Problems and Solutions in

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Jul 9, 2010 - you prepare to take the comprehensive exams. Please email comments, suggestions, and corrections to [email protected]. Contents.
Problems and Solutions in R EAL AND C OMPLEX A NALYSIS 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 [email protected]. Abstract The pages that follow contain “unofficial” solutions to problems appearing on the comprehensive exams in analysis given by the Mathematics Department at the University of Hawaii over the period from 1991 to 2007. I have done my best to ensure that the solutions are clear and correct, and that the level of rigor is at least as high as that expected of students taking the ph.d. exams. In solving many of these problems, I benefited enormously from the wisdom and guidance of professors Tom Ramsey and Wayne Smith. Of course, some typos and mathematical errors surely remain, for which I am solely responsible. Nonetheless, I hope this document will be of some use to you as you prepare to take the comprehensive exams. Please email comments, suggestions, and corrections to [email protected].

Contents 1

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Real Analysis 1.1 1991 November 21 1.2 1994 November 16 1.3 1998 April 3 . . . . 1.4 2000 November 17 1.5 2001 November 26 1.6 2004 April 19 . . . 1.7 2007 November 16

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3 3 10 12 18 20 27 32

Complex Analysis 2.1 1989 April . . . . . . . . . . . . 2.2 1991 November 21 . . . . . . . 2.3 1995 April 10 . . . . . . . . . . 2.4 2001 November 26 . . . . . . . 2.5 2004 April 19 . . . . . . . . . . 2.6 2006 November 13 . . . . . . . 2.7 2007 April 16 . . . . . . . . . . 2.8 2007 November 16 . . . . . . . 2.9 Some problems of a certain type

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1

A Miscellaneous Definitions and Theorems A.1 Real Analysis . . . . . . . . . . . . . . . . . . . . . . . . A.1.1 Metric Spaces . . . . . . . . . . . . . . . . . . . . A.1.2 Measurable Functions . . . . . . . . . . . . . . . A.1.3 Integration . . . . . . . . . . . . . . . . . . . . . A.1.4 Approximating Integrable Functions1 . . . . . . . A.1.5 Absolute Continuity of Measures . . . . . . . . . A.1.6 Absolute Continuity of Functions . . . . . . . . . A.1.7 Product Measures and the Fubini-Tonelli Theorem A.2 Complex Analysis . . . . . . . . . . . . . . . . . . . . . . A.2.1 Cauchy’s Theorem2 . . . . . . . . . . . . . . . . . A.2.2 Maximum Modulus Theorems . . . . . . . . . . . B List of Symbols

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71 71 71 71 72 73 74 75 75 77 77 78 79

2

1

1

REAL ANALYSIS

Real Analysis

1.1

1991 November 21

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  > 0 and let N0 ∈ N be such that n ≥ N0 implies |fn (x) − f (x)| < /2, for all x ∈ [0, 1]. Let δ > 0 be such that |f (x) − f (y)| < /2, for all x, y ∈ [0, 1] with |x − y| < δ. Finally, let 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)| < /2 + /2 = . t u (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, f (x) = 2nx − (n − 1),   1,

1 if 0 ≤ x < 21 − 2n , 1 1 if 2 − 2n ≤ x < 12 , if 12 ≤ x ≤ 1.

(1)

1 That is, fn (x) is constantly zero for x less than 21 − 2n , then it increases linearly until it reaches one at x = 1/2, and then it remains constantly one for x bigger than 1/2. Now define the sequence xn = 12 − n1 . Then fn (xn ) = 0 for all n ∈ N and xn → 1/2, while the sequence fn approaches the characteristic function f , χ[ 12 ,1] which is one on [ 12 , 1] and zero elsewhere. Therefore, f (1/2) = 1 6= 0 = limn fn (xn ). t u

2. Let f : R → R be differentiable and assume there is no x ∈ R such that f (x) = f 0 (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 0 (x) = 0, which will give us our desired contradiction. Since |xn − x| → 0, we can write the derivative of f as follows: f 0 (x) = lim

n→∞

f (x + (xn − x)) − f (x) f (xn ) − f (x) = lim = 0. n→∞ xn − x xn − x

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

t u

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

3

1.1

1991 November 21

1

REAL ANALYSIS

Solution: For n = 1, 2, . . ., define An = {x ∈ X : 1/n ≤ |f (x)| < n}. Clearly,

∞ [

A1 ⊆ A2 ⊆ · · · ↑ A ,

An

n=1

and each An is measurable (why?).3 Next, define A0 = {x ∈ X : f (x) = 0}

and

A∞ = {x ∈ X : |f (x)| = ∞}.

Then X = A0 ∪ A ∪ A∞ is a disjoint union, and Z Z Z Z |f | = |f | + |f | + X

A0

A

Z |f | =

A∞

|f |.

(2)

A

The first term in the middle expression is zero since f is zero on A0 , and the third term is zeroR since f ∈ L1 (µ) implies µ(A∞ ) = 0. To prove the result, then, we must find a measurable set E such that A\E |f | < , and µ(E) < ∞. Define 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). R Since RAn ⊆ An+1 , we have 0 ≤ f1 (x) ≤ f2 (x) ≤ · · · , so the monotone convergence theorem4 implies X fn → A |f |, and, by (2), Z Z Z lim |f | dµ = |f | dµ = |f | dµ. n→∞

An

Therefore, there is some N > 0 for which

A

X

Z |f | dµ < . X\AN

Finally, note that 1/N ≤ |f | < N on AN , so Z µ(AN ) ≤ N

Z |f | dµ ≤ N

AN

|f | dµ < ∞. X

t u

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.

R 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. 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). 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. 3 Answer:

4 Alternatively,

4

1.1

1991 November 21

1

REAL ANALYSIS

Then, Z

Z

ν(∪n En ) =

f dµ = S

=

n

En

Z X ∞

f χSn En dµ

f χEn dµ

(∵ the En are disjoint)

f χEn dµ

(∵ f χEn ≥ 0, n = 1, 2, . . .)

n=1

= =

∞ Z X n=1 ∞ X

ν(En ).

n=1

The penultimate equality follows Pm from the monotone convergence theorem applied to the sequence of non-negative t u measurable functions gm = n=1 f χEn (m = 1, 2, . . .). (See also: April ’98, problem A.3.)

5. Suppose f is a bounded, real valued function on [0, 1]. Show that f is Lebesgue measurable if and only if Z Z sup ψ dm = inf φ dm where m is Lebesgue measure on [0, 1], and ψ and φ range over all simple functions, ψ ≤ f ≤ φ. Solution: This is proposition 4.3 of Royden, 3ed. [6].

6.6 If f is Lebesgue integrable on [0, 1] and  > 0, show that there is δ > 0 such that for all measurable sets E ⊂ [0, 1] with m(E) < δ, Z f dm < . 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, An = {x ∈ X : 1/n ≤ |f (x)| < n}. Here, X = [0, 1]. As we saw in problem 3, Z Z Z lim |f | dm = |f | dm = |f | dm. n→∞

6 See

An

A

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

5

X

1.1

1991 November 21

1

Let n > 0 be such that

REAL ANALYSIS

Z |f | dm < /2. X\An

Define δ = (2n)−1 , and suppose E ⊂ [0, 1] is a measurable set with mE < δ. We must show | Z Z f dm ≤ |f | dm E ZE Z = |f | dm + |f | dm (X\An )∩E An ∩E Z Z ≤ |f | dm + |f | dm

R E

f dm| < .

An ∩E

X\An

< /2 + n m(An ∩ E)  = . < /2 + n 2n The penultimate inequality holds because |f | < n on An .

t u

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  > 0 there is a δ > 0 such that |ν(E)| <  whenever µ(E) < δ. The signed measure defined by Z ν(E) =

f dµ E

is finite iff8 f ∈ L1 (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  > 0, there is a δ > 0 such that Z |ν(E)| = f dµ < , whenever µ(E) < δ. E

t u

7.9 Suppose f is a bounded, real valued, measurable function on [0, 1] such that with m Lebesgue measure. Show that f (x) = 0 a.e.

R

xn f dm = 0 for n = 0, 1, 2, . . .,

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 kφ − pn k∞ → 0 as n → ∞. Then, since all functions involved are 7 See

also Folland [4], page 89. 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 ∈ L1 , but that assumption makes the problem even easier than the others. 8 For

6

1.1

1991 November 21

1

REAL ANALYSIS

integrable, Z Z f φ = f (φ − pn + pn ) Z Z ≤ |f ||φ − pn | + f pn Z ≤ kf k1 kφ − pn k∞ + f pn = kf k1 kφ − pn k∞ .

(3)

The last equality holds since xn f = 0 for all n = 0, 1, 2, . . ., which implies that f pn = 0 for all polynomials pn . Finally, note that kf k1 < ∞, since f is bounded and Lebesgue measurable on the bounded interval [0, 1]. Therefore, the right-hand side of R (3) tends to zero as n tends to infinity. Since the left-hand side of (3) is independent of n, we have thus shown that f φ = 0 for any φ ∈ C[0, 1]. R

R

Now, since C[0, 1] is dense in L1 [0, 1], let {φn } ⊂ C[0, 1] satisfy kφn − f k1 → 0 as n → ∞. Then Z Z Z Z 2 0 ≤ f = f (f − φn + φn ) ≤ |f ||f − φn | + f φn . The second term on the right is zero by R R what we proved above. Therefore, R if M is the bound on |f |, we have 0 ≤ f 2 ≤ M kf − φn k1 → 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. t u Alternative Solution: Quinn Culver suggests shortening the proof by using the fact that polynomials are dense in L1 [0, 1]. Simply start from the line, “Now, since C[0, 1] is dense in L1 [0, 1], let {φn } ⊂ C[0, 1] satisfy...” but instead write, “Since Pol[0, 1] is dense in L1 [0, 1], let {φn } ⊂ Pol[0, 1] satisfy...” This is a nice observation and disposes of the problem quickly and efficiently. However, I have left the original, somewhat clumsy proof intact because it provides a nice demonstration of the Stone-Weierstrass theorem (which appears on the exam syllabus), and because everyone should know how to apply this fundamental theorem to problems of this sort.

8.10 If µ and ν are finite measures on the measurable space (X, Σ), show that there is a nonnegative measurable function f on X such that for all E in Σ, Z Z (1 − f ) dµ = f dν. (4) E

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 R positive measures on the measurable space (X, Σ). By the linearity property of the integral, and since µ(E) = E dµ, we have Z Z Z Z (1 − f ) dµ = dµ − f dµ = µ(E) − f dµ. E

E

E

E

Therefore, (4) is equivalent to Z µ(E) =

f dµ + E

10 See

Z

Z f dν =

E

f d(µ + ν)

(∀ E ∈ Σ)

E

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

11 Note

7

(5)

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 Z 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 Z µ(E) = f d(µ + ν) ∀ E ∈ Σ, E

t u

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 Z Z Z 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. 12 See

also: November ’97 (2), and others. 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. 13 I’m

8

1.1

1991 November 21

1

REAL ANALYSIS

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, Z Z Z Z |f (x)g(y)| dν(y) dµ(x) = |f (x)||g(y)| dν(y) dµ(x) X Y Z  ZX Y = |f (x)| |g(y)| dν(y) dµ(x) Y ZX Z = |f (x)| dµ(x) |g(y)| dν(y) < ∞, X

Y

which holds since f ∈ L1 (µ) and g ∈ L1 (ν). The Fubini-Tonelli theorem then implies that F (x, y) ∈ L1 (µ × ν). R R R Finally, we must prove that RF 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, Z Z Z F d(µ × ν) = F (x, y) dν(y) dµ(x). X×Y

X

Y

Therefore, Z

Z Z F d(µ × ν) =

X×Y

f (x)g(y) dν(y) dµ(x) Z = f (x) g(y) dν(y) dµ(x) Y ZX Z = f (x) dµ(x) g(y) dν(y). ZX

X

Y

Y

t u

9

1.2

1994 November 16

1.2

1

REAL ANALYSIS

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 converges to an s ∈ E. Solution: Suppose E is complete. Let {xn } ⊂ E be absolutely convergent; i.e., ∞ X

xn := lim

n=1

Let SN =

PN

n=1

N →∞

N X

P∞

P

1

kxn k < ∞, then

P∞ 1

xn

kxn k < ∞. We must

xn = s ∈ E.

(6)

n=1

xn . Then, for any j ∈ N, kSN +j

N +j

N +j

X

X

xn ≤ kxn k → 0 − SN k =

n=N +1

n=N +1

P

as N → ∞, since kxn k < ∞. Therefore, {SN } is a Cauchy sequence. Since E is complete, there is an s ∈ E P∞ such that n=1 xn = lim SN = s. N →∞ P∞ P∞ Conversely, suppose whenever 1 kxn k < ∞, then 1 xn converges to an s ∈ E. Let {yn } ⊂ E be a Cauchy sequence. That is, kyn − ym k → 0 as n, m → ∞. Let n1 < n2 < · · · be a subsequence such that n, m ≥ nj ⇒ kyn − ym k < 2−j . Next observe, for k > 1, ynk = yn1 + (yn2 − yn1 ) + (yn3 − yn2 ) + · · · + (ynk − ynk−1 ) = yn1 +

k−1 X

(ynj+1 − ynj ),

j=1

and

∞ X

kynj+1 − ynj k
0. Consider the change of variables, y = x − n. Then dy = dx and x = y + n, so Z ∞ Z ∞ x y+n f (x − n) f (y) dx = dy 1 + |x| 1 + |y + n| −∞ −∞ Z −n Z ∞ y+n y+n f (y) f (y) dy + dy. = 1 + y + n 1 − (y + n) −∞ −n Note that, when y ≥ −n,

y+n 1+y+n

x 1+|x|



dx.

