Problems and Solutions in EAL AND COMPLEX ANALYSIS

1 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 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 exams. In solving many of these Problems , I benefited enormously from the wisdom and guidance of professors Tom Ramsey and Wayne Smith.

2 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 Contents 1 Real ANALYSIS 3. 1991 November 21 .. 3. 1994 November 16 .. 10. 1998 April 3 .. 12. 2000 November 17 .. 18. 2001 November 26 .. 20. 2004 April 19 .. 27. 2007 November 16 .. 32. 2 COMPLEX ANALYSIS 38. 1989 April .. 39. 1991 November 21 .. 43. 1995 April 10 .. 48. 2001 November 26 .. 52. 2004 April 19 .. 59. 2006 November 13 .. 60. 2007 April 16 .. 64. 2007 November 16 .. 68. Some Problems of a certain type .. 69. 1. A Miscellaneous Definitions and Theorems 71.

3 Real ANALYSIS .. 71. Metric Spaces .. 71. Measurable Functions .. 71. Integration .. 72. Approximating Integrable Functions1 .. 73. Absolute Continuity of Measures .. 74. Absolute Continuity of Functions .. 75. Product Measures and the Fubini-Tonelli Theorem .. 75. COMPLEX ANALYSIS .. 77. Cauchy's Theorem2 .. 77. Maximum Modulus Theorems .. 78. B List of Symbols 79. 2. 1 REAL ANALYSIS . 1 Real ANALYSIS 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.

4 Fix > 0 and let N0 N be such that n N0 implies |fn (x) f (x)| < /2, for all x [0, 1]. Let > 0 be such that |f (x) f (y)| < /2, for all x, y [0, 1] with |x y| < . Finally, let N1 N be such that n N1 implies |xn 1/2| < . 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, if 0 x < 21 2n 1. , f (x) = 2nx (n 1), 1. if 2 2n x < 12 , 1. (1). if 12 x 1.. 1, . That is, fn (x) is constantly zero for x less than 21 2n 1. , then it increases linearly until it reaches one at x = 1/2, and then it remains constantly one for x bigger than 1/2.

5 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 , [ 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 ) there is a limit point x S; , there is a sequence {xn } of distinct points in S.

6 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 (x + (xn x)) f (x) f (xn ) f (x). f 0 (x) = lim = lim = 0. n xn x n xn x The last equality holds since f (x) = f (xn ) = 0 holds for all n N. t u 3. If (X, , ) is a measure space and if f is integrable, show that for every > 0 there is E such that (E) < and Z. |f | d < . X\E. 3. 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)| = }.]

7 Then X = A0 A A is a disjoint union, and Z Z Z Z Z. |f | = |f | + |f | + |f | = |f |. (2). X A0 A A 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 ( ). implies (A ) = 0. To prove the result, then, we must find a measurable set E such that A\E |f | < , and (E) < . Define fn = |f | An . Then {fn } is a sequence of non-negative measurable functions and, for each x X, limn fn (x) = |f (x)| A (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 | d = |f | d = |f | d . n An A X. Therefore, there is some N > 0 for which Z. |f | d < . X\AN. Finally, note that 1/N |f | < N on AN , so Z Z.

8 (AN ) N |f | d N |f | d < . AN X. Therefore, the set E = AN meets the given criteria. t u R. If (X, , ) is a measure space, f is a non-negative measurable function, and (E) = E. f d , show that is a measure. Solution: Clearly (E) = 0 (E) = 0. Therefore, ( ) = ( ) = 0. In particular is not identically infinity, so we need only check countable additivity. Let {E1 , E2 , ..} be a countable collection of disjoint measurable sets. 3 Answer: f is measurable and x |x| is continuous, so g = |f | is measurable. Therefore, An = g 1 ([1/n, n)) is measurable (theorem ). 4 Alternatively,we could have cited the dominated convergence theorem here since fn (x) |f (x)| (x X; n = 1, 2, ..). 5 See also: Rudin [8], chapter 1.]

9 Thanks to Matt Chasse for pointing out a mistake in my original solution to this problem . I believe the solution given here is correct, but the skeptical reader is encouraged to consult Rudin. 4. 1991 November 21 1 REAL ANALYSIS . Then, Z Z. ( n En ) = S. f d = f Sn En d . En n Z . X. = f En d ( the En are disjoint). n=1. Z. X. = f En d ( f En 0, n = 1, 2, ..). n=1. X . = (En ). n=1. The penultimate equality follows Pm from the monotone convergence theorem applied to the sequence of non-negative measurable functions gm = n=1 f En (m = 1, 2, ..). (See also: April '98, problem ) t u 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 dm = inf dm where m is Lebesgue measure on [0, 1], and and range over all simple functions, f.

10 Solution: This is proposition of Royden, 3ed. [6]. If f is Lebesgue integrable on [0, 1] and > 0, show that there is > 0 such that for all measurable sets E [0, 1]. with m(E) < , 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.