A Complete Solution Guide to Real
and Complex Analysis I
by Kit-Wing Yu, PhD
kitwing@hotmail.com
Copyright c 2019 by Kit-Wing Yu. All rights reserved. No part of this publication may be
reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic,
mechanical, photocopying, recording, or otherwise, without the prior written permission of the
author.
ISBN: 978-988-78797-8-7 (eBook)
ISBN: 978-988-78797-9-4 (Paperback)
ii
About the author
Dr. Kit-Wing Yu received his B.Sc. (1st Hons), M.Phil. and Ph.D. degrees in Math. at the
HKUST, PGDE (Mathematics) at the CUHK. After his graduation, he has joined United Christian College to serve as a mathematics teacher for at least seventeen years. He has also taken
the responsibility of the mathematics panel since 2002. Furthermore, he was appointed as a
part-time tutor (2002 – 2005) and then a part-time course coordinator (2006 – 2010) of the
Department of Mathematics at the OUHK.
Besides teaching, Dr. Yu has been appointed to be a marker of the HKAL Pure Mathematics
and HKDSE Mathematics (Core Part) for over thirteen years. Between 2012 and 2014, Dr. Yu
was invited to be a Judge Member by the World Olympic Mathematics Competition (China). In
the research aspect, he has published over twelve research papers in international mathematical
journals, including some well-known journals such as J. Reine Angew. Math., Proc. Roy. Soc.
Edinburgh Sect. A and Kodai Math. J.. His research interests are inequalities, special functions
and Nevanlinna’s value distribution theory.
iii
iv
Preface
Professor Walter Rudina is the author of the classical and famous textbooks: Principles of Mathematical Analysis, Real and Complex Analysis, and Functional Analysis. (People commonly call
them “Baby Rudin”, “Papa Rudin” and “Grandpa Rudin” respectively.) Undoubtedly, they
have produced important and extensive impacts to the study of mathematical analysis at university level since their publications.
As far as you know, Real and Complex Analysis keeps the features of Principles of Mathematical Analysis which are well-organized and expositions of theorems are clear, precise and
well-written. Therefore, Real and Complex Analysis is always one of the main textbooks or
references of graduate real analysis course in many universities. Actually, some universities will
request their Ph.D. students to study this book for their qualifying examinations.
After the publication of the book A Complete Solution Guide to Principles of Mathematical
Analysis, some purchasers have suggested me to write a solution book of Real and Complex Analysis. (To the best of the author’s knowledge, Papa Rudin has no solution manual.) I was afraid
of doing so at the beginning because the exercises are at graduate level and they are much more
difficult than those in Baby Rudin. Fortunately, I was used to keeping solutions of mathematics
exercises done by me in my undergraduate and graduate study. In fact, I have kept at least 25%
of the exercises in the first six chapters of Papa Rudin. Therefore, after thorough consideration, I decided to start the next project of a solution book to Rudin’s Real and Complex Analysis.
Since Papa Rudin consists of two components, I plan to write the solutions for “Real Analysis” part first. In fact, the present book A Complete Solution Guide to Real and Complex
Analysis I covers all the exercises of Chapters 1 to 9 and its primary aim is to help every
mathematics student and instructor to understand the ideas and applications of the theorems
in Rudin’s book. To accomplish this goal, I have adopted the way I wrote the book A Complete
Solution Guide to Principles of Mathematical Analysis. In other words, I intend writing the
solutions as comprehensive as I can so that you can understand every detailed part of a proof
easily. Apart from this, I also keep reminding you what theorems or results I have applied by
quoting them repeatedly in the proofs. By doing this, I believe that you will become fully aware
of the meaning and applications of each theorem.
Before you read this book, I have two gentle reminders for you. Firstly, as a mathematics
instructor at a college, I understand that the growth of a mathematics student depends largely
on how hard he/she does exercises. When your instructor asks you to do some exercises from
Rudin, you are not suggested to read my solutions unless you have tried your best to prove them
yourselves. Secondly, when I prepared this book, I found that some exercises require knowledge
that Rudin did not cover in his book. To fill this gap, I refer to some other analysis or topology
a
https://en.wikipedia.org/wiki/Walter_Rudin.
v
vi
books such as [9], [10], [22], [42] and [47]. Other useful references are [1], [12], [24], [28], [29],
[59], [64], [66] and [67]. Of course, it is not a surprise that we will regard the exercises in Baby
Rudin as some known facts and if you want to read proofs of them, you are strongly advised to
read my book [63].
The features of this book are as follows:
• It covers all the 176 exercises from Chapters 1 to 9 with detailed and complete solutions.
As a matter of fact, my solutions show every detail, every step and every theorem that I
applied.
• There are 11 illustrations for explaining the mathematical concepts or ideas used behind
the questions or theorems.
• Sections in each chapter are added so as to increase the readability of the exercises.
• Different colors are used frequently in order to highlight or explain problems, lemmas,
remarks, main points/formulas involved, or show the steps of manipulation in some complicated proofs. (ebook only)
• Necessary lemmas with proofs are provided because some questions require additional
mathematical concepts which are not covered by Rudin.
• Many useful or relevant references are provided to some questions for your future research.
Since the solutions are written solely by me, you may find typos or mistakes. If you really
find such a mistake, please send your valuable comments or opinions to
kitwing@hotmail.com.
Then I will post the updated errata on my website
https://sites.google.com/view/yukitwing/
irregularly.
Kit Wing Yu
April 2019
List of Figures
2.1
The graph of gn on [−1, 1]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
2.2
2.3
The graphs of gn,k on [0, 1]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The pictures of V, E ′ , E ′ ∩ V and (E ′ ∩ V )c . . . . . . . . . . . . . . . . . . . . . .
31
57
2.4
The set (E ′ ∩ V )c \ U . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57
3.1
The distribution of x and ǫn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
85
3.2
The geometric interpretation of a special case. . . . . . . . . . . . . . . . . . . . .
89
5.1
5.2
The unit circle in different p-norm. . . . . . . . . . . . . . . . . . . . . . . . . . . 144
The square K. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
6.1
The graphs of gn,1 (x) and gn,2 (x).
7.1
The closed intervals En and En+1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
7.2
Construction of the sequence Ei (p). . . . . . . . . . . . . . . . . . . . . . . . . . 242
. . . . . . . . . . . . . . . . . . . . . . . . . . 194
vii
List of Figures
viii
Contents
Preface
v
List of Figures
vii
1 Abstract Integration
1.1 Problems on σ-algebras and Measurable Functions . . . . . . . . . . . . . . . . .
1.2
1
1
Problems related to the Lebesgue’s MCT/DCT . . . . . . . . . . . . . . . . . . .
7
2 Positive Borel Measures
2.1 Properties of Semicontinuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
17
2.2
Problems on the Lebesgue Measure on R . . . . . . . . . . . . . . . . . . . . . . .
25
2.3
2.4
Integration of Sequences of Continuous Functions . . . . . . . . . . . . . . . . . .
Problems on Borel Measures and Lebesgue Measures . . . . . . . . . . . . . . . .
30
36
2.5
Problems on Regularity of Borel Measures . . . . . . . . . . . . . . . . . . . . . .
43
2.6
Miscellaneous Problems on
L1
and Other Properties . . . . . . . . . . . . . . . .
59
3 Lp -Spaces
3.1 Properties of Convex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
69
69
3.2
Relations among Lp -Spaces and some Consequences . . . . . . . . . . . . . . . .
71
3.3
3.4
Applications of Theorems 3.3, 3.5, 3.8, 3.9 and 3.12 . . . . . . . . . . . . . . . . .
Hardy’s Inequality and Egoroff’s Theorem . . . . . . . . . . . . . . . . . . . . . .
87
91
3.5
3.6
Convergence in Measure and the Essential Range of f ∈ L∞ (µ) . . . . . . . . . . 106
A Converse of Jensen’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
3.7
The Completeness/Completion of a Metric Space . . . . . . . . . . . . . . . . . . 112
3.8
Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
4 Elementary Hilbert Space Theory
123
4.1 Basic Properties of Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 123
4.2
Application of Theorem 4.14
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
4.3
Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
5 Examples of Banach Space Techniques
143
5.1 The Unit Ball in a Normed Linear Space . . . . . . . . . . . . . . . . . . . . . . . 143
5.2
5.3
Failure of Theorem 4.10 and Norm-preserving Extensions . . . . . . . . . . . . . 146
The Dual Space of X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
ix
Contents
x
5.4
Applications of Baire’s and other Theorems . . . . . . . . . . . . . . . . . . . . . 157
5.5
Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
6 Complex Measures
183
6.1
6.2
Properties of Complex Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
Dual Spaces of Lp (µ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
6.3
6.4
Fourier Coefficients of Complex Borel Measures . . . . . . . . . . . . . . . . . . . 191
Problems on Uniformly Integrable Sets . . . . . . . . . . . . . . . . . . . . . . . . 196
6.5
Dual Spaces of Lp (µ) Revisit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
7 Differentiation
207
7.1
7.2
Lebesgue Points and Metric Densities . . . . . . . . . . . . . . . . . . . . . . . . 207
Periods of Functions and Lebesgue Measurable Groups . . . . . . . . . . . . . . . 210
7.3
7.4
The Cantor Function and the Non-measurability of f ◦ T . . . . . . . . . . . . . 216
Problems related to the AC of a Function . . . . . . . . . . . . . . . . . . . . . . 218
7.5
Miscellaneous Problems on Differentiation . . . . . . . . . . . . . . . . . . . . . . 232
8 Integration on Product Spaces
249
8.1
8.2
Monotone Classes and Ordinate Sets of Functions . . . . . . . . . . . . . . . . . . 249
Applications of the Fubini Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 252
8.3
8.4
The Product Measure Theorem and Sections of a Function . . . . . . . . . . . . 268
Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273
9 Fourier Transforms
281
9.1 Properties of The Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . . . . 281
9.2
9.3
The Poisson Summation Formula and its Applications . . . . . . . . . . . . . . . 297
Fourier Transforms on Rk and its Applications . . . . . . . . . . . . . . . . . . . 301
9.4
Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310
Index
315
Bibliography
317
CHAPTER
1
Abstract Integration
1.1
Problems on σ-algebras and Measurable Functions
Problem 1.1
Rudin Chapter 1 Exercise 1.
Proof. Let X be a set. Assume that M was an infinite σ-algebra in X which has only countably
many members. Then we have M = {∅, X, A1 , A2 , . . .}. Let x ∈ X be fixed, we define the set
\
Sx =
A,
(1.1)
x∈A
where x ∈ A ∈ M. (By this definition (1.1), it is clear that x ∈ Sx .) Since M is countable, the
intersection in (1.1) is actually at most countable. Thus it follows from Comment 1.6(c) that
Sx ∈ M. Next we define the set
S = {Sx | x ∈ X}
so that S ⊆ M. We need a lemma about the set S:
Lemma 1.1
If Sx , Sy ∈ S and Sx ∩ Sy 6= ∅, then Sx = Sy .
Proof of Lemma 1.1. Let z ∈ Sx ∩ Sy . We want to show that Sz = Sx . On the one
hand, since z ∈ Sx , we know from the definition (1.1) that z ∈ A for all A containing
x. Now if w ∈ Sx , then w ∈ A for all A containing x. By the previous observation,
such A must contain z, so we get w ∈ Sz and then
Sx ⊆ Sz .
(1.2)
On the other hand, if w ∈ Sz , then w ∈ A for all A containing z. Thus the hypothesis
z ∈ Sx ∩ Sy implies that w ∈ Sx ∩ Sy , i.e., Sz ⊆ Sx ∩ Sy . By this result and the
definition (1.1), we obtain
Sz ⊆ Sx ∩ Sy ⊆ Sx .
(1.3)
Combining the observations (1.2) and (1.3), we have the desired result that Sz = Sx .
Similarly, we can show that Sz = Sy . Hence we conclude that Sx = Sy .
1
Chapter 1. Abstract Integration
2
Now we return to the proof of the problem. By Lemma 1.1, we may assume that all elements
in S are distinct. If B ∈ M, then the definition (1.1) certainly gives y ∈ Sy for every y ∈ B so
that
[
B⊆
Sy .
(1.4)
y∈B
Therefore, if the cardinality of S is N , then the cardinality of its power set 2S is 2N and so
there are at most 2N elements in M by the inclusion (1.4), a contradiction. Thus the set S must
be infinite.
Recall that S ⊆ M and M is countable, S is also countable by [49, Theorem 2.8, p. 26].
Since M is a σ-algebra, it must contain the power set of S. However, it is well-known that the
power set of an infinite countable set is uncountable [47, Problem 22, p. 16]. We conclude from
these facts that M is uncountable, a contradiction. This completes the proof of the problem. Problem 1.2
Rudin Chapter 1 Exercise 2.
Proof. Put f (x) = (f1 (x), . . . , fn (x)). Since f1 , . . . , fn : X → R, f maps the measurable space
X into Rn . Let Y be a topological space and Φ : Rn → Y be continuous. Define
h(x) = Φ (f1 (x), . . . , fn (x)) .
Since h = Φ ◦ f , Theorem 1.7(b) shows that it suffices to prove the measurability of f .
To this end, we first consider f −1 (R) for an open rectangle R in Rn . By the definition,
R = I1 × · · · × In , where Ii is an open interval in R for 1 ≤ i ≤ n. Now we know from [42,
Exercise 3, p. 21] that
f −1 (R) = {x ∈ X | f (x) ∈ R}
= {x ∈ X | fi (x) ∈ Ii for 1 ≤ i ≤ n}
= {x ∈ X | f1 (x) ∈ I1 } ∩ · · · ∩ {x ∈ X | fn (x) ∈ In }
= f1−1 (I1 ) ∩ · · · ∩ fn−1 (In ).
Since each fi is measurable and each Ii is open in R, the set fi−1 (Ii ) is measurable in X by
Definition 1.3(c). Thus this implies that f −1 (R) is measurable in X by Comment 1.6(c).
Our proof will be complete if we can show that every open set V in Rn can be written as a
countable union of open rectangles Rj in Rn . We need some topology. By [42, Exercise 8(a), p.
83], the countable collection
B = {I = (a, b) | a < b and a, b ∈ Q}
is a basis that generates the standard topology on R. Then we follow from [42, Theorem 15.1,
p. 86] that the collection
C = {I1 × · · · × In | Ii = (ai , bi ), ai , bi ∈ Q, 1 ≤ i ≤ n}
(1.5)
is a basis for the (product) topology of Rn .a Since elements in the collection (1.5) are open
rectangles and it is countable by [49, Theorem 2.13, p. 29], every open set V in Rn is, in fact, a
countable union of open rectangles Rj , i.e,
f
−1
(V ) = f
−1
∞
[
j=1
a
For details, please read [42, §13 and §15].
Rj =
∞
[
j=1
f −1 (Rj ).
3
1.1. Problems on σ-algebras and Measurable Functions
Hence f −1 (V ) is also a measurable set in X by Definition 1.3(a)(iii). This completes the proof
of the problem.
Problem 1.3
Rudin Chapter 1 Exercise 3.
Proof. Let f : X → R ⊂ [−∞, ∞] and M be the σ-algebra of X. Let α be real. By [49, Theorem
1.20(b), p. 9], we can find a sequence {rn } of rational numbers such that α < rn for all n ∈ N,
rn → α as n → ∞. In other words, we have
(α, ∞] =
∞
[
[rn , ∞]
n=1
which implies that
f
−1
((α, ∞]) = f
−1
∞
[
∞
[
f −1 ([rn , ∞]).
[rn , ∞] =
n=1
n=1
Recall that
f −1 ([rn , ∞]) = {x | f (x) ≥ rn }
is assumed to be measurable for each n, so we have f −1 ([rn , ∞]) ∈ M and then we follow from
Definition 1.3(a)(iii) that f −1 ((α, ∞]) ∈ M. Since α is arbitrary, we conclude from Theorem
1.12(c) that f is measurable. This finishes the proof of the problem.
Problem 1.4
Rudin Chapter 1 Exercise 4.
Proof.
(a) Let αk = sup{−ak , −ak+1 , . . .} for k = 1, 2, . . .. By the definition, we have αk ≥ −an for
all n ≥ k and if α ≥ −an for all n ≥ k, then α ≥ αk . Note that this is equivalent to the
fact that −αk ≤ an for all n ≥ k and if −α ≤ an for all n ≥ k, then −α ≤ −αk . In other
words, we have
−αk = inf{ak , ak+1 , . . . , }
for k = 1, 2, . . .. Thus this implies that
sup(−an ) = sup{−ak , −ak+1 , . . .} = − inf{ak , ak+1 , . . .} = − inf (an ).
n≥k
n≥k
(1.6)
Similarly, we have
inf (−cn ) = − sup(cn )
n≥k
(1.7)
n≥k
for a sequence {cn } in [−∞, ∞]. By applying the equality (1.6) and then the equality
(1.7), we achieve that
n
o
o
o
n
n
lim sup(−an ) = inf sup(−an ) = inf − inf (an ) = − sup inf (an ) = − lim inf (an )
n→∞
k≥1
k≥1 n≥k
n≥k
n→∞
k≥1 n≥k
| {z }
cn
which is our desired result.
Chapter 1. Abstract Integration
4
(b) This part is proven in [63, Problem 3.5, pp. 32, 33].
(c) Since an ≤ bn for all n = 1, 2, . . ., we must have
αk = inf (an ) ≤ inf (bn ) = βk
n≥k
n≥k
(1.8)
for all k = 1, 2, . . .. Thus {αn } and {βn } are two sequences in [−∞, ∞] such that αn ≤ βn
for all n = 1, 2, . . .. By a similar argument, we also have
sup αk ≤ sup βk
k≥m
(1.9)
k≥m
for all m = 1, 2, . . .. Combining the inequalities (1.8) and (1.9), we have
n
o
n
o
sup inf (an ) ≤ sup inf (bn )
k≥m
n≥k
k≥m
n≥k
for all m = 1, 2, . . .. By Definition 1.13, we have the desired result.
For a counterexample to part (b), we consider an = (−1)n and bn = (−1)n+1 for all n =
1, 2, . . .. On the one hand, we have an + bn = 0 for all n = 1, 2, . . . so that
lim sup(an + bn ) = 0.
(1.10)
n→∞
On the other hand, we have
lim sup an = lim a2k = 1 and
n→∞
k→∞
lim sup bn = lim b2k+1 = 1
n→∞
k→∞
which imply that
lim sup an + lim sup bn = 2.
n→∞
(1.11)
n→∞
Hence we obtain from the results (1.10) and (1.11) that the strict inequality can hold in part
(b). This completes the proof of the problem.
Problem 1.5
Rudin Chapter 1 Exercise 5.
Proof.
(a) Let MX be a σ-algebra of X. Further, suppose that Sf (±∞) = {x ∈ X | f (x) = ±∞},
and Sg (±∞) = {x ∈ X | g(x) = ±∞}. We see that
Sf (∞) =
∞
\
{x ∈ X | f (x) > n}.
n=1
Since f is measurable and (n, ∞] is openb in [−∞, ∞], f −1 ((n, ∞]) = {x | f (x) > n} ∈ MX
by Theorem 1.12(b) for each n ∈ N. Thus Sf (∞) ∈ MX by Comment 1.6(c). Similarly,
all the other sets Sf (−∞), Sg (∞) and Sg (−∞) belong to MX too.
Next, we let X ′ = X\(Sf (∞)∪Sf (−∞)∪Sg (∞)∪Sg (−∞)). Since Sf (∞), Sf (−∞), Sg (∞)
and Sg (−∞) are measurable, X ′ ∈ M by Definition 1.3(a)(ii) and Comment 1.6(b). By
b
See the proof of Theorem 1.12(c) in [51, p. 13]
5
1.1. Problems on σ-algebras and Measurable Functions
the comment following Proposition 1.24, we know that X ′ is itself a measure space and if
we let MX ′ be a σ-algebra of X ′ , then
MX ′ ⊆ MX .
(1.12)
Furthermore, the restricted mappings fX ′ : X ′ → R and gX ′ : X ′ → R are also measurable
because of Definition 1.3(c) and the fact that any open set V in R is a countable union
of segments of the type (α, β) so that V is also open in the extended number system
[−∞, ∞].c In fact, we have
−1
−1
fX
(V ) ∈ MX ′
′ (V ) = f
and
−1
−1
gX
(V ) ∈ MX ′ .
′ (V ) = g
We need to prove one more thing: the mapping −g : X → [−∞, ∞] is measurable. Since
g is measurable, we know from [49, Definition 11.13] that {x ∈ X | g(x) > −a} ∈ MX for
every real a. It is obvious that {x ∈ X | − g(x) < a} = {x ∈ X | g(x) > −a} for every real
a, so we deduce from [49, Theorem 11.15, p. 311] that −g is measurable.
Now we are ready to prove the desired results. Notice that
{x ∈ X | f (x) = g(x)} = {x ∈ X | h(x) = 0} = h−1 (0),
where h = f − g. Since f and −g are measurable, the new (real) function h = f − g is also
measurable by Theorem 1.9(c). Since
−1
h
(0) =
∞
\
n=1
h−1
−
1 1 ,
n n
and h−1 ((− n1 , n1 )) ∈ MX ′ for every n ∈ N, we yield from this and the relation (1.12) that
h−1 (0) ∈ MX ′ ⊆ MX . This shows the second assertion. For the first assertion, we note
that
{x ∈ X | f (x) < g(x)} = {x ∈ X | h(x) = f (x) − g(x) < 0}
= h−1 (−∞, 0) ∪ [Sf (−∞) \ Sg (−∞)] ∪ [Sg (∞) \ Sf (∞)]
\ [Sf (∞) ∩ Sg (∞)] ∪ [Sf (−∞) ∩ Sg (−∞)] .
Recall that Sf (∞), Sf (−∞), Sg (∞), Sg (−∞) ∈ MX , so we have
Sf (−∞) \ Sg (−∞), Sg (∞) \ Sf (∞), Sf (∞) ∩ Sg (∞), Sf (−∞) ∩ Sg (−∞) ∈ MX . (1.13)
By the measurability of h and the relation (1.12), we have
h−1 ((−∞, 0)) ∈ MX .
(1.14)
Hence the facts (1.13) and (1.14) show that {x | f (x) < g(x)} ∈ MX .
(b) This part is proven in [63, Problem 11.3, p. 339].
This completes the proof of the problem.
Problem 1.6
Rudin Chapter 1 Exercise 6.
c
This can be seen, again, from the proof of Theorem 1.12(c) or from the comment following Proposition 1.24.
Chapter 1. Abstract Integration
6
Proof. We prove the assertions one by one.
• M is a σ-algebra in X. We check Definition 1.3(a). Since X c = ∅, it is at most
countable. Thus we have X ∈ M. Let A ∈ M. If Ac is at most countable, then Ac ∈ M.
Similarly, if A is at most countable, then since (Ac )c = A, we have Ac ∈ M. Suppose
that An ∈ M for n = 1, 2, . . .. Then we have either An or Acn is at most countable for
n = 1, 2, . . .. If all An are at most countable, then we deduce from the corollary following
[49, Theorem 2.12, p. 29] that the set
A=
∞
[
An
(1.15)
n=1
is also at most countable so that A ∈ M. Otherwise, without loss of generality, we suppose
that A1 is uncountable but Ac1 is at most countable. Then we consider
c
A =
∞
[
An
n=1
c
=
∞
\
n=1
Acn ⊆ Ac1
which means that Ac is at most countable too. Hence Ac ∈ M and then M is a σ-algebra
in X.
• µ is a measure on M. We check Definition 1.18(a). Since ∅ is at most countable,
we have µ(∅) = 0 so that µ is not identically ∞. In fact, it is clear that we have
µ : M → {0, 1} ⊂ [0, ∞]. Let {An } be a disjoint countable collection of members of M.
Case (i): All An are at most countable. Recall that the set A given by (1.15)
is at most countable. By the definition of µ, we have µ(A) = µ(An ) = 0 for all
n = 1, 2, . . .. Thus we have
∞
X
µ(An )
(1.16)
µ(A) =
n=1
in this case.
Case (ii): There is at least one Ak is uncountable. Since Ak ∈ M, Ack must
be at most countable. Since An ∩ Ak = ∅ for all n 6= k, we have An ⊆ Ack for all
n 6= k. In other words, the measurable sets An are at most countable for all n 6= k.
Therefore, we have µ(Ak ) = 1 and µ(An ) = 0 for all n 6= k. Since
c
A =
∞
\
n=1
Acn ⊆ Ack ,
Ac is at most countable and then µ(A) = 1. Hence the equality (1.16) also holds in
this case.
This completes the proof that µ is a measure on M.
• The determination of measurable functions and their integrals. Let f : X → R
be a measurable function. Since X is uncountable, we have µ(X) = 1 which is the only
thing that we know and start with. For every n ∈ Z, we know that
[n, n + 1) = R \ (−∞, n) ∪ [n + 1, ∞) .
d
By the fact thatd f −1 (A \ B) = f −1 (A) \ f −1 (B), we obtain
f −1 [n, n + 1) = f −1 (R) \ f −1 (−∞, n) ∪ f −1 [n + 1, ∞) .
See [42, Exercise 2(d), p. 20].
(1.17)
7
1.2. Problems related to the Lebesgue’s MCT/DCT
By [49, Theorem 11.15, p. 311], each set on the right-hand side in (1.17) is measurable.
This implies that
f −1 [n, n + 1) ∈ M.
Let En = f −1 [n, n + 1) , where n ∈ Z. By the definition of M and then the definition of
µ, we have either µ(En ) = 0 or µ(En ) = 1. For every x ∈ X, we must have f (x) ∈ [n, n+1)
for some n ∈ Z, i.e., x ∈ En for some n ∈ Z. Therefore, we have
∞
[
En = X.
n=−∞
It is clear that R =
∞
[
[n, n + 1), so {En } is a disjoint countable collection of members
n=−∞
of M. By Definition 1.18(a), we see that
µ(X) = µ
∞
[
n=−∞
∞
X
µ(En ).
En =
(1.18)
n=−∞
The fact µ(X) = 1 and the equality (1.18) force that there exists an integer n0 such that
µ(En0 ) = 1. Without loss of generality, we may assume that n0 = 0, i.e.,
µ(E0 ) = µ f −1 [0, 1) = 1.
(1.19)
1
1
Next if we write [0, 1) = [0, 2 ) ∪ [ 2 , 1), then the above
argument and the value (1.19)
1
1
−1
−1
imply that either µ f
[0, 2 ) = 1 or µ f
[ 2 , 1) = 1. This process can be done
continuously so that a sequence of intervals {[an , bn )} is constructed such that
µ f −1 [an , bn ) = 1,
0 ≤ bn − a n ≤
1
2n
and
lim (bn − an ) = 0,
n→∞
where n = 1, 2, . . .. In other words, it means that
µ f −1 (a) = 1
and µ f −1 (b) = 0
for some a ∈ R and all other real numbers b 6= a, but this is equivalent to saying that f −1 (a)
is uncountable and f −1 (b) is at most countable for all b 6= a. Now we have completely
characterized every measurable function on X.
Finally, it is clear that X \ f −1 (a) is a set of measure 0. Therefore, we have
Z
Z
f dµ = a.
f dµ =
X
f −1 (a)
Hence, this completes the proof of the problem.
1.2
Problems related to the Lebesgue’s MCT/DCT
Problem 1.7
Rudin Chapter 1 Exercise 7.
Chapter 1. Abstract Integration
8
Proof. Let Ek = {x ∈ X | f1 (x) > k} = f1−1 (k, ∞] and E = {x ∈ X | f1 (x) = ∞}, where
k = 1, 2, . . .. It is clear that each (k, ∞] is a Borel set in [0, ∞], so each Ek is measurable by
Theorem 1.12(b). Since
∞
\
E=
Ek ,
k=1
E is also measurable by Comment 1.6(c). If µ(E) > 0, then we know from the definition that
Z
f1 dµ = ∞.
(1.20)
E
By using the result (1.20), Proposition 1.24(b) and Theorem 1.33, we conclude that
Z
X
|f1 | dµ ≥
Z
X
f1 dµ ≥
Z
E
f1 dµ = ∞
which contradicts the hypothesis that f1 ∈ L1 (µ). In other words, we must have µ(E) = 0 and
thus f1 ∈ L1 (µ) on X \ E.
Here we may assume that the form of Theorem 1.34 (Lebesgue’s Dominated Convergence
Theorem) is also valid for measurable functions defined a.e. on X.e Now we see that the
measurable function f1 in the problem plays the role of g in the theorem and hence our desired
result follows from Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) immediately.
For a counterexample, we consider X = R, In = (n, ∞), µ = m the Lebesgue measure (see
[49, Definition 11.5, pp. 302, 303]) and define for each n = 1, 2, . . .,
fn (x) = χIn (x) =

 1, if x ∈ In ;

0, if x ∈
/ In .
It is clear that f1 ≥ f2 ≥ · · · ≥ 0 on X and
Z
Z
Z
fn dm =
|fn | dm =
R
R
In
dm = ∞
(1.21)
for every n = 1, 2, . . .. In particular, the integrals (1.21) show that f1 ∈
/ L1 (m). Furthermore,
we have
f (x) = lim fn (x) = lim χIn (x) = 0
(1.22)
n→∞
n→∞
for every x ∈ R. Thus we deduce from the expression (1.22) and Proposition 1.24(d) that
Z
f dm = 0.
(1.23)
R
Hence the inconsistence of the integrals (1.21) and (1.23) show that the condition “f1 ∈ L1 (µ)”
cannot be omitted. This completes the proof of the problem.
Problem 1.8
Rudin Chapter 1 Exercise 8.
e
Actually, Rudin [51, p. 29] assumed this fact in the proof of Theorem 1.38 or the reader may refer to the
comment between Theorem 11.32 and its proof in [49, p. 321].
9
1.2. Problems related to the Lebesgue’s MCT/DCT
Proof. If x ∈ E, then for all k ∈ N, we have
lim f2k (x) = lim 1 − χE (x) = 0.
k→∞
k→∞
Similarly, if x ∈
/ E, then for all k ∈ N, we have
lim f2k+1 (x) = lim χE (x) = 0.
k→∞
k→∞
Thus we have
lim inf fn (x) = 0
n→∞
for all x ∈ X. However, we have
Z
fn dµ =
X
=
 Z

χ dµ,


 X E
if n is odd;
Z




(1 − χE ) dµ, if n is even.
X

if n is odd;
 µ(E),

(1.24)
µ(X \ E), if n is even.
Therefore we obtain from the results (1.24) that
Z
fn dµ = min(µ(E), µ(X \ E)) 6= 0
lim inf
n→∞
X
if we assume that µ(E) > 0 and µ(X \E) > 0. Hence the example here shows that the inequality
in Fatou’s lemma can be strict, completing the proof of the problem.f
Problem 1.9
Rudin Chapter 1 Exercise 9.
Proof. If µ(X) = 0, then Proposition 1.24(e) implies that c =
Z
f dµ = 0, a contradiction. Let
X
E = {x ∈ X | f (x) = ∞}. We claim that µ(E) = 0. Otherwise, Proposition 1.24(b) implies that
Z
Z
f dµ = ∞
f dµ ≥
c=
X
E
which is a contradiction. Therefore, in the following discussion, we may assume that x ∈ X \ E
so that 0 ≤ f (x) < ∞.
For each n = 1, 2, . . ., we defineg fn : X \ E → [0, ∞) by
h
f (x) α i
.
fn (x) = n log 1 +
n
Since f is measurable on X \ E, g(x) = [1 + ( nx )α ] and h(x) = n log(1 + x) are continuous on
[0, ∞), Theorem 1.7(b) implies that
f
g
h
f (x) α i
fn (x) = n log 1 +
= h g f (x)
n
There is another example in [63, Problem 11.5, p. 340].
X \ E is itself a measure space by the remark following Proposition 1.24.
Chapter 1. Abstract Integration
10
is also measurable on X \ E.
Now we are going to show that when α ≥ 1, there exists a function g ∈ L1 (µ) such that
|fn (x)| ≤ g(x)
holds for all n = 1, 2, . . . and all x ∈ X \ E. We first show the following lemma:
Lemma 1.2
For each n ∈ N and α ≥ 1, we have
h
x α i
≤ αx
n log 1 +
n
(1.25)
on [0, ∞).
Proof of Lemma 1.2. For x ∈ [0, ∞), we let
Then basic calculus gives
Since
h
x α i
F (x) = αx − n log 1 +
.
n
x α−1
α
α−1
nxα−1 n = α − αnx
=
α
1
−
.
F ′ (x) = α −
x α
n α + xα
n α + xα
1+
n
(1.26)
nxα−1
xα
≤ α
<1
α
α
n +x
n + xα
for n ≤ x and since
nα
nxα−1
<
<1
n α + xα
n α + xα
for n > x and α ≥ 1, we deduce from the derivative (1.26) that
F ′ (x) < 0
for all x ∈ [0, ∞). By [49, Theorem 5.11, p. 108] and the continuity of F on [0, ∞), we
establish that F (x) is decreasing on [0, ∞) which implies the validity of the inequality
(1.25).
Let’s return to the proof of the problem. By Lemma 1.2, it is evident that for all n ∈ N and
α ≥ 1, we have
h
f α i
≤ αf
fn = n log 1 +
n
on X \ E. Furthermore, we apply Proposition 1.24(c) to get
Z
Z
f dµ = αc < ∞.
|αf | dµ = α
X
X
This means that αf ∈ L1 (µ) and then g = αf is the desired function.
Next, there are two cases for consideration:
(1.27)
11
1.2. Problems related to the Lebesgue’s MCT/DCT
Case (i): α = 1. In this case, we haveh
on X \ E.
h
f (x) α i
f (x) n
= lim log 1 +
= log ef (x) = f (x)
lim n log 1 +
n→∞
n→∞
n
n
Case (ii): α > 1. In this case, we apply L’Hospital’s rule [49, Theorem 5.13, p. 109] to
conclude that
h
f α i
αf α y α−1
0
log(1 + f α y α )
lim n log 1 +
=
= lim
= lim
=0
(1.28)
α
α
+
+
n→∞
n
y
1+0
y→0 1 + f y
y→0
on X \ E.
Thus it follows from Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) that
 Z

f dµ, if α = 1;


Z
 X
h
f α i
n log 1 +
dµ =
lim
Z
n→∞ X

n



0 dµ, if α > 1,
=
 X
 c, if α = 1;

0, if α > 1.
It remains the case that 0 < α < 1. In this case, Theorem 1.28 (Fatou’s Lemma) can be
applied directly to get
)
Z
Z (
h
f α i
n
h
f α io
n log 1 +
dµ ≤ lim inf
lim inf n log 1 +
dµ.
(1.29)
n→∞ X
n→∞
n
n
X
Since 0 < α < 1, we have y α−1 =
1
y 1−α
so that the limit (1.28) becomes
h
f α i
αf α
= ∞.
= lim 1−α
lim n log 1 +
n→∞
n
(1 + f α y α )
y→0+ y
(1.30)
Since lim xn = ∞ if and only if lim sup xn = lim inf xn = ∞, the inequality (1.29) and the limit
n→∞
n→∞
n→∞
(1.30) combine to imply that
n
h
f α io
lim inf n log 1 +
=∞
n→∞
n
and then it certainly gives
lim
n→∞
Z
h
f α i
dµ = ∞.
n log 1 +
n
X
This completes the proof of the problem.
Problem 1.10
Rudin Chapter 1 Exercise 10.
h
We use the definition ex = lim
n→∞
1+
x n
here.
n
Chapter 1. Abstract Integration
12
Proof. Since fn → f uniformly on X, there exists a positive integer N such that n ≥ N implies
|fn (x) − f (x)| ≤ 1
for all x ∈ X. By this, we have
|fn (x)| ≤ |fn (x) − f (x)| + |f (x)| ≤ |f (x)| + 1
(1.31)
|f (x)| ≤ |f (x) − fN (x)| + |fN (x)| ≤ |fN (x)| + 1
(1.32)
and
for all n ≥ N and x ∈ X. Combining the inequalities (1.31) and (1.32), we see that
|fn (x)| ≤ |fN (x)| + 2
(1.33)
g(x) = max{|f1 (x)|, . . . , |fN −1 (x)|, |fN (x)| + 2}.
(1.34)
for all n ≥ N and x ∈ X.
We define g : X → [0, ∞) by
Then it is easy to see from the inequality (1.33) and the definition (1.34) that
|fn (x)| ≤ g(x)
for x ∈ X and n = 1, 2, . . ..
Next we want to show that g ∈ L1 (µ). By Theorem 1.9(b), |f1 (x)|, . . . , |fN −1 (x)|, |fN (x)|
are measurable. Thus it follows from the corollaries following Theorem 1.14 that g is also
measurable. Furthermore, we know from the hypothesis that |fn (x)| ≤ Mn on X for some
constants Mn , where n = 1, 2, . . . , N . Therefore, this and the definition (1.34) certainly imply
that
g(x) = |g(x)| ≤ M
on X for some constant M . Since it is obvious that g ≥ 0 on X, we get from Proposition 1.24(a)
that
Z
Z
M dµ = M µ(X) < ∞
|g| dµ ≤
0≤
X
X
L1 (µ).
so that g ∈
In conclusion, our sequence of functions {fn } satisfies the hypotheses of Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) and hence the desired result follows
immediately from this.
For a counterexample, consider X = R and µ = m so that m(R) = ∞. For each n ∈ N,
define fn : R → R by
1
fn (x) = .
n
Then it is easy to prove that fn → f ≡ 0 uniformly on R. Since
Z
Z
m(R)
=∞
fn dm =
f dm = 0 and
n
R
R
for n = 1, 2, . . ., we conclude that
lim
Z
n→∞ R
This finishes the proof of the problem.
fn dm 6=
Z
f dm.
R
13
1.2. Problems related to the Lebesgue’s MCT/DCT
Problem 1.11
Rudin Chapter 1 Exercise 11.
Proof. For each n ∈ N, define
Bn =
∞
[
Ek .
k=n
Recall that A is the set of all x ∈ X which lie in infinitely many Ek . On the one hand, if x ∈ A,
∞
\
Bn . In other words, we have
then x ∈ Bn for all n ∈ N so that x ∈
n=1
A⊆
On the other hand, if x ∈
∞
\
n=1
∞
\
Bn .
(1.35)
n=1
Bn , then x ∈ Bn for each positive integer n and this is equivalent
to the condition that x belongs to infinitely many Ek , i.e.,
∞
\
n=1
Bn ⊆ A.
(1.36)
Hence the set relations (1.35) and (1.36) imply the desired result that
A=
∞
\
Bn =
∞ [
∞
\
Ek .
n=1 k=n
n=1
We have to show that µ(A) = 0. By the definition of Bn , we have
B1 ⊇ B2 ⊇ B3 ⊇ · · · .
Furthermore, our hypothesis and the subadditive property of a measure (see , for example, [54,
Corollary 4.6, p. 26]) show that
µ(B1 ) = µ
∞
[
k=1
Ek ≤
∞
X
k=1
µ(Ek ) < ∞.
Thus it follows from Theorem 1.19(e) and the subadditive property of µ again that
0 ≤ µ(A) = µ
∞
\
n=1
∞
∞
[
X
Bn = lim µ(Bn ) = lim µ
Ek ≤ lim
µ(Ek ) = 0.
n→∞
n→∞
k=n
n→∞
Hence we must have µ(A) = 0, completing the proof of the problem.
Problem 1.12
Rudin Chapter 1 Exercise 12.
k=n
Chapter 1. Abstract Integration
14
Proof. Let n be a positive integer. Define fn : X → [0, ∞] by
fn (x) = min |f (x)|, n
(1.37)
for every x ∈ X. This definition (1.37) clearly satisfies
0 ≤ fn (x) ≤ n
(1.38)
for every x ∈ X. Since f ∈ L1 (µ), it is obviously measurable. By the corollaries following
Theorem 1.14, each fn is also measurable on X. Furthermore, if x0 ∈ X such that |f (x0 )| ≤ n,
then the definition (1.37) implies that
fn (x0 ) = |f (x0 )|
and fn+1 (x0 ) = |f (x0 )|;
(1.39)
if n < |f (x0 )| ≤ n + 1, then we know again from the definition (1.37) that
fn (x0 ) = n
and fn+1 (x0 ) = |f (x0 )|;
(1.40)
and fn+1 (x0 ) = n + 1.
(1.41)
if n + 1 < |f (x0 )|, then we must have
fn (x0 ) = n
Thus we can conclude from the computations (1.39), (1.40) and (1.41) that the inequality
0 ≤ fn (x) ≤ fn+1 (x)
(1.42)
holds for all n ∈ N and x ∈ X. By Theorem 1.26 (Lebesgue’s Monotone Convergence Theorem),
we have
Z
Z
F dµ,
fn dµ =
lim
n→∞ X
X
where F = lim fn . By the definition (1.37), we have F = |f | and so
n→∞
lim
Z
n→∞ X
or equivalently,
lim
fn dµ =
Z
n→∞ X
Z
X
|f | dµ
|fn − f | dµ = 0.
Thus for every ǫ > 0, there exists a positive integer N such that
Z
ǫ
|fn − f | dµ <
2
X
(1.43)
for all n ≥ N .
We fix this N . Since 0 ≤ |f | ≤ |f −fN |+|fN | by the triangle inequalityi , we apply Proposition
1.24(a) and then Theorem 1.27 and the property (1.38) to the inequality (1.43) to derive
Z
Z
Z
Z
ǫ
ǫ
|fN | dµ < +
|f − fN | dµ +
N dµ = + N µ(E)
|f | dµ ≤
(1.44)
2
2
E
E
E
E
ǫ
for every E ∈ M. Therefore, if we take δ = 2N
and µ(E) < δ, then our inequality (1.44) becomes
Z
|f | dµ < ǫ
E
as desired. Hence we complete the proof of the problem.
i
See [49, Definition 2.15, p. 30].
15
1.2. Problems related to the Lebesgue’s MCT/DCT
Problem 1.13
Rudin Chapter 1 Exercise 13.
Proof. Let f : X → [0, ∞] ⊂ [−∞, ∞] be a measurable function. For each n = 1, 2, . . ., we
define fn : X → [0, ∞] ⊂ [−∞, ∞] by
fn (x) = nf (x).
It is clear that
n
α
α
io
i
fn−1 (α, ∞] = {x ∈ X | fn (x) ∈ (α, ∞]} = x ∈ X f (x) ∈
, ∞ = f −1
,∞
n
n
for all real α. Since f is measurable, we have f −1 ( αn , ∞] ∈ M for every real α, where M is a σalgebra in X. Thus we obtain from Theorem 1.12(c) that each fn is measurable for n = 1, 2, . . ..
Furthermore, we have
0 ≤ f1 (x) ≤ f2 (x) ≤ · · · ≤ ∞
on X and fn (x) → ∞ · f (x) as n → ∞. By Proposition 1.24(c), we have
Z
Z
Z
f dµ
nf dµ = n
fn dµ =
X
(1.45)
X
X
for n = 1, 2, . . .. Hence we conclude from the equality (1.45) and Theorem 1.26 (Lebesgue’s
Monotone Convergence Theorem) that
Z
Z
Z
Z
f dµ = ∞ ·
fn dµ = lim n
f dµ.
∞ · f dµ = lim
X
n→∞ X
n→∞
X
X
Thus Proposition 1.24(c) also holds when c = ∞ and so we complete the proof of the problem.
Chapter 1. Abstract Integration
16
CHAPTER
2
Positive Borel Measures
2.1
Properties of Semicontinuity
Problem 2.1
Rudin Chapter 2 Exercise 1.
Proof. We have fn : R → [0, ∞) for all n ∈ N. In Proposition 2.5, we will prove a property
which is equivalent to Definition 2.8 and our proof of Statement (a) below becomes simpler.
However, we choose to apply Definition 2.8 to prove the statements here.
• Statement (a): For any real α, β1 and β2 , let
E(α) = {x ∈ R | f1 (x) + f2 (x) < α}
and
Fi (βi ) = {x ∈ R | fi (x) < βi },
(2.1)
where i = 1, 2. If α ≤ 0, then we see that E(α) = ∅ which is open in R. Similarly,
Fi (βi ) = ∅ if βi ≤ 0. So in the following discussion, we assume that α > 0, βi > 0 and
furthermore, E(α) 6= ∅ and Fi (βi ) 6= ∅.
Now we claim that
[
E(α) =
β1 +β2 ≤α
β1 ,β2 >0
To prove the claim, on the one hand, if
[
x∈
[F1 (β1 ) ∩ F2 (β2 )].
(2.2)
[F1 (β1 ) ∩ F2 (β2 )],
β1 +β2 ≤α
β1 ,β2 >0
then there exist real β1 and β2 with β1 + β2 ≤ α and β1 , β2 > 0 so that x ∈ F1 (β1 )∩ F2 (β2 ).
By the definition (2.1), this x satisfies f1 (x) < β1 and f2 (x) < β2 and their sum implies
f1 (x) + f2 (x) < β1 + β2 ≤ α.
Thus we have x ∈ E(α), i.e.,
[
[F1 (β1 ) ∩ F2 (β2 )] ⊆ E(α).
β1 +β2 ≤α
β1 ,β2 >0
17
Chapter 2. Positive Borel Measures
18
On the other hand, if x ∈ E(α), then we let η = f1 (x) + f2 (x). Define the two numbers
β1 and β2 by
η+α
η+α
β1 =
− f2 (x) and β2 =
− f1 (x).
2
2
By direct computation, we know that
βi >
f1 (x) + f2 (x) + f1 (x) + f2 (x)
− fi (x) = f1 (x) + f2 (x) − fi (x) ≥ 0,
2
where i = 1, 2. Since η = f1 (x) + f2 (x), it is easy to check that f1 (x) + f2 (x) < α if and
only if [f1 (x) + f2 (x)] + [f1 (x) + f2 (x)] < α + η if and only if f1 (x) + f2 (x) < η+α
2 if and
only if
f1 (x) < β1 .
(2.3)
Similarly, the above argument can be used to show that f1 (x) + f2 (x) < α if and only if
f2 (x) < β2 .
(2.4)
Now we deduce from the inequalities (2.3) and (2.4) that x ∈ F1 (β1 )∩F2 (β2 ). Furthermore,
since β1 + β2 = η + α − f1 (x) − f2 (x) = α, we have
[
[F1 (β1 ) ∩ F2 (β2 )],
x∈
β1 +β2 ≤α
β1 ,β2 >0
i.e.,
E(α) ⊆
[
β1 +β2 ≤α
β1 ,β2 >0
[F1 (β1 ) ∩ F2 (β2 )].
Hence the claim (2.2) holds.
Since F1 (β1 ) and F2 (β2 ) are open in R, F1 (β1 ) ∩ F2 (β2 ) is also open in R. Since the
union of any collection of open sets is open ([49, Theorem 2.24, p. 34]), we follow from
this and the equality (2.2) that E(α) is open in R. By Definition 2.8, f1 + f2 is upper
semicontinuous.
• Statement (b): By a similar argument as in part (a), we can show that the sum of two
lower semicontinuous functions is lower semicontinuous.
• Statement (c): This is not true in general. We use Proposition A.7 to give a counterex1
, n1 ]. Then each χFn is upper semicontinuous because
ample. For each n ∈ N, let Fn = [ n+1
Fn is closed in R. Consider the set
n
E= x∈R
∞
X
χFn (x) <
n=1
1o
.
2
If x ∈ Fn for some n ∈ N, then we have
1≤
∞
X
n=1
χFn (x) ≤ 2.
In other words, we have Fn * E for n = 1, 2, . . .. Since
(2.5)
∞
[
Fn = (0, 1],a we see that
n=1
(0, 1] * E.
a
If x ∈ (0, 1], then there exists a positive integer k such that
1
k+1
(2.6)
< x which implies that x ∈ F1 ∪ F2 ∪ · · · ∪ Fk .
19
2.1. Properties of Semicontinuity
However, if x ≤ 0 or x > 1, then x ∈
/ Fn for every n ∈ N which means that χFn (x) = 0,
i.e.,
(−∞, 0] ∪ (1, ∞) ⊆ E.
(2.7)
Hence, by combining the set relations (2.6) and (2.7), we conclude that
E = (−∞, 0] ∪ (1, ∞)
which is not open in R. By Definition 2.8,
∞
X
χFn is not upper semicontinuous.
n=1
• Statement (d): The set
Fn (α) = {x ∈ R | fn (x) > α}
is open for every real α. By applying part (b) repeatedly, we know that
N
X
fn is lower
n=1
semicontinuous for every positive integer N . For every x ∈ [0, ∞), let
f (x) = lim
N →∞
N
X
fn (x) =
∞
X
fn (x),
n=1
n=1
E(α) = {x ∈ R | f (x) > α},
N
o
n
X
fn (x) > α
FN (α) = x ∈ R
(2.8)
n=1
for real α and N ∈ N. We claim that
∞
[
E(α) =
FN (α)
(2.9)
N =1
for every real α. Similar to the proof of part (a), we suppose that α > 0, E(α) 6= R and
FN (α) 6= R.
Suppose that x ∈ E(α), i.e., f (x) > α. Then there exists a ǫ > 0 such that
f (x) > α + ǫ.
Since each fn is nonnegative,
N
nX
n=1
o
fn is an increasing sequence. By this and the definition
′
of f in (2.8), there exists a positive integer
N′
such that
N
X
fn (x) > α + ǫ which implies
n=1
that x ∈ FN ′ (α), i.e.,
E(α) ⊆
To prove the other side, if x ∈
′
N ′ , i.e.,
that
N
X
∞
[
N =1
∞
[
FN (α).
(2.10)
N =1
FN (α), then x ∈ FN ′ (α) for some positive integer
fn (x) > α. Again, the fact that
N
nX
n=1
n=1
fn
o
N′
f (x) ≥
X
n=1
fn (x) > α,
is an increasing sequence implies
Chapter 2. Positive Borel Measures
i.e., x ∈ E(α) and then
20
∞
[
N =1
FN (α) ⊆ E(α).
(2.11)
Hence the set relations (2.10) and (2.11) definitely imply the claim (2.9) is true and since
each FN (α) is open in R by Definition 2.8, E(α) is also open in R. Since α is arbitrary, f
is lower semicontinuous by Definition 2.8.
In the proof of Statements (a) and (b) above, we don’t use the property that f1 and f2 are
nonnegative. Therefore, they remain valid even if the word “nonnegative” is omitted. However,
Statement (c) cannot hold anymore if the word “nonnegative” is omitted. In fact, we consider
the sequence of real functions {fn } defined by
f1 = χ[−1,1]
and
fn = −χ[ 1 ,
1
]
n n−1
(n = 2, 3, . . .).
1
Since [−1, 1 and [ n1 , n−1
] are closed in R, we know that f1 , f2 , . . . are upper semicontinuous. It
is clear that f1 (0) = −1, so f1 is not a nonnegative function on R. By definition, we have
f=
∞
X
n=1
fn = χ[−1,1] −
∞
X
n=2
χ[ 1 ,
1
]
n n−1
= χ[−1,0] + χ(0,1] −
∞
X
χ[
n=1
1
,1]
n+1 n
.
(2.12)
By the inequalities (2.5), we see that

−2, if x = 21 , 13 , . . .;




∞

X
χ[ 1 , 1 ] (x) =
−
−1, if x ∈ (0, 1] \ { 21 , 13 , . . .};
n+1 n


n=1



0,
if x ≤ 0 or x > 1.
Thus we follow from the expressions (2.12) and (2.13) that

1 1

 −1, if x = 2 , 3 , . . .;



f (x) =
0,
if x ∈ (0, 1] \ { 12 , 13 , . . .} or x < −1 or x > 1;





1,
if x ∈ [−1, 0].
(2.13)
(2.14)
Therefore, the expression (2.14) of f gives
n
1o
E = x ∈ R f (x) >
= [−1, 0]
2
which is not open in R. By Definition 2.8, f is not upper semicontinuous. By Definition 2.8, a
function f is lower semicontinuous if and only if −f is upper semicontinuous. This observation
indicates that we can deduce a counterexample to Statement (d) from the counterexample
(2.12).
Finally, the truths of the Statements (a), (b) and (d) depend only on the range of f and
the fact that the union of any collection of open sets in a topological space X is open in X (see
the set equalities (2.2) and (2.9)). This completes the proof of the problem.
Problem 2.2
Rudin Chapter 2 Exercise 2.
21
2.1. Properties of Semicontinuity
Proof. We formulate and prove the general statement: Let X be a topological space, U an open
set in X containing the point x and f : X → C. Define
ϕ(x, U ) = sup{|f (s) − f (t)| | s, t ∈ U } and ϕ(x) = inf{ϕ(x, U ) | U is open, x ∈ U }. (2.15)
We claim that ϕ is upper semicontinuous, f is continuous at x ∈ X if and only if ϕ(x) = 0 and
the set of points of continuity of an arbitrary complex function is a Gδ .
• ϕ is upper semicontinuous. Let E = {x ∈ X | ϕ(x) < α}, where α ∈ R. We have to
show that E is open in X. If E = ∅, then there is nothing to prove. Thus we suppose
that E 6= ∅. In this case, we have p ∈ E so that
ϕ(p) < α.
This fact shows that there exists an open set U containing p andb
ϕ(p, U ) < α.
Pick q ∈ U \ {p}. Since U is open, there exists an open set V containing q such that
q ∈ V ⊆ U . Since p, q ∈ U , we know from the definition (2.15) that
ϕ(p, U ) = ϕ(q, U ).
(2.16)
Furthermore, we observe from the definition (2.15) that
ϕ(x, U ′ ) ≤ ϕ(x, U ) for every open sets U, U ′ with U ′ ⊆ U .
(2.17)
Now these facts (2.16) and (2.17) imply that ϕ(q, V ) ≤ ϕ(q, U ) = ϕ(p, U ) < α and then
ϕ(q) ≤ ϕ(q, V ) < α.
In other words, q ∈ E. Since q ∈ U \ {p} is arbitrary, we have shown that p ∈ U ⊆ E, i.e.,
E is open for every α ∈ R and hence ϕ is upper semicontinuous by Definition 2.8.
• f is continuous at x if and only if ϕ(x) = 0. Suppose that f is continuous at x. Recall
from the definition of continuity ([42, Theorem 18.1, p. 104] or [51, p. 9]) that for every
ǫ > 0, the neighborhood
n
ǫo
(2.18)
B(f (x), ǫ) = z ∈ C |f (x) − z| <
2
has a neighborhood Uǫ of x such that f (Uǫ ) ⊆ B(f (x), ǫ), i.e., f (y) ∈ B(f (x), ǫ) for all
y ∈ Uǫ . Now we take this Uǫ in the definition (2.15):
ϕ(x, Uǫ ) = sup{|f (s) − f (t)| | s, t ∈ Uǫ }
and it follows from the definition (2.18) thatc
ϕ(x, Uǫ ) ≤ ǫ.
(2.19)
By the definition (2.15), we have ϕ(x) ≤ ϕ(x, U ) for every open set U containing x.
Therefore, we deduce from this and the inequality (2.19) that ϕ(x) ≤ ǫ. Since ǫ is arbitrary,
we have the desired result that ϕ(x) = 0.
b
Otherwise, we have ϕ(p, U ) ≥ α for every open set U containing p and this implies that α ≤ ϕ(p), a
contradiction.
c
Note that |f (s) − f (t)| ≤ |f (s) − f (x)| + |f (x) − f (t)| < ǫ for every s, t ∈ Uǫ .
Chapter 2. Positive Borel Measures
22
Conversely, suppose that ϕ(x) = 0. Thus for every ǫ > 0, there exists a neighborhood
Uǫ of x such that ϕ(x, Uǫ ) < ǫ. By the definition (2.15), this implies that
|f (s) − f (t)| < ǫ
(2.20)
for every s, t ∈ Uǫ . In particular, if we take s = x to be fixed and t = y vary, then the
inequality (2.20) can be rewritten as |f (x) − f (y)| < ǫ for every y ∈ Uǫ . By the definition
([42, Theorem 18.1, p. 104] or [51, p. 9]), f is continuous at x.
• The set of points of continuity of an arbitrary complex function is a Gδ . By the
previous analysis, we establish that
G = {x ∈ X | f is continuous at x} = {x ∈ X | ϕ(x) = 0} =
∞ n
\
n=1
x ∈ X ϕ(x) <
1o
.
n
Since ϕ is upper semicontinuous, each set {x ∈ X | ϕ(x) < n1 } is open in X. By Definition
1.11, we see that G is actually a Gδ .
This completes the proof of the problem.
Problem 2.3
Rudin Chapter 2 Exercise 3.
Proof. The first part is proven in [63, Problem 4.20, pp. 73, 74]. The function in the question
can be used to prove Urysohn’s Lemma for any metric space X directly.
Lemma 2.1 (Urysohn’s Lemma)
Suppose that X is a metric space with metric ρ, A and B are disjoint nonempty
closed subsets of X. Then there exists a continuous function f : X → [0, 1] such
that 0 ≤ f (x) ≤ 1 for all x ∈ X, f (x) = 0 precisely on A and f (x) = 1 precisely
on B.
Proof of Lemma 2.1. This lemma is proven in [63, Problem 4.22, pp. 75, 76]. Readers
are recommended to read [42, Theorem 32.2, p. 202; Theorem 33.1, pp. 207 – 210]. This ends the proof of the problem.
Problem 2.4
Rudin Chapter 2 Exercise 4.
Proof.
(a) Recall from [51, Eqn. (2), p. 42] that
µ(E) = inf{µ(V ) | E ⊆ V and V is open}
(2.21)
which is defined for every subset E of X.d By the result [51, Eqn. (4), p. 42], we have
µ(E1 ∪ E2 ) ≤ µ(E1 ) + µ(E2 ).
d
Here we don’t require E to be a measurable set.
(2.22)
23
2.1. Properties of Semicontinuity
We need a lemma:
Lemma 2.2
For disjoint open sets V1 and V2 , we have
µ(V1 ∪ V2 ) = µ(V1 ) + µ(V2 ).
Proof of Lemma 2.2. The proof of the inequality µ(V1 ∪ V2 ) ≤ µ(V1 ) + µ(V2 ) can be
found in [51, STEP I, p. 42], so we only prove the other side. Suppose that g ∈ Cc (X)
and g ≺ V1 . Similarly, suppose that h ∈ Cc (X) and h ≺ V2 . By Definition 2.9, since
Cc (X) is a vector space, f = g + h ∈ Cc (X). Furthermore, we know from Definition
2.9(a) that
supp (f ) = supp (g + h) ⊆ supp (g) + supp (h) ⊆ V1 ∪ V2 .
(2.23)
By assumption, we have V1 ∩V2 = ∅ which means that g(x) = 0 on X \V1 and h(x) = 0
on X \ V2 . Thus we deduce from these facts that

g(x), if x ∈ V1 ;





f (x) =
h(x), if x ∈ V2 ;





0,
if x ∈ X \ (V1 ∪ V2 ).
In other words, we have
0 ≤ f (x) ≤ 1
(2.24)
on X. Therefore, we can conclude from the set relation (2.23) and the inequalities
(2.24) that
f ≺ V1 ∪ V2 .
(2.25)
Now, by Definition 2.1, the relation (2.25) and then [51, Eqn. (1), p.41], we obtain
that
Λ(g) + Λ(h) = Λ(g + h) = Λ(f ) ≤ µ(V1 ∪ V2 ).
(2.26)
We first fix the h in the inequality (2.26) and since it holds for every g ≺ V1 , the
definition of supremum gives
µ(V1 ) + Λ(h) ≤ µ(V1 ∪ V2 ).
(2.27)
Next, the inequality (2.27) holds for every h ≺ V2 , we establish from the definition of
supremum that
µ(V1 ) + µ(V2 ) ≤ µ(V1 ∪ V2 ).
Hence we have µ(V1 ) + µ(V2 ) = µ(V1 ∪ V2 ).
Let’s go back to the original proof. We want to compute µ(E1 ∪ E2 ). By the definition
(2.21), it needs to consider all open sets containing E1 ∪ E2 . Since V1 ⊆ V2 certainly
implies µ(V1 ) ≤ µ(V2 ), we can restrict our attention to any open set W such that
E1 ∪ E2 ⊆ W ⊆ V1 ∪ V2 .
Since V1 ∩ V2 = ∅, we have W = (W ∩ V1 ) ∪ (W ∩ V2 ), where (W ∩ V1 ) ∩ (W ∩ V2 ) = ∅.
Chapter 2. Positive Borel Measures
24
Furthermore, since W ∩ V1 and W ∩ V2 are open sets in X, it yields from Lemma 2.2 that
µ(W ) = µ(W ∩ V1 ) + µ(W ∩ V2 ) ≥ µ(E1 ) + µ(E2 ).
(2.28)
Since W is arbitrary, µ(E1 ) + µ(E2 ) is a lower bound of the set in the definition (2.21).
Hence we follow from this fact and the inequality (2.28) that
µ(E1 ∪ E2 ) ≥ µ(E1 ) + µ(E2 ).
(2.29)
By the inequalities (2.22) and (2.29), we have the desired result that
µ(E1 ∪ E2 ) = µ(E1 ) + µ(E2 ).
(b) Define E1 = E. By [51, STEP V, p. 44], there exists a compact set K1 and an open set
V1 such that
1
K1 ⊆ E1 ⊆ V1 and µ(V1 \ K1 ) < 2 .
2
By [51, STEP II, p. 43], we have K1 ∈ MF . Then we follow from [51, STEP VI, p. 44]
that E1 \ K1 ∈ MF . Define E2 = E1 \ K1 so that
E = E1 = E2 ∪ K1 ,
where µ(E2 ) = µ(E1 \ K1 ) ≤ µ(V1 \ K1 ) < 212 . By induction, we can show that there exists
a sequence of compact sets {Kn } and a sequence of open sets {Vn } such that
Kn ⊆ En ⊆ Vn ,
where En+1 = En \ Kn , En+1 ∈ MF and µ(En+1 ) <
the following relations:
E = En+1 ∪ K1 ∪ K2 ∪ · · · ∪ Kn
(2.30)
1
.
2n+1
Therefore, the set E satisfies
and µ(En+1 ) <
1
2n+1
(2.31)
for every positive integer n. By the set relations (2.30), we deduce that
Kn+1 ⊆ En+1 = En \ Kn = En−1 \ (Kn−1 ∪ Kn ) = E \ (K1 ∪ K2 ∪ · · · ∪ Kn )
(2.32)
for all n ∈ N. In other words, Kn+1 is disjoint with all K1 , K2 , . . . , Kn and thus {K1 , . . . , Kn }
is a disjoint collection of compact sets in X. Now, by letting n → ∞ in the relations (2.31),
we achieve that E is given by
E = N ∪ K1 ∪ K2 ∪ · · · ,
where {Kn } is a disjoint countable collection of compact sets in X and
N =E\
∞
[
Kn .
n=1
By the equalities in (2.32), N ⊆ En+1 holds for all n ∈ N and the inequality in (2.31)
shows that
1
µ(N ) = lim µ(En+1 ) = lim n+1 = 0.
n→∞
n→∞ 2
This completes the proof of the problem.
2.2. Problems on the Lebesgue Measure on R
25
2.2
Problems on the Lebesgue Measure on R
In Problems 2.5 to 2.8, m stands for the Lebesgue measure on R.
Problem 2.5
Rudin Chapter 2 Exercise 5.
Proof. This problem is proven in [49, Remarks 11.11(f), p. 309].
Problem 2.6
Rudin Chapter 2 Exercise 6.
Proof. Recall from [49, Theorem 2.47, p. 42] that a subset F of R is connected if and only if
(x, y) ⊆ F for every x, y ∈ F with x < y. Therefore, if F is a connected subset of a totally
disconnected set K ⊂ R and F consists of more than one point, then K must contain a segment,
a contradiction. By this observation, we know that a totally disconnected set K ⊂ R contains
no segment and this gives a hint to our construction: It is well-known that the Cantor set E is
compact and contains no segment. However, Problem 2.5 shows that m(E) = 0, so we have to
“modify” the construction of E so as to fit our problem.
Let K0 = [0, 1]. The idea of the construction is to remove the “middle
fact, we first remove the segment ( 38 , 58 ) of length 212 from K0 and let
h 3i h5 8i
K1 = 0,
∪ , .
8
8 8
5 7
27
Next, we remove the middle segments ( 32
, 32 ) and ( 25
32 , 32 ) each of length
component of K1 to get
1
24
1
th”
22n
segments. In
from each connected
h 5 i h 7 12 i h 20 25 i h 27 32 i
∪
∪
∪
.
,
,
,
K2 = 0,
32
32 32
32 32
32 32
Continuing in this way, we obtain a sequence of compact sets Kn such that
K1 ⊃ K2 ⊃ K3 ⊃ · · · .
Therefore, we have
n
n
X
X
1
1
k−1
m(Kn ) = 1 −
·2
=1−
.
2k
k+1
2
2
k=1
Define the set
K=
(2.33)
k=1
∞
\
Kn .
n=1
We know that K 6= ∅ because of the corollary [49, Theorem 2.36, p. 38]. Applying an argument
similar to that used in [49, §2.44, pp. 41, 42], we can show that K is a totally disconnected
compact set in R. Furthermore, since m(K1 ) is finite, Theorem 1.19(e) and the Lebesgue measure
(2.33) ensure that
m(K) = lim m(Kn ) = lim
n→∞
n→∞
n
X
1−
k=1
1 2k+1
=1−
∞
X
k=1
1
1
= > 0.
2k+1
2
Chapter 2. Positive Borel Measures
26
For the second assertion, let v be lower semicontinuous and v ≤ χK . Assume that there was
a p ∈ R such that v(p) > 0. Since v is lower semicontinuous, Definition 2.8 implies that the set
A = {x ∈ R | v(x) > 0} is open in R. By the assumption, we have p ∈ A, so there exists a δ > 0
such that
(p − δ, p + δ) ⊆ A.
(2.34)
Now we follow from the hypothesis v ≤ χK and the relation (2.34) that
χK (x) > 0
on (p − δ, p + δ), therefore we have (p − δ, p + δ) ∈ K which contradicts to the fact that K
contains no segment. Hence we have v ≤ 0 and this completes the proof of the problem.
Problem 2.7
Rudin Chapter 2 Exercise 7.
Proof. By the idea used in Problem 2.6, it is not hard to see that the construction of the compact
ǫ
sets Kn and K still work for removing the “middle 22n−1
th” segments. In this case, instead of
the Lebesgue measure (2.33), we have
m(Kn ) = 1 −
and so
n
X
k=1
ǫ
22k−1
k−1
·2
n
X
1
=1−ǫ
2k
(2.35)
k=1
m(K) = lim m(Kn ) = 1 − ǫ.
n→∞
Since K is closed in [0, 1], the complement
E = [0, 1] \ K
(2.36)
is definitely open in [0, 1].
Now it remains to show that E is dense in [0, 1]. We prove an equivalent definition of a dense
set: Let A ⊆ B ⊆ R. We say A is dense in B means that for every x, y ∈ B with x < y, we
can find z ∈ A such that x < z < y. In particular, suppose that x, y ∈ [0, 1] with x < y. If
(x, y) ∩ E = ∅, then the definition (2.36) says that (x, y) ⊆ K which is impossible. Therefore,
we have
(x, y) ∩ E 6= ∅
and this means that there exists z ∈ E such that x < z < y. Hence E is dense in [0, 1],
completing the proof of the problem.
Problem 2.8
Rudin Chapter 2 Exercise 8.
Proof. Let {rn } be the enumeration of all rationals in R. Let Fn be the segment given by
1
1 Fn = rn − 2n , rn + 2n ,
(2.37)
2
2
where n = 1, 2, . . .. Define the sets
Gn = Fn \ (Fn+1 ∪ Fn+2 ∪ · · · ) = Fn \
We divide the proof into several steps.
∞
[
k=1
Fn+k .
(2.38)
2.2. Problems on the Lebesgue Measure on R
27
• Step 1: m(Gn ) > 0. By the definition (2.37), we know that m(Fn+1 ) =
n = 1, 2, . . .. By the definition (2.38), for every n ∈ N, we have
Fn = Gn ∪
∞
[
m(Fn )
22
for every
Fn+k
k=1
so that the subadditive property of a measure (see the proof of Problem 1.11) implies that
m(Fn ) ≤ m(Gn ) +
∞
X
m(Fn+k ) = m(Gn ) + m(Fn )
k=1
∞
X
1
m(Fn )
= m(Gn ) +
.
22k
3
k=1
Therefore, we have
m(Gn ) ≥
2m(Fn )
>0
3
(2.39)
for every positive integer n.
n)
• Step 2: Existence of a Borel subset An ⊂ Fn with m(An ) = m(F
2 . It is easy to
1
. By Definition 2.19, we know thate
check from the definition (2.37) that m(Fn ) = 22n−1
Fn′ =
1 h
1 i
Fn − rn − 2n
= (0, 1).
m(Fn )
2
Thus, by the proof of Problem 2.7f , there exists a (totally disconnected) compact set
A′n ⊂ Fn′ such that m(A′n ) = 21 . Then Theorem 2.20(c) implies that the set An given by
An =
m(Fn )A′n
1 + rn − 2n
2
is a (totally disconnected) compact subset of Fn with measure
1 m(Fn )
′
m(An ) = m m(Fn )An + rn − 2n
. (2.40)
= m m(Fn )A′n = m(Fn )m(A′n ) =
2
2
Since An is compact, it is closed by the Heine–Borel theorem [49, Theorem 2.41, p. 40].
By Definition 1.11, it is also a Borel set, completing the proof of Step 2.
Now we can construct a Borel set with the desired properties. The construction is as
follows: For every positive integer n, we define
En = Gn ∩ An
and E =
∞
[
En .
(2.41)
n=1
By the definition (2.38), since each Fn is Borel, each Gn is also Borel. Thus each En and
their countable union E are also Borel sets. It remains to show that the E satisfies the
requirements of the problem.
• Step 3: 0 < m(E ∩ Fn ). We first show that m(En ∩ Fn ) > 0 for every positive integer n.
To prove this claim, we note from the definitions (2.38) and (2.41) that En ⊆ Gn ⊆ Fn , so
we have
m(En ∩ Fn ) = m(En ) = m(Gn ∩ An ).
Furthermore, since Gn ⊆ Fn and An ⊂ Fn (by Step 2), we have
m(Gn ∪ An ) ≤ m(Fn ).
e
f
It is just a translation and an enlargement.
Take ǫ = 12 in the measure (2.35).
(2.42)
Chapter 2. Positive Borel Measures
28
Combining the inequality (2.39) and the expression (2.40), we can reduce the inequality
(2.42) to
m(Gn ∪ An ) < m(Gn ) + m(An ).
(2.43)
Now we apply a property of additive function [49, Eqn. (7), p. 302] to the inequality
(2.43) to obtain
m(Gn ∪ An ) < m(Gn ∪ An ) + m(Gn ∩ An )
which means that m(Gn ∩ An ) > 0 for every positive integer n. Therefore, we follow from
this and the fact that En ∩ Fn ⊆ E ∩ Fn
0 < m(Gn ∩ An ) = m(En ∩ Fn ) ≤ m(E ∩ Fn )
for every positive integer n.
• Step 4: m(E ∩ Fn ) < m(Fn ). Fix an integer n. For m < n, we apply the identity
A \ B = B c ∩ A to the definition (2.38) to get
c
Gcm = [Fm \ (Fm+1 ∪ Fm+2 ∪ · · · )]c = (Fm+1 ∪ Fm+2 ∪ · · · ) ∪ Fm
which implies that Fn ⊆ Gcm for m = 1, 2, . . . , n − 1. Recall from the definition (2.41) that
Enc = Gcn ∪ Acn , so we must have
c
Fn ⊆ Em
for m = 1, 2, . . . , n − 1. In other words, it says that Fn ∩ Em = ∅ for m = 1, 2, . . . , n − 1.
Therefore, this fact implies
m(E ∩ Fn ) = m
∞
[
m=1
∞
∞
[
[
Am ∩ Fn .
Em ∩ Fn ≤ m
Em ∩ Fn = m
(2.44)
m=n
m=n
By the subadditive property of a measure again, we further deduce the inequality (2.44)
to
∞
∞
X
X
m(Am ).
(2.45)
m(Am ∩ Fn ) ≤
m(E ∩ Fn ) ≤
m=n
Recall the result (2.40) and the fact m(Fn+1 ) =
(2.45) that
m=n
m(Fn )
,
22
so we derive from the inequality
∞
∞
X
m(Fn ) X 1
2m(Fn )
m(Fn+m )
=
< m(Fn )
=
m(E ∩ Fn ) ≤
2m
2
2
2
3
m=0
m=0
for every positive integer n.
• Step 5: 0 < m(E ∩ I) < m(I) for every nonempty segment I. By Step 3 and Step
4, the inequalities
0 < m(E ∩ Fn ) < m(Fn )
(2.46)
hold for every positive integer n, where Fn and E are given by (2.37) and (2.41) respectively.
Lemma 2.3
Let I = (α, β) with α < β. Then there is a positive integer n0 such that Fn0 ⊆ I.
2.2. Problems on the Lebesgue Measure on R
29
γ+β
Proof of Lemma 2.3. Let γ = α+β
2 and δ = 2 . By the density of Q in R, there exists
a rational in (γ, δ). Let it be rn for some positive integer n. If Fn ⊆ I, then we are
done. Otherwise, we may find another rational in (rn , δ), namely rn+1 . In fact, this
process can be repeated m times, i.e., there is a rational rn+m in (rn+m−1 , δ). It is
clear that
1
1
1
(2.47)
rn+m + 2(n+m) ≤ δ + 2(n+m) < δ + 2m
2
2
2
and
1
1
1
rn+m − 2(n+m) > γ − 2(n+m) > γ − 2m .
(2.48)
2
2
2
Now we may pick the m large enough such that
1
<β−δ
22m
and
1
β−α
<
22m
2
simultaneously. Then it follows from the inequalities (2.47) and (2.48) that
rn+m +
1
22(n+m)
<β
(2.49)
and
α+β β−α
1
1
−
= α.
(2.50)
> γ − 2m >
2
2
2
22(n+m)
Hence the inequalities (2.49) and (2.50) together imply that Fn0 ⊆ I, where n0 = n+m,
completing the proof of the lemma.
rn+m −
We return to the proof of the problem. By Lemma 2.3, we have E ∩ Fn0 ⊆ E ∩ I and the
left-hand side of the inequalities (2.46) shows that
0 < m(E ∩ Fn0 ) ≤ m(E ∩ I).
(2.51)
Since I = Fn0 ∪ (I \ Fn0 ) and Fn0 ∩ (I \ Fn0 ) = ∅, we have
E ∩ I = (E ∩ Fn0 ) ∪ [E ∩ (I \ Fn0 )].
Hence, by applying the right-hand side of the inequalities (2.46) and Theorem 1.19(b)
twice, we obtain
m(E ∩ I) = m(E ∩ Fn0 ) + m E ∩ (I \ Fn0 ) < m(Fn0 ) + m(I \ Fn0 ) = m(I).
(2.52)
Now our desired result follows immediately if we combine the inequalities (2.51) and (2.52).
• Step 6: m(E) < ∞. By the subadditive property of the measure m, the expression (2.40)
and the definition (2.41), we get
m(E) ≤
∞
X
n=1
m(En ) ≤
∞
X
m(An ) =
n=1
This completes the proof of the problem.g
g
∞
∞
n=1
n=0
1X
m(F1 ) X 1
2m(F1 )
m(Fn ) =
=
< ∞.
2n
2
2
2
3
Instead of Borel sets of the real number line R, Rudin proved a similar result for measurable set A in [0, 1],
see [50]. Furthermore, there are two interesting results ([14] and [35]) related to this problem and some of its
applications can be found in [13], [18], [25], [39], [46] and [60].
Chapter 2. Positive Borel Measures
2.3
30
Integration of Sequences of Continuous Functions
Problem 2.9
Rudin Chapter 2 Exercise 9.
Proof. For every n ∈ N, we consider the function gn : [−1, 1] → [0, 1] defined by
gn (x) =

 0,

if x 6∈ [− n1 , n1 ];
(2.53)
1 − n|x|, if x ∈ [− n1 , n1 ].
It is clear that each gn is continuous on [−1, 1] and 0 ≤ gn ≤ 1 on [−1, 1]. The graph of gn is
shown in Figure 2.1 below.
Figure 2.1: The graph of gn on [−1, 1].
Next, we define gn,k : [0, 1] → [0, 1] by
k gn,k (x) = gn x −
,
2n
(2.54)
k
where x ∈ [0, 1] and k = 0, 1, . . . , 2n. It is easy to check that x − 2n
∈ [−1, 1] so that each gn,k
is well-defined by (2.54). We claim that if we define the sequence {f1 , f2 , . . .} to be
{g1,0 , g1,1 , g1,2 , g2,0 , g2,1 , g2,2 , g2,3 , g2,4 , . . .},
then {fn } satisfies the hypotheses of the problem.
To this end, we first note that since each gn,k is continuous on [0, 1] and 0 ≤ gn,k ≤ 1 on
[0, 1], each fn is continuous on [0, 1] and 0 ≤ fn ≤ 1. To see why
lim
Z
1
n→∞ 0
we have to check the behaviour of
Z
0
fn (x) dx = 0,
1
gn,k (x) dx
(2.55)
31
2.3. Integration of Sequences of Continuous Functions
for every positive integer n and k = 0, 1, . . . , 2n. In fact, we know from the definitions (2.53)
and (2.54) that
 1
1

2 + nx, if x ∈ [0, 2n );



1


 1 − n|x|, if x ∈ [0, n );
3
1
3
(2.56)
gn,0 (x) =
and gn,1 (x) =
2 − nx, if x ∈ [ 2n , 2n ];



0,
if x ∈ [ n1 , 1]



3
, 1].
0,
if x ∈ ( 2n
Similarly, we have
gn,2n−1 (x) =
and


 0,
if x ∈ [0, 2n−3
2n );

 1−n x−
gn,2n (x) =
2n−1
2n

 0,

(2.57)
, if x ∈ [ 2n−3
2n , 1]
if x ∈ [0, n−1
n );
(2.58)
1 + n(x − 1), if x ∈ [ n−1
n , 1].
Finally, for k = 2, 3, . . . , 2n − 2, we have

0,






1−n x−
gn,k (x) =






0,
if x ∈ [0, k−2
2n );
k
2n
k+2
, if x ∈ [ k−2
2n , 2n ];
(2.59)
if x ∈ ( k+2
2n , 1].
The graphs of these gn,k are shown as follows:
Figure 2.2: The graphs of gn,k on [0, 1].
In other words, it is clear from Figure 2.2 that {gn,0 , gn,1 . . . , gn,2n } is a family of tent functions
1
0
h
, 2n
, . . . , 2n
centered at { 2n
2n }. Now we have the following cases.
h
We use the fact that each gn,k is a part or the whole of an isosceles triangle with height 1 and base less than
or equal to n2 .
Chapter 2. Positive Borel Measures
• When k = 0, we have
Z
1
gn,0 (x) dx =
0
• When k = 1, we have
Z
1
gn,1 (x) dx =
Z
1
2n
0
0
• When 2 ≤ k ≤ 2n − 2, we have
Z 1
Z
32
1
gn (x) dx =
0
1
2
+ nx dx +
gn,k (x) dx =
• When k = 2n − 1, we have
Z
Z 1
gn,2n−1 (x) dx =
• When k = 2n, we have
Z
Z 1
gn,2n (x) dx =
0
1
2n−3
2n
Z
(1 − nx) dx =
3
2n
1
2n
1
.
2n
(2.60)
7
− nx dx =
.
2
8n
3
1 2
1
× ×1= .
2 n
n
1−n x−
1
1
1− n
1
n
0
0
0
Z
gn (x − 1) dx =
Z
(2.62)
2n − 1 7
.
dx =
2n
8n
1
1
1− n
(2.61)
[1 + n(x − 1)] dx =
(2.63)
1
.
2n
(2.64)
Combining the expressions (2.60) to (2.64), we may conclude that the limit (2.55) holds.
Now it remains to show that {fn (x)} converges for no x ∈ [0, 1]. If x = 0, then we obtain
from the definitions (2.56) and (2.57) that
gn,0 (0) = 1 and
gn,2n−1 (0) = 0.
In other words, we can find subsequences {fnk (0)} and {fnl (0)} such that fnk (0) → 1 and
fnl (0) → 0 as k → ∞ and l → ∞ respectively. By Definition 1.13, they imply that
lim sup fn (0) = 1 and
n→∞
lim inf fn (0) = 0.
(2.65)
n→∞
Similarly, if x = 1, then the definitions (2.56) and (2.58) show that
gn,0 (1) = 0 and gn,2n (1) = 1
which then imply the limits (2.65). If x ∈ (0, 1) ∩ Q, then we have x =
Thus it follows from the definitions (2.56) and (2.59) that
p
p
= 0 and gnq,2np
=1
gn,0
q
q
p
q
=
2p
2q ,
where 0 < p < q.
(2.66)
for n ≥ q. Therefore, the limits (2.65) also hold for x ∈ (0, 1) ∩ Q. For irrational x in (0, 1), the
Archimedean Property shows that there exists N ∈ N such that x > N1 . Thus it implies that
gn,0 (x) = 0
(2.67)
for all n ≥ N . By the density of rationals, there exists a sequence { pqnn } of rationals in (0, 1)
such that
pn
→x
qn
33
2.3. Integration of Sequences of Continuous Functions
as n → ∞, where 0 < pn < qn for all n ∈ N. Since we always have
pn h npn − 1 npn + 1 i
,
∈
,
qn
nqn
nqn
we derive from the second expression (2.66) that
p n
gnqn ,2npn
=1
qn
(2.68)
for all n ∈ N. Combining the two results (2.67) and (2.68), we have shown that the limits (2.65)
remain valid for irrationals x, completing the proof of the problem.
Problem 2.10
Rudin Chapter 2 Exercise 10.
Proof. Given that ǫ > 0. For each positive integer n, we define
En = {x ∈ [0, 1] | fk (x) > ǫ for some k ≥ n}.
(2.69)
It is clear that each En is bounded and
En ⊇ En+1 ⊇ · · · .
If En0 = ∅ for some positive integer n0 , then we know from the definition (2.69) that
fk (x) ≤ ǫ
for every x ∈ [0, 1] and every positive integer k ≥ n0 . Therefore it implies that
0≤
Z
0
1
fk (x) dx ≤ ǫ
for every positive integer k ≥ n0 . Since ǫ is arbitrary, we obtain from this that
Z 1
fn (x) dx = 0.
lim
n→∞ 0
Without loss of generality, we assume that En 6= ∅ for all n ∈ N. Let’s prove two properties of
En first:
• Property 1: En is open. Let p ∈ En . Then we have fk (p) > ǫ for some k ≥ n. Define
ǫ′ = 12 (fk (p) − ǫ) > 0. Since fk is continuous on [0, 1], it is continuous at p. Thus for this
particular ǫ′ , there is a δ > 0 such that
|fk (x) − fk (p)| < ǫ′
for all points x ∈ [0, 1] with |x − p| < δ. Now the inequality (2.70) implies that
fk (x) > fk (p) − ǫ′ =
fk (p) + ǫ
> ǫ.
2
In other words, we have (p − δ, p + δ) ⊆ En as desired.
(2.70)
Chapter 2. Positive Borel Measures
• Property 2:
∞
\
n=1
34
En = ∅. Otherwise, there was a p ∈ [0, 1] such that p ∈ En for all
n ∈ N. Therefore, for each positive integer n, we have
fk (p) > ǫ
(2.71)
for some k ≥ n. If n → ∞, then k → ∞ and the inequality (2.71) implies that
0 = lim fn (p) = lim fk (p) ≥ ǫ > 0,
n→∞
k→∞
a contradiction.
To finish our proof, we have to study the lengths of certain subsets of En . Let F be a finite
union of bounded (open or closed) intervals of [0, 1]. Then we may write
F =
m
[
[ak , bk ] ∪
k=1
s
[
(ar , br )
(2.72)
r=1
where 0 ≤ ak < bk ≤ 1 and 0 ≤ ar < br ≤ 1 for k = 1, 2, . . . , m and r = 1, 2, . . . , s. Define the
length of F , denoted by ℓ(F ), to be
ℓ(F ) =
m
X
k=1
m
s
s
X
X
X
(br − ar ).
ℓ (ar , br ) =
(bk − ak ) +
ℓ [ak , bk ] +
r=1
k=1
r=1
Given a nonempty finite union of bounded interval F expressed in the form (2.72) and assume
ǫ
that F has at least one open interval. In addition, we suppose that ℓ(F ) > ǫ. Let δ = 4s
. Then
the set
m
s
[
[
[ar + δ, br − δ]
G=
[ak , bk ] ∪
r=1
k=1
is clearly a nonempty finite union of bounded and closed intervals of [0, 1]. Now it is easy to see
that
m
s
X
X
ǫ
(br − ar − 2δ) = ℓ(F ) − 2sδ = ℓ(F ) −
ℓ(G) =
(bk − ak ) +
2
r=1
k=1
implying that
ℓ(G) > ℓ(F ) − ǫ.
Here we need a lemma about the sequence of bounded subsets (2.69):
Lemma 2.4
Let Sn = {F | F is a finite union of bounded intervals of En } for every positive integer n and
Ln = sup{ℓ(F ) | F ∈ Sn }.
(2.73)
Then we have
lim Ln = 0.
n→∞
35
2.3. Integration of Sequences of Continuous Functions
Proof of Lemma 2.4. By Property 1, we know that each Sn is nonempty. It is obvious
that {Ln } is a bounded decreasing sequence. By the Monotone Convergence Theorem
[49, Theorem 3.14, p. 55], {Ln } converges in R. Assume that this limit was nonzero.
• Step 1: The construction of a compact set Kn . There exists a δ > 0 such
that Ln ≥ δ for every positive integer n. By the definition (2.73), Ln − 2δn is not
an upper bound of {ℓ(F ) | F ∈ Sn }. In other words, there exists a Fn ∈ Sn such
that
ℓ(Fn ) > Ln − δ · 2−n
(2.74)
for every positive integer n. By the observation preceding Lemma 2.4, we may
n
\
Fk ⊆ Fn . Since each Fk
assume further that each Fn is closed. Let Kn =
k=1
is closed in R, Kn is closed in R too. By the Heine-Borel Theorem, the set Kn
must be compact. Furthermore, we have Kn ⊇ Kn+1 for each n = 1, 2, . . ..
• Step 2: Each Kn is nonempty. There exists a F ∈ Sn such that
ℓ(F ) ≥ δ
(2.75)
for each n ∈ N. Otherwise, there is a N ∈ N such that ℓ(F ) < δ for all F ∈ SN
which means that δ is an upper bound of LN , but this is a contradiction. Now
we are going to show that if Kn = ∅, then it is impossible to have the inequality
(2.75). This contradiction bases on the following two facts:
– Fact 1: Suppose that G is a finite union of bounded intervals of En \ Fn
for every n ∈ N, where Fn are those sets considered in Step 1. Then we
have G ∩ Fn = ∅ and G ∪ Fn ∈ Sn so that ℓ(G) + ℓ(Fn ) = ℓ(G ∪ Fn ) ≤ Ln .
This and the inequality (2.74) give, for every n ∈ N,
ℓ(G) < δ · 2−n .
(2.76)
– Fact 2: Suppose that F is a finite union of bounded intervals of En \ Kn
for every n ∈ N. By De Morgan’s law (see [42, p. 11]), we have
(F \ F1 ) ∪ · · · ∪ (F \ Fn ) = F \ (F1 ∩ · · · ∩ Fn ) = F \ Kn = F.
(2.77)
Note that F \ Fk is also a finite union of bounded intervals of Ek (and hence
of Ek \ Fk ) for k = 1, 2, . . . , n, so it follows from the inequality (2.76) that
ℓ(F \ Fk ) < δ · 2−k
(2.78)
for k = 1, 2, . . . , n. Hence it follows from the expression (2.77) and the
inequality (2.78) that
ℓ(F ) < δ.
(2.79)
If Km = ∅ for some m ∈ N, then the F considered in Fact 2 is a subset of Em ,
but the inequality (2.79) definitely contradicts the inequality (2.75).
• Step 3: A contradiction to Property 2. By Step 1 and Step 2, we
∞
\
Kn 6= ∅. Since Fk ⊆ Ek for
deduce from [49, Theorem 2.36, p. 38] that
n=1
k = 1, 2, . . . , n, we must have Kn ⊆ En . However, these two facts contradict
Property 2.
Hence the limit must be 0 which completes the proof of the lemma.
Chapter 2. Positive Borel Measures
36
We return to the proof of the problem. By Lemma 2.4, for ǫ > 0, there is a positive integer
N such that for n ≥ N , we have
ℓ(F ) ≤ Ln < ǫ,
(2.80)
where F is any finite union of bounded intervals of En . Now one may think that we can derive
Z 1
fn (x) dx < 2ǫ
0
for n ≥ N directly from the uniform boundedness of fn , the definition (2.69) and the inequality
(2.80). However, it fails because we have no way to estimate the integral
Z
fn (x) dx.
En \F
To overcome this problem, we play the trick that since fn is Riemann integrable on [0, 1], the
integral must be equal to its lower Riemann integral, see [2, §1.17, p. 74]. Thus we have to find
nZ 1
o
sup
sn (x) dx 0 ≤ sn (x) ≤ fn (x) and sn is a step function ,
0
where the sup is taken over all step functions sn below fn on [0, 1].
For every n ≥ N , we define
En = {x ∈ [0, 1] | sn (x) > ǫ} and Fn = [0, 1] \ En .
Since sn is a step function, it is clear that En and Fn are finite unions of bounded intervals.
Furthermore, since 0 ≤ sn (x) ≤ fn (x), we have En ⊆ En and En ∈ Sn . Then we deduce from the
inequality (2.80) that
ℓ(En ) < ǫ
for all n ≥ N . Hence we obtain from this and repeated uses of [49, Theorem 6.12(c), p. 128]
that for all n ≥ N , we have
Z
Z
Z
Z
Z 1
ǫ dx ≤ ℓ(En ) + ǫ < 2ǫ.
dx +
sn (x) dx ≤
sn (x) dx +
sn (x) dx =
0≤
0
En
Fn
En
Fn
Since ǫ is arbitrary, we have shown that
lim
Z
n→∞ 0
1
fn (x) dx = 0.
This completes the proof of the problem.
Remark 2.1
There are many mathematicians who have provided different proofs of Problem 2.10. See,
for examples, [37], [38] and [53].
2.4
Problems on Borel Measures and Lebesgue Measures
Problem 2.11
Rudin Chapter 2 Exercise 11.
37
2.4. Problems on Borel Measures and Lebesgue Measures
Proof. We follow the given hint. Suppose that Ω is the family of all compact subsets Kα of X
with µ(Kα ) = 1. Since X is compact and µ(X) = 1, Ω is not empty. Now we define
\
K=
Kα .
(2.81)
Kα ∈Ω
Since X is Hausdorff, each Kα is closed in X by the corollaries following Theorem 2.5. Thus K
is closed in X and then K is Borel (see Definition 1.11), i.e., K ∈ M. Since K ⊆ X, Theorem
2.4 says that K is also compact.
Suppose that K ⊆ V and V is open in X. Then V c is closed in X and so it is compact in X
by Theorem 2.4. As we have mentioned in the previous paragraph that each Kα is closed in X,
so each Kαc is open in X. By the definition (2.81) and the fact that K ⊆ V , we have
[
V c ⊆ Kc =
Kαc .
Kα ∈Ω
In other words, {Kαc } forms an open cover of the compact set V c . Thus we have
V c ⊆ Kαc 1 ∪ Kαc 2 ∪ · · · ∪ Kαc n
(2.82)
for some positive integer n. Since µ(Kαm ) = µ(X) = 1 for m = 1, 2, . . . , n, we have µ(Kαc m ) = 0
for m = 1, 2, . . . , n. Therefore, it follows from these and the set relation (2.82) that
c
µ(V ) ≤
n
X
µ(Kαc m ) = 0,
m=1
i.e., µ(V c ) = 0 or equivalent µ(V ) = 1. Since µ is regular and K ∈ M, Definition 2.15 implies
that
µ(K) = inf{µ(V ) | K ⊆ V and V is open} = 1
which proves the first assertion.
For the second assertion, let H be a proper compact subset of K, i.e., H ⊂ K. Assume
that µ(H) = 1. Then it means that H ∈ Ω and the definition (2.81) shows that K ⊆ H, a
contradiction. Hence we must have µ(H) < 1. This completes the proof of the problem.
Problem 2.12
Rudin Chapter 2 Exercise 12.
Proof. Let B be the σ-algebra of all Borel sets in R and K be a nonempty compact subset of R.
We have to show that there exists a measure µ on B such that K is the smallest closed subset
with the property µ(K c ) = 0.
Since every compact set K has a countable base (see [49, Exercise 25, Chap. 2, p. 45]), there
exists a countable sequence F = {x1 , x2 , . . .} ⊆ K such that
K = F.
Next we define µ : F → [0, ∞] by
µ(xn ) =
1
2n
and for any E ∈ B, we define µ : B → [0, ∞] by
∞
X
1
χ
(E),
µ(E) =
2n {xn }
n=1
(2.83)
Chapter 2. Positive Borel Measures
38
where χ{xn } is the characteristic function of the set {xn }. Suppose that {Ei } is a disjoint
countable collections of members of B and
E=
∞
[
Ei .
i=1
If xn ∈ Ei for some i, then xn ∈
/ Ej for all j 6= i. In other words, each element of F belongs to
i
at most one element of {Ei }. Thus it follows from the definition (2.83) that
µ(E) =
X 1
χ
(E),
2n {xn }
n
(2.84)
where the summation runs through all n such that xn ∈ Ei for some i. Since the series
∞
X
1
2n
n=1
converges absolutely, its rearrangement converges to the same limit (see [49, Theorem 3.55, p.
78]). Therefore, we can rewrite the expression (2.84) as
∞ X
X
1
χ
(Ei ),
µ(E) =
2ik {xik }
(2.85)
i=1 ik
where the inner summation runs through all ik such that xik ∈ Ei . It may happen that the
set F ∩ Ei is finite for some i = 1, 2, . . .. Since the inner summation in the expression (2.85) is
exactly µ(Ei ), we deduce from the expression (2.85) that
µ
∞
[
i=1
Ei = µ(E) =
∞
X
µ(Ei ).
i=1
By Definitions 1.18(a) and 2.15, µ is a Borel measure on R.
By the definition (2.83), it is easy to see that for every E ∈ B, we have
µ(E) > 0 if and only if
K ∩ E 6= ∅
(2.86)
so that µ(K c ) = 0. Assume that there was a proper closed subset H ⊂ K in X such that
µ(H c ) = 0. Since H ⊂ K, we have K c ⊂ H c . If H c ∩ K = ∅, then H c ⊆ K c , a contradiction.
Thus H c ∩ K 6= ∅. Since H c is open in X, we have H c ∈ B. Therefore, we can conclude from
the condition (2.86) that
µ(H c ) > 0
which contradicts to our hypothesis. In other words, K is the smallest closed subset in R such
that µ(K c ) = 0, or equivalently,
K = supp (µ).
This completes the analysis of the problem.
Problem 2.13
Rudin Chapter 2 Exercise 13.
i
However, Ei may contain more than one element of F .
39
2.4. Problems on Borel Measures and Lebesgue Measures
Proof. It is clear that the point set {0} is a compact subset of R. Assume f : R → R was
a continuous function such that {0} = supp (f ). Let V = (−∞, 0) ∪ (0, ∞). We know from
Definition 2.9 that supp (f ) = f −1 (V ), so
{0} = f −1 (V )
(2.87)
so that f −1 (V ) 6= ∅. Since V is open in R and f is continuous on R, f −1 (V ) is open in R by
Definition 1.2(c). Thus, if p ∈ f −1 (V ), then there exists a δ > 0 such that
(p − δ, p + δ) ⊆ f −1 (V ).
(2.88)
Thus we deduce from the set relations (2.87) and (2.88) that
[p − δ, p + δ] ⊆ f −1 (V ) = {0},
a contradiction. Hence there is no continuous function f such that {0} = supp (f ).
For the second assertion, we suppose that K is a compact subset of R. We claim that K is
the support of a continuous function if and only if K is the closure of an open set V in X. If K
is the support of a continuous function f : R → R, then Definition 2.9 shows that
K = {x ∈ R | f (x) 6= 0} = f −1 (−∞, 0) ∪ (0, ∞) = f −1 (−∞, 0) ∪ f −1 (0, ∞) .
Since f is continuous on R, f −1 (−∞, 0) and f −1 (0, ∞) are open in R. Thus this proves one
direction. Conversely, if we have
K=V
(2.89)
for some open set V , then we consider F = V c which is closed in R. Define ρF : R → R by
ρF (x) = inf{|x − y| | y ∈ F }.
By Problem 2.3, we know that ρF is uniformly continuous on R and ρF (x) = 0 if and only if
x ∈ F . Therefore, we must have
ρF (x) > 0
if and only if
x ∈ V.
(2.90)
Applying the relation (2.90) to the expression (2.89), we have
K = {x ∈ R | ρF (x) 6= 0},
i.e., K is the support of the continuous function ρF .
We note that the third assertion is not valid in other topological spaces. We consider the
three-element set X = {a, b, c}. By [42, Example 1, p. 76], we see that {∅, {b}, X} is a topology
of X. Let K = {a, b}. Then it is easy to check from the definition that K is compact. Now
if K = V for some open set V , then K must be closed in X (see [42, p. 95]). However, since
X \ K = {c} which is not open in X, K is not closed in X. This gives a counter-example to
explain that the description “K is the closure of an open set V in X” cannot guarantee that K
is the support of a continuous function in an arbitrary topological space and hence we complete
the proof of the problem.
Problem 2.14
Rudin Chapter 2 Exercise 14.
Chapter 2. Positive Borel Measures
40
Proof. Let f : Rk → R. We follow the proof of Theorem 2.24 (Lusin’s Theorem). Suppose that
0 ≤ f < 1. Attach a sequence {sn } of simple measurable functions to f , as in the proof of
Theorem 1.17 (The Simple Function Approximation Theorem). Put t1 = s1 and tn = sn − sn−1
for n = 2, 3, . . .. Then 2n tn is the characteristic function of a measurable set Tn ⊆ Rk and
f (x) =
∞
X
tn (x) =
∞
X
2−n χTn (x)
(2.91)
n=1
n=1
on Rk . By Theorem 2.20(b), there exist An , Bn ∈ B in Rk such that
An ⊆ Tn ⊆ Bn
and
m(Bn \ An ) = 0
(2.92)
for every positive integer n.j
We define g : Rk → R and h : Rk → R by
g(x) =
∞
X
−n
2
χAn (x)
and h(x) =
∞
X
2−n χBn (x).
(2.93)
n=1
n=1
We claim that g and h are Borel measurable functions on Rk . It is easy to check that

∅, if α ≤ 0;





Acn , if 0 < α ≤ 1;
{x ∈ Rk | χAn (x) < α} =





R, if α > 1.
Thus each χAn is a Borel measurable function by Definition 1.11. Similarly, each χBn is also a
Borel measurable function. Since the partial sums gn and hn of g and h are just simple functions
and An , Bn ∈ B, they are Borel measurable functions by the comment following Definition 1.16.
Furthermore, for each n = 1, 2, . . ., we have
|2−n χAn (x)| ≤ 2−n
and
|2−n χBn (x)| ≤ 2−n
∞
X
2−n converges, it follows from Weierstrass M -test (see [49, Theorem 7.10, p.
Since
Xn=1
X
147]) that
2−n χAn and
2−n χBn converge uniformly on Rk to g and h respectively. By
the Corollary (a) following Theorem 1.14, g and h are Borel measurable. This proves our claim.
on
Rk .
By the measure (2.92), we know that g = h a.e. [m] on Rk . Obviously, if F ⊆ E, then we
must have χF (x) ≤ χE (x). Therefore, we observe from the set relations (2.92) that
χAn (x) ≤ χTn (x) ≤ χBn (x)
(2.94)
on Rk . Hence we apply the inequalities (2.94) to functions (2.91) and (2.93), we obtain
g(x) ≤ f (x) ≤ h(x)
(2.95)
on Rk .
Next, if 0 ≤ f < M for some positive constant M on Rk , then the above argument can
f
be applied to M
so that the inequalities (2.95) hold with g and h replaced by M g and M h
respectively.
j
In fact, each An is an Fσ and each Bn is a Gδ , see Definition 1.11.
41
2.4. Problems on Borel Measures and Lebesgue Measures
In the general case, we consider the measurable sets EN = {x ∈ Rk | − N ≤ f (x) ≤ N },
where N ∈ N. Now we have χEN f → f as N → ∞ on Rk . In fact, the above argument shows
that we can find Boreal functions gN and hN such that gN (x) = hN (x) a.e. [m] on Rk and
gN (x) ≤ χEN f (x) ≤ hN (x)
on Rk . By Theorem 1.14, we see that the functions g and h defined by
g(x) = lim sup gN (x)
and
h(x) = lim sup hN (x)
N →∞
(2.96)
N →∞
are Borel measurable functions on Rk . Since gN (x) = hN (x) a.e. [m] on Rk , we have g(x) = h(x)
a.e. [m] on Rk . Furthermore, we note that
g(x) = lim sup gN (x) ≤ lim sup χEN f (x) ≤ lim sup hN (x) = h(x)
N →∞
N →∞
(2.97)
N →∞
on Rk . Since
lim χEN f (x) = lim sup χEN f (x) = f (x)
N →∞
N →∞
on Rk , the inequalities (2.95) follow immediately from the inequalities (2.96) and the fact (2.97).
Hence our desired results also hold in this general case, completing the proof of the problem. Problem 2.15
Rudin Chapter 2 Exercise 15.
Proof. For each positive integer n, we define fn : [0, ∞) → [0, ∞] by
 x n x

e 2 , if x ∈ [0, n];
 1−
n
fn (x) =


0,
if x > n.
It is easy to see that each fn is continuous on [0, ∞) and so Definition 1.11 shows that it is Borel
measurable. Fix a x ∈ R. Then the Archimedean Property ensures the existence of a positive
integer N such that N ≥ x. By this fact and the fact that (1 + nx )n → ex as n → ∞, we have
x
x
x n x
e 2 = e−x · e 2 = e− 2 .
lim fn (x) = lim 1 −
n→∞
n→∞
n
x
Since e− 2 ∈ L1 (R) and
x
|fn (x)| ≤ e− 2
(2.98)
for all n = 1, 2, . . . and x ∈ [0, ∞), it follows from Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) that
Z ∞
Z ∞
x
(2.99)
lim
fn dx =
e− 2 dx = 2.
n→∞ 0
0
By the inequality (2.98), we get
Z
Z n
Z ∞
fn dx =
fn dx −
0
0
∞
n
which implies that
lim
Z
n→∞ 0
∞
fn dx ≤
Z
∞
n
fn dx = lim
Z
|fn | dx ≤
n→∞ 0
Z
∞
x
n
e− 2 dx = 2e− 2
n
n
fn dx.
(2.100)
Chapter 2. Positive Borel Measures
42
Combining the results (2.99) and (2.100), we obtain
Z n
fn dx = 2.
lim
n→∞ 0
For the second integral, we define gn : [0, ∞) → [0, ∞] by
 x n −2x

e , if x ∈ [0, n];
1
+


n
gn (x) =
x


 2 ,
if x > n.
e2x
Therefore each gn is continuous on [0, ∞). Furthermore, by using a similar argument as above,
we have for x ∈ R,
x n −2x
e
= ex · e−2x = e−x .
lim gn (x) = lim 1 +
n→∞
n→∞
n
Since e−x ∈ L1 (R) and |gn (x)| ≤ e−x for all n = 1, 2, . . . and x ∈ [0, ∞), it follows from Theorem
1.34 (Lebesgue’s Dominated Convergence Theorem) that
Z ∞
Z ∞
e−x dx = 1.
gn dx =
lim
n→∞ 0
0
Now we apply the same trick as obtaining the result (2.100), we are able to show that
Z n
gn dx = 1.
lim
n→∞ 0
We have completed the proof of the problem.
Problem 2.16
Rudin Chapter 2 Exercise 16.
Proof. Let T : Rk → Rk be linear, Y = T (Rk ) with dim Y = r < k. Let B = {b1 , . . . , br } be a
basis of Y . Since B is a finite set of linearly independent vectors of Rk , it can be enlarged to a
basis {b1 , . . . , br , br+1 , . . . , bk } for Rk .k Define the linear transformation T : Rk → Rk by
T (α1 b1 + · · · + αk bk ) = (α1 , . . . , αk ).
Then it is easy to see that T is one-to-one and so det T 6= 0. Furthermore, we have
T (Y ) = {(α1 , . . . , αr , 0, . . . , 0) | α1 , . . . , αr ∈ R}
so that
T (Y ) =
∞
[
Wn ,
(2.101)
n=1
where Wn = {(ξ1 , . . . , ξr , 0, . . . , 0) | −n ≤ ξi ≤ n and i = 1, 2, . . . , r}. By the definition, we have
Wn ⊆ Wn+1
k
See, for example, [23, Theorem 30.19, p. 279]. Sometimes, this result is called the Basis Extension
Theorem.
43
2.5. Problems on Regularity of Borel Measures
for every n ∈ N. It is trivial to see from Theorem 2.20(a) that
m(Wn ) = vol (Wn ) = 0
(2.102)
for each n = 1, 2, . . .. In addition, since Rk is a Borel set, Y is also a Borel set. Hence we follow
from this fact, the expressions (2.101), (2.102) and Theorem 1.19(d) that
m T (Y ) = lim m(Wn ) = lim vol (Wn ) = 0.
(2.103)
n→∞
n→∞
When we read the proof of Theorem 2.20(e) and §2.23 carefully, we see that the validity of
the formula
m T (E) = | det T |m(E)
(2.104)
when T is an one-to-one map of Rk onto Rk is independent of whether T (Rk ) is a subspace of
lower dimension or not. As a consequence, we
may apply the formula (2.104) here. Hence we
can conclude from m(Y ) = | det T |−1 m T (Y ) and the result (2.103) that m(Y ) = 0, completing
the proof of the problem.
2.5
Problems on Regularity of Borel Measures
Problem 2.17
Rudin Chapter 2 Exercise 17.
Proof. Let d be the mentioned distance in the problem, p = (x1 , y1 ) and q = (x2 , y2 ). We are
going to prove the assertions one by one.
• X is a metric space with metric d. We check the definition of a metric space, see [49,
Definition 2.15, p. 30]:
– If p = q, then d(p, q) = |y1 − y2 | = 0. Otherwise, we have
|y1 − y2 |,
if x1 = x2 and y1 6= y2 ;
d(p, q) =
1 + |y1 − y2 |, if x1 6= x2 ,
6= 0.
– It is clear that d(p, q) = d(q, p) holds because |y1 − y2 | = |y2 − y1 |.
– Let r = (x3 , y3 ). Suppose that r 6= p and r 6= q. Otherwise, the triangle inequality is
trivial. There are two cases:
∗ Case (i): x1 = x2 . Then we have d(p, q) = |y1 − y2 | and
|y1 − y3 |,
if x3 = x1 and y3 6= y1 ;
d(p, r) =
1 + |y1 − y3 |, if x3 6= x1 .
(2.105)
Similarly, we have
d(q, r) =
|y2 − y3 |,
if x3 = x2 and y3 6= y2 ;
1 + |y2 − y3 |, if x3 6= x2 .
(2.106)
Since |y1 − y2 | ≤ |y1 − y3 | + |y3 − y2 |, any combination of the distances (2.105)
and (2.106) imply that
d(p, q) ≤ d(p, r) + d(r, q).
(2.107)
Chapter 2. Positive Borel Measures
44
∗ Case (ii): x1 6= x2 . In this case, we have d(p, q) = 1 + |y1 − y2 |. If x3 = x1 ,
then x3 6= x2 so that d(p, r) = |y1 − y3 | and d(q, r) = 1 + |y2 − y3 |. The situation
for x3 = x1 is similar. If x3 6= x1 and x3 6= x2 , then we see from the distances
(2.105) and (2.106) again that d(p, r) = 1 + |y1 − y3 | and d(q, r) = 1 + |y2 − y3 |.
Hence the triangle inequality (2.107) remains true.
By the above analysis, we conclude that d is a metric.
• X is locally compact. Let X = (R2 , d) and p = (x, y), q = (u, v) ∈ R2 . Fix p first. By
the definition, d(p, q) < 1 implies that x = u and |y − v| < 1. Thus the neighborhood
B(p, 1) of p is given by
B(p, 1) = {q ∈ X | d(p, q) < 1}
= {q ∈ X | x = u and |y − v| < 1}
= {x} × (y − 1, y + 1).
(2.108)
Geometrically, B(p, 1) is the vertical line segment with half length less than 1 and (x, y)
as its midpoint. In addition, this line segment is open in R with the usual metric. Furthermore, it is clear from the definition (2.108) that
B(p, 1) = {x} × [y − 1, y + 1].
(2.109)
We want to show that B(p, 1) is compact with respect to the metric d. To this end, let
{Vα } be an open cover of B(p, 1), i.e.,
[
B(p, 1) ⊆
Vα .
α
Let a ∈ B(p, 1), where a = (s, t) with s = x and t ∈ [y − 1, y + 1]. Then (x, t) ∈ Vα for
some α. Since Vα is open in X, there exists a δt ∈ (0, 1) such that
B(a, δt ) = {q = (u, v) | u = s = x and |v − t| < δt } = {x} × (t − δt , t + δt ) ⊆ Vα (2.110)
so that
B(p, 1) ⊆
[
a∈B(p,1)
a=(s,t)
B(a, δt ) ⊆
[
Vα .
(2.111)
α
Applying the expressions (2.109) and (2.110) to the set relation (2.111), we see that
[
[
B(p, 1) = {x} × [y − 1, y + 1] ⊆
{x} × (t − δt , t + δt ) ⊆
Vα
(2.112)
t∈[y−1,y+1]
α
which means that {Ut }, where Ut = (t − δt , t + δt ) with t ∈ [y − 1, y + 1], is an open cover
of [y − 1, y + 1]. Since [y − 1, y + 1] is compact with respect to the usual metric | · | by the
Heine-Borel Theorem, there are finitely many indices t1 , . . . , tk such that
[y − 1, y + 1] ⊆ (t1 − δt1 , t1 + δt1 ) ∪ (t2 − δt2 , t2 + δt2 ) ∪ · · · ∪ (tk − δtk , tk + δtk ).
Suppose that (x, ti ) ∈ Vαi for i = 1, 2, . . . , k. Hence we follow immediately from the set
relation (2.112) that
B(p, 1) ⊆ Vt1 ∪ Vt2 ∪ · · · ∪ Vtk .
In other words, the closure B(p, 1) is compact with respect to the metric d and since p is
an arbitrary point in X, Definition 2.3(f) shows that X is locally compact.l
l
We remark that we cannot use the Heine-Borel Theorem directly to the closure (2.109) because we are working
in the space X with metric d, not R2 with the usual metric.
45
2.5. Problems on Regularity of Borel Measures
• Construction of a positive linear functional. Let f ∈ Cc (X), i.e., f : (R2 , d) → C is
a continuous function and supp (f ) is compact. First of all, we notice that the set
E = {x ∈ R | f (x, y) 6= 0 for some y}
is finite. To see this, let K = supp (f ) and 0 < δ < 1. Since
[
K⊆
B(p, δ),
p∈K
where B(p, δ) = {x} × (y − δ, y + δ). Now the compactness of K ensures that there exists
some positive integer n such that
K⊆
n
[
B(pi , δ) =
n
[
{q ∈ X | u = xi and |yi − v| < δ}
i=1
i=1
which forces E = {x1 , x2 , . . . , xn }, as claimed.
Next, we claim that the mapping Λ : Cc (X) → R defined by
Λ(f ) =
n Z
X
j=1
∞
f (xj , y) dy
−∞
is a positive linear functional. To this end, let f, g ∈ Cc (X). Since Cc (X) is a vector space,
we have αf + βg ∈ Cc (X) for any α, β ∈ C. Furthermore, let f (xi , y) 6= 0 and g(xj , y) 6= 0
for at least one y, where 1 ≤ i ≤ n and n + 1 ≤ j ≤ m, and f (zk , y)g(zk , y) 6= 0 for at least
one y, where m + 1 ≤ k ≤ r. By the definition, we have
f (xi , y) = g(xj , y) = 0,
where n + 1 ≤ i ≤ m, 1 ≤ j ≤ n and all y, so we get
Λ(αf + βg) =
r Z
X
s=1
"
=α
∞
n Z
X
s=1
+β
"
[αf (xs , y) + βg(xs , y)] dy
−∞
∞
f (xs , y) dy +
−∞
m Z
X
∞
s=n+1 −∞
= αΛ(f ) + βΛ(g).
Z
r
X
∞
f (zs , y) dy
s=m+1 −∞
Z ∞
r
X
g(xs , y) dy +
s=m+1 −∞
#
g(zs , y) dy
#
By Definition 2.1, Λ is a linear functional on Cc (X). If f ≥ 0, then Λ(f ) ≥ 0. Thus Λ is
positive. This proves the claim.
• The measures of the x-axis and its compact subsets. In order to apply Theorem
2.14 (The Riesz Representation Theorem), we have to show that X is Hausdorff. Given
p, q ∈ X and p 6= q. Let p = (x, y) and q = (u, v). There are two cases:
– Case (i): x = u but y 6= v. Then we define δ = |y − x| and it obtains from the
definition (2.108) that
δ h
δ
δ
δ
δ i h
δ i
∩ B q,
= {x} × y − , y +
∩ {x} × v − , v +
= ∅.
B p,
2
2
2
2
2
2
Chapter 2. Positive Borel Measures
46
– Case (ii): x 6= u but y = v. Then the geometric property of the neighborhoods
B(p, 1) and B(q, 1) ensures that their intersection must be empty.
In conclusion, we have shown that X is Hausdorff.
By Theorem 2.14 (The Riesz Representation Theorem), there exists a unique positive
measure µ on M associated with Λ. Let
E = {(x, 0) | x ∈ R}.
Let δ ∈ (0, 1) and
Uδ =
[
[
B(p, δ) =
x∈R
p∈E
{x} × (−δ, δ).
Then Uδ is clearly an open set in X and E ⊆ Uδ . Take (x1 , 0), . . . , (xn , 0) ∈ E and
construct
K = B (x1 , 0), δ ∪ B (x2 , 0), δ ∪ · · · ∪ B (xn , 0), δ
= {x1 } × [−δ, δ] ∪ {x2 } × [−δ, δ] ∪ · · · ∪ {xn } × [−δ, δ] .
Since each B (xi , 0), δ is compact, K is also compact and K ⊆ Uδ . By Theorem 2.12
(Urysohn’s Lemma), there exists an f ∈ Cc (X) such that
K ≺ f ≺ Uδ .
Since µ(Uδ ) = sup{Λ(f ) | f ≺ Uδ } (see [51, Eqn. (1), p. 41]), we have
µ(Uδ ) ≥ Λ(f ) =
m Z
X
j=1
∞
−∞
f (x′j , y) dy,
(2.113)
where x′1 , . . . , x′m are those values of x for which f (x, y) 6= 0 for at least one y and for
some positive integer m. Since K ≺ f , we have f (xi , y) = 1 for all i = 1, 2, . . . , n and
y ∈ (−δ, δ). Thus we have {x1 , x2 , . . . , xn } ⊆ {x′1 , x′2 , . . . , x′m } and then the inequality
(2.113) reduces to
n Z δ
X
dy = nδ
µ(Uδ ) ≥
i=1
−δ
so that µ(Uδ ) → ∞ as n → ∞. If V is an open set containing E, then V must contain the
open set Uδ for some small δ and thus µ(V ) = ∞ for every open set V containing E. By
Theorem 2.14(c), we have
µ(E) = inf{µ(V ) | E ⊆ V , V is open} = ∞.
Next, let K ⊆ E be compact. Choose δ ∈ (0, 1). Since
[
K⊆W =
B (a, 0), δ ,
(2.114)
(a,0)∈K
the compactness of K implies that
K⊆
m
[
i=1
B((ai , 0), δ) =
m
[
({ai } × (−δ, δ))
i=1
for some positive integer m. In other words, we know from the definition (2.114) that
K = {(a1 , 0), (a2 , 0), . . . , (am , 0)}.
47
2.5. Problems on Regularity of Borel Measures
Besides, since W is open in X, Theorem 2.12 (Urysohn’s Lemma) again ensures there is a
g ∈ Cc (X) such that
K ≺ g ≺ W.
Recall from [51, Eqn. (7), p. 43] that
µ(K) = inf{Λ(f ) | K ≺ f } ≤ Λ(g) =
m Z
X
i=1
∞
g(ai , y) dy = 2mδ.
(2.115)
−∞
Since m is fixed and δ is arbitrary, it follows from the inequality (2.115) that µ(K) = 0.
This completes the proof of the problem.
Problem 2.18
Rudin Chapter 2 Exercise 18.
Proof. Recall that X has an order relation ‘<’ and every nonempty subset A of X has a
first/smallest element a, i.e., a ≤ x for all x ∈ A.m If ω0 = min X (ω0 is a smallest element in
X), then we have
X = [ω0 , ω1 ].
Next, we characteristic the topology on X as follows:
The order topology of X: For α ∈ X, we define
Pα = {x ∈ X | x < α} = [ω0 , α)
and
Sα = {x ∈ X | α < x} = (α, ω1 ].
A subset E of X is open in X if it is in one of the following classes
[ω0 , α),
(β, ω1 ],
[ω0 , α) ∩ (β, ω1 ] = (β, α)
or
a union of such sets.
(2.116)
See [42, §14, pp. 84 – 86].
Now we prove the assertions one by one.
(a) X is a compact Hausdorff space. Suppose that V is an open cover of [ω0 , ω1 ]. Let
α0 = ω1 . Then there is an open set V0 ∈ V such that α0 ∈ V0 . If V0 = X, then X is
compact. Therefore, in the following discussion, we may assume that V0 6= X.
Since V0 is open in X, it follows from the uncountability of X and the order topology
(2.116) that there exists an α1 ∈ X such that (α1 , α0 ] ⊆ V0 . We note that α1 < α0 . Now
this α1 can be chosen such that α1 ∈
/ V0 . Otherwise, α1 ∈ V0 and we may choose another
α2 ∈ X such that (α2 , α1 ] ⊆ V0 . Here we have α2 < α1 . Similarly, this α2 can be chosen
such that α2 ∈
/ V0 . If not, then we continue this process in the same manner, but this must
stop in a finite number of steps. Otherwise, we can find an infinite decreasing sequence
α0 > α1 > α2 > · · · .
(2.117)
in X, which is a contradiction to the hint. Therefore, we may assume that
α1 ∈
/ V0
m
See [42, p. 63].
and
(α1 , α0 ] ⊆ V0 .
(2.118)
Chapter 2. Positive Borel Measures
48
Since V is a cover of [ω0 , ω1 ], there exists a V1 ∈ V containing α1 . By the above argument,
we may find an α2 ∈ X such that
α2 ∈
/ V1
and
(α2 , α1 ] ⊆ V1 .
(2.119)
If α2 ∈ V0 , then since [α2 , α0 ] ⊆ V0 and α2 < α1 < α0 , we have α1 ∈ V0 which contradicts
the result (2.118). Thus α2 ∈
/ V0 and then we deduce from this fact, the results (2.118)
and (2.119) that
α2 ∈
/ V0 ∪ V1 and (α2 , α0 ] ⊆ V0 ∪ V1 .
Inductively, we can find αn+1 ∈ X such that
αn+1 ∈
/ V0 ∪ V1 ∪ · · · ∪ Vn
and
(αn+1 , α0 ] ⊆ V0 ∪ V1 ∪ · · · ∪ Vn .
If this process continues infinitely, then we can find an infinite decreasing sequence (2.117)
again, a contradiction. Therefore, the sequence must stop eventually at the N th step and
it must stop at αn+1 = ω0 for all n ≥ N , i.e.,
ω0 ∈
/ V0 ∪ V1 ∪ · · · ∪ VN
and
(ω0 , α0 ] ⊆ V0 ∪ V1 ∪ · · · ∪ VN .
Since ω0 ∈ V for some V ∈ V, we conclude that
[ω0 , ω1 ] = [ω0 , α0 ] ⊆ V0 ∪ V1 ∪ · · · ∪ VN ∪ V.
In other words, X is compact.
Next, we prove that X is Hausdorff and we need the help of the following lemma:
Lemma 2.5
Define the mapping γ : [ω0 , ω1 ] → [ω0 , ω1 ] by

 min(Sα ), if α ∈ [ω0 , ω1 );
γ(α) =

ω1 ,
if α = ω1 .
Then we have X \ Sα = Pγ(α) = [ω0 , α].
Proof of Lemma 2.5. For every α ∈ [ω0 , ω1 ), Sα 6= ∅ so that the mapping γ is welldefined. By the definition, we have
Sα = [γ(α), ω1 ]
with α < γ(α). Furthermore, there is no element between α and γ(α). Otherwise,
γ(α) is not a smallest element of Sα anymore. Hence we deduce from this that
X \ Sα = [ω0 , α] = [ω0 , γ(α)) = Pγ(α) .
If α = ω1 , then Sω1 = ∅ and Pγ(ω1 ) = [ω0 , ω1 ] = X so that X \ Sω1 = Pγ(ω1 ) in this
case, completing the proof of the lemma.
Let α, β ∈ X and we may assume that α < β. Then we see that α ∈ Pγ(α) = [ω0 , α] and
β ∈ Sα . Thus Pγ(α) and Sα are neighborhoods of α and β respectively. By Lemma 2.5,
they are disjoint and this shows that X is Hausdorff, as required.
49
2.5. Problems on Regularity of Borel Measures
(b) X \ {ω1 } is open but not σ-compact. Let X \ {ω1 } = [ω0 , ω1 ) and α ∈ [ω0 , ω1 ). As we
have proven in part (a) that Pγ(α) = [ω0 , γ(α)) is a neighborhood of α. Hence X \ {ω1 } is
open in X.
Assume that X \ {ω1 } = [ω0 , ω1 ) was σ-compact. Let K be a compact subset of [ω0 , ω1 ).
We know that the family
{Px = [ω0 , x) | x < ω1 }
is an open cover of K so that there are finitely many indices x1 , x2 , . . . , xk such that
K ⊆ Px1 ∪ Px1 ∪ · · · ∪ Pxk .
Without loss of generality, we may assume that x1 < x2 < · · · < xk . Thus we have
K ⊆ [ω0 , xk ) ⊂ [ω0 , xk ].
Since xk < ω1 , our hypothesis shows that K is at most countable. Since X \ {ω1 } is
σ-compact, it is a countable union of compact sets so that X \ {ω1 } is countable. However,
we have
X = X \ {ω1 } ∪ {ω1 }
which means that X is countable. Evidently, this result contradicts our hypothesis. Hence
X \ ω1 is not σ-compact.
(c) Every f ∈ C(X) is constant on Sα for some α 6= ω1 . Suppose that x = f (ω1 ) and
B(x, n1 ) = {z ∈ C | |z − x| < n1 }, where n
on X
is a positive integer. Since f is continuous
1
1
−1
−1
B(x, n ) is an open subset of X and ω1 ∈ f
B(x, n1 ) .
and B(x, n ) is open in C, f
By the topology (2.116), there exists an α1 ∈ X \ {ω1 } such that
1 .
(2.120)
(α1 , ω1 ] ⊆ f −1 B x,
n
By Lemma 2.5, we note that (α1 , ω1 ] is uncountable, so a sequence {αn } exists such that
α1 < α2 < · · · < ω1 .
(2.121)
Now the results (2.120) and (2.121) together give
1 Sαn ⊆ Sαn+1 ⊆ f −1 B x,
n
for all n ∈ N. Before we proceed further, we need the following result:n
Lemma 2.6
Every well-ordered set A has the least upper bound property.
Proof of Lemma 2.6. Let B be a nonempty subset of A having an upper bound in
A. Therefore, the set U of upper bounds of B is nonempty. Since U ⊆ A and A is
well-ordered, U has a least element, completing the proof of Lemma 2.6.
By the relation (2.121), we see that the nonempty subset {αn } of X is bounded above
by ω1 . Therefore, it follows from Lemma 2.6 that sup{αn } exists in X. Call it α. It is
n∈N
clear that α ≤ ω1 and
n
See [42, Exercise 1, p. 66]
1 Sα ⊆ Sαn ⊆ f −1 B x,
n
(2.122)
Chapter 2. Positive Borel Measures
50
for all n = 1, 2, . . .. In fact, it is impossible to have α = ω1 because the set
Sω1 = {x ∈ Y | ω1 < x}
is uncountable, where Y is the well-ordered set as described in the “Construction” in
the question. As ω1 = α, we have Sω1 = Sα ⊆ Sαn so that Sαn is uncountable too, but
it contradicts to the fact that αn is a predecessor of ω1 . Hence we obtain α < ω1 and we
deduce from the set relations (2.122) that f (Sα ) ⊆ B(x, n1 ) for all n ∈ N and this means
that
∞
1
\
B x,
f (Sα ) ⊆
= {x}.
n
n=1
Now this is exactly our desired result.
(d) The intersection of {Kn } of uncountable compact subsets of X is also uncountable compact. We first prove a lemma which indicates a relationship between the cardinality of a compact set K and the topological property of ω1 in K.
Lemma 2.7
Suppose that K is a nonempty compact subset of X. Then K is uncountable if
and only if ω1 is a limit point of K.
Proof of Lemma 2.7. Let K be uncountable. By the proof of part (b), we know that
K cannot lie inside [ω0 , α] for every predecessor α < ω1 . Otherwise, K will be at most
countable, a contradiction. Hence this implies that
K ∩ (α, ω1 ] 6= ∅
for every α < ω1 . By the definition (see [42, p. 97]), it follows that ω1 is a limit point
of K.
Conversely, suppose that ω1 is a limit point of K. By [42, Theorem 17.9, p. 99], K
must be infinite. Assume that K was countable. Then K has the following representation
K = {αn },
where α1 ≤ α2 ≤ · · · . Since K is compact and X is Hausdorff, K is closed in X by
Corollary (a) following Theorem 2.5. Thus it must be ω1 ∈ K, i.e., every neighborhood
Sα of ω1 satisfies
K ∩ Sα \ {ω1 } 6= ∅.
Let αN ∈ K ∩ Sα \ {ω1 } for some positive integer N , i.e., α < αN < ω1 . Recall that
{αn } is increasing, so we must have
αn ∈ Sα
for all n ≥ N . By the definition (see [42, p. 98]), {αn } converges to ω1 . However,
by an argument similar to the proof of part (c), this fact will imply the contradiction
that Sαn is uncountable. Hence K is uncountable and we complete the proof of the
lemma.
Let’s go back to the proof of part (d). Denote the intersection of {Kn } by K. Since
X is Hausdorff, each Kn is closed in X. Since K ⊆ Kn , Theorem 2.4 implies that K is
51
2.5. Problems on Regularity of Borel Measures
compact. For each n ∈ N, let Kn′ = K1 ∩ K2 ∩ · · · ∩ Kn . It is easy to check that
n
\
Ki′
=
n
\
(K1 ∩ · · · ∩ Ki ) =
i=1
i=1
n
\
Ki ,
i=1
so we may assume that
K1 ⊇ K2 ⊇ · · · .
(2.123)
Therefore, any finite subcollection of {Kn } is nonempty and Theorem 2.6 shows that
K 6= ∅. By Lemma 2.7, ω1 is a limit point of every Kn . Then, using [42, Theorem 17.9, p.
99] again, it means that every neighborhood (α, ω1 ] of ω1 contains infinitely many points
of every Kn . Combining this fact and the sequence (2.123), every neighborhood (α, ω1 ] of
ω1 must contain infinitely many points of the compact set K. Hence ω1 is a limit point of
K and we finally conclude from Lemma 2.7 that K is uncountable.
(e) M is a σ-algebra containing all Borel sets in X. Suppose that
M = {E ⊆ X | either E ∪ {ω1 } or E c ∪ {ω1 } contains an uncountable compact set}.
– M is a σ-algebra. We check Definition 1.3(a). In fact, it is obvious that X ∈ M
because X is itself compact uncountable. Let E ∈ M. Then either E ∪ {ω1 } or
E c ∪ {ω1 } contains an uncountable compact set. Since (E c )c = E, it must be true
that E c ∈ M.
∞
[
En . If there exists an
To verify Definition 1.3(a)(iii), let En ∈ M and E =
n=1
n0 ∈ N such that En0 ∪ {ω1 } contains an uncountable compact set K, then
K ⊆ En0 ∪ {ω1 } ⊆
∞
[
n=1
En ∪ {ω1 } = E ∪ {ω1 }
so that E ∈ M. Otherwise, all En ∪ {ω1 } contain no uncountable compact sets. This
forces that every Enc ∪ {ω1 } contains an uncountable compact set Kn . We note that
c
E ∪ {ω1 } =
∞
\
Enc
n=1
∪ {ω1 } =
∞
\
n=1
Enc ∪ {ω1 } ,
so it is true that
K=
∞
\
n=1
Kn ⊆
∞
\
n=1
Enc ∪ {ω1 } = E c ∪ {ω1 }.
Now the result of part (d) illustrates that K is uncountable compact, thus E c ∈ M
which implies that E ∈ M.
– M contains all Borel sets in X. If α < ω1 , then (α, ω1 ) 6= ∅. Let β ∈ (α, ω1 ).
Since X = Pβ ∪ [β, ω1 ] and Pβ is countable, [β, ω1 ] must be uncountable. Since
[β, ω1 ] = X \Pβ , [β, ω1 ] is closed in X and Theorem 2.4 verifies that [β, ω1 ] is compact.
Furthermore, we note that
[β, ω1 ] ⊆ Sα ,
(2.124)
so these facts show that Sα ∈ M.
Similarly, for every α > ω0 , we have X \ Pα = [α, ω1 ] and then X \ Pα ∈ M. Since
M is a σ-algebra, M also contains Pα . Hence M contains the order topology τ of X
which proves that B(X) ⊆ M as desired.
Chapter 2. Positive Borel Measures
52
(f) λ is a measure on M but not regular. Suppose that λ : M → [0, ∞] is defined by

 1, if E ∪ {ω1 } contains an uncountable compact set;
λ(E) =
(2.125)

0, if E c ∪ {ω1 } contains an uncountable compact set.
If E ∈ M, then we know from the definition of M in part (e) that it is impossible for
both E ∪ {ω1 } and E c ∪ {ω1 } containing uncountable compact sets, so this function is
well-defined.
– λ is a measure on M. Suppose that {Ei } is a disjoint countable collection of
members of M. We claim that at most one Ei ∪ {ω1 } contains an uncountable
compact set. To see this, assume that both E1 ∪ {ω1 } and E2 ∪ {ω1 } contained
uncountable compact sets K1 and K2 respectively. Since E1 ∩ E2 = ∅, we have
K1 ∩ K2 ⊆ E1 ∪ {ω1 } ∩ E2 ∪ {ω1 }
= E1 ∩ {ω1 } ∪ E2 ∩ {ω1 } ∪ {ω1 }
= {ω1 }.
(2.126)
However, part (d) tells us that K1 ∩ K2 is uncountable so that the result (2.126) is
impossible. This proves the claim.
By the previous analysis, if E1 ∪ {ω1 } is the only set containing an uncountable
compact set K, then we get
∞
∞
[
[
Ei ∪ {ω1 } =
Ei ∪ {ω1 },
K⊆
i=1
i=1
so we follow from the definition (2.125) that
∞
[
λ
Ei = 1.
i=1
On the other hand, we know from the facts λ(E1 ) = 1 and λ(Ei ) = 0 for i = 2, 3, . . .
that
∞
X
λ(Ei ) = 1.
i=1
Therefore, we have
∞
∞
X
[
λ(Ei )
λ
Ei =
(2.127)
i=1
i=1
in this case. Similarly, if there is no Ei ∪ {ω1 } contains an uncountable compact set,
then each Eic ∪ {ω1 } contains an uncountable compact set Ki so that
λ(Ei ) = 0
for every i = 1, 2, . . .. Furthermore, we have
∞
∞
∞
∞
c
[
\
\
\
Ei ∪ {ω1 }
Eic ∪ {ω1 } =
Eic ∪ {ω1 } =
Ki ⊆
K=
i=1
i=1
i=1
i=1
and part (d) ensures that K is uncountable compact. Therefore, we obtain
∞
[
Ei = 0
λ
i=1
and then the formula (2.127) still holds in this case. Finally, since λ(X) = 1 < ∞.
By Definition 1.18(a), λ is a positive measure.
53
2.5. Problems on Regularity of Borel Measures
– λ is not regular. For every α < ω1 , we know from the set relation (2.124) that
Sα ∪ {ω1 } must contain an uncountable compact set, so the definition (2.125) gives
that λ(Sα ) = 1. If E = {ω1 }, then E c ∪ {ω1 } = {ω1 }c ∪ {ω1 } = X so that {ω1 } ∈ M.
Therefore, we derive from the definition (2.125) that
λ({ω1 }) = 0.
However, these facts give
λ({ω1 }) 6= inf{λ(Sα ) | {ω1 } ⊆ Sα }.
By Definition 2.15, λ is not regular.
(g) The validity of the integral. Let f ∈ Cc (X). By part (c), there exists an α0 < ω1 such
that f (x) = f (ω1 ) on Sα0 . Recall from part (e) that Sα0 ∈ M. By the set relation (2.124),
every Sα ∪ {ω1 } contains an uncountable compact set and thus, in particular, λ(Sαc 0 ) = 0.
By this, we gain
Z
f dλ =
X
Z
f dλ +
c
Sα
0
Z
f dλ =
Sα0
Z
Sα0
f dλ = f (ω1 )λ(Sα0 ) = f (ω1 )
which is the required result.
(h) The regular µ associates with the linear functional in part (f ). It is clear that
Λ : Cc (X) → C defined by
Λ(f ) = f (ω1 )
(2.128)
is a linear functional on Cc (X). By Theorem 2.14 (The Riesz Representation Theorem)
and Theorem 2.17(b), we have a unique regular positive Borel measure µ on X. By the
result (2.128), we know that
Z
f dµ = f (ω1 ).
X
By [51, Eqn. (1), p. 41], we have µ(V ) = sup{Λ(f ) | f ≺ V } = f (ω1 ) for every open set V
in X. Thus this implies that
µ(E) = inf{µ(V ) | E ⊆ V and V is open} = f (ω1 )
for every E ∈ M.
We have completed the proof of the problem.
Remark 2.2
The measure considered in Problem 2.18 is called the Dieudonné’s measure, see [10,
Exampl 7.1.3, pp. 68, 69] and [17].
Problem 2.19
Rudin Chapter 2 Exercise 19.
Chapter 2. Positive Borel Measures
54
Proof. Suppose that X is a compact metric space with metric ρ and Λ is a positive linear
functional on Cc (X), the space of all continuous complex functions on X with compact support.
To begin with the construction of the class µ, we use Λ to define a set function µ∗ on every open
set V in X by
µ∗ (V ) = sup{Λ(f ) | f ≺ V }
= sup{Λ(f ) | f ∈ Cc (X), 0 ≤ f ≤ 1 on X and supp (f ) ⊆ V }
(2.129)
and for every subset E ⊆ X that
µ∗ (E) = inf{µ∗ (V ) | V is open in X and E ⊆ V }.
(2.130)
Now we are going to present the proof by quoting several facts (some are with proofs and some
are not). In fact, the idea of the following proof is stimulated by Feldman’s online article [20].
• Fact 1: The set function µ∗ is an outer measure. We check the definition [47, p.
346].
– Since supp (f ) ⊆ ∅ if and only if supp (f ) = ∅, we get from Definition 2.9 that f ≡ 0
on X which implies that
µ∗ (∅) = sup{Λf | f ∈ Cc (X), 0 ≤ f ≤ 1 on X and supp (f ) ⊆ ∅} = Λ(0) = 0.
– Suppose that E, F ⊆ X and E ⊆ F . Since any open set V containing F must also
contain E, we obtain from the definition (2.130) that
µ∗ (E) = inf{µ∗ (W ) | W is open in X and E ⊆ W }
≤ inf{µ∗ (V ) | V is open in X and F ⊆ V }
= µ∗ (F ).
– The proof of the subadditivity
µ∗
∞
[
i=1
∞
X
µ∗ (Ei )
Ei ≤
(2.131)
i=1
follows exactly the same as the proof of Step I.
Hence µ∗ is an outer measure on X.
• Fact 2: µ∗ (K) < ∞ for every compact set K ⊆ X. Since X is open in X, we follow
from the definition (2.129) that µ∗ (X) ≤ Λ(1). By Fact 1, we have µ∗ (K) ≤ µ∗ (X) ≤ Λ(1)
for every compact set K ⊆ X.
• Fact 3: A topological result in metric spaces. Now we need the following topological
result about metric spaces which will be used in Fact 4:
Lemma 2.8
Let X be a metric space with metric ρ and E a nonempty proper subset of X.
Then the set V = {x ∈ X | ρE (x) < ǫ} is open in X, where ǫ > 0 and ρE (x) is
defined in Problem 2.3.
55
2.5. Problems on Regularity of Borel Measures
Proof of Lemma 2.8. If V = ∅, then there is nothing to prove. Thus we assume that
V 6= ∅ and pick x0 ∈ V so that ρE (x0 ) < ǫ. Then there exists a δ > 0 such that
ǫ − δ > 0 and ρE (x0 ) < ǫ − δ. We consider the ball
B(x0 , δ) = {x ∈ X | ρ(x, x0 ) < δ}
and we claim that B(x0 , δ) ⊆ V . To see this, let x ∈ B(x0 , δ). For every y ∈ E, we
have
ρ(x, y) ≤ ρ(x, x0 ) + ρ(x0 , y) < δ + ρ(x0 , y)
and therefore we obtain
ρE (x) = inf{ρ(x, y) | y ∈ E}
≤ inf{ρ(x, x0 ) + ρ(x0 , y) | y ∈ E}
< δ + inf{ρ(x0 , y) | y ∈ E}
= δ + ρE (x0 )
< ǫ.
In other words, x ∈ V . Since x is arbitrary, B(x0 , δ) ⊆ V and then V is open in X,
proving Lemma 2.8.
• Fact 4: If V is open in X, then V is measurable with respect to µ∗ . We recall
from [47, p. 347] that V ⊆ X is measurable with respect to the outer measure µ∗ if for
every subset E of X, we have
µ∗ (E) = µ∗ (E ∩ V ) + µ∗ (E ∩ V c ).
(2.132)
It suffices to show the inequality
µ∗ (E) ≥ µ∗ (E ∩ V ) + µ∗ (E ∩ V c ) − ǫ
(2.133)
holds for every ǫ > 0 because the other side follows directly from the subadditivity (2.131).
By the definition (2.130), there is an open set E ′ such that E ⊆ E ′ and
ǫ
µ∗ (E) ≥ µ∗ (E ′ ) − .
2
(2.134)
By the property in Fact 1, we have
µ∗ (E ∩ V ) ≤ µ∗ (E ′ ∩ V ) and µ∗ (E ∩ V c ) ≤ µ∗ (E ′ ∩ V c ).
(2.135)
Thus it is easy to see from the inequalities (2.134) and (2.135) that one can obtain the
inequality (2.133) if we can show the following result holds
ǫ
µ∗ (E ′ ) ≥ µ∗ (E ′ ∩ V ) + µ∗ (E ′ ∩ V c ) − .
2
(2.136)
To this end, we first find bounds of µ∗ (E ′ ∩ V ) and µ∗ (E ′ ∩ V c ). Since E ′ ∩ V is open
in X, the definition (2.129) implies the existence of a continuous function f1 : X → [0, 1]
such that supp (f1 ) ⊆ E ′ ∩ V and
ǫ
µ∗ (E ′ ∩ V ) ≤ Λ(f1 ) + .
8
(2.137)
Let ǫ ∈ 0, 8Λ(1) and δ > 0 be a constant such that δ ≤ ǫ[8Λ(1) − ǫ]−1 . Then we have
ǫ
δ
Λ(1) ≤
1+δ
8
and F1 =
f1
.
1+δ
(2.138)
Chapter 2. Positive Borel Measures
56
Recall the fact that the positivity of Λ implies the monotonicity of Λ. In particular,
Λ(f ) ≤ Λ(1) for every f ≺ E ′ ∩ V . Thus we apply the results (2.138) and the monotonicity
of Λ to the inequality (2.137) to get
µ∗ (E ′ ∩ V ) ≤
1
δ
ǫ
δ
ǫ
ǫ
Λ(f1 )+
Λ(f1 )+ ≤ Λ(F1 )+
Λ(1)+ = Λ(F1 )+ . (2.139)
1+δ
1+δ
8
1+δ
8
4
We replace E by the closed set (E ′ ∩V )c in Lemma 2.8 to get the open set U . Given δ > 0.
Since f1 is uniformly continuous on X, there exists a η > 0 such that |f1 (x) − f1 (y)| ≤ δ
for all x, y ∈ X with ρ(x, y) < η. Recall that f1 vanishes on the closed set (E ′ ∩ V )c , so if
we take y ∈ (E ′ ∩ V )c , then we have
f1 (x) ≤ δ
for all x ∈ X with ρ(x, y) < η. Since ρ(E ′ ∩V )c (x) < η, Lemma 2.8 ensures that the set
U = {x ∈ X | ρ(E ′ ∩V )c (x) < η}
is open in X. In other words, we have established from Fact 3 that there exists an open
set U such that
(2.140)
U ⊃ (E ′ ∩ V )c ⊇ V c and f1 (x) ≤ δ on U .
Since E ′ ∩ U is open in X, the definition (2.129) again shows that we can find a continuous
function f2 : X → [0, 1] such that supp (f2 ) ⊆ E ′ ∩ U and furthermore, by using similar
argument as in the proof of the inequality (2.139), we obtain that
µ∗ (E ′ ∩ V c ) ≤ µ∗ (E ′ ∩ U ) ≤ Λ(f2 ) +
where F2 =
ǫ
ǫ
≤ Λ(F2 ) + ,
4
4
(2.141)
f2
1+δ .
Now we have found the bounds (2.139) and (2.141) of µ∗ (E ′ ∩ V ) and µ∗ (E ′ ∩ V c )
respectively. Next, we want to show that F3 ∈ Cc (X), where F3 = F1 + F2 . To this
1
end, we recall the facts that F1 , F2 : X → [0, 1+δ
] because 0 ≤ f1 ≤ 1 and 0 ≤ f2 ≤ 1.
Furthermore, we note from the set relation (2.140) that
V ∩ U 6= ∅.
Therefore, we have the following facts:
– Since supp (f1 ) ⊆ E ′ ∩ V , supp (F1 ) ⊆ E ′ ∩ V , i.e., F1 (x) = 0 on (E ′ ∩ V )c . Therefore,
we conclude that
1
F3 = F2 ≤
≤1
1+δ
on (E ′ ∩ V )c . Similarly, since supp (F2 ) ⊆ E ′ ∩ U , we have F2 (x) = 0 on (E ′ ∩ U )c
and then
1
≤1
F3 = F1 ≤
1+δ
on (E ′ ∩ U )c .
– On (E ′ ∩ V ) ∩ U , we immediately follow from the inequality in (2.140) that f1 (x) ≤ δ
on (E ′ ∩ V ) ∩ U ⊆ U . It is clear that f2 (x) ≤ 1 on (E ′ ∩ V ) ∩ U , so these facts imply
that
f2
1
δ
f1
+
≤
+
= 1.
F3 = F1 + F2 =
1+δ 1+δ
1+δ 1+δ
– On (E ′ ∩ V ) \ U , we have F1 ≤ 1 and F2 = 0 so that F3 ≤ 1.
57
2.5. Problems on Regularity of Borel Measures
– On U \ (E ′ ∩ V ), we have F1 = 0 and F2 ≤ 1 so that F3 ≤ 1.
To have a better understanding of the above facts, we can draw some pictures. For
examples, Figure 2.3(a) shows the sets V, E ′ and E ′ ∩ V , the shaded blue part in Figure
2.3(b) indicates the closed set (E ′ ∩ V )c and the part inside the circle in Figure 2.4 is the
set (E ′ ∩ V )c \ U .
(a) The sets V, E ′ and E ′ ∩ V .
(b) The set (E ′ ∩ V )c .
Figure 2.3: The pictures of V, E ′ , E ′ ∩ V and (E ′ ∩ V )c .
Figure 2.4: The set (E ′ ∩ V )c \ U .
Thus we have shown that 0 ≤ F3 (x) ≤ 1 on X. Next, since F1 and F2 are continuous
on X, F3 is also continuous on X. By the definition, we know that x0 ∈ supp (F3 ) if and
only if x0 ∈ supp (F1 ) ⊆ E ′ ∩ V or x0 ∈ supp (F2 ) ⊆ E ′ ∩ U . In addition, we note that
(E ′ ∩ V ) ∪ (E ′ ∩ U ) ⊆ E ′ . On the other hand, if y ∈ E ′ but y ∈
/ E ′ ∩ V , then we have
′
c
y ∈ (E ∩ V ) and we follow from the set relation (2.140) that y ∈ U . Thus this implies
that y ∈ E ′ ∩ U and so
E ′ = (E ′ ∩ V ) ∪ (E ′ ∩ U ).
In other words, it gives x0 ∈ supp (F3 ) if and only if x0 ∈ supp (F1 ) ∪ supp (F2 ) ⊆ E ′ , i.e.,
supp (F3 ) ⊆ E ′ .
Finally, it yields from the inequalities (2.139) and (2.141) as well as the definition (2.129)
Chapter 2. Positive Borel Measures
58
that
µ∗ (E ′ ∩ V ) + µ∗ (E ′ ∩ V c ) ≤ Λ(F1 ) + Λ(F2 ) +
ǫ
ǫ
ǫ
= Λ(F3 ) + ≤ µ∗ (E ′ ) +
2
2
2
which is exactly the inequality (2.136). Hence we obtain the desired result that (2.132).
• Fact 5: The set M∗ of all measurable sets (with respect to µ∗ ) is a σ-algebra
and µ = µ∗ |B (X) is a (complete) measure. The first assertion follows directly from
Carathéodory’s Theorem [47, Theorem 8, p. 349]. By Fact 4, we have
B(X) ⊆ M∗
which is exactly Step VII, where B(X) denotes the set of all Borel sets in X. By
Carathéodory’s Theorem again, it shows that µ is a (complete) measure on B(X) and
this is the same as Step IX. Consequently, Steps III, IV, V, VI and VIII can all be
skipped.o
• Fact 6: Deduction of parts (b) and (d). Let K be a compact set of X. Since X is a
metric space, K is also Hausdorff. By Corollary (a) following Theorem 2.5, we know that
K is closed in X and so K ∈ B(X). Therefore, we deduce from Facts 2 and 5 that
µ(K) = µ∗ (K) ≤ µ∗ (X) ≤ Λ(1)
which is exactly part (b) of Theorem 2.14 (The Riesz Representation Theorem).
If E is open in X, then E ∈ B(X). Thus we observe from the equation (2.132) and
then the definition (2.130) that
µ(E) = µ∗ (E)
= µ∗ (X) − µ∗ (E c )
= µ∗ (X) − inf{µ∗ (V ) | V is open in X and E c ⊆ V }
= µ∗ (X) − inf{µ∗ (X) − µ∗ (V c ) | V c is closed in X and V c ⊆ E}
= sup{µ∗ (V c ) | V c is closed in X and V c ⊆ E}
= sup{µ(V c ) | V c is closed in X and V c ⊆ E}.
(2.142)
Since X is compact, V c is compact by Theorem 2.4 so that the expression (2.142) can be
expressed as
µ(E) = sup{µ(K) | K compact and K ⊆ E }
which is the same as part (d) of Theorem 2.14 (The Riesz Representation Theorem).
• Fact 7: Deduction of part (a). As Rudin pointed out in Step X that it suffices to
prove
Z
f dµ
(2.143)
Λ(f ) ≤
X
o
holds for f ∈ Cc (X) because the opposite inequality follows by changing the sign of f . In
fact, the proof of our inequality (2.143) for compact metric X is essentially the same as
that of Step X, except that Rudin applied Step II to show that
X
µ(K) ≤
Λ(hi ),
We cannot say at this stage that Step II can be omitted because the proof of Step X still needs it, but we
will see very soon in Fact 7 that it is eventually allowable to do so.
2.6. Miscellaneous Problems on L1 and Other Properties
59
P
where hi ≺ Vi and
hi = 1 on K.p However, this kind of argument can be replaced
completely by using Fact 6 that µ(K) ≤ Λ(1) holds for every compact set K ⊆ X.
Therefore, we finally arrive at the expected inequality (2.143) and furthermore, Step II
can also be skipped.
Hence we conclude that Steps II to IX can be replaced by a simpler argument (outer measure
in the sense of Carathéodory). In other words, only Steps I and X should be kept. Of course,
the proof of the uniqueness of µ cannot be omitted. This completes the proof of the problem. Remark 2.3
(a) We remark that a Radon measure is a measure µ defined on the σ-algebra M containing
all Borel sets B(X) of the Hausdforff space X and it satisfies µ(K) < ∞ for all compact
subsets K, outer regular on M and inner regular on open sets in X, see [22, p. 212]
or [47, p. 455]. Hence the unique positive measure µ in Theorem 2.14 (The Riesz
Representation Theorem) is in fact a Radon measure when X is Hausdforff.
(b) You are also advised to read the paper [52] for a similar proof of Problem 2.19.
2.6
Miscellaneous Problems on L1 and Other Properties
Problem 2.20
Rudin Chapter 2 Exercise 20.
Proof. Let f = sup fn : [0, 1] → [0, ∞]. By Definition 1.30, f ∈
/ L1 on [0, 1] is equivalent to
n
Z
0
1
|f (x)| dx = ∞.
In fact, the inspiration of the construction of {fn } comes from the idea of Problem 2.9. For each
positive integer n, we consider the continuous functions gn : [−1, 1] → [0, ∞) given by

if x ∈
/ [− n12 , n12 ];
 0,
gn (x) =

n − n3 |x|, if x ∈ [− n12 , n12 ].
Next, we define fn : [0, 1] → [0, ∞) by

0,
if x ∈
/ [ n1 −
1 
fn (x) = gn x −
=

n
n − n2 |nx − 1|, if x ∈ [ n1 −
1 1
,
n2 n
1 1
,
n2 n
+
1
];
n2
+
1
].
n2
(2.144)
Recall that fn (x) is the tent function centered at n1 and the graph of it is an isosceles triangle
with height n and base n22 so that each fn is continuous on [0, 1] and its area is n1 which implies
that
Z 1
1
fn (x) dx = → 0
n
0
p
See Step X for the meanings of the notations used here.
Chapter 2. Positive Borel Measures
60
as n → ∞. Furthermore, if x = 0, then x ∈
/ [ n1 −
1 1
,
n2 n
+
1
]
n2
for all n ∈ N. and thus
fn (0) = 0
for all n ∈ N by the definition (2.144). If x ∈ (0, 1], then there is a positive integer N such that
1
1
/ [ n1 − n12 , n1 + n12 ] for all n ≥ N . Therefore,
n + n2 < x for all n ≥ N . In other words, we have x ∈
the definition (2.144) certainly shows that
fn (x) → 0
as n → ∞ for all x ∈ (0, 1].
Now it remains to show that f ∈
/ L1 on [0, 1]. To this end, we study the behaviour of f as
x → 0. For every n ∈ N, we consider every x in [ n1 − n12 , n1 + n12 ]. If n is sufficiently large enough,
we have
1
1
,
(2.145)
x= +o
n
n
where o(x) is the little-o notation.q Therefore, it follows from the definition (2.144) and the
relation (2.145) that
1
fn (x) = n + o(1) = + o(1)
(2.146)
x
for large enough n and small positive x. By the definition and the expression (5.102), we have
f (x) = sup fn (x) ≥ fn (x) =
n
1
+ o(1)
x
(2.147)
/ L1 on [0, 1], so we conclude from the estimate that (2.147)
for small positive x. Obviously, x1 ∈
1
that f ∈
/ L on [0, 1]. This completes the proof of the problem.
Problem 2.21
Rudin Chapter 2 Exercise 21.
Proof. Let α ∈ f (X) ⊆ R and
Eα = {x ∈ X | f (x) ≥ α} = f −1 [α, ∞) .
By the definition, it is clear that Eα is nonempty. Furthermore, since
Eα = X \ f −1 (−∞, α) ⊆ X,
the set Eα is closed in X. Now we consider the collection {Eα } of subsets of X. Suppose that
α1 , α2 , . . . , αn ∈ f (X) and we define α = max(α1 , α2 , . . . , αn ). Then we have
n
\
k=1
Eαk =
n
\
{x ∈ X | f (x) ≥ αk } = Eα 6= ∅.
k=1
Therefore, {Eα } has the finite intersection property and then since X is compact, we conclude
thatr
\
Eα 6= ∅.
Thus suppose that x0 ∈
\
α
α
Eα which means that f (x0 ) ≥ α for all α ∈ f (X), i.e., f (x0 ) ≥ f (x)
for all x ∈ X. Hence this shows that f attains its maximum at some point of X, completing the
proof of the problem.
q
r
Recall that g(n) = o(f (n)) means fg(n)
→ 0 as n → ∞.
(n)
See, for instances, [42, Theorem 26.9, p. 169] or [64, Problem 4.23, p. 43].
2.6. Miscellaneous Problems on L1 and Other Properties
61
Remark 2.4
Problem 2.21 can be treated as the Extreme Value Theorem for upper semicontinuous
functions. Similarly, we can show that if f : X → R is lower semicontinuous and X is
compact, then f attains its minimum at some point of X.
Problem 2.22
Rudin Chapter 2 Exercise 22.
Proof. By the definition and the triangle inequality, for all p ∈ X, we have
gn (x) ≤ f (p) + nd(x, p) ≤ [f (p) + nd(y, p)] + nd(x, y).
In other words, gn (x) − nd(x, y) is a lower bound of {f (p) + nd(y, p) | p ∈ X}. Thus, by the
definition of infimum, we must have gn (x) − nd(x, y) ≤ gn (y) or equivalently,
gn (x) − gn (y) ≤ nd(x, y).
(2.148)
gn (y) − gn (x) ≤ nd(x, y).
(2.149)
Likewise, we have
Therefore, we conclude from the inequalities (2.148) and (2.149) that
|gn (x) − gn (y)| ≤ nd(x, y).
Since f (x) ≥ 0 on X and d(x, y) ≥ 0 for every x, y ∈ X, we have gn (x) ≥ 0 for all positive
integers n. Fix x ∈ X, we observe that
f (p) + nd(x, p) ≥ f (p) + (n − 1)d(x, p) ≥ gn−1 (x)
for every p ∈ X and n = 2, 3, . . .. In other words, this implies that
gn (x) ≥ gn−1 (x)
for every x ∈ X and n = 2, 3, . . .. Besides, it is trivial that gn (x) ≤ f (x) holds for all n ∈ N
if f (x) = ∞. Without loss of generality, we may assume that f (x) < ∞. In this case, since
f (p) ≤ f (p) + nd(x, p) for all p ∈ X, if we take p = x in the definition of gn , then we have
gn (x) ≤ f (x)
for every n = 1, 2, . . .. This proves property (ii).
To prove property (iii), we need the following lemma:
Lemma 2.9
A function f : X → R is lower semicontinuous on X if and only if given x ∈ X, for
every {xn } ⊆ X \ {x} converging to x and for every ǫ > 0, there exists a positive
integer N such that n ≥ N implies
f (x) < f (xn ) + ǫ.
(2.150)
Chapter 2. Positive Borel Measures
62
Proof of Lemma
2.9. For every α ∈ R, denote Eα = {x ∈ X | f (x) ≤ α}. Recall that
f −1 (α, ∞) is open in X if and only if X \ f −1 (α, ∞) = Eα is closed in X. In other
words, f is lower semicontinuous on X if and only if Eα is closed in X.
Let Eα be closed in X. Assume that the inequality (2.150) was invalid. Then it
means that there exists a x0 ∈ X and a sequence {xn } in X converging to x0 such that
for some ǫ > 0, the inequality
f (x0 ) − ǫ ≥ f (xnk )
(2.151)
holds for infinitely many k. Take α ∈ (f (x0 ) − ǫ, f (x0 )). On the one hand, since
f (x0 ) > α, x0 ∈
/ Eα . On the other hand, we gain from the inequality (2.151) that
f (xnk ) < α for infinitely many k and this implies that
{xnk } ⊆ Eα .
Since the set Eα is closed and xnk → x0 as k → ∞, we have x0 ∈ Eα , a contradiction.
Hence the inequality (2.150) must hold for a lower semicontinuous function f .
Conversely, we prove that the inequality (2.150) implies Eα is closed in X by showing
that Eα contains all its limit points. To this end, pick α ∈ R such that Eα 6= ∅. Let
{xn } ⊆ Eα \ {x} and xn → x as n → ∞. Given ǫ > 0. By the inequality (2.150), there
exists a positive integer N such that
f (x) − ǫ < f (xn )
(2.152)
for all n ≥ N . Since f (xn ) ≤ α, we deduce immediately from the inequality (2.152)
that
f (x) < α + ǫ.
Since ǫ is arbitrary, we must have f (x) ≤ α or equivalently x ∈ Eα . This ends the
proof of the lemma.
Remark 2.5
(a) Similarly, a function f : X → R is upper semicontinuous on X if and only if given
x ∈ X, for every {xn } ⊆ X \ {x} converging to x and for every ǫ > 0, there exists a
positive integer N such that n ≥ N implies f (x) > f (xn ) − ǫ.
(b) Except the equivalent definition stated in Lemma 2.9, a lower semicontinuous function
f on X can also be reformulated in terms of the lower limit of f : lim inf f (y) ≥ f (x)
y→x
for every y → x in X \ {x}, see [61, pp. 69, 70] and [65, Definition 3.62, p. 64].
Now it is time to go back to the proof of the problem. On the one hand, since f is lower
semicontinuous on X, Lemma 2.9 implies that for every ǫ > 0, there exists a δ > 0 such that
f (y) > f (x) − ǫ
(2.153)
for all y ∈ B(x, δ). In this case, we always have
f (y) + nd(x, y) > f (x) − ǫ
(2.154)
for every positive integer n. Since ǫ > 0 is arbitrary, the inequality (2.154) becomes
f (y) + nd(x, y) ≥ f (x).
(2.155)
2.6. Miscellaneous Problems on L1 and Other Properties
63
On the other hand, we consider y ∈
/ B(x, δ) so that d(x, y) ≥ δ. It is clear that nδ − ǫ > 0 holds
for large enough n,s thus we get from the inequality (2.153) that
f (y) + nd(x, y) ≥ f (y) + nδ > f (x) − ǫ + nδ > f (x)
for large enough n. Therefore, we conclude that the inequality (2.155) always holds for all y ∈ X
and for all large enough n. By the definition of limit inferior, we have
lim inf gn (x) ≥ f (x).
(2.156)
lim sup gn (x) ≤ f (x)
(2.157)
n→∞
By property (ii), we see that
n→∞
holds for every x ∈ X. Hence it deduces from the inequalities (2.156) and (2.157) that
lim gn (x) = f (x)
n→∞
on X.
By property (i), each gn is continuous on X. Next, property (ii) says that {gn } is increasing
and bounded by f . Finally, property (iii) ensures that {gn } converges to f on X pointwisely.
This ends the proof of the problem.
Remark 2.6
With the aid of Lemma 2.9, we can prove Problem 2.1(b) easily. In fact, given x ∈ X, for
every {xn } ⊆ X \ {x} converging to x and for every ǫ > 0, there exists a positive integer N1
such that n ≥ N1 implies that
ǫ
(2.158)
f1 (x) < f1 (xn ) + .
2
Similarly, there exists a positive integer N2 such that n ≥ N2 implies
ǫ
f2 (x) < f2 (xn ) + .
2
(2.159)
Take N = max(N1 , N2 ). If n ≥ N , then we derive from the inequalities (2.158) and (2.159)
that
f1 (x) + f2 (x) < f1 (xn ) + f2 (xn ) + ǫ.
Hence it follows from Lemma 2.9 that f is lower semicontinuous on X.
Problem 2.23
Rudin Chapter 2 Exercise 23.
Proof. Let x ∈ Rk and f : Rk → R be defined by
f (x) = µ(V + x).
Since V + x is just a translation of V , it is also open in Rk , i.e., V + x ∈ B.
s
Obviously, the n depends on δ and ǫ.
Chapter 2. Positive Borel Measures
64
We first construct a counter example which is not upper semicontinuous or continuous. By
Example 1.20(b), we consider the unit mass concentrated at 0, i.e.,

 1, if 0 ∈ E;
µ(E) =

0, if 0 ∈
/ E.
If V = B(0, 1), then we have V + x = B(x, 1) and
µ(V + x) =
=

 1, if 0 ∈ B(x, 1);

0, if 0 ∈
/ B(x, 1)

 1, if |x| < 1;

(2.160)
0, if |x| ≥ 1.
Thus it establishes from the result (2.160) that
f −1 (−∞, 1) = {x ∈ Rk | f (x) < 1} = {x ∈ Rk | µ(V + x) < 1} = Rk \ B(x, 1)
which is closed in Rk . By Definition 2.8, f is not upper semicontinuous or continuous.
However, we claim that f is always lower semicontinous. Let {xn } ⊆ Rk \ {x} be such that
xn → x as n → ∞. Recall that both V + x and V + xn are open in Rk . We claim that y ∈ V + x
implies that y ∈ V + xn for all but finitely many n. Let y = a + x for some a ∈ V . Since V is
open in Rk , there exists a ǫ > 0 such that
B(a, ǫ) ⊆ V.
By the hypothesis, there is a positive integer N such that n ≥ N implies
xn − x =
ǫ
2
or
ǫ
xn − x = − .
2
In the first case, we have
y =a+x=a−
ǫ
ǫ
ǫ
+ x + = a − + xn .
2
2
2
(2.161)
Since a − 2ǫ ∈ V , the expression (2.161) guarantees that y ∈ V + xn for all n ≥ N . The other
case can be done similarly, so we omit the details here. This proves our claim.
Now if χV +x (y) = 1, then we have χV +xn (y) = 1 for all but finitely many n. In other words,
it means that
lim inf χV +xn (y) ≥ χV +x (y)
(2.162)
n→∞
for every y ∈ Rk . By Proposition 1.9(d), each χV +xn : Rk → [0, ∞] is measurable, so we may
apply Theorem 1.28 (Fatou’s Lemma) to conclude that
Z Z
(2.163)
lim inf
lim inf χV +xn dµ.
χV +xn dµ ≥
n→∞
Rk
Rk
By Proposition 1.24(f), we have
Z
Z
χV +xn dµ =
Rk
V +xn
n→∞
dµ = µ(V + xn ).
(2.164)
2.6. Miscellaneous Problems on L1 and Other Properties
65
In addition, it follows from the inequality (2.162) and the application of Proposition 1.24(f)
again that
Z
Z
Z dµ = µ(V + x).
(2.165)
χV +x (y) dµ =
lim inf χV +xn dµ ≥
Rk
n→∞
Rk
V +x
Thus, by substituting the results (2.164) and (2.165) into the inequality (2.163), we achieve that
lim inf µ(V + xn ) ≥ µ(V + x)
n→∞
or equivalently,
lim inf µ(V + y) ≥ µ(V + x)
y→x
Rk \{x}.
for every y → x in
Hence we conclude from Remark 2.5(b) that f is lower semicontinuos
k
on R , completing the proof of the problem.
Problem 2.24
Rudin Chapter 2 Exercise 24.
Proof. Suppose that S1 and S2 are the sets of simple functions and step functions on R respectively. We divide the proof into several steps.
• Step 1: S1 is dense in L1 (R). Let f = f + − f 1 . By the Corollary (b) following
Theorem 1.14, both f + and f − are measurable. Since f + : R → [0, ∞], we follow from
Theorem 1.17 (The Simple Function Approximation Theorem) that there is a sequence
{sn } of nonnegative increasing simple functions such that for every x ∈ R, sn (x) → f + (x)
as n → ∞. Thus Theorem 1.26 (Lebesgue’s Monotone Convergence Theorem) implies that
Z ∞
Z ∞
f + dx
(2.166)
sn dx →
−∞
−∞
as n → ∞. It is clear that |f + (x) − sn (x)| = f + (x) − sn (x) for every x ∈ R, so we havet
Z ∞
Z ∞
+
[f + (x) − sn (x)] dx
|f (x) − sn (x)| dx =
−∞
−∞
Z ∞
Z ∞
+
sn (x) dx.
(2.167)
f (x) dx −
=
−∞
−∞
If we apply the limit (2.166) to the expression (2.167), then we gain
Z ∞
|f + (x) − sn (x)| dx = 0.
lim
n→∞ −∞
Likewise, the same is true for f − , so we have
Z ∞
|f − (x) − tn (x)| dx = 0
lim
n→∞ −∞
where {tn } is a sequence of nonnegative increasing simple functions such that for every
x ∈ R, tn (x) → f − (x) as n → ∞. Hence {gn = sn − tn } is a sequence of simple functions
such that
Z ∞
Z ∞
|f + − f − − sn + tn | dx
|f − gn | dx =
−∞
−∞
Since |f (x)| ≤ |f (x)| on R and f ∈ L (R), we have f + ∈ L1 (R). Thus this makes the second equality holds
because the right-hand side is not in the form ∞ − ∞.
t
+
1
Chapter 2. Positive Borel Measures
≤
66
Z
∞
−∞
+
|f − sn | dx +
→0
Z
∞
−∞
|f − − tn | dx
as n → ∞.
• Step 2: S2 is dense in S1 . To this end, it suffices to prove that for every measurable
set E ⊂ R with m(E) < ∞ and every ǫ > 0, there exists a step function g on R such that
Z ∞
|χE − g| dx < ǫ.
(2.168)
−∞
By Theorem 2.20(b), there exists an open set V in R such that E ⊆ V and m(V \ E) < 2ǫ .
By [49, Exercise 29, Chap. 2, p. 45], we have
V =
∞
[
(an , bn ),
(2.169)
n=1
where (ai , bi ) ∩ (aj , bj ) = ∅ if i 6= j. Assume that there was an unbounded segment in the
expression (2.169). Then we have m(V ) = ∞. However, since m(V ) = m(V \ E) + m(E),
the facts m(V ) = ∞ and m(V \ E) < 2ǫ will force that m(E) = ∞, a contradiction.
Furthermore, the expression (2.169) implies that
m(V ) =
∞
X
n=1
m (an , bn ) ,
so there exists a positive integer N such that
m(V ) −
N
X
ǫ
m (an , bn ) .
<
2
(2.170)
n=1
Define g : R → R by
g(x) =
N
X
χ(an ,bn ) (x)
n=1
and VN = (a1 , b1 ) ∪ · · · ∪ (aN , bN ). Let x ∈ (E \ VN ) ∪ (VN \ E). If x ∈ E \ VN , then
we have χE (x) = 1 and g(x) = 0. Similarly, if x ∈ VN \ E, then we have χE (x) = 0 and
g(x) = 1. Therefore, we have
|g − χE | = 1
(2.171)
on (E \ VN ) ∪ (VN \ E). On the other hand, we let
x ∈ (E \ VN )c ∩ (VN \ E)c = (VN ∪ E c ) ∩ (E ∪ VNc ) = (VN ∩ E) ∪ (VNc ∩ E c ).
If x ∈ VN ∩ E, then we have χE (x) = g(x) = 1. Similarly, x ∈ VNc ∩ E c implies that
χE (x) = g(x) = 0. Both cases give
|g − χE | = 0
(2.172)
on (E \ VN )c ∩ (VN \ E)c . Now the facts E, VN ⊆ V and (2.170) definitely imply that
ǫ
ǫ
m(VN \ E) ≤ m(V \ E) <
and m(E \ VN ) ≤ m(V \ VN ) < .
(2.173)
2
2
Hence we deduce from the results (2.171), (2.172) and the estimates (2.173) that
Z
Z
Z ∞
dx = m(VN \ E) + m(E \ VN ) < ǫ.
dx +
|g − χE | dx =
−∞
E\VN
This is the desired result (2.168).
VN \E
2.6. Miscellaneous Problems on L1 and Other Properties
67
• Step 3: S2 is dense in L1 (R). We need the following lemma:
Lemma 2.10
Suppose that A, B and C are subsets of a metric space X. If A ⊆ B ⊆ C, A is
dense in B and B is dense in C, then A is dense in C.
Proof of Lemma 2.10. Recall the definition that A is dense in S if A ⊆ S ⊆ A, where
A denotes the closure of A in X. Then we have
A ⊆ B ⊆ A and B ⊆ C ⊆ B
which show immediately that A ⊆ C. Let p ∈ B. We claim that p ∈ A. To this end,
we know from the assumption that
B(p, δ) ∩ (B \ {p}) 6= ∅
for every δ > 0. Let q ∈ B(p, δ) ∩ (B \ {p}). Since B ⊆ A, A also contains q. If q ∈ A,
then we have
B(p, δ) ∩ (A \ {p}) 6= ∅
(2.174)
for every δ > 0. Therefore, the definition implies that p ∈ A. Next, suppose that
q ∈ A′ . Since B(p, δ) is open in X, there exists a ǫ > 0 such that B(q, ǫ) ⊆ B(p, δ).
Since q is a limit point of A, the definition shows that B(q, ǫ) ∩ A 6= ∅ and thus the
set relation (2.174) still holds in this case. Therefore, we have proven our claim that
p ∈ A and this means that B ⊆ A. Hence we establish the set relations
A⊆C⊆A
which implies that A is dense in C, completing the proof of Lemma 2.10.
Let’s return to the original proof. By applying Lemma 2.10 to the results in Step 1
and Step 2, we conclude easily that S2 is dense in L1 (R) which is our desired result.
Hence we have completed the proof of the problem.
Remark 2.7
By Definition 1.16, we remark that a simple function s is a finite linear combination of
characteristic functions of an arbitrary set Ai and when all Ai are intervals, then s becomes
a step function. In fact, the class of step functions is one of the building blocks for the theory
of Riemann integration, see [49, Chap. 6].
Problem 2.25
Rudin Chapter 2 Exercise 25.
Proof.
(a) We note that et > 0 and log(1 + et ) > 0 for every t > 0. Furthermore, we know that the
functions ex and log x are increasing in their corresponding domains respectively. Thus
Chapter 2. Positive Borel Measures
68
the inequality log(1 + et ) < c + t is equivalent to 1 + et < ec+t and finally, it is equivalent
to
0 < 1 + e−t < ec .
(2.175)
Apply log to both sides of the inequality (2.175), we have
log(1 + e−t ) < c
(2.176)
for every t > 0. Since the left-hand side of the inequality (2.176) is a decreasing function
of t, the smallest value of c for the validity of the inequality (2.176) on (0, ∞) is found by
lim log(1 + e−t ) = log 2.
t→0
t>0
(b) Let f ∈ L1 on [0, 1]. Define E = {x ∈ [0, 1] | f (x) > 0} = f −1 (0, ∞) . By Theorem
1.12(b), E is a measurable set in [0, 1]. Thus we follow from part (a) that, on E,
nf (x) = log enf (x) < log 1 + enf (x) < log 2 + nf (x)
on (0, ∞) and this implies that
Z
Z
Z
log 2
1
f dµ.
log(1 + enf ) dµ <
+
f dµ <
n E
n
E
E
(2.177)
Next, we have 1 ≤ 1 + enf (x) ≤ 2 on [0, 1] \ E. Thus we have
0 ≤ log(1 + enf ) ≤ log 2
(2.178)
on [0, 1]\E. Since [0, 1]\E is also a measurable set in [0, 1], we deduce from the inequalities
(2.178) that
Z
log 2
1
log(1 + enf ) dµ ≤
.
0≤
n [0,1]\E
n
Now the sum of the inequalities (2.177) and (2.178) give
Z
1
f dµ <
n
E
Z
1
0
2 log 2
+
log(1 + e ) dµ <
n
nf
Z
f dµ.
(2.179)
E
Hence, by taking n → ∞ in the inequalities (2.179), we get that
1
lim
n→∞ n
Z
1
nf
log(1 + e ) dµ =
0
This completes the proof of the problem.
Z
f dµ.
E
CHAPTER
3
Lp-Spaces
3.1
Properties of Convex Functions
Problem 3.1
Rudin Chapter 3 Exercise 1.
Proof. Let {ϕα } be a collection of convex functions on (a, b). Define
ϕ = sup{ϕα }
(3.1)
and assume that it is finite. Suppose that x, y ∈ (a, b) and 0 ≤ λ ≤ 1. By the definition (3.1)
and the convexity of ϕα , we have
ϕα (1 − λ)x + λy ≤ (1 − λ)ϕα (x) + λϕα (y) ≤ (1 − λ)ϕ(x) + λϕ(y)
for all α. In other words, we have
ϕ (1 − λ)x + λy ≤ (1 − λ)ϕ(x) + λϕ(y).
By Definition 3.1, ϕ is convex on (a, b).
Suppose that {ϕn } is a sequence of convex functions on (a, b). For x ∈ (a, b), we define
ϕ : (a, b) → R to be
ϕ(x) = lim ϕn (x).
(3.2)
n→∞
Now for x, y ∈ (a, b) and 0 ≤ λ ≤ 1, we follow from the limit (3.2) that
lim [(1 − λ)ϕn (x) + λϕn (y)] = (1 − λ)ϕ(x) + λϕ(y)
n→∞
and
lim ϕn (1 − λ)x + λy = ϕ (1 − λ)x + λy .
n→∞
(3.3)
(3.4)
Since ϕn (1 − λ)x + λy ≤ (1 − λ)ϕn (x) + λϕn (y) for every n = 1, 2, . . ., we take limits to both
sides and apply the results (3.3) and (3.4) to conclude that
ϕ (1 − λ)x + λy ≤ (1 − λ)ϕ(x) + λϕ(y).
Hence ϕ is convex on (a, b) by Definition 3.1.
69
Chapter 3. Lp -Spaces
70
For each positive integer n, suppose that fn : (a, b) → R is convex. Define f : (a, b) → R by
f (x) = lim sup fn (x),
n→∞
where x ∈ (a, b). By Definition 1.13, we have
f (x) = lim gk (x),
k→∞
(3.5)
where gk = sup{fk , fk+1 , . . .} for k = 1, 2, . . .. Thus if each fn is convex on (a, b), then the
first assertion shows that each gk is convex on (a, b). By the definition (3.5), since f is the
pointwise limit of the sequence of convex functions {g1 , g2 , . . .} defined on (a, b), we deduce from
the second assertion that f is also convex on (a, b). However, the lower limit of a sequence of
convex functions may not be convex. For example, consider the functions fn : R → R defined
by
fn (x) = (−1)n x.
It is clear that each fn is convex on R. However, for each x ∈ R, we have
f (x) = lim inf fn (x) = −|x|.
n→∞
This shows that f is not convex on R and we complete the proof of the problem.
Problem 3.2
Rudin Chapter 3 Exercise 2.
Proof. For every x, y ∈ (a, b) and 0 ≤ λ ≤ 1, since ϕ is convex on (a, b), we have
ϕ (1 − λ)x + λy ≤ (1 − λ)ϕ(x) + λϕ(y).
(3.6)
Since ψ is convex and nondecreasing on ϕ (a, b) , we obtain from the inequality (3.6) that
ψ ϕ (1 − λ)x + λy ≤ ψ (1 − λ)ϕ(x) + λϕ(y) ≤ (1 − λ)ψ ϕ(x) + λψ ϕ(y) .
By Definition 3.1, ψ ◦ ϕ is convex on (a, b).
Let ϕ > 0 on (a, b). By [51, Theorem (c), p. 2], the function exp is a monotonically increasing
positive function on R. Since (ex )′ = ex , it is convex on R by the paragraph following Definition
3.1. Hence, if log ϕ is convex on (a, b), then we conclude from the first assertion that ϕ = elog ϕ
is also convex on (a, b).
However, the converse of the second assertion is false. For instance, we know that
ϕ(x) = x
is convex on (0, ∞), but log x is not convex on (0, ∞) because (log x)′′ = −x−2 < 0 on (0, ∞),
i.e., (log x)′ is not a monotonically increasing function on (0, ∞). We have completed the proof
of the problem.
Problem 3.3
Rudin Chapter 3 Exercise 3.
Proof. This problem is proven in [63, Problem 4.24, pp. 79 – 81].
3.2. Relations among Lp -Spaces and some Consequences
71
Relations among Lp -Spaces and some Consequences
3.2
Problem 3.4
Rudin Chapter 3 Exercise 4.
Proof.
(a) Since 0 < r < p < s, we can find λ ∈ (0, 1) such that p = λr + (1 − λ)s. By Theorem 3.5
(Hölder’s Inequality), we have
Z
|f |p dµ
ϕ(p) =
ZX
|f |λr × |f |(1−λ)s dµ
=
X
nZ
o1−λ
oλ n Z
1
1
≤
|f |λr λ dµ ×
[|f |(1−λ)s ] 1−λ dµ
X
X
= [ϕ(r)]λ × [ϕ(s)]1−λ .
(3.7)
Since r, s ∈ E, ϕ(r) and ϕ(s) are finite. Hence the inequality (3.7) ensures that ϕ(p) is
also finite, i.e., p ∈ E.
(b) We prove the assertion one by one.
– Case (i): ln ϕ is convex in E ◦ . If E ◦ = ∅, then there is nothing to prove. Assume
that E ◦ 6= ∅. By [49, Theorem 2.47, p. 42] and part (a), the set E is connected.
Since E ⊆ (0, ∞), E is either an interval in one of the forms [a, b], [a, b), (a, b] or (a, b)
for some positive a and b with a < b. In each of the cases, we always have E ◦ = (a, b).
Let x, y ∈ (a, b) and λ ∈ [0, 1]. Since (a, b) is a convex set, λx + (1 − λ)y ∈ (a, b).
Thus it follows from the inequality (3.7) that
ϕ λx + (1 − λ)y ≤ [ϕ(x)]λ × [ϕ(y)]1−λ .
(3.8)
If ϕ(p) = 0 for some p ∈ (0, ∞), then we have
Z
|f |p dµ = 0
X
so that |f (x)| = 0 almost everywhere on X. By the remark following Definition 3.7, it
implies that kf k∞ = 0, a contradiction. Hence we have ϕ(p) > 0 and we are allowed
to take the logarithm to both sides of the inequality (3.8) to get
ln ϕ λx + (1 − λ)y ≤ ln [ϕ(x)]λ × [ϕ(y)]1−λ ≤ λ ln ϕ(x) + (1 − λ) ln ϕ(y).
By Definition 3.1, ln ϕ is convex in (a, b).
– Case (ii): ϕ is continuous on E. By Theorem 3.2, ln ϕ is continuous on (a, b).
Thus ϕ is also continuous on (a, b), so it remains to show that ϕ is continuous at the
endpoints. Let a ∈ E. Then E is either [a, b) or [a, b], but no matter which case E is,
a is a limit point of E so that we can find a decreasing sequence {pn } ⊆ (a, b) such
that pn → a as n → ∞. By this, there exists a ǫ > 0 and a positive integer N such
that pn ∈ (a, a + ǫ) ⊂ E for all n ≥ N . Fix this ǫ and then n ≥ N implies that

 |f (x)|a+ǫ , if |f (x)| ≥ 1;
|f (x)|pn ≤

|f (x)|a ,
if |f (x)| < 1.
Chapter 3. Lp -Spaces
72
Hence we obtain that
|f (x)|pn ≤ |f (x)|a+ǫ + |f (x)|a
on X.
We recall that a, a + ǫ ∈ E, so the definition implies that ϕ(a) and ϕ(a + ǫ) are
finite. Since f is measurable, |f |a and |f |a+ǫ are measurable by Proposition 1.9(b) and
Theorem 1.7(b). Thus these two facts show that |f |a , |f |a+ǫ ∈ L1 (µ). By Theorem
1.32, we have |f |a+ǫ + |f |a ∈ L1 (µ). For each x ∈ X, |f (x)| ≥ 0. Since an exponential
function with nonnegative base is continuous on its domain, we have
lim |f (x)|pn = |f (x)|a
n→∞
on X. In conclusion, we have shown that the sequence {|f |pn } satisfies the hypotheses
of Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) and then
Z
Z
|f |a dµ = ϕ(a).
|f |pn dµ =
lim ϕ(pn ) = lim
n→∞
n→∞ X
X
By definition, ϕ is continuous at a. Similarly, ϕ is continuous at b if b ∈ E. Consequently, ϕ is continuous on E.
(c) We claim that E can be any open or closed connected subset of (0, ∞). We consider
X = (0, ∞), µ = m and 0 < a ≤ b. We need the following result (see [2, Examples 1 & 5,
pp. 417, 419]):
Lemma 3.1
Let x > 0 and b > 1. Then we have