(7)

∈ [0, 1), and increases to 1 as n tends to infinity. Thus, 0 ≤ |f (y)|

y+n ≤ |f (y)|, 1+y+n

for all y ≥ −n. Define the function14 gn (y) = f (y)

y+n 1[−n,∞) (y). 1+y+n

Then |gn | ≤ |f | and lim gn = f . Therefore, by the dominated convergence theorem, n→∞

Z



lim

n→∞

f (y) −n

y+n dy = lim n→∞ 1+y+n

Z



Z



gn (y) dy = −∞

f (y) dy. −∞

Next, consider the second term in (7). Define the function hn (y) = f (y) It is not hard to check that

y+n 1(−∞,−n] (y). 1 − (y + n)

|y + n| 1(−∞,−n] (y) ∈ [0, 1), |1 − (y + n)|

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

n→∞

n→∞

y+n 1(−∞,−n] (y) = 0. 1 − (y + n)

Therefore, the dominated convergence theorem implies that Z −n y+n lim f (y) dy = 0. n→∞ −∞ 1 − (y + n) Combining the two results above, we see that lim

n→∞

R∞

f (x − n) −∞



x 1+|x|



dx =

R∞ −∞

f (x) dx.

t u

Remark. Intuitively, this is the result we expect because the translation f (x − n) = Tn f (x) is merely shifting the x support of f to the right tail of the measure dµ := 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

1.3

1998 April 3

1

REAL ANALYSIS

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 x ˆn = sup xk k≥n

(a) Explain why the limit ` = limn→∞ x ˆn exists. (b) Prove that, for any  > 0 and positive integer N , there exists an integer k such that k ≥ N and |xk − `| < . 2. Let C be a collection of subsets of the real line R, and define \ Aσ (C) = {A : C ⊂ A and A is a σ-algebra of subsets of R}. (a) Prove that Aσ (C) is a σ-algebra, that C ⊂ Aσ (C), and that Aσ (C) ⊂ A for any other σ-algebra A containing all the sets of C. (b) Let O be the collection of all finite open intervals in R, and F the collection of all finite closed intervals in R. Show that Aσ (O) = Aσ (F ).

3. Let (X, A, µ) be a measure space, and suppose X = ∪n Xn , 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, Z XZ f dµ = f dµ. X

n

Xn

Pn Solution: 15 Define fn = k=1 f χXk = f χ∪n1 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). 15 Note

R 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

1.3

1998 April 3

1

REAL ANALYSIS

Therefore, ∞ Z X k=1

Xk

f dµ = lim

n→∞

= lim

n→∞

n Z X k=1

f χXk dµ

X

Z X n

f χXk dµ

(by linearity of the integral)

X k=1

Z = lim fn dµ n→∞ X Z = lim fn dµ X n→∞ Z = f dµ.

(by definition of fn ) (by the monotone convergence theorem)

X

t u

4. Using the Fubini/Tonelli theorems to justify all steps, evaluate the integral Z 1Z 1  πy  dx dy. x−3/2 cos 2x 0 y RR Solution: By Tonelli’s theorem, if f (x, y) ≥ 0 is measurable and one of the iterated integrals f (x, y) dx dy or RR 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) ∈ L1 (dx, dy). Fubini’s theorem states: if f (x, y) ∈ L1 (dx, dy), then the iterated integrals exist and are equal. Now let g(x, y) = x−3/2 cos(πy/2x), and apply the Tonelli theorem to the non-negative measurable function |g(x, y)| as follows: Z 1Z x Z 1Z x Z 1 Z 1Z x  πy  −3/2 −3/2 x · 1 dy dx = x−1/2 dx = 2. |g(x, y)| dy dx = |x| dy dx ≤ cos 2x 0 0 0 0 0 0 0 Thus one of the iterated integrals of |g(x, y)| is finite which, by the Tonelli theorem, implies g(x, y) ∈ L1 (dx, dy). Therefore, the Fubini theorem applies to g(x, y), and gives the first of the following equalities: Z 1Z 1 Z 1Z x  πy   πy  −3/2 x cos dx dy = dy dx x−3/2 cos 2x 2x 0 0 y 0 Z 1 2x h  πy iy=x = x−3/2 · sin dx π 2x y=0 0 Z 1 2 −1/2 = x dx π 0 2 h 1/2 ix=1 = 2x π x=0 4 = . π t u 13

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 X f (x, y) = φi (x)ψi (y), i

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 Z 1 dm(x). µ(E) = 2 E 1+x Show that m is absolutely continuous with respect to µ, and compute the Radon-Nikodym derivative dm/dµ. Solution: Obviously both measures are non-negative. We must first prove m  µ. To this end, suppose m(E) > 0, where E ∈ M, the σ-algebra of Lebesgue measurable sets. Then, if we can show µ(E) > 0, this will establish that the implication µ(E) = 0 ⇒ m(E) = 0 holds for all E ∈ M; i.e., m  µ. For n = 1, 2, . . ., define 

1 1 1 < ≤ An = x ∈ R : 2 n+1 1+x n Then Ai ∩ Aj = ∅ for all i 6= j in N, and, for all n = 1, 2, . . ., µ(An ) ≥ Also, R = ∪An , since 0
0 implies the existence of an n ∈ N such that m(An ∩E) > 0. Therefore,

µ(E) ≥ µ(An ∩ E) ≥

1 m(An ∩ E) > 0, n+1

which proves that m  µ. By the Radon-Nikodym theorem (A.1), there is a unique h ∈ L1 (µ) such that Z Z Z m(E) = h dµ, and f dm = f h dµ ∀ f ∈ L1 (m). In particular, if E ∈ M and f (x) =

1 1+x2 χE ,

Z µ(E) = E

then

1 dm(x) = 1 + x2

Z E

h(x) dµ(x). 1 + x2

h(x) That is, E dµ = dµ(x) holds for all measurable sets E, which implies16 that, 1+x 2 = 1 holds for µ-almost every x ∈ R. Therefore, dm (x) = h(x) = 1 + x2 . dµ One final note: h is uniquely defined only up to an equivalence class of functions that are equal to 1 + x2 , µ-a.e. t u

R

h(x) E 1+x2

R

16 Recall

R R 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

1.3

1998 April 3

1

REAL ANALYSIS

PART B 7. Let φ(x, y) = x2 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 µ(E) = m(φ−1 (E)). (a) Show that µ is a Borel measure on R. (b) Let χE denote the characteristic function of the set E. Show that Z Z χE dµ = χE ◦ φ dm. S

R

(c) Evaluate the integral



Z

t2 dµ(t).

−∞

8. Let f be a real valued and increasing function on the real line R, such that f (−∞) = 0 and f (∞) = 1. Prove that f is absolutely continuous on every closed finite interval if and only if Z f 0 dm = 1. R

Solution: 17 First note that f is increasing, so f 0 exists for a.e. x ∈ R, and f 0 (x) ≥ 0 wherever f 0 exists. Also, f 0 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 0 exists. Then m(R \ E) = 0, and f 0 = g on E (by the definition of derivative), so f 0 is measurable. R (⇐) Suppose R f 0 dm = 1. We must show f ∈ AC[a, b] for all −∞ < a < b < ∞. First, check that f 0 ∈ L1 (R), since Z Z Z Z 1= f 0 dm = f 0 dm + f 0 dm = f 0 dm, R

E

R\E

E

and, since f is increasing, f 0 ≥ 0 on E, so Z Z Z Z Z 0 0 0 0 |f |dm = |f |dm + |f |dm = |f |dm = f 0 dm = 1. R

R\E

E

E

E

Thus, f 0 ∈ L1 (R) as claimed. A couple of lemmas will be needed to complete the ⇐ direction of the proof. The first is proved in the appendix (sec. A), while the second can be found in Royden [6] on page 100. Lemma 1.2 Let f : R → R be a function. If f is differentiable on [a, b], f 0 ∈ L1 ([a, b]), and f (x) − f (a) for a ≤ x ≤ b, then f ∈ AC[a, b].

Rx a

f 0 (t)dt =

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 0 ∈ L1 ([a, b]), then

Rx a

f 0 (t)dt ≤ f (x) − f (a).

R Rb To finish the ⇐ direction of the proof, by lemma 1.2, it suffices to show that R f 0 dm = 1 implies a f 0 (t)dt = Rb f (b) − f (a) holds for all −∞ < a < b < ∞. By lemma 1.3, we have a f 0 (t)dt ≤ f (b) − f (a), so we need only Rb show that strict inequality cannot hold. Suppose, by way of contradiction, that a f 0 (t)dt < f (b) − f (a) holds for some −∞ < a < b < ∞. Then, Z Z a Z b Z ∞ 1= f 0 dm = f 0 dm + f 0 dm + f 0 dm −∞

R

a

b

< [f (a) − f (−∞)] + [f (b) − f (a)] + [f (∞) − f (b)] = f (∞) − f (−∞) = 1. This contradiction proves that desired.

R R

f 0 dm = 1 implies

Rb a

f 0 (t)dt = f (b) − f (a) holds for all −∞ < a < b < ∞, as

(⇒) Now assume f ∈ AC[a, b] for all −∞ < a < b < ∞. We must show f (∞) − f (−∞) = 1, so this is equivalent to showing Z x lim f 0 (t)dm(t) = lim [f (x) − f (−x)]. x→∞

R R

f 0 dm = 1. By assumption

x→∞

−x

Rx Let x ∈ R, x > 0, and f ∈ AC[−x, x]. Then we claim f (x) − f (−x) = −x f 0 dm. Assuming the claim is true (see Royden [6], p. 110 for the proof), we have Z x Z 1 = lim [f (x) − f (−x)] = lim f 0 (t)dm(t) = f 0 dm. x→∞

x→∞

−x

R

t u

9. Let F be a continuous linear functional on the space L1 [−1, 1], with the property that F (f ) = 0 for all odd functions f in L1 [−1, 1]. Show that there exists an even function φ such that Z

1

F (f ) =

f (x)φ(x) dx,

for all f ∈ L1 [−1, 1].

−1

[Hint: One possible approach is to use the fact that any function in Lp [−1, 1] is the sum of an odd function and an even function.] Solution: Since F ∈ L1 [−1, 1]∗ , then by the Riesz representation theorem19 there is a unique h ∈ L∞ [−1, 1] such that Z 1 F (f ) = f (x)h(x) dx (∀f ∈ L1 [−1, 1]) −1 18 See 19 See

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

16

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

ψ(x) =

and

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, 1

Z

1

Z

−1

−1

Now note that fe ψ is an odd function (since it’s an even times an odd), so R1 Similarly, −1 fo φ = 0. Therefore, Z

1

F (f ) = F (fe ) =

Z

1

fe φ = −1

Z

1

fe φ + −1

Z

Z

R1

fe ψ.

−1

−1

fe ψ = 0, since [−1, 1] is symmetric.

1

fo φ = −1

1

fe φ +

fe h =

F (f ) = F (fe ) + F (fo ) = F (fe ) =

Z

1

(fe + fo )φ = −1

f φ. −1

t u

17

1.4

2000 November 17

1.4

1

REAL ANALYSIS

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 : R3 → R3 is an invertible linear map and that g : R3 → R3 has continuous first order partial derivatives and satisfies kg(x)k ≤ Ckxk2 for some constant C and all x ∈ R3 . Here kxk denotes the usual Euclidean norm on R3 . Prove that f (x) = L(x) + g(x) is locally invertible near 0. Solution: (a) (Inverse function theorem (IFT) of calculus)21 Let f : E → Rn be a C 1 -mapping of an open set E ⊂ Rn . Suppose that f 0 (a) is invertible for some a ∈ E and that f (a) = b. Then, (i) there exist open sets U and V in Rn such that a ∈ U , b ∈ V , and f maps U bijectively onto V , and (ii) if g is the inverse of f (which exists by (i)), defined on V by g(f (x)) = x, for x ∈ U , then g ∈ C 1 (V ).

