Problems and Solutions in - Mathematics - University of Hawaii

Problems and Solutions in
REAL 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 reproduce
this document for other purposes may be obtained by emailing the author at williamdemeo@gmail.com.
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
williamdemeo@gmail.com.
Contents
1 Real Analysis 3
1.1 1991 November 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
2.6 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
A.1.3 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
A.1.4 Approximating Integrable Functions 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 Theorem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
A.2.2 Maximum Modulus Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
B List of Symbols 79
2

1 Real Analysis
1.1 1991 November 21
1 REAL ANALYSIS
1. (a) Let fn be a sequence of continuous, real valued functions on [0, 1] which converges uniformly to f. Prove that
limn→∞ fn(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 ɛ > 0 and let N0 ∈ N be such
that n ≥ N0 implies |fn(x) − f(x)| < ɛ/2, for all x ∈ [0, 1]. Let δ > 0 be such that |f(x) − f(y)| < ɛ/2,
for all x, y ∈ [0, 1] with |x − y| < δ. Finally, let N1 ∈ N be such that n ≥ N1 implies |xn − 1/2| < δ. Then
n ≥ max{N0, N1} implies |fn(xn) − f(1/2)| ≤ |fn(xn) − f(xn)| + |f(xn) − f(1/2)| < ɛ/2 + ɛ/2 = ɛ. ⊓⊔
(b) Suppose the convergence is only point-wise. Then the conclusion is false, as the following counterexample
demonstrates:
Define fn(x) to be the function
⎧
⎪⎨ 0, if 0 ≤ x <
f(x) =
⎪⎩
1
2
2nx − (n − 1), if 1 1
2 − 2n
≤ x ≤ 1.
1, if 1
2
1 − 2n ,
1 ≤ x < 2 ,
That is, fn(x) is constantly zero for x less than 1 1
2 − 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 = 1 1
2 − n . Then fn(xn) = 0
for all n ∈ N and xn → 1/2, while the sequence fn approaches the characteristic function f χ [ 1
2 ,1] which is one
on [ 1
2 , 1] and zero elsewhere. Therefore, f(1/2) = 1 = 0 = limn fn(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 ′ f(x + (xn − x)) − f(x) f(xn) − f(x)
(x) = lim
= lim
= 0.
n→∞ xn − x
n→∞ xn − x
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 ɛ > 0 there is E ∈ Σ such that
µ(E) < ∞ and
X\E
|f| dµ < ɛ.
3
(1)

1.1 1991 November 21 1 REAL ANALYSIS
Solution: For n = 1, 2, . . ., define
Clearly,
and each An is measurable (why?). 3 Next, define
An = {x ∈ X : 1/n ≤ |f(x)| < n}.
A1 ⊆ A2 ⊆ · · · ↑ A
∞
n=1
A0 = {x ∈ X : f(x) = 0} and A∞ = {x ∈ X : |f(x)| = ∞}.
Then X = A0 ∪ A ∪ A∞ is a disjoint union, and
|f| = |f| +
X
A0
A
|f| +
A∞
An
|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
|f| < ɛ, and
A\E
µ(E) < ∞.
Define fn = |f|χAn . Then {fn} is a sequence of non-negative measurable functions and, for each x ∈ X,
limn→∞ fn(x) = |f(x)|χA(x). Since An ⊆ An+1, we have 0 ≤ f1(x) ≤ f2(x) ≤ · · · , so the monotone
convergence theorem4 implies
X fn →
|f|, and, by (2),
A
lim
n→∞
An
Therefore, there is some N > 0 for which
Finally, note that 1/N ≤ |f| < N on AN, so
µ(AN) ≤ N
|f| dµ =
X\AN
AN
A
|f| dµ =
|f| dµ < ɛ.
|f| dµ ≤ N
X
X
|f| dµ.
|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) =
measure.
E
f dµ, show that ν is a
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 {E1, E2, . . .} be a countable collection of disjoint measurable sets.
3 Answer: 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).
4 Alternatively, we could have cited the dominated convergence theorem here since fn(x) ≤ |f(x)| (x ∈ X; n = 1, 2, . . .).
5 See 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

1.1 1991 November 21 1 REAL ANALYSIS
Then,
ν(∪nEn) =
n En
f dµ =
fχ n
En dµ
∞
= fχEn dµ (∵ the En are disjoint)
n=1
n=1
∞
= fχEn dµ (∵ fχEn ≥ 0, n = 1, 2, . . .)
=
∞
ν(En).
n=1
The penultimate equality follows from the monotone convergence theorem applied to the sequence of non-negative
measurable functions gm = m fχEn n=1 (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 ɛ > 0, show that there is δ > 0 such that for all measurable sets E ⊂ [0, 1]
with m(E) < δ,
f dm
< ɛ.
E
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,
Here, X = [0, 1]. As we saw in problem 3,
lim
n→∞
An = {x ∈ X : 1/n ≤ |f(x)| < n}.
An
6 See also: April ’92 (4), November ’97 (6), April ’03 (4).
|f| dm =
5
A
|f| dm =
X
|f| dm.

1.1 1991 November 21 1 REAL ANALYSIS
Let n > 0 be such that
X\An
|f| dm < ɛ/2.
Define δ = (2n) −1 ɛ, and suppose E ⊂ [0, 1] is a measurable set with mE < δ. We must show |
E
f dm
≤
=
≤
E
|f| dm
(X\An)∩E
X\An
|f| dm +
|f| dm +
< ɛ/2 + n m(An ∩ E)
< ɛ/2 + n ɛ
= ɛ.
2n
An∩E
An∩E
|f| dm
|f| dm
E
f dm| < ɛ.
The penultimate inequality holds because |f| < n on An. ⊓⊔
Proof 2: 7 This proof relies on the following lemma about absolute continuity of measures:
Lemma 1.1 Let ν be a finite signed measure and µ a positive measure on a measurable space (X, M). Then ν ≪ µ
if and only if for every ɛ > 0 there is a δ > 0 such that |ν(E)| < ɛ whenever µ(E) < δ.
The signed measure defined by
ν(E) =
is finite iff 8 f ∈ L 1 (µ). It is also clearly absolutely continuous with respect to µ. Therefore, lemma 1.1 can be
applied to the real and imaginary parts of any complex-valued f ∈ L1 (µ). It follows that, for every ɛ > 0, there is
a δ > 0 such that
|ν(E)| =
f dµ
< ɛ, whenever µ(E) < δ.
E
7. 9 Suppose f is a bounded, real valued, measurable function on [0, 1] such that x n f dm = 0 for n = 0, 1, 2, . . .,
with m Lebesgue measure. Show that f(x) = 0 a.e.
Solution: Fix an arbitrary continuous function on [0, 1], say, φ ∈ C[0, 1]. By the Stone-Weierstrass theorem, there
is a sequence {pn} of polynomials such that φ − pn∞ → 0 as n → ∞. Then, since all functions involved are
7 See also Folland [4], page 89.
8 For what follows we only need that ν is finite if f is integrable, but the converse is also true.
9 This question appears very often in varying forms of difficulty. cf. November ’92 (7b, very easy version), November ’96 (B2, very easy),
November ’91 (this question, easy), April ’92 (6, moderate), November ’95 (6, hard–impossible?). I have yet to solve the November ’95 version.
One attempted solution (which I think is the one given in the black notebook), seems to assume f ∈ L 1 , but that assumption makes the problem
even easier than the others.
6
E
f dµ
⊓⊔

1.1 1991 November 21 1 REAL ANALYSIS
integrable,
fφ
=
f(φ − pn + pn)
≤
|f||φ − pn| +
fpn
≤ f1φ − pn∞ +
fpn
= f1φ − pn∞. (3)
The last equality holds since x n f = 0 for all n = 0, 1, 2, . . ., which implies that fpn = 0 for all polynomials
pn. Finally, note that f1 < ∞, 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 fφ = 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 − f1 → 0 as n → ∞. Then
0 ≤ f 2
=
f(f − φn + φn)
≤
|f||f − φn| +
fφn
.
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 ≤ Mf − 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
L 1 [0, 1]. Simply start from the line, “Now, since C[0, 1] is dense in L 1 [0, 1], let {φn} ⊂ C[0, 1] satisfy...” but
instead write, “Since Pol[0, 1] is dense in L 1 [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,
(1 − f) dµ = f dν. (4)
E
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) =
dµ, we have
E
(1 − f) dµ = dµ − f dµ = µ(E) − f dµ.
E
E E
E
Therefore, (4) is equivalent to
µ(E) =
E
E
f dµ + f dν = f d(µ + ν) (∀ E ∈ Σ) (5)
E
E
10 See also: November ’97 (7).
11 Note that a measure µ is called “positive” when it is, in fact, nonnegative; that is, µE ≥ 0 for all E ∈ Σ.
7

1.1 1991 November 21 1 REAL ANALYSIS
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) = g dm, ∀E ∈ Σ.
E
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) = f d(µ + ν) ∀ E ∈ Σ,
E
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 g dν.
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 into R×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,
|f(x)g(y)| dν(y) dµ(x) = |f(x)||g(y)| dν(y) dµ(x)
X Y
X
Y
= |f(x)| |g(y)| dν(y) dµ(x)
X
Y
= |f(x)| dµ(x) |g(y)| dν(y) < ∞,
12 See also: November ’97 (2), and others.
13 I’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
X
Y

1.1 1991 November 21 1 REAL ANALYSIS
which holds since f ∈ L 1 (µ) and g ∈ L 1 (ν). The Fubini-Tonelli theorem then implies that F (x, y) ∈ L 1 (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,
F d(µ × ν) = F (x, y) dν(y) dµ(x).
Therefore,
X×Y
X×Y
F d(µ × ν) =
=
=
X
X
X
X
9
Y
f(x)g(y) dν(y) dµ(x)
Y
f(x) g(y) dν(y) dµ(x)
Y
f(x) dµ(x) g(y) dν(y).
Y
⊓⊔

1.2 1994 November 16 1 REAL ANALYSIS
1.2 1994 November 16
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 ∞ converges to an s ∈ E.
Solution: Suppose E is complete. Let {xn} ⊂ E be absolutely convergent; i.e., xn < ∞. We must
∞
xn := lim
n=1
Let SN = N
n=1 xn. Then, for any j ∈ N,
SN+j − SN =
N→∞
n=1
N+j
n=N+1
1 xn
N
xn = s ∈ E. (6)
xn
≤
N+j
n=N+1
xn → 0
as N → ∞, since xn < ∞. Therefore, {SN } is a Cauchy sequence. Since E is complete, there is an s ∈ E
such that ∞ n=1 xn = lim
N→∞ SN = 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
Next observe, for k > 1,
and
ynk
n, m ≥ nj ⇒ yn − ym < 2 −j .
k−1
= yn1 + (yn2 − yn1 ) + (yn3 − yn2 ) + · · · + (ynk − ynk−1 ) = yn1 +
By hypothesis, this implies that
ynk
∞
j=1
ynj+1
− yn1 =
− ynj <
k−1
j=1
(ynj+1
∞
2 −j = 1
j=1
− ynj ) → s ∈ E,
j=1
(ynj+1
− ynj ),
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 fn be a sequence of real continuous functions on a compact Hausdorff space X. Show that if f1 ≥ f2 ≥ f3 ≥
· · · , and fn(x) → 0 for all x ∈ X, then fn → 0 uniformly.
10