1 − x1−α


Z 1

, if α 6= 1;
1−α
t−α dt =

x


− ln x
if α = 1
and
 1−β
b
−1


, if β =
6 1;

b
1−β
t−β dt =

1


ln b,
if β = 1.
Z
Furthermore, the improper integrals
Z 1
x−α dx and
0
Z
∞
x−β dx
1
are convergent if and only if α < 1 and β > 1 respectively.
– Case (i): E = (a, b). For x ∈ (0, ∞), let

1

 x− b , if x ∈ (0, 1);
f (x) =

 x− a1 , if x ∈ [1, ∞).
We have to find p ∈ (0, ∞) such that
Z
ϕ(p) =
0
∞
|f (x)|p dx < ∞.
3.2. Relations among Lp -Spaces and some Consequences
73
We assume that the integral can be split as follows:
Z 1
Z ∞
p
p
x− b dx +
ϕ(p) =
x− a dx.
0
(3.9)
1
Apply Lemma 3.1 to the two improper integrals in the expression (3.9), we know that
ϕ(p) < ∞ if and only if pb < 1 and ap > 1 if and only if a < p < b. In other words, we
have E = (a, b) and such split is allowable.
– Case (ii): E = [a, b]. In this case, we need another lemma:a
Lemma 3.2
The improper integral
Z
e−1
dx
xα | ln x|β
0
converges if and only if (i) α < 1 or (ii) α = 1 and β > 1. The improper integral
Z ∞
dx
α
x (ln x)β
e
converges if and only if (i) α > 1 or (ii) α = 1 and β > 1.
We consider

1

,
if x ∈ (0, e−1 );

1

1+ 1b

b
x | ln x|






1
1

1
e− a − e b
f (x) =
b
(x − e−1 ), if x ∈ (e−1 , e);
e +
−1

e
−
e







1


if x ∈ [e, ∞).
 1
1 ,
x a (ln x)1+ a
(3.10)
It is clear that f is continuous on (0, ∞) and so it is measurable. Similar to Case
(i), we assume that the integral can be split as follows:
Z ∞
|f (x)|p dx
ϕ(p) =
=
Z
0
e−1
dx
p
1
x b | ln x|p(1+ b )
Z ∞
dx
0
+
p
e
1
+
x a (ln x)p(1+ a )
Z
e
e−1
h
1
1
ip
e− a − e b
−1
(x
−
e
)
dx
e +
e − e−1
1
b
.
(3.11)
By Lemma 3.2, the first integral in the expression (3.11) converges if and only if (i)
p
p
1
b < 1 or (ii) b = 1 and p(1 + b ) > 1. The first condition gives p < b. If p = b, then
p(1 + 1b ) = 1 + b > 1 so that the second condition is actually p = b. In this case, we
have p ≤ b.
Similarly, we apply Lemma 3.2 to the third integral in the expression (3.11), so it
converges if and only if (i) ap > 1 or (ii) ap = 1 and p(1 + a1 ) > 1. The first condition
a
The integrals in Lemma 3.2 are called Bertrand’s integrals, see the following webpage
https://fr.wikipedia.org/wiki/Int%C3%A9grale_impropre.
Chapter 3. Lp -Spaces
74
shows p > a. If p = a, then p(1 + a1 ) = 1 + a > 1 so that the second condition implies
that p = a. In this case, we have p ≥ a.
Since the second integral in the expression (3.11) is finite, we combine these observations to conclude that ϕ(p) < ∞ if and only if p ∈ [a, b].
– Case (iii): E = {a}. If we take b = a in the definition (3.10), then we obtain p ≥ a
and p ≤ a. Hence we establish E = {a}.
– Case (iv): E = ∅. Consider f ≡ 1 on (0, ∞). Since the integral
Z ∞
|1|p dx
0
is obviously divergent for every p ∈ (0, ∞), we conclude that E = ∅ in this case.
(d) Let r < p < s. By the proof of part (a), we see that p = λr + (1 − λ)s for some λ ∈ (0, 1)
so that the inequality (3.7) holds. In other words, we have
λ
1−λ
kf kpp ≤ kf krr × kf kss
.
(3.12)
It is clear that
kf krr
λ
× kf kss
1−λ
≤
=

λ
1−λ

, if kf kr ≤ kf ks ;
 kf krs × kf kss

 kf kr λ × kf ks 1−λ , if kf k ≤ kf k
s
r
r
r

p
 kf ks , if kf kr ≤ kf ks ;

kf kpr ,
(3.13)
if kf ks ≤ kf kr .
Hence, by combining the inequalities (3.12) and (3.13), we get
kf kp ≤ max(kf kr , kf ks ).
(3.14)
Let f ∈ Lr (µ) ∩ Ls (µ). By the definition, we have
kf kr < ∞
and
kf ks < ∞.
Hence we follow from these and the inequality (3.14) that kf kp < ∞, i.e., f ∈ Lp (µ) and
then Lr (µ) ∩ Ls (µ) ⊆ Lp (µ).
(e) Recall that kf k∞ > 0.
– Case (i): kf k∞ < ∞. Let Eα = {x ∈ X | |f (x)| ≥ α}, where α ∈ (0, kf k∞ ). It is
clear that Eα is measurable because |f | is measurable. By Definition 3.7, kf k∞ is the
smallest number such that
µ({x ∈ X | |f (x)| > kf k∞ }) = 0.
Thus we have µ(Eα ) > 0 for every α ∈ (0, kf k∞ ). Assume that µ(Eα ) = ∞. Then
we have
Z
Z
Z
r
r
r
|f |r dµ ≥ αr µ(Eα ) = ∞
|f | dµ +
|f | dµ =
kf kr =
X\Eα
Eα
X
which contradicts to our hypothesis. Thus we have 0 < µ(Eα ) < ∞ and then Proposition 1.24(b) implies that
Z
Z
|f |p dµ ≥ αp µ(Eα )
|f |p dµ ≥
kf kpp =
X
Eα
3.2. Relations among Lp -Spaces and some Consequences
75
which shows that
1
kf kp ≥ α[µ(Eα )] p .
(3.15)
1
Since lim [µ(Eα )] p = 1, we obtain from the inequality (3.15) that
p→∞
lim inf kf kp ≥ α.
p→∞
In particular, since α is arbitrary in (0, kf k∞ ), we have
lim inf kf kp ≥ kf k∞ .
p→∞
Next, if p > r, then we have
Z
Z
p
p
|f | dµ =
kf kp =
|f |p−r × |f |r dµ.
(3.16)
(3.17)
X
X
By Definition 3.7, since |f (x)| ≤ kf k∞ for almost all x ∈ X, we get from the expression (3.17) and Proposition 1.24(b) that
Z
Z
p
p−r
r
p−r
r
|f |r dµ = kf kp−r
kf kp ≤
kf k∞ × |f | dµ = kf k∞ ×
(3.18)
∞ × kf kr ,
X\E
X\E
where E = {x ∈ X | |f (x)| > kf k∞ } is of measure 0. Rewrite the inequality (3.18) in
the form
r
1− r
kf kp ≤ kf k∞ p kf krp
which, by using the fact that kf kr < ∞, implies
lim sup kf kp ≤ kf k∞ .
(3.19)
p→∞
Hence our desired result follows immediately from the inequalities (3.16) and (3.19).
– Case (ii): kf k∞ = ∞. Then we deduce immediately from the inequality (3.16) that
lim kf kp = lim inf kf kp = ∞ = kf k∞ .
p→∞
p→∞
Hence we have completed the proof of the problem.
Problem 3.5
Rudin Chapter 3 Exercise 5.
Proof.
(a) Suppose that s = ∞. Since |f (x)| ≤ kf k∞ holds for almost all x ∈ X, we have
µ(E) = µ({x ∈ X | |f (x)| > kf k∞ }) = 0
so that
kf krr
=
Z
X\E
r
|f | dµ ≤
Z
X\E
kf kr∞ dµ = kf kr∞ · µ(X \ E) = kf kr∞ .
Hence we have kf kr ≤ kf k∞ in this case.
(3.20)
Chapter 3. Lp -Spaces
76
Next, we suppose that s < ∞. Since f is measurable, |f |r is also measurable on X by
Proposition 1.9(b) and Theorem 1.7(b) for every r > 0. We apply Theorem 3.5 (Hölder’s
Inequality) to the measurable functions |f |r and 1 to obtain that
Z
X
|f |r dµ ≤
=
=
nZ
Z
|f |r
X
s
s
r
|f | dµ
X
kf krs .
dµ
r
s
or
s
×
nZ
X
dµ
o1− r
s
(3.21)
Hence we conclude that kf kr ≤ kf ks .
(b) Suppose that 0 < r < s < ∞. Since kf kr = kf ks , it means that the equality holds in the
inequality (3.21). Recall that the inequality (3.21) is derived from Theorem 3.5 (Hölder’s
Inequality), so there are constants α and β, not both 0, such that
s
α(|f |r ) r = β
a.e. on X.
If α = 0, then it is trivial that β = 0, a contradiction. Thus α 6= 0 and then we have
|f | =
β 1
s
α
a.e.
In other words, |f | is a constant a.e. on X.
Next, suppose that s = ∞ and kf kr = kf k∞ < ∞, where 0 < r < ∞. Then the equality
holds in the inequality (3.20) so that |f |r = kf kr∞ < ∞ a.e. on X, i.e.,
|f | = kf k∞
a.e. on X.
In other words, we achieve that |f | is a constant a.e. on X in this case.
Consequently, we conclude that 0 < r < s ≤ ∞ and the conditions kf kr = kf ks < ∞ hold
if and only if |f | is a constant a.e. on X
(c) If f ∈ Ls (µ), then kf ks < ∞ and we know from part (a) that kf kr < ∞, i.e., f ∈ Lr (µ).
Thus it is true that Ls (µ) ⊆ Lr (µ). Now the other inclusion Lr (µ) ⊆ Ls (µ) is a consequence
of the following lemmab :
Lemma 3.3
Suppose that (X, M, µ) is a measure space and M0 = {E ∈ M | µ(E) > 0}. Then
the following conditions are equivalent:
– Condition (1). inf{µ(E) | E ∈ M0 } > 0,
– Condition (2). Lr (µ) ⊆ Ls (µ) for all 0 < r < s ≤ ∞.
b
This lemma is part of the statements in [62, Theorem 1] and our proof follows part of the argument there.
3.2. Relations among Lp -Spaces and some Consequences
77
Proof of Lemma 3.3. Suppose that Condition (1) is true. Suppose that f ∈ Lr (µ)
and En = {x ∈ X | |f (x)| ≥ n} for every n ∈ N. We claim that there exists an N ∈ N
such that µ(EN ) = 0. Otherwise, there exists ǫ > 0 such that µ(En ) ≥ ǫ for each
∞
\
En . By the construction of En , we have E1 ⊇ E2 ⊇ · · · . Since
n ∈ N. Let E =
n=1
µ(E1 ) ≤ µ(X) < ∞, it yields from Theorem 1.19(e) that
µ(E) = lim µ(En ) ≥ ǫ > 0.
(3.22)
n→∞
However, we deduce from the inequality (3.22) that
1
1 Z
Z
1
r
r
r
|f |r dµ = ∞ · µ(E) r = ∞,
|f | dµ ≥
kf kr =
E
X
a contradiction. Therefore, we have our claim that µ(EN ) = 0 for some N ∈ N. In
other words, |f (x)| ≤ N holds for almost all x ∈ X and we obtain from Definition 3.7
that kf k∞ ≤ N , i.e., f ∈ L∞ (µ) and so f ∈ Ls (µ). Hence Condition (2) holds.
Assume that Condition (1) was false. Thus there exists a sequence of {En } ⊆ M0
such that
1
(3.23)
0 < µ(En ) < n .
2
Put F1 = E1 and Fn = En \ (En−1 ∪ · · · ∪ E1 ), where n = 2, 3, . . .. Then it is easy to
check that Fi ∩ Fj = ∅ for i 6= j and
1
0 < µ(Fn ) = µ En \ (En−1 ∪ · · · ∪ E1 ) ≤ µ(En ) < n
2
for every n = 1, 2, . . .. Therefore, we may assume that {En } is a sequence of disjoint
measurable sets.
∞
[
En . Suppose first that
Next, we define αn = µ(En ) for n = 1, 2, . . . and E =
n=1
s < ∞. Then we define f : X → R by
∞
X
f (x) =
−1
αn s χEn (x).
(3.24)
n=1
Since r < s, we have 1 − rs < 1. Note that f (x) = 0 if x ∈ X \ E. Thus it reduces from
the bounds (3.23) and Theorem 1.29 that
kf krr
=
Z
r
X\E
|f | dµ +
Z
r
E
|f | dµ =
∞ Z
X
−r
αn s
dµ =
n=1 En
∞
X
1− r
αn s
n=1
r
21− s
<
r < ∞
1 − 21− s
which means f ∈ Lr (µ). Similarly, it is easy to derive that
kf kss
=
Z
s
E
|f | dµ =
∞ Z
X
n=1 En
α−1
n dµ =
∞
X
n=1
1=∞
which implies that f ∈
/ Ls (µ). Hence we have shown that
Lr (µ) * Ls (µ).
−1
−
1
If s = ∞, then we replace the coefficients αn s by αn 2r in the expression (3.24). By
−
1
n
similar argument, since αn 2r > 2 2r for all n ∈ N, we conclude kf k∞ = ∞ and then
Lr (µ) * L∞ (µ).
Chapter 3. Lp -Spaces
78
(d) If 0 < p < q, then part (a) says that kf kp ≤ kf kq < ∞. Let E0 = {x ∈ X | |f (x)| = 0}.
By Problem 1.5, E0 is measurable.
– Case (i): µ(E0 ) > 0. Now the set E∞ = {x ∈ X | |f (x)| = ∞} is measurable
by Problem 1.5. Pick p ∈ (0, r), so part (a) indicates that kf kp ≤ kf kr < ∞. If
µ(E∞ ) > 0, then we have
nZ
o1
nZ
o1
1
p
p
p
kf kp =
|f | dµ
≥
|f |p dµ
= [∞ · µ(E∞ )] p = ∞,
X
E∞
a contradiction. Thus we have µ(E∞ ) = 0, i.e., |f (x)| < ∞ a.e. on X. Consequently,
we have
Z
Z
Z
Z
log |f | dµ
(3.25)
log |f | dµ +
log |f | dµ =
log |f | dµ =
X\(E∞ ∪E0 )
X\E∞
X
E0
Since µ(X) = 1 and |f (x)| is nonzero finite on X \ (E∞ ∪ E0 ), the first integral on
the right-hand side of the equation (3.25) is finite. However, the second integral on
the right-hand side of the equation (3.25) is in the form −∞ so that the right-hand
side of our desired result is in the form exp{−∞} which is defined to be 0. On the
other hand, we notice that
o1 n Z
nZ
o1
p
p
p
|f | dµ
kf kp =
|f |p χX\E0 dµ .
=
(3.26)
X
X
Apply Theorem 3.5 (Hölder’s Inequality) with u > 1 and v > 1 as fixed conjugate
exponents to the right-hand side of the expression (3.26), we see that
o1
o1
o 1 nZ
nZ
nZ
p
pu
pv
p u
p
(χX\E0 )v dµ
(|f | ) dµ
|f | χX\E0 dµ
≤
X
X
X
1
pv
= µ(X \ E0 ) kf kpu .
(3.27)
Combining the expression (3.26) and the inequality (3.27), we gain
1
kf kp ≤ µ(X \ E0 ) pv kf kpu .
Since µ(X) = 1 and µ(E0 ) > 0, we have µ(X \ E0 ) < 1 and then
1
lim µ(X \ E0 ) pv = 0.
(3.28)
p→0
In addition, when p is chosen to be small enough such that pu < r, part (a) implies
that kf kpu ≤ kf kr < ∞. Hence we observe from the limit (3.28) and this fact that
1
1
lim kf kp ≤ lim µ(X \ E0 ) pv kf kpu ≤ kf kr lim µ(X \ E0 ) pv = 0.
p→0
p→0
p→0
In other words, the desired result holds in this case.
– Case (ii): µ(E0 ) = 0. Then |f (x)| > 0 a.e. on X. By the fact used in Case (i), we
may assume that
0 < |f (x)| < ∞ a.e. on X.
(3.29)
This fact allows us to apply [51, Eqn. (7), p. 63] to the almost positive function |f |p
to obtain
o
n1 Z
o
nZ
o1
nZ
p
p
p
log |f | dµ
≥ exp
log |f | dµ = exp
|f | dµ
kf kp =
p X
X
X
3.2. Relations among Lp -Spaces and some Consequences
79
for every p ∈ (0, r). In particular,
lim kf kp ≥ exp
p→0+
nZ
X
o
log |f | dµ .
(3.30)
For the reverse direction, we consider the function φ : (0, ∞) → R defined by
φ(x) = x − 1 − log x.
Since φ′ (x) = 1 − x1 = 0 and φ′′ (x) = x12 , φ has the absolute minimum at x = 1, i.e.,
φ(x) ≥ φ(1) on (0, ∞) or equivalently
log x ≤ x − 1
(3.31)
on (0, ∞). Recall the fact (3.29), so we replace x by kf kpp in the inequality (3.31) and
then using the fact that µ(X) = 1 to obtain
Z
Z
Z |f |p − 1
kf kpp − 1
1
p
log kf kp ≤
dµ =
|f | dµ −
=
dµ.
(3.32)
p
p X
p
X
X
Let E = {x ∈ X | |f (x)| = 1} and F = {x ∈ X | |f (x)| =
6 1}. Then we get
Z
Z
Z
|f |p − 1
|f |p − 1
|f |p − 1
dµ =
dµ +
dµ.
p
p
p
E
X
F
(3.33)
It is clear from Proposition 1.24(d) that
Z
|f |p − 1
dµ = 0,
p
E
so it suffices to evaluate
Z
|f |p − 1
dµ.
p→0+ F
p
To this end, we need the following lemma:
lim
Lemma 3.4
Let |f | > 0 and kf kr < ∞ for some r > 0. Define ψ : R → R by ψ(p) = |f |p . Then
we have
|f |p − 1
∈ L1 (µ)
p
for all p ∈ (0, r).
Proof of Lemma 3.4. Since ψ ′′ = |f |p (log |f |)2 > 0 on R, ψ is convex on R. Thus we
see from [49, Exercise 23, p. 101] that, for 0 < p < r,
ψ(p) − ψ(0)
ψ(r) − ψ(0)
≤
p−0
r−0
which reduces to
and so
|f |p − 1
|f |r − 1
≤
p
r
Z
X
|f |p − 1
dµ ≤
p
In other words, we have
lemma.
|f |p −1
p
Z
X
|f |r − 1
1
1
dµ = kf krr − < ∞.
r
r
r
∈ L1 (µ) for all p ∈ (0, r), completing the proof of the
Chapter 3. Lp -Spaces
80
Let’s return to the proof of the problem. By Theorem 1.34 (Lebesgue’s Dominated
Convergence Theorem), we obtain that
Z
Z
|f |p − 1 |f |p − 1
lim
dµ.
(3.34)
lim
dµ =
p→0+ F
p
p
F p→0+
By L’ Hôspital Rule, we derive
|f |p − 1
= lim |f |p log |f | = log |f |
p→0+
p→0+
p
lim
(3.35)
on X. Therefore, when we combine the results (3.32), (3.33), (3.34) and (3.35), we
establish that
Z
Z
Z
|f |p − 1 lim
log |f | dµ.
(3.36)
lim log kf kp ≤
log |f | dµ =
dµ =
p→0+
p
F p→0+
X
F
Since log x is continuous for x > 0, we get from the inequality (3.36) that
nZ
o
lim kf kp ≤ exp
log |f | dµ .
p→0+
(3.37)
X
Hence the required result follows immediately from the inequalities (3.30) and (3.37).
This ends the proof of the problem.
Problem 3.6
Rudin Chapter 3 Exercise 6.
Proof. Let x > 0 and 0 ≤ c ≤ 1. We claim that the function Φ : [0, ∞) → R satisfies the relation
cΦ(x) + (1 − c)Φ(1) = Φ(xc ).
(3.38)
Obviously, the relation (3.38) holds for c = 0 and c = 1. Without loss of generality, we may
assume that 0 < c < 1 in the following discussion.
Since f is bounded measurable and positive on [0, 1], there exists a positive constant M such
that
Z 1
|f | dx ≤ M < ∞.
kf k1 =
0
Thus we get from Problem 3.5(d) that
lim kf kp = exp
p→0
nZ
1
0
o
log |f (t)| dt .
(3.39)
By putting the expression (3.39) into the relation in question, we obtain
nZ
Φ exp
0
1
log |f (t)| dt
We define f : [0, 1] → R by
f (t) =
o
=
Z
0
1
Φ f (t) dt.

 x, if x ∈ [0, c];

1, if x ∈ (c, 1].
(3.40)
(3.41)
3.2. Relations among Lp -Spaces and some Consequences
81
Since x > 0, the f is bounded measurable and positive on [0, 1]. Therefore, we substitute the
function (3.41) into the relation (3.40) to yield
Φ(xc ) = Φ exp(c log x)
Z c
= Φ exp
log x dt
0
Z 1
Z c
Φ(1) dt
Φ(x) dt +
=
0
c
= cΦ(x) + (1 − c)Φ(1).
This proves the claim (3.38).
Next, we prove a lemma which is useful of determining the explicit form of the function Φ:
Lemma 3.5
The relation (3.38) holds for all x > 0 and c ≥ 0.
Proof of Lemma 3.5. The case 0 ≤ c ≤ 1 is clear by the above analysis. Let c > 1 so
that 0 < 1c < 1. Then the relation (3.38) becomes
1
1
1
1
Φ x c = Φ(x) + 1 −
Φ(1).
c
c
(3.42)
Take y = x c . Since 1c > 0, y will take all positive real numbers. Therefore, it follows
from the expression (3.42) that
1
1
Φ(1)
Φ(y) = Φ(y c ) + 1 −
c
c
and after simplification, we have
Φ(y c ) = cΦ(y) + (1 − c)Φ(1).
This proves the lemma.
We return to the proof of the problem. Define Ψ : [0, ∞) → R by Ψ(x) = Φ(x) − Φ(1), then
the relation (3.38) can be rewritten as
Ψ(xc ) = cΨ(x),
(3.43)
where x > 0 and c ≥ 0. Take x = e and c = ln y, where y ≥ 1. Then the expression (3.43)
becomes
Ψ(y) = Ψ(e) ln y,
(3.44)
where y ≥ 1. For 0 < y < 1, we put x = e−1 and c = ln(y −1 ) > 0 into the expression (3.43) so
that
−1 Ψ(y) = Ψ e− ln(y ) = −Ψ(e−1 ) ln y.
(3.45)
Combining the results (3.44) and (3.45), we establish that

if x ≥ 1;
 Ψ(e) ln x,
Ψ(x) =

−Ψ(e−1 ) ln x, if 0 < x < 1
Chapter 3. Lp -Spaces
82
which imply that
Φ(x) =

 Φ(1) + [Φ(e) − Φ(1)] ln x,

if x ≥ 1;
Φ(1) − [Φ(e−1 ) − Φ(1)] ln x, if 0 < x < 1,
completing the proof of the problem.
Problem 3.7
Rudin Chapter 3 Exercise 7.
Proof. In this problem, we assume that µ is a positive measure. We give examples one by one:
Example 3.1. Let X = N with the counting measure µ. By Definition 3.6, we have
∞
o
1
n
X
p
<∞ .
ℓp (N) = x = {ξn } kxkp =
|ξn |p
n=1
Let r < s. Thus if {ξn } ∈ ℓr (N), then {|ξn |r } is a convergent real sequence so that there exists
a positive integer N such that |ξn | ≤ 1 for all n ≥ N , see [49, Theorem 3.23, p. 60]. Therefore,
we must have |ξn |s ≤ |ξn |r for all n ≥ N and this implies that
∞
X
n=N
s
|ξn | ≤
∞
X
n=N
|ξn |r < ∞.
In other words, we prove that {ξn } ∈ ℓs (N) and then ℓr (N) ⊆ ℓs (N).
Example 3.2. We consider the Lebesgue measure m on X = [0, 1]. Since m(X) = 1, Problem
3.5(c) shows that Ls ([0, 1]) ⊆ Lr ([0, 1]) if 0 < r < s.
Example 3.3. Consider X = (0, ∞) and the Lebesgue measure m. Let 1 ≤ r < s ≤ ∞. Take
1
1
s < α < r and define f : (0, ∞) → R by
f (x) = x−α χ(0,1) (x).
Then we have
nZ ∞
o1 n Z
r
r
|f | dx
=
0
1
x
−αr
dx
0
o1
r
and
nZ
∞
0
s
|f | dx
o1
s
=
nZ
1
x−αs dx
0
o1
s
.
(3.46)
By Lemma 3.1, the integrals on the right-hand sides of the two expressions (3.46) converge if
and only if αr < 1 and αs < 1 respectively. Since 1s < α < 1r , we conclude that f ∈ Lr ((0, ∞))
but f ∈
/ Ls ((0, ∞)), i.e., Lr ((0, ∞)) * Ls ((0, ∞)).
Similarly, we define g : (0, ∞) → R by
g(x) = x−α χ(1,∞) (x).
Then we have
nZ ∞
o1 n Z
r
r
=
|f | dx
0
1
∞
x
−αr
dx
o1
r
and
nZ
∞
0
s
|f | dx
o1
s
=
nZ
∞
1
x−αs dx
o1
s
.
Thus Lemma 3.1 again shows that g ∈ Ls ((0, ∞)) but g ∈
/ Lr ((0, ∞)). In other words, it means
that Ls ((0, ∞)) * Lr ((0, ∞)).
3.2. Relations among Lp -Spaces and some Consequences
83
Next, we are going to find conditions on µ under which these situations will occur:
• Lr (µ) ⊆ Ls (µ) for 0 < r < s ≤ ∞. Recall from Lemma 3.3 that it is equivalent to the
condition that inf{µ(E) | E ∈ M0 } > 0, where M0 = {E ∈ M | µ(E) > 0}.
• Ls (µ) ⊆ Lr (µ) for 0 < r < s < ∞. Similar to the proof of Lemma 3.3, the following result
is part of the statements in [62, Theorem 2]:
Lemma 3.6
Suppose that (X, M, µ) is a measure space and M∞ = {E ∈ M | µ(E) is finite}.
Then the following conditions are equivalent:
– Condition (1). sup{µ(E) | E ∈ M∞ } < ∞,
– Condition (2). Ls (µ) ⊆ Lr (µ) for all 0 < r < s < ∞.
Proof of Lemma 3.6. Suppose that Condition (1) holds. To begin with, let f ∈ Ls (µ)
1
≤ |f (x)| < n1 } for each n = 1, 2, . . .. Then it is clear that En ∈ M.
and En = {x ∈ X | n+1
Now for each fixed n, we see that
Z
Z
Z
µ(En )
1
s
s
dµ =
|f | dµ ≥
|f | dµ ≥
s
(n + 1) En
(n + 1)s
En
X
which implies that
µ(En ) ≤ (n + 1)s
Z
X
|f |s dµ = (n + 1)s kf kss < ∞.
Therefore, we conclude that En ∈ M∞ . Note that Ei ∩ Ej = ∅ if i 6= j and µ is a positive
measure, if Fn = E1 ∪ E2 ∪ · · · ∪ En , then we deduce from Theorem 1.19(b) that
µ(Fn ) = µ
n
[
Ek =
k=1
n
X
k=1
µ(Ek ) < ∞.
In other words, each Fn is an element of M∞ . Next, we know that F1 ⊆ F2 ⊆ · · · and
n
n
[
[
Fk =
Ek , so Theorem 1.19(d) implies that
k=1
k=1
lim µ(Fn ) = µ
n→∞
∞
[
n=1
∞
∞
X
[
µ(En ).
En =
Fn = µ
n=1
(3.47)
n=1
Given a positive integer n, it is clear that
µ(Fn ) ≤ sup{µ(Fk ) | k ∈ N} ≤ sup{µ(E) | E ∈ M∞ }.
(3.48)
Furthermore, by combining Condition (1), the expression (3.47) and the inequality
(3.48), it yields that
∞
X
n=1
µ(En ) = lim µ(Fn ) ≤ sup{µ(E) | E ∈ M∞ } < ∞.
n→∞
Chapter 3. Lp -Spaces
On X \
∞
[
n=1
84
En , we claim that |f (x)| ≥ 1. Otherwise, there is a x0 ∈ X \
∞
[
En such that
n=1
|f (x0 )| < 1. By the definition, x0 ∈ EN for some N ∈ N, a contradiction. This observation
implies that
|f (x)|r ≤ |f (x)|s
on X \
Z
∞
[
En . Finally, we have
n=1
r
X
|f | dµ =
Z
S
X\ ∞
n=1 En
r
|f | dµ +
∞ Z
X
r
n=1 En
|f | dµ ≤
Z
X
|f |s dµ +
∞
X
µ(En )
n=1
nr
< ∞.
This proves that Condition (2) holds.
Conversely, since Ls (µ) ⊆ Lr (µ) implies that Lαs (µ) ⊆ Lαr (µ) for every α ∈ (0, ∞),
we may assume that r ≥ 1. It is known that if r, s ∈ [1, ∞] and Ls (µ) ⊆ Lr (µ), then the
mapping
T : Ls (µ) → Lr (µ)
is continuous, see [62, Lemma 1]. This result guarantees the existence of a positive constant
k such that kf kr ≤ kkf ks for all f ∈ Ls (µ), so
rs
µ(E) ≤ k s−r
for every E ∈ M∞ and this implies Condition (1).
We have completed the proof of the problem.
Problem 3.8
Rudin Chapter 3 Exercise 8.
Proof. For every n ∈ N, since g(x) → ∞ as x → 0, there exists a sequence {ǫn } of positive
numbers such that
g(x) ≥ n
(3.49)
for every x ∈ (0, ǫn ). Without loss of generality, we may assume that {ǫn } is decreasing and
ǫn+1 <
ǫn
2
(3.50)
for every n = 1, 2, . . .. Now we define hn : (0, 1) → R and h : (0, 1) → R by
x
hn (x) = n 1 −
χ
(x) and h(x) = sup hn (x)
ǫn (0,ǫn )
n
respectively.
Let x ∈ (0, 1) be fixed but arbitrary. Since ǫn → 0 as n → ∞, there exists a positive integer
Nx such that x ≤ ǫn for all 1 ≤ n ≤ Nx and x ≥ ǫn for all n > Nx . Thus we have x ∈ (0, ǫn )
for all 1 ≤ n ≤ Nx and x ∈
/ (0, ǫn ) for all n > Nx . See Figure 3.1 for an illustration of the
distribution of x and ǫn .
3.2. Relations among Lp -Spaces and some Consequences
85
Figure 3.1: The distribution of x and ǫn .
These facts imply that
so that
 x

n
1
−
, if 1 ≤ n ≤ Nx ;

ǫn
hn (x) =


0,
if n > Nx
(3.51)
x
h(x) = sup hn (x) = sup{h1 (x), h2 (x), . . .} = max n 1 −
< ∞,
1≤n≤Nx
ǫn
n
i.e., h is finite. Since each hn is convex on (0, 1), Problem 3.1 shows that the h is also convex
on (0, 1).
Next, we want to compare the magnitudes of hn (x) and g(x) at the fixed point x ∈ (0, 1).
On the one hand, for 1 ≤ n ≤ Nx , we have x ∈ (0, ǫn ), so n(1 − ǫxn ) ≤ n and we deduce from
the inequality (3.49) and the definition (3.51) that
hn (x) ≤ g(x)
(3.52)
for 1 ≤ n ≤ Nx . On the other hand, if n > Nx , then x ∈
/ (0, ǫn ) and we have hn (x) = 0, but the
inequality (3.49) still holds for n = Nx (i.e., x ∈ (0, ǫNx )) so that the inequality (3.52) remains
valid in this case. Hence we have established that
h(x) ≤ g(x)
(3.53)
for the fixed point x ∈ (0, 1). Since x is arbitrary, the inequality (3.53) holds on (0, 1).
Take n = Nx − 1 in the definition (3.51) to get
hNx −1 (x) = (Nx − 1) 1 −
x
ǫNx −1
.
(3.54)
By the hypothesis (3.50), we have 2ǫNx < ǫNx −1 . Then since x < ǫNx , we have
x
1
x
< .
<
ǫNx −1
2ǫNx
2
(3.55)
By substituting the estimate (3.55) into the expression (3.54), we obtain
hNx −1 (x) >
Nx − 1
.
2
Since x → 0 if and only if Nx → ∞, we conclude that hNx (x) → ∞ as x → 0 and the definition
implies that h(x) → ∞ as x → 0 as required.
1
The second assertion is false. To see this, consider g(x) = x 3 which satisfies g(x) → ∞ as
x → ∞. Assume that there was a convex function h : (0, ∞) → R such that
1
h(x) ≤ x 3
(3.56)
Chapter 3. Lp -Spaces
86
on (0, ∞) and h(x) → ∞ as x → ∞. We know from [66, Exercise 6(c), p. 262]c that if
h : (0, ∞) → R is convex on (0, ∞), then the ratio
h(x)
x
tends to a finite limit or to infinity as x → ∞. If this limit is finite, since h(x) → ∞ as x → ∞,
this limit must be positive. Thus it must be true that
h(x) ≥ kx
(3.57)
as x → ∞ for some positive constant k. However, the inequality (3.57) definitely contradicts
the inequality (3.56). This completes the proof of the problem.
Problem 3.9
Rudin Chapter 3 Exercise 9.
Proof. Let Φ : (0, ∞) → (0, ∞) be such that Φ(p) → ∞ as p → ∞. Suppose that E1 = (0, 21 )
and
h1
1
1
1 1
1 h
1
1
+ 2 + · · · + n−1 , + 2 + · · · + n = 1 − n−1 , 1 − n ,
En =
2 2
2
2 2
2
2
2
where n ≥ 2. Obviously, we have En ⊂ (0, 1) for every n ∈ N and Ei ∩ Ej = ∅ for i 6= j.
Furthermore, each En is measurable and
m(En ) =
1
> 0.
2n
Suppose that x ∈ (0, 1). If x < 21 , then x ∈ E1 . Otherwise, 12 ≤ x < 1 and there exists a δ > 0
such that x < δ < 1. Take N to be least positive integer such that
N >−
Then we have x < 1 −
1
2N
log(1 − δ)
> 0.
log 2
< 1 so that x ∈ E2 ∪ E3 ∪ · · · ∪ EN . In other words, we get
(0, 1) =
∞
[
En .
n=1
Finally, we define fn , f : (0, 1) → (0, ∞) by
fn (x) =
n
X
kχEk (x)
and f (x) = lim fn (x) =
k=1
n→∞
∞
X
nχEn (x)
(3.58)
n=1
respectively. Since each En is measurable, each χEn is measurable by Proposition 1.9(d). Thus
each function fn is measurable and then Theorem 1.14 implies that f is also measurable. Now
it remains to show that the function f satisfies the required properties of the problem.
• Property 1: kf kp → ∞. Assume that f ∈ L∞ ((0, 1)). Then there exists a positive
constant M such that kf k∞ < M . By Definition 3.7, we see that
|f (x)| < M
c
a.e. on (0, 1).
See also [31, Theorem 126, p. 99] and [51, Eqn. (2), p. 62].
(3.59)
87
3.3. Applications of Theorems 3.3, 3.5, 3.8, 3.9 and 3.12
Since Ei ∩ Ej = ∅ for i 6= j, we have
|f (x)| = f (x) ≥ fn (x) ≥ n
(3.60)
for every x ∈ En . Since n → ∞, there exists a positive integer N such that N ≥ M .
Recall that m(EN ) = 21N > 0, so we obtain from the inequality (3.60) that
|f (x)| ≥ M
on EN , but this contradicts our assumption (3.59). Therefore, we must have f ∈
/ L∞ ((0, 1))
and in fact kf k∞ = ∞. Furthermore, it follows from the Ratio Test (see [49, Theorem
3.34]) that
Z 1
∞
∞
X
X
n
n × m(En ) =
f (x) dx =
kf k1 =
< ∞,
n
2
0
n=1
n=1
so we have kf k1 < ∞. Hence we deduce from Problem 3.4(e) that
kf kp → kf k∞ = ∞
as p → ∞.
• Property 2: kf kp ≤ Φ(p) for sufficiently large p. Fix a positive integer n, we consider
h n ip
Φ(p)
for p > 0. Since Φ(p) → ∞ as p → ∞, when p is sufficiently large, we have
that
h n ip
1
× m(En ) ≤ n .
Φ(p)
2
n
Φ(p)
≤ 1 so
(3.61)
Therefore, we see that
kf kpp
=
Z
0
∞
1hX
n=1
∞
ip
X
np × m(En ).
nχEn (x) dx =
(3.62)
n=1
Consequently, we follow from the inequality (3.61) and the expression (3.62) that
∞
∞ h
∞
X
X
X
kf kpp
n ip
1
1
p
× m(En ) ≤
n × m(En ) =
=
=1
n
[Φ(p)]p
[Φ(p)]p n=1
Φ(p)
2
n=1
n=1
for sufficiently large p, or equivalently,
kf kp ≤ Φ(p)
for sufficiently large p.
We end the proof of the problem.
3.3
Applications of Theorems 3.3, 3.5, 3.8, 3.9 and 3.12
Problem 3.10
Rudin Chapter 3 Exercise 10.
Chapter 3. Lp -Spaces
88
Proof. First of all, we have to show that f ∈ Lp (µ). Recall that Lp (µ) is a metric space with
metric k · k. By the triangle inequality
kfn − fm kp ≤ kfn − f kp + kf − fm kp
and the hypothesis kfn − f kp → 0 as n → ∞ together imply that {fn } is a Cauchy sequence in
Lp (µ). Now the completeness of Lp (µ) ensures that {fn } converges to an element of Lp (µ). By
the uniqueness of limits, this element must be f so that f ∈ Lp (µ). By Theorem 3.12, {fn } has
a subsequence {fnk } such that fnk → f a.e. on X. Since fn → g a.e. on X as n → ∞, every
subsequence of {fn } converges to g a.e. on X. In particular, we take this subsequence {fnk }
and the uniqueness of limits again imply that
f =g
a.e. on X. We have completed the proof of the problem.
Problem 3.11
Rudin Chapter 3 Exercise 11.
Proof. Since f, g : Ω → [0, ∞], the hypothesis f g ≥ 1 implies that
(Hölder’s Inequality), we see that
Z
Z p
2 Z
f g dµ ≤
g dµ.
f dµ ×
Ω
By Proposition 1.24(a), we have
Z p
Ω
f g dµ ≥
(3.63)
Ω
Ω
Z
√
f g ≥ 1. By Theorem 3.5
dµ = µ(Ω) = 1.
(3.64)
Ω
Now our desired result follows immediately by combining the inequalities (3.63) and (3.64). This
completes the analysis of the problem.
Problem 3.12
Rudin Chapter 3 Exercise 12.
√
Proof. Suppose that A = ∞. Since h ≤ 1 + h2 on Ω, we have
Z p
1 + h2 dµ = ∞,
Ω
so the
√ equalities
√ definitely hold in this case. Next, suppose that A = 0. By the fact that
1 ≤ 1 + h2 ≤ 1 + 2h + h2 = 1 + h, we achieve that
Z
Z
Z p
Z
2
dµ.
(3.65)
1 + h dµ ≤ (1 + h) dµ =
dµ ≤
Ω
Ω
Ω
Since µ(Ω) = 1, we conclude from the inequalities (3.65) that
Z p
1 + h2 dµ = 1.
Ω
Thus the equalities also hold in this case.
Ω
89
3.3. Applications of Theorems 3.3, 3.5, 3.8, 3.9 and 3.12
Suppose that 0 < A < ∞ which means h(x) ∈ (0, ∞) a.e. on Ω. We consider the function
ϕ : R → (0, ∞) defined by
p
ϕ(x) = 1 + x2
which implies that
ϕ′ (x) = √
x
1 + x2
and ϕ′′ (x) =
1
3
(1 + x2 ) 2
.
(3.66)
Since ϕ′′ (x) ≥ 0 on R, we know from [49, Exercise 14, p. 115] that ϕ is convex on R. In
particular, ϕ is convex on (0, ∞). By Theorem 3.3 (Jensen’s Inequality), we obtain that
p
1+
A2
=
s
1+
Z
Ω
Z p
h dµ ≤
1 + h2 dµ
(3.67)
Ω
which is exactly
√ the left-hand side inequality. The right-hand side inequality is easy because we
always have 1 + h2 ≤ 1 + h. Hence we have verified the validity of the inequalities.
For the second assertion, we suppose that µ = m on Ω = [0, 1], h = f ′ and h is continuous
on Ω. Then we have
Z 1
f ′ (x) dx = f (1) − f (0).
A=
0
In addition, we notice that
Z p
Ω
1+
h2 dµ
=
Z
0
1p
1 + [f ′ (x)]2 dx
which is the arc length of the curve from (0, f (0)) to (1, f (1)).d Thus the inequalities give
bounds of such arc length. Geometrically, the upper bound consists of the length from (0, f (0))
to (1, f (0)) (which is 1) and the length from (1, f (0)) to (1, f (1)) (which is A). The lower bound
is just the hypotenuse of the right triangle with vertices (0, f (0)), (0, f (1)) and (1, f (1)), see
Figure 3.2 below:
Figure 3.2: The geometric interpretation of a special case.
d
See, for examples, [2, p. 535] and [49, Theorem 6.27, p. 137].
Chapter 3. Lp -Spaces
90
Our first inequality (3.67) comes from the application of Theorem 3.3 (Jensen’s Inequality),
so the equality of the first inequality (3.67) holds if and only if the equality in [51, Eqn. (2), p.
62] holds if and only if
ϕ(A) = ϕ h(x) .
(3.68)
We see from the first derivative in (3.66) that ϕ′ (x) > 0 in (0, ∞) so that it is injective in (0, ∞).
Recall that A ∈ (0, ∞) and h(x) ∈ (0, ∞) a.e. on Ω, so we deduce that the equality (3.68) holds
if and only if
h = A a.e. on Ω.
√
Next we recall the second inequality is derived from the fact that 1 + h2 ≤ 1 + h and some
simple computations show that the equality holds if and only if
h≡0
a.e. on Ω.
Hence we have ended the proof of the problem.
Problem 3.13
Rudin Chapter 3 Exercise 13.
Proof. There are two cases.
• Case (i): 1 < p < ∞. Then the inequality in Theorem 3.8 is just Hölder’s inequality. If
kf kp = 0 or kgkp = 0, then we have f = 0 a.e. on X or g = 0 a.e. on X by Theorem
1.39(a). In either case, the equality holds trivially. Therefore, we may assume that both
kf kp > 0 and kgkp > 0. Furthermore, by the remark following the proof of Theorem 3.5
(Hölder’s Inequality), we may further assume that
0 < kf kp < ∞ and
0 < kgkp < ∞.
In this case, we see that the equality in Theorem 3.8 holds if and only if there are positive
constants α and β such that
αf p = βgq
a.e. on X.
In Theorem 3.9, the inequality is in fact Minkowski’s inequality. By simple observation,
it is clear that the equality holds if f = 0 a.e. on X or g = 0 a.e. on X. Therefore, we may
assume that f 6= 0 and g 6= 0 on measurable sets E and F with µ(E) > 0 and µ(F ) > 0
respectively. By examining the proof of [51, Eqn. (2), Theorem 3.5, pp. 64, 65], we know
that it applies Hölder’s inequality to the functions f and (f + g)p−1 as well as the functions
g and (f + g)p−1 . Thus we deduce from the previous paragraph that the equality in
Z
X
f · (f + g)p−1 dµ ≤
nZ
f p dµ
X
o1
p
×
nZ
(f + g)(p−1)q dµ
X
o1
q
holds if and only if there are positive constants α and β such that
αf p = β(f + g)q
a.e. on X.
Similarly, the equality in
Z
o1
o1 n Z
nZ
p
q
(p−1)q
p
p−1
(f + g)
dµ
×
g dµ
g · (f + g)
dµ ≤
X
X
X
(3.69)
91
3.4. Hardy’s Inequality and Egoroff ’s Theorem
holds if and only if there are positive constants λ and ν such that
λg p = ν(f + g)q
a.e. on X.
(3.70)
By combining the equations (3.69) and (3.70), we conclude that
fp =
β
βλ p
(f + g)q =
g
α
αν
a.e.
which means that f = Kg a.e. for some positive constant K.
• Case (ii): p = ∞. The inequality in Theorem 3.8 comes from the inequalitye
|f (x)g(x)| ≤ kf k∞ · |g(x)|
for almost all x. Therefore, it is easy to see that the equality in Theorem 3.8 holds if and
only if |f (x)g(x)| = kf k∞ · |g(x)| for almost all x if and only if g(x) = 0 for almost x or
f (x) = kf k∞ for almost x.f
Now the triangle inequality
|f + g| ≤ |f | + |g|
implies the inequality in Theorem 3.9. Thus the equality in Theorem 3.9 holds if and only
if f and g are either nonnegative functions or nonpositive functions for almost all x.
This completes the proof of the problem.
3.4
Hardy’s Inequality and Egoroff’s Theorem
Problem 3.14
Rudin Chapter 3 Exercise 14.
Proof. We follow the suggestions given by Rudin.
(a) We divide the proof into two cases:
– Special Case: f ≥ 0 and f ∈ Cc (0, ∞) . By the First Fundamental Theorem of
Calculus, the function F (x) is differentiable on (0, ∞) and
xF ′ (x) = f (x) − F (x)
(3.71)
for every x ∈ (0, ∞). By Integration by Partsg and the expression (3.71), we have
Z ∞
Z ∞
p
p
∞
pxF p−1 (x)F ′ (x) dx.
(3.72)
F (x) dx = [xF (x)]0 −
0
0
Now we have to evaluate the limits:
lim xF p (x) and
x→0
e
lim xF p (x).
x→∞
This is [51, Eqn. (2), Theorem 3.8, p. 66].
The latter case means that f is a nonnegative constant for almost all x.
g
Here we assume that the formula for integration by parts is also valid for improper integrals, see [3, p. 278].
f
Chapter 3. Lp -Spaces
92
Since supp (f ) = {x ∈ (0, ∞) | f (x) 6= 0} is compact, the Heine-Borel Theorem guarantees that supp (f ) = [a, b] ⊂ (0, ∞). By this and the continuity of f , we conclude
that f is bounded on (0, ∞), i.e., |f (x)| ≤ M on (0, ∞) for some positive constant
M . Thus we follow from this that
Z
1 x
|F (x)| ≤
|f (t)| dt ≤ M
x 0
for every x ∈ (0, ∞), so we see that
lim xF p (x) = 0.
(3.73)
x→0
Again, the boundedness of f implies that
Z b
Z ∞
Z x
|f (t)| dt ≤ M (b − a) < ∞
|f (t)| dt =
|f (t)| dt ≤
|xF (x)| ≤
0
0
on (0, ∞). Since xF p =
that
(xF )p
xp−1
(3.74)
a
and p > 1, it reduces from these and the bound (3.74)
lim xF p (x) = 0.
(3.75)
x→∞
Therefore, we deduce from the formula (3.72) and the two limits (3.73) and (3.75)
that
Z ∞
Z ∞
p
F p−1 (x)[f (x) − F (x)] dx
F (x) dx = −p
0
Z ∞
Z0 ∞
F p (x) dx
F p−1 (x)f (x) dx + p
= −p
0
0
so that
(p − 1)
Z
∞
F p (x) dx = p
Z
∞
F p−1 (x)f (x) dx.
(3.76)
0
0
Note that 1 < p < ∞, let q be its conjugate exponent. Obviously, our hypotheses
make sure that f and F p−1 have range [0, ∞], so we may apply Theorem 3.5 (Hölder’s
Inequality) to them. In fact, it yields from the fact (p − 1)q = p and the expression
(3.76) that
Z ∞
p
F p (x) dx
(p − 1)kF kp = (p − 1)
0
o1
nZ ∞
o1n Z ∞
p
q
F (p−1)q (x) dx
≤p
f p (x) dx
0
0
p
q
= pkf kp · kF kp
which is exactly
p− pq
kF kp = kF kp
≤
p
kf kp .
p−1
(3.77)
– General Case: f ∈ Lp (0, ∞) . By Definition 3.6, we always have
kf kp = |f | p ,
so we may assume that f ≥ 0 on (0, ∞). Since 1 < p < ∞, Theorem 3.14 says that
Cc (0, ∞) is dense in Lp (0, ∞) . Thus there exists a sequence {fn } ⊆ Cc (0, ∞)
such that
kfn − f kp → 0
(3.78)
93
3.4. Hardy’s Inequality and Egoroff ’s Theorem
as n → ∞. Notice that fn may not be nonnegative, but since |fn | − |f | ≤ |fn − f |
and f ≥ 0, we have
|fn (x)| − f (x) ≤ |fn (x) − f (x)|
(3.79)
for every x ∈ (0, ∞). Certainly, the inequality (3.79) implies that
|fn | − f
p
≤ kfn − f kp
(3.80)
for every n = 1, 2, . . .. Now the result (3.78) and the inequality (3.80) ensure that
|fn | − f
p
→0
as n → ∞. In other words,
we may also assume that fn ≥ 0 for each n = 1, 2, . . ..
Recall that fn ∈ Cc (0, ∞) , so we derive from the Special Case that
kFn kp ≤
where
p
kfn kp ,
p−1
1
Fn (x) =
x
Z
1
F (x) =
x
Z
and n = 1, 2, . . .. Suppose that
(3.81)
x
fn (t) dt
0
x
f (t) dt.
0
Since f ≥ 0 and fn ≥ 0, it is easily seen from the definition that F ≥ 0 and Fn ≥ 0.
We claim that {Fn } converges pointwise to F on (0, ∞). To see this, fix x ∈ (0, ∞),
then their definitions give
Z
1 x
|fn (t) − f (t)| dt.
(3.82)
|Fn (x) − F (x)| ≤
x 0
By the result (3.78), we know that for every ǫ > 0, there exists a positive integer N
such that n ≥ N implies
kfn − f kp < ǫ.
Apply Theorem 3.5 (Hölder’s Inequality) with 1p + 1q = 1 to the inequality (3.82), we
get immediately that
Z
o1 n Z x
o1
1n x
p
q
p
|fn (t) − f (t)| dt ×
1q dt
|Fn (x) − F (x)| ≤
x 0
0
1
1
≤ kfn − f kp × x q
x
1
= x− p kfn − f kp
1
< ǫx− p
(3.83)
for all n ≥ N . Since x is fixed and independent of ǫ, the estimate (3.83) verifies the
truth of the claim.
Next, recall that each fn is continuous on [a, x] for every a > 0, so the First
Fundamental Theorem of Calculus shows that each Fn is continuous on [a, x]. As a
continuous function, every Fn is measurable on [a, x].h Hence, it follows from the
h
See §1.11 or [49, Example 11.14].
Chapter 3. Lp -Spaces
94
pointwise convergence of {Fn }, Theorem 1.28 (Fatou’s Lemma), the inequality (3.81)
and then the result (3.78) that
Z ∞
p
F p (x) dx
kF kp =
Z0 ∞
lim inf Fnp (x) dx
=
n→∞
0
Z ∞
Fnp (x) dx
≤ lim inf
n→∞ 0
Z ∞
Fnp (x) dx
≤ lim
n→∞ 0
= lim kFn kpp
n→∞
p p
≤
lim kfn kpp
p − 1 n→∞
p p
kf kpp
≤
p−1
which is equivalent to the desired result
kF kp ≤
p
kf kp .
p−1
Hence the mapping T : Lp (0, ∞) → Lp (0, ∞) given by T (f ) = F is continuous.
(b) Suppose that f ∈ Lp (0, ∞) and
kF kp =
p
kf kp < ∞.
p−1
(3.84)
Recall that kf kp = |f | p , so we may suppose further that f ≥ 0 on (0, ∞). By the
General Case of the proof in part (a), there is a nonnegative sequence {fn } ⊆ Cc (0, ∞)
such that kfn −f kp → 0 as n → ∞. Now we replace F and f by Fn and fn in the expression
(3.76) respectively, we gain
Z ∞
Z ∞
p
p
p
Fn (x) dx =
kFn kp =
Fnp−1 (x)fn (x) dx
(3.85)
p−1 0
0
for each n = 1, 2, . . .. Since fn ∈ Cc (0, ∞) , T (fn ) = Fn ∈ Lp (0, ∞) , where T is the
continuous mapping considered in part (a). By Theorem 3.5 (Hölder’s Inequality), it is
easy to see that fn Fnp−1 ∈ Lp (0, ∞) for each n = 1, 2, . . .. Since the mapping T is
continuous, it follows from the expression (3.85) that
Z ∞
Z ∞
p
p
F p−1 (x)f (x) dx.
(3.86)
F (x) dx =
p
−
1
0
0
In other words, the formula (3.76) holds for f ∈ Lp (0, ∞) . We apply Theorem 3.5
(Hölder’s Inequality) to the right-hand side of the inequality (3.86) and then using our
hypothesis (3.84) to conclude that
Z ∞
p
p
kF kp =
F p−1 (x)f (x) dx
p−1 0
Z
o1 n Z ∞
o1
p n ∞ p
p
q
p
f (x) dx
×
(3.87)
F (x) dx
≤
p−1
0
0
95
3.4. Hardy’s Inequality and Egoroff ’s Theorem
p
p
kf kp × kF kpq
p−1
= kF kpp .
=
Hence the equality in the inequality (3.87) must hold. Consequently, there are constants
α and β, not both 0, such that αf p = β(F p−1 )q = βF p a.e. on (0, ∞) and so
1
1
αpf = β pF
a.e. on (0, ∞).
(3.88)
Here we have several cases:
– Case (i): β = 0. Then we have α 6= 0 and f = 0 a.e. on (0, ∞). Thus we are done.
– Case (ii): α = 0. Then we have β 6= 0 and F = 0 a.e. on (0, ∞). By the hypothesis
(3.84), we see that kf kp = 0 which gives f = 0 a.e. on (0, ∞) and we are done again.
– Case (iii): αβ 6= 0. By Theorem 3.8, we see that f ∈ L1 ([a, b]), where [a, b] is any
bounded interval of (0, ∞). By the comment following the proof of Theorem 11.3 on
[49, p. 324], we know that F is differentiable a.e. on [a, b] and
′
xF (x) = f (x)
(3.89)
a.e. on [a, b]. Since [a, b] is arbitrary, the formula (3.89) holds a.e. on (0, ∞). Combining the equations (3.88) and (3.89), we see immediately that xF ′ + xF = f which
implies
xf ′ = (c − 1)f a.e. on (0, ∞)
(3.90)
for some nonzero constant c. By solving the differential equation (3.90), we obtain
that
f (x) = γxc−1 a.e. on (0, ∞)
for some constant γ. Since γxc−1 ∈ Lp ((0, ∞)), it forces that γ = 0. Hence we have
shown that the equality holds only if f = 0 a.e. on (0, ∞).
1
(c) Take f (x) = x− p χ[1,A] (x) for large A. Then we have
kf kp =
nZ
0
∞
|x
− p1
p
χ[1,A] (x)| dx
o1
p
=
nZ
A
x
1
−1
dx
o1
p
1
= (log A) p .
Next, we know from the definition that
Z
1 x − p1
t χ[1,A] (t) dt
F (x) =
x 0

0,
if x ∈ (0, 1);






Z x


1
 1
t− p dt, if x ∈ [1, A];
=
x 1




Z


1 A − p1



t dt, if x ∈ (A, ∞);
x 1

0,
if x ∈ (0, 1);








 p (x− p1 − x−1 ), if x ∈ [1, A];
p−1
=





1− 1

A p −1
p



·
, if x ∈ (A, ∞).
p−1
x
(3.91)
(3.92)
Chapter 3. Lp -Spaces
96
Note that A is large and 1 −
1
p
> 0, so we have
1
A1− p − 1 > 0.
(3.93)
Therefore, we deduce from the expressions (3.92) and the fact (3.93) that
Z ∞
p
|F (x)|p dx
kF kp =
0
Z ∞
Z A
|F (x)|p dx
|F (x)|p dx +
=
1
A
p p Z ∞ A1− p1 − 1 p
p p Z A
− p1
−1 p
|x − x | dx +
dx
=
p−1
p−1
x
1
A
p p Z A
p 1−p
1
pp
1− p1
(x− p − x−1 )p dx +
A
=
A
−
1
p−1
(p − 1)p+1
1
p p Z A
1
(x− p − x−1 )p dx.
>
p−1
1
Fix p first. Assume that the constant
for some δ < 1, i.e.,
p
p−1
kF kp ≤
(3.94)
could be replaced by a smaller number
δp
kf kp .
p−1
δp
p−1
1
Take ǫ > 0 such that δ < ǫ < 1. Then it is clear that for large A, we have x p −1 < 1 − ǫ
for x ≥ A2 or equivalently,
x
for x ≥
A
2.
Z
− p1
− x−1 > ǫx
− p1
>0
(3.95)
By this estimate (3.95), we obtain
A
(x
− p1
1
−x
−1 p
) dx >
Z
A
A
2
A
> ǫp log A.
ǫp x−1 dx = ǫp log A − log
2
(3.96)
By substituting the inequality (3.96) into the inequality (3.94) and then using the fact
(3.91), we can show that
ǫp p
δp p
kF kpp >
log A >
kf kpp ≥ 0,
p−1
p−1
but this implies
kF kp >
a contradiction.
δp
kf kp ,
p−1
(d) Since f (x) > 0 on (0, ∞), there exists a δ > 0 such that we can find an α ∈ (0, ∞) with
Z α
f (t) dt > δ > 0.
0
By the definition, we have F (x) > 0 on (0, ∞) so that
Z ∞
Z ∞
Z ∞
Z ∞ Z x
δ
1
f (t) dt dx >
dx = ∞.
kF k1 =
F (x) dx ≥
F (x) dx =
0
α x
0
α
α x
Hence F ∈
/ L1 . In other words, this shows that Hardy’s inequality does not hold when
p = 1.
97
3.4. Hardy’s Inequality and Egoroff ’s Theorem
Hence we have completed the proof of the problem.
Problem 3.15
Rudin Chapter 3 Exercise 15.
Proof. We follow the hint. Suppose that {an } is decreasing. Let f : (0, ∞) → (0, ∞) be given
by
∞
X
an χ(n−1,n] (x).
(3.97)
f (x) =
n=1
Firstly, we note that
Z
∞
p
f (x) dx =
0
which implies that f ∈
∞ Z
X
n
n=1 n−1
apn dx
=
∞
X
apn
(3.98)
n=1
∞
X
apn < ∞. This shows that the desired
(0, ∞) if and only if
Lp
n=1
∞
X
p
apn < ∞ in
inequality holds if f ∈
/ L (0, ∞) . Without loss of generality, we assume that
n=1
the following discussion.
Secondly, let N be a positive integer and x ∈ (N − 1, N ]. Then we deduce from the definition
(3.97) that
Z
1 x
F (x) =
f (t) dt
x 0
Z x
N −1 Z
i
1h X n
f (t) dt
f (t) dt +
=
x
N −1
n=1 n−1
Z
Z
N −1
i
x
1h X n
=
aN dt
an dt +
x n=1 n−1
N −1
1
[a1 + a2 + · · · + aN −1 + aN (x − N + 1)].
x
=
(3.99)
It is clear from the expression (3.99) and the fact {an } is decreasing that
N −1
N
i 1X
1h X
an + aN (x − N + 1) =
an − N aN + xaN
x
x
n=1
=
1
x
n=1
N
X
n=1
an − N aN + aN
N
1 X
≥
an − N aN + aN
N n=1
N
1 X
an .
=
N
(3.100)
n=1
Thus we obtain from the expression (3.99) and the inequality (3.100) that
kF kp =
nZ
∞
0
F p (x) dx
o1
p
=
∞ Z
nX
N
N =1 N −1
F p (x) dx
o1
p
≥
∞ N
nX
1 X p o 1p
an
.
N n=1
N =1
(3.101)
Chapter 3. Lp -Spaces
98
Furthermore, by Problem 3.14(a) and the expression (3.98), we derive that
p
p n
kF kp ≤
kf kp =
p−1
p−1
Z
∞
p
f (x) dx
0
o1
p
p n X p o p1
a
=
.
p − 1 n=1 n
∞
(3.102)
Hence our desired inequality follows immediately from the inequalities (3.101) and (3.102).
∞
X
For the general case, recall the assumption that an > 0 for all n ∈ N and
apn < ∞, so the
n=1
sequence {apn } converges absolutely and every rearrangement converges to the same sum (see
[49, Theorem 3.55, p. 78]). On the other hand, if we fix a positive integer N , then
N X
k
X
1
k=1
k
n=1
an
p
=
a p
a + a p a + a + a p
1
2
1
2
3
+
1
2
3
a + a + · · · + a p
1
2
N
,
+ ··· +
N
1
+
so it follows from this form that the sum attains its maximum if and only if a1 ≥ a2 ≥ · · · ≥ aN .
Therefore, we have
∞ ∞ ∞
∞
N
N
X
X
1 X p p p X p p p X p
1 X p
αn =
an ,
an ≤
αn ≤
N
N
p−1
p−1
n=1
n=1
n=1
n=1
N =1
N =1
{z
}
|
By the special case.
where {αn } is the decreasing rearrangement of {an }. This completes the proof of the problem.
Problem 3.16
Rudin Chapter 3 Exercise 16.
Proof. Let’s prove the assertions one by one.
• A proof of Egoroff ’s Theorem. Put
\ n
1o
.
S(n, k) =
x ∈ X |fi (x) − fj (x)| <
k
(3.103)
i,j>n
For each k ∈ N, we follow from the definition (3.103) that
\
S(n, k) =
i,j>n+1
∩
n
\ n
1o
1o
∩
x ∈ X |fi (x) − fj (x)| <
x ∈ X |fi (x) − fj (x)| <
k
k
\ n
1o
x ∈ X |fi (x) − fj (x)| <
k
i=n+1
j>n
j=n+1
i>n
which implies that S(n, k) ⊆ S(n + 1, k).
Next, for each fixed k ∈ N, we claim that
X=
∞
[
n=1
S(n, k).
(3.104)
99
3.4. Hardy’s Inequality and Egoroff ’s Theorem
To this end, we see that the set inclusion
∞
[
S(n, k) ⊆ X
n=1
is evident. To prove the reverse direction, we note that for each k ∈ N and x ∈ X, we
yield from our hypothesis fn (x) → f (x) that there exists a positive integer N such that
i, j > N implies that
1
|fi (x) − fj (x)| < .
k
In other words, we must have x ∈ S(n, k), i.e.,
X⊆
∞
[
S(n, k).
n=1
Thus these prove the validity of our claim (3.104) and Theorem 1.19(d) gives
lim µ S(n, k) = µ(X)
n→∞
for every k = 1, 2, . . ..
Given that ǫ > 0. For each k ∈ N, this fact allows us to choose nk such that
ǫ
|µ S(nk , k − µ(X)| < k .
2
Define
E=
∞
\
S(nk , k).
(3.105)
k=1
Then it is clear that
X \E =
which implies that
µ(X \ E) ≤
Pick a k with
1
k
∞
X
k=1
∞
[
k=1
X \ S(nk , k)
∞
∞
X
X
ǫ
µ(X) − µ S(nk , k) <
µ X \ S(nk , k) ≤
= ǫ.
2k
k=1
k=1
< ǫ. If x ∈ S(nk , k), then we have
|fi (x) − fj (x)| < ǫ
(3.106)
for all i, j > nk . By the definition (3.105), we know that E ⊆ S(nk , k) for every k ∈ N,
so the inequality (3.106) holds on E. By the definition, fn → f uniformly on E, proving
Egoroff’s Theorem.
• A counterexample on a σ-finite space. Recall from Definition 2.16 that X is said to
be a σ-finite space if X is a countable union of sets Ei with µ(Ei ) < ∞. It is clear that
R=
∞
[
[n − 1, n) ∪
n=1
∞
[
(−n, 1 − n],
n=1
so R is σ-finite. Let fn (x) = nx on R. It is clear that the sequence converges pointwise to 0
at every point of R. Furthermore, we know that a measurable set E ⊆ R with m(R\E) < 1
must be unbounded. Otherwise, E ⊆ [−M, M ] for some M > 0 and this implies that
m(R \ E) ≥ m R \ [−M, M ] = ∞,
Chapter 3. Lp -Spaces
100
a contradiction. However, {fn } cannot converge uniformly to 0 on the unbounded set E.
Thus this counterexample shows that Egoroff’s Theorem cannot be extended to σ-finite
spaces.
• An extension of Egoroff ’s Theorem. Suppose that {ft } is a family of complex measurable functions such that
(i) lim ft (x) = f (x) and
t→∞
(ii) Fix a x ∈ X. Then the function F : (0, ∞) → C defined by
F (t) = ft (x)
is continuous.
For each n ∈ N, we consider the real function gn : X → R given by
gn (x) = sup{|ft (x) − f (x)|}.
(3.107)
t≥n
Then, for every x ∈ X, gn (x) → 0 as n → ∞. Now it remains to show that each gn is
measurable. To this end, let a ∈ R and
Ea (n) = {x ∈ X | gn (x) < a}
= {x ∈ X | |ft (x) − f (x)| < a for all t ≥ n}
\
=
{x ∈ X | |fr (x) − f (x)| < a}.
(3.108)
r≥n
r∈Q
Since fr is measurable, f is also measurable by Corollary (a) following Theorem 1.14. Thus
|fr − f | is measurable by Proposition 1.9(b) and then the set
{x ∈ X | |fr (x) − f (x)| < a}
is measurable for every a ∈ R by Definition 1.3(c). By the expression (3.108), since there
are countable such measurable sets, Comment 1.6(c) implies that Ea (n) is measurable for
every real a. By Definition 1.3(c) again, each gn is measurable and thus Egoroff’s Theorem
can be applied to conclude that for every ǫ > 0, there exists a measurable set E ⊆ X such
that µ(X \ E) < ǫ and {gn } converges uniformly to 0 on E. By the definition (3.107), {ft }
converges to f uniformly on E.
This completes the proof of the problem.
Problem 3.17
Rudin Chapter 3 Exercise 17.
Proof.
(a) Since the inequality is clearly holds if either α = 0 or β = 0, we assume, without loss of
generality, that α and β cannot be both zero in the following discussion.
Let 0 < p ≤ 1. We claim that the continuous function ϕ : [0, 1] → R defined by
ϕ(x) = xp + (1 − x)p
101
3.4. Hardy’s Inequality and Egoroff ’s Theorem
satisfies ϕ(x) ≥ 1 on [0, 1]. To this end, we note that
ϕ′ (x) = pxp−1 − p(1 − x)p−1 = 0
if and only if x = 21 . Since p − 1 ≤ 0, we know that
ϕ′ (x) ≥ 0 and
ϕ′ (x) ≤ 0
on (0, 12 ) and ( 21 , 1) respectively. In other words, ϕ is increasing and decreasing on (0, 21 )
and ( 21 , 1) respectively. By the continuity of ϕ, these are also valid on [0, 12 ] and [ 21 , 1]
respectively. Thus we observe
ϕ(x) ≥ ϕ(0) = 1 on [0, 21 ] and
ϕ(x) ≥ ϕ(1) = 1 on [ 12 , 1]
which mean that
xp + (1 − x)p ≥ 1
(3.109)
on [0, 1]. For arbitrary nonzero complex numbers α and β, we put x =
inequality (3.109) to get
|α|p
|β|p
+
≥1
(|α| + |β|)p
(|α| + |β|)p
|α|
|α|+|β|
into the
which implies
(|α| + |β|)p ≤ |α|p + |β|p .
(3.110)
By the triangle inequality, it is true that |α − β| ≤ |α| + |β|, so this and the inequality
(3.110) give
|α − β|p ≤ |α|p + |β|p .
(3.111)
Next, let 1 < p < ∞. The function ψ : [0, 1] → R given by
ψ(x) = xp
has second derivative p(p − 1)xp−2 > 0 on (0, 1). By Definition 3.1, it is convex on (0, 1),
i.e.,
[(1 − λ)x + λy]p ≤ (1 − λ)xp + λy p ,
(3.112)
where x, y ∈ (0, 1) and λ ∈ [0, 1]. Put λ = 12 , x =
(3.112), we obtain
|α|
|α|+|β|
and y =
|β|
|α|+|β|
into the inequality
1 |α| + |β| p 1
|α|p
|β|p
1
≤ ·
·
·
+
2 |α| + |β|
2 (|α| + |β|)p
2 (|α| + |β|)p
which reduces to
(|α| + |β|)p ≤ 2p−1 (|α|p + |β|p ).
Again, the triangle inequality and this show that
|α − β|p ≤ 2p−1 (|α|p + |β|p ).
(3.113)
Hence the desired inequality follows from combining the inequalities (3.111) and (3.113).
(b)
(i) If µ(X) = 0, then Proposition 1.24(e) implies that kf kp = 0 for every f ∈ Lp (µ)
and there is nothing to prove. Thus, without loss of generality, we may assume that
µ(X) > 0. We divide the proof into several steps:
Chapter 3. Lp -Spaces
102
∗ Step 1: Lemma 3.7 and its application.
Lemma 3.7
For every ǫ > 0, there exists a δ > 0 such that for every E ∈ M with µ(E) < δ, we
have
Z
|f |p dµ ≤ ǫ.
(3.114)
E
Proof of Lemma 3.7. Note that |f | must be bounded a.e. on X. Otherwise, there
exists an E ∈ M such that µ(E) > 0 and |f | = ∞ on E. Then Proposition 1.24(b)
shows that
Z
Z
|f |p dµ = ∞,
|f |p dµ ≥
E
X
a contradiction. Thus there exists a positive constant M such that |f | ≤ M a.e.
on X. Given ǫ > 0. Now for every E ∈ M with µ(E) < Mǫ p , we have
Z
|f |p dµ ≤ M p µ(E) ≤ ǫ a.e. on X
E
which is exactly what we want.
By Lemma 3.7, there exists a δ > 0 such that the inequality (3.114) holds
for every E ∈ M with µ(E) < δ. For this δ > 0, suppose that we can find a
measurable set F such that fn → f a.e. on F and µ(F ) < ∞. (The existence
of such a F will be clear at the end of Step 2 below.) Then Egoroff’s Theorem
guarantees that there exists a measurable set B ⊆ F such that µ(F \ B) < δ and
{fn } converges uniformly to f on B. Since F \ B ∈ M and µ(F \ B) < δ, we also
have
Z
ǫ
(3.115)
|f |p dµ ≤ .
2
F \B
Furthermore, since fn → f uniformly on B, there is a positive integer N such
that n ≥ N implies that
|fn (x) − f (x)| ≤
ǫ 1
p
µ(B)
for all x ∈ B and this means that
Z
B
|fn − f |p dµ ≤ ǫ
(3.116)
for n ≥ N .
∗ Step 2: Lemma 3.8 and its application.
Lemma 3.8
If f ∈ L1 (µ), then for every ǫ > 0, there exists an E ∈ M such that F = X \ E,
µ(F ) < ∞ and
Z
|f | dµ ≤ ǫ.
E
103
3.4. Hardy’s Inequality and Egoroff ’s Theorem
Proof of Lemma 3.8. Since |f | is measurable, Theorem 1.17 (The Simple Function
Approximation Theorem) ensures that there exists a sequence of simple measurable functions {sk } on X such that
0 ≤ s1 ≤ s2 ≤ · · · ≤ |f |
(3.117)
and sk converges to |f | pointwisely. By Theorem 1.26 (Lebesgue’s Monotone
Convergence Theorem), we have
Z
Z
|f | dµ.
sk dµ =
lim
k→∞ X
X
Hence, for every ǫ > 0, there exists a positive integer N such that
Z
Z
(|f | − sk ) dµ ≤ ǫ.
(|f | − sk ) dµ =
(3.118)
X
X
By Definition 1.16, we have
sk =
nk
X
αi χFi ,
i=1
where each αi is positive, Fi is measurable and Fi ∩ Fj = ∅ for all i 6= j. Recall
that |f | ∈ L1 (µ), so the hypothesis (3.117) implies that sk ∈ L1 (µ) for every
n ∈ N. Since each αi is positive, µ is positive on X (i.e., µ(E) ≥ 0 for every
E ∈ M) and
Z
nk
X
sk dµ < ∞,
αi µ(Fi ) =
X
i=1
they force that µ(Fi ) < ∞ for every i = 1, 2, . . . , nk .
We suppose that
E = {x ∈ X | sk (x) = 0}
and F =
nk
[
Fi .
i=1
Then the previous analysis shows that µ(F ) < ∞. Furthermore, it is clear that
sk (x) > 0 if and only if x ∈ Fi for some i, so this fact implies that
X \ E = F.
Thus we deduce from the estimate (3.118) and the definition of E that
Z
Z
Z
Z
Z
sk dµ ≤ ǫ.
(|f | − sk ) dµ +
sk dµ ≤
|f | dµ = (|f | − sk ) dµ +
E
E
E
X
E
This proves Lemma 3.8.
p
p
1
Since f ∈ L (µ), we have |f | ∈ L (µ). Then Lemma 3.8 implies that there
exists an E ∈ M such that F = X \ E, µ(F ) < ∞ and
Z
ǫ
(3.119)
|f |p dµ ≤ .
2
E
∗ Step 3: Constructions of A and B satisfying the hypotheses. Let E and
F be defined as in Lemma 3.8 so that the estimate (3.119) holds. We remark
that
X = E ∪ F = E ∪ (F \ B) ∪ B = A ∪ B,
Chapter 3. Lp -Spaces
104
where B is the measurable set guaranteed by Egoroff’s Theorem in Step 1 and
A = E ∪ (F \ B). Recall that B ⊆ F , so it is easy to see that
µ(B) ≤ µ(F ) < ∞
and since E ∩ (F \ B) = ∅, we obtain from Theorem 1.29 and the two estimates
(3.119) and (3.115) that
Z
Z
Z
p
p
|f |p dµ ≤ ǫ.
(3.120)
|f | dµ +
|f | dµ =
F \B
E
A
Since fn → f a.e. on X and particularly on B, this and Theorem 1.28 (Fatou’s
Lemma) together show that
Z
Z
Z
p
p
|fn |p dµ
(3.121)
lim inf |fn | dµ ≤ lim inf
|f | dµ =
B n→∞
B
n→∞
B
Next, we know from the definition that A = (E ∪ F ) \ B = X \ B so that
A ∩ B = ∅. Since kfn kp → kf kp as n → ∞, it deduces from this, Problem 1.4
and the inequalities (3.120) and (3.121) that
Z
Z
Z
p
p
|fn |p dµ
lim sup
|fn | dµ ≤ lim sup
|fn | dµ + lim sup −
n→∞
n→∞
n→∞
B
A
Z
ZX
p
|fn |p dµ
= lim sup
|fn | dµ − lim inf
n→∞
B
Zn→∞ X Z
p
p
|f | dµ
|f | dµ −
≤
B
ZX
|f |p dµ
=
A
≤ ǫ.
(3.122)
∗ Step 4: The establishment of kfn − f kp → 0 as n → ∞. By part (a), the
estimates (3.116) and (3.122), we see that
Z
Z
Z
lim sup
|fn − f |p dµ ≤ lim sup
|fn − f |p dµ + lim sup
|fn − f |p dµ
n→∞
n→∞
n→∞
X
AZ
B
p
p
≤ γp lim sup (|f | + |fn | ) dµ + ǫ
n→∞
A
≤ 2γp ǫ + ǫ.
Since ǫ is arbitrary, we conclude that
lim kfn − f kp = lim
n→∞
n→∞
nZ
X
|fn − f |p dµ
o1
p
= 0.
(ii) Put hn = γp (|f |p + |fn |p ) − |f − fn |p . By part (a), we have hn ≥ 0 for all n ∈ N.
Since f and fn are measurable, each hn is also measurable. By Theorem 1.28 (Fatou’s
Lemma), we get
Z
Z hn dµ.
(3.123)
lim inf hn dµ ≤ lim inf
X
n→∞
f |p
n→∞
X
Since fn → f a.e. on X, |fn −
→ 0 and |fn → |f |p a.e. on X. Thus we must
have
lim hn = 2γp |f |p a.e. on X.
(3.124)
n→∞
|p
105
3.4. Hardy’s Inequality and Egoroff ’s Theorem
Putting the limit (3.124) into the inequality (3.123), we obtain from Problem 1.4 that
Z
p
X
2γp |f | dµ =
Z ZX lim hn dµ
n→∞
lim inf hn dµ
n→∞
X
Z
hn dµ
= lim inf
n→∞ X
Z
i
hZ
|fn − f |p dµ
γp (|f |p + |fn |p ) dµ −
≤ lim inf
n→∞
X
X
Z
hZ
i
p
p
(|f | + |fn | ) dµ − lim sup
|fn − f |p dµ
≤ γp lim inf
n→∞
n→∞
X
X
Z
|fn − f |p dµ.
(3.125)
= γp lim inf kfn kpp + kf kpp − lim sup
=
n→∞
n→∞
X
Since kfn kp → kf kp as n → ∞, we can further reduce the inequality (3.125) to
0 ≤ lim sup
n→∞
Z
X
|fn − f |p dµ ≤ γp
lim kfn kpp + kf kpp − 2γp kf kpp = 0.
n→∞
Hence this leads to the following
lim kfn − f kp = lim sup kfn − f kp = 0
n→∞
n→∞
as desired.
(c) Consider X = (0, 1) and µ = m the Lebesgue measure. For each n = 1, 2, . . ., we let
En = (0, n1p ) and fn = nχEn : (0, 1) → R. If x ∈ (0, 1), then there exists a positive integer
N such that x ∈
/ En for all n ≥ N . In this case, we have fn (x) = 0 for all n ≥ N . In other
words, we have
fn (x) → f (x) ≡ 0
for every x ∈ (0, 1). It is clear that f ∈ Lp (0, 1) and kf kp = 0. Besides, we have
kfn kp =
nZ
0
1
p
|fn (x)| dx
o1
p
=
nZ
En
np dx
o1
p
=1
for every n ∈ N. Thus we know that
kfn kp 9 kf kp
as n → ∞. Finally, since kfn − f kp = kfn kp = 1, we see that kfn − f kp 9 0 as n → ∞,
i.e., the conclusion of part (b) is false.
We have completed the proof of the problem.
Remark 3.1
We remark that there is a short and elementary proof of Problem 3.17(b) in [45].
Chapter 3. Lp -Spaces
106
Convergence in Measure and the Essential Range of f ∈ L∞ (µ)
3.5
Problem 3.18
Rudin Chapter 3 Exercise 18.
Proof.
(a) Given small ǫ > 0. For each positive integer k, we let
Ek = {x ∈ X | |fn (x) − f (x)| ≤ ǫ for all n ≥ k}
and E =
∞
[
Ek .
k=1
The definitions of Ek and E guarantee that
X \E =
∞
\
(X \ Ek ) =
k=1
∞
\
k=1
{x ∈ X | |fn (x) − f (x)| > ǫ for some n ≥ k}.
(3.126)
Since fn (x) → f (x) a.e. on X, we deduce from the expression (3.126) that
µ(X \ E) = 0.
Furthermore, we have E1 ⊆ E2 ⊆ · · · , so Theorem 1.19 implies that µ(Ek ) → µ(E) as
k → ∞. Thus it follows from this fact and the result µ(X \ E) = 0 that
lim µ(Ek ) = µ(E) = µ(X) < ∞.
k→∞
This result ensures that there exists a N ∈ N such that µ(X \ EN ) < ǫ.
{x ∈ X | |fn (x) − f (x)| > ǫ for all n > N } ⊆ X \ EN .
Hence we must have
µ({x ∈ X | |fn (x) − f (x)| > ǫ for all n > N }) < ǫ,
i.e., fn → f in measure.
(b) Suppose that 1 ≤ p < ∞. Since kfn − f kp → 0 as n → ∞, we must have fn − f ∈ Lp (µ)
for every n ∈ N. Since fn ∈ Lp (µ), we also have f = (f − fn ) + fn ∈ Lp (µ) by Theorem
3.9. Given ǫ > 0, there exists a positive integer N such that n ≥ N implies that
kfn − f kp < ǫp+1 .
(3.127)
Let Fn = {x ∈ X | |fn (x) − f (x)| > ǫ} for each n = 1, 2, . . .. Now it is easy to check thati
Z
Z
p
ǫp dµ = ǫp µ(Fn ).
(3.128)
|fn (x) − f (x)| dµ ≥
Fn
i
Fn
The inequality (3.128) is a consequence of the so-called Chebyshev’s Inequality: If f is a nonnegative,
extended real-valued measurable function on X with measure µ, 0 < p < ∞ and ǫ > 0, then
Z
1
µ({x ∈ X | f (x) ≥ ǫ}) ≤ p
f p dµ.
ǫ X
See [22, Theorem 6.17, p. 193].
107
3.5. Convergence in Measure and the Essential Range of f ∈ L∞ (µ)
Hence we establish from the inequalities (3.127) and (3.128) that
µ({x ∈ X | |fn (x) − f (x)| > ǫ}) = µ(Fn ) < ǫ
for n ≥ N . By the definition, fn → f in measure.
Next, we suppose that p = ∞. Now kfn − f k∞ → 0 means the existence of a positive
integer N such that n ≥ N implies |fn (x) − f (x)| < ǫ a.e. on X. By the definition, we
have µ(Fn ) = 0 < ǫ for all n ≥ N , i.e., fn → f in measure in this case.
(c) Suppose that fn → f in measure, i.e., for every ǫ > 0, there exists a positive integer N
such that n ≥ N implies that
µ({x ∈ X | |fn (x) − f (x)| > ǫ}) < ǫ.
In particular, for each positive integer k, there exists a positive integer Nk such that
µ({x ∈ X | |fn (x) − f (x)| > 2−k }) < 2−k
(3.129)
for all n ≥ Nk . Now we may choose nk > Nk freely in the estimate (3.129) and consider
the subsequence {fnk }. We define
−k
Ek = {x ∈ X | |fnk (x) − f (x)| > 2
}
and E =
∞ [
∞
\
Ek .
(3.130)
m=1 k≥m
Then it follows from the estimate (3.129) that
µ(E) ≤ µ
∞
[
k≥m
∞
∞
X
X
1
1
Ek ≤
µ(Ek ) <
= m−1 ,
k
2
2
k=m
k=m
where m = 1, 2, . . .. As a result, we must have µ(E) = 0. Take any p ∈ X \ E, the
definitions (3.130) say that
p∈
∞
\
(X \ Ek ) =
k≥m
for some m ∈ N and then
∞
\
{x ∈ X | |fnk (x) − f (x)| ≤ 2−k }
k≥m
|fnk (p) − f (p)| ≤ 2−k
for all k ≥ m. In other words, we have
lim fnk (p) = f (p)
k→∞
which is our desired result.
Here we propose two examples which say that the converses of parts (a) and (b) fail:
• Failure of the converse of part (a). By Problem 2.9, there exists a sequence of
continuous functions fn : [0, 1] → R such that 0 ≤ fn ≤ 1 and
kfn − 0k1 → 0
as n → ∞, but there is no x ∈ [0, 1] such that {fn (x)} converges. In other words, fn (x)
does not converge to 0 a.e. on [0, 1]. However, since each fn is continuous [0, 1], it is R
on [0, 1] so that fn ∈ L1 [0, 1] by [49, Theorem 11.33, p. 323]. Thus the sequence of
functions {fn } satisfies the hypotheses of part (b) which shows that fn → 0 in measure.
Hence this example implies the the converse of part (a) is false.
Chapter 3. Lp -Spaces
108
• Failure of the converse of part (b). Suppose that X = [0, 1], µ is the Lebesgue
measure and
fn = en χ[0, 1 ]
n
for each n = 1, 2, . . .. Then fn → 0 a.e. on [0, 1] and part (a) implies that fn → 0 in
measure. However, we note that
n Z n1
o1
nZ
o1
en
p
p
p
=
= 1 →∞
enp dx
kfn kp =
|fn (x)| dµ
0
X
np
as n → ∞, so the sequence {fn } fails to converge in Lp [0, 1] .
Finally, by examining the proofs of parts (b) and (c), we see that they remain to be true in
the case µ(X) = ∞. However, part (a) does not hold in this case. For instance, let X = [0, ∞)
and En = [0, n] for each n ∈ N. Consider the functions fn = χEn . Then for every x ∈ [0, ∞),
there exists a positive integer N such that x ∈ En for all n ≥ N , so it is true that fn (x) = 1 for
all n ≥ N , i.e., fn (x) → f (x) ≡ 1 on [0, ∞). However, for every ǫ ∈ (0, 1), since fn (x) = 0 if and
only if x ∈ (n, ∞), we see that
{x ∈ [0, ∞) | |fn (x) − 1| > ǫ} = (n, ∞)
for every n ∈ N. Hence it leads that
m({x ∈ [0, ∞) | |fn (x) − 1| > ǫ}) = ∞
for every n ∈ N, i.e., fn does not converge in measure to f .
Hence we end the proof of the problem.
Problem 3.19
Rudin Chapter 3 Exercise 19.
Proof. For every ǫ > 0, define Eǫ (w) = {x ∈ X | |f (x) − w| < ǫ} and
\
Rf = {w ∈ C | µ Eǫ (w) > 0 for every ǫ > 0} =
{w ∈ C | µ Eǫ (w) > 0}.
(3.131)
ǫ>0
We are going to prove the assertions one by one:
• Rf is compact. Given {wn } ⊆ Rf and w ∈ C such that wn → w as n → ∞. For every
ǫ > 0, there exists a positive integer N such that |wn − w| < 2ǫ . By the triangle inequality,
we have
ǫ
|f (x) − w| ≤ |f (x) − wn | + |wn − w| < |f (x) − wn | +
2
ǫ
which means that E 2 (wn ) ⊆ Eǫ (w) or equivalently
(3.132)
µ Eǫ (w) > µ E 2ǫ (wn ) > 0.
Since ǫ is arbitrary, we know from the estimate (3.132) that w ∈ Rf . In other words, Rf
is closed in C.
Recall that kf k∞ < ∞, so we pick a complex number w0 such that |w0 | > kf k∞ and
then consider the positive number ǫ = |w0 | − kf k∞ . If x ∈ Eǫ (w0 ), then the triangle
inequality shows that
|w0 | − |f (x)| < |f (x) − w0 | < ǫ = |w0 | − kf k∞
109
3.5. Convergence in Measure and the Essential Range of f ∈ L∞ (µ)
which implies that |f (x)| > kf k∞ . Thus we have Eǫ (w0 ) ⊆ {x
∈ X | |f (x)| > kf k∞ }. By
Definition 3.7, µ({x ∈ X | |f (x)| > kf k∞ }) = 0, so µ Eǫ (w0 ) = 0 too. By the definition
(3.131), we have w0 ∈
/ Rf and hence the set Rf is bounded by kf k∞ . By the Heine–Borel
Theorem, we conclude that Rf is compact.
• A relation between Rf and kf k∞ . By the previous analysis, it is obvious that the
relation
Rf ⊆ {w ∈ C | |w| ≤ kf k∞ }
(3.133)
holds, i.e., Rf lies in B(0, kf k∞ ). We claim that
kf k∞ = max{|z| | z ∈ Rf }.
(3.134)
Given w ∈ C and ǫ > 0. Let B(w, ǫ) = {z ∈ C | |z − w| < ǫ}. Then we have
f −1 B(w, ǫ) = {x ∈ X | f (x) ∈ B(w, ǫ)} = {x ∈ X | |f (x) − w| < ǫ} = Eǫ (w).
By the definition, we see that
\
Rf =
{w ∈ C | µ f −1 B(w, ǫ) > 0} = C \ V,
(3.135)
ǫ>0
where V = {w ∈ C | µ f −1 B(w, ǫ) = 0 for some ǫ > 0}. Now we have the following two
facts:
– Fact 1: It is easy to show that V is the largest open subset of C such that
µ f −1 (V ) = µ({x ∈ X | f (x) ∈ V }) = 0,
i.e., V is the largest open subset of C such that f (x) ∈
/ V for almost all x ∈ X. By
this fact and the relation (3.135), we conclude that Rf is the smallest closed subset
of C such that f (x) ∈ Rf for almost all x ∈ X.
– Fact 2: Recall from Definition 3.7 that the number kf k∞ is the minimum of the set
{α ≥ 0} such that µ({x ∈ X | |f (x)| > α}) = 0, so we see from the relation (3.133)
that it is the minimum radius of the closed disc centered at 0 containing the set Rf .
Hence we follow immediately from these two observations that the equality actually holds
in the set relation (3.133). Since they are equal, our claim (3.134) is established.
• Relations between Af and Rf . By the definition, we have
o
n 1 Z
f dµ E ∈ M and µ(E) > 0 .
Af =
µ(E) E
(3.136)
We claim that Rf ⊆ Af . Given ǫ > 0 and w ∈ Rf . If w ∈ Af , then there is nothing to
prove. Therefore, we may assume that w ∈
/ Af . Consider Eǫ = {x ∈ X, | |f (x) − w| < ǫ}.
By the definition of Rf , we have µ(Eǫ ) > 0. If µ(Eǫ ) = ∞ for every ǫ, then f (x) = w a.e.
on X. In this case, we have
Z
1
f dµ = w ∈ Af ,
µ(Eǫ ) Eǫ
a contradiction. Therefore, we may assume that µ(Eǫ ) < ∞ for infinitely many ǫ > 0.
In fact, we may take ǫ = n1 and let E(n) = E 1 . Then, since f ∈ L∞ (µ), we have
f − w ∈ L1 (E(n)), where
n
L E(n) = f : X → C
1
n
Z
E(n)
o
|f | dµ < ∞ .
Chapter 3. Lp -Spaces
110
Consequently, it follows from Theorem 1.33 and the definition of E that
Z
Z
1
1
1
f dµ − w ≤
|f − w| dµ < .
n
µ E(n) E(n)
µ E(n) E(n)
In other words, w is a limit point of Af and this means that w ∈ Af . Hence this proves
the claim that Rf ⊆ Af .
• Af is not always closed. For example, we consider
X = [0, 1], f (x) = x with µ = m the
Lebesque measure. By Definition 3.7, f ∈ L∞ [0, 1] . In addition, for a measurable set E
in [0, 1] with m(E) > 0, we have
Z
1
f (x) dx.
(3.137)
w(E) =
m(E) E
We claim that w(E) can take any value in (0, 1). Indeed, if a, b ∈ (0, 1) and E = (a, b),
then we have m(E) = b − a > 0 and it follows from the definition (3.137) that
Z
Z b
1
1
b2 − a 2
a+b
1
w(E) =
f dx =
×
=
x dx =
m(E) E
b−a a
b−a
2
2
which implies the claim.
Next, we claim that w(E) 6= 0. Assume that w(E0 ) = 0 for a measurable set E0 ⊆ [0, 1]
with m(E0 ) > 0. Then we have
Z
f (x) dx = 0,
E0
but Theorem 1.39(a) implies that f (x) = 0 a.e. on E0 which contradicts the fact that
f (x) = 0 only at x = 0. Hence 0 is not a limit point of Af and Af is not closed in R.
• A measure µ such that Af is convex for every f ∈ L∞ (µ). Consider X = {0}
and µ the counting measure (see Example 1.20(a)). Then we have M = {∅, X} and
µ(X) = µ({0}) = 1 > 0. By the definition, kf k∞ = |f (0)| < ∞ for every f ∈ L∞ (µ) so
that f (0) is either kf k∞ or −kf k∞ . Now for every f ∈ L∞ (µ), we have
Z
1
w(X) =
f dµ = f (0).
µ(X) X
Therefore, the definition (3.136) gives either Af = {kf k∞ } or Af = {−kf k∞ }, but both
cases are also convex sets.
• A measure µ such that Af is not convex for some f ∈ L∞ (µ). We consider the set
X = {0, 1}, µ the counting measure and f (x) = x. Then we have M = {∅, {0}, {1}, X},
µ({0}) = µ({1}) = 1 > 0 and µ(X) = 2 > 0. Now we have
Z
Z
1
1
w({0}) =
f dµ = f (0) = 0, w({1}) =
f dµ = f (1) = 1,
µ({0}) {0}
µ({1}) {1}
Z
1
1
1
1
w(X) =
f dµ = [f (0)µ({0}) + f (1)µ({1})] = [f (0) + f (1)] = .
µ(X) X
2
2
2
Thus we have Af = {0, 1, 21 } is not convex.
• The situations when L∞ (µ) is replaced by L1 (µ). We consider f : (0, ∞) → C by

1


 √ , if x ∈ (0, 1];
x
f (x) =


 0,
if x > 1
111
3.6. A Converse of Jensen’s Inequality
and take µ = m the Lebesgue measure. By Lemma 3.1, we have f ∈ L1 (0, ∞) . For any
large positive integer N , if x ∈ (0, N12 ), then we have f (x) > N and
1 1
0, 2
= 2 > 0.
N
N
= ∞, i.e., f ∈
/ L∞ (0, ∞) . Given that ǫ > 0 and
m({x ∈ (0, ∞) | f (x) > N }) = m
By Definition 3.7, we have kf k∞
sufficiently large N ∈ N.
– Rf is unbounded and not compact. If x ∈
have |f (x) − N | < ǫ so that
m({x ∈ (0, ∞) | |f (x) − N | < ǫ}) = m
1
, 1
(N +ǫ)2 (N −ǫ)2
⊆ (0, 1), then we
1
1
,
> 0.
(N + ǫ)2 (N − ǫ)2
Thus N ∈ Rf and then N ⊆ Rf , i.e., Rf is unbounded and not compact.
– Af is unbounded. Next, if E = (0, N42 ), then m(E) =
1
w(E) =
m(E)
Z
N2
f (x) dx =
4
E
Z
0
4
N2
x
− 12
4
N2
> 0 and
√
N2
×2 x
dx =
4
4
N2
0
= N.
Therefore, N ⊆ Af , i.e., Af is unbounded.
– Af is not always closed. In fact, the
previous∞example
that X = [0, 1], f (x) = x
1
and µ = m satisfies both f ∈ L [0, 1] and f ∈ L [0, 1] . Thus the same conclusion
that 0 is not a limit point of Af is achieved.
– A measure µ such that Af is convex for every f ∈ L1 (µ). The example that
X = {0} and µ the counting measure also works for this case because if f ∈ L1 (µ),
then |f (0)| = kf k1 < ∞. Thus we have either Af = {kf k1 } or Af = {−kf k1 }.
– A measure µ such that Af is not convex for some f ∈ L1 (µ). The example we
considered for the case L∞ (µ) also works in this case.
We complete the proof of the problem.
3.6
A Converse of Jensen’s Inequality
Problem 3.20
Rudin Chapter 3 Exercise 20.
Proof. We check Definition 3.1. Let p, q ∈ R and λ ∈ [0, 1]. By the trick of Proposition 1.24(f),
we see that
Z
Z
χ[λ,1] (x) dx
χ[0,λ] (x) dx + q
λp + (1 − λ)q = p
=
Z
[0,1]
1
0
[0,1]
[pχ[0,λ] (x) + qχ[λ,1] (x)] dx.
(3.138)
Suppose that
f (x) = pχ[0,λ] (x) + qχ[λ,1] (x)
(3.139)
which is clearly a real and bounded function. Besides, since [0, λ] and [λ, 1] are Borel sets in
[0, 1], χ[0,λ] and χ[λ,1] are measurable by Proposition 1.9(d). By the last paragraph in [51, §1.22,
Chapter 3. Lp -Spaces
112
p. 19], the function f , as the sum of two measurable functions, is also measurable. Hence, by
substituting the expressions (3.138) and (3.139) into the inequality in question, we obtain
Z
1
ϕ pχ[0,λ] (x) + qχ[λ,1] (x) dx
0
Z 1
Z λ
ϕ(q) dx
ϕ(p) dx +
=
ϕ λp + (1 − λ)q ≤
0
λ
= λϕ(p) + (1 − λ)ϕ(q).
Hence ϕ is convex in R, completing the proof of the problem.
Remark 3.2
We note that Problem 3.20 is a converse of Theorem 3.3 (Jensen’s Inequality).
3.7
The Completeness/Completion of a Metric Space
Problem 3.21
Rudin Chapter 3 Exercise 21.
Proof. Let (X, d) and (X ∗ , ρ) be two metric spaces. An isometry of X into X ∗ is a mapping
ϕ : X → Y such that
ρ ϕ(p), ϕ(q) = d(p, q)
(3.140)
for all p, q ∈ X. We notice immediately that an isometry ϕ is necessarily injective and continuous.
If ϕ is surjective, then we call ϕ an isomorphism. Two metric spaces X and Y are called
isomorphic if there is an isomorphism between them. Before stating the statement, we need:
Lemma 3.9
Let (X, d) be a metric space with metric d and p ∈ X. Then the function f : X → R
defined by f (x) = d(x, p) is continuous.
Proof of Lemma 3.9. Given ǫ > 0. Let x ∈ X. If y ∈ X satisfies d(x, y) < ǫ, then the
triangle inequality implies that
|f (x) − f (y)| = |d(x, p) − d(y, p)| ≤ d(x, y) < ǫ.
Hence f is continuous, completing the proof of Lemma 3.9.
Lemma 3.10
Suppose that (X, d) and (Y, ρ) are metric spaces and Y is complete. If E is dense
in X and f : E → Y is an isometry, then there exists an isometry g : X → Y such
that g|E = f .
113
3.7. The Completeness/Completion of a Metric Space
Proof of Lemma 3.10. If x ∈ E, we define g(x) = f (x). Let x ∈ X \ E. Since E is
dense in X, we can choose a sequence {xn } ⊆ E such that xn → x. By [49, Theorem
3.11(a)], {xn } is Cauchy in X. Thus, given ǫ > 0, there exists a positive integer N
such that n, m ≥ N imply that
d(xm , xn ) < ǫ.
Since f is an isometry, it follows from the expression (3.140) that
ρ f (xm ), f (xn ) < ǫ
for all n, m ≥ N . In other words, {f (xn )} is Cauchy in Y . Since Y is complete,
{f (xn )} converges to a limit y in Y and we define g(x) = y in this case.
Now this y is uniquely determined by x. Indeed, let {x′n } ⊆ E be another sequence
converging to x and y ′ be the limit of {f (x′n )} in Y . For each m ∈ N, it follows from
Lemma 3.9 and the expression (3.140) that
ρ y, f (x′m ) = lim ρ f (xn ), f (x′m ) = lim d(xn , x′m ) = d(x, x′m ).
n→∞
n→∞
Then this implies that
ρ(y, y ′ ) = lim ρ y, f (x′m ) = lim d(x, x′m ) = 0.
m→∞
m→∞
Since ρ is a metric, we must have y = y ′ and so we obtain a mapping g : X → Y given
by


if x ∈ E;
 f (x),
(3.141)
g(x) =

 lim f (xn ), if x ∈ X \ E, {xn } ⊆ X and xn → x.
n→∞
Next, we show that g is an isometry and we consider the following situations:
• Case (i): x, y ∈ E. In this case, we know from the definition (3.141) that
g(x) = f (x) and g(y) = f (y) and then the use of the expression (3.140) implies
that
ρ g(x), g(y) = ρ f (x), f (y) = d(x, y).
• Case (ii): x ∈ X \ E and y ∈ E. In this case, we have
ρ g(x), g(y) = ρ g(x), f (y) .
Choose a sequence {xn } in E such that xn → x. By Lemma 3.9 and the expression (3.140), we establish that
(3.142)
ρ g(x), f (y) = lim ρ f (xn ), f (y) = lim d(xn , y) = d(x, y).
n→∞
n→∞
• Case (iii): x, y ∈ X \ E. Let {xn }, {yn } ⊆ E be sequences converging to x and
y respectively. Then we deduce from the definition (3.141) and the expression
(3.142) that
ρ g(x), f (yn ) = d(x, yn ).
(3.143)
Apply Lemma 3.9 to (3.143), we see that
ρ g(x), g(y) = lim ρ g(x), f (yn ) = lim d(x, yn ) = d(x, y).
n→∞
By the definition, g is an isometry and g|E = f .
n→∞
Chapter 3. Lp -Spaces
114
Lemma 3.11
The mapping g in Lemma 3.9 is unique up to isometry.
Proof of Lemma 3.11. Let g ′ : X → Y be an isometry such that g′ |E = f . If x ∈ E,
then
g(x) = f (x) = g ′ (x).
If x ∈ X \ E, then there is a sequence {xn } ⊆ E such that xn → x as n → ∞. The
triangle inequality, the expressions (3.140) and (3.142) indicate that
ρ g(x), g ′ (x) ≤ ρ g(x), g(xn ) + ρ g(xn ), g′ (xn ) + ρ g′ (xn ), g′ (x)
= ρ g(x), f (xn ) + ρ f (xn ), f ′ (xn ) + ρ f ′ (xn ), g′ (x)
= ρ g(x), f (xn ) + d(xn , xn ) + ρ f ′ (xn ), g′ (x)
= d(x, xn ) + d(xn , x)
Hence when n → ∞, it is clear that ρ(g(x), g′ (x)) = 0, so g(x) = g′ (x) on X \ E and
our result follows from this.
Let’s return to the proof of the problem. Let (Y1 , ρ1 ) and (Y2 , ρ2 ) be complete metric spaces,
ϕ1 : X → Y1 and ϕ2 : X → Y2 be isometries such that ϕ1 (X) is dense in Y1 and ϕ2 (X) is dense
in Y2 . We claim that (Y1 , ρ1 ) and (Y2 , ρ2 ) are isomorphic. Define f : ϕ1 (X) → Y2 by
f = ϕ2 ◦ ϕ−1
1 .
(3.144)
If p, q ∈ X, x = ϕ1 (p) and y = ϕ1 (q), then we have
−1
ρ2 f (x), f (y) = ρ2 ϕ2 ϕ−1
1 (x) , ϕ2 ϕ1 (y)
= ρ2 ϕ2 (p), ϕ2 (q)
= d(p, q)
= ρ1 ϕ1 (p), ϕ1 (q)
= ρ1 (x, y).
Consequently, f is an isometry. Since ϕ1 (X) is dense in Y1 and Y2 is complete, Lemma 3.10
ensures that there exists an isometry g : Y1 → Y2 such that
g|ϕ1 (X) = f.
It remains to prove that g is surjective. To this end, let y ∈ Y2 . Since ϕ2 (X) = Y2 , we
can select a sequence {yn } ⊆ ϕ2 (X) such that yn → y as n → ∞. Furthermore, there exists a
sequence {xn } ⊆ X such that
yn = ϕ2 (xn )
for all n ∈ N. Consider the corresponding sequence
zn = ϕ1 (xn )
(3.145)
for n ∈ N. Since {yn } is Cauchy in ϕ2 (X) (and hence in Y2 ) and ϕ2 is an isometry, the expression
(3.140) forces that {xn } is Cauchy in X. Then, since ϕ1 is an isometry, the expression (3.140)
again forces that {zn } is Cauchy in ϕ1 (X) (and hence in Y1 ). Since Y1 is complete, we can find
z ∈ Y1 such that zn → z as n → ∞. We claim that
g(z) = y.
115
3.8. Miscellaneous Problems
Thus we deduce from Lemma 3.10, the definition (3.145) and the fact g|ϕ1 (X) = f that
(3.146)
ρ2 g(z), y = lim ρ2 g(zn ), y = lim ρ2 g ϕ1 (xn ) , y = lim ρ2 f ϕ1 (xn ) , y .
n→∞
n→∞
n→∞
Finally, the definition (3.144) reduces the limit (3.146) to
ρ2 g(z), y = lim ρ2 f ϕ1 (xn ) , y = lim ρ2 ϕ2 (xn ), y = lim ρ2 (yn , y) = ρ2 (y, y) = 0.
n→∞
n→∞
n→∞
Therefore, g(z) = y and g is surjective. Hence g is an isomorphism and our claim follows,
completing the proof of the problem.
Problem 3.22
Rudin Chapter 3 Exercise 22.
Proof. This problem is proven in [63, Problem 3.20, p. 50].
3.8
Miscellaneous Problems
Problem 3.23
Rudin Chapter 3 Exercise 23.
Proof. If αn0 = 0 for some n0 , then we must have f = 0 a.e. on X by Theorem 1.39(a). However,
this implies that the measure of
f −1 (α, ∞] = {x ∈ X | f (x) > α}
is zero for every α > 0. By Definition 3.7, it means that kf k∞ = 0 which contradicts the
hypothesis. Thus we have αn 6= 0 for all n ∈ N so that the limit in question is well-defined.
By Definition 3.7, we see that |f (x)| ≤ kf k∞ holds for almost all x ∈ X. Therefore, we have
αn+1 ≤ kf k∞ · αn
a.e. on X
(3.147)
for all n ∈ N. Since kf k∞ > 0, we can find a ǫ > 0 such that kf k∞ > ǫ > 0. By the definition,
the measurable set
F = {x ∈ X | |f (x)| ≥ kf k∞ − ǫ > 0}
satisfies µ(F ) > 0. Apply Theorem 3.5 (Hölder’s Inequality) to the measurable functions f n
and 1 with p = n+1
n , we obtain
αn =
Z
n
X
|f | dµ ≤
nZ
X
n
(|f | )
n+1
n
o
n
n+1
nZ
1n+1 dµ
X
o
1
n+1
so that
−1
1
αn+1
≥ (αn+1 ) n+1 × µ(X) n+1
αn
nZ
o 1
−1
n+1
≥
|f |n+1 dµ
× µ(X) n+1
F
1
−1
≥ (kf k∞ − ǫ)n+1 µ(F ) n+1 × µ(X) n+1
n
1
= (αn+1 ) n+1 × µ(X) n+1
Chapter 3. Lp -Spaces
116
=
µ(F ) 1
n+1
µ(X)
× (kf k∞ − ǫ).
(3.148)
Thus we follow from the inequalities (3.147) and (3.148) that
µ(F ) 1
n+1
µ(X)
× (kf k∞ − ǫ) ≤
αn+1
≤ kf k∞
αn
a.e. on X
(3.149)
1
µ(F ) n+1
)
tends to 1 as n → ∞. By
for all n ∈ N. Since 0 < µ(F ) ≤ µ(X) < ∞, the number ( µ(X)
taking limit to both sides of the inequality (3.149), we see immediately that
kf k∞ − ǫ ≤ lim inf
n→∞
αn+1
αn+1
≤ lim sup
≤ kf k∞
αn
αn
n→∞
a.e. on X.
Remember that ǫ is arbitrary, so this implies that
lim
n→∞
αn+1
= kf k∞
αn
a.e. on X
and we complete the proof of the problem.
Problem 3.24
Rudin Chapter 3 Exercise 24.
Proof. Let x ≥ 0 and y ≥ 0. Put α − β = x and α = y into the equality (3.111), we gain
|x|p − |y|p ≤ |x − y|p .
(3.150)
Next, we substitute α − β = y and α = x into the equality (3.111) to get
−|x − y|p ≤ |x|p − |y|p .
(3.151)
Obviously, the inequalities (3.150) and (3.151) together imply that
|xp − y p | = |x|p − |y|p ≤ |x − y|p
(3.152)
if 0 < p < 1.
If x = y, then the inequality
|xp − y p | ≤ p|x − y|(xp−1 + y p−1 )
certainly holds. Without loss of generality, we may assume that x < y. For p > 1, we consider
the function ϕ(t) = tp on [x, y]. Since ϕ is differentiable in (x, y), the Mean Value Theorem
implies that
|y p − xp | ≤ |x − y||ϕ′ (ξ)| ≤ p|x − y|ξ p−1
(3.153)
for some ξ ∈ (x, y). Clearly, we have ξ p−1 ≤ xp−1 + y p−1 , so it reduces from the inequality
(3.153) that
|xp − y p | ≤ p|x − y|(xp−1 + y p−1 ).
(3.154)
Hence we establish from the inequalities (3.152) and (3.154) that

if 0 < p < 1;
 |x − y|p ,
p
p
|x − y | ≤

p|x − y|(xp−1 + y p−1 ), if 1 ≤ p < ∞.
117
3.8. Miscellaneous Problems
(a) We prove the assertions one by one:
– The truth of the inequality. Put x = |f | ≥ 0 and y = |g| ≥ 0 into the inequality
(3.152) and then take integration, we get
Z
Z
p
p
p
|f | − |g| dµ.
(3.155)
|f | − |g| dµ ≤
Since |a| − |b| ≤ |a − b| holds for every a, b ∈ R, we follow immediately from the
inequality (3.155) that
Z
Z
Z
p
p
p
|f | − |g| dµ ≤
|f | − |g| dµ ≤ |f − g|p dµ,
(3.156)
where 0 < p < 1.
– ∆ is a metric. Define ∆ : Lp (µ) × Lp (µ) → C by
Z
∆(f, g) = |f − g|p dµ.
By Remark 3.10, Lp (µ) is a complex vector space, so f − g ∈ Lp (µ) if f, g ∈ Lp (µ).
Thus we have
Z
0 ≤ ∆(f, g) = |f − g|p dµ < ∞
for all f, g ∈ Lp (µ). Next, Theorem 1.39(a) ensures that ∆(f, g) = 0 if and only if
|f − g|p = 0 a.e. on X, i.e., f = g a.e. on X. It is clear that
Z
Z
∆(f, g) = |f − g|p dµ = |g − f |p dµ = ∆(g, f ).
Finally, since |f |p − |g|p ≤ |f |p − |g|p , it is clear from the inequality (3.156) that
Z
Z
Z
|f |p − |g|p dµ ≤ |f − g|p dµ.
(|f |p − |g|p ) dµ ≤
(3.157)
By replacing f and g by f − g and h − g in the inequality (3.157), we deduce that
Z
∆(f, g) − ∆(h, g) = (|f − g|p − |h − g|p ) dµ
Z
≤ |(f − g) − (h − g)|p dµ
Z
= |f − h|p dµ
= ∆(f, h).
After rearrangement, we have ∆(f, g) ≤ ∆(f, h) + ∆(h, g). Hence, by the definition,
∆ is in fact a metric.
– (Lp (µ), ∆) is a complete metric space. Suppose that {fn } is a Cauchy sequence
in Lp (µ) with respect to the metric ∆, i.e., for every ǫ > 0, there exists a positive
integer N such that n, m ≥ N implies that
∆(fn , fm ) < ǫ.
(3.158)
Although ∆(f, g) = kf −gkpp here, we cannot apply Theorem 3.11 directly to conclude
that Lp (µ) is complete with respect to ∆ because p ≥ 1 in Theorem 3.11.
Chapter 3. Lp -Spaces
118
We imitate the proof of Theorem 3.11. To start with, there is a subsequence {fni }
such that
∆(fni+1 , fni ) < 2−i
(3.159)
for each i = 1, 2, . . .. We define gk : X → [0, ∞] and g : X → [0, ∞] by
gk =
k
X
i=1
|fni+1 − fni | and g =
∞
X
i=1
|fni+1 − fni |
respectively. It is clear that gk ≥ 0 and g ≥ 0 and furthermore, it follows from
Theorem 3.5 (Minkowski’s Inequality) that
∆(gk , 0) = ∆
k
X
i=1
=
|fni+1 − fni |, 0
Z X
k
i=1
≤
=
p
|fni+1 − fni | dµ
k nZ
hX
i=1
k
hX
|fni+1 − fni |p dµ
1
∆(fni+1 , fni ) p
i=1
ip
o 1 ip
p
(3.160)
for each k = 1, 2, . . ..
To proceed further, we need the following result:
Lemma 3.12
Let p ∈ (0, 1). For nonnegative real numbers a1 , a2 , . . . , ak , we have
(a1 + a2 + · · · + ak )p ≤ ap1 + ap2 + · · · + apk .
Proof of Lemma 3.12. Replace β by −β in the inequality (3.111), we get
(α + β)p ≤ |α|p + |β|p
for p ∈ (0, 1). In particular, if α ≥ 0 and β ≥ 0, then we have
(α + β)p ≤ αp + β p
(3.161)
for p ∈ (0, 1). By applying the inequality (3.161) repeatedly to a1 , a2 , . . . , ak , we
derive the desired inequality. This completes the proof of Lemma 3.12.
Apply Lemma 3.9 to the inequality (3.160) and then using the inequality (3.159)
to get
k
k
X
X
2−i ≤ 1
(3.162)
∆(fni+1 , fni ) <
∆(gk , 0) ≤
i=1
for each k = 1, 2, . . ..
i=1
119
3.8. Miscellaneous Problems
Since gk (x) → g(x) as k → ∞ for every x ∈ X, we have gk (x)p → g(x)p as k → ∞
on X. We note that since each fni is measurable, each gkp is also measurable and we
may apply Theorem 1.28 (Fatou’s Lemma) to {gkp } and the inequality (3.162) to get
Z
Z
Z
Z
p
p
p
g dµ =
lim gk dµ = lim inf gk dµ ≤ lim inf gkp dµ = lim inf ∆(gk , 0) ≤ 1,
k→∞
k→∞
k→∞
k→∞
implying that g ∈ Lp (µ). In particular, g < ∞ a.e. on X, so the series
∞
X
[fni+1 (x) − fni (x)]
fn1 (x) +
(3.163)
i=1
converges absolutely for almost every x ∈ X. Denote the sum (3.163) by f (x) for
those x at which the sum (3.163) converges and put f (x) = 0 on the remaining set
of measure zero.
Since we have
k−1
X
[fni+1 (x) − fni (x)] = fnk (x),
fn1 (x) +
i=1
it implies that fni (x) → f (x) as i → ∞ a.e. on X.
Now it remains to show that f ∈ Lp (µ) and ∆(fn , f ) → 0 as n → ∞. Given ǫ > 0.
Since {fn } is Cauchy in Lp (µ) with respect to ∆, there exists a N ∈ N such that the
estimate (3.158) holds for n, m ≥ N . Take m ≥ N , since fni → f as i → ∞ a.e. on
X, Theorem 1.28 (Fatou’s Lemma) again implies that
Z
Z lim inf |fni − fm |p dµ
|f − fm |p dµ =
i→∞
Z
≤ lim inf
|fni − fm |p dµ
i→∞
= lim inf ∆(fni , fm ) < ǫ.
i→∞
(3.164)
so we conclude that f − fm ∈ Lp (µ), hence that f ∈ Lp (µ).
Finally, according to the estimate (3.164), it follows that ∆(f, fm ) < ǫ for all
m ≥ N . Since ǫ is arbitrary, we conclude that
lim ∆(f, fm ) = 0.
m→∞
This shows the completeness of the space Lp (µ) with respect to ∆.
(b) Similar to part (a), a direct substitution of x = |f | and y = |g| into the inequality (3.154)
and the use of the fact |x| − |y| ≤ |x − y| give
Z
Z
p
p
|f | − |g| × (|f |p−1 + |g|p−1 ) dµ
|f | − |g| dµ ≤ p
Z
(3.165)
≤ p |f − g|(|f |p−1 + |g|p−1 ) dµ.
By Theorem 3.5 (Hölder’s Inequality) and the facts that kf kp ≤ R and kgkp ≤ R, we
observe that
Z
Z
Z
p−1
p−1
p−1
dµ + |f − g| · |g|p−1 dµ
|f − g|(|f |
+ |g| ) dµ ≤ |f − g| · |f |
nZ
o p−1
o1 n Z
p
p
p
p−1 p−1
p
(|f | )
≤
dµ
|f − g| dµ
Chapter 3. Lp -Spaces
120
+
nZ
p
|f − g| dµ
o1 n Z
p
p
(|g|p−1 ) p−1 dµ
= kf − gkp · (kf kp−1
+ kgkp−1
p
p )
o p−1
≤ 2Rp−1 kf − gkp .
p
(3.166)
Combining the inequalities (3.165) and (3.166), we discover that
Z
|f |p − |g|p dµ ≤ 2pRp−1 kf − gkp
which is our expected result.
We have completed the proof of the problem.
Problem 3.25
Rudin Chapter 3 Exercise 25.
Proof. Suppose that E ⊆ X and 0 < µ(E) < ∞. Let ME be a σ-algebra in E. Then (E, ME , µ)
is a measure space.j Define
Z
1
dµ (F ∈ ME ).
ϕ(F ) =
µ(E)
F
Then we follow from Theorem 1.29 that ϕ is a measure on ME and
Z
Z
1
f dµ
f dϕ =
µ(E) E
E
(3.167)
for every measurable f on E with range in [0, ∞]. It is trivial that ϕ(E) = 1 and
Z
Z
1
1
f dϕ ≤
f dµ =
<∞
0≤
µ(E)
µ(E)
E
X
which means f ∈ L1 (ϕ). Since 0 < f < ∞ on E, log f is well-defined on E.
Define φ : (0, ∞) → R by φ(x) = − log x which is convex on (0, ∞). Apply Theorem 3.3
(Jensen’s Inequality) with φ and measure ϕ to conclude that
Z
Z
log f dϕ
f dϕ ≤ −
− log
E
E
or equivalently,
Z
E
log f dϕ ≤ log
Z
E
f dϕ .
Now, by using the fact (3.167) to the inequality (3.168), we gain
Z
1 Z
1 Z
1
1
log f dµ ≤ log
f dµ ≤ log
f dµ = log
µ(E) E
µ(E) E
µ(E) X
µ(E)
which implies the first inequality.
For the second inequality, we define ψ : (0, ∞) → R by
ψ(x) = −xp ,
j
See the paragraphs following Proposition 1.24.
(3.168)
121
3.8. Miscellaneous Problems
where 0 < p < 1. Since ψ ′′ (x) = p(1 − p)xp−2 > 0 on (0, ∞), we apply Theorem 3.3 (Jensen’s
Inequality) with ψ and measure ϕ to conclude that
Z
op
nZ
f p dϕ.
(3.169)
f dϕ ≤ −
−
E
E
Next, we use the fact (3.167) to reduce the inequality (3.169) to
Z
n 1 Z
op n 1 Z
op
1
1
f p dµ ≤
f dµ ≤
f dµ =
µ(E) E
µ(E) E
µ(E) X
µ(E)p
so that
Z
E
completing the proof of the problem.
f p dµ ≤ µ(E)1−p ,
Problem 3.26
Rudin Chapter 3 Exercise 26.
Proof. Let M be a σ-algebra in [0, 1]. Suppose that there exists an E ∈ M such that f (x) = ∞
and m(E) > 0, where m is the Lebesgue measure. Then
Z
1
f (x) log f (x) dx =
0
Z
f (x) log f (x) dx +
E
Z
f (x) log f (x) dx.
(3.170)
[0,1]\E
On the one hand, since f > 0 on [0, 1] \ E, we know from Proposition 1.24(a) that the second
integral on the right-hand side of the expression (3.170) is nonnegative. Since f (x) log f (x) = ∞
on E and m(E) > 0, we get from the expression (3.170) that
Z
0
1
f (x) log f (x) dx = ∞.
(3.171)
On the other hand, by using similar argument, we can show that
Z 1
Z 1
log f (t) dt = ∞.
f (s) ds = ∞ and
(3.172)
0
0
Therefore, we conclude from the results (3.171) and (3.172) that
Z
1
f (x) log f (x) dx =
0
Z
1
f (s) ds
0
Z
1
log f (t) dt
0
in this case.
Next, we suppose that 0 < f (x) < ∞ a.e. on [0, 1]. Define ϕ : (0, ∞) → R and ψ : (0, ∞) → R
by
ϕ(x) = x log x and ψ(x) = − log x
respectively. Since ϕ′′ (x) = x1 > 0 and ψ ′′ (x) = x−2 > 0 on (0, ∞), they are convex on (0, ∞).
Since 0 < f (x) < ∞ a.e. on [0, 1] and m([0, 1]) = 1, we apply Theorem 3.3 (Jensen’s Inequality)
with ϕ to f to obtain
o Z 1
nZ 1
on Z 1
f (x) log f (x) dx
(3.173)
≤
f (x) dx
f (x) dx log
0
0
0
Chapter 3. Lp -Spaces
122
Similarly, we apply Theorem 3.3 (Jensen’s Inequality) with ψ to f to obtain
−log
Z
1
Z
f (x) dx ≤ −
1
log f (x) dx.
(3.174)
0
0
Combining the inequalities (3.173) and (3.174), we see that
nZ
1
f (x) dx
0
on Z
0
1
o
log f (x) dx ≤
We have ended the proof of the problem.
Z
1
f (x) log f (x) dx.
0
CHAPTER
4
Elementary Hilbert Space Theory
In the following problems, we use hx, yi to denote the inner product of the complex vectors x
and y and k · k the norm with respect to the Hilbert space H. Besides, we use span (S) to denote
the span of a set S.
4.1
Basic Properties of Hilbert Spaces
Problem 4.1
Rudin Chapter 4 Exercise 1.
Proof. Let x ∈ M . Recall that M ⊥ = {y ∈ H | y ⊥ x for all x ∈ M }, so x ⊥ y for every y ∈ M ⊥
which means that x ∈ (M ⊥ )⊥ , i.e.,
M ⊆ (M ⊥ )⊥ .
Conversely, suppose that x ∈ (M ⊥ )⊥ . Since M is a closed subspace of H, it follows from
Theorem 4.11(a) that x = y + z, where y ∈ M and z ∈ M ⊥ . On the one hand, since x ∈ (M ⊥ )⊥
and z ∈ M ⊥ , we have hx, zi = 0. On the other hand, we deduce from Definition 4.1 that
hx, zi = hy + z, zi = hy, zi + hz, zi = hz, zi.
Since hz, zi = 0, it must be z = 0 and then x ∈ M , i.e, (M ⊥ )⊥ ⊆ M . In conclusion, we obtain
M = (M ⊥ )⊥ .
Suppose that M is a subspace of H which may not be closed. We claim that
M = (M ⊥ )⊥ .
To see this, we recall from §4.8 that M is a closed subspace of H, so the first assertion implies
that
⊥
M = (M )⊥ .
(4.1)
⊥
Let x ∈ M . By §4.9, we see that x ⊥ M so that x ⊥ M particularly. In other words, we have
M
⊥
⊆ M ⊥.
(4.2)
⊥
For the other direction, suppose that x ∈ M ⊥ , we want to show that x ∈ M , i.e., x ⊥ M . To
this end, given y ∈ M , there exist a sequence {yn } ⊆ M such that yn → y as n → ∞. Since
123
Chapter 4. Elementary Hilbert Space Theory
124
x ∈ M ⊥ and yn ∈ M , we must have hyn , xi = 0. By Theorem 4.6, the mapping f : H → C
defined by
f (y) = hy, xi
is continuous on H so that
hy, xi = f (y) = lim f (yn ) = lim hyn , xi = 0.
n→∞
In other words, x ⊥ M , i.e.,
n→∞
⊥
M⊥ ⊆ M .
(4.3)
Now the claim is derived by combining the set relations (4.2) and (4.3) and then using the result
(4.1), completing the proof of the problem.
Problem 4.2
Rudin Chapter 4 Exercise 2.
Proof. We note that this is actually the Gram-Schmidt Process. Suppose that n = 1. Then
the set {u1 } is clearly orthonormal and span (u1 ) = span (x1 ). Thus the statement is true for
n = 1.
Assume that the statement is true for n = k for some k ∈ N, i.e., {u1 , u2 , . . . , uk } is an
orthonormal set and
span (u1 , u2 , . . . , uk ) = span (x1 , x2 , . . . , xk ).
(4.4)
Let n = k + 1. Note that xk+1 ∈
/ span (x1 , x2 , . . . , xk ) because {x1 , x2 , . . .} is linearly independent. By the assumption (4.4), it implies that xk+1 ∈
/ span (u1 , u2 , . . . , uk ) so that vk+1 6= 0.
By this fact and Definition 4.1, for j = 1, 2, . . . , k, we derive that
vk+1
, uj
huk+1 , uj i =
kvk+1 k
+
*
k
X
1
=
hxk+1 , ui i ui , uj
xk+1 −
kvk+1 k
i=1
=
=
1
kvk+1 k
1
kvk+1 k
= 0.
hxk+1 , uj i −
hxk+1 , uj i −
1
kvk+1 k
1
kvk+1 k
k
X
i=1
hxk+1 , ui i hui , uj i
hxk+1 , uj i huj , uj i
Therefore, {u1 , u2 , . . . , uk+1 } is orthonormal. Next, by the assumption (4.4), we know that
x ∈ span (x1 , x2 , . . . , xk , xk+1 ) if and only if
x ∈ span (u1 , u2 , . . . , uk , xk+1 ).
Since
vk+1 = xk+1 −
k
X
i=1
hxk+1 , ui i ui
and uk+1 =
(4.5)
vk+1
,
kvk+1 k
the xk+1 in the result (4.5) can be replaced by vk+1 and ultimately by uk+1 . Thus we have
span (u1 , u2 , . . . , uk+1 ) = span (x1 , x2 , . . . , xk+1 ).
125
4.1. Basic Properties of Hilbert Spaces
Hence the statement is true for n = k + 1 if it is true for n = k. By induction, the construction
yields an orthonormal set {u1 , u2 , . . .} with
span (u1 , u2 , . . . , un ) = span (x1 , x2 , . . . , xn )
for all n ∈ N. This completes the proof of the problem.
Problem 4.3
Rudin Chapter 4 Exercise 3.
Proof. Recall the definition that a space is separable if it contains a countable dense subset.
Suppose that 1 ≤ p < ∞. Since T is compact, the second paragraph following Definition
3.16 says that Cc (T ) = C0 (T ) = C(T ). Thus we deduce from this and Theorem 3.14 that C(T )
is dense in Lp (T ) in the norm k · kp . Let P be the set of all trigonometric polynomials on T .
By Theorem 4.25 (The Weierstrass Approximation Theorem) and Definitions 4.23 (or Remark
3.15), P is dense in C(T ) in the norm k · k∞ . By Definition 3.7, we know that |f (x)| ≤ kf k∞
a.e. on T and then
o1 n 1 Z π
o1
n 1 Z π
p
p
|f (t)|p dt
kf kp∞ dt
≤
= kf k∞ .
(4.6)
kf kp =
2π −π
2π −π
In other words, the inequality (4.6) implies that P is also dense in C(T ) in the norm k · kp . By
Lemma 2.10, we conclude that P is dense in Lp (T ) in the norm k · kp .
Next we suppose that PQ is the set of all trigonometric polynomials with rational coefficients.
Note thata
n
n
o
X
PQ = P (t) =
(ak + ibk )eikt a−n , . . . , an , b−n , . . . , bn ∈ Q and n ∈ N
k=−n
=
∼
∞
[
n=1
∞
[
n
P (t) =
n
X
(ak + ibk )eikt a−n , . . . , an , b−n , . . . , bn ∈ Q
k=−n
o
Q2(2n+1) .
(4.7)
n=1
For each n ∈ N, since each Q2(2n+1) is countable, it follows from the set relation (4.7) that PQ
is also countable. Given f ∈ P. Since the set {p + iq | p, q ∈ Q} is dense in C and f must be in
the form
n
X
ck eikt
f (t) =
k=−n
for some positive integer n, there exists a sequence {fnk } ⊆ PQ such that kfnk − f k∞ → 0 as
nk → ∞. By the inequality (4.6), it is true that
kfnk − f kp → 0
as nk → ∞. Consequently, PQ is dense in P in the norm k · kp and an application of Lemma 2.10
shows that PQ is dense in Lp (T ) in the norm k · kp . Since PQ is countable, Lp (T ) is separable.
By Definition 4.23, every function defined on T can be identified with a 2π-periodic function
on R, so we may identity L∞ (T ) with L∞ [0, 2π] . Consider the set
E = {fθ = χ[0,θ] | θ ∈ [0, 2π]}.
a
We write A ∼ B if there is a bijection between the two sets A and B, see [49, Definition 2.3, p. 25].
Chapter 4. Elementary Hilbert Space Theory
126
It is clear that χ[0,θ] ∈ L∞ [0, 2π] so that E ⊆ L∞ [0, 2π] . Since χ[0,θ1] 6= χ[0,θ2 ] if θ1 6= θ2 ,
E ∼ [0, 2π] and then it is uncountable. In fact, we have
kχ[0,θ1 ] − χ[0,θ2 ] k∞ = 1
(4.8)
if θ1 < θ2 .
Assume that L∞ [0, 2π] was separable. Let F = {fn } be a countable dense subset of
L∞ [0, 2π] . Then we must have
[ 1
E ⊆ L∞ [0, 2π] ⊆
B fn , ,
2
(4.9)
fn ∈F
where B(fn , 12 ) = {f ∈ L∞ ([0, 2π]) | kf − fn k∞ < 12 }. If χ[0,θ1 ] , χ[0,θ2 ] ∈ B(fn , 21 ) with θ1 6= θ2 ,
then it follows from the result (4.8) that
1 = kχ[0,θ1 ] − χ[0,θ2 ] k∞ ≤ kχ[0,θ1 ] − fn k∞ + kfn − χ[0,θ2 ] k∞ < 1,
a contradiction. Thus each B(fn , 12 ) contains at most one element of E. Since F is countable
but E is uncountable, we have
[ 1
E*
B fn ,
2
fn ∈F
which definitely contradicts the set relation (4.9). Hence L∞ [0, π] is not separable and we
have completed the proof of the problem.
Problem 4.4
Rudin Chapter 4 Exercise 4.
Proof. Suppose that H is separable. By the definition, it has a countable dense subset {un }.
By Problem 4.2, we may assume that {un } is also an orthonormal set. Let {unk } be a maximal
linearly independent subset of {un }. Note that {unk } is at most countable. Since we have
span ({unk }) = {un }, span ({unk }) is also dense in H. By Theorem 4.18, the set {unk } is in fact
an at most countable maximal orthonormal set in H.
Conversely, we suppose that E = {un } is an at most countable maximal orthonormal set in
H. By Theorem 4.18 again, span (E) is dense in H. Let span Q (E) be the span of the set E
with rational coefficients. It is easy to see that
span Q (E) =
n
nX
k=1
=
∼
(ak + ibk )uk a1 , . . . , an , b1 , . . . , bn ∈ Q and n ∈ N
∞ nX
n
[
n=1
∞
[
k=1
(ak + ibk )uk a1 , . . . , an , b1 , . . . , bn ∈ Q
o
o
Q2n .
n=1
As a result, span Q (E) must be countable. Next, since Q is dense in R, we conclude that
span Q (E) is dense in span (E). By Lemma 2.10, span Q (E) is also dense in H and hence it is
separable. This completes the proof of the problem.
127
4.1. Basic Properties of Hilbert Spaces
Problem 4.5
Rudin Chapter 4 Exercise 5.
Proof. Suppose that M 6= H. If L = 0, then M = H so that L 6= 0. By Theorem 4.12 (The
Riesz Representation Theorem for Hilbert Spaces), there is a unique z ∈ H such that
L(x) = hx, zi
for all x ∈ H. This z must be nonzero, otherwise we have L = 0 which contradicts our
assumption. It is clear that
M = {x ∈ H | hx, zi = 0} = L−1 (0),
(4.10)
so M is closed in H. Apart from this, if x, y ∈ M , then since L is a linear functional, Definition
2.1 shows that
L(x + y) = L(x) + L(y) = 0 and L(αx) = αL(x) = 0
for a scalar α. Thus M is a closed subspace of H.
Next, we note that span (z) = {αz | α ∈ C} and so
⊥
span (z) = {x ∈ H | hx, αzi = 0 for every α ∈ C}.
(4.11)
Since hx, αzi = α hx, zi, the two expressions (4.10) and (4.11) give
⊥
M = span (z)
and then
M⊥ =
⊥ ⊥
span (z)
.
(4.12)
We claim that span (z) is a closed subspace of H. In fact, it is a subspace of H follows easily
from the definition of span (z). Suppose that {xn } is a sequence in span (z). Then we know that
each xn has the form xn = αn z, where αn ∈ C. Let α = lim αn and x = αz. We want to show
n→∞
that xn → x in H with respect to the norm k · k. Indeed, we have
kxn − xk2 = h(αn − α)z, (αn − α)zi = |αn − α2 | × kzk
so that kxn − xk → 0 as n → ∞, i.e.,
lim xn = x
n→∞
as desired. In conclusion, span (z) contains all its limit points and therefore it is closed in H.
⊥ ⊥
By Problem 4.1, we have span (z)
= span (z). Combining this fact and the expression
(4.12), we have
M ⊥ = span (z)
and since {z} is a basis of span (z), we have dim span (z) = 1 and our required result follows.
We end the proof of the problem.
Problem 4.6
Rudin Chapter 4 Exercise 6.
Chapter 4. Elementary Hilbert Space Theory
128
Proof. We prove the assertions one by one:
• E = {un } is closed and bounded, but not compact. Since E is orthonormal, kun k ≤ 1
for every n ∈ N. Thus E is bounded by 1. If n 6= m, then it follows from Definition 4.1
that
kun − um k2 = hun − um , un − um i = hun , un i + hum , um i = 2.
(4.13)
Assume that u was a limit point√ of E. Then there exists a positive integer N such that
n ≥ N implies that kun − uk < 22 . If n, m ≥ N , then we have
kun − um k ≤ kun − uk + ku − xm k <
√
2
which contradicts the result (4.13). In other words, E has no limit point and then it is
closed in H.
Suppose that B(un , 1) = {x ∈ H | kun −xk < 1} which is clearly open in H. Furthermore,
it is obvious that {B(un , 1)} is an open cover of E. By the fact (4.13), we conclude that
each B(un , 1) contains exactly one element of E, so it is impossible to have
E⊆
k
[
B(uni , 1)
i=1
for any finite k, i.e., E is not compact.
• S is compact if and only if
∞
X
δn2
n=1
< ∞. Suppose that
∞
X
n=1
δn2 < ∞. It is well-known
that a metric space X is compact if and only if X is sequentially compact.b As a subset
of a metric space, S is itself a metric space. Let {xk } ⊆ S be a sequence of the form
xk =
∞
X
ckn un ,
(4.14)
n=1
where k = 1, 2, . . .. Note that we have |ckn | ≤ δn for all k ∈ N. Since {ck1 } ⊆ B(0, δ1 ) ⊆ C
for all k ∈ N and B(0, δ1 ) is compact, B(0, δ1 ) is sequentially compact and then there
exists a subsequence k1 (m) such that
k (m)
c11
→ c1 ∈ B(0, δ1 )
k (m)
as m → ∞. Next, we consider the sequence {c21
subsequence {k2 (m)} ⊆ {k1 (m)} such that
k (m)
c22
} ⊆ B(0, δ2 ). Then we can find a
→ c2 ∈ B(0, δ2 )
as m → ∞. Furthermore, we also have
k (m)
c12
→ c1 ∈ B(0, δ1 )
as m → ∞. By this observation, for every positive integer N , we can select a sequence
{kN +1 (m)} ⊆ {kN (m)} such that for all n ≤ N ,
cnkN (m) → cn ∈ B(0, δn )
b
(4.15)
We say that the space X is sequentially compact if every sequence of points of X has a convergent subsequence,
see [42, Theorem 28.2, p. 179].
129
4.1. Basic Properties of Hilbert Spaces
as m → ∞. Now the limit (4.15) makes sure that
x=
∞
X
n=1
cn un ∈ S.
We claim that the sequence {xkm (m) }, which is clearly a subsequence of {xk } converges
to the point x. To this end, given ǫ > 0, there is a N ′ ∈ N such that
X
ǫ
δn2 < .
(4.16)
8
′
n>N
Then we obtain from the definition (4.14), the fact |ckn | ≤ δn for all k and the estimate
(4.16) that
kx
km (m)
2
− xk =
=
∞
X
[cknm (m) − cn ]un
n=1
∞
X
n=1
|cknm (m) − cn |2
′
=
N
X
n=1
|cknm (m) − cn |2 +
N′
≤
X
n=1
|cknm (m) − cn |2 +
N′
≤
X
n=1
N′
<
2
X
n=1
∞
X
n=N ′ +1
∞
X
|cnkm (m) − cn |2
(|cknm (m) |2 + 2|cknm (m) | · |cn | + |cn |2 )
n=N ′ +1
|cknm (m) − cn |2 + 4 ×
∞
X
δn2
n=N ′ +1
ǫ
|cknm (m) − cn |2 + .
2
(4.17)
To finish our proof, we have to find an upper bound of the summation in the inequality
(4.17). Recall from the limit (4.15) that for each n ≤ N ′
k
cnN
′ (m)
→ cn ∈ B(0, δn )
as m → ∞. In other words, for each n ≤ N ′ , there is a Mn ∈ N such that m ≥ Mn implies
k
|cnN ′
k (m)
k
(m)
− cn |2 <
ǫ
.
2N ′
′ (m)
(4.18)
Since {cnm } ⊆ {cnN
} for m ≥ N ′ , this means that the inequality (4.18) also holds
kN ′ (m)
k (m)
when cn
is replaced by cnm . Let M = max(M1 , Mn , . . . , MN ′ ). If m ≥ M , then we
conclude from the inequalities (4.17) and (4.18) that
′
kx
km (m)
N
X
ǫ
ǫ
+ = ǫ,
− xk <
′
2N
2
n=1
2
proving our claim. Hence S is sequentially compact and then compact.
Conversely, we suppose that S is compact. Consider the point
xN =
N
X
n=1
δn un ,
Chapter 4. Elementary Hilbert Space Theory
2
where N ∈ N. Then we have kxN k =
N
X
130
δn2 . Since S is a compact metric space, it is
n=1
bounded. Thus there exists a positive constant M such that
N
X
n=1
for all N ∈ N and hence
∞
X
n=1
δn2 ≤ M
δn2 < ∞.
• Q is compact. Obviously, we havec
∞
X
n=1
∞
X
1
|cn | ≤
< ∞.
n2
2
n=1
Therefore, the previous assertion implies that Q is compact.
• H is not locally compact. Recall that H is itself a vector space, so it contains the zero
element 0. Assume that H was locally compact. Then the closure
B(0, r) = {x ∈ H | kxk ≤ r}
must be compact for every r > 0. Note that {run } ⊆ B(0, r). Since Er = {run } is closed
in H, Theorem 2.4 implies that Er is compact. Similar to part (a), {B(run , r)} is an open
cover of Er . A direct computation shows that
√
(4.19)
krun − rum k = 2r.
If rum ∈ B(run , r) for n 6= m, then we have
krun − rum k < r
which contradicts the fact (4.19). Therefore, each B(run , r) contains exactly one element
of Er so that {B(run , r)} does not have a finite subcover for Er . In other words, Er is not
compact, a contradiction. Hence we establish the fact that H is not locally compact.
We have completed the proof of the problem.
Problem 4.7
Rudin Chapter 4 Exercise 7.
Proof. We follow the suggestion given by Rudin. Assume that
k ∈ N, there are disjoint sets E1 , E2 , . . . such that
X
a2n > k > 0.
n∈Ek
Now for n ∈ Ek , we define
1
an
bn = n X o 1 × .
k
2
a2n
n∈Ek
c
See [49, Theorem 3.28, p. 62].
P
a2n = ∞. Therefore, for every
(4.20)
131
4.1. Basic Properties of Hilbert Spaces
On the one hand, we have
∞
X
b2n =
n=1
∞ X
X
b2n =
∞
X
k=1
k=1 n∈Ek
1
X
a2n
×
X
n∈Ek
n∈Ek
1
a2n × 2
k
!
=
∞
X
1
< ∞.
k2
k=1
On the other hand, we obtain from the inequality (4.20) that
!
∞ n X
∞ X
∞
∞
∞
o1
X
X
X
X
X
X
1
a2n
1
2 21
√
×
=
a
an bn =
a n bn =
>
n
n X o1
k
k
2
k
n=1
n∈Ek
k=1 n∈Ek
k=1 n∈Ek
k=1
k=1
a2n
n∈Ek
which is divergent, a contradiction. Hence we must have
of the problem.
P
a2n < ∞ and it completes the proof
Problem 4.8
Rudin Chapter 4 Exercise 8.
Proof. By Problem 4.2, H contains an orthonormal set A and then Theorem 4.22 (The Hausdorff
Maximality Theorem) implies that A is contained in a maximal orthonormal set in H. Suppose
that {uα | α ∈ A} and {vβ | β ∈ B} are maximal orthonormal sets in H1 and H2 respectively.
Without loss of generality, we may assume that |A| ≤ |B|. Then we take a subset B ′ of B which
is of the same cardinality as A. Now we consider
H = span ({vβ | β ∈ B ′ }) ⊆ H2 .
By Definition 4.13, span ({vβ | β ∈ B ′ }) is a subspace of H2 . Thus we follow from the last
paragraph in §4.7 that H is a closed subspace of H2 . By the given hint, H is also Hilbert.
Finally, we know from §4.19 that H1 and H are Hilbert space isomorphic to ℓ2 (A) and ℓ2 (B ′ )
respectively. Since A and B have the same cardinality, ℓ2 (A) is Hilbert space isomorphic to
ℓ2 (B ′ ). In other words, H1 is Hilbert space isomorphic to H ⊆ H2 , completing the proof of the
problem.
Problem 4.9
Rudin Chapter 4 Exercise 9.
Proof. Since A ⊆ [0, 2π] and A is measurable, we have χA ∈ L2 (T ). By Theorem 3.11, L2 (T )
is a complete metric space, so it is a Hilbert space by Definition 4.4. Since {un | n ∈ Z} is a
maximal orthonormal set in the Hilbert space L2 (T ), Theorem 4.18(iii) implies that
X
n∈Z
hun , χA i
2
2
= kχA k =
Z
2π
0
|χA (x)|2 dx = m(A) < ∞.
By [49, Theorem 3.23, p. 60], we know that
lim hun , χA i = lim
n→∞
n→−∞
hun , χA i = 0.
(4.21)
Chapter 4. Elementary Hilbert Space Theory
132
By Definition 4.1, we have
un + u−n
hcos nx, χA i =
, χA =
2
un − u−n
hsin nx, χA i =
, χA =
2i
1
(hun , χA i + hu−n , χA i),
2
1
(hun , χA i − hu−n , χA i).
2i
(4.22)
By applying the limits (4.21) to the inner products (4.22), we obtain that
Z π
Z
(cos nt) · χA (t) dt = 2π lim hcos nx, χA i = 0
cos nt dt = lim
lim
n→∞ A
and
lim
Z
n→∞ A
n→∞
n→∞ −π
sin nt dt = lim
Z
π
n→∞ −π
(sin nt) · χA (t) dt = 2π lim hsin nx, χA i = 0.
n→∞
Thus we have completed the proof of the problem.
Problem 4.10
Rudin Chapter 4 Exercise 10.
Proof. This problem is proven in [63, Problem 11.16, p. 352].
Problem 4.11
Rudin Chapter 4 Exercise 11.
Proof. For each n = 1, 2, . . ., we define xn =
of L2 (T ) because, by Definition 4.23,
n+1
n un
kxn k = kxn k2 =
and E = {xn } which is obviously a subset
n+1
<∞
n
for every n ∈ N. It is clear that E 6= ∅. Now we are going to apply a similar argument as in
the proof of the first assertion in Problem 4.6. In fact, we note that hxn , xm i = 0 for n 6= m, so
we follow from Definition 4.1 that
kxm − xn k2 = hxm − xn , xm − xn i
= hxm , xm − xn i − hxn , xm − xn i
= hxm , xm i − hxm , xn i − hxn , xm i + hxn , xn i
m + 1 2
n + 1 2
=
hum , um i +
hun , un i
m
n
m + 1 2 n + 1 2
+
=
m
n
≥2
which implies that
kxm − xn k ≥
√
2.
(4.23)
Assume that x was a√ limit point of E. Then there exists a positive integer N such that n ≥ N
implies kxn − xk < 22 . If n, m ≥ N , then we have
√
kxn − xm k ≤ kxn − xk + kx − xm k < 2
133
4.1. Basic Properties of Hilbert Spaces
which contradicts the result (4.23). Thus E has no limit point and hence it is closed in L2 (T ).
Furthermore, assume that xN was an element of E such that kxN k ≤ kxn k for all xn ∈ E.
However, since we have
N +1
N +2
kxN k =
>
= kxN +1 k,
N
N +1
this contradiction shows that E has no element of smallest norm. Hence we end the proof of
the problem.
Remark 4.1
Problem 4.11 indicates that the condition “convexity” in Theorem 4.10 cannot be omitted.
Problem 4.12
Rudin Chapter 4 Exercise 12.
Proof. By the half-angle formula for cosine function, we have
ck
1=
π
Z
1 + cos t k
π
2
0
Z
2ck
dt =
π
π
2
cos2k x dx.
(4.24)
0
A further substitution with z = sin x reduces the expression (4.24) to
2ck
1=
π
Z
1
0
1
(1 − z 2 )k− 2 dz.
(4.25)
If we substitute y = z 2 and use the integral form of the beta function (see [49, Theorem 8.20,
p. 193]), then the expression (4.25) can be written as
ck
1=
π
Z
1
0
1
1
(1 − y)k− 2 y − 2 dy =
ck Γ(k + 21 )Γ( 12 )
ck Γ(k + 21 )
·
=√ ·
.
π
Γ(k + 1)
kΓ(k)
π
(4.26)
Finally, we recall from [49, Exercise 30, p. 203] that
Γ(k + c)
=1
k→∞ k c Γ(k)
lim
(4.27)
for every real constant c. Take c = 12 in the limit (4.27) and then substitute the corresponding
result into the expression (4.26), we get
lim k
k→∞
− 12
ck = lim
This completes the proof of the problem.
Problem 4.13
Rudin Chapter 4 Exercise 13.
k→∞
√
1
√
k 2 Γ(k)
π·
π.
1 =
Γ(k + 2 )
Chapter 4. Elementary Hilbert Space Theory
134
Proof. We follow the hint. Consider the function f (t) = e2πikt , where k = 0, ±1, ±2, . . .. When
k = 0, we have f ≡ 1 and then the formula holds trivially. Suppose that k 6= 0. Since α is
irrational, kα ∈
/ Z and then e2πikα 6= 1. By this, we have
N
N
1 X 2πiknα
1 X 2πikα n e2πikα e2πikN α − 1
e
=
(e
) =
· 2πikα
N n=1
N n=1
N
e
−1
so that
lim
N →∞
On the other hand, we have
N
2
1 X 2πiknα
e
≤ lim
= 0.
N →∞ N |e2πikα − 1|
N
(4.28)
n=1
Z
1
e2πikt dt =
0
e2πikt
2πik
1
0
= 0.
(4.29)
Hence we deduce from the inequality (4.28) and the integral (4.29) that
Z 1
N
1 X 2πiknα
lim
e
=
e2πikt dt
N →∞ N
0
(4.30)
n=1
for every k = 0, ±1, ±2, . . .. Therefore the formula (4.30) holds for every trigonometric polynomials in the form
N
X
P (t) =
ck e2πikt ,
(4.31)
k=−N
where c−N , . . . , cN ∈ C.
Suppose that f : R → R is continuous and f (x + 1) = f (x) for every x ∈ R. By Theorem
4.25 (The Weierstrass Approximation Theorem), for every ǫ > 0, there exists a trigonometric
polynomial P in the form (4.31) such that
|f (t) − P (t)| <
ǫ
3
(4.32)
for every t ∈ R. Next, the analysis in the previous paragraph shows that there is a positive
integer N such that n ≥ N implies
Z 1
N
1 X
ǫ
P (t) dt < .
P (nα) −
N n=1
3
0
(4.33)
Hence we follow from the estimates (4.32) and (4.33) that n ≥ N implies
Z 1
N
N
N
N
1 X
1 X
1 X
1 X
f (t) dt =
f (nα) −
f (nα) −
P (nα) +
P (nα)
N
N
N
N
0
n=1
n=1
n=1
n=1
Z 1
Z 1
Z 1
f (t) dt
P (t) dt −
P (t) dt +
−
0
N
X
0
1
1
≤
f (nα) − P (nα) +
N
N
n=1
Z 1
+
|f (t) − P (t)| dt
0
N
X
n=1
P (nα) −
Z
1
P (t) dt
0
0
< ǫ.
Hence the formula (4.30) holds for every continuous function on R with period 1, completing
the proof of the problem.
135
4.2. Application of Theorem 4.14
Remark 4.2
A question similar to Problem 4.13 is also proven in [63, Problem 8.19, pp. 194 – 196].
In fact, Problem 4.13 is a very important tool in the study of the Weyl’s Equidistribution
Theorem, see [57, §4.2, pp. 105 – 113].
4.2
Application of Theorem 4.14
Problem 4.14
Rudin Chapter 4 Exercise 14.
Proof. Let H be the Hilbert space L2 [−1, 1] . Define the inner product of f, g ∈ H by
hf, gi =
Z
1
f (x)g(x) dx.
(4.34)
−1
It is clear that xn ∈ H for every n = 0, 1, 2, . . . and the set {1, x, x2 , . . .} is a linearly independent
set of vectors in H. By Problem 4.2, we can obtain an orthonormal set {u1 (x), u2 (x), . . .} such
that {1, x, . . . , xN −1 } and {u1 (x), u2 (x), . . . , uN (x)} have the same span for all positive integers
N . Next, we let F = {0, 1, 2} and we consider the closed subspace
MF = span ({1, x, x2 })
(4.35)
of H. Then the orthonormal set {u1 (x), u2 (x), u3 (x)} obtained by {1, x, x2 } is given by
r
r
3 5 2 1
1
3
x −
.
x and u3 (x) =
u1 (x) = √ , u2 (x) =
2
2 2
3
2
Let x3 ∈ H. Define
sF (x3 ) = x3 , u1 (x) · u1 (x) + x3 , u2 (x) · u2 (x) + x3 , u3 (x) · u3 (x).
By Theorem 4.14(b) and the inner product (4.34), we derive
3
h
3
kx − sF k = min{kx − sk | s ∈ MF } = min
Since x3 , u1 (x) = 0, x3 , u2 (x) =
2
5
q
3
2
a,b,c
h
3
x3 − x =
5
1
−1
3
2 2
|x − a − bx − cx | dx
2
.
(4.36)
3
x
5
2 i 1 r 8
3
2
x3 − x dx =
.
5
175
−1
Z
i1
and x3 , u3 (x) = 0, we have
sF (x3 ) =
and then
Z
1
(4.37)
Combining the results (4.36) and (4.37), it yields that
min
a,b,c
Z
1
−1
|x3 − a − bx − cx2 |2 dx =
8
.
175
(4.38)
Chapter 4. Elementary Hilbert Space Theory
136
For the second assertion, we first notice that the definite integral must be a real number
because we are considering its maximum. Secondly, the conditions on g actually mean that
h1, gi = hx, gi = x2 , g = 0 and kgk = 1.
Thus, by the discussion in §4.9, we have
g ∈ MF⊥
and
kgk = 1,
where MF is defined by (4.35). Now we follow from these facts that
Z 1
Z 1
3
max
x3 g(x) dx = max{| x3 , g | | g ∈ MF⊥ and kgk = 1}.
x g(x) dx ≤ max
g∈MF⊥
kgk=1
g∈MF⊥
kgk=1
−1
(4.39)
−1
By Problem 4.16d , we see that
max{| x3 , g | | g ∈ MF⊥ and kgk = 1} = min{kx3 − sk | s ∈ MF }.
(4.40)
Therefore, we combine the expressions (4.39) and (4.40) and then using the value (4.38) to get
r
Z 1
8
3
x g(x) dx ≤
max
.
(4.41)
⊥
175
g∈MF −1
If g(x) =
q
kgk=1
175
8
x3 − 35 x , then we have
Z
1
3
x g(x) dx =
−1
r
8
175
so that the equality in the inequality (4.41) is attainable. Hence we have completed the proof
of the problem.
Problem 4.15
Rudin Chapter 4 Exercise 15.
Proof. Now we work on the space H = L2 [0, ∞] and it is easy to check that the formula given
by
Z ∞
f ge−x dx
hf, gi =
(4.42)
0
satisfies Definition 4.1, so it is indeed an inner product for the space H.
Now we follow the idea of proof in Problem 4.14. Evidently, we have xn ∈ H for every
n = 0, 1, 2, . . . and the set {1, x, x2 , . . .} is a linearly independent set of vectors in H. By
Problem 4.2, we can obtain an orthonormal set {u1 (x), u2 (x), . . .} such that {1, x, . . . , xN −1 }
and {u1 (x), u2 (x), . . . , uN (x)} have the same span for all positive integers N . Next, we let
F = {0, 1, 2} and we consider the closed subspace
MF = span ({1, x, x2 })
of H. Then the orthonormal set {u1 (x), u2 (x), u3 (x)} obtained by {1, x, x2 } is given by
u1 (x) = 1,
d
We assume its truth here.
1
u2 (x) = x − 1 and u3 (x) = (x2 − 4x + 2).
2
137
4.3. Miscellaneous Problems
Let x3 ∈ H. Define
sF (x3 ) = x3 , u1 (x) · u1 (x) + x3 , u2 (x) · u2 (x) + x3 , u3 (x) · u3 (x).
By Theorem 4.14(b) and the inner product (4.42), we derive
Z 1
h
i1
2
3
3
kx − sF k = min{kx − sk | s ∈ MF } = min
|x3 − a − bx − cx2 |2 e−x dx .
a,b,c
(4.43)
−1
Since x3 , u1 (x) = 6 and x3 , u2 (x) = x3 , u3 (x) = 18, we have
sF (x3 ) = 6 + 18(x − 1) + 9(x2 − 4x + 2) = 9x2 − 18x + 6
so that
3
2
kx − 9x + 18x − 6k =
hZ
1
−1
(x3 − 9x2 + 18x − 6)2 e−x dx
Combining the results (4.43) and (4.44), it yields that
Z 1
|x3 − a − bx − cx2 |2 e−x dx = 36.
min
a,b,c
i1
2
= 6.
(4.44)
(4.45)
−1
For the second assertion, we observe again that the definite integral is a real number and the
conditions on g imply that
g ∈ MF⊥ and kgk = 1.
Therefore, it yields that
Z
Z 1
x3 g(x)e−x dx ≤ max
max
g∈MF⊥
kgk=1
g∈MF⊥
−1
kgk=1
1
x3 g(x)e−x dx
−1
= max{| x3 , g | | g ∈ MF⊥ and kgk = 1}.
(4.46)
max{| x3 , g | | g ∈ MF⊥ and kgk = 1} = min{kx3 − sk | s ∈ MF }.
(4.47)
By Problem 4.16 again, we see that
By combining the expressions (4.46) and (4.47) and then using the value (4.45), we gain
Z 1
max
x3 g(x)e−x dx ≤ 6.
(4.48)
g∈MF⊥
kgk=1
−1
If g(x) = 16 (x3 − 9x2 + 18x − 6), then we have
Z
1
x3 g(x)e−x dx = 6
−1
so that the equality in the inequality (4.48) is attainable and we end the proof of the problem. 4.3
Miscellaneous Problems
Problem 4.16
Rudin Chapter 4 Exercise 16.
Chapter 4. Elementary Hilbert Space Theory
138
Proof. Since M is a closed linear subspace of H, we follow from Theorem 4.11(a) that x0 has
the unique decomposition
x0 = P x0 + Qx0 ,
where P x0 ∈ M and Qx0 ∈ M ⊥ . In addition, Theorem 4.11(b) ensures that
kP x0 − x0 k = min{kx − x0 k | x ∈ M }.
(4.49)
On the other hand, if y ∈ M ⊥ and kyk = 1, then we have
hx0 , yi = hP x0 + Qx0 , yi = hQx0 , yi
and then Theorem 4.2 (The Schwarz Inequality) implies that
| hx0 , yi | = | hQx0 , yi | ≤ kQx0 k × kyk = kQx0 k.
(4.50)
In other words, the inequality (4.50) shows that
max{| hx0 , yi | | y ∈ M ⊥ and kyk = 1} ≤ kQx0 k.
(4.51)
If Qx0 = 0, then it means that x0 = P x0 ∈ M . In this case, we deduce from the expression
(4.49) and the inequality (4.51) that
min{kx − x0 k | x ∈ M } = kP x0 − x0 k = max{| hx0 , yi | | y ∈ M ⊥ and kyk = 1} = 0.
If Qx0 6= 0, since Qx0 ∈ M ⊥ , we may consider y0 =
and ky0 k = 1. A direct computation shows that
| hx0 , y0 i | =
Qx0
kQx0 k
which satisfies the conditions y0 ∈ M ⊥
1
1
× | hx0 , Qx0 i | =
× | hQx0 , Qx0 i | = kQx0 k.
kQx0 k
kQx0 k
Thus the inequality (4.51) actually becomes
max{| hx0 , yi | | y ∈ M ⊥ and kyk = 1} = kQx0 k.
(4.52)
Hence our desired result follows from the comparison between the expressions (4.49) and (4.52)
and the fact that kP x0 − x0 k = kQx0 k. We have ended the proof of the problem.
Problem 4.17
Rudin Chapter 4 Exercise 17.
Proof. Take H = L2 [0, 1] which is a Hilbert space with the inner product given by
hf, gi =
Z
1
f (t)g(t) dt,
(4.53)
0
where f, g ∈ H, see Example 4.5(b). For t ∈ [0, 1], we note that χ[0,t] ∈ H because
kχ[0,t] k =
nZ
0
1
2
|χ[0,t] (x)| dx
o1
2
=
nZ
t
dx
0
Thus we can consider the mapping γ : [0, 1] → H defined by
γ(t) = χ[0,t]
o1
2
=
√
t ≤ 1.
139
4.3. Miscellaneous Problems
which is well-defined. It is trivial that γ is injective. Furthermore, for every p ∈ [0, 1], if t > p,
then we have
nZ 1
o1 n Z t
o1
1
2
2
2
kγ(t) − γ(p)k =
|χ(p,t] (x)| dx
=
dx
= (t − p) 2 .
(4.54)
0
p
Similarly, if t < p, then we have
kγ(t) − γ(p)k =
nZ
p
dx
t
o1
2
1
= (p − t) 2 .
(4.55)
By combining the two expressions (4.54) and (4.55), we conclude that
1
kγ(t) − γ(p)k = |t − p| 2
and thus γ is continuous on [0, 1].
Let 0 ≤ a ≤ b ≤ c ≤ d ≤ 1. If a = b, then γ(a) − γ(b) = 0 and
Z 1
0 × χ(c,d] (x) dx = 0.
hγ(b) − γ(a), γ(d) − γ(c)i = h0, γ(d) − γ(c)i =
0
Similarly, if c = d, then hγ(b) − γ(a), γ(d) − γ(c)i = 0. Without loss of generality, we may
assume that a < b and c < d. Then (a, b] ∩ (c, d] = ∅ and so we have
Z 1
hγ(b) − γ(a), γ(d) − γ(c)i = χ(a,b] , χ(c,d] =
χ(a,b] (x)χ(c,d] (x) dx = 0.
0
Hence the desired result holds in the special case that H = L2 [0, 1] .
To treat the general case, we first consider the function un (t) = e2πint for every n ∈ Z. It is
clear that un ∈ H and

Z 1
 1, if n = m;
2πi(n−m)t
e
dt =
hun , um i =

0
0, if n 6= m.
Thus {un | n ∈ Z} is an orthonormal set in H. By Theorem 3.14, C [0, 1] is dense in H with
respect to the norm induced by the inner product (4.53). By Theorem 4.25 (The Weierstrass
Approximation Theorem) and Definition 4.23, we know that the set P of all trigonometric
polynomials in the form
N
N
X
X
cn e2πint
cn un (t) =
P (t) =
n=−N
n=−N
is dense in C [0, 1] with respect to the norm k · k∞ . Note that P = span ({un | n ∈ Z}). By
Definition 3.7, it can be shown easily that P is also dense in C [0, 1] with respect to the norm
induced by the inner product (4.53). Therefore, we derive from Lemma 2.10 that P is also dense
in H in the norm induced by the inner product (4.53). Consequently, we conclude from Theorem
4.18(i) that {un | n ∈ Z} is in fact maximal and thene
H∼
= ℓ2 (N).
(4.56)
Suppose that H ′ is a Hilbert space with an infinite maximal orthonormal set {vα | α ∈ A},
where A is countable or uncountable infinite. Then we have
H′ ∼
= ℓ2 (A).
e
We say A ∼
= B if A is Hilbert space isomorphic to B.
(4.57)
Chapter 4. Elementary Hilbert Space Theory
140
Applying Problem 4.8 to the Hilbert space isomorphisms (4.56) and (4.57), we conclude that H
must be Hilbert space isomorphic to a subspace of H ′ and then the special case is applicable to
this general case. Indeed, let Λ : H → K be a Hilbert space isomorphism, where K is a Hilbert
subspace of H ′ . If we define λ : [0, 1] → H ′ by
λ = Λ ◦ γ,
then λ is injective and continuous. Furthermore, since hΛ(f ), Λ(g)i = hf, gi for all f, g ∈ H, we
deduce from the special case above that
hλ(b) − λ(a), λ(d) − λ(c)i = Λ γ(b) − Λ γ(a) , Λ γ(d) − Λ γ(c)
= Λ γ(b) , Λ γ(d) − Λ γ(b) , Λ γ(c)
− Λ γ(a) , Λ γ(d) + Λ γ(a) , Λ γ(c)
= hγ(b), γ(d)i − hγ(b), γ(c)i − hγ(a), γ(d)i + hγ(a), γ(c)i
= hγ(b) − γ(a), γ(c) − γ(d)i
= 0.
This completes the proof of the problem.
Problem 4.18
Rudin Chapter 4 Exercise 18.
Proof. If r, s ∈ R and r 6= s, then we have
Z A
2i sin[(r − s)A]
1
= 0.
ei(r−s)t dt = lim
hur , us i = lim
A→∞
A→∞ 2A −A
2A(r − s)
If r = s, then
1
A→∞ 2A
hus , us i = lim
Z
A
1
A→∞ 2A
eist e−ist dt = lim
−A
Z
(4.58)
A
dt = 1.
(4.59)
−A
Thus {us | s ∈ R} is an orthonormal set of H by Definition 4.13. Since f and g are finite
combinations of us , we may suppose that
f (t) =
n
X
isp t
cp e
and g(t) =
p=1
m
X
dq eirq t .
(4.60)
q=1
Therefore, we obtain
Z A X
Z AX
m
n
n X
m
X
1
1
lim
dq e−irq t dt = lim
cp eisp t
cp dq ei(sp −rq )t dt
A→∞ 2A −A
A→∞ 2A −A
q=1
p=1
p=1 q=1
which is undoubtedly finite by the values (4.58) and (4.59). As a consequence, we have shown
that hf, gi is well-defined for all f, g ∈ X.
Next we are going to show that h · i satisfies Definition 4.1. Firstly, it is clear that
1
hf, gi = lim
A→∞ 2A
Z
A
1
f (t)g(t) dt = lim
A→∞ 2A
−A
Z
A
−A
Secondly, for f, g, h ∈ X, we observe that
Z A
1
[f (t) + g(t)]h(t) dt
hf + g, hi = lim
A→∞ 2A −A
g(t)f (t) dt = hg, f i.
141
4.3. Miscellaneous Problems
1
= lim
A→∞ 2A
Z
A
1
f (t)h(t) dt + lim
A→∞ 2A
−A
= hf, hi + hg, hi .
Z
A
g(t)h(t) dt
−A
Thirdly, for any scalar α, we have
Z A
Z A
1
1
αf (t)h(t) dt = α lim
f (t)h(t) dt = α hf, gi .
hαf , gi = lim
A→∞ 2A −A
A→∞ 2A −A
Fourthly, if f takes the representation (4.60), then we establish from the results (4.58) and (4.59)
that
Z A
Z A
n
X
1
1
2
|cp |2 ≥ 0.
(4.61)
f (t)f (t) dt = lim
|f (t)| dt =
hf, f i = lim
A→∞ 2A −A
A→∞ 2A −A
p=1
Finally, if hf, f i = 0, then the result (4.61) definitely shows that c1 = c2 = · · · = cp = 0. In
other words, we have f ≡ 0 in this case. Hence this inner product certainly makes X into a
unitary space.
Let H be the completion of X, i.e., X is dense in H and H = X. Since span ({us | s ∈ R}) =
X, Theorem 4.18(i) says that the set {us | s ∈ R} is in fact a maximal orthonormal in H. By
the discussion in §4.19, we have
H∼
(4.62)
= ℓ2 (R).
Assume that H was separable. By Problem 4.4, H has an at most countable maximal orthonormal system {vn | n ∈ N}. By the discussion in §4.19 again, we know that
H∼
= ℓ2 (N).
(4.63)
Now the two isomorphic relations (4.62) and (4.63) imply that ℓ2 (R) ∼
= ℓ2 (N), but then R
and N have the same cardinal number, a contradiction. Hence we complete the proof of the
problem.
Problem 4.19
Rudin Chapter 4 Exercise 19.
Proof. If N = 1, then ω = e2πi = 1 and k can only take the value 0. Thus the orthogonality
relation holds in this special case. Suppose that N > 1. If k = 0, then ω 0 = 1 so that
N
1 X 0
ω = 1.
N
n=1
Let k = 1, 2, . . . , N − 1. We remark that
z N − 1 = (z − 1)(z N −1 + · · · + z + 1).
Putting z = ω k into the expression (4.64), we obtain
(ω k )N − 1 = (ω k − 1) (ω k )N −1 + · · · + ω k + 1 = (ω k − 1) ω k(N −1) + · · · + ω k + 1 .
(4.64)
(4.65)
It is clear that (ω k )N = (ω N )k = 1. Furthermore, since 1 ≤ k ≤ N − 1, k is evidently not
divisible by N which implies that ω k 6= 1. Therefore, it deduces from the expression (4.65) that
ω k(N −1) + · · · + ω k + 1 = 0
Chapter 4. Elementary Hilbert Space Theory
142
or equivalently
ω kN + ω k(N −1) + · · · + ω k = 0.
In conclusion, we have

N
1, if k = 0;
1 X nk 
ω =

N
n=1
0, if 1 ≤ k ≤ N − 1.
Let N ≥ 3. It is clear from the orthogonality relations and the properties in Definition 4.1
that
N
N
1 X
1 X
n 2 n
kx + ω yk ω =
hx + ω n y, x + ω n yi ω n
N
N
n=1
n=1
N
N
N
X
X
1 hX
hω n y, xi ω n
hx, ω n yi ω n +
hx, xi ω n +
=
N
n=1
n=1
+
N
X
n=1
hω n y, ω n yi ω n
n=1
i
X
X
X
1h
=
hy, xi ω n × ω n
hx, yi ω −n × ω n +
ωn +
hx, xi
N
+ hy, yi
N
n=1
N
X i
n
N
n=1
n=1
ω
n=1
N
X
1h
=
hx, xi
N
= hx, yi
N
n=1
n
ω + N hx, yi + hy, xi
N
X
n=1
ω
2n
+ hy, yi
N
X
ωn
n=1
i
(4.66)
as desired.
Finally, Definition 4.1 gives
D
E
kx + eiθ yk2 = x + eiθ y, x + eiθ y = hx, xi + e−iθ hx, yi + eiθ hy, xi + hy, yi
so that
1
2π
Z
Z π
1
kx + e yk e dθ =
kxk2 + kyk2 + e−iθ hx, yi + eiθ hy, xi eiθ dθ
2π −π
−π
Z π
Z π
1
1
hx, yi dθ +
e2iθ hy, xi dθ
=
2π −π
2π −π
π
iθ
2 iθ
= hx, yi .
This completes the proof of the problem.
(4.67)
Remark 4.3
We note that the polarization identity (see [51, p. 86]) can be written in the form
3
hx, yi =
1 X −n
i kx + in yk2 .
4
n=0
Thus the two identities (4.66) and (4.67) in Problem 4.19 are actually generalizations of this.
CHAPTER
5
Examples of Banach Space Techniques
5.1
The Unit Ball in a Normed Linear Space
Problem 5.1
Rudin Chapter 5 Exercise 1.
Proof. For 0 < p ≤ ∞, we suppose that
B = {f ∈ Lp (µ) | kf kp ≤ 1}.
There are two cases:
• Case (i): 0 < p < ∞. In this case, we have
o1
nZ
p
|f |p dµ
kf kp =
X
1
= |f (a)|p × µ({a}) + |f (b)|p × µ({b}) p
1
1
= 1 {|f (a)|p + |f (b)|p } p .
2p
Consequently, kf kp ≤ 1 if and only if
|f (a)|p + |f (b)|p ≤ 2.
(5.1)
Therefore, it follows from the inequality (5.1) that the unit ball B is a circle if and only if
p = 2. Besides, the unit ball B becomes a square if and only if p = 1.
• Case (ii): p = ∞. By Definition 3.7, we know that
kf k∞ = max(|f (a)|, |f (b)|)
(5.2)
max(|f (a)|, |f (b)|) ≤ 1
(5.3)
so that kf k∞ ≤ 1 if and only if
which is a square.
See Figure 5.1 for an illustration.
143
Chapter 5. Examples of Banach Space Techniques
144
Figure 5.1: The unit circle in different p-norm.
If µ({a}) 6= µ({b}), then the expression (5.1) is replaced by
µ({a})|f (a)|p + µ({b})|f (b)|p ≤ 1
which cannot be a circle or a square for any p ∈ (0, ∞). However, we note that the expression
(5.2) is still valid even in the case µ({a}) 6= µ({b}), so we have the inequality (5.3) and then B
is a square when p = ∞. This ends the analysis of the problem.
Problem 5.2
Rudin Chapter 5 Exercise 2.
Proof. Let X be a normed linear space with norm k · k and B(0, 1) = {x ∈ X | kxk < 1} be the
open unit ball. For every x, y ∈ B(0, 1) and t ∈ [0, 1], we follow from Definition 5.2 that
k(1 − t)x + tyk ≤ (1 − t)kxk + tkyk < 1
so that (1 − t)x + ty ∈ B(0, 1). By §4.8, B(0, 1) is convex. Similarly, the convexity of the closed
unit ball B(0, 1) = {x ∈ X | kxk ≤ 1} can be proven similarly. Hence we complete the proof of
the problem.
Problem 5.3
Rudin Chapter 5 Exercise 3.
Proof. Suppose that kf kp = kgkp = 1 and f 6= g. By Theorem 3.9, we always have
khkp =
1
(f + g)
2
p
1
1
≤ kf kp + kgkp = 1.
2
2
Assume that khkp = 1. Then it follows from Problem 3.13 that f = λg a.e. on X for some
λ > 0. Since λkgkp = kf kp = kgkp , we have λ = 1 implying f = g a.e. on X, a contradiction.
However, strictly convexity does not hold for L1 (µ), L∞ (µ) and C(X). Ignore trivialities, we
suppose that X contains more than one point. In fact, we can pick E, F ∈ M \ {∅} such that
E ∩ F = ∅ and µ(E), µ(F ) ∈ (0, ∞). Define f : X → C and g : X → C by
f (x) =
1
χE (x) and
µ(E)
g(x) =
1
χF (x)
µ(F )
respectively. It is obvious that f 6= g. Furthermore, we have
Z
Z
1
|f (x)| dx =
kf k1 =
χE (x) dx = 1,
µ(E) E
X
(5.4)
145
5.1. The Unit Ball in a Normed Linear Space
kgk1 =
f +g
2
1
=
Z
ZX
X
Z
1
χF (x) dx = 1,
µ(F ) F
Z
Z
1
f (x) + g(x)
1
f (x) dx +
g(x) dx = 1.
dx =
2
2 E
2 F
|g(x)| dx =
These show that the invalidity of strictly convexity in L1 (µ). For L∞ (µ), we replace the functions
f and g in the definition (5.4) by
f (x) = χE (x)
and g(x) = χE∪F (x)
respectively. Then it is also true that f 6= g. By Definition 3.7, we have
f +g
= 1.
2
∞
Therefore, the strictly convexity does not hold in L∞ (µ).
Finally, we consider the space C(X), where X is Hausdorff. By Definition 3.16, we see that
X is assumed to be compact and so the norm k · k∞ of C(X) is given by
kf k∞ = kgk∞ =
kf k∞ = sup |f (x)|.
(5.5)
x∈X
Take f ∈ C(X) and suppose that f is nonconstant. Thus one can find a x0 ∈ X such that
|f (x0 )| < kf k∞ .
(5.6)
Otherwise, we have |f (x)| ≥ kf k∞ on X, but this and the definition (5.5) will imply that
|f (x)| = kf k∞ on X, i.e., f is a constant which is a contradiction. Define
1
3
a = kf k∞ + |f (x0 )| and
4
4
Consider the sets
o
n
K = x ∈ X |f (x)| ≥ a = |f |−1 [a, ∞)
and
1
3
b = kf k∞ + |f (x0 )|.
4
4
o
n
E = x ∈ X |f (x)| ≤ b = |f |−1 (−∞, b] .
It is clear that x0 ∈ E, i.e., E 6= ∅. Since |f | : X → R is continuous, the Extreme Value
Theorem ([42, Theorem 27.4, p. 174]) implies that kf k∞ = max |f (x)| = |f (p)| for some p ∈ X.
x∈X
Consequently, we have
p ∈ K.
(5.7)
Furthermore, both K and E are closed in X by [42, Theorem 18.1, p. 104]. If K ∩ E 6= ∅, then
we have
1
1
3
3
kf k∞ + |f (x0 )| ≤ kf k∞ + |f (x0 )|
4
4
4
4
but these imply that kf k∞ ≤ |f (x0 )| which contradicts the hypothesis (5.6). Hence we have
K ∩ E = ∅.
What we have done in the previous paragraph is that V = E c is open in X and K ⊆ V .
Since X is compact, Theorem 2.4 ensures that K is also compact. By Theorem 2.12 (Urysohn’s
Lemma), there exists a h ∈ Cc (X) such that K ≺ h ≺ V , i.e., 0 ≤ h(x) ≤ 1 on X, h(x) = 1 on
K and supp (h) ⊆ V (or equivalently h(x) = 0 on E). If we define g = h × f : X → C, then we
must have g 6≡ f and furthermore, the fact (5.7) shows that
kgk∞ = sup |h(x)f (x)| = |h(p)| · |f (p)| = |f (p)| = kf k∞
x∈X
and
f +g
2
∞
=
1
1
1
kf + hf k∞ = sup [1 + h(x)]f (x) = |1 + h(p)| · |f (p)| = |f (p)| = kf k∞ .
2
2 x∈X
2
Hence strictly convexity does not hold for C(X), completing the proof of the problem.
Chapter 5. Examples of Banach Space Techniques
5.2
146
Failure of Theorem 4.10 and Norm-preserving Extensions
Problem 5.4
Rudin Chapter 5 Exercise 4.
Proof. Define f : [0, 1] → C by