(b) First note that L and g both have continuous first order partial derivatives; i.e., L, g ∈ C 1 (R3 ). Therefore, the derivative of f = L + g,  3 ∂fi f 0 (x) , Jf (x) , ∂xj i,j=1 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 (R3 ). By the IFT, then, we need only show that f 0 (0) is invertible. Since f 0 (x) = L + g 0 (x), we must show f 0 (0) = L + g 0 (0) is invertible. Consider the matrix g 0 (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 gi (0 + hxj ) − gi (0) gi (hxj ) (0) = lim = lim . h→0 h→0 ∂xj h h

(8)

The second equality follows by the hypothesis that g is continuous and satisfies kg(x)k ≤ Ckxk2 , which implies that g(0) = 0. Finally, to show that (8) is zero, consider |gi (hxj )| ≤ kg(hxj )k ≤ Ckhxj k2 = C|h|2 , which implies |gi (hxj )| |hxj |2 ≤C = C|h| → 0, as h → 0. |h| |h| This proves that f 0 (0) = L, which is invertible by assumption, so the IFT implies that f (x) is locally invertible near 0. t u 2. Let f be a differentiable real valued function on the interval (0, 1), and suppose the derivative of f is bounded on this interval. Prove the existence of the limit L = limx→0+ f (x). 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 Z x Z x F (x) = f (t) dt, G(x) = g(t) dt. 0

0

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

1

Z F (x)g(x) dx = F (1)G(1) −

1

f (x)G(x) dx. 0

0

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 h(X,iA) that is absolutely continuous with respect to µ. Prove that the norm of the Radon-Nikodym derivative f =

dν dµ

is the same in L∞ (µ) as it is in

L∞ (ν). R1 5. Suppose that {fn } is a sequence of Lebesgue measurable functions on [0, 1] such that limn→∞ 0 |fn | dx = 0 and R 1 there is an integrable function g on [0, 1] such that |fn |2 ≤ 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) for all x. Determine, with proof, the closure of Pe in L1 [−1, 1]. You may use without proof the fact 2 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 Z a · m(U ) ≤ f dm ≤ b · m(U ), U

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

19

1.5

1.5

2001 November 26

1

REAL ANALYSIS

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 Z Z f (x) dx = lim fn (x) dx. n→∞

E

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  > 0 such that |f (x0 )| = M + . By the triangle inequality, then, for all n ∈ N, |f (x0 ) − fn (x0 )| ≥ ||f (x0 )| − |fn (x0 )|| = |M +  − |fn (x0 )|| ≥ , which contradicts fn (x0 ) → f (x0 ). Thus, |f (x)| ≤ M for all x ∈ E. Next, fix  > 0. By Egoroff’s theorem (A.8), there is a G ⊂ E such that µ(E \ G) <  and fn → f uniformly on G. Furthermore, since |fn | ≤ M and |f | ≤ M and µ(E) < ∞, it’s clear that {fn } ⊂ L1 and f ∈ L1 , so the following inequalities make sense (here we’re using the notation kf kG = sup{|f (x)| : x ∈ G}): Z Z Z Z Z f dµ − ≤ |f − f | dµ = |f − f | dµ + |f − fn | dµ f dµ n n n E E\G G E E Z Z ≤ |f | dµ + |fn | dµ + kf (x) − fn (x)kG µ(G) E\G

E\G

≤ 2M µ(E \ G) + kf (x) − fn (x)kG µ(G) <  + kf (x) − fn (x)kG µ(G). Finally, µ(G) ≤ µ(E) < ∞ and kf (x) − fn (x)kG → 0, which proves that

R E

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 Z f (x) dx > 0. F

(b) Prove that if Z

x

f (x) dx = 0,

for each x ∈ [0, 1],

0

then f (x) = 0 for almost all x ∈ [0, 1]. Solution: (a) Define Fn = {x ∈ F : f (x) > 1/n}. Then F1 ⊆ F2 ⊆ · · · ↑

[ n

20

Fn = F,

fn dµ →

R E

f dµ.

t u

1.5

2001 November 26

1

REAL ANALYSIS

and m(F ) > 0 implies 0 < m(F ) = m(∪n Fn ) ≤

X

m(Fn ).

n

Therefore, m(Fk ) > 0 for some k ∈ N, and then it follows from the definition of Fk that Z Z 1 0 < m(Fk ) ≤ f dm ≤ f dm. k Fk F t u R(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 ofR positive measure. (That such a closed subset exists follows from E Prop. 3.15 of Royden [6].) Then, again by (a), F f dm > 0. Now consider the set G = [0, 1] \ F , which is open in [0, 1], and hence22 is a countable union of disjoint open intervals; i.e., G = ∪· n (an , bn ). Therefore, Z Z XZ 0= f dm = f dm + f dm, [0,1]

so

R F

n

F

f dm > 0 implies XZ R (ak ,bk )

f dm < 0.

(an ,bn )

n

Thus,

(an ,bn )

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

Z

(ak ,bk )

bk

Z f (x) dm(x) −

f dm = 0

ak

f (x) dm(x). 0

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

t u

3. State each of the following: (a) The Stone-Weierstrass theorem (b) The Lebesgue (dominated) convergence theorem (c) H¨older’s inequality (d) The Riesz representation theorem for Lp (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 ∈R L1 (X, M, Rµ) 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 kfn − f k1 → 0. 22 Every

open set of real numbers is the union of a countable collection of disjoint open intervals (Royden [6], Prop. 8, page 42). 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]. 24 See theorem A.7 for a more general version. 23 The

21

1.5

2001 November 26

1

REAL ANALYSIS

(c) (H¨older’s inequality) Let f and g be measurable functions. (i) If 1 < p < ∞ and

1 p

1 q ∞

+

(ii) If p = ∞ and if f ∈ L

= 1, then kf gk1 = kf kp kgkq . Thus, if f ∈ Lp and g ∈ Lq , then f g ∈ L1 and g ∈ L1 , then |f g| ≤ kf k∞ |g|, so kf gk1 ≤ kf k∞ kgk1 .

(d) (Riesz representation theorem for Lp )25 Suppose 1 < p < ∞ and p1 + 1q = 1. If Λ is a linear functional on Lp , then there is a unique g ∈ Lq such that Z Λf = f g dµ (∀f ∈ Lp ). (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 kT¯kX = kT kY , where kT¯kX and kT kY are the usual operator norms, kT¯kX = sup{|T¯x| : x ∈ X, kxk ≤ 1}

and

kT kY = sup{|T x| : x ∈ Y, kxk ≤ 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) T∞ 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 G¯n 6= ∅ for at least one n ∈ N.

S∞

n=1

Gn then

Corollary 2. A nonempty complete metric space is not a countable union of nowhere dense sets. S∞ (b) Define Am = {x ∈ X : |f (x)| ≤ m, ∀f ∈ F }. Then X = m=1 T 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 7→ |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 6= ∅, so the set G = A◦m and the number M = m satisfy the given criteria. t u 25 Note

to self: add case p = ∞ 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. 26 See

22

1.5

2001 November 26

1

REAL ANALYSIS

5. Prove or disprove: (a) L2 convergence implies pointwise convergence. (b)



Z lim

n→∞

0

sin(xn ) dx = 0. xn

(c) Let {fn } be a sequence of measurable functions defined on [0, ∞). If fn → 0 uniformly on [0, ∞), as n → ∞, then Z Z lim fn (x) dx = lim fn (x) dx. [0,∞)

[0,∞)

Solution: (a) This is false, as the following example demonstrates: For each k ∈ N, define fk,j = χ[ j−1 , j ) for j = 1, . . . , k, k 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 , . . . √ R Then |fk,j |2 dµ = 1/k for each j = 1, . . . , k, so kfk,j k2 = 1/ k → 0, as k → ∞. Therefore kgn k2 → 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 0 ∈ N and j 0 ∈ {1, . . . , k 0 } such that gn0 (x) = fk0 ,j 0 (x) = 0 with n0 ≥ N . t u (b) For any fixed 0 < x < 1, limn→∞ xn = 0. Also, Together, these two facts yield

sin t t

→ 1, as t → 0, which can be proved by L’Hopital’s rule.

sin xn = 1. n→∞ xn Rθ Now, recall that | sin θ| ≤ |θ| for all real θ. Indeed, since sin θ = 0 cos x dx, we have, for θ ≥ 0, lim

Z | sin θ| ≤

θ

θ

Z | cos x| dx ≤

1 dx = θ,

0

0

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

sin xn xn ,

| sin xn | |xn |

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

to obtain Z lim

n→∞

0

1

sin xn dx = xn

Z

1

1 dx = 1.

(9)

0

Next consider the part of the integral over 1 ≤ x < N , for any real N > 1. Fix n ≥ 2. The change of variables 1 u = xn results in du = nxn−1 dx, and, since u1/n = x, we have xn−1 = u1− n . Therefore, Z N Z Nn Z n sin u du sin xn 1 N sin u dx = = 1 du. xn u nu1− n1 n 1 u2− n 1 1 23

1.5

2001 November 26

Now,

1

REAL ANALYSIS

Z n N n Z n 1 n −1 1 1 1 N sin u 1 N u 1 n −2 du = lim = lim u . 1 du ≤ lim 1 2− N →∞ n 1 N →∞ N →∞ n 1 n n −1 n−1 u n 1

Therefore,

Z



1

sin xn 1 dx ≤ n − 1, xn

and so,



Z lim

n→∞

1

Combining results (9) and (10) yields Z lim

n→∞

0



sin xn dx = 0. xn

(10)

sin xn dx = 1. xn t u

R(c) This is false, as the followingR lim fn = 0. On the other hand,

example demonstrates: Let fn = n1 Rχ[0,n) . Then fn fn = 1 for all n ∈ N. Therefore, lim fn = 1 6= 0 =

R→ 0 uniformly and so lim fn . t u

6. Let f : H → H be a bounded linear functional on a separable Hilbert space H (with inner product denoted by h·, ·i). Prove that there is a unique element y ∈ H such that f (x) = hx, yi for all x ∈ H

and kf k = kyk.

Hint. You may use the following facts: A separable Hilbert space, H, contains a complete orthonormal sequence, {φk }∞ ∈ H and if hx, φk i = hy, φk i for all k, then x = y. (2) k=1 , satisfying the following properties: (1) If x, y P ∞ Parseval’s equality holds; that is, for all x ∈ H, hx, xi = k=1 a2k , where ak = hx, φk i. P∞ Solution: Define y = k=1 f (φk )φk , and check that this y ∈ H has the desired properties. P∞ First observe that, by properties (1) and (2) given the hint, any x ∈ H can be written as x = k=1 ak φk , where ak = hx, φk i, for each k ∈ N. Therefore, by linearity of f , X X f (x) = f ( ak φk ) = ak f (φk ). (11) k

Now

k

X X hx, yi = h ak φk , yi = ak hφk , yi, k

(12)

k

and, by definition of y, hφk , yi = hφk ,

X

f (φj )φj i =

j

X

f (φj )hφk , φj i = f (φk ).

(13)

j

The last equality holds by orthonormality; i.e., hφk , φj i is 1 when j = k and 0 otherwise. Putting it all together, we see that, for every x ∈ H, X f (x) = ak f (φk ) ∵ (11) k

=

X

ak hφk , yi

∵ (13)

k

= hx, yi

∵ (12) 24

1.5

2001 November 26

1

REAL ANALYSIS

Moreover, this y is unique. For, suppose there is another y 0 ∈ H such that f (x) = hx, y 0 i for all x ∈ X. Then hx, yi = f (x) = hx, y 0 i for all x ∈ X. In particular, hφk , yi = f (φk ) = hφk , y 0 i for each k ∈ N , which, by property (1) of the hint, proves that y = y 0 . Finally, we must show kf k = kyk. Observe, kf k = sup {|f (x)| : kxk ≤ 1} = sup x∈X

x∈X

|f (x)| |(x, y)| = sup , kxk x∈X kxk

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

(14)

|(y, y)| kyk2 |(x, y)| ≥ = = kyk. kxk kyk kyk

(15)

kf k = sup x∈X

On the other hand, kf k = sup x∈X

Together, (14) and (15) give kf k = kyk, as desired.

t u

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 kAk = supkxk≤1 kAxk, 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., kTn − Tm k → 0 as m, n → ∞. Fix x ∈ X. Then, kTn x − Tm xkY ≤ kTn − Tm kkxkX → 0, as n, m → ∞. Therefore, the sequence {Tn x} ⊂ Y is a Cauchy sequence in (Y, k · kY ). Since the latter is complete, the limit limn→∞ Tn x = y ∈ Y exists. Define T : X → Y by T x = limn→∞ Tn x, for each x ∈ X. To complete the proof, we must check that T is linear, bounded, and satisfies limn→∞ kTn − T k = 0. • T is linear: For x1 , x2 ∈ X, T (x1 + x2 ) = lim Tn (x1 + x2 ) n→∞

= lim (Tn x1 + Tn x2 )

(∵ Tn is linear)

= lim Tn x1 + lim Tn x2

(∵ both limits exist)

n→∞

n→∞

n→∞

= T x1 + T x2 . • T is bounded: First, note that {kTn k} is a Cauchy sequence of real numbers, since | kTn k − kTm k | ≤ kTn − Tm k → 0, as n, m → ∞. Therefore, there is a c ∈ R such that kTn k → c, as n → ∞. For some N ∈ N, then, kTn k ≤ c + 1 for all n ≥ N . Thus, kTn xkY ≤ kTn kkxkX ≤ (c + 1)kxkX 25

(∀x ∈ X).

(16)

1.5

2001 November 26

1

REAL ANALYSIS

Now, by definition, Tn x → T x, for all x ∈ X and, since the norm k · kY is uniformly continuous,27 kTn xkY → kT xkY

(∀x ∈ X).

(17)

Taken together, (16) and (17) imply kT xkY ≤ (c + 1)kxkX , for all x ∈ X. Therefore, T is bounded. • limn→∞ kTn − T k = 0: Fix  > 0 and choose N ∈ N such that n, m ≥ N implies kTn − Tm k < . Then, kTn x − Tm xk ≤ kTn − Tm kkxkX < kxkX holds for all n, m ≥ N , and x ∈ X. Letting m go to infinity, then, kTn x − T xk = lim kTn x − Tm xk ≤ kxkX . m→∞

That is, kTn x − T xk ≤ kxkX , for all n ≥ N and x ∈ X. Whence, kTn − T k ≤  for all n ≥ N . t u

27 Proof:

| kakY − kbkY | ≤ ka − bkY (∀a, b ∈ Y ).