1.2 1994 November 16 1 REAL ANALYSIS
3. Let f be integrable on the real line with respect to Lebesgue measure. Evaluate lim
Justify all steps.
n→∞
∞
x
f(x − n)
−∞ 1+|x| dx.
Solution: Fix n > 0. Consider the change of variables, y = x − n. Then dy = dx and x = y + n, so
∞
−∞
Note that, when y ≥ −n,
x
f(x − n) dx =
1 + |x|
for all y ≥ −n. Define the function 14
=
∞
−∞
∞
−n
y + n
f(y)
1 + |y + n| dy
y + n
f(y) dy +
1 + y + n
−n
−∞
y+n
1+y+n ∈ [0, 1), and increases to 1 as n tends to infinity. Thus,
0 ≤ |f(y)|
gn(y) = f(y)
y + n
≤ |f(y)|,
1 + y + n
y + n
1 + y + n 1 [−n,∞)(y).
Then |gn| ≤ |f| and lim
n→∞ gn = f. Therefore, by the dominated convergence theorem,
∞
lim
n→∞
−n
y + n
f(y)
1 + y + n
Next, consider the second term in (7). Define the function
It is not hard to check that
hn(y) = f(y)
∞
∞
dy = lim
n→∞
gn(y) dy =
−∞
f(y) dy.
−∞
y + n
1 − (y + n) 1 (−∞,−n](y).
|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,
lim
n→∞ hn(y)
y + n
= f(y) lim
n→∞ 1 − (y + n) 1 (−∞,−n](y) = 0.
Therefore, the dominated convergence theorem implies that
−n
lim
n→∞
−∞
y + n
f(y)
dy = 0.
1 − (y + n)
y + n
f(y)
dy. (7)
1 − (y + n)
∞
x
Combining the two results above, we see that lim f(x − n)
n→∞ −∞ 1+|x| dx = ∞
f(x) dx. ⊓⊔
−∞
Remark. Intuitively, this is the result we expect because the translation f(x − n) = Tnf(x) is merely shifting the
support of f to the right tail of the measure dµ := x
1+|x| dx, and in the tail this measure looks like dx.
14 Here 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
(a) Explain why the limit ℓ = limn→∞ ˆxn exists.
ˆxn = sup xk
k≥n
(b) Prove that, for any ɛ > 0 and positive integer N, there exists an integer k such that k ≥ N and |xk − ℓ| < ɛ.
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 = ∪nXn, where {Xn} ∞ 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,
f dµ =
f dµ.
X
n
Solution: 15 Define fn = n k=1 fχXk = fχ∪n 1
Xk . Then it is clear that the hypotheses of the monotone convergence
theorem are satisfied. That is, for all x ∈ X,
(i) 0 ≤ f1(x) ≤ f2(x) ≤ · · · ≤ f(x), and
(ii) limn→∞ fn(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
Xn

1.3 1998 April 3 1 REAL ANALYSIS
Therefore,
∞
k=1
Xk
f dµ = lim
n
n→∞
k=1
X
fχXk dµ
n
= lim fχXk dµ
n→∞
X
k=1
(by linearity of the integral)
= lim
n→∞
fn dµ
X
(by definition of fn)
= lim
X
n→∞ fn dµ
(by the monotone convergence theorem)
= f dµ.
X
4. Using the Fubini/Tonelli theorems to justify all steps, evaluate the integral
1 1
0
y
x −3/2 cos
πy
dx dy.
2x
Solution: By Tonelli’s theorem, if f(x, y) ≥ 0 is measurable and one of the iterated integrals f(x, y) dx dy or
f(x, y) dy dx exists, then they both exist and are equal. Moreover, if one of the iterated integrals is finite, then
f(x, y) ∈ L 1 (dx, dy). Fubini’s theorem states: if f(x, y) ∈ L 1 (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 x
0
0
|g(x, y)| dy dx =
1 x
0
0
|x| −3/2
πy
cos dy dx ≤
2x
1 x
0
0
x −3/2 · 1 dy dx =
1
0
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) ∈ L 1 (dx, dy).
Therefore, the Fubini theorem applies to g(x, y), and gives the first of the following equalities:
1 1
0
y
x −3/2 cos
πy
dx dy =
2x
=
=
1 x
0
1
0
1
0
= 2
π
= 4
π .
13
0
x −3/2 cos
x −3/2 · 2x
π
2
π x−1/2 dx
2x 1/2 x=1
x=0
πy
dy dx
2x
πy
y=x sin
dx
2x y=0
⊓⊔
⊓⊔

1.3 1998 April 3 1 REAL ANALYSIS
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) =
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
1
µ(E) = dm(x).
1 + x2 E
Show that m is absolutely continuous with respect to µ, and compute the Radon-Nikodym derivative dm/dµ.
i
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 <
Then Ai ∩ Aj = ∅ for all i = j in N, and, for all n = 1, 2, . . .,
µ(An) ≥ 1
n + 1 m(An).
1
1
≤ .
1 + x2 n
Also, R = ∪An, since 0 < 1
1+x2 ≤ 1 holds for all x ∈ R. Therefore,
µ(E) = µ(E ∩ (∪nAn)) = µ(∪n(An ∩ E)) =
µ(An ∩ E).
The last equality might need a bit of justification: Since f(x) = 1
1+x 2 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 + 1 m(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) = h dµ, and f dm = fh dµ ∀ f ∈ L 1 (m).
In particular, if E ∈ M and f(x) = 1
1+x2 χE, then
1
h(x)
µ(E) = dm(x) = dµ(x).
E 1 + x2 E 1 + x2 That is, h(x)
dµ = E E 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+x 2 , 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
n

1.3 1998 April 3 1 REAL ANALYSIS
PART B
7. Let φ(x, y) = x 2 y 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
(a) Show that µ is a Borel measure on R.
µ(E) = m(φ −1 (E)).
(b) Let χE denote the characteristic function of the set E. Show that
χE ◦ φ dm.
χE dµ =
R
(c) Evaluate the integral ∞
S
t
−∞
2 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
f ′ dm = 1.
R
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) = lim sup [f(x + 1/n) − f(x)] n.
n→∞
As a lim sup of a sequence of measurable functions, g is measurable (Rudin [8], theorem 1.14?). Let E be the set
on 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
and, since f is increasing, f ′ ≥ 0 on E, so
|f ′
|dm = |f ′
|dm +
R
R\E
f ′
dm +
E
E
|f ′
|dm =
f ′
dm =
E
E
|f ′
|dm =
f ′ dm,
E
f ′ dm = 1.
Thus, f ′ ∈ L 1 (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 ′ ∈ L 1 ([a, b]), and x
a f ′ (t)dt =
f(x) − f(a) for a ≤ x ≤ b, then f ∈ AC[a, b].
17 I 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

1.3 1998 April 3 1 REAL ANALYSIS
The converse of this lemma is also true. 18
Lemma 1.3 If f : R → R is increasing and f ′ ∈ L 1 ([a, b]), then x
a f ′ (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 b
a f ′ (t)dt =
f(b) − f(a) holds for all −∞ < a < b < ∞. By lemma 1.3, we have b
a f ′ (t)dt ≤ f(b) − f(a), so we need only
show that strict inequality cannot hold. Suppose, by way of contradiction, that b
a f ′ (t)dt < f(b) − f(a) holds for
some −∞ < a < b < ∞. Then,
1 =
R
f ′ dm =
a
−∞
f ′ dm +
b
a
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 b
a 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
f(∞) − f(−∞) = 1, so this is equivalent to showing
x
lim f
x→∞
−x
′ (t)dm(t) = lim [f(x) − f(−x)].
x→∞
R f ′ dm = 1. By assumption
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
x
1 = lim [f(x) − f(−x)] = lim
x→∞ x→∞
f ′
(t)dm(t) = f ′ dm.
9. Let F be a continuous linear functional on the space L 1 [−1, 1], with the property that F (f) = 0 for all odd
functions f in L 1 [−1, 1]. Show that there exists an even function φ such that
F (f) =
1
−1
−x
f(x)φ(x) dx, for all f ∈ L 1 [−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 ∈ L 1 [−1, 1] ∗ , then by the Riesz representation theorem 19 there is a unique h ∈ L ∞ [−1, 1]
such that
F (f) =
1
−1
f(x)h(x) dx (∀f ∈ L 1 [−1, 1])
18 See Folland [4] for a nice, concise treatment of the fundamental theorem of calculus for Lebesgue integration.
19 See problem 3 of section 1.5.
16
R
⊓⊔

1.3 1998 April 3 1 REAL ANALYSIS
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 = fe + fo be the decomposition of f into a sum of even and odd functions. Then, by linearity of
F , and since F (fo) = 0 by hypothesis,
F (f) = F (fe) + F (fo) = F (fe) =
1
−1
feh =
1
−1
1
feφ + feψ.
−1
Now note that feψ is an odd function (since it’s an even times an odd), so 1
−1 feψ = 0, since [−1, 1] is symmetric.
Similarly, 1
−1 foφ = 0. Therefore,
F (f) = F (fe) =
1
−1
feφ =
1
−1
1
feφ + foφ =
−1
17
1
(fe + fo)φ =
−1
1
−1
fφ.
⊓⊔

1.4 2000 November 17 1 REAL ANALYSIS
1.4 2000 November 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 : R 3 → R 3 is an invertible linear map and that g : R 3 → R 3 has continuous first order partial
derivatives and satisfies g(x) ≤ Cx 2 for some constant C and all x ∈ R 3 . Here x denotes the usual
Euclidean norm on R 3 . 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 → R n be a C 1 -mapping of an open set E ⊂ R n . 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 R n 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 (R 3 ). Therefore, the
derivative of f = L + g,
f ′
∂fi
(x) Jf (x)
∂xj
exists. Furthermore, Jf (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 (R 3 ). 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) = Jg(0). We claim, Jg(0) = 0. Indeed, if x1, x2, x3 are the elementary unit vectors (also known
as i, j, k), then the elements of Jg(0) are
∂gi
∂xj
3
i,j=1
gi(0 + hxj) − gi(0)
(0) = lim
= lim
h→0 h
h→0
gi(hxj)
. (8)
h
The second equality follows by the hypothesis that g is continuous and satisfies g(x) ≤ Cx 2 , which implies
that g(0) = 0. Finally, to show that (8) is zero, consider
which implies
|gi(hxj)| ≤ g(hxj) ≤ Chxj 2 = C|h| 2 ,
|gi(hxj)|
|h|
2 |hxj|
≤ C = C|h| → 0, as h → 0.
|h|
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 = lim x→0 + f(x).
20 The 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.
21 See Rudin [7].
18

1.4 2000 November 17 1 REAL ANALYSIS
3. Let f and g be Lebesgue integrable functions on [0, 1], and let F and G be the integrals
F (x) =
x
Use Fubini’s and/or Tonelli’s theorem to prove that
1
0
0
f(t) dt, G(x) =
F (x)g(x) dx = F (1)G(1) −
1
0
x
0
g(t) dt.
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 =
L ∞ (ν).
dν
dµ
is the same in L ∞ (µ) as it is in
1
5. Suppose that {fn} is a sequence of Lebesgue measurable functions on [0, 1] such that limn→∞ 0 |fn| dx = 0 and
there is an integrable function g on [0, 1] such that |fn| 2 1
≤ g, for each n. Prove that limn→∞ 0 |fn| 2 dx = 0.
6. Denote by Pe the family of all even polynomials. Thus a polynomial p belongs to Pe if and only if p(x) =
p(x)+p(−x)
2 for all x. Determine, with proof, the closure of Pe 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) ≤ f dm ≤ b · m(U),
for all open sets U in R. Prove that a ≤ f(x) ≤ b a.e.
U
19

