Selected Solutions to
Walter Rudin’s
Real and Complex Analysis
Prepared by Richard G. Ligo
Chapter 1
Exercise 1.1: Does there exist an infinite σ-algebra which has only countably many members?
Proof: We claim that the answer is no. By way of contradiction, suppose that A is a countable σ-algebra on a space X. Further, note that X must not be finite, or else A would be
T
finite. We can then define a map f : X → A by x 7→ x∈A∈A A. As countable intersections
T
of elements of A must be contained in A and A is countable, we have that x∈A∈A A ∈ A for
all x ∈ X. Thus, f is well-defined.
We now claim that the image of f partitions A, that is, f (x) ∩ f (y) = ∅ or f (x) = f (y) for
distinct x, y ∈ X. Suppose that f (x) ∩ f (y) 6= ∅. If x ∈
/ f (y), then x ∈ f (x) \ f (y) ⊆ f (x),
but this contradicts the definition of f . Thus, we know that x ∈ f (y). Similarly, we have
that y ∈ f (x). Then the definition of f gives us that f (x) ⊆ f (y) and f (y) ⊆ f (x), so it
follows that f (x) = f (y). Let P represent the set of distinct elements of the form f (x),
where x ∈ X. Finally, observe that as f is defined at every x ∈ X, P does cover all of X.
We know from our definition of f that every P ∈ P will be minimal, that is, there does not
exist some Q ∈ A such that Q ( P for any P ∈ P. Combining this with the fact that P
covers X allows us to conclude that any A ∈ A may be generated by a union of sets from
P. As P partitions X, we know that the elements of the power set of P will be in bijection
with all possible unions of elements from P. We then have three cases for the cardinality of
P:
Case 1: P is finite. Then we know that the powerset of P will be finite, so it follows that A
will be finite, which is a contradiction.
Case 2: P is countably infinite. Then we know that the powerset of P will be uncountably
infinite, so it follows that A will be uncountably infinite, which is a contradiction.
Case 3: P is uncountably infinite. Then we know that the powerset of P will be uncountably
infinite, so it follows that A will be uncountably infinite, which is a contradiction.
Thus, we have that a countable σ-algebra cannot exist. Exercise 1.9: Suppose µ is a positive measure on X, f : X → [0, ∞] is measurable,
2
R
X
f dµ = c, where 0 < c < ∞, and α is a constant. Prove



Z
 ∞
α
lim
n log[1 + (f /n) ] dµ =
c
n→∞ X


 0
Proof:
that
if 0 < α < 1,
if α = 1,
if 1 < α < ∞.
As we will eventually show that the limit can be pushed into the integral, we first
determine the pointwise limit of n log[1 + (f /n)α ]. Fix x ∈ X and consider limn→∞ n log[1 +
(f (x)/n)α ]. As we can observe that this limit gives the indeterminant form ∞ · 0, we may
rewrite this expression and use l’Hôpital’s rule, as shown:
α log[1 +
f
lim n log 1 +
= lim
1
n→∞
n→∞
n
n
fα
]
nα
= lim
n→∞
−αf α
nα+1
1
fα
1+ n
α
−1
n2
1
= lim
α
n→∞ 1 + f
nα
−αf α
nα+1
n2
−1
αn1−α f α
α .
n→∞ 1 + f
nα
= lim
We then have separate cases for α.
Case 1: 0 < α < 1. Then we have that 1 − α = > 0, so it follows that
αnε f α
αn1−α f α
=
lim
α
α = ∞.
n→∞ 1 + f
n→∞ 1 + f
α
n
nα
lim
As this is true for all x ∈ X, it then follows that
Z
lim n log[1 + (f /n)α ]dµ = ∞
X n→∞
when 0 < α < 1. We have from Fatou’s lemma that
Z
Z
α
lim n log[1 + (f /n) ]dµ =
lim inf n log[1 + (f /n)α ]dµ
n→∞
X n→∞
X
Z
Z
α
≤ lim inf
n log[1 + (f /n) ]dµ ≤ lim
n log[1 + (f /n)α ]dµ.
n→∞
n→∞
X
X
And thus it follows from our work above that
Z
lim
n log[1 + (f /n)α ]dµ = ∞,
n→∞
X
as desired.
Case 2: α ≥ 1. Recall the fact that for p ≥ 1 we have that 1 + xp ≤ (1 + x)p for all x ≥ 0.
As f is a positive, real-valued function and α ≥ 1 in this case, we have that
1 + (f /n)α ≤ (1 + f /n)α =⇒ n log[1 + (f /n)α ] ≤ n log[(1 + f /n)α ].
3
Furthermore, we may observe via properties of logarithms that
n log[(1 + f /n)α ] = α log[(1 + f /n)n ].
We now claim that limn→∞ (1 + nx )n is an increasing sequence whose limit is ex . To see this,
consider the following chain of implications for all n ∈ N and real x ≥ 0:
n+1
n+1
ex
x
ex
x
ex
≤ e =⇒
≤
=⇒
≤
=⇒ 1 + ≤ e +
n
n
x
n
n+1
n
n+1
x
x
x
x
n
n+1
=⇒ 1 + ≤ e(1 +
) =⇒ e (1 + ) ≤ e (1 +
)
n
n+1
n
n+1
x
x
x
x n+1
=⇒ n log(1 + ) ≤ (n + 1) log(1 +
) =⇒ (1 + ) ≤ (1 +
) .
n
n+1
n
n+1
Thus, we have that the sequence increasing. Let L represent the limit of this sequence; we
then apply l’Hôpital’s rule to determine this limit:
log(1 + nx )
x n
x
x n
L = lim (1 + ) =⇒ log L = lim log(1 + ) = lim n log(1 + ) = lim
1
n→∞
n→∞
n→∞
n→∞
n
n
n
n
1
−x
x
x
1+ n
n2
x
=⇒ log L = lim
= lim
x = x =⇒ L = e .
−1
n→∞
n→∞
1
+
n
n2
It then follows from our claim, the monotonicity of log, and our above work that
n log[1 + (f /n)α ] ≤ α log[ef ] = αf
for all n. As f is integrable, αf will certainly be integrable, so we may apply the Lebesgue
Dominated Convergence Theorem to observe that
Z
Z
α
lim n log[1 + (f /n)α ]dµ
n log[1 + (f /n) ]dµ =
lim
n→∞
X n→∞
X
when α ≥ 1. We now consider two subcases to conclude this problem.
Subcase i: α = 1. Then we have that 1 − α = 0, so it follows that
αn1−α f α
f
= lim
α
α = f.
f
n→∞ 1 +
n→∞ 1 + f
α
n
nα
lim
As this is true for all x ∈ X, it then follows that
Z
Z
α
lim n log[1 + (f /n) ]dµ =
f dµ = c
X n→∞
X
when α = 1. Thus, it follows from our work above that
Z
lim
n log[1 + (f /n)α ]dµ = c,
n→∞
X
4
as desired.
Subcase ii: 1 < α < ∞. Then we have that 1 − α = δ < 0, so it follows that
αn1−α f α
αnδ f α
=
lim
α
α = 0.
n→∞ 1 + f
n→∞ 1 + f
nα
nα
lim
As this is true for all x ∈ X, it then follows that
Z
lim n log[1 + (f /n)α ]dµ = 0
X n→∞
when 1 < α < ∞. Thus, it follows from our work above that
Z
lim
n log[1 + (f /n)α ]dµ = 0
n→∞
X
as desired. Exercise 1.12: Suppose f ∈ L1 (µ). Prove that to each > 0 there exists a δ > 0 such that
R
|f | dµ < whenever µ(E) < δ.
