A Complete Solution Guide to
Principles of Mathematical Analysis
by Kit-Wing Yu, PhD
kitwing@hotmail.com
Copyright c 2018 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: 978-988-78797-0-1 (eBook)
ISBN: 978-988-78797-1-8 (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 wellknown 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.
In my memory, Principles of Mathematical Analysis was the standard textbook when I was a year 2
undergraduate mathematics student many years ago. I read Chapters 1-7 plus 11 again when I prepared
my Ph.D. qualifying examination. In my personal experience, I think that chapters in these books are
well-organized and expositions of theorems are clear, precise and well-written. I believe that you will
agree with me after reading them.
Although Principles of Mathematical Analysis stopped to have any update after 1976, many instructors nowadays still like to select this book as the standard textbook or one of the main reference books
in their analysis courses. I guess the “beauty” of this book that I described above is one of the reasons
for this kind of choice. Furthermore, many people believe that the best way to study mathematics is
by working through examples and exercises. Thus the excellent and well-designed exercises provided in
the book is another main factor that attracts us. This can be justified if you type some keywords in
google, then you will see many people are discussing and finding solutions of those exercises in different
platforms. For example, the Math Stack Exchange:
https://math.stackexchange.com/
Actually, Professor Roger Cooke wrote “Solutions Manual to Walter Rudin’s Principles of Mathematical
Analysis” in 1976, see
https://minds.wisconsin.edu/handle/1793/67009
However, a glance at the files will reveal that the readability of the manual is a bit low because of the
lacks of equation numbers and supporting illustrations. Besides, the most important point is that it
seems that some solutions are incomplete!
Therefore, these give “birth” to this book: I decide to write the solutions myself. In other words, this
book is not a collection of solutions from others, but rather I use my own words and own ways to solve
and prove the problems. In fact, I have written a complete solution guide so that it helps every mathematics student and instructor to understand the ideas and applications of the theorems in Rudin’s book.
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.
a https://en.wikipedia.org/wiki/Walter_Rudin.
v
vi
The features of this book are as follows:
• It covers all the 285 exercises with detailed and complete solutions. As a matter of fact, my solutions
show every detail, every step and every theorem that I applied. That’s why my book has over 390
pages!
• There are 55 illustrations and 3 tables 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 and references are provided because some questions require additional
mathematical concepts which are not covered by Rudin.
• Three appendices are included which further explain and supplement some theories in Chapters 10
and 11.
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 new website
https://sites.google.com/view/yukitwing/.
Finally, if the sales of this book are good (I hope so), then I will start to write a solution guide of the
book Real and Complex Analysis soon.
Kit Wing Yu
February 2018
List of Figures
2.1
2.2
2.3
2.4
The neighborhoods Nh (q) and Nr (p). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Convex sets and nonconvex sets. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The sets Nh (x), N h (x) and Nqm (xk ). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
The construction of the shrinking sequence. . . . . . . . . . . . . . . . . . . . . . . . . . .
13
23
25
29
3.1
The Cantor set. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
4.1
4.2
4.3
4.4
4.5
4.6
4.7
The graph of g on [an , bn ]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The sets E and Ini . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The graphs of [x] and√(x). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
An example for α = 2 and n = 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The distance from x ∈ X to E. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The graph of a convex function f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The positions of the points p, p + κ, q − κ and q. . . . . . . . . . . . . . . . . . . . . . . .
59
63
70
72
74
76
77
5.1
5.2
5.3
5.4
5.5
5.6
The zig-zag path of the process in (c). . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The zig-zag path induced by the function f in Case (i). . . . . . . . . . . . . . . . . . . .
The zig-zag path induced by the function g in Case (i). . . . . . . . . . . . . . . . . . . .
The zig-zag path induced by the function f in Case (ii). . . . . . . . . . . . . . . . . . .
The zig-zag path induced by the function g in Case (ii). . . . . . . . . . . . . . . . . . .
The geometrical interpretation of Newton’s method. . . . . . . . . . . . . . . . . . . . . .
105
108
109
109
110
111
8.1
8.2
8.3
8.4
8.5
8.6
The graph of the continuous function y = f (x) = (π − |x|)2 on [−π, π]. . . . . . . . . . . .
The graphs of the two functions f and g. . . . . . . . . . . . . . . . . . . . . . . . . . . .
A geometric proof of 0 < sin x ≤ x on (0, π2 ]. . . . . . . . . . . . . . . . . . . . . . . . . . .
The graph of y = | sin x|. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The winding number of γ around an arbitrary point p. . . . . . . . . . . . . . . . . . . . .
The geometry of the points z, f (z) and g(z). . . . . . . . . . . . . . . . . . . . . . . . . . .
186
197
199
199
202
209
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
An example of the range K of f . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The set of q ∈ K such that (∇f3 )(f −1 (q)) = 0. . . . . . . . . . . . . . . . . . . . . . . . .
Geometric meaning of the implicit function theorem. . . . . . . . . . . . . . . . . . . . . .
The graphs around the four points. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The graphs around (0, 0) and (1, 0). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The graph of the ellipse X 2 + 4Y 2 = 1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The definition of the function ϕ(x, t). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The four regions divided by the two lines αx1 + βx2 = 0 and αx1 − βx2 = 0. . . . . . . .
219
220
232
233
236
239
243
252
10.1 The compact convex set H and its boundary ∂H. . . . . . . . . . . . . . . . . . . . . . . .
10.2 The figures of the sets Ui , Wi and Vi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.3 The mapping T : I 2 → H. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.4 The mapping T : A → D. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.5 The mapping T : A◦ → D0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.6 The mapping T : S → Q. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
256
264
269
270
271
277
vii
List of Figures
10.7 The open sets Q0.1 , Q0.2 and Q. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.8 The mapping T : I 3 → Q3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.9 The mapping τ1 : Q2 → I 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.10The mapping τ2 : Q2 → I 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.11The mapping τ2 : Q2 → I 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.12The mapping Φ : D → R2 \ {0}. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.13The spherical coordinates for the point Σ(u, v). . . . . . . . . . . . . . . . . . . . . . . . .
10.14The rectangles D and E. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.15An example of the 2-surface S and its boundary ∂S. . . . . . . . . . . . . . . . . . . . . .
10.16The unit disk U as the projection of the unit ball V . . . . . . . . . . . . . . . . . . . . . .
10.17The open cells U and V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.18The parameter domain D. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.19The figure of the Möbius band. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10.20The “geometric” boundary of M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
viii
278
280
288
289
289
296
300
302
304
325
326
332
333
335
11.1 The open square Rδ ((p, q)) and the neighborhood N√2δ ((p, q)). . . . . . . . . . . . . . . . 350
B.1 The plane angle θ measured in radians. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365
B.2 The solid angle Ω measured in steradians. . . . . . . . . . . . . . . . . . . . . . . . . . . . 366
B.3 A section of the cone with apex angle 2θ. . . . . . . . . . . . . . . . . . . . . . . . . . . . 366
List of Tables
6.1
The number of intervals & end-points and the length of each interval for each En . . . . . 121
9.1
9.2
Expressions of x around four points. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
Expressions of y around four points. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235
ix
List of Tables
x
Contents
Preface
v
List of Figures
vii
List of Tables
ix
1 The Real and Complex Number Systems
1.1 Problems on rational numbers and fields . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Properties of supremums and infimums . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 An index law and the logarithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Properties of the complex field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5 Properties of Euclidean spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6 A supplement to the proof of Theorem 1.19 . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1
2
2
5
7
9
2 Basic Topology
2.1 The empty set and properties of algebraic numbers . . . . . . . . . . . . . . . . . . . . . .
2.2 The uncountability of irrational numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Limit points, open sets and closed sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4 Some metrics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 Compact sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6 Further topological properties of R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7 Properties of connected sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8 Separable metric spaces and bases and a special case of Baire’s theorem . . . . . . . . . .
11
11
12
12
15
17
18
21
24
3 Numerical Sequences and Series
3.1 Problems on sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Problems on series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Recursion formulas of sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4 A representation of the Cantor set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5 Cauchy sequences and the completions of metric spaces . . . . . . . . . . . . . . . . . . .
31
31
33
45
49
50
4 Continuity
4.1 Properties of continuous functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 The extension, the graph and the restriction of a continuous function . . . . . . . . . . . .
4.3 Problems on uniformly continuous functions . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Further properties of continuous functions . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5 Discontinuous functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.6 The distance function ρE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.7 Convex functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.8 Other properties of continuous functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57
57
58
63
68
69
73
76
81
xi
Contents
5 Differentiation
85
5.1 Problems on differentiability of a function . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
5.2 Applications of Taylor’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
5.3 Derivatives of higher order and iteration methods . . . . . . . . . . . . . . . . . . . . . . . 102
5.4 Solutions of differential equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
6 The Riemann-Stieltjes Integral
117
6.1 Problems on Riemann-Stieltjes integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
6.2 Definitions of improper integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
6.3 Hölder’s inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
6.4 Problems related to improper integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
6.5 Applications and a generalization of integration by parts . . . . . . . . . . . . . . . . . . . 133
6.6 Problems on rectifiable curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
7 Sequences and Series of Functions
141
7.1 Problems on uniform convergence of sequences of functions . . . . . . . . . . . . . . . . . 141
7.2 Problems on equicontinuous families of functions . . . . . . . . . . . . . . . . . . . . . . . 157
7.3 Applications of the (Stone-)Weierstrass theorem . . . . . . . . . . . . . . . . . . . . . . . . 164
7.4 Isometric mappings and initial-value problems . . . . . . . . . . . . . . . . . . . . . . . . . 167
8 Some Special Functions
173
8.1 Problems related to special functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
8.2 Index of a curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201
8.3 Stirling’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
9 Functions of Several Variables
213
9.1 Linear transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
9.2 Differentiable mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
9.3 Local maxima and minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219
9.4 The inverse function theorem and the implicit function theorem . . . . . . . . . . . . . . . 226
9.5 The rank of a linear transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237
9.6 Derivatives of higher order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241
10 Integration of Differential Forms
255
10.1 Integration over sets in Rk and primitive mappings . . . . . . . . . . . . . . . . . . . . . . 255
10.2 Generalizations of partitions of unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263
10.3 Applications of Theorem 10.9 (Change of Variables Theorem) . . . . . . . . . . . . . . . . 267
10.4 Properties of k-forms and k-simplexes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284
10.5 Problems on closed forms and exact forms . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
10.6 Problems on vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330
11 The Lebesgue Theory
337
11.1 Further properties of integrable functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 337
11.2 The Riemann integrals and the Lebesgue integrals . . . . . . . . . . . . . . . . . . . . . . 340
11.3 Functions of classes L and L 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345
Appendix
355
A A proof of Lemma 10.14
355
B Solid angle subtended by a surface at the origin
365
C Proofs of some basic properties of a measure
369
Index
377
Bibliography
379
CHAPTER
1
The Real and Complex Number Systems
Unless the contrary is explicitly stated, all numbers that are mentioned in these exercises are understood
to be real.
1.1
Problems on rational numbers and fields
Problem 1.1
Rudin Chapter 1 Exercise 1.
Proof. Assume that r + x was rational. Then it follows from Definition 1.12(A1), (A4) and (A5) that
x = (r + x) − x
is also rational, a contradiction. Similarly, if rx was rational, then it follows from Definition 1.12(M1),
M(4) and M(5) that
rx
x=
x
is also rational, a contradiction. This ends the proof of the problem.
Problem 1.2
Rudin Chapter 1 Exercise 2.
√
√
Proof. Assume that 12 was rational so that 12 = m
n , where m and n are co-prime integers. Then we
have m2 = 12n2 and thus m is divisible by 3. Let m = 3k for some integer k. Then we have m2 = 9k 2
and this shows that
4n2
k2 =
,
3
so n is divisible by 3. This contradicts the fact that m and n are co-prime, completing the proof of the
problem.
Problem 1.3
Rudin Chapter 1 Exercise 3.
Proof. Since x 6= 0, there exists x1 ∈ F such that x · x1 = 1.
(a) Therefore, it follows from Definition 1.12(M2), (M3), (M4) and (M5) that xy = xz implies that
y = z.
1
Chapter 1. The Real and Complex Number Systems
2
(b) Similarly, it follows from Definition 1.12(M2), (M3), (M4) and (M5) that xy = x implies that y = 1.
(c) Similarly, it follows from Definition 1.12(M2), (M3), (M4) and (M5) that xy = 1 implies that y = x1 .
(d) Since x1 ∈ F , we have 11 ∈ F such that x1 · 11 = 1. Now we have x1 · 11 = x1 · x(= 1), then Proposition
x
x
1.15(a) implies that 11 = x.
x
x
This completes the proof of the problem.
1.2
Properties of supremums and infimums
Problem 1.4
Rudin Chapter 1 Exercise 4.
Proof. Since E ⊂ S, the definitions give α ≤ x and x ≤ β for all x ∈ E. Thus Definition 1.5(ii) implies
that α ≤ β. This completes the proof of the problem.
Problem 1.5
Rudin Chapter 1 Exercise 5.
Proof. Theorem 1.19 says that R is an ordered set with the least-upper-bound property. Since A is a
non-empty subset of R and A is bounded below, inf A exists in R by Definition 1.10. Furthermore, −A is
a non-empty subset of R. Let y be a lower bound of A, i.e. y ≤ x for all x ∈ A. Then we have −x ≤ −y
for all x ∈ A. Thus −y is an upper bound of −A and sup(−A) exists in R by Definition 1.10.
Let α = inf A and β = sup(−A). By definition, we have y ≤ β for all y ∈ −A, where y = −x for
some x ∈ A. It implies that x = −y ≥ −β for all x ∈ A, so −β is a lower bound of A and then −β ≤ α.
Similarly, we have α ≤ x for all x ∈ A so that −α ≥ −x for all x ∈ A. It implies that −α is an upper
bound of −A, so β ≤ −α and then −β ≥ α. Hence we have α = −β, i.e. inf A = − sup(−A). This
completes the proof of the problem.
1.3
An index law and the logarithm
Problem 1.6
Rudin Chapter 1 Exercise 6.
Proof.
(a) Since bm > 0 and n ∈ N, Theorem 1.21 implies that there exists one and only one real y such that
y n = bm . Similarly, there exists one and only one real z such that z q = bp . We have
y nq = (y n )q = (bm )q = bmq = bpn = (bp )n = (z q )n = z qn
1
1
which implies that y = z, i.e., (bm ) n = (bp ) q .
p
m
(b) Let br = b n and bs = b q . Without loss of generality, we may assume that n and q are positive.
Then the corollary of Theorem 1.21 implies that
br+s = b
mq+np
nq
1
1
1
1
m
p
= (bmq+np ) nq = (bmq × bnp ) nq = (bmq ) nq × (bnp ) nq = b n × b q = br × bs .
3
1.3. An index law and the logarithm
(c) By definition, B(r) = {bt | t ∈ Q, t ≤ r}, where r ∈ Q. It is clear that br ∈ B(r), so it is a nonempty subset of R. Since b > 1, we have bt ≤ br for all t ≤ r so that br is an upper bound of B(r).
Therefore, Theorem 1.19 and Definition 1.10 show that sup B(r) exists in R.
Now we show that br = sup B(r). If 0 < γ < br , then γ is obviously not an upper bound of B(r)
because br ∈ B(r). By Definition 1.8, we have br = sup B(r).
(d) By part (c), we know that bx , by and bx+y are all well-defined in R. By definition, we have
B(x) = {br | r ∈ Q, r ≤ x},
B(y) = {bs | s ∈ Q, s ≤ y},
B(x + y) = {bt | t ∈ Q, t ≤ x + y}.
Before continuing the proof, we need to show several results:
Lemma 1.1
For every real x and y, we define
B(x, y) = B(x) × B(y) = {br × bs | r, s ∈ Q, r ≤ x, s ≤ y}.
Then we have bx × by = sup B(x, y).
Proof of Lemma 1.1. By definition, bx and by are upper bounds of B(x) and B(y) respectively,
so we have br ≤ bx and bs ≤ by for every br ∈ B(x) and bs ∈ B(y). Therefore, we have
br × bs ≤ bx × by for every br × bs ∈ B(x, y). In other words, bx × by is an upper bound of
B(x, y).
Let 0 < α < bx × by . Then we have bαx < by . We define the number p = 21 ( bαx + by ). It is
obvious from this definition that bαx < p < by . By bαx < p, we have αp < bx and so there exists
br ∈ B(x) such that
α
< br .
(1.1)
p
Similarly, the inequality p < by implies that there exists bs ∈ B(y) such that
p < bs .
(1.2)
Now inequalities (1.1) and (1.2) show that
α < br × bs
for some br × bs ∈ B(x, y). Hence α is not an upper bound of B(x, y) and we have
bx × by = sup B(x, y),
completing the proof of the lemma.
Lemma 1.2
Let S be a set of positive real numbers and bounded above and S −1 = {x−1 | x ∈ S}. Then
1
we have sup S =
.
inf S −1
Proof of Lemma 1.2. Suppose that α is an upper bound of S, i.e., 0 < x ≤ α for all x ∈ S.
Then we have 0 < α−1 ≤ x−1 for all x−1 ∈ S −1 . Hence the result follows directly from the
definitions of the least upper bound and the greatest lower bound.
Chapter 1. The Real and Complex Number Systems
4
Lemma 1.3
For every real x, we have b−x =
1
.
bx
Proof of Lemma 1.3. We have two facts:
1
1
– Fact 1. If b > 1, then b n > 1 for every positive integer n. Otherwise, 0 < b n < 1 implies
1
that 0 < b = (b n )n < 1n = 1 by Theorem 1.21, a contradiction.
1
1
– Fact 2. If m and n are positive integers such that n > m, then b m > b n . Otherwise, it
1
1
follows from Fact 1 that 1 < b m < b n and so it implies that
1
1
1
b = (b m )m < (b n )m < (b n )n = b,
a contradiction.
Let r and s be rational. Define A(x) = {bs | s ∈ Q, s ≥ x}. We next want to prove that
sup B(x) = inf A(x). In fact, it is clear that sup B(x) ≤ inf A(x) by definitions. Suppose that
1
D = inf A(x) − sup B(x). Assume that D > 0. By Fact 2 above, b n − 1 is decreasing as n is
increasing, so there exists a positive integer N such that
1
bx (b n − 1) < D
1
for all n ≥ N . By Theorem 1.20(b), we see that there exist r, s ∈ Q such that x − 2n
<s<x
1
1
1
1
and x < r < x + 2n . Thus we have r − s < x + 2n − (x − 2n ) = n and this implies that
D < br − bs = bs (br−s − 1) ≤ bx (br−s − 1) < D,
a contradiction. Hence we must have sup B(x) = inf A(x).
Define B ′ (−x) = {b−t | t ∈ Q, t ≤ −x}. Let s = −t. Then we have b−t = bs and the inequality
t ≤ −x is equivalent to s ≥ x. Thus we have B ′ (−x) = A(x). Now it follows from Lemma 1.2
and the above facts that
b−x = sup B(−x) =
1
1
1
1
=
=
= x,
inf B ′ (−x)
inf A(x)
sup B(x)
b
completing the proof of the lemma.
We can continue our proof of the problem. Let br and bs be any elements of B(x) and B(y)
respectively. Then we have r and s are rational numbers such that r ≤ x and s ≤ y. Since the sum
t = r + s is also rational and t ≤ x + y, part (b) implies that br × bs = br+s ≤ bx+y . Thus bx+y is
an upper bound of the set B(x, y) and Lemma 1.1 implies that
bx × by ≤ bx+y .
Assume that bx by < bx+y . We deduce from Lemma 1.3 that
by = (b−x bx )by = b−x (bx by ) < b−x bx+y ≤ b−x+(x+y) = by ,
a contradiction. Hence we have the desired result that bx by = bx+y .
This completes the proof of the problem.
Problem 1.7
Rudin Chapter 1 Exercise 7.
Proof.
5
1.4. Properties of the complex field
(a) Since b > 1, it is obvious that
bn − 1 = (b − 1)(bn−1 + bn−2 + · · · + 1) ≥ (b − 1)(1 + 1 + · · · + 1) = n(b − 1)
for any positive integer n.
1
1
(b) It is obvious that b n > 1; otherwise, we have b = (b n )n < 1 which is impossible. The result follows
1
by replacing b by the real number b n in part (a) and Problem 1.6(a).
(c) If t > 1 and n > b−1
t−1 , then part (b) implies that
1
b − 1 ≥ n(b n − 1) >
1
b−1
× (b n − 1)
t−1
1
and so b n < t.
(d) Let w be a number such that bw < y. Let t = y · b−w . It is easily to check that t > 1. If n
is sufficiently large enough, then we have n > b−1
t−1 . Hence it follows from parts (c) and (b) that
1
1
−w
w+ n
n
and thus b
< y for sufficiently large n.
b <t=y·b
(e) Let w be a number such that bw > y. Let t = y −1 · bw . It is obvious that t > 1. If n is sufficiently
large enough, then we have n > b−1
t−1 . Hence it follows from part (c) and then part (b) that
1
n(y −1 bw − 1) > b − 1 ≥ n(b n − 1)
1
and thus bw− n > y for sufficiently large n.
(f) We have A = {w ∈ R | bw < y}. Since x is the least upper bound of A, we have w ≤ x for all w ∈ A.
1
If bx < y, then part (d) implies that bx+ n < y for sufficiently large n and so x + n1 ∈ A. Therefore,
we have x + n1 ≤ x and then n1 ≤ 0, a contradiction. Similarly, if bx > y, then x ∈
/ A and so w < x
1
for all w ∈ A. Now part (e) implies that bx− n > y for sufficiently large n, so we have
w <x−
1
<x
n
for some sufficiently large n. This means that x − n1 is an upper bound of A, contradicting to the
fact that x is the least upper bound of A. Hence, we must have
bx = y.
(g) The uniqueness of x follows from the uniqueness of the least upper bound of the set A.
We finish the proof of the problem.
1.4
Properties of the complex field
Problem 1.8
Rudin Chapter 1 Exercise 8.
Proof. Assume that i > 0. Then we have i · i > i · 0 which implies that −1 > 0, a contradiction. Similarly,
the case i < 0 is impossible.
Problem 1.9
Rudin Chapter 1 Exercise 9.
Chapter 1. The Real and Complex Number Systems
6
Proof. We check Definition 1.5 in this case. Let z = a + bi, w = c + di ∈ C. If a 6= c, then we have either
z < w or z > w. If a = c and b 6= d, then we have either z < w or z > w. If a = c and b = d, then we
have z = w. Therefore, this relation satisfies Definition 1.5(i).
Let z = a + bi, w = c + di and q = e + f i be complex numbers such that z < w and w < q. Since
z < w, we have either a < c or a = c and b < d. Similarly, since w < q, we have either c < e or c = e and
d < f . Combining the above two results, we get either a < e or a = e and b < f . This proves Definition
1.5(ii). Hence this turns C into an ordered set.
Let S = {z = a + bi | b ∈ R, a < 0} ⊂ C. Since −1 ∈ S, it is clear that S is not empty. We will
show that sup S does not exist in C. Assume that w = c + di was the least upper bound of S for some
c, d ∈ R. That is, z ≤ w for all z ∈ S. If c > 0, then the definition of the dictionary order implies that
z < ζ = 0 + bi < w for all z ∈ S, contradicting to the fact that w = sup S. If c < 0, then we have c < 2c
so that ζ = 2c + di ∈ S and w < ζ, contradicting to the fact that w = sup S. Hence this order set does
not have the least-upper-bound property, completing the proof of the problem.
Problem 1.10
Rudin Chapter 1 Exercise 10.
Proof. If v ≥ 0, then we have
|w|2 − u2
z = a + 2abi − b = u + 2
4
2
2
2
12
1
1
i = u + (|w|2 − u2 ) 2 i = u + (v 2 ) 2 i = u + vi = w.
If v ≤ 0, then we have
1
1
z 2 = a2 − 2abi − b2 = u − (|w|2 − u2 ) 2 i = u − (v 2 ) 2 i = u − (−v)i = u + vi = w.
By the above results, we see that every non-zero complex number w has two complex square roots z, z
which are defined as in the question. However, when w = 0, then u = v = 0 which imply that a = b = 0
and thus z = z = 0. Hence, the complex number 0 has only one complex square root which is 0 itself,
completing the proof of the problem.
Problem 1.11
Rudin Chapter 1 Exercise 11.
Proof. Let |z| = r > 0. We define z = rw, where w = zr . Then it is easily to see that this expression
satisfies the required conditions. Assume r1 w1 = r2 w2 , where r1 > 0, r2 > 0 and |w1 | = |w2 | = 1. Then
1
we have rr21 = w
w2 which leads to
r1
w1
|w1 |
=| |=
= 1.
r2
w2
|w2 |
Hence we have r1 = r2 and then w1 = w2 . This completes the proof of the problem.
Problem 1.12
Rudin Chapter 1 Exercise 12.
Proof. This result follows from induction and Theorem 1.33(e).
Problem 1.13
Rudin Chapter 1 Exercise 13.
7
1.5. Properties of Euclidean spaces
Proof. Since
|x| = |x − y + y| ≤ |x − y| + |y|
by Theorem 1.33(e), we have |x| − |y| ≤ |x − y|. Similarly, since
|y| = |y − x + x| ≤ |y − x| + |x|
by Theorem 1.33(e), we have −|x − y| ≤ |x| − |y|. Hence these two results together imply that the desired
result, completing the proof of the problem.
Problem 1.14
Rudin Chapter 1 Exercise 14.
Proof. It follows from Definition 1.32 that
|1 + z|2 + |1 − z|2 = (1 + z)(1 + z) + (1 − z)(1 − z) = 2(1 + |z|2 )
holds. This completes the proof of the problem.
Problem 1.15
Rudin Chapter 1 Exercise 15.
Proof. By the proof of the Schwarz inequality (Theorem 1.35), the equality holds if and only if each term
n
X
|Baj − Cbj |2 is zero, i.e.,
in the sum
j=1
|Baj − Cbj |2 = 0
2
2
(Baj − Cbj )(Baj − Cbj ) = 0
B |aj | + C 2 |bj |2 − 2BCRe (aj bj ) = 0.
(1.3)
By Theorem 1.33(d), we have Re (aj bj ) ≤ |aj bj | and so the relation (1.3) implies that
0 ≥ B 2 |aj |2 + C 2 |bj |2 − 2BC|aj ||bj | = (B|aj | − C|bj |)2 .
|a |
Hence we have the equality holds if and only if B|aj | = C|bj | if and only if |bjj | is a constant for
j = 1, 2, . . . , n. This completes the proof of the problem.
1.5
Properties of Euclidean spaces
Problem 1.16
Rudin Chapter 1 Exercise 16.
Proof. Let m be the mid-point of x and y and u = z − m = (u1 , u2 , . . . , uk ). Geometrically, these
conditions say that the three vectors x, y and u form an isosceles triangle with sides r, r and d. In other
words, u must satisfy the equations
u · (x − y) = 0 and |u|2 = r2 −
d2
.
4
(1.4)
Chapter 1. The Real and Complex Number Systems
8
(a) Since |x − y| = d > 0, we have from Theorem 1.37(b) that x and y are distinct and then we may
assume without loss of generality that x1 6= y1 . If u2 , u3 , . . . , uk are arbitrary, then we define
−1
[u2 (x2 − y2 ) + · · · + uk (xk − yk )].
x1 − y1
u1 =
Then it is clear that this vector u = (u1 , u2 , . . . , uk ) satisfies the equation u · (x − y) = 0. Since
u2 , u3 , . . . , uk are arbitrary, there are infinitely many u1 and then u.
Next, suppose that u is a vector satisfying the equation u · (x − y) = 0. Since 2r > d, we must have
2
2
r2 − d4 > 0. If |u|2 6= r2 − d4 , then we consider the vector u defined by
u=
q
2
r2 − d4
|u|
u.
Thus it is easy to check that u satisfies both equations (1.4). In fact, this proves that there are
infinitely many u satisfying both equations (1.4) and hence there are infinitely many z such that
|z − x| = |z − y| = r.
2
(b) If 2r = d, then we have r2 − d4 = 0 and so |u|2 = 0. It follows from Theorem 1.37(b) that u = 0,
i.e., z = m. This prove the uniqueness of such z.
2
(c) If 2r < d, then we have r2 − d4 < 0 and it is clear that there is no u such that |u|2 < 0. Hence
there is no such z in this case.
When k = 1 or 2, the results will be different from those when k ≥ 3 and the analysis is given as
follows:
• Case (i): k = 2. If 2r > d, then
u=
q
2
r2 − d4
|u|
u
are the only vectors satisfying the equations (1.4), where
x −y
2
2
u2 , u2 .
u=± −
x1 − y1
If 2r = d, then u = 0 is the only vector such that u · (x − y) = 0 and |u|2 = 0. If 2r < d, then
there is no such u. Hence, there are two such z if 2r > d, exactly one such z if 2r = d and no such
z if 2r < d.
• Case (ii): k = 1. Then there is no point u satisfying both
u(x1 − y1 ) = 0
and |u|2 = r2 −
d2
4
if 2r 6= d. However, u = 0 is the only point such that u(x1 − y1 ) = 0 and |u|2 = 0 if 2r = d. Hence,
there is no such z if 2r 6= d and only one such z if 2r = d.
This completes the proof of the problem.
Problem 1.17
Rudin Chapter 1 Exercise 17.
9
1.6. A supplement to the proof of Theorem 1.19
Proof. Let x = (x1 , x2 , . . . , xk ) and y = (y1 , y2 , . . . , yk ). Then we have
2
2
|x + y| + |x − y| =
k
X
j=1
2
2
[(xj + yj ) + (xj − yj ) ] = 2
k
X
j=1
x2j + 2
k
X
yj2 = 2|x|2 + 2|y|2 .
j=1
Suppose that x and y are sides of a parallelogram. Then x + y and x − y are the diagonals of the
parallelogram and the above result can be interpreted as follows: the sum of the squares of the lengths of
the diagonals (left-hand side) is double to the sum of the squares of the lengths of the sides (right-hand
side). This completes the proof of the problem.
Problem 1.18
Rudin Chapter 1 Exercise 18.
Proof. We define x = (x1 , x2 , . . . , xk ), where xj ∈ R for j = 1, 2, . . . , k. If x1 = x2 = · · · = xk = 0, then
the element y = (1, 0, . . . , 0) satisfies the requirements that y 6= 0 and x · y = 0. Otherwise, without loss
of generality, we may assume that x1 6= 0. If we define
x2 + x3 + · · · + xk
, 1, . . . , 1 ,
y= −
x1
then we still have y 6= 0 and x · y = 0. Let x be a non-zero real number. If y 6= 0, then Proposition
1.16(b) implies that x · y 6= 0. Hence this is not true if k = 1, finishing the proof of the problem.
Problem 1.19
Rudin Chapter 1 Exercise 19.
Proof. Let 3c = 4b − a and 3r = 2|b − a|. Since |x|2 = x · x (Definition 1.36), we have the following
relations:
|x − a| = 2|x − b| ⇔ |x|2 − 2a · x + |a|2 = 4|x|2 − 8b · x + 4|b|2
⇔ 3|x|2 − 8b · x + 2a · x + 4|b|2 − |a|2 = 0
⇔ 9|x|2 − 24b · x + 6a · x + 12|b|2 − 3|a|2 = 0
⇔ (3x − 4b + a) · (3x − 4b + a) = 4(b − a)(b − a)
4
1
4
1
4
⇔ x − b + a · x − b + a = (b − a)(b − a)
3
3
3
3
9
1
2
4
⇔ x − b + a = |b − a|
3
3
3
⇔ |x − c| = r,
completing the proof of the problem.
1.6
A supplement to the proof of Theorem 1.19
Problem 1.20
Rudin Chapter 1 Exercise 20.
Proof. Let us recall the first two properties now:
(I) α 6= ∅ and α 6= Q.
Chapter 1. The Real and Complex Number Systems
10
(II) If p ∈ α and q ∈ Q such that q < p, then q ∈ α.
Suppose that “(III) If p ∈ α, then p < r for some r ∈ α” is deleted in the definition of a cut, see p. 17.
Then it is easy to see that Step 2 and Step 3 are still satisfied. For Step 4, we still have the definition
of addition of cuts: if α, β ∈ R, then α + β = {r + s | r ∈ α, s ∈ β}. We define
0∗ = {p ∈ Q | p ≤ 0}.
It is clear that 0∗ satisfies (I) and (II), but it has the maximum element 0. It is also obvious that the
addition so defined satisfies axioms (A1) to (A3).
Let α ∈ R. If r ∈ α and s ∈ 0∗ , then either r + s < r or r + s = r which imply that r + s ∈ α, i.e.,
α + 0∗ ⊆ α. For any p ∈ α, we always have p + 0 = p so that p ∈ α + 0∗ , i.e., α ⊆ α + 0∗ . Hence we have
α = α + 0∗ and the addition satisfies (A4).
Let α = {p ∈ Q | p < 0} ∈ R. Assume that there was β ∈ R such that α + β = 0∗ . Since α 6= ϕ, we
have p ∈ α and then β contains an element q ∈ Q such that p + q = 0. Since p is a negative rational, q
must be a positive rational. By Theorem 1.20(b), we have 0 < q1 < q for some q1 ∈ Q. By (II), we have
q1 ∈ β. However, we have p + q1 > 0 so that p + q1 ∈
/ 0∗ by definition. This contradicts the assumption
and hence we have the fact that the addition does not satisfy (A5). This completes the proof of the
problem.
CHAPTER
2
Basic Topology
2.1
The empty set and properties of algebraic numbers
Problem 2.1
Rudin Chapter 2 Exercise 1.
Proof. Assume that there was a set A such that ∅ is not a subset of it. Then ∅ contains an element
which is not in A. However, ∅ has no element by definition. Hence no such element exists and ∅ must
be a subset of every set. This completes the proof of the problem.
Problem 2.2
Rudin Chapter 2 Exercise 2.
Proof. We don’t use the hint to prove the result. For each positive integer n, we let Pn be the set of all
polynomials of degree less than or equal to n with integer coefficients. Then it is easy to check that the
mapping f : Pn → Zn+1 defined by
f (a0 z n + a1 z n−1 + · · · + an−1 z + an ) = (a0 , a1 , . . . , an )
is bijective. By Example 2.5, it is clear that Z is countable. Thus it follows from Theorem 2.13 that Zn+1
is countable and thus Pn is also countable. For each p(z) ∈ Pn , we let Bp(z) be the set of all roots of
p(z). Since a polynomial p(z) ∈ Pn has at most n (distinct) roots, Bp(z) is a finite set. By the corollary
to Theorem 2.12, the set
[
Bp(z)
S=
p(z)∈Pn
is at most countable. Since every positive integer is an algebraic number (consider the polynomial z − n),
the set of all algebraic numbers A is infinite. Since A is a subset of S, we have A is countable, completing
the proof of the problem.
Problem 2.3
Rudin Chapter 2 Exercise 3.
Proof. Assume that all real numbers were algebraic. Then Problem 2.2 implies that R is countable which
contradicts the corollary of Theorem 2.43. Hence there exist real numbers which are not algebraic. This
completes the proof of the problem.
11
Chapter 2. Basic Topology
2.2
12
The uncountability of irrational numbers
Problem 2.4
Rudin Chapter 2 Exercise 4.
Proof. The set of all irrational real numbers R \ Q is uncountable. Otherwise the corollary of Theorem
2.12 implies that
R = (R \ Q) ∪ Q
is countable, a contradiction to the corollary of Theorem 2.43. This completes the proof of the problem.
2.3
Limit points, open sets and closed sets
Problem 2.5
Rudin Chapter 2 Exercise 5.
Proof. We let
E0 =
and
n1
n
o
n = 1, 2, . . . ,
n
o
1
E1 = 2 +
n = 1, 2, . . . ,
n
n
o
1
E2 = 4 +
n = 1, 2, . . .
n
E = E0 ∪ E1 ∪ E2 .
It is clear that E0 , E1 and E2 are bounded. Furthermore, by Example 2.21(e), we know that each of
E0 , E1 and E2 has exactly one limit point, namely 0, 2 and 4 respectively. Hence the E satisfies the
required conditions. This completes the proof of the problem.
Problem 2.6
Rudin Chapter 2 Exercise 6.
Proof. Let p ∈ (E ′ )′ and Nr (p) be a neighborhood of p for some r > 0. Since p is a limit point of E ′ ,
Definition 2.18(b) implies that there exists a point q 6= p in Nr (p) such that q ∈ E ′ . By the definition of
E ′ , q is a limit point of E and thus there exists a point s 6= q in Nh (q) such that s ∈ E for every h > 0.
If we take h = 21 min(d(p, q), r − d(p, q)), then we have
Nh (q) ⊂ Nr (p).
Furthermore, we must have s 6= p. Otherwise, we have p ∈ Nh (q). If h = 12 d(p, q), then this fact implies
that
1
d(p, q) < h = d(p, q),
2
a contradiction. If h = 21 (r − d(p, q)), then we have r − d(p, q) ≤ d(p, q), but d(p, q) < h = 21 (r − d(p, q))
which implies that
2d(p, q) < r − d(p, q),
a contradiction again. Thus what we have shown is that every neighborhood Nr (p) of p contains a point
s 6= p such that s ∈ E. By Definition 2.18(b), p is a limit point of E and hence p ∈ E ′ and E ′ is closed
by Definition 2.18(d). See Figure 2.1 for the neighborhoods Nh (q) and Nr (p) below.
13
2.3. Limit points, open sets and closed sets
Figure 2.1: The neighborhoods Nh (q) and Nr (p).
We first show that (E)′ ⊆ E ′ . Suppose that p ∈ (E)′ . Then p is a limit point of E = E ′ ∪ E, so for
every r > 0, there exists q ∈ Nr (p) and q 6= p such that q ∈ E = E ′ ∪ E. If q ∈ E, then p is already a limit
point of E and thus p ∈ E ′ . If q ∈ E ′ , then the argument in the previous paragraph can be applied to
obtain that p ∈ E ′ . So we have (E)′ ⊆ E ′ . Conversely, suppose that p ∈ E ′ and Nr (p) is a neighborhood
of p for some r > 0. Since p is a limit point of E, Definition 2.18(b) implies that there exists a point
q 6= p in Nr (p) such that q ∈ E. Recall that E = E ∪ E ′ , we must have q ∈ E and thus p is a limit point
of E. Hence E ′ ⊆ (E)′ and then E ′ = (E)′ .
The sets E and E ′ may have different limit points. For example, we consider the set E = {1, 12 , 13 , . . .}
whose only limit point is 0, i.e., E ′ = {0}. Since E ′ has only one element, Theorem 2.20 implies that
(E ′ )′ = ∅. This completes the proof of the problem.
Problem 2.7
Rudin Chapter 2 Exercise 7.
Proof.
(a) If x ∈ Bn , then x ∈ Bn or x ∈ Bn′ . If x ∈ Bn , then x ∈ Ai for some i ∈ {1, 2, . . . , n}, so x ∈ Ai .
Therefore we have
n
[
x∈
Ai .
i=1
If x ∈ Bn′ , then x is a limit point of Bn . We claim that x ∈ A′i for some i ∈ {1, 2, . . . , n}. Assume
that x 6∈ A′i for all i ∈ {1, 2, . . . , n}. Then for each i ∈ {1, 2, . . . , n}, there exists a neighborhood
Nri (x) of x for some ri > 0 such that Nri (p) ∩ Ai = ∅. Let r = min {ri } > 0.a Then we must
1≤i≤n
have Nr (x) ∩ Ai = ∅ for all i ∈ {1, 2, . . . , n}, so Nr (x) ∩ Bn = ∅ by Remarks 2.11, contradicting
the fact that x is a limit point of Bn . Hence we must have x ∈ A′i for some i ∈ {1, 2, . . . , n} and
a We can define this r because there are only finitely many subsets, namely A , A , . . . , A , of a metric space.
n
1
2
Chapter 2. Basic Topology
14
then
Bn ⊆
Conversely, if x ∈
n
[
i=1
n
[
Ai .
(2.1)
i=1
Ai , then x ∈ Ai and thus x ∈ Ai or x ∈ A′i for some i ∈ {1, 2, . . . , n}. If
x ∈ Ai , then we have x ∈ Bn ⊆ Bn . If x ∈ A′i , then x is a limit point of Ai . Thus there exists
y ∈ Nr (x) for every r > 0 and y 6= x such that y ∈ Ai . Since Ai ⊆ Bn by definition, x is also a
limit point of Bn , i.e., x ∈ Bn′ ⊆ Bn . Both cases together imply that
n
[
i=1
Ai ⊆ Bn .
(2.2)
Hence the above set relations (2.1) and (2.2) imply that
Bn =
n
[
Ai ,
i=1
for n = 1, 2, . . ..
(b) The result follows from a similar argument of part (a). However, the inclusion may be proper. For
example, we consider Ai = { 1i } for all i ∈ N, so we have B = {1, 21 , 13 , . . . , }. By the corollary of
Theorem 2.20, we have A′i = ∅ and thus Ai = { 1i } for all i ∈ N. Thus we have
o
n 1 1
Ai = 1, , , . . . ,
2 3
i=1
∞
[
but B = {0, 1, 12 , 13 , . . .}. Hence we have B ⊃
This completes the proof of the problem.
∞
[
Ai .
i=1
Problem 2.8
Rudin Chapter 2 Exercise 8.
Proof. Let p = (p1 , p2 ) ∈ E. Since E is open in R2 , we have Ns (p) ⊆ E for some s > 0. In other words,
there exists q = (q1 , q2 ) ∈ Ns (p) such that (q1 , q2 ) 6= (p1 , p2 ). Let r > 0. If s ≤ r, then it is obvious that
(q1 , q2 ) ∈ Ns (p) ⊆ Nr (p) in this case. If r < s, then we consider the point p′ = (p1 + 21 r, p2 ). Since
0<
r
1
1
(p1 + r − p1 )2 + (p2 − p2 )2 = r < r,
2
2
we have p′ ∈ Nr (p) and p′ 6= p. Hence p is a limit point of E.
However, the case is not true for closed sets. For example, let E = {(1, 0), (2, 0)}. Then E is closed
by Example 2.21(c) but E ′ = ∅ by the corollary of Theorem 2.20. This completes the proof of the
problem.
Problem 2.9
Rudin Chapter 2 Exercise 9.
Proof.
15
2.4. Some metrics
(a) Let p ∈ E ◦ . To show that p is an interior point of E ◦ , it is equivalent to show that there exists a
neighborhood N of p such that N ⊆ E ◦ and this is equivalent to show that every point of such N
is an interior point of E. By definition, p is an interior point of E, so there is a neighborhood Np
of p such that Np ⊆ E. Now for each q ∈ Np , since Np is an open set by Theorem 2.19, there exists
a neighborhood Nq of q such that
Nq ⊆ Np ⊆ E.
Thus every point of Np is an interior point of E by Definition 2.18(e). Hence we have p is an interior
point of E ◦ , i.e., E ◦ is always open.
(b) By Definition 2.18(e), we always have E ◦ ⊆ E. Now it follows from Definition 2.18(f) that E is
open if and only if every point of E is an interior point of E if and only if E ⊆ E ◦ . Hence we have
E = E ◦ if and only if E is open.
(c) Let p ∈ G. Since G is open, there exists a neighborhood N of p such that
N ⊆ G ⊆ E.
Hence p is also an interior point of E, i.e., p ∈ E ◦ and then G ⊆ E ◦ .
(d) If x ∈ (E ◦ )c , then x 6∈ E ◦ . This means x is not an interior point of E. If x 6∈ E, then x ∈ E c ⊆ E c .
If x ∈ E, then Definition 2.18(e) implies that every neighborhood N of x must satisfy N ∩ E c 6= ∅.
Hence x is a limit point of E c by Definition 2.18(b) and then x ∈ (E c )′ ⊆ E c . Hence we have
(E ◦ )c ⊆ E c .
Conversely, if x ∈ E c , then we have x ∈ E c or x ∈ (E c )′ . If x ∈ E c , then x ∈
/ E. Since E ◦ ⊆ E,
◦
◦ c
c ′
c
we have x ∈
/ E and thus x ∈ (E ) . If x ∈ (E ) , then x is a limit point of E . If x ∈
/ (E ◦ )c , then
◦
x ∈ E . By definition, x is an interior point of E so that there exists a neighborhood N of x such
that N ⊆ E which gives N ∩ E c = ∅, a contradiction to the fact that x is a limit point of E c .
Hence we have x ∈ (E ◦ )c and then E c ⊆ (E ◦ )c .
Combining the two parts, we must have (E ◦ )c = E c .
(e) It is not always true. For example, take E = (−1, 0) ∪ (0, 1) in R. Then it is easy to see that
◦
◦
E ◦ = E by part (b) but E = [−1, 1] so that E = (−1, 1). Hence we have the fact that E ◦ 6= E .
(f) It is not always true. For example, take E = {1} in R. Then the corollary of Theorem 2.20 yields
that E is closed so that E = E. However, we have E ◦ = ∅ by definition so that E ◦ = ∅. Hence
we have the fact that E ◦ 6= E.
This completes the proof of the problem.
2.4
Some metrics
Problem 2.10
Rudin Chapter 2 Exercise 10.
Proof. It is clear that Definition 2.15(a) and (b) are true for this d. Since X is an infinite set, there exists
r ∈ X such that r 6= p and r 6= q. Thus for r ∈ X, we have the following cases:
• Case (i): p = q = r. Then d(p, q) = d(p, r) = d(r, q) = 0 so that d(p, q) = d(p, r) + d(r, q).
• Case (ii): p = q but r 6= p. Then d(p, q) = 0 and d(p, r) = d(r, q) = 1 so that
d(p, q) < d(p, r) + d(r, q).
• Case (iii): p 6= q and r = p. Then d(p, q) = d(r, q) = 1 and d(p, r) = 0 so that
d(p, q) = d(p, r) + d(r, q).
Chapter 2. Basic Topology
16
• Case (iv): p 6= q and r = q. Then d(p, q) = d(p, r) = 1 and d(r, q) = 0 so that
d(p, q) = d(p, r) + d(r, q).
• Case (v): p 6= q, r 6= p and r 6= q. Then d(p, q) = d(p, r) = d(r, q) = 1 so that
d(p, q) < d(p, r) + d(r, q).
Hence Definition 2.15(c) is also true for this d and then it is a metric.
Let x ∈ X. We consider its neighborhood
Nr (x) = {y ∈ X | d(x, y) < r}.
Since Nr (x) ⊂ {x} for any 0 < r ≤ 1, the set {x} is open in X for every x ∈ X. By Theorem 2.24(a),
every subset of X is open in X. Next, it follows from Theorem 2.23 that every subset of X is also closed
in X. It is clear that every finite subset of X is compact. Assume K ⊆ X was compact and infinite. For
x ∈ K, since Gx = {x} is open in X and
[
K⊆
Gx ,
x∈K
the collection {Gx } is an open cover of K. Since K is compact, there exists a positive integer n such that
K⊆
n
[
i=1
Gi = {x1 , . . . , xn }
which contradicts the hypothesis that K is infinite. Hence K is compact in X if and only if K is a finite
subset of X. This completes the proof of the problem.
Problem 2.11
Rudin Chapter 2 Exercise 11.
Proof. We answer the questions one by one:
• For d1 : It is clear that d1 (3, 1) = 4, d1 (3, 2) = 1 and d1 (2, 1) = 1, so we have d1 (3, 1) > d1 (3, 2) +
d1 (2, 1). Hence d1 is not a metric.
• For d2 : It is easy to check that the function d2 satisfies √
Definition 2.15(a) and (b). For any
ab + b. This certainly implies that
non-negative
real
numbers
a
and
b,
we
have
a
+
b
≤
a
+
2
√
√
√
a + b ≤ a + b and so for any x, y, z ∈ R,
p
p
p
p
d2 (x, y) = |x − y| ≤ |x − z| + |z − y| ≤ |x − z| + |z − y| = d2 (x, z) + d2 (z, y).
Hence d2 is a metric.
• For d3 and d4 : The function d3 is not a metric because d3 (1, −1) = |12 − (−1)2 | = 0. Similarly,
the function d4 is not a metric too because d4 (1, 1) = |1 − 2| = 1.
• For d5 : It is clear that d5 satisfies Definition 2.15(a) and (b). To show that d5 also satisfies
Definition 2.15(c), we need a lemma first:
Lemma 2.1
Suppose that a, b and c are non-negative real numbers. If a ≤ b + c, then we have
b
c
a
≤
+
.
1+a
1+b 1+c
17
2.5. Compact sets
Proof of Lemma 2.1. Since 0 ≤ a ≤ b + c, we have 1 ≤ 1 + a ≤ 1 + b + c and then
1+a−1
1
1
b
c
b
c
a
=
= 1−
≤1−
=
+
≤
+
,
1+a
1+a
1+a
1+b+c
1+b+c 1+b+c
1+b 1+c
completing the proof of the lemma.
Now if we put a = |x − y|, b = |x − z| and c = |y − z| into Lemma 2.1, we immediately have the
result that
d5 (x, y) ≤ d5 (x, z) + d5 (z, y).
Hence d5 is a metric.
Now we end the proof of the problem.
2.5
Compact sets
Problem 2.12
Rudin Chapter 2 Exercise 12.
S
Proof. Let K = {0, 1, 21 , 13 , . . .} and {Gα } be a collection of open subsets of R such that K ⊆ α Gα .
Then we must have 0 ∈ Gα1 for some α1 . Since Gα1 is open in R, 0 is an interior point of Gα1 by
Definition 2.18(f). Thus there exists an interval (a, b), where a < 0 < b, such that (a, b) ⊆ Gα1 . By
Theorem 1.20(a) (the Archimedean property), there exists a positive integer N such that N b > 1, i.e.,
b > N1 . Therefore, we have n1 ∈ (a, b) ⊆ Gα1 for all positive integers n ≥ N .
We rewrite K = {1, 12 , . . . , N1−1 } ∪ {0, N1 , N1+1 , . . .}. Now it follows from the previous paragraph that
n
0,
o
1
1
,
, . . . ⊆ Gα1 .
N N +1
In addition, since {1, 21 , . . . , N1−1 } is a finite set, there are finitely many Gα2 , Gα3 , . . . , Gαm such that
Hence we have
n
1
1 o
1, , . . . ,
⊆ Gα2 ∪ Gα3 ∪ · · · ∪ Gαm .
2
N −1
K⊆
m
[
Gαi ,
i=1
i.e., K is compact by Definition 2.32, completing the proof of the problem.
Problem 2.13
Rudin Chapter 2 Exercise 13.
Proof. We define
n
o
n1
o
1 1
1
K0 = 0, 1, , , . . .
and Kn =
+
m = n, n + 1, . . . ,
2 3
n m
where n = 1, 2, . . .. We also define K =
∞
[
Kn . By definition, K0 and each Kn have limit point 0 and
n=0
1 1
1
′
n respectively. Therefore we have {0, 1, 2 , 3 , . . .} ⊆ K .
We claim that K ′ = K0 . Let p ∈ R be a limit point of K. If p < 0, then we define δ = |p|
2 and thus
(p − δ, p + δ) ∩ K = ∅. If p > 2, then we define δ = p−2
and
thus
(p
−
δ,
p
+
δ)
∩
K
=
∅.
If
1 < p ≤ 2,
2
Chapter 2. Basic Topology
18
then we define δ = 12 min(p − 1, 2 − p) and thus the set (p − δ, p + δ) ∩ K contain only finitely many points
of K. Hence we have shown that
K ′ ⊆ [0, 1].
Next, we suppose that p ∈ [0, 1]\K0 . By this assumption, there exists a positive integer k such that
1
1
1
1
1
1
1
1
1
1
2
k+1 < p < k . Since n + n ≥ n + m and n + n ≥ m + m for all m ≥ n, k+m is the maximum of the set
1
1
1
1
, k1 ). If δ = 12 ( k1 − p),
Kk+m ∪ Kk+m+1 ∪ · · · . Define δ = 2 min(p − k+1 , k − p), so (p − δ, p + δ) ⊂ ( k+1
then we have
p−δ =
2
2k − 1
3p
2
1
1 3
1
=
>
−
>
−
2
2
2k
2 k+1 k
2(k + 1)
k+m
for all m > 4(k+1)
2k−1 − k. In this case, the interval (p − δ, p + δ) contains only finitely many points of
1
K1 ∪ K2 ∪ · · · ∪ Km−1 . Similarly, if δ = 21 (p − k+1
), then we have
p−δ =
1
1
2
p
+
>
>
2 2(k + 1)
k+1
k+m
for all m > k+2. In this case, the interval (p−δ, p+δ) contains only finitely many points of K1 ∪· · ·∪Km−1 .
Now both cases show that the interval (p − δ, p + δ) can only contain finitely many points of K and thus
it is not a limit point of K by Theorem 2.20. Therefore we have K ′ = K0 ⊂ K so that K is closed.
Since |x| ≤ 2 for all x ∈ K, the set K is a bounded set. Hence it follows from Theorem 2.41 (the
Heine-Borel theorem) that K is compact, completing the proof of the problem.
Problem 2.14
Rudin Chapter 2 Exercise 14.
Proof. For each n = 2, 3, . . ., we consider the interval Gn = ( n1 , 1). If x ∈ (0, 1), then it follows from
Theorem 1.20(a) (the Archimedean property) that there exists a positive integer n such that nx > 1, i.e.,
x ∈ Gn . Furthermore, we have
∞
[
Gn ,
(0, 1) ⊆
i=2
i.e., {G2 , G3 , . . .} is an open cover of the segment (0, 1).
Assume that {Gn1 , Gn2 , . . . , Gnk } was a finite subcover of (0, 1), where n1 , n2 , . . . , nk are positive
integers and 2 ≤ n1 < n2 < · · · < nk . By definition, we have Gn1 ⊆ Gn2 ⊆ · · · ⊆ Gnk and so
(0, 1) ⊆
k
[
i=1
Gni ⊆ Gnk ,
contradicting to the fact that 2n1k ∈ (0, 1) but 2n1k ∈
/ ( n1k , 1). Hence {G2 , G3 , . . .} does not have a finite
subcover for (0, 1). This completes the proof of the problem.
2.6
Further topological properties of R
Problem 2.15
Rudin Chapter 2 Exercise 15.
Proof. We take the metric space to be the real number line R.
• Example 1. Let En = [n, ∞), where n = 1, 2, 3, . . .. It follows from the corollary of Theorem 2.23
that each En is closed in R. Furthermore, if 1 ≤ n1 < n2 < · · · < nk , then we have
k
\
i=1
Eni = [nk , ∞) 6= ∅,
2.6. Further topological properties of R
19
but
∞
\
n=1
En = ∅ because if x is a real number such that x ∈
contradiction.
∞
\
n=1
En , then x > n for all n ∈ N, a
• Example 2. Let Fn = (0, n1 ), where n = 1, 2, 3, . . .. It is clear that each Fn is bounded. If
1 ≤ n1 < n2 < · · · < nk , then we have
1 Fni = 0,
6= ∅,
nk
i=1
k
\
but
∞
\
n=1
Fn = ∅ because if x is a real number such that x ∈
contradiction.
∞
\
n=1
Fn , then x < n1 for all n ∈ N, a
We note that the above examples satisfy the condition that En+1 ⊆ En and Fn+1 ⊆ Fn for n = 1, 2, . . ..
Hence they show that Theorem 2.36 and its corollary become false if the word “compact” is replaced by
“closed” or by “bounded”. This completes the proof of the problem.
Problem 2.16
Rudin Chapter 2 Exercise 16.
√
√
√ √
Proof. We have E = {p ∈ Q | 2 < p2 < 3} = (− 3, − 2) ∪ ( 2, 3). It is easy to check the boundedness
of E from Definition 2.18(i) because we have 0 ∈ Q and |p| < 10 for all p ∈ E. Let p ∈ Q be a limit point
of E. Thus for every r > 0, there exists q ∈ E and q 6= p such that |p − q| < r. In particular, if we take
r = n1 , where n is a large positive integer, then we have
0<q−
1
1
<p<q+ .
n
n
1
Since 2 < q 2 < 3, we have 2 < q 2 − δ and q 2 + δ < 3 for some δ > 0. Now we choose n > max{ 3q
δ , q }.
We can deduce from this that
p2 < q +
and
2q
1 2
3q
1
= q2 +
+ 2 < q2 +
< q2 + δ < 3
n
n
n
n
(2.3)
2q
1 2
2q
1
= q2 −
+ 2 > q2 −
> q 2 − δ > 2.
(2.4)
n
n
n
n
Thus the inequalities (2.5) and (2.6) imply that p ∈ E and hence E is closed by Definition 2.18(d).
Assume that E was compact in Q. For each n = 1, 2, . . ., we let the sets Vn = {p ∈ Q | 2− n1 < p2 < 3},
q
q
√
√ Vn− = − 3, − 2 − n1 and Vn+ =
2 − n1 , 3 . Then it is easy to see that
p2 > q −
Vn = Vn− ∪ Vn+ .
√
√
√
√
If p ∈ E, then either − 3 < p < − 2 or 2 < p < 3 which implies that either
r
r
√
√
1
1
− 3<p<− 2−
2− <p< 3
or
n
n
(2.5)
for some n ∈ N. Thus we have p ∈ Vn− ∪ Vn+ = Vn for some n ∈ N so that
E⊆
Let x− and x+ be the mid-points of the intervals
∞
[
Vn .
(2.6)
n=1
−
q
q
√
√ 3, − 2 − n1 and
2 − n1 , 3 respectively. If
q ∈ Vn , then we have either q ∈ Vn− or q ∈ Vn+ . If q ∈ Vn− , then there exists a positive real number h > 0
Chapter 2. Basic Topology
20
such that d(x− , q) = r − h, where r =
Definition 2.15(c) that
q
√
3 − 2 − n1 . For all s such that d(q, s) < h, we have from
d(x− , s) ≤ d(x− , q) + d(q, s) < r − h + h = r
so that s ∈ Vn− , i.e., Nh (q) ⊆ Vn− .
By similar argument, we also have Nh (q) ⊆ Vn+ if q ∈ Vn+ . Hence
it follows from the definition (2.5) that q is an interior point of Vn (i.e., Vn is open in Q) and then the
relation (2.6) implies that {Vn } is an open cover of E. Since E is compact, we have
E ⊆ Vn1 ∪ Vn2 ∪ · · · ∪ Vnk ,
where n1 , n2 , . . . , nk are positive integers and n1 < n2 < · · · < nk . Since Vn+1 ⊆ Vn for all positive
integers n, we have E ⊆ Vn1 , a contradiction.
The argument of proving the openness of the set Vn in the previous paragraph can also be applied to
show that E is open in Q. Hence, this completes the proof of the problem.
Problem 2.17
Rudin Chapter 2 Exercise 17.
Proof. By definition, the element of E must be in the representation 0.x1 x2 x3 . . ., where xn ∈ {4, 7} for
all n = 1, 2, . . .. We answer the questions one by one:
• Is E countable? By a similar argument as the proof of Theorem 2.14, we can show that E is
uncountable.
• Is E dense in [0, 1]? Let x = 0.1 and y = 0.2. By definition, we have x < y < z for all z ∈ E.
Hence E is not dense in [0, 1].
• Is E compact? Since 0.4 ≤ x < 0.8 and [0, 1] is bounded, E is bounded. Let x be a limit point of
E. It is clear that 0.4 < x < 0.8. Assume that x ∈
/ E. If we let xn be the first digit in x which is
not 4 or 7 and
h = 0. 00
. . . 0} 1,
| {z
n terms
then it is not difficult to see that the nth or (n + 1)th digit of every number in the interval
(x − h, x + h) is not 4 or 7. This implies that (x − h, x + h) ∩ E = ∅ and x is not a limit point of
E, a contradiction. Hence we have x ∈ E so that E is closed and then compact by Theorem 2.41
(the Heine-Borel theorem).
• Is E perfect? We know that E is closed, i.e., every limit point of E is a point of E. Let
x = 0.x1 x2 x3 . . . and Nh (x) be an element of E and a neighborhood of E respectively, where h > 0.
Let 0 < h < 1, i.e., h = 0.h1 h2 . . ..b Let hn be the first non-zero digit in h. Let y = 0.y1 y2 . . . be
the number such that if the (n + 1)th digit of x is 4 (resp. 7), then the (n + 1)th digit of y is 7
(resp. 4) and all other digits of y are kept the same as those of x. It follows from this definition
that y 6= x and
|x − y| = 0. 0| .{z
. . 0} 3 < h
n terms
so that y ∈ Nh (x). It is obvious that y ∈ E and so x is a limit point of E. Hence E is perfect by
Definition 2.18(h).
This completes the proof of the problem.
Problem 2.18
Rudin Chapter 2 Exercise 18.
b We note that the argument of proving the result when 0 < h < 1 can be applied completely to prove the case when
h ≥ 1.
21
2.7. Properties of connected sets
Proof. By Problem 2.17, we know that E is a nonempty perfect set. However, E may contain a rational
47
number, e.g. 99
= 0.47474747 · · · ∈ E. In fact, we have the following result about Q:
Lemma 2.2
Every rational number in (0, 1) has a terminating or repeating decimal expansion.
Proof of Lemma 2.2. Let pq be a rational number, where p and q are positive integers and p < q.
To find the first decimal digit of pq , we consider 10p ÷ q and the division algorithm (i.e., given
two integers a and b, with b 6= 0, there exist unique integers q and r such that a = bq + r, where
0 ≤ r < |b|) gives 10p = q1 q + p1 , where q1 and p1 are unique integers such that 0 ≤ p1 < q. To
find the second decimal digit of pq , we consider 10p1 ÷ q and the division algorithm again gives
10p1 = q2 q + p2 , where q2 and p2 are unique integers such that 0 ≤ p2 < q. We repeat this
process to get the sequence of integers {p1 , p2 , . . .} such that 0 ≤ p1 , p2 , . . . < q. Since there
are only finite number of integers between 0 and q, this sequence will terminate or repeat. Now we can go back to the proof of the problem. We let F = a + E = {a + x | x ∈ E}, where
a = 0.1 01 001 0001 · · ·. Since F is just a translation of the perfect set E, F is also perfect. The chosen
constant a ensures that elements in F will not have terminating or repeating decimal expansion and thus
are irrational by Lemma 2.2. Hence F is the desired set. This completes the proof of the problem.
2.7
Properties of connected sets
Problem 2.19
Rudin Chapter 2 Exercise 19.
Proof.
(a) Since A and B are closed, we have A = A and B = B by Theorem 2.27(b). Since A and B are
disjoint, we always have A ∩ B = A ∩ B = ∅. Hence they are separated.
(b) Let A and B be disjoint open sets. Without loss of generality, we may assume that A ∩ B 6= ∅. Let
x ∈ A ∩ B. Since A ∩ B = ∅ and A = A ∪ A′ , we must have x ∈ A′ . Since x ∈ B and B is open,
there is a neighborhood N of x such that N ⊆ B. Since x is a limit point of A, there exits y ∈ N
and y 6= x such that y ∈ A. Hence y ∈ A ∩ B, a contradiction.
(c) Fix p ∈ X and δ > 0. By definition, we have A = {q ∈ X | d(p, q) < δ} and B = {q ∈ X | d(p, q) > δ}
which are obviously disjoint open sets by Theorem 2.19. By the result of part (b), A and B are
separated.
(d) Let X be a connected metric space with at least two (distinct) points. Let the two points be a and
b. Assume that X was countable, i.e., X = {a, b, x3 , x4 , . . .}. We define r = d(a, b). Since a and b
are distinct, it follows from Definition 2.15(a) that r > 0. Since [0, r] = (0, r) ∪ {0, r}, (0, r) must
be uncountable by the corollary of Theorem 2.43. Therefore there exists δ ∈ (0, r) such that
d(a, x) 6= δ
for all x ∈ X. Now if we define A = {x ∈ X | d(a, x) < δ} and B = {x ∈ X | d(a, x) > δ}, then they
are nonempty (a ∈ A and b ∈ B) and the result of (c) shows that A and B are separated. However,
X = A∪B
so that X is not connected by Definition 2.45. This contradiction proves the problem.
We complete the proof of the problem.
Chapter 2. Basic Topology
22
Problem 2.20
Rudin Chapter 2 Exercise 20.
Proof. Let E be a connected set. Assume that E = A ∪ B, where A ∩ B = A ∩ B = ∅. Since A ∩ E ⊆ A
and B ∩ E ⊆ B, we have A ∩ E ⊆ A and B ∩ E ⊆ B. Since A ⊆ A, we have A ∩ B ⊆ A ∩ B = ∅. Since
E ⊆ E = A ∪ B, we have
E = E ∩ E = E ∩ (A ∪ B) = (E ∩ A) ∪ (E ∩ B).
Since (A ∩ E) ∩ (B ∩ E) ⊆ A ∩ B = ∅ and (B ∩ E) ∩ (A ∩ E) ⊆ B ∩ A = ∅, we have E is disconnected
which is a contradiction. Hence E must be connected.
However, the interior of a connected set may not be connected. We prove the following lemma first:
Lemma 2.3
Suppose that A and B are connected sets and A ∩ B 6= ∅. Then the set E = A ∪ B is also
connected.
Proof of Lemma 2.3. Assume that E was not connected. By definition, there are separated
sets U and V such that U ∪ V = E = A ∪ B. Let UA = U ∩ A and VA = V ∩ A. Since U and
V are separated, we have
UA ∩ VA ⊆ (U ∩ A) ∩ (V ∩ A) = (U ∩ V ) ∩ (A ∩ A) = ∅ ∩ A = ∅
and similarly UA ∩ VA = ∅. Since UA ∪ VA = A, A is not connected which contradicts the
hypothesis. Therefore we have either
UA = ∅ or VA = ∅.
(2.7)
Similarly, the sets UB = U ∩ B and VB = V ∩ B will imply that B is not connected, a
contradiction again. Therefore we have either
UB = ∅ or VB = ∅.
(2.8)
Now we verify that the cases (2.7) and (2.8) will induce a contradiction. For examples, if
UA = ∅ and UB = ∅, then U ∩ A = ∅ and U ∩ B = ∅ but they imply that
U = U ∩ E = U ∩ (A ∪ B) = (U ∩ A) ∪ (U ∩ B) = ∅
which contradicts U 6= ∅; if UA = ∅ and VB = ∅, then A ∩ U = ∅ and B ∩ V = ∅ which give
(A ∩ B) ∩ U = ∅ and (A ∩ B) ∩ V = ∅ respectively. Thus these imply that
∅ = [(A ∩ B) ∩ U ] ∪ [(A ∩ B) ∩ V ] = (A ∩ B) ∩ (U ∪ V ) = (A ∩ B) ∩ E = A ∩ B
which contradicts the fact that A ∩ B 6= ∅. Other cases can be done similarly, so we have the
desired result that E is connected.
Let’s go back to the proof of the problem. For example, we consider the disks
A = {x ∈ R2 | |x| ≤ 1} and B = {x ∈ R2 | |x − (2, 0)| ≤ 1}.
Define E = A ∪ B. By Definition 2.17, both A and B are convex so that we obtain from Problem 2.21(c)
that A and B are connected. It is clear that x = (1, 0) is the (only) common point of A and B. Hence it
follows from Lemma 2.3 that E is connected. However, it is easy to check from the definition that
E ◦ = A◦ ∪ B ◦ ,
where A◦ = {x ∈ R2 | |x| < 1} and B ◦ = {x ∈ R2 | |x − 2| < 1}. Since A◦ and B ◦ are clearly separated,
E ◦ is not connected. We end the proof of the problem.
23
2.7. Properties of connected sets
Problem 2.21
Rudin Chapter 2 Exercise 21.
Proof.
(a) By definition, we have A0 = {t ∈ R | p(t) ∈ A} and B0 = {t ∈ R | p(t) ∈ B}. If t ∈ A0 ∩ B0 , then
t ∈ A0 and t ∈ B0 . Since t ∈ B0 , we have p(t) ∈ B. Since t ∈ A0 , we have t ∈ A0 or t ∈ A′0 .
If t ∈ A0 , then p(t) ∈ A. Thus p(t) ∈ A ∩ B ⊆ A ∩ B and then A and B are not separated, a
contradiction.
Suppose that t ∈ A′0 . Then t is a limit point of A0 . Therefore we have s ∈ Nh (t) and s 6= t such
that s ∈ A0 for every h > 0. Since s ∈ Nh (t), we have |s − t| < h and thus
|p(s) − p(t)| = |(−s + t)a + (s − t)b| ≤ |s − t|(|a| + |b|) < h(|a| + |b|).
In fact, this implies that p(Nh (t)) = Nh(|a|+|b|)(p(t)) so that
p(s) ∈ Nh(|a|+|b|)(p(t)).
If p(s) = p(t), then we have a = b which contradicts the fact that A and B are separated. Thus
we have p(s) 6= p(t). Recall that s ∈ A0 so that p(s) ∈ A. Hence we have actually shown that p(t)
is a limit point of A, i.e., p(t) ∈ A which gives p(t) ∈ A ∩ B and then A and B are not separated,
a contradiction.
(b) Assume that p(t) ∈ A ∪ B for all t ∈ (0, 1). Since p(0) = a and p(1) = b, we have p(t) ∈ A ∪ B
for all t ∈ [0, 1] and thus [0, 1] ⊆ A0 ∪ B0 by definition. Let C0 = [0, 1] ∩ A0 and D0 = [0, 1] ∩ B0 .
Then C0 and D0 are nonempty because p(0) = a ∈ A and p(1) = b ∈ B. Furthermore, since A0
and B0 are separated, we have
C0 ∩ D0 = ∅ and C0 ∩ D0 = ∅.
Therefore [0, 1] is not connected which contradicts Theorem 2.47. Hence we have p(t0 ) ∈
/ A ∪ B for
some t0 ∈ (0, 1).
(c) By Definition 2.17, a set S ⊆ Rk is said to be convex if, for all x, y ∈ S and all t ∈ (0, 1), we have
(1 − t)x + ty ∈ S, see Figure 2.2.
Figure 2.2: Convex sets and nonconvex sets.
Chapter 2. Basic Topology
24
Assume that S was not connected. Then we have S = A ∪ B, where A and B are separated. Pick
a ∈ A and b ∈ B. By part (b), there exists t0 ∈ (0, 1) such that p(t0 ) ∈
/ A ∪ B = S, so S is not
convex which is contrary to our hypothesis. Hence S must be connected.
This completes the proof of the problem.
2.8
Separable metric spaces and bases and a special case of Baire’s theorem
Problem 2.22
Rudin Chapter 2 Exercise 22.
Proof. Let Qk = {(q1 , q2 , . . . , qk ) | q1 , q2 , . . . , qk ∈ Q}. Since Q is countable, Theorem 2.13 shows that Qk
is countable. To prove that Rk is separable, we must show that every nonempty open subset of Rk contains
at least one element in Qk .c Let S be an open set of Rk and p = (p1 , p2 , . . . , pk ) ∈ S. By Definition
2.18(f), there exists h > 0 such that B(p, h) ⊆ S. By definition, B(p, h) = {x ∈ Rk | d(x, p) < h}. If we
h
h
h
take x = (p1 − 2k
, p2 − 2k
, . . . , pk − 2k
), then we have
d(x, p) =
k
X
i=1
pi −
h
h
− pi = < h
2k
2
h
so that x ∈ B(p, h). Since pi − 2k
< pi , where i = 1, 2, . . . , k, Theorem 1.20(b) shows that there exist
h
qi ∈ Q such that pi − 2k < qi < pi , where i = 1, 2, . . . , k. Let q = (q1 , q2 , . . . , qk ). By Definition 2.15(c),
we have
h h
d(q, p) ≤ d(q, x) + d(x, p) < + = h.
2
2
Hence we have q ∈ B(p, h) ⊆ S and Qk is dense in Rk . This ends the proof of the problem.
Problem 2.23
Rudin Chapter 2 Exercise 23.
Proof. Let X be a separable metric space. Since X is separable, it contains a countable dense subset.
Let C = {x1 , x2 , . . .} be a countable dense subset of X. Let B = {Nqm (xk )} be a collection of subsets of
X, where qm ∈ Q+ and xk ∈ C. (Here Q+ is the set of all positive rational numbers.) By Theorem 2.19,
each Nqm (xk ) is open in X. Since C and Q+ are countable, the set B is also countable by Theorem 2.13.
Suppose that x ∈ X and G is an open subset of X containing x. Since G is open, x is an interior
point of G and thus we have Nh (x) ⊆ G for some h > 0, see the black dotted circle in Figure 2.3.
Since C is dense in X, there exists xk ∈ C such that xk ∈ N h (x), see the red dotted circle in Figure
2
2.3. It is clear from Theorem 1.20(b) that d(x, xk ) < qm < h2 for some qm ∈ Q+ so that x ∈ Nqm (xk ).
Since d(x, xk ) < qm < h2 , for every y ∈ Nqm (xk ), we have
d(y, x) ≤ d(y, xk ) + d(xk , x) < qm + qm <
h h
+ =h
2
2
so that y ∈ Nh (x), see the green dotted circle in Figure 2.3. Hence we have
x ∈ Nqm (xk ) ⊆ Nh (x) ⊆ G,
i.e., B is a countable base for X. This completes the proof of the problem.
c That is, every point of Rk is a limit point of Qk . Since Qk ⊆ Rk and Rk is closed in Rk , it follows from Theorem 2.27(c)
that Qk ⊆ Rk . By Definition 2.26, (Qk )′ is the set of all limit points of Qk . Since every point of Rk is a limit point of Qk ,
we have Rk ⊆ (Qk )′ ⊆ Qk . Hence we have Rk = Qk .
25
2.8. Separable metric spaces and bases and a special case of Baire’s theorem
Figure 2.3: The sets Nh (x), N h (x) and Nqm (xk ).
2
Problem 2.24
Rudin Chapter 2 Exercise 24.
Proof. By definition (see Problem 2.22), we have to prove that X has a countable dense subset. Fix δ > 0,
since X 6= ∅, we can pick x1 ∈ X. Then we can choose x2 ∈ X such that d(x1 , x2 ) ≥ δ. Otherwise, we
have d(x1 , x) < δ for all x ∈ X which means that
X ⊆ Nδ (x1 ).
In this case, we replace δ by 2δ and choose x2 ∈ X such that d(x1 , x2 ) ≥ δ2 . Otherwise, we have
X ⊆ N δ (x1 ). Then the process can be repeated but it must stop after a finite number of steps because
2
X ⊆ N δn (x1 )
2
for all nonnegative integers n. However, this implies that X = ∅, a contradiction.
Let j be a positive integer. Having chosen x1 , . . . , xj ∈ X, choose xj+1 ∈ X, if possible, so that
d(xi , xj+1 ) ≥ δ for i = 1, . . . , j. In other words, x1 , x2 . . . , xj+1 are distinct elements. This process must
stop after a finite number of steps. Otherwise, we have the infinite subset E = {x1 , x2 , . . .} of X and
the hypothesis guarantees that E has a limit point x ∈ X. By Theorem 2.20, every neighborhood of x
contains infinitely many points of E. Therefore we have xj , xj+1 , . . . ∈ N δ (x) for some positive integer j
2
so that d(xi , x) < δ2 for all i = j, j + 1, . . .. By this, we have
d(xj , xj+1 ) ≤ d(xj , x) + d(x, xj+1 ) <
δ
δ
+ = δ,
2 2
a contradiction.
Therefore X can be covered by finitely many neighborhoods of radius δ. For each n = 1, 2, . . ., we
take δ = n1 and consider the set En = {xn1 , xn2 , . . . , xnmn }, where xn1 , xn2 , . . . , xnmn are the centers of
such neighborhoods for some positive integer mn , i.e.,
X ⊆ N n1 (xn1 ) ∪ N n1 (xn2 ) ∪ · · · ∪ N n1 (xnmn ).
(2.9)
Chapter 2. Basic Topology
26
Since the set En is at most countable (in fact, it is finite) for each n, the corollary of Theorem 2.12 shows
that
∞
[
E=
En
n=1
is countable.
We claim that E is a countable dense subset of the metric space X. In other words, we have to verify
that every nonempty open subset V of X contains at least one element of E. Let x ∈ V ⊆ X. Since x
is an interior point of V , we have Nh (x) ⊆ V for some h > 0. By Theorem 1.20(a) (the Archimedean
property), there exists a positive integer n such that n1 < h. Fix this n and we follow from the relation
(2.9) that
x ∈ N n1 (xnk )
for some k ∈ {1, 2, . . . , mn }. By definition, we have d(x, xnk ) < n1 < h so that
xnk ∈ Nh (x) ⊆ V.
Hence the claim is true, i.e., E is a countable dense subset of X. This completes the proof of the
problem.
Problem 2.25
Rudin Chapter 2 Exercise 25.
Proof. Since K is compact, it follows from Theorem 2.37 that every infinite subset E of K has a limit
point in K. Since K is a metric space, we have K is separable by Problem 2.24. Hence we deduce from
Problem 2.23 that K has a countable base, completing the proof of the problem.
Problem 2.26
Rudin Chapter 2 Exercise 26.
Proof. By Problem 2.24, X is separable. Then by Problem 2.23, it has a countable base. It follows that
every open cover of X has a countable subcover {Gn }, n = 1, 2, 3, . . .. If no finite subcollection of {Gn }
covers X, then the complement
Fn = X \ (G1 ∪ G2 ∪ · · · ∪ Gn )
T
T
is nonempty for each n. However, Fn is an empty set. Otherwise, x ∈ Fn implies that x ∈ Fn for
every n and then
x∈
/ G1 ∪ G2 ∪ · · · ∪ Gn
for each n. This contradicts the fact that {Gn } is an open cover of X.
Let E = {x1 , x2 , . . .}, where xi ∈ Fi and i = 1, 2, . . .. We note that Fn+1 ⊆ Fn , so it may happen that
xn+1 = xn . Assume that E was finite, i.e., E = {xn1 , . . . , xnk }. Since {Gn } is an open cover of X, we
must have xni ∈ Gmi , where mi ∈ N for i = 1, 2, . . . , k.d Let m = max(m1 , m2 , . . . , mk ). Since xm ∈ Fm
by definition of E, we have
(2.10)
xm ∈
/ Gm1 ∪ Gm2 ∪ · · · ∪ Gmk .
However, since xm = xni for some i = 1, 2, . . . , k, we must have xm ∈ Gmi which contradicts the
hypothesis (2.10). Therefore E must be an infinite set and it follows from the hypothesis that E has a
limit point x in X.
Again we recall that X is covered by G1 , G2 , . . . so that x ∈ Gn for some n. Since Gn is open, there
is a δ > 0 such that Nδ (x) ⊆ Gn . Let m be a positive integer greater than n. If xm ∈ Nδ (x), then we
have xm ∈ Gn and so
xm ∈ G1 ∪ · · · ∪ Gn ∪ · · · ∪ Gm
which implies that xm ∈
/ Fm , a contradiction. Therefore x is not a limit point of E which is a contradiction.
Hence we have the desired result that X is compact. This completes the proof of the problem.
d We don’t require that n
i = mi in general.
27
2.8. Separable metric spaces and bases and a special case of Baire’s theorem
Problem 2.27
Rudin Chapter 2 Exercise 27.
Proof. Suppose that E ⊆ Rk , E is uncountable, and let P be the set of all condensation points of E.
• At most countably many points of E are not in P . This statement is equivalent to the
statement that
“P c ∩ E is at most countable.”
By Problem 2.22, we know that Rk is separable and then Problem 2.23 implies that it has a
countable base. Let {Vn } be a countable base of Rk , let W be the union of those Vn for which
E ∩ Vn is at most countable.
Let p ∈ P . Assume that p ∈ W . Since W is open by Theorem 2.24(a), we have p ∈ Vk ⊆ W for
some positive integer k because {Vn } is a base of X (see Problem 2.23). Since Vk is open, we have
Nh (p) ⊆ Vk for some h > 0. By definition of W , E ∩ Vk is at most countable and so is E ∩ Nh (p),
but this contradicts the fact that Nh (p) ∩ E is uncountable. Hence we have p ∈ W c so that
P ⊆ W c.
If p ∈ W c , then p ∈
/ W . Assume that Nh (p) was a neighborhood of p such that Nh (p) ∩ E has
at most countably many points for some h > 0. Since {Vn } is a base of X and Nh (p) is open,
there exists Vk such that p ∈ Vk ⊆ Nh (p) for some positive integer k. Since Nh (p) ∩ E is at most
countable, Vk ∩ E is at most countable too and this implies that
p ∈ Vk ⊆ W,
a contradiction. Hence Nh (p) ∩ E must be uncountable for all h > 0 so that p ∈ P , i.e., W c ⊆ P .
Now the above two paragraphs show our desired result that
P = W c.
(2.11)
• P is perfect. Since W is open, P = W c is closed by the relation (2.11) and Theorem 2.23.e Next,
we let x ∈ P and Nh (x) be a neighborhood of x, where h > 0. Assume that Nh (x) ∩ P = {x}. This
means that if y ∈ Nh (x) \ {x}, then y ∈ P c and it is easy to get from the relation (2.11) that y ∈ W .
Thus we have Nh (x) \ {x} ⊆ W . Since Nh (x) = (Nh (x) \ {x}) ∪ {x}, we have Nh (x) ⊆ W ∪ {x}
and so
Nh (x) ∩ E ⊆ (W ∪ {x}) ∩ E ⊆ (W ∩ E) ∪ {x}.
(2.12)
By definition, W is the union of those Vn for which E ∩ Vn is at most countable. By the corollary
of Theorem 2.12, we have W ∩ E is at most countable and then (2.12) implies that Nh (x) ∩ E is at
most countable too. This implies that x ∈
/ P , a contradiction. Hence there exists y ∈ Nh (x) and
y 6= x such that y ∈ P for every h > 0. In other words, x is a limit point of P and we obtain from
Definition 2.18(h) that P is perfect.
This completes the proof of the problem.
Problem 2.28
Rudin Chapter 2 Exercise 28.
e We can prove that P is closed directly without using the relation (2.11): Let p be a limit point of P . Then for every
r > 0, the neighborhood Nr (p) of p contains a point q 6= p such that q ∈ P . By definition, q is a condensation point of E,
so the neighborhood Nh (q) contains uncountably many points of E, where h = 12 min(d(p, q), r − d(p, q)). Since we have
Nh (q) ⊂ Nr (p) (see Figure 2.1 for clarification), p is also a condensation point of E. That is p ∈ P and so P is closed.
Chapter 2. Basic Topology
28
Proof. Let E be a closed set in a separable metric space X. If E is at most countable, then we have
E = ∅ ∪ E. By Theorem 2.23, the empty set is closed. Since ∅ contains no point, it contains no isolated
point. Thus the empty set must be perfect and we are done in this case.
Suppose that E is uncountable. If we read the proof of Problem 2.27 in detail, then it can be seen
that the proof does not depend on the metric space in which P is embedded. In fact, what the proof
requires is that the space X has a countable base and this is automatically satisfied because of Problem
2.23. Therefore the set P of all condensation points of E is perfect. By definition, a condensation point
of E must be a limit point of E so that P ⊆ E ′ . Since E is closed, we have
P ⊆E
by Theorem 2.27(b). Let F = E \ P . By Problem 2.27 again, the set F must be at most countably many
points. Hence we have
E =P ∪F
and this completes the proof of the problem.
Problem 2.29
Rudin Chapter 2 Exercise 29.
Proof. Let G be an open set in R and x ∈ G. We shall construct the segments with the required
properties. The idea of the construction is that for each x ∈ G, we have to find the maximal segment
containing x and show that such segment is a subset of G. Next, we have to show that two maximal
segments are disjoint. The construction is divided into several steps.
• Step 1: Let Ex = {y ∈ R | (x, y) ⊆ G} and Fx = {y ∈ R | (y, x) ⊆ G}. Since G is open, we have
x ∈ (x − h, x + h) ⊆ G for some h > 0 so that x + h ∈ Ex , x − h ∈ Fx and thus Ex and Fx are
nonempty. We define ax = inf Fx , bx = sup Ex and the segment
Ix = (ax , bx ).f
(2.13)
• Step 2: We show that Ix ⊆ G. By definition, we have x ∈ Ix . Let p ∈ Ix \ {x}. Then we have
either ax < p < x or x < p < bx . Suppose that ax < p < x. Then it follows from Theorem 1.20(b)
(or by Problem 2.22) that
ax < q < p
for some q ∈ Q. Since ax is the greatest lower bound of Fx , we must have y < q for some y ∈ Fx
so that
(q, x) ⊂ (y, x) ⊆ G.
(2.14)
Since q < p < x, we obtain from the relation (2.14) that p ∈ G so that (ax , x) ⊆ G. The other side
is similar and we omit the details here. By these, we have proven Step 2 that Ix ⊆ G. For each
x ∈ G, we have x ∈ Ix ⊆ G which implies that
[
G=
Ix .
x∈G
• Step 3: The segment defined in (2.13) is the maximal segment containing x. To show this property,
we must have
ax ∈
/ G.
Otherwise, we have (ax − h, ax + h) ⊆ G for some h > 0 as G is an open set. This implies that
(ax − h, x) ⊆ G so that ax − h ∈ Fx and then we have ax < ax − h, but ax is the greatest lower
bound of Fx . This is clearly a contradiction. Hence we have ax ∈
/ G. Similarly, we must have
bx ∈
/ G. Now if Ix was not the maximal segment containing x, then we have
Ix ⊆ (a, b) ⊆ G,
where a < ax or bx < b. These imply that ax ∈ G or bx ∈ G, but both lead to a contradiction.
f Here a and b can be possibly −∞ and +∞ respectively.
x
x
29
2.8. Separable metric spaces and bases and a special case of Baire’s theorem
• Step 4: If r, s ∈ G, then we have either
Ir = Is
or Ir ∩ Is = ∅.
Let p ∈ Ir ∩ Is . Then it follows from Theorem 2.24(a) that Ir ∪ Is is an open set containing the
point p. By Step 3 above, we have Ir = Ir ∪ Is and Is = Ir ∪ Is which imply that Ir = Is .
We have shown from Step 1 to Step 4 that the open set G is a union of disjoint segments. By
Theorem 1.20(b), each Ix contains at least one rational number. Since Q is countable, the union of
disjoint segments is at most countable. This finishes the proof of the problem.
Problem 2.30
Rudin Chapter 2 Exercise 30.
Proof. Assume that S
the interior of every Fn was empty. That is, Fn◦ = ∅ for every n. Then we have
S
∞ ◦ c
∞ ◦
k
F
=
∅
and
so
(
n
1 Fn ) = R . By Theorem 2.22, we have
1
∞
\
1
(Fn◦ )c =
∞
[
1
Fn◦
c
= Rk .
By Problem 2.9(d), we have (Fn◦ )c = Fnc . By this and the relation (2.15), we have
implies that
Rk = Fnc
(2.15)
T∞
1
Fnc = Rk which
(2.16)
for every n. For every n, since Fn is closed, Fnc is open and thus this and the relation (2.16) deduce that
each Fnc is a dense open subset of Rk .
As suggested by Rudin, we can “imitate” the proof of Theorem 2.43 to obtain the required result.
When we read the proof of Theorem 2.43 closely, we see that the core idea of it is to construct a shrinking
sequence of nonempty compact sets Kn so that the corollary of Theorem 2.36 can be applied to obtain a
contradiction. Now we follow this idea to construct such a shrinking sequence of nonempty compact sets
in the following paragraph, see also Figure 2.4 for the idea of constructing the shrinking sequence.
Figure 2.4: The construction of the shrinking sequence.
Chapter 2. Basic Topology
30
Let G be an open set of Rk .g Since F1c is a dense subset of Rk , there exists p ∈ F1c such that p ∈ G,
see Problem 2.22 and its footnote. Thus we have
G1 = F1c ∩ G 6= ∅.
Since F1c is open in Rk , G1 must be open in Rk by Theorem 2.24(c). Let p1 ∈ G1 .h Since G1 is open, we
have Nr1 (p1 ) ⊆ G1 for some r1 > 0. Without loss of generality, we may assume that
Nr1 (p1 ) ⊆ G1 .
Since F2c is a dense subset of Rk , the set G2 = F2c ∩ Nr1 (p1 ) is a nonempty open subset of Rk . Let
p2 ∈ G2 . Then we can choose r2 > 0 small enough such that
Nr2 (p2 ) ⊆ G2 .
By definition, we have
Nr2 (p2 ) ⊆ G2 = F2c ∩ Nr1 (p1 ) ⊂ Nr1 (p1 ).
Now we can continue this process to obtain the following shrinking sequence
· · · ⊂ Nr3 (p3 ) ⊂ Nr2 (p2 ) ⊂ Nr1 (p1 ).
(2.17)
By Theorem 2.27(a), each Nrn (pn ) is closed. It is clear that each Nrn (pn ) is bounded, so Theorem
2.41 (the Heine-Borel theorem)
T∞implies that each member in (2.17) is compact. Therefore the corollary
of Theorem 2.36 shows that 1 Nrn (pn ) is nonempty. Since it is true that Nrn (pn ) ⊆ Fnc for every n,
we have
∞
∞
\
\
Nrn (pn ) ⊆
Fnc
which says that the set
T∞
1
1
c
1 Fn is nonempty. However, this and Theorem 2.22 lead the result that
∞
[
1
Fn
c
6= ∅.
S∞
In other words, we have 1 Fn 6= Rk which contradicts our hypothesis. Hence there is at least one Fn
having a nonempty interior. This completes the proof of the problem.
g The construction of the required shrinking sequence can be “seen” from Figure 2.4.
h Here we don’t assume that p = p .
1
CHAPTER
3
Numerical Sequences and Series
3.1
Problems on sequences
Problem 3.1
Rudin Chapter 3 Exercise 1.
Proof. Suppose that {sn } converges to s. By Definition 3.1, for every ε > 0, there is an integer N such
that n ≥ N implies that |sn − s| < ε. By Problem 1.13, we have
||sn | − |s|| ≤ |sn − s| < ε
for all n ≥ N . Hence the sequence {|sn |} converges to |s|. However, the converse is not true. For
example, we let sn = (−1)n so that {sn } is divergent, but |sn | = 1 → 1. This completes the proof of the
problem.
Problem 3.2
Rudin Chapter 3 Exercise 2.
Proof. We have
p
n2 + n − n2
1
1
lim ( n2 + n − n) = lim √
= lim q
= .
n→∞
n→∞
2
n2 + n + n n→∞ 1 + 1 + 1
n
This completes the proof of the problem.
Problem 3.3
Rudin Chapter 3 Exercise 3.
Proof. It follows from s1 =
√
2 < 2 and the induction that
q
√
√
sk+1 = 2 + sk < 2 + 2 < 2
when sk < 2. Hence we have sn < 2 for all n = 1, 2, 3, . . .. By similar argument, we can show that sn > 0
for all n = 1, 2, 3, . . .. Therefore we have
0 < sn < 2
31
Chapter 3. Numerical Sequences and Series
32
for all n = 1, 2, 3, . . . so that {sn } is bounded. Since
√
√
q
−( sn − 2)( sn + 1)
√
p
and sn+1 − sn = 2 + sn − sn =
,
√
2 + sn + sn
√
√
sn − 2 < 2 − 2
we have sn+1 − sn > 0 so that {sn } is strictly increasing. By Theorem 3.14 (Monotone Convergence
Theorem), we have {sn } converges. This completes the proof of the problem.
Problem 3.4
Rudin Chapter 3 Exercise 4.
Proof. By checking the first few terms, we see that
s1 = 0, s2 = 0, s3 =
1
3
3
7
1
, s4 = , s5 = , s6 = , s7 = , . . . .
2
4
4
8
8
Therefore we can show by induction that
s2m =
2m−1 − 1
2m
and s2m+1 =
2m − 1
,
2m
where m = 1, 2, . . .. By Definition 3.16, we have E = { 21 , 1}. Hence we have
lim sup sn = 1 and
n→∞
lim inf sn =
n→∞
1
,
2
completing the proof of the problem.
Problem 3.5
Rudin Chapter 3 Exercise 5.
Proof. Let s = lim sup(an + bn ), a = lim sup an and b = lim sup bn . Suppose that the sum on the right is
n→∞
n→∞
not of the form ∞ − ∞.
n→∞
• Case (i): When a = +∞ or b = +∞. The inequality holds trivially in this case.
• Case (ii): When a < +∞ and b < +∞. By Theorem 3.17(a), there exists a subsequence {ank +bnk }
such that s = lim (ank + bnk ).
k→∞
– If any one of {ank } and {bnk } converges, say {ank }, then the equation bnk = (ank + bnk ) − ank
implies that {bnk } also converges and it follows from Theorem 3.3(a) and Definition 3.16 that
s = lim (ank + bnk ) = lim ank + lim bnk ≤ a + b.
k→∞
k→∞
k→∞
– If both {ank } and {bnk } diverge but {ank + bnk } convergea, then we consider
α = lim sup ank
k→∞
and β = lim sup bnk .
k→∞
By Definition 3.16, we have
lim sup ank ≤ lim sup an
k→∞
n→∞
and
lim sup bnk ≤ lim sup bn
k→∞
n→∞
a For example, a
k
k+1 and b
nk = (−1) so that ank + bnk = 0 for all positive integers k.
nk = (−1)
33
3.2. Problems on series
so that α and β are finite. By Theorem 3.17(a) again, there exists a subsequence {ankj } of
the divergent sequence {ank } such that
lim ankj = α.
j→∞
We note that {bnk } diverges, so it may happen that its subsequence also diverges. However, we
recall from Definition 3.5 that a sequence {pn } converges to p if and only if every subsequence
of {pn } converges to p. Since s = lim (ank + bnk ), the subsequence {ankj + bnkj } converges
k→∞
to this s too. Thus it follows from Theorem 3.3(a) that
lim bnkj = lim [(ankj + bnkj ) − ankj ] = lim (ankj + bnkj ) − lim ankj = s − α.
j→∞
j→∞
j→∞
j→∞
By this, we know that the subsequence {bnkj } of the divergent sequence {bnk } is convergent.
Hence we obtain from the definition of s and Theorem 3.3(a) that
lim sup(an + bn ) = lim (ank + bnk )
k→∞
n→∞
= lim (ankj + bnkj )
j→∞
= lim ankj + lim bnkj
j→∞
j→∞
≤α+β
≤a+b
= lim sup an + lim sup bn .
n→∞
n→∞
This completes the proof of the problem.
3.2
Problems on series
Problem 3.6
Rudin Chapter 3 Exercise 6.
Proof.
(a) We have
an =
√
√
√ √
√
√
1
1
( n + 1 − n)( n + 1 + n)
1
√
=√
n + 1− n =
=
√
√ ≥ √
1 ≥ 0.
n+1+ n
n+1+ n
2 n+1
2 · (n + 1) 2
By Theorem 3.28,
(b) We have
P
1
1
(n+1) 2
diverges. By Theorem 3.25 (Comparison Test),
P
an diverges.
√
√
√
√ √
√
( n + 1 − n)( n + 1 + n)
1
1
1
n+1− n
√
=
= √
an =
√
√ ≤ √ =
3 .
n
2n n
n( n + 1 + n)
n( n + 1 + n)
2n 2
Since
P 1
3
n2
converges by Theorem 3.28, the comparison test shows that
P
an converges.
(c) Since 23 > 1, we have ( 32 )n ≥ ( 32 )2 > 2 for n ≥ 2. Then it can be shown by induction that
n<
3 n
2
Chapter 3. Numerical Sequences and Series
34
P 1
√
for n ≥ 2. By this, we have n n − 1 < 21 which gives an < 21n for n ≥ 2. Since
2n converges by
Theorem 3.26, it follows from Theorem 3.25 (Comparison Test) that
∞
X
an
n=2
converges. Hence
P
(d) The convergence of
an converges because a1 = 0.
P
an depends on the value of |z|.
It is clear that if |z| ≤ 1, then the triangle inequality implies that |1 + z n | ≤ 1 + |z|n so that
|an | =
By Theorem 3.23, we acquire
P
1
1
1
≥
≥ 6= 0.
1 + zn
1 + |z|n
2
an diverges if |z| ≤ 1.
On the other hand, if |z| > 1, then the triangle inequality implies that |z n | ≤ |1 + z n | + 1 so that
1
1
≤ n
.
1 + zn
|z| − 1
|an | =
Since |z| > 1, we have |z| = 1 + δ for some δ > 0. Let N be the least positive integer such thatb
N≥
log 2
.
log(1 + δ)
Then for all positive integers n such that n > N , we have |z|n = (1 + δ)n > 2 so that
|an | ≤
1
2
≤ n.
|z|n − 1
|z|
Hence it follows from Theorem 3.25 (Comparison Test) that
This completes the proof of the problem.
P
an converges.
Problem 3.7
Rudin Chapter 3 Exercise 7.
P
√
Proof. Suppose that an ≥ 0 for all positive integers n and
an converges. Since 0 ≤ ( an − n1 )2 , we
have
√
an
1
1
≤ (an + 2 ).
n
2
n
P 1
By Theorem 3.28,
n2 converges. By the assumption and Theorem 3.25 (Comparison Test), we see that
X √an
n
converges. This completes the proof of the problem.
Problem 3.8
Rudin Chapter 3 Exercise 8.
b The base of the logarithm here is supposed to be 10.
35
3.2. Problems on series
Proof. Suppose that {bn } is increasing.c Since {bn } is bounded, there exists a positive number M such
that −M ≤ bn ≤ M for all positive integers n. Therefore we have −M |an | ≤ |an |bn ≤ M |an | and then
|an bn | ≤ M |an |
P
for all positive integers n. Since
an converges,
M |an | also converges and it follows from Theorem
P
3.25 (Comparison Test) that
an bn converges. This completes the proof of the problem.d
P
Problem 3.9
Rudin Chapter 3 Exercise 9.
Proof.
3
1 3
(a) Since | (n+1)
n3 | = (1 + n ) , we have
α = lim sup
n→∞
Hence we have R = 1.
(b) Since
2n+1
(n+1)!
2n
n!
(n + 1)3
1 3
= 1.
= lim sup 1 +
3
n
n
n→∞
2
, we have
= n+1
α = lim sup
n→∞
2n+1
(n+1)!
2n
n!
= lim sup
n→∞
2
= 0.
n+1
Hence we have R = ∞.
(c) Since
2n+1
(n+1)2
2n
n2
n 2
) , we have
= 2( n+1
α = lim sup
n→∞
2n+1
(n+1)2
2n
n2
= lim sup 2(
(n+1)3
3n+1
n3
3n
1 n+1 3
1
= lim sup (
) = .
n
3
n→∞ 3
n→∞
n 2
) = 2.
n+1
Hence we have R = 12 .
(d) Since
(n+1)3
3n+1
n3
3n
3
= 31 ( n+1
n ) , we have
α = lim sup
n→∞
Hence we have R = 3.
We complete the proof of the problem.
Problem 3.10
Rudin Chapter 3 Exercise 10.
Proof. Suppose that {ank } is the subsequence of {an } such that ank 6= 0, where nk are positive integers
such
p that n1 < n2 < · · · . The subsequence is infinite by the given condition. Since |ank | ≥ 1, we have
nk
|ank | ≥ 1 so that
p
α = lim sup n |an | ≥ 1.
n→∞
Hence we have R = α1 ≤ 1 which is the desired result.
c The case for decreasing sequences is similar, so we omit the details here.
d This result is well-known as Abel’s Test.
Chapter 3. Numerical Sequences and Series
36
Problem 3.11
Rudin Chapter 3 Exercise 11.
Proof.
(a) If the sequence {an } is not bounded, then
1
an
=
→ 1.
1 + an
1 + a1n
P an
It follows from Theorem 3.23 that
1+an diverges. If the sequence {an } is bounded, then an > 0
implies that there is a positive real number M such that an < M for all positive integers n and
thus 1 + an ≤ 1 + M . Therefore we have
an
an
≥
>0
1 + an
1+M
and Theorem 3.25 (Comparison Test) yields that
X
diverges.
an
1 + an
(b) Since an > 0, we have sN +k ≥ sN +j for any fixed positive integer N and for all j = 1, 2, . . . , k.
Therefore we have sN1+k ≤ sN1+j for j = 1, 2, . . . , k. Hence we have
aN +1
aN +k
aN +1
aN +2
aN +k
+ ···+
≥
+
+ ··· +
sN +1
sN +k
sN +k
sN +k
sN +k
1
(aN +1 + aN +2 + · · · + aN +k )
=
sN +k
1
=
(sN +k − sN )
sN +k
sN
=1−
.
sN +k
(3.1)
Since {sk } diverges, {sN +k } diverges. Since an > 0 for all n ∈ N, we have
lim sN +k = +∞.
k→∞
(3.2)
Combining the inequality (3.1) and the limit (3.2), we have
lim
k→∞
Assume that
a
P an
sn
N +1
sN +1
+ ···+
sN 1
aN +k ≥ lim 1 −
= 1 − sN lim
= 1.
k→∞
k→∞ sN +k
sN +k
sN +k
(3.3)
was convergent. By Theorem 3.22, there exists an integer N such that
aN +1
am
1
+ ···+
≤
sN +1
sm
2
if m ≥ N + 1, but this contradicts the result (3.3) if we take m → ∞ in the above inequality. Hence
∞
X
an
n=1
diverges.
sn
37
3.2. Problems on series
(c) Since sn ≥ sn−1 for every positive integer n ≥ 2, we have
1
1
sn − sn−1
an
−
=
≥ 2.
sn−1
sn
sn−1 sn
sn
(3.4)
Since an > 0, we have sn > 0. This and the inequality (3.4) together imply that
2 =
k=1
X 1
1
1
1
2
a1 X ak
1
1
+
+
−
+
−
< .
≤
=
2
2
s1
sk
a1
sk−1
sk
a1
s1
sn
a1
n
n
X
ak
sk
n
k=2
k=2
Hence the partial sums of the series {
P an
s2n } are bounded and it follows from Theorem 3.24 that
X an
s2n
converges.
(d) Since an > 0, we have n2 an < 1 + n2 an and thus
1
an
≤ 2.
1 + n2 an
n
By Theorem 3.28,
P 1
n2
converges. By Theorem 3.25 (Comparison Test), the series
converges.
However, the convergence of the series
X
an
1 + n2 an
X
an
1 + nan
depends on the choice of the sequence {an }.
P
– If an = 1, then we have
an diverges and it follows from Theorem 3.28 that the series
X
diverges.
– To construct a sequence {an } so that
terms
P
1
n
1
1+n
an diverges but
and
an
1 + nan
P
an
1+nan converges, we note that the
an
are of the same magnitude. Therefore, if there are “too many” terms 1+na
in the series, then
n
P1
the series must be divergent because
n diverges.
This observation motivates the construction of such sequence {an } as follows: Define {an } by
k
2 , if n = 2k , where k = 0, 1, 2, . . .;
an =
0, otherwise.
For examples, a1 = 1, a2 = 2, a3 = 0, a4 = 4, a5 = a6 = a7 = 0, a8 = 8 and
a9 = · · · = a15 = 0, a16 = 16, . . . .
It is clear that
∞
X
an =
n=1
By Theorem 3.26, the series
∞
X
k=0
2k
and
∞
X
∞
X 2k
an
=
.
1 + nan
1 + 22k
n=1
X
2k
k=0
Chapter 3. Numerical Sequences and Series
38
P 1
2k
1
diverges. Since 1+2
is convergent by Theorem 3.26, we obtain from Theorem
2k ≤ 2k and
2k
3.25 (Comparison Test) that the series
X
converges.
2k
1 + 22k
This completes the proof of the problem.
Problem 3.12
Rudin Chapter 3 Exercise 12.
Proof.
(a) If m < n, then rn < rm so that the sequence {rn } is strictly decreasing and r1m < r1n . It follows
from this that
am
am+1
an
am
am+1
an
+
+ ···+
>
+
+ ···+
rm
rm+1
rn
rm
rm
rm
rm − rn+1
=
rm
rn+1
=1−
rm
rn
>1−
.
(3.5)
rm
P an
We use similar argument as in the proof of Problem 3.11(b). Assume that
rn was convergent.
Since an > 0 for every n ∈ N, it is easy to see that rn > 0 for all n ∈ N. By Theorem 3.22, there is
an integer N such that
am
an
1
+ ···+
≤
(3.6)
rm
rn
2
if n > m ≥ N . Therefore, by putting the inequality (3.6) into the inequality (3.5), we obtain that
1−
rn
1
≤
rN
2
(3.7)
if n > N . Recall that {rn } is strictly decreasing and rn > 0 for all n ∈ N. Therefore, we apply
Theorem 3.14 (Monotone Convergence Theorem) to {rn } to get the result that
lim rn = 0.
n→∞
Hence we deduce from this and the inequality (3.7) that
rn 1
≤ ,
1 = lim 1 −
n→∞
rN
2
a contradiction and so the required result follows.
(b) Since
√
rn −
√
√
√
√
( rn − rn+1 )( rn + rn+1 )
rn − rn+1
an
√
=√
> √ ,
rn+1 =
√
√
√
rn + rn+1
rn + rn+1
2 rn
the desired inequality follows. Since
n
X
√
√
ak
√
√ < 2( r1 − rn+1 ) < 2 r1 ,
rk
k=1
P an
√
it follows from Theorem 3.24 that
rn converges.
This completes the proof of the problem.
39
3.2. Problems on series
Problem 3.13
Rudin Chapter 3 Exercise 13.
P
bn be two absolutely convergent series. Let
cn be the Cauchy product of the
n
X
P
two series, where cn =
ak bn−k . We have to show that
|cn | converges and we follow part of the idea
Proof. Let
P
an and
P
k=0
of the proof of Theorem 3.50.
n
X
P
P
P
P
Since
an and
bn converge absolutely, we let A =
|an | and B =
|bn |. Put An =
|ak |,
n
X
Bn =
k=0
|bk | and Cn =
n
X
k=0
k=0
|ck |. Now for all n ≥ 0, |an | and |bn | are nonnegative terms of An and Bn
respectively. This implies that An ≤ A and Bn ≤ B for all n ≥ 0 and then we have
|Cn | =
n
X
k=0
|ck |
≤ |a0 ||b0 | + |a0 ||b1 | + |a1 ||b0 | + · · · + |a0 ||bn | + |a1 ||bn−1 | + · · · + |an ||b0 |
= |a0 |Bn + |a1 |Bn−1 + · · · + |an |B0
≤ |a0 |B + |a1 |B + · · · + |an |B
= An B
≤ AB.
Hence {Cn } is a bounded sequence (bounded by AB) and we deduce from Theorem 3.24 that
converges. We finish the proof of the problem.
P
|cn |
Problem 3.14
Rudin Chapter 3 Exercise 14.
Proof.
(a) Given ǫ > 0, there exists a positive integer N such that |sn − s| < 2ǫ for all n ≥ N . Fix this N , we
have
(s0 − s) + · · · + (sN −1 − s) + (sN − s) + · · · + (sn − s)
n+1
1
1
≤
(|s0 − s| + · · · + |sN −1 − s|) +
(|sN − s| + · · · + |sn − s|)
{z
}
n+1
n+1 |
|σn − s| =
(n − N + 1) terms
1
n−N +1 ǫ
≤
(|s0 − s| + · · · + |sN −1 − s|) +
·
n+1
n+1
2
1
ǫ
<
(|s0 − s| + · · · + |sN −1 − s|) + .
n+1
2
(3.8)
Let M = max(|s0 − s|, . . . , |sN −1 − s|) and N ′ be the least integer such that N ′ > 2M
ǫ − 2. Since N
M
< 2ǫ so that inequality
is fixed, M and then N ′ are fixed too. Then for all n ≥ N ′ + 1, we have n+1
(3.8) implies that
|σn − s| < ǫ
for all n ≥ N ′ + 1. This shows that lim σn = s.
1
if n is
(b) Let sn = (−1)n . Then {sn } is obviously divergent. Since σn = 0 if n is odd and σn = n+1
even, we have lim σn = 0.
n→∞
Chapter 3. Numerical Sequences and Series
40
(c) Such sequence {sn } must satisfy two conditions:
– Condition (1): It must contain a divergent subsequence.
1
– Condition (2): The growth of the sum s0 + s1 + · · · + sn = O(n k ) for some k ≥ 2 as n → ∞e
so that
s0 + s1 + · · · + sn
→0
σn =
n+1
as n → ∞.
Now we define the sequence {sn } by
1
n 3 = k, if n = k 3 , where k = 1, 2, . . .;
sn =
1
otherwise.
n2 ,
This sequence {sn } satisfies Condition (1) because
sk 3 = k → ∞
as k → ∞ so that lim sup sn = ∞. Next, for any positive integer n, let k be the largest positive
integer such that k 3 ≤ n < (k + 1)3 . Then we have
(k+1)3
3
(k+1)
X 1
k(k + 1)
1
=
+
0 ≤ s0 + s1 + · · · + sn ≤
m+
m2
2
m2
m=1
m=1
m=1
k
X
X
so that
"
#
(k+1)3
X 1
k(k + 1)
1
s0 + s1 + · · · + sn
.
(3.9)
≤ 3
+
0 ≤ σn =
n+1
k +1
2
m2
m=1
P 1
By Theorem 3.28,
m2 converges. Since the right-hand side of the inequality (3.9) tends to 0 as
k → ∞, it follows from the remarkf above Theorem 3.20 that
lim σn = 0.
n→∞
That is, this sequence {sn } satisfies Condition (2). Hence this prove part (c).
(d) It is clear that the equation is true for n = 1. Assume that it is also true for n = k, where k is a
positive integer. Then for n = k + 1 we have
k
k+1
i
1 X
1 hX
jaj + (k + 1)ak+1
jaj =
k + 2 j=1
k + 2 j=1
1
[(k + 1)(sk − σk ) + (k + 1)ak+1 ]
k+2
1
[(k + 1)sk+1 − (k + 1)σk ]
=
k+2
1
=
[(k + 1)sk+1 + sk+1 − (k + 2)σk+1 ]
k+2
= sk+1 − σk+1 .
=
Hence the induction shows that the expression is true for all positive integers.
Since lim(nan ) = 0, we obtain from part (a) that
n
1 X
kak = 0.
n→∞ n + 1
lim
k=1
e Let f (n) and g(n) be two functions defined on N. One writes f (n) = O(g(n)) as n → ∞ if and only if there is a positive
constant M such that for all sufficiently large values of n, we have |f (n)| ≤ M |g(n)|.
f That is, if 0 ≤ x ≤ s for n ≥ N , where N is some fixed number, and if s → 0, then x → 0.
n
n
n
n
41
3.2. Problems on series
Let lim σn = σ. Then it follows from Theorem 3.3(a) that
lim sn = lim
n→∞
n→∞
σn +
1 X
kak = σ + 0 = σ.
n+1
n
k=1
Hence {sn } converges to σ.
(e) If m < n, then we have
(m + 1)(σn − σm ) +
n
X
i=m+1
(sn − si )
s + s + · · · + s
s0 + s1 + · · · + sm 0
1
n
= (m + 1)
+ [(sn − sm+1 ) + · · · + (sn − sn )]
−
n+1
m+1
m+1
(s0 + s1 + · · · + sn ) + (n − m)sn − (s0 + s1 + · · · + sn )
=
n+1
m+1−n−1
=
(s0 + s1 + · · · + sn ) + (n − m)sn
n+1
= (n − m)(sn − σn )
1
1
which yields the desired formula. Since i ≥ m + 1, we have n − i ≤ n − m − 1 and i+1
≤ m+2
so
that
|sn − si | = |sn − sn−1 + sn−1 − sn−2 + · · · + si+1 − si |
≤ |sn − sn−1 | + |sn−1 − sn−2 | + · · · + |si+1 − si |
M
M
M
≤
+
+ ···+
n
n
−
1
i
+ 1}
|
{z
(n − i) terms
M
M
+ ···+
≤
i+1
i+1
(n − i)
=
M
i+1
(n − m − 1)
≤
M.
m+2
Fix ǫ > 0. Since (m + mǫ + ǫ + 1 + ǫ) − (m + mǫ + ǫ) = 1 + ǫ > 1, there exists an integer n such
that m + mǫ + ǫ ≤ n < m + mǫ + ǫ + 1 + ǫ and the inequalities are equivalent to
m≤
n−ǫ
< m + 1.
1+ǫ
It is easy to see that the above inequalities imply that
m+1
1
≤
n−m
ǫ
and
n−m−1
< ǫ.
m+2
Since |sn − si | ≤ n−m−1
m+2 M , we have |sn − si | < M ǫ and hence
lim sup |sn − σ| ≤ M ǫ.
n→∞
Since ǫ was arbitrary, lim sn = σ.
This completes the proof of the problem.
Problem 3.15
Rudin Chapter 3 Exercise 15.
Chapter 3. Numerical Sequences and Series
42
Proof. We prove the theorems one by one:
• Proof of generalized Theorem 3.22: Let a = (a1 , a2 , . . . , ak ). We further let, for each positive
n
X
ai converges to a ∈ Rk if and
integer n, an = (an1 , an2 , . . . , ank ), where an1 , . . . , ank ∈ R. Now
only if
n
X
i=1
g
aij converges to aj for each j = 1, 2, . . . , k. By Theorem 3.22, we have
i=1
aij <
i=n
aij converges
i=1
to aj if and only if for every ǫ there is an integer Nj such that
m
X
n
X
ǫ
k
(3.10)
if m ≥ n ≥ Nj .
If
n
X
ai converges, then it follows from the inequality (3.10) and
i=1
m
X
i=n
ai ≤
that
m
X
i=n
m
X
i=n
ai <
i=n
for m ≥ n ≥ N = max(N1 , . . . , Nk ).
m
X
ai1 +
ai2 + · · · +
m
X
aik
i=n
ǫ
ǫ
+ ···+ ≤ ǫ
k
k
(3.11)
Conversely, if the inequality (3.11) holds for m ≥ n ≥ N , then since
m
X
i=n
aij ≤
m
X
ai ,
i=n
for each j = 1, 2, . . . , k, we have
m
X
aij < ǫ
i=n
for m ≥ n ≥ N . By Theorem 3.22 again,
n
X
n
X
aij converges to aj for each j = 1, 2, . . . , k so that
i=1
ai converges to a.
i=1
• Proof of generalized Theorem 3.23: We take m = n in the inequality (3.11), then it becomes
|an | < ǫ for all n ≥ N . Since an = (an1 , an2 , . . . , ank ), we have
q
a2n1 + a2n2 + · · · + a2nk < ǫ
for all n ≥ N . By Definition 3.1, we have
lim (a2n1 + a2n2 + · · · + a2nk ) = 0.
n→∞
Since a2nj ≥ 0 for 1 ≤ j ≤ k, we have lim anj = 0 for 1 ≤ j ≤ k, i.e.,
n→∞
lim an = 0.
n→∞
g This holds because we have
n
X
i=1
aij − aj ≤
n
X
i=1
ai − a ≤
n
X
i=1
ai1 − a1 + · · · +
n
X
i=1
aik − ak , where j = 1, 2, . . . , k.
43
3.2. Problems on series
• Proof of generalized Theorem 3.25(a): Given ǫ > 0, there exists N ≥ N0 such that m ≥ n ≥ N
implies
m
X
ck ≤ ǫ
k=n
by the Cauchy criterion. It follows from Theorem 1.37(e) that
m
X
k=n
ak ≤
m
X
k=n
|ak | ≤
m
X
ck < ǫ
k=n
for all m ≥ n ≥ N . By the generalized Theorem 3.22, we have
P
ak converges.
• Proof of generalized
p Theorem 3.33: If α < 1, then we can choose β so that α < β < 1 and an
integer N such that n |an | < β for n ≥ N . That is, for all n ≥ N , we have
|an | < β n .
P n
Since 0 < β <
β converges by Theorem 3.26. Hence it follows from the generalized Theorem
P1,
3.25(a) that
|an | converges.
If α > 1, then we obtain from Theorem 3.17(a) that there is a sequence {nk } such that
p
nk
|ank | → α.
Hence |an | > 1 for infinitely many values of n which contradicts the generalized Theorem 3.23.
• Proof of generalized Theorem 3.34: If part (a) holds, we can find β < 1 and an integer N such
that
an+1
<β
an
for n ≥ N . In particular, we have
|aN +p | < β p |aN |,
(3.12)
where p is a positive integer. It follows from the inequality (3.12) that
|an | < |aN |β −N · β n
for n ≥ N . Hence part (a) follows from the generalized Theorem 3.25(a) because
P
β n converges.
If |an+1 | ≥ |an | for n ≥ n0 , it is easily seen that the condition an → 0 does not hold, and part (b)
follows.
• Proof of generalized
P Theorem 3.42: For every integer n, let an = (an1 , . . . , ank ). Since the
partial sums An of
an form a bounded sequence, we choose
P a positive constant M such that
|An | ≤ M for all n. Let Anj be the partial sum of the series
anj for 1 ≤ j ≤ k. Then we have
|Anj | ≤
for 1 ≤ j ≤ k.
q
A2n1 + A2n2 + · · · + A2nk = |An | ≤ M
(3.13)
Given ǫ > 0, we follow from the facts b0 ≥ b1 ≥ b2 ≥ · · · and lim bn = 0 that there is an integer
n→∞
N such that
ǫ
bN ≤ √
.
(3.14)
2 kM
Let q ≥ p ≥ N . Now we have bn − bn+1 ≥ 0 for all nonnegative integers n. It follows from this,
inequalities (3.13), (3.14) and the facts that (a + b)2 ≤ (|a| + |b|)2 for every a, b ∈ R, we have
q
X
n=p
an bn
Chapter 3. Numerical Sequences and Series
q
X
=
an1 bn , . . . ,
=
n=p
ank bn
n=p
n=p
q−1
X
q
X
44
!
An1 (bn − bn+1 ) + Aq1 bq − A(p−1)1 bp , . . . ,
q−1
X
n=p
Ank (bn − bn+1 ) + Aqk bq − A(p−1)k bp
v
#2
u k " q−1
uX X
u
=u
Anj (bn − bn+1 ) + Aqj bq − A(p−1)j bp
uj=1 n=p
{z
}
|
t
This is b.
{z
}
|
!
This is a.
v
#2
u k " q−1
uX X
t
≤
Anj (bn − bn+1 ) + Aqj bq − A(p−1)j bp
j=1
n=p
v
#2
" q−1
u k
uX
X
t
2
(bn − bn+1 ) + bq + bp
M
≤
j=1
n=p
v
u k
uX
= t (4M 2 b2 )
p
j=1
√
≤ 2M bN k
≤ ǫ.
Now the convergence of the series
P
an bn follows immediately from the generalized Theorem 3.22.
• Proof of generalized Theorem 3.45: The assertion follows from the inequality
m
X
k=n
ak ≤
m
X
k=n
|ak |
plus the generalized Theorem 3.22.
• Proof of generalized Theorem 3.47: Let An =
An + Bn =
n
X
n
X
ak and Bn =
k=0
n
X
bk . Then we acquire
k=0
(ak + bk ).
(3.15)
k=0
Since lim An = A and lim Bn = B, we see from the expression (3.15) that
n→∞
n→∞
lim (An + Bn ) = A + B.
n→∞
The proof of the second assertion is similar.
P ′
• Proof of generalized Theorem 3.55: Let
an be a rearrangement with partial sums s′n . Given
ǫ > 0, there exists an integer N such that m ≥ n ≥ N implies that
m
X
i=n
|ai | < ǫ.
(3.16)
Now choose a positive integer p such that the integers 1, 2, . . . , N are all contained in the set
k1 , k2 , . . . , kp (here we use the notation of Definition 3.52). Then if n > p, the vectors a1 , . . . , aN
will cancel in the difference sn − s′n so that the inequality (3.16) implies that
|sn − s′n | < ǫ.
Hence {s′n } converges to the same sum as {sn }.
This completes the proof of the problem.
45
3.3. Recursion formulas of sequences
3.3
Recursion formulas of sequences
Problem 3.16
Rudin Chapter 3 Exercise 16.
Proof.
√
(a) Since α > 0, it can be shown by induction that xn > α > 0 for all positive integers n.h By this,
we have
−α + x2 x2 + α
n
≤ 0.
− xn = −
xn+1 − xn = n
2xn
2xn
√
Thus {xn } decreases monotonically. Since α < xn ≤ x1 for all positive integers n, the sequence
{xn } is bounded and Theorem 3.14 (Monotone Convergence Theorem) implies that {xn } converges.
Let x = lim xn . Then Theorem 3.3 implies that
α
1
lim xn+1 = lim
xn +
n→∞
n→∞ 2
xn
1
α
x=
x+
2√
x
x = ± α.
Since xn > 0 for all positive integers n, we have x > 0 and hence x =
√
α, as desired.
(b) We have
√
√
1
1 (xn − α)2
ǫ2
α √
= n .
xn +
− α= ·
ǫn+1 = xn+1 − α =
2
xn
2
xn
2xn
√
√
As shown in part (a) that xn > α > 0 for all positive integers n. Thus if β = 2 α, then we have
ǫn+1 =
By this, it is clear that ǫ2 < β ǫβ1
n = k + 1, we have
2
ǫk+2 <
ǫ2n
ǫ2
ǫ2
< √n = n .
2xn
2 α
β
. Assume that ǫk+1 < β ǫβ1
2k
for some positive integer k. For
k
ǫ 2k+1
ǫ2k+1
1 h ǫ1 2 i 2
1
β
.
=β
<
β
β
β
β
Hence it follows from induction that
ǫn+1 < β
for n = 1, 2, . . ..
ǫ 2n
1
β
√
√
√
√
√
√ 3 < 1 . Since ǫ = x1 −
3 = 2 − 3 and β = 2 3, we
(c) Since 25 < 27, we have 5 < 3 3 and so 2−
10
2 3
have
ǫ1
1
<
.
β
10
By part (b), we have
√
ǫ5 < 2 3 · 10−16 < 4 · 10−16
√
and ǫ6 < 2 3 · 10−32 < 4 · 10−32 .
We complete the proof of the problem.
α 2
1
h Note that x2
n+1 − α = 4 (xn − xn ) for every positive integer n.
Chapter 3. Numerical Sequences and Series
46
Problem 3.17
Rudin Chapter 3 Exercise 17.
Proof.
We need two results to prove (a) and (b). Lemma 3.1 is about the magnitude between xn and
√
α. In Lemma 3.2, we obtain the equations of x2n+1 − x2n−1 and x2n − x2n−2 .
Lemma 3.1
For every positive integer n, we have
x2n−1 >
√
α and x2n <
√
α.
Proof of Lemma 3.1. It is clear from
√ the definition and induction that xn > 0 for all positive
integers n. Since α > 1, we have α − 1 > 0 and then
√
√
α−1 √
(3.17)
xn+1 − α =
( α − xn )
1 + xn
for all positive integers n. Therefore the expression (3.17) implies that
√
√
√
( α − 1)2
(x2n−1 − α),
x2n+1 − α =
(1 + x2n )(1 + x2n−1 )
√
√
√
( α − 1)2
x2n+2 − α =
(x2n − α).
(1 + x2n+1 )(1 + x2n )
(3.18)
for all positive integers n. Since x1 > α and x2 < α, we can show by induction and the
expressions (3.18) that
√
√
x2n−1 > α and x2n < α
for all positive integers n.
Lemma 3.2
For every positive integer n, we have
x2n+1 − x2n−1 =
2(α − x22n−1 )
1 + α + 2x2n−1
and x2n+2 − x2n =
2(α − x22n )
.
1 + α + 2x2n
(3.19)
Proof of Lemma 3.2. We note that
α + xn
− xn−1
1 + xn
n−1
α + α+x
1+xn−1
− xn−1
=
n−1
1 + α+x
1+xn−1
xn+1 − xn−1 =
α(1 + xn−1 ) + α + xn−1
− xn−1
1 + xn−1 + α + xn−1
2(α − x2n−1 )
=
.
1 + α + 2xn−1
=
Hence it is easily seen that the expressions (3.19) follows from this.
It is time to return to the proof of Problem 3.17.
47
3.3. Recursion formulas of sequences
(a) Let n be a positive integer. By Lemma 3.1, we have x22n−1 − α > 0. Hence it follows from this and
Lemma 3.2 that x2n+1 < x2n−1 .
(b) Similarly, Lemma 3.1 implies that x22n − α < 0 and we obtain from Lemma 3.2 that x2n+1 > x2n .
(c) By Lemma 3.1 and part (a), {x2n−1 } is monotonically decreasing, so Theorem 3.14 (Monotone
Convergence Theorem) shows that {x2n−1 } converges. Similarly, since Lemma 3.1 and part (b)
imply that {x2n } is monotonically increasing, we obtain from Theorem 3.14 (Monotone Convergence
Theorem) that {x2n } converges. Furthermore, it can be deduced easily from the expressions (3.19)
that
√
lim x2n−1 = lim x2n = α.
(3.20)
n→∞
n→∞
By the triangle inequality, we have
|xm − xn | ≤ |xm −
√
√
α| + | α − xn |
(3.21)
for positive integers m and n. Hence it follows from the limits (3.20) and the inequality (3.21) that
{xn } is a Cauchy sequence. By Theorem 3.11(c), the sequence {xn } converges. Let lim xn = x.
Then we have
α + x n
lim xn+1 = lim
n→∞
n→∞ 1 + xn
α+x
x=
1+x
√
x = ± α.
√
Since xn > 0 for every positive integer n, we have x > 0 and hence x = α.
√
(d) Put ǫn = |xn − α|. We know from the expression (3.17) that
√
√
(1 − α)2
|1 − α|
ǫn =
ǫn−1 .
ǫn+1 =
1 + xn
(1 + xn )(1 + xn−1 )
By the definition of xn , we have (1 + xn+1 )(1 + xn ) = 1 + α + 2xn > 1 + α so that
√
√
√
(1 − α)2
α−1
(1 − α)2
ǫn+1 =
ǫn−1 <
ǫn−1 = √
ǫn−1
(1 + xn )(1 + xn−1 )
1+α
α+1
which implies that
√α − 1 n
ǫ1
ǫ2n+1 < √
α+1
√α − 1 n
and ǫ2n+2 < √
ǫ2 ,
α+1
(3.22)
where n is a positive integer.
To compare the rapidity of convergence of the process with the√one described in Problem 3.16, we
√
3
3
< 10
take the same example that α = 3 and x1 = 2. Then we have √3−1
, ǫ1 = |2 − 3| < 10
and
3+1
√
1
5
ǫ2 = | 3 − 3| < 10 . By these and the inequalities (3.22), we have
ǫ2n+1 <
3 n+1
10
and ǫ2n+2 <
1 3 n
,
·
10
10
where n is a positive integer. Put n = 2 in the above estimates, we have
ǫ5 <
3 3
10
and ǫ6 <
1
3 2
.
·
10 10
Now it is easy to see that the rapidity of convergence in Problem 3.16 is much faster than the one
in this problem.
This completes the analysis of the problem.
Chapter 3. Numerical Sequences and Series
48
Problem 3.18
Rudin Chapter 3 Exercise 18.
Proof. Firstly, it is trivial to check that p ≥ 2 because if p = 1, then xn+1 = α which is meaningless.
Secondly, the recursion formula here reduces to the recursion formula in Problem 3.16 when p =√ 2.
Thirdly, if lim xn exists and let it be x, then it can be shown from the recursion formula that x = p α.
n→∞
√
To prove that {xn } converges, we show that xn > p α for all positive integers n and {xn } is monotonically decreasing in the following two lemmae:
Lemma 3.3
√
√
Suppose that α > 0 and x1 > p α. For every positive integer n, we have xn > p α.
Proof of Lemma 3.3. Assume that xk >
theorem, if 0 < x < 1, then we have
√
p
α for some positive integer k. By the binomial
(1 − x)p = 1 − px +
p(p − 1) 2
x − · · · > 1 − px.
2
(3.23)
Let y = 1 − x. The inequality (3.23) becomes
y p > 1 − p(1 − y).
Next, we put y =
(3.24)
√
pα
xk into the inequality (3.24) to get
√
p
α
α
>
1
−
p
1
−
p
xk
xk
√
p
α
α
>1− p
p 1−
xk
xk
√
1
α
xk − p α > xk − p−1
p
pxk
√
p−1
α
xk + p−1 > p α
p
pxk
√
xk+1 > p α,
completing the proof of the lemma.
Lemma 3.4
The sequence {xn } is monotonically decreasing.
Proof of Lemma 3.4. It is clear from the definition of xn that
1
1
α
xn+1 − xn = − xn + p−1 = p−1 (α − xpn )
p
pxn
pxn
for every positive integer n. By Lemma 3.3, we have α − xpn < 0 so that
xn+1 − xn < 0
for every positive integer n. In other words, the sequence {xn } is monotonically decreasing. 49
3.4. A representation of the Cantor set
Now we can continue our proof of Problem 3.18.
By Lemmae 3.3 and 3.4, the sequence {xn } is
√
monotonically decreasing and bounded below by p α. Hence it follows from Theorem 3.14 (Monotone
Convergence
√ Theorem) that {xn } converges and the analysis preceding Lemma 3.3 gives the limit of it
must be p α. This completes the proof of the problem.
3.4
A representation of the Cantor set
Problem 3.19
Rudin Chapter 3 Exercise 19.
∞
o
n
X
αn
, αn ∈ {0, 2} . Recall that the Cantor set P is defined by
Proof. Let E = x(a) x(a) =
n
3
n=1
m−1
P = [0, 1] \
∞ 3 [−1 [
3k + 1 3k + 2 ,
,
3m
3m
m=1
k=0
see equation (2.24) on [21, p. 42], see Figure 3.1.i
Figure 3.1: The Cantor set.
3k+2
m−1
− 1.
In other words, x ∈
/ P if and only if x ∈ ( 3k+1
3m , 3m ) for some m = 1, 2, . . . and k = 0, 1, . . . , 3
∞
X βn
Suppose that b = {βn } and x(b) =
, where βn ∈ {0, 1, 2}. We want to show that x(b) ∈
/ P if
3n
n=1
and only if βn = 1 for some positive integer n. Then the previous paragraph says that x(b) ∈
/ P if and
only if
∞
X
βn 3k + 1 3k + 2 (3.25)
∈
,
3n
3m
3m
n=1
for some positive integer m and k = 0, 1, . . . , 3m−1 − 1. Fix this m and it is obvious that the relation
(3.25) is equivalent to
∞
X
βn
∈ (3k + 1, 3k + 2).
(3.26)
n−m
3
n=1
Since
∞
m−1
∞
X
X
X
βn
βn
m−n
=
β
3
+
β
+
n
m
n−m
n−m
3
3
n=1
n=1
n=m+1
and each m − n is a positive integer for n = 1, 2, . . . , m − 1, we have each 3m−n is divisible by 3 and then
we have
m−1
X
βn 3m−n = 3N
(3.27)
n=1
i The figure can be found in https://en.wikipedia.org/wiki/Cantor_set.
Chapter 3. Numerical Sequences and Series
for some positive integer N . Let γm = βm +
expression (3.27) that
50
∞
X
βn
, so we have from the relation (3.26) and the
n−m
3
n=m+1
3N + γm ∈ (3k + 1, 3k + 2).
(3.28)
Since 0 ≤ βn ≤ 2 for all n, we get from Theorem 3.26 that
0 ≤ γm ≤ 2 +
1
1
=
2
1
+
+
+
·
·
·
= 3.
3n−m
3 32
n=m+1
∞
X
2
If γm = 0 or 3, then 3N + γm = 3N or 3N + 3 which contradicts the relation (3.28). In fact, the bounds
of γm force that N = k and 1 < γm < 2. Since
0≤
∞
X
βn
≤ 1,
n−m
3
n=m+1
we must have 0 < βm < 2 for this fixed positive integer m. However, we acquire βm ∈ {0, 1, 2} which
implies that βm = 1 for a positive integer m.
Hence we have shown that x(b) ∈
/ P if and only if βn = 1 for some positive integer n and this means
that
E=P
as required. We finish the proof of the problem.
3.5
Cauchy sequences and the completions of metric spaces
Problem 3.20
Rudin Chapter 3 Exercise 20.
Proof. Given ǫ > 0, there is an integer N1 such that d(pni , p) < 2ǫ for ni ≥ N1 . Since {pn } is a Cauchy
sequence, there is an integer N2 such that d(pn , pm ) < 2ǫ for m, n ≥ N2 . Put N = max(N1 , N2 ). Then
for all n ≥ N , we have
ǫ
ǫ
d(pn , p) ≤ d(pn , pni ) + d(pni , p) < + = ǫ.
2 2
Hence the full sequence {pn } converges to p, completing the proof of the problem.
Problem 3.21
Rudin Chapter 3 Exercise 21.
T∞
Proof. Let E = 1 En . Since each En is nonempty, we can construct a sequence {pn }, where pn ∈ En .
Since Em ⊆ En if m ≥ n, we have
pm ∈ En
(3.29)
if m ≥ n.
We first show that the sequence {pn } is convergent. Given that ǫ > 0. Since each En is bounded, we
know from Definition 3.9 that each diam En is well-defined. Besides, since diam En → 0 as n → ∞, there
is an integer N such that
diam En < ǫ
for all n ≥ N . In particular, we take n = N and we obtain from Definition 3.9 that Sn = {d(p, q) | p, q ∈
En } and
d(p, q) ≤ sup SN = diam EN < ǫ
(3.30)
51
3.5. Cauchy sequences and the completions of metric spaces
for p, q ∈ EN .j Since En ⊇ En+1 , we have Em ⊆ En ⊆ EN so that
pm , pn ∈ EN
(3.31)
for any integers m, n with m ≥ n ≥ N . Therefore it follows from the inequality (3.30) and the relation
(3.31) that d(pn , pm ) < ǫ for all m ≥ n ≥ N which shows that {pn } is a Cauchy sequence by Definition
3.8. Since X is a complete metric space, Definition 3.12 ensures that the sequence {pn } converges to a
point p ∈ X.
Next, we prove that E = {p}. To this end, we first show that p ∈ E. Define A to be the subset of N
such that n ∈ A if and only if pn = p. We also define B to be the complement of A so that n ∈ B if and
only if pn 6= p. By definition, we know that n ∈ A if and only if p ∈ En . (However, n ∈ B does not imply
that p ∈
/ En .) Now there are two cases for consideration:
• Case (i): B is finite. Then we have B = {n1 , n2 , . . . , nk }, where n1 < n2 < · · · < nk . This
implies that pn = p for all n ≥ nk + 1 and thus p ∈ En for all n ≥ nk + 1. Since
Enk+1 ⊆ Enk ⊆ Enk−1 ⊆ · · · ⊆ En1 ,
we have p ∈ En for all n ∈ B. Hence we have p ∈ E in this case.
• Case (ii): B is countable. We want to show that p is a limit point of each En , so we fix the integer
n first. Since pm → p as m → ∞, we get from Theorem 3.2(a) that if Nr (p) is any neighborhood
of p, then there is an integer N such that
pm ∈ Nr (p)
for all m ≥ N . Since B is countable, we can pick an element m ∈ B with m ≥ n and m ≥ N such
that the relation (3.29) yields pm ∈ En . By definition, we have pm 6= p and hence it follows from
Definition 2.18(b) that p is a limit point of each En . Since En is closed for each positive integer n,
we have p ∈ En for each positive integer n and thus p ∈ E in this case too.
Now we have shown that p ∈ E and we shall prove the uniqueness of p. Assume that p′ ∈ E but
p 6= p. Then we have d(p, p′ ) > 0 and p, p′ ∈ En for every positive integer n. By Theorem 1.20(b), there
is q ∈ Q which does not depend on n such that 0 < q < d(p, p′ ). Therefore Definition 3.9 implies that
′
diam En ≥ q > 0
for every positive integer n, a contradiction. Hence we must have p = p′ , i.e., E = {p}, completing the
proof of the problem.
Problem 3.22
Rudin Chapter 3 Exercise 22.
Proof. Basically, we follow the idea of proof of Problem 2.30. Let G be an open set of X. Since G1 is a
dense subset of X, there exists p ∈ G1 such that p ∈ G. Thus the set
F1 = G1 ∩ G
is nonempty. Since G1 is open in X, F1 must be open in X by Theorem 2.24(c). Let p1 ∈ F1 . (Here we
don’t assume that p = p1 .) Since F1 is an open subset, we have
E1 = Nr1 (p1 ) ⊆ F1 ⊆ G1
for some r1 > 0. Without loss of generality, we may assume that
E1 = Nr1 (p1 ) ⊆ F1 ⊆ G1 .
j In fact, the inequality (3.30) holds for all n ≥ N with E
N replaced by En .
Chapter 3. Numerical Sequences and Series
52
Since G2 is a dense subset of X, the set
F2 = G2 ∩ E1
is a nonempty open subset of X. Let p2 ∈ F2 . Then we can choose r2 > 0 small enough such that
E2 = Nr2 (p2 ) ⊆ F2 ⊆ G2 .
By definition, we have
E2 = Nr2 (p2 ) ⊆ F2 = G2 ∩ E1 = G2 ∩ Nr1 (p1 ) ⊂ Nr1 (p1 ) = E1 .
Now we can continue this process to obtain the following shrinking sequence
· · · ⊂ E3 ⊂ E2 ⊂ E1 .
(3.32)
Since each En is actually a neighborhood, En is closed and bounded. Furthermore, the sequence (3.32)
implies that {rn } is a strictly decreasing sequence of positive real numbers (here 0 is the greatest lower
bound of {rn }) and it follows from Theorem 3.14 (Monotone Convergence Theorem) that lim rn = 0.
n→∞
Now it is clear that, for each positive integer n,
diam En = sup{d(p, q) | p, q ∈ En } = 2rn ,
so we must have
lim diam En = lim (2rn ) = 0.
n→∞
n→∞
Since X is a nonempty complete metric space, Problem 3.21 implies that
∞
\
n=1
En = {p}
for some p ∈ X. Since p ∈ En ⊆ Gn for each positive integer n, we have p ∈
proof of the problem.
T∞
1
Gn , completing the
Problem 3.23
Rudin Chapter 3 Exercise 23.
Proof. We follow the hint. For any positive integers m and n, we have
d(pn , qn ) ≤ d(pn , pm ) + d(pm , qm ) + d(qm , qn )
or d(pm , qm ) ≤ d(pn , pm ) + d(pn , qn ) + d(qm , qn ) (3.33)
which implies that
|d(pn , qn ) − d(pm , qm )| ≤ d(pn , pm ) + d(qm , qn ).
Since {pn } and {qn } are Cauchy sequences in a metric space X, there is an integer N such that
d(pn , pm ) <
ǫ
2
and d(qn , qm ) <
ǫ
2
(3.34)
for all m, n ≥ N . It follows from the inequalities (3.33) and (3.34) that
|d(pn , qn ) − d(pm , qm )| <
ǫ
ǫ
+ =ǫ
2 2
for m, n ≥ N . Therefore the sequence {d(pn , qn )} is a Cauchy sequence in R. By Theorem 3.11(c), the
sequence {d(pn , qn )} is convergent. This ends the proof of the problem.
Problem 3.24
Rudin Chapter 3 Exercise 24.
53
3.5. Cauchy sequences and the completions of metric spaces
Proof.
(a) We check that this satisfies Definition 2.3.
– Reflexive: It is clear that {pn } ∼ {pn } because d(pn , pn ) = 0 by definition.
– Symmetric: Since d(pn , qn ) = d(qn , pn ), we have {pn } ∼ {qn } implies {qn } ∼ {pn }.
– Transitive: Suppose that {pn }, {qn } and {rn } are Cauchy sequences such that {pn } ∼ {qn }
and {qn } ∼ {rn }. Then we have lim d(pn , qn ) = 0 and lim d(qn , rn ) = 0. The triangle
n→∞
n→∞
inequality gives
d(pn , rn ) ≤ d(pn , qn ) + d(qn , rn )
which yields lim d(pn , rn ) = 0. Thus we have {pn } ∼ {rn }.
n→∞
Hence this is an equivalence relation.
(b) Let {pn }, {p′n } ∈ P and {qn }, {qn′ } ∈ Q. By the triangle inequality, we have
d(p′n , qn′ ) ≤ d(p′n , pn ) + d(pn , qn ) + d(qn , qn′ )
which gives
lim d(p′n , qn′ ) ≤ lim d(p′n , pn ) + lim d(pn , qn ) + lim d(qn , qn′ ).
n→∞
n→∞
n→∞
Since {pn } ∼ {p′n } and {qn } ∼ {qn′ }, we have
have
lim
n→∞
d(p′n , pn ) =
n→∞
lim d(qn′ , qn ) = 0. Therefore, we
n→∞
lim d(p′n , qn′ ) ≤ lim d(pn , qn ).
n→∞
n→∞
(3.35)
Now the roles of pn , qn and p′n , qn′ can be exchanged in the inequality (3.35), so we have
lim d(pn , qn ) ≤ lim d(p′n , qn′ ).
n→∞
n→∞
Hence we have lim d(pn , qn ) = lim d(p′n , qn′ ), i.e., ∆(P, Q) is unchanged.
n→∞
n→∞
Next, we check ∆ is a metric in X ∗ .
– Condition 1: If P 6= Q, then we have {pn } ≁ {qn } for all {pn } ∈ P and {qn } ∈ Q. By the
definition from part (a), {pn } and {qn } are not equivalent so that lim d(pn , qn ) > 0. By the
n→∞
definition of ∆, we have
∆(P, Q) > 0.
Since d(pn , pn ) = 0 for every {pn } ∈ P , we have
∆(P, P ) = lim d(pn , pn ) = 0.
n→∞
– Condition 2: Since d(pn , qn ) = d(qn , pn ) for every {pn } ∈ P and {qn } ∈ Q, we have
∆(P, Q) = ∆(Q, P ).
– Condition 3: Now it is easily seen that the inequality
∆(P, Q) ≤ ∆(P, R) + ∆(R, Q)
follows directly from the triangle inequality d(pn , qn ) ≤ d(pn , rn )+d(rn , qn ) for every {pn } ∈ P ,
{qn } ∈ Q and {rn } ∈ R.
(c) Let {Pn } be a Cauchy sequence in X ∗ with the metric ∆. To show that {Pn } converges in X ∗ , it
is equivalent to showing that there exists a P ∈ X ∗ such that ∆(Pn , P ) → 0 as n → ∞.
Chapter 3. Numerical Sequences and Series
54
– Existence of P . For each positive integer n, recall that elements of each equivalence class Pn
are Cauchy sequences in X with the metric d, so we can choose
{pn1 , pn2 , pn3 , . . .} ∈ Pn .
(1)
Given that ǫ > 0. Since {Pn } is a Cauchy sequence in X ∗ , there exists an integer Nǫ
that
ǫ
∆(Pr , Ps ) <
4
(1)
(2)
for every r, s ≥ Nǫ . By the definition of ∆, there exists a positive integer Nǫ
such
such that
ǫ
d(prt , pst ) <
4
(3.36)
(2)
for t ≥ Nǫ .
Now for each positive integer k, the sequence {pk1 , pk2 , . . .} ∈ Pk is a Cauchy sequence in X,
so there exists an integer Nk such that
d(pkn , pkm ) <
1
2k
(3.37)
if m, n ≥ Nk .
Lemma 3.5
Suppose that pk = pkNk for each positive integer k. We define the sequence {p1 , p2 , . . .}
which is a Cauchy sequence in X.
Proof of Lemma 3.5. We note that the integer Nk depends solely on k (not the arbitrary
log 1
(3)
(1)
constant ǫ), so the sequence {p1 , p2 , . . .} is well-defined. Let Nǫ = max(Nǫ , ⌈2 + log 2ǫ ⌉) and
(2)
n = max(Nr , Ns , Nǫ ). (We denote ⌈x⌉ to be the function which gives the smallest integer
greater than or equal to x.)
(3)
If r and s are fixed positive integers such that r, s ≥ Nǫ , then we have 21r ≤ 4ǫ , 21s ≤ 4ǫ and it
follows from the inequalities (3.36), (3.37) and the triangle inequality that
d(pr , ps ) = d(prNr , psNs )
≤ d(prNr , prn ) + d(prn , psn ) + d(psn , psNs )
1
1
< r + s + d(prn , psn )
2
2
1
1
ǫ
< r + s+
2
2
4
ǫ
ǫ
ǫ
< + +
4 4 4
< ǫ.
Hence {p1 , p2 , . . .} is a Cauchy sequence in X, completing the proof of Lemma 3.5.
By Lemma 3.5, we can define P to be the equivalent class containing the Cauchy sequence
{p1 , p2 , . . .} and it is obvious that P ∈ X ∗ .
– Convergence of {Pm }. We want to show that
∆(Pm , P ) → 0
(3)
as m → ∞. Given that ǫ > 0. By the proof of Lemma 3.5, we see that if m, t ≥ Nǫ , then we
have
3ǫ
(3.38)
d(pm , pt ) < .
4
55
3.5. Cauchy sequences and the completions of metric spaces
(3)
(3)
(1)
log 1
We put m ≥ Nǫ . Since Nǫ = max(Nǫ , ⌈2 + log 2ǫ ⌉), we have 21m ≤ 4ǫ . Then we obtain
from the triangle inequality and the inequality (3.37) that
d(pmt , pt ) ≤ d(pmt , pm ) + d(pm , pt )
= d(pmt , pmNm ) + d(pm , pt )
1
< m + d(pm , pt )
2
(3.39)
if t ≥ Nm .
(4)
(3)
(4)
(3)
(4)
(3)
Let Nǫ = max(Nm , Nǫ ). Now for every t ≥ Nǫ , since m ≥ Nǫ and t ≥ Nǫ ≥ Nǫ , it
follows from the inequality (3.38) that
3ǫ
.
4
d(pm , pt ) <
(4)
Recall that this chosen m gives 21m ≤ 4ǫ and t ≥ Nǫ
inequality (3.39) that
d(pmt , pt ) <
(3)
≥ Nm , so we obtain from these and the
1
ǫ
3ǫ
+ d(pm , pt ) < +
=ǫ
2m
4
4
(3.40)
(4)
if m ≥ Nǫ and t ≥ Nǫ . Therefore we can easily see from the inequality (3.40) and the
definition ∆(Pm , P ) = lim d(pmt , pt ) that ∆(Pm , P ) → 0 as m → ∞.
t→∞
Hence we have shown that a Cauchy sequence {Pn } in X ∗ converges to P ∈ X ∗ . By Definition
3.12, X ∗ is complete.
(d) The sequence {p, p, . . .} is the required Cauchy sequence. Since {p, p, . . .} ∈ Pp and {q, q, . . .} ∈ Pq
for all p, q ∈ X, we get from the part (b) that
∆(Pp , Pq ) = lim d(p, q) = d(p, q).
n→∞
In other words, this shows that ϕ : X → X ∗ defined by ϕ(p) = Pp is an isometry.
(e) To prove that ϕ(X) is dense in X ∗ , we must show that every neighborhood of P ∈ X ∗ contains
ϕ(p) ∈ ϕ(X). To this end, let P ∈ X ∗ and {pn } ∈ P . Given that ǫ > 0. Since {pn } is a Cauchy
sequence in X, there exists a positive integer N such that d(pm , pn ) < 2ǫ for all m, n ≥ N . By part
(d), we have
ϕ(pN ) = PpN .
Recall that PpN contains the Cauchy sequence all of whose terms are pN , so it follows from this
and part (b) that
ǫ
∆(P, ϕ(pN )) = ∆(P, PpN ) = lim d(pn , pN ) ≤ < ǫ.
n→∞
2
In other words, it means that the neighborhood of P with radius ǫ contains an element ϕ(pN ) ∈
ϕ(X). Hence ϕ(X) is dense in X ∗ .
Suppose that X is complete. Let P ∈ X ∗ and {pn } ∈ P . Since {pn } is a Cauchy sequence in the
complete metric space X, we have pn → p for some p ∈ X. Thus we have
∆(P, Pp ) = lim d(pn , p) = 0.
n→∞
Hence we have P = Pp and then ϕ(X) = X ∗ .
This completes the proof of the problem.
Problem 3.25
Rudin Chapter 3 Exercise 25.
Chapter 3. Numerical Sequences and Series
56
Proof. Now we have X = Q. By Problem 3.24(e), Q is dense in Q∗ . By the footnote c in Chapter 2, we
have
Q∗ = Q = R,
finishing the proof of the problem.
CHAPTER
4
Continuity
4.1
Properties of continuous functions
Problem 4.1
Rudin Chapter 4 Exercise 1.
Proof. The answer is no! For example, let f : R → R be defined by f (0) = 1 and f (x) = 0 for x ∈ R \ {0}.
Then it is easy to check that
lim [f (x + h) − f (x − h)] = 0
h→0
for every x ∈ R. However, f is not continuous at 0. To see this, we pick ǫ = 21 and for every δ > 0, those
x ∈ R with 0 < |x − 0| < δ always imply that
|f (x) − f (0)| = |0 − 1| = 1 >
1
.
2
By Definition 4.5, f is not continuous at 0. We finish the proof of the problem.
Problem 4.2
Rudin Chapter 4 Exercise 2.
Proof. Since x ∈ E implies f (x) ∈ f (E) ⊆ f (E), we have
E ⊆ f −1 (f (E)).
By Theorem 2.27(a), f (E) is closed in Y . Thus the corollary of Theorem 4.8 ensures that f −1 (f (E)) is
closed in X. By Theorem 2.27(c), we must have E ⊆ f −1 (f (E)) and hence we have
f (E) ⊆ f (E)
for every set E ⊆ X.
To prove the second assertion, we consider the example f : R \ {0} → R defined by
f (x) =
1
.
x2
Take E = Z, so E = Z and f (E) = f (Z) = {1, 212 , 312 . . .}, but f (E) = f (Z) = {0} ∪ {1, 212 , 312 . . .}. Hence
we have the desired result that
f (E) ⊂ f (E),
completing the proof of the problem.
57
Chapter 4. Continuity
58
Problem 4.3
Rudin Chapter 4 Exercise 3.
Proof. By definition, we have
Z(f ) = {p ∈ X | f (p) = 0} = f −1 ({0}).
Since {0} is a closed set in R, the corollary of Theorem 4.8 implies that Z(f ) is closed in X. We complete
the proof of the problem.
Problem 4.4
Rudin Chapter 4 Exercise 4.
Proof. By definition, the statement “f (E) is dense in f (X)” is equivalent to the statement “every point
of f (X) is a limit point of f (E)” which is also equivalent to f (X) ⊆ (f (E))′ ⊆ f (E). Since it is clear
that f (E) ⊆ f (X), the last statement is equivalent to
f (E) = f (X).
(4.1)
The direction f (E) ⊆ f (X) is obvious. For the other direction, we note from Problem 4.2 that
f (E) ⊆ f (E)
for every set E ⊆ X. If E is a dense subset of X, then we have E = X and this implies that f (X) ⊆ f (E)
which is the relation (4.1).
Let p ∈ X. Thus it is a limit point of E. Since E is dense in X, there exists a sequence {pn } in E
such that lim pn = p. Since f and g are continuous at p, it follows from Theorem 4.2 that
n→∞
lim f (pn ) = f (p) and
n→∞
lim g(pn ) = g(p).
n→∞
Since f (pn ) = g(pn ) for all positive integers n, we have
f (p) = lim f (pn ) = lim g(pn ) = g(p),
n→∞
n→∞
completing the proof of the problem.
4.2
The extension, the graph and the restriction of a continuous function
Problem 4.5
Rudin Chapter 4 Exercise 5.
Proof. We have f : E ⊆ R → R and E is closed in R. By Theorem 2.23, E c is open in R and Problem 2.29
tells us that E c is the union of an at most countable collection of disjoint segments, possibly including
(−∞, a) or (b, ∞) for some a, b ∈ R.a We define g : R → R to be the function such that
g(x) = f (x)
for all x ∈ E. To define g on E c , we let (an , bn ) be one of the disjoint segments of E c , where an < bn .
There are two cases:
a For examples, if E = [0, 1], then we have E c = (−∞, 0) ∪ (1, ∞); if E = (−∞, 0] ∪ [1, +∞), then we have E c = (0, 1).
59
4.2. The extension, the graph and the restriction of a continuous function
• Case (i): E c does not contain (−∞, a) or (b, ∞). We define g on [an , bn ] by