 −8x + 4, if x ∈ [0, 12 ];
f (x) =

if x ∈ ( 21 , 1].
0,
Then it is easy to see that f ∈ M , i.e., M is nonempty. Suppose that f, g ∈ M , t ∈ [0, 1] and
h = (1 − t)f + tg. Then we have
Z
1
2
0
h(x) dx −
Z
1
1
2
h(x) dx =
Z
1
2
0
[(1 − t)f (x) + tg(x)] dx −
hZ
= (1 − t)
+t
=1
hZ
f (x) dx −
0
1
g(x) dx −
0
Z
1
Z
1
1
1
1
2
[(1 − t)f (x) + tg(x)] dx
f (x) dx
1
2
g(x) dx
1
2
Z
i
i
so that h ∈ M , i.e., M is convex.
Next, we note that C and M are metric spaces. By [49, Theorem 7.15, p. 151], we see that
C is a complete metric space. Since M ⊆ C, M is also a complete metric space. Suppose that
{fn } ⊆ M and
kfn − f k = sup |fn (x) − f (x)| → 0
x∈[0,1]
as n → ∞. Then it follows froma [49, Theorem 7.9, p. 148] that fn → f uniformly on [0, 1].
Since each fn is continuous on [0, 1], f is also continuous on [0, 1]. Furthermore, we have
lim
n→∞
Z
1
2
0
hZ
0
1
2
fn (t) dt −
Z
lim fn (t) dt −
n→∞
Z
1
2
0
1
2
1
Z
1
1
2
i
fn (t) dt = 1
lim fn (t) dt = 1
n→∞
f (t) dt −
Z
1
1
2
f (t) dt = 1.
In other words, f ∈ M and M is closed in the space C.
Finally, we want to show that M contains no element of minimal norm. To this end, we
notice that
1=
a
Z
1
2
0
f (t) dt −
Z
1
1
2
f (t) dt ≤
Z
0
1
2
|f (t)| dt +
Or we can see the rephrased Theorem 7.9 on p. 151 in [49].
Z
1
1
2
|f (t)| dt = kf k1 ≤ kf k∞
(5.8)
147
5.2. Failure of Theorem 4.10 and Norm-preserving Extensions
for every f ∈ M . Assume that there was a f ∈ M such that kf k∞ = 1. Then we deduce from
the inequality (5.8) that
Z 1
Z 1
(1 − |f (t)|) dt = 0.
(5.9)
|f (t)| dt = 1 or
0
0
Since kf k∞ = 1, we have |f (x)| ∈ [0, 1] for all x ∈ [0, 1] so that the function g(x) = 1 − |f (x)| is
continuous and nonnegative on [0, 1]. Applying Theorem 1.39(a) to the second integral in (5.9),
we obtain
g(x) = 0 a.e. on [0, 1].
Now the continuity of g forces that g(x) = 0 on [0, 1], i.e., |f (x)| = 1 on [0, 1]. In particular, we
have
−1 ≤ Re f (x) ≤ 1
(5.10)
on [0, 1].
On the other hand, since f ∈ M , the definition gives
1 = Re
Z
1
2
f (t) dt −
0
or equivalently,
Z
1
2
0
Z
1
1
2
Z
f (t) dt =
Re f (t) − 1 dt +
Z
1
2
0
1
1
2
Re f (t) dt −
Z
1
1
2
Re f (t) dt
−1 − Re f (t) dt = 0.
(5.11)
Observe from the inequalities (5.10) that Re f (x) − 1 ≤ 0 and
−1 − Re f (x) ≤ 0 on
[0, 1].
Therefore, the equation (5.11) and the continuities of Re f (x) − 1 and −1 − Re f (x) imply
that

if [0, 12 ];
 1,
Re f (x) =

−1, if [ 21 , 1].
However, this says that Re f (x) is discontinuous at x = 21 , a contradiction. Hence no such f
exists and this completes the proof of the problem.
Problem 5.5
Rudin Chapter 5 Exercise 5.
Proof. It is clear that 1 ∈ M , so M 6= ∅. Suppose that f, g ∈ M ⊆ L1 [0, 1] , t ∈ [0, 1] and
h = (1 − t)f + tg. Then we have
Z 1
Z 1
Z 1
|g(x)| dx < ∞
|f (x)| dx + t
|h(x)| dx ≤ (1 − t)
0
0
0
which means that h ∈ L1 [0, 1] . Besides, we also have
Z
1
h(x) dx =
Z
0
0
1
[(1 − t)f (x) + tg(x)] dx = (1 − t)
Z
0
1
f (x) dx + t
Z
1
g(x) dx = 1
0
so that h ∈ M , i.e., M is convex. Next, suppose that {fn } ⊆ M and
there exists a function f on
1
[0, 1] such that kfn − f k1 → 0 as n → ∞. Thus fn − f ∈ L [0, 1] and it yields from Theorem
1.33 that
Z 1
Z 1
Z 1
|fn (t) − f (t)| dt = kfn − f k1
f (t) dt ≤
fn (t) dt −
0
0
0
Chapter 5. Examples of Banach Space Techniques
which implies that
lim
Z
n→∞ 0
1
fn (t) dt =
Z
148
1
f (t) dt.
0
In other words, we have f ∈ M and M is closed in L1 [0, 1] .
For every f ∈ M ⊆ L1 [0, 1] , we see from Theorem 1.33 that kf k1 ≥ 1. For each n = 1, 2, . . .,
we define fn : [0, 1] → C by
fn (x) = nxn−1 .
By direct checking, it is clear that
kfn k1 =
Z
1
ntn−1 dt = 1.
0
In addition, fn 6= fm if n 6= m. Hence M contains infinitely many elements of minimal norm,
completing the proof of the problem.
Problem 5.6
Rudin Chapter 5 Exercise 6.
Proof. We note that M is not necessarily closed in H. Let f : M ⊆ H → C be a bounded
linear functional. If we check the proof of Theorem 5.4 carefully, we see that the bounded linear
transformation Λ : X → Y is in fact uniformly continuous on X. Thus f : M → C is uniformly
continuous on M . Since H is a metric (in fact Hilbert) space, M is also a metric space. Since
C is a complete metric space and M is dense in M , we follow from [49, Exercies 4 & 13, pp.
98, 99] that f can be uniquely extended to a continuous function fe : M → C.b Without loss of
generality, we may assume that M is closed in H.
As a closed subspace of a Hilbert space H, M is itself a Hilbert space (see the note in
Problem 4.8). By Theorem 4.12 (The Riesz Representation Theorem for Hilbert Spaces), there
corresponds a unique x0 ∈ M such that
f (x) = hx, x0 i
(5.12)
for all x ∈ M . Recall from Definition 5.3 that kf k is the smallest number such that
|f (x)| ≤ kf k · kxk
for every x ∈ M . In particular, we must have kf (x0 )k ≤ kf k · kx0 k. On the other hand, the
representation (5.12) gives
|f (x0 )| = f (x0 ) = hx0 , x0 i = kx0 k2 = kx0 k · kx0 k,
so we must have
kx0 k = kf k.
(5.13)
By Theorem 4.11, every x ∈ H can be expressed uniquely as x = P x + Qx, where P x ∈ M
and Qx ∈ M ⊥ . This means that H = M ⊕ M ⊥ .c Now if we define F : H → C by
F (x) = hx, x0 i
In fact, such fe is uniformly continuous on M . See also [42, Exercise 2, p. 270].
Let U and W be two vector subspaces of the vector space V . Then V is said to be the direct sum of U and
W , denoted by V = U ⊕ W , if V = U + W and U ∩ W . See [4, pp. 95, 96].
b
c
149
5.2. Failure of Theorem 4.10 and Norm-preserving Extensions
for every x ∈ H, then we obtain
F (x) = hx, x0 i =

 f (x), if x ∈ M ;

if x ∈ M ⊥ .
0,
Furthermore, by the same analysis in obtaining the representation (5.13), we get kF k = kx0 k
and then using the representation (5.13) again, we conclude that
kF k = kf k = kx0 k,
(5.14)
i.e., F is a norm-preserving extension on H of f which vanishes on M ⊥ .
Suppose that F ′ : H → C is another norm-preserving extension of f that vanishes on M ⊥ .
Then F ′ is bounded and thus continuous by Theorem 5.4. By Theorem 4.12 (The Riesz Representation Theorem for Hilbert Spaces), there corresponds a unique x1 ∈ H such that
F ′ (x) = hx, x1 i
for all x ∈ H and also
Since F ′ (x) = f (x) on M , we have
kF ′ k = kf k = kx1 k.
(5.15)
hx, x1 i = hx, x0 i
on M . By Definition 4.1, hx, x1 − x0 i = 0 and then x1 − x0 ∈ M ⊥ . Since x1 = x0 + (x1 − x0 ),
where x0 ∈ M and x1 − x0 ∈ M ⊥ , Theorem 4.11 shows that x1 and x1 − x0 are unique and
kx0 k2 = kx1 k2 + kx0 − x1 k2 .
(5.16)
By putting the numbers (5.14) and (5.15) into the formula (5.16), we see that
kf k2 = kf k2 + kx0 − x1 k2
and this reduces to kx0 − x1 k2 = 0. By Definition 4.1, we must have x0 = x1 , i.e., F ′ = F . This
proves the uniqueness part and so we have completed the proof of the problem.
Problem 5.7
Rudin Chapter 5 Exercise 7.
Proof. Consider
X = [−1, 1] with µ = m, the Lebesgue measure. Then the vector space
L1 [−1, 1] consists of all complex measurable function on [−1, 1] such that
kf k1 =
We define
Z
1
−1
|f (x)| dx < ∞.
M = {f ∈ L1 [−1, 1] | f is real-valued and f (x) = 0 on [−1, 0]}.
Now the function χ[0,1] is obviously an element
of M , so M 6= ∅ and also kχ[0,1] k1 = 1. It is
also clear that M is a subspace of L1 [−1, 1] . Next, we define the functional Λ : M → C by
Λ(f ) =
Z
1
−1
f (x) dx =
Z
1
f (x) dx
0
(5.17)
Chapter 5. Examples of Banach Space Techniques
150
which is linear. By Definition 5.3, the definition (5.17) and the fact that χ[0,1] ∈ M , we have
kΛk = sup{|Λ(f )| | f ∈ M and kf k1 = 1}
Z
Z 1
nZ 1
= sup
|f (x)| dx =
f (x) dx f ∈ M and
−1
0
= 1.
Thus Λ is bounded.
Let δ ∈ (0, 1). If we define Λδ : L1 [−1, 1] → C by
Λδ (f ) =
Z
1
−1
Re [f (x)]χ(−δ,1) (x) dx =
Z
1
0
o
f (x) dx = 1
1
Re [f (x)] dx,
−δ
then it satisfies Definition 2.1 so that Λδ is linear on L1 [−1, 1] . It is also clear that Λδ |M = Λ.
A direct computation also shows that
Z 1
Z 1
Z 1
|f (x)| dx
|Re [f (x)]| dx ≤
Re [f (x)] dx ≤
|Λδ (f )| =
−δ
−δ
−1
which implies that
kΛδ k = sup{|Λδ (f )| | f ∈ L1 [−1, 1] and kf k1 = 1} = 1 = kΛk.
Thus Λδ is a norm-preserving extension on L1 [−1, 1] of Λ. Hence if α 6= β, then we have
Λα 6= Λβ . This completes the proof of the problem.
5.3
The Dual Space of X
Problem 5.8
Rudin Chapter 5 Exercise 8.
Proof.
(a) We first show that X ∗ is a normed linear space by checking Definition 5.2. For all f, g ∈ X ∗ ,
since |f (x) + g(x)| ≤ |f (x)| + |g(x)| for all x ∈ X, it must be true that
kf + gk ≤ kf k + kgk.
Next, for a scalar α and f ∈ X ∗ , we have |αf (x)| = |α| · |f (x)| for all x ∈ X which implies
that
kαf k = |α| · kf k.
Finally, suppose that f ∈ X ∗ is such that kf k = 0. By [51, Eqn. (3), p. 96], we have
|f (x)| ≤ kf k · kxk = 0
holds for every x ∈ X. Thus f ≡ 0 on X. By Definition 5.2, X ∗ is a normed linear space.
It remains to show that X ∗ is complete. To this end, let {Λn } ⊆ X ∗ be Cauchy. Then
given ǫ > 0, there corresponds a positive integer N such that n, m ≥ N imply that
kΛn − Λm k < ǫ
151
5.3. The Dual Space of X
and for arbitrary but fixed x ∈ X, this and [51, Eqn. (3), p. 96] together imply that
|Λn (x) − Λm (x)| = |(Λn − Λm )(x)| ≤ kΛn − Λm k · kxk < ǫkxk.
(5.18)
Thus {Λn (x)} ⊆ C is Cauchy. Since C is complete, {Λn (x)} converges to a point in C and
it is reasonable to define the pointwise limit Λ : X → C by
Λ(x) = lim Λn (x).
n→∞
Now our target is to show that Λ ∈ X ∗ and {Λn } converges to Λ in X ∗ , i.e., kΛn −Λk → 0
as n → ∞. Let α, β ∈ C and x, y ∈ X. Since X is a normed linear space, αx + βy ∈ X
and thus
Λ(αx + βy) = lim Λn (αx + βy)
n→∞
= lim [αΛn (x) + βΛn (y)]
n→∞
= α lim Λn (x) + β lim Λn (y)
n→∞
n→∞
= αΛ(x) + βΛ(y).
By Definition 2.1, Λ is linear. By the hypothesis, {Λn } is a Cauchy sequence in X ∗ so that
it is bounded, i.e., there exists a positive constant M such that
kΛn k ≤ M
(5.19)
for every n = 1, 2, . . .. Therefore, it follows from the triangle inequality, then [51, Eqn.
(3), p. 96] and finally the inequalities (5.19) that
|Λ(x)| ≤ |Λn (x)| + |Λ(x) − Λn (x)|
≤ kΛn k · kxk + |Λ(x) − Λn (x)|
≤ M kxk + |Λ(x) − Λn (x)|
(5.20)
for all n > N and for all x ∈ X. Recall that if we take m → ∞ in the inequality (5.18),
then we have
|Λn (x) − Λ(x)| ≤ ǫkxk
(5.21)
for all n > N . Thus, by putting the inequality (5.19) into the inequality (5.20), we get
|Λ(x)| ≤ (M + ǫ)kxk
for every x ∈ X. Hence we obtain kΛk < ∞, i.e., Λ ∈ X ∗ . Since the inequality (5.21) is
true for all x ∈ X, it must be true for those x with kxk ≤ 1 and so n > N implies that
kΛn − Λk = sup{|Λn (x) − Λ(x)| | x ∈ X and kxk ≤ 1} ≤ ǫ.
Since ǫ is arbitrary, we have {Λn } converges to Λ in X ∗ . In other words, X ∗ is complete
and we conclude that X ∗ is a Banach space.
(b) Fix the x ∈ X, define Λ∗x : X ∗ → C by
Λ∗x (f ) = f (x).
If x = 0, then it is clear that Λ∗0 (f ) = f (0) = 0 on X ∗ . Therefore, Λ∗0 must be linear and of
norm 0. Without loss of generality, we may assume that x 6= 0 in the following discussion.
For any f, g ∈ X ∗ and α, β ∈ C, we have
Λ∗x (αf + βg) = (αf + βg)(x) = αf (x) + βg(x) = αΛ∗x (f ) + βΛ∗x (g).
Chapter 5. Examples of Banach Space Techniques
152
By Definition 2.1, Λ∗x is a linear functional on X ∗ . On the one hand, by [51, Eqn. (3), p.
96], we know that
|Λ∗x (f )| = |f (x)| ≤ kf k · kxk ≤ kxk
(5.22)
for all f ∈ X ∗ with kf k = 1. Thus it means that kΛ∗x k ≤ kxk. On the other hand, since
X is a normed linear space, and x 6= 0, Theorem 5.20 ensures the existence of a g ∈ X ∗
such that kgk = 1 and g(x) = kxk, so we have
kΛ∗x k ≥ |Λ∗x (g)| = |g(x)| ≥ kxk,
(5.23)
where kgk = 1. Combining the inequalities (5.22) and (5.23), we get the desired result
that
kΛ∗x k = kxk.
(c) Suppose that {xn } ⊆ X and {f (xn )} is bounded for every f ∈ X ∗ . We consider the
mapping Λ∗xn : X ∗ → C given by
Λ∗xn (f ) = f (xn ).
By part (a), X ∗ is a Banach space. By part (b), kΛ∗xn k = kxn k for every n ∈ N. By
the hypothesis, for each f ∈ X ∗ , the set {Λ∗xn (f )} = {f (xn )} is bounded by a positive
constant Mn , so you can’t find a f ∈ X ∗ such that
sup |Λ∗xn (f )| = ∞.
n∈N
Therefore, we deduce from Theorem 5.8 (The Banach-Steinhaus Theorem) that
kΛ∗xn k ≤ M
for all n ∈ N, where M is a positive constant. Hence the sequence {kxn k} is bounded.
This completes the proof of the problem.
Problem 5.9
Rudin Chapter 5 Exercise 9.
Proof.
(a) We are going to prove the assertions one by one.
– kΛk = kyk1 . Fix y = {ηi } ∈ ℓ1 . Define Λ : c0 → C by
Λ(x) =
∞
X
ξi ηi ,
(5.24)
i=1
for every x = {ξi } ∈ c0 ⊆ ℓ∞ . Since ξi → 0 as i → ∞, the sequence {ξi } is bounded
by a positive constant M (see [49, Theorem 3.2(c), p. 48]). Since y ∈ ℓ1 , we have
∞
X
i=1
ξi ηi ≤
∞
X
i=1
|ξi ηi | ≤ M
∞
X
i=1
|ηi | < ∞
so that the series (5.24) converges (absolutely) and thus the map Λ is well-defined.
153
5.3. The Dual Space of X
Let z = {ωi } ∈ c0 and α, β ∈ C. Since c0 is a vector space, we have αx + βz ∈ c0 .
By the definition, we have
∞
X
(αξi + βωi )ηi
Λ(x + z) =
(5.25)
i=1
By the analysis in the previous paragraph, the series (5.25) is allowed to split into
two series so that
Λ(x + z) = α
∞
X
ξi ηi + β
∞
X
ωi ηi = αΛ(x) + βΛ(z),
i=1
i=1
i.e., Λ is linear. By the definition (5.24), it is easy to see that
|Λ(x)| ≤ kxk∞ · kyk1
for every x ∈ c0 . Thus it is true that kΛk ≤ kyk1 .
For the reverse direction, since y ∈ ℓ1 , we know that |ηi | → 0 as i → ∞. Next, for
each n ∈ N, we consider the sequence xn = {ξn,i } defined by

ηi


, if i = 1, 2, . . . , n and ηi 6= 0;

|ηi |
(5.26)
ξn,i =


 0,
if i > n or η = 0.
i
Since | |ηηii | | ≤ 1 for all i ∈ N, we have xn ∈ ℓ∞ for each n ∈ N. In fact, we have
kxn k∞ ≤ 1. Furthermore, for every fixed n ∈ N, one can find i ∈ N such that i > n,
thus we deduce easily from the definition (5.26) that
ξn,i → 0
as i → ∞, i.e., xn ∈ c0 . Now we derive from Definition 5.2 that
kΛk = sup{|Λ(x)| | x ∈ c0 and kxk∞ ≤ 1} ≥ |Λ(xn )| =
n
X
i=1
ξn,i ηi =
n
X
i=1
|ηi |
for every n ∈ N, i.e., kΛk ≥ kyk1 . In conclusion, we have kΛk = kyk1 < ∞ and thus
Λ ∈ (c0 )∗ .
– (c0 )∗ is isometric isomorphic to ℓ1 . In the first assertion, we have shown that
given a y ∈ ℓ1 , if Λ is defined in the form (5.24), then Λ ∈ (c0 )∗ . More precisely, the
mapping Φ : ℓ1 → (c0 )∗ given by
Φ(y) = Λy
is well-defined, where Λy is in the form (5.24).
∗ Step 1: Φ is linear. For every y = {ηi } ∈ ℓ1 , z = {θi } ∈ ℓ1 and α, β ∈ C, we
have αy + βz = {αηi + βθi } ∈ ℓ1 . If x = {ξi } ∈ c0 , then we deduce from the
definition (5.24) that
Λαy+βz (x) =
∞
X
i=1
ξi (αηi + βθi ) = α
∞
X
ξi ηi + β
i=1
Consequently, Φ(αy + βz) = αΦ(y) + βΦ(z).
∞
X
i=1
ξi θi = αΛy (x) + βΛz (x).
Chapter 5. Examples of Banach Space Techniques
154
∗ Step 2: Φ is surjective. Given Λ ∈ (c0 )∗ . We want to show that there exists
a y ∈ ℓ1 such that Φ(y) = Λ.
It is well-known that the sequence
{ei }∞
i=1 of standard unit vectors of c0 is a
P
d In fact, we have
basis for c0 and that {ξi } =
ξi ei whenever {ξi }∞
i=1 ∈ c0 .
∞
ei = {δi,k }k=1 , where δi,k is the Kronecker delta function. Then it is routine to
check that ei ∈ c0 and kei k∞ = 1.
Next, we define the sequence y = {ηi }∞
i=1 by
ηi = Λ(ei )
(5.27)
for every i = 1, 2, . . .. Our aim is to show that y ∈ ℓ1 . If the sequence xn =
{ξn,i }ni=1 is given by the form (5.26), then we immediately know that xn ∈ c0 and
kxn k∞ = 1. Since
n
∞
X
X
ξn,i ei ,
ξn,i ei =
xn =
i=1
i=1
the linearity of Λ shows that
Λ(xn ) =
n
X
ξn,i Λ(ei ) =
n
X
ξn,i ηi =
i=1
i=1
n
X
i=1
|ηi |.
(5.28)
Therefore, we apply [51, Eqn. (3), p. 96] to the expression (5.28) to obtain
n
X
i=1
|ηi | = |Λ(xn )| ≤ kΛk · kxn k∞ = kΛk
holds for every n = 1, 2, . . .. In particular, we have
kyk1 =
∞
X
i=1
|ηi | ≤ kΛk < ∞,
i.e, y ∈ ℓ1 .
Finally, for every x = {ξi } ∈ c0 , we have
x=
∞
X
ξi ei
i=1
and the application of the expressions (5.27) and the linearity of Λ indicate that
∞
∞
∞
X
X
X
ξi ηi
ξi Λ(ei ) =
ξi ei =
Λ(x) = Λ
i=1
i=1
i=1
which is exactly in the form (5.24). Hence the map Φ is surjective.
∗ Step 3: Φ is injective. Let y = {ηi } ∈ ℓ1 , z = {θi } ∈ ℓ1 . Suppose further that
Φ(y) = Φ(z). Since Φ is linear, we have Λy−z = Φ(y − z) = 0. By this, for every
j = 1, 2, . . ., we get
Λy−z (ej ) =
∞
X
i=1
δj,i (ηi − θi ) = ηj − θj = 0
so that ηj = θj . In other words, we conclude that y = z, as desired.
The sequence {ei } is a Schauder basis for c0 (and also for ℓ1 ). In fact, a sequence {xi }∞
i=1 in a Banach
space X is called a Schauder basis if for each x ∈ X, there is a unique sequence {αi } of scalars such that
∞
n
X
X
x=
αi xi = lim
αi xi . See, for instance, [40, Example 4.1.3, p. 351].
d
i=1
n→∞
i=1
155
5.3. The Dual Space of X
∗ Step 4: Φ is an isometry. This follows from the first assertion directly because
kΦ(y)k = kΛy k = kyk1 .
Hence we have (c0 )∗ = ℓ1 .
(b) For each y = {ηi } ∈ ℓ∞ , we define Θ : ℓ1 → C by
Θ(x) =
∞
X
ξi ηi
(5.29)
i=1
for every x = {ξi } ∈ ℓ1 . Similar to the proof of the first assertion in part (a), we can show
that the series (5.29) converges (absolutely) and Θ is linear. It is clear from the definition
that |Θ(x)| ≤ kyk∞ · kxk1 , so we have
kΘk ≤ kyk∞ < ∞.
(5.30)
This means that Θ ∈ (ℓ1 )∗ .
On the other hand, suppose that Θ ∈ (ℓ1 )∗ , i.e., kΘk < ∞. If we consider the Schader
basis {ei } for ℓ1 and the relationships (5.27) with Λ replaced by Θ, then we have
|ηi | = |Θ(ei )| ≤ kΘk · kei k1 = kΘk < ∞
(5.31)
for every i = 1, 2, . . .. Put z = {ηi }. Then the inequality (5.31) immediately implies that
z ∈ ℓ∞ . In fact, it shows further that kzk∞ ≤ kΘk. Now for every x = {ξi } ∈ ℓ1 , we have
x=
∞
X
ξi ei
i=1
and then
Θ(x) =
∞
X
ξi ηi
i=1
which is exactly in the form (5.29), so the inequality (5.30) holds for z. Therefore, we have
kΘk = kzk∞ , i.e., Θ ∈ ℓ∞ .
If we define the mapping Ψ : ℓ∞ → (ℓ1 )∗ by
Ψ(y) = Θy ,
where Θy is given by the form (5.29), then the previous paragraph actually shows that Ψ
is surjective. By using similar argument as in the proofs of Step 1, Step 3 and Step 4
in part (a), it is easy to conclude that Ψ is an isometric vector space isomorphism.
(c) Let y = {ηi } ∈ ℓ1 . Define the functional Λ : ℓ∞ → C by
Λ(x) =
∞
X
ξi ηi ,
i=1
where x = {ξi } ∈ ℓ∞ . Since this series converges (absolutely), it is a linear functional.
Since |Λ(x)| ≤ kxk∞ · kyk1 , we know from [51, Eqn. (2), p. 96] that kΛk ≤ kyk1 , i.e.,
Λ ∈ (ℓ∞ )∗ .
Assume that ℓ1 was isometric isomorphic to (ℓ∞ )∗ . It is clear that x = {1, 1, . . .} ∈
ℓ∞ \ c0 . By [40, Example 1.2.13, p. 14], we know that c0 is a proper closed subspace of
Chapter 5. Examples of Banach Space Techniques
156
ℓ∞ so that ℓ∞ \ c0 6= ∅. Take x0 ∈ ℓ∞ \ c0 . Then Theorem 5.19 implies that there is a
f ∈ (ℓ∞ )∗ such that kf k = 1, f (x) = 0 on c0 and
f (x0 ) 6= 0.
(5.32)
If f ∈ ℓ1 , then we have f ∈ (c0 )∗ by part (a). Since f (x) = 0 on c0 , f is in fact the trivial
bounded functional on c0 . By the assumption and part (a), (c0 )∗ is isometric isomorphic
to (ℓ∞ )∗ so that the trivial bounded functional on c0 corresponds to the trivial bounded
functional on ℓ∞ . In other words, f (x) = 0 on ℓ∞ which contradicts the result (5.32).
Hence we conclude that ℓ1 is not isometric isomorphic to (ℓ∞ )∗ .
(d) We prove the assertions one by one.
– c0 is separable. For each k ∈ N, we define
Sk = {{ξ1 , . . . , ξk , 0, 0, . . .} | ξ1 , . . . , ξk ∈ Q} and S =
∞
[
k=1
Sk .
Since each Sk is countable, S is also countable. We claim that S is dense in c0 . Pick
y = {ηi } ∈ c0 . Given ǫ > 0. Since ηi → 0 as i → ∞, there exists a positive integer N
such that i ≥ N implies
|ηi | < ǫ.
(5.33)
By the density of Q in R, we can take x = {ξ1 , . . . , ξN , 0, 0, . . .} ∈ SN such that
|ξi − ηi | < ǫ
(5.34)
for all i = 1, 2, . . . , N . Thus we obtain from the definition of k · k∞ , the inequalities
(5.33) and (5.34) that
kx − yk∞ = sup |ξi − ηi | ≤ ǫ.
i∈N
Hence this proves the claim and then c0 is separable.
–
ℓ1
is separable. Take y = {ηi } ∈
ℓ1 .
Given that ǫ > 0. Since
∞
X
i=1
|ηi | < ∞, the
Cauchy Criterion (see [49, Theorem 3.22, p. 59]) shows that one can find a positive
integer N ′ such that i ≥ N ′ implies
∞
X
i=N ′
|ηi | <
ǫ
.
2
(5.35)
By the density of Q in R again, we can take x = {ξ1 , . . . , ξN ′ , 0, 0, . . .} ∈ SN ′ such
that
N′
X
ǫ
(5.36)
|ξi − ηi | < .
2
i=1
Combining the inequalities (5.35) and (5.36), we get
kx − yk1 =
∞
X
i=1
′
|ξi − ηi | ≤
N
X
i=1
|ξi − ηi | +
∞
X
i=N ′
Hence S is also dense in ℓ1 and thus ℓ1 is separable.
|0 − ηi | <
ǫ
ǫ
+ = ǫ.
2 2
157
5.4. Applications of Baire’s and other Theorems
– ℓ∞ is not separable. Let A be the set of all sequences x = {ξi } such that ξi ∈ {0, 1}.
Clearly, A is uncountable (see [49, Theorem 2.14]) and A ⊆ ℓ∞ . Let x, y ∈ A be such
that x = {ξi } and y = {ηi }. If x 6= y, then there is at least one i such that ξi 6= ηi
which implies that
kx − yk∞ = 1.
By this and the fact that B(x, 21 ) ∩ B(y, 21 ) = ∅ if x 6= y, we conclude that the
collection of open balls {B(x, 12 ) | x ∈ A} is uncountable. If S was a dense subset of
ℓ∞ , then we have
1
6= ∅
S ∩ B x,
2
for every x ∈ A so that S must be uncountable. Hence ℓ∞ is not separable.
Remark 5.1
We say two normed linear spaces X and Y isometric isomorphism if there is a bijective
linear map T : X → Y such that
kT (x)k = kxk
for every x ∈ X.
5.4
Applications of Baire’s and other Theorems
Problem 5.10
Rudin Chapter 5 Exercise 10.
Proof. For each positive integer n, we define Λn : c0 → C by
Λn (x) =
n
X
αi ξi ,
(5.37)
i=1
where x = {ξi } ∈ c0 . By the hypotheses of Problem 5.9, we know that c0 is a Banach space. On
the one hand, we note from the definition (5.37) that Λn is linear and the analysis in Problem
5.9(a) first assertion shows that
|Λn (x)| ≤ kxk∞ ·
n
X
i=1
|αi |
for every x ∈ c0 ⊆ ℓ∞ and therefore,
kΛn k ≤
n
X
i=1
|αi |.
(5.38)
On the other hand, if we define the sequence xn = {ξn,i } by (5.26) with ηi replaced by αi , then,
using the argument as in the proof of Problem 5.9(a), we conclude that xn ∈ c0 . Thus we obtain
from Definition 5.2 that
kΛn k = sup{|Λn (x)| | x ∈ c0 and kxk∞ ≤ 1} ≥ |Λn (xn )| =
n
X
i=1
|αi |
(5.39)
Chapter 5. Examples of Banach Space Techniques
158
for every n ∈ N. Combining the inequalities (5.38) and (5.39), we conclude that
kΛn k =
n
X
i=1
|αi |
(5.40)
which is bounded for each n ∈ N. By the hypothesis and the definition (5.37), there is no x ∈ c0
such that sup |Λn (x)| = ∞. Thus Theorem 5.8 (The Banach-Steinhaus Theorem) implies that
n∈N
there exists a positive constant M such that
kΛn k ≤ M
(5.41)
for every n ∈ N. Consequently, we deduce from the two results (5.40) and (5.41) that
∞
X
i=1
|αi | < ∞,
as required. We have completed the proof of the problem.
Problem 5.11
Rudin Chapter 5 Exercise 11.
Proof. The proof will be divided into several steps:
• Step 1: Lip α is a vector space. For every f, g ∈ Lip α and µ, ν ∈ C, we have
|µf (s) + νg(s) − µf (t) − νg(t)|
|s − t|α
|g(s) − g(t)|
|f (s) − f (t)|
+ |ν| · sup
≤ |µ| · sup
α
|s − t|
|s − t|α
= |µ| · Mf + |ν| · Mg
Mµf +νg = sup
(5.42)
<∞
which implies that µf + νg ∈ Lip α, i.e., Lip α is a vector space.
• Step 2: Lip α is a normed linear space with the mentioned norms. If we define
kf k1 = |f (a)| + Mf , then for f, g ∈ Lip α, it is easy to see from the inequality (5.42) that
kf + gk1 = |f (a) + g(a)| + Mf +g ≤ |f (a)| + |g(a)| + Mf + Mg = kf k1 + kgk1 .
Next, if µ ∈ C, then we have
kµf k1 = |µf (a)| + Mµf = |µ| · |f (a)| + |µ| · Mf = |µ| · kf k1 .
Finally, if kf k1 = 0, then |f (a)| = Mf = 0 so that |f (s) − f (a)| = 0 for all s ∈ (a, b]. Thus
we establish that f (x) = 0 on [a, b] and we can say that Lip α is a normed linear space
with respect to the first norm.
For the second norm, if f, g ∈ Lip α, then we see that
kf +gk2 = Mf +g + sup |f (x)+g(x)| ≤ Mf +Mg + sup |f (x)|+ sup |g(x)| = kf k2 +kgk2 .
x∈[a,b]
x∈[a,b]
x∈[a,b]
159
5.4. Applications of Baire’s and other Theorems
Next, if µ ∈ C, then we have
kµf k2 = Mµf + sup |µf (x)| = |µ| · Mf + |µ| · sup |f (x)| = |µ| · kf k2 .
x∈[a,b]
x∈[a,b]
Finally, if kf k2 = 0, then we have Mf = sup |f (x)| = 0 implying that f (x) = 0 on [a, b].
x∈[a,b]
Hence Lip α is also a normed linear space with respect to the second norm.
• Step 3: Lip α is complete. Suppose that {fn } ⊆ Lip α is Cauchy with respect to the
norm k · k2 . Given ǫ > 0. There exists a positive integer N such that n, m ≥ N imply that
ǫ
sup |fn (x) − fm (x)| + Mfn −fm = kfn − fm k2 < .
2
x∈[a,b]
(5.43)
Particularly, we have
ǫ
(5.44)
2
for all x ∈ [a, b]. Thus {fn (x)} is a Cacuhy sequence in C. Since C is complete, there
exists a function f : [a, b] → C such that
|fn (x) − fm (x)| <
f (x) = lim fn (x)
(5.45)
n→∞
for each x ∈ [a, b].
We claim that kfn − f k2 → 0 as n → ∞ and f ∈ Lip α. On the one hand, we take
m → ∞ in the inequality (5.44) to get
|fn (x) − f (x)| ≤
ǫ
2
(5.46)
for all x ∈ [a, b]. Thus we have
sup |fn (x) − f (x)| ≤
x∈[a,b]
ǫ
.
2
(5.47)
On the other hand, we know from the inequality (5.43) that Mfn −fm < 2ǫ , or explicitly
ǫ
|fn (s) − fm (s) − fn (t) + fm (t)|
<
|s − t|α
2
(5.48)
for all s, t ∈ [a, b] with s 6= t. Taking m → ∞ in the inequality (5.48), we observe that
|fn (s) − f (s) − fn (t) + f (t)|
ǫ
≤
α
|s − t|
2
for all s, t ∈ [a, b] with s 6= t. In other words, we have
ǫ
Mfn −f ≤ .
2
(5.49)
Hence we combine the inequalities (5.47) and (5.49) to establish that
kfn − f k2 = sup |fn (x) − f (x)| + Mfn −f ≤ ǫ,
x∈[a,b]
i.e., kfn − f k2 → 0 as n → ∞. Furthermore, for s, t ∈ [a, b] with s 6= t, since we have
|f (s) − fN (s) + fN (t) − f (t) + fN (s) − fN (t)|
|f (s) − f (t)|
=
|s − t|α
|s − t|α
Chapter 5. Examples of Banach Space Techniques
160
|f (s) − fN (s) + fN (t) − f (t)| |fN (s) − fN (t)|
+
|s − t|α
|s − t|α
≤ MfN −f + MfN ,
≤
where N is the positive integer which makes the inequality (5.43) holds. Therefore, we
deduce from the definition and the inequality (5.49) that
Mf ≤ MfN −f + MfN ≤
ǫ
+ MfN < ∞,
2
i.e., f ∈ Lip α. In other words, Lip α is complete with respect to the norm k · k2 .
For the first norm k · k1 , we note that
kf k1 = |f (a)| + Mf ≤ sup |f (x)| + Mf = kf k2 .
(5.50)
x∈[a,b]
If {fn } ⊆ Lip α is Cauchy with respect to the norm k · k1 , then we deduce from the
inequality (5.50) that
kfn − f k1 ≤ kfn − f k2 < ǫ
for large enough n, where the f is defined by the limit (5.45). Since f ∈ Lip α, we have
shown that Lip α is complete with respect to the norm k · k1 .
Hence Lip α is also a Banach space with respect to the two norms and this completes the proof
of the problem.
Problem 5.12
Rudin Chapter 5 Exercise 12.
Proof. We have A = {f : K → R | f (x, y) = αx + βy + γ, where α, β, γ ∈ R}.
• Case (i): K is a triangle. Suppose that P = (x1 , y1 ), Q = (x2 , y2 ), R = (x3 , y3 ) and
H = {P, Q, R}. Then it is well-known that
K = Conv (P, Q, R)
= {λP + τ Q + νR | λ, τ, ν ≥ 0 and λ + τ + ν = 1}
= {(λx1 + τ x2 + νx3 , λy1 + τ y2 + νy3 ) | λ, τ, ν ≥ 0 and λ + τ + ν = 1}.
Therefore, for each (x0 , y0 ) ∈ K, we have the representation
(x0 , y0 ) = λP + τ Q + νR = (λx1 + τ x2 + νx3 , λy1 + τ y2 + νy3 )
(5.51)
and thus
f (x0 , y0 ) = f (λx1 + τ x2 + νx3 , λy1 + τ y2 + νy3 )
= α(λx1 + τ x2 + νx3 ) + β(λy1 + τ y2 + νy3 ) + (λ + τ + ν)γ
= λ(αx1 + βy1 + γ) + τ (αx2 + βy2 + γ) + ν(αx3 + βy3 + γ)
= λf (x1 , y1 ) + τ f (x2 , y2 ) + νf (x3 , y3 ).
(5.52)
If we define µ(P ) = µ(x1 , y1 ) = λ, µ(Q) = µ(x2 , y2 ) = τ and µ(R) = µ(x3 , y3 ) = ν, then
the expression (5.52) becomes
Z
f dµ.
(5.53)
f (x0 , y0 ) =
H
161
5.4. Applications of Baire’s and other Theorems
To prove the uniqueness of µ, we suppose that the integral representation (5.53) holds
for the measure µ′ . Then the integral (5.53) implies that
f (x0 , y0 ) = µ′ (P )f (x1 , y1 ) + µ′ (Q)f (x2 , y2 ) + µ′ (R)f (x3 , y3 )
= f µ′ (P )x1 + µ′ (Q)x2 + µ′ (R)x3 , µ′ (P )y1 + µ′ (Q)y2 + µ′ (R)y3 .
(5.54)
By the definition of f , the expression (5.54) means that
(x0 , y0 ) = µ′ (P ) · P + µ′ (Q) · Q + µ′ (R) · R.
(5.55)
Comparing the expressions (5.51) and (5.55), we derive that
[µ′ (P ) − λ]P + [µ′ (Q) − τ ]Q + [µ′ (R) − ν]R = 0.
(5.56)
Since the set {P, Q, R} is linearly independent, it follows from the equation (5.56) that
µ′ (P ) = λ,
µ′ (Q) = τ
and
µ′ (R) = ν.
In other words, we establish the fact that µ′ = µ.
• Case (ii): K is a square. Without loss of generality, we may assume that the vertices
of K are P (0, 0), Q(1, 0), R(1, 1) and S(0, 1). Let A(a, b) be a point of K, see Figure 5.2
below.
Figure 5.2: The square K.
Suppose that (a, b) = λP + τ Q + νR + θS, where λ, τ, ν, θ ∈ R. Then we have
(a, b) = τ (1, 0) + ν(1, 1) + θ(0, 1) = (τ + ν, ν + θ).
Thus we have ν + θ = b and τ + ν = a. Solving the equations, we have
ν =b−θ
Now we have five subcases below:
and
τ = a − b + θ.
(5.57)
Chapter 5. Examples of Banach Space Techniques
162
– Subcase (i): A belongs to Region I. Here we have 0 < b ≤ a < 1 and a+b−1 ≤ 0.
If θ = 0, then we obtain from the equations (5.57) that ν = b and τ = a − b ≥ 0. If
λ = 1 − a ≥ 0, then it is easy to see that
(a, b) = (1 − a)P + (a − b)Q + bR + 0 · S.
Similarly, θ = b, ν = 0, τ = a and λ = 1 − a − b ≥ 0 imply that
(a, b) = (1 − a − b)P + aQ + 0 · R + bS.
Hence, every point in Region I has two different convex combinations of P, Q, R and
S.
– Subcase (ii): A belongs to Region II. In this case, we have 0 < b ≤ a < 1 and
a + b − 1 ≥ 0. Since we have
(a, b) = (1 − a)P + (a − b)Q + bR + 0 · S
= 0 · P + (1 − b)Q + (a + b − 1)R + (1 − a)S,
every point in Region II has two distinct convex combinations of P, Q, R and S.
– Subcase (iii): A belongs to Region III. In this case, we have 0 < a ≤ b < 1 and
a + b − 1 ≥ 0. It is easy to see that
(a, b) = 0 · P + (1 − b)Q + (a + b − 1)R + (1 − a)S
= (1 − b)P + 0 · Q + aR + (b − a)S,
so every point in Region III also has two different convex combinations of P, Q, R and
S.
– Subcase (iv): A belongs to Region IV. Now we have 0 < a ≤ b < 1 and
a + b − 1 ≤ 0. Then we certainly have
(a, b) = (1 − b)P + 0 · Q + aR + (b − a)S
= (1 − a − b)P + aQ + 0 · R + bS
which means that it has two different convex combinations of P, Q, R and S in Region
IV.
– Subcase (v): A belongs to the boundary of K. Suppose that A = (a, b). We
note that
(a, 0) = (1 − a) · P + aQ + 0 · R + 0 · S = 1 · P + aQ + 0 · R + 0 · S,
thus every point on P Q can be expressed as two distinct linear (but not convex)
combinations of P, Q, R and S. Similarly, all points lie on QR, RS and SP also have
two different linear combinations of P, Q, R and S.
In conclusion, every point (x0 , y0 ) in the square K has at least two distinct linear
combinations of P, Q, R and S. Now, instead of the expression (5.52), we have
f (x0 , y0 ) = λf (P ) + τ f (Q) + νf (R) + θf (S) = λ′ f (P ) + τ ′ f (Q) + ν ′ f (R) + θ ′ f (S),
where {λ, τ, ν, θ} and {λ′ , τ ′ , ν ′ , θ ′ } are two sets of distinct numbers corresponding to two
distinct linear combinations of P, Q, R and S. Thus {λ, τ, ν, θ} and {λ′ , τ ′ , ν ′ , θ ′ } give
distinct measures µ and µ′ such that the representation (5.53) holds.
163
5.4. Applications of Baire’s and other Theorems
• Case (iii): The general situation. Suppose that H = {v0 , v1 , . . . , vn } and
K = {λ0 v0 + λ1 v1 + · · · + λn vn ⊆ Rn+1 | λ0 , λ1 , . . . , λn ≥ 0 and λ0 + · · · + λn = 1}.
Let f : K → R be defined by
f (v0 , v1 , . . . , vn ) = α0 v0 + α1 v1 + · · · + αn vn + γ,
where α0 , α1 , . . . , αn , γ are real. For each (v0 , v1 , . . . , vn ) ∈ K, there corresponds a measure
µ on H such that
Z
f dµ.
f (v0 , v1 , . . . , vn ) =
H
Furthermore, the measure µ is unique if K is a n-simplex, i.e., {v1 − v0 , v2 − v0 , . . . , vn − v0 }
is linearly independent, see [32, pp. 102, 103].
We have completed the proof of the problem.
Problem 5.13
Rudin Chapter 5 Exercise 13.
Proof.
(a) For each N = 1, 2, . . ., we define
EN = {x ∈ X | |fn (x)| ≤ N for all n ∈ N}.
Since each fn is continuous on X, each EN is closed in X. On the one hand, we have
EN ⊆ X.
(5.58)
On the other hand, for each fixed x ∈ X, since f (x) is well-defined as a complex number,
it must be true that
|f (x)| ≤ m
for some m > 0. By the convergence of {fn (x)}, there is a positive integer N0 such that
n ≥ N0 implies that
|f (x) − fn (x)| < 1
Thus, for n ≥ N0 , the triangle inequality shows that
|fn (x)| ≤ |f (x)| + |f (x) − fn (x)| < m + 1.
(5.59)
Take m′ = max f1 (x), f2 (x), . . . , fN0 −1 (x) . Combining this and the inequality (5.59), we
obtain that
|fn (x)| < max(m′ , m + 1)
for all n ∈ N. Hence we have x ∈ EN1 for some N1 ≥ max(m′ , m + 1) and then we follow
from the set relation (5.58) that
∞
[
EN .
X=
N =1
By Theorem 5.6 (Baire’s Theorem), X is not of the first category, i.e., X is not a countable
union of nowhere dense sets. Therefore, some EN ′ contains a nonempty open subset V of
X, i.e., for all x ∈ V and n = 1, 2, . . ., we have
|fn (x)| ≤ N ′ ,
where N ′ < ∞.
Chapter 5. Examples of Banach Space Techniques
164
(b) We follow the hint. For each N = 1, 2, 3, . . ., we put
AN = {x ∈ X | |fm (x) − fn (x)| ≤ ǫ for all n, m ≥ N }.
Since fm and fn are continuous on X, fm − fn is also continuous on X. Thus for each pair
n, m ∈ N, the set
S(m, n) = (fm − fn )−1 [−ǫ, ǫ] = {x ∈ X | |fm (x) − fn (x)| ≤ ǫ}
is closed in X. Since
AN =
\
S(m, n),
n,m≥N
AN is definitely closed in X.
It is clear that
AN ⊆ X
(5.60)
for each N = 1, 2, . . .. Now for every x ∈ X, by the convergence of {fn (x)}, it is Cauchy
in C, i.e., there exists a N0 ∈ N such that n, m ≥ N0 imply that
|fm (x) − fn (x)| ≤ ǫ
or equivalently, x ∈ AN0 . In other words, we have
X⊆
∞
[
AN .
(5.61)
N =1
Combining the set relations (5.60) and (5.61), we conclude that
X=
∞
[
AN .
N =1
By Theorem 5.6 (Baire’s Theorem), X is not of the first category, i.e., X is not a countable
union of nowhere dense sets. Therefore, some AN ′ contains a nonempty open subset V of
X. Particularly, if x ∈ V and n, m ≥ N ′ , then we have
|fm (x) − fn (x)| ≤ ǫ.
(5.62)
Taking m → ∞ in the inequality (5.62), we have established that
|f (x) − fn (x)| ≤ ǫ
for all x ∈ V and n ≥ N ′ .
This has completed the proof of the problem.
Problem 5.14
Rudin Chapter 5 Exercise 14.
Proof. Let C = {f : [0, 1] → R | f is continuous on [0, 1]} and kf k = sup |f (x)|. Furthermore,
x∈[0,1]
for a positive integer n, let
Xn = {f ∈ C | there exists a t ∈ I such that |f (s) − f (t)| ≤ n|s − t| for all s ∈ I}.
165
5.4. Applications of Baire’s and other Theorems
To prove the first assertion, we fix n ∈ N. We follow the hint given by Rudin and the construction
of g will be shown in several steps below.
Let V be an open set in C and f ∈ V . Then there exists a ǫ > 0 such that B(f, 2ǫ) ⊆ V .
Since f is continuous on the compact [0, 1], it is uniformly continuous on [0, 1], i.e., there exists
a k ∈ N such that
|f (x) − f (y)| < ǫ
(5.63)
for all x, y ∈ [0, 1] with |x − y| ≤ k1 . In particular, we consider the partition P = {0 =
x0 , x1 , . . . , xk = 1} of [0, 1], where xi = ki for i = 0, 1, . . . , k. Then we know from the inequality
(5.63) that
|f (x) − f (xi )| < ǫ
(5.64)
for all x ∈ [xi−1 , xi+1 ]. To prove the first assertion, we are going to divide the proof into several
steps:
• Step 1: Construction of a g ∈ C with kg − f k < ǫ. Suppose that
gi (x) = gi λx xi + (1 − λx )xi+1 = λx f (xi ) + (1 − λx )f (xi+1 )
(5.65)
on [xi , xi+1 ], where λx ∈ [0, 1]. Since x = λx xi + (1 − λx )xi+1 , we have
λx =
x − xi+1
= −k(x − xi+1 ) = −kx + (i + 1).
xi − xi+1
(5.66)
Thus λx is a continuous function of x. Therefore, each zigzag function gi is continuous on
[xi , xi+1 ] and furthermore, we have
gi (xi+1 ) = f (xi+1 ) = gi+1 (xi+1 ).
Thus if we “glue” all the zigzag functions g0 , g1 , . . . , gk together and let it be g, then we
have g ∈ C. In addition, if x ∈ [xi , xi+1 ], then we see from the definition (5.65) and the
inequality (5.64) that
|g(x) − f (x)| ≤ λx · |f (xi ) − f (x)| + (1 − λx ) · |f (x) − f (xi+1 )| < ǫ.
In other words, kg − f k < ǫ as required.
• Step 2: Construction of a continuous function with large one-sided derivatives.
To complete the construction, we have to consider the function ϕ : R → [0, 1] defined by
ϕ(x + 1) = ϕ(x), where

if x ∈ [0, 12 ];
 2x,
ϕ(x) =

2 − 2x, if x ∈ [ 21 , 1].
Clearly, ϕ is a zigzag continuous function on R.e Now our desired function is constructed
in the following lemma:
Lemma 5.1
For each n ∈ N, if we define
ϕn (x) = 2−n ϕ(4n x),
then ϕn is a zigzag continuous function on R such that |ϕ′n (x+)| ≥ 2n and
|ϕ′n (x−)| ≥ 2n for all x ∈ R.
e
Such zigzag function ϕ is something looks like the graph in Figure 2.1.
Chapter 5. Examples of Banach Space Techniques
166
Proof of Lemma 5.1. Obviously, ϕn is a zigzag continuous function on R. Suppose
that x ∈ [N, N + 1] for some N ∈ N. Then we have 4n x ∈ [4n N, 4n (N + 1)]. Since ϕ
is a function of period 1, we have
ϕ(4n x) = ϕ(4n x − 4n N ),
(5.67)
where 4n x − 4n N ∈ [0, 4n ]. It is clear that there exists N ′ ∈ N ∪ {0} such that
4n x − 4n N − N ′ ∈ [0, 1], then the expression (5.67) and the periodicity of ϕ again
shows that
ϕ(4n x) = ϕ(4n x − 4n N ) = ϕ(4n x − 4n N − N ′ ).
In other words, we may assume that 4n x ∈ [0, 1). Let h > 0 be so small such that
4n (x + h) ∈ [0, 1]. Then we obtain
ϕn (x + h) − ϕn (x)
h
1 ϕ 4n (x + h) − ϕ(4n x)
= n·
2
h

n (x + h) − 2 · 4n x
2
·
4
1

 n·
,
if 4n x, 4n (x + h) ∈ [0, 12 ];


2
h





1 1 − 2 · 4n h − 1
=
·
,
if 4n x = 12 and 4n (x + h) =
n

2
h





n
n


 1 · −2 · 4 (x + h) + 2 · 4 x , 4n x, 4n (x + h) ∈ [ 1 , 1],
2
n
h
 2 n
1
n
n
2·2 ,
if 4 x, 4 (x + h) ∈ [0, 2 ];





−2 · 2n , if 4n x = 12 and 4n (x + h) = 21 + 4n h;
=





−2 · 2n , 4n x, 4n (x + h) ∈ [ 12 , 1]
1
2
+ 4n h;
which implies that |ϕ′n (x+)| ≥ 2n . Similarly, |ϕ′n (x−)| ≥ 2n , completing the proof of
Lemma 5.1.
• Step 3: Construction of a e
g ∈ C with large slopes and ke
g − f k < 2ǫ. We define
ge = g + ϕk ,
where the k will be determined later. We claim that gb satisfies the required properties.
To this end, let
Mk = max k|f (xi+1 ) − f (xi )|.
0≤i≤k−1
Now for every x ∈ [0, 1], if we take h > 0 small enough such that x + h ∈ [xi , xi+1 ], then
we see from the definition (5.65) and the expression (5.66) that
g(x + h) − g(x)
|λx+h − λx | · |f (xi+1 ) − f (xi )|
= k|f (xi+1 ) − f (xi )| ≤ Mk
=
h
h
(5.68)
which implies that |g′ (x+)| ≤ Mk on [0, 1]. Similarly, we have |g′ (x−)| ≤ Mk for every
x ∈ [0, 1].f
f
If x = 1, then g ′ (1+) does not exist. Similarly, if x = 0, then g ′ (0−) does not exist.
167
5.4. Applications of Baire’s and other Theorems
Suppose that N ′ ∈ N. Since f is bounded on [0, 1], we must have Mk ≤ kM for some
constant M > 0 and then k can be chosen so large that
2k ≥ kM ≥ Mk + N ′
and
ǫ
1
< .
k
2
2
(5.69)
Since g and ϕk are continuous on [0, 1], ge is also continuous on [0, 1]. Next, for x ∈ [0, 1)
and h > 0 so small such that x + h ∈ [0, 1], we derive from Lemma 5.1 and the inequality
(5.68) that
g(x + h) − g(x) ϕ(x + h) − ϕ(x)
ge(x + h) − ge(x)
+
=
h
h
h
ϕ(x + h) − ϕ(x)
g(x + h) − g(x)
≥
−
h
h
k
≥ 2 − Mk
> N′
(5.70)
which yields |e
g ′ (x+)| > N ′ for all x ∈ [0, 1). Similarly, we have |e
g′ (x−)| > N ′ for all
x(0, 1]. Since N ′ can be chosen arbitrary large, our e
g is continuous function on [0, 1] with
large slopes.
Finally, it remains to show that ke
g − f k < ǫ. On [0, 1], we know from Steps 1 and 2
that
|e
g(x) − f (x)| ≤ |g(x) − f (x)| + |ϕk (x)| < ǫ + 2−k |ϕ(4k x)| < ǫ +
ǫ
< 2ǫ.
2
Thus we conclude that ke
g − f k < 2ǫ. By the definition of Xn and the lower bound (5.70),
we note that ge ∈
/ Xn if we pick N ′ ≥ n.
• Step 4: Xn is closed in C. Let {fk } ⊆ Xn and fk → f in C. Then one can find a
sequence {tk } ⊆ [0, 1] such that
|fk (s) − fk (tk )| ≤ n|s − tk |
(5.71)
for every s ∈ [0, 1] and k ∈ N. Since [0, 1] is compact, we may assume that tk → t ∈ [0, 1].
By Theorem 3.17, C is a metric space and then we recall from the rephrased Theorem 7.9
[49, p. 151] that fk → f uniformly on [0, 1]. Hence it follows from [49, Exercise 9, p. 166]
that
lim fk (tk ) = f (t)
k→∞
and we deduce from the inequality (5.71) that
|f (s) − f (t)| ≤ n|s − t|,
for every s ∈ [0, 1]. In other words, we conclude that f ∈ Xn which means Xn is closed in
C.
• Step 5: Construction of B(e
g , ǫ′ ) ⊆ C such that B(e
g, ǫ′ ) ∩ Xn = ∅. By Step 3 and
our choice of ǫ > 0, we know that e
g ∈ B(f, 2ǫ) ⊆ V and ge ∈
/ Xn . Since ge ∈ Xnc and Xnc is
open in C by Step 4. There exists a δ > 0 such that B(e
g, δ) ⊆ Xnc , i.e., B(e
g, δ) ∩ Xn = ∅.
1
′
g − f k, 2ǫ − ke
g − f k) and h ∈ B(e
g , ǫ′ ), then we have
If we take ǫ = 2 min(ke
kh − f k ≤ kh − gek + ke
g − f k < ǫ′ + ke
g − fk ≤
which means B(e
g, ǫ′ ) ⊆ B(f, 2ǫ) ⊆ V .
1
(2ǫ − ke
g − f k) + ke
g − f k < 2ǫ
2
Chapter 5. Examples of Banach Space Techniques
168
Hence we have shown the first assertion.
To prove the second assertion, we notice that Xn◦ = ∅; otherwise, since Xn◦ is open in C, the
first assertion shows the existence of a non-empty open set G ⊆ Xn◦ ⊆ Xn such that G ∩ Xn = ∅,
a contradiction. Hence each Xnc is open and dense in C. Since C is a complete metric space, we
establish from Theorem 5.6 (Baire’s Theorem) that the set
X=
∞
\
Xnc
n=1
is a dense Gδ in C. Suppose that f ∈ C and f is differentiable at p ∈ [0, 1). Then there exists
a δ > 0 such that for all 0 < |h| < δ, we haveg
f (p + h) − f (p)
− f ′ (p) ≤ 1
h
which implies that
f (p + h) − f (p)
≤ 1 + |f ′ (p)|.
h
(5.72)
f (p + h) − f (p)
2kf k
|f (p + h)| + |f (p)
≤
.
≤
h
h
δ
(5.73)
Obviously, if |h| ≥ δ, then
Combining the inequalities (5.72) and (5.73), we obtain that
|f (x) − f (p)| ≤ N |x − p|
if N = max(1 + |f ′ (p)|, 2δ−1 kf k). In the case that p = 1, we will replace f (p + h) by f (p − h)
in the inequalities (5.72) and (5.73).
c
Consequently, f ∈ Xn for every n ≥ N . By this, we must have f ∈
/ X. Otherwise, f ∈ XN
or equivalently, f ∈
/ XN which is a contradiction. Hence we have completed the proof of the
problem.
Problem 5.15
Rudin Chapter 5 Exercise 15.
Proof. Recall from Problem 5.9 that c0 is the subspace of ℓ∞ consisting of all x = {ξi } ∈ ℓ∞ for
which ξi → 0 as i → ∞. By the definition, we have
{σi } = A(x)
whenever the series converges (but ξi not necessarily converging to 0). We are going to prove
the results one by one:
• Necessity part: Suppose that A transforms every convergent sequence {sj } to a sequence
{σi } which converges to the same limit.
To prove Condition (a), it suffices to prove the case when both {sj } and {σi } converge
to 0 because we may replace sj and σi by sj − L and σi − L respectively, where L is the
g
Of course, only one-sided derivative f ′ (0+) will be considered if p = 0.
169
5.4. Applications of Baire’s and other Theorems
common limit of {sj } and {σi }. In other words, we work on the space c0 . In fact, for each
fixed i ∈ N, we define Λi : c0 → C by
Λi (x) = σi =
∞
X
aij sj
j=0
for all x ∈ c0 so that A(x) = {Λi (x)}. Since {σi } is convergent, it is bounded so that
|Λi (x)| ≤ M
(5.74)
for all x ∈ c0 and i = 1, 2, . . . for some positive constant M . For every k = 0, 1, 2, . . ., we
take sj = δjk (the Kronecker delta function). Then we have sj → 0 as j → ∞ and we
obtain
∞
X
0 = lim σi = lim
aij sj = lim aik
i→∞
i→∞
i→∞
j=0
which is Condition (a).
If sj = 1 for all j ∈ N, then sj → 1 as j → ∞ and thus
1 = lim σi = lim
i→∞
i→∞
∞
X
aij
j=0
which is Condition (c).
To prove Condition (b), we first show that
∞
X
j=0
|aij | < ∞ for every i = 1, 2, . . .. Assume
that it was not the case. Then one is able to find a strictly increasing sequence of integers
{nk } such that
nk+1
X
|aij | > k.
j=nk +1
Let k ∈ N and define
sj =

0,


if j = 1, 2, . . . , nk ;

 sgn (aij ) , if j = n + 1, n + 2, . . . , n .
k
k
k+1
k
(5.75)
Obviously, j → ∞ if and only if k → ∞. Then we have sj → 0 as j → ∞. Thus
x = {sj } ∈ c0 and so lim σi = 0. However, we notice from the construction (5.75) that
i→∞
|Λi (x)| = |σi | =
∞
X
aij sj
j=0
∞ nX
k+1
X
|aij |
=
=∞
k
k=1 j=nk +1
which contradicts the inequality (5.74). Thus we must have
∞
X
j=0
|aij | < ∞.
By the inequality (5.74) again, we have Λi ∈ c∗0 and since c0 is Banach, we may apply
Theorem 5.8 (The Banach-Steinhaus Theorem) to conclude that
kΛi k = sup{|Λi (x)| | x ∈ c0 ⊆ ℓ∞ and kxk∞ = 1} ≤ M
(5.76)
Chapter 5. Examples of Banach Space Techniques
for all i ∈ N. Recall that
∞
X
j=0
170
|aij | < ∞ for every i = 1, 2, . . ., so yi = {aij } ∈ ℓ1 for every
i = 1, 2, . . . and Problem 5.9(a) implies that
kΛi k = kyi k1 =
∞
X
j=0
|aij |
(5.77)
for every i = 1, 2, . . .. Hence Condition (b) follows from the inequality (5.76) and the
equality (5.77).
• Sufficiency part: Let x = {sj } and sj → L as j → ∞. By Condition (b), the series
∞
X
aij sj converges absolutely for every i ∈ N. We write
σi =
j=0
σi =
∞
X
j=0
aij (sj − L) + L ·
By Condition (c), we have
lim L ·
i→∞
∞
X
∞
X
aij .
(5.78)
j=0
aij = L.
(5.79)
j=0
Let the supremum in Condition (b) be M . Given ǫ > 0, we choose a N ∈ N such that
|sj − L| < Mǫ for all n ≥ N . Thus we see that
∞
X
j=0
aij (sj − L) ≤
N
X
j=0
|aij | · |sj − L| +
∞
X
n=N +1
|aij | · |sj − L| ≤
N
X
j=0
|aij | · |sj − L| + ǫ.
By Condition (a), we know that
lim
i→∞
so that
lim
i→∞
N
X
j=0
|aij | · |sj − L| = 0
∞
X
j=0
aij (sj − L) ≤ ǫ.
Since ǫ is arbitrary, we conclude from the representation (5.78) and the limit (5.79) that
lim σi = L.
i→∞
• Two examples. It is easy to check that the examples satisfy the conditions of the
problem.
To show the last assertion, we consider the sequence {sj } defined by
 √
if j = 2k;
 k,
sj =
 √
− k, if j = 2k − 1.
Then it is clear that {sj } is unbounded and direct computation shows that, for each
i = 1, 2, . . .,
σi =
∞
X
j=0
aij sj
171
5.5. Miscellaneous Problems
=
i
X
j=0
=
=
1
sj
i+1

2k

1 X



sj , if i = 2k;



 2k + 1 j=0


2k−1

1 X



sj ,
if i = 2k − 1;

 2k
j=0


0,
if i = 2k;


1


 − √ , if i = 2k − 1.
2 k
Since i → ∞ if and only if k → ∞, we have σi → 0 as i → ∞, i.e., {σi } is convergent.
For the other example, pick δ to be a number such that ri < δ < 1 for every i ∈ N. If
sj = (−δ)−j , then {sj } is divergent and we have
σi =
∞
X
j=0
aij sj = (1 − ri )
∞ X
j=0
ri j
1 − ri
−
=
→0
δ
1 + δ−1 ri
as i → ∞.
We complete the proof of the problem.
Remark 5.2
Classically, Problem 5.15 is called the Silverman-Toepliotz Theorem. For more information
or other proofs about this theorem, please refer to [30, §3.2], [44, Chap. 4] and [68, §1.2].
5.5
Miscellaneous Problems
Problem 5.16
Rudin Chapter 5 Exercise 16.
Proof. We follow Rudin’s hint. Let X ⊕Y = {(x, y) | x ∈ X and y ∈ Y } with addition and scalar
multiplication defined componentwise. We define the norm k · k on X ⊕ Y byh
k(x, y)k = kxkX + kykY .
We check Definition 5.2. For (x1 , y1 ), (x2 , y2 ) ∈ X ⊕ Y , we have
k(x1 , y1 ) + (x2 , y2 )k = k(x1 + x2 , y1 + y2 )k
= kx1 + x2 kX + ky1 + y2 kY
≤ kx1 kX + kx2 kX + ky1 kY + ky2 kY
= k(x1 , y1 )k + k(x2 , y2 )k.
h
Of course, k · kX and k · kY denote the norms in X and Y respectively.
(5.80)
Chapter 5. Examples of Banach Space Techniques
172
Next, if (x, y) ∈ X ⊕ Y and α is a scalar, then we have
kα(x, y)k = k(αx, αy)k = kαxkX + kαykY = |α| · kxkX + |α| · kykY = |α| · k(x, y)k.
Finally, if k(x, y)k = 0, then kxkX + kykY = 0. Since kxkX and kykY are nonnegative for every
x ∈ X and y ∈ Y , kxkX + kykY = 0 implies that kxkX = kykY = 0 and thus (x, y) = (0, 0).
Hence X ⊕ Y is a normed linear space.
Suppose that {(xn , yn )} ⊆ X ⊕ Y is Cauchy. Given ǫ > 0, Then there exists a positive integer
N such that n, m ≥ N imply that
k(xn , yn ) − (xm , ym )k < ǫ.
By the definition (5.80), we must have kxn − xm kX < ǫ and kyn − ym kY < ǫ for all n, m ≥ N .
In other words, {xn } ⊆ X and {yn } ⊆ Y are also Cauchy. Since X and Y are Banach, there
exist x ∈ X and y ∈ Y such that
kxn − xkX → 0
and kyn − ykY → 0
as n → ∞. Now these limits and the definition (5.80) show that
k(xn , yn ) − (x, y)k = k(xn − x, yn − y)k = kxn − xkX + kyn − ykY → 0
as n → ∞. By Definition 5.2, X ⊕ Y ia Banach.
Suppose
that
G
=
{
x,
Λ(x)
| x ∈ X} ⊆ X ⊕ Y . Furthermore, we let α, β ∈ C and
x, Λ(x) , y, Λ(y) ∈ G. Then the linearity of Λ says that
α x, Λ(x) + β y, Λ(y) = αx + βy, αΛ(x) + βΛ(y) = αx + βy, Λ αx + βy ∈ G.
Thus G is a linear subspace of X ⊕ Y and it is also a metric space. We claim that G is closed
in X ⊕ Y . To see this, let { xn , Λ(xn ) } ⊆ G and k xn , Λ(xn ) − (x, y)k → 0 as n → ∞. By the
definition (5.80), we have kxn − xkX → 0 and kΛ(xn ) − ykY → 0 as n → ∞, i.e.,
x = lim xn
n→∞
and y = lim Λ(xn ).
n→∞
By the hypothesis, we gain that y = Λ(x) and so (x, y) = x, Λ(x) ∈ G by the definition.
Therefore, we have shown the claim that G is closed in X ⊕ Y . Since X ⊕ Y is complete, we
deduce from [49, Theorem 3.11, p. 53] that G is also complete and thus Banach.
Consider the mapping Φ : G → X defined by
Φ x, Λ(x) = x.
Then it is clear that Φ is linear and bijective. Furthermore, for every x ∈ X, we see that
kΦ x, Λ(x) kX = kxkX ≤ kxkX + kΛ(x)kY = k x, Λ(x) k
which implies that
kΦk = sup{kΦ x, Λ(x) kX | x, Λ(x) ∈ G and k x, Λ(x) k = 1} ≤ 1,
i.e., Φ is bounded. By Theorem 5.10, Φ−1 is also a bounded linear transformation of X onto G
and it follows from Theorem 5.10’s proof that there exists a positive constant δ such that
1
kΦ−1 k ≤ .
δ
173
5.5. Miscellaneous Problems
Besides, Theorem 5.4 shows that Φ−1 is continuous. Combining these two facts and [51, Eqn.
(3), p. 96], we obtain
1
kxkX + kΛ(x)kY = k x, Λ(x) k = kΦ−1 (x)k ≤ kΦ−1 k · kxkX ≤ kxkX < ∞
(5.81)
δ
for every x ∈ X. Since kxkX and kΛ(x)kY are nonnegative, we get from the inequality (5.81)
that δ < 1 and
1
0 ≤ kΛ(x)kY ≤
− 1 · kxkX
(5.82)
δ
on X. Hence we conclude from the inequality (5.82) that
1
< ∞ − 1.
δ
Consequently, Λ is bounded and it is continuous on X by Theorem 5.4, completing the proof of
the problem.
kΛk = sup{kΛ(x)kY | x ∈ X and kxkX = 1} ≤
Problem 5.17
Rudin Chapter 5 Exercise 17.
Proof. We prove the assertions one by one.
• kMf k ≤ kf k∞ . Given f ∈ L∞ (µ). Define Mf : L2 (µ) → L2 (µ) by
Mf (g) = f g.
By Definition 3.7, we have |f g| ≤ kf k∞ · |g| for all f ∈ L2 (µ) and then Remark 3.10 shows
that
o1
nZ
o1 n Z
2
2
2
≤
kf k2∞ · |g|2 dµ
= kf k∞ · kgk2 . (5.83)
kMf (g)k2 = kf gk2 =
|f g| dµ
X
X
By the inequality (5.83) and Definition 5.3, we obtain immediately that
kMf k = sup{kMf (g)k2 | g ∈ L2 (µ) and kgk2 = 1} ≤ kf k∞ .
(5.84)
• Measures µ with kMf k = kf k∞ for all f ∈ L∞ (µ). We call the measure µ semifinite
if for each E ∈ M with µ(E) = ∞ one can find a F ∈ M with F ⊂ E such that
0 < µ(F ) < ∞, see [22, p. 25].
Now we claim that kMf k = kf k∞ for all f ∈ L∞ (µ) if and only if the measure µ is
semifinite.
– Suppose that µ is semifinite. Since f ∈ L∞ (µ), we have kf k∞ < ∞. Let α = kf k∞ .
If α = 0, then the inequality (5.84) forces that
kMf k = kf k∞ = 0
and we are done. If α > 0, then we see from Problem 3.19 that α = max{|z| | z ∈ Rf },
where Rf denotes the essential range of f . Thus there exists a z0 ∈ C such that
α = |z0 |. Without loss of generality, we may assume that z0 = α and so
µ{x ∈ X | |f (x) − α| < ǫ} > 0
(5.85)
for every ǫ > 0. Combining Definition 3.7 and the result (5.85), we establish that the
measure of the set
E = {x ∈ R | |f (x)| > α − ǫ}
is nonzero.
Chapter 5. Examples of Banach Space Techniques
174
∗ Case (i): µ(E) < ∞. Now it is trivial that the function
g=p
1
µ(E)
χE
satisfies the conditions that g ∈ L2 (µ) and kgk2 = 1. Furthermore, we derive
from the definition that
o 1 n (α − ǫ)2 Z
o1
n 1 Z
2
2
2
|f χE | dµ
dµ
>
= α − ǫ.
(5.86)
kMf (g)k2 =
µ(E) E
µ(E) E
Since ǫ is arbitrary, the estimate (5.86) implies that
kMf k = sup{kMf (g)k2 | g ∈ L2 (µ) and kgk2 = 1} ≥ α
and hence kMf k = kf k∞ by the inequality (5.84).
∗ Case (ii): µ(E) = ∞. Then the hypothesis ensures that there exists a F ∈ M
with F ⊂ E and 0 < µ(F ) < ∞. Now the function
1
g = √ χF
F
also satisfies the estimate (5.86) and thus kMf k = kf k∞ holds in this case.
– Suppose that µ is not semifinite. Then there is a E ∈ M such that µ(E) = ∞ and
every F ⊂ E satisfies either µ(F ) = 0 or µ(F ) = ∞. Take g ∈ L2 (µ) so that
Z
Z
2
|g(x)|2 dµ < ∞.
|g(x)| dµ ≤
E
X
If there is a F ⊂ E such that |g(x)| > 0 on F and g(x) = 0 on E \ F , then µ(F ) 6= ∞
by Proposition 1.24(a). Recall that µ is not semifinite, we have µ(F ) = 0. In other
words, it must be true that g = 0 a.e. on E. Let f = χE . Then we have
f g = 0 a.e. on E
for every g ∈ L2 (µ). By the definition, we obtain
kMf k = 0.
On the other hand, kf k∞ = 1 so that kMf k < kf k∞ .
Hence we have proven the claim.
• Functions f ∈ L∞ (µ) such that Mf is onto. Suppose that µ satisfies the condition
that every measurable set E of positive measure contains a measurable subset F with
0 < µ(F ) < ∞.i We claim that the map Mf is onto if and only if f1 ∈ L∞ (µ).
– If Mf is onto, then we let
E = {x ∈ X | f (x) = 0}.
i
(5.87)
We claim that µ(E) = 0. Assume that µ(E) > 0. By the hypothesis, there exists
a F ⊆ E with 0 < µ(F ) < ∞. Thus χF ∈ L2 (µ). Since f g = 0 on F for every
g ∈ L2 (µ), it implies that
χF ∈
/ Mf L2 (µ)
Some books take this as the definition of a semifinite measure µ. See, for example, [7, Exercise 25.9]
175
5.5. Miscellaneous Problems
which contradicts the surjective property of Mf . Thus we conclude that µ(E) = 0.
If g ∈ L2 (µ) is such that Mf (g) = f g = 0, then it follows from the definition (5.87)
that g = 0 a.e. on E c . Since µ(E) = 0, we obtain fact that
g = 0 a.e. on X
and this means that Mf is one-to-one. Since Mf is assumed to be onto, it is bijective.
On the one hand, recall that L2 (µ) is Banach and kMf k ≤ kf k∞ < ∞ by the first
assertion, Theorem 5.10 ensures that there corresponds a δ > 0 such that
kMf (g)k2 ≥ δkgk2
(5.88)
for every g ∈ L2 (µ). On the other hand, we consider F = {x ∈ X | |f (x)| < 2δ }. If
µ(F ) 6= 0, then our hypothesis tells us that there is a G ⊂ F such that 0 < µ(G) < ∞
and thus
1
kχG k2 = [µ(G)] 2
which verifies that
kMf (χG )k2 = kf χG k2 =
nZ
G
|f |2 dµ
o1
2
<
n Z δ2 o 1
1
δ
δ
2
dµ
= [µ(G)] 2 = kχG k2 ,
2
2
G 4
but it contradicts the inequality (5.88). Hence we have µ(F ) = 0 if and only if
1
≤ 2δ a.e. on X if and only if f1 ∈ L∞ (µ).
|f (x)| ≥ 2δ > 0 a.e. on X if and only if |f (x)|
– Suppose that f1 ∈ L∞ (µ). Then it is clear that M 1 is the inverse operator of Mf so
f
that Mf is bijective. In particular, Mf is surjective.
Hence we have completed the proof of the problem.
Problem 5.18
Rudin Chapter 5 Exercise 18.
Proof. Let x ∈ X and ǫ > 0. Since E is dense in the normed linear space X, we can find y ∈ E
such that
ǫ
.
(5.89)
kx − yk <
2M
Since {Λn (y)} converges in the Banach space Y , there exists a positive integer N such that
n, m ≥ N imply that
ǫ
kΛn (y) − Λm (y)k < .
(5.90)
2
Therefore, if n, m ≥ N , then we deduce immediately from the inequalities (5.89) and (5.90) that
kΛn (x) − Λm (x)k ≤ kΛn (x) − Λn (y)k + kΛn (y) − Λm (y)k + kΛm (y) − Λm (x)k
ǫ
< kΛn (x − y)k + + kΛm (y − x)k
2
ǫ
≤ kΛn k · kx − yk + kΛm k · kx − yk +
2
ǫ
ǫ
< +
2 2
= ǫ.
This means that {Λn (x)} is Cauchy in Y . Since Y is Banach, {Λn (x)} converges in Y and we
complete the analysis of the problem.
Chapter 5. Examples of Banach Space Techniques
176
Problem 5.19
Rudin Chapter 5 Exercise 19.
Proof. Given f ∈ C(T ), i.e., f is a continuous complex function on T . Recall that
1
sn (f ; x) =
2π
Z
π
−π
f (t)Dn (x − t) dt
and Dn (t) =
n
X
k=−n
eikt =
sin(n + 21 )t
.
sin 2t
(5.91)
By §5.11, C(T ) is Banach relative to the supremum norm kf k∞ .j Next, since f ∈ C(T ), we
follow from the equations (5.91) that sn (f ; x) ∈ C(T ) for each n ∈ N.
Let X = Y = C(T ) and for each n = 2, 3, . . ., we define Λn : C(T ) → C(T ) by
Λn (f ) =
sn (f ; x)
.
log n
Since f ∈ C(T ), there exists a M > 0 such that |f (x)| ≤ M for all x ∈ T so that
Z π
Z
1
M π
|sn (f ; x)| ≤
|f (t)| · |Dn (x − t)| dt ≤
|Dn (x − t)| dt.
2π −π
2π −π
(5.92)
Since Dn (x − t) = eikx Dn (t) and Dn (t) is an even function, the estimate (5.92) can be replaced
by
Z
Z
Z
M π
M π sin(n + 21 )t
M π
|Dn (t)| dt =
|Dn (t)| dt =
dt.
(5.93)
|sn (f ; x)| ≤
2π −π
π 0
π 0
sin 2t
Next, we consider f (x) = sin x2 − x4 on [0, π]. Using differentiation, we can show that sin x2 ≥ x4
on [0, π]. Thus the estimate (5.93) can be further written as
Z
4M π | sin(n + 12 )t|
|sn (f ; x)| ≤
dt
π 0
t
1
Z
4M (n+ 2 )π | sin t|
dt
=
π 0
t
Z
n Z (k+1)π
X
4M π sin t
| sin t| (5.94)
≤
dt +
dt .
π
t
t
0
kπ
k=1
1
t
On the interval [kπ, (k + 1)π], it is easy to see that
n Z
X
k=1
(k+1)π
kπ
X 1 | sin t|
dt ≤
t
kπ
n
k=1
Z
0
is bounded by
Z
j
π
0
π
2
sin t
dt =
t
π
2
1
kπ ,
(k+1)π
kπ
2
π
Besides, we get from [49, Problem 8.6, p. 197] that
Z
≤
where k = 1, 2, . . . , n, so we have
n
2X
1
| sin t| dt =
.
π
k
(5.95)
k=1
<
sin t
t
< 1 for 0 < t < π2 , so the integral
sin t
dt
t
and thus
Z
0
π
2
sin t
dt +
t
Z
π
π
2
sin t
π
dt ≤ +
t
2
Z
π
π
2
dt
π
π
π
= + ln π − ln = + ln 2.
t
2
2
2
(5.96)
To see this, since T is compact, we follow from Definition 3.16 that C(T ) = C0 (T ) and Theorem 3.17 implies
that C(T ) is complete. By Definition 5.2, we see that C(T ) is Banach.
177
5.5. Miscellaneous Problems
Now, by putting the estimates (5.95) and (5.96) into the estimate (5.94), we derive that
|Λn (f )| =
n
X
1
π
8M
sn (f ; x)
4M
× ( + ln 2) + 2
·
.
≤
log n
π log n
2
π log n
k
(5.97)
k=1
When kf k∞ = 1, we may take M = 1 in the estimate (5.97). It is well-known (see [63, Eqn.
n
X
1
(6.27), p. 124]) that log n and
are of the same growth as n → ∞, so we follow from the
k
k=1
estimate (5.97) that there exists a M ′ > 0 such that
kΛn k ≤ M ′
for all n = 2, 3, . . .. That is, {Λn } satisfies the first hypothesis of Problem 5.18.
By Theorem 4.25 (The Weierstrass Approximation Theorem), the set of all trigonometric
polynomials, namely P, is dense in C(T ). Let Pm (t) = eimt for some m ∈ Z. If n ≥ m, then we
follow from the result [51, Eqn. (8), p. 89] that
1
sn (Pm ; x) =
2π
Therefore, if P (t) =
N
X
Z
π
−imt
e
−π
n
X
ikt
e
Z π
n
X
1
dt =
ei(k−m)t dt = 1.
2π −π
(5.98)
k=−n
k=−n
cm Pm (t), then we get immediately from the result (5.98) that
m=−N
sn (P ; x) =
N
X
cm sn (Pm ; x) =
N
X
cm
m=−N
m=−N
for every n ≥ N and this implies that
N
X
1
sn (P ; x)
cm → 0
=
·
Λn (P ) =
log n
log n
m=−N
as n → ∞. Thus {Λn } satisfies the second hypothesis of Problem 5.18. Hence we establish from
Problem 5.18 that {Λn (f )} converges to 0 for each f ∈ C(T ) and as a consequence, we have
lim
n→∞
ksn k∞
= lim kΛn k = 0.
n→∞
log n
For the second assertion, we note that if we define
Λn f =
sn (f ; 0)
,
λn
then we may apply an argument similar to that used in §5.11 to show that
kΛn k =
kDn k1
|λn |
holds. Furthermore, we follow from the hypothesis
λn
log n
→ 0 as n → ∞ thatk
4 X1
4
4 log n
γ kDn k1
→∞
≥ 2
≥ 2
(log n + γ) = 2
+
|λn |
π |λn |
k
π |λn |
π |λn |
|λn |
n
k=1
k
We have applied the estimate of kDn k1 used in [51, p. 102] in the first inequality in (5.99).
(5.99)
Chapter 5. Examples of Banach Space Techniques
178
as n → ∞, where γ is the famous Euler constant. Hence Theorem 5.8 (the Banach-Steinhaus
Theorem) ensures that the sequence
n s (f ; 0) o
n
λn
is unbounded for every f in some dense Gδ set in C(T ), completing the proof of the problem. Problem 5.20
Rudin Chapter 5 Exercise 20.
Proof.
(a) Assume that such a sequence of continuous positive functions {fn } existed. Let x ∈ R and
n, k ∈ N. Furthermore, let
U = {x ∈ R | {fn (x)} is unbounded} and
We claim that
U=
Uk = {x ∈ R | fn (x) > k for some n ∈ N}.
∞
\
Uk .
(5.100)
k=1
On the one hand, if x ∈ Uk for all k ∈ N, then for every k ∈ N, there exists a n ∈ N such
that fn (x) > k. In other words, we have x ∈ U . On the other hand, if x ∈ U , then since
{fn (x)} is unbounded, for every k ∈ N, there exists a n ∈ N such that fn (x) > k which
is equivalent to saying that x ∈ Uk for all k ∈ N. Thus this proves the claim. Next, it is
easy to see that
∞
[
{x ∈ R | fn (x) > k}.
Uk =
n=1
Since each fn is continuous, the set {x ∈ R | fn (x) > k} is open in R and therefore, each Uk
is open in R. By the definition (5.100), U is a Gδ set. Thus we follow from the assumption
that
Q=U
is also a Gδ set. We deduce from the definition (5.100) that
R = Q ∪ Qc = Q ∪
∞
[
k=1
Ukc =
∞
[
{qk } ∪
k=1
∞
[
Ukc
(5.101)
k=1
Clearly, each qk is a nowhere dense subset of R. Furthermore, it is trivial that each Ukc is
closed in R. Assume that Ukc0 was not a nowhere dense subset of R for some k0 . Let Vk0
be a nonempty open subset of Ukc0 . Then there exists a p ∈ Vk0 and a δ > 0 such that
(p − δ, p + δ) ⊆ Vk0 ⊆ Ukc0 ⊆ Qc ,
but this means that Q ∩ Qc 6= ∅, a contradiction. Thus every Ukc is also a nowhere dense
subset of R and the representation (5.101) shows that R is a set of the first category which
contradicts Theorem 5.6 (Baire’s Theorem) that no complete metric space is of the first
category.
(b) Suppose that Q = {q1 , q2 , . . .} and for each n ∈ N, we define fn : R → R by
fn (x) = min(n|x − q1 | + 1, n|x − q2 | + 2, . . . , n|x − qn | + n).
(5.102)
179
5.5. Miscellaneous Problems
It is clear that fn is continuous on R and fn (x) ≥ 1 for all x ∈ R, i.e., {fn } is a sequence
of continuous positive functions on R.
Suppose that θ is irrational and N ∈ N. Consider the number
α = min(|θ − q1 |, |θ − q2 |, . . . , |θ − qN |) > 0.
Then the Archimedean Property ([49, Theorem 1.20(a)]) implies that there is a positive
integer n such that nα > N and thus
fn (θ) = min(n|θ − q1 | + 1, n|θ − q2 | + 2, . . . , n|θ − qn | + n) > N + 1.
Since N is arbitrary, we have fn (θ) → ∞ as n → ∞ so that {fn (θ)} is unbounded.
To prove the other direction (i.e., {fn (x)} is unbounded implies x is irrational), we prove
its contrapositive. Let qk be a rational number. If n ≥ k, then we have
fn (qk ) = min(n|qk − q1 | + 1, . . . , n|qk − qk | + k, . . . , n|qk − qn | + n) ≤ k
so that fn (qk ) ≤ min f1 (qk ), . . . , fk−1 (qk ), k . In other words, the set {fn (qk )} is bounded.
(c) The sequence of the functions given by (5.102) shows that the assertion is true for irrationals. For the rational numbers, we first prove the case on [0, 1].l To begin with, suppose
that Qn = {q1 , q2 , . . . , qn } ⊆ [0, 1] ∩ Q, where q1 = 0. Furthermore, we let
δn =
1
2n+1
min{|qi − qj | | 1 ≤ i < j ≤ n} > 0.
Clearly, we have
lim δn = 0 and
n→∞
(qi − δn , qi + δn ) ∩ (qj − δn , qj + δn ) = ∅
(5.103)
for all 1 ≤ i < j ≤ n. (If qi = 0 or qi = 1, then (qi − δn , qi + δn ) are replaced by [0, δn ) or
(1 − δn , 1] respectively.)
n
[
(qi − δn , qi + δn ) and fn : [0, 1] → R is defined by
Suppose that En =
i=1
 n

(x − qi + δn ),
if x ∈ (qi − δn , qi ] and 1 ≤ i ≤ n;


δn




n
fn (x) =
− (x − qi − δn ), if x ∈ [qi , qi + δn ) and 1 ≤ i ≤ n;

δn






0,
if x ∈ [0, 1] \ En .
Thus fn is a continuous function on [0, 1], zig-zag on (qi − δn , qi + δn ) and fn (qi ) = n for
each i = 1, 2, . . . , n.m Therefore, if x ∈ [0, 1] ∩ Q, then x = qk for some k ∈ N and thus we
obtain
lim fn (x) = ∞.
n→∞
Next, suppose that x is irrational. Assume that there was an N ∈ N such that x ∈ En
for all n ≥ N . By the definition, it means that
x ∈ (qk − δN , qk + δN )
l
m
The following argument is stimulated by the papers of Fabrykowski [19] and Myerson [43].
The graph of the fn looks like the graph shown in Figure 2.1.
(5.104)
Chapter 5. Examples of Banach Space Techniques
180
for some k ∈ {1, 2, . . . , N }. Since x ∈ En for all n ≥ N , the set relation (5.104) shows that
x ∈ (qk − δn , qk + δn )
for all n ≥ N . By the limit (5.103), we know that x = qk ∈ Q, a contradiction. Hence
we must have x ∈
/ En for infinitely many n, i.e., fn (x) = 0 for infinitely many n so that
lim fn (x) = ∞ is impossible.
n→∞
In conclusion, we have constructed the sequence of continuous functions fn : [0, 1] → R
which satisfies fn (x) → ∞ as n → ∞ if and only if x ∈ Q. If we suppose that fn is
a function of period 1, then the domain of fn can be extended to R and we obtain the
desired result.
Hence we have completed the proof of the problem.
Problem 5.21
Rudin Chapter 5 Exercise 21.
Proof. Since Q is a countable union of closed
√ sets of R, we have Q ∈ B. It is well-known that
m(Q) = 0. Obviously, the translate Q + 2 does not intersect Q. This gives an affirmative
answer to the first assertion.
Assume that there was a homeomorphismn h : R → R such that
e = h(E) ∩ E 6= ∅
E
for every measurable E ⊆ R with m(E) = 0. Given Q = {qk } and ǫ > 0. We first construct
a particular measurable set E with m(E) = 0. To this end, for all qk ∈ Q, we consider the
neighborhoods (qk − 2−k ǫ, qk + 2−k ǫ) and their union
E(ǫ) =
∞
[
(qk − 2−k ǫ, qk + 2−k ǫ).
k=1
It is easy to see that every E(ǫ) is nonempty open in R. In addition, since Q ⊆ E(ǫ), each E(ǫ)
is dense in R with m E(ǫ) ≤ 2ǫ. Hence it follows from Theorem 5.6 (Baire’s Theorem) that
the set
∞
1
\
(5.105)
E
E=
n
n=1
is a dense Gδ set in R and m(E) = 0.
Suppose that E is the set (5.105). Since h is a homeomorphism, it is an open map which
implies that
∞ \
1
h E
h(E) =
n
n=1
e is also a dense Gδ set
is also a dense Gδ set in R with m h(E) = 0. Thus the intersection E
1
e ⊆E
of measure zero. Since E n is countable for all n ∈ N, E is also countable and thus E
is a countable dense Gδ in R of measure zero. However, R is a complete metric space which
e certainly contradicts Theorem 5.13. Hence no such
has no isolated points, the existence of E
homeomorphism h exists and this completes the proof of the problem.
n
By the definition, h is a continuous bijection and h−1 is continuous.
181
5.5. Miscellaneous Problems
Problem 5.22
Rudin Chapter 5 Exercise 22.
Proof. Suppose that f ∈ C(T ), f ∈ Lip α and f (0) = 0. We need to show that
lim sn (f ; x) = f (x).
(5.106)
n→∞
To achieve the goal, we first quote the following stronger form of the Riemann-Lebesgue Lemma
whose proof can be found in [3, Theorem 11.16, p. 313].
Lemma 5.2 (Riemann-Lebesgue Lemma)
For every f ∈ L1 (T ), we define
1
fb(n) =
2π
Z
π
f (t)e−i(n+β)t dt,
−π
where β ∈ R. Then we have fb(n) → 0 as |n| → ∞.
To begin with, since f ∈ C(T ), we have f ∈ L1 (T ). Furthermore, the hypothesis f ∈ Lip α
implies that |f (s) − f (t)| ≤ Mf |s − t|α for all s, t ∈ [−π, π], where Mf is finite and α ∈ (0, 1].
Particularly, take s = 0 so that |f (t)| ≤ Mf |t|α for all t ∈ [−π, π] which implies
Z
π
f (t)
dt ≤ Mf
t
−π
Z
π
−π
In other words, we have
|t|α−1 dt = Mf
f (t)
t
hZ
0
(−1)α−1 tα−1 dt +
−π
Z
π
0
i 2M π α
f
tα−1 dt =
< ∞.
α
∈ L1 (T ). Since we have
i
1
1 h i(n+ 1 )t
1
2
t=
e
sin n +
− e−i(n+ 2 )t ,
2
2i
it yields
1
π
Z
π
−π
f (t) ·
Z π
sin(n + 12 )t
f (t) i(n+ 1 )t
1
2
dt =
e
dt −
t
2πi −π t
Z π
1
f (t) i(n+ 1 )t
2
dt +
≤
e
2π −π t
Z π
f (t) −i(n+ 1 )t
1
2
e
dt
2πi −π t
Z π
1
f (t) −i(n+ 1 )t
2
dt .
e
2π −π t
(5.107)
By Lemma 5.2, each of the integrals on the right-hand side of the inequality (5.107) tends to 0
as |n| → ∞. Thus it is true that
1
π
Z
π
−π
f (t) ·
sin(n + 12 )t
dt → 0 = f (0)
t
as |n| → ∞.o
Next, we claim that
o
Integrals of the form
Z
0
sn (f ; 0) → f (0)
b
g(t)
(5.108)
sin αt
dt are called Dirichlet integrals, where α > 0 and g is defined on [0, b].
t
Chapter 5. Examples of Banach Space Techniques
182
as n → ∞. To this end, we consider
sn (f ; 0) −
1
π
Z
π
−π
f (t) ·
Z π
Z
sin(n + 12 )t
sin(n + 12 )t
1 π
1
dt =
dt
f (t)Dn (t) dt −
f (t)
t
2π −π
π −π
t
Z π
1
1
1
1
=
−
f
(t)
sin
n
+
t dt
t
2π −π sin 2t
2
2
Z π
1
1
t dt ,
(5.109)
F (t)f (t) sin n +
=
2π −π
2
where F : [0, π] → R is defined by
F (t) =



1
sin 2t
− 1t , if t ∈ [−π, π] \ {0};

 0,
2
if t = 0.
Clearly, F is continuous on [−π, π] and so F f ∈ L1 (T ). Using Lemma 5.2 to the right-hand side
of the equation (5.109), we see that
Z π
1
1
t dt = 0
F (t)f (t) sin n +
lim
n→∞ 2π −π
2
and this guarantees the validity of the claim (5.108).
For the general case, we consider the function g : [−π, π] → R defined by
g(t) = f (x + t) − f (x)
(5.110)
for every x ∈ R. Clearly, the real function g satisfies the conditions g ∈ C(T ), g ∈ Lip α and
g(0) = 0. By the above argument, we obtain
lim sn (g; 0) = 0.
n→∞
(5.111)
By the definition (5.110), we gain
sn (g; 0) = sn f (x + t) − f (x); 0
Z π
Z π
1
1
=
f (x + t)Dn (−t) dt −
f (x)Dn (−t) dt
2π −π
2π −π
Z π
1
f (x + t)Dn (−t) dt − f (x)
=
2π −π
Z π
1
=
f (x − t)Dn (t) dt − f (x)
2π −π
= sn (f ; x) − f (x).
(5.112)
By applying the limit (5.111) to the equation (5.112), we obtain our desired result (5.106),
completing the proof of the problem.
CHAPTER
6
Complex Measures
6.1
Properties of Complex Measures
Problem 6.1
Rudin Chapter 6 Exercise 1.
Proof. By the definition, we have
λ(E) = sup
n
nX
k=1
|µ(Ei )| E1 , E2 , . . . , Ek are mutually disjoint and E =
k
[
j=1
o
Ej .
By Definition 6.1, we have
|µ|(E) = sup
∞
nX
k=1
|µ(Ei )| E1 , E2 , . . . are mutually disjoint and E =
∞
[
j=1
o
Ej .
Obviously, we have λ(E) ≤ |µ|(E) for every E ∈ M.
For the other direction, we suppose that {Ei } is a partition of E ∈ M. Given ǫ > 0. Since µ
is a complex measure, we get from Definition 6.1 that the series
∞
X
µ(Ei )
i=1
converges absolutely. Thus there exists a N ∈ N such that
∞
X
n=N
|µ(En )| < ǫ.
(6.1)
′
′
Define E1′ = E1 , E2′ = E2 , . . . , EN
−1 = EN −1 and EN = EN ∪ EN +1 ∪ · · · . By Definition 1.3(a),
′
we have EN ∈ M. Then we have
∞
[
i=1
′
′
Ei = E1′ ∪ E2′ ∪ · · · ∪ EN
−1 ∪ EN
which implies, with the aid of the estimate (6.1), that
∞
X
i=1
|µ(Ei )| =
N
−1
X
i=1
|µ(Ei′ )| +
∞
X
i=N
|µ(Ei )| <
N
−1
X
i=1
183
′
|µ(Ei′ )| + |µ(EN
)| + ǫ ≤ λ(E) + ǫ.
(6.2)
Chapter 6. Complex Measures
184
Since the inequality (6.2) holds for every partition of E, we have |µ|(E) ≤ λ(E) + ǫ. Since ǫ is
arbitrary, we obtain |µ|(E) ≤ λ(E). Hence we conclude that λ = |µ| which completes the proof
of the problem.
Problem 6.2
Rudin Chapter 6 Exercise 2.
Proof. Let µ be Lebesgue measure on (0, 1) and λ be the counting measure on the σ-algebra M
of all Lebesgue measurable sets in (0, 1). Assume that λ was σ-finite. Then we have
(0, 1) =
∞
[
En ,
(6.3)
n=1
where λ(En ) < ∞ for each n ∈ N. Clearly, we have λ(E) < ∞ if and only if E is finite.
Therefore, the representation (6.3) implies that (0, 1) is countable, a contradiction. Hence λ is
not σ-finite.
Next, we check µ ≪ λ. Suppose that λ(E) = 0, where E ∈ M. By the definition of λ, we
know that E = ∅ which
implies definitely that µ(E) = µ(∅) = 0. In addition, the definition of
µ ensures that µ (0, 1) = 1, i.e., µ is a bounded measure.
Assume that there was a h ∈ L1 (λ) such that dµ = h dλ, i.e.,
Z
h dλ
(6.4)
µ(E) =
E
for every E ∈ M. Note that µ(E) ≥ 0, so h(x) ≥ 0 a.e. [λ] on E.a Particulary, if E = (0, 1),
then h(x) ≥ 0 a.e. [λ] on (0, 1). Suppose that En = {x ∈ (0, 1) | h(x) ≥ n1 } for every n ∈ N.
Since h ∈ L1 (λ), Proposition 1.24(a) implies that
Z
1
h dλ ≥ λ(En ) ≥ 0.
∞>
n
En
Thus we must have λ(En ) ∈ [0, ∞) or equivalently, En is finite or En = ∅. Let
E=
∞
[
n=1
En = {x ∈ (0, 1) | h(x) > 0}.
(6.5)
Therefore, the set E is countable. There are two cases for consideration:
• Case (i): E 6= ∅. Let E = {x1 , x2 , . . .}. On the one hand, we know from [49, Remark
11.11(f), p. 309] that µ(F ) = 0 for every countable subset F ⊂ (0, 1), so we deduce from
the representation (6.4) and the definition (6.5) that
Z
∞
X
h(xn ) 6= 0,
h dλ =
0 = µ(E) =
E
n=1
a contradiction.
• Case (ii): E = ∅. In this case, we have h(x) = 0 a.e. [λ] on (0, 1) and the representation
(6.4) again shows that
Z
h dλ = 0,
1 = µ (0, 1) =
(0,1)
a contradiction.
a
See Definition 1.35 for the meaning of the notation a.e. [λ].
185
6.1. Properties of Complex Measures
Hence no such h exists and this ends the proof of the problem.
Problem 6.3
Rudin Chapter 6 Exercise 3.
Proof. We first show that if µ and λ are complex regular Borel measures, then both µ + λ and
αµ are complex regular Borel measures too. By §6.18, it is equivalent to show that both |µ + ν|
and |αµ| are regular Borel measures. To this end, we follow from §6.18 again that both |µ| and
|λ| are regular Borel measures. Let E ∈ B and ǫ > 0. By Definition 2.15, there exist open sets
V1 , V2 ⊇ E such that
|µ|(V1 ) < |µ|(E) +
ǫ
2
ǫ
and |λ|(V2 ) < |λ|(E) + .
2
(6.6)
Similarly, there are compact sets K1 , K2 ⊆ E such that
|µ|(E) < |µ|(K1 ) +
ǫ
2
ǫ
and |λ|(E) < |λ|(K2 ) + .
2
(6.7)
The triangle inequality certainly implies |µ(E) + λ(E)| ≤ |µ(E)| + |λ(E)| and then Definition
6.1 gives
∞
∞
X
X
|µ(E)| + |λ(E)| ,
(6.8)
|µ(E) + λ(E)| ≤ sup
|µ + λ|(E) = sup
i=1
i=1
where the supremum being taken over all partitions {Ei } of E. Since the series
∞
X
i=1
∞
X
i=1
|µ(E)| and
|λ(E)| converge, we deduce from the expression (6.8) that
|µ + λ|(E) ≤ sup
∞
X
i=1
|µ(E)| + sup
∞
X
i=1
|λ(E)| = |µ|(E) + |λ|(E).
Thus we have
|µ + λ| ≤ |µ| + |λ|.
(6.9)
Let V = V1 ∩ V2 and K = K1 ∪ K2 . Then K is compact and V is open in X. Now we follow
from the estimates (6.6) and (6.7) and the inequality (6.9) that
|µ + λ|(V ) = |µ + λ|(E) + |µ + λ|(V \ E)
≤ |µ + λ|(E) + |µ|(V \ E) + |λ|(V \ E)
≤ |µ + λ|(E) + |µ|(V1 \ E) + |λ|(V2 \ E)
< |µ + λ|(E) + ǫ.
Since ǫ and E are arbitrary, the measure |µ + λ|(E) is in fact outer regular. Similarly, we have
|µ + λ|(K) = |µ + λ|(E) − |µ + λ|(E \ K)
≥ |µ + λ|(E) − |µ + λ|(E \ K1 ) − |µ + λ|(E \ K2 )
> |µ + λ|(E) − ǫ.
In other words, |µ + λ| is also inner regular. By Definition 2.15, µ + λ is a regular complex Borel
measure, i.e., µ + λ ∈ M (X). Now the regularity of |αµ| is easy to prove, so we omit the details
here.
Chapter 6. Complex Measures
186
Now it is time to prove the assertion in the question. Since X is a locally compact Hausdorff space, Theorem 6.19 (The Riesz Representation Theorem) ensures every Φ ∈ C0 (X)∗ is
represented by a unique µΦ ∈ M (X) in the sense that
Z
f dµΦ
(6.10)
Φ(f ) =
X
for every f ∈ C0 (X). By [51, Eqn. (3), p. 130], we see that
Z
Z
Z
Z
f d(µΦ + µΨ )
f dµΨ =
f dµΦ +
f dµΦ+Ψ = (Φ + Ψ)(f ) = Φ(f ) + Ψ(f ) =
X
X
X
and
Therefore, they imply that
Z
f dµαΦ = αΦ(f ) =
Z
X
f d(αµΦ ).
X
X
µΦ+Ψ = µΦ + µΨ
and
µαΦ = αµΦ .
(6.11)
By the previous analysis, we see that µΦ+Ψ , µαµ ∈ M (X), so we may define the mapping
F : C0 (X)∗ → M (X) by
F (Φ) = µΦ
and it is easy to see from the results (6.11) that
F (Φ + Ψ) = F (Φ) + F (Ψ) and
F (αΦ) = αF (Φ).
Furthermore, Theorem 6.19 (The Riesz Representation Theorem) also implies that F is a bijection and kΦk = |µΦ |(X) = kµΦ k = kF (Φ)k. Consequently, F is actually an isometric vector
space isomorphism, i.e.,
C0 (X)∗ ∼
= M (X).
Since C0 (X) is a Banach space with the supremum norm, Problem 5.8 guarantees that C0 (X)∗
is Banach. Hence M (X) is Banach and we have completed the proof of the problem.
6.2
Dual Spaces of Lp (µ)
Problem 6.4
Rudin Chapter 6 Exercise 4.
Proof. Since µ is positive and σ-finite, we can write
X=
∞
[
Xn ,
(6.12)
n=1
where {Xn } is an increasing sequence of measurable sets and µ(Xn ) < ∞ for all n ∈ N, see [54,
Definition 4.22, p. 22]. Let A = {x ∈ X | |g(x)| = ∞} and An = {x ∈ Xn | |g(x)| = ∞}, where
n ∈ N. Since g is measurable, every An and A are measurable by Problem 1.5. Besides, it is
evident that A1 ⊆ A2 ⊆ · · · and if x ∈ A, then x ∈ XN0 for some N0 ∈ N so that x ∈ AN0 . As
a result, we get
∞
[
An .
A=
n=1
Now we are going to divide the proof into several steps:
6.2. Dual Spaces of Lp (µ)
187
• Step 1: µ(A) = 0. Otherwise, it follows from the construction and Theorem 1.19(c) that
lim µ(An ) = µ(A) > 0.
n→∞
This means that one can find a N1 ∈ N such that 0 < µ(AN1 ) ≤ µ(XN1 ) < ∞. We know
that χAN1 ∈ Lp (µ), so our hypothesis implies that χAN1 g ∈ L1 (µ), but
Z
Z
|g| dµ = µ(AN1 ) · ∞ = ∞
|χAN1 g| dµ =
kχAN1 gk1 =
AN1
X
which is a contradiction. Hence µ(A) = 0, i.e., g is finite a.e. on X.
• Step 2: g ∈ Lq (µ) when 1 < p < ∞. In this case, we have q ∈ (1, ∞). Here we define
En = {x ∈ Xn | |g(x)| ≤ n}.
Obviously, {En } is also an increasing sequence of measurable sets. Furthermore, if x0 ∈
X \ A, then x0 ∈ XN2 for some N2 ∈ N by the definition (6.12) so that x0 ∈ Xn for all
n ≥ N2 . Since g is finite a.e. on X, there exists a positive integer N3 such that |g(x)| ≤ N3
for almost all x ∈ X. Take N4 = max(N2 , N3 ), then x0 ∈ XN4 and |g(x0 )| ≤ N4 which
imply that x0 ∈ EN4 . Consequently, we have shown that
∞
[
X=
En .
n=1
Let gn = χEn g : X → C, where n = 1, 2, . . .. Then each gn is measurable on X and
Z
Z
q
q
|g|q dµ ≤ nq µ(En ) ≤ nq µ(Xn ) < ∞
(6.13)
|gn | dµ =
kgn kq =
En
X
for each n = 1, 2, . . .. Thus we have gn ∈ Lq (µ) for all n ∈ N. Next, we define Λn : Lp (µ) →
C by
Z
f gn dµ
Λn (f ) =
X
which is linear for n ∈ N. Since f gn = (f χEn )g and f χEn ∈ Lp (µ), we have f gn ∈ L1 (µ).
By Theorem 1.33 and Theorem 3.8, we obtain
Z
Z
|f gn | dµ = kf gn k1 ≤ kf kp × kgn kq ,
(6.14)
f gn dµ ≤
X
X
where n = 1, 2, . . .. By Definition 5.3 and the inequality (6.14), we know that
kΛn k = sup{|Λn (f ) | f ∈ Lp (µ) and kf kp = 1} ≤ kgn kq ,
where n = 1, 2, . . .. Recall that 1 < q < ∞, so we may consider
−q
f0 = kgn kq p |gn |q−2 gn .
(q−1)p so that
Then we have |f0 |p = kgn k−q
q |gn |
Z
Z
Z
(q−1)p
−q
|f0 |p dµ = kgn k−q
|g
|
dµ
=
kg
k
|gn |q dµ = 1
kf0 kp =
n
n q
q
X
X
and
Λn (f0 ) =
Z
X
X
−q
kgn kq p |gn |q−2 gn · gn dµ
(6.15)
Chapter 6. Complex Measures
188
− pq
= kgn kq
=
−q
kgn kq p
= kgn kq
Z
X
|gn |q dµ
× kgn kqq
(6.16)
for all n ∈ N. Combining the results (6.13), (6.15) and (6.16), we have established the fact
that
kΛn k = kgn kq < ∞,
(6.17)
where n ∈ N. Hence {Λn } is a family of bounded linear transformations of Lp (µ) into C.
Since Lp (µ) is Banach (see Definition 5.2) and
Z
Z
Z
Z
|f g| dµ = kf gk1 < ∞
|f g| dµ ≤
|f gn | dµ =
f gn dµ ≤
|Λn (f )| =
X
X
En
X
for every f ∈ Lp (µ) and n ∈ N, Theorem 5.8 (The Banach-Steinhaus Theorem) implies
that there is a M > 0 such that
kΛn k ≤ M
(6.18)
for all n ∈ N. Using the results (6.17) and (6.18), we conclude that
kgn kq ≤ M
(6.19)
for all n ∈ N.
By Step 1 and the definition of gn , we have |gn (x)| ≤ |g(x)| < ∞ a.e. on X for all
n ∈ N and
|g1 (x)| ≤ |g2 (x)| ≤ · · ·
for every x ∈ X. Now the Monotone Convergence Theorem [49, Theorem 3.14, p. 55]
implies that |gn (x)| → |g(x)| < ∞ a.e. on X as n → ∞. Hence we gain from Theorem
1.26 (Lebesgue’s Monotone Convergence Theorem) that
Z
Z
|g|q dµ = kgkqq .
(6.20)
|gn |q dµ =
lim kgn kqq = lim
n→∞
n→∞ X
X
Hence it follows from the inequality (6.19) and the limit (6.20) that
kgkq ≤ M < ∞,
i.e. g ∈ Lq (µ).
• Step 3: g ∈ L∞ (µ) when p = 1. In this case, q = ∞. We recall from Step 1 that g is
finite a.e. on X, so there exists a M > 0 such that |g(x)| ≤ M a.e. on X. By Definition
3.7, we assert that kgk∞ ≤ M , i.e., g ∈ L∞ (µ).
• Step 4: g ∈ L1 (µ) when p = ∞. Define f1 : X → C by

g(x)



, if g(x) 6= 0;
|g(x)|
f1 (x) =



0,
otherwise.
Since |f1 (x)| ≤ 1 for all x ∈ X, we get from Definition 3.7 that kf1 k∞ ≤ 1, i.e., f1 ∈ L∞ (µ).
By the hypothesis, we have f1 g ∈ L1 (µ). Since f1 g = |g|, we conclude that g ∈ L1 (µ).
6.2. Dual Spaces of Lp (µ)
189
Now we have completed the proof of the problem.
Problem 6.5
Rudin Chapter 6 Exercise 5.
Proof. The answer is negative. On the one hand, if f1 , f2 : X → C are defined by
f1 (a) = 0,
f1 (b) = 1 and f2 (a) = 1,
f2 (b) = 0,
(6.21)
then we have f1 , f2 ∈ L∞ (µ) and f1 6≡ f2 . For every f ∈ L∞ (µ), we have f (a) = A + Bi
and f (b) = C + Di for some A, B, C, D ∈ R. Obviously, we get from the definition (6.21) the
representation
f = (C + Di)f1 + (A + Bi)f2 .
As a result, L∞ (µ) is a two-dimensional space spanned by f1 and f2 , i.e., dim L∞ (µ) = 2.
On the other hand, if f ∈ L1 (µ), then kf k1 < ∞ and we follow from Definition 3.6 that
f (b) = 0 and kf k1 = |f (a)| = |f (a)|f2 (a). Therefore, L1 (µ) is an one-dimensional space spanned
by f2 . As a vector space (see Remark 3.10), we have
dim L1 (µ)∗ = dim L1 (µ) = 1.
Hence, L∞ (µ) 6= L1 (µ)∗ , completing the proof of the problem.
Problem 6.6
Rudin Chapter 6 Exercise 6.
Proof. We want to show that Lp (µ)∗ ∼
= Lq (µ) for 1 < p < ∞. Equivalently, we have to show
that for each Λ ∈ Lp (µ)∗ , there exists a unique g ∈ Lq (µ) such that
Z
f g dµ
(6.22)
Λf =
X
for all f ∈ Lp (µ). The following proof follows mainly Follan’s argument ([22, p. 190]):
• Step 1: µ is finite. Let s be a simple function on X. Since µ is finite, we have
µ({x ∈ X | s(x) 6= 0}) ≤ µ(X) < ∞ and then it follows from Theorem 3.13 that s ∈ Lp (µ).
Let Λ ∈ Lp (µ)∗ , E be measurable and λ(E) = Λ(χE ). For every partition {Ei } of E, if we
let Fn = E1 ∪ E2 ∪ · · · ∪ En , then we obtain
Z
Z
Z
p
p
χE\Fn dµ = µ(E \ Fn )
(6.23)
|χE\Fn | dµ =
|χE − χFn | dµ =
X
X
X
for all n ∈ N. In fact, the expression (6.23) can be rewritten as
1
kχE − χFn kp = µ(E \ Fn ) p .
(6.24)
Since E \ F1 ⊇ E \ F2 ⊇ · · · and µ(E \ F1 ) is finite, it establishes from Theorem 1.19(e)
that
∞
\
lim µ(E \ Fn ) = µ
(E \ Fn ) = µ(∅) = 0.
(6.25)
n→∞
n=1
Chapter 6. Complex Measures
190
Combining the results (6.24) and (6.25) and using the fact p ∈ (1, ∞), we derive that
lim kχFn − χE kp = 0.
n→∞
Since χE ∈ Lp (µ) and each χEi g is measurable, we deduce from the representation (6.22)
and then Theorem 1.27 that
Z X
Z
∞
∞ Z
∞
X
X
λ(χEi ).
χEi g dµ =
(χEi g) dµ =
χE g dµ =
λ(E) = Λ(χE ) =
X i=1
X
i=1
X
i=1
By Definition 6.1, λ is a complex measure.
If E ∈ M satisfies µ(E) = 0, then χE = 0 in Lp (µ) and so λ(E) = Λ(0) = 0. By
Definition 6.7, we have λ ≪ µ. Thus we know from Theorem 6.10 (The Lebesgue-RadonNikodym Theorem) that there exists a unique h ∈ L1 (µ) such that
Z
Z
χE g dµ
g dµ =
Λ(χE ) = λ(E) =
E
X
for every E ∈ M. Recall that Λ ∈ Lp (µ)∗ , Λ is linear and bounded so that
Z
sg dµ
Λ(s) =
X
for every simple function s ∈
Z
X
Lp (µ).
By [51, Eqn. (3), p. 96], we know that
sg dµ = kΛ(s)k ≤ kΛk · kskp < ∞.
By [22, Theorem 6.14, p. 189], we conclude that g ∈ Lq (µ). Given that f ∈ Lp (µ). By
Theorem 3.8, we have f g ∈ L1 (µ). By Theorem 3.13, there exists a sequence of simple
functions {sn } such that |sn | ≤ |f | for every n ∈ N and sn → f in Lp (µ) as n → ∞.
Since |sn g| ≤ |f g| and f g ∈ L1 (µ), we obtain from Theorem 1.34 (Lebesgue’s Dominated
Convergence Theorem) and Theorem 5.4 that
Z
Z
f g dµ.
sn g dµ =
Λ(f ) = lim Λ(sn ) = lim
n→∞
n→∞ X
X
• Step 2: µ is σ-finite. This is exactly the hypothesis of Theorem 6.16, so we have
Lp (µ)∗ ∼
= Lq (µ) in this case.
• Step 3: µ is any measure and p > 1. In this case, q ∈ (1, ∞). As in Step 2, for each
σ-finite subset E of X, there corresponds a unique gE ∈ Lq (E) such that
Z
f gE dµ
Λ(f ) =
X
for all f ∈ Lp (E), where Lp (E) = {f ∈ Lp (µ) | f (x) = 0 for all x ∈
/ E} and Lq (E) is
defined similarly. By Theorem 6.16, we know that
kgE kq = Λ|Lp (E) ≤ kΛk.
(6.26)
If F is a σ-finite subset of X containing E, then the uniqueness of gE implies that gF = gE
a.e. on E so that
kgE kq ≤ kgF kq .
(6.27)
Now we define
M = sup{kgE kq | E a σ-finite subset of X}.
191
6.3. Fourier Coefficients of Complex Borel Measures
Then it follows from the inequality (6.26) that
M ≤ kΛk
holds. By the definition, we may select a sequence {En } of σ-finite subsets of X such that
∞
[
En which is obviously σ-finite subset of X so
kgEn kq → M as n → ∞. Define F =
n=1
that kgF kq ≤ M . Furthermore, we deduce from the inequality (6.27) that
kgEn kq ≤ kgF kq
for all n ∈ N. Therefore, the definition of M implies that M ≤ kgF kq and then
M = kgF kq .
Finally, if A is a σ-finite subset of X containing F , we get
Z
Z
Z
Z
q
q
q
q
q
|gF |q dµ
|gA | dµ = kgA kq ≤ M =
|gA\F | dµ =
|gF | dµ +
X
X
X
X
which means that gA\F = 0 a.e. on X or equivalently,
gA = gF a.e. on X.
(6.28)
If f ∈ Lp (µ), then the set A = F ∪ {x ∈ X | f (x) 6= 0} is clearly σ-finite. Therefore, we
observe from the result (6.28) that
Z
Z
f gF dµ.
f gA dµ =
Λ(f ) =
X
X
Hence we may pick g = gF and the above argument makes sure this g is unique.
We complete the proof of the problem.
6.3
Fourier Coefficients of Complex Borel Measures
Problem 6.7
Rudin Chapter 6 Exercise 7.
Proof. If µ is a real measure, then we have
Z
Z
int
b(n)
µ
b(−n) = e dµ(t) = e−int dµ(t) = µ
for every n ∈ Z. Therefore, our result follows in this case.
Since µ is a complex Borel measure, Theorem 6.12 implies the existence of a Borel measurable
function h such that |h(t)| = 1 on [0, 2π) and dµ = h d|µ|. Thus we have
d|µ| = 1 · d|µ| = h · h d|µ| = h dµ.
Since |µ| is real, the previous paragraph and the expression (6.29) imply that
Z
Z
Z
−int
int
µ
b(−n) = e h d|µ| = e
h(t) d|µ|(t) = e−int [h(t)]2 dµ(t)
(6.29)
Chapter 6. Complex Measures
192
for every n ∈ Z. By Theorem 3.14, C(T ) is dense in L1 (|µ|). Using Theorem 4.25 (The
Weierstrass Approximation Theorem) and then Lemma 2.10, we deduce that the set of all
trigonometric polynomials P is dense in L1 (|µ|).
We want to apply Problem 5.18. For any f ∈ L1 (|µ|), we define Λn : L1 (|µ|) → C by
Z
Λn (f ) = f (t)e−int dµ(t)
for all n ∈ N. Then we have
Z
Z
Z
|Λn (f )| ≤ |f | · | dµ(t)| = |f | · |h| d|µ(t)| = |f | d|µ(t)| = kf k1 .
(6.30)
By Definition 5.3, we see from the inequality (6.30) that
kΛn k = sup{|Λn (f )| | f ∈ L1 (|µ|) and kf k1 = 1} ≤ 1
for every n ∈ N. Next, if g(t) = eikt , then we have
Z
Z
−int
Λn (g) = g(t)e
dµ(t) = e−i(n−k)t dµ(t).
Now our hypothesis ensures that Λn (g) → 0 as n → ∞ so that
Λn (f ) → 0
(6.31)
as n → ∞ for each f ∈ P. Therefore, Problem 5.18 says that the limit (6.31) also holds for
2
every f ∈ L1 (|µ|). Obviously, we have h ∈ L1 (|µ|) and so
Z
2
µ
b(−n) = e−int [h(t)]2 dµ(t) = Λn (h ).
Hence we conclude from the result (6.31) that
2
lim µ
b(−n) = lim Λn (h ) = 0.
n→∞
n→∞
This completes the proof of the problem.
Problem 6.8
Rudin Chapter 6 Exercise 8.
Proof. Suppose that X = [0, 2π), B is the collection of all Borel sets in X and µ
b is periodic
−ikt
with period k ∈ Z. Denote dλ(t) = (e
− 1) dµ(t). Clearly, we have
Z
Z
b
(6.32)
µ
b(n + k) − µ
b(n) = e−int (e−ikt − 1) dµ(t) = e−int dλ(t) = λ(n)
for all n ∈ Z. Particularly, take n = 0 in the result (6.32) and use the periodicity of µ
b and
Theorem 6.12 to get
Z
Z
h d|λ| =
dλ = 0,
(6.33)
where h is a measurable function such that |h(x)| = 1 on X. If |λ|(X) 6= 0, then Theorem
1.39(a) and the form (6.33) together imply that h = 0 a.e. on X, a contradiction. This means
that |λ|(X) = 0 and we assert from the fact |λ(E)| ≤ |λ|(X) = 0 that
λ(E) = 0
(6.34)
193
6.3. Fourier Coefficients of Complex Borel Measures
for every E ∈ B.
Using Theorem 6.12 to write dµ = H d|µ|, where H is a measurable function with |H(x)| = 1
on X. Next, we recall the meaning of the notation dλ(t) = (e−ikt − 1) dµ(t) and we observe
from the result (6.34) that
Z
Z
−ikt
0 = λ(E) = (e
(6.35)
− 1) dµ = (e−ikt − 1)H d|µ|
E
E
for every E ∈ B. Let A1 = {t ∈ X | cos kt − 1 = 0}, A2 = {t ∈ X | sin kt = 0} and
A = {t ∈ X | e−ikt − 1 = 0} = A1 ∩ A2 .
Since f1 (t) = cos kt − 1, f2 (t) = sin kt and g(t) = 0 are continuous on X, they are Borel
measurable.b By Problem 1.5(a), we have A1 , A2 ∈ B and also A ∈ B.
We claim that µ is concentrated on A. Assume that it was not the case, i.e., there is E0 ∈ B
such that a E0 ∩ A = ∅ but µ(E0 ) 6= 0. We recall from the result (6.35) that (e−ikt − 1)H = 0
a.e. on any E with |µ|(E) > 0. Since |H| = 1 on X, we have e−ikt = 1 a.e. on any E with
|µ|(E) > 0. Particularly, since µ(E0 ) 6= 0, we have |µ|(E0 ) > 0 and then
e−ikt = 1
(6.36)
a.e. on E0 . However, E0 ∩ A = 0 = ∅ means that e−ikt 6= 1 on E0 which contradicts the
expression (6.36). Hence we have proven our claim that µ is concentrated on A.
On the other hand, if µ is concentrated on A, then for every E ∈ B, we write E = (E \ A) ∪ A
and then
Z
Z
Z
−ikt
−ikt
(6.37)
(e
− 1) dµ + (e−ikt − 1) dµ.
(e
− 1) dµ =
E
E\A
A
Notice that e−ikt − 1 = 0 on A and (E \ A) ∩ A = ∅ so that µ(E \ A) = 0. Thus we deduce
from Proposition 1.24(d) and (e) that all the integrals in the expression (6.37) are zero. This
certainly implies that λ(E) = 0 for every E ∈ B and then
Z
b
λ(n) = e−int dλ = 0
(6.38)
for every n ∈ Z. Combining the expression (6.32) and the result (6.38), we may conclude that
µ
b(n + k) = µ
b(n)
for all n ∈ Z.
Hence we have shown that µ is a complex Borel measure with periodic Fourier coefficient µ
b
with period k if and only if µ is concentrated on
o
n
2nπ
n ∈ N ∪ {0} ,
A = {t ∈ X | e−ikt − 1 = 0} = t =
k
completing the proof of the problem.
Problem 6.9
Rudin Chapter 6 Exercise 9.
b
See §1.11.
Chapter 6. Complex Measures
194
Proof. The assertion is false. Take µ = m the Lebesgue measure on I. Since m ≪ m, if m ⊥ m,
then Proposition 6.8(g) implies that m = 0, a contradiction. Now it remains to construct a
sequence of {gn } satisfying the required properties.
2
. Define gn,k : [0, 1] → R and
Suppose that n, k ∈ N and k = 1, 2, . . . , n. Let δn = n(2n+3)
gn : [0, 1] → R by

h k
δn
k
δn i


n
+
1,
if
x
∈
;
−
,
+


n+1
2 n+1
2






k

δn
δn
k
δn 

;
−
−
,
−
 linear, if x ∈
n+1
2
2n + 2 n + 1
2
gn,k (x) =


k

δn
k
δn
δn 


;
+
,
+
+
linear,
if
x
∈


n+1
2 n+1
2
2n + 2






0,
otherwise
and
gn (x) =
n
X
gn,k (x)
k=1
respectively. By direct computation, it is easy to check that
0<
1
δn
δn
−
−
,
n+1
2
2n + 2
n
δn
δn
+
+
<1
n+1
2
2n + 2
and
δn
δn
k + 1 δn
δn
k
+
+
<
−
−
n+1
2
2n + 2
n+1
2
2n + 2
for k = 1, 2, . . . , n − 1. See Figure 6.1 for gn,1 (x) and gn,2 (x) below:
Figure 6.1: The graphs of gn,1 (x) and gn,2 (x).
Now the Lebesgue measure of the support of each gn,k is given by
k
2(n + 2)
δn
δn
k
δn
δn
1 δn =
+
+
−
+
+
= 1+
n+1
2
2n + 2 n + 1
2
2n + 2
n+1
n(n + 1)(2n + 3)
195
6.3. Fourier Coefficients of Complex Borel Measures
so that the Lebesgue measure of the support of gn is given by
n
X
k=1
2(n + 2)
2(n + 2)
=
→0
n(n + 1)(2n + 3)
(n + 1)(2n + 3)
as n → ∞. This shows that condition (i) holds for this gn . Since each gn,k is in fact a trapezium
(see Figure 6.1 again) with area
Z 1
2n + 3
(n + 1) n + 2
1
gn,k (x) dx =
δn + δn =
δn =
2
n+1
2
n
0
which implies that
Z
1
gn (x) dx =
0
n Z
X
k=1
1
gn,k (x) dx = 1.
0
Thus this gn satisfies condition (ii). Finally, note that for large n, we have
1
n(2n + 3)
=
≈ 1.
(n + 1)2 δn
2(n + 1)2
Thus for every f ∈ C(I) and sufficiently large n, we have
Z
1
f gn dx −
0
n Z 1
n+1
n
X
1 X k f (1)
1 X k f gn,k dx −
+
=
f
f
n+1
n+1
n
+
1
n
+
1
n
+1
k=1 0
k=1
k=1
"
#
Z 1
n
k X
1
|f (1)|
≤
f gn,k dx −
+
f
. (6.39)
n+1 n+1
n+1
0
k=1
Since f ∈ C(I), there is a M > 0 such that |f (x)| ≤ M on I. On the linear part, we have
gn,k = ak x + bk for some ak , bk ∈ R. Then it asserts that, for example,
Z
k
− δ2n
n+1
δn
k
− δ2n − 2n+2
n+1
(ak x + bk )f (x) dx ≤ M |ak | ·
Z
k
− δ2n
n+1
δn
k
− δ2n − 2n+2
n+1
x dx + M |bk | ·
δn
2n + 2
(6.40)
for every k = 1, 2, . . . , n. By routine computation, it can be shown easily that the integral on
′
, so the integral on the left-hand
the right-hand side of the inequality (6.40) is of the growth M
n4
′
side of the inequality (6.40) is bounded by M
for
some
positive
constant M ′ . Using this fact,
n3
the inequality (6.39) can be further reduced to
Z
0
1 X k f
n+1
n+1
n+1
1
f gn dx −
"Z
n
X
k=1
#
k f kf k∞
1
M
(n + 1)f (x) dx −
≤
+ 2 +
f
k
n+1 n+1
n
n+1
− δ2n
n+1
k=1
#
"
δ
δ
k
k
Z
Z
n
+ 2n
+ 2n X
f kf k∞
n+1
n+1
1
k M
(n + 1)f (x) dx −
f
dx + 2 +
=
k
(n + 1)δn k − δn
n+1
n
n+1
− δ2n
n+1
n+1
2
k=1
#
"
Z k + δn
n
k X
f kf k∞
n+1
2
1
M
(n + 1) f (x) −
dx
+
f
+
≤
k
(n + 1)2 δn n + 1
n2
n+1
− δ2n
n+1
k=1
#
"
Z k + δn
n
k X
f kf k∞
n+1
2
M
dx + 2 +
,
(6.41)
f (x) − f
(n + 1)
≤
k
n+1
n
n+1
− δn
k=1
k
+ δ2n
n+1
n+1
2
Chapter 6. Complex Measures
196
f is a positive constant.
where M
Applying the Mean-Value Theorem for Integrals (see [3, Theorem 7.30, pp. 160, 161]) to the
integral in the inequality (6.39), we see that
Z
k
where ξk ∈ [ n+1
−
we get
Z
0
1
k
+ δ2n
n+1
f (x) − f
k
− δ2n
n+1
δn
k
2 , n+1
+
δn
2 ].
k k dx = f (ξ) − f
δn ,
n+1
n+1
(6.42)
Substituting the expression (6.42) into the inequality (6.41),
k X
1 X k kf k∞
f gn dx −
(n + 1) f (ξk ) − f
≤
δn +
f
n+1
n+1
n+1
n+1
n+1
n
k=1
k=1
≤ n(n + 1)δn Mn +
where
Mn = max
f (ξ1 ) − f
f kf k∞
M
,
+
n2
n+1
(6.43)
2 n 1 , f (ξ2 ) − f
, . . . , f (ξn ) − f
.
n+1
n+1
n+1
k
for each k = 1, 2, . . . , n so that Mn → 0 as n → ∞. By this observation
When n → ∞, ξk → n+1
and the fact lim n(n + 1)δn = 1, we derive from the inequality (6.43) that
n→∞
lim
Z
1
n→∞ 0
f gn dx =
Z
1
f dx
0
which is exactly condition (iii).
We have completed the proof of the problem.
6.4
Problems on Uniformly Integrable Sets
Problem 6.10
Rudin Chapter 6 Exercise 10.
Proof.
(a) Let Φ = {f1 , f2 , . . . , fN } ⊆ L1 (µ) for some N ∈ N and ǫ > 0. By Problem 1.12, there
exists a δk > 0 such that
Z
|fk | dµ < ǫ
(6.44)
Ek
whenever µ(Ek ) < δk , where k = 1, 2, . . . , N . Suppose that δ = min(δ1 , δ2 , . . . , δk ). Then
it yields from Theorem 1.33 and the estimates (6.44) that if µ(E) < δ ≤ δk , then
Z
E
fk dµ ≤
Z
E
|fk | dµ < ǫ
for each k = 1, 2, . . . , N . By the definition, Φ is uniformly integrable.
197
6.4. Problems on Uniformly Integrable Sets
(b) We first show that
Lemma 6.1
Suppose that Φ ⊆ L1 (µ) is uniformly integrable. Given ǫ > 0, there exists a δ > 0
such that
Z
|f | dµ < ǫ
E
whenever µ(E) < δ and f ∈ Φ.
Proof of Lemma 6.1. Suppose that Φ is a collection of real measurable functions in
L1 (µ) and E ∈ M. Given ǫ > 0. Since Φ is uniformly integrable, there corresponds a
δ > 0 such that
Z
ǫ
(6.45)
f dµ <
4
E
whenever f ∈ Φ and µ(E) < δ.
Define E+ = {x ∈ E | f (x) ≥ 0} and E− = {x ∈ E | f (x) < 0}. It is clear that
E = E+ ∪ E− , E+ ∩ E− = ∅ and µ(E± ) ≤ µ(E) < δ. Thus we gain from Theorem
1.29 that
Z
Z
Z
Z
Z
f dµ
f dµ −
|f | dµ =
|f | dµ +
|f | dµ =
and then the estimate (6.45) yields that
Z
Z
Z
|f | dµ ≤
|f | dµ =
E
E
E−
E+
E−
E+
E
f dµ +
Z
f dµ <
E−
E+
ǫ
2
(6.46)
for all f ∈ Φ and µ(E) < δ. In other words, Lemma 6.1 holds for subsets of real
measurable functions in L1 (µ).
Next, suppose that Φ is a collection of complex measurable functions in L1 (µ). If
f ∈ Φ, then f = u + iv, where u and v are real measurable functions in L1 (µ). Let
Ψ = {Re (f ) | f ∈ Φ} and
Ω = {Im (f ) | f ∈ Φ}.
Since 0 < |u| ≤ |f | and 0 < |v| ≤ |f |, we see that u, v ∈ L1 (µ) and then Ψ, Ω ⊆ L1 (µ).
Besides, by the estimate (6.46), we have
Z
Z
ǫ
ǫ
|v| dµ <
and
(6.47)
|u| dµ <
2
2
E
E
when µ(E) < δ and for all u ∈ Ψ and v ∈ Ω. By the definition, the sets Ψ and Ω are
uniformly integrable. Since |f | ≤ |u| + |v|, we deduce from the estimates (6.47) that
Z
Z
Z
|v| dµ < ǫ
|u| dµ +
|f | dµ ≤
E
E
E
whenever µ(E) < δ and for all f ∈ Φ. This completes the proof of Lemma 6.1.
Now it is time to return to the proof of the problem.
– Proof of kf k1 < ∞. Since {fn } is uniformly integrable, it follows from Lemma 6.1
that given ǫ > 0, there is a δ > 0 such that
Z
|fn | dµ < ǫ
(6.48)
E
Chapter 6. Complex Measures
198
whenever µ(E) < δ and for all n ∈ N. Employing Theorem 1.28 (Fatou’s Lemma) to
the estimate (6.48), we know that
Z
Z
|fn | dµ ≤ ǫ
(6.49)
|f | dµ ≤ lim inf
n→∞
E
E
whenever µ(E) < δ.
By hypothesis (iii), there exists a E1 ∈ M such that µ(X\E1 ) = 0 and fn (x) → f (x)
pointwisely a.e. on E1 . Since µ(X) < ∞ and hypothesis (iii) again, Egoroff’s Theorem
asserts the existence of a measurable set E2 ⊆ X with µ(X \ E2 ) < δ such that
fn (x) → f (x)
(6.50)
uniformly a.e. on E2 as n → ∞. Since uniform convergence implies pointwise convergence, we may assume that E2 ⊆ E1 . Otherwise, we can replace E2 by the set
E2 ∩ E1 . In this case, De Morgan’s Laws give
µ X \ (E1 ∩ E2 ) = µ (X \ E1 ) ∪ (X \ E2 ) ≤ µ(X \ E1 ) + µ(X \ E2 ) < δ
so that the limit (6.50) also holds on E2 ∩ E1 .
By hypothesis (iv), there exists a E3 ∈ M such that µ(X \ E3 ) = 0 and |f (x)| < ∞
on E3 . Take E = E2 ∩ E3 ⊆ E2 so that the uniform convergence (6.50) also holds on
E. This means that there exists a positive integer N such that n ≥ N implies
|fn (x) − f (x)| < ǫ
a.e. on E and then hypothesis (i) implies
Z
|fn − f | dµ < ǫµ(E) ≤ ǫµ(X) < ∞
(6.51)
E
for all n ≥ N . By De Morgan’s Laws again, we obtain
µ(X \ E) ≤ µ(X \ E2 ) + µ(X \ E3 ) < δ,
so we deduce from the estimates (6.49) that
Z
|f | dµ ≤ ǫ.
(6.52)
X\E
Now using the estimates (6.51), (6.52) and the fact fN ∈ L1 (µ) that
Z
Z
Z
Z
Z
Z
|fN | dµ +
|fN − f | dµ +
|f | dµ ≤
|f | dµ +
|f | dµ =
X
E
E
E
X\E
In other words, we have f ∈ L1 (µ).
X\E
|f | dµ < ∞.
– Proof of kfn − f k1 → 0 as n → ∞. Again the estimates (6.48), (6.51) and (6.52)
together show that
Z
Z
Z
|fn − f | dµ
|fn − f | dµ +
|fn − f | dµ =
X\E
E
X
Z
Z
|f | dµ
|fn | dµ +
< ǫµ(X) +
X\E
X\E
< [2 + µ(X)] · ǫ
for all n ≥ N . Hence we conclude that kfn − f k1 → 0 as n → ∞.
199
6.4. Problems on Uniformly Integrable Sets
(c) For each n ∈ N, define fn : R → R by

1

 , if x ∈ [0, n];
n
fn (x) =


0, otherwise.
Then it is easy to see that
kfn k1 =
Z
R
|fn | dm =
Z
n
0
1
dm = 1,
n
(6.53)
i.e., {kfn k1 } is bounded. Next, for every x ∈ R, we have fn (x) → 0 as n → ∞. So we let
f (x) = 0. Since |fn (x)| ≤ 1 for all n ∈ N and x ∈ R, for each ǫ > 0, if δ = ǫ, then for every
E ∈ B with m(E) < δ, we obtain
Z
Z
|fn | dm ≤ m(E) < ǫ.
fn dm ≤
E
E
Thus {fn } is uniformly integrable. However, the expression (6.53) indicates that
Z
|fn − 0| dm 6= 0.
lim
n→∞ R
Hence hypothesis (i) cannot be omitted in part (b).
(d) We construct two examples showing that hypothesis (iv) is redundant in some cases, but
not in other cases.
– Example 1. Let B be the collection of Borel sets of [0, 1]. We employ the concept
of atomicless measures . A set E ∈ M is called an atom of the measure µ if
µ(E) > 0 and every measurable subset F ⊂ E has measure either 0 or µ(E). If µ has
no atoms, then it is called atomless. Lebesgue measure m on [0, 1] is atomless. In
other words, for every ǫ ∈ (0, 1), if E ∈ B satisfies m(E) = ǫ > 0, then there exists
a F ∈ B and F ⊂ E such that 0 < m(F ) < ǫ. See [9, Definition 1.12.7, Example
1.12.8, p. 55].
We claim that Vitali’s Convergence Theorem holds for m without the hypothesis
(iv). To this end, it suffices to show that hypotheses (i) to (iii) imply hypothesis (iv).
Suppose that
E = {x ∈ [0, 1] | |f (x)| = ∞}.
If m(E) = 0, then there is nothing to prove. So we assume that m(E) > 0. Since
{fn } is uniformly integrable, Lemma 6.1 implies the existence of a δ > 0 such that
Z
|fn | dx < 1
E
whenever m(E) < δ and all n ∈ N. Since m is atomless and m(E) > 0, there exists
a F ∈ B such that 0 < m(F ) < δ and thus
Z
|fn | dx < 1
(6.54)
F
for all n ∈ N. By Theorem 1.28 (Fatou’s Lemma) and the fact that fn (x) → f (x)
a.e. on [0, 1] as n → ∞, we deduce from the inequality (6.54) that
Z
Z
|fn | dx ≤ 1.
(6.55)
|f | dx ≤ lim inf
F
n→∞
F
However, since F ⊂ E and m(F ) > 0, the left-most integral in the inequality (6.55)
is actually ∞, a contradiction. Hence m(E) = 0 or equivalently, |f (x)| < ∞ a.e. on
[0, 1].
Chapter 6. Complex Measures
200
– Example 2. Let M be the collection of all sets E ⊆ R such that either E or E c is
at most countable and define µ(E) = 0 in the first case, µ(E) = 1 in the second. By
Problem 1.6, we see that M is a σ-algebra in R and µ is a measure on M. Clearly, µ
is a finite measure. We define f (x) ≡ ∞ on R and fn : R → R by fn (x) = n for every
n ∈ N so that fn (x) → f (x) as n → ∞ on R, but hypothesis (iv) does not hold.
To each ǫ > 0, we pick δ = 1, so if E ∈ M satisfies µ(E) < 1, then it must be
µ(E) = 0 and thus Proposition 1.24(e) implies that
Z
fn dµ = 0 < ǫ
E
for all n ∈ N. By the definition, {fn } is uniformly integrable. However, it is clear
that f ∈
/ L1 (µ) because Rc = ∅ is at most countable. Consequently, conclusion of
part (b) is false.
(e) Suppose that the hypotheses of Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) hold. Then hypotheses (i) and (iii) are true definitely. On the one hand, since
g ∈ L1 (µ), Problem 1.12 says that each ǫ > 0 there exists a δ > 0 such that
Z
|g| dµ < ǫ
(6.56)
E
whenever µ(E) < δ. On the other hand, |fn (x)| ≤ g(x) for all n ∈ N will imply that
Z
Z
|g| dµ < ∞.
(6.57)
|fn | dµ ≤
X
X
Therefore, {fn } ⊆ L1 (µ) and then we gain from this, Theorem 1.33 and the estimate (6.56)
that
Z
Z
|fn | dµ < ǫ
fn dµ ≤
E
E
whenever µ(E) < δ and all n ∈ N. Hence {fn } is uniformly integrable, i.e., hypothesis
(ii) is true. Since |fn | are measurable by Proposition 1.9(b), it follows from Theorem 1.28
(Fatou’s Lemma) and the estimate (6.57) that
Z
Z
|fn | dµ < ∞.
|f | dµ ≤ lim inf
n→∞
X
X
Thus |f (x)| < ∞ a.e. on X, i.e., hypothesis (iv) is valid. Hence part (b) verifies that
Vitali’s Convergence Theorem implies Theorem 1.34 (Lebesgue’s Dominated Convergence
Theorem) if µ(X) < ∞.
Now we are going to construct an example in which Vitali’s Theorem applies, but
not Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem). For every n ∈ N, we
consider fn : (0, 1) → R defined by
fn (x) =
1
χ 1 1 ≥ 0.
x ( n+1 , n )
Evidently, we have m (0, 1) = 1 and |f (x)| ≤ on (0, 1) which are hypotheses (i) and (iv)
respectively. For each x ∈ (0, 1), we have x ∈
/ χ( 1 , 1 ) for all sufficiently large n so that
n+1 n
fn (x) → 0 = f (x) as n → ∞. Thus we have hypothesis (iii). It is trivial to check that
Z
1
0
|fn | dx =
Z
1
n
1
n+1
1
1
<∞
dx = ln 1 +
x
n
(6.58)
201
6.4. Problems on Uniformly Integrable Sets
so that {fn } ∈ L1 (0, 1) . Given ǫ > 0. One can find a positive integer N such that
ln(1 + n1 ) < ǫ for all n ≥ N . By this result, we observe from the representation (6.58) that
Z
E
fn dx ≤
Z
1
fn dx ≤
0
Z
1
0
|fn | dx < ǫ
for every n ≥ N and every measurable subset E of (0, 1). By part (a), the finite set
{f1 , f2 , . . . , fN −1 } is uniformly integrable. Therefore, these two facts confirm that {fn } is
uniformly integrable, i.e., hypothesis (ii). Hence the set {fn } satisfies all the hypotheses
of Vitali’s Convergence Theorem.
However, for every x ∈ (0, 1), we note from Definition 1.13 that
sup(fn (x)) =
n∈N
1
,
x
/ L1 , g ∈
/ L1 too and this
so if |fn (x)| ≤ g(x) for all n ∈ N, then x1 ≤ g(x). Since x1 ∈
means that {fn } does not satisfy the hypotheses of Theorem 1.34 (Lebesgue’s Dominated
Convergence Theorem).
(f) For each n ∈ N, we consider fn : [0, 1] → R defined by
fn (x) = nχ(0, 1 ) (x) − nχ(1− 1 ,1) (x).
n
n
Notice that each fn is measurable and satisfies fn (0) = fn (1) = 0. If x ∈ (0, 1), then there
exists a positive integer N such that x ∈
/ (0, n1 ) and x ∈
/ (1 − n1 , 1) for all n ≥ N . Thus we
obtain fn (x) = 0 for all n ≥ N . Next,
Z
1
fn (x) dx =
Z
0
0
1
n
n dx −
Z
1
1
1− n
n dx = n ×
1
1
−n× = 0
n
n
for each n ∈ N.
However, {fn } is not uniformly integrable. To see this, assume that there corresponds
a δ > 0 such that
Z
(6.59)
fn dx < 1
E
whenever m(E) < δ and for all n ∈ N. In particular, fix n = N >
so that m(En ) = N1 < δ and
Z
EN
fN dx =
Z
1
N
fN dx =
0
Z
1
N
1
δ
and take EN = (0, N1 )
N dx = 1
0
which contradicts to the inequality (6.59).
(g) If µ(X) = 0, then there is nothing to prove. Without loss of generality, we may assume
that µ(X) > 0. We follow the hint and prove it into several steps.
– Step 1: ρ is a metric (modulo sets of measure 0)c . Define
Z
|χA − χB | dµ ≥ 0.
ρ(A, B) =
X
c
Remember that sets A and B with A = B a.e. on X are treated to be identical.
(6.60)
Chapter 6. Complex Measures
202
Since µ(X) > 0, ρ(A, B) = 0 if and only if χA = χB a.e. on X if and only if A = B
a.e. on X. It is trivial that ρ(A, B) = ρ(B, A). Finally, since
|χA − χB | ≤ |χA − χC | + |χC − χB |
for every A, B, C ∈ M, we must have
ρ(A, B) ≤ ρ(A, C) + ρ(C, B).
By the definition, ρ is a metric (modulo sets of measure 0).
– Step 2: (M, ρ) is a complete metric space (modulo sets of measure 0). Let
{En } ⊆ M be a Cauchy sequence, i.e., given ǫ > 0, there exists a positive integer N
such that n, m ≥ N imply that
ρ(En , Em ) < ǫ.
We note from the definition (6.60) that ρ(A, B) = kχA − χB k1 , so we obtain
kχEn − χEm k1 < ǫ
for all n, m ≥ N , i.e., {χEn } is Cauchy in L1 (µ). By Theorem 3.11, there exists
f ∈ L1 (µ) such that {χEn } converges to f in L1 (µ). By Theorem 3.12, {χEn } has a
subsequence {χEnk } such that
χEnk (x) → f (x)
pointwise a.e. on X as n → ∞. By the definition of χEnk , we get immediately that
either f (x) = 0 or 1 for almost every x ∈ X. Now if we define E = {x ∈ X | f (x) = 1},
then E ∈ M, f = χE a.e. on X and therefore
ρ(En , E) = kχEn − χE k1 → 0
as n → ∞. Hence (M, ρ) is a complete metric space (modulo sets of measure 0).
R
– Step 3: The map E → E fn dµ is continuous for each n. Denote this map
by ϕn . Recall that fn ∈ L1 (µ), so Problem 1.12 asserts that for every ǫ > 0, there
corresponds a δ > 0 such that
Z
E
|fn | dµ < ǫ
(6.61)
whenever µ(E) < δ. Suppose that E, F ∈ M such that ρ(E, F ) < δ. We deduce from
the fact E ∪ F = (E \ F ) ∪ (E ∩ F ) ∪ (F \ E) that
Z
|χE − χF | dµ
ρ(E, F ) =
X
Z
|χE − χF | dµ
=
Z
ZE∪F
|χE − χF | dµ
|χE − χF | dµ +
=
E\F
= µ(E \ F ) + µ(F \ E)
F \E
(6.62)
so that µ(E \ F ) < δ and µ(F \ E) < δ. Since (χE − χF )fn ∈ L1 (µ), Theorem 1.33
and the estimate (6.61) imply that
Z
Z
fn dµ
fn dµ −
|ϕn (E) − ϕn (F )| =
E
F
203
6.4. Problems on Uniformly Integrable Sets
Z
=
ZX
=
Z
≤
X
ZX
=
χE fn dµ −
χF fn dµ
X
(χE − χF )fn dµ
|χE − χF | · |fn | dµ
Z
|fn | dµ
|fn | dµ +
E\F
< 2ǫ.
Z
F \E
Hence ϕn is continuous on M for every n ∈ N.
– Step 4: Completion of the proof. Now {ϕn } is a sequence of continuous complex
functions on the complete metric space (M, ρ). Denote
ϕ(E) = lim ϕn (E)
n→∞
as a complex number for every E ∈ M. If ǫ > 0, it follows from Problem 5.13
(particularly the inequality (5.62)) that there exist E0 ∈ M, δ > 0 and N ∈ N such
that
Z
(6.63)
(fn − fN ) dµ = |ϕn (E) − ϕN (E)| < ǫ
E
for all E ∈ M with ρ(E, E0 ) < δ and n ≥ N .
If µ(A) < δ and B = E0 \ A, then the expression (6.62) indicates that
ρ(B, E0 ) = µ (E0 \ A) \ E0 + µ(E0 \ (E0 \ A)) = µ(E0 ∩ A) ≤ µ(A) < δ.
Similarly, if C = E0 ∪ A, then we know that
ρ(C, E0 ) = µ (E0 ∪ A) \ E0 + µ(E0 \ (E0 ∪ A)) = µ(A) < δ.
Thus the estimate (6.63) holds with B and C in place of E.
Since E0 ∪ A = (E0 \ A) ∪ A and (E0 \ A) ∩ A = ∅, we get
Z
Z
Z
(fn − fN ) dµ
(fn − fN ) dµ = (fn − fN ) dµ +
E0 ∪A
E0 \A
A
which implies that
Z
Z
Z
(fn − fN ) dµ
(fn − fN ) dµ −
(fn − fN ) dµ =
E0 \A
E0 ∪A
A
Z
Z
(fn − fN ) dµ
(fn − fN ) dµ +
≤
< 2ǫ
E0 ∪A
E0 \A
(6.64)
for all n > N . By part (a), the set {f1 , f2 , . . . , fN } is uniformly integrable, i.e., there
corresponds a δ′ > 0 such that
Z
fn dµ < ǫ
(6.65)
A
whenever µ(A) < δ′ and n = 1, 2, . . . , N . Define δ′′ = min(δ, δ′ ). If µ(A) < δ′′ ,
then since µ(A) < δ′ , the inequality (6.65) still holds for n = 1, 2, . . . , N . Next, for
Chapter 6. Complex Measures
204
n = N + 1, N + 2, . . ., since µ(A) < δ, it yields from the estimates (6.64) and (6.65)
that
Z
Z
Z
fN dµ < 3ǫ.
(6.66)
(fn − fN ) dµ +
fn dµ ≤
A
A
A
By combining the results (6.65) and (6.66), we conclude that
Z
fn dµ < 3ǫ
A
if µ(A) < δ′′ and for all n ∈ N, i.e., {fn } is uniformly integrable.
We have completed the proof of the problem.
Problem 6.11
Rudin Chapter 6 Exercise 11.
Proof. We note that if µ(X) = 0, then Proposition 1.24(e) shows that the result holds trivially.
Therefore, without loss of generality, we may assume that µ(X) 6= 0. Now we want to apply
Problem 6.10(b). By our hypotheses, it suffices to show that Φ = {fn } is uniformly integrable
and |f (x)| < ∞ a.e. on X.
Since fn ∈ L1 (µ) for every n ∈ N, we apply Theorem 1.33 and then Theorem 3.5 (Hölder’s
Inequality) to obtain
Z
E
fn dµ ≤
Z
E
|fn | dµ ≤
nZ
E
|fn |p dµ
o1 n Z
p
E
dµ
o1
q
1
1
< C p × µ(E) q
(6.67)
−q
for every E ∈ M. Since p > 1, q < ∞. Take δ = ǫq C p . If µ(E) < δ, then it follows from the
inequality (6.67) that
Z
fn dµ < ǫ.
E
Thus the set Φ = {fn } is uniformly integrable.
It remains to show that |f (x)| < ∞ a.e. on X. Since fn (x) → f (x) a.e. on X, |fn (x)| → |f (x)|
a.e. on X. By Theorem 1.28 (Fatou’s Lemma) and the inequality (6.67), we know that
Z
Z
1
1
|fn | dµ < C p × µ(X) q < ∞.
(6.68)
|f | dµ ≤ lim inf
X
n→∞
X
Since µ(X) 6= 0, the inequality (6.68) ensures that |f (x)| < ∞ a.e. on X. Hence we deduce
from Problem 6.10(b) that
Z
|f − fn | dµ = 0
lim
n→∞ X
holds, completing the proof of the problem.
6.5
Dual Spaces of Lp (µ) Revisit
Problem 6.12
Rudin Chapter 6 Exercise 12.
6.5. Dual Spaces of Lp (µ) Revisit
205
Proof. We note that Problem 1.6 establishes that M is a σ-algebra. Let E = ( 14 , 12 ) which is open
in [0, 1]. It is clear that g −1 (E) = E. Since E and E c are uncountable, we have g−1 (E) ∈
/ M
and thus g is not M-measurable by Definition 1.3(c). We have completed the proof of the
problem.
Problem 6.13
Rudin Chapter 6 Exercise 13.
Proof. Let C(I) be the space of all continuous functions on I. It is clear that χ[0,δ] ∈ L∞ but
χ[0,δ] ∈
/ C(I) for some δ > 0. We need the following result:
Lemma 6.2
The space C(I) is closed in the metric space L∞ .
Proof of Lemma 6.2. Let {fn } ⊆ C(I) and {fn } converges to f in L∞ . Given ǫ > 0.
It means that there exists a positive integer N such that n ≥ N implies
|fn (x) − f (x)| ≤ sup |fn (x) − f (x)| = kfn − f k∞ < ǫ
x∈I
for all x ∈ I. In other words, {fn } converges uniformly to f on I and thus f is
continuous on I (see [49, Theorem 7.12, p. 150]), i.e., f ∈ C(I). Hence C(I) is closed
in L∞ , completing the proof of the lemma.
We return to the proof of the problem. Since L∞ is a Banach space, we get from Lemma 6.2
and the equivalent form of Theorem 5.19 that there exists a bounded linear functional Λ on L∞
such that
Λ(f ) = 0
(6.69)
for every f ∈ C(I), but Λ(χ[0,δ] ) 6= 0.
Assume that there was a g ∈ L1 (µ) such that
Z
Λ(f ) = f g dm
(6.70)
I
for every f ∈ L∞ . In particular, the result (6.69) and the representation (6.70) imply that
Z
f g dm = 0
(6.71)
I
on C(I). Since f ∈ C(I) and g ∈ L1 (µ), we have f g ∈ L1 (µ) and then we apply Theorem
1.39(b) to the integral (6.71) to conclude that f g = 0 a.e. on I. If we take f (x) = ex on I which
belongs to C(I), then we obtain the fact that
g=0
a.e. on I. Hence we follow from the representation (6.70) that the result (6.69) actually holds
on the whole L∞ , a contradiction to the fact that Λ(χ[0,δ] ) 6= 0. This completes the proof of the
problem.
Chapter 6. Complex Measures
206
CHAPTER
7
Differentiation
7.1
Lebesgue Points and Metric Densities
Problem 7.1
Rudin Chapter 7 Exercise 1.
Proof. Let x be a Lebesgue point of f . Since f ∈ L1 (Rk ), we deduce from the triangle inequality
and Theorem 1.33 that
Z
Z
1
1
f (y) dy +
f (y) dy
|f (x)| ≤ f (x) −
m(Br ) Br
m(Br ) Br
Z
Z
1
1
[f (x) − f (y)] dy +
|f (y)| dy
≤
m(Br ) Br
m(Br ) Br
Z
1
≤
|f (y) − f (x)| dy + (M f )(x).
(7.1)
m(Br ) Br
Since x is a Lebesgue point of f , if we let r → 0+ in the inequality (7.1), then we obtain
|f (x)| ≤ (M f )(x)
as required, completing the proof of the problem.
Problem 7.2
Rudin Chapter 7 Exercise 2.
Proof. We have I = (−δ, δ). Now we are going to prove the existence of a measurable set E ⊆ R
such that
m E ∩ I(δ)
m E ∩ I(δ)
= α and lim sup
= β.
(7.2)
lim inf
δ→0
2δ
2δ
δ→0
In fact, it suffices to prove that
m E ∩ [0, δ)
=α
lim inf
δ→0
δ
and
m E ∩ [0, δ)
lim sup
= β,
δ
δ→0
(7.3)
where E ⊆ [0, ∞) because we can get the desired result (7.2) by reflecting about the origin. To
start our proof, we consider five cases:
207
Chapter 7. Differentiation
208
• Case (i): 0 < α < β < 1. In this case, we consider the numbers
θ=
α(1 − β)
α
< <1
β(1 − α)
β
β−α n
·θ >0
1−α
and dn =
for every n ∈ N. Then it is easy to see that
h
β − α n dn
α(1 − β) i
=
θ =
> dn .
θ n − θ n+1 = θ n 1 −
β(1 − α)
β(1 − α)
β
(7.4)
Next, for each n = 1, 2, . . ., we define the closed interval
En = [θ n − dn , θ n ].
We claim that the measurable set
E=
∞
[
En
n=1
satisfies the requirement (7.3). To see this, we first note from the inequality (7.4) that
dn → 0 as n → ∞ and En ∩ En+1 = ∅ for every n ∈ N, see Figure 7.1 below:
Figure 7.1: The closed intervals En and En+1 .
Next, choose δ′ > δ > 0 as shown in Figure 7.1, then it is easy to know that
n
[0, θ − dn ) ∩ E =
∞
[
k=n+1
Ek ,
n
[0, θ ) ∩ E =
∞
[
Ek ,
k=n
and
[0, δ′ ) ∩ E = [θ n − dn , δ′ ) ∪
∞
[
[0, δ) ∩ E =
∞
[
k=n+1
Ek
k=n+1
so that
∞
X
m [0, θ n − dn ) ∩ E
θ
β−α
1−α
·
= α,
dk =
=
θ n − dn
(1 − β)θ n
1−β 1−θ
k=n+1
∞
n
X
m [0, θ ) ∩ E
1
β−α
1
·
= β,
dk =
= n
n
θ
θ
1−α 1−θ
k=n
∞
m [0, θ n+1 ) ∩ E)
m [0, δ) ∩ E
1 X
βθ n+1
=
=
dk =
δ
δ
δ
δ
k=n+1
Ek
209
7.1. Lebesgue Points and Metric Densities
and
m [0, δ′ ) ∩ E
m [0, θ n ∩ E − m [δ′ , θ n ]
βθ n + δ′ − θ n
θn
=
=
=
1
−
(1
−
β)
.
δ′
δ′
δ′
δ′
Therefore, the above analysis shows trivially that the function f : (0, ∞) → R defined by
f (δ) =
m([0, δ) ∩ E)
δ
(7.5)
attains the values α and β at the end-points of each En , decreases from β to α in the
interval [θ n+1 , θ n − dn ] and increases from α to β in En . Hence, we have
α = min f (x) and
x∈(0,δ)
β = max f (x)
x∈(0,δ)
which mean that the limits (7.3) hold.
• Case (ii): α = 0 and β = 1. In this case, we defined the measurable set
∞ h
i
[
1
1
E=
.
,
(2n)! (2n − 1)!
n=1
By similar analysis as in Case (i), we see that
1
h
)
m E ∩ [0, (2n)!
1
1
1
≤
(2n)!
×
m
0,
= (2n)! ×
=
0≤
1
(2n + 1)!
(2n + 1)!
2n + 1
(2n)!
and
m E ∩ [0, 1 )
h 1
2n − 1
1
(2n−1)!
≤ 1.
≤
= (2n − 1)! × m
,
1
2n
(2n)! (2n − 1)!
(2n−1)!
Hence we have proven the limits (7.3) in this case.
• Case (iii): 0 ≤ α = β ≤ 1. For each n ∈ N, we take
h α
1 − α 1i
En =
+
, .
n+1
n
n
Thus it is easy to see that En ∩ En+1 = ∅ for all n ∈ Z. We define E =
δ = n1 , then we follow similar argument as in Case (i) that
∞
∞
h 1 X
X
1
m(Ek ) = α
=
.
m E ∩ 0,
n
k(k + 1)
k=n
Since
N
X
k=n
∞
[
En . If we take
n=1
(7.6)
k=n
1
1
1
= − , we have
k(k + 1)
n N
∞
X
k=n
1
1
= .
k(k + 1)
n
Now we put the infinite series (7.7) into the expression (7.6) to get
m E ∩ [0, n1 )
=α
1
n
which induces exactly the limits (7.3).
(7.7)
Chapter 7. Differentiation
210
• Case (iv): 0 = α < β < 1. Similar to the construction in Case (i), we suppose that
En =
h1 − β
(2n)!
β
1 i
,
(2n + 2)! (2n)!
+
where n ∈ N. Similarly, we have En ∩ En+1 = ∅ for all n ∈ N. Let E =
one hand, since
(7.8)
∞
[
En . On the
n=1
∞ h
h 1−β
i
X
1
β
β
1
m E ∩ 0,
=β
=
+
−
,
(2n)!
(2n + 2)!
(2k)! (2k + 2)!
(2n + 2)!
k=n+1
we have
1−β
m E ∩ [0, (2n)!
+
1−β
(2n)!
+
β
)
(2n+2)!
β
(2n+2)!
=
β
(2n+2)!
β
1−β
(2n)! + (2n+2)!
=
β
→0
(1 − β)(2n + 2)(2n + 1) + β
as n → ∞. On the other hand, we have
h
m E ∩ 0,
we have
i
Xh 1
1 β
1
=β
=
−
,
(2n)!
(2k)! (2k + 2)!
(2n)!
∞
k=n
1
)
m E ∩ [0, (2n)!
1
(2n)!
=
β
(2n)!
1
(2n)!
= β.
1−β
β
1
, (2n)!
+ (2n+2)!
] and
Besides, the function (7.5) decreases from β to 0 in the interval [ (2n+2)!
increases from 0 to β in En . Hence these imply the results (7.3).
• Case (v): 0 < α < β ≤ 1. Instead of the intervals (7.8), we consider the intervals
En =
h 1
α
1−α i
,
,
+
(2n)! (2n)! (2n + 2)!
where n ∈ N. Thus it can be shown similar to Case (iv) that the limits (7.3) hold for the
∞
[
En and we omit the details here.
measurable set E =
n=1
This completes the proof of the problem.
Remark 7.1
The limits defined in (7.2) are called the upper and lower densities of E at 0 respectively.
7.2
Periods of Functions and Lebesgue Measurable Groups
Problem 7.3
Rudin Chapter 7 Exercise 3.
211
7.2. Periods of Functions and Lebesgue Measurable Groups
Proof. We follow Rudin’s hint. Pick α ∈ R and put
F (x) = m(E ∩ [α, x])
for x > α. If y > x > α + pi , then we deduce from Theorem 2.20(c) and the fact
[α − pi , α + pi ] = [α − pi , x] \ (α + pi , x]
that
F (x + pi ) − F (x − pi ) = m(E ∩ [α, x + pi ]) − m(E ∩ [α, x − pi ])
= m(E ∩ [α, x + pi ] − pi ) − m(E ∩ [α, x − pi ] + pi )
= m(E ∩ [α − pi , x]) − m(E ∩ (α + pi , x])
= m(E ∩ [α − pi , α + pi ])
= F (y + pi ) − F (y − pi ).
(7.9)
If F is differentiable at x, then we have
1 h F (x + pi ) − F (x) F (x − pi ) − F (x) i
F (x + pi ) − F (x − pi )
= lim
+
= F ′ (x). (7.10)
i→∞ 2
i→∞
2pi
pi
−pi
lim
Combining the results (7.9) and (7.10), we see that F ′ (x) = F ′ (y) for all x, y > α and thus
F ′ (x) = cα
(7.11)
for all x > α, where cα is a constant. Next, we recall from the definition of F that
Z x
χE (t) dt.
F (x) = m(E ∩ [α, x]) =
α
Since χE ∈ L1 (R), it establishes from Theorem 7.11 that F is differentiable and F ′ (x) = χE (x)
a.e. for x > α. Using this and the fact (7.11), we get
χE (x) = cα
(7.12)
a.e. for x > α. Since χE takes only 0 or 1, we see that cα ∈ {0, 1}. If x > β > α, then
cβ = F ′ (x) = χE (x) = F ′ (x) = cα ,
so cα does not depend on α and we can simply replace cα by c in the results (7.11) and (7.12).
Finally, if c = 0, then we conclude from the result (7.12) that
m(E) =
Z
R
χE dm = 0.
Otherwise, c = 1 and the result (7.12) implies that χE c (x) = 0 and hence m(E c ) = 0. We have
completed the proof of the problem.
Problem 7.4
Rudin Chapter 7 Exercise 4.
Chapter 7. Differentiation
212
Proof. Our proof does not follow Rudin’s hint. In fact, we mainly follow the argument of Cignoli
and Hounie [15] and now we first quote one of their results that will be used very soon:
Lemma 7.1
Let D be a dense subset of R and µ a Borel measure on R such that
µ d + [a, b) = µ [a, b)
for every d ∈ D and a, b ∈ R. Then we have µ = km, where k = µ (0, 1] .
Next, it is clear that the set G(f ) = {ns + mt | n, m ∈ Z} is an additive group of R. Without
loss of generality, we assume that t > 0. Since α = s/t is irrational, it follows from the
Kronecker’s Approximation Theorema that for every θ ∈ R and every ǫt−1 > 0, there exists
n, m ∈ Z such that
θ
ǫ
nα + m −
<
t
t
which is equivalent to saying that
|ns + mt − θ| < ǫ.
Hence G(f ) is dense in R.
We first consider the special case that f is nonnegative and bounded a.e. on R. Note that
if F (x) = f (xt), then F (x + 1) = f (xt + t) = f (xt) = F (x), so F is a function of period 1 and
we may assume that f has period 1. Consequently, it suffices to prove the problem when f is a
nonnegative, bounded a.e. function on [0, 1] having period 1. Suppose that µ : B → [0, ∞) is
defined by
Z
f (x) dx
µ(E) =
(7.13)
E
for all E ∈ B. By Theorem 1.29, µ is a Borel measure. Let p be a period of f . Then we deduce
from the representation (7.13) that
Z
Z
Z
f (x) dx = µ(E).
f (x + p) dx =
f (x) dx =
µ(E + p) =
E
E
E+p
In other words, µ has the same periods as f . Furthermore, if p ∈ G(f ), then we obtain from
Theorem 7.26 (The Change-of-variables Theorem)b that
Z b+p
Z b
Z b
µ p + [a, b) = µ [a + p, b + p) =
f (x) dx =
f (x + p) dx =
f (x) dx = µ [a, b) .
a+p
a
a
By Lemma 7.1, we have
µ = km,
(7.14)
where 0 < k = µ (0, 1] < ∞. By combining the results (7.13) and (7.14), we establish that
Z
[f (x) − k] dx = 0
E
for every measurable
subset E of [0, 1]. Since f is bounded a.e. on [0, 1], it is clear that
f − k ∈ L1 [0, 1] . Now Theorem 1.39(b) implies that f = k a.e. on [0, 1] and the periodicity of
f shows that
f =k
a
b
See [63, Lemma 4.6, p. 82]
It is applied to the third equality. Recall that we are discussing Lebesgue integral, not Riemann integral.
213
7.2. Periods of Functions and Lebesgue Measurable Groups
a.e. on R in this special case.
For the general case, the composite function g(x) = π2 + tan−1 f (x) is clearly a measurable
function on R. Furthermore, g is also
a.e. on R. If f (x + 1) = f (x),
nonnegative and bounded
then g(x + 1) = π2 + tan−1 f (x + 1) = π2 + tan−1 f (x) = g(x) so that g is also a function of
period 1. Applying the previous part to g to conclude that g = k a.e. on R for a constant k and
this implies that
π
f (x) = tan k −
2
a.e. on R.
Finally, if we define G = {ns + mt | m, n ∈ Z} and f : R → R by

 0, if x ∈ G;
f (x) =

1, otherwise,
then it is trivial that f has periods s and t as well as f = 1 a.e. on R.c This completes the
proof of the problem.
Problem 7.5
Rudin Chapter 7 Exercise 5.
Proof. We prove the assertions one by one.
• A + B contains a segment. Since m(A) > 0 and m(B) > 0, A and B are Lebesgue
measurable. By the comment in §7.12, the metric densities of A and B are 1 at almost
every point of A and B. Thus we pick a0 ∈ A and b0 ∈ B such that
m B ∩ B(b0 , r)
m A ∩ B(a0 , r)
= lim
= 1.
(7.15)
lim
r→0+
r→0+
m B(a0 , r)
m B(b0 , r)
Since χA , χB ∈ L1 (R), we may assume with the aid of Theorem 7.7 that a0 and b0 are
Lebesgue points of χA and χB respectively.
Set c0 = a0 + b0 . For each ǫ > 0, define
Bǫ = {c0 + ǫ − b | b ∈ B and |b − b0 | < δ},
(7.16)
where δ > 0 is small. For every c0 + ǫ − b, we have
|c0 + ǫ − b − (a0 + ǫ)| = |c0 + ǫ − b − (c0 − b0 + ǫ)| = |b − b0 | < δ
which implies that Bǫ ⊆ (a0 + ǫ − δ, a0 + ǫ + δ). If we take δ = 2ǫ and ǫ is sufficiently
small, then we have
Bǫ ⊆ (a0 + ǫ − δ, a0 + ǫ + δ) = (a0 − ǫ, a0 + 3ǫ) ⊆ (a0 − 4ǫ, a0 + 4ǫ) = B(a0 , 4ǫ). (7.17)
By the definition (7.16), we note that
Bǫ − (c0 + ǫ) = {−b | b ∈ B and |b − b0 | < 2ǫ} = B ∩ B(b0 , 2ǫ).
c
Since Bǫ − (c0 + ǫ) is just a translate of Bǫ , Theorem 2.20(c) says that
m(Bǫ ) = m Bǫ − (c0 + ǫ) = m B ∩ B(b0 , 2ǫ) .
By Problem 7.6, we have m(G) = 0.
(7.18)
Chapter 7. Differentiation
214
By the metric density of B at b0 in (7.15), we know that
m B(b0 , 2ǫ)
m B ∩ B(b0 , 2ǫ) >
= 2ǫ.
2
(7.19)
m(Bǫ ) > 2ǫ > 0.
(7.20)
Combining the expression (7.18) and the inequality (7.19), we arrive at
We claim that A ∩ Bǫ 6= ∅ for all small enough ǫ > 0. Otherwise, no matter how small
ǫ > 0 is, we can find a η ∈ (0, ǫ) such that A ∩ Bη = ∅. In fact, we may pick η small
enough so that the metric density of A at a0 in (7.15) implies
7
m A ∩ B(a0 , 4η) > m B(a0 , 4η) = 7η.
8
(7.21)
Since A ∩ Bη = ∅, we have A ∩ [B(a0 , 4η) \ Bη ] = [A ∩ B(a0 , 4η)] \ [A ∩ Bη ] = A ∩ B(a0 , 4η)
so that
m A ∩ B(a0 , 4η) = m A ∩ [B(a0 , 4η) \ Bη ] ≤ m B(a0 , 4η) \ Bη .
(7.22)
Recall from the set relation (7.17), we have the fact B(a0 , 4η) = [B(a0 , 4η) \ Bη ] ∪ Bη . By
this fact and the inequality (7.20), we may further reduce the inequality (7.22) to
m A ∩ B(a0 , 4η) ≤ m B(a0 , 4η) − m(Bη ) < 8η − 2η = 6η.
(7.23)
However, the inequalities (7.21) and (7.23) are contradictory and this proves our claim
that A ∩ Bǫ 6= ∅ for all small ǫ > 0.
Pick x ∈ (c0 − ǫ′ , c0 + ǫ′ ) for sufficiently small ǫ′ and without loss of generality, suppose
that ǫ = x − c0 > 0. Then we have 0 < ǫ < ǫ′ and the above claim implies that
A ∩ Bǫ 6= ∅.
Choose a ∈ A ∩ Bǫ . By the definition (7.16) and then using the fact ǫ = x − c0 , we have
a = c0 + ǫ − b = x − b
for some b ∈ B with |b − b0 | < 2ǫ. In other words, we have the representation
x = a + b ∈ A + B,
i.e., (c0 − ǫ′ , c0 + ǫ′ ) ⊆ A + B.
•
C
2
+
C
2
= [0, 1]. Recall from [49, Exercise 19, p. 81] that if c ∈ C, then
c=
∞
X
αn
n=1
3n
,
where αn = 0 or 2. On the one hand, let x ∈ [0, 1] and we have
x=
∞
X
an
n=1
3n
,
where an ∈ {0, 1, 2} for all n = 1, 2, . . .. Define


 0, if an ∈ {0};
 0, if an ∈ {0, 1};
and γn =
βn =


1, otherwise.
1, otherwise,
(7.24)
215
7.2. Periods of Functions and Lebesgue Measurable Groups
Then it is always true that an = βn + γn for all n = 1, 2, . . .. Thus the x given in (7.24)
can be represented as
x = y + z.
Here y and z are given by
y=
∞
X
βn
n=1
3n
and
z=
∞
X
γn
n=1
3n
,
(7.25)
where βn , γn ∈ {0, 1} for all n = 1, 2, . . ..d In other words, the analysis implies that
[0, 1] ⊆ 21 C + 21 C.
On the other hand, let y, z ∈ 12 C be expressed by the series (7.25), where βn , γn ∈ {0, 1}.
Then we have
∞
X
βn + γn
y+z =
,
3n
n=1
where βn + γn ∈ {0, 1, 2} for all n = 1, 2, . . .. If βn + γn = 0 for all n = 1, 2, . . ., then
y + z = 0. If βn + γn = 2 for all n = 1, 2, . . ., then
∞
X
2
= 1.
y+z =
3n
n=1
Otherwise, y + z ∈ (0, 1). Therefore, we have 12 C + 12 C ⊆ [0, 1]. Hence we have obtained
the required result that 21 C + 12 C = [0, 1].
We have completed the proof of the problem.
Remark 7.2
The result in Problem 7.5 is called the Steinhaus Theorem. It has a generalization in Rk
which says that if A ∈ B k and mk (A) > 0, then 0 is an interior point of A − A, see [5,
Theorem 26.6, p. 163] for details.
Problem 7.6
Rudin Chapter 7 Exercise 6.
Proof. Assume that m(G) > 0. By Problem 7.5, G + G contains a segment (a, b). Since G is
a group with addition, we have G + G ⊆ G so that (a, b) ⊆ G. Let m = a+b
2 ∈ (a, b) so that
m ∈ G. Since G is a group, −m ∈ G so that (a − m, b − m) ⊆ G. Since m is the mid-point of a
and b, the segment (a − m, b − m) is in the form (−c, c) for some fixed c ∈ R \ {0}. Recall that
G + G ⊆ G, so we have (−2c, 2c) ⊆ G. Repeat this kind of argument, we may conclude that
(−nc, nc) ⊆ G
for all n ∈ N, but this implies that G = R, a contradiction. Hence m(G) = 0 and this completes
the proof of the problem.
d
Since the series (7.25) converge, we can add them together.
Chapter 7. Differentiation
7.3
216
The Cantor Function and the Non-measurability of f ◦ T
Problem 7.7
Rudin Chapter 7 Exercise 7.
Proof. The construction has been done in [26, Example 15, pp. 96 – 98]. We won’t repeat it
here. See also [26, Example 30, pp. 105, 195].
Problem 7.8
Rudin Chapter 7 Exercise 8.
Proof. We have T : V = (a, b) → R. Since V is a bounded segment, we define ϕ(a) = ϕ(b) = 0
so that the domain of T can be extended to [a, b] easily.
(a) Obviously, V is open in R and Lebesgue measurable. Since 0 ≤ ϕn (x) < 2−n in V and
∞
X
2−n < ∞, the Weierstrass M -test [49, Theorem 7.10, p. 148] indicates that
n=1
ϕ=
∞
X
ϕn
(7.26)
n=1
converges uniformly on V . Since each ϕn is continuous on V , ϕ is continuous on V .
Furthermore, 0 ≤ ϕ(x) ≤ 1 on V so that ϕ is also bounded on [a, b]. By [49, Theorem
11.33, p. 323], we see that ϕ ∈ R on [a, x] and
Z x
Z x
ϕ(t) dt
ϕ(t) dt = R
T (x) =
a
a
for every x ∈ [a, b]. Therefore, it follows from the First Fundamental Theorem of Calculus
that T is differentiable at every point of the continuity of ϕ in [a, x]. Consequently, T is
differentiable (and hence continuous) at every point of V .
Next, if T (x) = T (y) for x < y, then we have
Z
Z y
ϕ(t) dt = T (x) +
T (y) =
ϕ(t) dt
x
a
which implies that
y
Z
y
ϕ(t) dt = 0.
(7.27)
x
Since W is dense in V , (x, y) ∩ W 6= ∅ so that (x, y) ∩ WN 6= ∅ for some positive integer
N . Let p ∈ (x, y) ∩ WN . By the definition, we have ϕ(p) > 0 and the sign-preserving
property of continuous functions (see [64, Problem 7.15, p. 112]) assures that there exists
a δ > 0 such that (p − δ, p + δ) ⊆ (x, y) ∩ Wn so that ϕ(q) > 0 for all q ∈ (p − δ, p + δ), but
this implies that
Z p+δ
Z y
ϕ(t) dt > 0,
ϕ(t) dt ≥
x
p−δ
a contradiction to the result (7.27). Thus T is one-to-one and condition (ii) is satisfied.
Since T is differentiable at every point of V , it is continuous on V and condition (i) is also
satisfied. Condition (iii) is obvious because X \ X = ∅.
217
7.3. The Cantor Function and the Non-measurability of f ◦ T
(b) By the First Fundamental Theorem of Calculus, we know that T ′ = ϕ which is continuous
on V . If p ∈ K, then p ∈
/ Wn for all n ∈ N which implies that ϕn (p) = 0 for all n ∈ N.
Therefore, we have
∞
X
′
ϕn (p) = 0.
(7.28)
T (p) = ϕ(p) =
n=1
For the third assertion, we first show that T (K) is a Borel set. To this end, since T is
injective, we note from [42, Exercise 2(h), p. 21] that
T (K) = T (V \ W ) = T (V ) \ T (W ) = T (V ) \
∞
[
T (Wn ).
n=1
Since T is continuous and injective on V , the property of Invariance of Domain [32,
Theorem 2B.3, p. 172] shows that T is an open map. Thus T (V ) and each T (Wn ) is open
in R so that T (K) ∈ B, as desired.
By the previous result, the function f = χT (K) is Lebesgue measurable. By part (a), T
satisfies the hypotheses of Theorem 7.26 (The Change-of-variables Theorem), so we have
Z
Z
Z
(7.29)
f dm = (f ◦ T )|JT | dm,
χT (K) dm =
m T (K) =
T (V )
T (V )
V
where |JT (x)| denotes the Jacobian of T at x. Recall that T is injective, so T (x) ∈ T (K)
if and only if x ∈ K and then we reduce from the expression (7.29) that
Z
Z
χK (x)|JT | dm.
(7.30)
χT (K) T (x) |JT | dm =
m T (K) =
V
K
Notice from the result (7.28) that |JT (x)| = | det T ′ (x)| = 0 on K, so we obtain from the
integral (7.30) that m T (K) = 0.
(c) Since m(K) > 0, it follows from Corollary to Theorem 2.22 that K contains a nonmeasur
able subset E. Let A = T (E). Since E ⊂ K, we have A ⊂ T (K). Since m T (K) = 0, we
recall from Theorem 1.36 that all subsets of sets of measure 0 are measurable, thus A is a
measurable set and χA is a measurable function. By the definition, we know that