E
Proof: Let X be our measure space. As f ∈ L1 (µ), we have that there exists some sequence
of simple functions {sn } such that {sn } → f and 0 ≤ s1 ≤ s2 ≤ · · · ≤ f . Let ε > 0; then we
R
R
may pick s ∈ {sn } such that X f dµ ≤ X s dµ + 2ε .
R
R
We claim that for any E ⊆ X, we have E f dµ ≤ E s dµ + 2ε . By way of contradiction
R
R
suppose that X f dµ > X s dµ + 2ε . As we know that f ≥ s everywhere it follows that
R
R
R
R
R
R
f
dµ
≥
s
dµ.
Then
we
have
that
f
dµ
+
f
dµ
>
s
dµ
+
s dµ + 2ε , but this
c
c
c
E
E
E
E
E
Ec
R
R
implies that X f dµ > X s dµ + 2ε , which is a contradiction. Thus, we have our claim.
R
As s is simple, we have that s is bounded above by some M . Thus, we have that E s dµ ≤
R
ε
M dµ for all E ⊆ X. Now set δ = 2M
and observe that for E ⊆ X with µ(E) < δ we
E
R
ε
have that E M dµ = M µ(E) < M 2M
= 2ε . It then follows from our work above that for
R
R
R
such E we have E f dµ ≤ E s dµ + 2ε ≤ E M dµ + 2ε < 2ε + 2ε = ε, as desired. 5
Chapter 2
Exercise 2.7: Show that for each ε, 0 < ε < 1, there is an open subset E ⊆ [0, 1] whose
closure is [0, 1] such that the Lebesgue measure of E equals ε.
Proof: To accomplish the desired task, we first construct a “fat Cantor set.” This is done
by sucessively removing 2n−1 open intervals of length tn from the unit interval [0, 1], where
0 < t ≤ 13 . Approaching this iteratively, where Fn is the nth iteration gives the sets
F0 = [0, 1]
1−t
1+t
F1 = [0,
]∪[
, 1]
2
2
1 − t + 2t2 1 − t
1 + t 3 + t − 2t2
3 + t + 2t2
1 − t − 2t2
]∪[
,
]∪[
,
]∪[
, 1]
F2 = [0,
4
4
2
2
4
4
..
.
We can then define our desired set to be F =
T∞
n=0
Fn . As each Fn is a closed set, it follows
that F will also be closed, hence measurable. We also define E = [0, 1] \ F , and note that
E is an open set (this can be seen either by the fact that F is closed or that E is a union
of the open intervals, specifically, the ones removed from each Fn ). We claim that E is the
desired set.
Let ε > 0. As E is constructed from a disjoint countable union of open intervals (hence
measurable), we know that the measure of E will be equal to the sum of the lengths of these
intervals. Note that 0 < t ≤
1
3
implies that 2t < 1; it then follows that
m(E) =
∞
X
n−1 n
2
t =t
ε
1+2ε
(2t)n−1 =
n=1
n=1
Now set t =
∞
X
t
.
1 − 2t
and observe that
ε
t
1+2ε
=
= ε,
ε
1 − 2t
1 − 2( 1+2ε
)
so by choosing such t we have m(E) = ε, as desired.
It then remains to show that E = [0, 1], that is, that E is dense in [0, 1]. Recall an equivalent
definition of dense: “A subset D is a space X is said to be dense in X if for any open subset
U in X we have D ∩ U 6= ∅.” In an effort to use this equivalent definition, let U be an
open set in [0, 1]. Recall now that any open set in R can be constructed from a disjoint
union of intervals. As a result, it suffices to show that some interval I ⊆ U intersects E.
If I ∩ E = ∅, then we have that I ⊆ F . However, we know that F is totally disconnected
6
by construction, as any neighborhood of a point in F will contain some removed interval.
As a result, we know that I 6⊆ F , so I ∩E 6= ∅, implying that E is dense in [0, 1], as desired. Exercise 2.9: Construct a sequence of continuous functions {fn }n≥1 on [0, 1]such that
0 ≤ fn (x) ≤ 1, for all n and x, such that
Z
lim
n→∞
1
fn (x) dx = 0,
0
but such that the sequence {fn (x)}n≥1 converges for no x ∈ [0, 1].
Proof: Recall the enumeration of the positive rationals {Qn }∞
n=1 corresponding to traversing
the integer ordered pairs in quadrant I of the coordinate plane and the x-axis, that is the
enumeration
0 1 0 0 1 2 3 2 1 0 0 1 2 3 4
y
{Qn }∞
n=1 = , , , , , , , , , , , , , , , . . . , , . . .
1 1 2 3 2 1 1 2 3 4 5 4 3 2 1
x
corresponding to the following diagram:
..
.
..
.
..
.
..
.
..
.
··
(1, 4)
(2, 4)
(3, 4)
(4, 4)
(5, 4)
···
(1, 3)
(2, 3)
(3, 3)
(4, 3)
(5, 3)
···
(1, 2)
(2, 2)
(3, 2)
(4, 2)
(5, 2)
···
(1, 1)
(2, 1)
(3, 1)
(4, 1)
(5, 1)
···
(1, 0)
(2, 0)
(3, 0)
(4, 0)
(5, 0)
···
·
∞
∞
We can then define the subsequence {qk }∞
k=1 = {Qnk }k=1 , where {nk }k=1 = {n | Qn ≤ 1}.
We then have that {qk }∞
k=1 is an enumeration of the rationals in the interval [0, 1]; note also
that this enumeration visits each rational infinitely many times, as we will visit all integer
multiples of the ordered pairs previously traversed, which in turn represent the same rational
numbers. The sequence {qk }∞
k=1 can then be pictured as the following diagram:
7
..
.
..
.
..
.
..
.
..
.
··
(1, 4)
(2, 4)
(3, 4)
(4, 4)
(5, 4)
···
(1, 3)
(2, 3)
(3, 3)
(4, 3)
(5, 3)
···
(1, 2)
(2, 2)
(3, 2)
(4, 2)
(5, 2)
···
(1, 1)
(2, 1)
(3, 1)
(4, 1)
(5, 1)
···
(1, 0)
(2, 0)
(3, 0)
(4, 0)
(5, 0)
···
·
We now define a sequence of functions on [0, 1] by {fk }∞
k=1 : [0, 1] → R where fk is a triangle
spike function with a spike of height 1 and base width
4
10blog10 xc
(here x corresponds to the diagram above)
centered at qk and zero elsewhere (if part of this spike happens to leave [0, 1], we may simply
truncate it at 0 or 1, as our function will still be continuous on [0, 1]).
It follows from this definition that the integrals of the fk ’s will indeed converge to zero, as
Z
1
4
2
fk dm =
(1) = blog xc ,
blog
xc
10
2 10
10 10
and we can see from our diagram representing {qk }∞
k=1 that
k → ∞ =⇒ blog10 xc → ∞ =⇒
2
10blog10 xc
→ 0.
It the remains to show that for any x ∈ [0, 1] the sequence {fk (x)}∞
k=1 does not converge.
We have two cases.
