FACULTY OF SCIENCE FINAL EXAMINATION MATHEMATICS MATH 355. Analysis 4. Examiner: Professor S. W. Drury Date: Tuesday, April 18, 2006 INSTRUCTIONS

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1 FACULTY OF SCIENCE FINAL EXAMINATION MATHEMATICS MATH 355 Analysis 4 Examiner: Professor S. W. Drury Date: Tuesday, April 18, 26 Associate Examiner: Professor K. N. GowriSankaran Time: 2: pm. 5: pm. INSTRUCTIONS Attempt six questions for full credit. This is a closed book examination. Write your answers in the booklets provided. All questions are of equal weight, each is alloted 2 marks. This exam has 7 questions and 9 pages 1

2 1. (i (5 points Define the concept σ-field. (ii (5 points Define the concept measure on a σ-field. (iii (5 points Let Y be a subset of X. For any family M of subsets of X, denote by M Y the family M Y {M Y ; M M} of subsets of Y. If M is a σ-field of subsets of X, show that M Y is σ-field on Y. Note that you are not allowed to assume that Y M. (iv (5 points If µ is a measure on (Y, M Y, does ν(m µ(m Y define a measure on (X, M? Justify your answer. (i Let X be a set. Then a collection F of subsets of X is a σ-field if and only if (a X F. (b A F X \ A F. (c A k F for k K, K countable k K A k F. (ii A measure on a σ-field F of subsets of X as a function µ : F [, ] such that (a µ(. (b µ ( k K A k k K µ(a k whenever K is a countable index set and A k are pairwise disjoint subsets of X with A k F and k K A k F. (iii (a Since X M, Y X Y M Y. (b Let B M Y. Then there exists A M such that B A Y. But Y \B (X\A Y M Y, since (X \ A M. (c Let B k M Y for k N, then there exist A k M with B k A k Y. But now we find B k ( A k Y and it follows that B k M Y since A k M. (iv (a ν( µ(. (b Let A k be pairwise disjoint subsets of X in M, then B k A k Y are pairwise disjoint subsets of Y in M Y. Further, as in (c above B k ( A k Y and consequently ( (( ( ν A k µ A k Y µ B k µ(b k ν(a k. 2

3 2. Let (X, M, µ be a measure space. (i (5 points Under what conditions can one define f(xdµ(x for a signed M-measurable function f on X? In this case give the definition in terms of the integral of nonnegative M- measurable functions on X. Let g be a nonnegative M-measurable function on X satisfying g(xdµ(x <. (ii (5 points Prove Tchebychev s inequality µ({x; g(x > t} 1 g(xdµ(x for t >. t (iii (5 points If f L p (X, M, µ where 1 p <, show that the inequality µ({x; f(x > t} t p f p p holds for every t >. (iv (5 points For every p with 1 p <, find a measure space (X, M, µ and a measurable function f on (X, M such that µ({x; f(x > t} t p holds for every t > but f / L p (X, M, µ. (i The integral is only defined if f(x dµ(x < and in this case we set f(xdµ(x f + (xdµ(x f (xdµ(x where f ± (x max(, ±f(x. (ii Let E t {x; g(x > t}, then we have g(xdµ(x t11 Et dµ(x tµ(e t, so that µ(e t 1 g(xdµ(x as required. t (iii Applying the previous part of the question to g f p we have, since {x; g(x > t p } {x; f(x > t} that µ({x; f(x > t} µ({x; g(x > t p } t p g(xdµ(x f p pt p. (iv Put f(x x 1 p measure exactly t p, but on ], [ with Lebesgue measure and then {x; f(x > t} ], t p [ has f(x p dx x 1 dx, so f / L p (], [. 3

4 3. (2 points State whether the following statement is true or false and give either a proof or an explicit counterexample. Every closed subset of the real line with empty interior has zero Lebesgue measure. This is false. The easiest solution is as follows. Let (q k be an enumeration of the points of Q [, 1]. Let X ]q k 2 2 k, q k k [. Then X is a union of open intervals and hence is open. The measure of X is bounded by 2 1 k 1. Hence [, 1] \ X is a closed set of measure 2 at least 1. But if [, 1] \ X contains an interval of strictly positive length, there is a rational 2 number q [, 1] \ X. But this is impossible since then q q k X for some k N. 4

5 4. Let (X, M, µ be a measure space. (i (1 points Let (f n n1 be a sequence of real-valued M-measurable functions on X. Let A {x X; lim n f n (x exists}. Show from first principles that A M. (ii (1 points If A n M for n 1, 2, 3,... and if µ(a n <, show that µ-almost every point of X lies in only finitely many of the A n. n1 (i We show first that if (f n n1 is a sequence of M-measurable functions on X taking values in the extended real line [, ], then sup n f n is also M-measurable. This is a consequence of the identity {x; sup n1 f n (x > t} {x; f n (x > t} and the fact that the intervals ]t, ] generate all Borel subsets of [, ]. An exactly similar argument shows that inf n f n is M-measurable. It now follows that lim sup n f n (x inf sup m1 nm n1 f n (x and lim inf n f n(x sup inf f n(x nm are also M-measurable. Now, interpreting the phrase lim n f n (x exists to mean that the limit exists in R (as opposed to [, ], we see that X \ A {x; lim sup f n (x } {x; lim inf f n(x } {x; lim inf n n n m1 f n(x < lim sup f n (x}. n The first two of these sets are seen to be measurable and the third can be written {x; lim inf f n(x < lim sup f n (x} ( {x; lim inf f n(x < q} {x; q < lim sup f n (x} n n n n q Q and hence is also measurable. It follows that X \ A and hence A is measurable. (ii The set E of x that belong to infinitely many of the A n can be written as Thus, for all m N, since µ(e µ ( nm A n nm µ(a n m, m1 nm µ(a n <. It follows that µ(e. So, for almost all x, x lies in only finitely many n1 of the A n. 5 A n.