1.5 2001 November 26 1 REAL ANALYSIS
1.5 2001 November 26
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 {fn} be a sequence of Lebesgue measurable functions on a set E ⊂ R, where E is of finite Lebesgue measure.
Suppose that there is M > 0 such that |fn(x)| ≤ M for n ≥ 1 and for all x ∈ E, and suppose that
limn→∞ fn(x) = f(x) for each x ∈ E. Use Egoroff’s theorem to prove that
f(x) dx = lim fn(x) dx.
n→∞
E
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 ɛ > 0 such that |f(x0)| = M + ɛ. By the triangle inequality, then, for all n ∈ N,
|f(x0) − fn(x0)| ≥ ||f(x0)| − |fn(x0)|| = |M + ɛ − |fn(x0)|| ≥ ɛ,
which contradicts fn(x0) → f(x0). Thus, |f(x)| ≤ M for all x ∈ E.
Next, fix ɛ > 0. By Egoroff’s theorem (A.8), there is a G ⊂ E such that µ(E \ G) < ɛ and fn → f uniformly
on G. Furthermore, since |fn| ≤ M and |f| ≤ M and µ(E) < ∞, it’s clear that {fn} ⊂ L 1 and f ∈ L 1 , so the
following inequalities make sense (here we’re using the notation fG = sup{|f(x)| : x ∈ G}):
f dµ − fn dµ
E
E
≤
|f − fn| dµ = |f − fn| dµ + |f − fn| dµ
E
E\G
G
≤ |f| dµ + |fn| dµ + f(x) − fn(x)Gµ(G)
E\G
E\G
≤ 2Mµ(E \ G) + f(x) − fn(x)Gµ(G)
< ɛ + f(x) − fn(x)Gµ(G).
Finally, µ(G) ≤ µ(E) < ∞ and f(x) − fn(x)G → 0, which proves that
E fn 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(x) dx > 0.
(b) Prove that if x
f(x) dx = 0, for each x ∈ [0, 1],
then f(x) = 0 for almost all x ∈ [0, 1].
Solution: (a) Define Fn = {x ∈ F : f(x) > 1/n}. Then
and m(F ) > 0 implies
0
F
F1 ⊆ F2 ⊆ · · · ↑
Fn = F,
0 < m(F ) = m(∪nFn) ≤
m(Fn).
20
n
E
n
E
f dµ. ⊓⊔

1.5 2001 November 26 1 REAL ANALYSIS
Therefore, m(Fk) > 0 for some k ∈ N, and then it follows from the definition of Fk that
0 < 1
k m(Fk)
≤
f dm ≤ f dm.
Fk
(b)
Suppose there is a subset E ⊂ [0, 1] of positive measure such that f > 0 on E. Then part (a) implies
f dm > 0. Let F ⊂ E be a closed subset of positive measure. (That such a closed subset exists follows from
E
Prop. 3.15 of Royden [6].) Then, again by (a),
f dm > 0. Now consider the set G = [0, 1] \ F , which is open
F
in [0, 1], and hence22 is a countable union of disjoint open intervals; i.e., G = ·∪n(an, bn). Therefore,
0 = f dm =
f dm + f dm,
so
f dm > 0 implies
F
[0,1]
n
n
(an,bn)
(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 =
bk
0
ak
f(x) dm(x) − f(x) dm(x).
0
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) Hölder’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 {fn} be a sequence of measurable functions on (X, M, µ)
such that fn → f a.e.. If there exists g ∈ L1 (X, M, µ) such that |fn(x)| ≤ g(x) holds for all x ∈ X and
n = 1, 2, . . .. Then {fn} ⊂ L1 , f ∈ L1 , lim
X fn dµ =
X f dµ, and fn − f1 → 0.
(c) (Hölder’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
F
F
⊓⊔

1.5 2001 November 26 1 REAL ANALYSIS
(i) If 1 < p < ∞ and 1 1
p + q = 1, then fg1 = fpgq. Thus, if f ∈ Lp and g ∈ Lq , then fg ∈ L1 (ii) If p = ∞ and if f ∈ L ∞ and g ∈ L 1 , then |fg| ≤ f∞|g|, so fg1 ≤ 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 Lp , then there is a unique g ∈ L q such that
Λf = fg dµ (∀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 Mx such that
|f(x)| ≤ Mx 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 = ∞ o
= ∅ for at least one n ∈ N.
¯
Gn
Corollary 2. A nonempty complete metric space is not a countable union of nowhere dense sets.
n=1 Gn then
(b) Define Am = {x ∈ X : |f(x)| ≤ m, ∀f ∈ F }. Then X = ∞ m=1 Am, since for every x there is a finite
number Mx such that |f(x)| ≤ Mx 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 A◦ m = ∅, so the set G = A◦ m and
the number M = m satisfy the given criteria. ⊓⊔
25 Note to self: add case p = ∞
26 See 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

1.5 2001 November 26 1 REAL ANALYSIS
5. Prove or disprove:
(a) L 2 convergence implies pointwise convergence.
(b)
∞
sin(x
lim
n→∞
0
n )
xn dx = 0.
(c) Let {fn} be a sequence of measurable functions defined on [0, ∞). If fn → 0 uniformly on [0, ∞), as n → ∞,
then
lim fn(x) dx = lim fn(x) dx.
Solution:
[0,∞)
[0,∞)
(a) This is false, as the following example demonstrates: For each k ∈ N, define fk,j = χ [ j−1
k
and let {gn} be the sequence defined by
g1 = f1,1,
g2 = f2,1, g3 = f2,2,
g4 = f3,1, g5 = f3,2, g6 = f3,3,
g7 = f4,1, . . .
j
, k ) for j = 1, . . . , k,
Then |fk,j| 2 dµ = 1/k for each j = 1, . . . , k, so fk,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) = fk,j(x) = 1 with n ≥ N, and we can also find a k ′ ∈ N
and j ′ ∈ {1, . . . , k ′ } such that gn ′(x) = fk ′ ,j ′(x) = 0 with n′ ≥ N. ⊓⊔
sin t
t
(b) For any fixed 0 < x < 1, limn→∞ xn = 0. Also,
Together, these two facts yield
sin x
lim
n→∞
n
xn = 1.
→ 1, as t → 0, which can be proved by L’Hopital’s rule.
Now, recall that | sin θ| ≤ |θ| for all real θ. Indeed, since sin θ = θ
cos x dx, we have, for θ ≥ 0,
0
and, for θ < 0,
In particular, for any 0 < x < 1, | sin xn |
|xn |
function
sin xn
x n , to obtain
| sin θ| ≤
θ
0
| cos x| dx ≤
θ
0
1 dx = θ,
| sin θ| = | sin(−θ)| ≤ | − θ| = |θ|.
≤ 1. Therefore, we can apply the dominated convergence theorem to the
1
lim
n→∞
0
sin xn dx =
xn 1
0
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 1 1− = u n . Therefore,
N
1
sin xn n
N
sin u
dx =
xn 1 u
23
du
1
nu1− n
= 1
n
N
n 1
sin u
1
u2− n
du.

1.5 2001 November 26 1 REAL ANALYSIS
Now,
1
lim
N→∞ n
N n
1
sin u
1
u2− n
du
Therefore, ∞
and so,
Combining results (9) and (10) yields
n
N
1
≤ lim u
N→∞ n 1
1
n −2 1
du = lim
N→∞ n
1
dx
n
sin x n
x
∞
lim
n→∞
1
∞
lim
n→∞
0
≤ 1
n − 1 ,
u 1
n −1
− 1
1
n
N n
1
= 1
n − 1 .
sin xn dx = 0. (10)
xn sin xn dx = 1.
xn (c) This is false, as the following example demonstrates: Let fn = 1
n χ [0,n). Then fn → 0 uniformly and so
lim fn = 0. On the other hand, fn = 1 for all n ∈ N. Therefore, lim fn = 1 = 0 = lim fn. ⊓⊔
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 a2 k , 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=1 akφk, where
ak = 〈x, φk〉, for each k ∈ N. Therefore, by linearity of f,
f(x) = f(
akφk) =
akf(φk). (11)
Now
and, by definition of y,
k
k
〈x, y〉 = 〈
akφk, y〉 =
ak〈φk, y〉, (12)
k
k
〈φk, y〉 = 〈φk,
f(φj)φj〉 =
f(φj)〈φk, φj〉 = f(φk). (13)
j
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) =
akf(φk) ∵ (11)
k
=
ak〈φk, y〉 ∵ (13)
k
= 〈x, y〉 ∵ (12)
24
j
⊓⊔

1.5 2001 November 26 1 REAL ANALYSIS
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(x)| |(x, y)|
f = sup{|f(x)|
: x ≤ 1} = sup = sup
x∈X
x∈X x x∈X x ,
and recall that |(x, y)| ≤ xy holds for all x, y ∈ X. Whence,
On the other hand,
|(x, y)|
f = sup ≤ y. (14)
x∈X x
|(x, y)| |(y, y)| y2
f = sup ≥ = = y. (15)
x∈X x y y
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 = sup x≤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 {Tn} ⊂ B(X, Y ) be a Cauchy sequence; i.e., Tn − Tm → 0 as m, n → ∞. Fix x ∈ X. Then,
Tnx − TmxY ≤ Tn − TmxX → 0, as n, m → ∞.
Therefore, the sequence {Tnx} ⊂ Y is a Cauchy sequence in (Y, · Y ). Since the latter is complete, the limit
limn→∞ Tnx = y ∈ Y exists. Define T : X → Y by T x = limn→∞ Tnx, for each x ∈ X. To complete the proof,
we must check that T is linear, bounded, and satisfies limn→∞ Tn − T = 0.
• T is linear:
For x1, x2 ∈ X,
T (x1 + x2) = lim
n→∞ Tn(x1 + x2)
= lim
n→∞ (Tnx1 + Tnx2) (∵ Tn is linear)
= lim
n→∞ Tnx1 + lim
n→∞ Tnx2
= T x1 + T x2.
(∵ both limits exist)
• T is bounded:
First, note that {Tn} is a Cauchy sequence of real numbers, since | Tn − Tm | ≤ Tn − Tm → 0,
as n, m → ∞. Therefore, there is a c ∈ R such that Tn → c, as n → ∞. For some N ∈ N, then,
Tn ≤ c + 1 for all n ≥ N. Thus,
TnxY ≤ TnxX ≤ (c + 1)xX (∀x ∈ X). (16)
25

1.5 2001 November 26 1 REAL ANALYSIS
Now, by definition, Tnx → T x, for all x ∈ X and, since the norm · Y is uniformly continuous, 27
TnxY → 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→∞ Tn − T = 0:
Fix ɛ > 0 and choose N ∈ N such that n, m ≥ N implies Tn − Tm < ɛ. Then,
Tnx − Tmx ≤ Tn − TmxX < ɛxX
holds for all n, m ≥ N, and x ∈ X. Letting m go to infinity, then,
Tnx − T x = lim
m→∞ Tnx − Tmx ≤ ɛxX.
That is, Tnx − T x ≤ ɛxX, for all n ≥ N and x ∈ X. Whence, Tn − T ≤ ɛ for all n ≥ N.
27 Proof: | aY − bY | ≤ a − bY (∀a, b ∈ Y ).
26
⊓⊔

1.6 2004 April 19 1 REAL ANALYSIS
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 ɛ > 0 there is a positive number M such that |f| ≤ M, except on a
set of measure less than ɛ.
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} =
∞
{x ∈ S : f(x) < α − qi} ∩ {x ∈ S : g(x) < qi}.
i=1
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 ɛ > 0. For n ∈ N, define An = {x ∈ [a, b] : |f(x)| ≤ n}. Then
[a, b] =
∞
An ∪ A∞, (18)
n=1
where 28 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 ) < ɛ. 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 ) < ɛ.
The second equality holds since we assumed f(x) = ±∞ only on a set of measure zero; i.e., µ(A∞) = 0. ⊓⊔
28 Since 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

