Problems and Solutions in Complex and Real Analysis

8/2/2019 Problems and Solutions in Complex and Real Analysis

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Problems and Solutions inREAL AND COMPLEX ANALYSIS

William J. DeMeo

July 9, 2010

c William J. DeMeo. All rights reserved. This document may be copied for personal use. Permission to reproducethis document for other purposes may be obtained by emailing the author at [email protected].

Abstract

The pages that follow contain “unofficial” solutions to problems appearing on the comprehensive exams in

analysis given by the Mathematics Department at the University of Hawaii over the period from 1991 to 2007. I have

done my best to ensure that the solutions are clear and correct, and that the level of rigor is at least as high as that

expected of students taking the ph.d. exams. In solving many of these problems, I benefited enormously from the

wisdom and guidance of professors Tom Ramsey and Wayne Smith. Of course, some typos and mathematical errors

surely remain, for which I am solely responsible. Nonetheless, I hope this document will be of some use to you as

you prepare to take the comprehensive exams. Please email comments, suggestions, and corrections to

[email protected].

Contents

1 Real Analysis 3

1.1 1991 November 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 1994 November 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.3 1998 April 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.4 2000 November 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.5 2001 November 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.6 2004 April 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.7 2007 November 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2 Complex Analysis 38

2.1 1989 April . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2.2 1991 November 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

2.3 1995 April 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

2.4 2001 November 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

2.5 2004 April 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 592.6 2006 November 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

2.7 2007 April 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

2.8 2007 November 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

2.9 Some problems of a certain type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

1



A Miscellaneous Definitions and Theorems 71

A.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

A.1.1 Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

A.1.2 Measurable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71A.1.3 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

A.1.4 Approximating Integrable Functions1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

A.1.5 Absolute Continuity of Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

A.1.6 Absolute Continuity of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

A.1.7 Product Measures and the Fubini-Tonelli Theorem . . . . . . . . . . . . . . . . . . . . . . . 75

A.2 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

A.2.1 Cauchy’s Theorem2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

A.2.2 Maximum Modulus Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

B List of Symbols 79

2



1 REAL ANALYSIS

1 Real Analysis

1.1 1991 November 21

1. (a) Let f n be a sequence of continuous, real valued functions on [0, 1] which converges uniformly to f . Prove that

limn→∞ f n(xn) = f (1/2) for any sequence xn which converges to 1/2.

(b) Must the conclusion still hold if the convergence is only point-wise? Explain.

Solution: (a) Let xn be a sequence in [0, 1] with xn → 1/2 as n → ∞. Fix > 0 and let N 0 ∈ N be such

that n ≥ N 0 implies |f n(x) − f (x)| < /2, for all x ∈ [0, 1]. Let δ > 0 be such that |f (x) − f (y)| < /2,

for all x, y ∈ [0, 1] with |x − y| < δ. Finally, let N 1 ∈ N be such that n ≥ N 1 implies |xn − 1/2| < δ. Then

n ≥ maxN 0, N 1 implies |f n(xn) − f (1/2)| ≤ |f n(xn) − f (xn)| + |f (xn) − f (1/2)| < /2 + /2 = . (b) Suppose the convergence is only point-wise. Then the conclusion is false, as the following counterexample

demonstrates:

Define f n(x) to be the function

f (x) =

0, if 0 ≤ x < 12 − 1

2n ,

2nx − (n − 1), if 12 − 12n ≤ x < 1

2 ,

1, if 12 ≤ x ≤ 1.

(1)

That is, f n(x) is constantly zero for x less than 12 − 1

2n , then it increases linearly until it reaches one at x = 1/2,

and then it remains constantly one for x bigger than 1/2. Now define the sequence xn = 12 − 1

n . Then f n(xn) = 0

for all n ∈ N and xn → 1/2, while the sequence f n approaches the characteristic function f χ[ 12,1] which is one

on [12 , 1] and zero elsewhere. Therefore, f (1/2) = 1 = 0 = limn f n(xn).

2. Let f : R → R be differentiable and assume there is no x ∈ R such that f (x) = f (x) = 0. Show that

S = x | 0 ≤ x ≤ 1, f (x) = 0 is finite.

Solution: Consider f −1(0). Since 0 is closed and f continuous, f −1(0) is closed. Therefore S = [0, 1] ∩f −1(0) is a closed and bounded subset of R. Hence, S is compact. Assume, by way of contradiction, that S is

infinite. Then (by theorem A.1) there is a limit point x ∈ S ; i.e., there is a sequence xn of distinct points in S which converges to x. Also, as all points are in S , f (xn) = f (x) = 0 for all n ∈ N.

We now show that f (x) = 0, which will give us our desired contradiction. Since |xn − x| → 0, we can write the

derivative of f as follows:

f (x) = limn→∞

f (x + (xn − x)) − f (x)

xn − x= limn→∞

f (xn) − f (x)

xn − x= 0.

The last equality holds since f (x) = f (xn) = 0 holds for all n ∈ N.

3. If (X, Σ, µ) is a measure space and if f is µ integrable, show that for every > 0 there is E ∈ Σ such that

µ(E ) < ∞ and X\E

|f | dµ < .

3



1.1 1991 November 21 1 REAL ANALYSIS

Solution: For n = 1, 2, . . ., define

An = x ∈ X : 1/n ≤ |f (x)| < n.

Clearly,

A1 ⊆ A2 ⊆ · · · ↑ A

∞n=1

An

and each An is measurable (why?).3 Next, define

A0 = x ∈ X : f (x) = 0 and A∞ = x ∈ X : |f (x)| = ∞.

Then X = A0 ∪ A ∪ A∞ is a disjoint union, and X

|f | =

A0

|f | +

A

|f | +

A∞

|f | =

A

|f |. (2)

The first term in the middle expression is zero since f is zero on A0, and the third term is zero since f ∈ L1(µ)implies µ(A∞) = 0. To prove the result, then, we must find a measurable set E such that

A\E

|f | < , and

µ(E ) < ∞.

Define f n = |f |χAn . Then f n is a sequence of non-negative measurable functions and, for each x ∈ X ,limn→∞ f n(x) = |f (x)|χA(x). Since An ⊆ An+1, we have 0 ≤ f 1(x) ≤ f 2(x) ≤ · · · , so the monotone

convergence theorem4 implies X

f n → A

|f |, and, by (2),

limn→∞

An

|f | dµ =

A

|f | dµ =

X

|f | dµ.

Therefore, there is some N > 0 for which

X\AN

|f | dµ < .

Finally, note that 1/N ≤ |f | < N on AN , so

µ(AN ) ≤ N

AN

|f | dµ ≤ N

X

|f | dµ < ∞.

Therefore, the set E = AN meets the given criteria.

4.5 If (X, Σ, µ) is a measure space, f is a non-negative measurable function, and ν (E ) = E

f dµ, show that ν is a

measure.

Solution: Clearly µ(E ) = 0

⇒ν (E ) = 0. Therefore, ν (

∅) = µ(

∅) = 0. In particular ν is not identically infinity,

so we need only check countable additivity. Let E 1, E 2, . . . be a countable collection of disjoint measurable sets.

3Answer: f is measurable and x → |x| is continuous, so g = |f | is measurable. Therefore, An = g−1([1/n,n)) is measurable (theorem A.2).4Alternatively, we could have cited the dominated convergence theorem here since f n(x) ≤ |f (x)| (x ∈ X; n = 1, 2, . . .).5See also: Rudin [8], chapter 1. Thanks to Matt Chasse for pointing out a mistake in my original solution to this problem. I believe the solution

given here is correct, but the skeptical reader is encouraged to consult Rudin.

4




Then,

ν (

∪nE n) =

n En

f dµ = f χn En

dµ

=

∞n=1

f χEn dµ ( the E n are disjoint)

=∞n=1

f χEn dµ ( fχEn ≥ 0, n = 1, 2, . . .)

=

∞n=1

ν (E n).

The penultimate equality follows from the monotone convergence theorem applied to the sequence of non-negative

measurable functions gm =

mn=1 f χEn (m = 1, 2, . . .). (See also: April ’98, problem A.3.)

5. Suppose f is a bounded, real valued function on [0, 1]. Show that f is Lebesgue measurable if and only if

sup

ψ dm = inf

φ dm

where m is Lebesgue measure on [0, 1], and ψ and φ range over all simple functions, ψ ≤ f ≤ φ.

Solution: This is proposition 4.3 of Royden, 3ed. [6].

6.6 If f is Lebesgue integrable on [0, 1] and > 0, show that there is δ > 0 such that for all measurable sets E ⊂ [0, 1]

with m(E ) < δ , E

f dm

< .

Solution: This problem appears so often, I think it’s worth giving two different proofs. The first relies on the

frequently useful technique, employed in problem 3, in which the domain is written as a union of the nested sets

An = x ∈ X : 1/n ≤ |f (x)| < n. The second is a shorter proof, but it relies on a result about absolute

continuity of measures, which is almost equivalent to the original problem statement. I recommend that you learn

the first proof. The second proof is also worth studying, however, as it connects this result to the analogous result

about absolutely continuous measures.

Proof 1: Let An, n = 1, 2, . . . be the sequence of measurable sets defined in problem 3. That is,

An = x ∈ X : 1/n ≤ |f (x)| < n.

Here, X = [0, 1]. As we saw in problem 3,

limn→∞

An

|f | dm =

A

|f | dm =

X

|f | dm.

6See also: April ’92 (4), November ’97 (6), April ’03 (4).

5






integrable,

f φ = f (φ

− pn + pn)

≤

|f ||φ − pn| +

f pn

≤ f 1φ − pn∞ +

f pn

= f 1φ − pn∞. (3)

The last equality holds since

xnf = 0 for all n = 0, 1, 2, . . ., which implies that

f pn = 0 for all polynomials

pn. Finally, note that f 1 < ∞, since f is bounded and Lebesgue measurable on the bounded interval [0, 1].

Therefore, the right-hand side of (3) tends to zero as n tends to infinity. Since the left-hand side of (3) is independent

of n, we have thus shown that

f φ = 0 for any φ ∈ C [0, 1].

Now, since C [0, 1] is dense in L1[0, 1], let φn ⊂ C [0, 1] satisfy φn − f 1 → 0 as n → ∞. Then

0 ≤

f 2 =

f (f − φn + φn)

≤

|f ||f − φn| +

f φn

.

The second term on the right is zero by what we proved above. Therefore, if M is the bound on |f |, we have

0 ≤ f 2 ≤ M f − φn1 → 0. As

f 2 is independent of n, we have

f 2 = 0. Since f 2 ≥ 0, this implies

f 2 = 0 a.e., hence f = 0 a.e. Alternative Solution: Quinn Culver suggests shortening the proof by using the fact that polynomials are dense in

L1[0, 1]. Simply start from the line, “Now, since C [0, 1] is dense in L1[0, 1], let φn ⊂ C [0, 1] satisfy...” but

instead write, “Since Pol[0, 1] is dense in L1[0, 1], let φn ⊂ Pol[0, 1] satisfy...” This is a nice observation and

disposes of the problem quickly and efficiently. However, I have left the original, somewhat clumsy proof intact

because it provides a nice demonstration of the Stone-Weierstrass theorem (which appears on the exam syllabus),

and because everyone should know how to apply this fundamental theorem to problems of this sort.

8.10 If µ and ν are finite measures on the measurable space (X, Σ), show that there is a nonnegative measurable

function f on X such that for all E in Σ, E

(1 − f ) dµ =

E

f dν. (4)

Solution: There’s an assumption missing here: µ and ν must be positive measures.11 In fact, one can prove the

result is false without this assumption. So assume µ and ν are finite positive measures on the measurable space

(X, Σ). By the linearity property of the integral, and since µ(E ) = E

dµ, we have

E

(1

−f ) dµ =

E

dµ

− E

f dµ = µ(E )

− E

f dµ.

Therefore, (4) is equivalent to

µ(E ) =

E

f dµ +

E

f dν =

E

f d(µ + ν ) (∀ E ∈ Σ) (5)

10See also: November ’97 (7).11Note that a measure µ is called “positive” when it is, in fact, nonnegative; that is, µE ≥ 0 for all E ∈ Σ.

7




so this is what we will prove. The Radon-Nikodym theorem (A.12) says, if λ m are σ-finite positive measures

on a σ-algebra Σ, then there is a unique g ∈ L1(dm) such that

λ(E ) = E g dm, ∀E ∈ Σ.

In the present case, µ µ + ν (since the measures are positive), so the theorem provides an f ∈ L1(µ + ν ) such

that

µ(E ) =

E

f d(µ + ν ) ∀ E ∈ Σ,

which proves (5).

9.12 If f and g are integrable functions on (X, S , µ) and (Y, T , ν ), respectively, and F (x, y) = f (x) g(y), show that

F is integrable on X × Y and

F d(µ × ν ) =

f dµ

gdν.

Solution: 13 To show F (x, y) = f (x)g(y) is integrable, an important (but often overlooked) first step is to prove

that F (x, y) = f (x)g(y) is (S ⊗ T )-measurable. Define Ψ : X × Y → R×R by Ψ(x, y) = (f (x), g(y)), and let

Φ : R× R→ R be the continuous function Φ(s, t) = st. Then,

F (x, y) = f (x)g(y) = (Φ Ψ)(x, y).

Theorem A.2 states that a continuous function of a measurable function is measurable. Therefore, if we can show

that Ψ(x, y) is an (S ⊗T )-measurable function from X ×Y intoR×R, then it will follow that F (x, y) is (S ⊗T )-

measurable. To show Ψ is measurable, let R be an open rectangle in R×R. Then R = A × B for some open sets

A and B in R, and

Ψ−1(R) = Ψ−1(A × B)

= (x, y) : f (x) ∈ A, g(y) ∈ B

= (x, y) : f (x) ∈ A ∩ (x, y) : g(y) ∈ B= (f −1(A) × Y ) ∩ (X × g−1(B))

= f −1(A) × g−1(B).

Now, f −1(A) ∈ S and g−1(B) ∈ T , since f and g are S - and T -measurable, resp. Therefore, Ψ−1(R) ∈ S ⊗ T ,which proves the claim.

Now that we know F (x, y) = f (x)g(y) is (S⊗T )-measurable, we can apply part (b) of the Fubini-Tonelli theorem

(A.13) to prove that F (x, y) = f (x)g(y) is integrable if one of the iterated integrals of |F (x, y)| is finite. Indeed, X

Y

|f (x)g(y)| dν (y) dµ(x) =

X

Y

|f (x)||g(y)| dν (y) dµ(x)

=

X

|f (x)| Y

|g(y)| dν (y)

dµ(x)

= X

|f (x)| dµ(x) Y

|g(y)| dν (y) < ∞,

12See also: November ’97 (2), and others.13I’m not sure if the claim is true unless the measure spaces are σ-finite, so I’ll assume all measure spaces σ-finite.

In my opinion, the most useful version of the Fubini-Tonelli theorem is the one in Rudin [8], which assumes σ-finite measure spaces. There is a

version appearing in Royden [6] that does not require σ-finiteness. Instead it begins with the assumption that f is integrable. To me, the theorem

in Rudin is much easier to apply. All you need is a function that is measurable with respect to the product σ-algebra S ⊗ T , and from there, in a

single theorem, you get everything you need to answer any of the standard questions about integration with respect to a product measure.

8




which holds since f ∈ L1(µ) and g ∈ L1(ν ). The Fubini-Tonelli theorem then implies that F (x, y) ∈ L1(µ × ν ).

Finally, we must prove that

F d(µ×ν ) =

f dµ

g dν . Since F (x, y) ∈ L1(µ×ν ), part (c) of the Fubini-Tonelli

theorem asserts that φ(x) = Y F (x, y) dν (y) is defined almost everywhere, belongs to L1(µ), and, moreover, X×Y

F d(µ × ν ) =

X

Y

F (x, y) dν (y) dµ(x).

Therefore, X×Y

F d(µ × ν ) =

X

Y

f (x)g(y) dν (y) dµ(x)

=

X

f (x)

Y

g(y) dν (y) dµ(x)

=

X

f (x) dµ(x)

Y

g(y) dν (y).

9





Masters students: Do any 5 problems.

Ph.D. students: Do any 6 problems.

1. Let E be a normed linear space. Show that E is complete if and only if, whenever∞

1 xn < ∞, then∞

1 xnconverges to an s ∈ E .

Solution: Suppose E is complete. Let xn ⊂ E be absolutely convergent; i.e., xn < ∞. We must

∞n=1

xn := limN →∞

N n=1

xn = s ∈ E. (6)

Let S N =N n=1 xn. Then, for any j ∈ N,

S N +j − S N =N +jn=N +1

xn ≤

N +jn=N +1

xn → 0

as N → ∞, since xn < ∞. Therefore, S N is a Cauchy sequence. Since E is complete, there is an s ∈ E

such that∞n=1 xn = lim

N →∞S N = s.

Conversely, suppose whenever∞

1 xn < ∞, then∞

1 xn converges to an s ∈ E . Let yn ⊂ E be a Cauchy

sequence. That is, yn − ym → 0 as n, m → ∞. Let n1 < n2 < · · · be a subsequence such that

n, m ≥ nj ⇒ yn − ym < 2−j .

Next observe, for k > 1,

ynk = yn1 + (yn2 − yn1) + (yn3 − yn2) + · · · + (ynk − ynk−1) = yn1 +

k−1

j=1(ynj+1 − ynj ),

and∞j=1

ynj+1 − ynj <

∞j=1

2−j = 1

By hypothesis, this implies that

ynk − yn1 =

k−1j=1

(ynj+1 − ynj ) → s ∈ E,

as k → ∞. We have thus found a subsequence ynk ⊆ yn having a limit in E . Finally, since yn is Cauchy,

it is quite easy to verify that

yn

must converge to the same limit. This proves that every Cauchy sequence in E converges to a point in E .

2. Let f n be a sequence of real continuous functions on a compact Hausdorff space X . Show that if f 1 ≥ f 2 ≥ f 3 ≥· · · , and f n(x) → 0 for all x ∈ X , then f n → 0 uniformly.

10




3. Let f be integrable on the real line with respect to Lebesgue measure. Evaluate limn→∞

∞−∞

f (x − n)

x1+|x|

dx.

Justify all steps.

Solution: Fix n > 0. Consider the change of variables, y = x − n. Then dy = dx and x = y + n, so ∞−∞

f (x − n)x

1 + |x| dx =

∞−∞

f (y)y + n

1 + |y + n| dy

=

∞−n

f (y)y + n

1 + y + ndy +

−n−∞

f (y)y + n

1 − (y + n)dy. (7)

Note that, when y ≥ −n, y+n1+y+n ∈ [0, 1), and increases to 1 as n tends to infinity. Thus,

0 ≤ |f (y)| y + n

1 + y + n≤ |f (y)|,

for all y ≥ −n. Define the function

14

gn(y) = f (y)y + n

1 + y + n1[−n,∞)(y).

Then |gn| ≤ |f | and limn→∞

gn = f . Therefore, by the dominated convergence theorem,

limn→∞

∞−n

f (y)y + n

1 + y + ndy = lim

n→∞

∞−∞

gn(y) dy =

∞−∞

f (y) dy.

Next, consider the second term in (7). Define the function

hn(y) = f (y)y + n

1

−(y + n)

1(−∞,−n](y).

It is not hard to check that |y + n||1 − (y + n)| 1(−∞,−n](y) ∈ [0, 1),

from which it follows that |hn| ≤ |f |. Also, it is clear that, for all y,

limn→∞

hn(y) = f (y) limn→∞

y + n

1 − (y + n)1(−∞,−n](y) = 0.

Therefore, the dominated convergence theorem implies that

limn→∞

−n−∞

f (y)y + n

1 − (y + n)dy = 0.

Combining the two results above, we see that limn→∞

∞−∞

f (x − n)

x1+|x|

dx =

∞−∞

f (x) dx. Remark. Intuitively, this is the result we expect because the translation f (x − n) = T nf (x) is merely shifting the

support of f to the right tail of the measure dµ := x1+|x| dx, and in the tail this measure looks like dx.

14Here 1A(x) denotes the indicator function of the set A, which is 1 if x ∈ A and 0 if x /∈ A.

11



1.3 1998 April 3 1 REAL ANALYSIS

1.3 1998 April 3

Instructions Do at least four problems in Part A, and at least two problems in Part B.

PART A

1. Let xn∞n=1 be a bounded sequence of real numbers, and for each positive n define

xn = supk≥n

xk

(a) Explain why the limit = limn→∞ xn exists.

(b) Prove that, for any > 0 and positive integer N , there exists an integer k such that k ≥ N and |xk − | < .

2. Let C be a collection of subsets of the real line R, and define

Aσ(C ) =A : C ⊂ A and A is a σ-algebra of subsets of R.

(a) Prove that Aσ(C ) is a σ-algebra, that C ⊂ Aσ(C ), and that Aσ(C ) ⊂ A for any other σ-algebra A containing

all the sets of C .

(b) Let O be the collection of all finite open intervals in R, and F the collection of all finite closed intervals in R.

Show that

Aσ(O) = Aσ(F ).

3. Let (X,A,µ) be a measure space, and suppose X = ∪nX n, where X n∞n=1 is a pairwise disjoint collection of

measurable subsets of X . Use the monotone convergence theorem and linearity of the integral to prove that, if f is

a non-negative measurable real-valued function on X ,

X

f dµ =n

Xn

f dµ.

Solution: 15 Define f n =nk=1 f χXk

= f χ∪n1Xk

. Then it is clear that the hypotheses of the monotone conver-

gence theorem are satisfied. That is, for all x ∈ X ,

(i) 0 ≤ f 1(x) ≤ f 2(x) ≤ · · · ≤ f (x), and

(ii) limn→∞ f n(x) = f (x)χX(x) = f (x).