26

1.6

2004 April 19

1.6

1

REAL ANALYSIS

2004 April 19

Instructions. Use a separate sheet of paper for each new problem. Do as many problems as you can. Complete solutions to five problems will be considered as an excellent performance. Be advised that a few complete and well written solutions will count more than several partial solutions. Notation: f ∈ C(X) means that f is a real-valued, continuous function defined on X. 1. (a) Let S be a (Lebesgue) measurable subset of R and let f, g : S → R be measurable functions. Prove that (i) f + g is measurable and (ii) if φ ∈ C(R), then φ(f ) is measurable. (b) Let f : [a, b] → [−∞, ∞] be a measurable function. Suppose that f takes the value ±∞ only on a set of (Lebesgue) measure zero. Prove that for any  > 0 there is a positive number M such that |f | ≤ M, except on a set of measure less than . Solution: (a) Proof 1: Since f and g are real measurable functions of S, and since the mapping Φ : R × R → R defined by Φ(x, y) = x + y is continuous, theorem A.3 implies that the function f + g = Φ(f, g) is measurable. If φ ∈ C(R), then φ(f ) is measurable by part (b) of theorem A.2. Proof 2: Let {qi }∞ i=1 be an enumeration of the rationals. Then, for any α ∈ R, {x ∈ S : f (x) + g(x) < α} =

∞ [

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

∞ [

An ∪ A∞ ,

(18)

n=1

where28 A∞ = {x : f (x) = ±∞}. Also, A1 ⊆ A2 ⊆ · · · and, since f is measurable, each An is measurable. Therefore, µ(An ) ↑ µ(∪n An ), as n → ∞. Note that all sets are contained in [a, b] and thus have finite measure. Let M ∈ N be such that µ(∪n An ) − µ(AM ) < . Then |f | ≤ M except on [a, b] \ AM , and by (18), µ([a, b] \ AM ) = µ(∪n An ∪ A∞ \ AM ) ≤ µ(∪n An \ AM ) + µ(A∞ ) = µ(∪n An \ AM ) < . 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

t u

1.6

2004 April 19

1

REAL ANALYSIS

2. (a) State Egorov’s theorem. (b) State Fatou’s lemma. (c) Let {fn } ⊂ Lp [0, 1], where 1 ≤ p < ∞. Suppose that fn → f a.e., where f ∈ Lp [0, 1]. Prove that kfn − f kp → 0 if and only if kfn kp → kf kp . Solution: (a) See theorem A.8. (b) See theorem A.6. (c) (⇒) By the Minkowsky inequality, kfn kp = kfn − f + f kp ≤ kfn − f kp + kf kp . Similarly, kf kp ≤ kfn − f kp + kfn kp . Together, the two inequalities yield |kfn kp − kf kp | ≤ kfn − f kp . Therefore, kfn − f kp → 0 implies |kfn kp − kf kp | → 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 Lp (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 ≤ 2p−1 (αp + β p ). Proof: When p ≥ 1, φ(x) = xp is convex on [0, ∞). Thus, for all α, β ∈ [0, ∞), p    α+β 1 1 α+β =φ ≤ [φ(α) + φ(β)] = (αp + β p ). 2 2 2 2

When α, β ∈ R, the triangle inequality followed by the lemma yields |α − β|p ≤ ||α| + |β||p ≤ 2p−1 (|α|p + |β|p ).

(19)

Proof 1: By (19), |fn − f |p ≤ 2p−1 (|fn |p + |f |p ). In particular, fn − f ∈ Lp , for each n ∈ N. Moreover, the functions gn = 2p−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, Z Z Z Z  p−1 2p |f |p = lim gn ≤ lim gn = lim 2 (|fn |p + |f |p ) − |fn − f |p . Since kfn kp → kf kp , this implies p

2 Equivalently 0 ≤ − lim 29 Disclaimer:

R

Z

p

p

|f | ≤ 2

Z

p

Z

|f | − lim

|fn − f |p .

|fn − f |p . This proves kfn − f kp → 0.

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

28

t u

1.6

2004 April 19

1

REAL ANALYSIS

Proof 2: By (19), |fn − f |p ≤ 2p−1 (|fn |p + |f |p ).

(21)

In particular, fn − f ∈ Lp , for each n ∈ N. Define the functions gn = 2p−1 (|fn |p + |f |p ) and g = 2p |f |p . R R p Then gn → g a.e., and kfn kp → kf kp implies gn → g. R Also, gnp ≥ |fn − f | → 0 a.e., by (21). Therefore, the dominated convergence theorem (theorem A.7) implies |fn − f | → 0. t u

Proof 3: Since fR ∈ Lp , for all  > 0, there R is a number δ > 0 and a set B ∈ M of finite measure such that f is bounded on B, X\B |f |p dµ < /2, and E |f |p dµ < /2, for all E ∈ M with µE < δ. By Egoroff’s theorem, there is a set A ⊆ B such that µ(B \ A) < δ and fn → f uniformly on A. Therefore, Z Z Z Z |f |p = |f |p + |f |p + |f |p X X\B B\A A Z p < /2 + /2 + |f | A Z |fn |p , (22) ≤  + lim A

lim |fn |p ≤ lim (Z Z Z p lim |fn |p − |fn | = lim

since, by Fatou’s lemma,

R

A

|f |p =

R

A

X

A

R A

|fn |p . By hypothesis, kfn kp → kf kp . Therefore ) Z Z

|fn |p

|f |p − lim

=

X\A

X

|fn |p .

X\A

By (22), then, Z

p

Z

p

|f | <  + X

Therefore, lim

R X\A

Z

|fn |p .

|f | − lim X

X\A

|fn |p <  (since f ∈ Lp ). Finally, note that kfn − f kp = k(fn − f )χA + (fn − f )χX\A kp ≤ k(fn − f )χA kp + k(fn − f )χX\A kp

(Minkowsky)

≤ k(fn − f )χA kp + kfn χX\A kp + kf χX\A kp . Therefore, lim kfn − f kp ≤ lim{|fn (x) − f (x) : x ∈ A}µ(A)1/p + lim

Z X\A

!1/p |fn |p

Z +

!1/p |f |p

.

X\A

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

29

t u

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 kfn kp ≤ M for all n, prove that for each g ∈ Lq (S), p1 + 1q = 1, we have Z lim

n→∞

Z fn g =

S

f g. S

(b) Show by means of an example that this result is false for p = 1. Solution: (a) Since g ∈ Lq (S), for all  > 0 there exists δ > 0 such that if A is a measurable set with µA < δ then Z |g|q dµ < . 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, Z Z Dn := |fn g − f g| dµ = |fn − f ||g| dµ S ZS Z = |fn − f ||g| dµ + |fn − f ||g| dµ S\B

B

≤ k(fn − f )χS\B kp kgkq + kfn − f kp kgχB kq . The final inequality holds because fn , f ∈ Lp implies |fn − f | ∈ Lp and by H¨older’s inequality. Now, by Minkowski’s inequality, kfn − f kp ≤ kfn kp + kf kp , so Dn ≤ sup{|fn (x) − f (x)| : x ∈ / B}µ(S \ B)

1/p

Z kgkq + (kfn kp + kf kp )

q

|g| dµ

1/q .

B

Now we are free to choose the δ > 0 above so that Z 1/q q |g| dµ < B

 2(M + kf kp )

holds whenever µB < δ. Also, fn → f uniformly on S \ B, so let N be such that sup{|fN (x) − f (x)| : x ∈ / B}
0 such that 1/N < x and thus fn (x) R= 0 for all n ≥ N . Therefore, {x ∈ [0, 1] : fRn (x) 9 0} = {0}. Now, if g is Rthe constant function g ≡ 1, then fn g dµ = 1 for all R n = 1, 2, . . . . Therefore, fn g dµ → 1, while f g dµ = 0g dµ = 0. Finally, note that kfn k1 = nµ([0, n1 ]) = 1 for n = 1, 2, . . . , so {fn } satisfies the hypothesis kfn k1 ≤ some constant M . t u

30

1.6

2004 April 19

1

REAL ANALYSIS

4. State and prove the closed graph theorem.

5. Prove or disprove: P∞ (a) For 1 ≤ p < ∞, let `p := {x = {xk } | kxkp = ( k=1 |xk |p )1/p < ∞}. Then for p 6= 2, `p is a Hilbert space. R1 (b) Let X = (C[0, 1], k · k1 ), where the linear space C[0, 1] is endowed with the L1 -norm: kf k1 = 0 |f (x)| dx. Then X is a Banach space. (c) Every real, separable Hilbert space is isometrically isomorphic to `2 .

6. (a) Give a precise statement of some version of Fubini’s theorem that is valid for non-negative functions. (b) Let f, g ∈ L1 (R). (i) Prove that the integral Z f (x − t)g(t) dt

h(x) = R

exists for almost all x ∈ R and that h ∈ L1 (R). (ii) Show that khk1 ≤ kf k1 kgk1 . 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 Z 1 f dµ ∈ S AE (f ) = µ(E) E for every E ∈ B with µ(E) > 0, prove that f (x) ∈ S for almost all x ∈ X.

31

1.7

2007 November 16

1.7

1

REAL ANALYSIS

2007 November 16

Notation: R is the set of real numbers and Rn is n-dimensional Euclidean space. Denote by m Lebesgue measure on R and mn n-dimensional Lebesgue measure. Be sure to give a complete statement of any theorems from analysis that you use in your proofs below. 1. Let µ be a positive measure on a measure space X. Assume that E1 , E2 , . . . are measurable subsets of X with the property that for n 6= m, µ(En ∩ Em ) = 0. Let E be the union of these sets. Prove that µ(E) =

∞ X

µ(En )

n=1

Solution: Define F1 = E1 , F2 = E2 \ E1 , F3 = E3 \ (E1 ∪ E2 ), . . . , and, in general, Fn = En \

n−1 [

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 6= j, and F1 ∪ F2 ∪ · · · ∪ Fn = E1 ∪ E2 ∪ · · · ∪ En for all n ∈ N. Thus, ∞ [

∞ [

Fn =

n=1

En , E,

n=1

and, by σ-additivity of µ, µ(E) = µ(

∞ [

∞ X

Fn ) =

n=1

µ(Fn ).

n=1

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, . . . , n−1 [

Fn = E n ∩ (

Ei )c

(23)

i=1

and

n−1 [

µ(En ) = µ(En ∩ (

Ei )c ) + µ(En ∩ (

i=1

Equation (24) holds because

Sn−1 i=1

n−1 [

Ei )).

(24)

i=1

Ei is a measurable set for each n = 2, 3, . . . . Finally, note that n−1 [

En ∩ (

Ei ) =

n−1 [

i=1

(En ∩ Ei ),

i=1

which implies µ(En ∩ (

n−1 [

Ei )) ≤

i=1

n−1 X

µ(En ∩ Ei ),

i=1

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

t u

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 ∈ L1 (m) for n = 1, 2, . . . . Assuming that kfn − f k1 → 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 kfn − f k1 → 0, then fn → f in measure. (Proof: m({x : |fn (x) − f (x)| > }) ≤ 1 kfn − f k1 → 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 } ⊂ L1 (m) and kfn − f k1 → 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) 9 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) 9 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) 6= 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. t u 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] : fP (x) 6= g(x)}. Then mBn = 0 for all n ∈ N, so that if B = {x ∈ R : f (x) 6= 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 R the fact that f, g ∈ L1 ([−n, n], m), which we prove below.) Then, for all  > 0 there is a δ > 0 such that E |f − g| dm <  for all measurable E ⊆ [−n, n] with mE < δ. Now apply Egoroff’s theorem to find a set A ⊆ [−n, n] such that m([−n, n] \ A) < δ and fn → g uniformly on A. Then Z n Z Z |f − g| dm = |f − g| dm + |f − g| dm −n [−n,n]\A A Z Z ≤+ |f − fn | dm + |fn − g| dm A

A

≤  + kf − fn k1 + 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

1.7

2007 November 16

1

REAL ANALYSIS

where kf − fn k1 → 0, by assumption, and supx∈A |fn (x) − g(x)| → 0 since fn → g uniformly on A. Since  > 0 Rn was arbitrary, it follows that −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 kf k1 ≤ kf − fn k1 + kfn k1 < ∞. 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, Z Z Z kgk1 = |g| dm = lim |fn | dm ≤ lim |fn | dm = lim kfn k1 = kf k1 < ∞. The last equality holds because, by the triangle inequality, |kfn k1 − kf k1 | ≤ kfn − f k1 → 0.

t u

3. Let K be a compact subset in R3 and let f (x) = dist(x, K). (a) Prove that f is a continuous function and that f (x) = 0 if and only if x ∈ K. RRR 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) = inf k∈K |x − k|. Clearly, for all k ∈ K, f (x) ≤ |x − k|. Therefore, by the triangle inequality, for any x, y ∈ R3 , f (x) ≤ |x − y| + |y − k|,

∀k ∈ K,

and so, taking the infimum over k ∈ K on the right, f (x) ≤ |x − y| + f (y).

(26)

f (y) ≤ |x − y| + f (x).

(27)

Similarly, 3

Obviously, for any given x ∈ R , f (x) is finite. Therefore, (26) and (27) together imply that |f (x) − f (y)| ≤ |x − y|,

∀ x, y ∈ R3 .