 1, if x ∈ E;
χA T (x) = χT (E) T (x) =

0, otherwise.
In other words, we have χA ◦ T = χE . Since E is nonmeasurable, χE is not measurable.
(d) By part (b), we know that T ′ (x) = ϕ(x) on V . Thus if we can construct a ϕ which
is infinitely differentiable, then we are done. To see this, we are going to construct an
infinitely differentiable function ϕn on Wn for each positive integer n. Let’s recall the
bump function ψ : R → R defined by

1 
, if x ∈ (−1, 1);
 exp −
1 − x2
ψ(x) =


0,
otherwise.
The ψ has derivatives of all orders in R, i.e., ψ is an element of C ∞ (R) the space of all
infinitely differentiable functions on R. Since each Wn is a segment, we let Wn = (an , bn )
for some an , bn ∈ (a, b) with an < bn . Therefore the bump function ϕn : V → R given by
ϕn (x) = 2−n ψ
x − an +bn 2
2
(x − bn ) + 1 = 2−n ψ
bn −an
bn − a n
2
Chapter 7. Differentiation
218
is what we want. Put U1 = W1 and Un = Wn \ (W1 ∪ W2 ∪ · · · ∪ Wn−1 ) for n = 2, 3, . . ..
Then we have Ui ∩ Uj = ∅ for all i 6= j and their union is also W . Since each Wn is a
segment, each Un is a union of at most countable disjoint segments. In other words, we
may assume that Wi ∩ Wj = ∅ for all i 6= j. Therefore, the function ϕ : V → R defined
by (7.26) satisfies

 ϕn (x), if x ∈ Wn ⊆ W ;
ϕ(x) =

0,
if x ∈ V \ W .
Hence, as we have mentioned above, ϕ is infinitely differentiable in V and we complete the
construction of the required integral T .
This completes the proof of the problem.
7.4
Problems related to the AC of a Function
Problem 7.9
Rudin Chapter 7 Exercise 9.
Proof. The function f discussed in §7.16 is called a singular function (see [8, p. 161] or [34,
Definition 7.1.44, p. 170]). Let 0 < α < 1. Pick t so that tα = 2. Let δn = ( 2t )n = 2n t−n for all
n ∈ N.
We remark that comprehensive materials about the relations between singular functions and
Cantor sets can be found in [33, Chap. 4] and [34, Chap. 8]. In fact, our proof here relies
heavily on the known results there.
We fix some notations first. Let n ≥ 0. At the nth step, we have
n
En =
2
[
In (k),
(7.31)
k=1
where In (1), In (2), . . . , In (2n ) are the 2n disjoint closed intervals, each of length 2−n δn = t−n .
Next the deleted open segment between the intervals In (k) and In (k +1) is denoted by J(k ·2−n ),
where k = 1, 2, . . . , 2n − 1. Suppose that
D = {m · 2−n | 0 ≤ m ≤ 2n and n ≥ 0}
which is the set of dyadic rationals in [0, 1].
We claim that the function f discussed in §7.16 satisfies
f (x) = k · 2−n
(7.32)
for all x ∈ J(k · 2−n ) and k · 2−n ∈ D. To this end, the hypothesis x ∈ J(k · 2−n ) implies that
there are k closed disjoint intervals In (1), In (2), . . . , In (k) on its left-hand side, so we derive from
this and the fact [51, Eqn. (5), p. 145] that
fn (x) =
Z
0
x
gn (t) dt =
k Z
X
m=1 In (m)
gn (t) dt = k · 2−n .
(7.33)
Since x ∈
/ En by the definition (7.31), we see from [51, Eqn. (6), p. 145] and the values (7.33)
that
f (x) = lim fn (x) = k · 2−n .
n→∞
219
7.4. Problems related to the AC of a Function
This proves our claim. By [51, Eqn. (3), p. 145], we have E c =
∞
[
n=1
Enc . Since x ∈
/ En , we have
x ∈ Enc ⊆ E c . Therefore, the f is well-defined on E c by the values (7.32).
By [33, p. 84] or [34, p. 202], E is nowhere dense in [0, 1] so that E c is a dense subset of
[0, 1]. Next, since f is uniformly continuous on [0, 1] and particularly on E c , it follows from [49,
Exercise 13, pp. 99, 100] that f has a unique continuous extension on [0, 1] which agrees with
f on E c . Hence the uniqueness shows that our f considered in §7.16 is the so-called Lebesgue
function associated with the Cantor-like set E. Let h : [0, 1] → R be defined by
h(x) = xα .
Since α ∈ (0, 1), it is immediate from basic calculus that h is strictly concave on [0, 1]. Clearly,
we know that
1
(7.34)
h(2−n δn ) = (2−n δn )α = nα = 2−n ,
t
where n = 0, 1, 2, . . .. Thus it gives
h 2−(n−1) δn−1 = 2−(n−1) = 2 · 2−n = 2h(2−n δn ) > h(2−n+1 δn )
for all n = 1, 2, . . .. Since h is strictly increasing on [0, 1], we obtain 2−(n−1) δn−1 > 2−n+1 δn
which is equivalent to δn−1 > δn for each n = 1, 2, . . .. By the formula (7.34), we know that
h(δ0 ) = 1 which implies δ0 = 1. In other words, the Cantor-like set E in §7.16 is constructed
by means of our fixed concave function h. By [33, p. 97] or [34, Theorem 8.5.12, p. 205], it
concludes that
|f (x) − f (y)| ≤ h |x − y| = |x − y|α
for all x, y ∈ [0, 1]. Hence f belongs to Lip α on [0, 1], completing the proof of the problem. Problem 7.10
Rudin Chapter 7 Exercise 10.
Proof. By Problem 5.11, there exists a constant M > 0 such that
|f (s) − f (t)| ≤ M |s − t|
for all s, t ∈ [a, b]. For every ǫ > 0, if we take δ =
of segments (α1 , β1 ), . . . , (αn , βn ) in [a, b] with
ǫ
M,
(7.35)
then for any n and any disjoint collection
n
X
ǫ
(βi − αi ) <
,
M
i=1
the inequality (7.35) implies that
n
X
i=1
|f (βi ) − f (αi )| ≤ M
n
X
i=1
n
X
(βi − αi ) < ǫ.
|βi − αi | = M
i=1
By Definition 7.17, f is AC on [a, b].
By Theorem 7.20, f ′ exists a.e. on [a, b]. Let f be differentiable at x ∈ [a, b]. Then we follow
from the inequality (7.35) that
|f ′ (x)| = lim
h→0
(x + h) − x
f (x + h) − f (x)
≤ lim M
= M,
h→0
h
h
i.e., |f ′ | ≤ M a.e. on [a, b]. Hence f ′ ∈ L∞ and we have completed the proof of the problem. Chapter 7. Differentiation
220
Problem 7.11
Rudin Chapter 7 Exercise 11.
Proof. Since f is AC on [a, b], we get from Theorem 7.20 that
Z
Z x
′
f (t) dt and f (y) − f (a) =
f (x) − f (a) =
a
for any a ≤ y ≤ x ≤ b. Now their difference gives
f (x) − f (y) =
Z
x
y
f ′ (t) dt
a
f ′ (t) dt.
y
By Theorem 7.20 again, we have f ′ ∈ L1 (m) so that Theorem 1.33 implies
Z x
Z x
′
|f ′ (t)| dt.
f (t) dt ≤
|f (x) − f (y)| =
(7.36)
y
y
Since f ′ ∈ Lp , we apply Theorem 3.5 (Hölder’s Inequality) to the right-hand side of the inequality
(7.36) to obtain
nZ x
o1 n Z x o1
p
q
′
p
|f (x) − f (y)| ≤
|f (t)| dt
dt
≤ kf ′ kp · |x − y|α
y
and thus
y
|f (x) − f (y)|
≤ kf ′ kp < ∞.
α
|x
−
y|
x,y∈[a,b]
sup
x6=y
By the definition (see Problem 5.11), we conclude that f ∈ Lip α, completing the proof of the
problem.
Problem 7.12
Rudin Chapter 7 Exercise 12.
Proof.
(a) Since ϕ is nondecreasing, we follow from [49, Theorems 4.29 & 4.30, pp. 95, 96] that ϕ(x+)
and ϕ(x−) exist for every point x of [a, b] and the set of discontinuities of ϕ is at most
countable. Let this set be A = {x1 , x2 , . . .} and x1 < x2 < · · · . We define f : [a, b] → R by
f (x) = sup ϕ(t). In fact, we have
t<x
f (x) =

ϕ(x),



if x ∈ [a, b] \ A;

lim ϕ(xk − t) = ϕ(xk −), if x = xk for some k ∈ N.

 t→0
t>0
Let x < y and x, y ∈ [a, b]. Then the nondecreasing property of ϕ ensures that
f (x) = sup ϕ(t) ≤ sup ϕ(t) = f (y)
t<x
t<y
which implies that f is nondecreasing on [a, b].
(7.37)
221
7.4. Problems related to the AC of a Function
Clearly, {x ∈ [a, b] | f (x) 6= ϕ(x)} ⊆ A is at most countable. Given x ∈ (a, b] and ǫ > 0.
By the definition (7.37), if x = xk , then there is a δ > 0 such that
|f (xk ) − ϕ(xk − t)| < ǫ
(7.38)
for all t ∈ (0, δ). By the construction of A, the number δ can be chosen so small such that
xk − t ∈
/ A for all t ∈ (0, δ), i.e., xk − t ∈ [a, b] \ A for all t ∈ (0, δ). Then ϕ(x) is continuous
at every xk − t and ϕ(xk − t) = f (xk − t). Thus the inequality (7.38) reduces to
|f (xk ) − f (xk − t)| < ǫ
for all t ∈ (0, δ). In other words, f is a left-continuous function.
(b) Consider the function g : [a, b] → R defined by

ϕ(t+) − ϕ(t−), if t ∈ (a, b);





0,
if t = a;
g(t) =





0,
if t = b.
(7.39)
Since ϕ is nondecreasing, we have g(t) ≥ 0 on [a, b].
Firstly, we claim that
X
t∈[a,b]
g(t) ≤ ϕ(b) − ϕ(a).
Since g(t) = 0 on [a, b] \ A, we have
X
g(t) =
X
(7.40)
g(t).
t∈A
t∈[a,b]
Let AN = {x1 , x2 , . . . , xN } ⊆ A for a positive integer N . Insert {y0 , y1 , y2 , . . . , yN } into
AN in such a way that
a ≤ y0 ≤ x1 < y1 < x2 < · · · < yN −1 < xN ≤ yN ≤ b.
(7.41)
Since ϕ is nondecreasing, we have
ϕ(xk +) ≤ ϕ(yk ) ≤ ϕ(xk+1 −)
for k = 1, 2, . . . , N − 1. For the end points x1 and xN , if a ≤ y0 < x1 , then ϕ(x1 −) ≥
ϕ(y0 ) ≥ ϕ(a); if a = y0 = x1 , then we have obviously ϕ(x1 ) = ϕ(y0 ) = ϕ(a). Similarly,
if xN < yN ≤ b, then ϕ(xN +) ≤ ϕ(yN ) ≤ ϕ(b); if xN = yN = b, then we certainly have
ϕ(xN ) = ϕ(yN ) = ϕ(b). Hence this analysis implies that
N
N
X
X
ϕ(b) − ϕ(a) ≥ ϕ(yN ) − ϕ(y0 ) ≥
[ϕ(xk +) − ϕ(xk −)] =
g(xk ).
k=1
(7.42)
k=1
Now our claim (7.40) follows directly by taking limit N → ∞ in the inequality (7.42).
Secondly, we define h : [a, b] → R by
h(x) =
X
t∈A
g(t)χ(t,b] (x) =
∞
X
k=1
g(xk )χ(xk ,b] (x).
(7.43)
Chapter 7. Differentiation
222
By the claim (7.40) (or the second inequality in (7.42)) and the Weierstrass M -test, we
know that the sum (7.43) actually converges uniformly on [a, b]. If a ≤ x < y ≤ b, then
we have χ(t,b] (x) ≤ χ(t,b] (y) and this shows that
h(x) =
X
t∈A
g(t)χ(t,b] (x) ≤
X
g(t)χ(t,b] (y) = h(y),
t∈A
i.e., h is nondecreasing on [a, b].
Thirdly, we need two lemmas about the properties of the function F : [a, b] → R defined
by F (x) = f (x) − h(x).
Lemma 7.2
The function F is continuous on [a, b].
Proof of Lemma 7.2. If x ∈
/ A, then f = ϕ is continuous at x. Furthermore, if xm <
x < xm+1 for some m ∈ N, then for all small enough δ, we have (x − δ, x + δ) ⊂
(xm , xm+1 ). In this case, we obtain
lim h(x ± δ) = lim
δ→0
δ→0
∞
X
k=1
g(xk )χ(xk ,b] (x ± δ) = lim
δ→0
m
X
k=1
g(xk )χ(xk ,b] (x ± δ) = h(x),
i.e., h is continuous at x. By the definition, F is also continuous at x.
If x ∈ A, then x = xm for some m ∈ N. On the one hand, the definition (7.37) gives
f (xm +) = lim f (xm + δ) = ϕ(xm +) and f (xm −) = ϕ(xm −).
δ→0
δ>0
On the other hand, the definitions (7.39) and (7.43) show
h(xm +) = lim h(xm + δ) = lim
δ→0
δ>0
∞
X
δ→0
δ>0 k=1
g(xk )χ(xk ,b] (xm + δ) =
m
X
[ϕ(xk +) − ϕ(xk −)]
k=1
and
h(xm −) = lim h(xm + δ) = lim
δ→0
δ<0
∞
X
δ→0
δ<0 k=1
g(xk )χ(xk ,b] (xm + δ) =
m−1
X
k=1
[ϕ(xk +) − ϕ(xk −)].
Consequently, they imply that
F (xm +) = f (xm +) − h(xm +)
m
X
= ϕ(xm +) −
[ϕ(xk +) − ϕ(xk −)]
= ϕ(xm −) −
k=1
m−1
X
k=1
[ϕ(xk +) − ϕ(xk −)]
= f (xm −) − h(xm −)
= F (xm −),
i.e., F is continuous at xm and this ends the proof of the lemma.
223
7.4. Problems related to the AC of a Function
Lemma 7.3
The function F is nondecreasing on [a, b].
Proof of Lemma 7.3. Let x, y ∈ [a, b] be x < y. Suppose that xm < x ≤ xm+1 for some
m ∈ N. If xm < x < y ≤ xm+1 , then the definition (7.43) gives
h(y) − h(x) =
=
∞
X
k=1
m
X
k=1
g(xk )χ(xk ,b] (y) −
g(xk )χ(xk ,b] (y) −
=0
∞
X
k=1
m
X
g(xk )χ(xk ,b] (x)
g(xk )χ(xk ,b] (x)
k=1
≤ f (y) − f (x).
Suppose that
xm < x ≤ xm+1 < · · · < xm+p ≤ y
(7.44)
for some p ∈ N, then we get from the definition (7.39) that
h(y) − h(x) =
=
=
≤
∞
X
k=1
m
X
g(xk )χ(xk ,b] (y) −
g(xk )χ(xk ,b] (y) +
∞
X
g(xk )χ(xk ,b] (x)
k=1
m+p
X
k=m+1
k=1
m+p
X
g(xk )χ(xk ,b] (y) −
m
X
g(xk )χ(xk ,b] (x)
k=1
g(xk )χ(xk ,b] (y)
k=m+1
m+p
X
[ϕ(xk +) − ϕ(xk −)].
(7.45)
k=m+1
If we apply the same trick (the sequence (7.41) is replaced by the sequence (7.44) with
a and b are replaced by x and y respectively) as in proving the inequality (7.42), we
can easily deduce from the inequality (7.45) that h(y) − h(x) ≤ f (y) − f (x).
In conclusion, we have shown that
h(y) − h(x) ≤ f (y) − f (x)
for all x, y ∈ [a, b] with x < y. Equivalently, it means that
F (x) = f (x) − h(x) ≤ f (y) − h(y) ≤ F (y)
and so F is nondecreasing, as desired.
Fourthly, we express F in terms of a finite and positive Borel measure. By Lemmas 7.2
and 7.3, we know that the function
g(x) = x + F (x)
is continuous and strictly increasing on [a, b]. Then g is one-to-one on [a, b] so that its inverse function g −1 : [g(a), g(b)] → [a, b] is a continuous and bijective function on [g(a), g(b)]
Chapter 7. Differentiation
224
by [49, Theorem 4.17, p. 90]. For every E ⊆ [a, b] and E ∈ B, since g−1 is continuous on
[g(a), g(b)], Definition 1.2(c) tells us that g(E) = (g −1 )−1 (E) ∈ B so that we are able to
define
ν(E) = m g(E) .
(7.46)
We have to show that ν is a positive Borel measure, but it follows exactly the same
reasons as in [51, p. 147]. Furthermore, since m is a finite measure, the construction
(7.46) implies that ν is also a finite measure. By Theorem 6.10 (The Lebesgue-RadonNikodym Theorem), there exists a unique pair of positive and finite Borel measures νac
and νsc such that
νac ≪ m, νsc ⊥ m and ν = νac + νsc .
Now the function F and the measures ν and m are linked up in the following lemma:
Lemma 7.4
Let µ′ = ν − m. Then µ′ is finite and positive Borel measure. Furthermore, we
have
µ′ [a, x) = F (x) − F (a).
(7.47)
Proof of Lemma 7.4. Clearly, the measure µ′ is a complex Borel measure. The finiteness of µ′ is also clear. If x ∈ [a, b], then we deduce from the definition (7.46) that
µ′ [a, x) = m g[a, x) − m([a, x)) = g(x) − g(a) − x + a = F (x) − F (a).
It remains to verify that µ′ is positive. To this end, it is obvious from the definition
and Lemma 7.3 that for x, y ∈ [a, b] with x < y, we have
ν (x, y) = m g (x, y) = g(y) − g(x) = y + F (y) − x − F (x) ≥ y − x = m (x, y) .
Suppose that V is an open set in [a, b]. By [49, Exercise 29, p. 45], V can be expressed
as a union of an at most countable collection of disjoint segments {Vi }. Since µ and
m are measures, we have
ν(V ) = ν
∞
[
i=1
∞
∞
∞
[
[
[
Vi = m(V ).
m(Vi ) = m
ν(Vi ) ≥
Vi =
i=1
i=1
(7.48)
i=1
Next, it is well-known that m is regular. Since [a, b] is compact, it is σ-compact.
Recall that ν is a finite positive Borel measure, so ν(K) < ∞ for every compact set
K ⊆ [a, b]. Therefore, ν is regular by Theorem 2.18. Thus if E ∈ B and E ⊆ [a, b],
then Theorem 2.14(c) implies that given ǫ > 0, there exists an open set V containing
E such that
ν(V ) < ν(E) + ǫ.
(7.49)
Combining the inequalities (7.48) and (7.49), we obtain
ν(E) + ǫ > ν(V ) ≥ m(V ) ≥ m(E).
Since ǫ is arbitrary, we establish the fact that ν(E) ≥ m(E) and then µ′ (E) ≥ 0 for
all E ∈ B and E ⊆ [a, b]. This proves the positivity of the measure µ′ and ends the
proof of Lemma 7.4.
225
7.4. Problems related to the AC of a Function
We come to the final step of the proof of part (b). Recall that a complex Borel measure
µ′′ is called a P
discrete measure if there
P is a countable set {xj } ⊂ R and complex numbers
cj such that
|cj | < ∞ and µ′′ =
cj δxj , where δx is the point mass at x, see [22, p.
106] or [51, Example 1.20(b), p. 17]. Now if we let
µ′′ (E) =
∞
X
k=1
g(xk )δxk (E) ≥ 0
for any E ⊆ [a, b] and E ∈ B, then it follows from the fact (7.40) that µ′′ is finite. Besides,
we deduce from this and the definition (7.43) easily that
∞
∞
X
X
µ′′ [a, x) =
g(xk )δxk [a, x) =
g(xk )χ(xk ,b] (x) = h(x) − h(a).
k=1
(7.50)
k=1
Therefore, if we further define µ = µ′ + µ′′ , then µ must be a finite positive Borel measure
on [a, b] and it yields from the representations (7.47) and (7.50) that
µ([a, x)) = µ′ ([a, x)) + µ′′ ([a, x)) = F (x) − F (a) + h(x) − h(a) = f (x) − f (a).
(7.51)
This proves that µ is the desired measure.
(c) By Theorem 6.10 (The Lebesgue-Radon-Nikodym Theorem), we have
µ = µac + µsc ,
where µac ≪ m and µsc ⊥ m. Suppose that ω is the Radon-Nikodym derivative of µac
with respect to m. Since ω ∈ L1 (m) on [a, b], almost every x ∈ [a, b] is a Lebesgue point
of ω by Theorem 7.7.
Let x ∈ (a, b) be a Lebesgue point of ω. Define r = min(x − a, b − x) and we can choose
{ri } ⊆ (0, r) such that ri → 0 as i → ∞. Then the two sequences of sets {Ei } and {Ei′ }
given by
Ei = [x, x + ri ) and Ei′ = [x − ri , x)
satisfy the inequalities
m(Ei ) = ri ≥
1
1
· 2ri = m (x − ri , x + ri )
2
2
and m(Ei′ ) ≥
1
m (x − ri , x + ri )
2
for all i = 1, 2, . . .. In other words, they mean that {Ei } and {Ei′ } shrink to x nicely.
Since µ is a complex Borel measure on [a, b] and dµ = ω dm + dµs , Theorem 7.14 and the
formula (7.51) imply that
µ [x, x + ri )
f (x + ri ) − f (x)
′
= ω(x)
f+ (x) = lim
(7.52)
= lim
i→∞
i→∞ m [x, x + ri )
ri
a.e. [m] on [a, b]. Similarly, we have
f−′ (x)
µ [x − ri , x)
f (x − ri ) − f (x)
= ω(x)
= lim
= lim
i→∞ m [x − ri , x)
i→∞
−ri
(7.53)
a.e. [m] on [a, b]. Now the results (7.52) and (7.53) definitely give f ′ (x) exists a.e. [m] on
[a, b] and f ′ = ω ∈ L1 (m) on [a, b].
Finally, since dµac = ω dm, we have
Z
Z x
ω dm =
µac [a, x) =
a
x
′
f dm =
a
Z
a
x
f ′ (t) dt
Chapter 7. Differentiation
226
so that
f (x) − f (a) = µ [a, x) = µac [a, x) + µsc [a, x) =
Z
x
f ′ (t) dt + s(x),
a
where s(x) = µsc [a, x) for all x ∈ [a, b]. Now µsc [a, x) is positive finite because µ is
positive finite, so s is nondecreasing on [a, b]. Since f ′ ∈ L1 (m) on [a, b], Theorem 7.11
shows that
Z x
′
′
′
s (x) = [f (x) − f (a)] −
f ′ (t) dt = f ′ (x) − f ′ (x) = 0
a
a.e. [m] on (a, b).
(d) By Theorem 6.8(e), µ ⊥ m if and only if µac = 0 if and only if
Z
f ′ dm = 0
(7.54)
E
for every E ∈ B and E ⊆ [a, b]. Recall that f ′ ∈ L1 (m). By Theorem 1.39(b), the result
(7.54) is equivalent to f ′ (x) = 0 a.e. [m] on [a, b].
Similarly, µ ≪ m if and only if µsc = 0 if and only if s = 0 on [a, b] if and only if
Z x
f ′ dm
(7.55)
f (x) − f (a) =
a
for all x ∈ [a, b]. As Rudin pointed out in the discussion preceding Definition 7.17 (on p.
145), one may apply Theorem 6.11 to show that f is AC on [a, b] if the formula (7.55)
holds. Conversely, if f is AC on [a, b], then f is automatically continuous on [a, b] and
since f is nondecreasing, Theorem 7.18 implies that the formula (7.55) holds.
(e) Suppose that f ′ (p) exists at p ∈ (a, b) \ A. Then f is continuous at p and the definition
(7.37) implies the existence of a δ > 0 such that
f (x) = ϕ(x)
for all x ∈ (p − δ, p + δ). If h ∈ (0, δ), then p + h ∈ (p − δ, p + δ) and so
ϕ(p + h) − ϕ(p)
f (p + h) − f (p)
− f ′ (p) =
− f ′ (p) .
h
h
(7.56)
By taking h → 0+ in the expression (7.56), we get
|ϕ′+ (p) − f ′ (p)| = |f+′ (p) − f ′ (p)| = 0,
i.e., ϕ′+ (p) = f ′ (p). The other side ϕ′− (p) = f ′ (p) can be done similarly. Thus we conclude
that ϕ′ (p) = f ′ (p). Since f ′ (x) exists a.e. [m] on [a, b], we have ϕ′ (x) = f ′ (x) a.e. [m] on
[a, b].
Hence we have completed the proof of the problem.
Remark 7.3
By the result of part (b) in Problem 7.12, we have
singular
continuous
µ = µ′ + µ′′ = v − m + µ′′ = (νa − m) +
| {z }
absolutely
continuous
See also the decomposition in [22, Theorem 3.35, p. 106].
z}|{
νs
+ µ′′ .
|{z}
discrete
227
7.4. Problems related to the AC of a Function
Problem 7.13
Rudin Chapter 7 Exercise 13.
Proof. By the definition,
n
BV = f : [a, b] → C F (b) = sup
N
X
i=1
o
|f (ti ) − f (ti−1 )| < ∞ ,
where the supremum is taken over all N and over all choices of {ti } such that
a = t0 < t1 < · · · < tN = b.
(7.57)
(a) Let f : [a, b] → C be a bounded nondecreasing function. Then we have
N
X
i=1
|f (ti ) − f (ti−1 )| =
N
X
i=1
[f (ti ) − f (ti−1 )] = f (b) − f (a) < ∞,
where {ti } satisfies the requirement (7.57). Hence we have f ∈ BV . If f is a bounded
nonincreasing function, then we consider −f which is a bounded nondecreasing function.
(b) Let f ∈ BV be real. Define
f1 (x) =
F (x) + f (x)
2
and f2 (x) =
F (x) − f (x)
,
2
(7.58)
where F is given by [51, Eqn. (1), p. 147]. It is easy to see that f = f1 − f2 , so it remains
to prove that they are bounded and monotonic.
Suppose first that f is nondecreasing. By Theorem 7.19, F +f and F −f are nondecreasing,e so are f1 and f2 . By the definition of the space BV , we have 0 ≤ F (x) ≤ F (b) < ∞
on [a, b]. If x ∈ [a, b], then the triangle inequality gives
|f (x)| ≤ |f (a)| + |f (x) − f (a)| ≤ |f (a)| + F (b).
Thus f is bounded on [a, b]. By the definition (7.59), f1 and f2 are also bounded on
[a, b]. If f is nonincreasing, then the function −f is nondecreasing but the functions
F −(−f )
F +(−f )
f1 = F +f
and f2 = F −f
remain nondecreasing.f
2 =
2
2 =
2
(c) Suppose that f ∈ BV and it is left-continuous. It suffices to prove that F is left-continuous
at p, i.e., if a < p ≤ b and ǫ > 0, then there is a δ > 0 so that
|F (p) − F (p − h)| < ǫ
whenever 0 < h < δ. To start with, we define
F (x − h; x) = sup
N
X
i=1
|f (ti ) − f (ti−1 )|,
where x − h = t0 < t1 < · · · < tN = x. We claim that
F (x) − F (x − h) = F (x − h; x)
e
f
The conclusion of this assertion does not depend on the assumption that f is AC on [a, b].
If we write f = (−f2 ) − (−f1 ), then −f1 and −f2 are nonincreasing.
(7.59)
Chapter 7. Differentiation
228
for every small h > 0 such that x − h ∈ [a, b]. By the definition, if {ti } and {sj } are
partitions of [a, x − h] and [x − h, x] respectively, then {ti } ∪ {sj } is certainly a partition
of [a, b] and so
F (x) ≥ F (x − h) + F (x − h; x).
(7.60)
On the other hand, let {t0 , t1 , . . . , tN } be a partition of [a, x]. If tM < x − h < tM +1 , then
we may insert a number t′ such that
a = t0 < t1 < · · · ≤ tM < t′ = x − h < tM +1 < · · · < tN = x.
Clearly, the triangle inequality implies
N
X
i=1
|f (ti ) − f (ti−1 )| ≤
M
hX
i=1
i
|f (ti ) − f (ti−1 )| + |f (t′ ) − f (tM )|
N
i
h
X
|f (ti ) − f (ti−1 )|
+ |f (tM +1 ) − f (t′ )| +
i=M +2
≤ F (x − h) + F (x − h; x).
(7.61)
Since the partition {t0 , t1 , . . . , tN } is arbitrary, the inequality (7.61) asserts that
F (x) ≤ F (x − h) + F (x − h; x).
(7.62)
Hence our desired claim follows immediately from the inequalities (7.60) and (7.62).
Since f is left-continuous at p ∈ (a, b], given ǫ > 0, there exists a δ > 0 such that
h ∈ (0, δ) implies
ǫ
|f (p) − f (p − h)| < .
2
For this same ǫ, we can find a partition {ti } of [a, p], say
a = t0 < t1 < · · · < tN = p,
such that
N
F (a; p) −
ǫ X
|f (ti ) − f (ti−1 )|
<
2
(7.63)
i=1
Since adding more points to the partition {ti } results in an increase of the summation
(7.63), we may assume that tN − tN −1 = h ∈ (0, δ) and this implies that
|f (tN ) − f (tN −1 )| = |f (tN ) − f (tN − h)| = |f (p) − f (p − h)| <
ǫ
2
and the inequality (7.63) becomes
F (a; p) −
N −1
ǫ
ǫ
ǫ X
|f (ti ) − f (ti−1 )| ≤ + F (a; p − h)
< +
2
2
2
i=1
or equivalently,
F (a; p) − F (a; p − h) < ǫ.
However, we note that
0 ≤ F (p) − F (p − h) = F (p − h; p) = F (a; p) − F (a; p − h) < ǫ
which is exactly the expected result (7.59). Hence F and then f1 as well as f2 are leftcontinuous on [a, b].
229
7.4. Problems related to the AC of a Function
(d) By parts (b) and (c), f = f1 − f2 , where f1 and f2 are left-continuous and nondecreasing
on [a, b]. Now Problem 7.12(b) shows that there exist positive and finite Borel measures
λ and ν on [a, b] such that
f1 (x) − f1 (a) = λ [a, x) and f2 (x) − f2 (a) = ν [a, x) ,
where x ∈ [a, b]. Then the difference µ = λ − ν is easily seen to be a Borel measure on
[a, b] satisfying the equation in the question. In fact, the finiteness of λ and ν imply the
finiteness of µ. (However, µ may not be positive.)
As Theorem 6.10 (The Lebesgue-Radon-Nikodym Theorem) tells us that there exists a
unique pair of positive and finite measures µa ≪ m and µs ⊥ m such that µ = µa + µs .
Furthermore, there is a unique h ∈ L1 (m) on [a, b] such that
Z
h dm
(7.64)
µa (E) =
E
for every E ∈ B.
If µ ≪ m, then Proposition 6.8(e) implies that µs = 0 and thus the µa in the representation (7.64) can be replaced by µ. Now we put E = [a, x) into the representation and use
the result f (x) − f (a) = µ([a, x)) to obtain
Z x
Z
h dm.
h dm =
f (x) − f (a) = µ([a, x)) =
[a,x)
a
Since h ∈ L1 (m) on [a, b], we follow from [9, Theorem 5.3.6, p. 339] or [22, Theorem 3.35,
p. 106] that f is AC on [a, b].g Conversely, suppose that f is AC on [a, b]. By Theorem
7.19, F, f1 and f2 are AC and nondecreasing on [a, b]. By Problem 7.12(d), we conclude
that
λ ≪ m and ν ≪ m.
Since µ = λ − ν, Proposition 6.8(c) ensures that µ ≪ m.
(e) By part (b), we have f = f1 − f2 , where f1 and f2 are nondecreasing. By Problem 7.12(e),
f1′ and f2′ exist a.e. [m] on [a, b]. By Problem 7.12(c), we further know that f1′ , f2′ ∈ L1 (m)
on [a, b]. Hence we conclude that f ′ exists a.e. [m] on [a, b] and f ′ ∈ L1 (m).
We complete the proof of the problem.
Problem 7.14
Rudin Chapter 7 Exercise 14.
Proof. Suppose that f and g are AC on [a, b]. By Definition 7.17, they are continuous on [a, b],
so the Extreme Value Theorem ([49, Theorem 4.16, p. 89]) ensures that we may assume that
there exists a M > 0 such that
|f (x)| ≤ M
and
|g(x)| ≤ M
(7.65)
on [a, b]. Given ǫ > 0, there corresponds a δ > 0 such that
n
X
k=1
g
ǫ
|f (βk ) − f (αk )| <
2M
and
n
X
k=1
|g(βk ) − g(αk )| <
ǫ
2M
(7.66)
These theorems do not need the assumptions of Theorem 7.18: f is continuous and nondecreasing on [a, b].
Chapter 7. Differentiation
230
for every n and any disjoint collection of segments (α1 , β1 ), . . . , (αn , βn ) in [a, b] with
n
X
(βk − αk ) < δ.
k=1
It is clear that
n
X
k=1
|f (βk )g(βk ) − f (αk )g(αk )| =
≤
n
X
k=1
n
X
k=1
≤M
|f (βk )g(βk ) − f (βk )g(αk ) + f (βk )g(αk ) − f (αk )g(αk )|
|f (βk )g(βk ) − f (βk )g(αk )| +
n
X
k=1
|g(βk ) − g(αk )| + M
n
X
k=1
n
X
k=1
|f (βk )g(αk ) − f (αk )g(αk )|
|f (βk ) − f (αk )|.
(7.67)
By substituting the inequalities (7.66) into the inequality (7.67), we gain
n
X
k=1
|f (βk )g(βk ) − f (αk )g(αk )| < ǫ
which implies that f g is AC on [a, b].
Now Theorem 7.20 shows that f, g and f g are differentiable a.e. on [a, b] and f ′ , g′ , (f g)′ ∈ L1
on [a, b]. Apply Theorem 7.20 to f g, we have
f (b)g(b) − f (a)g(a) =
Z
b
a
′
f (x)g(x) dx.
(7.68)
By the bounds (7.65), we see that f ′ g, f g′ ∈ L1 on [a, b]. Furthermore, since (f g)′ = f ′ g + f g′
a.e. on [a, b], we deduce from the formula (7.68) that
Z
b
a
f ′ (x)g(x) dx = f (b)g(b) − f (a)g(a) −
Z
b
f (x)g′ (x) dx,
a
completing the proof of the problem.
Problem 7.15
Rudin Chapter 7 Exercise 15.
Proof. It is well-known [49, Example 5.6(b), p. 106] that the function f : R → R defined by
has derivative given by

1

 x2 sin , if x 6= 0;
x
f (x) =


0,
otherwise

1
1

 2x sin − cos , if x 6= 0;
x
x
′
f (x) =


0,
otherwise.
231
7.4. Problems related to the AC of a Function
Thus f is differentiable at all points but f ′ is not a continuous function. However, such function
oscillates around 0 so that it is not monotonic and we have to “modify” it. To do this, we notice
that since
sin( x1 )
1
lim x sin = lim
= 1,
x→±∞
x x→±∞ x1
f ′ (x) is bounded on R. Let M > 0 be a bound of f ′ on R, i.e.,
−M ≤ f ′ (x) ≤ M
(7.69)
for all x ∈ R. We consider the function F : R → R given by
F (x) = f (x) + M x.
Now F ′ (x) = f ′ (x) + M which exists finitely on R. In addition, the inequalities (7.69) implies
that
F ′ (x) ≥ −M + M ≥ 0
for all x ∈ R. Hence [49, Theorem 5.11, p. 108] ensures that F is monotonically increasing on R.
Since f ′ is not continuous at 0, F ′ is also discontinuous at 0 and thus F is our desired function.
This completes the proof of the problem.
Problem 7.16
Rudin Chapter 7 Exercise 16.
Proof. We follow the hint given by Rudin. Since m is outer regular (see Theorem 2.14(c) and
Theorem 2.20(b)) and m(E) = 0, we can find a sequence {Vn } of open subsets of R such that
E ⊆ · · · ⊆ V2 ⊆ V1 and m(Vn ) < 2−n . This construction gives
E⊆
∞
\
Vn .
(7.70)
n=1
Consider F : [a, b] → R defined by
F (x) =
∞
X
n=1
χVn (x) ≥ 0.
Since each χVn is measurable, it follows from Theorem 1.27 that
0≤
Z
a
b
F (x) dx =
∞ Z
X
b
χVn (x) dx =
∞ Z
X
dx =
n=1 Vn
n=1 a
∞
X
n=1
In other words, F ∈ L1 [a, b] . Next we define f : [a, b] → R by
Z x
F (t) dt ≥ 0
f (x) =
m(Vn ) <
∞
X
n=1
2−n < ∞.
(7.71)
a
which is clearly nondecreasing. By Theorem 7.11, f is differentiable a.e. on [a, b] and then we
follow from Theorem 7.18 that f is AC on [a, b].
Now it remains to show that f ′ (x) = ∞ on E. To see this, if x ∈ E, then the set relation
(7.70) tells us that x ∈ Vn for all n ∈ N. Since each Vn is open, there exists a δn > 0 such that
(x − δn , x + δn ) ⊆ Vn
Chapter 7. Differentiation
232
for all n ∈ N. In fact, we may assume that {δn } is a decreasing sequence of positive numbers.
Pick y ∈ (x − δn , x + δn ) so that
Z
|f (y) − f (x)| =
Z
y
x
F (t) dt ≥
n
yX
x k=1
[χV1 (t) + · · · + χVn (t)] dt
(7.72)
Since Vn ⊆ Vn−1 ⊆ · · · ⊆ V2 ⊆ V1 and t ∈ [x, y] ⊆ (x − δn , x + δn ) ⊆ Vn ,h the inequality (7.72)
becomes
|f (y) − f (x)| ≥ n|y − x|.
Notice that y → x if and only if n → ∞, so we have
|f ′ (x)| = lim
y→x
f (y) − f (x)
= ∞.
y−x
By the definition (7.71), we conclude that f ′ (x) = ∞ on E. We have completed the proof of the
problem.
7.5
Miscellaneous Problems on Differentiation
Problem 7.17
Rudin Chapter 7 Exercise 17.
Proof. We prove the assertions one by one:
• µ is a Borel measure. Suppose that {Ei } is a collection of mutually disjoint Borel
∞
[
k
Ei . Since µn (Ei ) ≥ 0 for all i ∈ N, we know from [49, Exericse
subsets of R . Let E =
i=1
3, p. 196] that
µ(E) =
∞
X
µn (E) =
∞ X
∞
X
µn (Ei ) =
µn (Ei ) =
i=1 n=1
n=1 i=1
n=1
∞
∞ X
X
∞
X
µ(Ei ).
i=1
Clearly, we have µ(E) ≥ 0 for all E ∈ B. By Definition 1.18(a), µ is a positive Borel
measure on Rk .
• The relation between the Lebesgue decompositions of the µn and that of µ.
Since m is a positive σ-finite measure on B, it follows from Theorem 6.10 (The LebesgueRadon-Nikodym Theorem) that
µ = µa + µs
and µn = µn,a + µn,s ,
where
µa , µn,a ≪ m
for all n ∈ N. Let
λa =
∞
X
n=1
h
Or t ∈ [y, x].
µn,a
and µs , µn,s ⊥ m
and
λs =
∞
X
n=1
µn,s .
(7.73)
233
7.5. Miscellaneous Problems on Differentiation
Since µ and µn are positive, so are µa , µn,a , µs and µn,s and it deduces from the first
assertion that both λa and λs are positive Borel measures of Rk . Furthermore, we have
µ=
∞
X
µn =
∞
X
(µn,a + µn,s ) =
µn,a +
∞
X
µn,s = λa + λs .
n=1
n=1
n=1
n=1
∞
X
By Theorem 6.8 and the relations (7.73), we assert that λa ≪ m and λs ⊥ m. Recall that
the Lebesgue decomposition of µ relative to m is unique, it must be true that
µ a = λa =
∞
X
µn,a
and µs = λs =
n=1
∞
X
µn,s .
n=1
• The truth of the formula. By Theorem 6.10 (The Lebesgue-Radon-Nikodym Theorem)
again, there exists a unique h ∈ L1 (m) such that
Z
h dm
(7.74)
µa (E) =
E
for all E ∈ B. Similarly, for each n ∈ N, there exists a unique hn ∈ L1 (m) such that
Z
hn dm
(7.75)
µn,a (E) =
E
for all E ∈ B. Since µa , µn,a ≪ m for all n ∈ N, we observe from Theorem 7.8 that
Dµa = h and
Dµn,a = hn
(7.76)
a.e. [m] for all n ∈ N. Since µs ⊥ m and µn,s ⊥ m for all n ∈ N, Theorem 7.14 ensures
that
Dµs = 0 and Dµn,s = 0
(7.77)
a.e. [m] for all n ∈ N. Hence the two results (7.76) and (7.77) say that
Dµ = h and
Dµn = hn
a.e. [m] for all n ∈ N.
Recall from the second assertion and the application of the integrals (7.74) and (7.75)
that
Z
∞
∞ Z
X
X
h dm = µa (E) =
µn,a (E) =
hn dm
(7.78)
E
n=1 E
n=1
for all E ∈ B. Since hn ∈ L1 (m), each hn is measurable and Theorem 1.27 implies that
the expression (7.78) can be further reduced to
Z
for all E ∈ B, or equivalently,
h dm =
E
hn dm
E n=1
Z ∞
X
hn dm = 0
h−
E
for all E ∈ B.
Z X
∞
n=1
(7.79)
Chapter 7. Differentiation
Since µa (E) ≥
N
X
n=1
234
µn,a(E) for every N ∈ N, the integrals (7.74) and (7.75) show that
Z N
X
hn dm ≥ 0
h−
E
(7.80)
n=1
for every E ∈ B. Assume that h <
N
X
hn on E ′ with m(E ′ ) > 0. For this Borel set E ′ ,
n=1
we derive from Proposition 1.24(a) that
Z
E′
h−
N
X
n=1
hn dm < 0
which contradicts the result (7.80). Consequently, we always have h ≥
Rk and since N is arbitrary, this implies that
h−
∞
X
n=1
N
X
hn a.e. [m] on
n=1
hn ≥ 0
a.e. [m] on Rk . Now we apply Theorem 1.39(a) to the integral (7.79) to conclude that
∞
X
h(x) =
hn (x)
n=1
a.e. [m] on Rk , then this shows the formula
Dµ(x) =
∞
X
(Dµn )(x)
n=1
a.e. [m] on Rk .
• A corresponding theorem for a sequence {fn } of positive nondecreasing func∞
X
fn (x) converges on R. We claim that f ′ (x) exists
tions on R. Suppose that f (x) =
and
n=1
f ′ (x) =
∞
X
fn′ (x)
(7.81)
n=1
a.e. [m] on R. By the definition, it is trivial that f is nondecreasing and 0 < f (x) < ∞
on R. Let [a, b] be a compact interval of R. Suppose that
E = {x ∈ (a, b) | f ′ (x) and fn′ (x) exist}.
By Problem 7.12(e), we see that f and each fn are differentiable a.e. [m] on [a, b] so that
[a, b] \ E is of measure zero.
Let x ∈ E and h > 0 be sufficiently small so that x + h ∈ [a, b]. Then for any N ∈ N,
we have
N
f (x + h) − f (x) X fn (x + h) − fn (x)
≥
h
h
n=1
235
7.5. Miscellaneous Problems on Differentiation
so if we let h → 0, then we have
f ′ (x) ≥
N
X
fn′ (x).
(7.82)
n=1
Now we take N → ∞ in the inequality (7.82) to conclude that
′
f (x) ≥
∞
X
fn′ (x)
(7.83)
n=1
a.e. [m] on [a, b]. Recall that x ∈ E so that the inequality (7.83) ensures that the series of
its right-hand side is convergent a.e. [m] on [a, b]. Consequently, it implies that
lim f ′ (x)
n→∞ n
=0
(7.84)
a.e. [m] on [a, b].
Next we define
sN (x) =
N
X
fn (x).
n=1
Since sN (b) → f (b) as N → ∞, there is a sequence {Nk } ⊆ N such that
0 ≤ f (b) − sNk (b) <
1
.
2k
(7.85)
Obviously, we have
gk (x) = f (x) − sNk (x) =
∞
X
fn (x),
n=Nk +1
so each gk is a positive nondecreasing function on [a, b]. To apply the first part of the
∞
X
proof, we have to show that the series
gk (x) converges on [a, b]. Combining this fact
k=1
and the inequality (7.85), we obtain
0 ≤ gk (x) <
1
2k
on [a, b]. As a result, the series
∞
X
k=1
∞
X
(f − sNk )
gk =
k=1
converges (uniformly) to g(x) on [a, b] by the Weierstrass M -test. Now we are able to apply
the result (7.84) to the sequence {gk } of positive nondecreasing functions and conclude
that
lim gk (x) = lim [f ′ (x) − s′Nk (x)] = 0
k→∞
k→∞
a.e. [m] on [a, b] or equivalently,
lim
k→∞
Nk
X
n=1
fn′ (x) = f ′ (x)
(7.86)
Chapter 7. Differentiation
236
a.e. [m] on [a, b]. Since each fn is nondecreasing on [a, b], fn′ (x) ≥ 0 for all x ∈ E. Hence
the limit (7.86) implies that
∞
X
fn′ (x)
f ′ (x) =
n=1
a.e. [m] on [a, b]. Since any real number is contained in some compact interval [a, b], our
desired result (7.81) follows from this.
It completes the proof of the problem.
Remark 7.4
The last assertion in Problem 7.17 is called Fubini’s Theorem on Differentiation. See
[33, Sec. C, pp. 527 – 529].
Problem 7.18
Rudin Chapter 7 Exercise 18.
Proof. First of all, we show that {ϕn } is orthonormal on [0, 1]. Suppose that m, n ∈ N and
m ≥ n. If m = n, then ϕn (t)ϕn (t) = ϕ2n (t) = 1 and so
Z 1
ϕn (t)ϕm (t) dt = 1.
0
If m > n, then we deduce from Theorem 7.26 (The Change-of-variables Theorem) that
Z 1
Z 1
ϕ0 (2n t)ϕ0 (2m t) dt
ϕn (t)ϕm (t) dt =
0
0
Z 2n
1
ϕ0 (y)ϕ0 (2m−n y) dy
= n
2 0
2n−1 Z
1 X 2i
ϕ0 (y)ϕ0 (2m−n y) dy.
= n
2
2(i−1)
(7.87)
i=1
Since ϕ0 (y) is periodic with period 2, we have ϕ0 (y) = 1 on [2i − 2, 2i − 1) and ϕ0 (y) = −1 on
[2i − 1, 2i). Therefore, we can reduce the expression (7.87) to
Z
0
1
Z 2i
2n−1 Z
i
1 X h 2i−1
ϕ0 (2m−n y) dy .
ϕn (t)ϕm (t) dt = n
ϕ0 (2m−n y) dy −
2
2i−1
2(i−1)
(7.88)
i=1
Note that the interval [2m−n+1 i − 2m−n , 2m−n+1 i) is of length 2m−n , so the integrals on the
right-hand side of the equation (7.88) are 0 and this gives
Z 1
ϕn (t)ϕm (t) dt = 0
0
if m 6= n. By Definition 4.16, {ϕn } is orthonormal on [0, 1].
P
Since
|cn |2 < ∞, it yields from the Riesz-Fischer Theorem [49, Theorem 11.43, p. 330]
that
∞
X
cn ϕn
(7.89)
f∼
n=1
237
7.5. Miscellaneous Problems on Differentiation
for some f ∈ L2 [0, 1] , i.e., the series in (7.89) is the Fourier series of f .
Consider a = j · 2−N , b = (j + 1) · 2−N , a < t < b and sN = c1 ϕ1 + · · · + cN ϕN , where
j = 0, 1, 2, . . .. On the one hand, if 1 ≤ k ≤ N , then a ≤ t ≤ b implies that
j · 2k−N ≤ 2k t ≤ (j + 1)2k−N and m [2k a, 2k b] = 2k−N ≤ 1
so that each ϕk (t) = ϕ0 (2k t) (of course, it is 1 or −1) is constant in [a, b]. This implies that
1
b−a
Z
b
a
1
sN (t) dt = sN (t) ·
b−a
Z
b
dt = sN (t).
a
On the other hand, if k > N , then 2k−N > 1 and we apply Theorem 7.26 (The Change-ofvariables Theorem) to get
Z
b
ϕk (t) dt =
a
Z
b
ϕ0 (2k t) dt =
a
1
2k
Z
2k b
2k a
ϕ0 (y) dy =
1
2k
Z
(j+1)·2k−N
j·2k−N
ϕ0 (y) dy.
Using the same argument as in evaluating the integrals in the equation (7.88), we can show that
1
2k
Z
(j+1)·2k−N
ϕ0 (y) dy = 0.
j·2k−N
In conclusion, we arrive at
1
sN (t) =
b−a
Z
b
a
Z
1
sN dm =
b−a
b
sn dm
(7.90)
a
for every n = 1, 2, . . .. By the relation (7.89), sn → f in L2 on [0, 1]. By Theorem 3.5 (Hölder’s
Inequality), we obtain
Z
a
b
|sn − f | dm ≤
nZ
b
a
2
|sn − f | dm
o1 n Z
b
2
dm
a
o1
2
1
= (b − a) 2 ksn − f k2 → 0
as n → ∞. Thus this implies that sn − f ∈ L1 on [0, 1] for large enough n and so we derive from
Theorem 1.33 that
Z b
Z b
f dm.
(7.91)
sn dm =
lim
n→∞ a
a
By combining the representation (7.90) and the limit (7.91), we establish
sN (t) =
1
b−a
Z
b
f dm.
(7.92)
a
Since f ∈ L2 on [0, 1], we have f ∈ L1 on [0, 1] by Theorem 3.8. By Theorem 7.7, almost every
x ∈ [0, 1] is a Lebesgue point of f . Let p ∈ [0, 1] be such a point and for each N ∈ N, we define
EN (p) = [2−N , 2−N +1 ]. Clearly, we have
m EN (p) = 2−N
and
∞
X
2−N = 1.
N =1
Therefore, one can find a N ∈ N such that p ∈ [2−N , 2−N +1 ]. Furthermore, it is easy to check
that the sequence {EN (p)} shrinks to p nicely because
EN (p) ⊆ B(p, 2−N +1 )
1
and m EN (p) = m B(p, 2−N +1 )
4
Chapter 7. Differentiation
238
for every N ∈ N. Using Theorem 7.10 (Lebesgue Differentiation Theorem) and then the representation (7.92), we find that
1
f (p) = lim
−N
N →∞ m [2
, 2−N +1 ]
In other words, the series
∞
X
Z
f dm = lim sN (p) =
N →∞
[2−N ,2−N+1 ]
∞
X
cN ϕN (p).
N =1
cn ϕn converges to f a.e. in [0, 1].
n=1
Finally, we know that
ϕn (t + 1) = ϕ0 2n (t + 1) = ϕ0 (2n t + 2n ) = ϕ0 (2n t) = ϕn (t)
for every n ∈ N, the series and f are functions with period 1. Now for every t ∈ R, we can write
t = t′ + m with t′ ∈ [0, 1] and m ∈ Z. Furthermore, if t is a Lebesgue point of f , then it can be
shown from the definition and the periodicity of f that t′ is also a Lebesgue point of f . By this
fact (for the third equality below), we obtain
∞
X
cn ϕn (t) =
n=1
∞
X
′
cn ϕn (t + m) =
∞
X
cn ϕn (t′ ) = f (t′ ) = f (t′ + m) = f (t),
n=1
n=1
completing the proof of the problem.
Problem 7.19
Rudin Chapter 7 Exercise 19.
Proof. Suppose that c ∈ R. Since f is continuous on R, it is Boreal measurable. Thus the
function fn (x) = nc f (nx) is also Borel measurable for every n ∈ N. By Theorem 1.14, hc (x) is
Borel measurable.
(a) By the hypotheses, it is true that
max f (x) = max f (x) = max f (x)
x∈R
x∈[0,1]
x∈(0,1)
(7.93)
exists and it is positive. Let this number be M . If x ≤ 0, then nx ≤ 0 for all n ∈ N.
Similarly, if x ≥ 1, then nx ≥ 1 for all n ∈ N. In other words, we have
hc (x) = 0
for all x ∈ R \ (0, 1).
Fix x ∈ (0, 1), then 0 < nx < 1 if and only if
0<n<
1
.
x
(7.94)
In this case, there are only finitely many n satisfying the inequalities (7.94), so we have
n
h 1 io M
hc (x) = sup{nc f (nx) | n = 1, 2, 3, . . .} = max nc f (np) n ≤
≤ c.
(7.95)
x
x
Since c ∈ (0, 1), direct computation shows that
Z 1
M
dx = M x1−c |10 = M
c
x
0
239
7.5. Miscellaneous Problems on Differentiation
and then
0≤
Z
R
hc (x) dx =
Z
1
hc (x) dx ≤ M.
0
In conclusion, we have hc ∈ L1 (R) for c ∈ (0, 1).
(b) By the inequality (7.95), we see that
h1 (x) ≤
M
x
(7.96)
for x ∈ (0, 1) and h1 (x) = 0 otherwise. Take λ > 0. If the real number x satisfies
|h1 (x)| = h1 (x) > λ, then we must have x ∈ (0, 1) and furthermore 0 < x < M
λ by the
inequality (7.96). In other words, we have
and so
n
Mo
{x ∈ R | h1 (x) > λ} ⊆ x ∈ R 0 < x <
λ
n
M
Mo
=λ·
= M.
λ · m{|h1 | > λ} ≤ λ · m x ∈ R 0 < x <
λ
λ
By the definition in §7.5, we see that h1 is in weak L1 .
Recall the fact (7.93), so the Extreme Value Theorem [49, Theorem 4.16, p. 89] ensures
that M = f (p) for some p ∈ (0, 1). Given ǫ > 0. Consider the function g : R → R defined
by
g(x) = f (x) − M + ǫ
which is continuous on R. Since g(p) = f (p) − M + ǫ = ǫ > 0, the sign-preserving property
for continuous functions [64, Problem 7.15, p. 112] says that there exists a δ > 0 small
enough such that g(x) > 0 on (p−δ, p+δ) ⊂ (0, 1), i.e., f (x) > M −ǫ on (p−δ, p+δ) ⊂ (0, 1).
Particularly, we may take ǫ = M
2 so that
f (x) >
M
2
on (p − δ, p + δ). Fix this δ. If 0 < y < 2δ, then
Then it is easy to check that
(7.97)
2δ
y
> 1. Suppose that Ny = [ p−δ
y ] + 1.
p−δ
p−δ
p−δ+y
p+δ
< Ny ≤
+1=
<
y
y
y
y
(7.98)
which is equivalent to saying that Ny y ∈ (p−δ, p+δ). Hence it follows from the inequalities
(7.97) and (7.98) that
h1 (y) ≥ Ny f (Ny y) >
However, we know that
Z
2δ
0
Ny M
(p − δ)M
>
.
2
2y
dt
= ∞, so we conclude that h1 ∈
/ L1 (R).
t
(c) Let c > 1 and δ > 0 be the number such that the inequality (7.97) holds on the segment
p+δ
(p − δ, p + δ) ⊂ (0, 1). Now for each k ∈ N, if x ∈ ( p−δ
k , k ), then kx ∈ (p − δ, p + δ) and
thus
M kc
hc (x) ≥ kc f (kx) >
2
Chapter 7. Differentiation
240
c
c
p+δ
Mk
Mk
which means ( p−δ
k , k ) ⊂ {x ∈ R | hc (x) >
2 }. Consequently, if we take λk =
2 ,
then we obtain
p − δ p + δ M k c 2δ
λk · m{hc > λk } > λk · m
=
,
·
= M δkc−1 .
(7.99)
k
k
2
k
As M and δ are fixed as well as c > 1, the inequality (7.99) implies that
λk · m{hc > λk } → ∞
as k → ∞. Then we have proved that hc is not in weak L1 if c > 1.
We have ended the proof of the problem.
Problem 7.20
Rudin Chapter 7 Exercise 20.
Proof. By the definition, we have ∂E = E \ E ◦ .
(a) If m(∂E) = 0, then we have m(E \E ◦ ) = 0. Since E ◦ is the union of all open sets contained
in E and E is the intersection of all closed sets containing E, we have E ◦ ⊆ E ⊆ E and
E ◦ , E ∈ B. In fact, we have
m(E) = m(E ◦ )
by Theorem 1.36.i
measurable.j
Furthermore, since m is a complete measure, E is also Lebesgue
(b) To prove this part, we need stronger versions of Theorems 7.7 and 7.10 (The Lebesgue
Differentiation Theorem). Roughly speaking, the hypothesis f ∈ L1 (Rk ) can be replaced
by f ∈ L1loc (Rk ), where L1loc (Rk ) is the space of all locally integrable functions:
Z
|f | dm < ∞
K
for every bounded measurable set K ⊆ Rk , see [22, p. 95]. For convenience, we state the
stronger versions of these theorems in a single lemma:
Lemma 7.5
(a) If f ∈ L1loc (Rk ), then almost every x ∈ Rk is a Lebesgue point of f .
(b) Associate to each x ∈ Rk a sequence {Ei (x)} that shrinks to x nicely, and
let f ∈ L1loc (Rk ). Then
Z
1
f (x) = lim
f dm
i→∞ m Ei (x)
Ei (x)
at every Lebesgue point of f , hence a.e. [m] on Rk .
i
See also [49, Theorem 2.27, p. 35; Exercise 9, p. 43].
Notice the difference between a Borel measurable set and a Lebesgue measurable set: Not every subset of a
Borel set with measure 0 is also Borel measurable, but Lebesgue measure is obtained by enlarging B to include
all subsets of sets of Borel measure 0.
j
241
7.5. Miscellaneous Problems on Differentiation
Let A be a (possibly uncountable) set. Suppose that
E=
[
B(xα , rα ),
(7.100)
α∈A
where B(xα , rα ) is a closed disc in R2 and rα ≥ 1.
Assume that m(∂E) > 0. Clearly, E ◦ is an open subset of R2 , so it is a Borel subset in
As in §7.12, we consider the characteristic function f = χE ◦ . Denote DA (x) to be the
density of the Lebesgue measurable set A at x. Since f ∈ L1loc (R2 ), we apply Lemma 7.5
to conclude that
Z
m E ◦ ∩ B(x, r)
1
DE ◦ (x) = lim
= lim
χE ◦ dm = χE ◦ (x) = 0
r→0
r→0 m B(x, r)
m B(x, r)
B(x,r)
R2 .
for almost every point x ∈ ∂E ⊆ (E ◦ )c . In particular, we choose a p ∈ ∂E such that
DE ◦ (p) = 0.
(7.101)
To continue our proof, we need the following result:
Lemma 7.6
There is a sequence {Ei (p)} that shrinks to p nicely.
Proof of Lemma 7.6. For every i ∈ N, we have E ∩ B(p, 1i ) 6= ∅. (In fact, this is an
equivalent definition of a boundary point of E, see [3, Definition 3.40, p. 64].) Pick
{qi } to be an arbitrary sequence temporarily such that qi ∈ E ∩ B(p, 1i ). By the
1
definition (7.100), we know that qi ∈ B(xαi , rαi ) for some αi ∈ A. If |p − qi | ≤ 2i
,
then we have
1
1
⊂ B p,
(7.102)
B qi ,
2i
i
1
and we define Ei (p) = B(qi , 2i
). Otherwise, we consider the point
1
s2i ∈ B(xβ2i , rβ2i ) ∩ B p,
2i
1
which satisfies |p − s2i | < 2i
so that if we let qi = s2i and αi = β2i , then we gain
the set inclusion (7.102) in this case. (See Figure 7.2 below.) Now this process can be
done continually and then we obtain a sequence {qi } satisfying
1
1
1
qi ∈ E ∩ B p,
and Ei (p) = B qi ,
⊂ B p, .
i
2i
i
Clearly, we have
1 1 = m B p,
(7.103)
2i
4
i
for every i ∈ N, therefore {Ei (p)} shrinks to p nicely by the definition in §7.9, com
pleting the proof of the lemma.
m(Ei (p)) = π ×
1 2
We go back to the proof of part (b). Recall that qi ∈ B(xαi , rαi ). There are two cases:
Chapter 7. Differentiation
242
Figure 7.2: Construction of the sequence Ei (p).
– Case (i): qi ∈ ∂B(xαi , rαi ). Geometrically, the area of B(qi , r) ∩ B(xαi , rαi ) is
always greater than that of 14 B(qi , r) whenever 0 < r < rαi so that
1
(7.104)
m B(qi , r) ∩ B(xαi , rαi ) ≥ m B(qi , r) .
4
Since B(α, rα ) ⊆ E for all α ∈ A, we have B(α, rα ) ⊆ E ◦ for all α ∈ A. Furthermore,
1
< rαi for all i. Hence these
since rα ≥ 1 for all α ∈ A, we certainly have 2i
observations allow us to apply the equality (7.103) and the inequality (7.104) to get
m E ◦ ∩ B(p, 1i )
DE ◦ (p) = lim
i→∞
m B(p, 1i )
m E ◦ ∩ B(p, 1i )
= lim
i→∞
4m Ei (p)
1
)
m B(xαi , rαi ) ∩ B(qi , 2i
1
≥ · lim
1
4 i→∞
m B(qi , 2i
)
1
≥
16
which contradicts the result (7.101).
1
and qi ∈ B(xαi , rαi ) ∩ B(p, 1i ),
– Case (ii): qi ∈ B(xαi , rαi ). Since |p − qi | < 2i
1
so that the inequality (7.104) holds in this case.
the geometry tells us that rαi > 2i
1
Finally, we have DE ◦ (p) ≥ 16 , a contradiction again.
In conclusion, we have proved that m(∂E) = 0.
1
(c) The crux of the analysis in part (b) is the inequalities rαi > 2i
. For unrestricted radii
rα , we can pick a sequence {rαi } satisfying this condition for sufficiently large i and thus
1
DE ◦ (p) ≥ 16
(d) By the comments in §2.21, there exists a non-Borel subset of R. Let this set be A. Suppose
that
[
B (x, 0), 1 .
E=
x∈A
243
7.5. Miscellaneous Problems on Differentiation
Assume that E was Borel in R2 . If we can show that R × {1} is Borel in R2 , then since
E ∩ (R × {1}) = A × {1},
A is also Borel in R, a contradiction. Hence E must be non-Borel in R2 . Now it remains
to show that R × {1} is Borel in R2 and this is the content of the following lemma:
Lemma 7.7
Let A and B be Boreal subsets of R. Then A × B is also Borel in R2 .
Proof of Lemma 7.7. Suppose that M1 = {A ⊆ R | A × R is Borel in R2 }. It is obvious that R ∈ M1 . Let A ∈ M1 , i.e., A × R is Borel in R2 . Since Ac × R = R2 \ (A × R),
Ac × R is Borel in R2 by Comment 1.6(d). Suppose that An ∈ M1 for all n = 1, 2, . . .,
∞
[
2
(An × R) is Borel in R2 . Since
i.e., each An × R is Borel in R so that
n=1
∞
[
n=1
An
∞
o
n
[
An , y ∈ R
× R = (a, y) a ∈
n=1
= {(a1 , y) | a1 ∈ A1 , y ∈ R} ∪ {(a2 , y) | a2 ∈ A1 , y ∈ R} ∪ · · ·
= (A1 × R) ∪ (A2 × R) ∪ · · ·
∞
[
(An × R),
=
n=1
we establish that
∞
[
n=1
An ∈ M1 .
By Definition 1.3(a), M1 is a σ-algebra in R. Next, let V be open in R. Since V × R
is open in R2 (see [42, p. 86]), it is Borel in R2 and thus V ∈ M1 . Hence B ⊆ M1 .
Similarly, if we define M2 = {B ⊆ R | R × B is Borel in R2 }, then in the same way
as above we can show that M2 is also a σ-algebra in R containing B. Finally, if
A, B ∈ B, then A × R and R × B are Borel in R2 . Since (A × R) ∩ (R × B) = A × B,
our desired result follows.
(e) As Balcerzak and Kharazishvili [6, p. 205] pointed out that the union of an arbitrary
family of convex bodies (compact convex sets with non-empty interiors) in R2 is Lebesgue
measurable. Therefore, the closed discs can be replaced by arbitrary closed polygons
provided that they are convex.k
Hence we have completed the proof of the problem.
Problem 7.21
Rudin Chapter 7 Exercise 21.
k
More generally, it is known that if E =
[
Xα and for each Xα , there exists an open ball Bα such that
Bα ⊆ Xα ⊆ Bα , then E is Lebesgue measurable. See [33, Problem 22, p. 482].
Chapter 7. Differentiation
244
Proof. Notice that the triangle inequality implies that f, g ∈ BV on [a, b] if and only if f +g ∈ BV
on [a, b]. Clearly, the function g(t) = t belongs to BV on [0, 1], so γ ∈ BV on [0, 1] if and only
if f ∈ BV on [0, 1].
By the definition, the length of the graph of f is the total variation of γ on [0, 1], say Vγ (1).
We want to show that
Z 1p
(7.105)
Vγ (1) =
1 + [f ′ (t)]2 dt.
0
Suppose that f is AC on [0, 1]. Then γ is also AC on [0, 1]. Without loss of generality, we may
assume that f (0) = 0; otherwise, we consider the function fe : [0, 1] → R defined by fe = f − f (0).
This assumption implies that γ(0) = 0. By Theorem 7.18,l we know that γ ′ ∈ L1 a.e. [m] on
[0, 1] and
Z
x
γ(x) =
γ ′ (t) dt.
0
Therefore, if x, y ∈ [0, 1] and x < y, then the above fact and Theorem 1.33 establish
Z yp
Z y
Z y
′
′
1 + f ′ (t) dt.
|γ (t)| dt =
γ (t) dt ≤
|γ(y) − γ(x)| =
(7.106)
x
x
x
Finally, it follows from the definition and the inequality (7.106) that if 0 = t0 < t1 < · · · < tN =
1, then we have
Vγ (1) = sup
N
X
i=1
|γ(ti ) − γ(ti−1 | ≤ sup
N Z
X
i=1
ti
ti−1
p
1+
f ′ (t) dt
=
Z
0
1p
1 + f ′ (t) dt.
This proves one direction.
For the other direction, since γ is AC on [0, 1], Vγ is also AC on [0, 1] by Theorem 7.19. Thus
Theorem 7.18 and the fact Vγ (0) = 0 imply that
Z
1
0
Vγ′ (t) dt = Vγ (1).
(7.107)
Employing the notation in the proof of Problem 7.13(c), if 0 ≤ x < y ≤ 1, then we have
Vγ (y) − Vγ (x) = Vγ (x; y) = sup
N
X
i=1
|γ(ti ) − γ(ti−1 )|,
(7.108)
where the supremum is taken over all positive integer N and over all partitions {ti } such that
x = t0 < t1 < · · · < tN = y. By repeated use of the triangle inequality, we certainly have
p
(y − x)2 + [f (y) − f (x)]2 = |γ(y) − γ(x)| ≤
N
X
i=1
|γ(ti ) − γ(ti−1 )| ≤ Vγ (x; y).
(7.109)
Therefore, it follows from the expressions (7.108) and (7.109) that
p
(y − x)2 + [f (y) − f (x)]2 ≤ Vγ (y) − Vγ (x)
and then
l
s
1+
h f (y) − f (x) i2
y−x
≤
Vγ (y) − Vγ (x)
.
y−x
Monotonicity of γ is not necessary for the implication (a) → (c).
(7.110)
245
7.5. Miscellaneous Problems on Differentiation
Recall that f and Vγ are AC on [0, 1], f ′ and Vγ′ exist a.e. [m] on [0, 1] by Theorem 7.18, so the
inequality (7.110) ensures that
p
1 + [f ′ (x)]2 ≤ Vγ′ (x)
a.e. [m] on [0, 1]. Substituting this into the expression (7.107), we get
Z 1p
1 + [f ′ (t)]2 dt ≤ Vγ (1)
0
which proves the other direction. Hence we have completed the proof of the problem.
Problem 7.22
Rudin Chapter 7 Exercise 22.
Proof.
(a) Assume that f 6= 0 on a measurable set E ⊆ Rk of positive measure. We claim that there
corresponds a constant c = c(f ) > 0 such that
(M f )(x) ≥ c|x|−k
for sufficiently large |x|. If it was not the case, then for each n ∈ N, there exists a xn ∈ Rk
with |xn | > n such that
|xn |−k
(M f )(xn ) <
.
n
Recall the basic fact that the Lebesgue measure of the ball B(x, r) in Rk is given by
k
π2
rk .
k
Γ( 2 + 1)
Thus it follows from the definition [51, Eqn. (4), p. 138] and this fact that
Z
k
π2
|xn |−k
k
|f | dm ≤ m(Br ) × (M f )(xn ) <
(7.111)
r
×
n
Γ( k2 + 1)
B(xn ,r)
√
for all r > 0. Particularly, we pick the sequence {rn }, where rn = 2k n · |xn |, and the
inequality (7.111) becomes
Z
Z
k
π2
1
χB(xn ,rn ) |f | dm =
|f | dm < k
×√ .
n
Γ( 2 + 1)
Rk
B(xn ,rn )
(7.112)
By the Archimedean Property [49, Theorem 1.20(a), p. 9], for every x ∈ Rk , there is a
positive integer n such that n ≥ |x|, so we have |x| < n < |xn | and then
√
|x − xn | ≤ |x| + |xn | < 2|xn | < 2k n · |xn |
for large enough n (remember that k is fixed). In other words, this means that x ∈
√
B(xn , 2k n · |xn |) and thus
(7.113)
lim χB(xn , 2k√n|xn |) (x) = 1.
n→∞
Let |fn | = χB(xn ,rn ) |f | for all n ∈ N. Then each |fn | is measurable, |fn | ≤ |f | and the
limit (7.113) implies
|f |(x) = lim |fn |(x)
n→∞
Chapter 7. Differentiation
246
for every x ∈ Rk . Since f ∈ L1 (Rk ), the sequence {|fn |} satisfies the hypotheses of
Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem). Thus we have
Z
Z
Z
|f | dm.
(7.114)
|f | dm = lim
|fn | dm =
lim
n→∞ B(x ,r )
n n
n→∞ Rk
Rk
Now we combine the inequality (7.112) and the result (7.114) to get
Z
k
π2
1
|f | dm ≤ lim
× √ = 0.
k
n→∞
n
Γ( 2 + 1)
Rk
By Theorem 1.39(a), f = 0 a.e. on Rk , a contradiction. Hence there exists a constant
c = c(f ) > 0 and r > 0 such that for all |x| ≥ r, we have
(M f )(x) ≥ c|x|−k
which implies that kM f k1 = ∞, a contradiction. Hence we must have f = 0 a.e. on Rk .
(b) For every x ∈ (0, 41 ), if r ∈ (0, x], then we have x − r ≥ 0 and x + r < 12 . Particularly, we
take r = x so that
Z
Z x+r
Z 2x
1
1
1
1
|f | dm =
dt.
(7.115)
|f (t)| dt =
m(Br ) B(x,r)
2r x−r
2x 0 t(log t)2
Since t(log t)2 > 0 on (0, 21 ) and t−1 (log t)−2 ∈ R on [ǫ, 2x], we apply [49, Theorem 11.33,
p. 323] to obtain
Z
2x
0
1
dt ≥
t(log t)2
Z
2x
ǫ
1
dt = R
t(log t)2
Z
2x
ǫ
for every ǫ > 0. As ǫ → 0+, we know that
Z
2x
0
1
1
dt = −
t(log t)2
log t
1
log ǫ
2x
ǫ
=−
1
1
+
log(2x) log ǫ
→ 0− and so
1
1
.
dt ≥ −
2
t(log t)
log(2x)
(7.116)
1
> 0 and then we substitute the inequality (7.116) into
Since 2x ∈ (0, 21 ), we have − log(2x)
the expression (7.115) to obtain
Z
1
1
1
(M f )(x) ≥ sup
|f | dm ≥ −
=
,
2x log(2x)
2x log(2x)
0<r<∞ m(Br ) B(x,r)
as required.
Furthermore, direct computation gives
Z
1
4
0
dx
= ∞.
|2x log(2x)|
Recall the basic fact that (M f )(x) ≥ 0, so it implies that
Z
0
1
(M f )(x) dx ≥
Z
0
1
4
dx
= ∞,
|2x log(2x)|
as desired.
Hence we have completed the proof of the problem.
247
7.5. Miscellaneous Problems on Differentiation
Problem 7.23
Rudin Chapter 7 Exercise 23.
Proof. We first show that (SF )(x) is well-defined. To this end, let α1 and α2 be two complex
numbers such that
Z
Z
1
1
|f1 − α1 | dm = lim
|f2 − α2 | dm = 0
(7.117)
lim
r→0+ m(Br ) B(x,r)
r→0+ m(Br ) B(x,r)
for some f1 , f2 ∈ F . Then we see that
Z
1
|α1 − α2 | =
|α1 − α2 | dm
m(Br ) B(x,r)
Z
1
≤
(|α1 − f | + |f − g| + |g − α2 |) dm
m(Br ) B(x,r)
Z
Z
1
1
|α1 − f | dm +
|g − α2 | dm
=
m(Br ) B(x,r)
m(Br ) B(x,r)
Z
1
+
|f − g| dm.
m(Br ) B(x,r)
(7.118)
Since f and g belong to the same equivalence class F , f = g a.e. on Rk and the last integral in
the inequality (7.118) is actually 0. By taking r → 0+ to the integrals in the inequality (7.118)
and using the hypotheses (7.117), we obtain α1 = α2 , i.e., (SF )(x) is unique if x is a Lebesgue
point of F .
Next, if x is a Lebesgue point of f and f ∈ F , then Definition 7.6 gives
Z
1
|f (y) − f (x)| dm(y) = 0
lim
r→0+ m(Br ) B(x,r)
so that x is also a Lebesgue point of F and the uniqueness of Lebesgue points of F in the
previous paragraph shows immediately that (SF )(x) = f (x). This completes the proof of the
problem.
Chapter 7. Differentiation
248
8
CHAPTER
Integration on Product Spaces
8.1
Monotone Classes and Ordinate Sets of Functions
Problem 8.1
Rudin Chapter 8 Exercise 1.
Proof. Suppose that
M = {∅, R, (−∞, a), (−∞, a], (b, ∞), [b, ∞) | a, b ∈ R}.
We claim that M is a monotone class. To see this, let Ai ∈ M, Ai ⊂ Ai+1 and A =
have several cases to consider.
∞
[
Ai . We
i=1
• Case (i): Ai = (−∞, ai ). Then the condition Ai ⊂ Ai+1 forces that Ai+1 is in one of the
following forms:
(−∞, ai ], (−∞, ai+1 ) and (−∞, ai+1 ]
(8.1)
for some ai+1 > ai . Thus A is in one of the following forms:
(−∞, a),
and R
(−∞, a]
(8.2)
for some a ∈ R. Both cases say that A ∈ M.
• Case (ii): Ai = (−∞, ai ]. Now Ai+1 is still in one of the forms (8.1), so similar analysis
shows that A is in one of the forms (8.2). Therefore, A ∈ M.
• Case (iii): Ai = (bi , ∞). In this case, Ai+1 is in one of the forms:
[bi , ∞),
(bi+1 , ∞) and
[bi+1 , ∞)
(8.3)
for some bi+1 < bi . Then it can be shown easily that A is in one of the following forms:
(b, ∞),
[b, ∞)
and R
(8.4)
for some b ∈ R. Consequently, we have A ∈ M.
• Case (iv): Ai = [b, ∞). The Ai+1 is expressed in one of the forms (8.3) so that A is in
one of the forms (8.4). In conclusion, we have A ∈ M.
249
Chapter 8. Integration on Product Spaces
Next, let Bi ∈ M, Bi ⊃ Bi+1 and B =
∞
[
i=1
250
Bi . Similar to the sequence of sets {Ai }, there are
also four cases for {Bi }. Since the analysis of these cases is very similar to those above, so we
omit the details here. By Definition 8.1, we see that M is a monotone class.
Clearly, R ∈ M. Since R \ (−∞, a) = [a, ∞), R \ (−∞, a] = (a, ∞), R \ (b, ∞) = (−∞, b] and
R \ [b, ∞) = (−∞, b), we immediately have R \ A ∈ M for every A ∈ M. However, M is not
a σ-algebra because (−∞, 0), (1, ∞) ∈ M but (−∞, 0) ∪ (1, ∞) ∈
/ M which violates Definition
1.3(a). This completes the proof of the problem.
Problem 8.2
Rudin Chapter 8 Exercise 2.
Proof. We have f : R → [0, ∞) and
A(f ) = {(x, y) ∈ R2 | 0 < y < f (x)}.
(8.5)
Now all the answers are affirmative.
(a) Since f is Lebesgue measurable, Theorem 1.17 (The Simple Function Approximation Theorem) ensures the existence of simple measurable functions sn on R such that 0 ≤ s1 ≤
s2 ≤ · · · ≤ f and sn (x) → f (x) as n → ∞ for every x ∈ R.
Suppose that s is a nonnegative simple function on R. Then, by Definition 1.6, we have
s=
n
X
αi χAi ,
i=1
where A1 , A2 , . . . , An can be assumed to be mutually disjoint Lebesgue measurable sets of
R. Therefore, we obtain
A(s) = {(x, y) ∈ R2 | 0 < y < s(x)}
n
[
{(x, y) ∈ R2 | x ∈ Ai and 0 < y < αi }
=
=
i=1
n
[
[Ai × (0, αi )].
i=1
By Definition 8.1, A(s) must be a Lebesgue measurable subset of R2 . It is easy to check
from the definition of the ordinate set of a function that the following lemma holds:
Lemma 8.1
If f and g are functions on R such that f (x) ≤ g(x) for every x ∈ R, then
A(f ) ⊆ A(g).
Now we claim that
A(f ) =
∞
[
A(sn ).
n=1
The direction
∞
[
n=1
A(sn ) ⊆ A(f )
(8.6)
251
8.1. Monotone Classes and Ordinate Sets of Functions
holds trivially because of the fact that sn ≤ f for every n = 1, 2, . . . and Lemma 8.1. For
the other direction, let (x, y) ∈ A(f ). This implies that x ∈ R and 0 < y < f (x). Since
sn (x) → f (x) as n → ∞, there exists a N ∈ N such that 0 < y < sN (x) ≤ f (x) which
means (x, y) ∈ A(sN ) so that
∞
[
A(sn )
A(f ) ⊆
n=1
also holds. Hence we have established the claim (8.6). Finally, since each A(sn ) is a
Lebesgue measurable subset of R2 , the relation (8.6) implies that A(f ) is also a Lebesgue
measurable subset of R2 .
(b) By the definition (8.5), the x-section of A(f ) is given by
[A(f )]x = {y ∈ R | (x, y) ∈ A(f )} = {y ∈ R | 0 < y < f (x)} = 0, f (x) .
Since R is obviously σ-finite, it follows from Theorem 8.6 (see also Definition 8.7) that
Z
Z
f (x) dx.
m [A(f )]x dx =
m2 A(f ) = (m × m) A(f ) =
R
R
(c) Recall that the graph of f , namely G, is given by
G = {(x, y) ∈ R2 | y = f (x)}.
For each i ∈ N, the set A(f + 1i ) is Lebesgue measurable in R2 by part (a). Then the set
A=
∞
\
1
A f+
= {(x, y) ∈ R2 | 0 < y ≤ f (x)}
i
i=1
is also Lebesgue measurable by Comment 1.6(c). Therefore, the set
A \ A(f ) = {(x, y) ∈ R2 | y = f (x) but y > 0}
is Lebesgue measurable in R2 . Obviously, we have
{(x, 0) ∈ R2 | 0 = f (x)} = {x ∈ R | f (x) = 0} × {0} = f −1 (0) × {0}.
Since f is Lebesgue measurable, f −1 (0) is a Lebesgue measurable set in R by Definition
1.3(c).a Since {0} is a Lebesgue measurable set in R, we use Definition 8.1 to conclude
that f −1 (0) × {0} is Lebesgue measurable in R2 . Since
G = A \ A(f ) ∪ [f −1 (0) × {0}],
the preceding analysis shows that G is a Lebesgue measurable subset of R2 .
(d) For each fixed x ∈ R, we consider the x-section of G:
Gx = {y ∈ R | (x, y) ∈ G} = {y ∈ R | y = f (x)} = {f (x)}.
By a similar argument as in part (b), we conclude easily from this that
Z
m(Gx ) dx = 0
m2 (G) = (m × m)(G) =
R
as required.
This completes the proof of the problem.
a
Of course, we replace “for every open set V in Y ” by “for every closed set V in Y ” here.
Chapter 8. Integration on Product Spaces
8.2
252
Applications of the Fubini Theorem
Problem 8.3
Rudin Chapter 8 Exercise 3.
Proof. Note that
ϕ(x) =
for each x ∈ (0, 1). Suppose that
Z
1
f (x, y) dy
0
1 2
−1
f (x, y) = y (x− 2 )
in (0, 1) × (0, 1). It is clear that f is a positive continuous function in (0, 1) × (0, 1). If x = 21 ,
then f ( 12 , y) = y −1 so that
1 Z 1
y −1 dy = ln y|10 = ∞.
=
ϕ
2
0
If x 6= 21 , then we have
ϕ(x) =
Z
1
1 2
−1
y (x− 2 )
dy =
0
1
(x − 21 )2
and this implies that
Z
1
ϕ(x) dx =
0
Z
1
0
1
1 −1
dx
=
−
x
−
2
(x − 21 )2
1
0
= −4.
This is a required example, completing the proof of the problem.
Problem 8.4
Rudin Chapter 8 Exercise 4.
Proof.
(a) As Rudin pointed out, we may assume that f and g are Borel functions on R. If p = 1,
then it is in fact Theorem 8.14. If p = ∞, then kgk∞ < ∞ and so
|(f ∗ g)(x)| =
Z
Z
∞
−∞
∞
f (x − t)g(t) dt
|f (x − t)| · |g(t)| dt
Z ∞
|f (x − t)| dt
≤ kgk∞
Z−∞
∞
|f (t)| dt
= kgk∞
≤
−∞
−∞
= kgk∞ · kf k1 .
Thus we have kf ∗ gk∞ ≤ kf k1 · kgk∞ .
253
8.2. Applications of the Fubini Theorem
1
Suppose that 1 < p < ∞. Since f and g are measurable on R, the functions |f (x − t)| q
1
and |f (x − t)| p · |g(y)| are also measurable on R by Theorem 1.7, where q is the conjugate
exponent of p. By Theorem 3.5 (Hölder’s Inequality), we derive that
Z ∞
Z ∞
1
1
|f (x − y)g(y)| dy =
|f (x − y)| q · |f (x − y)| p |g(y)| dy
−∞
−∞
nZ ∞
o1 n Z ∞
o1
q
p
p
≤
|f (x − y)| dy
×
|f (x − y)| · |g(y)| dy
−∞
−∞
nZ ∞
o1
1
p
|f (x − y)| · |g(y)|p dy
= kf k1q ·
−∞
which gives
nZ
∞
−∞
|f (x − y)g(y)| dy
Define F : R2 → C by
op
p
q
≤ kf k1
Z
∞
−∞
|f (x − y)| · |g(y)|p dy.
(8.7)
F (x, y) = f (x − y)g(y).
Using similar argument as in the proof of Theorem 8.14, we are able to show that the
function F ∈ L1 (R2 ). Since R is a σ-finite measure space, we get from Theorem 8.8 (The
Fubini Theorem) that
Z ∞
Z ∞
f (x − y)g(y) dy = (f ∗ g)(x)
Fx (y) dy =
ϕ(x) =
−∞
−∞
L1 (R)
a.e. is in
so that the integral defining (f ∗ g)(x) exists for almost all x. This proves
the first assertion.
To verify the second and the third assertions, we recall that |f |, |g|p ∈ L1 (R), so Theorem
8.14 ensures that the function h : R → [0, ∞] given by
Z ∞
|f (x − y)| · |g(y)|p dy
h(x) =
−∞
L1 (R)
belongs to
which means that h is a measurable function on R. By Theorem 1.33,
we observe from the inequality (8.7) that
Z ∞
p
f (x − y)g(y) dy
|(f ∗ g)(x)|p =
−∞
nZ ∞
op
≤
|f (x − y)g(y)| dy
−∞
Z ∞
p
q
≤ kf k1
|f (x − y)| · |g(y)|p dy.
(8.8)
−∞
Now we apply Theorem 8.8 (The Fubini Theorem) to the right-hand side of the inequality
(8.8) to get
Z ∞
Z ∞Z ∞
p
q
p
|(f ∗ g)(x)| dx ≤ kf k1
|f (x − y)| · |g(y)|p dy dx
−∞
−∞
Z−∞
∞ Z ∞
p
= kf k1q
|f (x − y)| · |g(y)|p dx dy
−∞ −∞
Z ∞
p
q
= kf k1 · kf k1
|g(y)|p dy
=
kf kp1
· kgkpp
which implies that kf ∗ gkp ≤ kf k1 · kgkp < ∞.
−∞
Chapter 8. Integration on Product Spaces
254
(b) Let p = 1. If f ≥ 0 and g ≥ 0, then the definition gives
Z ∞
f (x − y)g(y) dy ≥ 0.
h(x) = (f ∗ g)(x) =
(8.9)
−∞
Therefore we deduce from Theorem 8.8 (The Fubini Theorem) that
kf ∗ gk1 = khk1
Z ∞
h(x) dx
=
−∞
Z ∞Z ∞
f (x − y)g(y) dy dx
=
−∞ −∞
Z ∞
Z ∞
f (x − y) dx
g(y) dy
=
−∞
= kgk1 · kf k1
−∞
as required.
Let p = ∞. If f ≥ 0 and g = c ≥ 0, then the inequality (8.9) holds in this case. Recall
that f ∈ L1 (R) and g ∈ L∞ (R), so we have
Z ∞
f (x − y)c dy = ckf k1 < ∞
|h(x)| =
−∞
for almost all x ∈ R. By Definition 3.7, we obtain
kf ∗ gk∞ = khk∞ = kf k1 × c = kf k1 · kgk∞ .
(c) Suppose that 1 < p < ∞ and kf ∗ gkp = kf k1 · kgkp . By the definition of f ∗ g and the
application of part (a), we observe that
kf ∗ gkp ≤ |f | ∗ |g|
which forces |f | ∗ |g|
p
= |f |
1
p
≤ |f |
· |g|
p
1
· |g|
p
= kf k1 · kgkp = kf ∗ gkp
. Thus we may assume that f, g ≥ 0.
Recall that the inequality kf ∗ gkp ≤ kf k1 · kgkp derives from (see the deduction of the
inequality (8.8))
nZ ∞
op
(f ∗ g)(x)p =
f (x − y)g(y) dy
−∞
nZ ∞
op
1
1
=
f (x − y) q · [f (x − y) p g(y)] dy
−∞
op n Z ∞
nZ ∞
o
q
(8.10)
f (x − y) dy
≤
f (x − y)g(y)p dy
×
−∞
−∞
nZ ∞
o
p
q
= kf k1 ×
f (x − y)g(y)p dy
−∞
and so the equality kf ∗ gkp = kf k1 · kgkp forces that the equality (8.10) holds, but this
means that the equality of Theorem 3.5 (Hölder’s Inequality) holds there i.e., there are
constants (with respect to y) α(x) and β(x), not both zero, such that
q
1
α(x)f (x − y) = α(x)f (x − y) q = β(x)[f (x − y) p g(y)]p = β(x)f (x − y)g(y)p
(8.11)
255
8.2. Applications of the Fubini Theorem
for almost all y ∈ R. Let E = {z ∈ R | f (−z) > 0}. If kf k1 = 0, then since m(R) = ∞,
f = 0 a.e. on R and we are done. Therefore, without loss of generality, we may assume
that kf k1 = 1 which gives m(E) > 0.
On the one hand, if β(x) = 0 for some x ∈ R, then α(x) 6= 0 and so the equation (8.11)
implies that f (x − y) = 0 for almost all y ∈ R which is equivalent to f = 0 a.e. on R. On
the other hand, if α(x) = 0 for some x ∈ R, then β(x) 6= 0 and so
f (x − y)g(y)p = 0
(8.12)
for almost all y ∈ R. Now if y ∈ Ex = x+E = {x+z | z ∈ E}, then y = x+z or −z = x−y
for some z ∈ E so that
f (x − y) > 0
for all y ∈ Ex . Therefore, it follows from the equation (8.12) that g(y) = 0 for almost all
y ∈ Ex . Assume that m(Ex ) < ∞. Then m(Exc ) = ∞. Furthermore, we have g(y) > 0 for
all y ∈ Exc and this implies that
Z
Z
g(y)p dy = ∞
g(y)p dy =
kgkpp =
R
Exc
which contradicts g ∈ Lp (R).
(d) Suppose that p = ∞. We simply take f (x) = χ[0,1] and g(x) = 1 on R. Then we have
kf k1 = kgk∞ = 1. Since f ≥ 0 and g ≥ 0 on R, we deduce from Theorem 7.26 (The
Change-of-variables Theorem) that
Z 1
Z ∞
dy = 1.
f (y)g(x − y) dy =
(f ∗ g)(x) =
0
−∞
Hence we always have
kf ∗ gk∞ = 1 > (1 − ǫ)kf k1 · kgk∞ .
Next, we suppose that p < ∞. Let 0 < α < 1 − (1 − ǫ)p < 1 and define
f (x) =
1
χ
(x)
2α [−α,α]
and g(x) =
1
χ
(x).
2p [−1,1]
It is easy to see that
kf k1 =
Z
kgkp =
Z
and
∞
−∞
∞
−∞
|f (x)| dx =
p
Z
|g(x)| dx =
α
−α
Z
1
χ
(x) dx = 1
2α [−α,α]
1
−1
1
χ
(x) dx = 1.
2 [−1,1]
Since f ≥ 0 and g ≥ 0 on R, we deduce from Theorem 7.26 (The Change-of-variables
Theorem) that
Z ∞
Z α
1
f (y)g(x − y) dy =
(f ∗ g)(x) =
χ[−1,1] (x − y) dy.
(8.13)
1
−∞
2α · 2 p −α
Since −1 ≤ x − y ≤ 1 if and only if x − 1 ≤ y ≤ x + 1, we follow from the expression (8.13)
that
Z α
1
χ[x−1,x+1](y) dy
(f ∗ g)(x) =
1
2α · 2 p −α
Chapter 8. Integration on Product Spaces
1
=
=
1
p
256
m [−α, α] ∩ [x − 1, x + 1]
2α · 2

0,








x+1+α



1


 2α · 2 p




 2α
1

2α · 2 p







1+α−x



1


2α · 2 p






0,
if x ∈ (−∞, −1 − α];
if x ∈ (−1 − α, −1 + α];
if x ∈ (−1 + α, 1 − α];
if x ∈ (1 − α, 1 + α];
if x ∈ (1 + α, ∞).
Hence direct computation shows that
kf
∗ gkpp
=
=
=
Z
∞
|(f ∗ g)(x)|p dx
−∞
Z 1+α
|(f ∗ g)(x)|p dx
−1−α
Z −1+α
−1−α
Z 1+α
+
=2
Z
1
(x + 1 + α)p dx +
2(2α)p
1
(1 + α − x)p dx
2(2α)p
1−α
−1+α
−1−α
Z
1−α
−1+α
(2α)p
dx
2(2α)p
1
(x + 1 + α)p dx + (1 − α).
2(2α)p
(8.14)
Since (x+1+α)p ≥ 0 on [−1−α, −1+α], the integral in the expression (8.14) is nonnegative
so that
kf ∗ gkpp ≥ 1 − α > (1 − ǫ)p
and this is equivalent to saying that kf ∗ gkp > (1 − ǫ)kf k1 · kgkp .
This completes the proof of the problem.
Remark 8.1
The result in Problem 8.4(a) is sometimes called Minkowski’s Inequality which is a special
case of the Young’s Convolution Inequality.
Problem 8.5
Rudin Chapter 8 Exercise 5.
Proof.
(a) Clearly, the function α : R2 → R defined by α(x, y) = x + y is continuous. We want to
show the following result first:
257
8.2. Applications of the Fubini Theorem
Lemma 8.2
It is true that α−1 (E) ∈ B2 if E ∈ B1 . (See Problem 8.11 for the meaning of the
notation Bk for any k ∈ N.)
Proof of Lemma 8.2. Let M = {E ⊆ R | α−1 (E) ∈ B2 }. Since α is continuous,
α−1 (V ) ∈ B2 for every open set V in R. In other words, M contains the standard
topology of R. It is easy to check that
– R ∈ M.
– If E ∈ M, then E c ∈ M because α−1 (E c ) = α−1 (R)\α−1 (E) = R2 \α−1 (E) ∈ B2 .
– If En ∈ M so that α−1 (En ) ∈ B2 for every n ∈ N, then we have
α−1
∞
[
n=1
∞
[
En =
α−1 (En ) ∈ B2 .
n=1
By Definition 1.3(a), M is a σ-algebra and so M = B1 which completes the proof of
Lemma 8.2
Let’s return to the proof of the problem. We want to check Definition 6.1:
(µ ∗ λ)(E) =
∞
X
i=1
(µ ∗ λ)(E i )
(8.15)
for every partition {E i } ⊆ B1 of E ∈ B1 . For each i = 1, 2, . . . , we denote
E2i = {(x, y) ∈ R2 | x + y ∈ E i } ⊆ R2
and E2 = {(x, y) ∈ R2 | x + y ∈ E} ⊆ R2 .
Then we have
E2 =
∞
[
E2i .
i=1
E2j
∩
for i 6= j, then x0 + y0 ∈ E i ∩ E j but E i ∩ E j = ∅, a
Next, if (x0 , y0 ) ∈
contradiction. Thus the set {E21 , E22 , . . .} must be a partition of E2 and since µ × λ is a
measure by Definition 8.7, we get
E2i
(µ ∗ λ)(E) = (µ × λ)(E2 ) =
∞
∞
X
X
(µ ∗ λ)(E i ),
(µ × λ)(E2i ) =
i=1
(8.16)
i=1
i.e., µ ∗ ν ∈ M and this proves the first assertion.
For the second assertion, since µ and λ are complex measures on R, Theorem 6.12
implies that
dµ = h1 d|µ| and dλ = h2 d|λ|,
where |h1 | = |h2 | = 1 on R. For any A ∈ B(R2 ), we observe from Definition 8.7 that
Z
λ(Ax ) dµ(x)
(µ × λ)(A) =
X
Z
λ(Ax )h1 (x) d|µ|(x)
=
X
Chapter 8. Integration on Product Spaces
=
=
Z nZ
ZX n ZY
Y
X
258
o
χAx (y) dλ(y) h1 (x) d|µ|(x)
o
χA (x, y)h2 (y) d|λ|(y) h1 (x) d|µ|(x).
(8.17)
Since |µ| and |λ| are finite by Theorem 6.4, we obtain
Z
Z
∗
ϕ∗ d|µ| = |λ|(Y )|µ|(X) < ∞.
|g|x d|λ| = |λ|(Y ) and
ϕ (x) =
X
Y
Thus, by Theorem 8.8 (The Fubini Theorem), the integral (8.17) can be reduced to
Z Z
χA (x, y)h2 (y)h1 (x) d|λ|(y) d|µ|(x)
(µ × λ)(A) =
X
Y
Z
χA (x, y)h2 (y)h1 (x) d(|µ| × |ν|)
=
X×Y
Z
f (x, y) d(|µ| × |ν|),
(8.18)
=
A
where f (x, y) = h2 (y)h1 (x). Obviously, Theorem 8.8 (The Fubini Theorem) also says that
f (x, y) ∈ L1 (|µ| × |λ|) so that Theorem 6.13 may be applied to the result (8.18) to get
Z
Z
d(|µ| × |λ|) = (|µ| × |λ|)(A),
(8.19)
|f (x, y)| d(|µ| × |λ|) =
|µ × λ|(A) =
A
A
i.e., the total variation of µ × λ is |µ| × |λ|.
Since µ ∗ λ ∈ M , the series (8.16) converges absolutely and then the fact (8.16) implies
|(µ ∗ λ)|(R) ≤
∞
X
i=1
|µ ∗ λ|(E i ) ≤
∞
X
i=1
|(µ × λ)|(E2i )
(8.20)
where {E i } is a partition of R. By Theorem 6.2, |µ × λ| is a positive measure. Recall that
{E2i } is a partition of E2 , so if we apply these facts and the result (8.19) to the inequality
(8.20), then it becomes
|(µ ∗ λ)|(R) ≤ |(µ × λ)|(E2 ) = (|µ| × |λ|)(E2 ) ≤ (|µ| × |λ|)(R2 ).
Hence we follow from the definition of the norm in M and Definition 8.7 (R is σ-finite)
that
kµ ∗ λk = |µ ∗ λ|(R) ≤ (|µ| × |λ|)(R × R) = |µ|(R) × |λ|(R) = kµk · kλk.
(8.21)
(b) Firstly, for E ∈ B1 , we know from the definition of µ ∗ λ, Definition 8.7 and then [51, Eqn.
(3), p. 163] that
Z
χE (x) d(µ ∗ λ) = (µ ∗ λ)(E)
R
= (µ × λ)(E2 )
Z
λ(E2y ) dµ(x)
=
ZR Z
χE (x + y) dµ(x) dλ(y).
=
R
R
259
8.2. Applications of the Fubini Theorem
Thus the formula holds for χE and then it also holds for any simple function s =
R
αi χAi ,
i=1
where each Ai is a Borel set in R, i.e.,
Z Z
Z
s(x + y) dµ(x) dλ(y).
s(x) d(µ ∗ λ) =
R
n
X
(8.22)
R
Next, we let f ∈ C0 (R) so that f : R → C is continuous. By Definition 3.16, f is
bounded by a positive constant M on R and Borel measurable. By [22, Theorem 2.10(b),
p. 47], there exist simple Borel measurable functions sn : R → C such that
0 ≤ |s1 | ≤ |s2 | ≤ · · · ≤ |f |
(8.23)
and sn (x) → f (x) as n → ∞ for every x ∈ R. Now we replace s by sn in the formula
(8.22) to get
Z Z
Z
sn (x + y) dµ(x) dλ(y).
(8.24)
sn (x) d(µ ∗ λ) =
R
R
R
By Theorem 6.12, there is a Borel measurable function h such that |h(x)| = 1 on R and
d(µ ∗ λ) = h d|µ ∗ λ|. If we consider the sequence {sn h} of Borel measurable functions,
then the inequalities (8.23) are still valid. Furthermore, we have sn (x)h(x) → f (x)h(x) as
n → ∞ for every x ∈ R. By the inequality (8.21), we see that
Z
|f (x)h(x)| d|µ ∗ λ| ≤ M |(µ ∗ λ)|(R) ≤ M |µ|(R)| · λ|(R) < ∞.
kf hk1 =
R
Therefore, it means f h ∈ L1 (µ ∗ λ) and we apply Theorem 1.34 (Lebesgue’s Dominated
Convergence Theorem) to the left-hand side of the formula (8.24) to obtain
Z
Z
Z
Z
f d(µ ∗ λ).
f (x)h(x) d|µ ∗ λ| =
sn (x)h(x) d|µ ∗ λ| =
sn (x) d(µ ∗ λ) = lim
lim
n→∞ R
n→∞ R
R
R
Now we may apply similar analysis to show that the limit of the right-hand side of the
formula (8.24) is exactly
Z Z
f (x + y) dµ(x) dλ(y).
R
Hence we conclude that
Z
R
R
f d(µ ∗ λ) =
Z Z
R
f (x + y) dµ(x) dλ(y).
(8.25)
R
Since µ ∗ λ ∈ M , §6.18 on p. 130 indicates that the mapping
Z
f d(µ ∗ λ)
f 7→
R
is a bounded linear functional on C0 (R). Consequently, Theorem 6.19 (The Riesz Representation Theorem) ensures that µ ∗ ν is the unique complex Borel measure satisfying the
formula (8.25).
(c) Since R is commutative, x + y ∈ E if and only if y + x ∈ E. Thus it follows from the
formula (8.25) and Theorem 8.8 (The Fubini Theorem) that
Z
Z Z
Z Z
Z
f d(µ ∗ λ) =
f (x + y) dµ(x) dλ(y) =
f (y + x) dλ(y) dµ(x) = f d(λ ∗ µ).
Chapter 8. Integration on Product Spaces
260
Therefore, the uniqueness property in part (b) implies that µ ∗ λ = λ ∗ µ.
On the one hand, it is clear from the formula (8.25) and Theorem 8.8 (The Fubini
Theorem) that
Z Z Z
Z
f (x + y + z) dµ(x) dλ(y) dν(z).
(8.26)
f d[(µ ∗ λ) ∗ ν] =
R
R
R
R
On the other hand, we also have
Z nZ
Z
o
f (x + Y ) dµ(x) d(λ ∗ ν)(Y )
f d[µ ∗ (λ ∗ ν)] =
R
R
| R
{z
}
F
Z Z Z
f (x + y + z) dµ(z) dλ(y) dν(z).
=
R
R
(8.27)
R
Thus the associativity of the convolution in M follows immediately from the formulas
(8.26) and (8.27), i.e., (µ ∗ λ) ∗ ν = µ ∗ (λ ∗ ν).
By the formula (8.25) again, we see that
Z Z
Z
f (x + y) dµ(x) d[λ(y) + ν(y)]
f d µ ∗ (λ + ν) =
R
Z Z
ZR ZR
f (x + y) dµ(x) dν(y)
f (x + y) dµ(x) dλ(y) +
=
R R
Z
ZR R
f d(µ ∗ ν)
f d(µ ∗ λ) +
=
R
ZR
f d(µ ∗ ν + µ ∗ ν)
=
R
which shows immediately that the convolution in M is distributive with respect to addition,
i.e., µ ∗ (λ + ν) = µ ∗ λ + µ ∗ ν.
(d) By Definition 8.7, we have
(µ × λ)(Q) =
Z
y
µ(Q ) dλ(y) =
R
Z nZ
Qy
o
dµ(x) dλ(y),
(8.28)
where Qy = {x ∈ R | (x, y) ∈ Q}. Let E ∈ B1 and y ∈ R. Then we have
E2y = {x ∈ R | (x, y) ∈ E2 } = {x ∈ R | x + y ∈ E} = {x − y ∈ R | x ∈ E} = E − y.
By the formula (8.28), we have
Z nZ
Z nZ
o
dµ(x) dλ(y) =
(µ × λ)(E2 ) =
E2y
E−y
Z
o
dµ(x) dλ(y) = µ(E − y) dλ(y)
as required.
(e) Suppose that µ and λ are concentrated on the countable sets A and B respectively, i.e.,
µ(E) = λ(F ) = 0
whenever E ∩ A = ∅ and F ∩ B = ∅, where A, B ∈ B1 . Let Q ∈ B2 and Q ∩ (A × B) = ∅.
Then Theorem 8.2 implies that Qx ∈ B1 for every x ∈ R. Assume that y0 ∈ Qx ∩ B. Then
(x, y0 ) ∈ Q for every x ∈ R. However, this means that (x0 , y0 ) ∈ Q ∩ (A × B) for some
261
8.2. Applications of the Fubini Theorem
x0 ∈ R, a contradiction. As a result, it must be true that Qx ∩ B = ∅. Similarly, we can
show that Qy ∩ A = ∅. Therefore, we have
λ(Qx ) = 0
which definitely give
(µ × λ)(Q) =
Z
R
and µ(Qy ) = 0
λ(Qx ) dµ(x) =
Z
µ(Qy ) dλ(y) = 0
(8.29)
R
by Definition 8.7. In other words, µ × λ is concentrated on A × B.b
To continue the proof, we need the following lemma:
Lemma 8.3
E2 ∩ (A × B) = ∅ if and only if E ∩ (A + B) = ∅.
Proof of Lemma 8.3. It is easy to see that x = a + b ∈ E ∩ (A + B) if and only if
(a, b) ∈ E2 ∩ (A × B), so our expected result follows.
By the definition, if E ∩ (A + B) = ∅, then Lemma 8.3 implies that E2 ∩ (A × B) = ∅
and the analysis in the preceding paragraph shows that
(µ ∗ λ)(E) = (µ × λ)(E2 ) = 0.
Since A + B is countable, we have established the result that µ ∗ λ is discrete.
Next, suppose that µ is continuous and λ ∈ M . By part (d), if x ∈ R, then we have
Z
Z
Z
0 dλ(t) = 0.
µ({x − t}) dλ(t) =
µ({x} − t) dλ(t) =
(µ ∗ λ)({x}) =
R
R
R
Thus µ ∗ λ is continuous.
Finally, let µ ≪ m and E ∈ B1 satisfy m(E) = 0. By Theorem 2.20(c), we have
m(E − t) = m(E) = 0 for every t ∈ R and then µ(E − t) = 0 for every t ∈ R. By this fact
and part (d), we achieve
Z
Z
0 dλ(t) = 0.
µ(E − t) dλ(t) =
(µ ∗ λ)(E) =
R
R
By Definition 6.7, we conclude that µ ∗ λ ≪ m.
(f) By part (b) and the assumptions, we note that
Z Z
Z Z
Z
f (x)g(y) dm dm.
dµ(x) dλ(y) =
d(µ ∗ λ) =
R
R
R
R
By Theorem 8.14, since
f ∗g =
Z
R
(8.30)
R
f (x − y)g(y) dm,
we may further reduce the expression (8.30) to
Z
Z Z
Z
f (x − y)g(y) dm dm. = (f ∗ g) dm
d(µ ∗ λ) =
R
R
R
R
and this is equivalent to d(µ ∗ λ) = (f ∗ g) dm.
b
It seems that it is enough to assume either µ or λ is discrete for the validity of the result (8.29). However, as
Definition 8.7 indicates, (µ × λ)(Q) can be computed by using two different integrals, so both µ and λ must be
assumed to be discrete in order that the two integrals there are equal and thus (µ × λ)(Q) makes sense.
Chapter 8. Integration on Product Spaces
262
(g) To being with, suppose that X1 = {µ ∈ M | µ is discrete}, X2 = {µ ∈ M | µ is continuous}
and X3 = {µ ∈ M | µ ≪ m}.
Recall from [40, Definition 3.3.1, p. 305] that a set X with binary operations + and ∗
from X × X to X, and that · is a binary operation from F × X to X is called an algebra
if
– (X, +, ·) is a vector space;
– for all x, y, z ∈ X and every scalar α ∈ F, we have
Condition (1). x ∗ (y ∗ z) = (x ∗ y) ∗ z;
Condition (2). x ∗ (y + z) = (x ∗ y) + (x ∗ z) and (x + y) ∗ z = (x ∗ z) + (y ∗ z);
Condition (3). α · (x ∗ y) = (α · x) ∗ y = x ∗ (α · y).
A subset of X is called a subalgebra if it is an algebra.
Now let’s prove the assertions one by one:
– Case(1): (X1 , +, ∗, ·) is an algebra. If µ and λ are concentrated on countable
(Borel) sets A and B respectively, then µ(E) = 0 and λ(F ) = 0 whenever E ∩ A = ∅
and F ∩ B = ∅. Let C = A ∪ B. If E ∩ (A ∪ B) = ∅, then we must have E ∩ A = ∅
and E ∩ B = ∅ which imply immediately that
(αµ + βλ)(E) = αµ(E) + βλ(E) = 0,
where α and β are real. Thus (X1 , +, ·) is a vector space. Next, by part (e), we know
that the map X1 × X1 → X1 given by
(µ, λ) 7→ µ ∗ λ
is well-defined. Thus it can be seen that Conditions (1) to (3) come directly from
the results of part (c). By the definition, (X1 , +, ∗, ·) is an algebra.
– Case (2): X2 is an ideal in M . Recall that Y is called an ideal of an algebra X
if Y is a vector subspace of X and xY, Y x ⊆ Y for every x ∈ X. Here
xY = {x ∗ y | y ∈ Y }
and Y x = {y ∗ x | y ∈ Y }.
Let µ, λ ∈ X2 so that µ({x}) = λ({x}) = 0 for every x ∈ R, but these imply trivially
that
(αµ + βλ)({x}) = αµ({x}) + βλ({x}) = 0
for every real α and β. In other words, (X2 , +, ·) is a vector space. Since M is
commutative by part (c), it suffices to prove X2 λ ⊆ X2 for every λ ∈ M , but this
fact follows immediately from part (e). Hence we may conclude that X2 is an ideal
in M .
– Case(3): X3 is an ideal in M . Let µ, λ ∈ X3 . Then µ(E) = λ(E) = 0 whenever
m(E) = 0. For every α, β ∈ R, if m(E) = 0, then
(αµ + βλ)(E) = αµ(E) + βλ(E) = 0.
Therefore, it means that (αµ + βλ) ≪ m and so (X3 , +, ·) is a vector space. Next,
for any µ ∈ X3 and λ ∈ M , µ ∗ λ ≪ m by part (e), so we have µλ ∈ X3 and X3 is an
ideal in M by the definition.
Recall that m is a positive σ-finite measure, thus for every µ ∈ X3 , we haveµ ≪ m
so that Theorem 6.10 (The Lebesgue-Radon-Nikodym Theorem) ensures that there
263
8.2. Applications of the Fubini Theorem
is a unique f ∈ L1 (R) such that dµ = f dm. Therefore, it is reasonable to define
Φ : (X3 , ∗) → L1 (R), ∗ by
Φ(µ) = f.
We claim that Φ is in fact an isomorphism. Let λ be another element in X3 . Then
there is a unique g ∈ L1 (R) such that dλ = g dm. By part (f), we have
d(µ ∗ λ) = (f ∗ g) dm.
(8.31)
By Theorem 6.10 (The Lebesgue-Radon-Nikodym Theorem) again, the h ∈ L1 (R)
satisfying the formula (8.31) must be unique. Hence we have
Φ(µ ∗ ν) = f ∗ g,
i.e., Φ is a homomorphism. Next, if Φ(µ) = Φ(λ), then dµ = dλ which implies
exactly µ = λ. Thus Φ is injective. Next, suppose that f ∈ L1 (R). By Definition
1.31, if f = u + iv, where u and v are real measurable functions, then we have
Z
Z
Z
Z
Z
+
−
+
v − dm.
(8.32)
v dm − i
u dm + i
u dµ −
f dm =
µ(E) =
E
E
E
E
E
where E ∈ B1 . Since u± and v ± are measurable and nonnegative, Theorem 1.29
guarantees that every integral in the expression (8.32) is a measure on B1 . Thus, if
m(E) = 0, then Proposition 1.24(e) tells us that µ(E) = 0. By Definition 6.7, µ ≪ m
or equivalently µ ∈ X3 and Φ is surjective.
In conclusion, Φ is an isomorphism, as required.
(h) Consider the Dirac measure δ0 : B1 → [0, ∞], that is

 1, if 0 ∈ E;
δ0 (E) =

0, otherwise.
It is clear that δ0 ∈ M and part (d) shows that
Z
µ(E − t) dδ0 (t) = µ(E)
(δ0 ∗ µ)(E) =
R
for all µ ∈ M . Hence δ0 is a unit of M .
(i) Suppose that M is defined to be the Banach space of all complex Borel measures on Rk
with norm kµk = |µ|(Rk ) and associate to each Borel set E ⊆ Rk the set
E2 = {(x, y) | x + y ∈ E} ⊆ Rk × Rk .
If µ, λ ∈ M , we define
(µ ∗ λ)(E) = (µ × λ)(E2 )
for every Borel set E ⊆ Rk . Then all the assertions from parts (a) to (h) also hold when we
replace R and B1 by Rk and Bk respectively because Rk is always a commutative group
and mk is a translation invariant Borel measure on Rk . The proofs go exactly the same,
so we omit the details here.
For the k-dimensional torus T k , we need to seek a measure similar to the Lebesgue
measure mk on Rk . Note that T k is a locally compact group with the σ-algebra B(T k ),
so the Haar measure denoted by σk is what we want. (For the existence, uniqueness
and basic properties of such a measure, the reader is suggested to read, for examples, [16,
Chapter 8. Integration on Product Spaces
264
Chap. 9], [22, §11.1, pp. 339 – 348] and [48, pp. 128 – 132, 211].) Once we have such a
measure, we can follow the same line as above to obtain the corresponding results. Again,
we omit the details here.
This completes the analysis of the problem.
Problem 8.6
Rudin Chapter 8 Exercise 6.
Proof. Let x ∈ Rk \ {0} and r = |x|. Then we have r > 0 so that it is meaningful to define
u = xr which certainly gives the representation
x = ru and
|u| = 1.
(8.33)
Suppose that x = r ′ u′ , where r ′ > 0 and |u′ | = 1. Then the equality ru = r ′ u′ implies that
r = |ru| = |r ′ u′ | = r ′
and so u = u′ . In other words, the representation (8.33) is unique. Therefore, we may “identify”
Rk \ {0} as (0, ∞) × Sk−1 , i.e., the map
ϕ : Rk \ {0} → (0, ∞) × Sk−1
is a homeomorphism with the continuous inverse
ϕ−1 : (0, ∞) × Sk−1 → Rk \ {0}
and they are given by
x
and ϕ−1 (r, u) = ru
ϕ(x) = |x|,
|x|
(8.34)
respectively. To proceed further, we need the following result:c
Lemma 8.4
Let MX and MY be σ-algebras in X and Y respectively. Suppose that ϕ : X → Y
is measurable and µ : MX → [0, ∞] is a measure. Define ν = ϕ∗ µ by
ν(E) = µ ϕ−1 (E)
for E ∈ MY . Then the set function ν is a measure on MY and for every measurable
function g : Y → [0, ∞], we have
Z
Z
(g ◦ ϕ) dµ.
g dν =
Y
X
Let ρ be the measure on (0, ∞) given by dρ = r k−1 dr. Now it suffices to prove that
(8.35)
ν(E) = mk ϕ−1 (E) = (ρ × σk−1 )(E)
for every E ∈ B (0, ∞) × Sk−1 , where σk−1 is the measure defined on Sk−1 in the question.
This is because if the formula (8.35) holds, then for every nonnegative Borel function f on Rk ,
c
The measure ν in Lemma 8.4 is called the push-forward measure of µ. See [9, §3.6].
265
8.2. Applications of the Fubini Theorem
we deduce immediately from Lemma 8.4 (with X = Rk \ {0}, Y = (0, ∞) × Sk−1 , ϕ is given by
the formula (8.34), µ = mk and g = f ◦ ϕ−1 ) and Theorem 8.8 (The Fubini Theorem) that
Z
Z
f dmk =
(f ◦ ϕ−1 ) d(ρ × σk−1 )
k
R \{0}
(0,∞)×Sk−1
Z
f ϕ−1 (r, u) dσk−1 dρ
=
(0,∞)×S
Z ∞ Z k−1
r k−1 f (ru) dσk−1 dr.
(8.36)
=
0
Sk−1
Since {0} is a Borel set in Rk and mk (0) = 0, the integral on the left-hand side of the equation
(8.36) can be replaced by
Z
Rk
f dmk
and we have what we want. To prove the formula (8.35), we have several steps:
• Step 1: The formula (8.35) holds for (r1 , r2 ) × A, where A ∈ B(Sk−1 ). On the one
hand, it is clear from Definition 8.7 that
(ρ × σk−1 ) (r1 , r2 ) × A = ρ (r1 , r2 ) σk−1 (A)
Z r2
r k−1 dr
= σk−1 (A)
r1
r k − r1k
= 2
σk−1 (A).
k
(8.37)
On the other hand, we see from the definition (8.34) that
e \ [(r1 A)
e ∪ {r1 u | u ∈ A}]
ϕ (r1 , r2 ) × A = {ru | 0 < r1 < r < r2 and u ∈ A} = (r2 A)
so that
ν (r1 , r2 ) × A = mk ϕ (r1 , r2 ) × A
e \ [(r1 A)
e ∪ {r1 u | u ∈ A}]
= mk (r2 A)
e − mk (r1 A)
e −0
= mk (r2 A)
e
= (r k − r k )mk (A)
2
1
r k − r1k
= 2
σk−1 (A).
k
(8.38)
Hence our claim follows by combining the two expressions (8.37) and (8.38).
• Step 2: The formula (8.35) holds for B × A, where A ∈ B(Sk−1 ) and B ∈
B (0, ∞) . For any open V in (0, ∞), we know from [49, Exercise 29, Chapter 2] that it
can be represented as an at most countable union of disjoint segments in the form
(a, b).
Since both ν (by Lemma 8.4) and ρ × σk−1 are measures on B (0, ∞) × Sk−1 , Step 1
ensures that the formula (8.35) also holds for V × A and hence for all K × A, where K is
closed in (0, ∞).
Next, let U be a bounded Borel set in (0, ∞). The proof of Theorem 2.17(c) says that
you can find an increasing sequence of closed sets Kn and a decreasing sequence of bounded
open sets Vn such that
K1 ⊆ K2 ⊆ · · · ⊂ U ⊂ · · · ⊂ V2 ⊂ V1
(8.39)
Chapter 8. Integration on Product Spaces
266
and m(Vn \ Kn ) → 0 as n → ∞. Clearly, it is true that V1 \ K1 ⊇ V2 \ K2 ⊇ · · · and V1 \ K1
is a bounded set. Let r k−1 ≤ M on V1 \ K1 for some positive constant M . Then we have
Z
Z
k−1
dm = M · m(Vn \ Kn ) → 0
(8.40)
r
dm ≤ M
ρ(Vn \ Kn ) =
Vn \Kn
Vn \Kn
as n → ∞. Therefore, we apply the set relations (8.39), then the fact that ϕ−1 sends Borel
sets in (0, ∞) × Sk−1 to Borel sets in Rk \ {0} and finally Definition 8.7 to get
ν (Vn \ U ) × A ≤ ν (Vn \ Kn ) × A
= (ρ × σk−1 ) (Vn \ Kn ) × A
= ρ(Vn \ Kn )σk−1 (A).
(8.41)
By using the inequality (8.40), the inequality (8.41) shows that ν(Vn \ U ) × A) → 0 as
n → ∞. In other words, it means that
ν(Vn × A) → ν(U × A)
(8.42)
as n → ∞. Furthermore, if we suppose
K=
∞
[
Kn
and
∞
\
V =
Vn
n=1
n=1
which are an Fσ set and a Gδ set respectively, then Theorem 1.19(d) and (e) imply that
ρ(V ) − ρ(K) = lim [ρ(Vn ) − ρ(Kn )] = lim ρ(Vn \ Kn ) = 0
n→∞
n→∞
so that ρ(V ) = ρ(K). Since K, V ∈ B (0, ∞) and K ⊂ U ⊂ V , Theorem 1.36 asserts
that ρ is complete and then
ρ(U ) = ρ(K) = ρ(U ) = lim ρ(Vn ).
n→∞
(8.43)
Thus we establish from Definition 8.7 and the limits (8.42) and (8.43) that
(ρ × σk−1 )(U × A) = ρ(U )σk−1 (A)
= lim [ρ(Vn )σk−1 (A)]
n→∞
= lim (ρ × σk−1 )(Vn × A)
n→∞
= lim ν(Vn × A)
n→∞
= ν(U × A).
Finally, if U is any bounded or unbounded Borel set in (0, ∞), then we consider the sets
Un = U ∩ (n − 1, n]
for n ∈ N. It is trivial that each Un is a bounded Borel set, so the preceding paragraph
gives
(ρ × σk−1 )(Un × A) = ν(Un × A)
(8.44)
for each n = 1, 2, . . .. Since Ui ∩ Uj = ∅ if i 6= j, (Ui × A) ∩ (Uj × A) = ∅ for all i 6= j and
then the result (8.44) shows that
∞
[
(ρ × σk−1 )(U × A) = (ρ × σk−1 )
Un × A
n=1
267
8.2. Applications of the Fubini Theorem
∞
[
= (ρ × σk−1 )
(Un × A)
n=1
∞
X
=
(ρ × σk−1 )(Un × A)
n=1
∞
X
=
n=1
=ν
ν(Un × A)
∞
[
n=1
Un × A
= ν(U × A),
i.e., the formula (8.35) holds for every U × A, where U ∈ B (0, ∞) and A ∈ B(Sk−1 ).
• Step 3: The formula (8.35) holds for E ∈ B (0, ∞) × Sk−1 . Suppose that
M = {E ∈ B (0, ∞) × Sk−1 | the formula (8.35) holds for E} ⊆ B (0, ∞) × Sk−1 .
(8.45)
By using similar argument as in the proof of Problem 8.11, we can prove that
B (0, ∞) × B(Sk−1 ) = B (0, ∞) × Sk−1 .
By Step 2, we see that B (0, ∞) × B(Sk−1 ) ⊆ M. Thus it concludes from the set
inclusion (8.45) that
M = B (0, ∞) × Sk−1 .
When k = 2, for a nonnegative Borel function f on R2 , the formula becomes
Z ∞ n Z 2π
Z
o
r
f (r cos θ, r sin θ) dθ dr.
f (x) dm2 =
R2
0
We have to show that
Z
0
f (ru) dσ1 (u) =
S1
Z
2π
f (r cos θ, r sin θ) dθ
(8.46)
0
for every r ∈ (0, ∞). To this end, let u ∈ S1 . Then u is a point on the unit circle so that
u = ( √ 2x 2 , √ 2y 2 ). We define F : B(0, 1) \ {0} → [0, ∞] by
x +y
x +y
rx
ry
F (u) = f (ru) = f p
= f (r cos θ, r sin θ),
,p
x2 + y 2
x2 + y 2
where θ ∈ [0, 2π). Recall the definition of σ1 so that
Z
Z
F (u) dm2
f (ru) dσ1 = 2
S1
=2
Z
=
Z
B(0,1)\{0}
1 Z 2π
0
2π
f (r cos θ, r sin θ) dθ da
a
0
f (r cos θ, r sin θ) dθ
0
which is exactly the formula (8.46). When k = 3, the formula becomes
Z ∞ Z 2π Z π
Z
f (x) dm3 =
r 2 f (r cos ϕ sin θ, r sin ϕ sin θ, r cos θ) sin θ dθ dϕ dr,
R3
0
0
0
where θ ∈ [0, π) and ϕ ∈ [0, 2π). This completes the proof of the problem.
Chapter 8. Integration on Product Spaces
268
Remark 8.2
We note that it can be further shown that the Borel measure σk−1 on Sk−1 is unique, see
[22, Theorem 2.49, pp. 78, 79] for a proof of this. For similar or other proofs of Problem
8.6, you are suggested to read [36, pp. 172, 173, 322], [54, Theorem 15.13, pp. 154, 155] or
[58, §3.2, pp. 279 – 281].
8.3
The Product Measure Theorem and Sections of a Function
Problem 8.7
Rudin Chapter 8 Exercise 7.
Proof. Since X and Y are σ-finite, X is the union of countably many disjoint sets Xn with
µ(Xn ) < ∞, and that Y is the union of countably many disjoint sets Ym with λ(Ym ) < ∞. For
each Xn × Ym , we have
ψ(Xm × Ym ) = µ(Xn )λ(Ym ) < ∞.
(8.47)
Suppose that Ω is the class of all E ∈ S × T for which the conclusion of the problem holds,
i.e.,
Ω = {E ∈ S × T | ψ(E) = (µ × λ)(E)}.
If E = A × B for some A ∈ S and B ∈ T , then Ex = B if x ∈ A and Ex = ∅ if x ∈ B. Thus
we deduce from Definition 8.7 that
Z
Z
λ(B) dµ(x) = µ(A)λ(B) = ψ(A × B) = ψ(E).
(8.48)
λ(Ex ) dµ(x) =
(µ × λ)(E) =
A
X
In other words, A × B ∈ Ω. Besides, if Ai ∈ Ω, Ai ∩ Aj = ∅ for i 6= j and A =
ψ and µ × µ are measures,
ψ(A) = ψ
∞
[
i=1
∞
[
Ai , then since
i=1
∞
∞
∞
[
X
X
Ai = (µ × λ)(A).
(µ × λ)(Ai ) = (µ × λ)
ψ(Ai ) =
Ai =
i=1
(8.49)
i=1
i=1
Suppose that
M = {E ∈ S × T | E ∩ (Xn × Ym ) ∈ Ω for all n, m ∈ N} ⊆ S × T .
(8.50)
Our goal is to show that M is a monotone class containing E so that Theorem 8.3 (The Monotone
Class Theorem) can be applied to obtain the reverse direction of the set inclusion (8.50).
∞
[
Ai .
We claim that M is a monotone class. To see this, let Ai ∈ M, Ai ⊆ Ai+1 and A =
i=1
By the definition (8.50), we know that
ψ Ai ∩ (Xn × Ym ) = (µ × λ) Ai ∩ (Xn × Ym )
(8.51)
for all n, m ∈ N. Since Ai ∩ (Xn × Ym ) ∈ Ω and S × T is a σ-algebra in X × Y , we have
A ∩ (Xn × Ym ) =
∞
[
[Ai ∩ (Xn × Ym )] ∈ S × T .
i=1
269
8.3. The Product Measure Theorem and Sections of a Function
Furthermore, since Ai ∩ (Xn × Ym ) ⊆ Ai+1 ∩ (Xn × Ym ) for all n, m ∈ N, Theorem 1.19(d) and
the equality (8.51) imply that
∞
[
[Ai ∩ (Xn × Ym )]
ψ A ∩ (Xn × Ym ) = ψ
i=1
= lim ψ Ai ∩ (Xn × Ym )]
i→∞
= lim (µ × λ) Ai ∩ (Xn × Ym )
i→∞
∞
[
= (µ × λ)
[Ai ∩ (Xn × Ym )]
i=1
= (µ × λ) A ∩ (Xn × Ym ) .
In other words, A ∩ (Xn × Ym ) ∈ Ω so that A ∈ M. Next, if Bi ∈ M, Bi ⊇ Bi+1 and B =
By the definition (8.50) again, we have
ψ Bi ∩ (Xn × Ym ) = (µ × λ) Bi ∩ (Xn × Ym )
∞
\
Bi .
i=1
(8.52)
for all n, m ∈ N. Since Bi ∩ (Xn × Ym ) ∈ Ω and S × T is a σ-algebra in X × Y , we have
B ∩ (Xn × Ym ) =
∞
\
[Bi ∩ (Xn × Ym )] ∈ S × T
i=1
by Comment 1.6(c). In addition, we follow from the inequality (8.47) that
and
ψ B1 ∩ (Xn × Ym ) ≤ ψ(Xn × Ym ) < ∞
(µ × λ) B1 ∩ (Xn × Ym ) ≤ (µ × λ)(Xn × Ym ) < ∞.
Since Bi ∩ (Xn × Ym ) ⊇ Bi+1 ∩ (Xn × Ym ) for all n, m ∈ N, Theorem 1.19(e) and the equality
(8.52) imply that
∞
\
ψ B ∩ (Xn × Ym ) = ψ
[Bi ∩ (Xn × Ym )]
i=1
= lim ψ Bi ∩ (Xn × Ym )]
i→∞
= lim (µ × λ) Bi ∩ (Xn × Ym )
i→∞
∞
\
= (µ × λ)
[Bi ∩ (Xn × Ym )]
i=1
= (µ × λ) B ∩ (Xn × Ym ) .
In other words, B ∩ (Xn × Ym ) ∈ Ω so that B ∈ M. By Definition 8.1, M is a monotone class
as claimed.
We claim that E ⊆ M. Denote Q = R1 ∪ R2 ∪ · · · ∪ Rn , where each Ri is a measurable
rectangle and Ri ∩ Rj = ∅ for i 6= j. Then each Ri is in the form Ai × Bi for some Ai ∈ S and
Bi ∈ T , so it is clear from the equality (8.48) that Ri ∈ Ω. Since
Ri ∩ (Xn × Ym ) = (Ai ∩ Xn ) × (Bi ∩ Ym ),
Chapter 8. Integration on Product Spaces
270
we have Ri ∩ (Xn × Ym ) ∈ Ω and then Ri ∈ M. Since Ri 6= Rj for i 6= j, we observe from the
fact (8.49) that Q ∈ M. Hence we have our claim that E ⊆ M.
Now we have proven our goal that M is a monotone class containing E . These fact imply
that
S × T ⊆ M.
(8.53)
Consequently, our proof is completed if we compare the set inclusions (8.50) and (8.53).
Remark 8.3
Problem 8.7 is sometimes called the Product Measure Theorem.
Problem 8.8
Rudin Chapter 8 Exercise 8.
Proof. We follow the hint given by Rudin.
(a) Given x ∈ R. We take
αn (x) =
[nx]
,
n
where [x] denotes the greatest integer function. Since x ∈ [ ni , i+1
n ) if and only if nx ∈
[i, i + 1) so that
i
αn (x0 ) =
n
if x0 ∈ [ ni , i+1
n ). Thus this implies that
0 ≤ x − αn (x) <
1
n
for every n ∈ N and αn (x) → x as n → ∞. Therefore, it is illegal to say that there are
n ∈ N and i(n) ∈ Z such that
i(n) − 1
i(n)
= ai(n)−1 ≤ x ≤ ai(n) =
n
n
and ai(n) → x as n → ∞. Now we simply write ai = ai(n) and define fn : R2 → R by
x − ai−1
ai − x
f (ai−1 , y) +
f (ai , y)
ai − ai−1
ai − ai−1
ai − x
x − ai−1
=
fai−1 (y) +
fa (y).
ai − ai−1
ai − ai−1 i
fn (x, y) =
(8.54)
By the hypotheses, fai−1 (y) and fai (y) are Borel measurable. Obviously, both
ai − x
ai − ai−1
and
x − ai−1
ai − ai−1
are continuous in x so that they are Borel measurable. By Proposition 1.19(c), each
fn (x, y) is Borel measurable. Furthermore, it satisfies
|fn (x, y) − f (x, y)| =
x − ai−1
ai − x
ai − x
f (ai−1 , y) +
f (ai , y) −
f (x, y)
ai − ai−1
ai − ai−1
ai − ai−1
271
8.3. The Product Measure Theorem and Sections of a Function
x − ai−1
f (x, y)
ai − ai−1
≤ |f (ai−1 , y) − f (x, y)| + |f (ai , y) − f (x, y)|
−
= |f y (ai−1 ) − f y (x)| + |f y (ai ) − f y (x)|,
(8.55)
where y ∈ R. Since ai and ai−1 tend to x as n → ∞ and f y is continuous for all y ∈ R,
we deduce from the inequality (8.55) that
f (x, y) = lim fn (x, y).
n→∞
By Theorem 1.14, f is also Borel measurable in R2 .
(b) We prove it by induction on k. The case for k = 1 is obviously true. Assume that the
statement is also true for k = m, i.e., if g(x1 , x2 , . . . , xm ) is continuous in each of the m
variables separately, then g is a Borel function in Rm . For k = m+1, g(x1 , x2 , . . . , xm+1 ) is
continuous in each of the m + 1 variables separately. Now for each xm+1 ∈ R, the function
gxm+1 (x1 , x2 , . . . , xm )
is continuous separately in x1 , . . . , xm . Thus we get from our assumption that every
gxm+1 is a Borel function in Rm . By our hypothesis of the question, for every choice
y = (x2 , x3 , . . . , xm+1 ), the function
g y (x1 ) = g(x1 , x2 , . . . , xm+1 )
is continuous in x1 .
Now if we define
i−1
n
= ai−1 ≤ x1 ≤ ai =
gn (x, y) =
i
n
and gn : Rm+1 → R by
x1 − ai−1
ai − x 1
g(ai−1 , y) +
g(ai , y),
ai − ai−1
ai − ai−1
then we use a similar argument as part (a) to conclude that each gn is Borel measurable
in Rm+1 and the limit
g(x, y) = lim gn (x, y)
n→∞
implies that g is a Borel function in Rm+1 . Hence our desired result follows from induction.
This completes the proof of the problem.
Problem 8.9
Rudin Chapter 8 Exercise 9.
Proof. For each x ∈ R, let {pn } ⊆ E such that pn → x. By the hypotheses, let N be the
Lebesgue measurable set such that m(N ) = 0 and f y is continuous on R for all y ∈ R \ N . Now
for every y ∈ R \ N , these facts imply that
lim fpn (y) = lim f (pn , y) = lim f y (pn ) = f y (x) = f (x, y) = fx (y).
n→∞
n→∞
n→∞
Since each fpn is Lebesgue measurable, Theorem 1.14 ensures that fx is also Lebesgue measurable. In other words, what we have shown is that for each x ∈ R, the function fx is Lebesgue
measurable on R \ N .
Chapter 8. Integration on Product Spaces
272
Using the same idea of the proof of Problem 8.8, we define fn : R × (R \ N ) → R by
fn (x, y) =
ai − x
x − ai−1
f (ai−1 , y) +
f (ai , y),
ai − ai−1
ai − ai−1
where
i(n) − 1
i(n)
= ai(n)−1 ≤ x ≤ ai(n) =
.
n
n
Imitate Problem 8.8’s proof, we conclude that each fn (x, y) is Lebesgue measurable and the
inequality (8.55) holds for every y ∈ R \ N which further imply that
f (x, y) = lim fn (x, y)
n→∞
for every y ∈ R \ N . By Theorem 1.14, the function f : R × R \ N → R is clearly Lebesgue
measurable. Eventually, the completeness of m2 guarantees that f : R2 → R is Lebesgue
measurable which ends the analysis of the proof.
Problem 8.10
Rudin Chapter 8 Exercise 10.
Proof. Define each fn : R2 → R by the representation (8.54), putd
ai − g(y)
g(y) − ai−1
hn (y) = fn g(y), y =
f (ai−1 , y) +
f (ai , y)
ai − ai−1
ai − ai−1
if g(y) ∈ [ai−1 , ai ]. Similar to the proof of Problem 8.8,
fai−1 (y) = f (ai−1 , y)
and fai (y) = f (ai , y)
are Lebesgue measurable on R by the hypotheses. Furthermore, since g is Lebesgue measurable
on R, both
g(y) − ai−1
ai − g(y)
and
ai − ai−1
ai − ai−1
are also Lebesgue measurable on R. By Proposition 1.19(c), each hn (y) is Lebesgue measurable
on R.
Now our hn and h satisfy
ai − g(y)
g(y) − ai−1
ai − g(y)
f (ai−1 , y) +
f (ai , y) −
f g(y), y
ai − ai−1
ai − ai−1
ai − ai−1
g(y) − ai−1
f g(y), y
−
ai − ai−1
≤ |f (ai−1 , y) − f g(y), y | + |f (ai , y) − f g(y), y |
= |f y (ai−1 ) − f y g(y) | + |f y (ai ) − f y g(y) |.
|hn (y) − h(y)| =
By using similar argument as in the proof of Problem 8.8, we can show that
h(y) = lim hn (y)
n→∞
which implies that h is Lebesgue measurable on R by Theorem 1.14, completing the proof of
the problem.
d
Remember that i actually depends on n.
273
8.4. Miscellaneous Problems
8.4
Miscellaneous Problems
Problem 8.11
Rudin Chapter 8 Exercise 11.
Proof. Recall from the definition on [42, p. 86] that the product topology on Tm+n = Rm+n is
Rm × Rn which is the topology having as basis the collection
Bm+n = {U × V | U ∈ Tm and V ∈ Tn }.
Furthermore, the result [42, Theorem 15.1, p. 86] shows that the basis Bm+n can be replaced
by the basis
′
Bm+n
= {U × V | U ∈ Bm and V ∈ Bn }.
For every k ≥ 1, since Rk is second-countable, it has a countable basis.e Thus we may assume
′
is also countable. Obviously, we have
that Bm and Bn are countable and consequently Bm+n
U ∈ Bm ⊆ Tm ⊆ Bm and V ∈ Bn ⊆ Tn ⊆ Bn so that
U × V ∈ B m × B n ⊆ Tm × Tn ⊆ B m × B n .
(8.56)
′
′
If W ∈ T m+n , then it can be expressed as a union of basis elements of Bm+n
. Since Bm+n
is countable, this union must be countable too. Therefore, it follows from this fact, the fact
Bm × Bn is an σ-algebra and the set relation (8.56) that Tm+n ⊆ Bm × Bn and this means
that
Bm+n ⊆ Bm × Bn .
(8.57)
For the other direction, the projections π1 : Rm × Rn → Rm and π2 : Rm × Rn → Rn are
continuous mappings because π1−1 (U ) = U × Rn and π2−1 (V ) = Rm × V which are open in
Rm × Rn = Rm+n if U and V are open in Rm and Rn respectively. If A ∈ Bm , then A × Rn is
a Borel set in Rm × Rn = Rm+n . Similarly, if B ∈ Bn , then Rm × B ia a Borel set in Rm+n .
Therefore, we get A × Rn , Rm × B ∈ Bm+n and these imply that
A × B = (A × Rn ) ∩ (Rm × B) ∈ Bm+n .
(8.58)
We recall from Definition 8.1 that Bm × Bn is the smallest σ-algebra in Rm × Rn = Rm+n
containing every measurable rectangle A × B, where A ∈ Bm and B ∈ Bn . Then we deduce
from the set relation (8.58) that
Bm × Bn ⊆ Bm+n .
(8.59)
Hence our desired result follows if we combine the set relations (8.57) and (8.59), completing
the proof of the problem.
Problem 8.12
Rudin Chapter 8 Exercise 12.
Proof. Let f (x, y) = (sin x)e−xt . Since f is continuous in R2 , f is a m2 -measurable function on
R2 . It is trivial that (0, A) and (0, ∞) are σ-finite, where A > 0. Furthermore, we have
Z A
Z ∞
Z A
Z ∞
Z A
| sin x|
−xt
e
dt =
| sin x| dx
|f (x, t)| dt =
dx.
(8.60)
dx
x
0
0
0
0
0
e
Read [42, §30] for details.
Chapter 8. Integration on Product Spaces
274
By [49, Exercise 7, p. 197], we have π2 < sinx x < 1 on (0, π2 ). Therefore, the integral (8.60)
becomes
Z A
Z ∞
Z A
| sin x|
|f (x, t)| dt =
dx
dx
x
0
0
0
Z π
Z A
2 sin x
| sin x|
≤
dx +
dx
π
x
x
0
2
Z A
π
dx
< +
π
2
x
2
π
π
= + ln A − ln
2
2
<∞
for large A > 0.
By Theorem 8.8 (The Fubini Theorem)f and the given relation, we see that
Z
0
A
Z
Z
∞
sin x dx
e−xt dt
0
0
Z ∞
Z A
(sin x)e−xt dt
dx
=
0
0
Z ∞ Z A
(sin x)e−xt dx
dt
=
0
0
Z ∞h
i
e−At
1
−
(cos
A
+
t
sin
A)
dt
=
1 + t2 1 + t2
0
Z ∞ −At
e
π
= −
(cos A + t sin A) dt.
2
1
+ t2
0
sin x
dx =
x
A
(8.61)
For A ≥ 1, we notice that
e−At
1
t
(cos A + t sin A) ≤
e−At +
e−At ≤ 2e−At ≤ 2e−t .
2
2
1+t
1+t
1 + t2
(8.62)
Then the given relation implies that e−t ∈ L1 (m) on (0, ∞). Obviously, for t ≥ 0, we note from
the second inequality in (8.62) that
e−At
(cos A + t sin A) = 0,
A→∞ 1 + t2
lim
so we deduce from this and Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) that
lim
Z
A→∞ 0
∞
e−At
(cos A + t sin A) dt = 0.
1 + t2
(8.63)
Hence our desired result follows from the representation (8.61) and the limit (8.63), completing
the proof of the problem.
Remark 8.4
The improper integral in Problem 8.12 is known as Dirichlet integral.
f
See the Notes on [51, p. 165]
275
8.4. Miscellaneous Problems
Problem 8.13
Rudin Chapter 8 Exercise 13.
Proof. We follow Rudin’s hint. Since µ is a complex measure on a σ-algebra M in X, Theorem
6.12 implies that there is a measurable function h such that |h(x)| = 1 on X and dµ = h d|µ|.
This asserts that there exists a real measurable function θ such that h = eiθ and so
dµ = eiθ d|µ|.
Given E ∈ M and we define Aα to be the subset of E where cos(θ − α) > 0. Then we have
Z
Z
Z
χAα eiθ d|µ|
eiθ d|µ| =
dµ =
µ(Aα ) =
E
Aα
Aα
and this gives
−iα
|µ(Aα )| = |e
−iα
µ(Aα )| ≥ Re e
µ(Aα ) = Re
Z
E
χAα ei(θ−α) d|µ| .
(8.64)
Since χAα cos(θ − α) = cos+ (θ − α) by the Corollary following Theorem 1.14, the inequality
(8.64) becomes
Z
Z
cos+ (θ − α) d|µ|.
(8.65)
χAα cos(θ − α) d|µ| =
|µ(Aα )| ≥
E
E
As in the proof of Lemma 6.3 from Rudin, we may choose α0 so as to maximize the integral
in the inequality (8.65). Since this maximum is at least as large as the average of the sum over
[−π, π], we obtain
Z π Z
1
|µ(Aα0 )| ≥
cos+ (θ − α) d|µ| dθ.
2π −π E
Since |µ|(E) < ∞ by Theorem 6.4, E is σ-finite. It is clear that [−π, π] is also σ-finite. Thus
we deduce from Theorem 8.8 (The Fubini Theorem) and the integral in Lemma 6.3 that
Z h Z π
Z π Z
i
1
1
1
+
cos (θ − α) d|µ| dθ =
cos+ (θ − α) dθ d|µ| = |µ|(E).
|µ(Aα0 )| ≥
2π −π E
π
E 2π −π
By [16, Example 9.2.1(c), p. 285], if T is the circle group and f : [0, 2π] → T is given by
f (t) = (cos t, sin t), then the Haar measure µ on T is given by
µ(S) =
1
m f −1 (S) .
2π
(8.66)
In particular, we have µ(T ) = 1. Since f is continuous and 2 < 2π, if we consider f ([0, 2]) = A,
then we see from the definition (8.66) that
µ(A) =
1
1
1
1
m f −1 (A) =
m([0, 2]) = = µ(S).
2π
2π
π
π
Since µ is a positive measure, we have |µ| = µ so that the Haar measure considered here gives an
example showing that π1 is the beast constant. We have completed the proof of the problem. Problem 8.14
Rudin Chapter 8 Exercise 14.
Chapter 8. Integration on Product Spaces
276
Proof. Since f (t)tα and t−α satisfy the hypotheses of Theorem 3.5 (Hölder’s Inequality), we
have
Z x
o1 n Z x
o 1 x−αq+1 1
nZ x
o1 n Z x
p
q
p
q
tαq dt
=
f p tαp dt ×
f (t)tα t−α dt ≤
xF (x) =
f ptαp dt
−αq
+
1
0
0
0
0
which gives
Z
1−p −1−αp
p
F (x) ≤ (1 − αq)
x
x
f p tαp dt
0
and then
Z
∞
0
1−p
p
F (x) dx ≤ (1 − αq)
= (1 − αq)1−p
Z
∞
Z 0∞
x
−1−αp
x−1−αp
0
Z
Z
x
0
∞
0
f p tαp dt dx
(8.67)
χ(0,x) (t)f p tαp dt dx.
(8.68)
As the space (0, ∞) is σ-finite, χ(0,x) x−1−αp f p tαp ≥ 0 and measurable on (0, ∞) × (0, ∞),
Theorem 8.8 (The Fubini Theorem) allows the inequality (8.67) can be rewritten as
Z ∞
Z ∞
Z ∞
p αp
p
1−p
f t
χ(0,x) (t)x−1−αp dx dt
F (x) dx ≤ (1 − αq)
0
Z0 ∞
Z0 ∞
1−p
p αp
= (1 − αq)
f t
x−1−αp dx dt
t
Z0 ∞
t−αp
= (1 − αq)1−p
f p tαp ×
dt
αp
0
Z ∞
= (1 − αq)1−p (αp)−1
f p dt.
0
If we define g(α) = (1 − αq)1−p (αp)−1 on (0, 1q ), then we can apply basic calculus to verify
1
that g will attain the absolute minimum at α = pq
and its minimum value is
p p
.
p−1
Hence we have again proven the result of Problem 3.14 and this completes the proof of the
problem.
Problem 8.15
Rudin Chapter 8 Exercise 15.
Proof. Notice that
ϕ(x) =
For all x ∈ R, we have
(f ∗ g)(x) =
Z
∞
−∞
This proves Statement (i).

 1 − cos x, if x ∈ [0, 2π];

0,
f (x − y)g(y) dy =
(8.69)
otherwise.
Z
∞
−∞
sin y dy =
Z
2π
sin y dy = 0.
0
277
8.4. Miscellaneous Problems
Clearly, if x ∈ R, then it is easy to see that
Z
Z ∞
Z ∞
g(y)h(x − y) dy =
g(x − y)h(y) dy =
(g ∗ h)(x) =
−∞
−∞
2π
0
ϕ′ (y)h(x − y) dy.
(8.70)
Applying the Mean Value Theorem to ϕ(t) = 1 − cos t, we get
|ϕ(x) − ϕ(y)| ≤ |x − y|
for all x, y ∈ [0, 2π]. Next, for x ∈ [0, 2π], we follow from the definition (8.69) that
Z x
Z x
(1 − cos t) dt = x − sin x.
ϕ(t) dt =
h(x) =
0
−∞
Using the Mean Value Theorem to h, we obtain
|h(x) − h(y)| ≤ 2|x − y|
for all x, y ∈ [0, 2π]. Hence we have
ϕ, h ∈ Lip 1
on [0, 2π]. By Problem 7.10, ϕ and h are AC on [0, 2π] and then we may apply Problem 7.14 to
the integral (8.70) to establish
Z 2π
ϕ′ (y)h(x − y) dy
(g ∗ h)(x) =
0
Z 2π
ϕ(y)h′ (x − y) dy
= ϕ(2π)h(x − 2π) − ϕ(0)h(x) +
0
Z 2π
ϕ(y)ϕ(x − y) dy
=
0
Z ∞
ϕ(y)ϕ(x − y) dy
=
−∞
= (ϕ ∗ ϕ)(x)
for all x ∈ R. Suppose that x ∈ (0, 2π]. Then we have
Z x
ϕ(y)ϕ(x − y) dy.
(ϕ ∗ ϕ)(x) =
0
It is obvious that ϕ(y) = 1 − cos y > 0 on (0, x). Similarly, if y ∈ (0, x), then x − y ∈ (0, x) and
so ϕ(x − y) = 1 − cos(x − y) > 0 for all y ∈ (0, x). Consequently, we deduce that
(ϕ ∗ ϕ)(x) > 0
(8.71)
on (0, 2π]. Suppose that x ∈ (2π, 4π). Then the definition (8.69) implies that
Z x−2π
Z x
ϕ(y)ϕ(x − y) dy.
ϕ(y)ϕ(x − y) dy =
(ϕ ∗ ϕ)(x) =
0
0
Obviously, ϕ(y) = 1 − cos y > 0 on (0, x − 2π). If 0 < y < x − 2π, then 2π < x − y < x and so
ϕ(x − y) = 1 − cos(x − y) > 0 for all y ∈ (0, x − 2π). Therefore, the inequality (8.71) is till valid
whenever x ∈ (2π, 4π). In conclusion, we establish that
(g ∗ h)(x) = (ϕ ∗ ϕ)(x) > 0
on (0, 4π). This is Statement (ii).
Chapter 8. Integration on Product Spaces
278
On the one hand, Statement (i) says that f ∗ g = 0 which implies that (f ∗ g) ∗ h = 0. On
the other hand, if x ∈ R, since f (x) = 1 for all x ∈ R, we have
Z ∞
Z ∞
(g ∗ h)(x − y) dy.
(8.72)
f (y)(g ∗ h)(x − y) dy =
[f ∗ (g ∗ h)](x) =
−∞
−∞
By Theorem 7.26 (The Change-of-variables Theorem) (or its special case on p. 156), we see that
the last integral in the expression (8.72) becomes
Z ∞
(g ∗ h)(y) dy.
(8.73)
−∞
Recall from the definition (8.69) that ϕ(y) = 0 outside [0, 2π] and ϕ(x−y) = 0 outside [x−2π, x].
If x ≤ 0 or x ≥ 4π, then [0, 2π] ∩ [x − 2π, x] is of measure 0 so that (ϕ ∗ ϕ)(x) = 0 outside (0, 4π).
Hence it follows from this fact and the expression (8.73) that
Z 4π
Z 4π
Z ∞
(ϕ ∗ ϕ)(y) dy > 0
(g ∗ h)(y) dy =
(g ∗ h)(y) dy =
[f ∗ (g ∗ h)](x) =
−∞
0
0
by Statement (ii).
Now the inconsistence in Statement (iii) is due to the fact that f ∈
/ L1 (R). This completes
the proof of the problem.
Problem 8.16
Rudin Chapter 8 Exercise 16.
Proof. Suppose that (X, L , µ) and (Y, T , ν) are σ-finite measure spaces. Let f be an (L × T )measurable function on X × Y . If 0 ≤ f ≤ ∞ and 1 ≤ p < ∞, then we have the analogy of
Minkowski’s inequality in the question.
If p = 1, then it is exactly Theorem 8.8 (The Fubini Theorem). Suppose that 1 < p < ∞ and
q is the conjugate exponent of p. Since the inequality holds trivially when
Z hZ
i1
p
f p (x, y) dµ(x) dλ(y) = ∞,
so without loss of generality, we assume further that this integral is finite. Let g ∈ Lq (µ). By
Theorem 8.8 (The Fubini Theorem) and Theorem 3.5 (Hölder’s Inequality), we have
Z hZ
Z Z
i
f (x, y) dλ(y) · |g(x)| dµ(x) =
f (x, y) · |g(x)| dµ(x) dλ(y)
Z nZ
o1
o1n Z
p
q
p
|g|q dµ(x) dλ(y)
f (x, y) dµ(x)
≤
Z hZ
i1
p
= kgkq
f p (x, y) dµ(x) dλ(y).
(8.74)
Thus if we define Φ : Lq (µ) → [0, ∞] by
Z hZ
i
Φ(g) =
f (x, y) dλ(y) · |g(x)| dµ(x),
(8.75)
then the inequality (8.74) and the assumption imply that Φ is a bounded linear functional on
Lq (µ). Applying Theorem 6.16 to Φ, there exists a unique h ∈ Lp (µ) such that
Z
Φ(g) = hg dµ(x).
(8.76)
279
8.4. Miscellaneous Problems
Since Φ(|g|) = Φ(g), the uniqueness of h and the comparison of the representations (8.75) and
(8.76) show that
Z
h = f (x, y) dλ(y).
Furthermore, Theorem 6.16 gives that kΦk = khkp and we deduce from this and the inequality
(8.74) that
nZ hZ
ip
o1
p
f (x, y) dλ(y) dµ(x)
= khkp
= kΦk
= sup{|Φ(g)| | g ∈ Lq (µ) and kgkq = 1}
Z hZ
i1
p
≤
f p (x, y) dµ(x) dλ(y)
which completes the proof of the problem.
Chapter 8. Integration on Product Spaces
280
CHAPTER
9
Fourier Transforms
9.1
Properties of The Fourier Transforms
Problem 9.1
Rudin Chapter 9 Exercise 1.
Proof. For every y ∈ R, |f (x)e−ixy | = |f (x)| = f (x) so that f (x)e−ixy ∈ L1 . By Definition 9.1
and Theorem 1.33, we have
Z ∞
Z ∞
1
1
−ixy
b
f (x)e
dx ≤ √
f (x) dx = fb(0)
|f (y)| = √
2π −∞
2π −∞
for every y ∈ R. Assume that |fb(y)| = fb(0) for some y 6= 0, i.e.,
Z ∞
Z ∞
Z ∞
−ixy
|f (x)e−ixy | dx.
f (x) dx =
f (x)e
dx =
−∞
−∞
−∞
By Theorem 1.39(c), there is a constant α such that
αf (x)e−ixy = |f (x)e−ixy | = |f (x)| = f (x)
(9.1)
a.e. on R. Since f (x) > 0 on R, the expression (9.1) implies that αe−ixy = 1 a.e. on R. Thus it
gives y = 0, a contradiction. This completes the proof of the problem.
Problem 9.2
Rudin Chapter 9 Exercise 2.
Proof. We prove the assertions one by one.
• The Fourier transform of χ[a,b] . By Definition 9.1, we have
Z ∞
1
√
χ
[
χ[a,b] (x)e−ixt dx
[a,b] (t) =
2π −∞
Z b
1
√
e−ixt dx
=
2π a
281
Chapter 9. Fourier Transforms
282
=

(e−ibt − e−iat )



√
, if t 6= 0;
−


2πit


b−a


 √ ,
2π
(9.2)
otherwise.
By similar argument, we see that χ[
[
[
(a,b) , χ
(a,b] and χ
[a,b) also give the same answer (9.2).
• The expression of χ[−n,n] ∗ χ[−1,1] . We have gn = χ[−n,n] and h = χ[−1,1] . By Definition
9.1, we have
Z ∞
1
(gn ∗ h)(x) = √
χ[−n,n](x − y)χ[−1,1] (y) dy
2π −∞
Z 1
1
χ[−n,n] (x − y) dy
=√
2π −1
1
= √ m [x − n, x + n] ∩ [−1, 1] .
(9.3)
2π
If x ≤ −n − 1, then x + n ≤ −1 so that
m [x − n, x + n] ∩ [−1, 1] = 0.
If −n − 1 ≤ x ≤ −n + 1, then −1 ≤ x + n ≤ 1 and x − n ≤ −2n + 1 ≤ −1 so that
m [x − n, x + n] ∩ [−1, 1] = x + n − (−1) = x + n + 1.
If −n + 1 ≤ x ≤ n − 1, then x + n ≥ 1 and x − n ≤ −1 so that [−1, 1] ⊆ [x − n, x + n] and
m [x − n, x + n] ∩ [−1, 1] = 2.
If n − 1 ≤ x ≤ n + 1, then −1 ≤ x − n ≤ 1 and x + n ≥ 2n − 1 ≥ 1 so that
m [x − n, x + n] ∩ [−1, 1] = 1 − (x − n) = 1 − x + n.
If x ≥ n + 1, then x − n ≥ 1 so that
m [x − n, x + n] ∩ [−1, 1] = 0.
2c
fn = gn ∗ h. We first claim that fn ∈ L1 for each n = 1, 2, . . .. Recall the
π
sin x
basic fact that
→ 1 as x → 0. Thus if we define fn (0) = n, then fn is continuous on
x
R. Particularly, fn is bounded by a positive constant Mn on [−1, 1]. Since |fn (x)| ≤ x12 if
|x| ≥ 1, we have
Z
Z
Z
|fn (x)| dx
|fn (x)| dx +
|fn (x)| dx =
• Proof of
R
≤
Z
|x|≥1
∞
1
dx
+
x2
= 2 + Mn .
Therefore, our claim is true.
Z
−1
−∞
|x|≤1
dx
+ Mn
x2
283
9.1. Properties of The Fourier Transforms
b
Next, since gn , h ∈ L1 , Theorem 9.2(c) implies that g\
n ∗ h = gbn · h. By the formulas
(9.2), we see that
 r
 r


2 sin nx
2 sin x




·
, if x 6= 0;
·
, if x 6= 0;




π
x
π
x
gbn (x) =
and b
h(x) =
r
r






2
2




,
otherwise
,
otherwise.
π
π
Therefore, we have
(g\
n ∗ h)(x) =
and then

2 sin x sin nx


, if x 6= 0;

 π·
x2


2

 ,
π
otherwise
2
· fn (x)
(g\
n ∗ h)(x) =
π
for all x 6= 0. Now the equation (9.4) clearly gives
Z ∞
Z ∞
2
2c
ixt
f (t)eixt dm(x) =
dm(x)
fn (−x) =
(g\
n ∗ h)(t) e
|
{z
}
π
π
−∞
−∞
| {z }
(9.4)
It is f in the
theorem
It is g in the
theorem
1
Furthermore, the relation (9.4) also gives (g\
n ∗ h) ∈ L . Recall the expression (9.3) ensures
that gn ∗ h ∈ L1 . In other words, the function gn ∗ h satisfies the hypotheses of Theorem
9.11 (The Inversion Theorem), so we conclude that
2c
fn (−x) = (gn ∗ h)(x)
π
(9.5)
a.e. on R. Since fn is an odd function, the equation (9.5) reduces to
a.e. on R, as required.
2c
fn (x) = (gn ∗ h)(x)
π
• kfn k1 → ∞ as n → ∞. By [49, Exercise 7, p. 197], we have
sin x
2
>
x
π
(9.6)
if 0 < x < π2 . Again the fact sinx x → 1 as x → 0 implies that the inequality (9.6) also holds
at x = 0. By the definition, we have
|fn (x)| >
2 | sin nx|
·
π
x
π
2
and so
Z π
Z nπ
Z ∞
2 | sin x|
2
2 2 | sin nx|
dm(x) =
dm(x).
|fn (x)| dm(x) ≥
kfn k1 =
π 0
x
π 0
x
−∞
if 0 ≤ x <
(9.7)
It is well-known that
Z ∞
n Z (k+1)π
n Z (k+1)π
n
X
X
| sin x|
| sin x|
| sin x|
2X 1
dx ≥
dx ≥
dx =
→∞
x
x
(k + 1)π
π
k+1
0
kπ
kπ
k=0
k=0
k=0
as n → ∞, so the right-most integral in the inequality (9.7) tends to ∞ as n → ∞.
Chapter 9. Fourier Transforms
284
• The mapping f 7→ fb maps L1 into a proper subset of C0 . Suppose that F : L1 → C0
is given by
F(f ) = fb.
By Definition 9.1, F is linear. If f ∈ L1 and kf k1 = 1, then Theorem 9.6 gives kfbk∞ ≤ 1
so that
kFk = sup{kfbk∞ | f ∈ L1 and kf k1 = 1} ≤ 1,
i.e., F is bounded. If F(f ) = F(g), then the linearity of F gives
f[
− g = F(f − g) = F(f ) − F(g) = 0.
By Theorem 9.12 (The Uniqueness Theorem), we obtain f (x) = g(x) for almost all x ∈ R
which means that F is injective.
Assume that F was surjective. Since both L1 and C0 are Banach, we know from Theorem
5.10 that there is a δ > 0 such that
kfbk∞ ≥ δkf k1
for all f ∈ L1 . Recall the facts from the above assertions that
1
fn ∈ L ,
1
gn ∗ h ∈ L ,
2c
fn = gn ∗ h and
π
so we get from the inequality (9.8) that
1
kfn k1 ≤
δ
r
π
π
kc
fn k∞ = kgn ∗ hk∞ = ·
2
2
(9.8)
r
2
=
π
r
2
,
π
2
<∞
π
but this contradicts the fact kfn k1 → ∞. Hence F is not surjective.
• The set B = {fb| f ∈ L1 } is dense in C0 . To solve this part, we need the locally compact
Hausdorff version of the Stone-Weierstrass Theorema . In fact, it is
Lemma 9.1 (The Stone-Weierstrass Theorem)
If A is a subalgebra of C0 that separates points on R and vanishes at no point of
R, then A is dense in C0 .
It is clear that B is a vector subspace of C0 . For every pair of f, g ∈ L1 , Theorem 8.14
says that L1 is closed under convolution, i.e., f ∗ g ∈ L1 . By Theorem 9.2(c), we get
f[
∗ g = fb · b
g ∈ B.
By the definition, B is an algebra. Suppose that p, q ∈ R and we have two cases for
consideration.
– Case (i): p < q and p 6= −q. Consider the function
f (x) = e−|x| .
(9.9)
By [51, Eqn. (3), p. 183], since f is even in R, we have
r
Z ∞
Z ∞
1
2
ixt
−ixt
f (x)e dm(x) = h1 (t) =
f (x)e
dm(x) =
fb(t) =
·
. (9.10)
π
1
+
t2
−∞
−∞
a
Since fb(x) = fb(y) if and only if x = ±y, we obtain fb(p) 6= fb(q) in our case.
See [49, pp. 159 – 165] for the compact cases and [22, §4.7] for the locally compact Hausdorff cases
285
9.1. Properties of The Fourier Transforms
– Case (ii): p < q and p = −q. We consider g(x) = f (x)eix , where f is the function
(9.9). By Theorem 9.2(a), we have
r
1
2
b
gb(t) = f (t − 1) =
·
.
π 1 + (t − 1)2
If gb(p) = b
g(−q) = b
g(q), then we have (q − 1)2 = (q + 1)2 which implies that q = 0, a
contradiction. Hence we have b
g(−q) 6= b
g(q).
In other words, B separates points on R. Furthermore, we see from the expression (9.10)
that fb(t) 6= 0 for each t ∈ R. Thus B vanishes at no point of R. Hence our desired result
follows immediately from Lemma 9.1.
This completes the proof of the problem.
Remark 9.1
The result in Problem 9.2 that F(L1 ) ⊂ C0 is called the Riemann-Lebesgue Lemma
([58, Theorem 1.4, p. 80]). See also §5.14 from Rudin.
Problem 9.3
Rudin Chapter 9 Exercise 3.
Proof. By similar analysis as in Problem 9.2, if gλ = χ[−λ,λ] , then for t 6= 0, we have
gbλ =
r
2 sin λt
·
.
π
t
Note that b
gλ ∈
/ L1 , so we can’t apply any inversion theorems (Theorems 9.11 or 9.14) directly.
To solve this problem, we apply Theorem 9.13.
It is clear that gλ , gbλ ∈ L2 . Using the same notation as in Theorem 9.13(d), we have
r Z A
Z A
Z
1
2
1 A sin λt ixt
sin λt ixt
ixt
gbλ (t)e dm(t) = √ ·
ψA (x) =
e dt =
e dt
π −A t
π −A t
2π
−A
By Theorem 9.13(d) again,
kψA − gλ k2 → 0
(9.11)
as A → ∞. If we restrict A to be a positive integer, then the result (9.11) means that {ψA } is
a Cauchy sequence in L2 with limit gλ .b Thus it follows from Theorem 3.12 that {ψA } has a
subsequence which converges pointwise almost everywhere to gλ , i.e.,

Z Ak
Z A
 π, if x ∈ [−λ, λ];
sin λt ixt
sin λt ixt
e dt = lim
e dt = πχ[−λ,λ] (x) =
(9.12)
lim

A→∞ −A
k→∞ −Ak
t
t
0, otherwise
for almost all real x, where λ > 0.
Since the result (9.12) only holds for almost all real x, if we want to evaluate the limit for
all x ∈ R, we need an inversion theorem which is applicable to gλ . Fortunately, there is such a
b
Here we use the basic fact [49, Theorem 3.11(a)].
Chapter 9. Fourier Transforms
286
theorem for piecewise smooth functionsc . Indeed, we have (see [21, Theorem 7.6, p. 220] or
[55, Theorem 6.6.2, p. 330])
Lemma 9.2
If f is a piecewise smooth function on R such that f ∈ L1 , then we have
Z A
π
lim
fb(t)eixt dm(t) = [f (x−) + f (x+)]
A→∞ −A
2
for every x ∈ R.
Now the gλ has discontinuities only at x = ±λ, so it is piecewise smooth on R and gλ ∈ L1 .
Apply Lemma 9.2 to gλ , we establish at once that

π, if x ∈ (−λ, λ);




Z A

π
sin λt ixt
π
if x = ±λ;
e dt = [gλ (x−) + gλ (x+)] =
lim
2

A→∞ −A
t
2




0, otherwise.
Hence we have ended the analysis of the proof.
Problem 9.4
Rudin Chapter 9 Exercise 4.
Proof. Suppose that there exists a g ∈ L1 ∩ L2 such that gb ∈
/ L1 and
f (x) = b
g(−x)
(9.13)
on R. Then the hypotheses immediately give f ∈
/ L1 . Since g ∈ L2 , Theorem 9.13 (The
Plancherel Theorem) implies that
kb
g k2 = kgk2 < ∞.
Thus we have b
g ∈ L2 and we deduce immediate from the hypothesis (9.13) that f ∈ L2 .
Consequently, we must have f ∈ L2 \ L1 . Now it remains to show that fb ∈ L1 . To this end, we
need the following property of Fourier transform ([55, Theorem 6.5.1, p. 324]):
Lemma 9.3
b
b
If f ∈ L2 , then we have fb = f − , where fb denotes the Fourier transform of fb and
f − (x) = f (−x).
Now the expression (9.13) can be written as f − = b
g. Since g ∈ L2 , this and Lemma 9.3 imply
that
− =g
b = g− .
fc
b
− ∈ L1 . Recall from Definition 9.1 that
Since g ∈ L1 , we have g− ∈ L1 and then fc
Z ∞
Z ∞
Z ∞
−ixt
−
−ixt
c
−
f (x)eixt dm(x) = fb(t).
f (−x)e
dm(x) =
f (x)e
dm(x) =
f (t) =
−∞
c
−∞
For the definition, see [55, Definition 1.4.2, p. 29].
−∞
287
9.1. Properties of The Fourier Transforms
Hence, we may conclude that fb ∈ L1 .
Finally, we give examples of the problem. Consider gn = χ[−n,n] which is clearly an element
of L1 ∩ L2 for every n ∈ N. Since
r
Z ∞
Z n
1
2 2 sin nt
−ixt
−ixt
gn (x)e
dm(x) = √
gbn (t) =
·
,
e
dx =
π
t
2π −n
−∞
we have gbn ∈
/ L1 . Therefore, the expression (9.13) ensures that each function
r
2 2 sin nx
fn (x) =
·
π
x
satisfies the hypotheses, completing the proof of the problem.
Problem 9.5
Rudin Chapter 9 Exercise 5.
Proof. By Theorem 9.6, we have fb is continuous on R which vanishes at infinity. Thus fb is
bounded by a positive constant, say M , on [−1, 1]. In addition, it is obvious that
|fb(t)| ≤ |tfb(t)|
if |t| > 1 and so our hypothesis implies that
Z
Z
Z
b
b
|f (t)| dt +
|f (t)| dt =
R
|fb(t)| dt ≤ M +
|t|>1
|t|≤1
Z
|t|>1
|tfb(t)| dt < ∞.
Therefore, fb ∈ L1 and Theorem 9.11 (The Inversion Theorem) ensures that
Z ∞
f (x) = g(x) =
fb(t)eixt dm(t)
(9.14)
−∞
a.e. on R.
It remains to show that g is differentiable with the mentioned derivative in the question. To
this end, we define F (t) = fb(t) and G(t) = −itfb(t) so that
G(x) = −ixF (x)
for all x ∈ R. Since |G(t)| = |tfb(t)|, the hypotheses give G ∈ L1 . Thus we deduce from Theorem
9.2(f) that Fb is differentiable and
b
Fb′ (t) = G(t)
(9.15)
for all t ∈ R. By the integral (9.14) and Definition 9.1, we see that
Z ∞
F (t)eixt dm(t) = Fb(−x).
g(x) =
−∞
Using the equation (9.15) and the definition of G, we conclude that
Z ∞
Z ∞
b
tfb(t)eixt dm(t).
G(t)eixt dm(t) = i
g′ (x) = −Fb′ (−x) = −G(−x)
=−
−∞
Hence we have completed the proof of the problem.
−∞
Chapter 9. Fourier Transforms
288
Problem 9.6
Rudin Chapter 9 Exercise 6.
Proof. Let f = χ[−1,1] . Clearly, f ∈ L1 and f ′ = 0 a.e. on R. Thus we have f ′ ∈ L1 too.
However, we see from Definition 9.1 that
Z ∞
′
b
f ′ (x)e−ixt dm(x) = 0
f (t) =
−∞
a.e. on R and
tifb(t) = ti
Z
∞
−ixt
f (x)e
−∞
ti
dm(x) = √
2π
Z
1
−ixt
e
−1
r
2
2 sin t
ti
=i
sin t.
dx = √ ·
t
π
2π
d
′ )(t) 6= tifb(t) which completes the proof of the problem.
Hence we conclude that (f
Problem 9.7
Rudin Chapter 9 Exercise 7.
Proof. Let f ∈ S. Denote F[f (x)](t) = fb(t), or simply F[f (x)] = fb, to be the Fourier transform
of f . We want
to show that, for every m and n = 0, 1, 2, . . ., there are positive numbers
Amn F[f (x)] < ∞ such that
|xn D m F[f (x)]| ≤ Amn (F[f (x)]).
Since |x2 f (x)| ≤ A02 (f ) for every x ∈ R, it means that
|f (x)| ≤
A02
x2
for every x 6= 0. By this, we have f ∈ L1 and thus Theorem 9.6 gives
F[f (x)](t) ≤ F[f (x)]
∞
≤ kf k1 < ∞.
It is easy to prove
that if f ∈ S, then f ′ (x) ∈ S and xf (x) ∈ S. These facts show that
D n (−ix)m f (x) belongs to S for every m, n = 0, 1, 2, . . . and so
F D n (−ix)m f (x) < ∞
(9.16)
for every m, n = 0, 1, 2, . . ..
By Theorem 9.2(f), we observe that
d
{F[f (x)](t)} = F[−ixf (x)](t)
dt
Therefore, we see that
and
d2
{F[f (x)](t)} = F[(−ix)2 f (x)](t).
dt2
dm
{F[f (x)](t)} = F[(−ix)m f (x)](t)
dtm
for every m = 0, 1, 2, . . .. To continue the proof, we need the following lemma:
D m F[f (x)](t) =
Lemma 9.4
Let f ∈ S. Then we have F[f ′ (x)](t) = itF[f (x)](t).
(9.17)
289
9.1. Properties of The Fourier Transforms
Proof of Lemma 9.4. By Integration by Parts, we obtain
Z N
Z N
N
f (x) d(e−ixt )
−
f ′ (x)e−ixt dx = f (x)e−ixt
−N
−N
−N
= f (N )e−iN t − f (−N )eiN t + it
Z
N
f (x)e−ixt dx.
(9.18)
−N
Since f ∈ S, f (x) → 0 as x → ±∞. Thus we follow from the expression (9.18) by
letting N → ∞ that
F[f ′ (x)](t) = itF[f (x)](t),
completing the proof of the lemma.
Using Lemma 9.4 repeatedly, we establish the relation
F[D n f (x)](t) = (it)n F[f (x)](t)
(9.19)
for every n = 0, 1, 2, . . .. Particularly, by replacing f (x) by (−ix)m f (x) in the formula (9.19),
we gain
F[D n (−ix)m f (x) ](t) = (it)n F[(−ix)m f (x)](t)
or equivalently,
F[(−ix)m f (x)](t) = (it)−n F[D n (−ix)m f (x) ](t).
(9.20)
Now we combine formulas (9.17) and (9.20) to conclude
(it)n D m F[f (x)](t) = F[D n (−ix)m f (x) ](t)
and we use the result (9.16) to get
tn D m F[f (x)](t) = F[D n (−ix)m f (x) ] < ∞
for every m, n = 0, 1, 2, . . .. Hence we have proven that F[f (x)] = fb ∈ S.
2
The function f (x) = e−x is called a Gaussian function which is an element of S. Similarly,
2
the function f (x) = P (x)e−x belongs to S for any polynomial P (x). This completes the proof
of the problem.
Remark 9.2
The class S in Problem 9.7 is called the Schwartz space or space of rapidly decreasing
functions on R.
Problem 9.8
Rudin Chapter 9 Exercise 8.
Proof. We are going to show the assertions one by one.
• h = f ∗ g is uniformly continuous on R. We claim that (f ∗ g)z = fz ∗ g. To see this,
we note from the definition that
Z ∞
Z ∞
fz (x − y)g(y) dy = (fz ∗ g)(x).
f (x − z − y)g(y) dy =
(f ∗ g)z (x) = (f ∗ g)(x − z) =
−∞
−∞
Chapter 9. Fourier Transforms
290
It is clear that either p or q must be finite. Suppose that 1 ≤ p < ∞. We need the
following result:
Lemma 9.5
Let p, q ∈ [1, ∞] be conjugate exponents. If f ∈ Lp and g ∈ Lq , then f ∗ g ∈ L∞
and
kf ∗ gk∞ ≤ kf kp · kgkq .
(9.21)
Proof of Lemma 9.5. Suppose that σ(f )(x) = f (−x). By Theorem 3.8 and the fact
that σ(f−x )(y) = f−x (−y) = f (−y + x) for every y ∈ R, if x ∈ R, then
Z ∞
f (x − y)g(y) dm(y) = kσ(f−x ) × gk1 ≤ kσ(f−x )kp · kgkq . (9.22)
|(f ∗ g)(x)| =
−∞
In addition, we have
nZ
kσ(fx )kp =
∞
−∞
p
|f (−y − x)| dy
o1
p
=
nZ
∞
−∞
p
|f (y)| dy
o1
p
= kf kp .
(9.23)
Hence we derive (9.23) by simply putting the result (9.23) back into the inequality
(9.22). This proves Lemma 9.5.
Now it is easily seen that the convolution of functions f and g is distributive, so for every
x ∈ R, it follows from Lemma 9.5 that
|(f ∗ g)(x − δ) − (f ∗ g)(x)| = |(f ∗ g)δ (x) − (f ∗ g)(x)|
≤ k(f ∗ g)δ − f ∗ gk∞
= k(fδ − f ) ∗ gk∞
≤ kfδ − f kp · kgkq
which implies that |(f ∗ g)(x − δ) − (f ∗ g)(x)| → 0 as δ → 0. Hence f ∗ g is uniformly
continuous on R.
• h ∈ C0 if 1 < p < ∞. Recall from Theorem 3.14 that Cc (R) is dense in Lp and Lq . Choose
{fn } ⊆ Cc (R) ⊂ Lp and {gn } ⊆ Cc (R) ⊂ Lq such that kfn − f kp → 0 and kgn − gkq → 0
as n → ∞.
Let M be a positive constant such that supp (fn ), supp (gn ) ⊆ [−M, M ]. Thus for every
|x| > 2M and every y ∈ R, we have either y ∈ R \ [−M, M ] or x − y ∈ R \ [−M, M ].
Otherwise, it will give the contradiction that |x| = |x − y + y| ≤ 2M . Thus this fact gives
fn (x − y)gn (y) = 0 and then
(fn ∗ gn )(x) = 0
for every x such that |x| > 2M , i.e., supp (fn ∗ gn ) ⊆ [−2M, 2M ] which means that each
fn ∗ gn is an element of Cc (R).
Next, by Lemma 9.5 and then Theorem 3.9, we obtain
|(fn ∗ gn )(x) − (fm ∗ gm )(x)| ≤ |(fn ∗ gn )(x) − (fn ∗ gm )(x)|
+ |(fn ∗ gm )(x) − (fm ∗ gm )(x)|
= [fn ∗ (gn − gm )](x) + [(fn − fm ) ∗ gm ](x)
≤ kfn kp · kgn − gm kq + kfn − fm kp · kgm kq
291
9.1. Properties of The Fourier Transforms
≤ (kfn − f kp + kf kp ) × kgn − gm kq
+ kfn − fm kp × (kgm − gkq + kgkq ).
Given ǫ > 0. Now there exists a N ∈ N such that n, m ≥ N imply that
kfn −f kp ≤ 1,
kgm −gkp ≤ 1,
kgn −gm kq <
ǫ
2(1 + kf kp )
and kfn −fm kp <
(9.24)
ǫ
.
2(1 + kgkq )
Hence if n, m ≥ N , then we deduce from the inequality (9.24) that
|(fn ∗ gn )(x) − (fm ∗ gm )(x)| < ǫ
for every x ∈ R. In other words, {fn ∗gn } is Cauchy in Cc (R) relative to the metric induced
by the supremum norm.
Finally, using similar argument as in proving the inequality (9.24), we can show that
one can find a positive integer N ′ such that n ≥ N ′ implies
|(fn ∗ gn )(x) − (f ∗ g)(x)| < ǫ.
Consequently, it means that
kfn ∗ gn − f ∗ gk∞ → 0
(9.25)
as n → ∞. By Theorem 3.17, we know that C0 (R) = Cc (R) so the limit (9.25) establishes
the fact that f ∗ g ∈ C0 (R).
• A counter-example. Consider f = χ[−1,1] and g = 1 so that f ∈ L1 and g ∈ L∞ .
However, for any x ∈ R, we note that
Z ∞
Z ∞
χ[x−1,x+1](y) dy = m [x − 1, x + 1] = 2.
f (x − y)g(y) dy =
h(x) =
−∞
−∞
Thus h does not vanish at ∞, i.e., h ∈
/ C0 (R).
This completes the proof of the problem.
Problem 9.9
Rudin Chapter 9 Exercise 9.
Proof. We have
g(x) =
Z
R
χ[x,x+1] (t)f (t) dt =
Z
R
χ[0,1] (t − x)f (t) dt.
(9.26)
By the change of variable formula [51, p. 156], the integral (9.26) reduces to
Z
Z
χ[0,1] (t)f−x (t) dt.
χ[0,1] (t)f (t + x) dt =
g(x) =
R
R
For every x ∈ R, by Theorem 9.5, the mapping x 7→ f−x is continuous at x. In other words,
given ǫ > 0, there exists a δ > 0 such that kf−x − f−y kp < ǫ for all y ∈ R with |x − y| < δ.
Since f−x − f−y ∈ Lp and χ[0,1] ∈ Lq , where q be the conjugate exponent of p, we deduce from
Theorem 3.8 and then Theorem 1.33 that
Z
χ[0,1] (t)[f−x (t) − f−y (t)] dt
|g(x) − g(y)| =
R
= χ[0,1] · (f−x − f−y )
1
Chapter 9. Fourier Transforms
292
≤ kχ[0,1] kq · kf−x − f−y kp
<ǫ
for all y ∈ R with |x − y| < δ. Thus g is continuous at x.
Next, suppose that a and b are real numbers with a < b. Since f ∈ L1 , we have f ∈ Lp ([a, b])
and then Theorem 3.5 (Hölder’s Inequality) implies that
Z b
nZ b
o1
n Z b o1 n Z b
o1
1
q
p
p
p
q
|f (t)|p dt .
|f (t)| dt ≤
dt
|f (t)| dt
= (b − a) ·
(9.27)
a
a
a
a
In particular, if a = n and b = n + 1, then we know from the inequality (9.27) that
Z n+1
n Z n+1
o1
p
|f (t)| dt ≤
|f (t)|p dt .
n
(9.28)
n
Given ǫ > 0. Again f ∈ Lp gives
Z ∞
∞ Z
X
p
|f (t)| dt =
n
n=−∞ −n
−∞
|f (t)|p dt < ∞.
Thus there is a positive integer N such that for all n ∈ Z, |n| ≥ N implies that
Z n+1
|f (t)|p dt < ǫp .
(9.29)
n
Combining the inequalities (9.28) and (9.29), we get
Z n+1
|f (t)|p dt < ǫ
n
if |n| ≥ N . In other words, we have
lim
Z
lim
Z
n+1
|n|→∞ n
and equivalently,
x+1
|x|→∞ x
|f (t)|p dt = 0
|f (t)|p dt = 0.
(9.30)
Finally, we use Theorem 1.33 and then apply the limit (9.30) to the inequality (9.27) with a = x
and b = x + 1 to establish
Z x+1
o1
n Z x+1
p
|f (t)|p dt
|f (t)| dt ≤ lim
lim |g(x)| ≤ lim
= 0.
|x|→∞
|x|→∞
|x|→∞ x
x
Hence we conclude that g vanishes at infinity and then g ∈ C0 (R).
Suppose that f ∈ L∞ . Therefore, we can say that |f (x)| ≤ kf k∞ < ∞ for almost all x ∈ R.
Thus if x < y, then we have
Z y+1
Z x+1
f (t) dt
f (t) dt −
|g(x) − g(y)| =
=
=
Z
Z
x
0
f (t) dt +
x
y
x
f (t) dt −
Z
y
x+1
f (t) dt −
0
Z
y+1
f (t) dt
x+1
Z
y
0
f (t) dt −
Z
0
y+1
f (t) dt
293
9.1. Properties of The Fourier Transforms
≤
Z
x
y
|f (t)| dt +
Z
y+1
x+1
|f (t)| dt
≤ 2(y − x)kf k∞ .
Hence this means that g is uniformly continuous on R and furthermore,
Z x+1
Z x+1
Z x+1
kf k∞ dt = kf k∞
|f (t)| dt ≤
f (t) dt ≤
|g(x)| =
x
x
x
for almost every x ∈ R so that kgk∞ ≤ kf k∞ . This completes the proof of the problem.
Problem 9.10
Rudin Chapter 9 Exercise 10.
Proof. Let’s prove the results one by one.
• Cc∞ does not consist of 0 alone. In Problem 7.8(d), we define the bump function
ψ : R → [0, ∞) by