15Note that the hypotheses imply ν (E ) = E

f dµ is a measure (problem 4, Nov. ’91), from which the desired conclusion immediately follows.

Of course, this does not answer the question as stated, since the examiners specifically require the use of the MCT and linearity of the integral.

12




Therefore,

∞

k=1

Xk

f dµ = limn→∞

n

k=1

X

f χXkdµ

= limn→∞

X

nk=1

f χXkdµ (by linearity of the integral)

= limn→∞

X

f n dµ (by definition of f n)

=

X

limn→∞

f n dµ (by the monotone convergence theorem)

=

X

f dµ.

4. Using the Fubini/Tonelli theorems to justify all steps, evaluate the integral 10

1y

x−3/2 cosπy

2x

dxdy.

Solution: By Tonelli’s theorem, if f (x, y) ≥ 0 is measurable and one of the iterated integrals

f (x, y) dxdy or f (x, y) dydx exists, then they both exist and are equal. Moreover, if one of the iterated integrals is finite, then

f (x, y) ∈ L1(dx,dy). Fubini’s theorem states: if f (x, y) ∈ L1(dx, dy), then the iterated integrals exist and are

equal.

Now let g(x, y) = x−3/2 cos(πy/2x), and apply the Tonelli theorem to the non-negative measurable function

|g(x, y)

|as follows: 1

0

x0

|g(x, y)| dydx =

10

x0

|x|−3/2cos

πy

2x

dydx ≤ 10

x0

x−3/2 · 1 dydx =

10

x−1/2 dx = 2.

Thus one of the iterated integrals of |g(x, y)| is finite which, by the Tonelli theorem, implies g(x, y) ∈ L1(dx,dy).

Therefore, the Fubini theorem applies to g(x, y), and gives the first of the following equalities: 10

1y

x−3/2 cosπy

2x

dxdy =

10

x0

x−3/2 cosπy

2x

dydx

=

10

x−3/2 · 2x

π

sin

πy

2x

y=xy=0

dx

= 10

2

π x

−1/2

dx

=2

π

2x1/2

x=1

x=0

=4

π.

13




5. Let I be the interval [0, 1], and let C (I ), C (I × I ) denote the spaces of real valued continuous functions on I and

I × I , respectively, with the usual supremum norm on these spaces. Show that the collection of finite sums of the

form

f (x, y) =i

φi(x)ψi(y),

where φi, ψi ∈ C (I ) for each i, is dense in C (I × I ).

6. Let m be Lebesgue measure on the real line R, and for each Lebesgue measurable subset E of R define

µ(E ) =

E

1

1 + x2dm(x).

Show that m is absolutely continuous with respect to µ, and compute the Radon-Nikodym derivative dm/dµ.

Solution: Obviously both measures are non-negative. We must first prove m µ. To this end, suppose m(E ) >0, where E ∈M, the σ-algebra of Lebesgue measurable sets. Then, if we can show µ(E ) > 0, this will establish

that the implication µ(E ) = 0⇒

m(E ) = 0 holds for all E ∈M; i.e., m

µ.

For n = 1, 2, . . ., define

An =

x ∈ R :

1

n + 1<

1

1 + x2≤ 1

n

.

Then Ai ∩ Aj = ∅ for all i = j in N, and, for all n = 1, 2, . . .,

µ(An) ≥ 1

n + 1m(An).

Also, R = ∪An, since 0 < 11+x2 ≤ 1 holds for all x ∈ R. Therefore,

µ(E ) = µ(E ∩ (∪nAn)) = µ(∪n(An ∩ E )) =n

µ(An ∩ E ).

The last equality might need a bit of justification: Since f (x) = 11+x2

is continuous, hence measurable, the sets

An are measurable. Therefore, the last equality holds by countable additivity of disjoint measurable sets.Now note that m(E ) =

m(An∩E ) > 0 implies the existence of an n ∈ N such that m(An∩E ) > 0. Therefore,

µ(E ) ≥ µ(An ∩ E ) ≥ 1

n + 1m(An ∩ E ) > 0,

which proves that m µ. By the Radon-Nikodym theorem (A.1), there is a unique h ∈ L1(µ) such that

m(E ) =

hdµ, and

f dm =

fhdµ ∀ f ∈ L1(m).

In particular, if E ∈M and f (x) = 11+x2 χE , then

µ(E ) =

E

1

1 + x2dm(x) =

E

h(x)

1 + x2dµ(x).

That is, E dµ =

E h(x)1+x2 dµ(x) holds for all measurable sets E , which implies16 that, h(x)1+x2 = 1 holds for

µ-almost every x ∈ R. Therefore,dm

dµ(x) = h(x) = 1 + x2.

One final note: h is uniquely defined only up to an equivalence class of functions that are equal to 1+ x2, µ-a.e. 16Recall the standard result: if f and g are integrable functions such that

E

f = E

g holds for all measurable sets E , then f = g, µ-a.e.

This is an exam problem, but I can’t remember on which exam it appears. When I come across it again I’ll put a cross reference here.

14




PART B

7. Let φ(x, y) = x2y be defined on the square S = [0, 1]

×[0, 1] in the plane, and let m be two-dimensional Lebesgue

measure on S . Given a Borel subset E of the real line R, define

µ(E ) = m(φ−1(E )).

(a) Show that µ is a Borel measure on R.

(b) Let χE denote the characteristic function of the set E . Show that R

χE dµ =

S

χE φdm.

(c) Evaluate the integral

∞

−∞

t2 dµ(t).

8. Let f be a real valued and increasing function on the real line R, such that f (−∞) = 0 and f (∞) = 1. Prove that

f is absolutely continuous on every closed finite interval if and only if R

f dm = 1.

Solution: 17 First note that f is increasing, so f exists for a.e. x ∈ R, and f (x) ≥ 0 wherever f exists. Also, f

is measurable. To see this, define

g(x) = limsupn→∞

[f (x + 1/n) − f (x)] n.

As a lim sup of a sequence of measurable functions, g is measurable (Rudin [8], theorem 1.14?). Let E be the seton which f exists. Then m(R \ E ) = 0, and f = g on E (by the definition of derivative), so f is measurable.

(⇐) Suppose R

f dm = 1. We must show f ∈ AC [a, b] for all −∞ < a < b < ∞. First, check that

f ∈ L1(R), since

1 =

R

f dm =

R\E

f dm +

E

f dm =

E

f dm,

and, since f is increasing, f ≥ 0 on E , so R

|f |dm =

R\E

|f |dm +

E

|f |dm =

E

|f |dm =

E

f dm = 1.

Thus, f

∈L1(R) as claimed. A couple of lemmas will be needed to complete the

⇐direction of the proof. The

first is proved in the appendix (sec. A), while the second can be found in Royden [6] on page 100.

Lemma 1.2 Let f : R → R be a function. If f is differentiable on [a, b], f ∈ L1([a, b]), and xa f (t)dt =

f (x) − f (a) for a ≤ x ≤ b, then f ∈ AC [a, b].

17I have worked this problem a number of times, and what follows is the clearest and most instructive proof I’ve come up with. It’s by no means

the shortest, most elegant solution, and probably not the type of detailed answer one should give on an actual exam. However, some of the facts that

I prove in detail have appeared as separate questions on other exams, so the proofs are worth knowing.

15




The converse of this lemma is also true.18

Lemma 1.3 If f : R→ R is increasing and f ∈ L1([a, b]), then x

af (t)dt ≤ f (x) − f (a).

To finish the ⇐ direction of the proof, by lemma 1.2, it suffices to show that R

f dm = 1 implies ba f (t)dt =

f (b) − f (a) holds for all −∞ < a < b < ∞. By lemma 1.3, we have ba f (t)dt ≤ f (b) − f (a), so we need only

show that strict inequality cannot hold. Suppose, by way of contradiction, that ba f (t)dt < f (b) − f (a) holds for

some −∞ < a < b < ∞. Then,

1 =

R

f dm =

a−∞

f dm +

ba

f dm +

∞b

f dm

< [f (a) − f (−∞)] + [f (b) − f (a)] + [f (∞) − f (b)]

= f (∞) − f (−∞) = 1.

This contradiction proves that R f dm = 1 implies ba f (t)dt = f (b)

−f (a) holds for all

−∞< a < b <

∞, as

desired.

(⇒) Now assume f ∈ AC [a, b] for all −∞ < a < b < ∞. We must show R

f dm = 1. By assumption

f (∞) − f (−∞) = 1, so this is equivalent to showing

limx→∞

x−x

f (t)dm(t) = limx→∞

[f (x) − f (−x)].

Let x ∈ R, x > 0, and f ∈ AC [−x, x]. Then we claim f (x) − f (−x) = x−x

f dm. Assuming the claim is true

(see Royden [6], p. 110 for the proof), we have

1 = limx→∞

[f (x) − f (−x)] = limx→∞

x

−x

f (t)dm(t) = R f dm.

9. Let F be a continuous linear functional on the space L1[−1, 1], with the property that F (f ) = 0 for all odd

functions f in L1[−1, 1]. Show that there exists an even function φ such that

F (f ) =

1−1

f (x)φ(x) dx, for all f ∈ L1[−1, 1].

[Hint: One possible approach is to use the fact that any function in L p[−1, 1] is the sum of an odd function and an

even function.]

Solution: Since F ∈ L1[−1, 1]∗, then by the Riesz representation theorem19 there is a unique h ∈ L∞[−1, 1]such that

F (f ) =

1−1

f (x)h(x) dx (∀f ∈ L1[−1, 1])

18See Folland [4] for a nice, concise treatment of the fundamental theorem of calculus for Lebesgue integration.19See problem 3 of section 1.5.

16




Now (using the hint) write h = φ + ψ, where φ and ψ are the even and odd functions

φ(x) =h(x) + h(−x)

2

and ψ(x) =h(x) − h(−x)

2

.

Similarly, let f = f e + f o be the decomposition of f into a sum of even and odd functions. Then, by linearity of

F , and since F (f o) = 0 by hypothesis,

F (f ) = F (f e) + F (f o) = F (f e) =

1−1

f eh =

1−1

f eφ +

1−1

f eψ.

Now note that f eψ is an odd function (since it’s an even times an odd), so 1−1 f eψ = 0, since [−1, 1] is symmetric.

Similarly, 1−1

f oφ = 0. Therefore,

F (f ) = F (f e) =

1−1

f eφ =

1−1

f eφ +

1−1

f oφ =

1−1

(f e + f o)φ =

1−1

f φ.

17





Do as many problems as you can. Complete solutions to five problems would be considered a good performance.

1. (a)20 State the inverse function theorem.

(b) Suppose L : R3 → R3 is an invertible linear map and that g : R3 → R3 has continuous first order partial

derivatives and satisfies g(x) ≤ C x2 for some constant C and all x ∈ R3. Here x denotes the usual

Euclidean norm on R3. Prove that f (x) = L(x) + g(x) is locally invertible near 0.

Solution:

(a) (Inverse function theorem (IFT) of calculus)21

Let f : E → Rn be a C 1-mapping of an open set E ⊂ Rn. Suppose that f (a) is invertible for some a ∈ E and

that f (a) = b. Then,

(i) there exist open sets U and V in Rn such that a ∈ U , b ∈ V , and f maps U bijectively onto V , and

(ii) if g is the inverse of f (which exists by (i)), defined on V by g(f (x)) = x, for x ∈ U , then g ∈ C 1(V ).

(b) First note that L and g both have continuous first order partial derivatives; i.e., L, g ∈ C 1(R3). Therefore, the

derivative of f = L + g,

f (x) J f (x)

∂f i∂xj

3

i,j=1

exists. Furthermore, J f (x) is continuous in a neighborhood of the zero vector, because this is true of the partials of

g(x), and the partials of L(x) are the constant matrix L. Therefore, f ∈ C 1(R3). By the IFT, then, we need only

show that f (0) is invertible. Since f (x) = L + g(x), we must show f (0) = L + g(0) is invertible. Consider

the matrix g(0) = J g(0). We claim, J g(0) = 0. Indeed, if x1, x2, x3 are the elementary unit vectors (also known

as i, j, k), then the elements of J g(0) are

∂gi∂xj(0) = lim

h→0gi(0 + hxj) − gi(0)h = lim

h→0gi(hxj)h . (8)

The second equality follows by the hypothesis that g is continuous and satisfies g(x) ≤ C x2, which implies

that g(0) = 0. Finally, to show that (8) is zero, consider

|gi(hxj)| ≤ g(hxj) ≤ C hxj2 = C |h|2,

which implies|gi(hxj)|

|h| ≤ C |hxj |2

|h| = C |h| → 0, as h → 0.

This proves that f (0) = L, which is invertible by assumption, so the IFT implies that f (x) is locally invertible

near 0.

2. Let f be a differentiable real valued function on the interval (0, 1), and suppose the derivative of f is bounded on

this interval. Prove the existence of the limit L = limx→0+ f (x).

20The inverse function theorem does not appear on the syllabus and, as far as I know, this is the only exam problem in which it has appeared. The

implicit function theorem does appear on the syllabus, but I have never encountered an exam problem that required it.21See Rudin [7].

18




3. Let f and g be Lebesgue integrable functions on [0, 1], and let F and G be the integrals

F (x) = x

0

f (t) dt, G(x) = x

0

g(t) dt.

Use Fubini’s and/or Tonelli’s theorem to prove that 10

F (x)g(x) dx = F (1)G(1) − 10

f (x)G(x) dx.

Other approaches to this problem are possible, but credit will be given only to solutions based on these theorems.

4. Let (X,A,µ) be a finite measure space and suppose ν is a finite measure on (X, A) that is absolutely continuous

with respect to µ. Prove that the norm of the Radon-Nikodym derivative f =dνdµ

is the same in L∞(µ) as it is in

L∞(ν ).

5. Suppose that f n is a sequence of Lebesgue measurable functions on [0, 1] such that limn→∞

10

|f n| dx = 0 and

there is an integrable function g on [0, 1] such that |f n|2 ≤ g, for each n. Prove that limn→∞

10

|f n|2 dx = 0.

6. Denote by P e the family of all even polynomials. Thus a polynomial p belongs to P e if and only if p(x) = p(x)+ p(−x)

2 for all x. Determine, with proof, the closure of P e in L1[−1, 1]. You may use without proof the fact

that continuous functions on [−1, 1] are dense in L1[−1, 1].

7. Suppose that f is real valued and integrable with respect to Lebesgue measure m on R and that there are real

numbers a < b such that

a · m(U ) ≤ U

f dm ≤ b · m(U ),

for all open sets U in R. Prove that a ≤ f (x) ≤ b a.e.

19





Instructions Masters students do any 4 problems Ph.D. students do any 5 problems. Use a separate sheet of paper for

each new problem.

1. Let f n be a sequence of Lebesgue measurable functions on a set E ⊂ R, where E is of finite Lebesgue mea-

sure. Suppose that there is M > 0 such that |f n(x)| ≤ M for n ≥ 1 and for all x ∈ E , and suppose that

limn→∞ f n(x) = f (x) for each x ∈ E . Use Egoroff ’s theorem to prove that E

f (x) dx = limn→∞

E

f n(x) dx.

Solution: First note that |f (x)| ≤ M for all x ∈ E. To see this, suppose it’s false for some x0 ∈ E , so that

|f (x0)| > M . Then there is some > 0 such that |f (x0)| = M + . By the triangle inequality, then, for all n ∈ N,

|f (x0) − f n(x0)| ≥ ||f (x0)| − |f n(x0)|| = |M + − |f n(x0)|| ≥ ,

which contradicts f n(x0)

→f (x0). Thus,

|f (x)

| ≤M for all x

∈E .

Next, fix > 0. By Egoroff’s theorem (A.8), there is a G ⊂ E such that µ(E \ G) < and f n → f uniformly

on G. Furthermore, since |f n| ≤ M and |f | ≤ M and µ(E ) < ∞, it’s clear that f n ⊂ L1 and f ∈ L1, so the

following inequalities make sense (here we’re using the notation f G = sup|f (x)| : x ∈ G): E

f dµ − E

f n dµ

≤ E

|f − f n| dµ =

E\G

|f − f n| dµ +

G

|f − f n| dµ

≤ E\G

|f | dµ +

E\G

|f n| dµ + f (x) − f n(x)Gµ(G)

≤ 2M µ(E \ G) + f (x) − f n(x)Gµ(G)

< + f (x) − f n(x)Gµ(G).

Finally, µ(G)

≤µ(E ) <

∞and

f (x)

−f n(x)

G

→0, which proves that E f n dµ

→ E f dµ.

2. Let f (x) be a real-valued Lebesgue integrable function on [0, 1].(a) Prove that if f > 0 on a set F ⊂ [0, 1] of positive measure, then

F

f (x) dx > 0.

(b) Prove that if x0

f (x) dx = 0, for each x ∈ [0, 1],

then f (x) = 0 for almost all x ∈ [0, 1].

Solution: (a) Define F n = x ∈ F : f (x) > 1/n. Then

F 1 ⊆ F 2 ⊆ · · · ↑n

F n = F,

and m(F ) > 0 implies

0 < m(F ) = m(∪nF n) ≤n

m(F n).

20




Therefore, m(F k) > 0 for some k ∈ N, and then it follows from the definition of F k that

0 <1

k

m(F k)

≤ F k

f dm

≤ F

f dm.

(b) Suppose there is a subset E ⊂ [0, 1] of positive measure such that f > 0 on E . Then part (a) implies E

f dm > 0. Let F ⊂ E be a closed subset of positive measure. (That such a closed subset exists follows from

Prop. 3.15 of Royden [6].) Then, again by (a), F

f dm > 0. Now consider the set G = [0, 1] \ F , which is open

in [0, 1], and hence22 is a countable union of disjoint open intervals; i.e., G = ·∪n(an, bn). Therefore,

0 =

[0,1]

f dm =n

(an,bn)

f dm +

F

f dm,

so F f dm > 0 implies

n

(an,bn)

f dm < 0.

Thus, (ak,bk)

f dm < 0 for some (ak, bk) ⊂ [0, 1]. On the other hand,

(ak,bk)

f dm =

bk0

f (x) dm(x) − ak0

f (x) dm(x).

By the initial hypothesis, both terms on the right are zero, which gives the desired contradiction.

3. State each of the following:

(a) The Stone-Weierstrass theorem

(b) The Lebesgue (dominated) convergence theorem

(c) Holder’s inequality(d) The Riesz representation theorem for L p

(e) The Hahn-Banach theorem.

Solution: 23

(a) (Stone-Weierstrass theorem)

Let X be a compact Hausdorff space and let A be a closed subalgebra of functions in C (X,R) which separates

points. Then either A = C (X,R), or A = f ∈ C (X,R) : f (x0) = 0 for some x0 ∈ X . The first case occurs iff

A contains the constant functions.

(b) (Lebesgue dominated convergence theorem)24 Let f n be a sequence of measurable functions on (X,M, µ)such that f n → f a.e.. If there exists g ∈ L1(X,M, µ) such that |f n(x)| ≤ g(x) holds for all x ∈ X and

n = 1, 2, . . .. Then

f n

⊂L1, f

∈L1, lim X f n dµ = X f dµ, and

f n

−f

1

→0.

(c) (Holder’s inequality)

Let f and g be measurable functions.

22Every open set of real numbers is the union of a countable collection of disjoint open intervals (Royden [6], Prop. 8, page 42).23The presentations of (a) and (c) in Folland [4] are especially nice. For (b) and (e), as well as (c), I like Rudin [8]. A version of (d) appears in

Royden [6].24See theorem A.7 for a more general version.

21




(i) If 1 < p < ∞ and 1 p + 1

q = 1, then f g1 = f pgq . Thus, if f ∈ L p and g ∈ Lq , then f g ∈ L1

(ii) If p = ∞ and if f ∈ L∞ and g ∈ L1, then |f g| ≤ f ∞|g|, so f g1 ≤ f ∞g1.

(d) (Riesz representation theorem for L p)25

Suppose 1 < p < ∞ and 1 p + 1

q = 1. If Λ is a linear functional on L p, then there is a unique g ∈ Lq such that

Λf =

f g d µ (∀f ∈ L p).

(e) (Hahn-Banach theorem)

Suppose X is a normed linear space, Y ⊆ X is a subspace, and T : Y → R is a bounded linear functional.

Then there exists a bounded linear functional T : X → R such that T (y) = T (y) for all y ∈ Y , and such that

T X = T Y , where T X and T Y are the usual operator norms,

T X = sup|T x| : x ∈ X, x ≤ 1 and T Y = sup|T x| : x ∈ Y, x ≤ 1.

4. (a) State the Baire category theorem.

(b) Prove the following special case of the uniform boundedness theorem: Let X be a (nonempty) complete metric

space and let F ⊆ C (X ). Suppose that for each x ∈ X there is a nonnegative constant M x such that

|f (x)| ≤ M x for all f ∈ F.

Prove that there is a nonempty open set G ⊆ X and a constant M > 0 such that

|f (x)| ≤ M holds for all x ∈ G and for all f ∈ F .

Solution: 26

(a) (Baire category theorem)

If X is a complete metric space and An is a collection of open dense subsets, then∞n=1 An is dense in X .

Corollary 1. If X is a complete metric space and G ⊆ X is a non-empty open subset and G =∞n=1 Gn then

Gno = ∅ for at least one n ∈ N.

Corollary 2. A nonempty complete metric space is not a countable union of nowhere dense sets.