1.6 2004 April 19 1 REAL ANALYSIS
2. (a) State Egorov’s theorem.
(b) State Fatou’s lemma.
(c) Let {fn} ⊂ L p [0, 1], where 1 ≤ p < ∞. Suppose that fn → f a.e., where f ∈ L p [0, 1]. Prove that
fn − fp → 0 if and only if fnp → fp.
Solution:
(a) See theorem A.8.
(b) See theorem A.6.
(c) (⇒) By the Minkowsky inequality, fnp = fn − f + fp ≤ fn − fp + fp. Similarly, fp ≤
fn − fp + fnp. Together, the two inequalities yield |fnp − fp| ≤ fn − fp. Therefore, fn − fp → 0
implies |fnp − fp| → 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) = x p is convex on [0, ∞). Thus, for all α, β ∈ [0, ∞),
p
α + β α + β
= φ ≤
2
2
1
1
[φ(α) + φ(β)] =
2 2 (αp + β p ).
When α, β ∈ R, the triangle inequality followed by the lemma yields
Proof 1: By (19),
|α − β| p ≤ ||α| + |β|| p ≤ 2 p−1 (|α| p + |β| p ). (19)
|fn − f| p ≤ 2 p−1 (|fn| p + |f| p ).
In particular, fn − f ∈ L p , for each n ∈ N. Moreover, the functions
gn = 2 p−1 (|fn| p + |f| p ) − |fn − f| p . (20)
are non-negative. Now notice that lim gn = 2p |f| p . Applying Fatou’s lemma to (20), then,
2 p |f| p
=
lim gn ≤ lim
2p−1 gn = lim (|fn| p + |f| p ) − |fn − f| p .
Since fnp → fp, this implies
2 p
|f| p ≤ 2 p
|f| p
− lim
|fn − f| p .
Equivalently 0 ≤ − lim |fn − f| p . This proves fn − f p → 0. ⊓⊔
29 Disclaimer: I made up the first proof, so you should check it carefully for yourself and decide whether you believe me.
28

1.6 2004 April 19 1 REAL ANALYSIS
Proof 2: By (19),
In particular, fn − f ∈ L p , for each n ∈ N. Define the functions
|fn − f| p ≤ 2 p−1 (|fn| p + |f| p ). (21)
gn = 2 p−1 (|fn| p + |f| p ) and g = 2 p |f| p .
Then gn → g a.e., and fnp → fp implies gn → g. Also, gn ≥ |fn − f| p → 0 a.e., by (21). Therefore,
the dominated convergence theorem (theorem A.7) implies |fn − f| p → 0. ⊓⊔
Proof 3: Since f ∈ Lp , for all ɛ > 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 < ɛ/2, and
E |f|p dµ < ɛ/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 fn → f uniformly on A. Therefore,
|f| p
= |f| p
+ |f| p
+ |f| p
X
X\B
since, by Fatou’s lemma,
A |f|p =
A lim |fn| p ≤ lim
lim |fn|
A
p = lim
By (22), then,
Therefore, lim
Therefore,
X
B\A
A
< ɛ/2 + ɛ/2 + |f|
A
p
≤ ɛ + lim |fn| p , (22)
X
A
|fn| p
− |fn|
X\A
p
A |fn| p . By hypothesis, fnp → fp. Therefore
=
|f| p
< ɛ + |f|
X
p
− lim |fn|
X\A
p .
X\A |fn| p < ɛ (since f ∈ L p ). Finally, note that
fn − f p = (fn − f)χA + (fn − f)χ X\Ap
≤ (fn − f)χAp + (fn − f)χ X\Ap
≤ (fn − f)χAp + fnχ X\Ap + fχ X\Ap.
lim fn − f p ≤ lim{|fn(x) − f(x) : x ∈ A}µ(A) 1/p + lim
X
|f| p
− lim |fn|
X\A
p .
|fn|
X\A
p
(Minkowsky)
1/p
+
|f|
X\A
p
The first term on the right goes to zero since fn → f uniformly on A. The other terms are bounded by 2ɛ 1/p . ⊓⊔
29
1/p
.

1.6 2004 April 19 1 REAL ANALYSIS
3. (a) Let S = [0, 1] and let {fn} ⊂ Lp (S), where 1 < p < ∞. Suppose that fn → f a.e. on S, where f ∈ Lp (S). If
there is a constant M such that fnp ≤ M for all n, prove that for each g ∈ Lq (S), 1 1
p + q = 1, we have
lim
n→∞
fng =
S
fg.
S
(b) Show by means of an example that this result is false for p = 1.
Solution:
(a) Since g ∈ Lq (S), for all ɛ > 0 there exists δ > 0 such that if A is a measurable set with µA < δ then
|g| q dµ < ɛ.
A
Let B0 ⊂ S denote the set on which fn does not converge to f. Let B ⊂ S be such that fn → 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 := |fng − fg| dµ = |fn − f||g| dµ
S
S
=
|fn − f||g| dµ +
S\B
|fn − f||g| dµ
B
≤ (fn − f)χ S\Bpgq + fn − fpgχBq.
The final inequality holds because fn, f ∈ L p implies |fn − f| ∈ L p and by Hölder’s inequality. Now, by
Minkowski’s inequality, fn − fp ≤ fnp + fp, so
Dn ≤ sup{|fn(x) − f(x)| : x /∈ B}µ(S \ B) 1/p
gq + (fnp + fp) |g|
B
q 1/q dµ .
Now we are free to choose the δ > 0 above so that
|g|
B
q dµ
1/q
<
ɛ
2(M + fp)
holds whenever µB < δ. Also, fn → f uniformly on S \ B, so let N be such that
ɛ
sup{|fN(x) − f(x)| : x /∈ B} <
2µ(S \ B) 1/p .
gq
Then DN < ɛ/2 + ɛ/2. ⊓⊔
(b) The result is not true for p = 1 as the following example shows: Let fn = nχ [0,1/n], n = 1, 2, . . . . First note
that fn → 0 a.e. For if x ∈ (0, 1], there exists N > 0 such that 1/N < x and thus fn(x) = 0 for all n ≥ N.
Therefore, {x ∈ [0, 1] : fn(x) 0} = {0}. Now, if g is the constant function g ≡ 1, then fng dµ = 1 for all
n = 1, 2, . . . . Therefore, fng dµ → 1, while fg dµ = 0g dµ = 0. Finally, note that fn1 = nµ([0, 1
n ]) = 1
for n = 1, 2, . . . , so {fn} satisfies the hypothesis fn1 ≤ some constant M. ⊓⊔
30

1.6 2004 April 19 1 REAL ANALYSIS
4. State and prove the closed graph theorem.
5. Prove or disprove:
(a) For 1 ≤ p < ∞, let ℓ p := {x = {xk} | xp = ( ∞
k=1 |xk| p ) 1/p < ∞}. Then for p = 2, ℓ 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: f1 = 1
|f(x)| dx.
0
Then X is a Banach space.
(c) Every real, separable Hilbert space is isometrically isomorphic to ℓ 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 ∈ L 1 (R). (ii) Show that h1 ≤ f1g1.
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
f dµ ∈ S
µ(E)
for every E ∈ B with µ(E) > 0, prove that f(x) ∈ S for almost all x ∈ X.
31
E

1.7 2007 November 16 1 REAL ANALYSIS
1.7 2007 November 16
Notation: R is the set of real numbers and R n 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 E1, E2, . . . are measurable subsets of X with the
property that for n = m, µ(En ∩ Em) = 0. Let E be the union of these sets. Prove that
µ(E) =
∞
µ(En)
Solution: Define F1 = E1, F2 = E2 \ E1, F3 = E3 \ (E1 ∪ E2), . . . , and, in general,
n=1
n−1
Fn = En \ Ei (n = 2, 3, . . . ).
i=1
If M is the σ-algebra of µ-measurable subsets of X, then Fn ∈ M for each n ∈ N, since M is a σ-algebra. Also,
Fi ∩ Fj = ∅ for i = j, and F1 ∪ F2 ∪ · · · ∪ Fn = E1 ∪ E2 ∪ · · · ∪ En for all n ∈ N. Thus,
and, by σ-additivity of µ,
∞
n=1
Fn =
n=1
∞
En E,
n=1
∞
∞
µ(E) = µ( Fn) = µ(Fn).
Therefore, if we can show µ(En) = µ(Fn) holds for all n ∈ N, the proof will be complete.
Now, for each n = 2, 3, . . . ,
and
n=1
n−1
Fn = En ∩ ( Ei) c
i=1
n−1
µ(En) = µ(En ∩ ( Ei) c n−1
) + µ(En ∩ ( Ei)). (24)
Equation (24) holds because n−1
i=1 Ei is a measurable set for each n = 2, 3, . . . . Finally, note that
which implies
i=1
n−1
En ∩ ( Ei) =
i=1
n−1
µ(En ∩ ( Ei)) ≤
i=1
(En ∩ Ei),
n−1
i=1
i=1
n−1
µ(En ∩ Ei),
by σ-subadditivity. By assumption, each term in the last sum is zero, and therefore, by (23) and (24),
n−1
µ(En) = µ(En ∩ ( Ei) c ) = µ(Fn) holds for each n = 2, 3, . . . .
i=1
For n = 1, we have F1 = E1, by definition. This completes the proof. ⊓⊔
32
i=1
(23)

1.7 2007 November 16 1 REAL ANALYSIS
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 fn ∈ L 1 (m) for n = 1, 2, . . . . Assuming that fn − f1 → 0 and fn → 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 {fn} ⊂ L1 (m) and
fn − f1 → 0, then fn → f in measure. (Proof: m({x : |fn(x) − f(x)| > ɛ}) ≤ 1
ɛ fn − f1 → 0.) Next recall
another important theorem31 which states that if fn → f in measure then there is a subsequence {fnj } ⊆ {fn}
which 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 {fn} ⊂ L 1 (m) and fn − f1 → 0 then there is a subsequence
{fnj } ⊆ {fn} with the property fnj (x) → f(x) for almost all x ∈ R.
Now, if fn(x) → g(x) for almost all x ∈ R, and if B1 be the set of measure zero where fn(x) g(x), then off of
B1 the sequence fn, as well as every subsequence of fn, converges to g. Let {fnj } be the subsequence mentioned
above which converges to f almost everywhere. Then
|f(x) − g(x)| ≤ |f(x) − fnj (x)| + |fnj (x) − g(x)|. (25)
Define B2 = {x ∈ R : fnj (x) f(x)}. Then the set B = B1 ∪ B2 has measure zero and, for all x ∈ R \ B,
fnj (x) → f(x) and fnj (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)}, then
B = ∪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 ɛ > 0 there is a δ > 0 such that
|f − g| dm < ɛ for
E
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 fn → g uniformly on A. Then
n
|f − g| dm = |f − g| dm + |f − g| dm
−n
[−n,n]\A
≤ ɛ + |f − fn| dm + |fn − g| dm
A
A
≤ ɛ + f − fn1 + mA sup |fn(x) − g(x)|,
x∈A
30 Note 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.
31 Folland [4], theorem 2.30 and its corollary.
32 Perhaps this statement is the version of the Littlewood principle dealing with a.e. convergence that we were meant to cite in part (a).
33
A

1.7 2007 November 16 1 REAL ANALYSIS
where f − fn1 → 0, by assumption, and supx∈A |fn(x) − g(x)| → 0 since fn → g uniformly on A. Since ɛ > 0
was arbitrary, it follows that n
−n |f − g| dm = 0 and, for functions f, g ∈ L1 ([−n, n], m), this implies that f = g
a.e. on [−n, n].
It remains to show that f, g ∈ L1 ([−n, n], m). It’s clear that f ∈ L1 since f1 ≤ f − fn1 + fn1 < ∞. To
a.e.
prove g ∈ L1 ([−n, n], m) note that fn −−→ g implies limn |fn(x)| = |g(x)| for almost all x, so by Fatou’s lemma,
g1 = |g| dm = lim |fn| dm ≤ lim |fn| dm = lim fn1 = f1 < ∞.
The last equality holds because, by the triangle inequality, |fn1 − f1| ≤ fn − f1 → 0. ⊓⊔
3. Let K be a compact subset in R 3 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.
n (b) Let g = max(1 − f, 0) and prove that limn→∞ g exists and is equal to m3(K).
Solution: (a) Define dist(x, K) = f(x) = infk∈K |x − k|. Clearly, for all k ∈ K, f(x) ≤ |x − k|. Therefore, by
the triangle inequality, for any x, y ∈ R 3 ,
and so, taking the infimum over k ∈ K on the right,
Similarly,
f(x) ≤ |x − y| + |y − k|, ∀k ∈ K,
f(x) ≤ |x − y| + f(y). (26)
f(y) ≤ |x − y| + f(x). (27)
Obviously, for any given x ∈ R 3 , f(x) is finite. Therefore, (26) and (27) together imply that
Whence f is (Lipschitz) continuous.
|f(x) − f(y)| ≤ |x − y|, ∀ x, y ∈ R 3 .
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 ɛ-neighborhood about x fully contained in K c , in which case f(x) > ɛ. 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 K1 to be a closed and bounded
set containing K on which f(x) ≤ 1. That is, K1 is the set of points that are a distance of not more than 1 unit
from the set K. In particular K ⊂ K1. Notice that g = max(1 − f, 0) = (1 − f)χK1 . Also, if x ∈ K1 \ K,
then 0 ≤ 1 − f(x) < 1, so g n → 0 on the set K1 \ K, while on the set K, g n = 1 for all n ∈ N. Therefore,
g n → χK. Finally, note that g n ≤ χK1 ∈ L 1 (R 3 ) so the dominated convergence theorem can be applied to yield
limn→∞
g n = χK = m3(K). ⊓⊔
4. Let E be a Borel subset of R 2 .
(a) Explain what this means.
(b) Suppose that for every real number t the set Et = {(x, y) ∈ E | x = t} is finite. Prove that E is a Lebesgue
null set.
34