1 
, if x ∈ (−1, 1);
 exp −
1 − x2
ψ(x) =
(9.31)


0,
otherwise.
This is an element of C ∞ . Since supp (ψ) = [−1, 1], it must be true that ψ ∈ Cc∞ .
• If f ∈ L1loc and g ∈ Cc∞ , then f ∗ g ∈ C ∞ . To begin with, we have to show that f ∗ g is
well-defined first. Since supp (g) is compact in R, let M be a positive number such that
supp (g) ⊆ [−M, M ]. Thus we have
Z M
Z
f (x − y)g(y) dy.
(9.32)
f (x − y)g(y) dy =
(f ∗ g)(x) =
R
−M
As g is continuous on [−M, M ], the Heine-Borel Theorem says that it is bounded. Let a
bound of g be K. Then |g(y)| ≤ K on [−M, M ]. Therefore, the boundedness of g and the
fact f ∈ L1loc indicate that the integral in the equation (9.32) is finite. Consequently, f ∗ g
is well-defined.
Next, we claim that (f ∗ g)′ = f ∗ g′ . In fact, if x ∈ R and 0 < δ < 1, then we write
(f ∗ g)(x + δ) − (f ∗ g)(x)
δ
Z ∞
Z
i
1h ∞
f (x − y)g(y) dy
f (x + δ − y)g(y) dy −
=
δ −∞
−∞
Z M
Z M
h
i
1
=
f (x + δ − y)g(y) dy −
f (x − y)g(y) dy
δ −M
−M
Z x+(M +δ)
Z x+M
h
i
1
=
f (y)g(x − y) dy .
f (y)g(x + δ − y) dy −
δ x−(M −δ)
x−M
(9.33)
Since x + δ − [x + (M + δ)] = −M , x + δ − [x − (M + δ)] = M + 2δ and supp (g) ⊆ [−M, M ],
the first integral in (9.33) can be replaced by
Z x+(M +1)
f (y)g(x + δ − y) dy.
x−(M +1)
Chapter 9. Fourier Transforms
294
Similarly, we may extend the lower limit and the upper limit of the second integral in
(9.33) to x − (M + 1) and x + (M + 1) respectively. Therefore, the expression (9.33) can
be reduced to
Z x+(M +1)
(f ∗ g)(x + δ) − (f ∗ g)(x)
g(x − y + δ) − g(x − y)
=
· f (y) dy.
(9.34)
δ
δ
x−(M +1)
Now we fix x ∈ R. Let g = g1 + ig2 . Since g ∈ Cc∞ , g1 and g2 are real differentiable
functions. It follows from the Mean Value Theorem [49, Theorem 5.10, p. 108] that there
exist ξ, η ∈ (x − y, x − y + δ) such that
g1 (x − y + δ) − g1 (x − y) = δg1′ (ξ) and
g2 (x − y + δ) − g2 (x − y) = δg2′ (η).
Then they imply that g(x − y + δ) − g(x − y) = δ[g1′ (ξ) + ig2′ (η)] which gives
q
p
g(x − y + δ) − g(x − y)
≤ [g1′ (ξ)]2 + [g2′ (η)]2 ≤ |g′ (ξ)|2 + |g ′ (η)|2 .
δ
(9.35)
Since g ∈ Cc∞ , g′ is continuous and thus bounded on [x − y, x − y + δ]. Recall that
y ∈ [x − (M + 1), x + (M + 1)] and 0 < δ < 1. Thus, there is an A > 0 such that |g ′ (t)| ≤ A
for every t ∈ [−M − 2, M + 2]d and then we observe from the inequality (9.35) that
√
g(x − y + δ) − g(x − y)
· f (y) ≤ 2A|f (y)|.
δ
Since f is integrable on [x − (M + 1), x + (M + 1)], Theorem 1.34 (Lebesgue’s Dominated
Convergence Theorem) ensures that the expression (9.34) gives
(f ∗ g)(x + δ) − (f ∗ g)(x)
δ→0
δ
Z x+(M +1)
g(x − y + δ) − g(x − y)
= lim
· f (y) dy
δ→0 x−(M +1)
δ
Z x+(M +1)
g(x − y + δ) − g(x − y)
· f (y) dy
lim
=
δ
x−(M +1) δ→0
Z x+(M +1)
g′ (x − y) · f (y) dy
=
(f ∗ g)′ (x) = lim
x−(M +1)
′
= (g ∗ f )(x).
Since x is arbitrary, we have f ∗ g ∈ C 1 and (f ∗ g)′ = g ′ ∗ f . Since g has compact support,
g (n) also has compact support for every n ∈ N. By induction, we have obtained
(f ∗ g)(n) = g(n) ∗ f.
• Existence of a sequence {gn } ⊆ Cc∞ such that kf ∗ gn − f k1 → 0 as n → ∞ for
every f ∈ L1 . If f ∈ L1 , then f ∈ L1loc and so f ∗ g ∈ C ∞ for any g ∈ Cc∞ . Let
Z 1
ψ(x) dx,
c=
−1
where ψ is the bump function (9.31). It is clear that c is a finite positive number because
ψ > 0. If we define ϕ : R → [0, ∞) by
ϕ(x) = c−1 ψ(x),
d
Notice that A does not depend on x, y or δ.
295
9.1. Properties of The Fourier Transforms
then ϕ satisfies the conditions that ϕ ∈ Cc∞ , supp (ϕ) = [−1, 1], ϕ ∈ L∞ and kϕk1 = 1.
Next, for ǫ > 0, we definee
ϕǫ (x) = ǫ−1 ϕ(ǫ−1 x).
Then it is easily checked that
ϕǫ ∈ Cc∞ ,
ϕǫ ∈ L∞
supp (ϕǫ ) = [−ǫ, ǫ],
and
kϕǫ k1 = kϕk1 = 1.
We observe that
f ∗ ϕǫ (x) − f (x) =
=
=
Z
ZR
f (x − y)ϕǫ (y) dy −
Z
R
f (x)ϕǫ (y) dy
R
[f (x − y) − f (x)]ϕǫ (y) dy
R
[f (x − y) − f (x)]ϕ(ǫ−1 y)ǫ−1 dy.
Z
If we let y = ǫz, then the expression (9.36) becomes
Z
[fǫz (x) − f (x)]ϕ(z) dz .
|f ∗ ϕǫ (x) − f (x)| =
(9.36)
(9.37)
R
Since f ∈ L1 and ϕ ∈ L∞ , Theorem 3.8 (with p = 1 and q = ∞) makes sure that
[fǫz (x) − f (x)]ϕ(z) ∈ L1 (with respect to z). By Theorem 1.33, the expression (9.37) gives
Z hZ
Z
i
|fǫz (x) − f (x)| · |ϕ(z)| dz dx. (9.38)
|f ∗ ϕǫ (x) − f (x)| dx ≤
kf ∗ ϕǫ − f k1 =
{z
}
R|
R
R
nonnegative integrand
Since the integrand in the expression (9.38) is nonnegative, we may apply Theorem 8.8 (The
Fubini Theorem) to change the order of integration in the expression (9.38) to conclude
that
Z
|ϕ(z)| · kfǫz − f k1 dz.
(9.39)
kf ∗ ϕǫ − f k1 ≤
R
By Theorem 9.5, we note that kfǫz − f k1 → 0 as ǫ → 0 for each z ∈ R. Furthermore,
since kfǫz − f k1 ≤ 2kf k1 , we apply Theorem 1.34 (Lebesgue’s Dominated Convergence
Theorem) to the inequality (9.39) to obtain
kf ∗ ϕǫ − f k1 → 0
as ǫ → 0. Now, if we let ǫ =
1
n
and gn = ϕ 1 , then we can obtain our desired sequence.
n
• Existence of a sequence {gn } ⊆ Cc∞ such that (f ∗ gn )(x) → f (x) a.e. for every
f ∈ L1loc . For each fixed n ∈ N, let Kn = [− n1 , n1 ] ⊂ Vn = (− n1 − 2−n , n1 + 2−n ). By [49,
Exercise 6, p. 289], there exists functions ψ1 , . . . , ψs ∈ C ∞ such thatf
– 0 ≤ ψi ≤ 1 for 1 ≤ i ≤ s;
– supp (ψi ) ⊂ Vn ;
– ψ1 (x) + ψ2 (x) + · · · + ψs (x) = 1 for every x ∈ Kn .
e
The collection {ϕǫ } is called an approximate identity or a sequence of mollifiers on R, see [48, Definition
6.31, p. 173] or [11, p. 108].
f
For a proof of this, please read [63, Problem 10.6, pp. 266, 267]. Of course, our s depends on n.
Chapter 9. Fourier Transforms
296
If we define gn : R → R by
n
(ψ1 + ψ2 + · · · + ψs ),
2
then the above conditions imply that gn ∈ Cc∞ , gn (x) = n2 on Kn , gn (x) = 0 outside Vn
and 0 ≤ gn (x) ≤ n2 for all x ∈ R.
gn =
Let x ∈ R and f (x) ≥ 0 on R. On the one hand, we have
Z
Z 1
1
n n
f (x − y) dy =
f (y) dy.
(f ∗ gn )(x) ≥
2 −1
m B 0, n1 ) B(x, 1 )
n
(9.40)
n
Since f ∈ L1loc , the integral (9.40) is finite. On the other hand, we have
Z 1 −n
n n +2
(f ∗ gn )(x) ≤
f (x − y) dy
2 − 1 −2−n
n
Z
n
1
= 1+ n ×
f (y) dy.
2
m B(0, n1 + 2−n ) B(x, n1 +2−n )
(9.41)
Similarly, the integral (9.41) is finite because of the fact that f ∈ L1loc . Now if x ∈ R is a
Lebesgue point of f , then we combine the inequalities (9.40) and (9.41) to get
Z
1
f (y) dy = f (x).
(9.42)
lim (f ∗ gn )(x) = lim
n→∞
n→∞ m B(0, 1 )
1
)
B(x, n
n
By Lemma 7.5, we see that almost every x ∈ R is a Lebesgue point of f . Hence we
conclude from this that (f ∗ gn )(x) → f (x) a.e. on R for every f ∈ L1loc and f (x) ≥ 0 on
R.
For arbitrary f ∈ L1loc , we can write f = f + − f − and apply the above analysis to f +
and f − individually so that the limit (9.42) also holds when f is replaced by f + and f − .
It is clear from Lemma 7.5 that almost every x ∈ R is a common Lebesgue point of f, f +
and f − , so
lim (f ∗ gn )(x) = lim (f + ∗ gn )(x) − lim (f − ∗ gn )(x) = f + (x) − f − (x) = f (x).
n→∞
n→∞
n→∞
Consequently, the result (9.42) is also true in this general case.
• (f ∗ hλ )(x) → f (x) a.e. if f ∈ L1 as λ → 0 and f ∗ hλ ∈ C ∞ . First of all, we notice that
the hλ is in the form of the so-called Poisson kernel on R, see [21, p. 228]. Next, it is
easy to see that
hλ (x) = λ−1 h1 (λ−1 x),
1
where h1 (x) = 1+x
2 . Furthermore, we obtain from the hypotheses of hλ that h1 is an even
function such that it is decreasing on (0, ∞) and h1 ∈ L1 . With the aid of [56, Theorem
1.25, p. 13], we see that
Z
lim (f ∗ hλ )(x) = f (x) h1 (t) dm(t) = f (x)
λ→0
R
for every Lebesgue point x of f . By Theorem 7.7, we obtain our desired result that
(f ∗ hλ )(x) → f (x) a.e. on R if f ∈ L1 as λ → 0.
It remains to prove that f ∗ hλ ∈ C ∞ . To see this, the definition of hλ shows clearly that
(n)
hλ ∈ C ∞ . In addition, direct computation gives easily that hλ is bounded on R for every
positive integer n. As a consequence of [22, Proposition 8.10, p. 242], we may conclude
immediately that f ∗ hλ ∈ C ∞ .
This completes the proof of the problem.
297
9.2. The Poisson Summation Formula and its Applications
9.2
The Poisson Summation Formula and its Applications
Problem 9.11
Rudin Chapter 9 Exercise 11.
Proof. Suppose that f is a Schwartz function. We are going to show that the Poisson summation
formula holds in this situation and consider the limiting case as α → 0 under the extra hypothesis
that fb ∈ L1 .
• The validity of the Poisson summation formula. We first assume that f ∈ S.
Consider
∞
X
f (x + 2kπ)
F (x) =
k=−∞
which is a periodic function of period 2π, with Fourier coefficients
1
Fb(n) =
2π
Z
π
F (x)e−inx dx =
−π
1
2π
Z
π
−π
∞
X
k=−∞
f (x + 2kπ)e−inx dx.
(9.43)
Since f ∈ S, there exists a positive constant M such that |f (x)| ≤ M and |x2 f (x)| ≤ M
for all x ∈ R. In particular, if k ≥ 1, then
|f (x + 2kπ)| ≤
M
M
≤
2
(x + 2kπ)
(2k − 1)2 π 2
for all x ∈ [−π, π]. By the Weierstrass M -test [49, Theorem 7.10, p. 148], we see that the
series
∞
X
f (x + 2kπ)e−inx
k=1
converges uniformly and absolutely on [−π, π]. Similarly, the series
∞
X
k=1
f (x − 2kπ)e−inx
also converges uniformly and absolutely on [−π, π]. If k = 0, then
Z π
Z π
−inx
|f (x)| dx ≤ 2πM.
f (x)e
dx ≤
−π
−π
Consequently, the series
∞
X
f (x + 2kπ)e−inx
(9.44)
k=−∞
converges uniformly and absolutely on [−π, π] so that the integral and summation in the
expression (9.43) can be interchanged (by [49, Corollary, p. 152]) and we get
∞ Z π
1 X
b
f (x + 2kπ)e−inx dx
F (n) =
2π
−π
k=−∞
∞ Z (2k+1)π
1 X
=
f (x)e−inx dx
2π
(2k−1)π
k=−∞
Chapter 9. Fourier Transforms
298
Z
1
=
2π
= ϕ(n)
∞
f (x)e−inx dx
−∞
for every n ∈ Z. In other words, we have
∞
X
∞
X
f (x + 2kπ) = F (x) =
ϕ(n)einx .
(9.45)
n=−∞
k=−∞
If we put x = 0 in the formula (9.45), then we get
∞
X
f (2kπ) =
∞
X
ϕ(n).
(9.46)
n=−∞
k=−∞
Let γ > 0 and g(x) = f (γx). Since g ∈ S, the preceding formula (9.46) also gives
∞
X
g(2kπ) =
∞
X
ψ(n),
(9.47)
n=−∞
k=−∞
where
ψ(t) =
1
2π
Z
∞
g(x)e−itx dx.
−∞
Direct computation shows
ψ(t) =
1
2π
Z
∞
f (γx)e−itx dx =
−∞
1 t
ϕ
γ
γ
and after substituting this back into the formula (9.47) and then using the definition of g
, we obtain
∞
∞
X
X
1 n
f (2γkπ) =
.
(9.48)
ϕ
γ
γ
n=−∞
k=−∞
Suppose that α =
becomes
1
γ
> 0 and β =
∞
X
2π
α
> 0. Then we have αβ = 2π and the formula (9.48)
f (kβ) = α
∞
X
ϕ(nα).
(9.49)
n=−∞
k=−∞
• The case when α → 0. Now we suppose further that the inequality (9.51) in Remark
9.3 below holds. On the one hand, we know that
Z ∞
Z ∞
∞
1 X n
ϕ(x) dx =
fb(x) dm(x)
=
ϕ
ϕ(nα) = lim
lim α
α→0
N →∞ N
N
−∞
−∞
n=−∞
n=−∞
∞
X
because ϕ(n) =
√1 fb(n).
2π
lim
α→0
∞
X
k=−∞
On the other hand, we have
f (kβ) = lim
α→0
Again the fact f ∈ S implies that
∞
X
k=−∞
f
2πk |f (x)| ≤
α
M
|x|2
= lim
N →∞
∞
X
k=−∞
f (2πkN ).
(9.50)
299
9.2. The Poisson Summation Formula and its Applications
holds for all x 6= 0, where M is a positive constant. Thus this shows that
∞
X
k=−∞
k6=0
f (2πkN ) ≤
∞
X
k=−∞
k6=0
|f (2πkN )| ≤
∞
X
k=−∞
k6=0
M
→0
|2πkN |2
as N → ∞ and then the limit (9.50) becomes
lim
α→0
∞
X
f (kβ) = f (0).
k=−∞
Hence we arrive at the result
f (0) =
Z
∞
−∞
fb(x) dm(x)
which is in agreement with Theorem 9.11 (The Inversion Theorem).
We have completed the proof of the problem.
Remark 9.3
If there exists a constant M and p > 1 such that
|f (x)| ≤
M
|x|p
(9.51)
for all sufficiently large x, then the series (9.44) still converges uniformly and absolutely on
[−π, π] so that the Poisson summation formula (9.49) also holds in this case.
Problem 9.12
Rudin Chapter 9 Exercise 12.
Proof. We note that f is not a Schwartz function, but the result [49, Theorem 8.6(f)] ensures
that f satisfies the inequality (9.51), so we may apply the first assertion of Problem 9.11. First
of all, since e−|x| cos tx and e−|x| sin tx are even and odd functions in x respectively, we have
Z ∞
Z
1
1 ∞ −x
−|x| −itx
ϕ(t) =
e
e
dx =
e cos tx dx.
2π −∞
π 0
By [27, §2.662 Eqn. 2, p. 228], we have
Z
(− cos xt + t sin xt)e−x
e−x cos tx dx =
1 + t2
so that
ϕ(t) =
1
.
(1 + t2 )π
Substituting the expression (9.52) and f (x) = e−|x| into the formula (9.49), we obtain
∞
X
k=−∞
e−|k|β = α
∞
X
1
(1 + n2 α2 )π
n=−∞
(9.52)
Chapter 9. Fourier Transforms
300
∞
1 X
α
1 + e−β
=
−β
1−e
π n=−∞ 1 + n2 α2
∞
eβ + 1
1 X
α
=
,
eβ − 1
π n=−∞ 1 + n2 α2
(9.53)
where αβ = 2π. If we take β = 2πγ, then α = γ1 . Hence we deduce from the formula (9.53) that
∞
e2πγ + 1
1 X
γ
=
e2πγ − 1
π n=−∞ γ 2 + n2
(9.54)
and the desired result follows if we replace γ by α in the formula (9.54). This completes the
proof of the problem.
Problem 9.13
Rudin Chapter 9 Exercise 13.
Proof.
(a) By Definition 9.1 and similar argument as in the proof of Problem 9.12, we have
r Z
Z ∞
Z ∞
2 −ixt
2 ∞ −cx2
−cx
−ixt
e
e
dm(x) =
fc (x)e
dm(x) =
fbc (t) =
e
cos tx dx. (9.55)
π 0
−∞
−∞
By [27, §3.922, Eqn. 4, p. 494], we see that
Z
∞
−cx2
e
0
1
cos tx dx =
2
r
t2 π
,
exp −
c
4c
so the formula (9.55) becomes
t2 1
fbc (t) = √ exp −
.
4c
2c
(b) By part (a), if fbc = fc , then we have
x2 1
√ exp −
= exp(−cx2 )
4c
2c
4c2 − 1 √
exp
x2 = 2c
4c
(9.56)
for every x ∈ R. Since the right-hand side of the equation (9.56) is constant, 4c2 − 1 = 0
and then c = 21 . Direct checking immediately shows that c = 21 satisfies the equation
(9.56).
(c) Let a, b ∈ (0, ∞). By the definition, we see that
Z ∞
fa (x − y)fb (y) dy
(fa ∗ fb )(x) =
−∞
Z ∞
exp[−a(x − y)2 − by 2 ] dy
=
−∞
9.3. Fourier Transforms on Rk and its Applications
301
=
Z
∞
−∞
exp{−[(a + b)y 2 − 2axy + ax2 ]} dy.
(9.57)
Using [27, §2.33, Eqn. 1, p. 108] and the facts thatg erf(∞) = 1 and erf(−∞) = −1, we
are able to represent the integral (9.57) as
r
√
π
1
ax ab 2 (fa ∗ fb )(x) =
a + by − √
exp −
x × erf
2 a+b
a+b
a+b
r
π
ab 2
=
exp −
x .
a+b
a+b
y=∞
y=−∞
Therefore, we conclude that
γ=
r
π
a+b
and
c=
ab
.
a+b
(d) By Problem 9.7, we know that fc ∈ S. Thus the definition and the formula [27, §2.33,
Eqn. 1, p. 108] together give
1
ϕ(t) =
2π
Z
∞
−cx2 −itx
e
e
−∞
1
dx =
2π
Z
t2 1
.
exp[−(cx2 + itx)] dx = √ exp −
2 πc
4c
−∞
∞
Hence we establish from the formula (9.49) that
∞
X
−ck 2 β 2
e
k=−∞
∞
h
2nπ 2 i
1
1
2π X
√ exp −
×
=
β n=−∞ 2 πc
4c
β
r
∞
n2 π 2 1 π X
exp −
.
=
β c n=−∞
cβ 2
We complete the proof of the problem.
9.3
Fourier Transforms on Rk and its Applications
Problem 9.14
Rudin Chapter 9 Exercise 14.
Proof. Recall that if f ∈ L1 (Rk ), then
Z
b
f (x) exp(−ix · y) dmk (x).
f (y) =
(9.58)
Rk
We first show the analogue of Theorem 9.6 that fb ∈ C0 (Rk ) and kfbk∞ ≤ kf k1 . To this end, we
imitate the proof of Theorem 9.6. The required inequality is obvious from the definition (9.58)
and the integrability of f .
To prove fb ∈ C0 (Rk ), we need the analogue of Theorem 9.5 for Rk :h
g
h
Here erf(x) is the error function.
Its proof is very similar to that of Theorem 9.5, so we omit the details here.
Chapter 9. Fourier Transforms
302
Lemma 9.6
For any function f on Rk and every y ∈ Rk , let fy be the translate of f defined by
fy (x) = f (x − y)
for x ∈ Rk . If 1 ≤ p < ∞ and f ∈ Lp (Rk ), then the mapping
y 7→ fy
is a uniformly continuous mapping of Rk into Lp (Rk ).
Let t ∈ Rk . If tn → t, then
|fb(tn ) − fb(t)| ≤
Z
Rk
|f (x)| · | exp(−itn · x) − exp(−it · x)| dmk (x).
(9.59)
Since the integrand of the integral (9.59) is bounded by 2|f (x)| ∈ L1 (Rk ), Theorem 1.34
(Lebesgue’s Dominated Convergence Theorem) implies that
fb(tn ) → fb(t)
as n → ∞. In other words, fb is continuous on Rk .
Next we have to prove that f vanishes at infinity. Suppose that y = (η1 , . . . , ηk ) and η1 6= 0.
Take
π
, 0, . . . , 0 .
y1 =
η1
Since e−πi = −1, we have
h
exp − iy · x + y1
i
= exp
n
k
h io
π X
− i η1 ξ1 +
ηj ξj
+
η1
= exp − i
j=2
k
X
j=1
ξj ηj × e−iπ
= − exp(−ix · y)
so that the definition (9.58) implies that
Z
Z
i
h
b
f (x) exp − iy · x + y1 dmk (x) = −
f (y) = −
Rk
Rk
f (x − y1 ) exp(−iy · x) dmk (x).
Thus we get from this that
2fb(y) =
=
Z
k
ZR
Rk
[f (x) − f (x − y1 )] exp(−iy · x) dmk (x)
[f (x) − fy1 (x)] exp(−iy · x) dmk (x).
(9.60)
Now we apply Lemma 9.6 to the integral (9.60), we assert that
2|fb(y)| ≤ kf − fy1 k1 = kf0 − fy1 k1
(9.61)
which tends to 0 as |y1 | → 0 or equivalently, as η1 → ±∞. Or we can say this way: Given ǫ > 0,
there exists a δ > 0 such that
kf0 − fy1 k1 < 2ǫ
9.3. Fourier Transforms on Rk and its Applications
303
for all y1 ∈ Rk for which |y1 | < δ or |η1 | > πδ . Notice also that η1 → ±∞ means that |y| → ∞.
By similar argument as the previous part, we note that the result (9.61) also holds if y1 is
replaced by any yj = (0, . . . , 0, ηπj , 0, . . . , 0) with ηj 6= 0, where j = 2, 3, . . . , k. Write it explicitly,
we have
kf0 − fyj k1 < 2ǫ
(9.62)
for all yj ∈ Rk for which |ηj | > πδ , where j = 1, 2, . . . , k. Now for arbitrary y ∈ Rk \ {0}, there
|y|
is at least one j such that |ηj | ≥ |y|
k . Otherwise, we have |ηj | < k for every j = 1, 2, . . . , k, but
2
this means that k < 1, a contradiction. Thus if we take
n
kπ o
K = y ∈ Rk |y| ≤
,
δ
then K is compact and for y ∈
/ K, we have
|ηj | ≥
π
|y|
>
k
δ
for some 1 ≤ j ≤ k. Hence the inequality (9.62) always holds for this j and then we conclude
from the inequality (9.61) that
|fb(y)| < ǫ
if y ∈
/ K. By Definition 3.16, fb vanishes at infinity, i.e., fb ∈ C0 (Rk ).
Our next target is to obtain the analogue of Proposition 9.8. Similar to the work in §9.7, we
put
H(t) = exp[−(|t1 | + |t2 | + · · · + |tk |)],
where t = (t1 , t2 , . . . , tk ). Let λ > 0. Then 0 < H(t) ≤ 1 and H(λt) → 1 as λ → 0. Define
Z
H(λt) exp(it · x) dmk (t).
hλ (x) =
Rk
A simple computation with repeated applications of Theorem 8.8 (The Fubini Theorem), it gives
hλ (x) =
k 1
Y
2 2
j=1
π
λ2
λ
.
+ ξj2
By Theorem 8.8 (The Fubini Theorem) and [51, Eqn. (4), p. 183], we get
#
"Z
k
2 1
Y
λ
2
dm(ξ1 ) = 1.
hλ (x) dmk (x) =
π
λ2 + ξj2
k
R
R
Z
j=1
By this result, we establish the analogue of Proposition 9.8: If f ∈ L1 (Rk ), then
Z
H(λt)fb(t) exp(ix · t) dmk (t).
(f ∗ hλ )(x) =
(9.63)
Rk
Similarly, we have the following analogue of Theorem 9.10: If 1 ≤ p < ∞ and f ∈ Lp (Rk ), then
lim kf ∗ hλ − f kp = 0.
λ→0
Now it is time to prove the required results.
(9.64)
Chapter 9. Fourier Transforms
304
• The Inversion Theorem for Rk . Suppose that f, fb ∈ L1 (Rk ). The integrand on the
right-hand side of the formula (9.63) are bounded by |fb(t)|. By Theorem 1.34 (Lebesgue’s
Dominated Convergence Theorem) and the fact that H(λt) → 1 as λ → 0, the right-hand
side of the formula (9.63) converges to
Z
fb(t) exp(ix · t) dmk (t).
g(x) =
Rk
for all x ∈ Rk . By the limit (9.64) and Theorem 3.12, there is a sequence {λn } such that
λn → 0 and
lim (f ∗ hλn )(x) = f (x)
n→∞
a.e. on
Consequently, we have f (x) = g(x) a.e. on Rk . Since fb ∈ L1 (Rk ) and g(−x)
is the Fourier transform of fb, the analogue of Theorem 9.6 above implies that g ∈ C0 (Rk )
as desired.
Rk .
• The Plancherel Theorem for Rk . Fix f ∈ L1 (Rk ) ∩ L2 (Rk ). Put fe(x) = f (−x) and
g = f ∗ fe. Then we have
Z
Z
f (x + y)f (y) dmk (y) = hf−x , f i .
f (x − y)f (−y) dmk (y) =
g(x) =
Rk
Rk
By the use of Lemma 9.6 and Theorem 4.6, it can be shown that g is a continuous function
on Rk and thus Theorem 4.2 (The Schwarz Inequality) gives
|g(x)| ≤ kf−x k2 · kf k2 = kf k22
so that g is bounded on Rk . Since f ∈ L1 (Rk ), we have fe ∈ L1 (Rk ) and then g ∈ L1 (Rk )
by the analogue of Theorem 8.14 for Rk .i By the formula (9.63), we have
Z
(g ∗ hλ )(0) =
(9.65)
H(λt)b
g (t) dmk (t).
Rk
Since g is continuous and bounded on Rk , the analogue of Theorem 9.9j shows that
lim (g ∗ hλ )(0) = g(0) = kf k22 .
λ→0
(9.66)
Now the analogues of Theorem 9.2(c) and (d) imply that
b
gb = fb × fe = fb × fb = |fb|2 ≥ 0.
Since H(λt) is measurable and increases to 1 as λ → 0, Theorem 1.26 (Lebesgue’s Monotone Convergence Theorem) implies that
Z
Z
H(λt)b
g (t) dmk (t) =
lim
|fb(t)|2 dmk (t).
(9.67)
λ→0 Rk
Rk
Hence it follows from the expression (9.65) and the two limits (9.66) and (9.67) that
Z
(9.68)
H(λt)b
g (t) dmk (t) = lim (g ∗ hλ )(0) = kf k22 < ∞.
kfbk22 = lim
λ→0 Rk
Suppose that
i
j
λ→0
Y = {fb| f ∈ L1 (Rk ) ∩ L1 (Rk )}.
The analogue of Theorem 8.14 is that f, g ∈ L1 (Rk ) implies f ∗ g ∈ L1 (Rk ), but we won’t give a proof here.
If g ∈ L∞ (Rk ) and g is continuous at a point x ∈ Rk , then (g ∗ hλ )(x) → g(x) as λ → 0.
9.3. Fourier Transforms on Rk and its Applications
305
The equality (9.68) implies that Y is a subspace of L2 (Rk ). We claim that Y is dense in
L2 (Rk ). To this end, we recall from Problem 4.1 that (Y ⊥ )⊥ = Y . Now we observe that
Y ⊥ = {0} if and only if
Y = {0}⊥ = {f ∈ L2 (Rk ) | hf, 0i = 0} = L2 (Rk ).
Thus it suffices to show that Y = {0}⊥ . Let a ∈ Rk and λ > 0. Denote the mapping
x 7→ exp(ia · x)H(λx) by F . It is clear that F (x) ∈ L1 (Rk ) ∩ L2 (Rk ) for all x ∈ Rk and
Z
Z
b
exp i(a − t) · x H(λx) dmk (x) = hλ (a − t)
F (x) exp(−ix · t) dmk (t) =
F (t) =
Rk
Rk
so that hλ ∈ Y . If w ∈ Y ⊥ = {f ∈ L2 (Rk ) | hf, gi = 0 for all g ∈ Y }, then it follows that
Z
(hλ ∗ w)(a) =
hλ (a − t)w(t) dmk (t) = 0
Rk
for all a ∈ Rk . By the limit (9.64), we see that w = 0, i.e., Y ⊥ = {0} and thus Y = L2 (Rk ).
Now we define Φ : L1 (Rk ) ∩ L2 (Rk ) → Y by
Φ(f ) = fb.
Then the equality (9.68) certainly shows that Φ is an L2 (Rk )-isometry. Since both L1 (Rk )∩
L2 (Rk ) and Y are dense in L2 (Rk ), as the paragraph before [51, Eqn. (10), p. 187] indicates
that Φ can be extended to an isometry
e : L2 (Rk ) → L2 (Rk ).
Φ
e ) (of course, it is the Plancherel transform), then the equality (9.68)
If we denote fb = Φ(f
holds for every f ∈ L2 (Rk ). Using a similar argument as in the proof of Theorem 4.18, we
can prove the Parserval formula
Z
Z
D
E
fb(t)b
g (t) dmk (t) = fb, gb
f (x)g(x) dmk (x) =
hf, gi =
Rk
Rk
for every f, g ∈ L2 (Rk ). Finally, for the symmetric relation in Theorem 9.13(d), the ϕA
and ψA should be replaced by
Z
Z
f
f (x) exp(−ix · t) dmk (x) and ψA (x) =
fb(t) exp(ix · t) dmk (t)
ϕ
fA (t) =
[−A,A]k
[−A,A]k
respectively. Employing a similar argument as shown in [51, p. 187], it can be verified
that
fA − f k2 → 0
kϕ
fA − fbk2 → 0 and kψ
as A → ∞.
• The analogue of Theorem 9.23. To every complex homomorphism ϕ on L1 (Rk ) (except
for ϕ = 0) there corresponds a unique t ∈ Rk such that
ϕ(f ) = fb(t).
Here it is seen that the content up to Eqn. (6) in §9.22 can be extended to Rk easily. If
x = ξe1 + ξ2 e2 + · · · + ξk ek , where {e1 , e2 , . . . , ek } is the usual basis of Rk , then we have
β(x) = β(ξe1 + ξ2 e2 + · · · + ξk ek ) = β(ξ1 e1 ) × β(ξ2 e2 ) × · · · × β(ξ1 e1 ).
(9.69)
Chapter 9. Fourier Transforms
306
Let βj : R → R be defined by βj (ξj ) = β(ξj ej ) for 1 ≤ j ≤ k. Since β is continuous on
Rk , each βj is continuous on R. Since β is not identity 0, each βj is not identically 0.
Furthermore, each βj satisfies
βj (x + y) = β (x + y)ej = β(xej )β(yej ) = βj (x)βj (y).
Thus each βj satisfies the hypotheses qon [51, p. 192] which implies βj (ξj ) = exp(−itj ξj )
for a unique tj ∈ R. By the formula (9.69), we obtain
β(x) = exp(−it1 ξ1 ) × exp(−it2 ξ2 ) × · · · × exp(−itk ξk ) = exp(−it · x),
where t = (t1 , t2 , . . . , tk ) and it is unique. Hence, we get from this that
Z
Z
f (x) exp(−it · x) dmk (x) = fb(t).
f (x)β(x) dmk (x) =
ϕ(f ) =
Rk
Rk
We have completed the proof of the problem.
Problem 9.15
Rudin Chapter 9 Exercise 15.
Proof. Let A : Rk → Rk be a linear operator with det A > 0. Thus A−1 exists. Let y = A−1 (x).
Then we follow from Theorem 7.26 (The Change-of-variables Theorem) (with X = Y = V = Rk ,
T = A−1 which is one-to-one and differentiable on Rk , T (Rk ) = A−1 (Rk ) = Rk and y = A−1 (x))
that
g (z) = f[
b
◦ A(z)
Z
f A(y) exp(−iy · z) dmk (y)
=
k
{z
}
|R
Z
This is
=
Z
Rk
f dm.
Y
f (x) exp[−iA−1 (x) · z] × | det A−1 | dmk (x)
{z
}
|
(9.70)
This is (f ◦ T )(x) = (f ◦ A−1 )(x).
for every z ∈ Rk . We need a result from linear algebra:
Lemma 9.7
If A is a k × k matrix, then for all x, y ∈ Ck , we have hA(x), yi = x, AT (y) ,
where AT is the transpose of A.
Proof of Lemma 9.7. Using [41, p. 109; Example 5.3.1, p. 286], it can be seen easily
that
hA(x), yi = [A(x)]T y = xT AT (y) = x, AT (y) ,
completing the proof of Lemma 9.7.
Thus we apply Lemma 9.7 to the expression (9.70) to deduce that
Z
f (x) exp[−ix · A−T (z)] dmk (x) = | det A| · fb A−T (z) .
g(z) = | det A| ·
b
Rk
9.3. Fourier Transforms on Rk and its Applications
307
Consequently, we conclude that
f\
◦ A = | det A| · (fb ◦ A−T ).
(9.71)
Particularly, let A be a rotation matrix of Rk . Then it preserves the distance of x from the
origin, i.e., hA(x), A(x)i = hx, xi. By [4, Proposition 5.1.13, p. 134], A is orthogonal which
means
AT = A−1 .
Furthermore, this implies that | det A| = 1, so the formula (9.71) can be simplified to
f\
◦ A = fb ◦ A.
(9.72)
In other words, if f is invariant under rotations, then f ◦A = f and the expression (9.72) implies
fb ◦ A = fb,
i.e., fb is also invariant under rotations. This completes the proof of the problem.
Problem 9.16
Rudin Chapter 9 Exercise 16.
Proof. We make the following assumption: Suppose that f ∈ S(Rk ), where S(Rk ) is the class
of all functions f : Rk → C such that f ∈ C ∞ and xβ ∂ α f is bounded for all multi-indices α
∂f
and β.k Let 1 ≤ j ≤ k, x = (xj , xj ) and F (x) =
, where x = (x1 , . . . , xj , . . . , xk ) and
∂xj
xj = (x1 , . . . , xj−1 , xj+1 , . . . , xk ). We note that
x · y = x1 y1 + x2 y2 + · · · + xk yk = xj · yj + xj yj .
For every y ∈ Rk , we follow from this, the definition in Problem 9.14 and Lemma 9.4 that
Z
∂f
b
F (y) =
(x) exp(−ix · y) dmk (x)
k ∂xj
ZR
i
h Z ∂f
(xj , xj ) exp(−ixj · yj ) dm(xj )
exp(−ixj · yj ) dmk−1 (xk )
=
∂xj
Rk−1
| R
{z
}
=
Z
Apply Lemma 9.4 to this.
Rk−1
Z
= iyj
i
h Z
f (x) exp(−ixj · yj ) dm(xj )
exp(−ixj · yj ) dmk−1 (xk ) iyj
Rk
R
f (x) exp(−ix · y) dmk (x)
= iyj fb(y)
which implies that
Therefore, we have
k
2f
∂d
(y) = −yj2 fb(y).
∂x2j
g(x) = −
b
k
X
j=1
x2j fb(x) = −|x|2 · fb(x)
In fact, S is the Schwartz space on Rk , see Remark 9.2.
(9.73)
Chapter 9. Fourier Transforms
308
for all x ∈ Rk .
Suppose that f has continuous second derivatives so that the Laplacian ∆f is well-defined.
Suppose further that supp (f ) is compact. In this case, f ∈ L1 (Rk ) so that fb is well-defined on
Rk . Furthermore, we suppose that A is a rotation of Rk about the origin 0 and gA = ∆(f ◦ A).
Then we deduce from the formulas (9.72) and (9.73) that
2 \
2
2 b
b
gc
b(Ax) = g[
◦ A(x).
A (x) = −|x| · f ◦ A(x) = −|Ax| · (f ◦ A)(x) = −|Ax| · f (Ax) = g
\
\
By the definition, we have ∆(f
◦ A) = (∆f
) ◦ A. To go further, we need the analogue of
Theorem 9.12 (The Uniqueness Theorem):
Lemma 9.8
If f ∈ L1 (Rk ) and fb(t) = 0 for all t ∈ Rk , then
f (x) = 0
(9.74)
a.e. on Rk . Furthermore, if f is continuous on Rk , then the expression (9.74) holds
everywhere on Rk .
Proof of Lemma 9.8. Since fb = 0 on Rk , we have fb ∈ L1 (Rk ) and the result (9.74)
follows from the Inversion Theorem for Rk (Problem 9.14). If f is continuous on Rk ,
the definition of continuity ensures that the relation (9.74) holds everywhere on Rk . Now we let F = ∆(f ◦ A) − (∆f ) ◦ A. It is clear that f ◦ A ∈ S(Rk ) if f ∈ S(Rk ). Similar
to the proof of Problem 9.7, we can show that
∂ α f ∈ S(Rk )
and ∂ α f ∈ L1 (Rk ).
Thus we have ∆(f ◦ A) ∈ L1 (Rk ) and (∆f ) ◦ A ∈ L1 (Rk ) so that F ∈ L1 (Rk ). Since Fb(t) = 0
for all t ∈ Rk and F is continuous on Rk , Lemma 9.8 implies that
∆(f ◦ A) = (∆f ) ◦ A
(9.75)
holds everywhere on Rk . It is well-known that any rotation about a point p is the composition of
a rotation about the origin 0 and a translation. By this fact, the analysis preceding Lemma 9.8
and the hypothesis that ∆ commutes with translations, we conclude that the expression (9.75)
is still valid for every rotation about every point p.
It is time to consider the general situation. Fix a p ∈ Rk . There exists a r > 0 such
that p ∈ Kr = B(0, r) ⊂ Vr = B(0, 2r). By [49, Exercise 6, p. 289], there exists functions
ψ1 , . . . , ψs ∈ C ∞ (Rk ) such thatl
• 0 ≤ ψi ≤ 1 for 1 ≤ i ≤ s;
• supp (ψi ) ⊂ Vr ;
• ψ1 (x) + ψ2 (x) + · · · + ψs (x) = 1 for every x ∈ Kr .
If we define Ψ : Rk → R by
l
Ψ = ψ1 + ψ2 + · · · + ψs ,
For a proof of this, please read [63, Problem 10.6, pp. 266, 267 ].
9.3. Fourier Transforms on Rk and its Applications
309
then the above conditions imply that Ψ ∈ Cc∞ (Rk ) and Ψ(x) = 1 on Kr . Thus we have f · Ψ = f
on Kr . Particularly, we have
∂Ψ
∂2Ψ
(p) = 0
(p) =
∂xj
∂x2j
(f · Ψ)(p) = f (p) and
for every 1 ≤ j ≤ k. As a consequence, these imply that
∂ 2 (f · Ψ)
∂Ψ
∂2Ψ
∂2f
∂f
∂2f
(p)
·
(p)
+
f
·
(p)
=
Ψ(p)
·
(p)
+
2
(p)
=
(p)
∂xj
∂xj
∂x2j
∂x2j
∂x2j
∂x2j
(9.76)
for 1 ≤ j ≤ k. By the definition of the Laplacian and the result (9.76), we yield
k
k
X
X
∂ 2 (f · Ψ)
∂2f
∆(f · Ψ) (p) =
(p)
=
(p) = (∆f )(p).
∂x2j
∂x2j
j=1
j=1
(9.77)
Note that f · Ψ is compactly supported in this case. If Ap = q, then q ∈ Kr . On the one hand,
the expressions (9.75) (with f replaced by f · Ψ) and (9.77) give
∆ f · Ψ) ◦ A (p) = [(∆(f · Ψ)) ◦ A](p) = ∆(f · Ψ) (q) = (∆f )(q) = [(∆f ) ◦ A](p). (9.78)
On the other hand, since
[(f · Ψ) ◦ A](p) = [(f ◦ A)(p)] · [(Ψ ◦ A)(p)] = (f · Ψ)(q) = f (q) = (f ◦ A)(p)
on Kr , we have (f · Ψ) ◦ A = f ◦ A on Kr so that
∆ (f · Ψ) ◦ A (p) = ∆(f ◦ A) (p)
(9.79)
Finally, by combining the expressions (9.78) and (9.79) and using the fact that p is arbitrary,
we may conclude that our desired formula (9.75) also holds everywhere on Rk in this general
case. This completes the proof of the problem.
Problem 9.17
Rudin Chapter 9 Exercise 17.
Proof. We prove the case for Rk directly. Suppose that ϕ : Rk → C is a Lebesgue measurable
character. We define Φ : L1 (Rk ) → C by
Z
f (x)ϕ(x) dmk (x).
(9.80)
Φ(f ) =
Rk
Now direct computation gives
Z
Z
(f ∗ g)(x)ϕ(x) dmk (x) =
Φ(f ∗ g) =
Rk
Rk
Z
Rk
f (x − y)g(y)ϕ(x) dmk (y) dmk (x).
(9.81)
Since ϕ is a character, we have ϕ(x) = ϕ(x − y)ϕ(y). Substituting this into the expression
(9.81), we see that
Z Z
f (x − y)g(y)ϕ(x − y)ϕ(y) dmk (y) dmk (x)
Φ(f ∗ g) =
k
k
R
R
Z hZ
i
f (x − y)ϕ(x − y) dmk (x) g(y)ϕ(y) dmk (y)
=
Rk
Rk
Chapter 9. Fourier Transforms
=
Z
Rk
310
Φ(f )g(y)ϕ(y) dmk (y)
= Φ(f ) · Φ(g).
Therefore, Φ is a linear functional on L1 (Rk ). In other words, Φ is a complex homomorphism of
the Banach algebra L1 (Rk ). Hence the analogue of Theorem 9.23 in Problem 9.14 shows that
one may find a β ∈ L∞ (Rk ) and a unique y ∈ Rk such that
Z
f (x) exp(−iy · x) dmk (x) = fb(y).
(9.82)
β(x) = exp(−iy · x) and Φ(f ) =
Rk
Next, we deduce from the formulas (9.80) and (9.82) that
Z
f (x)[ϕ(x) − exp(−iy · x)] dmk (x) = 0
(9.83)
Rk
holds for all f ∈ L1 (Rk ). Since f (x) = e−|x| is in L1 (Rk ) and f (x) 6= 0 on Rk , the result (9.83)
implies that
ϕ(x) = exp(−iy · x)
(9.84)
a.e. on Rk . To proceed further, we need the following result which is a generalization of Problem
7.6 and its proof uses the Steinhaus Theorem in Rk (see Remark 7.2), but we don’t present
it here.
Lemma 9.9
Suppose that G is a subgroup of Rk (relative to addition), G 6= Rk , and G is
Lebesgue measurable. Then mk (G) = 0.
To apply this lemma, we consider h(x) = ϕ(x) − exp(−iy · x) and G = {x ∈ Rk | h(x) = 0}.
For a, b ∈ G, if a + b ∈
/ G, then we have h(a + b) 6= 0 and
ϕ(a)ϕ(b) = ϕ(a + b) 6= exp[−iy · (a + b)] = exp(−iy · a) exp(−iy · b) = ϕ(a)ϕ(b)
which implies 1 6= 1, a contradiction. Thus a + b ∈ G which means G is a subgroup of Rk
(relative to vector addition). Besides, since h is obviously Lebesgue measurable by Proposition
1.9(c) and G = h−1 (0), G is a Lebesgue measurable set. Since mk (G) > 0, Lemma 9.9 forces
that
G = Rk
so that the equation (9.84) holds for all x ∈ Rk . Now the continuity of the exponential function
implies that ϕ is continuous on Rk . This completes the proof of the problem.
9.4
Miscellaneous Problems
Problem 9.18
Rudin Chapter 9 Exercise 18.
Proof. First of all, the equation
f (x + y) = f (x) + f (y)
for all x, y ∈ R is called the Cauchy functional equation.
(9.85)
311
9.4. Miscellaneous Problems
• Existence of real discontinuous functions f satisfying the equation (9.85). It
is well-known that the Hausdorff Maximality Theorem, the Axiom of Choice and Zorn’s
Lemma are equivalent to each other. To prove this part, we would like to apply Zorn’s
Lemma which says that
Lemma 9.10 (Zorn’s Lemma)
If X is a partially ordered set and every totally ordered subset of X has an upper
bound, then X has a maximal element.
As a consequence, Zorn’s Lemma can be used to prove that every vector space has a
basis. In fact, if A is a linearly independent subset of a vector space V over a field K, then
there is a basis B of V that contains A.m By this result, there exists a basis of R over Q
with the usual addition and scalar multiplication.n
Let x ∈ R and H be a basis of R over Q. Thus x has a unique representation
x = r 1 b1 + r 2 b2 + · · · + r n bn ,
where b1 , b2 , . . . , bn ∈ H and r1 , r2 , . . . , rn ∈ Q. Define f : R → R by
f (x) = r1 + r2 + · · · + rn .
It is obvious that f satisfies the equation (9.85). Assume that f was continuous on R.
Then f (R) must be connected (see [49, Theorem 4.22, p. 93]), but f (R) = Q which is a
contradiction.
• If f is Lebesgue measurable and satisfies the equation (9.85), then f is continuous. We use the following special form of Theorem 2.24 (Lusin’s Theorem):o
Lemma 9.11
Suppose that f : [a, b] → R is Lebesgue measurable. For every ǫ > 0, there is a
compact set K ⊆ [a, b] such that m([a, b] \ K) < ǫ and f |K is continuous.
By Lemma 9.11, there exists a compact set K ⊆ [0, 1] such that
m(K) >
2
3
(9.86)
and f |K is continuous. Since K is compact, f is actually uniformly continuous on K.
Given ǫ > 0. There is a δ > 0 such that |f (x) − f (y)| < ǫ for every x, y ∈ K with
|x − y| < δ. Without loss of generality, we may assume that δ < 13 . Let η ∈ (0, δ) and
K − η = {x − η | x ∈ K}. Then K − η ⊆ [−η, 1 − η]. Assume that K ∩ (K − η) = ∅. Since
K, K − η ⊆ [−η, 1], we have
m(K) + m(K − η) = m K ∪ (K − η) ≤ m([−η, 1]) = 1 + η.
(9.87)
By Theorem 2.20(c), m(K − η) = m(K), so we deduce from the inequalities (9.86) and
(9.87) that 1 + η ≥ 2m(K) > 43 and thus η > 13 , a contradiction. Hence we have
K ∩ (K − η) 6= ∅.
m
For a hint of this proof, please refer to [42, Exercise 8, p. 72]
This kind of basis is called a Hamel basis.
o
Read [22, Exercise 44, p. 64] or [47, p. 66]
n
Chapter 9. Fourier Transforms
312
Let p ∈ K ∩ (K − η). Then p, p + η ∈ K and |p + η − p| = η < δ so that
|f (η) − f (0)| = |f (p) + f (η) − f (p)| = |f (p + η) − f (p)| < ǫ.
(9.88)
In other words, f is continuous at 0. Now for any x ∈ [0, 1], the inequality (9.88) implies
clearly that
|f (x + η) − f (x)| = |f (x) + f (η) − f (x)| = |f (η) − f (0)| < ǫ.
Hence f is continuous on [0, 1].
Next, if z ∈ [1, 2], then we have z = x+y for some x, y ∈ [0, 1]. Since f (z) = f (x)+f (y),
we have
|f (z + η) − f (z)| = |f (x + η) − f (x)| < ǫ
and the above paragraph shows that f is also continuous on [1, 2]. Now this kind of
argument can be applied repeatedly and we conclude that f is continuous on R.
• If graph (f ) is not dense in R2 and satisfies the equation (9.85), then f is continuous. Suppose that f is discontinuous. Then f cannot be of the form f (x) = cx for
any constant c. We have
graph (f ) = { x, f (x) | x ∈ R}.
Choose a nonzero real number x1 . Then there exists another nonzero real number x2 such
that
f (x2 )
f (x1 )
6=
x1
x2
or equivalently,
x1 f (x1 )
6= 0.
x2 f (x2 )
In other words, the vectors u = x1 , f (x1 ) and v = x2 , f (x2 ) are linearly independent
and thus span the space R2 . Let p ∈ R2 . Given ǫ > 0. Since Q2 is dense in R2 , one can
find q1 , q2 ∈ Q such that
|p − q1 u − q2 v| < ǫ.
Clearly, we have
Therefore, the set
q1 u + q2 v = q1 x1 + q2 x2 , f (q1 x1 + q2 x2 ) .
G = { x, f (x) ∈ R2 | x = q1 x1 + q2 x2 and q1 , q2 ∈ Q}
is dense in R2 . Since G ⊆ graph (f ), graph (f ) is dense in R2 .
• The form of all continuous functions satisfying the equation (9.85). In fact, by
a similar argument as in the proof used in [63, Problem 8.6, p. 178], it can be shown that
the function f satisfies
f (r) = rf (1)
for every r ∈ Q. Now for every x ∈ R, since Q is dense in R, we can find a sequence
{rn } ⊆ Q such that rn → x as n → ∞. Since f is continuous, we have
f (x) = lim f (rn ) = lim rn f (1) = xf (1).
n→∞
n→∞
Hence every continuous solution of the equation (9.85) must be in the form
f (x) = f (1) · x.
313
9.4. Miscellaneous Problems
We have completed the proof of the problem.
Problem 9.19
Rudin Chapter 9 Exercise 19.
Proof. Let f = χA and g = χB . Since f, g ∈ L1 , their convolution h = f ∗ g is well-defined by
Theorem 8.14. Suppose that p and q are conjugate exponents. Direct computation gives
Z ∞
Z
m(A)
1
dx = √
<∞
|χA (x)|p dm(x) = √
kf kp =
2π A
2π
−∞
and
Z
∞
1
kgkq =
|χB (x)| dm(x) = √
2π
−∞
Thus we have f ∈
We consider
Lp
Z
and g ∈
R
Lq .
q
Z
B
m(B)
dx = √
< ∞.
2π
By Problem 9.8, h = χA ∗ χB is (uniformly) continuous on R.
(f ∗ g)(x) dm(x) =
Z Z
R
R
f (x − y)g(y) dm(y) dm(x).
(9.89)
Recall that R is σ-finite. If we set F (x, y) = f (x − y)g(y) = χA (x − y)χB (y) ≥ 0 on R2 , then F
satisfies the hypotheses of Theorem 8.8 (The Fubini Theorem). We notice that
Z
Z
Z
Z
y
f (x − y)g(y) dm(x) = g(y) f (x − y) dm(x) = g(y) f (x) dm(x),
F dm(x) =
ψ(y) =
R
R
R
R
so the integral on the right-hand side in the expression (9.89) can be simplified to
Z
Z
ψ(y) dm(y)
(f ∗ g)(x) dm(x) =
R
R
#
"
Z
Z
R
R
=
f (x) dm(x) dm(y)
g(y)
=
nZ
R
o
o nZ
g(y) dm(y)
f (x) dm(x) ×
R
1
=
m(A)m(B) > 0.
2π
Hence it is impossible that χA ∗ χB = 0.
Recall the definition that
(χA ∗ χB )(x) =
Z
∞
−∞
χA (x − y)χB (y) dm(y).
Since y ∈ B if and only if χB (y) = 1, the integral (9.90) reduces to
Z
χA (x − y) dm(y).
(χA ∗ χB )(x) =
(9.90)
(9.91)
B
Similarly, x − y ∈ A if and only if χA (x − y) = 1. Define x − A = {x − a | a ∈ A}. We notice
that x − y ∈ A implies that y ∈ x − A and then y ∈ B ∩ (x − A). These facts show that the
integral (9.91) can be further simplified to
Z
dm(y) = m B ∩ (x − A) ≥ 0.
(9.92)
(χA ∗ χB )(x) =
B∩(x−A)
Chapter 9. Fourier Transforms
314
By the previous paragraph, there exists a x0 ∈ R such that (χA ∗ χB )(x0 ) > 0, so the expression
(9.92) gives
m B ∩ (x0 − A) > 0.
In other words, we have B ∩ (x0 − A) 6= ∅ and we can find a b ∈ B such that x0 = a + b for
some a ∈ A. Consequently, we have
x0 ∈ A + B.
(9.93)
Finally, since χA ∗ χB is continuous at x0 , there exists a neighborhood I around x0 such that
(χA ∗ χB )(x) > 0 on I. Hence we conclude from the relation (9.93) that
I ⊆ A + B,
completing the proof of the problem.
Index
Gram-Schmidt Process, 124
A
algebra, 262
approximate identity, 295
arc length, 89
atom, 199
atomicless measures, 199
Axiom of Choice, 311
H
Haar measure, 263
Hamel basis, 311
Heine-Borel Theorem, 35
I
ideal, 262
Invariance of Domain, 217
isometric isomorphism, 157
isometry, 112
isomorphism, 112
B
basis, 2
Basis Extension Theorem, 42
Bertrand’s integrals, 73
bump function, 217, 293
K
Kronecker delta function, 154
Kronecker’s Approximation Theorem, 212
C
Cauchy functional equation, 310
Chebyshev’s Inequality, 106
connected sets, 25
convex combinations, 162
L
Lebesgue function associated with the
Cantor-like set E, 219
limit point, 50
little-o notation, 60
locally compact group, 263
locally integrable functions, 240
lower densities of E at 0, 210
lower Riemann integral, 36
D
De Morgan’s law, 35
dense, 67
Dieudonné’s measure, 53
direct sum, 148
Dirichlet integral, 274
Dirichlet integrals, 181
discrete measure, 225
dyadic rationals, 218
M
Mean Value Theorem, 116
Minkowski’s Inequality, 256
mollifiers, 295
E
error function, 301
Euler constant, 178
Extreme Value Theorem, 61
O
order topology, 47
outer measure, 54
F
finite intersection property, 60
Fubini’s Theorem on Differentiation, 236
P
Poisson kernel, 296
power set, 2
Product Measure Theorem, 270
push-forward measure, 264
G
Gaussian function, 289
315
Index
R
Radon measure, 59
Riemann-Lebesgue Lemma, 285
Riesz-Fischer Theorem, 236
S
Schauder basis, 154
Schwartz space, 289
second-countable, 273
semifinite, 173
sequentially compact, 128
Silverman-Toepliotz Theorem, 171
singular function, 218
standard topology, 2
Stone-Weierstrass Theorem, 284
subadditive, 13, 27
subalgebra, 262
316
T
tent function, 59
The Steinhaus Theorem, 215, 310
triangle inequality, 14, 43
U
upper densities of E at 0, 210
Urysohn’s Lemma, 22
usual metric, 44
W
Weierstrass Approximation Theorem, 125,
139
Weyl’s Equidistribution Theorem, 135
Y
Young’s Convolution Inequality, 256
Z
Zorn’s Lemma, 311
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A Complete Solution Guide to Real
and Complex Analysis II
by Kit-Wing Yu, PhD
kitwing@hotmail.com
Copyright c 2021 by Kit-Wing Yu. All rights reserved. No part of this publication may be
reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic,
mechanical, photocopying, recording, or otherwise, without the prior written permission of the
author.
ISBN: 978-988-74156-4-0 (eBook)
ISBN: 978-988-74156-5-7 (Paperback)
ii
About the author
Dr. Kit-Wing Yu received his B.Sc. (1st Hons), M.Phil. and Ph.D. degrees in Math. at the
HKUST, PGDE (Mathematics) at the CUHK. After his graduation, he joined United Christian
College (UCC) to serve as a mathematics teacher between 2000 and 2020. From 2002 to 2020,
he also took the responsibility of the mathematics panel at UCC. Starting from Sept. 2020,
Dr. Yu was promoted to be the Vice Principal (Academic) at Kiangsu-Chekiang College (Kwai
Chung). Furthermore, he was appointed as a part-time tutor (2002 – 2005) and then a part-time
course coordinator (2006 – 2010) of the Department of Mathematics at the OUHK.
Besides teaching, Dr. Yu has been appointed to be a marker of the HKAL Pure Mathematics
and HKDSE Mathematics (Core Part) for over thirteen years. Between 2012 and 2014, Dr. Yu
was invited to be a Judge Member by the World Olympic Mathematics Competition (China).
In the research aspect, he has published research papers in international mathematical journals,
including some well-known journals such as J. Reine Angew. Math., Proc. Roy. Soc. Edinburgh
Sect. A and Kodai Math. J.. His research interests are inequalities, special functions and
Nevanlinna’s value distribution theory. In the area of academic publication, he is the author of
the following seven books:
• A Complete Solution Guide to Complex Analysis
• A Complete Solution Guide to Real and Complex Analysis I
• A Complete Solution Guide to Principles of Mathematical Analysis
• Problems and Solutions for Undergraduate Real Analysis
• Problems and Solutions for Undergraduate Real Analysis I
• Problems and Solutions for Undergraduate Real Analysis II
• Mock Tests for the ACT Mathematics
iii
iv
Preface
This is the continuum of my book A Complete Solution Guide to Real and Complex Analysis I.
It covers the “Complex Analysis” part of Rudin’s graduate book. In fact, we study all exercises
of Chapters 10 to 20.
Same as A Complete Solution Guide to Real and Complex Analysis I, the primary aim of this
book is to help every mathematics student and instructor to understand the ideas and applications of the theorems in Rudin’s book. To accomplish this goal, I have adopted the way I wrote
the solution guides of Baby Rudin and the first part of Papa Rudin. In other words, I intend
writing the solutions as comprehensive as I can so that you can understand every detailed part
of a proof easily. Apart from this, I also keep reminding you what theorems or results I have
applied by quoting them repeatedly in the proofs. By doing this, I believe that you will become
fully aware of the meaning and applications of each theorem.
Before you read this book, I have two gentle reminders for you. Firstly, as a mathematics
instructor at a college, I understand that the growth of a mathematics student depends largely
on how hard he/she does exercises. When your instructor asks you to do some exercises from
Rudin, you are not suggested to read my solutions unless you have tried your best to prove them
seriously yourselves. Secondly, when I prepared this book, I found that some exercises require
knowledge that Rudin did not cover in his book. To fill this gap, I refer to some other analysis
or topology books such as [2], [9], [15], [18], [23], [42] and [65]. Other useful references are [3],
[27], [28], [37], [69], [79], [80], [82] and [83]. Of course, we will use the exercises in Baby Rudin
and the first part of Papa Rudin freely and if you want to read proofs of them, you are strongly
advised to read my books [77] and [78].
As you will expect, this book always keeps the main features of my previous books [77] and
[78]. In fact, its features are as follows:
• It covers all the 221 exercises from Chapters 10 to 20 with detailed and complete solutions.
As a matter of fact, my solutions show every detail, every step and every theorem that I
applied.
• There are 29 illustrations for explaining the mathematical concepts or ideas used behind
the questions or theorems.
• Sections in each chapter are added so as to increase the readability of the exercises.
• Different colors are used frequently in order to highlight or explain problems, lemmas,
remarks, main points/formulas involved, or show the steps of manipulation in some complicated proofs. (ebook only)
• Necessary lemmas with proofs are provided because some questions require additional
mathematical concepts which are not covered by Rudin.
v
vi
• Many useful or relevant references are provided to some questions for your future research.
Since the solutions are written solely by me, you may find typos or mistakes. If you really
find such a mistake, please send your valuable comments or opinions to
kitwing@hotmail.com.
Then I will post the updated errata on my website
https://sites.google.com/view/yukitwing/
irregularly.
Kit Wing Yu
April 2021
List of Figures
10.1 The closed contour ΓA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
10.2 The contour ΓA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
10.3 The contours ΓA , Γ1 and Γ2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
10.4 The annulus A(r1 , r2 ) and the circles γ1 , γ2 . . . . . . . . . . . . . . . . . . . . . .
23
10.5 A non null-homotopic closed path Γ = γ1 − γ3 − γ2 + γ4 in Ω. . . . . . . . . . . .
32
11.1 The I divides ∂∆ into several triangles. . . . . . . . . . . . . . . . . . . . . . . .
48
12.1 The boundary ∂∆. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
69
12.2 The sectors ∆1 , ∆2 and the ray Lα . . . . . . . . . . . . . . . . . . . . . . . . . .
76
13.1 The simply connected set Ω. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
84
13.2 The compact sets Dn and En . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
85
13.3 The compact sets An , Bn and Cn . . . . . . . . . . . . . . . . . . . . . . . . . . .
86
13.4 The disc ∆n , the arc Ln and its neighborhood Ωn . . . . . . . . . . . . . . . . . .
87
14.1 The region Ω bounded by C1 and C2 .
. . . . . . . . . . . . . . . . . . . . . . . . 119
14.2 The constructions of Ωn−1 , D(0; rn ) and αn . . . . . . . . . . . . . . . . . . . . . . 122
14.3 The construction of the symmetric point z ∗ of z. . . . . . . . . . . . . . . . . . . 134
14.4 The conformal mapping ψ(z) = φ ϕ(z) . . . . . . . . . . . . . . . . . . . . . . . 143
π
π
14.5 The conformal mapping f : U → A(e− 2 , e 2 ). . . . . . . . . . . . . . . . . . . . . 144
14.6 The locus of f (z) for t ∈ (0, π). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
14.7 The locus of f (z) for t ∈ (π, 2π). . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
14.8 The image of f (E) when r = 0.25. . . . . . . . . . . . . . . . . . . . . . . . . . . 146
15.1 The distribution of the zeros zk,n of exp(exp(z)). . . . . . . . . . . . . . . . . . . 151
15.2 The paths γ(z + h) and −γ(z). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
16.1 The paths βζ (I) and γζ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
16.2 The paths γz (I), γω (I) and γω,ζ (I). . . . . . . . . . . . . . . . . . . . . . . . . . . 182
16.3 The fundamental domain R of G. . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
16.4 The regions Ωα and Ωβ if α < β. . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
16.5 The regions of convergence of the two series. . . . . . . . . . . . . . . . . . . . . . 199
19.1 The closed contour Γr . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255
vii
viii
List of Figures
20.1 The compact set X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271
Contents
Preface
v
List of Figures
viii
10 Elementary Properties of Holomorphic Functions
10.1 Basic Properties of Holomorphic Functions
1
. . . . . . . . . . . . . . . . . . . . .
1
10.2 Evaluation of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
10.3 Composition of Holomorphic Functions and Morera’s Theorem . . . . . . . . . .
13
10.4 Problems related to Zeros of Holomorphic Functions . . . . . . . . . . . . . . . .
17
10.5 Laurent Series and its Applications . . . . . . . . . . . . . . . . . . . . . . . . . .
22
10.6 Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
11 Harmonic Functions
33
11.1 Basic Properties of Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . .
33
11.2 Harnack’s Inequalities and Positive Harmonic Functions . . . . . . . . . . . . . .
49
11.3 The Weak∗ Convergence and Radial Limits of Holomorphic Functions . . . . . .
57
11.4 Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
60
12 The Maximum Modulus Principle
69
12.1 Applications of the Maximum Modulus Principle . . . . . . . . . . . . . . . . . .
69
12.2 Asymptotic Values of Entire Functions . . . . . . . . . . . . . . . . . . . . . . . .
78
12.3 Further Applications of the Maximum Modulus Principle . . . . . . . . . . . . .
79
13 Approximations by Rational Functions
13.1 Meromorphic Functions on
S2
83
and Applications of Runge’s Theorem . . . . . . .
83
13.2 Holomorphic Functions in the Unit Disc without Radial Limits . . . . . . . . . .
87
13.3 Simply Connectedness and Miscellaneous Problems . . . . . . . . . . . . . . . . .
93
14 Conformal Mapping
99
14.1 Basic Properties of Conformal Mappings . . . . . . . . . . . . . . . . . . . . . . .
99
14.2 Problems on Normal Families and the Class S . . . . . . . . . . . . . . . . . . . 112
14.3 Proofs of Conformal Equivalence between Annuli . . . . . . . . . . . . . . . . . . 118
14.4 Constructive Proof of the Riemann Mapping Theorem . . . . . . . . . . . . . . . 122
15 Zeros of Holomorphic Functions
149
ix
x
Contents
15.1 Infinite Products and the Order of Growth of an Entire Function . . . . . . . . . 149
15.2 Some Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158
15.3 Problems on Blaschke Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
15.4 Miscellaneous Problems and the Müntz-Szasz Theorem . . . . . . . . . . . . . . . 172
16 Analytic Continuation
179
16.1 Singular Points and Continuation along Curves . . . . . . . . . . . . . . . . . . . 179
16.2 Problems on the Modular Group and Removable Sets . . . . . . . . . . . . . . . 183
16.3 Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
17 H p -Spaces
205
17.1 Problems on Subharmonicity and Harmonic Majoriants . . . . . . . . . . . . . . 205
17.2 Basic Properties of H p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
17.3 Factorization of f ∈ H p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217
17.4 A Projection of Lp onto H p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224
17.5 Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230
18 Elementary Theory of Banach Algebras
235
18.1 Examples of Banach Spaces and Spectrums . . . . . . . . . . . . . . . . . . . . . 235
18.2 Properties of Ideals and Homomorphisms . . . . . . . . . . . . . . . . . . . . . . 239
18.3 The Commutative Banach algebra H ∞ . . . . . . . . . . . . . . . . . . . . . . . . 249
19 Holomorphic Fourier Transforms
251
19.1 Problems on Entire Functions of Exponential Type . . . . . . . . . . . . . . . . . 251
19.2 Quasi-analytic Classes and Borel’s Theorem . . . . . . . . . . . . . . . . . . . . . 260
20 Uniform Approximation by Polynomials
271
Index
275
Bibliography
277
CHAPTER
10
Elementary Properties of Holomorphic Functions
10.1
Basic Properties of Holomorphic Functions
Problem 10.1
Rudin Chapter 10 Exercise 1.
Proof. In fact, this is [61, Exercise 21, p. 101]. For a solution of it, please refer to [77, p. 75].
This completes the proof of the problem.
Problem 10.2
Rudin Chapter 10 Exercise 2.
Proof. Let a ∈ C and
f (z) =
∞
X
n=0
cn (z − a)n .
(10.1)
By [62, Eqn. (8), p. 199], we have n!cn = f (n) (a) for all n = 0, 1, 2, . . . and so the hypothesis
implies that f (n) (a) = 0 for some n ∈ N ∪ {0}.
For every n ∈ N ∪ {0}, let Zn = {z ∈ C | f (n) (z) = 0} ⊆ C. Now the previous paragraph
implies that
∞
[
Zn .
(10.2)
C=
n=0
f (n)
Since f is entire, every
is also entire. If f (n) 6≡ 0 for every n ∈ N ∪ {0}, then Zn 6= C for
every n ∈ N ∪ {0} by Theorem 10.18. Furthermore, each Zn is at most countable and it deduces
from the set relation (10.2) that C is countable, a contradiction. Thus there exists an N ∈ N
such that f (n) ≡ 0 for all n > N and so the representation (10.1) implies that f is a polynomial
of degree at most N , completing the proof of the problem.
Problem 10.3
Rudin Chapter 10 Exercise 3.
1
2
Chapter 10. Elementary Properties of Holomorphic Functions
Proof. We claim that f (z) = cg(z) for some |c| ≤ 1. If g 6≡ 0 in C, then we follow from Theorem
10.18 that Z(g) has no limit point in C. Consider h : C \ Z(g) → C given by
h(z) =
f (z)
.
g(z)
For each a ∈ Z(g), we have h ∈ H(D ′ (a; r)) and |h(z)| ≤ 1 in D ′ (a, r) for some r > 0. Now we
follow from Theorem 10.20 that h has a removable singularity at a and then h can be defined at
a so that it is holomorphic in D(a; r). Since it is true for every a ∈ Z(g), h is in fact entire and
|h(z)| ≤ 1 in C. Therefore, Theorem 10.23 (Liouville’s Theorem) asserts that h(z) = c for some
constant c such that |c| ≤ 1. Consequently, we obtain f (z) = cg(z) as required. This proves the
claim and we end the proof of the problem.
Problem 10.4
Rudin Chapter 10 Exercise 4.
Proof. For every n = 0, 1, 2, . . ., we apply the hint given in Problem 10.2, Theorem 10.26 (The
Cauchy’s Estimates) and the hypothesis to get
|cn | =
f (n) (0)
A + Rk
,
≤
n!
Rn
(10.3)
where R > 0. If n > k, then we take R → ∞ to both sides of the inequality (10.3) to conclude
cn = 0 for all n > k. In other words, f is a polynomial of degree at most k. This completes the
proof of the problem.
Problem 10.5
Rudin Chapter 10 Exercise 5.
Proof. Since {fn } is uniformly bounded in an open subset Ω ⊆ C, there exists a M > 0 such
that |fn (z)| ≤ M for all z ∈ Ω and all n ∈ N. Let a ∈ Ω and f (z) = lim fn (z). Since Ω is open
n→∞
in C, one can find a δ > 0 such that D(a; 4δ) ⊆ Ω. Define γ : [0, 2π] → Ω by
γ(t) = a + 2δeit
(10.4)
which is clearly a circle centered at a with radius 2δ. If z ∈ D(a; δ), then we have |γ(t) − z| > δ
for all t ∈ [0, 2π]. Therefore, we get from this fact and the representation (10.4) that
γ ′ (t)
2δ
=2
<
γ(t) − z
δ
(10.5)
for all t ∈ [0, 2π].
Next, for every n ∈ N, we define Fn : [0, 2π] → C and F : [0, 2π] → C by
Fn (t) = fn (γ(t))
and
F (t) = f (γ(t))
respectively. Clearly, the pointwise convergence of {fn } on Ω implies the pointwise convergence
of {Fn } on [0, 2π]. Besides, since all fn and γ are continuous functions, every Fn is complex
10.1. Basic Properties of Holomorphic Functions
3
measurable on the measurable space [0, 2π]. Since 2M ∈ L1 (m) and |Fn (t)| ≤ M for all n ∈ N,
we deduce from Theorem 1.34 (The Lebesgue’s Dominated Convergence Theorem) that
Z 2π
Z 2π
F (t) dt.
(10.6)
Fn (t) dt =
lim
n→∞ 0
Now the bound (10.5) ensures that
certainly have
0
γ ′ (t)
γ(t)−z
Fn (t) ·
is continuous on [0, 2π] so that for all z ∈ D(a; δ), we
γ ′ (t)
γ ′ (t)
→ F (t) ·
γ(t) − z
γ(t) − z
′
γ (t)
pointwisely on [0, 2π] and each Fn (t)· γ(t)−z
is complex measurable on [0, 2π]. Since 2M ∈ L1 (m)
′
γ (t)
| ≤ 2M for all n ∈ N, further application of Theorem 1.34 (The Lebesgue’s
and |Fn (t) · γ(t)−z
Dominated Convergence Theorem) also gives
Z 2π
Z 2π
γ ′ (t)
γ ′ (t)
F (t) ·
dt =
dt.
Fn (t) ·
lim
(10.7)
n→∞ 0
γ(t) − z
γ(t) − z
0
Recall that z ∈
/ γ ∗ , so Ind γ (z) = 1. Then we obtain from Theorem 10.15 (The Cauchy’s
Formula in a Convex Set) and the limit (10.7) thata
f (z) = lim fn (z)
n→∞
Z
1
fn (ζ)
= lim
dζ
n→∞ 2πi γ ζ − z
Z 2π
γ ′ (t)
1
dt
fn (γ(t)) ·
= lim
n→∞ 2πi 0
γ(t) − z
Z 2π
γ ′ (t)
1
Fn (t) ·
dt
= lim
n→∞ 2πi 0
γ(t) − z
Z 2π
γ ′ (t)
1
F (t) ·
dt.
=
2πi 0
γ(t) − z
(10.8)
Given that ǫ > 0. The result (10.6) guarantees that there is an N ∈ N such that n ≥ N implies
Z 2π
|Fn (t) − F (t)| dt < ǫπ.
0
In this case, for every z ∈ D(a, δ), we obtain from the expression (10.8) that
Z 2π
Z 2π
1
1
γ ′ (t)
γ ′ (t)
|fn (z) − f (z)| =
dt −
dt
Fn (t) ·
F (t) ·
2πi 0
γ(t) − z
2πi 0
γ(t) − z
Z 2π
γ ′ (t)
1
|Fn (t) − F (t)| ·
dt
≤
2π 0
γ(t) − z
<ǫ
for all n ≥ N . In other words, {fn } converges uniformly to f on D(a; δ).
Finally, every compact K ⊆ Ω can be covered by D(a1 , δ), D(a2 , δ), . . . , D(am , δ) for some
m ∈ N. Thus the previous analysis shows immediately that for all z ∈ D(ak , δ), the inequality
|fn (z) − f (z)| < ǫ
a
(10.9)
We remark that we cannot apply Theorem 10.15 (The Cauchy’s Formula in a Convex Set) directly to f to
obtain the representation (10.8) because it is not clear whether f ∈ H(Ω) or not.
4
Chapter 10. Elementary Properties of Holomorphic Functions
holds if n ≥ Nk , where k = 1, 2, . . . , m. Set N = max(N1 , N2 , . . . , Nm ). Therefore, if n ≥ N ,
then the inequality (10.9) remains valid on D(a1 , δ) ∪ D(a2 , δ) ∪ · · · ∪ D(am , δ), so particularly
on K. Hence we have shown that the convergence is in fact uniform on every compact subset
of Ω. This completes the proof of the problem.
Problem 10.6
Rudin Chapter 10 Exercise 6.
Proof. We divide the proof into several parts.
• The existence of Ω. Since f (z) = 1z is continuous on D(1; 1), we follow from Theorem
10.14 (The Cauchy’s Theorem in a Convex Set) that one can find an F0 ∈ H(D(1; 1)) such
that F0′ (z) = z1 in D(1; 1). If F (z) = F0 (z) − F0 (1), then we also have F ′ (z) = F0′ (z) = z1
in D(1; 1) and F (1) = 0. Next, we define g : D(1; 1) → C by
g(z) =
eF (z)
.
z
(10.10)
Direct computation gives
g ′ (z) =
zeF (z) F ′ (z) − eF (z)
=0
z2
for all z ∈ D(1; 1). Using [9, Exercise 5, p. 42], we know that g is a constant in D(1; 1).
Obviously, we have g(1) = eF (1) = 1, so it is true that g(z) = 1 in D(1; 1). Hence it follows
from the form (10.10) that
eF (z) = z
(10.11)
for all z ∈ D(1; 1).
• The injection of exp. Set Ω = F (D(1; 1)). We claim that Ω is a region such that
exp(Ω) = D(1; 1).
(10.12)
To this end, since F ′ (z) = z1 , Ω is not a point set. Then Ω is a region follows from the Open
Mapping Theorem directly. Finally, the expression (10.11) implies that the set equality
(10.12). Furthermore, the expression (10.11) implies immediately that exp is one-to-one
in Ω.
• The number of such region Ω. For each n ∈ Z, we define Ωn = {z + 2nπi | z ∈ Ω}
which is also a region and satisfies
exp(Ωn ) = exp(Ω + 2nπi) = exp(Ω) = D(1; 1).
Since F is continuous on D(1; 1), Ω must be bounded. Therefore, there exists an N ∈ N
such that ΩN 6= Ω. Hence {ΩkN | k ∈ Z} is a set of distinct regions satisfying the set
equality (10.12).
• The derivative of log z. Define log z, for |z − 1| < 1, such that
elog z = z.
(10.13)
By differentiating the equation (10.13), we get log′ (z)elog z = 1 and so
log′ (z) =
1
elog z
=
1
.
z
10.1. Basic Properties of Holomorphic Functions
5
• The coefficients an and cn . It is easy to see that
∞
∞
X
X
1
1
=
=
(1 − z)n =
(−1)n (z − 1)n ,
z
1 − (1 − z) n=0
n=0
so we derive that an = (−1)n for all n = 0, 1, 2, . . .. To find cn , we first consider the power
series
∞
X
(−1)n−1
(z − 1)n .
h(z) =
n
n=1
1
Since lim sup √
= 1, the radius of convergence of the power series h is 1, so termwise
n
n
n→∞
differentiation can be performed to obtainb
h′ (z) =
∞
X
(−1)n−1 (z − 1)n−1 =
n=1
∞
X
(−1)n (z − 1)n =
n=0
1
.
z
Since log′ (z) = z1 , we have h(z) = log z + C for some constant C. Since h(1) = log 1 = 0,
we get C = 0 and
∞
X
(−1)n−1
log z = h(z) =
(z − 1)n .
n
n=1
Hence we establish

(−1)n−1



, if n ≥ 1;
n
cn =


 0,
if n = 0.
• What other discs can this be done? This can be done in every disc D(a; |a|), where
a ∈ C \ {0}. In fact, we pick b ∈ C such that eb = a. By similar argument as the proof of
the first assertion, there exists an F1 ∈ H(D(a; |a|)) such that F1′ (z) = z1 in D(a; |a|). If
we set F2 (z) = F1 (z) − F1 (a) + b, then we also have
F2′ (z) = F1′ (z) =
1
z
and F2 (a) = b
in D(a; |a|). Furthermore, the function G : D(a; |a|) → C given by
G(z) =
eF2 (z)
z
(10.14)
satisfies G′ (z) = 0 in D(a; |a|). Thus G is a constant in D(a; |a|) and since G(a) = 1, we
conclude that G(z) = 1 in D(a; |a|) and we get from the definition (10.14) that
eF2 (z) = z
in D(a; |a|). According to this construction, all the above assertions can be proven similarly
and we won’t repeat the argument here.
We have completed the analysis of the problem.
Problem 10.7
Rudin Chapter 10 Exercise 7.
b
Read [9, Theorem 2.9, pp. 28, 32] for details.
6
Chapter 10. Elementary Properties of Holomorphic Functions
Proof. The conditions should be that Γ is a cycle in the open set Ω and

/ Ω;
 0, if α ∈
Ind Γ (α) =

1, if α ∈ Ω \ Γ∗ .
In fact, with the above hypotheses, we get from Theorem 10.35 (Cauchy’s Theorem) that
Z
1
f (ζ)
f (z) =
dζ
2πi Γ ζ − z
for all z ∈ Ω \ Γ∗ . This proves the formula for the case n = 0. Assume that
Z
k!
f (ζ)
(k)
f (z) =
dζ
2πi Γ (ζ − z)k+1
(10.15)
on Ω \ Γ∗ for some k ∈ N ∪ {0}. Next, we define g : Ω \ {z} → C by
g(ζ) =
f (ζ)
,
(ζ − z)k+1
where ζ ∈ Ω \ Γ∗ . Then it is clear that
g′ (ζ) =
(k + 1)f (ζ)
f ′ (ζ)
−
.
k+1
(ζ − z)
(ζ − z)k+2
(10.16)
Since f ∈ H(Ω), we have f ′ ∈ H(Ω) and thus the formula (10.16) ensures that g ∈ H(Ω \ {z}).
Notice that Ω \ {z} is also an open set in C, so a combined application of Theorem 10.35
(Cauchy’s Theorem) and the formula (10.16) implies that
Z
g(ζ) dζ = 0
Γ
Z
Z
f (ζ)
f ′ (ζ)
dζ
=
(k
+
1)
dζ.
(10.17)
k+1
(ζ
−
z)
(ζ
−
z)k+2
Γ
Γ
Applying the induction step (10.15) to the left-hand side of the formula (10.17) (with f replaced
by f ′ ), we see immediately that
Z
2πi (k+1)
f (ζ)
f
(z) = (k + 1)
dζ
k+2
k!
Γ (ζ − z)
which gives the desired result for the case k + 1. By induction, we have completed the proof of
the problem.
10.2
Evaluation of Integrals
Problem 10.8
Rudin Chapter 10 Exercise 8.
Proof. We note that
Z
∞
−∞
R(x) dx = lim
Z
A
A→∞ −A
P (x)
dx.
Q(x)
(10.18)
Let ΓA be the closed contour consisting of the real segment [−A, A] and the upper semi-circle
CA = {z ∈ C | |z| = A and Im z > 0} positively oriented, see Figure 10.1
10.2. Evaluation of Integrals
7
Figure 10.1: The closed contour ΓA .
Clearly, if A is large enough, then ΓA will contain all zeros of Q(z) lying in the upper half
plane. Hence it follows from Theorem 10.42 (The Residue Theorem) that
Z
X
R(z) dz = 2πi
Res (R; zk ),
(10.19)
ΓA
k
where {zk } is the set of all zeros of Q in the upper half plane. In fact, we can write the expression
(10.19) in the form
Z A
Z
X
R(x) dx = 2πi
R(z) dz +
Res (R; zk ).
−A
CA
k
Since CA is a semi-circle of radius A, its length is πA. Using this fact and deg Q − deg P ≥ 2,
we obtain from the estimate [62, Eqn. (5), Definition 10.8, p. 202] that
Z
πM
M
(10.20)
R(z) dz ≤ 2 · πA =
A
A
CA
for some positive constant M . Taking A → ∞ in the inequality (10.20), we get
Z
R(z) dz = 0.
lim
A→∞ CA
Finally, we combine the results (10.18), (10.19) and (10.21) to conclude that
Z ∞
X
R(x) dx = 2πi
Res (R; zk ).
−∞
(10.21)
(10.22)
k
For the analogous statement for the lower half plane, the formula (10.22) will be replaced by
Z ∞
X
R(x) dx = −2πi
Res (R; zk ),
−∞
k
where the set {zk } now consists of all zeros of Q in the lower half plane.c To compute the
integral, we note from the formula (10.22) and some basic facts of calculating residuesd that
Z ∞
z2
h
z2
πi 3πi i
x2
+
Res
dx
=
2πi
Res
;
exp
;
exp
4
1 + z4
4
1 + z4
4
−∞ 1 + x
c
Here we have the negative sign in the formula because the corresponding semi-circle in the lower half plane
is negatively oriented.
d
See, for examples, [9, pp. 129, 130] or [65, pp. 75, 76].
8
Chapter 10. Elementary Properties of Holomorphic Functions
πi 3πi i
πi h
exp −
+ exp −
2
4
4
πi −2i
× √
=
2
2
π
=√ .
2
=
This completes the analysis of the problem.
Problem 10.9
Rudin Chapter 10 Exercise 9.
Proof. Let ΓA be the closed contour consisting of the real segment [−A, A] and the upper semicircle CA = {z ∈ C | |z| = A and Im z > 0}, see Figure 10.1. Furthermore, we let P and Q be
polynomials such that deg Q − deg P ≥ 1, Q(x) 6= 0 (except perhaps at zeros of cos x or sin x)
P (x)
. By the discussion of Type II integrals in [9, pp. 144 – 146], we see that
and R(x) = Q(x)
Z
∞
Z
ix
R(x)e dx = lim
A→∞ ΓA
−∞
R(z)eiz dz = 2πi
X
Res (R(z)eiz ; zk ),
(10.23)
k
where the points zk are the poles of R(z) in the upper half plane.
Suppose that t ≥ 0. By the substitution y = tx, we have
Z ∞
Z ∞
eitx
t
dx =
eiy dy.
2
2 + y2
1
+
x
t
−∞
−∞
Set R(z) =
t
t2 +z 2 .
Then it follows from the representation (10.23) that
Z
∞
−∞
t
eitx
π
iz
dx
=
2πiRes
e
;
ti
= t.
2
2
2
1+x
t +z
e
Next, if t = −u for some u > 0, then we have
Z ∞ −iux
Z ∞
e
eitx
dx =
dx
2
2
−∞ 1 + x
−∞ 1 + x
Z ∞
u
eiy dy
=
2
2
−∞ u + y
u
iz
= 2πiRes 2
e
;
ui
u + z2
π
= u.
e
(10.24)
(10.25)
Combining the two expressions (10.24) and (10.25), we conclude that
Z
∞
−∞
π
eitx
dx = |t| .
2
1+x
e
Using the theory of Fourier transforms, we notice that if f (t) =
Definition 9.1 that
Z ∞
1
fb(x) = √
f (t)e−ixt dt
2π −∞
(10.26)
pπ
−|t| ,
2e
then we know from
10.2. Evaluation of Integrals
1
=
2
=
Z
∞
9
e−|t| e−ixt dt
−∞
0
Z
1h
e(1−ix)t dt +
Z
∞
e−(1+ix)t dt
i
2 −∞
0
1 n exp[(1 − ix)t] 0
exp[−(1 + ix)t]
= ·
+
2
1 − ix
−1 + ix
−∞
1 1 1
+
=
2 1 − ix 1 + ix
1
=
.
1 + x2
∞o
0
Now it is clear that f, fb ∈ L1 (R), so we follow from Theorem 9.11 (The Inversion Theorem) that
g(t) = f (t), where
Z ∞
Z ∞
1
eixt
1
ixt
b
√
√
g(t) =
dx.
f (x)e dx =
2π −∞
2π −∞ 1 + x2
Consequently, we obtain
Z
∞
−∞
√
eixt
dx = 2π ·
2
1+x
r
π
π −|t|
e
= |t|
2
e
which is consistent with the result (10.26). Hence we have completed the proof of the problem.
Problem 10.10
Rudin Chapter 10 Exercise 10.
Proof. Let f (z) = (ez − e−z )z −4 . Then f is holomorphic in C \ {0}. By the power series
expansion of ez (see [62, Eqn. (1), p. 1]), we have
1
ez − e−z
1
z
=
2
+
+
+
·
·
·
z4
z 3 3!z 5!
so that f has a pole of order 3 at 0. By Theorem 10.21(b), the difference
1
1 f (z) − 2 3 +
z
3!z
has a removable singularity at 0. Thus there exists an entire function g such thate
1
1 f (z) − 2 3 +
= g(z)
z
3!z
which gives
Z
Z Z
1
1
1
1 1
dz
+
f (z) dz =
+
g(z) dz.
2πi γ
πi γ z 3 3!z
2πi γ
(10.27)
Applying Theorems 10.10 and 10.12 to the right-hand side of the expression (10.27), we establish
that
Z z
Z
Z
1
e − e−z
dz
1
1
1
dz
=
f
(z)
dz
=
= .
2πi γ
z4
2πi γ
6πi γ z
3
This ends the proof of the problem.
e
Obviously, we have
g(z) = 2
∞
X
n=2
z 2n−3
.
(2n + 1)!
10
Chapter 10. Elementary Properties of Holomorphic Functions
Problem 10.11
Rudin Chapter 10 Exercise 11.
Proof. Since |α| =
6 1, we have either |α| < 1 or |α−1 | < 1. If z = eiθ , then cos θ = 21 (z + z1 ).
Using [9, Eqn. (5), p. 150], we see that
Z 2π
Z
dθ
dz
1
·
=
1
1
2
2
1 − 2α cos θ + α
iz
0
|z|=1 1 − 2α · 2 (z + z ) + α
Z
dz
=i
2
2
|z|=1 αz − (1 + α )z + α

1

−2πRes
;
α
,
if |α| < 1;



αz 2 − (1 + α2 )z + α
=


1

−1
 −2πRes
, if |α−1 | < 1
;
α
αz 2 − (1 + α2 )z + α

2π



 − α2 − 1 , if |α| < 1;
=


2π


,
if |α−1 | < 1.
2
α −1
Hence we complete the analysis of the problem.
Problem 10.12
Rudin Chapter 10 Exercise 12.
Proof. Let ΓA be the path obtained by going from −A to −1 along the real axis, from −1 to
1 along the lower half of the unit circle C and from 1 to A along the real axis, see Figure 10.2
below.
Figure 10.2: The contour ΓA .
We note that
Z −1
Z 1
Z A
Z A
sin2 x itx
sin2 x itx
sin2 x itx
sin2 x itx
e
dx
=
e
dx
+
e
dx
+
e dx.
2
2
x2
x2
−A
−1 x
1
−A x
(10.28)
Since z −2 · sin2 z · eitz is entire for every t ∈ R, it follows from Theorem 10.14 (The Cauchy’s
Theorem in a Convex Set) that
Z −1
Z
sin2 x itx
sin2 z itz
e
dz
+
e dx = 0
2
x2
1
C z
10.2. Evaluation of Integrals
or equivalently,
Z
1
−1
11
sin2 x itx
e dx =
x2
Z
C
sin2 z itz
e dz.
z2
(10.29)
Combing the integral relations (10.28) and (10.29), we see immediately that
Z
A
−A
Next, we write 2i sin z = eiz − e−iz so that
Z
ΓA
sin2 z itz
e dz =
z2
Z
ΓA
Now we define
ϕA (s) =
Z
Z
sin2 z itz
e dz.
z2
(10.30)
e2iz − 2 + e−2iz itz
e dz.
−4z 2
(10.31)
sin2 x itx
e dx =
x2
ΓA
ΓA
eisz
dz,
z2
(10.32)
so the expression (10.31) becomes
Z
ΓA
sin2 z itz
1
1
e dz = − [ϕA (t + 2) + ϕA (t − 2)] + ϕA (t).
2
z
4
2
(10.33)
If we combine (10.30) and (10.33), then we have
Z
A
−A
sin2 x itx
1
1
e dx = ϕA (t) − [ϕA (t + 2) + ϕA (t − 2)].
2
x
2
4
(10.34)
Complete ΓA to a closed path in two different ways: Firstly, we consider the semi-circle Γ1
from A to −Ai and then to −A; secondly, we consider the semi-circle Γ2 from A to Ai and then
to −A, see Figure 10.3.
Figure 10.3: The contours ΓA , Γ1 and Γ2 .
12
Chapter 10. Elementary Properties of Holomorphic Functions
It is easily checked that the function eisz · z −2 has a pole of order 2 at 0, and so the residue
is is. Thus in the first way, we have
Z
Z
eisz
eisz
dz
+
dz = 0
2
2
ΓA z
Γ1 z
so that if z = Aeiθ , where θ ∈ [−π, 0], then we deduce from the definition (10.32) that
Z 0
Z
exp(isAeiθ )
eisz
dz
=
i
dθ
(10.35)
ϕA (s) =
2
Aeiθ
−π
ΓA z
and in the second way, Theorem 10.42 (The Residue Theorem) yields
Z
Z
eisz eisz
eisz
dz
+
dz
=
2πiRes
; 0 = −2πs
2
2
z2
ΓA z
Γ2 z
which implies that
ϕA (s) =
Since
Z
ΓA
eisz
dz = −2πs − i
z2
Z
π
0
exp(isAeiθ )
dθ.
Aeiθ
(10.36)
exp(isAeiθ )
exp(−As sin θ)
→0
≤
Aeiθ
A
as A → ∞ if s and sin θ have the same sign. Thus it follows from Theorem 1.34 (The Lebesgue’s
Dominated Convergence Theorem) that the integral (10.35) tends to 0 if s < 0 and the one in
(10.36) tends to 0 if s > 0. In other words, we obtain

 −2πs, if s > 0;
lim ϕA (s) =
(10.37)

A→∞
0,
if s < 0.
Finally, we apply the result (10.37) to the expression (10.34) to get

if |t| > 2;

Z ∞
 0,
2
sin x itx
(10.38)
e dx =
2

−∞ x
 π − πt , if −2 < t < 0 or 0 < t < 2.
2
When t = 0, we know from the integral (10.35) that ϕA (0) = − A2 , so we establish from the
expression (10.34) that

Z ∞
 π, if t = 0;
2
sin x itx
e dx =
(10.39)
2

−∞ x
0, if t = ±2.
By combining the results (10.38) and (10.39), we

0,





Z ∞

sin2 x itx
πt
e dx =
π− ,
2

x
2
−∞





π,
achieve
if |t| ≥ 2;
if −2 < t < 0 or 0 < t < 2;
if t = 0.
We have completed the proof of the problem.
Problem 10.13
Rudin Chapter 10 Exercise 13.
Proof. This has been solved on [76, p. 143] which completes the proof of the problem.
10.3. Composition of Holomorphic Functions and Morera’s Theorem
10.3
13
Composition of Holomorphic Functions and Morera’s Theorem
Problem 10.14
Rudin Chapter 10 Exercise 14.
Proof. Both cases can be negative. Let Ω1 = C \ {0} and Ω2 = C. Define f (z) = z in Ω1 and

 z, if z 6= 0;
g(z) =

1, otherwise.
Then f (Ω1 ) ⊆ Ω2 and h(z) = g(f (z)) = g(z) = z in Ω1 . Hence f and h are holomorphic in Ω1 ,
but g is discontinuous in Ω2 .
Next, we consider Ω1 = Ω2 = C. Define

 −1, if z 6= 0;
f (z) =

1,
otherwise
and g(z) = z 2 . Obviously, we have f (Ω1 ) = {±1} ⊆ Ω2 and h(z) = g(f (z)) = 1 for all z ∈ Ω1 .
It is clear that both g and h are holomorphic in Ω2 and Ω1 respectively, but f is not continuous
in Ω1 . This ends the analysis of the proof.
Remark 10.1
A problem similar to Problem 10.14 but for (uniform) continuity has been discussed in [61,
Exercise 26, p. 102].
Problem 10.15
Rudin Chapter 10 Exercise 15.
Proof. According to Theorem 10.18, we have f (z) = (z − ω0 )m h(z), where h ∈ H(ϕ(Ω)) and
h(ω0 ) 6= 0. Then we have
g(z) = f (ϕ(z)) = [ϕ(z) − ϕ(z0 )]m h(ϕ(z)).
(10.40)
Suppose that n ≥ 1 is the order of the zero of ϕ(z) − ϕ(z0 ). Now Theorem 10.18 implies that
ϕ(z) − ϕ(z0 ) = (z − z0 )n φ(z),
where φ ∈ H(Ω) and φ(z0 ) 6= 0. Assume that n ≥ 2. We follow from
ϕ′ (z) = (z − z0 )n−1 [(z − z0 )φ′ (z) + φ(z)]
(10.41)
that ϕ′ (z0 ) = 0, a contradiction. Consequently, n = 1 and we can write the expression (10.40)
as
g(z) = (z − z0 )m · φm (z)h(ϕ(z)).
(10.42)
Finally, since φm (z0 )h(ϕ(z0 )) = φm (z0 )h(ω0 ) 6= 0, the representation (10.42) ensures that g has
a zero of order m at z0 .
14
Chapter 10. Elementary Properties of Holomorphic Functions
If ϕ′ has a zero of order k at z0 , then the expression (10.41) will imply that n = k + 1 so the
representation (10.40) becomes
g(z) = (z − z0 )m(k+1) · φm (z)h(ϕ(z)).
In conclusion, g has a zero of order m(k + 1) at z0 . This completes the proof of the problem. Problem 10.16
Rudin Chapter 10 Exercise 16.
Proof. Since ϕ is bounded on Ω × X, there exists a M > 0 such that |ϕ(z, t)| ≤ M for all
(z, t) ∈ Ω × X. Let z0 ∈ Ω. Since Ω is open in C, there exists a ǫ > 0 such that D(z0 ; 3ǫ) ⊆ Ω.
Then we have D(z0 ; 2ǫ) ⊆ Ω.
We claim that for every pair z, ω ∈ D(z0 ; ǫ), z 6= ω and p ∈ X, we have
ϕ(z, p) − ϕ(ω, p)
2M
.
≤
z−ω
ǫ
(10.43)
To this end, we consider the closed curve γ(t) = z0 + 2ǫeit , where t ∈ [0, 2π]. Obviously, since
ϕ(z, t) ∈ H(D(z0 ; 3ǫ)) for each t ∈ X, we establish from Theorem 10.15 (The Cauchy’s Formula
in a Convex Set) that if z ∈ D(z0 ; 2ǫ) ⊆ D(z0 ; 3ǫ) and p ∈ X, then z ∈
/ γ ∗ and
Z
Z 2π
1
ϕ(ζ, p)
ϕ(γ(t), p) ′
1
ϕ(z; p) =
dζ =
· γ (t) dt.
(10.44)
2πi γ ζ − z
2πi 0
γ(t) − z
Note that z ∈ D(z0 ; ǫ) implies |γ(t) − z| > ǫ. Thus we follow from the formula (10.44) that
Z 2π
1
1
1
ϕ(z, p) − ϕ(ω, p)
·
−
≤
· |ϕ(γ(t), p)| · |γ ′ (t)| dt
z−ω
2π|z − ω| 0
γ(t) − z γ(t) − ω
Z 2π
z−ω
1
· 2M ǫ dt
·
≤
2π|z − ω| 0
γ(t) − z γ(t) − ω
Z 2π
Mǫ
dt
≤
·
π
|γ(t) − z| · |γ(t) − ω|
0
Z 2π
Mǫ
dt
≤ 2
ǫ π 0
2M
=
ǫ
which is exactly the inequality (10.43).
Recall that µ is a complex measure, so Theorem 6.12 tells us that there is a measurable
function h such that |h(x)| = 1 in X and dµ = h d|µ|. This fact ensures that
Z
2M
2M |µ|(X)
d|µ| =
< ∞,
ǫ X
ǫ
i.e.,
2M
ǫ
∈ L1 (|µ|). Suppose that {zn } ⊆ D(z0 ; ǫ) \ {z0 } satisfies zn → z0 . Define
gn (x) =
ϕ(zn , x) − ϕ(z0 , x)
· h(x).
zn − z0
By the hypotheses, we know that each gn is measurable of x and
lim gn (x) = ϕ′ (z0 , x) · h(x).
n→∞
10.3. Composition of Holomorphic Functions and Morera’s Theorem
15
In other words, the sequence {gn } satisfies the conditions of Theorem 1.34 (The Lebesgue’s
Dominated Convergence Theorem), so we conclude that ϕ′ (z0 , x) · h(x) ∈ L1 (|µ|) and furthermore,
Z
ϕ(zn , x) − ϕ(z0 , x)
f (zn ) − f (z0 )
= lim
· h(x) d|µ|
lim
n→∞
n→∞
zn − z0
zn − z0
ZX
gn (x) d|µ|
= lim
n→∞ X
Z
ϕ′ (z0 , x) · h(x) d|µ|.
=
X
Consequently, f ′ (z0 ) exists. Since z0 is arbitrary, we establish that f ∈ H(Ω), completing the
proof of the problem.
Problem 10.17
Rudin Chapter 10 Exercise 17.
Proof. We apply Problem 10.16 to the functions one by one.
• The function f (z). We write
f (z) =
Z
ϕ(z, t) dm,
X
were X = [0, 1] and ϕ(z, t) = (1 + tz)−1 . Notice that if z0 ∈ (−∞, −1], then ϕ(z0 , − z10 ) is
unbounded. Thus it is reasonable to take Ω = C\(−∞, −1]. Now the function ϕ(z, t) satisfies all hypotheses of Problem 10.16 except the boundedness condition because ϕ(z, 1) → ∞
as z → −1 in Ω. Instead, it is really bounded locally. In fact, for every z ∈ Ω, there exists
a δ > 0 such that D(z; δ) ⊂ D(z; 2δ) ⊆ Ω so that ϕ is bounded on D(z; δ) × X. Hence
Problem 10.16 implies that f ∈ H(D(z; δ)). Since z is arbitrary, it yields that f ∈ H(Ω).
• The function g(z). We have
g(z) =
Z
ϕ(z, t) dm,
X
where X = [0, ∞) and ϕ(z, t) = etz (1 + t2 )−1 . Since |etz | = etRe z , if Re (z0 ) > 0, then
etRe (z0 )
= ∞.
t→∞ 1 + t2
lim |ϕ(z0 , t)| = lim
t→∞
In other words, ϕ(z, t) is unbounded in any set containing a point of the right half plane.
Thus we may take Ω to be the left half plane which is an open set in C. We want to apply
Morera’s Theorem and Fubini’s Theorem.f
Let z0 , zn ∈ Ω for all n ∈ N, where zn → z0 as n → ∞. Fix t ∈ [0, ∞), then it is easy to
see that ϕ(z, t) is continuous at z0 . In addition, we know that
|ϕ(z, t)| ≤
f
1
1 + t2
(10.45)
Problem 10.16 cannot be applied directly in this case because m(X) = ∞, i.e., m is not a complex measure
on X.
16
Chapter 10. Elementary Properties of Holomorphic Functions
1
1
for all z ∈ Ω and 1+t
2 ∈ L (m). Then we deduce from Theorem 1.34 (The Lebesgue’s
Dominated Convergence Theorem) that
Z
Z
ϕ(z0 , t) dm = g(z0 ),
ϕ(zn , t) dm =
lim g(zn ) = lim
n→∞
n→∞ X
X
i.e., g is continuous at z0 . Since z0 is arbitrary, g is continuous in Ω.
Next, we suppose that ∆ ⊆ Ω is a closed triangle and ∂∆ is parameterized by a piecewise
continuously differentiable curve γ : [a, b] → Ω. We have
Z
g(z) dz =
∂∆
Z
b
′
g(γ(x))γ (x) dx =
a
Z bZ
a
∞
ϕ(γ(x), t)γ ′ (x) dt dx.
(10.46)
0
Define φ(x, t) = ϕ(γ(x), t) · γ ′ (x) on [a, b] × [0, ∞). Clearly, both [a, b] and [0, ∞) are σfinite measure spaces. Since γ is piecewise continuously differentiable on [a, b], there exists
a M > 0 such that
|γ ′ (x)| ≤ M
on [a, b]. Furthermore, φ(x, t) is piecewise continuous on [a, b] × [0, ∞) so that φ is a
measurable function on [a, b] × [0, ∞). Finally, for each x ∈ [a, b], we know from the
inequality (10.45) that
M
|φ|x = |ϕ(γ(x), t) · γ ′ (x)| ≤
1 + t2
which gives
∗
φ (x) =
Z
∞
0
and
|φ|x dt ≤ M
Z
b
a
Z
∞
0
Mπ
dt
=
2
1+t
2
φ∗ dx < ∞.
Consequently, we may apply Theorem 8.8 (The Fubini Theorem) to change the order of
integration in the integral (10.46) and get
Z
g(z) dz =
∂∆
Z
0
∞hZ b
a
Z
i
ϕ(γ(x), t)γ ′ (x) dx dt =
0
∞Z
ϕ(z, t) dz dt.
(10.47)
∂∆
Since ϕ(z, t) is holomorphic in Ω for every t ∈ [0, ∞), we conclude from Theorem 10.13
(The Cauchy’s Theorem for a Triangle) that
Z
ϕ(z, t) dz = 0,
∂∆
so the integral (10.47) reduces to
Z
g(z) dz = 0.
∂∆
Finally, we apply Theorem 10.17 (Morera’s Theorem) to obtain the desired conclusion that
g ∈ H(Ω).
• The function h(z). We have
h(z) =
Z
X
ϕ(z, t) dm,
10.4. Problems related to Zeros of Holomorphic Functions
17
where X = [−1, 1] and ϕ(z, t) = etz (1 + t2 )−1 . For every z ∈ C, we have z ∈ D(z; 1) ⊂ C.
We know that
|ϕ(z, t)| ≤ e|Re z| < ∞
in D(z; 1) × X. Thus the function ϕ satisfies all the requirements of Problem 10.16 in
D(z; 1) × X, so h ∈ H(D(z; 1)). Since z is arbitrary, we conclude that h ∈ H(C), i.e., h is
entire.
We complete the proof of the problem.
10.4
Problems related to Zeros of Holomorphic Functions
Problem 10.18
Rudin Chapter 10 Exercise 18.
Proof. By [76, Problem 10.11, p. 131], we know that
Z ′
X
1
f (z) p
z dz =
mk zkp ,
2πi γ f (z)
k
where the sum is taken over all the zeros of f inside γ and mk is the multiplicity of the zero zk
of f . This is the answer of the first assertion.
′
(z)
For the second assertion, let F (z) = ff (z)
ϕ(z). Then F is a meromorphic function in Ω. Let
A ⊆ Ω be the set of poles of F . Since f 6= 0 on γ ∗ , γ is a cycle in Ω \ A and Ind γ (α) = 0 for
all α ∈ Ω. In other words, our F satisfies Theorem 10.42 (The Residue Theorem) which implies
that
Z ′
X
f (z)
1
ϕ(z) dz =
Res (F ; zk ),
(10.48)
2πi γ f (z)
k
where the zk denote the isolated singularities of F inside γ which are exactly the zeros of f
inside γ. By Theorem 10.18, there exists a g ∈ H(Ω) and a unique positive integer mk such that
f (z) = (z − zk )mk g(z) and g(zk ) 6= 0. Clearly, we have
F (z) =
mk ϕ(z) g′ (z)
f ′ (z)
ϕ(z) =
+
ϕ(z).
f (z)
z − zk
g(z)
If ϕ(zk ) = 0, then F is holomorphic at zk . Otherwise, F has a simple pole at zk and it yields
from [9, Eqn. (1), p. 129] that
Res (F ; zk ) = lim (z − zk )F (z) = mk ϕ(zk ).
z→zk
(10.49)
Substituting the residues (10.49) into the formula (10.48), we get
Z ′
X
f (z)
1
ϕ(z) dz =
mk ϕ(zk ),
2πi γ f (z)
k
where the sum is taken over all the zeros of f which are not zeros of ϕ inside γ. This completes
the proof of the problem.
Problem 10.19
Rudin Chapter 10 Exercise 19.
18
Chapter 10. Elementary Properties of Holomorphic Functions
Proof. We claim that f = cg for some nonzero constant c. To this end, we consider h = fg in U .
Since f (z) 6= 0 on U , we know that h(z) 6= 0 on U . In addition, since f, g ∈ H(U ) and g(z) 6= 0
on U , we conclude that h ∈ H(U ). Direct differentiation gives
f ′ (z)g(z) − f (z)g ′ (z)
f (z) h f ′ (z) g′ (z) i
h (z) =
=
·
−
g 2 (z)
g(z)
f (z)
g(z)
′
so that h′ ( n1 ) = 0 on {1, 2, . . .}. By Theorem 10.18, we conclude that h′ (z) = 0 in U . Next, the
Fundamental Theorem of Calculus [9, Proposition 4.12, p. 51] shows that
Z
h′ (ζ) dζ = h(0) 6= 0
h(z) = h(0) +
[0,z]
for every z ∈ U , where [0, z] is a path connecting 0 and z in U . By the definition, we establish
that
f (z) = h(0)g(z)
on U . This ends the proof of the problem.
Problem 10.20
Rudin Chapter 10 Exercise 20.
Proof. Suppose that f (z0 ) = 0 for some z0 ∈ Ω. Assume that f 6≡ 0. Then there exists a circle
C(z0 ; R) for some R > 0 such that f (z) 6= 0 on C(z0 ; R). By Theorem 10.28, we know that
fn′ → f ′ uniformly on compact subsets of Ω. Thus the convergence
f′
fn′
→
fn
f
is also uniform on C(z0 ; R). Next, Theorem 10.43(a) gives
Z
Z
1
f ′ (z)
fn′ (z)
1
Nf =
dz and Nfn =
dz.
2πi C(z0 ;R) f (z)
2πi C(z0 ;R) fn (z)
Our hypotheses give Nf = 1 but Nfn = 0 which is a contradiction, so we establish the result
f ≡ 0 on Ω.
For the second assertion, suppose that
∞
[
n=1
fn (Ω) ⊆ Ω′ .
Choose a point a ∈ C \ Ω′ . Define Fn : Ω → C by Fn = fn − a for every n = 1, 2, . . .. Then each
Fn is holomorphic in Ω, Fn (z) 6= 0 for all z ∈ Ω and Fn → f − a uniformly on compact subsets
of Ω. Thus the first assertion and the fact that f is nonconstant imply that f (z) 6= a in Ω. In
other words, we must have
f (Ω) ⊆ Ω′
which completes the proof of the problem.
Remark 10.2
We note that Problem 10.20 is classically called Hurwitz’s Theorem, see [9, Theorem
10.13, p. 139] or [18, p. 152].
10.4. Problems related to Zeros of Holomorphic Functions
19
Problem 10.21
Rudin Chapter 10 Exercise 21.
Proof. Let g(z) = f (z) − z and h(z) = −z on Ω. Since D(0; 1) ⊆ Ω and |f (z)| < 1 on |z| = 1,
we have
|g(z) − h(z)| = |f (z)| < 1 = |h(z)|
on |z| = 1. By Theorem 10.43(b) (Rouché’s Theorem), we conclude that Ng = Nh = 1 in the
disc D(0; 1). This completes the analysis of the proof.
Problem 10.22
Rudin Chapter 10 Exercise 22.
Proof. Assume that f (z) 6= 0 for all z ∈ Ω. By the Corollary to Theorem 10.24 (The Maximum
Modulus Theorem), we see that
1 = |f (0)| ≥ min |f (reiθ )| > 2,
θ
a contradiction. Hence f has at least one zero in the unit disc, completing the proof of the
problem.
Problem 10.23
Rudin Chapter 10 Exercise 23.
Proof. Here we list some observations about the zeros of Pn and Qn .
• By Theorem 10.25 (The Fundamental Theorem of Algebra), both Pn and Qn have precisely
n zeros in C.
• The definitions of Pn and Qn guarantee that Pn and Qn cannot have any common zeros.
• Since Qn (0) = 0, Qn always has a zero at 0 for every n.
• By [9, Exercise 12, p. 142], we know that for every R > 0, if n is large enough, Pn (z) has
no zeros in |z| ≤ R, i.e., all zeros ζ of Pn satisfy |ζ| > R for large n. In fact, it has been
shown further in [33] that every zero ζ of Pn lie in the annulus
n
< |ζ| < n
e2
for large enough n.
• Applying [13, Corollary 1.2.3, p. 13] to Qn , all the zeros z 6= 0 of
Qn (z) = z +
lie inside the annulus
2=
zn
z
z n−1 z2
+ ··· +
= z 1 + + ··· +
2!
n!
2!
n!
min (k + 2) ≤ |z| <
0≤k≤n−2
max (k + 2) = n.
0≤k≤n−2
20
Chapter 10. Elementary Properties of Holomorphic Functions
• We imitate the proof of [76, Problem 10.12, pp. 131, 132]. First, we notice that ez − 1 = 0
if and only if z = ±2kπi for all k ∈ N, so if we take 2kπ < Rk < 2(k + 1)π, then there is a
ǫk > 0 such that
|ez − 1| > ǫk .
(10.50)
Since Pn (z) → ez uniformly in D(0; Rk ), there exists a Mk ∈ N such that n ≥ Mk implies
|ez − 1 − Qn | = |ez − Pn | < ǫk .
(10.51)
Thus the inequalities (10.50) and (10.51) give
|ez − 1 − Qn | < |ez − 1|
for all n ≥ Mk and for all z ∈ C(0; Rk ). By Theorem 10.43(b) (Rouché’s Theorem), we
conclude that
NQn = Nez −1 = 2k + 1
inside C(0; Rk ) for all n ≥ Mk . This also means that
NQn = n − 2k − 1
(10.52)
outside C(0; Rk ) for all n ≥ Mk .
Similarly, since 2(k + 1)π < Rk + 2π < 2(k + 2)π, Theorem 10.43(b) (Rouché’s Theorem)
tells us that
NQn = Nez −1 = 2k + 3
inside C(0; Rk + 2π) for all n ≥ Mk+1 , or equivalently,
NQn = n − 2k − 3
(10.53)
outside C(0; Rk + 2π) for all n ≥ Mk+1 . Now if we combine the results (10.52) and (10.53),
there exists a Mk′ ∈ N such that
NQ n = 2
in the annulus A = {z ∈ C | Rk < |z| < Rk + 2π} for all large enough n ≥ Mk′ , where
k = 1, 2, 3, . . ..
We complete the proof of the problem.
Remark 10.3
There are some books concerning the location of the zeros of polynomials. For instances,
[13, Chap. 1], [39] and [40, Chap. 3].
Problem 10.24
Rudin Chapter 10 Exercise 24.
Proof. We have Ω = K ◦ which is an open set in C. Put
ϕ(z) = |f (z) − g(z)| − |f (z)| − |g(z)|,
E = {z ∈ K | ϕ(z) = 0} and {zn } ⊆ E with zn → z0 as n → ∞. We divide the proof into several
steps.
10.4. Problems related to Zeros of Holomorphic Functions
21
• Step 1: E ⊆ Ω. Our hypothesis implies that
|f (z) − g(z)| < |f (z)| + |g(z)|
(10.54)
on ∂Ω = K \ Ω and this ensures that f and g cannot have any zero on ∂Ω. In addition,
the continuity of f and g imply the continuity of ϕ and then z0 ∈ E. Thus E is a closed,
hence compact, subset of K by Theorem 2.4. By the inequality (10.54), we know that
E ∩ ∂Ω = ∅ which means that E ⊆ Ω and
|f (z) − g(z)| < |f (z)| + |g(z)|
(10.55)
in K \ E.
• Step 2: The numbers of zeros of f and g are finite. Suppose that
Z(f ) = {a ∈ Ω | f (a) = 0}.
Assume that Z(f ) was infinite. Then [79, Problem 5.25, p. 68] leads to us that Z(f ) has
a convergent subsequence and Theorem 10.18 says that Z(f ) = Ω, but the continuity of
f implies immediately that f ≡ 0 on K which is impossible. Consequently, Z(f ) is finite.
Furthermore, since f (a) = 0 implies ϕ(a) = 0, all zeros of f lie in E. Similarly, Z(g) is
also finite and all zeros of g belong to E.
• Step 3: A lemma and its application. Here we need the following result whose proof
can be found in [63, IX.8 & 9, pp. 115 – 118] or [67, Lemma 5.8, pp. 61, 62]:
Lemma 10.1
Let G be an open subset of C and K a compact subset of G. Then there exists a
cycle Γ in G \ K such that K ⊆ int Γ ⊆ G and
Z
f (ζ)
1
dζ
(10.56)
f (z) =
2πi Γ ζ − z
for every f ∈ H(G) and z ∈ K.
By Step 1, we see that E and Ω satisfy the roles of K and G of Lemma 10.1 respectively,
so it ensures that there exists a cycle Γ in Ω \ E such that E ⊆ int Γ ⊆ Ω and the formula
(10.56) holds for every f ∈ H(Ω) and every z ∈ E.
As an application, let Fe ∈ H(Ω), z ∈ E and F (ζ) = (ζ − z)Fe(ζ). Obviously, we have
F ∈ H(Ω) and F (z) = 0, so the formula (10.56) gives
Z
Γ
Fe(ζ) dζ = 0.
(10.57)
• Step 4: The calculation of |Z(f )| and |Z(g)|. Recall from the hypotheses and Step
2 that f ∈ H(Ω) and f has only finitely many zeros a1 , a2 , . . . , aN with multiplicities
p1 , p2 , . . . , pN respectively in E. Then the function
N
f ′ (z) X pk
fe(z) =
−
f (z)
z − ak
k=1
22
Chapter 10. Elementary Properties of Holomorphic Functions
can be shown to have a removable singularity at each ak by Theorem 10.18 and Definition
10.19. Thus fe is holomorphic in Ω. Hence it follows from the result (10.57) that
Z
Γ
N
X
f ′ (ζ)
pk
dζ =
f (ζ)
k=1
Z
Γ
dζ
.
ζ − ak
Next, we apply Lemma 10.1 with f ≡ 1 and z = ak to get
Z
dζ
2πi =
.
ζ
−
ak
Γ
(10.58)
(10.59)
By combining the integrals (10.58) and (10.59), we conclude that
1
2πi
Z
N
Γ
X
f ′ (ζ)
dζ =
pk = |Z(f )|.
f (ζ)
(10.60)
k=1
Similarly, we obtain
1
2πi
Z
Γ
g′ (ζ)
dζ = |Z(g)|.
g(ζ)
(10.61)
By the strict inequality (10.55), it is impossible that f = −N ′ g for some N ′ ∈ N ∪ {0} on
Ω \ E. Therefore, the function h : Ω \ E → C \ (−∞, 0] given by
h(z) = log
f (z)
g(z)
is well-defined and holomorphic in the open set Ω \ E. Taking differentiation, we have
h′ =
f ′ g′
−
f
g
on Ω \ E. Since Z(f ), Z(g) * Ω \ E, h′ is continuous in Ω \ E. Since Γ is a cycle in Ω \ E,
Theorem 10.12 guarantees that
Z
h′ (ζ) dζ = 0
Γ
or equivalently,
Z
Γ
f ′ (ζ)
dζ =
f (ζ)
Z
Γ
g′ (ζ)
dζ.
g(ζ)
(10.62)
Finally, by substituting the results (10.60) and (10.61) into the expression (10.62), we have
established that
|Z(f )| = |Z(g)|.
Now we complete the proof of the problem.
10.5
Laurent Series and its Applications
Problem 10.25
Rudin Chapter 10 Exercise 25.
10.5. Laurent Series and its Applications
23
Proof. Define A(r1 , r2 ) = {z ∈ C | r1 < |z| < r2 }. Let ǫ > 0 be such that r1 + ǫ < r2 − ǫ.
Furthermore, we define ρr : [0, 2π] → C by
ρr (t) = reit ,
where r > 0. Particularly, we have γ1 = −ρr1 +ǫ and γ2 = ρr2 −ǫ .
(a) Note that γ1 and γ2 are negatively oriented and positively oriented circles respectively. By
Theorem 10.11, we have

 −1, if z ∈ D(0; r1 + ǫ);
Ind γ1 (z) =

0,
if z ∈
/ D(0; r1 + ǫ)
and
Ind γ2 (z) =

 1, if z ∈ D(0; r2 − ǫ);

0, if z ∈
/ D(0; r2 − ǫ).
Let Γ = γ1 + γ2 which is the sum of two circles in A(r1 , r2 ). Let α ∈
/ A(r1 , r2 ). If |α| ≤ r1 ,
then we have α ∈ D(0; r1 + ǫ) ⊆ D(0; r2 − ǫ). Thus we follow from [62, Eqn. (8), p. 218]
that
Ind Γ (α) = Ind γ1 (α) + Ind γ2 (α) = 0.
(10.63)
Similarly, if |α| ≥ r2 , then we still have the result (10.63). Next if z ∈ A(r1 + ǫ, r2 − ǫ),
then z ∈
/ D(0; r1 + ǫ) and z ∈ D(0; r2 − ǫ). Thus it is easy to check that
Ind Γ (z) = Ind γ1 (z) + Ind γ2 (z) = 1.
See Figure 10.4 for an illustration below.
Figure 10.4: The annulus A(r1 , r2 ) and the circles γ1 , γ2 .
Hence, Theorem 10.35 (Cauchy’s Theorem) and [62, Eqn. (5), p. 217] assert that
24
Chapter 10. Elementary Properties of Holomorphic Functions
1
f (z) =
2πi
Z
for every z ∈ A(r1 + ǫ, r2 − ǫ).
Γ
f (ζ)
1 dζ =
ζ −z
2πi
Z
Z f (ζ)
+
dζ
γ2 ζ − z
γ1
(10.64)
(b) Let R1 = r1 + ǫ and R2 = r2 − ǫ so that r1 < R1 < |z| < R2 < r2 . Define
Z
Z
1
f (ζ)
f (ζ)
1
f1 (z) =
dζ and f2 (z) =
dζ.
2πi −ρR ζ − z
2πi ρR ζ − z
1
(10.65)
2
Then the formula (10.64) simplifies to f = f1 + f2 .
– Step 1: f1 and f2 are well-defined. To see this, we first recall from the hypothesis
that f ∈ H(A(r1 , r2 )). Next, we fix z and take r1 < R1 < R1′ < |z| which means that
z∈
/ D(0; R1′ ). Let Γ1 = ρR1 − ρR′1 . By Theorem 10.11, we know that
Ind Γ1 (α) = Ind ρR1 (α) − Ind ρR′ (α) = 0
1
for every α ∈
/ A(r1 , r2 ). Thus it yields from Theorem 10.35 (Cauchy’s Theorem)g
that Ind Γ1 (z) = 0 − 0 = 0 and then
Z
Z
1
f (ζ)
1
f (ζ)
dζ =
dζ.
2πi ρR ζ − z
2πi ρR′ ζ − z
1
1
In other words, f1 (z) is uniquely determined by z, not by R1 . Since ǫ is arbitrary,
this means that f1 is actually well-defined in C \ D(0; r1 ).
Similarly, we take |z| < R2 < R2′ < r2 so that z ∈ D(0; R2 ) ⊆ D(0; R2′ ). Let
Γ2 = ρR2 − ρR′2 . By Theorem 10.11 again, we have
Ind Γ2 (α) = Ind ρR2 (α) − Ind ρR′ (α) = 0
2
for every α ∈
/ A(r1 , r2 ). Hence we have Ind Γ2 (z) = 1 − 1 = 0 and similar argument
shows that f2 (z) is well-defined in D(0; r2 ).
– Step 2: f2 ∈ H(D(0; r2 )). As suggested by the proof of Theorem 10.16, we may
apply Theorem 10.7 to the integral representation (10.65) of f2 with X = [0, 2π],
ϕ = ρR2 and dµ(t) = f (ρR2 (t))ρ′R2 (t) dt to establish the fact that f2 is representable
by a power series in D(0; R2 ). Since ǫ is arbitrary, f2 is representable by a power
series in D(0; r2 ) and hence Theorem 10.6 concludes that f2 ∈ H(D(0; r2 )).
– Step 3: f1 ∈ H(C \ D(0; r1 )). For f1 , we consider the function g(ζ) = ζ1 f ( 1ζ ) on
|ζ| = R11 . Tale X = [0, 2π], ϕ = ρ 1 and dµ(t) = g ρ 1 (t) ρ′ 1 (t) dt in Theorem 10.7.
R1
Obviously, we have ϕ([0, 2π]) = ρ
and thus the function
R1
1
R1
R1
([0, 2π]) = C(0; R11 ), so ϕ([0, 2π]) ∩ D(0; R11 ) = ∅
1
g(z) =
2πi
Z
ρ
1
R1
dµ(ζ)
ζ−z
is representable by power series in D(0; R11 ). Furthermore, for ρ
have
1
2πi
g
Z
ρ
1
R1
dµ(ζ)
1
=
ζ −z
2πi
Z
2π
R1 e−it f (R1 e−it )
0
In fact, we have applied the formula [62, Eqn. (2), p. 219] here.
i it
e ·
R1
1
R1
(t) =
dt
−z
1 it
R1 e
1 it
R1 e ,
we
10.5. Laurent Series and its Applications
1
=
2πi
Z
2π
0
25
if (R1 e−it ) ·
R1 e−it dt
( R11 eit − z)R1 e−it
Z 2π
1 1
if (R1 e−it )R1 e−it dt
·
1
−it
z 2πi 0
z − R1 e
Z
f (ζ)
1
=
dζ
z −ρ 1 ζ − z1
R1
1
1
.
= f1
z
z
=
Therefore, the function 1z f1 ( z1 ) is representable by power series in D(0; R11 ). Again,
since ǫ is arbitrary, z1 f1 ( 1z ) can be represented by power series in D(0; r11 ) and then
Theorem 10.6 ensures that z1 f1 ( 1z ) ∈ H(D(0; r11 )). Hence, this certainly shows that
f1 ∈ H(C \ D(0; r1 )) as required.
– Step 4: Uniqueness of the decomposition. Suppose that f = g1 + g2 , where
g1 ∈ H(C \ D(0; r1 )) and g2 ∈ H(D(0; r2 )). Then we have
g1 − f1 = f2 − g2
(10.66)
in A(r1 , r2 ). Define h : C → C by

 f2 (z) − g2 (z), if z ∈ D(0; r2 );
h(z) =

g1 (z) − f1 (z), if z ∈ C \ D(0; r1 ).
Now the equation (10.66) shows that the two definitions of h actually agree in
A(r1 , r2 ), so h is well-defined in C and in fact it is entire. Thus it suffices to prove
that h ≡ 0. Since f1 (z), g1 (z) → 0 as |z| → ∞, we have h(z) → 0 as |z| → ∞ and thus
h is bounded. By Theorem 10.23 (Liouville’s Theorem), we conclude that h(z) = 0
for all z ∈ C and this means that f1 (z) = g1 (z) and f2 (z) = g2 (z) in C \ D(0; r1 ) and
D(0; r2 ) respectively. This proves the uniqueness of the decomposition.
(c)
– Existence of a Laurent series. On γ2 , we have |ζ| > |z| so that
1
z
z2
1
1
=
+ 2 + 2 + ··· .
z =
ζ −z
ζ(1 − ζ )
ζ
ζ
ζ
(10.67)
On −γ1 , since |ζ| < |z|, we have
1
ζ
ζ2
1
1
=−
= − − 2 − 3 − ··· .
ζ
ζ−z
z z
z
z(1 − z )
(10.68)
Substituting the series (10.67) and (10.68) into the formula (10.64), we get
1
f (z) =
2πi
Z
∞
X
f (ζ)z n
γ2 n=0
ζ n+1
1
dζ +
2πi
Z
−∞
X
f (ζ)z n
dζ.
ζ n+1
(10.69)
−γ1 n=−1
Since the convergence of the series (10.67) and (10.68) are uniform, we can switch
the order of integration and summation in the expression (10.69) to obtain
f (z) =
∞
X
−∞
cn z n ,
26
Chapter 10. Elementary Properties of Holomorphic Functions
where
1
cn =
2πi
Z
γ
f (ζ)
dζ
ζ n+1
and γ =

 γ2 ,

if n = 0, 1, 2, . . .;
(10.70)
−γ1 , if n = −1, −2, −3, . . ..
– Uniqueness of the Laurent series. We claim that if
A, then an = cn for all n ∈ Z. To this end, since
on γ, we have
Z
γ
∞
X
f (z)
dz
=
z k+1
−∞
Z
∞
X
∞
X
an z n converges to f in
−∞
an z n converges uniformly to f
−∞
an z n−k−1 dz = 2πiak ,
γ
where k ∈ Z. Hence it asserts from the definitions (10.70) that cn = an hold for all
n ∈ Z. This proves the claim and then the uniqueness follows.
– Uniform convergence on compact subsets of A. Let K be a compact subset of
A. It is easy to check that C \ A is closed in C and (C \ A) ∩ K = ∅. Thus we deduce
from Problem 10.1 that there is a δ > 0 such that d(K, C \ A) = 2δ > 0. In addition,
we have
K ∩ C(0; r1 + δ) = ∅ and K ∩ C(0; r2 − δ) = ∅
which imply that K ⊆ A(r1 + δ, r2 − δ) ⊆ A(r1 , r2 ). For r1 + δ ≤ |z| ≤ r2 − δ, we
know that
∞
X
−∞
n
|cn z | ≤
≤
≤
−1
X
n=−∞
∞
X
n=1
∞
X
n=1
< ∞.
n
|cn z | +
|cn |
+
|z|n
∞
X
n=0
∞
X
n=0
|cn z n |
|cn |(r2 − δ)n
∞
X
|cn |
|cn |(r2 − δ)n
+
(r1 + δ)n
n=0
Hence it follows from the Weierstrass M -Test [9, Theorem 1.9, p. 15] or [61, Theorem
7.10, p. 148] that the series converges to f uniformly on K.
(d) Let r1 < s1 < s2 < r2 . Firstly, since f2 ∈ H(D(0; r2 )) and D(0; s2 ) is a compact subset of
D(0; r2 ), f2 is bounded in D(0; s2 ). Secondly, since f1 (z) → 0 as |z| → ∞, there exists a
M > s1 such that
|f1 (z)| < 1
(10.71)
on C \ D(0; M ). The set D(0; M ) \ D(0; s1 ) = {z ∈ C | s1 ≤ |z| ≤ M } is a closed and
bounded subset in C, so it is compact. Since f1 ∈ H(C \ D(0; r1 )), f1 is bounded in
D(0; M ) \ D(0; s1 ). This fact and the bound (10.71) combine to say that f1 is bounded in
C \ D(0; s1 ).
Now we write
f2 = f − f1
on D(0; r2 ) \ D(0; s2 ). Since D(0; r2 ) \ D(0; s2 ) ⊆ C \ D(0; s1 ), f1 is bounded there.
Since f is bounded in A, f2 is bounded in D(0; r2 ) \ D(0; s2 ). Using the first result in
the previous paragraph, we conclude that f2 is bounded in D(0; r2 ). Similarly, we write
10.5. Laurent Series and its Applications
27
f1 = f − f2 on D(0; s1 ) \ D(0; r1 ). Clearly, we know that D(0; s1 ) \ D(0; r1 ) ⊆ D(0; s2 ), so
f2 is bounded there and thus f1 is also bounded there. Using the second assertion in the
previous paragraph, we see that f1 is bounded in C \ D(0; r1 ).
(e) All the foregoing parts remain valid if r1 = 0, or r2 = ∞ or both. In the case r2 = ∞,
we note that f2 represents an entire function and thus it reduces to 0 by Theorem 10.23
(Liouville’s Theorem).
(f) Suppose that 0 < r1 < r2 < r3 < · · · < rn−1 < rn < ∞. Each A(rk , rk+1 ) is an annulus,
n−1
[
A(rk , rk+1 ) and f ∈ H(A). Then the foregoing
where k = 1, 2, . . . , n − 1. Let A =
k=1
results can be applied to f in each A(rk , rk+1 ) and obtain the corresponding Laurent series
in each A(rk , rk+1 ).
Hence we have completed the analysis of the problem.
Problem 10.26
Rudin Chapter 10 Exercise 26.
Proof. Let f be the function in the question. The function f is holomorphic in C \ {−1, 1, 3}.
By Problem 10.25, we have to consider the following three regions
D(0; 1) = {z ∈ C | |z| < 1},
A = {z ∈ C | 1 < |z| < 3}
Note that if |z| < 1, then we have
and
B = {z ∈ C | |z| > 3}.
∞
X
1
z 2n .
=
1 − z2
(10.72)
n=0
If |z| > 1, then | 1z | < 1 so that
∞
X 1
1
1
1
=− 2 ·
=−
.
2
−2
1−z
z 1−z
z 2n
(10.73)
n=1
Similarly, if |z| < 3, then we have
| z3 |
< 1 so that
1
1
1
= ·
3−z
3 1−
=
z
3
∞
1 X zn
3 n=0 3n
(10.74)
and if |z| > 3, then | 3z | < 1 which gives
1
1
1
=− ·
3−z
z 1−
3
z
=−
∞
1 X 3n
.
z
zn
n=0
Therefore, for every z ∈ D(0; 1), we follow from the series (10.72) and (10.74) that
f (z) =
∞
X
n=0
Thus we have
z 2n +
1 X z n X h 1 + (−1)n
1 i n
=
+
z .
3
3n
2
3n+1
∞
∞
n=0
n=0

1
1 + (−1)n


+ n+1 , if n = 0, 1, 2, . . .;
2
3
cn =


0,
otherwise.
(10.75)
28
Chapter 10. Elementary Properties of Holomorphic Functions
Next, if z ∈ A, then we deduce from the series (10.73) and (10.74) that
∞
∞
∞
X
X
1
1 X zn
f (z) = −
cn z n ,
+
=
2n
n
z
3
3
n=−∞
n=1
n=0
where
cn =

1



 3n+1 ,
if n ≥ 0;

n+1


 −1 + (−1)
, otherwise.
2
Finally, if z ∈ B, then the series (10.73) and (10.75) imply that
∞
∞
∞
X
X
1 X 3n
1
cn z n ,
−
=
f (z) = −
z 2n z
z n n=−∞
n=0
n=1
where
cn =

0,


if n ≥ 0;
h
i
n

 − 1 + (−1) + 3|n|−1 , otherwise.
2
Hence we have completed the analysis of the problem.
Problem 10.27
Rudin Chapter 10 Exercise 27.
Proof. Suppose that
ζ = e2πiz
or
z=
1
i
arg ζ −
log |ζ|,
2π
2π
(10.76)
where 0 < arg ζ < 2π. Set F (ζ) = f (z). We claim that F (ζ) does not depend on the choice of
arg ζ. In fact, it is easy to see from the second equation (10.76) that
i
arg ζ
i
arg ζ + 2π
−
log |ζ| =
+1−
log |ζ| = z + 1.
2π
2π
2π
2π
Since f is a function of period 1, we have F (ζ) = f (z) = f (z + 1) = F (ζe2πi ) which proves our
claim. Denote the annulus A = {ζ ∈ C | e−2πb < |ζ| < e−2πa }. Then the mapping G(z) = e2πiz
clearly maps the horizontal strip Ω = {z ∈ C | a < Im z < b} onto A. Furthermore, G is
holomorphic in Ω and f = F ◦ G. Here we need a positive result to Problem 10.14:
Lemma 10.2
Suppose that Ω1 and Ω2 are two regions, f and g are nonconstant complex functions
defined in Ω1 and Ω2 respectively. Put h = g ◦ f . If f and h are holomorphic in
Ω1 and f (Ω1 ) = Ω2 , then g is also holomorphic in Ω2 .
10.5. Laurent Series and its Applications
29
Proof of Lemma 10.2. We consider
S = {z ∈ Ω2 | every ω ∈ Ω1 with f (ω) = z implies f ′ (ω) = 0}.
(10.77)
Let β ∈ Ω2 \ S. Since f (Ω1 ) = Ω2 , there exists an α ∈ Ω1 such that f (α) = β. By
the definition (10.77), f ′ (α) 6= 0. Using Theorem 10.30, there exist neighborhoods
V ⊆ Ω1 and W ⊆ Ω2 of α and β respectively such that f : V → W is a bijection and a
holomorphic inverse f −1 : W → V exists. Thus we have g = g ◦ (f ◦ f −1 ) = h ◦ f −1 on
W which guarantees that g is holomorphic in W and thus g is holomorphic in Ω2 \ S.
Next, we want to show that every point b of S is isolated. This means that one can
find a neighborhood W ⊆ Ω2 of b such that W ∩ S = {b}. By the hypothesis, one can
find an a ∈ Ω1 such that f (a) = b. By the definition (10.77), we have f ′ (a) = 0. Since
f is nonconstant, f ′ is nonzero. Since f ′ ∈ H(Ω1 ), Theorem 10.18 implies that Z(f ′ )
has no limit point in Ω1 . In other words, Ω1 contains a neighborhood U of a such that
f ′ (z) 6= 0
(10.78)
for all z ∈ U \ {a}. By the Open Mapping Theorem, f (U ) ⊆ Ω2 is our wanted
neighborhood because if f (U ) contains a c ∈ S \ {b}, then we have f (ζ) = c for some
ζ ∈ U \ {a} but the definition (10.77) shows that f ′ (ζ) = 0 which contradicts the result
(10.78).
Given ǫ > 0. Since h is continuous at a, there exists a δ > 0 such that ω ∈ D(a; δ)
implies
h(ω) ∈ D(h(a); ǫ).
(10.79)
Since f is holomorphic in Ω1 , f (D(a; δ)) is open in C by the Open Mapping Theorem.
Since b ∈ f (D(a; δ)), there is a δ′ > 0 such that D(b; δ′ ) ⊆ f (D(a; δ)). Obviously, if
z ∈ D(b; δ′ ), then we have f (ω) = z for some ω ∈ D(a; δ). Combining this fact with
the property (10.79), we conclude that
g(z) = g(f (ω)) = h(ω) ∈ D(h(a); ǫ) = D(g(f (a)); ǫ) = D(g(b); ǫ).
Consequently, g is continuous on S. According to the Riemann’s Principle of Removable Singularities [9, Theorem 9.3, p. 118], every point of S is a removable singularity
of g. Hence g can be extended to a function holomorphic in Ω2 , as required.
Since f and G are holomorphic in Ω and G(Ω) = A, Lemma 10.2 ensures that F is analytic
in A. By Problem 10.25(c), F admits the Laurent series
F (ζ) =
∞
X
cn ζ n
(10.80)
−∞
in A and this implies
f (z) = F (ζ) =
∞
X
cn e2nπiz .
(10.81)
−∞
Finally, we denote
Ω′ = {z ∈ C | a + ǫ ≤ Im z ≤ b − ǫ} and A′ = {ζ ∈ C | e−2π(b−ǫ) ≤ |ζ| ≤ e−2π(a+ǫ) }.
Then we have G(Ω′ ) = A′ and A′ is a compact subset of A. By Problem 10.25(c), the series
(10.80) converges uniformly in A′ and hence the corresponding series (10.81) converges uniformly
in Ω′ . This completes the proof of the problem.
30
Chapter 10. Elementary Properties of Holomorphic Functions
10.6
Miscellaneous Problems
Problem 10.28
Rudin Chapter 10 Exercise 28.
Proof. If we consider γ = Γ − α, then we observe immediately from [62, Eqn. (2), p. 203] that
1
Ind Γ (α) =
2πi
Z
2π
0
Γ′ (s)
1
ds =
Γ(s) − α
2πi
Z
2π
0
γ ′ (s)
ds = Ind γ (0).
γ(s)
Thus, without loss of generality, we may assume that α = 0 in the following discussion. Now we
prove the assertions one by one.
• Ind Γn (0) = Ind Γm (0) if m and n are sufficiently large. Since Γ : [0, 2π] → C is a closed
curve, it is a continuous function with period 2π. Thus the Stone-Weierstrass Theorem [61,
Theorem 8.15, p. 190] asserts that there exists a sequence of trigonometric polynomials
{Γn } converges to Γ uniformly in [0, 2π]. In other words, choose |Γ(θ)| > δ > 0, there
exists an N ∈ N such that n ≥ N implies
|Γ(θ) − Γn (θ)| <
δ
4
for all θ ∈ [0, 2π]. According to [61, Exercise 26, p. 202]h , we conclude that
Ind Γn (0) = Ind Γm (0)
for all m, n ≥ N .
• The result is independent of the choice of {Γn }. Note that [77, Problem 8.26, pp.
204 – 206] also includes this part.
• Lemma 10.39 is true for closed curves. Again, [77, Problem 8.26, pp. 204 – 206]
contains this part.
• Another proof of Theorem 10.40. By the definition, there exists a continuous map
H : I 2 → Ω such that
H(s, 0) = Γ0 (s),
H(s, 1) = Γ1 (s) and
H(0, t) = H(1, t).
Put Γt (s) = H(s, t). According to [62, Eqn. (3) & (4), p. 223], we obtain
|Γt (s)| > 2ǫ
(10.82)
for all (s, t) ∈ I 2 and there exists a positive integer n such that
|Γt (s) − Γt′ (s)| < ǫ
(10.83)
for all t, t′ ∈ I with |t − t′ | ≤ n1 and all s ∈ I. Let t0 = 0, tn = 1 and tk = tk−1 + n1 , where
k = 1, 2, . . . , n. Then the set {t0 , t1 , . . . , tn } forms a partition of I and tk − tk−1 = n1 for
each k = 1, 2, . . . , n. Therefore, it follows from the inequalities (10.82) and (10.83) that
|Γtk (s) − Γtk−1 (s)| < ǫ < |Γtk−1 (s)|
h
The author has proven it in [77, Problem 8.26, pp. 204 – 206].
10.6. Miscellaneous Problems
31
for every s ∈ I, where k = 1, 2, . . . , n. By the improved version of Lemma 10.39, we see
that
Ind Γtk (0) = Ind Γtk−1 (0),
where k = 1, 2, . . . , n. Consequently, we have the desired result that
Ind Γ0 (0) = Ind Γt0 (0) = Ind Γt1 (0) = · · · = Ind Γtn (0) = Ind Γ1 (0).
This ends the proof of the problem.
Problem 10.29
Rudin Chapter 10 Exercise 29.
Proof. Suppose that z 6= 0, so we can write
1
f (z) =
π
1
π
=
1
π
=
Z
Z
Z
0
0
0
1Z π
−π
1Z π
−π
1Z π
Z
r
dθ dr
+z
reiθ
re−iθ
dθ dr
r + ze−iθ
r −iθ
ze
r
−iθ
−π z + e
Z
π
1 h
dθ dr
i
−ieiθ
r
dθ
dr
r
−iθ − (− )
−π e
0 z
z
Z π
i
1 h
r 1
−ieiθ
dθ
dr
r
−iθ − (− )
0 z 2πi −π e
z
Z 1
−r r
dr
= −2
· Ind γ
z
0 z
1
=−
πi
Z
= −2
(10.84)
where γ(θ) = e−iθ with −π ≤ θ ≤ π which is the negatively oriented circle with center at 0 and
radius 1. By Theorem 10.11, we know that

r
 −1, if |z| < 1;
−r 
Ind γ
=

z
r
 0,
if |z|
> 1.
If |z| < 1, then the expression (10.84) reduces to
hZ
f (z) = −2
=2
Z
|z|
0
|z|
0
Z
−r r
dr +
· Ind γ
z
z
r
dr
z
2 |z|
r dr
z 0
|z|2
=
z
= z.
=
Z
1
|z|
−r i
r
dr
· Ind γ
z
z
32
Chapter 10. Elementary Properties of Holomorphic Functions
Next, if |z| ≥ 1, then we always have Ind γ (− zr ) = −1 so that
2
f (z) =
z
Z
1
r dr =
0
1
.
z
Finally, if z = 0, then it is easy to see that
Z Z
1 1 π −iθ
e dθ dr = 0.
f (0) =
π 0 −π
We have completed the analysis of the problem.
Problem 10.30
Rudin Chapter 10 Exercise 30.
Proof. Without loss of generality, we may assume that Ω = C \ {−1, 1} which is clearly open in
C. Consider the boundaries γ1 , γ2 , γ3 and γ4 of the discs D(−1; 1), D(1; 1), D(−2; 2) and D(2; 2)
respectively. Each starts and ends at 0 with the orientation as shown in Figure 10.5. Then
Figure 10.5: A non null-homotopic closed path Γ = γ1 − γ3 − γ2 + γ4 in Ω.
Γ = γ1 − γ3 − γ2 + γ4 is a closed curve in Ω and we observe easily that
Ind Γ (±1) = 0.
Assume that Γ was null-homotopic to the constant map 0 in Ω. By §10.38, there exists a
continuous map Γt : [0, 1] → Ω connecting Γ and 0 such that Γ0 = 0 and Γ1 = Γ. However,
the continuity of Γt forces that it must pass through one of the omitted points 1 and −1, a
contradiction. This completes the proof of the problem.
CHAPTER
11
Harmonic Functions
11.1
Basic Properties of Harmonic Functions
Problem 11.1
Rudin Chapter 11 Exercise 1.
Proof. Let u, v : Ω → R. We prove the assertions one by one.
• uv is harmonic if and only if u + icv ∈ H(Ω) for some c ∈ R. Direct computation
gives
∆(uv) = (uv)xx + (uv)yy
= (uvxx + 2ux vx + uxx v) + (uvyy + 2uy vy + uyy v)
= 2(ux vx + uy vy )
= 2(ux , uy ) · (vx , vy ).
Thus uv is harmonic if and only if
(ux , uy ) · (vx , vy ) = 0.
(11.1)
If u + icv ∈ H(Ω), then it yields from the Cauchy-Riemann equations that ux = cvy and
uy = −cvx so that ux vx + uy vy = cvy vx − cvx vy = 0 in Ω. This means that ∆(uv) = 0.
Conversely, suppose that uv is harmonic in Ω. If u is constant, then Theorem 11.2 ensures
that ux = uy = 0 in Ω which gives the equation (11.1). The case for v being constant
is similar. Therefore, without loss of generality, we may assume that both u and v are
nonconstant. Clearly, the equation (11.1) implies that there exists a function c : R2 → R
such that (ux , uy ) = c(x, y)(vy , −vx ) in Ω, or equivalently,
ux = c(x, y)vy
and uy = −c(x, y)vx
(11.2)
in Ω. On the one hand, since ∆u = 0 in Ω, we obtain
cx vy − cy vx = 0
in Ω. On the other hand, uxy − uyx = 0 and ∆v = 0 imply that
cy vy + cvyy − (−cx vx − cvxx ) = 0
33
(11.3)
34
Chapter 11. Harmonic Functions
and thus
cy vy − cx vx = 0.
(11.4)
Eliminating vx and vy from the equations (11.3) and (11.4), we get
(c2x + c2y )vx = 0 and
(c2x + c2y )vy = 0
(11.5)
in Ω. Since v is harmonic in Ω, the function f = vx − ivy is holomorphic in Ω by Theorem
11.2. If Z(f ) = Ω, then vx = vy = 0 in Ω so that v is constant in Ω, a contradiction.
By Theorem 10.18, Z(f ) has no limit point in Ω. Pick z ∈ Ω \ Z(f ). Then there exists
a δ > 0 such that D(z; δ) ∩ Z(f ) = ∅. By the equations (11.5), we have (c2x + c2y )f = 0
which implies that c2x + c2y = 0 in D(z; δ). Since z is arbitrary, we conclude that
c2x + c2y = 0
(11.6)
in Ω \ Z(f ). Since f = vx − ivy ∈ H(Ω), f has continuous derivatives of all orders. Thus
both vx and vy have continuous partial derivatives of all orders in Ω. Similarly, both ux
and uy have continuous partial derivatives of all orders in Ω. Therefore, it follows from any
one of the equations (11.2) that both cx and cy are continuous on Ω. Hence the equation
(11.6) holds in Ω so that cx = cy = 0 in Ω. In conclusion, c ∈ R and hence u + icv satisfies
the Cauchy-Riemann equations.
• u2 cannot be harmonic in Ω unless u is constant. This part is shown in [76, Problem
16.3, pp. 199, 200].
• |f |2 is harmonic. Since f ∈ H(Ω), we know from Theorem 11.4 that both u and v are
harmonic in Ω. Note that |f |2 is harmonic in Ω if and only if u2 + v 2 is harmonic in Ω.
Clearly, we have
∆(u2 + v 2 ) = (u2 + v 2 )xx + (u2 + v 2 )yy
= (2uux + 2vvx )x + (2uuy + 2vvy )y
= (2uuxx + 2u2x + 2vvxx + 2vx2 ) + (2uuyy + 2u2y + 2vvyy + 2vy2 )
= 2(u2x + vx2 + u2y + vy2 ).
Therefore, ∆(u2 + v 2 ) = 0 in Ω if and only if ux = uy = vx = vy = 0 in Ω if and only if
both u(x, y) and v(x, y) are constant in Ω. In other words, |f |2 is harmonic in Ω if and
only if f is constant in Ω.
This completes the proof of the problem.
Problem 11.2
Rudin Chapter 11 Exercise 2.
Proof. Let f = u + iv. The result is obvious if f is constant, so without loss of generality, we
may assume that f is nonconstant. Since f is harmonic in Ω, both u and v are harmonic in Ω.
Similarly, since f 2 = u2 − v 2 + 2iuv is harmonic in Ω, both u2 − v 2 and uv are harmonic in Ω.
Thus ∆(u2 − v 2 ) = 0 and ∆(uv) = 0 imply that
u2x + u2y = vx2 + vy2
and ux vx + uy vy = 0
in Ω respectively. Consequently, they show that
(ux + ivx )2 + (uy + ivy )2 = 0
11.1. Basic Properties of Harmonic Functions
35
(ux + ivx )2 − [i(uy + ivy )]2 = 0
[ux + ivx + i(uy + ivy )][ux + ivx − i(uy + ivy )] = 0
(ux − vy + ivx + iuy )(ux + vy + ivx − iuy ) = 0
holds in Ω.
Next, the equation (ux − ivx )2 + (uy − ivy )2 = 0 gives
(ux − vy − ivx − iuy )(ux + vy − ivx + iuy ) = 0
(11.7)
g = ux − vy − i(vx + uy ) and h = ux + vy + i(uy − vx ).
(11.8)
in Ω. We denote
It is obvious that
(ux − vy )x = uxx − vyx = −uyy − vxy = −(uy + vx )y
and
(ux − vy )y = uxy − vyy = uyx + vxx = −[−(uy + vx )]x
hold in Ω. In other words, g ∈ H(Ω). Similarly, we have h ∈ H(Ω).
Assume that g 6≡ 0 and h 6≡ 0 in Ω. Then Theorem 10.18 ensures that Z(g) and Z(h) are
at most countable. Therefore, Z(gh) = Z(g)Z(h) is also at most countable. Note that zeros
of h and h are identical. Therefore, the equation (11.7) says that Z(gh) = Ω, a contradiction.
Hence we have g ≡ 0 or h ≡ 0 in Ω. If g ≡ h ≡ 0 in Ω, then we deduce from the definition
(11.8) that ux = uy = 0 in Ω, i.e., u is constant in Ω. By this and the fact f ∈ H(Ω), we obtain
immediately from Theorem 11.2 that v is also constant in Ω. Hence f is also constant in H(Ω),
a contradiction. In other words, we have either g ≡ 0 or h ≡ 0 in Ω.
• Case (i): g ≡ 0 in Ω. By the definition, g ≡ 0 implies that ux − vy = i(vx + uy ) in Ω.
Since ux , uy , vx and vy are real functions, we get ux = vy and uy = −vx in Ω. By Theorem
11.2, we conclude that f ∈ H(Ω).
• Case (ii): h ≡ 0 in Ω. Similarly, this shows that ux = −vy and uy = vx in Ω and these
mean that f = u − iv ∈ H(Ω).
We have completed the proof of the problem.
Problem 11.3
Rudin Chapter 11 Exercise 3.
Proof. Let u be a real function. Suppose that
V = {z ∈ Ω | grad u = 0} = {z ∈ Ω | ux (z) = uy (z) = 0}
and f = ux − iuy .
Since ux and uy are continuous in Ω, V is closed in Ω. By Theorem 11.2, f ∈ H(Ω). It is trivial
that V = Z(f ). By Theorem 10.18, either V = Ω or V has no limit point in Ω.a This ends the
analysis of the problem.
a
Note that the case V = Ω implies that u is constant in Ω.
36
Chapter 11. Harmonic Functions
Problem 11.4
Rudin Chapter 11 Exercise 4.
Proof. We prove the assertions one by one.
• Every partial derivative of a harmonic is harmonic. See [76, Problem 16.2, p. 199].
• Pr (θ − t) is a harmonic function of reiθ for a fixed t. It can be shown easily that the
Laplacian equation in polar form is given by
1
1
∆u = urr + ur + 2 uθθ .
r
r
(11.9)
Fix t. Write P = Pr (θ − t) and A = A(r, θ) = 1 − 2r cos(θ − t) + r 2 for convenience. Then
[62, Eqn. (2), §11.5, p. 233] can be written as P A = 1 − r 2 . Direct differentiation gives
Pr
2[r − cos(θ − t)]
2
=−
P− ,
r
rA
A
4[r − cos(θ − t)]
2
2
Prr = −
Pr − P − ,
A
A
A
8 sin2 (θ − t)
2 cos(θ − t)
Pθθ
=
P−
P.
r2
A2
rA
(11.10)
Put these equations (11.10) into the Laplacian equation (11.9) to get
4[r − cos(θ − t)]
2
2
2[r − cos(θ − t)]
2
Pr − P − −
P−
A
A
A
rA
A
2 cos(θ − t)
8 sin2 (θ − t)
P−
P
+
A2
rA
8r[r − cos(θ − t)]
4
4
8 sin2 (θ − t)
8[r − cos(θ − t)]2
P
+
−
P
−
+
P
=
A2
A2
A
A
A2
8r[r − cos(θ − t)] − 4A
8[r − cos(θ − t)]2 − 4A + 8 sin2 (θ − t)
P+
=
2
A
A2
2
2
2
8[r − 2r cos(θ − t) + 1] − 4A
4[r + r − 2r cos(θ − t)] − 4A
=
P+
2
A
A2
2
4(r + A − 1) − 4A
4P
+
=
A
A2
2
4(1 − r ) 4(r 2 − 1)
=
+
A2
A2
= 0.
∆P = −
Hence Pr (θ − t) is a harmonic function of reiθ for a fixed t.
• P [ dµ] is harmonic in U . Suppose that µ is a finite Borel measure on T and u = P [ dµ].
It suffices to show that
Z
Z
Z
∂
∂
Pr (θ − t) dµ(eit ) =
Pr (θ − t) dµ(eit ) = [Pr (θ − t)]r dµ(eit ) (11.11)
ur =
∂r T
T ∂r
T
and
∂
uθ =
∂θ
Z
T
it
Pr (θ − t) dµ(e ) =
Z
T
∂
Pr (θ − t) dµ(eit ) =
∂θ
Z
T
[Pr (θ − t)]θ dµ(eit ) (11.12)
11.1. Basic Properties of Harmonic Functions
37
because they certainly yield that
Z
Z
it
urr = [Pr (θ − t)]rr dµ(e ) and uθθ = [Pr (θ − t)]θθ dµ(eit )
T
T
so that
∆u =
Z
T
∆Pr (θ − t) dµ(eit ) = 0.
In other words, u = P [ dµ] is harmonic in U .
Fix r ∈ [0, 1). For very small h > 0 such that 1 − r − 2h > 0, we note that
Z
u(r + h, θ) − u(r, θ)
Pr+h (θ − t) − Pr (θ − t)
=
dµ(eit )
h
h
T
Z
Pr+h (θ − t) − Pr (θ − t)
ur = lim
dµ(eit ).
h→0 T
h
We observe that
lim
h→0
(11.13)
∂
Pr+h (θ − t) − Pr (θ − t)
=
Pr (θ − t)
h
∂r
and
Pr+h (θ − t) − Pr (θ − t)
h
i
1 − r2
1 − (r + h)2
1 h
−
= ·
h 1 − 2(r + h) cos(θ − t) + (r + h)2 1 − 2r cos(θ − t) + r 2
[1 − (r + h)2 ][1 − 2r cos(θ − t) + r 2 ] − (1 − r 2 )[1 − 2(r + h) cos(θ − t) + (r + h)2 ]
=
h[1 − 2(r + h) cos(θ − t) + (r + h)2 ] · [1 − 2r cos(θ − t) + r 2 ]
2h cos(θ − t) + 2r 2 − 2(r + h)2 + 2hr(r + h) cos(θ − t)
=
h[1 − 2(r + h) cos(θ − t) + (r + h)2 ][1 − 2r cos(θ − t) + r 2 ]
2h cos(θ − t) − 2h(2r + h) + 2hr(r + h) cos(θ − t)
=
h[1 − 2(r + h) cos(θ − t) + (r + h)2 ] · [1 − 2r cos(θ − t) + r 2 ]
2 cos(θ − t) − 2(2r + h) + 2r(r + h) cos(θ − t)
=
.
(11.14)
[1 − 2(r + h) cos(θ − t) + (r + h)2 ] · [1 − 2r cos(θ − t) + r 2 ]
Since 1 − r − 2h > 0 implies 1 − (r + h) >
1−r
2 ,
this and the expression (11.14) give
4(8 + 4h)
40
2 + 2(2 + h) + 2(1 + h)
Pr+h (θ − t) − Pr (θ − t)
≤
≤
≤
h
(1 − r)2 [1 − (r + h)]2
(1 − r)4
(1 − r)4
for every eit ∈ T . Since µ is a finite Borel measure on T , we have
Z
40
40
dµ(eit ) =
· µ(T ) < ∞
4
(1
−
r)
(1
−
r)4
T
40
1
Hence Theorem 1.34 (The Lebesgue’s Dominated
which means that (1−r)
4 ∈ L (T ).
Convergence Theorem) ensures that the order of integration and the limit in (11.13) can
be changed and this action shows that the formula (11.11) holds. Similarly, it is easily to
check that
−4r(1 − r 2 ) sin(θ − t + h2 ) sin h2
Pr (θ + h − t) − Pr (θ − t)
=
h
h[1 − 2r cos(θ + h − t) + r 2 ][1 − 2r cos(θ − t) + r 2 ]
≤
2r(1 − r 2 ) sin h2
· h
(1 − r)4
2
38
Chapter 11. Harmonic Functions
3
4
·
(1 − r)3 2
6
.
=
(1 − r)3
≤
6
1
Again, the finiteness of µ implies that (1−r)
3 ∈ L (T ) and we can apply Theorem 1.34
(The Lebesgue’s Dominated Convergence Theorem) to conclude that the formula (11.12)
holds.
This ends the proof of the problem.
Problem 11.5
Rudin Chapter 11 Exercise 5.
Proof. Let f = u + iv. Since f ∈ H(Ω), both u and v are harmonic in Ω. Since |f | =
6 0 in Ω,
1
2
2
2
log |f | is a well-defined real function and in fact log |f | = log(u + v ) . Using ∆u = ∆v = 0
and the Cauchy-Riemann equations, we get
1
1
∆(log |f |) = [log(u2 + v 2 ) 2 ]xx + [log(u2 + v 2 ) 2 ]yy
1
=
log(u2 + v 2 ) xx + log(u2 + v 2 ) yy
2
uu + vv uu + vv y
y
x
x
=
+
u2 + v 2 x
u2 + v 2 y
(u2 + v 2 )(u∆u + u2x + u2y + v∆v + vx2 + vy2 ) − 2(uux + vvx )2 − 2(uuy + vvy )2
=
(u2 + v 2 )2
2
2
2
2
2
2
(u + v )(ux + uy + vx + vy ) − 2[u2 (u2x + u2y ) + 2uv(ux vx + uy vy ) + v 2 (vx2 + vy2 )]
=
(u2 + v 2 )2
2
2
2
2
2
2
2
2(u + v )(vx + vy ) − 2(u + v )(vx + vy2 )
=
(u2 + v 2 )2
=0
in Ω. Hence log |f | is harmonic in Ω.
If f is holomorphic and non-vanishing in Ω, then
(log f )′
f′
f ,
1
′
f,f
∈ H(Ω). Thus we obtain
f′
f
∈ H(Ω).
we conclude log f ∈ H(Ω) so that its real part, which is log |f |, is harmonic
Since
=
in Ω, completing the proof of the problem.
Problem 11.6
Rudin Chapter 11 Exercise 6.
Proof. Let A(Ω) be the area of the region Ω = f (U ). Referring to [2, §2.4, pp. 75, 76],b we
know that
ZZ
|f ′ (z)|2 dx dy,
A(Ω) =
U
b
See also the discussion on [59, Eqn. (6), p. 11].
11.1. Basic Properties of Harmonic Functions
39
where f (z) = u(x, y) + iv(x, y). Put z = reiθ , where 0 ≤ r < 1 and 0 ≤ θ ≤ 2π. Then the
formula of A(Ω) becomes
Z
A(Ω) =
Z
2π
0
1
0
|f ′ (r cos θ, r sin θ)|2 r dr dθ.
(11.15)
Since |f ′ (z)|2 = f ′ (z) · f ′ (z), we see that
|f ′ (r cos θ, r sin θ)|2 =
=
∞
X
mcm r m−1 e−i(m−1)θ
ncn r n−1 ei(n−1)θ ×
∞
X
m=1
n=1
∞
X
nmcn cm r n+m−2 ei(n−m)θ .
m,n=1
Recall from Definition 4.23 that {einθ } forms an orthonormal set, the integral (11.15) becomes
A(Ω) =
Z
2π
0
′
Since f (z) =
∞
X
n=1
Z
0
∞
1X
n=1
2
2 2n−1
n |cn | r
dr dθ.
(11.16)
ncn z n−1 ∈ H(U ), we have
lim sup
n→∞
which implies that
lim sup
n→∞
p
n
n|cn | = 1
p
n
n2 |cn |2 = 1.
Therefore, the radius of convergence of the series
∞
X
n=1
n2 |cn |2 r 2n−1 is 1 and then the order of
integration in the integral (11.16) can be switched so that
A(Ω) =
Z
0
∞ Z 1
2π X
n=1
0
∞
∞
X
X
1
n2 |cn |2 ×
n2 |cn |2 r 2n−1 dr dθ = 2π
n|cn |2 ,
=π
2n
n=1
n=1
as required. This completes the proof of the problem.
Problem 11.7
Rudin Chapter 11 Exercise 7.
Proof. Suppose that f = u + iv.
(a) Let ψ be a twice differentiable function on (0, ∞). Then f f = u2 + v 2 = |f |2 . Using the
formulas [62, Eqn. (3), p. 231], we see that
∂ ∂
∂ 1 ∂
ψ(|f |2 )
−i
+i
4 ∂x
∂y ∂x
∂y
1 ∂
∂ ′
=
ψ (|f |2 ) · (uux + vvx ) + iψ ′ (|f |2 ) · (uuy + vvy )
−i
2 ∂x
∂y
1 n ′′
2ψ (|f |2 ) · (uux + vvx )2 + ψ ′ (|f |2 ) · (u2x + uuxx + vx2 + vvxx )
=
2
∂∂[ψ ◦ (f f )] =
40
Chapter 11. Harmonic Functions
+ iψ ′′ (|f |2 ) · (uux + vvx )(uuy + vvy )
+ iψ ′ (|f |2 ) · (ux uy + uuxy + vx vy + vvxy )
+ −iψ ′′ (|f |2 ) · (uuy + vvy )(uux + vvx )
−iψ ′ (|f |2 )(uy ux + uuxy + vx vy + vvxy )
o
+ 2ψ ′′ (|f |2 ) · (uuy + vvy )2 + ψ ′ (|f |2 ) · (u2y + uuyy + vy2 + vvyy ) . (11.17)
Since f ∈ H(Ω), both u and v are harmonic in Ω. Furthermore, we note from Theorem
11.2 that |f ′ |2 = u2x + vx2 = u2y + vy2 , so the expression (11.17) becomes
∂∂[ψ ◦ (f f )] =
1 n ′′
2ψ (|f |2 ) · u2 (u2x + u2y ) + v 2 (vx2 + vy2 )
2
o
+ ψ ′ (|f |2 ) · u2x + u2y + vx2 + vy2 + u(uxx + uyy ) + v(vxx + vyy )
1 ′′
2ψ (|f |2 ) · |f |2 · |f ′ |2 + 2ψ ′ (|f |2 ) · |f ′ |2
2
= [|f |2 ψ ′′ (|f |2 ) + ψ ′ (|f |2 )] · |f ′ |2
=
= (ϕ ◦ |f |2 ) · |f ′ |2
(11.18)
as required.
α
Now the function ψ(t) = t 2 is clearly twice differentiable on (0, ∞) with
ϕ(t) =
α
αα
α α
α2 α −1
− 1 t 2 −1 + t 2 −1 =
t2 ,
2 2
2
4
so we combine the formula (11.18) and [62, Eqn. (3), p. 232] to get
∆(|f |α ) = 4∂∂(|f |α ) = α2 |f |α−2 |f ′ |2 .
(b) Suppose that Φ : f (Ω) → C is defined by Φ(f ) = Φ(u + iv) = Φ(u(x, y), v(x, y)). Since
uxx + uyy = vxx + vyy = 0, we have
∆[Φ ◦ f ] = ∆Φ(f )
∂2
∂2
Φ(u,
v)
+
Φ(u, v)
∂x2
∂y 2
∂
∂
(Φu · ux + Φv · vx ) +
(Φu · uy + Φv · vy )
=
∂x
∂y
∂
∂
= ux Φu + Φu · uxx + vx Φv + Φv · vxx
∂x
∂x
∂
∂
+ uy Φu + Φu · uyy + vy Φv + Φv · vyy
∂y
∂y
= ux (Φuu · ux + Φuv · vx ) + Φu · uxx + vx (Φuv · ux + Φvv · vx ) + Φv · vxx
=
+ uy (Φuu · uy + Φuv · vy ) + Φu · uyy + vy (Φuv · uy + Φvv · vy ) + Φv · vyy
= Φuu · (u2x + u2y ) + Φvv · (vx2 + vy2 ) + 2Φuv · ux vx + 2Φuv · uy vy .
(11.19)
According to the Cauchy-Riemann equations, we can further reduce the expression (11.19)
to
∆[Φ ◦ f ] = Φuu · (u2x + vx2 ) + Φvv · (vx2 + u2x )
= (Φuu + Φvv ) · |f ′ |2
= [(∆Φ) ◦ f ] · |f ′ |2
(11.20)
11.1. Basic Properties of Harmonic Functions
41
as desired.
Finally, we take Φ(w) = |w| so that Φ(w) = Φ(|w|). Suppose that f = u + iv and
α
w = |f |α = (u2 + v 2 ) 2 . Direct differentiation implies that
α
α
α
Φuu + Φvv = 2α(u2 + v 2 ) 2 −1 + α(α − 2)u2 (u2 + v 2 ) 2 −2 + α(α − 2)v 2 (u2 + v 2 ) 2 −2
α
= α2 (u2 + v 2 ) 2 −1
and thus
(∆Φ) ◦ f = α2 |f |α−2 .
Substituting this into the formula (11.20), we have established that
∆(|f |α ) = α2 |f |α−2 · |f ′ |2 .
This ends the analysis of the problem.
Problem 11.8
Rudin Chapter 11 Exercise 8.
Proof. The proof of this problem will be divided into two steps as follows:
• Step 1: {fn (z)} converges at every point of Ω. Suppose that
S1 = {z ∈ Ω | {fn (z)} converges}
Let a ∈ Ω and R > 0 be such that
D(a; 2R) ⊆ Ω.
(11.21)
We consider the functions u
cn (z) = un (z + a) and c
fn (z) = fn (z + a) which are defined in
D(0; 2R) and in D(0; 2R) respectively. By [9, Theorem 16.9, p. 233]c , we have
Z π
2Reit + z 1
c
c
u
cn (2Reit ) dt
fn (z) = iIm fn (0) +
2π −π 2Reit − z
and thus
fn (z + a) = iIm fn (a) +
1
2π
Z
2Reit + z un (a + 2Reit ) dt,
it − z
2Re
−π
π
where z ∈ D(0; 2R). If we take z = Reiθ , then it is easy to see that
2Reit + z
2Reit + Reiθ
≤3
=
2Reit − z
2Reit − Reiθ
and this implies that
|fn (z + a) − fm (z + a)| ≤ Im [fn (a) − fm (a)]
Z π 2Reit + z 1
[un (a + 2Reit ) − um (a + 2Reit )] dt
+
2π −π 2Reit − z
Z π
3
≤ |fn (a) − fm (a)| +
un (a + 2Reit ) − um (a + 2Reit ) dt
2π −π
c
In fact, this is a generalization of Theorem 11.9 and it is called the Schwarz Integral Formula. See, for
example, [37, p. 408].
42
Chapter 11. Harmonic Functions
≤ |fn (a) − fm (a)| + 3 · max
−π≤t≤π
un (a + 2Reit ) − um (a + 2Reit ) .
Consequently, we conclude that
sup
z∈D(0;R)
|fn (z + a) − fm (z + a)| ≤ 3 · max
−π≤t≤π
un (a + 2Reit ) − um (a + 2Reit )
+ |fn (a) − fm (a)|
or equivalently,
sup
z∈D(a;R)
|fn (z) − fm (z)| ≤ 3 ·
max
ζ∈C(a;2R)
|un (ζ) − um (ζ)| + |fn (a) − fm (a)|.
(11.22)
Given ǫ > 0. Now our hypothesis asserts that there is an N1 ∈ N such that n, m ≥ N1
imply
ǫ
|un (ζ) − um (ζ)| <
6
for all ζ ∈ D(a; 2R). Thus it follows from the inequality (11.22) that
|fn (z) − fm (z)| ≤ |fn (a) − fm (a)| +
ǫ
2
(11.23)
for all z ∈ D(a; R).
Particularly, if a ∈ S1 , then there exists an N2 ∈ N such that n, m ≥ N2 imply
|fn (a) − fm (a)| <
ǫ
2
so that the inequality (11.23) gives
|fn (z) − fm (z)| < ǫ
(11.24)
for all z ∈ D(a; R) and all n, m ≥ N = max(N1 , N2 ). By [61, Theorem 7.8, p. 147], the
sequence {fn } converges uniformly in D(a; R) and this definitely implies
D(a; R) ⊆ S1 .
(11.25)
To finish the proof that S1 = Ω, we need a result from [15, Theorem 1.28, p. 28]:
Lemma 11.1 (The Basic Connectedness Lemma)
Suppose that Ω is open in C and A is a nonempty subset of Ω. If there exists
a θ > 0 such that for every a ∈ A and D(a; 2R) ⊆ Ω for some R > 0, we have
D(a; 2θR) ⊆ A, then A is relatively open and closed in Ω. Furthermore, if Ω is a
region, then we have
A = Ω.
Put A = S1 , the above set relations (11.21) and (11.25) mean that we can take θ = 21
in Lemma 11.1 (The Basic Connectedness Lemma). Hence we conclude immediately that
S1 = Ω, as desired.
• Step 2: {fn } converges uniformly on compact subsets of Ω. Let K be a compact
subset of Ω. Clearly,
[
K⊆
D(a; Ra ),
a∈K
11.1. Basic Properties of Harmonic Functions
43
where Ra > 0 is a number satisfying the set relation (11.21). Therefore, it follows from
Step 1 that the inequality (11.24) holds in D(a; Ra ) and all n, m ≥ N (a, Ra ) (of course,
the positive integer N (a, Ra ) depends on a and Ra ) for every a ∈ K. Since K is compact,
there exists a finite set {a1 , a2 , . . . , ap } ⊆ K such that
K⊆
p
[
j=1
D(aj ; Rj ) ⊆ Ω,
where Rj ∈ {Ra | a ∈ K} for 1 ≤ j ≤ p. In particular, if we take N = max N (aj , Rj ),
1≤j≤p
then it is true that n, m ≥ N imply
|fn (z) − fm (z)| < ǫ
for all z ∈ K. Using [61, Theorem 7.8, p. 147] again, the sequence {fn } converges
uniformly on K.
This completes the analysis of the problem.
Problem 11.9
Rudin Chapter 11 Exercise 9.
Proof. Let D(a; r) ⊆ Ω. Since u is locally in L1 , the double integral considered in the question
is well-defined. If u is harmonic in Ω, then Definition 11.12 says that it satisfies
Z π
1
u(a) =
u(a + reiθ ) dθ
(11.26)
2π −π
for every r > 0 with D(a; r) ⊆ Ω. Multiplying both sides of the expression (11.26) by ρ and
integrating from 0 to r, we obtain
Z r
r2
u(a)ρ dρ
u(a) =
2
Z0 r Z π
1
u(a + ρeiθ ) dθ ρ dρ
=
2π
0
Z r Z π−π
1
=
u(a + ρeiθ )ρ dθ dρ
2π 0 −π
which gives exactly
1
u(a) = 2
πr
Z
0
r
Z
π
1
u(a + ρe )ρ dθ dρ = 2
πr
−π
iθ
ZZ
u(x, y) dx dy.
(11.27)
D(a;r)
Conversely, suppose that the formula (11.27) holds for any D(a; r) ⊆ Ω. On the one hand,
applying the polar coordinates, we can write
Z rZ π
ZZ
u(x, y) dx dy =
u(a + ρeiθ )ρ dθ dρ.
(11.28)
0
D(a;r)
On the other hand, we have
2
r u(a) = 2
−π
Z
r
ρu(a) dρ.
0
(11.29)
44
Chapter 11. Harmonic Functions
Substituting the expressions (11.28) and (11.29) into the formula (11.27), we get
Z r
Z r
ρu(a) dρ
ρ dρ =
u(a)
0
0
Z rZ π
1
=
u(a + ρeiθ )ρ dθ dρ
2π 0 −π
Z r
1
=
U (a, ρ, θ)ρ dρ,
2π 0
where
U (a, ρ, θ) =
Z
π
(11.30)
(11.31)
u(a + ρeiθ ) dθ.
−π
As the integral on the left-hand side in the formula (11.30) is differentiable with respect to r,
so is the integral (11.31). Consequently, we derive from the First Fundamental Theorem of
Calculus [79, p. 161] that
Z π
r
1
· U (a, r, θ)r =
u(a + reiθ ) dθ.
(11.32)
ru(a) =
2π
2π −π
After cancelling the r in the expression (11.32), we find that
Z π
1
u(a) =
u(a + reiθ ) dθ
2π −π
whenever D(a; r) ⊆ Ω.
To finish the proof, we have to show that u is continuous in Ω. To this end, fix z ∈ Ω. Given
that ǫ > 0. Since u ∈ L1loc (Ω),d we must have u ∈ L1 (D(z; 2r ′ )) for some r ′ > 0 such that
D(z; 2r ′ ) ⊆ Ω. We fix this r ′ . By Problem 1.12, there exists a δz > 0 such that
Z
|u| dm < π(r ′ )2 ǫ
(11.33)
E
whenever m(E) < δz and E ⊆ D(z; 2r ′ ). Clearly, we may assume that δz < r ′ . For every
ω ∈ D(z; r ′ ), it is always true that D(ω; r ′ ) ⊆ D(z; 2r ′ ) and
m D(ω; r ′ ) \ D(z; r ′ ) = m D(z; r ′ ) \ D(ω; r ′ ) < δz
if ω is very close to z. Therefore, we follow from the formula (11.27) and the inequality (11.33)
that
ZZ
ZZ
1
u(x, y) dx dy
u(x, y) dx dy −
|u(z) − u(ω)| =
π(r ′ )2
D(ω;r ′ )
D(z;r ′ )
=
1
π(r ′ )2
1
=
2π(r ′ )2
1
≤
2π(r ′ )2
<
d
Z
D(z;r ′ )
Z
u dm −
D(z;r ′ )\D(ω;r ′ )
"Z
Z
u dm
D(ω;r ′ )
u dm −
D(z;r ′ )\D(ω;r ′ )
Z
|u| dm +
1 ′ 2
π(r ) ǫ + π(r ′ )2 ǫ
′
2
2π(r )
u dm
D(ω;r ′ )\D(z;r ′ )
Z
D(ω;r ′ )\D(z;r ′ )
The notation L1loc (Ω) is the set of all locally integrable functions on Ω.
|u| dm
#
11.1. Basic Properties of Harmonic Functions
45
= ǫ.
By the definition, u is continuous at z. Since z is arbitrary, u is actually continuous in Ω. Finally,
Theorem 11.13 asserts that u is harmonic in Ω, completing the proof of the problem.
Problem 11.10
Rudin Chapter 11 Exercise 10.
Proof.
• By the definition, we have
Z b
1
1
1
f (x + iǫ) − f (x − iǫ) =
dt
−
ϕ(t)
2πi a
t − x − iǫ t − x + iǫ
Z b
ǫ
ϕ(t)
=
·
dt
π
(x
−
t)2 + ǫ2
a
Z b
Pǫ (x − t)ϕ(t) dt
=
a
Z ∞
ϕ(x − t)Pǫ (t) dt,
=
(11.34)
−∞
where
Pǫ (t) =
ǫ
1
· 2
π t + ǫ2
relates to the formula [62, Eqn. (3), §9.7, p. 183] (in fact, it is the Poisson kernel for the
upper half-plane, see [66, p. 149]) and ϕ(t) = 0 if t ∈ R \ I. Using the convolution notation
introduced in Theorem 8.14, the expression (11.34) becomes
f (x + iǫ) − f (x − iǫ) = (ϕ ∗ Pǫ )(x).
Since |ϕ(x)| ≤ max |ϕ(t)| for every x ∈ R, we get ϕ ∈ L∞ (R).
t∈I
To proceed, we need to modify Theorem 9.9 and givese
Lemma 11.2
If g ∈ L∞ , then we have
lim (g ◦ hλ )(x) =
λ→0
g(x+) + g(x−)
,
2
where hλ is the formula [62, Eqn. (3), §9.7, p. 183].
e
If g is continuous at x, then g(x+) = g(x−) and it is exactly Theorem 9.9.
(11.35)
46
Chapter 11. Harmonic Functions
Proof of Lemma 11.2. Following the proof of Theorem 9.9, we have
Z 0 h
g(x+) + g(x−) i
g(x+) + g(x−)
g(x − λs) −
(g ∗ hλ )(x) −
h1 (s) dm(s)
=
2
2
−∞
Z ∞h
g(x+) + g(x−) i
g(x − λs) −
+
h1 (s) dm(s).
2
0
Since the integrands are dominated by 2kgk∞ h1 (s) and the integrals converge pointwise
for every s as λ → 0, we deduce from Theorem 1.34 (The Lebesgue’s Dominated
Convergence Theorem) that
h
g(x+) + g(x−) i
lim (g ∗ hλ )(x) −
λ→0
2
Z 0 h
g(x+) + g(x−) i
h1 (s) dm(s)
g(x − λs) −
= lim
λ→0 −∞
2
Z ∞h
g(x+) + g(x−) i
g(x − λs) −
+ lim
h1 (s) dm(s)
λ→0 0
2
Z 0
h
g(x+) + g(x−) i
lim g(x − λs) −
=
h1 (s) dm(s)
2
−∞ λ→0
Z ∞
h
g(x+) + g(x−) i
h1 (s) dm(s)
lim g(x − λs) −
+
2
0 λ→0
Z 0 h
g(x+) + g(x−) i
g(x+) −
=
h1 (s) dm(s)
2
−∞
Z ∞h
g(x+) + g(x−) i
g(x−) −
+
h1 (s) dm(s)
2
0
Z ∞
Z 0
g(x−) − g(x+)
g(x+) − g(x−)
h1 (s) dm(s) +
h1 (s) dm(s)
=
2
2
0
−∞
Z
Z
g(x+) − g(x−) ∞
g(x−) − g(x+) ∞
=
h1 (s) dm(s) +
h1 (s) dm(s)
2
2
0
0
=0
which implies the desired result. We complete the proof of Lemma 11.2.
By Lemma 11.2, we conclude at once that
lim [f (x + iǫ) − f (x − iǫ)] =
ǫ→0
ǫ>0
ϕ(x+) + ϕ(x−)
2
for every x ∈ R. Hence the formula (11.36) asserts that