if x = an ;
 f (an ),
m(x − an ) + f (an ) if an < x < bn ;
g(x) =

f (bn ),
if x = bn ,
where
m=
f (bn ) − f (an )
.
b n − an
In other words, the graph of g is a straight line connecting (an , f (an )) and (bn , f (bn )) with slope
m, see Figure 4.1 for an illustration.
Figure 4.1: The graph of g on [an , bn ].
• Case (ii): E c contains (−∞, a) or (b, ∞). We consider the case (−∞, a) only because the case
(b, +∞) can be done similarly. We define g on (−∞, a] by
g(x) = f (a)
for all x ∈ (−∞, a]. In other words, the graph of g is a constant function.
Since a straight line is continuous on its domain, the only points of uncertainty of continuity of g(x)
are at the endpoints x = an and x = bn . Given ǫ > 0 small enough and suppose that m 6= 0. Since f is
continuous at an , there exists a δ > 0 such that
|f (x) − f (an )| < ǫ
ǫ
if |x − an | < δ. Let δ1 = min(δ, |m|
).b Then, for |x − an | < δ1 , we have
|g(x) − g(an )| =
=
<
|m(x − an ) + f (an ) − f (an )|, if an ≤ x < an + δ1 ;
|f (x) − f (an )|,
if an − δ1 < x < an ,
|m| · |x − an |,
|f (x) − f (an )|,
if an ≤ x < an + δ1 ;
if an − δ1 < x < an ,
ǫ, if an ≤ x < an + δ1 ;
ǫ, if an − δ1 < x < an .
b We note that m can be negative, so we take its absolute value to make
ǫ
positive.
|m|
Chapter 4. Continuity
60
If m = 0, then we take δ1 = min(ǫ, δ) in the above analysis and we get
|f (an ) − f (an )|, if an ≤ x < an + δ1 ;
|g(x) − g(an )| =
|f (x) − f (an )|,
if an − δ1 < x < an ,
0,
if an ≤ x < an + δ1 ;
=
|f (x) − f (an )|, if an − δ1 < x < an ,
ǫ, if an ≤ x < an + δ1 ;
<
ǫ, if an − δ1 < x < an .
Hence we have
|g(x) − g(an )| < ǫ
if |x − an | < δ1 and then g is continuous at x = an by Definition 4.5. Since the continuity of g in the
case for the endpoint x = bn and the case for E c containing (−∞, a) or (b, +∞) can be done similarly as
above, we skip the details here.
Note that the word “closed” cannot be omitted in the above result. We modify Example 4.27(c) to
be
−x − 2, if x < 0;
f (x) =
x + 2,
if x > 0.
Then f has a simple discontinuity at x = 0 and is continuous at every other point of the open set
(−∞, 0) ∪ (0, +∞). Since f (0+) = 2 and f (0−) = −2, this makes f impossible to have a continuous
extension.
Suppose that E ⊆ R is a closed set and f : E → Rk is a vector-valued function defined by
f (x) = (f1 (x), f2 (x), . . . , fk (x)),
where each fi : E → R is a real function defined on E. By Theorem 4.10(a), f is continuous on E if
and only if each of the functions f1 , . . . , fk is continuous on E. By the previous analysis, each fi has a
continuous extension gi . Therefore, the vector-valued function g : R → Rk defined by
g(x) = (g1 (x), g2 (x), . . . , gk (x))
is continuous on R by Theorem 4.10(a). For each 1 ≤ i ≤ k, we have gi (x) = fi (x) for all x ∈ E and this
implies that
g(x) = f (x)
for all x ∈ E. This completes the proof of the problem.
Problem 4.6
Rudin Chapter 4 Exercise 6.
Proof. Let f : E → Y , E and Y are metric spaces. By definition, since E × f (E) = {(x, f (y)) | x, y ∈ E},
we have
graph (f ) = {(x, f (x)) | x ∈ E} ⊆ E × f (E) ⊆ E × Y.
Before we start to prove the result, there are a few points to note. We know that E and Y may not be
subsets of Rk for some positive integer k, so we cannot apply any result relevant to Rk (e.g. Theorems
2.41, 4.10, 4.15 and etc.) directly in this problem. Instead, the strategy we use here is that we define a
metric on the set E × Y and we consider the mapping g : E → E × f (E) defined by
g(x) = (x, f (x)).
If we can show that g is continuous, then g(E) = graph (f ) is compact by Theorem 4.14.
The first step is to define the metric in E × Y induced by the metrics dE and dY . Let p1 , p2 ∈ E × Y .
Thus we have p1 = (x1 , y1 ) and p2 = (x2 , y2 ) for some x1 , x2 ∈ E and y1 , y2 ∈ Y , We define
dE×Y (p1 , p2 ) = dE (x1 , x2 ) + dY (y1 , y2 ),
(4.2)
where dE and dY are the metrics in the spaces E and Y respectively. We must check Definition 2.15:
61
4.2. The extension, the graph and the restriction of a continuous function
• Since p1 = p2 if and only if x1 = x2 and y1 = y2 if and only if dE (x1 , x2 ) = 0 and dY (y1 , y2 ) = 0,
we have dE×Y (p1 , p2 ) = 0 if and only if p1 = p2 .
• Since dE (x1 , x2 ) = dE (x2 , x1 ) and dY (y1 , y2 ) = dY (y2 , y1 ), we have
dE×Y (p1 , p2 ) = dE×Y (p2 , p1 ).
• For any p1 , p2 , p3 ∈ E × Y , we have
dE×Y (p1 , p2 ) = dE (x1 , x2 ) + dY (y1 , y2 )
≤ dE (x1 , x3 ) + dE (x3 , x2 ) + dY (y1 , y3 ) + dY (y3 , y2 )
= dE×Y (p1 , p3 ) + dE×Y (p2 , p3 ).
These show that the definition (4.2) is indeed a metric in E × Y . By Example 2.16, we know that every
subset Y of a metric space X is a metric in its own right, with the same distance function. Hence the
expression (4.2) is also a metric in E × f (E).
The second step is to show that g is continuous on E. Let p be a point of E. Given that ǫ > 0. Since
f is continuous at E, it is continuous at p by Definition 4.5. Thus there exists a δ > 0 such that
ǫ
dY (f (x), f (p)) <
(4.3)
2
if dE (x, p) < δ and x ∈ E. We let δ ′ = min( 2ǫ , δ). Then we still have the inequality (4.3) if dE (x, p) < δ ′
and x ∈ E. Hence it follows from the definition (4.2) and the inequality (4.3) that
dE×Y (g(x), g(p)) = dE (x, p) + dY (f (x), f (p)) < δ ′ +
ǫ
ǫ
ǫ
≤ + <ǫ
2
2 2
if dE (x, p) < δ ′ and x ∈ E. Since p is arbitrary, g is continuous on E by Definition 4.5 and then
g(E) = graph (f ) is compact by Theorem 4.14.
Conversely, suppose that graph (f ) is compact and p ∈ E. By Theorem 2.34, we know that graph (f )
is a closed subset of E × f (E). By Theorem 4.6, f is continuous at p if and only if
lim f (x) = f (p).
x→p
The latter statement is equivalent to the fact that
lim f (pn ) = f (p)
n→∞
(4.4)
for every sequence {pn } in E such that pn 6= p and lim pn = p by Theorem 4.2.
n→∞
Let {pn } be a sequence in E such that pn 6= p and lim pn = p and let qn = (pn , f (pn )). Then it is
n→∞
clear that {qn } is a sequence in graph (f ). Since graph (f ) is a compact metric space, we obtain from
Theorem 3.6(a) that a subsequence
{qnk = (pnk , f (pnk ))}
converges to a point q = (q1 , f (q1 )) ∈ graph (f ). Since lim pn = p if and only if every subsequence of
n→∞
{pn } converges to p (Definition 3.5) and the uniqueness of the limit of a sequence (Theorem 3.2(b)), we
must have q1 = p and then
lim f (pnk ) = f (q1 ) = f (p).
(4.5)
k→∞
In fact, the limit (4.5) is true for every subsequence {pnj } converging to p. Otherwise, there was a
subsequence {pnj } converging to p but lim f (pnj ) = r 6= f (p).c Then this implies that
j→∞
lim qnj = (p, r) ∈
/ graph (f )
j→∞
which means the set graph (f ) does not contain its limit point (p, r), but this certainly contradicts the
fact that graph (f ) is closed (Definition 2.18(b)). Therefore the limit (4.5) is true for every subsequence
{pnj } converging to p so that Definition 3.5 induces the fact (4.4). Hence, since p is arbitrary, we have f
is continuous on E, finishing the proof of the problem.
c If r = f (p′ ) for some p′ ∈ E, then it is easily proven that p′ = p. So it is reasonable to assume that r ∈
/ f (E).
Chapter 4. Continuity
62
Problem 4.7
Rudin Chapter 4 Exercise 7.
Proof. We show the result steps by steps as follows:
2
4
• f is bounded on R2 : By the A.M. ≥ G.M., we have x +y
≥
2
|f (x, y)| ≤
p
x2 y 4 = |x|y 2 so that
xy 2
|x|y 2
1
=
≤
2
4
2
4
x +y
x +y
2
for every (x, y) ∈ R2 . Hence f is bounded on R2 .
• g is unbounded in every neighborhood of (0, 0): If we take x = y 3 , then we have
g(x, y) = g(y 3 , y) =
1
y5
=
6
6
y +y
2y
so that g(x, y) → ∞ as y → 0. Hence g is unbounded in every neighborhood of (0, 0).
• f is not continuous at (0, 0): If we take x = y 2 , then we have f (x, y) = f (y 2 , y) = 12 so that
|f (x, y) − f (0, 0)| =
1
2
for every y. Therefore, f does not satisfy Definition 4.5 when ǫ < 12 which means that f is not
continuous at (0, 0).
• The restrictions of both f and g to every straight line in R2 are continuous:
– Case (i): The straight line does not pass through the origin. We have y = mx + c for
some m, c ∈ R and c 6= 0. Let L : R → R2 \ {(0, 0)} be defined by
L(x) = (x, mx + c)
and E = {(x, mx + c) | x ∈ R} be the graph of L. Now we have fE = f ◦ L : R → R.
Since the mapping L is obviously continuous on R by Theorem 4.10(a) and f is continuous
on R2 \ {(0, 0)}, it follows from Theorem 4.7 that fE is continuous on R. The case for the
continuity of gE is similar to that of fE , so we omit its details here.
– Case (ii): The straight line passes through the origin. We have y = mx for some
m ∈ R. Let L′ : R → R2 be defined by
L′ (x) = (x, mx)
and E ′ = {(x, mx) | x ∈ R} be the graph of L′ . It is clear that f is continuous at every point
except at the origin, so we only need to check the continuity of fE ′ = f ◦ L′ : R → R at the
origin. By definition, we have
f (x, mx) =
x(mx)2
x2 + (mx)4
=
m2 x
.
1 + m4 x2
If m = 0, then we have f (x, 0) = 0 so that fE ′ is continuous at the origin because
|f (x, 0) − f (0, 0)| = 0
for every x ∈ R. Without loss of generality, we may assume that m 6= 0. Given that ǫ > 0.
We let δ = mǫ2 . If |x| < δ, then we have
|f (x, mx) − f (0, 0)| =
m2 x
≤ m2 |x| < m2 δ = ǫ.
1 + m4 x2
By Definition 4.5, fE ′ is continuous at the origin. The case for the continuity of gE ′ is similar
to that of fE ′ , so we omit its details here.
We complete the proof of the analysis.
63
4.3. Problems on uniformly continuous functions
4.3
Problems on uniformly continuous functions
Problem 4.8
Rudin Chapter 4 Exercise 8.
Proof. Since E is bounded in R, Theorem 1.19 guarantees that inf E and sup E exist in R. Let them be
a and b respectively so that E ⊆ [a, b]. Since f is uniformly continuous on E, we take ǫ = 1 there exists
a δ > 0 such that
|f (x) − f (y)| < 1
for all x, y ∈ E and |x − y| < δ. Let ∆ = 2(b−a)
and N be the least positive integer such that N ≥ 2(b−a)
.
δ
δ
Then we define the intervals
h
δ
δi
In = a + (n − 1) , a + n ,
2
2
where n = 1, 2, . . . , N . By definition, we have
h
δi
I1 = a, a +
2
where a + N 2δ ≥ b. See Figure 4.2.
h
δi
δ
and IN = a + (N − 1) , a + N ,
2
2
Figure 4.2: The sets E and Ini .
Since the width of each In is 2δ , we have
|f (x) − f (y)| < 1
(4.6)
for all x, y ∈ In ∩ E.d In the following discussion, we suppose that n1 , n2 , . . . , nk are positive integers
such that 1 ≤ n1 < n2 < · · · < nk ≤ N and Ini ∩ E 6= ∅ for i = 1, 2, . . . , k. Therefore, we can pick and
fix an element xni ∈ Ini ∩ E and it follows from the triangle inequality and the inequality (4.6) that
|f (x)| ≤ |f (x) − f (xni )| + |f (xni )| < 1 + |f (xni )|
(4.7)
for all x ∈ Ini ∩ E, where i = 1, 2, . . . , k. Let M = max(|f (xn1 )|, |f (xn2 )|, . . . , |f (xnk )|). It is clear that
(In1 ∩ E) ∪ (In2 ∩ E) ∪ · · · ∪ (Ink ∩ E) = E.
If x ∈ E, then we have x ∈ Ini ∩ E for some i = 1, 2, . . . , k and it follows from the inequality (4.7) that
|f (x)| < 1 + M.
Hence f is bounded on E.
We know that f (x) = x is a real uniformly continuous function on R (we just need to take δ = ǫ in
Definition 4.18). However, it is obviously not bounded on R, completing the proof of the problem.
d Note that it may happen that I
n ∩ E = ∅ for some n ∈ {1, 2, . . . , N }.
Chapter 4. Continuity
64
Problem 4.9
Rudin Chapter 4 Exercise 9.
Proof. Let us restate the two requirements first:
• Statement 1: To every ǫ > 0 there exists a δ > 0 such that diam f (E) < ǫ for all E ⊆ X with
diam E < δ.
• Statement 2: To every ǫ > 0 there exists a δ > 0 such that dY (f (p), f (q)) < ǫ for all p, q ∈ E ⊆ X
with dX (p, q) < δ.
Recall from Definition 3.9 that if E is a nonempty subset of a metric space X,
SE = {dX (p, q) | p, q ∈ E}
and
Then we have
Sf (E) = {dY (P, Q) | P, Q ∈ f (E)} = {dY (f (p), f (q)) | p, q ∈ E}.
diam E = sup SE
and diam f (E) = sup Sf (E) .
(4.8)
Suppose that Statement 1 is true. Let p, q ∈ E. If dX (p, q) < δ, then Definition 1.7 implies that
δ is greater than any element of SE and so the first expression (4.8) implies that diam E < δ. Since
Statement 1 is true, we have diam f (E) < ǫ. By the second expression (4.8) again, the latter inequality
implies that dY (f (p), f (q)) < ǫ. This shows Statement 2 is true. Next, we suppose that Statement 2
is true. If diam E < δ, then the first expression (4.8) implies that dX (p, q) < δ for all p, q ∈ E and thus
dY (f (p), f (q)) < ǫ for all p, q ∈ E. By this and the second expression (4.8), we have
diam f (E) < ǫ.
Therefore, Statement 1 is true. Hence Statement 1 and Statement 2 are equivalent, completing the
proof of the problem.
Problem 4.10
Rudin Chapter 4 Exercise 10.
Proof. Assume that f was not uniformly continuous on X. Then it follows from Definition 4.18 that
there exists a ǫ > 0 such that for every δ > 0, we have
dY (f (p), f (q)) ≥ ǫ
for some p, q ∈ X with dX (p, q) < δ. For each positive integer n, if we take δ = n1 , then there are
sequences {pn }, {qn } in X such that dX (pn , qn ) < n1 but
dY (f (pn ), f (qn )) ≥ ǫ.
(4.9)
Furthermore, we know from the inequality (4.9) that pn 6= qn for all positive integers n. Since X is a
compact metric space, Theorem 3.6(a)e implies that a subsequence of {pn }, namely {pnk }, converges to
p ∈ X. By the triangle inequality, we have
dX (qnk , p) ≤ dX (qnk , pnk ) + dX (pnk , p) <
1
+ dX (pnk , p).
nk
(4.10)
Thus this inequality (4.10) implies that the subsequence {qnk } of {qn } also converges to the point p.
Since p is a point of the domain of f , f is continuous at p. Therefore we can get from Theorems 4.2(a)
and 4.6 that there exists integers Np and Nq such that
dY (f (pnk ), p) <
ǫ
4
and dY (f (qnk ), p) <
e We can apply Theorem 2.37 here instead of Theorem 3.6(a).
ǫ
4
65
4.3. Problems on uniformly continuous functions
for all k ≥ Np and k ≥ Nq respectively. Hence if k ≥ N = max(Np , Nq ), then we get from the inequality
(4.9) that
ǫ
ǫ
ǫ
ǫ ≤ dY (f (pnk ), f (qnk )) ≤ dY (f (pnk ), f (p)) + dY (f (p), f (qnk )) < + = ,
4 4
2
a contradiction. Hence this proves Theorem 4.19 that f is uniformly continuous on the compact metric
space X. This completes the proof of the problem.
Problem 4.11
Rudin Chapter 4 Exercise 11.
Proof. Given that ǫ > 0. By Definition 4.18, there exists a δ > 0 such that dY (f (p), f (q)) < ǫ for all
p, q ∈ X and dX (p, q) < δ. Since {xn } is a Cauchy sequence in X, there is an integer N such that
dX (xn , xm ) < δ if m, n ≥ N . Hence we have
dY (f (xn ), f (xm )) < ǫ
if m, n ≥ N , i.e., {f (xn )} is a Cauchy sequence by Definition 3.8.
Alternative proof of Problem 4.13: Let p ∈ X \ E. Since E is a dense subset of X, there is a
sequence {pn } in E such that
lim pn = p.
n→∞
By Theorem 3.11(a), {pn } is a Cauchy sequence in X. Since f is uniformly continuous on E, our Problem
4.11 implies that {f (pn )} is also a Cauchy sequence in R and we follow from Theorem 3.11(c) that {f (pn )}
converges to P ∈ R. Suppose that there is another sequence {qn } in E which converges to p. By the
previous analysis, we know that {f (qn )} converges to Q ∈ R. Now we have the following result about P
and Q:
Lemma 4.1
We have P = Q.
Proof of Lemma 4.1. Given that δ > 0. Since pn → p and qn → p as n → ∞, there exist
positive integers N1 and N2 such that
dX (pn , p) <
δ
2
and dX (qn , p) <
δ
2
if n ≥ N1 and n ≥ N2 respectively. Let N = max(N1 , N2 ). Then the triangle inequality
implies that
δ
δ
dX (pn , qn ) ≤ dX (pn , p) + dX (qn , p) < + = δ
2 2
for n ≥ N .
Since f is uniformly continuous on E, for every ǫ > 0, there exists a δ > 0 such that
|f (a) − f (b)| < ǫ
for all a, b ∈ E and dX (a, b) < δ. Thus if we take a = pn and b = qn , then there exists a
positive integer N such that dX (pn , qn ) < δ for all n ≥ N . Therefore we have for all n ≥ N ,
|f (pn ) − f (qn )| < ǫ.
(4.11)
By Definition 3.1,
lim [f (pn ) − f (qn )] = 0
n→∞
and it follows from Theorem 3.3(a) that P = Q. This completes the proof of the lemma.
Chapter 4. Continuity
66
We can continue the proof of the problem. We define g : X → R by
(
f (p),
if p ∈ E;
g(p) =
lim f (pn ), if p ∈ X \ E, {pn } ⊆ E and lim pn = p.
n→∞
(4.12)
n→∞
By Lemma 4.1, the definition (4.12) is well-defined. It remains to show that g is continuous at every
p ∈ X. At the first glance, one may think that it is not necessary to consider points in E because f is
uniformly continuous on E. However, the definition only says that
|f (x) − f (p)| <
ǫ
3
(4.13)
is true for all x, p ∈ E (not x, p ∈ X) satisfying dX (x, p) < δ. If x or p is a point of X, then the validity
of the inequality (4.13) is not clear to us. Therefore, we have to check that g satisfies Definition 4.5 on
X. Given that ǫ > 0.
• Case (i): p ∈ E. In this case, g(p) = f (p). Thus it is easy to see that there exists a δ > 0 such
that
ǫ
|g(x) − g(p)| = |f (x) − f (p)| <
3
for all x ∈ E and dX (x, p) < δ. Thus, without loss of generality, we may assume that x ∈ X \ E.
Let dX (x, p) < δ. Since E is dense in X, we have {xn } ⊆ E such that lim xn = x. Since we have
n→∞
g(x) = lim f (xn ), there exists a positive integer N such that
n→∞
dX (xN , x) + dX (x, p) < δ
and |g(x) − f (xN )| <
2ǫ
.
3
(4.14)
Now it follows from the first inequality in (4.14) that
dX (xN , p) < δ.
(4.15)
Again, we follow from the inequality (4.13), the second inequality in (4.14) and the inequality (4.15)
that
ǫ
2ǫ
+ =ǫ
|g(x) − g(p)| = |g(x) − f (p)| ≤ |g(x) − f (xN )| + |f (xN ) − f (p)| <
3
3
for all x ∈ X \ E and dX (x, p) < δ.
• Case (ii): p ∈ X \ E. Similarly, we let dX (x, p) < δ. Basically, we apply the argument used in
Case (i). Since E is dense in X, we can find {xn }, {pn } ⊆ E such that
lim xn = x
n→∞
and
lim pn = p.
n→∞
Furthermore, since we have
g(x) = lim f (xn ) and g(p) = lim f (pn ),
n→∞
n→∞
there exists a positive integer M such that
2dX (pM , p) + dX (x, p) < δ
and |g(p) − g(pM )| <
ǫ
3
(4.16)
2dX (xM , x) + dX (x, p) < δ
and |g(x) − g(xM )| <
ǫ
.
3
(4.17)
dX (xM , pM ) ≤ dX (xM , x) + dX (x, p) + d(p, pM ) < δ.
(4.18)
and
Now it follows from the left-hand inequalities in (4.16) and (4.17) that
Again, we follow from the inequality (4.13), the right-hand inequalities in (4.16) and (4.17) as well
as the inequality (4.18) that
|g(x) − g(p)| ≤ |g(x) − g(xM )| + |g(xM ) − g(pM )| + |g(pM ) − g(p)|
67
4.3. Problems on uniformly continuous functions
ǫ
ǫ
+ |f (xM ) − f (pM )| +
3
3
ǫ
ǫ
ǫ
< + +
3 3 3
=ǫ
<
for all x ∈ X \ E and dX (x, p) < δ.
Since ǫ is arbitrary, we know from Definition 4.5 that g is continuous on X. Hence g is the continuous
extension of f and this extension is unique because of Problem 4.4. This completes the proof of the
problem
Problem 4.12
Rudin Chapter 4 Exercise 12.
Proof. The statement can be stated as follows: “Suppose that X, Y, Z are metric spaces, f : X → Y and
g : Y → Z and h : X → Z is a mapping defined by
h(x) = g(f (x))
for all x ∈ X. If f is uniformly continuous on X and g is uniformly continuous on Y , then h is uniformly
continuous on X.”
We follow the idea of proof in [21, Theorem 4.7, p. 86]. Given that ǫ > 0. Since g is uniformly
continuous on Y , there exists a η > 0 such that
dZ (g(f (p)), g(f (q))) < ǫ
(4.19)
for all f (p), f (q) ∈ Y and dY (f (p), f (q)) < η. Since f is uniformly continuous on X, there exists a δ > 0
such that
dY (f (p), f (q)) < η
(4.20)
for all p, q ∈ X and dX (p, q) < δ. It follows from the inequalities (4.19) and (4.20) that
dZ (h(p), h(q)) = dZ (g(f (p)), g(f (q))) < ǫ
for all p, q ∈ X and dX (p, q) < δ. This completes the proof of the problem.
Problem 4.13
Rudin Chapter 4 Exercise 13.
Proof. Let f : E → R. For each p ∈ X \ E and each positive integer n, let
n
1o
.
Vn (p) = q ∈ E d(p, q) <
n
We want to apply Problem 3.21 to {f (Vn (p))}, so we check the conditions as follows:
• Since f (p) ∈ f (Vn (p)) ⊆ f (Vn (p)) for each positive integer n, we have
f (Vn (p)) 6= ∅.
• Since f is uniformly continuous on E, it is also a real uniformly continuous function on each
boundedf Vn (p). By Problem 4.8, f is bounded on Vn (p). Therefore, Definition 4.13 implies that
f (Vn (p))
is bounded.
f See Definition 2.18(i).
Chapter 4. Continuity
68
• By Problem 4.9, we have diam f (Vn (p)) → 0 as n → ∞. By Theorem 3.10(a),
diam f (Vn (p)) = diam f (Vn (p))
so that diam f (Vn (p)) → 0 as n → ∞.
• It is clear that f (Vn (p)) is closed in R for every positive integer n. Since Vn+1 (p) ⊆ Vn (p), we have
f (Vn+1 (p)) ⊆ f (Vn (p))
and thus
f (Vn+1 (p)) ⊆ f (Vn (p)).
R,
Now what we have shown is that {f (Vn (p))} is a sequence of nonempty, closed and bounded sets in
f (Vn+1 (p)) ⊆ f (Vn (p)) and
lim diam f (Vn (p)) = 0.
n→∞
Since R is complete by Theorem 3.11(c), Problem 3.21 guarantees that
∞
\
f (Vn (p))
1
consists of exactly one point. Let this point be g(p). By similar argument as in the proof of Problem 4.11,
it can be shown that the function g so defined is the continuous extension of the function f . Therefore
we won’t repeat the proof here.
Next, we answer the questions one by one as follows:
• Could the range space R be replaced by Rk ? Since Problem 3.21 is valid for any complete
metric space and Rk is complete by Theorem 3.11(c), the range space R can be replaced by Rk .
• Could the range space R be replaced by any compact metric space? Since a compact
metric space must be complete by Theorem 3.11(b), the range space R can be replaced by any
compact metric space.
• Could the range space R be replaced by any complete metric space? The answer is
affirmative.
• Could the range space R be replaced by any metric space? The answer is no! For example,
we consider f : Q → Q ⊂ R defined by
f (x) = 2x.
It is obvious that f is a uniformly continuous real function on Q. (Take δ = 2ǫ in Definition 4.18.)
Assume that g : R → Q was a continuous extension of f from Q to R. Then Theorem 4.22 yields
that g(R) ⊆ Q is connected which is a contradiction because any subset of Q cannot be connected
by Theorem 2.47.
This completes the proof of the problem.
4.4
Further properties of continuous functions
Problem 4.14
Rudin Chapter 4 Exercise 14.
Proof. Let g(x) = x − f (x). If g(0) = 0 or g(1) = 0, then we are done. Suppose that g(0) 6= 0 and
g(1) 6= 0. Then we must have
f (0) > 0 and f (1) < 1
which imply that
g(0) = 0 − f (0) < 0 and g(1) = 1 − f (1) > 0.
By Theorem 4.23, there exists c ∈ (0, 1) such that g(c) = 0, i.e., f (c) = c. This completes the proof of
the problem.
69
4.5. Discontinuous functions
Problem 4.15
Rudin Chapter 4 Exercise 15.
Proof. Let f : R → R be a continuous open mapping. Assume that f was not monotonic. In other words,
there exists a < c < b such that either
f (a) > f (c) and f (c) < f (b)
(4.21)
f (a) < f (c) and f (c) > f (b).
(4.22)
or
If inequalities (4.21) hold, then we consider the restriction of f (see Problem 4.7) on the compact set
[a, b]. By Theorem 4.16, there exists p, q ∈ [a, b] such that f attains its maximum and minimum at p and
q respectively. Certainly, we have q ∈ (a, b) because of the inequalities (4.21).
• Case (i): p ∈ (a, b). By definition, we have
for all x ∈ (a, b) so that
f (q) ≤ f (x) ≤ f (p)
(4.23)
f ((a, b)) ⊆ [f (q), f (p)].
(4.24)
By Theorem 2.47, (a, b) is connected. Thus Theorem 4.22 implies that f ((a, b)) is a connected
subset of R. Since f (p), f (q) ∈ f ((a, b)) and the inequalities (4.23), Theorem 2.47 again implies
that
[f (q), f (p)] ⊆ f ((a, b)).
(4.25)
Now we obtain from the set relations (4.24) and (4.25) that
f ((a, b)) = [f (q), f (p)].
Since f is an open map and (a, b) is an open set in R, f ((a, b)) = [f (q), f (p)] is open in R which is
a contradiction.
• Case (ii): p ∈
/ (a, b). Then f attains its maximum only at the end points a or b. However, by
applying similar argument as obtaining the set relations (4.24) and (4.25), we can show that either
f ((a, b)) = [f (q), f (a)) or f ((a, b)) = [f (q), f (b)).
Since both [f (q), f (a)) and [f (q), f (b)) are not open, it contradicts the hypothesis that f is an open
map.
Therefore, inequalities (4.21) cannot hold. Similarly, inequalities (4.22) cannot hold. Hence, f must
be monotonic. This completes the proof of the problem.
4.5
Discontinuous functions
Problem 4.16
Rudin Chapter 4 Exercise 16.
Proof. Both the functions f (x) = [x] and g(x) = (x) have simple discontinuities at all integers. To see
this, let n ∈ Z, then we have
f (n+) = lim f (x) = n,
x→n+
f (n−) = lim f (x) = n − 1
x→n−
and
g(n+) = lim g(x) = 0,
x→n+
g(n−) = lim g(x) = 1.
x→n−
Chapter 4. Continuity
70
Let us show that f and g are continuous on R \ Z. Let x ∈ R \ Z. Then we have x ∈ (k − 1, k) for some
k ∈ Z so that
f (x) = [x] = k − 1 and g(x) = (x) = x − [x] = x − k + 1.
In other words, f (x) is constant on each (k − 1, k), so f is continuous on each (k − 1, k). Similarly, since
g(x) is a polynomial of x on each (k − 1, k), g is continuous on each (k − 1, k). See Figure 4.3 for the
graphs of the functions [x] (the blue line) and (x) (the red line):
Figure 4.3: The graphs of [x] and (x).
This completes the proof of the problem.
Problem 4.17
Rudin Chapter 4 Exercise 17.
Proof. By Definition 4.26, there are two ways in which a function can have a simple discontinuity, so we
consider the cases separately.
• Case (i): f (x+) 6= f (x−). Let E+ = {x | f (x−) < f (x+)} ⊆ (a, b). Since f (x−) < f (x+),
Theorem 1.20(b) implies the existence of a rational number p such that
f (x−) < p < f (x+).
Therefore this p satisfies condition (a). Fix this p, Theorem 1.20(b) again implies the existence of
a δ > 0 such that f (x−) + δ < p. By Definition 4.25, for every ǫ > 0, there exists a δ > 0 such that
|f (t) − f (x−)| < ǫ
for all a < t < x and x − t < δ. Since the ǫ can be chosen such as f (x−) + ǫ < p, we have f (t) < p
for all t such that x − δ < t < x. Since there exists a rational number q such that x − δ < q < t,
condition (b) is satisfied. Similarly, there exists a rational number r satisfying condition (c). Now
we have shown that for each x ∈ E+ , we have associated a triple (p, q, r) of rational numbers
satisfying conditions (a) to (c). By Theorem 2.13, the set of all such triples is countable.
Suppose that the triple (p, q, r) is associated with x and y in E+ . Assume that x < y. Then we have
x < t0 < y. Since x < t < r < b implies f (t) > p, we have f (t0 ) > p. However, since a < q < t < y
71
4.5. Discontinuous functions
implies f (t) < p, we have f (t0 ) < p. Thus a contradiction occurs and we cannot have x < y. Now
the inequality y < x can be shown to be impossible by similar argument. Therefore, we have x = y.
In other words, each triple is associated with at most one point of E+ . Hence the set E+ is at most
countable.
If we define E− = {x | f (x+) < f (x−)} ⊆ (a, b), then the above analysis can also be applied to
show that E− is also at most countable.
• Case (ii): f (x+) = f (x−) 6= f (x). Let F+ = {x | f (x+) = f (x−) < f (x)} ⊆ (a, b). By applying
the argument in Case (i), it can be shown that each point x ∈ F+ is associated a triple (p, q, r) of
rational numbers such that
(1) f (x−) = f (x+) < p < f (x) and a < q < t < x or x < t < r < b implies that
f (t) < p,
(2) each such triple (p, q, r) is associated with at most one point of F+ . By Theorem 2.13, F+ is
at most countable.
Next if we define F− = {x | f (x+) = f (x−) > f (x)} ⊆ (a, b), then it is very clear from the above
analysis that it is also at most countable.
Since the set of points at which f has a simple discontinuity, namely G, is the union of E+ , E− , F+
and F− and each set is at most countable, G is also at most countable by Theorem 2.13, completing the
proof of the problem.
Problem 4.18
Rudin Chapter 4 Exercise 18.
Proof. The function in question is called Thomae’s function, named after Carl Johannes Thomae. Read
the website https://en.wikipedia.org/wiki/Thomae%27s_function for more details.
Let α be irrational. We want to show that for every ǫ > 0, there exists a δ > 0 such that
|f (x) − f (α)| = |f (x)| < ǫ
for all x ∈ R with |x − α| < δ. Given that ǫ > 0. By Theorem 1.20(a) (the Archimedean property), we let
n be the least positive integer such that n1 < ǫ. Now it is clear from the definition that f (x) = 0 if x is
irrational and so the crux idea of the proof is that the δ is constructed in order that all rational numbers
in the interval (α − δ, α + δ) cannot have denominators ≤ n.
The construction of the number δ is as follows: Let k be a positive integer ≤ n. Since α is irrational,
kα is irrational too. By applying the proof of Theorem 1.20(b), we can find an integer mk such that
mk < kα < mk + 1 which implies that
α ∈ Ik =
m
k
k
,
mk + 1 .
k
Since the width of Ik is less than k1 , the interval Ik does not contain a rational number with k as its
denominator. For this k, we define δk = min(| mkk − α|, |α − mkk+1 |). Since (α − δk , α + δk ) ⊂ Ik ,
(α − δk , α + δk ) does not contain a rational number with k as its denominator too. Next, we define
δ = min(δ1 , δ2 , . . . , δn ) and we consider the interval (α − δ, α + δ). By the above analysis, this interval
does not contain a rational number whose denominator ≤ n. If x ∈ (α − δ, α + δ) and x = pq , then we
must have q ≥ n + 1 and thus
1
1
1
|f (x)| = ≤
< < ǫ,
q
n+1
n
as required. By definition, f is continuous at every irrational number, i.e., lim f (x) = f (α) = 0. See
x→α
√
Figure 4.4 for an example that α = 2 and n = 5.
Chapter 4. Continuity
72
Figure 4.4: An example for α =
√
2 and n = 5.
Let x = m
n . To prove that f has a simple discontinuity at x, we check Definitions 4.25 and 4.26
directly. We need the following result:
Lemma 4.2
The set R \ Q is also dense in R.
Proof of Lemma 4.2. Let x, y ∈ R and x < y. Suppose first that 0 ∈
/ (x, y). We consider
x
√
√y . By Theorem 1.20(b), we have
<
2
2
x
y
√ <r< √
2
2
√
√
for some r ∈ Q \ {0} and this means that x < r 2 < y. By Problem 1.1, r 2 is irrational.
Next, we suppose that x < 0 < y. Then we consider √x2 < 0. By Theorem 1.20(b) again, we
have
x
√ < r′ < 0
2
√
′
for some r ∈ Q \ {0} which implies that x < r′ 2 < 0 < y. This completes the proof of the
lemma.
We can continue our proof of the problem. By Lemma 4.2, there exists an irrational αk ∈ (x, x + k1 )
for each positive integer k. In other words, we have lim αk = x and then
k→∞
lim f (αn ) = 0.
k→∞
By this, it is reasonable to claim that f (x+) = 0. To prove this, we need the following lemma:
Lemma 4.3
m
Let {rk = pqkk } be a sequence of rational numbers such that rk > m
n for all k and rk → n .
Then we have
qk → +∞ as k → ∞.
73
4.6. The distance function ρE
Proof of Lemma 4.3. Assume that the set {qk } was bounded. Then there is a positive integer
N such that 1 ≤ qk ≤ N for all k. If {pk } is bounded, then {rk } is a finite sequence which
contradicts the hypothesis that rk → m
n as k → ∞. Thus we have pk → +∞ as k → ∞.
However, this means that rk → +∞ as k → ∞, a contradiction again. Hence we have
qk → +∞ as k → ∞.
By the definition of f , we have {f (rk )} = { q1k }.
Now we return to the proof of the problem.
Lemma 4.3, we have q1k → 0 as k → ∞. In other words, we have
By
lim f (rk ) = 0.
k→∞
1
By Definition 4.25, we have f (x+) = 0. Similarly, we also have f (x−) = 0. Since f (x) = f ( m
n ) = n , we
have
f (x+) = f (x−) 6= f (x).
By Definition 4.26, f has a simple discontinuity at every rational point. This competes the proof of the
problem.
Problem 4.19
Rudin Chapter 4 Exercise 19.
Proof. For every rational r, let Er = {x ∈ R | f (x) = r}. By the hypothesis, Er is closed in R. Assume
that f was not continuous at x0 . By Theorems 4.2 and 4.6, there exists a sequence {xn } in R such that
xn → x0 and xn 6= x0 for all n but f (xn ) 9 f (x0 ). Thus there exists a subsequence {xnk } of {xn } such
that
|f (xnk ) − f (x0 )| ≥ ǫ
for some ǫ > 0 and for all k. For the convenience of the discussion, we may rename this subsequence as
{xn } so that
f (xn ) − f (x0 ) ≥ ǫ
for all n, i.e., f (xn ) ≥ f (x0 ) + ǫ for all n. By Theorem 1.20(b), there is a rational r such that
f (xn ) ≥ f (x0 ) + ǫ > r > f (x0 )
for all n. By the assumption, we have f (tn ) = r for some tn between x0 and xn . Since xn → x0 as
n → ∞, we must have tn → x0 as n → ∞. By this and the fact that tn ∈ Er for all n, we know that x0
is a limit point of Er . Since Er is closed, x0 ∈ Er and thus f (x0 ) = r which contradicts our choice of r.
Hence f is continuous, completing the proof of the analysis.
4.6
The distance function ρE
Problem 4.20
Rudin Chapter 4 Exercise 20.
Proof.
(a) By definition, it is clear that ρE (x) ≥ 0.
Suppose that x ∈ E. Then x ∈ E or x ∈ E ′ . If x ∈ E, then it is clear that ρE (x) = 0. If x ∈ E ′ ,
then x is a limit point of E. Thus for every N n1 (x), there is a point zn ∈ N n1 (x) but zn 6= x such
that zn ∈ E. By the choice of N n1 (x), we have d(x, zn ) < n1 which gives
0 ≤ ρE (x) <
1
.
n
Chapter 4. Continuity
74
By the remark preceding Theorem 3.20, we must have ρE (x) = 0.
Conversely, we suppose that ρE (x) = 0. If x ∈ E, then we are done. Now without loss of generality,
we may assume that x ∈
/ E. We claim that for every positive integer n, there exists a z ∈ E such
that d(x, z) < n1 . Otherwise, there is a positive integer N such that
1
≤ d(x, z)
N
for all z ∈ E but this means that N1 is a lower bound of the set Ax = {d(x, z) | z ∈ E}. Thus we
have
1
ρE (x) ≥
> 0,
N
a contradiction. This proves our claim. In addition, it is easy to see that z 6= x because x ∈
/ E.
Next, we let Nδ (x) be a neighborhood of x for some δ > 0. Then there exists a positive integer n
such that n1 < δ so that
N n1 (x) ⊂ Nδ (x).
The preceding analysis makes sure that there exists a z ∈ Nδ (x) such that z 6= x and z ∈ E.
Hence, it follows from Definition 2.18(b) that x is a limit point of E. Hence we have x ∈ E ′ ⊆ E,
completing the proof of part (a).
Figure 4.5: The distance from x ∈ X to E.
(b) Since ρE (x) ≤ d(x, z) ≤ d(x, y) + d(y, z) for all x, y ∈ X, we have
ρE (x) ≤ d(x, y) + ρE (y).
(4.26)
Similarly, we have ρE (y) ≤ d(y, z) ≤ d(y, x) + d(x, z) for all x, y ∈ X which implies that
ρE (y) ≤ d(x, y) + ρE (x).
(4.27)
Combining inequalities (4.26) and (4.27), we have
|ρE (x) − ρE (y)| ≤ d(x, y)
for all x ∈ X, y ∈ X. Given ǫ > 0. Let δ = 2ǫ . Then for all x, y ∈ X with d(x, y) < δ, we have
|ρE (x) − ρE (y)| ≤ d(x, y) < δ =
ǫ
< ǫ.
2
By Definition 4.18, ρE is a uniformly continuous function on X.
This finishes the proof of the problem.
75
4.6. The distance function ρE
Problem 4.21
Rudin Chapter 4 Exercise 21.
Proof. We define the function ρF : K → R by
ρF (x) = inf d(x, z),
z∈F
where x ∈ K. Since F is closed, we have F = F by Theorem 2.27(b). Since F ∩ K = ∅, we have
F ∩K = ∅
and Problem 4.20(a) implies that ρF (x) 6= 0 for all x ∈ K. Therefore, we follow from Problem 4.20(b)
that ρF is a positive (uniformly) continuous function on K. By Theorem 4.16, there exists an a ∈ K
such that ρF (a) = min ρF (x). Since ρF is positive, we must have ρF (a) > 0 so that there exists a δ > 0
x∈K
such that
0 < δ < ρF (a) ≤ ρF (q) ≤ d(p, q)
for all p ∈ F and q ∈ K.
Suppose that K = {1, 2, 3, . . .} and F = {1 + 21 , 2 + 13 , . . .} Then both K and F are closed as well as
1
∈ F for every n ∈ N, we have
K ∩ F = ∅, but they are not compact. Since n ∈ K and n + n+1
d n, n +
1
1 = →0
n+1
n
as n → ∞. Thus our conclusion fails in this case and this ends the proof of the problem.
Problem 4.22
Rudin Chapter 4 Exercise 22.
Proof. By definition, we have ρA ≥ 0 and ρB ≥ 0 on X. We claim that
ρA (x) + ρB (x) > 0
(4.28)
for all x ∈ X. Since A and B are closed, it follows from Problem 4.20(a) that ρA (x) = 0 if and only if
x ∈ A and ρB (x) = 0 if and only if x ∈ B. Since A ∩ B = ∅, there is no point x ∈ X such that
ρA (x) = ρB (x) = 0
which yields the result (4.28). By Problem 4.20(b), both ρA and ρB are continuous functions on X. By
Theorem 4.9, we see immediately that f is a continuous function on X.
To find the range of f , we note that 0 ≤ ρA (x) ≤ ρA (x) + ρB (x). Combining this and the inequality
(4.28), we have
ρA (x)
≤ 1.
0≤
ρA (x) + ρB (x)
Since f (a) = 0 if a ∈ A and f (b) = 1 if b ∈ B, we have
f (X) = [0, 1].
Since f (p) = 0 if and only if ρA (p) = 0 and A is closed, we deduce from Problem 4.20(a) that
ρA (p) = 0 if and only if p ∈ A.
Hence f (p) = 0 precisely on A. The result that f (p) = 1 precisely on B can be proven similarly, so we
omit the details here.
Now the above result can be applied to obtain a converse of Problem 4.3: Let A ⊆ X be closed. Then
A = Z(f ) for some real continuous function f . To see why it is true, we consider two cases:
Chapter 4. Continuity
76
• Case (i): A = X. Then the function f : X → R defined by f (x) = 0 for all x ∈ X satisfies the
required conditions.
• Case (ii): A ⊂ X. Then we take any point y ∈ X \ A. Thus B = {y} is closed and A ∩ B = ∅.
So the f defined in this problem is the desired continuous real function with the property that
Z(f ) = A.
Let V = f −1 ([0, 12 )) and W = f −1 (( 21 , 1]). It is obvious that V and W are disjoint because if
x ∈ V ∩ W , then f (x) ∈ [0, 12 ) and f (x) ∈ ( 12 , 1] which is impossible. Since
1 1
h 1
= [0, 1] ∩ − ,
0,
2
2 2
and
1
i
1 3
, 1 = [0, 1] ∩
, ,
2
2 2
Theorem 2.30 implies that both [0, 12 ) and ( 12 , 1] are open sets in [0, 1]. By Theorem 4.8, we know that
V and W are disjoint open sets in X. Finally, if p ∈ A, then f (p) = 0 so that
p ∈ V,
i.e., A ⊆ V . Similarly, if p ∈ B, then f (p) = 1 so that p ∈ W , i.e., B ⊆ W . This completes the proof of
the problem.
4.7
Convex functions
Problem 4.23
Rudin Chapter 4 Exercise 23.
Proof. Before we prove the results, let’s look at the graph of a convex function first:
Figure 4.6: The graph of a convex function f .
From Figure 4.6, we see that the graph of a convex function is below the straight line connecting the
points (x, f (x)) and (y, f (y)). Furthermore, the graph indicates that the equality holds if and only if
x = y. Now we are going to prove the assertions one by one.
• f is continuous in (a, b). Suppose that p, q ∈ (a, b) and p < q. We first show that
77
4.7. Convex functions
Lemma 4.4
The function f is bounded in [p, q].
p−a b−q
Proof of Lemma 4.4. Let 0 < r < min( q−p
4 , 4 , 4 ), Mp,q = max(f (p), f (q)) and t ∈ (p, q).
t−p
It is obvious that [p+r, q −r] ⊂ (a, b). If λ = q−p
, then we have 0 < λ < 1 and λq +(1−λ)p = t.
Therefore, the definition implies that
f (t) = f (λq + (1 − λ)p)
≤ λf (q) + (1 − λ)f (p)
≤ λMp,q + (1 − λ)Mp,q
= Mp,q
for all t ∈ (p, q). In other words, f is bounded above in [p, q].
r
Since a < p − r < p < q, if t ∈ (p, q) and λ = t−p+r
, then 0 < λ < 1 and we have
p = (1 − λ)(p − r) + λt. By the definition of a convex function, we have
f (p) = f ((1 − λ)(p − r) + λt) ≤ (1 − λ)f (p − r) + λf (t)
so that
1
1−λ
f (p) −
f (p − r)
λ
λ
t−p+r
t−p
=
f (p) −
f (p − r)
r
r
q−p
f (p − r)
> f (p) −
r
f (t) ≥
for all t ∈ (p, q). In other words, f is bounded below in [p, q]. Hence there is a positive number
Np,q such that |f (x)| ≤ Np,q for all x ∈ [p, q], completing the proof of the lemma.
Let’s return to the proof of the problem. Given that ǫ > 0 and x is a fixed number in the interval
(a, b). Then Theorem 1.20(b) ensures that there exist p, q ∈ (a, b) such that a < p < x < q < b.
Suppose that 0 < κ < 12 min(x − p, q − x). Then we have
a < p < p + κ < x < q − κ < q < b.
See Figure 4.7 for the positions of the points p, p + κ, q − κ and q.
Figure 4.7: The positions of the points p, p + κ, q − κ and q.
Case (i): Let y be a real number such that p < p + κ < x < y < q − κ < q. If λ = x−p
y−p , then
it is easy to check that 0 < λ < 1 and x = (1 − λ)p + λy. Thus we have y − p ≥ κ and Lemma
4.4 implies that
Chapter 4. Continuity
78
f (x) ≤ (1 − λ)f (p) + λf (y)
= f ((1 − λ)p + λy)
f (x) − f (y) ≤ (1 − λ)[f (p) − f (y)]
y−x
[f (p) − f (y)]
=
y−p
2Np,q
(y − x)
<
κ
(4.29)
(4.30)
for some positive number Np,q . Similarly, if λ = y−x
q−x , then it is easy to check that 0 < λ < 1
and y = (1 − λ)x + λq. Thus we have q − x > κ and Lemma 4.4 implies that
f (y) ≤ (1 − λ)f (x) + λf (q)
= f ((1 − λ)x + λq)
f (y) − f (x) ≤ λ[f (q) − f (x)]
y−x
[f (q) − f (x)]
=
q−x
2Np,q
<
(y − x).
κ
(4.31)
By combining the inequalities (4.29) and (4.31), we always have
|f (y) − f (x)| <
2Np,q
(y − x)
κ
(4.32)
for all y ∈ [p + κ, q − κ] with x < y.
Case (ii): Let y be a real number such that p < p + κ < y < x < q − κ < q. Then, instead of
the inequalities (4.29) and (4.31), we have
f (y) − f (x) <
2Np,q
2Np,q
(x − y) and f (x) − f (y) <
(x − y)
κ
κ
(4.33)
for some positive number Np,q . Again, by combining the inequalities (4.33), we see that
|f (y) − f (x)| <
2Np,q
(x − y)
κ
(4.34)
for all y ∈ [p + κ, q − κ] with y < x.
Hence, if we let δ = 2Nκp,q ǫ, then it follows from the inequalities (4.32) and (4.34) that
|f (y) − f (x)| < ǫ
for all y ∈ [p + κ, q − κ] with |y − x| < δ. By Definition 4.5, f is continuous at x and then on (a, b).
• Every increasing convex function of a convex function is convex. Let f : (a, b) → R be a
convex function, g : f ((a, b)) → R be an increasing convex function and h : (a, b) → R be defined
by h = g ◦ f .
Further, let a < x, y < b and 0 < λ < 1. We first show that
λf (x) + (1 − λ)f (y) ∈ f ((a, b)).
(4.35)
To this end, if f (x) = f (y), then we have the (4.35) holds trivially. So, without loss of generality,
we may assume that f (x) < f (y).g Then we have
f (x) < λf (x) + (1 − λ)f (y) < f (y).
g The case for f (x) > f (y) is similar, so we omit the details here.
79
4.7. Convex functions
Since f is convex, it is continuous on (a, b). Since (a, b) is connected, Theorem 4.22 implies that
f ((a, b)) is connected. By Theorem 2.47, we must have the (4.35).
Since f is convex, we have f (λx + (1 − λ)y) ≤ λf (x) + (1 − λ)f (y). Since g is increasing and convex
plus λf (x) + (1 − λ)f (y) ∈ f ((a, b)), we have
h(λx + (1 − λ)y) = g(f (λx + (1 − λ)y))
≤ g(λf (x) + (1 − λ)f (y))
≤ λg(f (x)) + (1 − λ)g(f (y))
= λh(x) + (1 − λ)h(y).
By definition, h is convex.
• The validity of inequalities. Suppose that f is convex in (a, b) and a < s < t < u < b. If
u−t
λ = u−s
, then we have 0 < λ < 1 and λs + (1 − λ)u = t. Thus the definition implies that
f (λx + (1 − λ)y) ≤ λf (x) + (1 − λ)f (y)
u−t
u − t
f (t) ≤
f (u)
f (s) + 1 −
u−s
u−s
f (t) − f (s)
f (u) − f (s)
≤
.
t−s
u−s
(4.36)
t−s
Similarly, if we let λ = u−s
, then we have 0 < λ < 1 and λu + (1 − λ)s = t. Thus the definition
implies that
f (λx + (1 − λ)y) ≤ λf (x) + (1 − λ)f (y)
t−s
t−s
f (u) + 1 −
f (s)
f (t) ≤
u−s
u−s
f (u) − f (t)
f (u) − f (s)
≤
.
u−s
u−t
(4.37)
Hence, by combining the inequalities (4.36) and (4.37), we have the desired inequalities in the
question.
This completes the proof of the problem.
Problem 4.24
Rudin Chapter 4 Exercise 24.
Proof. We have a lemma first:
Lemma 4.5
Let f be a continuous real function defined in (a, b) satisfying the hypothesis of the problem.
For every x, y ∈ (a, b) and λ = 2mn , where m and n are positive integers such that 0 < m <
2n , we have
f (λx + (1 − λ)y) ≤ λf (x) + (1 − λ)f (y).
(4.38)
Chapter 4. Continuity
80
Proof of Lemma 4.5. We prove it by induction on n. If n = 1, then we see from the hypothesis
of the problem that the inequality (4.38) holds for λ = 12 . Thus it is true for the case n = 1.
Assume that it is also true for n = k for some positive integer k, i.e.,
m
m m
m
f k x + 1 − k y ≤ k f (x) + 1 − k f (y),
(4.39)
2
2
2
2
where m is a positive integer such that 0 < m < 2k . Let n = k + 1 and 0 < m < 2k+1 . We can
rewrite
1 2m
1m − 1 m + 1
m
=
=
+ k+1 .
·
2k+1
2 2k+1
2 2k+1
2
If we let µ = m−1
and ν = m+1
, then we have λ = 21 (µ + ν) and it is clear that
2k+1
2k+1
λx + (1 − λ)y =
1
[µx + (1 − µ)y] + [νx + (1 − ν)y]
{(µ + ν)x + [2 − (µ + ν)]y} =
.
2
2
By definition, we have µ ∈ [0, 1) and ν ∈ (0, 1]. Now we split the proof into two cases as
follows:
• Cases (i): µ = 0 or ν = 1. We know that µ = 0 if and only if m = 1. In this particular
case, we have
λx + (1 − λ)y =
x
+
1−
k+1
1
2
i
1 h1
1 1
y
=
y
+
y
.
x
+
1
−
·
2k+1
2 2k
2k
Since 21k x + (1 − 21k )y ∈ (a, b) for all x, y ∈ (a, b), the hypothesis of the problem and the
inequality (4.39) imply that
21k x + 1 − 21k y + y f (λx + (1 − λ)y) = f
2
i
1h 1
1 ≤ f k x + 1 − k y + f (y)
2
2
2
i
1h 1
1
≤
f (x) + 1 − k f (y) + f (y)
k
2 2
2
= λf (x) + (1 − λ)f (y).
In other words, the inequality (4.38) also holds in this particular case. Similarly, ν = 1
if and only if m = 2k+1 − 1. In this particular case, we have
λx + (1 − λ)y =
i
2k+1 − 1
1 h1
2k+1 − 1 1
y
=
x
+
x
.
x
+
1
−
y
+
1
−
·
2k+1
2k+1
2 2k
2k
Thus similar argument can be applied to show that the inequality (4.38) also holds in
this particular case.
• Case (ii): µ, ν ∈ (0, 1). By definition, we have 0 < m − 1 < m + 1 < 2k+1 . Besides, we
have µx + (1 − µ)y ∈ (a, b) and νx + (1 − ν)y ∈ (a, b) for all x, y ∈ (a, b). Therefore, it
follows from the hypothesis of the problem and the inequality (4.39) that
f (λx + (1 − λ)y) = f
[µx + (1 − µ)y] + [νx + (1 − ν)y] 2
1
≤ · [f (µx + (1 − µ)y) + f (νx + (1 − ν)y)]
2
[µf (x) + (1 − µ)f (y)] + [νf (x) + (1 − ν)f (y)]
≤
2
= λf (x) + (1 − λ)f (y).
Hence the inequality (4.39) is also true for n = k+1 and the induction shows that the inequality
(4.39) is true for all positive integers k, where m is a positive integer such that 0 < m < 2k .
This completes the proof of Lemma 4.5.
81
4.8. Other properties of continuous functions
Now we may continue our proof of the problem. Let us accept the fact that every real number
an
α ∈ (0, 1) has a unique binary representation in the sense that if λn = n , where an ∈ {0, 1}, then we
2
have
∞
X
λn .
α=
n=1
By Theorem 3.26, the series is convergent (absolutely). If we let αn =
n
X
k=1
λk , then we have αn → α as
n → ∞. It is easy to see that αn is in the form of 2mn , where m is a positive integer such that 0 < m < 2n ,
therefore Lemma 4.5 implies that
f (αn x + (1 − αn )y) ≤ αn f (x) + (1 − αn )f (y).
(4.40)
Since f is continuous on (a, b), we apply Theorems 3.3, 3.19, 4.2 and 4.6 to the inequality (4.40) to get
lim f (αn x + (1 − αn )y) ≤ lim [αn f (x) + (1 − αn )f (y)]
n→∞
n→∞
f (αx + (1 − α)y) ≤ lim αn f (x) + lim (1 − αn )f (y)
n→∞
n→∞
f (αx + (1 − α)y) ≤ αf (x) + (1 − α)f (y).
By Problem 4.23, we see that f is convex. This finishes the proof of the problem.
4.8
Other properties of continuous functions
Problem 4.25
Rudin Chapter 4 Exercise 25.
Proof.
(a) We follow the hint. Take z ∈
/ K + C, put F = z − C = {z − y | y ∈ C}. Assume that K ∩ F 6= ∅.
Then we have x ∈ K and x ∈ F . By definition, x = z − y for some y ∈ C which is equivalent to
z = x + y ∈ K + C,
a contradiction. Therefore K and F are disjoint. Since K is compact and C is closed, Problem 4.21
ensures that there exists a δ > 0 such that
|p − q| > δ
for all p ∈ K and q ∈ F .
Now we consider the open ball B(z, δ) = {x ∈ Rk | |x − z| < δ}. Assume that there was x ∈ Rk
such that x ∈ B(z, δ) ∩ (K + C). Then we have |x − z| < δ and x = p + y for some p ∈ K and
y ∈ C. For this y, we have q = z − y ∈ F which implies
x = p + y = p + z − q.
Hence, by the definition of the chosen δ, we have
δ < |p − q| = |x − z| < δ,
a contradiction.
(b) We
∈ Z, the set (k − 1, k) is open in R. By Theorem 2.24(a),
[ have C2 = {nα | n ∈ C1 }. For every k[
(k−1, k) is open in R. Since R\C1 =
(k−1, k), C1 is closed in R by Theorem 2.23. By similar
k∈Z
k∈Z
argument, it can be shown that C2 is closed. By definition, we have C1 + C2 = {h + kα | h, k ∈ C1 }.
Chapter 4. Continuity
82
Now the mapping f : C1 + C2 → C1 × C1 defined by f (h + kα) = (h, k) shows that C1 + C2 is
equivalent to a subset of C1 × C1 . Since C1 × C1 is countable and C1 ⊂ C1 + C2 , it follows from
Theorem 2.8 that C1 + C2 is countable.h
For a positive integer n, we consider the fractional part {nα} = nα − [nα]. To prove the density of
the set, we need a preliminary result first:
Lemma 4.6
Suppose that α is irrational and 0 < θ < 1. For every ǫ > 0, there exists a positive integer
k such that |{kα} − θ| < ǫ. Hence, if h = [kα], then we have |kα − h − θ| < ǫ.
Proof of Lemma 4.6. We note that if m 6= n, then {mα} 6= {nα}. Otherwise, we have mα −
[mα] = nα − [nα] which implies that
α=
[mα] − [nα]
m−n
is rational, a contradiction. Given that ǫ > 0 and choose θ such that 0 < θ < 1. Recall
Dirichlet’s theorem that for any irrational α, there exist integers h and k such that |kα−h| < ǫ.
Now we have either kα > h or kα < h. Suppose that kα > h. Since |kα − h| = kα − h =
{kα} + [kα] − h, h and [kα] are integers and {kα}, we have [kα] = h and
0 < {kα} < ǫ.
(4.41)
Now we consider the following sequence {kα}, {2kα}, {3kα}, . . .. Since kα = [kα] + {kα}, we
have mkα = m[kα] + m{kα} for every integer m. Thus we have
m{kα} = {mkα}
if and only if
{kα} <
1
.
m
(4.42)
By this observation, we let N be the largest integer satisfying {kα} < N1 . In other words, we
always have
1
1
< {kα} < .
(4.43)
N +1
N
By the relation (4.42) and the inequalities (4.43), we have {mkα} = m{kα} for m = 1, 2, . . . , N .
Therefore, the numbers {kα}, {2kα}, . . . , {N kα} form an increasing chain with equal spacing
{kα}. Now it follows from the inequalities (4.43) that
1−
1
< {N kα} < 1
N
and then this and the inequalities (4.41) imply that
|1 − {N kα}| <
1
< {kα} < ǫ.
1+N
Hence the numbers {kα}, {2kα}, . . . , {N kα} subdivide the interval (0, 1) into subintervals of
length less than ǫ. Recall that 0 < θ < 1, so θ lies in one of these subintervals and the result
follows, proving Lemma 4.6.
For a general real θ, we note that θ = [θ] + {θ}, where 0 < {θ} < 1. Thus Lemma 4.6 implies that
|kα − [kα] − {θ}| < ǫ
h This result is known as the one-dimensional case of the famous Kronecker’s Approximation Theorem. For details, please
read [2, §7.4, pp. 148, 149] or http://mathworld.wolfram.com/KroneckersApproximationTheorem.html . Here our proof is
based on the one presented in [2, §7.4, pp. 148, 149].
83
4.8. Other properties of continuous functions
or equivalently,
|kα − ([kα] − [θ]) − θ| < ǫ.
(4.44)
If we take h = [kα] − [θ] in the inequality (4.44), then we have the desired result that C1 + C2 is
dense in R.
This ends the proof of the problem.
Problem 4.26
Rudin Chapter 4 Exercise 26.
Proof. We have f : X → Y, g : Y → Z and h : X → Z. Since Y is compact and g is continuous on
Y , g(Y ) is compact by Theorem 4.14. Since g : Y → g(Y ) is continuous and bijective, we obtain from
Theorem 4.17 that g −1 : g(Y ) → Y is a bijective continuous mapping. By Theorem 4.19, g −1 is uniformly
continuous on g(Y ). Since f (x) = g −1 (h(x)) and h is uniformly continuous on X, it follows from Problem
4.12 that f is also uniformly continuous on X. If h is continuous at p ∈ X, then Theorem 4.7 says that
f is also continuous at p ∈ X. This completes the proof of the problem.
Chapter 4. Continuity
84
CHAPTER
5
Differentiation
5.1
Problems on differentiability of a function
Problem 5.1
Rudin Chapter 5 Exercise 1.
(x)
Proof. We have −(x − y)2 ≤ f (x) − f (y) ≤ (x − y)2 . If t > x, then −(t − x) ≤ f (t)−f
≤ t − x. If t < x,
t−x
(x)
≥ t − x. By these, we have
then we have −(t − x) ≥ f (t)−f
t−x
−|t − x| ≤
f (t) − f (x)
≤ |t − x|.
t−x
By Definition 5.1 and the remark preceding Theorem 3.20, we have
f ′ (x) = lim
t→x
f (t) − f (t)
=0
t−x
for all x ∈ R. By Theorem 5.11(b), f is constant which completes the proof of the problem.
Problem 5.2
Rudin Chapter 5 Exercise 2.
Proof. By Theorem 5.10, we have f (x2 ) − f (x1 ) = (x2 − x1 )f ′ (x) for every pair of numbers x1 , x2 in (a, b)
with x1 < x2 and for some x between x1 and x2 . Since f ′ (x) > 0 in (a, b), we have f (x2 ) − f (x1 ) > 0
which means that f is strictly increasing in (a, b).
To prove the second assertion, we prove the following result first:
Lemma 5.1
If f is strictly increasing in (a, b), then f is one-to-one in (a, b).
Proof of Lemma 5.1. If x1 , x2 ∈ (a, b) and x1 6= x2 , then either x1 < x2 or x1 > x2 . If x1 < x2 ,
then since f is strictly increasing in (a, b), we have f (x1 ) < f (x2 ). Similarly, if x1 > x2 ,
then since f is strictly increasing in (a, b), we have f (x1 ) > f (x2 ). In both cases, we have
f (x1 ) 6= f (x2 ). By Definition 2.2, f is one-to-one in (a, b), completing the proof of Lemma
5.1.
85
Chapter 5. Differentiation
86
Suppose that c and d are real numbers such that a < c < d < b. By Lemma 5.1, f is one-to-one in
[c, d]. Since f is differentiable in (a, b), it is also continuous in (a, b) by Theorem 5.2. In particular, f
is continuous on [c, d]. Let E = f ([c, d]). By Theorem 4.17, the inverse function g : E → [c, d] of f is
well-defined and
g(f (x)) = x.
(5.1)
Besides, g is continuous on E.
Let t, y ∈ E and t 6= y. Then there exist s, x ∈ [c, d] such that f (s) = t and f (x) = y. Now we have
from the expression (5.1) that
φ(t) =
g(t) − g(y)
g(f (s)) − g(f (x))
s−x
=
=
.
t−y
f (s) − f (x)
f (s) − f (x)
(5.2)
Since t → y if and only if s → x and f ′ (x) > 0 in (a, b), the expression (5.2) and Theorem 4.4(c) imply
that
g ′ (f (x)) = g ′ (y) = lim φ(t) = lim
t→y
t→y
1
g(t) − g(y)
s−x
1
= lim
= lim f (s)−f (x) = ′
,
s→x f (s) − f (x)
s→x
t−y
f (x)
s−x
where x ∈ [c, d]. For any x ∈ (a, b), it is obvious that we can find c, d such that x ∈ [c, d] ⊂ (a, b),
g ′ (f (x)) = f ′1(x) is true for every x ∈ (a, b). This completes the proof of the problem.
Problem 5.3
Rudin Chapter 5 Exercise 3.
Proof. Since g is differentiable in R, f is also differentiable in R by Theorem 5.3(a). In fact, we acquire
f ′ (x) = 1 + ǫg ′ (x).
Since −M ≤ g ′ (x) ≤ M in R, we have
1 − ǫM ≤ f ′ (x) ≤ 1 + ǫM.
Thus if ǫ is small enough, then we have 1 − ǫM > 0 so that f ′ (x) > 0 in R. By Problem 5.2, we have f
is strictly increasing in R and then it is one-to-one in R. This completes the proof of the problem.
Problem 5.4
Rudin Chapter 5 Exercise 4.
2
n+1
Proof. Let f (x) = C0 x + C1 x2 + · · · + Cn xn+1 be a polynomial defined in R. In particular, f is continuous
on [0, 1] and differentiable in (0, 1). By Theorem 5.10, there exists a α ∈ (0, 1) such that f (1) − f (0) =
f ′ (α). Since f (1) = f (0) = 0, the equation
f ′ (x) = C0 + C1 x + · · · + Cn−1 xn−1 + Cn xn = 0
has at least one real root in (0, 1) which is the desired result, completing the proof of the problem.
Problem 5.5
Rudin Chapter 5 Exercise 5.
87
5.1. Problems on differentiability of a function
Proof. Since f is differentiable in (0, +∞), it is continuous on (0, +∞) by Theorem 5.2. In particular,
f is continuous on [x, x + 1] and differentiable in (x, x + 1) for every x > 0. By Theorem 5.10 and the
definition of g, we have
g(x) = f (x + 1) − f (x) = (x + 1 − x)f ′ (y) = f ′ (y)
(5.3)
for some y ∈ (x, x + 1). If x → +∞, then we have y → +∞ and so f ′ (y) → 0. Hence it follows from the
expression (5.3) that g(x) → 0 as x → +∞. This finishes the proof of the problem.
Problem 5.6
Rudin Chapter 5 Exercise 6.
Proof. By Theorem 5.3 and condition (b), g is differentiable in (0, +∞) and
g ′ (x) =
xf ′ (x) − f (x)
.
x2
We note that
g ′ (x) ≥ 0 if and only if
xf ′ (x) − f (x) ≥ 0.
For every x > 0, conditions (a) and (b) imply that f is continuous in [0, x] and differentiable in (0, x).
By Theorem 5.10, we have
f (x) − f (0) = (x − 0)f ′ (ξ)
for some ξ ∈ (0, x). By condition (c), we have f (x) = xf ′ (ξ). By condition (d), we have f ′ (ξ) ≤ f ′ (x)
and thus f (x) ≤ xf ′ (x). Therefore, we have g ′ (x) ≥ 0 for every x > 0 and Theorem 5.11(a) implies that
g is monotonically increasing in (0, +∞).
Problem 5.7
Rudin Chapter 5 Exercise 7.
Proof. Since f (x), g(x) are differentiable, g ′ (x) 6= 0 and f (x) = g(x) = 0, it follows from Theorem 4.4(c)
that
f (t)−f (x)
f (t)
f (t) − f (x)
f ′ (x)
t−x
lim
= lim
= lim g(t)−g(x)
= ′
t→x g(t)
t→x g(t) − g(x)
t→x
g (x)
t−x
which is our desired result. This completes the proof of the problem.
Problem 5.8
Rudin Chapter 5 Exercise 8.
Proof. Since f ′ is continuous in [a, b] and [a, b] is compact, Theorem 4.19 implies that f ′ is uniformly
continuous in [a, b]. By Definition 4.18, there exists a δ > 0 such that
|f ′ (t) − f ′ (x)| < ǫ
(5.4)
for every x, t ∈ [a, b] with |t − x| < δ. Suppose that t < x. Then we deduce from Theorem 5.10 that there
exists a ξ ∈ (t, x) such that
f (x) − f (t)
= f ′ (ξ).
(5.5)
x−t
Since 0 < |ξ − x| < |t − x| < δ, the inequality (5.4) and the expression (5.5) together give
f (x) − f (t)
− f ′ (x) = |f ′ (ξ) − f (x)| < ǫ
x−t
Chapter 5. Differentiation
88
which is our desired result in the case t < x. Since the case for x < t is similar, we omit the details here.
We claim that this also holds for vector-valued functions: f ′ (x) is continuous on [a, b] and ǫ > 0. Then
there exists δ > 0 such that
f (t) − f (x)
− f ′ (x) < ǫ
t−x
whenever 0 < |t − x| < δ and x, t ∈ [a, b]. To prove this result, suppose that
f (x) = (f1 (x), f2 (x), . . . , fk (x)),
then Remark 5.16 implies that f is differentiable on [a, b] if and only if each f1 , . . . , fk is differentiable
on [a, b]. Furthermore, we apply Theorem 4.10 to get the result that f ′ (x) is continuous on [a, b] if and
only if each f1′ , . . . , fk′ is continuous on [a, b] and thus they are uniformly continuous on [a, b] by Theorem
4.19. In other words, for each i = 1, 2, . . . , k, there exists δi > 0 such that
fi (t) − fi (x)
ǫ
− fi′ (x) < √
t−x
k
(5.6)
whenever 0 < |t − x| < δi and t, x ∈ [a, b]. Let δ = min(δ1 , . . . , δk ). Hence we follow from the inequalities
(5.6) and the proof of Theorem 4.10 that for every 0 < |t − x| < δ and t, x ∈ [a, b], we have
f (t) − f (x)
f (t) − f (x)
fk (t) − fk (x)
1
1
− f ′ (x) =
− f1′ (x), . . . ,
− fk′ (x)
t−x
t−x
t−x
v
u k
uX fi (t) − fi (x)
2
≤t
− fi′ (x)
t−x
i=1
v
u k
uX ǫ2
<t
k
i=1
= ǫ.
This completes the proof of the problem.
Problem 5.9
Rudin Chapter 5 Exercise 9.
Proof. The answer is affirmative. By Definition 5.1, we have to show that the limit
f (x) − f (0)
x→0
x−0
lim
exists. Define the functions
F (x) = f (x) − f (0) and G(x) = x.
Then they are real and differentiable in (0, 1), and G′ (x) = 1 6= 0 on (0, 1). Since
F ′ (x)
= f ′ (x) → 3
G′ (x)
as x → 0, we apply Theorem 5.13 to the functions F (x) and G(x) to get lim
F (x)
x→0 G(x)
= 3. As
F (x)
f (x) − f (0)
= lim
,
x→0 G(x)
x→0
x−0
lim
the derivative f ′ (0) exists and its value must be 3.a We complete the proof of the problem.
a If one checks the proof carefully, then it can be seen easily that the number 3 can be replaced by any real number.
89
5.1. Problems on differentiability of a function
Problem 5.10
Rudin Chapter 5 Exercise 10.
Proof. By the hint, we acquire
1
1
f (x) f (x)
=
− A · g(x) + A g(x) .
g(x)
x
x
(5.7)
x
Let A = A1 + iA2 and B = B1 + iB2 , where A1 , A2 , B1 , B2 ∈ R. Let f = f1 + if2 and g = g1 + ig2 , where
f1 , f2 , g1 and g2 are real functions on (0, 1). By the hypothesis, we have
lim f1′ (x) = A1 ,
x→0
lim f2′ (x) = A2 ,
x→0
lim g1′ (x) = B1
x→0
lim g2′ (x) = B2 .
and
x→0
Then the formula (5.7) can be rewritten as
f (x) f1 (x)
f2 (x)
1
1
=
+i
− A · g (x) g (x) + A g (x) g (x) .
2
1
1
g(x)
x
x
+
+ 2
x
x
x
(5.8)
x
Obviously, we have the following facts:
• Fact 1. f (x) → 0 as x → 0 if and only if f1 (x) → 0 and f2 (x) → 0 as x → 0;
• Fact 2. f ′ (x) → A as x → 0 if and only if f1′ (x) → A1 and f2′ (x) → A2 as x → 0;
• Fact 3. g(x) → 0 as x → 0 if and only if g1 (x) → 0 and g2 (x) → 0 as x → 0;
• Fact 4. g ′ (x) → B as x → 0 if and only if g1′ (x) → B1 and g2′ (x) → B2 as x → 0.
We obtain from Theorem 5.13 that
lim
f1 (x)
= lim f1′ (x) = A1
x→0
x
and
lim
g1 (x)
= lim g1′ (x) = B1
x→0
x
and
x→0
lim
f2 (x)
= lim f2′ (x) = A2 .
x→0
x
lim
g2 (x)
= lim g2′ (x) = B2 .
x→0
x
x→0
Similarly, we have
x→0
x→0
Since B 6= 0, it is impossible that B1 = B2 = 0. Hence, it follows from the formula (5.8) that
f (x)
1
1
1
A
A
+A·
=0· +
= (A1 + iA2 − A) ·
= .
x→0 g(x)
B1 + iB2
B1 + iB2
B
B
B
lim
This completes the proof of the problem.
Problem 5.11
Rudin Chapter 5 Exercise 11.
Proof. Let E be the neighborhood of x in which f is defined. Suppose further that h is so small such
that x + h, x − h ∈ E. Then we can define
F (h) = f (x + h) + f (x − h) − 2f (x)
and G(h) = h2 .
Since f ′′ (x) exists, f is continuous at x by Theorem 5.2. It is clear that x is a limit point of E and so
Theorem 4.6 implies that
lim f (x + h) = f (x) and
h→0
lim f (x − h) = f (x).
h→0
Chapter 5. Differentiation
90
Therefore, we have F (h) → 0 and G(h) → 0 as h → 0. By Definition 5.14, since f ′′ (x) exists at x, f ′ (t)
must exist in (x − δ, x + δ) for some δ > 0 and then f must be differentiable in (x − δ, x + δ). Since our
h can be chosen arbitrarily so that
(x − h, x + h) ⊆ (x − δ, x + δ),
we can conclude that F (h) is differentiable as a function of h and Theorem 5.5b gives
F ′ (h) = f ′ (x + h) − f ′ (x − h).
By Theorem 5.13, we have
F (h)
F ′ (h)
f ′ (x + h) − f ′ (x − h)
f (x + h) + f (x − h) − 2f (x)
= lim
= lim ′
= lim
.
2
h→0 G(h)
h→0 G (h)
h→0
h→0
h
2h
(5.9)
lim
On the other hand, we follow from Definition 5.1 that
1 ′′
(f (x) + f ′′ (x))
2
1h
f ′ (x + h) − f ′ (x)
f ′ (x) − f ′ (x − h) i
=
lim
+ lim
h→0
2 h→0
h
h
f ′ (x + h) − f ′ (x − h)
.
= lim
h→0
2h
f ′′ (x) =
(5.10)
By comparing the limits (5.9) and (5.10), we have the desired result. This finishes the proof of the
problem.
Problem 5.12
Rudin Chapter 5 Exercise 12.
Proof. We have
f (x) =
so that
′
f (x) =
−3x2 , if x < 0;
3x2 ,
if x > 0,
−x3 , if x ≤ 0;
x3 ,
if x > 0
′′
and f (x) =
To compute f ′ (0), we consider f ′ (0+) and f ′ (0−) so that
−6x, if x < 0;
6x,
if x > 0.
f (t) − f (0)
t3
f (t) − f (0)
−t3
= lim
= 0 and f ′ (0−) = lim
= lim
= 0.
t→0
t→0 t
t→0
t→0 t
t−0
t−0
f ′ (0+) = lim
t>0
t>0
t<0
t<0
Hence we have f ′ (0) = f ′ (0+) = f ′ (0−) = 0.
Similarly, to compute f ′′ (0), we consider f ′′ (0+) and f ′′ (0−) so that
3t2
f ′ (t) − f ′ (0)
−3t2
f ′ (t) − f ′ (0)
= lim
= 0 and f ′′ (0−) = lim
= lim
= 0.
t→0 t
t→0
t→0
t→0
t−0
t−0
t
f ′′ (0+) = lim
t>0
t>0
′′
′′
t<0
t<0
′′
Hence we have f (0) = f (0+) = f (0−) = 0.
Finally, we consider f (3) (0+) and f (3) (0−) so that
f ′′ (t) − f ′′ (0)
6t
= lim
=6
t→0
t→0 t
t−0
f (3) (0+) = lim
t>0
t>0
f ′′ (t) − f ′′ (0)
−6t
= lim
= −6.
t→0
t→0 t
t−0
and f (3) (0−) = lim
t<0
t<0
Since f (3) (0+) 6= f (3) (0−), we conclude that f (3) (0) does not exist, finishing the proof of the problem. b By our construction, we know that the function a(h) = x + h is continuous on [0, δ ], a′ (h) = 1 exists on [0, δ ], the
2
2
interval (x − δ, x + δ) certainly contains a([0, 2δ ]) = [x, x + 2δ ], and f is differentiable on [x, x + 2δ ]. Hence the functions a
and f satisfy the conditions of Theorem 5.5.
91
5.1. Problems on differentiability of a function
Problem 5.13
Rudin Chapter 5 Exercise 13.
Proof. It is unavoidable to make two assumpations here. First of all, by Example 5.6(a) or [21, Eqn.
(49), Chap. 8], we know that the function sin x is differentiable in R. Secondly, when x < 0, the function
xa is a complex multi-valued function. In fact, we have xa = ea ln x , where ln x = ln |x| + i arg x is the
complex logarithm function. If we take the principal branch of the function ln x, then the power function
xa is single-valued. Here we assume without proof that xa is differentiable in [−1, 1] and
(xa )′ = axa−1 .
(a) By Theorem 4.9, we know that f is continuous for all x 6= 0. Therefore, only the continuity at the
point x = 0 should be considered.
– Case (i): a > 0. Then since −1 ≤ | sin(|x|−c )| ≤ 1 for all x ∈ [−1, 1], we have
−|xa | ≤ |f (x)| ≤ |xa |.
Since lim |xa | = 0, we have lim |f (x)| = 0 and thus lim f (x) = 0. Since f (0) = 0, we have
x→0
x→0
x→0
lim f (x) = f (0) and so Theorem 4.6 implies that f is continuous at x = 0.
x→0
– Case (ii): a = 0. Consider the sequence {xn } defined by xn =
have lim xn = 0. Now we have
1
1
c
(2nπ+ π
2)
. Since c > 0, we
n→∞
π
=1
(5.11)
f (xn ) = x0n sin(|xn |−c ) = sin 2nπ +
2
for all positive integers n so that lim f (xn ) = 1 6= f (0). Thus f is not continuous at x = 0 in
n→∞
this case.
1
– Case (iii): a < 0. Similarly, we consider the sequence {xn } defined by xn =
1 .
π
(2nπ+ 2 ) c
Instead of the expression (5.11), we have
π π − ac
π − ac
sin 2nπ +
(5.12)
f (xn ) = xan sin(|xn |−c ) = 2nπ +
= 2nπ +
2
2
2
for all positive integers n. Since a < 0 and c > 0, − ac > 0 and then the expression (5.12)
implies that lim f (xn ) = ∞ =
6 f (0). Thus f is not continuous at x = 0 in this case.
n→∞
Hence we establish the result that f is continuous if and only if a > 0.
(b) Note that we have
φ(x) =
xa sin (|x|−c ) − 0
f (x) − f (0)
=
= xa−1 sin (|x|−c )
x−0
x−0
(5.13)
and so f ′ (0) = lim φ(x) = lim xa−1 sin (|x|−c ).
x→0
x→0
– Case (i): a ≤ 1. We consider the sequence {xn } defined by xn =
have lim xn = 0. By a similar argument as in part (a), we have
1
1
c
(nπ+ π
2)
. Since c > 0, we
n→∞
a−1
a−1
π
π − c
π − c
sin nπ +
φ(xn ) = xna−1 sin(|xn |−c ) = nπ +
= (−1)n nπ +
2
2
2
for all positive integers n. Now if a = 1, then
(5.14)
φ(x2k ) = 1 and φ(x2k+1 ) = −1.
By Theorem 4.2, lim φ(xn ) does not exist in this case! If a < 1, then − a−1
> 0 and the
c
n→∞
expression (5.14) implies that φ(x2k ) → +∞ and φ(x2k+1 ) → −∞ as k → ∞. Thus lim φ(xn )
n→∞
does not exist in this case! Therefore, we have shown that if f ′ (0) exists, then a > 1.
Chapter 5. Differentiation
92
– Case (ii): a > 1. Then since −1 ≤ | sin(|x|−c )| ≤ 1 for all x ∈ [−1, 1], we have
0 ≤ |φ(x)| ≤ |x|a−1 .
Since lim |x|a−1 = 0, we have lim |φ(x)| = 0 and thus lim φ(x) = 0. In other words, we have
x→0
x→0
f ′ (0) = 0 in this case.
x→0
(c) Since xa and sin(|x|−c ) are differentiable in [−1, 1] \ {0}, Theorem 5.3(b) implies that f is differentiable in [−1, 1] \ {0}. Suppose that a > 1. By the result of part (b) and Theorem 5.5, we have
 a−1
[a sin(x−c ) − cx−c cos(x−c )],
if 0 < x ≤ 1;
 x
′
f (x) =
0,
if x = 0;
(5.15)
 a−1
x
[a sin(−x)−c − c(−x)−c cos(−x)−c ], if −1 ≤ x < 0.
– Case (i): a ≥ 1 + c. Then we have a > 1 and a − 1 − c ≥ 0. Therefore, we can deduce from
the expressions (5.15) that
|f ′ (x)| ≤ |axa−1 sin(|x|−c )| + |cxa−1−c | cos(|x|−c )| ≤ a|x|a−1 + c|x|a−1−c ≤ a + c
for all x ∈ [−1, 1]. This shows that f ′ is bounded by a + c in this case.
1
– Case (ii): a < 1 + c. Consider the sequence {xn } defined by xn = (2nπ)− c > 0. Then we
−c
have sin(x−c
n ) = sin 2nπ = 0, cos(xn ) = cos 2nπ = 1 and so
f ′ (xn ) = −c(xn )a−1−c = −c(2nπ)
c+1−a
c
→ −∞
as n → ∞. Therefore, f ′ is unbounded on [−1, 1] in this case.
Hence we have shown that f ′ is bounded if and only if a ≥ 1 + c.
(d) By the expressions (5.15), we know that f ′ is continuousc for all x ∈ [−1, 1] \ {0}, so we check
whether f ′ is continuous at x = 0 or not.
– Case (i): a > 1 + c. Then we have a > 1 and a − 1 − c > 0. Since
|f ′ (x)| ≤ |axa−1 sin(|x|−c )| + |cxa−1−c cos(|x|−c )| ≤ a|x|a−1 + c|x|a−1−c ,
we have lim |f ′ (x)| = 0 which implies that lim f ′ (x) = 0. By the expression (5.15), we have
x→0
x→0
lim f ′ (x) = f ′ (0), so f ′ is continuous on [−1, 1] by Theorem 4.6.
x→0
– Case (ii): a < 1 + c. Then we know from part (c) that f ′ is unbounded on [−1, 1]. Assume
that f ′ was continuous on the compact set [−1, 1]. By Theorem 4.15, f ′ is bounded on [−1, 1],
a contradiction. Thus it is not continuous on [−1, 1].
1
– Case (iii): a = 1 + c. Then we consider the sequence {xn } defined by xn = (2nπ)− c > 0.
−c
Then we have sin(x−c
n ) = sin(2nπ) = 0, cos(xn ) = cos(2nπ) = 1 and so
f ′ (xn ) = −c(xn )a−1−c = −c.
Since f ′ (0+) = lim f ′ (xn ) = −c 6= 0 = f ′ (0), f ′ is not continuous at x = 0 in this case.
n→∞
Hence we have shown that f ′ is continuous if and only if a > 1 + c.
(e) It follows from the expressions (5.15) that
f ′ (x) − f ′ (0)
x−0
xa−1 [a sin (|x|−c ) − c|x|−c cos(|x|−c )]
=
x−0
a−2
−c
=x
[a sin (|x| ) − c|x|−c cos(|x|−c )]
ϕ(x) =
and so f ′′ (0) = lim ϕ(x) = lim xa−2 [a sin (|x|−c ) − c|x|−c cos(|x|−c )].
x→0
x→0
c By [21, Eqn. (49), Chap. 8] again, we assume that the function cos x is differentiable in R.
93
5.1. Problems on differentiability of a function
– Case (i): a ≤ 2 + c. We consider the sequences {xn } and {yn } defined by xn =
yn =
1
1
[(2n+1)π] c
1
1
(2nπ) c
and
. Since c > 0, we have lim xn = 0 and lim yn = 0. By a similar argument as
n→∞
n→∞
in part (b), we have
ϕ(xn ) = xna−2 [a sin (|xn |−c ) − c|xn |−c cos(|xn |−c )]
a−2
= (2nπ)− c [a sin(2nπ) − c(2nπ) cos(2nπ)]
= −c(2nπ)
c+2−a
c
(5.16)
and
ϕ(yn ) = yna−2 [a sin (|yn |−c ) − c|yn |−c cos(|yn |−c )]
a−2
= [(2n + 1)π]− c {a sin(2n + 1)π − c[(2n + 1)π] cos(2n + 1)π}
= c[(2n + 1)π]
c+2−a
c
(5.17)
for all positive integers n. If a = c + 2, then the expressions (5.16) and (5.17) imply that
ϕ(xn ) = −c and ϕ(yn ) = c respectively. Since ϕ(xn ) 6= ϕ(yn ), f ′′ (0) does not exist by
Theorem 4.2. If a < c + 2, then we obtain from the expression (5.16) that lim ϕ(xn ) = −∞.
n→∞
Thus f ′′ (0) does not exist by Theorem 4.2.
– Case (ii): a > 2 + c. Then since −1 ≤ | sin(x−c )| ≤ 1 and −1 ≤ | cos(x−c )| ≤ 1 for all
x ∈ [−1, 1], we have
0 ≤ |ϕ(x)| ≤ a|x|a−2 + c|x|a−2−c .
Since lim |x|a−1 = 0 and lim |x|a−2−c = 0, we have lim |ϕ(x)| = 0 and thus lim ϕ(x) = 0. In
x→0
x→0
other words, we have f ′′ (0) = 0 in this case.
x→0
x→0
Hence we have shown that f ′′ (0) exists if and only if a > 2 + c.
(f) Suppose that a > 2 + c. By the result of part (e) and the expressions (5.15), we have

a(a − 1)xa−2 − c2 xa−2−2c sin(x−c )




if 0 < x ≤ 1;
 −c(2a − 1 − c)xa−2−c cos(x−c ),
′′
0,
if x = 0;
f (x) =
a−2

2
−2c
−c

a(a
−
1)
−
c
(−x)
x
sin(−x)



