Austin Mohr Math 704 Homework 6

Transcription

1 Austin Mohr Math 704 Homework 6 Problem 1 Integrability of f on R does not necessarily imply the convergence of f(x) to 0 as x. a. There exists a positive continuous function f on R so that f is integrable on R, but yet lim sup f(x) x. Proof. Define the function f to take on the value n if x [n, n + 1 n 3 ) for n. Elsewhere, the function is zero except for the line segments required to make the function continuous. We define these segments in such a that the graph resembles a sequence of trapezoids with height n and bases of length 1 n 3 and 3 n 3. By construction, f is positive and continuous on R. Now, fdx n( n n < n 1 n n 3 Hence, f is integrable. The fact that lim sup f(x) is also immediate (for any N, there are infinitely-many x x (N, ) such that f(x) > N), so the claim is proven. ) b. However, if we assume that f is uniformly continuous on R and integrable, then lim sup f(x) 0. x Proof. Suppose, for the sake of contradiction, that lim sup f(x) c > 0 (we consider first only the case x where x + ). Choose some d so that 0 < d < c. Then, there is a sequence x 1, x,... with each x i far apart (to be made precise later) so that f(x i ) d for all i. Choose ɛ 0 so that 0 < ɛ 0 < d. Since f is uniformly continuous, there is some δ 0 > 0 so that, for each x i, f(x i ) f(y) < ɛ 0 for all y N(x i, δ 0 ). Since f(x i ) d > ɛ 0, we have that f(y) > ɛ 0 for all y N(x i, δ 0 ). Hence, the area contributed by the function over the interval N(x i, δ 0 ) is at least δ 0 ɛ 0. Now, if we choose the x i far enough apart so that each of the N(x i, δ 0 ) are disjoint, we have that f(x) δ 0 ɛ 0 0 which contradicts the fact that f is integrable. We can force the same contradiction when x, and so we conclude that lim sup f(x)dx 0. x n1 1

2 Problem Suppose f 0, and let E k everywhere, then and the sets k are disjoint. {x f(x) > k } and k {x k < f(x) k+1 }. If f is finite almost k {f(x) > 0}, Proof. Since f is a function, it has a unique output for each input x. Hence, k < f(x) k+1 for a single value of k. That is, the k are disjoint. ( ) Let x k. Then k < f(x) k+1 for some k, so f(x) 0. Since f 0, we have that f(x) > 0. That is, x {f(x) > 0}. ( ) Let x {f(x) > 0}. Then f(x) > 0, and so k < f(x) k+1 for some k. That is, x Prove that f is integrable if and only if k m( k ) < if and only if k m(e k) < k. Proof. (i ii) Suppose f is integrable. Let x k for some k. By definition of k, k < f(x). Since the k are disjoint, it follows that > fdx > k χ k dx k χ k dx k m( k ) (ii i) Suppose k m( k ) <. Then k m( k ) <. Let x k for some k. By definition of k, k+1 f(x). Since the k are disjoint, it follows that fdx < k+1 χ k dx < k+1 χ k dx k+1 m( k )

3 (ii iii) Suppose k m( k ) <. Observe that E k n. n k measurable, it follows that Since the n are disjoint and m(e k) m( n k n ) n k m( n ) So k m(e k) k m( n ) n k n n n n n m( n ) k m( n ) n n+1 m( n ) n m( n ) k Hence, if either of k m(e k) or k m( k ) is finite, then the other is also finite. Use this result to verify the following assertions. Let and f(x) { x a if x 1 0 otherwise g(x) { x b if x > 1 0 otherwise Then f is integrable on R d if and only if a < d; also g is integrable on R d if and only if b > d. Proof. Let x k. Then Hence k < x a k+1 k a > x k 1 a m(b(0; k a )) > m(k ) m(b(0; k 1 a )) v d dk a > m( k ) v d d(k+1) a 3

4 where v d is the volume of the unit ball. Now k m( k ) Since x 1, m( k ) v d for all k. Hence So it suffices to show that 0 k m( k ) v d 0 k m( k ) converges. v d d a k (v d d(k+1) a ) k m( k ) + v d + k + k m( k ) k m( k ) k m( k ) k m( k ) < k(1 d a ) k m( k ) < v d k (v d dk a ) k(1 d a ) We see that the upper and lower bounds converge if and only if 0 < a < d, forcing the convergence of k m( k ), which in turn implies that f is integrable. Let x E k. Then k < x b k b > x Observe also that E k is empty for k 0. Hence, E k B(0; k b ) \ B(0; 1), and so m(e k) v d dk b v d. Now k m(e k) v d k m(e k) k (v d dk b v d ) k(1 d b ) v d which converges if and only if b > d, which in turn implies that f is integrable. k Problem 3 a. Prove that if f is integrable on R d, real-valued, and f(x)dx 0 for every measurable E, then E f(x) 0 a.e. x. Proof. Define to be the set {x f(x) < 0}. Since f is integrable, f is measurable, and so is measurable. We claim that m( ) 0. 4

5 Since is measurable, we have that f(x)dx 0 by hypothesis. Now, observe that, for any n 1, nfχ f. It follows that nfχ dx fdx n fdx fdx fdx 1 fdx n fdx 0 Hence, fdx 0. Since f(x) < 0 for all x, we conclude that m( ) 0. That is, f(x) 0 almost everywhere. b. As a result, if f(x)dx 0 for every measurable E, then f(x) 0 a.e. E Proof. rom the first part, we see that f(x) 0 almost everywhere. Let G be the set {x f(x) > 0}. It suffices to show that m(g) 0. As before, f is measurable, so G is measurable. By hypothesis, we have that fdx 0. Since f(x) > 0 G for all x G, we conclude that m(g) 0. Hence, the set of x so that f(x) 0 has measure 0. That is, f(x) 0 almost everywhere. Problem 4 a. Let a n, b n R such that a n a R. Prove that lim inf(a n + b n ) a + lim inf b n Proof. Since a n a, we have a lim a n lim sup a n lim inf a n. Since lim sup(a n ) lim inf( a n ), we n n n have lim inf(a n + b n ) lim sup( a n b n ) lim sup( a n ) lim sup( b n ) lim inf a n + lim inf b n a + lim inf b n Now, construct a subsequence (b nk ) of (b n ) with lim b n k k induced by the indices chosen for (b nk ). lim inf n b n. Let (a nk ) be the subsequence Therefore, lim inf(a n + b n ) a + lim inf b n. a + lim inf b n lim inf a n + lim inf b n lim a nk + lim b nk lim(a nk + b nk ) lim inf(a n + b n ) b. Let f, f n be integrable functions. Assume f n (x)arrowf(x) a.e. and f n dx f dx. Prove that fn f dx 0. Proof. Define the function g n to be f + f n f f n. Then g n f as n. By atou s lemma gdx lim inf gn dx 5

6 Hence Now, f dx lim inf ( f + f n f f n ) dx lim inf( f dx + f n dx f f n dx) f dx + lim inf( f n dx f f n dx) f dx + f dx + lim inf( f f n dx) (by part (a)) 0 lim inf( f f n dx) 0 lim sup f f n dx lim f f n dx n Therefore, as n, f f n dx 0. 6