ϕ(x),








0,




lim [f (x + iǫ) − f (x − iǫ)] =
ϕ(a+)
ǫ→0

,

ǫ>0


2







 ϕ(b−) ,
2
(11.36)
if x ∈ (a, b);
if x ∈ R \ I;
if x = a;
if x = b.
• The case when ϕ ∈ L1 . In this case, we apply Theorem 9.10 to the expression (11.35)
11.1. Basic Properties of Harmonic Functions
47
to get
lim
Z
∞
ǫ→0 −∞
ǫ>0
|f (x + iǫ) − f (x − iǫ) − ϕ(x)| dm(x) = lim kϕ ∗ Pǫ − ϕk1 = 0.
ǫ→0
ǫ>0
(11.37)
Denote f (x + iǫ) = fǫ+ (x) and f (x − iǫ) = fǫ− (x). Then the result (11.37) means that
lim kfǫ+ − fǫ− k1 = kϕk1 .
ǫ→0
ǫ>0
• The case when ϕ(x+) and ϕ(x−) exist at x. This case has been settled already in the
formula (11.36).
We have completed the proof of the problem.
Remark 11.1
The integral considered in Problem 11.10 is an example of the so-called Cauchy type
integrals. For more details of this subject, please refer to Muskhelishvili’s book [44].
Problem 11.11
Rudin Chapter 11 Exercise 11.
Proof. We note that the following proof uses only the techniques from Chapter 10, not from
Chapter 11. Actually, the author admits that he is not able to apply the theory of harmonic
functions to prove this problem.
• By Theorem 10.17 (Morera’s Theorem), it suffices to prove that
Z
f (z) dz = 0
(11.38)
∂∆
for every closed triangle ∆ ⊆ Ω. There are three cases.
– Case (i): ∆ ∩ I = ∅. Then ∆ ⊆ Ω \ I. Since Ind ∂∆ (x) = 0 for every x ∈ I, Theorem
10.35 (Cauchy’s Theorem) implies that the result (11.38) holds trivially.
– Case (ii): ∂∆ has a side lying on I. Since ∆ is compact, C \ Ω is closed in C
and ∆ ∩ (C \ Ω) = ∅, Problem 10.1 ensures that there exists a δ1 > 0 such that
d(∆, C \ Ω) = δ1 > 0. Therefore, it is true that
∆n = ∆ +
i
⊆Ω
n
(11.39)
for all n > δ11 . Thus we have ∆n ∩ I = ∅ so that the result (11.38) holds for every
∂∆n with n > δ11 . Suppose that ∂∆ = [a, b]+[b, c]+[c, a], where a, b and c are vertices
of the triangle ∆. Then we have
i i
i + [b, c] +
+ [c, a] +
.
∂∆n = [a, b] +
n
n
n
Using [62, Eqn. (4), p. 202], we know that
Z
Z
f (z) dz = (b − a)
[a,b]+ ni
1
0
i
f a + + (b − a)t dt
n
48
Chapter 11. Harmonic Functions
so that
Z
[a,b]
f (z) dz −
= |b − a| ·
≤ |b − a| ·
Z
0
1
Z
1h
0
Z
[a,b]+ ni
f (z) dz
i
i
f a + + (b − a)t − f a + (b − a)t dt
n
i
f a + + (b − a)t − f a + (b − a)t dt.
n
Since f is continuous on Ω, it is uniformly continuous on any compact subset of Ω.
Given ǫ > 0, there exists a δ2 > 0 such that n > δ12 implies
i
ǫ
f a + + (b − a)t − f a + (b − a)t <
n
3|b − a|
and then
Z
[a,b]
f (z) dz −
Z
[a,b]+ ni
f (z) dz <
ǫ
.
3
(11.40)
Thus, if n > max( δ11 , δ12 ), then both the conditions (11.39) and (11.40) hold simultaneously. Since the inequality (11.40) also holds for [b, c] and [c, a] for large enough n,
we obtain immediately that
Z
f (z) dz < ǫ
∂∆n
which means
Z
f (z) dz = lim
∂∆
Z
n→∞ ∂∆
n
f (z) dz = 0.
– Case (iii): ∆ intersects with I at only two points. Then I divides ∆ into a
triangle and a quadrangle or another triangle. If it is a quadrangle, then it can be
further divided into two triangles, see Figure 11.1 for an illustration. Since ∂∆ is a
sum of two or three boundaries of triangles, it follows from Case (i) and Case (ii)
that our result (11.38) remains true in this case.
(a) ∂∆ = ∂∆1 + ∂∆2 .
(b) ∂∆ = ∂∆1 + ∂∆2 + ∂∆3 .
Figure 11.1: The I divides ∂∆ into several triangles.
11.2. Harnack’s Inequalities and Positive Harmonic Functions
49
• Removable sets for holomorphic functions of class C (Ω). Let N ∈ N. It is clear
that the above argument can be applied to the compact set in the form
K = I1 ∪ I2 ∪ · · · ∪ IN
and f ∈ H(Ω \ K), where each Ij = [aj , bj ] is a subset of Ω for 1 ≤ j ≤ N .
We have completed the analysis of the problem.f
Remark 11.2
(a) Classically, Problem 11.11 is a topic of the so-called removable sets for holomorphic
functions. To say it more precisely, let F be a class of functions from Ω to C. Then
a compact set K ⊆ Ω is said to be removable for holomorphic functions of class F if
every f ∈ F such that f ∈ H(Ω \ E) can be extended to a holomorphic function in Ω.
Examples of F are L∞ (Ω), C (Ω) and Lip α (Ω), where 0 < α ≤ 1.
(b) It is clear that Theorem 10.20 is a positive result for bounded and holomorphic functions. Because of this, Painlevé was motivated and studied some more general problems: “Which subsets of C are removable? What geometric characterization(s) must
these subsets satisfy?” In fact, Painlevé proved a sufficient condition that if a compact
set K ⊆ Ω is of one-dimensional Hausdorff measure [12, pp. 215, 216], then it is
removable for bounded and holomorphic functions in Ω \ K.
(c) For the class Lip α (Ω) with 0 < α < 1, Dolženko [19] has shown in 1963 that a compact
set K is removable for holomorphic functions of this class if and only if the (1 + α)dimensional Hausdorff measure is zero. For the remaining case α = 1, Uy [74] verified
in 1979 that K is removable if and only if m(K) = 0.
(d) Besides the approach of Hausdorff measure, Ahlfors [1] introduced the analytic capacity (a purely complex-analytic concept) of a compact set K to study removable
compact sets for holomorphic functions of class L∞ . In fact, he proved that K is
removable for bounded analytic functions if and only if its analytic capacity vanishes.
11.2
Harnack’s Inequalities and Positive Harmonic Functions
Problem 11.12
Rudin Chapter 11 Exercise 12.
Proof.
• Proof of Harnack’s Inequalities. We first prove the special case that if u is harmonic
in D(a; R) and u > 0 in D(a; R), then for every 0 ≤ r < R and z = a + reiθ ∈ D(a; R), we
have
R+r
R−r
u(a) ≤ u(z) ≤
u(a).
(11.41)
R+r
R−r
f
See also Problem 16.10.
50
Chapter 11. Harmonic Functions
To see this, choose ρ > 0 such that r < ρ < R. Using the second set of inequalities on [62,
p. 236], we know that
ρ−r
ρ+r
u(a) ≤ u(z) = u(a + reiθ ) ≤
u(a)
ρ+r
ρ−r
(11.42)
for every θ ∈ R. Letting ρ → R in the inequalities (11.42), we obtain the desired results
(11.41).
Since Ω is a region and K ⊆ Ω, C \ Ω is closed in C and (C \ Ω) ∩ K = ∅. By Problem 10.1,
there exists a δ > 0 (depending on K and Ω) such that d(C \ Ω, K) = 2δ > 0. Clearly, we
have
[
[
K⊆
D(z; δz ) ⊆
D(z; δ) ⊆ Ω,
z∈K
z∈K
where 0 < δz < δ. Since K is compact, one can find a finite set {z1 , z2 , . . . , zm } with
positive numbers δ1 , δ2 , . . . , δm such that
K⊆
m
[
k=1
D(zk ; δk ) ⊆
m
[
k=1
D(zk ; δ) ⊆ Ω.
Now we apply the special case (11.41) to each disc D(zk ; δ) and consider only points
z ∈ D(zk ; δk ) to get
δ − δk
δ−r
δ+r
δ + δk
u(zk ) ≤
u(zk ) ≤ u(z) ≤
u(zk ) ≤
u(zk ).
δ + δk
δ+r
δ−r
δ − δk
Take positive numbers α and β such thatg
α=
δ − δk
1
· min
u(zk )
u(z0 ) 1≤k≤m δ + δk
and β =
1
δ + δk
· max
u(zk ).
u(z0 ) 1≤k≤m δ − δk
Then we establish
for every z ∈
m
[
k=1
αu(z0 ) ≤ u(z) ≤ βu(z0 )
(11.43)
D(zk ; δk ). In particular, the inequalities (11.43) are true for all z ∈ K.
• The behavior of {un } in Ω \ {z0 } if un (z0 ) → 0. Let un (z) → u(z) for every z ∈ Ω and
a ∈ Ω \{z0 }. Since Ω \{z0 } is open in C, there exists a R > 0 such that D(a; R) ⊆ Ω \{z0 }.
Obviously, D(a; R) is a compact subset of Ω. By the inequalities (11.43), we have
αun (z0 ) ≤ un (z) ≤ βun (z0 )
(11.44)
for every z ∈ D(a; R) and n = 1, 2, . . ., where α and β depend on z0 , K and Ω only. Take
n → ∞ in the inequalities (11.44) and then use the hypothesis, we get
u(z) = 0
(11.45)
for all z ∈ D(a; R). Particularly, u(a) = 0. Since a is arbitrary, we conclude that u ≡ 0 in
Ω \ {z0 }. Finally, the continuity of u ensures that u(z0 ) = 0 and then u ≡ 0 in Ω.
• The behavior of {un } in Ω \ {z0 } if un (z0 ) → ∞. Instead of the result (11.45), we
obtain
u(z) = ∞
for all z ∈ D(a; R). Hence, using similar argument as the previous assertion, we conclude
that u(z) = ∞ in Ω.
g
Since δ depends on K and Ω, α and β trivially depend on z0 , K and Ω.
11.2. Harnack’s Inequalities and Positive Harmonic Functions
51
• The positivity of {un } is essential. For each n ∈ N, we consider u(x, y) = nx − ny in C.
Then it is easily checked that ∆un = 0 so that un is harmonic in C. As un (0, 1) = − n1 < 0
and un (1, 0) = n > 0, each un is neither positive nor negative. Furthermore, un (0, 1) → 0
but un (1, 0) → ∞ as n → ∞. Hence this counterexample shows that the positivity of {un }
cannot be omitted for these results.
We have completed the analysis of the problem.
Problem 11.13
Rudin Chapter 11 Exercise 13.
Proof. By the Poisson formula, we have
Z π
1
1 − r2
u(reiθ ) =
u(eit ) dt,
2π −π 1 − 2r cos(θ − t) + r 2
where 0 ≤ r < 1 and −π ≤ θ ≤ π. Pick r =
u( 12 )
Since
3
5+4
≤
3
5−4 cos t
1
=
2π
≤
Z
3
5−4
π
−π
1
2
(11.46)
and θ = 0 in the formula (11.46) to get
1 − 41
1
it
1 u(e ) dt = 2π
1 − cos t + 4
Z
π
−π
3
u(eit ) dt.
5 − 4 cos t
(11.47)
on [−π, π], the integral (11.47) gives
1 1
·
3 2π
Z
π
−π
it
u(e ) dt ≤
u( 12 )
1
≤3·
2π
Z
π
u(eit ) dt.
(11.48)
−π
Put r = 0 in the Poisson formula (11.46), we obtain
Z π
1
u(eit ) dt,
u(0) =
2π −π
therefore, we conclude from the hypothesis u(0) = 1 and the inequalities (11.48) that
1
≤ u( 21 ) ≤ 3.
3
This completes the proof of the problem.
Problem 11.14
Rudin Chapter 11 Exercise 14.
Proof. By translation and/or rotation, we may assume that L1 is the real axis (i.e., y = 0) and
if L1 and L2 intersect, the intersection point is the origin.
• Case (i): L1 and L2 are parallel. Suppose that the equation of L2 is y = A for some
A ∈ R \ {0}. Then we consider the function
πx
u(x, y) = e A sin
πy
.
A
It is obvious that u(x, 0) = u(x, A) = 0 for all x ∈ R. Furthermore, direct computation
gives uxx + uyy = 0 in R2 so that u is a harmonic function in R2 .
52
Chapter 11. Harmonic Functions
• Case (ii): L1 and L2 are perpendicular. Suppose that L2 : x = 0. Then it is easy to
check that the function u(x, y) = xy satisfies all the requirements.
• Case (iii): L1 and L2 are non-parallel and non-perpendicular. Suppose that the
m
angle between L1 and L2 is a rational multiple of π, say mπ
n , where m, n ∈ N and n is
1
n
not a multiple of 2 . Since z = un (x, y) + ivn (x, y) is entire, its imaginary part vn (x, y) is
harmonic in R2 by Theorem 11.4. On L1 , we have
xn = (x + i · 0)2 = un (x, 0) + ivn (x, 0)
for all x ∈ R. This implies that vn (x, 0) = 0 on R. Similarly, since points on L2 are in the
mπ
form z = cos mπ
n + i sin n , so we have
(−1)m =
cos
mπ
mπ
mπ n
mπ mπ mπ
= un cos
+ ivn cos
+ i sin
, sin
, sin
n
n
n
n
n
n
which implies that vn (x, y) = 0 on L2 .
Now suppose that the angle between them is an irrational multiple of π, say απ for some
α ∈ R \ Q. Assume that u(x, y) was a harmonic function in R2 vanishing on L1 ∪ L2 . We
need to develop a harmonic version of Theorem 11.14 (The Schwarz reflection principle).
To this end, we recall the following concept: Let L be a straight line passing through
the origin. We say that a pair of points are symmetric with respect to L if L is the
perpendicular bisector of the line segment joining these points. For each z = (x, y) ∈ C,
it is easy to see that there exists a unique zL = (xL , yL ) ∈ C such that z and zL are
symmetric with respect to L. Next, the following result is taken from [7, Theorem 4.12,
p. 68]:
Lemma 11.3
Let z0 = (a, b) ∈ C and consider the line L = {(x, y) ∈ C | (x, y) · (a, b) = c} for
some a, b, c ∈ R. Define L+ = {(x, y) ∈ C | (x, y) · (a, b) > c}. Suppose that Ω ⊆ C
is a region symmetric with respect to L. If u is continuous on Ω∩L+, u is harmonic
on Ω ∩ L+ and u = 0 on Ω ∩ L, then the function

if (x, y) ∈ Ω ∩ L+ ;
 u(x, y),
U (x, y) =

−u(xL , yL ), if (x, y) ∈ Ω ∩ L−
is harmonic in Ω.
Applying Lemma 11.3 to our u with L = L2 , we see immediately that u vanishes on the
line L3 : y = (tan απ)x. In fact, repeated applications of Lemma 11.3 show that u vanishes
on lines in the form
y = (tan nαπ)x
(11.49)
for every n ∈ Z.
To finish the proof, we have to show that the collection of the straight lines (11.49), denoted
by L, is dense in R2 . To see this, let α > 0. Given that 0 < θ < 12 , δ > 0 and ǫ > 0. By
the Kronecker’s Approximation Theoremh , we find that there exist m, n ∈ N such that
|nαπ − mπ − θπ| < δ.
h
See, for example, [5, §7.4, pp. 148, 149].
11.2. Harnack’s Inequalities and Positive Harmonic Functions
53
Note that tan(nαπ − mπ) = (−1)m tan(nαπ). The continuity of tan x implies that
|(−1)m tan(nαπ) − tan(θπ)| < ǫ.
(11.50)
If m is odd, then we can replace (−1)m tan(nαπ) by tan(−nαπ) in the estimation (11.50).
If − 21 < θ < 0, then a similar argument gives the following estimation
|(−1)m+1 tan(nαπ) − tan(θπ)| < ǫ.
In this case, if m is even, then (−1)m+1 tan(nαπ) will be replaced by tan(−nαπ). In other
words, for every θ ∈ (− 12 , 12 ) \ {0}, we always have
| tan(nαπ) − tan(θπ)| < ǫ
(11.51)
for some n ∈ Z. If θ = 0, then we follow from the Dirichlet’s Approximation Theoremi
π
that for a positive integer N with N
< δ, there exist m, n ∈ N with 0 < n ≤ N such that
|nαπ − mπ| <
π
<δ
N
and again the continuity of tan x implies that
| tan(nαπ)| < ǫ.
Consequently, the inequality (11.51) actually holds for all − 21 < θ < 21 . Since the range of
tan x on (− π2 , π2 ) is R, our inequality (11.51) means that the collection L is dense in R2
for the case α > 0. For the case α < 0, we just consider −α > 0 and then the inequality
(11.51) remains valid with the integer −n.
Finally, since u is continuous on R2 and vanishes on L, we conclude that u ≡ 0 on R2 .
This completes the proof of the problem.
Problem 11.15
Rudin Chapter 11 Exercise 15.
Proof. We first prove the following lemma:
Lemma 11.4
Given ǫ ∈ (0, 1). There exists a constant M > 0 such that
P1−ǫ (t) ≥
for all t ∈ [−ǫ, ǫ].
i
Read [79, Problem 2.4, p. 12].
M
ǫ
(11.52)
54
Chapter 11. Harmonic Functions
Proof of Lemma 11.4. Recall from the series expansion of the Poisson kernel that
Pr (t) = Pr (−t). We know from [62, Eqn. (4), p. 233] that Pr (t) is decreasing on
[0, π] and [−π, 0]. Therefore, it suffices to prove that the inequality (11.52) holds for
t = ǫ. In fact, we have 1 − (1 − ǫ)2 = 2ǫ − ǫ2 ≥ ǫ and
1 − 2(1 − ǫ) cos ǫ + (1 − ǫ)2 = 2(1 − ǫ) − 2(1 − ǫ) cos ǫ + ǫ2
= 2(1 − ǫ)(1 − cos ǫ) + ǫ2
ǫ2 ǫ4
= 2(1 − ǫ)
− + · · · + ǫ2
2!
4!
2
ǫ
≤
M
for some constant M > 0. Thus they imply that
ǫP1−ǫ (ǫ) = ǫ ·
Mǫ
1 − (1 − ǫ)2
≥ǫ· 2 =M
2
1 − 2(1 − ǫ) cos ǫ + (1 − ǫ)
ǫ
which gives the desired result.
Let’s return to the proof of the problem. When u is considered in D(0; r) for 0 < r < 1,
[9, Theorem 16.5, p. 227] implies that u attains its maximum on C(0; r). This fact and the
positivity of u imply that
u(reiθ ) ≤ u(seiθ )
for every θ ∈ [−π, π] if 0 ≤ r ≤ s < 1. Using this result and the hypothesis, if θ 6= 0, then we
obtain
Z
Z
1
1
iθ
sup kur k1 =
sup u(re ) dθ ≤
lim u(reiθ ) dθ = 0.
2π [−π,π]\{0} 0<r<1
2π [−π,π]\{0} r→1
0<r<1
Hence it follows from Theorem 11.30 that one can find a unique positive Borel measure on T
such that u = P [ dµ]. Next, it is easy to see that
Z
Z
1
1
P1−ǫ (θ − t) dµ(eit ) =
P1−ǫ (t) dµ ei(θ−t) ,
(11.53)
u (1 − ǫ)eiθ =
2π T
2π I(eiθ ;2π)
where θ 6= 0 and I(eiθ ; 2π) ⊆ T denotes the open arc centred at eiθ 6= 1 with arc length 2π. We
notice that [−ǫ, ǫ] ⊆ I(eiθ ; 2π), so we apply Lemma 11.4 to the representation (11.53) and get
Z
Z
M
M
M
1
i(θ−t)
iθ
dµ e
dµ ei(θ−t) ≥
µ I(eiθ ; 2ǫ) . (11.54)
≥
u (1 − ǫ)e ≥
2π [−ǫ,ǫ] ǫ
ǫ [−ǫ,ǫ]
ǫ
Take ǫ → 0 in the inequality (11.54), the condition u(reiθ ) → 0 as r → 1 implies that
µ I(eiθ ; 2ǫ)
lim
=0
ǫ→0
ǫ
for every eiθ 6= 1. Consequently, we have µ(T \ {1}) = 0 so that µ is a positive point mass at 1
which means that
Z
1
iθ
Pr (θ − t) dµ(eit ) = µ(1)Pr (θ),
u(re ) =
2π T
completing the proof of the problem.
Problem 11.16
Rudin Chapter 11 Exercise 16.
11.2. Harnack’s Inequalities and Positive Harmonic Functions
55
Proof. Assume that u was the Poisson integral of a complex measure µ on T , i.e., u = P [ dµ].
We know from [62, Eqn. (2), p. 244] that
kur k1 ≤ kµk = |µ|(T ) < ∞
for every 0 < r < 1. Thus it deduces from Theorem 11.30(a) that µ is a (unique) complex Borel
measure. Next, we express u as
u(z) = −4 · Im (z) ·
1 − |z|2
.
|1 + z|4
(11.55)
On the one hand, as the first inequality of Theorem 11.20 only requires that µ is a Borel measure
on T , so we may apply it with a fixed 0 < α < 1 and one can find a constant cα > 0 such that
0 ≤ cα (Nα u)(−1) ≤ (Mrad u)(−1) = sup{|u(−r)| | 0 ≤ r < 1} ≤ lim |u(−r)|.
r→1
(11.56)
If z ∈ (−1, 0], then we have Im (z) = 0 and we see from the representation (11.55) that u(z) = 0.
Consequently, the inequalities (11.56) force that
(Nα u)(−1) = 0.
(11.57)
On the other hand, given 0 < ǫ < sin−1 α, if z = (−1 + ǫ) + iǫ2 , then it is easily checked that
z ∈ −Ωα and the expression (11.55) becomes
Therefore, we obtain
4 2 − ǫ − ǫ3
.
u(z) = u (−1 + ǫ) + iǫ2 = − ·
ǫ (1 + ǫ2 )2
4 2 − ǫ − ǫ3
lim u(z) = lim − ·
= −∞
ǫ→0
z→−1
ǫ (1 + ǫ2 )2
z∈−Ωα
which implies (Nα u)(−1) = ∞, a contradiction to the result (11.57). This proves the first
assertion that u cannot be a Poisson integral of any measure on T .
For the second assertion, if u = v − w, where both v and w are positive harmonic functions
in U , then we have |u(z)| ≤ |v(z)| + |w(z)| = v(z) + w(z) for all z ∈ U so that
Z π
Z π
Z π
1
1
1
iθ
iθ
kur k1 =
|u(re )| dθ ≤
v(re ) dθ +
w(reiθ ) dθ
(11.58)
2π −π
2π −π
2π −π
for every 0 < r < 1. Since v and w are harmonic in U , they satisfy the mean value property, so
the inequality (11.58) gives
Z π
1
|u(reiθ )| dθ ≤ v(0) + w(0)
kur k1 =
2π −π
for every 0 < r < 1. In other words, sup kur k1 is bounded and Theorem 11.30(a) shows that
0<r<1
u = P [ dµ] for a unique complex Borel measure on T . However, it is impossible by the first
assertion and hence u is not the difference of two positive harmonic functions in U . We have
completed the analysis of the problem.
Problem 11.17
Rudin Chapter 11 Exercise 17.
56
Chapter 11. Harmonic Functions
Proof. Set Φ = {u : U → R | u is positive, harmonic and u(0) = 1} and let C be the set whose
members are the positive Borel measures µ on T of |µ|(T ) = 1. We divide the proof into several
steps:
• Step 1: There is an isomorphism between Φ and C. On the one hand, for each
µ ∈ C, we know from §11.17 that
Z
P (z, eit ) dµ(eit )
(11.59)
u(z) =
T
is harmonic and also positive in U . In addition, it is easy to see that
Z
Z
it
it
dµ(eit ) = |µ|(T )
P (0, e ) dµ(e ) =
u(0) =
(11.60)
T
T
which implies u(0) = 1, i.e., u ∈ Φ. On the other hand, if u ∈ Φ, then the mean value
property shows that
Z π
Z π
1
1
iθ
kur k1 =
|u(re )| dθ =
u(reiθ ) dθ = u(0) < ∞
2π −π
2π −π
so that u = P [ dµ] for a unique positive Borel measure on T by Theorem 11.30(a). Besides,
the expression (11.60) gives |µ|(T ) = 1, so µ ∈ C. Consequently, the mapping f : Φ → C
defined by
f (u) = µ
is bijective. In fact, f is a homemorphism because for every u, v ∈ Φ, if their corresponding
positive Borel measures are µ and ν respectively, then
Z
Z
it
it
P (z, eit ) dν(eit )
P (z, e ) dµ(e ) + β
αu(z) + βv(z) = α
T
T
Z
P (z, eit ) d[αµ + βν](eit ),
=
T
i.e., f (αu + βv) = αµ + βν = αf (u) + βf (v).
• Step 2: Both Φ and C are convex. Suppose that u1 , u2 ∈ Φ and u = λu1 + (1 − λ)u2 ,
where 0 ≤ λ ≤ 1. Since u1 and u2 are positive and harmonic in U , u is also positive and
harmonic in U . Furthermore, we have u(0) = λu1 (0) + (1 − λ)u2 (0) = 1. In other words,
u ∈ Φ and thus Φ is a convex set. By Step 1, C is also convex.
• Step 3: Extreme points of C. Denote ext C to be the set of extreme points of C.j
Let µ ∈ ext C. Since u is positive in U , it follows from Step 1 that µ(E) > 0 for every
Borel subset E of T . Now we claim that supp µ = {eit ∈ T | µ(eit ) 6= 0} is a unit mass
concentrated at eit (see [62, Example 1.20(b), p. 17] for the definition). Otherwise, there
was a measurable set E ⊆ T such that 0 < µ(E) < 1. Then we have 0 < µ(T \ E) < 1.
We define
µ(F \ E)
µ(F ∩ E)
and µT \E (F ) =
.
(11.61)
µE (F ) =
µ(E)
µ(T \ E)
Since µ is a positive Borel measure on T , both µE and µT \E are also positive Borel measures
on T . Furthermore, it is clear that
|µE |(T ) =
j
|µ|(T \ E)
|µ|(T ∩ E)
= 1 and |µT \E |(T ) =
= 1.
|µ|(E)
|µ|(T \ E)
A point x of a convex set X is an extreme point of X if x cannot be written as a proper convex combination
x = λx1 + (1 − λ)x2 , where 0 < λ < 1, x1 , x2 ∈ X and x1 6= x2 .
11.3. The Weak∗ Convergence and Radial Limits of Holomorphic Functions
57
Thus µE and µT \E belong to C. If we set λ = µ(E), then for every measurable subset
F ⊆ T , we get from the definitions (11.61) that
µ(F ∩ E)
µ(F \ E)
+ [1 − µ(E)] ·
µ(E)
µ(T \ E)
= µ(F ∩ E) + µ(F \ E)
λµE (F ) + (1 − λ)µT \E (F ) = µ(E) ·
= µ(F )
which means µ ∈
/ ext C, a contradiction. Hence, we obtain
ext C = {δeit | eit ∈ T },
where δx is the Dirac delta function at x.
• Step 4: Extreme points of Φ. By Step 2, the restriction fext Φ is certainly an isomorphism between ext Φ and ext C. Using Step 3 and the definition (11.59), we see
that
ext Φ = {P (z, eit ) | eit ∈ T }.
This completes the proof of the problem.
11.3
The Weak∗ Convergence and Radial Limits of Holomorphic Functions
Problem 11.18
Rudin Chapter 11 Exercise 18.
Proof. Let X be a Banach space and X ∗ be its dual space.k Recall Definition 5.3 that
kΛn k = sup{|Λn (x)| | x ∈ X and kxk = 1}.
Suppose that {Λn } ⊆ X ∗ converges weakly to Λ ∈ X ∗ , so Λn (x) → Λ(x) for every x ∈ X and
then
sup |Λn (x)| = Mx < ∞
n∈N
for each x ∈ X. By the definition, {Λn } is a collection of bounded linear functionals. Hence
Theorem 5.8 (The Banach-Steinhaus Theorem) asserts the existence of a positive constant M
such that
sup kΛn k ≤ M,
n∈N
completing the proof of the problem.
Remark 11.3
We note that the convergence in Problem 11.18 is called the weak∗ convergence. In
fact, there are two important convergences in functional analysis. They are called strong
convergence and weak convergence. To be more precisely, let X be a Banach space and
fn , f ∈ X for every n ∈ N. If kfn − f k → 0, then {fn } is said to converge strongly to f .
Similarly, we say {fn } converges weakly to f if we have g(fn ) → g(f ) for every g ∈ X ∗∗ .
Read, for example, [7, p. 115, 116].
k
Recall from Remark 5.21 that X ∗ is the collection of all bounded linear functionals on X.
58
Chapter 11. Harmonic Functions
Problem 11.19
Rudin Chapter 11 Exercise 19.
Proof.
(a) Since 0 < r < 1, 1 − r > 0. By the power series of the cosine function, we have
cos(1 − r) > 1 −
(1 − r)2
2
which implies that
(1 − r)Pr (1 − r) =
(1 − r)(1 − r 2 )
1 − r − r2 + r3
>
= 1.
2
1 − 2r cos(1 − r) + r
1 + r(1 − r)2 − 2r + r 2
(b) Clearly, we have
δu(1 − δ) =
Z
T
it
it
δP (1 − δ, e ) dµ(e ) =
Z
δP1−δ (t) dµ(eit ).
(11.62)
T
We apply part (a) and the fact that Pr (t) is an even function of t to the integral (11.62)
to get
Z
δu(1 − δ) ≥
dµ(eiθ ) = µ(Iδ ).
Iδ
Since µ ≥ 0, we have u(1 − δ) ≥ 0. By Definition 11.19, we have
µ(Iδ )
µ(Iδ )
=
≤ u(1 − δ) ≤ (Mrad u)(1)
πσ(Iδ )
δ
(11.63)
for every δ > 0. Recall the definition [62, Eqn. (3), §11.19, p. 241], we conclude that
(M µ)(1) ≤ π(Mrad u)(1)
as desired.
(c) In fact, the results of part (b) are valid for every point eiθ on T if we apply the special case
to the rotated measure µθ (E) = µ(eiθ E). In particular, we deduce from the inequalities
(11.63) that
µ(Iδ )
≤ u (1 − δ)eiθ ≤ (Mrad u)(eiθ ),
(11.64)
πσ(Iδ )
where eiθ ∈ T and Iδ is the open arc with center eiθ and length 2δ.
Next, we know from [62, Eqn. (2), §11.17 p. 240] that
kur k1 ≤ kµk = |µ|(T ) < ∞
for every 0 < r < 1. Therefore, we follow from Theorem 11.30(A) that µ is a Borel measure
on T . By the hypothesis, µ is positive. Since µ ⊥ m, it follows from Theorem 7.15 that
(Dµ)(eiθ ) = ∞
a.e. [µ].
(11.65)
Finally, we observe from the inequality (11.64), the result (11.65) and [62, Eqn. (4), p.
241] that
lim u (1 − δ)eiθ = ∞ a.e. [µ]
δ→0
which is exactly the required result.
This ends the proof of the problem.
11.3. The Weak∗ Convergence and Radial Limits of Holomorphic Functions
59
Problem 11.20
Rudin Chapter 11 Exercise 20.
Proof. Since m(E) = 0, for each n ∈ N, we know from [61, Remark 11.11(b), p. 309] that there
exists an open set Vn ⊆ T containing E such that m(Vn ) ≤ 21n . Define
χVn (eiθ ) =
and ϕ : T → R by

 1, if eiθ ∈ Vn ;

ϕ(eiθ ) =
0, otherwise
∞
X
χVn (eiθ ).
n=1
Obviously, we have
∞ Z
X
iθ
χVn (e ) dm =
n=1 T
∞
X
n=1
∞
X
1
= 1,
m(Vn ) ≤
2n
n=1
so Theorem 1.38 implies that
Z
iθ
T
|ϕ(e )| dm =
∞ Z
X
χVn (eiθ ) dm = 1.
n=1 T
In other words, ϕ ∈ L1 (T ) and kϕk1 = 1.
eiθ
Define u : U → R by u = P [ϕ] which is harmonic in U by Theorem 11.7. Now, for each
∈ E, we have eiθ ∈ Vn for every n ∈ N, so it follows from the definition that
1
lim u(re ) = lim
r→1
r→1 2π
iθ
Z
π
−π
P (reiθ , eit )ϕ(eit ) dt = ∞.
Since U is simply connected, u is the real part of a holomorphic function g in U , see [9, Theorem
16.3, p. 226]. Let f = e−g . Then we have f ∈ H(U ) and |f | = e−Re g = e−u so that
lim f (reiθ ) = lim e−u = 0
r→1
r→1
for every eiθ ∈ E. By the definition of f , f (z) 6= 0 for every z ∈ U . Thus f is nonconstant.
Since ϕ ≥ 0, we have u ≥ 0 and consequently, f ∈ H ∞ . If f (0) 6= 1, then we can replace f by
e
fe(z) = ff (z)
(0) so that f (0) = 1. This completes the proof of the problem.
Remark 11.4
For further information, the reader can refer to [53, p. 295] and [84, pp. 105, 276].
Problem 11.21
Rudin Chapter 11 Exercise 21.
60
Chapter 11. Harmonic Functions
t+is−1
Proof. We first show that g ∈
/ H ∞ . In fact, fix s ∈ R, consider zt = t+is+1
for 0 < t < ∞. Then
t
it is easy to check that |zt | < 1 and 1+z
=
t
+
is.
Therefore,
we
have
1−zt
g(zt ) =
2
exp(−et+is )
(t + 1) + is
which gives
|g(zt )| = p
2
(t + 1)2 + s2
exp(−et cos s).
(11.66)
Now we put s = π in the expression (11.66) to get |g(zt )| → ∞ as t → ∞. Consequently,
g∈
/ H ∞.
Next, if eiθ ∈ T and 0 ≤ r < 1, then we have
1 − r 2 + 2i sin θ
1 + reiθ
=
1 − reiθ
1 − 2r cos θ + r 2
and
f (reiθ ) = exp
1 − r 2 + 2i sin θ .
1 − 2r cos θ + r 2
By the definition of g, we have
h
1 − r 2 + 2i sin θ i
g(reiθ ) = (1 − reiθ ) · exp − exp
1 − 2r cos θ + r 2
and so

h
i sin θ i

, if θ =
6 0;
 (1 − eiθ ) exp − exp
1 − cos θ
g∗ (eiθ ) = lim g(reiθ ) =
r→1


0,
if θ = 0.
(11.67)
Hence g∗ (eiθ ) exists for every eiθ ∈ T . By the representation (11.67), we know that g ∗ (eiθ ) is
continuous at every θ ∈ (0, 2π). Thus it suffices to show that g∗ (eiθ ) is continuous at θ = 0. To
see this, since
cos θ
− sin θ
sin θ
= lim
= lim
= 0,
lim
θ→0 sin θ
θ→0 cos θ
θ→0 1 − cos θ
the representation (11.67) gives
h
i sin θ i
lim g∗ (eiθ ) = lim (1 − eiθ ) exp − exp
θ→0
θ→0
1 − cos θ
h
i sin θ i
= lim (1 − eiθ ) lim exp − exp
θ→0
θ→0
1 − cos θ
−1
=0·e
= 0.
As a consequence, we establish the fact that g∗ ∈ C(T ) and this completes the proof of the
problem.
11.4
Miscellaneous Problems
Problem 11.22
Rudin Chapter 11 Exercise 22.
11.4. Miscellaneous Problems
61
Proof. For each 0 ≤ r < 1, the function ur : T → C is continuous, so u−1
r (R) is measurable.
− , where u+ and u− are the positive and negative parts of u , see
Suppose that ur = u+
−
u
r
r
r
r
r
Definition 1.15. Since {ur } is uniformly integrable, there exists a δ > 0 such that
Z
(11.68)
ur dm < 1
E
iθ
+
whenever 0 ≤ r < 1 and m(E) < 2δ. Let Er+ = {θ ∈ [0, 2π] | u+
r (e ) > 0}. Then Er is
measurable and
Z
Z π
+
ur dm.
(11.69)
ur dm =
Er+
−π
(k−1)δi to
l
Let N be the least positive integer such that 2π
N ≤ δ. Define Ik to be the arc from e
ekδi including e(k−1)δi but not ekδi , where k = 1, 2, . . . , N − 1. Similarly, we define IN to be the
arc from e(N −1)δi to e2πi = 1 including e(N −1)δi but not e2πi = 1. Clearly, the central angles of
I1 , I2 , . . . , IN −1 are exactly δ and the central angle of IN is less than or equal to δ. Next, we
denote
+
Er,k
= Er+ ∩ Ik
+
+
+
} forms a disjoint measurable subsets of Er+ and
so that {Er,1
, Er,2
, . . . , Er,N
+
) ≤ m(Ik ) ≤ δ < 2δ
m(Er,k
for k = 1, 2, . . . , N . Hence we follow from the inequality (11.68) and the result (11.69) that
Z
π
−π
u+
r dm =
N Z
X
k=1
+
Er,k
ur dm ≤
Z
N
X
k=1
+
Er,k
ur dm ≤ N
for every 0 ≤ r < 1. Similarly, it can be shown that
Z π
u−
r dm ≤ N
−π
for every 0 ≤ r < 1. Thus they imply that
Z π
Z
Z π
N
1
1 π + iθ
iθ
iθ
kur k1 =
u−
(e
)
dθ
≤
ur (e ) dθ +
|ur (e )| dθ ≤
r
2π −π
2π −π
π
−π
for every 0 ≤ r < 1.
Hence we apply Theorem 11.30(a) to obtain a unique complex Borel measure µ such that
u = P [ dµ]. Next, we know form Theorem 11.24 that there exists a f ∈ L1 (T ) such that
lim u(zj ) = f (eiθ )
j→∞
for almost all points eiθ ∈ T , where zj → eiθ and zj ∈ eiθ Ωα for α < 1. Particularly, this implies
that
urj (eiθ ) = u(rj eiθ ) → f (eiθ )
as j → ∞ a.e. on T , where {rj } ⊆ [0, 1) and rj → 1 as j → ∞. Furthermore, since σ(T ) < ∞
and {ur } ⊆ L1 (T ) is uniformly integrable, Problem 6.10(d) ensures that
Z
|urj (eiθ ) − f (eiθ )| dσ(eiθ ) → 0
kurj − f k1 =
T
l
It is obvious that N is independent of r.
62
Chapter 11. Harmonic Functions
as j → ∞ or equivalently, we have
Z
Z
iθ
iθ
f (eiθ ) dσ(eiθ ).
urj (e ) dσ(e ) =
lim
j→∞ T
(11.70)
T
Finally, we take g = 1 in [62, Eqn. (3), p. 247] to get
Z
Z
iθ
iθ
dµ(eiθ ).
urj (e ) dσ(e ) =
lim
j→∞ T
(11.71)
T
Combining the results (11.70) and (11.71), we establish
dµ = f dσ
and hence u = P [f ] for some f ∈ L1 (T ) which completes the proof of the problem.
Remark 11.5
We can also apply Theorem 17.13 (F. and M. Riesz Theorem) on [62, p. 341] to prove that
µ ≪ σ.
Problem 11.23
Rudin Chapter 11 Exercise 23.
Proof. Given z ∈ U , since |eiθ − z|2 ≥ |eiθ | − |z|
|P (z, eiθ )| =
2
= (1 − |z|)2 , we have
1 + |z|
1 − |z|2
1 − |z|2
=
< ∞.
≤
iθ
2
2
|e − z|
(1 − |z|)
1 − |z|
Thus the series
v(z) =
∞
X
n
−2
iθn
P (z, e
) and w(z) =
∞
X
n−2 P (z, e−iθn )
n=1
n=1
converge absolutely so that we may split the representation of u into the difference of v and w.
For each n ∈ N, we define
vn (z) =
n
X
k=1
k
−2
iθk
P (z, e
) and wn (z) =
n
X
k−2 P (z, e−iθk ).
k=1
Using Problem 11.4, we know that P (z, eiθk ) and P (z, e−iθk ) are harmonic in U for k = 1, 2, . . . , n
so that vn and wn are also harmonic in U . Fix 0 ≤ r < 1. If z ∈ D(0, r), then we have
1 − |z| > 1 − r and thus
∞
∞
X
P (z, eiθn )
1+r X 1
·
≤
< ∞.
n2
1 − r n=1 n2
n=1
By the Weierstrass M -test, we see that {vn } converges uniformly to v in D(0; r). Let K be a
compact subset of U . Then there exists a 0 < r < 1 such that K ⊆ D(0, r), so it follows from
Theorem 11.11 (Harnack’s Theorem) that v is harmonic in U . By a similar argument, we are
able to show that w is also harmonic in U .
11.4. Miscellaneous Problems
63
Since P (z, eit ) > 0 for every z ∈ U and eit ∈ T , we have v > 0 and w > 0. Now, for every
0 ≤ r < 1, we observe that
Z π
1
kvr k1 =
v(reiθ ) dθ
2π −π
Z π hX
∞
i
1
n−2 P (reiθ , eiθn ) dθ
=
2π −π
n=1
∞
h 1 Z π
i
X
n−2
=
P (reiθ , eiθn ) dθ
2π −π
n=1
=
∞
X
n−2
n=1
< ∞.
Similarly, we have sup kwr k1 < ∞ and hence v and w satisfy the requirements of Theorem
0<r<1
11.30. Then there exist unique positive Borel measures µ and ν on T such that
v = P [ dµ] and w = P [ dν].
Since u = v − w, we establish
u = P [ d(µ − ν)],
where µ − ν is clearly a measure on T .
Next, if x ∈ (−1, 1), then we get from [62, Eqn. (2), p. 233] that
P (x, eiθn ) = P (x, e−iθn ) =
1 − x2
.
1 − 2x cos θn + x2
Consequently, we conclude that u(x) = 0 if −1 < x < 1. Finally, suppose that z = 1 − ǫ + iǫ,
where ǫ = sin θ > 0. Then it is easy to check that
P (z, eiθ ) − P (z, e−iθ ) = P (1 − sin θ + i sin θ, eiθ ) − P (1 − sin θ + i sin θ, e−iθ )
1 − [(1 − sin θ)2 + sin2 θ]
1 − [(1 − sin θ)2 + sin2 θ]
+
| cos θ − 1 + sin θ|2
|(cos θ − 1 + sin θ) − 2i sin θ|2
i
h
1
1
−
= 2 sin θ(1 − sin θ) ·
(cos θ + sin θ − 1)2 (cos θ + sin θ − 1)2 + 4 sin2 θ
h
1
= 2 sin θ(1 − sin θ) ·
2 θ
4 sin 2 (cos 2θ − sin 2θ )2
i
1
−
.
(11.72)
4 sin2 θ2 (cos 2θ − sin 2θ )2 + 4 sin2 θ
=
Since (cos θ2 − sin 2θ )2 = 1 − sin θ, the expression (11.72) reduces to
P (z, eiθ ) − P (z, e−iθ ) = 2 sin θ ·
=
≥
4 sin2 θ2 (1 − sin θ) + 4 sin2 θ − 4 sin2 θ2 (1 − sin θ)
4 sin2 2θ [4 sin2 θ2 (1 − sin θ) + 4 sin2 θ]
sin3 θ
2 sin2 2θ [sin2 θ2 (1 − sin θ) + sin2 θ]
sin3 θ
2 sin2 2θ (sin2 θ2 + sin2 θ)
64
Chapter 11. Harmonic Functions
≥
sin θ
.
2 sin2 2θ
(11.73)
By the definition, ǫ is small if and only if θ > 0 is small. In this case, cos2
(11.73) becomes
P (z, eiθ ) − P (z, e−iθ ) ≥
θ
2
> 12 , so the inequality
2 cos2 θ2
1
1
sin2 θ
1
=
>
= .
·
2 θ sin θ
sin θ
sin θ
ǫ
2 sin 2
(11.74)
If z = x + iy ∈ U with x > 0 and y > 0, then it is easy to see from [62, Eqn. (6), p. 233] that
P (x + iy, eiθ ) − P (x + iy, e−iθ ) > 0
(11.75)
for every θ ∈ (0, π2 ). By the definition of u, if we put z = 1 − ǫ + iǫ, then we may apply the
estimate (11.75) to get
u(1 − ǫ + iǫ) > n−2 [P (1 − ǫ + iǫ, eiθn ) − P (1 − ǫ + iǫ, e−iθn )]
(11.76)
for every n ∈ N. Now we take ǫn = sin θn = sin 21n in the inequality (11.76) and then apply the
estimate (11.74) as well as the fact that sin θn ≤ θn for every n ∈ N to obtain
u(1 − ǫn + iǫn ) >
for every n ∈ N. Since
that
2n
n2
1
1
2n
=
≥
n 2 ǫn
n2
n2 sin 21n
(11.77)
→ ∞ as n → ∞, we conclude immediately from the inequality (11.77)
lim u(1 − ǫn + iǫn ) = ∞,
n→∞
completing the proof of the problem.
Problem 11.24
Rudin Chapter 11 Exercise 24.
Proof. Most assertions of this problem come from [61, Exercise 15, p. 199] and their solutions
are shown by the author in [77, Problem 8.15, pp. 187 – 190], so we just prove the necessary
assertions here.
• Proof of KN −1 (t) ≤ LN (t). Since sin 2θ =
1−cos θ
,
2
we know that
1 sin N2t 2
.
·
N
sin 2t
KN −1 (t) =
(11.78)
By [61, Exercise 8, p. 197], we always have | sin nθ| ≤ n| sin θ| for n = 0, 1, 2, . . . so that
sin N t 2
sin
for N2 |t| ≤
that
π
2.
2
t
2
=
| sin N2t |2
≤ N2
| sin 2t |2
If 0 ≤ t ≤ π, then the power series of sin x implies that sin 2t ≥
sin N t 2
sin
2
t
2
≤
1
sin2
t
2
≤
π2
.
t2
(11.79)
t
π
≥ 0 so
(11.80)
11.4. Miscellaneous Problems
65
If −π ≤ t ≤ 0, then we let t = −s so that 0 ≤ s ≤ π. In this case, we have
sin N t 2
sin
2
t
2
=
sin N s 2
sin
2
s
2
≤
1
sin2
s
2
≤
π2
π
= 2.
s2
t
(11.81)
Hence, by putting the inequalities (11.79), (11.80) and (11.81) into the expression (11.78),
π
, then
we conclude that if |t| ≤ N
KN −1 (t) ≤
if
π
N
1
· N2 = N;
N
(11.82)
≤ |t| ≤ π, then
π2
.
N t2
By the definition of LN , we see that KN −1 (t) ≤ LN (t) for every t ∈ [−π, π].
R
• Proof of T LN dσ ≤ 2. Using the estimates (11.82) and (11.83), we have
Z
Z π
1
LN dσ =
Ln (t) dt
2π −π
T
Z
Z
π
dt
1
N dt +
=
2π |t|≤ π
2N π ≤|t|≤π t2
KN −1 (t) ≤
N
(11.83)
N
1
=2−
N
≤2
for every N ∈ N.
• Proof of Fejér’s Theorem. The result about the convergence of the arithmetic means
is called Fejér’s Theorem.m Recall from [62, Eqn. (1), p. 101] that
Z
Z
it
f ei(θ−t) Dn (t) dσ,
f (e )Dn (θ − t) dσ =
sn (f ; θ) =
T
T
so it is easy to check that
σN (f ; θ) =
Z
i(θ−t)
f e
T
h
Z
N
i
1 X
f ei(θ−t) KN (t) dσ.
Dn (t) dσ =
N +1
T
(11.84)
n=0
Suppose that eiθ is a Lebesgue point of f ∈ L1 (T ). Imitating the proof of Theorem 11.23
(Fatou’s Theorem), we may assume without loss of generality that f (eiθ ) = 0. Thus it
suffices to prove that
lim σN (f ; θ) = 0.
N →∞
It is easy to check that the KN satisfies KN (t) ≥ 0, KN (t) is even and
Z
KN (t) dσ = 1
T
m
for every N ∈ N. Thus it follows from the expression (11.84) that
Z π
1
f ei(θ−t) KN (t) dt
|σN (f ; θ)| =
2π −π
See, for instances, [61, Exercises 15, 16, p. 199] or [84, Theorem 3.4, p. 89].
66
Chapter 11. Harmonic Functions
Z 0
Z π
1
i(θ−t)
f ei(θ−t) KN (t) dt
f e
KN (t) dt +
=
2π 0
−π
Z π
1
f ei(θ−t) + f ei(θ+t) · KN (t) dt.
≤
2π 0
Put g(t) = f ei(θ−t) + f ei(θ+t) and
Z x
g(t) dt,
G(x) =
0
where 0 ≤ x ≤ π. We have
0<
and since
that
eiθ
G(x)
1
≤
x
x
Z
0
x
1
|f ei(θ−t) | dt +
x
Z
0
x
|f ei(θ+t) | dt
is a Lebesgue point of f , we know from the definition [62, Eqn. (5), p. 241]
G(x)
= 0.
x→0 x
lim
(11.85)
Given ǫ > 0. Firstly, the result (11.85) means that we may choose a δ > 0 such that
G(x) < ǫx for all 0 < x < δ. If N > 1δ , then we deduce from (11.82) that
Z 1
Z 1
N
N
N +1
ǫ < 2ǫ.
(11.86)
g(t)KN (t) dt ≤ (N + 1)
g(t) dt <
N
0
0
Secondly, the property (11.83) actually holds for all 0 < |t| ≤ π so that
Z δ
Z
π 2 δ g(t)
dt
g(t)KN (t) dt <
1
N 1 t2
N
N
Z
π 2 δ dG(t)
=
N 1 t2
N
Z δ
h
1
2
G(t) i
π G(δ)
2
+
2
−
N
G
dt
=
1
N δ2
N
t3
N
Z δ
dt π2 ǫ
<
+ 2ǫ
1 t2
N δ
N
π2 ǫ
+ 2ǫN
<
N δ
< 3π 2 ǫ.
(11.87)
Thirdly, we note that
Z π
Z π
π2
g(t) dt
g(t)KN (t) dt ≤
N δ2 δ
δ
Z
Z
i
π2 h
i(θ+t)
i(θ−t)
≤
|f
e
| dt
|f
e
|
dt
+
N δ2 T
T
π2
· 2kf k1
=
N δ2
<ǫ
(11.88)
for large enough N . Finally, by combining the inequalities (11.86), (11.87) and (11.88),
we conclude immediately that
Z π
3(1 + π 2 )
1
1
(3ǫ + 3π 2 ǫ) =
ǫ
g(t)KN (t) dt <
|σN (f ; θ)| ≤
2π 0
2π
2π
11.4. Miscellaneous Problems
67
for sufficiently large N . Since ǫ is arbitrary, we have obtained the desired result that
σN (f ; θ) → 0 as N → ∞.
We have completed the proof of the problem.
Problem 11.25
Rudin Chapter 11 Exercise 25.
Proof. Let z = x + iλ, where λ > 0. Using [62, Eqn. (3), §9.7, p. 183], we find that
Z ∞
Z
1
1 ∞ λf (x − y)
u(z) = u(x, λ) = (f ∗ hλ )(x) = √
dy.
f (x − y)hλ (y) dy =
π −∞ y 2 + λ2
2π −∞
(11.89)
We prove the problem by showing the following steps:
• Step 1: ϕ(z, t) is harmonic in Π+ for every t ∈ R. By the change of variable t = x− y,
the expression (11.89) can be written as
Z
Z
1 ∞
λf (t)
1 ∞
u(z) = u(x, λ) =
dt
=
ϕ(x + iλ, t)f (t) dt,
π −∞ (x − t)2 + λ2
π −∞
where ϕ : Π+ × R is given by
ϕ(x + iλ, t) =
1
λ
.
=
Im
(x − t)2 + λ2
t − (x + iλ)
(11.90)
1
Since t−(x+iλ)
is holomorphic in Π+ for every t ∈ R, Theorem 11.4 ensures that ϕ is
harmonic in Π+ .n Hence ϕ(z, t) is continuous on Π+ and then it satisfies the mean value
property for every t ∈ R:
Z π
1
ϕ(z, t) =
ϕ(z + reiθ , t) dθ,
2π −π
where r is any positive number such that D(z, r) ⊆ Π+ .
• Step 2: ϕ(z, t) ≤
such that
Mz
1+t2
for some Mz > 0. We claim that there exists a constant Mz > 0
0 < ϕ(z, t) = ϕ(x + iλ, t) ≤
Mz
1 + t2
(11.91)
for all t ∈ R. To see this, simple calculation shows that
holds for all t ∈ R if and only if
λ
Mz
≤
2
2
(x − t) + λ
1 + t2
(λ − Mz )t2 + 2Mz xt + (λ − Mz x2 − Mz λ2 ) ≤ 0
for all t ∈ R if and only if
(2Mz x)2 − 4(λ − Mz )(λ − Mz x2 − Mz λ2 ) ≤ 0
λMz2 − (x2 + λ2 + 1)Mz + λ ≥ 0
and the inequality (11.92) is true for large Mz > 0. This proves the claim.
n
In fact, it can be shown directly from the definition (11.90) that, for each t ∈ R, we have
ϕxx =
which give ∆ϕ = 0.
6λ(x − t)2 − 2λ3
[(x − t)2 + λ2 ]3
and
ϕλλ =
[(x − t)2 + λ2 ](−2λ) − 4λ[(x − t)2 − λ2 ]
[(x − t)2 + λ2 ]3
(11.92)
68
Chapter 11. Harmonic Functions
• Step 3: ϕ(z, t) ∈ Lq (R) for every 1 ≤ q ≤ ∞. By Step 2, if 1 ≤ q < ∞, then we have
1 ≤ 1 + t2 ≤ (1 + t2 )q so that
Z ∞
Z ∞
Z ∞
Mzq
1
q
q
ϕ (z, t) dt ≤
dt ≤ Mz ·
dt < ∞.
(11.93)
2 )q
(1
+
t
1
+
t2
−∞
−∞
−∞
In other words, we have ϕ(z, t) ∈ Lq (R1 ). By the inequality (11.91), since ϕ(z, t) is
obviously bounded by Mz , we know that ϕ(z, t) ∈ L∞ (R1 ).
• Step 4: u satisfies the mean value property. Suppose that z ∈ Π+ and r is a positive
number such that D(z; r) ⊆ Π+ . If q is the conjugate exponent of p, then we obtain from
Theorem 3.8 and Step 3 that
Z ∞
iθ
|ϕ(z + reiθ , t)f (t)| dt ≤ kf kp × kϕ(z + reiθ , ·)kq < ∞.
kϕ(z + re , ·)f (·)k1 =
−∞
Next, we observe from the inequality (11.92) that we can pick Mζ in such a way that the
set {Mζ | ζ = z + reiθ and −π ≤ θ ≤ π} is bounded. Therefore, we get
Z π
kϕ(z + reiθ , ·)f (·)k1 dθ < ∞.
−π
Consequently, we apply Theorem 8.8 (The Fubini Theorem) and Step 1 to conclude that
Z π Z ∞
Z π
1
1
iθ
u(z + re ) dθ = 2
ϕ(z + reiθ , t)f (t) dt dθ
2π −π
2π −π −∞
Z
Z π
1 ∞ 1
ϕ(z + reiθ , t) dθ f (t) dt
=
π −∞ 2π −π
Z
1 ∞
ϕ(z, t)f (t) dt
=
π −∞
= u(z).
• Step 5: u is continuous on Π+ . Let {zn } be a sequence of Π+ converging to z0 ∈ Π+ .
Then we must have ϕq (zn , t) → ϕq (z0 , t) as n → ∞. Now the inequality (11.93) guarantees
that we may use Theorem 1.34 (The Lebesgue’s Dominated Convergence Theorem) to
conclude that
Z
1 ∞
lim u(zn ) = lim
ϕ(zn , t)f (t) dt
n→∞
n→∞ π −∞
Z ∞h
i
1
lim ϕ(zn , t) f (t) dt
=
π −∞ n→∞
Z
1 ∞
ϕ(z0 , t)f (t) dt
=
π −∞
= u(z0 ).
Thus u is continuous on Π+ .
• Step 6: u is harmonic in Π+ . In fact, this follows immediately from Steps 4, 5 and
Theorem 11.13.
Hence we have completed the analysis of the proof.
CHAPTER
12
The Maximum Modulus Principle
12.1
Applications of the Maximum Modulus Principle
Problem 12.1
Rudin Chapter 12 Exercise 1.
Proof. Let f (z) = (z − a)(z − b)(z − c). Then f is nonconstant and entire. By Theorem 10.24
(The Maximum Modulus Theorem), we know that
max |f (z)| = max |f (z)|.
z∈∆
z∈∂∆
Let L = |a − b| = |b − c| = |c − a| and t = t(z) = |z − a| ∈ [0, L], see Figure 12.1 below.
Figure 12.1: The boundary ∂∆.
If z ∈ [a, b], then we have
|f (z)|2 = |z − a|2 · |z − b|2 · |z − c|2
√
h
L 2 3L 2 i
2
2
+
= t (L − t) · t −
2
2
69
70
Chapter 12. The Maximum Modulus Principle
= t2 (L − t)2 (t2 − Lt + L2 ).
Define the function F : [0, L] → R by F (t) = t2 (L − t)2 (t2 − Lt + L2 ). Elementary differentiation
shows that
F ′ (t) = 6t5 − 15Lt4 + 16L2 t3 − 9L3 t2 + 2L4 t
= t(2t − L)(3t3 − 6Lt2 + 5L2 t − 2L3 )
= t(2t − L)(t − L)(3t2 − 3Lt + 2L2 ).
Thus F ′ (t) = 0 if and only if t = 0, L2 , L. By the First Derivative Test, it is easily seen that F
3 6
L at t = L2 . Hence we conclude that
attains its maximum 64
r √
p
L
3 3
max |f (z)| = max |f (z)| = max F (t) = F
=
L .
z∈∆
z∈∂∆
2
8
t∈[0,L]
This completes the analysis of the problem.
Problem 12.2
Rudin Chapter 12 Exercise 2.
Proof. Suppose that f (i) = α. If |α| = 1, then Theorem 10.24 (The Maximum Modulus Theorem) forces that f (z) = α in Π+ . In this case, we have |f ′ (i)| = 0. Suppose that α ∈ U . By
Definition 12.3, ϕα (α) = 0. Furthermore, we know from [9, Theorem 13.16, p. 183] that the
mapping h : Π+ → U given by
z − β ,
(12.1)
h(z) = eiθ
z−β
where Im β > 0 and θ ∈ R, is a bijection. Particularly, we take β = i in the definition (12.1).
Clearly, we have h(i) = 0. Next, we consider the mapping F = ϕα ◦ f ◦ h−1 : U → U . Then we
have F ∈ H ∞ , kF k∞ ≤ 1 and
F (α) = ϕα f h−1 (0) = ϕα f (i) = ϕα (α) = 0.
Hence it follows from Theorem 12.2 (Schwarz’s Lemma) that
|F ′ (0)| ≤ 1.
Since h−1 (z) = i ·
that
z+eiθ
,
eiθ −z
(12.2)
we have (h−1 )′ (0) = 2ie−iθ . Consequently, we see from Theorem 12.4
F ′ (0) = ϕ′α (α) × f ′ (i) × (h−1 )′ (0) =
so the inequality (12.2) implies that
|f ′ (i)| ≤
2ie−iθ f ′ (i)
,
1 − |α|2
(12.3)
1 − |α|2
.
2
Thus |f ′ (i)| attains the maximum 12 when α = 0. In this case, we observe from the expression
(12.3) that |F ′ (0)| = 1, so Theorem 12.2 (Schwarz’s Lemma) implies that
F (z) = λz for some
−1
constant λ with |λ| = 1. Since ϕ0 (z) = z, we conclude that f h (z) = λz. Now if we put
z = h(ζ), then it asserts that
ζ − i
.
(12.4)
f (ζ) = λh(ζ) = λeiθ
ζ +i
Since λ = eiφ for some φ ∈ R, we may simply replace λeiθ by eiθ in the representation (12.4)
which gives all extremal functions with |f ′ (i)| = 12 . This completes the proof of the problem. 12.1. Applications of the Maximum Modulus Principle
71
Problem 12.3
Rudin Chapter 12 Exercise 3.
Proof. If f is constant, then there is nothing to prove. Thus, without loss of generality, we may
assume that f is nonconstant. In this case, f has a local minimum in Ω if and only if f has a
zero in Ω. Assume that f was a non-vanishing function in Ω. Then f1 ∈ H(Ω) and |f | has a
local minimum at z0 ∈ Ω if and only if |f1 | has a local maximum at z0 . By Theorem 10.24 (The
Maximum Modulus Theorem), f1 is forced to be constant which is impossible by our hypothesis.
Hence f has a zero in Ω, completing the proof of the problem.
Problem 12.4
Rudin Chapter 12 Exercise 4.
Proof.
(a) Assume that f was non-vanishing in D. By Theorem 10.24 (The Maximum Modulus
Theorem), f 6= 0 on ∂D. Thus it is true that f 6= 0 in D. Denote M = |f (z)| on ∂D. On
the one hand, Theorem 10.24 (The Maximum Modulus Theorem) again implies that
|f (z)| ≤ M
for all z ∈ D. On the other hand, since f1 ∈ H(D) and
10.24 (The Maximum Modulus Theorem) that
(12.5)
1
f
∈ C(D), we follow from Theorem
1
1
≤
|f (z)|
M
(12.6)
for all z ∈ D. Combining the inequalities (12.5) and (12.6), we conclude that |f (z)| = M
in D, or equivalently, f (D) ∈ {−M, M } which contradicts the Open Mapping Theorem.
Hence f has at least one zero in D.
(b) This part is proven in [76, Problem 7.5, pp. 85 – 87].
We end the proof of the problem.
Problem 12.5
Rudin Chapter 12 Exercise 5.
Proof. Given that ǫ > 0. Since fn → f uniformly on ∂Ω, there exists an N ∈ N such that
m, n ≥ N imply that
|fn (z) − fm (z)| < ǫ
(12.7)
for every z ∈ ∂Ω. Since Ω is bounded, we may apply Theorem 10.24 (The Maximum Modulus
Theorem) to assert that the inequality (12.7) holds for every z ∈ Ω, i.e.,
max |fn (z) − fm (z)| = max |fn (z) − fm (z)| < ǫ
z∈Ω
z∈∂Ω
(12.8)
for m, n ≥ N . According to the Cauchy Criterion for Uniform Convergence [61, Theorem 7.8,
p. 147], the inequality (12.8) ensures that {fn } converges uniformly on Ω which ends the proof
of the problem.
72
Chapter 12. The Maximum Modulus Principle
Problem 12.6
Rudin Chapter 12 Exercise 6.
Proof. By the definitions, Γ∗ is closed in C and then C \ Γ∗ is open in C. By Definition 10.1,
C \ Γ∗ is a union of disjoint open connected sets, and hence components. Suppose that V is a
component of C \ Γ∗ such that Ind Γ (z) 6= 0 for every z ∈ V . Let Γ = γ1 +̇ · · · +̇γn , where each γj
is a closed path in Γ. Assume that V was unbounded. If z ∈ V , then it follows from Theorem
10.10 that Ind γj (z) = 0 for every 1 ≤ j ≤ n. Since
Ind Γ (z) =
n
X
Ind γj (z),
j=1
we have Ind Γ (z) = 0, a contradiction. Therefore, V must be bounded. Since Ind Γ (α) = 0 for
all α ∈
/ Ω, we get
V ⊆ Ω.
(12.9)
Let ζ ∈ ∂V . Then ζ ∈
/ V . If ζ lies in another component U , then there exists a δ > 0 such that
D(ζ; δ) ∩ V 6= ∅ and D(ζ; δ) ∩ U 6= ∅, but this is a contradiction by [42, Theorem 25.1, p. 159].
Hence ζ ∈
/ C \ Γ∗ , i.e., ζ ∈ Γ∗ and then ∂V ⊆ Γ∗ . Since Γ is a cycle in Ω, we have
Γ∗ ⊆ Ω.
(12.10)
Combining the set relations (12.9) and (12.10), we conclude that V ⊆ Ω. Recall that V is
bounded, V is compact. By the hypothesis, we have |f (ζ)| ≤ 1 for every ζ ∈ ∂V ⊆ Γ∗ . By
Theorem 10.24 (The Maximum Modulus Theorem), we observe that
|f (z)| ≤ 1
for every z ∈ V with Ind Γ (z) 6= 0. This completes the proof of the problem.
Problem 12.7
Rudin Chapter 12 Exercise 7.
Proof. Suppose that Ω = {x + iy | a < x < b, y ∈ R} and M (a) = 0. We have to show that
M (x) = 0 for all x ∈ (a, b). Consider the function g(x + iy) = f (y + a + ix) which is defined
in the horizontal strip Ω+
g = {x + iy | −∞ < x < ∞ and 0 < y < b − a}. By the definition, we
know that
|g(x)| = |f (a + ix)| = 0
for all x ∈ R. In particular, g is real on the segment L = (0, 1). By Theorem 11.14 (The
−
Schwarz Reflection Principle), there exists a function G holomorphic in Π = Ω+
g ∪ L ∪ Ωg such
that G(x) = g(x) = 0 for every x ∈ L. By Theorem 10.18, we have G(z) = 0 in Π. Since
G(z) = g(z) = 0 in Ω+
g , we obtain
f (z) = 0
in Ω which implies the required result that M (x) = 0 for all x ∈ (a, b) and hence Theorem 12.8
(The Hadamard’s Three-Line Theorem) is also true if M (a) = 0. This completes the proof of
the problem.
12.1. Applications of the Maximum Modulus Principle
73
Problem 12.8
Rudin Chapter 12 Exercise 8.
Proof. Suppose that Ω = {z = x + iy | c < x < d and y ∈ R} for some −∞ < c < d < ∞. By
the hypothesis, the exponential function ζ = ez maps Ω onto A(R1 , R2 ). We are given that
R1 < a < r < b < R2 . Then there exist c < α < β < d such that ζ = ez maps the closed strip
Ω′ = {z = x + iy | α ≤ x ≤ β and y ∈ R} onto the closed annulus A(a, b). Thus we have
eα = a
We define F : Ω′ → C by
and eβ = b.
(12.11)
F (z) = f (ez ) = f (ζ)
which is continuous on Ω′ and F ∈ H(Ω′ ). By Theorem 10.24 (The Maximum Modulus Theorem)
and the Extreme Value Theorem, there is a positive constant B > 0 such that |F (z)| < B for
all z ∈ Ω′ . In other words, we may apply Theorem 12.8 (The Hadamard’s Three-Line Theorem)
to our function F to get
M (x)β−α ≤ M (α)β−x × M (β)x−α ,
(12.12)
where the two expressions (12.11) give
M (x) = sup{|F (x + iy)| | α ≤ x ≤ β and y ∈ R}
= sup{|f (ex · eiy )| | α ≤ x ≤ β and y ∈ R}
= sup{|f (reiy )| | a ≤ r ≤ b and y ∈ R}
= M (r).
Therefore, it deduces from the inequality (12.12) that
(β − α) log M (r) ≤ (β − x) log M (a) + (x − α) log M (b)
log M (r) ≤
log rb
log ab
log M (a) +
log
log
r
a
b
a
log M (b)
(12.13)
which is the required result.
We claim that the equality (12.13) holds if and only if f (ζ) = Aζ λ for some A ∈ C and λ ∈ Z.
Obviously, the equality holds if f is of this form. Conversely, the we can rewrite the equality
(12.13) as
log
b
r
b
log M (r) = log log M (a) + log log M (b)
a
r
a
r
b
r
· [log M (a)] + log log M (b)
= log − log
a
a
a
b
r
M (b)
= log log M (a) + log log
a
a
M (a)
log ar
M (b)
.
log M (r) = log M (a) +
log
b
M
(a)
log a
If we denote λ = (log ab )−1 log
M (a)
M (b)
(12.14)
∈ R, then the expression (12.14) can be further simplified to
log M (r) = log M (a) − λ log
r
a
74
Chapter 12. The Maximum Modulus Principle
M (r) =
a λ
r
M (a).
(12.15)
If we combine the Extreme Value Theorem and the expression (12.15), then one can show that
there exists an |ζ0 | = r ∈ (a, b) such that |f (ζ0 )| = M (r) = ( ar )λ M (a) which can be rewritten as
|ζ0λ f (ζ0 )| = aλ M (a).
(12.16)
Notice that ζ λ f (ζ) may not be holomorphic because λ may not be an integer. However, remember that ζ = ez , so we can express the expression (12.16) as
|eλz0 f (ez0 )| = aλ M (a) = eλα M (α),
where ζ0 = ez0 ∈ Ω′ . Now eλz f (ez ) ∈ H(Ω′ ) and continuous on Ω′ , so we may apply Theorem
12.4 (The Maximum Modulus Theorem) to this function to conclude that
eλz f (ez ) = c
(12.17)
in Ω′ , where c is some constant. Since Ω′ is a region, Theorem 10.18 asserts that the result
(12.17) also holds in Ω. Using the substitution ζ = ez again, the result (12.17) becomes
f (ζ) = cζ −λ
in A(R1 , R2 ). Recall that f ∈ H A(R1 , R2 ) , it forces that λ ∈ Z. This ends the analysis of the
problem.
Problem 12.9
Rudin Chapter 12 Exercise 9.
Proof. We prove the assertions one by one.
• |f (z)| ≤ 1 in the open right half plane Π. If α = 0, then |f (z)| ≤ Ae in Π. Since
Π is unbounded and ∂Π is exactly the imaginary axis, we may apply Problem 12.11 to
conclude that |f (z)| ≤ 1 in Π. Next, if α < 0, then there exists a R > 0 such that |z| ≥ R
implies
|f (z)| ≤ 1.
(12.18)
Since D(0; R) ∩ Π is compact, it is bounded. Therefore, Theorem 10.24 (The Maximum
Modulus Theorem), the Extreme Value Theorem and the hypothesis |f (iy)| ≤ 1 for all
y ∈ R ensure that the inequality (12.18) is also valid for all z ∈ Π. Thus we may assume
that 0 < α < 1 in the following discussion.
Denote Ω = {x + iy | x ∈ R, |y| < π2 }. Consider the mapping ϕ : Ω → Π defined by
ϕ(z) = ez
(12.19)
which is clearly an isomorphism and ϕ ∈ H(Ω). Next, we define g : Ω → C by g = f ◦ ϕ.
By Problem 10.14, we know that g ∈ H(Ω). By the definition of (12.19), we see that
ϕ(∂Ω) = ∂Π. Since ϕ and f are continuous on Ω and Π respectively, g is continuous on
Ω. Furthermore, we have
πi πi g x±
= f exp x ±
= |f (±iex )| ≤ 1
2
2
12.1. Applications of the Maximum Modulus Principle
75
for all x ∈ R. Finally, if z ∈ Ω, then we have
|g(z)| = |f ϕ(z) | < A exp(|ϕ(z)|α ) < A exp(|ez |α ) = A exp(eαx ) ≤ A exp(eα|x| ). (12.20)
Choose B such that B − 1 ≥ log A. Note that eα|x| ≥ 1 for all x ∈ R. Then it is easy to
see that
log A ≤ (B − 1)eα|x|
log A + eα|x| ≤ Beα|x|
A exp(eα|x| ) ≤ exp(Beα|x| ).
(12.21)
Combining the inequalities (12.20) and (12.21), the inequality
|g(z)| < exp(Beα|x| )
holds for all z = x + iy ∈ Ω. Hence Theorem 12.9 (The Phragmen-Lindelöf Theorem)
asserts that |g(z)| ≤ 1 in Ω which means that the inequality (12.18) holds in Π.
z
• The conclusion is false for α = 1. It is easy to check that the function f (z) = ee gives
a counterexample to the result.
• The modified result. Suppose that ∆ is an open sector between two rays from the origin
with sectoral angle βπ < π for some β > 1. Suppose that f is continuous on ∆, f ∈ H(∆)
and there are constants A < ∞ and α ∈ (0, β) such that
|f (z)| < A exp(|z|α )
(12.22)
for all z ∈ ∆. Furthermore, if |f (z)| ≤ 1 on ∂∆, then we have |f (z)| ≤ 1 in ∆.
To see this, let θ be the angle between the real axis and the ray nearest to it. Then we see
that the mapping φ : ∆ → Π defined by
φ(z) = −i(e−iθ z)β = −ie−iβθ exp(β log z)
is clearly an isomorphism such that φ maps the boundary of ∆ onto the boundary of Π.
Since 0 ∈
/ ∆, we can define a branch for log z so that φ ∈ H(∆). By Theorem 10.33, we
−1
have φ ∈ H(Π). Next, the map F : Π → C defined by
F = f ◦ φ−1
1
1
1
is continuous on Π and F ∈ H(Π). Since φ−1 (ζ) = eiθ i β ζ β , we have |φ−1 (ζ)| = |ζ| β and
thus the hypothesis (12.22) implies
where
α
β
α
|F (ζ)| = |f φ−1 (ζ) | < A exp(|φ−1 (ζ)|α ) = A exp(|ζ| β ),
< 1. If ζ = iy for some y ∈ R, then since φ−1 (iy) lies on ∂∆, we get
|F (iy)| = |f φ−1 (ζ) | ≤ 1.
Hence we establish from our first assertion that |F (ζ)| ≤ 1 in Π or equivalently, |f (z)| ≤ 1
in ∆.
We have completed the analysis of the problem.
76
Chapter 12. The Maximum Modulus Principle
Problem 12.10
Rudin Chapter 12 Exercise 10.
Proof. For each n = 1, 2, 3, . . ., we define gn (z) = f (z)enz . Suppose that Lα = {z = reiα | r ≥ 0},
o
n
n
πo
π
.
and ∆2 = z ∈ Π α < arg z <
∆1 = z ∈ Π − < arg z < α
2
2
Certainly, ∆1 and ∆2 have the sectoral angles α +
Figure 12.2:
π
2
< π and
π
2
− α < π respectively. Refer to
Figure 12.2: The sectors ∆1 , ∆2 and the ray Lα .
In order to use Problem 12.9 as stated in the hint, it is necessary to assume that f is
continuous on the closure of Π. Since f ∈ H(Π) and ∆1 , ∆2 are proper subsets of Π, each gn is
2π
continuous on ∆j and gn ∈ H(∆j ) for j = 1, 2. Choose an γ ∈ (0, 2α+π
) and let Fn (x) = nx − xγ
for x ≥ 0. By elementary calculus, the Fn attains its maximum value
n 1 γ − 1
γ−1
Mn = n
γ
γ
1
at x = (nγ −1 ) γ−1 > 0. Therefore, if we pick An = 1 + eMn , then we have
exp(n|z| − |z|γ ) = eFn (|z|) ≤ eMn < An
(12.23)
for all z ∈ ∆1 . Since |f (z)| < 1 for all z ∈ Π, for each fixed n ∈ N, we note from the inequality
(12.23) that
|gn (z)| < exp(n|z|) < An exp(|z|γ )
for all z ∈ ∆1 . Consequently, each gn satisfies the inequality (12.22). Furthermore, we observe
from the definition of gn that
log |gn (reiα )|
log |f (reiα )|
=
+ n cos α → −∞
r
r
as r → ∞, so |gn (z)| is bounded on Lα . Without loss of generality, we may assume that the
bound is 1. Obviously, we know from the additional assumption that
π
π
(12.24)
|gn re±i 2 | = |f re±i 2 | < 1
12.1. Applications of the Maximum Modulus Principle
77
for all r ≥ 0. In other words, gn is bounded by 1 on ∂∆1 . By the modified result of Problem
12.9, we conclude that each gn is bounded by 1 in ∆1 . Similarly, each gn is also bounded by
1 on ∂∆2 and then in ∆2 . Since Π = ∆1 ∪ Lα ∪ ∆2 , what we have shown is that each gn is
bounded by 1 in Π, so for every z ∈ Π, this implies that
|f (z)| < e−nr cos θ
for all n ∈ N which means f (z) = 0 because cos θ > 0 for θ ∈ (− π2 , π2 ), i.e., f = 0 as required.
This completes the proof of the problem.
Problem 12.11
Rudin Chapter 12 Exercise 11.
Proof. Since the result is trivial if f is constant, we assume that f is nonconstant in the following
discussion. Besides, without loss of generality, we may assume that |f (z)| ≤ 1 on Γ = ∂Ω. It
suffices to prove that
|f (ω)| ≤ 1
(12.25)
for every ω ∈ Ω. We choose a ∈ Ω and consider
f (z) − f (a)
.
fe(z) =
z−a
Using Theorem 10.16 and then Theorem 10.6, we see that fe ∈ H(Ω). Furthermore, the continuity
of f on Ω ∪ Γ certainly implies that fb is also continuous on Ω ∪ Γ. Now the boundedness of f
guarantees that fe(z) → 0 as z → ∞. In other words, there is a positive constant C such that
|fe(z)| ≤ C
(12.26)
in Ω ∪ Γ. Next, let ΩR = D(0; R) ∩ Ω ⊆ Ω for R > 0 and Fe(z) = f N (z)fe(z) for some N ∈ N.
Clearly, we have
Fe ∈ H(Ω).
By the boundedness of f and the fact fe(z) → 0 as z → ∞, we can find R large enough such
that ω ∈ ΩR and |Fe(z)| ≤ C for all z ∈ ∂ΩR = C(0; R) ∩ Γ ⊆ Γ from the bound (12.26). By
Theorem 10.24 (The Maximum Modulus Theorem) and the Extreme Value Theorem, we assert
that
|Fe (ω)| ≤ C.
(12.27)
If fe(ω) 6= 0, then we deduce from the inequality (12.27) that
1
CN
|f (ω)| ≤
1 .
|fe(ω)| N
(12.28)
Taking N → ∞ in the inequality (12.28) which yields the required result (12.25) immediately.
Consequently, the inequality
|f (ω)| ≤ 1
holds for all ω ∈ Ω \ Zfe.
Assume that Zfe had a limit point in Ω. Now Theorem 10.18 ensures that fe ≡ 0 in Ω and then
f (z) = f (a) for all z ∈ Ω, a contradiction to our hypothesis that f is nonconstant. Therefore,
Zfe is discrete and hence the continuity of f forces definitely that the inequality (12.25) remains
valid in Ω which completes the proof of the problem.
78
Chapter 12. The Maximum Modulus Principle
12.2
Asymptotic Values of Entire Functions
Problem 12.12
Rudin Chapter 12 Exercise 12.
Proof. Let E1 = {z ∈ C | |f (z)| > 1}. Since f is nonconstant, E1 6= ∅. Let F1 be a component
of E1 . By Definition 10.1, F1 is an open set. By the continuity of f , it is true that |f (z)| ≥ 1
on ∂F1 . Assume that |f (z0 )| > 1 for some z0 ∈ ∂F1 . Obviously, z0 ∈ E1 . Furthermore, we
also have |f (ω)| > 1 for all ω ∈ D(z0 ; δ) for some δ > 0 by the Sign-preserving Property [79,
Problem 7.15, p. 112]. Therefore, D(z0 ; δ) ⊆ E1 . By [42, Theorem 25.1, p. 159], we know that
D(z0 ; δ) intersects only F1 which is impossible by the definition of a boundary point of a set
[4, Definition 3.40, p. 64]. Hence we have |f (z)| = 1 on ∂F1 . Assume that F1 was bounded.
Since F1 is compact, it follows from Theorem 10.24 (The Maximum Modulus Theorem) and
the Extreme Value Theorem that f attains its maximum on ∂F1 . Thus |f (z)| ≤ 1 on F1 ,
contradicting to the fact that F1 ⊆ E1 . Consequently, every component of E1 is unbounded.
Now for every n = 2, 3, . . ., we define
En = {z ∈ C | |f (z)| > n}.
By similar argument, it can be shown easily that every component of En is unbounded.
Since f is unbounded on F1 , En 6= ∅ for every n = 2, 3, . . .. Clearly, we have En+1 ⊆ En for
every n ∈ N. Let Fn+1 be a component of En+1 . By similar argument of the previous paragraph,
Fn+1 is open in C so that it is a region. By the definition, we have Fn+1 ⊆ En and then it must
lie entirely in a component of En , namely Fn . Hence, we obtain a sequence of regions
F1 ⊇ F2 ⊇ · · · .
(12.29)
For each k = 1, 2, . . ., pick zk ∈ Fk . Now the sequence (12.29) ensures that zn ∈ Fk for all
n ≥ k. Using [9, Proposition 1.7, p. 14], one can find a continuous mapping γk : [k − 1, k] → Fk
connecting zk and zk+1 . We note that the definition
of Ek asserts that |f (z)| > k for all z ∈ Fk , so
particularly, it is also true on γk , i.e., |f γ(t) | > k for every t ∈ [k −1, k]. Define γ : [0, ∞) → F1
by
γ = γ1 +̇ γ2 +̇ · · · .
t
), then γ
e is a
Then it is easy to see that f (γ(t)) → ∞ as t → ∞. Finally, if we set γ
e(t) = γ( 1−t
well-defined continuous function on [0, 1) and satisfies
lim f (e
γ (t)) = ∞,
t→1
completing the proof of the problem.
Problem 12.13
Rudin Chapter 12 Exercise 13.
Proof. If z = x < ∞, then |ez | = e−x → 0 as |z| → ∞. By the definition, 0 is an asymptotic
value of ez . Since ez is nonconstant entire, ∞ is also one of its asymptotic value by Problem
12.12. Hence it remains to show that if α is an asymptotic value of ez , then α is either 0 or
∞. Let γ : [0, 1) → C be a continuous curve such that γ(t) → ∞ and exp(γ(t)) → α as t → 1.
Let γ(t) = a(t) + ib(t), where a and b are continuous real-valued functions. Then we note that
a2 (t) + b2 (t) → ∞ as t → 1.
12.3. Further Applications of the Maximum Modulus Principle
79
If a(t) → +∞ as t → 1, then it is clear that | exp(γ(t))| = exp(a(t)) → ∞ as t → 1 so that
α = ∞. Next, if a(t) → −∞ as t → 1, then we | exp(γ(t)) = exp(a(t)) → 0 as t → 1 and so
α = 0 in this case. Finally, if a(t) → A as t → 1 for some finite A, then b(t) → ∞ as t → 1.
Since b is continuous, there exist sequences {tn } and {t′n } in [0, 1) such that b(tn ) = 2nπ and
b(t′n ) = (2n + 1)π respectively. Thus we have
eγ(tn ) = ea(tn ) × eib(tn ) → A and
′
′
′
eγ(tn ) = ea(tn ) × eib(tn ) → −A
as n → ∞. These imply that α = A = −A and so α = A = 0. In conclusion, we have shown
that exp has exactly two asymptotic values: 0 and ∞.
For the entire functions sin z and cos z, we notice that
sin z =
eiz − e−iz
2i
and
cos z =
eiz + e−iz
.
2
Therefore, ∞ is the only asymptotic value of sin z and cos z, completing the proof of the problem.
Problem 12.14
Rudin Chapter 12 Exercise 14.
Proof. Suppose that f is nonconstant. Since f (z) 6= α for all z ∈ C, the function
F (z) =
1
f (z) − α
is nonconstant entire. Now Problem 12.12 implies that F has ∞ as an asymptotic value which
means that α is an asymptotic value of f . This completes the proof of the problem.
12.3
Further Applications of the Maximum Modulus Principle
Problem 12.15
Rudin Chapter 12 Exercise 15.
Proof. The case is trivial if f is constant. So we may assume that f is nonconstant. Suppose
first that Z(f ) = ∅. Then we have f1 ∈ H(U ). For every n ∈ N, we have 0 < 1 − n1 < 1. By
combining Theorem 10.24 (The Maximum Modulus Theorem) and the Extreme Value Theorem,
we see that
1
1
≤
max
1
|f (0)| z∈C(0;1− n ) |f (z)|
or equivalently,
|f (0)| ≥
min
1
)
z∈C(0;1− n
|f (z)|.
We simply take zn = 1 − n1 which gives |f (zn )| ≤ |f (0)| for all n ∈ N. Obviously, |zn | → 1 as
n → ∞, so our assertion is true in this case.
Next, we suppose that Z(f ) 6= ∅. Then there are two cases.
80
Chapter 12. The Maximum Modulus Principle
• Case (i): Z(f ) is infinite. Since Z(f ) ⊆ U which is bounded, the Bolzano-Weierstrass
Theorem [79, Problem 5.25, pp. 68, 69] ensures that Z(f ) has a convergent subsequence
{ζk }. Now Theorem 10.18 forces that ζk → ζ ∈ C(0, 1). Since f (ζk ) = 0 for all k ∈ N, the
assertion remains true in this case.
• Case (ii): Z(f ) is finite. Suppose that Z(f ) = {ζ1 , ζ2 , . . . , ζN } for some N ∈ N. Suppose
further that mk is the order of zero of f at ζk , where 1 ≤ k ≤ N . Consider
g(z) =
f (z)
N
Y
.
(12.30)
(z − ζk )mk
k=1
Therefore, we know that g ∈ H(U ) and Z(g) = ∅. Thus the special case implies that there
is a sequence {zn } ⊆ U and a positive constant M such that |zn | → 1 and |g(zn )| ≤ M for
all n ∈ N. Hence it follows from the representation (12.30) that
|f (zn )| ≤ M
N
Y
k=1
mk
|zn − ζk |
≤M
N
Y
(1 + |ζk |) < ∞
k=1
for all n ∈ N.
Consequently, we have completed the proof of the problem.
Problem 12.16
Rudin Chapter 12 Exercise 16.
Proof. The result is always true if f is a constant function. Without loss of generality, we may
assume that f is nonconstant. Let α = sup{|f (z)| | z ∈ Ω}. If |f (ζ)| = α for some ζ ∈ Ω, then
f is constant by Theorem 10.24 (The Maximum Modulus Theorem), a contradiction. Thus we
always have
|f (z)| < α
(12.31)
for all z ∈ Ω. By the definition, there is a sequence {zn } ⊆ Ω such that |f (zn )| → α as n → ∞.
Since Ω is bounded, the Bolzano-Weierstrass Theorem ensures that {zn } contains a convergent
subsequence {znk }. Let znk → z0 . If z0 ∈ Ω, then α = |f (z0 )| which is impossible. Therefore,
we must have z0 ∈ ∂Ω and our hypothesis gives α ≤ M . By the inequality (12.31), we conclude
that
|f (z)| < α ≤ M
for all z ∈ Ω. This ends the proof of the analysis.
Problem 12.17
Rudin Chapter 12 Exercise 17.
Proof. By the definitions, we have
Φ = {f ∈ H(U ) | 0 < |f (z)| < 1 for all z ∈ U } and
where 0 < c < 1.
Φc = {f ∈ Φ | f (0) = c},
12.3. Further Applications of the Maximum Modulus Principle
81
• The value of M . Without loss of generality, we may assume that f (0) > 0. Otherwise,
we can consider the function fb = eiθ f , where θ = − arg f (0). Then fb(0) = |f (0)| > 0 and
fb ∈ Φ.
Let Ω− = {z ∈ C | Re z < 0}. Since 0 < |f (z)| < 1 for z ∈ U , the mapping f1 = log f
maps U into Ω− . Next, the mapping f2 (z) = −iz clearly maps Ω− onto the upper half
plane Π+ . Finally, for Im α > 0, we know that the mapping
f3 (z) =
z−α
z−α
maps Π+ into U . Hence the mapping F = f3 ◦ f2 ◦ f1 maps U into U . Since f1 ∈ H(U ),
f2 ∈ H(Ω− ) and f3 ∈ H(Π+ ), it is true that F ∈ H(U ). Clearly, the definition of F
implies that F ∈ H ∞ , kF k∞ ≤ 1 and
F (0) =
−i log f (0) − α
.
−i log f (0) − α
(12.32)
If we take α = −i log f (0), then Im α = − log f (0) > 0 because 0 < f (0) < 1. Thus the
expression (12.32) gives F (0) = 0 in this case. By Theorem 12.2 (Schwarz’s Lemma), we
have |F (z)| ≤ |z| for all z ∈ U and |F ′ (0)| ≤ 1.
Now the explicit formula of F is given by
(z)
log ff (0)
−i log f (z) + i log f (0)
=
F (z) =
−i log f (z) − i log f (0)
log[f (0)f (z)]
so that
F ′ (z) =
f ′ (z)
log f (0)2
·
.
f (z) {log[f (0)f (z)]}2
(12.33)
Since |F ′ (0)| ≤ 1, we see from the formula (12.33) that
|f ′ (0)| ≤ 2|f (0) log f (0)|.
(12.34)
Elementary calculus shows that the function g : (0, 1) → R defined by g(x) = x log x
attains its absolute minimum −e−1 at x = e−1 . Therefore, we have M ≤ 2e−1 .
Next, we claim that M = 2e−1 . To see this, we consider the function f (z) = e−2z−1 which
is holomorphic in U . Since e−2 < |e2z | = e2r cos θ < e2 for all z = reiθ ∈ U , it is easy to see
that
0 < e−3 < |f (z)| < e−1 < 1
in U so that f ∈ Φ. As f ′ (z) = −2e−2z−1 , we have |f ′ (0)| = 2e−1 = M as required.
• The value of M (c). Since f (0) = c, it follows from the inequality (12.34) that

 2c| log c|, if c < e−1 ;
M (c) =
 −1
2e ,
if e−1 ≤ c < 1.
Hence we complete the analysis of the problem.
Remark 12.1
The first assertion of Problem 12.17 is called Rogosinski’s Theorem. See, for instances,
[57] and [15, Exercise 6.36, pp. 213, 214] for a different proof.
82
Chapter 12. The Maximum Modulus Principle
CHAPTER
13
Approximations by Rational Functions
13.1
Meromorphic Functions on S 2 and Applications of Runge’s Theorem
Problem 13.1
Rudin Chapter 13 Exercise 1.
Proof. Suppose that f is meromorphic on S 2 and A ⊆ S 2 is the set of poles of f . If A is infinite,
then the compactness of S 2 implies that A has a limit point in S 2 . However, this contradicts
the note following Definition 10.41 and so A must be finite. Let A = {a1 , a2 , . . . , aN } for some
N ∈ N and mk be the order of ak for 1 ≤ k ≤ N . If we define
P (z) = f (z) ·
N
Y
(z − ak )mk ,
k=1
then this expression implies that P (z) is entire and has at most a pole at ∞. If ∞ is not a pole
of f , then P is a constant by Theorem 10.23 (Liouville’s Theorem). Otherwise, the function
P ( z1 ) has a pole at 0. If P (z) = c0 + c1 z + c2 z 2 + · · · , then we have P ( 1z ) = c0 + cz1 + zc22 + · · ·
and the nature of its singularity at 0 implies that cp = cp+1 = · · · = 0 for some p ∈ N. In other
words, P must be a polynomial. Since
f (z) =
P (z)
N
Y
,
(z − ak )mk
k=1
f must be rational which completes the proof of the problem.
Problem 13.2
Rudin Chapter 13 Exercise 2.
Proof.
(a) It is clear that Ω is simply connected, so S 2 \ Ω is connected by Theorem 13.11. Hence
Theorem 13.9 (Runge’s Theorem) implies that there exists a sequence {Pn } of polynomials
such that Pn → f uniformly on compact subsets of Ω. See Figure 13.1 for details.
83
84
Chapter 13. Approximations by Rational Functions
Figure 13.1: The simply connected set Ω.
(b) The answer is negative. Consider the function
f (z) =
1
z−
1
2
(13.1)
which belongs to H(Ω). Assume that there was a sequence {Pn } of polynomials such that
Pn → f uniformly in Ω. Take γ = C(− 34 ; 14 ). Then we have γ ⊆ Ω, see Figure 13.1 again.
On the one hand, we have
Z
1
= 2πi.
f (z) dz = 2πi · Ind γ
2
γ
Furthermore, Theorem 10.12 gives
Z
Pn (z) dz = 0
γ
for every n ∈ N. On the other hand, the uniform convergence shows that
Z
Z
Pn (z) dz = f (z) dz = 2πi,
0 = lim
n→∞ γ
γ
a contradiction. Hence no such sequence exists.
(c) The answer is still negative. The function (13.1) considered in part (b) is in fact holomorphic in C \ { 12 } which is open in C and it contains Ω.
We complete the proof of the problem.
Problem 13.3
Rudin Chapter 13 Exercise 3.
Proof. For every n ∈ N, let Dn = {z ∈ D(0; n) | |Im z| ≥ n1 } and En = [ n1 , n] × {0}, see Figure
13.2. It is clear that both Dn and En are compact, so the set Kn = Dn ∪ En ∪ {0} is compact
1
. Then the sets
too. Furthermore, we note that S 2 \ Kn is connected. Take 0 < δn < 2n
o
n
1
1
and En′ =
Dn′ = z ∈ D(0; n + δn ) |Im z| ≥ − δn
− δn , n + δn × (−δn , δn )
n
n
13.1. Meromorphic Functions on S 2 and Applications of Runge’s Theorem
85
are open sets containing Dn and En respectively. Obviously, the set Dn′ ∪ En′ is open in C and
is disjoint from Un = (−δn , δn ) × (−δn , δn ). Define the function fn : Ωn = Dn′ ∪ En′ ∪ Un → C by

 1, if z ∈ Un ;
fn (z) =

0, if z ∈ Dn′ ∪ En′ .
Then we have fn ∈ H(Ωn ) and Ωn is an open set containing Kn . According to Theorem 13.7,
one can find a polynomial Qn such that |Qn (z) − fn (z)| < n1 for all z ∈ Kn . In fact, we get

 |Qn (z) − 1|, if z ∈ Un ;
|Qn (z) − fn (z)| =
(13.2)

|Qn (z)|,
if z ∈ Dn′ ∪ En′ .
If we define Pn (z) = Qn (z) − Qn (0) + 1, then the definition (13.2) implies immediately that
Pn (0) = Qn (0) − Qn (0) + 1 = 1 for n = 1, 2, . . .. Since we have
C \ {0} =
∞
[
(Dn′ ∪ En′ ),
n=1
if z 6= 0, then there exists an N ∈ N such that z ∈ Dn′ ∪ En′ for all n ≥ N and thus the definition
(13.2) implies that
|Pn (z)| = |Qn (z) − Qn (0) + 1| ≤ |Qn (z)| + |Qn (0) − 1| <
2
n
for all n ≥ N . Consequently, we conclude from this that Pn (z) → 0 as n → ∞, completing the
proof of the problem.
Figure 13.2: The compact sets Dn and En .
86
Chapter 13. Approximations by Rational Functions
Problem 13.4
Rudin Chapter 13 Exercise 4.
Proof. For every n ∈ N, we consider the three sets An = [−n, n] × [ n1 , n], Bn = [−n, n] × {0}
and Cn = [−n, n] × [− n1 , −n], see Figure 13.3 for an illustration. Let
Kn = An ∪ Bn ∪ Cn .
1
Obviously, each Kn is compact and the set S 2 \ Kn is connected. Choose 0 < δn < 2n
. Then
1
′
′
the three sets An = (−n − δn , n + δn ) × ( n − δn , n + δn ), Bn = (−n − δn , n + δn ) × (−δn , δn ) and
Cn′ = (−n − δn , n + δn ) × (− n1 − δn , n + δn ) are open sets containing An , Bn and Cn respectively.
Besides, they are disjoint and Kn ⊆ Ωn = A′n ∪ Bn′ ∪ Cn′ . Define

1,
if z ∈ A′n ;





0,
if z ∈ Bn′ ;
fn (z) =





−1, if z ∈ Cn′ .
Since fn ∈ H(Ωn ), it follows from Theorem 13.7 that there
|Pn (z) − fn (z)| < n1 for all z ∈ Kn . In fact, we have

|Pn (z) − 1|, if





|Pn (z)|,
if
|Pn (z) − fn (z)| =





|Pn (z) + 1|, if
exists a polynomial Pn such that
z ∈ An ;
z ∈ Bn ;
z ∈ Cn .
Therefore, such sequence {Pn } of polynomials satisfy the requirements of the problem. This
completes the proof of the problem.
Figure 13.3: The compact sets An , Bn and Cn .
13.2. Holomorphic Functions in the Unit Disc without Radial Limits
13.2
87
Holomorphic Functions in the Unit Disc without Radial Limits
Problem 13.5
Rudin Chapter 13 Exercise 5.
1
1
), Cn = {(1 − 2n+1
)eiθ | π2 ≤ θ ≤ 2π},
Proof. For each n ∈ N, suppose that ∆n = D(0; 1 − 2n
1
1
1
π
iθ
Dn = [1 − 2n+1 , 1 − 2n+2 ] and En = {(1 − 2n+2 )e | 0 ≤ θ ≤ 2 }. Then the union
Ln = Cn ∪ Dn ∪ En
is an arc in U \ ∆n with the property that each Ln intersects every radius of U . Finally, we
1
suppose that Ωn is a tubular region of Ln with width 2(2n+2)
4 . The construction ensures that
Ωn ∩ ∆n = ∅, see Figure 13.4 for an illustration. Similar to Problem 13.4, we can select very
1
small δn > 0 such that the modified tubes Ω′n of Ln with widths 2(2n+2)
4 + δn and Πn =
D(0; 1 −
1
2n
+ δn ) also satisfy Ω′n ∩ Πn = ∅. Notice that Ω′n ⊆ Ωn .
Figure 13.4: The disc ∆n , the arc Ln and its neighborhood Ωn .
We apply induction to construct the sequence of polynomials {Qn } and a holomorphic function f such that Qn → f uniformly on U : Consider Q0 ≡ 0 and

 Q0 (z), if z ∈ Π◦1 ;
f1 (z) =

1,
if z ∈ Ω′1 .
Obviously, we have f1 ∈ H(Π◦1 ∪ Ω′1 ). Since C \ (Π◦1 ∪ Ω′1 ) is connected, it follows from Theorem
13.9 (Runge’s Theorem) that one can find a polynomial Q1 such that
|Q1 (z) − f1 (z)| <
1
2
(13.3)
88
Chapter 13. Approximations by Rational Functions
on compact subsets of Π◦1 ∪Ω′1 . Particularly, the estimate (13.3) holds for all z ∈ ∆1 ∪Ω1 because
∆1 ⊆ Π◦1 and Ω1 ⊆ Ω′1 .
Assume that the polynomial Qn is chosen with the property that
|Qn (z) − fn (z)| <
1
2n
on ∆n ∪ Ωn . We consider the function defined by

Q (z),
if z ∈ Π◦n+1 ;

 n
fn+1 (z) =
n

 1 + (−1) , if z ∈ Ω′ .
n+1
2
(13.4)
(13.5)
Then we have fn+1 ∈ H(Π◦n+1 ∩ Ω′n+1 ). Since C \ (Π◦n+1 ∩ Ω′n+1 ) is connected, Theorem 13.9
(Runge’s Theorem) guarantees that one can find a polynomial Qn+1 such that
|Qn+1 (z) − fn+1 (z)| <
1
2n+1
(13.6)
on compact subsets of Π◦n+1 ∩ Ω′n+1 and thus on ∆n+1 ∪ Ωn+1 because ∆n+1 ⊆ Π◦n+1 and
Ωn+1 ⊆ Ω′n+1 . Consequently, we have constructed the polynomial Qn+1 and then a sequence of
polynomials {Qn } satisfying the property (13.4).
Next, let N be a positive integer. Then we have ∆N ⊆ ∆n for all n ≥ N . Therefore, it
deduces from the definition (13.5) and the inequality (13.6) that
|Qn+1 (z) − Qn (z)| <
1
2n+1
(13.7)
holds for all z ∈ ∆N ⊆ ∆n ⊆ ∆n+1 and all n ≥ N . Given ǫ > 0. We pick the N large enough so
that 21N < ǫ. For all positive integers p and n ≥ N , we observe from the inequality (13.7) that
|Qn+p (z) − Qn (z)| ≤ |Qn+p (z) − Qn+p−1 (z)| + · · · + |Qn+1 (z) − Qn (z)|
1
1
1
< n+p + n+p−1 + · · · + n+1
2
2
2
1
< n
2
<ǫ
for all z ∈ ∆N . Hence it follows from the Cauchy Criterion for Uniform Convergence (see [61,
Theorem 7.8, p. 147]) that the sequence of polynomials {Qn } converges uniformly to a function
f on ∆N . By Theorem 10.28, we have f ∈ H(∆◦N ). Since it is true for every large N and
U=
∞
[
∆N ,
N =1
we obtain f ∈ H(U ) and Qn → f uniformly on compact subsets of U . Finally, we define
Pn = Qn − Qn−1 for every n = 1, 2, . . .. Each Pn is also a polynomial and
n
X
Pk = Qn .
k=1
Therefore, we have constructed a sequence of polynomials {Pn } such that the series f =
is holomorphic in U .
∞
X
n=1
Pn
13.2. Holomorphic Functions in the Unit Disc without Radial Limits
The above construction of {Qn } starts with the inequality (13.3), where
replaced by any small δ > 0 so that
1
2
89
can be actually
|P1 (z)| = |Q1 (z) − Q0 (z)| = |Q1 (z)| < δ
on ∆1 . By the inequality (13.7) again, we obtain
|Pn (z)| = |Qn (z) − Qn−1 (z)| < δn
on ∆n for every n = 2, 3, . . .. This means that the polynomials Pn are very small on ∆n .
Furthermore, we can also replace the values 0 and 1 by another Qn + g ∈ H(Ω′n+1 ) in the
definition (13.5), and hence Qn + g ∈ H(Ωn+1 ). As a result, we obtain from the inequality
(13.6) that
1
|Pn+1 (z) − g(z)| < n+1
2
in Ωn+1 . In other words, it means that the sequence {Pn } can be chosen more or less arbitrary
on Ln , i.e., any holomorphic g can be approximated by the sequence of polynomials {Pn } in a
neighborhood of Ln .
Now it remains to show that f has no radial limit at any point of T . To this end, by the
definitions of ∆n and Ωn , we see easily that Ω2n ⊆ ∆p for all p ≥ 2n + 2. On the one hand,
1
.
suppose that z ∈ Ω2n ⊆ Ω′2n . Then we recall from the definition (13.5) that |Q2n (z)| < 22n
Combining this fact, Qn → f in U and the inequality (13.7), we know that
|f (z)| ≤ |f (z) − Q2n (z)| + |Q2n (z)|
≤ |Q2n+1 (z) − Q2n (z)| + |f (z) − Q2n+1 (z)| + |Q2n (z)|
≤ |Q2n+1 (z) − Q2n (z)| + |Q2n+2 (z) − Q2n+1 (z)| + |f (z) − Q2n+2 (z)| + |Q2n (z)|
∞
X
≤
|Qk+1 (z) − Qk (z)| + |Q2n (z)|
<
k=2n
∞
X
k=2n
≤
1
2k+1
1
22n−1
+
1
22n
.
(13.8)
On the other hand, if z ∈ Ω2n+1 ⊆ Ω′2n+1 , then we have |Q2n+1 (z) − 1| <
|f (z) − 1| ≤
<
∞
X
k=2n+1
∞
X
k=2n+1
≤
1
22n+1
so that
|Qk+1 (z) − Qk (z)| + |Q2n+1 (z) − 1|
1
2k+1
+
1
22n+1
1
.
22n
(13.9)
Assume that lim f (reiθ0 ) existed for some θ0 ∈ [0, 2π). Since every Ln intersects the radius
r→1
ℓ0 = {reiθ0 | 0 ≤ r < 1}, every Ωn also intersects ℓ0 . In other words, we can choose a sequence
{zn = rn eiθ0 } tending to eiθ0 ∈ T with the property that z2n+1 ∈ Ω2n+1 and z2n ∈ Ω2n . Then
the above estimates (13.8) and (13.9) show that
lim f (z2n ) = 0 and
n→∞
lim f (z2n+1 ) = 1
n→∞
which are contrary. Hence this contradiction means that our f has no radial limit on T , com
pleting the analysis of the problem.
90
Chapter 13. Approximations by Rational Functions
Problem 13.6
Rudin Chapter 13 Exercise 6.
Proof. Let’s prove the assertions one by one.
• The series converges if |z| < 1. It is easily seen that
 k−1
 22
nk ≥

For large k, since
if 1 ≤ k ≤ 3;
(k − 1)! · 2k if k ≥ 4.
k
k
≤
,
nk
(k − 1)! · 2k
we have
k
nk
→ 0 as k → ∞ and then
k
lim sup 5 nk = 1.
k→∞
Hence it asserts from §10.5 that the radius of convergence of the series of h is 1, as required.
• There is a constant c > 0 such that |h(zm )| > c · 5m for all z with |zm | = 1 − n1m
and m ≥ 3. We remark that the hint is not true for the cases m = 1, 2. For example, take
n1 = 4 and n2 = 9, so we have
∞
3 X
4 4iθ
9 9iθ
iθ
= 5(0.75) e + 25(0.75) e +
5k (0.75)nk eiθnk .
h e
4
k=3
Since 0.754 < 0.32 and 5(0.75)9 > 0.375, the first term is not the dominant term in the
series defining h(z) which means the hint is not correct anymore. Examples for the case
m = 2 can be found similarly. However, we can show affirmative result if m ≥ 3. Suppose
that zm = (1 − n1m )eiθ .
1
nm
5m
h (1 −
eiθ )
=
m−1
X
k=1
5k 1 nm iθnm
1 nk iθnk e
+
1
−
1
−
e
5m
nm
nm
∞
X
+
k=m+1
5k 1 nk iθnk
e
1
−
5m
nm
i
h m−1
X 5k 1 nk iθnk
=
e
−
S
1
−
m−1 + (Sm−1 + am )
5m
nm
+
k=1
∞
X
k=m+1
where
Sm−1 =
m−1
X
k=1
5k 1 nk iθnk
,
e
1
−
5m
nm
5k iθnk
e
5m
1 nm iθnm
and am = 1 −
e
nm
for every m = 3, 4, . . .. By elementary calculus, we can show easily the following result:
13.2. Holomorphic Functions in the Unit Disc without Radial Limits
91
Lemma 13.1
If α ≥ 1, then the function
1 αx
fα (x) = 1 −
x
−α
is strictly increasing and 0 < fα (x) ≤ e
on [1, ∞). In addition, fα (x) → e−α as
x → ∞.
Particularly, Lemma 13.1 says that {|am |} is a strictly increasing sequence of positive
numbers such that
1
1
≤ |am | ≤
(13.10)
4
e
for all m = 1, 2, . . . and |am | → e−1 as m → ∞. In fact, if m ≥ 3, then |am | > 0.35. Simple
algebra gives
1 1 |Sm−1 | ≤ × 1 − m−1
4
5
for all m = 3, 4, . . .. Thus it is true that
|Sm−1 + am | ≥ |am | − |Sm | > 0.11
(13.11)
for all m = 3, 4, . . ..
Next, we know that
m−1
X
k=1
m−1
X 5k h
5k 1 nk iθnk
1 nk i
.
1−
− Sm−1 ≤
1− 1−
e
5m
nm
5m
nm
(13.12)
k=1
Using differentiation, we always have 1 − (1 − x)n ≤ nx for every x ∈ [0, 1] and n ∈ N. By
the definition of nm , we see that
nm > 2j+1 (m − 1)(m − 2) · · · (m − j − 1)nm−j−1
for j = 0, 1, . . . , m − 2. Applying these to the inequality (13.12) to obtain
m−1
X
k=1
m−1
1 nk iθnk
5k 1 X k nk
1−
− Sm−1 ≤ m
e
5 ·
5m
nm
5
nm
k=1
m−1
X
≤
1
5m
≤
1
m−1
5k
2m−k (m − 1)(m − 2) · · · k
k=1
m−1
X
k=1
1
2m−k
· 5k−m
1 1
· 1 − m−1
≤
8(m − 1)
5
< 0.06
(13.13)
for every m = 3, 4, . . ..
Clearly, we deduce from the upper bound (13.10) that
∞
X
k=m+1
1 nk iθnk
5k e
=
1
−
5m
nm
∞
X
∞
nk
5k
1 X 5k
5
nm
·a
≤ nm+1 + m
nk .
5m k
5
n
e nm
k=m+1
k=m+2 e m
(13.14)
92
Chapter 13. Approximations by Rational Functions
As m ≥ 3, we get
nm+1
nm
≥ 6 which implies that
5
e
≤
nm+1
nm
5
< 0.0124.
e6
(13.15)
Since nk > 2(k − 1)nm for k = m + 2, m + 3, . . ., we have
nk
e nm > e2(k−1)
and then the inequality (13.14) becomes
∞
X
k=m+2
∞
5k e2 X 5 k
1 nk iθnk
25
e
≤
< 0.026
1
−
= 2
5m
nm
5m
e2
(e − 5)e2m
(13.16)
k=m+2
for every m ≥ 3. Finally, by combining the bounds (13.11), (13.13), (13.15) and (13.16),
we conclude immediately that
m−1
X 5k h (1 − n1m )eiθ
1 nk iθnk
≥
1
−
− Sm−1
e
|S
+
a
|
−
m−1
m
5m
5m
nm
k=1
−
∞
X
5k
k=m+1
5m
1−
1 nk iθnk
e
nm
≥ 0.11 − 0.06 − 0.0124 − 0.026
> 0.
Consequently, there exists a constant C > 0 such that
|h(z)| > C · 5m
holds for every z with |z| = 1 −
1
nm
and m ≥ 3.
• Proof that h has no finite radial limits. If zm = (1 −
assertion shows that
1 iθ e
> C · 5m
h 1−
nm
1
iθ
nm )e ,
then the previous
for every m ≥ 3 and θ ∈ [0, 2π]. Therefore, it guarantees that
1 iθ = ∞.
e
lim h 1 −
m→∞
nm
In other words, h has no finite radial limits.
• The h has infinitely many zeros in U . Assume that h had only finitely many zeros
α1 , α2 , . . . , αp in U . Then the function
ϕ(z) = z
m
p
Y
k=1
ϕαk (z) = z
m
p
Y
z − αk
1 − αk z
(13.17)
k=1
has exactly the same zeros as h counted with multiplicity, where m is the order of zero of
h at the origin. If h has no zero in U , then we let ϕ = 1. Now the function f = ϕh satisfies
f ∈ H(U ). By the proof of Theorem 12.4, we know that |ϕαk (z)| < 1 if |z| < 1, so the
definition (13.17) implies that
|f (z)| > C · 5m
(13.18)
13.3. Simply Connectedness and Miscellaneous Problems
93
for every z with |z| = 1− n1m and m ≥ 3. Furthermore, f has no zero in U so that f1 ∈ H(U ).
Combining this fact, the inequality (13.18) and Theorem 10.24 (The Maximum Modulus
Theorem), we see immediately that
1
1
<
f (z)
C · 5m
for all z ∈ D(0; 1 − n1m ) and m ≥ 3. Since nm → ∞ as m → ∞, we conclude from this
1
that f (z)
= 0 in U which is impossible.
• The function h assumes every complex number α infinitely many times in U .
Define b
h(z) = f (z) − α. Then b
h ∈ H(U ). For large enough m, 5m > C2 |α| so that
C
|b
h(z)| = |h(z) − α| ≥ |h(z)| − |α| > C · 5m − |α| > · 5m
2
for |z| = 1 − n1m . Therefore, we can apply similar argument as above to obtain the desired
result.
This completes the analysis of the problem.
Remark 13.1
(a) A sequence {nk } of positive integers is said to be lacunary if there is a constant c > 1
such that nk+1 > cnk for all k ∈ N. A power series
∞
X
ak z n k
(13.19)
k=1
is called a lacunary power series or a power series with Hadamard gaps. Thus
our h is an example of this type of power series with c = 2. See, for instance, [84,
Chap. V].
(b) In [29, Problem 5.36 & Update 5.36, p. 113], it is pointed out that the best known
result concerning the number of zeros of a lacunary power series inside U is due to
Chang [16], who proved that if
∞
X
k=0
|ak |2+ǫ = ∞
for some ǫ > 0, then the series (13.19) has infinitely many zeros in any sector. See also
[26], [43] and [75]
13.3
Simply Connectedness and Miscellaneous Problems
Problem 13.7
Rudin Chapter 13 Exercise 7.
Proof. Suppose that A intersects each component of S 2 \ Ω. Choose a sequence of compact sets
Kn in Ω with the properties specified in Theorem 13.3. Fix a positive integer n. Let V be a
94
Chapter 13. Approximations by Rational Functions
component of S 2 \ Kn . By the proof of Theorem 13.9 (Runge’s Theorem), it suffices to prove
that V ∩ A 6= ∅.
Since every component of S 2 \ Kn contains a component of S 2 \ Ω, we have
U ⊆V
(13.20)
for at least one component U of S 2 \ Ω. Since A intersects every component of S 2 \ Ω, we have
A ∩ U 6= ∅.
(13.21)
Now the set relations (13.20) and (13.21) together imply that V ∩ A 6= ∅. If V ∩ A 6= ∅, then
we are done. Otherwise, p ∈ V ∩ A′ , where A′ is the set of limit points of A. By the openness
of V and the definition of limit points, there exists a δ > 0 such that q ∈ A ∩ D ′ (p; δ) and
D(p; δ) ⊆ V . This means that V ∩ A 6= ∅. Hence we have obtained the requirement and this
completes the analysis of the problem.
Problem 13.8
Rudin Chapter 13 Exercise 8.
Proof. Let Ω = C. For every n = 1, 2, . . ., we denote Kn = D(0; n) which is compact. Put
A1 = A ∩ K1 and An = A ∩ (Kn \ Kn−1 ) for n = 2, 3, . . .. Since An ⊆ Kn and A has no limit
point in C (hence none in Kn ), every An is a finite set. Put
X
Pα (z),
Qn (z) =
α∈An
where n = 1, 2, . . .. Since An is finite, each Qn is a rational function and the poles of Qn lie
in An for n ≥ 2. In particular, Qn is holomorphic in an open set V containing Kn−1 . By the
known fact given in Definition 10.5, the power series of Qn at 0 converges uniformly to Qn in
Kn−1 . This means that for each n = 2, 3, . . ., there exists a polynomial Rn such that
|Rn (z) − Qn (z)| <
1
2n
(13.22)
for all z ∈ Kn−1 .
By imitating the remaining part of the proof of Theorem 13.10 (The Mittag-Leffler Theorem),
it can be seen that the function
∞
X
[Qn (z) − Rn (z)],
f (z) = Q1 (z) +
n=2
where z ∈ C, has the desired properties. In fact, we fix a positive integer N first. On KN , we
have
∞
N
X
X
(Qn − Rn ).
(13.23)
(Qn − Rn ) +
f = Q1 +
n=2
n=N +1
Using the inequality (13.22), each term in the last sum in the expression (13.23) is less than 2−n
on KN , hence this last series converges uniformly on KN to a function gN +1 which is holomorphic
◦ . Since each R is a polynomial, the function
in KN
n
f − (Q1 + Q2 + · · · + QN ) = gN +1 −
N
X
Rn
n=2
◦ and therefore, f has precisely the prescribed principal parts in K ◦ . Since
is holomorphic in KN
N
N is arbitrary, it is actually true in C. This completes the analysis of the problem.
13.3. Simply Connectedness and Miscellaneous Problems
95
Problem 13.9
Rudin Chapter 13 Exercise 9.
Proof. As f (z) 6= 0 for all z ∈ Ω, we have f1 ∈ H(Ω). Since Ω is simply connected, Theorem
13.11 says that there exists an h ∈ H(Ω) such that
f = eh .
Take g = exp( nh ) which is obviously holomorphic in Ω. Hence we have
g n = eh = f
which ends the proof of the problem.
Problem 13.10
Rudin Chapter 13 Exercise 10.
Proof. We want to show that there exists a g ∈ H(Ω) such that f = exp g if and only if there
exists a ϕn ∈ H(Ω) such that f = ϕnn for every positive integer n.
Suppose that there exists a g ∈ H(Ω) such that f = exp g. Then the function ϕn = exp( ng )
satisfies ϕn ∈ H(Ω) and ϕnn = exp g = f for every positive integer n. Conversely, suppose that
there exists a ϕn ∈ H(Ω) such that f = ϕnn for every n = 1, 2, . . .. We claim that f (z) 6= 0 for
all z ∈ Ω. Assume that there was an a ∈ Ω such that f (a) = 0. Since f 6≡ 0, Theorem 10.18
asserts that there is a unique positive integer m such that
f (z) = (z − a)m g(z)
(13.24)
for some g ∈ H(Ω) and g(a) 6= 0. Particularly, we take n = m + 1. Since ϕm+1 (a) = 0, Theorem
10.18 ensures that
ϕm+1 (z) = (z − a)φm+1 (z)
(13.25)
for some φ ∈ H(Ω). If we combine the representations (13.24) and (13.25), then we conclude
that
g(z) = (z − a)φm+1
m+1 (z)
for all z ∈ Ω, but this implies that g(a) = 0, a contradiction which proves our claim. In other
words, we have f1 ∈ H(Ω). Next, we may take n = 2 so that f = ϕ22 for some ϕ2 ∈ H(Ω). Hence
it follows from Theorem 13.11 that f has a holomorphic logarithm g in Ω and thus we complete
the analysis of the problem.
Problem 13.11
Rudin Chapter 13 Exercise 11.
Proof. Put ϕ(z) = sup |fn (z)| in Ω. Now ϕ is well-defined because fn → f pointwise in Ω.
n∈N
Suppose that U is a nonempty open subset of Ω with U ⊆ Ω. Such a set exists because Ω is an
open set. For k = 1, 2, . . ., we define
Vk = {z ∈ U | |fn (z)| ≤ k for all n ∈ N} ⊆ U .
(13.26)
96
Chapter 13. Approximations by Rational Functions
By the hypothesis, we know that fn (z) → f (z) for every z ∈ U , the set {fn (z) | n ∈ N} must be
bounded for each fixed z. Thus each z ∈ U lies in some Vk , i.e.,
U=
∞
[
Vk .
k=1
Since U is a compact subset of the metric space C, it is a complete metric space. By Theorem
5.6 (Baire’s Theorem) (see also §5.7), it is impossible that all Vk are nowhere dense sets. Thus
there exists an N ∈ N such that the closure VN contains a nonempty disc DU of U , so it yields
from the definition (13.26) that
|fn (z)| ≤ N
for all n ∈ N and z ∈ DU . Equivalently, it means that ϕ(z) is bounded on DU and we may apply
Problem 10.5 to conclude that fn → f uniformly on every compact subset of DU . According to
Theorem 10.28, we have f ∈ H(DU ).
Now we let
V =
[
DU ,
U
where U is any arbitrary nonempty open subset of Ω with U ⊆ Ω. Obviously, we observe that
f ∈ H(V ). If D(z0 ; δ) ⊂ Ω and D(z0 ; δ) ∩ V = ∅ for some z0 ∈ Ω and δ > 0,a then the above
argument shows that there exists a nonempty open subset D of D(z0 ; δ) such that f ∈ H(D).
Therefore, D ⊆ V which implies the contradiction D(z0 ; δ) ∩ V 6= ∅. As a result, it means that
V is a dense open subset of Ω which ends the proof of the problem.
Remark 13.2
The result of Problem 13.11 is sometimes called Osgood’s Theorem [48]. In fact, Problem
10.5 is the well-known Vitali Convergence Theorem, see [72, p. 168].
Problem 13.12
Rudin Chapter 13 Exercise 12.
Proof. We regard C as R2 and consider the two-dimensional Lebesgue measure m2 . Suppose
that f : R2 → R2 is measurable. Let f = u + iv and RN = [−N, N ] × [−N, N ]. Consider the
measurable function
u|RN : RN → R,
where N ∈ N. By [67, Theorem 4.3, p. 32], there exists a sequence {ψN,k } of step functions
converging to u|RN for almost every z ∈ RN , i,e., there corresponds an pN ∈ N such that k ≥ pN
implies
1
(13.27)
u|RN (z) − ψN,k (z) < N +1
2
for almost every z ∈ RN .
Note that each ψN,k has the form
ψN,k (z) =
mk
X
αN,j χRN,j (z),
j=1
a
Without loss of generality, we may assume further that D(z0 ; δ) ⊆ Ω.
13.3. Simply Connectedness and Miscellaneous Problems
97
where {RN,j } forms a set of disjoint open rectangles and
mk
m
m
[k
[
[k
RN,j \ RN
m2 RN ∆
RN,j ∪
RN,j = m2 RN \
j=1
j=1
j=1
Suppose that
ΩN =
m
[k
!
1
<
2N +1
.
(13.28)
RN,j .
j=1
Then we can represent the inequality (13.28) as
m2 (RN ∆ΩN ) = m2 (RN \ ΩN ) ∪ (ΩN \ RN ) <
1
2N +1
.
By [67, Theorem 3.4, p. 21], we know that one can find a compact subset KN of ΩN such that
m2 (ΩN \ KN ) <
1
2N +1
.
Since we have RN \ KN ⊆ (RN \ ΩN ) ∪ (ΩN \ KN ) and KN \ RN ⊆ ΩN \ RN , we get
m2 (RN ∆KN ) = m2 (RN \ KN ) ∪ (KN \ RN )
≤ m2 [(RN \ ΩN ) ∪ (ΩN \ KN )] ∪ (ΩN \ RN )
≤ m2 (RN ∆ΩN ) + m2 (ΩN \ KN )
1
< N.
2
(13.29)
Obviously, each ΩN is open in C because each RN,j is an open set. Besides, as the complement
C \ ΩN is path-connected, it is also connected.b Since ψN,k is constant on each RN,j , it is
holomorphic in RN,j and hence in ΩN , i.e., ψN,k ∈ H(ΩN ). By Theorem 13.9 (Runge’s Theorem),
for each pair of fixed N and k, there is a polynomial sequence {QN,k,n } such that QN,k,n → ψN,k
as n → ∞ uniformly on compact subsets of ΩN . Fix the KN as above, there is an qN ∈ N such
that n ≥ qN implies that
1
(13.30)
|QN,k,n (z) − ψN,k (z)| < N +1
2
for all z ∈ KN . Combining the estimates (13.27) and (13.30), for k ≥ pN and n ≥ qN , we can
establish that
u|RN (z) − QN,k,n (z) ≤ u|RN (z) − ψN,k (z) + ψN,k (z) − QN,k,n (z) <
1
2N
(13.31)
for almost every RN ∩ KN . Recall that
C=
∞
[
RN .
N =1
By this fact and the inequality (13.29), we obtain
lim KN = C.
N →∞
If we pick UN (z) = QN,pN ,qN (z), then it can be established from the inequality (13.31) that
lim UN (z) = u(z)
N →∞
b
Refer to the paragraph following the definition in [42, p. 155].
(13.32)
98
Chapter 13. Approximations by Rational Functions
for almost every z ∈ C. Now the same result also holds for the imaginary part, i.e., there exists
a sequence of polynomials {VN } such that
lim VN (z) = v(z)
N →∞
(13.33)
for almost every z ∈ C. If we let PN = UN + iVN , then each PN is a polynomial and therefore,
our desired result follows immediately from the limits (13.32) and (13.33). Hence we have ended
the proof of the problem.
CHAPTER
14
Conformal Mapping
14.1
Basic Properties of Conformal Mappings
Problem 14.1
Rudin Chapter 14 Exercise 1.
Proof. The linear fractional transformation
f (z) =
az + b
cz + d
(14.1)
maps Π+ onto itself if and only if a, b, c and d are real numbers such that ad − bc > 0.
Suppose that a, b, c and d are real numbers such that ad − bc > 0. By §14.3, the f in the
form (14.1) is already a one-to-one mapping of S 2 onto S 2 . Since a, b, c, d ∈ R, f must map the
real axis onto itself. Furthermore, we have
Im f (i) =
ad − bc
> 0,
c2 + d2
so i is mapped into Π+ which proves that the transformation f maps Π+ onto itself. The
converse part can be found in [76, Problem 13.16, p. 181], completing the proof of the problem.
Problem 14.2
Rudin Chapter 14 Exercise 2.
Proof. Denote Π+ to be the upper half plane. Let z ∈ U . We have to make clear the meaning
of the reflection, namely z ∗ , of z with respect to the arc L. In fact, by the discussion in [9, pp.
102, 103]a , we know that z ∗ lies on the same ray as z and |z ∗ | = |z|−1 . In other words, we have
z∗ =
a
See also [18, pp. 50, 51]
99
1
.
z
100
Chapter 14. Conformal Mapping
(a) We have the following analogous reflection theorem for this part:
Lemma 14.1
Suppose that Ω ⊆ Π+ , L = R and every point t ∈ L is the center of an open disc
Dt such that Π+ ∩ Dt ⊆ Ω. Let Ω− be the reflection of Ω, i.e.,
Ω− = {z ∈ C | z ∈ Ω}.
If f ∈ H(Ω) and |f (zn )| → 1 for every {zn } in Ω which converges to a point of L,
then there exists a function F , holomorphic in Ω ∪ L ∪ Ω− , such that

f (z), if z ∈ Ω ∪ L;



F (z) =
(14.2)
1

−.

,
if
z
∈
Ω

f (z)
Proof of Lemma 14.1. Fix a point t ∈ L. By the hypothesis |f (z)| → 1 as z → t ∈ L,
it is legitimate to select a disc Dt so small that f (z) 6= 0 in Π+ ∩ Dt . Then the function
g(z) = i log f (z) is well-defined and holomorphic in Π+ ∩ Dt . Furthermore, we know
that
Im g(zn ) = log |f (zn )| → 0
for every {zn } in Ω converging to a point of L. By the application of Theorem 11.14
(The Schwarz Reflection Principle), we see that one can find a function G, holomorphic
in Ω ∪ L ∪ Ω− , such that G(z) = g(z) in Ω and satisfies
G(z) = G(z)
(14.3)
for all z ∈ Ω ∪ L ∪ Ω− . Define F (z) = e−iG(z) . Since f (z) = e−ig(z) , if z ∈ Ω, then we
have
F (z) = e−iG(z) = e−ig(z) = f (z).
Next, if z ∈ Ω− , then z ∈ Ω and we deduce from the equation (14.3) and the definition
of F that
1
1
F (z) = e−iG(z) = e−iG(z) = eiG(z) =
=
F (z)
f (z)
which is exactly the equation (14.2).
(b) Recall from [62, Eqn. (1), p. 281] that
ψ(z) =
z−i
z+i
is a conformal one-to-one mapping of Π+ onto U and ψ(R) ⊆ T . So the inverse
ψ −1 (ζ) =
i(1 + ζ)
1−ζ
is a conformal one-to-one mapping of U onto Π+ . For every θ ∈ [0, 2π], we know that
ψ −1 (eiθ ) =
sin θ
i(1 + eiθ )
∈ R.
=−
iθ
1−e
1 − cos θ
14.1. Basic Properties of Conformal Mappings
101
b = ψ −1 (L) and Ω
b = ψ −1 (Ω) are a segment of R and a region in Π+ respectively.
Thus L
Define the map
b ⊆ Π+ → C.
fb = f ◦ ψΩb : Ω
(14.4)
b converging to a z0 ∈ L,
b the points ζn = ψ(zn ) ∈ Ω ⊆ U converging
Then for every {zn } ∈ Ω
to ζ0 = ψ(z0 ) ∈ L ⊆ T , so the hypothesis guarantees that
|fb(zn )| = f ψ(zn ) = |f (ζn )| → 1
as n → ∞. Hence it follows from Lemma 14.1 that there exists a function Fb, holomorphic
b ∪L
b∪Ω
b − , such that
in Ω

b ∪ L;
b
fb(z), if z ∈ Ω



Fb(z) =
1

b −,

, if z ∈ Ω

b
f (z)

b ∪ L;
b
f ψ(z) , if z ∈ Ω



(14.5)
=
1

−.
b

,
if
z
∈
Ω

f ψ(z)
b − = {z ∈ C | z ∈ Ω}.
b If z ∈ Ω
b − , then z ∈ Ω
b and so we note from the definition of ψ
Here Ω
that
1
.
(14.6)
ψ(z) =
ψ(z)
Hence the formula (14.5) becomes
F (ζ) =
where Ω∗ = {ζ ∈ C | ζ
−1
∈ Ω}.


f (ζ),







if ζ ∈ Ω ∪ L;
1
f ζ
, if ζ ∈ Ω∗ ,
−1 (14.7)
(c) We have the following analogous reflection theorem for U :
Lemma 14.2 (The Schwarz Reflection Principle for U )
Every eit ∈ L ⊆ T is the center of an open disc Dt such that Dt ∩ U lies in Ω.
Denote Ω∗ to be the reflection of Ω, i.e.,
o
n
1
Ω∗ = z ∈ C z ∗ = ∈ Ω .
z
Suppose that f ∈ H(Ω) and Im f (zn ) → 0 for every sequence {zn } in Ω which
converges to a point of L. Then there exists a function F , holomorphic in the set
Ω ∪ L ∪ Ω∗ , such that


f (z),
if z ∈ Ω ∪ L;


F (z) =
(14.8)
1


∗
 f
, if z ∈ Ω .
z
102
Chapter 14. Conformal Mapping
Proof of Lemma 14.2. With the same function ψ : Π+ → U as in part (b), we see that
b and z → L”
b is equivalent to saying that “ψ(z) ∈ Ω and ψ(z) → L”. This
“z ∈ Ω
property implies that the function (14.4) satisfies
Im fb(z) = Im f ψ(z) → 0
b and z → L. According to Theorem 11.14 (The Schwarz Reflection Principle),
as z ∈ Ω
b ∪L
b∪Ω
b − , such that
there is a function Fb , holomorphic in Ω

b ∪ L;
b

 f ψ(z) , if z ∈ Ω
b
(14.9)
F (z) =

 f ψ(z ), if z ∈ Ω
b −.
Using the formula (14.6), the function (14.9) can be expressed in the following form:

f (ζ),
if ζ ∈ Ω ∪ L;



