Problems 1(a)(b)(c), 2(a)(b)(c), 3(a)(b)(c)(d)(e), 4(a)

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1 Problems 1(a)(b)(c), 2(a)(b)(c), 3(a)(b)(c)(d)(e), 4(a) Bob Lutz MATH 138 Final xamination 03/06/2012 Problem 1. Suppose that F and G are increasing functions on R. Recall that µ F is the unique Borel measure for which µ F (a, b] F(b + ) F(a + ) for all half-open intervals. Recall also that µ F {a} F(a + ) F(a ). (a) Show that F(x + ) dg(x) + G(x ) df(x) F(x ) dg(x) + G(x + ) df(x). Do this by applying Cavalieri s Theorem to the portion of the diagonal y x with x in [a, b]. (b) Show that F and G have no common points of discontinuity if and only if F(x+ ) dg F(x ) dg for all Borel sets. (c) Use part (b) to write out the simplified version of the integration by parts formula for closed intervals which holds when F and G have no common points of discontinuity. What if, in addition, F and G are everywhere right continuous? (a) Let {(x, x) : a x b}. By Cavalieri s theorem, we have µ F µ G ( ) µ G ( x ) df(x) µ G ({x}) df(x) ( G(x + ) G(x ) ) df(x) Similarly, We conclude that G(x + ) df(x) µ F µ G ( ) G(x ) df(x) µ F ( y ) dg(y) µ F ({y}) dg(y) ( F(y + ) F(y ) ) dg(y) F(x + ) dg(x) from which the desired inequality follows after rearranging terms. F(x ) dg(x), 1

2 (b) Observe that F and G have no common points of discontinuity iff µ F µ G ({(x, x) : x X}) 0, which occurs iff µ F µ G ({(x, x) : x }) 0 for all Borel X. Let {(x, x) : x }, so that µ F µ G ( ) 0. By Cavalieri s theorem, we have F(x + ) dg(x) F(x ) dg(x) (F(x + ) F(x )) dg(x) µ F ({x}) dg(x) F(( ) x ) dg(x) µ F µ G ( ) 0. Hence F(x+ ) dg(x) F(x ) dg(x), as desired. (c) Due to (b), if F and G have no common points of discontinuity, then since [a, b] B X, we may simplify the integration by parts formula for closed intervals to F(x) dg(x) + G(x) df(x) F(b + )G(b + ) F(a )G(a ). If F and G are everywhere right continuous, then we have F(x) F(x + ) and G(x) G(x + ) for all x X, whence the formula becomes F(x) dg(x) + G(x) df(x) F(b)G(b) F(a )G(a ). 2

3 Problem 2. The Fourier transform of a function f in L 1 (R) is defined by ˆf (t) f (x)e itx dx. (a) Prove that ˆf is a continuous function and that ˆf u f 1. (b) Prove that lim t ˆf (t) 0. (c) Suppose that f and g belong to L 1 (R), so that the convolution f g also belongs to L 1 (R). Prove that ( f g)(t) ˆf (t)ĝ(t). (a) Note that f (x)e itx f (x) e itx f (x) for a.e. x. If t t 0, then f (t) f (t 0 ) a.e.; thus, since f 0 is integrable, ˆf (t) ˆf (t 0 ) by dominated convergence. Hence f (t) is continuous. Moreover, for t R, ˆf (t) f (x)e itx dx f (x)e itx dx f 1, whence ˆf u f 1. (b) Let a < b and f χ. Then ˆf (t) b a e itx dx eita e itb it 0, as t. The result follows if f is a simple function with standard representation f n 1 c jχ [aj,b j ]. Since such functions are dense in L 1, for all ε > 0 and f L 1 (R), there exists a simple function g such that f g 1 < ε 2. Furthermore, since ˆf (t) 0 as t, there exists t 0 > 0 such that if t > t 0, then g(t) < ε 2. Hence if t > t 0, then ˆf (t) ˆf (t) ĝ(t) + ĝ(t) ( f (x) g(x))e itx dx + ĝ(t) f g 1 + ĝ(t) < ε. Since ε > 0 was arbitrary, the result follows. (c) Observe that [ ( f g)(t) ] f (y)g(x y) dy e itx dx. 3

4 Substituting u x y, we have ( f g)(t) ˆf (t)ĝ(t), f (y)g(u)e it(u+y) dy du f (y)e ity g(u)e itu dy du f (y)e ity dy g(u)e itu du as desired. 4