−c(2a − 1 − c)xa−2 (−x)−c cos(−x)−c , if −1 ≤ x < 0.
(5.18)
– Case (i): a ≥ 2 + 2c. Then we have a − 2 ≥ 2c > 0, a − 2 − c ≥ c > 0 and a − 2 − 2c ≥ 0.
Therefore, we can deduce from the expressions (5.18) that
|f ′′ (x)| ≤ [a(a − 1)xa−2 − c2 xa−2−2c ] sin(|x|−c ) + c(2a − 1 − c)xa−2−c cos(|x|−c )
≤ a(a − 1)|x|a−2 + c2 |x|a−2−2c + c(2a − 1 − c)|x|a−2−c
≤ a(a − 1) + c2 + c(2a − 1 − c)
= a(a − 1) + c(2a − 1)
for all x ∈ [−1, 1]. This shows that f ′′ is bounded by a(a − 1) + c(2a − 1) in this case.
– Case (ii): a < 2 + 2c and 2a − 1 − c 6= 0. If a ≥ 2, then we consider the sequence {xn }
1
defined by xn = (2nπ + π2 )− c > 0 so that we have
1
π = 1 and
sin(x−c
)
=
sin
2n
+
n
2
Thus we have
1
cos(x−c
)
=
cos
2n
+
π = 0.
n
2
a−2−c
sin(x−c
cos(x−c
f ′′ (xn ) = a(a − 1)xna−2 − c2 xa−2−2c
n ) − c(2a − 1 − c)xn
n )
n
2−a
2+2c−a
π c
π
c
= a(a − 1) 2nπ +
− c2 2nπ +
.
2
2
Chapter 5. Differentiation
94
2+2c−a
Since 2−a
> 0, we have f ′′ (xn ) → −∞ as n → +∞.
c ≤ 0 and
c
1
If a < 2, then we consider the sequence {yn } defined by yn = (2nπ)− c > 0 so that we have
sin(yn−c ) = sin(2nπ) = 0 and cos(yn−c ) = cos(2nπ) = 1. Thus we have
f ′′ (yn ) = a(a − 1)yna−2 − c2 yna−2−2c sin(yn−c ) − c(2a − 1 − c)yna−2−c cos(yn−c )
= −c(2a − 1 − c)(2nπ)
c+2−a
c
.
Since c+2−a
> 1 and 2a − 1 − c 6= 0, we have f ′′ (yn ) → −∞. Therefore, f ′′ is unbounded in
c
this case.
– Case (iii): a < 2 + 2c and 2a − 1 − c = 0. Then we have from the expressions (5.18) that

if 0 < x ≤ 1;
 a(a − 1)xa−2 − c2 xa−2−2c sin(x−c ),
0,
if x = 0;
f ′′ (x) =

a(a − 1) − c2 (−x)−2c xa−2 sin(−x)−c , if −1 ≤ x < 0.
If a ≥ 2, then we can apply the same sequence {xn } as in Case (ii) so that we have
2−a
2+2c−a
π c
π c
f ′′ (xn ) = a(a − 1) 2nπ +
− c2 2nπ +
.
2
2
(5.19)
2+2c−a
Since 2−a
> 0, we have f ′′ (xn ) → −∞ as n → +∞.
c ≤ 0 and
c
If a < 2, then the expression (5.19) can be rewritten as
2−a
π 2 i
π c h
a(a − 1) − c2 2nπ +
.
f ′′ (xn ) = 2nπ +
2
2
′′
′′
Since 2−a
c > 0, we have f (xn ) → −∞ as n → +∞. Therefore, f is also unbounded in this
case.
Hence we have shown that f ′′ is bounded if and only if a ≥ 2 + 2c.
(g) By the expressions (5.18), we know that f ′′ is continuous for all x ∈ [−1, 1] \ {0}, so we check
whether f ′′ is continuous at x = 0 or not.
– Case (i): a > 2 + 2c. Then we have a > 2 and a − 2 − c > a − 2 − 2c > 0. Since
|f ′′ (x)| ≤ |a(a − 1)xa−2 − c2 xa−2−2c sin(|x|−c )| + |c(2a − 1 − c)xa−2−c cos(|x|−c )|
≤ a(a − 1)|x|a−2 + c2 |x|a−2−2c + c(2a − 1 − c)|x|a−2−c ,
we have lim |f ′′ (x)| = 0 which implies that lim f ′′ (x) = 0. By the expressions (5.18) again,
x→0
x→0
we have lim f ′′ (x) = f ′′ (0), so f ′′ is continuous at 0 by Theorem 4.6.
x→0
– Case (ii): a < 2 + 2c. Then we know from part (f) that f ′′ is unbounded on [−1, 1]. Assume
that f ′′ was continuous on the compact set [−1, 1]. By Theorem 4.15, f ′′ is bounded on [−1, 1],
a contradiction. Thus it is not continuous on [−1, 1].
1
– Case (iii): a = 2+2c. Then we consider the sequence {xn } defined by xn = (2nπ + π2 )− c > 0.
−c
Then we have sin(x−c
n ) = 1, cos(xn ) = and so
π −2
f ′′ (xn ) = a(a − 1)xna−2 − c2 xa−2−2c
=
a(a
−
1)
2nπ
+
− c2 .
n
2
Since f ′′ (0+) = lim f ′′ (xn ) = −c2 6= 0 = f ′′ (0), f ′′ is not continuous at x = 0 in this case.
n→∞
Hence we have shown that f ′′ is continuous if and only if a > 2 + 2c.
This completes the proof of the problem.
95
5.1. Problems on differentiability of a function
Problem 5.14
Rudin Chapter 5 Exercise 14.
Proof. We prove the results one by one.
• f is convex if and only if f ′ is monotonically increasing.
– Suppose that f is convex in (a, b). Then it follows from Problem 4.23 that
f (t) − f (s)
f (u) − f (s)
f (u) − f (t)
≤
≤
t−s
u−s
u−t
(5.20)
whenever a < s < t < u < b. Similarly, we have
f (u) − f (t)
f (v) − f (t)
f (v) − f (u)
≤
≤
u−t
v−t
v−u
(5.21)
whenever a < t < u < v < b.
Combining the inequalities (5.20) and (5.21), we have
f (u) − f (t)
f (v) − f (u)
f (t) − f (s)
≤
≤
t−s
u−t
v−u
(5.22)
whenever a < s < t < u < v < b. Since f is differentiable in (a, b), we have from Definition
5.1 that f ′ (x) = f ′ (x+) = f ′ (x−) for every x ∈ (a, b). In particular, we have f ′ (s) = f ′ (s+)
and f ′ (u) = f ′ (u+). Thus these and the inequalities (5.22) together imply that
f ′ (s) = f ′ (s+) = lim
t→s
t>s
f (t) − f (s)
f (v) − f (u)
≤ v→u
lim
= f ′ (u+) = f ′ (u)
t−s
v
−
u
v>u
if a < s < u < b. Hence it means that f ′ is monotonically increasing in (a, b).
– Suppose that f ′ is monotonically increasing in (a, b). Let a < x < b, a < y < b and 0 < λ < 1,
we want to show that
f (λx + (1 − λ)y) ≤ λf (x) + (1 − λ)f (y).
(5.23)
If t = λx + (1 − λ)y, then it is easily to see that
0<λ=
t−x
y−t
< 1 and 1 − λ =
.
y−x
y−x
After doing some simple algebra, we can show that the inequality (5.23) is equivalent to the
following inequalities
y−t
t−x
f (x) +
f (y)
y−x
y−x
(y − x)f (t) ≤ (y − t)f (x) + (t − x)f (y)
[(y − t) + (t − x)]f (t) ≤ (y − t)f (x) + (t − x)f (y)
f (t) ≤
f (t) − f (x)
f (y) − f (t)
≤
.
t−x
y−t
(5.24)
Since f is differentiable in (a, b), f is differentiable in [x, y] particularly and then it is continuous
on [x, y]. By Theorem 5.10, we have
f (t) − f (x) = (t − x)f ′ (η)
and f (y) − f (t) = (y − t)f ′ (ξ),
(5.25)
where t is any real number such that x < t < y, η and ξ are some real numbers such that
η ∈ (x, t) and ξ ∈ (t, y) respectively. Now we follow from the inequality (5.24) and the
expressions (5.25) that
f ′ (η) ≤ f ′ (ξ).
(5.26)
Chapter 5. Differentiation
96
By our assumption, f ′ is monotonically increasing in (a, b), so the inequality
f ′ (r) ≤ f ′ (s)
must hold for any real numbers r, s with a < r < s < b. In particular, the inequality (5.26)
holds so that the inequality (5.24) and then the inequality (5.23) also hold. By Problem 4.23,
f is convex in (a, b).
• f is convex if and only if f ′′ (x) ≥ 0 for all x ∈ (a, b).
– Suppose that f is convex in (a, b). Then the previous part shows that f ′ is monotonically
increasing in (a, b). That is, if a < x < t < b, then we have f ′ (x) ≤ f ′ (t) and similarly, if
a < t < x < b, then we have f ′ (t) ≤ f ′ (x). In both cases, we always have
f ′ (t) − f ′ (x)
≥0
t−x
(5.27)
for every x, t ∈ (a, b). Thus it follows from the inequality (5.27) and Definition 5.1 that
f ′ (t) − f ′ (x)
≥0
t→x
t−x
f ′′ (x) = lim
for all x ∈ (a, b).
– Suppose that f ′′ (x) ≥ 0 for all x ∈ (a, b). By Theorem 5.11, f ′ must be monotonically
increasing in (a, b). Thus the previous part shows that f is convex in (a, b).
Hence, we complete the proof of the problem.
5.2
Applications of Taylor’s theorem
Problem 5.15
Rudin Chapter 5 Exercise 15.
Proof. The inequality M12 ≤ 4M0 M2 is called Landau’s inequality.
We note that the inequality holds trivially if “M0 = +∞ or M2 = +∞” or “M1 = 0”. Suppose that
M1 6= 0, 0 ≤ M0 < +∞ and 0 ≤ M2 < +∞. Furthermore, if M0 = 0, then we have |f (x)| ≤ 0 for all
x ∈ (a, +∞) so that
f (x) = f ′ (x) = f ′′ (x) = 0
for all x ∈ (a, +∞). Thus we have M0 = M1 = M2 = 0 which contradicts to the fact that M1 6= 0. Next,
if M2 = 0, then we have |f ′′ (x)| ≤ 0 for all x ∈ (a, +∞) so that
f ′′ (x) = 0
for all x ∈ (a, +∞). By Theorem 5.11(b), we have f ′ (x) = c and then f (x) = cx + d for some constants
c and d.d In this case, we have M1 = c and M0 = +∞ which contradicts to the fact that M0 < +∞.
Therefore, we may suppose further that
0 < M0 < +∞ and 0 < M2 < +∞.
Let x ∈ (a, +∞). Then it is true that [x, x + 2h] ⊂ (a, +∞) for every h > 0. By the hypothesis, f is
a twice-differentiable real function in [x, x + 2h]. In other words, the function f satisfies the conditions
of Theorem 5.15 (Taylor’s Theorem). Put P (t) = f (x) + f ′ (x)(t − x) and so Theorem 5.15 (α = x and
β = x + 2h) implies that
f (x + 2h) = P (x + 2h) +
f ′′ (ξ)
(x + 2h − x)2 = f (x) + f ′ (x) · (2h) + 2f ′′ (ξ)h2
2
d We apply the theory of Chap. 6 here to obtain that f (x) = cx + d.
(5.28)
97
5.2. Applications of Taylor’s theorem
for some ξ ∈ (x, x + 2h). Now we rewrite (5.28) to get
f ′ (x) =
1
[f (x + 2h) − f (x)] − hf ′′ (ξ)
2h
and then
1 |f (x + 2h)| + |f (x)| + h|f ′′ (ξ)|
(5.29)
2h
for some ξ ∈ (x, x + 2h). By the definitions of M0 , M1 and M2 , the inequality (5.29) implies that
|f ′ (x)| ≤
|f ′ (x)| ≤ hM2 +
M0
,
h
(5.30)
where x ∈ (a, +∞).
q
0
Since h is arbitrary and M0 , M2 are positive numbers, we may take h = M
M2 > 0 in the inequality
√
√
(5.30) to get |f ′ (x)| ≤ 2 M0 M2 on (a, +∞). Since it is true for all x ∈ (a, +∞), we have M1 ≤ 2 M0 M2
which gives
M12 ≤ 4M0 M2
(5.31)
as desired.
Consider the given example in the hint, we acquire
f ′ (x) =
4x
4x
(x2 +1)2
(−1 < x < 0),
(0 < x < ∞),
and f ′′ (x) =
(
4
4−12x2
(x2 +1)3
(−1 < x < 0),
(0 < x < ∞).
Since f ′ (x) → 0 and f ′′ (x) → 4 as x → 0, the footnote of Problem 5.9 shows that f ′ (0) = 0 and f ′′ (0) = 4.
Thus we have


(−1 < x < 0),
(−1 < x < 0),
 4
 4x
4
(x = 0),
0
(x = 0),
(5.32)
and f ′′ (x) =
f ′ (x) =
 4−12x2

4x
(0
<
x
<
∞),
(0
<
x
<
∞).
(x2 +1)2
(x2 +1)3
From the definition of f and the expressions (5.32), we have M0 = 1, M1 = 4 and M2 = 4 which give the
equality (5.31).
We claim that the inequality M12 ≤ 4M0 M2 also holds for vector-valued functions. Suppose that
f = (f1 , f2 , . . . , fn ), where f1 , f2 , . . . , fn are twice-differentiable real functions on (a, +∞). By Remarks
5.16, the vector-valued function f is also twice-differentiable on (a, +∞). Let, further, M0 , M1 and M2
be the least upper bounds of |f (x)|, |f ′ (x)| and |f ′′ (x)| on (a, +∞) respectively. For any c ∈ (a, +∞), we
consider the function F : R → R defined by
F (x) = f1′ (c)f1 (x) + f2′ (c)f2 (x) + · · · + fn′ (c)fn (x).
Since each fk is twice-differentiable real function on (a, +∞), Theorem 5.3 implies that F is also a
twice-differentiable real
√ function on (a, +∞). Therefore, the previous analysis shows that, instead of the
inequality |f ′ (x)| ≤ 2 M0 M2 , we have
p
|F ′ (x)| ≤ 2 M0 M2 ,
(5.33)
where M0 , M1 and M2 are the least upper bounds of |F (x)|, |F ′ (x)| and |F ′′ (x)|, respectively, on (a, +∞).
Next, let’s recall the Cauchy-Schwarz inequality for vectors: If u = (u1 , u2 , . . . , un ) and v = (v1 , v2 , . . . , vn ),
then we have
q
q
|u1 v1 + u2 v2 + · · · + un vn | ≤ u21 + u22 + · · · + u2n · v12 + v22 + · · · + vn2 .
(5.34)
Apply the result (5.34) to |F (x)| and |F ′′ (x)| to get
|F (x)| = |f1′ (c)f1 (x) + f2′ (c)f2 (x) + · · · + fn′ (c)fn (x)|
q
p
≤ [f1′ (c)]2 + [f2′ (c)]2 + · · · + [fn′ (c)]2 · [f1 (x)]2 + [f2 (x)]2 + · · · + [fn (x)]2
Chapter 5. Differentiation
98
≤ M1 · M0
and
|F ′′ (x)| = |f1′ (c)f1′′ (x) + f2′ (c)f2′′ (x) + · · · + fn′ (c)fn′′ (x)|
q
q
≤ [f1′ (c)]2 + [f2′ (c)]2 + · · · + [fn′ (c)]2 · [f1′′ (x)]2 + [f2′′ (x)]2 + · · · + [fn′′ (x)]2
≤ M1 · M2
on (a, +∞). Therefore, it follows from these and the inequality (5.33) that
|F ′ (x)| ≤ 2
p
p
p
M0 M2 ≤ 2 M1 · M0 · M1 · M2 = 2M1 M0 M2 ,
(5.35)
where x ∈ (a, +∞). In particular, we may take x = c in the inequality (5.35) to obtain
[f1′ (c)]2 + [f2′ (c)]2 + · · · + [fn′ (c)]2 ≤ 2M1
p
M0 M2 .
(5.36)
Since the right-hand side of the inequality
√ (5.36) is independent of the choice of c, we let c run through all
values in (a, +∞) so that M12 ≤ 2M1 M0 M2 which implies that the desired inequality (5.31). Finally,
the equality also holds for vector-valued functions if we consider f (x) = (f (x), 0, . . . , 0), where f (x) is the
given example in the hint. This completes the proof of the problem.
Problem 5.16
Rudin Chapter 5 Exercise 16.
Proof. By the hypothesis, we have |f ′′ (x)| ≤ K on (0, ∞) for some positive K. Let
M0 (a) =
sup |f (x)|,
M1 (a) =
x∈(a,∞)
sup |f ′ (x)|
and M2 (a) =
x∈(a,∞)
sup |f ′′ (x)|.
x∈(a,∞)
Thus Problem 5.15 implies that
M12 (a) ≤ 4M2 (a)M0 (a) ≤ 4KM0 (a).
(5.37)
Since f (x) → 0 as x → ∞, we have M0 (a) → 0 as a → ∞. Otherwise, there was a sequence {an }, a
positive integer N and a ǫ > 0 such that an → ∞ as n → ∞ but
M0 (an ) > ǫ
for all n ≥ N . Since M0 (an ) is the least upper bound of |f (x)| on (an , ∞), there exist xn ∈ (an , ∞) such
that |f (xn )| > ǫ for all n ≥ N , contradicting to the fact that f (x) → 0 as n → ∞. Therefore, it follows
from the inequality (5.37) that
lim M12 (a) ≤ 4K lim M0 (a) = 0.
a→∞
a→∞
(5.38)
It is clear that a → ∞ implies that x → ∞. Since |f (x)| ≤ M1 (a) for all x ∈ (a, ∞), we follow from this
and the inequality (5.38) that
lim |f ′ (x)| ≤ lim M1 (a) = 0.
x→∞
′
a→∞
Hence we have f (x) → 0 as x → ∞, completing the proof of the problem.
Problem 5.17
Rudin Chapter 5 Exercise 17.
99
5.2. Applications of Taylor’s theorem
Proof. By Theorem 5.15, we have
f (1) = P (1) +
f (3) (s)
(1 − 0)3
3!
and f (−1) = P (−1) +
f (3) (t)
(0 − 1)3
3!
(5.39)
for some s ∈ (0, 1) and t ∈ (−1, 0). By the hypothesis, we have
P (1) = f (0) + f ′ (0) · 1 +
f ′′ (0)
f ′′ (0) 2
·1 =
2
2
and
P (−1) = f (0) + f ′ (0) · (−1) +
f ′′ (0)
f ′′ (0)
· (−1)2 =
.
2
2
Therefore, the expressions (5.39) imply that
1=
f ′′ (0) f (3) (s)
+
2
6
and 0 =
f ′′ (0) f (3) (t)
−
2
6
so that f (3) (s) + f (3) (t) = 6. Assume that f (3) (x) < 3 for all x ∈ (−1, 1). Then we have f (3) (s) < 3 and
f (3) (t) < 3 so that f (3) (s) + f (3) (t) < 6, a contradiction. Hence we have
f (3) (x) ≥ 3
for some x ∈ (−1, 1). This end the proof of the problem.
Problem 5.18
Rudin Chapter 5 Exercise 18.
Proof. We can prove this version of Taylor’s theorem by induction. For n = 1, we have
P (β) +
0
X
Q(1−1) (α)
f (i) (α)
(β − α)1 =
(β − α)i + Q(α)(β − α) = f (α) − [f (α) − f (β)] = f (β).
(1 − 1)!
i!
i=0
Therefore, the statement is true for n = 1. Assume that it is true for n = k, i.e.,
f (β) = P (β) +
k−1 (i)
X
f (α)
Q(k−1) (α)
Q(k−1) (α)
(β − α)k =
(β − α)i +
(β − α)k .
(k − 1)!
i!
(k
−
1)!
i=0
(5.40)
For n = k + 1, since f (t) − f (β) = (t − β)Q(t), we have
f ′ (t) = (t − β)Q′ (t) + Q(t)
f ′′ (t) = (t − β)Q′′ (t) + 2Q′ (t)
..
..
.
.
f (k) (t) = (t − β)Q(k) (t) + kQ(k−1) (t).
(5.41)
Therefore, it follows from the assumption (5.40) and the expression (5.41) that
P (β) +
k
X
f (i) (α)
Q(k) (α)
Q(k) (α)
(β − α)k+1 =
(β − α)i +
(β − α)k+1
k!
i!
k!
i=0
k−1
X
f (k) (α)
Q(k) (α)
f (i) (α)
(β − α)i +
(β − α)k +
(β − α)k+1
i!
k!
k!
i=0
"
#
k−1
X f (i) (α)
Q(k) (α)
Q(k−1) (α)
i
k+1
k
(β − α) + −
(β − α)
+
(β − α)
=
i!
k!
(k − 1)!
i=0
=
Chapter 5. Differentiation
100
Q(k) (α)
(β − α)k+1
k!
k−1
X f (i) (α)
Q(k−1) (α)
(β − α)i +
(β − α)k
=
i!
(k
−
1)!
i=0
+
= f (β).
Thus the statement is also true for n = k + 1. Hence we follow from induction that it is true for all
positive integers n, completing the proof of the problem.
Problem 5.19
Rudin Chapter 5 Exercise 19.
Proof.
(a) Let αn < 0 < βn . Since f ′ (0) exists, |βn − αn | ≥ |βn | and |βn − αn | ≥ |αn |, we have
f (βn ) − f (αn )
− f ′ (0)
βn − αn
f (βn ) − f (0) − βn f ′ (0)
f (αn ) − f (0) − αn f ′ (0)
≤
+
βn − αn
βn − αn
f (βn ) − f (0)
f (αn ) − f (0)
≤
− f ′ (0) +
− f ′ (0) .
βn
αn
|Dn − f ′ (0)| =
(5.42)
(5.43)
Since f ′ (0) exists, Definition 5.1 and Theorem 4.2 imply that
f (pn ) − f (0)
= f ′ (0)
n→∞
pn − 0
lim
for every sequence {pn } in (−1, 1) such that pn 6= 0 and lim pn = 0. In particular, we have
n→∞
f (βn ) − f (0)
f (αn ) − f (0)
= lim
= f ′ (0).
n→∞
n→∞
αn
βn
lim
Hence we deduce from these and the inequality (5.43) that lim Dn = f ′ (0).
n→∞
(b)
n
n
Since 0 < αn < βn and { βnβ−α
} is bounded, we have βnβ−α
≤ M for some positive M so that
n
n
M
M
1
≤
<
.
βn − αn
βn
αn
Therefore, these and the inequality (5.42) imply that
f (αn ) − f (0) − βn f ′ (0)
f (βn ) − f (0) − βn f ′ (0)
+M
βn
αn
f (βn ) − f (0)
f (αn ) − f (0)
=M
− f ′ (0) + M
− f ′ (0) .
βn
αn
|Dn − f ′ (0)| < M
Hence, a similar argument as in part (a) gives the desired result that lim Dn = f ′ (0).
n→∞
′
(c) Since f is continuous in (−1, 1) and −1 < αn < βn < 1, f is continuous on [αn , βn ] and differentiable
in (αn , βn ). By Theorem 5.10, we have
f (βn ) − f (αn ) = (βn − αn )f ′ (ξn )
Dn = f ′ (ξn ),
(5.44)
101
5.2. Applications of Taylor’s theorem
where ξn ∈ (αn , βn ). Again, by the continuity of f ′ , we have lim f ′ (ξn ) = f ′ (0) and so the
n→∞
expression (5.44) implies that
lim Dn = f ′ (0).
n→∞
Recall from Example 5.6(b) that the function f : (−1, 1) → R defined by
2
x sin x1 , (x 6= 0);
f (x) =
0,
(x = 0)
1
is differentiable in (−1, 1), but f ′ is not continuous at 0, i.e., lim f ′ (x) 6= f ′ (0) = 0. Let αn = 2nπ+
π
x→0
2
1
. Then we have −1 < αn < βn < 1, αn → 0 and βn → 0 as n → ∞. Since
and βn = 2nπ
f (αn ) = α2n sin
sin(2nπ + π2 )
1
1
=
=
αn
(2nπ + π2 )2
(2nπ + π2 )2
and f (βn ) = βn2 sin
1
sin(2nπ)
=
= 0,
βn
(2nπ)2
we have
Dn =
π
−4n
−1
f (βn ) − f (αn )
=
=
π 2 × (4n) 2nπ +
βn − αn
(2nπ + 2 )
2
2nπ + π2
so that
lim Dn = −
n→∞
2
6= f ′ (0).
π
This completes the proof of the problem.
Problem 5.20
Rudin Chapter 5 Exercise 20.
Proof. When we read the Eqn. (24) of Theorem 5.15, we may modify it to the following form: Suppose
f = (f1 , f2 , . . . , fm ) : [a, b] → Rm , where f1 , f2 , . . . , fm are real functions on [a, b], n is a positive integer,
f (n−1) is continuous on [a, b], f (n) exists for every t ∈ (a, b). Let α, β be distinct points of [a, b], and define
P(t) = (P1 (t), P2 (t), . . . , Pm (t)) =
n−1
X
k=0
where Pi (t) =
n−1
X
k=0
f (k) (α)
(t − α)k ,
k!
(5.45)
(k)
fi (α)
(t − α)k for i = 1, 2, . . . , m. Then there exists a point x ∈ (α, β) such that
k!
|f (β) − P(β)| ≤
f (n) (x)
(β − α)n .
n!
(5.46)
We see that the inequality (5.46) follows immediately from Theorem 5.15 (Taylor’s Theorem) if m = 1.
For the general case, we let
1
z=
[f (β) − P(β)]
|f (β) − P(β)|
which is clearly a unit vector. Let, further, ϕ(t) = z · f (t), where t ∈ [α, β]. Then it is easy to see that
ϕ is a real-valued continuous function on [α, β] which is differentiable in (α, β). Thus it is the case when
m = 1, so we deduce from Theorem 5.15 that
|ϕ(β) − Q(β)| ≤
where
Q(β) =
n−1
X
k=0
ϕ(n) (x)
(β − α)n ,
n!
ϕ(k) (α)
(β − α)k .
k!
(5.47)
Chapter 5. Differentiation
102
Since ϕ(k) (t) = z · f (k) (t), where k = 0, 1, . . . , n, we have
Q(β) = z ·
n−1
X
f (k) (α)
(β − α)k = z · P(β).
k!
k=0
(5.48)
Hence, it follows from the inequality (5.47) and the expression (5.48) that
z · f (n) (x)
(β − α)n
n!
f (n) (x)
1
[f (β) − P(β)] · [f (β) − P(β)] ≤ |z|
(β − α)n
|f (β) − P(β)|
n!
|z · [f (β) − P(β)]| ≤
1
f (n) (x)
|f (β) − P(β)|2 ≤
(β − α)n
|f (β) − P(β)|
n!
|f (β) − P(β)| ≤
f (n) (x)
(β − α)n
n!
which is exactly our desired inequality (5.46). This finishes the proof of the problem.
5.3
Derivatives of higher order and iteration methods
Problem 5.21
Rudin Chapter 5 Exercise 21.
Proof. We prove the last assertion only because it implies all the other cases. We note from Theorem
8.6(b) that the exponential function ex has derivatives of all orders on R, so we construct the desired
function based on this property.
Let E be a non-empty closed subset of R. Then E = R\E is an open subset of R. Recall from Problem
2.29 that every open set in R is the union of an at most countable collection of disjoint segments. Thus
we suppose that
∞
[
E=
(ak , bk ),
k=1
where (a1 , b1 ), (a2 , b2 ), . . . are disjoint.
belong to E, i.e., ak , bk ∈ E.
We start with the function
e
By this definition, we know that all the endpoints ak and bk
g(x) =
1
e− x2 , if x > 0;
0,
if x ≤ 0.
(5.49)
By Theorem 8.6(c), we have g(x) = 0 if and only if x ≤ 0. Furthermore, it follows from Theorem 5.5
(Chain Rule) and Theorem 8.6(b) that g is differentiable at all non-zero x. We claim that g is also
differentiable at 0. To this end, we follow from the definition (5.49) and Theorem 8.6(f) that
1
g(x) − g(0)
e− x2
g (0+) = lim
= lim
=0
x→0
x→0
x−0
x
′
x>0
and
g ′ (0−) = lim
x→0
x<0
′
x>0
0
g(x) − g(0)
= lim = 0.
x→0
x−0
x
x<0
′
Since g (0+) = g (0−), we obtain our claim. In fact, it can be shown that g has derivatives of all orders
on R.
e It may happen that (a , b ) = (−∞, b ) or (a , ∞) for some positive integer k.
k k
k
k
103
5.3. Derivatives of higher order and iteration methods
Next, we construct functions which have derivatives of all orders in the segments (ak , bk ), (ak , ∞) and
(−∞, ak ) respectively. Firstly, for each segment (ak , bk ), we define
fk (x) = g((x − ak )(bk − x)).
If x ≤ ak , then (x − ak )(bk − x) ≤ 0 so that fk (x) = 0. Similarly, if x ≥ bk , then (x − ak )(bk − x) ≤ 0 so
that fk (x) = 0. By these observations and Theorem 8.6(c), we have
fk (x) = 0 if and only if x 6∈ (ak , bk ).
Since g has derivatives of all orders on R, it follows from Theorem 5.5 (Chain Rule) that the function fk
also has derivatives of all orders on R. In fact, the expression of fk is given as follows:
g((x − ak )(bk − x)), if ak < x < bk ;
fk (x) =
0,
if x ≤ ak or x ≥ bk ,
(
1
−
e (x−ak )2 (x−bk )2 , if ak < x < bk ;
=
(5.50)
0,
if x ≤ ak or x ≥ bk .
Secondly, for the segment (ak , ∞), we consider the function fk (x) = g(x − ak ) so that fk (x) = 0 if and
only if x ≤ ak and fk has derivatives of all orders. Finally, for the segment (−∞, ak ), we consider the
function fk (x) = g(ak − x) so that fk (x) = 0 if and only if x ≥ ak and fk has derivatives of all orders.f
Now it is time to define the function f which has derivative of all orders and f (x) = 0 if and only if
x ∈ E. In fact, we define
fk (x), if x ∈ (ak , bk ) ⊆ E;
f (x) =
0,
if x ∈ E.
Hence it is easy to check that f (x) = 0 if and only if x ∈ E and we follow from the definition (5.50) that
the expression of f is given by
fk (x), if x ∈ (ak , bk ) ⊆ E;
f (x) =
0,
if x ∈ E.
(
− (x−a )21(x−b )2
k
k
, if x ∈ (ak , bk ) ⊆ E;
e
=
0,
if x ∈ E.
We finish the proof of the problem.
Problem 5.22
Rudin Chapter 5 Exercise 22.
Proof.
(a) Let F (x) = f (x) − x. By Theorem 5.3(a), F is differentiable in (−∞, ∞). Since f ′ (x) 6= 1 for every
real x, we have
F ′ (x) 6= 0
for every real x. Assume that a < b were two fixed points of f . Then we have
F (a) = F (b) = 0.
By Theorem 5.10 (the Mean Value Theorem), we have
0 = F (b) − F (a) = (b − a)F ′ (ξ)
for some ξ ∈ (a, b), but this clearly implies that F ′ (ξ) = 0, a contradiction. Hence f has at most
one fixed point.
f Actually, it can be shown by induction that
−
(n)
fk (x) = p(x − ak , bk − x)e
1
where p(x − ak , bk − x) is a polynomial in x−a
k
and b 1−x .
k
1
(x−ak )2 (x−bk )2
,
Chapter 5. Differentiation
104
(b) If f (t) = t + (1 + et )−1 = t, then we have
1
=0
1 + et
which is impossible for a real t. Hence the function f has no fixed point in (−∞, ∞).
(c) Assume that xn+1 6= xn for all positive integers n. Otherwise, if xn0 +1 = xn0 for some positive
integer n0 , then we have
xn = xn0
for all n ≥ n0 + 1g which means that xn0 is a fixed point of f .
Suppose that there is a constant A < 1 such that
|f ′ (t)| ≤ A
for all real t. Let {xn } be a sequence defined by
xn+1 = f (xn )
and x1 is an arbitrary real number. Since f is differentiable in R, f is continuous on [xn , xn+1 ]
or [xn+1 , xn ] and differentiable in (xn , xn+1 ) or (xn+1 , xn ) for all positive integers n. By Theorem
5.10, both cases imply that
|xn+2 − xn+1 | = |f (xn+1 ) − f (xn )|
= |xn+1 − xn ||f ′ (x)|
≤ A|xn+1 − xn |
(5.51)
for all positive integers n. By considering the inequality (5.51) recursively, we have
|xn+2 − xn+1 | ≤ An |x2 − x1 |
(5.52)
for all positive integers n. Since 0 < A < 1, we have An → 0 as n → ∞. Therefore, if m ≥ n ≥ 1,
then it follows from the inequality (5.52) that
|xm − xn | ≤ |xm − xm−1 | + |xm−1 − xm−2 | + · · · + |xn+1 − xn |
≤ (Am−2 + Am−3 + · · · + An−1 )|x2 − x1 |
≤ An−1 (1 + A + A2 + · · · )|x2 − x1 |
≤
|x2 − x1 | n−1
A
.
1−A
(5.53)
Given ǫ > 0. If we take N to be a positive integer such that
l 1
ǫ(1 − A) m
N >1+
· log
log A
|x2 − x1 |
(note that log A < 0), then we have
|x2 − x1 | N −1
A
< ǫ,
1−A
so this and the inequality (5.53) imply that for m ≥ n ≥ N , we have
|xm − xn | < ǫ.
Thus {xn } is a Cauchy sequence so that lim xn = x exists. Since f is continuous on R, we have
n→∞
lim f (xn ) = f (x) and then
n→∞
x = lim xn+1 = lim f (xn ) = f (x).
n→∞
n→∞
Hence x is a fixed point of f .
(d) The process described in part (c) can be “visualized” by the zig-zag path, where the points
(xn , xn+1 ) are replaced by the points (pn−1 , pn ) (i.e., the red crosses on the blue curves).
g For example, we have x
n0 +2 = f (xn0 +1 ) = f (xn0 ) = xn0 +1 .
105
5.3. Derivatives of higher order and iteration methods
Figure 5.1: The zig-zag path of the process in (c).
Problem 5.23
Rudin Chapter 5 Exercise 23.
Proof. Since α, β and γ are fixed points of f , they are the solutions of the equation
x3 − 3x + 1 = 0.
By using a scientific calculator, we can see easily that α = −1.87939 . . . , β = 0.34730 . . . and γ =
1.53209 . . .. We define g(x) = f (x) − x. It is obvious that
g(x) =
x3 − 3x + 1
= 0 if and only if f (x) = x.
3
In other words, fixed points of f (x) are the only zeros of g(x).h Thus α, β and γ are the only zeros of
g(x).
(a) Suppose that x1 < α. Since g(−2) = − 31 < 0, the continuity of g implies that g(x) < 0 on (−∞, α).
In particular, we have g(x1 ) < 0. Now we apply induction to prove that {xn } is a strictly decreasing
sequence. For n = 1, we have
x2 − x1 = f (x1 ) − x1 = g(x1 ) < 0,
i.e., x2 < x1 . The statement is true for n = 1. Assume that it is true for n = k for some positive
integer k, i.e.,
xk+1 < xk .
For n = k + 1, we note that
g ′ (x) = x2 − 1 > 0
on (−∞, −1), so Theorem 5.10 implies that g is strictly increasing on (−∞, −1). Therefore, our
assumption gives the result that g(xk+1 ) < g(xk ) which is equivalent to
xk+2 − xk+1 < xk+1 − xk < 0.
Thus it is also true for n = k + 1. By induction, it is true for all positive integers n.
h The following proof in this problem uses mainly the continuity property of the function g and induction. In fact, one
may prove the problem by using mainly Theorem 5.11 and the differentiability and the convexity of the function f (Problems
4.23 and 5.14).
Chapter 5. Differentiation
106
Next, we show that {xn } is unbounded. Assume that {xn } was bounded. By Theorem 3.14, we
have {xn } is convergent. Let lim xn = θ. It is clear that θ < α. Furthermore, it follows from this
n→∞
and Theorem 4.2 that
θ = lim xn+1 = lim f (xn ) = f (θ).
(5.54)
n→∞
n→∞
Therefore, we arrive at a conclusion that θ is the fourth fixed point of f , a contradiction to our
hypothesis. Hence, we have xn → −∞ as n → ∞.
(b) Suppose that α < x1 < γ. We note that α < β < γ, so there are three cases:
– Case (i): x1 = β. Then we have
x2 = f (x1 ) =
x31 + 1
3β
=
= β.
3
3
Thus we have xn = β for all positive integers n and we are done in this case.
– Case (ii): x1 ∈ (β, γ). We claim that xn ∈ (β, γ) and xn+1 < xn for all positive integers n.
We prove the claim by induction.
For the first assertion, it follows from 0 < β < x1 that f (β) < f (x1 ) which is equivalent to
β < x2 .
For the second assertion, since g(β) = g(γ) = 0 and g(x) 6= 0 on (β, γ), we must have g(x1 ) 6= 0.
Since
13 + 1
1
g(1) =
−1=− ,
3
3
the continuity of g guarantees that g(x) < 0 on (β, γ). In particular, we have g(x1 ) < 0 and
then f (x1 ) < x1 which is equivalent to
x2 < x1 .
Thus the statements are true for n = 1.
Assume that they are true for n = k for some positive integer k, i.e., xk ∈ (β, γ) and xk+1 < xk .
For n = k + 1, it follows from 0 < β < xk that f (β) < f (xk ) so that
β < xk+1 .
Since xk+1 ∈ (β, γ), we have g(xk+1 ) < 0 so that f (xk+1 ) < xk+1 which is equivalent to
xk+2 < xk+1 .
Therefore, the statements are true for n = k + 1. Hence the claim follows from induction.
Since {xn } is bounded, Theorem 3.14 implies that {xn } converges. Let lim xn = η. By using
n→∞
similar argument as obtaining the result (5.54), we have
η = lim xn+1 = lim f (xn ) = f (η).
n→∞
n→∞
Therefore, η is also a fixed point of f and η ∈ [β, γ). Hence we have η = β in this case.
– Case (iii): x1 ∈ (α, β). We claim that xn ∈ (α, β) and xn < xn+1 for all positive integers
n. Since the proof of this part is very similar to that of Case (ii) above, we omit the details
here.
Since {xn } is bounded, Theorem 3.14 implies that {xn } converges. Let lim xn = τ . By using
n→∞
similar argument as obtaining the result (5.54), we have
τ = lim xn+1 = lim f (xn ) = f (τ ).
n→∞
n→∞
Therefore, τ is also a fixed point of f and τ ∈ (α, β]. Hence we have τ = β in this case.
107
5.3. Derivatives of higher order and iteration methods
(c) Suppose that γ < x1 . Since g(2) = 1 > 0, the continuity of g implies that g(x) > 0 on (γ, ∞).
In particular, we have g(x1 ) > 0. Now by applying similar argument as in part (a), we can show
that {xn } is a strictly increasing sequence and it is unbounded. Hence we must have xn → +∞ as
n → ∞.
This completes the proof of the problem.
Problem 5.24
Rudin Chapter 5 Exercise 24.
Proof. By Problem 3.16, we see that the convergence
of the function f is much more rapid than that of
√
the function g. Take the same starter x1 > α in the recursions of f and g. By the inequalities (5.51)
to (5.53), the rate of the convergence depends on the bound of the recursion formula. It is easy to check
that
√
1
α √
x+
− α
f (x) − α =
2
x
√
√ x
α α
α
=
−
−
+
2
2
2x
2
√ √
√
α
α
1
−1
= (x − α) +
2
x
√ 2
√
√
α
1
f (x) − α =
(5.55)
1−
(x − α)
2
x
and
√
√
√
√
√
α+x √
α+x− α−x α
1− α
g(x) − α =
− α=
=
(x − α).
1+x
1+x
1+x
(5.56)
Let {xn (f )} and {xn (g)} be the two sequences generated by f and g respectively, where x1 (f ) = x1 (g).
(Note that xn (f ) 6= xn (g) in general.) Then it follows from the formulas (5.55) and (5.56) that
√
√
√
α
1
|xn+1 (f ) − α| = 1 −
(5.57)
· |xn (f ) − α|
2
xn (f )
and
√
|xn+1 (g) − α| =
√
√
1− α
· |xn (g) − α|.
1 + xn (g)
(5.58)
• Case √
(i): α > 1. By Problem 3.16, Lemma 3.1, Problem 3.17(a) and (b) and the fact that
x1 > α > 1, we consider the functions f and g defined on the interval [1, x1 ]. Therefore, the
expressions (5.57) and (5.58) imply that
√
n
√
√
α
1 Y
|x1 − α|
|xn+1 (f ) − α| = n
1−
2
xk (f )
(5.59)
k=1
and
|xn+1 (g) −
n
Y
√
α| =
k=1
√
√
1− α
|x1 − α|
1 + xk (g)
(5.60)
for all positive integers n. Since {xn (f )} decreases monotonically (see Problem 3.16(a)), we have
from the expression (5.59) that
x − √α n
√
√
1
|xn+1 (f ) − α| ≤
(5.61)
|x1 − α|.
2x1
Chapter 5. Differentiation
108
Similarly, since xn (g) ∈ [1, x1 ], we have from the expression (5.60) that
√α − 1 n
√α − 1 n
√
√
√
|x1 − α|.
|x1 − α| ≤ |xn+1 (g) − α| ≤
1 + x1
2
(5.62)
We compare the magnitudes of the two constants
√
√
x1 − α
α−1
and
.
2x1
1 + x1
√
√
In fact, if x1 is chosen to be close to α,i then x1 − α is very small so that
√
√
x1 − α
α−1
<
< 1.
2x1
1 + x1
Therefore, we follow from the inequalities (5.61) and (5.62) that
√
√
|xn+1 (f ) − α| < |xn+1 (g) − α|.
This explains why the rate of convergence of {xn (f )} is much more rapid than that of {xn (g)} if n
is large enough.
For example, take α = 10, then we have
√ √
( α, α) = (3.162277660168379, 3.162277660168379).
Note that
7
89
15761
, x3 (f ) =
, x4 (f ) =
,...
2
28
4984
and its zig-zag path is shown in Figure 5.2j , where the green, orange and purple dots are
89 15761 7 7 89 ,
,
,
and
5,
2
2 28
28 4984
x1 (f ) = 5,
x2 (f ) =
respectively.
Figure 5.2: The zig-zag path induced by the function f in Case (i).
Similarly, we have
x1 (g) = 5,
x2 (g) =
15
,
6
x3 (g) =
75
,
21
x4 (g) =
286
,
96
x5 (g) =
1245
,...
381
and its zig-zag path is shown in Figure 5.3, where the green, orange, purple and black dots are
285 1245 15 15 75 75 285 ,
,
and
,
,
,
5,
6
6 21
21 96
96 381
109
5.3. Derivatives of higher order and iteration methods
Figure 5.3: The zig-zag path induced by the function g in Case (i).
respectively.
In conclusion,
√ √ the pattern of the zig-zag path induced by the function f is approaching to the fixed
point ( 10, 10) from one-side, but the one induced by the function g goes near the fixed point
“alternatively”.
1
• Case (ii): 0 < α < 1. Take α = 10
and x1 = 15 . Then we have
√ √
( α, α) = (0.4472135954999579, 0.4472135954999579).
Therefore, we have
x1 (f ) =
1
,
5
x2 (f ) =
7
,
20
x3 (f ) =
89
280
and x4 (f ) =
15761
49840
and its zig-zag path is shown in Figure 5.4, where the green, orange and purple dots are
respectively.
1 7 ,
,
5 20
7
89 20 280
,
and
89 15761 ,
280 49840
Figure 5.4: The zig-zag path induced by the function f in Case (ii).
i This can be done by an initial estimation.
j The figures are drawn by “desmos”.
Chapter 5. Differentiation
110
Similarly, we acquire
x1 (g) =
1
,
5
x2 (g) =
1
,
4
x3 (g) =
7
,
25
x4 (g) =
19
,
64
x5 (g) =
127
,...
415
and its zig-zag path is shown in Figure 5.3, where the green, orange, purple and black dots are
respectively.
1 1
,
,
5 4
1 7 ,
,
4 25
7 19 ,
25 64
and
19 127 ,
64 415
Figure 5.5: The zig-zag path induced by the function g in Case (ii).
In this case, both the
q of the zig-zag paths induced by the functions f and g are approaching
qpatterns
to the fixed point (
(ii).
1
10 ,
1
10 ) from one-side. This is main difference between Case (i) and Case
We complete the proof of the problem.
Problem 5.25
Rudin Chapter 5 Exercise 25.
Proof.
(a) Now the given formula can be rewritten as
f ′ (xn ) =
0 − f (xn )
.
xn+1 − xn
(5.63)
Therefore, it means that the slope of the graph of f at xn (the left-hand side of the formula (5.63))
equals to the slope of the straight line passing through
(xn , f (xn )) and (xn+1 , 0)
(the right-hand side of the formula (5.63)). See the following figure:
111
5.3. Derivatives of higher order and iteration methods
Figure 5.6: The geometrical interpretation of Newton’s method.
(b) We prove the result by induction. By the continuity of f and the fact that ξ is the unique point
in (a, b) at which f (ξ) = 0, we have f (x1 ) > 0. Otherwise, we have f (t) = 0 for some t ∈ [x1 , b)
contradicting to the uniqueness of ξ. Since f ′ (x) > 0 for all x ∈ [a, b],
f (x1 )
>0
f ′ (x1 )
and so x2 < x1 . This shows that the case for n = 1 is true.
Assume that it is also true for n = k for some positive integer k, i.e., xk+1 < xk . Note that
xk+1 < xk is equivalent to f (xk ) > 0. For n = k + 1, we have
xk+2 = xk+1 −
f (xk+1 )
.
f ′ (xk+1 )
(c) It is clear that the function f satisfies the conditions of Theorem 5.15 (Taylor’s theorem) with α = ξ
and β = xn , so we have
f (xn ) = f (ξ) + f ′ (xn )(xn − ξ) +
= f ′ (xn )(xn − ξ) +
f ′′ (t)
(xn − ξ)2
2
f ′′ (t)
(xn − ξ)2
2
f ′′ (t)
f ′ (xn )
= (xn − ξ) + ′
(xn − ξ)2
f (xn )
2f (xn )
(d) Since f ′ (x) ≥ δ > 0 and 0 ≤ f ′′ (x) ≤ M for all x ∈ [a, b], it follows from part (c) that
0 ≤ xn+1 − ξ ≤
M
(xn − ξ)2 = A(xn − ξ)2
2δ
which implies that
0 ≤ xn+1 − ξ ≤
n
1
[A(x1 − ξ)]2
A
(5.64)
as required.
When we compare these inequalities (5.64) with Problems 3.16 and 3.18, we see that the recursion
formula in Problem 3.18 which is
xn+1 =
α
p−1
xn + x−p+1
p
p n
Chapter 5. Differentiation
112
corresponds to applying Newton’s method with the function f (x) = xp − α. Actually, we have
f ′ (x) = pxp−1 so that
xn+1 = xn −
f (xn )
xpn − α
α
p−1
=
x
−
xn + x−p+1
.
n
p−1 =
f ′ (xn )
p
p n
pxn
(5.65)
Put p = 2, the formula (5.65) reduces to Problem 3.16 as a special case and the error estimation
ǫn+1 < β
ǫ 2n
1
β
is equivalent to the error estimation (5.64) deduced by Newton’s method.
(e) It is clear that g(ξ) = ξ if and only if f (ξ) = 0. In other words, the fixed point ξ of g is exactly
the zero of f . Hence, Newton’s method of computing ξ as described in parts (a) to (c) amounts to
finding the fixed point of the function g.
Since
g ′ (x) =
f (x)f ′ (x)
,
[f ′ (x)]2
we have
M
f (x)f ′′ (x)
≤ 2 |f (x)|
[f ′ (x)]2
δ
M
lim |g ′ (x)| ≤ lim 2 |f (x)|
x→ξ
x→ξ δ
|g ′ (x)| =
which implies that lim g ′ (x) = 0.
x→ξ
2
(f) It is obvious that f (x) = 0 if and only if x = 0. Since f ′ (x) = 13 x− 3 , f ′ (x) → 0 as x → ∞. This
shows that the function f does not satisfy one of the conditions of Newton’s method:
f ′ (x) ≥ δ > 0
for some positive constant δ and for all x ∈ (−∞, ∞). Therefore, Newton’s method fails in this
case. In fact, we have
f (xn )
= −2xn
xn+1 = xn − ′
f (xn )
so that xn+1 = (−2)n x1 . Since x1 ∈ (0, ∞), the sequence {xn } is divergent.
This ends the proof of the problem.
Problem 5.26
Rudin Chapter 5 Exercise 26.
Proof. We follow the hint. Fix x0 ∈ [a, b], let
M0 (f ) =
sup |f (x)|
and M0 (f ′ ) =
x∈[a,x0 ]
sup |f ′ (x)|.
x∈[a,x0 ]
Since |f ′ (x)| ≤ A|f (x)| on [a, b], we always have M0 (f ′ ) ≤ AM0 (f ). Now for any x ∈ [a, x0 ], since f is
differentiable in [a, b], it is clear that it is continuous on [a, x] and differentiable in (a, x). By Theorem
5.10, we have
|f (x) − f (a)| = |(x − a)f ′ (t)|
for some t ∈ (a, x). By our hypothesis, this implies that
|f (x)| ≤ |f ′ (t)|(x0 − a) ≤ M0 (f ′ )(x0 − a) ≤ A(x0 − a)M0 (f ).
(5.66)
113
5.4. Solutions of differential equations
Let x0 be chosen so that A(x0 − a) < 21 . Assume that M0 (f ) 6= 0. Then A(x0 − a) < 21 and the inequality
(5.66) imply that
M0 (f )
|f (x)| <
2
for all x ∈ [a, x0 ] and this gives the contradiction that M0 (f ) ≤ M02(f ) . Hence we must have M0 (f ) = 0
on [a, x0 ] which induces
f (x) = 0
on [a, x0 ].
Now we choose x0 , x1 , . . . , xn such that a < x0 < x1 < · · · < xn < b with
A(x0 − a) <
1
,
2
A(xk − xk−1 ) <
1
2
and A(b − xn ) <
1
,
2
where k = 1, 2, . . . , n. Next, we define
Mk (f ) =
sup
x∈[xk−1 ,xk ]
|f (x)|
and Mk (f ′ ) =
sup
x∈[xk−1 ,xk ]
|f ′ (x)|
for each k = 1, 2, . . . , n. Then the argument in the previous paragraph can be applied to each Mk (f ) to
get the result that
f (x) = 0
(5.67)
on [xk−1 , xk ], where k = 1, 2, . . . , n. The remaining case for [xn , b] can be considered similarly and finally
the equation (5.67) also holds on [xn , b].
Hence, by combining the above cases, we have f (x) = 0 on [a, b], completing the proof of the problem.
5.4
Solutions of differential equations
Problem 5.27
Rudin Chapter 5 Exercise 27.
Proof. Assume that the initial-value problem had two solutions f1 and f2 . Let f = f1 − f2 . Since f1 and
f2 are differentiable in [a, b], f is also differentiable in [a, b] by Theorem 5.3(a). Furthermore, we have
f (a) = f1 (a) − f2 (a) = c − c = 0.
Since α ≤ f1 (x) ≤ β and α ≤ f2 (x) ≤ β, we have (x, f1 (x)), (x, f2 (x)) ∈ R. Therefore, the hypothesis
implies that
|φ(x, f2 (x)) − φ(x, f1 (x))| ≤ A|f2 (x) − f1 (x)|
|f2′ (x) − f1′ (x)| ≤ A|f2 (x) − f1 (x)|
|f ′ (x)| ≤ A|f (x)|.
In other words, the function f satisfies the conditions of Problem 5.26 and it yields that f ≡ 0 on [a, b],
i.e., f1 ≡ f2 on [a, b].
2
Now it is easy to check that the functions f (x) = 0 and f (x) = x4 are two distinct solutions of the
initial-value problem. To find all other solutions, we suppose first that the interval under consideration
is I = [0, ∞). Next, we let f (x) be a (real) solution of the initial-value problem on I, f (x) 6≡ 0 and
2
f (x) 6≡ x4 . Now we must have
f (x) ≥ 0
1
on I. Otherwise, if f (x0 ) < 0 for some x0 ∈ I, then (f (x0 )) 2 is a complex-value function in a neighborhood
of x0 which is a contradiction. Since f (x) 6≡ 0, there exists x0 ∈ I such that f (x0 ) > 0. Since f is
continuous on I, we have f (x) > 0 in an interval (a, b) containing x0 with
f (a) = 0,
Chapter 5. Differentiation
114
where 0 ≤ a < ∞.kp
Define F (x) = f (x) on (a, b). Then it is easy to see that F ′ (x) = 12 so that F (x) = 12 (x + C) and
then f (x) = 14 (x + C)2 for some constant C. Since f (a) = 0, we have C = −a which gives
f (x) =
1
(x − a)2
4
on (a, b). The final task is to find the explicit form of f (x) on [0, a). By the continuity of f , the initial
1
condition f (0) = 0 and f (a+) = x→a
lim f (x) = x→a
lim (x − a)2 = 0, we have
x>a
x>a 4
f (x) ≡ 0
on [0, a). Hence, the other solutions must be in the form
0,
if 0 ≤ x < a;
f (x) =
1
2
(x
−
a)
,
if a ≤ x < b.
4
2
As a final remark, since f (x) 6≡ x4 , it is impossible that a = 0 and b = ∞.l This finishes the proof of the
problem.
Problem 5.28
Rudin Chapter 5 Exercise 28.
Proof. A solution of the initial-value problem for systems of differential equations
y′ = Φ(x, y),
y(a) = c = (c1 , c2 , . . . , ck ) (αi ≤ ci ≤ βi ; i = 1, 2, . . . , k)
(5.68)
is, by definition, a differentiable function f on [a, b] such that f (a) = c and
f ′ = Φ(x, f )
(a ≤ x ≤ b).
We claim that the initial-value problem (5.68) has at most one solution if there is a constant A such
that
|Φ(x, y2 ) − Φ(x, y1 )| ≤ A|y2 − y1 |
(5.69)
whenever (x, y1 ), (x, y2 ) ∈ R, where R is a (k + 1)-cell.
To prove the claim, let y1 and y2 be two solutions of the initial-value problem (5.68). Define the
function f (x) = y2 (x) − y1 (x). It is clear that f is differentiable on [a, b],
f (a) = y2 (a) − y1 (a) = c − c = 0
and
|f ′ (x)| = |y2′ (x) − y1′ (x)| = |Φ(x, y2 ) − Φ(x, y1 )| ≤ A|y2 − y2 | = A|f (x)|
on [a, b]. By using the vector-valued form of Problem 5.26, we have f (x) = 0 for all x ∈ [a, b]. Hence, we
have
y2 (x) = y1 (x)
for all x ∈ [a, b] and the uniqueness of solutions of the initial-value problem (5.68) follows immediately.
This completes the proof of the problem.
k Such an a must exist. Otherwise, if f (x) > ǫ > 0 for some ǫ > 0 and for all x > 0, then the continuity of f gives
0 = f (0) = lim f (x) ≥ ǫ,
x→0
a contradiction.
l We shall remark that Problem 5.27 is actually part of the so-called Picard’s existence theorem and the inequality in
the problem is called the Lipschitz condition. See [5, §2.4, §2.8].
115
5.4. Solutions of differential equations
Problem 5.29
Rudin Chapter 5 Exercise 29.
Proof. Let y be a solution of the differential equation with the initial conditions
(k)
y + gk (x)y (k−1) + · · · + g2 (x)y ′ + g1 (x)y = f (x),
y(a) = c1 , y ′ (a) = c2 , . . . , y (k−1) (a) = ck .
(5.70)
′
Suppose that y1 = y, y2 = y1′ , y3 = y2′ , . . . , yk = yk−1
, yk′ = y (k) . Then the differential equation (5.70) is
equivalent to
yk = f (x) − [g1 (x)y + g2 (x)y ′ + · · · + gk (x)y (k−1) ] = f (x) −
k
X
gj (x)yj .
j=1
Besides, the initial conditions y (j−1) (a) = cj are equivalent to the conditions yj (a) = cj respectively,
where j = 1, 2, . . . , k. Therefore, what we have shown is that the differential equation (5.70) is equivalent
to the initial-value problem

k
X

 ′
gj (x)yj , (j = 1, 2, . . . , k − 1);
yj = yj+1 , yk′ = f (x) −
(5.71)
j=1


yj (a) = cj
(j = 1, 2, . . . , k).
We note that the initial-value problem (5.71) is a special case of the system of differential equations
in Problem 5.28 with
φ1 = y2 ,
φ2 = y3 , . . . , φk−1 = yk
and φk = f (x) −
k
X
gj (x)yj .
j=1
Assume that u and v were two solutions of the differential equation (5.70). Then they induce two solutions
u = (u1 , u2 , . . . , uk ) = (u, u′ , . . . , u(k−1) ) and v = (v1 , v2 , . . . , vk ) = (v, v ′ , . . . , v (k−1) )
of the initial-value problem (5.71) with
Φ(x, y) = (φ1 , φ2 , . . . , φk ) =
y2 , y3 , . . . , yk , f (x) −
k
X
j=1
!
gj (x)yj .
To complete our problem, we have to show that the solutions u and v satisfy the inequality (5.69) for
some positive constant A.
Since g1 , g2 , . . . , gk are continuous real functions on [a, b], Theorem 4.16 ensures that maximums of
g1 , g2 , . . . , gk can be found in [a, b]. Let Mj = max gj (x) and define
x∈[a,b]
M = max(M1 , M2 , . . . , Mk ).
Therefore, it follows from the definition (5.71), Theorem 1.35 and Definition 1.36 that
|Φ(x, u) − Φ(x, v)| = |u′ − v′ |
= |(u′1 , u′2 , . . . , u′k ) − (v1′ , v2′ , . . . , vk′ )|
=
=
u2 , u3 , . . . , uk , f (x) −
u2 − v2 , . . . , uk − vk ,
k
X
j=1
k
X
j=1
gj (x)uj
!
−
v2 , v3 , . . . , vk , f (x) −
!
gj (x)(vj − uj )
k
X
j=1
gj (x)vj
!
Chapter 5. Differentiation
=
(
≤
(
≤
(
116
2
2
(u2 − v2 ) + · · · + (uk − vk ) +
2
2
" k
X
j=1
(u2 − v2 ) + · · · + (uk − vk ) + M
2
#2 ) 21
gj (x)(vj − uj )
" k
X
j=1
(u2 − v2 )2 + · · · + (uk − vk )2 + M 2 × k
#2 ) 21
(uj − vj )
k
X
j=1
(uj − vj )2
) 12
i 12
h
≤ (1 + kM 2 ) × (u1 − v1 )2 + (u2 − v2 )2 + · · · + (uk − vk )2
1
2
≤ A|u − v|,
1
where A = (1 + kM 2 ) 2 > 0. Hence, Problem 5.28 implies that u = v, as desired. This completes the
proof of the problem.
CHAPTER
6
The Riemann-Stieltjes Integral
6.1
Problems on Riemann-Stieltjes integrals
Problem 6.1
Rudin Chapter 6 Exercise 1.
Proof. It is clear that f is bounded on [a, b] and is discontinuous only at x0 ∈ [a, b]. Since α is continuous
at x0 , it follows from Theorem 6.10 that f ∈ R(α). Let P = {t0 , t1 , . . . , tn } be a partition of [a, b] and
Ii = [ti−1 , ti ], where i ∈ {1, . . . , n}. Let, further, x0 ∈ Ii for some i. By Definition 6.2, we have
mi = inf f (x) = 0 (i = 1, 2, . . . , n).
x∈Ii
Thus we have
L(P, f, α) =
n
X
mi ∆αi = 0
i=1
and then sup L(P, f, α) = 0 which gives
Z b
f dα = 0.
a
Since f ∈ R(α), we must have
Z b
f dα =
a
Z b
f dα = 0
a
as desired. We complete the proof of the problem.
Problem 6.2
Rudin Chapter 6 Exercise 2.
Proof. Assume that f (x0 ) 6= 0 for some x0 ∈ [a, b]. Since f ≥ 0 on [a, b], we have f (x0 ) > 0 and the
continuity of f implies that
f (x0 )
>0
f (x) >
2
on (x0 − δ, x0 + δ) ⊂ [a, b] for some δ > 0. Let [c, d] ⊂ (x0 − δ, x0 + δ) ⊂ [a, b], where c < d. By Theorem
6.21 (Second Fundamental Theorem of Calculus), we know that
Z b
0 dx = 0
and
a
Z b
a
117
1 dx = b − a.
Chapter 6. The Riemann-Stieltjes Integral
118
By Theorem 6.12(b) and (c), we have
Z b
Z c
Z d
Z b
Z d
(d − c)f (x0 )
f dx =
f dx +
f dx +
f dx ≥ 0 +
f dx + 0 ≥
>0
2
a
a
c
d
c
which is a contradiction. Hence we have f (x) = 0 for all x ∈ [a, b], completing the proof of the problem.
Problem 6.3
Rudin Chapter 6 Exercise 3.
Proof. Suppose that P = {x0 , x1 , . . . , xn } is a partition of [−1, 1]. By Theorem 6.4, we have for j = 1, 2, 3,
U (P ∗ , f, βj ) − L(P ∗ , f, βj ) ≤ U (P, f, βj ) − L(P, f, βj )
if P ∗ is a refinement of P . Thus, without loss of generality, we may assume that 0 ∈ P . Let xk = 0 for
some positive integer k, where 0 < k < n. Then we have

 β1 (xi ) − β1 (xi−1 ) = 0, if i = 1, 2, . . . , k;
1,
if i = k + 1;
(6.1)
∆(β1 )i = β1 (xi ) − β1 (xi−1 ) =

β1 (xi ) − β1 (xi−1 ) = 0, if i = k + 2, . . . , n,

 β2 (xi ) − β2 (xi−1 ) = 0, if i = 1, 2, . . . , k − 1;
1,
if i = k;
(6.2)
∆(β2 )i = β2 (xi ) − β2 (xi−1 ) =

β2 (xi ) − β2 (xi−1 ) = 0, if i = k + 1, . . . , n,

 β3 (xi ) − β3 (xi−1 ) = 0, if i = 1, 2, . . . , k − 1;
1
,
if i = k, k + 1;
(6.3)
∆(β3 )i = β3 (xi ) − β3 (xi−1 ) =
 2
β3 (xi ) − β3 (xi−1 ) = 0, if i = k + 2, . . . , n.
By the definitions (6.1), (6.2) and (6.3), we have
L(P, f, β1 ) =
L(P, f, β2 ) =
n
X
i=1
n
X
mi ∆(β1 )i = mk+1 ,
U (P, f, β1 ) =
n
X
Mi ∆(β1 )i = Mk+1 ,
i=1
mi ∆(β2 )i = mk ,
U (P, f, β2 ) =
n
X
Mi ∆(β2 )i = Mk ,
i=1
i=1
n
X
mk + mk+1
mi ∆(β3 )i =
L(P, f, β3 ) =
,
2
i=1
U (P, f, β3 ) =
n
X
Mi ∆(β3 )i =
i=1
Mk + Mk+1
2
which imply that
U (P, f, β1 ) − L(P, f, β1 ) = Mk+1 − mk+1 ,
U (P, f, β2 ) − L(P, f, β2 ) = Mk − mk ,
(6.4)
(6.5)
U (P, f, β3 ) − L(P, f, β3 ) =
(6.6)
(Mk+1 − mk+1 ) + (Mk − mk )
.
2
(a) By Theorem 6.6 and the expression (6.4), f ∈ R(β1 ) if and only if for every ǫ > 0, there exists a
partition P such that
Mk+1 − mk+1 < ǫ.
(6.7)
Suppose that such a partition P exists. If we take δ = xk+1 > 0, then for all x with x > 0 and
x − 0 < δ, we have
|f (x) − f (0)| ≤ Mk+1 − mk+1 < ǫ.
(6.8)
By Definition 4.25, this means that f (0+) = f (0).
Conversely, suppose that f (0+) = f (0). Then for every ǫ > 0, there exists a δ > 0 such that
|f (x) − f (0)| < ǫ
(6.9)
119
6.1. Problems on Riemann-Stieltjes integrals
for all x with 0 < x < δ. Let P = {−1, 0, δ, 1} be a partition of [−1, 1]. Thus the expression (6.4)
and inequality (6.9) give
U (P, f, β1 ) − L(P, f, β1 ) = sup f (x) − inf f (x)
x∈[0,δ]
x∈[0,δ]
≤
sup f (x) − f (0) +
x∈[0,δ]
inf f (x) − f (0)
x∈[0,δ]
≤ǫ+ǫ
= 2ǫ.
Since ǫ is arbitrary, we know from Theorem 6.6 that f ∈ R(β1 ). Hence we have shown that
f ∈ R(β1 ) if and only if f (0+) = f (0).
If the inequality (6.7) holds for the partition P , then it follows from Theorem 6.7(c) that
f (tk+1 ) −
Z
f dβ1 < ǫ,
(6.10)
where tk+1 ∈ [0, xk+1 ]. Hence we deduce from the inequalities (6.8) and (6.10) that, for all tk+1 ∈
[0, xk+1 ],
f (0) −
Z
f dβ1 ≤ f (0) − f (tk+1 ) + f (tk+1 ) −
Since ǫ is arbitrary, it means that
as desired.
Z
Z
f dβ1 < ǫ + ǫ = 2ǫ.
f dβ1 = f (0)
(b) We claim that f ∈ R(β2 ) if and only if f (0−) = f (0) and then
Z
f dβ2 = f (0).
Since the proof of this is almost identical to that in part (a) with applying the expressions (6.2)
and (6.5), we omit the details of its proof here.
(c) By the analysis in part (a), we see that Mk+1 − mk+1 < ǫ if and only if f (0+) = f (0). Similarly,
the analysis in part (b) implies that Mk − mk < ǫ if and only if f (0−) = f (0). By these, we obtain
from the expression (6.6) that
U (P, f, β3 ) − L(P, f, β3 ) <
ǫ
ǫ
+ =ǫ
2 2
if and only if f (0+) = f (0) = f (0−), i.e., f is continuous at 0 by Definition 4.25. Hence Theorem
6.6 shows that f ∈ R(β3 ) if and only if f is continuous at 0.
By similar argument as in proving part (a), we can show that
Z
f dβ3 = f (0).
(d) If f is continuous at 0, then we must have f (0) = f (0+) = f (0−) and the results of parts (a) - (c)
imply that
Z
Z
Z
f dβ1 =
This completes the proof of the problem.
f dβ2 =
f dβ3 = f (0).
Chapter 6. The Riemann-Stieltjes Integral
120
Problem 6.4
Rudin Chapter 6 Exercise 4.
Proof. Assume that f ∈ R on [a, b] for some a < b. By Theorem 6.6, for every ǫ > 0, there exists a
partition P = {x0 , x1 , . . . , xn } such that
U (P, f ) − L(P, f ) < ǫ.
(6.11)
By Definition 6.1, we have Mi = 1 and mi = 0 for all i = 1, 2, . . . , n. Therefore, we have
U (P, f ) =
n
X
i=1
Mi ∆xi = xn − x0 = b − a
and L(P, f ) =
n
X
mi ∆xi = 0.
i=1
However, their difference is b − a which contradicts the assumption (6.11). Hence we must have f ∈
/R
on [a, b] for any a < b, completing the proof of the problem.
Problem 6.5
Rudin Chapter 6 Exercise 5.
Proof. Define the function
f (x) =
1,
if x ∈ [a, b] and x is rational;
−1, if x ∈ [a, b] and x is irrational.
Then it is bounded on [a, b]. Now f 2 (x) = 1 for all x ∈ [a, b] so that f 2 ∈ R on [a, b]. However, using
similar argument as in the proof of Problem 6.4, we can show that f 6∈ R on [a, b].
For the second assertion, we know from the hypothesis that there exist constants M and m such that
m ≤ f (x) ≤ M
for all x ∈ [a, b]. Define φ : [m3 , M 3 ] → R by
1
φ(y) = y 3 .
Then the function φ is continuous on [m3 , M 3 ] and
φ(f 3 (x)) = f (x)
on [a, b]. By Theorem 6.11, we have f ∈ R(α) on [a, b]. This completes the proof of the problem.
Problem 6.6
Rudin Chapter 6 Exercise 6.
Proof. Recall the definition of the Cantor set P : Suppose that E0 , E1 , E2 , . . . are the intervals defined in
Sec. 2.44. Then we have
∞
\
P =
En .
n=1
Table 6.1 below shows the number of intervals, the number of the end-points and the length of each
interval for each En .
We apply Theorem 6.6 to show that f ∈ R on [0, 1]. In fact, our goal is to construct a suitable
partition P = {x0 , x1 , . . . , xn } of [0, 1] such that
U (P, f ) − L(P, f ) =
n
X
i=1
(Mi − mi )∆xi < ǫ
(6.12)
121
6.1. Problems on Riemann-Stieltjes integrals
Interval
E0
E1
E2
..
.
Number of intervals
20
21
22
..
.
Number of end-points
including 0 and 1
1
2
22
23
..
.
1
30
1
31
1
32
En
2n
2n+1
1
3n
Length of each interval
..
.
Table 6.1: The number of intervals & end-points and the length of each interval for each En .
for every ǫ > 0.
Since f is bounded on [0, 1], there are constants M and m such that
m ≤ f (x) ≤ M
for all x ∈ [0, 1]. Suppose that n is a large positive integer such that
1+
ǫ
log 8(M−m)
log 32
< n.
(6.13)
Now for each En , we let
0 = a1 < b1 < a2 < b2 < · · · < a2n < b2n = 1
be the end-points contained in En .a By the information given in Table 6.1, we know two facts about the
end-points:
• Fact 1. For k = 1, 2, . . . , 2n and j = 1, 2, . . . , 2n − 1, we have
b k − ak =
1
3n
and aj+1 − bj ≤
1
3n−1
(6.14)
• Fact 2. If x ∈ P , then it follows from the definition of P and the footnote a that
x ∈ [ak , bk ]
for some positive integer k. Furthermore, by the hypotheses, f is continuous at every point x ∈
[0, 1] \ P . However, it does not mean that f is discontinuous at every point of P . In fact, the only
thing that we can say is that
“f is possibly discontinuous at points of P .”
Now we can start to construct our partition based on Fact 1. Let δ > 0 be a number such that
δ < 2·31 n . By the observations (6.14), we define a partition Pn (δ) by
Pn (δ) = {0 = a1 < b1 + δ < a2 − δ < b2 + δ < · · · < a2n − δ < b2n = 1}.
(6.15)
Since there are totally 2n+1 distinct points in Pn (δ), they make the interval [0, 1] into 2n+1 − 1 intervals.b
We consider
2n+1
X−1
(Mi − mi )∆xi .
(6.16)
U (Pn (δ), f ) − L(Pn (δ), f ) =
i=1
Next, we estimate the magnitude of (6.16) by using Fact 2. In fact, we split the summation (6.16)
into two parts: One part consists of only intervals on which f is continuous and the other part consisting
of intervals that f might contain discontinuities. By the definition (6.15) of Pn (δ) and Fact 2, we know
that
[a1 , b1 ] ⊂ [a1 , b1 + δ], [a2n , b2n ] ⊂ [a2n − δ, b2n ] and [ak , bk ] ⊂ [ak − δ, bk + δ]
a We note that each [a , b ] is an interval in E , where k = 1, 2, . . . , n.
n
k k
b You should be careful that they are not the intervals in E !
n
Chapter 6. The Riemann-Stieltjes Integral
122
for k = 2, 3, . . . , 2n − 1. In other words, all possible discontinuities of f must fall into the intervals with
odd indices of the summation (6.16). Hence we can rewrite the summation (6.16) as
U (Pn (δ), f ) − L(Pn (δ), f ) =
n
2X
−1
s=1
|
n
(M2s − m2s )∆x2s +
{z
corresponding to the
continuity of f
n
2X
−1
≤ (M − m)
s=1
}
2
X
r=1
|
(M2r−1 − m2r−1 )∆x2r−1
{z
corresponding to possibly
the discontinuity of f
n
2
X
∆x2s + (M − m)
∆x2r−1 .
}
(6.17)
r=1
In order to find the estimate of the inequality (6.17), we must find the bounds of the summations in
(6.17). To this end, we obtain from the expressions (6.14) that
∆x2s = x2s − x2s−1 = (as+1 − δ) − (bs + δ) = as+1 − bs − 2δ ≤
1
− 2δ,
3n−1
where s = 1, 2, . . . , 2n − 1. Since δ is any number less than 2·31 n , we can further assume that
"
#
3 1
1
ǫ
0<
≤δ<
−
.
2 3n
2(2n − 1)(M − m)
2 · 3n
(6.18)
(6.19)
Then we get from the expression (6.18) that
n
2X
−1
(M − m)
s=1
∆x2s ≤ 3(M − m)
n
2X
−1 s=1
n
2X
−1
3ǫ
2δ 3ǫ
1
≤
−
= .
n − 1)
3n
3
2(2
2
s=1
(6.20)
Similarly, we note from the expressions (6.14) that
∆x1 = x1 − x0 = b1 − a1 + δ =
1
+ δ,
3n
∆x2r−1 = x2r−1 − x2r−2 = (br + δ) − (ar − δ) = br − ar + 2δ =
∆x2n+1 −1 = x2n+1 −1 − x2n+1 −2 = b2n − a2n + δ =
1
+ 2δ,
3n
(6.21)
1
+ δ,
3n
where r = 2, 3 . . . , 2n − 1. Therefore, it follows from the inequality (6.13), the upper bound of δ and the
expressions (6.21) that
2n 2n
2 n−1
X
X
1
ǫ
∆x2r−1 < (M − m)
< .
(M − m)
+
2δ
≤
4(M
−
m)
n
3
3
2
r=1
r=1
(6.22)
Hence it follows from the inequalities (6.20) and (6.22) that
U (Pn (δ), f ) − L(Pn (δ), f ) < 2ǫ
holds for every n satisfying the inequality (6.13) and every δ satisfying the inequality (6.19). By Theorem
6.6, we have f ∈ R on [0, 1]. This completes the proof of the problem.
6.2
Definitions of improper integrals
Problem 6.7
Rudin Chapter 6 Exercise 7.
Proof. We use a red symbol on the left-hand side of the definition because we want to emphasize that it
is a “new” definition of integral.
123
6.2. Definitions of improper integrals
(a) Since f ∈ R on [0, 1] and 0 < c < 1, Theorem 6.12(c) implies that f ∈ R on [0, c] and on [c, 1]. In
addition, f ∈ R on [0, 1] also implies that f is bounded on [0, 1] by Definition 6.1. Thus, we let
|f (x)| ≤ M
on [0, 1] for some M > 0. Therefore, we follow from Theorem 6.12(b) and (c) that
Z 1
Z 1
Z c
f dx =
f dx −
f dx ≤ M c
0
so that
Z 1
lim
Since
Z 1
0
c
c→0
c>0
0
f dx −
Z 1
f dx = lim M c = 0.
c
c→0
c>0
f dx is a constant, we have
0
Z 1
f dx = lim
0
c→0
c>0
Z 1
f dx =
Z 1
f dx.
0
c
Hence this “new” definition of integral agrees with the old one.
∞ [
1
1i
1
(b) Let (0, 1] =
, k1 ]. Then we have
, . Define f (x) = (−1)k+1 (k + 1) for all x ∈ ( k+1
k+1 k
k=1
Z k1
1
1 (−1)k+1
=
−
.
f (x) dx = (−1)k+1 (k + 1)
1
k k+1
k
k+1
(6.23)
1
It is clear that c ∈ ( k+1
, k1 ] for one and only one positive integer k, so this and expression (6.23)
imply that
1
Z 1
Z 1
Z 12
Z k−1
Z k1
f (x) dx =
f (x) dx +
f (x) dx + · · · +
f (x) dx +
f (x) dx
c
1
2
1
3
k
Z 1
c
f (x) dx −
Z k1
c
=1−
f (x) dx = 1 −
(−1)
1 1
+ − ··· +
+
2 3
k−1
(−1)k
1 1
+ − ··· +
.
2 3
k−1
Since c → 0 if and only if k → ∞, we have
Z k1
c
Z k1
1
k
c
f (x) dx
c
(6.24)
f (x) dx → 0
as c → 0 and then we deduce from the expression (6.24) and Theorem 3.43 that
Z 1
Z 1
1 1
f (x) dx = 1 − + − · · · = ln 2.
f (x) dx = lim
c→0 c
2 3
0
c>0
1
, k1 ], we follow from
However, we have |f (x)| = k + 1 ≥ 0 for any positive integer k. Since c ∈ ( k+1
the expression (6.23) that
Z 1
Z 1
Z 1
k
k−1
X
X1
1
|f (x)| dx ≥
|f (x)| dx ≥
|f (x)| dx =
=
.
1
1
j
j
c
k+1
k
j=1
j=1
By the fact that c → 0 if and only if k → ∞ and Theorem 3.28, we have
Z 1
lim
|f (x)| dx
c→0
c>0
does not exist.
c
Chapter 6. The Riemann-Stieltjes Integral
124
We end our analysis of the problem here.
Problem 6.8
Rudin Chapter 6 Exercise 8.
Proof. Let k be a positive integer such that 1 ≤ k < n. Since f decreases monotonically on [1, ∞), we
have
f (k + 1) ≤ f (x) ≤ f (k)
(6.25)
for all x ∈ [k, k + 1]. Recall that f ∈ R on [1, b] for every b > 1, so we know from Theorem 6.12(c) and
the inequalities (6.25) that f ∈ R on [k, k + 1] which certainly imply
f (k + 1) ≤
Z k+1
k
f (x) dx ≤ f (k).
(6.26)
By the inequalities (6.26) and the hypothesis f (x) ≥ 0,c we have
! n−1
n−1
n−1
X
X Z k+1
X
f (k + 1) ≤
f (x) dx ≤
f (k)
k=1
k
k=1
n
X
k=2
n
X
k=2
f (k) ≤
f (k) ≤
Since f (x) ≥ 0 on [1, ∞), the sequences
(Z
Z n
1
Z n
1
f (x) dx ≤
f (x) dx ≤
)
n
f (x) dx
1
k=1
and
n−1
X
k=1
n
X
f (k)
f (k).
(6.27)
k=1
( n
X
)
f (k)
k=1
(6.28)
are increasing monotonically. If the series converges, then we deduce from the right-hand inequality
in (6.27) that the sequence of integrals in (6.28) is bounded above. By Theorem 3.14, it converges.
Conversely, if the sequence of integrals converges, then the left-hand inequality in (6.27) guarantees that
the series in (6.28) is also bounded above. By Theorem 3.14 again, it converges. This completes the proof
of the problem.
Problem 6.9
Rudin Chapter 6 Exercise 9.
Proof. The result is stated as follows:
Integration by Parts for Improper Integrals. Suppose that F and
Z G are differentiable functions on
∞
[a, ∞), F ′ = f ∈ R and G′ = g ∈ R on [a, ∞). If lim F (b)G(b) and
b→∞
Z ∞
f (x)G(x) dx exist, then
a
F (x)g(x) dx
a
also exists and we have
Z ∞
a
F (x)g(x) dx = lim F (b)G(b) − F (a)G(a) −
b→∞
c The hypothesis f (x) ≥ 0 is used in the last set of inequalities.
Z ∞
a
f (x)G(x) dx.
(6.29)
125
6.2. Definitions of improper integrals
Proof of Integration by Parts for Improper Integrals. Let a be fixed and a < b. By Theorem 6.22 (Integration by parts), we have
Z b
Z b
F (x)g(x) dx = F (b)G(b) − F (a)G(a) −
f (x)G(x) dx.
(6.30)
a
a
Taking b → ∞ to the right-hand side of the formula (6.30), since both
Z ∞
lim F (b)G(b) and
f (x)G(x) dx
b→∞
a
exist, they guarantee that the right-hand side of the formula (6.30) exists. Therefore, this shows that
Z ∞
F (x)g(x) dx
a
exists and then the formula (6.29) holds.
1
into the formula
Now we return to the proof of the problem. Put a = 0, F (x) = sin x and G(x) = 1+x
(6.29), we get
Z ∞
Z ∞
sin x
sin b
cos x
dx = lim
−0−
dx
2
b→∞
−(1
+
x)
1
+
b
1
+x
0
0
Z ∞
Z ∞
sin x
cos x
dx.
dx =
1
+
x
(1
+ x)2
0
0
By Theorem 6.21 (Second Fundamental Theorem of Calculus), we have
Z b
Z b
1
1
1
1 sin x
=1−
dx
≤
dx = −
− −
2
2
(1
+
x)
(1
+
x)
1
+
b
1
+
0
1
+
b
0
0
so that
Z ∞
0
In other words,
sin x
dx = lim
b→∞
(1 + x)2
Z ∞
0
converges absolutely.
To show that the integral
0
sin x
dx ≤ 1.
(1 + x)2
sin x
dx
(1 + x)2
Z ∞
0
Z b
cos x
dx
1+x
does not converge absolutely, notice that
Z nπ+ π2
n Z (k+1)π
X
2
| cos x|
| cos x|
dx ≥
dx
(k−1)π
1
+
x
1+x
0
2
k=2
Z (k+1)π
n
X
2
2
| cos x| dx
≥
(k + 1)π + 2 (k−1)π
2
k=2
≥
≥
n
X
4
(k + 1)π + 2
k=2
n
X
4
π
k=2
1
k+2
for every positive integer n. By Theorem 3.28, we know that
n
X
k=2
1
k+2
diverges and hence this completes the proof of the problem.
Chapter 6. The Riemann-Stieltjes Integral
6.3
126
Hölder’s inequality
Problem 6.10
Rudin Chapter 6 Exercise 10.
Proof.
(a) The claim is certainly true if u = 0 or v = 0. Therefore, we assume that both u and v are positive.
By Theorem 8.6(b) and (c), we have (ex )′′ = ex ≥ 0. Then Problem 5.14 implies that ex is convex
on R and so
p
1
q
1
uv = elog(uv) = e p log(u )+ q log(v ) ≤
up
vq
1 log up 1 log vq
e
+ e
=
+ .
p
q
p
q
(6.31)
By Problem 4.23, we know that the equality of
f (λx + (1 − λ)y) ≤ λf (x) + (1 − λ)f (y)
holds if and only if x = y. Hence the equality of the inequality (6.31) holds if and only if
log(up ) = log(v q )
which is equivalent to up = v q .d
(b) Let u = f and v = g. By part (a), we have
fg ≤
1 p 1 q
f + g .
p
q
(6.32)
Since f, g ∈ R(α) on [a, b], Theorem 6.13(a) shows that f g ∈ R(α) on [a, b]. Furthermore, there
are constants mf , mg , Mf and Mg such that
mf ≤ |f (x)| ≤ Mf
and mg ≤ |g(x)| ≤ Mg
on [a, b]. Since the functions φp (x) = xp and φq (x) = xq are obviously continuous on [mf , Mf ] and
[mg , Mg ] respectively, Theorem 6.11 guarantees that
f p , g q ∈ R(α) on [a, b].
Therefore, it follows from these, the inequality (6.32) and Theorem 6.12(a) that
Z b
a
f g dα ≤
Z b
a
1 p 1 p
1
f + g dα =
p
q
p
Z b
f p dα +
a
1
q
Z b
g q dα =
a
1 1
+ = 1.
p q
By part (a), we know that the equality holds if and only if f p = g q .
(c) Suppose that f and g are complex-valued functions in R(α) on [a, b]. Let
F (x) = (
Z b
a
|f (x)|
) p1
|f (x)|p dα
and G(x) = (
Z b
Since f, g ∈ R(α), we deduce from Theorem 6.13(b) that
|f (x)|, |g(x)| ∈ R(α) on [a, b].
d Note that the inequality proven in (a) is called Young’s inequality.
a
|g(x)|
) q1 .
|g(x)|q dα
127
6.3. Hölder’s inequality
Then Theorem 6.12(a) implies that F, G ∈ R(α) on [a, b]. Now it is obvious that F (x) ≥ 0, G(x) ≥ 0
on [a, b] and
Z b
Z b
F p (x) dα =
Gq (x) dα = 1.
a
a
Hence we obtain from part (b) and then Theorem 6.13(b) that
Z b
1≥
F (x)G(x) dα
(Z
|f (x)|p dα
a
(Z
a
) p1 ( Z
b
p
|f (x)| dα
a
a
) q1
≥
) q1
Z b
≥
Z b
|g(x)|q dα
a
) p1 ( Z
b
b
b
q
|g(x)| dα
a
|f (x)||g(x)| dα
f (x)g(x) dα
a
which is our desired inequality. By part (b), the equality holds if and only if F p = Gq which is
equivalent to
|g(x)|q
|f (x)|p
= Z b
.
Z b
p
q
|f (x)| dα
|g(x)| dα
a
(d)
a
– Case (i): “Improper” integrals described in Problem 6.7. Suppose that f and g are
real functions on (0, 1] and f, g ∈ R on [c, 1] for every c > 0. Then Theorem 6.13(a) implies
that f g ∈ R on [c, 1]. By part (c), we have
Z 1
0≤
c
(Z
f (x)g(x) dα ≤
) p1 ( Z
1
c
|f (x)|p dα
c
) q1
1
|g(x)|q dα
.
(6.33)
If any integral on the right-hand side of the inequality (6.33) is divergent when c → 0, then
Hölder’s inequality certainly holds in this case. If
Z 1
Z 1
Z 1
Z 1
lim
|f (x)|p dα =
|f (x)|p dα and lim
|g(x)|q dα =
|g(x)|q dα,
c→0
c>0
0
c
then we obtain from Theorem 4.2 that
Z 1
Z 1
|f (x)|p dα
|f (x)|p dα =
lim
n→∞
and
0
cn
c→0
c>0
c
lim
Z 1
n→∞
0
|g(x)|q dα =
cn
Z 1
0
|g(x)|q dα
for every sequence {cn } in [c, 1] such that cn 6= c and lim cn = c. By the inequality (6.33),
n→∞
we have
) q1
(Z
) p1 ( Z
Z
1
0 ≤ lim
n→∞
cn
1
1
f (x)g(x) dα ≤ lim
n→∞
= lim
n→∞
=
(Z
cn
(Z
1
cn
|f (x)|p dα
p
) p1 ( Z
|f (x)| dα
Z 1
cn
) p1
1
0
so that Theorem 4.2 ensures that
|f (x)|p dα
0
|g(x)|q dα
× lim
n→∞
1
q
(Z
) q1
1
cn
|g(x)|q dα
) 1q
<∞
|g(x)| dα
f (x)g(x) dα
0
exists. Hence Hölder’s inequality also holds for “improper” integrals described in Problem 6.7.
Chapter 6. The Riemann-Stieltjes Integral
128
– Case (ii): “Improper” integrals described in Problem 6.8. The argument for proving
this part is very similar to that in Case (i), so we omit the details here.
We complete the proof of the problem.
Problem 6.11
Rudin Chapter 6 Exercise 11.
Proof. Since f, g, h ∈ R(α) on [a, b], we follow from Theorem 6.12(a) that
f − h, f − g, g − h ∈ R(α)
on [a, b]. By Problem 6.10(c), when p = q = 2, we have
Z b
a
|f − g||g − h| dα ≤
(Z
) 12
b
a
|f − g|2 dα
×
(Z
) 21
b
a
|g − h|2 dα
.
(6.34)
Hence, it follows from the inequality (6.34) and Theorem 1.33(e) that
Z b
kf − hk22 =
|f − h|2 dα
a
=
≤
=
Z b
a
|(f − g) + (g − h)|2 dα
a
(|f − g| + |g − h|)2 dα
a
|f − g|2 dα + 2
Z b
Z b
Z b
a
|f − g||g − h| dα +
≤ kf − gk22 + 2kf − gk2 kg − hk2 + kg − hk22
Z b
a
|g − h|2 dα
= (kf − gk2 + kg − hk2 )2
which implies that
kf − hk2 ≤ kf − gk2 + kg − hk2 .
This completes the proof of the problem.
Problem 6.12
Rudin Chapter 6 Exercise 12.
Proof. We follow the given hint. Let P = {x0 , . . . , xn } be a partition of [a, b], define
g(t) =
xi − t
t − xi−1
f (xi−1 ) +
f (xi )
∆xi
∆xi
(6.35)
if xi−1 ≤ t ≤ xi .e As a function of t, it is clear that g is continuous on [xi−1 , xi ]. Since
#
"
t − xi−1
xi − t
f (xi−1 ) +
f (xi ) = f (xi )
g(xi −) =
lim g(t) =
lim
t→xi
t→xi
∆xi
∆xi
t∈[xi−1 ,xi )
t∈[xi−1 ,xi )
and
g(xi +) =
lim
t→xi
t∈(xi ,xi+1 ]
g(t) =
lim
t→xi
t∈(xi ,xi+1 ]
"
#
xi+1 − t
t − xi
f (xi ) +
f (xi+1 ) = f (xi ),
∆xi+1
∆xi+1
e In fact, the geometric meaning of the function g is that if we consider the graph of y = f (x), then (t, g(t)) is a point on
the line segment joining the points (xi−1 , f (xi−1 )) and (xi , f (xi )).
129
6.3. Hölder’s inequality
we have
g(xi +) = g(xi −) = f (xi )
for all i = 1, 2, . . . , n − 1. Hence g is continuous on [a, b].
To show the result kf − gk2 < ǫ, it is convenient to rewrite the function (6.35) as
g(t) = f (xi−1 ) +
f (xi ) − f (xi−1 )
(t − xi−1 ).
∆xi
If f (xi ) ≥ f (xi−1 ), then we have
max
t∈[xi−1 ,xi ]
g(t) = f (xi ) and
min
t∈[xi−1 ,xi ]
g(t) = f (xi−1 ).
(6.36)
g(t) = f (xi ).
(6.37)
If f (xi ) < f (xi−1 ), then we have
max
t∈[xi−1 ,xi ]
g(t) = f (xi−1 ) and
min
t∈[xi−1 ,xi ]
Therefore, the observations (6.36) and (6.37) imply that
min(f (xi−1 ), f (xi )) ≤ g(t) ≤ max(f (xi−1 ), f (xi ))
(6.38)
on [xi−1 , xi ]. Since f ∈ R(α) on [a, b], we have from Theorem 6.12(c) that f ∈ R(α) on [xi−1 , xi ] and
then it is a bounded real function on [xi−1 , xi ], i.e.,
mi ≤ f (x) ≤ Mi
where Mi =
sup
f (x) and mi =
x∈[xi−1 ,xi ]
inf
x∈[xi−1 ,xi ]
(6.39)
f (x). Thus we follow from the inequalities (6.38) and
(6.39) that
mi ≤ g(x) ≤ Mi
(6.40)
f
on [xi−1 , xi ]. Then we obtain from the inequalities (6.39) and (6.40) that
|f (x) − g(x)| ≤ Mi − mi
(6.41)
on [xi−1 , xi ].
Since g is continuous on [a, b], Theorem 6.8 implies that g ∈ R(α) on [a, b] and then Theorem 6.12(a)
shows that f − g ∈ R(α) on [a, b]. Hence we have from the inequalities (6.41) and the definition that
kf − gk22 =
Z b
a
|f − g|2 dα =
n Z xi
X
i=1
xi−1
|f − g|2 dα ≤
n
X
i=1
(Mi − mi )2 ∆αi .
(6.42)
Again, since f ∈ R(α) on [a, b], it is a bounded real function on [a, b]. Therefore, there exist constants
M and m such that m ≤ f (x) ≤ M for all x ∈ [a, b]. It is obvious that
M i − mi ≤ M − m
for all i = 1, 2, . . . , n, so the inequality (6.42) implies that
n
X
kf − gk22 ≤ (M − m) (Mi − mi )∆αi .
(6.43)
i=1
By Theorem 6.6, for every ǫ > 0 there exists a partition P = {x0 , x1 , . . . , xn } such that
U (P, f, α) − L(P, f, α) <
f Here we change the independent variable of g from t to x.
ǫ2
.
M −m
(6.44)
Chapter 6. The Riemann-Stieltjes Integral
130
By Definition 6.2, we have
U (P, f, α) − L(P, f, α) =
n
X
i=1
(Mi − mi )∆αi .
Hence it follows from the inequalities (6.43) and (6.44) that
kf − gk22 ≤ (M − m)[U (P, f, α) − L(P, f, α)] < ǫ2
which implies that
kf − gk2 < ǫ.
This completes the proof of the problem.
6.4
Problems related to improper integrals
Problem 6.13
Rudin Chapter 6 Exercise 13.
Proof.
(a) Put t2 = u. It follows from Theorem 6.21 (Integration by Parts) that
f (x) =
=
Z x+1
sin(t2 ) dt
x
Z (x+1)2
x2
=−
du
sin u √
2 u
Z (x+1)2
x2
d(cos u)
√
2 u
cos(x2 ) cos(x + 1)2
−
−
=
2x
2(x + 1)
Z (x+1)2
x2
cos u
3
4u 2
du.
(6.45)
Hence, for x > 0, we get from the expression (6.45) and Theorem 6.13(b) that
Z (x+1)2
cos u
cos(x2 ) cos(x + 1)2
−
−
|f (x)| =
3 du
2x
2(x + 1)
2
4u 2
x
Z (x+1)2
1
cos u
1
+
+
≤
3 du
2x 2(x + 1)
4u 2
x2
Z (x+1)2
1
cos u
1
≤
du
+
+
3
2x 2(x + 1)
2
4u 2
x
Z (x+1)2
1
1
1
+
+
≤
3 du
2x 2(x + 1)
4u 2
x2
1
1 1
1
1
+
−
−
=
2x 2(x + 1) 2 x + 1 x
1
= .
x
Assume that cos x2 and cos(x + 1)2 attained 1 at the same time. Then we have
x2 = kπ
and (x + 1)2 = lπ,
(6.46)
131
6.4. Problems related to improper integrals
where l > k > 0. This leads 2x + 1 = (l − k)π and then x =
(l − k)π − 1
. Therefore, we have
2
[(l − k)π − 1]2
4
2 2
4kπ = (l − k) π − 2(l − k)π + 1
x2 =
(l − k)2 π 2 − 2(l + k)π + 1 = 0.
(6.47)
However, the equation (6.47) means that π is an algebraic number which is a contradiction. Hence
the inequality (6.46) is strict, i.e., |f (x)| < x1 for x > 0.
(b) If we define
Z (x+1)2
cos(x + 1)2
− 2x
r(x) =
x+1
cos u
du,
(6.48)
2xf (x) = cos(x2 ) − cos[(x + 1)2 ] + r(x).
(6.49)
x2
then it follows from the expression (6.45) that
3
4u 2
Hence we have from the expression (6.48) that
1
+ 2x
|r(x)| ≤
x+1
Z (x+1)2
x2
du
4u
3
2
=
1
1
2
2
1
−x
−
<
=
x+1
x+1 x
x+1
x
as required.
(c) Let’s rewrite the expression (6.49) as
1
1
sin x +
+ r(x).
2xf (x) = 2 sin x2 + x +
2
2
(6.50)
By the result of part (b), we know that r(x) → 0 as x → ∞. Thus it follows from the expression
(6.50) that
−1 ≤ xf (x) ≤ 1
for all x > 0. We claim that
lim sup xf (x) = 1
and
x→∞
lim inf xf (x) = −1.
x→∞
To prove this claim, we note from the result of part (b) that 2xf (x) has the same magnitude as
cos(x2 ) − cos[(x + 1)2 ]
(6.51)
√
as x → ∞. Put xn = n 2π into the term (6.51) and then apply the periodicity of cos x to get
√
√
cos(x2n ) − cos[(xn + 1)2 ] = cos(2n2 π) − cos 2n2 π + n 8π + 1 = 1 − cos n 8π + 1 .
q
It is clear that the number α = π2 is irrational. By Lemma 4.6, we know that the set
S = {kα − h | k ∈ N, h ∈ Z}
1
, then for every ǫ > 0 there exists an
is dense in R. In other words, we consider the number 12 − 2π
g
integer N , sequences {hm } ⊂ Z and {km } ⊂ N such that
km α − hm −
1
2
−
g It is obvious that
lim km = ∞.
m→∞
ǫ
1 <
2π
2π
(6.52)
Chapter 6. The Riemann-Stieltjes Integral
132
for all m ≥ N . It is clear that the inequality (6.52) is equivalent to
√
km 8π + 1 − (2hm π + π) < ǫ
for all m ≥ N . Therefore, we deduce from this, the periodicity and the continuity of cos x that
√
lim cos(km 8π + 1) = lim cos(2hm π + π) = cos π = −1.
m→∞
m→∞
In other words, we have
lim sup xf (x) = 1.
x→∞
The case for
lim inf xf (x) = −1
x→∞
is similar, so we omit the details here.
(d) Let N be a positive integer. We consider
Z N
2
sin t dt =
0
N
−1 Z k+1
X
k=0
=
sin t2 dt
k
N
−1
X
f (k)
k=0
#
cos(k + 1)2
r(k)
cos k 2
−
+
= f (0) +
2k
2k
2k
k=1
#
"
N −1
N −1
cos(k + 1)2
1 X r(k) 1 X cos k 2
.
+
−
= f (0) +
2
k
2
k
k
N
−1
X
"
k=1
Since
(6.53)
k=1
cos(k + 1)2
cos(k + 1)2
cos(k + 1)2
−
=−
,
k+1
k
k(k + 1)
we follow from the expression (6.53) that
Z N
0
N −1
1 X r(k) 1
+
sin t dt = f (0) +
2
k
2
2
cos N 2
cos 1 −
N −1
k=1
Recall from part (b) that |r(k)| <
!
−
N −1
1 X cos k 2
.
2
k(k − 1)
(6.54)
k=2
c
for a constant c, so we have
k
N
−1
X
k=1
N −1
N −1
k=1
k=1
X r(k)
X c
r(k)
.
≤
<
k
k
k2
Thus Theorems 3.25 and 3.28 imply that
∞
X
r(k)
k=1
k
converges. Similarly, since | cos k 2 | ≤ 1, by applying Theorems 3.25 and 3.28 again, we can show
that
∞
X
cos k 2
k(k − 1)
k=2
converges. By Theorem 3.20(a), it is easy to see that
cos N 2
= 0.
N →∞ N − 1
lim
133
6.5. Applications and a generalization of integration by parts
Hence every term in the right-hand side of the expression (6.54) converges as N → ∞ and then
Z ∞
sin t2 dt = lim
N →∞
0
Z N
sin t2 dt
0
converges.
This ends the proof of the problem.
Problem 6.14
Rudin Chapter 6 Exercise 14.
Proof. Put et = u. It follows from Theorem 6.21 (Integration by Parts) that
f (x) =
=
Z x+1
x
Z ex+1
ex
=−
sin(et ) dt
sin u
du
u
Z ex+1
ex
d(cos u)
u
cos(ex+1 ) cos(ex )
+
−
=−
ex+1
ex
Z ex+1
ex
cos u
du.
u2
(6.55)
Hence, we have
|f (x)| ≤
1
e
+
x+1
1
+
ex
Z ex+1
ex
1
2
du
1
1
1
= x.
=
+
−
−
u2
ex+1
ex
ex+1
ex
e
(6.56)
Since cos u < 1 for some u ∈ [ex , ex+1 ], the inequality (6.56) is strict, i.e., ex |f (x)| < 2. If we define
r(x) = −ex
Z ex+1
ex
cos u
du,
u2
(6.57)
then the expression (6.55) becomes
ex f (x) = cos(ex ) − e−1 cos(ex+1 ) + r(x).
Therefore, it follows from the definition (6.57) that
|r(x)| = − ex
Z ex+1
ex
cos u
du ≤ ex
u2
Z ex+1
ex
1 e−1
du
x 1
=
=
e
−
< 2e−1 .
u2
ex
ex+1
e
This finishes the proof of the problem.
6.5
Applications and a generalization of integration by parts
Problem 6.15
Rudin Chapter 6 Exercise 15.
Proof. Since f is a continuously differentiable function on [a, b], Theorem 6.8 and Theorem 6.13(a) imply
that
f, f 2 , f ′ , xf f ′ ∈ R
Chapter 6. The Riemann-Stieltjes Integral
134
on [a, b]. Let F (x) = xf (x) and G(x) = f (x). Then we have
F ′ (x) = xf ′ (x) + f (x)
and G′ (x) = f ′ (x)
which are also Riemann-integrable on [a, b]. Therefore, Theorem 6.22 (Integration by Parts) implies that
Z b
Z b
F (x)G′ (x) dx = F (b)G(b) − F (a)G(a) −
F ′ (x)G(x) dx
a
Z b
a
xf (x)f ′ (x) dx = −
=−
Z b
Z b
a
[xf ′ (x) + f (x)]f (x) dx
a
Z b
′
xf (x)f (x) dx −
a
1
xf (x)f ′ (x) dx = − .
2
a
Z b
f 2 (x) dx
a
Apply Problem 6.10(c) to the functions f ′ (x) and xf (x) with p = q = 2, we have
(Z
) 12
b
′
2
[f (x)] dx
a
Z b
a
×
′
(Z
b
2
) 12
≥
[xf (x)] dx
a
2
[f (x)] dx ·
Z b
a
x2 f 2 (x) dx ≥
Z b
xf (x)f ′ (x) dx
a
1
.
4
(6.58)
Next we show that the inequality (6.58) is strict. By Problem 6.10(c), the equality holds if and only if
x2 f (x)2
[f ′ (x)]2
=
,
A
B
where
A=
Z b
[f ′ (x)]2 dx and B =
a
Z b
(6.59)
x2 f (x)2 dx.
a
From the inequality (6.58), we have AB = 14 so that the expression (6.59) becomes either
f ′ (x)
= 2Ax or
f (x)
which yield either
f (x) = eAx
2
f ′ (x)
= −2Ax
f (x)
or f (x) = e−Ax
2
respectively. However, both cases contradict the hypothesis that f (a) = f (b) = 0. Hence the inequality
(6.58) must be strict. This completes the proof of the problem.
Problem 6.16
Rudin Chapter 6 Exercise 16.
Proof.
(a) As suggested by the hint, we consider
f (N ) = s
Z N
1
N
X
1
[x]
dx
−
.
s
xs+1
n
n=1
Since [x] is the greatest integer less than or equal to x, we have [x] = n on [n, n + 1) so that
f (N ) = s
Z N
1
N
X
1
[x]
dx
−
s
xs+1
n
n=1
135
6.5. Applications and a generalization of integration by parts
=s
N
−1 Z n+1
X
n=1
n
N
−1 Z n+1
X
N
X
1
[x]
dx
−
s
xs+1
n
n=1
N
X
1
dx
−
s+1
s
x
n
n=1 n
n=1
"
#
N
−1
N
X
X
n
1
n
=−
−
−
s
s
(n + 1)
n
ns
n=1
n=1
=s
=1+
=1+
=−
n
N
−1
X
N
N
X
n
n−1 X 1
−
−
ns n=2 ns
ns
n=2
n=1
N
−1
X
N
N
N
X
X
X
n
n
1
1
−
+
−
s
s
s
s
n
n
n
n
n=2
n=2
n=2
n=1
1
N s−1
.
(6.60)
Since s > 1, we get from Theorem 3.20(a) that
lim
1
N →∞ N s−1
= 0.
Hence it yields from the expression (6.60) that lim f (N ) = 0 and then
N →∞
ζ(s) = s
Z ∞
1
[x]
dx
xs+1
as required.
(b) By Problem 6.8 and Theorem 6.21 (Second Fundamental Theorem of Calculus), we have
Z ∞
s
dx
=
.
s
s
x
s
−
1
1
Therefore, the result of part (a) implies that we can split the integral of the right-hand side into
two integrals. Hence we have
Z ∞
Z N
s
s
x − [x]
x − [x]
dx =
dx
−s
− s lim
s+1
N
→∞
s−1
x
s
−
1
xs+1
1
1
!
Z N
Z N
[x]
s
dx
−
dx
− s lim
=
N →∞
s−1
xs
xs+1
1
1
Z N
Z N
s
[x]
dx
+
s
lim
dx
− s lim
=
N →∞ 1
N →∞ 1
s−1
xs
xs+1
Z ∞
[x]
dx
=s
xs+1
1
= ζ(s).
For the second assertion, we note that 0 ≤ x − [x] < 1 for all x > 1, so we have
Z ∞
Z N
Z N
1
1 1
x − [x]
x − [x]
1
0≤
dx
=
lim
dx
<
lim
dx = lim −
−1 =
s+1
s+1
s+1
N →∞ 1
N →∞ 1
N →∞
x
x
x
s N
s
1
for all s > 0. Hence the integral in part (b) converges for all s > 0. This ends the proof of the problem. Problem 6.17
Rudin Chapter 6 Exercise 17.
Chapter 6. The Riemann-Stieltjes Integral
136
Proof. We follow the hint. Take g to be real, without loss of generality. Given P = {x0 , x1 , . . . , xn }
which will be determined later. Since g is a real continuous function on [xi−1 , xi ], Theorem 5.10 (Mean
Value Theorem) implies that
G(xi ) − G(xi−1 ) = (xi − xi−1 )G′ (ti ) = g(ti )∆xi
for some ti ∈ (xi−1 , xi ). Therefore, we have
n
X
α(xi )g(ti )∆xi =
i=1
=
n
X
i=1
n
X
α(xi )[G(xi ) − G(xi−1 )]
G(xi )α(xi ) −
i=1
n
X
−
=
G(xi−1 )α(xi ) +
n
X
G(xi−1 )α(xi−1 )
i=1
i=1
G(xi−1 )α(xi−1 )
i=1
n
X
i=1
n
X
G(xi )α(xi ) −
n
X
i=1
G(xi−1 )α(xi−1 ) −
= [G(xn )α(xn ) − G(x0 )α(x0 )] +
= G(b)α(b) − G(a)α(a) −
n
X
n−1
X
i=1
n
X
G(xi−1 )∆αi
i=1
G(xi )α(xi ) −
n−1
X
i=1
G(xi−1 )∆αi .
G(xi )α(xi ) −
n
X
G(xi−1 )∆αi
i=1
(6.61)
i=1
Since g(x) = G′ (x) on [a, b], we have G is continuous on [a, b] and then Theorem 6.8 implies that
G ∈ R(α) on [a, b]. By Theorem 6.6, for every ǫ > 0, there exists a partition P1 such that
U (P1 , G, α) − L(P1 , G, α) < ǫ
and so we obtain from Theorem 6.7(c) that
n
X
i=1
G(ti )∆αi −
Z b
G dα < ǫ,
(6.62)
a
where ti ∈ [xi−1 , xi ]. Since ti is arbitrary, we may take ti = xi−1 in the inequality (6.62). Hence the
summation on the right-hand side of the expression (6.61) can be made arbitrarily close to
Z b
G dα.
a
Now it remains to show that
n
X
α(xi )g(ti )∆xi
i=1
can also be made arbitrarily close to
Z b
α(x)g(x) dx.
a
To this end, since α increases monotonically on [a, b], Theorem 6.9 implies that α ∈ R on [a, b] and then
it follows from Theorem 6.13(a) that αg ∈ R on [a, b]. By arguing as above, we know that for every
ǫ > 0, there exists a partition P2 such that
U (P2 , αg) − L(P2 , αg) < ǫ
and so we obtain from Theorem 6.7(c) that
n
X
i=1
α(ti )g(ti )∆xi −
Z b
a
α(x)g(x) dx <
ǫ
,
2
(6.63)
137
6.6. Problems on rectifiable curves
where ti ∈ [xi−1 , xi ]. Since ti is arbitrary, we may take ti = xi in the inequality (6.63) so that
n
X
i=1
α(xi )g(xi )∆xi −
Z b
α(x)g(x) dx <
a
ǫ
.
2
(6.64)
It is clear that
n
X
i=1
α(xi )g(xi )∆xi −
n
X
α(xi )g(ti )∆xi =
n
X
α(xi )[g(xi ) − g(ti )]∆xi
i=1
i=1
≤ | max(α(a), α(b))| ×
n
X
i=1
[g(xi ) − g(ti )]∆xi ,
(6.65)
where ti ∈ [xi−1 , xi ]. Since g is continuous on [xi−1 , xi ], for every ǫ > 0, there exists a δ > 0 such that
|g(xi ) − g(t)| <
ǫ
2(b − a)| max(α(a), α(b))|
for every t ∈ [xi−1 , xi ] such that |xi − t| < δ. Make ti be such t, so the inequality (6.65) reduces to
n
X
i=1
α(xi )g(xi )∆xi −
n
X
i=1
α(xi )g(ti )∆xi ≤ | max(α(a), α(b))|
n
X
ǫ
∆xi
2(b − a) i=1
ǫ
= .
2
n
X
i=1
|g(xi ) − g(ti )|∆xi
<
(6.66)
Therefore, it follows from the inequalities (6.64) and (6.66) that
n
X
i=1
α(xi )g(ti )∆xi −
Z b
a
α(x)g(x) dx ≤
n
X
α(xi )g(xi )∆xi −
i=1
n
X
+
i=1
n
X
α(xi )g(xi )∆xi −
ǫ
ǫ
+
2 2
= ǫ.
α(xi )g(ti )∆xi
i=1
Z b
α(x)g(x) dx
a
<
(6.67)
In other words, the summation on the left-hand side of the expression (6.61) can be made arbitrarily
close to
Z
b
α(x)g(x) dx.
a
Hence, if we take P to be the common refinement of P1 and P2 (i.e., P = P1 ∪P2 ), then the inequalities
(6.62) and (6.67) will hold simultaneously and they imply the desired formula that
Z b
Z b
α(x)g(x) dx = G(b)α(b) − G(a)α(a) −
G dα.
a
This completes the proof of the problem.
6.6
a
h
Problems on rectifiable curves
Problem 6.18
Rudin Chapter 6 Exercise 18.
h We remark that we can’t apply Theorem 6.17 directly because α may not be differentiable.
Chapter 6. The Riemann-Stieltjes Integral
138
Proof. Suppose that
1 1
− ,
t
2
where t ∈ [0, 2π]. It is clear that they are continuous functions on [0, 2π]. By Theorem 8.7(a), the
function eit is periodic with the period 2π. Since the lengths of F ([0, 2π]) and G([0, 2π]) are 2π and 4π
respectively, we have
γ1 ([0, 2π]) = γ2 ([0, 2π]).
F (t) = t,
G(t) = 2t
and H(t) = t sin
For the function H(t), since we have
√
3 3
1
π
1
3 3 1
1
= sin − =
H(0) = − < 0 and H
− > ,
2
π
π
3
2
2π
2
2
it follows from Theorem 4.23 (Intermediate Value Theorem) that H([0, π3 ]) is an interval of length at
least 1. Since H([0, π3 ]) ⊆ H([0, 2π]), 2πH([0, 2π]) is an interval of length at least 2π. This implies that
γ1 ([0, 2π]) = γ2 ([0, 2π]) = γ3 ([0, 2π]),
i.e., the three curves have the same range.
By Theorems 5.5 and 8.6(b), the functions γ1 (t), γ2 (t) and γ3 (t) are differentiable on [0, 2π] and we
have
1
1
1 1
(6.68)
γ1′ (t) = ieit , γ2′ (t) = 2ie2it and γ3′ (t) = 2πi sin − cos e2πit sin t .
t
t
t
By the first two equations in (6.68), we know that γ1′ (t) and γ2′ (t) are continuous on [0, 2π]. By Theorem
6.27, we conclude that γ1 (t) and γ2 (t) are rectifiable and then
Z 2π
Z 2π
Z 2π
Z 2π
′
′
2 dt = 4π.
|γ2 (t)| dt =
dt = 2π and Λ(γ2 ) =
|γ1 (t)| dt =
Λ(γ1 ) =
0
0
0
0
By the third equation in (6.68), we have
Λ(γ3 ) =
Z 2π
0
|γ3′ (t)| dt = 2π
Z 2π
0
sin
1 1
1
− cos dt.
t
t
t
(6.69)
By the triangle inequality that |a| − |b| ≤ |a − b| and the fact that | sin 1t | ≤ 1, we have
sin
1 1
1
1
1
1
1
1
− cos
cos − sin
cos − 1
≥
≥
t
t
t
t
t
t
t
t
so that the integral (6.69) reduces to
Λ(γ3 ) ≥ 2π
Z 2π
0
!
Z 2π
1
1
1
1
cos − 1 dt = 2π
cos dt − 4π 2 .
t
t
t
t
0
Now we know from Problem 6.7 that we must evaluate the limit
Z 2π
1
1
cos dt.
lim
c→0 c
t
t
(6.70)
(6.71)
By applying the substitution x = 1t to the integral in (6.71), we have
Z 2π
c
1
1
cos dt = −
t
t
1
Z 2π
1
c
cos x
dx =
x
Z 1c
1
2π
cos x
dx ≥
x
Z 1c
2π
cos x
dx.
x
Since c is assumed to be very small, let N be the largest positive integer such that 2N π ≤ 1c . Then we
have
Z 2N π
Z 1c
cos x
cos x
dx ≥
dx
x
x
2π
2π
139
6.6. Problems on rectifiable curves
=
Z 2kπ+ π2
N
−1
X
2kπ
k=1
N
−1 Z 2kπ+ π
X
2
≥
≥
=
k=1
2kπ
N
−1
X
1
2kπ + π2
k=1
N
−1
X
k=1
cos x
dx −
x
cos x
dx
x
Z 2kπ+ π2
Z 2kπ+ 3π
2
2kπ+ π
2
cos x
dx +
x
Z 2(k+1)π
2kπ+ 3π
2
!
cos x
dx
x
cos x dx
2kπ
1
.
2kπ + π2
If c → 0, then N → ∞ and we conclude from Theorem 3.28 that the series
N
−1
X
k=1
1
2kπ + π2
and then the integral (6.71) diverges. Hence γ3 is not rectifiable. This completes the proof of the
problem.
Problem 6.19
Rudin Chapter 6 Exercise 19.
Proof. We have γ1 : [a, b] → Rk . Since γ2 = γ1 ◦ φ, we get from Theorem 4.7 that γ2 is continuous and
thus a curve. By Theorem 4.17, the inverse mapping φ−1 : [a, b] → [c, d] is also bijective and continuous.
Here we need a lemma for proving one of our results:
Lemma 6.1
Suppose that f : [a, b] → [c, d] is a continuous and one-to-one function. Then f is monotonic.
Proof of Lemma 6.1. Assume that f was not monotonic. Then we have x, y, z ∈ [a, b] with
x < y < z such that either
f (x) ≤ f (y) and f (y) ≥ f (z)
or
f (x) ≥ f (y) and f (y) ≤ f (z).
Suppose that f (x) ≤ f (y) and f (y) ≥ f (z). In fact, it is impossible that f (x) = f (y) or
f (y) = f (z) because f is one-to-one. Thus we have f (x) < f (y) and f (y) > f (z). It is clear
that we can find a number c such that
f (x) < c < f (y) and f (y) > c > f (z).
For such number c, Theorem 4.23 (Intermediate Value Theorem) implies that there exist p ∈
(x, y) and q ∈ (y, z) such that f (p) = f (q) = c. However, this contradicts the hypothesis that
f is one-to-one. The case for f (y) ≥ f (z) or f (x) ≥ f (y) and f (y) ≤ f (z) can be proven
similarly, so we omit the details here. Hence we have the desired result that f is monotonic.
This completes the proof of the lemma.
Since the mapping φ : [c, d] → [a, b] is bijective, continuous and φ(c) = a, we conclude that φ(d) = b.
Now we prove the remaining results one by one.
Chapter 6. The Riemann-Stieltjes Integral
140
• γ1 is an arc if and only if γ2 is an arc. Suppose that γ1 is one-to-one. If γ2 (s) = γ2 (t), then
γ1 (φ(s)) = γ1 (φ(t)).
Since γ1 is one-to-one, we have φ(s) = φ(t). Since φ is one-to-one, we have s = t. Therefore, γ2 is
one-to-one.
Suppose that γ2 is one-to-one. Let γ1 (u) = γ1 (v), where u, v ∈ [a, b]. By the above observation,
there exist s, t ∈ [c, d] such that
φ(s) = u and φ(t) = v.
Since γ1 (u) = γ1 (v), we have γ1 (φ(s)) = γ1 (φ(t)) and then γ2 (s) = γ2 (t). Since γ2 is one-to-one,
we have s = t and then u = v. Therefore, γ1 is one-to-one.
• γ1 is a closed curve if and only if γ2 is a closed curve. We note that
γ1 (t) = γ2 (φ−1 (t))
for all t ∈ [a, b]. By Lemma 6.1, φ is monotonic. Since φ(c) = a, it is monotonically increasing and
also φ(d) = b. Thus we have γ1 (a) = γ1 (b) if and only if γ2 (φ−1 (a)) = γ2 (φ−1 (b)) if and only if
γ2 (c) = γ2 (d). Hence γ1 is a closed curve if and only if γ2 is a closed curve.
• γ1 is a rectifiable curve if and only if γ2 is a rectifiable curve. The result can be easily seen
from the formula (6.72) below.i
Before proving that they have the same length, we have an observation first. If P = {x0 , x1 , . . . , xn }
is a partition of [a, b], then
P (φ−1 ) = {φ−1 (x0 ), φ−1 (x1 ), . . . , φ−1 (xn )}
is a partition of [c, d]. Similarly, if P ∗ = {x0 , x1 , . . . , xn } is a partition of [c, d], then
P ∗ (φ) = {φ(x0 ), φ(x1 ), . . . , φ(xn )}
is a partition of [a, b]. In other words, we see that there is a bijective mapping between partitions of [a, b]
and partitions of [c, d].
Now let P ∗ = {x0 , x1 , . . . , xn } be a partition of [c, d]. By Definition 6.26 and the definition of γ2 , we
have
Λ(P, γ2 ) =
n
X
i=1
γ2 (xi ) − γ2 (xi−1 ) =
n
X
i=1
γ1 (φ(xi )) − γ1 (φ(xi−1 )) = Λ(P ∗ (φ), γ1 ).
(6.72)
Since there is a bijective mapping between partitions of [a, b] and partitions of [c, d], we conclude from
(6.72) that
Λ(γ2 ) = sup Λ(P, γ2 ) = sup Λ(P ∗ (φ), γ1 ) = Λ(γ1 ),
where the leftmost and rightmost supremums are taken over all partitions of [a, b] and [c, d] respectively.
Hence they have the same length and this completes the proof of the problem.
i Note that we can’t apply Theorem 6.27 directly because we are not sure whether γ is differentiable or not.
1
CHAPTER
7
Sequences and Series of Functions
7.1
Problems on uniform convergence of sequences of functions
Problem 7.1
Rudin Chapter 7 Exercise 1.
Proof. Suppose that {fn } is a uniformly convergent sequence of bounded functions. Let ǫ = 1. Since
{fn } is uniformly convergent, there exists an integer N1 such that m ≥ n ≥ N1 implies
|fm − fn | ≤ 1.
Let the limit function be f . Then there exists an integer N2 such that n ≥ N2 implies
|fn − f | ≤ 1.
Let N = max(N1 , N2 ). Then, for m ≥ n ≥ N , the triangle inequality shows that
|fn | ≤ |fn − fm | + |fm − f | + |f − fN +1 | + |fN +1 | = 3 + |fN +1 |.
(7.1)
Since fN +1 is bounded, we have |fN +1 (x)| ≤ M for all x ∈ R. Therefore, we follow from the inequality
(7.1) that
|fn | ≤ 3 + M
for all n ≥ N and x ∈ R. Since f1 , f2 , . . . , fN are bounded functions, we have |fk (x)| ≤ Mk for all x ∈ R
and k = 1, 2, . . . , N . Hence we have
|fn (x)| ≤ max(M1 , M2 , . . . , MN , M )
for all x ∈ R and all n ≥ 1, completing the proof of the problem.
Problem 7.2
Rudin Chapter 7 Exercise 2.
Proof. For every ǫ > 0, there exist integers N1 and N2 such that
if m ≥ n ≥ N1 and
|fm − fn | <
ǫ
2
|gs − gt | <
ǫ
2
141
Chapter 7. Sequences and Series of Functions
142
if s ≥ t ≥ N2 . Let N = max(N1 , N2 ). Then, for u, v ≥ N , we have
|fu + gu − fv − gv | ≤ |fu − fv | + |gu − gv | <
ǫ
ǫ
+ = ǫ.
2 2
Hence {fn + gn } converges uniformly on E.
Since {fn } and {gn } are uniformly convergent sequences of bounded functions, we conclude from
Problem 7.1 that {fn } and {gn } are uniformly bounded. Let A and B be non-negative numbers such
that
|fn (x)| ≤ A and |gn (x)| ≤ B
for all n = 1, 2, . . . and for all x ∈ E. Given that ǫ > 0. Then there exist integers N3 and N4 such that
m ≥ n ≥ N3 implies
ǫ
|fm − fn | <
2B
and s ≥ t ≥ N4 implies
ǫ
|gs − gt | <
.
2A
Let N ∗ = max(N3 , N4 ). Then, for u, v ≥ N ∗ , we have
|fu gu − fv gv | ≤ |fu gu − fu gv | + |fu gv − fv gv |
≤ |fu | × |gu − gv | + |gv | × |fu − fv |
ǫ
ǫ
<A·
+B·
2A
2B
= ǫ.
Hence {fn gn } converges uniformly on E. This completes the proof of the problem.
Problem 7.3
Rudin Chapter 7 Exercise 3.
Proof. Suppose that fn (x) = x and gn (x) = n1 are functions defined on E = N, where n = 1, 2, . . .. Let
f (x) = x and g(x) = 0. For every ǫ > 0, we have
|fn (x) − f (x)| = |x − x| = 0 ≤ ǫ
for all n ≥ 1 and x ∈ N. By Definition 7.7, {fn } converges uniformly to f on N. Similarly, for every
ǫ > 0, let N be a positive integer such that N ≥ 1ǫ . Then we have
|gn (x) − g(x)| =
1
1
−0 ≤
≤ǫ
n
N
for all n ≥ N and x ∈ N. Thus {gn } also converges uniformly to g on N.
Now we have
x
fn (x)gn (x) =
n
for all n = 1, 2, . . .. For each x ∈ N, it is obvious that
fn (x)gn (x) → 0
as n → ∞. Therefore, the limit function of {fn gn } must be F (x) = 0 for all x ∈ N. Assume that the
convergence was uniform. Then for 0 < ǫ < 1, there exists an integer N such that
x
|fn (x)gn (x) − 0| = ≤ ǫ
(7.2)
n
for all n ≥ N and x ∈ N. In particular, if we take n = x = N in the inequality (7.2), then we have
|fN (N )gN (N ) − 0| =
N
−0 =1≤ǫ<1
N
which is a contradiction. Hence the sequence {fn gn } does not converge uniformly on N. This completes
the proof of the problem.
143
7.1. Problems on uniform convergence of sequences of functions
Problem 7.4
Rudin Chapter 7 Exercise 4.
Proof. Suppose that
1
gn (x) =
1 + n2 x
and fn (x) =
n
X
gk (x).
k=1
We answer the questions one by one:
• For what values of x does the series converge absolutely?
– Case (i): x > 0. Then we always have |1 + n2 x| = 1 + n2 x ≥ n2 x which leads to
|gn (x)| =
Since
1
1
≤ 2 .
|1 + n2 x|
n x
(7.3)
∞
1X 1
converges by Theorem 3.28,
x n=1 n2
∞
X
1
|1
+
n2 x|
n=1
converges by Theorem 3.25(a).
– Case (ii): x = 0. Then we have
fn (0) =
n
X
k=1
1
= n.
1 + k2 · 0
(7.4)
Thus the series diverges in this case.
– Case (iii): x < 0. Since gn (x) is undefined at x = − n12 , the function f (x) is undefined on
the set
o
n
1
1
(7.5)
S = − 1, − 2 , − 2 , . . . .
2
3
So let x ∈
/ S. By this, we get x ∈ (−∞, −a] \ S for some positive a. Thus we have
1
1
≥
= |gn (x)|.
an2 − 1
1 + n2 x
By Theorems 3.25(a) and 3.28 again, we have
∞
X
1
and then
2−1
an
n=1
∞
X
1
1 + n2 x
n=1
converges.
In conclusion, the series converges absolutely only if x > 0 or x < 0 and x ∈
/ S and diverges in other
cases.
• On what intervals does it converge uniformly? Let I = [a, b] be an interval by Definition
2.17, where a < b. Thus it is obvious that the uniform convergence of {fn } on [a, b] depends solely
on the values of a and b.
Chapter 7. Sequences and Series of Functions
144
– Case (i): a = α > 0. Then we have I = [α, b] and it follows from the inequality (7.3) that
|gn (x)| =
1
1
≤
|1 + n2 x|
αn2
∞
X
1
converges by Theorem 3.28, we conclude from Theorem 7.10 (Weier2
αn
n=1
strass M -Test) that the series converges uniformly on I = [α, b].
holds on I. Since
– Case (ii): a = 0. Then we have
|gn (x)| =
1
≤1
1 + n2 x
for all x ∈ I. Assume that the series converged uniformly on I. Then Problem 7.1 implies that
the sequence {fn } is uniformly bounded on I. However, since I is an interval, there exists a
positive integer N such that N12 ∈ I. Thus we have n12 ∈ I for all n ≥ N and then
fn
1
n2
=
n
X
1
n
1
≥
= →∞
k2
2
2
1
+
n2
k=1
k=1
n
X
as n → ∞. This contradicts Definition 7.19.
– Case (iii): b = 0. Then we see from the expression (7.4) that the series diverges on [a, 0].
– Case (iv): b = β < 0. Then there are two subcases:
∗ I does not contain any element of the set (7.5). Then we have β < −1. Since
x < β < 0, we have 1 + n2 x ≤ 1 + βn2 < 0 for n ≥ 1. Therefore, we have
|gn (x)| =
1
1
≤
1 + n2 x
1 + βn2
on [a, β] for n = 1, 3, . . .. Let m and n be positive integers such that m ≥ n ≥ 2. Then we
have
|fm (x) − fn (x)| =
m
X
k=n+1
m
X
1
≤
1 + k2 x
k=n+1
m
X
1
≤
1 + k2 x
k=n+1
1
1 + βk 2
(7.6)
∞
X
1
converges and Theorem 3.22
1 + βk 2
k=1
implies that for every ǫ > 0, there exists an integer N such that
for all x ∈ [a, β]. By Theorem 3.28, the series
m
X
k=n+1
1
≤ǫ
1 + βk 2
(7.7)
for m ≥ n ≥ N . Hence we establish from the inequalities (7.6) and (7.7) that
|fm (x) − fn (x)| ≤ ǫ
for all m ≥ n ≥ N and x ∈ [a, β]. By Theorem 7.8, the sequence of functions {fn }
converges uniformly on [a, β].
∗ I contains an element of the set (7.5). Then we have I = [a, β], where −1 ≤ β < 0.
This means that − n12 ∈ I for some n ≥ 1 and so the function fn (x) is undefined at x = − n12
for such n ≥ 1. Hence the sequence of functions {fn } does not converge uniformly on [a, β].
In conclusion, {fn } converges uniformly to f only on [α, b] or [a, β], where α > 0 and β < −1.
• On what intervals does it fail to converge uniformly? By the above analysis, we see that it
fails to converge uniformly on the intervals [0, b] and [a, β], where −1 ≤ β ≤ 0.
145
7.1. Problems on uniform convergence of sequences of functions
• Is f continuous wherever the series converges? Using the same notations as above, we know
that {fn } converges uniformly to f on the intervals [α, b] or [a, β], where α > 0 and β < −1. Since
each fn is continuous on [α, b] or [a, β], then Theorem 7.12 implies that the function f is continuous
on [α, b] or [a, β].
• Is f bounded? We can see from the above argument that f is not bounded on any set containing
an element of the set S.
This completes the proof of the problem.
Problem 7.5
Rudin Chapter 7 Exercise 5.
Proof. We claim that for each x ∈ R, fn (x) → 0 as n → ∞. In other words, {fn } converges to 0 pointwise
on R. In fact, if x ≤ 0, then we have fn (x) = 0 for all positive integers n and thus we have
|fn (x) − 0| = |0 − 0| ≤ ǫ
for all n ≥ 1. If x > 0, then we let N be a positive integer such that N > x1 . For all n ≥ N , since x > n1 ,
we have
|fn (x) − 0| = |0 − 0| ≤ ǫ.
This proves our claim.
However, it does not converge to 0 uniformly on R. In fact, for every positive integer n, we take
1
, n1 ] and so
x = n+1 1 ∈ [ n+1
2
fn (x) = fn
1
1 2
n
+
=
sin
π = 1.
2
n + 12
Now this implies that
|fn (x) − 0| = |1 − 0| = 1.
Hence {fn } does not converge to 0 uniformly on R by Definition 7.7.
If x ≤ 0, then it is clear from the definition that fn (x) = 0 for all n ≥ 1. We note that

