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m5051mtmod - Ma 5051 Real Variables and Functional Analysis...

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Ma 5051 — Real Variables and Functional Analysis Solutions for Take-Home Midterm Prof. Sawyer — Washington University Let ( X, M , μ ) be a measure space. Recall R A f ( x ) = R I A ( x ) f ( x ) for A ∈ M and f L + , where I A ( x ) is the indicator function of A . 1. (a) The problem is to show that the integrand is dominated by an integrable function. Since 1 - e - x = R x 0 e - y dy R x 0 dy = x , it follows that 1 - x e - x , 1 - (1 /n ) x e - x/n , and ( 1 - (1 /n ) x ) n e - x . Thus the integrand is dominated by x k e - x , which is integrable on (0 , ). Then by the dominated convergence theorem lim n →∞ Z n 0 x k ( 1 - n - 1 x ) n dx = Z 0 lim n →∞ I (0 ,n ] ( x ) x k ( 1 - n - 1 x ) n dx = Z 0 x k e - x dx = k ! (b) The substitution x x/ n changes the integral to I ( n ) = Z n 0 1 + x 2 ( 1 + (1 /n ) x 2 ) n dx By the binomial theorem (see also the model solutions for Problem 4a of HW5), 1 + x 2 n n 1 + n x 2 n + n ( n - 1) 2 x 2 n 2 1 + 1 3 x 4 for x > 0 and n 3. Thus the integrand of I ( n ) is dominated by 3(1+ x 2 ) / (3+ x 4 ), which is integrable on (0 , ). Then by the dominated convergence theorem lim n →∞ I ( n ) = Z 0 lim n →∞ I (0 , n ] ( x ) 1 + x 2 ( 1 + (1 /n ) x 2 ) n dx = Z 0 1 + x 2 e x 2 dx = Z 0 e - x 2 dx + Z 0 1 2 x 2 xe - x 2 · dx = 3 2 Z 0 e - x 2 dx = 3 4 π 2. The integral is I ( a ) = Z 0 e - y 2 sin( ay ) dy = Z 0 e - y 2 X n =0 ( - 1) n a 2 n +1 y 2 n +1 (2 n + 1)! dy
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Ma 5051– Real Variables and Functional Analysis– Take-Home Midterm . . . . . . . . . 2 The partial sums of the integrand of I ( a ) are dominated for fixed a > 0 by e - y 2 X n =0 a 2 n +1 y 2 n +1 (2 n + 1)! e - y 2 X n =0 a n y n n ! = e - y 2 e ay where the second series is formed from the first by adding in the even terms a 2 n y 2 n / (2 n )!. Since the last expression above is integrable on (0 , ), dominated convergence allows us to interchange the integral and sum in I ( a ) and conclude I ( a ) = X n =0 ( - 1) n a 2 n +1 (2 n + 1)! Z 0 e - y 2 y 2 n +1 dy = X n =0 ( - 1) n a 2 n +1 (2 n + 1)! 1 2 Z 0 e - x x n dx = X n =0 ( - 1) n a 2 n +1 n ! 2(2 n + 1)! = a 2 X n =0 ( - a 2 / 2) n (2 n + 1)(2 n - 1) . . . (3)(1) since 2 n n ! = (2 n )(2 n - 2) . . . 2. One can use complex-variable techniques to show I ( a ) = R 0 e - y 2 sin( ay ) dy = e - a 2 / 4 R a/ 2 0 e x 2 dx , but that is not required. See the Appendix for two derivations of this identity using complex-variable methods, one based on analytic continuation and one using Cauchy’s theorem. 3. Let A = lim inf n →∞ R f n ( x ) . Then there exists a sequence n k ↑ ∞ such that lim k →∞ R f n k = A . Since f n k f in measure, there exists a further subsequence { f n k j } such that lim j →∞ f n k j ( x ) = f ( x ) a.e. μ . Since f n ( x ) 0, this implies Z f ( x ) lim inf j →∞ Z f n k j ( x ) = lim k →∞ Z f n k ( x ) = lim inf n →∞ Z f n ( x ) and R f ( x ) lim inf n →∞ R f n ( x ) .
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