(b) Define Am = x ∈ X : |f (x)| ≤ m, ∀f ∈ F . Then X =∞m=1 Am, since for every x there is a finite

number M x such that |f (x)| ≤ M x for all f ∈ F . Now note that Am =

f ∈F x ∈ X : |f (x)| ≤ m, and, since f

and a→ |

a|

are continuous functions, each

x

∈X :

|f (x)

| ≤m

is closed, so Am is closed. Therefore, corollary

2 of the Baire category theorem implies that there must be some m ∈ N such that Am = ∅, so the set G = A

m and

the number M = m satisfy the given criteria.

25Note to self: add case p = ∞26See Royden [6], § 7.8, for an excellent treatment of this topic. Part (b) of this problem appears there as theorem 32, and another popular exam

question is part c of problem 37.

22




5. Prove or disprove:

(a) L2 convergence implies pointwise convergence.

(b)

limn→∞

∞0

sin(xn)

xndx = 0.

(c) Let f n be a sequence of measurable functions defined on [0, ∞). If f n → 0 uniformly on [0, ∞), as n → ∞,

then

lim

[0,∞)

f n(x) dx =

[0,∞)

lim f n(x) dx.

Solution:

(a) This is false, as the following example demonstrates: For each k ∈ N, define f k,j = χ[ j−1k , jk )for j = 1, . . . , k,

and let gn be the sequence defined by

g1 = f 1,1,

g2 = f 2,1, g3 = f 2,2,

g4 = f 3,1, g5 = f 3,2, g6 = f 3,3,

g7 = f 4,1, . . .

Then |f k,j|2 dµ = 1/k for each j = 1, . . . , k, so f k,j2 = 1/

√k → 0, as k → ∞. Therefore gn2 → 0 as

n → ∞. However, gn does not converge pointwise since, for every x ∈ [0, 1] and every N ∈ N, we can always

find some k ∈ N and j ∈ 1, . . . , k such that gn(x) = f k,j(x) = 1 with n ≥ N , and we can also find a k ∈ Nand j ∈ 1, . . . , k such that gn(x) = f k,j(x) = 0 with n ≥ N .

(b) For any fixed 0 < x < 1, limn→∞ xn = 0. Also, sin tt → 1, as t → 0, which can be proved by L’Hopital’s rule.

Together, these two facts yield

limn→∞

sin xn

xn = 1.

Now, recall that | sin θ| ≤ |θ| for all real θ. Indeed, since sin θ = θ0 cos x dx, we have, for θ ≥ 0,

| sin θ| ≤ θ0

| cos x| dx ≤ θ0

1 dx = θ,

and, for θ < 0,

| sin θ| = | sin(−θ)| ≤ |− θ| = |θ|.In particular, for any 0 < x < 1,

| sin xn||xn|

≤ 1. Therefore, we can apply the dominated convergence theorem to the

function sinxn

xn , to obtain

limn→∞ 10

sin xn

xn dx = 10 1 dx = 1.

(9)

Next consider the part of the integral over 1 ≤ x < N , for any real N > 1. Fix n ≥ 2. The change of variables

u = xn results in du = nxn−1dx, and, since u1/n = x, we have xn−1 = u1− 1n . Therefore, N

1

sin xn

xndx =

N n1

sin u

u

du

nu1− 1n

=1

n

N n1

sin u

u2− 1n

du.

23




Now,

limN →∞

1

n N n

1

sin u

u2− 1n

du ≤ limN →∞

1

n N n

1

u1n−2 du = lim

N →∞

1

n

u1n−1

1

n −1 N n

1

=1

n−

1.

Therefore, ∞1

sin xn

xndx

≤ 1

n − 1,

and so,

limn→∞

∞1

sin xn

xndx = 0. (10)

Combining results (9) and (10) yields

limn→∞

∞0

sin xn

xndx = 1.

(c) This is false, as the following example demonstrates: Let f n = 1

nχ[0,n). Then f n → 0 uniformly and so lim f n = 0. On the other hand,

f n = 1 for all n ∈ N. Therefore, lim

f n = 1 = 0 =

lim f n.

6. Let f : H → H be a bounded linear functional on a separable Hilbert space H (with inner product denoted by

·, ·). Prove that there is a unique element y ∈ H such that

f (x) = x, y for all x ∈ H and f = y.

Hint. You may use the following facts: A separable Hilbert space, H , contains a complete orthonormal sequence,

φk∞k=1, satisfying the following properties: (1) If x, y ∈ H and if x, φk = y, φk for all k, then x = y. (2)

Parseval’s equality holds; that is, for all x ∈ H , x, x =∞k=1 a2k, where ak = x, φk.

Solution: Define y =

∞k=1 f (φk)φk, and check that this y ∈ H has the desired properties.

First observe that, by properties (1) and (2) given the hint, any x∈

H can be written as x = ∞

k=1ak

φk

, where

ak = x, φk, for each k ∈ N. Therefore, by linearity of f ,

f (x) = f (k

akφk) =k

akf (φk). (11)

Now

x, y = k

akφk, y =k

akφk, y, (12)

and, by definition of y,

φk, y = φk,j

f (φj)φj =j

f (φj)φk, φj = f (φk). (13)

The last equality holds by orthonormality; i.e., φk, φj is 1 when j = k and 0 otherwise. Putting it all together,

we see that, for every x ∈ H ,

f (x) =k

akf (φk) (11)

=k

akφk, y (13)

= x, y (12)

24




Moreover, this y is unique. For, suppose there is another y ∈ H such that f (x) = x, y for all x ∈ X . Then

x, y = f (x) = x, y for all x ∈ X . In particular, φk, y = f (φk) = φk, y for each k ∈ N , which, by

property (1) of the hint, proves that y = y.

Finally, we must show f = y. Observe,

f = supx∈X

|f (x)| : x ≤ 1 = supx∈X

|f (x)|x = sup

x∈X

|(x, y)|x ,

and recall that |(x, y)| ≤ xy holds for all x, y ∈ X . Whence,

f = supx∈X

|(x, y)|x ≤ y. (14)

On the other hand,

f = supx∈X

|(x, y)|x ≥ |(y, y)|

y =y2y = y. (15)

Together, (14) and (15) give f = y, as desired.

7. Let X be a normed linear space and let Y be a Banach space. Let

B(X, Y ) = A | A : X → Y is a bounded linear operator.

Then with the norm A = supx≤1 Ax, B(X, Y ) is a normed linear space (you need not show this). Prove

that B(X, Y ) is a Banach space; that is, prove that B(X, Y ) is complete.

Solution: Let T n ⊂ B(X, Y ) be a Cauchy sequence; i.e., T n − T m → 0 as m, n → ∞. Fix x ∈ X . Then,

T nx

−T mx

Y

≤ T n

−T m

x

X

→0, as n, m

→ ∞.

Therefore, the sequence T nx ⊂ Y is a Cauchy sequence in (Y, · Y ). Since the latter is complete, the limit

limn→∞ T nx = y ∈ Y exists. Define T : X → Y by T x = limn→∞ T nx, for each x ∈ X . To complete the proof,

we must check that T is linear, bounded, and satisfies limn→∞ T n − T = 0.

• T is linear:

For x1, x2 ∈ X ,

T (x1 + x2) = limn→∞

T n(x1 + x2)

= limn→∞

(T nx1 + T nx2) ( T n is linear)

= limn→∞

T nx1 + limn→∞

T nx2 ( both limits exist)

= T x1 + T x2.

• T is bounded:

First, note that T n is a Cauchy sequence of real numbers, since | T n − T m | ≤ T n − T m → 0,

as n, m → ∞. Therefore, there is a c ∈ R such that T n → c, as n → ∞. For some N ∈ N, then,

T n ≤ c + 1 for all n ≥ N . Thus,

T nxY ≤ T nxX ≤ (c + 1)xX (∀x ∈ X ). (16)

25




Now, by definition, T nx → T x, for all x ∈ X and, since the norm · Y is uniformly continuous,27

T nxY → T xY (∀x ∈ X ). (17)

Taken together, (16) and (17) imply T xY ≤ (c + 1)xX , for all x ∈ X . Therefore, T is bounded.

• limn→∞ T n − T = 0:

Fix > 0 and choose N ∈ N such that n, m ≥ N implies T n − T m < . Then,

T nx − T mx ≤ T n − T mxX < xXholds for all n, m ≥ N , and x ∈ X . Letting m go to infinity, then,

T nx − T x = limm→∞

T nx − T mx ≤ xX .

That is, T nx − T x ≤ xX , for all n ≥ N and x ∈ X . Whence, T n − T ≤ for all n ≥ N .

27Proof: | aY − bY | ≤ a − bY (∀a, b ∈ Y ).

26




1.6 2004 April 19

Instructions. Use a separate sheet of paper for each new problem. Do as many problems as you can. Complete

solutions to five problems will be considered as an excellent performance. Be advised that a few complete and well

written solutions will count more than several partial solutions.

Notation: f ∈ C (X ) means that f is a real-valued, continuous function defined on X .

1. (a) Let S be a (Lebesgue) measurable subset of R and let f, g : S → R be measurable functions. Prove that (i)

f + g is measurable and (ii) if φ ∈ C (R), then φ(f ) is measurable.

(b) Let f : [a, b] → [−∞, ∞] be a measurable function. Suppose that f takes the value ±∞ only on a set of

(Lebesgue) measure zero. Prove that for any > 0 there is a positive number M such that |f | ≤ M, except on a

set of measure less than .

Solution:

(a) Proof 1: Since f and g are real measurable functions of S , and since the mapping Φ : R× R → R defined by

Φ(x, y) = x + y is continuous, theorem A.3 implies that the function f + g = Φ(f, g) is measurable. If φ

∈C (R),

then φ(f ) is measurable by part (b) of theorem A.2.

Proof 2: Let qi∞i=1 be an enumeration of the rationals. Then, for any α ∈ R,

x ∈ S : f (x) + g(x) < α =∞i=1

x ∈ S : f (x) < α − qi ∩ x ∈ S : g(x) < qi.

Since each set on the right is measurable, and since σ-algebras are closed under countable unions and intersections,

x ∈ S : f (x) + g(x) < α is measurable. Since α was arbitrary, f + g is measurable.

The function φf is measurable if and only if, for any open subset U of R, the set (φf )−1(U ) is measurable. Let U be open in R. Then φ−1(U ) is open, since φ ∈ C (R), and so (φf )−1(U ) = f −1(φ−1(U )) is measurable, since f is measurable. Therefore, φf is measurable.

(b) Fix > 0. For n ∈ N, define An = x ∈ [a, b] : |f (x)| ≤ n. Then

[a, b] =

∞n=1

An ∪ A∞, (18)

where28 A∞ = x : f (x) = ±∞. Also, A1 ⊆ A2 ⊆ · · · and, since f is measurable, each An is measurable.

Therefore, µ(An) ↑ µ(∪nAn), as n → ∞. Note that all sets are contained in [a, b] and thus have finite measure.

Let M ∈ N be such that µ(∪nAn) − µ(AM ) < . Then |f | ≤ M except on [a, b] \ AM , and by (18),

µ([a, b] \ AM ) = µ(∪nAn ∪ A∞ \ AM )

≤ µ(∪nAn \ AM ) + µ(A∞)

= µ(∪nAn \ AM ) < .

The second equality holds since we assumed f (x) = ±∞ only on a set of measure zero; i.e., µ(A∞) = 0.

28Since f is an extended real valued function, we must not forget to include A∞, without which the union in (18) would not be all of [a, b].

27




2. (a) State Egorov’s theorem.

(b) State Fatou’s lemma.

(c)Let f n ⊂ L

p

[0, 1], where 1 ≤ p < ∞. Suppose that f n → f a.e., where f ∈ L

p

[0, 1]. Prove thatf n − f p → 0 if and only if f n p → f p.

Solution:

(a) See theorem A.8.

(b) See theorem A.6.

(c) (⇒) By the Minkowsky inequality, f n p = f n − f + f p ≤ f n − f p + f p. Similarly, f p ≤f n− f p+ f n p. Together, the two inequalities yield |f n p − f p| ≤ f n− f p. Therefore, f n− f p → 0implies |f n p − f p| → 0. This proves necessity.

(⇐) I know of three proofs of sufficiency. The second is similar to the first, only much shorter as it exploits the full

power of the general version of Lebesgue’s dominated convergence theorem, whereas the first proof merely relies

on Fatou’s lemma.29 The third proof uses both Fatou’s lemma and Egoroff’s theorem, so, judging from parts (a)

and (b), this may be closer to what the examiners had in mind. Note that none of the proofs use the assumption that

the measure space is finite, so we may as well work in the more general space L p(X,M, µ).

Both proofs 1 and 2 make use of the following:

Lemma 1.4 If α, β ∈ [0, ∞) and 1 ≤ p < ∞, then (α + β) p ≤ 2 p−1(α p + β p).

Proof: When p ≥ 1, φ(x) = xp is convex on [0, ∞). Thus, for all α, β ∈ [0, ∞),

α + β

2

p= φ

α + β

2

≤

1

2[φ(α) + φ(β)] =

1

2(α

p + βp).

When α, β ∈ R, the triangle inequality followed by the lemma yields

|α − β| p ≤ ||α| + |β|| p ≤ 2 p−1(|α| p + |β| p). (19)

Proof 1: By (19),

|f n − f | p ≤ 2 p−1(|f n| p + |f | p).

In particular, f n − f ∈ L p, for each n ∈ N. Moreover, the functions

gn = 2 p−1(|f n| p + |f | p) − |f n − f | p. (20)

are non-negative. Now notice that lim gn = 2 p|f | p. Applying Fatou’s lemma to (20), then, 2 p|f | p =

lim gn ≤ lim

gn = lim

2 p−1(|f n| p + |f | p) − |f n − f | p .

Since f n p → f p, this implies

2 p

|f | p ≤ 2 p

|f | p − lim

|f n − f | p.

Equivalently 0 ≤ − lim |f n − f | p. This proves f n − f p → 0.

29Disclaimer: I made up the first proof, so you should check it carefully for yourself and decide whether you believe me.

28




Proof 2: By (19),

|f n − f | p ≤ 2 p−1(|f n| p + |f | p). (21)

In particular, f n−

f ∈

L p, for each n∈N. Define the functions

gn = 2 p−1(|f n| p + |f | p) and g = 2 p|f | p.

Then gn → g a.e., and f n p → f p implies

gn → g. Also, gn ≥ |f n − f | p → 0 a.e., by (21). Therefore,

the dominated convergence theorem (theorem A.7) implies |f n − f | p → 0.

Proof 3: Since f ∈ L p, for all > 0, there is a number δ > 0 and a set B ∈ M of finite measure such that f is

bounded on B, X\B

|f | p dµ < /2, and E

|f | p dµ < /2, for all E ∈ M with µE < δ. By Egoroff’s theorem,

there is a set A ⊆ B such that µ(B \ A) < δ and f n → f uniformly on A. Therefore,

X |

f

| p =

X\B |f

| p +

B\A |f

| p +

A |f

| p

< /2 + /2 +

A

|f | p

≤ + lim

A

|f n| p, (22)

since, by Fatou’s lemma, A

|f | p = A

lim |f n| p ≤ lim A

|f n| p. By hypothesis, f n p → f p. Therefore

lim

A

|f n| p = lim

X

|f n| p − X\A

|f n| p

=

X

|f | p − lim

X\A

|f n| p.

By (22), then, X

|f | p < + X

|f | p − lim X\A

|f n| p.

Therefore, lim X\A

|f n| p < (since f ∈ L p). Finally, note that

f n − f p = (f n − f )χA + (f n − f )χX\A p≤ (f n − f )χA p + (f n − f )χX\A p (Minkowsky)

≤ (f n − f )χA p + f nχX\A p + f χX\A p.

Therefore,

lim f n − f p ≤ lim|f n(x) − f (x) : x ∈ Aµ(A)1/p + lim

X\A |f n| p

1/p

+

X\A |f | p

1/p

.

The first term on the right goes to zero since f n → f uniformly on A. The other terms are bounded by 21/p.

29




3. (a) Let S = [0, 1] and let f n ⊂ L p(S ), where 1 < p < ∞. Suppose that f n → f a.e. on S, where f ∈ L p(S ). If

there is a constant M such that f n p ≤ M for all n, prove that for each g ∈ Lq(S ), 1 p + 1

q = 1, we have

limn→∞

S

f ng = S

fg.

(b) Show by means of an example that this result is false for p = 1.

Solution:

(a) Since g ∈ Lq(S ), for all > 0 there exists δ > 0 such that if A is a measurable set with µA < δ then A

|g|q dµ < .

Let B0 ⊂ S denote the set on which f n does not converge to f . Let B ⊂ S be such that f n → f uniformly in

S \ B, and such that µB < δ. (Such a set exists by Egoroff’s theorem since µS < ∞.) Now throw the set B0 in

with B (i.e. redefine B to be B∪

B0). Then,

Dn :=

S

|f ng − f g| dµ =

S

|f n − f ||g| dµ

=

S\B

|f n − f ||g| dµ +

B

|f n − f ||g| dµ

≤ (f n − f )χS\B pgq + f n − f pgχBq.

The final inequality holds because f n, f ∈ L p implies |f n − f | ∈ L p and by Holder’s inequality. Now, by

Minkowski’s inequality, f n − f p ≤ f n p + f p, so

Dn ≤ sup|f n(x) − f (x)| : x /∈ Bµ(S \ B)1/pgq + (f n p + f p)

B

|g|q dµ

1/q

.

Now we are free to choose the δ > 0 above so that B

|g|q dµ

1/q

<

2(M + f p)

holds whenever µB < δ. Also, f n → f uniformly on S \ B, so let N be such that

sup|f N (x) − f (x)| : x /∈ B <

2µ(S \ B)1/pgq .

Then DN < /2 + /2.

(b) The result is not true for p = 1 as the following example shows: Let f n = nχ[0,1/n], n = 1, 2, . . . . First note

that f n → 0 a.e. For if x ∈ (0, 1], there exists N > 0 such that 1/N < x and thus f n(x) = 0 for all n ≥ N .Therefore, x ∈ [0, 1] : f n(x) 0 = 0. Now, if g is the constant function g ≡ 1, then

f ng dµ = 1 for all

n = 1, 2, . . . . Therefore,

f ng dµ → 1, while

f g d µ =

0g dµ = 0. Finally, note that f n1 = nµ([0, 1n ]) = 1

for n = 1, 2, . . . , so f n satisfies the hypothesis f n1 ≤ some constant M .

30




4. State and prove the closed graph theorem.

5. Prove or disprove:(a) For 1 ≤ p < ∞, let p := x = xk | x p = (

∞k=1 |xk| p)1/p < ∞. Then for p = 2, p is a Hilbert

space.

(b) Let X = (C [0, 1], · 1), where the linear space C [0, 1] is endowed with the L1-norm: f 1 = 10

|f (x)| dx.

Then X is a Banach space.

(c) Every real, separable Hilbert space is isometrically isomorphic to 2.

6. (a) Give a precise statement of some version of Fubini’s theorem that is valid for non-negative functions.

(b) Let f, g ∈ L1(R). (i) Prove that the integral

h(x) = R

f (x−

t)g(t) dt

exists for almost all x ∈ R and that h ∈ L1(R). (ii) Show that h1 ≤ f 1g1.

7. (a) State the Radon-Nikodym theorem.

(b) Let (X,B, µ) be a complete measure space, where µ is a positive measure defined on the σ-algebra, B, of

subsets of X . Suppose µ(X ) < ∞. Let S be a closed subset of R and let f ∈ L1(µ), where f is an extended

real-valued function defined on X . If

AE(f ) =1

µ(E )

E

f dµ ∈ S

for every E ∈B with µ(E ) > 0, prove that f (x) ∈ S for almost all x ∈ X .

31





Notation: R is the set of real numbers and Rn is n-dimensional Euclidean space. Denote by m Lebesgue measure on

R and mn n-dimensional Lebesgue measure. Be sure to give a complete statement of any theorems from analysis that

you use in your proofs below.

1. Let µ be a positive measure on a measure space X . Assume that E 1, E 2, . . . are measurable subsets of X with the

property that for n = m, µ(E n ∩ E m) = 0. Let E be the union of these sets. Prove that

µ(E ) =∞n=1

µ(E n)

Solution: Define F 1 = E 1, F 2 = E 2 \ E 1, F 3 = E 3 \ (E 1 ∪ E 2), . . . , and, in general,

F n = E n \n−1i=1

E i (n = 2, 3, . . . ).

If M is the σ-algebra of µ-measurable subsets of X, then F n ∈M for each n ∈ N, sinceM is a σ-algebra. Also,F i ∩ F j = ∅ for i = j, and F 1 ∪ F 2 ∪ · · · ∪ F n = E 1 ∪ E 2 ∪ · · · ∪ E n for all n ∈ N. Thus,

∞n=1

F n =∞n=1

E n E,

and, by σ-additivity of µ,

µ(E ) = µ(∞n=1

F n) =∞n=1

µ(F n).

Therefore, if we can show µ(E n) = µ(F n) holds for all n ∈ N, the proof will be complete.

Now, for each n = 2, 3, . . . ,

F n = E n ∩ (

n−1i=1 E i)

c

(23)

and

µ(E n) = µ(E n ∩ (n−1i=1

E i)c) + µ(E n ∩ (

n−1i=1

E i)). (24)

Equation (24) holds becausen−1i=1 E i is a measurable set for each n = 2, 3, . . . . Finally, note that

E n ∩ (n−1i=1

E i) =n−1i=1

(E n ∩ E i),

which implies

µ(E n ∩

(n−1

i=1

E i))

≤

n−1

i=1

µ(E n ∩

E i),

by σ-subadditivity. By assumption, each term in the last sum is zero, and therefore, by (23) and (24),

µ(E n) = µ(E n ∩ (

n−1i=1

E i)c) = µ(F n) holds for each n = 2, 3, . . . .

