Solutions: Problem Set 4 Math 201B, Winter 2007

Transcription

1 Solutions: Problem Set 4 Math 2B, Winter 27 Problem. (a Define f : by { x /2 if < x <, f(x otherwise. Use the monotone convergence theorem to show that f L (. (b Suppose that {r n Q n N} is an enumeration of the rational numbers. Define g : by g(x 2 f(x r n, n n where f is the function defined in (a. Show that g L (, even though it is unbounded on every interval. Define f n : by { x /2 if /n < x < /n, f n (x otherwise. Then (f n is a monotone increasing sequence of nonnegative, measurable functions (since f ((a, is open and therefore measurable for every a which converges pointwise to f on (so f is measurable. Each f n is iemann integrable on [, ], and /n f n dx x /2 dx /n [ 2x /2] /n /n 2 as n. The monotone convergence theorem implies that f dx lim f n dx 2 <, so f L (.

2 (b Let g N (x N n 2 n f(x r n. Then (g N is a monotone increasing sequence, since f, that converges pointwise to g. By the linearity of the integral and the translation invariance of Lebesgue measure, g N dx 2 N n f(x r 2 n n dx N n 2 n 2 as N. Hence, by the monotone convergence theorem g dx 2, so g is integrable. (In particular, the series defining g diverges to on at most a set of measure zero. The function g is unbounded in any neighborhood of r n Q, and therefore on any open interval since the rationals are dense in. emark. This example illustrates that integrable functions are not necessarily as nice as one might imagine; in particular they do not necessarily approach at infinity.

3 Problem 2. If f L (, prove that lim 2n n n f dx. Give an example to show that this result need not be true if f is not integrable on. Let f n 2n χ [ n,n]f, where χ [ n,n] is the characteristic function of the interval [ n, n]. Then f n dx 2n n n f dx. We have f n (x as n whenever f(x ±, so f n pointwise a.e. on. Also, for n, f n 2 f L (. The Lebesgue dominated convergence theorem implies that lim f n dx lim f n dx dx, which proves the result If f, then In this case the sequence lim n f dx. 2n n f n 2n χ [ n,n] converges pointwise (and even uniformly to on as n, but the integrals do not. Note that the convergence is not monotone and the sequence (f n is not dominated by any integrable function.

4 Problem 3. Define f : 2 by /x 2 if < y < x <, f(x, y /y 2 if < x < y <, otherwise. Compute the following integrals: ( f(x, y dxdy; f(x, y dx dy; 2 Are your results consistent with Fubini s theorem? If y or y, then f(x, y, and f(x, y dx. If < y <, then It follows that f(x, y dx y y dx + 2 y x dx 2 [ xy ] xy [ + ] x 2 x x y + y. ( f(x, y dx dy If x or x, then f(x, y, and f(x, y dy. ( f(x, y dy dx. xy dy.

5 If < x <, then f(x, y dy x x dy + 2 [ y ] [ yx + x 2 y y x ] y y 2 dy yx x + x. It follows that ( f(x, y dy dx dx. According to Fubini s theorem, we can evaluate the integral of f as an iterated integral (in either order: ( f(x, y dxdy f(x, y dx dy 2 ( y y dx + 2 y x dx dy 2 ( [ ] xy [ x + ] x dy y 2 x x xy ( 2 y dy ( 2 lim /n y dy lim [2 log y y] /n. This example shows that if one drops the assumption that f L in Fubini s theorem then the iterated integrals with different orders need not be equal. Also note that both f + dxdy and f dxdy are equal to, so f dxdy is undefined.

6 Problem 4. Define f : (, (, by f(x, y xe x2 (+y 2. Compute the iterated integrals with respect to x, y and y, x, and use Fubini s theorem to show that π e t2 dt 2. Integrating with respect to x followed by y, and using the monotone convergence theorem, we get ( ( n f(x, y dx dy lim xe x2 (+y 2 dx dy ( [ ] lim e x2 (+y2 xn dy 2 + y lim 2 lim π 4. + y dy 2 n + y dy 2 [ tan y ] n Integrating with respect to y followed by x, using the monotone convergence theorem, and making the change of variables t xy, we get ( ( n f(x, y dy dx lim xe x2 (+y 2 dy dx ( nx lim e (x2 +t 2 dt dx ( ( e x2 dx e t2 dt ( 2 e dt t2. x

7 The function f is non-negative, so Fubini s theorem implies that the iterated integrals of f are equal, and both are finite if one is finite. Equating the two iterated integrals and taking the square-root, we get the result.

8 Problem 5. In a normed space X, let B r (a {x X x a < r}. be the ball of radius r > centered at a X. Lebesgue measure m on d (with the Euclidean norm, say has the properties that every ball is measurable with finite, nonzero measure, and the measure of a ball is invariant under translations. That is, for every < r < and a X < m (B r (a <, m (B r (a m (B r (. Prove that it is not possible to define a measure with these properties on an infinite-dimensional Hilbert space. Since H is an infinite-dimensional Hilbert space, it contains a countably infinite orthonormal set {e n n N}. If k n then the Pythagorean theorem implies that e k e n e k 2 + e n 2 2. It follows that the balls B 2/2 (e k and B 2/2 (e n are disjoint. Suppose that m (B 2/2 ( ɛ. Then, by translation invariance, ( m B 2/2 (e n ɛ for every n N. Since the balls are disjoint, the countable additivity of m implies that if ɛ > ( m B 2/2 (e n ( m B 2/2 (e n n N n N n N ɛ.

9 If x B 2/2 (e n, then It follows that x e n + x e n < + B 2/2 (e n B + 2/2 (. n N 2 2. The additivity and non-negativity of m implies that ( ( m B 2/2 (e n m B + 2/2 (, n N so if m (B 2/2 ( > then ( m B + 2/2 (. Thus, if the measure of the smaller ball, with radius 2/2, is nonzero, the measure of the larger ball, with radius ( + 2/2, is infinite, so there is no translation invariant measure that assigns a non-zero, finite measure to every ball.