1.7 2007 November 16 1 REAL ANALYSIS
Solution:
(a) The Borel σ-algebra of R 2 , which we denote by B(R 2 ), is the smallest σ-algebra that contains the open subsets
of R 2 . The sets belonging to B(R 2 ) are called Borel subsets of R 2 .
(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 easy
to prove that if G is any countable subset, then mG = 0. (Just fix ɛ > 0 and cover each point xn ∈ G with a set
En of measure less than ɛ2 −n . Then mG ≤ mEn < ɛ.)
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 R 2 by 33 (X × Y, B(X) ⊗ B(Y ), µ × ν) = (R 2 , B(R 2 ), 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 = χE(x, y) dµ(x) dν(y) = χE(x, y) dν(y) dµ(x). (28)
Now, let
Y
X
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) ∈ Et = {(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 = χGt (y) dν(y) dµ(t) = νGt dµ(t) = 0.
X
Y
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 dν
d(µ+ν)
represent the Radon-Nikodym derivative of ν with respect to µ + ν. Show that
0 <
X
Y
X
dν
< 1 a.e. [µ].
d(µ + ν)
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) = f d(µ + ν) ∀E ∈ A.
E
Indeed, f is the Radon-Nikodym derivative; i.e., f = dν
d(µ+ν) . We want to show 0 < f(x) < 1 holds 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.
33 The notation B(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(R 2 ).
35

1.7 2007 November 16 1 REAL ANALYSIS
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 {bn}
satisfying 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 Tk : ℓq(N) → R by Tk(b) = k
n=1 anbn, for b ∈ ℓq(N). Then {Tk} is a
family of pointwise bounded linear functionals. That is, each Tk is a linear functional, and, for each b ∈ ℓq(N),
there is an Mb ≥ 0 such that |Tk(b)| ≤ Mb 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
Sk
k
∞
anbn → anbn = x ∈ R.
n=1
Thus, {Tk(b)} = {Sk} is a convergent sequence of reals, so, if N ∈ N is such that k ≥ N implies |Sk − x| < 1,
and if M ′ b is defined to be max{|Sk| : 1 ≤ k ≤ N}, then, for any k ∈ N,
n=1
|Tk(b)| ≤ Mb max{M ′ b, x + 1}.
Next note that ℓq is a Banach space, so the (Banach-Steinhauss) principle of uniform boundedness implies that
there is a single M > 0 such that Tk ≤ M for all k ∈ N. In other words,
(∃ M > 0) (∀b ∈ ℓq) (∀k ∈ N) |Tk(b)| ≤ Mb.
Define let T (b) ∞
n=1 anbn = limk→∞ Tk(b), which exists by assumption. Since |·| is continuous, we conclude
that limk→∞ |Tk(b)| = |T (b)|. Finally, since |Tk(b)| ≤ Mb for all k ∈ N, we have |T (b)| ≤ Mb. That is T is
a bounded linear functional on ℓq(N).
Now, by the Riesz representation theorem, if 1 ≤ q < ∞, then any bounded linear functional T ∈ ℓ ∗ q is uniquely
representable by some α = (α1, α2, . . . ) ∈ ℓp as
T (b) =
∞
αnbn. (29)
n=1
On the other hand, by definition, T (b) = ∞
n=1 anbn, for all b ∈ ℓq. Since the representation in (29) is unique,
a = α ∈ ℓp. That is,
n |an| p < ∞. ⊓⊔
34 On the original exam this question asked only about the special case p = q = 2.
36

1.7 2007 November 16 1 REAL ANALYSIS
(b) Consider the case p = 1 and q = ∞. First recall that the Riesz representation theorem says that every
T ∈ ℓ ∗ q (1 ≤ q < ∞) is uniquely representable by some α ∈ ℓp (where p = q/(q − 1), so 1 < p ≤ ∞). That is ℓp
is the dual of ℓq, when 1 ≤ q < ∞ and p = q/(q − 1). However, in the present case we have q = ∞ and p = 1
and ℓ1 is not the dual of ℓ∞. (Perhaps the easiest way to see this is to note that ℓ1 is separable but ℓ∞ is not. For
the collection of a ∈ ℓ∞ such that an ∈ {0, 1}, n ∈ N, is uncountable and, for any two distinct such sequences
a, b ∈ {0, 1} N , we have a − b∞ = 1, so there cannot be a countable base, so ℓ∞ 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
ān/|an|, for an = 0,
bn = sgn(an) =
(n ∈ N).
0, for an = 0,
Then
n |an| =
n anbn converges by the hypothesis, since |bn| ∈ {0, 1} implies b ∈ ℓ∞. Therefore, a ∈ ℓ1.
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 williamdemeo@gmail.com.
37

2 Complex Analysis
38
2 COMPLEX ANALYSIS

2.1 1989 April 2 COMPLEX ANALYSIS
2.1 1989 April
INSTRUCTIONS: Do at least four problems.
TIME LIMIT: 1.5 hours 35
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) = anz n
n=0
converges uniformly on any compact subset of Ω. The unit circle T = {z : |z| = 1} = {eiθ : θ ∈ R} is one such
compact subset, and here the series is
f(e iθ ∞
) = ane inθ .
If we write the coefficients as an = cn + ibn, where cn, bn ∈ R, then we have
n=0
ane inθ = (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(e iθ ) =
∞
[cn sin(nθ) + bn cos(nθ)]
n=0
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
lim f(z) = 0.
z→∞
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

2.1 1989 April 2 COMPLEX ANALYSIS
Show that if |z| > 2, then
Solution: By Cauchy’s formula, if |z| < R, then
f(z) = 1
2πi |ζ|=R
1 f(ζ)
dζ = −f(z). (30)
2πi |ζ|=2 ζ − z
f(ζ) 1
dζ −
ζ − z 2πi |ζ|=2
f(ζ)
dζ. (31)
ζ − z
Note that this holds for all R > |z| > 2. Fix ɛ > 0. Let Rɛ be such that |f(ζ)| < ɛ/2 for all |ζ| = Rɛ and
Rɛ > 2|z|. Then |ζ − z| > Rɛ/2 for all |ζ| = Rɛ, so
1 |f(ζ)| ɛ/2 2πRɛ
|dζ| < = ɛ.
2π |ζ|=Rɛ |ζ − z| 2π Rɛ/2
Therefore, by (31),
1
2πi
|ζ|=2
f(ζ)
ζ − z
dζ + f(z) < ɛ.
This holds for any ɛ, 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−z
1+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 and
arguing 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.
Thus, ϕ0 : U + → Ω − .
ϕ0(i/2) =
1 − i/2
1 + i/2 =
1 − i/2
1 + i/2
1 − i/2
=
1 − i/2
3 − 4i
∈ Ω
5
− .
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 −1
0 (z) = ϕ0(z) maps Ω onto U. Putting it all
together, we see that a map satisfying the requirements is
37 This problem also appears in April ’95 (5) and November ’06 (2).
ϕ(z) = (ϕ0 ◦ ϕ3 ◦ ϕ2 ◦ ϕ1 ◦ ϕ0)(z).
40
⊓⊔

2.1 1989 April 2 COMPLEX ANALYSIS
4. Let {fn} be a sequence of analytic functions in the unit disk U, and suppose there exists a constant M such that
|fn(z)| |dz| ≤ M
C
for each fn and for all circles C lying in U. Prove that {fn} 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) =
n
k=1
P (ak)
Q ′ (ak)(z − ak) .
6. Use contour integration and the residue method to evaluate the integral
∞
0
cos x
(1 + x2 dx.
) 2
Solution: Denote the integral by I. Since the integrand is even,
2I =
∞
−∞
cos x
(1 + x2 dx.
) 2
Consider the simple closed contour ΓR = γR ∪ [−R, R], where the trace of γR is the set {Re iθ : 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 cos z
=
) 2 (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) = lim
z→i
d
dz [(z − i)2 f(z)] = lim
z→i
d
dz
cos z
(z + i) 2
−(z + i)
= lim
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,
f(z) dz = 2πi Res(f, i) = π
2e .
ΓR
41

2.1 1989 April 2 COMPLEX ANALYSIS
It remains to check that
2I =
∞
−∞
Indeed,
γR
f(z) dz → 0, as R → ∞, which will allow us to conclude that
f(x) dx = lim
R→∞
ΓR
f(z) dz −
f(z) dz
γR
=
cos z
γR (1 + z2
dz
) 2 ≤
This inequality holds for all R > 1, so, letting R → ∞, we have
I =
∞
0
γR
f(z) dz = lim f(z) dz =
R→∞ ΓR
π
. (32)
2e
1
(R2 − 1) 2 ℓ(γR)
πR
=
(R2 .
− 1) 2
γR
cos x
(1 + x2 π
dx =
) 2 4e .
42
f(z) dz → 0. Therefore, by (32),
⊓⊔

2.2 1991 November 21 2 COMPLEX ANALYSIS
2.2 1991 November 21
INSTRUCTIONS: In each of sections A, B, and C, do all but one problem.
TIME LIMIT: 2 hours
1. Where does the function
SECTION A
(Do 3 of the 4 problems.)
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
= x 2 + xy + x + i(xy − y 2 − y)
= u(x, y) + iv(x, y),
where u(x, y) = x 2 + xy + x and v(x, y) = xy − y 2 − 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
f ∈ H(Ω), then the derivative of f is given by f ′ (z) = (∂f)(z), z ∈ Ω. In the present case,
∂f
= 2x + y + 1 + iy,
∂x
∂f
∂y
= x + i(x − 2y − 1).
∂ ∂
∂x − i ∂y , and recall that, if
Therefore, ∂f(x + iy) = 1
1
2 [(2x + y + 1 + iy) − i(x + i(x − 2y − 1))] = 2 [(3x − y) + i(y − x)], and finally,
f ′ (1 − i) = 1
(4 − 2i) = 2 − i.
2
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
(z − λn),
n=1
43
⊓⊔

2.2 1991 November 21 2 COMPLEX ANALYSIS
where a and each root λn are complex constants. Prove that if P has only real coefficients, then P has a factorization
K
M
P (z) = a (z − rk) (z 2 − bmz + cm),
k=1
where a and each rk, bm, cm are real constants.
m=1
3. Use complex residue methods to compute the integral
π
1
5 + 3 cos θ dθ.
Solution: Let I = π
0
For z = e iθ ,
0
1
5+3 cos θ dθ. Note that cos θ is an even function (i.e., cos(−θ) = cos θ), so
2I =
and dz = ieiθdθ, from which it follows that
2I =
π
−π
cos θ = eiθ + e −iθ
2
= 1
i
= 2
3i
|z|=1
|z|=1
|z|=1
1
5 + 3 cos θ dθ.
= 1 1
(z +
2 z ),
1
5 + 3 1
2 (z + z )
dz
iz
dz
5z + 3
2 (z2 + 1)
dz
z2 + 10
3 z + 1.
Let p(z) = z 2 + 10
3 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
Now,
which implies
Therefore,
2I = 2
1
· 2πi · Res , z1 .
3i p(z)
1
Res , z1 = lim
p(z) z→z1
1
p(z) =
1
(z − z1)(z − z2) ,
1
=
z − z2
1
− 1
3
=
3 − (−3) 8 .
2I = 2 3 π
· 2πi · =
3i 8 2 ,
so I = π
4 . ⊓⊔
44

2.2 1991 November 21 2 COMPLEX ANALYSIS
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
(Do 3 of the 4 problems.)
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 representation
f(z) =
∞
an(z − z0) n ,
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
f(z)
dz, (n ∈ Z; 1 < R < 2).
2πi (z − z0) n+1
|z|=R
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,
pm(z)(z − z0) −n−1 dz = 0, (n = −1, −2, . . . ).
Therefore,
|z|=R
|an| = 1
f(z)
pm(z)
dz −
dz
2π |z|=R (z − z0) n+1
|z|=R (z − z0) n+1
≤ 1
|f(z) − pm(z)|
2π |z|=R |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, what
is
(Verify your answer.)
38 See also: April ’96 (8).
|z|=1
f ′ (z)g(z)
f(z)
45
dz ?