For n = 1, we have F 1 = E 1, by definition. This completes the proof.

32




2. (a) State a theorem that illustrates Littlewood’s Principle for pointwise a.e. convergence of a sequence of functions

on R.

(b) Suppose that f n∈

L1(m) for n = 1, 2, . . . . Assuming that

f n

−f

1

→0 and f n

→g a.e. as n

→ ∞, what

relation exists between f and g? Make a conjecture and then prove it using the statement in Part (a).

Solution:

(a) I think it’s generally accepted that the Littlewood principle dealing with a.e. convergence of a sequence of

functions on R is Egoroff’s theorem, which is stated below in section A.8.

(b) Conjecture: f = g a.e.

Proof 1:30 First recall that L1 convergence implies convergence in measure. That is, if f n ⊂ L1(m) and

f n − f 1 → 0, then f n → f in measure. (Proof: m(x : |f n(x) − f (x)| > ) ≤ 1f n − f 1 → 0.) Next recall

another important theorem31 which states that if f n → f in measure then there is a subsequence f nj ⊆ f nwhich converges a.e. to f as j → ∞. Combining these two results in the present context (Lebesgue measure on

the real line), we can say the following:32 If f n ⊂ L1(m) and f n − f 1 → 0 then there is a subsequence

f n

j ⊆ f n

with the property f n

j

(x)→

f (x) for almost all x∈R.

Now, if f n(x) → g(x) for almost all x ∈ R, and if B1 be the set of measure zero where f n(x) g(x), then off of

B1 the sequence f n, as well as every subsequence of f n, converges to g. Let f nj be the subsequence mentioned

above which converges to f almost everywhere. Then

|f (x) − g(x)| ≤ |f (x) − f nj (x)| + |f nj (x) − g(x)|. (25)

Define B2 = x ∈ R : f nj (x) f (x). Then the set B = B1 ∪ B2 has measure zero and, for all x ∈ R \ B,

f nj (x) → f (x) and f nj (x) → g(x) . Therefore, by (25), |f (x) − g(x)| = 0 for all x ∈ R \ B. It follows that the

set x ∈ R : f (x) = g(x) ⊂ B, as a subset of a null set, must itself be a null set (since m is complete). That is,

f = g a.e. and the conjecture is proved.

Proof 2: First, we claim that if f = g a.e. on [

−n, n] for every n

∈N, then f = g a.e. in R. To see this, let

Bn = x ∈ [−n, n] : f (x) = g(x). Then mBn = 0 for all n ∈ N, so that if B = x ∈ R : f (x) = g(x), thenB = ∪Bn and mB ≤

mBn = 0, as claimed. Thus, to prove the conjecture, it is enough to show that f = g for

almost every −n ≤ x ≤ n, for an arbitrary fixed n ∈ N.

Fix n ∈ N, and suppose we know that f − g ∈ L1([−n, n], m). (This will follow from the fact that f, g ∈L1([−n, n], m), which we prove below.) Then, for all > 0 there is a δ > 0 such that

E

|f − g| dm < for

all measurable E ⊆ [−n, n] with mE < δ. Now apply Egoroff’s theorem to find a set A ⊆ [−n, n] such that

m([−n, n] \ A) < δ and f n → g uniformly on A. Then n−n

|f − g| dm =

[−n,n]\A

|f − g| dm +

A

|f − g| dm

≤ +

A

|f − f n| dm +

A

|f n − g| dm

≤ + f − f n1 + mA supx∈A

|f n(x) − g(x)|,30Note that Proof 1, which seems to me the more natural one, doesn’t use Egoroff’s theorem, so either the examiners were looking for a different

proof, or a different conjecture, or perhaps Egoroff’s theorem was not the Littlewood principle they had in mind. In any event, I have found a way

to prove the conjecture which does make use of Egoroff’s theorem, and this appears here as Proof 2.31Folland [4], theorem 2.30 and its corollary.32Perhaps this statement is the version of the Littlewood principle dealing with a.e. convergence that we were meant to cite in part (a).

33




where f −f n1 → 0, by assumption, and supx∈A |f n(x)−g(x)| → 0 since f n → g uniformly on A. Since > 0was arbitrary, it follows that

n−n

|f − g| dm = 0 and, for functions f, g ∈ L1([−n, n], m), this implies that f = ga.e. on [−n, n].

It remains to show that f, g ∈ L1([−n, n], m). It’s clear that f ∈ L1 since f 1 ≤ f − f n1 + f n1 < ∞. To

prove g ∈ L1([−n, n], m) note that f na.e.−−→ g implies limn |f n(x)| = |g(x)| for almost all x, so by Fatou’s lemma,

g1 =

|g| dm =

lim |f n| dm ≤ lim

|f n| dm = lim f n1 = f 1 < ∞.

The last equality holds because, by the triangle inequality, |f n1 − f 1| ≤ f n − f 1 → 0.

3. Let K be a compact subset in R3 and let f (x) = dist(x, K ).

(a) Prove that f is a continuous function and that f (x) = 0 if and only if x ∈ K .

(b) Let g = max(1 − f, 0) and prove that limn→∞ gn exists and is equal to m3(K ).

Solution: (a) Define dist(x, K ) = f (x) = inf k∈K |x − k|. Clearly, for all k ∈ K , f (x) ≤ |x − k|. Therefore, by

the triangle inequality, for any x, y ∈ R3,

f (x) ≤ |x − y| + |y − k|, ∀k ∈ K,

and so, taking the infimum over k ∈ K on the right,

f (x) ≤ |x − y| + f (y). (26)

Similarly,

f (y) ≤ |x − y| + f (x). (27)

Obviously, for any given x ∈ R3, f (x) is finite. Therefore, (26) and (27) together imply that

|f (x)

−f (y)

| ≤ |x

−y|,

∀x, y

∈R3.

Whence f is (Lipschitz) continuous.

Now, if x ∈ K , then it’s clear that f (x) = 0. Suppose x /∈ K ; that is, x is in the complement K c of K . Since K is

closed, K c is open and we can find an -neighborhood about x fully contained in K c, in which case f (x) > . We

have thus proved that f (x) = 0 if and only if x ∈ K .

(b) First observe that f (x) = 0 for all x ∈ K , and f (x) > 0 for all x /∈ K . Define K 1 to be a closed and bounded

set containing K on which f (x) ≤ 1. That is, K 1 is the set of points that are a distance of not more than 1 unit

from the set K . In particular K ⊂ K 1. Notice that g = max(1 − f, 0) = (1 − f )χK1. Also, if x ∈ K 1 \ K ,

then 0 ≤ 1 − f (x) < 1, so gn → 0 on the set K 1 \ K , while on the set K , gn = 1 for all n ∈ N. Therefore,

gn → χK . Finally, note that gn ≤ χK1∈ L1(R3) so the dominated convergence theorem can be applied to yield

limn→∞ gn = χK = m3(K ).

4. Let E be a Borel subset of R2.

(a) Explain what this means.

(b) Suppose that for every real number t the set E t = (x, y) ∈ E | x = t is finite. Prove that E is a Lebesgue

null set.

34




Solution:

(a) The Borel σ-algebra of R2, which we denote byB(R2), is the smallest σ-algebra that contains the open subsets

of R2. The sets belonging toB(R2) are called Borel subsets of R2.

(b) First observe that if G is a finite subset of R, then G is a Lebesgue null set. That is, mG = 0. In fact, it is easyto prove that if G is any countable subset, then mG = 0. (Just fix > 0 and cover each point xn ∈ G with a set

E n of measure less than 2−n. Then mG ≤ mE n < .)

In problems involving 2-dimensional Lebesgue measure, distinguishing x and y coordinates sometimes clarifies

things. To wit, let (X,B(X ), µ) = (Y,B(Y ), ν ) be two identical copies of the measure space (R,B(R), m), and

represent Lebesgue measure on R2 by33 (X × Y,B(X ) ⊗B(Y ), µ × ν ) = (R2,B(R2), m2).

Our goal is to prove that m2E = 0. First note that

m2E = (µ × ν )(E ) =

X×Y

χE d(µ × ν ).

The integrand χE is non-negative and measurable (since E is Borel). Therefore, by Tonelli’s theorem (A.13),

m2E = Y

X

χE(x, y) dµ(x) dν (y) = X

Y

χE(x, y) dν (y) dµ(x). (28)

Now, let

Gt = y ∈ R : (x, y) ∈ E and x = t.

This is the so called “x-section” of E at the point x = t. It is a subset of R, but we can view it as a subset of R2 by

simply identifying each point y ∈ Gt with the point (t, y) ∈ E t = (x, y) ∈ E : x = t. It follows that, for each

t ∈ R, Gt is a finite subset of R. Therefore, mGt = 0. Finally, by (28),

m2E =

X

Y

χGt(y) dν (y) dµ(t) =

X

νGt dµ(t) = 0.

since νGt mGt = 0.

5. Let µ and ν be finite positive measures on the measurable space (X, A) such that ν µ ν , and let dνd(µ+ν)

represent the Radon-Nikodym derivative of ν with respect to µ + ν . Show that

0 <dν

d(µ + ν )< 1 a.e. [µ].

Solution: First note that ν µ implies ν µ + ν , so, by the Radon-Nikodym theorem (A.12), there is a unique

f ∈ L1(µ + ν ) such that

ν (E ) =

E

f d(µ + ν ) ∀E ∈ A.

Indeed,f

is the Radon-Nikodym derivative; i.e.,f =

dν

d(µ+ν). We want to show

0 < f (x) < 1holds for

µ-almost

every x ∈ X . Since we’re dealing with positive measures, we can assume f (x) ≥ 0 for all x ∈ X .

If B0 = x ∈ X : f (x) = 0, then

ν (B0) =

B0

f d(µ + ν ) = 0.

33The notationB(X) ⊗B(Y ) denotes the σ-algebra generated by all sets A × B ⊆ X × Y with A ∈ B(X) and B ∈ B(Y ). See A.1.7. In

this case,B(X) ⊗B(Y ) is the same as B(R2).

35




Therefore, µ(B0) = 0, since µ ν , which proves that f (x) > 0, [µ]-a.e.

If B1 = x ∈ X : f (x) ≥ 1, then

ν (B1) = B1

f d(µ + ν ) ≥ (µ + ν )(B1) = µ(B1) + ν (B1).

Since ν is finite by assumption, we can subtract ν (B0) from both sides to obtain µ(B1) = 0. This proves f (x) < 1,

[µ]-a.e.

6.34 Suppose that 1 < p < ∞ and that q = p/( p − 1).

(a) Let a1, a2, . . . be a sequence of real numbers for which the series

anbn converges for all real sequences bnsatisfying the condition

|bn|q < ∞. Prove that |an| p < ∞.

(b) Discuss the cases of p = 1 and p = ∞. Prove your assertions.

Solution: (a) For each k ∈ N, define T k : q(N) → R by T k(b) =kn=1 anbn, for b ∈ q(N). Then T k is a

family of pointwise bounded linear functionals. That is, each T k is a linear functional, and, for each b ∈ q(N),

there is an M b ≥ 0 such that |T k(b)| ≤ M b holds for all k ∈ N. To see this, simply note that a convergent sequence

of real numbers is bounded, and, in the present case, we have

S k kn=1

anbn →∞n=1

anbn = x ∈ R.

Thus, T k(b) = S k is a convergent sequence of reals, so, if N ∈ N is such that k ≥ N implies |S k − x| < 1,

and if M b is defined to be max|S k| : 1 ≤ k ≤ N , then, for any k ∈ N,

|T k(b)| ≤ M b maxM b, x + 1.

Next note that q is a Banach space, so the (Banach-Steinhauss) principle of uniform boundedness implies that

there is a single M > 0 such that T k ≤ M for all k ∈ N. In other words,

(∃ M > 0) (∀b ∈ q) (∀k ∈ N) |T k(b)| ≤ M b.

Define let T (b) ∞n=1 anbn = limk→∞ T k(b), which exists by assumption. Since |·| is continuous, we conclude

that limk→∞ |T k(b)| = |T (b)|. Finally, since |T k(b)| ≤ M b for all k ∈ N, we have |T (b)| ≤ M b. That is T is

a bounded linear functional on q(N).

Now, by the Riesz representation theorem, if 1 ≤ q < ∞, then any bounded linear functional T ∈ ∗q is uniquely

representable by some α = (α1, α2, . . . ) ∈ p as

T (b) =

∞n=1 αnbn. (29)

On the other hand, by definition, T (b) =∞n=1 anbn, for all b ∈ q. Since the representation in (29) is unique,

a = α ∈ p. That is,n |an| p < ∞.

34On the original exam this question asked only about the special case p = q = 2 .

36




(b) Consider the case p = 1 and q = ∞. First recall that the Riesz representation theorem says that every

T ∈ ∗q (1 ≤ q < ∞) is uniquely representable by some α ∈ p (where p = q/(q − 1), so 1 < p ≤ ∞). That is pis the dual of q , when 1 ≤ q < ∞ and p = q/(q − 1). However, in the present case we have q = ∞ and p = 1

and 1 is not the dual of ∞. (Perhaps the easiest way to see this is to note that 1 is separable but ∞ is not. Forthe collection of a ∈ ∞ such that an ∈ 0, 1, n ∈ N, is uncountable and, for any two distinct such sequences

a, b ∈ 0, 1N, we have a − b∞ = 1, so there cannot be a countable base, so ∞ is not second countable, and a

metric space is separable iff it is second countable.)

So we can’t use the same method of proof for this case. However, I believe the result still holds by the following

simple argument: Define b = (b1, b2, . . . ) by

bn = sgn(an) =

an/|an|, for an = 0,

0, for an = 0,(n ∈ N).

Thenn |an| =

n anbn converges by the hypothesis, since |bn| ∈ 0, 1 implies b ∈ ∞. Therefore, a ∈ 1.

Finally, in case p = ∞ and q = 1, the Riesz representation theorem can be applied as in part (a).

Please email comments, suggestions, and corrections to [email protected].

37



2 COMPLEX ANALYSIS

2 Complex Analysis

38



2.1 1989 April 2 COMPLEX ANALYSIS

2.1 1989 April

INSTRUCTIONS: Do at least four problems.

TIME LIMIT: 1.5 hours35

1. (a) Let U be the unit disk in the complex plane C, U = z ∈ C : |z| < 1, and let f be an analytic function in a

neighborhood of the closure of U . Show that if f is real on all the boundary of U , then f must be constant.

(b) Let u be a real harmonic function in all the complex plane C. Show that if u(z) ≥ 0 for all z ∈ C, then

u must be constant.

Solution: (a) The hypotheses imply that Imf (eiθ) = 0 for all θ ∈ R. Since f is holomorphic in a neighborhood

Ω of U , the series

f (z) =

∞n=0

anzn

converges uniformly on any compact subset of Ω

. The unit circle T= z : |z| = 1 = e

iθ

: θ ∈R

is one such

compact subset, and here the series is

f (eiθ) =∞n=0

aneinθ.

If we write the coefficients as an = cn + ibn, where cn, bn ∈ R, then we have

aneinθ = (cn + ibn)[cos(nθ) + i sin(nθ)] = [cn cos(nθ) − bn sin(nθ)] + i[cn sin(nθ) + bn cos(nθ)].

Thus, by the hypothesis, the series

Imf (eiθ) =∞n=0

[cn sin(nθ) + bn cos(nθ)]

converges uniformly to zero for all θ ∈ [0, 2π]. Therefore, with the possible exception of c0, we have cn = bn = 0,for all n, so f ≡ c0.

(b) Since C is simply connected, there is a real-valued harmonic conjugate v(z) such that the function f (z) =u(z) + iv(z) is entire. Now, since u(z) ≥ 0, f (z) maps the complex plane into the right half-plane, Ref (z) ≥ 0.

It follows immediately from Picard’s theorem that f must be constant.36 However, an elementary argument using

only Liouville’s theorem is probably preferable, so let f be as above, and define g(z) = f (z) + 1. Then g is

entire and maps C into w ∈ C : Rew ≥ 1. In particular, g(z) is bounded away from zero, so the function

h(z) = 1/g(z) is a bounded entire function. (In fact, |h(z)| ≤ 1.) Therefore, by Liouville’s theorem, h is constant,

so f is constant, so u = Ref is constant.

2. Let f be an analytic function in the region z : |z| > 1, and suppose that

limz→∞

f (z) = 0.

35In 1989 there was a single three hour test covering both real and complex analysis. Students were required to do nine problems, with at least

four from each part.36Picard’s theorem states that a non-constant entire function can omit at most one value of C from its range. This is a very powerful theorem, but

if you use it for an easy problem like this one, you might be accused of killing a fly with a sledge hammer!

39




Show that if |z| > 2, then1

2πi

|ζ|=2

f (ζ )

ζ

−z

dζ = −f (z). (30)

Solution: By Cauchy’s formula, if |z| < R, then

f (z) =1

2πi

|ζ|=R

f (ζ )

ζ − zdζ − 1

2πi

|ζ|=2

f (ζ )

ζ − zdζ. (31)

Note that this holds for all R > |z| > 2. Fix > 0. Let R be such that |f (ζ )| < /2 for all |ζ | = R and

R > 2|z|. Then |ζ − z| > R/2 for all |ζ | = R, so

1

2π

|ζ|=R

|f (ζ )||ζ − z| |dζ | <

/2

2π

2πRR/2

= .

Therefore, by (31), 12πi

|ζ|=2

f (ζ )ζ − z

dζ + f (z) < .

This holds for any , which proves (30).

3.37 Let U be the open unit disk in C, and let U + be the top half of this disk,

U + = z ∈ C : Imz > 0, |z| < 1.

Exhibit a one-to-one conformal mapping from U + onto U .

Solution: Consider ϕ0(z) = 1−z1+z

, a linear fractional transformation which takes U onto the right half-plane

Ω = z ∈ C : Rez > 0. (This property of ϕ0 can be seen by considering ϕ0(0) = 1 and ϕ0(∞) = −1 andarguing by symmetry.)

Note that ϕ0(1) = 0 and ϕ0(x) ∈ R, for all x ∈ R. Also, ϕ0 is conformal, so it preserves the right angle formed

by the intersection of the circle and the real axis at the point z = 1. Therefore, ϕ0 takes U + onto either the first

quadrant, Ω+ = z ∈ Ω : Imz > 0, or the fourth quadrant, Ω− = z ∈ Ω : Imz < 0. To see which, consider

z = i/2.

ϕ0(i/2) =1 − i/2

1 + i/2=

1 − i/2

1 + i/2

1 − i/2

1 − i/2

=

3 − 4i

5∈ Ω−.

Thus, ϕ0 : U + → Ω−.

Let ϕ1(z) = iz, so that ϕ1 : Ω− → Ω+, let ϕ2(z) = z2, so that ϕ2 : Ω+ → Imz > 0, and let ϕ3(z) = −iz,

so that ϕ3 : Imz > 0 → Ω = Rez > 0. Finally, note that ϕ−10 (z) = ϕ0(z) maps Ω onto U . Putting it all

together, we see that a map satisfying the requirements is

ϕ(z) = (ϕ0 ϕ3 ϕ2 ϕ1 ϕ0)(z).

37This problem also appears in April ’95 (5) and November ’06 (2).

40




4. Let f n be a sequence of analytic functions in the unit disk U , and suppose there exists a constant M such that

C |

f n(z)

| |dz

| ≤M

for each f n and for all circles C lying in U . Prove that f n has a subsequence converging uniformly on compact

subsets of U .

Solution: See the solution to problem 7 of November ’91.

5. Let Q be a complex polynomial with distinct simple roots at the points a1, a2, . . . , an, and let P be a complex

polynomial of degree less than that of Q. Show that

P (z)

Q(z)=

nk=1

P (ak)

Q(ak)(z − ak).

6. Use contour integration and the residue method to evaluate the integral ∞0

cos x

(1 + x2)2dx.

Solution: Denote the integral by I . Since the integrand is even,

2I =

∞−∞

cos x

(1 + x2)2dx.

Consider the simple closed contour ΓR = γ R∪

[−

R, R], where the trace of γ R is the set

Reiθ : 1≤

θ≤

π

,

oriented counter-clockwise. Note that, if R > 1, then i is inside the region bounded by ΓR.

The function

f (z) =cos z

(1 + z2)2=

cos z

(z + i)2(z − i)2

is holomorphic inside and on ΓR, except for a double pole at z = i, where the residue is computed as follows:

Res(f, i) = limz→i

d

dz[(z − i)2f (z)] = lim

z→i

d

dz

cos z

(z + i)2

= limz→i

−(z + i)2 sin z − 2(z + i)cos z

(z + i)4

=−(2i)2 sin(i) − 4i cos(i)

(2i)4

=sin(i) − i cos(i)

4=

−iei·i

4=

−i

4e.

By the residue theorem, it follows that, for all R > 1, ΓR

f (z) dz = 2πi Res(f, i) =π

2e.

41




It remains to check that γR

f (z) dz → 0, as R → ∞, which will allow us to conclude that

2I = ∞

−∞

f (x) dx = limR→∞

ΓR

f (z) dz−

γR

f (z) dz = limR→∞

ΓR

f (z) dz =π

2e. (32)

Indeed, γR

f (z) dz

=

γR

cos z

(1 + z2)2dz

≤ 1

(R2 − 1)2(γ R) =

πR

(R2 − 1)2.

This inequality holds for all R > 1, so, letting R → ∞, we have γR

f (z) dz → 0. Therefore, by (32),

I =

∞0

cos x

(1 + x2)2dx =

π

4e.

42



2.2 1991 November 21 2 COMPLEX ANALYSIS


INSTRUCTIONS: In each of sections A, B, and C, do all but one problem.