Whence f is (Lipschitz) continuous. Now, if x ∈ K, then it’s clear that f (x) = 0. Suppose x ∈ / K; that is, x is in the complement K c of K. Since K is c closed, K is open and we can find an -neighborhood about x fully contained in K c , in which case f (x) > . We have thus proved that f (x) = 0 if and only if x ∈ K. t u (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 → χKRRR . Finally, RRR note that g n ≤ χK1 ∈ L1 (R3 ) so the dominated convergence theorem can be applied to yield n limn→∞ g = χK = m3 (K). t u

4. Let E be a Borel subset of R2 . (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 R2 , which we denote by B(R2 ), is the smallest σ-algebra that contains the open subsets of R2 . The sets belonging to B(R2 ) are called Borel subsets of R2 . (b) First observe that if G is a finite subset of R, then G is a Lebesgue null set. That is, mG = 0. In fact, it is easy to prove that if G is any countable subset, then P mG = 0. (Just fix  > 0 and cover each point xn ∈ G with a set En of measure less than 2−n . Then mG ≤ mEn < .) In problems involving 2-dimensional Lebesgue measure, distinguishing x and y coordinates sometimes clarifies things. To wit, let (X, B(X), µ) = (Y, B(Y ), ν) be two identical copies of the measure space (R, B(R), m), and represent Lebesgue measure on R2 by33 (X × Y, B(X) ⊗ B(Y ), µ × ν) = (R2 , B(R2 ), m2 ). Our goal is to prove that m2 E = 0. First note that Z m2 E = (µ × ν)(E) =

χE d(µ × ν). X×Y

The integrand χE is non-negative and measurable (since E is Borel). Therefore, by Tonelli’s theorem (A.13), Z Z Z Z m2 E = χE (x, y) dµ(x) dν(y) = χE (x, y) dν(y) dµ(x). (28) Y

X

X

Y

Now, let Gt = {y ∈ R : (x, y) ∈ E and x = t}. This is the so called “x-section” of E at the point x = t. It is a subset of R, but we can view it as a subset of R2 by simply identifying each point y ∈ Gt with the point (t, y) ∈ 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), Z Z Z m2 E = χGt (y) dν(y) dµ(t) = νGt dµ(t) = 0. X

Y

X

t u

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 represent the Radon-Nikodym derivative of ν with respect to µ + ν. Show that 0
0 such that kTk k ≤ M for all k ∈ N. In other words, (∃ M > 0) (∀b ∈ `q ) (∀k ∈ N) |Tk (b)| ≤ M kbk. P∞ Define let T (b) , n=1 an bn = limk→∞ Tk (b), which exists by assumption. Since |·| is continuous, we conclude that limk→∞ |Tk (b)| = |T (b)|. Finally, since |Tk (b)| ≤ M kbk for all k ∈ N, we have |T (b)| ≤ M kbk. That is T is a bounded linear functional on `q (N). Now, by the Riesz representation theorem, if 1 ≤ q < ∞, then any bounded linear functional T ∈ `∗q is uniquely representable by some α = (α1 , α2 , . . . ) ∈ `p as T (b) =

∞ X

αn bn .

n=1 34 On

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

36

(29)

1.7

2007 November 16

1

On the other hand, by Pdefinition, T (b) = a = α ∈ `p . That is, n |an |p < ∞.

P∞

n=1

REAL ANALYSIS

an bn , for all b ∈ `q . Since the representation in (29) is unique, t u

(b) Consider the case p = 1 and q = ∞. First recall that the Riesz representation theorem says that every T ∈ `∗q (1 ≤ q < ∞) is uniquely representable by some α ∈ `p (where p = q/(q − 1), so 1 < p ≤ ∞). That is `p is the dual of `q , when 1 ≤ q < ∞ and p = q/(q − 1). However, in the present case we have q = ∞ and p = 1 and `1 is not the dual of `∞ . (Perhaps the easiest way to see this is to note that `1 is separable but `∞ is not. For the collection of a ∈ `∞ such that an ∈ {0, 1}, n ∈ N, is uncountable and, for any two distinct such sequences a, b ∈ {0, 1}N , we have ka − bk∞ = 1, so there cannot be a countable base, so `∞ is not second countable, and a metric space is separable iff it is second countable.) So we can’t use the same method of proof for this case. However, I believe the result still holds by the following simple argument: Define b = (b1 , b2 , . . . ) by ( a ¯n /|an |, for an 6= 0, (n ∈ N). bn = sgn(an ) = 0, for an = 0, Then

P

n

|an | =

P

n

an bn converges by the hypothesis, since |bn | ∈ {0, 1} implies b ∈ `∞ . Therefore, a ∈ `1 .

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

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

37

t u

2

2

Complex Analysis

38

COMPLEX ANALYSIS

2.1

2.1

1989 April

2

COMPLEX ANALYSIS

1989 April

INSTRUCTIONS: Do at least four problems. TIME LIMIT: 1.5 hours35 1. (a) Let U be the unit disk in the complex plane C, U = {z ∈ C : |z| < 1}, and let f be an analytic function in a neighborhood of the closure of U . Show that if f is real on all the boundary of U , then f must be constant. (b) Let u be a real harmonic function in all the complex plane C. Show that if u(z) ≥ 0 for all z ∈ C, then u must be constant. Solution: (a) The hypotheses imply that Imf (eiθ ) = 0 for all θ ∈ R. Since f is holomorphic in a neighborhood Ω of U , the series ∞ X f (z) = an z 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 ∞ X f (eiθ ) = an einθ . n=0

If we write the coefficients as an = cn + ibn , where cn , bn ∈ R, then we have an einθ = (cn + ibn )[cos(nθ) + i sin(nθ)] = [cn cos(nθ) − bn sin(nθ)] + i[cn sin(nθ) + bn cos(nθ)]. Thus, by the hypothesis, the series Imf (eiθ ) =

∞ X

[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 . t u (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. t u

2. Let f be an analytic function in the region {z : |z| > 1}, and suppose that lim f (z) = 0.

z→∞

35 In 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. 36 Picard’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 1 2πi

Z |ζ|=2

f (ζ) dζ = −f (z). ζ −z

Solution: By Cauchy’s formula, if |z| < R, then Z Z 1 f (ζ) 1 f (ζ) f (z) = dζ − dζ. 2πi |ζ|=R ζ − z 2πi |ζ|=2 ζ − z

(30)

(31)

Note that this holds for all R > |z| > 2. Fix  > 0. Let R be such that |f (ζ)| < /2 for all |ζ| = R and R > 2|z|. Then |ζ − z| > R /2 for all |ζ| = R , so Z |f (ζ)| 1 /2 2πR |dζ| < = . 2π |ζ|=R |ζ − z| 2π R /2 Therefore, by (31), 1 Z f (ζ) dζ + f (z) < . 2πi |ζ|=2 ζ − z t u

This holds for any , which proves (30).

3.37 Let U be the open unit disk in C, and let U + be the top half of this disk, U + = {z ∈ C : Imz > 0, |z| < 1}. Exhibit a one-to-one conformal mapping from U + onto U . Solution: Consider ϕ0 (z) = 1−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.    1 − i/2 1 − i/2 1 − i/2 3 − 4i ϕ0 (i/2) = = = ∈ Ω− . 1 + i/2 1 + i/2 1 − i/2 5 Thus, ϕ0 : U + → Ω− . Let ϕ1 (z) = iz, so that ϕ1 : Ω− → Ω+ , let ϕ2 (z) = z 2 , 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 ϕ(z) = (ϕ0 ◦ ϕ3 ◦ ϕ2 ◦ ϕ1 ◦ ϕ0 )(z). t u 37 This

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

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 Z |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 n

P (z) X P (ak ) = . Q(z) Q0 (ak )(z − ak ) k=1

6. Use contour integration and the residue method to evaluate the integral Z ∞ cos x dx. (1 + x2 )2 0

Solution: Denote the integral by I. Since the integrand is even, Z ∞ cos x dx. 2I = (1 + x2 ) 2 −∞ Consider the simple closed contour ΓR = γR ∪ [−R, R], where the trace of γR is the set {Reiθ : 1 ≤ θ ≤ π}, oriented counter-clockwise. Note that, if R > 1, then i is inside the region bounded by ΓR . The function f (z) =

cos z cos z = 2 2 (1 + z ) (z + i)2 (z − i)2

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

z→i

d d cos z [(z − i)2 f (z)] = lim z→i dz (z + i)2 dz

−(z + i)2 sin z − 2(z + i) cos z z→i (z + i)4

= lim =

−(2i)2 sin(i) − 4i cos(i) (2i)4

=

sin(i) − i cos(i) −iei·i −i = = . 4 4 4e

By the residue theorem, it follows that, for all R > 1, Z π f (z) dz = 2πi Res(f, i) = . 2e ΓR 41

2.1

1989 April

2

It remains to check that Z

R

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



Z

Z f (z) dz −

f (x) dx = lim

2I = −∞

Indeed,

γR

COMPLEX ANALYSIS

R→∞

ΓR



Z

f (z) dz = lim γR

R→∞

f (z) dz = ΓR

π . 2e

(32)

Z f (z) dz =

cos z 1 πR dz ≤ `(γR ) = . 2 2 2 2 2 (R − 1) (R − 1)2 γR (1 + z ) γR R This inequality holds for all R > 1, so, letting R → ∞, we have γR f (z) dz → 0. Therefore, by (32), Z

Z I= 0



cos x π dx = . 2 2 (1 + x ) 4e t u

42

2.2

2.2

1991 November 21

2

COMPLEX ANALYSIS

1991 November 21

INSTRUCTIONS: In each of sections A, B, and C, do all but one problem. TIME LIMIT: 2 hours SECTION A (Do 3 of the 4 problems.) 1. Where does the function f (z) = zRez + z¯Imz + z¯ have a complex derivative? Compute the derivative wherever it exists. Solution: Writing f in terms of the real and imaginary parts of z = x + iy, we have f (x + iy) = (x + iy)x + (x − iy)y + x − iy = x2 + xy + x + i(xy − y 2 − y) = u(x, y) + iv(x, y), where u(x, y) = x2 + 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 ∂ = 21 ∂x , and recall that, if − i ∂y f ∈ H(Ω), then the derivative of f is given by f 0 (z) = (∂f )(z), z ∈ Ω. In the present case, ∂f = 2x + y + 1 + iy, ∂x

∂f = x + i(x − 2y − 1). ∂y

Therefore, ∂f (x + iy) = 21 [(2x + y + 1 + iy) − i(x + i(x − 2y − 1))] = 12 [(3x − y) + i(y − x)], and finally, f 0 (1 − i) =

1 (4 − 2i) = 2 − i. 2 t u

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 Y

(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 Y

P (z) = a

(z − rk )

M Y

(z 2 − bm z + cm ),

m=1

k=1

where a and each rk , bm , cm are real constants. 3. Use complex residue methods to compute the integral Z π

1 dθ. 5 + 3 cos θ

0

Solution: Let I =



1 0 5+3 cos θ

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

Z

1 dθ. 5 + 3 cos θ

2I = −π

For z = eiθ ,

eiθ + e−iθ 1 1 = (z + ), 2 2 z

cos θ = and dz = ieiθ dθ, from which it follows that Z

1

2I = |z|=1

=

=

1 i

5+

Z |z|=1

3 2 (z

dz iz

dz 5z + 32 (z 2 + 1)

Z

2 3i

+

1 z)

dz

|z|=1

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   2 1 2I = · 2πi · Res , z1 . 3i p(z) Now,

1 1 = , p(z) (z − z1 )(z − z2 )

which implies  Res

1 , z1 p(z)

 = lim

z→z1

Therefore, 2I = so I =

1 1 3 = 1 = . z − z2 8 − 3 − (−3)

2 3 π · 2πi · = , 3i 8 2

π 4.

t u 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 ∞ X 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, Z f (z) 1 dz, (n ∈ Z; 1 < R < 2). an = 2πi |z|=R (z − z0 )n+1 Let {pm } be the sequence of polynomials mentioned in the problem statement. Of course, pm ∈ H(C), so Cauchy’s R theorem implies |z|=R pm (z) dz = 0, and, more generally, Z pm (z)(z − z0 )−n−1 dz = 0, (n = −1, −2, . . . ). |z|=R

Therefore, Z Z f (z) pm (z) dz − dz n+1 |z|=R (z − z0 )n+1 |z|=R (z − z0 ) Z 1 |f (z) − pm (z)| ≤ |dz|. 2π |z|=R |z − z0 |n+1

1 |an | = 2π

(35)

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

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 Z f 0 (z)g(z) dz ? f (z) |z|=1 (Verify your answer.) 38 See

also: April ’96 (8).