2.2 1991 November 21 2 COMPLEX ANALYSIS
7. Let {fn} be a sequence of analytic functions in the unit disk D, and suppose there exists a positive constant M
such that
|fn(z)| |dz| ≤ M
C
for each fn and for every circle C lying in D. Prove that {fn} has a subsequence converging uniformly on compact
subsets of D.
Solution: We must show that F = {fn} 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 MK > 0 such that
|fn(z)| ≤ MK 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 Mα and a
neighborhood B(zα, rα) ⊂ D such that |fn(z)| ≤ Mα 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
|fn(z)| ≤ 1
|fn(ζ)|
2π |ζ−zα|=Rα |ζ − z| |dζ|
≤ 1 1
2π Rα/2
≤ M
.
πRα
|ζ−zα|=Rα
|fn(ζ)| |dζ|
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 |fn(z)| |dz| ≤ M for any circle C in D. Letting Mα = M
πRα , and rα = Rα/2, we
have |fn(z)| ≤ Mα 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.
39 Answer: If K ⊂ D is compact, we could select a finite covering of K by such neighborhoods B(zαj , rα j ) (j = 1, . . . , J) and then
|fn(z)| ≤ maxj Mα j MK, for all z ∈ K and n = 1, 2, . . . .
46

2.2 1991 November 21 2 COMPLEX ANALYSIS
SECTION C
(Do 2 of the 3 problems.)
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) = anz n 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
for every R > 0. Therefore,
|an| ≤ 1
2π
|ζ|=R
an = f (n) (0)
n!
1 f(ζ)
= dζ,
2πi |ζ|=R ζn+1 |f(ζ)| 1 A R
|dζ| ≤
|ζ| n+1 2π
N + B
Rn+1 2πR = A R N−n + B R −n .
Again, this holds for every R > 0. Thus, for any n > N and ɛ > 0, taking R large enough forces |an| < ɛ
(n = N + 1, N + 2, . . . ). Since ɛ was arbitrary, we have an = 0 for all n = N + 1, N + 2, . . . . Therefore,
f(z) = N
n=0 anz n . ⊓⊔
(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 singularity
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) = anz n 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=−∞
be the Laurent series expansion of the function g about z = 0. Suppose the pole at z = 0 is of order m. Clearly m
is 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
bnz n
bnz n = 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 anz n , 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) =
48
m
anz n .
n=0

2.3 1995 April 10 2 COMPLEX ANALYSIS
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)
. (37)
(z − α1) · · · (z − αN)
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,
min |g(z)| ≥ min |g(z)| = ɛ > 0,
z∈DR
|z|=R
for some ɛ > 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. Let 40 f(z) = ∞
n=0 anz n be analytic and one-to-one on |z| < 1. Suppose that |f(z)| < 1 for all |z| < 1.
(a) Prove that
(b) Is the constant 1 the best possible?
∞
n|an| 2 ≤ 1.
n=1
Solution: (a) This is a special case of the following area theorem:
Theorem 2.1 Suppose f(z) = ∞
n=0 anz n is a holomorphic function which maps the unit disk D = {|z| < 1}
bijectively onto a domain f(D) = G having area A. Then
A = π
∞
n|an| 2 .
n=1
40 On 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

2.3 1995 April 10 2 COMPLEX ANALYSIS
Proof: The area of the image of D under f is the integral over D of the Jacobian of f. That is,
A = |f ′ (z)| 2 dx dy.
Compute |f ′ (z)| by differentiating the power series of f(z) term by term,
f ′ ∞
(z) = nanz n−1 .
Next, take the squared modulus,
This gives,
Letting z = re iθ ,
A =
|f ′ (z)| 2 =
A =
∞
m,n=1
∞
D
m,n=1
∞
D m,n=1
m n aman
n=1
m n amanz m−1 z n−1 .
m n amanz m−1 z n−1 dx dy.
1 2π
0
0
r m+n−1 e i(m−n)θ dθ dr.
Now, for all k = 0, the integral of e ikθ 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 2 |an| 2
1
r 2n−1 ∞
dr = π n 2 |an| 2 . (38)
n=1
0
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 2 |an| 2 ,
n=1
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) > 0
for 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 > 0
such 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. ⊓⊔
41 Recall problem 1(b), April ’89, where we proved that a real valued harmonic function u(z) satisfying u(z) ≥ 0 for all z ∈ C must be constant.
50
n=1
⊓⊔

2.3 1995 April 10 2 COMPLEX ANALYSIS
5. 42 Find an explicit conformal mapping of the semidisk
onto the unit disk.
H = {z : |z| < 1, Rez > 0}
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:
(a) State the Schwarz lemma, as applied to f.
(b) If f(0) = 1
2 , how large can |f ′ (0)| be?
Solution: (a) See theorem A.22.
|f(z)| < 1 all |z| < 1.
(b) Assume f satisfies the given hypotheses. In particular, f(0) = 1
2 . Consider the map
ϕ(z) =
1
2
− z
1 − z .
2
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
Now,
1 ≥ |g ′ (0)| = |ϕ ′ (1/2)||f ′ (0)|. (39)
ϕ ′ (z) = − 1 − z
1 1
2 + 2 2 − z
z 2 .
1 − 2
Therefore, ϕ ′ (1/2) = −4/3, and it follows from (39) that
|f ′ (0)| ≤
42 This problem also appears in April ’89 (3) and November ’06 (2).
43 A very similar problem appeared in November ’06 (3).
1
|ϕ ′ = 3/4.
(1/2)|
51
⊓⊔

2.4 2001 November 26 2 COMPLEX ANALYSIS
2.4 2001 November 26
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 C1 functions defined on a neighborhood
of the closure of a bounded region G ⊂ C with boundary which is parametrized by a properly oriented, piecewise
C1 curve γ. If u and v obey the Cauchy-Riemann equations, show that Cauchy’s theorem
f(z) dz = 0 follows
γ
from Green’s theorem, namely
∂Q ∂P
P dx + Q dy = − dx dy for C
∂x ∂y
1 functions P and Q. (40)
γ
G
(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 ′ f(z) − f(z0)
(z0) = lim
z→z0 z − z0
exists at some (possibly only one!) point z0 ∈ G. Show that given any ɛ > 0, we can find a triangular region ∆
containing z0, such that if T is the boundary curve of ∆, then
f(z) dz
1
=
2 ɛL2 ,
T
where L is the length of the perimeter of ∆.
Hint for (b) Note that part (a) yields
(az + b) dz = 0 for a, b ∈ C, which you can use here in (b), even if you
T
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
u(x, y) dx − v(x, y) dy = (vx + uy) dx dy = 0. (41)
γ
G
Similarly, if P = v and Q = u in Green’s theorem, then the Cauchy-Riemann equations imply
v(x, y) dx + u(x, y) dy = (ux − vy) dx dy = 0. (42)
γ
G
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),
f(z) dz =
44 These are ux = vy and uy = −vx.
γ
γ
u(x, y) dx − v(x, y) dy + i v(x, y) dx + u(x, y) dy = 0.
γ
52
⊓⊔

2.4 2001 November 26 2 COMPLEX ANALYSIS
(b) Suppose u and v are continuous and f ′ (z) exists at the point z0 ∈ G. Then, for any ɛ > 0 there is a δ > 0 such
that B(z0, δ) ⊆ G, and
f ′
f(z) − f(z0)
(z0) −
z − z0
< ɛ, 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
R(z) dz = T
Finally, note that
R(z)
z − z0
=
f ′
f(z) − f(z0)
(z0) −
z − z0
< ɛ, for all |z − z0| < δ.
Therefore,
f(z) dz
T
=
R(z) dz
T
≤
|R(z)| |dz|
T
=
R(z)
T z − z0
|z − z0| |dz|
≤ ɛ |z − z0| |dz| ≤ ɛrL.
T
T
f(z) dz.
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
2 ɛL2 .
2. Give two quite different proofs of the fundamental theorem of algebra, that if a polynomial with complex coefficients
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, Rouché’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 + a2z 2 + · · · + anz n with an = 0 for some n ≥ 1. Both proofs also rely on the following observation:
If {aj} n j=0 ⊂ C with an = 0, then for all 1 ≤ R ≤ |z| < ∞,
a0
an
z −n + · · · + an−1
z −1
|a0|
≤
|an| |z|−n + · · · + |an−1|
|z|
|an|
−1
an
|aj|
≤ n max
0≤j

2.4 2001 November 26 2 COMPLEX ANALYSIS
In particular, if we choose45 R = 1 + 2 n max0≤j

2.4 2001 November 26 2 COMPLEX ANALYSIS
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 suitable
singularity.
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, ɛ) \ {z0} ⊂ G such that |f(z) − w0| > δ > 0 for all z ∈ ¯ D0. Define
g(z) = 1/(f(z) − w0) on D0. Then
lim sup
z→z0
z∈D0
|g(z)| = lim sup
z→z0
z∈D0
1 1
≤ < ∞.
|f(z) − w0| δ
Thus, by lemma 2.1 (Nov. ’06, prob. 1), z0 is a removable singularity of g(z). Therefore, g(z) ∈ H(B(z0, ɛ)). In
particular, g is continuous and non-zero at z = z0, so it is non-zero in a neighborhood B(z0, ɛ0) of z0. Therefore,
f(z)−w0 = 1/g(z) is holomorphic in B(z0, ɛ0), which implies that the singularity of f(z) at z = z0 is removable.
This contradiction proves the theorem. ⊓⊔
(b) Consider f(z) = e z . This function has an essential singularity at ∞, and, for every horizontal strip,
Sα = {x + iy : x ∈ R, α ≤ y < α + 2π},
of width 2π, f(z) maps Sα onto C \ {0}. (In particular, f(z) comes arbitrarily close to every w ∈ C.) Now let
NR = {z ∈ C : |z| > R} be any neighborhood of ∞. There is clearly a strip Sα contained in NR (e.g., with
α = R + 1). Therefore, f(z) = e z maps points in NR 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
f(z) dz = 1. Be careful with minus signs and
γ
factors of 2πi.
(b) From the theory of Laurent expansions, it is known that there are constants an such that, for 1 < |z| < 4,
Find a−10 and a10 by the method of your choice.
1
z 2 − 5z + 4 =
∞
n=−∞
anz n .
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,
n
f(z) dz = 2πi Res(f, zj).
γ
55
j=1