TIME LIMIT: 2 hours

SECTION A

(Do 3 of the 4 problems.)

1. Where does the function

f (z) = zRez + zImz + z

have a complex derivative? Compute the derivative wherever it exists.

Solution: Writing f in terms of the real and imaginary parts of z = x + iy, we have

f (x + iy) = (x + iy)x + (x − iy)y + x − iy

= x2 + xy + x + i(xy−

y2

−y)

= u(x, y) + iv(x, y),

where u(x, y) = x2 + xy + x and v(x, y) = xy − y2 − y are the real and imaginary parts of f . Therefore,

ux = 2x + y + 1 vy = x − 2y − 1 (33)

uy = x vx = y. (34)

If f is holomorphic in some region, the Cauchy-Riemann equations (ux = vy, uy = −vx) must hold there. By (33)

and (34), this requires 2x + y + 1 = x − 2y − 1 and x = −y. Substituting the second equation into the first yields

−y + 1 = −3y − 1, or y = −1. Then, since x = −y, we must have x = 1. Therefore, f has a complex derivative

at (x, y) = (1, −1), or z = 1 − i.

For any region Ω

⊆C, we define the linear functional ∂ : H (Ω)

→C by ∂ = 1

2 ∂ ∂x −

i ∂

∂y, and recall that, if

f ∈ H (Ω), then the derivative of f is given by f (z) = (∂f )(z), z ∈ Ω. In the present case,

∂f

∂x= 2x + y + 1 + iy,

∂f

∂y= x + i(x − 2y − 1).

Therefore, ∂f (x + iy) = 12 [(2x + y + 1 + iy) − i(x + i(x − 2y − 1))] = 1

2 [(3x − y) + i(y − x)], and finally,

f (1 − i) =1

2(4 − 2i) = 2 − i.

2. (a) Prove that any nonconstant polynomial with complex coefficients has at least one root.

(b) From (a) it follows that every nonconstant polynomial P has the factorization

P (z) = a

N n=1

(z − λn),

43




where a and each root λn are complex constants. Prove that if P has only real coefficients, then P has a factorization

P (z) = aK

k=1

(z−

rk

)M

m=1

(z2

−bm

z + cm

),

where a and each rk, bm, cm are real constants.

3. Use complex residue methods to compute the integral π0

1

5 + 3 cosθdθ.

Solution: Let I =

π

01

5+3cos θ dθ. Note that cos θ is an even function (i.e., cos(−θ) = cos θ), so

2I = π−π

15 + 3 cosθ

dθ.

For z = eiθ,

cos θ =eiθ + e−iθ

2=

1

2(z +

1

z),

and dz = ieiθdθ, from which it follows that

2I =

|z|=1

1

5 + 32 (z + 1

z )

dz

iz

=1

i |z|=1

dz

5z + 32(z2 + 1)

=2

3i

|z|=1

dz

z2 + 103 z + 1

.

Let p(z) = z2 + 103 z + 1. Then the roots of p(z) are z1 = −1/3 and z2 = −3. Only z1 = −1/3 is inside the circle

|z| = 1, so the residue theorem implies

2I =2

3i· 2πi · Res

1

p(z), z1

.

Now,1

p(z)=

1

(z − z1)(z − z2),

which implies

Res

1

p(z), z1

= limz→z1

1

z − z2=

1

−13 − (−3)

=3

8.

Therefore,

2I =2

3i· 2πi · 3

8=

π

2,

so I = π4 .

44




4. (a) Explain how to map an infinite strip (i.e., the region strictly between two parallel lines) onto the unit disk by a

one-to-one conformal mapping.

(b) Two circles lie outside one another except for common point of tangency. Explain how to map the region

exterior to both circles (including the point at infinity) onto an infinite strip by a one-to-one conformal mapping.

SECTION B


5.38 Suppose that f is analytic in the annulus 1 < |z| < 2, and that there exists a sequence of polynomials converging

to f uniformly on every compact subset of this annulus. Show that f has an analytic extension to all of the disk

|z| < 2.

Solution: Note that the function f , being holomorphic in the annulus 1 < |z| < 2, has Laurent series representa-

tion

f (z) =

∞n=−∞

an(z − z0)n

,

converging locally uniformly for 1 < |z| < 2, where z0 is any point in the disk |z| < 2. I claim that an = 0 for all

negative integers n. To see this, first recall the formula for the coefficients in the Laurent series,

an =1

2πi

|z|=R

f (z)

(z − z0)n+1dz, (n ∈ Z; 1 < R < 2).

Let pm be the sequence of polynomials mentioned in the problem statement. Of course, pm ∈ H (C), so Cauchy’s

theorem implies |z|=R pm(z) dz = 0, and, more generally,

|z|=R

pm(z)(z − z0)−n−1 dz = 0, (n = −1, −2, . . . ).

Therefore,

|an| =1

2π

|z|=R

f (z)

(z − z0)n+1dz −

|z|=R

pm(z)

(z − z0)n+1dz

≤ 1

2π

|z|=R

|f (z) − pm(z)||z − z0|n+1

|dz|. (35)

Finally, pm → f uniformly on |z| = R, so (35) implies |an| = 0 for n = −1, −2, . . . . This proves that f (z) =∞n=0 an(z − z0)n, converging locally uniformly in |z| < 2. Whence f ∈ H (|z| < 2).

6. Let f be analytic in |z| < 2, with the only zeros of f being the distinct points a1, a2, . . . , an, of multiplicities

m1, m2, . . . , mn, respectively, and with each aj lying in the disk |z| < 1. Given that g is analytic in |z| < 2, whatis |z|=1

f (z)g(z)

f (z)dz ?

(Verify your answer.)

38See also: April ’96 (8).

45




7. Let f n be a sequence of analytic functions in the unit disk D, and suppose there exists a positive constant M such that

C |

f n(z)

| |dz

| ≤M

for each f n and for every circle C lying in D. Prove that f n has a subsequence converging uniformly on compact

subsets of D.

Solution: We must show that F = f n is a normal family. If we can prove that F is a locally bounded family

of holomorphic functions – that is, F ⊂ H (D) and, for any compact set K ⊂ D, there is an M K > 0 such that

|f n(z)| ≤ M K for all z ∈ K and all n = 1, 2, . . . – then the Montel theorem (corollary 2.2) will give the desired

result.

To show F is locally bounded, it is equivalent to show that, for each point zα ∈ D, there is a number M α and a

neighborhood B(zα, rα) ⊂ D such that |f n(z)| ≤ M α for all z ∈ B(zα, rα) and all n = 1, 2, . . .. (Why is this

equivalent?)39 So, fix zα ∈ D. Let Rα > 0 be such that B(zα, Rα) = z ∈ C : |z − zα| ≤ Rα ⊂ D. Then, for

any z ∈ B(zα, Rα/2), Cauchy’s formula gives

|f n(z)| ≤ 1

2π

|ζ−zα|=Rα

|f n(ζ )||ζ − z| |dζ |

≤ 1

2π

1

Rα/2

|ζ−zα|=Rα

|f n(ζ )| |dζ |

≤ M

πRα.

The second inequality holds since |ζ − zα| = Rα and |z − zα| < Rα/2 imply |ζ − z| > Rα/2. The last inequality

follows from the hypothesis C

|f n(z)| |dz| ≤ M for any circle C in D. Letting M α = M πRα

, and rα = Rα/2, we

have |f n(z)| ≤ M α for all z ∈ B(zα, rα) and all n = 1, 2, . . . , as desired.

8. State and prove:

(a) the mean value property for analytic functions

(b) the maximum principle for analytic functions.

39Answer: If K ⊂ D is compact, we could select a finite covering of K by such neighborhoods B(zαj , rαj ) ( j = 1, . . . , J ) and then

|f n(z)| ≤ maxj M αj M K , for all z ∈ K and n = 1, 2, . . . .

46




SECTION C


9. Let X be a Hausdorff topological space, let K be a compact subset of X , and let x be a point of X not in K . Show

that there exist open sets U and V such that

K ⊂ U, x ∈ V, U ∩ V = ∅.

10. A topological space X satisfies the second axiom of countability. Prove that every open cover of X has a countable

subcover.

11. Let X be a topological space, and let U be a subset of X .

(a) Show that if an open set intersects the closure of Y then it intersects Y .

(b) Show that if Y is connected and if Y ⊂ Z ⊂ Y , then Z is connected.

47



2.3 1995 April 10 2 COMPLEX ANALYSIS

2.3 1995 April 10

Instructions. Work as many of the problems as you can. Each solution should be clearly written on a separate sheet

of paper.

1. Let f (z) =

anzn be an entire function.

(a) Suppose that |f (z)| ≤ A|z|N + B for all z ∈ C where A, B are finite constants. Show that f is a polynomial

of degree N or less.

(b) Suppose that f satisfies the condition: |f (zn)| → ∞ whenever |zn| → ∞. Show that f is a polynomial.

Solution: (a) By Cauchy’s formula, we have

an =f (n)(0)

n!=

1

2πi

|ζ|=R

f (ζ )

ζ n+1dζ,

for every R > 0. Therefore,

|an| ≤ 12π |ζ|=R

|f (ζ )||ζ |n+1 |dζ | ≤ 12π A R

N

+ BRn+1 2πR = A RN −n + B R−n.

Again, this holds for every R > 0. Thus, for any n > N and > 0, taking R large enough forces |an| < (n = N + 1, N + 2, . . . ). Since was arbitrary, we have an = 0 for all n = N + 1, N + 2, . . . . Therefore,

f (z) =N n=0 anzn.

(b) We give three different proofs. The first is the shortest, but relies on the heaviest machinery.

Proof 1: If we take for granted that any transcendental (i.e. non-polynomial) entire function has an essential singu-

larity at infinity, then the Casorati-Weierstrass theorem (see 3 of Nov. ’01) implies that, for any complex number

w, there is a sequence zn with zn → ∞ and f (zn) → w as n → ∞. Since this contradicts the given hypotheses,

f (z) cannot be a transcendental function. That is, f (z) must be a polynomial.

Proof 2: Since f ∈ H (C), the series f (z) =

anzn converges locally uniformly in C. The hypotheses imply that

the function f (1/z) has a pole at z = 0. Let

g(z) = f (1/z) =∞

n=−∞

bnzn

be the Laurent series expansion of the function g about z = 0. Suppose the pole at z = 0 is of order m. Clearly mis finite, by the criterion for a pole (i.e., limz→0 f (1/z) = ∞). Therefore, we can write

g(z) = f (1/z) =

∞n=−m

bnzn = b−mz−m + b−m+1z−m+1 + · · · b−1z−1 + b0 + b1z + · · · (36)

Now f is entire, so it has the form f (z) =∞

n=0 anzn

, which implies that f (1/z) = a0 + a1z−1

+ a2z−2

+ · · · .Compared with (36),

a0 + a1z−1 + a2z−2 + · · · = f (1/z) = b−mz−m + b−m+1z−m+1 + · · · b−1z−1 + b0 + b1z + · · ·That is, 0 = am+1 = am+2 = · · · , so

f (z) =

mn=0

anzn.

48




Proof 3: By the hypotheses, there is an R > 0 such that |f (z)| > 0 for all |z| > R. Therefore, the zeros of f are

confined to a closed disk DR = |z| ≤ R. Since the zeros of f are isolated, there are at most finitely many of

them in any compact subset of C. In particular, DR contains only finitely many zeros of f . This proves that f has

only finitely many zeros in C.Let α1, . . . , αN be the collections of all zeros of f (counting multiplicities). Consider the function

g(z) =f (z)

(z − α1) · · · (z − αN ). (37)

This is defined and holomorphic in C\α1, . . . , αN , but the αi’s are removable singularities, so g(z) is a nonzero

entire function. In particular, for any R > 0,

minz∈DR

|g(z)| ≥ min|z|=R

|g(z)| = > 0,

for some > 0. Therefore, 1/g is a bounded entire function, hence constant, by Liouville’s theorem. What we

have shown is that the left hand side of (37) is constant, and this proves that f (z) is a polynomial.

Remark: A nice corollary to part (b) is the following:

Corollary 2.1 If f is an injective entire function, then f (z) = az + b for some constants a and b.

The proof appears below in section 2.9.

2. (a) State a form of the Cauchy theorem.

(b) State a converse of the Cauchy theorem.

Solution: (a) See theorem A.16.

(b) See theorem A.18.

3. Let40 f (z) =∞n=0 anzn be analytic and one-to-one on |z| < 1. Suppose that |f (z)| < 1 for all |z| < 1.

(a) Prove that∞n=1

n|an|2 ≤ 1.

(b) Is the constant 1 the best possible?

Solution: (a) This is a special case of the following area theorem:

Theorem 2.1 Suppose f (z) =∞

n=0 anzn

is a holomorphic function which maps the unit disk D = |z| < 1bijectively onto a domain f (D) = G having area A. Then

A = π∞n=1

n|an|2.

40On the original exam, the power series representation was given as f (z) =∞

n=1 anzn. However, the problem can be solved without

assuming a0 = 0 a priori.

49




Proof: The area of the image of D under f is the integral over D of the Jacobian of f . That is,

A =

D

|f (z)|2 dxdy.

Compute |f (z)| by differentiating the power series of f (z) term by term,

f (z) =

∞n=1

nanzn−1

.

Next, take the squared modulus,

|f (z)|2 =

∞m,n=1

m n amanzm−1

zn−1

.

This gives,

A =

D

∞m,n=1

m n amanzm−1z

n−1 dxdy.

Letting z = reiθ,

A =

∞m,n=1

m n aman 10

2π0

rm+n−1

ei(m−n)θ

dθ dr.

Now, for all k = 0, the integral of eikθ over 0 ≤ θ < 2π vanishes, so the only non-vanishing terms of the series are those for

which m = n. That is,

A = 2π

∞n=1

n2|an|2

10

r2n−1

dr = π

∞n=1

n2|an|2. (38)

To apply this theorem to the problem at hand, note that the hypotheses of the problem imply that f maps the unit

disk bijectively onto its range f (D), which is contained inside D and, therefore, has area less or equal to π. This

and (38) together imply

π ≥ π∞

n=1

n2|an|2,

which gives the desired inequality.

(b) The identity function f (z) = z satisfies the given hypotheses and its power series expansion has coefficients

a1 = 1 and 0 = a0 = a2 = a3 = · · · . This shows that the upper bound of 1 is obtained and is therefore the best

possible.

4. Let u(z) be a nonconstant, real valued, harmonic function on C. Prove there exists a sequence zn with |zn| → ∞for which u(zn) → 0.

Solution: Suppose, by way of contradiction, that there is no such sequence. Then u(z) is bounded away from zero

for all z in some neighborhood of infinity, say, |z| > R, for some R > 0. Since u is continuous, either u(z) > 0for all |z| > R, or u(z) < 0 for all |z| > R. Assume without loss of generality that u(z) > 0 for all |z| > R.

Since u is continuous on the compact set |z| ≤ R, it attains its minimum on that set. Thus, there is an M > 0such that −M ≤ u(z) for all |z| ≤ R. Consider the function U (z) = u(z) + M . By construction, U (z) ≥ 0 for

all z ∈ C, and U is harmonic in C. But this implies U (z), hence u(z), must be constant.41 This contradicts the

hypothesis that u(z) be nonconstant and completes the proof. 41Recall problem 1(b), April ’89, where we proved that a real valued harmonic function u(z) satisfying u(z) ≥ 0 for all z ∈ Cmust be constant.

50




5.42 Find an explicit conformal mapping of the semidisk

H = z : |z| < 1, Rez > 0

onto the unit disk.

Solution: See the solution to (3) of April ’89, or (2) of November ’06.

6.43 Suppose f (z) is a holomorphic function on the unit disk which satisfies:

|f (z)| < 1 all |z| < 1.

(a) State the Schwarz lemma, as applied to f .

(b) If f (0) = 12 , how large can |f (0)| be?


(b) Assume f satisfies the given hypotheses. In particular, f (0) = 12 . Consider the map

ϕ(z) =12 − z

1 − z2

.

This is a holomorphic bijection of the unit disk, with φ(1/2) = 0. Therefore, g = ϕ f satisfies the hypotheses of

Schwarz’s lemma. In particular, |g(0)| ≤ 1. Since g(z) = ϕ(f (z))f (z), we have

1 ≥ |g(0)| = |ϕ(1/2)||f (0)|. (39)

Now,

ϕ(z) = − 1−z

2 + 1

2 1

2 −z

1 − z2

2 .

Therefore, ϕ(1/2) = −4/3, and it follows from (39) that

|f (0)| ≤ 1

|ϕ(1/2)| = 3/4.

42This problem also appears in April ’89 (3) and November ’06 (2).43A very similar problem appeared in November ’06 (3).

51





Instructions. Make a substantial effort on all parts of the following problems. If you cannot completely answer Part

(a) of a problem, it is still possible to do Part (b). Partial credit is given for partial progress. Include as many details as

time permits. Throughout the exam, z denotes a complex variable, and C denotes the complex plane.

1. (a) Suppose that f (z) = f (x + iy) = u(x, y) + iv(x, y) where u and v are C 1 functions defined on a neighborhood

of the closure of a bounded region G ⊂ C with boundary which is parametrized by a properly oriented, piecewise

C 1 curve γ . If u and v obey the Cauchy-Riemann equations, show that Cauchy’s theorem γ f (z) dz = 0 follows

from Green’s theorem, namely γ

P dx + Q dy =

G

∂Q

∂x− ∂P

∂y

dxdy for C 1 functions P and Q. (40)

(b) Suppose that we do not assume that u and v are C 1, but merely that u and v are continuous in G and

f

(z0) = limz→z0

f (z)

−f (z0)

z − z0

exists at some (possibly only one!) point z0 ∈ G. Show that given any > 0, we can find a triangular region ∆containing z0, such that if T is the boundary curve of ∆, then

T

f (z) dz

=1

2L2,

where L is the length of the perimeter of ∆.

Hint for (b) Note that part (a) yields T (az + b) dz = 0 for a, b ∈ C, which you can use here in (b), even if you

could not do Part (a). You may also use the fact that T g(z) dz

≤ L · sup|g(z)| : z ∈ T for g continuous on T .

Solution: (a) Let P = u and Q =

−v in (40). Then, by the Cauchy-Riemann equations,44

γ

u(x, y) dx − v(x, y) dy =

G

(vx + uy) dxdy = 0. (41)

Similarly, if P = v and Q = u in Green’s theorem, then the Cauchy-Riemann equations imply γ

v(x, y) dx + u(x, y) dy =

G

(ux − vy) dxdy = 0. (42)

Next, note that

f (z) dz = [u(x, y) + iv(x, y)] d(x + iy) = u(x, y) dx − v(x, y) dy + i[v(x, y) dx + u(x, y) dy].

Therefore, by (41) and (42), γ

f (z) dz =

γ

u(x, y) dx − v(x, y) dy + i

γ

v(x, y) dx + u(x, y) dy = 0.

44These are ux = vy and uy = −vx.

52




(b) Suppose u and v are continuous and f (z) exists at the point z0 ∈ G. Then, for any > 0 there is a δ > 0 such

that B(z0, δ) ⊆ G, and

f (z0)−

f (z) − f (z0)

z − z0 < , for all

|z

−z0

|< δ .

Pick a triangular region ∆ ⊂ B(z0, δ) with z0 ∈ ∆, and let T be the boundary. Define

R(z) = f (z) − [f (z0) + f (z0)(z − z0)].

Then, by Cauchy’s theorem (part (a)), T [f (z0) + f (z0)(z − z0)] dz = 0, whence

T R(z) dz =

T f (z) dz.

Finally, note that R(z)

z − z0

=

f (z0) − f (z) − f (z0)

z − z0

< , for all |z − z0| < δ .

Therefore,

T

f (z) dz = T

R(z) dz ≤ T |

R(z)

| |dz

|=

T

R(z)

z − z0

|z − z0| |dz|

≤

T

|z − z0| |dz| ≤ rL.

where L denotes the length of the perimeter of ∆ (i.e., the length of T ), and r denotes the length of one side of T ,which must, of course, be greater than |z − z0| for all z ∈ T . Also, the length of one side of ∆ is surely less than

half the length of the perimeter (i.e., r < L/2). Therefore, T

f (z) dz

≤ 1

2L2.

2. Give two quite different proofs of the fundamental theorem of algebra, that if a polynomial with complex coeffi-

cients has no complex zero, then it is constant. You may use independent, well-known theorems and principles

such as Liouville’s theorem, the argument principle, the maximum principle, Rouche’s theorem, and/or the open

mapping theorem.

Solution: In the two proofs below, we begin by supposing p(z) is not constant and thus has the form p(z) =a0 + a1z + a2z2 + · · · + anzn with an = 0 for some n ≥ 1. Both proofs also rely on the following observation:

If ajnj=0 ⊂ C with an = 0, then for all 1 ≤ R ≤ |z| < ∞,

a0an

z−n + · · · +an−1

anz−1

≤ |a0|

|an

||z|−n + · · · +

|an−1|

|an

||z|−1

≤ n max0≤j<n

|aj ||an| |z|−1

≤ n max0≤j<n

|aj ||an|R−1.

53




In particular, if we choose45 R = 1 + 2 n max0≤j<n |aj |/|an|, then

a0

anz−n +

· · ·+

an−1

anz−1 ≤

1/2, for all

|z

| ≥R. (43)

Proof 1: Assume p(z) = a0+a1z+· · ·+anzn with an = 0 for some n ≥ 1, and let R = 1+2 n max0≤j<n |aj |/|an|,as above. We claim that

| p(z) − anzn| < |anzn|, for all |z| = R. (44)

To see this, check that

| p(z) − anzn||anzn| =

a0an

z−n + · · · +an−1

anz−1

< 1, for all |z| = R.