45

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 Z |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 ¯ α , Rα ) = {z ∈ C : |z − zα | ≤ Rα } ⊂ D. Then, for equivalent?)39 So, fix zα ∈ D. Let Rα > 0 be such that B(z any z ∈ B(zα , Rα /2), Cauchy’s formula gives Z |fn (ζ)| 1 |dζ| |fn (z)| ≤ 2π |ζ−zα |=Rα |ζ − z| Z 1 1 |fn (ζ)| |dζ| ≤ 2π Rα /2 |ζ−zα |=Rα ≤

M . πRα

The second inequality holds Rsince |ζ − zα | = Rα and |z − zα | < Rα /2 imply |ζ − z| > Rα /2. The last inequality M follows from the hypothesis C |fn (z)| |dz| ≤ M for any circle C in D. Letting Mα = πR , and rα = Rα /2, we α have |fn (z)| ≤ Mα for all z ∈ B(zα , rα ) and all n = 1, 2, . . . , as desired. t u

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 , rαj ) (j = 1, . . . , J) and then |fn (z)| ≤ maxj Mαj , 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

2.3

1995 April 10

2

COMPLEX ANALYSIS

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. P 1. Let f (z) = an z 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 an =

f (n) (0) 1 = n! 2πi

Z |ζ|=R

f (ζ) dζ, ζ n+1

for every R > 0. Therefore, |an | ≤

1 2π

Z |ζ|=R

|f (ζ)| 1 A RN + B |dζ| ≤ 2πR = A RN −n + B R−n . n+1 |ζ| 2π Rn+1

Again, this holds for every R > 0. Thus, for any n > N and  > 0, taking R large enough forces |an | <  (n = NP+ 1, N + 2, . . . ). Since  was arbitrary, we have an = 0 for all n = N + 1, N + 2, . . . . Therefore, N f (z) = n=0 an z n . t u (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. t u Proof 2: Since f ∈ H(C), the series f (z) = the function f (1/z) has a pole at z = 0. Let

P

an z n converges locally uniformly in C. The hypotheses imply that

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

∞ X

bn z n

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) =

∞ X

bn z n = b−m z −m + b−m+1 z −m+1 + · · · b−1 z −1 + b0 + b1 z + · · ·

(36)

n=−m

Now f is entire, so it has the form f (z) = Compared with (36),

P∞

n=0

an z n , which implies that f (1/z) = a0 + a1 z −1 + a2 z −2 + · · · .

a0 + a1 z −1 + a2 z −2 + · · · = f (1/z) = b−m z −m + b−m+1 z −m+1 + · · · b−1 z −1 + b0 + b1 z + · · · That is, 0 = am+1 = am+2 = · · · , so f (z) =

m X n=0

48

an z n .

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) . (z − α1 ) · · · (z − αN )

(37)

This is defined and holomorphic in C \ {α1 , . . . , αN }, but the αi ’s are removable singularities, so g(z) is a nonzero entire function. In particular, for any R > 0, min |g(z)| ≥ min |g(z)| =  > 0,

z∈DR

|z|=R

for some  > 0. Therefore, 1/g is a bounded entire function, hence constant, by Liouville’s theorem. What we have shown is that the left hand side of (37) is constant, and this proves that f (z) is a polynomial. t u Remark: A nice corollary to part (b) is the following: Corollary 2.1 If f is an injective entire function, then f (z) = az + b for some constants a and b. The proof appears below in section 2.9.

2. (a) State a form of the Cauchy theorem. (b) State a converse of the Cauchy theorem. Solution: (a) See theorem A.16. (b) See theorem A.18.

3. Let40 f (z) =

P∞

n=0

an z n be analytic and one-to-one on |z| < 1. Suppose that |f (z)| < 1 for all |z| < 1.

(a) Prove that

∞ X

n|an |2 ≤ 1.

n=1

(b) Is the constant 1 the best possible? Solution: (a) This is a special case of the following area theorem: P∞ Theorem 2.1 Suppose f (z) = n=0 an z 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

∞ X

n|an |2 .

n=1 40 On

the original exam, the power series representation was given as f (z) = assuming a0 = 0 a priori.

49

P∞

n=1

an z n . However, the problem can be solved without

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, ZZ A= |f 0 (z)|2 dx dy. D 0

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

f 0 (z) =

nan z n−1 .

n=1

Next, take the squared modulus, |f 0 (z)|2 =

∞ X

m n am an z m−1 z n−1 .

m,n=1

This gives, ∞ X

ZZ A=

m n am an z m−1 z n−1 dx dy.

D m,n=1

Letting z = reiθ , A=

∞ X

Now, for all k 6= 0, the integral of e which m = n. That is,

Z

0

m,n=1 ikθ

1

Z m n am an



rm+n−1 ei(m−n)θ dθ dr.

0

over 0 ≤ θ < 2π vanishes, so the only non-vanishing terms of the series are those for

A = 2π

∞ X n=1

n2 |an |2

1

Z

r2n−1 dr = π

0

∞ X

n2 |an |2 .

(38)

n=1

u t

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 ∞ X π≥π n2 |an |2 , n=1

t u

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. t u

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. t u 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

2.3

1995 April 10

2

COMPLEX ANALYSIS

5.42 Find an explicit conformal mapping of the semidisk H = {z : |z| < 1, Rez > 0} onto the unit disk. Solution: See the solution to (3) of April ’89, or (2) of November ’06.

6.43 Suppose f (z) is a holomorphic function on the unit disk which satisfies: |f (z)| < 1 all |z| < 1. (a) State the Schwarz lemma, as applied to f . (b) If f (0) = 21 , how large can |f 0 (0)| be? Solution: (a) See theorem A.22. (b) Assume f satisfies the given hypotheses. In particular, f (0) = 12 . Consider the map 1 2

−z . 1 − z2

ϕ(z) =

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 (0)| ≤ 1. Since g 0 (z) = ϕ0 (f (z))f 0 (z), we have 1 ≥ |g 0 (0)| = |ϕ0 (1/2)||f 0 (0)|. Now, 0

ϕ (z) =

− 1−

z 2



+

1−

1 2  z 2 2



1 2

−z

(39)  .

Therefore, ϕ0 (1/2) = −4/3, and it follows from (39) that |f 0 (0)| ≤

1 = 3/4. |ϕ0 (1/2)| t u

42 This 43 A

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

51

2.4

2.4

2001 November 26

2

COMPLEX ANALYSIS

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 C 1 functions defined on a neighborhood of the closure of a bounded region G ⊂ C with boundary which is parametrized by a properly R oriented, piecewise C 1 curve γ. If u and v obey the Cauchy-Riemann equations, show that Cauchy’s theorem γ f (z) dz = 0 follows from Green’s theorem, namely  Z Z  ∂Q ∂P − dx dy for C 1 functions P and Q. P dx + Q dy = (40) ∂x ∂y 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 0 (z0 ) = lim

z→z0

f (z) − f (z0 ) z − z0

exists at some (possibly only one!) point z0 ∈ G. Show that given any  > 0, we can find a triangular region ∆ containing z0 , such that if T is the boundary curve of ∆, then Z f (z) dz = 1 L2 , 2 T where L is the length of the perimeter of ∆. R Hint for (b) Note that part (a) yields T (az + b) dz = 0 for a, b ∈ C, which you can use here in (b), even if you R 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 Z Z u(x, y) dx − v(x, y) dy = (vx + uy ) dx dy = 0. γ

(41)

G

Similarly, if P = v and Q = u in Green’s theorem, then the Cauchy-Riemann equations imply Z Z v(x, y) dx + u(x, y) dy = (ux − vy ) dx dy = 0. γ

(42)

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), Z Z Z f (z) dz = u(x, y) dx − v(x, y) dy + i v(x, y) dx + u(x, y) dy = 0. γ

γ

γ

t u 44 These

are ux = vy and uy = −vx .

52

2.4

2001 November 26

2

COMPLEX ANALYSIS

(b) Suppose u and v are continuous and f 0 (z) exists at the point z0 ∈ G. Then, for any  > 0 there is a δ > 0 such that B(z0 , δ) ⊆ G, and 0 f (z0 ) − f (z) − f (z0 ) < , for all |z − z0 | < δ. z − z0 Pick a triangular region ∆ ⊂ B(z0 , δ) with z0 ∈ ∆, and let T be the boundary. Define R(z) = f (z) − [f (z0 ) + f 0 (z0 )(z − z0 )]. R R R Then, by Cauchy’s theorem (part (a)), T [f (z0 ) + f 0 (z0 )(z − z0 )] dz = 0, whence T R(z) dz = T f (z) dz. Finally, note that R(z) 0 = f (z0 ) − f (z) − f (z0 ) < , for all |z − z0 | < δ. z − z0 z − z0 Therefore, Z Z Z f (z) dz = R(z) dz ≤ |R(z)| |dz| T T T Z R(z) |z − z0 | |dz| = T z − z0 Z ≤ |z − z0 | |dz| ≤ rL. T

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, Z f (z) dz ≤ 1 L2 . 2 T t u 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, Rouch´e’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 + a1 z + a2 z 2 + · · · + an z n with an 6= 0 for some n ≥ 1. Both proofs also rely on the following observation: If {aj }nj=0 ⊂ C with an 6= 0, then for all 1 ≤ R ≤ |z| < ∞, a0 −n z + · · · + an−1 z −1 ≤ |a0 | |z|−n + · · · + |an−1 | |z|−1 an |an | an |an | ≤ n max

|aj | −1 |z| |an |

≤ n max

|aj | −1 R . |an |

0≤j R/2 and |ζ − w2 | > R/2. Therefore, Z |w1 − w2 | |f (ζ)| |f (w1 ) − f (w2 )| ≤ |dζ| 2 2π |ζ−a|=R (R/2) ≤

|w1 − w2 | supγ |f (ζ)| `(γ) 2π R2 /4



4M |w1 − w2 |, R

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

t u

(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 ¯ R) ⊂ G. Then, by local boundedness, there is equicontinuity constant49 δ. So, fix a ∈ G and  > 0, and let B(a, ¯ 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 |w1 − w2 |, R

for all w1 , w2 ∈ {|w − a| ≤ R/2}.

R If δ = 4M  and r = R/2, then |f (w1 ) − f (w2 )| <  whenever w1 , w2 ∈ B(a, r) and |w1 − w2 | < δ. Therefore, F is equicontinuous in B(a, r). We have thus shown that local boundedness implies conditions (i) and (ii) of the Arzela-Ascoli theorem and thereby implies normality. t u

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(a , r ) : j = 1, . . . , n} by j j 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

2.5

2004 April 19

2

COMPLEX ANALYSIS

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.) R ϕ(w) Let F (z) = γ (w−z) dw, z ∈ C \ γ ∗ . R ϕ(w) Prove that F 0 (z) = γ (w−z) z ∈ C \ γ ∗ , without using Leibniz’s Rule. 2 dw, 2. (a) State the Casorati-Weierstrass theorem. (b) Evaluate the integral I=

1 2πi



Z (z − 3) sin |z|=R

1 z+2

 dz

where R ≥ 4.

3. Let f (z) be an entire function such that f (0) = 1, f 0 (0) = 0 and 0 < |f (z)| ≤ e|z|

for all z ∈ C

Prove that f (z) = 1 for all z ∈ C. Solution: I know of two ways to prove this. One can be found in Rudin’s Functional Analysis ([9], p. 250). The other goes as follows:50 By the Hadamard factorization theorem (see, e.g., [11]), an entire function f with zeros at {an } ⊂ C \ {0} and m zeros at z = 0 has the form  ∞  Y z P (z) m f (z) = e z ez/an , (46) 1− a n 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) = eBz+C , for some constants B, C. We are given that f (0) = 1 and f 0 (0) = 0, so eC = 1, and f 0 (0) = BeC = B = 0. It follows that f (z) = 1. t u

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

2.6

2006 November 13

2.6

2

COMPLEX ANALYSIS

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  > 0 and an M > 0 such that the deleted neighborhood B o , {z ∈ C : 0 < |z − z0 | ≤ } is contained in G and such that |f (z)| ≤ M for all z ∈ B o . Let

∞ X

f (z) =

an (z − z0 )n

n=−∞ o

be the Laurent expansion of f for z ∈ B , where an =

Z

1 2πi

f (ζ) dζ. (ζ − z0 )n+1

C

Here C denotes the positively oriented circle |ζ − z0 | = . Changing variables, ζ = z0 + eiθ the coefficients are an =

1 2π i

Z

dζ = i eiθ dθ



f (z0 + eiθ ) ieiθ dθ. (z0 + eiθ − z0 )n+1

1 2π

Z

0

Therefore, |an | ≤





0

M M d|θ| = n , n+1 

which makes it clear that, if n < 0, then |an | can be made arbitrarily small, by choosing a sufficiently small . This proves that an = 0 for negative n, and so f (z) =

∞ X

an (z − z0 )n .

n=0

Thus, f ∈ H(G).

t u

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 )m f (z)| remains bounded, then z0 is a pole of order m. t u

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 7→ 1 and ∞ 7→ −1, the positive imaginary axis is mapped to either the upper half-circle or the lower half-circle, and similarly for the negative imaginary axis. Checking that φ0 (i) = −i, it is clear that the positive imaginary axis is mapped to the lower half-circle {eiθ : −π < θ < 0}. Therefore, in mapping the right half-plane onto the unit disk, φ0 maps the first quadrant to the lower half-disk D ∩ Π− , and must therefore map the fourth quadrant to the upper half-disk. That is, φ0 : Q4 → D ∩ Π+ , where Q4 = {z ∈ C : Rez > 0, Imz < 0}. Next construct a mapping of the unit disk onto the fourth quadrant as follows: If φ1 (z) = iz, then φ1 ◦ φ0 : D → Π+ . Let φ2 (z) = 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+ . t u (b) (Riemann)52 Let Ω ⊂ C be a simply connected region such that Ω 6= 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 0 (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). up the precise statement of the Riemann mapping theorem. 53 A very similar problem appeared in April ’95 (6). 52 Look

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(b) In order to apply the Schwarz lemma, map the disk to the upper half-plane with the M¨oebius map φ : D → Π+ defined by 1−z φ(z) = i . 1+z φ

f

Then, φ(0) = i. Therefore, the function g = f ◦ φ : D − → Π+ − → D satisfies |g(z)| ≤ 1 and g(0) = f (φ(0)) = 0 f (i) = 0. By Schwarz’s lemma, then, |g (0)| ≤ 1. Finally, observe that g 0 (z) = f 0 (φ(z))φ0 (z), and then check that φ0 (0) = −2i. Whence, g 0 (0) = f 0 (φ(0))φ0 (0) = f 0 (i)(−2i), which implies 1 ≥ |g 0 (0)| = 2|f 0 (i)|. Therefore |f 0 (i)| ≤ 1/2. t u

4. (a) State Cauchy’s theorem and its converse. (b) Suppose that f is a continuous function defined on the entire complex plane. Assume that (i) f ∈ H(Π+ ∪ Π− ) (ii) f (¯ z ) = f (z) all z ∈ C. Prove that f is an entire function. Solution: (a) See theorems A.16 and A.18. (b) See Marsden and Hoffman.