2.4 2001 November 26 2 COMPLEX ANALYSIS
Therefore, if we were to find a function f(z) ∈ H(C) with exactly two isolated singularities in G (e.g., at z1 = i
and z2 = −i), and such that Res(f, zj) = −i
4π , then
f(z) dz = 2πi
−i −i
Res(f, zj) = 2πi + = 1,
4π 4π
and the problem would be solved. Clearly,
γ
f(z) = −i
4π
j
1 1
− =
z − i z + i
1
2πi
z
z2 + 1
is such a function. ⊓⊔
(b) Expand the function in partial fractions:
Then, note that
converges for |z| < 4, while
converges for |z| > 1. Therefore,
1
z2 − 5z + 4 =
1 1/3 1/3
= −
(z − 4)(z − 1) z − 4 z − 1 .
1/3
z − 4
1/3
z − 1
1
z2 = −1
− 5z + 4 3
1 −1
=
3 4(1 − z/4)
1
= −1
3 z(1 − 1/z)
−1
n=−∞
∴ a−10 = − 1
3
z n − 1
12
= − 1
12
∞
n=0
= − 1
3z
∞
n=0
z
n 4
∞
z −n
n=0
n 1
z
4
n , for 1 < |z| < 4.
and a10 = − 1
12
10 1
.
4
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| ≤ 1
2R}, 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
f(ζ)
2πi ζ − w dζ.
|ζ−a|=R
56
⊓⊔

2.4 2001 November 26 2 COMPLEX ANALYSIS
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
f(ζ)
−
2πi |ζ−a|=R ζ − w1
f(ζ)
dζ
ζ − w2
= w1
− w2
f(ζ)
2πi
(ζ − w1)(ζ − w2) dζ.
W 2
W 1
.
.
a
|ζ−a|=R
. R/2
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| |f(ζ)|
|dζ|
2π
(R/2) 2
.
w
≤ |w1 − w2|
2π
≤ 4M
R |w1 − w2|,
|ζ−a|=R
R
. ζ
supγ |f(ζ)|
R2 ℓ(γ)
/4
where γ denotes the positively oriented circle {ζ : |ζ − a| = R}, and ℓ(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.
47 The best treatment of normal families and the Arzela-Ascoli theorem is Ahlfors [1].
57

2.4 2001 November 26 2 COMPLEX ANALYSIS
Recall that a family F of functions is called locally bounded on G iff for all compact K ⊂ G there is a constant
MK such that |f(z)| ≤ MK 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 and
F ⊂ 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 constant 49 δ. So, fix a ∈ G and ɛ > 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 = R
4M ɛ and r = R/2, then |f(w1) − f(w2)| < ɛ 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. ⊓⊔
48 Answer: 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.
49 The 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 2 COMPLEX ANALYSIS
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) =
Prove that F ′ (z) =
γ
ϕ(w)
(w−z) dw, z ∈ C \ γ∗ .
2. (a) State the Casorati-Weierstrass theorem.
(b) Evaluate the integral
γ
ϕ(w)
(w−z) 2 dw, z ∈ C \ γ ∗ , without using Leibniz’s Rule.
I = 1
1
(z − 3) sin dz where R ≥ 4.
2πi |z|=R
z + 2
3. Let f(z) be an entire function such that f(0) = 1, f ′ (0) = 0 and
Prove that f(z) = 1 for all z ∈ C.
0 < |f(z)| ≤ e |z|
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 m
zeros at z = 0 has the form
f(z) = e P (z) z m
∞
1 − z
e z/an , (46)
n=0
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) = e Bz+C ,
for some constants B, C. We are given that f(0) = 1 and f ′ (0) = 0, so e C = 1, and f ′ (0) = Be C = B = 0. It
follows that f(z) = 1. ⊓⊔
50 This 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
an

2.6 2006 November 13 2 COMPLEX ANALYSIS
2.6 2006 November 13
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
lim sup |f(z)| < ∞,
z→z0
z∈G
then z0 is a removable singularity and f may be extended holomorphically to all of G.
Proof: Under the stated hypotheses, there is an ɛ > 0 and an M > 0 such that the deleted neighborhood B o
{z ∈ C : 0 < |z − z0| ≤ ɛ} is contained in G and such that |f(z)| ≤ M for all z ∈ B o .
Let
f(z) =
be the Laurent expansion of f for z ∈ B o , where
an = 1
2πi
∞
n=−∞
C
an(z − z0) n
f(ζ)
dζ.
(ζ − z0) n+1
Here C denotes the positively oriented circle |ζ − z0| = ɛ. Changing variables,
the coefficients are
Therefore,
ζ = z0 + ɛe iθ ⇒ dζ = i ɛe iθ dθ
an = 1
2π
2π i 0
f(z0 + ɛe iθ )
(z0 + ɛe iθ − z0) n+1 iɛeiθ dθ.
|an| ≤ 1
2π
M M
ɛd|θ| = ,
2π 0 ɛn+1 ɛn which makes it clear that, if n < 0, then |an| can be made arbitrarily small, by choosing a sufficiently small ɛ. This
proves that an = 0 for negative n, and so
f(z) =
∞
an(z − z0) n .
n=0
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

2.6 2006 November 13 2 COMPLEX ANALYSIS
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 D
onto 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−z
1+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 {e iθ : −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) = z 1/2 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/2 z = −iz, which takes the first quadrant to the
fourth 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.
Solution: (a) See theorem A.22.
51 This problem also appears in April ’89 (3) and April ’95 (5).
52 Look up the precise statement of the Riemann mapping theorem.
53 A very similar problem appeared in April ’95 (6).
61
⊓⊔

2.6 2006 November 13 2 COMPLEX ANALYSIS
(b) In order to apply the Schwarz lemma, map the disk to the upper half-plane with the Möebius map φ : D → Π +
defined by
1 − z
φ(z) = i
1 + z .
Then, φ(0) = i. Therefore, the function g = f ◦ φ : D φ −→ 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
Hint. Consider the functions ft(z) = f(tz) where t > 0.
Γ = {z ∈ Π + : |Rez| ≤ Imz}.
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.
54 Despite the wording of the problem, the family need not satisfy F ⊂ H(Ω) in order to be normal.
55 Problem 5 (b) of the November 2001 exam asks for a proof of Montel’s theorem using the Arzela-Ascoli theorem.
62

2.6 2006 November 13 2 COMPLEX ANALYSIS
Recall that a family F of functions is called locally bounded on Ω iff for all compact K ⊂ Ω there is a constant
MK such that |f(z)| ≤ MK for all f ∈ F and z ∈ K.
Corollary 2.3 (little Montel theorem 56 ) Assume the set-up of the Arzela-Ascoli theorem, and suppose S = C
and 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 fn(z) = f(|zn|z).
Then, since z ∈ Γ ⇒ |zn|z ∈ Γ, we have
|fn(z)| = |f(|zn|z)| < 1, for all z ∈ Γ and n ∈ N.
Therefore, F is a normal family in Γ. Also note that each fn 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 {fn}; i.e., there is some subsequence nk such that, as k → ∞, fnk → g locally uniformly
in Γ.
Consider the point z = i. Since f(it) → 0 as t ↓ 0,
g(i) = lim fnk (i) = lim f(|znk |i) = 0.
k→∞ k→∞
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
fn
zn
= f |zn|
|zn|
zn
= f(zn). (47)
|zn|
The numbers zn/|zn| lie in the compact set γ = {z ∈ Γ : |z| = 1}. Since fnk → g uniformly in γ, for any ɛ > 0,
there is a K > 0 such that |fnk (z) − g(z)| = |fnk (z)| < ɛ, for all k ≥ K and all z ∈ γ. That is,
lim sup{|fnk (z)| : z ∈ γ} lim
k→∞ k→∞ fnkγ = 0,
and, since znk /|znk | ∈ γ,
znk
fnk
|znk |
≤ fnkγ. ∴ lim
k→∞ fnk
znk
|znk |
= 0.
By (47), then, limk→∞ f(znk ) = 0.
Finally, recall that f(zn) → 0 iff every subsequence znj has a further subsequence znj k such that f(znj k ) → 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 znj k such that f(znj k ) → 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 this
result, 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 {znj }
such 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 2 COMPLEX ANALYSIS
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
1. Give the Laurent series expansion of z(z−1) in the region A ≡ {z ∈ C : 2 < |z + 2| < 3}.
Solution:
f(z) =
1 1 − z + z 1 1
= = −
z(z − 1) z(z − 1) z − 1 z .
Let u = z + 2. Then z = u − 2 and A = {u ∈ C : 2 < |u| < 3}. Therefore,
1 1 1 1
= =
z u − 2 u (1 − 2/u)
= 1
u
∞
n=0
n 2
u
converges for |u| > 2 and, substituting u = z + 2 in the last expression, we have
1 1
=
z z + 2
∞
n=0
n 2
=
z + 2
converging for 2 < |z + 2|. Next, consider that
∞
n=0
2 n (z + 2) −n−1 =
1 1
=
z − 1 u − 3 =
−1 −1
=
3(1 − u/3) 3
−1
n=−∞
∞
n=0
n+1 1
(z + 2)
2
n ,
u
n 3
converges for |u| < 3 and, substituting u = z + 2 in the last expression, we have
converging for |z + 2| < 3. Therefore,
f(z) = 1 1
−
z − 1 z
1
z − 1
= −
∞
n=0
= −
∞
n=0
n+1 1
3
n+1 1
(z + 2)
3
n ,
(z + 2) n −
−1
n=−∞
n+1 1
(z + 2)
2
n ,
for z ∈ A. ⊓⊔
2. (i) Prove: Suppose that for all z ∈ D and all n ∈ N we have that fn is holomorphic in D and |fn(z)| < 1. Also
suppose that limn→∞ Imfn(x) = 0 for all x ∈ (−1, 0). Then limn→∞ Imfn(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 ∞ 1
−∞ 1+x4 dx.
64

2.7 2007 April 16 2 COMPLEX ANALYSIS
Solution: Note that
f(z) =
1
=
1 + x4 1
(z 2 + i)(z 2 − i) =
1
(z + e iπ/4 )(z − e iπ/4 )(z + e i3π/4 )(z − e i3π/4 ) ,
which reveals that the poles of f in the upper half plane are at e iπ/4 and e i3π/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, e iπ/4 ) + Res(f, e i3π/4
) . (48)
The other two poles of f are in the lower half-plane, so both e iπ/4 and e i3π/4 are simple poles. Therefore,
Res(f, e iπ/4 ) = lim
z→eiπ/4 (z − e iπ/4 1
)f(z) =
2eiπ/4 (eiπ/4 − ei3π/4 )(eiπ/4 + ei3π/4 = −1
) 4 ie−iπ/4 ,
Res(f, e i3π/4 ) = lim
z→ei3π/4 (z − e i3π/4 1
)f(z) =
2ei3π/4 (ei3π/4 − eiπ/4 )(ei3π/4 + eiπ/4 1
=
) 4 ie−i3π/4 .
Plugging these into (48) yields
1
f(z)dz = 2πi
4 ie−i3π/4 − 1
4 ie−iπ/4
= π
2 (e−iπ/4 − e −i3π/4 ) = π √ .
2
It remains to show
ΓR
lim
R→∞
Changing variables via z = Reiθ (0 ≤ θ ≤ π),
f(z)dz
=
π
g(R)
0
g(R)
iRe iθ
1 + (Re iθ ) 4
f(z)dz
= 0.
πR
≤
R4 → 0, as R → ∞.
− 1
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
⊓⊔

2.7 2007 April 16 2 COMPLEX ANALYSIS
Solution: First consider 57 φ1(z) = 1−z
1+z , which maps D onto the right half-plane P + = {z ∈ C : Rez > 0}.
Let φ2(z) = e iπ/2 z = iz, which maps P + onto the upper half-plane Π + = {z ∈ C : Imz > 0}. Next,
let φ3(z) = z 1/2 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) =
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−z
1−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) = e iθ z 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.
57 This is my favorite Möebius 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 −1
1 = 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.
58 See Rudin [8] page 254-5 (in particular, theorem 12.4) for a nice discussion of functions of the form φα(z) = z−α
. In addition to 12.4,
1− ¯αz
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
1 + i
√ 2
⊓⊔