6 5. Let µ be Lebesgue measure on [, 1]. (i (6 points Show that the functions f k (t 2 cos(kπt form an orthonormal set in the Hilbert space H L 2 ([, 1], B [,1], µ as k runs over the positive integers. (ii (4 points Let M be the closed linear span of {f k ; k 1, 2, 3...}. Find g H, such that g f k for all k N and deduce that M is not the whole of H. (iii (6 points Let f(t 2 sin(πt and P the orthogonal projection onto M. Find a series expansion (convergent in H for P (f in the form P (f a k f k computing the a k explicitly. (iv (4 points By comparing the norms of f and P (f, show that f / M. (i We have and for k l f k f k, f l (cos(kπt 2 dt 1 cos(kπt cos(lπtdt [ sin((k lπt + (k lπ ( [ 1 + cos(2kπt dt t + sin(2kπt ] 1 1 2kπ 1 sin((k + lπt (k + lπ showing that (f k is an orthonormal set. (ii Let g 11 [,1] H and then it is clear that f k, g 2 1 ( cos((k lπt + cos((k + lπt dt ] 1 cos(kπtdt [ sin(kπt 2 kπ so that g f k for all k N. Extending by linearity and continuity, we see that g M, that g / M and that M H. 6 ] 1

7 (iii We obtain a k f, f k 2 1 cos(kπt sin(πtdt 1 [ cos((k 1πt cos((k + 1πt (k 1π (k + 1π { 4 if k is even (k 2 1π if k is odd ( sin((k + 1πt sin((k 1πt dt ] 1 2(1 + ( 1k (k 2 1π Now we also have 1 1 ( [ f (sin(πt 2 dt 1 cos(2πt dt t sin(2πt ] 1 1 2π So, that if f M then f P (f and f P (f. We can compute (using Maple l1 P (f 2 a k 2 l1 16 π 2 (4l 2 1 π2 8 < 1 f 2 2 π 2 and it follows that f / M. The exact summation is difficult to arrive at and is easily circumvented with estimates. We have 4l 2 1 3l 2 for l 1, leading to 16 π 2 (4l l 4 16 ( 1 + x 4 dx π 2 9π 2 27π < 1. 2 l1 using the integral test to make the estimation (i (5 points State Fubini s Theorem. (ii (5 points State the Dominated Convergence Theorem. In the remaining part of this question, you may assume without proof that for t > and a > 2π. a 2π e tx cos(xdx (sin(a t cos(ae ta + te 2πt t (iii (1 points By considering the iterated integral x2π a x2π cos(x te 2πt dx x t t dt cos(x e tx dt dx, show that t where the integral on the left is treated as an improper integral. Be sure to justify all steps. 7

8 (i Let (X, S, µ and (Y, T, ν be σ-finite measure spaces. If any one of the three quantities f(x, y dν(ydµ(x f d(µ ν f(x, y dµ(xdν(y. is finite, then f(x, ydν(ydµ(x fd(µ ν f(x, ydµ(xdν(y. (ii Let f n be a sequence of measurable functions and suppose that f n f pointwise. Further suppose that there is a (nonnegative function g such that f n g pointwise for every n N. If gdµ <, then necessarily f n dµ fdµ. n (iii We leave a fixed for the moment. Then a x2π cos(x so from Fubini s Theorem we have a x2π cos(x dx x t a e tx dt dx x2π a t t a cos(x x2π x2π t cos(x dx < x e tx dt dx cos(xe tx dx dt (sin(a t cos(ae ta + te 2πt dt t cos(x Therefore, bearing in mind that dx is interpreted as an improper integral (note that x2π x it is not defined as a lebesgue integral, it suffices only to show that t (sin(a t cos(ae ta t dt a. We could use dominated convergence here, but it is quite easy to use hard estimates (sin(a t cos(ae ta dt t t + 1 t e ta dt t since 2(t (t + 1 2(t t t 2e ta dt 2a 1 a.

9 7. Let f be an L 1 function on the circle group T for the normalized linear measure η. For each of the following properties of f determine an equivalent condition on the sequence ( ˆf(n n Z of Fourier coefficients and briefly justify your answer. (i (7 points f is real-valued. (ii (6 points f L 2 (T, η. (iii (7 points f has an infinitely differentiable version. (i We have f ˆ(n f(te int dη(t [ ˆf( n ] f(te int dη(t So, if f is real, then f f and ˆf(n ˆf( n. Conversely, if ˆf(n ˆf( n then f and f have the same Fourier transform and by the uniqueness theorem, they must agree (almost everywhere. Hence f is real (almost everywhere. (ii Since the exponential system e n (t e int (n Z is an orthonormal basis for L 2 (T, we have f 2 2 n Z ˆf(n 2. A necessary and sufficient condition for f L 2 is then n ˆf(n 2 <. (iii We have from integration by parts (f ˆ(n f (te int dη(t in f(te int dη(t in ˆf(n provided that f and f are continuous. Thus, if f is k times continuously differentiable, we have ( ( k d ˆ f (n (in k ˆf(n dt and since the k th derivative is in L 2 this gives n 2k ˆf(n 2 <. It follows that for every k N, n Z sup n Z n k ˆf(n <. Conversely, if for every k N, sup n Z n k ˆf(n <, then replacing k with k + 2 we find n k ˆf(n <. Consequently all the Fourier series n Z(in l ˆf(ne int for l, 1,..., k converge n Z absolutely and uniformly and standard results from MATH 255 allow one to conclude that is continuous. ( k d f dt 9