In fact, (43) implies that the sum is no greater than 1/2, for all

|z

| ≥R, which is more than we need. Now (44)

and Rouche’s theorem imply that the function g(z) = anzn has the same number of zeros in |z| < R as does thefunction p(z). Clearly z = 0 is a zero of g(z) (of multiplicity n). Therefore, p(z) has a zero in |z| < R.

Proof 2:46 Assume p(z) = a0 + a1z + · · · + anzn with an = 0 for some n ≥ 1, and consider

| p(z)| = |anzn| a0

anz−n + · · · +

an−1

anz−1 + 1

≥ |an||z|n1 − |

n−1j=0

ajan

z−n+j | . (45)

If we choose R = 1 + 2 n max0≤j<n |aj |/|an| as above, then

0

≤ n−1

j=0

aj

anz−n+j = a0

anz−n +

· · ·+

an−1

anz−1 ≤

1/2, for all

|z

| ≥R,

and (45) becomes | p(z)| ≥ |an||z|n/2, for all |z| ≥ R. Therefore, the function f (z) 1/p(z) satisfies

|f (z)| =1

| p(z)| ≤ 2

|an||z|n , for all |z| ≥ R.

Now suppose p(z) has no complex zero. Then f (z) ∈ H (C). In particular, f (z) is continuous, hence bounded

on the compact set |z| ≤ R. Therefore f (z) is a bounded entire function, so, by Liouville’s theorem, it must be

constant, but then p(z) must be constant. This contradicts our initial assumption and proves that p(z) must have a

complex zero. In fact, we have proved a bit more: If p(z) = a0 + a1z + · · · + anzn with an = 0 for some n ≥ 1, and R is

either 1 or R = 2 n max0≤j<n

|aj

|/

|an

|(whichever is greater), then p(z) vanishes for some

|z

|< R, while for all

|z| ≥ R, | p(z)| is bounded from below by |an||z|n/2. Thus all the zeros of p(z) are contained in the disk |z| < R.

45Note, we add 1 here just to be sure R is safely over 1.46Conway [3] (p. 77) presents a similar, but more elegant proof.

54




3. (a) State and prove the Casorati-Weierstrass theorem concerning the image of any punctured disk about a certain

type of isolated singularity of an analytic function. You may use the fact that if a function g is analytic and bounded

in the neighborhood of a point z0, then g has a removable singularity at z0.

(b) Verify the Casorati-Weierstrass theorem directly for a specific analytic function of your choice, with a suitablesingularity.

Solution:

Theorem 2.2 (Casorati-Weierstrass) If f is a holomorphic function in a region G ∈ C except for an essential

singularity at the point z = z0, then for any w ∈ C there is a sequence zn ⊂ G approaching z0 such that

f (zn) → w as n → ∞.

Proof: Fix w0 ∈ C and suppose there is no sequence zn ⊂ G approaching z0 such that f (zn) → w0 as n → ∞.

Then there is a punctured disk D0 B(z0, ) \ z0 ⊂ G such that |f (z) − w0| > δ > 0 for all z ∈ D0. Define

g(z) = 1/(f (z) − w0) on D0. Then

limsupz→z0z∈D0

|g(z)| = limsupz→z0z∈D0

1

|f (z) − w0| ≤1

δ < ∞.

Thus, by lemma 2.1 (Nov. ’06, prob. 1), z0 is a removable singularity of g(z). Therefore, g(z) ∈ H (B(z0, )). In

particular, g is continuous and non-zero at z = z0, so it is non-zero in a neighborhood B(z0, 0) of z0. Therefore,

f (z)−w0 = 1/g(z) is holomorphic in B(z0, 0), which implies that the singularity of f (z) at z = z0 is removable.

This contradiction proves the theorem.

(b) Consider f (z) = ez . This function has an essential singularity at ∞, and, for every horizontal strip,

S α = x + iy : x ∈ R, α ≤ y < α + 2π,

of width 2π, f (z) maps S α onto C \ 0. (In particular, f (z) comes arbitrarily close to every w ∈ C.) Now let

N R = z ∈ C : |z| > R be any neighborhood of ∞. There is clearly a strip S α contained in N R (e.g., withα = R + 1). Therefore, f (z) = ez maps points in N R to points arbitrarily close (in fact equal when w = 0) to all

points w ∈ C.

4. (a) Define γ : [0, 2π] → C by γ (t) = sin(2t) + 2i sin(t). This is a parametrization of a “figure 8” curve, traced

out in a regular fashion. Find a meromorphic function f such that γ

f (z) dz = 1. Be careful with minus signs and

factors of 2πi.

(b) From the theory of Laurent expansions, it is known that there are constants an such that, for 1 < |z| < 4,

1

z2 − 5z + 4=

∞n=−∞

anzn.

Find a−10 and a10 by the method of your choice.

Solution: (a) Let G be the region whose boundary is the curve γ , and suppose f (z) ∈ H (C) except for isolated

singularities at the points z1, . . . , zn ⊂ G. By the residue theorem, γ

f (z) dz = 2πinj=1

Res(f, zj).

55




Therefore, if we were to find a function f (z) ∈ H (C) with exactly two isolated singularities in G (e.g., at z1 = iand z2 = −i), and such that Res(f, zj) = −i

4π , then

γ

f (z) dz = 2πij

Res(f, zj) = 2πi−i

4π+ −i

4π

= 1,

and the problem would be solved. Clearly,

f (z) =−i

4π

1

z − i− 1

z + i

=

1

2πi

z

z2 + 1

is such a function.

(b) Expand the function in partial fractions:

1

z2 − 5z + 4=

1

(z − 4)(z − 1)=

1/3

z − 4 −1/3

z − 1.

Then, note that

1/3

z − 4=

1

3

−1

4(1 − z/4)= − 1

12

∞n=0

z

4

nconverges for |z| < 4, while

1/3

z − 1= −1

3

1

z(1 − 1/z)= − 1

3z

∞n=0

z−n

converges for |z| > 1. Therefore,

1

z2

− 5z + 4

=

−1

3

−1

n=−∞

zn

−1

12

∞

n=0

1

4n

zn, for 1 <

|z

|< 4.

∴ a−10 = −1

3and a10 = − 1

12

1

4

10

.

5. (a) Suppose that f is analytic on a region G ⊂ C and z ∈ C : |z − a| ≤ R ⊂ G. Show that if |f (z)| ≤ M for

all z with |z − a| = R, then for any w1, w2 ∈ z ∈ C : |z − a| ≤ 12R, we have

|f (w1) − f (w2)| ≤ 4M

R|w1 − w2|

(b) Explain how Part (a) can be used with the Arzela-Ascoli theorem to prove Montel’s theorem asserting the

normality of any locally bounded family F of analytic functions on a region G.

Solution: (a) By Cauchy’s formula (A.14), if w is any point in the disk |w − a| < R, then

f (w) =1

2πi

|ζ−a|=R

f (ζ )

ζ − wdζ.

56




In particular, if w1, w2 are any two points inside the “half-disk” |w − a| < R/2 (see figure 1), then

f (w1)

−f (w2) =

1

2πi |ζ−a|=R

f (ζ )

ζ − w1 −f (ζ )

ζ − w2 dζ

=w1 − w2

2πi

|ζ−a|=R

f (ζ )

(ζ − w1)(ζ − w2)dζ.

R

.w

.a

W1

W2

.

. R/2

.ζ

Figure 1: Note that, if ζ is any point on the outer radius, |ζ − a| = R , and if w is any point in the disk |w − a| < R/2 ,

then |ζ − w| > R/2.

Now, for all ζ on the outer radius in figure 1, it is clear that |ζ − w1| > R/2 and |ζ − w2| > R/2. Therefore,

|f (w1) − f (w2)| ≤ |w1 − w2|2π

|ζ−a|=R

|f (ζ )|(R/2)2

|dζ |

≤ |w1 − w2|2π

supγ |f (ζ )|R2/4

(γ )

≤ 4M

R|w1 − w2|,

where γ denotes the positively oriented circle ζ : |ζ − a| = R, and (γ ) denotes its length, 2πR.

(b)47 We must explain how part (a) can be used with the Arzela-Ascoli theorem to prove Montel’s theorem asserting

the normality of any locally bounded family

F ⊂H (G).

Theorem 2.3 (Arzela-Ascoli) Let F ⊂ C (G, S ) be a family of continuous functions from an open set G ⊆ C into

a metric space (S, d). Then F is a normal family if and only if

(i) F is equicontinuous on each compact subset of G, and

(ii) for each z ∈ G, the set f (z) : f ∈ F is contained in a compact subset of S .

47The best treatment of normal families and the Arzela-Ascoli theorem is Ahlfors [1].

57




Recall that a family F of functions is called locally bounded on G iff for all compact K ⊂ G there is a constant

M K such that |f (z)| ≤ M K for all f ∈ F and z ∈ K .

Corollary 2.2 (little Montel theorem) Assume the set-up of the Arzela-Ascoli theorem, and suppose S = C andF ⊂ H (G). Then F is a normal family if and only if it is locally bounded.

Because of the way the problem is stated, it is probably enough to prove just one direction of Montel’s theorem;

i.e., local boundedness implies normality. For a proof of the other direction, see Conway [3], page 153.

Let S = C in the Arzela-Ascoli theorem. In that case, K ⊂ C is compact if and only if K is closed and bounded.

Therefore, if F is locally bounded, condition (ii) of the theorem is clearly satisfied. To check that local boundedness

also implies condition (i), we use part (a).

It suffices (why?)48 to prove that for any a ∈ G there is a neighborhood B(a, r) in which F is equicontinuous with

equicontinuity constant49 δ. So, fix a ∈ G and > 0, and let B(a, R) ⊂ G. Then, by local boundedness, there is

an M > 0 such that |f (z)| ≤ M for all z ∈ B(a, R) and all f ∈ F . Therefore, by part (a),

|f (w1) − f (w2)| ≤4M

R |w1 − w2|, for all w1, w2 ∈ |w − a| ≤ R/2.

If δ = R4M and r = R/2, then |f (w1) − f (w2)| < whenever w1, w2 ∈ B(a, r) and |w1 − w2| < δ. Therefore,

F is equicontinuous in B(a, r). We have thus shown that local boundedness implies conditions (i) and (ii) of the

Arzela-Ascoli theorem and thereby implies normality.

48Answer: If, instead of a single point a ∈ G, we are given a compact set K ⊂ G, then there is a finite cover B(aj, rj) : j = 1, . . . , n by

such neighborhoods with equicontinuity constants δ1, . . . , δn. Then, δ = minj δj , is a single equicontinuity constant that works for all of K .49The careful reader might note the distinction between this type of “uniform” equicontinuity, which is taken for granted in complex analysis

texts, e.g., Ahlfors [1] and Rudin [8], and the “pointwise” equicontinuity discussed in topology books like the one by Munkres [5]. To make peace

with this apparent discrepancy, check that the two notions coincide when the set on which a family of functions is declared equicontinuous is

compact.

58




2.5 2004 April 19

Instructions. Use a separate sheet of paper for each new problem. Do as many problems as you can. Complete

solutions to five problems will be considered as an excellent performance. Be advised that a few complete and well

written solutions will count more than several partial solutions.

Notation. D(z0, R) = z ∈ C : |z − z0| < R R > 0. For an open set G ⊆ C, H (G) will denote the set of

functions which are analytic in G.

1. Let γ be a rectifiable curve and let ϕ ∈ C (γ ∗). (That is, ϕ is a continuous complex function defined on the trace,

γ ∗, of γ .)

Let F (z) = γϕ(w)(w−z) dw, z ∈ C \ γ ∗.

Prove that F (z) = γϕ(w)

(w−z)2 dw, z ∈ C \ γ ∗, without using Leibniz’s Rule.

2. (a) State the Casorati-Weierstrass theorem.

(b) Evaluate the integral

I =1

2πi

|z|=R

(z − 3)sin

1

z + 2

dz where R ≥ 4.

3. Let f (z) be an entire function such that f (0) = 1, f (0) = 0 and

0 < |f (z)| ≤ e|z| for all z ∈ CProve that f (z) = 1 for all z ∈ C.

Solution: I know of two ways to prove this. One can be found in Rudin’s Functional Analysis ([9], p. 250). The

other goes as follows:50

By the Hadamard factorization theorem (see, e.g., [11]), an entire function f with zeros at an ⊂ C \ 0 and mzeros at z = 0 has the form

f (z) = eP (z)zm∞n=0

1 − z

an

ez/an , (46)

where P (z) is a polynomial of degree ρ, the “order of growth,” and k ≤ ρ < k + 1. For the function in question,

we have |f (z)| > 0 so an = ∅ and m = 0. Also, since |f (z)| ≤ e|z|, the order of growth is ρ = 1, which implies

that P (z) is a polynomial of degree 1. Therefore, (46) takes the simple form,

f (z) = eBz+C ,

for some constants B, C . We are given that f (0) = 1 and f (0) = 0, so eC = 1, and f (0) = BeC = B = 0. It

follows that f (z) = 1.

50This proof came to me by sheer lucky coincidence – I worked on this exam after having just read a beautiful treatment of the Hadamard

factorization theorem in Stein and Sharkachi’s new book [11]. If you need convincing that this theorem is worth studying, take a look at how easily

it disposes of this otherwise challenging exam problem. Also, Stein and Sharkachi seem to have set things up just right, so that the theorem is very

easy to apply.

59





Notation: C is the set of complex numbers, D = z ∈ C : |z| < 1 is the open unit disk, Π+ and Π− are the upper

and lower half-planes, respectively, and, given an open set G⊂C, H (G) is the set of holomorphic functions on G.

1. (a) Suppose that f ∈ H (D \ 0) and that |f (z)| < 1 for all 0 < |z| < 1. Prove that there is F ∈ H (D) with

F (z) = f (z) for all z ∈ D \ 0.

(b) State a general theorem about isolated singularities for holomorphic functions.

Solution:

Lemma 2.1 Suppose G ⊂ C is an open set and f is holomorphic in G except for an isolated singularity at z0 ∈ G.

If

limsupz→z0z∈G

|f (z)| < ∞,

then z0 is a removable singularity and f may be extended holomorphically to all of G.

Proof: Under the stated hypotheses, there is an > 0 and an M > 0 such that the deleted neighborhood Bo z ∈ C : 0 < |z − z0| ≤ is contained in G and such that |f (z)| ≤ M for all z ∈ Bo.

Let

f (z) =∞

n=−∞

an(z − z0)n

be the Laurent expansion of f for z ∈ Bo, where

an =1

2πi

C

f (ζ )

(ζ − z0)n+1dζ.

Here C denotes the positively oriented circle

|ζ

−z0

|= . Changing variables,

ζ = z0 + eiθ ⇒ dζ = i eiθdθ

the coefficients are

an =1

2π i

2π0

f (z0 + eiθ)

(z0 + eiθ − z0)n+1ieiθdθ.

Therefore,

|an| ≤ 1

2π

2π0

M

n+1d|θ| =

M

n,

which makes it clear that, if n < 0, then |an| can be made arbitrarily small, by choosing a sufficiently small . This

proves that an = 0 for negative n, and so

f (z) =

∞n=0

an(z − z0)n.

Thus, f ∈ H (G). The lemma solves part (a) and is also an example of a general theorem about isolated singularities of holomorphic

functions, so it answers part (b). Here is another answer to part (b):

60




Theorem 2.4 (Criterion for a pole) Let G ⊂ C be open. and suppose f (z) is holomorphic for all z ∈ G except

for an isolated singularity at z = z0 ∈ G. Then

(i) z0 is a pole of f if and only if limz→z0 |f (z)| = ∞;(ii) if m > 0 is the smallest integer such that limz→z0 |(z − z0)mf (z)| remains bounded, then z0 is a pole of

order m.

2. (a) Explicitly construct, through a sequence of mappings, a one-to-one holomorphic function mapping the disk Donto the half disk D ∩ Π+.

(b) State a general theorem concerning one-to-one mappings of D onto domains Ω ⊂ C.

Solution: (a)51 Let φ0(z) = 1−z1+z

. Our strategy will be to show that φ0 maps the fourth quadrant onto D ∩ Π+,

and then to construct a conformal mapping, f , of the unit disk onto the fourth quadrant. Then the composition

φ0 f will have the desired properties.

Consider the boundary of the first quadrant. Note that φ0 maps the real line onto itself. Furthermore, φ0 takes 0 to

1 and takes ∞ to -1. Since φ0(1) = 0, we see that the positive real axis (0, ∞) is mapped onto the segment (−1, 1).

Now, since φ0 maps the right half-plane P + onto the unit disk, it must map the boundary of P + (i.e., the imaginary

axis) onto the boundary of the unit disk. Thus, as 0 → 1 and ∞ → −1, the positive imaginary axis is mapped to

either the upper half-circle or the lower half-circle, and similarly for the negative imaginary axis. Checking that

φ0(i) = −i, it is clear that the positive imaginary axis is mapped to the lower half-circle eiθ : −π < θ < 0.

Therefore, in mapping the right half-plane onto the unit disk, φ0 maps the first quadrant to the lower half-disk

D ∩ Π−, and must therefore map the fourth quadrant to the upper half-disk. That is, φ0 : Q4 → D ∩ Π+, where

Q4 = z ∈ C : Rez > 0, Imz < 0.

Next construct a mapping of the unit disk onto the fourth quadrant as follows: If φ1(z) = iz, then φ1 φ0 : D →Π+. Let φ2(z) = z1/2 be a branch of the square root function on Π+. Then φ2 maps Π+ onto the first quadrant,

Q1 = z ∈ C : Rez > 0, Imz > 0. Finally, let φ3(z) = e−iπ/2z = −iz, which takes the first quadrant to thefourth quadrant. Then, since all of the mappings are conformal bijections, f = φ3 φ2 φ1 φ0 is a conformal

bijection of D onto Q4. Therefore, φ0 f is a conformal bijection of D onto D ∩ Π+.

(b) (Riemann)52 Let Ω ⊂ C be a simply connected region such that Ω = C. Then Ω is conformally equivalent

to D. That is, there is a conformal bijection, φ, of Ω onto the unit disk. Moreover, if we specify that a particular

z0 ∈ Ω must be mapped to 0, and we specify the value of arg φ(z0), then the conformal mapping is unique.

3.53 (a) State the Schwarz lemma.

(b) Suppose that f ∈ H (Π+) and that |f (z)| < 1 for all z ∈ Π+. If f (i) = 0 how large can |f (i)| be? Find the

extremal functions.


51This problem also appears in April ’89 (3) and April ’95 (5).52Look up the precise statement of the Riemann mapping theorem.53A very similar problem appeared in April ’95 (6).

61




(b) In order to apply the Schwarz lemma, map the disk to the upper half-plane with the M oebius map φ : D → Π+

defined by

φ(z) = i1 − z

1 + z

.

Then, φ(0) = i. Therefore, the function g = f φ : Dφ−→ Π+ f −→ D satisfies |g(z)| ≤ 1 and g(0) = f (φ(0)) =

f (i) = 0. By Schwarz’s lemma, then, |g(0)| ≤ 1. Finally, observe that g(z) = f (φ(z))φ(z), and then check

that φ(0) = −2i. Whence, g(0) = f (φ(0))φ(0) = f (i)(−2i), which implies 1 ≥ |g(0)| = 2|f (i)|. Therefore

|f (i)| ≤ 1/2.

4. (a) State Cauchy’s theorem and its converse.

(b) Suppose that f is a continuous function defined on the entire complex plane. Assume that

(i) f ∈ H (Π+ ∪ Π−)

(ii) f (z) = f (z) all z∈C.

Prove that f is an entire function.

Solution: (a) See theorems A.16 and A.18.

(b) See Marsden and Hoffman.

5. (a) Define what it means for a family F ⊂ H (Ω) to be a normal family. State the fundamental theorem for normal

families.

(b) Suppose f ∈ H (Π+) and |f (z)| < 1 all z ∈ Π+. Suppose further that

lim t

→0+f (it) = 0.

Prove that f (zn) → 0 whenever the sequence zn → 0 and zn ∈ Γ where

Γ = z ∈ Π+ : |Rez| ≤ Imz.

Hint. Consider the functions f t(z) = f (tz) where t > 0.

Solution: (a) Let Ω be an open subset of the plane. A family F of functions in Ω is called a normal family if

every sequence of functions in F has a subsequence which converges locally uniformly in Ω. (The same definition

applies when the family F happens to be contained in H (Ω).)54

I think of the Arzela-Ascoli theorem as the fundamental theorem for normal families. However, since the examiners

asked specifically about the special case when F is a family of holomorphic functions, they probably had in mind

the version of Montel’s theorem stated below, which is an easy consequence of the Arzela-Ascoli theorem.55

Theorem 2.5 (Arzela-Ascoli) Let F ⊂ C (Ω, S ) be a family of continuous functions from an open set Ω ⊆ C into

a metric space (S, d). Then F is a normal family if and only if

(i) F is equicontinuous on each compact subset of Ω, and

(ii) for each z ∈ Ω, the set f (z) : f ∈ F is contained in a compact subset of S .

54Despite the wording of the problem, the family need not satisfy F ⊂ H (Ω) in order to be normal.55Problem 5 (b) of the November 2001 exam asks for a proof of Montel’s theorem using the Arzela-Ascoli theorem.

62




Recall that a family F of functions is called locally bounded on Ω iff for all compact K ⊂ Ω there is a constant

M K such that |f (z)| ≤ M K for all f ∈ F and z ∈ K .

Corollary 2.3 (little Montel theorem56

) Assume the set-up of the Arzela-Ascoli theorem, and suppose S = Cand F ⊂ H (Ω). Then F is a normal family if and only if it is locally bounded.

