Problem Set 5. 2 n k. Then a nk (x) = 1+( 1)k

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1 Problem Set 5 1. (Folland 2.43) For x [, 1), let 1 a n (x)2 n (a n (x) = or 1) be the base-2 expansion of x. (If x is a dyadic rational, choose the expansion such that a n (x) = for large n.) Then the sequence {a n (x)} n=1 in {, 1}[,1) has no pointwise convergence subsequence. (Hence {, 1} [,1), with the product topology arising from the discrete topology on {, 1} is not sequentially compact. It is, however, compact. Proof. Suppose that a nk, k = 1, 2... is a subsequence of a n (x). Let x = (1+( 1) k ) k 2 2 n k. Then a nk (x) = 1+( 1)k 2. Therefore a nk (x) does not converge and hence a nk is not pointwise convergent. 2. (Folland 2.49) Let X be a compact Hausdorff space and E X. Lemma.1. A E is compact relative to E if and only if it is compact in X. Proof. Suppose that A E is compact in X. Suppose that {W α } is an open cover of A relative to E. Then A α W α = α V α E α V α for some V α open in X with W α = V α E. Thus, there is a finite subcover V αi and hence a finite subcover W αi. Suppose that A E is compact relative to X. Then suppose that V α is an open cover of A in X. Then A α V α A α V α E and V α E is open relative to E. Hence, there is a finite sub cover V αi E and in particular a finite subcover V αi. Lemma.2. For A E, the closure of A relative to E, A E is equal to E A. Proof. Notice that F is closed relative to E if and only if F = W E for some W closed in X. Thus, since A = W, A E = A W W closed A W W closed W E = A F F closed in E F = A E. (a) If E is open, then E is locally compact in the relative topology. Proof. Let x E. Then since E c is a closed subset of a compact space, it is compact and hence there exist disjoint open sets of X, U, V with E c V and x U. Then U E and so U E = U is open in the relative topology of E. Now, V c E is closed in X U V c. Therefore, U V c E and in particular, U E = U E = U. Thus, U E is closed in X and hence is compact in X. Thus, U E is a compact neighborhood of x in E. 1

2 2 (b) If E is dense in X and locally compact in the relative topology, then E is open. Proof. Suppose E is not open. Then there exists x E so that for all neighborhoods U of x, U E c. Now, E is locally compact. Therefore, there exists K compact K E and U open in X so that x U E K. Now, E is dense in X and U \ K is open. Therefore, if U \K is nonempty, then E (U \K). But, this is a contradiction since U E K. (c) E is locally compact in the relative topology if and only if E is relatively open in E. Proof. First, observe that if X is compact Hausdorff, then so is E in the relative topology. Furthermore, observe that U is open in the relative topology on E if and only if there exists W open in X so that U = W X. On the other hand, U is open in the relative topology on E with respect to E if and only if there exists W 1 open in the relative topology on E so that U = W 1 E if and only if there exists W open in X so that W 1 = W E. But then, U = W E E = W E. Therefore, the relative topology on E with respect to X and that with respect to E are the same and we may take X = E without loss of generality.then the statement follows from (a) and (b) since the relative topology on E in E is the same is the relative topology on E in X. 3. (Folland 2.51) If X and Y are topological space, φ C(X, Y ) is called proper if π 1 (K) is compact in X for every compact K Y. Suppose that X and Y are LCH spaces and X and Y Are their one point compactificiations. If φ C(X, Y ), then φ is proper if and only if φ extends continuously to a map from X to Y by setting φ( X ) = Y. (Please read about one point compactifications on page 132 of Folland) Proof. Let φ(x) = { φ(x) x X Y x = X. Suppose φ is proper. Let U Y be open. We consider two cases. First, suppose that Y / U. Then φ 1 (U) = φ 1 (U) is open in X and hence is open in X. Next, suppose Y U. Then U c is compact by the definition of the topology on Y. Therefore, φ 1 (U c ) = φ 1 (U c ) is compact and hence closed. In particular, φ 1 (U) = φ 1 (U c ) c is open. Thus, φ is continuous. Now, suppose that φ is continuous and suppose that K Y is compact. Then Y \ K is open and Y \ K = { Y } Y \ K. In particular, Y \ K is open. Therefore, φ 1 (Y \ K) is open and X φ 1 (Y \ K). Therefore, is compact. Notice that so, X \ ( φ 1 (Y \ K) \ { X }) c φ 1 (Y \ K) = φ 1 (Y \ K) \ { X } φ 1 (K) = X \ φ 1 (Y \ K) = X \ ( φ 1 (Y \ K) \ { X }) is compact and φ is proper.