5. (a) Define what it means for a family F ⊂ H(Ω) to be a normal family. State the fundamental theorem for normal families. (b) Suppose f ∈ H(Π+ ) and |f (z)| < 1 all z ∈ Π+ . Suppose further that lim t → 0+f (it) = 0. Prove that f (zn ) → 0 whenever the sequence zn → 0 and zn ∈ Γ where Γ = {z ∈ Π+ : |Rez| ≤ Imz}. Hint. Consider the functions ft (z) = f (tz) where t > 0. Solution: (a) Let Ω be an open subset of the plane. A family F of functions in Ω is called a normal family if every sequence of functions in F has a subsequence which converges locally uniformly in Ω. (The same definition applies when the family F happens to be contained in H(Ω).)54 I think of the Arzela-Ascoli theorem as the fundamental theorem for normal families. However, since the examiners asked specifically about the special case when F is a family of holomorphic functions, they probably had in mind the version of Montel’s theorem stated below, which is an easy consequence of the Arzela-Ascoli theorem.55 Theorem 2.5 (Arzela-Ascoli) Let F ⊂ C(Ω, S) be a family of continuous functions from an open set Ω ⊆ C into a metric space (S, d). Then F is a normal family if and only if (i) F is equicontinuous on each compact subset of Ω, and (ii) for each z ∈ Ω, the set {f (z) : f ∈ F} is contained in a compact subset of S. 54 Despite

the wording of the problem, the family need not satisfy F ⊂ H(Ω) in order to be normal. 5 (b) of the November 2001 exam asks for a proof of Montel’s theorem using the Arzela-Ascoli theorem.

55 Problem

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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 theorem56 ) 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 |zn |



  zn = f |zn | = f (zn ). |zn |

(47)

The numbers zn /|zn | lie in the compact set γ = {z ∈ Γ : |z| = 1}. Since fnk → g uniformly in γ, for any  > 0, there is a K > 0 such that |fnk (z) − g(z)| = |fnk (z)| < , for all k ≥ K and all z ∈ γ. That is, lim sup{|fnk (z)| : z ∈ γ} , lim kfnk kγ = 0,

k→∞

and, since znk /|znk | ∈ γ,

k→∞

  znk fn k |zn | ≤ kfnk kγ . k   znk ∴ lim fnk = 0. k→∞ |znk |

By (47), then, limk→∞ f (znk ) = 0. Finally, recall that f (zn ) → 0 iff every subsequence znj has a further subsequence znjk such that f (znjk ) → 0, as k → ∞. Now, if znj is any subsequence, then {f (znj )} is a normal family, and, repeating the argument above, there is, indeed, a further subsequence znjk such that f (znjk ) → 0. This completes the proof. t u 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.

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2007 April 16

Notation: C is the set of complex numbers, D = {z ∈ C : |z| < 1}, and, for any open set G ⊂ C, H(G) is the set of holomorphic functions on G. 1. Give the Laurent series expansion of

1 z(z−1)

Solution: f (z) =

in the region A ≡ {z ∈ C : 2 < |z + 2| < 3}.

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, ∞  n 1 1 1 1X 2 1 = = = z u−2 u (1 − 2/u) u n=0 u

converges for |u| > 2 and, substituting u = z + 2 in the last expression, we have n X ∞  ∞ −1  n+1 X 1 1 X 2 1 n −n−1 = = 2 (z + 2) = (z + 2)n , z z + 2 n=0 z + 2 2 n=−∞ n=0 converging for 2 < |z + 2|. Next, consider that ∞ 1 1 −1 −1 X  u n = = = z−1 u−3 3(1 − u/3) 3 n=0 3

converges for |u| < 3 and, substituting u = z + 2 in the last expression, we have ∞  n+1 X 1 1 =− (z + 2)n , z−1 3 n=0

converging for |z + 2| < 3. Therefore, ∞  n+1 −1  n+1 X X 1 1 1 1 n f (z) = − =− (z + 2) − (z + 2)n , z−1 z 3 2 n=−∞ n=0

for z ∈ A.

2.

t u

(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

R∞

1 dx. −∞ 1+x4

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Solution: Note that f (z) =

1 1 1 = 2 = , 1 + x4 (z + i)(z 2 − i) (z + eiπ/4 )(z − eiπ/4 )(z + ei3π/4 )(z − ei3π/4 )

which reveals that the poles of f in the upper half plane are at eiπ/4 and ei3π/4 . Let ΓR be the contour shown in the figure below; i.e., ΓR = g(R) ∪ [−R, R], where R > 1. Then, by the residue theorem, Z h i f (z)dz = 2πi Res(f, eiπ/4 ) + Res(f, ei3π/4 ) . (48) ΓR

The other two poles of f are in the lower half-plane, so both eiπ/4 and ei3π/4 are simple poles. Therefore, Res(f, eiπ/4 ) = Res(f, ei3π/4 ) =

lim (z − eiπ/4 )f (z) = z→eiπ/4

1 2eiπ/4 (eiπ/4

lim (z − ei3π/4 )f (z) = z→ei3π/4



ei3π/4 )(eiπ/4

+

ei3π/4 )

1 = − ie−iπ/4 , 4

1 1 = ie−i3π/4 . 4 2ei3π/4 (ei3π/4 − eiπ/4 )(ei3π/4 + eiπ/4 )

Plugging these into (48) yields   Z 1 −i3π/4 1 −iπ/4 π π f (z)dz = 2πi ie − ie = (e−iπ/4 − e−i3π/4 ) = √ . 4 4 2 2 ΓR It remains to show

Z lim f (z)dz = 0. R→∞ g(R)

Changing variables via z = Reiθ (0 ≤ θ ≤ π), Z Z π iReiθ πR f (z)dz = → 0, ≤ 4 iθ )4 g(R) 1 + (Re R −1 0

as R → ∞. t u

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.

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+ Solution: First consider57 φ1 (z) = 1−z = {z ∈ C : Rez > 0}. 1+z , which maps D onto the right half-plane P iπ/2 + + Let φ2 (z) = e 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): 1+i (φ3 ◦ φ2 ◦ φ1 )(0) = φ3 (i) = √ 2 √ √ so apparently we’re off by a factor of 2. This is easy to fix: let φ4 (z) = 2z. Then the holomorphic function f , φ4 ◦ φ maps D bijectively onto Q1 and f (0) = 1 + i, as desired. t u φ1 (0) = 1

5.



(φ2 ◦ φ1 )(0) = φ2 (1) = i



(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). 58 ¯ ¯ Solution: (i) Define φ(z) = 1/2−z 1−z/2 . This is a holomorphic bijection 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 (0)| ≤ 1, with equality in (a) for some z ∈ D or equality in (b) iff g(z) = eiθ z for some constant θ ∈ R. By condition (a),

1/2 ≥ |g(1/2)| = |f (φ(1/2))| = |f (0)|. t u (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 M¨ oebius 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 φ (z) = z−α . In addition to 12.4, α 1−αz ¯ sec. 12.5 and theorem 12.6 are popular exam questions. 59 There 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.

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Solution: First, recall that if A ⊂ G ⊂ C, then A is said to be open relative to G, or simply open in G, if for any a ∈ A there is a neighborhood B(a, ) = {z ∈ C : |z − a| < } such that B(a, ) ∩ G ⊂ A.60 Next, recall that a subset G ⊂ C is connected iff the only subsets of G that are both open and closed relative to G are the empty set and G itself. Equivalently, if there exist non-empty disjoint subsets A, B ⊂ G that are open in 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 , ) ⊂ G for some  > 0, and clearly B(z0 , ) ⊂ Ω. In particular, Ω 6= ∅ . If we can show Ω is both open and closed in G, then it will follow by connectedness that Ω = G, and the problem will be solved. (Ω is open) Let w ∈ Ω be connected to z0 by a parametric curve γ ⊂ G. Since G is open, ∃ > 0 such that B(w, ) ⊂ G. Clearly any w1 ∈ B(w, ) can be connected to z0 by a parametric curve (from w1 to w, then from w to z0 via γ) that remains in G. This proves that B(w, ) ⊂ Ω, so Ω is open. (Ω is closed) We show G \ Ω is open (and thus, in fact, empty). If z ∈ G \ Ω, then, since G is open, ∃δ > 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. t u

60 For

example, the set A = [0, 1], although closed in C, is open in G = [0, 1] ∪ {2}. To see the equivalence note that, in this case, A is open in G, as is Ac = G \ A = B, so A is both open and closed in G. Also it is instructive to check, using either definition, that G = [0, 1] ∪ {2} is disconnected. 61

67

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2007 November 16

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COMPLEX ANALYSIS

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 Z 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 (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 [email protected].

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Some problems of a certain type

2

COMPLEX ANALYSIS

Some problems of a certain type

Collected in this section are miscellaneous problems about such things as what can be said of a holomorphic (or harmonic) function when given information about how it behaves near a boundary or near infinity. Behavior near infinity 1. If f (z) is an entire function which tends to infinity as z tends to infinity, then f (z) is a polynomial. 2. If f (z) is an injective entire function, then f (z) = az + b for some constants a and b. 3. If u(z) is a nonconstant real valued harmonic function of C, then there is a sequence {zn } ⊂ C with zn → ∞ and u(zn ) → 0 as n → ∞. Behavior on or near the unit circle 4. If f (z) is holomorphic in an open set containing the closed unit disk, and if f (eiθ ) is real for all θ ∈ R, then f (z) is constant. 5. Prove or disprove: There exists a function f (z) holomorphic on the unit disk D such that |f (zn )| → ∞ whenever {zn } ⊂ D and |zn | → 1. 6. Prove or disprove: There exists a function u(z) harmonic on the unit disk D such that |u(zn )| → ∞ whenever {zn } ⊂ D and |zn | → 1. Other Problems 7. If f is holomorphic in the punctured disk {0 < |z| < R} and if Ref ≤ M for some constant M , then 0 is a removable singularity. 8. If f is holomorphic in the unit disk, with |f (z)| ≤ 1, f (0) = 0, and f (r) = f (−r) for some r ∈ (0, 1), then 2 z − r2 . |f (z)| ≤ |z| 1 − r2 z 2

Solutions 1. See (1b) of April ’95. 2. Suppose f ∈ H(C) is injective. Then f −1 is a continuous function in C which maps compact sets to compact sets. Therefore, if {zn } ⊂ C is any sequence tending to infinity, then the image f ({zn }) cannot remain inside any closed disk (since f −1 maps all such disks to closed bounded sets in C). Thus f (z) → ∞ whenever z → ∞. By the previous problem, f is a polynomial. Finally, if f has degree greater than one, or if f is constant, then f would not be injective. Therefore, f is a polynomial of degree one. t u

3. See (4) of April ’95. 4. See (1a) of April ’89. 69

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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 − } ⊂ 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. t u Please email comments, suggestions, and corrections to [email protected].

70

A

A

MISCELLANEOUS DEFINITIONS AND THEOREMS

Miscellaneous Definitions and Theorems

A.1

Real Analysis

A.1.1

Metric Spaces

The following theorem is found in Conway [3]. Theorem A.1 Let (X, d) be a metric space; then the following are equivalent statements: (a) X is compact; (b) Every infinite set in X has a limit point (in X); (c) X is sequentially compact (d) X is complete and for all  > 0 there exist {x1 , . . . , xn } ⊂ X such that X=

n [

B(xk , )

k=1

(The last property is called total boundedness.) A.1.2

Measurable Functions

Definition A.1 A complex function s on a measurable space X whose range consists of only finitely many points is called a simple function. If α1 , . . . , αn are the distinct values of a simple function, and if we set Ai = {x ∈ X : s(x) = αi }, then clearly s=

n X

αi χAi ,

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.262 Let Y and Z be topological spaces, and let g : Y → Z be continuous. (a) If X is a topological space, if f : X → Y is continuous, and if h = g ◦ f , then h : X → Z is continuous. (b) If X is a measurable space, if f : X → Y is measurable, and if h = g ◦ f , then h : X → Z is measurable. Proof: If V is open in Z, then g −1 (V ) is open in Y , and h−1 (V ) = (g ◦ f )−1 (V ) = f −1 (g −1 (V )). If f is continuous, then h−1 (V ) is open, proving (a). If f is measurable, then h−1 (V ) is measurable, proving (b). u t

Note, however, that measurable functions of continuous functions need not be measurable. Theorem A.363 Let u and v be real measurable functions on a measurable space X, let Φ be a continuous mapping of the plane into a topological space Y , and define h(x) = Φ(u(x), v(x)) Then h : X → Y is measurable. 62 Theorem 63 Theorem

1.7 of Rudin [8]. 1.8 of Rudin [8].

71

(x ∈ X).