(b) Fix a sequence zn ⊂ Γ with zn → 0 as n → ∞. We must prove f (zn) → 0. Define f n(z) = f (|zn|z).

Then, since z ∈ Γ ⇒ |zn|z ∈ Γ, we have

|f n(z)| = |f (|zn|z)| < 1, for all z ∈ Γ and n ∈ N.

Therefore, F is a normal family in Γ. Also note that each f n is holomorphic in Γ since f (tz) ∈ H (Γ) for any

constant t > 0. Thus, F is a normal family of holomorphic functions in Γ.

Let g be a normal limit of f n; i.e., there is some subsequence nk such that, as k → ∞, f nk → g locally uniformly

in Γ.

Consider the point z = i. Since f (it)

→0 as t

↓0,

g(i) = limk→∞

f nk(i) = limk→∞

f (|znk |i) = 0.

In fact, for any point z = iy with y > 0, we have g(z) = 0. Since g is holomorphic in Γ, the identity theorem

implies that g ≡ 0 in Γ.

Next, consider

f n

zn|zn|

= f

|zn| zn

|zn|

= f (zn). (47)

The numbers zn/|zn| lie in the compact set γ = z ∈ Γ : |z| = 1. Since f nk → g uniformly in γ , for any > 0,

there is a K > 0 such that |f nk(z) − g(z)| = |f nk(z)| < , for all k ≥ K and all z ∈ γ . That is,

limk→∞

sup|f nk(z)| : z ∈ γ limk→∞

f nkγ = 0,

and, since znk/|znk | ∈ γ , f nk

znk|znk |

≤ f nkγ .

∴ limk→∞

f nk

znk|znk |

= 0.

By (47), then, limk→∞ f (znk) = 0.

Finally, recall that f (zn) → 0 iff every subsequence znj has a further subsequence znjk such that f (znjk ) → 0,

as k → ∞. Now, if znj is any subsequence, then f (znj ) is a normal family, and, repeating the argument above,

there is, indeed, a further subsequence znjk such that f (znjk ) → 0. This completes the proof.

Remark: In the last paragraph, we made use of the fact that a sequence converges to zero iff any subsequence has,

in turn, a further subsequence that converges to zero. An alternative concluding argument that doesn’t rely on thisresult, but proceeds by way of contradiction, runs as follows: Assume we have already shown limk→∞ f (znk) = 0,

as above, and suppose f (zn) does not converge to 0 as n → ∞. Then there is a δ > 0 and a subsequence znjsuch that |f (znj )| > δ for all j ∈ N. Relabel this subsequence zn. Then f (zn) is itself a normal family and

we can repeat the argument above to get a further subsequence znk with limk→∞ f (znk) = 0. This contradicts

the assumption that |f (zn)| > δ for all n ∈ N. Therefore, f (zn) → 0, as desired.

63




2.7 2007 April 16

Notation: C is the set of complex numbers, D = z ∈ C : |z| < 1, and, for any open set G ⊂ C, H (G) is the set of

holomorphic functions on G.

1. Give the Laurent series expansion of 1z(z−1) in the region A ≡ z ∈ C : 2 < |z + 2| < 3.

Solution:

f (z) =1

z(z − 1)=

1 − z + z

z(z − 1)=

1

z − 1− 1

z.

Let u = z + 2. Then z = u − 2 and A = u ∈ C : 2 < |u| < 3. Therefore,

1

z=

1

u − 2=

1

u

1

(1 − 2/u)=

1

u

∞n=0

2

u

n

converges for |u| > 2 and, substituting u = z + 2 in the last expression, we have

1

z=

1

z + 2

∞n=0

2

z + 2

n=

∞n=0

2n(z + 2)−n−1 =−1

n=−∞

1

2

n+1

(z + 2)n,

converging for 2 < |z + 2|. Next, consider that

1

z − 1=

1

u − 3=

−1

3(1 − u/3)=

−1

3

∞n=0

u

3

n

converges for |u| < 3 and, substituting u = z + 2 in the last expression, we have

1

z

−1

= −∞

n=0

1

3n+1

(z + 2)n,

converging for |z + 2| < 3. Therefore,

f (z) =1

z − 1− 1

z= −

∞n=0

1

3

n+1

(z + 2)n −−1

n=−∞

1

2

n+1

(z + 2)n,

for z ∈ A.

2. (i) Prove: Suppose that for all z ∈ D and all n ∈ N we have that f n is holomorphic in D and |f n(z)| < 1. Also

suppose that limn→∞ Imf n(x) = 0 for all x ∈ (−1, 0). Then limn→∞ Imf n(1/2) = 0.

(ii) Give a complete statement of the convergence theorem that you use in part (2i).

Solution: (i)

(ii)

3. Use the residue theorem to evaluate ∞−∞

11+x4 dx.

64




Solution: Note that

f (z) =1

1 + x4=

1

(z2 + i)(z2

−i)

=1

(z + eiπ/4)(z−

eiπ/4)(z + ei3π/4)(z−

ei3π/4),

which reveals that the poles of f in the upper half plane are at eiπ/4 and ei3π/4. Let ΓR be the contour shown in

the figure below; i.e., ΓR = g(R) ∪ [−R, R], where R > 1. Then, by the residue theorem, ΓR

f (z)dz = 2πi

Res(f, eiπ/4) + Res(f, ei3π/4)

. (48)

The other two poles of f are in the lower half-plane, so both eiπ/4 and ei3π/4 are simple poles. Therefore,

Res(f, eiπ/4) = limz→eiπ/4

(z − eiπ/4)f (z) =1

2eiπ/4(eiπ/4 − ei3π/4)(eiπ/4 + ei3π/4)= −1

4ie−iπ/4,

Res(f, ei3π/4) = limz→ei3π/4

(z − ei3π/4)f (z) =1

2ei3π/4(ei3π/4 − eiπ/4)(ei3π/4 + eiπ/4)=

1

4ie−i3π/4.

Plugging these into (48) yields ΓR

f (z)dz = 2πi

1

4ie−i3π/4 − 1

4ie−iπ/4

=

π

2(e−iπ/4 − e−i3π/4) =

π√2

.

It remains to show

limR→∞

g(R)

f (z)dz

= 0.

Changing variables via z = Reiθ (0 ≤ θ ≤ π),

g(R)

f (z)dz

=

π0

iReiθ

1 + (Reiθ)4

≤ πR

R4 − 1→ 0, as R → ∞.

4. Present a function f that has all of the following properties: (i) f is one-to-one and holomorphic on D. (ii)

f (z) : z ∈ D = w ∈ C : Rew > 0 and Imw > 0. (iii) f (0) = 1 + i.

65




Solution: First consider57 φ1(z) = 1−z1+z , which maps D onto the right half-plane P + = z ∈ C : Rez > 0.

Let φ2(z) = eiπ/2z = iz, which maps P + onto the upper half-plane Π+ = z ∈ C : Imz > 0. Next,

let φ3(z) = z1/2 be a branch of the square root function on Π+. Then φ3 maps Π+ onto the first quadrant

Q1 = z ∈ C : 0 < arg(z) < π/2.

The function φ = φ3 φ2 φ1 satisfies the first two conditions, so we check whether it satisfies condition (iii):

φ1(0) = 1 ⇒ (φ2 φ1)(0) = φ2(1) = i ⇒ (φ3 φ2 φ1)(0) = φ3(i) =1 + i√

2

so apparently we’re off by a factor of √

2. This is easy to fix: let φ4(z) =√

2z. Then the holomorphic function

f φ4 φ maps D bijectively onto Q1 and f (0) = 1 + i, as desired.

5. (i) Prove: If f : D → D is holomorphic and f (1/2) = 0, then |f (0)| ≤ 1/2.

(ii) Give a complete statement of the maximum modulus theorem that you use in part (i).

Solution: (i) Define φ(z) =1/2−z1−z/2 . This is a holomorphic bijection58 of D onto D. Therefore, g = f φ ∈ H (D),

|g(z)| ≤ 1 for all z ∈ D, and g(0) = f (φ(0)) = f (1/2) = 0. Thus g satisfies the hypotheses of Schwarz’s lemma

(theorem A.22), which allows us to conclude the following:

(a) |g(z)| ≤ |z|, for all z ∈ D, and

(b) |g(0)| ≤ 1,

with equality in (a) for some z ∈ D or equality in (b) iff g(z) = eiθz for some constant θ ∈ R. By condition (a),

1/2 ≥ |g(1/2)| = |f (φ(1/2))| = |f (0)|.

(ii) In part (i) I used Schwarz’s lemma, a complete statement of which appears in the appendix (theorem A.22).

This is sometimes thought of as a version of the maximum modulus principle since it is such an easy corollary of what is usually called the maximum modulus principle. Here is a complete statement of the latter:

(max modulus principle, version 1)

Suppose G ⊂ C is open and f ∈ H (G) attains its maximum modulus at some point a ∈ G. Then f is constant.

That is, if there is a point a ∈ G with |f (z)| ≤ |f (a)| for all z ∈ G, then f is constant.59

6. Prove: If G is a connected open subset of C, any two points of G can be connected by a parametric curve in G.

57This is my favorite Moebius map. Not only does it map the unit disk onto the right half-plane, but also it maps the right half-plane onto the

unit disk. This feature makes φ1 an extremely useful tool for conformal mapping problems, where you’re frequently required to map half-planes to

the unit disk and vice-versa. Another nice feature of this map is that φ−11 = φ1. (Of course this must be the case if φ1 is to have the first feature.)

Also note that, like all linear fractional transformations, φ1 is a holomorphic bijection of C. Therefore, if φ1 is to map the interior of the unit disk

to the right half-plane, it must also map the exterior of the unit disk to the left half-plane.58See Rudin [8] page 254-5 (in particular, theorem 12.4) for a nice discussion of functions of the form φα(z) = z−α1−αz

. In addition to 12.4,

sec. 12.5 and theorem 12.6 are popular exam questions.59There are a couple of other versions of the maximum modulus principle you should know, though for most problems on the comprehensive

exams, the version above usually suffices. The other two versions are stated and proved clearly and concisely in Conway [3], but they also appear

as theorems A.20 and A.21 of the appendix.

66




Solution: First, recall that if A ⊂ G ⊂ C, then A is said to be open relative to G, or simply open in G, if for any

a ∈ A there is a neighborhood B(a, ) = z ∈ C : |z − a| < such that B(a, ) ∩ G ⊂ A.60

Next, recall that a subset G

⊂C is connected iff the only subsets of G that are both open and closed relative to G

are the empty set and G itself. Equivalently, if there exist non-empty disjoint subsets A, B ⊂ G that are open in Gand have the property G = A ∪ B, then G is not connected, or disconnected .61

Now, suppose G is a connected open subset of C. Fix z0 ∈ G and let Ω ⊂ G be the subset of points that can

be connected to z0 by a parametric curve in G. Since G is open, ∃B(z0, ) ⊂ G for some > 0, and clearly

B(z0, ) ⊂ Ω. In particular, Ω = ∅ . If we can show Ω is both open and closed in G, then it will follow by

connectedness that Ω = G, and the problem will be solved.

(Ω is open) Let w ∈ Ω be connected to z0 by a parametric curve γ ⊂ G. Since G is open, ∃ > 0 such that

B(w, ) ⊂ G. Clearly any w1 ∈ B(w, ) can be connected to z0 by a parametric curve (from w1 to w, then from wto z0 via γ ) that remains in G. This proves that B(w, ) ⊂ Ω, so Ω is open.

(Ω is closed) We show G \ Ω is open (and thus, in fact, empty). If z ∈ G \ Ω, then, since G is open, ∃δ > 0such that B(z, δ) ⊂ G. We want B(z, δ) ⊂ G \ Ω. This must be true since, otherwise, there would be a point

z1∈

B(z, δ)∩

Ω which could be connected to both z and z0 by parametric curves in G. But then a parametric

curve in G connecting z to z0 could be constructed, which would put z in Ω – a contradiction.

We have thus shown that Ω is both open and closed in G, as well as non-empty. Since G is connected, Ω = G.

60For example, the set A = [0 , 1], although closed in C, is open in G = [0, 1] ∪ 2.61To see the equivalence note that, in this case, A is open in G, as is Ac = G \ A = B, so A is both open and closed in G. Also it is instructive

to check, using either definition, that G = [0, 1] ∪ 2 is disconnected.

67





Do as many problems as you can. Complete solutions (except for minor flaws) to 5 problems would be considered an

excellent performance. Fewer than 5 complete solutions may still be passing, depending on the quality.

1. Let G be a bounded open subset of the complex plane. Suppose f is continuous on the closure of G and analytic

on G. Suppose further that there is a constant c ≥ 0 such that |f | = c for all z on the boundary of G. Show that

either f is constant on G or f has a zero in G.

2. (a) State the residue theorem.

(b) Use contour integration to evaluate ∞0

x2

(x2 + 1)2dx.

Important: You must carefully: specify your contours, prove the inequalities that provide your limiting arguments,

and show how to evaluate all relevant residues.

3. (a) State the Schwarz lemma.

(b) Suppose f is holomorphic in D = z : |z| < 1 with f (D) ⊂ D. Let f n denote the composition of f with

itself n times (n = 2, 3, . . . ). Show that if f (0) = 0 and |f (0)| < 1, then f n converges to 0 locally uniformly

on D.

4. Exhibit a conformal mapping of the region common to the two disks |z| < 1 and |z − 1| < 1 onto the region inside

the unit circle |z| = 1.

5. Let

f n

be a sequence of functions analytic in the complex plane C, converging uniformly on compact subsets of

Cto a polynomial p of positive degree m. Prove that, if n is sufficiently large, then f n has at least m zeros (countingmultiplicities).

Do not simply refer to Hurwitz’s theorem; prove this version of it.

6. Let (X, d) be a metric space.

(a) Define what it means for a subset K ⊂ X to be compact.

(b) Prove (using your definition in (a)) that K ⊂ X is compact implies that K is both closed and bounded in X .

(c) Give an example that shows the converse of the statement in (b) is false.


68



2.9 Some problems of a certain type 2 COMPLEX ANALYSIS

2.9 Some problems of a certain type

Collected in this section are miscellaneous problems about such things as what can be said of a holomorphic (or

harmonic) function when given information about how it behaves near a boundary or near infinity.

Behavior near infinity

1. If f (z) is an entire function which tends to infinity as z tends to infinity, then f (z) is a polynomial.

2. If f (z) is an injective entire function, then f (z) = az + b for some constants a and b.

3. If u(z) is a nonconstant real valued harmonic function of C, then there is a sequence zn ⊂ C with zn → ∞ and

u(zn) → 0 as n → ∞.

Behavior on or near the unit circle

4. If f (z) is holomorphic in an open set containing the closed unit disk, and if f (eiθ) is real for all θ ∈ R, then f (z)is constant.

5. Prove or disprove: There exists a function f (z) holomorphic on the unit disk D such that |f (zn)| → ∞ whenever

zn ⊂ D and |zn| → 1.

6. Prove or disprove: There exists a function u(z) harmonic on the unit disk D such that |u(zn)| → ∞ whenever

zn ⊂ D and |zn| → 1.

Other Problems

7. If f is holomorphic in the punctured disk 0 < |z| < R and if Ref ≤ M for some constant M , then 0 is a

removable singularity.

8. If f is holomorphic in the unit disk, with |f (z)| ≤ 1, f (0) = 0, and f (r) = f (−r) for some r ∈ (0, 1),

then|f (z)| ≤ |z|

z2 − r2

1 − r2z2

.

Solutions

1. See (1b) of April ’95.

2. Suppose f ∈ H (C) is injective. Then f −1 is a continuous function in C which maps compact sets to compact

sets. Therefore, if zn ⊂ C is any sequence tending to infinity, then the image f (zn) cannot remain inside any

closed disk (since f −1 maps all such disks to closed bounded sets in C). Thus f (z)

→ ∞whenever z

→ ∞. By the

previous problem, f is a polynomial. Finally, if f has degree greater than one, or if f is constant, then f would not be

injective. Therefore, f is a polynomial of degree one.

3. See (4) of April ’95.

4. See (1a) of April ’89.

69



2.9 Some problems of a certain type 2 COMPLEX ANALYSIS

5. & 6. That both of these statements are false is a corollary of the next two lemmas.

Lemma 1: If f ∈ H (D), then there is a sequence zn ⊂ D with |zn| → 1 such that the sequence |f (zn)| isbounded.

Lemma 2: If u is harmonic in D, then there is a sequence zn ⊂ D with |zn| → 1 such that the sequence u(zn)is bounded.

Proof of Lemma 1: First suppose that f has infinitely many zeros in D. Then, in any closed disk |z| ≤ 1 − ⊂ D,

the zeros of f must be isolated (otherwise f ≡ 0). Since such a disk is compact, it contains only finitely many zeros

of f . We conclude that there must be a sequence of zeros of f tending to the boundary of D.

Now suppose f has finitely many zeros in D. Let α1, . . . , αN bethe set ofall zeros of f (counting multiplicities).

Then

f (z) = (z − α1) · · · (z − αN )g(z),

where g is holomorphic and non-zero in D. Therefore, the function 1/g is also holomorphic in D. By the maximummodulus principal, in each compact disk Dn = |z| ≤ 1 − 1/n (n ≥ 2), the function |1/g(z)| attains its maximum

in Dn on the boundary at, say, the point zn, where |zn| = 1 − 1/n. The reciprocals of these maxima must satisfy

|g(x2)| ≥ |g(x3)| ≥ ·· · . Of course, the product (z − α1) · · · (z − αN ) is bounded in D, so the sequence |f (zn)| is

bounded.


70



A MISCELLANEOUS DEFINITIONS AND THEOREMS

A Miscellaneous Definitions and Theorems

A.1 Real Analysis

A.1.1 Metric Spaces

The following theorem is found in Conway [3].

Theorem A.1 Let (X, d) be a metric space; then the following are equivalent statements:

(a) X is compact;

(b) Every infinite set in X has a limit point (in X );

(c) X is sequentially compact

(d) X is complete and for all > 0 there exist x1, . . . , xn ⊂ X such that

X =n

k=1

B(xk, )

(The last property is called total boundedness.)

A.1.2 Measurable Functions

Definition A.1 A complex function s on a measurable space X whose range consists of only finitely many points is

called a simple function.

If α1, . . . , αn are the distinct values of a simple function, and if we set Ai = x ∈ X : s(x) = αi, then clearly

s =

ni=1

αiχAi ,

where χAi

is the characteristic function of the set Ai. Note that the definition assumes nothing about the sets A

i(in

particular, they may or may not be measurable). Thus, a simple function, as defined here, is not necessarily measurable.

Continuous functions of continuous functions are continuous, and continuous functions of measurable functions

are measurable. We state this as

Theorem A.262 Let Y and Z be topological spaces, and let g : Y → Z be continuous.

(a) If X is a topological space, if f : X → Y is continuous, and if h = g f , then h : X → Z is continuous.

(b) If X is a measurable space, if f : X → Y is measurable, and if h = g f , then h : X → Z is measurable.

Proof: If V is open in Z , then g−1(V ) is open in Y , and h−1(V ) = (g f )−1(V ) = f −1(g−1(V )). If f is continuous, then

h−1(V ) is open, proving (a). If f is measurable, then h−1(V ) is measurable, proving (b).

Note, however, that measurable functions of continuous functions need not be measurable.

Theorem A.363 Let u and v be real measurable functions on a measurable space X , let Φ be a continuous mapping of

the plane into a topological space Y , and define

h(x) = Φ(u(x), v(x)) (x ∈ X ).

Then h : X → Y is measurable.62Theorem 1.7 of Rudin [8].63Theorem 1.8 of Rudin [8].

71



A.1 Real Analysis A MISCELLANEOUS DEFINITIONS AND THEOREMS

A.1.3 Integration

There are a handful of results that are the most essential, and lay the foundation on which everything else is built.

Rudin [8] gives a beautifully succinct and clear presentation of these in just seven pages (pp. 21–27). 64 Some of these

results are presented below, but do yourself a favor and read the master [8].

The following theorem65 is an essential ingredient of many proofs (e.g. the proof that simple functions are dense

in L p, presented below).

Theorem A.4 (Rudin Theorem 1.17) Let f : X → [0, ∞] be measurable. There exist simple measurable functions

sn on X such that

(a) 0 ≤ s1 ≤ s2 ≤ · · · ≤ f ,

(b) sn(x) → f (x) as n → ∞, for every x ∈ X .

Theorem A.5 (Lebesgue’s Monotone Convergence Theorem) Let f n be a sequence of measurable functions on

X , and suppose that, for every

x ∈ X ,

(a) 0 ≤ f 1(x) ≤ f 2(x) ≤ ·· · ≤ ∞,

(b) f n(x) → f (x) as n → ∞.

Then f is measurable, and X

f n dµ → X

f dµ as n → ∞.

Theorem A.6 (Fatou’s lemma) If f n : X → [0, ∞] (n = 1, 2, . . . ) is a sequence of positive measurable functions,

then liminf f n ≤ liminf

f n

Theorem A.7 (Lebesgue’s dominated convergence theorem) Let f n be a sequence of measurable functions on

(X,M, µ) such that f n → f a.e. If there is another sequence of measurable functions gn satisfying

(i) gn → g a.e.,

(ii)

gn → g < ∞, and

(iii) |f n(x)| ≤ gn(x) (x ∈ X ; n = 1, 2, . . .),

then f ∈ L1(X,M, µ),

f n → f , and f n − f 1 → 0.

Theorem A.8 (Egoroff) If (X,M, µ) is a measure space, E ∈ M a set of finite measure, and f n a sequence of

measurable functions such that f n(x) → f (x) for almost every x ∈ E , then for all > 0 there is a measurable subsetA ⊆ E such that f n → f uniformly on A and µ(E \ A) < .