F (ζ) =
1


 f
, if ζ ∈ Ω∗ .
ζ
This completes the proof of Lemma 14.2.
Now, since ( α1 )−1 = α, it is easy to conclude from the expression (14.7) that if f (α) = 0 for
some α ∈ Ω, then α1 ∈ Ω∗ so that
1
1
F
= ∞,
=
α
f (α)
i.e., F has a pole at α1 . By the expression (14.8), the analogue of part (c) is that F ( α1 ) = f (α) = 0.
For part (a), if f (α) = 0 for some α ∈ Ω, then the expression (14.2) implies that
F (α) =
1
f (α)
= ∞,
i.e., F has a pole at α. This finishes the proof of the problem.
Problem 14.3
Rudin Chapter 14 Exercise 3.
b
Proof. If |z| = 1, then it is easy to see that z = z1 and the rational function R(z)
= R(z) · R( z1 )
satisfies
1
b
= R(z) · R(z) = |R(z)|2 = 1
R(z)
= R(z) · R
z
b = P , where P and Q are polynomials, we have P (z) = Q(z) on the unit
if |z| = 1. Since R
Q
circle. Now the Corollary following Theorem 10.18 guarantees that P (z) ≡ Q(z) in C and this
implies that R(z) = 1 for every z ∈ C, i.e.,
1
1
=
R
z
R(z)
for every z ∈ C. Therefore, ω is a zero of order m of R if and only if ω1 is a pole of order m of
R. This fact shows that the zeros and poles of R inside U completely determines all zeros and
poles of R in C.
14.1. Basic Properties of Conformal Mappings
103
Next, suppose that R has a zero at z = 0 of order m. Suppose, further, that {0, α1 , α2 , . . . , αk }
are the distinct zeros and poles of R inside U . We consider the product
k
Y
z − αn
B(z) = z ·
1 − αn z
n=1
m
which is a rational function having the same zeros and poles of the same order as R. Recall
z−αn
| = 1 for |z| = 1 and thus |B(z)| = 1 for
from the proof of Theorem 12.4, we know that | 1−α
nz
|z| = 1. Consequently, the quotient
R(z)
f (z) =
B(z)
|R(z)|
=1
is a rational function without zeros or poles in D(0; r) for some r > 1. Since |f (z)| = |B(z)|
on |z| = 1, we get from the Corollary following Theorem 10.18 that f (z) = c for some constant
c with |c| = 1 in D(0; r) and hence in C \ { α11 , α12 , . . . , α1n }, i.e.,
R(z) = cB(z) = cz m ·
k
Y
z − αn
1 − αn z
n=1
as desired. This ends the proof of the problem.
Problem 14.4
Rudin Chapter 14 Exercise 4.
Proof. We have R(z) > 0 on |z| = 1. By the hint, R must have the same number of zeros as
poles in U . Let α1 , α2 , . . . , αN and β1 , β2 , . . . , βN be the zeros and poles of R inside U , where
N is a positive integer. Next, we consider the rational function
f (α, β, z) =
(z − α)(1 − αz)
,
(z − β)(1 − βz)
(14.10)
where α, β ∈ U . Obviously, if |z| = 1, then z · z = |z|2 = 1 and 1 − αz 6= 0 which imply that
f (α, β, z) =
(1 − αz)(1 − αz)
|1 − αz|2
z (z − α)(1 − αz)
·
=
=
> 0.
z (z − β)(1 − βz)
|1 − βz|2
(1 − βz)(1 − βz)
Now the representation (14.10) indicates that α and β are the only zeros and poles of f (α, β, z)
inside U respectively. Therefore, the rational function
Q(z) =
N
Y
f (αn , βn , z)
n=1
has the same numbers of zeros and poles as those of R inside U .
Consequently, the quotient
F (z) =
R(z)
Q(z)
is a rational function without zeros or poles in U , i.e., F ∈ H(U ). Since f (αn , βn , z) > 0 on
|z| = 1 for every n = 1, 2, . . . , N , we also have Q(z) > 0 on |z| = 1 and hence F (z) > 0 on
|z| = 1. In other words, Im F ≡ 0 on |z| = 1. Recall that Im F is a continuous real-valued
104
Chapter 14. Conformal Mapping
function on U and is harmonic in U , so we follow from [7, Corollary 1.9, p. 7] that Im F ≡ 0 in
U . Finally, using [9, Proposition 3.6, p. 39], there is a positive constant c such that
F (z) = c
(14.11)
in U . By the Corollary following Theorem 10.18, we conclude that the result (14.11) holds in
Ω = C \ {β1 , β2 , . . . , βN , β1 , β1 , . . . , β1 } which means that
1
2
N
N
Y
(z − αn )(1 − αn z)
R(z) = cQ(z) = c
(z − βn )(1 − βn z)
n=1
(14.12)
in Ω, completing the analysis of the problem.
Problem 14.5
Rudin Chapter 14 Exercise 5.
Proof. Suppose that g(ζ) =
n
X
ak ζ n which is a rational function. Then g(eiθ ) = f (θ), so g is
k=−n
positive on T . By the representation (14.12), one can find a positive constant c such that
g(ζ) = c
n
Y
(ζ − αj )(1 − αj ζ)
(ζ − βj )(1 − βj ζ)
j=1
(14.13)
in Ω = C \ {β1 , β2 , . . . , βn , β1 , β1 , . . . , β1 }. Here {α1 , α2 , . . . , αn } and {β1 , β2 , . . . , βn } are sets of
n
1
2
zeros and poles of g inside U respectively. Since ζ n g(ζ) is a polynomial, the expression (14.13)
implies that β1 = β2 = · · · = βn = 0. Hence we obtain
f (θ) = g(eiθ )
n
Y
(eiθ − αj )(1 − αj eiθ )
=c
eiθ
=c
=c
=c
j=1
n
Y
(eiθ − αj )(e−iθ − αj )
j=1
n
Y
(eiθ − αj )(eiθ − αj )
j=1
n
Y
j=1
=
|eiθ − αj |2
n
√ Y
(eiθ − αj )
c
2
j=1
where P (z) =
√
= |P (eiθ )|2 ,
c(z − α1 )(z − α2 ) · · · (z − αn ). We have completed the proof of the problem. Problem 14.6
Rudin Chapter 14 Exercise 6.
14.1. Basic Properties of Conformal Mappings
105
Proof. If α = 0, then ϕ0 (z) = z, so the fixed points of ϕ0 are U . Next, we suppose that α 6= 0.
By Definition 12.3, ϕα (z) = z if and only if z 2 = αα if and only if
z=±
α
.
|α|
This gives our first assertion.
For the second assertion, we note that since ϕα is a special case of the linear fractional
transformation ϕ(z) = az+b
cz+d , we consider the general case for ϕ. By a suitable rotation, we may
assume that the straight line is the real axis. We claim that ϕ maps R ∪ {∞} into R ∪ {∞}
if and only if a, b, c and d are real. It is easy to see that if a, b, c and d are real, then we have
ϕ(R∪{∞}) ⊆ R∪{∞}. Conversely, suppose that ϕ(R∪{∞}) ⊆ R∪{∞}. Since ϕ(0) ∈ R∪{∞},
we have either d = 0 or db ∈ R.
• Case (i): d = 0. Since ϕ(z) → ac as z → ∞ along R, we have ac ∈ R or c = 0. In the
latter case, the transformation is ϕ(z) = ∞ for all z ∈ R ∪ {∞} and we can rewrite it as
1
. For the case ac ∈ R, since ϕ(1) = ac + bc ∈ R with c 6= 0, we have bc ∈ R.
ϕ(z) = 0·z+0
Therefore, we obtain
a
z + bc
az + b
= c
.
ϕ(z) =
cz
1·z+0
• Case (ii):
b
d
∈ R. Note that d 6= 0. Let B = db . Then we have
az + Bd
.
cz + d
ϕ(z) =
(14.14)
Using ϕ(z) → ac as z → ∞ along R again, we know that A = ac ∈ R or c = 0. In the latter
case, we have ϕ(z) = ad z + db . Since ϕ(1) = ad + db ∈ R and db ∈ R, we have ad ∈ R so that
ϕ(z) =
For the case A =
a
c
+ db
.
0z + 1
a
dz
∈ R, the representation (14.14) becomes
ϕ(z) =
Acz + Bd
.
cz + d
(14.15)
If c + d = 0, then c = −d 6= 0 so that the representation (14.15) reduces to
ϕ(z) =
−Adz + Bd
−Az + B
=
.
−dz + d
−z + 1
d
d
Otherwise, ϕ(1) = Ac+Bd
c+d = A + (B − A) · c+d ∈ R if and only if c+d ∈ R. Since d 6= 0,
c
d
c+d ∈ R if and only if d ∈ R. Let c = Cd for some C ∈ R. Now the representation (14.15)
reduces to
ACdz + Bd
ACz + B
ϕ(z) =
=
.
Cdz + d
Cz + 1
This proves our claim.
Return to our original problem. Suppose that ϕα maps a straight line L into itself. It is clear
that eiθ L = R for some θ ∈ [0, 2π], so we assume that eiθ ϕα maps R into R. Write
eiθ ϕα (z) =
eiθ z − eiθ α
.
(eiθ α)z + eiθ
Then the above claim says that eiθ , eiθ α and eiθ α are real, or equivalently, both eiθ and α are
real. Hence we end the analysis of the problem.
106
Chapter 14. Conformal Mapping
Problem 14.7
Rudin Chapter 14 Exercise 7.
Proof. Suppose that z, ω ∈ U . By the definition, fα (z) = fα (0) = 0 so that z = 0 and α is
arbitrary in this case. Thus, without loss of generality, we may assume that z, ω 6= 0. Now
fα (z) = fα (ω) if and only if (z − ω)(1 − αzω) = 0 if and only if
z=ω
or
α=
1
.
zω
(14.16)
1
Since |z| < 1 and |ω| < 1, we have |zω|
> 1. Suppose that |α| ≤ 1. Then the result (14.16) leads
to us that z = ω, i.e., fα is one-to-one in U .
Let |α| ≤ 1. It is easy to check that fα (0) = 0 and fα′ (0) = 1. Besides, we have f ∈ H(U ).
Combining these facts and the previous result, we know that fα ∈ L and
f (z) = z − αz 3 + α2 z 5 + · · · .
Therefore, it deduces from Theorem 14.14 that D(0; 14 ) ⊆ fα (U ). This ends the analysis of the
problem.
Problem 14.8
Rudin Chapter 14 Exercise 8.
Proof. Let z = reiθ , where r > 0 and 0 ≤ θ < 2π. Then we have
e−iθ
1
1
f (reiθ ) = reiθ +
cos θ + i r −
sin θ.
= r+
r
r
r
Suppose that
1
1
cos θ and y = r −
sin θ.
(14.17)
x= r+
r
r
If r = 1, then x = 2 cos θ and y = 0 so that f C(0; 1) = [−2, 2]. Suppose, otherwise, that r 6= 1,
so we obtain
y2
x2
+
=1
(r + 1r )2 (r − 1r )2
which is an ellipse.
Next, suppose that θ is a fixed number. Denote Lθ = {reiθ | 0 ≤ r < ∞}. If θ = 0, then
cos θ = 1 and sin θ = 0 and thus it easily follows from the representations (14.17) that
f (L0 ) = (0, ∞).
Similarly, if θ = π, then cos θ = −1 and sin θ = 0 so that
f (Lπ ) = (−∞, 0).
If θ = π2 , 3π
2 , then cos θ = 0 and sin θ = ±1. Simple computation verifies that
f L π2 = f L 3π = iR.
2
Finally, if θ ∈ [0, 2π) \ {0,
representations (14.17) that
π
3π
2 , π, 2 },
then we have cos θ · sin θ 6= 0 and we deduce from the
y2
x2
−
=4
cos2 θ sin2 θ
which is trivially a hyperbola. This completes the analysis of the problem.
14.1. Basic Properties of Conformal Mappings
107
Problem 14.9
Rudin Chapter 14 Exercise 9.
Proof. Define Ω1 = {z ∈ C | 0 < Im z < π} and Π+ = {z ∈ C | Im z > 0}.
πi
′
(a) Notice that the map f1 : Ω → Ω1 defined by f1 (z) = πi
2 (z + 1) satisfies f (z) = 2 6= 0
in Ω. By Theorem 14.2, it is a one-to-one conformal mapping of Ω onto Ω1 . Next, we
know from [9, p. 176] that f2 (z) = ez is a one-to-one conformal mapping of Ω1 onto the
upper half plane Π+ . Recall from [62, Eqn. (1), p. 281] that f3 (z) = z−i
z+i is a one-to-one
conformal mapping of Π+ onto U . Hence the mapping f = f3 ◦ f2 ◦ f1 : Ω → U is the
required mapping. Explicitly, we have
f (z) =
exp( iπ
2 z) − 1
.
iπ
exp( 2 z) + 1
(b) The inverse f −1 : U → Ω is given by
f −1 (z) = −
1 + z 2i
2i
1+z
1+z
2
− log
log
= arg
.
π
1−z
π
1−z
π
1−z
(14.18)
If f −1 = u + iv, then we have
u(z) =
1 + z 2
2
1+z
and v(z) = − log
arg
.
π
1−z
π
1−z
Since f −1 ∈ H(U ), u must be bounded and harmonic in U . Its harmonic conjugate v
is unbounded in U because |v(z)| → ∞ as z → ±1. It remains to prove that u can
be extended continuously to U . The definition of f shows that it can be extended to a
continuous function of Ω onto U . Hence its inverse, which is u, can also be extended to a
continuous function on U .
(c) Since our f −1 and g satisfy the hypotheses of Problem 14.10, we establish immediately
that
g D(0; r) ⊆ f −1 D(0; r)
(14.19)
for all 0 < r < 1. Of course, we observe from the definition (14.18) that
π
1+z
− Im f −1 (z) = log
2
1−z
for all z ∈ U . Let f −1 (z) =
∞
X
n=0
(14.20)
cn z n , where z ∈ U . Since n!cn = (f −1 )(n) (0) for every
n = 0, 1, 2, . . ., it is easy to see that f −1 has the form
f −1 (z) = −
∞
4i X z 2n−1
π
2n − 1
n=1
for all z ∈ U . Thus for every z ∈ D(0; r), we deduce from the expression (14.20) that
|f −1 (z)| ≤
∞
1 + |z|
2
1+r
2
4 X |z|2n−1
= −Im f −1 (|z|) = log
.
= log
π
2n − 1
π
1 − |z|
π
1−r
n=1
Finally, the set relation (14.19) asserts that
|g(z)| ≤
for all z ∈ D(0; r).
2
1+r
log
π
1−r
108
Chapter 14. Conformal Mapping
(d) Now suppose that Ω = {x + iy | − π2 < y < π2 } and h : Ω → Ω is a conformal bijective
mapping such that h(a + iβ) = 0. It is known from Problem 14.32 that the mapping
ψ(z) = log
1+z
1−z
sends U conformally, one-to-one and onto the horizontal strip Ω with ψ(0) = 0. Thus its
inverse
ez − 1
ψ −1 (z) = z
e +1
is a conformal one-to-one mapping of Ω onto U and ψ −1 (0) = 0. Denote ψ −1 (α + iβ) = A
so that
eα+iβ − 1
= A.
(14.21)
eα+iβ + 1
It is well-known [9, Theorem 13.15, p. 183] that the conformal mapping ϕ of U onto itself
and ϕ(A) = 0 is represented by
ϕ(z) = eiθ ·
z−A
1 − Az
for some θ ∈ R. Therefore, the composition h = ψ ◦ ϕ ◦ ψ −1 is a conformal one-to-one
mapping from Ω onto itself and
h α + iβ = ψ ϕ ψ −1 (α + iβ) = ψ ϕ(A) = ψ(0) = 0.
Notice that
ψ ′ (z) =
2
,
1 − z2
(ψ −1 )′ (z) =
2ez
(ez + 1)2
and ϕ′ (z) = eiθ ·
1 − |A|2
.
(1 − Az)2
By the Chain Rule, we have
h′ (α + iβ) = ψ ′ ϕ ψ −1 (α + iβ) · ϕ′ ψ −1 (α + iβ) · (ψ −1 )′ (α + iβ)
2eα+iβ
(eα+iβ + 1)2
2eα+iβ
eiθ
×
.
=2×
1 − |A|2 (eα+iβ + 1)2
= ψ ′ (0) × ϕ′ (A) ×
(14.22)
According to the value (14.21) and the expression (14.22), we see that
|eα+iβ − 1|2 −1
2eα
|h′ (α + iβ)| = 2 1 − α+iβ
×
|e
+ 1|2
|eα+iβ + 1|2
4eα
= α+iβ
|e
+ 1|2 − |eα+iβ − 1|2
1
=
cos β
which is the desired result.
Consequently, we have completed the proof of the problem.
Remark 14.1
Problem 14.9 contributes to the theory of the Principle of Subordination, read [15, Chap.
VI, §5, pp. 207 - 215 ], [21, §1.5, pp. 10 – 13] or [47, Chap V, §9, pp. 226 – 236].
14.1. Basic Properties of Conformal Mappings
109
Problem 14.10
Rudin Chapter 14 Exercise 10.
Proof. Since f is one-to-one, f −1 : Ω → U exists. Let h = f −1 ◦ g. Then it is clear that
h ∈ H(U ), h(U ) = f −1 g(U ) ⊆ U and h(0) = 0. Thus Theorem 12.2 (Schwarz’s Lemma)
ensures that
|h(z)| ≤ |z|
(14.23)
for all z ∈ U . For every 0 < r < 1, if z ∈ D(0; r), then the inequality (14.23) implies that
h(z) ∈ D(0; r). Equivalently, this means that g(z) ∈ f D(0; r) and henceb
g(D(0; r)) ⊆ f (D(0; r)),
completing the proof of the problem.
Problem 14.11
Rudin Chapter 14 Exercise 11.
Proof. Now we have Ω = {z ∈ U | Im z > 0}. Using [9, Example 1, p. 180; Theorem 13.16, p.
183], if f : Ω → U is a conformal bijective mapping, then f has the representation
f (z) = eiθ ·
(z − 1)2 + 4α(z + 1)2
,
(z − 1)2 + 4α(z + 1)2
(14.24)
(z − 1)2 + i(z + 1)2
(z − 1)2 − i(z + 1)2
(14.25)
where θ ∈ [0, 2π] and Im α > 0. Putting f (−1) = −1, f (0) = −i and f (1) = 1 into the formula
(14.24), we obtain that θ = π and α = 4i , so
f (z) = −
is the desired conformal mapping.√By the representation (14.25), it is easily seen that if z ∈ Ω
satisfies f (z) = 0, then z = (−1 + 2)i. Furthermore, simple algebra gives f ( 2i ) = 7i which ends
the proof of the problem.
Problem 14.12
Rudin Chapter 14 Exercise 12.
Proof. For convenience, we let u(z) = Re f ′ (z) : C → R. We prove the assertions as follows:
• f is one-to-one in Ω when u(z) > 0 for all z ∈ Ω. Choose a, b ∈ Ω and a 6= b. Since
Ω is convex, the path γ(t) = a + (b − a)t for all t ∈ [0, 1] is in Ω. Then we know from the
Fundamental Theorem of Calculus that
Z 1
Z
′
f ′ a + (b − a)t dt = f (b) − f (a)
f (z) dz = (b − a)
0
γ
so that
Re
h f (b) − f (a) i
=
Z
1
u a + (b − a)t dt > 0.
b−a
0
Consequently, Re f (a) 6= Re f (b) which implies that f (a) 6= f (b). Since the pair of points
{a, b} is arbitrary, we assert that f is one-to-one in Ω.
b
Rudin used the notation “⊂” to mean “⊆”, see [61, Definition 1.3, p. 3].
110
Chapter 14. Conformal Mapping
• f is either one-to-one or constant in Ω when u(z) ≥ 0 for all z ∈ Ω. Since f ∈ H(Ω),
u has continuous derivative of all orders. Thus the set
S = {z ∈ Ω | u(z) = 0} = u−1 (0)
(14.26)
is closed in C. Let p ∈ S. Since u is harmonic in Ω, it follows from the mean value property
that
Z 2π
1
0 = u(p) =
u(p + Reiθ ) dθ
(14.27)
2π 0
for some R > 0 such that D(p; R) ⊆ Ω. If there exists a measurable set E ⊆ [0, 2π] such
that m(E) > 0 and u(p + Reiθ ) > 0 for every θ ∈ E, then we have
Z
Z
Z 2π
iθ
iθ
u(p + Reiθ ) dm > 0
u(p + Re ) dm +
u(p + Re ) dθ =
0
[0,2π]\E
E
which contradicts the result (14.27). Therefore, no such E exists and then u(p + Reiθ ) = 0
a.e. on [0, 2π]. Now the continuity of u on Ω forces that u(z) = 0 for every z ∈ D(p; R).
In other words, D(p; R) ⊆ S which means that S is open in C.
Since S is both open and closed in C, we have either S = ∅ or S = C. Suppose that S = ∅,
then Re f ′ (z) > 0 in Ω so that f is one-to-one in Ω by the previous assertion. Next, we
suppose that S = C, then the definition (14.26) implies f ′ (Ω) is purely imaginary. Since
f ′ ∈ H(Ω), the Open Mapping Theorem ensures that f ′ (Ω) = A for some constant A.
Finally, if we consider g(z) = f (z) − Az, then g ∈ H(Ω) and g ′ ≡ 0 there. Thus it follows
from [9, Exercise 5, p. 42] that g is a constant B. Consequently, we have f (z) = Az + B.
If A 6= 0, then f is linear and so it is one-to-one. Otherwise, f ≡ B in Ω.
• The condition “convex” cannot be replaced by “simply connected”. The following example can be found in [30].c Take β = π2 + δ for very small δ > 0. Define
1+ π
Ω = {z ∈ C | − β < arg z < β} and f (z) = z 2β . Then it is clear that f ∈ H(Ω) and
π
π
f ′ (z) = (1 + 2β
)z 2β so that
π arg z π
π 2β
Re f ′ (z) = 1 +
|z| cos
.
(14.28)
2β
2β
z
π
Since −β < arg z < β, we have − π2 < π arg
2β < 2 and it follows from the expression (14.28)
that Re f ′ (z) > 0 for every z ∈ Ω. Suppose that z1 = reiθ1 and z2 = reiθ2 are points of Ω,
where θ1 6= θ2 . Clearly,
1+
π
1+
π
z1 2β = z2 2β .
π
) = 1 if and only if
if and only if exp i(θ1 − θ2 )(1 + 2β
θ1 − θ2 =
If − π2 − δ < θ2 < − π2 +
θ1 = θ2 +
δ2
π+δ
<
π
2
4πβ
π
=
· (π + 2δ).
2β + π
π+δ
(14.29)
(14.30)
+ δ, then the inequality (14.30) yields
π
δ2
π
π
π
· (π + 2δ) < − +
+
· (π + 2δ) = + δ.
π+δ
2 π+δ π+δ
2
Now we have found distinct θ1 , θ2 ∈ (−β, β), and hence distinct z1 and z2 of Ω such that
the equation (14.29) holds. Consequently, our f is not one-to-one in Ω.
c
In fact, it was shown by Tims [71, Theorem 2] that for any simply connected non-convex region Ω whose
boundary contains more than one point, there exists distinct points α, β ∈ Ω and an f ∈ H(Ω) such that Re f ′ (z)
is nonzero in Ω and f (α) = f (β).
14.1. Basic Properties of Conformal Mappings
This completes the proof of the problem.
111
Problem 14.13
Rudin Chapter 14 Exercise 13.
Proof. Assume that z1 6= z2 but f (z1 ) = f (z2 ). Pick a δ > 0 such that z2 ∈
/ D(z1 ; δ) and
D(z1 ; δ) ⊆ Ω. Since every fn is one-to-one in D(z1 ; δ), none of the functions fn (z) − fn (z2 )
has a zero in D(z1 ; δ). By the hypotheses, fn (z) − fn (z2 ) → f (z) − f (z2 ) uniformly on every
compact subset in D(z1 ; δ), so we deduces from Problem 10.20 (Hurwitz’s Theorem) that either
f (z) − f (z2 ) 6= 0 for all z ∈ D(z1 ; δ) or f (z) − f (z2 ) = 0 in D(z1 ; δ). Since f (z1 ) = f (z2 ), we
have f (z) = f (z2 ) in D(z1 ; δ) and this means that
D(z1 ; δ) ⊆ Z f − f (z2 ) .
By Theorem 10.18, we conclude immediately that f is constant in Ω.
If fn (z) = n1 ez for every n ∈ N, then each fn is entire and one-to-one in C. Since fn (z) → 0
pointwise in C, we have f ≡ 0. On each compact set K of C, since ez is bounded on K, fn → f
1
uniformly on K. In this case, our limit function f is a constant. Next, we consider fn (z) = ez+ n
in C, then each fn is entire and one-to-one in C. Besides, it is easily checked that fn → f = ez
uniformly on every compact subset of C. In this case, we have f (z) = ez which is one-to-one in
C. This completes the proof of the problem.
Problem 14.14
Rudin Chapter 14 Exercise 14.
Proof. Let’s answer the questions step by step.
• f (x + iy) → 0 as x → ∞ for all y ∈ (−1, 1). Assume that one could find an y ′ ∈
(−1, 1) \ {0} such that the limit lim f (x + iy ′ ) is nonzero. Since |f | < 1, the Bolzanox→∞
Weierstrass Theorem ensures the existence of a sequence {xn } and a nonzero complex
number L such that xn → ∞ as n → ∞ and
lim f (xn + iy ′ ) = L.
n→∞
(14.31)
Consider the family F = {fn } ⊆ H(Ω), where fn (z) = f (z + xn ). Now the boundedness
of f implies that F is uniformly bounded on each compact subset of Ω. By Theorem 14.6
(Montel’s Theorem), it is a normal family and then there exists a subsequence {nk } and
an F ∈ H(Ω) such that fnk → F uniformly on every compact subset of Ω. It is clear that
{x, iy ′ } is compact. On the one hand, the hypothesis gives
F (x) = lim fnk (x) = lim f (x + xnk ) = 0.
k→∞
k→∞
On the other hand, it follows from the limit (14.31) that
F (iy ′ ) = lim fnk (iy ′ ) = lim f (xnk + iy ′ ) = L.
k→∞
k→∞
Thus it is a contradiction and we obtain our desired result.
112
Chapter 14. Conformal Mapping
• The passage to the limit is uniform if y is confined to [−α, α], where α < 1.
Assume that the limit was not uniform in Kα = {x + iy | x ∈ R and y ∈ [−α, α]} for some
α < 1. Then there exists some ǫ > 0 so that for all N ∈ N, one can find xN ≥ N and
yN ∈ [−α, α] such that
|fN (iyN )| = |f (xN + iyN )| > ǫ.
(14.32)
As we have shown above that {fN } has a subsequence {fNk } which converges uniformly
on compact subsets of Ω to a holomorphic function g, and g 6≡ 0 in view of the inequality
(14.32). By the previous assertion, we have
fNk (x + iy) = f xNk + (x + iy) → 0
as k → ∞ for every (x, y) ∈ [−α, α]2 ⊆ Kα ⊆ Ω, so this means that g(z) = 0 for all
z ∈ [−α, α]2 and then the Corollary following Theorem 10.18 implies that g(z) = 0 in Ω,
a contradiction. Hence the limit must be uniform in Kα .
• Boundary behavior of a function g ∈ H ∞ with a radial limit.d Let g ∈ H ∞ .
Without loss of generality, we may assume that |g(z)| < C1 for all z ∈ U and g(reiθ ) → C2
as r → 1 for some θ, where C1 and C2 are some constants. Using the mapping (14.124),
we know that
π
e2z − 1
κ(z) = π z
e2 + 1
is a conformal one-to-one mapping of Ω onto U . Since κ(x) → 1 as x → ∞, the composite
g κ(z)eiθ − C2
h(z) =
C1 + C2
is a mapping from Ω into U satisfying h ∈ H(Ω), |h(z)| < 1 for all z ∈ Ω and
g κ(x)eiθ − C2
→0
h(x) =
C1 + C2
as x → ∞. Hence the first assertion implies that
lim h(x + iy) = 0 or
x→∞
lim g κ(x + iy)eiθ = C2
x→∞
(14.33)
for every y ∈ (−1, 1). We observe from the definition that |κ(x+iy)| < 1 and κ(x+iy) → 1
as x → ∞. By §11.21, the limit (14.33) means that g has non-tangential limit C2 at eiθ .
The analogue of the second assertion can be stated similarly and we omit the details here.
We complete the proof of the problem.
14.2
Problems on Normal Families and the Class S
Problem 14.15
Rudin Chapter 14 Exercise 15.
Proof. Let Π be the right half plane. Recall that the ϕ given by [62, Eqn. (6), p. 281] is a
conformal one-to-one mapping of U onto Π. Thus ϕ−1 : Π → U is conformal and bijective. Since
−1 ◦ f : U → U is holomorphic and
ϕ−1 (z) = z−1
z+1 and f : U → Π is holomorphic, we have g = ϕ
g(0) = ϕ−1 (f (0)) = ϕ−1 (1) = 0.
d
Recall from Theorem 11.32 (Fatou’s Theorem) that our g has radial limits almost everywhere on T .
14.2. Problems on Normal Families and the Class S
113
According to Theorem 12.2 (Schwarz’s Lemma), we always have
|g(z)| ≤ |z|
(14.34)
for all z ∈ U . Thus the auxiliary family G = {g = ϕ−1 ◦ f | f ∈ F } is uniformly bounded on
each compact subset of U . Particularly, Theorem 14.6 (Montel’s Theorem) implies that G is a
normal family.
Let K be a compact subset of U . Then there exists a constant 0 < R < 1 such that
K ⊆ D(0; R), so the inequality (14.34) gives |g(z)| ≤ R forevery g ∈ G and all z ∈ K. Since
ϕ is conformal, it is continuous on U . Therefore, ϕ D(0; R) is bounded by a positive constant
M and then
f (z) = ϕ g(z) ∈ ϕ D(0; R) ⊆ D(0; M )
(14.35)
for all z ∈ K. As g runs through G , f runs through F . Hence, it yields from the result (14.35)
that F is uniformly bounded on K. Again, Theorem 14.6 (Montel’s Theorem) implies that F
is normal.
The condition “f (0) = 1” can be omitted or replaced by “|f (0)| ≤ 1”. In fact, we suppose
that F ′ = {f ∈ H(U ) | Re f > 0} and the auxiliary family G ′ = {g = e−f | f ∈ F ′ }. It is
evident that
1
|g(z)| = |e−f (z) | = Re f (z) ≤ 1
e
for all z ∈ U , the family G ′ is uniformly bounded on U . Using similar argument as the previous
paragraph, it can be shown that the family F ′ is also normal. This completes the proof of the
problem.
Problem 14.16
Rudin Chapter 14 Exercise 16.
Proof. Let p ∈ U . Then there exists a R > 0 such that D(p; 2R) ⊆ U . For every z ∈ D(p; R),
we have D(p; r) ⊆ U for all 0 < r ≤ R. Since f ∈ H(U ), it is harmonic in U by Theorem 11.4.
By the mean value property, we have
1
f (z) =
2π
Z
2π
f (z + reit ) dt
0
which implies
Z 2π
1
rf (z + reit ) dt
rf (z) =
2π 0
Z R
Z R Z 2π
1
rf (z) dr =
rf (z + reit ) dt dr
2π
0
0
0
Z R Z 2π
1
R2
f (z) =
rf (z + reit ) dt dr.
2
2π 0 0
Applying Theorem 3.5 (Hölder’s Inequality) to the expression (14.36), we obtain
Z R Z 2π
1
|f (z)| =
rf (z + reit ) dt dr
πR2 0 0
Z Z
o 1 n Z R Z 2π √
o1
2
1 n R 2π √ 2
2
2
it
r|f (z + re )| dr dt
( r) dr dt ·
≤
πR2
0
0
0
0
(14.36)
114
Chapter 14. Conformal Mapping
n
1 √
=
πR
·
·
πR2
ZZ
|f (z)|2 dx dy
U
o1
2
1
≤√
πR
(14.37)
for all z ∈ D(p; R). Let K ⊆ Ω be compact and {D(p; 2R)} be an open cover of K, where
D(p; 2R) ⊆ U . Then there exist finitely many points p1 , p2 , . . . , pN and positive numbers
R1 , R2 , . . . , RN such that
K ⊆ D(p1 ; R1 ) ∪ D(p2 ; R2 ) ∪ · · · ∪ D(pN ; RN ).
If R = min(R1 , R2 , . . . , RN ), then we conclude from the inequality (14.37) that F is uniformly
1
bounded by √πR
on K. Hence Theorem 14.6 (Montel’s Theorem) shows that F is a normal
family and we complete the proof of the problem.
Problem 14.17
Rudin Chapter 14 Exercise 17.
Proof. The conclusion is affirmative. To this end, we need the following version of Hurwitz’s
Theorem [18, p. 152]:
Lemma 14.3 (Hurwitz’s Theorem)
Let Ω be a region and fn ∈ H(Ω) for n = 1, 2, . . .. If {fn } converges to f uniformly
on compact subsets of Ω, f 6≡ 0, D(a; R) ⊆ Ω and f (z) 6= 0 on C(a; R), then there
is an N ∈ N such that f and fn have the same number of zeros in D(a; R) for all
n ≥ N.
Fix an z0 ∈ Ω and let a ∈ Ω. Then the functions gn = fn − fn (z0 ) converge uniformly to
g = f − f (z0 ) on compact subsets of Ω. Since f is one-to-one in Ω, g(z) 6= 0 on C(a; r) for
every r > 0 such that D(a; r) ⊆ Ω. By Lemma 14.3 (Hurwitz’s Theorem), there corresponds an
N (a, r) ∈ N such that gn and g have the same number of zeros in D(a; r) for all n ≥ N (a, r). In
other words, we obtain
fn (z) 6= fn (z0 )
(14.38)
in D(a; r) \ {z0 } and for all n ≥ N (a, r). Suppose that K ⊆ Ω is compact, p ∈ K and {D(a; r)}
is an open covering of K, where each D(a; r) is a subset of Ω. Then we have
K ⊆ D(a1 ; r1 ) ∪ D(a2 ; r2 ) ∪ · · · ∪ D(am ; rm )
for some positive integer m. Let N (K) = max{N (a1 , r1 ), N (a2 , r2 ), . . . , N (am , rm )}. It follows
from the result (14.38) that
fn (z) 6= fn (p)
(14.39)
in K \ {p} for all n ≥ N (K). Since p is an arbitrary point of K, the result (14.39) means that
fn is one-to-one in K for all n ≥ N (K), completing the proof of the problem.
Problem 14.18
Rudin Chapter 14 Exercise 18.
14.2. Problems on Normal Families and the Class S
115
Proof. Suppose that f, g : Ω → U and f (z0 ) = g(z0 ) = 0. Then F = g ◦ f −1 : U → U is a
bijective conformal mapping such that
g′ (z0 )
.
F (0) = g f −1 (0) = g(z0 ) = 0 and F ′ (0) = ′
f (z0 )
(14.40)
By [9, Theorem 13.15, p. 183], we know that
z−α 1 − αz
F (z) = eiθ ·
(14.41)
for some |α| < 1 and θ ∈ [0, 2π]. By the conditions (14.40), we have α = 0 and eiθ =
which imply that
g′ (z0 )
g′ (z0 )
f (z) = ′
ϕ0 f (z)
g(z) = ′
f (z0 )
f (z0 )
g ′ (z0 )
f ′ (z0 )
for all z ∈ Ω.
For the case that f (z0 ) = g(z0 ) = a, the conditions (14.40) are replaced by
F (a) = a
and
F ′ (a) =
By the form (14.41) again, we can show that eiθ =
g(z) =
g ′ (z0 )
f ′ (z0 )
g ′ (z0 )
.
f ′ (z0 )
·
(1−αa)2
1−|α|2
which gives
g′ (z0 ) (1 − αa)2
·
ϕα f (z)
′
2
f (z0 ) 1 − |α|
for all z ∈ Ω. This ends the proof of the problem.
Problem 14.19
Rudin Chapter 14 Exercise 19.
Proof. We claim that the mapping given by
f (z) = z exp
i 1 − |z|
(14.42)
is a homeomorphism of U onto U . If we consider z = reiθ and represent the function (14.42) as
1 ,
(14.43)
f (r, θ) = r, θ +
1−r
then the inverse function f −1 is given by
f −1 (r, θ) = r, θ −
1 .
1−r
Now it is easy to see that both f and f −1 are continuous and bijective. In other words, f
is a homeomorphism. However, this homeomorphism cannot be extended to U continuously.
Otherwise, we assume that F : U → U was a continuous extension to f , i.e., F = f on U .
′
Therefore, we have F C(0; 1) = C(0; 1). Take F (eiθ ) = (1, 0) for some θ ′ ∈ [0, 2π] and V any
′
neighborhood of (1, 0). Suppose that i ∈
/ V . Then there exists a neighborhood W of eiθ such
′
that F (W ) ⊆ V .e Indeed, we can find a sequence {rn eiθ } in W such that rn < 1 for all n ∈ N,
′
rn → 1 and F (rn eiθ ) = f (rn , θ ′ ) → (1, 0)
e
See, for example, [42, Theorem 18.1, p. 104].
116
Chapter 14. Conformal Mapping
as n → ∞. Using the formula (14.43), we have
f (rn , θ ′ ) = rn , θ ′ +
h
1 1 i
= rn exp i θ ′ +
.
(14.44)
1 − rn
1 − rn
1
is continuous on [0, 1) and s [0, 1) = [θ ′ + 1, ∞), if we take
Since the function s(r) = θ ′ + 1−r
1
the points θ ′ + 1−r
= (2n + 1)π for all large n, then we deduce from the formula (14.44) that
n
h
f (rn , θ ′ ) = i 1 −
i
1
(2n + 1)π − θ ′
which means that f (rn , θ ′ ) cannot be contained in the neighborhood V of (1, 0) for all large n.
Hence no such continuous extension exists and we complete the proof of the problem.
Problem 14.20
Rudin Chapter 14 Exercise 20.
Proof. Write f (z) = zϕ(z). Then ϕ ∈ H(U ), ϕ(0) = 1 and ϕ has no zero in U . By Problem
13.9, there exists an h ∈ H(U ) such that hn (z) = ϕ(z) and h(0) = 1. Put
g(z) = zh(z n )
(14.45)
g n (z) = z n hn (z n ) = z n ϕ(z n ) = f (z n )
(14.46)
in U . Then we know that
for every z ∈ U . It is clear that g(0) = 0 and g ′ (0) = h(0) = 1.
To prove g ∈ L , it suffices to show that g is one-to-one in U . Suppose that z, ω ∈ U and
g(z) = g(ω). Since f is one-to-one in U , the formula (14.46) ensures that z n = ω n or equivalently,
z=e
2kπi
n
ω,
where k = 0, 1, . . . , n − 1. Put this into the equation (14.45) to get
g(z) = e
2kπi
n
ωh(e2kπi ω n ) = e
2kπi
n
2kπi
· ωh(ω n ) = e n g(ω).
(14.47)
Recall that g(z) = g(ω), so it follows from the equation (14.47) that we have g(z) = g(ω) = 0
2kπi
or e n = 1. In the latter case, we have z = w Otherwise, since g(z) = 0 if and only if z = 0 in
U , we have z = w = 0, completing the proof of the problem.
Problem 14.21
Rudin Chapter 14 Exercise 21.
Proof. By Definition 14.10, we have
f (z) = z +
∞
X
n=2
an z n .
14.2. Problems on Normal Families and the Class S
117
(a) Since f ∈ S , it is one-to-one. By Theorems 10.33 and 14.2, f is conformal and its inverse
f −1 : f (U ) → U exists and is conformal.f As U ⊆ f (U ), we consider g = f −1 |U : U → U .
Clearly, we have
g(0) = f −1 (0) = 0 and g ′ (0) =
1
1
= ′
= 1.
f (0)
f ′ f −1 (0)
By Theorem 12.2 (Schwarz’s Lemma), we conclude that g(z) = z and consequently, f (z) =
z in U .
(b) We claim that there is no element f ∈ S with U ⊆ f (U ). Assume that f was such a
function. By part (a), we know that f (z) = z so that f (U ) = U which contradicts our
assumption that U ⊆ f (U ).
(c) Consider the function F given in [62, Eqn. (1), p. 286]. If |α1 | = 1, then Theorem 14.13
(The Area Theorem) implies that αn = 0 for all n = 2, 3, . . .. In this case, we have
F (z) =
1
+ α0 + eiθ z.
z
Now we know from the proof of Theorem 14.14 that |a2 | = 2 is equivalent to |α1 | = 1, so
we have
1
1
= G(z) = + α0 + eiθ z.
g(z)
z
By Theorem 14.12, we have f (z 2 ) = g 2 (z) which implies definitely that α0 = 0 and then
f (z 2 ) =
z2
.
(1 + eiθ z 2 )2
Consequently, f must be in the form
f (z) =
z
.
(1 + eiθ z)2
We complete the proof of the problem.
Problem 14.22
Rudin Chapter 14 Exercise 22.
Proof. Let f : U → S be a one-to-one conformal mapping, where S is a square with center at 0.
Now it is clear that both if : U → S and f −1 : S → U are conformal. We consider
g = f −1 ◦ (if ) : U → U
which is also a one-to-one conformal mapping of U onto itself. Clearly, g(0) = f −1 if (0) = 0.
By Remark 10.3, we have
if ′ (z)
g′ (z) = ′
f if (z)
which gives |g′ (0)| = |i| = 1. Hence Theorem 12.2 (Schwarz’s Lemma) implies that the formula
if (z) = f (λz)
f
Or it can be seen directly from [9, Theorem 13.8, p. 174].
(14.48)
118
Chapter 14. Conformal Mapping
holds in U for some constant λ with |λ| = 1. By the note following Theorem 14.2, we have
f ′ (0) 6= 0. Now we observe from this fact and the expression (14.48) that if ′ (z) = λf ′ (λz) so
that λ = i if we put z = 0 into this equation. Thus we get what we want
if (z) = f (iz)
for every z ∈ U . Suppose that f (z) =
P
∞
X
n=1
(14.49)
cn z n . By the expression (14.49), we have
icn (1 − in−1 )z n = 0
for all z ∈ U . If n − 1 is not a multiple of 4, then in−1 6= 1, so it yields from Theorem 10.18 that
cn = 0 for such n.
A generalization is as follows: Let S be a simply connected region with rotational symmetry
of order N and center at 0, i.e., exp( 2πi
N )S = S. Let f : U → S be a one-to-one conformal
−1 : S → U are conformal. By
mapping with f (0) = 0. Thus both exp( 2πi
N )f : U → S and f
considering g = f −1 ◦ [exp( 2πi
N )f ] : U → U which is clearly a one-to-one conformal mapping of
U onto itself. Obviously, we have
2πi g(0) = f −1 exp
f (0) = 0.
N
We observe from Remark 10.3 that
g′ (z) =
′
exp( 2πi
N )f (z)
,
f ′ exp( 2πi
N )f (z)
so |g ′ (0)| = 1. Hence Theorem 12.2 (Schwarz’s Lemma) implies that
e
2πi
N
f (z) = f (λz)
(14.50)
holds for some constant λ with |λ| = 1. Since f ′ (0) 6= 0, we take differentiation to both sides of
the formula (14.50) to conclude that λ = exp( 2πi
N ) and hence
e
2πi
N
f (z) = f e
2πi
N
z
for every z ∈ U . Next, if n − 1 is not a multiple of N and f (z) =
(14.51) gives
∞
X
2π(n−1)i 2πi
e N cn 1 − e N
zn = 0
(14.51)
P
cn z n , then the formula
n=1
2π(n−1)i
N
6= 1 if n − 1 is not a multiple of N , Theorem 10.18 implies that
for all z ∈ U . Since e
cn = 0 for such n. This completes the analysis of the problem.
14.3
Proofs of Conformal Equivalence between Annuli
Problem 14.23
Rudin Chapter 14 Exercise 23.
Proof. Suppose that ∂Ω = C1 ∪C2 , where C1 lies in the inside of C2 . By appropriate translation,
rotation and homothety, we assume that C2 = T (the unit circle) and the center of C1 lies on
the real axis with x-intercepts a and b, where |a| < b < 1. See Figure 14.1 below:
14.3. Proofs of Conformal Equivalence between Annuli
119
Figure 14.1: The region Ω bounded by C1 and C2 .
By Theorem 12.4, ϕα carries T onto itself and U onto U , where |α| < 1. Since ϕα is a linear
fractional transformation, §14.3 ensures that it will send C1 ⊂ U onto a circle or a line. Now
the condition ϕα (U ) = U implies that ϕα (C1 ) must be a circle.
Suppose that α ∈ R. Since C1 intersects the real axis at a and b perpendicularly and ϕα is
a conformal map, ϕα (a) and ϕα (b) are the end-points of a diameter of ϕα (C1 ). Furthermore,
since ϕα (R) = R, both ϕα (a) and ϕα (b) are real. Thus if ϕα (C1 ) is a circle centered at 0, then
we must have
ϕα (a) = −ϕα (b)
which gives
b−α
a−α
=−
1 − αa
1 − αb
2(1
+
ab)
α2 −
α + 1 = 0.
a+b
Solving this equation to get
1 + ab
α± =
±
a+b
r
1 + ab 2
− 1.
a+b
(14.52)
Since |a| < b < 1, we always have a(1 − b) < 1 − b or equivalently, 1 + ab > a + b > 0. Combining
this and the formulas (14.52), it follows that
1 + ab
α+ =
+
a+b
r
1 + ab 2
− 1 > 1.
a+b
Since α+ · α− = 1, we conclude that 0 < α− < 1. Takethis α− . Then the conformal map
ϕα− carries U onto U , T onto T and C1 onto
C 0; ϕα− (a) . In other words, it is a one-to-one
conformal mapping of Ω onto A ϕα− (a), 1 . This completes the proof of the problem.
120
Chapter 14. Conformal Mapping
Remark 14.2
An example of this kind of conformal mappings can be found in [76, Problem 13.20, pp. 182,
183].
Problem 14.24
Rudin Chapter 14 Exercise 24.
Proof. Since 1 < R2 < R1 , we have A(1, R2 ) ⊆ A(1, R1 ). Assume that f : A(1, R1 ) → A(1, R2 )
was a bijective conformal mapping. By the first half of the proof of Theorem 14.22, we may
assume without loss of generality that |f (z)| → 1 as |z| → 1 and |f (z)| → R2 as |z| → R1 .
Consider the family of holomorphic functions F = {fn }, where
f1 = f
and
fn+1 = f ◦ fn : A(1; R1 ) → A(1; R2 )
for all n ∈ N. Thus each fn is bijective.
By the definition, we have
f A(1, R1 ) = A(1, R2 ) ⊆ A(1, R1 ) and
fn A(1, R1 ) ⊆ A(1, R1 )
(14.53)
for every n = 1, 2, . . ., so the family F is uniformly bounded on compact subsets of A(1, R1 ).
Therefore, it follows from Theorem 14.6 (Montel’s Theorem) that F is normal and then there
exists a subsequence {fnk } converging uniformly on compact subsets of A(1, R1 ) to a holomorphic
function
g : A(1, R1 ) → A(1, R2 ).
Denote Ω = A(R2 , R1 ). We claim that
fn (Ω) ∩ fm (Ω) = ∅,
(14.54)
where n, m ∈ N and n 6= m. Fix m. Assume that ω ∈ fm (Ω) ∩ fn (Ω). Then we can find p, q ∈ Ω
such that fn (p) = ω and fm (q) = ω. If n < m, then we have
fn (p) = fm (q) = fn fm−n (q)
which gives fm−n (q) = p ∈ Ω, but it contradicts the fact that fm−n A(1, R1 ) ∩ Ω = ∅. If
n > m, then we have
fm (q) = fn (p) = fm fn−m(p)
which gives fn−m (p) = q ∈ Ω, a contradiction again. Consequently, we prove the claim (14.54).
Assume that the range of g contained a nonempty open set. This means that g is not
constant and so we pick
an p ∈ Ω ⊆ A(1, R1 ) such that g(p) = ω. By the Open Mapping
Theorem, g A(1, R1 ) is an open set so that there exists a δ > 0 such that 0 ∈
/ g D(p; δ) .
Suppose that
hk (z) = fnk (z) − ω and h(z) = g(z) − ω.
We observe that hk , h ∈ H A(1, R1 ) , {hk } converges to h uniformly on compact subsets of
A(1, R1 ) and h 6≡ 0. Furthermore, we can select δ if necessary so that h(z) 6= 0 on C(p; δ).g
Hence it follows from Lemma 14.3 (Hurwitz’s Theorem) that there corresponds an N ∈ N such
g
Otherwise, h(z) = 0 for all z ∈ A(1, R1 ) by Theorem 10.18 which means that g is constant.
14.3. Proofs of Conformal Equivalence between Annuli
121
that if k ≥ N , then hk and h have the same number of zeros in D(p; δ). Since h(p) = g(p)−ω = 0,
there exists an zk ∈ D(p; δ) ⊆ Ω such that k ≥ N implies
fnk (zk ) = ω
but this contradicts the fact (14.54). Therefore, the range of g cannot contain any nonempty
open set and the Open Mapping Theorem ensures that g is constant.
√
On the other hand, g cannot be constant on the circle C(0; R1 ). Otherwise, Theorem√10.24
(The Maximum Modulus Theorem)√and its Corollary establish that g is constant in A(1, R1 ).
Let K be a compact subset of A(1, R1 ). Since fnk → g uniformly on K, fnk is also constant in
K for large enough k. However, this contradicts
the fact that fn is injective for every n = 1, 2, . . ..
√
Therefore, g is not constant on C(0; R1 ).
Now the above two results are contrary, so they force that no such f exists and we have
completed the proof of the problem.
Problem 14.25
Rudin Chapter 14 Exercise 25.
Proof. We have f : A(1, R1 ) → A(1, R2 ), where 1 < R2 < R1 . By the first half of the proof of
Theorem 14.22, we may assume without loss of generality that
lim |f (z)| = 1
|z|→1
and
lim |f (z)| = R2 .
|z|→R1
Applying Problem 14.2(b), the reflection across the inner circle (i.e., the unit circle) extends f
to a conformal mapping
f1 : A(R1−1 , R1 ) → A(R2−1 , R2 )
and f1 satisfies
lim
|z|→R−1
1
|f1 (z)| = R2−1 ,
lim |f1 (z)| = 1 and
|z|→1
lim |f1 (z)| = R2 .
|z|→R1
Next, by considering the function
g1 (z) =
f1 (R1 z)
,
R2
we see that g1 is holomorphic in the region Ω = {z ∈ C | R1−2 < |z| < 1} such that |g1 (z)| → 1
as |z| → 1 and |g(z)| → R2−2 as |z| → R1−2 . Thus Problem 14.2(b) may be applied to extend g1
to a conformal mapping f2 : A(R1−2 , R12 ) → A(R2−2 , R22 ) and f2 satisfies
lim
|z|→R−2
1
|f2 (z)| = R2−2 ,
lim |f2 (z)| = 1 and
|z|→1
lim |f2 (z)| = R22 .
|z|→R21
This process can be repeated infinitely many times and finally we obtain a conformal mapping
F of the punctured plane C \ {0}. If F has a pole of order m at the origin, then z m F (z) is
entire and |F (z)| = 1 whenever |z| = 1. Now Problem 12.4(b) asserts that F (z) = αz n for some
|α| = 1 and some n ∈ Z \ {0}. Assume that n ≥ 2. Let a ∈ A(1, R1 ). Take ζ 6= 1 to be an n-root
of unity. Then we have
F (a) = αan = α(ζa)n = F (ζa),
(14.55)
but a 6= ζa. This contradicts the fact that F is one-to-one in A(1, R1 ). Assume that n ≤ −1.
The equation (14.55) also holds in this case which in turn contradicts the injectivity of F again.
In other words, F (z) = αz which implies that R1 = R2 , a contradiction. Next, if F has a
removable singularity at the origin, then F is actually entire and we use the same argument as
above to show that no such F exists. Hence no such mapping f exists which completes the proof
of the problem.
122
Chapter 14. Conformal Mapping
Remark 14.3
(a) Theorem 14.22 is sometimes called Schottky’s Theorem. Besides the proofs given
in the text (Problems 14.24 and 25), you can also find a simple and elegant proof ofs
this theorem in [6].
(b) Besides the analytical proofs provided in the text and the problems, one can find a
pure algebraic proof in [58].
14.4
Constructive Proof of the Riemann Mapping Theorem
Problem 14.26
Rudin Chapter 14 Exercise 26.
Proof.
(a) Suppose that the regions Ω0 , Ω1 , . . . , Ωn−1 and functions f1 , f2 , . . . , fn are constructed such
that Ωk = fj (Ωk−1 ), where k = 1, 2, . . . , n. Define
rn = inf{|z| | z ∈ C \ Ωn−1 }.
(14.56)
Then rn is the largest number such that D(0; rn ) ⊆ Ωn−1 and the definition shows that
there is an αn ∈ ∂Ωn−1 with |αn | = rn .h See Figure 14.2 below.
Figure 14.2: The constructions of Ωn−1 , D(0; rn ) and αn .
h
Geometrically, αn is a point on ∂Ωn−1 nearest the origin.
14.4. Constructive Proof of the Riemann Mapping Theorem
123
Choose βn2 = −αn and put
Fn = ϕ−αn ◦ s ◦ ϕ−βn : U → U,
where
ϕα (z) =
By the Chain Rule, we have
z−α
1 − αz
(14.57)
and s(ω) = ω 2 .
Fn′ (z) = ϕ′−αn s ϕ−βn (z) × s′ ϕ−βn (z) × ϕ′−βn (z)
= 2ϕ′−αn s ϕ−βn (z) × ϕ′−βn (z) × ϕ−βn (z).
By Theorems 10.33 and 12.4, ϕ′−αn (z) 6= 0 and ϕ′−βn (z) 6= 0 for every z ∈ Ωn−1 . Thus
Fn′ (z) 6= 0 for all z ∈ Ωn−1 and it follows from Theorem 10.30(c) that Fn has a holomorphic
inverse Gn in Ωn−1 .
(b) By the definition and Theorem 12.4, we may write
−1
−1
◦ ϕαn : Ωn−1 → U,
Gn = ϕ−1
◦ ϕ−1
−αn = ϕβn ◦ s
−βn ◦ s
where s−1 (z) =
(14.58)
√
z. Combining the Chain Rule and Theorem 12.4, we get
G′n (0) = ϕ′βn s−1 ϕαn (0) × (s−1 )′ ϕαn (0) × ϕ′αn (0)
1
= ϕ′βn s−1 (−αn ) × √
× (1 − |αn |2 )
2 −αn
1 − rn2
=
× ϕ′βn (βn )
2βn
1 − rn2
.
=
2(1 − |βn |2 )βn
Put fn = λn Gn : Ωn−1 → U , where λn =
fn′ (0) = λn G′n (0) = |G′n (0)| =
|G′n (0)|
G′n (0) .
(14.59)
Now the formula (14.59) implies that
1 + rn
1 − rn2
1 − rn2
=
√ = √ .
2
2(1 − |βn | ) · |βn |
2(1 − rn ) rn
2 rn
Using the A.M. ≥ G.M., it is easy to see that fn′ (0) > 1.
(c) We prove the assertions one by one.
– Each ψn : Ω → Ωn is bijective. By the definition, we have ψn = fn ◦ fn−1 ◦ · · · ◦ f1 .
Since Ω = Ω0 and fn (Ωn−1 ) = Ωn ⊆ U , we have
ψn (Ω) = fn (fn−1 (· · · f1 (Ω0 ))) = fn (fn−1 (· · · f2 (Ω1 ))) = fn (Ωn−1 ) = Ωn ⊆ U.
By the representation (14.58) and Theorem 12.4, each Gn and hence each fn is injective on Ωn−1 . Consequently, each
ψn : Ω → Ωn ⊆ U
is injective on Ω.
– {ψn′ (0)} is bounded. Since ϕ−αn , ϕ−βn , s ∈ H(U ), the definition (14.57) gives Fn ∈
H(U ). Furthermore, it is clear that
Fn (0) = ϕ−αn ϕ2−βn (0) = ϕ−αn (βn2 ) = ϕ−αn (−αn ) = 0,
(14.60)
124
Chapter 14. Conformal Mapping
so we know from Theorem 12.2 (Schwarz’s Lemma) that
|Fn (ω)| ≤ |ω| and |Fn′ (0)| ≤ 1
(14.61)
for all ω ∈ U . Put ω = Gn (z) into the inequality (14.61), we get that
|G′n (0)| ≥ 1
|Gn (z)| ≥ |z| and
(14.62)
hold for all z ∈ Ωn−1 .
Lemma 14.4
For every n = 1, 2, . . ., we have 0 < r1 ≤ r2 ≤ · · · ≤ 1.
Proof of Lemma 14.4. Notice that we have fn (Ωn−1 ) = Ωn , D(0; rn ) ⊆ Ωn−1 and
D(0; rn+1 ) ⊆ Ωn . By the definition (14.56), there exists a boundary point α ∈ ∂Ωn
such that |α| = rn+1 . Select {zn−1,k } ⊆ Ωn−1 such that fn (zn−1,k ) → α as k → ∞.
Assume that {zn−1,k } had a limit point β in Ωn−1 . Since fn is obviously continuous on
Ωn−1 , we have α = fn (β) which means that Ωn ∩ ∂Ωn 6= ∅, a contradiction. Therefore,
the sequence {zn−1,k } cannot have a limit point in Ωn−1 . Since D(0; rn ) ⊆ Ωn−1 , we
must have
lim sup |zn−1,k | ≥ rn .
k→∞
Since |fn (z)| = |Gn (z)|, we derive from the first inequality (14.62) that
rn+1 = |α| = lim |fn (zn−1,k )| = lim |Gn (zn−1,k )| ≥ lim sup |zn−1,k | ≥ rn
k→∞
k→∞
k→∞
as desired. This ends the proof of the lemma.
z
r1
Now Lemma 14.4 ensures that D(0; r1 ) lies in every Ωn . Since φ(z) = maps D(0; r1 )
one-to-one and onto U , the map Ψn = ψn ◦ φ−1 : U → U satisfies the hypotheses of
Theorem 12.2 (Schwarz’s Lemma) so that |Ψ′n (0)| ≤ 1 which means that
|ψn′ (0)| ≤
1
r1
(14.63)
for every n = 1, 2, . . .. In other words, {ψn′ (0)} is bounded.
– A formula of ψn′ (0). By the aid of the Chain Rule and part (b), we establish easily
that
n
Y
1 + rk
′
′
′
′
ψn (0) = fn (0) × fn−1 (0) × · · · × f1 (0) =
(14.64)
√ .
2 rk
k=1
– The sequence {rn } converges to 1. For m > n ≥ 1, we define
ψm,n = fm ◦ fm−1 ◦ · · · ◦ fn+1
which is holomorphic in Ωn and hence on D(0; rn+1 ). In view of the value (14.60),
we know that Gn (0) = 0 and then fn (0) = 0 for every n ∈ N. Recalling the fact
ψn = fn ◦ fn−1 ◦ · · · ◦ f1 , we therefore have ψn (0) = 0 for every n ∈ N which implies
ψm,n (0) = 0. Using similar attack as in proving the inequality (14.63),i we can obtain
′
|ψm,n
(0)| ≤
i
1
rn+1
.
That is, D(0; rn+1 ) lies in every Ωm for all m ≥ n + 1, the map Ψm,n = ψm,n ◦ φ−1 : U → U satisfies the
−1
hypotheses of Theorem 12.2 (Schwarz’s Lemma), where φ(z) = rn+1
z maps D(0; rn+1 ) one-to-one and onto U .
14.4. Constructive Proof of the Riemann Mapping Theorem
125
Now the Chain Rule and part (b) assert that
′
|ψm,n
(0)| =
so that
m
Y
k=n+1
|fk′ (0)| =
m
Y
1 + rk
√
2 rk
k=n+1
m
Y
1 + rk
1
1<
≤
√
2 rk
rn+1
(14.65)
k=n+1
which implies the convergence of
∞
Y
1 + rn
√
2 rn
n=1
by taking m → ∞. By a basic fact about the convergence theory of infinite products
[9, Note 1, p. 242], we see that
lim
n→∞
1 + rn
= 1.
√
2 rn
Combining this fact and the identity
√
(1 − rn )2
1 + rn
,
√ −1=
√
2 rn
2 rn
we conclude immediately that rn → 1 as n → ∞.
(d) We verify the parts one by one.
– |hn | ≤ |hn+1
| for each n ∈ N. By the definition of ψn , we have ψn+1 (z) =
fn+1 ψn (z) so that
(14.66)
zhn+1 (z) = fn+1 zhn (z) .
Applying the first inequality (14.62) to the right-hand side of the expression (14.66),
we assert that
|zhn+1 (z)| ≥ |zhn (z)|
which implies our expected result immediately.
– ψn → ψ converges uniformly on compact subsets of Ω. Recall from part
(c) that each ψn is injective on Ω, so every hn is zero-free in Ω and we can define
gn = log hn as a holomorphic function in Ω. Define un = Re gn = log |hn | which is
harmonic in Ω. Since each ψn (Ω) is a subset of U , |hn (z0 )| < ∞ for some z0 ∈ Ω \ {0}
and all n ∈ N. Now the inequality |hn | ≤ |hn+1 | guarantees that
u1 ≤ u2 ≤ · · · .
According to Theorem 11.11 (Harnack’s Theorem), we see immediately that {un }
converges uniformly on compact subsets of Ω. Since ψn′ (z) = hn (z) + zh′n (z), the
second fact in part (c) assures us that {hn (0)} is bounded and therefore {hn (0)}
converges. Consequently, {gn (0)} converges and we conclude from Problem 11.8 that
{gn } converges uniformly on compact subsets of Ω to g. By the definition, we know
that
ψn (z) = zegn (z) ,
so {ψn } also converges uniformly on compact subsets of Ω to ψ. In view of Theorem
10.28, it is true that ψ ∈ H(Ω).
126
Chapter 14. Conformal Mapping
– The map ψ : Ω → U is surjective. Let ω ∈ U . By Lemma 14.4 and the fact
rn → 1 as n → ∞, we may select an N ∈ N such that
|ω| < rn
for all n ≥ N . Therefore, for p ≥ 1, we have
ω ∈ D(0, rN +p )
⊆ ΩN +p−1
= fN +p−1 (ΩN +p−2 )
= ···
= fN +p−1 ◦ fN +p−2 ◦ · · · ◦ fN +1 ◦ fN (ΩN −1 )
= fN +p−1 ◦ fN +p−2 ◦ · · · ◦ fN +1 ◦ ψN (Ω)
which means that there is a zp ∈ Ω such that
ω = fN +p−1 ◦ fN +p−2 ◦ · · · ◦ fN +1 ◦ ψN (zp )
= ψN +p−1 (zp ).
(14.67)
(14.68)
Applying the inequality (14.62) repeatedly to the expression (14.67), we see that
|ω| ≥ |ψN (zp )|
(14.69)
lim ψN +pj −1 (zpj ) = ψ(z).
(14.70)
−1
for all p ≥ 1. Clearly, D(0; |ω|) is a compact subset of ΩN , so ψN
D(0; |ω|) is also a
compact subset of Ω. Denote this set by K. By the result (14.69), we establish that
{zp } ⊆ K. Then the Bolzano-Weierstrass Theorem [79, Problem 5.25, p. 68] suggests
that there corresponds a subsequence {zpj } such that zpj → z ∈ K. Since ψn → ψ
uniformly on K, it observes that
j→∞
Combining the expression (14.68) and the limit (14.70), we conclude at once that
ω = ψ(z)
which means that ψ is surjective.
– The map ψ : Ω → U is injective. Since ψ is surjective, it is not constant. Recall
that each ψn is injective, so Problem 14.13 indicates easily that ψ is also injective.
Hence we have completed the analysis of the problem.
Problem 14.27
Rudin Chapter 14 Exercise 27.
Proof. Taking logarithms in the inequality (14.64) with m = 2n and then using the hint to get
2n
X
k=n+1
√
2n
h
X
(1 − rk )2 i
1 + rk
1
=
log 1 +
log √
≤ log
= − log rn+1 .
√
2 rk
2 rk
rn+1
k=n+1
(14.71)
14.4. Constructive Proof of the Riemann Mapping Theorem
Using the Mean Value Theorem for Derivatives, one can show that log(1 + x) >
Since 0 <
√
(1− rk )2
2
<
1
2
127
x
1+x
for x > 0.
for all k ∈ N, the inequality (14.71) reduces to
− log rn+1 >
2n
X
k=n+1
√
√
2n
X
(1 − rk )2
(1 − rk )2 i
>
> 0.
log 1 +
√
2 rk
3
h
(14.72)
k=n+1
By elementary calculus again, we know that log(1+x)
is strictly decreasing for x > −1. Combining
x
this fact and Lemma 14.4, it is true that for all n ≥ 1,
log rn+1
log[1 + (rn+1 − 1)]
log[1 + (r1 − 1)]
log r1
=
≤
=
rn+1 − 1
rn+1 − 1
r1 − 1
r1 − 1
which implies
− log rn+1 ≤ (1 − rn+1 ) ·
As (1 +
√
rk )2 ≤ 4, we have
(1 −
log r1
log r1−1
≤ (1 − rn ) ·
.
r1 − 1
1 − r1
√ 2
(1 − rk )2
(1 − rk )2
rk ) =
.
√ 2 ≥
(1 + rk )
4
(14.73)
(14.74)
log r −1
Let A = 1−r11 . Now we observe by substituting the inequalities (14.73) and (14.74) into the
inequality (14.72) that
√
2n
2n
X
(1 − rk )2
1 X 1 − rk 2
≤
< − log rn+1 < A(1 − rn )
0<
3
2
3
k=n+1
k=n+1
or equivalently,
2n X
1 − rk 2 1 − rn
<
,
2B
B
0<
k=n+1
where B = 3A. Using Lemma 14.4, the inequality (14.75) gives
n
1 − r
2n
2B
2
<
1 − rn
.
B
After some algebra, we can show that it is actually equivalent to
h
2n
1 − r
Particularly, for p = 1, 2, . . ., we obtain
2n
16B
i2
<n
1 − r n
.
16B
1 − r p−1 h 1 − r p i2
2
2
< 2p−1
2p
16B
16B
1 − r p−1 i2
h
1 − r p−2 2
2
2p−1
< 2p−2
16B
16B
..
.
h 1 − r 1 i2 1 − r
1
2
<
21
16B
16B
and they certainly imply that
h 1 − r p i2p
1 − r1
2
<
.
2p
16B
16B
(14.75)
128
Chapter 14. Conformal Mapping
Since
that
1−r1
16B
1
is constant and a n → 1 as n → ∞ if a > 0, there exists a positive constant M such
0 < 1 − r2p <
M
2p
for every p = 1, 2, . . .. Thus it gives
0 < 2p (1 − r2p )2 <
for every p = 1, 2, . . .. Obviously, the series
∞
X
M2
p=1
[79, Theorems 6.5, 6.6, p. 76] that
∞
X
2p
M2
2p
converges, so we deduce from the results
(1 − rn )2 < ∞.
n=1
This completes the proof of the problem.
Remark 14.4
(a) The proofs of Problems 14.26 and 14.27 are more or less the same as Ostrowski’s paper
[49].
(b) For more properties of the so-called Koebe mapping and details of the constructive
proof of the Riemann Mapping Theorem, please refer to [52, §3, pp. 15, 16, 183 – 186]
and [10, §1.7, pp. 180 – 214] respectively.
Problem 14.28
Rudin Chapter 14 Exercise 28.
Proof. Since αn ∈ U \ Ωn−1 , we have |αn | ≥ rn , see Figure 14.2 again. By Problem 14.26, we
may assume further that
1 + rn
rn < |αn | ≤
.
(14.76)
2
In this case, we also have β 2 = −αn . Furthermore, recall from the definition (14.57) that
Fn′ (z) 6= 0 for all z ∈ Ωn−1 so that Fn has a holomorphic inverse Gn in Ωn−1 . Next, we observe
from the formula (14.59) and the A.M. ≥ G.M. that
|G′n (0)| =
1 + |βn |2
1 + |αn |
1 − |αn |2
=
= p
> 1.
2
2(1 − |βn | )|βn |
2|βn |
2 |αn |
In conclusion, we always have |fn′ (0)| > 1 for n = 1, 2, . . .. Now it is easy to see that Lemma
14.4 remains valid, each ψn : Ω → Ωn is bijective and {ψn′ (0)} is bounded by r11 . It remains to
prove that rn → 1 as n → ∞.
Instead of the formula (14.64) and the inequalities (14.65), we obtain
ψn′ (0)
=
fn′ (0)
×
and
1<
′
fn−1
(0)
× ···
× f1′ (0)
n
Y
1 + |αk |
p
=
2 |αk |
k=1
m
Y
1
1 + |αk |
p
,
≤
rn+1
2
|α
|
k
k=n+1
14.4. Constructive Proof of the Riemann Mapping Theorem
129
where m > n ≥ 1. By the right-most inequality (14.76), we may write
1 + rn
= |αn | + δn ,
2
where 0 ≤ δn <
1−rn
2 .
Since we have
1 + |αk |
1 + rk + δk
p
,
= √
2 rk + δk
2 |αk |
we can show that the infinite product
∞
Y
1 + rn + δn
√
2
rn + δn
n=1
converges and so
1 = lim
n→∞
1 + rn + δn
√
.
2 rn + δn
(14.77)
By Lemma 14.4, {rn } converges to a positive number r. Suppose that δn → δ as n → ∞, where
0 ≤ δ ≤ 1−r
2 . Then it follows from the limit (14.77) that
1=
1+r+δ
√
2 r+δ
which implies r + δ = 1. If r 6= 1, then δ 6= 0. Now the expression δ = 1 − r contradicts the fact
0 < δ ≤ 1−r
2 . Hence we have r = 1 and we complete the proof of the problem.
Problem 14.29
Rudin Chapter 14 Exercise 29.
Proof.
(a) By translating Ω by −a, we may assume without loss of generality that a = 0. We claim
that
fn′ (0) = [f ′ (0)]n
(14.78)
for all n = 1, 2, . . .. We use induction and the case n = 1 is trivial. Assume that
fk′ (0) = [f ′ (0)]k
(14.79)
for some k ∈ N. By the definition, we have
′
(0) = f ′ fk (0) · fk′ (0).
fk+1
(14.80)
The hypothesis f (0) = 0 implies that fn (0) = 0 for all n ∈ N, so the expression (14.80)
′
(0) = f ′ (0) · fk′ (0). By the inductive step (14.79), we conclude that
reduces to fk+1
fk+1 (0) = [f ′ (0)]k+1 .
Hence the claim follows from induction.
Since Ω is bounded, there exists a positive constant M such that |z| ≤ M for all z ∈ Ω.
Since Ω is a region, one can find a r > 0 such that D(0; r) ⊆ Ω. By the hypothesis
130
Chapter 14. Conformal Mapping
f (Ω) ⊆ Ω, it is true that fn (Ω) ⊆ Ω for every n = 1, 2, . . .. Hence we have |fn (z)| ≤ M
for all z ∈ D(0; r) and n ∈ N, and then we yield from Theorem 10.26 (Cauchy’s Estimate)
that
k!M
(14.81)
|fn(k) (0)| ≤ k
r
for k = 1, 2, . . .. Combining the expression (14.78) and the estimate (14.81), we get
′
|f (0)| =
1
[fn′ (0)] n
≤
M 1
n
r
for every n = 1, 2, . . .. By taking n → ∞, we establish that |f ′ (0)| ≤ 1 as required.
(b) If f (k) (0) = 0 for all k ≥ 2, then since f ∈ H(Ω), Theorem 10.16 implies that f (z) = a + bz
for some a, b ∈ C. Since f (0) = 0 and f ′ (0) = 1, we have a = 0 and b = 1. Thus f (z) = z
as desired. Suppose that N ≥ 2 is the smallest positive integer such that f (N ) (0) 6= 0 and
let ck =
f (k) (0)
k!
for k ≥ N . In particular, we have cN 6= 0. Since f ′ (0) = 1, we have
f (z) = z +
∞
X
ck z k = z + cN z N g(z),
(14.82)
k=N
where g(0) = 1.
We claim that for each n ∈ N, there is a rn > 0 and an gn ∈ H D(0; rn ) such that for all
z ∈ D(0; rn ), we have
fn (z) = z + cN z N gn (z)
and gn (0) = n.
(14.83)
The expression (14.82) is just the case n = 1, so we assume that the result (14.83)
holds for some positive
integer n. Since fn (0) = 0, we can pick rn+1 ∈ (0, rn ) such
that fn D(0; rn+1 ) ⊆ D(0; rn ). Then it follows from the expressions (14.82) and (14.83)
that if z ∈ D(0; rn+1 ), then
fn+1 (z) = f fn (z)
= fn (z) + cN [fn (z)]N g fn (z)
N
= z + cN z N gn (z) + cN z + cN z N gn (z) g fn (z)
N
= z + cN z N gn (z) + 1 + cN z N −1 gn (z) g fn (z) .
N
Suppose that gn+1 (z) = gn (z) + 1 + cN z N −1 gn (z) g fn (z) . Recall that N − 1 ≥ 1, so
we obtain
gn+1 (0) = gn (0) + g fn (0) = n + 1.
By induction, our claim follows. Next, we differentiate the expression (14.83) N times and
put z = 0, we have
fn(N ) (0) = N !cN gn (0) = nN !cN
(N )
for every n = 1, 2, . . .. Since cN 6= 0, we have fn (0) → ∞ as n → ∞ which contradicts
the fact (14.81). Consequently, this means that f (k) (0) = 0 for all k ≥ 2 which implies
f (z) = z in D(0; r). By the Corollary following Theorem 10.18, it is actually true in Ω.
(c) By the hypothesis, we know that f ′ (0) = eiθ for some θ ∈ [0, 2π]. If eiθ is an N -root of
unity, then the integer nk = kN satisfies
[f ′ (0)]nk = (eiN θ )k = 1.
14.4. Constructive Proof of the Riemann Mapping Theorem
131
Otherwise, we need the following form of the Kronecker’s Approximation Theorem [5,
Theorem 7.8, p. 149]:
Lemma 14.5 (The Kronecker’s Approximation Theorem)
Given any real α, any irrational β and any ǫ > 0, there exist integers m and n with
n > 0 such that
|nβ − m − α| < ǫ.
θ
Take α = 0 and β = 2π
. For every k ∈ N, we obtain from Lemma 14.5 that there exist
integers nk and mk with nk > 0 such that
1
nk θ
− mk <
2π
2kπ
or equivalently
|nk θ − 2mk π| <
1
k
which implies that |nk θ − 2mk π| → 0 as k → ∞. Consequently, we have eink θ → 1 as
k → ∞.
Next, we consider the family F = {fnk }. Recall the fact that fn (Ω) ⊆ Ω, so F is bounded
on Ω and Theorem 14.6 (Montel’s Theorem) ensures that F is normal and thus it has
a convergent subsequence which we also call {fnk } for convenience. Let g be its limit
function. By Theorem 10.28, g ∈ H(Ω) and fn′ k (0) → g′ (0) as k → ∞. By the formula
(14.78) and the hypothesis f ′ (0)]nk → 1 as k → ∞, we get fn′ k (0) → 1 as k → ∞ and
thus g′ (0) = 1. In other words, g is not constant and it follows from Problem 10.20 that
g(Ω) ⊆ Ω. By part (b), we see that
g(z) = z
(14.84)
in Ω.
Finally, if z, ω ∈ Ω and f (z) = f (ω), then we have
z = g(z) = lim fnk (z) = lim fnk (ω) = g(ω) = ω.
k→∞
k→∞
Thus f is one-to-one. If p ∈ Ω \ f (Ω), then p ∈ Ω \ fn (Ω) for every n ∈ N because we
always have fn (Ω) ⊆ f (Ω). Particularly, the function fnk (z) − p 6= 0 in Ω. However,
fnk (p) − p → g(p) − p = 0 as k → ∞. By Problem 10.20 again, we establish that g(z) = p
for all z ∈ Ω which certainly contradicts the result (14.84). Hence we have f (Ω) = Ω, i.e.,
f is onto.
We end the proof of the problem.
Problem 14.30
Rudin Chapter 14 Exercise 30.
Proof. We have
n
o
az + b
Λ = ϕ(z) =
a, b, c, d ∈ C and ad − bc 6= 0 .
cz + d
The number ad − bc is called the determinant of ϕ.
132
Chapter 14. Conformal Mapping
(a) If α, β and γ are distinct, then the definition gives
ϕ(z) = [z, α, β, γ] =
(z − α)(β − γ)
(β − γ)z − α(β − γ)
=
.
(z − γ)(β − α)
(β − α)z − γ(β − α)
(14.85)
By direct computation, we see that
−γ(β − γ)(β − α) + α(β − γ)(β − α) = (β − γ)(β − α)(α − γ) 6= 0
which means ϕ ∈ Λ in this case. Next, if α = ∞ and β 6= γ, then we have
ϕ(z) = [z, ∞, β, γ] =
0 · z + (β − γ)
z−γ
(14.86)
so that 0 · (−γ) − 1 · (β − γ) 6= 0. Thus we also have ϕ ∈ Λ in this case.j
Furthermore, the representations (14.85) and (14.86) imply easily that ϕ maps {α, β, γ}
to {0, 1, ∞}.
(b) Suppose that {α, β, γ} and {a, b, c} are two groups of distinct complex numbers. By the
definition (14.85), we have
(β − γ)z − α(β − γ)
(b − c)ω − a(b − c)
=
(b − a)ω − c(b − a)
(β − α)z − γ(β − α)
which implies
a(b − c)(β − α)(z − γ) − c(b − a)(β − γ)(z − α)
(b − c)(β − α)(z − γ) − (b − a)(β − γ)(z − α)
a(b − c)(β − α) − c(b − a)(β − γ) z + cα(b − a)(β − γ) − aγ(b − c)(β − α)
= (b − c)(β − α) − (b − a)(β − γ) z + α(b − a)(β − γ) − γ(b − c)(β − α)
ab(β − α) + ac(α − γ) + bc(γ − β) z + bcα(β − γ) + acβ(γ − α) + abγ(α − β)
= b(γ − α) + a(β − γ) + c(α − β) z + bβ(α − γ) + aα(γ − β) + cγ(β − α)
ω=
= ϕ(z).
(14.87)
Direct checking gives the determinant of ϕ as follows
ab(β − α) + ac(α − γ) + bc(γ − β) · bβ(α − γ) + aα(γ − β) + cγ(β − α)
− bcα(β − γ) + acβ(γ − α) + abγ(α − β) · b(γ − α) + a(β − γ) + c(α − β)
= ab2 β(α − γ)(β − α) + abcβ(α − γ)2 + b2 cβ(γ − β)(α − γ) + a2 bα(γ − β)(β − α)
+ ac2 α(γ − β)(α − γ) + abcα(γ − β)2 + abcγ(β − α)2 + ac2 γ(β − α)(α − γ)
+ bc2 γ(γ − β)(β − α) − b2 cα(γ − α)(β − γ) − abcβ(γ − α)2 ab2 γ(γ − α)(α − β)
− abcα(β − γ)2 − a2 cβ(β − γ)(γ − α) − a2 bγ(α − β)(β − γ) − bc2 α(β − γ)(α − β)
− ac2 β(α − β)(γ − α) − abcγ(α − β)2
= (α − β)(β − γ)(γ − α)(ab2 − b2 c − a2 b + a2 c − ac2 + bc2 )
= (α − β)(β − γ)(γ − α)(a − b)(b − c)(c − a)
6= 0
so that ϕ ∈ Λ in this case.
j
The cases for β = ∞ or γ = ∞ can be done similarly, so we omit the details here.
14.4. Constructive Proof of the Riemann Mapping Theorem
133
Next, if {∞, β, γ} and {a, b, c} are two groups of distinct complex numbersk , then we follow
from the definition (14.86) (or by taking α → ∞ in the definition (14.87) that
(b − c)ω − a(b − c)
β−γ
=
(b − a)ω − c(b − a)
z−γ
which gives
ω=
a(b − c)z − aγ(b − c) − c(b − a)(β − γ)
= ϕ(z).
(b − c)z − γ(b − c) − (β − γ)(b − a)
(14.88)
Now simple algebra shows that
a(b − c)[−γ(b − c) − (β − γ)(b − a)] − (b − c)[−aγ(b − c) − c(b − a)(β − γ)]
= −aγ(b − c)2 − a(b − a)(b − c)(β − γ) + aγ(b − c)2 + c(b − a)(b − c)(β − γ)
= (b − a)(b − c)(c − a)(β − γ)
6= 0
which means ϕ ∈ Λ in this case.
Furthermore, if {∞, β, γ} and {∞, b, c} are two groups of distinct complex numbersl , then
we have
β−γ
b−c
=
ω−c
z−γ
and so
ω=
(b − c)z + cβ − bγ
= ϕ(z).
β−γ
(14.89)
Since (b − c)(β − γ) 6= 0, we also have ϕ ∈ Λ.
Finally, by the formulas (14.87), (14.88) and (14.89), it is easy to see that ϕ maps {α, β, γ}
to {a, b, c}.
(c) Suppose that φ(z) = [z, β, γ, δ] which sends {β, γ, δ} to {0, 1, ∞} by part (a). Then the
mapping φ ◦ ϕ−1 carries {ϕ(β), ϕ(γ), ϕ(δ)} to {0, 1, ∞}. By §14.3, this map φ ◦ ϕ−1 is
unique, so (φ ◦ ϕ−1 )(z) = [z, ϕ(β), ϕ(γ), ϕ(δ)] and then we have
[ϕ(α), ϕ(β), ϕ(γ), ϕ(δ)] = (φ ◦ ϕ−1 ) ϕ(α) = φ(α) = [α, β, γ, δ].
(d) This part has been solved in [76, Problem 13.18, p. 182].
(e) Now we have
[z ∗ , α, β, γ] = [z, α, β, γ].
(14.90)
If C is a straight line, then we choose γ = ∞ in the equation (14.90) to get
z∗ − β
z−β
.
=
α−β
α−β
(14.91)
This certainly gives |z ∗ − β| = |z − β|. Since β is an arbitrary point on C, z and z ∗ are in
fact equidistance from each point on C. Furthermore, the equation (14.91) implies that
Im
z−β
z∗ − β
z−β
= Im
= −Im
α−β
α
−β
α−β
which means that z and z ∗ lie in different half planes determined by C.
k
l
The other cases can be done similarly, so we only consider this case and omit the others.
Again, we have omitted other similar cases.
134
Chapter 14. Conformal Mapping
Next, we suppose that C = {z ∈ C | |z| = 1}. By the equation (14.90) and applications of
the invariance property of ϕ ∈ Λ as verified in part (c), we see that
i
h 1 1 1 i h1
[z ∗ , α, β, γ] = [z, α, β, γ] = [z, α, β, γ] = z, , ,
= , α, β, γ
α β γ
z
which implies z ∗ =
from the ratio
1
z
or z ∗ · z = 1. Thus we have |z ∗ | = |z|−1 and furthermore, it deduces
z∗
1
= 2 >0
z
|z|
that z ∗ lies on the ray L = {tz | t ∈ R}. Geometrically, see Figure 14.3 for the construction
of the point z ∗ .
Figure 14.3: The construction of the symmetric point z ∗ of z.
(f) Let α, β, γ ∈ C. Then it follows from part (c) and the definition that
[ϕ(z ∗ ), ϕ(α), ϕ(β), ϕ(γ)] = [z ∗ , α, β, γ] = [z, α, β, γ] = [ϕ(z), ϕ(α), ϕ(β), ϕ(γ)].
Hence ϕ(z ∗ ) and ϕ(z) are symmetric with respect to ϕ(C).
This finishes the analysis of the problem.
Problem 14.31
Rudin Chapter 14 Exercise 31.
Proof.
(a) Given ϕ, ψ, φ ∈ Λ by
ϕ(z) =
az + b
,
cz + d
ψ(z) =
αz + β
γz + δ
and φ(z) =
– Composition as group operation. It is easy to see that
(αa + βc)z + αb + βd
ψ ϕ(z) =
(γa + δc)z + γb + δd
Az + B
.
Cz + D
14.4. Constructive Proof of the Riemann Mapping Theorem
135
and its determinant is
(αa + βc)(γb + δd) − (γa + δc)(αb + βd) = (ad − bc)(αδ − βγ) 6= 0
so that ψ ◦ ϕ ∈ Λ.
– Associativity. Simple algebra verifies
[ϕ(z) + φ(z)] + ψ(z) = ϕ(z) + [φ(z) + ψ(z)].
– The identity element. Now the usual identity map id : C → C is the identity
element of Λ because its determinant is 1 and
id ◦ ϕ = ϕ ◦ id = ϕ.
– The inverse of ϕ. The equation ω = ϕ(z) has exactly one solution and indeed, it is
z = ϕ−1 (ω) =
dω − b
.
cω − a
Since the determinant of ϕ−1 is −ad + bc 6= 0, we have ϕ−1 belongs to Λ. Clearly, we
know that ϕ−1 ◦ ϕ = ϕ ◦ ϕ−1 = id.
By the definition (see [25, Definition 4.1, pp. 37, 38]),
Λ is indeed a group.
If1 we take
which
ϕ(z) = z + 1 and ψ(z) = z1 , then we see that ϕ ψ(z) = z1 + 1 and ψ ϕ(z) = z+1
imply that Λ is not commutative.
(b) Let ϕ ∈ Λ be given by
az + b
cz + d
and ϕ 6= id. Define ∆ = ad − bc 6= 0. Since we may write
ϕ(z) =
ϕ(z) =
√a z
∆
√c z
∆
+
+
√b
∆
√d
∆
,
we may assume without loss of generality that ad − bc = 1. Now the equation z = ϕ(z) is
equivalent to saying that
cz 2 + (d − a)z − b = 0.
(14.92)
– Case (i): c = 0. Thus we have ad 6= 0 and ϕ(∞) = ∞ so that ∞ is a fixed point of
ϕ. If a 6= d, then by solving the equation (14.92), we get one more (finite) fixed point
which is
b
.
z=
d−a
Otherwise, a = d implies that
b
(14.93)
ϕ(z) = z +
d
whose fixed point is also ∞. Since ϕ 6= id, b 6= 0 so that ϕ has only a unique (infinite)
fixed point in this case.
– Case (ii): c 6= 0. Then the equation (14.92) has two roots
p
p
a − d ± (d − a)2 + 4bc
a − d ± (a + d)2 − 4
z=
=
.
(14.94)
2c
2c
Since ϕ(∞) = ac , ∞ is not transformed into itself. This means that ϕ has either one
or two finite fixed points on S 2 depending on whether a + d = ±2 or not. In the case
of the unique finite fixed point, it is given by
z=
a−d
.
2c
136
Chapter 14. Conformal Mapping
In conclusion, ϕ has either one or two fixed points on S 2 .
(c) We consider two cases.
– Case (i): ϕ has a unique fixed point. Given ϕ1 (z) = z + 1 which is obviously an
element of Λ.
If c = 0, then we follow from part (b) that ϕ has a unique (infinite) fixed point if and
only if it takes the form (14.93). Define ψ(z) = db z. Recall that bd 6= 0, so ψ ∈ Λ and
ψ −1 (z) = db z. Furthermore, it is clear that
d b
b
ψ −1 ϕ ψ(z) = ·
= z + 1 = ϕ1 (z).
z+
b
d
d
Hence we have shown that ϕ is conjugate to ϕ1 in this case.
Next, if c 6= 0, then we observe from the roots (14.94) that ϕ has a unique fixed point
z1 = a−d
2c if and only if a + d = ±2. Define the linear fractional transformation
S(z) =
1
z − z1
which carries z1 to ∞. Therefore, the linear fractional transformation
T = S ◦ ϕ ◦ S −1
(14.95)
has ∞ as its only fixed point because if p is a fixed point of T , then S −1 (p) will be a
fixed point of ϕ so that
p = S(z1 ) = ∞.
Hence it follows from part (b) that T (z) = z + B for some B ∈ C \ {0}. If we take
P (z) = Bz, then P −1 (z) = Bz and so
ϕ1 = P −1 ◦ T ◦ P.
(14.96)
By combining the expressions (14.95) and (14.96), we conclude that
ϕ1 = (P −1 ◦ S) ◦ ϕ ◦ (S −1 ◦ P ) = (S −1 ◦ P )−1 ◦ ϕ ◦ (S −1 ◦ P ).
Hence we have ψ = S −1 ◦ P . Since P, S ∈ Λ, we have ψ ∈ Λ.
– Case (ii): ϕ has two distinct fixed points. Consider the linear fractional transformation
φα (z) = αz,
where α is a non-zero complex number which will be determined soon. By part (b),
we have either “c = 0 and a 6= d” or “c 6= 0 and a + d 6= ±2”.
∗ Subcase (i): c = 0 and a 6= d. We notice that
ϕ(z) =
az + b
.
d
b
which is obviously a linear fractional transformation and
Take ψ(z) = z + d−a
b
−1
ψ (z) = z − d−a . Direct computation gives
a
b b
a
b
ϕ ψ(z) =
z+
+ = z+
d
d−a
d
d
d−a
14.4. Constructive Proof of the Riemann Mapping Theorem
137
and then
a
b
b
a
−
= z = φ ad (z).
ψ −1 ϕ ψ(z) = z +
d
d−a d−a
d
Consequently, ϕ is conjugate to φα with
α=
a
.
d
(14.97)
Particularly, 0 is the finite fixed point if and only if b = 0, so we have ϕ(z) = ad z
and ψ(z) = z.
∗ Subcase (ii): c 6= 0 and a + d 6= ±2. The two distinct finite fixed points
are given by (14.94). Let z1 and z2 be the roots corresponding to the negative
square root and the positive square root respectively. Now the linear fractional
transformation
z − z1
S(z) =
z − z2
maps the ordered pair {z1 , z2 } into {0, ∞}. Then the linear fractional transformation T = S ◦ ϕ ◦ S −1 fixes 0 and ∞. By the particular case of Subcase (i),
we know that
φα = ψ −1 ◦ T ◦ ψ = T
for some complex α. Since S −1 (z) =
z2 z−z1
z−1 ,
we obtain from the definition that
(az2 + b)z − (az1 + b)
ϕ S −1 (z) =
.
(cz2 + d)z − (cz1 + d)
and
T (z) =
(az − dz1 + 2b)z + cz12 + (d − a)z1 − b
22
.
− cz2 + (d − a)z2 − b z − (az1 − dz2 + 2b)
(14.98)
Since z1 , z2 are roots of the equation (14.92), the formula (14.98) simplifies to
T (z) = −
(az2 − dz1 + 2b)
z.
(az1 − dz2 + 2b)
(14.99)
Using the formula (14.94), the expression (14.99) can further reduce to
p
(a + d)2 − 4 + (a + d)
z.
T (z) = − p
(a + d)2 − 4 − (a + d)
Hence we obtain the formula
p
α = −p
(a + d)2 − 4 + (a + d)
(a + d)2 − 4 − (a + d)
.
(14.100)
Finally, α is determined by either (14.97) or (14.100).
(d) We prove the assertions one by one.
– The existence of β. Since ϕ has only a unique finite fixed point, the analysis of
part (b) leads us to the result that c 6= 0 and a + d = ±2. In this case, we have
α = a−d
2c . Then we have
d = a − 2αc.
(14.101)
Since α is a root of the equation (14.92), we have
b − dα = cα2 − aα = α(cα − a).
(14.102)
138
Chapter 14. Conformal Mapping
Obviously, cα − a 6= 0. Otherwise, put cα = a into the equation (14.101) will give
a + d = 0 which is impossible. Now we note that
1
=
ϕ(z) − α
1
az+b
cz+d
−α
=
cz + d
cz + d
=
az + b − cαz − dα
(a − cα)z + (b − dα)
(14.103)
Substituting the values (14.101) and (14.102) into the expression (14.103) to get
cz + a − 2αc
c(z − α) + a − αc
1
c
1
=
=
=
+
(14.104)
ϕ(z) − α
(a − cα)z + α(cα − a)
(z − α)(a − αc)
z − α a − αc
which means that
β=
With the aid of α =
a−d
2c
c
∈ C.
a − αc
and a + d = ±2, we can further show that
β = c.
(14.105)
– Gα is a subgroup of Λ. Let Gα = {ϕ ∈ Λ | ϕ(α) = α} ∪ {id} ⊆ Λ. Let ϕ and φ be
elements of Gα with the corresponding constant βϕ and βφ respectively. Therefore,
we see that
1
1
1
=
+ βϕ =
+ βϕ + βφ
(14.106)
φ(z)
−
α
z
−
α
ϕ φ(z) − α
so that ϕ ◦ φ ∈ Gα . By the definition, we have id ∈ Gα . Furthermore, since ϕ has α
as its only finite fixed point, so is ϕ−1 . Thus we have
1
ϕ−1 (z)
−α
=
1
− β.
z−α
In other words, it means that ϕ−1 ∈ Gα . By [25, Theorem 5.14, p. 52], Gα is a
subgroup of Λ.
– Gα is isomorphic to (C, +). Define f : Gα → C by f (ϕ) = βϕ , where βϕ is the
complex number satisfying the equation
1
1
=
+ βϕ .
ϕ(z) − α
z−α
(14.107)
As the expression (14.106) shows definitely that
f (ϕ ◦ φ) = βϕ + βφ ,
1
1
if and only
= z−α
so f is a homomorphism. Next, suppose that β = 0. Then ϕ(z)−α
if ϕ = id. In other words, the kernel of f is {id}. Finally, given β ∈ C \ {0}. we
consider a = 1 + αβ, b = −α2 β, c = β and d = 1 − αβ. Direct computation gives
ad − bc = 1. Besides, the linear fractional transformation
ϕβ (z) =
(1 + αβ)z − α2 β
βz + (1 − αβ)
(14.108)
fixes α only and satisfies the equation (14.107).m Consequently, we have ϕβ ∈ Gα
and f (ϕβ ) = β, i.e., f is surjective. Hence f is in fact an isomorphismn .
m
In fact, we establish from the value (14.105) that the representation (14.108) becomes
ϕ(z) =
n
See, for instance, [25, p. 132].
(1 + cα)z − cα2
.
cz + (1 − cα)
14.4. Constructive Proof of the Riemann Mapping Theorem
139
(e) Now we have Gα,β = {ϕ ∈ Λ | ϕ(α) = α and ϕ(β) = β}. This refers to the case c 6= 0 and
a + d 6= ±2.
– Every ϕ ∈ Gα,β satisfies the required equation. Since α and β are roots of
the equation (14.92), the formula (14.102) also holds forpβ. Obviously, a − βc 6= 0.
Otherwise, it implies the contradiction that a + d = ± (a + d)2 − 4. We observe
that
ϕ(z) − α
=
ϕ(z) − β
az+b
cz+d
az+b
cz+d
−α
−β
(a − αc)z + (b − dα)
(a − βc)z + (b − dβ)
(a − αc)z + α(cα − a)
=
(a − βc)z + β(cβ − a)
z−α
,
=γ·
z−β
=
where
γ=
a − αc
∈ C.o
a − βc
(14.109)
– Gα,β is a subgroup of Λ. Since id fixes α and β, we have id ∈ Gα,β . For every
ϕ, φ ∈ Gα,β , let γϕ and γφ be their corresponding complex numbers respectively.
Since
φ(z) − α
z−α
ϕ(φ(z)) − α
= γϕ ·
= γϕ · γφ ·
,
(14.110)
ϕ(φ(z)) − β
φ(z) − β
z−β
we have ϕ ◦ φ ∈ Gα,β . Assume that ϕ ∈ Gα,β was a constant map. Then it implies
that α = β, a contradiction. In addition, γϕ 6= 0. Otherwise, ϕ(z) = α for all z ∈ S 2
which is impossible. Next, if ϕ ∈ Gα,β , then ϕ−1 also fixes α and β, and we have
ϕ−1 (z) − α
1 z−α
=
·
.
−1
ϕ (z) − β
γϕ z − β
Consequently, these imply that ϕ−1 ∈ Gα,β . Hence Gα,β is a subgroup of Λ.
– Gα,β is isomorphic to (C \ {0}, ×). Define g : Gα,β → C \ {0} by g(ϕ) = γϕ , where
γϕ is the complex number satisfying the equation
z−α
ϕ(z) − α
= γϕ ·
.
ϕ(z) − β
z−β
(14.111)
The equation (14.110) implies that g(ϕ ◦ φ) = γϕ × γφ so that g is a homomorphism.
If g(ϕ) = 1, then we have a − αc = a − βc so that c = 0 and ϕ takes the form
ϕ(z) =
az + b
.
d
Put this into the equation (14.111) with γϕ = 1 and after simplification, we conclude
that ϕ(z) = z, i.e., the kernel of g is {id}. Let γ ∈ C \ {0}. By changing the subject
of the formula (14.109) to c and using the formula α + β = a−d
c , we can show that
d=
o
αγ − β
a.
α − βγ
This number is called the multipler of the transformation ϕ, read [24, pp. 15, 16].
140
Chapter 14. Conformal Mapping
Next, we apply the fact
√
1
γ+ √ =a+d
γ
to represent a, c and d in terms of α, β and γ as follows:
√
α − βγ
1 γ+√ ,
a=
(1 + γ)(α − β)
γ
αγ − β
1 √
d=
γ+√ ,
(1 + γ)(α − β)
γ
√
1−γ
1 c=
γ+√ .
(1 + γ)(α − β)
γ
(14.112)
Finally, we employ the formula αβ = − cb to obtain
b=−
αβ(1 − γ) √
1 γ+√ .
(1 + γ)(α − β)
γ
(14.113)
Now it is a routine task to check that the linear fractional transformation ϕ with
the coefficients given by the formulas (14.112) and (14.113) has α and β as its fixed
points, ad − bc = 1 and satisfies
g(ϕ) = γ.
In other words, the map g is surjective and hence, an isomorphism.
(f) By the hypothesis, the fixed points are finite. The following proof is due to Drazin [20]
az+b
has invariant circles if
who verified that the linear fractional transformation ϕ(z) = cz+d
and only if its determinant ∆ 6= 0 and
(a+d)2
∆
is real.
– Case (i): ϕ has a unique finite fixed point. Recall from the explicit form
(14.108) that
(1 + βα)z − βα2
,
(14.114)
ϕ(z) =
βz + (1 − βα)
where β = c 6= 0. Simple algebra gives
βϕ(z) = 1 + βα −
1
.
βz + (1 − βα)
(14.115)
Define
ϕ∗ = βϕ + (1 − βα)
and
ζ = βz + (1 − βα).
(14.116)
Then the equation (14.115) becomes
1
ϕ∗ (ζ) = 2 − .
ζ
(14.117)
Consequently, this change of variables establishes a one-to-one correspondence between the invariant circles of the linear fractional transformations (14.114) and (14.117).
Let C ∗ = {ζ ∈ C | |ζ − ρeiθ | = R} be an invariant circle of ϕ∗ , where ρ, θ, R are
real and R > 0. Consider the two points (ρ − R)eiθ and (ρ + R)eiθ which are the
endpoints of a diameter of C ∗ . Since ϕ∗ is conformal, the points P = ϕ∗ (ρ − R)eiθ
and Q = ϕ∗ (ρ + R)eiθ are also endpoints of a diameter of C ∗ so that |P − Q| = 2R
and 21 (P + Q) = ρeiθ . Notice that
P =2−
1
(ρ − R)eiθ
and Q = 2 −
1
,
(ρ + R)eiθ
(14.118)
14.4. Constructive Proof of the Riemann Mapping Theorem
so we have
ρ2 − R2 = ±1 and
2 = ρeiθ +
141
ρe−iθ
.
ρ2 − R2
If ρ2 − R2 = −1, then we have 1 = iρ sin θ which is impossible. Therefore, we must
have
R2 = ρ2 − 1 and 1 = ρ cos θ.
In this case, C ∗ are circles with centers 1 ± iR. By the transformation (14.116), the
invariant circles of ϕ satisfy the equations
R
R
z− α±i
=
,
β
|β|
where R > 0.
– Case (ii): ϕ has two finite fixed points. By the expressions (14.112) and (14.113),
the explicit form of ϕ (after the cancellation of the common coefficient) is given by
ϕ(z) =
(α − βγ)z − αβ(1 − γ)
(1 − γ)z + (αγ − β)
(14.119)
and ∆ = (α − βγ)(αγ − β) + αβ(1 − γ)2 = γ(α − β)2 6= 0. Now we have
√
∆
(α − βγ)[(1 − γ)z + (αγ − β)] − ∆
= (α − βγ) −
,
(1 − γ)ϕ(z) =
(1 − γ)z + (αγ − β)
ζ
where
ζ=
(1 − γ)z + (αγ − β)
√
.
∆
(14.120)
1
Define ϕ∗ = ∆− 2 [(1 − γ)ϕ + (αγ − β)]. Then we have
∗
ϕ (ζ) = ∆
− 12
h
√ i
1+γ
1
∆
= √ − .
(α − β)(1 + γ) −
ζ
γ
ζ
(14.121)
Similar to Case (i), this change of variables establishes a one-to-one correspondence
between the invariant circles of the linear fractional transformations (14.119) and
(14.121). Instead of the expressions (14.118), we have
1
1+γ
P = √ −
γ
(ρ − R)eiθ
so that
ρ2 − R2 = ±1
Denote χ =
1+γ
√
2 γ.
and
1+γ
1
and Q = √ −
γ
(ρ + R)eiθ
1+γ
ρe−iθ
.
√ = ρeiθ + 2
γ
ρ − R2
Thus we have either
R2 = ρ2 + 1 and χ = iρ sin θ
(14.122)
R2 = ρ2 − 1 and χ = ρ cos θ.
(14.123)
or
Since ρ and θ are real, the expressions involving χ in (14.122) and (14.123) show that
it is either purely real or purely imaginary.
142
Chapter 14. Conformal Mapping
√ |. By the equations (14.122),
∗ Subcase (i): χ2 < 0. Here χ = it, where t = | 21+γ
γ
∗
the invariant circle of ϕ has the form
p
ζ − ± R2 − t2 − 1 + it = R,
where R2 ≥ 1 − χ2 = 1 + t2 . Transforming back to the original system (using
(14.120)), we get
p
(1 − γ)z + (αγ − β)
√
− ± R2 − t2 − 1 + it = R
∆
√
√
p
(β − αγ) + ∆ ± R2 − t2 − 1 + it
R|α − β| |γ|
z−
=
1−γ
|1 − γ|
√ q
p
√ |2 − 1 + i| 1+γ
√
(β − αγ) + ∆ ± R2 − | 21+γ
γ
2 γ|
R|α − β| |γ|
z−
.
=
1−γ
|1 − γ|
∗ Subcase (ii): χ2 = 0. In this subcase, we know that γ = −1. Furthermore, it
can be seen from the expressions involving χ in (14.122) and (14.123) that ρ = 0
and then R = 1. Consequently, we have |ζ| = 1 which gives
z−
|α − β|
α+β
.
=
2
2
√ is real and we get from the expression
∗ Subcase (iii): χ2 > 0. Then χ = 21+γ
γ
(14.123) that
p
ζ − χ ± i R2 + 1 − χ2 = R,
where R2 ≥ χ2 − 1. Hence, after transforming back to the original system, we
assert that
q
√ p
√ ± i R2 + 1 − ( 1+γ
√ )2
(β − αγ) + ∆ 21+γ
γ
2 γ
R|α − β| |γ|
z−
,
=
1−γ
|1 − γ|
where R2 ≥
(1+γ)2
4γ
−1 =
(1−γ)2
4γ .
Now we have completed the analysis of the problem.
Remark 14.5
(a) The expressions (14.107) and (14.111) are called the normal forms of the linear
fractional transformation ϕ.
(b) The number of finite fixed points can be used to classify the linear fractional transformations ϕ. In fact, we rewrite the multiplier (14.109) as ρeiθ . If ρ 6= 1 and θ = 2nπ,
then ϕ is called a hyperbolic transformation. If ρ = 1 and θ 6= 2nπ, then it is
called an elliptic transformation. If ρ > 0 but ρ 6= 1 and θ 6= 2nπ, then it is
called a loxodromic transformation. The case for one finite fixed point is called
a parabolic transformation and it can be thought as corresponding to ρ = 1 and
θ = 2nπ. See [1, §3.5, pp. 84 – 89] and [24, pp. 15 – 23] for further details.
Problem 14.32
Rudin Chapter 14 Exercise 32.
14.4. Constructive Proof of the Riemann Mapping Theorem
143
Proof. We notice from §14.3 that the linear fractional transformation
ω = ϕ(z) =
1+z
1−z
is a conformal one-to-one mapping of U onto the open right half plane
Π = {ω = X + iY ∈ C | X > 0}.
The images of the upper semi-circle and the lower semi-circle under ϕ are the positive Y -axis
and the negative Y -axis respectively. Next, the mapping
ζ = φ(ω) = log ω = log
maps Π conformally onto the horizontal strip
1+z
1−z
S = {ζ = u + iv ∈ C | u ∈ R and − π2 < v < π2 }.
Furthermore, the positive Y -axis is mapped onto the line v =
mapped onto the line v = − π2 . Consequently, the mapping
π
2
and the negative Y -axis is
1+z
ψ(z) = φ ϕ(z) = log
1−z
(14.124)
sends U conformally onto the horizontal strip S, the images of the upper semi-circle and the
lower semi-circle under ψ are the lines v = π2 and v = − π2 respectively. See Figure 14.4 for the
illustration.
Figure 14.4: The conformal mapping ψ(z) = φ ϕ(z)
Finally, it is easy to see that the mapping ζ 7→ iζ carries the horizontal strip S conformally
onto the vertical strip
H = {ω = X + iY ∈ C | − π2 < X <
π
2
and Y ∈ R}
and the lines v = ± π2 onto the lines X = ∓ π2 respectively.
(a) Since ez maps the horizontal strip {z ∈ C | a < Re z < b} conformally onto the annulus
π
π
A(ea , eb ), our conformal mapping f carries U onto the annulus A(e− 2 , e 2 ).p Furthermore, the images of the upper semi-circle and the lower semi-circle under f are the circles
π
π
C 0, e− 2 and C 0, e 2 respectively, see Figure 14.5 for details below.
p
By direct differentiation, we see that
f ′ (z) =
n
2i
1 +zo
6= 0
exp i log
2
1−z
1−z
in U , so f is conformal in U by Theorem 14.2.
144
Chapter 14. Conformal Mapping
π
π
Figure 14.5: The conformal mapping f : U → A(e− 2 , e 2 ).
(b) Refer to part (a).
(c) Refer to part (a).
(d) The circular arc in U from −1 to 1 can be parametrized as
x = cos t
and y = b sin t,
where b ∈ (−1, 0) ∪ (0, 1) and t ∈ (0, π) ∪ (π, 2π). Straightforward computation gives
1 + cos t + ib sin t 1 − cos t + ib sin t
1+z
=
×
1−z
1 − cos t − ib sin t 1 − cos t + ib sin t
sin2 t − b2 sin2 t + 2ib sin t
=
(1 − cos t)2 + b2 sin2 t
b cos 2t
(1 − b2 ) cos2 2t
+i·
.
=
sin2 2t + b2 cos2 2t
sin 2t (sin2 2t + b2 cos2 2t )
which implies
log
1+z
1
= log
1−z
2
1
= log
2
1+z 2
1+z
+ i arg
1 − z 2 2 2 t 1 −2z t
(1 − b ) sin 2 cos 2 + b2 cos2
sin2 2t (sin2 2t + b2 cos2 2t )2
t
2
+ i tan−1
2b
.
(1 − b2 ) sin t
Consequently, we have
h
i
2b
f (z) = exp − tan−1
(1 − b2 ) sin t
ni
(1 − b2 )2 sin2 2t cos2 2t + b2 cos2 2t o
log
× exp
2
sin2 2t (sin2 2t + b2 cos2 2t )2
so that
h
|f (cos t, b sin t)| = exp − tan−1
Furthermore, suppose that
(14.125)
i
2b
.
(1 − b2 ) sin t
(1 − b2 )2 sin2 2t cos2 2t + b2 cos2
F (b, t) =
sin2 2t (sin2 2t + b2 cos2 2t )2
t
2
which is continuous on [(−1, 0) ∪ (0, 1)] × [(0, π) ∪ (π, 2π)]. Now for any fixed b, we have
F (b, t) → ∞ as t → 0. This implies that arg f (z) takes any value in [0, 2π].
14.4. Constructive Proof of the Riemann Mapping Theorem
For example, put b =
1
2
145
into the equation (14.125), we see that
f (z) = exp − tan−1
hi
(9 sin2 2t cos2
4 × exp
log
3 sin t
2
sin2 2t (4 sin2
t
2
t
2
+ 4) cos2 2t i
,
+ cos2 2t )2
where t ∈ (0, π) ∪ (π, 2π). For t ∈ (0, π), the locus of f (z) is pictured in Figure 14.6:
Figure 14.6: The locus of f (z) for t ∈ (0, π).
Similarly, Figure 14.7 shows the locus of f (z) for t ∈ (π, 2π):
Figure 14.7: The locus of f (z) for t ∈ (π, 2π).
(e) On the radius [0, 1), z is real so that log 1+z
1−z is also real. In this case, |f (z)| = 1 for every
z ∈ [0, 1), so f [0, 1) starts at z = 1 and runs through the unit circle T anticlockwise
infinitely many times.
146
Chapter 14. Conformal Mapping
(f) Suppose that z = r + (1 − r)eiθ , where 0 < r < 1 and −π ≤ θ ≤ π. Then it is easy to
check that
cot θ2
1+z
r
ω = ϕ(z) =
=
+i
1−z
1−r
1−r
r
so that ϕ maps the circle C(r; 1 − r) onto the vertical line Re ω = 1−r
. Hence ϕ maps the
r
}. Next, we
disc E = D(r; 1 − r) conformally onto the vertical strip {ω | 0 < Re ω < 1−r
have
r 2 + cot2 2θ
cot 2θ
1
log ϕ(z) = log
+
i
arg
2
(1 − r)2
r
so that
hi
cot 2θ r 2 + cot2 2θ i
f (z) = exp − arg
× exp
log
.
r
2
(1 − r)2
cot
θ
will run through [− π2 , π2 ] so that arg r 2 runs
cot θ
through [− π2 , π2 ]. Since |f (z)| = exp(− arg r 2 ), this means that f C(r; 1 − r) is a curve
π
π
connecting the two circles C(0; e− 2 ) and C(0; e 2 ), and thus f maps E onto the annulus
π
π
A(e− 2 , e 2 ), see Figure 14.8 when r = 0.25.
Fix an r. As θ runs through [−π, π],
θ
2
Figure 14.8: The image of f (E) when r = 0.25.
(g) Suppose that γ : [0, 1] → U is a curve in U such that γ(t) → 1 as t → 1. Generally
speaking, f γ(t) is also a curve in C and we have to study the behaviour of f γ(t) as
t → 1. The fact γ(t) → 1 as t → 1 implies that γ(t) ∈ D(r; 1 − r) for any 0 < r < 1 and
for all t sufficiently close to 1. Furthermore, γ(t) → 1 as t → 1 also implies that θ(t) → 0
so that
cot θ(t)
π
2
→ e− 2
exp − arg
r
14.4. Constructive Proof of the Riemann Mapping Theorem
147
as t → 1. Therefore, it follows from part (f) that
π
lim f γ(t) = e− 2 .
t→1
We have completed the analysis of the problem.
Remark 14.6
In the literature, there are two ways to define conformal maps. The first definition says that
a holomorphic function f to be conformal at all points with f ′ (z) 6= 0. This is the only
Rudin uses in his book, see also [2, p. 73]. The second way to define a conformal map f is
that it is one-to-one and holomorphic on an open set in the plane, see for example [65, p.
206].
Problem 14.33
Rudin Chapter 14 Exercise 33.
Proof.
(a) We may write ϕα = u + iv = (u, v), so we may think of ϕα : U → U is given by
ϕα (x, y) = u(x, y), v(x, y)
which is an invertible C 2 mapping. Now the Jacobian of ϕα is
 ∂u ∂u 
∂y 

.
∂v 
∂y
 ∂x

Jϕα (x, y) = 
 ∂v
∂x
Therefore, it follows from this and the Cauchy-Riemann equations that
∂u ∂v
∂v ∂u ∂u 2 ∂v 2
det Jϕα (x, y) =
+
= ϕ′α (x, y)
·
−
·
=
∂x ∂y ∂x ∂y
∂x
∂x
2
6= 0
for all (x, y) ∈ U . Finally, we follow from the Change of Variables Theorem [61, Theorem
10.9, pp. 252, 253] that
Z
Z
Z
Z
2
2
′
ϕ′α dm
ϕα (x, y) dx dy =
det Jϕα (x, y) dx dy =
1 dx dy =
π=
U
U
U
U
which is exactly our required result.
(b) An easy computation gives
1 − |α|2
|1 − αz|2
1 − |α|2
=
(1 − αz)(1 − αz)
∞
∞
o nX
o
nX
(αz)k
= (1 − |α|2 ) ·
(αz)n ·
ϕ′α (z) =
n=0
k=0
(14.126)
148
Chapter 14. Conformal Mapping
holds for all z ∈ U . Since |α| < 1, the radii of convergence of the power series in the
1
> 1. Consequently, they converge absolutely and uniformly in
expression (14.126) are |α|
U so that an interchange of integration and summation is legitimate. In terms of polar
coordinates, this means that
1
π
Z
U
ϕ′α
1 − |α|2
dm =
π
=
Z
∞
X
(αz)n (αz)k dm
U n,k=0
∞ Z
|α|2 X
1−
π
(αz)n (αz)k dm
n,k=0 U
∞ Z 2π
|α|2 X
1−
=
π
=
n,k=0 0
∞
|α|2 X
1−
π
If n 6= k, then
·
Z
Thus the expression becomes
1
π
Z
U
ϕ′α
2π
0
Z
1
(αn αk )
n,k=0
Z
1
(αreiθ )n (αre−iθ )k r dr dθ
0
Z
2π
0
Z
1
r n+k+1 ei(n−k)θ dr dθ.
(14.127)
0
r n+k+1 ei(n−k)θ dr dθ = 0.
0
Z
Z
∞
1 − |α|2 X 2n 2π 1 2n+1
dm =
·
|α|
r
dr dθ
π
0
0
n=0
∞
1 − |α|2 X 2n π
|α|
=
·
π
n+1
n=0
∞
X
|α|2n
2
.
= 1 − |α|
n+1
(14.128)
n=0
Consider the power series expansion of log(1 + z) about z = 0, we know that
log(1 + z) =
∞
X
(−1)n
n=0
z n+1
n+1
which has radius of convergence 1. Substituting this with z = −|α|2 into the right-hand
side of the formula (14.128), we obtain finally that
Z
1 − |α|2
1
1
ϕ′α dm =
log
.
2
π U
|α|
1 − |α|2
We have ended the proof of the problem.
Remark 14.7
Problem 14.33(a) is a special case of the classical result Lusin Area Integral, see [34, p.
150].
CHAPTER
15
Zeros of Holomorphic Functions
15.1
Infinite Products and the Order of Growth of an Entire Function
Problem 15.1
Rudin Chapter 15 Exercise 1.
Proof. Let S be the set in which the infinite product converges uniformly. Define un (z) =
Note that
bn − a n
z − an
=1+
= 1 + un (z).
z − bn
z − bn
Suppose that
bn −an
z−bn .
δ = d S, {bn } = inf{|z − ω| | z ∈ S and ω ∈ {bn }} > 0.
Then it is easy to see that
1
|un (z)| ≤ |bn − an | < ∞
δ
for every n ∈ N and z ∈ S. Furthermore, we know that
∞
X
n=1
|un (z)| =
∞
X
|bn − an |
n=1
|z − bn |
≤
∞
1X
|bn − an | < ∞
δ
n=1
for every z ∈ S. By Theorem 15.4, the infinite product
f (z) =
∞
Y
z − an
z − bn
n=1
(15.1)
converges uniformly on S.
Clearly, S ◦ is an open set in C, fn (z) =
component of S ◦ . By the above paragraph,
∞
X
n=1
z−an
z−bn
∈ H(S ◦ ) for n = 1, 2, . . . and fn 6≡ 0 in any
|1 − fn (z))| =
∞
X
n=1
|un (z)|
converges uniformly on every compact subset of S ◦ . By Theorem 15.6, the infinite product
(15.1) is holomorphic in S ◦ . This ends the proof of the problem.
149
150
Chapter 15. Zeros of Holomorphic Functions
Problem 15.2
Rudin Chapter 15 Exercise 2.
Proof. Denote λ to be the order of the entire function f . By the definition, we have
λ = inf{ρ | |f (z)| < exp(|z|ρ ) holds for all large enough |z|}.
Using the fact an =
f (n) (0)
n!
and Theorem 10.26 (Cauchy’s Estimates), we have
λ
er
|an | ≤ n
r
(15.2)
for large enough r. Let g(r) = r −n exp(r λ ), where r > 0. Applying elementary differentiation,
we can show that g attains its minimum
λn
λ
n
1
1
exp
n
λ
at r = n λ λ− λ . Note that r is large if and only if n is large. Thus it follows from the inequality
(15.2) that
n eλ n
λn
λ
λ
=
exp
|an | ≤
n
λ
n
holds for all large enough n.
Consider the entire functions f (z) = exp(z k ), where k = 1, 2, . . .. It is clear that λ = k. By
k
1
the power series expansion of ez , we have ank = n!
. By induction, we obtain
|ank | =
e n
1
<
n!
n
for every large enough n. Consequently, the above bound is not close to best possible. This
completes the analysis of the proof.
Problem 15.3
Rudin Chapter 15 Exercise 3.
z
Proof. The part of finding solutions of ee = 1 has been solved in [76, Problem 3.19, pp. 44 –
45]. In fact, they are given by

π

,
if k > 0 and n ∈ Z;
ln(2kπ)
+
i
2nπ
+


2
(15.3)
z=

3π

 ln(−2kπ) + i 2nπ +
, if k < 0 and n ∈ Z.
2
Denote the zeros (15.3) by zk,n , where k, n ∈ Z and k 6= 0, see Figure 15.1.
Assume that f was an entire function of finite order having a zero at each (15.3). Suppose
further that f 6≡ 0. Consider the disc D(0, RN ), where RN = N π and N is a sufficiently large
positive integer. Then we have zk,n ∈ D(0, RN ), where
exp(N π)
exp(N π)
−1≤k ≤
.
2π
2π
15.1. Infinite Products and the Order of Growth of an Entire Function
151
Figure 15.1: The distribution of the zeros zk,n of exp(exp(z)).
Therefore, we gain
so that
exp(N π)
n RN ≥
4π
log n RN
N π − log 4π
→∞
≥
log RN
log N π
as N → ∞, but it means that f is of infinite order, a contradiction. Hence no such entire
function exists and we finish the proof of the problem.
Problem 15.4
Rudin Chapter 15 Exercise 4.
Proof. We prove the assertions one by one.
152
Chapter 15. Zeros of Holomorphic Functions
• Both functions have a simple pole with residue 1 at each integer. Since sin πz
has a simple zero at every integer N , we have
Res (π cot πz; N ) = Res
π cos πz
π cos πN
;N =
=1
sin πz
π cos πN
which shows that π cot πz has a simple pole with residue 1 at each integer.
We claim that
X
n∈Z
n6=0
z
1 X
1
=
+
z−n n
n(z − n)
(15.4)
n∈Z
n6=0
converges absolutely and uniformly on compact subsets of C \ Z. To see this, let |z| ≤ R
with R > 0. Then we have
X
|n|≥2R
X
X
X 1
|z|
R
R
≤
≤
<∞
n ≤ 2R
|n| · |n − z|
|n| · (|n| − R)
|n| · | 2 |
n2
|n|≥2R
n∈Z
n6=0
|n|≥2R
which gives the desired claim. Since
z
z
2z
+
= 2
,
n(z − n) (−n)(z + n)
z − n2
the function f can be expressed as
f (z) =
1 X
1 X 1
1
1 X 2z
z
.
+
=
+
=
+
+
z
z 2 − n2
z
n(z − n)
z
z−n n
∞
n=1
n∈Z
n6=0
n∈Z
n6=0
By Theorem 10.28, we see that f ∈ H(C \ Z). Furthermore, for each N ∈ N, we see that
lim (z − N )f (z) = lim
z→N
z→N
hz − N
z
X 1
z−N
1 i
=1
+1+
+ (z − N )
+
N
z−n n
n∈Z
n6=0,N
and
lim (z−N )2 f (z) = lim
z→N
z→N
h (z − N )2
z
+(z−N )+
X 1
(z − N )2
1 i
= 0,
+(z−N )2
+
N
z−n n
n∈Z
n6=0,N
so it follows from [9, Theorem 9.5, p. 118] that f also has a simple pole with residue 1 at
each integer N . Simple computation gives
N X
n=−N
n6=0
N
X
1
1
1
1
=− +
+
,
z−n n
z
z−n
n=−N
so the function f also has the form
f (z) =
N
N X
X
1
1
1
1
= lim
+ lim
+
.
N →∞
z N →∞
z−n n
z−n
n=−N
n6=0
n=−N
15.1. Infinite Products and the Order of Growth of an Entire Function
153
• Both functions are periodic. If z ∈
/ Z, then it is obvious to check that
π cot π(z + 1) = πi ·
eπi(z+1) + e−πi(z+1)
eπiz + e−πiz
= π cot πz.
=
πi
·
eπiz − e−πiz
eπi(z+1) − e−πi(z+1)
By the representation (15.4), we have
Xh
1
1i
1
f (z + 1) =
+
+
z+1
z − (n − 1) n
n∈Z
n6=0
=
X 1
1
1 +
+
z+1
z−n n+1
n∈Z
n6=−1
=
X 1
1
1
1
1
1 +1+ +
+ − +
z+1
z
z−n n n n+1
n∈Z
n6=−1,0
=
X 1
X 1
1
1
1
1 −
+1+ +
+
−
z+1
z
z−n n
n n+1
n∈Z
n6=−1,0
=
X 1
1
1
1
−1+ +
+
z+1
z
z−n n
n∈Z
n6=−1,0
n∈Z
n6=−1,0
=
1 X 1
1
+
+
z
z−n n
n∈Z
n6=0
= f (z).
Hence both functions are periodic.
• Their difference is a constant. Now the function ∆(z) = π cot πz − f (z) must be
entire and periodic (i.e., ∆(z + 1) = ∆(z)). To show that it is bounded, it is enough to
show that ∆(z) is bounded in the strip |Re z| ≤ 21 . By Theorem 10.24 (The Maximum
Modulus Theorem), ∆ is bounded in the rectangle {z ∈ C | |Re z| ≤ 21 and |Im z| ≤ 1}. Let
z = x + iy, where |x| ≤ 12 and |y| > 1. On the one hand, we have
cot πz = i ·
so that
|e−2πy | + 1
< ∞.
|e−2πy | − 1
| cot πz| ≤
On the other hand, we write
f (z) =
e−2πy + e−2πix
e−2πy − e−2πix
∞
X
2(x + iy)
1
+
.
2
x + iy
x − y 2 − n2 + 2ixy
n=1
√
If y > 1 and |x| ≤ 21 , we have |x + iy| ≤ 2y and
|x2 − y 2 − n2 + 2ixy| =
Thus they imply that
p
[x2 − (y 2 + n2 )]2 + 4x2 y 2 ≥ (y 2 + n2 ) −
|f (z)| ≤ 1 +
∞
X
n=1
2|x + iy|
|x2 − y 2 − n2 + 2ixy|
1
y 2 + n2
.
≥
4
2
154
Chapter 15. Zeros of Holomorphic Functions
∞
√ X
≤1+4 2
y2
n=1
∞
√ Z
≤1+4 2
y
+ n2
y dx
.
y 2 + x2
0
(15.5)
By the change of variable x = yt, it is easily checked that the integral in the inequality
(15.5) becomes
Z ∞
Z ∞
dt
y dx
=
2
2
y +x
1 + t2
0
0
which implies that |f (z)| ≤ M for some M > 0 and for all |x| ≤ 12 and y > 1. Similarly,
f (z) is also bounded for all |x| ≤ 21 and y < −1. Consequently, we have shown that ∆(z)
is a bounded entire function and Theorem 10.23 (Liouville’s Theorem) says that it is in
fact a constant.
To find this constant, we note that
lim f (iy) = lim
y→∞
y→∞
h1
iy
= −2i lim
y→∞
= −2i
= −πi
Z
∞
0
+
∞
X
n=1
∞
X
n=1
y2
2iy i
−y 2 − n2
y
+ n2
dt
1 + t2
and
lim π cot iπy = −πi.
y→∞
Thus ∆(z) ≡ 0 and then
∞
1 X 2z
π cot πz = +
.
z
z 2 − n2
(15.6)
n=1
• The product representation of
sin πz
. If g(z) = sin πz, then it is clear that
πz
g′ (z)
= π cot πz.
g(z)
Consequently, we observe from the representation (15.6) that
∞
g′ (z)
1 X 2z
= +
.
g(z)
z
z 2 − n2
(15.7)
n=1
Next, we consider the infinite product
P (z) = πz
∞ Y
z2 1− 2 .
n
n=1
2
(15.8)
Now each Pn (z) = 1 − nz 2 ∈ H(C \ Z) and Pn 6≡ 0 in C \ Z. In addition, if K is a compact
subset of C \ Z, then the Weierstrass M -test [61, Theorem 7.10, p. 148] guarantees that
∞
X
n=1
|1 − Pn (z)| =
∞
X
|z|2
n=1
n2
15.1. Infinite Products and the Order of Growth of an Entire Function
155
converges uniformly on K. By Theorem 15.6, the product (15.8) is holomorphic in C \ Z.
Using [18, Exercise 10, p. 174], we have
∞
∞
∞
P ′ (z)
1 X Pn′ (z)
1 X − n2z2
1 X 2z
= +
= +
+
=
P (z)
z n=1 Pn (z)
z n=1 1 − z 22
z n=1 z 2 − n2
(15.9)
n
for every z ∈ C \ Z. Substituting the result (15.9) into the formula (15.7), we see that
g′ (z)
P ′ (z)
=
g(z)
P (z)
and then it gives
h P (z) i′
g(z)
= 0.
Therefore, we have P (z) = cg(z) for some constant c in C \ Z. Since
we have c = 1 and eventually,
P (z)
z
→ 1 as z → 0,
∞ Y
z2 1− 2
sin πz = πz
n
n=1
in C \ Z. Since sin πz and πz
whole plane.
∞
Y
(1 −
n=1
(15.10)
z2
) agree on Z, the formula (15.10) holds in the
n2
Hence we have completed the analysis of the problem.
Remark 15.1
Another way to prove Problem 15.4 is to consider the square CN with vertices (N + 12 )(±1±i)
for N ∈ N. According to Theorem 10.42 (The Residue Theorem), we have
Z
cot πz
X
cot πz
1
dz =
; zk ,
(15.11)
Res
2πi CN z − ζ
z−ζ
k
πz
where ζ ∈ C and the zk denotes a pole of g(z) = cot
/ Z and ζ to be
z−ζ inside CN . Take ζ ∈
a point inside CN . Then it is clear that the poles of the function g(z) occur at z = ζ and
at z = n ∈ {−N, −N + 1, . . . , 0, 1, . . . , N }, and they are all simple. By the basic method of
evaluating residue [9, p. 129], we know that
Res
cot πζ
;ζ =
= cot πζ
z−ζ
1
cot πz
and Res
cot πz
z−ζ
;n =
1
.
π(n − ζ)
By putting these values into the equation (15.11), we get
1
2i
Z
CN
N
X
cot πz
1
dz = π cot πζ −
.
z−ζ
ζ −n
n=−N
It can be shown that
lim
Z
N →∞ CN
cot πz
dz = 0
z−ζ
which implies the desired result. For details, please refer to [54, Lemma 7.22, pp. 181, 182].
156
Chapter 15. Zeros of Holomorphic Functions
Problem 15.5
Rudin Chapter 15 Exercise 5.
Proof. By Theorem 15.9, our f is an entire function having a zero at each point zn . We claim
that
M (r) < exp(|z|k+1 )
for sufficiently large enough |z|. If |z| < 12 , then
h
zk i
z2
+ ··· +
log |Ek (z)| = Re log(1 − z) + z +
2
k
1
1
= Re −
z k+1 −
z k+2 − · · ·
k+1
k+1
1
|z|
|z|2
≤ |z|k+1
+
+
+ ···
k+1 k+2 k+3
1
1
≤ |z|k+1 1 + + 2 + · · ·
2 2
≤ 2|z|k+1 .
Since |Ek (z)| ≤ 1 + |z| exp |z| +
which gives
|z|2
2
+ ··· +
|z|k k ,
(15.12)
we have
|z|k
|z|2
+ ··· +
log |Ek (z)| ≤ log 1 + |z| + |z| +
2
k
lim
z→∞
log |Ek (z)|
=0
|z|k+1
so that if M1 > 0, then there exists a R > 0 such that
log |Ek (z)| < M1 |z|k+1
(15.13)
for all |z| > R. On the set S = {z ∈ C | 12 ≤ |z| ≤ R}, the function g(z) = |z|−(k+1) log |Ek (z)| is
continuous except at z = 1, where g(z) → −∞ as z → 1. Thus there is a constant M2 > 0 such
that
log |Ek (z)| ≤ M2 |z|k+1
(15.14)
for all z ∈ S.
Let M = max{2, M1 , M2 }. Now we combine the inequalities (15.12), (15.13) and (15.14) to
get
log |Ek (z)| ≤ M |z|k+1
(15.15)
for every z ∈ C. By the hypothesis, one can find an N ∈ N such that
∞
X
n=N +1
1
1
<
k+1
|zn |
2M
and
N
X
n=1
1
≤ N.
|zn |k+1
Using the inequality (15.15), we obtain
∞
X
n=N +1
log Ek
∞
z
X
≤M
zn
n=N +1
z
zn
k+1
<
|z|k+1
.
2
(15.16)
15.1. Infinite Products and the Order of Growth of an Entire Function
157
Since M1 can be chosen arbitrary, we deduce from the inequality (15.13) that there exists a
R1 > 0 such that
|z|k+1
log |Ek (z)| ≤
2N
for all |z| > R1 . Let R2 = max{|z1 |R1 , |z2 |R1 , . . . , |zN |R1 }. Then it is obvious that
N
X
log Ek
n=1
z
|z|k+1
≤
zn
2
(15.17)
for all |z| > R2 . Finally, the inequalities (15.16) and (15.17) give
log |f (z)| =
∞
X
log Ek
n=1
z
≤ |z|k+1
zn
for all |z| > R2 , and it is equivalent to saying that
|f (z)| < exp(|z|k+1 )
for all |z| > R2 . This proves our claim and thus f is of finite order. This completes the analysis
of the problem.
Problem 15.6
Rudin Chapter 15 Exercise 6.
Proof. Given ǫ > 0. Notice that
X
|zn |≥1
|zn |−p−ǫ =
∞ X
k=0
X
2k ≤|z
n
|<2k+1
∞
X
|zn |−p−ǫ ≤
2−k(p+ǫ) n 2k+1 .
(15.18)
k=0
Since |f (z)| < exp(|z|p ), it follows from [62, Eqn. (2) & (3), p. 309] that
n(r) ≤ Cr p
(15.19)
for some constant C > 0 and all sufficiently large enough r. Combining the inequalities (15.18)
and (15.19), we see that
X
|zn |≥1
|zn |−p−ǫ < C
Hence we have
∞
X
k=0
2−k(p+ǫ) · 2(k+1)p = C · 2p
∞
X
n=1
Rudin Chapter 15 Exercise 7.
k=0
2ǫ
< ∞.
|zn |−p−ǫ < ∞
for every ǫ > 0, completing the proof of the problem.
Problem 15.7
∞ X
1 k
158
Chapter 15. Zeros of Holomorphic Functions
Proof. Without loss of generality, we may assume that
f 6≡ 0. By the definition (see Problem
15.2), f is of finite order. In the disc D 0; N + 21 for large enough positive integer N , the
number of zeros of f inside D 0; N + 12 is at least N 2 , i.e., n(N + 21 ) ≥ N 2 . By the hypothesis,
we know that M (r) < exp(|z|α ), so it follows from [62, Eqn. (4), p. 309] that
2 ≤ lim sup
N →∞
log n(N + 21 )
≤ α.
log N
Therefore, if 0 < α < 2, then no entire function can satisfy the hypotheses of the problem. In
other words, f (z) = 0 for all z ∈ C if 0 < α < 2, completing the proof of the problem.
15.2
Some Examples
Problem 15.8
Rudin Chapter 15 Exercise 8.
Proof. Let A = {zn }. We are going to verify the results one by one.
• f is independent of the choice of γ(z). Suppose that γ, η : [0, 1] → C are paths from
0 to z and they pass through none of the points zn . Then Γ = γ − η is a simple closed
path. Let {z1 , z2 , . . . , zN } be the set of points which are surrounded by Γ for some N ∈ N.
Now we follow from Theorem 10.42 (The Residue Theorem) that
1
2πi
which gives
Z
Z
g(ζ) dζ =
Γ(z)
N
X
Res (g; zn ) =
γ(z)
N
X
mn
n=1
n=1
g(ζ) dζ = 2πi
N
X
mn +
n=1
Z
g(ζ) dζ.
η(z)
Since the summation is a positive integer, it is true that
exp
nZ
g(ζ) dζ
γ(z)
o
n
= exp 2πi
N
X
mn +
n=1
Z
g(ζ) dζ
η(z)
o
= exp
nZ
η(z)
g(ζ) dζ
o
and this means that f is independent of the choice of γ(z).
• f ∈ H(C \ A). The definition of f shows that the holomorphicity of f depends on the
holomorphicity of the integral. Let z ∈ C \ A and denote
Z
g(ζ) dζ.
G(z) =
γ(z)
Now there exists a disc D(0; R) containing z. Let h be so small that z + h ∈ D(0; R)
and [z, z + h] ∩ A = ∅. By Definition 10.41, D(0; R) contains only finitely many points
of A. Without loss of generality, we may assume that {z1 , z2 , . . . , zN } ⊆ D(0; R) for some
positive integer N . Then both γ(z) and γ(z + h) must lie in Ω = D(0; R) \ {z1 , z2 , . . . , zN }.
Suppose that γ(z) consists of only horizontal or vertical line segments in Ω and γ(z + h)
shares the same path with γ(z) until the point z. Since [z, z +h]∩A = ∅, we can connect z
and z + h by another set of horizontal or vertical line segments in a way that only triangles
are produced. Figure 15.2 illustrates this setting.
15.2. Some Examples
159
Figure 15.2: The paths γ(z + h) and −γ(z).
Now we consider the difference
G(z + h) − G(z) =
=
Z
γ(z+h)
m
XZ
g(ζ) dζ −
Z
g(ζ) dζ +
g(ζ) dζ
γ(z)
Z
g(ζ) dζ
(15.20)
[z,z+h]
k=1 ∂∆k
where ∆1 , ∆2 , . . . , ∆m are the triangles produced. Obviously, the compactness of the set
∆1 ∪ ∆2 ∪ · · · ∪ ∆m ensures that there corresponds an open set Ω′ in Ω such that ∆k ⊆ Ω′
for every k = 1, 2, . . . , m. Since g ∈ H(Ω), Theorem 1013 (The Cauchy’s Theorem for a
Triangle) implies that each integral in the summation (15.20) is 0. As g is continuous at
z, we can write g(ζ) = g(z) + ǫ(ζ), where ǫ(ζ) → 0 as ζ → z. Therefore, the expression
(15.20) becomes
Z
Z
Z
ǫ(ζ) dζ. (15.21)
ǫ(ζ) dζ = hg(z) +
g(z) dζ +
G(z + h) − G(z) =
Since
[z,z+h]
[z,z+h]
[z,z+h]
Z
[z,z+h]
ǫ(ζ) dζ ≤
sup
ζ∈[z,z+h]
|ǫ(ζ)| · |h|,
the last integral actually tends to 0 as h → 0. Consequently, we have proved that
lim
h→∞
G(z + h) − G(z)
= g(z),
h
i.e., G is holomorphic at z. Since z is arbitrary, G ∈ H(C \ A) which implies the desired
result that f ∈ H(C \ A).
• f has a removable singularity at each zn . It suffices to verify that
lim (z − zn )f (z) = 0.
z→zn
Since zn is a simple pole of g with residue mn , there exists a δn > 0 such that
mn
+ h(z)
g(z) =
z − zn
(15.22)
(15.23)
160
Chapter 15. Zeros of Holomorphic Functions
for all |z − zn | < 2δn and h ∈ H D(zn ; 2δn ) .a In fact, we can choose δn small enough
such that D(zn ; 2δn ) contains only the pole zn . Suppose that z lies on the line segment
joining 0 and zn , z ∈ D(zn ; δn ) and θn = arg zn . We split the path γ(z) into two paths
γ1 (z) and γ2 (z) as follows: The path γ1 (z) is the line segment from 0 to zn − δn eiθn . If it
passes through a pole of g, then we can make a small circular arc around that pole. The
path γ2 (z) is the line segment from zn − δn eiθn to z.
On γ1 (z), since δn is fixed and g is continuous on γ1 (z), there exists a positive constant
M1 such that
Z
g(ζ) dζ ≤ M1 .
(15.24)
γ1 (z)
By the definition, we parameterize γ2 (z) : [0, 1] → C as
γ2 (z; t) = z + (1 − t)(zn − δn eiθn − z)
so that γ2 (z; 0) = zn − δn eiθn and γ2 (z; 1) = z. Substitute γ2 (z; t) into the Laurent series
(15.23) to getb
Z 1n
Z
o Z
z − zn − δn eiθn dt
g(ζ) dζ = mn
h(ζ) dζ
+
t z − zn − δn eiθn − δn eiθn
0
γ2 (z)
γ2 (z)
Z
1
iθn
iθn
h(ζ) dζ
= mn ln{t[z − (zn − δn e )] − δn e } +
0
γ2 (z)
Z
h(ζ) dζ.
(15.25)
= mn ln(z − zn ) − i(θn + π) − ln δn +
γ2 (z)
Since h ∈ H D(zn ; 2δn ) , there exists a positive constant M2 such that
Z
h(ζ) dζ ≤ M2
γ2 (z)
which gives
exp
nZ
g(ζ) dζ
γ2 (z)
o
≤ |z − zn |mn eM2 × exp(−mn ln δn ) .
(15.26)
Combining the estimate (15.24) and the expression (15.26), we establish
o
o
nZ
nZ
|f (z)| = exp
g(ζ) dζ × exp
g(ζ) dζ
γ1 (z)
M1 +M2
≤e
· exp(−mn ln δn ) × |z
γ2 (z)
− zn |mn
which implies the limit (15.22).
• The extension of f has a zero of order mn at zn . Since f has a removable singularity
at each zn , it follows from Definition 10.19 that f can be extended to be holomorphic at
zn . The analysis in the previous part further shows that
nZ
o
nZ
o
f (z) = exp
g(ζ) dζ × exp
g(ζ) dζ
γ1 (z)
γ2 (z)
nZ
o
nZ
o
= exp
g(ζ) dζ × exp
h(ζ) dζ × δn−mn e−imn (θn +π) (z − zn )mn .
γ1 (z)
γ2 (z)
In other words, f has a zero of order mn at zn .
a
b
See, for example, [9, Corollary 9.11, p. 124].
If zn is real, then θn = 0 and z is also real. In this case, the integrated result (15.25) becomes ln |z −zn |−ln δn .
15.2. Some Examples
161
Thus we have completed the analysis of the problem.
Problem 15.9
Rudin Chapter 15 Exercise 9.
Proof. Suppose that z1 , z2 , . . . , zn are the zeros of f , listed according to their multiplicities, such
that z1 , z2 , . . . , zn ∈ D(0; β). Define
n
Y
z − zk
.
g(z) =
1 − zk z
k=1
Then g is clearly holomorphic in a neighbourhood V containing U and |g(z)| = 1 on T by
Theorem 12.4. Now the function
f (z)
h(z) =
g(z)
must be holomorphic in U . Since f (U ) ⊆ U , we deduce from Theorem 10.24 (The Maximum
Modulus Theorem) that
|h(z)| ≤ 1
for all z ∈ U . If z = 0, then we get
α
≤1
|z1 | × |z2 | × · · · × |zn |
which implies 0 < α ≤ β n . Consequently, we obtain
n≤
(a) Put α = β =
1
2
log α
.
log β
(15.27)
into the inequality (15.27), we conclude that n ≤ 1.
(b) Similarly, by the inequality (15.27) again, we know that n ≤ 2.
(c) In this case, we have n = 0.
(d) In this case, we have n ≤ 3.
This completes the analysis of the problem.
Problem 15.10
Rudin Chapter 15 Exercise 10.
Proof. Let I be the ideal generated by the set {gN | N ∈ N}. By Definition 15.14, every element
of I is of the form
fN1 gN1 + fN2 gN2 + · · · + fNk gNk
(15.28)
for some increasing sequence {Nk } of positive integers, where fN1 , fN2 , . . . , fNk are entire. By
the definition of gN , if N < M , then there exists an entire function hM such that gN = hM gM .
Consequently, the element (15.28) can be expressed as
fN1 gN1 + fN2 hN2 gN1 + · · · + fNk hNk gN1 = f gN1 ,
162
Chapter 15. Zeros of Holomorphic Functions
where f is entire. Thus we have
I = {f gN | f is entire and N ≥ 1}.
(15.29)
Assume that I was principal. One can find an entire function g ∈ I such that I = [g]. By the
representation (15.29), we know that g = f gN for some entire f and some N ∈ N. Thus we
may assume that there exists some positive integer N such that I = {f gN | f is entire}. Then
we have
gN +1 = f gN
(15.30)
for some entire f . However, since gN (N ) = 0 but gN +1 (N ) 6= 0, the equation (15.30) is a
contradiction. Hence I is not principal and we have completed the proof of the problem.
Problem 15.11
Rudin Chapter 15 Exercise 11.
z−1
Proof. Recall from [62, Eqn. (6), p. 281] that ϕ−1 (z) = z+1
is a conformal one-to-one mapping
of Π = {z ∈ C | Re z > 0} onto U . Therefore, we can reduce the problem to the existence of a
bounded holomorphic function f in U which is not identically zero and its zeros are precisely at
αn =
iyn
,
2 + iyn
where n = 1, 2, . . .. In this case, §15.22 shows that {αn } satisfies
∞
X
(1 − |αn |) < ∞.
n=1
(a) We have
∞ X
1−
n=1
∞ p
i log n X 4 + (log n)2 − log n
p
=
2 + i log n
4 + (log n)2
n=1
=
∞
X
n=1
For n ≥ 3, we have
p
p
4
+ (log n)2
·
4 + (log n)2 ≤ 2 log n so that
∞ X
1−
n=3
p
4
4 + (log n)2 + log n
.
(15.31)
∞
∞
i log n X
2
2X1
.
≥
≥
2
2 + i log n
3(log
n)
3
n
n=3
n=3
Hence the series (15.31) diverges and thus no such bounded holomorphic function in Π.
(b) In this case, we know from the A.M. ≥ G.M. that
√
√
∞ √
∞ X
i n X 4+n− n
√
√
=
1−
2+i n
4+n
n=1
n=1
=
≤
∞
X
n=1
∞
X
n=1
√
√
4
√
√ 4+n·
4+n+ n
4+n·
2
p
4
(4 + n)n
15.2. Some Examples
163
=2
<2
∞
X
(n + 4) 4 · n 4
n=1
n2
n=1
∞
X
1
5
1
1
3
< ∞,
so the answer is affirmative.
(c) The answer to this part is affirmative because
∞ X
1−
n=1
∞ √
in X 4 + n2 − n
√
=
2 + in
4 + n2
=
n=1
∞
X
4
√
√
2
2+n
4
+
n
·
4
+
n
n=1
∞
X
2
<
2
n
n=1
< ∞.
(d) The answer to this part is affirmative because
∞ X
1−
n=1
∞
∞
∞
in2 X |2 + in2 | − n2 X 2 + n2 − n2 X 2
≤
=
<∞
=
2 + in2
|2 + in2 |
n2
n2
n=1
n=1
We complete the analysis of the problem.
n=1
Problem 15.12
Rudin Chapter 15 Exercise 12.
Proof. Let E = { α1n } and Ω = C \ E. The Blaschke condition implies that αn → eiθ as n → ∞
for some real θ, so eiθ ∈ E. Let K be a compact subset of Ω. Since E is closed and K ∩ E = ∅,
Problem 10.1 shows that δ = d(K, E) > 0. In particular, we have |eiθ − z| ≥ δ for every z ∈ K
or equivalently
|1 − e−iθ z| ≥ δ
(15.32)
for all z ∈ K. It is easy to see from the representation
∞
Y
αn − z |αn |
B(z) =
·
1
− αn z αn
n=1
that B has a pole at each point
1
αn .
Next, the nth term of the series
∞
X
n=1
is given by
1−
αn − z |αn |
·
1 − αn z αn
1 + |z|
αn + |αn |z
.
· 1 − |αn | ≤ 1 − |αn | ·
(1 − αn z)αn
|1 − αn z|
(15.33)
(15.34)
164
Chapter 15. Zeros of Holomorphic Functions
As K is compact, it is bounded by a positive constant M so that |z| ≤ M for all z ∈ K. Now
there exists an N ∈ N such that n ≥ N implies
|αn − e−iθ | ≤
δ
.
2M
(15.35)
Combining the estimates (15.32) and (15.35), if n ≥ N , then we gain
δ ≤ |1 − e−iθ z| ≤ |1 − αn z| + |αn − e−iθ | · |z| ≤ |1 − αn z| +
δ
2
so that
δ
>0
2
for all z ∈ K. Thus the inequality (15.34) reduces to
|1 − αn z| ≥
2(1 + M )
αn + |αn |z
.
· 1 − |αn | ≤ 1 − |αn | ·
(1 − αn z)αn
δ
for all z ∈ K and all n ≥ N . Using the Blaschke condition and the Weierstrass M -test, we
conclude immediately that the series (15.33) converges uniformly on K. Eventually, Theorem
10.28 ensures that B ∈ H(Ω), and we end the analysis of the problem.
15.3
Problems on Blaschke Products
Problem 15.13
Rudin Chapter 15 Exercise 13.
Proof. Since αn are real, we have
B(z) = z k
∞
Y
αn − z
1 − αn z
n=1
for some k ∈ N. Suppose that αN −1 < r < αN . Since 0 < αn < 1 and αn < αn+1 for all positive
integers n, we obtain r k < 1 and
αn − r
αn − r
<1
=
1 − αn r
1 − αn r
for every n = N, N + 1, . . .. Consequently, we have
N
−1
∞
N
−1
∞
Y
Y
Y
Y
αn − r
αn − r
r − αn
αn − r
.
<
×
<
|B(r)| = r
1 − αn r
1 − αn r
1 − αn r
1 − αn r
n=1
n=1
n=1
k
(15.36)
n=N
Simple algebra shows that αN −αn > r−αn > 0 and 1−αn r > 1−αn for each n = 1, 2, . . . , N −1,
the inequality (15.36) reduces to
|B(r)| <
Since
αN −αn
1−αn
=1−
n2
N2
N
−1
Y
n=1
αN − αn
.
1 − αn
and log x ≤ x − 1 for x > 0, the inequality (15.37) becomes
|B(r)| <
N
−1 Y
n=1
1−
n2 N2
(15.37)
15.3. Problems on Blaschke Products
h
165
≤ exp log
= exp
N
−1 Y
n=1
1−
n2 i
N2
n2 i
log 1 − 2
N
n=1
−1
h NX
−1 2 N
X
n
≤ exp −
N2
n=1
h (N − 1)(2N − 1) i
≤ exp −
6N
h (N − 1)2 i
< exp −
3N
2
N
−
< e3 e 3
N
< 2e− 3 .
(15.38)
Hence, by combining the fact r → 1 if and only if N → ∞ and the estimate (15.38), we have
established that B(r) → 0 as r → 1 and r ∈ (0, 1), as required. This completes the proof of the
problem.
Problem 15.14
Rudin Chapter 15 Exercise 14.
Proof. Consider αn = 1 − e−n and xn = 1 − 12 e−n for n = 1, 2, . . .. Now we are going to modify
the sequence {αn } and select a subsequence of {xn } so that a Blaschke product with zeros at
the modified sequence satisfies the requirement.
We note from [8, Eqn. (15), p. 12] that we can write
|B(x2p )| =
p
4p−1
∞
Y
Y αn − x2p
Y
x2p − αn
αn − x2p
×
×
1 − αn x2p
1 − αn x2p
1 − αn x2p
n=1
n=p+1
n=4p
= T1 (p) · T2 (p) · T3 (p).
For n = 1, 2, . . . , p, since
(15.39)
x2p − αn
x2p − αp
≥
≥ 1 − e−p ,
1 − αn x2p
1 − αp x2p
we obtain (1 − e−p )p < T1 (p) < 1 which implies that
lim T1 (p) = 1.
p→∞
(15.40)
Furthermore, for sufficiently large p, the inequalities
∞
X
n
e− 2 < T3 (p) < 1
exp − 8
n=4p
give
lim T3 (p) = 1.
n→∞
However, the function T2 (p) only satisfies
lim T2 (p) > 0.
p→∞
(15.41)
166
Chapter 15. Zeros of Holomorphic Functions
Now here is the trick: For a positive integer p, we first replace αp+1 , αp+2 , . . . , α4p−1 by α4p .
Then we have
1 > S2 (p) =
4p−1
Y
n=p+1
α − x 3p−1 4 3p−1
α4p − x2p
4p
2p
=
≥ 1 − 2p
1 − α4p x2p
1 − α4p x2p
e +2
so that
lim S2 (p) = 1.
(15.42)
p→∞
Next, we select a subsequence {pk } of positive integers such that 4pk − 1 < pk+1 + 1 for each
k = 1, 2, . . .. This makes sure that
{αpk +1 , αpk +2 , . . . , α4pk −1 } ∩ {αpk+1 +1 , αpk+1 +2 , . . . , α4pk+1 −1 } = ∅
for each k = 1, 2, . . .. Then the modified sequence {α′n } will be the one that replaces only the
terms αpk +1 , αpk +2 , . . . , α4pk −1 by α4pk from the original sequence {αn } for k = 1, 2, . . .. Since
∞
X
(1
n=1
− |α′n |)
≤
∞
X
(1 − |αn |) < ∞,
n=1
Thus it follows from Theorem 15.21 that
B(z) =
∞
Y
α′k − z
1 − α′k z
(15.43)
k=1
is an element of
so
H∞
and has no zeros except at α′k . It is no doubt that |B(α′n )| → 0 as n → ∞,
lim inf |B(r)| = 0.
(15.44)
r→1
Besides, we apply the representation (15.39) to our Blaschke product (15.43) to obtain
|B(x2pk )| =
pk
Y
x2pk − α′n
×
1 − α′n x2pk
n=1
4pY
k −1
n=pk +1
= T1 (pk ) · S2 (pk ) · T3 (pk ).
∞
Y
α4pk − x2pk
α′n − x2pk
×
1 − α4pk x2pk
1 − α′n x2pk
n=4pk
Combining the limits (15.40), (15.41) and (15.42), we conclude immediately that
lim |B(x2pk )| = 1
k→∞
which means
lim sup |B(r)| = 1.
(15.45)
r→1
Hence the two results (15.44) and (15.45) imply that the function (15.43) has no radial limit at
z = 1, completing the analysis of the problem.
Problem 15.15
Rudin Chapter 15 Exercise 15.
Proof. As a linear fractional transformation, ϕ is one-to-one and ϕ ∈ H(U ). Since ϕ(U ) = U ,
it follows from Theorem 12.6 that
z−α
(15.46)
ϕ(z) = λ ·
1 − αz
for some α ∈ U and |λ| = 1.
15.3. Problems on Blaschke Products
167
(a) If α = 0, then λ 6= 1 because ϕ is not the identity function. In this case, 0 is the unique
fixed point and ϕ(z) = λz. However, it implies that
∞
X
n=1
1<∞
which is impossible. Thus α 6= 0 and we deduce from the expression (15.46) that ϕ(z) = z
is equivalent to
αz 2 − (1 − λ)z − λα = 0.
(15.47)
It is clear from the equation (15.47) that any fixed point must be of modulus 1 because if
z is a root of the equation (15.47), then the fact λ = λ1 implies that z 6= 0 and 1z is also
one of its roots.
– Case (i): Suppose that ϕ has a unique fixed point on T . Let it be b. Consider
−1
the linear fractional transformation µ−1 (z) = 1+bz
z . Clearly, µ (∞) = b, so the
−1
conjugate ψ = µ ◦ ϕ ◦ µ fixes ∞ and then it can be expressed asc
ψ(z) = z + A
for some nonzero constant A. Now ψn (z) = z + nA for every z ∈ U , so ψn (z) → ∞
as n → ∞. Since
ϕn = µ−1 ◦ ψn ◦ µ,
it establishes that
ϕn (z) =
for every z ∈ U . Put z = 0 to get
ϕn (0) =
so for sufficiently large n, we have
(1 + nAb)z − nAb2
nAz + (1 − nAb)
b
−nAb2
=b−
,
1 − nAb
1 − nAb
|ϕn (0)| ≈ 1 −
1
,
n|A| − 1
1 − |ϕn (0)| ≈
1
.
n|A| − 1
but it implies that
Therefore, such ϕ cannot satisfy the Blaschke condition by [79, Theorem 6.10, p. 77].
– Case (ii): Suppose that ϕ has two distinct fixed points on T . Call them b1 and b2
respectively. Using Problem 14.31(c), we have ψ = ν ◦ ϕ ◦ ν −1 , where
ν(z) =
z − b1
.
z − b2
Since ψ fixes 0 and ∞, it can be expressed as ψ(z) = kz for some nonzero complex
constant k. Since ψn (z) = kn z, we have
ϕn = ν −1 ◦ ψn ◦ ν.
Note that ν −1 (z) =
c
b2 z−b1
z−1 ,
so
z − b1 ϕn (z) = ν −1 kn ·
z − b2
Or we may apply Problem 14.31(c) directly here.
168
Chapter 15. Zeros of Holomorphic Functions
=
=
b2 k n ·
kn ·
z−b1
z−b2 − b1
z−b1
z−b2 − 1
(b2 kn − b1 )z + (1 − kn )b1 b2
.
(kn − 1)z + (b2 − b1 kn )
(15.48)
∗ Subcase (i): |k| < 1. Then for sufficiently large enough n, the expression
(15.48) gives
|ϕn (z)| ≈
(kn − b1 )z + (b1 − kn )b2
≈ 1 − |k|n
−z + b2
so that 1 − |ϕn (z)| ≈ |k|n . In other words, ϕ satisfies the Blaschke condition by
[79, Theorem 6.9, p. 77].
∗ Subcase (ii): |k| > 1. In this case, for sufficiently large enough n, the expression
(15.48) implies that
b1
1
k n )z + ( k n − 1)b1 b2
(1 − k1n )z + ( kbn2 − b1 )
(b2 − k1n )z + ( k1n − b2 )b1
(b2 −
|ϕn (z)| =
≈
≈1−
1
|k|n
z − b1
which means 1 − |ϕn (z)| ≈ |k|1n . Hence the ϕ also satisfies the Blaschke condition
in this case.
∗ Subcase (iii): |k| = 1 and k is an N th root of unity for some N . Put
z = 21 into the expression (15.48), we have
ϕN
so that
∞ h
X
n=1
1 − ϕn
1 i
2
1
≥
=
2
∞ h
X
p=1
1
2
1 − ϕpN
1 i
2
= ∞.
Thus ϕ does not satisfy the Blaschke condition in this case.
∗ Subcase (iv): |k| = 1 and k is not an nth root of unity for all n. We
claim that the set S = {kn | n ≥ 0} is dense on T . To see this, we write k = eiθ ,
where θ is an irrational multiple of 2π.d Therefore, we obtain
S = {einθ | n ∈ N}.
Given ǫ > 0. Let ℓ be an arc on T with angle ǫ. Choose N ∈ N such that 2π
N < ǫ.
By the hypothesis, the (N + 1) points 1, eiθ , e2iθ , . . . , eN iθ are all distinct. As a
result, two of them must have a counterclockwise angle less than 2π
N , i.e.,
e(p−q)iθ <
2π
<ǫ
N
for some p < q. This means that the points en(p−q)iθ with n ≥ 0 are distributed
on T at successive angles less then ǫ. Consequently, we have en(p−q)iθ ∈ ℓ for
some n ≥ 0.
d
Otherwise, we have θ =
2πp
q
for some p ∈ Z and q ∈ N. This implies that kq = 1, a contradiction.
15.3. Problems on Blaschke Products
169
Put z = 0 into the expression (15.48) to get
|ϕn (0)| =
|1 − kn |
.
|b2 − b1 kn |
Now the claim ensures that there exists a sequence {nm } such that |b1 − b2 knm | <
1
m and then
m
|ϕnm (0)| > m · |1 − knm | >
2
for large enough m. This shows that ϕ does not satisfy the Blaschke condition.
(b) We claim that ϕ satisfies
ϕn (z) = z
(15.49)
for some n ∈ N. To this end, if f ◦ ϕ = f for some nonconstant f ∈ H ∞ , then we must
have
f (ϕn (z)) = f (ϕ(ϕn−1 (z))) = f (ϕn−1 (z)) = · · · = f (z)
for every z ∈ U and every n = 1, 2, . . ..
– Case (i): ϕ has a unique fixed point in U . Recall from the equation (15.46) that
ϕ has a unique fixed point in U if and only if α = 0. In this case, we have ϕ(z) = λz
for some λ such that |λ| = 1. Obviously, we know that
ϕn (z) = λn z
for every n ∈ N and z ∈ U . If λ is an N th root of unity, then the expected result
(15.49) holds immediately for N . Otherwise, we fix z = z0 ∈ U \ {0} so that
f (λn z0 ) = f (z0 )
for every n ≥ 0. By Subcase (iv), the set S ′ = {λn | n ≥ 0} is dense on T which
implies that
f (z) = f (z0 )
(15.50)
on C(0; |z0 |). By Theorem 10.18, the result (15.50) shows that f (z) = f (z0 ) for every
z ∈ U and this contradicts the Open Mapping Theorem.
– Case (ii): ϕ has a unique fixed point b on T . In this case, it follows from the
equation (15.48) that ϕn (z) → b as n → ∞ for every z ∈ U . This means that
f (z) = lim f (ϕn (z)) = f (b)
n→∞
for every z ∈ U , but it also contradicts the Open Mapping Theorem.
– Case (iii): ϕ has two distinct fixed points b1 and b2 on T . Then it yields from
the equation (15.48) that for every z ∈ U , we have

 b1 , if |k| < 1;
lim ϕn (z) =
n→∞

b2 , if |k| > 1.
By similar argument to Case (ii), we can show that it is impossible for these two
cases so that |k| = 1.
Suppose that k is an N th root of unity for some N ∈ N. Then we see from the
equation (15.48) again that ϕN (z) = z. Otherwise, Subcase (iv) ensures that the
set S = {kn | n ≥ 0} is dense on T . Using similar argument as Case (i), we conclude
that this is impossible.
170
Chapter 15. Zeros of Holomorphic Functions
Consequently, the linear fractional transformation ϕ must satisfy the claim (15.49).
Hence we have completed the analysis of the problem.
Problem 15.16
Rudin Chapter 15 Exercise 16.
Proof. Note that
1 − |αj | =
so we have
Z
1
dr,
|αj |
∞ Z
∞
X
X
(1 − |αj |) =
j=1
j=1
1
dr.
(15.51)
|αj |
Define the characteristic function
χj (r) =
Then formula (15.51) becomes

 1, if r ≥ |αj |;

0, otherwise.
∞ Z
∞
X
X
(1 − |αj |) =
j=1
j=1
Observe that
∞
X
1
χj (r) dr.
(15.52)
0
χj (r) = n(r).
(15.53)
j=1
Since each χj : [0, 1] → [0, ∞] is measurable for every j = 1, 2, . . ., we apply Theorem 1.27 to
interchange the summation and the integration in the equation (15.52) and then use the formula
(15.53) to gain
Z 1
Z 1 X
∞
∞
X
n(r) dr.
χj (r) dr =
(1 − |αj |) =
0
j=1
j=1
0
This ends the proof of the problem.
Problem 15.17
Rudin Chapter 15 Exercise 17.
Proof. Assume that B(z) =
∞
X
k=0
ck z k was a Blaschke product with ck ≥ 0 for all k = 0, 1, 2, . . .
and B(α) = 0 for some α ∈ U \ {0}. Combining Theorem 15.24 and the facte that |f (x)| ≤ λ
holds for almost all x if and only if kf kL∞ ≤ λ, we have
kB ∗ kL∞ (T ) ≤ 1 or
e
Refer to [62, p. 66]
B(eiθ ) ∈ L∞ (T ).
15.3. Problems on Blaschke Products
171
Furthermore, we also have An (eiθ ) = e−inθ B(eiθ ) ∈ L2 (T ) for every n = 1, 2, . . .. Clearly, we
have L∞ (T ) ⊆ L2 (T ).
Recall that L2 (T ) is an inner product space. On the one hand, we may apply [62, Eqn. (7)
& (8), p. 89] to get
Z π
1
hB, An i =
B(eiθ ) · An (eiθ ) dθ
2π −π
Z π X
∞
∞
X
1
ikθ
×
ck e
ck ei(n−k)θ dθ
=
2π −π
k=0
=
∞
X
k=0
ck cn+k .
(15.54)
k=0
On the other hand, the fact |B(eiθ )| = 1 a.e. on T gives
h 1 Z π
2
i2 1 Z
inθ
iθ 2
hB, An i =
einθ dθ ,
e |B(e )| dθ =
2π −π
2π T \N
(15.55)
where |B(eiθ )| =
6 1 on N with m(N ) = 0. Recall from Remark 3.10 that L2 (T ) is a space whose
elements are equivalence classes of functions, so we can express (15.55) by
2
1 Z
einθ dθ = 0.
(15.56)
hB, An i =
2π T
Combining the two results (15.54) and (15.56), we conclude that ck = 0 for all k ∈ N, which is
impossible. Hence no such Blaschke product exists and we complete the proof of the problem. Problem 15.18
Rudin Chapter 15 Exercise 18.
Proof. Obviously, we have
f ′ (z) = 2(z − 1)B(z) + (z − 1)2 B ′ (z).
Thus it suffices to show that (z − 1)B ′ (z) is bounded in U . Suppose that {αn } is the sequence
of zeros of B. Then we know that {αn } ⊆ (0, 1) and it satisfies the Blaschke condition
∞
X
(1 − α2n ) < ∞.
(15.57)
n=1
Without loss of generality, we may assume that
0 < α1 ≤ α2 ≤ · · · < 1.
(15.58)
Furthermore, suppose that
∞
Y
αn − z
B(z) =
1 − αn z
n=1
and
∞
Y
αk − z
Bn (z) =
.
1 − αk z
k=1
k6=n
We yield from [18, Exercise 10, p. 174] that
B ′ (z) X αn − z ′ αn − z −1
=
B(z)
1 − αn z
1 − αn z
∞
n=1
172
Chapter 15. Zeros of Holomorphic Functions
B ′ (z) = −
∞
X
n=1
1 − α2n
Bn (z).
(1 − αn z)2
It is well-known that |Bn (z)| < 1 for every z ∈ U . Now we observe from the assumption (15.58)
that
δ = min(α1 , α2 , . . .) > 0.
Geometrically, if z ∈ U and p > 1, then
|p − z| > |1 − z|.
(15.59)
Put p = α1n into the estimate (15.59) to get |1 − αn z|2 ≥ δ2 |1 − z|2 for every z ∈ U . Hence, the
condition (15.57) implies
|B ′ (z)| ≤
∞
X
n=1
∞
X
1 − α2n
1
(1 − α2n ) < ∞
·
|B
(z)|
≤
n
(1 − αn z)2
δ2 |1 − z|2 n=1
holds for all z ∈ U . Consequently, the modulus |(z − 1)2 B ′ (z)| is bounded in U which shows the
boundedness of f ′ in U , completing the proof of the problem.
Remark 15.2
Blaschke products serve as an important subclass of H(U ). If you are interested in the
literature of Blaschke products, you are suggested to read the book by Colwell [17].
15.4
Miscellaneous Problems and the Müntz-Szasz Theorem
Problem 15.19
Rudin Chapter 15 Exercise 19.
Proof. Let 0 < r < 1. On the one hand, we notice that
reiθ + 1 r2 − 1
= 2
.
log |f (reiθ )| = log exp Re iθ
re − 1
r − 2r cos θ + 1
Thus [62, Eqn, (3), §11.5, p. 233] asserts that
Z π
Z π
1
r2 − 1
1
µr (f ) =
dθ = −
Pr (θ) dθ = −1.
2π −π r 2 − 2r cos θ + 1
2π −π
(15.60)
On the other hand, we have
f ∗ (eiθ ) = lim f (reiθ ) = lim exp
r→1
r→1
so that
log |f ∗ (eiθ )| = Re
Therefore, we obtain
reiθ + 1 reiθ − 1
eiθ + 1 eiθ − 1
µ∗ (f ) = 0.
= exp
eiθ + 1 eiθ − 1
= 0.
(15.61)
15.4. Miscellaneous Problems and the Müntz-Szasz Theorem
173
Hence it follows from the results (15.60) and (15.61) that
lim µr (f ) < µ∗ (f ),
r→1
completing the proof of the problem.
Problem 15.20
Rudin Chapter 15 Exercise 20.
Proof. If λN < 0 for some N ∈ N, then there exists a δ > 0 such that λN < δ < 0. By the
hypothesis, we actually have λn < δ < 0 for all n ≥ N , but this contradicts another hypothesis
that λn → 0 as n → ∞. Therefore, we have λ1 > λ2 > · · · > 0 which implies that
0<
1
1
<
< ··· .
λ1
λ2
Suppose that X is the closure of the in C(I) of the set of all finite linear combinations of the
functions
1
1
1
1, t λ1 , t λ2 , t λ3 , . . . .
By Theorem 15.26 (The Müntz-Szasz Theorem), we have the following results:
• If
• If
∞
X
λn = ∞, then X = C(I).
∞
X
λn < ∞, λ ∈
/ {λn } and λ 6= ∞, then X does not contain the function t λ .
n=1
n=1
1
This ends the analysis of the problem.
Problem 15.21
Rudin Chapter 15 Exercise 21.
Proof. Let {λn } be a sequence of distinct real numbers and λn > − 12 . Then the set of all finite
linear combinations of the functions
tλ0 , tλ1 , tλ2 , . . .
if dense in L2 (I) if and only if
∞
X
2λn + 1
= ∞.
2+1
(2λ
+
1)
n
n=0
To this end, let m ∈ N and m ∈
/ {λn }. By [13, p. 173], we know that
m
min t −
ak ∈C
n−1
X
k=0
ak t
λk
2
=√
n−1
Y
1
m − λk
.
m
+ λk + 1
1 + 2m k=0
Thus we see that
tm ∈ span {tλ0 , tλ1 , . . .}
(15.62)
174
Chapter 15. Zeros of Holomorphic Functions
if and only if
lim sup
n→∞
if and only if
lim sup
n→∞
n−1
Y
k=0
λk >m
n−1
Y
m − λk
=0
m + λk + 1
k=0
2m + 1
×
1−
m + λk + 1
n−1
Y
k=0
− 12 <λk ≤m
1−
2λk + 1
= 0.
m + λk + 1
Hence the set relation (15.62) is true if and only if
n−1
Y
lim sup
n→∞
k=0
λk >m
1−
2m + 1
= 0 or
m + λk + 1
lim sup
n→∞
Since m + λk + 1 > 0 and λk > − 12 , we have
results (15.63) hold if and only if
∞
X
k=0
λk >m
Since λk +
1
2
2m + 1
=∞
m + λk + 1
or
n−1
Y
k=0
− 21 <λk ≤m
1−
2λk +1
2m+1
m+λk +1 , m+λk +1
∞
X
k=0
− 12 <λk ≤m
2λk + 1
= 0.
m + λk + 1
(15.63)
∈ (0, 1). By Theorem 15.5, the
2λk + 1
= ∞.
m + λk + 1
(15.64)
< m + λk + 1 < 2λk + 1, we get
1
1
2
<
<
2λk + 1
m + λk + 1
2λk + 1
which means that the first summation (15.64) is equivalent to
∞
X
k=0
λk >m
1
= ∞.
2λk + 1
(15.65)
Since 1 < m + λk + 1 ≤ 2m + 1, the second summation (15.64) is equivalent to
∞
X
(2λk + 1) = ∞.
(15.66)
k=0
− 21 <λk ≤m
If λk > m, then it is clear that
2m + 1
2m + 1
2λk + 1
<
<
< 2λk + 1.
2
2λk + 1
(2λk + 1) + 1
(2λk + 1)2 + 1
Similarly, if − 12 < λk ≤ m, then it is easy to see that
2λk + 1
2m + 1
2m + 1
2λk + 1
≤
≤
<
.
2
2
2
(2m + 1) + 1
(2λk + 1) + 1
(2λk + 1) + 1
2λk + 1
Therefore, we establish that one of the summations (15.65) and (15.66) holds if and only if
∞
X
k=0
2λk + 1
=∞
(2λk + 1)2 + 1
holds. Finally, our desired result is proven if we combine Theorem 3.14 and the Weierstrass
Approximation Theorem [61, Theorem 7.26, p. 159]. This completes the proof of the problem.
15.4. Miscellaneous Problems and the Müntz-Szasz Theorem
175
Remark 15.3
The proof of Problem 15.21 follows basically the proof of Theorem 2.2 in [14]. For the version
of a complex sequence {λn }, please refer to [50, §12, pp. 32 – 36].
Problem 15.22
Rudin Chapter 15 Exercise 22.
Proof. Let M be the set of all finite linear combinations of the functions fn . It is clear that M
is a subspace of L2 (0, ∞). Let g ∈ L2 (0, ∞) be orthogonal to each fn , i.e.,
Z ∞
fn (t)g(t) dt = 0
hfn , gi =
0
for each n = 1, 2, . . .. Define
F (z) =
Z
∞
e−tz g(t) dt
0
on Π = {z ∈ C | Re z > 0}. For every fixed z ∈ Π, since
Z ∞
Z ∞
1
< ∞,
e−2tRe z dt = √
|e−tz |2 dt =
2 Re z
0
0
we have e−tz ∈ L2 (0, ∞). Using Theorem 3.8, we see that
|F (z)| = ke−tz gk1 ≤ ke−tz k2 · kgk2 < ∞.
In other words, F is well-defined in Π.
Next, we have to compute F ′ (z) in Π. To this end, we consider
Z ∞ −tω
e
− e−tz
F (ω) − F (z)
=
· g(t) dt.
ω−z
ω−z
0
(15.67)
Given ǫ > 0. Let t > 0 and
ω = z − ζ,
(15.68)
where |ζ| < ǫ. Then we have
etζ − 1
e−t(z−ζ) − e−tz
= −e−tz ·
.
z−ζ−z
ζ
Suppose that h(ζ) =
etζ −1
ζ .
The power series expansion of h is given by
h(ζ) =
∞ n
X
t
n=1
n!
ζ n−1 ,
(15.69)
so we have h ∈ H(D(0; ǫ)) by Theorem 10.6. Since h is continuous on D(0; ǫ), we obtain from
Theorem 10.24 (The Maximum Modulus Theorem) and the Extreme Value Theorem that the
maximum of |h(ζ)| occurs on the boundary |ζ| = ǫ. By the expansion (15.69), we see that
max |h(ζ)| ≤
|ζ|=ǫ
∞ n
X
t
n=1
n!
· |ζ|n−1 =
etǫ − 1
.
ǫ
(15.70)
176
Chapter 15. Zeros of Holomorphic Functions
By the Mean Value Theorem, we know that etǫ − 1 = ǫtetξ for some ξ ∈ (0, ǫ), so we induce from
the estimate (15.70) that
etζ − 1
≤ tetξ < tetǫ ,
ζ
where |ζ| < ǫ and t > 0. If z ∈ Π, then Re z > 2ǫ for some ǫ > 0. With this ǫ > 0, we may pick
ω ∈ Π satisfying the condition (15.68). Therefore, we have
e−tω − e−tz
etζ − 1
= e−tz ·
< te−tRe z · etǫ = te−t(Re z−ǫ) ≤ te−tǫ
ω−z
ζ
for t > 0. Since te−tǫ , g ∈ L2 (0, ∞), Theorem 3.8 implies that te−tǫ g ∈ L1 (0, ∞). Consequently,
Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) can be applied to show the limit
and the integral can be interchanged in the following deduction:
F (ω) − F (z)
ω→z
ω−z
Z ∞ −ωt
e
− e−tz
· g(t) dt
= lim
ω→z 0
ω−z
Z ∞
e−ωt − e−tz
lim
=
· g(t) dt
ω→z
ω−z
Z0 ∞
−tetz g(t) dt.
=
F ′ (z) = lim
0
Since z is arbitrary, we have proven that F ∈ H(Π).
In fact, this process can be repeated to get
Z ∞
(n)
n
F (z) = (−1)
tn e−tz g(t) dt,
(15.71)
0
where n ∈ N and z ∈ Π. By the hypothesis and the formula (15.71), we have
Z ∞
(n)
n
F (1) = (−1)
tn e−t g(t) dt = (−1)n hfn , gi = 0
0
so that F ≡ 0. Particularly, if we denote G(t) = χ(0,∞) (t)e−t g(t), then we see that
b
G(y)
=
√
2π
Z
∞
−ity
G(t)e
−∞
√ Z
dt = 2π
∞
e−t g(t)e−ity dt =
√
2πF (1 + iy) = 0,
0
where y ∈ R. As G ∈ L1 (0, ∞), we observe from Theorem 9.12 (The Uniqueness Theorem) that
G(t) = 0 a.e. on R which implies that
g(t) = 0
(15.72)
a.e. on (0, ∞). Recall that L2 (0, ∞) is Hilbert and M is a subspace of L2 (0, ∞). If we have
M 6= L2 (0, ∞), according to the Corollary of Theorem 4.11, there corresponds a g ∈ L2 (0, ∞)
such that g 6= 0 and hfn , gi = 0 for every n = 1, 2, . . .. However, this definitely contradicts the
above conclusion (15.72). Hence M = L2 (0, ∞), as desired. This completes the analysis of the
problem.
Problem 15.23
Rudin Chapter 15 Exercise 23.
15.4. Miscellaneous Problems and the Müntz-Szasz Theorem
177
Proof. Since f (0) = 0, λn 6= 0 for all n = 1, 2, . . . , N . Suppose that
B(z) =
Since
N
X
N
Y
λn − z |λn |
·
.
1 − λn z λn
n=1
(1 − |λn |) < ∞, Theorem 15.21 implies that B ∈ H ∞ and B has no zeros except at
n=1
the points λn . Consider the function g =
continuous on U and
B
1−f .
Then we have g ∈ H(U ). By Theorem 12.4, g is
λn − eiθ
=1
1 − λn eiθ
for every n = 1, 2, . . . , N . Therefore, we see that
|g(eiθ )| =
1
1
|B(eiθ )|
≤
≤ .
|1 − f (eiθ )|
|f (eiθ )| − 1
2
Consequently, it yields from Theorem 10.24 (The Maximum Modulus Theorem) that
|λ1 λ2 · · · λN | = |g(0)| <
We end the proof of the problem.
1
.
2
178
Chapter 15. Zeros of Holomorphic Functions
CHAPTER
16
Analytic Continuation
16.1
Singular Points and Continuation along Curves
Problem 16.1
Rudin Chapter 16 Exercise 1.
Proof. By Theorem 16.2, f has a singularity at some point eiθ . If we consider the power series
for f about the point 21 , then the representation
f (z) =
∞
X
bk
k=0
∞
1 k
1 k X f (k) ( 21 ) z−
=
z−
2
k!
2
(16.1)
k=0
holds in D( 21 ; 21 ). Thus the radius of convergence of the power series (16.1) must be 12 . Otherwise,
the power series would define a holomorphic extension of f beyond eiθ , a contradiction.
Assume that f was regular at z = 1. Then the series
∞
X
f (k) ( 21 ) 1 k
x−
k!
2
k=0
converges at some x > 1. For every k ≥ 1, we derive from the representation f (z) =
that
∞
1 X
n(n − 1) · · · (n − k + 1)an
=
.
f (k)
2
2n−k
P
an z n
n=k
Now we deduce from the binomial theorem that
∞
X
f (k) ( 1 ) 2
k=0
k!
x−
∞
∞
1 k X h X n(n − 1) · · · (n − k + 1)
an i 1 k
=
× n−k × x −
2
k!
2
2
k=0
=
n=k
∞
∞ X
X
k=0 n=k
Ckn ·
an 1 k
.
·
x
−
2n−k
2
(16.2)
an
Since an ≥ 0, Ckn · 2n−k
·(x− 12 )k ≥ 0. Consequently, the order of the summation in the expression
(16.2) can be switched (see [61, Exercise 3, p. 196]) so that
n
∞
∞
∞
n−k X
f (k) ( 12 ) 1 k X h X n 1
1 k i X
an xn
an
x−
=
Ck
=
−0
· x−
k!
2
2
2
k=0
n=0
n=0
k=0
179
180
Chapter 16. Analytic Continuation
P
which means that the radius of convergence of f (z) =
an z n is greater than 1. This is a
contradiction to our hypothesis and we have completed the analysis of the problem.
Problem 16.2
Rudin Chapter 16 Exercise 2.
Proof. Suppose that (f, D) and (g, D) can be analytically continued along γ to (fn , Dn ) and
(gm , Dm ) respectively. According to Definition 16.9, there exist chains Cf = {D0 , D1 , . . . , Dn }
′ }, where D = D ′ = D. Furthermore, there are numbers
and Cg = {D0′ , D1′ , . . . , Dm
0
0
0 = s0 < s1 < · · · < sn = 1 and
0 = t0 < t1 < · · · < tm = 1
′ ,
such that γ(0) is the center of D0 = D0′ , γ(1) is the center of Dn and Dm
γ [sj , sj+1 ] ⊆ Dj and γ [tk , tk+1 ] ⊆ Dk′
for j = 0, 1, . . . , n − 1 and k = 0, 1, . . . , m − 1. We notice that there are function elements
′
) for j = 0, 1, . . . , n−1 and k = 0, 1, . . . , m−1,
(fj , Dj ) ∼ (fj+1 , Dj+1 ) and (gk , Dk′ ) ∼ (gk+1 , Dk+1
where f0 = f and g0 = g.
Since P f0 (z), g0 (ζ) = 0 for all z, ζ ∈ D0 , we have P f1 (z), g0 (ζ) = 0 for all z ∈ D0 ∩ D1
and ζ ∈ D0 . For each fixed ζ ∈ D0 , P is a polynomial in z so that P ∈ H(D1 ). By the Corollary
to Theorem 10.18, we have
P f1 (z), g0 (ζ) = 0
(16.3)
for all z ∈ D1 . Since ζ is arbitrary, the equation (16.3) is actually true for all ζ ∈ D0 . Repeat
this process, we conclude that P (fn (z), g0 (ζ)) = 0 for all z ∈ Dn and ζ ∈ D0 . Next, we fix a
z ∈ Dn and now P is a polynomial in ζ so that P ∈ H(D1′ ). Similar argument shows that
P fn (z), g1 (ζ) = 0
holds for all ζ ∈ D1′ and hence also for every z ∈ Dn . Repeat the process also implies that the
equation
P fn (z), gm (ζ) = 0
(16.4)
holds in Dn and Dm . Finally, we can establish the required equation of the problem if we replace
fn and gm by f1 and g1 in the equation (16.4).
Obviously, this can be extended to n function elements (f1 , D), (f2 , D), . . . , (fn , D) provided
that P (z1 , z2 , . . . , zn ) is a polynomial in n variables, f1 , f2 , . . . , fn can be analytically continued
along a curve γ to g1 , g2 , . . . , gn and P (f1 , f2 , . . . , fn ) = 0 in D. In fact, our above proof only
uses the holomorphicity of the polynomial P , so similar results can be established if we only
require that P is a function of n variables such that P (. . . , zj , . . .) is holomorphic in each variable
zj for j = 1, 2, . . . , n. This completes the analysis of the problem.
Problem 16.3
Rudin Chapter 16 Exercise 3.
Proof. By Theorem 11.2, the function f = ux − iuy is holomorphic in Ω. Since Ω is simply
connected, Theorem 13.11 ensures that there corresponds an F ∈ H(Ω) such that F ′ = f . If
F = A + iB, then we have
F ′ (z) = Ax + iBx = Ax − iAy = ux − iuy
16.1. Singular Points and Continuation along Curves
181
so that A(x, y) = u(x, y) + C for some constant C. Hence u is the real part of F − C ∈ H(Ω).
Next, we suppose that Ω is a region, but not simply connected. Let f ∈ H(Ω) and f (z) 6= 0
for every z ∈ Ω. Then log |f | is harmonic in Ω by Problem 11.5. If it has harmonic conjugate,
then there corresponds an F ∈ H(Ω) such that
|eF (z) | = eRe F (z) = elog |f (z)| = |f (z)|
which means that
|f e−F | = 1
in Ω. According to [9, Proposition 3.7, p. 39], f (z)e−F (z) = eiθ for some constant θ in Ω. Now
we can write
f (z) = eg(z)
for some g ∈ H(Ω). By Theorem 13.11, Ω is simply connected, a contradiction. Hence this shows
that the statement of the problem fails in every region that is not simply connected, completing
the proof of the problem.
Problem 16.4
Rudin Chapter 16 Exercise 4.
Proof. Denote I = [0, 1]. Let α : I → C \ {0} be an arbitrary path from 1 to f (0). By Definition
10.8, without loss of generality, we may assume further that α′ is continuous on I. Define
Z 1 ′
α (t) dt
.
(16.5)
g(0) =
α(t)
0
We note that
g(0) =
Z
1
0
d ln α(t) = log f (0)
and it means that f (0) = eg(0) . Let ζ ∈ X \ {0} and βζ : I → X be the line segment joining 0
and ζ. Next, we define γζ : I → C \ {0} by

if t ∈ [0, 21 ];
 α(2t),
γζ (t) =
(16.6)

f βζ (2t − 1) , otherwise.
Finally, we define
g(ζ) =
Z
0
1
γζ′ (t) dt
γζ (t)
.
Clearly, we know that
Z 1
d log γζ (t) = log γζ (1) − log γζ (0) = log f βζ (1) − log α(0) = log f (ζ)
g(ζ) =
(16.7)
0
so that f (ζ) = eg(ζ) . See Figure 16.1 for the paths βζ (I) and γζ .
Now it suffices to prove that the function g : X → C defined by the equations (16.5) and
(16.7) is continuous on X. To this end, let z, ω ∈ X and suppose that βω,z : I → X is the line
segment from ω to z. We also define γω,z : I → C \ {0} by
γω,z (t) = f βω,z (t) .
182
Chapter 16. Analytic Continuation
Figure 16.1: The paths βζ (I) and γζ .
If γ is a path in X from x0 to x1 , and if λ is a path in X from x1 to x2 , then we define the
product γ ∗ λ of γ and λ to be the path η given by the equations

if t ∈ [0, 21 ];
 γ(2t),
η(t) =

λ(2t − 1), otherwise.
Thus both βz and βω ∗ βω,z are paths in X from 0 to z. Since X is simply connected, we follow
from Theorem 16.14 or [42, p. 323] that βz and βω ∗ βω,z are (path) homotopic in
X, i.e.,
βz ≃p βω ∗ βω,z in X. Since f is continuous on X, we must have f (βz ) ≃p f βω ∗ βω,z in f (X).
Recall that α is arbitrary in the definition (16.6), it establishes that
γz ≃p γω ∗ γω,z
in f (X) ∪ α(I), see Figure 16.2 for the paths γz (I), γω (I) and γω,z (I)
Figure 16.2: The paths γz (I), γω (I) and γω,ζ (I).
Since X is compact and f (z) 6= 0 for all z ∈ X, there exists a constant m > 0 such that
κ
< ǫ. By the continuity of f , there
m = min |f (z)|. Given ǫ > 0. Choose κ > 0 such that m
z∈X
corresponds a δ > 0 such that for z, ω ∈ X and |z − ω| < δ, the length of γω,z is less than κ. Let
16.2. Problems on the Modular Group and Removable Sets
ℓ(z, ω) be the length of γω,z . Hence, if |z − ω| < δ, then we obtain
Z 1 ′
Z 1
γz (t) dt
γω (t) dt
|g(z) − g(ω)| =
−
γ
(t)
γω (t)
z
0
0
Z 1 ′
Z 1
′
γω (t) dt
(γω ∗ γω,z ) (t) dt
−
=
(γω ∗ γω,z )(t)
γω (t)
0
Z0
Z
dt
dt
=
−
t
γ ∗γ
γω t
Z
Z ω ω,z Z
dt
dt
dt
+
−
=
t
t
γω,z
γω t
γ
Z ω
dt
=
.
γω,z t
183
(16.8)
Applying the definition of m to the integral (16.8), we get
ℓ(z, ω)
κ
<
< ǫ.
m
m
In other words, g is continuous at ω so that it is actually continuous on X, as required. This
completes the proof of the problem.
|g(z) − g(ω)| ≤
16.2
Problems on the Modular Group and Removable Sets
Problem 16.5
Rudin Chapter 16 Exercise 5.
Proof. Let τ (z) = z + 1, σ(z) = − z1 and
ϕ(z) =
az + b
∈ G.
cz + d
Then we have a, b, c, d ∈ Z and ad − bc = 1.a Furthermore, we notice that τ −1 (z) = z − 1 and
σ −1 (z) = − z1 .
• Case (i): a = 0. We have bc = −1 so that b = −c. Obviously, b = ±1 if and only if
c = ∓1. If b = 1, then c = −1 and we have
1
1
1
=−
= − −d
= σ τ −d (z) ,
ϕ(z) =
−z + d
z−d
τ (z)
i.e., ϕ = σ ◦ τ −d . Similarly, we have ϕ = σ ◦ τ d if b = −1 and c = 1.
az+b
• Case (ii): a = ±1. Since −az−b
−cz−d = cz+d , we may only consider the case that a = 1. Thus
z+b
. We note that
we have d − bc = 1 and ϕ(z) = cz+d
−cz − d −cz − d
−cz − d
−(c − 1)z − (d − b)
σ ϕ(z) =
=
and τ
+1=
z+b
z+b
z+b
z+b
which imply that
−(d − bc)
1
τ c σ ϕ(z) =
=−
= σ τ b (z) .
z+b
z+b
a
Hence we have ϕ = σ −1 ◦ τ −c ◦ σ ◦ τ b .
With the aid of Problem 16.7, we remark that G = SL2 (Z).
184
Chapter 16. Analytic Continuation
• Case (iii): |a| > 1. We may take |a| > |c|. Otherwise, we consider σ ϕ(z) = − cz+d
az+b
az+b
. Now it is easy to see that one can find an N ∈ Z satisfying
instead of ϕ(z) = cz+d
0 ≤ |a − N c| < |c| < |a|. Since
we establish that
az + b
(a − c)z + (b − d)
τ −1 ϕ(z) =
−1=
,
cz + d
cz + d
ϕ1 (z) = σ τ −N ϕ(z) =
−cz − d
a1 z + b1
=
.
(a − N c)z + (b − N d)
c1 z + d1
Simple algebra shows that ϕ1 ∈ G, |a1 | < |a| and 0 ≤ |c1 | < |c|. If |c1 | = 0, then a1 d1 = 1
so that a1 = ±1 which goes back to Case (ii). Otherwise, we can repeat the above process
finitely many times, say m times, to get
ϕm (z) =
am z + bm
∈ G,
cm z + dm
where either am = 0 or am = ±1. Therefore, the ϕm , and hence ϕ, is generated by τ and
σ.
Consequently, this proves the first assertion that τ and σ generate the modular group G.
For the second assertion, suppose that
R1 = {z = x + iy | |x| < 12 , y > 0 and |z| > 1},
R2 = {z = x + iy | − 12 ≤ x ≤ 0 and |z| = 1},
n
o
1
R3 = z = − + iy y > 0 and |z| ≥ 1 .
2
(16.9)
Then we have R = R1 ∪ R2 ∪ R3 , see Figure 16.3.
Figure 16.3: The fundamental domain R of G.
We check Theorem 16.19(a) and (b). Based on Apostol’s description [5, p. 30], two points
ω, ω ′ ∈ Π+ = {z ∈ C | Re z > 0} are said to be equivalent under G if ω ′ = ϕ(ω) for some
ϕ ∈ G. With this terminology, property (a) means that no two distinct points of R are equivalent
under G and property (b) implies that for every ω ∈ Π+ , there exists a z ∈ R such that z is
equivalent to ω.
16.2. Problems on the Modular Group and Removable Sets
185
Lemma 16.1
Suppose that z1 , z2 ∈ R, z1 6= z2 and z2 = ϕ(z1 ) for some ϕ ∈ G. Then we have
Re z1 = ± 21 and z2 = z1 ∓ 1 or
|z1 | = 1 and z2 = − z11 .
Proof of Lemma 16.1. Without loss of generality, we may assume that Im z2 ≥ Im z1
by symmetry. Let ϕ(z) = az+b
cz+d . Combing the assumption and the relation
Im ϕ(z) =
Im z
|cz + d|2
(16.10)
to get the condition
|cz1 + d|2 ≤ 1.
(16.11)
√
Since z1 ∈ R, it is easy to see that Im z1 ≥ 23 . Thus |c| ·
As c ∈ Z, this forces that either c = 0 or |c| = 1.
√
3
2
≤ |c|Im z1 ≤ |cz1 + d| ≤ 1.
• Case (i): c = 0. Then ad = 1 and since a, d ∈ Z, we have a = d = ±1. In this
case, the relation (16.10) shows that Im z2 = Im z1 . Furthermore, ϕ(z) = z ± b
so that Re z2 = Re z1 ± b. Since b is an integer, the definition (16.9) shows b = 1
which implies that Re z1 = ± 21 and hence z2 = z1 ∓ 1.
• Case (ii): |c| = 1. Then the condition (16.11) becomes |z1 ± d|2 ≤ 1 or equivalently,
(Re z1 ± d)2 + (Im z1 )2 ≤ 1.
(16.12)
Further reduction implies that
(Re z1 ± d)2 ≤ 1 − (Im z1 )2 ≤ 1 −
so that
|Re z1 ± d| ≤
3
1
= ,
4
4
1
.
2
(16.13)
Since − 21 ≤ Re z1 ≤ 12 , we have |d| ≤ 1 which means either d = 0 or |d| = 1.
– Subcase (i): |d| = 1. Using the inequality (16.13), we have |Re z1 ± 1| = 21
and then Re z1 = ± 21 . Next, it follows from the inequality (16.12) that
0 ≤ Im z1 ≤√
z1 =
± 12
+
√
i 3
2
3
2 ,
so actually we have Im z1 =
√
3
2 .
Consequently, we have
and then both our results hold in this case.
– Subcase (ii): d = 0. Now the condition (16.11) implies that |z1 | ≤ 1.
Since z1 ∈ R by the definition (16.9), we actually have |z1 | = 1. Simple
calculation gives
1
z2 = ±a − ,
z1
where a ∈ Z. Let z1 = x + iy. Thus z2 = ±a − x + iy. If a 6= 0, then since
z1 , z2 ∈ R and z1 6= z2 , we get z1 = − 21 + iy and z2 = 21 + iy. Otherwise,
a = 0 so that z1 = x + iy and z2 = −x + iy for every 0 < x ≤ 12 . Obviously,
this case satisfies |z1 | = 1 and z2 = − z11 .
This completes the proof of the lemma.
186
Chapter 16. Analytic Continuation
Combining the definition (16.9) of R and Lemma 16.1, we see immediately that no two
distinct points of R are equivalent under G which is property (a). For proving property (b), we
need the following result whose proof can be found in [5, Lemma 1, pp. 31, 32]:
Lemma 16.2
ω′
Given ω1′ , ω2′ ∈ C with ω2′ not real. Let Ω = {mω1′ + nω2′ | m, n ∈ Z}. Then there
1
exist ω1 , ω2 ∈ C such that ω2 = aω2′ + bω1′ and ω1 = cω2′ + dω1′ , where ad − bc = 1,
|ω2 | ≥ |ω1 | and |ω1 ± ω2 | ≥ |ω2 |.
Now we go back to the proof of our problem. If ω1′ = 1 and ω2′ = ω ∈ Π+ , then it is easy to
ω′
see that ω2′ ∈
/ R. By Lemma 16.2, there exist ω1 and ω2 with |ω2 | ≥ |ω1 | and |ω1 ± ω2 | ≥ |ω2 |
1
such that
ω2 = aω + b and ω1 = cω + d.
Let z =
ω2
ω1 .
These relations give
z=
aω + b
= ϕ(ω)
cω + d
(16.14)
with ad − bc = 1, |z| ≥ 1 and |z ± 1| ≥ |z|. The relation (16.14) means that there exists a point
z ∈ R equivalent to ω ∈ Π+ under G which is exactly property (b). Hence we obtain the result
that R is a fundamental domain of G and we end the proof of the problem.
Problem 16.6
Rudin Chapter 16 Exercise 6.
Proof. Since ψ ϕ(z) = z + 1, it follows from Problem 16.5 that G is also generated by ϕ and
ψ. It is easy to see that
ϕ2 (z) = ϕ ϕ(z) = z
and
ψ 2 (z) = −
1
z−1
and ψ 3 (z) = z.
Hence ϕ has period 2 and ψ has period 3. This completes the proof of the problem.
Problem 16.7
Rudin Chapter 16 Exercise 7.
Proof. For each linear fractional transformation ϕ(z) =
Mϕ =
a b
c d
az+b
cz+d ,
we associate the 2 × 2 matrix
.
Here we identify each matrix with its negative because Mϕ and −Mϕ represent the same transformation. If Mϕ and Mψ are the matrices associated with the linear fractional transformations
ϕ and ψ respectively, then it is easy to see that the matrix product Mϕ Mψ is associated with
the function composition ϕ ◦ ψ.
16.2. Problems on the Modular Group and Removable Sets
187
• An algebraic proof of Theorem 16.19(c). Now the group Γ is generated by the
matrices
1 0
1 2
A=
and B =
.
2 1
0 1
If M ∈ Γ, then we have
M = An1 Bm1 An2 Bm2 · · · Anp Bmp ,
(16.15)
where the nk , mk are integers. Direct computation gives
1 0
1 −2
−1
−1
A =
and B =
−2 1
0 1
so that
A
nk
=
1
(−2)nk
0
1
=
1
(−2)nk
(−2)mk
(−2)mk +nk + 1
and
mk
B
=
1 (−2)mk
0
1
=
1
2Nk
2Mk 2Lk + 1
and then
nk
A B
mk
,
where Nk , Mk and Lk are integers. Thus we obtain
1 + 2Nk,j
2Mk,j
Ank Bmk Anj Bmj =
,
2Pk,j
1 + 2Lk,j
where Nk,j , Mk,j , Lk,j and Pk,j are integers. Hence we apply this to the expression (16.15)
to conclude immediately that if
a b
M=
,
c d
then a and d are odd, b and c are even.
• Proof of the first part of Problem 16.5. Note that the transformations z 7→ z + 1
and z 7→ − z1 correspond to the matrices
T=
1 1
0 1
and S =
0 −1
1 0
respectively. We claim that if M ∈ G, then it has the form
M = Tn1 STn2 S · · · Tnp S,
(16.16)
where the nk are integers. To this end, we first notice that S2 = I,b so this explains why
only the S appears in the form (16.16). Next, it suffices to prove those matrices
a b
M=
c d
with c ≥ 0. If c = 0, then ad = 1 or equivalently, a = d = ±1 so that
±1 b
1 ±b
M=
=
= T±b .
0 ±1
0 1
b
Remember that we have identified I = −I.
188
Chapter 16. Analytic Continuation
Next, if c = 1, then ad − b = 1 so that b = ad − 1 and
a ad − 1
1 a
0 −1
1 d
M=
=
= Td STd .
1
d
0 1
1 0
0 1
Assume that the form (16.16) is true for all matrices with lower left-hand element less
than c for some c ≥ 1. Since ad − bc = 1, c and d must be coprime so that d = cq + r for
some q ∈ Z and 0 < r < c. Since
a b
1 −q
a −aq + b
MT−q =
=
,
c d
0 1
c
r
we have
MT
−q
S=
−aq + b −a
r
−c
.
(16.17)
By the hypothesis, the matrix (16.17) has the form (16.16) which implies that M can be
expressed in the form (16.17).
This completes the proof of the problem.
Problem 16.8
Rudin Chapter 16 Exercise 8.
Proof. Since E ⊆ R is compact, we can define
A = max x
x∈E
and B = min x.
x∈E
Notice that Ω = C \ E. Denote R = max(|A|, |B|).
(a) Let x, y ∈ R such that x < y < A. Then x 6= y and
Z Z
1
dt
1 f (x) − f (y) =
dt = (x − y)
−
.
t
−
x
t
−
y
(t
−
x)(t
− y)
E
E
(16.18)
Since x, y ∈ R and E ⊂ R, the integrals in the equation (16.18) are real integrals. Clearly,
1
1
1
1
1
1
t−x ≥ A−x > 0 and t−y ≥ A−y > 0 for all t ∈ E. Let δx = A−x and δy = A−y . Then it
follows from the expression (16.18) that
Z
δx δy dt = (x − y)δx δy m(E) > 0,
f (x) − f (y) ≥ (x − y)
E
i.e., f (x) 6= f (y). Consequently, f is nonconstant.
(b) The answer is negative. Assume that f could be extended to an entire function. Since
1
|t| ≤ R on E, we have | t−z
| → 0 as |z| → ∞ for every t ∈ E. In other words, we see
that f (z) → 0 as |z| → ∞ which means f is bounded in C. By Theorem 10.23 (Liouville’s
Theorem), f is constant which contradicts part (a). Hence we conclude that f cannot be
extended to an entire function.
z
on E. We claim that
(c) Given ǫ > 0 and z ∈ C \ D(0; R + Rǫ ). Define g(z, t) = − z−t
g(z, t) → −1 uniformly on E. In fact, we see that
|g(z, t) + 1| =
|t|
|z − t|
16.2. Problems on the Modular Group and Removable Sets
189
on E. Since |z| > R, we have |z − t| ≥ |z| − |t| ≥ |z| − R > 0 so that
|g(z, t) + 1| =
on E. Since |z| > R +
R
ǫ,
|t|
R
≤
|z − t|
|z| − R
(16.19)
the inequality (16.19) implies that
|g(z, t) + 1| < ǫ
for all t ∈ E. This proves our claim which asserts that
Z
g(z, t) dt + m(E)
|zf (z) + m(E)| =
E
Z
[g(z, t) + 1] dt
=
E
Z
|g(z, t) + 1| dt
≤
E
< ǫm(E)
for every |z| ≥ R +
R
ǫ.
(16.20)
Since ǫ is arbitrary, we conclude from the estimate (16.20) that
lim zf (z) = −m(E).
z→∞
(d) The compactness of E implies that Ω is open in C. We have to show that Ω is connected.
Since E ⊂ R, we have C \ R ⊆ Ω. Thus the upper half plane Π+ lies in a component of
Ω. Similarly, the lower half plane Π− must lie in a component of Ω. Since E 6= R, one
can have a real number a lying in Ω. Since Π+ is connected, it follows from [42, Theorem
23.4, p. 150] that Π+ ∪ {a} is also connected. Similarly, the set Π− ∪ {a} is also connected.
According to [42, Theorem 23.3, p. 150], the union Π+ ∪ {a} ∪ Π− = (C \ R) ∪ {a} is
connected. Finally, since
(C \ R) ∪ {a} ⊆ Ω ⊆ C,
the connectedness of Ω can be deduced again from [42, Theorem 23.4, p. 150].
Assume that f had a holomorphic square root in Ω. By the definition, Ω is a region.
Furthermore, we observe from Theorem 13.11 that Ω is simply connected. However, the
closed curve C(0; 2R) is not null-homotopic in Ω because E lies inside C(0; 2R). By the
definition, Ω is not simply connected and hence f has no holomorphic square root in Ω.
(e) Assume that Re f was bounded in Ω. We use part (f) in advance that f will be bounded
in Ω. This implies that f can be extended to a bounded entire function because E is
compact. Hence it contradicts part (a) and then Re f is unbounded in Ω.
(f) Suppose that z = a + ib ∈ Ω. Then it is easy to see that
Z
Z
Z
(t − a) dt
b dt
dt
=
+i
.
f (z) =
2
2
2
2
E (t − a) + b
E (t − a) + b
E (t − a) − ib
For every z ∈ Ω, we deduce from the expression (16.21) that
Z
b dt
|Im f (z)| =
(t − a)2 + b2
ZE∞
b dt
≤
(t
−
a)2 + b2
−∞
(16.21)
190
Chapter 16. Analytic Continuation
= tan−1
π
−
2
= π.
=
t − a
−
b
π
∞
−∞
2
(g) Suppose that γ is a positively oriented circle which has E in its interior. Since γ ∗ (the
range of γ) is closed in C and γ ∗ ∩ E = ∅, we have δ = inf∗ |t − z| > 0 so that
z∈γ
t∈E
Z Z
γ
E
dt dz ≤
t−z
Z Z
γ
E
1
ℓ(γ)m(E)
dt dz =
< ∞,
δ
δ
where ℓ(γ) is the circumference of the circle γ. Hence Theorem 8.8 (The Fubini Theorem)
and Theorem 10.11 together assert that
Z
Z Z
Z
Z Z
1
1
dt = 2πm(E)i.
2πiInd γ (t) dt = 2πi
dt dz =
dz dt =
γ t−z
E
E
E
γ E t−z
(h) By parts (e) and (f), we see that
f (Ω) ⊆ {z = x + iy | x ∈ R and −π ≤ y ≤ π}.
Define g(z) = eiz and ϕ = g ◦ f : Ω → C. Now part (a) ensures that ϕ is not constant.
Furthermore, it is clear that
ϕ(Ω) = g f (Ω) ⊆ {reiθ | e−π ≤ r ≤ eπ and θ ∈ [0, 2π]}
so that ϕ is bounded on Ω. Thus it remains to show that ϕ ∈ H(Ω) and this follows from
the result f ∈ H(Ω). To see this, we write
Z
ψ(z, t) dt,
f (z) =
E
1
where ψ(z, t) = t−z
. For each fixed z ∈ Ω, ψ(z, t) is measurable. For each fixed t ∈ E,
we have ψ(z, t) ∈ H(Ω). Furthermore, for each z0 ∈ Ω, we have inf |t − z0 | > 0. Let this
t∈E
number be 2δ. Then we have
δ=
inf
z∈D(z0 ;δ)
t∈E
|t − z|
which implies that |t − z| ≥ δ for every z ∈ D(z0 ; δ) and t ∈ E. Therefore, we get
Z
Z
dt
1
m(E)
sup
dt ≤
=
< ∞.
|t
−
z|
δ
δ
E
z∈D(z0 ;δ) E
In other words,
Z
E
|ψ(z, t)| dt is locally bounded. Hence we conclude that f ∈ H(Ω).c
We end the proof of the problem.
Problem 16.9
Rudin Chapter 16 Exercise 9.
c
See the online paper http://www.nieuwarchief.nl/serie5/pdf/naw5-2001-02-1-032.pdf or [22].
16.2. Problems on the Modular Group and Removable Sets
191
Proof.
(a) It is easy to see that
f (−2) =
Z
1
−1
dt
= log(t + 2)
t+2
1
−1
= log 3
and f (−4) = log(t + 4)
1
−1
5
= log .
3
Thus f is not constant in Ω.
(b) Similar to Problem 16.8(b), f cannot be extended to an entire function.
(c) According to Problem 16.8(c), the value of the limit is −2 because m(E) = 2.
(d) Similar to Problem 16.8(d), f has no holomorphic square root in Ω.
(e) Similar to Problem 16.8(e), Re f is unbounded in Ω.
(f) Since E = [−1, 1], we see that
Im f =
Z
1
−1
b
−1 t − a
dt
=
tan
(t − a)2 + b2
b
1
−1
= tan−1
1 − a
b
− tan−1
−1 − a .
b
(g) We have the exact result
Z Z
γ
1
−1
1
dt dz = 2πm(E)i = 4πi.
t−z
(h) By Problem 16.18(h), the nonconstant bounded holomorphic function ϕ in Ω is given by
Z
ϕ(z) = exp i
1
−1
z − 1
dt = exp i log
.
t−z
z+1
This completes the proof of the problem.
Problem 16.10
Rudin Chapter 16 Exercise 10.
Proof.
(a) We first need the following lemma:
Lemma 16.3
Suppose that E is compact and has no interior, and K satisfies the following two
conditions:
– Condition (1): K ⊆ E is compact (K can possibly be empty) and
– Condition (2): Each f ∈ H(C \ E) can be extended to an fK ∈ H(C \ K).
Let E ′ be the intersection of all such compact subsets K of E. Then E ′ also
satisfies the conditions.
192
Chapter 16. Analytic Continuation
Proof of Lemma 16.3. Obviously, we have E ′ ⊆ K and E ′ is compact. This means
that E ′ satisfies Condition (1). To show that E ′ also satisfies Condition (2), let
f ∈ H(C \ E) and z ∈ C \ E ′ . Then z ∈
/ K (or equivalently z ∈ C \ K) for some
compact K ⊆ E satisfying Condition (2). Thus our f ∈ H(C \ E) can be extended
to an fK ∈ H(C \ K). Since E has no interior, there exists a sequence {zn } ⊆ C \ E
such that zn → z and
fK (z) = lim f (zn )
n→∞
which implies that the value fK (z) is uniquely determined by the limit, i.e., all the
values fK (z) must be equal for all compact K ⊆ E satisfying z 6∈ K and the conditions.
Hence we may define fb : C \ E ′ → C by
fb(z) = fK (z)
(16.22)
for any such compact K. Recall that fK ∈ H(C \ K) and E ′ ⊆ K, so the expression
(16.22) ensures that fb ∈ H(C \ E ′ ), completing the proof of Lemma 16.3
Suppose that E is countable compact, i.e., E = {z1 , z2 , . . .}. Since {zn } has no interior, it
is nowhere dense and we observe from the Baire Category Theorem (see §5.7) that E has
no interior. Let E ′ be the set in Lemma 16.3. Assume that E ′ 6= ∅. Notice that
E′ =
∞
[
n=1
E ′ ∩ {zn } .
Since E ′ is compact, E ′ is closed in C so that it is a complete metric space by [61,
Theorem 3.11, p. 53]. If each E ′ ∩ {zn } has no interior, then it is nowhere dense and
the Baire Category Theorem shows that E ′ is of the first category, a contradiction. Thus
{zN } = E ′ ∩ {zN } has a nonempty interior for some N ∈ N which is impossible. Hence we
have E ′ = ∅ and we deduce from Lemma 16.3 that every f ∈ H(C \ E) can be extended to
an entire function and we denote it by the same notation f . Particularly, if f is bounded,
then the corresponding entire function f is also bounded and Theorem 10.23 (Liouville’s
Theorem) forces that it is a constant. By the definition, E is removable.
(b) Let E ⊆ R be compact and m(E) = 0. Let f ∈ H(C \ E) be bounded by a positive
constant M . By the proof of Theorem 13.5, there exists a cycle Γ in C \ E such that
Ind Γ (z) = 1
for every z ∈ E. Suppose that V is the union of the collection of those components of
C \ Γ intersecting E. Thus we have E ⊆ V . Define g : V → C by
Z
f (ζ)
1
dζ.
F (z) =
2πi Γ ζ − z
for every z ∈ V . By an argument similar to part (c) below, we see that F ∈ H(V ).
Fix α ∈ V \ E. Since
the set (C \ V ) ∪ {α} is closed in C and [(C \ V ) ∪ {α}] ∩ E = ∅,
d E, (C\V )∪{α} > 0. Let 0 < ǫ < d E, (C\V )∪{α} . Since E is compact and m(E) = 0,
E can be covered by a finite number of open intervals I1 = (a1 , b1 ), I2 = (a2 , b2 ), . . .,
In = (an , bn ) whose total length is less than ǫ. Without loss of generality, we may assume
that I1 , I2 , . . . , In are pairwise disjoint and intersect E. Let γk denote the counterclockwise
circle having Ik as its diameter and
Γǫ =
n
[
k=1
γk .
16.2. Problems on the Modular Group and Removable Sets
193
We notice that the length of Γǫ is less than πǫ. By applying Theorem 10.35 (Cauchy’s
Theorem) to the cycle Γ − Γǫ in C \ E, we obtain
Z
Z
1
1
f (ζ)
f (ζ)
dζ = F (α) −
dζ.
(16.23)
f (α) =
2πi Γ−Γǫ ζ − α
2πi Γǫ ζ − α
It is easy to see that
1
2πi
Z
Γǫ
1
M πǫ
f (ζ)
dζ ≤
×
.
ζ −α
2π d(α, E)
Since ǫ is arbitrary small, the second integral in the equation (16.23) is actually zero, we
get
f (α) = F (α)
for every α ∈ V \ E. Since F ∈ H(V ), we conclude immediately that f ∈ H(C) which
implies that it is a constant. By the definition, E is removable.
(c) Suppose thatf : Ω \ E → C is bounded by M . We fix z0 ∈ Ω \ E. Let Γ1 be a cycle in
Ω \ E ∪ {z0 } with winding number 1 around
E ∪ {z0 } and zero around C \ Ω. Similarly,
suppose that Γ2 is a cycle in Ω \ E ∪ {z0 } with winding number 1 around E and zero
around (C \ Ω) ∪ {z0 }. Since Ind Γ1 (α) = Ind Γ2 (α) = 0 for every α ∈
/ Ω, Theorem 10.35
(Cauchy’s Theorem) asserts that our construction guarantees
Z
Z
f (ζ)
f (ζ)
1
1
dζ −
dζ.
f (z0 ) =
2πi Γ1 ζ − z0
2πi Γ2 ζ − z0
Define f1 , f2 : Ω \ E → C by
Z
f1 (z) =
Γ1
f (ζ)
dζ
ζ−z
and f2 (z) =
Z
Γ2
f (ζ)
dζ.
ζ −z
We claim that f1 ∈ H(Ω) and f2 ∈ H(C \ E). To this end, we first note from Theorem
10.35 (Cauchy’s Theorem) that f1 is independent of Γ1 . Next, we take z ∈ Ω and fix the
cycle Γ1 as constructed above. Denotethe length of Γ1 to be ℓ(Γ1 ). Since E ∪ {z} lies
entirely inside Γ1 , we have Γ∗1 ∩ E ∪ {z} = ∅. Recall that E is compact, so is E ∪ {z} and
then d Γ∗1 , E ∪{z} > 0. Let this number be 2δ. If h is very small such that z +h ∈ D(z; δ),
then we have
Z
f (ζ)
f1 (z + h) − f1 (z)
=
dζ.
(16.24)
h
Γ1 (ζ − z)(ζ − z − h)
Clearly, for every ζ ∈ Γ1 , we have
f (ζ)
M
≤ 2.
(ζ − z)(ζ − z − h)
2δ
Using this and the fact that ℓ(Γ1 ) < ∞, we may apply Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) to the expression (16.24) to conclude that
Z
f (ζ)
dζ.
f1′ (z) =
2
Γ1 (ζ − z)
Since z ∈ Ω is arbitrary, we get the desired result that f1 ∈ H(Ω). Using a similar
argument, we can show that f2 ∈ H(C \ E) which proves the desired claim.
Therefore, we have f = f1 − f2 on Ω \ E. Now the boundedness of f certainly implies
the boundedness of f2 . Since E is removable, f2 is a constant. Consequently, we obtain
f ∈ H(Ω).
194
Chapter 16. Analytic Continuation
(d) Suppose that E ⊂ C is compact and m2 (E) = 0.d Then E is removable. Here we need
the following lemma to prove this result.
Lemma 16.4
E ⊆ C is removable if and only if every bounded holomorphic function f on C \ E
satisfies f ′ (∞) = 0.
Proof of Lemma 16.4. By the definition, we know that E is removable if and only if
every bounded holomorphic function f on C\E is constant. Obviously, if f ∈ H(C\E)
is constant, then f ′ (∞) = 0. Conversely, let g : C \ E → C be nonconstant and
bounded. Then there exists an z0 ∈ C \ E such that g(z0 ) 6= g(∞). Define
f (z) =
g(z) − g(z0 )
z − z0
on C \ E. Obviously, f is also a bounded and nonconstant function on C \ E and
f (∞) = lim f (z) = 0.
z→∞
Consequently, we establish
f ′ (∞) = lim z[f (z) − f (∞)]
z→∞
z[g(z) − g(z0 )]
z − z0
z
z
g(z) − g(z0 ) lim
= lim
z→∞ z − z0
z→∞ z − z0
= g(∞) − g(z0 ) 6= 0,
= lim
z→∞
completing the proof of Lemma 16.4.
We return to the proof of the problem. Let f be a bounded holomorphic function on C \E,
i.e., |f (z)| ≤ M on C \ E for some positive constant M . Given ǫ > 0. Then E can be
covered by open discs D1 , D2 , . . . , Dn of radii r1 , r2 , . . . , rn respectively such that
n
X
rk < ǫ.
k=1
Let Γ = ∂D1 ∪ ∂D2 ∪ · · · ∪ ∂Dn . Using [73, Eqn. (1.2), p. 16], we have
1
|f (∞)| =
2πi
′
Z
Γ
f (z) dz ≤ M
n
X
rk < M ǫ.
(16.25)
k=1
Since ǫ is arbitrary, the inequality (16.25) guarantees that f ′ (∞) = 0. Now we conclude
from Lemma 16.4 that E is in fact removable.
(e) Suppose first that E ⊂ C is compact and removable. If F ⊆ E is a connected component
of E containing more than one point, then it follows from Theorem 14.8 (The Riemann
Mapping Theorem) that there exists a conformal mapping f : C \ F → U which is nonconstant. Thus f |C\E ∈ H(C \ F ) and f |C\E must be bounded. By the definition, E is
d
Here m2 denotes the Lebesgue measure in two dimensional space C.
16.3. Miscellaneous Problems
195
non-removable, a contradiction. Hence connected components of E are one-point sets, i.e.,
E is totally disconnected, see [42, Exercise 5, p. 152].
Now if E ⊂ C is a connected subset with more than one point, then the above paragraph
ensures that E must be non-removable.e
We have completed the analysis of the proof of the problem.
Remark 16.1
Recall that we have studied the special case of removable sets in Problem 11.11. See also
Remark 11.2.
16.3
Miscellaneous Problems
Problem 16.11
Rudin Chapter 16 Exercise 11.
Proof. By the definition, we have Ωα ⊂ Ωβ if α < β. In Figure 16.4, Ωβ is the union of Ωα and
the region shaded by straight lines.
Figure 16.4: The regions Ωα and Ωβ if α < β.
e
Recall from point-set topology [45, p. 3] that a nonempty compact connected metric space is a continuum.
196
Chapter 16. Analytic Continuation
• fβ is an analytic continuation of fα if α < β. Let ζ ∈ Ωα . Then there exists a δ > 0
such that D(ζ; δ) ⊆ Ωα and D(ζ; δ) ∩ Γ∗α = ∅. Define ψα : D(ζ; δ) × Γ∗α → C by
ψα (z, ω) =
exp(eω )
.
ω−z
If ω = t + πi for t ∈ [α, ∞), then we have eω = −et so that
| exp(eω )| = exp(−et ) ≤
1
<∞
eeα
(16.26)
for all t ∈ [α, ∞). In fact, this bound (16.26) remains true on −t − πi for all t ∈ (−∞, −α].
If ω = α + πit
α for t ∈ [−α, α], then we see that
πt ≤ exp(eα ) < ∞.
| exp(eω )| = exp eα cos
α
In other words, the function exp(eω ) is bounded on Γ∗α . Next, it is obvious that ψα (z, ω)
is a measurable function of ω, for each fixed z ∈ D(ζ; δ), and ψα (z, ω) is holomorphic in
D(ζ; δ), for each fixed ω ∈ Γ∗α . By Problem 10.18, we conclude immediately that fα is
holomorphic at ζ. Since ζ is arbitrary, we obtain fα ∈ H(Ωα ).
Recall that Ωα ⊂ Ωβ if α < β, so fβ is an analytic continuation of fα if α < β.
• Existence of an entire function f such that f = fα on Ωα . We fix an α > 0 and an
ζ ∈ Ωα . Since Ωα is open in C, one can find a δζ > 0 such that
D(ζ; δζ ) ⊆ Ωα .
Let γ be a curve in C with parameter interval [0, 1] that starts at the center of D(ζ; δζ ).
Note that this may happen that γ([0, 1]) * Ωα . However, the compactness of γ([0, 1])
ensures that there corresponds
an β > α such that γ([0, 1]) ⊆ Ωβ . Hence the first assertion
guarantees that fα , D(ζ; δ) can be analytically continued along the curve γ in C. By
Theorem 16.15 (The Monodromy Theorem), there exists an entire function f such that
f (z) = fα (z)
for all z ∈ D(ζ; δζ ). By the Corollary to Theorem 10.18, we have f = fα on Ωα .
• f (reiθ ) → 0 as r → ∞ for every eiθ 6= 1. Suppose that r > 0 and θ is real. By the second
assertion, we know that f (z) = f1 (z) on Ω1 . By the assumption, we have reiθ ∈ Ω1 for
large enough r > 0. Write Γα = γα− + Lα + γα+ , where γα− = −t − πi for t ≤ −α, γα+ = t + πi
for t ≥ α and Lα = α + πit
α for t ∈ [−α, α]. Therefore, we see that
Z
1
exp(eω )
|f (reiθ )| =
dω
2π Γ1 ω − z
Z
Z
Z
exp(eω )
exp(eω )
exp(eω )
1
1
1
≤
dω +
dω +
dω
2π γ1− ω − z
2π L1 ω − z
2π γ1+ ω − z
Z −1
Z 1
exp(−e−t )
exp(e cos πt) · exp(ie sin πt)
1
1
dt +
πi dt
=
iθ
2π −∞ −t − πi − re
2π −1
1 + πit − reiθ
Z ∞
1
exp(−et )
+
dt .
(16.27)
2π 1 t + πi − reiθ
Since |ω − reiθ | ≥ r sin θ − 1 for large enough r > 0 and for every ω ∈ Γ∗1 , the inequality
(16.27) reduces to
Z ∞
Z 1
i
h Z −1
1
exp(−et ) dt
exp(e cos πt) dt +
exp(−e−t ) dt +
|f (reiθ )| ≤
2π(r sin θ − 1) −∞
1
−1
16.3. Miscellaneous Problems
h
1
≤
2
2π(r sin θ − 1)
197
Z
∞
1
i
exp(−et ) dt + 2ee .
(16.28)
Since et ≤ exp(et ) for every t ≥ 0, the inequality (16.28) further reduces to
|f (reiθ )| ≤
1
(ee + e−1 ).
π(r sin θ − 1)
Since eiθ 6= 1, sin θ 6= 0 which implies that
lim f (reiθ ) = 0.
r→∞
• f is not constant. Fix r > 0. Let 0 < r < α < R and Γ = Γα ∪ LR , where LR = R + πit
R
for t ∈ [−R, R]. Assume that f was constant. Since f is entire, the third assertion forces
that f (z) = 0 in C. In particular, we have
1
0 = f (r) =
2πi
Z
Γα
exp(eω )
dω
ω−r
(16.29)
for every α > r. It is clear that Γ is closed and Ind Γ (r) = 0. Using Theorem 10.35
(Cauchy’s Theorem), we know that
1
2πi
Z
Γα
1
exp(eω )
dω +
ω−r
2πi
which implies
Z
Z
LR
LR
exp(eω )
1
dω =
ω−r
2πi
Z
Γ
exp(eω )
dω = 0
ω−r
exp(eω )
dω = 0
ω−r
(16.30)
for every R > r. Write
Z
1
exp(eω )
f (r) =
dω
2πi ΓR ω − r
Z
Z
Z
1
exp(eω )
1
1
exp(eω )
exp(eω )
=
dω +
dω +
dω
2πi γR+ ω − r
2πi γR− ω − r
2πi LR ω − r
Z ∞
Z ∞
Z
exp(−et )
exp(−et )
exp(eω )
1
1
1
dt −
dt +
dω
=
2πi R t + πi − r
2πi R t − πi − r
2πi LR ω − r
Z
Z ∞
exp(eω )
1
exp(−et )
dt
+
dω.
=−
2
2
2πi LR ω − r
R (t − r) + π
Using the results (16.29) and (16.30), we immediately see that
Z
∞
R
− exp(−et )
dt = 0
(t − r)2 + π 2
for every R > r. Since exp(−et ) ≤ e−t and
Z
∞
R
1
(t−r)2 +π 2
1
− exp(−et )
dt ≥ −
(t − r)2 + π 2
(R − r)2 + π 2
Z
∞
≤
1
,
(R−r)2 +π 2
e−t dt =
R
which contradicts the result (16.31). Hence f (r) 6= 0.
(16.31)
we get
1
>0
eR [(R − r)2 + π 2 ]
198
Chapter 16. Analytic Continuation
• g(reiθ ) → 0 as r → ∞ for every eiθ . If eiθ 6= 1, then the third assertion implies that
lim g(reiθ ) = lim f (reiθ ) exp[−f (reiθ )] = 0 · 1 = 0.
r→∞
r→∞
Next, suppose that eiθ = 1. Since f (r) → ∞ as r → ∞, we see immediately that
f (r)
= 0.
r→∞ exp[f (r)]
lim g(r) = lim
r→∞
This gives the fifth assertion.
• Existence of an entire function h with the required properties. By the fourth and
the fifth assertions, we know that g is a nonconstant entire function such that g(reiθ ) → 0
as r → ∞ for every eiθ . If g has a zero of order N at z = 0, then we write g(z) = z N G(z).
Thus G is nonconstant entire, G(0) 6= 0 and
lim G(reiθ ) = 0
r→∞
for every eiθ . Define h(z) =
G(z)
G(0) .
(16.32)
Therefore, h is nonconstant entire and h(0) = 1.
Furthermore, if z 6= 0, then we write z = reiθ for some r > 0. Since
h(nz) = h(nreiθ ) =
G(nreiθ )
,
G(0)
we follow from the limit (16.32) that h(nz) → 0 as n → ∞. If g(0) 6= 0, then we consider
the nonconstant entire function h(z) = g(z)
g(0) which satisfies h(0) = 1 and h(nz) → 0 as
n → ∞. In conclusion, there exists an entire function h such that