3 4. Suppose K C(R R; C) is properly supported in the sense that the maps π L : supp K X and π R : supp K Y are proper where π L and π R are respectively the projection maps onto the right and left copy of R. Consider the map T K : C c (R) C R given by f K(x, y)f(y)dy. Show that in fact T K : C c (R) C c (R). Proof. Suppose that f C c (R) and consider T K f(x) = K(x, y)f(y)dy. Let F = supp f. Then F is compact and hence πr 1 (F ). In particular, then π L(πR 1 (F )) is compact. But supp T K f π L (πr 1 (supp f)). Therefore, T Kf is compactly supported. To see that f is continuous, let x n x. Then T K f(x n ) = K(x n, y)f(y)dy = 1 supp f (y)k(x n, y)f(y)dy. Now, by continuity of K, sup K(w, y) = M < w [x ɛ,x+ɛ],y supp f and by continuity of f, together with the fact that it is compactly supported, sup f = N <. Therefore, by the dominated convergence theorem, since 1 supp f (y) K(x n, y) f(y) 1 supp f (y)mn L 1, T K f(x n ) lim K(x n, y)f(y)dy = K(x, y)f(y)dy = T K f(x). n 5. (Folland 2.58) If {X α } α A is a family of topological spaces of which infinitely many are noncompact, then every closed compact subset of α A X α is nowhere dense. Proof. Suppose that K α A X α =: X is closed and compact. Recall that the sets of the form { n } (1) B := πα 1 i (U αi ) U αi X αi open and n N i=1 is a base for the topology on X. Suppose that K has nonempty interior. Then there exists B B with B K. In particular, n B = πα 1 i (V αi ). i=1 Let β A \ {α 1,... α n } such that X β is noncompact. Such a β exists since infinitely X α s are noncompact. now, since K is compact, and π β is continuous, π β (K) is compact. But π β (K) π β (B) = X β is not compact, a contradiction. Thus K has nonempty interior and since K is closed, having nonempty interior is equivalent to being nowhere dense. 3

4 4 6. (Folland 2.63) Let K C([, 1] [, 1]). For f C([, 1]), let T K f(x) = 1 K(x, y)f(y)dy. Then T f C([, 1]), and {T f f 1} is precompact in C([, 1]). Proof. By exercise 4, T f C([, 1]). Then, T f(x) 1 1 K(x, y) f(y) dy f K(x, y) dy M f where M = sup (x,y) [,1] 2 K(x, y) <. Therefore, the collection F := {T f f 1} is pointwise bounded. We next show that it is equicontinuous. Since K(x, y) is continuous on compact set, it is uniformly continuous. Hence, there exists δ > so that if x w < δ, then K(x, y) K(w, y) < ɛ. In particular, then T f(x) T f(w) = (K(x, y) K(w, y))f(y)dy K(x, y) K(w, y) f(y) dy f ɛ. Therefore, F is equicontinuous and hence by the Arzela-Ascoli theorem, it is precompact. 7. (Folland 2.65) Let U be an open subset of C, and let {f n } be a sequence of holomorphic functions on U. If {f n } is uniformly bounded on compact subsets of U, there is a subsequence that converges uniformly to a holomorphic function on compact subsets of U. (Use the Cauchy integral formula to obtain equicontinuity). Proof. Let K U be compact. Then there exists V open and precompact with K V V U. In particular, then there exists ɛ > so that Now, for x K, we can write inf x y > 3δ. y K,x V c f n (x) = 1 2πi γ δ (x) z x dz where γ δ (x) is the circle of radius 3δ around x. Moreover, for x y < δ, y B(x, 3δ) and hence we can write f n (x) f n (y) = 1 ( 1 2πi γ δ (x) z x 1 ) dz z y = 1 y x 2πi (z x)(z y) dz γ δ (x)

5 Now, sup n N, x V f n (x) = M <. Now, for z γ δ (x), min( z x, z y ) > 2δ. Therefore, f n (x) f n (y) M 8πδ 2 x y 6πδ M x y δ Hence, for x y < min(ɛδm 1, δ), f n (x) f n (y) < ɛ. In particular, {f n } is equicontinuous on K. Lemma.3. Suppose that X is a second countable LCH space and U X is open. Then U is σ-compact. Proof. Let B be a countable base for X. Then for x U, there exists K x U a compact neighborhood of x. Hence, there exists x B K x with B B. In particular, then there exists a countable collection {B n } B with B n K n U so that n B n = U. But then n K n = U and hence U is σ-compact. Now, let K m U with K m compact. Then by the Arzela-Ascoli theorem, since {f n Km } is pointwise bounded and equicontinuous, there exists a subsequence fn m converging uniformly on K m. Letting g n = fn n, g n converges uniformly on compact subsets of U to g. Now, with K and γ δ (x) as before and x K, by the dominated convergence theorem, g(x) = lim g n n n(x) = lim n In particular, g(x) is holomorphic on K. g n (z) γ δ z x dz = γ δ g(z) z x dz. 5

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