64Study these seven pages until you can recite all seven theorems and their proofs in your sleep. Also, pay attention to the details. Rudin is

careful to choose definitions and hypotheses that lend themselves to a succinct exposition, usually without too much loss of generality. For example,

he sometimes takes the range of a “real-valued” function to be [−∞, ∞], rather than R. It is instructive to pause occasionally to consider how his

arguments depend on such choices.65I label it “Rudin Theorem 1.17” because I cited it as such so often when practicing to take my comprehensive exams that the number stuck with

me.

72




A.1.4 Approximating Integrable Functions66

Let X be a locally compact Hausdorff space. (See Rudin [8], section 2.3 for the definition.) Let C c(X ) denote

the vector space of complex valued continuous functions on X with compact support (i.e. the closure of the set

x ∈ X : f (x) = 0 is compact). First, we present an important application of Rudin Theorem 1.17 (theorem A.4

above) and the dominated convergence theorem (DCT).

Theorem A.9 Let S be the set of all complex, measurable, simple functions on X such that

µ(x ∈ X : s(x) = 0) < ∞.

If 1 ≤ p < ∞, then S is dense in L p(µ).

Proof: Let 1 ≤ p < ∞, and f ∈ Lp(µ).

Case 1. f ≥ 0.

Let sn be as in Theorem 1.17. Then sn ≤ f implies sn ∈ Lp(µ) for all n = 1, 2, . . . . Define gn = |f − sn|p. Then gn ∈ Lp(µ),

n = 1, 2, . . . , gn → 0 as n → ∞, and gn ≤ |f |p ∈ L1(µ). Therefore, by the DCT, gn → 0 as n → ∞. That is,

f − snp → 0 as n → ∞.

Case 2. f : X → [−∞, ∞].

Let f = f + − f − where f + = max0, f and f − = min0, −f . Let sn and tn be simple functions such that

0 ≤ s1(x) ≤ s2(x) ≤ · · · ≤ f +(x)

0 ≤ t1(x) ≤ t2(x) ≤ · · · ≤ f −(x).

Then, by Case 1, we have

f + − snp → 0 and f

− − tnp → 0

as n → ∞. Finally, note that

f − (sn − tn)pp = f + − f

− − (sn − tn)pp ≤ f + − snpp + f

− − tnpp.

Case 3. f : X → C.

This case follows from Case 2 once we split f up into real and complex parts.

The following is a deep result whose proof depends on Urysohn’s lemma. (For the proof, see Rudin [8], section

2.4.)

Theorem A.10 (Lusin’s Theorem) Fix > 0. Let f be a complex measurable function which vanishes off a set of

finite measure. Then there exists g ∈ C c(X ) such that

µ(x ∈ X : f (x) = g(x)) < .

Furthermore, we may arrange it so that

sup

x ∈X

|g(x)

| ≤sup

x∈X |f (x)

|Lusin’s theorem is a key ingredient in the following. The short, elegant proof is lifted directly from Rudin. (I won’t

pretend I can improve on his masterpiece.)

Theorem A.11 For 1 ≤ p < ∞, C c(X ) is dense in L p(µ).

66This topic is very important and appears on the comprehensive exam syllabus.

73




Proof: Define S as in theorem A.9. If s ∈ S and > 0, then by Lusin’s Theorem there exists g ∈ C c(X) such that g(x) = s(x)except on a set of measure < , and |g| ≤ s∞. Hence,

g − sp ≤ 21/ps∞.

Since S is dense in Lp(µ), this completes the proof.

We close out this subsection by giving a careful solution to a basic but important exercise from Rudin [8], Chapter

2. First, define a step function to be a finite linear combination of characteristic functions of bounded intervals in R1.

(Notice how much more special these functions are than the simple functions, defined above.)

Lemma A.1 (Rudin Exercise 2.24) If f ∈ L1(R) then there exists a sequence gn of step functions on R such that

limn→∞

|f − gn| = 0.

Proof: We proceed in steps of successively greater generality. (I’ll fill in the details soon!)

Case 1. f = χA for some measurable set A with µ(A) < ∞.

Case 2. f =n

i=1 αiχAi ∈ L1

(R).N.B. the assumption that f is integrable implies, in particular, that each Ai is measurable with µ(Ai) < ∞.

Case 3. f ∈ L1(R), with f ≥ 0.

Case 4. f ∈ L1(R).

(Details coming soon!)

A.1.5 Absolute Continuity of Measures

Two excellent sources for the material appearing in this section are Rudin [8] (§ 6.7, 6.10) and Folland [4] (§ 3.2).

Let µ be a positive measure on a σ-algebraM, and let λ be an arbitrary complex measure on M. (Recall that

the range of a complex measure is a subset of C, while a positive measure takes values in [0, ∞]. Thus the positive

measures do not form a subclass of the complex measures.)

Suppose, for any E ∈M, that µ(E ) = 0

⇒λ(E ) = 0. In this case, we say that λ is absolutely continuous with

respect to µ, and write λ µ. If there is a set A ∈M such that, for all E ∈M, λ(E ) = λ(A ∩ E ), then we say that

λ is concentrated on A. Suppose λ1 and λ2 are measures onM and suppose there exists a pair of disjoint sets A and

B such that λ1 is concentrated on A and λ2 is concentrated on B. Then we say that λ1 and λ2 are mutually singular ,

and write λ1 ⊥ λ2.

Theorem A.12 (Lebesgue-Radon-Nikodym)67 Let µ be a positive σ-finite measure on a σ-algebraM in a set X , and

let λ be a complex measure onM.

(a) There is then a unique pair of complex measures λa and λs onM such that

λ = λa + λs, λa µ, λs ⊥ µ.

If λ is positive and finite, then so are λa and λs.

(b) There is a unique h ∈ L1(µ) such that

λa(E ) =

E

h dµ ∀E ∈M.

67Rudin[8], 6.10.

74




The pair (λa, λs) is called the Lebesgue decomposition of λ relative to µ. We call h the Radon-Nikodym derivative of

λa with respect to µ, and write h = dλa/dµ and

dλa =

dλadµ dµ.

Strictly speaking, dλa/dµ should be viewed as the equivalence class of functions that are equal to h µ-a.e.

Corollary A.168 Suppose ν is a σ-finite complex measure and µ, λ are σ-finite measures on (X,M) such that ν µ λ. Then

(a) If g ∈ L1(ν ), then g dνdµ ∈ L1(µ) and g dν =

g

dν

dµdµ.

(b) ν λ, anddν

dλ=

dν

dµ

dµ

dλλ-a.e.

A.1.6 Absolute Continuity of Functions

Lemma 1.2 Let f : R → R be a function. If f is differentiable on [a, b], f ∈ L1([a, b]), and xa

f (t)dt =f (x) − f (a) for a ≤ x ≤ b, then f ∈ AC [a, b].

Proof Assuming the stated hypotheses, by a standard theorem,69 f ∈ L1 implies that for all > 0 there is a δ > 0such that, if E ⊂ R is measurable mE < δ, then

E

|f |dm < . (49)

Let A = ∪ni=1(ai, bi) be a finite union of disjoint open intervals in [a, b] such thatni=1(bi − ai) < δ. Then

mA ≤ni=1(bi − ai) < δ , so

ni=1

|f (bi) − f (ai)| =ni=1

bi

ai

f dm ≤ n

i=1

bi

ai

|f |dm = A

|f |dm < (50)

by (49). Thus, f ∈ AC [a, b].

A.1.7 Product Measures and the Fubini-Tonelli Theorem

Let (X, S , µ) and (Y, T , ν ) be measure spaces. If we want to construct a measurable space out of X × Y , it is natural

to start by considering the collection of subsets S × T = A × B ⊆ X × Y : A ∈ S , B ∈ T . Note, however, that

this collection is not, in general, an algebra of sets. To get an adequate collection on which to define product measure,

then, define70 S ⊗T = σ(S ×T ); that is, S ⊗T is the σ-algebra generated by S ×T .In my opinion, the most useful version of the Fubini and Tonelli theorems is the one in Rudin [8]. It begins

by assuming only that the function f (x, y) is measurable with respect to the product σ-algebra

S ⊗ T . Then, in a

single, combined Fubini-Tonelli theorem, you get everything you need to answer any of the standard questions aboutintegration with respect to product measure. Here it is:

68Folland [4], Prop. 3.9.69The “standard theorem” cited here appears often on the comprehensive exams (cf. Nov. ’91 #6), but in a slightly weaker form in which the

conclusion is that | E

f dm| < . In the present case we need E |f |dm < to get the sum in (50) to come out right.

70This notation is not completely standard. In Aliprantis and Burkinshaw [2] (p. 154), for example, S ⊗ T denotes what we call S × T , while

σ(S ⊗ T ) denotes what we have labeled S ⊗ T . At the opposite extreme, I believe Rudin[8] simply takes S × T to be the σ-algebra generated by

the sets A × B : A ∈ S , B ∈ T .

75




Theorem A.13 (Fubini-Tonelli)

Assume (X, S , µ) and (Y, T , ν ) are σ-finite measure spaces, and f (x, y) is a (S ⊗T )-measurable function on X × Y .

(a) If f (x, y) ≥ 0, and if φ(x) = Y f (x, y) dν (y) and ψ(y) =

X f (x, y) dµ(x), then φ is S -measurable, ψ isT -measurable, and

X

φ dµ =

X×Y

f (x, y) d(µ × ν ) =

Y

ψ dν. (51)

(b) If f : X × Y → C and if one of Y

X

|f (x, y)| dµ(x) dν (y) < ∞ or

X

Y

|f (x, y)| dν (y) dµ(X ) < ∞

holds, then so does the other, and f ∈ L1(µ × ν ).

(c) If f ∈ L1(µ × ν ), then,

(i) for almost every x∈

X, f (x, y)∈

L1(ν ),

(ii) for almost every y ∈ Y, f (x, y) ∈ L1(µ),

(iii) φ(x) = Y f dν is defined almost everywhere (by (i)), moreover φ ∈ L1(µ),

(iv) ψ(y) = X

f dµ is defined almost everywhere (by (ii)), moreover ψ ∈ L1(ν ), and

(v) equation (51) holds.

76



A.2 Complex Analysis A MISCELLANEOUS DEFINITIONS AND THEOREMS

A.2 Complex Analysis

A.2.1 Cauchy’s Theorem71

A continuous function γ : [a, b] → C, where [a, b] ⊂ R, is called a path in C, and such a path is called rectifiable if itis of bounded variation, i.e., if there is a constant M > 0 such that, for any partition a = t1 < t2 < · · · < tn = b of

[a, b],i |γ (ti) − γ (ti−1)| ≤ M . In particular, if γ is a piecewise smooth path, it is rectifiable.

If γ : [a, b] → C is a path in C, the set of points γ (t) : a ≤ t ≤ b is called the trace of γ . Some authors denote

this set by γ ∗, and others by γ . We will write γ ∗ if clarity demands it. Otherwise, if we simply write z ∈ γ , it should

be obvious that we mean γ (t) = z for some a ≤ t ≤ b. Finally, if γ is a closed rectifiable path in C, we call the region

which has γ as its boundary the interior of γ .A curve is an equivalence class of paths that are equal modulo a change of parameter. If a path γ has some (non-

parametric) property that interests us (e.g., it is closed or smooth or rectifiable), then invariably that property is shared

by every path in the equivalence class of parametrization of γ . Therefore, when parametrization has no relevance to

the discussion, we often speak of the “curve” γ , by which we mean any one of the paths that represent the curve.

Definition A.2 If γ is a closed rectifiable curve in C then, for w /

∈γ ∗, the number

n(γ ; w) =1

2πi

γ

dz

z − w

is called the index of γ with respect to the point w. It is also sometimes called the winding number of γ around w.

Theorem A.14 (Cauchy’s formula, ver. 1) Let G ⊆ C be an open subset of the plane and suppose f ∈ H (G). If γ is a closed rectifiable curve in G such that n(γ ; w) = 0 for all w ∈ C \ G, then for all z ∈ G \ γ ∗,

f (z)n(γ ; z) =1

2πi

γ

f (ζ )

ζ − zdζ.

A number of important theorems include a hypothesis like the one above concerning γ – i.e., a closed rectifiable

curve with n(γ ; w) = 0 for all w ∈ C \ G (where G is some open subset of the plane). This simply means that γ is contained with its interior in G. In other words, γ does not wind around any points in the complement of G (e.g.,

“holes” in G, or points exterior to G). Such a curve γ is called homologous to zero in G, denoted γ ≈ 0, and a version

of a theorem with this as one of its hypotheses may be called the “homology version” of the theorem.

More generally, if G ⊆ C is open and γ 1, . . . , γ m are closed rectifiable curves in G, then the curve γ = γ 1 + · · · +γ m is homologous to zero in G provided n(γ 1, w) + · · ·+ n(γ m, w) = 0 for all w ∈ C\ G. Thus, either theorem A.14,

or the following generalization, might be called “the homology version of Cauchy’s formula:”

Theorem A.15 (Cauchy’s formula, ver. 2) Let G ⊆ C be an open subset of the plane and suppose f ∈ H (G). If

γ 1, . . . , γ m are closed rectifiable curves in G with γ = γ 1 + · · · + γ m ≈ 0, then for all z ∈ G \ γ ∗,

f (z)m

j=1

n(γ j , z) =1

2πi

m

j=1 γjf (ζ )

ζ

−z

dζ.

The next theorem (or its generalization below) might be called “the homology version of Cauchy’s theorem:”

Theorem A.16 (Cauchy’s theorem, ver. 1) Let G ⊆ C be an open set and suppose f ∈ H (G). If γ is a closed

rectifiable curve that is homologous to zero in G, then γ f (z)dz = 0.

71Most of the material in this section can be found in Conway [3].

77



A.2 Complex Analysis A MISCELLANEOUS DEFINITIONS AND THEOREMS

Theorem A.17 (Cauchy’s theorem, ver. 2) Let G ⊆ C be an open set and suppose f ∈ H (G). If γ 1, . . . , γ m are

closed rectifiable curves in G such that γ = γ 1 + · · · + γ m ≈ 0, then

mj=1

γj

f (z) dz = 0.

A partial converse of Cauchy’s theorem is the following:

Theorem A.18 Let G be an open set in the plane and f ∈ C (G,C). Suppose, for any triangular contour T ⊂ G with

T ≈ 0 in G, that T f (z)dz = 0. Then f ∈ H (G).

This theorem is still valid (and occasionally easier to apply) if we replace “any triangular contour” with “any rectan-

gular contour with sides parallel to the real and imaginary axes.” This stronger version is sometimes called Morera’s

theorem, and the exercise on page 81 of Sarason [10] asks you to prove it using theorem A.18.

A.2.2 Maximum Modulus Theorems

Theorem A.19 (max mod principle, ver. 1) Suppose G ⊂ C is open and f ∈ H (G) attains its maximum modulus

at some point a ∈ G. Then f is constant.

That is, if there is a point a ∈ G with |f (z)| ≤ |f (a)| for all z ∈ G, then f is constant.72

Theorem A.20 (max mod principle, ver. 2) If G ∈ C is open and bounded, and if f ∈ C (G) ∩ H (G), then

max|f (z)| : z ∈ G = max|f (z)| : z ∈ ∂G.

That is, in an open and bounded region, if a holomorphic function is continuous on the boundary, then it attains its

maximum modulus there.

Theorem A.21 (max mod principle, ver. 3) Let G ⊂ C = C∪∞ be open, let f ∈ H (G), and suppose there is an

M > 0 such that

limz→a

|f (z)| ≤ M, for every a ∈ ∂ ∞G.

Then |f (z)| ≤ M for all z ∈ G.

Theorem A.22 (Schwarz’s lemma) Let f ∈ H (D), |f (z)| ≤ 1 for all z ∈ D, and f (0) = 0. Then

(a) |f (z)| ≤ |z|, for all z ∈ D,

(b) |f (0)| ≤ 1,

with equality in (a) for some z ∈ D \ 0 or equality in (b) iff f (z) = eiθz for some constant θ ∈ R.


72This version of the maximum modulus principle is an easy consequence of the open mapping theorem, which itself can be proved via the local

mapping theorem, which in turn can be proved using Rouch e’s theorem. Of course, you should know the statements of all of these theorems and,

since proving them in this sequence is not hard, you might as well know the proofs too! Two excellent references giving clear and concise proofs

are Conway [3] and Sarason [10].

78



REFERENCES

B List of Symbols

F an arbitrary field

Q the rational numbersZ the integers

N the natural numbers, 1, 2, . . .C the complex numbers (a.k.a. the complex plane)

C the extended complex plane, C ∪∞R the real numbers (a.k.a. the real line)

T the unit circle, z ∈ C : |z| = 1R the extended real line, [−∞, ∞]Rez the real part of a complex number z ∈ CImz the imaginary part of a complex number z ∈ CD or U the open unit disk, z ∈ C : |z| < 1H (G) the holomorphic functions on an open set G ⊂ C

Π+ the upper half-plane,

z

∈C : Imz > 0

Π− the lower half-plane, z ∈ C : Imz < 0P + the right half-plane, z ∈ C : Rez > 0P − the left half-plane, z ∈ C : Rez < 0∂ ∞G the extended boundary of a set G ⊂ C

C [0, 1] the space of continuous real valued functions on [0, 1].

L1 the space of integrable functions; i.e., measurable f such that |f | < ∞.

L p for 0 < p < ∞, the space of measurable functions f such that |f | p < ∞.

L∞ the space of essentially bounded measurable functions;

i.e., measurable f such that x : |f (x)| > M has measure zero for some M < ∞.

S ⊗T the product σ-algebra generated by S and T .

References

[1] Lars Ahlfors. Complex Analysis. McGraw-Hill, New York, 3rd edition, 1968.

[2] Charalambos D. Aliprantis and Owen Burkinshaw. Principles of Real Analysis. Academic Press, New York, 3rd

edition, 1998.

[3] John B. Conway. Functions of One Complex Variable I . Springer-Verlag, New York, 2nd edition, 1978.

[4] Gerald B. Folland. Real Analysis: Modern Techniques and Their Applications. John Wiley & Sons Ltd, New

York, 1999.

[5] James R. Munkres. Topology: A First Course. Prentice Hall International, Englewood Cliffs, NJ, 1975.

[6] H. L. Royden. Real Analysis. Macmillan, New York, 3rd edition, 1988.

[7] Walter Rudin. Principles of Mathematical Analysis. McGraw-Hill, New York, 3rd edition, 1976.

[8] Walter Rudin. Real and Complex Analysis. McGraw-Hill, New York, 3rd edition, 1987.

[9] Walter Rudin. Functional Analysis. McGraw-Hill, New York, second edition, 1991.

[10] Donald J. Sarason. Notes on Complex Function Theory. Henry Helson, 1994.

[11] Elias Stein and Rami Shakarchi. Complex Analysis. Princeton University Press, 2003.

79



Index

absolute continuityof functions, 15–16, 73

of measures, 6, 14, 72

approximating integrable functions, 7

area theorem, 48

Arzela-Ascoli theorem, 57 , 62

Baire category theorem, 22

Banach space

p, 35

of bounded linear operators, 26

Banach-Steinhauss theorem, 22, 35

Borel σ-algebra, 34

Borel set, 34

Casorati-Weierstrass theorem, 54

applied, 47, 58

Cauchy’s formula, 75

applied, 39, 47, 56

Cauchy’s theorem, 75–76

converse of, see Morera’s theorem

problems, 44, 45, 48, 52, 61

proof by Green’s theorem, 51–52

Cauchy-Riemann equations, 42, 51

closed graph theorem, 30

conformal mapping

problems, 39, 44, 50, 60, 66, 68connected, 67

criterion for a pole, 59

curve, 75

disconnected, 67

dominated convergence theorem

applied, 11, 23, 28, 33

general version, 72

standard version, 21

Egoroff’s theorem, 72

problems, 20, 28–29, 32–33

equicontinuitypointwise vs. uniform, 57

equicontinuous, 62

even functions, 16

Fatou’s lemma, 72

Fubini-Tonelli theorem, 73–74

applied, 8, 13fundamental theorem of algebra, 52

fundamental theorem of calculus, 16

Green’s theorem, 51

Holder’s inequality, 22

Hadamard factorization theorem

applied, 58

Hahn-Banach theorem, 22

homologous to zero, 75

implicit function theorem, 18

index, 75inverse function theorem

of calculus, 18

Laurent expansion, 55, 59, 64

Lebesgue decomposition, 73

Liouville’s theorem

applied, 38, 48, 53

maximum modulus theorem, 76

applied, 66

monotone convergence theorem

applied, 4, 5, 12

Montel’s theorem, 57 , 62

applied, 45

Morera’s theorem, 76

problems, 48, 61

mutually singular, 72

normal family, 40, 45, 57, 62–63

odd functions, 16

path, 75

Picard’s theorem, 38

product measures, 8, 73

Radon-Nikodymderivative, 14, 73

problems, 7–8, 14, 19, 30, 34

theorem, 72–73

removable singularity theorem, 59

residue theorem, 68

applied, 40, 43, 55, 65, 68

80



INDEX INDEX

Riemann mapping theorem, 60

Riesz representation theorem, 35–36

applied, 16–17

for L p

, 22Rouche’s theorem, 53

Schwarz’s lemma, 76

applied, 50, 61, 66, 68

Stone-Weierstrass theorem, 21

applied, 7

Tonelli’s theorem, see Fubini-Tonelli theorem

uniform boundedness principle, 22, 35

winding number, 75

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Problems and Solutions in Complex and Real Analysis