Case 1: x ∈ Q ∩ [0, 1]. Then we know from our work above that x will be visited infinitely
∞
many times by {qk }∞
k=1 , so {fk (x)}k=1 will have a constant subsequence of 1. However, as
the triangle spikes of the fk are becoming abitrarily narrow and we must visit every other
rational in [0, 1] infinitely many times, we know that fk (x) must be zero for infinitely many
∞
k, so {fk (x)}∞
k=1 will also have a constant subsequence of 0. Thus, we have that {fk (x)}k=1
is not convergent.
Case 2: x ∈ (R \ Q) ∩ [0, 1]. As x is irrational in [0, 1], we know that x can be represented by
some nonterminating, never-repeating decimal expansion, i.e. x = .x1 x2 x3 x4 x5 · · · , where xi
8
is in the ith digit in the decimal expansion of x. We can then define a sequence of rational
numbers converging to x by taking
.x1 , .x1 x2 , .x1 x2 x3 , .x1 x2 x3 x4 , . . . .
Note xn ∈ N for all n, as xn is simply a digit in the decimal expansion of x. Further, note
that x − .x1 < .1, x − .x1 x2 < .01, x − .x1 x2 x3 < .001, and so on. This generalizes to
x − .x1 x2 . . . xn <
1
.
10n
Also, we can alternatively represent this sequence as
x1 x1 x2 x1 x2 x 3 x1 x2 x3 x4
,
,
,
,....
10 100 1000
10000
It follows from this alternative representation and the definition of the fk that the function
fk1 , where k1 corresponds to
4
10
x1
10
will give a triangle spike centered at .x1 =
x1
10
of width
= .4. As we know that x − .x1 < .1, we know that x will be within .1 of the center of
this triangle, hence closer to the center of the spike than the outside. Thus, we have that
fk1 (x) ≥ 21 .
Similarly, we have that the width of the spike given by fk2 , where k2 corresponds to
and centered at .x1 x2 =
x1 x2
100
will have width
4
.
100
x1 x2
100
Again, as we know that x − .x1 x2 < .01,
we know that x will be within .01 of the center of this triangle, hence closer to the center of
the spike than the outside. Again, we may conclude that fk2 (x) ≥ 21 .
We can then extend this idea to fkn , where kn corresponds to
x1 x2 ...xn
,
10n
so that x will be
closer to the middle of the the triangle spike corresponding to the function fkn than to the
outside of this spike. Thus, we again have that fkn (x) ≥
1
.
2
As a result, we know that
the sequence {fk (x)} has infinitely many values above 12 , so we know that {fk (x)} has a
subsequence not converging to zero. However, we also know from the same reasons stated
in case 1 that {fk (x)} does have a subsequence converging to 0, so it follows that {fk (x)}
cannot be convergent. 9
Chapter 3
Exercise 3.3: Assume that φ is a continuous real function on (a, b) such that
x+y
1
1
φ
≤ φ(x) + φ(y)
2
2
2
for all x, y ∈ (a, b). Prove that φ is convex.
Proof: Let λ ∈ [0, 1]; then it follows that λ can be represented as some (possibly nonterminating) binary decimal number. As a result, we may write λ as
λ=
∞
X
dk
k=1
2k
,
where dk is either 1 or 0 for each k. Now define the sequence {an } by
( n
)
n
na o
X
X
d
d
k
n
k
=⇒
=
→ λ.
an = 2n
k
n
k
2
2
2
k=1
k=1
We shall now show via induction that
a
an an
an n
φ n x + 1 − n y ≤ n φ(x) + 1 − n φ(y)
2
2
2
2
for all n. Let x, y ∈ (a, b) with x < y. We first handle the base case, in which n = 1 and two
cases.
Case 1: a1 = d1 = 1. Then we immediately have that
a
a1 x+y
1
1
a1
a1 1
φ
x+ 1−
y =φ
≤ φ(x) + φ(y) = φ(x) + 1 −
φ(y)
2
2
2
2
2
2
2
by the given information.
Case 2: a1 = d1 = 0. Then we immediately have that
a
a1 a1
a1 1
y = φ (y) = φ(x) + 1 −
φ(y),
φ
x+ 1−
2
2
2
2
so the base case holds.
Now assume that
φ
a
n
x
n
2
an an
an + 1 − n y ≤ n φ(x) + 1 − n φ(y).
2
2
2
Note that we may without loss of generality assume that
an+1
2n
< 1. Now consider the following:
a
an+1 n+1
φ n+1 x + 1 − n+1 y
2
2
10
an+1
2n+1
n+1
< 1 − a2n+1
, which implies that
an+1
1
1
an+1 =φ
x + 1 − n+1 y − y + y
2n+1
2
2
2
1
an+1
1 an+1
− n+1 y + y
=φ
x+
n+1
2
2 2
2
1 an+1
1
an+1
=φ
x+ 1− n y + y
2 2n
2
2
1
an+1
1
an+1
≤ φ
x + 1 − n y + φ(y)
n
2
2
2
2
1
1 an+1
an+1 ≤
φ(y)
+ φ(y)
φ(x)
+
1
−
2 2n
2n
2
an+1
an+1
= n+1 φ(x) + 1 − n+1 φ(y)
2
2
Add and subtract
Combine terms
Factor
Given
Inductive hypothesis
Simplify
So our induction holds (note that the inductive hypothesis does apply here because we
< 1). By the continuity of φ we then have that
without loss of generality had an+1
2n
a
an+1 an+1
an+1 n+1
lim φ n+1 x + 1 − n+1 y ≤ lim n+1 φ(x) + 1 − n+1 φ(y)
n→∞
n→∞ 2
2
2
2
implies that
lim φ (λx + (1 − λ) y) ≤ lim λφ(x) + (1 − λ) φ(y).
n→∞
n→∞
Thus, we have that φ is convex. Exercise 3.19: Define the essential range of a function f ∈ L∞ (µ) to be the set Rf consisting
of all complex numbers w such that
µ({x : |f (x) − w| < }) > 0
for every > 0. Prove that Rf is compact. What relation exists between the set Rf and the
number kf k∞ ? Let Af be the set of all averages
Z
1
f dµ
µ(E) E
where E ∈ M and µ(E) > 0. What relations exist between Af and Rf ? Is Af always closed?
Are there measures µ such that Af is convex for every f ∈ L∞ (µ)? Are there measures µ
such that Af fails to be convex for some f ∈ L∞ (µ)? How are these results affected if L∞ (µ)
is replaced by L1 (µ), for instance?
Proof: It is helpful to note that an alternative definition for the essential range of a function
f is
Rf =
\
{w | µ({x : |f (x) − w| < }) > 0}.
>0
11
As f ∈ L∞ , we have that f is essentially bounded by some M ∈ R.
We first show that Rf is bounded, so let w ∈ C such that |w| > M . Then there exists > 0
such that for all z ∈ B (w) ∩ f (X) we have |z| > M . But as we know that M is the essential
bound of f , it follows that µ(f −1 (B (w))) = 0, so then z 6∈ Rf .
We now show that Rf is closed. Let {wn } be some convergent sequence in Rf with limit
w and let > 0. Consider f −1 (B (w)) = {x : |f (x) − w| < }. Choose N such that for
all n > N we have wn ∈ B (w). But then B (w) open implies that there exists some δ
such that Bδ (wn ) ⊆ B (w), so f −1 (Bδ (wn )) ⊆ f −1 (B (w)). But then wn ∈ Rf implies that
0 < m(f −1 (Bδ (wn ))) ≤ m(f −1 (B (w))). As this is true for all , we have that w ∈ Rf .