 1, if z = 0;
lim h(nz) =
n→∞

0, if z 6= 0.
We have ended the analysis of the problem.
Problem 16.12
Rudin Chapter 16 Exercise 12.
Proof. Suppose that f is represented by the series
∞ X
z − z 2 3k
k=1
2
.
2
(16.33)
Evidently, if |z − z 2 | < 2, then | z−z
2 | < 1 so that the series (16.33) converges by [61, Theorem
3.26, p. 61]. Furthermore, if |z − z 2 | > 2, then the series (16.33) diverges. The red shaded part
in Figure 16.5 indicates the regionf of convergence of the power series (16.33).
k
Next, suppose that Pk (z) = [z(1 − z)]3 , so
∞
X
1
f (z) =
k Pk (z).
23
k=1
f
This is (x − x2 + y)2 + (y − 2xy)2 < 4.
16.3. Miscellaneous Problems
199
Figure 16.5: The regions of convergence of the two series.
Note that the highest power and the lowest power of z in Pk (z) and in Pk+1 (z) are 2 · 3k and
3k+1 respectively. Since 3k+1 − 2 · 3k = 3k > 0 for every k ≥ 1, the polynomial Pk (z) contains
no power of z that appear in any other Pj (z) for all j 6= k. If we replace every Pk (z) by its
expansion in powers of z, then we get the power series
f (z) =
∞
X
an z n
(16.34)
n=1
with the property that a1 = a2 = 0 and for each positive integer k, we have

3k
(−1)n+1 Cn−3

k


, if n = 3k , 3k + 1, . . . , 2 · 3k ;
k
3
2
an =



0,
if 2 · 3k < n < 3k+1 .
(16.35)
If both n and r tend to infinity, then it follows from Stirling’s formula [61, Eq. (103), p. 194]
that
r
nn
n
· r
.
Crn ∼
2πr(n − r) r (n − r)n−r
k
3
Recall that Cn−3
k takes its maximum value when nk =
3k+1 +1
,
2
so it is true that
3k
3k
=
2π(nk − 3k )[3k − (nk − 3k )]
2π(nk − 3k )(2 · 3k − nk )
3k
=
k
k
2π · 3 2+1 · 3 2−1
2
π
2
∼
π
=
3k
32k − 1
1
· k
3
·
(16.36)
200
Chapter 16. Analytic Continuation
and
k
(3k )3
(
3k +1
2
)
3k +1
2
k
3k −1
(
2
1
)
3k −1
2
∼ 23 .
(16.37)
1
Since g(x) = x x is decreasing for x > e and x x → 1 as x → ∞ and nk ∼ 1.5 × 3k for large k, it
yields from the estimates (16.36) and (16.37) that
1
C 33k +1 ! 1.5×3
k
k
lim |ank |
1
nk
k→∞
2
= lim
23k
k→∞
1
= lim
k→∞
= lim
k→∞
= 1.
× C 33k +1
2
3
·
2
1
2
k
2
3
2
2
1
3k+1
π
1
1.5×3k
·
1
1 k+1
2
3
· 23
k
3
In other words, the radius of convergence of the power series is 1. Check the blue shaded part
in Figure 16.5.
Let λ = 3. Let pk = 2 · 3k and qk = 3k+1 for k = 1, 2, . . .. Then they satisfy λqk > (λ + 1)pk
and an = 0 for pk < n < qk for all positive integers k. Now the power series (16.33) and (16.34)
assert that there exists a δ > 0 such that
∞ X
z − z 2 3k
2
k=1
=
∞
X
an z n
n=1
for all z ∈ D(0; 1) ∩ D(1; δ). By Definition 16.1, it means that 1 is a regular point of f . Finally,
it concludes from Theorem 16.5 that the sequence {spk (z)} converges in a neighborhood of 1,
where sp (z) is the pth partial sum of the power series (16.34).
By Figure 16.5 again, we know that all boundary points of T , except z = −1, are regular
points of f . Observe from the representation (16.34) and the definition (16.35) that
−f (−z) =
∞
X
bn z n ,
n=1
where bn ≥ 0 for every n ≥ 1. Thus Problem 16.1 ensures that −f (−z) has a singularity at
−z = 1. Hence z = −1 is the singular point of f which is nearest to the origin, so we have
completed the proof of the problem.
Problem 16.13
Rudin Chapter 16 Exercise 13.
Proof. For each positive integer n, we have
Xn = {f ∈ H(Ω) | f = g(n) for some g ∈ H(Ω)}.
(a) If f ∈ X1 , then f = g′ for some g ∈ H(Ω). Since γ is a closed path lying in Ω, Theorem
10.12 implies that
Z
Z
(16.38)
f (z) dz = g′ (z) dz = 0.
γ
γ
16.3. Miscellaneous Problems
201
Conversely, suppose that the integral (16.38) holds. Since f ∈ H(Ω) and Ω is an annulus,
Problem 10.25 shows that f admits the Laurent series
f (z) = f1 (z) + f2 (z) =
−1
X
cn z n + f2 (z),
−∞
where f1 ∈ H C \ D(0; 21 ) and f2 ∈ H D(0; 2) . Since
1
cn =
2πi
Z
γ
f (z)
dz,
z n+1
the integral (16.38) implies that c−1 = 0. Thus we obtain
f1 (z) =
∞
X
n=2
c−n z −n .
If we let ω = z1 , then the function
F1 (ω) = f1
1
ω
∞
X
=
n=2
c−n ω n
is holomorphic in {ω ∈ C | |ω| < 21 }. Therefore, it is true that lim sup
implies that the radius of convergence of the series
n→∞
1
√
n c
which
−n ≥
2
∞
X
c−n n−1
G(ω) =
ω
1
−n
n=2
is at least 21 too. Next, it is clear from Theorem 10.6 that G′ (ω) = −ω −2 F1 (ω) in the
disc{ω ∈ C | |ω| < 12 }. By transforming back to the variable z, we see that
d X c−n 1−n z
f1 (z) =
dz n=2 1 − n
∞
(16.39)
holds in C \ D(0; 21 ). Similarly, we can show that
f2 (z) =
d X cn n+1 z
dz n=0 n + 1
∞
(16.40)
holds in D(0; 2). Finally, if we define
g(z) =
∞
∞
X
c−n 1−n X cn n+1
z
+
z
1−n
n+1
n=0
n=2
for all z ∈ Ω, then the facts (16.39) and (16.40) combine to imply immediately that
f (z) = g ′ (z) in Ω, i.e., f ∈ X1 .
(b) Since f ∈ H(Ω) and Ω is an annulus, Problem 10.25 shows that f admits a representation
f (z) =
∞
X
n=−∞
an z n .
(16.41)
202
Chapter 16. Analytic Continuation
(m)
If f ∈ Xm , then f = gm
i.e.,
for some gm ∈ H(Ω). Again, gm has the Laurent series in Ω,
gm (z) =
∞
X
bn,m z n
n=−∞
which gives
∞
X
(m)
an z n = gm
(z)
n=−∞
=
=
∞
X
n=−∞
∞
X
n(n − 1) · · · (n − m + 1)bn,m z n−m
(n + m)(n + m − 1) · · · (n + 1)bn+m,m z n .
n=−∞
Therefore, it means that a−1 = a−2 = · · · = a−m = 0. As m runs through all positive
integers, the Laurent series (16.41) reduces to
f (z) =
∞
X
an z n
n=0
which implies that f ∈ H D(0; 2) .
Conversely, suppose that there exists an g ∈ H D(0; 2) such that f (z) = g(z) for all
z ∈ Ω. By Theorem
13.11, the simply connectedness of D(0; 2) ensures that one can find
an g1 ∈ H D(0; 2) such that g1′ = g. In fact, this argument can be repeated to achieve
(n)
the existence of an gn ∈ H D(0; 2) with gn = g for each positive integer n. Hence we
obtain
f (z) = gn(n) (z)
for all z ∈ Ω and this means that f ∈ Xn for every positive integer n.
We have completed the proof of the problem.
Problem 16.14
Rudin Chapter 16 Exercise 14.
Proof. Our proof here basically follows that in [64, §2.7, pp. 54 – 56]. Since normality is a local
property, we may assume that Ω is the unit disc U . Suppose that
F = {f ∈ H(U ) | |f (p)| ≤ R and 0, 1 ∈
/ f (U )}.
Recall from Theorem 16.20 that the modular function λ is invariant under Γ (i.e., λ ◦ ϕ = λ for
all ϕ ∈ Γ) and maps Π+ onto C \ {0, 1}. Since f (U ) ⊆ C \ {0, 1}, the function λ−1 ◦ f has a local
branch defined in a sufficiently small neighbourhood of f (0). Then this function element may
be analytically continued in U , so we assert from Theorem 16.15 (The Monodromy Theorem)
that there exists a holomorphic function fb : U → Π+ such that
λ ◦ fb = f.
(16.42)
Let {fn } ⊆ F . Since |fn (p)| ≤ R for all n ∈ N, the Bolzano-Weierstrass Theorem [79, Problem
5.25, pp. 68, 69] ensures that there is a convergent subsequence {fnk (p)}. Let this limit be ℓ.
16.3. Miscellaneous Problems
203
• Case (i): ℓ 6= 0 and ℓ 6= 1. Then we can fix a branch of λ−1 in a neighborhood of ℓ and
c
use this to define the functions fc
nk by the equation (16.42). Since Im fnk > 0, we have
c
c
Re (−ifc
nk ) = Im fnk > 0 so that the family {−ifnk | k ∈ N} is normal by Problem 14.15.
Consequently, the family
c = {fc
F
nk | k ∈ N}
is also normal. For simplicity, we may assume that fc
nk converges normally to g ∈ H(U ).
Clearly, we have g(U ) ⊆ Π+ . By the equation (16.42) again, we conclude that
g(p) = lim fbnk (p) = lim λ−1 fnk (p) = λ−1 (ℓ).
k→∞
k→∞
Recall that the domain of λ is Π+ , so the Open Mapping Theorem implies that g(U ) ⊆ Π+
and λ ◦ g : U → C is well-defined such that
lim fnk (z) = lim λ fc
nk (z) = λ g(z)
k→∞
k→∞
for all z ∈ U . Hence {fnk } is the required subsequence.
• Case (ii): ℓ = 1. Since fnk ∈ H(U ) and 0 ∈
/ fnk (U ), we have fn1 ∈ H(U ). Since U is
k
simply connected, we deduce from Theorem 13.11 that each fnk has a holomorphic square
root hk in U . We choose the branch such that
lim hk (p) = −1.
(16.43)
k→∞
/ hk (U ) and |hk (p)| ≤
Since fnk = h2k , we have 0, 1 ∈
√
R. Consider the family
H = {hk | k ∈ N}.
Now the limit (16.43) guarantees that we can apply Case (i) to H to obtain a convergent
subsequence hkj in U . In conclusion, the limit
lim fnkj (z)
j→∞
exists for all z ∈ U .
• Case (iii): ℓ = 0. In this case, we may apply Case (ii) to the sequence {1 − fnk | k ∈ N}.
Hence we have completed the proof of the problem.
Remark 16.2
Problem 16.14 is classically called the Fundamental Normality Test.
Problem 16.15
Rudin Chapter 16 Exercise 15.
Proof. Without loss of generality, we may assume that D is the unit disc and D ⊆ Ω = D(0; R)
for some R > 1. Let (f, D) be analytically continued along every curve in Ω that starts at the
origin 0. Since f ∈ H(D), f has a power series expansion at 0, i.e.,
f (z) =
∞
X
n=0
an z n .
204
Chapter 16. Analytic Continuation
Suppose that r is the radius of convergence of this power series. Clearly, we have 1 ≤ r. If r = 1,
then we know from Theorem 16.2 that f has at least one singular point on the unit circle T . Let
ω be a singular point of f on T and γ be a curve in Ω starting at 0 and passing through ω. Now
the assumption ensures that (f, D) can be analytically continued along γ in Ω, so ω is a regular
point of f , a contradiction. As a result, 1 < r ≤ R. Next, if r < R, then similar argument can
be applied to conclude that no point on C(0; r) is a singular point, but this contradicts Theorem
16.2. Hence we must have r ≥ R and it means that Theorem 16.5 holds in this special case.
This proves the first assertion.
Let Ω be any simply connected region (other than the complex plane itself), D ⊆ Ω and
(f, D) be analytically continued along every curve in Ω that starts at the center of D. Let
z0 ∈ D. By Theorem 14.8 (The Riemann Mapping Theorem) or [9, §14.2, pp. 200 – 204], one
can find a (unique) conformal mapping F : Ω → U such that F (z0 ) = 0 and F ′ (z0 ) > 0. Let
S = F (D). Obviously, we have G = F −1 |S : S ⊂ U → D so that G(0) = z0 . We consider the
mapping
g = f ◦ G : S ⊂ U → C.
(16.44)
Since S is an open set containing the origin, we may assume that it is an open disc centered at
0 so that S and U are concentric. Since f ∈ H(D), we conclude that g ∈ H(S). Furthermore,
since (f, D) can be analytically continued along every curve in Ω, the function element (g, S) can
also be analytically continued along every curve in U . Therefore, the first assertion guarantees
that there corresponds a h ∈ H(U ) such that h(z) = g(z) for all z ∈ S. Using the definition
(16.44), we have h ◦ F ∈ H(Ω) and for all z ∈ D,
h F (z) = h G−1 (z) = h g −1 f (z) = f (z).
This proves the second assertion and we end the analysis of the problem.
CHAPTER
17
H p-Spaces
17.1
Problems on Subharmonicity and Harmonic Majoriants
Problem 17.1
Rudin Chapter 17 Exercise 1.
Proof. Let u : Ω → R be an upper semicontinuous subharmonic function. By Definition 2.8, for
every real α, the set {z ∈ C | u(z) < α} is open in C.
• Let K ⊂ Ω be compact and h : K → R be continuous such that h is harmonic in V = K ◦
and u(z) ≤ h(z) for all boundary points of K. Put u1 = u − h. Assume that u1 (ζ) > 0
for some ζ ∈ V . Since h is continuous on K, −h is upper semicontinuous on K. Thus u1
is also upper semicontinuous on K by [78, Problem 2.1, pp. 17, 18]. Since K is compact,
[78, Problem 2.21, p. 60] ensures that u1 attains its maximum m on K. Since u1 ≤ 0 on
the boundary of K, the set E = {z ∈ K | u1 (z) = m} is a nonempty compact subset of V .
Let z0 be a boundary point of E. Since E is compact, V is open in C and E ⊂ V , there
exists an r > 0 such that D(z0 ; r) ⊂ V . Now some subarc of the boundary of D(z0 ; r) lies
in V \ E. Hence we have
Z π
1
u1 (z0 + reiθ ) dθ
u1 (z0 ) = m >
2π −π
which means that u1 is not subharmonic in V . However, since u and h are subharmonic
and harmonic in V respectively, it follows from the mean value property that u1 is also
subharmonic in V , a contradiction. This proves that no such ζ exists and then u(z) ≤ h(z)
for all z ∈ K.
• By Definition 17.1, it suffices to prove that Theorem 17.5 is true for subharmonic function
in U . Let 0 ≤ r < 1. Then K = D(0; r) ⊂ U and u : K → R is subharmonic. In particular,
u is an upper semicontinuous function. We need the following result:a
Lemma 17.1 (Baire’s Theorem on Semicontinuous Functions)
Let K ⊂ C be compact and −∞ ≤ u(z) < ∞ on K. If u is upper semicontinuous,
then it is the limit of a monotone decreasing sequence of continuous functions {un }
on K.
a
Read https://encyclopediaofmath.org/wiki/Baire_theorem#Baire.27s_theorem_on_semi-continuous_functions .
205
Chapter 17. H p -Spaces
206
Let 0 ≤ r1 < r2 < 1. Using this lemma, we know that u(z) ≤ un (z) on C(0; r2 ). By
Theorem 11.8, Hun ∈ C(K), Hun is harmonic in D(0; r2 ) and (Hun )|C(0;r2 ) = un . Clearly,
we have u(z) ≤ un (z) = (Hun )(z) on C(0; r2 ), so the first assertion implies that
u(z) ≤ (Hun )(z)
for all z ∈ K. Furthermore, the mean value property gives
Z π
Z π
1
1
(Hun )(r2 eit ) dt =
un (r2 eit ) dt.
(Hun )(0) =
2π −π
2π −π
Combining the inequality (17.1) and the formula (17.2) we obtain
Z π
Z π
Z π
1
1
1
u(r1 eit ) dt ≤
(Hun )(r1 eit ) dt = (Hun )(0) =
un (r2 eit ) dt.
2π −π
2π −π
2π −π
(17.1)
(17.2)
(17.3)
Finally, we apply Problem 1.7b to the inequality (17.3) to get
Z π
Z π
Z π
1
1
1
u(r1 eit ) dt ≤ lim
un (r2 eit ) dt =
u(r2 eit ) dt.
n→∞ 2π −π
2π −π
2π −π
This completes the proof of the problem.
Problem 17.2
Rudin Chapter 17 Exercise 2.
Proof. Let u(z) = log 1 + |f (z)| = log 1 + elog |f (z)| . Define ϕ : R → R by ϕ(x) = log(1 + ex ).
Then we can write
u(z) = ϕ log |f (z)| .
(17.4)
Since f ∈ H(Ω), it follows from Theorem 17.3 that log |f (z)| is subharmonic in Ω. Evidently,
ex
′
′
ϕ′ (x) = 1+e
x > 0 for all x ∈ R and ϕ (s) < ϕ (t) if s < t, so ϕ is a monotonically increasing
convex function on R. By applying Theorem 17.2 to the function (17.4), we conclude that u is
subharmonic in Ω, as required. This completes the analysis of the problem.
Problem 17.3
Rudin Chapter 17 Exercise 3.
Proof. It seems that the hypothesis 0 < p ≤ ∞ should be replaced by 0 < p < ∞ because if
p = ∞, then the function f (z) ≡ 2 belongs to H ∞ . In this case, we have |f (z)|∞ = ∞ for any
z ∈ U.
• f ∈ H p if and only if |f (z)|p ≤ u(z) for some harmonic function u in U . Suppose
that there exists a harmonic function u in U such that |f (z)|p ≤ u(z) for all z ∈ U .
Combining this and the mean value property, we get
nZ
o1 n Z
o1
1
p
p
iθ p
iθ
kfr kp =
|f (re )| dσ
≤
u(re ) dσ
= [u(0)] p < ∞
T
T
for every 0 ≤ r < 1. By Definition 17.7, we have f ∈ H p .
b
In fact, it is
17.1. Problems on Subharmonicity and Harmonic Majoriants
207
Conversely, suppose that f ∈ H p . Now it is easy to see that
Z
Z
p
p
|fr |p dσ = kfr kpp
|fr | dσ =
kfr k1 =
T
T
which means kf p k1 = kf kpp < ∞. As a consequence, we have f p ∈ H 1 . Applying [58, Eqn.
(5), p. 344] directly to f p , we have
|f (z)|p ≤ |Qf p (z)|
n 1 Z eit + z
o
p ∗ it
= exp
log
|(f
)
(e
)|
dt
2π T eit − z
io
n h 1 Z eit + z
p ∗ it
log
|(f
)
(e
)|
dt
= exp Re
2π T eit − z
n 1 Z
o
= exp
Pr (θ − t) log |(f p )∗ (eit )| dt
2π T
(17.5)
for all z ∈ U . Using the same argument as in proving the inequality in the proof of
Theorem 17.16(c), we obtain from the inequality (17.5) that
Z
1
Pr (θ − t)|(f p )∗ (eit )| dt = P [(f p )∗ ](z)
|f (z)|p ≤
2π T
in U .c By Theorem 17.11(b), (f p )∗ ∈ L1 (T ), so Theorem 11.7 ensures that P [(f p )∗ ] is
harmonic in U .
• The existence of a least harmonic majorant. Let uf = P [(f p )∗ ]. In fact, this is
a least harmonic majorant. To see this, let u be a harmonic majorant in U , i.e., u is
harmonic in U and |f (z)|p ≤ u(z) for every z ∈ U . For any ρ < 1, we have
Z
Z
1
1
it p
Pr (θ − t)|f (ρe )| dt ≤
Pr (θ − t)u(ρeit ) dt = u(ρz),
2π T
2π T
where 0 ≤ r < 1. As ρ → 1, we getd
Z
1
Pr (θ − t)|(f p )∗ (eit )| dt
uf (z) =
2π T
Z
1
Pr (θ − t)lim |f p (ρeit )| dt
=
ρ→1
2π T
h 1 Z
i
Pr (θ − t)|f (ρeit )|p dt
= lim
ρ→1 2π T
≤ u(z)
for every z ∈ U .
1
• kf kp = uf (0) p . Let 0 < p < ∞. By the previous assertion, we know that
Z
P (z, eit )|f (eit )|p dσ.
uf (z) =
T
1
When z = 0, we have P (0, eit ) = 1 so that uf (0) = kf kpp , i.e., kf kp = uf (0) p .
c
Recall that P [f ] is the Poisson integral of f .
We can interchange the limit and the integral because Theorem 10.24 (The Maximum Modulus Theorem)
asserts that Fρ (t) = Pr (θ − t)|f (ρeit )|p is increasing with respect to ρ, so we may apply Theorem 1.26 (The
Lebesgue’s Monotone Convergence Theorem).
d
Chapter 17. H p -Spaces
208
We have completed the proof of the problem.
Problem 17.4
Rudin Chapter 17 Exercise 4.
Proof. On the one hand, if log+ |f | has a harmonic majorant in U , then there exists a harmonic
function u in U such that 0 ≤ log+ |f (z)| = log+ |f (z)| ≤ u(z). Combining this and the mean
value property, we obtain
Z
Z
+
it
u(reit ) dσ = u(0) < ∞
0 ≤ kfr k0 = exp
log |fr (e )| dσ ≤
T
T
for all z = reit ∈ U and all 0 ≤ r < 1. By Definition 17.7, we get kf k0 < ∞ so that f ∈ N .
On the other hand, suppose that f ∈ N . If f ≡ 0 so that log+ |f (z)| = 0 in U , then
there is nothing to prove. With the aid of the result in §17.19, there correspond two functions
b1 , b2 ∈ H ∞ such that b2 has no zero in U and
f=
b1
.
b2
Without loss of generality, we may assume that kb1 k∞ ≤ 1 and kb2 k∞ ≤ 1. Since U is simply
connected and b12 ∈ H(U ), we deduce from Theorem 13.11 that there exists an g ∈ H(U ) such
that b2 = eg . Since kb2 k∞ = eRe g ≤ 1, the function u(z) = Re g(z) is less than or equal to zero
in U and this implies
log |f (z)| = log |b1 (z)| − log |b2 (z)| ≤ − log eu(z) = −u(z)
(17.6)
for all z ∈ U . Since −u(z) ≥ 0, the inequality (17.6) and the definition in §15.22 yield
log+ |f (z)| ≤ −u(z)
in U . By Theorem 11.4 and g ∈ H(U ), −u is harmonic in U . Hence, log+ has a harmonic
majorant in U and this completes the proof of the problem.
17.2
Basic Properties of H p
Problem 17.5
Rudin Chapter 17 Exercise 5.
Proof. Since f ∈ H(U ), it is true that f ◦ ϕ ∈ H(U ). Since f ∈ H p , Problem 17.3 ensures that
there is a harmonic function u in U such that |f (z)|p ≤ u(z) for all u ∈ U . Thus this shows that
p
≤ u ϕ(z)
f ϕ(z)
for all z ∈ U . Applying Problem 11.7(b) with Φ = u and f = ϕ there, we see that
∆[u ◦ ϕ] = [(∆u) ◦ ϕ] × |ϕ′ |2 = 0.
By the definition, u ◦ ϕ is harmonic in U and then Problem 17.3 asserts that f ◦ ϕ ∈ H p .
17.2. Basic Properties of H p
209
The assertion is also true when we replace H p by N . To see this, Problem 17.4 ensures that
there exists a harmonic function u in U such that log+ |f (z)| ≤ u(z) holds for all z ∈ U . Then
we have
(17.7)
log+ f ϕ(z) ≤ u ϕ(z)
in U . Applying Problem 11.7 to u ◦ ϕ and then using the fact that ∆u = 0 in U , we see
immediately that
∆[u ◦ ϕ] = [(∆u) ◦ ϕ] · |ϕ′ |2 = 0.
In other words, u ◦ ϕ is harmonic in U . Combining the inequality (17.7) and Problem 17.4, we
conclude that f ◦ ϕ ∈ N , completing the proof of the problem.
Problem 17.6
Rudin Chapter 17 Exercise 6.
Proof. Let α > 0 and consider the functione
fα (z) =
1
.
(1 − z)α
Put z = reit , we have
Iα (r) =
Z
T
it
|fα (re )| dt =
Z
π
−π
1
dt.
|1 − reit |α
We want to estimate Iα (r) as r → ∞. Of course, it depends on the value of α.
Since r → 1, we may assume that r > 21 . By considering the triangle formed by 1, r and reit .
Clearly, this is an obtuse triangle. If t ∈ [−π, π], then we have
|1 − reit | > max{1 − r, r|1 − eit |}
1
≥ 1 − r + r|1 − eit |
2
1
1
>
1 − r + |1 − eit |
2
2
1
t =
1 − r + sin
.
2
2
(17.8)
Recall the fact [61, Exercise 7, p. 197] that | sin x| ≥ π2 |x| for every x ∈ [− π2 , π2 ], so the inequality
(17.8) can further reduce to
|1 − reit | ≥
|t| 1
1
1−r+
≥
(1 − r + |t|).
2
π
2π
(17.9)
On the other hand, the triangle inequality gives
|1 − reit | ≤ |1 − eit | + |eit − reit | = |1 − eit | + 1 − r ≤ |t| + 1 − r.
(17.10)
Combining the inequalities (17.9) and (17.10), we obtain
Here we note that 1 − z 6= 0 in U , so we may take the branch such that (1 − z)α = exp α log(1 − z) , where
< arg(1 − z) < π2 .
e
− π2
1
1
(2π)α
α ≤
α
≤
|1 − reit |α
|t| + 1 − r
|t| + 1 − r
Chapter 17. H p -Spaces
210
Z
π
−π
dt
α ≤ Iα (r) ≤ (2π)α
|t| + 1 − r
Z
π
−π
dt
α .
|t| + 1 − r
(17.11)
Since |t| is an even function in t, we get
Z π
Z π
dt
dt
α = 2
α
t+1−r
−π |t| + 1 − r
0
 if α = 1;
 2 log(1 − r + π) − log(1 − r) ,
=
 2 1−α − (1 − r)1−α , otherwise.
1−α (1 − r + π)
1−α
If α < 1, then the integral in the inequalities (17.11) tends to 2π1−α as r → 1 so that Iα (r)
is bounded as r → 1. If α = 1, then we observe from the inequalities (17.11) that
0 < m1 ≤
Iα (r)
≤ M1
log(1 − r)
as r → 1 for some positive constants m1 and M1 . Similarly, if α > 1, then the inequalities
(17.11) tells us that
Iα (r)
0 < m2 ≤
≤ M2
(1 − r)1−α
as r → 1 for some positive constants m2 and M2 .
We notice that
n 1 Z
o 1 I (r) 1
p
p
αp
=
,
k(fα )r kp =
|fα (reit )|p dt
2π T
2π
so the previous paragraph indicates that fα ∈ H p if and only if αp < 1. Thus, for 0 < r < s < ∞,
if we take α ∈ ( 1s , 1r ), then it is easy to see that fα ∈ H r but fα ∈
/ H s , i.e., H s ⊂ H r . The
case for s = ∞ is obvious because we have fα ∈
/ H ∞ for every α > 0. In particular, if we take
1
r
∞
/ H . This completes the analysis of the problem. α = 2r , then we have f 1 ∈ H but f 1 ∈
2r
2r
Problem 17.7
Rudin Chapter 17 Exercise 7.
Proof. Now Problem 17.6 guarantees that H ∞ ⊂ H p for every 0 < p < ∞. By Definition 17.7,
H p ⊆ N holds for every 0 < p < ∞, so we have H ∞ ⊂ N . Hence we conclude easily that
\
\
H∞ ⊂ N ∩
Hp ⊆
H p.
0<p<∞
This completes the proof of the problem.
0<p<∞
Problem 17.8
Rudin Chapter 17 Exercise 8.
Proof. If p = 1, then there is nothing to prove. If 0 < p < 1, then Problem 17.6 implies that
H 1 ⊂ H p so that the result is trivial. Thus we may assume that 1 < p < ∞. Since f ∈ H 1 ,
Theorem 17.11 guarantees that f is the Poisson integral of f ∗ , i.e.,
Z
P (z, eit )f ∗ (eit ) dσ
f (z) =
T
17.2. Basic Properties of H p
211
which shows
iθ
iθ
fr (e ) = f (re ) =
Z
T
Pr (θ − t)f ∗ (eit ) dσ,
where 0 ≤ r < 1. Since Pr (t) > 0 for all t ∈ T , we may apply Theorem 3.5 (Hölder’s Inequality)
to obtain
Z
p
iθ p
Pr (θ − t)f ∗ (eit ) dσ
|fr (e )| =
ZT
p
Pr (θ − t)|f ∗ (eit )| dσ
≤
ZT h p−1
i h 1
i
p
Pr p (θ − t) × Prp (θ − t)|f ∗ (eit )| dσ
=
T
o n Z h p−1
op−1
nZ
i p
p−1
∗ it p
Pr p (θ − t)
Pr (θ − t)|f (e )| dσ ×
dσ
≤
T
T
nZ
o nZ
op−1
∗ it p
=
Pr (θ − t)|f (e )| dσ ×
Pr (θ − t) dσ
T
Z T
∗ it p
=
Pr (θ − t)|f (e )| dσ
T
which implies
kfr kpp
Z
1
|fr |p dt
=
2π T
Z Z
1
≤
Pr (θ − t)|f ∗ (eit )|p dσ dθ
2π T T
Z h Z
i
1
Pr (θ − t) dθ |f ∗ (eit )|p dσ
≤
2π
T
ZT
|f ∗ (eit )|p dσ
=
T
= kf ∗ kp ,
where 0 ≤ r < 1. By the hypothesis, kf ∗ kp < ∞ so that f ∈ H p as required. We complete the
analysis of the problem.
Problem 17.9
Rudin Chapter 17 Exercise 9.
Proof. Since f (U ) is not dense in the complex plane, there exists a point α ∈ C and r > 0 such
1
. Then F is
that D(α; r) ⊆ C \ f (U ), i.e., |f (z) − α| > r for all z ∈ U . Let F (z) = f (z)−α
∞
a bounded holomorphic function in U so that F ∈ H . Thus it follows from Theorem 11.32
(Fatou’s Theorem) that
F ∗ (eiθ ) = lim F (reiθ )
r→1
exists at almost all points of T . This implies that the same conclusion also holds for the function
f , completing the proof of the problem.
Problem 17.10
Rudin Chapter 17 Exercise 10.
Chapter 17. H p -Spaces
212
Proof. Define Φα : H 2 → C by
Φα (f ) = f (α).
It is obvious that Φα is a linear functional on H 2 . Let
f (z) =
∞
X
an z n .
n=0
The Cauchy-Schwarz inequality and Theorem 17.12 combine to yield
|f (α)| =
∞
X
n
an α
n=0
≤
∞
X
2
|an |
n=0
1
2
×
∞
X
n=0
2n
|α|
1
2
=
∞
X
n=0
|α|2n
1
2
· kf k2 < ∞.
By Definition 5.3, Φα is a bounded linear functional. By Theorem 5.4, Φα is continuous and
since H 2 is a Hilbert space, Theorem 4.12 (The Riesz Representation Theorem) ensures the
existence of an g ∈ H 2 such thatf
Z
Z
f ∗ (reiθ )g∗ (reiθ ) dσ,
(17.12)
f (α) = Φα (f ) = hf, gi = lim
f (reiθ )g(reiθ ) dσ =
r→1 T
T
where f ∗ , g ∗ ∈ L2 (T ) by Theorem 17.11. If we let
g(z) =
∞
X
bn z n ,
n=0
then Theorem 17.12 implies
∞
X
n=0
formula (17.12) becomes
|bn |2 < ∞. By the Parseval Theorem [62, Eqn. (6), p. 91], the
∞
X
n
an α = f (α) =
∞
X
an bn .
n=0
n=0
As a result, we have bn = αn for all n = 0, 1, 2, . . .. Hence g has the representation
g(z) =
∞
X
αn z n .
n=0
This ends the proof of the problem.
Problem 17.11
Rudin Chapter 17 Exercise 11.
Proof. Suppose that
f (z) =
∞
X
an z n
n=0
in U . By the Parseval Theorem [62, Eqn. (6), p. 91], we have
Z
f
it 2
T
|f (re )| dσ =
Or you may apply [62, Eqn. (6), p. 347] directly.
∞
X
n=0
|an |2 r 2n ,
17.2. Basic Properties of H p
213
where 0 ≤ r < 1, so that
∞
X
n=0
2
|an |
1
2
= sup
r<1
nZ
|f (reit )|2 dσ
T
o1
2
= kf k2 ≤ 1.
By the Cauchy-Schwarz inequality, we can establish
′
∞
X
2
|f (α)| =
≤
≤
The identity
2
nan αn−1
n=1
∞
X
n=1
∞
X
n=1
∞
X
2
|an |
×
∞
X
n=1
n2 |α|2(n−1)
n2 |α|2(n−1) .
n2 xn−1 =
n=1
(17.13)
1+x
(1 − x)3
(17.14)
is valid for |x| < 1, so we may apply this to the inequality (17.13) to obtain
′
|f (α)| ≤
s
1 + |α|2
.
(1 − |α|2 )3
(17.15)
We claim that the estimation (17.15) is sharp. To see this, let α 6= 0 and Cα be the number
in the estimation (17.15). Consider the function
Fα (z) =
∞
eiθ X
n|α|n−1 z n ,
Cα
(17.16)
n=1
where z ∈ U . Direct computation indicates that the radius of convergence of the series (17.16)
1
is |α|
> 1 so that Fα ∈ H(U ). Next, using the identity (17.14), we see that
kFα k22
∞
X
1
=
Cα
n=0
2
· n2 |α|2(n−1)
∞
(1 − |α|2 )3 X 2 2(n−1)
=
n |α|
·
1 + |α|2
n=0
1 + |α|2
(1 −
×
1 + |α|2
(1 − |α|2 )3
= 1.
|α|2 )3
=
Finally, we know that
|Fα′ (α)|
∞
1 X 2 2(n−1)
n |α|
=
=
|Cα |
n=1
s
1 + |α|2
(1 − |α|)3
×
=
1 + |α|2
(1 − |α|)3
If α = 0, then instead of the function (17.16), we consider
F0 (z) = eiθ z
s
1 + |α|2
.
(1 − |α|)3
Chapter 17. H p -Spaces
214
which satisfies kF0 k22 = 1 and |F0′ (0)| = 1. Hence this proves the claim and the functions (17.16)
are extremal.
Next, we consider the general n. Firstly, we have
∞
X
f (n) (α) =
m=n
m(m − 1) · · · (m − n + 1)am αm−n
so that
|f
(n)
∞
X
2
(α)| =
m=n
≤
≤
m(m − 1) · · · (m − n + 1)am αm−n
∞
X
m=n
∞
nX
m=n
2
|am |
2
×
∞
nX
m=n
2
[m(m − 1) · · · (m − n + 1)]2 |α|2(m−n)
|m(m − 1) · · · (m − n + 1)αm−n |2
o2
o2
which implies
|f
(n)
We denote
Cα,n
(α)| ≤
∞
X
m=n
|m(m − 1) · · · (m − n + 1)αm−n |2 .
v
u ∞
uX
=t
|m(m − 1) · · · (m − n + 1)αm−n |2 > 0.
m=n
Then we claim that
∞
eiθ X
m(m − 1) · · · (m − n + 1)|α|m−n z m
Fα (z) =
Cα,n m=n
are the extremal functions for α 6= 0. In fact, we know that
kFα k22
=
1
|Cα,n
|2
∞
X
m=n
|m(m − 1) · · · (m − n + 1)αm−n |2 =
1
|Cα,n |2
× |Cα,n |2 = 1.
Furthermore, direct computation gives
|Fα(n) (α)|
=
1
Cα,n
×
∞
X
m=n
|m(m − 1) · · · (m − n + 1)αm−n |2 = Cα,n
and we prove the claim. If α = 0, then it is easily seen that
F0 (z) = eiθ n!z n
(n)
are the extremal functions in this case because |F0 (α)| = (n!)2 . Therefore, we have completed
the proof of the problem.
Problem 17.12
Rudin Chapter 17 Exercise 12.
17.2. Basic Properties of H p
215
Proof. Since p ≥ 1, we know that f ∈ H 1 and Theorem 17.11 tells us that
Z π
1
f (z) =
P (z, eit )f ∗ (eit ) dt
2π −π
for all z ∈ U . Now the hypothesis guarantees that f is real a.e. in U . By the Open Mapping
Theorem, f must be constant.
1+z
Consider f (z) = i 1−z
in U . It is easy to see that f ∈ H p for every 0 < p < 1 and
f ∗ (eit ) = lim f (reit ) = − cot
r→1
t
2
is real a.e. on T , but f is not constant. Hence we have completed the proof of the problem. Problem 17.13
Rudin Chapter 17 Exercise 13.
Proof. Since |f (reit )| = |γr (t)| ≤ M for every 0 ≤ r < 1 and t ∈ [−π, π], f is bounded in U .
Thus f ∈ H ∞ so that f ∈ H 1 . Furthermore, we also have the fact that f ∗ is bounded on T .
Let f ∗ (eit ) = µ(t). Since f (reπi ) = f (re−πi ), it is easy to see that
µ(π) = f ∗ (eπi ) = lim f (reπi ) = lim f (re−πi ) = f ∗ (e−πi ) = µ(−π).
r→1
r→1
On the one hand, if
f (z) =
∞
X
(17.17)
an z n ,
n=0
then we have
an =
1
2πi
Z
C(0;r)
f (ζ)
1
dζ =
ζ n+1
2π
Z
π
r −n e−int f (reit ) dt
(17.18)
−π
for every n ≥ 0 and 0 < r < 1. On the other hand, it follows from Theorem 17.11 that
f ∗ ∈ L1 (T ), so the Fourier coefficients of f ∗ are given by
Z π
1
e−int f ∗ (eit ) dt
(17.19)
fc∗ (n) =
2π −π
for every n ∈ Z. By observing the coefficients (17.18) and (17.19), we know that
Z
Z
−int
it
∗ it
n
∗
c
|f (reit ) − f (eit )| dσ = kfr − f ∗ k1 → 0
e
[f (re ) − f (e )] dσ ≤
|r an − f (n)| =
T
T
as r → 1 by Theorem 17.11. In other words, we get

 an , if n ≥ 0;
∗
c
f (n) =

0, if n < ∞.
This implies that
for every n = 1, 2, . . ..
Z
π
int
e
−π
µ(t) dt =
Z
π
−π
eint f ∗ (eit ) dt = 0
(17.20)
Chapter 17. H p -Spaces
216
Using the two facts (17.17) and (17.20), we find
Z
Z π
Z π
int
int
int
π
µ(t) d(e ) = −in
e dµ(t) = [e µ(t)]−π −
−π
−π
π
eint µ(t) dt = 0
−π
holds for every n = 1, 2, . . .. Next, Theorem 17.13 (The F. and M. Riesz Theorem) shows that
µ is absolutely continuous with m, i.e., µ(E) = 0 if m(E) = 0. Recall that f ∗ (eit ) = µ(t), so
f ∗ ∈ C(T ). In fact, f ∗ is AC on T because of Theorem 7.18 because f maps sets of measure 0
to sets of measure 0. Finally, according to Theorem 17.11, we have
Z π
Z π
1
1
it
f (z) =
P (z, e )µ(t) dt =
P (z, eit )f ∗ (eit ) dt = P [f ∗ ](z)
2π −π
2π −π
for all z ∈ U . Thus if we define the function Hf ∗ : U → C as [62, Eqn. (1), p. 234], then
Theorem 11.8 shows that Hf ∗ ∈ C(U ) and the restriction of Hf ∗ to T is exactly f ∗ . This is a
required extension of f and we end the analysis of the problem.
Problem 17.14
Rudin Chapter 17 Exercise 14.
Proof. Recall from Problem 2.11 that the support of a measure µ is the smallest closed set
K ⊆ T such that µ(T \ K) = 0. Without loss of generality, we may assume that µ 6≡ 0. Given
that K is a proper closed subset of T . Note that T \ K is nonempty open in T , so σ(T \ K) > 0.
Our goal is to show that
µ(T \ K) > 0.
(17.21)
Now we observe from the proof of Theorem 17.13 (The F. and M. Riesz Theorem) that
dµ = f ∗ (eit ) dσ, i.e.,
Z
1
f ∗ (eit ) dt
µ(E) =
2π E
holds for every measurable subset E of T , where f = P [f ∗ ] ∈ H 1 and f ∗ ∈ L1 (T ). Thusit
suffices to show that f ∗ doesn’t vanish on T \ K. Let g(t) = f ∗ (eit ). Assume that g(T \ K) = 0.
Then it means that log |g| = ∞ on T \ K, but it contradicts Theorem 17.17 (The Canonical
Factorization Theorem) that
log |g| = log |f ∗ | ∈ L1 (T ).
Hence f ∗ does not vanish on T \ K with σ(T \ K) > 0 which implies the result (17.21). This
completes the analysis of the problem.
Problem 17.15
Rudin Chapter 17 Exercise 15.
Proof. Denote CK to be the set of all continuous functions on K.g Then the problem is equivalent
to show that the set of polynomials PK on K is dense in CK with respect to the norm
kf k∞ = sup{|f (z)| | z ∈ K}.
This is well-defined because f is continuous on the compact set K. Assume that PK was not
dense in CK . In other words, there exists an g ∈ CK such that g ∈
/ PK . It is clear that PK
g
See Definition 3.16, p. 70.
17.3. Factorization of f ∈ H p
217
is a linear subspace of the normed linear space CK . Then Theorem 5.19 implies that there is a
bounded linear functional Φ on CK such that
Φ(P ) = 0
for all P ∈ PK and Φ(g) 6= 0. According to Theorem 6.19, there exists a unique regular complex
Borel measure ν on K such that
Z
f dν
(17.22)
Φ(f ) =
K
for every f ∈ CK . This measure ν must be nonzero because of Φ(g) 6= 0.
Define µ(E) = ν(E ∩ K) for E ∈ MT . Then it is easily checked that µ is also a complex
Borel measure on T . Combining this and the representation (17.22), we get
Z
Z
Z
P dµ = Φ(P ) = 0
P dν +
P dµ =
T
T \K
K
for every P ∈ PK . In particular, we have
Z
e−int dµ = 0
T
for every n = 1, 2, . . .. Since µ 6≡ 0, Problem 17.14 asserts that the support of µ is exactly all of
T so that µ(T \ K) > 0 because K is a proper compact (hence closed) subset of T . However, the
definition of µ implies that µ(T \ K) = ν(∅) = 0, a contradiction. This completes the analysis
of the problem.
17.3
Factorization of f ∈ H p
Problem 17.16
Rudin Chapter 17 Exercise 16.
Proof. Suppose that 0 < p < 1. Recall from the first paragraph of the proof of Theorem 7.17
that we may assume that f has no zeros in U . Therefore, we deduce from Theorem 17.10 that
2
one can find a zero-free function h ∈ H 2 such that f = h p . Note that h = Mh Qh by Theorem
17.17, so we have
2
2
f = Mhp Qhp .
By the definition of an inner function, we know that
n 2 Z eit + z
n Z eit + z
o
o
2
2
2
p
p
p
exp
−
dµ
(t)
=
c
dµ
(t)
,
Mh (z) = c exp −
h
f
it
p T eit − z
T e −z
(17.23)
(17.24)
where µh is a finite positive Borel measure on T and µf = p2 µh . Clearly, µf is also a finite
positive Borel measure on T , so it follows from Theorem 17.15 that the right-most function is
in fact an inner function. Let it be Mf .
Next, according to [62, Eqn. (1), p. 344],
n 1 Z
2
p
Qh (z) = exp
2π T
n 1 Z
= exp
2π T
we see that
eit + z
eit − z
eit + z
eit − z
o
2
· log |h∗ (eit )| dt
p
o
2
log (h∗ ) p (eit ) dt
·
Chapter 17. H p -Spaces
218
n 1 Z eit + z
o
∗ it
= exp
log
|f
(e
)|
dt
.
2π T eit − z
(17.25)
Since log |h∗ | ∈ L1 (T ), we immediately have log |f ∗ | ∈ L1 (T ). By Definition 17.14, the function
(17.25) is an outer function and we let it be Qf . Since Qh ∈ H 2 , Theorem 17.16 implies that
|h∗ | ∈ L2 (T ) and thus |f ∗ | ∈ Lp (T ). Using Theorem 17.16 again, we conclude that Qf ∈ H p .
By substituting the expressions (17.24) and (17.25) into the formula (17.23), we obtain
f = Mf Qf .
Finally, we note that the inequality
1
log |h(0)| ≤
2π
Z
(17.26)
T
log |h∗ (eit )| dt
1
2π
Z
(17.27)
T
log |f ∗ (eit )| dt.
is equivalent to the inequality
log |f (0)| ≤
Hence equality holds in (17.27) if and only if equality holds in (17.26) if and only if Mh is
constant if and only if Mf is constant too. Consequently, we have completed the proof of the
problem.
Problem 17.17
Rudin Chapter 17 Exercise 17.
Proof.
(a) Assume that ϕ1 ∈ H p for some p > 0. Since ϕ1 has no zero in U and nonconstant, Theorem
17.10 (The Riesz Factorization Theorem) implies that there is a zero-free function f ∈ H 2
such that
2
1
= f p.
ϕ
Therefore,
1
p
ϕ2
= f is in H 2 . Let
f (z) =
∞
X
an z n .
n=0
By the Parseval Theorem [62, Eqn. (6), p. 91] and also the proof of Theorem 17.12, we
get
Z
Z
Z
∞
X
1
1
2
2
|an | = lim
|fr | dσ = lim
dσ =
dσ.
p
∗ p
r→1 T
r→1 T |ϕ|
T |ϕ |
n=0
Since ϕ is an inner function in U , Definition 17.14 implies that |ϕ∗ | = 1 a.e. on T and
then
∞
X
|an |2 = 1.
n=0
In particular, we have |f (0)| = |a0 | ≤ 1 or equivalently, |ϕ(0)| ≥ 1. By Theorem 17.15,
every inner function M satisfies |M (z)| ≤ 1 in U . Combining this fact and Theorem
10.24 (The Maximum Modulus Theorem), we establish that ϕ is constant, a contradiction.
Consequently, ϕ1 ∈
/ H p for all p > 0.
17.3. Factorization of f ∈ H p
219
(b) By the hypotheses, we have
ϕ(z) = c exp
n
−
Z
T
o
eit + z
dµ(t)
,
eit − z
where |c| = 1, µ is a finite positive Borel measure on T , and µ ⊥ m. By Theorem 17.15
and Theorem 10.24 (The Maximum Modulus Theorem), we know that log |ϕ| is always
negative, i.e., 0 < |ϕ(z)| < 1 for every z ∈ U . Since ϕ ∈ H(U ) and ϕ(z) 6= 0 for all z ∈ U ,
it follows from Problem 11.5 that log |ϕ| is harmonic in U . Recall from [62, Eqn. (2),
§11.5, p. 233] that
u(z) = − log |ϕ(z)| =
Z
Re
T
eit + z eit − z
dµ(t) =
Z
P (z, eit ) dµ(t)
T
which means u = P [ dµ]. Since µ ⊥ m, it follows from Problem 11.19 that u(reiθ ) → ∞
a.e. [µ]. Consequently, there exists an eiθ ∈ T such that
lim ϕ(reiθ ) = 0.
r→1
We have completed the analysis of the problem.
Problem 17.18
Rudin Chapter 17 Exercise 18.
Proof. Suppose that
ϕα (z) =
ϕ(z) − α
1 − αϕ(z)
for every z ∈ U . It is clear that ϕα has no zero in U because α ∈
/ ϕ(U ). If we can show that ϕα
is nonconstant and inner, then it follows directly from Problem 17.17(b) that there is at least
one eiθ ∈ T such that ϕα (reiθ ) → 0 as r → 1. Equivalently, it means that
lim ϕ(reiθ ) = α.
r→1
(17.28)
To this end, we directly apply the following result from [68, p. 323]:
Lemma 17.2
If ψ and ϕ are inner functions in U , then ψ ◦ ϕ is also inner.
z−α
Since ψα (z) = 1−αz
is clearly inner and ϕα = ψα ◦ ϕ, we follow from Lemma 17.1 that ϕα is
also inner. Now ϕα is nonconstant because ϕ is also nonconstant. Therefore, we conclude that
the result (17.28) holds and we have completed the proof of the problem.
Problem 17.19
Rudin Chapter 17 Exercise 19.
Chapter 17. H p -Spaces
220
Proof. Let g = f1 . Since f, g ∈ H 1 and f, g are not identically 0, we deduce from Theorem 17.17
(The Canonical Factorization Theorem) that
f = Mf Qf
and g = Mg Qg
which imply that 1 = (Mf Mg )(Qf Qg ). By Definition 17.14 and Theorem 17.15, finite products
of inner functions and outer functions remain inner and outer respectively. Thus we can write
1 = MQ
(17.29)
for some inner and outer functions M and Q, where M has no zero in U .
We claim that this factorization (17.29) is unique up to a constant of modulus 1. Suppose
that we have 1 = M1 Q1 = M2 Q2 . By Theorem 17.15, M1 and M2 can be expressed in the
form [58, Eqn. (1), p. 342] which gives |M1 (z)| = |M2 (z)| = 1 on T . Therefore, we also have
|Q1 (z)| = |Q2 (z)| = 1 on T . Since
Q1
M2
=
Q2
M1
Q1
Q2
Q2 and Q1
Q1 (z)
| =
Since | Q
2 (z)
both
Theorem) that
and
Q2
M1
=
,
Q1
M2
are inner functions without zero in U . In other words, we have
Q2 (z)
|Q
| =
1 (z)
Q1 (z)
| Q2 (z) | ≤ 1
Q1 Q2
Q2 , Q1
∈ H(U ).
1 on T , it follows from Theorem 10.24 (The Maximum Modulus
2 (z)
and | Q
Q1 (z) | ≤ 1 in U which imply that
Q1 (z) = cQ2 (z)
in U for some constant c with |c| = 1. Since Q is unique up to a constant of modulus 1, M is
also unique up to a constant of modulus 1 in the factorization (17.29), as required.
By the definition of g and Theorem 17.17 (The Canonical Factorization Theorem), we know
that Qg = Q1f . Consequently, this fact and the above claim show immediately that Qf Qg =
Mf Mg = 1. By Theorem 17.15 again, Mf Mg = 1 implies that Mf = 1 and hence f = Qf ,
completing the proof of the problem.
Problem 17.20
Rudin Chapter 17 Exercise 20.
Proof. Given ǫ > 0. Define fǫ (z) = f (z) + ǫ in U . Then
Clearly, Problem 17.18 implies that fǫ = Qfǫ , i.e.,
1
fǫ
is bounded in U so that
o
n 1 Z π eit + z
∗ it
log
(f
)
(e
)
dt
fǫ (z) = c exp
ǫ
2π −π eit − z
1
fǫ
∈ H 1.
(17.30)
for some constant c with |c| = 1. By the definition, the functions log |(fǫ )∗ | decrease to log |f ∗ |
as ǫ → 0. Next, we know from Theorem 17.17 (The Canonical Factorization Theorem) that
log |f ∗ | ∈ L1 (T ). Finally, we apply Problem 1.7 to the expression (17.30) to conclude that
o
n 1 Z π eit + z
∗ it
log
|f
(e
)|
dt
= Qf (z)
f (z) = c exp
2π −π eit − z
for all z ∈ U . This completes the proof of the problem.
17.3. Factorization of f ∈ H p
221
Problem 17.21
Rudin Chapter 17 Exercise 21.
Proof. Suppose first that f = hg , where g, h ∈ H ∞ . There is no loss of generality to assume that
|g(z)| ≤ 1 and |h(z)| ≤ 1. By the definition of log+ , it is easy to see that
Z π
Z π
Z π
+
it
+
log |h(reit )| dt
(17.31)
log |f (re )| dt ≤
log |fr | dt =
−π
−π
−π
for 0 ≤ r < 1. Since h(z) 6= 0 in U , Theorem 15.18 (Jensen’s Formula) gives
Z π
1
log |h(reit )| dt = log |h(0)|.
2π −π
(17.32)
Combining the inequality (17.31) and the result (17.32), we conclude that f ∈ N .
Conversely, let f ∈ N and f 6≡ 0. According to Theorem 17.9, we may assume that f has no
zero in U . By Problem 11.5, u = log |f | is harmonic in U . Thus the mean value property gives
Z
1
u(r) dt
u(0) =
2π T
Z
1
log |fr (eit )| dt
=
2π T
Z
Z
1
1
+
it
log |fr (e )| dt −
log− |fr (eit )| dt
=
2π T
2π T
Z
1
log− |fr (eit )| dt
(17.33)
≤ log kf k0 −
2π T
for all 0 ≤ r < 1. Since f ∈ N and the left-hand side of the equation (17.33) is independent of
r, we know immediately that
Z
1
sup
log− |fr (eit )| dt < ∞
2π
0≤r<1
T
so that
sup kur k1 = sup log |f | 1
r<1
Z
1
= sup
log |fr (eit )| dt
2π
0≤r<1
T
Z
o
n 1 Z
1
+
it
log |fr (e )| dt +
log− |fr (eit )| dt
= sup
2π T
0≤r<1 2π T
0≤r<1
<∞
By Theorem 11.30(a), there is a unique complex Borel measure µ on T such that
Z
P (z, eit ) dµ(eit ).
u(z) = P [ dµ] =
(17.34)
T
By the definition, u is real, so µ is in fact real. Using Theorem 6.14 (The Hahn Decomposition
Theorem), we can write µ = µ+ −µ− . Put u± = P [ dµ± ]. Then both u+ and u− are nonnegative
harmonic functions in U .
Chapter 17. H p -Spaces
222
Since U is simply connected and f has no zero in U , it follows from Theorem 13.11 that
f = eg for some g ∈ H(U ). We may assume that g(0) = log |f (0)| so that g is unique. Recall
that u = log |f |, so we have u = Re g and then the expression (17.34) implies that
Z it
e +z
g(z) =
dµ(eit ).
it
T e −z
If we denote
±
g (z) =
Z
T
then we have g =
g+
−
g−
and
Re g ±
eit + z ± it
dµ (e ),
eit − z
= u± ≥ 0. Suppose that
g1 = e−g
−
+
and g2 = e−g .
Clearly, g1 , g2 ∈ H(U ) and they have no zero in U . In addition, we have
−
|g1 | = e−Re g = e−u− ≤ 1.
Similarly, we have |g2 | = e−u+ ≤ 1. These mean that g1 , g2 ∈ H ∞ and furthermore,
g1
exp(−g− )
=
= exp(g+ − g − ) = eg = f.
g2
exp(−g+ )
This completes the proof of the problem.
Problem 17.22
Rudin Chapter 17 Exercise 22.
Proof. Since log+ t ≤ | log t| for every 0 < t < ∞, the hypothesis gives
Z π
lim
log+ |f (reiθ )| dθ = 0.
r→1 −π
(17.35)
Since f ∈ H(U ), Theorem 17.3 says that log+ |f | is subharmonic in U . Since log+ |f | is continuous and nonnegative in U , it follows from Theorem 17.5 that
Z π
Z π
+
iθ
log+ |f (r2 eiθ )| dθ
(17.36)
log |f (r1 e )| dθ ≤
0≤
−π
−π
if 0 ≤ r1 < r2 < 1. Combining the limit (17.35) and the inequality (17.36), we deduce that
Z π
log+ |f (reiθ )| dθ = 0
−π
for every 0 ≤ r < 1. In other words, we have log+ |f (z)| = 0 and so |f (z)| ≤ 1 for every z ∈ U ,
i.e., f ∈ H ∞ .
Recall from Definition 17.7 that H ∞ ⊆ N . The hypothesis yields that f ≡ 0, so Theorem
17.9 asserts that
f = Bg,
(17.37)
where B is the Blaschke product formed with the zeros of f and g ∈ N . Clearly, we always have
|g(z)| ≤ 1 in U . Furthermore, we see from the representation (17.37) that
Z π
Z π
Z π
1
iθ
iθ
0 = lim
log |f (re )| dθ = lim
log |g(re )| dθ = lim
log
dθ.
r→1 −π
r→1 −π
r→1 −π
|g(reiθ )|
1
| ≤ 1 in U so that |g(z)| = 1 in U . This
Therefore, the previous paragraph implies that | g(z)
means that g is a constant of modulus 1 and then f is a Blaschke product. We have completed
the analysis of the problem.
17.3. Factorization of f ∈ H p
223
Problem 17.23
Rudin Chapter 17 Exercise 23.
Proof. Let M1 6= 0 and M2 6= 0 be two closed S-invariant subspaces of H 2 . By Theorem 17.21
(Beurling’s Theorem), there exist inner functions ϕ1 and ϕ2 such that
M1 = ϕ1 H 2
and M2 = ϕ2 H 2 .
By Theorem 17.15, we have
ϕ1 (z) = c1 B1 (z) exp
and
ϕ2 (z) = c2 B2 (z) exp
n
n
−
Z
−
Z
π
−π
π
−π
o
eit + z
dµ
(t)
1
eit − z
(17.38)
o
eit + z
dµ
(t)
2
eit − z
(17.39)
where c1 and c2 are constants such that |c1 | = |c2 | = 1, B1 and B2 are Blaschke products, and
µ1 and µ2 are finite positive Borel measures on T such that µ1 , µ2 ⊥ m.
ϕ1
is an inner function. This is
We claim that M1 ⊆ M2 if and only if the quotient ϕ = ϕ
2
equivalent to saying that every zero of B2 is also a zero of B1 with at least the same multiplicity
and µ2 (E) ≤ µ1 (E) for every Borel subset E of T .
To prove the claim, we first suppose that every zero of B2 is also a zero of B1 with at least
B1
the same multiplicity and µ2 (E) ≤ µ2 (E) for every Borel subset E of T . Clearly, B
is another
2
Blaschke product B. Next, it follows from the representations (17.38) and (17.39) that
n Z π eit + z
o
c1
ϕ1 (z)
.
= B(z) exp −
d
µ
(t)
−
µ
(t)
ϕ(z) =
1
2
it
ϕ2 (z)
c2
−π e − z
Define the measure µ on T by µ(E) = µ1 (E) − µ2 (E) for every Borel subset E of T . Since
µ2 (E) ≤ µ1 (E), µ is a finite positive measure. Recall that µ1 , µ2 ⊥ m, so Proposition 6.8 gives
µ ⊥ m. By Theorem 17.15, the function ϕ is inner.
Conversely, let ϕ1 H 2 ⊆ ϕ2 H 2 . Since 1 ∈ H 2 , it follows that ϕ1 ∈ ϕ2 H 2 and then ϕ1 = ϕ2 h
for some h ∈ H 2 . Since |ϕ∗1 | = |ϕ∗2 | = 1 a.e. on T , we have |h∗ | = 1 a.e. on T . Consequently,
h∗ ∈ L∞ (T ). By Problem 17.6, we know that H 2 ⊂ H 1 . Then we deduce from Problem 17.8
that h ∈ H ∞ . By Definition 17.14, h is an inner function, so we may write
n Z π eit + z
o
h(z) = cB(z) exp −
dµ(t)
,
it
−π e − z
where c, B and µ satisfy the conditions of Theorem 17.15. Then we have
B1 (z) = B2 (z)B(z)
and
exp
n
−
Z
π
−π
o
n Z π eit + z
o
eit + z
.
dµ
(t)
=
exp
−
d
µ(t)
+
µ
(t)
1
2
it
eit − z
−π e − z
(17.40)
(17.41)
Now the relation (17.40) implies that every zero of B2 is also a zero of B1 with at least the same
multiplicity. Next, the expression (17.41) implies that there exists a real number C such that
Z π it
Z π it
e +z
e +z
dµ
(t)
=
d µ(t) + µ2 (t) + iC
1
it
it
−π e − z
−π e − z
Chapter 17. H p -Spaces
224
for all z ∈ U . Put z = 0 and recall that the measures are finite positive, we get C = 0. Recall
from the formula following [62, Eqn. (8), p. 111] that
∞
X
eit + z
(ze−it )n ,
=
1
+
2
eit − z
n=1
where z ∈ U . Hence we obtain
Z
Z π
∞
X
n −int
1+2
dµ1 (t) =
z e
Z
−π
π
dµ1 (t) + 2
−π
n=1
∞ hZ π
X
i
e−int dµ1 (t) z n =
−π
n=1
π
−π
Z
π
−π
+2
(17.42)
∞
X
1+2
z n e−int d µ(t) + µ2 (t)
n=1
d µ(t) + µ2 (t)
∞ hZ
X
n=1
π
−π
i
e−int d µ(t) + µ2 (t) z n .
It is easy to see that the power series converge in U , so they are holomorphic in U by Theorem
10.6. Since the power series representation of any f ∈ H(U ) is unique, we have
Z π
Z π
d µ(t) + µ2 (t)
(17.43)
dµ1 (t) =
−π
−π
and
Z
π
−int
e
dµ1 (t) =
Z
π
−π
−π
e−int d µ(t) + µ2 (t)
(17.44)
for every n ∈ N. Since all the measures are positive, it yields from taking complex conjugates
to the expression (17.44) that it also holds for all nonzero integers n. Combining this with the
expression (17.43), we may conclude that
Z π
eint d[µ(t) + µ2 (t) − µ1 (t)] = 0
−π
for every n ∈ Z. By Theorem 17.13 (The F. and M. Riesz Theorem), we have µ + µ2 − µ1 ≪ m.
We know that µ + µ2 − µ1 ⊥ m, so Proposition 6.8 implies that µ + µ2 − µ1 = 0 or µ1 = µ + µ2 .
This means that
µ2 (E) ≤ µ1 (E)
for every Borel subset E of T , as desired. This completes the analysis of the problem.
17.4
A Projection of Lp onto H p
Problem 17.24
Rudin Chapter 17 Exercise 24.
Proof. By the explanation in §17.24, Theorem 17.26 (M. Riesz’s Theorem) is equivalent to saying
thath
sup kvr kp < ∞.
0≤r<1
h
See also [21,Z Theorem 4.1, p. 54]. More precisely, it says that if u is harmonic in U and it satisfies
π
1
kukp = sup
|u(reiθ )|p dθ < ∞, then there exists a constant Ap , depending only on p, such that
0≤r<1 2π −π
kvkp ≤ Ap kukp ,
where v is the harmonic conjugate of u.
17.4. A Projection of Lp onto H p
225
Now we consider the function
f (z) = i log
1+z
= u(z) + iv(z)
1−z
which maps U conformally onto the vertical strip {z ∈ C | − π2 < Re z < π2 }. In other words, u
is bounded, but v is not so that M. Riesz’s Theorem fails in the case p = ∞.
For p = 1, recall the facts [62, Eqn. (2) & (3), §11.5, p. 233] that
Z π
1 + z 1
Pr (θ) = Re
and
sup kPr (θ)k1 =
Pr (θ) dθ < ∞.
1−z
2π −π
0≤r<1
Denote f (z) =
1+z
1−z .
kf k1 = kf ∗ k1 =
If f ∈ H 1 , then Theorem 17.11 gives
1
2π
Z
π
−π
1
1 + eit
dt =
it
1−e
2π
Z
π
−π
dt
1
t = π
| tan 2 |
Z
π
0
t iπ
dt
2h
=∞
ln
sin
=
π
2 0
tan 2t
which is impossible. Hence we have f ∈
/ H 1 and Theorem 17.26 (M. Riesz’s Theorem) also fails
in the case. This completes the proof of the problem.
Remark 17.1
Although Theorem 17.26 (M. Riesz’s Theorem) fails to say that sup kvr k1 < ∞, it is true
0≤r<1
that sup kvr kp < ∞ for all p < 1. In fact, this is the content of the so-called Kolmogorov’s
0≤r<1
Theorem, see [21, Theorem 4.2, p. 57].
Problem 17.25
Rudin Chapter 17 Exercise 25.
Proof.
(a) Let z ∈ U . Note that
∞
1
1 X z n
=
,
eit − z
eit n=0 eit
so we obtain
Z
∞
eit + z X z n it
f (e ) dt
eit n=0 eit
−π
Z π
Z π
∞ h
∞ h
i
i
X
X
1
1
it −int
n
=
f (e )e
dt · z +
f (eit )e−int dt · z n
2π −π
2π −π
(ψf )(z) =
=
=
1
2π
n=0
∞
X
n=0
∞
X
n=0
Define
π
n=1
fb(n)z n +
∞
X
n=1
fb(n)z n
2fb(n)z n − fb(0).
(17.45)
∞
(ψf )(z) + fb(0) X b
=
f (n)z n .
F (z) =
2
n=0
Chapter 17. H p -Spaces
226
Now Theorem 17.26 (M. Riesz’s Theorem) implies that F ∈ H p . By Theorem 17.11,
g = F ∗ ∈ Lp (T ). Direct computation gives
Z π
1
gb(n) =
g(eit )e−int dt
2π −π
Z π
1
F ∗ (eit )e−int dt
=
2π −π
Z π
∞
X
1
b
ei(m−n)t dt
=
f (m) ·
2π −π
m=0

 fb(n), for all n ≥ 0;
=

0,
for all n < 0.
Next, we notice that Theorem 3.5 (Hölder’s Inequality) gives
Z π
1
b
|f (n)| =
f (eit )e−int dt ≤ kf k1 ≤ kf kp
2π −π
(17.46)
for every n ∈ Z. Finally, we obtain from the inequality (17.46), Theorem 17.11 and
Theorem 17.26 (M. Riesz’s Theorem) that
kgkp = kF ∗ kp = kF kp ≤
|fb(0)| kψf kp
kf kp
+
≤
+ Ap kf kp = Cp kf kp ,
2
2
2
where Cp = 21 + Ap which is a constant depending only on p. In addition, this means that
the mapping Φ : Lp (T ) → Lp (T ) defined by
Φ(f ) = g
is a bounded linear projection.
(b) Let k > 0. It follows from the representation (17.45) that we may define
1
(ψf )(z) + fb(0) − 2fb(0) − 2fb(1)z − · · · − 2fb(k − 1)z k−1
2
∞
X
=
fb(n)z n .
F (z) =
n=k
Similarly, we have F ∈ H p , g = F ∗ ∈ Lp (T ) and then

Z π
∞
 fb(n), for all n ≥ k;
X
1
i(m−n)t
b
gb(n) =
f (m) ·
e
dt =

2π −π
m=k
0,
for all n < k.
(17.47)
Finally, if we combine the inequality (17.46), Theorem 17.11 and Theorem 17.26 (M.
Riesz’s Theorem), then we get
kgkp = kF kp
≤ Ap kf kp +
≤ Ap kf kp +
k−1
|fb(0)| X b
|f (n)|
+
2
kf kp
+
2
n=0
k−1
X
n=0
kf kp
17.4. A Projection of Lp onto H p
227
= Cp,k · kf kp ,
where Cp,k = Ap + k + 21 , a constant depending only on p and k.
Similarly, if k < 0, then we define
1
(ψf )(z) + fb(0) + 2fb(−1)z −1 + 2fb(−2)z −2 + · · · + 2fb(k)z k
2
∞
X
fb(n)z n .
=
F (z) =
n=k
Now if we define g = F ∗ , then it is easy to check that g satisfies the condition (17.47) so
that
k
|fb(0)| X b
kgkp ≤ Ap kf kp +
|f (n)| = Cp,k · kf kp ,
+
2
n=0
where Cp,k = Ap + k +
3
2
in this case.
(c) We have to show that there exists a constant M > 0 such that ksn kp ≤ M for n = 1, 2, . . ..
Recall that
n
X
fb(k)eikt = g1 (t) − g2 (t),
sn (t) =
k=−n
where
g1 (t) =
∞
X
k=−n
fb(k)eint
and g2 (t) =
∞
X
k=n+1
fb(k)eint .
Using part (b), there exists a constant Cp , depending only on p, such that
kg1 kp ≤ Cp kf kp
and
kg2 kp ≤ Cp kf kp
so that
ksn kp ≤ kg1 kp + kg2 kp ≤ 2Cp kf kp
(17.48)
for all n ∈ N. Since f ∈ Lp (T ), we may take M = 2Cp kf kp and we are done.
Given ǫ > 0. Recall from Theorems 3.14 and 4.25 (Weierstrass Theorem) that there exists
a trigonometric polynomial g such that
kf − gkp <
ǫ
.
1 + 2Cp
(17.49)
Clearly, if n ≥ deg g, then we have
sn (g) =
n
X
−n
g(n)eint = g(t)
b
(17.50)
for every t ∈ T . By Theorem 3.5 (Minkowski’s Inequality), we see that
ksn (f ) − f kp ≤ kf − gkp + kg − sn (g)kp + ksn (g) − sn (f )kp .
For sufficiently large enough n, we get from the inequalities (17.48), (17.49) and the expression (17.50) that
ksn (f ) − f kp <
ǫ
ǫ
+ ksn (f − g)kp ≤
+ 2Cp kf − gkp < ǫ
1 + 2Cp
1 + 2Cp
which implies the desired result
lim kf − sn (f )kp = 0.
n→∞
Chapter 17. H p -Spaces
228
(d) Suppose that f ∈ Lp (T ) and set fb(n) = 0 for all n < 0. Using the estimate (17.46), we
know that
1
1
lim sup |fb(n)| n ≤ lim sup kf kpn = 1
n→∞
n→∞
which implies that F ∈ H D(0; R) for some R ≥ 1, where F is the function defined in
part (a). In fact, F is the Poisson integral of f because
Z π
1
Pr (θ − t)f (eit ) dt
P [f ](z) =
2π −π
Z π
∞
X
1
|n|
r ×
=
f (eit )ein(θ−t) dt
2π
−π
n=−∞
=
=
∞
X
n=0
∞
X
n=0
fb(n)r n einθ
fb(n)(reiθ )n
= F (z).
Thus it follows from Theorem 11.16 that kFr kp ≤ kf kp for all 0 ≤ r < 1. By Definition
17.7, we conclude that F ∈ H p .
Conversely, we suppose that F ∈ H p . By Theorem 17.11, we have f = F ∗ ∈ Lp (T ). Since
1 < p < ∞, we have H p ⊂ H 1 so that F is the Cauchy integral of f , i.e.,
Z
π
f (eiθ )
dθ
−iθ z
−π 1 − e
Z π X
∞
1
=
e−inθ z n f (eiθ ) dθ
2π −π n=0
Z π
∞ h
i
X
1
=
f (eiθ )e−inθ dθ z n
2π −π
1
F (z) =
2π
=
n=0
∞
X
n=0
We end the proof of the problem.
fb(n)z n .
Remark 17.2
(a) Problem 17.25(a) can be treated as an equivalent form of Theorem 17.26 (M. Riesz
Theorem), see the first paragraph of the proof of the theorem in [31, p. 152].
(b) We may regard H p as a subspace of Lp (T ) by Theorem 17.11(d), where 1 ≤ p ≤ ∞.
In addition, H p is closed in Lp (T ). To see this, let f be a limit point of H p . Given
ǫ > 0. There exists an N ∈ N such that kfn − f kp < 2ǫ , where fn ∈ H p . For m, n ≥ N ,
the inequality
kfm − fn kp ≤ kfm − f kp + kf − fn kp < ǫ
implies that {fn } is Cauchy. Note that H p is Banach by Remark 17.8(c), so f ∈ H p
and then H p is closed in Lp (T ).
17.4. A Projection of Lp onto H p
229
Problem 17.26
Rudin Chapter 17 Exercise 26.
Proof. Take p = 2. Denote H = ψh which is holomorphic in U . Therefore, we follow from the
expression (17.42) that
1
H(z) =
2π
1
=
2π
Z
π
−π
Z
π
n=1
z n e−int h(eit ) dt
it
h(e ) dt + 2
∞
X
n=1
∞
X
n=1
H(re ) = b
h(0) +
iθ
where
1+2
∞
X
−π
=b
h(0) +
and then
h 1 Z π
i
z
h(eit )e−int dt
2π −π
n
2b
h(n)z n
∞
X
n=1
2b
h(n)(reiθ )n =
∞
X
n=−∞
iθ n
b
H(n)(re
) ,
(17.51)
 b
2h(n), if n = 1, 2, . . .;





b
b
H(n)
=
h(0)
if n = 0;





0,
if n = −1, −2, . . ..
Using the Parseval Theorem [62, Eqn. (6), p. 91] and the Fourier series (17.51), we see that
kHk22
Z π
1
=
|H(reiθ )|2 dθ
2π −π
∞
X
2
b
|H(n)|
=
n=−∞
= |b
h(0)|2 +
≤4
∞
X
n=−∞
1
=4·
2π
Z
∞
X
n=1
|2b
h(n)|2
|b
h(n)|2 + |b
h(0)|2
π
−π
(17.52)
|h(eit )|2 dt + |b
h(0)|2
= 4khk22 + |b
h(0)|2 .
When b
h(n) = 0 for n = 0, −1, −2, . . .,i the equality (17.52) holds and we get kHk2 = 2khk2 .
Hence the best possible value of A2 is 2 and we complete the proof of the problem.
i
This is equivalent to
for n = 0, −1, −2, . . ..
Z
π
h(eiθ )e−inθ dθ = 0
−π
Chapter 17. H p -Spaces
230
17.5
Miscellaneous Problems
Problem 17.27
Rudin Chapter 17 Exercise 27.
Proof. Define g : U → C by
g(z) =
∞
X
n=0
|an |z n .
By Theorem 10.6, we have f, g ∈ H(U ) so that
∞
X
′
|f (z)| =
nan z
n−1
n=1
≤
∞
X
n=1
n|an | · |z|n−1 = g ′ (|z|)
and thus for every x ∈ (0, 1), we have
Z
x
′
iθ
|f (re )| dr ≤
0
Z
x
0
g′ (r) dr = g(x) − g(0) ≤
∞
X
n=0
|an | < ∞.
Since this is true for every 0 < x < 1, we actually have
Z 1
|f ′ (reiθ )| dr < ∞,
0
completing the proof of the problem.
Problem 17.28
Rudin Chapter 17 Exercise 28.
Proof. Suppose that {nk } is a sequence of positive integers.
• |f ′ (z)| >
and
nk
10k
for all 1 −
1
nk
< |z| < 1 −
1
2nk .
Let n1 be a sufficiently large positive integer
√
nk > 4k(k − 1) e · exp(nk−1 )
for k ≥ 2. Then we have nk > k which implies nk > k >
1
1
n
nk k
1
<
1
k nk
1
k
>
(17.53)
1
nk
and
1
n
< nk k .
Thus the radius of convergence of the power series will be 1 and Theorem 10.6 implies that
′
f (z) =
∞
X
nk z nk −1
k=1
k
=
k−1
X
np z np −1
p=1
p
∞
X
np z np −1
nk z nk −1
+
+
k
p
p=k+1
which implies
|f ′ (z)| ≥
k−1
X
nk |z|nk −1
np z np −1
−
−
k
p
p=1
∞
X
np z np −1
p
p=k+1
(17.54)
17.5. Miscellaneous Problems
for all z satisfying 1 −
1
nk
231
< |z| < 1 −
1
2nk .
Since nk is sufficiently large enough, we have
nk |z|nk −1
1 nk −1 nk
nk 1−
≈
>
k
k
nk
ek
(17.55)
and the property (17.53) shows that
k−1
X
np z np −1
p
p=1
≤
≤
≤
k−1
X
p=1
np |z|np −1
k−1
X
np p=1
p
k−1
X
np h
p=1
1 np −1
2np
1−
p
1−
1 2np −2 i 12
2np
k−1
X
n
√p
≈
ep
p=1
(k − 1)nk−1
√
e
nk
.
<
4ek
≤
(17.56)
If p ≥ k + 1, then np − 1 = exp(exp · · · exp nk ) − 1. Therefore, we obtain
|
{z
}
(p − k) iterations
np −1
∞
∞
X
X
1 2nk i 2nk
np h
np z np −1
1−
≤
p
p
2nk
p=k+1
≈
p=k+1
∞
X
p=k+1
np
p exp(
np −1 .
2nk )
(17.57)
Besides the property (17.53), we require that the sequence {nk } satisfies
exp
np − 1 4eknp
> 2p−k ×
2nk
pnk
for every p ≥ k + 1. Then the estimate (17.57) becomes
∞
∞
X
np z np −1
nk
nk X 1
=
.
<
p
4ek
2p−k
4ek
p=k+1
(17.58)
p=k+1
Substituting the estimates (17.55), (17.56) and (17.58) into the inequality (17.54), we get
|f ′ (z)| >
in 1 −
1
nk
< |z| < 1 −
nk
nk
nk
nk
nk
−
−
=
>
ek
4ek 4ek
2ek
10k
1
2nk .
• The divergence of the integral. It is clear from the first assertion that
Z
1
0
|f ′ (reiθ )| dr ≥
∞ Z
X
k=1
1− 2n1
k
1− n1
k
|f ′ (reiθ )| dr
Chapter 17. H p -Spaces
232
≥
∞ Z
X
k=1
1− 2n1
k
1− n1
k
nk
dr
10k
∞
X
nk 1
1 =
−
10k nk 2nk
k=1
∞
X
1
=
20k
k=1
for every θ. Hence we have
Z
for every θ.
1
0
|f ′ (reiθ )| dr = ∞
(17.59)
• The convergence of the limit. Let 0 < R < 1. By the definition of f , we have
Z RX
Z R
∞
nk r nk −1 i(nk −1)θ
′
iθ
f (re ) dr =
e
dr
k
0 k=1
0
Z
∞
X
nk R nk −1 i(nk −1)θ
r
dr e
=
k
0
k=1
=
∞
X
R nk
k=1
−iθ
=e
As
k
ei(nk −1)θ
f (Reiθ ).
(17.60)
∞
X
1
< ∞, Theorem 17.12 implies that f ∈ H 2 and then we follow from Theorem
k2
k=1
17.11 that f ∗ (eiθ ) exists a.e. on T . Thus we deduce from the expression (17.60) that
Z R
f ′ (reiθ ) dr = e−iθ f ∗ (eiθ )
lim
R→1 0
exists for almost all θ.
• The geometrical meaning of the integral (17.59). If f ∈ H(U ), we denote
Z 1
V (f ; θ) =
|f ′ (reiθ )| dr
0
which is the total variation of f on the radius of U terminating at the point eiθ . Hence
the result (17.59) tells us that the length of the curve which is the image of the radius
with angle θ under f is always infinite.
We complete the proof of the problem.
Problem 17.29
Rudin Chapter 17 Exercise 29.
Proof. Let g ∈ Lp (T ). Suppose that g(eit ) = f ∗ (eit ) a.e. for some f ∈ H p . Let
f (z) =
∞
X
n=0
an z n
17.5. Miscellaneous Problems
233
and fc∗ (n) be the Fourier coefficients of its boundary function f ∗ (eit ), i.e.,
Z π
1
∗
c
f (n) =
e−int f ∗ (eit ) dt
2π −π
for all n ∈ Z. The Taylor coefficients of f can be expressed in the form
Z π
Z π
1
f (reit )
1
an =
dt
=
r −n e−int f (reit ) dt
2π −π r n eint
2π −π
(17.61)
(17.62)
for every 0 < r < 1. Combining the coefficients (17.61) and (17.62), we get
Z π
Z π
1
1
n
−int
it
∗ it
∗
c
r an − f (n) =
e
f (reit ) − f ∗ (eit ) dt
f (re ) − f (e ) dt ≤
2π −π
2π −π
for every n ∈ Z. Applying Theorem 17.11, we obtain
as r → 1. Therefore, we have
r n an − fc∗ (n) ≤ kfr − f ∗ k1 → 0
fc∗ (n) =

 an , if n ≥ 0;

0,
if n < 0.
By the hypothesis, we conclude that
Z π
Z π
1
1
−int
it
e
g(e ) dt =
e−int f ∗ (eit ) dt = 0
2π −π
2π −π
(17.63)
for all negative integers n.
Conversely, we suppose that the formula (17.63) holds for all negative integers n. In other
words, gb(n) = 0 for all negative integers n. Let
Z π
1
iθ
Pr (θ − t)g(eit ) dt,
(17.64)
f (re ) =
2π −π
where 0 < r < 1. Since the Poisson kernel has the expansion
Pr (t) = 1 +
∞
X
r n (eint + e−int ),
n=1
it follows from the formula (17.63) that
Z π
1
f (reiθ ) =
Pr (θ − t)g(eit ) dt
2π −π
Z π
Z π
∞
∞
X
X
1
1
n in(θ−t)
it
r e
g(e ) dt +
e−in(θ−t) g(eit ) dt
=
2π
2π
−π
−π
n=1
n=0
Z
∞
h 1
i
π
X
(reiθ )n
=
e−int g(eit ) dt
2π −π
=
n=0
∞
X
n=0
which means
gb(n)(reiθ )n
f (z) =
∞
X
n=0
g(n)z n
b
Chapter 17. H p -Spaces
234
for every z ∈ U and then f ∈ H(U ). By the integral (17.64) and [62, Eqn. (3), p. 233], it is
easy to see that
Z π
1
kf k1 =
|f (reiθ )| dθ
2π −π
Z πh Z π
i
1
1
Pr (θ − t)|g(eit )| dt dθ
≤
2π −π 2π −π
Z πh Z π
i
1
1
Pr (θ − t) dθ · |g(eit )| dt
=
2π −π 2π −π
Z π
1
=
|g(eit )| dt
2π −π
so that f ∈ H 1 .
By Theorem 17.11, we see that f ∗ ∈ L1 (T ). Now we consider
Z π
1
Φ(z) =
Pr (θ − t)f ∗ (eit ) dt = P [f ∗ ](z).
2π −π
Let z ∈ U . For any fixed 0 < ρ < 1, since f ∈ H(U ), it follows from Theorem 11.4 and the
mean value property that
Z π
1
Pr (θ − t)f (ρeit ) dt.
f (ρz) =
2π −π
We observe from Theorem 17.11 that
Z π
lim
|f (ρeit ) − f ∗ (eit )| dt → 0,
ρ→1 −π
so we obtain
f (z) = lim f (ρz) = Φ(z)
ρ→1
for every z ∈ U , i.e.,
f (reiθ ) =
1
2π
Z
π
−π
Pr (θ − t)f ∗ (eit ) dt.
(17.65)
Comparing the two integrals (17.64) and (17.65), we conclude immediately that g(z) = f ∗ (z)
a.e. on T . Since g ∈ Lp (T ), we know that f ∗ ∈ Lp (T ) and it yields from Problem 17.8 that
f ∈ H p , as required. Now we have completed the proof of the problem.
CHAPTER
18
Elementary Theory of Banach Algebras
18.1
Examples of Banach Spaces and Spectrums
Problem 18.1
Rudin Chapter 18 Exercise 1.
Proof. Let X be a Banach space and
B(X) = {A : X → X | A is linear and bounded with mentioned conditions}.
(18.1)
Denote k · kX to be the norm of the Banach space X. The hypotheses ensure that an associative
and distributive multiplication is well-defined in B(X). For every α ∈ C, A1 , A2 ∈ B(X) and
x ∈ X, we see that
α(A1 A2 )(x) = αA1 (A2 x) = A1 (αA2 x) = (αA1 )(A2 x),
i.e., α(A1 A2 ) = A1 (αA2 ) = (αA1 )A2 . Thus B(X) is a complex algebra.
We check Definition 5.2. By the definition, kAk must be nonnegative. By the definition
(18.1), there exists a positive constant M such that kAxkX ≤ M kxkX for all x ∈ X. Therefore,
we have
kAxkX
≤ M,
kAk = sup
kxkX
i.e., kAk is a real number. Next, for all A1 , A2 ∈ B(X), we apply the fact that X is a Banach,
so we have
k(A1 + A2 )(x)kX
kxkX
kA1 x + A2 xkX
= sup
kxkX
kA1 xkX + kA2 xkX
≤ sup
kxkX
= kA1 k + kA2 k.
kA1 + A2 k = sup
Furthermore, if α ∈ C and A ∈ B(X), then we obtain
kαAk = sup
kAxkX
k(αA)(x)kX
= |α| sup
= |α| · kAk.
kxkX
kxkX
235
236
Chapter 18. Elementary Theory of Banach Algebras
Let kAk = 0. This means that kAxkX = 0 for all x ∈ X so that Ax = 0 for all x ∈ X.
Consequently, it must be the case A = 0 and then B(X) is a normed linear space.
For A1 , A2 ∈ B(X), we see that
k(A1 A2 )(x)kX
kxkX
kA2 xkX
kA1 (A2 x)kX
×
= sup
kA2 xkX
kxkX
kA1 (A2 x)kX
kA2 xkX
≤ sup
× sup
kA2 xkX
kxkX
≤ kA1 k · kA2 k.
kA1 A2 k = sup
Hence B(X) is also a normed complex algebra.
We claim that B(X) is a complete metric space. Fix x ∈ X. Given ǫ > 0. Let {An } be
Cauchy with respect to the norm k · k. Then it suffices to show that there exists an A ∈ B(X)
such that
kAn − Ak → 0
as n → ∞. Since there exists an N ∈ N such that kAm − An k < 1 for all n, m ≥ N . Using the
triangle inequality of the norm k · k, we see that
kAn k < 1 + kAN k
for all n ≥ N . Denote M = max{kA1 k, kA2 k, . . . , kAN −1 k, 1 + kAN k}. Thus we get
kAn k < 1 + kAN k
for all n ∈ N. This means that
kAn xkX ≤ M kxkX
(18.2)
for all x ∈ X and n ∈ N. Now, for all x ∈ X, we can establish
kAm x − An xkX = k(Am − An )xkX ≤ kAm − An k · kxkX → 0
as n, m → ∞. In other words, {An x} is a Cauchy sequence in X. Since X is Banach, the sequence
converges to an element in X, namely Ax. Then we can define the operator A : X → X by
this and the linearity of taking limits implies the linearity of A. Besides, we follow from the
inequality (18.2) that
kAxkX = lim kAn xkX ≤ M kxkX
n→∞
for every x ∈ X. Consequently, it means that A ∈ B(X).
Now it remains to verify that kAm − Ak → 0 as m → ∞. Recall that {An } is Cauchy, so
given ǫ > 0, there exists an N ∈ N such that
kAm − An k ≤ ǫ
whenever n, m ≥ N . Therefore, for every x ∈ X, we have
kAm x − An xkX ≤ kAm − An k · kxkX < ǫ · kxkX
whenever n, m ≥ N . If n → ∞, then we see that
kAm x − AxkX ≤ ǫ · kxkX
for all m ≥ N and all x ∈ X. By the definition, we conclude that kAm − Ak ≤ ǫ for all m ≥ N .
Hence this proves our claim that B(X) is complete and then it is a Banach algebra, completing
the proof of the problem.
18.1. Examples of Banach Spaces and Spectrums
237
Problem 18.2
Rudin Chapter 18 Exercise 2.
Proof. Suppose that we have
X = {v = (z1 , z2 , . . . , zn ) | z1 , z2 , . . . , zn ∈ C} = Cn
is equipped with the norm k · kCn and B(Cn ) is the algebra of all bounded linear operators on
Cn . By Problem 18.1, B(Cn ) is also a Banach algebra with the norm k · k given by
kAk = sup
kAvkCn
.
kvkCn
Our target is to find
σ(A) = {λ ∈ C | A − λI is not invertible}.
According to the explanation in [35, pp. 96, 97], every bounded linear operator A can be
represented by a matrix with entries in C. We also denote this matrix by A. Therefore, A − λI
is not invertible if and only if
det(A − λI) = 0.
(18.3)
Since det(A−λI) = 0 is an equation in λ of order n, the Fundamental Theorem of Algebra ensures
that it has at most n complex roots. Hence σ(A) consists of at most n complex numbers and
they are exactly the solutions of the equation (18.3), completing the proof of the problem. Problem 18.3
Rudin Chapter 18 Exercise 3.
Proof. Suppose that C is a positive constant such that |ϕ(x)| ≤ C a.e. on R. Define the mapping
Mϕ : L2 → L2 by
Mϕ (f ) = ϕ × f.
Of course, it is true that ϕf ∈ L2 , so Mϕ is well-defined. Furthermore, the linearity of Mϕ is
clear. Recall that L2 is a normed linear space with the norm kf k2 . Since we have
Z
|ϕ(x)f (x)|2 dx ≤ Ckf k2 ,
kϕf k2 =
R
so we get
kMϕ k = sup kMϕ (f )k | f ∈ L2 and kf k2 = 1
= sup kϕf k2 | f ∈ L2 and kf k2 = 1
≤ C.
By Definition 5.3, Mϕ is bounded.
For the second assertion, recall from [62, Exercise 19, p. 74] that
Rϕ = λ ∈ C m {x | |ϕ(x) − λ| < ǫ} > 0 for every ǫ > 0 .
(18.4)
Let I be the identity operator on L2 . We are required to prove that
σ(Mϕ ) = {λ ∈ C | Mϕ − λI is not invertible} = Rϕ .
(18.5)
238
Chapter 18. Elementary Theory of Banach Algebras
On the one hand, let λ ∈
/ Rϕ . Then |ϕ(x) − λ| ≥ ǫ for some ǫ > 0 a.e. on R, so we have
1
∞ (R) and this implies that the operator M
is bounded. Furthermore, it is easy to
∈
L
1
ϕ−λ
ϕ−λ
see that
M 1 Mϕ (f ) − λf = M 1 (ϕf − λf ) = f.
ϕ−λ
Thus we have M
1
ϕ−λ
ϕ−λ
is the inverse of Mϕ − λI which means λ ∈
/ σ(Mϕ ).
On the other hand, let λ ∈ Rϕ . For any n ∈ N, we denote
Sn = {x | |ϕ(x) − λ| < 2−n }.
The definition (18.4) reveals that m(Sn ) > 0. Suppose that there exists an N ≥ n such that
0 < m(SN ) < ∞. Otherwise, m(Sn ) = ∞ for all n ≥ 1 and this means that ϕ(x) = λ for almost
all x ∈ R. In this case, we know that Rλ = {λ}. Clearly, Mλ f − λf = 0, so λ ∈ σ(Mλ ). If
µ 6= λ, then Mλ (f ) − µf = (λ − µ)f so that
M 1 Mλ (f ) − µf = f.
λ−µ
Consequently, we obtain σ(Mλ ) = {λ} and then the equality (18.5) holds. Let 0 < m(SN ) < ∞.
Take φn = χSN . Then we have
Z
|ϕ(x) − λ|2 · |φn (x)|2 dx ≤ 2−2n kφn k2
k(Mϕ − λI)(φn )k2 = kϕφn − λφn k2 =
SN
so that the operator (Mϕ − λI)−1 is not bounded, i.e., Mϕ − λI is not invertible. Hence we
conclude that λ ∈ σ(Mϕ ) and we have established the equality (18.5), completing the proof of
the problem.
Problem 18.4
Rudin Chapter 18 Exercise 4.
Proof. Recall that ℓ2 =
given by
n
x = {ξ0 , ξ1 , ξ2 , . . .} kxk =
∞
nX
n=0
|ξn |2
o1
2
< ∞
o
and S : ℓ2 → ℓ2 is
Sx = {0, ξ0 , ξ1 , . . .}
which is a bounded linear operator on ℓ2 and kSk = 1.a We want to determine
σ(S) = {λ ∈ C | S − λI is not invertible}.
Since kSk = 1, Corollary 3 to Theorem 18.4 implies that
σ(S) ⊆ U .
(18.6)
Take 0 < |λ| < 1. Assume that λ ∈
/ σ(S). Then S − λI is invertible so that the equation
(S − λI)x = y
(18.7)
has a unique solution x ∈ ℓ2 for every y ∈ ℓ2 . If y = (1, 0, 0, . . .), then the equation (18.7) takes
the system
−λξ0 = 1 and ξn − λξn+1 = 0
a
In fact, S is called the right-shift operator.
18.2. Properties of Ideals and Homomorphisms
239
for every n = 0, 1, 2, . . .. Solving it, we obtain ξn = −λ−(n+1) for every n ∈ N. Since |λ| < 1, we
have |ξn | = |λ|−(n+1) > and thus x ∈
/ ℓ2 which is a contradiction. Hence we must have λ ∈ σ(S),
i.e.,
{λ ∈ C | 0 < |λ| < 1} ⊆ σ(S).
(18.8)
Finally, we observe from Theorem 18.6 that σ(S) is a closed set, so we conclude from the set
relations (18.6) and (18.8) that
σ(S) = U .
We end the analysis of the problem.
18.2
Properties of Ideals and Homomorphisms
Problem 18.5
Rudin Chapter 18 Exercise 5.
Proof. Let M be an ideal of the commutative complex algebra A. Then M is a vector space and
we note from §4.7 that M is also a vector space. Thus it remains to show that aM ⊆ M and
M a ⊆ M for all a ∈ A. Fix a ∈ A and consider b ∈ M . Then there exists a sequence {bn } ⊆ M
such that bn → b as n → ∞. By considering the sequences {abn } and {bn a}, since M is an ideal,
it is true that
abn ∈ M and bn a ∈ M
for all n = 1, 2, . . .. Since a is fixed, the mapping x →
7 ax and x 7→ xa are both continuous on
M . Therefore, abn → ab and bn a → ba as n → ∞. In other words, it is true that ab, ba ∈ M .
This completes the proof of the problem.
Problem 18.6
Rudin Chapter 18 Exercise 6.
Proof. Let C(X) be the algebra of all continuous complex functions on X with pointwise addition
multiplication and the supremum norm. The constant function 1 is the unit element. Let I be
an ideal in C(X). We claim that either I = C(X) or there exists a p ∈ X such that
I = {f ∈ C(X) | f (p) = 0}.
(18.9)
Assume that, for every p ∈ X, there exists a continuous function f ∈ I such that f (p) 6= 0.
Since X is compact Hausdorff, the continuity of f implies that there exists an open set Vp
containing p such that f (x) 6= 0 for all x ∈ Vp . Therefore, the collection {Vp } forms an open
covering of X. Since X is compact, there must exist a finite subcover. Call this subcover
V1 , V2 , . . . , VN and the corresponding functions f1 , f2 , . . . , fN for some N ∈ N. Define
2
F (x) = f12 (x) + f22 (x) + · · · + fN
(x).
(18.10)
Since fk ∈ I and I is an ideal, it follows from Definition 18.12 that F ∈ I. For every p ∈ X,
we have fk (p) 6= 0 for some k ∈ {1, 2, . . . , N } so that F (p) 6= 0. Since F is continuous on the
compact set X, it must attain a minimum. By the form (18.10), it is trivial that F (x) > 0 for
1
is the inverse of F in C(X). However, we note
all x ∈ X. This implies that F −1 (x) = F (x)
from Definition 18.12 that no proper ideal contains an invertible element, so we have I = C(X).
Consequently, we have obtained our claim.
240
Chapter 18. Elementary Theory of Banach Algebras
If I is maximal, then it has the form (18.9) for some p ∈ X. Assume that I ⊂ J for some
ideal J in C(X), where I 6= J. Then there corresponds an f ∈ J such that f (p) 6= 0. Since f is
continuous, one can find a neighborhood Vp of p such that f (x) 6= 0 for all x ∈ Vp . The point
set {p} is compact by Theorem 2.4. According to Urysohn’s Lemma, there exists an g ∈ C(X)
such that
{p} ≺ g ≺ Vp ,
i.e., g(p) = 1 and g(x) = 0 for all x ∈ X \ Vp . Take h = 1 − g which is also an element of C(X)
and it satisfies h(p) = 0 and h(x) = 1 for all x ∈ X \ Vp . As J has the unit, we have h ∈ J.
Next, we define
H(x) = f 2 (x) + h2 (x)
which is an element of J. Obviously, it is easy to check that H(x) > 0 on X. Since X is compact,
H attains its minimum in X and thus H is bounded from below by a positive number. Therefore,
its inverse H1 belongs to C(X) which asserts that J = C(X) by the previous paragraph. Hence
we have the expected conclusion that the ideal in the form (18.9) is maximal. This completes
the analysis of the problem.
Problem 18.7
Rudin Chapter 18 Exercise 7.
Proof. Let e be the unit of A. Given λ ∈
/ σ(x). Then x − λe is invertible so that (x − λe)−1 ∈ A.
Since A is generated by a single element x, this means that there are polynomials Pn such that
Pn (x) → (x − λe)−1
(18.11)
as n → ∞ in A. If z ∈ σ(x), then Theorem 18.17(b) ensures that h(x) = z for some h ∈ ∆.
Since h is a complex homomorphism of A, we have h(xm ) = z m for every m ∈ Z. By this and
Theorem 18.17(e), we establish that
|Pn (z) − (λ − z)−1 | = h Pn (x) − (λ − x)−1 ≤ kPn (x) − (λ − x)−1 k.
(18.12)
Applying the result (18.11) to the inequality (18.12), we obtain the result that Pn (z) → (λ−z)−1
uniformly on σ(x).
Assume that C \ σ(x) was disconnected. Let Ω be a (non-empty) bounded component of it.
Fix λ ∈ Ω, i.e., λ ∈
/ σ(x). Choose {Pn } as above. For every n ∈ N and z ∈ σ(x), we have
|(z − λ)Pn (z) − 1| = |z − λ| · Pn (z) − (z − λ)−1 ≤ ℓ · Ln ,
(18.13)
where
ℓ = sup |z − λ| and
z∈σ(x)
Ln = sup |Pn (z) − (z − λ)−1 |.
z∈σ(x)
The compactness of σ(x) by Theorem 18.6 asserts that both ℓ and Ln are finite. Furthermore,
∂Ω ⊆ σ(x). Next, we use Theorem 10.24 (The Maximum Modulus Theorem) to see that the
inequality (18.13) also holds on Ω. In particular, we get
|Pn (z) − (z − λ)−1 | ≤
ℓ · Ln
|z − λ|
(18.14)
for all z ∈ Ω \ {λ}. Since Ω is a component, one can find a δ > 0 small enough such that the
circle C(λ; δ) lies in Ω. Therefore, we conclude from the estimate (18.14) that
Z
Z
dz
ℓLn
−1
Pn (z) − (z − λ)
dz ≤ ℓ · Ln
2πi =
=
· 2πδ = 2πℓLn . (18.15)
δ
C(λ;δ) |z − λ|
C(λ;δ)
18.2. Properties of Ideals and Homomorphisms
241
Notice that Ln → 0 as n → ∞, so the inequality (18.15) implies a contradiction. Hence the set
C \ σ(x) must be connected, as required. This completes the proof of the problem.
Problem 18.8
Rudin Chapter 18 Exercise 8.
Proof. Since
∞
X
n=0
that
|cn | < ∞, there exists a positive constant M such that |cn | ≤ M . This implies
1
lim sup |cn | n ≤ 1,
n→∞
i.e., the radius of convergence R of the power series satisfies R ≥ 1. By Theorem 10.6, both f
and then f1 are holomorphic in a region containing U .
Define Cn = cn for all n ≥ 0 and Cn = 0 for all n < ∞. It is clear that
f (eit ) =
∞
X
cn eint =
∞
X
Cn eint
∞
X
and
n=−∞
n=−∞
n=0
|Cn | =
∞
X
n=0
|cn | < ∞.
Now the hypothesis |f (z)| > 0 for every z ∈ U implies that f (eit ) 6= 0 for every real t, so
Theorem 18.21 (Wiener’s Theorem) guarantees that f satisfies
∞
X
1
γn eint
=
f (eit ) n=−∞
Since
1
f
∞
X
and
n=∞
|γn | < ∞.
(18.16)
is holomorphic in a region containing U , we have γn = 0 for all n < 0 and
∞
∞
n=0
n=0
X
X
1
int
γn eint
a
e
=
=
n
f (eit )
which implies immediately that an = γn for all n ≥ 0 by the Corollary following Theorem 10.18.
Hence the second condition (18.16) gives the desired result that
∞
X
n=0
|an | < ∞.
This completes the proof of the problem.
Problem 18.9
Rudin Chapter 18 Exercise 9.
Proof. We note from Example 9.19(d) that we define the multiplication in L1 (R) by convolution.
Let f, g ∈ L1 (R) and φ ∈ L∞ (R). Here we employ the proof of [60, pp. 157, 158]: If I is a
translation invariant subspace of L1 (R), then we say that φ annihilates I if
Z
f (x − y)φ(y) dm(y) = 0
(f ∗ φ)(x) =
R
for all f ∈ I and x ∈ R. On the one hand, we note that
Z
(f ∗ g ∗ φ)(0) = [(f ∗ g) ∗ φ](0) = (f ∗ g)(0 − y)φ(y) dm(y).
R
(18.17)
242
Chapter 18. Elementary Theory of Banach Algebras
On the other hand, recall from [62, Example 9.19(d)] that f ∗ g = g ∗ f , so we have
Z
g(0 − y)(f ∗ φ)(y) dm(y).
(f ∗ g ∗ φ)(0) = [g ∗ (f ∗ φ)](0) =
(18.18)
R
In other words, the two integrals (18.17) and (18.18) are equal, i.e.,
Z
Z
g(0 − y)(f ∗ φ)(y) dm(y).
(f ∗ g)(0 − y)φ(y) dm(y) =
(18.19)
R
R
Let I be a closed translation invariant subspace of the Banach space L1 (R) and φ ∈ L1 (R)
annihilate f ∈ I. Then f ∗ φ = 0 and we see from the right-hand side of the expression (18.19)
that
(f ∗ g) ∗ φ = 0
(18.20)
for every g ∈ L1 (R).b Recall the basic fact from Theorem
6.16 that L1 (R) is isometrically
∗
isomorphic to the dual space of L∞ (R), i.e., L1 (R) ∼
= L∞ (R). By Remark 5.21, every
∞
1
φ ∈ L (R) is a bounded linear functional on L (R). Assume that f ∗ g ∈
/ I. Since I = I, there
corresponds an ϕ ∈ L∞ (R) such that f ∗ ϕ = 0 and (f ∗ g) ∗ ϕ 6= 0 by Theorem 5.19, but this
contradicts the result (18.20). Hence we conclude that f ∗ g ∈ I so that I is an ideal.
Conversely, let I be a closed ideal and f ∗ φ = 0 for all f ∈ I. Assume that F = fx0 ∈
/I=I
for some x0 ∈ R.c By Theorem 5.19, there exists an ϕ ∈ L∞ (R) such that f ∗ ϕ = 0 for all f ∈ I
but
F ∗ ϕ 6= 0.
(18.21)
Now the hypotheses show that we have f ∗ g ∈ I for every g ∈ L1 (R), so the left-hand side of
the expression (18.19) gives f ∗ φ annihilates every g ∈ L1 (R). This implies that f ∗ φ = 0 or
Z
f (x − y)φ(y) dm(y)
0 = (f ∗ φ)(x) =
R
for x ∈ R. In other words, φ annihilates every translate of f . Particularly, this implies that
F ∗ ϕ = 0 which contradicts the result (18.21). Consequently, fx ∈ I for every x ∈ R which
means it is translation invariant. Hence we have completed the analysis of the problem.
Remark 18.1
As [60, Theorem 7.1.2, p. 157] indicates, the result of Problem 18.9 is also valid if we replace
R by any locally compact abelian group. In particular, Problem 18.9 remains true for
the unit circle T .
Problem 18.10
Rudin Chapter 18 Exercise 10.
Proof. We prove the assertions one by one.
• L1 (T ) is a commutative Banach algebra. Suppose that f, g, h ∈ L1 (T ). It is clear
that f ∗ (g + h) = f ∗ g + f ∗ h, (f + g) ∗ h = f ∗ h + g ∗ h and α(f ∗ g) = f ∗ (αg) = (αf ) ∗ g
for every α ∈ C. It also satisfies the associative law by an application of Theorem 8.8 (The
Fubini Theorem)d Thus L1 (T ) is a complex algebra.
By Theorem 8.14, we know that f ∗ g ∈ L1 (R).
Recall that fx0 (y) = f (y − x0 ), see Theorem 9.5.
d
See also [62, Example 9.19(d), p. 190].
b
c
18.2. Properties of Ideals and Homomorphisms
243
Recall from [62, p. 96] that L1 (T ) is a Banach space normed by kf k1 . Using Theorem 8.8
(The Fubini Theorem) again, we know that
Z π
1
kf ∗ gk1 =
|(f ∗ g)(t)| dt
2π −π
Z π
Z π
1
1
=
f (t − s)g(s) ds dt
2π −π 2π −π
Z π
Z π
1
1
≤
|f (t − s)| · |g(s)| ds dt
2π −π 2π −π
Z π Z π
1
1
=
|f (t − s)| dt · |g(s)| ds
2π −π 2π −π
Z π
1
=
kf k1 · |g(s)| ds
2π −π
= kf k1 · kgk1 .
In other words, L1 (T ) is a Banach algebra by Definition 18.1. The definition implies
immediately that f ∗ g = g ∗ f . Hence L1 (T ) is commutative.
• L1 (T ) does not have a unit. Assume that e ∈ L1 (T ) was a unit. Then e∗f = f for every
f ∈ L1 (T ). Using similar argument as in the proof of Theorem 9.2(c) (The Convolution
Theorem), we can show that
b
h(n) = fb(n) · gb(n)
(18.22)
if f, g ∈ L1 (T ). Therefore, we have
eb(n) = 1
for every n ∈ Z, but it contradicts the Riemann-Lebesgue Lemma [62, §5.14, p. 103].
• Complex homomorphisms of L1 (T ). Denote ∆T to be the set of all complex homomorphisms of L1 (T ). Let ϕ ∈ ∆T and ϕ 6= 0. By Theorem 18.17(e), ϕ is bounded by 1,
so it follows from Theorem 6.16 that there is a unique β ∈ L∞ (T ) such that
Z π
1
f (t)β(t) dt.
(18.23)
ϕ(f ) =
2π −π
On the one hand, we have
Z π Z π
Z π
1
1
(f ∗ g)(t)β(t) dt =
f (t − s)g(s)β(t) ds dt.
ϕ(f ∗ g) =
2π −π
(2π)2 −π −π
On the other hand, we obtain
i h 1 Z π
i
h 1 Z π
f (t)β(t) dt ×
g(s)β(s) ds
2π −π
2π −π
Z πZ π
1
f (t − s)g(s)β(t − s)β(s) ds dt.
=
(2π)2 −π −π
ϕ(f )ϕ(g) =
The fact ϕ(f ∗ g) = ϕ(f )ϕ(g) asserts that β(t) = β(t − s)β(s) a.e. on T or equivalently,
β(x + y) = β(x)β(y)
a.e. on T . Employing similar analysis as in the proof of Theorem 9.23, since β is periodic
with period 1, it has the form
β(x) = e−iαx
244
Chapter 18. Elementary Theory of Banach Algebras
for a unique α ∈ R.e Since we must have β(x + 2π) = β(x), α must be an integer. Put
α = n. Substituting this back into the integral (18.23), we get
Z π
1
ϕ(f ) =
f (t)e−int dt = fb(n)
2π −π
for a unique integer n.
• IE is a closed ideal in L1 (T ). Let E ⊆ Z and
IE = f ∈ L1 (T ) fb(n) = 0 for all n ∈ E .
(18.24)
For any f ∈ IE and g ∈ L1 (T ), if we write h = f ∗ g, then the formula (18.22) implies that
b
h(n) = fb(n) · gb(n) = 0
for every n ∈ E. Thus we have f ∗ g ∈ IE and IE is an ideal in L1 (T ). For every α ∈ R,
we follow from the definition that
fb(n − α) = fb(n)e−iαn = 0
for every n ∈ E, so f (x − α) ∈ IE . This means that IE contains every translate of f and
Remark 18.1 ensures that IE is closed.
• Every closed ideal I in L1 (T ) has the form (18.24). Let I be a closed ideal of L1 (T ).
We have to prove that I = IE for some set E ⊆ Z. Suppose
each n ∈ Z, there
−1that for
int
1
c
c
exists an fn ∈ I such that fn (n) 6= 0. Put gn (t) = e fn (n)
∈ L (T ). Then we have
1
1
·
(gn ∗ fn )(t) =
c
fn (n) 2π
Z
π
−π
in(t−s)
fn (s)e
eint
1
ds =
·
c
fn (n) 2π
Z
π
fn (s)e−ins ds = eint .
−π
Since I is an ideal, we have eint ∈ I for every n ∈ Z. Using Theorems 3.14, 4.25 and
the fact that I is closed, we know that the set {eint | n ∈ Z} is dense in L1 (T ). Thus we
conclude that I = L1 (T ).
Without loss of generality, we may assume that I 6= L1 (T ). Then there exists an n ∈ Z
such that fb(n) = 0 for all f ∈ I. Let the collection of such integers be E, i.e.,
E = {n ∈ Z | fb(n) = 0 for all f ∈ I}.
(18.25)
Thus it is easy to see that I ⊆ IE . If n ∈
/ E, then there exists an g ∈ I such that b
g(n) 6= 0.
For simplicity, we may assume that this number is 1. Regarding eint as an element of
L1 (T ), we see that (g ∗ ein )(t) = −eint . Since I is an ideal, we have g ∗ ein ∈ I and then
eint ∈ I. Thus I contains every trigonometric polynomial of the form
X
an eint
provided that an = 0 for all n ∈ E. Suppose that f ∈ IE and we consider the set
Z(f ) = n ∈ Z fb(n) = 0 .
Now the definitions (18.24) and (18.25) imply that E ⊆ Z(f ), so it follows from Theorem
9.2(c) that if P is a trigonometric polynomial on T and n ∈ E, then we have
e
See also [23, Theorem 8.19, p. 247].
(f[
∗ P )(n) = fb(n) × Pb(n) = 0,
18.2. Properties of Ideals and Homomorphisms
245
i.e., E ⊂ Z(f ∗ g). Using [60, Theorem 2.6.6, p. 51], we see that kf − f ∗ P k1 can be made
as small as we want. Therefore, the closeness of I asserts that f ∈ I which implies IE ⊆ I.
Hence we conclude that
I = IE
as desired.
We end the proof of the problem.
Remark 18.2
For other classes of complex homomorphisms of specific Banach algebras, please refer to [81,
§9, pp. 39 – 43].
Problem 18.11
Rudin Chapter 18 Exercise 11.
Proof. Notice that λ, µ ∈ C \ σ(x). On the one hand, we have
(x − λe) R(λ, x) − R(µ, x) (x − µe) = (x − λe)(λe − x)−1 − (x − λe)(µe − x)−1 (x − µe)
= −e(x − µe) − (x − λe)(µe − x)−1 (x − µe)
= −x + µe + x − λe
= (µ − λ)e.
(18.26)
On the other hand, we see that
(x − λe) (µ − λ)R(λ, x)R(µ, x) (x − µe) = (µ − λ) (x − λe)R(λ, x)R(µ, x)(x − µe)
= (µ − λ)e.
(18.27)
It yields from the results (18.26) and (18.27) that
R(λ, x) − R(µ, x) = (µ − λ)R(λ, x)R(µ, x)
(18.28)
holds for all λ, µ ∈ C \ σ(x).
We follow from the identity (18.28) that
(x − µe)−1 − (x − λe)−1
R(λ, x) − R(µ, x)
=
= R(λ, x)R(µ, x) → R(λ, x)2 = (x − λe)−2
µ−λ
µ−λ
as µ → λ. This is exactly [62, Eqn. (3), p. 359], so the argument in the proof of Theorem 18.5
can be applied directly. This completes the proof of the problem.
Remark 18.3
The result in Problem 18.11 is called Hilbert’s identity.
Problem 18.12
Rudin Chapter 18 Exercise 12.
246
Chapter 18. Elementary Theory of Banach Algebras
Proof. Denote M be the set of maximal ideals of A. Let M be a maximal ideal of A. By
Theorem 18.17, we have M = ker h for some h ∈ ∆. Conversely, if h ∈ ∆, then it follows from
the First Isomorphism Theorem [25, Theorem 16.2, p. 145] that
A/ ker h ∼
= h(A) = C.
Since C is a field, ker h is a maximal ideal of A. In other words, there exists an one-to-one
correspondence between M and ∆.
Let rad A be the radical of A. Suppose that x ∈ rad A, i.e.,
\
x∈
M.
(18.29)
M ∈M
Now the previous paragraph yields that the set relation (18.29) is equivalent to the condition
\
x∈
h−1 (0)
h∈∆
which means that h(x) = 0 for every h ∈ ∆. Consequently, statements (a) and (c) are equivalent.
Next, Theorem 18.17(b) means that the spectrum σ(x) is exactly the set {h(x) | h ∈ ∆}.
Since x ∈ rad A if and only if h(x) = 0 for every h ∈ ∆, this implies that x ∈ rad A if and
1
only if σ(x) = {0} if and only if kxn k n → 0 as n → ∞ by Theorem 18.9 (The Spectral Radius
Formula). Hence we have shown the three statements are equivalent, completing the proof of
the problem.
Problem 18.13
Rudin Chapter 18 Exercise 13.
Proof. Let X = C([0, 1]). Then X is a Hilbert space (and hence a Banach space). For each
f ∈ X, we define
Z t
f (s) ds,
T (f )(t) =
0
where t ∈ [0, 1]. Since f ∈ X, T (f ) ∈ X so that T ∈ B(X), the algebra of all bounded linear
operations on X. By Problem 18.1, B(X) is a Banach space.
Particularly, we take f (x) = 1. Then we observe
Z t
Z tZ s
t2
2
s ds = .
du ds =
T (f )(t) =
2
0
0
0
More generally, for n = 1, 2, . . ., we obtain
T n (f )(t) =
tn
n!
which implies that T n (f ) 6= 0 for all n > 0 and
kT n k = sup
kT n (f )k∞
1
= .
kf k∞
n!
(18.30)
1
Since (n!) n → ∞ as n → ∞, we conclude immediately from the result (18.30) that
1
lim kT n k n = 0.
n→∞
This completes the analysis of the problem.
18.2. Properties of Ideals and Homomorphisms
247
Problem 18.14
Rudin Chapter 18 Exercise 14.
Proof. By the definition, the function x
b : ∆ → C is given by x
b(h) = h(x) and we have the set
b = {b
b by G(x) = x
A
x | x ∈ A}. Denote the surjective mapping G : A → A
b.
• The mapping G is a homomorphism. Suppose that x, y ∈ A, α ∈ C and h ∈ ∆. Then
we see that
d
(αx)(h)
= h(αx) = αh(x) = (αb
x)(h)
and
\
(x
+ y)(h) = h(x + y) = h(x) + h(y) = x
b(h) + yb(h) = (b
x + yb)(h)
x
cy(h) = h(xy) = h(x)h(y) = x
b(h)b
y (h) = (b
xyb)(h).
Therefore, the map G is a homomorphism. Its kernel consists of those x ∈ A such that
h(x) = 0 for every h ∈ ∆. By Problem 18.12, it is exactly the radical of A, i.e., ker x
b=
rad A.
Combining the First Isomorphism Theorem and the previous result, we see that
b
A/rad A = A/ ker x
b∼
= A.
b and G becomes an isomorphism.
Thus if rad A = {0},f then we get A ∼
=A
• ρ(x) = kb
xk∞ = sup{|b
x(h)| | h ∈ ∆}. By Theorem 18.17(e), we have ρ(x) ≥ |h(x)| = |b
x(h)|
for every h ∈ ∆ which means that
ρ(x) ≥ kb
xk∞ .
For the other direction, λ belongs to the range of x
b means that λ = x
b(h) = h(x) for some
h ∈ ∆ and it follows from Theorem 18.17(b) that this happens if and only if λ ∈ σ(x), so
Definition 18.8 establishes
ρ(x) = sup{|λ| | λ ∈ σ(x)} ≤ sup{|b
x(h)| | h ∈ ∆} = kxk∞ .
• The range of x
b is σ(x). The analysis in the previous part also implies that the range of
the function x
b is exactly the spectrum σ(x).
We have completed the proof of the problem.
Problem 18.15
Rudin Chapter 18 Exercise 15.
Proof. Let A1 = {(x, λ) | x ∈ A and λ ∈ C} and k(x, λ)k = kxk + |λ|. For any (x, λ), (y, µ) ∈ A1 ,
we define the multiplication in A1 by
(x, λ)(y, µ) = (xy + µx + λy, λµ).
f
In this case, A is called semisimple.
(18.31)
248
Chapter 18. Elementary Theory of Banach Algebras
• A1 is a commutative Banach algebra with unit. It is easily checked that this is
associative and distributive. Thus A1 is a complex algebra. Furthermore, the element
(0, 1) is a unit for this multiplication because
(x, λ)(0, 1) = (x · 0 + 1 · x + λ · 0, λ · 1) = (x, λ) = (0, 1)(x, λ).
For any (x, λ), (y, µ) ∈ A1 , we see that
k(x, λ) + (y, µ)k = k(x + y, λ + µ)k
= kx + yk + |λ + µ|
≤ kxk + kyk + |λ| + |µ|
= k(x, λ)k + k(y, µ)k.
If α ∈ C, then we have
kα(x, λ)k = k(αx, αλ)k = kαxk + |αλ| = |α| · kxk + |α| · |λ| = |α| · k(x, λ)k.
As k(x, λ)k = 0 if and only if kxk + |λ| = 0 if and only if kxk = 0 if and only if x = 0 and
λ = 0, A1 is a normed linear space by Definition 5.2. Since kxyk ≤ kxk · kyk, we see that
k(x, λ)(y, µ)k = k(xy + µx + λy, λµ)k
= kxy + µx + λyk + |λµ|
≤ kxyk + kµxk + kλyk + |λ| · |µ|
≤ kxkkyk + |µ| · kxk + |λ| · kyk + |λ| · |µ|
= kxk + |λ| · kyk + |µ|
= k(x, λ)k · k(y, µ)k.
Since the spaces A and C are complete, A1 is obviously complete and then it is a Banach
algebra with unit by Definition 18.1. It is commutative because A and C are commutative
so that (y, µ)(x, λ) equals to the right-hand side of the expression (18.31).
• The mapping x 7→ (x, 0) is an isometric isomorphism of A onto a maximal ideal
of A1 . Let Φ be this mapping. It is trivial surjective. If Φ(x) = Φ(y), then (x, 0) = (y, 0)
which means that (x − y, 0) = 0. Since k(x − y, 0)k = 0, we get x = y and thus Φ is
injective. Is is easily checked that Φ satisfies
Φ(x + y) = Φ(x) + Φ(y),
so Φ is an isomorphism onto Φ(A). It is also isometric because
kΦ(x) − Φ(y)k = kΦ(x − y)k = k(x − y, 0)k = kx − yk.
Now we may identify A with Φ(A) ⊆ A1 . Since (x, λ)(y, 0) = (xy + λy, 0) ∈ Φ(A) for
every (x, λ) ∈ A1 and (y, 0) ∈ Φ(A), Φ(A) is an ideal of A1 by Definition 18.12. Since
A1 = A ⊕ C, we have A1 /Φ(A) ∼
= A1 /A ∼
= C which implies that Φ(A) is a maximal ideal
of A1 as required.
This completes the proof of the problem.
18.3. The Commutative Banach algebra H ∞
18.3
249
The Commutative Banach algebra H ∞
Problem 18.16
Rudin Chapter 18 Exercise 16.
Proof. It is clear that H ∞ is a commutative complex algebra. Recall from §11.31 that its norm
is defined by
kf k∞ = sup{|f (z)| | z ∈ U }.
This norm makes H ∞ satisfy Definition 5.2. Thus H ∞ is a normed linear space, so it is a normed
complex algebra. The fact that H ∞ is complete has been shown in [62, Remark 17.8(c), p. 338],
so H ∞ is a commutative Banach algebra by Definition 18.1. The element 1 ∈ H ∞ is easily seen
to be its unit which gives the first assertion.
Suppose that |α| < 1. Define Φα : H ∞ → C by
Φα (f ) = f (α).
(18.32)
Then it satisfies Φα (f g) = f (α)g(α) = Φα (f )Φα (g). For every constant a, the function f (z) = a
gives Φα (f ) = a so it is surjective. In other words, Φα ∈ ∆. To see that there are complex
homomorphisms of H ∞ other than the point homomorphisms (18.32), we let I be the set of
functions f ∈ H ∞ such that f (α) → 0 as α → 1 and α > 0. Then it is easy to see that I is
a proper ideal of H ∞ . By Theorem 18.13, I is contained in a maximal ideal J of H ∞ which
means that there exists a complex homomorphism of H ∞ , say ϕ ∈ ∆ such that ϕ(f ) = 0 for all
f ∈ I by Theorem 18.17(a). However, ϕ 6= Φα for all α ∈ U because there is no α such that
Φα (f ) = f (α) = 0 for every f ∈ I. We have finished the proof of this problem.
Problem 18.17
Rudin Chapter 18 Exercise 17.
Proof. By Problem 18.16, H ∞ is a commutative Banach algebra. Let I = {(z − 1)2 f | f ∈ H ∞ }.
For any f, g ∈ H ∞ , we know that f g ∈ H ∞ . Thus if (z − 1)2 f ∈ I and g ∈ H ∞ , then we have
g · (z − 1)2 f = (z − 1)2 f g ∈ I.
By Definition 18.12, I is an ideal of H ∞ .
Given ǫ > 0. The function fǫ (z) = (1 + ǫ − z)−1 belongs to H ∞ because
|1 + ǫ − z| ≥ |1 + ǫ| − |z| > ǫ
so that |fǫ (z)| < ǫ−1 for all z ∈ U . We observe that
(1 − z)2 (1 + ǫ − z)−1 − (1 − z) =
ǫ(1 − z)
<ǫ
1−z+ǫ
for all z ∈ U . This means that (z − 1)2 fǫ (z) converges uniformly to 1 − z in U . However, we
know that 1− z ∈
/ H ∞ which means that I is not closed. This ends the proof of the problem. Problem 18.18
Rudin Chapter 18 Exercise 18.
250
Chapter 18. Elementary Theory of Banach Algebras
Proof. Denote I = {ϕf | f ∈ H ∞ }. Of course, we have ϕf ∈ H ∞ . Since f g ∈ H ∞ for any
f, g ∈ H ∞ , the space I is an ideal of H ∞ by Definition 18.12. Let {fn } be a sequence in H ∞
such that
kϕfn − gk∞ → 0
as n → ∞, where g ∈ H ∞ . Thus {ϕfn } is a Cauchy sequence by [61, Theorem 3.11(a), p. 53]
and hence so is {fn }. Observing from Remark 17.8(c) that H ∞ is Banach, so it is complete.
Then we have fn → f ∈ H ∞ as n → ∞ and this means that ϕf ∈ I. Hence we complete the
proof of the problem.
CHAPTER
19
Holomorphic Fourier Transforms
19.1
Problems on Entire Functions of Exponential Type
Problem 19.1
Rudin Chapter 19 Exercise 1.
Proof. By the hypothesis, we know that there exist some constants A and C such that
|f (z)| ≤ CeA|z|
for all z ∈ C. Suppose for simplicity that ϕ(0) < ∞, i.e.,
Z ∞
|f (x)|2 dx < ∞.
−∞
By Theorem 19.3 (The Paley and Wiener Theorem), there exists an F ∈ L2 (−A, A) such that
Z A
f (z) =
F (t)eitz dt
−A
for all z ∈ C. Define F1 and F2 by

 F (t), if 0 ≤ t < A;
F1 (t) =

0,
if t ≥ A
and F2 (t) =
Then we may express f as
f (z) = f1 (z) + f2 (z) =
Z
∞
itz
F1 (t)e

 F (t), if −A < t ≤ 0;

dt +
0,
Z
0
if t ≤ −A.
F2 (t)eitz dt.
(19.1)
−∞
0
By the definition, it is clear that F1 ∈ L2 (0, ∞), so we know from [62, Eqn. (3), p. 372] that
Z ∞
Z ∞
1
|F1 (t)|2 dt < ∞
(19.2)
|f1 (x + iy)|2 dx ≤
2π −∞
0
for every y > 0. For the second integral of the equation (19.1), we write
Z ∞
Z 0
f2 (t)e−itz dt,
F
F2 (t)eitz dt =
f2 (z) =
0
−∞
251
252
Chapter 19. Holomorphic Fourier Transforms
f2 (t) = F2 (−t). Since F2 ∈ L2 (−∞, 0), we have F
f2 ∈ L2 (0, ∞). By similar argument as
where F
in [62, pp. 371, 372], we can show that f2 is holomorphic in the lower half plane Π− and if we
write
Z ∞
f2 (t)ety · e−itx dt,
f2 (x + iy) =
F
0
regard y as fixed, then Theorem 9.13 (The Plancherel Theorem) implies that
Z ∞
Z ∞
Z ∞
1
2 2ty
2
f2 (t) 2 dt < ∞
f
F
F2 (t) e dt ≤
|f (x + iy)| dx =
2π −∞
0
0
(19.3)
for every y < 0. Now we substitute the estimates (19.2) and (19.3) into the expression (19.1),
we see from Theorem 3.8 that
Z ∞
1
1
ϕ(y) =
|f (x + iy)|2 dx
2π
2π −∞
Z ∞
Z
1 ∞
1
|f1 (x + iy)|2 dx +
|f1 (x + iy)| · |f2 (x + iy)| dx
≤
2π −∞
π −∞
Z ∞
1
|f2 (x + iy)|2 dx
+
2π −∞
Z
Z ∞
o1 n Z ∞
o1
1n ∞
2
2
2
2
|f1 (x + iy)| dx ·
|F1 (t)| dt +
≤
|f2 (x + iy)|2 dx
π
−∞
0
−∞
Z ∞
f2 (t) 2 dt
F
+
0
f2
= kF1 k22 + 2kF1 k2 · F
2
f2
= F1 k2 + kF
2
2
<∞
f2
+ F
2
2
for every real y. This proves the first assertion.
For the second assertion, we suppose that ϕ is a bounded function. By Theorem 19.3 (The
Paley and Wiener Theorem), there exists an F ∈ L2 (−A, A) such that
f (x + iy) = f (z) =
Z
A
F (t)eitz dt =
Z
A
−A
−A
F (t)e−ty · e−itx dt
(19.4)
for all z = x + iy ∈ C. If we extend F (t) = 0 for all t ≤ −A and t ≥ A, then the integral (19.4)
becomes
Z ∞
[F (t)e−ty ] · e−itx dt.
f (x + iy) =
−∞
Observing from Theorem 9.13 (The Plancherel Theorem), we have
ϕ(y) =
Z
∞
−∞
2
|f (x + iy)| dx = 2π
Z
∞
−∞
−ty 2
|F (t)e
| dt = 2π
Z
A
−A
e−2ty |F (t)|2 dt.
If we consider y → −∞, then the boundedness of ϕ implies that F (t) = 0 a.e. on [0, ∞).
Similarly, if y → ∞, then we have F (t) = 0 a.e. on (−∞, 0]. Thus we have F (t) = 0 a.e. on R,
so the integral representation (19.4) implies that f = 0 which ends the proof of the problem. Problem 19.2
Rudin Chapter 19 Exercise 2.
19.1. Problems on Entire Functions of Exponential Type
253
Proof. Since f is of exponential type, there exist some constants A and C such that
|f (z)| ≤ C exp(A|z|)
(19.5)
for all z ∈ C. Using rotation, we may assume that one of the nonparallel lines is the real line,a
so we also have
Z ∞
|f (x)|2 dx < ∞.
−∞
Now Theorem 19.3 (The Paley and Wiener Theorem) asserts that there exists an F ∈ L2 (−A, A)
such that
Z
A
f (z) =
F (t)eitz dt
−A
for all z. Particularly, we take z = x ∈ R so that
Z A
nZ
1
2
|F (t)| dt ≤ (2A) ·
|f (x)| ≤
−A
A
−A
|F (t)|2 dt
o1
2
<∞
by Theorem 3.8. Thus f is bounded on the real line. Similarly, it can be shown that f is also
bounded on the other line. Then we may apply Problem 19.3 to conclude that f is actually a
constant. Since f ∈ L2 , this constant is in fact zero. Hence we obtain our desired result and
this ends the proof of the problem.
Problem 19.3
Rudin Chapter 19 Exercise 3.
Proof. Without loss of generality, we assume that the two nonparallel lines pass through the
origin and f is bounded by 1 on these lines. Suppose that ∆ is the open sector between the two
lines with sectoral angle πβ < π for some β > 1. Since f is entire, it is continuous on ∆ and
holomorphic in ∆. Since f is of exponential type, the inequality (19.5) holds for all complex z.
Take 1 < α < β. Then there exists a positive number R such that A|z| ≤ |z|α for all z with
|z| > R. For |z| ≤ R, we let M = CeAR so that
|f (z)| ≤ M exp(|z|α )
for all z ∈ ∆. Thus Problem 12.9 reveals that |f (z)| is bounded in ∆ which implies that f is a
bounded entire function. Hence it follows from Theorem 10.23 (Liouville’s Theorem) that f is
constant, as desired. This ends the proof of the problem.
Problem 19.4
Rudin Chapter 19 Exercise 4.
Proof. Now f is an entire function of exponential type.
• The series converges if |w| > A. By the hypothesis, it is true that |f (z)| < exp(|z|λ )
for all large enough |z|, where λ > 1. According to Problem 15.2, f is of order 1. We note
from [11, Eqns. (2.1.6) & (2.2.12), pp. 8, 12] that
1
lim sup |f (n) (0)| n = A.
n→∞
a
Notice that f (eiθ z) is also an entire function satisfying the inequality (19.5).
254
Chapter 19. Holomorphic Fourier Transforms
This means that
1
lim sup |n!an | n = A
n→∞
and so the power series Φ converges if
1
1
1
<
,
1 =
|w|
A
lim sup |n!an | n
n→∞
i.e., |w| > A.b
• The function f can be expressed as an integral. By the power series of Φ, we see
that
Z
Z X
∞
∞
n!an X z k wk 1
1
dw
Φ(w)ewz dw =
×
2πi Γ
2πi Γ
wn+1
k!
n=0
k=0
Z X
∞
m
1
am z
=
dw
2πi Γ
w
m=0
Z
∞
1 X
dw
m
=
am z
2πi
Γ w
m=0
=
∞
X
am z m
m=0
= f (z),
where the term-by-term integration being justified by the uniform convergence of the series
on Γ.c
• Φ is the function which occurred in the proof of Theorem 19.3. Recall from
Remark 19.4 that the functions Φα are restrictions of a function holomorphic in the complement of the interval [−iA, iA]. Thus it suffices to prove that Φ is such function.
Our first assertion and Theorem 10.6 ensure that the Borel transform Φ is holomorphic in
the complement of [−iA, iA]. It remains to show that
Φ|Πα = Φα
(19.6)
for every real α.d If Re (weiα ) > 3A, then we have
| exp(−wseiα )| = exp − sRe (weiα ) < e−3As .
Furthermore, if we define Mf (r) =
(19.7)
max |f (z)|, then we follow from Theorem 10.26
z∈D(0;r)
(Cauchy’s Estimates) that
|an | ≤
and the remainder
Rn (s) =
Mf (r)
rn
∞
X
ak s k
k=n+1
b
The function Φ(w) in question is called the Borel transform of the function f (z), see [11, §5.3, p. 73] or
[36, §20, p. 84].
c
The integral is sometimes called the Pólya representation of the function f .
d
Indeed, the half plane Πα is given by {w = x + iy | x cos α − y sin α > A}.
19.1. Problems on Entire Functions of Exponential Type
255
of the power series of f satisfies
∞ X
Mf (r) s n+1
s k
|Rn (s)| ≤ Mf (r)
=
.
·
r
1 − rs
r
k=n+1
By putting r = 2s, we get
e2s
2n
Therefore, we deduce from the inequalities (19.7) and (19.8) that the series
|Rn (s)| ≤
∞
X
(19.8)
an e−wz z n
n=0
converges uniformly on the ray Γα = {seiα | s ≥ 0}. Consequently, for every w ∈ Πα , we
obtain
Z
Z
∞
∞
X
X
n!an
−wz
an
z n e−wz dz =
e
f (z) dz =
Φα (ω) =
= Φ(w)
wn+1
Γα
Γα
n=0
n=0
which is exactly the expression (19.6).
We have completed the proof of the problem.
Problem 19.5
Rudin Chapter 19 Exercise 5.
Proof. Suppose that f ∈ H(Π+ ) and
1
sup
0<y<∞ 2π
Z
∞
−∞
|f (x + iy)|2 dx = C < ∞.
(19.9)
• The Cauchy formula holds in Π+ . Let 0 < ǫ < y and z = x + iy. Let r > 0 be large
and γr be the semicircle in Π+ with radius r and centered at iǫ. Define Γr to be the union
of γr and the line segment [−r + iǫ, r + iǫ]. Since r is large, we may assume that z ∈
/ Γ∗r .
See Figure 19.1 below:
Figure 19.1: The closed contour Γr .
256
Chapter 19. Holomorphic Fourier Transforms
Using Theorem 10.15 (The Cauchy’s Formula in a Convex Set), we have
Z
1
f (ω)
f (z) =
dω
2πi Γr ω − z
Z r
Z π
f (ξ + iǫ)
ireiθ f (reiθ + iǫ)
1
1
dξ +
dθ.
=
2πi −r ξ + iǫ − z
2πi 0
reiθ + iǫ − z
(19.10)
By Theorem 19.2 (The Paley and Wiener Theorem), there exists an F ∈ L2 (0, ∞) such
that
Z ∞
F (t)eitz dt,
f (z) =
0
Π+ .
reiθ
where z ∈
Take z =
+ iǫ = r cos θ + i(r sin θ + ǫ), where 0 < θ < π. Since
itz = −t(r sin θ + ǫ) + itr cos θ, we have
|eitz | = e−t(r sin θ+ǫ) ≤ e−tr sin θ ,
(19.11)
where 0 < θ < π. According to Theorem 3.8 and the estimate (19.11), we obtain
Z ∞
e−2tr sin θ ∞
kF k22
iθ
2
2
e−2tr sin θ dt = kF k22 ×
|f (re + iǫ)| ≤ kF k2 ×
=
.
(19.12)
−2r sin θ 0
2r sin θ
0
Substituting the bound (19.12) into the second integral in the formula (19.10), we obtain
Z π
Z π
ireiθ f (reiθ + iǫ)
rkF k2
dθ
dθ ≤ √
1
iθ + iǫ − z
re
2r(r − |iǫ − z|) 0 sin 2 θ
0
√
Z π
2
2rkF k2
dθ
=
.
(19.13)
r − |iǫ − z| 0 sin 21 θ
By the inequality π2 ≤ sinθ θ for 0 < θ ≤ π2 , we can show that the integral in the estimate
(19.13) is bounded so that
Z π
ireiθ f (reiθ + iǫ)
dθ = 0
lim
r→∞ 0
reiθ + iǫ − z
and consequently, the expression (19.10) gives
Z ∞
1
f (ξ + iǫ)
f (z) =
dξ
2πi −∞ ξ + iǫ − z
for every 0 < ξ < y.
• f ∗ (x) = lim f (x + iy) exists for almost all x. Notice that the linear fractional transy→0
1+z
formation ϕ(z) = i 1−z
maps U conformally onto Π+ and ϕ(T ) = R ∪ {∞}. If we consider
g = f ◦ ϕ ∈ H(U ), then the condition (19.9) implies definitely that
Z
|g(z)|2 dθ dr < ∞.
U
Therefore, g is bounded for almost all z ∈ U which means kgk∞ < ∞, i.e., g ∈ H ∞ . By
Theorem 11.32 (Fatou’s Theorem), the limit
lim g(reiθ )
r→1
(19.14)
exists for almost everywhere on [−π, π]. After transforming back to f , the result (19.14)
means that
f ∗ (x) = lim f (x + iy)
y→1
exists for almost all x.
19.1. Problems on Entire Functions of Exponential Type
• The relation between f ∗ and F . The second assertion implies
Z ∞
∗
f (x) = lim f (x + iy) = lim
F (t)e−ty+itx dt.
y→0
y→0 0
257
(19.15)
t
Let fn (t) = F (t)e− n +itx . Then we have |f1 | ≥ |f2 | ≥ · · · ≥ 0 and fn (t) → F (t)eitx as
n → ∞ for almost every t ∈ (0, ∞). It is clear from Theorem 3.8 that
Z ∞
Z ∞
nZ ∞
o1
kF k2
2
−t
|F (t)|e dt ≤ kF k2 ×
e−2t dt
|f1 (t)| dt =
= √ < ∞,
2
0
0
0
i.e., |f1 | ∈ L1 (0, ∞). Hence we may apply Problem 1.7 to the limit (19.15) to get
Z ∞
∗
F (t)eitx dt
(19.16)
f (x) =
0
for almost all real x. Since we may assume that F vanishes on (−∞, 0) (see §19.1), we
follow from the expression (19.16) that
√
f ∗ (x) = 2π · Fb(−x)
for almost all x ∈ R.
• The case when ǫ = 0. Let z = x + iy with y > 0. Notice that
f (ξ + iǫ)
|f (ξ + iǫ)|
≤
ξ + iǫ − z
|ξ − z|
for every ξ ∈ (−∞, ∞) and 0 < ǫ < y. Thus it follows from Theorem 3.8 that
Z ∞
Z ∞
|f (ξ + iǫ)|
f (ξ + iǫ)
dξ
dξ ≤
ξ
+
iǫ
−
z
|ξ − x|
−∞
−∞
nZ ∞
o1 n Z ∞
dξ o 12
2
≤
×
|f (ξ + iǫ)|2 dξ
2
−∞
−∞ (ξ − x)
Z
n ∞
√
dξ o 21
≤ 2Cπ ·
2
−∞ (x − ξ)
< ∞,
so Theorem 1.34 (The Lebesgue’s Dominated Convergence Theorem) ensures that
Z ∞ ∗
f (ξ)
1
dξ.
f (z) =
2πi −∞ ξ − z
We complete the proof of the problem.
Problem 19.6
Rudin Chapter 19 Exercise 6.
Proof. Here we follow mainly [50, Theorem XII, pp. 16 – 20]. Since 0 < ϕ < eϕ , we have
log ϕ < ϕ. Combining the hypothesis and Theorem 3.5 (Hölder’s Inequality), we obtain
Z ∞
Z ∞
nZ ∞
o1 n Z ∞
ϕ(x)
dx o 21
log ϕ(x)
2
2
×
< ∞.
dx
<
dx
≤
|ϕ(x)|
dx
−∞ <
2
2
2 2
−∞ 1 + x
−∞ (1 + x )
−∞
−∞ 1 + x
258
Chapter 19. Holomorphic Fourier Transforms
In other words, we have
Z
∞
−∞
| log ϕ(x)|
dx < ∞.
1 + x2
We write z = x + iy with y > 0 and consider
Z
y
1 ∞
log ϕ(t) dt.
u(z) =
π −∞ (x − t)2 + y 2
(19.17)
Using the half-plane version of Fatou’s Theorem [55, Theorem 5.5, pp. 86, 87], we see that the
function (19.17) is harmonic in Π+ and
lim u(x + iy) = log ϕ(x)
y→0
or
lim |f (x + iy)| = ϕ(x)
y→0
(19.18)
holds for almost all x ∈ R.e Let v(z) be its harmonic conjugate and write
f (z) = exp(u(z) + iv(z)).
Because of [62, Eqn. (7), p. 63], we see that
|f (x + iy)| = eu(z) ≤
1
π
Z
∞
−∞
ϕ(t)y
dt,
(x − t)2 + y 2
it follows from Theorem 3.5 (Hölder’s Inequality) that
Z
Z ∞
ϕ(t)y
ϕ(s)y
1 ∞
2
|f (x + iy)| ≤ 2
dt ×
ds
2
2
2
2
π −∞ (x − t) + y
−∞ (x − s) + y
Z
o 1 nZ ∞
o1
1n ∞
|ϕ(t)|2 y
y
2
2
≤ 2
dt
dt
2 + y2
2 + y2
π
(x
−
t)
(x
−
t)
−∞
−∞
nZ ∞
o 1 nZ ∞
o1
|ϕ(s)|2 y
y
2
2
ds
ds
×
2 + y2
2 + y2
(x
−
s)
(x
−
s)
−∞
−∞
Z
|ϕ(t)|2 y
1 ∞
dt
=
π −∞ (x − t)2 + y 2
which implies
Z
∞
−∞
Z
Z
1 ∞ ∞
|ϕ(t)|2 y
dt dx
π −∞ −∞ (x − t)2 + y 2
Z ∞
Z
y
1 ∞
2
dx
|ϕ(t)| dt ·
=
2
2
π
−∞ (x − t) + y
Z ∞−∞
|ϕ(t)|2 dt.
=
|f (x + iy)|2 dx ≤
−∞
In other words, we have established
sup
y>0
Z
∞
−∞
|f (x + iy)|2 dx < ∞.
According to Theorem 19.2 (The Paley-Wiener Theorem), there exists an F ∈ L2 (−∞, ∞)
vanishing on (−∞, 0) such that
Z ∞
F (t)eitz dt
f (z) =
−∞
e
Here the function
y
(x − t)2 + y 2
is the Poisson kernel in the upper half plane Π+ . See also [7, pp. 145 – 147; Theorem 7.28, pp. 160, 161].
19.1. Problems on Entire Functions of Exponential Type
for all z ∈ Π+ . In particular, we have
lim f (x + iy) = f (x) =
y→0
Z
∞
itx
F (t)e
dt =
Z
∞
−∞
0
259
F (t)eitx dt = Fb(−x)
(19.19)
for x ∈ R. If we denote G(x) = Fb (−x), then we combine the results (19.18) and (19.19) to get
|G(x)| = ϕ(x).
b
Since G ∈ L2 (−∞, ∞), we derive from [78, Lemma 9.3, p. 286] that G(x)
= F (−x) which
vanishes on [0, ∞).
Conversely, we suppose that there exists an f with fb = ϕ such that f (x) = 0 for all x ≤ 0.
Let us write
Z ∞
1
fb(x) = √
f (t)e−ixt dt
(19.20)
2π −∞
and
1
ψ(z) = √
2π
Z
∞
f (t)e−izt dt,
(19.21)
−∞
where z ∈ Π+ and the integral in (19.21) is taken along a horizontal line in the z-plane. Certainly,
we have ψ ∈ H(Π+ ) by §19.1. Suppose that we map Π+ (conformally) onto U by z = i ζ+1
ζ−1 .
Write
ζ + 1
and K(eiθ ) = fb(x),
k(ζ) = ψ(z) = ψ i
ζ −1
iθ
+1
where ζ = reiθ and 0 ≤ r < 1 and x = i eeiθ −1
. Then it is easily seen from Theorem 12.12 (The
Hausdorff-Young Theorem) that
Z
π
−π
|K(eiθ )|2 dθ = 2
Z
∞
−∞
|fb(x)|2
dx ≤ 2
1 + x2
Z
∞
−∞
|fb(x)|2 dx = 2 fb
2
2
≤ 2kf k22 = 2kϕk22 < ∞.
Therefore, we have K ∈ L2 (T ). On the other hand, if z = x + iy, then the integral (19.20)
implies that
Z π
Z
1
y
1 ∞
fb(t) dt
K(eiθ )Pr (θ − φ) dθ =
2π −π
π −∞ (x − t)2 + y 2
Z
Z ∞
y
1
1 ∞
×√
f (ξ)e−itξ dξ dt
=
π −∞ (x − t)2 + y 2
2π 0
Z ∞
1 Z ∞
e−itξ y
1
dt
dξ
f (ξ) ×
=√
π −∞ (x − t)2 + y 2
2π 0
Z ∞
1
=√
f (ξ)e−ixξ+yξ dξ
2π 0
Z ∞
1
√
f (ξ)e−izξ dξ
=
2π 0
= ψ(z)
= k(reiθ ).
By Definition 11.6, k is the Poisson integral of K, i.e., k = P [K]. Since K ∈ L2 (T ) and
k = P [K], it follows from Theorem 11.16 that
Z π
Z π
Z π
|K(reiθ )|2 dθ.
(19.22)
|k(reiθ )|2 dθ ≤
log+ |k(reiθ )| dθ ≤
−π
−π
−π
260
Chapter 19. Holomorphic Fourier Transforms
If k(0) 6= 0, then we apply Theorem 15.18 (Jensen’s Formula) to obtain
Z π
1
log |k(0)| ≤
log |k(reiθ )| dθ.
2π −π
(19.23)
Now the formula
Z π
Z π
Z π
1
1
1
log |k(reiθ )| dθ =
log+ |k(reiθ )| dθ +
log− |k(reiθ )| dθ
2π −π
2π −π
2π −π
clearly implies
Z π
Z π
Z π
1
1
1
log |k(reiθ )| dθ =
log+ |k(reiθ )| dθ −
log− |k(reiθ )| dθ
2π −π
2π −π
2π −π
Z
Z π
1
1 π
+
iθ
log |k(re )| dθ −
log |k(reiθ )| dθ.
=
π −π
2π −π
Substituting the inequalities (19.22) and (19.23) into the formula (19.24), we obtain
Z π
Z
1
1 π
|K(reiθ )|2 dθ − log |k(0)|
log |k(reiθ )| dθ ≤
2π −π
π −π
(19.24)
(19.25)
for all 0 ≤ r < 1. If k has zero at 0 with multiplicity m, then the inequality (19.25) becomes
Z π
Z
1
k(ζ)
1 π
iθ
log |k(re )| dθ ≤
|K(reiθ )|2 dθ − log m
− m log r.
(19.26)
2π −π
π −π
ζ
ζ=0
for all 0 ≤ r < 1. By the definition, log |k(reiθ )| → log |K(eiθ )| as r → 1 almost everywhere, so
we conclude from the inequality (19.26) that
1
π
Z
∞
−∞
| log ϕ(x)|
1
dx =
2
1+x
π
Z
∞
−∞
Consequently, this ensures that
Z
∞
−∞
log |fb(x)|
1
dx =
2
1+x
2π
log ϕ(x)
Z
−π
log |K(eiθ )| dθ < ∞.
dx
> −∞.
1 + x2
Hence we have completed the proof of the problem.
19.2
π
Quasi-analytic Classes and Borel’s Theorem
Problem 19.7
Rudin Chapter 19 Exercise 7.
Proof. It is trivial that Condition (a) implies Condition (b). Conversely, suppose that Condition
(b) holds. For each α ∈ E, we fix a neighborhood Vα of α and put
[
Ω=
Vα .
α∈E
It is clear that E ⊂ Ω and Ω is an open set. Now we define F : Ω → C as follows: Given z ∈ Ω.
Then we have z ∈ Vα for some Vα and we define
F (z) = Fα (z).
(19.27)
19.2. Quasi-analytic Classes and Borel’s Theorem
261
We claim that F ∈ H(Ω) and F (z) = f (z) for z ∈ E. If z ∈ Vβ for β 6= α, then we have
Vα ∩ Vβ 6= ∅. Since Vα ∩ Vβ is an open set, there exists a δ > 0 such that z ∈ D(0; δ) ⊆ Vα ∩ Vβ ,
so Theorem 10.18 says that Fα ≡ Fβ in Vα ∩ Vβ and thus the formula (19.27) is well-defined.
Furthermore, it is clear that F is holomorphic at every point of Ω. Finally, if ζ ∈ E, then ζ ∈ Vζ .
Since Fζ (z) = f (z) for all z ∈ Vζ ∩ E, we must have
F (ζ) = Fζ (ζ) = f (ζ).
This ends the proof of the problem.
Problem 19.8
Rudin Chapter 19 Exercise 8.
Proof. Since n! ≤ nn for every n ≥ 1, we have kD n f k∞ ≤ βf Bfn n! ≤ βf Bfn nn . By Definition
19.6, we have C{n!} ⊆ C{nn }. Recall the approximation to Stirling’s formula [61, Exercise 20,
p. 200] that
1
nn
∼√
en n!
2πn
for large n. Thus there exists a constant M > 0 such that nn ≤ M en n! for all n ≥ 1. If
f ∈ C{nn }, then we have
kD n f k∞ ≤ βf Bfn nn ≤ (M βf )(eBf )n n!
which means f ∈ C{n!}. Consequently, we obtain the desired result that C{n!} = C{nn },
completing the proof of the problem.
Problem 19.9
Rudin Chapter 19 Exercise 9.
Proof. Let M0 = 1, M1 = 1, M2 = 2 and Mn = n!(log n)n for every n ≥ 3. Thus we always have
Mn2 ≤ Mn−1 Mn+1 for all n = 1, 2, . . .. Using Theorem 19.11 (The Denjoy-Carleman Theorem),
we know that C{Mn } is quasi-analytic. If f ∈ C{n!}, then there exist positive constants βf and
Bf such that
kD n f k∞ ≤ βf Bfn n!
for all n ≥ 0. By the definition of {Mn }, we also have
kD n f k∞ ≤ βf Bfn Mn
for every n ≥ 0. Thus we have
C{n!} ⊆ C{Mn }.
The construction of an example f belonging to C{Mn }, but f ∈
/ C{n!} is basically motivated
for
every
n
=
0,
1,
2,
.
.
.. Since
by [70, Theorem 1, p. 4]. Put mn = MMn+1
n
mn − mn−1 =
Mn
Mn+1 Mn−1 − Mn2
Mn+1
−
=
≥0
Mn
Mn−1
Mn Mn−1
for every n ≥ 0. Thus {mn } is a positive increasing sequence. It is clear that f ∈ C ∞ . For every
n, k ∈ N, if k ≤ n, then we see that
1
mnn−k
=
1
1
1
×
× ··· ×
mn mn
mn
262
Chapter 19. Holomorphic Fourier Transforms
1
1
1
×
× ··· ×
mn−1 mn−2
mk
Mk
Mn−1 Mn−2
×
× ··· ×
=
Mn
Mn−1
Mk+1
Mk
=
.
Mn
≤
If k > n, then we have
1
mnn−k
= mnk−n
≤ mn × mn+1 × · · · × mk−1
Mn+1 Mn+2
Mk
=
×
× ··· ×
Mn
Mn+1
Mk−1
Mk
=
.
Mn
In other words, we obtain the estimate
1
mnn−k
≤
Mk
.
Mn
(19.28)
Define
fn (x) =
Mn 2mn ix
e
(2mn )n
and
f (x) =
∞
X
fn (x)
n=0
for x ∈ R. We first show that f ∈ C{Mn }. For every k ≥ 1, we have
fn(k) (x) =
ik Mn
e2mn ix .
(2mn )n−k
Combining this and the estimate (19.28), we get
|f
(k)
(x)| ≤
∞
X
n=0
|fn(k) (x)|
=
∞
X
n=0
∞
∞
n=0
n=0
X Mn Mk
X 1
Mn
≤
·
=
M
·
≤ 2 · 2k Mk (19.29)
k
(2mn )n−k
2n−k Mn
2n−k
for every k ≥ 1 and x ∈ R. Furthermore, we also have
|f (x)| ≤
∞
X
n=0
By induction, we can show that
|fn (x)| =
Mnn+1
n
Mn+1
∞
X
∞
X 1 M n+1
Mn
=
· n .
n
n Mn
(2m
)
2
n
n+1
n=0
n=0
≤ 1 for each n ≥ 0. Therefore, the estimate (19.30) gives
|f (x)| ≤ 2 = 2 · 20 M0
for every x ∈ R. Now we conclude from the estimates (19.29) and (19.30) that
kD k f k∞ ≤ 2 · 2k Mk
holds for every k ≥ 0 so that f ∈ C{Mn } as required.
Next, we want to show that |f (k) (0)| ≥ Mk for every k ≥ 0. It is obvious that
f (0) =
∞
X
n=0
(19.30)
Mn
M1
= M0 +
+ · · · ≥ 1 = M0 .
(2mn )n
2m1
19.2. Quasi-analytic Classes and Borel’s Theorem
In addition, if k ≥ 1, then since every term
|f
(k)
Mn
(2mn )n−k
k
(0)| = |i |
∞
X
n=0
263
is positive for every n = 0, 1, 2, . . ., we have
Mn
≥ Mk .
(2mn )n−k
(19.31)
Hence we have obtained what we want. If f ∈ C{n!}, then it must be true that
|f (k) (0)| ≤ βf Bfk k!
for some positive constants βf and Bf . However, if k is sufficiently large so that (log k)k ≥ βf Bfk ,
then this will certainly contradict the estimate (19.31). Hence we conclude that f ∈
/ C{n!} and
then we complete the proof of the problem.
Problem 19.10
Rudin Chapter 19 Exercise 10.
Proof. Suppose that λ =
∞
X
λn is positive finite. Recall from Definition 2.9 that Cc (R) is the
n=1
collection of all continuous complex functions on R whose support is compact. Now we let g0
to be the function modified from the (19.38) in such the way that g0 (x) = 1 for −λ ≤ x ≤ λ,
g0 (x) = 0 for |x| ≥ 2λ and 0 ≤ g0 (x) ≤ 1 if x ∈ [−2λ, −λ] ∪ [λ, 2λ]. Then g0 ∈ Cc (R) and g0 is
integrable in R. Write
gn (x) = g λ1 , λ2 , . . . , λn ; g0 (x)
Z λ1
Z λn
Z λ2
1
dt1
g0 (x − t1 − t2 − · · · − tn ) dtn .
(19.32)
dt2 · · ·
= n
2 λ1 λ2 · · · λn −λ1
−λn
−λ2
Since |g0 (x)| ≤ 1 for every x ∈ R, the definition (19.32) ensures that |gn (x)| ≤ 1 for every
n = 0, 1, 2, . . . and x ∈ R. In other words, the family {gn } is (uniform) bounded in R. Besides,
if |x| ≥ 3λ, then |x − λ1 − λ2 − · · · − λn | ≥ 2λ so that gn (x) = 0 there.
Obviously, we have
g λ1 , λ2 , . . . , λn ; g0 (x) = g λ1 , λ2 , . . . , λk ; g λk+1 , λk+2 , . . . , λn ; g0 (x) .
(19.33)
By the definition (19.32) again, we see that
g1 (x) =
1
2λ1
Z
λ1
−λ1
g0 (x − t1 ) dt1 =
1
2λ1
Z
x+λ1
g0 (t) dt,
x−λ1
so the Fundamental Theorem of Calculus yields that g1 (x) is differentiable in R and
g1′ (x) =
1
[g0 (x + λ1 ) − g0 (x − λ1 )].
2λ1
Next, we assume that the function gn−1 (x) is continuous
for any n ≥2 in R and if n ≥ 2, then
the function gn (x) = g λn ; g(λ1 , λ2 , . . . , λn−1 ; g0 (x) = g λn ; gn−1 (x) is differentiable in R and
it follows from the formula (19.33) that
gn′ (x) =
d
g λ1 , λ2 , . . . , λn ; g0 (x)
dx
264
Chapter 19. Holomorphic Fourier Transforms
d
g λ1 ; g λ2 , λ3 , . . . , λn ; g0 (x)
dx
1 =
g λ2 , λ3 , . . . , λn ; g0 (x + λ1 ) − g λ2 , λ3 , . . . , λn ; g0 (x − λ1 )
2λ1
1
=
g λ2 , λ3 , . . . , λn ; g0 (x + λ1 ) − g0 (x − λ1 ) .
2λ1
=
(19.34)
This also implies that gn (x) has continuous derivatives of order 0, 1, . . . , n−1 in R. Furthermore,
if we combine the formula (19.34) and the Mean Value Theorem for Derivatives, we get, for every
x ∈ R, that
′
|gn′ (x)| ≤ max |g′ λ2 , λ3 , . . . , λn ; g0 (x) | = max |gn−1
(x)| ≤ · · · ≤ max |g2′ (x)| < ∞. (19.35)
x∈R
x∈R
x∈R
For every n ≥ 2 and any x, y ∈ R, the bound (19.35) asserts that
|gn (x) − gn (y)| = |x − y| · |gn′ (ξ)| ≤ |x − y| · max |g2′ (x)|
x∈R
which shows that the family {gn } is equicontinuous on R. Recall that gn (x) = 0 outside [−3λ, 3λ],
so {gn } is actually equicontinuous on [−3λ, 3λ]. The fact |gn (x)| ≤ 1 in R guarantees that {gn }
converges pointwise on R. Hence it asserts from [61, Exercise 16, p. 168] that {gn } converges
uniformly to a continuous function g on [−3λ, 3λ], i.e.,
g(x) = lim gn (x)
n→∞
for every x ∈ [−3λ, 3λ]. Since the argument also applies to any compact interval of R, the
function g is also continuous at the end points ±3λ. It is trivial that if x ∈
/ [−3λ, 3λ], then
g(x) = 0. Hence we must have g(±3λ) = 0.
Denote g(x) = g λ1 , λ2 , . . . ; g0 (x) . Then we know that
g(x) = lim g λ1 , λ2 , . . . , λn ; g0 (x)
n→∞
Z λ1
1
g λ2 , λ3 , . . . , λn ; g0 (x − t) dt
= lim
n→∞ 2λ1 −λ
1
Z λ1
1
=
g λ2 , λ3 , . . . ; g0 (x − t) dt
2λ1 −λ1
is true for all x ∈ [−3λ, 3λ] which means that
1 g λ2 , λ3 , . . . ; g0 (x + λ1 ) − g λ2 , λ3 , . . . ; g0 (x − λ1 )
2λ1
g0 (x + λ1 ) − g0 (x − λ1 ) = g λ2 , λ3 , . . . ;
2λ1
g′ (x) =
(19.36)
holds in [−3λ, 3λ]. Using similar reasoning as the previous paragraph, it can be shown that g is
also differentiable at the end points ±3λ and the formula (19.36) holds in R. In conclusion, we
have g ∈ C ∞ .
Since |λ1 + λ2 + · · · + λn | < λ, we obtain g0 (−t1 − t2 − · · · − tn ) = 1 for all −λk ≤ tk ≤ λk ,
where 1 ≤ k ≤ n. Thus we note that
Z λn
Z λ2
Z λ1
1
dtn = 1,
dt2 · · ·
dt1
g(0) = lim n
n→∞ 2 λ1 λ2 · · · λn −λ
−λn
−λ2
1
so g is not identically zero in R.
19.2. Quasi-analytic Classes and Borel’s Theorem
265
Put
G0 (x) = g0 (x)
G0 (x + λ1 ) − G0 (x − λ1 )
G1 (x) =
,
2λ1
..
.
Gn−1 (x + λn ) − Gn−1 (x − λn )
Gn (x) =
.
2λn
Thus the formula (19.36) can be written as
In fact, it is true that
g′ (x) = g λ2 , λ3 , . . . ; G1 (x) .
g (n) (x) = g λn+1 , λn+2 , . . . ; Gn (x)
(19.37)
for every n = 0, 1, 2, . . . and x ∈ R. Recall that |g0 (x)| ≤ 1 on R, so we have
|Gn (x)| ≤
1
= Mn .
λ1 λ2 · · · λn
on R. Thus it follows from the formula (19.37) that
|g(n) (x)| ≤ Mn
for all n = 0, 1, 2, . . . and x ∈ R. Consequently, g ∈ C{Mn } which completes the proof of the
problem.
Remark 19.1
The construction in Problem 19.10 follows basically the unpublished work of H. E. Bray
which was quited in Mandelbrojt’s article [38, pp. 79 – 84]. See also [32].
Problem 19.11
Rudin Chapter 19 Exercise 11.
Proof. An example of a function ϕ ∈ C ∞ with the required properties can be found in [77,
Problem 10.6, pp. 266, 267]. In fact, we start with
 −1
 e x , if x > 0;
g(x) =

0,
if x ≤ 0.
It is known that g ∈ C ∞ and g(m) (0) = 0 for all m = 1, 2, . . .. Define ϕ : R → R by
ϕ(x) =
g(2 − |x|)
.
g(2 − |x|) + g(|x| − 1)
(19.38)
Now it is easy to see that ϕ ∈ C ∞ . Furthermore, we have ϕ(x) = 1 for −1 ≤ x ≤ 1, ϕ(x) = 0
for |x| ≥ 2 and 0 ≤ ϕ(x) ≤ 1 if x ∈ [−2, −1] ∪ [1, 2] so that
supp ϕ ⊆ [−2, 2].
This completes the proof of the problem.
266
Chapter 19. Holomorphic Fourier Transforms
Problem 19.12
Rudin Chapter 19 Exercise 12.
Proof. Let ϕ be as in Problem 19.11. Set β =
fn (x) =
αn
n!
and gn (x) = βn xn ϕ(x). Take
gn (λn x)
= βn xn ϕ(λn x),
λnn
where λn is large enough. Fix the non-negative integer n, we notice that
(D k fn )(x) = βn
k
X
n!
k k−m n−m (m)
Cm
λn x
ϕ (λn x),
(n
−
m)!
m=0
(19.39)
where k = 0, 1, 2, . . . , n − 1. Recall from the definition of ϕ in Problem 19.11 that
supp ϕ(m) ⊆ supp ϕ ⊆ [−2, 2]
holds for every m = 0, 1, . . . , k. If λn x ∈
/ [−2, 2], then ϕ(m) (λn x) = 0 so that (D k f )(x) = 0. If
2
(m)
is continuous on [−2, 2], there is a positive constant M
λn x ∈ [−2, 2], then |x| ≤ λn . Since ϕ
(m)
such that |ϕ (λn x)| ≤ M . Thus we obtain
|(D k fn )(x)| ≤ |βn |M
k
X
n−m
n!
k k−m 2
Cm
λn
· n−m
(n − m)!
λn
m=0
k
X
2n (n!)2
≤ |βn |M
λnn−k
m=0
≤
n2n (n!)2 |βn |M
λn
(19.40)
for all x ∈ R and k = 0, 1, . . . , n − 1. Since λn can be chosen large enough, we observe from the
estimate (19.40) that
1
(19.41)
kD k fn k∞ < n
2
for all k = 0, 1, . . . , n − 1. Take f = f0 + f1 + · · · .
It is clear that f0 (0) + f1 (0) + · · · = α0 . Besides, the result (19.41) ensures that the series
{f0′ + f1′ + · · · + fn′ } converges uniformly on R. Using [61, Theorem 7.17, p. 152], termwise
differentiation is legitimate so that f ′ = f0′ + f1′ + · · · . Now this argument can be applied
repeatedly to show that f ∈ C ∞ . Next, it follows from the expression (19.39) that (D k fn )(0) = 0
for k = 0, 1, . . . , n − 1. Since ϕ(x) = 1 on [−1, 1], ϕ(n) (0) = 0 for all n = 1, 2, . . . and this implies
that (D n fm )(0) = 0 for m = 0, 1, . . . , n − 1. Hence we have
(D n f )(0) = (D n f0 )(0) + (D n f1 )(0) + · · · + (D n fn−1 )(0) + (D n fn )(0)
+ (D n fn+1 )(0) + (D n fn+2 )(0) + · · ·
= n!βn
= αn
for every n = 0, 1, 2, . . ., as required. This completes the proof of the problem.f
f
Please also read [41] and [56].
19.2. Quasi-analytic Classes and Borel’s Theorem
267
Remark 19.2
Problem 19.12 is called Borel’s Theorem which says that every power series is the Taylor
series of some smooth function, see, for examples, [46, Theorem 1.5.4, p. 30] and [51].
Problem 19.13
Rudin Chapter 19 Exercise 13.
Proof. It suffices to prove that
lim sup
n→∞
|(D n f )(a)| 1
n
n!
= ∞.
(19.42)
We follow the suggestion. Let ck = λ1−k
k , where the sequence {λk } satisfies
ck λkk
= λk > 2
k−1
X
j=1
cj λkj = 2(λk1 + λ2k−1 + · · · + λ2k−1 ) and
λk > k2k > 1.
Fix the non-negative integer n, we have
∞
X
k=1
ck λnk
=
n+1
X
λkn+1−k
+
k=1
∞
X
λkn+1−k
k=n+2
<
n+1
X
λkn+1−k
+
k=1
∞
X
k=n+2
1
< ∞.
k2k
We put
f (z) =
∞
X
ck eiλk x ,
k=1
where z = x + iy. Let fk (x) = ck eiλk x , where k ∈ N. For each n = 0, 1, 2, . . ., we observe that
|(D n fk )(a)| = ck λnk = λkn+1−k
for every a ∈ R. Using similar argument as in the proof of Problem 19.12, it is easy to show
that f ∈ C ∞ . The choices of our {ck } and {λk } reveal that
|(D n f )(a)| ≥ |cn λnn | −
∞
X
k=1
k6=n
ck λnk = λn −
∞
X
ck λnk >
k=1
k6=n
n2n
.
2
Combining this and Stirling’s formula, for large enough n, we get
|(D n f )(a)| 1
n
n!
>
n2n 1
n
2n!
≈
en n2n 1
en
n
√
=
2
2
n
2n 2πn
(8π) n · n n
which implies the result (19.42). Hence the power series
∞
X
(D n f )(a)
n=0
n!
(x − a)n
has radius of convergence 0 for every a ∈ R, completing the proof of the problem.
268
Chapter 19. Holomorphic Fourier Transforms
Remark 19.3
Let S = {2n | n ∈ N}. Define
g(x) =
X
e−
√
k
cos(kx).
k∈S
Then it can be shown that g also satisfies the requirements of Problem 19.13.
Problem 19.14
Rudin Chapter 19 Exercise 14.
Proof. Suppose that f ∈ C{Mn } has infinitely many zeros {xn } in [0, 1]. Then {xn } has a
convergent subsequence by the Bolzano-Weierstrass Theoremg . Without loss of generality, we
may assume that {xn } is itself convergent, distinct, increasing and its limit is α. The continuity
of f gives
f ′ (α) = 0.
By the Mean Value Theorem for Derivatives, we see that f ′ (ξn ) = 0 for some ξn ∈ (xn , xn+1 )
for all n = 1, 2, . . .. The fact xn → α as n → ∞ ensures that ξn → α as n → ∞. Since f ∈ C ∞ ,
the continuity of f ′ implies that
f ′ (α) = lim f ′ (ξn ) = 0.
n→∞
This argument can be repeated to show that f (n) (α) = 0 for all n = 0, 1, 2, . . .. Since C{Mn }
is quasi-analytic, Definition 19.8 gives f (x) ≡ 0 for all x ∈ R, completing the proof of the
problem.
Problem 19.15
Rudin Chapter 19 Exercise 15.
Proof. Suppose that
X = f ∈ H(C) |f (z)| ≤ Ceπ|z| for some C > 0, f ∈ L2 [−π, π] .
Recall from Definition 3.6 that the sequence space ℓ2 is given by
n
ℓ2 = {f (n)}
Define the map Φ : X → ℓ2 by
∞
X
n=−∞
o
|f (n)|2 < ∞ .
Φ(f ) = {f (n)}.
For any α, β ∈ C and f, g ∈ X, it is clear that |f (z)| ≤ C1 eπ|z| and |g(z)| ≤ C2 eπ|z|
for some
π|z|
positive constants C1 and C2 . Therefore, we have |αf (z) + βg(z)| ≤ |α|C1 + |β|C2 e
and
kαf + βgk2 ≤ |α| · kf k2 + β · kgk2 < ∞
g
See [79, Problem 5.25, pp. 68, 69]
19.2. Quasi-analytic Classes and Borel’s Theorem
269
by Theorem 3.9. This means that αf + βg ∈ X. Besides, the fact
∞
X
n=−∞
|αf (n) + βg(n)|2 ≤ 2|α|2
∞
X
n=−∞
|f (n)|2 + 2|β|2
∞
X
n=−∞
|g(n)|2 < ∞
implies that {αf (n) + βg(n)} ∈ ℓ2 . Thus the relation
Φ(αf + βg) = {αf (n) + βg(n)} = α{f (n)} + β{g(n)} = αΦ(f ) + βΦ(g)
holds, i.e, Φ is linear.
For every f ∈ L2 (T ), we define
F (z) =
1
2π
Z
π
f (t)e−izt dt.
−π
Using the analysis in §19.2, one can show that F is entire, |F (z)| ≤ Ceπ|z| for all z ∈ C and
F ∈ L2 (−∞, ∞). In other words, this means that F ∈ X. Furthermore, we note that F = fb.
Next, for each n ∈ Z, recall from Definition 4.23 that {un (t) = eint | n ∈ Z} forms an orthonormal
set in L2 (T ). Furthermore, we have
Z π
ei(n−z)π − e−i(n−z)π
sin[(z − n)π]
1
1
×
=
ei(n−z)t dt =
Un (z) = u
bn (z) =
2π −π
2π
i(n − z)
(z − n)π
holds for every n ∈ Z. Recall from [62, Example 4.5(b), p. 78] that
Z π
1
hf, giT =
f (t)g(t) dt
2π −π
defines an inner product in L2 (T ). Then it is easily checked that we can induce a norm k · k to
X by defining
p
p
(19.43)
kF kX = hF, F iX = hf, f iT = kf kT ,
so it makes X Hilbert. Of course, it follows from the expression (19.43) that

 1, if n = m;
hUn , Um iX = hun , um iT =

0, otherwise.
In other words, {Un | n ∈ Z} forms an orthonormal set in X. Since {un | n ∈ Z} is maximal in
L2 (T ), {Un | n ∈ Z} is also maximal in X. By Theorem 4.18 (The Riesz-Fischer Theorem), it is
true that
X
F (z) =
ck Uk (z),
k∈Z
where c1 , c2 , . . . are some scalars. Since F is entire, we have
X
F (n) = lim
ck Uk (z) = cn Un (n) = cn
z→n
k∈Z
for every n ∈ Z. Finally, as a Hilbert space X with a maximal orthonormal set {Un | n ∈ Z}, we
conclude from §4.19 that the mapping
F 7→ hF, Un i = cn = F (n)
is a Hilbert space isomorphism of X onto ℓ2 (Z). This means that our map Φ is a bijection,
completing the analysis of the problem.
270
Chapter 19. Holomorphic Fourier Transforms
Problem 19.16
Rudin Chapter 19 Exerci