2.7 2007 April 16 2 COMPLEX ANALYSIS
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, ɛ) = {z ∈ C : |z − a| < ɛ} such that B(a, ɛ) ∩ 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 G
and 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, ɛ) ⊂ G for some ɛ > 0, and clearly
B(z0, ɛ) ⊂ 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, ∃ɛ > 0 such that
B(w, ɛ) ⊂ G. Clearly any w1 ∈ B(w, ɛ) can be connected to z0 by a parametric curve (from w1 to w, then from w
to z0 via γ) that remains in G. This proves that B(w, ɛ) ⊂ 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 > 0
such 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. ⊓⊔
60 For example, the set A = [0, 1], although closed in C, is open in G = [0, 1] ∪ {2}.
61 To see the equivalence note that, in this case, A is open in G, as is A c = 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

2.8 2007 November 16 2 COMPLEX ANALYSIS
2.8 2007 November 16
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 dx.
+ 1) 2
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 fn denote the composition of f with
itself n times (n = 2, 3, . . . ). Show that if f(0) = 0 and |f ′ (0)| < 1, then {fn} 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 {fn} 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 fn has at least m zeros (counting
multiplicities).
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.
Please email comments, suggestions, and corrections to williamdemeo@gmail.com.
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(e iθ ) 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|
z
2 − r2
.
1. See (1b) of April ’95.
Solutions
1 − r 2 z 2
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)|} is
bounded.
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 − ɛ} ⊂ 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} be the set of all 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 maximum
modulus 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. ⊓⊔
Please email comments, suggestions, and corrections to williamdemeo@gmail.com.
70

A Miscellaneous Definitions and Theorems
A.1 Real Analysis
A.1.1 Metric Spaces
The following theorem is found in Conway [3].
A MISCELLANEOUS DEFINITIONS AND THEOREMS
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 ɛ > 0 there exist {x1, . . . , xn} ⊂ X such that
n
X = B(xk, ɛ)
(The last property is called total boundedness.)
A.1.2 Measurable Functions
k=1
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
n
s = αiχAi,
i=1
where χAi is the characteristic function of the set Ai. Note that the definition assumes nothing about the sets Ai (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.2 62 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.3 63 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
Then h : X → Y is measurable.
62 Theorem 1.7 of Rudin [8].
63 Theorem 1.8 of Rudin [8].
h(x) = Φ(u(x), v(x)) (x ∈ X).
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 theorem 65 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 {fn} be a sequence of measurable functions on
X, and suppose that, for every x ∈ X,
(a) 0 ≤ f1(x) ≤ f2(x) ≤ · · · ≤ ∞,
(b) fn(x) → f(x) as n → ∞.
Then f is measurable, and
X
fn dµ →
X
f dµ as n → ∞.
Theorem A.6 (Fatou’s lemma) If fn : X → [0, ∞] (n = 1, 2, . . . ) is a sequence of positive measurable functions,
then
lim inf fn ≤ lim inf
Theorem A.7 (Lebesgue’s dominated convergence theorem) Let {fn} be a sequence of measurable functions on
(X, M, µ) such that fn → f a.e. If there is another sequence of measurable functions {gn} satisfying
(i) gn → g a.e.,
(ii) gn → g < ∞, and
(iii) |fn(x)| ≤ gn(x) (x ∈ X; n = 1, 2, . . .),
then f ∈ L 1 (X, M, µ), fn → f, and fn − f1 → 0.
Theorem A.8 (Egoroff) If (X, M, µ) is a measure space, E ∈ M a set of finite measure, and {fn} a sequence of
measurable functions such that fn(x) → f(x) for almost every x ∈ E, then for all ɛ > 0 there is a measurable subset
A ⊆ E such that fn → f uniformly on A and µ(E \ A) < ɛ.
64 Study 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.
65 I 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
fn

A.1 Real Analysis A MISCELLANEOUS DEFINITIONS AND THEOREMS
A.1.4 Approximating Integrable Functions 66
Let X be a locally compact Hausdorff space. (See Rudin [8], section 2.3 for the definition.) Let Cc(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
If 1 ≤ p < ∞, then S is dense in L p (µ).
Proof: Let 1 ≤ p < ∞, and f ∈ L p (µ).
µ({x ∈ X : s(x) = 0}) < ∞.
Case 1. f ≥ 0.
Let sn be as in Theorem 1.17. Then sn ≤ f implies sn ∈ L p (µ) for all n = 1, 2, . . . . Define gn = |f − sn| p . Then gn ∈ L p (µ),
n = 1, 2, . . . , gn → 0 as n → ∞, and gn ≤ |f| p ∈ L 1 (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 + = max{0, f} and f − = min{0, −f}. Let sn and tn be simple functions such that
Then, by Case 1, we have
as n → ∞. Finally, note that
0 ≤ s1(x) ≤ s2(x) ≤ · · · ≤ f + (x)
0 ≤ t1(x) ≤ t2(x) ≤ · · · ≤ f − (x).
f + − snp → 0 and f − − tnp → 0
f − (sn − tn) p p = f + − f − − (sn − tn) p p ≤ f + − sn p p + f − − tn p p.
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 ɛ > 0. Let f be a complex measurable function which vanishes off a set of
finite measure. Then there exists g ∈ Cc(X) such that
Furthermore, we may arrange it so that
µ({x ∈ X : f(x) = g(x)}) < ɛ.
sup
x
∈ X|g(x)| ≤ sup |f(x)|
x∈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 < ∞, Cc(X) is dense in L p (µ).
66 This topic is very important and appears on the comprehensive exam syllabus.
73

A.1 Real Analysis A MISCELLANEOUS DEFINITIONS AND THEOREMS
Proof: Define S as in theorem A.9. If s ∈ S and ɛ > 0, then by Lusin’s Theorem there exists g ∈ Cc(X) such that g(x) = s(x)
except on a set of measure < ɛ, and |g| ≤ s∞. Hence,
g − sp ≤ 2ɛ 1/p s∞.
Since S is dense in L p (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 R 1 .
(Notice how much more special these functions are than the simple functions, defined above.)
Lemma A.1 (Rudin Exercise 2.24) If f ∈ L 1 (R) then there exists a sequence {gn} of step functions on R such that
lim
n→∞ |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 ∈ L 1 (R), with f ≥ 0.
Case 4. f ∈ L 1 (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-algebra M, 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 on M 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-algebra M in a set X, and
let λ be a complex measure on M.
(a) There is then a unique pair of complex measures λa and λs on M such that
If λ is positive and finite, then so are λa and λs.
(b) There is a unique h ∈ L 1 (µ) such that
67 Rudin[8], 6.10.
λ = λa + λs, λa ≪ µ, λs ⊥ µ.
λa(E) =
E
74
h dµ ∀E ∈ M.

A.1 Real Analysis A MISCELLANEOUS DEFINITIONS AND THEOREMS
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 = dλa
dµ dµ.
Strictly speaking, dλa/dµ should be viewed as the equivalence class of functions that are equal to h µ-a.e.
Corollary A.1 68 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 ∈ L 1 (ν), then g dν
dµ ∈ L1 (µ) and
(b) ν ≪ λ, and
A.1.6 Absolute Continuity of Functions
g dν =
g dν
dµ dµ.
dν dν dµ
=
dλ dµ dλ λ-a.e.
Lemma 1.2 Let f : R → R be a function. If f is differentiable on [a, b], f ′ ∈ L 1 ([a, b]), and x
a 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 ɛ > 0 there is a δ > 0
such that, if E ⊂ R is measurable mE < δ, then
|f ′ |dm < ɛ. (49)
E
Let A = ∪n i=1 (ai, bi) be a finite union of disjoint open intervals in [a, b] such that n i=1 (bi − ai) < δ. Then
mA ≤ n i=1 (bi − ai) < δ, so
n
n
bi
|f(bi) − f(ai)| = f
′
dm
≤
n
bi
|f ′
|dm = |f ′ |dm < ɛ (50)
i=1
i=1
ai
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, define 70 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 about
integration with respect to product measure. Here it is:
68 Folland [4], Prop. 3.9.
69 The “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| < ɛ. In the present case we need
E |f ′ |dm < ɛ to get the sum in (50) to come out right.
70 This 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
i=1
ai
A

A.1 Real Analysis A MISCELLANEOUS DEFINITIONS AND THEOREMS
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) =
f(x, y) dν(y) and ψ(y) = f(x, y) dµ(x), then φ is S-measurable, ψ is
Y X
T -measurable, and
φ dµ = f(x, y) d(µ × ν) = ψ dν. (51)
X
X×Y
(b) If f : X × Y → C and if one of
|f(x, y)| dµ(x) dν(y) < ∞ or
Y
X
holds, then so does the other, and f ∈ L 1 (µ × ν).
(c) If f ∈ L 1 (µ × ν), then,
(i) for almost every x ∈ X, f(x, y) ∈ L 1 (ν),
X
Y
Y
|f(x, y)| dν(y) dµ(X) < ∞
(ii) for almost every y ∈ Y, f(x, y) ∈ L 1 (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 Theorem 71
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 it
is 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 (nonparametric)
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
dz
2πi 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
f(ζ)
2πi γ ζ − z dζ.
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 γ1, . . . , γm are closed rectifiable curves in G, then the curve γ = γ1 + · · · +
γm is homologous to zero in G provided n(γ1, w) + · · · + n(γm, 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
γ1, . . . , γm are closed rectifiable curves in G with γ = γ1 + · · · + γm ≈ 0, then for all z ∈ G \ γ ∗ ,
f(z)
m
j=1
n(γj, z) = 1
2πi
γ
m
j=1
γj
f(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
f(z)dz = 0.
71 Most of the material in this section can be found in Conway [3].
γ
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 γ1, . . . , γm are
closed rectifiable curves in G such that γ = γ1 + · · · + γm ≈ 0, then
m
j=1
A partial converse of Cauchy’s theorem is the following:
γj
f(z) dz = 0.
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
f(z)dz = 0. Then f ∈ H(G).
T
This theorem is still valid (and occasionally easier to apply) if we replace “any triangular contour” with “any rectangular
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 ⊂ Ĉ = C ∪ {∞} be open, let f ∈ H(G), and suppose there is an
M > 0 such that
lim |f(z)| ≤ M, for every a ∈ ∂∞G.
z→a
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) = e iθ z for some constant θ ∈ R.
Please email comments, suggestions, and corrections to williamdemeo@gmail.com.
72 This 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 Rouché’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

B List of Symbols
F an arbitrary field
Q the rational numbers
Z the integers
N the natural numbers, {1, 2, . . .}
C the complex numbers (a.k.a. the complex plane)
Ĉ the extended complex plane, C ∪ {∞}
R the real numbers (a.k.a. the real line)
T the unit circle, {z ∈ C : |z| = 1}
ˆR the extended real line, [−∞, ∞]
Rez the real part of a complex number z ∈ C
Imz the imaginary part of a complex number z ∈ C
D or U the open unit disk, {z ∈ C : |z| < 1}
H(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 < 0}
P + the right half-plane, {z ∈ C : Rez > 0}
P − the left half-plane, {z ∈ C : Rez < 0}
∂∞G the extended boundary of a set G ⊂ Ĉ
C[0, 1] the space of continuous real valued functions on [0, 1].
L 1 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.
REFERENCES
[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 continuity
of 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
ℓp, 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, 68
connected, 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
equicontinuity
pointwise vs. uniform, 57
equicontinuous, 62
even functions, 16
Fatou’s lemma, 72
Fubini-Tonelli theorem, 73–74
80
applied, 8, 13
fundamental theorem of algebra, 52
fundamental theorem of calculus, 16
Green’s theorem, 51
Hölder’s inequality, 22
Hadamard factorization theorem
applied, 58
Hahn-Banach theorem, 22
homologous to zero, 75
implicit function theorem, 18
index, 75
inverse 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-Nikodym
derivative, 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

INDEX INDEX
Riemann mapping theorem, 60
Riesz representation theorem, 35–36
applied, 16–17
for L p , 22
Rouché’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
81