As we have that Rf ⊆ C and Rf is closed and bounded, it follows that Rf is compact.
Additionally, we have that ess sup(f ) = max{|w| : w ∈ Rf }.
We claim that the closure of Af is contained in the convex hull of Rf (that is, the smallest
convex set containing Rf ). First let f ∈ L∞ with f ∈ Cc (X) and define m = inf f and
M = sup f , which exist because f ∈ L∞ . Further, we know m, M ∈ Rf because Rf is
compact and f is continuous. Now let x ∈ Af , and note m ≤ x ≤ M , so there exists some
λ such that x = λm + (1 − λ)M . As a result, we have confirmed that Af is contained in
the convex hull of Rf . It follows from the compactness of Rf that the convex hull of Rf will
be closed, so we have that the closure of Af must be contained in the convex hull of Rf .
Finally, it follows from the definition of complex integration and the desity of Cc (X) in L∞
that this result can be generalized to any f ∈ L∞ .
Af is not always closed. Consider the example f (x) =
1
x
on (1, ∞) with the Lebesgue mea-
sure. The averages of this function over larger and larger intervals will only asymptotically
approach zero, as any integral of this function on an unbounded set is itself unbounded.
If we let X be the one point space with the counting measure, any function on X is necessarily
constant, and Rf and Af will equal that constant. It is then immediate that Af is convex.
However, if we let X = {0, 1} (the two-point space) with the counting measure and f (x) = x,
then Af = {0, 21 , 1}, so Af is certainly not convex.
Replacing L∞ with L1 allows for functions such as

x −1
2
0<x≤1
f (x) =
.
0
x>1
This immediately gives that both Rf and Af are unbounded, which deteriorates many of
our earlier results, including the compactness of Rf . 12
Chapter 4
Exercise 4.2: Let {xn : n = 1, 2, 3, . . .} be a linearly independent set of vectors in H. Show
that the following construction yields an orthonormal set {un } such that {x1 , . . . , xN } and
{u1 , . . . , uN } have the same span for all N . Put u1 = x1 /kx1 k. Having u1 , . . . , un−1 define
vn = xn −
n−1
X
(xn , ui )ui ,
un = vn /kvn k.
i=1
Note that this leads to a proof of the existence of a maximal orthonormal set in separable
Hilbert spaces which makes no appeal
to the Hausdorff maximality principle.
vn kvn k
Proof: First note that kun k = kvn k = kvn k = 1, so we have that un is normal.
Note that it suffices to show (un , um ) = 0 for all m < n to prove that {un } is an orthogonal
set. Also note that
(un , um ) = 0 ⇐⇒ kvn k(un , um ) = 0 ⇐⇒ (kvn kun , um ) = 0 ⇐⇒ (vn , um ) = 0,
so it suffices to show that (vn , um ) = 0 for all m < n, which we shall shall do inductively.
First consider the case in which n = 2 (as there is nothing to show in the n = 1 case). Then
we need only observe
x1
x1
x1
x1
(v2 , u1 ) = (x2 − (x2 , u1 )u1 , u1 ) = x2 − x2 ,
,
= x2 − 0,
= 0.
kx1 k kx1 k kx1 k
kx1 k
Now assume that for all j ≤ k we have m < j implies (vj , um ) = 0. Let j ≤ k + 1 and
consider the following:
(vk + 1, uj ) =
k
X
xk+1 −
(xk+1 , ui )ui , uj
!
i=1
= (xk+1 , uj ) −
k
X
(xk+1 , ui )(ui , uj ) Properties of an inner product
i=1
= (xk+1 , uj ) − (xk+1 , uj )(uj , uj )
Inductive hypothesis: i 6= j =⇒ (ui , uj ) = 0
= (xk+1 , uj ) − (xk+1 , uj )kuj k
Definition of the norm
= (xk+1 , uj ) − (xk+1 , uj )
uj is normal
=0
Simplify
Thus, our induction holds.
As uN is constructed from the xi with 1 ≤ i ≤ N , we have that uN ∈
definition of the ui also gives that
N
−1
−1
N
X
X
(xn , ui )ui +
(xN , ui )ui = xN ,
xN −
i=1
i=1
13
W
{x1 , . . . , xN } . The
so we may conclude that xN ∈
W
{u1 , . . . , uN }. As a result, we have that {x1 , . . . , xN } and
{u1 , . . . , uN } have the same span for all N . Exercise 4.4: Show that H is separable if and only if H contains a maximal orthonormal
system which is at most countable.
Proof: (⇒) Assume that H is separable, which by definition implies the existence of some
countable dense subset D ⊆ H. Let X be the maximal linearly independent subset of D.
Then it follows from the previous exercise that there exists some maximal orthonormal set
U corresponding to X that is at most countable.
(⇐) Assume that H contains a maximal orthonormal system which is at most countable,
call it {uα }. Then it follows from Theorem 4.18 that the set P of all finite linear combinations of elements in {uα } is dense in H. Now define P 0 to be the set of all finite linear
combinations of elements in {uα } with rational coefficients in C, and note that P 0 is dense
in P . Additionally, P 0 is a countable set, as it is the set of all finite linear combinations of
a countable set with coefficients in a countable set. Finally, P 0 is dense in P and P is dense
in H, so we may conclude that P 0 is dense in H. Hence, H is separable. Exercise 4.6: Let {un } (n = 1, 2, 3, . . .) be an orthonormal set in H. Show that this gives
an example of a closed and bounded set which is not compact. Let Q be the set of all x ∈ H
of the form
x=
∞
X
cn u n
(where |cn | ≤
1
1
).
n
Prove that Q is compact. (Q is called the Hilbert cube.) More generally, let {δn } be a
sequence of positive numbers, and let S be the set of all x ∈ H of the form
x=
∞
X
cn u n
(where |cn | ≤ δn ).
1
Prove that S is compact if and only if
P∞
1
δn2 < ∞. Prove that H is not locally compact.
Proof: As {un } is an normal set, it is clear that {un } is bounded (in particular, every element
has norm 1). Let d represent the metric on H, and let P be the set spanned by finite linear
combinations of {un }. It then follows from the Riesz-Fisher theorem that P isometric to
√
`2 (N), so we have from the definition of the `2 (N) metric that d(un , um ) = 2 for all n 6= m.
As H is a metric space, we can then construct the disjoint set B = {B√2/3 (un )}, which
we note also covers H. Since each set in B contains exactly one un , it follows that {un } is
14
necessarily closed. Note that this fact also implies that no proper subset of B covers {un }, so
we may conclude from the fact that {un } is countably infinite that B has no finite subcover.
Thus, we have that {un } is not compact.
We shall handle the compactness of the usual Hilbert cube as a special case of the generalized
Hilbert cube, see the work below.
(⇒) Assume that S is compact, which implies that S is bounded. Riesz-Fisher allows us to
isometrically embed S into `2 , so we may once again utilize the norm in `2 . Note that the
P
point x = ∞
n=1 δn un is contained in S. Then we have that
kxk =
∞
X
! 21
|δn |2
.
n=1
But as we know that S must be bounded, it follows that
∞
X
! 21
|δn |2
< ∞ =⇒
n=1
∞
X
δn < ∞,
n=1
as desired.