1

0,
x
<
;


1 2
π
f1 (x) =
2
sin
,
≤
x
≤
1
;


x
2

0,
(1 < x).
Thus if x ≥ 1, then we also have fn (x) = 0 for all n ≥ 1. Therefore, if x ≤ 0 or x ≥ 1, we have
X
fn = 0.
If 0 < x < 1, then Theorem 1.20(a) (Archimedean property) guarantees that there is a positive integer
N ≥ 2 such that x ≥ N1 . In fact, we may assume without loss of generality that this N is the smallest
integer with this property. This implies that we must have x ≤ N 1−1 . Thus we have
1
1
≤x≤
N
N −1
and then
∞
X
n=1
Now we have
∞
X
n=1
fn (x) = fN −1 (x) = sin2
fn (x) =
(
π
.
x
0,
if x ≤ 0 or x ≥ 1;
2 π
sin , if 0 < x < 1.
x
Chapter 7. Sequences and Series of Functions
146
P
Since sin2 πx is clearly non-negative on (0, 1), the series must converge absolutely. Assume that
fn
converged uniformly to some function f . In other words, for every ǫ > 0, there is an integer N such that
n
X
k=1
ǫ
2
fk (x) − f (x) ≤
for all n ≥ N and x ∈ R. By the fact that
fn =
n
X
k=1
n−1
X
fk −
fk ,
k=1
we have
|fn (x) − 0| =
≤
n
X
k=1
n
X
k=1
fk (x) −
n−1
X
k=1
fk (x) − f (x) + f (x)
fk (x) − f (x) +
ǫ
ǫ
+
2 2
=ǫ
n−1
X
k=1
fk (x) − f (x)
≤
for all n ≥ N + 1 and x ∈ R. Hence this means that {fn } converges to 0 uniformly on R which is a
contradiction. In conclusion, we have the result that absolute convergence on R does not imply uniformly
convergence. This ends the proof of the problem.
Problem 7.6
Rudin Chapter 7 Exercise 6.
Proof. For every bounded interval [a, b], where −∞ < a < b < ∞, we have |x2 | ≤ M = max(a2 , b2 ).
Suppose that
m
X
x2 + k
,
fm (x) =
(−1)k
k2
k=1
where m = 1, 2, . . .. Then we have
fm (x) = x2
m
X
(−1)k
k=1
Given ǫ > 0. By Theorem 3.43, we know that
k2
m
X
(−1)k
k=1
∞
X
(−1)k
k=1
there is an integer N1 such that
+
k
k
.
converges. Then Theorem 3.22 implies that
m
X
(−1)k
ǫ
≤
k
2
(7.8)
k=n+1
if m ≥ n ≥ N1 . By Theorem 3.28, we have
that
∞
X
(−1)k
k=1
k2
converges. Similarly, there is an integer N2 such
m
X
ǫ
(−1)k
≤
k2
2M
(7.9)
k=n+1
if m ≥ n ≥ N2 . Let N = max(N1 , N2 ). Then it yields from the inequalities (7.8) and (7.9) that
|fm (x) − fn (x)| = x2
m
X
(−1)k
k=1
k2
+
m
X
(−1)k
k=1
k
− x2
n
X
(−1)k
k=1
k2
−
n
X
(−1)k
k=1
k
147
7.1. Problems on uniform convergence of sequences of functions
≤M
m
X
(−1)k
+
k2
k=n+1
m
X
(−1)k
k
k=n+1
ǫ
ǫ
+
≤M·
2M
2
=ǫ
for all m ≥ n ≥ N and x ∈ [a, b]. By Theorem 7.8, the series converges uniformly on every bounded
interval.
∞
X
2
1
≥ n1 > 0 for all positive integers n and the series
However, since x n+n
diverges by Theorem
2
n
n=1
3.28, the series
∞
X
x2 + n
(−1)n
n2
n=1
does not converge absolutely for any x by Theorem 3.25(b). We complete the proof of the problem.
Problem 7.7
Rudin Chapter 7 Exercise 7.
Proof. For each x ∈ R, we have
lim fn (x) = lim
x
n→∞ 1 + nx2
n→∞
= 0.
Thus the limit function of the sequence of functions {fn } must be f (x) = 0. Given that ǫ > 0. Then
there is a positive integer N such that N ≥ 4ǫ12 and so we have
√
1
n|x|
x
=√ ·
|fn (x) − 0| =
2
1 + nx
n 1 + nx2
(7.10)
for all n ≥ N and x ∈ R. By applying A.M. ≥ G.M. to 1 and nx2 , we have
√
1 + nx2
≥ n|x|.
2
Therefore we follow from this and the expression (7.10) that
1
1
|fn (x) − 0| ≤ √ ≤ √ ≤ ǫ
2 n
2 N
for all n ≥ N and x ∈ R. By Definition 7.7, we have the desired result that {fn } converges uniformly to
a function f = 0 on R.
Since f (x) = 0 on R, we have f ′ (x) = 0 for every x ∈ R. Now fn is clearly differentiable and we have
fn′ (x) =
which gives
lim fn′ (x) =
n→∞
1 − nx2
(1 + nx2 )2
1 − nx2
=
(1 + nx2 )2
Hence we have
0, if x 6= 0;
1, if x = 0.
f ′ (x) = lim fn′ (x)
n→∞
is correct if x 6= 0, but false if x = 0, finishing the proof of the problem.
Problem 7.8
Rudin Chapter 7 Exercise 8.
Chapter 7. Sequences and Series of Functions
148
Proof. Suppose that
gn (x) = cn I(x − xn ) and fn (x) =
where x ∈ [a, b]. By the definition of I(x), we know that
n
X
gk (x),
k=1
|gn (x)| = |cn I(x − xn )| ≤ |cn |.
P
P
Since
|cn | converges, we conclude from Theorem 7.10 that fn =
gn converges uniformly to f on
[a, b].
Let x 6= xn . If each fn is continuous at x, then Theorem 7.12 implies that f is also continuous at x.
We check Definition 4.5 for
n
X
fn (x) =
ck I(x − xk ).
k=1
Since x 6= xk for k = 1, 2, . . . , n, we let δ = min(|x − x1 |, |x − x2 |, . . . , |x − xn |) > 0. Let x − xk > 0 for
some k so that x − xk ≥ δ > 0. If x − δ < t < x + δ, then we have
t − xk > x − δ − xk = x − xk − δ ≥ δ − δ = 0.
This implies that
I(x − xk ) = I(t − xk ) = 1.
Similarly, let x − xk < 0 for some k so that x − xk ≤ −δ < 0. If x − δ < t < x + δ, then we have
t − xk < x + δ − xk = x − xk + δ ≤ δ − δ = 0.
In this case, we have
I(x − xk ) = I(t − xk ) = 0.
By these, if |x − t| < δ, then we have
|fn (x) − fn (t)| =
n
X
k=1
ck [I(x − xk ) − I(t − xk )] = 0 < ǫ.
By Definition 4.5, each fn is continuous at every x 6= xn . Hence our desired result follows from Theorem
7.12. This completes the proof of the problem.
Problem 7.9
Rudin Chapter 7 Exercise 9.
Proof. By Theorem 7.12, we have f is continuous on E. Let x ∈ E and {xn } be a sequence in E such
that xn 6= x and lim xn = x. Given ǫ > 0. Then there is an integer N1 such that for all n ≥ N1 , we
n→∞
have
ǫ
|f (xn ) − f (x)| < .
(7.11)
2
Since {fn } converges uniformly to f on E, there is an integer N2 such that
|fn (x) − f (x)| ≤
ǫ
2
(7.12)
for all n ≥ N2 and x ∈ E. In particular, we put x = xn into the inequality (7.12) so that
|fn (xn ) − f (xn )| ≤
ǫ
2
(7.13)
for all n ≥ N2 . Let N = max(N1 , N2 ). Then for n ≥ N , we follow from the inequalities (7.11) and (7.13)
that
ǫ
ǫ
|fn (xn ) − f (x)| ≤ |fn (xn ) − f (xn )| + |f (xn ) − f (x)| < + = ǫ.
2 2
149
7.1. Problems on uniform convergence of sequences of functions
Hence we have the required result that
lim fn (xn ) = f (x)
n→∞
for every sequence of points xn ∈ E such that xn → x, and x ∈ E.
1
The converse is false. For example, let fn (x) = nx
and E = (0, 1). It is clear that each fn is a
continuous function on E. Fix x ∈ E. Then we have
lim fn (x) = 0.
n→∞
In other words, {fn } converges pointwise to f = 0 on E.
Given that ǫ > 0. Let {xn } be a sequence of E such that xn → x. Since 0 < x < 1 and x is fixed,
there is an integer N1 such that |x − xn | < x2 for all n ≥ N1 and this inequality is equivalent to
0<
x
3x
< xn <
2
2
for all n ≥ N1 . Besides, we also have 0 < xn < 1 for all positive integers n. If we take N to be an integer
2
), then for all n ≥ N , we have
such that N > max(N1 , ǫx
|fn (xn ) − 0| =
1
2
2
−0<
≤
< ǫ.
nxn
nx
Nx
This means that
lim fn (xn ) = 0
n→∞
holds for every sequence of points xn ∈ E such that xn → x and x ∈ E. However, {fn } does not converge
uniformly to f = 0 on E. In fact, if m > n ≥ 1, then we have
|fn (x) − fm (x)| =
1
1
(m − n)
1
−
≥
≥1>ǫ
=
nx mx
mnx
mnx
as x → 0. Hence the desired result follows from Theorem 7.8 and this completes the proof of the
problem.
Problem 7.10
Rudin Chapter 7 Exercise 10.
Proof. Suppose that
gn (x) =
(nx)
n2
and fn (x) =
n
X
gk (x),
k=1
where n is a positive integer. By the definition of (x), we know that gn (x) ≥ 0 for every positive integer
n and x ∈ R. Furthermore, the fact |(nx)| ≤ 1 for every positive integer n implies that
|gn (x)| =
1
(nx)
≤ 2.
n2
n
(7.14)
P
Thus we follow from Theorems 3.28 and 7.10 that the series
gn converges uniformly to f on R which
is equivalent to saying that fn → f uniformly on R.
By Figure 4.3, we know that the function g(x) = (x) is continuous on R \ Z and discontinuous on Z.
Therefore, each function gn (x) is discontinuous only at every point x ∈ R such that nx ∈ Z. Since nx ∈ Z
if and only if x ∈ Q, we have gn (x) is discontinuous only on Q. Then this means that the function fn (x)
is discontinuous on Q and is continuous on R \ Q (the set of irrational numbers). Since fn → f uniformly
on R, we have fn → f uniformly on R \ Q and then Theorem 7.12 implies that f is continuous on R \ Q.
Now f may be discontinuous on Q.
Chapter 7. Sequences and Series of Functions
150
Let x = pq ∈ Q, where p ∈ Z, q ∈ N, p and q are relative prime (the only positive integer which divides
them is 1). We check Definition 4.25 at x = pq . In other words, we want to show that
f (x+) 6= f (x−).
To this end, we notice that if n is an integer, then
lim [t] = n
and
t→n
t>n
lim [t] = n − 1.
(7.15)
t→n
t<n
Therefore, it follows from the limits (7.15) and the definition of gn that
gq (x+) = lim gq (t) = lim
t→x
t>x
t→x
t>x
and
gq (x−) = lim gq (t) = lim
t→x
t<x
t→x
t<x
(r)
r − [r]
(qt)
= r→p
lim 2 = r→p
lim
=0
2
q
q
q2
r>p
r>p
(r)
r − [r]
1
(qt)
= r→p
lim 2 = r→p
lim
= 2.
2
q2
q
q
q
r<p
r<p
Thus we always have gq (x−) − gq (x+) > 0 for every positive integer q. Suppose that N1 is an integer
such that
∞
X
1
1
< 2.
(7.16)
2
n
2q
n=N1
Let N = max(N1 , q). On the one hand, we have
fN (x−) − fN (x+) = lim fN (t) − lim fN (t)
t→x
t<x
= lim
t→x
t>x
N
X
t→x
t<x k=1
=
N
X
k=1
=
N
X
k=1
=
N
X
t→x
t>x k=1
gk (t)
#
lim gk (t) − lim gk (t)
t→x
t<x
t→x
t>x
[gk (t−) − gk (t+)]
N
X
1
k=1
>
"
gk (t) − lim
k2
1
.
q2
(7.17)
On the other hand, we follow from the inequalities (7.14) and (7.16) that
0≤
∞
X
k=N +1
gk (t) ≤
∞
X
k=N1
gk (t) ≤
∞
X
1
1
< 2.
2
k
2q
(7.18)
k=N1
Since |t − pq | < δ if and only if pq − δ < t < pq + δ, we conclude from the inequalities (7.17) and (7.18) that
1
1
− lim f x +
f (x−) − f (x+) = lim f x −
s→∞
s→∞
s
" ∞ s
#
∞
X X
1
1
= lim
gk x −
−
gk x +
s→∞
s
s
k=1
k=1
#
" N
N
X 1
1 X −
gk x +
= lim
gk x −
s→∞
s
s
k=1
k=1
151
7.1. Problems on uniform convergence of sequences of functions
"
#
∞
X
1
1
+ lim
gk x −
−
gk x +
s→∞
s
s
k=N +1
k=N +1
" N
#
N
∞
X
X 1
1
1 X > lim
−
gk x +
− lim
gk x +
gk x −
s→∞
s→∞
s
s
s
∞
X
k=N +1
k=1
k=1
= [fN (x−) − fN (x+)] − lim
s→∞
1
1
> 2− 2
q
2q
1
= 2
2q
> 0.
∞
X
k=N +1
1
gk x +
s
This means that f is discontinuous at x = pq and then on Q. Since Q is a countable dense subset of R,
the second assertion follows immediately.
For the third assertion, let [a, b] be a bounded interval. Recall that (nx) = nx − [nx] so that we may
rewrite gn (x) as
x [nx]
gn (x) = − 2 .
n
n
For every positive integer n, we have hn (x) = [nx] is clearly monotonic on [a, b] (see Figure 4.3 again).
By Theorem 6.9, we have hn ∈ R on [a, b]. It is obvious that nx ∈ R on [a, b], so we follow from Theorem
6.12(a) that gn ∈ R on [a, b] and then fn ∈ R on [a, b]. Recall the fact that fn converges uniformly to f
on R, so we conclude from Theorem 7.16 that
f ∈R
on [a, b] which is our desired result. This completes the proof of the problem.
Problem 7.11
Rudin Chapter 7 Exercise 11.
Proof. We follow the idea of the proof of Theorem 3.42. Suppose that
hn (x) =
n
X
fk (x).
k=1
By the hypothesis (a), {hn } is uniformly bounded on E. By Definition 7.19, there is a positive number
M such that
|hn (x)| < M
for all x ∈ E and n = 1, 2, . . .. Given ǫ > 0. By the hypothesis (b), there is an integer N such that
|gn (x)| = |gn (x) − 0| ≤
ǫ
2M
(7.19)
for all n ≥ N and all x ∈ E. If m ≥ n ≥ N and x ∈ E, then it follows from Theorem 3.41 and the
inequality (7.19) that
m
X
k=n
fk gk = hm gm − hn−1 gn +
≤M
m−1
X
k=n
= 2M gn
m−1
X
k=n
hk (gk − gk+1 )
(gk − gk+1 ) + gn + gm
Chapter 7. Sequences and Series of Functions
By Theorem 7.8, the series
Problem 7.12
P
152
≤ 2M gN
≤ ǫ.
fn gn converges uniformly on E. This finishes the proof of the problem. Rudin Chapter 7 Exercise 12.
Proof. Given that ǫ > 0. Our goal is to prove that there is an integer N such that
Z ∞
0
fn (x) dx −
Z ∞
f (x) dx < ǫ
(7.20)
0
for all n ≥ N . To this end, we divide the proof into a few steps.
• Step 1: Convergence of the integrals in (7.20). We first show that both integrals
Z ∞
Z ∞
fn (x) dx and
f (x) dx
0
(7.21)
0
are well-defined in the sense of Problems 6.7 and 6.8. Since |fn | ≤ g on [t, T ] for every positive
integer n, we have from Theorem 6.13(b) that
Z T
t
fn (x) dx ≤
Z T
t
|fn (x)| dx ≤
Z T
g(x) dx
(7.22)
t
for every positive integer n. Similarly, for every x ∈ [t, T ] we have
|f (x)| =
so that
Z T
t
lim fn (x) ≤ lim |fn (x)| ≤ g(x)
n→∞
f (x) dx ≤
(7.23)
n→∞
Z T
t
|f (x)| dx ≤
Z T
g(x) dx.
(7.24)
t
Then the convergence of the two integrals (7.21) follow immediately from the fact that
Z ∞
g(x) dx < ∞
0
and by letting t → 0 and T → ∞ one by one in the inequalities (7.22) and (7.24). In particular, we
have the following formulas:
Lemma 7.1
Let 0 < t < T < ∞. Then we have
Z t
0
and
Z t
0
f (x) dx ≤
Z t
g(x) dx,
fn (x) dx ≤
Z t
g(x) dx,
where n = 1, 2, 3, . . ..
0
0
Z ∞
f (x) dx ≤
Z ∞
fn (x) dx ≤
T
T
Z ∞
g(x) dx
(7.25)
T
Z ∞
T
g(x) dx
(7.26)
153
7.1. Problems on uniform convergence of sequences of functions
• Step 2: Establish a “temporary” upper bound of (7.20) by other integrals. For any
0 < t < T < ∞ and every positive integer n, the triangle inequality shows that the left-hand side
of the inequality (7.20) can be reduced as
Z ∞
0
fn (x) dx −
Z ∞
0
Z T
Z ∞
fn (x) dx −
+
f (x) dx −
f (x) dx ≤
0
Z T
t
fn (x) dx +
t
Z T
t
Z ∞
[fn (x) − f (x)] dx
f (x) dx .
(7.27)
0
Thus we show the inequality (7.20) is valid by investigating the absolute value of each term in the
right-hand side of the inequality (7.27). Before this, we need a lemma:
Lemma 7.2
Theorem 6.12(c) is also valid for improper integrals in the sense of Problems 6.7 and 6.8.
Proof of Lemma 7.2. Let b > c > a > 0. Then it follows from Theorem 6.12(c) that
"Z
# Z
Z
Z
Z
Z
b
b
f dα = lim
a→0
a
f dα = lim
Z b
0
c
f dα = lim
b
f dα +
a→0
a
f dα = lim
"Z
c
f dα =
b
f dα +
f dα
0
c
Z c
Z ∞
c
and
Z ∞
b→∞
a
b→∞
a
c
f dα +
a
Z b
c
#
f dα =
f dα +
a
f dα.
c
• Step 3: Estimate the absolute value of each term in (7.27). By Lemmas 7.1 and 7.2 as well
as the fact that |fn | ≤ g for every positive integer n, we have
Z ∞
0
fn (x) dx −
Z T
t
fn (x) dx ≤
≤
Z t
fn (x) dx +
0
Z t
Z ∞
fn (x) dx
T
g(x) dx +
0
Z ∞
g(x) dx.
(7.28)
T
Let 0 < t ≤ a < b ≤ T < ∞. Since g ∈ R on [t, T ], it follows from Theorem 6.20 (First Fundamental
Theorem of Calculus) that the function G : [a, b] → R defined by
Z u
G(u) =
g(x) dx
t
is continuous on [t, T ]. Furthermore, since 0 ≤ |fn (x)| ≤ g(x) for all x ∈ [a, b], we deduce from
Theorem 6.12(b) that G(x) ≥ 0 for all x ∈ [a, b] and G is monotonically increasing on [a, b]. By
Problem 6.7, we know that
Z
Z
a
a
g(x) dx = lim
0
t→0
g(x) dx
t
which means that for every ǫ > 0, there exists a δ1 > 0 such that
Z t
Z a
Z a
ǫ
g(x) dx =
g(x) dx −
g(x) dx <
6
0
0
t
(7.29)
for all t with 0 < t < δ1 . Similarly, we see from Problem 6.8 that there exists a M1 > 0 such that
Z ∞
Z ∞
Z T
ǫ
(7.30)
g(x) dx =
g(x) dx −
g(x) dx <
6
T
t
t
Chapter 7. Sequences and Series of Functions
154
for all T > M1 . Therefore, we apply the inequalities (7.29) and (7.30) to the inequality (7.28) to
establish
Z T
Z ∞
ǫ
ǫ
ǫ
(7.31)
fn (x) dx < + =
fn (x) dx −
6 6
3
t
0
for 0 < t < δ1 and T > M1 . By a similar argument, there exist δ2 > 0 and M2 > 0 such that
Z ∞
0
f (x) dx −
Z T
f (x) dx <
t
ǫ
ǫ
ǫ
+ =
6 6
3
(7.32)
for 0 < t < δ2 and T > M2 . Now it remains to evaluate the middle integral in (7.27). Since fn → f
uniformly on [t, T ], there is an integer N such that n ≥ N impliesa
|fn (x) − f (x)| ≤
for all x ∈ [t, T ] so that
Z T
t
[fn (x) − f (x)] dx ≤
Z T
t
ǫ
3(T − t)
|fn (x) − f (x)| dx ≤
ǫ
.
3
(7.33)
• Step 4: Obtain the desired inequality (7.20). Let δ = min(δ1 , δ2 ) and M = max(M1 , M2 ).
We pick a t and a T satisfying 0 < t < δ and T > M so that both the inequalities (7.31) and (7.32)
hold. After selecting the values of t and T , we can choose an integer N such that the inequality
(7.33) holds. Hence, our desired inequality (7.20) follows from the inequality (7.27) by combining
the three inequalities (7.31), (7.32) and (7.33).
This completes the proof of the problem.
Problem 7.13
Rudin Chapter 7 Exercise 13.
Proof.
(a) We follow the given hint.
(i) By the hypothesis, we have |fn (x)| ≤ 1 for all n and all x ∈ R. In particular, it is also true
that |fn (x)| ≤ 1 for all n and all x ∈ Q. By Definition 7.19, {fn } is uniformly bounded (thus
pointwise bounded) on Q. Recall that Q is countable, so Theorem 7.23 implies that {fn }
contains a subsequence {fni } such that {fni (r)} converges for every r ∈ Q.
(ii) Define f : R → R to be
f (x) = sup f (r),
(7.34)
r≤x
r∈Q
where f (r) = lim fni (r). By (i), the value f (r) is well-defined. Thus the supremum (7.34) is
i→∞
well-defined because R has the least-upper-bound property (see Definition 1.10 and Theorem
1.19).
(iii) Given that ǫ > 0. Since f is continuous at x, there exists a δ > 0 such that
|f (p) − f (x)| <
ǫ
2
(7.35)
for all |p − x| < δ. By Theorem 1.20(b), we may take p to be rational. Now the inequality
(7.35) can be rewritten as
ǫ
ǫ
f (x) − < f (p) < f (x) +
(7.36)
2
2
a Obviously, the integer N depends on t and T .
155
7.1. Problems on uniform convergence of sequences of functions
for all p ∈ (x − δ, x + δ). If we choose r, s ∈ Q and x − δ < r < x < s < x + δ, then since f is
a monotonically increasing function on R, we obtain from the inequality (7.36) that
f (x) −
ǫ
ǫ
< f (r) ≤ f (x) ≤ f (s) < f (x) + .
2
2
(7.37)
Besides, since every fni is also a monotonically increasing function on R, we have
0 ≤ fni (r) ≤ fni (x) ≤ fni (s).
(7.38)
Since f (p) = lim fni (p), there exists an integer Np such that i ≥ Np implies
i→∞
|fni (p) − f (p)| <
ǫ
.
2
(7.39)
If we take N = max(Nr , Ns ), then it follows from the inequality (7.38) and then the inequality
(7.39) that for all i ≥ N , we have
f (r) −
ǫ
ǫ
< fni (r) ≤ fni (x) ≤ fni (s) < f (s) + .
2
2
(7.40)
Combining the inequalities (7.37) and (7.40), we conclude that
f (x) − ǫ < f (r) −
ǫ
ǫ
< fni (r) ≤ fni (x) ≤ fni (s) < f (s) + < f (x) + ǫ
2
2
for all i ≥ N . In other words, we have
|fni (x) − f (x)| < ǫ
for all i ≥ N and this means that fni (x) → f (x) at every x at which f is continuous.
(iv) Since every fni is a monotonically increasing function on R, the definition (7.34) says that
f is also a monotonically increasing function on R. By Theorem 4.30, the set of points of
R at which f is discontinuous is at most countable. Let this set be E = {p1 , p2 , . . .} ⊆ R.
By Theorem 7.23 again, since {fni } is still a uniformly bounded (thus pointwise bounded)
sequence on the countable set E, {fni } has a subsequence {fnik } such that {fnik } converges
for every x ∈ E.
By the above steps, we define a function f : R → R and obtain a sequence {nik } such that
f (x) = lim fnik (x)
k→∞
for every x ∈ R.
(b) Suppose that K is a compact subset of R, each fnk : K → R is monotonically increasing and
f : K → R is continuous. We want to show that for every ǫ > 0, there exists an integer N such
that nk ≥ N implies
|fnk (x) − f (x)| ≤ ǫ
for all x ∈ K.
Since f is continuous on K, we know from Theorem 4.14 that f is uniformly continuous on K, i.e.,
for every ǫ > 0, there exists a δ > 0 such that
|f (x) − f (y)| <
ǫ
4
(7.41)
for all x, y ∈ K for which |x − y| < δ. It is clear that {(x − δ, x + δ) | x ∈ K} is an open cover of K.
Since K is compact, we have
m
[
(xi − δ, xi + δ)
K⊆
i=1
for some x1 , . . . , xm ∈ K. Since K is bounded, we may assume further that xi+1 − xi < δ for
1 ≤ i ≤ m − 1.
Chapter 7. Sequences and Series of Functions
156
To simplify our notations, we replace fnik by fnk in the following discussion. Let x ∈ K. Then we
must have x ∈ [xi , xi+1 ] for some 1 ≤ i ≤ m − 1. We fix this i and consider the following inequality
|fnk (x) − f (x)| ≤ |fnk (x) − fnk (xi )| + |fnk (xi ) − f (xi )| + |f (xi ) − f (x)|.
(7.42)
Now we are going to find an estimate of each modulus in the right-hand side of the inequality (7.42).
– Since {fnk } converges to f pointwise on K, for each xi , there exists an integer Ni such that
nk ≥ Ni implies that
ǫ
|fnk (xi ) − f (xi )| ≤ .
(7.43)
5
Take N = max(N1 , N2 , . . . , Nm ). Then the inequality (7.43) certainly holds for all nk ≥ N
and all 1 ≤ i ≤ m.
– Next, since xi+1 − xi < δ and x ∈ [xi , xi+1 ], we have |xi − x| < δ and thus the inequality (7.41)
implies that
ǫ
(7.44)
|f (xi ) − f (x)| < .
5
– By the hypothesis, each fnk is a monotonically increasing function on K and x ∈ [xi , xi+1 ] for
some 1 ≤ i ≤ m − 1, we have
fnk (xi ) ≤ fnk (x) ≤ fnk (xi+1 )
which implies
0 ≤ fnk (x) − fnk (xi ) ≤ fnk (xi+1 ) − fnk (xi ).
(7.45)
Now we apply the triangle inequality to (7.45) and then the inequalities (7.43) and (7.44) come
into play, we get
|fnk (x) − fnk (xi )| ≤ |fnk (xi+1 ) − fnk (xi )|
≤ |fnk (xi+1 ) − f (xi+1 )| + |f (xi+1 ) − f (xi )| + |fnk (xi ) − f (xi )|
ǫ
ǫ
ǫ
< + +
5 5 5
3ǫ
= .
(7.46)
5
Hence, by substituting the results (7.43), (7.44) and (7.46) into the inequality (7.42), we derive that
|fnk (x) − f (x)| ≤
ǫ
3ǫ ǫ
+ + =ǫ
5
5 5
for all nk ≥ N and x ∈ K. This is our expected inequality and so fnk → f uniformly on compact
sets.
Hence, this completes the proof of the problem.
Problem 7.14
Rudin Chapter 7 Exercise 14.
Proof. By the property 0 ≤ f (t) ≤ 1 for every t ∈ R, we have
0 ≤ x(t) =
0 ≤ y(t) =
∞
X
n=1
∞
X
n=1
2−n f (32n−1 t) ≤
2−n f (32n t) ≤
1
1
+ · · · = 1,
+
2 22
1
1
+ ··· = 1
+
2 22
for every t ∈ R. Furthermore, since
|2−n f (32n−1 t)| ≤ 2−n
and |2−n f (32n t)| ≤ 2−n
157
7.2. Problems on equicontinuous families of functions
for every t ∈ R, Theorems 7.10 and 3.26 imply that x(t) and y(t) are well-defined by the uniform
convergence of both series. Since f is continuous on R, we have {2−n f (32n−1 t)} and {2−n f (32n t)} are
sequences of continuous functions on R. By Theorem 7.12, the functions x(t) and y(t) are continuous on
R. Hence Φ(t) is continuous on R.
We follow the given hint. Let
t0 =
∞
X
3−i−1 (2ai ) =
i=1
2a1
2a2
+ 3 + ··· .
32
3
(7.47)
By Problem 3.19, we see that the set of all points t0 defined in (7.47) is exactly the Cantor set. If a1 = 0,
then we have
2
2
1
0 ≤ 3t0 ≤ 2 + 3 + · · · =
3
3
3
so that f (3t0 ) = 0 in this case. If a1 = 1, then we have
2
2
2
≤ 3t0 ≤ 2 + 3 + · · · = 1
3
3
3
so that f (3t0 ) = 1. Thus they mean that
f (3t0 ) = a1 .
(7.48)
For k = 2, 3, . . ., the definition (7.47) gives
3k t0 = 2(a1 + a2 + · · · + ak−1 ) +
2ak
2ak+1
+ ··· .
+
3
32
Therefore, a similar argument shows that
2a
2ak
2ak+1
2ak+1
k
f (3k t0 ) = f 2(a1 + a2 + · · · + ak−1 ) +
+
·
·
·
=
f
+
·
·
·
= ak .
+
+
3
32
3
32
(7.49)
Hence we follow from the expressions (7.48) and (7.49) that
x(t0 ) = x0
and y(t0 ) = y0 .
This completes the proof of the problem.
7.2
Problems on equicontinuous families of functions
Problem 7.15
Rudin Chapter 7 Exercise 15.
Proof. Let ǫ > 0. Since {fn } is equicontinuous on [0, 1], there exists a δ > 0 such that
|fn (t) − fn (y)| < ǫ
(7.50)
for all |t − y| < δ and fn , where t, y ∈ [0, 1]. In particular, we may choose y = 0 so that the inequality
(7.50) becomes
|fn (t) − fn (0)| < ǫ
(7.51)
for all 0 ≤ t < δ, t ∈ [0, 1] and fn . By definition, the inequality (7.51) is equivalent to
|f (nt) − f (0)| < ǫ
for all t ∈ [0, 1] with 0 ≤ t < δ and for all positive integers n. Let x ∈ R. By Theorem 1.20(a)
(Archimedean property), there exists a positive integer n such that nδ > x and this implies that we have
nt = x for some t ∈ [0, 1] with 0 ≤ t < δ.b Hence we obtain from the inequality (7.51) that
|f (x) − f (0)| < ǫ
for all x ∈ R. Since ǫ is arbitrary, we have f (x) = f (0) on R, i.e., f is a constant function.
b In fact, we have t = x < δ.
n
Chapter 7. Sequences and Series of Functions
158
Problem 7.16
Rudin Chapter 7 Exercise 16.
Proof. Given that ǫ > 0. Suppose that X is the metric space containing K. Since {fn } is equicontinuous
on K, there is a δ > 0 such that
ǫ
(7.52)
|fn (x) − fn (y)| <
3
whenever dX (x, y) < δ, x ∈ K, y ∈ K and every positive integer n. For any x ∈ K, let
Kx = {y ∈ K | dX (x, y) < δ}.
Then Kx is an open subset of X and K =
[
Kx . Since K is compact, we have
x∈K
K ⊆ Kx1 ∪ Kx2 ∪ · · · ∪ K xs
for some positive integer s. Since fn converges pointwise on K, {fn (x)} is a Cauchy sequence for every
x ∈ K. Therefore, for each xi , there is an integer Ni such that m, n ≥ Ni implies
|fm (xi ) − fn (xi )| <
ǫ
,
3
(7.53)
where i = 1, 2, . . . , s. Put N = max(N1 , N2 , . . . , Ns ). Thus if m, n ≥ N , then we obtain from (7.53) that
|fm (xi ) − fn (xi )| <
ǫ
,
3
(7.54)
where i = 1, 2, . . . , s.
Let x ∈ K. We know that x ∈ Kxi for some 1 ≤ i ≤ s. Then we have dX (x, xi ) < δ and it follows
from the inequality (7.52) that
ǫ
|fn (x) − fn (xi )| <
(7.55)
3
for every positive integer n. Now if m, n ≥ N , then we see from the inequalities (7.54) and (7.55) that
|fm (x) − fn (x)| ≤ |fm (x) − fm (xi )| + |fm (xi ) − fn (xi )| + |fn (xi ) − fn (x)|
ǫ
ǫ
ǫ
< + +
3 3 3
=ǫ
for all x ∈ K. Hence we obtain from Theorem 7.8 (Cauchy Criterion for Uniform Convergence) that {fn }
converges uniformly on K. This completes the proof of the problem.
Problem 7.17
Rudin Chapter 7 Exercise 17.
Proof. Let X and Y be metric spaces and fn : E ⊆ X → Y for n = 1, 2, 3, . . .. We state the definitions
of uniform convergence and equicontinuity for the sequence of mappings {fn }.
Definition 1. We say that {fn } converges uniformly on E ⊆ X to a mapping f : X → Y if for every
ǫ > 0, there is an integer N such that n ≥ N implies that
dY (fn (x), f (x)) < ǫ
for all x ∈ E.
159
7.2. Problems on equicontinuous families of functions
Definition 2. A family F of mappings f : E ⊆ X → Y is said to be equicontinuous on E if for every
ǫ > 0, there exists a δ > 0 such that
dY (f (x), f (y)) < ǫ
whenever dX (x, y) < δ, x ∈ E, y ∈ E and f ∈ F .
• Generalized Theorems 7.9 and 7.12. Given that ǫ > 0. Define
Mn = sup dY (fn (x), f (x)).
x∈E
By Definition 1, we see that dY (fn (x), f (x)) < ǫ if and only if Mn → 0 as n → ∞. This proves the
generalized Theorem 7.9.
Suppose that p ∈ E. Since there is nothing to prove if p is an isolated point of E, we may assume
that p is a limit point of E. By the triangle inequality, we have
dY (f (x), f (p)) ≤ dY (f (x), fn (x)) + dY (fn (x), fn (p)) + dY (fn (p), f (p))
(7.56)
for every x ∈ E, where the positive integer n will be determined very soon. Since fn → f uniformly
on E, there is an integer N such that n ≥ N implies
dY (fn (x), f (x)) ≤
ǫ
3
(7.57)
ǫ
.
3
(7.58)
for all x ∈ E. In particular, we also have
dY (fn (p), f (p)) ≤
Now we fix n = N in the inequalities (7.56), (7.57) and (7.58), i.e.,
dY (f (x), f (p)) ≤ dY (f (x), fN (x)) + dY (fN (x), fN (p)) + dY (fN (p), f (p))
2ǫ
≤
+ dY (fN (x), fN (p)).
3
(7.59)
It remains to find an estimate of dY (fN (x), fN (p)). In fact, since fN is continuous on E, there is a
δ > 0 such that
ǫ
dY (fN (x), fN (p)) <
(7.60)
3
for all x ∈ E with 0 < dX (x, p) < δ. Hence, for these x ∈ E with 0 < dX (x, p) < δ, we may
substitute the inequality (7.60) into the inequality (7.59) to get
dY (f (x), f (p)) <
2ǫ ǫ
+ = ǫ.
3
3
Hence f is continuous at p. This proves the generalized Theorem 7.12.c
• Generalized Theorems 7.8 and 7.11. Suppose that the metric space Y is complete and ǫ > 0.
It is easy to see the the part of proof on [21, p. 147] remains valid when the absolute values are
replaced by the metric dY . Conversely, suppose that N is an integer such that m, n ≥ N implies
that
dY (fn (x), fm (x)) ≤ ǫ
(7.61)
for every x ∈ E. Since Y complete, it follows from Definition 3.12 that the sequence {fn (x)}
converges for every x ∈ E, to a limit which we may call f (x). Thus the sequence {fn } converges
on E, to f . Next, we have to prove that the convergence is uniform. To this end, fix n(≥ N )
temporary and let m → ∞ in the inequality (7.61). Since fm (x) → f (x) as m → ∞, this gives
dY (fn (x), f (x)) ≤ ǫ
c We note that Rudin used Theorem 7.11 to prove Theorem 7.12, but we haven’t applied any generalized Theorem 7.11
to prove our generalized Theorem 7.12 here because Theorem 7.11 is not valid for any metric space.
Chapter 7. Sequences and Series of Functions
160
for every n ≥ N and every x ∈ E. This proves the generalized Theorem 7.8.
One of the core parts of the proof of Theorem 7.11 (see [21, p. 149]) is that the sequence {An },
where
An = lim fn (t),
t→x
is Cauchy in the complete metric space R and therefore converges to a real number A. By this, one
can prove the inequalities (19) to (22) in [21, p. 149] and then finally the result of Theorem 7.11.
Since our Y is supposed to be complete, the Cauchy sequence {An } still converges to A ∈ Y . Hence
all the inequalities (19) to (22) in [21, p. 149] remain valid when the absolute values are replaced
by the metric dY and this shows the generalized Theorem 7.12 is true.
• Generalized Theorems 7.10, 7.16, 7.17, 7.24 and 7.25. Let k be a positive integer,
fn = (fn1 , fn2 , . . . , fnk ) : E ⊆ X → Rk
be vector-valued functions for every positive integer n. Here each fni : E ⊆ X → R is a real-valued
function, where 1 ≤ i ≤ k. Suppose that
|fn (x)| ≤ Mn
P
for all x ∈ E and all positive integers n. If
Mn converges, then for m and n sufficiently large
enough, we have
m
m
m
X
X
X
Mj ≤ ǫ
|fj (x)| ≤
fj (x) ≤
j=n
j=n
j=n
P
for all x ∈ E. Therefore, the sequence { Pfn } satisfies the condition (7.61) on E. Hence it follows
from the generalized Theorem 7.8 that
fn converges uniformly on E. This is the generalized
Theorem 7.10.
We put ǫn = sup |fn (x) − f (x)|. By Definition 1.36, we know that
a≤x≤b
|fni (x) − fi (x)| ≤ |fn (x) − f (x)| ≤ ǫn
for all i = 1, 2, . . . , k and for all x ∈ [a, b]. Then, for each i = 1, 2, . . . , k, we have
fni − ǫn ≤ fi ≤ fni + ǫn .
Therefore, by applying the argument as in the proof of Theorem 7.16 to each function fni , we can
show that fi ∈ R(α) on [a, b] and
Z b
fi dα = lim
a
n→∞
Z b
fni dα.
(7.62)
a
Hence we deduce from Definition 6.23 and the expression (7.62) that f ∈ R(α) on [a, b] and
Z b
a
f dα = lim
n→∞
Z b
fn dα
a
as required. This proves the generalized Theorem 7.16.
Given that ǫ > 0. Since {fn (x0 )} converges for some x0 ∈ [a, b] and {fn′ } converges uniformly on
[a, b], we choose an integer N such that m ≥ N and n ≥ N imply that
ǫ
2
(7.63)
ǫ
2(b − a)
(7.64)
|fn (x0 ) − fm (x0 )| <
and
′
|fn′ (t) − fm
(t)| <
for all t ∈ [a, b].
161
7.2. Problems on equicontinuous families of functions
If we apply Theorem 5.19 to the vector-valued function fn − fm , then the inequality (7.64) becomes
|fn (x) − fm (x) − fn (t) + fm (t)| ≤ |x − t|
sup
y∈(x,t) or
y∈(t,x)
′
|fn′ (y) − fm
(y)| ≤ |x − t| ·
ǫ
ǫ
≤
2(b − a)
2
(7.65)
for any x, t ∈ [a, b] and n, m ≥ N . By the inequalities (7.63) and (7.65), we have
|fn (x) − fm (x)| ≤ |fn (x) − fm (x) − fn (x0 ) + fm (x0 )| + |fn (x0 ) − fm (x0 )|
ǫ
ǫ
< +
2 2
=ǫ
for any x ∈ [a, b], n ≥ N and m ≥ N . This means that {fn } converges uniformly on [a, b]. Let
f (x) = lim fn (x),
n→∞
where x ∈ [a, b]. We now fix a point x in [a, b] and define
Φn (t) =
fn (t) − fn (t)
t−x
and Φ(t) =
f (t) − f (t)
t−x
(7.66)
for t ∈ [a, b] but t 6= x. Then we have
lim Φn (t) = fn′ (x)
t→x
(7.67)
for n = 1, 2, . . .. Now the second inequality in (7.65) shows that
|Φn (t) − Φm (t)| ≤
ǫ
2(b − a)
if n ≥ N and m ≥ N so that the sequence {Φn } converges uniformly for t 6= x. Since {fn } converges
(uniformly) to f on [a, b], we conclude from the definition (7.66) that
lim Φn (t) = Φ(t)
n→∞
(7.68)
uniformly for t ∈ [a, b] but t 6= x.
If we apply Theorem 7.11 to {Φn }, then the limits (7.67) and (7.68) imply that
lim Φ(t) = lim fn′ (x)
t→x
n→∞
which is our desired result
f ′ (x) = lim fn′ (x).
n→∞
This completes the proof of the generalized Theorem 7.17.
We follow the proof of Theorem 7.24 to prove the generalized Theorem 7.24. Let K be a compact
metric space and fn ∈ C (K) for n = 1, 2, 3, . . .. Suppose that {fn } converges uniformly on K. Then
there exists an integer N such that
kfn − fN k < ǫ
(7.69)
for n > N . Since fn is continuous on the compact set K, Theorem 4.19 implies that fn is uniformly
continuous on K. Thus there is a δ > 0 such that
|fi (x) − fi (y)| < ǫ
(7.70)
if 1 ≤ i ≤ N and x, y ∈ K with d(x, y) < δ. If n > N and x, y ∈ K with d(x, y) < δ, we obtain
from the inequality (7.69) that
|fn (x) − fn (y)| ≤ |fn (x) − fN (x)| + |fN (x) − fN (y)| + |fN (y) − fn (y)| < 3ǫ.
Combining this with the inequality (7.70), we see that the sequence {fn } satisfies Definition 2.
Hence it is equicontinuous on K. This shows the generalized Theorem 7.24.
Finally, we prove the generalized Theorem 7.25 by following the proof of Theorem 7.25:
Chapter 7. Sequences and Series of Functions
162
(a) Let ǫ > 0 be given and choose a δ > 0 such that
|fn (x) − fn (y)| < ǫ
(7.71)
for n = 1, 2, . . . and x, y ∈ K with d(x, y) < δ.
Since K is compact, there are finitely many points p1 , p2 , . . . , pr in K such that to every x ∈ K
corresponds at least one pi with d(x, pi ) < δ. Since {fn } is pointwise bounded on K, there
exists Mi < ∞ such that
|fn (pi )| < Mi
for n = 1, 2, . . .. If M = max(M1 , M2 , . . . , Mr ), then we have
|fn (x)| = |fn (x) − fn (pi ) + fn (pi )| ≤ |fn (x) − fn (pi )| + |fn (pi )| < Mi + ǫ ≤ M + ǫ
for all x ∈ K. This shows that {fn } is uniformly bounded on K.
(b) By Problem 2.25, let E be a countable dense subset of K. Then Theorem 7.23 shows that
{fn } has a subsequence {fni } such that {fni (x)} converges for every x ∈ E.d
Put gi = fni . By the assumptions in the proof of part (a), we let
V (x, δ) = {y ∈ K | d(x, y) < δ}.
Since E is dense in K and K is compact, there are finitely many points x1 , . . . , xm in E such
that
K ⊆ V (x1 , δ) ∪ V (x2 , δ) ∪ · · · ∪ V (xm , δ).
(7.72)
Since {gi (x)} converges for every x ∈ E, there is an integer N such that
|gi (xs ) − gj (xs )| < ǫ
(7.73)
whenevery i ≥ N, j ≥ N and 1 ≤ s ≤ m. If x ∈ K, then the relation (7.72) implies that
x ∈ V (xs , δ) for some s so that the inequality (7.71) gives
|gi (x) − gi (xs )| < ǫ
(7.74)
for every positive integer i. Now if i ≥ N and j ≥ N , then it follows from the inequalities
(7.73) and (7.74) that
|gi (x) − gj (x)| ≤ |gi (x) − gi (xs )| + |gi (xs ) − gj (xs )| + |gj (xs ) − gj (x)| < 3ǫ
for all x ∈ K. Hence we obtain from the generalized Theorem 7.8 that {gi } converges uniformly
on K.
This completes the proof of the generalized Theorem 7.25.e
Hence we finish the proof of the problem.
Problem 7.18
Rudin Chapter 7 Exercise 18.
Proof. Since {fn } is uniformly bounded on [a, b], we have |fn (x)| < M for all x ∈ [a, b] and n = 1, 2, . . .,
where M is a positive number. Since fn ∈ R on [a, b], each Fn is well-defined on [a, b]. By Theorem
6.12(d), we have
Z x
|Fn (x)| =
fn (t) dt ≤ M (b − a).
a
d Here we assume without proof that Theorem 7.23 is also valid for vector-valued functions.
e This version of Theorem 7.25 will be applied to prove results in Problem 7.26.
163
7.2. Problems on equicontinuous families of functions
This means that {Fn } is uniformly bounded on [a, b]. By Theorem 6.20 (First Fundamental Theorem of
Calculus), we know that Fn is continuous on [a, b] so that Fn ∈ C ([a, b]) for every positive integer n.
ǫ
Given that ǫ > 0. Let δ = M
> 0. Thus if x, y ∈ [a, b] and |x − y| < δ, then we have
|Fn (x) − Fn (y)| =
Z x
a
fn (t) dt −
Z y
a
fn (t) dt ≤
Z x
y
fn (t) dt ≤ M |y − x| < ǫ
which means that {Fn } is equicontinuous on [a, b] by Definition 7.22. Hence it contains a subsequence
{Fnk } converges uniformly on [a, b] by Theorem 7.25. This completes the proof of the problem.
Problem 7.19
Rudin Chapter 7 Exercise 19.
Proof. We prove the necessity and the sufficiency of the problem as follows:
• Necessity. Suppose that S is compact.
– Uniformly closed. By Theorem 2.34, S is closed in C (K), so S is uniformly closed in C (K)
by Definition 7.14.
– Pointwise bounded. Let x ∈ K. Define the set
Vn = {f ∈ C (K) | kf k < n},
where n = 1, 2, 3, . . .. By Definition 2.18, each Vn is a neighborhood of the zero function 0.
Then Theorem 2.19 implies that the Vn is open in C (K). Since
S ⊆ C (K) ⊆
∞
[
Vn ,
n=1
the collection {Vn } forms an open cover of S. Since S is compact, we have
S ⊆ Vn1 ∪ Vn2 ∪ · · · ∪ Vnk
(7.75)
for some positive integers n1 , n2 , . . . , nk . Without loss of generality, we may assume that
n1 < n2 < · · · < nk so that the relation (7.75) reduces to
S ⊆ Vnk .
In other words, we have kf k < nk for all f ∈ S. By Definition 7.19, S is uniformly bounded
and thus pointwise bounded.
– Equicontinuous. We follow the hint to prove the last assertion. Assume that S was not
equicontinuous on K. Then there is a ǫ > 0 such that for every δ > 0, we have
|f (x) − f (y)| ≥ ǫ
for some f ∈ S and some x, y ∈ K with d(x, y) < δ. Now we take δ = n1 , where n = 1, 2, . . .
in the above consideration so that a sequence of functions {fn } ⊆ S plus sequences of points
{xn } and {yn } in K with d(xn , yn ) < n1 are constructed. In other words, we have
|fn (xn ) − fn (yn )| ≥ ǫ,
(7.76)
where xn , yn ∈ K and d(xn , yn ) < n1 for n = 1, 2, . . ..
To complete this part, we need a lemma:
Lemma 7.3
The above sequence of functions {fn } does not contain any equicontinuous subsequence
on K.
Chapter 7. Sequences and Series of Functions
164
Proof of Lemma 7.3. Otherwise, the family of functions
F = {fnk } ⊆ {fn }
was equicontinuous on K. By Definition 7.22, there exists a δ ′ > 0 such that
|fnk (x) − fnk (y)| < ǫ
(7.77)
whenever x, y ∈ K with d(x, y) < δ ′ and all fnk ∈ F . However, if we choose nk large enough
so that n1k < δ ′ , then the points xnk , ynk satisfy
d(xnk , ynk ) < δ ′
and thus we obtain from the inequality (7.76) that
|fnk (xnk ) − fnk (ynk )| ≥ ǫ
which contradicts the inequality (7.77).
If {fn } ⊆ S contains a uniformly convergent subsequence on K, then Theorem 7.24 implies
that such subsequence must be equicontinuous on K which obviously contradicts our Lemma
7.3. Therefore, no subsequence of {fn } converges in C (K) by the rephrased Theorem 7.9 on
[21, p. 151]. However, this contradicts to the fact that S is compact.f Hence we conclude that
S is equicontinuous on K.
• Sufficiency. Now we suppose that the subset S is uniformly closed, poinwise bounded and equicontinuous on K. If S is a finite set, then there is nothing to prove by Definition 2.32. Thus we may
assume without loss of generality that S is infinite. Let {fn } ⊆ S. Then {fn } must be pointwise
bounded and equicontinuous on K. By Theorem 7.25, we know that {fn } contains a uniformly
convergent subsequence {fnk } on K. By the rephrased Theorem 7.9 on [21, p. 151] again, such
subsequence {fnk } must converge to a function f in C (K). Since S is closed in C (K), we have
f ∈ S. By Problem 2.26, S is compact.
This completes the proof of the problem.
7.3
Applications of the (Stone-)Weierstrass theorem
Problem 7.20
Rudin Chapter 7 Exercise 20.
Proof. Let P (x) = a0 xn + a1 xn−1 + · · · + an−1 x + an , where a0 , a1 , . . . , an ∈ R. Now it follows from the
hypotheses and Theorem 6.12(a) that
Z 1
f (x)P (x) dx = 0.
(7.78)
0
Since f is continuous on [0, 1], Theorem 7.26 implies that there exists a sequence of polynomial Pn
converging uniformly to f on [0, 1]. In particular, we have from the integral equation (7.78) that
Z 1
f (x)Pn (x) dx = 0
(7.79)
0
for all n = 1, 2, . . .. Since f, Pn ∈ R on [0, 1] for all n = 1, 2, . . ., we have
f Pn ∈ R
f In fact, since S is compact, we deduce from Theorem 3.6(a) that {f
C (K).
n } has a subsequence converging in S and then in
165
7.3. Applications of the (Stone-)Weierstrass theorem
on [0, 1] from Theorem 6.13(a). Since Pn → f uniformly on [0, 1], f Pn → f 2 uniformly on [0, 1]. Therefore,
we conclude from Theorem 7.16 and the integral equation (7.79) that
Z 1
f 2 (x) dx = lim
n→∞
0
Z 1
f (x)Pn (x) dx = 0.
0
Since f 2 ≥ 0 and f 2 is continuous on [0, 1], Problem 6.2 implies that f 2 (x) = 0 on [0, 1]. Hence f (x) = 0
for all x ∈ [0, 1], finishing the proof of the problem.
Problem 7.21
Rudin Chapter 7 Exercise 21.
Proof. We have K = {z ∈ C | |z| = 1}. It is clear that f (z) = z ∈ A because we have
f (eiθ ) = eiθ = 0 · e0 + eiθ .
Now the algebra A separates points on K by this f . Furthermore, we see that f (z) 6= 0 for all z ∈ K so
that A vanishes at no point of K. We notice that for every f ∈ A , we have
Z 2π
iθ
iθ
f (e )e dθ =
0
Z 2π X
N
0
cn e
inθ
n=0
iθ
e dθ =
N Z 2π
X
cn ei(n+1)θ dθ = 0.
(7.80)
0
n=0
By Definition 7.28, the closure B of A is the set of all functions which are limits f of uniformly convergent
sequences {fn } in A . Define
gn (θ) = fn (eiθ ) and g(θ) = f (eiθ ).
Since fn → f uniformly on K, we have gn → g uniformly on [0, 2π]. By Theorem 7.16, we can conclude
that the integral equation (7.80) is also true for every f in the closure of A . It is obvious that the
function F (z) = z1 is a continuous function on K.
However, F ∈
/ B because
Z 2π
iθ
iθ
F (e )e dθ =
0
Z 2π
e−iθ eiθ dθ = 2π.
(7.81)
0
This completes the proof of the problem.
Problem 7.22
Rudin Chapter 7 Exercise 22.
Proof. Suppose that ǫ > 0. Since f ∈ R(α) on [a, b], Problem 6.12 ensures that there exists a continuous
function g on [a, b] such that
ǫ
(7.82)
kf − gk2 < .
2
Since g is continuous on [a, b], we follow from Theorem 7.26 (Weierstrass’s Theorem) that there exists a
sequence of polynomials Pn converging uniformly to g on [a, b]. By Definition 7.7, there is an integer N
such that n ≥ N implies that
ǫ2
|Pn (x) − g(x)| ≤
4[α(b) − α(a)]
for all x ∈ [a, b]. By this and Theorem 6.12(d), we have
kPn − gk2 =
(Z
b
2
a
) 21
|Pn (x) − g(x)| dx
Chapter 7. Sequences and Series of Functions
≤
=
(
ǫ
2
166
ǫ2
[α(b) − α(a)]
4[α(b) − α(a)]
) 21
(7.83)
for all n ≥ N . Hence it follows from the inequalities (7.82), (7.83) and Problem 6.11 that
kf − Pn k2 ≤ kf − gk2 + kg − Pn k2 < ǫ
for all n ≥ N which is exactly what we want. We complete the proof of the problem.
Problem 7.23
Rudin Chapter 7 Exercise 23.
Proof. We follow the given hint. For n = 0, we know that
P1 (x) = P0 (x) +
x2 − P02 (x)
x2
=
.
2
2
Since |x| ≤ 1, we have 0 ≤ P0 (x) ≤ P1 (x) ≤ |x|. This means that the statement is true for n = 0. Assume
that the statement is also true for n = k for some non-negative integer k, i.e.,
0 ≤ Pk (x) ≤ Pk+1 (x) ≤ |x|
if |x| ≤ 1. Now for n = k + 1, we note from the assumption that |x|− Pk+1 (x) ≥ 0 if |x| ≤ 1. Furthermore,
if |x| ≤ 1, then the assumption also implies that
0≤
|x| + |x|
|x| + Pk+1 (x)
≤
= |x| ≤ 1,
2
2
so we have
|x| + Pk+1 (x)
≥ 0.
2
Therefore, we obtain from these and the given identity that
#
"
|x| + Pk+1 (x)
≥0
|x| − Pk+2 (x) = [|x| − Pk+1 (x)] 1 −
2
1−
if |x| ≤ 1. By definition and the assumption, we have
Pk+2 (x) − Pk+1 (x) =
2
x2 − Pk+1
(x)
[|x| − Pk+1 (x)][|x| + Pk+1 (x)]
=
≥0
2
2
if |x| ≤ 1. Therefore, the statement is true for n = k + 1 if it is true for n = k. Hence it follows from
induction that it is true for all non-negative integers n.
By using the identity repeatedly, we have
#
"
|x| + Pn−1 (x)
|x| − Pn (x) = [|x| − Pn−1 (x)] 1 −
2
"
#
|x| + Pn−2 (x)
|x| − Pn−1 (x) = [|x| − Pn−2 (x)] 1 −
(7.84)
2
..
.
..
.
#
|x| + P0 (x)
.
|x| − P1 (x) = [|x| − P0 (x)] 1 −
2
"
167
7.4. Isometric mappings and initial-value problems
We remark that if |x| ≤ 1, then the inequalities 0 ≤ Pn (x) ≤ Pn+1 (x) ≤ |x| certainly imply that
1−
|x| + Pn (x)
|x|
≤1−
2
2
(7.85)
for all non-negative integer n. Therefore, if |x| ≤ 1 and n is a non-negative integer, then the relations
(7.84) and the inequalities (7.85) give
# "
#
"
|x| + Pn−2 (x)
|x| |x| + Pn−1 (x)
× 1−
× ···× 1 −
|x|
0 ≤ |x| − Pn (x) = 1 −
2
2
2
|x| n
≤ |x| 1 −
2
2
<
.
n+1
Given that ǫ > 0. There exists an integer N ≥ 2ǫ − 1 such that for n ≥ N , we have
||x| − Pn (x)| <
2
2
≤
≤ǫ
n+1
N +1
for all x ∈ [−1, 1]. By Definition 7.7, we have
lim Pn (x) = |x|,
n→∞
uniformly on [−1, 1], completing our proof of the problem.
7.4
Isometric mappings and initial-value problems
Problem 7.24
Rudin Chapter 7 Exercise 24.
Proof. If x ∈ X, then it deduces from Definition 2.15(c) that fp (x) = d(x, p) − d(x, a) ≤ d(a, p) which
gives the desired result that |fp (x)| ≤ d(a, p). This proves our first assertion.
For the second assertion, since a and p are fixed, we note from the first assertion that fp is bounded
on X. Next, let q ∈ X and ǫ > 0. We take δ = 2ǫ . Thus if x ∈ X and d(x, q) < δ, then the triangle
inequality implies that
|fp (x) − fp (q)| ≤ |d(x, p) − d(q, p)| + |d(q, a) − d(x, a)|
≤ d(x, q) + d(q, x)
= 2d(x, q)
< 2δ
= ǫ.
By Definition 4.5, fp is continuous at q and thus on X. By Definition 7.14, we have fp ∈ C (X).
For all p, q, x ∈ X, we have |fp (x) − fq (x)| = |d(x, p) − d(x, q)| ≤ d(p, q) and the equality holds if x = p
or x = q. Therefore, we have
kfp − fq k = sup |fp (x) − fq (x)| = d(p, q).
x∈X
This proves the third assertion.
If we define Φ : X → C (X) by Φ(p) = fp , then it follows from the third assertion that Φ is an
isometry of X onto Φ(X). Suppose that Y = Φ(X). By Theorem 2.27(a), Y is closed in C (X). By the
paragraph following Definition 3.12, we know that every closed subset E of a complete metric space X is
also complete. Thus Y must be complete which completes the proof of the fourth assertion. We end the
proof of the problem.
Chapter 7. Sequences and Series of Functions
168
Problem 7.25
Rudin Chapter 7 Exercise 25.
Proof. We follow the given hint.
(a) We prove the assertions as follows:
– It is clear that 0 = x0 < x1 < · · · < xn = 1. If t = xi for some 0 ≤ i ≤ n, then we have
∆n (xi ) = 0 which means that
fn′ (xi ) = φ(xi , fn (xi )).
Since |φ(x, y)| ≤ M for all x ∈ [0, 1] and y ∈ R, we have |fn′ (xi )| ≤ M for all i = 0, 1, 2, . . . , n.
If t 6= xi for all i = 0, 1, 2, . . . , n, then t ∈ (xi , xi+1 ) for some i = 0, 1, . . . , n − 1. In this case,
we have
|fn′ (t)| = |φ(xi , fn (xi ))| ≤ M.
In conclusion, we have
|fn′ (t)| ≤ M
for all t ∈ [0, 1].
– Next the second assertion |∆n (t)| ≤ 2M , where t ∈ [0, 1], follows immediately from the first
assertion and the hypothesis that |φ(x, y)| ≤ M for all x ∈ [0, 1] and y ∈ R.
– Since φ is a continuous bounded real function on [0, 1] × R and
fn′ (t) = φ(xi , fn (xi ))
if xi < t < xi+1 , the function fn′ is bounded and continuous on [0, 1] \ {x0, x1 , . . . , xn }. By the
definition of ∆n , we know that ∆n is bounded on [0, 1] and continuous on [0, 1]\{x0, x1 , . . . , xn }.
By Theorem 6.10, we have ∆n ∈ R on [0, 1].
– Since φ(t, fn (t)) + ∆n (t) = fn′ (t) on [0, 1] \ {x0 , x1 , . . . , xn }, we have for every x ∈ [0, 1],
|fn (x)| = c +
Z x
0
[φ(t, fn (t)) + ∆n (t)] dt ≤ |c| +
Z x
0
|fn′ (t)| dt ≤ |c| + M x ≤ |c| + M.
ǫ
> 0. Now for all x, y ∈ [0, 1] with |x − y| < δ, since |fn′ | ≤ M on [0, 1],
(b) Given that ǫ > 0. Let δ = M
we obtain from Theorem 6.12(c) that
|fn (x) − fn (y)| =
Z x
0
fn′ (t) dt −
Z y
0
fn′ (t) dt =
Z x
y
fn′ (t) dt ≤ M |x − y| < M δ = ǫ,
where n = 1, 2, . . ..g By Definition 7.22, {fn } is equicontinuous on [0, 1].
(c) By part (a), {fn } is uniformly bounded (and thus pointwise bounded) on [0, 1]. Since {fn } is
equicontinuous on [0, 1], Theorem 7.25(b) implies that it contains a uniformly convergent subsequence on [0, 1].
(d) By Definition 2.17, the rectangle
K = {(x, y) | 0 ≤ x ≤ 1, |y| ≤ M1 }
is a 2-cell. Then it follows from Theorem 2.40 that K is compact. Since φ is continuous on K, we
deduce from Theorem 4.19 that φ is uniformly continuous on K. Thus, given ǫ > 0 there exists a
δ > 0 such that
|φ(x, y) − φ(x′ , y ′ )| < ǫ
(7.86)
g Here we assume that x > y. If x < y, then we have the integral
Z y
x
fn′ (t) dt instead of
Z x
y
fn′ (t) dt.
169
7.4. Isometric mappings and initial-value problems
p
for all (x, y), (x′ , y ′ ) ∈ K with (x − x′ )2 + (y − y ′ )2 < δ. In particular, we take x = x′ = t ∈ [0, 1]
in the inequality (7.86) so that
|φ(t, y) − φ(t, y ′ )| < ǫ
(7.87)
for all |y| ≤ M1 and |y ′ | ≤ M1 with |y − y ′ | < δ. By part (c) with the fixed δ, there exists an integer
N such that nk ≥ N implies that
|fnk (t) − f (t)| < δ
(7.88)
for all t ∈ [0, 1]. If we put y = fnk (t) and y ′ = f (t), then the inequality (7.88) and part (a) show
that y and y ′ satisfy the hypotheses |y| ≤ M1 , |y ′ | ≤ M1 and |y − y ′ | < δ as required in (7.87).
Therefore, for nk ≥ N , we have
|φ(t, fnk (t)) − φ(t, f (t))| < ǫ
for all t ∈ [0, 1]. Hence, by Definition 7.7, the sequence {φ(t, fnk (t))} converges uniformly to
φ(t, f (t)) on [0, 1].
(e) Our goal is to show that for every ǫ > 0, there exists an integer N such that n ≥ N implies that
|∆n (t)| ≤ ǫ
(7.89)
for all t ∈ [0, 1].
Since fn′ (t) = φ(xi , fn (xi )) on (xi , xi+1 ) and ∆n (t) = fn′ (t) − φ(t, fn (t)), we have
∆n (t) = φ(xi , fn (xi )) − φ(t, fn (t))
for all t ∈ (xi , xi+1 ), where i = 0, 1, . . . , n − 1.
For every t ∈ [0, 1], we know that t ∈ [xi , xi+1 ] for some 0 ≤ i ≤ n − 1 so that
|t − xi | ≤
1 h
.
n
Recall that φ is uniformly continuous on K. In particular, if we put x = xi , x′ = t, y = fn (xi ) and
y ′ = fn (t) into the inequality (7.86), then there is a δ > 0 such that
ǫ
(7.90)
|φ(xi , fn (xi )) − φ(t, fn (t))| <
2
for all t ∈ [0, 1] with
p
(xi − t)2 + [fn (xi ) − fn (t)]2 < δ.
(7.91)
Now it is easy to see that the inequality (7.89) follows immediately from the inequality (7.90) if we
can find an integer N such that the inequality (7.91) holds for all n ≥ N .i
By part (b), since {fn } is equicontinuous on [0, 1], for the fixed δ > 0, there is a η > 0 such that
δ
|fn (x) − fn (t)| < √
2
(7.92)
for all x, t ∈√[0, 1] with |x − t| < η and n = 1, 2, . . .. We take N to be an integer such that
N > max( η1 , δ2 ). Then for all n ≥ N , we get
|t − xi | ≤
1
1
≤
<η
n
N
δ
and |t − xi | < √ .
2
(7.93)
Therefore, for n ≥ N , we follow from the inequalities (7.92) and (7.93) that
r
p
δ2
δ2
2
2
+
=δ
(xi − t) + [fn (xi ) − fn (t)] <
2
2
which is exactly (7.91). Hence we obtain our desired inequality (7.89) and then
∆n (t) → 0
uniformly on [0, 1].
h This means that x
i → t as n → ∞.
i This corresponds a partition such that every t must be a point in some interval with length less than or equal to 1 .
N
Chapter 7. Sequences and Series of Functions
170
(f) Combining parts (d) and (e), we have
φ(t, fnk (t)) + ∆nk (t) → φ(t, f (t))
uniformly on [0, 1]. Hence we establish from Theorem 7.16 that
f (x) = lim fnk (x)
k→∞
c+
= lim
k→∞
=c+
=c+
Z x
Z0 x
Z x
[φ(t, fnk (t)) + ∆nk (t)] dt
0
!
lim [φ(t, fnk (t)) + ∆nk (t)] dt
k→∞
φ(t, f (t)) dt,
(7.94)
0
where x ∈ [0, 1]. It is clear that f (0) = c. Since φ(t, f (t)) is continuous on [0, x], Theorem 6.20
(First Fundamental Theorem of Calculus) implies that the integral on the right-hand side in the
expression (7.94) is differentiable on [0, x] and
!
Z x
d
′
f (x) =
φ(t, f (t)) dt = φ(x, f (x)).
dx
0
Hence the function f is a solution of the given problem.
This completes the proof of the problem.
Problem 7.26
Rudin Chapter 7 Exercise 26.
Proof. Basically, we follow the setting of the proof of Problem 7.25. Fix n. For i = 0, 1, . . . , n, put
xi = ni . Let fn be a continuous mapping on [0, 1] such that fn (0) = c,
fn′ (t) = Φ(xi , fn (xi ))
if xi < t < xi+1 ,
and put
∆n (t) = fn′ (t) − Φ(t, fn (t)),
except at the points xi , where ∆n (t) = 0. Then
Z x
fn (x) = c +
[Φ(t, fn (t)) + ∆n (t)] dt.
0
Choose M < ∞ so that |Φ| ≤ M .
(a) Then by using similar argument as in the proof of Problem 7.25(a), we have
|fn′ | ≤ M, |∆n | ≤ M
and |fn | ≤ |c| + M = M1
on [0, 1] for all positive integers n. For ∆n ∈ R on [0, 1], we note that ∆n is continuous on
[0, 1] \ {x0 , x1 , . . . , xn }. If we write
∆n = (∆n1 , . . . , ∆nk ),
then it follows from Theorem 4.10(a) that each ∆nj : [0, 1] → R is continuous on [0, 1]\{x0 , x1 , . . . , xn }
so that ∆nj ∈ R on [0, 1] by Theorem 6.10, where j = 1, 2, . . . , k. Thus we have ∆n ∈ R on [0, 1]
by Definition 6.23.
(b) By applying similar argument as in the proof of Problem 7.25(b), we have {fn } is equicontinuous
on [0, 1].
171
7.4. Isometric mappings and initial-value problems
(c) By the vector-valued version of Theorem 7.25 (see Problem 7.17), we have the result that a subsequence {fnk } converges to some f uniformly on [0, 1].
(d) Similarly, the argument in the proof of Problem 7.25(d) can be repeated to show that
Φ(t, fnj (t)) → Φ(t, f (t))
uniformly on [0, 1].
(e) It can be shown, by an argument similar to the proof of Problem 7.25(e), that
∆n (t) → 0
uniformly on [0, 1]. As a remark, we note that the inequality (7.91) is replaced by
p
(xi − t)2 + |fn (xi ) − fn (t)|2 < δ.
(f) Suppose that
Φ(t, fnj (t)) = (φ1 (t, fnj (t)), . . . , φk (t, fnj (t))),
Φ(t, f (t)) = (φ1 (t, f (t)), . . . , φk (t, f (t))),
∆nj (t) = (∆nj 1 (t), . . . , ∆nj k (t)).
By using part (d), part (e) and the proof of Theorem 4.10 on [21, p. 88], we see that
Φ(t, fnj (t)) + ∆nj (t) → Φ(t, f (t))
uniformly on [0, 1] if and only if
φi (t, fnj (t)) + ∆nj i (t) → φi (t, f (t))
(7.95)
uniformly on [0, 1], where i = 1, 2, . . . , k. Applying Theorem 7.16 to each equation in (7.95), we get
Z x
fi (x) = ci +
φi (t, f (t)) dt,
0
where f = (f1 , . . . , fk ), c = (c1 , . . . , ck ), ci ∈ [0, 1] and i = 1, 2, . . . , k. Hence these give
Z x
Φ(t, f (t)) dt
f (x) = c +
0
and it is easily shown that it is a solution of the given initial-value problem.
Hence, we have completed the proof of the problem.
Chapter 7. Sequences and Series of Functions
172
CHAPTER
8
Some Special Functions
8.1
Problems related to special functions
Problem 8.1
Rudin Chapter 8 Exercise 1.
Proof. We complete the proof by proving two claims:
• Claim 1: If x 6= 0, then we have
p(x) − 12
e x
q(x)
f (n) (x) =
for every positive integer n, where p(x) and q(x) are polynomials with q(x) 6≡ 0.a We prove the
claim by induction. For x 6= 0, since
1
2
f ′ (x) = 3 e− x2 ,
x
the statement is true for n = 1. Assume that the statement is true for n = k, where k is a positive
integer, i.e.,
f (k) (x) =
p(x) − 12
e x ,
q(x)
(8.1)
where x 6= 0, p(x) and q(x) are polynomials with q(x) 6≡ 0. Now, for x 6= 0, it follows from the
assumption (8.1), Theorem 5.3(b) and (c) that
#
"
p(x) 1
p(x)
d
2
e− x2
·
+
f (k+1) (x) =
x3 q(x)
dx q(x)
#
"
2 p(x) q(x)p′ (x) − q ′ (x)p(x) − 12
e x
·
+
=
x3 q(x)
q 2 (x)
=
2p(x)q(x) + x3 [q(x)p′ (x) − q ′ (x)p(x)] − 12
e x .
x3 q 2 (x)
(8.2)
Since q(x) 6≡ 0, q 2 (x) 6≡ 0 too. Thus it is also true for n = k + 1 if it is true for n = k. By induction,
the statement is true for all positive integers n.
• Claim 2: Suppose that p(x) and q(x) are polynomials with q(x) 6≡ 0. Then we have
p(x) − 12
e x = 0.
x→0 q(x)
lim
a The degrees of p(x) and q(x) are not important in the proof.
173
(8.3)
Chapter 8. Some Special Functions
174
Let p(x) = p0 xs +p1 xs−1 +· · ·+ps and q(x) = q0 xt +q1 xt−1 +· · ·+qt , where p0 , p1 , . . . , ps , q0 , q1 , . . . , qt
are all real and p0 q0 6= 0. If x = y1 , then the left-hand side of the limit (8.3) can be rewritten as
p( y1 ) −y2
p(x) − 12
x
e
e
lim
= lim
y→+∞ q( 1 )
x→0 q(x)
y
p1
p0
y s + y s−1 + · · · + ps −y 2
e
= lim q0
q1
y→+∞ t + t−1
+ · · · + qt
y
y
p0 + p1 y + · · · + ps y s t−s −y2
= lim
y e .
y→+∞ q0 + q1 y + · · · + qt y t
Now it is clear that
1
lim
y→+∞ q0 + q1 y + · · · + qt y t
(8.4)
= 0.
(8.5)
2
Furthermore, we know from Theorem 8.6(c) that e−y > e−y > 0 for large positive y so that
2
0 < y i e−y < y i e−y
for large positive y, where i = 0, 1, . . . , s. Therefore, we deduce from Theorem 8.6(f) that
2
lim pi y i e−y = 0,
(8.6)
y→+∞
where i = 0, 1, . . . , s. Hence we obtain from the limits (8.5) and (8.6) that the right-hand side of
the limit (8.4) is 0 and this is exactly our desired result (8.3).
For every positive integer n, it follows from the limit (8.3) that
p(x) − 12
f (n−1) (x) − f (0)
= lim
e x = 0.
x→0 xq(x)
x→0
x−0
f (n) (0) = lim
Hence this shows that f has derivatives of all orders at x = 0 and f (n) (0) = 0 for n = 1, 2, 3, . . .. This
completes the proof of the problem.
Problem 8.2
Rudin Chapter 8 Exercise 2.
Proof. We note that
∞ X
∞
X
i=1 j=1
aij =
∞ i−1
∞ ∞
X
X
X
X
1
1
1 − 1 + + · · · + i−1 = −
2j−i =
aii +
= −2
i−1
2
2
2
i=1
j=1
i=1
i=1
and
∞ X
∞
X
j=1 i=1
aij =
∞ X
j=1
ajj +
∞
X
i=j+1
∞
X
(−1 + 1) = 0.
2j−i =
j=1
This completes the proof of the problem.
Problem 8.3
Rudin Chapter 8 Exercise 3.
Proof. Since aij ≥ 0 for all i and j, we have |aij | = aij . For each i = 1, 2, . . ., let
two cases for consideration.
∞
X
j=1
aij = bi . There are
175
8.1. Problems related to special functions
Case (i):
P
bi converges. Then it follows from Theorem 8.3 that we have
∞ X
∞
X
aij =
∞ X
∞
X
aij .
(8.7)
j=1 i=1
i=1 j=1
We note that the double series in the expression (8.7) are finite.
P
Case (ii):
bi diverges. Let
n
X
bi .
sn =
i=1
We claim that lim sn = +∞. Otherwise, {sn } is bounded. Since aij ≥ 0, we have bi ≥ 0 and thus
n→∞
the definition implies that {sn } is a monotonic increasing sequence. P
By Theorem 3.14 (Monotone
Convergence Theorem), the sequence {sn } converges. By definition,
bi also converges which is a
contradiction. Therefore, we have
∞ X
∞
X
aij = +∞
(8.8)
i=1 j=1
in this case.
Next, for each j = 1, 2, . . ., we define
n
X
aij = cj . If
i=1
P
cj converges, then we may apply Theorem
8.3 to obtain the expression (8.7) and
P both double series are finite. This certainly contradicts the
result (8.8) and thus we must have
cj diverges. By an argument in the previous paragraph, we
can show that
∞ X
∞
X
aij = +∞.
j=1 i=1
Hence the expression (8.7) also holds in this subcase.
This finishes the proof of the problem.
Problem 8.4
Rudin Chapter 8 Exercise 4.
Proof.
(a) We follow from Theorem 5.13 (L’Hospital’s Rule) that, for b > 0,
bx − 1
ex log b − 1
log bex log b
= lim
= lim
= log b.
x→0
x→0
x→0
x
x
1
lim
(b) By Theorem 5.13 (L’Hospital’s Rule) and [21, Eqn. (38), p. 180], we have
1
log(1 + x)
= lim 1+x = 1.
x→0 1
x→0
x
lim
1
(c) Let y = (1 + x) x . Then we have
y=e
log(1+x)
x
.
By part (b), we have
lim y = lim e
x→0
x→0
log(1+x)
x
= exp
log(1 + x) = e.
x→0
x
lim
Chapter 8. Some Special Functions
(d) Notice that
176
"
#x
x n
x nx1
1+
= 1+
.
n
n
Since n → ∞ if and only if nx → 0, it follows from part (c) that
"
#x "
#x
1
1
x nx
x nx
x n
1+
lim 1 +
= xlim 1 +
= ex .
= xlim
n→∞
→0
→0
n
n
n
n
n
We end the proof of the problem.
Problem 8.5
Rudin Chapter 8 Exercise 5.
Proof.
1
(a) If y = (1 + x) x , then we have
log(1 + x)
.
x
We deduce from [21, Eqn. (38), p.180] and Theorem 5.5 (Chain Rule) that
log y =
x − (1 + x) log(1 + x)
y′
=
.
y
x2 (1 + x)
By Theorem 5.13 (L’Hospital’s Rule) twice, we have
1
− 1+x
1
x − (1 + x) log(1 + x)
− log(1 + x)
y′
=
lim
= lim
=
lim
=−
x→0 2 + 6x
x→0
x→0 2x + 3x2
x→0 y
x2 (1 + x)
2
lim
so that
1
e
lim y = − .
2 x→0
2
Hence Theorem 5.13 (L’Hospital’s Rule) implies that
lim y ′ = −
x→0
1
e − (1 + x) x
e
= lim −y ′ = .
x→0
x→0
x
2
lim
1
log n
(b) Since n n = e n , we have
log n
e n −1
1
n
[n n − 1] =
.
log n
log n
n
Since n → ∞ if and only if logn n → 0, it follows from the identity (8.9) that
1
n
ex − 1
ex − e0
[n n − 1] = lim
= lim
= e0 = 1,
n→∞ log n
x→0
x→0 x − 0
x
lim
where x = logn n .
(c) By applying Theorem 5.13 (L’Hospital’s Rule) three times, we have
sec2 x − 1
tan x − x
= lim
x→0 1 − cos x + x sin x
x→0 x(1 − cos x)
2 sec2 x tan x
= lim
x→0 sin x + sin x + x cos x
2 sec3 x tan x
= lim
x→0 2 tan x + x
lim
(8.9)
177
8.1. Problems related to special functions
2(3 sec3 x tan2 x + sec5 x)
x→0
2 sec2 x + 1
2
= .
3
= lim
(d) By Theorem 5.13 (L’Hospital’s Rule), we have
1 − cos x
1 − cos x
cos2 x
1
x − sin x
= lim
= lim
· cos2 x = lim
= .
2
2
x→0 sec x − 1
x→0 1 − cos x
x→0 1 + cos x
x→0 tan x − x
2
lim
We complete the proof of the problem.
Problem 8.6
Rudin Chapter 8 Exercise 6.
Proof.
(a) Put x = y = 0 into the given equation, we have f 2 (0) = f (0) which means that
f (0) = 0
or f (0) = 1.
Since f is not zero, we have f (0) = 1. Since f is differentiable on R, we haveb
f (x + h) − f (x)
f (x)f (h) − f (x)
f (h) − f (0)
= lim
= f (x) lim
= cf (x),
h→0
h→0
h→0
h
h
h−0
f ′ (x) = lim
where c = f ′ (0). Let F (x) = e−cx f (x). Then we have
F ′ (x) = e−cx f ′ (x) − ce−cx f (x) = ce−cx f (x) − ce−cx f (x) = 0
for every x ∈ R. By Theorem 5.11(b), we have F (x) = C for some constant C. Since we have
F (0) = f (0) = 1, C = 1 and thus
f (x) = ecx
as required.
(b) Suppose that we can prove that
f (r) = [f (1)]r ,
(8.10)
where r ∈ Q.
For any x ∈ R, since Q is dense in R (Theorem 1.20(b)), there exists a sequence {rn } ⊆ Q such
that rn → x as n → ∞. Since f is continuous on R, this and Theorem 4.2 imply that
f (x) = lim f (rn ) = lim [f (1)]rn = [f (1)]x .
n→∞
n→∞
(8.11)
By the hypothesis, we have
f (x) = f
x
2
+
x
x x
x
=f
f
= f2
>0
2
2
2
2
for all real x. In particular, we have f (1) > 0 so that elog f (1) = f (1) by [21, Eqn. (36), p.180]. If
we let c = log f (1), then we deduce from the expression (8.11) that
f (x) = ecx .
Thus we have transformed our problem into the question of the validity of the formula (8.10).
To this end, we divide the proof into several steps:
b This is an equivalent definition of the derivative of f at x (by letting h = t − x in [21, Eqn. (2), p. 103]).
Chapter 8. Some Special Functions
178
– Step 1: f (n) = [f (1)]n for all n ∈ N. The result follows from induction and the hypothesis
f (x)f (y) = f (x + y), so we skip its proof here.
– Step 2: f (n) = [f (1)]n for all n ∈ Z. Since f (−1)f (1) = f (1 − 1) = f (0) = 1, we have
f (−1) = [f (1)]−1 .
Assume that f (−k) = [f (1)]−k for some positive integer k. Since
f (−k − 1)f (1) = f (−k − 1 + 1) = f (−k) = [f (1)]−k ,
we have f (−(k + 1)) = [f (1)]−(k+1) . By induction, we have the claim.
– Step 3: The formula (8.10) holds. Suppose that r = pq , where p, q ∈ Z and q 6= 0. It is
clear that
1
1
1 h 1 ip
p
1
1
=f
= f
.
(8.12)
+ ···+
+ ···+
=f
×f
f
q
q
q
q
q
q
q
|
|
{z
}
{z
}
p terms
(p − 1) terms
Furthermore, since
1
1
1
1 1
h 1 iq
1
=f
+ + ··· +
×f
× ···× f
=f
= f (1),
f
q
q
q
q
q
q
q
we have
f
1
q
1
= [f (1)] q .
(8.13)
Combining the expressions (8.12) and (8.13), we get
p
p
= [f (1)] q
f
q
which is exactly formula (8.10).
Hence we have completed the proof of our problem.
Problem 8.7
Rudin Chapter 8 Exercise 7.
Proof. We consider the function f (x) = sinx x defined on (0, π2 ). Then we have
f ′ (x) =
x cos x − sin x
.
x2
Let g(x) = x cos x − sin x be a function defined on (0, π2 ). Since
g ′ (x) = −x sin x + cos x − cos x = −x sin x < 0
(8.14)
on (0, π2 ), g is strictly decreasing on (0, π2 ) by Problem 5.2. Since g is continuous on [0, π2 ], it is also
strictly decreasing on [0, π2 ].c Since g(0) = 0, we always have g(x) < 0 on (0, π2 ] and thus f ′ (x) < 0 on
(0, π2 ). Therefore, f is strictly decreasing on (0, π2 ). By the continuity of f , it yields that f is also strictly
decreasing on (0, π2 ]. Hence we have f (x) > f ( π2 ) which gives
sin π
2
sin x
> π2 =
x
π
2
(8.15)
c If g(0) = g(p) for some p ∈ (0, π ), then Theorem 5.10 (Mean Value Theorem) implies that g ′ (x) = 0 for some
2
x ∈ (0, p) ⊆ (0, π2 ) which contradicts the inequality (8.14).
179
8.1. Problems related to special functions
for every x ∈ (0, π2 ).
For the second inequality, we consider another function h(x) = sin x − x on (0, π2 ). Now we have
h′ (x) = cos x − 1 < 0
on (0, π2 ) so that h is strictly decreasing on (0, π2 ) by Problem 5.2. By the continuity of h, we know that
h is also strictly decreasing on [0, π2 ] and this implies that h(x) < h(0) which is equivalent to
sin x
<1
x
(8.16)
for every x ∈ (0, π2 ). Hence our desired result follows immediately by combining the inequalities (8.15)
and (8.16), completing the proof of the problem.
Problem 8.8
Rudin Chapter 8 Exercise 8.
Proof. The inequality holds obviously if n = 0. Assume that
| sin k| ≤ k| sin x|
for some non-negative integer k. If n = k + 1, then, by basic properties of trigonometric functions, we
have
| sin(k + 1)x| = | sin kx cos x + cos kx sin x|
≤ | sin kx|| cos x| + | cos kx|| sin x|
≤ | sin kx| + | sin x|
≤ k| sin x| + | sin x|
= (k + 1)| sin x|.
Hence the inequality also holds in this case. By induction, the inequality holds for all non-negative
integers n. This completes the proof of the problem.
Problem 8.9
Rudin Chapter 8 Exercise 9.
Proof.
(a) Let {pN } = {sN − log N }, where N ≥ 1. Then we follow from [21, Eqn. (39), p.180] that
Z N +1
1
dx
−
.
pN +1 − pN = sN +1 − log(N + 1) − sN + log N =
N +1
x
N
(8.17)
On [N, N + 1], we have N1+1 ≤ x1 so that the expression (8.17) yields
1
−
pN +1 − pN ≤
N +1
Z N +1
N
1
1
dx
=
−
= 0.
N +1
N +1 N +1
Thus the sequence {pN } is monotonic decreasing. Furthermore, we have
sN =
N −1 Z k+1
X
1
+
N
k=1
k
N −1 Z k+1
X
1
dx
≥
+
k
N
k=1
k
1
dx
=
+
x
N
so that
pN = sN − log N ≥
1
> 0.
N
Z N
1
1
dx
=
+ log N
x
N
(8.18)
Chapter 8. Some Special Functions
180
In other words, {pN } is bounded below. By Theorem 3.14 (Monotone Convergence Theorem), we
have
lim pN = lim (sN − log N )
N →∞
converges.
N →∞
(b) When N = 10m , the inequality (8.18) givesd
sN = s10m > log 10m =
m
log10 10m
=
.
log10 e
log10 e
Therefore, if m ≥ 44, then we have
m
44
≥
≈ 101.3137441 > 100.
log10 e
0.434294481
Hence the value of m is at least 44 to make sure that sN > 100.
This ends the analysis of the problem.
Problem 8.10
Rudin Chapter 8 Exercise 10.
Proof. We follow the hint. Let p1 , . . . , pk be those primes that divide at least one integer ≤ N . Then
p1 , . . . , pk are all primes less than or equal to N and each integer ≤ N must have the form
mk
1 m2
pm
1 p2 · · · pk
for some non-negative integers m1 , m2 , . . . , mk . This fact allows us to have the inequality
N
X
1
n=1
n
≤
k Y
1
1
1+
+ 2 + ··· .
pj
pj
j=1
(8.19)
Since pj ≥ 2, it follows from Theorem 3.26 that
1+
1
1
1
+ 2 + ··· =
.
pj
pj
1 − p1j
By [21, Eqn. (25), p.178], we know that
e2x ≥ 1 + 2x
for all x ≥ 0 so that
(1 − x)e2x ≥ (1 − x)(1 + 2x) = 1 + x(1 − 2x) ≥ 1
(8.20)
on [0, 21 ]. Therefore, combining inequalities (8.19) and (8.20), we have
N
X
1
n=1
n
≤
k Y
j=1
1−
k
k
X
1
1 −1 Y p2j
e = exp 2
.
≤
pj
p
j=1 j
j=1
(8.21)
Since N → ∞ if and only if k → ∞, we deduce immediately from the inequality (8.21), Theorem 3.25(b)
(Comparison Test) and Theorem 3.28 that
X
all primes p
1
p
diverges. This completes the proof of the problem.
d We have applied the change of base formula for logarithmic in the second equality.
181
8.1. Problems related to special functions
Problem 8.11
Rudin Chapter 8 Exercise 11.
Proof. We prove the result by considering two steps:
• Step 1: The improper integral in the question is well-defined. Since f (x) → 1 as x → ∞,
for every ǫ > 0, there exists an integer N > 0 such that
ǫ
ǫ
< f (x) < 1 +
4
4
0<1−
(8.22)
for all x > N . By the inequalities (8.22), we have
Z n
Z n
Z n
ǫ −tx
ǫ −tx
−tx
1+
1−
e f (x) dx <
e
dx <
e
dx
0<
4
4
N
N
N
Z
n
ǫ e−nt
ǫ e−nt
e−N t e−N t 0< 1−
<
e−tx f (x) dx < 1 +
.
+
+
4
−t
t
4
−t
t
N
(8.23)
By the most left-hand inequality in (8.23) the sequence
(Z
)
n
e−tx f (x) dx
N
is increasing monotonically. Fix t > 0. By the right-hand side inequality in (8.23), we know that
Z n
ǫ e−N t
e−tx f (x) dx < 1 +
4
t
N
so that the sequence is bounded above. By Theorem 3.14 (Monotone Convergence Theorem), the
limit
Z n
e−tx f (x) dx
lim
n→∞
N
exists. Since f (x), e−tx ∈ R on [0, N ], Theorem 6.13(a) implies that e−tx f (x) ∈ R on [0, N ]. Hence
the limit
!
Z
Z
Z
n
n→∞
N
e−tx f (x) dx = lim
lim
n→∞
0
=
Z N
n
e−tx f (x) dx +
0
N
e−tx f (x) dx + lim
n→∞
0
e−tx f (x) dx
Z n
e−tx f (x) dx
N
certainly exists.
• Step 2: The limit holds. Next, we are going to show the desired limit holds. When n → ∞, the
inequalities (8.23) become
Z ∞
ǫ −N t
ǫ −N t
e
<t
e−tx f (x) dx < 1 +
e
(8.24)
1−
4
4
N
for all t > 0. Since f ∈ R on [0, N ], f is clearly bounded on [0, N ]. In other words, there is a
positive number M such that
−M ≤ f (x) ≤ M
for all x ∈ [0, N ]. By this, we have
−M t
Z N
0
e−tx dx ≤ t
Z N
0
e−tx f (x) dx ≤ M t
Z N
0
e−tx dx
Chapter 8. Some Special Functions
182
−M (1 − e−N t ) ≤ t
Z N
0
e−tx f (x) dx ≤ M (1 − e−N t ).
(8.25)
We note from Theorem 8.6(c) that ex is a strictly increasing function of x and ex ≥ 1 for all x ≥ 0,
so we have e−N t ≤ 1 for all t ≥ 0 and there exist δ1 > 0 and δ2 > 0 such that
−
ǫ
ǫ
< 1 − e−N t <
4M
4M
(8.26)
for 0 < t < δ1 and
ǫ
ǫ
< 1 − e−N t <
4
4
for 0 < t < δ2 respectively. By inequalities (8.26), the inequalities (8.25) imply that
−
−
ǫ
<t
4
Z N
e−tx f (x) dx <
0
ǫ
4
(8.27)
(8.28)
for 0 < t < δ1 . Similarly, we apply the inequalities (8.27) to the inequalities (8.24) to get
Z ∞
ǫ 2
ǫ 2
1−
e−tx f (x) dx < 1 +
<t
4
4
N
for all 0 < t < δ2 . Rewrite the inequalities (8.29) as
Z ∞
ǫ2
ǫ2
ǫ
ǫ
<t
e−tx f (x) dx < 1 + +
1− +
2 16
2 16
N
(8.29)
(8.30)
for all 0 < t < δ2 .
Therefore, if we take δ = min(δ1 , δ2 ), then both the inequalities (8.26) and (8.27) hold for 0 < t < δ
which imply that we can add the inequalities (8.28) and (8.30) together to get
Z ∞
3ǫ
ǫ2
3ǫ
ǫ2
1−
+
<t
e−tx f (x) dx < 1 +
+
4
16
4
16
0
Z ∞
ǫ2
3ǫ
ǫ2
3ǫ
<t
e−tx f (x) dx − 1 <
+
<ǫ
(8.31)
−ǫ < − +
4
16
4
16
0
for 0 < t < δ. Hence we deduce from the inequality (8.31) that
Z ∞
lim t
e−tx f (x) dx = 1 (t > 0).
t→0
0
This completes the proof of the problem.
Problem 8.12
Rudin Chapter 8 Exercise 12.
Proof.
(a) It is clear that f ∈ R on [−π, π]. By Definition 8.9 and [21, Eqn. (46), p. 182], we have
cm =
1
2π
Z π
f (x)e−imx dx =
−π
1
2π
Z δ
e−imx dx =
−δ
sin mδ
i
(e−imδ − eimδ ) =
2mπ
mπ
(8.32)
for all non-zero integers m. If m = 0, then we have
c0 =
1
2π
Z π
−π
f (x) dx =
1
2π
Z δ
−δ
f (x) dx =
δ
.
π
(8.33)
183
8.1. Problems related to special functions
(b) Since
|f (0 + t) − f (0)| = |f (t) − f (0)| = 1 − 1 ≤ |t|
for all t ∈ (−δ, δ), it follows from Theorem 8.14 that
f (0) = lim sN (f ; 0).
(8.34)
N →∞
By the formulas (8.32) and (8.33), we have
sN (f ; 0) =
N
X
cn = c0 +
−N
1
X
cn +
N
X
cn =
1
−N
N
δ X 2 sin nδ
+
.
π n=1 nπ
Therefore, the limit (8.34) implies that
1 = f (0) =
∞
X
2 sin nδ
n=1
∞
X
nπ
=
π−δ
π
∞
δ X 2 sin nδ
+
π n=1 nπ
sin nδ
π−δ
=
n
2
n=1
which is our desired formula.
(c) Since f ∈ R on [−π, π] with period 2π, we deduce from Theorem 8.16 (Parseval’s Theorem), the
formulas (8.32) and (8.33) that
1
2π
Z π
−π
|f (x)|2 dx =
∞
δ 2 X 2 sin2 (nδ)
+
π 2 n=1 n2 π 2
∞
δ
δ 2 X 2 sin2 (nδ)
= 2+
π
π
n2 π 2
n=1
∞
X
πδ − δ 2
2 sin2 (nδ)
=
2
π
n2 π 2
n=1
∞
X
sin2 (nδ)
n=1
n2 δ
=
π−δ
2
(8.35)
as required.
(d) Given that ǫ > 0. Let b > 0 be a fixed large number such that
ǫ
1
< .
b
3
We consider δn = nb and the partition {x0 , x1 , . . . , xn } of [0, b]e such that xj = jδn for 0 ≤ j ≤ n.
Then we have
∆xi = xi − xi−1 = iδn − (i − 1)δn = δn
for 1 ≤ i ≤ n. We want to show that there exists an integer N such that
n
X
sin2 (iδn )
i=1
for all n ≥ N .
e Thus we must have b < n.
i2 δ
n
−
Z ∞
0
sin2 x
dx < ǫ
x2
Chapter 8. Some Special Functions
184
To this end, suppose that
sin2 x
,
x2
f (x) =
Mi =
sup
f (x) and mi =
xi−1 ≤x≤xi
inf
xi−1 ≤x≤xi
f (x),
where i = 1, 2, . . . , n. Then we have
L(P, f ) ≤
L(P, f ) ≤
n
X
sin2 (iδn )
· δn ≤ U (P, f )
i2 δn2
i=1
n
X
sin2 (iδn )
≤ U (P, f )
i 2 δn
i=1
(8.36)
Since f is bounded on [0, b]f , Theorem 6.10 implies that f ∈ R on [0, b]. By the inequalities (8.36),
we have
Z b
n
X
sin2 (iδn )
sin2 x
dx
=
lim
(8.37)
n→∞
x2
i 2 δn
0
i=1
which means that there exists an integer N such that
n
X
sin2 (iδn )
i=1
i 2 δn
−
Z b
0
sin2 x
ǫ
dx <
x2
3
(8.38)
for all n ≥ N .
Furthermore, since 1b < 3ǫ , we have
Z ∞
∞
∞
∞
X
X
sin2 (iδn )
1 X 1
1
1
1
ǫ
1
sin2 (iδn )
≤
≤
≤
dt =
= <
2δ
2δ
2
2
i
i
δ
i
δ
t
nδ
b
3
n
n
n
n
n
n
i=n+1
i=n+1
i=n+1
and
Z ∞
0
sin2 x
dx −
x2
Z b
0
Z ∞
sin2 x
dx =
x2
b
sin2 x
dx ≤
x2
Z ∞
b
1
ǫ
1
dx = < .
x2
b
3
Thus, for n ≥ N , combining the inequalities (8.38), (8.39) and (8.40), we have
∞
X
sin2 (iδn )
i=1
i 2 δn
−
Z ∞
0
Z b
n
X
sin2 (iδn )
sin2 x
sin2 x
dx
≤
−
dx
x2
i 2 δn
x2
0
i=1
Z ∞
Z b
∞
X
sin2 (iδn )
sin2 x
sin2 x
+
+
dx
−
dx
2
2
x
x
i 2 δn
b
0
i=n+1
Z b
n
∞
X
X
sin2 x
sin2 (iδn )
sin2 (iδn )
−
dx
≤
+
i 2 δn
x2
i 2 δn
0
i=1
i=n+1
Z ∞
Z b
sin2 x
sin2 x
+
dx
−
dx
x2
x2
b
0
ǫ
ǫ
ǫ
< + +
3 3 3
=ǫ
which is our required inequality. By part (b), we have
π − δn
−
2
for all n ≥ N .
f Note the classical result that
lim
x→0
sin x
= 1.
x
Z ∞
0
sin2 x
dx < ǫ
x2
(8.39)
(8.40)
185
8.1. Problems related to special functions
(e) Put δ = π2 in the formula (8.35), we have
∞
X
sin2 ( nπ )
2
n=1
n2
=
π2
.
8
(8.41)
2 nπ
Since sin2 ( nπ
2 ) = 1 if n is odd and sin ( 2 ) = 0 if n is even, we deduce from the formula (8.41) that
∞
X
π2
1
=
.
2
(2n + 1)
8
n=0
We complete the proof of the problem.
Problem 8.13
Rudin Chapter 8 Exercise 13.
Proof. Since the f is not a function with period 2π, we can’t apply Parseval’s Theorem directly and
we have to find a function based on f which satisfies the hypotheses of Parseval’s Theorem. Define the
function F : R → R by
F (x) = f (x − 2nπ) = x − 2nπ
(2nπ ≤ x < (2n + 2)π),
where n is an integer. Now if 2nπ ≤ x < (2n + 2)π, then
2(n + 1)π ≤ x + 2π < [2(n + 1) + 2]π.
This implies that
F (x + 2π) = f x + 2π − 2(n + 1)π = x + 2π − 2(n + 1)π = x − 2nπ = F (x).
Thus F is a function with period 2π. Since F is bounded on [−π, π] and is discontinuous only at 0, we
follow from Theorem 6.10 that F ∈ R on [−π, π]. By Definition 8.9 and Theorem 6.22 (Integration by
Parts), we have
Z π
1
cn =
F (x)e−inx dx
2π −π
"Z
#
Z π
0
1
−inx
−inx
=
F (x)e
dx +
F (x)e
dx
2π −π
0
"Z
#
Z π
0
1
xe−inx dx
(x + 2π)e−inx dx +
=
2π −π
0
" Z
#
Z π
0
1
−inx
−inx
=
2π
e
dx +
xe
dx
2π
−π
−π
"
#
0
π
Z π
xe−inx
1
e−inx
1
−inx
+
+
2π
=
e
dx
2π
−in
−in
in −π
−π
−π
"
#
2π
2π(−1)n
2π(−1)n
1
−
−
+
+0
=
2π
in
−in
−in
=−
1
,
in
where n is a non-zero integer. If n = 0, then we have
"Z
#
Z π
Z π
0
1
1
1
× 2π 2 = π.
(x + 2π) dx +
x dx =
F (x) dx =
c0 =
2π −π
2π −π
2π
0
Chapter 8. Some Special Functions
186
Hence it follows from Theorem 8.16 (Parseval’s Theorem) that
1
2π
Z π
−π
|F (x)|2 dx = π 2 + 2
∞
X
1
n=1
∞
X
n2
1
1
(2π)3
×
= π2 + 2
2
2π
3
n
n=1
∞
X
4π 2
1
= π2 + 2
2
3
n
n=1
∞
X
1
n=1
n
=
2
π2
.
6
This completes the proof of the problem.
Problem 8.14
Rudin Chapter 8 Exercise 14.
Proof. Since f (−π) = f (π) = 0, we can extend the function f to be a continuous function on R:
f (x) = (π − |x − (2n + 2)π|)2 ,
where x ∈ [(2n + 1)π, (2n + 3)π] and n ∈ Z, see Figure 8.1 for the illustration.
Figure 8.1: The graph of the continuous function y = f (x) = (π − |x|)2 on [−π, π].
Since f is integrable on [−π, π], we apply Theorem 6.22 (Integration by Parts) to get the Fourier
coefficients of f :
Z π
1
cn =
(π − |x|)2 e−inx dx
2π −π
"Z
#
Z π
0
1
2 −inx
2 −inx
=
(π − x) e
dx
(π + x) e
dx +
2π −π
0