A.1

A.1.3

Real Analysis

A

MISCELLANEOUS DEFINITIONS AND THEOREMS

Integration

There are a handful of results that are the most essential, and lay the foundation on which everything else is built. Rudin [8] gives a beautifully succinct and clear presentation of these in just seven pages (pp. 21–27).64 Some of these results are presented below, but do yourself a favor and read the master [8]. The following theorem65 is an essential ingredient of many proofs (e.g. the proof that simple functions are dense in Lp , 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 Z

Z fn dµ →

X

f dµ

as n → ∞.

X

Theorem A.6 (Fatou’s lemma) If fn : X → [0, ∞] (n = 1, 2, . . . ) is a sequence of positive measurable functions, then Z Z lim inf fn ≤ lim inf fn

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., R R (ii) gn → g < ∞, and (iii) |fn (x)| ≤ gn (x)

(x ∈ X; n = 1, 2, . . .), R R then f ∈ L1 (X, M, µ), fn → f , and kfn − f k1 → 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  > 0 there is a measurable subset A ⊆ E such that fn → f uniformly on A and µ(E \ A) < . 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

A.1

Real Analysis

A.1.4

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MISCELLANEOUS DEFINITIONS AND THEOREMS

Approximating Integrable Functions66

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) 6= 0} is compact). First, we present an important application of Rudin Theorem 1.17 (theorem A.4 above) and the dominated convergence theorem (DCT). Theorem A.9 Let S be the set of all complex, measurable, simple functions on X such that µ({x ∈ X : s(x) 6= 0}) < ∞. If 1 ≤ p < ∞, then S is dense in Lp (µ). Proof: Let 1 ≤ p < ∞, and f ∈ Lp (µ). Case 1. f ≥ 0. Let sn be as in Theorem 1.17. Then sn ≤ f implies sn ∈ Lp (µ) for all n = 1, 2, . . .R. Define gn = |f − sn |p . Then gn ∈ Lp (µ), n = 1, 2, . . . , gn → 0 as n → ∞, and gn ≤ |f |p ∈ L1 (µ). Therefore, by the DCT, gn → 0 as n → ∞. That is, kf − sn kp → 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 0 ≤ s1 (x) ≤ s2 (x) ≤ · · · ≤ f + (x) 0 ≤ t1 (x) ≤ t2 (x) ≤ · · · ≤ f − (x). Then, by Case 1, we have kf + − sn kp → 0

and

kf − − tn kp → 0

as n → ∞. Finally, note that kf − (sn − tn )kpp = kf + − f − − (sn − tn )kpp ≤ kf + − sn kpp + kf − − tn kpp . Case 3. f : X → C. This case follows from Case 2 once we split f up into real and complex parts.

u t

The following is a deep result whose proof depends on Urysohn’s lemma. (For the proof, see Rudin [8], section 2.4.) Theorem A.10 (Lusin’s Theorem) Fix  > 0. Let f be a complex measurable function which vanishes off a set of finite measure. Then there exists g ∈ Cc (X) such that µ({x ∈ X : f (x) 6= g(x)}) < . Furthermore, we may arrange it so that sup ∈ X|g(x)| ≤ sup |f (x)| 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 Lp (µ). 66 This

topic is very important and appears on the comprehensive exam syllabus.

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MISCELLANEOUS DEFINITIONS AND THEOREMS

Proof: Define S as in theorem A.9. If s ∈ S and  > 0, then by Lusin’s Theorem there exists g ∈ Cc (X) such that g(x) = s(x) except on a set of measure < , and |g| ≤ ksk∞ . Hence, kg − skp ≤ 21/p ksk∞ . Since S is dense in Lp (µ), this completes the proof.

u t

We close out this subsection by giving a careful solution to a basic but important exercise from Rudin [8], Chapter 2. First, define a step function to be a finite linear combination of characteristic functions of bounded intervals in R1 . (Notice how much more special these functions are than the simple functions, defined above.) Lemma A.1 (Rudin Exercise 2.24) If f ∈ L1 (R) then there exists a sequence {gn } of step functions on R such that lim |f − gn | = 0.

n→∞

Proof: We proceed in steps of successively greater generality. (I’ll fill in the details soon!) Case 1. f = χ set A with µ(A) < ∞. PA for some measurable 1 Case 2. f = n i=1 αi χAi ∈ L (R). N.B. the assumption that f is integrable implies, in particular, that each Ai is measurable with µ(Ai ) < ∞. Case 3. f ∈ L1 (R), with f ≥ 0. Case 4. f ∈ L1 (R). (Details coming soon!)

A.1.5

Absolute Continuity of Measures

Two excellent sources for the material appearing in this section are Rudin [8] (§ 6.7, 6.10) and Folland [4] (§ 3.2). Let µ be a positive measure on a σ-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 λa  µ,

λ = λa + λs ,

λs ⊥ µ.

If λ is positive and finite, then so are λa and λs . (b) There is a unique h ∈ L1 (µ) such that Z h dµ ∀E ∈ M.

λa (E) = E 67 Rudin[8],

6.10.

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Real Analysis

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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.168 Suppose ν is a σ-finite complex measure and µ, λ are σ-finite measures on (X, M) such that ν  µ  λ. Then dν (a) If g ∈ L1 (ν), then g dµ ∈ L1 (µ) and

Z

Z g dν =

g

dν dµ. dµ

(b) ν  λ, and dν dν dµ = dλ dµ dλ A.1.6

λ-a.e.

Absolute Continuity of Functions

Lemma 1.2 Let f : R → R be a function. If f is differentiable on [a, b], f 0 ∈ L1 ([a, b]), and f (x) − f (a) for a ≤ x ≤ b, then f ∈ AC[a, b].

Rx a

f 0 (t)dt =

Proof Assuming the stated hypotheses, by a standard theorem,69 f 0 ∈ L1 implies that for all  > 0 there is a δ > 0 such that, if E ⊂ R is measurable mE < δ, then Z |f 0 |dm < . (49) E

Let A mA ≤

= ∪ni=1 (ai , bi ) be a finite P n i=1 (bi − ai ) < δ, so n X i=1

− ai ) < δ. Then

Z n Z bi n Z bi X X 0 |f 0 |dm = |f 0 |dm <  f dm ≤ |f (bi ) − f (ai )| = ai ai A

(50)

i=1

i=1

by (49). Thus, f ∈ AC[a, b]. A.1.7

Pn

i=1 (bi

union of disjoint open intervals in [a, b] such that

t u

Product Measures and the Fubini-Tonelli Theorem

Let (X, S, µ) and (Y, T , ν) be measure spaces. If we want to construct a measurable space out of X × Y , it is natural to start by considering the collection of subsets S × T = {A × B ⊆ X × Y : A ∈ S, B ∈ T }. Note, however, that this collection is not, in general, an algebra of sets. To get an adequate collection on which to define product measure, then, define70 S ⊗ T = σ(S × T ); that is, S ⊗ T is the σ-algebra generated by S × T . In my opinion, the most useful version of the Fubini and Tonelli theorems is the one in Rudin [8]. It begins by assuming only that the function f (x, y) is measurable with respect to the product σ-algebra S ⊗ T . Then, in a single, combined Fubini-Tonelli theorem, you get everything you need to answer any of the standard questions about integration with respect to product measure. Here it is: 68 Folland

[4], Prop. 3.9. “standard Rtheorem” cited here appears often on the comprehensive exams (cf. Nov. ’91 #6), but in a slightly weaker form in which the R conclusion is that | E f 0 dm| < . In the present case we need E |f 0 |dm <  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 }. 69 The

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Real Analysis

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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 . R R (a) If f (x, y) ≥ 0, and if φ(x) = Y f (x, y) dν(y) and ψ(y) = X f (x, y) dµ(x), then φ is S-measurable, ψ is T -measurable, and Z Z Z φ dµ = f (x, y) d(µ × ν) = ψ dν. (51) X

X×Y

(b) If f : X × Y → C and if one of Z Z |f (x, y)| dµ(x) dν(y) < ∞ Y

Y

Z Z |f (x, y)| dν(y) dµ(X) < ∞

or

X

X

Y

holds, then so does the other, and f ∈ L1 (µ × ν). (c) If f ∈ L1 (µ × ν), then, (i) for almost every x ∈ X, f (x, y) ∈ L1 (ν), (ii) for almost every y ∈ Y, f (x, y) ∈ L1 (µ), R (iii) φ(x) = Y f dν is defined almost everywhere (by (i)), moreover φ ∈ L1 (µ), R (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.2

Complex Analysis

A.2.1

Cauchy’s Theorem71

A

MISCELLANEOUS DEFINITIONS AND THEOREMS

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 P [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 Z dz 1 n(γ; w) = 2πi γ 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 ∗ , Z f (ζ) 1 dζ. f (z)n(γ; z) = 2πi γ ζ − z 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 X

m

n(γj , z) =

j=1

1 X 2πi j=1

Z γj

f (ζ) dζ. ζ −z

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 ⊆ RC be an open set and suppose f ∈ H(G). If γ is a closed rectifiable curve that is homologous to zero in G, then γ f (z)dz = 0. 71 Most

of the material in this section can be found in Conway [3].

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Complex Analysis

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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 Z X j=1

f (z) dz = 0.

γj

A partial converse of Cauchy’s theorem is the following: Theorem A.18 Let R G be an open set in the plane and f ∈ C(G, C). Suppose, for any triangular contour T ⊂ G with T ≈ 0 in G, that T f (z)dz = 0. Then f ∈ H(G). This theorem is still valid (and occasionally easier to apply) if we replace “any triangular contour” with “any 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 ¯ ∩ H(G), then Theorem A.20 (max mod principle, ver. 2) If G ∈ C is open and bounded, and if f ∈ C(G) ¯ 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. ˆ = C ∪ {∞} be open, let f ∈ H(G), and suppose there is an Theorem A.21 (max mod principle, ver. 3) Let G ⊂ C 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 (0)| ≤ 1, with equality in (a) for some z ∈ D \ {0} or equality in (b) iff f (z) = eiθ z for some constant θ ∈ R. Please email comments, suggestions, and corrections to [email protected].

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 Rouch´e’s theorem. Of course, you should know the statements of all of these theorems and, since proving them in this sequence is not hard, you might as well know the proofs too! Two excellent references giving clear and concise proofs are Conway [3] and Sarason [10].

78

REFERENCES

B

List of Symbols

F Q Z N C ˆ C R T ˆ R Rez Imz D or U H(G) Π+ Π− P+ P− ∂∞ G C[0, 1] L1 Lp L∞ S ⊗T

an arbitrary field the rational numbers the integers the natural numbers, {1, 2, . . .} the complex numbers (a.k.a. the complex plane) the extended complex plane, C ∪ {∞} the real numbers (a.k.a. the real line) the unit circle, {z ∈ C : |z| = 1} the extended real line, [−∞, ∞] the real part of a complex number z ∈ C the imaginary part of a complex number z ∈ C the open unit disk, {z ∈ C : |z| < 1} the holomorphic functions on an open set G ⊂ C the upper half-plane, {z ∈ C : Imz > 0} the lower half-plane, {z ∈ C : Imz < 0} the right half-plane, {z ∈ C : Rez > 0} the left half-plane, {z ∈ C : Rez < 0} ˆ the extended boundary of a set G ⊂ C the space of continuous real valued functions on [0, 1]. R the space of integrable functions; i.e., measurable f such that R|f | < ∞. for 0 < p < ∞, the space of measurable functions f such that |f |p < ∞. the space of essentially bounded measurable functions; i.e., measurable f such that {x : |f (x)| > M } has measure zero for some M < ∞. the product σ-algebra generated by S and T .

References [1] Lars Ahlfors. Complex Analysis. McGraw-Hill, New York, 3rd edition, 1968. [2] Charalambos D. Aliprantis and Owen Burkinshaw. Principles of Real Analysis. Academic Press, New York, 3rd edition, 1998. [3] John B. Conway. Functions of One Complex Variable I. Springer-Verlag, New York, 2nd edition, 1978. [4] Gerald B. Folland. Real Analysis: Modern Techniques and Their Applications. John Wiley & Sons Ltd, New York, 1999. [5] James R. Munkres. Topology: A First Course. Prentice Hall International, Englewood Cliffs, NJ, 1975. [6] H. L. Royden. Real Analysis. Macmillan, New York, 3rd edition, 1988. [7] Walter Rudin. Principles of Mathematical Analysis. McGraw-Hill, New York, 3rd edition, 1976. [8] Walter Rudin. Real and Complex Analysis. McGraw-Hill, New York, 3rd edition, 1987. [9] Walter Rudin. Functional Analysis. McGraw-Hill, New York, second edition, 1991. [10] Donald J. Sarason. Notes on Complex Function Theory. Henry Helson, 1994. [11] Elias Stein and Rami Shakarchi. Complex Analysis. Princeton University Press, 2003.

79

Index absolute 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 `p , 35 of bounded linear operators, 26 Banach-Steinhauss theorem, 22, 35 Borel σ-algebra, 34 Borel set, 34 Casorati-Weierstrass theorem, 54 applied, 47, 58 Cauchy’s formula, 75 applied, 39, 47, 56 Cauchy’s theorem, 75–76 converse of, see Morera’s theorem problems, 44, 45, 48, 52, 61 proof by Green’s theorem, 51–52 Cauchy-Riemann equations, 42, 51 closed graph theorem, 30 conformal mapping problems, 39, 44, 50, 60, 66, 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

applied, 8, 13 fundamental theorem of algebra, 52 fundamental theorem of calculus, 16 Green’s theorem, 51 H¨older’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 80

INDEX

INDEX

Riemann mapping theorem, 60 Riesz representation theorem, 35–36 applied, 16–17 for Lp , 22 Rouch´e’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