(⇐) Assume that
P∞
n=1 δn
< ∞. We then define the map φ : CN → H by
(c1 , c2 , . . .) 7→
∞
X
cn u n
n=1
Define the set R = {(c1 , c2 , . . .) ∈ CN | |cn | ≤ δn }. Using the notation Dδn to represent the
closed disk of radius δn in C, it follows from the definition of R that
R=
∞
Y
D δn .
n=1
As each Dδn is closed and bounded in C, we have that each Dδn is compact. Then it follows
that R is a product of compact sets, so Tychonoff’s theorem implies that R will be compact
in the product topology on CN . It is clear from the definition of φ that φ maps R bijectively
to S, and we claim that φ is continuous on this subset of CN . Let c = (c1 , c2 , . . .) ∈ R and
let {c(k) } be a sequence of points in R converging to c (it is worth noting that I am using
upper-scripts here to denote the elements of my sequence, as I have reserved the subscripts
to represent the components in each point). Then we have that c(k) → c component-wise,
that is
(k)
{c1 } → c1 ,
(k)
{c2 } → c2 ,
15
(k)
{c3 } → c3 , . . . .
Note that the image of this sequence will be given by
)
(∞
X
.
c(k)
n un
n=1
So then our component-wise convergence implies that
(∞
)
∞
X
X
(k)
cn un →
cn un .
n=1
n=1
provided that the limit point exists. However, we know that this sum (hence the limit point)
exists, as the cn were bounded above by the square-summable sequence {δn }. Further, we
can see that
φ(c1 , c2 , . . .) =
∞
X
n=1
cn un =⇒ lim f ({c(k) }) = f ( lim {c(k) }),
k→∞
k→∞
so we have that φ is continuous on R. As R is a compact set, we then have that f (R) = S
must be compact. As Q is simply a special case for S in which δn = n1 , we have also shown
that Q is compact.
Finally, we show that H is not locally compact. Let U be an open neighborhood in H
containing the origin, and we claim that U is not compact. As U is open, there exists some
open ball B (0) ⊆ U with > 0. Note that B (0) ⊆ U and that B (0) contains the sequence
√
{xn } = {u1 , u2 , u3 , . . .}. It again follows from our metric that d(ui , uj ) = 2 for all
i 6= j. As a result, {xn } cannot contain any Cauchy subsequences, so {xn } cannot have a
convergent subsequence. Thus, U contains a sequence with no convergent subsequence, so
U is not sequentially compact. But sequential compactness and compactness are equivalent
because H is a metric space, so it follows that U cannot be compact. Hence, H is not locally
compact. Exercise 4.13: Suppose f is a continuous function on R1 , with period 1. Prove that
Z 1
N
1 X
lim
f (nα) =
f (t) dt
N →∞ N
0
n=1
for every irrational real number α.
Proof: My initial attack on this problem was founded upon some recent intuition acquired
in my differential equations class. While it ultimately lead to somewhat of an impasse, I feel
like it’s worth taking a brief moment to detail my approach:
16
As f (x + 1) = x, we may assume that 0 ≤ α < 1. In a previous differential equations
assignment required us to show that X = {nα mod 1 | n ∈ Z} was dense in [0, 1). From this,
it is not challenging to show that f (X) is dense in f ([0, 1]). (It is also worth noting that
a values of any continuous function on a dense set necessarily determine the values of the
function on the space containing the dense set.) Next, we took {xn } to be an enumeration
of X and hoped that this would quickly lead us to our result. Unfortunately, this requires
that X be equidistributed in [0, 1], which is defined to be
|{x1 , x2 , . . . , xn } ∩ [a, b]|
=b−a
n→∞
n
lim
however, proving appeared to be impossible, as we do not seem to have much control over
the orbit of α. An equivalent definition called uniformly distributed could be substituted,
but this is actually the result Rudin is asking us to show. As a result, we abandoned this
approach in favor of one using things we actually learned in this chapter.
We begin by performing a change of variables. Let 2πt = s; then we may define f˜(s) =
f ( s ) = f (t). Note that this implies that f˜(s) has a period of 2π, and dt = ds . Applying
2π
2π
this change to the right-hand side of our original expression allows us to assemble all these
pieces as
Z
1
Z
f (t) dt =
0
0
2π
ds
1
f˜(s)
=
2π
2π
Z
π
f˜(s) ds.
−π
Note that the last step is legimiate because f˜ has period 2π. Additionally, we can use the
definition of f˜ to write
N
N
1 X˜
1 X
lim
f (2πnα) = lim
f (nα).
N →∞ N
N →∞ N
n=1
n=1
We now move to the hint, so let f (t) = e2πikt , with k ∈ Z; note that then f˜(s) = eiks . There
are two cases:
Case 1: k = 0.
We first consider the right-hand side:
Z π
Z π
1
1
1
iks
e ds =
1 ds =
2π = 1.
2π −π
2π −π
2π
And we now turn our attention to the left-hand side:
N
N
1 X 2πiknα
1 X
1
e
= lim
1 = lim N = 1.
n→∞ N
n→∞ N
N →∞ N
n=1
n=1
lim
17
So the desired result holds for f (t) = e2πikt when k = 0.
Case 2: k 6= 0.
We first consider the right-hand side:
π
Z π
1
eiks 1
iks
e ds =
(1 − 1) = 0.
=
2π −π
2πik −π 2πik
And we now turn our attention to the left-hand side, where using the formula for a finite
geometric sum yields
N
N
1 X 2πikαn
1 X 2πikα n
e2πikα 1 − (e2πikα )N
lim
e
= lim
(e
) = lim
.
N →∞ N
N →∞ N
N →∞
N
1 − e2πikα
n=1
n=1
It is crucial here to note that if kα = j ∈ Z then we have that α = kj , which is a contradiction. As a result, we know that the denominator in the above expression must be nonzero.
Continuing this manipulation:
e2πikα 1 − (e2πikα )N
e2πikα e2πikαN − 1
lim
= lim
N →∞
N →∞
N
1 − e2πikα
N
e2πikα − 1
e2πikα
1
lim
(1 − e2πikαN )
= 2πikα
e
− 1 N →∞ N
We then take the norm to observe the following:
2πikα
2πikα e
e
1
2πikαN lim 1 1 − e2πikαN lim
(1
−
e
)
=
e2πikα − 1 N →∞ N
e2πikα − 1 N →∞ N
2πikα e
lim 1 (1 + |e2πikαN |)
≤ 2πikα
e
− 1 N →∞ N
2πikα e
lim 1 (1 + 1)
= 2πikα
e
− 1 N →∞ N
2πikα 2πikα e
e
2
= 2πikα
lim
=
e2πikα − 1 · 0 = 0
e
− 1 N →∞ N
So we have that the result does indeed hold in this case as well, for both sides equal 0.
If f˜ is a trignometric polynomial, we are have our result via the hint from earlier (as our
change of variable shows that we have in fact proved this for the trigonometric system).