2



 n2 , if n 6= 0;
(8.42)
=

2

π


, if n = 0.
3
For every x ∈ [−π, π] and t ∈ (−π, π), we have
|f (x + t) − f (x)| = (π − |x + t|)2 − (π − |x|)2
187
8.1. Problems related to special functions
= π 2 − 2π|x + t| + |x + t|2 − π 2 + 2π|x| − |x|2
= 2π|x| − 2π|x + t| + |x + t|2 − |x|2
≤ 2π|x| − 2π(|x| − |t|) + (|x| + |t|)2 − |x|2
≤ 2π|t| + 2|x||t| + |t|2
≤ 2π|t| + 2π|t| + π|t|
= 5π|t|.
By Theorem 8.14 and the definitions (8.42), we have
f (x) = lim sN (f ; x)
=
N →∞
−1
X
cn einx + c0 +
−∞
=
∞
X
cn einx
1
−1
X
∞
X 2
2 inx π
e
+
einx
+
2
2
n
3
n
−∞
1
2
∞
π 2 X 2 inx
(e
+ e−inx )
+
2
3
n
n=1
=
∞
π2 X 4
=
cos nx.
+
3
n2
n=1
(8.43)
To prove the two identities in the problem, we first put x = 0 into the expression (8.43), since
f (0) = π 2 , we have
π2 =
∞
X
1
n=1
n
=
2
∞
X
1
π2
+4
2
3
n
n=1
π2
.
6
Next, Theorem 8.16 (Parseval’s Theorem) implies that
1
2π
Z π
−π
(π − |x|)4 dx =
−1
X
−∞
|cn |2 + |c0 |2 +
−1
X
4
4
4
π
π
+
=
+
4
5
n
9
−∞
∞
X
π4
8
π4
−
=
5
9
n4
n=1
∞
X
1
n=1
n4
=
∞
X
1
|cn |2
∞
X
4
1
n4
π4
.
90
This completes the proof of the problem.
Problem 8.15
Rudin Chapter 8 Exercise 15.
Proof. By the definition of the Dirichlet kernel, we have
Dn (x) =
sin(n + 21 )x
sin x2
Chapter 8. Some Special Functions
188
so that
N
N
1 X
1 X sin(n + 21 )x
KN (x) =
.
Dn (x) =
N + 1 n=0
N + 1 n=0
sin x2
We claim that
N
X
sin(n + 1 )x
2
sin x2
n=0
=
1 − cos(N + 1)x
1 − cos x
(8.44)
for every non-negative integer N . This can be proven by induction. The case for N = 0 is easy. Assume
that for some non-negative integer k, we have
k
X
sin(n + 1 )x
sin x2
n=0
2
=
1 − cos(k + 1)x
.
1 − cos x
If N = k + 1, then we have
k
sin(n + 21 )x X sin(n + 12 )x sin(k + 32 )x
=
+
sin x2
sin x2
sin x2
n=0
n=0
k+1
X
=
1 − cos(k + 1)x sin(k + 23 )x
+
1 − cos x
sin x2
[1 − cos(k + 1)x] sin x2 + (1 − cos x) sin(k + 32 )x
sin x2 (1 − cos x)
x
1
sin 2 − 2 sin(k + 23 )x − sin(k + 12 )x + sin(k + 32 )x − 21 sin(k + 25 )x + sin(k + 21 )x
=
sin x2 (1 − cos x)
sin x2 + 12 sin(k + 23 )x − sin(k + 52 )x
=
sin x2 (1 − cos x)
x
sin 2 − cos(k + 2)x sin x2
=
sin x2 (1 − cos x)
1 − cos(k + 2)x
.
=
1 − cos x
=
Thus the statement is true for N = k + 1 if it is true for N = k. By induction, the formula (8.44) holds
for every non-negative integer N . Hence we have
KN (x) =
1 − cos(N + 1)x
1
·
.
N +1
1 − cos x
(8.45)
(a) Since −1 ≤ cos x < 1 for every x ∈ R \ {0, ±2π, ±4π, . . .}, we have
KN (x) ≥ 0
for all x ∈ R \ {0, ±2π, ±4π, . . .}. If x ∈ {0, ±2π, ±4π, . . .}, then cos x = 1 and thus Theorem 5.13
(L’Hospital’s Rule) implies that
1
KN (x) = lim
t→x N + 1
·
1 − cos(N + 1)t
sin(N + 1)t
(N + 1) cos(N + 1)t
= lim
= lim
= N + 1.
t→x
t→x
1 − cos t
sin t
cos t
Hence we have
KN (x) ≥ 0
for every x ∈ R.
(b) By definition, we have
1
2π
Z π
−π
KN (x) dx =
1
2π
Z π h
−π
N
i
1 X
Dn (x) dx
N + 1 n=0
189
8.1. Problems related to special functions
N Z π
X
1
Dn (x) dx
=
2(N + 1)π n=0 −π
"Z
#
N Z π
π
X
1
=
D0 (x) dx +
Dn (x) dx
2(N + 1)π −π
n=1 −π
! #
"
n
N Z π
X
X
1
ikx
dx
e
2π +
=
2(N + 1)π
n=1 −π
k=−n
"
!#
Z π
n Z π
N
X
X
1
ikx
=
dx +
2π +
e dx
2(N + 1)π
−π
−π
n=1
"
N
X
1
=
2π +
2(N + 1)π
n=1
k=−n
k6=0
Z π
dx +
−π
n Z π
X
k=−n
k6=0
e
ikx
!#
dx
−π
!#
"
n
N
X
X
1
2π +
2 sin kx
=
2π +
2(N + 1)π
n=1
N
X
1
=
2π +
2π
2(N + 1)π
n=1
!
k=−n
k6=0
= 1.
(c) Let δ > 0. If δ ≤ |x| ≤ π, then we have cos x ≤ cos δ < 1 which implies that
1 − cos x ≥ 1 − cos δ > 0.
(8.46)
By the equivalent definition of (8.45), the inequality (8.46) and part (a), we have
0 ≤ KN (x) = |KN (x)| =
1
1
1 − cos(N + 1)x
1
≤
·
·
.
N +1
1 − cos x
N + 1 1 − cos δ
By [21, Eqn. (78), p. 189], we have
sN (f ; x) =
1
2π
Z π
−π
f (x − t)DN (t) dt
so that
N
1 X
sn (f ; x)
N + 1 n=0
Z π
N
1
1 X
=
Dn (t) dt
f (x − t)
2π −π
N + 1 n=0
Z π
1
=
f (x − t)KN (t) dt.
2π −π
σN (f ; x) =
(8.47)
Proof of Fejér’s theorem. Suppose that f : R → R is a continuous function with period 2π. We proceed
as in Theorem 7.26. Given that ǫ > 0, by the continuity of the function f , we choose a δ > 0 such that
|y − x| < δ implies
ǫ
(8.48)
|f (y) − f (x)| < .
2
Let M = sup |f (x)|. By properties (a), (b) and (c), we see that for x ∈ [−π, π],
x∈[−π,π]
|σn (f ; x) − f (x)| =
1
2π
Z π
−π
f (x − t)Kn (t) dt − f (x) ·
1
2π
Z π
−π
Kn (t) dt
Chapter 8. Some Special Functions
190
Z π
1
[f (x − t) − f (x)]Kn (t) dt
2π −π
Z π
1
|f (x − t) − f (x)|Kn (t) dt
≤
2π −π
Z −δ
Z δ
1
1
|f (x − t) − f (x)|Kn (t) dt +
|f (x − t) − f (x)|Kn (t) dt
≤
2π −π
2π −δ
Z π
1
|f (x − t) − f (x)|Kn (t) dt
+
2π δ
Z −δ
Z δ
Z π
1
ǫ 1
1
< 2M ·
Kn (t) dt + ·
Kn (t) dt + 2M ·
Kn (t) dt
2π −π
2 2π −δ
2π δ
2
ǫ
4M
·
+
≤
n + 1 1 − cos δ
2
<ǫ
=
for all large enough n, which proves Fejér’s theorem.
This completes the proof of the problem.
Problem 8.16
Rudin Chapter 8 Exercise 16.
Proof. We follow the argument as in the proof of Fejér’s theorem in Problem 8.15. Suppose that x is a
real number such that both f (x+) and f (x−) exist. Given that ǫ > 0. We have
σN (f ; x) −
1
f (x+) + f (x−)
=
2
2π
Z π h
f (x+) + f (x−) i
KN (t) dt .
f (x − t) −
2
−π
Now we may split the integral on the right-hand side of (8.49) into two parts:
Z π h
1
f (x+) + f (x−) i
f (x − t) −
KN (t) dt
2π −π
2
Z 0 h
f (x+) + f (x−) i
1
KN (t) dt
f (x − t) −
=
2π −π
2
Z πh
1
f (x+) + f (x−) i
+
KN (t) dt.
f (x − t) −
2π 0
2
(8.49)
(8.50)
Let t = −y. If t = −π, then y = π. If t = 0, then y = 0. Therefore, the first integral on the right-hand
side in the expression (8.50) can be rewritten asg
Z 0 h
Z 0h
f (x+) + f (x−) i
f (x+) + f (x−) i
1
1
KN (t) dt =
KN (−y)(− dy)
f (x − t) −
f (x + y) −
2π −π
2
2π π
2
Z πh
f (x+) + f (x−) i
1
KN (t) dt. (8.51)
f (x + t) −
=
2π 0
2
Substitute the expression (8.51) back into the expression (8.50) and apply property (a) in Problem 8.15,
we get
1
2π
g Note that K
Z π h
Z π
1
f (x+) + f (x−) i
KN (t) dt =
f (x − t) −
F (x, t)KN (t) dt
2
2π 0
−π
Z π
1
|F (x, t)|KN (t) dt,
≤
2π 0
N (x) is an even function in x.
(8.52)
191
8.1. Problems related to special functions
where F (x, t) = f (x + t) + f (x − t) − f (x+) − f (x−).
By definitions of f (x+) and f (x−), there exists a δ > 0 such that 0 < t < δ implies
|f (x − t) − f (x−)| <
ǫ
2
and |f (x + t) − f (x+)| <
ǫ
2
which show that
|F (x, t)| ≤ |f (x − t) − f (x−)| + |f (x + t) − f (x+)| < ǫ.
(8.53)
Now we deduce from property (b) of Problem 8.15 and the inequality (8.53) that
Z δ
Z δ
ǫ
ǫ
ǫ
1
·π = .
|F (x, t)|KN (t) dt <
KN (t) dt ≤
2π 0
2π 0
2π
2
Since f ∈ R on [a, b], f is bounded on [a, b]. Therefore, we may define M =
(8.54)
sup
x∈[−π,π]
|f (x)|. Then it
follows from property (c) of Problem 8.15 that
Z π
Z π
1
1
· 4M
KN (t) dt
|F (x, t)|KN (t) dt ≤
2π δ
2π
δ
Z π
1
2
2M
·
·
dt
≤
π N + 1 1 − cos δ δ
1
4M
<
·
N + 1 1 − cos δ
ǫ
<
2
(8.55)
for large enough N . Hence after putting the inequalities (8.54) and (8.55) into the inequality (8.52), we
deduce from the expression (8.49) that for large enough N , we have
Z π
f (x+) + f (x−)
1
σN (f ; x) −
≤
|F (x, t)|KN (t) dt
2
2π 0
Z δ
Z π
1
1
|F (x, t)|KN (t) dt +
|F (x, t)|KN (t) dt
=
2π 0
2π δ
ǫ
ǫ
< +
2 2
= ǫ.
In other words, we have the desired result that
lim σN (f ; x) =
N →∞
1
[f (x+) + f (x−)].
2
This completes the proof of the problem.
Problem 8.17
Rudin Chapter 8 Exercise 17.
Proof.
(a) Here we must assume that f ∈ R on [−π, π] so that the coefficients cn are well-defined. By
definition, we have
Z π
n
f (x)e−inx dx.
ncn =
2π −π
1 −inx ′
Since (− in
e
) = e−inx , we deduce from Problem 6.17 and then Theorem 6.13(b) that
|ncn | =
n
2π
Z π
−π
f (x)e−inx dx
Chapter 8. Some Special Functions
192
#
"
Z π
1
−inx
−inπ
inπ
= −
e
df
e
f (π) − e f (−π) −
2πi
−π
"
#
Z π
1
−inx
f (π) + f (−π) +
e
df
≤
2π
−π
"
#
Z π
1
−inx
≤
f (π) + f (−π) +
|e
| df
2π
−π
≤
f (π)
π
for every positive integer n. Hence {ncn } is a bounded sequence.
(b) For every positive integer N , let aN = sN (f ; x) − sN −1 (f ; x). Then we have
aN =
N
X
n=−N
cn einx −
N
−1
X
n=−(N −1)
cn einx = c−N e−iN x + cN eiN x
so that
|N aN | ≤ |N c−N | + |N cN |.
By part (a), {N cN } is a bounded sequence which implies that there is M < ∞ such that |N aN | ≤ M
for all positive integers N . Since f is monotonic on [−π, π), Theorem 4.29 guarantees that f (x+)
and f (x−) exist for every x ∈ [−π, π). By Problem 8.16, we have
lim σN (f ; x) =
N →∞
1
[f (x+) + f (x−)].
2
Therefore, Problem 3.14(e) shows that
lim sN (f ; x) =
N →∞
1
[f (x+) + f (x−)]
2
for every x ∈ [−π, π).
(c) Since f ∈ R on [−π, π], it is a bounded function on [−π, π] so that f (α) and f (β) are finite. Define
a real function g : [−π, π] → R as follows:

 f (α), if x ∈ [−π, α];
f (x), if x ∈ (α, β);
g(x) =
(8.56)

f (β), if x ∈ [β, π].
Next, we define h : [−π, π] → R by
h(x) = f (x) − g(x).
Since h(x) = 0 for all x ∈ (α, β), the corollary of Theorem 8.14 (Localization Theorem) implies that
lim sN (h; x) = 0
N →∞
(8.57)
for every x ∈ (α, β). Since sN (h; x) = sN (f ; x) − sN (g; x) for every x ∈ (α, β), the limit (8.57) can
be rewritten as
lim sN (f ; x) = lim sN (g; x)
(8.58)
N →∞
N →∞
on (α, β).
If g is monotonic on [−π, π], then Theorem 6.9 implies that g ∈ R on [−π, π].h Since it is bounded
and monotonic on [−π, π], part (b) implies that
lim sN (g; x) =
N →∞
1
[g(x+) + g(x−)]
2
h Therefore we can apply the pointwise version of Fejér’s theorem (Problem 8.16) to g.
(8.59)
193
8.1. Problems related to special functions
for every x ∈ [−π, π]. Thus we substitute the expression (8.59) into the limit relation (8.58) to get
lim sN (f ; x) =
N →∞
1
1
[g(x+) + g(x−)] = [f (x+) + f (x−)]
2
2
(8.60)
for every x ∈ (α, β).
If g is not monotonic on [−π, π], then we may modify the definition (8.56) as

f (α)


,
if x ∈ [−π, α];



 M
g(x) =
f (x),
if x ∈ (α, β);






M f (β), if x ∈ [β, π],
where M is a positive constant such that
f (α)
< f (x) < M f (β)
M
for all x ∈ (α, β). In this case, this modified function g will be monotonic on [−π, π] and the above
argument can be repeated to show that the expression (8.60) still holds on (α, β).
This completes the proof of the problem.
Problem 8.18
Rudin Chapter 8 Exercise 18.
Proof. Since tan x → +∞ as x → π2 and x ∈ (0, π2 ), we have f (x) < 0 and g(x) < 0 as x → π2 and
x ∈ (0, π2 ). Thus it is reasonable to conjecture that
f (x) < 0 and g(x) < 0
on (0, π2 ). By applying [21, Eqn. (49), p. 182] repeatedly, it is clear that
f ′ (x) = 3x2 − (tan2 x + 2 sin2 x)
f ′′ (x) = 6x − 2(tan x + tan3 x + sin 2x)
f ′′′ (x) = 6 − 2(sec2 x + 3 tan2 x sec2 x + 2 cos 2x)
= 6 − 2(3 tan4 x + 4 tan2 x + 1 + 2 cos 2x)
f (4) (x) = −2(12 tan x3 sec2 x + 8 tan x sec2 x − 4 sin 2x)
= −8(3 tan3 x + 5 tan3 x + 2 tan x − sin 2x)
= −8(3 tan3 x + 5 tan3 x + 2 tan x sin2 x)
= −8 tan x(3 tan4 x + 5 tan2 x + 2 sin2 x).
(8.61)
Furthermore, it is easy to check that
f (0) = f ′ (0) = f ′′ (0) = f ′′′ (0) = f (4) (0) = 0.
(8.62)
Since tan x and sin x are positive on (0, π2 ), the expression (8.61) yields that
f (4) (x) < 0
for all x ∈ (0, π2 ). By Problem 5.2, we have f ′′′ (x) is strictly decreasing in (0, π2 ). Thus we follow from
the values (8.62) that f ′′′ (x) < 0 for all x ∈ (0, π2 ). By repeating this kind of argument, we can show that
f ′′ (x) < 0,
f ′ (x) < 0
Chapter 8. Some Special Functions
194
and then
f (x) < 0
for all x ∈ (0, π2 ), as desired.
Similarly, we have
g ′ (x) = 4x − (sin 2x + tan x + x + x tan2 x)
g ′′ (x) = 4 − (2 cos 2x + 2 + 2 tan2 x + 2x tan x + 2x tan3 x)
g ′′′ (x) = −(−4 sin 2x + 6 tan x + 6 tan3 x + 2x + 8x tan2 x + 6x tan4 x)
g (4) (x) = −(−8 cos 2x + 6 sec2 x + 18 tan2 x sec2 x + 2
+ 8 tan2 x + 16x tan x sec2 x + 6 tan4 x + 24x tan3 x sec2 x)
= −(−8 cos 2x + 8 + 32 tan2 x + 24 tan4 x + 16x tan x
3
(8.63)
5
+ 40x tan x + 24x tan x)
g
(5)
(x) = −(16 sin 2x + 16x + 80 tan x + 200 tan3 x + 120 tan5 x
+ 136x tan2 x + 240x tan4 x + 120x tan6 x).
(8.64)
g(0) = g ′ (0) = g ′′ (0) = g ′′′ (0) = g (4) (0) = g (5) (0) = 0.
(8.65)
In addition, we also have
Since tan x and sin 2x are clearly positive on (0, π2 ), the expression (8.63) yields that
g (5) (x) < 0
for all x ∈ (0, π2 ). Thus, by using similar argument as above and the values (8.65), we obtain that
g (4) (x) < 0,
g ′′′ (x) < 0,
g ′′ (x) < 0,
g ′ (x) < 0
and then
g(x) < 0
for all x ∈ (0, π2 ), completing the proof of the problem.
Problem 8.19
Rudin Chapter 8 Exercise 19.
Proof. Suppose that f (x) = eikx . If k = 0, then we have f (x) = 1 which means that
N
N
1 X
1 X
f (x + nα) = lim
1 = 1 and
N →∞ N
N →∞ N
n=1
n=1
lim
1
2π
Z π
f (t) dt =
−π
1
2π
Z π
dt = 1.
−π
Therefore, we have the desired result in this special case. Next, suppose that k 6= 0. Since α
π is irrational,
kα cannot be a multiple of π. In other words, we have eiknα 6= 1 for every positive integer n. By this, we
have
N
N
1 X ik(x+nα)
1 X
f (x + nα) = lim
e
N →∞ N
N →∞ N
n=1
n=1
lim
N
=e
ikx
1 X iknα
lim
e
N →∞ N
n=1
1 ikα 1 − eikN α
·e
·
N →∞ N
1 − eikα
= eikx lim
=0
195
and
8.1. Problems related to special functions
1
2π
Z π
1
f (t) dt =
2π
−π
Z π
eikt dt =
−π
1 eikπ − e−ikπ sin kπ
= 0.
=
2π
ik
kπ
Therefore, we also have the desired result in this case.
For the general case, since f is continuous with period 2π, we obtain from Theorem 8.15 (Stone–Weierstrass
theorem) that for every ǫ > 0, there is a trigonometric polynomial P such that
|P (x) − f (x)| <
ǫ
3
(8.66)
for all real x. Therefore, we have
N
N
1 X
1 X
lim
f (x + nα) = lim
{[f (x + nα) − P (x + nα)] + P (x + nα)}
N →∞ N
N →∞ N
n=1
n=1
N
N
1 X
1 X
[f (x + nα) − P (x + nα)] + lim
P (x + nα)
N →∞ N
N →∞ N
n=1
n=1
= lim
(8.67)
and
1
2π
Z π
Z π
1
{[f (t) − P (t)] + P (t)} dt
2π −π
Z π
Z π
1
1
[f (t) − P (t)] dt +
P (t) dt.
=
2π −π
2π −π
f (t) dt =
−π
(8.68)
By Definition 8.9, we have
P (x) =
N
X
cn einx ,
n=−N
where cn (−N ≤ n ≤ N ) are some constants. Thus the analysis for the special case f (x) = eikx implies
that
Z π
N
1 X
1
lim
P (x + nα) =
P (t) dt
N →∞ N
2π −π
n=1
for every x. In other words, given ǫ > 0, there exists an integer N such that
N
1 X
1
P (x + nα) −
N n=1
2π
Z π
−π
P (t) dt <
ǫ
3
(8.69)
for all n ≥ N .
Therefore, we deduce from the expressions (8.67), (8.68) and the inequality (8.69) with the integer N
defined in the inequality (8.69) that for all n ≥ N , we have
N
1 X
1
f (x + nα) −
N n=1
2π
Z π
−π
f (t) dt
Z π
N
N
1 X
1
1 X
P (t) dt
=
[f (x + nα) − P (x + nα)] +
P (x + nα) −
N n=1
N n=1
2π −π
Z π
1
−
[f (t) − P (t)] dt
2π −π
Z π
N
N
1
1 X
1 X
[f (x + nα) − P (x + nα)] +
P (x + nα) −
P (t) dt
≤
N n=1
N n=1
2π −π
Z π
1
+
[f (t) − P (t)] dt
2π −π
Chapter 8. Some Special Functions
196
N
ǫ
1 X
1
<
|f (x + nα) − P (x + nα)| + +
N n=1
3 2π
Z π
−π
|f (t) − P (t)| dt.
(8.70)
Finally, we apply the inequality (8.66) to the inequality (8.70), so we establish
N
1 X
1
f (x + nα) −
N n=1
2π
Z π
−π
f (t) dt <
ǫ
ǫ
ǫ
+ + = ǫ.
3 3 3
Hence we have the desired result and this completes the proof of the problem.
Problem 8.20
Rudin Chapter 8 Exercise 20.
Proof. For examples, if m = 1, 2, 3, 4, then we have

(x − 1) log 2,
if 1 ≤ x ≤ 2;








 (3 − x) log 2 + (x − 2) log 3, if 2 ≤ x ≤ 3;
f (x) =


(4 − x) log 3 + (x − 3) log 4, if 3 ≤ x ≤ 4;







(5 − x) log 4 + (x − 4) log 5, if 4 ≤ x ≤ 5

(x − 1) log 2,
if 1 ≤ x ≤ 2;








3
23



log
x + log 2 , if 2 ≤ x ≤ 3;


2
3

=

34
4


x
+
log
log
, if 3 ≤ x ≤ 4;


3
43






5


 log 5 x + log 4 , if 4 ≤ x ≤ 5
4
4
5
and
g(x) =
Therefore, their graphs are given by

x − 1,
if 21 ≤ x < 32 ;






 x


− 1 + log 2, if 23 ≤ x < 52 ;


 2
x


− 1 + log 3, if 25 ≤ x < 72 ;


3







 x − 1 + log 4, if 7 ≤ x < 9 .
2
2
4
197
8.1. Problems related to special functions
Figure 8.2: The graphs of the two functions f and g.
Now for a positive integer n, we have
Z n
f (x) dx =
1
=
=
n−1
X Z m+1
m=1 m
n−1
X Z m+1
m=1
n−1
X
m
f (x) dx
[(m + 1 − x) log m + (x − m) log(m + 1)] dx
1
[log m + log(m + 1)]
2
m=1
1
1
log(n − 1)! + log(n!)
2
2
1
= log(n!) − log n.
2
=
(8.71)
Chapter 8. Some Special Functions
198
Furthermore, we have
Z n
g(x) dx =
1
Z 32
g(x) dx +
Z n
g(x) dx +
n− 21
1
n−1
X Z m+ 1
2
g(x) dx
m− 21
m=2
n−1
X Z m+ 12 x
1
1
1
+ log n +
− 1 + log m dx
= −
1
8 8n 2
m
m=2 m− 2
=
n−1
X
1
1
1
−
+ log n +
log m
8 8n 2
m=2
1
1
1
−
+ log n + log(n − 1)!
8 8n 2
1
1
1
= log(n!) − log n + −
.
2
8 8n
=
(8.72)
Combining the expressions (8.71) and (8.72), we have for x ≥ 1,
Z n
Z n
1
1
g(x) dx.
f (x) dx = log(n!) − log n > − +
2
8
1
1
By Theorem 6.22, we have
Z n
Z n
Z n
log x dx = n log n − 1 · log 1 −
x d(log x) = n log n −
dx = n log n − n + 1.
1
1
(8.73)
(8.74)
1
Since f (x) ≤ log x ≤ g(x) if x ≥ 1, we follow from the inequality (8.73) and the expression (8.74) that
Z n
Z n
Z n
f (x) dx ≤
log x dx ≤
g(x) dx
1
1
1
1
1
1
log(n!) − log n ≤ n log n − n + 1 < log(n!) − log n +
2
2
8
which deduce the inequalities
7
1
log n + n < 1
< log(n!) − n +
8
2
(8.75)
for n = 2, 3, 4, . . .. Hence, by taking exponential to each part of the inequalities (8.75), we have eventually
the desired formula
7
n!
e 8 < n n √ < e.
(e) n
This completes the proof of the problem.
Problem 8.21
Rudin Chapter 8 Exercise 21.
Proof. Since |Dn (t)| is an even function on [−π, π], Theorem 6.12(c) and Theorem 6.19 give
"Z
#
Z π
Z
Z
0
1
1 π
1 π sin(n + 21 )t
Ln =
|Dn (t)| dt +
|Dn (t)| dt =
|Dn (t)| dt =
dt.
2π −π
π 0
π 0
sin 2t
0
(8.76)
By Theorem 6.19 again with t = 2x, the expression (8.76) implies that
2
Ln =
π
Z π2
0
sin(2n + 1)x
dx.
sin x
We know that 0 < sin x ≤ x on (0, π2 ], see Figure 8.3 for a geometric proof of this.
(8.77)
199
8.1. Problems related to special functions
Figure 8.3: A geometric proof of 0 < sin x ≤ x on (0, π2 ].
Thus we can deduce from the formula (8.77) that
Ln ≥
2
π
Z π2
sin(2n + 1)x
dx.
x
0
(8.78)
Apply Theorem 6.19 with y = (2n + 1)x to the inequality (8.78), we yield
Ln ≥
2
π
Z (2n+1) π2
0
2
sin y
dy ≥
y
π
It is clear that
Z nπ
0
n−1 Z (k+1)π
2X
sin y
dy =
y
π
k=0
kπ
sin y
dy.
y
(8.79)
1
1
≥
y
(k + 1)π
on [kπ, (k + 1)π], where k = 0, 1, . . . , n − 1. Therefore, the inequality (8.79) implies that
n−1
Ln ≥
2 X 1
π2
k+1
k=0
Z (k+1)π
kπ
For each k ∈ {0, 1, 2, . . . , n − 1}, we acquire
Z (k+1)π
kπ
| sin x| dx =
Z π
| sin y| dy.
(8.80)
sin x dx,
(8.81)
0
see Figure 8.4 for this geometric meaning.
Figure 8.4: The graph of y = | sin x|.
Chapter 8. Some Special Functions
200
By substituting the expression (8.81) into the inequality (8.80) and applying Problem 8.9(a), we obtain
n−1
2 X 1
π2
k+1
Ln ≥
k=0
Z π
n
4
4 X1
≥ 2 log n.
π2
k
π
sin x dx =
0
k=1
This finishes the analysis of the problem.
Problem 8.22
Rudin Chapter 8 Exercise 22.
Proof. We follow the given hint. Let
f (x) = 1 +
∞
X
α(α − 1) · · · (α − n + 1)
n!
n=1
xn .
(8.82)
Since
lim sup
n→∞
n!
α−n
α(α − 1) · · · (α − n)
|x| = |x|,
×
x = lim
n→∞
(n + 1)!
α(α − 1) · · · (α − n + 1)
n+1
we follow from Theorem 3.34(a) that the series converges if |x| < 1. By Theorem 8.1, the function defined
by the series (8.82) is differentiable in (−1, 1) and
f ′ (x) =
∞
X
α(α − 1) · · · (α − n + 1)
(n − 1)!
n=1
xn−1 .
(8.83)
Thus we follow from the definition (8.82) and the derivative (8.83) that
(1 + x)f ′ (x) = (1 + x)
∞
X
α(α − 1) · · · (α − n + 1)
n=1
=
(n − 1)!
∞
X
α(α − 1) · · · (α − n + 1)
(n − 1)!
n=1
=α+
=α+
=α+
xn−1 +
∞
X
α(α − 1) · · · (α − n + 1)
n=2
∞
X
(n − 1)!
xn−1
∞
X
α(α − 1) · · · (α − n + 1)
(n − 1)!
n=1
n−1
x
+
xn
∞
X
α(α − 1) · · · (α − n + 1)
n=1
(n − 1)!
∞
α(α − 1) · · · (α − n) n X α(α − 1) · · · (α − n + 1) n
x +
x
n!
(n − 1)!
n=1
n=1
∞
X
α(α − 1) · · · (α − n + 1) α − n
n=1
∞
X
=α+α
(n − 1)!
n
+ 1 xn
α(α − 1) · · · (α − n + 1) n
x
n!
n=1
= αf (x).
By [21, Eqn. (38), p. 180] and Theorem 5.5, we have
f ′ (x)
d
(log f (x)) =
dx
f (x)
d
α
(log f (x)) =
dx
1+x
Z x
α
dt
log f (x) − log f (0) =
0 1+t
log f (x) = α log(1 + x)
xn
201
8.2. Index of a curve
f (x) = (1 + x)α .
(8.84)
Hence, by the definition (8.82) and the expression (8.84), we have the desired result that
α
(1 + x) = 1 +
∞
X
α(α − 1) · · · (α − n + 1)
n!
n=1
xn .
(8.85)
By replacing x and α by −x and −α respectively in the expression (8.85), we have
(1 − x)−α = 1 +
=1+
∞
X
(−α)(−α − 1) · · · (−α − n + 1)
n!
n=1
∞
X
(−x)n
α(α + 1) · · · (α + n − 1) n
x .
n!
n=1
(8.86)
In particular, we suppose that −1 < x < 1 and α > 0 in the equation (8.86). By Theorem 8.18(a), we
know that
Γ((n + α − 1) + 1)
Γ(n + α)
=
Γ(α)
Γ(α)
(α + n − 1)Γ(α + n − 1)
=
Γ(α)
= ···
(α + n − 1) · · · (α + 1)αΓ(α)
Γ(α)
= (α + n − 1) · · · (α + 1)α.
=
(8.87)
By substituting the identity (8.87) into the expression (8.86), we get
(1 − x)−α = 1 +
∞
X
Γ(n + α)
n=1
n!Γ(α)
xn =
∞
X
Γ(n + α)
n=0
n!Γ(α)
where −1 < x < 1 and α > 0. This completes the proof of the problem.
8.2
xn ,
Index of a curve
Problem 8.23
Rudin Chapter 8 Exercise 23.
Proof. We follow the given hint. Since γ is continuously differentiable and γ(t) 6= 0 on [a, b], the function
γ′
γ
′
is well-defined and continuous on [a, b]. Thus γγ ∈ R on [a, b] by Theorem 6.8 and we can define
ϕ : [a, b] → C to be the function given by
Z x ′
γ (t)
dt.
(8.88)
ϕ(x) =
a γ(t)
By Theorem 6.20 (First Fundamental Theorem of Calculus), the function ϕ is differentiable on [a, b]
and
γ ′ (t)
ϕ′ (t) =
.
(8.89)
γ(t)
Chapter 8. Some Special Functions
202
Furthermore, we have ϕ(a) = 0. Let f (t) = γ(t)e−ϕ(t) , where t ∈ [a, b]. It is clear that f is differentiable
on [a, b] and we deduce from the expression (8.89) that
f ′ (t) = γ ′ (t)e−ϕ(t) + γ(t)[−ϕ′ (t)]e−ϕ(t) = γ ′ (t)e−ϕ(t) − γ ′ (t)e−ϕ(t) = 0
for all t ∈ (a, b). By Theorem 5.11(b), f is a constant on (a, b). Since f is continuous on [a, b], it must
be a constant on [a, b]. Since γ(a) = γ(b), we have
eϕ(b) = eϕ(a) = e0 = 1.
(8.90)
By [21, Eqn. (52), p. 183], Theorem 8.7(a) and (c), the expression (8.90) implies that
ϕ(b) = 2nπi
for some integer n. We note that ϕ(b) = 2πiInd (γ), so we have
2πiInd (γ) = 2nπi
which is equivalent to
Ind (γ) = n
for some integer n. This proves the first assertion.
For the second assertion, if γ(t) = eint , a = 0 and b = 2π, then we have
Z 2π
Z 2π
1
n
ineint
Ind (γ) =
dt
=
dt = n.
2πi 0
eint
2π 0
Now the number Ind (γ) is called the winding number of γ around 0 because it counts the total number
of times that the curve γ travels counterclockwise around the origin 0. This number certainly depends on
the orientation of the curve, so it is negative if the curve travels around the point clockwise. See Figure
8.5 about the winding number of γ around an arbitrary point p.
Figure 8.5: The winding number of γ around an arbitrary point p.
We end the analysis of the problem here.
Problem 8.24
Rudin Chapter 8 Exercise 24.
Proof. We follow the given hint. Let 0 ≤ c < ∞. It is clear that γ + c : [a, b] → C is still a continuously
differentiable closed curve. Since γ does not intersect the negative real axis, we must have
γ(t) + c 6= 0
203
8.2. Index of a curve
for every t ∈ [a, b]. Thus γ + c satisfies all the hypotheses of Problem 8.23 and it is meaningful to talk
about Ind (γ + c).
We define f : [0, ∞) → Z by
f (c) = Ind (γ + c) =
1
2πi
Z b
γ ′ (t)
dt,
a γ(t) + c
where c ∈ [0, ∞). We claim that f is a continuous function of c. To this end, we check the definition of
continuity (Definition 4.5). Given that ǫ > 0 and c ∈ [0, ∞). It is easy to see that
#
Z b" ′
1
γ ′ (t)
γ (t)
|f (x) − f (c)| =
dt
−
2πi a γ(t) + x γ(t) + c
Z b
1
γ ′ (t)
γ ′ (t)
≤
−
dt
2π a γ(t) + x γ(t) + c
Z b
1
γ ′ (t)(c − x)
≤
dt.
(8.91)
2π a (γ(t) + x)(γ(t) + c)
Since γ and γ ′ are continuous on [a, b], we know from Theorem 4.16 (Extreme Value Theorem) that there
exist real numbers m, m′ , M and M ′ such that
0 < m ≤ |γ(t)| ≤ M
and m′ ≤ |γ ′ (t)| ≤ M ′
(8.92)
for all t ∈ [a, b].i
If M ′ = 0, then γ ′ (t) = 0 for all t ∈ [a, b] which imply that γ = A for some constant A on (a, b). Since
γ is continuous on [a, b], we have γ = A on [a, b]. However, γ cannot be a closed curve in this case, a
contradiction. Thus we must have
M ′ > 0.
Furthermore, we note that c > 0 and x ≥ 0 imply that
m+c>m
and m + x ≥ m,
so we deduce easily from the inequalities (8.91) and (8.92) that
1
|f (x) − f (c)| ≤
2π
Z b
M ′ (b − a)
M′
|c − x|.
|c − x| dt <
2πm2
a (m + c)(m + x)
(8.93)
2
If we take δ = M2πm
′ (b−a) ǫ > 0, then we obtain from the inequality (8.93) that
|f (x) − f (x)| < ǫ
for all x ∈ [0, ∞) with |x − c| < δ. Hence we prove our claim that f is continuous on [0, ∞).
Next, the definition of f and the inequalities (8.92) also imply that
|f (c)| ≤
1
2π
Z b
a
M ′ (b − a)
γ ′ (t)
.
dt ≤
γ(t) + c
2π(m + c)
Thus we have
lim |f (c)| = 0.
c→∞
Since the range of f is Z, we have f (c) = 0 for all c ∈ [0, ∞). In particular, we take c = 0 to get
f (0) = Ind (γ) = 0,
as required. This completes the proof of the problem.
i Note that it is possible that m′ = 0, but m > 0 because γ(t) 6= 0 on [a, b].
Chapter 8. Some Special Functions
204
Problem 8.25
Rudin Chapter 8 Exercise 25.
Proof. We follow the given hint. Let γ = γγ21 . Since γ1 : [a, b] → C and γ2 : [a, b] → C are continuously
differentiable closed curves and γ1 (t)γ2 (t) 6= 0 for every t ∈ [a, b], the function γ : [a, b] → C is also a
continuously differentiable closed curve and γ(t) 6= 0 for every t ∈ [a, b].
Now the inequality |γ1 (t) − γ2 (t)| < |γ1 (t)| shows that
|1 − γ(t)| < 1
on [a, b], so we have 0 < γ(t) < 2 for all t ∈ [a, b]. Next, by Problem 8.24, we have Ind (γ) = 0. In
addition, a direct computation shows that
γ′
γ′
γ′
= 2− 1
γ
γ2
γ1
which gives
Ind (γ) =
1
2πi
Z b
1
γ ′ (t)
dt =
2πi
a γ(t)
Z b
a
1
γ2′ (t)
dt −
γ2 (t)
2πi
Z b
γ1′ (t)
dt = Ind (γ2 ) − Ind (γ1 ).
a γ1 (t)
Hence we have
Ind (γ1 ) = Ind (γ2 )
as required, completing the proof of the problem.
Problem 8.26
Rudin Chapter 8 Exercise 26.
Proof. For all t ∈ [0, 2π], we obtain from the triangle inequality that
δ < |γ(t)| ≤ |γ(t) − P1 (t)| + |P1 (t)| <
so that
|P1 (t)| >
δ
+ |P1 (t)|
4
3δ
δ
> >0
4
2
on [0, 2π]. By this and the triangle inequality, we have
|P1 (t) − P2 (t)| = |P1 (t) − γ(t) + γ(t) − P2 (t)| ≤ |P1 (t) − γ(t)| + |P2 (t) − γ(t)| <
δ
< |P1 (t)|
2
for all t ∈ [0, 2π]. Therefore, it follows from Problem 8.25 that
Ind (P1 ) = Ind (P2 ).
Define this common value to be Ind (γ). In other words, the winding number of the closed curve γ can
be defined in terms of that of any trigonometric polynomials P (t) satisfying the inequality
|P (t) − γ(t)| <
on [0, 2π].
δ
4
205
8.2. Index of a curve
• Extension of Problem 8.24. Suppose that the range of γ does not intersect the negative real axis.
Then there exists a η > 0 such that |γ(t) − x| > η for all t ∈ [0, 2π] and x ≤ 0. Put κ = min(δ, η).
Then both
|γ(t)| > κ and |γ(t) − x| > κ
are valid on [0, 2π] and on x ≤ 0. If P (t) is a trigonometric polynomial such that
κ
|P (t) − γ(t)| < ,
4
then the triangle inequality implies that
|γ(t) − x| ≤ |γ(t) − P (t)| + |P (t) − x|
and then it gives
κ
3κ
=
>0
4
4
for all t ∈ [0, 2π]. Therefore, the range of P (t) does not intersect the negative real axis. By Problem
8.24, we have Ind (P ) = 0. By definition, we have
|P (t) − x| ≥ |γ(t) − x| − |γ(t) − P (t)| > κ −
Ind (γ) = 0
which extends the result of Problem 8.24 to any closed curve (not necessarily differentiable) in C
with domain [0, 2π] and γ(t) 6= 0 for every t ∈ [0, 2π].
• Extension of Problem 8.25. Suppose that γ1 and γ2 are two closed curves in C with domain
[0, 2π], and γ1 (t)γ2 (t) 6= 0 on [0, 2π]. Then there exist δ1 > 0 and δ2 > 0 such that
|γ1 (t)| > δ1
and |γ2 (t)| > δ2
on [0, 2π]. By Theorem 8.15 (Stone–Weierstrass theorem), there exists trigonometric polynomials
P1 and P2 such that
δ2
δ1
and |γ2 (t) − P2 (t)| <
|γ1 (t) − P1 (t)| <
4
4
on [0, 2π]. Furthermore, we suppose that
|γ1 (t) − γ2 (t)| < |γ1 (t)|
for every t ∈ [0, 2π]. Thus there exists a δ3 > 0 such that
|γ1 (t)| − |γ1 (t) − γ2 (t)| > δ3
for all t ∈ [0, 2π]. Let δ = min(δ1 , δ2 , δ3 ). Then we have
δ
δ
, |γ2 (t) − P2 (t)| <
and |γ1 (t)| − |γ1 (t) − γ2 (t)| > δ
4
4
for all t ∈ [0, 2π]. By these and the triangle inequality, we have
|γ1 (t) − P1 (t)| <
|P1 (t) − P2 (t)| ≤ |P1 (t) − γ1 (t)| + |γ1 (t) − γ2 (t)| + |γ2 (t) − P2 (t)|
δ
δ
< + + |γ1 (t)| − δ
4 4
δ
= |γ1 (t)| −
2
δ
≤ |γ1 (t) − P1 (t)| + |P1 (t)| −
2
δ
δ
< |P1 (t)| + −
4 2
< |P1 (t)|
for every t ∈ [0, 2π]. By Problem 8.25, we have Ind (P1 ) = Ind (P2 ) which implies that
Ind (γ1 ) = Ind (γ2 ).
This definitely extends the result of Problem 8.25 to any closed curves γ1 and γ2 (not necessarily
differentiable) in C with domain [0, 2π] and γ1 (t)γ2 (t) 6= 0 for every t ∈ [0, 2π].
Chapter 8. Some Special Functions
206
This completes the proof of the problem.
Problem 8.27
Rudin Chapter 8 Exercise 27.
Proof. Assume that f (z) 6= 0 for all z ∈ C. Define γr (t) = f (reit ) for 0 ≤ r < ∞ and 0 ≤ t ≤ 2π.
(a) Since γ0 (t) = f (0) 6= 0 for all t ∈ [0, 2π], we have γ0′ (t) = 0 and thus
Ind (γ0 ) =
1
2πi
Z b
a
γ0′ (t)
dt = 0.
γ0 (t)
(b) Since
lim z −n f (z) = c,
|z|→∞
for every ǫ > 0 with ǫ < |c| there exists a R > 0 such that
|z −n f (z) − c| < ǫ
for all |z| ≥ R which is equivalent to
|γr (t) − crn eint | = |f (reit ) − crn eint | < ǫ|rn eint | < |crn eint |
(8.94)
for all r ≥ R and 0 ≤ t ≤ 2π. Now we apply Problem 8.25 to the inequality (8.94) and then the
definition, we get
Ind (γr (t)) = Ind (crn eint ) =
1
2πi
Z 2π
0
cinrn eint
dt = n.
crn eint
Therefore, we have Ind (γr (t)) = n for all sufficiently large r.
(c) Let p, r ∈ [0, ∞). Define d = |p − r| ≥ 0 and I(p, d) = [min(0, p − d), p + d]. Now we want to show
that for every ǫ > 0, there exists a δ > 0 such that r ∈ I(p, d) and |p − r| < δ imply thatj
|Ind (γp ) − Ind (γr )| < ǫ.
(8.95)
Next, we define the set
K(r, d) = {aeit | a ∈ I(r, d), 0 ≤ t ≤ 2π}.
Since f : C → C is a continuous function and the set K(p, 0) is compact, Theorem 4.16 (Extreme
Value Theorem) ensures that the value
m = min |f (peit )|
t∈[0,2π]
is finite.
Given that ǫ > 0 with m > ǫ. By Theorem 4.19, f is uniformly continuous on K(p, d). Thus there
exists a δ > 0 such that for all z1 , z2 ∈ K(p, d) with |z1 − z2 | < δ implies
|f (z1 ) − f (z2 )| < ǫ.
(8.96)
In particular, we may assume that z1 = peit and z2 = reit , where p is fixed and r varies. Thus the
inequality (8.96) implies that
|f (peit ) − f (reit )| < ǫ < m ≤ |f (peit )|
j The introduction of the set I(p, d) makes sure that if r ∈ I(p, d), then r ≥ 0.
(8.97)
207
8.2. Index of a curve
for all r ∈ I(p, d) with |p − r| < δ and 0 ≤ t ≤ 2π. By definition, the inequality (8.97) can be
rewritten as
|γp (t) − γr (t)| < |γp (t)|
for all r ∈ I(p, d) with |p − r| < δ and 0 ≤ t ≤ 2π. By Problem 8.25 or Problem 8.26, we have
Ind (γp ) = Ind (γr )
(8.98)
for all r ∈ I(p, d) with |p − r| < δ. In other words, the identity (8.98) means that the inequality
(8.94) holds, i.e., the function Ind (γp ) is continuous at p ∈ [0, ∞).
Recall from Problem 8.23 that Ind (γr ) ∈ Z. Since [0, ∞) is connected by Theorem 2.47, we can see from
Theorem 4.22 and part (c) that
Ind (γr )([0, ∞))
is also connected. By part (b), we know that Ind (γr ) = n for all sufficiently large r and thus we must
have Ind (γr ) = n for every r ∈ [0, ∞) and for some n ∈ Z. In particular, we have Ind (γ0 ) = n > 0 which
contradicts the result of (a). Hence we have f (z) = 0 for at least one complex number z. This completes
the proof of the problem.
Problem 8.28
Rudin Chapter 8 Exercise 28.
Proof. For 0 ≤ r ≤ 1, 0 ≤ t ≤ 2π, we put
γr (t) = g(reit ) and ψ(t) = e−it γ1 (t).
(8.99)
Assume that g(z) 6= −z for every z ∈ T . Then we have
ψ(t) 6= −1
(8.100)
for every t ∈ [0, 2π]. Since |g(z)| = 1 for every z ∈ D, we have
|ψ(t)| = |e−it γ1 (t)| = |e−it g(eit )| = 1 6= 0
(8.101)
for every t ∈ [0, 2π], i.e., ψ maps [0, 2π] into the unit circle T . By these two facts (8.100) and (8.101), we
have the result that the range of ψ does not intersect the negative real axis. Furthermore, we have
ψ(0) = γ1 (0) = g(1) and ψ(2π) = e−2πi γ1 (2π) = g(e2πi ) = g(1)
so that it is a closed curve. In conclusion, the curve ψ satisfies the hypotheses of Problem 8.24 or Problem
8.26. Therefore, we must have
Ind (ψ) = 0.
(8.102)
By a similar argument as in Problem 8.27(c), we know that Ind (γr ) is a continuous function of r, on
[0, 1]. Since [0, 1] is connected, Theorem 4.22 implies that Ind (γr )([0, 1]) is also connected. Recall that
Ind (γr ) ∈ Z, so it must be a fixed integer for all r ∈ [0, 1]. To derive a contradiction from this result, we
compute the values of Ind (γ0 ) and Ind (γ1 ).
We note that γ0 (t) = g(0) 6= 0 which is a non-zero constant, the definition of the winding number of
a closed curve (see Problem 8.23) gives
Z 2π
0
1
dt = 0.
Ind (γ0 ) =
2πi 0 g(0)
To find Ind (γ1 ), we need a lemma first.
Lemma 8.1
Let α, β : [a, b] → C be closed curves, α(t) 6= 0 and β(t) 6= 0 for every t ∈ [a, b]. Let
γ : [a, b] → C be defined by γ = α × β. Then we have Ind (γ) = Ind (α) + Ind (β).
Chapter 8. Some Special Functions
208
Proof of Lemma 8.1. We suppose that α and β are continuously differentiable. Then it is easy
to check that γ is also a continuously differentiable closed curve and γ(t) 6= 0 for every t ∈ [a, b].
By definition and the fact that γ ′ = α′ β + β ′ α, we have
Ind (γ) =
1
2πi
Z b
1
γ′
dt =
2πi
a γ
Z b
1
α′
dt +
2πi
a α
Z b
β′
dt = Ind (α) + Ind (β).
a β
Next, suppose that α and β are not differentiable so that γ may not be differentiable. However,
the numbers Ind (α), Ind (β) and Ind (γ) are still well-defined by Problem 8.26. Let
M1 = max |α(t)|
t∈[0,2π]
and M2 = max |β(t)|.
t∈[0,2π]
Since α, β and γ are non-zero on [0, 2π], there exists a small δ > 0 such that
|α(t)| > δ,
|β(t)| > δ
and |γ(t)| > δ
for every t ∈ [0, 2π].
• Case (i): M1 + M2 ≥ 1. By Theorem 8.15 (Stone–Weierstrass theorem), there are
trigonometric polynomials P1 and P2 such that
|P1 (t) − α(t)| <
δ
δ
<
8(M1 + M2 )
4
and |P2 (t) − β(t)| <
δ
δ
<
8(M1 + M2 )
4
(8.103)
for every t ∈ [0, 2π]. By Problem 8.26, we have Ind (α) = Ind (P1 ) and Ind (β) = Ind (P2 ),
so we derive from the triangle inequality and the inequalities (8.103) that
|P1 (t)P2 (t) − γ(t)| = |P1 (t)P2 (t) − P2 (t)α(t) + P2 (t)α(t) − α(t)β(t)|
≤ |P2 (t)||P1 (t) − α(t)| + |α(t)||P2 (t) − β(t)|
δ
< (|P2 (t)| + |α(t)|)
8(M1 + M2 )
δ
≤ |P2 (t) − β(t)| + |β(t)| + |α(t)|
8(M1 + M2 )
δ
+ M1 + M2 δ
·
< 4
2(M1 + M2 ) 4
δ
≤
4
for every t ∈ [0, 2π]. In this case, we have the formula
Ind (γ) = Ind (P1 P2 ) = Ind (P1 ) + Ind (P2 ) = Ind (α) + Ind (β).
(8.104)
• Case (ii): 0 < M1 + M2 < 1. Instead of the inequalities (8.103), we have
|P1 (t) − α(t)| <
δ
<1
8
and |P2 (t) − β(t)| <
δ
<1
8
(8.105)
for every t ∈ [0, 2π]. Similarly, we can derive from the triangle inequality and the inequalities (8.105) that
|P1 (t)P2 (t) − γ(t)| < (|P2 (t)| + |α(t)|)
δ
δ
δ
≤ |P2 (t) − β(t)| + |β(t)| + |α(t)| <
8
8
4
for every t ∈ [0, 2π]. Thus we also have the formula (8.104).
This completes the proof of Lemma 8.1.
209
8.2. Index of a curve
Now we may go back to the proof of Problem 8.28. Apply Lemma 8.1 to (8.102), we obtain
0 = Ind (ψ) = Ind (e−it γ1 ) = Ind (e−it ) + Ind (γ1 ) = −1 + Ind (γ1 )
which means that Ind (γ1 ) = 1. As a result, the consequences
Ind (γ0 ) = 0
and Ind (γ1 ) = 1
certainly contradict to the fact that Ind (γr ) is a fixed integer for every r ∈ [0, 1]. Hence g(z) = −z for
at least one z ∈ T and this ends the proof of the problem.
Problem 8.29
Rudin Chapter 8 Exercise 29.
Proof. Let T = {z ∈ C | |z| = 1} and f : D → D, where D = {z ∈ C | |z| ≤ 1}. Assume that f (z) 6= z for
every z ∈ D. Then we have
|z − f (z)| 6= 0
for every z ∈ D. Now we associate to each z ∈ D the point g(z) ∈ T which lies on the ray that starts at
f (z) and passes through the point z. See Figure 8.6 for an illustration.
Figure 8.6: The geometry of the points z, f (z) and g(z).
In fact, we have
g(z) =
z − f (z)
.
|z − f (z)|
(8.106)
By the definition (8.106), we have g(D) ⊆ T and we follow from Theorem 4.4 that g is continuous on D.
Thus g is a continuous mapping of D into the unit circle T . By Problem 8.28,
g(z0 ) = −z0
(8.107)
for some z0 ∈ T . However, the construction of g implies that
g(z) = z
(8.108)
if z ∈ T . Therefore, we deduce a contradiction from the expressions (8.107) and (8.108) that
z0 = 0 6∈ T.
Hence we have f (z) = z for some z ∈ D. This completes the proof of the problem.
Chapter 8. Some Special Functions
8.3
210
Stirling’s formula
Problem 8.30
Rudin Chapter 8 Exercise 30.
Proof. Let c ∈ R and y = x + c. By Theorem 8.18(a), we have
lim
x→∞
Γ(x + c)
1
Γ(y)
Γ(y + 1)
= lim
= lim
·
.
y→∞ (y − c)c Γ(y − c)
y→∞
xc Γ(x)
y
(y − c)c−1 Γ(y − c + 1)
It is easy to see that
p
√
y−c
( ye )y 2πy
( y−c
2π(y − c)
1
Γ(y + 1)
Γ(y + 1)
e )
√
p
·
=
·
·
.
y−c
y
(y − c)c−1 Γ(y − c + 1)
( ye )y 2πy y(y − c)c−1 Γ(y − c + 1) ( y−c
2π(y − c)
e )
Since
Stirling’s formula implies that
lim
x→∞
√
( ye )y 2πy
p
= 1,
lim
y→∞ ( y−c )y−c
2π(y − c)
e
1
Γ(y + 1)
Γ(x + c)
= lim
·
y→∞
xc Γ(x)
y
(y − c)c−1 Γ(y − c + 1)
p
y−c
( y−c
2π(y − c)
Γ(y + 1)
e )
= lim y y √
× lim
c−1
y→∞ ( )
y→∞
y(y − c) Γ(y − c + 1)
2πy
e
√
( ye )y 2πy
p
× lim y−c
y−c 2π(y − c)
y→∞ (
e )
p
y−c
2π(y − c)
( y−c
e )
= lim
y−c→∞ y(y − c)c−1 Γ(y − c + 1)
= 1.
This completes the proof of the problem.
Problem 8.31
Rudin Chapter 8 Exercise 31.
Proof. Since (1 − x2 )n is an even function in x, we have
Z 1
−1
2 n
(1 − x ) dx = 2
Z 1
0
(1 − x2 )n dx.
(8.109)
By the change of variable t = x2 , the integral on the right-hand side of (8.109) becomes
Z 1
0
1
t− 2 (1 − t)n dt.
(8.110)
Apply Theorem 8.20 with x = 12 and y = n + 1 to the integral (8.110), we have
Z 1
0
1
t− 2 (1 − t)n dt =
Γ( 21 )Γ(n + 1)
.
Γ(n + 23 )
(8.111)
211
8.3. Stirling’s formula
Substituting the identity (8.111) into the integral equation (8.109) and applying [21, Eqn. (99), p. 194]
and Problem 8.30, we have
√
Z 1
πΓ(n + 1)
2 n
(1 − x ) dx =
Γ(n + 32 )
−1
Z 1
1
√
√
n 2 Γ(n + 1)
(1 − x2 )n dx = π lim
lim n
n→∞ Γ(n + 3 )
n→∞
−1
2
√
= π.
This completes the proof of the problem.
Chapter 8. Some Special Functions
212
CHAPTER
9
Functions of Several Variables
9.1
Linear transformations
Problem 9.1
Rudin Chapter 9 Exercise 1.
Proof. Let span (S) be the span of S. Suppose that x, y ∈ span (S) and c is a scalar. By Definition
9.1(b), we have
x = c1 s 1 + c2 s 2 + · · · + cp s p
and y = d1 s′1 + d2 s′2 + · · · + dp s′q ,
where c1 , c2 , . . . , cp , d1 , d2 , . . . , dq are scalars and s1 , s2 , . . . , sp , s′1 , s′2 , . . . , s′q ∈ span (S). Then we have
x + y = c1 s1 + c2 s2 + · · · + cp sp + d1 s′1 + d2 s′2 + · · · + dp s′q ∈ span (S)
and
cx = (cc1 )s1 + (cc2 )s2 + · · · + (ccp )sp ∈ span (S).
By Definition 9.1(a), span (S) is a vector space. This completes the proof of the problem.
Problem 9.2
Rudin Chapter 9 Exercise 2.
Proof. Suppose that A ∈ L(X, Y ) and B ∈ L(Y, Z). Let x1 , x2 ∈ X and c be a scalar. Then we have
(BA)(x1 + x2 ) = (BA)x1 + (BA)x2 = B(Ax1 ) + B(Ax2 ) = (BA)x1 + (BA)x2
and
(BA)(cx) = B(A(cx)) = B(cAx) = cB(Ax) = c(BA)x.
By Definition 9.4, we obtain BA ∈ L(X, Z).
For y, y1 , y2 ∈ X, there exist unique x, x1 , x2 ∈ X such that Ax1 = y1 , Ax2 = y2 and Ax = y. By
Definition 9.4, we have
A(x1 + x2 ) = Ax1 + Ax2 ,
A(cx) = cAx,
A−1 y1 = x1 ,
A−1 y2 = x2 ,
A−1 y = x.
Thus we deduce from the expressions (9.1) that
A−1 (y1 + y2 ) = A−1 (Ax1 + Ax2 ) = A−1 (A(x1 + x2 )) = x1 + x2 = A−1 y1 + A−1 y2
213
(9.1)
Chapter 9. Functions of Several Variables
214
and
A−1 (cy) = A−1 (cAx) = A−1 (A(cx)) = cx = cA−1 y.
By Definition 9.4, A−1 is linear. If A−1 y1 = A−1 y2 , then we have
A(A−1 y1 ) = A(A−1 y2 )
so that y1 = y2 . Therefore, A−1 is one-to-one. For every x ∈ X, there exists a y ∈ X such that Ax = y
which means
A−1 y = x
by the expressions (9.1). Thus A−1 is onto. Hence A−1 is invertible. This completes the proof of the
problem.
Problem 9.3
Rudin Chapter 9 Exercise 3.
Proof. If Ax = Ay, then we have
A(x − y) = 0
which implies that x − y = 0, i.e., x = y. Hence A is 1-1, completing the proof of the problem.
Problem 9.4
Rudin Chapter 9 Exercise 4.
Proof. Let X and Y be vector spaces and A ∈ L(X, Y ). Furthermore, let N (A) = {x ∈ X | Ax = 0} be
the null space of A. If x, y ∈ N (A) and c is a scalar, then we follow from Definition 9.4 that
A(x + y) = Ax + Ay = 0 + 0 = 0
and A(cx) = cAx = 0.
Hence N (A) is a vector space in X.
Let R(A) = {Ax ∈ Y | x ∈ X} be the range of A. If x, x1 , x2 ∈ X and c is a scalar, then we have
Ax = y ∈ Y,
Ax1 = y1 ∈ Y
and Ax2 = y2 ∈ Y.
Thus it follows from Definitions 9.1 and 9.4 that
A(x1 + x2 ) = Ax1 + Ax2 = y1 + y2 ∈ Y
and A(cx) = cAx = cy ∈ Y.
Hence R(A) is also a vector space in Y . We finish the proof of the problem.
Problem 9.5
Rudin Chapter 9 Exercise 5.
Proof. Suppose that {e1 , . . . , en } is the standard basis of Rn and x =
j = 1, 2, . . . , n. By Definition 9.4, we have
X
X
cj Aej = x · y,
Ax = A
cj e j =
P
cj ej , where cj ∈ R for all
(9.2)
P
where y = (Aej )ej . This proves the existence of the vector y. In fact, the uniqueness of such vector
follows immediately from the form
X
y=
(Aej )ej
because for every A ∈ L(Rn , R), the real numbers Aej (1 ≤ j ≤ n) are uniquely determined.
215
9.2. Differentiable mappings
By Theorem 1.37(d) (Schwarz inequality) and the expression (9.2), we have
|Ax| = |x · y| ≤ |y||x|
for all x ∈ Rn . Then Definition 9.6(c) implies that
kAk ≤ |y|.
(9.3)
y
, then we have |x| = 1 so that
If x = |y|
Ax = x · y =
y·y
= |y|.
|y|
By Definition 9.6(c), we have
kAk ≥ |Ax| = |y|.
(9.4)
Hence our desired result follows immediately from the inequalities (9.3) and (9.4). This completes the
proof of the problem.
9.2
Differentiable mappings
Problem 9.6
Rudin Chapter 9 Exercise 6.
Proof. By Definition 9.16 and Theorem 5.13 (L’Hospital’s Rule), for all (x, y) 6= (0, 0), we have
(D1 f )(x, y) = lim
t→0
= lim
f ((x, y) + t(1, 0)) − f (x, y)
t
(x+t)y
xy
(x+t)2 +y 2 − x2 +y 2
t
[(x + t)2 + y 2 ]y − (x + t)y · 2(x + t)
= lim
t→0
[(x + t)2 + y 2 ]2
3
2
y − 2x y
= 2
(x + y 2 )2
t→0
and
f ((x, y) + t(0, 1)) − f (x, y)
t→0
t
x(y+t)
xy
−
x2 +(y+t)2
x2 +y 2
= lim
t→0
t
[x2 + (y + t)2 ]x − x(y + t) · 2(y + t)
= lim
t→0
[x2 + (y + t)2 ]2
x3 − 2y 2 x
.
= 2
(x + y 2 )2
(D2 f )(x, y) = lim
For (x, y) = (0, 0), we note that
f ((0, 0) + t(1, 0)) − f (0, 0)
0
= lim = 0
t→0
t→0 t
t
(D1 f )(0, 0) = lim
and
(D2 f )(0, 0) = lim
t→0
0
f ((0, 0) + t(0, 1)) − f (0, 0)
= lim = 0.
t→0 t
t
Chapter 9. Functions of Several Variables
216
Therefore, (D1 f )(x, y) and (D2 f )(x, y) exist at every point of R2 . However, if x = y 6= 0, then
|f (x, y) − f (0, 0)| = |f (x, x) − (0, 0)| =
1
,
2
so f is not continuous at (0, 0). This completes the proof of the problem.
Problem 9.7
Rudin Chapter 9 Exercise 7.
Proof. We have f : E ⊆ Rn → R. Fix x ∈ E and ǫ > 0. Since the partial derivatives are bounded in E,
there exists a positive number M such that
|(Dj f )(y)| ≤ M
(9.5)
for all y ∈ E and 1 ≤ j ≤ n. Since E is open, there is an open ball S ⊆ E, with center at x and radius
ǫ
nM . Suppose that y ∈ S. Then we have
y−x=h=
X
hj e j ,
ǫ
. Put v0 = 0 and vk = h1 e1 + · · · + hk ek , where 1 ≤ k ≤ n. Then we have
where |h| < nM
f (y) − f (x) = f (x + h) − f (x) =
n
X
j=1
[f (x + vj ) − f (x + vj−1 )].
(9.6)
ǫ
Since |vk | < nM
for 1 ≤ k ≤ n and S is convex by Definition 2.17, the segments with end points
x + vj−1
and x + vj
must lie in S. Since vj = vj−1 + hj ej , Theorem 5.10 (Mean Value Theorem) implies that
f (x + vj ) − f (x + vj−1 ) = hj (Dj f )(x + vj−1 + θj hj ej )
(9.7)
for some θj ∈ (0, 1). By the bounds (9.5), the expression (9.7) gives
|f (x + vj ) − f (x + vj−1 )| ≤ M |hj |.
Therefore, the expression (9.6) reduces to
|f (y) − f (x)| ≤ M
n
X
j=1
|hj | ≤ nM |h| < ǫ.
By Definition 4.5, f is continuous at x. Since x is arbitrary, f is continuous on E and this completes the
proof of the problem.
Problem 9.8
Rudin Chapter 9 Exercise 8.
Proof. We have f : E ⊆ Rn → R. Since E is open and f is differentiable at x, it follows from Theorem
9.17 that
f ′ (x)ej = (Dj f )(x)
for 1 ≤ j ≤ n so that its corresponding matrix is
[f ′ (x)] =
(D1 f )(x)
(D2 f )(x)
· · · (Dn f )(x)
(9.8)
217
9.2. Differentiable mappings
Since f has a local maximum at x ∈ E, there exists a δ > 0 such that
f (y) ≤ f (x)
for all y ∈ Nδ (x). Note that, for each 1 ≤ j ≤ n, if y = x + tej with |t| < δ, then we have y ∈ Nδ (x).
Suppose that Fj : (−δ, δ) → R is defined by
Fj (t) = f (x + tej ),
where 1 ≤ j ≤ n. By the previous analysis, we see that x + tej ∈ Nδ (x) so that each Fj is well-defined.
By the hypotheses, the real-valued function Fj is differentiable on (−δ, δ) and has a local maximum at
0. By Theorem 5.8, we see that
Fj′ (0) = 0
for 1 ≤ j ≤ n. By Definition 5.1 and [21, Eqn. (25), p. 215], we have
(Dj f )(x) = lim
t→0
f (x + tej ) − f (x)
Fj (t) − Fj (0)
= lim
= Fj′ (0) = 0
t→0
t
t
for all 1 ≤ j ≤ n. Hence we obtain from the matrix representation (9.8) that
[f ′ (x)] = 0 0 · · · 0
which is equivalent to f ′ (x) = 0. This completes the proof of the problem.
Problem 9.9
Rudin Chapter 9 Exercise 9.
Proof. We have f : E ⊆ Rn → Rm . We borrow a result from general topology, see [18, §23, p. 148].
Lemma 9.1
A space E is connected if and only if the only subsets of E that are both open and closed
in E are the empty set and E itself.
Pick and fix x ∈ E. Consider the set
A = {y ∈ E | f (y) = f (x)}.
Now for every y ∈ A, since E is open, there is an open ball Gy such that y ∈ Gy ⊆ E. It is clear from the
hypothesis that f ′ (x) = 0 for every x ∈ Gy . Since Gy is convex and open, we follow from the corollary
of Theorem 9.19 that
f (z) = f (x)
for all z ∈ Gy , i.e., Gy ⊆ A. By Definition 2.18(f), A is open in E.
Let u ∈ Ac . Then we have f (u) 6= f (x). Again, the openness of E implies that
u ∈ Gu ⊆ E
for some open ball Gu . By the corollary of Theorem 9.19, we have
f (v) = f (u) 6= f (x)
for all v ∈ Gu . Therefore, Gu ⊆ Ac and then Ac is open. By Theorem 2.23, A is closed in E. Hence
it follows from Lemma 9.1 that A = E which means f is constant in E, finishing the proof of the
problem.
Chapter 9. Functions of Several Variables
218
Problem 9.10
Rudin Chapter 9 Exercise 10.
Proof. We have f : E ⊆ Rn → R. Let x, x′ ∈ E and they differ only at the first coordinate, i.e.,
x = (a, x2 , . . . , xn ) and x′ = (b, x2 , . . . , xn ). Let, further, b > a and consider the function g : [a, b] → R
defined by
g(t) = f (t, x2 , . . . , xn ).
(9.9)
We first show that g is well-defined. To this end, note that
(1 − λ)x + λx′ ∈ E
(9.10)
for all λ ∈ [0, 1] because E is convex. Thus this implies that
((1 − λ)a + λb, x2 , . . . , xn ) ∈ E
for all λ ∈ [0, 1] which means that (t, x2 , . . . , xn ) ∈ E for t ∈ [a, b] and then the function g given by (9.9)
is well-defined. Since D1 f exists on E, the function g is differentiable on [a, b]. Apply Theorem 5.5 (Mean
Value Theorem) to g, we have
g(b) − g(a) = (b − a)g ′ (θ)
(9.11)
for some θ ∈ (a, b). By the hypothesis, we have
g ′ (θ) = (D1 f )(θ, x2 , . . . , xn ) = 0
so that we follow from the expression (9.11) that g(b) = g(a) which is equivalent to f (x) = f (x′ ), i.e.,
f (x) depends only on x2 , . . . , xn .
It is easy to see that the relation (9.10) still holds by a weaker condition that E is convex in the first
coordinate. Let W1 = {(x, y) ∈ R2 | x > 0, y > 0}, W2 = {(x, y) ∈ R2 | x < 0, y > 0} and E = W1 ∪ W2 .a
Define the real-valued function f : E → R by
y,
if (x, y) ∈ W1 ;
f (x, y) =
−y, if (x, y) ∈ W2 .
Since W1 and W2 are open, E is open. Obviously, E is not convex. Furthermore, we have
(D1 f )(x, y) = 0
for every (x, y) ∈ E. However, we have f (1, 1) = 1 and f (−1, 1) = −1 so that f depends on the first
coordinate. This completes the proof of the problem.
Problem 9.11
Rudin Chapter 9 Exercise 11.
Proof. By [21, Eqn. (34), p. 217], we have for x ∈ Rn ,
∇(f g)(x) =
n
X
(Di (f g))(x)ei .
i=1
Since Di (f g)(x) = f (Di g)(x) + g(Di f )(x), the expression (9.12) reduces to
∇(f g)(x) =
n
X
[f (x)(Di g)(x) + g(x)(Di f )(x)]ei
i=1
a In other words, E is the union of the first and second quadrants which is shaped like a horseshoe.
(9.12)
219
9.3. Local maxima and minima
= f (x)
n
X
(Di g)(x)ei + g(x)
i=1
n
X
(Di f )(x)ei
i=1
= f (x)∇(g)(x) + g(x)∇(f )(x).
Thus we have
∇(f g) = f ∇g + g∇f.
(9.13)
Suppose that f 6= 0 in Rn . Since 1 = f · f1 and ∇(1) = 0, we put g = f1 in the identity (9.13) to obtain
1 1
+ ∇(f )
0 = f∇
f
f
1
∇
= f −2 ∇(f ).
f
Hence, we complete the proof of the problem.
9.3
Local maxima and minima
Problem 9.12
Rudin Chapter 9 Exercise 12.
Proof. The range K of f is in fact the (ring) torus generated by rotating the circle given by
(x − b)2 + z 2 = a2
about the z-axis, see Figure 9.1 for an example produced by WolframAlphab with a = 1 and b = 2.
Figure 9.1: An example of the range K of f .
(a) By definition, we have
(∇f1 )(x) = (D1 f1 )(x)e1 + (D2 f1 )(x)e2 = (−a sin s cos t, −(b + a cos s) sin t),
where x = (s, t). Thus (∇f1 )(x) = 0 if and only if
sin s cos t = 0 and (b + a cos s) sin t = 0.
b See https://www.wolframalpha.com/ .
(9.14)
Chapter 9. Functions of Several Variables
220
Since b > a > 0, we have (b + a cos s) > 0 for any s, so the second equation in (9.14) implies that
t = 0 or π. In both cases, the first equation in (9.14) shows that s = 0 or s = π. Hence there are
exactly four points p ∈ K such that (∇f1 )(f −1 (p)) = 0. In fact, they are the images of the four
points (0, 0), (0, π), (π, 0) and (π, π) which are
f (0, 0) = (a + b, 0, 0),
f (0, π) = (−a − b, 0, 0),
f (π, 0) = (b − a, 0, 0)
and
f (π, π) = (−(b − a), 0, 0).
(b) Again, by definition, we have
(∇f3 )(x) = (D1 f3 )(x)e1 + (D2 f3 )(x)e2 = (a cos s, 0),
where x = (s, t). Thus (∇f3 )(x) = 0 if and only if cos s = 0 if and only if s = π2 or s = 3π
2 .
Therefore, the required set is given by
f1 ( π2 , t) = b cos t,
f2 ( π2 , t) = b sin t,
f3 ( π2 , t) = a
or
f1 ( 3π
2 , t) = b cos t,
f2 ( 3π
2 , t) = b sin t,
f3 ( 3π
2 , t) = −a.
(9.15)
Geometrically, the locus of the left-hand side in (9.15) is the circle center at z = a with radius b and
the locus of the right-hand side in (9.15) is the circle center at z = −a with radius b. The graph of
the loci can be seen in Figure 9.2.
Figure 9.2: The set of q ∈ K such that (∇f3 )(f −1 (q)) = 0.
(c) For any (s, t) ∈ [0, 2π] × [0, 2π], we have
−(a + b) ≤ f1 (s, t) ≤ a + b.
221
9.3. Local maxima and minima
Since
f1 (0, 0) = a + b,
f1 (0, π) = −(a + b),
f1 (π, 0) = b − a
and f1 (π, π) = −(b − a),
the points (0, 0) and (0, π) correspond to the local maximum (a + b, 0, 0) and the local minimum
(−(a + b), 0, 0) of f1 respectively. Finally, it is easy to see that any of the remaining two points is
neither a local maximum or a local minimum.
For f3 (s, t), we know that
−a ≤ f3 (s, t) ≤ a
for all (s, t) ∈ [0, 2π] × [0, 2π]. Since
3π f3
, t = −a
2
and f3
π ,t = a
2
for every t ∈ [0, 2π], the points a and −a are obviously the local maximum and the local minimum
of f3 respectively.
(d) We have g : R → R3 . By definition, we have
g(t) = f (t, λt) = ((b + a cos t) cos λt, (b + a cos t) sin λt, a sin t)
(9.16)
which implies that
g′ (t) = − λ(b + a cos t) sin λt − a sin t cos λt, λ(b + a cos t) cos λt − a sin t sin λt, a cos t .
Therefore, we have
|g′ (t)|2 = g′ (t) · g′ (t)
= [−λ(b + a cos t) sin λt − a sin t cos λt]2
+ [λ(b + a cos t) cos λt − a sin t sin λt]2 + a2 cos2 t
= λ2 (b + a cos t)2 + a2 sin2 t + a2 cos2 t
= a2 + λ2 (b + a cos t)2 .
This proves the second assertion of this part.
For the first assertion, we note from the definition (9.16) that if g(u) = g(v) for u, v ∈ R, then we
have sin u = sin v which means that
u = v + 2kπ
for some k ∈ Z. Since b + a cos t ≥ b − a > 0 for every t ∈ [0, 2π], we have
(b + a cos u) cos λu = (b + a cos v) cos λv
(b + a cos v) cos λ(v + 2kπ) = (b + a cos v) cos λv
cos λ(v + 2kπ) = cos λv
λ(v + 2kπ) = λv + 2mπ
kλ = m
for some m ∈ Z. If k 6= 0, then λ = m
k ∈ Q, a contradiction. Thus k = 0 and m = 0, so we have
u = v, i.e., g is 1-1.
Finally, we show that g(R) is dense in K and we divide its proof into several steps:
– Step 1: Rephrasing the problem. Pick x ∈ K. For every ǫ > 0, there exists a t ∈ R such
that
|g(t) − x| < ǫ.
To this end, suppose that x = ((b + a cos p) cos q, (b + a cos p) sin q, a sin p), where p, q ∈ R. By
the definition (9.16), we have
|g(t) − x|2 = [(b + a cos t) cos λt − (b + a cos p) cos q]2
+ [(b + a cos t) sin λt − (b + a cos p) sin q]2
+ (a sin t − a sin p)2 .
(9.17)
Chapter 9. Functions of Several Variables
222
– Step 2: Simplification of the expression (9.17). Now the expression in the right-hand
side of (9.17) is too complicated for computation, so we need to simplify it. Since sin t and
cos t are periodic functions, if we take
t = p + 2nπ
for some integer n, then the expression (9.17) reduces to
|g(p + 2nπ) − x|2 = [(b + a cos p) cos λ(p + 2nπ) − (b + a cos p) cos q]2
+ [(b + a cos p) sin λ(p + 2nπ) − (b + a cos p) sin q]2
n
2
≤ (b + a)2 cos λ(p + 2nπ) − 2mπ − cos q
2 o
+ sin λ(p + 2nπ) − 2mπ − sin q
(9.18)
for some m ∈ Z. (The reason why the term 2mπ is inserted in the inequality (9.18) will be
clear very soon in Step 3 and Step 4 below.)
– Step 3: Transformation of the problem. The expression in the right-hand side of the
inequality (9.18) is quite simple now, but we can do much better. In fact, by [21, Eqn. (49),
p. 182] and Definition 5.1, we have
| cos t − cos x| ≤ |t − x| and | sin t − sin x| ≤ |t − x|
as t → x. Thus we can further reduce the inequality (9.18) to
|g(p + 2nπ) − x|2 ≤ 2(b + a)2 |λ(p + 2nπ) − 2mπ − q|2 .
(9.19)
If we can show that for every ǫ > 0, there exist m, n ∈ Z such that
ǫ
,c
|2nλπ − 2mπ + λp − q| < √
2(b + a)
(9.20)
then the inequalities (9.19) and (9.20) imply that
|g(p + 2nπ) − x| < ǫ
(9.21)
for some integer n.
Let’s summarize what we have done so far. We have shown that our original problem
|g(t) − x| < ǫ
follows immediately from the validity of the inequality (9.20) for a sequence of integers, so
Step 4 comes into play.
– Step 4: Application of the Kronecker’s Approximation Theorem (Lemma 4.6).
To show that (9.20) holds, we need a previous result in Chapter 4. That is the Kronecker’s
Approximation Theorem (Lemma 4.6): Let 0 < θ < 1. Given ǫ > 0, there exists n ∈ N such
that
ǫ
,
(9.22)
|nλ − h − θ| < √
2 2π(b + a)
where a, b are constants with 0 < a < b and h = [nλ] ∈ Z. We notice that the θ in the
inequality (9.22) can be taken to be any real number. In particular, we substitute θ = q−λp
2π
in the inequality (9.22) to get
ǫ
|2nλπ − 2hπ + (λp − q)| < √
2(b + a)
which is exactly the inequality (9.20) with m = [nλ].
c This inequality may not hold without the existence of the term 2mπ. For example, if p = q = π
2
√
√
have |2 2nπ + π2 ( 2 − 1)| which cannot be arbitrary small.
and λ =
√
2, then we
223
9.3. Local maxima and minima
Hence the set g(R) is dense in K.
This completes the proof of the problem.d
Problem 9.13
Rudin Chapter 9 Exercise 13.
Proof. Suppose that f : R → R3 is given by
f (t) = (f1 (t), f2 (t), f3 (t)).
Now the differentiability of f implies the differentiability of f1 , f2 and f3 by Remark 5.16. Since f (t)·f (t) =
1, we have f1 (t)f1 (t) + f2 (t)f2 (t) + f3 (t)f3 (t) = 1 and thus
d
[f (t) · f (t)] = 0
dt
d
[f1 (t)f1 (t) + f2 (t)f2 (t) + f3 (t)f3 (t)] = 0
dt
2f1′ (t)f1 (t) + 2f2′ (t)f2 (t) + 2f3′ (t)f3 (t) = 0
(f1′ (t), f2′ (t), f3′ (t)) · (f1 (t), f2 (t), f3 (t)) = 0
f ′ (t) · f (t) = 0
as required, completing the proof of the problem.
Problem 9.14
Rudin Chapter 9 Exercise 14.
Proof.
(a) By simple computation, we have
(D1 f )(x, y) =
x2 (x2 + 3y 2 )
(x2 + y 2 )2
and (D2 f )(x, y) =
−2x3 y
,
(x2 + y 2 )2
where (x, y) 6= (0, 0). Thus we have
0 ≤ |(D1 f )(x, y)| ≤
x2 (3x2 + 3y 2 )
3x2
= 2
≤3
2
2
2
(x + y )
x + y2
and the A.M. ≥ G.M. implies that
p
x2 (2|x||y|)
x2 (2 x2 y 2 )
x2 (x2 + y 2 )
x2
0 ≤ |(D2 f )(x, y)| = 2
=
≤
=
= 1.
(x + y 2 )2
(x2 + y 2 )2
(x2 + y 2 )2
x2 + y 2
If (x, y) = (0, 0), then we have
(D1 f )(0, 0) = lim
t→0
f (0, t) − f (0, 0)
f (t, 0) − f (0, 0)
= 1 and (D2 f )(0, 0) = lim
= 0.
t→0
t
t
Hence both D1 f and D2 f are bounded in R2 .
d This problem can also be found in [6, Exericses 11.5, pp. 265, 266].
(9.23)
Chapter 9. Functions of Several Variables
224
(b) Let u = u1 e1 + u2 e2 , where u21 + u22 = 1. By [21, Eqn. (39), p. 217], we know that
f ((0, 0) + t(u1 , u2 )) − f (0, 0)
t
f (tu1 , tu2 )
= lim
t→0
t
t3 u31
= lim 2 2
t→0 t(t u1 + t2 u2
2)
(Du f )(0, 0) = lim
t→0
= u31 .
Since u is a unit vector, |u31 | ≤ |u|3 ≤ 1 and the result follows.
(c) We have γ : R → R2 with γ(0) = (0, 0) and |γ ′ (0)| > 0. Let γ(t) = (γ1 (t), γ2 (t)). Then we have
g(t) = f (γ(t)) =
γ13 (t)
.
2
γ1 (t) + γ22 (t)
(9.24)
Since γ is differentiable in R, γ1 and γ2 are differentiable in R. Thus it is easy to see from the righthand side in the expression (9.24) that g is differentiable at every point t with (γ1 (t), γ2 (t)) 6= (0, 0).
Now the remaining case is the differentiability of g at the point a where
γ1 (a) = γ2 (a) = 0.
In this case, we have
g(a) = f (γ(a)) = f (0, 0) = 0
which gives
h γ (t) − γ (a) i3
1
1
γ13 (t)
g(t) − g(a)
t−a
,
=h
=
γ1 (t) − γ1 (a) i2 h γ2 (t) − γ2 (a) i2
t−a
(t − a)[γ12 (t) + γ22 (t)]
+
t−a
t−a
(9.25)
where t 6= a. Here we must impose an addition assumption that
γ1 (a) 6= 0
or γ2 (a) 6= 0.
Thus it follows from the expression (9.25) that
g ′ (a) =
[γ1′ (a)]3
.
[γ1′ (a)]2 + [γ2′ (a)]2
(9.26)
This proves our first assertion.
For the second assertion, suppose that γ is continuously differentiable on R, i.e., γ ′ is continuous
on R. Then both γ1′ and γ2′ are continuous on R. By the expression (9.24) again, we know that g ′
is continuous at every point t with (γ1 (t), γ2 (t)) 6= (0, 0). It remains to check the continuity of g ′
at every point a such that
γ1 (a) = γ2 (a) = 0.
To this end, we establish from the expression (9.24) that
g ′ (t) =
γ14 (t)γ1′ (t) + 3γ12 (t)γ22 (t)γ1′ (t) − 2γ13 (t)γ2 (t)γ2′ (t)
.
[γ12 (t) + γ22 (t)]2
Then we follow from the equation (9.26) that
γ14 (t)γ1′ (t) + 3γ12 (t)γ22 (t)γ1′ (t) − 2γ13 (t)γ2 (t)γ2′ (t)
t→a
[γ12 (t) + γ22 (t)]2
(
h γ (t) − γ (a) i2 h γ (t) − γ (a) i2
h γ (t) − γ (a) i4
2
2
1
1
1
1
γ1′ (t) + 3
γ1′ (t)
= lim
t→a
t−a
t−a
t−a
lim g ′ (t) = lim
t→a
225
9.3. Local maxima and minima
)
h γ (t) − γ (a) i3 h γ (t) − γ (a) i
nh γ (t) − γ (a) i2 h γ (t) − γ (a) i2 o−2
1
1
2
2
1
1
2
2
′
−2
γ2 (t) ×
+
t−a
t−a
t−a
t−a
[γ1′ (a)]5 + 3[γ1′ (a)]3 [γ2′ (a)]2 − 2[γ1′ (a)]3 [γ2′ (a)]2
{[γ1′ (a)]2 + [γ2′ (a)]2 }2
[γ1′ (a)]3
= ′
[γ1 (a)]2 + [γ2′ (a)]2
=
= g ′ (a).
Hence g ′ is continuous at a and our desired result follows.e
(d) Assume that f was differentiable at (0, 0). Let u = u1 e1 + u2 e2 . By [21, Eqn. (40), p. 218] and
the limits (9.23), we acquire
(Du f )(0, 0) = (D1 f )(0, 0)u1 + (D2 f )(0, 0)u2 = u1
which contradicts the result of part (b).
Hence, we end the analysis of the problem.
Problem 9.15
Rudin Chapter 9 Exercise 15.
Proof.
(a) If x = 0 or y = 0, then the inequality clearly holds. Suppose that x 6= 0 and y 6= 0. Apply the A.M.
≥ G.M. to the positive numbers x4 and y 2 , we get the desired result.
(b) By direct computation, we have
gθ (t) = f (t cos θ, t sin θ) = t2 − 2t3 cos2 θ sin θ −
4t4 cos6 θ sin2 θ
,
(t2 cos4 θ + sin2 θ)2
gθ′ (t) = 2t − 6t2 cos2 θ sin θ −
16t3 cos6 θ sin4 θ
,
(t2 cos4 θ + sin2 θ)3
gθ′′ (t) = 2 − 12t cos2 θ sin θ −
48t2 cos6 θ sin4 θ(sin2 θ − t2 cos4 θ)
.
(t2 cos4 θ + sin2 θ)4
Therefore, we have
gθ (0) = 0,
gθ′ (0) = 0
and gθ′′ (0) = 2.
Hence, for each θ ∈ [0, 2π], the function gθ has a strict local minimum at t = 0.
(c) By direct computation, it is clear that
f (x, x2 ) = −x4 < 0 = f (0, 0)
so that the point (0, 0) is not a local minimum for f .
This completes the proof of the problem.
e We remark that the condition |γ ′ (0)| > 0 has not been used in our argument, so the author wonders that this may
happen to be a typo and the correct condition should be |γ ′ (t)| > 0 for all t ∈ R. However, the author can’t find a counterexample to show that the condition |γ ′ (t)| > 0 for all t ∈ R is necessary for the first assertion. Thus, can someone find a
differentiable curve γ in R2 with γ(0) = 0, |γ ′ (0)| > 0 and γ(1) = 0 (take a = 1 for example), but g is differentiable at 1?
Chapter 9. Functions of Several Variables
9.4
226
The inverse function theorem and the implicit function theorem
Problem 9.16
Rudin Chapter 9 Exercise 16.
Proof. We remark that this exercise was also been discussed in Hardy’s book [10, p. 236]. In this book,
Hardy used the function φ(x) = αx + x2 sin x1 instead of the function f (t) Rudin used here.
By [21, Example 5.6(b), p. 106] and the fact that
1
lim t sin ,
t
t→0
we have
f (0) = 0
1
f (t) − f (0)
= 1.
= lim 1 + 2t sin
t→0
t→0
t
t
and f ′ (0) = lim
Furthermore, for all t 6= 0, we have
f ′ (t) = 1 + 4t sin
1
1
− 2 cos .
t
t
(9.27)
1
Therefore, we have |f ′ (t)| ≤ 7 for all t ∈ (−1, 1). Now we put t = 2kπ
into the derivative (9.27), where
k ∈ N and k → ∞, so
1 4
f′
=1+
sin(2kπ) − 2 cos(2kπ) = −1
(9.28)
2kπ
2kπ
and then
1 lim f ′
= −1 6= f ′ (0),
k→∞
2kπ
i.e., f ′ is not continuous at 0.
Assume that f was one-to-one in (−δ, δ) for some δ > 0. Since f is continuous on (−δ, δ), it follows
from Lemma 6.1 that f is monotonic on any [a, b] ⊆ (−δ, δ), where a < b.
• f is monotonically increasing in (−δ, δ). Suppose that t is an arbitrary point in (−δ, δ). Then
we have
≥ 0, if x − t > 0;
f (x) − f (t)
≤ 0, if x − t < 0.
Both cases imply that
f (x) − f (t)
≥0
x−t
for all x, t ∈ (−δ, δ) with x 6= t. Therefore, we must have
(9.29)
f ′ (t) ≥ 0
1
∈ (0, δ), then the inequality (9.29) is
on (−δ, δ). If the positive integer k is chosen such that 2kπ
contradictory to the result (9.28).
• f is monotonically decreasing in (−δ, δ). Instead of the inequality (9.29), we have
f (x) − f (t)
≤0
x−t
(9.30)
for all x, t ∈ (−δ, δ) with x 6= t. In this case, we have
f ′ (t) ≤ 0
1
∈ (0, δ), then
on (−δ, δ). Now if we take the positive integer k to be large enough so that (2k+1)π
we derive from the expression (9.27) that
1
4
f′
sin(2k + 1)π − 2 cos(2k + 1)π = 1 + 2 = 3
(9.31)
=1+
(2k + 1)π
(2k + 1)π
which gives a contradiction.
227
9.4. The inverse function theorem and the implicit function theorem
Hence f is not one-to-one in any neighborhood of 0 and this completes the proof of the problem.
Problem 9.17
Rudin Chapter 9 Exercise 17.
Proof.
(a) Since f12 (x, y) + f22 (x, y) = e2x (cos2 y + sin2 y) = e2x , the range of f is R2 \ {(0, 0)}.
(b) Now we have
D1 f1 = ex cos y,
D2 f1 = −ex sin y,
so that
[f ′ (x, y)] =
D1 f2 = ex sin y
ex cos y
ex sin y
−ex sin y
ex cos y
and D2 f2 = ex cos y
,
(9.32)
where (x, y) ∈ R2 . Hence we have
Jf (x, y) = det[f ′ (x, y)] =
ex cos y
ex sin y
−ex sin y
ex cos y
= e2x 6= 0
for every (x, y) ∈ R2 .
By Theorem 9.36, the linear operator f ′ (x, y) is invertible for every (x, y) ∈ R2 and then we deduce
from Theorem 9.24 (The Inverse Function Theorem) that there exists a neighborhood of (x, y) such
that f is one-to-one. However, we note that
f (x, y + 2π) = (ex sin(y + 2π), ex cos(y + 2π)) = (ex sin y, ex cos y) = f (x, y),
so f is not one-to-one on R2 .
√
(c) We have b = f (0, π3 ) = (cos π3 , sin π3 ) = ( 12 , 23 ). Now we want the formula
g(ex cos y, ex sin y) = (x, y)
holds in a neighborhood of a. It is easy to see that if p = ex cos y and q = ex sin y, then we have
p
q
x = log p2 + q 2 and y = tan−1 ,
p
where − π2 < y < π2 . Thus the mapping g is given by
p
q
.
g(p, q) = log p2 + q 2 , tan−1
p
On the one hand, we have from the matrix (9.32) that
h p
q i
[f ′ (g(p, q))] = f ′ log p2 + q 2 , tan−1
p
 p

p
p
2 + q2 × √ q
2
2
√
−
p +q ×
p
p2 +q2
p2 +q2 
p
= p 2
p + q 2 × √ 2q 2
p2 + q 2 × √ 2p 2
p +q
p +q
p −q
=
q p
which implies that
[f ′ (g(p, q))]−1 =
1
p2 + q 2
p
−q
q
p
.
(9.33)
Chapter 9. Functions of Several Variables
228
On the other hand, we have
D1 g 1 =
p
p2 + q 2
,
D2 g 1 =
q
p2 + q 2
,
q
D1 g 2 = − 2
p + q2
and D2 g2 =
p
p2 + q 2
so that
′
[g (p, q)] =
p
p2 +q2
q
− p2 +q
2
q
p2 +q2
p
p2 +q2
1
= 2
p + q2
p
−q
q
p
(9.34)
which is exactly the matrix (9.33).
Finally, it is easy to check from the matrices (9.32) and (9.34) that
√ !
h 1 √3 i
h π i
1
3
−
′
′
′
2
2
√
and [g (b)] = g′ ,
[f (a)] = f 0,
=
=
3
1
3
2 2
2
2
1
2√
− 23
√
3
2
1
2
!
.
(d) For the vertical line x = c for some constant c, we have
f (c, y) = (ec cos y, ec sin y).
Let X = ec cos y and Y = ec sin y. Then we have
X 2 + Y 2 = e2c
which is the circle centered at the origin with radius ec in the XY -plane.
For the horizontal line y = c for some constant c, we have
f (x, c) = ex (cos c, sin c).
Let X = ex cos c and Y = ex sin c. If c 6= nπ
2 , where n ∈ Z, then the locus of the image is given by
Y = (tan c)X
which is the straight line but not passing through the origin in the XY -plane. If c is a multiple of
π
2 , then there are two cases:
– Case (i): c = (2m+1)π
for some m ∈ Z. Then we have
2
X =0
and Y = (−1)m ex
which is the positive or negative Y -axis depending on the value of m.
– Case (ii): c = mπ for some m ∈ Z. In this case, we have
X = (−1)m ex
and Y = 0
which is the positive or negative X-axis depending on the value of m.
We finish the proof of the problem.
Problem 9.18
Rudin Chapter 9 Exercise 18.
Proof. Let f = (f1 , f2 ) be the mapping of R2 into R2 given by
f1 (x, y) = x2 − y 2
and f2 (x, y) = 2xy.
(a) Given that u = x2 − y 2 and v = 2xy. If u = v = 0, then we have x = y = 0. In this case, we have
f (0, 0) = (0, 0).
Suppose that (u, v) 6= (0, 0). There are two cases for consideration.
229
9.4. The inverse function theorem and the implicit function theorem
v
– Case (i): v 6= 0. Then both x and y are non-zero. Let x = 2y
. By putting this into the first
equation, we have
v2
u = 2 − x2
4y
which implies that
4y 4 + 4uy 2 − v 2 = 0.
By the quadratic formula, we have
y2 =
−u ±
√
u2 + v 2
.
2
(9.35)
Next, by taking the positive value on the right-hand side in the solutions (9.35), we have
s
√
u + u2 + v 2
y=
(9.36)
2
and then
q√
q√
s√
u2 +v 2 −u
u2 +v 2 −u
u2 + v 2 − u
v
v
1
2
2
× q√
= × q
= sgn (v)
x= ×q √
. (9.37)
2
2
2
2
2
2
2
2
2
2
u +v −u
u+ u +v
u +v −u
2
4
2
Therefore, if x and y are given by the equations (9.36) and (9.37), then we must have
f (x, y) = (u, v).
– Case (ii): v = 0. Then we have
√
if u > 0;
f ( u,
√ 0) = (u, 0),
f (0, −u) = (u, 0), if u < 0.
Hence we have shown that f (R2 ) = R2 .
(b) Since
D1 f1 = 2x,
D2 f1 = −2y,
D1 f2 = 2y
and D2 f2 = 2x,
Jf (x, y) = det[f ′ (x, y)] =
2x −2y
2y 2x
= 4(x2 + y 2 ).
we have
(9.38)
Thus Jf (x, y) 6= 0 if and only if (x, y) 6= (0, 0), and then it follows from Theorem 9.24 (Inverse
Function Theorem) that every point of R2 \ {(0, 0)} has a neighborhood in which f is one-to-one.
However, it is clear that f is not one-to-one because
f (1, −1) = (0, 2) = f (−1, 1).
(c) Suppose that a = (2, 1). Then we have b = f (2, 1) = (3, 4). Let g be the continuous inverse of
f , defined in a neighborhood of b, such that g(b) = a. By the analysis of part (a), we know the
explicit formula for g:
s√
s√
!
u2 + v 2 + u
u2 + v 2 − u
.
,
g(u, v) =
2
2
On the one hand, we note from the partial derivatives (9.38) that
2x −2y
[f ′ (x, y)] =
2y 2x
Chapter 9. Functions of Several Variables
so that
 s
230
√
u2 + v 2 + u


2
s√
[f ′ (g(u, v))] = 2 


u2 + v 2 − u
2
and then

√
2
2
u +v −u 
−

2

s√

u2 + v 2 + u 
s
2
 s√
u2 + v 2 + u


1
2
 s
[f ′ (g(u, v))]−1 = √
√
2 u2 + v 2 
2

u + v2 − u
−
2
On the other hand, we have
s
2
1
u
√
√
D1 g 1 =
+1 ,
4
u2 + v 2 + u
u2 + v 2
s
2
1
u
√
√
D1 g 2 =
−1 ,
4
u2 + v 2 − u
u2 + v 2
1
D2 g 1 =
4
1
D2 g 2 =
4
s√

2
2
u +v −u 

2
.
s√

u2 + v 2 + u 
(9.39)
2
s
s
2
v
√
,
×√
2
2
2
u +v +u
u + v2
2
v
√
.
×√
2
2
2
u +v −u
u + v2
After simplification, these expressions become
s√
s√
1
1
u2 + v 2 + u
u2 + v 2 − u
, D2 g 1 = √
,
D1 g 1 = √
2
2
2 u2 + v 2
2 u2 + v 2
s√
s√
u2 + v 2 − u
u2 + v 2 + u
1
1
D1 g 2 = − √
, D2 g 2 = √
.
2
2
2 u2 + v 2
2 u2 + v 2
Therefore, we have
 s√
u2 + v 2 + u


1
2
 s
[g′ (u, v)] = √
√
2 u2 + v 2 
2

u + v2 − u
−
2
s√

u2 + v 2 − u 

2

s√

2
2
u +v +u 
2
which is exactly the matrix (9.39).
Finally, it is easy to see that
[f ′ (a)] = [f ′ (2, 1)] =
4 −2
2 4
 q
13
1
 q2
and [g′ (b)] = [g′ (3, 4)] =
10
− 7
2
This completes the proof of the problem.
q
7
q2
13
2

.
Problem 9.19
Rudin Chapter 9 Exercise 19.
Proof. The system of equations is equivalent to the mapping f : R3+1 → R3 defined by
f (x, y, z, u) = (3x + y − z + u2 , x − y + 2z + u, 2x + 2y − 3z + 2u).
(9.40)
231
9.4. The inverse function theorem and the implicit function theorem
If a = (x, y, z) = (0, 0, 0) and b = u = 0, then we have f (a, b) = (0, 0, 0). Therefore, its corresponding
matrix at (0, 0, 0, 0) is given by


3 1 −1 0
[f ′ (0, 0, 0, 0)] =  1 −1 2 1  .
(9.41)
2 2 −3 2
The determinants of the submatrices




3 1 0
3 −1 0
 1 −1 1  ,  1 2 1  ,
2 2 2
2 −3 2
are
−12,

1 −1
 −1 2
2 −3
21,

0
1 
2
and

3 1
 1 −1
2 2

−1
2 
−3
3 and 0
respectively. Hence, it follows from Theorem 9.28 (Implicit Function Theorem) that the system of equations can be solved for x, y, u in terms of z; for x, z, u in terms of y; for y, z, u in terms of x; but not for
x, y, z in terms of u. This completes the proof of the problem.
Problem 9.20
Rudin Chapter 9 Exercise 20.
Proof. Let us restate the implicit function theorem in the case n = m = 1 first. Suppose that E ⊆ R2
is an open set, f : E → R is a C ′ -mapping and (a, b) is a point in R2 such that f (a, b) = 0. Suppose,
further, that ∂x f (a, b) 6= 0.f Then there exists an open set U ⊆ R2 and an interval I ⊆ R with (a, b) ∈ U
and b ∈ I, having the following property: For every y ∈ I corresponds a unique x ∈ R such that
(x, y) ∈ U
and f (x, y) = 0.
If this x is defined to be g(y), where g : I ⊆ R → R, then the function g is C ′ , g(b) = a, f (g(y), y) = 0
for y ∈ I and
∂y f (a, b)
.
(9.42)
g ′ (b) = −
∂x f (a, b)
We can interpret the implicit function theorem in two approaches:
• Approach 1: By the explanation of Apostol [1, pp. 373, 374], the expression f (x, y) = 0 does
not necessarily represent a function. Then one may ask when the relation can be solved explicitly
for x in terms of y. The implicit function theorem solves this problem locally. Geometrically, it
means that given a point (a, b) such that f (a, b) = 0, if ∂x f (a, b) 6= 0, then there will be an interval
I of b such that the relation f (x, y) = 0 is in fact a function in I. In other words, we can find a
continuously differentiable function g : I → R implicitly such that
f (g(y), y) = 0,
i.e., x can be solved explicitly in terms of g in this neighborhood.
• Approach 2: Another way to look at the theorem is that the level curve
S = {(x, y) ∈ R2 | f (x, y) = 0}
(9.43)
is locally a graph of a function. Here the word “locally” means that for every (a, b) ∈ S, there exist
an interval I of b and an open set U ⊆ R2 of (a, b) such that U ∩ S is the graph of a continuously
differentiable function x = g(y), i.e.,
U ∩ S = {(x, y) = (g(y), y) | y ∈ I}.
Chapter 9. Functions of Several Variables
232
Figure 9.3: Geometric meaning of the implicit function theorem.
Furthermore, the slope of the tangent to the curve at the point (a, b) is given by the derivative
(9.42). Since ∂x f (a, b) 6= 0, we see that the tangent is not vertical.
For instance, if f (x, y) = x2 + y 2 − 1, then the level curve S defined by (9.43) is the unit circle in
R2 , see Figure 9.3. Around the point A(x1 , y1 ), x can be expressed in terms of y, i.e., x = g(y) =
p
1 − y 2 . Furthermore, the slope of the tangent at A is given by
g ′ (x1 ) = −
y1
2y1
=− .
2x1
x1
However, there is no such function around the point B(0, 1) because ∂x f (0, 1) = 0 so that g ′ is not
well-defined at B.
We complete the analysis of the problem.
Problem 9.21
Rudin Chapter 9 Exercise 21.
Proof.
(a) By [21, Eqn. (34), p. 217], we have ∇f (x, y) = (6x2 −6x)e1 +(6y 2 +6y)e2 . Therefore, ∇f (x, y) = 0
if and only if x2 − x = 0 and y 2 + y = 0 if and only if x = 0, 1 and y = 0, −1. Hence the four points
are (0, 0), (0, −1), (1, 0) and (1, −1). To find the local extreme of the function f , we need a result
from calculus of several variables, see [1, Theorem 13.11, p. 379].
Lemma 9.2
Let f be a real-valued function with continuous second-order partial derivatives at a stationary point (a, b) ∈ R2 . Let A = ∂xx f (a, b), B = ∂xy f (a, b), C = ∂yy f (a, b) and
A B
∆ = det
= AC − B 2 .
B C
(a) If ∆ > 0 and A > 0, then f has a locally minimum at (a, b).
(b) If ∆ > 0 and A < 0, then f has a locally maximum at (a, b).
(c) If ∆ < 0, then f has a saddle point at (a, b).
f Here the notation ∂ f means ∂f .
x
∂x
233
9.4. The inverse function theorem and the implicit function theorem
Now we have ∂xx f = 12x − 6, ∂yy f = 12y + 6 and ∂xy f = 0.
– At (0, 0): A = −6, B = 0 and C = 6. Since ∆ = −36 < 0, (0, 0) is a saddle point by Lemma
9.2(c).
– At (0, −1): A = −6, B = 0 and C = −6. Since ∆ = 36 > 0 and A < 0, (0, −1) is a local
maximum by Lemma 9.2(b).
– At (1, 0): A = 6, B = 0 and C = 6. Since ∆ = 36 > 0 and A > 0, (0, −1) is a local minimum
by Lemma 25(a).
– At (1, −1): A = 6, B = 0 and C = −6. Since ∆ = −36 < 0, (1, −1) is a saddle point by
Lemma 9.2(c).
The behaviours of the four points are shown in Figures 9.4(a) to (d) below.g
(a) The saddle point (0, 0).
(b) The local maximum point (0, −1).
(c) The local minimum point (1, 0).
(d) The saddle point (1, −1).
Figure 9.4: The graphs around the four points.
(b) By computation, we have
f (x, y) = (x + y)(2x2 − 2xy + 2y 2 − 3x + 3y).
By definition, we have
S = {(x, y) ∈ R2 | f (x, y) = 0}
= {(x, y) ∈ R2 | (x + y)(2x2 − 2xy + 2y 2 − 3x + 3y) = 0}.
g The graphs in Figures 9.4 and 9.5 are produced by using the free online software “3D Surface Plotter”, see
https://academo.org/demos/3d-surface-plotter/.
Chapter 9. Functions of Several Variables
234
Therefore, f (x, y) = 0 if and only if
or 2x2 − 2xy + 2y 2 − 3x + 3y = 0.
x+y =0
(9.44)
In other words, points of S are exactly the zeros of one of the equations in (9.44).
– Case (i): x as a function of y. For every y, we have from the leftmost equation in (9.44)
always gives the solution
x = −y.
(9.45)
Recall the fact that
D1 f (x, y) = 6x(x − 1),
so Theorem 9.28 (Implicit Function Theorem) implies that for every x ∈ R \ {0, 1}, x can be
expressed as a function of y locally. When x = 0, it follows from the rightmost equation in
(9.44) that
3
y = 0 or y = − ;
2
when x = 1, we obtain
1
y = −1 or y = .
2
Therefore, at points
1
3
1, ,
(9.46)
(0, 0),
0, − , (1, −1) and
2
2
x might not possibly be expressed as a function of y.
To check whether x can be solved in terms of y around these four points, we rewrite the
rightmost equation in (9.44) as
2x2 − (3 + 2y)x + (2y 2 + 3y) = 0
so that
x=
(3 + 2y) ±
p
(3 + 2y)2 − 8(2y 2 + 3y)
.
4
(9.47)
By the forms of solutions (9.45) and (9.47), we have the following table:
Expressions of x
p −y
(3 + 2y) + 3(1 − 2y)(3 + 2y)
p 4
(3 + 2y) − 3(1 − 2y)(3 + 2y)
4
y→0
0
y → − 23
3
2
y → −1
1
y → 21
− 21
3
2
0
1
1
0
0
− 12
1
Table 9.1: Expressions of x around four points.
Hence we can conclude from Table 9.1 that x cannot be expressed uniquely as a function of y
around these four points. For instance, as y → 0, both
p
(3 + 2y) − 3(1 − 2y)(3 + 2y)
x = −y and x =
4
tend to 0. This means that x has two different expressions around the point (0, 0).
– Case (ii): y as a function of x. Similarly, for every x, we have
y = −x
(9.48)
p
3(3 − 2x)(1 + 2x)
.
4
(9.49)
and for − 21 ≤ x ≤ 32 , we have
y=
(2x − 3) ±
235
9.4. The inverse function theorem and the implicit function theorem
Since
D2 f (x, y) = 6y(y + 1),
we know from Theorem 9.8 (Implicit Function Theorem) that for every y ∈ R \ {−1, 0}, y can
be expressed as a function of x locally and the only points of uncertainly are
1
3 − , −1 , (1, −1), (0, 0) and
,0 .
(9.50)
2
2
By the forms of solutions (9.48) and (9.49), we have the following table:
Expressions of y
p −x
(2x − 3) + 3(3 − 2x)(1 + 2x)
p 4
(2x − 3) − 3(3 − 2x)(1 + 2x)
4
x → − 12
1
2
x→1
−1
x→0
−1
x → 23
− 23
−1
1
2
−1
−1
−1
−1
− 32
−1
Table 9.2: Expressions of y around four points.
Hence, by similar analysis as in Case (i), we can conclude from Table 9.2 that y cannot be
expressed uniquely as a function of x around the four points.
Finally, we see easily from the sets of points (9.46) and (9.50) that there is no neighborhoods around
the points (0, 0) and (1, −1) such that f (x, y) = 0 cannot be solved for y in terms of x or for x in
terms of y.
This completes the proof of the problem.
Problem 9.22
Rudin Chapter 9 Exercise 22.
Proof. We follow the flow of the proof of Problem 9.21.
(a) By [21, Eqn. (34), p.217], we have
∇f (x, y) = 6(x2 + y 2 − x)e1 + 6y(2x + 1)e2 .
Therefore, ∇f (x, y) = 0 if and only if
x2 + y 2 − x = 0
and y(2x + 1) = 0
if and only ifh
x2 − x = 0
and y = 0
if and only if
x = 0, 1 and y = 0.
Hence ∇f (x, y) = 0 only at the two points (0, 0) and (0, 1). Since
∂xx f = 12x − 6,
∂yy f = 12x + 6 and ∂xy f = 12y.
– At (0, 0): A = −6, B = 0 and C = 6. Since ∆ = −36 < 0, (0, 0) is a saddle point by Lemma
9.2(c).
– At (1, 0): A = 6, B = 0 and C = 18. Since ∆ = 108 > 0 and A > 0, (1, 0) is a local minimum
by Lemma 9.2(a).
The behaviours of the four points are shown in Figures 9.5(a) and (b) below.
h It is impossible that x = − 1 because y 2 + 3 = 0 gives non-real y.
2
4
Chapter 9. Functions of Several Variables
236
(b) The local minimum point (1, 0).
(a) The saddle point (0, 0).
Figure 9.5: The graphs around (0, 0) and (1, 0).
(b) By definition, we have
S = {(x, y) ∈ R2 | f (x, y) = 2x3 − 3x2 + 6y 2 x + 3y 2 = 0}.
– Case (i): x as a function of y. We note from Theorem 9.28 (Implicit Function Theorem)
that if D1 f (x, y) 6= 0, then x in terms of y. Therefore, points of S that have no neighborhoods
in which the equation f (x, y) = 0 cannot be solved for x in terms of y must satisfy
D1 f (x, y) = 0.
(9.51)
By the result of part (a), we know that D1 f (x, y) = 0 is equivalent to y 2 = x − x2 . Put this
into the equation f (x, y) = 0 to reduce it to
3x − 4x3 = 0
which gives
x = 0 or
If x = 0, then y = 0. If x =
√
√
3
2 , then we have
x=±
√
3
.
2
p √
2 3−3
y=±
.
2
√
Now the point x = − 23 is rejected because y 2 = −2 43−3 < 0. In conclusion, x might possibly
be expressed as a function of y around the points
p √
√3
2 3 − 3
.
(9.52)
,±
(0, 0) and
2
2
It is well-known that the discriminant ∆ of the cubic equation
2x3 − 3x2 + 6y 2 x + 3y 2 = 0
(9.53)
is given byi
h i
3 2
∆ = −108y 2(16y 4 + 24y 2 − 3) = −108y 2 16 y 2 +
− 12 < 0,
4
so the equation (9.53) has one real root and a pair of complex conjugate roots for every real
y. Hence x cannot be expressed uniquely as a function of y at the points (9.52).
i See https://en.wikipedia.org/wiki/Cubic_function.
237
9.5. The rank of a linear transformation
– Case (ii): y as a function of x. Similarly, we consider D2 f (x, y) = 0 which is equivalent to
6y(2x + 1) = 0. If x = − 12 , then we have
1 f − , y = −1 6= 0,
2
so (− 12 , y) 6∈ S for every real y. If y = 0, then f (x, 0) = 0 if and only if x = 0 or x = 23 . As a
result, y might possibly be expressed as a function of x around the points
(0, 0) and
For fixed x, we have
y2 =
3
2
,0 .
(9.54)
x2 (3 − 2x)
3(2x + 1)
so that − 12 < x ≤ 32 for real solutions y. Hence, this shows that at the points (9.54), y cannot
be solved in terms of x.
We complete the analysis of the problem.
Problem 9.23
Rudin Chapter 9 Exercise 23.
Proof. It is obvious that f (0, 1, −1) = 0. By definition, we have
D1 f = 2xy1 + ex ,
D2 f = x2
and D3 f = 1
and then
D1 f (0, 1, −1) = 1 6= 0,
D2 f (0, 1, −1) = 0 and D3 f (0, 1, −1) = 1.
(9.55)
Therefore, we deduce from Theorem 9.28 (Implicit Function Theorem) that there exists a differentiable
function g in a neighborhood of (1, −1) in R2 such that g(1, −1) = 0 and
f (g(y1 , y2 ), y1 , y2 ) = 0.
To find (D1 g)(1, −1) and (D2 g)(1, −1), we derive from the note following [21, Eqn. (65), p.226] that
(D1 f )(0, 1, −1)(D1g)(1, −1) = −(D2 f )(0, 1, −1)
and
(D1 f )(0, 1, −1)(D2 g)(1, −1) = −(D3 f )(0, 1, −1).
Hence we obtain from the values (9.55) that
(D1 g)(1, −1) = 0 and (D2 g)(1, −1) = −1.
This completes the proof of the problem.
9.5
The rank of a linear transformation
Problem 9.24
Rudin Chapter 9 Exercise 24.
Chapter 9. Functions of Several Variables
238
Proof. We have the mapping f : R2 \ {(0, 0)} → R2 . Thus for every (x, y) 6= (0, 0), since
D1 f1 =
4xy 2
,
(x2 + y 2 )2
we have
D2 f1 =
−4x2 y
,
(x2 + y 2 )2
D1 f2 =

