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Unformatted text preview: Ma 5051 — Real Variables and Functional Analysis Solutions for TakeHome Midterm Prof. Sawyer — Washington University Let ( X, M ,μ ) be a measure space. Recall R A f ( x ) dμ = R I A ( x ) f ( x ) dμ 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 e y dy ≤ R x 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 x k ( 1 n 1 x ) n dx = Z ∞ lim n →∞ I (0 ,n ] ( x ) x k ( 1 n 1 x ) n dx = Z ∞ x k e x dx = k ! (b) The substitution x → x/ √ n changes the integral to I ( n ) = Z √ n 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 ∞ lim n →∞ I (0 , √ n ] ( x ) 1 + x 2 ( 1 + (1 /n ) x 2 ) n dx = Z ∞ 1 + x 2 e x 2 dx = Z ∞ e x 2 dx + Z ∞ 1 2 x ‡ 2 xe x 2 · dx = 3 2 Z ∞ e x 2 dx = 3 4 √ π 2. The integral is I ( a ) = Z ∞ e y 2 sin( ay ) dy = Z ∞ e y 2 ∞ X n =0 ( 1) n a 2 n +1 y 2 n +1 (2 n + 1)! dy Ma5051– Real Variables and Functional Analysis– TakeHome 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 ∞ e y 2 y 2 n +1 dy = ∞ X n =0 ( 1) n a 2 n +1 (2 n + 1)! 1 2 Z ∞ 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 complexvariable techniques to show I ( a ) = R ∞ e y 2 sin( ay ) dy = e a 2 / 4 R a/ 2 e x 2 dx , but that is not required. See the Appendix for two derivations of this identity using complexvariable methods, one based on analytic continuation and one using Cauchy’s theorem. 3. Let A = liminf n →∞ R f n ( x ) dμ . Then there exists a sequence n k ↑ ∞ such that lim k →∞ R f n k dμ = 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 ) dμ ≤ liminf j →∞ Z f n k j ( x...
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This note was uploaded on 11/02/2010 for the course MATH 205 taught by Professor Junying during the Fall '08 term at NYU.
 Fall '08
 JUNYING

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