Fortunately, as any f˜ is both continuous and has period 2π, we can write f˜ as the uniform
limit of some sequence {fi } of trigonometric polynomials. It then follows from our initial
work that and our change of variables
Z π
N
1 X
1
lim
fi (2πnα) =
fi (s) ds
N →∞ N
2π
−π
n=1
18
for all i. Then we can apply this uniform convergence as follows:
N
N
N
N
1 X˜
1 X
1 X
1 X
f (nα) = lim
f (2πnα) = lim
lim fi (2πnα) = lim lim
fi (2πnα)
i→∞
i→∞ N →∞ N
N →∞ N
N →∞ N
N →∞ N
n=1
n=1
n=1
n=1
Z 1
Z π
Z π
Z π
1
1
1
= lim
f (t) dt
fi (s) ds =
lim fi (s) ds =
f˜(s) ds =
i→∞ 2π −π
2π −π i→∞
2π −π
0
lim
Thus, we have the desired result. Exercise 4.14: Compute
Z
1
|x3 − a − bx − cx2 |2 dx
min
a,b,c
−1
and find
Z
1
x3 g(x) dx,
max
−1
where g is subject to the restrictions
Z
Z 1
Z 1
xg(x) dx =
g(x) dx =
2
Z
1
x g(x) dx = 0;
|g(x)|2 dx = 1.
−1
−1
−1
−1
1
Proof: We begin by manipulating the given integral:
Z 1
min
|x3 − a − bx − cx2 |2 dx
a,b,c −1
Z 1
= min
(x3 − a − bx − cx2 )2 dx
a,b,c −1
Z 1
x6 − 2cx5 + (−2b + c2 )x4 + (−2a + 2bc)x3 + (2ac + b2 )x2 + 2abx + a2 dx
= min
a,b,c
−1
= min 2a2 +
a,b,c
4ac 2b2 4b 2c2 2
+
−
+
+
3
3
5
5
7
We can then define the function
F (a, b, c) = 2a2 +
4ac 2b2 4b 2c2 2
+
−
+
+ .
3
3
5
5
7
This allows us to calculate
∂F
4c
= 4a + ,
∂a
3
∂F
4b 4
=
− , and
∂b
3
5
∂F
4a 4c
=
+ .
∂c
3
5
We can then solve the system
0 = 4a +
4c
,
3
0=
4b 4
4a 4c
− , and 0 =
+
3
5
3
5
19
to determine that the only relative extremum of F occurs at a = 0, b = 53 , and c = 0. We
then compute

∂2F
2
 ∂a2
∂ F
 ∂b∂a
∂2F
∂c∂a
∂2F
∂a∂b
∂2F
∂b2
∂2F
∂c∂b

∂2F
∂a∂c 
∂2F 
∂b∂c 
∂2F
∂c2

4 0

=
0
4
3
4
3
0

4
3
0
,
4
5
which has eigenvalues
√
36 + 4 61
,
λ1 =
15
√
4
36 − 4 61
λ2 = , and λ3 =
.
3
15
As λ1 , λ2 , λ3 > 0, we may conclude that the extrema at (0, 53 , 0) is a relative minimum. As
a result, we have that
Z
min
a,b,c
3 3 2
x − x dx = 8 .
|x − a − bx − cx | dx =
5 175
−1
−1
1
3
Z
2 2
1
We now define H = L2 ([−1, 1]) with the inner product
Z 1
(f, g) =
f (x)g(x)dx.
−1
Also define M =
W
{x3 , x2 , x, 1}, which implies that H = M ⊕ M ⊥ . As a result, we have
without loss of generality that g(x) will be of the form
g(x) = α[x3 − a − bx − cx2 ] + βf (x)
for some scalars a, b, c, α and f (x) ∈ M ⊥ . Note that this implies
(1, f (x)) = (x, f (x)) = (x2 , f (x)) = (x3 , f (x)) = 0.
We then wish to maximize the following:
(g(x), x3 ) = (α[x3 − a − bx − cx2 ] + βf (x), x3 )
= α[(x3 , x3 ) − (a, x3 ) − (bx, x3 ) − (cx2 , x3 )] + β(f (x), x3 )
Recall that (f (x), x3 ) = 0 and note that
Z 1
Z
2
3
5
3
(cx , x ) =
cx dx = 0 and (a, x ) =
−1
1
ax3 dx = 0.
−1
So we may combine these facts with the work above to observe that
Z 1
Z 1
2 2
3
3
3
3
6
4
(g(x), x ) = α[(x , x ) − (bx, x )] = α
x dx −
bx dx = α
− b .
7 5
−1
−1
20
It is important to note that this is then the expression we wish to maximize. As f (x) is
orthogonal to M and our additional constraints on g(x) are all based on elements of M , we
may assume that β = 0. Then g(x) must be of the form
g(x) = α[x3 − a − bx − cx2 ].
We can then transform these constraints by evaluating the inner products
0 = (g(x), 1) = −2a −
2c
,
3
0 = (g(x), x) =
2 2b
−2a 2c
− , and 0 =
− .
5
3
3
5
Solving this system of equations gives a = c = 0 and b = 35 . We have now determined that
g(x) = α[x3 − 35 x]. Applying the final condition that g(x) is normal gives
Z
1
1 = (g(x), g(x)) =
α
2
−1
Thus, we have that g(x) =
q
175
(x3
8
3
x −
5
3
2
8α2
=⇒ α =
dx =
175
r
175
.
8
− 53 x). As the satisfaction of our constraints has
completely determined g(x), we have that this particular g(x) must give the max of (g(x), x3 ).
For the sake of completeness, we calculate it here:
√
!
Z 1 r
2
175
3
14
(g(x), x3 ) =
x3 dx =
.
x3 − x
8
5
35
−1
21
Chapter 16
Exercise 16.1: Suppose f (z) =
P
an z n , an ≥ 0, and that the radius of convergence is 1.
Show that f has a singularity at z = 1.
Proof: Choose {bn }∞
n=0 such that f (z) =
bn (z − 21 )n . For clarity’s sake, let A(z) be the
P
original power series representation of f (z), and let B(z) be the new power series representation of f (z). Note that every point of the interior of D( 21 , 21 ) is contained in the interior of
D(0, 1), so we know that the series B(z) converges everywhere inside D( 12 , 12 ). As a result,
we have that the radius of convergence of B(z) is at least 21 . Additionally, we note that
∂D( 21 , 12 ) \ {1} is contained in the interior of D(0, 1), so we know that B(z) also converges
everywhere on ∂D( 21 , 12 ) \ {1}. There are then two distinct possibilities for the radius of
convergence of B(z).
Case 1: The radius of convergence of B(z) is equal to 12 . This then implies that B(z) has a
singularity at z = 1, so A(z) has a singularity at z = 1 and we are finished.
Case 2: The radius of convergence of B(z) is
1
2
+ for some > 0. This implies that B(z)
converges at z = 1, so we have that B(1) = c < ∞. We can also analytically continue f (z)
onto D(0, 1) ∪ D( 12 , 12 + ), which implies that z = 1 is also a regular point of f (z). Define
the series {zn } = {1 − n1 }, and note that {zn } → 1 and {zn } is contained in the interior of
D(0, 1). We can then define the sequence {B(zn )}, and note that {B(zn )} → c. Further, as
every zn is in the interior of D(0, 1), we have that B(zn ) = A(zn ) for all n, which implies
that {A(zn )} → c. We can then observe that
c = lim A(zn ) = lim
n→∞
n→∞
∞
X
k=0
ak znk = lim lim
n→∞ N →∞
N
X
ak znk = lim lim
k=0
= lim
N →∞
Hence,
so
P
P
N →∞ n→∞
N
X
k=0
ak lim
n→∞
znk
N
X
ak znk
k=0
= lim
N →∞
N
X
ak =
k=0
∞
X
ak .
k=0
an converges. But then we have from the fact that an ≥ 0 that
X
X
|an | =
an = c < ∞,
an is absolutely convergent. Now let z ∈ ∂D(0, 1) and observe that
X
X
X
X
|f (z)| = |
an z n | ≤
|an ||z|n =
|an | · 1 =
an = c < ∞,
so then
P
an z n is convergent at all z ∈ ∂D(0, 1). But then Theorem 16.2 implies that
the radius of convergence of A(z) must be strictly greater than 1, contradicting our given
information. Thus, we may conclude that Case 1 is the only possibility.
