m5051mtmod

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 ) 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– 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 ∞ 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 complex-variable 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 complex-variable 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...
View Full Document

This note was uploaded on 11/02/2010 for the course MATH 205 taught by Professor Junying during the Fall '08 term at NYU.

Page1 / 7

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online