Now it is easy to see that
4xy 2

2
(x + y 2 )2
[f ′ (x, y)] = 
 y(y 2 − x2 )
(x2 + y 2 )2
det[f ′ (x, y)] =
1
(x2 + y 2 )4
y(y 2 − x2 )
(x2 + y 2 )2
and D2 f2 =
x(x2 − y 2 )
,
(x2 + y 2 )2

−4x2 y
(x2 + y 2 )2 
.
x(x2 − y 2 ) 
(x2 + y 2 )2
(9.56)
[(4xy 2 )x(x2 − y 2 ) + (4x2 y)y(y 2 − x2 )] = 0
for any (x, y) 6= (0, 0). By Theorem 9.36, the matrix (9.56) is not invertible. Then we deduce from [15,
Theorem 8, p. 112] that the column vectors of [f ′ (x, y)] are dependent. Since R([f ′ (x, y)]) is a vector
space in R2 , we have
rank ([f ′ (x, y)]) = dim R([f ′ (x, y)]) ≤ 2.
Since R([f ′ (x, y)]) is spanned by the column vectors of [f ′ (x, y)] (see [21, p. 210]), we follow from
Theorem 9.3(a) that rank ([f ′ (x, y)]) = 2 is impossible. Therefore, we have either
rank ([f ′ (x, y)]) = 0 and rank ([f ′ (x, y)]) = 1.
Suppose that rank ([f ′ (x, y)]) = 0. Then Definition 9.30 says that
[f ′ (x, y)](u, v) = (0, 0)
for all (u, v) ∈ R2 \ {(0, 0)}. If x 6= −y, then we choose u = 1 and v = −1 so that the matrix (9.56)
implies that


−4x2 y
4xy 2
 (x2 + y 2 )2 (x2 + y 2 )2 
1
4xy(x + y)
0
1
′


6=
.
[f (x, y)](1, −1) = 
= 2
y(y 2 − x2 ) x(x2 − y 2 )  −1
(x − y)(x2 − y 2 )
0
(x + y 2 )2
(x2 + y 2 )2 (x2 + y 2 )2
If x = −y, then we choose u = v = 1 so that the matrix (9.56) gives
1
[f (x, −x)](1, −1) = 4
4x
′
4x3
0
4x3
0
1
1
=
2 ! 0
6=
.
x
0
0
Hence it is impossible to have rank ([f ′ (x, y)]) = 0 and so rank ([f ′ (x, y)]) = 1. This answers the first
assertion.
For the second assertion, let X = f1 (x, y) and Y = f2 (x, y). Since
X 2 + 4Y 2 = f12 (x, y) + 4f22 (x, y) =
x4 − 2x2 y 2 + y 4
4x2 y 2
x4 + 2x2 y 2 + y 4
+
=
= 1,
(x2 + y 2 )2
(x2 + y 2 )2
(x2 + y 2 )2
the range of f is a subset of the ellipse with radii 1 and 21 on the X and Y axes respectively, see Figure
9.6.
239
9.5. The rank of a linear transformation
Figure 9.6: The graph of the ellipse X 2 + 4Y 2 = 1.
Now we show that the range of f is exactly the ellipse X 2 + 4Y 2 = 1. Suppose that (X, Y ) is a point
on the ellipse such that
x2 − y 2
xy
X= 2
and Y = 2
(9.57)
2
x +y
x + y2
2
1−y
for some x and y with (x, y) 6= (0, 0). Fix x = 1. Then we have X = 1+y
2 which implies that
y=±
r
1−X
,
1+X
(9.58)
where X 6= −1. By the expressions (9.58) and the fact that X 2 + 4Y 2 = 1, when X 6= −1, we have
f 1, ±
r
r
r
1−X 1−X 1 − X = f1 1, ±
, f2 1, ±
1+X
1+X
1+X
q
1−X !
1−X ±
1 − 1+X
1+X
=
,
1−X
1 + 1+X
1 + 1−X
1+X
1p
= X, ±
1 − X2
2
= (X, Y ).
If X = −1, then we deduce from the definition (9.57) that x = 0 and thus Y = 0. In this case, we have
f (0, 2) = (f1 (0, 2), f2 (0, 2)) =
02 − 22
02 + 2
,
2
0·2 = (−1, 0).
02 + 22
Hence the range of f is exactly the graph of the ellipse X 2 + 4Y 2 = 1. This shows the second assertion
and thus completes the proof of the problem.
Problem 9.25
Rudin Chapter 9 Exercise 25.
Proof.
Chapter 9. Functions of Several Variables
240
(a) Recall the definitions from the proof of Theorem 9.32 that Y1 = R(A), {y1 , . . . , yr } is a basis of
Y1 , zi ∈ Rn is defined by
Azi = yi
(9.59)
for 1 ≤ i ≤ r, and a linear mapping S : Y1 → Rn is given by
S(c1 y1 + · · · + cr yr ) = c1 z1 + · · · + cr zr
(9.60)
for all scalars c1 , . . . , cr .
We note from Theorem 9.3(c) that Rn has a basis containing {y1 , . . . , yr }. Let such a basis be
{y1 , . . . , yr , xr+1 , . . . , xn }. Then we have
x = c1 y1 + · · · + cr yr + cr+1 xr+1 + · · · + cn xn
for some scalars c1 , . . . , cn . By Definition 9.30, we have Ax ∈ Y1 = R(A) so that
Ax = c1 y1 + · · · + cr yr .
(9.61)
By using the expressions (9.59) and (9.61), we have
ASAx = AS(c1 y1 + · · · + cr yr ) = A(c1 z1 + · · · + cr zr ) = c1 y1 + · · · + cr yr = Ax.
(9.62)
Thus it deduces from the expression (9.62) that
SASAx = SAx
for every x ∈ Rn . Hence SA is a projection in Rn .
For the second assertion, since SA is a projection in Rn , we follow from the property [21, p. 228]
that every x ∈ Rn has a unique representation of the form
x = x1 + x2 ,
(9.63)
where x1 ∈ R(SA) and x2 ∈ N (SA).
To finish the proof of this part, we have to prove two steps:
– Step 1: R(SA) = R(S). On the one hand, if z ∈ R(S), then we have z = Sy for some
y ∈ Y1 = R(A). Since y = Aw for some x ∈ Rn , we have
z = Sy = SAx.
Thus z ∈ R(SA), i.e., R(S) ⊆ R(SA). On the other hand, if z ∈ R(SA), then z = SAy for
some y ∈ Rn . By definition, Ay ∈ Y1 so that z ∈ R(S), i.e., R(SA) ⊆ R(S) and then
R(SA) = R(S).
– Step 2: N (SA) = N (A). On the one hand, if x ∈ N (A), then Ax = 0 which shows clearly
that
SAx = S0 = 0.
In other words, x ∈ N (SA), i.e., N (A) ⊆ N (SA). On the other hand, if x ∈ N (SA), then
we have SAx = 0 and it follows from the expression (9.62) that
Ax = ASAx = A0 = 0.
Thus, x ∈ N (A), i.e., N (SA) ⊆ N (A) and then
N (SA) = N (A).
Hence the vectors x1 and x2 in the representation (9.63) are elements of R(S) and N (A) respectively.
241
9.6. Derivatives of higher order
(b) We divide the proof into two steps:
– Step 1: dim R(S) = dim R(A). Let y, y′ ∈ Y1 be such that S(y) = S(y′ ). Then we have
S(c1 y1 + · · · + cr yr ) = S(c′1 y1 + · · · + c′r yr )
(9.64)
for some scalars c1 , . . . , cr , c′1 , . . . , c′r . By definition, we have from the expression (9.64) that
c1 z1 + · · · + cr zr = c′1 z1 + · · · + c′r zr .
(9.65)
Apply A to both sides of the expression (9.65), we have
c1 y1 + · · · + cr yr = c′1 y1 + · · · + c′r yr .
Since {y1 , . . . , yr } is a basis of Y1 , we have
c1 = c′1 , . . . , cr = c′r
so that S is an one-to-one linear mapping. Therefore, the mapping S : Y1 → S(Y1 ) is actually
an isomorphism j and we have
dim R(S) = dim S(Y1 ) = dim Y1 = dim R(A) = r.
(9.66)
– Step 2: dim N (A) = n − r. We know from part (a) that {y1 , . . . , yr , xr+1 , . . . , xn } is a basis
of Rn . For every x ∈ Rn , we have
x = c1 y1 + · · · + cr yr + cr+1 xr+1 + · · · + cn xn
for some scalars c1 , . . . , cr . In particular, if x ∈ N (A), then we have Ax = 0 and by the
expression (9.61), we have
c1 y1 + · · · + cr yr = Ax = 0.
Since {y1 , . . . , yr } is a basis of Y1 , it gives
c1 = c2 = · · · = cr = 0.
Thus every x ∈ N (A) is a linear combination of xr+1 , . . . , xn , i.e., N (A) is spanned by
{xr+1 , . . . , xn }.
Next, it is clear that the set {xr+1 , . . . , xn } is linearly independent, so it is a basis of N (A)
and then
dim N (A) = n − r,
(9.67)
as desired.
Combining the numbers (9.66) and (9.67), we have
dim N (A) + dim R(A) = n − r + r = n.
This completes the proof of the problem.
9.6
Derivatives of higher order
Problem 9.26
Rudin Chapter 9 Exercise 26.
j A linear mapping f : V → W between two vector spaces is called an isomorphism if it is one-to-one and onto. See, for
example, [15, p. 155].
Chapter 9. Functions of Several Variables
242
Proof. By Theorem 7.18, there exists a real continuous function on R which is nowhere differentiable.
Let this function be g. Define f : R2 → R by
f (x, y) = g(x).
Then D1 f (x, y) does not exist for every x(, y) ∈ R2 , but D12 f (x, y) = 0. We have completed the proof
of the problem.
Problem 9.27
Rudin Chapter 9 Exercise 27.
Proof. We have
f (x, y) =
(a) If (x, y) 6= (0, 0), then we have
D1 f (x, y) =

0,



if (x, y) = (0, 0);
xy(x2 − y 2 )


, if (x, y) 6= (0, 0).

x2 + y 2
y(x4 + 4x2 y 2 − y 4 )
(x2 + y 2 )2
and D2 f (x, y) =
x(x4 − 4x2 y 2 − y 4 )
.
(x2 + y 2 )2
(9.68)
(9.69)
Thus it is obvious that f, D1 f and D2 f are continuous at every point (x, y) 6= (0, 0).
Next we have to check their continuity at the point (0, 0). By the A.M. ≥ G.M., we have
|f (x, y) − f (0, 0)| =
|x2 − y 2 |
xy(x2 − y 2 )
≤
.
2
2
x +y
2
We observe that
|x2 − y 2 |
→0
2
as (x, y) → (0, 0), so this means that f is continuous at (0, 0).
By the definition (9.68) and the A.M. ≥ G.M., we have
D1 f (0, 0) = lim
t→0
f (t, 0) − f (0, 0)
f (0, t) − f (0, 0)
= 0 and D2 f (0, 0) = lim
= 0.
t→0
t
t
Therefore, we have from the expressions (9.69) that
|D1 f (x, y) − D1 f (0, 0)| =
(x2 + y 2 )2 − 2y 4
y(x4 + 4x2 y 2 − y 4 )
≤ |y|
≤ |y| → 0
2
2
2
(x + y )
(x2 + y 2 )2
|D2 f (x, y) − D2 f (0, 0)| =
(x2 + y 2 )2 − 2x4
x(x4 − 4x2 y 2 − y 4 )
≤ |x|
≤ |x| → 0
2
2
2
(x + y )
(x2 + y 2 )2
and
as (x, y) → (0, 0). Thus D1 f and D2 f are continuous at (0, 0).
(b) We have
and

if (x, y) = (0, 0);
 0,
y(x4 + 4x2 y 2 − y 4 )
D1 f (x, y) =
, if (x, y) 6= (0, 0).

(x2 + y 2 )2

if (x, y) = (0, 0);
 0,
x(x4 − 4x2 y 2 − y 4 )
D2 f (x, y) =
, if (x, y) 6= (0, 0).

(x2 + y 2 )2
(9.70)
(9.71)
243
9.6. Derivatives of higher order
For (x, y) 6= (0, 0), we see that
D12 f (x, y) = D21 f (x, y) =
(x2 − y 2 )(x4 + 10x2 y 2 + y 4 )
x6 + 9x4 y 2 − 9x2 y 4 − y 6
=
.
(x2 + y 2 )3
(x2 + y 2 )3
(9.72)
For (x, y) = (0, 0), we have
D2 f (t, 0) − D2 f (0, 0)
t−0
= lim
= 1,
t→0
t
t
D1 f (0, t) − D1 f (0, 0)
−t − 0
D21 f (0, 0) = lim
= lim
= −1.
t→0
t→0
t
t
D12 f (0, 0) = lim
t→0
(9.73)
(9.74)
Therefore, both D12 f and D21 f exist at every point of R2 . In addition, we deduce from the
expressions (9.72) that they are continuous at every point except possibly the origin (0, 0). However,
if x = y = t, then we have
D12 f (t, t) = D21 f (t, t) = 0
so that
|D12 f (t, t) − D12 f (0, 0)| = 1 and |D21 f (t, t) − D21 f (0, 0)| = 1,
i.e., they are discontinuous at (0, 0).
(c) The results have already been shown in the limits (9.73) and (9.74).
We end the proof of the problem.
Problem 9.28
Rudin Chapter 9 Exercise 28.
Proof. For t < 0, we have
p

(0p≤ x ≤ |t|);
 −x, p
p
ϕ(x, t) = −ϕ(x, |t|) =
x − 2 |t|, ( |t| ≤ x ≤ 2 |t|);

0,
(otherwise).
Now we can divide the xt-plane into six regions, see Figure 9.7:
Figure 9.7: The definition of the function ϕ(x, t).
(9.75)
Chapter 9. Functions of Several Variables
In fact, we have
244

x,


√


 −x + 2 t
−x, p
ϕ(x, t) =



x − 2 |t|,


0,
Region 1;
Region 2;
Region 3;
Region 4;
Regions 5 and 6.
(9.76)
Therefore, it is easy to see from the definition (9.76)
ϕ(x, t) is continuous on each
so we only
p region, p
√ that √
need to check its continuity on the curves x = t, x = 2 t when t ≥ 0 and x = |t|, x = 2 |t| when
t < 0.k
√
• Case (i): On the curve x = t. Since
√
ϕ( t+, t) =
lim
ϕ(x, t) =
√
√ x→ t√
t<x≤2 t
and
√
ϕ( t−, t) =
√
√
(−x + 2 t) = t
lim
√
√ x→ t√
t<x≤2 t
lim
√ ϕ(x, t) =
t
x→ √
0≤x< t
lim
√ x =
√
t,
x→ √
t
0≤x< t
we have
√
√
ϕ( t+, t) = ϕ( t−, t)
√
and so ϕ(x, t) is continuous on the curve x = t.
√
• Case (i): On the curve x = 2 t. Since
√
ϕ(2 t+, t) = lim√ ϕ(x, t) = lim√ 0 = 0
x→2
t
√
2 t<x
and
√
ϕ(2 t−, t) =
lim√
√ x→2 t√
t≤x<2 t
ϕ(x, t) =
x→2
t
√
2 t<x
lim√
√
(−x + 2 t) = 0,
√ x→2 t√
t≤x<2 t
we have
√
√
ϕ(2 t+, t) = ϕ(2 t−, t)
√
and thus ϕ(x, t) is continuous on the curve x = 2 t.
p
p
• Case (i): p
On the curves
x
=
|t|
and
x
=
2
|t|. Similar argument can be applied to the
p
curves x = |t| and x = 2 |t| when t < 0, so ϕ(x, t) is also continuous on them.
In conclusion, ϕ is continuous on R2 . This proves the first assertion.
For the second assertion, we have three cases as follows:
• Case (i): x = 0. Then we observe that
(D2 ϕ)(0, 0+) = lim
h→0
h>0
and
(D2 ϕ)(0, 0−) = lim
h→0
h<0
ϕ(0, h) − ϕ(0, 0)
ϕ(0, h) − 0
= lim
=0
h→0
h
h
h>0
ϕ(0, h) − ϕ(0, 0)
−ϕ(0, |h|) − 0
= lim
= 0.
h→0
h
h
h<0
Thus we have (D2 ϕ)(0, 0) = 0.
k They are the boundaries between Regions 1 and 2, Regions 2 and 5, Regions 3 and 4, as well as Regions 4 and 5
respectively.
245
9.6. Derivatives of higher order
√
• Case (ii): x > 0. If h > 0 and h is so small such that 0 < 2 h ≤ x, then we have
ϕ(x, h) = 0
which implies that
ϕ(x, h) − 0
0
ϕ(x, h) − ϕ(x, 0)
= lim
= lim = 0.
h→0
h→0 h
h→0
h
h
(D2 ϕ)(x, 0+) = lim
h>0
h>0
h>0
p
If h < 0 and |h| is so small such that 0 < 2 |h| ≤ x, then we have
ϕ(x, h) = −ϕ(x, |h|) = 0
from the definition (9.75). Therefore, this implies that
(D2 ϕ)(x, 0−) = lim
h→0
h<0
ϕ(x, h) − ϕ(x, 0)
−ϕ(x, |h|) − 0
0
= lim
= lim = 0.
h→0
h→0 h
h
h
h<0
h<0
Thus we have (D2 ϕ)(x, 0) = 0 if x > 0.
• Case (iii): x < 0. If h > 0, then we have
ϕ(x, h) = 0
by definition. Thus we have
ϕ(x, h) − ϕ(x, 0)
= 0.
h→0
h
(D2 ϕ)(x, 0+) = lim
h>0
If h < 0, then we also have
ϕ(x, h) = −ϕ(x, |h|) = 0
from the definition (9.75) which implies that
ϕ(x, h) − ϕ(x, 0)
= 0.
h→0
h
(D2 ϕ)(x, 0−) = lim
h<0
Thus we have (D2 ϕ)(x, 0) = 0 if x < 0.
For the third assertion, if 0 ≤ t < 14 , then we know from the definition of ϕ(x, t) that
Z 1
ϕ(x, t) dx
f (t) =
−1
=
Z 0
−1
=0+
=
ϕ(x, t) dx +
Z √t
Z √t
0
Z 1
ϕ(x, t) dx
0
ϕ(x, t) dx +
0
x dx +
Z 2√t
√
t
Z 2√t
√
t
ϕ(x, t) dx +
√
(−x + 2 t) dx
Z 1
√ ϕ(x, t) dx
2 t
√
√ i2 t
t h x2
= + −
+ 2 tx √
2
2
t
3t
t
= + 2t −
2
2
= t.
Similarly, if − 41 < t < 0, then we have
Z 1
Z 1
f (t) =
ϕ(x, t) dx = −
ϕ(x, |t|) dx = −f (|t|) = −f (−t) = t.
−1
−1
(9.77)
(9.78)
Chapter 9. Functions of Several Variables
246
Hence we follow from the expressions (9.77) and (9.78) that f (t) = t if |t| < 14 . Since (D2 ϕ)(x, 0) = 0 for
all x, we obtain that
Z 1
(D2 ϕ)(x, 0) dx = 0.
′
−1
′
However, since f (t) = 1, we have f (0) = 1. In other words,
Z 1
′
f (0) 6=
(D2 ϕ)(x, 0) dx.
−1
This completes the proof of the problem.
Problem 9.29
Rudin Chapter 9 Exercise 29.
Proof. This result is sometimes called Schwarz’s theorem, Clairaut’s theorem or Young’s theorem.
We have f : E ⊆ Rn → R. Since f ∈ C (k) (E), the kth-order derivative
Di1 i2 ...ik f = Di1 Di2 · · · Dik f
is well-defined and continuous on E, where i1 , i2 , . . . , ik = 1, 2, . . . , n. Here we provide two solutions and
both use induction.
• Method 1: We prove the assertion by induction. If k = 2, then the corollary of Theorem 9.41
shows that D21 f = D12 f .l Thus the statement is true for k = 2.
Assume that the statement is also true for k = m ≥ 2. That is, if f ∈ C (m) (E), then we have
Di1 i2 ...im f = Di1 Di2 · · · Dim f
is unchanged if the subscripts i1 , . . . , im are permuted, i.e.,
Dϕ(i1 ) Dϕ(i2 ) · · · Dϕ(im ) f = Dϕ(i1 )ϕ(i2 )...ϕ(im ) f = Di1 i2 ...im f = Di1 Di2 · · · Dim f
(9.79)
if ϕ is a permutation m on {i1 , . . . , im } and i1 , i2 , . . . , im = 1, 2, . . . , n.
Let k = m + 1 and A = {i1 , i2 , . . . , im+1 }, where m ≥ 2 and i1 , i2 , . . . , im+1 = 1, 2, . . . , n. Suppose
that f ∈ C (m+1) (E) and ϕ : A → A is a permutation. By the hypothesis, we know that
D1 f, . . . , Dn f ∈ C (m) (E).
– Case (i): ϕ(im+1 ) = im+1 . Then it is clear that ϕ is also a permutation on {i1 , i2 , . . . , im }.
Thus we obtain from the expression (9.79) and Theorem 9.41 that
Dϕ(i1 )ϕ(i2 )...ϕ(im+1 ) f = [Dϕ(i1 ) Dϕ(i2 ) · · · Dϕ(im ) ]Dϕ(im+1 ) f
= (Di1 Di2 · · · Dim )Dim+1 f
= Di1 i2 ···im im+1 f.
– Case (ii): ϕ interchanges the indices im and im+1 . In this case, ϕ is also a permutation on
the restricted set {i1 , i2 , . . . , im−1 } and we follow from the induction hypothesis and Theorem
9.41 that
Dϕ(i1 )ϕ(i2 )...ϕ(im+1 ) f = [Dϕ(i1 ) Dϕ(i2 ) · · · Dim−1 ][Dϕ(im ) Dϕ(im+1 ) f ]
= [Dϕ(i1 ) Dϕ(i2 ) · · · Dϕ(im−1 ) ](Dim+1 Dim f )
= [Dϕ(i1 ) Dϕ(i2 ) · · · Dϕ(im−1 ) ](Dim Dim+1 f )
= Di1 Di2 · · · Dim−1 Dim Dim+1 f
= Di1 i2 ...im+1 f.
l As mentioned by Rudin [21, p. 235], the results of Theorem 9.41 and its corollary are also true when R2 is replaced by
Rn .
m A bijective mapping of a set S to itself is called a permutation of the set.
247
9.6. Derivatives of higher order
– Case (iii): ϕ is a permutation such that ϕ(im ) = im+1 . Let α, β : A → A be permutations
such that
α(im ) = im+1 ,
α(im+1 ) = im
and α(is ) = is
for s 6= m, m + 1
and
β(im+1 ) = im+1 ,
β(im ) = ϕ(im+1 ) and β(is ) = ϕ(is ) for s 6= m, m + 1.
We claim that ϕ = β ◦ α on A, i.e.,
ϕ(is ) = β(α(is ))
(9.80)
for all s = 1, 2, . . . , m + 1. To prove this claim, we consider three subcases:
∗ Subcase (i): s = m. In this case, we have β(α(im )) = β(im+1 ) = im+1 = ϕ(im ).
∗ Subcase (ii): s = m + 1. We acquire β(α(im+1 )) = β(im ) = ϕ(im+1 ).
∗ Subcase (iii): s 6= m, m + 1. This implies that β(α(is )) = β(is ) = ϕ(is ).
Therefore, the formula (9.80) holds for all s = 1, 2, . . . , m + 1 and our claim is proven. Since
α and β are permutations as described in Case (ii) and Case (i) respectively, we have
Dα(i1 )α(i2 )...α(im+1 ) f = Di1 i2 ...im+1 f,
(9.81)
Dβ(α(i1 ))β(α(i2 ))...β(α(im+1 )) f = Dα(i1 )α(i2 )...α(im+1 ) f .
(9.82)
After combining the two expressions (9.81) and (9.82), we apply formula (9.80) to deduce that
Dϕ(i1 )ϕ(i2 )...ϕ(im+1 ) f = Di1 i2 ...im+1 f.
– Case (iv): ϕ is a permutation such that ϕ(ir ) = im+1 , where r 6= m, m + 1. Let
ρ, σ, τ : A → A be permutations such that
ρ(ir ) = im ,
ρ(im ) = ir ,
ρ(im+1 ) = im+1
and ρ(is ) = is
for s 6= r, m, m + 1,
and
σ(im ) = im+1 ,
σ(im+1 ) = im
and σ(is ) = is
and
τ (im+1 ) = im+1 ,
τ (ir ) = ϕ(im ),
τ (im ) = ϕ(im+1 )
for s 6= m, m + 1
and τ (is ) = ϕ(is ) for s 6= r, m, m + 1.
We claim that ϕ = τ ◦ σ ◦ ρ on A, i.e.,
ϕ(is ) = τ (σ(ρ(is )))
(9.83)
for all s = 1, 2, . . . , m + 1. To prove this claim, we consider four subcases:
∗ Subcase (i): s = r. Then we have τ (σ(ρ(ir ))) = τ (σ(im )) = τ (im+1 ) = im+1 = ϕ(ir ).
∗ Subcase (ii): s = m. In this case, we obtain that
τ (σ(ρ(im ))) = τ (σ(ir )) = τ (ir ) = ϕ(im ).
∗ Subcase (iii): s = m + 1. We get τ (σ(ρ(im+1 ))) = τ (σ(im+1 )) = τ (im ) = ϕ(im+1 ).
∗ Subcase (iv): s 6= r, m, m + 1. By definition of ρ, we have ρ(is ) = ϕ(is ) 6= im , im+1 .
This implies that
τ (σ(ρ(is ))) = τ (σ(is )) = τ (is ) = ϕ(is ).
Therefore, the formula (9.83) holds for all s = 1, 2, . . . , m + 1 and our claim is proven. Since
σ is a permutation as described in Case (ii) and ρ and τ are permutations as described in
Case (i), we have
Dρ(i1 )ρ(i2 )...ρ(im+1 ) f = Di1 i2 ...im+1 f,
(9.84)
Dσ(ρ(i1 ))σ(ρ(i2 ))...σ(ρ(im+1 )) f = Dρ(i1 )ρ(i2 )...ρ(im+1 ) f ,
(9.85)
Dτ (σ(ρ(i1 )))τ (σ(ρ(i2 )))...τ (σ(ρ(im+1 ))) f = Dσ(ρ(i1 ))σ(ρ(i2 ))...σ(ρ(im+1 )) f .
(9.86)
After combining the expressions (9.84), (9.85) and (9.86), then we use the formula (9.83) to
acquire
Dϕ(i1 )ϕ(i2 )...ϕ(im+1 ) f = Di1 i2 ...im+1 f.
Chapter 9. Functions of Several Variables
248
Hence it follows from induction that the kth-order derivative
Di1 i2 ...ik f = Di1 Di2 · · · Dik f
is unchanged if the subscripts i1 , . . . , ik are permuted.n
• Method 2: Besides the above argument, the author discovered that there is a shorter and nicer
solution provided by Christian Blatter on webo Here we repeat his argument with a slight modification.
Denote by Wk the set of words α = α1 α2 · · · αk of length k ≥ 1, where α1 , α2 , . . . , αk = 1, 2, . . . , n.
See, for example, [12, §39, p. 341]. For every α ∈ Wk , denote by α
b : {1, 2, . . . , n} → N ∪ {0} the
function giving the multiplicity with which each letter j occurs in α, i.e.,
α
b(j) = #{r ∈ {1, 2, . . . , n} | αr = j}.
We say that two words α, β ∈ Wk with α
b = βb are equivalent.
For α ∈ Wk , we denote by Dα f the kth partial derivative of f with respect to xαk , xαk−1 , . . . , xα1
in this order, i.e.,
Dα f = Dα1 Dα2 · · · Dαk f.
Now we have to prove that if α, β ∈ Wk and α ∼ β, then we must have
Dα f = Dβ f.
The case for k = 2 is obviously true, so we may assume that k ≥ 3 and that the statement is true
for m < k. Take two equivalent words α, β ∈ Wk . Then we have
α = α′ αk
and β = β ′ βk ,
where α′ , β ′ ∈ Wk−1 . If αk = βk , then we have α′ ∼ β ′ and the induction hypothesis implies that
Dα f =
′
′
∂
∂
Dα f =
Dβ f = Dβ f.
∂xαk
∂xαk
If αk 6= βk , then βk must occur in α′ and there is a word α′′ ∈ Wk−2 such that
α′ ∼ α′′ βk .
Similarly, there is a word β ′′ ∈ Wk−2 such that
β ′ ∼ β ′′ αk .
Since α ∼ β, we deduce from this and the corollary of Theorem 9.41 that
α′′ ∼ β ′′ .
Therefore, it follows from our induction hypothesis that
′
∂
Dα f
∂xαk
′′
∂2
=
Dα f
∂xαk ∂xβk
Dα f =
=
′′
∂2
Dβ f
∂xαk ∂xβk
n Here we haven’t applied the hint given by Rudin: repeated application of Theorem 9.41. If someone wants to use this
hint, he/she needs some results from abstract algebra first: By definition, a transposition on {1, 2, . . . , n} is a permutation
that leaves all elements but two fixed and it maps those elements onto the other, see [23, p. 13]. Then every permutation
on {1, 2, . . . , n} is a product of adjacent transpositions (1, 2), (2, 3), . . . , (n − 1, n), see [23, Exercise 38(b), p. 79]. Thus, if
ϕ is an adjacent transposition, then a repeated application of the corollary of Theorem 9.41 implies our result (9.79).
o See https://math.stackexchange.com/questions/2234894/proof-of-the-generalized-clairauts-theorem.
249
9.6. Derivatives of higher order
′′
∂2
Dβ f
∂xβk ∂xαk
′
∂
Dβ f
=
∂xβk
=
= Dβ f.
By induction, we have the desired result.
This completes the proof of the problem.
Problem 9.30
Rudin Chapter 9 Exercise 30.
Proof. Now we have p : [0, 1] → E ⊆ Rn and h : S ⊆ R → R, where S is a set containing the closed
interval [0, 1].
(a) Since p is differentiable on [0, 1] and f ∈ C (k) (E), we deduce from Theorem 9.15 (Chain Rule) that
h′ (t) = [f ′ (p(t))][p′ (t)],
(9.87)
where [f ′ (p(t))] and [p′ (t)] are the matrices of the linear transformations f ′ (p(t)) and p′ (t) respectively. By the result of Theorem 9.17, we have


x1
 x2 


[f ′ (p(t))] = D1 f (p(t)) D2 f (p(t)) · · · Dn f (p(t))
and [p′ (t)] =  .  ,
 .. 
xn
where x = (x1 , x2 , . . . , xn ). By these, we conclude from the expression (9.87) that
h′ (t) = D1 f (p(t))x1 + D2 f (p(t))x2 + · · · + Dn f (p(t))xn .
(9.88)
Since Dj f (p(t)) ∈ C (m−1) (E) for each j = 1, 2, . . . , n, we apply Theorem 9.15 (Chain Rule) to each
Dj f (p(t)) to acquire that
(Dj f (p(t)))′ = [(Dj f )′ (p(t))][p′ (t)]
=
D1 Dj f (p(t))
D2 Dj f (p(t))

x1


 x2 
· · · Dn Dj f (p(t))  . 
 .. 

xn
= D1 Dj f (p(t))x1 + D2 Dj f (p(t))x2 + · · · + Dn Dj f (p(t))xn
for j = 1, 2, . . . , n. Therefore, we follow from the expression (9.88) that
h′′ (t) = D1 D1 f (p(t))x1 x1 + D2 D1 f (p(t))x2 x1 + · · · + Dn D1 f (p(t))xn x1
+ D1 D2 f (p(t))x1 x2 + D2 D2 f (p(t))x2 x2 + · · · + Dn D2 f (p(t))xn x2
+ · · · + D1 Dn f (p(t))x1 xn + D2 Dn f (p(t))x2 xn + · · · + Dn Dn f (p(t))xn xn
n
X
(Dij f )(p(t))xi xj .
(9.89)
=
i,j=1
Hence, if k ≥ 3, then by applying Theorem 9.15 (Chain Rule) (k − 2)-times to the expression (9.89),
we obtain our desired result.
Chapter 9. Functions of Several Variables
250
(b) Since h(1) = f (p(1)) = f (a + x), it follows from Theorem 5.15 (Taylor’s Theorem) that
f (a + x) =
m−1
X
k=0
h(k) (0) h(m) (t)
+
k!
m!
(9.90)
for some t ∈ (0, 1). By the result of part (a), we have
X
X
h(k) (0) =
(Di1 i2 ...ik f )(p(0))xi1 · · · xik =
(Di1 i2 ...ik f )(a)xi1 · · · xik ,
where the sum extends over all ordered k-tuples (i1 , . . . , ik ) in which each ij is one of the integers
1, . . . , n. Thus the expression (9.90) can be rewritten asp
f (a + x) =
m−1
X
k=0
1 X
h(m) (t)
(Di1 i2 ...ik f )(a)xi1 · · · xik +
k!
m!
(9.91)
(m)
for some t ∈ (0, 1). Let r(x) = h m!(t) . Then we have
X
r(x)
1
=
(Di1 i2 ...im f )(p(t))xi1 · · · xim
|x|m−1
|x|m−1
(9.92)
for some t ∈ (0, 1). The sum extends over all ordered m-tuples (i1 , . . . , im ) in which each ij is one of
the integers 1, . . . , n. By the definition given in Problem 9.29, we see that each Di1 i2 ...im f ∈ C (E),
i.e., Di1 i2 ...im f is a real-valued, continuous and bounded function on E. Since p is continuous on
[0, 1], the set p([0, 1]) is compact by Theorem 4.14. By Theorem 4.16 (Extreme Value Theorem),
there exists a positive integer M such that
|(Di1 i2 ...im f )(p(t))| ≤ M
for all m-tuples (i1 , . . . , im ) and t ∈ [0, 1]. Therefore, the expression (9.92) gives
X
1
|r(x)|
(Di1 i2 ...im f )(p(t))xi1 · · · xim
≤
m−1
m−1
|x|
|x|
X
1
≤
M
xi1 · · · xim
|x|m−1
≤ M |x|
which implies that
lim
r(x)
x→0 |x|m−1
= 0.
(c) Fix the positive integer n. For each nonnegative k, we have the k letters i1 , i2 , . . . , ik which are
elements of the set of numbers {1, 2, . . . , n}. It is clear that there are k! ways to permute i1 , i2 , . . . , ik .
However, some letters may repeat. Let s1 , s2 , . . . , sn be the multiplicities of the numbers 1, 2, . . . , n
in the set {i1 , i2 , . . . , ik } respectively. Then they are nonnegative and
s1 + s2 + · · · + sn = k
so that the numberq
k!
s1 !s2 ! · · · sn !
represents the number of distinct ways to permute the numbers 1, 2, . . . , n.
Apply this formula (9.93) and Problem 9.29 to the inner summation (9.91), we have
f (a + x) =
m−1
X
k=0
1 X
(Di1 Di2 · · · Dik f )(a)xi1 · · · xik + r(x)
k!
p As mentioned in the problem, we may simply define (D
i1 i2 ...ik f )(a) = f (a) and xi1 · · · xik = 1 when k = 0.
q This number is the multinomial coefficient in the Multinomial Theorem.
(9.93)
251
9.6. Derivatives of higher order
m−1
X
1 X
k!
(Ds1 Ds2 · · · Dnsn f )(a)xs11 · · · xsnn + r(x)
k!
s1 !s2 ! · · · sn ! 1 2
k=0
X
(D1s1 D2s2 · · · Dnsn f )(a) s1
x1 · · · xsnn + r(x).
=
s1 !s2 ! · · · sn !
=
s1 +···+sn ≤m−1
Hence this completes the proof of the problem.
Problem 9.31
Rudin Chapter 9 Exercise 31.
Proof. We have f : E ⊆ R2 → R, where E is a neighborhood of a. By definition, we have
∇f (a) = (D1 f )(a)e1 + (D2 f )(a)e2 = 0
which is equivalent to the fact thatr
(D1 f )(a) = (D2 f )(a) = 0.
By these and the result of Problem 9.30, the Taylor polynomial of f at a is given by
f (a + x) − f (a) =
X
1≤s1 +s2 ≤2
(D1s1 D2s2 f )(a) s1 s2
x1 x2
s1 !s2 !
= (D1 f )(a)x1 + (D2 f )(a)x2
1
+ 2(D1 D2 f )(a)x1 x2 + (D12 f )(a)x21 + (D22 f )(a)x22 + r(x)
2
1
= 2(D1 D2 f )(a)x1 x2 + (D12 f )(a)x21 + (D22 f )(a)x22 + r(x),
2
(9.94)
where x = (x1 , x2 ) is close to 0 = (0, 0) and r(x) is the remainder such that
r(x)
→0
|x|2
as x → 0. Therefore, the sign of f (a + x) − f (a) is determined by that of the brackets in the right-hand
side of the expression (9.94).
By the hypothesis, we know that not all (D1 D2 f )(a), (D12 f )(a) and (D22 f )(a) are zero. Let
(D12 f )(a) (D12 f )(a)
Ha =
(D21 f )(a) (D22 f )(a)
be the Hessian matrix at the point a. Now we can rewrite the expression (9.94) as
f (a + x) − f (a) = Q(x) + r(x),
(9.95)
where Q(x) is defined bys
Q(x) =
1 2
(D1 f )(a)x21 + 2(D1 D2 f )(a)x1 x2 + (D22 f )(a)x22 .
2
(9.96)
Suppose that (D12 f )(a) 6= 0. Then we can rewrite the expression (9.96) as
Q(x) =
o
n
2
1
2
2
.
+
(det
H
)x
(D
f
)(a)x
+
(D
D
f
)(a)x
a
1
1
2
2
2
1
2(D12 f )(a)
r By definition, the point a is a critical point of f .
s Note that (D
12 f )(a) = (D21 f )(a) because of the corollary of Theorem 9.41.
(9.97)
Chapter 9. Functions of Several Variables
252
• Case (i): det Ha > 0. In this case, the brackets in the right-hand side of the expression (9.97) is the
sum of two squares. In this case, we have Q(x) > 0 if (D12 f )(a) > 0 and Q(x) < 0 if (D12 f )(a) < 0.
Hence it follows from this and the expression (9.95) that
– f has a local minimum at a if (D12 f )(a) > 0 and det Ha > 0,
– f has a local maximum at a if (D12 f )(a) < 0 and det Ha > 0.
• Case (ii): det Ha < 0. By the form (9.97), Q(x) can be expressed in the form
Q(x) = γ(αx1 + βx2 )(αx1 − βx2 ),
(9.98)
where α, β, γ are some constants and γ 6= 0. Now it is clear from the form (9.98) that Q(x) = 0 if
and only if
αx1 + βx2 = 0 or αx1 − βx2 = 0.
Besides, the lines αx1 + βx2 = 0 and αx1 − βx2 = 0 divides the x1 x2 -plane into four regions. See
Figure 9.8 for details.t
Figure 9.8: The four regions divided by the two lines αx1 + βx2 = 0 and αx1 − βx2 = 0.
By direct computation, we have
Q(1, 0) = Q(−1, 0) = α2 γ
and Q(0, 1) = Q(0, −1) = −β 2 γ.
When γ > 0, then Q(x) > 0 in Regions I and III and Q(x) < 0 in Regions II and IV; when γ < 0,
then Q(x) < 0 in Regions I and III and Q(x) > 0 in Regions II and IV. Hence f has a saddle point
at a.
• Case (iii): det Ha = 0. Then the point a may be a local maximum, a local minimum or a saddle
point.
This finishes the discussion of the first assertion.
We start to prove the second assertion. We consider the function f : E ⊆ Rn → R, where E is a
neighborhood of a. Suppose that
n
X
(Di f )(a)ei = 0
∇f (a) =
i=1
t Here we suppose that the slope of the line αx
1 + βx2 = 0 is positive. The other case can be done similarly.
253
9.6. Derivatives of higher order
and not all second-order derivatives of f are 0 at a. By these and the results from Problem 9.30, we have
f (a + x) − f (a) =
=
X
1≤s1 +···+sn ≤2
n
X
(D1s1 · · · Dnsn f )(a) s1
x1 · · · xsnn + r(x)
s1 ! · · · sn !
1
(Di Dj f )(a)xi xj + r(x).
2 i,j=1
(9.99)
Similar as in the case of R2 , we define
Ha = [(Dij f )(a)]
(9.100)
to be the Hessian matrix at the point a, where (Dij f )(a) is the entry appears in the ith row and jth
column of the matrix (9.100). By this, the expression (9.99) can be written in the form
f (a + x) − f (a) =
where
1 T
x Ha x + r(x),
2
(9.101)


x1
 x2 


x= . 
 .. 
xn
and xT is the transpose of the vector x. Recall that a symmetric matrix A (i.e., AT = A) is called positive
definite if the quadratic form
xT Ax > 0
for all x 6= 0. To finish our proof, we need to find bounds of the number xT Ax when A is symmetric.
For this purpose, some basic results about eigenvalues of a matrix are needed, see [15, Theorem 5, §7.2,
Chap. 7, p. 405] or [3, Exercise 7, §5, Chap. 7, p. 266]:
Lemma 9.3
Let A be an n × n symmetric matrix. Then a quadratic form xT Ax is
(a) positive definite if and only if the eigenvalues of A are all positive,
(b) negative definite if and only if the eigenvalues of A are all negative, or
(c) indefinite if and only if A has both positive and negative eigenvalues.
Lemma 9.3 provides a quick way to check whether a symmetric matrix A is positive definite or negative
definite. Another simple way to show that a matrix is positive definite is by using its submatrices: A real
symmetric n × n matrix A is positive definite if and only if
det Ai > 0
for each i = 1, 2, . . . , n, where Ai is the upper left i × i submatrix of A. We need one more lemma, see
[3, Proposition 5.7, §5, Chap. 7, p. 255]:
Lemma 9.4 Spectral Theorem (real case)
Let T be a symmetric operator on a real vector space V with a positive definite bilinear
form. Then there is an orthonormal basis of V consisting of eigenvectors of T .
Let λ1 , λ2 , . . . , λn be eigenvalues of an n× n symmetric and positive definite matrix A with real entries
and x1 , . . . , xn be their corresponding eigenvectors. By Lemma 9.3(a), λ1 , λ2 , . . . , λn are real and positive.
Chapter 9. Functions of Several Variables
254
By Lemma 9.4, without loss of generality, we may assume that the set {x1 , . . . , xn } is an orthonormal
basis of Rn . Given non-zero x ∈ Rn , we have
x = c1 x1 + · · · + cn xn
for some scalar c1 , . . . , cn . It follows that
xT Ax =
n
X
i=1
ci xi
n
n
n
n
X
T X
X
T X
λi c2i ≥ λ|x|2 ,
ci λi xi =
ci xi
ci xi =
A
i=1
i=1
i=1
(9.102)
i=1
where λ = min (λi ) > 0. Now it is time to determine whether f has a local maximum, or a local
1≤i≤n
minimum, or a saddle point, at a as follows:
• Case (i): Ha is positive definite. By the corollary of Theorem 9.41, the matrix Ha is symmetric.
Put A = Ha into the inequality (9.102) so that the expression (9.101) becomes
f (a + x) − f (a) ≥
1
λ|x|2 + r(x),
2
(9.103)
λ
where x is non-zero and so close to 0. Since r(x)
|x|2 → 0 as x → 0, for 2 > 0 there exists a δ > 0 such
that
r(x)
λ
<
(9.104)
|x|2
2
for 0 < |x| < δ. By the equivalent form
λ
λ
− |x|2 < r(x) < |x|2
2
2
of the inequality (9.104), the inequality (9.103) can be reduced to
f (a + x) − f (a) > 0
for 0 < |x| < δ. This means that f has a local minimum at the point a.
• Case (ii): Ha is negative definite. In this case, Lemma 9.3(b) says that λ1 , . . . , λn are all
negative. Instead of the inequalities (9.102) and (9.103), we have
xT Ha x ≤ ρ|x|2
and
f (a + x) − f (a) ≤
ρ 2
|x| + r(x),
2
(9.105)
where x is non-zero and so close to 0, and ρ = max (λi ) < 0. Again, since r(x)
|x|2 → 0 as x → 0, for
1≤i≤n
|ρ|
2 > 0 there exists a δ > 0 such that
r(x)
|ρ|
<
|x|2
2
(9.106)
for 0 < |x| < δ. By the equivalent form
−
|ρ| 2
|ρ| 2
|x| < r(x) <
|x|
2
2
of the inequality (9.106), we see that the inequality (9.105) induces
f (a + x) − f (a) ≤
ρ + |ρ| 2
ρ 2
|x| + r(x) <
|x| = 0
2
2
for 0 < |x| < δ. In other words, f has a local maximum at the point a.
• Case (iii): Ha is indefinite. In this case, the point a may be a local maximum, a local minimum
or a saddle point in this case.
This proves our second assertion, completing the proof of the problem.
CHAPTER
10
Integration of Differential Forms
10.1
Integration over sets in Rk and primitive mappings
Problem 10.1
Rudin Chapter 10 Exercise 1.
Proof. Let H be a compact convex set in Rk , supp (f ) ⊆ H and H ◦ 6= ∅, where H ◦ denotes the interior
of H (see Problem 2.9). If f ∈ C (H), we extend f to a function on I k containing H by setting f (x) = 0
for all x ∈ I k \ H, and define
Z
Z
f,
f=
(10.1)
Ik
H
where
I k = {(x1 , . . . , xk ) | 0 ≤ xi ≤ 1, i = 1, 2, . . . , k}.
As suggested by the hint, we are going to show that f can be approximated by functions F that are
continuous on Rk and supp (F ) ⊆ H. Before that, we need to show the existence of the integral on the
right-hand side in (10.1).
Denote ∂H to be the boundary of the compact convex set H. Since H ◦ 6= ∅, we define the function
ρ : I k → R by
ρ∂H (x), if x ∈ H;
ρ(x) =
0,
if x ∈ I k \ H,
inf{|x − y| | y ∈ ∂H}, if x ∈ H;
=
(10.2)
0,
if x ∈ I k \ H.
By Problem 4.20(a),
ρ∂H (x) = 0
if and only if x ∈ ∂H = ∂H.
By this and Problem 4.20(b), ρ is (uniformly) continuous on I k . Suppose 0 < δ < 1, put


 1, if δ ≤ t < +∞;
t
ϕ(t) =
, if 0 < t < δ;

δ

0, if t = 0.
Now it is clear that ϕ is continuous [0, +∞). Define F : I k → R by F (x) = ϕ(ρ(x))f (x). Since f ∈ C (H),
F is obviously continuous on H. Since f (x) = 0 on I k \ H, F is continuous on I k \ H too. However, it is
not clear whether F is continuous on ∂H or not. In fact, the answer to this question is affirmative: Let
p ∈ ∂H and {xn } be a sequence in I k \ ∂H converging to p. By the continuity of the functions ϕ and ρ,
we observe that
lim ϕ(ρ(xn )) = ϕ(ρ(p)) = 0.
n→∞
255
Chapter 10. Integration of Differential Forms
256
Since H is compact, we have p ∈ H so that f (p) is well-defined. Furthermore, Theorem 4.15 implies
that f is bounded by a positive constant M on H. Therefore, we have
0 ≤ lim |F (xn )| = lim |ϕ(ρ(xn ))f (xn )| ≤ M lim |ϕ(ρ(xn ))| = M |ϕ(ρ(p))| = 0
n→∞
n→∞
n→∞
which means that
lim F (xn ) = 0 = ϕ(ρ(p))f (p) = F (p).
n→∞
Hence F is also continuous on ∂H and then F ∈ C (I k ).a
Lemma 10.1
Put y = (x1 , . . . , xk−1 ) ∈ I k−1 . Let S = {xk ∈ [0, 1] | F (y, xk ) 6= f (y, xk )}. Then the set
S is either empty or is a line segment whose length does not exceed δ.
Proof of Lemma 10.1. Suppose that S 6= ∅. Then we have F (y, xk ) 6= f (y, xk ) for some
xk ∈ [0, 1]. By definition, this means that
0 ≤ ϕ(ρ(y, xk )) < 1
and f (y, xk ) 6= 0.
In addition, these conditions are equivalent to
0 ≤ ρ(y, xk ) < δ
and f (y, xk ) 6= 0.
(10.3)
By the second condition in (10.3), we must have (y, xk ) 6∈ I k \ H. Thus, (y, xk ) ∈ H so that
the first condition in (10.3) reduces to
0 < ρ(y, xk ) = ρ∂H (y, xk ) < δ.
Since H is convex, the line segment joining the point (y, xk ) and any point on ∂H is still inside
H. See Figure 10.1 for details. Therefore, S must be a line segment whose length is less than
δ. This completes the proof of the lemma.
Figure 10.1: The compact convex set H and its boundary ∂H.
a It should be noted that since f may not be continuous on I k , we can’t conclude that
lim f (xn ) = f (p).
n→∞
10.1. Integration over sets in Rk and primitive mappings
257
Let return to the proof of the problem. Since 0 ≤ ϕ ≤ 1, it follows that
Z 1
|Fk (y, xk ) dxk − fk (y, xk )| dxk ≤ δkf k,
|Fk−1 (y) − fk−1 (y)| ≤
(10.4)
0
where kf k = max |f (x)|. As δ → 0, the inequality (10.4) exhibits fk−1 as a uniform limit (with respect
x∈I k
to δ) of a sequence of continuous functions {Fk−1 } on I k−1 . This proves that fk−1 ∈ C ( I k−1 ) and also
the existence of the integral on the right-hand side in (10.1).
Furthermore, if F = Fk and f = fk , then we can rewrite the inequality (10.4) as
Z
Ik
F−
Z
Ik
f ≤ δkf k
which is true no matter what the order of the k single integrations is. As we have shown above, F ∈ C (I k ),
so
L(F ) = L′ (F )
Z
by Theorem 10.2. Hence the same is true for f , i.e.,
f is unaffected by any change of the order of
integration. This completes the proof of the problem.
Problem 10.2
Rudin Chapter 10 Exercise 2.
Proof. By Definition 10.3, we have
supp (f ) = {(x, y) ∈ R2 | f (x, y) 6= 0} ⊆ R2 .
Let Si = supp (ϕi ) ⊆ (2−i , 21−i ) ⊂ (0, 1) for each i ∈ N and
S=
∞
[
Si .
i=1
We need a lemma:
Lemma 10.2
We have f (x, y) 6= 0 if and only if x ∈ Si and y ∈ Si ∪ Si−1 for some i ∈ N.
Proof of Lemma 10.2. By the hypothesis, we note that Si ∩ Sj = ∅ for all i, j ∈ N and i 6= j.
In other words, the supports {Si } are mutually disjoint. Suppose that x ∈ Si and y ∈ Si ∪ Si−1
for some i ∈ N. Then we have ϕi (x) 6= 0, ϕj (x) = 0 for all j 6= i and either ϕi (y) 6= 0 or
ϕi−1 (y) 6= 0 so that
f (x, y) = [ϕi−1 (x) − ϕi (x)]ϕi−1 (y) + [ϕi (x) − ϕi+1 (x)]ϕi (y)
= −ϕi (x)ϕi−1 (y) + ϕi (x)ϕi (y)
= [ϕi (y) − ϕi−1 (y)]ϕi (x)
6= 0.
Next, suppose that x 6∈ Si or y 6∈ Si for all i ∈ N. Then we have
ϕi (x) = 0
or ϕi (y) = 0
for all i ∈ N. In this case, the definition of f gives f (x, y) = 0. This completes the proof of the
lemma.
Chapter 10. Integration of Differential Forms
258
By Lemma 10.2, if f (x, y) 6= 0, then (x, y) ∈ Si × (Si ∪ Si−1 ) ⊂ (0, 1) × (0, 1) for some i ∈ N. This
means that
S = {(x, y) ∈ R2 | f (x, y) 6= 0} ⊂ (0, 1) × (0, 1).
(10.5)
Since S ⊂ (0, 1) × (0, 1) ⊂ [0, 1] × [0, 1] and [0, 1] × [0, 1] is closed in R2 , Theorem 2.27(c) implies that
S ⊆ [0, 1] × [0, 1]. Since S = supp (f ), we have
supp (f ) ⊆ [0, 1] × [0, 1]
which means supp (f ) is a bounded set in R2 . By Theorem 2.27(a), supp (f ) is a closed set in R2 . Hence
we conclude from Theorem 2.41 (Heine-Borel Theorem) that supp (f ) is compact in R2 . This proves our
first assertion.
For the second assertion, we consider two cases:
• Case (i): f is continuous at (x, y) 6= (0, 0). Let (x, y) 6= (0, 0). If both x and y are nonzero,
then it is clear that
x ∈ Si ⊆ (2−i , 21−i ) and y ∈ Sj ⊆ (2−j , 21−j )
for some i, j ∈ N. Denote
η = min(|x − 2−i |, |x − 21−i |, |y − 2−j |, |y − 21−j |).
Then we have
Nη ((x, y)) ⊆ (2−i , 21−i ) × (2−j , 21−j ).
Furthermore, for every (p, q) ∈ Nη ((x, y)), we see that p 6∈ Sr for all r 6= i and q 6∈ St for all t 6= j.
These mean that ϕr (p) = 0 for all r 6= i and ϕt (q) = 0 for all t 6= j, and they imply that

0,
if i < j;



ϕi (p)ϕi (q),
if i = j;
(10.6)
f (p, q) =
−ϕi (p)ϕi−1 (q), if i = j + 1;



0,
if i > j + 1
for all (p, q) ∈ Nη ((x, y)).b By the definition (10.6), we always have

0,



|ϕi (x)ϕi (y) − ϕi (p)ϕi (q)|,
|f (x, y) − f (p, q)| =
|ϕi (x)ϕi−1 (y) − ϕi (p)ϕi−1 (q)|,



0,
if i < j;
if i = j;
if i = j + 1;
if i > j + 1
(10.7)
for all (x, y), (p, q) ∈ Nη ((x, y)). Since each ϕi is continuous on R, it is also continuous on (2−i , 21−i ).
Hence, given ǫ > 0, it is easily seen from the expressions in (10.7) that we can always find 0 < δ < η
small enough so that
|f (x, y) − f (p, q)| < ǫ
if |(x, y) − (p, q)| < δ. In other words, f is continuous at every (x, y) whenever x and y are nonzero.
The remaining cases x 6= 0 and y = 0 or x = 0 and y 6= 0 can be done similarly, so we don’t repeat
the details here and we simply conclude that f is continuous at every point (x, y) 6= (0, 0).
• Case (ii): f is discontinuous at (0, 0). Next we show that f is discontinuous at (0, 0). Consider
the sequence of points {(2−k , 2−k )}. It is obvious that
(2−k , 2−k ) → (0, 0)
as k → ∞. For each k ∈ N, since 2k 6∈ (2−i , 21−i ) for all i ∈ N, we have 2k 6∈ supp (ϕi ) for all i ∈ N.
In other words, we have ϕi (2k ) = 0 for all i ∈ N and then
f (2−k , 2−k ) = 0
(10.8)
b By the hypothesis, we just require that S = supp (ϕ ) is in (2−k , 21−k ) for each k ∈ N, not the whole interval. Thus
k
k
it may happen that p ∈ (2−i , 21−i ) but not in Si or q ∈ (2−j , 21−j ) but not in Sj . As a result, it is still possible that
f (p, q) = 0 in the second or third case in the definition (10.6). However, this does not give any trouble to our argument.
10.1. Integration over sets in Rk and primitive mappings
259
for all k ∈ N. Since
1=
Z
ϕk =
R
Z 21−k
2−k
R
R
ϕk = 1 for every k ∈ N, we have
ϕk (x) dx ≤
Z 21−k
2−k
max ϕk (x) dx = max ϕk (x)
x∈Sk
x∈Sk
Z 21−k
2−k
dx = 2−k × max ϕk (x).
x∈Sk
Thus we have
max ϕk (x) ≥ 2k
x∈Sk
for every k ∈ N. By Theorem 2.27(a), Sk is closed. Since Sk ⊂ (0, 1) for all k ∈ N, each Sk
is bounded and Theorem 2.41 (Heine-Borel Theorem) implies that each Sk is compact. Since
ϕk : Sk → R is continuous, Theorem 4.16 (Extreme Value Theorem) ensures that ϕk attains its
maximum value 2k at some point in the interval Sk . Let one of these points be pk ∈ Sk , i.e.,
ϕk (pk ) = 2k . Consider another sequence of points {(pk , pk )}. By the construction, it is clear that
pk ∈ (2−k , 21−k ) and so
(pk , pk ) → (0, 0)
as k → ∞. By the definition of f , we have
f (pk , pk ) = [ϕk (pk ) − ϕk+1 (pk )]ϕk (k) = [ϕk (k)] = 2k+1
giving
f (pk , pk ) → ∞
(10.9)
as k → ∞. By comparing the two results (10.8) and (10.9), we conclude that f is discontinuous at
(0, 0). This shows the second assertion.c
To prove the equations in the third assertion, we basically follow the consideration of Definition 10.1.
If y 6∈ Si for all i ∈ N, then we have ϕi (y) = 0 for all i ∈ N so that f (x, y) = 0 which gives
Z
f (x, y) dx = 0
and thus
Z
dy
Z
f (x, y) dx =
dy
0
R
R
Z 1
Z
f (x, y) dx =
R
∞ Z 21−i
X
i=1
dy
2−i
Z
f (x, y) dx.
On each Si , we know that f (x, y) = [ϕi (x) − ϕi+1 (x)]ϕi (y) and thus
Z 21−i
dy
2−i
Z
Z 21−i
f (x, y) dx =
ϕi (y)
2−i
R
R
Z 21−i
=
Z
ϕi (x) dx −
(10.10)
R
Z
!
ϕi+1 (x) dx dy
R
ϕi (y)(1 − 1) dy
2−i
= 0.
(10.11)
Combining the expressions (10.10) and (10.11), we get
Z
Z
dy f (x, y) dx = 0.
R
R
Similarly, if x 6∈ Si for all i ∈ N, then we have ϕi (x) = 0 for all i ∈ N so that f (x, y) = 0 which gives
Z
f (x, y) dy = 0
and thus
Z
R
dx
Z
R
f (x, y) dy =
Z 1
0
dx
Z
R
f (x, y) dy =
∞ Z 21−i
X
i=1
dx
2−i
c The limit (10.9) also shows that f is unbounded in every neighborhood of (0, 0).
Z
f (x, y) dy.
R
(10.12)
Chapter 10. Integration of Differential Forms
On each Si , we have
f (x, y) =
260
ϕi (x)[ϕi (y) − ϕi−1 (y)],
ϕ1 (x)ϕ1 (y),
if i 6= 1;
if i = 1.
Thus we obtain
Z


[ϕi (y) − ϕi−1 (y)] dy, if i 6= 1;
ϕ
(x)


Z
 i
R
f (x, y) dy =
Z

R


 ϕ1 (x) ϕ1 (y) dy,
if i = 1,
R
0,
if i 6= 1;
=
ϕ1 (x), if i = 1.
(10.13)
Combining the expressions (10.12) and (10.13), we obtain
Z
R
dx
Z
f (x, y) dy =
R
∞ Z 21−i
X
i=1
=
Z 1
dx
1
2
=
Z 1
1
2
dx
2−i
Z
Z
f (x, y) dy
R
f (x, y) dy +
R
∞ Z 21−i
X
i=2
ϕ1 (x) dx +
∞ Z 21−i
X
i=2
= 1.
2−i
dx
Z
f (x, y) dy
R
0 dx
2−i
This completes the proof of the problem.
Problem 10.3
Rudin Chapter 10 Exercise 3.
Proof.
(a) We modify the proof of Theorem 10.7 to obtain the result. In fact, put F1 = F. Since F′1 (0) = I,
we also have F′ (0) = I. By the assumption, we know that F1 ∈ C ′ (V1 ) for some neighborhood
V1 ⊆ E of 0. Then we still have [21, Eqn. (19), p. 249] in the case m = 1:
F′1 (0)e1 =
n
X
(D1 Fi )(0)ei ,
(10.14)
i=1
where F1 , . . . , Fn are real C ′ -functions in V1 . Recall from the proof of Theorem 10.7 that there is a
k such that 1 ≤ k ≤ n, (D1 Fk )(0) 6= 0 and B1 is defined to be the flip that interchanges 1 and this
k. Next, we apply the fact F′1 (0) = I to the formula (10.14) to obtain
e1 = (D1 F1 )(0)e1 + (D1 F2 )(0)e2 + · · · + (D1 Fn )(0)en .
Thus, we have (D1 F1 )(0) = 1 and (D1 Fi )(0) = 0 for all i = 2, 3, . . . , n. This means that k = m = 1,
so B1 = I. Since (D1 F1 )(0) = 1, we obtain from the expression
G1 (x) = x + [F1 (x) − x1 ]e1
that G1 ∈ C ′ (V1 ), G1 is primitive and
G′1 (0) = I.
Therefore, we have
F′1 (0) = G′1 (0) = B1 = I.
10.1. Integration over sets in Rk and primitive mappings
261
Assume that F′i (0) = G′i (0) = Bi = I, Gi ∈ C ′ (Vi ) and Gi is primitive for all 1 ≤ i ≤ m. By [21,
Eqn. (21), p. 250], we still have
−1
Fm+1 (y) = Bm Fm ◦ G−1
m (y) = Fm ◦ Gm (y)
(y ∈ Vm+1 ).
(10.15)
Apply Theorem 9.15 (Chain Rule) and the induction hypothesis to the expression (10.15) to get
−1
′
′
−1
′
−1
′
F′m+1 (0) = F′m (G−1
m (0))(Gm (0)) = Fm (0)(Gm (0)) = (Gm (0)) .
(10.16)
By using [21, Eqn . (52), p. 223] in the proof of Theorem 9.24 (Inverse Function Theorem) and
then the induction hypothesis, we acquire that
′
′
−1
−1
(G−1
= (G′m (0))−1 = I−1 = I.
m (0)) = (Gm (Gm (0)))
Thus the expression (10.16) implies that F′m+1 (0) = I. By Theorem 9.17, we have
F′m+1 (0)em+1 =
n
X
(Dm+1 Fi )(0)ei .
(10.17)
i=m+1
Recall from the proof of Theorem 10.7 that there is a k such that m + 1 ≤ k ≤ n, (Dm+1 Fk )(0) 6= 0
and Bm+1 is defined to be the flip that interchanges m + 1 and this k. Since F′m+1 (0) = I, we
deduce from the formula (10.17) that
em+1 = (Dm+1 Fm+1 )(0)em+1 + · · · + (Dm+1 Fn )(0)en
so that
(Dm+1 Fm+1 )(0) = 1 and (Dm+1 Fi )(0) = 0
for i = 2, . . . , n. As a result, we have k = m+1 and then Bm+1 = I. Similarly, since (Dm+1 Fm+1 )(0) =
1, we obtain from the expression
Gm+1 (x) = x + [Fm+1 (x) − xm+1 ]em+1
that
G′m+1 (0) = I.
To sum up, what we have shown is that F′i (0) = G′i (0) = Bi = I, Gi ∈ C ′ (Vi ) and Gi is primitive
for all 1 ≤ i ≤ n. The expression (10.15), with y = Gm (x), is equivalent to
Fm (x) = Fm+1 (Gm (x))
(x ∈ Um ).
(10.18)
By applying this with m = 1, 2 . . . , n − 1, we establish that
F1 = F2 ◦ G1 = F3 ◦ G2 ◦ G1 = · · · = Fn ◦ Gn−1 ◦ · · · ◦ G1
in some neighborhood of 0. By [21, Eqn. (18), p. 249], we have
Fn (x) = Pn−1 x + αn (x)en ,
(10.19)
so Fn is also primitive by Definition 10.5. Hence we just rename Fn by Gn in the formula (10.19)
so as to obtain our desired result.
(b) Let F : R2 → R2 be the mapping defined by
F(x, y) = (y, x).
Assume that F = G2 ◦ G1 , where each Gi : Vi ⊆ R2 → R2 is a primitive C ′ -mapping in some
neighborhood of 0, Gi (0) = 0 and G′i (0) is invertible. By Theorem 9.15 (Chain Rule), we have
F′ (0) = G′2 (G1 (0))G′1 (0) = G′2 (0)G′1 (0).
(10.20)
Chapter 10. Integration of Differential Forms
Now we have
′
F (0) =
262
0
1
1
0
.
(10.21)
Since G1 is a primitive, we have either
G1 (x, y) = (x, g1 (x, y))
or G1 (x, y) = (g1 (x, y), y),
where g1 : V1 → R is a real function and it is differentiable at 0. By direct computation, we have
either
1 0
a b
G′1 (0) =
and G′1 (0) =
,
a b
0 1
where a and b are real numbers. Similarly, we have either
1 0
c d
′
′
G2 (0) =
and G2 (0) =
,
c d
0 1
where c and d are real numbers. Thus the right-hand side of the expression (10.20) is in one of the
following forms:
1
0
c + ad bd
a
b
ac bc + d
,
,
,
,
c + ad bd
0
b
ac bc + d
0
1
but none of them is the matrix (10.21). Hence, the mapping F is not the composition of any two
primitive mappings.
We complete the proof of the problem.
Problem 10.4
Rudin Chapter 10 Exercise 4.
Proof. By direct computation, we have
(G2 ◦ G1 )(x, y) = G2 (G1 (x, y))
= G2 (ex cos y − 1, y)
= (ex cos y − 1, (1 + ex cos y − 1) tan y)
= (ex cos y − 1, ex sin y)
= F(x, y)
for every (x, y) ∈ R2 . Suppose that g1 , g2 : R2 → R are given by
g1 (x, y) = ex cos y − 1 and g2 (x, y) = (1 + x) tan y.
Then we have
G1 (x, y) = (0, y) + (ex cos y − 1, 0) = ye2 + g1 (x, y)e1
and
G2 (x, y) = (x, 0) + (0, (1 + x) tan y) = xe1 + g2 (x, y)e2 .
By Definition 10.5, G1 and G2 are primitive. By Definition 9.38, we have
x
e cos y −ex sin y
JG1 (x, y) =
,
0
1
1
0
JG2 (x, y) =
,
tan y (1 + x) sec2 y
x
e cos y −ex sin y
JF (x, y) =
.
ex sin y ex cos y
263
10.2. Generalizations of partitions of unity
Thus we have
JG1 (0, 0) = JG2 (0, 0) = JF (0, 0) =
1
0
0
1
.
Let D = {(u, v) ∈ R2 | u2 + v 2 ≤ 1} be the unit disk. Then we have e2u − v 2 ≥ 0 for all (u, v) ∈ D, so
p
H1 (u, v) = ( e2u − v 2 − 1, v)
is well-defined. Then we have
(H1 ◦ H2 )(x, y) = H1 (H2 (x, y))
= H1 (x, ex sin y)
p
= ( e2x − (ex sin y)2 − 1, ex sin y)
= (ex cos y − 1, ex sin y)
= F(x, y)
in the unit disk D, completing the proof of the problem.
10.2
Generalizations of partitions of unity
Problem 10.5
Rudin Chapter 10 Exercise 5.
Proof. Here is the analogue of Theorem 10.8: Suppose K is a compact subset of a metric space X with
metric d, and {Vα } is an open cover of K. Then there exists functions ψ1 , . . . , ψs ∈ C (X) such that
(a) 0 ≤ ψi ≤ 1 for 1 ≤ i ≤ s;
(b) each ψi has its support in some Vα , and
(c) ψ1 (x) + · · · + ψs (x) = 1 for every x ∈ K.
When we read Rudin’s proof carefully, we see that the key step of the construction there are relations
[21, Eqns. (26) & (27), p.251]. More precisely, the relation [21, Eqn. (26), p.251] tells us that
W (xi ) \ B(xi ) 6= φ
and B(xi ) ∩ W (xi )c = φ.
(10.22)
How do these relations (10.22) motivate ideas of the construction of functions in our question? We observe
a few points first. Functions of the type in Problem 4.22 are constructed on disjoint nonempty closed
sets A and B. Since
B(xi ) and W (xi )c
are disjoint nonempty closed subsets of Rn , they satisfy this condition. By constructing each ϕi ∈ C (Rn ),
the relation [21, Eqn. (27), p.251] plays the role of “gluing” them together to give functions satisfying
the unity requirement. Thus these give us a “direction” of proving our result. In fact, we first construct
sets having properties similar to relations [21, Eqns. (26) & (27), p.251]. Since K is compact, {Vα } has
a finite subcover and then there is a positive integer s such that
K ⊆ V1 ∪ V2 ∪ · · · ∪ Vs .
We need some results from topology:
Lemma 10.3
If X is a metric space, then it is Hausdroff.
Chapter 10. Integration of Differential Forms
264
Proof of Lemma 10.3. See [18, p. 98] for the definition of a Hausdroff space. Let x, y ∈ X
with x 6= y. Let δ = d(x, y), U = N δ (x) and V = N δ (y). Then it is easy to check that
2
2
U ∩ V = ∅,
proving the lemma.
Since K is a subset of X, it is also Hausdroff by [18, Theorem 17.11, p. 100]. Furthermore, it follows
from Lemma 10.3 and [18, Theorem 32.3, p. 202] that K is normal.d Now, by using Step 1 of the proof
of Theorem 36.1 in [18, p. 225], we can show that it is possible to find open coverings {U1 , . . . , Us } and
{W1 , . . . , Ws } of K such that
U i ⊂ Wi ⊂ W i ⊂ Vi
(10.23)
for each i = 1, . . . , s. Therefore, the sets Ui and Wi are what we need, see Figure 10.2 for the sets Ui , Wi
and Vi .
Figure 10.2: The figures of the sets Ui , Wi and Vi .
Next, we apply Problem 4.20 to construct the functions satisfying the requirements of the partitions
of unity. To do this, let
Ai = U i and Bi = Wic ,
where i = 1, . . . , s. Then the relations (10.23) show that Ai and Bi are disjoint nonempty closed sets.
By Problem 4.20, we consider the continuous function ϕi : X → [0, 1] defined by
ϕi (x) =
ρBi (x)
.
ρAi (x) + ρBi (x)
Then we have ϕi (x) = 0 precisely on Bi , ϕi (x) = 1 precisely on Ai and 0 ≤ ϕi (x) ≤ 1 for all x ∈ X. In
c
addition, since ϕ−1
i ((0, 1]) = Bi = Wi , Definition 10.3 and relations (10.23) imply that
supp (ϕi ) = ϕ−1
i ((0, 1]) = W i ⊂ Vi .
d See [18, p. 195] for the definition of a normal space.
265
10.2. Generalizations of partitions of unity
Let W = W1 ∪ W2 ∪ · · · ∪ Ws . Define ϕ : W → R by
ϕ(x) =
s
X
ϕi (x).
i=1
If x ∈ W , then x ∈ Wi for some i which means that ϕi (x) > 0. As a result, we have
ϕ(x) > 0
(10.24)
for all x ∈ W . Let Y = W c . By Theorem 2.24(a), W is open in X and then Theorem 2.23 ensures that
Y is closed in X. Since K and Y are disjoint nonempty closed subsets of X, Problem 4.20 again implies
that the function f : X → [0, 1] defined by
f (x) =
ρY (x)
ρK (x) + ρY (x)
is a continuous function on X, 0 ≤ f (x) ≤ 1 for all x ∈ X, f (x) = 0 precisely on Y and f (x) = 1 precisely
on K.
Finally, for each i = 1, . . . , s, we define the function ψi : X → [0, 1] by

ϕi (x)f (x)


, if x ∈ W ;

ϕ(x)
(10.25)
ψi (x) =



0,
if x ∈ X \ W .
We claim that the functions ψ1 , . . . , ψs satisfy conditions (a) to (c):e
• Proof of condition (a). To this end, we follow from the facts 0 ≤ ϕi (x) ≤ 1, 0 ≤ f (x) ≤ 1 for all
x ∈ X, the inequality (10.24) and the definition (10.25) that
0 ≤ ψi (x) ≤ 1
on X. This proves condition (a).
• Proof of condition (b). The definition (10.25) and the inequality (10.24) imply that ψ(x0 ) 6= 0
0 )f (x0 )
if and only if ϕi (xϕ(x
6= 0 if and only if ϕi (x0 )f (x0 ) 6= 0 if and only if ϕi (x0 ) 6= 0 and f (x0 ) 6= 0.
0)
By the definition of ϕi , we have
ϕi (x0 ) 6= 0
if and only if
x0 ∈ Bic = Wi .
(10.26)
Similarly, the definition of f implies that f (x0 ) 6= 0 if and only if x0 6∈ Y . Since Y = W c , we have
f (x0 ) 6= 0 if and only if
x0 ∈ W.
(10.27)
In conclusion, statements (10.26) and (10.27) tell us that
ψi (x) 6= 0 if and only if
x ∈ Wi .
Therefore, we deduce from this and the relations (10.23) that
supp (ψi ) = {x ∈ X | ψi (x) 6= 0} = W i ⊂ Vi .
This is exactly condition (b).
• Proof of condition (c). Since {W1 , . . . , Ws } is an open cover of K, if x ∈ K, then x ∈ W and it
follows from the definition (10.25) that
s
X
s
ψi (x) =
i=1
This shows condition (c).
e Such functions are called bump functions in X.
1
f (x) X
ϕi (x) =
· ϕ(x) = 1.
ϕ(x) i=1
ϕ(x)
Chapter 10. Integration of Differential Forms
266
To finish our proof, we have to show that each function ψi is continuous on X. It is trivial to see that
ψi is continuous on W and on X \ W by its definition (10.25). Thus what is left is to show that it is
continuous on the boundary of W .
Recall that the boundary of a set A, denoted by ∂A, is defined to be
A \ A◦ .
Let a ∈ ∂W . Since W is open, a ∈ X \ W f and then ψi (a) = 0. Let {an } be a sequenceg in W such that
lim an = a.
n→∞
Since f is continuous on X and f (x) = 0 on Y = W c , we have
lim f (an ) = f (a) = 0.
n→∞
i (x)
is bounded by 1 on W . Then we obtain from the remark
By the inequality (10.24), we know that ϕϕ(x)
preceding Theorem 3.20 that
lim ψi (an ) = 0 = ψi (a).
n→∞
Hence, by Theorem 4.2 and Definition 4.5, ψi is also continuous on ∂W and then on X, i.e., ψi ∈ C (X)
for i = 1, 2, . . . , s. We end the proof of the problem.
Problem 10.6
Rudin Chapter 10 Exercise 6.
Proof. As a remark, we note that a function f is called smooth if it has derivatives of all orders in its
domain.
Firstly, we recall the function in Problem 8.1:
f (x) =
1
e− x2 ,
0,
(x 6= 0);
(x = 0).
This is a function satisfying the conditions that f (x) > 0 for all x ∈ R, f ∈ C ∞ (R) and f (m) (0) = 0 for
all m = 1, 2, . . ..h Secondly, we define the function g : R → R by
√
f ( x), if x > 0;
g(x) =
0,
if x ≤ 0,
−1
e x , if x > 0;
=
0,
if x ≤ 0.
This function is nonnegative on R. Furthermore, it can be shown in the same way as the proof of Problem
8.1 that this function g has derivatives of all orders for all real x (i.e., g ∈ C ∞ (R)) and g (m) (0) = 0 for
all m = 1, 2, . . .). Thirdly, we define the function h : Rn → R by
h(x) =
g(2 − |x|)
.
g(2 − |x|) + g(|x| − 1)
(10.28)
If |x| < 2, then we have g(2 − |x|) > 0. If |x| ≥ 2, then we have g(2 − |x|) > 0 but g(|x| − 1) > 0. Thus
we always have
g(2 − |x|) + g(|x| − 1) > 0
f See [18, Exercise 19(a), p. 102]
g Such a sequence exists because of the equivalent definition of the boundary of a set: a ∈ ∂A if and only if N (a) ∩ A 6= ∅
δ
and Nδ (a) ∩ Ac 6= ∅ for every δ > 0.
h The notation C ∞ (Rn ) denotes the set of all infinitely differentiable functions on Rn for some positive integer n.
267
10.3. Applications of Theorem 10.9 (Change of Variables Theorem)
on Rn . Apart from this, since g(|x| − 1) = 0 for |x| ≤ 1, we have h(x) = 1 for |x| ≤ 1. Similarly, we
have h(x) = 0 for |x| ≥ 2 and 0 ≤ h(x) ≤ 1 for 1 ≤ |x| ≤ 2. By repeated applications of Theorems 9.15
(Chain Rule), 9.21 and the fact that g has derivatives of all orders, we conclude that h ∈ C ∞ (Rn ).
It is time to construct our desired function based on the function (10.28). We follow the proof of
Theorem 10.8. Without loss of generality, we may assume further that
Bri (xi ) ⊂ W2ri (xi ) ⊂ W2ri (xi ) ⊂ Vα(xi )
and K ⊆ Br1 (x1 ) ∪ · · · ∪ Brs (xs ),
(10.29)
where xi ∈ K, Bri (xi ) and W2ri (xi ) are open balls, centered at xi , with rational radii ri and 2ri
respectively, 1 ≤ i ≤ s. Now we define ϕi : Rn → R and ϕ : Rn → R by
ϕi (x) = h
x
ri
and ϕ(x) =
s
X
ϕi (x)
(10.30)
i=1
respectively. By the construction of the function h, we know that ϕi (x) = 1 if x ∈ Bri (xi ), ϕi (x) = 0
if x ∈ (W2ri (xi ))c and 0 ≤ ϕi (x) ≤ 1 on Rn . Besides, if x ∈ K, then x ∈ Bri (xi ) for some i so that
ϕi (x) > 0. Thus we always have
ϕ(x) > 0
on K and we can further define the function ψi : Rn → R by
ψi (x) =
ϕi (x)
.
ϕ(x)
(10.31)
We check that the functions ψ1 , . . . , ψs satisfy conditions (a) to (c):
• Proof of condition (a). It follows from the definitions (10.30) and (10.31) trivially.
• Proof of condition (b). Since ϕi (x) = 0 on (W2ri (xi ))c , we have
{x ∈ Rn | ϕi (x) 6= 0} ⊂ W2ri (x).
Therefore, we follow from the left-most relation in (10.29) that
supp (ϕi ) ⊂ W2ri (x) ⊂ Vα(xi ) .
This verifies condition (b).
• Proof of condition (c). For every x ∈ K, we have
s
X
s
1 X
ϕi (x) = 1
ψi (x) =
ϕ(x) i=1
i=1
so that condition (c) is satisfied.
This completes the proof of the problem.
10.3
Applications of Theorem 10.9 (Change of Variables Theorem)
Problem 10.7
Rudin Chapter 10 Exercise 7.
Proof.
Chapter 10. Integration of Differential Forms
268
(a) Recall that
Qk = {x = (x1 , . . . , xk ) ∈ Rk | x1 + · · · + xk ≤ 1
and x1 , . . . , xk ≥ 0}.
(10.32)
Let S = {0, e1 , . . . , ek }. Then the smallest convex subset of Rk containing S is exactly the convex
hull of S. If we denote the convex hull of S by Conv (S), then we have
Conv (S) = {c0 0 + c1 e1 + · · · + ck ek | c0 + c1 + · · · + ck = 1
and c0 , c1 , . . . , ck ≥ 0}.
(10.33)
We have to prove that Qk = Conv (S). To this end, we suppose that x ∈ Conv (S). Then we have
x = c0 0 + c1 e 1 + · · · + ck e k ,
where c0 + c1 + · · · + ck = 1 and c0 , c1 , . . . , ck ≥ 0. If x1 = c1 , x2 = c2 , . . . , xk = ck , then since
c0 0 = 0, we have
x = c1 e1 + · · · + ck ek = (c1 , . . . , ck ) = (x1 , . . . , xk ),
where x1 + · · · + xk ≤ 1 and x1 , . . . , xk ≥ 0. By definition (10.32), we have x ∈ Qk . Conversely, we
suppose that x ∈ Qk , so
x = (x1 , . . . , xk ) = x1 e1 + · · · + xk ek ,
where x1 + · · · + xk ≤ 1 and x1 , . . . , xk ≥ 0. Let c0 = 1 − (x1 + · · · + xk ), c1 = x1 , . . . , ck = xk . We
can see from these that
c0 + c1 + · · · + ck = 1, c0 , c1 . . . , ck ≥ 0
and
x = x1 e1 + · · · + xk ek = c0 0 + c1 e1 + · · · + ck ek .
By definition (10.33), we have x ∈ Conv (S). Hence, we establish
Qk = Conv (S).
(b) Let f : X → Y , where X is a convex set. Without loss of generality, we may assume that Y = f (X).
Our purpose is to show that Y is convex. Suppose that u, v ∈ Y . By Definition 10.26, we know
that
f (x) = f (0) + Ax
for some A ∈ L(X, Y ). Thus there are a, b ∈ X such that
u = f (a) = f (0) + Aa
and v = f (b) = f (0) + Ab.
Let 0 < λ < 1. Since X is convex, we have λa + (1 − λ)b ∈ X and then
λu + (1 − λ)v = f (0) + A[λa + (1 − λ)b] = f (λa + (1 − λ)b) ∈ Y.
In other words, Y is also convex which is our desired result.
This finishes the proof of the problem.
Problem 10.8
Rudin Chapter 10 Exercise 8.
Proof. Before proving the result, it is believed that Rudin missed the few words “and (1, 1) to (4, 5)” at
the end of the second sentence in the question.
By Definition 10.26, since T (0, 0) = (1, 1), the affine map T : I 2 → H has the form
T (x) = T (0) + Ax = (1, 1) + Ax,
where A is a 2 × 2 matrix and I 2 = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}. Let
a b
A=
.
c d
269
10.3. Applications of Theorem 10.9 (Change of Variables Theorem)
Then we have
T
x
y
=
1
1
+
a
c
b
d
x
y
=
ax + by + 1
cx + dy + 1
.
(10.34)
In particular, since T (1, 0) = (3, 2) and T (0, 1) = (2, 4), we obtain from the expression (10.34) that
a+1
3
b+1
2
=
and
=
c+1
2
d+1
4
which imply that a = 2, b = c = 1 and b = 3. Therefore, the affine map T is given by
x
1
2 1
x
2x + y + 1
T
=
+
=
.
y
1
1 3
y
x + 3y + 1
(10.35)
See 10.3 for the mapping T .
Figure 10.3: The mapping T : I 2 → H.
By the representation (10.35) and Definition 9.38, we have
2 1
′
JT (x) = det[T (x)] = det
=6−1=5
1 3
(10.36)
for every x ∈ I 2 .
Let I ◦ and H ◦ be the interiors of the square I 2 and the parallelogram H respectively. To evaluate the
integral α, one may think that Theorem 10.9 should be applied to the function f : H → R defined by
f (x, y) = ex−y .
Although the map T satisfies the hypotheses of the theorem (i.e., T ∈ C (I ◦ ) by Theorem 9.21 and the
fact that T ∈ C (I 2 ); T : I ◦ → H ◦ is bijective by the fact that A is invertible; and JT (x) 6= 0 on I ◦ ), the
function f fails in this case: f (x, y) 6= 0 for every (x, y) ∈ H ◦ so that
supp (f ) = H ◦ = H
which is compact in R2 , but
supp (f ) * T (I ◦ ) = H ◦ .
Therefore, we have to seek a variation of Theorem 10.9 which is applicable to this situation. In fact, the
version considered by Fitzpatrick ([8, Theorem 19.9, p. 506]) serves this purpose:i
Lemma 10.4
Suppose that U is open in Rn and the mapping T : U → Rn is a smooth change of variables.
Let D be an open Jordan domain such that K = D ∪ ∂D ⊆ U . Then T (K) is a Jordan
domain with the property that for any continuous function f : T (K) → R, the following
formula holds:
Z
Z
f (y) dy =
f (T (x))|JT (x)| dx.
T (K)
K
i T is called smooth if it is one-to-one and J (x) 6= 0 on U . Besides, a bounded subset of Rn is said to be a Jordan
T
domain if its boundary has Jordan content 0, see [8, p. 492].
Chapter 10. Integration of Differential Forms
270
To apply Lemma 10.4, we need to extend the domain of T from the unit square I 2 to the open 2-cell
R = (0, 2) × (0, 2). Now it is easy to check that T is also one-to-one and JT (x) 6= 0 on R2 . By [8,
Exercise 1, p. 496], the unit square I 2 is a Jordan domain. By definition, it means that ∂I 2 has Jordan
content 0. Since ∂I ◦ = ∂I 2 , the set I ◦ is an open Jordan domain. Furthermore, we have
2
K = I ◦ ∪ ∂I ◦ = I 2 ⊂ R2
and then T (K) = T (I 2 ) = H. Since f is obviously continuous on H, we follow from Lemma 10.4 that
Z
Z
Z
Z
f (T (x))|JT (x)| dx.
(10.37)
f (T (x))|JT (x)| dx =
f (y) dy =
ex−y dx dy =
α=
I2
T −1 (H)
H
H
By the formula (10.35), we have
f (T (x)) = f (2u + v + 1, x + 3v + 1) = eu−2v
and then it deduces from the formula (10.37) that
α=5
Z 1Z 1
0
eu−2v du dv.
(10.38)
0
Since eu−2v ∈ C (I 2 ), Theorem 10.2j says that the order of the integration on the right-hand side of
(10.38) does not matter. Hence we obtain from the integral (10.38) that
α=5
Z 1Z 1
0
eu−2v du dv = 5
0
Z 1
0
eu du
Z 1
5
e−2v dv = (e − 1)(e−2 + 1),
2
0
completing the proof of the problem.
Problem 10.9
Rudin Chapter 10 Exercise 9.
Proof. Rudin said that the interval is from (0, 0) to (0, a), but the end point is a typo, see [1, Example
1, p. 418]. Let A = {(r, θ) | 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π} and D = {(x, y) | x2 + y 2 = r} be the rectangle in the
rθ-plane and the closed disc with center (0, 0) and radius a respectively. See Figure 10.4 below:
Figure 10.4: The mapping T : A → D.
2
If f (x, y)
! I = [a, b] × [c, d], where a, b, c and d are some constants,
! = Zg(x)h(y) and
Z b
d
h(y) dy . See, for instance, [24, Theorem 3.10, p. 58].
g(x) dx
j It is a kind of Fubini’s Theorem:
then we have
Z
I2
f (x, y) dx dy =
a
c
271
10.3. Applications of Theorem 10.9 (Change of Variables Theorem)
Let T : A → D be defined by
T (r, θ) = (r cos θ, r sin θ).
Since (r cos θ) + (r sin θ) = r , T maps A into D. Given any (x, y) ∈ D. If x 6= 0, then let θ = tan−1 xy
p
(thus θ 6= π2 , 3π
x2 + y 2 which imply that
2 ) and r =
2
2
2
T (r, θ) = (x, y).
If x = 0 and y = r, then
π
T r,
= (0, r).
2
If x = 0 and y = −r, then
3π T r,
= (0, −r).
2
Therefore, the mapping T is onto.
Next, suppose that A◦ = {(r, θ) | 0 < r < a, 0 < θ < 2π} which is open by Problem 2.9(a). We
consider the mapping
T : A◦ → D0 ,
(10.39)
where D0 = {(x, y) | x2 + y 2 < a2 } \ {(r, 0) | 0 ≤ r ≤ a}.k See Figure 10.5 for the illustration.
Figure 10.5: The mapping T : A◦ → D0 .
We check the hypotheses of Theorem 10.9 for the map (10.39):
• Let (r, θ), (r′ , θ′ ) ∈ A◦ . If T (r, θ) = T (r′ , θ′ ), then we have (r cos θ, r sin θ) = (r′ cos θ′ , r′ sin θ′ )
which gives
r cos θ = r′ cos θ′ and r sin θ = r′ sin θ′ .
Thus tan θ = tan θ′ and either θ′ = θ or θ′ = π + θ. However, the relation θ′ = π + θ implies that
r cos θ = r′ cos(π + θ) = −r′ cos θ
and r sin θ = r′ sin(π + θ) = −r′ sin θ.
(10.40)
Since r > 0 and r′ > 0, there is no θ such that the two equations in (10.40) hold simultaneously.
Therefore, we have θ′ = θ and then r = r′ . In other words, the mapping T : A◦ → D0 is one-to-one.
• For the Jacobian of T , since
′
[T (r, θ)] =
we have
for every (r, θ) ∈ A◦ .
cos θ
sin θ
−r sin θ
r cos θ
,
JT (r, θ) = det[T ′ (r, θ)] = r 6= 0
k We notice that T (0, θ) = (0, 0) for every 0 ≤ θ ≤ 2π; T (r, 0) = T (r, 2π) = (r, 0) for every 0 ≤ r ≤ a; and T (a, θ) =
(a cos θ, a sin θ) for every 0 ≤ θ ≤ 2π which is the boundary of D.
Chapter 10. Integration of Differential Forms
272
Suppose that f ∈ C (D). Then f is a continuous and bounded function on the compact set D. By
Theorem 4.15, f (D) is bounded too. Since
supp (f ) = {(x, y) ∈ D | f (x, y) 6= 0} ⊆ f (D),
the set supp (f ) is also bounded. Since supp (f ) is closed, Theorem 2.41 (Heine-Borel Theorem) shows
that it is compact. In addition, since D0 ⊂ D, we know from Definition 4.5 that f is also continuous on
D0 . Let fD0 be the restriction of f to D0 (see Problem 4.7 for the definition). Then
supp (fD0 ) = {(x, y) ∈ D0 | f (x, y) 6= 0}
is a closed subset of the compact set supp (f ), so supp (fD0 ) is also compact by Theorem 2.35. Furthermore, it is clear that supp (fD0 ) lies in D0 = T (A◦ ). By Theorem 10.9 with y = (x, y) and x = (r, θ), we
must have
Z
Z
Z
(10.41)
f (T (x))|JT (x)| dx.
f (y) dy =
f (x, y) dx dy =
D0
A◦
D0
Since f ∈ C (D) and T, JT ∈ C (A), we have
(f ◦ T ) × JT ∈ C (A)
and then Theorem 10.2 implies that the mapping (f ◦ T ) × JT : A → R is integrable. By [17, Theorem
13.6, p. 110], the restriction
(f ◦ T ) × JT : A◦ → R
is also integrable and
Z
f (T (x))|JT (x)| dx =
Z
f (T (x))|JT (x)| dx.
(10.42)
A
A◦
Combining formulas (10.41) and (10.42), we obtain
Z
Z a Z 2π
f (x, y) dx dy =
f (T (r, θ))r dr dθ.
0
D◦
(10.43)
0
To summary, what we have shown in the previous paragraph is that if f ∈ C (D), then the equality
(10.43) holds on the interior D0 of D. To remove this restriction, we proceed as in Example 10.4. In
other words, we want to show something like [21, Eqn. (8), p. 248]. Since f ∈ C (D), we extend f to a
function on I 2 by setting f (x, y) = 0 on I 2 \ D, and define
Z
Z
(10.44)
f (x, y) dx dy =
f (x, y) dx dy.
D
I2
2
Here I is the 2-cell defined by
2
−1 ≤ x, y ≤ 1.
Since f may be discontinuous on I , the existence of the integral on the right-hand side of (10.44) needs
proof. To do this, suppose 0 < δ < 1 and letl

if 0 ≤ t ≤ 1 − δ;

 1,
1−t
ϕδ (t) =
(10.45)
, if 1 − δ < t ≤ 1;

 δ
0,
if t > 1.
It is easy to see that the function (10.45) is continuous on [0, ∞). Define Fδ : I 2 → R by
Fδ (x, y) = ϕδ (x2 + y 2 )f (x, y).
(10.46)
Here we have a result about this Fδ :
Lemma 10.5
The function Fδ : I 2 → R defined by (10.46) is continuous and bounded on I 2 .
l The functions ϕ and F defined in [21, Eqn. (5) & (6)] depend on δ, but Rudin didn’t mention this point clearly in the
text.
273
10.3. Applications of Theorem 10.9 (Change of Variables Theorem)
Proof of Lemma 10.5. If (x, y) ∈ D◦ , then we have Fδ (x, y) = f (x, y) which is clearly continuous on D◦ . Similarly, if (x, y) ∈ I 2 \ D, then we have Fδ (x, y) = 0 so that it is also continuous
on D \ I 2 . Suppose that (x, y) ∈ ∂D. Then we have
lim
2
2
x +y →1
x2 +y 2 <1
Fδ (x, y) =
lim
2
2
x +y →1
x2 +y 2 <1
ϕδ (x2 + y 2 )f (x, y) =
1 − (x2 + y 2 )
f (x, y) = 0
δ
x +y →1
lim
2
2
x2 +y 2 <1
and
lim
x2 +y 2 →1
x2 +y 2 >1
Fδ (x, y) =
lim
x2 +y 2 →1
x2 +y 2 >1
ϕδ (x2 + y 2 )f (x, y) =
lim
x2 +y 2 →1
x2 +y 2 >1
0 × f (x, y) = 0.
These mean that Fδ is continuous on ∂D. Since D is compact, we conclude from Theorem 4.15
that Fδ ∈ C (I 2 ).
Let’s return to the proof of the problem. For each x0 ∈ [−1, 1], let
Sx0 = {y ∈ [−1, 1] | Fδ (x0 , y) 6= f (x0 , y)}.
If Sx0 6= ∅, then Fδ (x0 , y) 6= f (x0 , y) for some y ∈ [−1, 1] which amounts to ϕ(x20 + y 2 ) 6= 1 and
f (x0 , y) 6= 0 and then it is equivalent to
1 − δ ≤ x20 + y 2 ≤ 1 and f (x0 , y) 6= 0.
√
In this case, the set Sx0 is a segment of length does not exceed 2(1 − 1 − δ). By definition (10.45), we
have 0 ≤ ϕ ≤ 1 which shows that
√
(10.47)
|F1δ (x) − f1 (x)| ≤ 2(1 − 1 − δ) · max 2 |f (x, y)|
(x,y)∈I
for all x ∈ [−1, 1], where
F1δ (x) =
Z 1
Fδ (x, y) dy
and f1 (x) =
−1
Z 1
f (x, y) dy.
−1
As δ → 0, the inequality (10.47) implies that f1 is a uniform limit of a sequence of continuous functions
{F1δ } on [−1, 1]. By Theorem 7.12, we have f1 ∈ C ([−1, 1]). This proves the existence of the integral on
the right-hand side of (10.44). Furthermore, we follow from the inequality (10.47) that
Z
I2
Fδ (x, y) dx dy −
Z
I2
f (x, y) dx dy ≤ 22 (1 −
Since Fδ ∈ C (I 2 ), the order of integration of
Z
√
1 − δ) · max 2 |f (x, y)|.
(x,y)∈I
Fδ
I2
is irrelevant and then the same is true for
Z
f.
I2
Hence, by the definition (10.44), f is integrable over D.
To get our final conclusion, we need a lemma (see [17, Theorem 13.6, p. 110]):
Lemma 10.6
Let S be a bounded set in Rn ; let f : S → R be a bounded continuous function. If f is
integrable over S, then f is integrable over S ◦ , and
Z
Z
f.
f=
S
S◦
Chapter 10. Integration of Differential Forms
274
Now the closed disk D is bounded and f : D → R is a bounded continuous function. By the above
analysis, f is integrable over D, so Lemma 10.6 shows that
Z
f (x, y) dx dy =
D
Z
f (x, y) dx dy.
(10.48)
D◦
Hence, we establish from the equalities (10.43) and (10.48) that
Z
f (x, y) dx dy =
Z a Z 2π
0
D
f (T (r, θ))r dr dθ.
(10.49)
0
This completes the proof of the problem.
Problem 10.10
Rudin Chapter 10 Exercise 10.
Proof. Let’s make clear what is the meaning of “f decreases sufficiently rapidly as |x| + |y| → ∞”. In
fact, it means that
lim
|x|+|y|→∞
|(x2 + y 2 )f (x, y)| = 0
(10.50)
or equivalently
|f (x, y)| ≤
A
(10.51)
(x2 + y 2 )1+c
for some positive constants A and c and for all large |x| + |y|. If we put x = r cos θ and y = r sin θ, then
the inequality (10.51) becomes
A
|f (r cos θ, r sin θ)| ≤ 2+2c
(10.52)
r
for all large enough r. Define fb : [0, b] → R by
fb(r) = r
so that
Z 2π
f (T (r, θ)) dθ = r
0
Z b Z 2π
0
Z 2π
f (r cos θ, r sin θ) dθ
0
f (T (r, θ))r dr dθ =
0
Z b
0
where b > 0. Now we show that the improper integral
Z ∞
0
fb(r) dr,
fb(r) dr
exists. To this end, we need a comparison result of improper integrals.
Lemma 10.7
Suppose that F : (a, ∞) → R is increasing. If there exists a constant M such that
F (x) ≤ M for all x ∈ (a, ∞), then lim F (x) exists and
x→∞
lim F (x) ≤ M.
x→∞
275
10.3. Applications of Theorem 10.9 (Change of Variables Theorem)
Proof of Lemma 10.7. Suppose that
E = F ((a, ∞)) = {y = F (x) | x > a} ⊆ R.
By the hypothesis, we know that E is bounded by M . By Theorem 1.19, R is an ordered field
with the least-upper-bound property. By Definition 1.10, E has a least upper bound in R.
Suppose that S = sup E. Given ǫ > 0, then the number S − ǫ is not an upper bound of E, so
we have
S − ǫ < F (x0 )
(10.53)
for some x0 > a. Since F is increasing, the inequality (10.53) shows that
S − ǫ < F (x0 ) ≤ F (x) ≤ S
for all x > x0 . In other words, we have
|F (x) − S| < ǫ
for all x > x0 . By Definition 4.33, we have
lim F (x) = S.
x→∞
Since S = sup E, it is trivial that S ≤ M which is our desired result.
Lemma 10.8
Let f (x) and g(x) be continuous functions and a be a constant. Suppose that
0 ≤ g(x) ≤ f (x)
Z ∞
Z ∞
for x ≥ a. If
f (x) dx exists, then the limit
g(x) dx also exists and
a
a
Z ∞
a
g(x) dx ≤
Proof of Lemma 10.8. Suppose that
Z y
F (y) =
f (x) dx
Z ∞
f (x) dx.
a
and G(y) =
a
Z y
g(x) dx,
a
where y ≥ a. Since f (x) ≥ 0 and g(x) ≥ 0 for all x ≥ a, the functions F and G are increasing.
Since 0 ≤ g(x) ≤ f (x) for all x ≥ a, we have
G(y) ≤ F (y)
(10.54)
for all y ≥ a. By the hypothesis, the number
Z ∞
M=
f (x) dx
a
is finite. Since we always have F (y) ≤ M for all y ≥ a, we deduce from the inequality (10.54)
that
G(y) ≤ M
for all y ≥ a. Hence the desired results follow immediately from Lemma 10.8.
Chapter 10. Integration of Differential Forms
276
Lemma 10.9
The improper integral
Z ∞
0
exists.
|fb(r)| dr
Proof of Lemma 10.9. Let N be a fixed positive integer such that the inequality (10.52) holds
for all r ≥ N . If b > N , then we get from Theorems 6.12(c) and 6.13(b) that
Z b
0
|fb(r)| dr =
≤
≤
Z N
0
Z N
0
Z N
0
|fb(r)| dr +
|fb(r)| dr +
|fb(r)| dr +
Z b
N
Z b
|fb(r)| dr
r
N
Z b
Z 2π
f (r cos θ, r sin θ) dr
0
2πA
dr
1+2c
N r
(10.55)
Since fb is continuous on the compact interval [0, N ] and N is fixed, Theorem 4.14 implies that
|fb(r)| ≤ m on [0, N ] for a positive constant m. Thus the inequality (10.55) becomes
0≤
Since
Z b
0
Z b
|fb(r)| dr ≤ mN +
Z b
2πA
dr.
1+2δ
N r
πA 1
1 2πA
dr
=
−
1+2δ
δ r2N
r2b
N r
and
πA
1 πA 1
− 2b = 2N ,
2N
b→∞ δ
r
r
δr
Lemma 10.8 implies that the limit
Z ∞
|fb(r)| dr
lim
0
exists. This completes the proof of this lemma.
Since 0 ≤ |fb(r)| − fb(r) ≤ 2|fb(r)| for every 0 ≤ r ≤ b, it follows from Lemmas 10.8 and 10.9 that
Z ∞
fb(r) dr
0
exists. Next, since D is the closed disk centered at the origin with radius a, D becomes the whole plane
R2 as a → ∞. Hence, we deduce from the equality (10.49) that
Z
f (x, y) dx dy =
R2
Z ∞ Z 2π
0
f (T (r, θ))r dr dθ.
0
Now the function f : R2 → R given by
f (x, y) = exp(−x2 − y 2 )
is clearly a continuous and bounded function on the closed disk D with center at (0, 0) and radius a so
that f ∈ C (D). Furthermore, f is a function satisfying the condition (10.50). Therefore, on the one
hand, we have
Z ∞ Z 2π
Z
exp(−x2 − y 2 ) dx dy =
f (T (r, θ))r dr dθ
R2
0
0
277
10.3. Applications of Theorem 10.9 (Change of Variables Theorem)
=
Z ∞ Z 2π
0
1
=
2
2
e−r r dr dθ
0
Z 2π " Z ∞
0
2
#
e−r d(r2 ) dθ
0
Z
2 ∞
1 2π − e−r
dθ
=
2 0
0
= π.
(10.56)
On the other hand, we have
Z
R2
exp(−x2 − y 2 ) dx dy =
Z ∞
2
!
e−x dx
−∞
×
Z ∞
2
e−y dy
−∞
!
=
Z ∞
−∞
2
!2
e−x dx
.
(10.57)
Combining the expressions (10.56) and (10.57), we have
Z ∞
√
2
e−x dx = π
−∞
which is exactly formula (101) of Chap. 8 (see [21, p. 194]). This completes the proof of our problem. Problem 10.11
Rudin Chapter 10 Exercise 11.
Proof. Let S = {(s, t) | 0 < s < ∞, 0 < t < 1} and Q = {(u, v) | u > 0, v > 0} be the strip in the
(s, t)-plane and the positive quadrant in the (u, v)-plane respectively. Define T : S → R by
T (s, t) = (s − st, st).
(10.58)
Since s > 0 and 0 < t < 1, s − st = s(1 − t) > 0 and st > 0. Thus we have T (S) ⊆ Q. If T (s, t) = T (s′ , t′ ),
then the definition (10.58) gives s − st = s′ − s′ t′ and st = s′ t′ which show that s = s′ and t = t′
v
.
immediately. Thus T is one-to-one. Besides, given u > 0 and v > 0, we define s = u + v and t = u+v
Then we have (s, t) ∈ S and
v
v T (s, t) = (s − st, st) = u + v − (u + v)
= (u, v).
, (u + v)
u+v
u+v
Therefore, we have T (S) = Q, i.e., the map T : S → Q is onto. See Figure 10.6 for the mapping T .
Figure 10.6: The mapping T : S → Q.
It is obvious from the definition (10.58) that


∂u ∂u
1−t
 ∂s ∂t 
JT (s, t) = det  ∂v
=
det

∂v
t
∂s ∂t
−s
s
= s(1 − t) + st = s.
Chapter 10. Integration of Differential Forms
278
This proves our second assertion.
For x > 0, y > 0, we consider the function f : Q → R and the integral given by
Z
Z
ux−1 v y−1
f (u, v) = u · v
and α =
f (u, v) du dv =
ux−1 e−u v y−1 e−v du dv
e
e
Q
Q
respectively. Obviously, S is an open set and T is an one-to-one (in fact, bijective) C ′ -mapping. Although
the function f (u, v) is clearly continuous on Q, its support is
supp (f ) = {(u, v) ∈ Q | f (u, v) 6= 0} = Q ∪ {(u, 0) | u ≥ 0} ∪ {(0, v) | v ≥ 0}
which consists of the positive quadrant plus the nonnegative u and v axes. Thus the set supp (f ) is
unbounded (and hence not compact in R2 by Theorem 2.41 (Heine-Borel Theorem)) and does not lie in
T (S) = Q. In other words, the hypotheses of Theorem 10.9 do not satisfy in this case.
To overcome these two problems, we try to consider subsets of S and Q and then we apply Lemma
10.4 to such subsets. To this end, for every small enough ǫ > 0, we consider the set
Sǫ = {(s, t) | ǫ < s < ǫ−2 , ǫ < t < 1 − ǫ} ⊂ S.
Let Aǫ = T (ǫ, ǫ), Bǫ = T (ǫ, 1 − ǫ), Cǫ = T (ǫ−2 , ǫ) and Dǫ = T (ǫ−2 , 1 − ǫ). Then direct computation gives
Aǫ = (ǫ(1 − ǫ), ǫ2 ),
Bǫ = (ǫ2 , ǫ(1 − ǫ)),
Cǫ = (ǫ−2 (1 − ǫ), ǫ−1 ) and Dǫ = (ǫ−1 , ǫ−2 (1 − ǫ)).
Now let Qǫ be the interior of the convex hull of the points Aǫ , Bǫ , Cǫ and Dǫ in the (u, v)-plane. That is,
if
Hǫ = {λ1 Aǫ + λ2 Bǫ + λ3 Cǫ + λ4 Dǫ | λi ≥ 0 for i = 1, 2, 3, 4 and λ1 + λ2 + λ3 + λ4 = 1},
then we have
Qǫ = Hǫ◦ ⊂ Q.
For examples, if ǫ = 0.1, then the four points are
A0.1 = (0.09, 0.01),
B0.1 = (0.01, 0.09),
C0.1 = (90, 10) and D0.1 = (10, 90);
B0.2 = (0.04, 0.16),
C0.2 = (20, 5) and D0.2 = (5, 20).
if ǫ = 0.2, then they are
A0.2 = (0.16, 0.04),
See Figure 10.7 for the open sets Q0.1 , Q0.2 and Q.
Figure 10.7: The open sets Q0.1 , Q0.2 and Q.
(10.59)
279
10.3. Applications of Theorem 10.9 (Change of Variables Theorem)
By definition, we know that both Sǫ and Qǫ are open and bounded in R2 . Next, we have
Sǫ ⊂ Sδ
and Qǫ ⊂ Qδ
if 0 < δ < ǫ. As ǫ → 0, we know from the representations of Sǫ and Qǫ that
Sǫ → S
and Qǫ → Q.
Furthermore, by the construction of the sets Sǫ and Qǫ , it is easily seen that the restriction T : Sǫ → Qǫ
is a bijective C ′ -mapping. Since Sǫ is a rectangle in R2 , it follows from [8, Exercise 1, p. 496] that it is
an open Jordan domain. Since Kǫ = Sǫ ∪ ∂Sǫ = Sǫ ⊂ S, we know from the definition (10.59) of Qǫ that
T (Kǫ ) = Qǫ = Hǫ .
Now our function f is obviously continuous on Hǫ , so we may apply Theorem 10.9 to the restriction
f : Hǫ → R to obtain
Z
Z
f (u, v) du dv =
f (T (s, t))|JT (s, t)| ds dt.
(10.60)
Hǫ
Sǫ
On the one hand, we know that
Z
f (u, v) du dv =
Z ǫ−2 (1−ǫ)
u
x−1 −u
e
!
(10.61)
sx+y−1 ty−1 (1 − t)x−1 e−s ds dt.
(10.62)
du
ǫ2
Hǫ
! Z −2
ǫ (1−ǫ)
v
y−1 −v
e
dv
ǫ2
and on the other hand, we have
Z
f (T (s, t))|JT (s, t)| ds dt =
Z 1−ǫ Z ǫ−1
ǫ
Sǫ
ǫ
By substituting the integrals (10.61) and (10.62) into the integral relation (10.60), we have
! Z −2
! "Z
#
Z −2
ǫ
(1−ǫ)
ǫ
ux−1 e−u du
ǫ2
(1−ǫ)
v y−1 e−v dv
1−ǫ
ǫ2
ǫ
×
and since
Z ǫ−1
ty−1 (1 − t)x−1 dt
=
Z ǫ−1
sx+y−1 e−s ds
ǫ
!
sx+y−1 e−s ds > 0
ǫ
for every ǫ > 0, we must have
Z 1−ǫ
ǫ
t
y−1
x−1
(1 − t)
dt =
Z ǫ−2 (1−ǫ)
u
x−1 −u
e
du
ǫ2
! Z −2
ǫ (1−ǫ)
v
y−1 −v
ǫ2
Z ǫ−1
s
x+y−1 −s
e
ds
ǫ
for every ǫ > 0. Recall from Definition 8.17 that the integral
Z ∞
tx−1 e−t dt
0
converges for all x > 0, so we can conclude that
lim
ǫ→0
Z ǫ−2 (1−ǫ)
ǫ2
ux−1 e−u du =
Z ∞
0
ux−1 e−u du = Γ(x),
e
dv
!
(10.63)
Chapter 10. Integration of Differential Forms
lim
ǫ→0
Z ǫ−2 (1−ǫ)
v
y−1 −v
e
dv =
ǫ2
lim
ǫ→0
Z ǫ−1
280
Z ∞
v x−1 e−v dv = Γ(y),
0
sx+y−1 e−s ds =
Z ∞
sx+y−1 e−s ds = Γ(x + y).
0
ǫ
Since Γ(x) is nonzero on (0, ∞), they imply that the integral on the left-hand side of the relation (10.63)
also converges and we have
Z 1
0
ty−1 (1 − t)x−1 dt = lim
ǫ→0
Z 1−ǫ
ǫ
ty−1 (1 − t)x−1 dt =
Γ(x)Γ(y)
Γ(x + y)
which is exactly formula (96) of Chap. 8. This completes the proof of the problem.
As a remark to Problem 10.11, Rudin mentioned that Theorem 10.9 has to be extended to cover the
case of improper integrals. The proof we present here does not use this approach. If the reader wants to
apply such extension to prove the problem, we recommend the following version of Theorem 10.9 which
can be found in [25, Theorem 1, p. 156]:
Lemma 10.10
Suppose that T : U → V is a diffeomorphism of the open set U ⊆ Rn onto the open set
V ⊆ Rn and f : U → R is integrable on all measurable compact subsets of U . If the
improper integral
Z
f (y) dy
V
converges, then the improper integral
Z
f (T (x))|JT (x)| dx
U
converges and has the same value.
Problem 10.12
Rudin Chapter 10 Exercise 12.
Proof. See Figure 10.8 for the example T : I 3 → Q3 .
Figure 10.8: The mapping T : I 3 → Q3 .
We divide the proof into several steps:
281
10.3. Applications of Theorem 10.9 (Change of Variables Theorem)
• Proof of the formula. The formula can be shown by induction. The case k = 1 is trivial. Assume
that the formula is true for k = n, i.e.,
n
X
i=1
xi = 1 −
n
Y
i=1
(1 − ui ).
(10.64)
If k = n + 1, then we follow from the formula (10.64) and the definition that
n+1
X
n
X
xi = xn+1 +
xi
i=1
i=1
= xn+1 + 1 −
n
Y
(1 − ui )
i=1
= 1 + (1 − u1 ) · · · (1 − un )un+1 −
= 1 + un+1
n
Y
i=1
=1+
=1−
n
Y
n
Y
(1 − ui )
i=1
n
Y
(1 − ui )
(1 − ui ) −
i=1
(1 − ui )(un+1 − 1)
i=1
n+1
Y
i=1
(1 − ui ).
Thus the formula is also true in the case k = n+ 1. Hence we follow from induction that the formula
is true for all positive integers k.
• The surjectivity of the map T : I k → Qk . Since 0 ≤ ui ≤ 1 for all i = 1, 2, . . . , k, we have
xi ≥ 0 for all i = 1, 2, . . . , k. By the formula (10.64), we have
k
X
i=1
xi ≤ 1,
so T maps I k into Qk . Given (x1 , . . . , xk ) ∈ Qk . If
points
m
X
i=1
xi 6= 1 for all m = 1, 2, . . . , k − 1, then the
u1 = x1 ,
x2
u2 =
,
1 − x1
x3
x3
u3 =
,
=
(1 − x1 )(1 − x2 )
1 − x1 − x2
·················· ,
xk
xk
uk =
=
(1 − x1 ) · · · (1 − xk−1 )
1 − x1 − · · · − xk−1
imply that
T (u1 , . . . , uk ) = (x1 , . . . , xk ).
If m is the least positive integer in the set {1, 2, . . . , k − 1} such that
m
X
xi = 1,
i=1
then we have
m−1
X
i=1
xi < 1 and xm+1 = xm+2 = · · · = xk = 0.
(10.65)
Chapter 10. Integration of Differential Forms
282
By these and the formula (10.64), the point (u1 , . . . , uk ) defined by
u1 = x1 ,
u2 =
m−1
X x2
xi ,
, . . . , um = xm ÷ 1 −
1 − x1
i=1
ui = xi ,
where i = m + 1, . . . , k, implies that
T (u1 , . . . , um , um+1 , . . . , uk ) = (x1 , x2 , . . . , xm , 0, . . . , 0).
If we have
k
X
xi = 1 and
i=1
k−1
X
xi < 1,
i=1
then the point (u1 , . . . , uk ) defined by
k−1
X x2
u1 = x1 , u2 =
xi
, . . . , uk = xk ÷ 1 −
1 − x1
i=1
implies that
T (u1 , . . . , uk ) = (x1 , x2 , . . . , xk ).
Thus it means that the mapping T is onto.
• The injectivity of the mapping T : I → Q. Let I and Q be the interiors of I k and Qk
respectively, i.e.,
I = {(u1 , . . . , uk ) ∈ Rk | 0 < u1 , . . . , uk < 1}
(10.66)
and
k
o
n
X
xi < 1 .
Q = (x1 , . . . , xk ) ∈ Rk x1 , . . . , xk > 0,
(10.67)
i=1
We first show that the mapping T is one-to-one in I. To this end, let u = (u1 , . . . , uk ) and
v = (v1 , . . . , vk ) be points in I such that
T (u) = x = (x1 , . . . , xk ) and T (v) = y = (y1 , . . . , yk ).
If T (u) = T (v), then we have x1 = y1 , x2 = y2 , . . . , xk = yk . Note that 0 < ui < 1 and 0 < vi < 1
for 1 ≤ i ≤ k. By definition, the relation x1 = y1 implies trivially that u1 = v1 . Similarly, the
inequality 0 < u1 = v1 < 1 and the relation x2 = y2 show that u2 = v2 . By using the inequalities
0 < ui = vi < and the relations xi = yi for 1 ≤ i ≤ k − 1, it deduces immediately from the definition
that xk = yk . Hence T is one-to-one in I.
Next, we show that T (I) = Q. Let x ∈ Q so that x = (x1 , . . . , xk ), where x1 , . . . , xk > 0 and
k
X
xk < 1. If we define u1 , . . . , uk by the equalities (10.65), then we obtain immediately that
i=1
T (u1 , . . . , uk ) = (x1 , . . . , xk ).
Thus T (I) = Q, as required.
In addition, it is clearly that the inverse S : Q → I is well-defined by the formulas (10.65).
• Computation of the JT (u) and JS (x). To find JT (u), we note that
T (u1 , . . . , uk ) = (u1 , (1 − u1 )u2 , . . . , (1 − u1 ) · · · (1 − uk−1 )uk ).
(10.68)
283
10.3. Applications of Theorem 10.9 (Change of Variables Theorem)
Apply Theorem 9.17 to (10.68), we get

1
0



−u2
1 − u1

..
..