22
As a result, we have that f (z) has a singularity at z = 1. Exercise 16.4: Suppose X is the closed unit square in the plane and that f is a continuous
function on X that does not vanish at any point. Show that there is a continuous function
g on X such that f = eg . For what class of spaces X (other than the above square) is this
also true?
Proof: First define the path α : [0, 1] → C \ {0} such that α(0) = 1 and α(1) = f (0). That
is, α is an arbitrary path in C \ {0} from 1 to f (0). Our goal is to define some g : X → C
satisfying the requirements stated above. To start this, we let
Z 1
1
g(0) =
dt,
0 α(t)
and note that this implies that f (0) = eg(0) . For all ω ∈ X we let βω : [0, 1] → X be such
that βω (0) = 0 and βω (1) = ω; specifically, let βω be the straight-line path. Now define
γω : [0, 1] → C \ {0} by
γω (t) =

α(2t)
t≤
f (β (2t − 1))
ω
1
2
1
2
,
≤t
and note this is a path whose “second half” is contained entirely in f (X). We define g for
all ω ∈ X by
Z
1
g(ω) =
0
1
dt,
γω (t)
and it follows immediately from this definition that f (ω) = eg(ω) for all ω ∈ X. As a result,
it only remains to show that g is continuous.
Now let y, z ∈ X and define βy,z : [0, 1] → X be the straight-line path from y to z. We
then also define γy,z : [0, 1] → C \ {0} by γy,z = f (βy,z ). Let ∗ be the operator indicating
concatenation of paths. Note that our definitions then give that βz and βy ∗ βy,z will be
paths from 0 to z contained in X. As X is simply connected, we may conclude that βz
and βy ∗ βy,z are homotopic on X, which we indicate as βz ∼ βy ∗ βy,z . This means that
the homotopy from βz to βy ∗ βy,z takes place entirely in X. Combining this fact with the
continuity of f then implies that the image of the homotopic paths must be homotopic;
specifically, f (βz ) ∼ f (βy ∗ βy,z ) in f (X). As the first half of γω (that is, α) is equal for any
choice of ω, we may then conclude that γz ∼ γy ∗ γy,z in f (X) ∪ α([0, 1]). In particular, this
implies that any integral over the path γz is equal to any integral over the path γy ∗ γy,z .
As X is compact and f is nowhere zero, there exists some m = minz∈X |f (z)| > 0. Let η > 0,
and let > 0 such that < ηm, which implies
23
m
< η. We have from the continuity f there
exists some δ > 0 such that for y, z ∈ X with |y − z| < δ that length(γy,z ) < . Consider the
following for |y − z| < δ:
Z 1
Z 1
1
1
|g(z) − g(y)| = dt −
dt
γz (t)
0 γy (t)
Z0 1
Z 1
1
1
dt −
dt
=
γ ∗ γy (t)
0 γy (t)
Z0 y,z
Z
1
1
=
dt −
dt
γy,z ∗γy t
t
γy
Z
Z
Z
1
1
1
dt +
dt −
dt
=
γy,z t
t
t
γy
γy
Z
1 =
dt
γy,z t 1
≤ length(γy,z )
m
≤
<η
m
Definition of g
γz ∼ γy ∗ γy,z
Definition of a curve integral
Properties of curve integrals
Simplify
Properties of curve integrals
Substitute and simplify
Thus, we may conclude that g is continuous, as desired. We can expand this result to any
space that is compact and simply connected, as those were the only properties of X used
above. Exercise 16.8: Let E be a compact subset of the real axis, of positive Lebesgue measure,
let Ω be the complement of E, relative to the plane, and define
Z
dt
(z ∈ Ω).
f (z) =
E t−z
Answer the following questions:
1. Is f constant?
2. Can f be extended to an entire function?
3. Does lim zf (z) exist as z → ∞?
4. Does f have a holomorphic square root in Ω?
5. Is the real part of f bounded in Ω? If so, give a bound.
6. Is the imaginary part of f bounded in Ω? If so, give a bound.
24
7. What is
R
γ
f (z) dz if γ is a positively oriented circle which has E on its interior?
8. Does there exist a bounded holomorphic function ϕ in Ω which is not constant?
Proof: (a) We claim that f (z) is nonconstant, and we aim to show this by demonstrating
that f (z1 ) − f (z2 ) for some z1 , z2 . Begin by defining
M = max t,
and m = min t,
t∈E
t∈E
which both exist by the compactness of E. Next, pick z1 , z2 ∈ R such that z1 < z2 < m, and
note that z1 − z2 6= 0. (It is helpful to note here that all the inequalities appearing in (a)
are indeed legitimate, as we have chosen z1 , z2 to be real.) Then we observe that
Z
Z
Z
1
z1 − z2
1
1
f (z1 )−f (z2 ) =
−
dt =
dt = (z1 −z2 )
dt.
t − z2
E t − z1
E (t − z1 )(t − z2 )
E (t − z1 )(t − z2 )
Now note that
1
1
≥
t − z1
M − z1
and
1
1
≥
,
t − z2
M − z2
so we define
d1 =
Then we have that
Z
(z1 − z2 )
E
1
1
6= 0 and d2 =
6= 0.
M − z1
M − z2
1
dt ≥ (z1 − z2 )
(t − z1 )(t − z2 )
Z
d1 d2 dt = (z1 − z2 )d1 d2 m(E) > 0,
E
as E was chosen to have positive Lebesgue measure. Thus, it follows that f (z) is nonconstant.
(b) We claim that f (z) cannot be extended to an entire function, and we aim to show this
by demonstrating that f (z) blows up as z approaches M from above on the real axis. Via
the monotone convergence theorem we may observe that
Z
Z
1
1
lim+ f (z) = lim+
dt =
lim+
dt → −∞.
z→M
z→M
E t−z
E z→M t − z
Although we have that M ∈
/ Ω, we can see that f (z) blows up as we approach M from above
on the real axis. As a result, we may conclude that any entire extension F (z) of f would
blow up identically on the same limit, implying that F (z) is undefined at z = M , which
contradicts the fact that F (z) is entire. Thus, we have f cannot be extended to an entire
function.
(c) We claim that the limit does indeed exist, and is −m(E). To show this we simply observe
that
Z
lim zf (z) = lim
z→∞
z→∞
E
−z
dt =
z−t
−z
lim
dt =
E z→∞ z − t
Z
25
Z
−1 dt = −m(E).
E
It is crucial here to note that we may only take the limit into the integral because E is
compact and
−z
z−t
→ −1 uniformly. To confirm the second statement we must show that
−z
→ 0.