5 Problem 3. Prove the following. (a) If f and g are measurable functions on X, then f g 1 f 1 g. If f L 1 and g L, f g 1 f 1 g iff g(x) g a.e. on the set where f (x) 0. (b) is a norm on L. (c) f n f 0 if and only if there exists M such that µ( c ) 0 and f n f uniformly on. (d) L is a Banach space. (e) The simple functions are dense in L. (a) Suppose that f and g are measurable functions on X. Since {x : g(x) > g } is null, we have f g 1 f g f g sup g(x) f g f f 1 g. c x c A c Suppose that f L 1 and g L. Note that f g 1 f 1 g if and only if f g f g. (1) Thus f (x) g(x) f (x) g for a.e. x, so that if f (x) 0, then dividing through yields the result. If, on the other hand, g(x) g a.e. on the set where f (x) 0, then certainly f g f g a.e., so (1) follows. (b) Since f (x) + g(x) f (x) + g(x) for all x X, we have f + g ess sup f (x) + g(x) x X ess sup f (x) + ess sup g(x) x X x X f + g, proving the triangle inequality. If f 0, then certainly f 0. If g 0, then g 0 a.e., so f 0 and g define the same element of L. Hence is nondegenerate, and is therefore a norm. (c) Suppose that f n f 0. Define n {x : f n (x) f (x) f n f } and n N n. Notice that µ( c ) µ( n N n) c n N n c 0, so µ( c ) 0. Since f n f 0, given ε > 0, there exists N N such that if n N, then f n f < ε. Thus for all x and n N, we have f n (x) f (x) f n f < ε, proving that f n f uniformly on. Suppose on the other hand that f n f uniformly on M such that µ( c ) 0. Given ε > 0, there exists N N such that if n N, then f n (x) f (x) < ε for all x. Thus µ({x : f n (x) f (x) ε}) 0, so f n f < ε. Hence f n f 0. 5

6 (d) Let { f n } n1 L be Cauchy with respect to. We may assume that f n (x) f n for all x X, since we identify f n and f n χ, where {x : f n (x) f n }. Then, fixing x X, we have f m (x) f n (x) ( f m f n )(x) f m f n for all m, n N. Thus { f n (x)} R is Cauchy and converges to a point p x R. Define f : X R by f (x) p x for all x X. Note that f (x) f n (x) f (x) + f n (x) for all n N and x X. Since f n (x) f (x) and f n is bounded for all n N, it follows that f is bounded as well. Given ε > 0, since { f n } is Cauchy, there exists N N such that if m, n N, then f m f n < 2 ε. Thus f m(x) f n (x) < 2 ε for all x X. Since the maps (a, b) a b and c c are continuous, we have lim m f m (x) f n (x) f (x) f n (x) 2 ε < ε for all x X, whence f n f < ε as well. Thus f n f with respect to, proving that L is complete. (e) Let f L be bounded and A {x : f (x) f }. Since f is measurable, A f 1 ([ f, f ]) is measurable. Thus, given ε > 0, for all k Z the set k A f 1 (((k 1)ε, kε]) is measurable as well. Hence the simple function ϕ kε f kεχ k has the property that ϕ(x) f (x) < ε for all x A. Since µ(a c ) 0, we have ϕ f < ε, proving that the simple functions are dense in L. 6

7 Problem 4. (a) Make reasonable definitions of almost everywhere Cauchy and of almost uniformly Cauchy. In each case prove that a sequence converges if and only if it is a Cauchy sequence. (b) Recall that convergence in a metric space satisfies the following condition: If every subsequence of { f n } has a subsubsequence which converges to f, then { f n } converges to f. Investigate whether this condition is true for almost everywhere convergence, almost uniform convergence, and convergence in measure. If you find a counterexample, or can t decide, for general measures, look at finite and σ-finite measures. (a) A sequence { f n } n1 of measurable complex-valued functions on (X, M, µ) is almost everywhere Cauchy if there exists M such that µ( c ) 0 and { f n (x)} is Cauchy for all x. Since f n (x) C for all x X, it follows that { f n } is Cauchy on if and only if f n (x) p x for some p x C and all x, i.e. f n f a.e., where f (x) p x for all x. The sequence { f n } is almost uniformly Cauchy if for all δ > 0, there exists M such that µ( c ) < δ and { f n } is uniformly Cauchy on. That is, given ε > 0, there exists N N such that if m, n N, then f m (x) f n (x) < ε for all x. Since uniform Cauchy iis a necessary and sufficient condition for uniform convergence in C, there exists a measurable function f such that f n f uniformly on. (b) We claim that the condition is not true in general for almost everywhere or almost uniform convergence. The following counterexample is adapted from [Folland, (iv) p. 61]. For each n N, write n 2 k + m, where 0 m < 2 k and k Z +. Consider the sequence { f n } L 1 ([0, 1]) defined by f n χ [ m 2 k, m+1 ]. 2 k We have f n (x) dx 2 k 0 as n, so f n f 0 in L 1. If { f n(k) } is a subsequence of { f n }, then f n(k) 0 in L 1 as well. By the all-ways theorem, there exists a subsequence of { f n(k) } which converges to f almost everywhere and almost uniformly. However, f n (x) does not converge for any x [0, 1], so the convergence of { f n } is neither almost everywhere nor almost uniform. Since m([0, 1]) 1, this counterexample addresses finite measures, and therefore σ-finite measures as well. We claim that the condition is true for convergence in measure. Suppose that { f n } does not converge to f in measure. Then there exists ε > 0 and a subsequence { f n(k) } such that µ({x : f n(k) (x) f (x) > ε}) ε for all k N. Any further subsequence also has this property, and hence does not converge to f in measure. 7