Math541, Spring 2013, Homework #6 Solution Due to Friday, April 12, 2013 # 1. Let (V; k·k) be a Banach space. Show that there exists a compact topological space X and there exists an isometric inclusion ι :(V; k · k) ! (C(X); k · k1): This shows that every Banach space can be identified with a norm closed subspace of some C(X) space. Proof: Let us first recall that we can define the weak* topology on the dual space ∗ − V , and by Banach-Alaoglu theorem that the closed unit ball X = BV ∗ (0; 1) of V ∗ is compact with respect to the weak* topology on V ∗. Let us consider this compact topological space (X; τw∗ ). Next, we consider the canonical isometric inclusion ^ : x 2 V ! x^ 2 V ∗∗, where we definex ^(f) = f(x) for all f 2 V ∗. According to the definition of weak* topology on V ∗, for every x 2 V ,x ^ is a weak* continuous linear functional on V ∗, and thus the restriction ofx ^ to X, which is still denoted byx ^, is a continuous function on X (with respect to the weak* topology). This defines a map ι : x 2 V ! x^ 2 (C(X); k · k1): It is clear that this is a well-defined linear map, and it is an isometry injection by the Hahn-Banach extension theorem. # 2. Find all extreme points of the closed unit ball of C([0; 1]) = ff : [0; 1] ! R; continuous functionsg: Proof: Let K be the closed unit ball of C([0; 1]). It is easy to see that the constant function 1 and −1 are extreme points of K. Indeed, if there exist continuous functions f; g : [0; 1] ! [−1; 1] in K and 0 < α < 1 such that 1 = αf(x) + (1 − α)g(x) for all x 2 [0; 1], we must have f = 1 = g. Now let f be an extreme point of K. We must have kfk1 = supfjf(x)j : x 2 [0; 1]g = 1 and thus there exists a point c 2 [0; 1] such that jf(c)j = 1. If f(c) = 1, we claim that f(x) = 1 for all x 2 [0; 1]. Suppose not. We can find a point x0 2 (0; 1) such that 0 < f(x0) < 1. Let " = 1 − f(x0) > 0. By the continuity, there exists a sufficiently small δ > 0 such that (x0 − δ; x0 + δ) ⊆ (0; 1) and " " 1 − < f(x) < 1 + 2 2 for all x 2 (x0 − δ; x0 + δ). Now it is known from Urysohn's lemma, there exists a continuous function h : [0; 1] ! [0; 1] such that h(x0) = 1 and h(x) = 0 for all " x2 = (x0 − δ; x0 + δ). In this case, we can get two distinct functions f1 = f + 2 h " 1 1 and f2 = f − 2 h in K such that f = 2 f1 + 2 f2. This contradicts to f to be an extreme point of K. Therefore, if f(c) = 1, we must have f = 1. Similarly, we can prove that if f(c) = −1 for some c 2 [0; 1], we must have f = −1. These are the only extreme points of K. 1 2 # 3. Show that 1. there exists no extreme point in the closed unit ball of L1([0; 1]), 2. L1([0; 1]) is not a dual Banach space. Proof of Part 1: Let K be the closed unit ball of L1([0; 1]) and let f be an arbitrary R function in K with kfk1 = [0;1] jf(x)jdx = 1. By the (absolute) continuity of R R 1 F (t) = [0;t] jf(x)jdx, we can find c 2 [0; 1] such that F (c) = [0;c] jf(x)jdx = 2 . We can consider the distinct integrable functions g(x) = 2f(x)χ[0;c] and h(x) = 1 2f(x)χ[c;1]. It is easy to see that kgk1 = khk1 = 1 and f = 2 (g + h). This shows that any function in the closed unit ball K can not be an extreme point. Therefore, K has no extreme point at all. Proof of Part 2: Since the closed unit ball of a dual Banach space is compact with respect to the weak* topology, it must contain some extreme points by the Krein-Milman theorem. Since the closed unit ball of L1([0; 1]) has no extreme point, L1([0; 1]) can not be a dual Banach space. # 4. Let (X; τ) be a compact topological space. It is known from the Riesz representation theorem that we have the isometric isomorphism M(X) = C(X)∗, where M(X) denotes the sapce of all real valued Radon measures on X. Therefore, we obtain the weak* topology on M(X) under the representation Z µ(f) = f(x)dµ(x): X 1. Show that the set P (X) of all positive Radon measures of norm 1 (i.e. all Radon probability measures) on X is a weak* closed convex subset of M(X). 2. Show that the extreme points of P (X) are exactly Dirac measures δx with x 2 X. 3. The map D : x 2 X ! δx 2 Ext(P (X)) is a homeomorphism from topo- logical space (X; τ) onto the topological space (Ext(P (X)); τw∗ ). Proof of Part 1: Let fµαg be a net of positive Radon measure in P (X) such that µα ! µ 2 M(X) in weak* topology. We have Z Z fdµ = lim fdµα ≥ 0 X α X for all positive functions f in C(X). This shows that µ is a positive Radon measure. Since µα(X) = kµαk = 1 for all index α, we have Z Z µ(X) = 1dµ = lim 1dµα = lim µα(X) = 1: X α X α This shows that µ is a positive Radon measure with kµk = µ(X) = 1. Therefore P (X) is a weak* closed subset of M(X). Proof of Part 2: Let us fix a point x 2 X and let δx be the Dirac measure at x. Then δx is a Radon measure on X. Suppose µ = αµ1 +(1−α)µ2 for some 1 < α < 1 1 and probability Radon measures µ1 and µ2 in P (X). Since for arbitrary open set O with x2 = O, we have µ(O) = 0, this implies that µi(O) = 0 and thus µi = δx for i = 1; 2. This shows that δx is an extreme point in P (X) 3 Now we want to show that if µ is an extreme point of P (X), it must be a Dirac measure µ = δx for some point x. Suppose not. Then there exist two distinct points x and y in the support of µ, i.e. there exist two disjoint open sets x 2 U and y 2 V such that µ(U) > 0 and µ(V ) > 0. In this case, we have 0 < α = µ(U) < 1, µjU µjUc and µ1 = α and µ2 = (1−α) are positive Radon measures on X (see Homework 3, #1) with kµik = µi(X) = 1. It is easy to see that µ = αµ1 + (1 − α)µ2 is a convex combination of µ1 and µ2. Therefore, µ is not an extreme point, a contradiction. Proof of Part 3: It is known from the above discussion that D : x 2 X ! δx 2 Ext(P (X)) is a well-defined bijection. If a net xα ! x in X, then for every continuous function f 2 C(X), we have δxα (f) = f(xα) ! f(x) = δx(f): This shows that the map D is continuous from X onto Ext(P (X)). To show D−1 is continuous, we notice that for every closed subset F of X, F is compact and thus D(F ) is compact, and thus closed. It follows that if O is an open subset in X, then D(O) is open in Ext(P (X)). This implies that D−1 is also continuous. Therefore, D is a homeomorphism from X onto Ext(P (X))..