.
.
[T ′ (u)] = 


k−1
 k−1
Y
Y

(1 − ui )uk
(1 − ui )uk −
 −
i=1
i6=2
i=2
−
0
···
0
0
..
.
···
..
.
0
..
.









k−1

Y

···
(1 − ui ) 
k−1
Y
i=1
i6=3

(1 − ui )uk
i=1
which is a lower triangular matrix. It is well-known that the determinant of a triangular matrix is
the product of the entries on the diagonal (see [15, Theorem 2, 167]), so we have
JT (u) = det[T ′ (u)]
= 1 × (1 − u1 ) × [(1 − u1 )(1 − u2 )] × · · · ×
= (1 − u1 )k−1 (1 − u2 )k−2 · · · (1 − uk−1 ),
k−1
Y
i=1
(1 − ui )
(10.69)
where u ∈ I. To find JS (x), we notice that
S(x1 , . . . , xk ) =
!
x2
xk
x1 ,
.
,...,
1 − x1
1 − x1 − · · · − xk−1
Similarly, we apply Theorem 9.17 to the expression (10.70), we get

1
0
0



1
x2

0
2

(1
−
x
)
1
−
x1
1

..
..
..

.
.
.
[S ′ (x)] = 



xk
xk
xk


k−1
k−1
k−1
 2 2 2
X
X
X

xi
xi
xi
1−
1−
1−
i=1
i=1
i=1
(10.70)
···
0
···
0
..
..
.
.
···

1
1−
k−1
X
i=1
and therefore,
xi













2 

JS (x) = det[S ′ (x)]
1
1
× ···×
=1×
1 − x1
1 − x1 − · · · − xk−1
= [(1 − x1 )(1 − x1 − x2 ) · · · (1 − x1 − x2 · · · − xk−1 )]−1 ,
where x ∈ Q.
We end the proof of the problem.
Problem 10.13
Rudin Chapter 10 Exercise 13.
Proof. Let T : I k → Qk be the mapping as defined in Problem 10.12. Then I is an open set, T : I → Q
is bijective and we deduce from the definition (10.66) and the formula (10.69) that
JT (u) 6= 0
Chapter 10. Integration of Differential Forms
284
for all u ∈ I. Define the function f : Q → R by
f (x) = xr11 xr22 · · · xrkk
which is clearly continuous on Q. By definition (10.67), we know that
supp (f ) = {x ∈ Q | f (x) 6= 0} = Q = Qk
which is compact by Theorem 2.41 (Heine-Borel Theorem).
Hence, it follows from Theorem 10.9 and the formula (10.69) that
Z
Z
xr11 xr22 · · · xrkk dx = f (T (u))|JT (u)| du
Q
I
Z
= f (T (u))|(1 − u1 )k−1 (1 − u2 )k−2 · · · (1 − uk−1 )| du1 du2 · · · duk .
(10.71)
I
By the definition of x, the integral (10.71) reduces to
Z
Z
xr11 xr22 · · · xrkk dx = f (T (u))|(1 − u1 )k−1 (1 − u2 )k−2 · · · (1 − uk−1 )| du1 du2 · · · duk
Q
I
Z
= ur11 [(1 − u1 )u2 ]r2 [(1 − u1 )(1 − u2 )u3 ]r3 · · · [(1 − u1 ) · · · (1 − uk−1 )uk ]rk
I
× |(1 − u1 )k−1 (1 − u2 )k−2 · · · (1 − uk−1 )| du1 du2 · · · duk
Z
o
n
= ur11 · · · urkk (1 − u1 )r2 +···+rk (1 − u2 )r3 +···+rk · · · (1 − uk−1 )rk
I
(1 − u1 )k−1 (1 − u2 )k−2 · · · (1 − uk−1 ) du1 du2 · · · duk
"Z
# "Z
1
=
0
ur11 (1 − u1 )k+r2 +···+rk −1 du1
× ··· ×
"Z
0
1
1
×
rk−1
uk−1
(1 − uk−1 )1+rk duk−1
0
#
×
ur22 (1 − u2 )k+r3 ···+rk −2 du2
Z 1
0
urkk duk
!
.
#
(10.72)
Finally, we apply Theorem 8.20 to each integral on the right-hand side of (10.72) and then Theorem
8.18(b), cancellations happen and it further reduces to
Z
Γ(r1 + 1)Γ(k + r2 + · · · + rk ) Γ(r2 + 1)Γ(k − 1 + r3 + · · · + rk )
×
xr11 xr22 · · · xrkk dx =
Γ(k + r1 + · · · + rk + 1)
Γ(k + r2 + · · · + rk )
Q
Γ(rk−1 + 1)Γ(2 + rk ) Γ(rk + 1)Γ(1)
×
Γ(3 + rk−1 + rk )
Γ(2 + rk )
Γ(r1 + 1)Γ(r2 + 1) · · · Γ(rk + 1)
=
Γ(k + r1 + · · · + rk + 1)
r1 !r2 ! · · · rk !
.
=
(k + r1 + · · · + rk )
× ···×
(10.73)
In particular, if we take r1 = · · · = rk = 0 into the integral (10.73), then we get
Z
1
dx =
k!
Qk
which gives the volume of the k-simplex Qk . This completes the proof of the problem.
10.4
Properties of k-forms and k-simplexes
Problem 10.14
Rudin Chapter 10 Exercise 14.
285
10.4. Properties of k-forms and k-simplexes
Proof. Recall from Definition 9.33 that if {j1 , . . . , jk } is an ordered k-tuple of distinct integers, then we
have
k q−1
Y
Y
sgn (jq − jp ).
(10.74)
s(j1 , . . . , jk ) =
q=2 p=1
Let k = 2. Then we have
s(j1 , j2 ) = sgn (j2 − j1 ) =
1,
if j2 > j1 ;
−1, if j2 < j1 .
If j2 > j1 , then dxj1 ∧ dxj2 is an increasing 2-index so that ε(j1 , j2 ) = 1. Similarly, if j2 < j1 , then
dxj1 ∧ dxj2 = − dxj2 ∧ dxj1 so that ε(j1 , j2 ) = −1 by [21, Eqn. (42), p. 256]. Therefore, we have
ε(j1 , j2 ) = s(j1 , j2 ).
Assume that we have
ε(j1 , . . . , jm ) = s(j1 , . . . , jm ),
(10.75)
where 2 ≤ m < k. Let k = m + 1. By definition (10.74) and the equality (10.75), we have
s(j1 , . . . , jm , jm+1 ) =
m+1
Y
Y q−1
q=2 p=1
=
m
Y
p=1
sgn (jq − jp )
sgn (jm+1 − jp ) ×
= ε(j1 , . . . , jm ) ×
m
Y
p=1
m q−1
Y
Y
q=2 p=1
sgn (jq − jp )
sgn (jm+1 − jp ).
(10.76)
To make the (m + 1)-tuple {j1 , . . . , jm , jm+1 } of distinct integers into an increasing (m + 1)-index, we
can first make the m-tuple {j1 , . . . , jm } into an increasing m-index, namely {jr1 , . . . , jrm } where
jr1 < · · · < jrm ,
and then add the integer jm+1 to that increasing m-index to produce the increasing (m + 1)-index.
Suppose that
jr1 < · · · < jrs < jm+1 < jrs+1 < · · · < jrm .
(10.77)
Then we have
ε(j1 , . . . , jm , jm+1 ) = ε(j1 , . . . , jm ) × (−1)m−s−1 .
(10.78)
By the inequalities (10.77), we have
m
Y
p=1
sgn (jm+1 − jp ) =
m
Y
p=1
sgn (jm+1 − jrp ) = (−1)m−s−1 .
(10.79)
Hecnce we follow from the equalities (10.76), (10.78) and (10.79) that
s(j1 , . . . , jm , jm+1 ) = ε(j1 , . . . , jm , jm+1 )
which implies that the statement is true for k = m + 1. By induction, formula (46) is true for all integers
k. This completes the proof of the problem.
Problem 10.15
Rudin Chapter 10 Exercise 15.
Chapter 10. Integration of Differential Forms
286
Proof. Suppose that ω and λ are represented in the standard presentation:
X
X
ω=
aI (x) dxI and λ =
bJ (x) dxJ ,
I
(10.80)
J
where the summations in (10.80) extend over all increasing k-indices I and m-indices J respectively. By
Definition 10.17, we have the (k + m)-form in an open set E ⊆ Rn
X
X
ω∧λ=
aI (x)bJ (x) dxI ∧ dxJ and λ ∧ ω =
bJ (x)aI (x) dxJ ∧ dxI ,
I,J
I,J
where I and J range independently over their
aI (x)bJ (x) = bJ (x)aI (x)
for every x ∈ E. Therefore, our result follows immediately if we can show that
dxI ∧ dxJ = (−1)km dxJ ∧ dxI
(10.81)
for each increasing k-indices I and increasing m-indices J.
Suppose that I and J have an element in common, then we know from [21, Eqn. (43), p. 256] that
dxI ∧ dxJ = dxJ ∧ dxI = 0
so that the formula (10.81) holds. Next, we suppose that {i1 , . . . , ik } and {j1 , . . . , jm } are increasing
k-indices and m-indices respectively with no element in common. By repeated application of the anticommutative relation [21, Eqn. (42), p. 256], we have
dxI ∧ dxJ = ( dxi1 ∧ · · · ∧ dxik ) ∧ ( dxj1 ∧ · · · ∧ dxjm )
= ( dxi1 ∧ · · · ∧ dxik−1 ∧ (−1) dxj1 ) ∧ ( dxik ∧ dxj2 ∧ · · · ∧ dxjm )
= ( dxi1 ∧ · · · ∧ dxik−2 ∧ (−1)2 dxj1 ∧ dxik−1 ) ∧ ( dxik ∧ dxj2 ∧ · · · ∧ dxjm )
= ((−1)k dxj1 ∧ dxi1 ∧ · · · ∧ dxik−1 ) ∧ ( dxik ∧ dxj2 ∧ · · · ∧ dxjm )
= ((−1)2k dxj1 ∧ dxj2 ∧ dxi1 ∧ · · · ∧ dxik−2 )
∧ ( dxik−1 ∧ dxik ∧ dxj3 ∧ · · · ∧ dxjm )
..
.
= (−1)km ( dxj1 ∧ dxj2 ∧ · · · ∧ dxjm ) ∧ ( dxi1 ∧ dxi2 ∧ · · · ∧ dxik )
= (−1)km dxJ ∧ dxI
which is exactly the equality (10.81). This completes the proof of the problem.
Problem 10.16
Rudin Chapter 10 Exercise 16.
Proof. Let k = 2. Then we get from Definition 10.29 that
∂σ = [p1 , p2 ] − [p0 , p2 ] + [p0 , p1 ]
which gives
∂ 2 σ = ∂[p1 , p2 ] − ∂[p0 , p2 ] + ∂[p0 , p1 ] = [p2 ] − [p1 ] − ([p2 ] − [p0 ]) + [p1 ] − [p0 ] = 0.
Let k = 3. Similarly, we obtain from Definition 10.29 that
∂σ = [p1 , p2 , p3 ] − [p0 , p2 , p3 ] + [p0 , p1 , p3 ] − [p0 , p1 , p2 ]
287
10.4. Properties of k-forms and k-simplexes
which implies that
∂ 2 σ = ∂[p1 , p2 , p3 ] − ∂[p0 , p2 , p3 ] + ∂[p0 , p1 , p3 ] − [p0 , p1 , p2 ]
= [p2 , p3 ] − [p1 , p3 ] + [p1 , p2 ] − ([p2 , p3 ] − [p0 , p3 ] + [p0 , p2 ])
+ [p1 , p3 ] − [p0 , p3 ] + [p0 , p1 ] − ([p1 , p2 ] − [p0 , p2 ]) + [p0 , p1 ])
= 0.
For the general case, let σi and σij be the (k − 1)-simplex and (k − 2)-simplex obtained by deleting
pi and pi plus pj from σ respectively, where i < j. That is
σi = [p0 , . . . , pi−1 , pi+1 , . . . , pk ] and σij = [p0 , . . . , pi−1 , pi+1 , . . . , pj−1 , pj+1 . . . , pk ]
where i, j = 0, . . . , k and i < j. Now each σij occurs exactly twice in ∂ 2 σ, one from deleting the pj first
and then the pi next, and the other one from deleting the pi first and then the pj next. We claim that
the resulting (k − 2)-simplex have opposite sign. To this end, we notice that the positions of pi and pj
in the oriented affine k-simplex σ first:
σ = [ p0 , . . . , pi−1 , pi , pi+1 , . . . , pj−1 , pj , pj+1 , . . . , pk ].
|
{z
}
i terms before pi
|
{z
j terms before pj
(10.82)
}
• Case (i): Delete the pj from the expression (10.82). We have
σj = [ p0 , . . . , pi−1 , pi , pi+1 , . . . , pj−1 , pj+1 , . . . , pk ]
|
{z
}
(10.83)
i terms before pi
and this contributes a factor (−1)j . We observe from the expression (10.83) that there are i terms
before the pi , so when we delete the pi from the expression (10.83), we have
σij = [p0 , . . . , pi−1 , pi+1 , . . . , pj−1 , pj+1 , . . . , pk ]
and this contributes another factor (−1)i . Thus we obtain
(−1)i+j σij
in this way.
• Case (ii): Delete the pi from (10.82). We get
σi = [p0 , . . . , pi−1 , pi+1 , . . . , pj−1 , pj , pj+1 , . . . , pk ]
|
{z
}
(10.84)
(j − 1) terms before pj
and this contributes a factor (−1)i . Now we notice from the form (10.84) that there are (j − 1)
terms before the pj , so if we delete the pj from the expression (10.84), then we have
σij = [p0 , . . . , pi−1 , pi+1 , . . . , pj−1 , pj+1 , . . . , pk ]
and this contributes another factor (−1)j−1 . Thus we obtain
(−1)i+j−1 σij
in this way.
This proves our claim and then completes the proof of our problem.
Problem 10.17
Rudin Chapter 10 Exercise 17.
Chapter 10. Integration of Differential Forms
288
As a remark to Definition 10.28, we recall from Rudin’s explanation [21, p. 268] that the notation
“+” used in the chain J 2 = τ1 + τ2 does not mean the addition of mappings. In fact, if we denote Ωk (E)
to be the collection of all k-forms in E and Φ : D ⊂ Rk → E ⊆ Rn is a k-surface in E, then we define
Φ : Ωk (E) → R to bem
Z
Φ(ω) =
ω.
Φ
Hence the “+” in the expression J 2 = τ1 + τ2 means that
2
J (ω) = τ1 (ω) + τ2 (ω) =
Z
ω+
τ1
Z
ω
(10.85)
τ2
for every 2-form ω. To avoid any ambiguity, we write τ1 +c τ2 to replace the original affine chain τ1 + τ2
with the meaning shown in the integrals (10.85).
Proof. We first find the explicit representations of τ1 and τ2 . Note that τ1 is characterized by
τ1 (0) = 0,
τ1 (e1 ) = e1
and τ1 (e2 ) = e1 + e2 .
By [21, Eqn. (80), p. 267], we know that τ2 = [0, e2 + e1 , e2 ] and then it is characterized by
τ2 (0) = 0,
τ2 (e1 ) = e1 + e2
and τ2 (e2 ) = e2 .
By [21, Eqn. (78), p. 266], we acquire the mappings τ1 : Q2 → R2 and τ2 : Q2 → R2 by
for all u ∈ Q2 , where
A=
τ1 (u) = Au
and τ2 (u) = Bu
and B =
1
0
1
1
1
1
0
1
(10.86)
.
(10.87)
Next, we find τ1 (Q2 ) and τ2 (Q2 ). By Definition 10.26,
Q2 = {ae1 + be2 | a, b ≥ 0, a + b ≤ 1}.
Therefore, by this and the mappings (10.86) with the matrices (10.87), we deduce that
1 1
a
a+b
τ1 (ae1 + be2 ) =
=
= (a + b)e1 + be2
0 1
b
b
(10.88)
which imply that
τ1 (Q2 ) = {(a + b)e1 + be2 | a, b ≥ 0, a + b ≤ 1}
(10.89)
and it is the “lower right” half of the unit square I 2 , see Figure 10.9:
Figure 10.9: The mapping τ1 : Q2 → I 2 .
m We use different colors for Φ so as to make clear that it has “different” meanings in “different” situations: For Φ, it
means a k-surface and for Φ, it means a function.
289
10.4. Properties of k-forms and k-simplexes
Similarly, we have
τ2 (ae1 + be2 ) =
1 0
1 1
a
b
=
a
a+b
= ae1 + (a + b)e2
(10.90)
which imply that
τ2 (Q2 ) = {ae1 + (a + b)e2 | a, b ≥ 0, a + b ≤ 1}
(10.91)
and it is the “upper left” half of the unit square I 2 , see Figure 10.10:
Figure 10.10: The mapping τ2 : Q2 → I 2 .
Thus we follow from the ranges (10.89) and (10.91) that
I 2 = τ1 (Q2 ) ∪ τ2 (Q2 )
and the interiors of τ1 (Q2 ) and τ2 (Q2 ) are disjoint. By the matrices (10.87), since
det A = det B = 1 > 0,
τ1 and τ2 are obviously one-to-one mappings of class C ′′ . Furthermore, we know from Definition 10.31
that J 2 has the positively oriented boundary. Hence it is reasonable to say that J 2 the positively oriented
square in R2 . See Figure 10.11 for the orientation of J 2 , where the red and green arrows connecting the
points (0, 0) and (1, 1) have opposite orientation so that they cancel each other.
Figure 10.11: The mapping τ2 : Q2 → I 2 .
Finally, we compute ∂J 2 and ∂(τ1 − τ2 ). By routine computation and [21, Eqn. (80), p. 267], we have
∂J 2 = ∂τ1 + ∂τ2
= [e1 , e1 + e2 ] − [0, e1 + e2 ] + [0, e1 ] − ([e2 , e1 + e2 ] − [0, e1 + e2 ] + [0, e2 ])
= [e1 , e1 + e2 ] + [0, e1 ] − [e2 , e1 + e2 ] − [0, e2 ]
= [0, e1 ] + [e1 , e1 + e2 ] + [e1 + e2 , e2 ] + [e2 , 0]
Chapter 10. Integration of Differential Forms
290
which is the sum of 4 oriented affine 1-simplexes.n Similarly, we have
∂(τ1 − τ2 ) = ∂τ1 − ∂τ2
= [e1 , e1 + e2 ] − [0, e1 + e2 ] + [0, e1 ]
+ ([e2 , e1 + e2 ] − [0, e1 + e2 ] + [0, e2 ])
= [e1 , e1 + e2 ] + [0, e1 ] + [e2 , e1 + e2 ] + [0, e2 ] − 2[0, e1 + e2 ].
Therefore, we have
∂J 2 6= ∂(τ1 − τ2 )
and this completes the proof of the problem.
Problem 10.18
Rudin Chapter 10 Exercise 18.
Proof. Recall from Definition 10.26 that
Q3 = {ae1 + be2 + ce3 | a, b, c ≥ 0, a + b + c ≤ 1}
and σ1 is characterized by
σ1 (0) = 0,
σ1 (e1 ) = e1 ,
σ1 (e2 ) = e1 + e2
and σ1 (e3 ) = e1 + e2 + e3 .
By [21, Eqn. (78), p. 266], we have the mapping σ1 : Q3 → R3 defined by
σ1 (u) = A1 u
for all u ∈ Q3 , where


1 1
1 1 .
0 1
1
A1 =  0
0
(10.92)
Since det A1 = 1 > 0, σ1 is positively oriented. The five permutations of (1, 2, 3) other than (1, 2, 3) are
(1, 3, 2),
(2, 1, 3),
(2, 3, 1),
(3, 1, 2) and (3, 2, 1).
By Problem 10.14, we have
s(1, 3, 2) = −1,
s(2, 1, 3) = −1,
s(2, 3, 1) = 1,
s(3, 1, 2) = 1
and s(3, 2, 1) = −1.
Therefore, we have
σ2 = −[0, e1 , e1 + e3 , e1 + e2 + e3 ] = [0, e1 , e1 + e2 + e3 , e1 + e3 ],
σ3 = −[0, e2 , e1 + e2 , e1 + e2 + e3 ] = [0, e2 , e1 + e2 + e3 , e1 + e2 ],
σ4 = [0, e2 , e2 + e3 , e1 + e2 + e3 ],
σ5 = [0, e3 , e1 + e3 , e1 + e2 + e3 ],
σ6 = −[0, e3 , e2 + e3 , e1 + e2 + e3 ] = [0, e3 , e1 + e2 + e3 , e2 + e3 ]
so that
and

1
A2 =  0
0

1 1
1 0 ,
1 1

0
A3 =  1
0

1 1
1 1 ,
1 0

0
A6 =  0
1

0
A4 =  1
0

1 0
1 1 .
1 1

0 1
1 1 ,
1 1

0 1
A5 =  0 0
1 1

1
1 ,
1
n Actually, the ∂J 2 is exactly the positively oriented boundary ∂I 2 of the unit square I 2 given in [21, p. 271].
291
10.4. Properties of k-forms and k-simplexes
Since
det A2 = det A3 = det A4 = det A5 = det A6 = 1 > 0,
σ2 , . . . , σ6 are positively oriented by Definition 10.26.
Put J 3 = σ1 + · · · + σ6 . By Definition 10.29, we have
∂J 3 =
6
X
∂σm
m=1
= [e1 , e1 + e2 , e1 + e2 + e3 ] − [0, e1 + e2 , e1 + e2 + e3 ] + [0, e1 , e1 + e2 + e3 ] − [0, e1 , e1 + e2 ]
− [e1 , e1 + e3 , e1 + e2 + e3 ] − [0, e1 + e3 , e1 + e2 + e3 ] + [0, e1 , e1 + e2 + e3 ] − [0, e1 , e1 + e3 ]
+ [e2 , e2 + e3 , e1 + e2 + e3 ] − [0, e2 + e3 , e1 + e2 + e3 ] + [0, e2 , e1 + e2 + e3 ] − [0, e2 , e2 + e3 ]
− [e3 , e2 + e3 , e1 + e2 + e3 ] − [0, e2 + e3 , e1 + e2 + e3 ] + [0, e3 , e1 + e2 + e3 ] − [0, e3 , e2 + e3 ]
= [e1 , e1 + e2 , e1 + e2 + e3 ] − [0, e1 , e1 + e2 ] − [e1 , e1 + e3 , e1 + e2 + e3 ] + [0, e1 , e1 + e3 ]
− [e2 , e1 + e2 , e1 + e2 + e3 ] − [0, e1 + e2 , e1 + e2 + e3 ] + [0, e2 , e1 + e2 + e3 ] − [0, e2 , e1 + e2 ]
+ [e3 , e1 + e3 , e1 + e2 + e3 ] − [0, e1 + e3 , e1 + e2 + e3 ] + [0, e3 , e1 + e2 + e3 ] − [0, e3 , e1 + e3 ]
− [e2 , e1 + e2 , e1 + e2 + e3 ] + [0, e2 , e1 + e2 ] + [e2 , e2 + e3 , e1 + e2 + e3 ]
− [0, e2 , e2 + e3 ] + [e3 , e1 + e3 , e1 + e2 + e3 ] − [0, e3 , e1 + e3 ]
− [e3 , e2 + e3 , e1 + e2 + e3 ] + [0, e3 , e2 + e3 ]
(10.93)
which consists exactly 12 oriented affine 2-simplexes.
By the matrix representation (10.92), A1 u = x if and only if


 

1 1 1
a
x1
 0 1 1   b  =  x2 
0 0 1
c
x3
if and only if
a + b + c = x1 ,
b + c = x2
and c = x3 .
3
Thus x ∈ σ1 (Q ) if and only if 0 ≤ x3 ≤ x2 ≤ x1 ≤ 1. Similarly, we have
A2 u = x if and only if 0 ≤ x2 ≤ x3 ≤ x1 ≤ 1,
A3 u = x if and only if 0 ≤ x3 ≤ x1 ≤ x2 ≤ 1,
A4 u = x if and only if 0 ≤ x1 ≤ x3 ≤ x2 ≤ 1,
A5 u = x if and only if 0 ≤ x2 ≤ x1 ≤ x3 ≤ 1,
A6 u = x if and only if 0 ≤ x1 ≤ x2 ≤ x3 ≤ 1.
These mean that
x ∈ σ2 (Q3 ) if and only if 0 ≤ x2 ≤ x3 ≤ x1 ≤ 1,
x ∈ σ3 (Q3 ) if and only if 0 ≤ x3 ≤ x1 ≤ x2 ≤ 1,
x ∈ σ4 (Q3 ) if and only if 0 ≤ x1 ≤ x3 ≤ x2 ≤ 1,
x ∈ σ5 (Q3 ) if and only if 0 ≤ x2 ≤ x1 ≤ x3 ≤ 1,
x ∈ σ6 (Q3 ) if and only if 0 ≤ x1 ≤ x2 ≤ x3 ≤ 1.
In other words, we have
σ1 (Q3 ) = {(x1 , x2 , x3 ) | 0 ≤ x3 ≤ x2 ≤ x1 ≤ 1},
σ2 (Q3 ) = {(x1 , x2 , x3 ) | 0 ≤ x2 ≤ x3 ≤ x1 ≤ 1},
σ3 (Q3 ) = {(x1 , x2 , x3 ) | 0 ≤ x3 ≤ x1 ≤ x2 ≤ 1},
σ4 (Q3 ) = {(x1 , x2 , x3 ) | 0 ≤ x1 ≤ x3 ≤ x2 ≤ 1},
σ5 (Q3 ) = {(x1 , x2 , x3 ) | 0 ≤ x2 ≤ x1 ≤ x3 ≤ 1},
σ6 (Q3 ) = {(x1 , x2 , x3 ) | 0 ≤ x1 ≤ x2 ≤ x3 ≤ 1}.
(10.94)
Chapter 10. Integration of Differential Forms
292
Hence it follows from the ranges (10.94) that they have disjoint interiors and
I 3 = σ1 (Q3 ) ∪ σ2 (Q3 ) ∪ · · · ∪ σ6 (Q3 ).
This completes the proof of the problem.
Problem 10.19
Rudin Chapter 10 Exercise 19.
Proof. By rewriting the expression (10.93), we get the following new expression for ∂J 3 :
n
∂J 3 = [e1 , e1 + e2 , e1 + e2 + e3 ] − [e1 , e1 + e3 , e1 + e2 + e3 ] − [0, e2 , e2 + e3 ]
o n
+[0, e3 , e2 + e3 ] + [0, e1 , e1 + e3 ] − [0, e3 , e1 + e3 ] − [e2 , e1 + e2 , e1 + e2 + e3 ]
o n
+[e2 , e2 + e3 , e1 + e2 + e3 ] + [e3 , e1 + e3 , e1 + e2 + e3 ] − [e3 , e2 + e3 , e1 + e2 + e3 ]
o
−[0, e1 , e1 + e2 ] + [0, e2 , e1 + e2 ] .
(10.95)
For r = 0, 1, i = 1, 2, 3, Bri : R2 → R3 are C ′′ -mappings so that each βri is an oriented 3-simplex of class
C ′′ . Since J 2 = τ1 + τ2 , we deduce from [21, Eqn. (88, p. 270] that
βri = Bri (τ1 + τ2 ) = Bri (τ1 ) + Bri (τ2 ).
(10.96)
Let b1ri = Bri (τ1 ) and b2ri = Bri (τ2 ), where r = 0, 1, i = 1, 2, 3. Then the expression (10.96) can be
rewritten as
βri = b1ri + b2ri .
(10.97)
Now we must find the explicit forms of the oriented affine 2-simplexes b1ri and b2ri in the definition
(10.97) in order to compute each (−1)i (β0i − β1i ). To this end, we recall the representations (10.88) and
(10.90) first:
τ1 (ae1 + be2 ) = (a + b)e1 + be2 = ((a + b), b) and τ2 (ae1 + be2 ) = ae1 + (a + b)e2 = (a, a + b).
• Computation of −(β01 − β11 ). For b101 and b201 , since
b101 (ae1 + be2 ) = (0, a + b, b) and b201 (ae1 + be2 ) = (0, a, a + b),
we have
b101 (0) = 0,
b101 (e1 ) = e2
and b101 (e2 ) = e2 + e3 ,
b201 (0) = 0,
b201 (e1 ) = e2 + e3
and b201 (e2 ) = e3 .
By [21, Eqn. (77), p. 266], we have
b101 = [0, e2 , e2 + e3 ] and b202 = [0, e2 + e3 , e3 ],
so the definition (10.97) gives
β01 = [0, e2 , e2 + e3 ] + [0, e2 + e3 , e3 ].
Similarly, since
b111 (ae1 + be2 ) = (1, a + b, b) and b211 (ae1 + be2 ) = (1, a, a + b),
we have
b111 (0) = e1 ,
b111 (e1 ) = e1 + e2
and b111 (e2 ) = e1 + e2 + e3 ,
(10.98)
293
10.4. Properties of k-forms and k-simplexes
b211 (0) = e1 ,
b211 (e1 ) = e1 + e2 + e3
and b211 (e2 ) = e1 + e3 .
By [21, Eqn. (77), p. 266] again, we have
b111 = [e1 , e1 + e2 , e1 + e2 + e3 ] and b212 = [e1 , e1 + e2 + e3 , e1 + e3 ],
so the definition (10.97) gives
β11 = [e1 , e1 + e2 , e1 + e2 + e3 ] + [e1 , e1 + e2 + e3 , e1 + e3 ].
(10.99)
Therefore, we deduce from the two expressions (10.98) and (10.99) that
(−1)1 (β01 − β11 ) = [e1 , e1 + e2 , e1 + e2 + e3 ] + [e1 , e1 + e2 + e3 , e1 + e3 ]
− [0, e2 , e2 + e3 ] − [0, e2 + e3 , e3 ]
= [e1 , e1 + e2 , e1 + e2 + e3 ] − [e1 , e1 + e3 , e1 + e2 + e3 ]
−[0, e2 , e2 + e3 ] + [0, e3 , e2 + e3 ]
which is exactly the first brackets in the expression (10.95).
• Computation of β02 − β12 . For b102 and b202 , since
b102 (ae1 + be2 ) = (a + b, 0, b) and b202 (ae1 + be2 ) = (a, 0, a + b),
we have
b102 (0) = 0,
b102 (e1 ) = e1
and b102 (e2 ) = e1 + e3 ,
b202 (0) = 0,
b202 (e1 ) = e1 + e3
and b202 (e2 ) = e3 .
Since
b102 = [0, e1 , e1 + e3 ] and b202 = [0, e1 + e3 , e3 ],
we have
β02 = [0, e1 , e1 + e3 ] + [0, e1 + e3 , e3 ].
(10.100)
Now we have
b112 (ae1 + be2 ) = (a + b, 1, b) and b212 (ae1 + be2 ) = (a, 1, a + b)
which imply that
b112 = [e2 , e1 + e2 , e1 + e2 + e3 ] and b212 = [e2 , e1 + e2 + e3 , e2 + e3 ].
Thus we have
β12 = [e2 , e1 + e2 , e1 + e2 + e3 ] + [e2 , e1 + e2 + e3 , e2 + e3 ].
(10.101)
Combining the expressions (10.100) and (10.101), we obtain
β02 − β12 = [0, e1 , e1 + e3 ] + [0, e1 + e3 , e3 ] − [e2 , e1 + e2 , e1 + e2 + e3 ]
− [e2 , e1 + e2 + e3 , e2 + e3 ]
= [0, e1 , e1 + e3 ] − [0, e3 , e1 + e3 ] − [e2 , e1 + e2 , e1 + e2 + e3 ]
+[e2 , e2 + e3 , e1 + e2 + e3 ]
which is exactly the second brackets in the expression (10.95).
• Computation of −(β03 − β13 ). The computation of (−1)3 (β03 − β13 ) can be done similarly as
above. We have
b103 (ae1 + be2 ) = (a + b, b, 0) and b203 (ae1 + be2 ) = (a, a + b, 0)
which imply that
b103 (0) = 0,
b103 (e1 ) = e1
and b103 (e2 ) = e1 + e2 ,
Chapter 10. Integration of Differential Forms
b203 (0) = 0,
294
b203 (e1 ) = e1 + e2
and b203 (e2 ) = e2 .
These mean that
b103 = [0, e1 , e1 + e2 ] and b203 = [0, e1 + e2 , e2 ],
so we have
β03 = [0, e1 , e1 + e2 ] + [0, e1 + e2 , e2 ].
(10.102)
β13 = [e3 , e1 + e3 , e1 + e2 + e3 ] + [e3 , e1 + e2 + e3 , e2 + e3 ].
(10.103)
Similarly, we have
Therefore we follow from the expressions (10.102) and (10.103) that
−(β03 − β13 ) = −[0, e1 , e1 + e2 ] − [0, e1 + e2 , e2 ] + [e3 , e1 + e3 , e1 + e2 + e3 ]
+ [e3 , e1 + e2 + e3 , e2 + e3 ]
= [e3 , e1 + e3 , e1 + e2 + e3 ] − [e3 , e2 + e3 , e1 + e2 + e3 ]
−[0, e1 , e1 + e2 ] + [0, e2 , e1 + e2 ]
which is the third brackets in the expression (10.95).
Hence, by the above computations, we get our desired result that
3
∂J =
3
X
i=1
(−1)i (β0i − β1i ).
This ends the proof of the problem.
10.5
Problems on closed forms and exact forms
Problem 10.20
Rudin Chapter 10 Exercise 20.
Proof. Suppose that E is an open set in Rn , f ∈ C ′ (E), ω is a k-form of class C ′ in E and Φ is a
(k + 1)-chain of class C ′′ in E. By Theorem 10.20(a), we have
d(f ω) = ( df ) ∧ ω + (−1)0 f dω = ( df ) ∧ ω + f dω.
(10.104)
Apply Theorem 10.33 (Stokes’ Theorem) to the left-hand side in the expression (10.104), we get
Z
Z
Z
Z
fω =
d(f ω) = ( df ) ∧ ω +
f dω
Φ
∂Φ
Φ
which implies that the desired equality
Z
Z
f dω =
Φ
∂Φ
fω −
Φ
Z
Φ
( df ) ∧ ω.
Let n = 1 and k = 0 in the above consideration. Let, further, E be an open set in R containing the
interval [a, b], where a and b are real numbers with a < b. Now we consider the oriented affine 1-simplex
Φ : [0, 1] → R, where
Φ(0) = a and Φ(1) = b.
By Definition 10.29, we have
∂Φ = [b] − [a]
295
10.5. Problems on closed forms and exact forms
which is an oriented 0-simplex. If ω = g is a 0-form of class C ′ in E, then f g is also a 0-form of class C ′
in E. Thus, by the equation just preceding Theorem 10.27 (see [21, p. 267]), we acquire
Z
Z
Z
f g = f (b)g(b) − f (a)g(a).
(10.105)
fg +
fω =
−a
+b
∂Φ
On the other hand, it is clear that f dg and ( df )g are 1-forms by Definition 10.18. In addition, we obtain
from [21, Eqn. (59)] that
dg = g ′ (x) dx,
where x = Φ(t), thus we see from [21, Eqn. (35), p. 254] that
Z
Z
f (x)g ′ (x)
dx
f dg =
| {z }
Φ
Φ
=
=
This is the a(x) in
[21, Eqn. (34), p. 254]
Z
f (Φ(t))g ′ (Φ(t))Φ′ (t) dt
[0,1]
Z 1
f (Φ(t))g ′ (Φ(t))Φ′ (t) dt.
0
If Φ is supposed to be strictly increasing on [0, 1], then [21, Eqn. (39), p. 133)] implies that
Z 1
Z b
f (u)g ′ (u) du.
f (Φ(t))g ′ (Φ(t)) Φ′ (t) dt =
|
{z
}
0
a
(10.106)
This is the f (ϕ(y)) in
[21, Eqn. (39), p. 133]
Similarly, we have df = f ′ (x) dx, where x = Φ(t). Then it deduces from [21, Eqn. (35), p. 254] that
Z 1
Z
Z
Z
Z
′
′
′
g(Φ(t))f ′ (Φ(t))Φ′ (t) dt.
g(Φ(t))f (Φ(t))Φ (t) dt =
g(x)f (x) dx =
g( df ) =
( df ) ∧ ω =
0
[0,1]
Φ
Φ
Φ
By [21, Eqn. (39), p. 133)] again, we obtain
Z 1
Z b
′
′
g(Φ(t))f (Φ(t))Φ (t) dt =
g(u)f ′ (u) du.
0
(10.107)
a
Combining the equalities (10.105) to (10.107), we establish that
Z b
Z b
′
f (u)g (u) du = f (b)g(b) − f (a)g(a) −
f (u)g ′ (u) du
a
a
which is exactly Theorem 6.22 (Integration of Parts). This completes the proof of the problem.
Problem 10.21
Rudin Chapter 10 Exercise 21.
Proof.
(a) Since x = r cos t and y = r sin t, the direct computation of the formula [21, Eqn. (113), p. 277] is
given as follows:
Z
Z 2π 2
Z 2π
r cos2 t dt + r2 sin2 t dt
η=
=
dt = 2π.
r2
γ
0
0
However, it follows from Theorem 10.20 and [21, Eqn. (59), p. 260] that
x y y
x ∧ dy +
∧ d2 y − d 2
∧ dx −
∧ d2 x
dη = d 2
2
2
2
2
2
2
x +y
x +y
x +y
x +y
Chapter 10. Integration of Differential Forms
=
h (x2 + y 2 ) − x(2x)
i
−x(2y)
dy
∧ dy
(x2 + y 2 )2
(x2 + y 2 )2
h −y(2x)
(x2 + y 2 ) − y(2y) i
−
dx
+
dy ∧ dx.
(x2 + y 2 )2
(x2 + y 2 )2
296
dx +
(10.108)
By the anticommutative relation ( dy ∧ dx = − dx ∧ dy) and [21, Eqn. (43), p. 256], we deduce
from the expression (10.108) that
h (x2 + y 2 ) − x(2x)
i
(x2 + y 2 ) − y(2y)
dx ∧ dy
2
2
2
(x + y )
i
h −x(2y)
−y(2x)
dy
∧
dy
−
dx
∧
dx
+
(x2 + y 2 )2
(x2 + y 2 )2
1
= 2
(−2xy dy ∧ dy + 2xy dx ∧ dx)
(x + y 2 )2
= 0.
dη =
(x2 + y 2 )2
dx ∧ dy −
(b) Let D = {(t, u) | 0 ≤ t ≤ 2π, 0 ≤ u ≤ 1} and Φ : D → R2 \ {0} be given as in the hint. Since
γ : [0, 2π] → R2 \ {0} and Γ : [0, 2π] → R2 \ {0} are C ′′ -mappings, Φ is a 2-surface in R2 \ {0} by
Definition 10.10. The geometric interpretation of the mapping Φ is given in Figure 10.12, where a
purple arrow is an interval [γ(t), Γ(t)] for some t ∈ [0, 2π] which does not contain the origin 0.o
Figure 10.12: The mapping Φ : D → R2 \ {0}.
By a similar analysis as in Example 10.32, we know that
∂Φ = Φ(∂D) = σ1 + σ2 + σ3 + σ4 ,
(10.109)
where
σ1 (t) = Φ(t, 0) = Γ(t),
σ2 (u) = Φ(2π, u) = (1 − u)Γ(2π) + uγ(2π),
σ3 (t) = Φ(2π − t, 1) = γ(2π − t),
σ4 (u) = Φ(0, 1 − u) = uΓ(0) + (1 − u)γ(0).
Since γ(0) = γ(2π) and Γ(0) = Γ(2π), we follow from [21, Eqn. (77) & (80), pp. 266, 267] in
Definition 10.26 that
σ2 = [Γ(0), γ(0)] = −[γ(0), Γ(0)] = −σ4 .
o In fact, the Φ is a homotopy between γ and Γ, see [18, p. 323].
297
10.5. Problems on closed forms and exact forms
Similarly, by direct application of [21, Eqn. (35), p. 254], we obtain
Z
Z
ω=− ω
γ
σ3
for every 1-form ω. In other words, we get from the relation (10.109) that
∂Φ = Γ − γ
(10.110)
By Theorem 10.33 (Stokes’ Theorem) and part (a), we get
Z
Z
Z
η=
dη =
0 = 0.
Φ
∂Φ
(10.111)
Φ
Hence it follows from the expressions (10.110) and (10.111) that
Z
Z
Z
Z
η=0
η+
η+
η+
σ4
σ3
σ2
σ1
Z
Z
Z
Z
η=0
η− η+
η−
σ4
γ
σ4
Γ
Z
Z
η.
η=
Γ
(10.112)
γ
Now, by the result of part (a), we have the desired result from (10.112)
Z
η = 2π.
Γ
(c) Suppose that
Γ(t) = (a cos t, b sin t),
where a > 0, b > 0 are fixed. Now Γ is a C ′′ -curve in R2 \ {0} with parameter interval [0, 2π] and
Γ(0) = Γ(2π). Let [γ(t), Γ(t)] be the interval joining the points γ(t) and Γ(t) for each t ∈ [0, 2π].
Since [γ(t), Γ(t)] does not contain the origin 0, we deduce from part (b) that
Z
η = 2π
Γ
Z
x dy − y dx
= 2π
x2 + y 2
[0,2π]
Z 2π
ab cos2 t dt + ab sin2 t dt
= 2π
a2 cos2 t + b2 sin2 t
0
Z 2π
ab
dt = 2π.
2
2
a cos t + b2 sin2 t
0
(d) Recall from [21, Remark 10.35(a), p. 275] that a 1-form
ω=
n
X
fi (x) dxi
i=1
is exact in an open set E ⊆ Rn if and only if there is a function (0-form) g ∈ C ′ (E) such that
(Di g)(x) = fi (x)
(x ∈ E, 1 ≤ i ≤ n).
(10.113)
Consider n = 2 and E to be any convex open set in which x 6= 0.p By the definition of η, we have
f1 (x, y) =
−y
x2 + y 2
and f2 (x, y) =
x
x2 + y 2
.
p In other words, E does not interest with the y-axis and this means that E lies either entirely in the left or right half
plane.
Chapter 10. Integration of Differential Forms
Let g : E → R be defined by
298
y
g(x, y) = arctan .
x
d
1
Since dx
(arctan x) = 1+x
2 , we can easily see that
y −y
1
∂ y
arctan
= 2 ·
=− 2
y2
∂x
x
x
x
+
y2
1+ 2
and
x
y
x
∂ arctan
= 2
.
∂y
x
x + y2
Now the function g satisfies the conditions (10.113), so the definition gives
x
η = dg = d arctan
y
in E.
Similarly, we suppose that F is any convex open set in which y 6= 0.q A direct computation shows
that
∂ x
y
x
x
∂ − arctan
=− 2
− arctan
= 2
and
,
2
∂x
y
x +y
∂y
y
x + y2
thus we have
∂ x
∂ x
x
=
− arctan
dx +
− arctan
dy
d − arctan
y
∂x
y
∂y
y
x
y
dx + 2
dy
=− 2
x + y2
x + y2
x dy − y dx
=
x2 + y 2
= η.
By the result of Example 10.36, η is not exact in R2 \ {0}, but the analysis verifies that we can say
that η is exact locally in R2 \ {0}, so it is reasonable to denote
η = dθ
for some 0-form θ.
(e) We write
[0, 2π] = I1 ∪ I2 ∪ · · · ∪ I6 ,
where Ii = [ (i−1)π
, iπ
3
3 ] and i = 1, 2, . . . , 6. Now
γ
lie in E, while
h π i
0,
,
3
γ
h 2π
γ
h π 2π i
,
3 3
,π
3
i
,
γ
h 4π i
π,
3
and γ
h 4π 5π i
,
3 3
Z
Z
lie in F . Therefore, it follows from (d) that
Z
η=
γ
6 Z
X
i=1
=
Z
and γ
h 5π
3
, 2π
i
η
γ(Ii )
Z
y
d arctan
x
γ(I1 )
γ(I3 )
γ(I4 )
γ(I6 )
Z
Z
x
.
d − arctan
+
+
y
γ(I5 )
γ(I2 )
+
+
+
q Now F does not interest with the x-axis and so F lies either entirely in the upper or lower half plane.
(10.114)
299
10.5. Problems on closed forms and exact forms
We apply Theorem 10.33 (Stokes’ Theorem) to the integrals on the right hand side in the expression
(10.114) and then [21, Eqn. (62), p. 261], we obtain
Z
η = tan−1 tan t
γ
π
3
0
π
+ tan−1 tan t 2π + tan−1 tan t
3
2π
3
5π
3
3
3
4π
3
π
2π
+ tan−1 tan t 5π
2
− tan−1 cot t π − tan−1 cot t 4π
π
2π
5π
π
π
3
3
− t π − tan−1
− t 4π
= × 4 − tan−1 tan
3
2
2
3
3
2π
5π
π
4π π
3
3
−
−t π −
− t 4π
=
3
2
2
3
3
4π 2π
+
=
3
3
= 2π.
This means that part (d) implies part (b).
(f) Given that Γ(t) = (Γ1 (t), Γ2 (t)), where t ∈ [0, 2π]. We also write
Γ = Γ1 + iΓ2
in the complex plane C. Since Γ is assumed to be a closed C ′ -curve in R2 \ {0}, it is a continuously
differentiable closed curve in C and Γ(t) 6= 0 for every t ∈ [0, 2π]. Thus, by the definition of the
index of Γ in Problem 8.23, we have
Ind (Γ) =
1
2πi
Z 2π
0
Γ′1 + iΓ′2
dt
Γ1 + iΓ2
Z 2π ′
1
Γ1 + iΓ′2 Γ1 − iΓ2
=
·
dt
2πi 0 Γ1 + iΓ2 Γ1 − iΓ2
Z 2π
Z 2π
1
Γ1 Γ′2 − Γ′1 Γ2
Γ1 Γ′1 + Γ2 Γ′2
1
dt
+
dt.
=
2
2
2π 0
Γ1 + Γ2
2πi 0
Γ21 + Γ22
(10.115)
dx
Since η = x dy−y
x2 +y 2 , Definition 10.11 implies that
1
2π
Z
Γ
η=
1
2π
Z 2π
0
Γ1 Γ′2 − Γ2 Γ1
dt
Γ21 + Γ22
(10.116)
which is exactly the real part of the complex number (10.115). Furthermore, since Γ(2π) = Γ(0),
we have
Z 2π
Z 2π
2π
Γ1 Γ′1 + Γ2 Γ′2
d(ln(Γ21 + Γ22 )) = ln[Γ21 (x) + Γ22 (x)]
dt
=
= 0.
(10.117)
2
2
Γ1 + Γ2
0
0
0
Hence we reach the desired result by comparing the expressions (10.115) to (10.117).
This completes the proof of the problem.
Problem 10.22
Rudin Chapter 10 Exercise 22.
Proof. It should be noted that the equations of x, y and z expressed in terms of u and v are actually the
spherical coordinates of the point (x, y, z) on the unit sphere.
Let’s “see” the spherical coordinates for the point Σ(u, v) in Figure 10.13.
Chapter 10. Integration of Differential Forms
300
Figure 10.13: The spherical coordinates for the point Σ(u, v).
(a) By Definition 10.18, we see that
y
z
x
dζ = d 3 dy ∧ dz + d 3 dz ∧ dx + d 3 dx ∧ dy.
r
r
r
(10.118)
Since
x
∂ x
∂ x
∂ x
d 3 =
dx
+
dy
+
dz
r
∂x r3
∂y r3
∂z r3
r2 − 3x2
3xy
3xz
=
dx − 5 dy − 5 dz,
6
r
r
r
y
∂ y
∂ y
∂ y
dx
+
dy
+
dz
d 3 =
r
∂x r3
∂y r3
∂z r3
3xy
r2 − 3y 2
3yz
= − 5 dx +
dy − 5 dz,
6
r
r
r
z
∂ z
∂ z
∂ z
dx
+
dy
+
dz
d 3 =
r
∂x r3
∂y r3
∂z r3
3xz
3yz
r2 − 3z 2
= − 5 dx − 5 dy +
dz,
r
r
r6
we apply these, the anticommutative relation and the facts dx ∧ dx = dy ∧ dy = dz ∧ dz = 0 to
the expression (10.118) to obtain
dζ =
3r2 − 3(x2 + y 2 + z 2 )
3r2 − 3r2
·
dx
∧
dy
∧
dz
=
· dx ∧ dy ∧ dz = 0
r6
r6
in R3 \ {0}.
(b) We remark that the first place where the concept of the area of a 2-surface in R3 occurs is Sec.
10.46, not Sec. 10.43., so it is believed that this is a typo.
301
10.5. Problems on closed forms and exact forms
Suppose that E ⊆ D is a compact set and S = ΣE : E → R3 . Then S is also a 2-surface in R3 , of
class C ′′ . Notice that
∂(y, z)
=
∂(u, v)
cos u sin v
− sin u
sin u cos v
0
∂(z, x)
=
∂(u, v)
− sin u
cos u cos v
0
− sin u sin v
= sin2 u sin v,
∂(x, y)
=
∂(u, v)
cos u cos v
cos u sin v
− sin u sin v
sin u cos v
= sin u cos u.
= sin2 u cos v,
(10.119)
Therefore, it follows from the Jacobians (10.119) and Definition 10.11 that
Z
Z
Z
∂(z, x)
∂(y, z)
sin u sin v ·
du ∧ dv +
du ∧ dv
sin u cos v ·
ζ=
∂(u,
v)
∂(u, v)
S
S
S
Z
∂(x, y)
du ∧ dv
cos u ·
+
∂(u,
v)
S
Z
Z
Z
∂(y, z)
∂(z, x)
∂(x, y)
=
sin u cos v ·
du dv +
sin u sin v ·
du dv +
cos u ·
du dv
∂(u,
v)
∂(u,
v)
∂(u, v)
E
E
E
Z
Z
Z
=
sin3 u cos2 v du dv +
sin3 u sin2 v du dv +
sin u cos2 u du dv
E
E
E
Z
sin u du dv.
(10.120)
=
E
By Definition 10.46, since
N(u, v) = (sin2 u cos v)e1 + (sin2 u sin v)e2 + (sin u cos u)e3 ,
we have
A(S) =
Z
S
|N(u, v)| du dv =
Z
sin u du dv.
(10.121)
E
Hence our result follows immediately from the expressions (10.120) and (10.121).
(c) By a bit algebra, we know that
∂(y, z)
=
∂(t, s)
g ′ (t)h2 (s) g(t)h′2 (s)
g ′ (t)h3 (s) g(t)h′3 (s)
= g ′ (t)g(t)[h2 (s)h′3 (s) − h′2 (s)h3 (s)],
∂(z, x)
=
∂(t, s)
g ′ (t)h3 (s) g(t)h′3 (s)
g ′ (t)h1 (s) g(t)h′1 (s)
= g ′ (t)g(t)[h3 (s)h′1 (s) − h1 (s)h′3 (s)],
∂(x, y)
=
∂(t, s)
g ′ (t)h1 (s) g(t)h′1 (s)
g ′ (t)h2 (s) g(t)h′2 (s)
= g ′ (t)g(t)[h1 (s)h′2 (s) − h′1 (s)h2 (s)].
Therefore, it follows from the Jacobians (10.122) and Definition 10.11 that
Z
Z
n
1
g 2 (t)g ′ (t)h1 (s)[h2 (s)h′3 (s) − h′2 (s)h3 (s)]
ζ=
3 ×
2
2
2
3
Φ
I 2 g (t)[h1 (s) + h2 (s) + h3 (s)] 2
+ g 2 (t)g ′ (t)h2 (s)[h3 (s)h′1 (s) − h1 (s)h′3 (s)]
o
+ g 2 (t)g ′ (t)h3 (s)[h1 (s)h′2 (s) − h′1 (s)h2 (s)] dt ds
Z
n
g ′ (t)
h1 (s)[h2 (s)h′3 (s) − h′2 (s)h3 (s)]
=
3 ×
2 (s) + h2 (s) + h2 (s)] 2
2
g(t)[h
I
1
2
3
o
′
+ h2 (s)[h3 (s)h1 (s) − h1 (s)h′3 (s)] + h3 (s)[h1 (s)h′2 (s) − h′1 (s)h2 (s)] dt ds
(10.122)
Chapter 10. Integration of Differential Forms
=
Z
g ′ (t)
302
′
′
3 × h1 (s)h2 (s)h3 (s) − h1 (s)h2 (s)h3 (s)
g(t)[h21 (s) + h22 (s) + h23 (s)] 2
+ h′1 (s)h2 (s)h3 (s) − h1 (s)h2 (s)h′3 (s) + h1 (s)h′2 (s)h3 (s) − h′1 (s)h2 (s)h3 (s) dt ds
I2
= 0.
(d) We follow the given hint. Since E is a closed rectangle, we have
E = [a, b] × [c, d]
for some constants a, b, c and d with 0 < a < b < π and 0 < c < d < 2π, see Figure 10.14 below:
Figure 10.14: The rectangles D and E.
Consider the 3-surface Ψ : [0, 1] × E → R3 \ {0} given by
Ψ(t, u, v) = [1 − t + tf (u, v)]Σ(u, v),
where (u, v) ∈ E, 0 ≤ t ≤ 1. For fixed v, define
Eu = {u ∈ [0, π] | (u, v) ∈ E} = [a, b]
and the mapping Φ : [0, 1] × [a, b] → R3 \ {0} given by
Φ(t, u) = Ψ(t, u, v).
Now [0, 1] × [a, b] is a 2-cell, thus it is compact by Theorem 2.40 and Φ is a 2-surface of class C ′′
with parameter domain [0, 1] × [a, b]. Since v is fixed, we have Φ(t, u) = (x, y, z), where
x = g(t, u) cos v sin u,
y = g(t, u) sin v sin u,
z = g(t, u) cos u
and
g(t, u) = [1 − t + tf (u, v)].
Since 0 ≤ t ≤ 1 and f (u, v) > 0 on D, g(t, u) > 0 on D.
By a similar argument as in part (c), instead of the Jacobians (10.122), we haver
r Here we denote g
t =
∂(y, z)
=
∂(t, s)
gt sin v sin u sin v(g cos u + gu sin u)
gt cos u
gu cos u − g sin u
= −ggt sin v,
∂(z, x)
=
∂(t, s)
gt cos u
gu cos u − g sin u
gt cos v sin u cos v(g cos u + gu sin u)
= ggt cos v,
∂(x, y)
=
∂(t, s)
gt cos v sin u cos v(g cos u + gu sin u)
gt sin v sin u sin v(g cos u + gu sin u)
= 0.
∂g(t,u)
and gu = ∂g(t,u)
.
∂t
∂u
(10.123)
303
10.5. Problems on closed forms and exact forms
Thus, by the Jacobians (10.123), we have
Z
Z
Z
ζ=
g 2 gt sin v cos v sin u du dv −
Φ
[0,1]×[a,b]
g 2 gt sin v cos v sin u du dv
[0,1]×[a,b]
= 0.
We notice that the same thing holds when u is fixed.
By the definition of Ψ, we get
∂Ψ = Ψ(0, u, v) − Ψ(1, u, v) + Ψ(t, a, v) − Ψ(t, b, v) + Ψ(t, u, c) − Ψ(t, u, d)
= S(u, v) − Ω(u, v) + Φa (t, v) − Φb (t, v) + Φc (t, u) − Ψd (t, u),
(10.124)
where Φc and Φd are the mappings defined by
(t, u) 7→ Ψ(t, u, c) and (t, u) 7→ Ψ(t, u, d)
respectively, while Φa and Φb are the mappings defined by
(t, v) 7→ Ψ(t, a, v) and (t, v) 7→ Ψ(t, b, v)
respectively. By the above analysis, we know that
Z
Z
Z
η=
η=
Z
η = 0.
(10.125)
Φd
Φc
Φb
Φa
η=
Since f ∈ C ′′ (D), Ψ is a 3-chain of class C ′′ in R3 \ {0}. In addition, since ζ is a 2-form of class C ′
in R3 \ {0}, Theorem 10.33 (Stokes’ Theorem) implies that
Z
Z
dζ =
ζ.
(10.126)
Ψ
∂Ψ
By part (a), dζ = 0 so that the integral (10.126) shows that
Z
ζ = 0.
(10.127)
∂Ψ
Hence we follow from the relations (10.124), the integrals (10.125) and the expression (10.127) that
Z
ζ
0=
Z
Z
Z
Z
Z∂Ψ Z
ζ
ζ−
ζ+
ζ−
ζ+
ζ−
=
Φd
Φc
Φb
Φa
Ω
S
Z
Z
ζ
ζ−
=
S
Ω
and it is equivalent to
Z
Ω
ζ=
Z
ζ = A(S).
S
For a better illustration of the set S = ΣE , by the analysis in Example 10.32, we have
∂S = ∂(ΣE ) = Σ(∂E) = γ1 + γ2 + γ3 + γ4 ,
where
γ1 (u) = Σ(u, c)
= (sin u cos c, sin u sin c, cos u),
γ2 (v) = Σ(b, v)
Chapter 10. Integration of Differential Forms
304
= (sin b cos v, sin b sin v, cos b),
γ3 (u) = Σ(b + a − u, d)
= (sin(b + a − u) cos d, sin(b + a − u) sin d, cos(b + a − u)),
γ4 (v) = Σ(a, c + d − v)
= (sin a cos(c + d − v), sin a sin(c + d − v), cos a),
with a ≤ u ≤ b and c ≤ v ≤ d. In particular, we consider the example that E = [ π4 , π2 ] × [ π2 , π].
Then we have
γ1 (u) = (0, sin u, cos u),
γ2 (v) = (cos v, sin v, 0),
3π
3π
γ3 (u) = − sin
− u , 0, cos
−u ,
4
4
3π
√2
3π
√2 √2
,
cos
−v ,
sin
−v ,
γ4 (v) =
2
2
2
2
2
where π4 ≤ u ≤ π2 and π2 ≤ v ≤ π. The corresponding 2-surface S and its boundary ∂S is shown in
Figure 10.15 below:
Figure 10.15: An example of the 2-surface S and its boundary ∂S.
305
10.5. Problems on closed forms and exact forms
(e) Let
V = {(x, y, z) | x2 + y 2 > 0, z ∈ R} = R3 \ {(0, 0, z) | z ∈ R}
which is an open set in R3 . In other words, V is the 3-dimensional space with deleted z-axis. Now
−
z
r
and η
are 0- and 1-forms respectively, we follow from Theorem 10.20(a), Definition 10.18 and Problem
10.21(a) that
h z i
z
dλ = − d
∧ η + (−1)0 ∧ dη
r
r
xz
yz
r2 − z 2 x dy − y dx dx + 3 dy −
dz ∧
=
r3
r
r3
x2 + y 2
2
2
z
x +y
x dy − y dx = 3 dx ∧ dy −
dz
∧
r
r3
x2 + y 2
z
x
y
= 3 dx ∧ dy − 3 dz ∧ dy + 3 dz ∧ dx
r
r
r
x dy ∧ dz + y dz ∧ dx + z dx ∧ dy
=
r3
= ζ.
Hence ζ is exact in the open set V .
(f) Note that
Ω(u, v) = f (u, v)Σ(u, v) = (f (u, v) sin u cos v, f (u, v) sin u sin v, f (u, v) cos u),
where (u, v) ∈ E. By the hypothesis E = [a, b] × [c, d] made in part (b), if we further assume that
0 < a < b < π, then it is easy to see that
[f (u, v) sin u cos v]2 + [f (u, v) sin u sin v]2 = f 2 (u, v) sin2 u > 0
which means
Ω ⊆ V.
(10.128)
Recall that f ∈ C ′′ (D) and Σ is a 2-surface of class C ′′ in R3 \ {0}, we have Ω is a 2-surface of
class C ′′ in R3 \ {0} too. Since λ is clearly a 1-form of class C ′ in V ,
ζ = dλ
in V by part (d) and the subset relation (10.128), the exactness of ζ still hold in Ω. Hence it follows
from Theorem 10.33 (Stokes’ Theorem) that
Z
Z
Z
Z
z
(10.129)
η.
dλ =
λ=
ζ=
∂Ω r
Ω
∂Ω
Ω
Similarly, the definition of S implies that S ⊆ V . Since S is a 2-surface of class C ′′ in R3 \ {0},
Theorem 10.33 (Stokes’ Theorem) again implies that
Z
Z
Z
Z
z
ζ=
dλ =
(10.130)
η.
λ=
S
S
∂S
∂S r
Now we want to show that the two right-most integrals in the relations (10.129) and (10.130) are
equal. Such a proof is presented in two steps below.
– Step 1: Analysis of zr on ∂Ω and ∂S. We notice that if (x, y, z) ∈ ΣE (u, v), then
x = sin u cos v,
y = sin u sin v,
z = cos u
(10.131)
Chapter 10. Integration of Differential Forms
so that
306
z
cos u
=
= cos u.
r
1
Similarly, if (x, y, z) ∈ Ω(u, v), then
x = f (u, v) sin u cos v,
y = f (u, v) sin u sin v,
z = f (u, v) cos u,
where f (u, v) > 0 so that
f (u, v) cos u
z
=
= cos u.
r
f (u, v)
In other words, zr is the same at ΣE (u, v) as at Ω(u, v). By part (b), we know that
S = ΣE
and thus zr is the same at ∂S as at ∂Ω.
– Step 2: Analysis of ζ on ∂Ω and ∂S. By Figure 10.14, we know that E does not intersect
u = 0, therefore we have
y
η = d arctan
x
by Problem 10.21(d).s On ∂S, we have
arctan
Similarly, on ∂Ω, we have
arctan
sin u sin v y
= arctan(tan v) = v.
= arctan
x
sin u cos v
f (u, v) sin u sin v y
= arctan(tan v) = v.
= arctan
x
f (u, v) sin u cos v
Thus the 1-form η is the same at ∂S as at ∂Ω too.
Hence we deduce from the above analysis and part (b) that
Z
Z
Z
Z
Z
Z
z
z
ζ=
ζ = A(S)
η=
η=
λ=
λ=
Ω
S
∂S
∂Ω
∂Ω r
∂S r
which is our expected result.
(g) The answer is affirmative. Let L be a straight line through the origin. Recall that the spherical
coordinates for a unit vector x = (x, y, z) are given by the formulas (10.131). It is well-knownt that
the matrices


cos u − sin u 0
Rxy (u) =  sin u cos u 0  ,
0
0
1


cos v 0 sin v
0
1
0 ,
Rzx (v) = 
− sin v 0 cos v


1
0
0
Ryz (w) =  0 cos w − sin w 
0 sin w
cos w
represent rotations by angle u, v and w counterclockwise in the (x, y)-plane, the (z, x)-plane and
the (y, z)-plane respectively.
s Note that E does not intersect v = 0, so the formula η = d
t See, for instance, [16, pp. 328 - 332].
is also applicable to get the same result.
− arctan x
y
307
10.5. Problems on closed forms and exact forms
Lemma 10.11
Let T : R3 → R3 be the transformation with matrix M defined by
M = Rzx (−u)Rxy (−v).
Then T transforms the straight line L onto the z-axis.
Proof of Lemma 10.11. We have
M = Rzx (−u)Rxy (−v)


cos u 0 − sin u
cos v
sin v
  − sin v cos v
1
0
= 0
sin u 0 cos u
0
0


cos u cos v cos u sin v − sin u
,
cos v
0
=  − sin v
sin u cos v sin u sin v
cos u

0
0 
1
(10.132)
where 0 ≤ u ≤ π and 0 ≤ v ≤ 2π. Then direct computation shows that
T (x) = Mx

cos u cos v
=  − sin v
sin u cos v
 
0
=  0 .
1
cos u sin v
cos v
sin u sin v


− sin u
sin u cos v
  sin u sin v 
0
cos u
cos u
In other words, the mapping T transforms L onto the z-axis, completing the proof of the
lemma.
Let E = R3 \ L. Since T is bijectiveu , we consider the mapping
TE = T : E → V
which is also bijective. Since λ is of class C ′ in V and the formula (10.132) of M implies that T
and then TE is of class C ′′ , we deduce from Theorem 10.22(c) and part (e) that
d(λTE ) = ( dλ)TE = ζTE .
Since ζ is a 2-form in V , ζTE is a 2-form in E. We claim that
ζ = ζTE .
To this end, we get from the formula (10.132) that



cos u cos v cos u sin v − sin u
x
 y 
cos v
0
T (x) =  − sin v
sin u cos v sin u sin v
cos u
z


(cos u cos v)x + (cos u sin v)y + (− sin u)z
.
(− sin v)x + (cos v)y
=
(sin u cos v)x + (sin u sin v)y + (cos u)z
Following the notations used in Definition 10.21, we have
t1 (x) = (cos u cos v)x + (cos u sin v)y + (− sin u)z,
u This can be easily checked by the matrix form (10.132).
Chapter 10. Integration of Differential Forms
308
t2 (x) = (− sin v)x + (cos v)y,
t3 (x) = (sin u cos v)x + (sin u sin v)y + (cos u)z
which imply that
dt1 = (cos u cos v) dx + (cos u sin v) dy − sin u dz,
dt2 = − sin u dx + cos v dy,
(10.133)
dt3 = (sin u cos v) dx + (sin u sin v) dy + cos u dz.
Thus we obtain from [21, Eqn. (67), p.262] that
ζTE = [(cos u cos v)x + (cos u sin v)y + (− sin u)z] dt2 ∧ dt3
+ [(− sin u)x + (cos v)y] dt3 ∧ dt1
+ [(sin u cos v)x + (sin u sin v)y + (cos u)z] dt1 ∧ dt2 .
(10.134)
We need to compute
dt2 ∧ dt3 ,
dt3 ∧ dt1
and
dt1 ∧ dt2 .
To do this, we know from the formulas (10.133) that
dt2 ∧ dt3 = [− sin v dx + cos v dy] ∧ [(sin u cos v) dx + (sin u sin v) dy + cos u dz]
= −(sin u sin2 v + sin u cos2 v) dx ∧ dy − (sin v cos u) dx ∧ dz
+ (cos u cos v) dy ∧ dz
= (− sin u) dx ∧ dy − (sin v cos u) dx ∧ dz + (cos u cos v) dy ∧ dz,
dt3 ∧ dt1 = [(sin u cos v) dx + (sin u sin v) dy + cos u dz]
(10.135)
∧ [(cos u cos v) dx + (cos u sin v) dy − sin u dz]
= (sin u cos u sin2 v) dx ∧ dy + (sin u2 cos v) dz ∧ dx
− (sin u cos u sin2 v) dx ∧ dy − (sin2 u sin v) dy ∧ dz
+ (cos2 u cos v) dz ∧ dx − (cos2 u sin v) dy ∧ dz
= (cos v) dz ∧ dx − (sin v) dy ∧ dz
(10.136)
and
dt1 ∧ dt2 = [(cos u cos v) dx + (cos u sin v) dy − sin u dz] ∧ [− sin v dx + cos v dy]
= (cos u cos2 v + cos u sin2 v) dx ∧ dy + (sin u sin v) dz ∧ dx
+ (sin u cos v) dy ∧ dz
= (cos u) dx ∧ dy + (sin u sin v) dz ∧ dx + (sin u cos v) dy ∧ dz.
Now we show that r is invariant under the rotation Rzx (−u)Rxy (−v):
Lemma 10.12
1
If we denote rTE = (t21 + t22 + t23 ) 2 , then we have
rTE = r.
(10.137)
309
10.5. Problems on closed forms and exact forms
Proof of Lemma 10.12. By direct computation, we have
rT2 E = t21 + t22 + t23
= [(cos u cos v)x + (cos u sin v)y + (− sin u)z]2 + [(− sin v)x + (cos v)y]2
+ [(sin u cos v)x + (sin u sin v)y + (cos u)z]2
= (cos2 u cos2 v)x2 + 2(cos u cos v)[(cos u sin v)y + (− sin u)z]x + [(cos u sin v)y + (− sin u)z]2
+ (sin2 v)x2 − (2 sin v cos v)xy + (cos2 v)y 2 + (sin2 u cos2 v)x2
+ (2 sin u cos v)[(sin u sin v)y + (cos u)z]x + [(sin u sin v)y + (cos u)z]2
= x2 + 2(cos u cos v)[(cos u sin v)y + (− sin u)z]x + [(cos u sin v)2 y 2 − (2 cos u sin u sin v)yz
+ (− sin u)2 z 2 ] − (2 sin v cos v)xy + (cos2 v)y 2 + (sin2 u cos2 v)x2
+ (2 sin u cos v)[(sin u sin v)y + (cos u)z]x + [(sin u sin v)2 y 2
+ (2 sin u cos u sin v)yz + (cos u)2 z 2 ]
= x2 + y 2 + z 2 + (2 cos2 u sin v cos v − 2 sin v cos v + 2 sin2 u cos v sin v)xy
(−2 cos u sin u sin v + 2 cos u sin u sin v)yz + (−2 sin u cos u cos v + 2 sin u cos u cos v)zx
= x2 + y 2 + z 2
= r2
which certainly gives
rTE = r
and this completes the proof of the lemma.
Now we return to the proof of the problem. After putting the identities (10.135), (10.136) and
(10.137) into the 2-form (10.134) and using Lemma 10.11, we have
rT3 E ζTE = [(cos u cos v)x + (cos u sin v)y + (− sin u)z] × [−(sin u) dx ∧ dy + (sin v cos u) dz ∧ dx
+ (cos u cos v) dy ∧ dz]
+ [(− sin v)x + (cos v)y] × [(cos v) dz ∧ dx + (− sin v) dy ∧ dz]
+ [(sin u cos v)x + (sin u sin v)y + (cos u)z] × [(cos u) dx ∧ dy + (sin u sin v) dz ∧ dx
+ (sin u cos v) dy ∧ dz]
n
= (cos u cos v)x + (cos u sin v)y + (− sin u)z × (− sin u)
o
+ [(sin u cos v)x + (sin u sin v)y + (cos u)z] × (cos u) dx ∧ dy
n
+ (cos u cos v)x + (cos u sin v)y + (− sin u)z × (cos u cos v)
+ (− sin v)x + (cos v)y × (− sin v)
o
+ (sin u cos v)x + (sin u sin v)y + (cos u)z × (sin u cos v) dy ∧ dz
n
+ (cos u cos v)x + (cos u sin v)y + (− sin u)z × (sin v cos u)
+ (− sin v)x + (cos v)y × (cos v)
o
(sin u cos v)x + (sin u sin v)y + (cos u)z × (sin u sin v) dz ∧ dx
= z dx ∧ dy + x dy ∧ dz + y dz ∧ dx
= r3 ζ
which implies that
ζTE = ζ.
(10.138)
As we have shown that d(λTE ) = ζTE holds in E = R3 \ L, this and the identity (10.138) imply
that
ζ = d(λTE ),
Chapter 10. Integration of Differential Forms
310
i.e., ζ is exact E.
This completes the proof of the problem.
Problem 10.23
Rudin Chapter 10 Exercise 23.
Proof.
(a) By applying Theorem 10.20 repeatedly, we have
d[(−1)i−1 (rk )−k xi dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk ]
= d[(−1)i−1 (rk )−k xi ] ∧ dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk
+ (−1)0 (−1)i−1 (rk )−k xi ∧ d( dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk )
= d[(−1)i−1 (rk )−k xi ] ∧ dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk
+ (−1)i−1 (rk )−k xi [( d2 x1 ) ∧ dx2 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk
| {z }
It is 0.
+(−1)1 dx1 ∧ d( dx2 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk )]
= d[(−1)i−1 (rk )−k xi ] ∧ dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk
+ (−1)i (rk )−k xi dx1 ∧ d( dx2 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk )
= ···
= d[(−1)i−1 (rk )−k xi ] ∧ dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk .
(10.139)
By [21, Eqn. (59), p.260], we have
d[(−1)i−1 (rk )−k xi ] = (−1)i−1
k
X
∂ xi dxj .
∂xj rkk
j=1
(10.140)
It is clear that

k
∂rk
k−1 ∂rk
k−1 xj


 −xi ∂xj = −xi krk ∂xj = −xi krk rk = −kxi xj , if j 6= i;



rk2k
rk2k
rk2k

rkk+2
∂ xi =

∂xj rkk

∂r k


rkk − xi ∂xki
rk − kx2i rkk−2
r2 − kx2


= k
= k k+2 i ,

2k
2k
rk
rk
rk
if j = i.
Therefore, it follows from the summation (10.140) that
i−1
d[(−1)
−k
(rk )
i−1
xi ] = (−1)
k
X
rk2 − kx2i
−kxi xj
dx
+
dxj
i
k+2
rk
rkk+2
j=1
!
j6=i
and then the expression (10.139) with an application of the anticommutative relation shows that
d[(−1)i−1 (rk )−k xi dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk ]
!
k
2
2
X
−kxi xj
i−1 rk − kxi
= (−1)
dxi +
dxj ∧ dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk
rkk+2
rkk+2
j=1
j6=i
r2 − kx2
= (−1)i−1 k k+2 i dxi ∧ dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk
rk
311
10.5. Problems on closed forms and exact forms
r2 − kx2
= (−1)2i−2 k k+2 i dx1 ∧ · · · ∧ dxk
rk
=
rk2 − kx2i
dx1 ∧ · · · ∧ dxk
rkk+2
(10.141)
Hence we deduce from the definition of ωk and the relation (10.141) that
dωk =
k
X
i=1
=
d[(−1)i−1 (rk )−k xi dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk ]
k
X
r2 − kx2
i
k
dx1 ∧ · · · ∧ dxk
i=1
kr2 − k(x21 + · · · + x2k )
dx1 ∧ · · · ∧ dxk
= k
rkk+2
rkk+2
=0
in Ek .
(b) We have fk : Ek → R. Note that the gradient of fk at x is given by
(∇fk )(x) =
k
X
(Di fk )(x)ei .
(10.142)
i=1
– Step 1: fk satisfies the equations given in the hint. By Theorem 6.20 (First Fundamental
Theorem of Calculus), if i 6= k, then we have
Z xk
k−3
∂ rk
(Di fk )(x) = (−1)
(1 − s2 ) 2 ds
∂xi −1
xk k−3
∂(
)
x2 2
= (−1)k rk 1 − 2k
∂xi
rk
k−3
2
−xi xk x1 + · · · + x2k−1 2
= (−1)k ·
rk3
rk2
k
= (−1)k+1
k−3
xi xk rk−1
.
rkk
(10.143)
Similarly, if i = k, then we have
(Dk fk )(x) = (−1)k
∂rk k−3
rk − xk ∂x
x2k 2
k
1
−
rk2
rk2
2
2
2
2 k−3
2
k rk − xk (rk − xk )
= (−1)
·
k−3
rk3
rk
k−1
r
.
= (−1)k k−1
rkk
(10.144)
By substituting the relations (10.143) and (10.144) into the definition (10.142) and then consider its dot product with x, we get
x · (∇fk )(x) =
=
k
X
xi (Di fk )(x)
i=1
k−1
X
i=1
(−1)k+1
k−3
x2i xk rk−1
x rk−1
k k k−1
+
(−1)
rkk
rkk
Chapter 10. Integration of Differential Forms
= (−1)k+1
312
k−1
x rk−1
xk rk−1
k k k−1
+
(−1)
rkk
rkk
= 0.
(10.145)
– Step 2: ωk = d(fk ωk−1 ) for k = 2, . . . , n. We notice from Theorem 10.20, part (a) and
Definition 10.17 that
d(fk ωk−1 ) = ( dfk ) ∧ ωk−1 + (−1)0 fk ∧ dωk−1
= ( dfk ) ∧ ωk−1
#
" k
X
1
= k−1
(Di fk )(x) dxi
rk−1 i=1
#
" k−1
X
(−1)i−1 xi dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk−1
∧
i=1
=
1 h
k−1
rk−1
(−1)1−1 x1 (D1 fk )(x) dx1 ∧ dx2 ∧ · · · ∧ dxk−1
{z
}
|
There are (k − 2) terms.
+(−1)2−1 x2 (D2 fk )(x) dx2 ∧ dx1 ∧ dx3 ∧ · · · ∧ dxk−1
|
{z
}
There are (k − 2) terms.
k−2
+ · · · + (−1)
+
(
xk−1 (Dk−1 fk )(x) dxk−1 ∧ dx1 ∧ · · · ∧ dxk−2
{z
}
|
There are (k − 2) terms.
1
k−1
rk−1
i
(Dk fk )(x) dxk
#)
" k−1
X
i−1
.
(−1) xi dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk−1
∧
{z
}
|
i=1
(10.146)
There are (k − 2) terms.
By applying the anticommutative relation (k − 2)-times to the red brackets in the expression
(10.146) and then the formula (10.144), it becomes
X
rk−1 k−1
1
(−1)i−1 xi dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk
(−1)2k−2 k−1 · k−1
k
rk i=1
rk−1
k−1
=
1 X
(−1)i−1 xi dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk .
rkk i=1
(10.147)
Now it remains to simplify the blue brackets in the expression (10.146). In fact, by the equation
(10.145), the blue brackets are equivalent to
xk
xk
− k−1 (Dk fk )(x) dx1 ∧ · · · ∧ dxk−1 = (−1)k+1 k dx1 ∧ · · · ∧ dxk−1 .
rk
rk−1
(10.148)
Hence we follow from substituting the expressions (10.147) and (10.148) back into the expression (10.146) that
xk
d(fk ωk−1 ) = (−1)k+1 k dx1 ∧ · · · ∧ dxk−1
rk
k−1
+
1 X
(−1)i−1 xi dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk
rkk i=1
k
=
1 X
(−1)i−1 xi dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxk
rkk i=1
313
10.5. Problems on closed forms and exact forms
= ωk
which is our desired property.
(c) We know from Problems 10.22 and 10.23 that ω2 = η and ω3 = ζ. By Examples 10.36 and 10.37,
we know that
Z
Z
ζ = 4π 6= 0,
η = 2π 6= 0 and
Σ
γ
where γ and Σ are parametrizations of the unit circle and the unit sphere in R2 and R3 respectively.
Furthermore, we conclude from the discussion parts of Examples 10.36 and 10.37 and the facts
∂γ = ∂Σ = 0 that
η and ζ
are not exact in R2 \ {0} and R3 \ {0} respectively. Thus it is reasonable to say that the answer in
this part is negative.
Let n ≥ 2. We basically follow a part of the argument as in [9, §6.1]. Consider the (n − 1)-sphere
S n−1 of radius 1 in Rn defined by
S n−1 = {x ∈ Rn | kxk = 1}.
Let ω = rnn ωn . Then we have ω = ωn on S n−1 so that
ω(S n−1 ) = ωn (S n−1 )
by Definition 10.11, i.e.,
Z
ωn =
S n−1
Z
ω.
S n−1
Since ωn is, by definition, a (n − 1)-form defined in En , ω is also a (n − 1)-form in En . We know
that
S n−1 = ∂Dn ,
where Dn = {x ∈ Rn | kxk ≤ 1} is the closed unit ball in Rn (see, for example, [11, p. 253]). Since
S n−1 can be treated as an k-simplex in Rn of class C ′′ in En , we apply Theorem 10.33 (Stokes’
Theorem) to S n−1 and ω to get
Z
Z
Z
Z
dω.
(10.149)
ω=
ω=
ωn =
S n−1
S n−1
∂Dn
Dn
Lemma 10.13
We have
dω = n dx1 ∧ · · · ∧ dxn .
Proof of Lemma 10.13. By Definition 10.18, we have
n
X
(−1)i−1 xi dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxn
dω = d
i=1
=
n
X
i=1
=
n
X
i=1
(−1)i−1 dxi ∧ dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxn
(−1)i−1 (−1)i−1 dx1 ∧ · · · ∧ dxi−1 ∧ dxi ∧ dxi+1 ∧ · · · ∧ dxn
= n dx1 ∧ · · · ∧ dxn
which is the required result.
Chapter 10. Integration of Differential Forms
314
Now we apply Lemma 10.13 to the relation (10.149) and then using Definition 10.44 (Volume
elements), we obtain
Z
S n−1
ω=n
Z
n
Dn
dx1 ∧ · · · ∧ dxn = Vol (Dn ) =
π2
6= 0,
n
Γ( 2 + 1)
(10.150)
where Γ is the Gamma function.
For the second assertion, assume that λ was a (n − 2)-form defined in En ⊆ Rn such that
ωn = dλ.
(10.151)
By Problem 10.16 or Remarks 10.35(c), we know that
∂S n−1 = ∂ 2 Dn = 0.
Therefore, we follow from this fact, Theorem 10.33 (Stokes’ Theorem) and the exactness (10.151)
that
Z
Z
Z
Z
∂S n−1
S n−1
λ=0
λ=
dλ =
ωn =
S n−1
∂ 2 Dn
which contradicts the result (10.150). Hence ωn is not exact in En .v
(d) We have the following generalizations:
– A generalization of Problem 10.22(c). Suppose that g ∈ C ′′ ([0, 1]), hi ∈ C ′′ ([0, 1]n−2 )
and g > 0. Suppose further that
Φ(s1 , . . . , sn−2 , t) = (x1 , . . . , xn )
is a (n − 1)-surface with parameter domain I n−1 given by
xi = g(t)hi (s1 , . . . , sn−2 ) (1 ≤ i ≤ n).
(10.152)
Then we have
Z
ωn = 0.
(10.153)
Φ
To prove the formula (10.153), we need to compute the Jacobians

∂x1
∂s1




..


.

 ∂xi−1

 ∂s1

∂(x1 , . . . , xi−1 , xi+1 , . . . , xn )

= det  ∂x
i+1

∂(s1 , . . . , sn−2 , t)

 ∂s1



..

.



 ∂xn
∂s1
···
∂x1
∂sn−2
···
..
.
∂xi−1
∂sn−2
···
∂xi+1
∂sn−2
..
.
..
.
···
∂xn
∂sn−2
..
.
∂x1
∂t





..


.

∂xi−1 

∂t 


.
∂xi+1 


∂t 



..

.



∂xn 
∂t
v One can prove the same result by generalizing the argument used in part (b), but the steps are very cumbersome.
315
10.5. Problems on closed forms and exact forms
By the definition (10.152), we have

g(t)
∂h1
∂s1
· · · g(t)
∂h1
∂sn−2
g ′ (t)h1









..
..
..
..


.


.
.
.






∂hi−1
∂hi−1


· · · g(t)
g ′ (t)hi−1 
 g(t)


∂s
∂s
1
n−2
∂(x1 , . . . , xi−1 , xi+1 , . . . , xn )


= det 



∂(s1 , . . . , sn−2 , t)

 g(t) ∂hi+1 · · · g(t) ∂hi+1 g ′ (t)h

i+1 
∂s1
∂sn−2








.
.
.
..


..
.
.
.
.
.








∂hn
∂hn
′
· · · g(t)
g (t)hn
g(t)
∂s1
∂sn−2


∂h1
∂h1
···
h1
 ∂s1

∂sn−2





..
..
.. 
..


.

.
.
. 






∂hi−1
 ∂hi−1

···
hi−1 

 ∂s1

∂sn−2


= g n−2 (t)g ′ (t) det 

 ∂h

∂hi+1
i+1

···
hi+1 


∂sn−2
 ∂s1







.
.
.
..

..
.
.
.
.
. 






 ∂hn

∂hn
···
hn
∂s1
∂sn−2
so that
Z
Φ
ωn =
Z
n
X
g ′ (t)
(−1)i−1 hi
n
n2
X
I n−1
i=1
h2j
g(t)

j=1
∂h1
∂s1




..


.



 ∂hi−1

 ∂s1

× det 
 ∂h
i+1


∂s

1



..

.



 ∂hn
∂s1
···
∂h1
∂sn−2
..
.
..
.
···
∂hi−1
∂sn−2
···
∂hi+1
∂sn−2
..
.
..
.
···
∂hn
∂sn−2
h1




.. 

. 




hi−1 


 ds1 · · · dsn−2 dt.

hi+1 




.. 

. 



hn
Chapter 10. Integration of Differential Forms
316
Since the summation in the above integral is just the expansion of the following determinant
along the first column


∂h1
∂h1
···
h1
h1


∂s1
∂sn−2




 .

.
.
.
.

..
..
..
.. 
det  ..







∂hn
∂hn
···
hn
hn
∂s1
∂sn−2
and Theorem 9.34(d) implies that it is actually 0. Hence we have
Z
Z
g ′ (t)
ωn =
n
n × (0) ds1 · · · dsn−2 dt = 0
X
Φ
I n−1
2 2
hj
g(t)
j=1
which proves the formula (10.153).
– A generalization of Problem 10.22(d). Our first step is to construct a (n − 1)-surface in
En whose role is similar to that of Σ in Example 10.32. Let Dn be the (n − 1)-cell
[0, π]n−2 × [0, 2π].
Suppose that Σn−1 : Dn → En ⊆ Rn \ {0} is the (n − 1)-surface defined by
Σn−1 (φ1 , . . . , φn−1 ) = (x1 , . . . , xn−1 , xn ),
where
x1 = cos φ1 ,
x2 = sin φ1 cos φ2 ,
x3 = sin φ1 sin φ2 cos φ3 ,
..
.
(10.154)
xn−2 = sin φ1 · · · sin φn−3 cos φn−2 ,
xn−1 = sin φ1 · · · sin φn−2 cos φn−1 ,
xn = sin φ1 · · · sin φn−2 sin φn−1
and 0 ≤ φ1 , . . . , φn−2 ≤ π, 0 ≤ φn−1 ≤ 2π. (See [4] for further details of the derivation of
Σn−1 .) By direct computation, we know from the definitions (10.154) that
x21 + x22 + · · · + x2n = 1.
Thus the range of Σn−1 is the (n − 1)-sphere S n−1 , i.e.,
Σn−1 (Dn ) = S n−1 .
(10.155)
Next, suppose E is a closed rectangle in Dn with edges parallel to those of Dn . In other words,
we have
E = [a1 , b1 ] × · · · × [an−1 , bn−1 ],
where ai and bi are some constants with 0 < ai < bi < π for 1 ≤ i ≤ n − 2 and 0 < an−1 <
bn−1 < 2π. Let f ∈ C ′′ (Dn ) and f > 0 on Dn . Let, further, that Ω be the (n − 1)-surface
with parameter domain E, defined by
Ω(φ1 , . . . , φn−1 ) = f (φ1 , . . . , φn−1 )Σn−1 (φ1 , . . . , φn−1 ).
Now we want to prove
317
10.5. Problems on closed forms and exact forms
Lemma 10.14
We have
Z
ωn =
Ω
Z
ωn = An−1 (S),
S
n−1
where S and An−1 (S) denote the restriction ΣE
and the “area” of S.
Since the proof of Lemma 10.14 is quite lengthy, we present its proof in Appendix A.
– A special case of Lemma 10.14. We claim that
Lemma 10.15
For n ≥ 2, we have
Z
n
ωn =
Σn−1
where
n
2π 2
Γ( n
2)
2π 2
,
Γ( n2 )
is the surface area of the (n − 1)-sphere of radius 1, see [4, p. 66].
Proof of Lemma 10.15. The case for n = 2 is done in Example 10.36. So we prove the case for
n ≥ 3 by induction. By the formula (A.16), we have
Z
ωn = An−1 (Σn−1 )
Σn−1
=
Z π
0
···
Z π Z 2π
0
0
sin φn−2 sin2 φn−3 · · · sinn−2 φ1 dφ1 · · · dφn−1
= (2π)I1 × · · · × In−2 ,
where
In−2 =
(10.156)
Z π
sinn−2 x dx.
0
When n = 3, we obtain from the formula (A.9) that
Z π
Z
sin x dx = 4π.
ω3 = 2πI1 = 2π
Σ2
0
√
By Theorem 8.18(a) and the fact that Γ( 12 ) = π, we have
3
3
3
2π 2
4π 2
2π 2
√ = 4π.
1 =
3 = 1
π
Γ( 2 )
2 Γ( 2 )
Thus the statement is true for n = 3. Assume that it is also true for n = k ≥ 3, i.e.,
Z
k
2π 2
ωk =
.
Γ( k2 )
Σk−1
For n = k + 1, we follow from the formula (A.9), the assumption and the properties of Γ(x)
that
Z
ωk+1 = [(2π)I1 × · · · × Ik−2 ] × Ik−1
Σk
Z
=
ωk × Ik−1
Σk−1
k
=
2π 2
× Ik−1
Γ( k2 )
Chapter 10. Integration of Differential Forms
318

2π m


if k = 2m;
 Γ(m) × I2m−1 ,
=
m+ 1
 2π 2

× I2m , if k = 2m + 1,

Γ(m + 21 )
√

2π m
2(m − 1)! π


,
×
 Γ(m)
Γ(m + 12 )
=
1
2m Γ(m + 21 )
π
2π m+ 2



√
× m ,
×
1
π
2 m!
Γ(m + 2 )

1
m+

2π 2


, if k = 2m;
Γ(m + 21 )
=

 2π m+1

,
if k = 2m + 1,
m!

1
 2π m+ 2

, if k = 2m;