−
(−1)
z − t
∞
First observe that
−z
+
1
z − t
∞
−z + (z − t) −t .
= sup = sup
z−t
t∈E
t∈E z − t
To find this supremum we define
gz (t) =
−t
−z
=⇒ gz0 (t) =
< 0.
z−t
(z − t)2
Thus, we may conclude that for sufficiently large z (specifically, z > M ) we have that
sup gz (t) = gz (M ).
t∈E
Substitution then yields
−t −M → 0,
=
sup z
−
t
z
−
M
t∈E
so we have that the convergence is indeed uniform. As a result, our work above with the
integral is justified, which gives that the limit in question is −m(E).
(d) We claim that f does not have a holomorphic square root in Ω, and we aim to show this
by building a contradiction with Theorem 13.11.
1
f
We claim that f ∈ H(Ω) and
d
d
f (z) =
dz
dz
Z
E
∈ H(Ω). This can be seen via the Leibniz rule as
1
dt =
t−z
Z
E
∂
(t − z)−1 dt =
∂z
Z
E
−1
dt.
(t − z)2
Note that this manipulation is legitimate, as the partial derivative of the integrand is still
continuous (smooth, in fact) and the integral still exists, all because Ω ∩ E = ∅. Thus,
we may conclude that f (z) ∈ H(Ω). As it is clear that f (z) 6= 0, we then have that both
f, f1 ∈ H(Ω). By way of contradiction suppose that a holomorphic square root exists for f .
Then Theorem 13.11 implies that Ω is simply connected, which is certainly false, hence a
contradiction. Thus, we have that f cannot have a holomorphic square root.
(e) We claim that the real part of f is not bounded in Ω, and while we have essentially shown
this in (b), we will provide an alternate method of showing this here, as it will be helpful in
(f).
26
Let z = a + bi, then we have that
Z
Z
1
1
(t − a) + bi
f (z) =
dt =
dt
E t − (a + bi)
E (t − a) − bi (t − a) + bi
Z
Z
Z
t−a
b
(t − a) − bi
=
dt =
dt + i
dt.
2
2
2
2
2
2
E (t − a) + b
E (t − a) + b
E (t − a) + b
Thus, we may conclude that
Z
Re(f (z)) =
E
t−a
dt.
(t − a)2 + b2
It follows from the definition of E that there exists some t0 ∈ E such that m((t0 −, t0 +)) > 0
for all > 0. Now define the sequence of points in C given by {zn } = {t0 + n1 i} and define
the sequence of intervals {In } = (t0 − n1 , t0 + n1 ). We can then observe that
Z
Z
t
−
t
t − t0
0
≤ lim lim Re(f (zn )) = lim dt
1
2
2
n→∞
n→∞
n→∞
E (t − t0 ) + n2
E∩In (t − t0 ) +
Note that t → t0 because In → {t0 }. Then we may conclude that (t − t0 )2 +
1
n2
1
n2
dt
→ 0 faster
than t − t0 → 0, so we have that the integrand approaches infinity. As a result, we have that
Re(f ) is unbounded.
(f) We claim that the imaginary part of f is bounded in Ω, and we will show this using some
of our work from (e). First note that for z = a + bi our work above gives
Z
b
Im(f (z)) =
dt.
2
2
E (t − a) + b
Then we observe the following:
Z ∞
Z
b
b
dt
(t − a)2 + b2 dt ≤ 2
2
(t − a) + b
E
−∞ ∞ t
−
a
= arctan
b
−∞
t−a
t−a
= lim arctan
− lim arctan
t→−∞
t→−∞
b
b
π −π
= −
=π
2
2
Thus, we may conclude that |Im(f (z))| ≤ π.
(g) We claim that such an integral evaluates to m(E).
Let γ represent the circle, and note that there exists c = minz∈γ,t∈E |t−z| by the compactness
of γ and E. Then we have that
Z Z
Z Z
1 1
m(E)`(γ)
1
1
< ∀z, t =⇒ =
dt,
dz
≤
dt
dz
< ∞.
t − z c
t
−
z
c
c
γ E
γ E
27
As a result, we may apply Fubini’s theorem to see that
Z Z
Z
Z
Z Z
1
1
dt dz =
dz dt =
Indγ (t) dt =
1 dt = m(E),
γ E t−z
E γ t−z
E
E
as claimed.
(h) We claim yes, and we shall show this via construction.
It follows from our work in (e) and (f) that
f (Ω) ⊆ {a + bi ∈ C | a, b ∈ R, |b| < π}.
Define the map g(z) = iz, and note g is simply a rotation by π2 . We can thus conclude that
g(f (Ω)) ⊆ {a + bi ∈ C | a, b ∈ R, |a| < π}.
Now define the map h(z) = ez , and note that
h(g(f (Ω))) ⊆ {reθi ∈ C | r, θ ∈ R, e−π ≤ r ≤ eπ }.
That is, h maps the vertical strip between −π and π to the annulus with inner radius e−π
and outer radius eπ . Define the composition function F (z) = h ◦ g ◦ f . It follows from our
work in (a) and the definitions of g and h that F is nonconstant. Note that we showed
f ∈ H(Ω) in (d). As h and g are clearly holomorphic via their definitions, we also have that
F is holomorphic on Ω. Exercise 16.9: Check your answers in Exercise 8 against the special case
E = [−1, 1].
Proof: (a) Observe that
Z
1
f (−2) =
−1
and
Z
1
f (−3) =
−1
1
1
dt = ln t + 2 = ln 3 − ln 1 = ln 3,
t+2
−1
1
1
dt = ln t + 3 = ln 4 − ln 2 = ln 2.
t+3
−1
Clearly f (−2) = ln 3 6= ln 2 = f (−3), so we have that f (z) is nonconstant.
(b) Here we use the same argument as in 8(b), simply specifying to
Z 1
1
lim+ f (z) = lim+
dt → ∞.
z→1
z→1
−1 t − z
28
(c) To check our work here, we can numerically evaluate the limit with a tool such as a TI-89
(which is my preference) or Wolfram Alpha, which yields
Z 1
z
lim zf (z) = limz → ∞
dt = −2.
z→∞
−1 z − t
We note here that −m([−1, 1]) = −2, confirming our findings from 8(c).
(d) Our general result from 8(d) still holds here, so I’m not sure what Rudin intends for us
to do more concretely show a holomorphic square root does not exist in this specific case.
(e) Our work in (b) has already demonstrated that Re(f (z)) is unbounded.
(f) The expansion to an integral over the entirety of R is unnecessary in this special case, so
picking up in the middle of our work in 8(f) gives
1
Z 1
b
t − a 1−a
−1 − a
|Im(f (z))| ≤
dt = arctan
= arctan
− arctan
2
2
b −1
b
b
−1 (t − a) + b
1 a
−1 a
−
−
= arctan
− arctan
.
b
b
b
b
Note that as b → 0 we then have
1 a
−1 a
π −π
−
−
= π,
arctan
− arctan
= −
b
b
b
b
2
2
so our bound is exactly the same as the general one found in 8(f).
(g) A specific calculation here would look exactly like our work in 8(g), simply replacing
every E with [−1, 1]. That being said, the value of the integral will be m([−1, 1]) = 2.
(h) Again, our work here would exactly match that in 8(h). We would be able to construct a
nonconstant holomorphic function F on Ω by using precisely the same g and h and defining
F as F = h ◦ g ◦ f . 29