Γ(m + 21 )
=

2π m+1


, if k = 2m + 1,
Γ(m + 1)
if k = 2m;
if k = 2m + 1,
k+1
2π 2
.
=
Γ( k+1
2 )
Hence the statement is still true for n = k + 1 and this finishes the proof of the lemma.
This completes the proof of the problem.
Problem 10.24
Rudin Chapter 10 Exercise 24.
Proof. Let x, y ∈ E and x 6= y. Since E is convex, the affine-oriented 2-simplexes σ = [p, x, y] is in E.
Furthermore, we know from Definition 10.30 that σ is of class C ′′ because the identity mapping is of class
C ′′ . By Theorem 10.33 (Stokes’ Theorem) and the fact that dω = 0, we have
Z
Z
ω.
(10.157)
dω =
0=
∂σ
σ
By Theorem 10.29, we have
∂σ = [x, y] − [p, y] + [p, x],
so the integral (10.157) implies that
Z
[x,y]
ω−
Z
ω+
Z
ω=0
Z
f (y) − f (x) =
[p,y]
[p,x]
ω.
(10.158)
[x,y]
By definition, [x, y] is the straight line segment in E joining the points x and y. Let γ : [0, 1] → E be
the 1-surface in E ⊆ Rn defined by
γ(t) = (1 − t)x + ty = ((1 − t)x1 + ty1 , . . . , (1 − t)xn + tyn ).
By Definition 10.11 (or Example 10.12(a)), we have
Z
ω=
[x,y]
=
Z 1X
n
0 i=1
n
XZ 1
i=1
0
ai (γ(t))
∂
((1 − t)xi + tyi ) dt
∂t
ai ((1 − t)x + ty)(yi − xi ) dt
319
10.5. Problems on closed forms and exact forms
Z 1
n
X
(yi − xi )
ai ((1 − t)x + ty) dt.
=
(10.159)
0
i=1
Combining the expressions (10.158) and (10.159), we have
f (y) − f (x) =
n
X
i=1
(yi − xi )
Z 1
0
ai ((1 − t)x + ty) dt.
(10.160)
Next, by [21, Eqn. (25), p. 215] and the expression (10.160), we have
f (x + sej ) − f (x)
s
Z
1 1
= lim (xj + s − xj )
aj ((1 − t)x + t(x + sej )) dt
s→0
s 0
Z 1
= lim
aj (x + tsej ) dt
(Dj f )(x) = lim
s→0
s→0
=
Z 1
0
aj (x) dt
0
= aj (x)
for j = 1, . . . , n. Hence if we define the real function f : E ⊆ Rn → R by
Z
ω,
f (x) =
[p,x]
then it is of class C ′ in E and
df =
n
X
(Di f )(x) dxi =
n
X
ai (x) dxi = ω
i=1
i=1
in E. This completes our proof of the problem.
Problem 10.25
Rudin Chapter 10 Exercise 25.
Proof. This problem relates the concepts “exactness” and “independence of path” of 1-forms. Suppose
that
X
ω=
ai (x) dxi .
We need some results from topology. Given a space X, define an equivalence relation (Definition 2.3)
on X by setting x ∼ y if there exists a connected subspace of X containing x and y. The equivalent
classes are called the connected component of X. Let’s recall some well-known results about connected
components of a space X ([18, Theorem 25.1, p. 159]):
Lemma 10.16
Let {Xα } be the collection of all connected components of X. Then we have
[
X=
Xα .
α
Furthermore, a space X is said to be locally connected at x if for every neighborhood U of x, there
exists a connected neighborhood V such that
x ∈ V ⊆ U.
Chapter 10. Integration of Differential Forms
320
If X is locally connected at each x, then X is called a locally connected space. We have the following
result about locally connected spaces ([18, Theorem 25.3, p. 161]):
Lemma 10.17
If X is a locally connected space, then each component of an open set U of X is also open
in X.
We know, by checking the definition of locally connectedness directly, that Rn is locally connected.
Since the set E in our question is supposed to be open in Rn , Lemmas 10.16 and 10.17 imply that
[
E=
Eα ,
α
where each Eα is an open (in Rn ) connected component of E. We claim that ω is exact in Eα . To this
end, fix pα ∈ Eα . Similar to Problem 10.24, we define a function fα : Eα → R by
Z
ω.
fα (x) =
[pα ,x]
For any x, y ∈ Eα , let γ = [x, y] − [pα , y] + [pα , x]. Then γ is a closed curve in Eα and the hypothesis
shows that
Z
Z
Z
Z
Z
0=
ω=
ω=
ω−
ω+
ω.
γ
[x,y]−[pα ,y]+[pα ,x]
[x,y]
Thus we have
fα (x) − fα (y) =
Z
[pα ,y]
[pα ,x]
ω
[x,y]
which is exactly the relation (10.158). By imitating the remaining part of the argument in Problem 10.24,
we conclude that
ω = dfα
in Eα . Finally, if we define the function f : E → R such that the restriction of f to Eα is fα , i.e.,
f |Eα = fα , then we have
ω = df
in E. This ends the proof of the problem.
Problem 10.26
Rudin Chapter 10 Exercise 26.
Proof. We follow the given hint. Let E = R3 \ {0}. Then E is obviously open in R3 . Define γ : [0, 1] → E
to be a closed curve in E, of class C ′ . Therefore, there is a 2-surface Φ : D → E such that
∂Φ = γ,
where D is a compact subset of R2 . Since ω is a 1-form in E of class C ′ and Φ is of class C ′′ in E, it
follows from Theorem 10.33 (Stokes’ Theorem) and then the hypothesis dω = 0 that
Z
Z
Z
ω=
ω=
dω = 0.
γ
∂Φ
Φ
By Problem 10.25, we establish that ω is exact in E = R3 \ {0}, finishing the proof of the problem.
Problem 10.27
Rudin Chapter 10 Exercise 27.
321
10.5. Problems on closed forms and exact forms
Proof. Let E = (p1 , q1 ) × (p2 , q2 ) × (p3 , q3 ) be the open 3-cell in R3 . By Theorem 10.20 and then [21,
Eqn. (59), p. 260], we have
dλ = d(g1 dx + g2 dy)
= ( dg1 ) ∧ dx + (−1)0 g1 ∧ d2 x + ( dg2 ) ∧ dy + (−1)0 g2 ∧ d2 y
∂g
∂g
∂g1
∂g1 ∂g2
∂g2 1
2
=
dx +
dy +
dz ∧ dx +
dx +
dy +
dz ∧ dy.
∂x
∂y
∂z
∂x
∂y
∂z
(10.161)
Since the anticommutative relation and dx ∧ dx = dy ∧ dy = 0 (see [21, Eqn. (42) & (43), p.256]), we
obtain from the expression (10.161) that
∂g1
∂g2
∂g1
∂g2
dy ∧ dx +
dx ∧ dy +
dz ∧ dx +
dz ∧ dy
∂y
∂x
∂z
∂z
∂g
∂g1
∂g1 ∂g2
2
dx ∧ dy +
=
−
dz ∧ dx +
dz ∧ dy.
∂x
∂y
∂z
∂z
dλ =
(10.162)
When x and y are fixed, since f2 ∈ C ′ (E), we have f2 ∈ C ′ ((p3 , q3 )). In particular, f2 ∈ R on [c, z]
and f2 is continuous at z, so Theorem 6.20 (First Fundamental Theorem of Calculus) implies that
Z z
Z y
∂g1
∂
∂
=
f2 (x, y, s) ds −
f3 (x, t, c) dt = f2 (x, y, z) − 0 = f2 (x, y, z).
(10.163)
∂z
∂z c
∂z b
Similarly, we have
∂g2
∂
=−
∂z
∂z
and
∂
∂y
−
Z z
Z y
c
f1 (x, y, s) ds = −f1 (x, y, z)
f3 (x, t, c) dt
b
!
(10.164)
= −f3 (x, y, c).
(10.165)
Finally, we have to evaluate
∂
∂x
−
Z z
c
f1 (x, y, s) ds
!
and
∂
∂y
Z z
!
f2 (x, y, s) ds .
c
To do this, we need Theorem 9.42, so we have to check its hypotheses. For the first integral, we fix y and
let F1 (x, s) = f1 (x, y, s).
• Since F1 (x, s) is defined for a ≤ x ≤ r1 , c ≤ s ≤ r3 for some r1 < q1 and r3 < q3 . (This is the
hypothesis (a) in Theorem 9.42.)
• For every (fixed) x ∈ [a, r1 ], the condition f1 ∈ C ′ (E) implies that F1 (x, s) ∈ R on [c, r3 ]. (This is
the hypothesis (c) in Theorem 9.42.)
1
• Since f1 ∈ C ′ (E), ∂F
∂s is a uniformly continuous function on [a, r1 ] × [c, r3 ]. By Definition 4.18, for
every ǫ > 0, there exists a δ > 0 such that
∂
∂
<ǫ
F1 (x, s)
−
F1 (x, s)
∂x
∂x
(p,q)
(u,v)
p
for all (p, q), (u, v) ∈ [a, r1 ] × [c, r3 ] with (p − u)2 + (q − v)2 < δ. In particular, if we fix u such
that a < u < r1 and put q = v = s, then the last inequality gives
∂
∂
<ǫ
F1 (x, s)
F1 (x, s)
−
∂x
∂x
x=p
x=u
for all s ∈ [c, r3 ] with |p − u| < δ. This is exactly the hypothesis (d) in Theorem 9.42.
Chapter 10. Integration of Differential Forms
322
Hence, by Theorem 9.42, we have
∂
∂x
−
c
!
Z z
!
Z z
f1 (x, y, s) ds
=−
Z z
c
∂
f1 (x, y, s) ds.
∂x
(10.166)
Similarly, we have
∂
∂y
f2 (x, y, s) ds
c
=
Z z
c
∂
f2 (x, y, s) ds.
∂y
(10.167)
Therefore, we follow from the partial derivatives (10.165) - (10.167) that
Z zh
i
∂g2
∂
∂g1
∂
−
= f3 (x, y, c) −
f1 (x, y, s) +
f2 (x, y, s) ds.
∂x
∂y
∂x
∂y
c
(10.168)
Now we derive from Theorem 10.20 and [21, Eqn. (45), p. 257] that
dω = d(f1 dy ∧ dz) + d(f2 dz ∧ dx) + d(f3 dx ∧ dy)
= df1 ∧ dy ∧ dz + f1 d( dy ∧ dz) + df2 ∧ dz ∧ dx + f2 d( dz ∧ dx)
{z
}
{z
}
|
|
It is 0.
It is 0.
+ df3 ∧ dx ∧ dy + f3 d( dx ∧ dy)
|
{z
}
It is 0.
∂f1
∂f2
∂f3
=
dx ∧ dy ∧ dz +
dy ∧ dz ∧ dx +
dz ∧ dx ∧ dy
∂x
∂y
∂z
∂f
∂f2
∂f3
1
dx ∧ dy ∧ dz.
=
+
+
∂x
∂y
∂z
(10.169)
By the assumption dω = 0, we obtain from the relation (10.169) that
∂f1 ∂f2
∂f3
=−
−
∂z
∂x
∂y
so that the relation (10.168) implies that
Z z
∂f3
∂g2 ∂g1
−
= f3 (x, y, c) +
ds
∂x
∂y
∂s
c
= f3 (x, y, c) + f3 (x, y, z) − f3 (x, y, c)
= f3 (x, y, z).
(10.170)
By substituting the relations (10.163), (10.164) and (10.170) into the relation (10.162), we reach the
desired result that
dλ = ω
in E. This proves our first assertion.
For the second assertion, if ω = ζ and E = R3 \ {0}, then, by Problem 10.22, we have
f1 (x, y, z) =
x
3
(x2 + y 2 + z 2 ) 2
,
f2 (x, y, z) =
To evaluate the integral
Z z
c
we apply the substitution s =
Z z
c
y
(x2 + y 2 + s2 )
y
and f3 (x, y, z) =
3
(x2 + y 2 + z 2 ) 2
y
3
(x2 + y 2 + s2 ) 2
ds,
3
2
ds = y
x2 +y2
arctan √ 2z
x +y2
3
.
(10.171)
p
x2 + y 2 tan θ, so the integral (10.171) becomes
Z arctan √ z
z
(x2 + y 2 + z 2 ) 2
1
(x2 + y 2 ) 2 sec2 θ dθ
3
(x2 + y 2 ) 2 sec3 θ
323
10.5. Problems on closed forms and exact forms
=y
Z arctan √ z
arctan √ 2z
x +y2
=
dθ
x2 +y2
y
x2 + y 2
(x2 + y 2 ) sec θ
Z arctan √ z
x2 +y2
cos θ dθ
arctan √ 2z
x +y2
y
= 2
sin θ
x + y2
arctan √ 2z
x +y2
arctan √ 2z
x +y2
y
= 2
x + y2
c
− sin arctan p
x2 + y 2
#
"
y
c
z
= 2
−
.
1
x + y 2 (x2 + y 2 + z 2 ) 12
(x2 + y 2 + c2 ) 2
Similarly, we have
Z y
sin arctan p
z
x2 + y 2
!
(10.172)
(10.173)
b
#
"
c
b
y
−
3 dt =
1
x2 + c2 (x2 + y 2 + c2 ) 12
(x2 + t2 + c2 ) 2
(x2 + b2 + c2 ) 2
Z z
"
#
z
x
c
−
.
3 ds =
1
x2 + y 2 (x2 + y 2 + z 2 ) 21
(x2 + y 2 + s2 ) 2
(x2 + y 2 + c2 ) 2
(10.174)
and
c
c
x
Now we put the integrals (10.172) - (10.174) into the definitions of g1 and g2 to acquire
"
"
#
#
z
y
c
c
b
y
−
− 2
−
g1 (x, y, z) = 2
1
1
x + y2 r
x + c2 (x2 + y 2 + c2 ) 12
(x2 + y 2 + c2 ) 2
(x2 + b2 + c2 ) 2
and
#
"
x
z
c
g2 (x, y, z) = − 2
−
1
x + y2 r
(x2 + y 2 + c2 ) 2
so that
λ = g1 dx + g2 dy
#
#
"
"
c
b
y
z
y
c
dx − 2
−
dx
= 2
−
1
1
x + y2 r
x + c2 (x2 + y 2 + c2 ) 12
(x2 + y 2 + c2 ) 2
(x2 + b2 + c2 ) 2
"
#
z
c
x
−
− 2
dy
1
x + y2 r
(x2 + y 2 + c2 ) 2
#
"
x dy − y dx c
z
−
=−
1
r
x2 + y 2
(x2 + y 2 + c2 ) 2
"
#
y
c
b
− 2
−
dx
1
x + c2 (x2 + y 2 + c2 ) 12
(x2 + b2 + c2 ) 2
#
#
"
"
b
y
c
z
c
η− 2
−
dx.
(10.175)
−
=−
1
1
r
x + c2 (x2 + y 2 + c2 ) 12
(x2 + y 2 + c2 ) 2
(x2 + b2 + c2 ) 2
Recall that (a, b, c) is a fixed point in E = R3 \ {0}, so we put b = c = 0 into the expression (10.175) to
get
z
λ=− η
r
which is exactly the form that occurs in part (e) of Problem 10.22. This completes the proof of the
problem.
Chapter 10. Integration of Differential Forms
324
Problem 10.28
Rudin Chapter 10 Exercise 28.
Proof. We have Φ : [a, b] × [0, 2π] → R2 . Since ω = x3 dy, we follow from Theorem 10.20 and then [21,
Eqn. (59), p. 260] that dω = d(x3 dy) = ( dx3 ) ∧ dy − x3 d2 y = 3x2 dx ∧ dy. Therefore, it follows from
Definition 10.11 that
Z
Z
∂(x, y)
3(r cos θ)2
dω =
dr dθ
∂(r, θ)
[a,b]×[0,2π]
Φ
Z b Z 2π
cos θ −r sin θ
=
3r2 cos2 θ
dr dθ
sin θ r cos θ
a
0
Z b Z 2π
=
3r3 cos2 θ dr dθ
0
a
3π 4
(b − a4 ).
=
4
By the hint in Problem 10.21(b), we know that ∂Φ = Γ − γ, where Γ : [0, 2π] → R2 \ {0} and
γ : [0, 2π] → R2 \ {0} are given by Γ(t) = (b cos θ, b sin θ) and γ(t) = (a cos θ, a sin θ). By Definition 10.11
again, we have
Z
Z
Z
ω=
ω− ω
Γ
∂Φ
=
Z 2π
γ
(b cos θ)3
0
= (b4 − a4 )
= (b4 − a4 )
Z 2π
∂(b sin θ)
dθ −
∂θ
Z 2π
(a cos θ)3
0
∂(a sin θ)
dθ
∂θ
cos4 θ dθ
0
Z 2π
0
cos 4θ + 4 cos 2θ + 3
dθ
8
3π 4
(b − a4 ).
=
4
Hence we see that the two integrals are equal and this finishes the proof of the problem.
Problem 10.29
Rudin Chapter 10 Exercise 29.
Proof. Let’s recall the hypotheses of the problem first: Suppose that E is a convex open set in Rn ,
f ∈ C ′ (E), p is an integer such that 1 ≤ p ≤ n, and
(Dj f )(x) = 0
b ∈ E).
(p < j ≤ n, x
(10.176)
b = (x, xp , x′ ), where x = (x1 , . . . , xp−1 ) and x′ = (xp+1 , . . . , xn ). Furthermore, we define
Denote x
V = {(x, xp ) ∈ Rp | (x, xp , x′ ) ∈ E for some x′ }.
(10.177)
Now V is a convex open subset in Rp . If we define U to be the projection of V onto its first (p − 1)coordinates, i.e.,
U = {x = (x1 , . . . , xp−1 ) ∈ Rp−1 | y = (x, xp ) ∈ V for some xp ∈ R},
(10.178)
then U is also a convex open subset in Rp−1 .
Claim 1: graph (α) ⊆ V . We show that there is a function α ∈ C ′ (U ) such that its graph lies in V ,
i.e.,
graph (α) ⊆ V.
(10.179)
325
10.5. Problems on closed forms and exact forms
• Case (i): E = Bn (a, r) for some a = (a1 , . . . , an ) and r > 0. For any (x1 , . . . , xp ) satisfying
(x1 − a1 )2 + · · · + (xp − ap )2 < r2 ,
we may choose (xp+1 , . . . , xn ) with the property that
(xp+1 − ap+1 )2 + · · · + (xn − an )2 <
1 2
{r − [(x1 − a1 )2 + · · · + (xp − ap )2 ]}
2
and this implies that
(x1 − a1 )2 + · · · + (xn − an )2 < r2 .
By the definition (10.177), we have
V = {(x1 , . . . , xp−1 , xp ) ∈ Rp | (x1 − a1 )2 + · · · + (xp − ap )2 < r2 } = Bp (a′ , r)
(10.180)
which is the ball centered at the fixed point a′ = (a1 , . . . , ap ) and of radius r. If we put xp = ap ,
then we get from the set (10.180) and the definition (10.178) that
U = {(x1 , . . . , xp−1 ) ∈ Rp | (x1 − a1 )2 + · · · + (xp−1 − ap−1 )2 < r2 }
We define α : U → R to be α(x) = ap for all x ∈ U . It is clear that α ∈ C ′ (U ) and we have
graph (α) = {(x, α(x)) | x ∈ U }
= {(x, ap ) | x ∈ U }
= {(x1 , . . . , xp−1 , ap ) | (x1 − a1 )2 + · · · + (xp−1 − ap−1 )2 < r2 }
⊂ V.
To have a better understanding between V and its projection U , we consider the example that
n ≥ 4, p = 3, a = 0 and r = 1, then the sets E, V and U are the unit balls in R4 , R3 and R2
respectively, see Figure 10.16 below:
Figure 10.16: The unit disk U as the projection of the unit ball V .
Notice that x21 + x22 + x23 = 1 and the point (x1 , x2 , a3 ) on the section x3 = a3 is projected onto the
point (x1 , x2 , 0).
Chapter 10. Integration of Differential Forms
326
• Case (ii): E = (a1 , b1 ) × · · · × (an , bn ). Then we follow from the definitions (10.177) and (10.178)
that
V = {(x1 , . . . , xp−1 , xp ) ∈ Rp | (x1 , . . . , xp−1 , xp , xp+1 , . . . , xn ) ∈ E for some (xp+1 , . . . , xn )}
= {(x1 , . . . , xp−1 , xp ) ∈ Rp | xi ∈ (ai , bi ), where 1 ≤ i ≤ p}
= (a1 , b1 ) × · · · × (ap , bp )
and
U = (a1 , b1 ) × · · · × (ap−1 , bp−1 ).
In this case, we define α : U → R by
α(x) = c
for some c ∈ (ap , bp ) and for all x ∈ U . The definition shows clearly that α ∈ C ′ (U ) and
graph (α) = {(x, α(x)) | x ∈ U }
= {(x, c) | x ∈ U }
= {(x1 , . . . , xp−1 , c) | xi ∈ (ai , bi ), where 1 ≤ i ≤ p − 1}
= (a1 , b1 ) × · · · × (ap−1 , bp−1 ) × {c}
⊂ V.
For instance, we take n ≥ 4, p = 3 and E = (0, 1) × (0, 1) × (0, 1) × (0, 1). Then we have
V = (0, 1) × (0, 1) × (0, 1) and U = (0, 1) × (0, 1),
see Figure 10.17 below:
Figure 10.17: The open cells U and V .
327
10.5. Problems on closed forms and exact forms
• Case (iii): E is an arbitrary open convex set in Rn . Since the set E is not compact, we can’t
apply Problem 10.6 directly to show our result. In fact, we need a variation of partitions of unity
which is suitable for arbitrary sets in Rn and the following result (see [24, pp. 63, 64]) serves this
purpose:
Lemma 10.18
Let A ⊆ Rn and let O be an open cover of A. Then there is a collection Φ = {ϕµ } of
infinitely differentiable functions ϕµ : W → R, where W is an open set containing A, with
the following properties:
(a) For each x ∈ A, we have 0 ≤ ϕµ (x) ≤ 1.
(b) For each x ∈ A, there is an open set V containing x (V may not be an element in
the open cover O) such that all but finitely many ϕµ ∈ Φ are 0 on V .
(c) For each x ∈ A, we have
X
ϕµ (x) = 1.
ϕµ ∈Φ
(d) For each ϕµ ∈ Φ, there is an open set U ∈ O such that ϕµ = 0 outside of some closed
set contained in U .
Now we are ready for the construction:
– Step 1: Construction of an open cover of U . Since V is open, for every x ∈ U , we may
take y ∈ V and a ǫ(y) > 0 such that the projection of y onto its first (p − 1)-coordinates is x
(i.e., y = (x, ap ) for some ap ∈ R) and Bp (y, ǫ(y)) ⊆ V . Now we see that
U=
[
Bp−1 (x, ǫ(y)).
x∈U
In other words, O = {Bp−1 (x, ǫ(y))}x∈U is an open cover of U .
– Step 2: Construction of the required function α. By Lemma 10.18, we can find a
collection Φ = {ϕµ } of C ∞ functions ϕµ : U → R (assume W = U for simplicity) subordinate
to the cover O, i.e., each ϕµ satisfies the condition
supp (ϕµ ) ⊆ Uµ = Bp−1 (xµ , ǫ(yµ ))
for some xµ ∈ U , where yµ = (xµ , aµp ) for some aµp ∈ R.
Lemma 10.19
Suppose that
Vµ = Bp (yµ , ǫ(yµ ))
and Uµ = Bp−1 (xµ , ǫ(yµ )).
Then the projection of Vµ is Uµ .
Proof of Lemma 10.19. Recall from the definition (10.178) that
Uµ = Bp−1 (xµ , ǫ(yµ )) = {x ∈ Rp−1 | |x − xµ | < ǫ(yµ )}.
By using the analysis as in proving the set relation (10.180) in Case (i), we know that the
projection of the ball Vµ = Bp (yµ , ǫ(yµ )) is exactly Uµ = Bp−1 (xµ , ǫ(yµ )), see Figure 10.16 for
an example in a lower dimension. This completes the proof of the lemma.
Chapter 10. Integration of Differential Forms
328
We return to the proof of Problem 10.29. Suppose that the constant aµp ∈ R is chosen to be
fixed. Then we have
y = (x, aµp ) ∈ Vµ
for every x ∈ Uµ .
(10.181)
In fact, for x ∈ Uµ , we have |x − xµ | < ǫ(y) so that Lemma 10.19 shows that
|y − yµ | = |(x, aµp ) − (xµ , aµp )| = |x − xµ | < ǫ(yµ ).
This proves the relation (10.181).
We define the function α : U → R by
α(x) =
X
ϕµ (x)aµp
µ
for every x ∈ U . Now, for each x ∈ U , Lemma 10.18(b) implies that there is an open ball
Bp−1 (x, δ) ⊆ U
such that all but finitely many ϕµ ∈ Φ are 0 on Bp−1 (x, δ), where δ > 0. Let m be a finite
positive integer such that ϕµj (x) 6= 0 on Bp−1 (x, δ) for 1 ≤ j ≤ m. Thus we have
α(x) =
X
ϕµ (x)aµp =
µ
m
X
ϕµj (x)aµj p .
(10.182)
j=1
Since ϕµj ∈ C ∞ (U ) for every 1 ≤ j ≤ m, each ϕµj is continuously differentiable at x and we
conclude from the expression (10.182) that α is continuously differentiable at x. Since x is
arbitrary, we have α ∈ C ′ (U ) (in fact, C ∞ (U )). Finally, by the convexity of the set V and
Lemma 10.19(c), we have
(x, α(x)) =
m
X
j=1
|
ϕµj (x)x,
{z
}
m
X
j=1
m
X
ϕµj (x)(x, aµj p ) ∈ V.
ϕµj (x)aµj p =
j=1
Its value is 1.
Hence we have graph (α) ⊆ V in this case. This completes the proof of Claim 1.
Claim 2: F ∈ C ′ (E). When p = 1, we have
Z x1
ϕ(x) dx (x ∈ E).
F (x) =
(10.183)
c
Since V is convex in R, Problem 2.21(c) implies that it is connected and then it follows from Theorem
2.47 that V is an interval. Pick c ∈ V . We know that
ϕ : [c, x1 ] → R,
ϕ(x) = f (x, x2 , . . . , xn ) = f (x) for x ∈ [c, x1 ] and f ∈ C ′ (E),
so ϕ is continuous at x1 . By Theorem 6.20 (First Fundamental Theorem of Calculus), F is differentiable
at x1 and
Z x1
∂
(D1 F )(x) =
ϕ(x) dx = ϕ(x1 ) = f (x).
(10.184)
∂x1 c
Now the continuity of (D1 F )(x) follows from the relation (10.184). By the definition (10.183), we see
that
(Dj F )(x) = 0
for j = 2, . . . , n and for all x ∈ E. Hence we follow from Theorem 9.21 that F ∈ C ′ (E).
Next, we suppose that p > 1. By the definition of the real function α, we have
(x′ , α(x′ )) ∈ V.
329
10.5. Problems on closed forms and exact forms
Since (x′ , xp ) ∈ V and V is convex, we must have
(x′ , λα(x′ ) + (1 − λ)xp ) = λ(x′ , α(x′ )) + (1 − λ)(x′ , xp ) ∈ V
for every λ ∈ (0, 1). By this, instead of the definition (10.183), the integral
Z xp
ϕ(x′ , x) dx (x ∈ E)
F (x) =
(10.185)
α(x′ )
is well-defined. Similarly, (Dj F )(x) = 0 for j = p + 1, . . . , n and for all x ∈ E. If α(x′ ) > xp , then we
can rewrite the integral (10.185) as
Z α(x′ )
ϕ(x′ , x) dx (x ∈ E).
F (x) = −
xp
Thus, without loss of generality, we may assume that α(x′ ) < xp .
We recall from the proof of Theorem 10.38 that since E is convex and (Dj f )(x) = 0 for all p < j ≤ n
and x ∈ E, f (x) does not depend on x′′ = (xp+1 , . . . , xn ). Thus the function ϕ : V → R is given by
ϕ(x′ , xp ) = f (x) = f (x1 , . . . , xp , xp+1 , . . . , xn ).
{z
}
|
They can be taken
any values.
For 1 ≤ i ≤ p − 1, we let xi vary and keep other components of x′ fix so that ϕ can be thought as a
function of the two variables xi and xp . In this case, we define ψ : [α(x′ ), xp ] × [ai , bi ] → R to be given by
ψ(x, xi ) = ϕ(x′ , x) = f (x),
(10.186)
where x = (x1 , . . . , xp−1 , x, xp+1 , . . . , xn ) ∈ E.
Lemma 10.20
The function ψ defined in (10.186) satisfies the hypotheses of Theorem 9.42.
Proof of Lemma 10.20. By the definition of ϕ, we yield from the relation (10.186) that ψ is
well-defined on [α(x′ ), xp ] × [ai , bi ]. For each fixed xi in [ai , bi ], we have
ψ xi (x) = ψ(x, xi ) = f (x)
is a function of one variable. Since f ∈ C ′ (E), it is continuous on E and, in particular, on
[α(x′ ), xp ]. Thus, by Theorem 6.8, we have
ψ xi ∈ R
on [α(x′ ), xp ] for every xi ∈ [ai , bi ]. Furthermore, by Theorem 9.21, D2 f is continuous on E
and, in particular, on the compact set [α(x′ ), xp ] × [ai , bi ]. Hence, condition (d) of Theorem
9.42 is satisfied and this completes the proof of lemma.
Let’s return to the proof of the problem. Now we apply Theorem 9.42 to the function ψ to get
Z xp
Z xp
Z xp
∂
∂
∂
′
(Di F )(x) =
ψ(x, xi ) dx.
ϕ(x , x) dx =
ψ(x, xi ) dx =
(10.187)
∂xi α(x′ )
∂xi α(x′ )
α(x′ ) ∂xi
In other words, the integral (10.187) shows that (Di F )(x) exists for all x ∈ E and 1 ≤ i ≤ p − 1. Finally,
we are going to verify
(Dp F )(x) = f (x)
(10.188)
for all x ∈ E. Since ϕ(x′ , x) is continuous at xp , we deduce from Theorem 6.20 (First Fundamental
Theorem of Calculus) that, for every x ∈ E,
Z xp
∂
(Dp F )(x) =
ϕ(x′ , x) dx = ϕ(x′ , xp ) = f (x)
∂xp α(x′ )
which is exactly the relation (10.188). This proves Claim 2 and so completes the proof of the problem. Chapter 10. Integration of Differential Forms
10.6
330
Problems on vector fields
Problem 10.30
Rudin Chapter 10 Exercise 30.
Proof. Let A be the matrix in the question. As N = (α2 β3 − α3 β2 )e1 +(α3 β1 − α1 β3 )e2 +(α1 β2 − α2 β1 )e3 ,
we have
|N|2 = (α2 β3 − α3 β2 )2 + (α3 β1 − α1 β3 )2 + (α1 β2 − α2 β1 )2 .
(10.189)
On the other hand, direct computation shows that
det A = α1 [β2 (α1 β2 − α2 β1 ) − β3 (α3 β1 − α1 β3 )] − α2 [β1 (α1 β2 − α2 β1 ) − β3 (α2 β3 − α3 β2 )]
+ α3 [β1 (α3 β1 − α1 β3 ) − β2 (α2 β3 − α3 β2 )]
= (α1 β2 − α2 β1 )(α1 β2 − α2 β1 ) + (α3 β1 − α1 β3 )(α3 β1 − α1 β3 ) + (α2 β3 − α3 β2 )(α2 β3 − α3 β2 )
= (α1 β2 − α2 β1 )2 + (α3 β1 − α1 β3 )2 + (α2 β3 − α3 β2 )2 .
(10.190)
Thus the first assertion follows immediately from the expressions (10.189) and (10.190).
Since T (e1 ) = α1 e1 + α2 e2 + α3 e3 and T (e2 ) = β1 e1 + β2 e2 + β3 e3 , it yeilds
N · T (e1 ) = α1 α2 β3 − α1 α3 β2 + α2 α3 β1 − α2 α1 β3 + α3 α1 β2 − α3 α2 β1 = 0
and
N · T (e2 ) = β1 α2 β3 − β1 α3 β2 + β2 α3 β1 − β2 α1 β3 + β3 α1 β2 − β3 α2 β1 = 0.
Hence we have N · T (e1 ) = N · T (e2 ) = 0. We complete the proof of the problem.
Problem 10.31
Rudin Chapter 10 Exercise 31.
Proof.
(a) Let ∇h = (D1 h)e1 + (D2 h)e2 + (D3 h)e3 so that
F = g∇h = g(D1 h)e1 + g(D2 h)e2 + g(D3 h)e3 .
Thus the divergence of F is given by
∇ · F = D1 F1 + D2 F2 + D3 F3
= D1 (gD1 h) + D2 (gD2 h) + D3 (gD3 h)
= (D1 gD1 h + gD12 h) + (D2 gD2 h + gD22 h) + (D3 gD3 h + gD32 h)
= g(D12 h + D22 + D32 h) + (D1 gD1 h + D2 gD2 h + D3 gD3 h)
= g∇2 h + (∇g) · (∇h).
(b) As a remark, we call the first formula as the Green’s first identity and the second one as the Green’s
second identity. Besides, there is a third identity, namely the Green’s third identity, see [14, Eqn.
(5.198), p. 362].
Since F is a vector field of class C ′ in the open set E and Ω is a closed subset of E with positively
oriented boundary, Theorem 10.51 (Divergence Theorem) together with part (a) imply that
Z
Z
Z
Z
∂h
dA.
(10.191)
[g∇2 h + (∇g) · (∇h)] dV =
(F · n) dA =
[(g∇h) · n] dA =
g
Ω
∂Ω
∂Ω
∂Ω ∂n
331
10.6. Problems on vector fields
By interchanging the roles of g and h in the formula (10.191), we obtain
Z
Z
∂g
2
[h∇ g + (∇h) · (∇g)] dV =
h
dA.
∂n
Ω
∂Ω
Next, by subtracting the formula (10.192) from the formula (10.191), we yield
Z Z
∂h
∂g dA.
g
−h
(g∇2 h − h∇2 g) dV =
∂n
∂n
∂Ω
Ω
(10.192)
(10.193)
(c) Suppose that ∇2 h = 0. If g = 1, then it follows from the Green identity (10.193) that
Z
Z ∂(1) ∂h
−h
dA = [1 × ∇2 h − h∇2 (1)] dV
1×
∂n
∂n
∂Ω
Z
ZΩ
∂h
∇2 h dV
dA =
∂Ω ∂n
Ω
= 0.
To complete the last part, put g = h into the Green’s first identity (10.192), we have
Z
Z
∂h
2
2
dA
h
[h∇ h + k∇hk ] dV =
∂n
Ω
Z
Z∂Ω
Z
∂h
h∇2 h dV.
dA −
h
k∇hk2 dV =
∂n
Ω
∂Ω
Ω
Since ∇2 h = 0 in E and h = 0 on ∂Ω, we have
Z
k∇hk2 dV = 0
Ω
which implies that k∇2 hk = 0 on Ω. Thus we obtain that
(Dh1 )(x) = (D2 h)(x) = (D3 h)(x) = 0
(10.194)
for every x ∈ Ω.
so every component of the interior Ω◦ of Ω is also open in R3 by Lemma 10.17. By the relations
(10.194) and Theorem 9.17, we know that
h′ (x) = 0
on each connected component Ω◦α of Ω◦ . Then we follow from Problem 9.9 that
h(x) = cα
for every x ∈ Ω◦α for some constant cα . Since h ∈ C ′′ (E), the continuity of h and the hypothesis
that h = 0 on ∂Ω ensures that the constant cα must be 0. Hence we have
h(x) = 0
for every x ∈ Ω.
(d) Let E ⊆ R2 be open and Ω be a closed subset of E with positively oriented boundary ∂Ω in
R2 . Suppose that F : E → R3 is a vector field defined by F = αe1 + βe2 . We have to find a
two-dimensional version of the divergence theorem. In fact,
Lemma 10.21
We have
Z
∂Ω
F · n ds =
Z
Ω
∇ · F dA,
where n is the outward normal vector to ∂Ω and ds is the element of arc length. See [20,
pp. 422, 423].
Chapter 10. Integration of Differential Forms
332
Proof of Lemma 10.21. Now we apply Theorem 10.45 (Green’s Theorem) with α = −f2 , β =
f1 and F = f1 e1 + f2 e2 , we see that
Z
Z ∂f1
∂f2 ∇ · F dA =
+
dA
∂x
∂y
Ω
ZΩ
(−f2 dx + f1 dy)
=
∂Ω
Z
(f1 , f2 ) · ( dy, − dx)
=
Z∂Ω
F · n ds.
=
∂Ω
This completes the proof of Lemma 10.21.
Let f1 = gD1 h and f2 = gD2 h. Then F = g∇h and we have
∇ · F = D1 (gD1 h) + D2 (gD2 h)
= (D1 gD1 h + gD12 h) + (D2 gD2 h + gD22 h)
= g∇2 h + (∇g) · (∇h).
By Lemma 10.21, we have
Z
Z
[g∇2 h + (∇g) · (∇h)] dA =
Ω
∂Ω
g∇h · n ds =
Z
∂Ω
g
∂h
ds
∂n
(10.195)
which is the Green’s first identity in the plane. Similar as the analysis in part (b), the Green’s
second identity in the plane can be derived from the formula (10.195):
Z Z
∂h
∂g 2
2
g
(g∇ h − h∇ g) dA =
dA.
−h
∂n
∂n
∂Ω
Ω
We have completed the proof of the problem.
Problem 10.32
Rudin Chapter 10 Exercise 32.
Proof. We remark that for a detailed deduction of a parametrization a Möbius band, readers please refer
to [7, pp. 108-110]. Besides, we also emphasize again that −Γ3 (θ, t) means that reverse of the curve
Γ3 (θ, t), not the vector −Γ3 (θ, t), see the last two paragraphs in [21, pp. 268, 269].
Now we can start to prove the assertions. The figure of the parameter domain D is given in Figure
10.18:
Figure 10.18: The parameter domain D.
Furthermore, the figure of a Möbius band is shown in 10.19.w
w The figure is taken from http://brookstonbeerbulletin.com/mobius-beer/.
333
10.6. Problems on vector fields
Figure 10.19: The figure of the Möbius band.
We prove the assertions as follows:
• Computation of ∂Φ. By definition, D is a 2-cell with boundary γ1 + γ2 + γ3 + γ4 . By the analysis
on the bottom on [21, p. 271], we know that
∂Φ = Φ(γ1 ) + Φ(γ2 ) + Φ(γ3 ) + Φ(γ4 ) = Γ1 + Γ2 + Γ3 + Γ4 .
(10.196)
• Computation of Φ(p1 ), Φ(p2 ), Φ(p3 ) and Φ(p4 ). By direct computation, we have
Φ(p1 ) = Φ(0, −δ) = ((1 + δ sin 0) cos 0, (1 + δ sin 0) sin 0, −δ) = (1, 0, −δ) = a,
Φ(p2 ) = Φ(π, −δ) = ((1 + δ sin π) cos 2π, (1 + δ sin π) sin 2π, −δ cos π) = (1, 0, δ) = b,
Φ(p3 ) = Φ(π, δ) = ((1 − δ sin π) cos 2π, (1 − δ sin π) sin 2π, δ cos π) = (1, 0, −δ) = a,
Φ(p4 ) = Φ(0, δ) = ((1 − δ sin 0) cos 0, (1 − δ sin 0) sin 0, δ cos 0) = (1, 0, δ) = b.
• Description of ∂Φ. Since γ1 = [p1 , p2 ], we have γ1 = {(θ, −δ) | 0 ≤ θ ≤ π}. Therefore, if
(x, y, z) ∈ Γ1 , then we have
x = (1 + δ sin θ) cos 2θ, y = (1 + δ sin θ) sin 2θ, z = −δ cos θ
and so the projection of Γ1 , denoted by λ1 (θ, δ) whose domain is [0, π], into the (x, y)-plane is given
by
λ1 (θ, δ) = (1 + δ sin θ) cos 2θ + i(1 + δ sin θ) sin 2θ = (1 + δ sin θ)(cos 2θ + i sin 2θ).
Now λ1 is obviously a continuously differentiable closed curve in C, so Problem 8.23 implies that
Z π ′
1
λ1 (θ, δ)
Ind (λ1 ) =
dθ
2πi 0 λ1 (θ, δ)
Z π
1
2(1 + δ sin θ)(− sin 2θ + i cos 2θ) + δ cos θ(cos 2θ + i sin 2θ)
=
dθ
2πi 0
(1 + δ sin θ)(cos 2θ + i sin 2θ)
Z π
Z π
1
1
− sin 2θ + i cos 2θ
δ cos θ
=
dθ +
dθ
πi 0 cos 2θ + i sin 2θ
2πi 0 1 + δ sin θ
Z π
Z π
1
1
=
i dθ +
d(ln(1 + δ sin θ))
πi 0
2πi 0
π
= 1 + ln(1 + δ sin θ)
0
= 1.
Since γ2 = [p2 , p3 ], γ2 = {(π, t) | − δ ≤ t ≤ δ}. Thus we have
Γ2 = Φ(γ2 )
Chapter 10. Integration of Differential Forms
334
= Φ(π, t)
= {((1 − t sin π) cos 2π, (1 − t sin π) sin 2π, t cos π) | − δ ≤ t ≤ δ}
= {(1, 0, −t) | − δ ≤ t ≤ δ}
= [b, a].
For Γ3 , we note that γ3 = [p3 , p4 ], so
γ3 = {(θ, δ) | θ decreases from π to 0} = {(π − ϕ, δ) | ϕ ∈ [0, π]}.
Therefore, if (x, y, z) ∈ Γ3 , then we have
x = [1 − δ sin(π − ϕ)] cos 2(π − ϕ), y = [1 − δ sin(π − ϕ)] sin 2(π − ϕ), z = δ cos(π − ϕ)
and so the projection of Γ3 , denoted by λ3 (ϕ, δ) with domain [0, π], into the (x, y)-plane is given by
λ3 (ϕ, δ) = [1 − δ sin(π − ϕ)] cos 2(π − ϕ) + i[1 − δ sin(π − ϕ)] sin 2(π − ϕ)
= [1 − δ sin(π − ϕ)][cos 2(π − ϕ) + i sin 2(π − ϕ)]
= (1 − δ sin ϕ)(cos 2ϕ − i sin 2ϕ).
Now λ3 is obviously a continuously differentiable closed curve in C, so Problem 8.23 implies that
Z π ′
1
λ3 (ϕ, δ)
Ind (λ3 ) =
dϕ
2πi 0 λ3 (ϕ, δ)
Z π
2(1 − δ sin ϕ)(− sin 2ϕ − i cos 2ϕ) − δ cos ϕ(cos 2ϕ − i sin 2ϕ)
1
dϕ
=
2πi 0
(1 − δ sin ϕ)(cos 2ϕ − i sin 2ϕ)
Z π
Z π
1
1
− sin 2ϕ − i cos 2ϕ
δ cos ϕ
=
dϕ −
dϕ
πi 0 cos 2ϕ − i sin 2ϕ
2πi 0 1 − δ sin ϕ
Z π
Z π
1
1
(−i) dϕ +
d(ln(1 − δ sin ϕ))
=
πi 0
2πi 0
π
= −1 + ln(1 − δ sin ϕ)
0
= −1.
Since γ4 = [p4 , p1 ], γ4 = {(0, −t) | − δ ≤ t ≤ δ}. Thus we have
Γ4 = Φ(γ4 )
= Φ(0, −t)
= {((1 + t sin 0) cos 0, (1 + t sin 0) sin 0, −t cos 0) | − δ ≤ t ≤ δ}
= {(1, 0, −t) | − δ ≤ t ≤ δ}
= [b, a].
Hence we deduce from the above analysis and the expression (10.196) that
∂Φ = Γ1 + Γ3 + 2Γ2 .
• The “geometric” boundary of M . Recall that γ1 = {(θ, −δ) | 0 ≤ θ ≤ π}, so
Γ1 (0, −δ) = Φ(0, −δ) = (1, 0, −δ) = a and Γ1 (π, −δ) = Φ(π, −δ) = (1, 0, δ) = b.
By the previous analysis, we notice that Γ1 and Γ3 spiral up from a to b. Furthermore, we have
Γ1 (θ, −δ) = ((1 + δ sin θ) cos 2θ, (1 + δ sin θ) sin 2θ, −δ cos θ)
(10.197)
Γ3 (θ, δ) = ((1 − δ sin θ) cos 2θ, (1 − δ sin θ) sin 2θ, δ cos θ),
(10.198)
for θ ∈ [0, π] and
where θ decreases from π to 0. Therefore,
Γ = Γ1 − Γ3
is a curve that spirals up from a to b and then back to a, see Figure 10.20 below where the blue
and green loci refer to Γ1 and −Γ3 respectively.
335
10.6. Problems on vector fields
Figure 10.20: The “geometric” boundary of M .
• A representation of Γ. We claim that a parametrization of Γ : [0, 2π] → R3 \ {0} is given by
x = (1 + δ sin θ) cos 2θ,
y = (1 + δ sin θ) sin 2θ,
(10.199)
z = −δ cos θ.
In fact, when θ ∈ [0, π], the parametrization (10.199) is exactly the trace of Γ1 (see the parametrization (10.197)). To see why the loci of the parametrization (10.199) for θ ∈ [π, 2π] is the “reverse”
of Γ3 , we put θ = π + ϕ into the parametrization (10.199) to get
x = (1 − δ sin ϕ) cos 2ϕ,
y = (1 − δ sin ϕ) sin 2ϕ,
z = δ cos ϕ,
(10.200)
where ϕ increases from 0 to π. Thus the parametrization (10.200) is exactly the reverse of the
parametrization (10.198).
• Proof of Γ 6= ∂Φ. If η is the 1-form discussed in Problems 10.21 and 10.22, then since η is closed
in R2 \ {0} by Example 10.36 (i.e., dη = 0), we get from Theorem 10.33 (Stokes’ Theorem) that
Z
Z
η=
dη = 0.
(10.201)
Φ
∂Φ
Since −δ ≤ t ≤ δ and −1 ≤ sin θ ≤ 1, we have
−δ ≤ −t sin θ ≤ δ
and then
x2 + y 2 = [(1 − t sin θ) cos 2θ]2 + [(1 − t sin θ) sin 2θ]2 = (1 − t sin θ)2 ≥ (1 − δ)2 > 0.
Thus η is well-defined on the projection of M into the (x, y)-plane. In this case, we follow from the
previous calculation of the winding numbers of the projections of Γ1 and Γ3 into the (x, y)-plane
around the origin that
Z
Γ
η=
Z
Γ1 −Γ3
η=
Z
Γ1
usual
substrac
-tion
η
↓
−
Z
Γ3
η = 2π − (−2π) = 4π.
(10.202)
Hence integrals (10.201) and (10.202) shows that Γ 6= ∂Φ.
This completes the proof of the problem.
Chapter 10. Integration of Differential Forms
336
CHAPTER
11
The Lebesgue Theory
11.1
Further properties of integrable functions
Problem 11.1
Rudin Chapter 11 Exercise 1.
Proof. We follow the hint. Let En ⊆ E and En = {x ∈ E | f (x) > n1 } for all n = 1, 2, . . .. We write
∞
[
A=
En .
(11.1)
n=1
Suppose that µ(A) = 0. Since En ⊂ A for every n, we follow from [21, Eqn. (18), p. 304] that
0 ≤ µ(En ) ≤ µ(A) = 0 for every n. In other words, µ(En ) = 0 for every n. Conversely, suppose that
µ(En ) = 0 for every positive integer n. We apply Theorem 11.8(b) to the definition (11.1) to get
∞
X
0 ≤ µ(A) ≤
µ(En ) = 0.
n=1
Thus we have µ(A) = 0. To prove our result, we need the following lemma:
Lemma 11.1
Let X be a measurable space and E be a measurable set. Let KF be the characteristic
function of F , where F ⊆ E is measurable. Then we have IE (KF ) = µ(E ∩ F ) and
Z
KF dµ = µ(F ).
E
Furthermore, if f is a nonnegative simple measurable function defined on E and given by
f (x) =
m
X
dk KFk (x),
k=1
where dk > 0 and Fk ⊆ E are measurable sets for k = 1, . . . , m, then we have
Z
E
f dµ =
m
X
k=1
337
dk µ(Fk ).
Chapter 11. The Lebesgue Theory
338
Lemma of 11.1. We remark that some books call the KF the indicator function of F .
By Definition 11.19, KF is also a simple function so that Definition 11.21 gives
IE (KF ) = µ(F ).
Suppose that
s(x) =
n
X
(11.2)
ci KEi (x),
i=1
where x ∈ X and c1 , . . . , cn > 0. By Definition 11.19, we know that Ei ∩ Ej = ∅ for all i 6= j.
Let
0 ≤ s ≤ KF .
If x ∈ Ei but x ∈
/ F , then we have 0 < ci ≤ 0, a contradiction. Thus x ∈ F so that Ei ⊆ F for
every i = 1, . . . , n which imply that
n
[
i=1
Ei ⊆ F ⊆ E.
In addition, for every x ∈ Ei ⊆ F , we have 0 < ci ≤ 1. Since each Ei is measurable by
Proposition A.7, we follow from this, the additive property of µ, the inequality [21, Eqn. (8),
p. 302] and the relation (11.2) that
IE (s) =
n
X
i=1
ci µ(E ∩ Ei ) =
n
X
i=1
ci µ(Ei ) ≤
n
X
µ(Ei ) = µ
n
[
i=1
i=1
Ei ≤ µ(F ) = IE (KF ).
Hence, by Definition 11.21, we obtain
Z
KF dµ = sup IE (s) = IE (KF ) = µ(F ).
(11.3)
E
This proves the first assertion.
For the second assertion, we let f be the simple function given by the lemma. By the first
assertion, we have
KFi ∈ L (µ)
on E. Then we deduce from Theorem 11.29, Remark 11.23(c) and the identity (11.3) that
Z
E
f dµ =
Z X
m
E k=1
dk KFk dµ =
m
X
dk
k=1
Z
KFk dµ =
E
m
X
dk µ(Fk ).
k=1
This proves the second assertion, completing the proof of the lemma.
We go back to the proof of Problem 11.1. By the definition (11.1), we have
A = {x ∈ E | f (x) > 0}.
We have to show that
µ(A) = 0.
Since En ⊂ E for every n, we must have A ⊂ E. Assume that µ(En ) > 0 for some n. By the result of
the first part, we have µ(A) > 0. By Definition 11.21 and Remark 11.23(b), we note the following
Z
Z
µ(En )
>0
f dµ ≥
f dµ = sup IE (s) ≥ sup IEn (s) =
n
E
En
which contradicts the hypothesis. This completes the analysis of the problem.
Problem 11.2
Rudin Chapter 11 Exercise 2.
339
11.1. Further properties of integrable functions
Proof. Let
A+ = {x ∈ E | f (x) ≥ 0} and A− = {x ∈ E | f (x) < 0}.
It is clear that E = A+ ∪ A− . Since f is measurable, Theorem 11.15 implies that A+ and A− are
measurable. By the hypothesis, we have
Z
Z
f dµ =
(−f ) dµ = 0.
A+
A−
Since f ≥ 0 on A+ and −f > 0 on A− , Problem 11.1 shows that f (x) = 0 almost everywhere on A+ and
−f (x) = 0 almost everywhere on A− . Hence, these facts imply that
f (x) = 0
almost everywhere on E, completing the proof of the problem.
Problem 11.3
Rudin Chapter 11 Exercise 3.
Proof. Suppose that X is the measurable space we are considering. Let A be the set of points x at which
the sequence {fn (x)} converges. Recall from the Cauchy criterion for convergence (see [21, p. 54]) that a
sequence converges in R if and only if it is a Cauchy sequence. Therefore, if x ∈ A, then for every n ∈ N,
there exists an integer N such that
1
|fp (x) − fq (x)| <
n
for all p, q ≥ N so that the set A is equal to
∞ [
∞
\
\ n
n=1 N =1 p,q≥N
x ∈ X |fp (x) − fq (x)| <
1o
.
n
(11.4)
By the proof of Theorem 11.16, we have
n
1o n
1o n
1o
x ∈ X |fp (x) − fq (x)| <
= x ∈ X fp (x) − fq (x) <
∩ x ∈ X fp (x) − fq (x) > −
. (11.5)
n
n
n
Since fp and fq are measurable, fp − fq is also measurable by Theorem 11.18. Thus the two sets in the
right-hand side of (11.5) are measurable too. Since there exists a σ-ring M of subsets of X, we see from
Definition 11.1 that the set (11.4) is measurable. This completes the proof of the problem.
Problem 11.4
Rudin Chapter 11 Exercise 4.
Proof. Suppose that M is a positive constant such that
|g(x)| ≤ M
for all x ∈ E. Since f ∈ L (µ) on E, Remark 11.23(d) and Theorem 11.26 imply that
M f ∈ L (µ)
and |f | ∈ L (µ)
on E respectively. Since f and g are measurable, f g is also measurable by Theorem 11.18. It is obvious
that
|f g| ≤ M |f |
on E, so we follow from Theorem 11.27 that
f g ∈ L (µ)
on E. This completes the proof of the problem.
Chapter 11. The Lebesgue Theory
340
Problem 11.5
Rudin Chapter 11 Exercise 5.
Proof. For the first assertion, we note that if x ∈ [0, 12 ], then {f2k } is a subsequence of {fn } such that
lim f2k (x) = g(x) = 0.
k→∞
If x ∈ ( 21 , 1], then {f2k+1 } is a subsequence of {fn } such that
lim f2k+1 (x) = g(1 − x) = 0.
k→∞
By definition, we have the desired result that
lim inf fn (x) = 0
n→∞
(0 ≤ x ≤ 1).
For the second assertion, when n = 2k, we have
Z 1
fn (x) dx =
0
Z 1
f2k (x) dx =
0
Z 1
g(x) dx =
0
Z 12
0 dx +
Z 1
1 dx =
1
2
0
1
.
2
Similarly, when n = 2k + 1, we have
Z 1
fn (x) dx =
0
Z 1
f2k+1 (x) dx =
0
Z 1
0
g(1 − x) dx =
Z 21
1 dx +
0
Z 1
0 dx =
1
2
This completes the proof of the problem.a
1
.
2
Problem 11.6
Rudin Chapter 11 Exercise 6.
Proof. Given that ǫ > 0. Let N be a positive integer such that N1 < ǫ. On the one hand, for all x ∈ R
and all integers n ≥ N , we have
|fn (x) − 0| = |fn (x)| ≤
1
1
≤
< ǫ.
n
N
By Definition 7.7, fn (x) → 0 uniformly on R. On the other hand, we have
Z n
Z ∞
Z n
Z n
Z ∞
1
dx = 2.
fn dx =
fn dx +
fn dx +
fn dx =
n
−n
n
−n
−∞
−∞
This completes the proof of the problem.b
11.2
The Riemann integrals and the Lebesgue integrals
Problem 11.7
Rudin Chapter 11 Exercise 7.
a This problem says that the strict inequality in Theorem 11.31 (Fatou’s Theorem) can happen.
b As the problem mentioned, the uniform convergence does not imply the dominated convergence in the case of infinite
measure. Compare with [21, Eqn. (84), p.321].
341
11.2. The Riemann integrals and the Lebesgue integrals
Proof. Suppose that f is bounded on [a, b] and α is a monotonically increasing function on [a, b]. As the
hint says, we consider the regular measure µα induced by α in Example 11.6(b):
µα ([a, b)) = α(b−) − α(a−),
µα ([a, b]) = α(b+) − α(a−),
(11.6)
µα ((a, b]) = α(b+) − α(a+),
µα ((a, b)) = α(b−) − α(a+).
However, the domain of the monotonically increasing function considered there is R, so we have to extend
the function α in question as follows:

 α(a), if x < a;
α(x) =
α(x), if a ≤ x ≤ b;

α(b), if x > b.
Then such extended α : R → R is now monotonically increasing on R and therefore, the regular measure
µα is well-defined as in (11.6).
Motivating from Theorem 6.10, we assume that α is continuous on [a, b].c By this, the measure in the
definition (11.6) unifies to the same value:
µα ([a, b)) = µα ([a, b]) = µα ((a, b]) = µα ((a, b)) = α(b) − α(a).
Now we claim that
“f ∈ R(α) on [a, b] if and only if f is continuous a.e. on [a, b] with respect to µα .”
To prove this claim, we mainly modify the proof of Theorem 11.33 and divide our proof into the following
steps:
• Step 1: Upper and lower Riemann-Stieltjes integrals. By Theorem 3.17 and Theorem 6.4,
there exists a sequence {Pk } of partitions of [a, b] such that Pk+1 is a refinement of Pk , the distance
between adjacent points of Pk is less than k1 and
lim L(Pk , f, α) = R
k→∞
Z b
f dα
a
and
lim U (Pk , f, α) = R
k→∞
Z b
f dα.
(11.7)
a
• Step 2: Representations of lim L(Pk , f, α) and lim U (Pk , f, α). Let Pk = {x0 , x1 , . . . , xn }
k→∞
k→∞
be a partition of [a, b], where x0 = a and xn = b. For 1 ≤ i ≤ n, let Ei = (xi−1 , xi ] and
Mi = sup f (x)
x∈Ei
and mi = inf f (x).
x∈Ei
Next, we define the functions Lk , Uk : [a, b] → R by
0,
if x = a;
Lk (x) =
mi , if x ∈ Ei for 1 ≤ i ≤ n,
n
X
mi KEi (x)
=
(11.8)
i=1
and
0,
if x = a;
Mi , if x ∈ Ei for 1 ≤ i ≤ n,
n
X
Mi KEi (x).
=
Uk (x) =
(11.9)
i=1
c We remark that this is a very restrictive requirement and one can find more general and less restrictive necessary and
sufficient conditions for f ∈ R(α) on [a, b] in [13].
Chapter 11. The Lebesgue Theory
342
Thus the representations (11.8) and (11.9) of Lk and Uk imply that
Z xi
Z xi
Z b
Z a
n
n
n
X
X
X
mi ∆αi = L(Pk , f, α)
dα =
mi R
Lk dα =
R
R
Lk dα = R
0 dα +
a
a
Z b
Z a
xi−1
i=1
xi−1
i=1
i=1
and
R
Uk dα = R
0 dα +
a
a
n
X
R
Z xi
Uk dα =
xi−1
i=1
n
X
Mi R
Z xi
dα =
xi−1
i=1
n
X
Mi ∆αi = U (Pk , f, α),
i=1
where ∆αi = α(xi ) − α(xi−1 ) (since α is continuous on [a, b]) for 1 ≤ i ≤ n. In conclusion, we have
Z b
Z b
L(Pk , f, α) = R
Lk dα and U (Pk , f, α) = R
Uk dα.
a
a
Putting these into the limits (11.7), we have
Z b
Z b
lim R
Lk dα = R
f dα and
k→∞
a
a
lim R
k→∞
Z b
Uk dα = R
a
Z b
f dα.
a
• Step 3: Computations of lim Lk and lim Uk . By the construction of the sequence of partitions
k→∞
k→∞
{Pk }, we know that Pk+1 refines Pk , so it leads from the definitions (11.8) and (11.9) that
L1 (x) ≤ L2 (x) ≤ · · · ≤ f (x) ≤ · · · ≤ U2 (x) ≤ U1 (x)
(11.10)
for all x ∈ [a, b]. Apply Theorem 3.14 (Monotone Convergence Theorem) to the inequalities (11.10),
there exist functions L, U : [a, b] → R such that
L(x) = lim Lk (x)
k→∞
and U (x) = lim Uk (x)
k→∞
and
for all x ∈ [a, b].
Z
• Step 4: Computations of lim
k→∞
L(x) ≤ f (x) ≤ U (x)
(11.11)
Z
Uk dµα . Since each Ei is a Borel set
Lk dµα and lim
k→∞
[a,b]
[a,b]
of [a, b] (see [22, Sec. 1.11, p. 12]), it is measurable. By Definition 11.19, all Lk and Uk are simple
functions. Thus they are measurable by Proposition A.7 and so the functions L and U are also
(bounded) measurable by Theorem 11.17. By Theorem 11.28 (Lebesgue’s Monotone Convergence
Theorem)d , we have
Z
Z
Z
Z
U dµα .
Uk dµα =
L dµα and lim
Lk dµα =
lim
k→∞
k→∞
[a,b]
[a,b]
[a,b]
[a,b]
• Step 5: Representations of U (Pk , f, α) and L(Pk , f, α) in terms of Lebesgue integrals.
Since Lk and Uk are simple functions, it follows from the representation (11.8) and [21, Eqn. (52)
& (54), p. 314] thate
Z
Z
n X
n
n Z
n
X
X
X
mj µ(Ei ∩ Ej ).
(11.12)
IEi (Lk ) =
Lk dµα =
Lk dµα +
Lk dµα =
[a,b]
{a}
i=1
Ei
i=1
i=1 j=1
Since Ei ∩ Ej = ∅ if i 6= j, we further deduce from the expression (11.12) that
Z
n
n
X
X
mi ∆α = L(Pk , f, α).
mi µ(Ei ) =
Lk dµα =
[a,b]
Similarly, we can show from the representation (11.9) of Uk that
Z
n
X
Mi ∆α = U (Pk , f, α).
Uk dµα =
[a,b]
i=1
d The constant “zero” in [21, Eqn. (64), p.318] can be replaced by any constant.
e Notice that µ (a) = α(a) − α(a) = 0.
α
(11.13)
i=1
i=1
(11.14)
343
11.2. The Riemann integrals and the Lebesgue integrals
• Step 6: Relations between Riemann-Stieltjes integrals and Lebesgue integrals. Now by
substituting the Lebesgue integrals (11.13) and (11.14) into the limits (11.7), we obtain
Z
Z
Z b
Z b
Lk dµα = R
lim
Uk dµα = R
f dα and lim
f dα.
(11.15)
k→∞
[a,b]
k→∞
a
[a,b]
a
By the result of Step 3, the expressions (11.15) can be written as
Z b
Z
Z b
Z
f dα and
U dµα = R
f dα.
L dµα = R
[a,b]
[a,b]
a
(11.16)
a
• Step 7: Equivalent statement of f ∈ R(α) on [a, b]. We note that f ∈ R(α) on [a, b] if and
only if
Z
(U − L) dµα = 0
(11.17)
[a,b]
by the integrals (11.16). Since U (x) ≥ L(x) on [a, b], the integral (11.17) holds if and only if
U (x) = L(x) a.e. on [a, b] respect to µα by Problem 11.1. In other words, we have
“f ∈ R(α) on [a, b] if and only if U (x) = L(x) a.e. on [a, b] with respect to µα .”
(11.18)
To finish our proof, we have to show that
“U (x) = L(x) a.e. on [a, b] w.r.t. µα if and only if f is continuous a.e. on [a, b] w.r.t. µα .”
(11.19)
This equivalent relation will be shown in the remaining three steps.
• Step 8: A lemma. We need the following lemma:f
Lemma 11.2
Suppose that x0 is chosen such that x0 6∈ Pk for every k ∈ N and U (x0 ) = L(x0 ). Then f
is continuous at x0 .
Proof of Lemma 11.2. Assume that f was discontinuous at x0 . Then there exists a ǫ > 0 and
a sequence {xp } ⊂ [a, b] such that |xp − x0 | < 1p but
|f (xp ) − f (x0 )| ≥ ǫ
for all p ∈ N. Now for every k ∈ N, we can find an i such that x0 ∈ Ei◦ . (Recall that Ei◦
denotes the interior of Ei , see Problem 2.9.) Thus it implies that xp ∈ Ei for all large enough
p. Hence we have
Uk (x0 ) − Lk (x0 ) =
n
X
i=1
(Mi − mi )KEi (x0 ) = Mi − mi ≥ sup |f (a) − f (b)| ≥ |f (xp ) − f (x)| ≥ ǫ,
a,b∈Ei
but this implies that
U (x0 ) − L(x0 ) ≥ ǫ
which contradicts the hypothesis U (x0 ) = L(x0 ). This completes the proof of our lemma.
By Lemma 11.2, if f is discontinuous at x ∈ [a, b], then x ∈ S1 ∪ S2 , where
S1 = {x ∈ [a, b] | U (x) 6= L(x)}
and S2 = {x ∈ [a, b] | x ∈ Pk for some k ∈ N}.
Therefore, f is continuous a.e. on [a, b] with respect to µα if and only if
µα (S1 ) = µα (S2 ) = 0.
f Lemma 11.2 and its proof are basically motivated by Step 7 of the proof of Theorem C in [13].
Chapter 11. The Lebesgue Theory
344
• Step 9: Proofs of µα (S1 ) = µα (S2 ) = 0. Since α is continuous on [a, b], we always have
µα ({x}) = 0
(11.20)
for every x ∈ [a, b]. Since each Pk is at most countable, their union
∞
[
Pk
k=1
is countable by Theorem 2.12. Since S2 ⊆
11.7 and Theorem 11.8(b) that
∞
[
Pk , we know from this, the measure (11.20), Definition
k=1
0 ≤ µα (S2 ) ≤ µα
∞
[
k=1
∞
X
Pk ≤
µα (Pk ) = 0.
k=1
In other words, we always have µα (S2 ) = 0.
By the definition of S1 , U (x) = L(x) a.e. on [a, b] with respect to µα if and only if µα (S1 ) = 0.
• Step 10: Final step. By the first part of Step 9, we always have µα (S2 ) = 0. Then, by the
result of Step 8 and the second part of Step 9, we know that f is continuous a.e. on [a, b] with
respect to µα if and only if U (x) = L(x) a.e. on [a, b] with respect to µα . Hence our claim follows
from the relations (11.18) and (11.19) in Step 7.
This completes the proof of the problem.
Problem 11.8
Rudin Chapter 11 Exercise 8.
Proof. Since f ∈ R on [a, b], we know from Definition 6.1 that f is supposed to be bounded on [a, b].
By Theorem 11.33(b), f is continuous almost everywhere on [a, b]. By Theorem 6.20, F is differentiable
almost everywhere on [a, b] and
F ′ (x) = f (x),
where f is continuous at x ∈ [a, b], finishing the proof of the problem.
Problem 11.9
Rudin Chapter 11 Exercise 9.
Proof. Given that f ∈ L on [a, b]. Let x0 ∈ [a, b] and {xn } be a sequence in [a, b] converging to x0 , where
xn 6= x0 for all n ∈ N. We want to show that
lim F (xn ) = F (x0 ).
n→∞
Now for each n ∈ N, we define the sequences of intervals {En } and functions {fn } given by En = [a, xn ]
and
f (x), x ∈ En ;
fn (x) =
(11.21)
0,
x ∈ [a, b] \ En
respectively. By the definition (11.21), we can write
f n = f × K En
11.3. Functions of classes L and L 2
345
on [a, b]. Since f and KEn are real-valued measurable functions defined on [a, b], it follows from Theorem
11.18 that fn is also measurable on [a, b]. Furthermore, we have
lim fn (x) = f (x)K[a,x0 ] (x)
n→∞
for all x ∈ [a, b]. Therefore, this shows that
|fn (x)| ≤ |f (x)|
for all n = 1, 2, 3, . . . and x ∈ [a, b]. Hence, we deduce easily from Theorem 11.32 (Lebesgue’s dominated
convergence theorem) that
Z x0
Z
Z
Z xn
f (x) dx = F (x0 )
f K[a,x0 ] dm =
fn dm =
f (x) dx = lim
lim F (xn ) = lim
n→∞
n→∞
n→∞
a
[a,b]
[a,b]
a
as desired. Hence F is continuous at x0 and since x0 is arbitrary, F is continuous on [a, b]. This completes
the proof of the problem.
11.3
Functions of classes L and L 2
Problem 11.10
Rudin Chapter 11 Exercise 10.
Proof. Suppose that g(x) = 1 for every x ∈ X. Then we have
X, if a ≤ 1;
E = {x ∈ X | g(x) > a} =
∅, if a > 1.
(11.22)
Since X is a measurable space, we know from Definition 11.12 that X ∈ M. Since M is a σ-ring, we
follow from Definition 11.1 that ∅ = X \ X ∈ M. Thus X and ∅ are measurable and then the set E
defined in (11.22) is also measurable. Since
Z
Z
|1|2 dµ =
dµ = µ(X) < +∞,
X
X
2
we get from Definition 11.34 that 1 ∈ L (µ) on X. By Theorem 11.35, we have
f = f · 1 ∈ L (µ)
on X.
On the other hand, this is false if µ(X) = +∞. For example, we consider X = R and the function
f (x) =
Then we have µ(R) = +∞ and
Z
R
|f |2 dm =
Z ∞
dx
≤
(1
+
|x|)2
−∞
so that f ∈ L on R. However, since
Z
R
+
f dm =
Z ∞
0
1
.
1 + |x|
dx
= lim
1 + x N →∞
Z N
0
Z ∞
dx
=0
2
x
−∞
dx
= lim ln(1 + N ) → ∞,
1 + x N →∞
it follows from Definition 11.22 that f 6∈ L on R. This ends the proof of the problem.
Chapter 11. The Lebesgue Theory
346
Problem 11.11
Rudin Chapter 11 Exercise 11.
Proof. By definition, we have
d(f, g) =
Z
X
|f − g| dµ.
Recall from Remarks 11.25 that two functions f, g ∈ L (µ) on X are supposed to be the same if f = g
almost everywhere on X. We first show that L (µ) is a metric space by checking Definition 2.15.
(a) If f 6= g on E ⊆ X with µ(E) > 0, then we have from Remark 11.23(f) that f, g ∈ L (µ) on E and
X \ E. By Theorem 11.29, we have f − g ∈ L (µ) on E and X \ E. Thus Theorem 11.26 implies
that
|f − g| ∈ L (µ)
on E and X \ E. Now we get from Theorem 11.24(b) that
Z
Z
Z
d(f, g) =
|f − g| dµ =
|f − g| dµ +
|f − g| dµ.
X
X\E
(11.23)
E
Since |f − g| ≥ 0 on X and |f − g| ≥ δ > 0 on E for some fixed δ, we apply Remark 11.23(c) to the
relation (11.23) to get
Z
d(f, g) ≥
E
|f − g| dµ > δµ(E) > 0.
Furthermore, if f = fˆ almost everywhere on X, then we deduce from Remark 11.23(e) that
Z
Z
Z
ˆ
ˆ
ˆ
|f − fˆ| dµ = 0,
|f − f | dµ +
|f − f | dµ =
d(f, f ) =
E
X\E
X
where E = {x ∈ X | f (x) 6= fˆ(x)} which has measure zero.
(b) Now d(f, g) = d(g, f ) is obvious because of the identity |f − g| = |g − f |.
(c) Suppose that f, g, h ∈ L (µ) on X. Then it follows from the triangle inequality
|f − g| ≤ |f − h| + |h − g|
that the inequality d(f, g) ≤ d(f, h) + d(h, g) always holds.
Hence L (µ) is a metric space.
Next, we want to show that L (µ) is complete. By Definition 3.12, we have to show that every Cauchy
sequence converges in L (µ): Suppose that {fn } is a Cauchy sequence in L (µ). We must show that {fn }
converges to a function f in L (µ). We mainly follow the proof of Theorem 11.42 here and we prove our
result in two steps.
• Step 1: Construction of the function f . Now we can find a sequence {nk }, k = 1, 2, . . ., such
that
Z
1
|fnk − fnk+1 | dµ = d(fnk , fnk+1 ) < k (k = 1, 2, 3, . . .)
2
X
which implies that
∞ Z
X
k=1
X
|fnk − fnk+1 | dµ ≤
∞
X
1
k=1
2k
= 1.
(11.24)
By Theorem 11.30, we may interchange the integration and summation on the left-hand side in the
inequality (11.24) to get
Z X
∞
|fnk − fnk+1 | dµ ≤ 1.
X k=1
11.3. Functions of classes L and L 2
347
Therefore, we must have
∞
X
k=1
|fnk − fnk+1 | < +∞
(11.25)
almost everywhere on X. Apply Theorem 3.45 to the inequality (11.25), we know that the series
∞
X
(fnk − fnk+1 )
k=1
converges almost everywhere on X. Since the kth partial sum of the series
∞
X
(fnk − fnk+1 )
k=1
is fnk+1 (x) − fn1 (x), we see that the limit
f (x) = lim fnk (x)
k→∞
defines f (x) for almost all x ∈ X.
• Step 2: Convergence of the sequence of functions {fn }. We claim that {fn } converges to f
in L (µ). Given that ǫ > 0. Since {fn } is Cauchy, there exists a positive integer N such that
d(fn , fm ) < ǫ
(11.26)
for all m, n ≥ N . If nk > N , then we may apply Theorem 11.31 (Fatou’s Theorem) to the sequence
{Fj = |fnj − fnk |}
of nonnegative measurable functions to get
|f − fnk | = lim inf |fnj − fnk |
j→∞
and
d(f, fnk ) ≤ lim inf d(fnj , fnk ).
j→∞
By the assumption (11.26), we deduce from the inequality (11.27) that
Z
|f − fnk | dµ ≤ ǫ.
d(f, fnk )) =
(11.27)
(11.28)
X
By the result of the inequality (11.28), we have
|f − fnk | ∈ L (µ).
Since {fnk } is a sequence of measurable functions, f is also measurable by the corollary of Theorem
11.17 and then Theorem 11.18 ensures that
f − fn−k
is also a measurable function. Thus (f − fnk )+ and (f − fnk )− are measurable by the corollary of
Theorem 11.17 again. Furthermore, we see that
|(f − fnk )+ | = (f − fnk )+ ≤ |(f − fnk )| and |(f − fnk )− | = (f − fnk )− ≤ |(f − fnk )| (11.29)
hold almost everywhere on X. If we apply Theorem 11.27 to the inequalities (11.29), then we can
conclude that
(f − fnk )+ , (f − fnk )− ∈ L (µ).
Chapter 11. The Lebesgue Theory
348
Therefore, since
f − fnk = (f − fnk )+ − (f − fnk )− ,
it follows from Theorem 11.29 that
f − fnk ∈ L (µ).
By this and the fact that f = (f − fnk ) + fnk , we see that
f ∈ L (µ).
Since ǫ is arbitrary, we can derive from the inequality (11.28) that
lim d(f, fnk ) = 0.
k→∞
Now we obtain from the triangle inequality
d(f, fn ) ≤ d(f, fnk ) + d(fnk , fn )
(11.30)
that the sequence {fn } converges to f in L (µ) if we take both n and nk become large enough on
the right-hand side in the equality (11.30).
This completes the proof of the problem.
Problem 11.12
Rudin Chapter 11 Exercise 12.
Proof. The answer is affirmative. It is clear that the set [0, 1] is Lebesgue measurable. Let α ∈ [0, 1] and
{xn } ⊆ [0, 1] be a sequence such that xn → α as n → ∞. By condition (b), for each fixed xn ∈ [0, 1], the
function fn defined by
fn (y) = f (xn , y)
is continuous in y. Thus, by Example 11.14, each fn (y) is Lebesgue measurable.
Furthermore, the continuity of f (x, y) for fixed y ∈ [0, 1] (condition (c)) shows that
lim fn (y) = lim f (xn , y) = f (α, y).
n→∞
n→∞
By condition (a), we know that
|fn (y)| = |f (xn , y)| ≤ 1
for all xn , y ∈ [0, 1]. Since 1 ∈ L on [0, 1], we follow from Theorem 11.32 (Lebesgue’s dominated
convergence theorem) that
lim g(xn ) = lim
n→∞
n→∞
Z 1
f (xn , y) dy = lim
n→∞
0
Z 1
fn (y) dy =
0
Z 1
f (α, y) dy = g(α).
0
Since the sequence {xn } is arbitrary, Theorem 4.6 implies that g is continuous at α. This completes the
proof of the problem.
Problem 11.13
Rudin Chapter 11 Exercise 13.
Proof. Let E be the set of these points. By Definition 11.34,
kf k =
(Z
π
−π
2
)2
|f (x)| dx
11.3. Functions of classes L and L 2
349
defines a norm in L 2 . By direct computation, we have
Z π Z π
Z π
1 cos 2x dx = π
sin2 nx dx =
−
|f (x)|2 dx =
kfn k2 =
2
−π 2
−π
−π
for all n = 1, 2, 3, . . .. In other words, we have
kfn k =
√
π
for all n = 1, 2, 3, . . .. Therefore, E is bounded. Furthermore, it is also easy to see that if n 6= m, then
we have
Z π
kfn − fm k2 =
|fn (x) − fm (x)|2 dx
−π
Z π
=
(sin nx − sin mx)2 dx
−π
Z π
=
(sin2 nx − 2 sin nx sin mx + sin2 mx) dx
−π
Z π
sin nx sin mx dx
= 2π − 2
= 2π.
−π
(11.31)
This means that the neighborhood N√π (fn ) contains only the point fn itself, so any point fn of E is not
a limit point of E by Theorem 2.20. In other words, E ′ = ∅. By Problem 2.1, ∅ ⊂ E. Hence E is closed
by Definition 2.18(d).
Assume that E was compact. By Theorem 11.42, L 2 is a complete metric space. By the identity
(11.31), E has no Cauchy sequence. Thus E cannot be complete by Definition 3.12. However, since E is
assumed to be compact, we deduce from Theorem 3.11(b) that E is complete, a contradiction. Hence E
is not compact and this ends the proof of the problem.
Problem 11.14
Rudin Chapter 11 Exercise 14.
Proof. Let f : X → C and V be an open set in C. By definition, we have f = u + iv, where u and v are
real. Recall that f is measurable if and only if u and v are measurable.
Suppose that f −1 (V ) is measurable for every open set V in C. It is easy to see that Vx = (a, ∞) × R
and Vy = R × (a, ∞) are open sets in C for every a ∈ R, so f −1 (Vx ) and f −1 (Vy ) are measurable. By
direct checking, we know that
f −1 (Vx ) = f −1 ((a, ∞) × R) = {x ∈ X | f (x) = u(x) + iv(x) ∈ (a, ∞) × R} = {x ∈ X | u(x) > a}
and
f −1 (Vy ) = f −1 (R × (a, ∞)) = {x ∈ X | f (x) = u(x) + iv(x) ∈ R × (a, ∞)} = {x ∈ X | v(x) > a}.
Since a is arbitrary, we obtain from Definition 11.13 that u and v are measurable. Hence f is measurable
too.
Conversely, suppose that f is measurable so that u and v are measurable. Let (p, q) ∈ V . Since V is
open in C, there exists a δ > 0 such that N√2δ ((p, q)) ⊆ V .g By definition, we have
√
N√2δ ((p, q)) = {(x, y) ∈ C | |(x, y) − (p, q)| < 2δ}.
Now the open square
Rδ ((p, q)) = {(x, y) ∈ X | x ∈ (p − δ, p + δ) and y ∈ (q − δ, q + δ)}
g Notice that the number δ depends on the point (p, q).
Chapter 11. The Lebesgue Theory
350
is obviously a subset of N√2δ ((p, q)) so that Rδ ((p, q)) ⊆ V and then
V =
[
Rδ ((p, q)),
(11.32)
(p,q)∈V
see Figure 11.1 for details.
Figure 11.1: The open square Rδ ((p, q)) and the neighborhood N√2δ ((p, q)).
By the definition in Problem 2.22, we see that the set of points
Q = {pn + iqn | pn , qn ∈ Q and n ∈ N}
is clearly a countable dense subset of C and so C is separable. Thus, by Problem 2.23, it has a countable
base {Vm }. Since Rδ ((p, q)) is an open subset of V , there exists a m ∈ N such that
(p, q) ∈ Vm ⊆ Rδ ((p, q)) ⊆ V.
(11.33)
Therefore, we know from the representation (11.32) that V is a union of subcollection of {Vm }. Since
every infinite subset of a countable set is also countable (see Theorem 2.8), there exists a subsequence
{mk } such that
∞
[
V =
Vmk
k=1
and furthermore, we deduce from the relation (11.33) that
V =
∞
[
Rδ ((pk , qk )),
k=1
where {(pk , qk )} ⊆ V . Now we follow from the representation (11.34) that
f −1 (V ) = f −1
∞
[
k=1
=
=
∞
[
k=1
∞
[
k=1
Rδ ((pk , qk ))
{x ∈ X | f (x) ∈ Rδ ((pk , qk ))}
{x ∈ X | u(x) ∈ (pk − δ, pk + δ) and v(x) ∈ (qk − δ, qk + δ)}
(11.34)
11.3. Functions of classes L and L 2
351
=
∞
[
k=1
{x ∈ X | pk − δ < u(x) < pk + δ} ∩ {x ∈ X | qk − δ < v(x) < qk + δ}.
Since u and v are measurable, the set
{x ∈ X | pk − δ < u(x) < pk + δ} ∩ {x ∈ X | qk − δ < v(x) < qk + δ}
is measurable for each k ∈ N. Since the set of all measurable sets of R is a σ-ring, Definition 11.1 implies
that f −1 (V ) is also measurable. This completes the proof of the problem.
Problem 11.15
Rudin Chapter 11 Exercise 15.
Proof. By Definition 11.4, an interval in (0, 1] is one of the following forms:
(0, b), (0, b], [a, b], [a, b), (a, b], (a, b),
(11.35)
where 0 < a ≤ b ≤ 1. If A ∈ R, then by a similar argument as in the proof of Proposition A.2(c), A
can be written as a finite union of disjoint intervals of (0, 1]. By a similar argument as in the proof of
Proposition A.3(d), any two different decompositions of A into disjoint intervals of (0, 1] give the same
value of φ(A).
Suppose that A, B ∈ R. By the previous paragraph, we have
A = I1 ∪ · · · ∪ Im
and B = J1 ∪ · · · ∪ Jn ,
where {I1 , . . . , Im } and {J1 , . . . , Jn } are collections of disjoint intervals of (0, 1]. Let A ∩ B = ∅. Since
A∩B =
n
m [
[
p=1 q=1
(Ip ∩ Jq ),
we must have
Ip ∩ Jq = ∅
for all p = 1, . . . , m and q = 1, . . . , n. In other words, {I1 , . . . , Im , J1 , . . . , Jn } is a set of disjoint intervals
of (0, 1]. Therefore, we have
φ(A ∪ B) = φ(I1 ∪ · · · ∪ Im ∪ J1 ∪ · · · ∪ Jn )
= [φ(I1 ) + · · · + φ(Im )] + [φ(J1 ) + · · · + φ(Jn )]
= φ(A) + φ(B).
Thus φ is an additive set function on R.
Let A = (0, 21 ) ∈ R so that φ(A) = 1 + 12 = 23 . Given that ǫ = 13 . If F ⊂ A and F is closed, then F
does not contain the end point 0 and thus
φ(F ) + ǫ ≤ 1 +
1
< φ(A).
3
1
By Definition 11.5, φ cannot be regular. For each k ∈ N, we suppose that Ak = ( 2k+1
, 21k ]. Then we have
Ai ∩ Aj = ∅
for i 6= j and
We notice that
1i
= A1 ∪ A2 ∪ · · · .
0,
2
φ(Ak ) =
1
1
− k+1
k
2
2
Chapter 11. The Lebesgue Theory
so that
∞
X
352
φ(Ak ) =
k=1
∞ X
1
k=1
but φ((0, 12 ]) = 1 + 12 = 23 . Therefore, we have
2
−
k
1 2k+1
=
1
,
2
∞
∞
[
X
φ
φ(Ak ),
Ak 6=
n=1
k=1
i.e., φ cannot be extended to a countably additive set function on a σ-ring, completing the proof of the
problem.
Problem 11.16
Rudin Chapter 11 Exercise 16.
Proof. Let fk (x) = sin nk x. By definition, we have
E = {x ∈ (−π, π) | {fk (x)} converges}.
By Example 11.14, since each fk is continuous, {fk } is a sequence of measurable functions. Thus, it
follows from Problem 11.3 that E is measurable. Let A be a measurable subset of E. By the hypothesis,
we let f : E → R be the function such that
lim fk (x) = f (x)
k→∞
on E. Since each fk is continuous on (−π, π), Example 11.14 says that each fk is measurable on (−π, π).
By Remark 11.23(f), fk is measurable on E and also on A. Since
|fk (x)| = | sin nk x| ≤ 1
for all k ∈ N and x ∈ A and 1 ∈ L on A, we deduce from Theorem 11.32 (Lebesgue’s dominated
Convergence Theorem) and the first part of the given hint that
Z
Z
Z
f dx = lim
fk dx = lim
sin nk x dx = 0.
(11.36)
A
k→∞
k→∞
A
A
By Problem 11.2, the relation (11.36) implies that f (x) = 0 almost everywhere on E and then
f 2 (x) = 0
almost everywhere on E. Now, on the one hand, we apply Theorem 11.32 (Lebesgue’s dominated Convergence Theorem) to the sequence of measurable functions {fk2 } (the measurability of fk2 follows from
Theorem 11.18) to obtain that
Z
Z
fk2 dx =
lim
k→∞
A
f 2 dx = 0.
(11.37)
A
On the other hand, since 2 sin2 nk x = 1 − cos 2nk x, we obtain from the second part of the hint that
Z
Z
Z
m(A)
1
lim
.
(11.38)
(1 − cos 2nk x) dx =
lim
fk2 dx = lim
sin2 nk dx =
k→∞ A
k→∞ A
2 k→∞ A
2
Combining the limits (11.37) and (11.38), we derive
Z
Z
m(A)
2
f 2 dx = 0
fk dx =
= lim
k→∞ A
2
A
which is equivalent to m(A) = 0. Since A is arbitrary, we have m(E) = 0 in particular. This finishes the
proof of the problem.
11.3. Functions of classes L and L 2
353
Problem 11.17
Rudin Chapter 11 Exercise 17.
Proof. Here we must have Theorem 8.12 (Bessel inequality) in terms of Lebesgue’s sense. If {φn } is
orthonormal on a measurable set X ⊆ R and f ∈ L 2 on X, then we have
f (x) ∼
∞
X
cn φn (x)
and
n=1
∞
X
n=1
where
cn =
Z
|cn |2 ≤
Z
X
|f (x)|2 dx = kf k2 ,
f φn dx (n = 1, 2, 3, . . .).
(11.39)
X
In particular, we have
lim cn = 0.
(11.40)
n→∞
Assume that there were infinitely may integers n such that
sin nx ≥ δ
for all x ∈ E. Consider the function KE which is clearly measurable. By Remark 11.23(c), since
|KE (x)|2 ≤ 1 on [−π, π], we have
Z π
Z π
dx = 2π.
|KE (x)|2 dx ≤
0 < m(E) =
−π
−π
Thus we must have KE ∈ L 2 on [−π, π] by Definition 11.34.
For each n ∈ N, we define
1
φn (x) = √ sin nx.
π
Then we can see easily that {φn } forms an orthonormal set of functions on [−π, π]. Now we apply the
definition (11.39) to the function f = KE and the orthonormal system {φn } to obtain
Z
Z
1
1
KE sin nx dx = √
sin nx dx.
(11.41)
cn = √
π X
π E
By our hypothesis, we can deduce from the identity (11.41) that
Z
δm(E)
1
>0
δ dx = √
cn ≥ √
π E
π
for infinitely many integers n, but this obviously contradicts the special case (11.40) of the Bessel inequality. Hence there are at most finitely many integers n such that sin nx ≥ δ on E, completing the
proof of the problem.
Problem 11.18
Rudin Chapter 11 Exercise 18.
Proof. We remark that there is a mistake in the statement. It still holds if we take f (x) ≡ 0 and g(x) ≡ 1
on X. Furthermore, the equality holds trivially if µ(X) = 0 by Remark 11.23(e). Therefore, we suppose
that f 6≡ 0 almost everywhere and µ(X) > 0 so that
Z
|f |2 dµ 6= 0.
Chapter 11. The Lebesgue Theory
354
We consider the (complex) constant given by
R
f g dµ
R
c=
.
|f |2 dµ
Now it is easy to see that
Z
Z
|g − cf |2 dµ = (g − cf )(g − cf ) dµ
Z
= (|g|2 − cf g − cf g + |c|2 |f |2 ) dµ
R
Z
Z
Z
| f g dµ|2
2
2
R
= |g| dµ +
g
dµ
·
|f
|
dµ
−
2Re
cf
( |f |2 dµ)2
R
Z
Z
| f g dµ|2
2
= |g| dµ + R
g
dµ
.
−
2Re
cf
|f |2 dµ
(11.42)
(11.43)
If g(x) = cf (x) almost everywhere, then g(x) − cf (x) = 0 and thus cf g = |g|2 almost everywhere. By
the identity (11.43), we establish that
R
Z
Z
Z
Z
| f g dµ|2
2
|g|2 dµ + R
g
dµ
=
2Re
=
2Re
|g|
dµ
=
2
|g|2 dµ
cf
|f |2 dµ
which implies that
R
Z
| f g dµ|2
R
=
|g|2 dµ
|f |2 dµ
Z
Z
Z
2
f g dµ = |f |2 dµ |g|2 dµ.
(11.44)
On the other hand, if the identity (11.44) holds, then we derive from the definition (11.42) and the
identity (11.43) that
Z
Z
Z
2
2
|g − cf | dµ = 2 |g| dµ − 2Re c f g dµ
!
Z
Z
Z
2
2
= 2 |g| dµ − R
Re
f g dµ · f g dµ
|f |2 dµ
!
Z
Z
2
2
= 2 |g|2 dµ − R
· Re
f g dµ
|f |2 dµ
Z
Z
2
2
= 2 |g|2 dµ − R
·
f g dµ
2
|f | dµ
Z
Z
= 2 |g|2 dµ − 2 |g|2 dµ
= 0.
Since |g − cf |2 ≥ 0, we find from Problem 11.1 that
|g − cf |2 = 0
almost everywhere. Hence we have our desired result that g(x) = cf (x) almost everywhere. This
completes the proof of the problem.
APPENDIX
A
A proof of Lemma 10.14
We see from the definition (10.154) that Σn−1 is of class C ′′ , so is S. We consider the n-surface Ψ :
[0, 1] × E → Rn \ {0} defined by
Ψ(t, φ1 , . . . , φn−1 ) = [1 − t + tf (φ1 , . . . , φn−1 )]Σn−1 (φ1 , . . . , φn−1 ),
(A.1)
where (φ1 , . . . , φn−1 ) ∈ E and 0 ≤ t ≤ 1. For each fixed φi , define
E(φ1 ,...,φi−1 ,φi+1 ,...,φn−1 )
= {(φ1 , . . . , φi−1 , φi+1 , . . . , φn−1 ) ∈ [0, π]n−2 × [0, 2π] | (φ1 , . . . , φi−1 , φi+1 , . . . , φn−1 ) ∈ E}
=
n−1
Y
[aj , bj ]
j=1
j6=i
and the mapping
Φ : [0, 1] ×
n−1
Y
j=1
j6=i
[aj , bj ] → Rn \ {0}
defined by
fixed
↓
Φ(t, φ1 , . . . , φi−1 , φi+1 , . . . φn−1 ) = Ψ(t, φ1 , . . . , φi , . . . φn−1 ).
Following the idea as in the proof of “A generalization of Problem 10.22(c)”, if we write
Φ(t, φ1 , . . . , φi−1 , φi+1 , . . . φn−2 ) = (x1 , . . . , xn ),
then we put
g(t, φ1 , . . . , φi−1 , φi+1 , . . . φn−1 ) = 1 − t + tf (t, φ1 , . . . , φi−1 , φi+1 , . . . φn−1 )
into the right-hand side of the definition (A.1) to get
xi = g(t, φ1 , . . . , φi−1 , φi+1 , . . . φn−1 )hi (φ1 , . . . , φi−1 , φi+1 , . . . φn−1 ),
where 1 ≤ i ≤ n. Since f ∈ C ′′ (Dn ) and f > 0, we must have g ∈ C ′′ (Dn ) and g > 0. Furthermore, since
[0, 1] ×
n−1
Y
[aj , bj ]
(A.2)
j=1
j6=i
is clearly a (n − 1)-cell, it is compact by Theorem 2.40 and then the mapping Φ is a (n − 1)-surface of
class C ′′ with (A.2) as the parameter domain to which the formula (10.153) holds. Since f ∈ C ′′ (Dn ),
355
Appendix A. A proof of Lemma 10.14
356
Ψ is a n-surface of class C ′′ in Rn \ {0}. Since each rxni is of class C ′ in Rn \ {0}, the same is true for
n
the (n − 1)-form ωn by Definition 10.11 and thus it follows from Theorem 10.33 (Stokes’ Theorem) and
Problem 10.23(a) that
Z
Z
ωn =
dωn = 0.
(A.3)
Ψ
∂Ψ
To finish the proof of the lemma, we need to compute ∂Ψ by generalizing the relation (10.124). In
fact, the definition of Ψ shows that
∂Ψ = Φ(0, φ1 , . . . , φn−1 ) − Φ(1, φ1 , . . . , φn−1 ) +
− Ψ(t, φ1 , . . . , bi , φi+1 , . . . , φn−1 )
= S(φ1 , . . . , φn−1 ) − Ω(φ1 , . . . , φn−1 ) +
n−1
X
i=1
− Φbi (t, φ1 , . . . , φi−1 , φi+1 , . . . , φn−1 ) .
The above analysis implies that
Z
ωn =
Φai
Z
ωn = 0
Φbi
n−1
X
i=1
Ψ(t, φ1 , . . . , ai , φi+1 , . . . , φn−1 )
Φai (t, φ1 , . . . , φi−1 , φi+1 , . . . , φn−1 )
(1 ≤ i ≤ n)
so that we derive from these and the formula (A.3) that
Z
Z
ωn −
ωn = 0
S
ZΩ
Z
ωn =
ωn .
Ω
(A.4)
S
To compute the integral on the right-hand side in the identity (A.4), we consider the following steps:
• Step 1: Observations from Examples 10.36 and 10.37. By Examples 10.36 and 10.37 (see
also Problem 10.21(a) and Problem 10.22(b)), we have
Z
Z
Z
Z 2π
η=
ω2 =
(1) dt =
dt = 2π
and
Thus we have
Z
Z
ω3 =
ζ=
S
S
S
Z
sin u du ∧ dv =
(ω2 )γ = dt
0
γ
γ
γ
Z π Z 2π
0
sin u du dv = 4π.
0
and (ω3 )Σ = sin u du ∧ dv.
In other words, we need to find a way to express ωn in terms of the parameters of a parametrization
of the (n − 1)-sphere. Fortunately, this has been done in (10.154) for the (n − 1)-surface (don’t
forget it is also a C ′ -mapping) Σn−1 : Dn → En already.
By Definition 10.21 with carefully chosen n, m, ti and xj or the deduction up to [21, Eqn. (72), pp.
264, 265], we havea
(ωn )Σn−1 =
n
hX
i=1
=
(−1)i−1 xi dx1 ∧ · · · ∧ dxi−1 ∧ dxi+1 ∧ · · · ∧ dxn
" n
X
i=1
i−1
(−1)
i
Σn−1
#
∂(x1 , . . . , xi−1 , xi+1 , . . . , xn )
xi (φ1 , . . . , φn−1 ) det
×
∂(φ1 , . . . , φn−1 )
dφ1 ∧ · · · ∧ dφn−1 .
(A.5)
Therefore, if we can compute the Jacobians in the formula (A.5), then an explicit formula of
(ωn )Σn−1 in terms of dφ1 , . . . , dφn−1 can be derived.
a The formula (A.5) can also be found in [25, Eqn. (12.31), p. 206].
357
• Step 2: Statement of the formula (ωn )Σn−1 .
Lemma A.1
For every n ≥ 2, we have
(ωn )Σn−1 = sin φn−2 sin2 φn−3 · · · sinn−2 φ1 dφ1 ∧ · · · ∧ dφn−1 .
To prove this lemma, we have to compute the Jacobians in the formula (A.5), so we need to consider
the following n × (n − 1) Jacobian matrix
M=
∂(x1 , . . . , xn )
∂(φ1 , . . . , φn−1 )


− sin φ1




cos φ1 cos φ2




..

.



n−2
=
Y

 cos φ1
sin φj cos φn−1

j=1


j6=1



n−1
Y


sin φj
cos
φ
1


j=1

j6=1
0
..
.
cos φ2
n−2
Y
sin φj cos φn−1
j=1
j6=2
cos φ2
n−1
Y





···
0




..
..

.
.



n−2

Y

sin φj (− sin φn−1 ) 
···

j=1





n−1
Y

sin φj 
· · · cos φn−1


j=1

j6=n−1
···
− sin φ1 sin φ2
sin φj
j=1
j6=2

0
so that
det
∂(x1 , . . . , xi−1 , xi+1 , . . . , xn )
= det Mi
∂(φ1 , . . . , φn−1 )
(1 ≤ i ≤ n),
where Mi is the lower triangular (n − 1) × (n − 1) matrix obtain by deleting the i row from M.
To get the idea of the computation of det Mi , we first consider a special case in Step 3. As a
result, a “guess” of the formula of det Mi can be obtained in the expression (A.8). A proof of such
formula is provided in Step 4 and finally the proof of Lemma A.1 is presented in Step 5.
• Step 3: The special case when n = 5. In this step, we consider the special case that n = 5.
Then we have
x1 = cos φ1 ,
x2 = sin φ1 cos φ2 ,
x3 = sin φ1 sin φ2 cos φ3 ,
x4 = sin φ1 sin φ2 sin φ3 cos φ4 ,
x5 = sin φ1 sin φ2 sin φ3 sin φ4
(A.6)
Appendix A. A proof of Lemma 10.14
358
andb
det
∂(x2 , x3 , x4 , x5 )
= sin3 φ1 sin2 φ2 sin φ3 cos φ1
∂(φ1 , φ2 , φ3 , φ4 )
= (−1)1−1 sin3 φ1 sin2 φ2 sin φ3 x1 ,
det
∂(x1 , x3 , x4 , x5 )
= − sin4 φ1 sin2 φ2 sin φ3 cos φ2
∂(φ1 , φ2 , φ3 , φ4 )
= (−1)2−1 sin3 φ1 sin2 φ2 sin φ3 × cos φ2
= (−1)2−1 sin3 φ1 sin2 φ2 sin φ3 x2 ,
det
= (−1)3−1 sin3 φ1 sin2 φ2 sin φ3 x3 ,
3−1
Y
sin φj
4−1
Y
sin φj
(A.7)
j=1
∂(x1 , x2 , x3 , x5 )
= − sin4 φ1 sin3 φ2 sin2 φ3 cos φ4
∂(φ1 , φ2 , φ3 , φ4 )
4−1
= (−1)
3
2
sin φ1 sin φ2 sin φ3 × cos φ4
= (−1)4−1 sin3 φ1 sin2 φ2 sin φ3 x4 ,
det
sin φj
j=1
∂(x1 , x2 , x4 , x5 )
= sin4 φ1 sin3 φ2 sin φ3 cos φ3
∂(φ1 , φ2 , φ3 , φ4 )
= (−1)3−1 sin3 φ1 sin2 φ2 sin φ3 × cos φ3
det
2−1
Y
j=1
5−1
Y
∂(x1 , x2 , x3 , x4 )
sin φj
= (−1)5−1 sin3 φ1 sin2 φ2 sin φ3 ×
∂(φ1 , φ2 , φ3 , φ4 )
j=1
= (−1)5−1 sin3 φ1 sin2 φ2 sin φ3 x5 .
Substituting the definitions (A.6) and the determinants (A.7) into the formula (A.5) (with n = 5)
and using the fact that
(x1 , x2 , x3 , x4 , x5 )
is a point of the 4-sphere, we obtain that
"
∂(x2 , x3 , x4 , x5 )
∂(x1 , x3 , x4 , x5 )
(ω5 )Σ4 = x1 det
− x2 det
∂(φ1 , φ2 , φ3 , φ4 )
∂(φ1 , φ2 , φ3 , φ4 )
∂(x1 , x2 , x3 , x5 )
∂(x1 , x2 , x4 , x5 )
− x4 det
∂(φ1 , φ2 , φ3 , φ4 )
∂(φ1 , φ2 , φ3 , φ4 )
#
∂(x1 , x2 , x3 , x4 )
× dφ1 ∧ dφ2 ∧ dφ3 ∧ dφ4
+ x5 det
∂(φ1 , φ2 , φ3 , φ4 )
+ x3 det
= sin3 φ1 sin2 φ2 sin φ3 (x21 + x22 + x23 + x24 + x25 ) dφ1 ∧ dφ2 ∧ dφ3 ∧ dφ4
= sin3 φ1 sin2 φ2 sin φ3 dφ1 ∧ dφ2 ∧ dφ3 ∧ dφ4
which is consistent with Lemma A.1. Hence, it is reasonable to make the following claim:
det
∂(x1 , . . . , xi−1 , xi+1 , . . . , xn )
= (−1)i−1 sinn−2 φ1 sinn−3 φ2 · · · sin φn−2 xi
∂(φ1 , . . . , φn−1 )
where 1 ≤ i ≤ n and n ≥ 2.
b The explicit form of the Jacobian matrix ∂(x1 , x2 , x3 , x4 , x5 ) can be found on the next page.
∂(φ1 , φ2 , φ3 , φ4 )
(A.8)
∂x1
∂φ3
∂x2
∂φ3
∂x3
∂φ3
∂x4
∂φ3
∂x5
∂φ3
∂x1
∂φ2
∂x2
∂φ2
∂x3
∂φ2
∂x4
∂φ2
∂x5
∂φ2
∂x1 

∂φ4 


∂x2 

∂φ4 



∂x3 

∂φ4 



∂x4 

∂φ4 


∂x5 


∂φ4


− sin φ1




cos φ1 cos φ2




=
cos φ1 sin φ2 cos φ3



 cos φ1 sin φ2 sin φ3 cos φ4



 cos φ sin φ sin φ sin φ
1
2
3
4

 ∂x1

 ∂φ1


 ∂x2

 ∂φ

1


 ∂x3
=
 ∂φ1



 ∂x4

 ∂φ1


 ∂x

5

∂φ1
∂(x1 , x2 , x3 , x4 , x5 )
∂(φ1 , φ2 , φ3 , φ4 )

0
− sin φ1 sin φ2 sin φ3
− sin φ1 sin φ2
sin φ1 cos φ2 cos φ3
sin φ1 cos φ2 sin φ3 sin φ4 sin φ1 sin φ2 cos φ3 sin φ4
sin φ1 cos φ2 sin φ3 cos φ4 sin φ1 sin φ2 cos φ3 cos φ4
0
0





0




.
0



− sin φ1 sin φ2 sin φ3 sin φ4 



sin φ1 sin φ2 sin φ3 cos φ4 
0

359
Appendix A. A proof of Lemma 10.14
360
• Step 4: A proof of the formula (A.8). We apply induction to prove the formula (A.8). To do
so, we rewrite the equations (10.154) as follows:
x1 = cos φ1 ,
x2
= cos φ2 ,
sin φ1
x3
= sin φ2 cos φ3 ,
sin φ1
..
.
xk−1
= sin φ2 · · · sin φk−2 cos φk−1 ,
sin φ1
xk
= sin φ2 · · · sin φk−2 sin φk−1 ,
sin φ1
(A.9)
where 0 ≤ φ1 , . . . , φk−2 ≤ π, 0 ≤ φk−1 ≤ 2π. For j = 1, . . . , k − 2, let ϕj = φj+1 so that
0 ≤ ϕ1 , . . . , ϕk−3 ≤ π
and 0 ≤ ϕk−2 ≤ 2π.
Furthermore, for j = 1, . . . , k − 2, we let
xj+1
Xj =
= sin φ2 · · · sin φj cos φj+1 = sin ϕ1 · · · sin ϕj−1 cos ϕj
sin φ1
and
xk
= sin φ2 · · · sin φk−2 sin φk−1 = sin ϕ1 · · · sin ϕk−3 sin ϕk−2 .
sin φ1
Thus the last (k − 1) equations of (10.154) can be reduced to
Xk−1 =
X1 = cos ϕ1 ,
X2 = sin ϕ1 cos ϕ2 ,
..
.
(A.10)
Xk−2 = sin ϕ1 · · · sin ϕk−3 cos ϕk−2 ,
Xk−1 = sin ϕ1 · · · sin ϕk−3 sin ϕk−2 .
Note that
2
X12 + · · · + Xk−1
= 1.
Assume the induction is true for n = k − 1 for some k ≥ 3, i.e.,
det
∂(X1 , . . . , Xi−1 , Xi+1 , . . . , Xk−1 )
= (−1)i−1 sink−3 ϕ1 sink−4 ϕ2 · · · sin ϕk−3 Xi ,
∂(ϕ1 , . . . , ϕk−2 )
where 1 ≤ i ≤ k − 1. Since
x1 = cos φ1
and xj+1 = Xj sin φ1
(1 ≤ j ≤ k − 1),
we have

 ∂x1

 ∂φ1



..


.

∂(x1 , . . . , xk )

=
∂(φ1 , . . . , φk−1 ) 
 ∂xk−1

 ∂φ1


 ∂x
k

 ∂φ
1
∂x1
∂φ2
···
..
.
..
∂xk−1
∂φ2
···
∂xk
∂φ2
···
.

∂x1 

∂φk−1 



..


.


∂xk−1 


∂φk−1 


∂xk 


∂φ
k−1
(A.11)
361

 − sin φ1




 X1 x1




..

.
=




 Xk−2 x1




 X
 k−1 x1
0
···
∂X1
sin φ1
∂φ2
···
..
.
..
∂Xk−2
sin φ1
∂φ2
···
∂Xk−1
sin φ1
∂φ2
···

 − sin φ1




 X1 x1




..

.
=




 Xk−2 x1




 X
 k−1 x1
.
0
···
∂X1
sin φ1
∂ϕ1
···
..
.
..
∂Xk−2
sin φ1
∂ϕ1
···
∂Xk−1
sin φ1
∂ϕ1
···
.





∂X1

sin φ1 

∂φk−1



..


.



∂Xk−2

sin φ1 

∂φk−1



∂Xk−1
sin φ1 

∂φ
0
k−1





∂X1

sin φ1 

∂ϕk−2



..


.



∂Xk−2

sin φ1 

∂ϕk−2



∂Xk−1
sin φ1 

∂ϕ
0
.
(A.12)
k−2
k×(k−1)
By deleting the first row of the matrix (A.12), we obtain
det
∂(x2 , . . . , xk )
=
∂(φ1 , . . . , φk−1 )
X1 x1
∂X1
sin φ1
∂ϕ1
···
∂X1
sin φ1
∂ϕk−2
..
.
..
.
..
.
..
.
Xk−2 x1
∂Xk−2
sin φ1
∂ϕ1
···
∂Xk−2
sin φ1
∂ϕk−2
Xk−1 x1
∂Xk−1
sin φ1
∂ϕ1
···
∂Xk−1
sin φ1
∂ϕk−2
X1
∂X1
∂ϕ1
···
∂X1
∂ϕk−2
..
.
..
.
..
.
..
.
Xk−2
∂Xk−2
∂ϕ1
···
∂Xk−2
∂ϕk−2
Xk−1
∂Xk−1
∂ϕ1
···
∂Xk−1
∂ϕk−2
= x1 sink−2 φ1
.
(A.13)
Appendix A. A proof of Lemma 10.14
362
Now, by expanding the determinant (A.13) along the first column and then using the assumption
(A.11), we have
k−1
X
∂(X1 , . . . , Xi−1 , Xi+1 , . . . , Xk−1 )
∂(x2 , . . . , xk )
k−2
(−1)i−1 det
= x1 sin
φ1
Xi
det
∂(φ1 , . . . , φk−1 )
∂(ϕ1 , . . . , ϕk−2 )
i=1
= x1 sink−2 φ1
k−1
X
i=1
sink−3 ϕ1 sink−4 ϕ2 · · · sin ϕk−3 Xi2
= x1 sink−2 φ1 sink−3 ϕ1 sink−4 ϕ2 · · · sin ϕk−3
= sink−2 φ1 sink−3 φ2 · · · sin φk−2 x1 .
k−1
X
Xi2
i=1
(A.14)
Next, for 2 ≤ i ≤ k, we first delete the i row from the matrix (A.12), then expand its corresponding
determinant along the first row and finally apply the assumption (A.11) to obtain that
∂(x1 , . . . , xi−1 , xi+1 , . . . , xk )
det
=
∂(φ1 , . . . , φk−1 )
− sin φ1
0
···
0
X1 x1
∂X1
sin φ1
∂ϕ1
···
∂X1
sin φ1
∂ϕk−2
..
.
..
.
..
.
..
.
Xi−2 x1
∂Xi−2
sin φ1
∂ϕ1
···
∂Xi−2
sin φ1
∂ϕk−2
Xi x1
∂Xi
sin φ1
∂ϕ1
···
∂Xi
sin φ1
∂ϕk−2
..
.
..
.
..
.
..
.
Xk−1 x1
∂Xk−1
sin φ1
∂ϕ1
···
∂Xk−1
sin φ1
∂ϕk−2
∂X1
sin φ1
∂ϕ1
..
.
= − sin φ1
··· ···
..
.
..
.
∂X1
sin φ1
∂ϕk−2
..
.
∂Xi−2
sin φ1
∂ϕ1
··· ···
∂Xi−2
sin φ1
∂ϕk−2
∂Xi
sin φ1
∂ϕ1
··· ···
∂Xi
sin φ1
∂ϕk−2
..
.
∂Xk−1
sin φ1
∂ϕ1
= − sink−1 φ1 det
..
.
..
.
··· ···
..
.
∂Xk−1
sin φ1
∂ϕk−2
∂(X1 , . . . , Xi−2 , Xi , . . . , Xk−1 )
∂(ϕ1 , . . . , ϕk−2 )
= (−1)i−1 sink−1 φ1 sink−3 ϕ1 sink−4 ϕ2 · · · sin ϕk−3 Xi−1
363
= (−1)i−1 sink−2 φ1 sink−3 φ2 · · · sin φk−2 xi .
(A.15)
Combining the results (A.14) and (A.15), we have shown that
det
∂(x1 , . . . , xi−1 , xi+1 , . . . , xk )
= (−1)i−1 sink−2 φ1 sink−3 φ2 · · · sin φk−2 xi
∂(φ1 , . . . , φk−1 )
for 1 ≤ i ≤ k. Hence, by induction, our claim (A.8) holds for every positive integer n ≥ 2.
• Step 5: Completion of the proof of Lemma A.1. By substituting the determinant (A.8) into
the formula (A.5), we establish easily that
(ωn )Σn−1 =
sinn−2 φ1 sinn−3 φ2 · · · sin φn−2
n−2
= sin
n−3
φ1 sin
n
X
i=1
x2i dφ1 ∧ · · · ∧ dφn−1
φ2 · · · sin φn−2 dφ1 ∧ · · · ∧ dφn−1 ,
completing the proof of Lemma A.1.
If we apply Lemma A.1 to the integral on the right-hand side in (A.4), then we acquire
Z
Z
sin φn−2 sin2 φn−3 · · · sinn−2 φ1 dφ1 ∧ · · · ∧ dφn−1 = An−1 (S),
ωn =
S
(A.16)
S
n−1
where An−1 (S) is the “area” of the restriction S = ΣE
. This completes the proof of Lemma 10.14.
Appendix A. A proof of Lemma 10.14
364
APPENDIX
B
Solid angle subtended by a surface at the origin
Roughly speaking, a solid angle is the two-dimensional “angle” in three-dimensional space. In fact, it is
an analogy to the case that a plane angle θ measured in radians equals its corresponding arc length s of
a circle of radius r. Recall that the angle θ can be found by the formula
s
θ= .
r
(B.1)
See Figure B.1 for an illustration.
Figure B.1: The plane angle θ measured in radians.
To extend this concept to our three dimensional space, we consider the solid angle Ω subtended by an
object to be the surface area A of its radical projection onto the unit sphere (the unit is steradians). In
other words, you have an object and you project it onto the the unit sphere. After this, you compute the
surface area of the projection and this number is called the solid angle. Similar to the definition (B.1),
we define the solid angle to be
A
Ω = 2,
(B.2)
r
where A is the surface area of the projection and r is the radius of the sphere. The idea can be seen in
Figure B.2:
365
Appendix B. Solid angle subtended by a surface at the origin
366
Figure B.2: The solid angle Ω measured in steradians.
To find the formula (B.2), we know from the spherical coordinates system that dA = r2 sin θ dθ dφ so
that
dA
(B.3)
dΩ = 2 = sin θ dθ dφ.
r
By integration, we deduce from the expression (B.3) that
Z Z
sin θ dθ dφ,
Ω=
(B.4)
S
where S denotes the projection of the object onto the sphere. Here the variables θ and φ in the formula
(B.4) correspond to the variables u and v in Problem 10.22 (see Figure 10.13).
When we look at Figure B.2 again, it may happen that two similar objects (two discs in Figure B.2)
may have the same solid angle. This coincides with our daily experience that a small nearby object and
a large far away object may “look” the same. Since the objects OCD or OEF are similar cones, let’s
calculate the solid angle of a cone with apex angle 2θ. First, we have a section of the cone in Figure B.3:
Figure B.3: A section of the cone with apex angle 2θ.
367
In this case, our formula (B.4) becomes
Ω=
Z 2π Z θ
0
0
sin ϕ dϕ dφ = 2π
Z θ
0
sin ϕ dϕ = 2π(1 − cos θ).
(B.5)
Now the formula (B.5) says that the solid angle Ω of a cone (a three-dimensional angle) can be found
by the plane angle θ (a two-dimensional angle). For instance, we know that the sun is much larger than
the moon and also much far away from the earth. However, the “sizes” of both objects (the sun and the
moon) are approximately the same because they have approximately the same solid angle. In fact, the
angular diameters of the sun and the moon seen from earth are approximately 9.35 × 10−3 and 9.2 × 10−3
in radians respectively. By substituting these values into the formula (B.5), we obtain the solid angle
subtended by the sun and the moon are given by
Ωsun = 6.87 × 10−5 sr and Ωmoon = 6.67 × 10−5 sr
respectively. Hence, these values explain why the moon can almost cover the whole sun in the sky during
a solar eclipse.
Appendix B. Solid angle subtended by a surface at the origin
368
APPENDIX
C
Proofs of some basic properties of a measure
Let p be a positive integer and Rp denote the p-dimensional Euclidean space. We call I an interval a in
Rp if
I = {x = (x1 , . . . , xp ) | − ∞ ≤ ai ≤ xi ≤ bi ≤ +∞ (i = 1, . . . , p)}
(C.1)
= [a1 , b1 ] × · · · × [ap , bp ],
where any or all of the “≤” signs can be replaced by “<” and the notation “≤ bi ” (resp. “ai ≤”) is
replaced by “< bi ” if bi = +∞ (resp. “ai <” if ai = −∞) in (C.1). Let I1 , . . . , In be intervals for some
positive integer n. If
A=
n
[
Ii ,
i=1
then A is called an elementary set .
We obtain some fundamental results about intervals in the following proposition, see [19, pp. 4-7] for
reference.
Proposition C.1
Suppose that I and J are intervals in the form (C.1).
(a) If I ⊆ J, then there exist intervals I1 , . . . , In such that
I = I0 ⊆ I1 ⊆ · · · ⊆ In = J.
(C.2)
(b) If I ⊆ J, then for such a sequence of intervals in the relation (C.2), the set-theoretic differences
Im \ Im−1
(m = 1, . . . , n)
form a sequence of intervals of type (C.1).
(c) The difference J \ I is a union of finite number of intervals.
Proof.
(a) We consider the special case J = Rp first, i.e., a1 = · · · = ap = −∞ and b1 = · · · = bp = +∞ in the
a Some books call it a rectangle or a box .
369
Appendix C. Proofs of some basic properties of a measure
370
definition (C.1). Suppose that I0 = I. Then we define I1 and I2 as follows:
I1 = {(x1 , . . . , xp ) | x1 ∈ [a1 , ∞), ai ≤ xi ≤ bi (i = 2, . . . , p)},
I2 = {(x1 , . . . , xp ) | x1 ∈ R, ai ≤ xi ≤ bi (i = 2, . . . , p)},
I2k−1 = {(x1 , . . . , xp ) | x1 , . . . , xk−1 ∈ R, xk ∈ [ak , ∞), ai ≤ xi ≤ bi (i = k + 1, . . . , p)},
I2k = {(x1 , . . . , xp ) | x1 , . . . , xk ∈ R, ai ≤ xi ≤ bi (i = k + 1, . . . , p)},
(C.3)
I2p−1 = {(x1 , . . . , xp ) | x1 , . . . , xp−1 ∈ R, xp ∈ [ap , ∞)},
I2p = {(x1 , . . . , xp ) | x1 , . . . , xp ∈ R} = Rp ,
where k = 2, . . . , p − 1. Hence it is easy to see from our construction that
I = I0 ⊆ I1 ⊆ · · · ⊆ I2p = Rp .
(C.4)
For the general case (i.e., J ⊂ Rp ), our special case shows that there exists intervals I0 , I1 , . . . , I2p
such that the relation (C.3) holds. Let
Ii′ = Ii ∩ J,
where i = 0, 1, . . . , 2p. It is obvious from the definition (C.1) that the intersection of two intervals
must be an interval. Thus each Ii′ is an interval and then the relation (C.3) implies that
′
I = I0′ ⊆ I1′ ⊆ · · · ⊆ I2p
= J.
(b) Similar as part (a), we prove the result for the special case J = Rp first. Note that we haveb
I2k \ I2k−1 = {(x1 , . . . , xp ) | x1 , . . . , xk−1 ∈ R, xk ∈ (−∞, ak ),
ai ≤ xi ≤ bi (i = k + 1, . . . , p)},
I2k−1 \ I2k−2 = {(x1 , . . . , xp ) | x1 , . . . , xk−1 ∈ R, xk ∈ (bk , ∞),
ai ≤ xi ≤ bi (i = k + 1, . . . , p)},
I2p−1 \ I2p−2 = {(x1 , . . . , xp ) | x1 , . . . , xp−1 ∈ R, xp ∈ (bp , +∞)},
(C.5)
I2p \ I2p−1 = {(x1 , . . . , xp ) | x1 , . . . , xp−1 ∈ R, xp ∈ (−∞, ap )},
where k = 1, . . . , p − 1. By the definition (C.1), all the above differences are certainly intervals and
so our result follows in this special case.
For the general case (i.e., J ⊂ Rp ), we deduce from the identity
A \ B = A ∩ Bc
(C.6)
that
′
′
′
′
Im
\ Im−1
= Im
∩ (Im−1
)c
= (Im ∩ J) ∩ (Im−1 ∩ J)c
c
= (Im ∩ J) ∩ (Im−1
∪ J c)
c
= (Im ∩ J ∩ Im−1
) ∪ (Im ∩ J ∩ J c )
c
= [(Im ∩ Im−1
) ∩ J] ∪ (Im ∩ ∅)
= (Im \ Im−1 ) ∩ J,
′
′
where m = 1, . . . , n. Since both Im \ Im−1 and J are intervals, the difference Im
\ Im−1
must be an
interval too.
(c) If J \ I is an interval, then there is nothing to prove. Suppose that J \ I is not an interval. We
consider the set Ib = I ∩ J. Then we have
Ib ⊂ J
b
and J \ I = J \ I.
b Notice that the intervals (−∞, a ) and (b , +∞) are replaced by (−∞, a ] and [b , +∞) if the interval [a , +∞) and
k
k
k
k
k
the inequality xk ≤ bk in (C.3) are replaced by (ak , +∞) and xk < bk respectively.
371
By part (a), there exist intervals I1 , . . . , In such that
Ib = I0 ⊆ I1 ⊆ · · · ⊆ In = J.
(C.7)
By part (b), the sequence in the relation (C.7) satisfies the condition that
Jk = Ik \ Ik−1
(k = 1, . . . , n)
form a set of intervals and
n
[
k=1
which is our desired result.
Jk =
n
[
(Ik \ Ik−1 ) = In \ I0 = J \ Ib = J \ I
k=1
Proposition C.2
Denote E to be the collection of all elementary sets of Rp .
(a) The set E is a ring, but not a σ-ring.
(b) If A ∈ E , then Ac ∈ E .
(c) If A ∈ E , then A is the union of a finite number of disjoint intervals.
(d) If A ∈ E and
A = I1 ∪ · · · ∪ Im = J1 ∪ · · · ∪ Jn ,
where {I1 , . . . , Im } and {J1 , . . . , Jn } are disjoint intervals, then we have
m(A) = m(I1 ) + · · · + m(Im ) = m(J1 ) + · · · + m(Jn ).
Proof.
(a) Suppose that A, B ∈ E . Then we have
A = I1 ∪ · · · ∪ Im
and B = J1 ∪ · · · ∪ Jn ,
where I1 , . . . , Im , J1 , . . . , Jn are intervals. Since
A ∪ B = I1 ∪ · · · ∪ Im ∪ J1 ∪ · · · ∪ Jn
is also a union of a finite number of intervals, we have A ∪ B ∈ E . Furthermore, we have
A\B =
=
n
[
Is \
s=1
n
[
s=1
=
n
[
s=1
=
=
m
[
Jt
t=1
m
c
[
Jt
Is ∩
t=1
Is ∩
m
n \
[
s=1 t=1
n \
m
[
m
\
t=1
Is ∩ Jtc
(Is \ Jt ).
s=1 t=1
Jtc
Appendix C. Proofs of some basic properties of a measure
372
By Proposition A.1(c), each Is \ Jt is a union of finite number of intervals. As we note in the proof
of Proposition A.1(a), intersections of intervals is still an interval, so the intersection
m
\
(Is \ Jt )
t=1
is a finite union of intervals. Hence the difference A \ B is a finite union of intervals. In other words,
we have A \ B ∈ E . By Definition 11.1, E is a ring.
However, E is not a σ-ring. Let p = 1 and An = (n − 12 , n + 12 ) ⊂ R for n = 1, 2, . . .. Then we have
An ∈ E for n = 1, 2, . . . and
A=
∞
[
An =
n=1
1
n1 3
o
, +∞ \
, ,...
2
2 2
is not elementary. Otherwise, we have
A=
N
[
Ik ,
k=1
where I1 , . . . , IN are some intervals. Since A is unbounded, Ik is unbounded for some k ∈ {1, . . . , N }
and this implies that such Ik contains n2 for some positive integer n, a contradiction.
(b) By definition, we have Ac = Rp \ A. Since A, Rp ∈ E and E is a ring, part (a) implies that Ac ∈ E .
(c) Let A =
n
[
k=1
Ik with I1 , . . . , In ∈ E . Write A1 = I1 and
Ak = (I1 ∪ · · · ∪ Ik ) \ (I1 ∪ · · · ∪ Ik−1 ),
where k = 1, . . . , n. Then we have
A=
n
[
(C.8)
Ak .
k=1
We claim that A1 , . . . , An are pairwise disjoint. Let s > t. Apply the identity (C.6) to the definition
(C.8), we get
As = Is \ (I1 ∪ · · · ∪ It ∪ · · · ∪ Is−1 ) and At = It \ (I1 ∪ · · · ∪ It−1 ) ⊂ It
which shows that
As ∩ At = ∅.
(d) For each p = 1, . . . , m, we have
m(Ip ) = m(Ip ∩ J1 ) + · · · + m(Ip ∩ Jn ).
Similarly, for each q = 1, . . . , n, we have
m(Jq ) = m(Jq ∩ I1 ) + · · · + m(Jq ∩ Im ).
Since
m
X
p=1
m(Ip ) =
m X
n
X
p=1 q=1
m(Ip ∩ Jq ) =
this means that m(A) is well-defined.
This finishes the proof of the proposition.
n X
m
X
q=1 p=1
m(Jq ∩ Ip ) =
n
X
m(Jq ),
q=1
Proposition C.3
Suppose that A ∈ M(µ) and ǫ > 0. Then there exist a closed set F and an open set G such that
F ⊆ A ⊆ G, where
µ(G \ A) < ǫ and µ(A \ F ) < ǫ.
373
Proof. Since A ∈ M(µ), we deduce from Definition 11.9 that A can be represented as the union of a
countable collection of µ-measurable sets {An } of MF (µ), i.e.,
A=
∞
[
An .
n=1
By the definition of MF (µ), for each n ∈ N, there exists a sequence {Ank } ⊆ E such that Ank → An as
k → ∞. Now for each k, since µ∗ is regular, there exist Fnk , Gnk ∈ E such that Fnk is closed and Gnk is
open,
Fnk ⊆ Ank ⊆ Gnk
(C.9)
and
µ∗ (Gnk ) − 2−n 2−k 2−1 ǫ ≤ µ∗ (Ank ) ≤ µ∗ (Fnk ) + 2−n 2−k 2−1 ǫ.
(C.10)
By the left-hand side of the inequality (C.10), we have
µ∗ (Gnk ) − µ∗ (Ank ) ≤ 2−n 2−k 2−1 ǫ.
(C.11)
Since each Gnk is open, Theorem 2.24(a) shows that the sets
Gn =
∞
[
Gnk
∞
[
and G =
k=1
Gn
(C.12)
k=1
are open. Now, by Remark 11.11(a), we have Gnk , Gn , G ∈ M(µ).c We define
An =
∞
[
Ank
and A =
k=1
∞
[
An .
(C.13)
k=1
Combining the relations (C.9), the definitions (C.12) and (C.13), we obtain
An ⊆ Gn
and A ⊆ G.
For each n, if x ∈ Gn \ An , then x ∈ Gn but x 6∈ An so that x ∈ Gnk for some k and x 6∈ Ani for all
i. Therefore, we have x ∈ Gnk \ Ank for some k which implies that
Gn \ An ⊆
(Gnk \ Ank ).
(C.14)
∞
[
(Gn \ An ).
(C.15)
k=1
Similarly, we can show that
G\A⊆
∞
[
k=1
By applying [21, Eqn. (18), p. 304] and Theorem 11.8(b) to the relation (C.15) first and then the relation
(C.14), we conclude that
µ∗ (G \ A) ≤ µ∗
≤
≤
≤
∞
[
∞
X
n=1
∞
X
n=1
(Gn \ An )
µ∗ (Gn \ An )
µ∗
∞
[
n=1
k=1
∞
∞ X
X
∗
n=1 k=1
(Gnk \ Ank )
µ (Gnk \ Ank ).
c Note that these also follow directly from the fact that M(µ) is a σ-ring.
(C.16)
Appendix C. Proofs of some basic properties of a measure
374
Since µ∗ is a nonnegative additive set function and Gnk , Ank ∈ E , we further apply the monotonic
property [21, Eqn. (8), p. 302] and the inequality (C.11) to the inequality (C.16) to obtain
µ∗ (G \ A) ≤
∞ X
∞
X
n=1 k=1
[µ∗ (Gnk ) − µ∗ (Ank )] ≤
∞ X
∞
X
2−n 2−k 2−1 ǫ = 2−1 ǫ < ǫ.
(C.17)
n=1 k=1
As Rudin commented on [21, p. 308], µ∗ (A) can be replaced by µ(A) if A ∈ M(µ). Recall that
G \ A ∈ M(µ), so our first half of the proposition follows from the inequality (C.17).
For the second half of the proposition, we note that F is a closed set contained in A if and only if its
open complement F c contained in Ac , i.e., Ac ⊆ F c . Next, we follow from the identity (C.2) that
A \ F = A ∩ F c = F c ∩ (Ac )c = F c \ Ac .
Therefore, we have µ(A \ F ) < ǫ if and only if µ(F c \ Ac ) < ǫ. If we let
An = Ank = (n, n) × · · · × (n, n)
|
{z
}
p terms
for every n, k ∈ N, then we have Ank ∈ E . By Eqn. (5) and (24) [21, pp. 301, 306], we obtain Ank → An
∞
[
as k → ∞ and thus An ∈ MF (µ). Since Rp =
An , we have Rp ∈ M(µ). Since Ac = Rp \ A and M(µ)
n=1
is a ring, Definition 11.1 implies that Ac ∈ M(µ). Now we apply the previous analysis to the set Ac to
get an open set G containing Ac such that µ(G \ Ac ) < ǫ and this implies that
µ(A \ Gc ) < ǫ,
where F = Gc is our desired closed set contained in A. This completes the proof of the proposition.
Proposition C.4
The collection B of all Borel sets in Rp is a σ-ring and it is the smallest σ-ring containing all open
sets of Rp .
Proof. Let O be the collection of all open subsets of Rp and Ω be the family of all σ-rings R in Rp which
p
contain the collection O. Since the power set of Rp (denoted by 2R ) is a σ-ring, Ω 6= ∅. Suppose that
\
R∗ =
R.
R∈Ω
By definition, it is clear that O ⊂ R ∗ and that R ∗ must lie in every σ-ring containing O. If A, B ∈ R ∗ ,
then A, B ∈ R for every R ∈ Ω. Since R is a σ-ring, we have A ∪ B ∈ R and A \ B ∈ R which imply
that A ∪ B ∈ R ∗ and A \ B ∈ R ∗ . By Definition 11.1, R ∗ is a σ-ring and thus it is the smallest σ-ring
containing the collection of all open subsets of Rp . Hence we have R ∗ = B.
Proposition C.5
For every measure µ, the sets of measure zero form a σ-ring.
Proof. Suppose that M0 is the class of all sets of measure zero under the measure µ. If A, B ∈ M0 , then
µ(A) = µ(B) = 0 and we follow from [21, Eqn. (7), p. 302] that
0 ≤ µ(A ∪ B) ≤ µ(A ∪ B) + µ(A ∩ B) = µ(A) + µ(B) = 0.
Thus we have A ∪ B ∈ M0 . By the identity (C.6), we have
0 ≤ µ(A \ B) = µ(A ∩ B c ) ≤ µ(A) = 0
375
which means that A \ B ∈ M0 . Let An ∈ M0 for n = 1, 2, . . .. By Theorem 11.8(b), we have
0≤µ
which means that
∞
[
n=1
∞
[
n=1
∞
X
An ≤
µ(An ) = 0
n=1
An ∈ M0 . By Definition 11.1, M0 is a σ-ring.
Proposition C.6
Every countable subset of Rp has Lebesgue measure zero.
Proof. Let x = (x1 , . . . , xp ) ∈ Rp . Given ǫ > 0. Since {xi } ⊂ (xi − ǫ, xi + ǫ) for i = 1, . . . , p, we have
{x} ⊂ I =
By Definition 11.4, we have
m(x) ≤
p
Y
i=1
p
Y
(xi − ǫ, xi + ǫ).
(2ǫ) = 2p ǫp → ǫ
i=1
as ǫ → 0. In other words, we obtain m(x) = 0. Let E = {x1 , x2 , . . .} be a countable subset of Rp .
By Proposition A.5, we see that E ∈ M0 which is a subset of M(m). By Theorem 11.10, the Lebesgue
measure m is countably additive on M(µ) and in particular, on M0 . Hence we have
m(E) = m
∞
[
n=1
∞
X
m(xn ) = 0
xn =
n=1
which is our desired result.
Proposition C.7
Let s be a real-valued function defined on the measurable space X, the range of s consist of the
distinct numbers c1 , . . . , cn and
Ei = {x | s(x) = ci }
(i = 1, . . . , n).
Then every simple function s is measurable if and only if the sets E1 , . . . , En are measurable.
Proof. We claim that the characteristic function KE is measurable if and only if E is measurable. Suppose
−1
that E is measurable. It is obvious that x 6∈ {0, 1} implies that KE
(x) = ∅. Let a ∈ R. By Theorem
11.15, we have

 ∅, if a > 1;
{x ∈ X | KE (x) ≥ a} =
E, if 0 < a < 1;
(C.18)

X, if a ≤ 0.
Since X is a measurable space, we have, by Definition 11.12, X is measurable. Since ∅ and E are
measurable, it follows from (C.18) that {x ∈ X | KE (x) ≥ a} is measurable. By Definition 11.13, the
characteristic function KE is measurable. Conversely, for every x ∈ X, we have
{x ∈ X | KE (x) > 0} = E.
Since KE is measurable, the set {x | KE (x) > 0} is measurable by Definition 11.13. This proves the claim.
Suppose that E1 , . . . , En are measurable. Then each characteristic function is measurable. Since
s=
n
X
i=1
ci K E i ,
Appendix C. Proofs of some basic properties of a measure
376
Theorem 11.18 implies that s is also measurable. Conversely, suppose that s is measurable. Then the
set s−1 (ci ) = {x ∈ X | s(x) = ci } is measurable, where i = 1, . . . , n. Since Ei = {x ∈ X | s(x) = ci }, Ei is
measurable for i = 1, . . . , n.
Index
A
A.M. ≥ G.M., 62, 147, 223, 225, 242
algebraic numbers, 11
angular diameter, 367
anticommutative relation, 286, 296
L
Landau’s inequality, 96
level curve, 231
Lipschitz condition, 114
locally connected, 319
lower triangular matrix, 283
B
binary representation, 81
Borel sets, 374
boundary of a set, 266
box, 369
bump functions, 265
O
orientation of the curve, 202
P
partitions of unity, 327
Picard’s existence theorem, 114
positive definite, 253
power set, 374
C
Carl Johannes Thomae, 71
Cauchy criterion for convergence, 339
Cauchy-Schwarz inequality, 97
characteristic function, 337
Clairaut’s theorem, 246
connected component, 319
convex hull, 268, 278
Q
quadratic form, 253
R
rectangle, 369
relative prime, 150
D
Dirichlet kernel, 187
S
saddle points, 252
Schwarz’s theorem, 246
set-theoretic differences, 369
simple discontinuity, 60
solid angle, 365
steradians, 365
E
eigenvalues, 253
elementary set, 369
G
Gamma function, 314
H
Hessian matrix, 251
homotopy, 296
T
Thomae’s function, 71
torus, 219
transposition, 248
I
indicator function, 338
initial-value problem, 114
interior, 255
interval, 369
isomorphism, 241
W
WolframAlpha, 219
word, 248
Y
Young’s inequality, 126
Young’s theorem, 246
K
Kronecker’s Approximation Theorem, 82, 222
377
Index
378
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