BS V c Gabriel Nagy Banach Spaces V: A Closer Look at the w- and the w∗-Topologies Notes from the Functional Analysis Course (Fall 07 - Spring 08) In this section we discuss two important, but highly non-trivial, results concerning the weak topology (w) on Banach spaces and the weak dual topology (w∗) on dual Banach spaces. Convention. Throughout this note K will be one of the fields R or C, and all vector spaces are over K. A. Convex w∗-closed sets in X ∗ The main result in this sub-section (Theorem 1) concerns the w∗-topology on dual Banach spaces. Recall that, given a topological vector space X , the w∗-topology on the topological dual space X ∗ is defined to be the weakest topology that makes all evaluation maps X ∗ 3 φ 7−→ φ(x) ∈ K, x ∈ X continuous. Equivalently, for a net (φλ)λ∈Λ and a functional φ in ∗ ∗ w X , one has φλ −→ φ, if and only if φλ(x) → φ(x), ∀ x ∈ X . In the case when X is a normed vector space, by Alaoglu’s Theorem (see BS II), we know that, for every r > 0, the ball ∗ ∗ ∗ ∗ (X )r = {φ ∈ X : kφk ≤ r} is w -compact in X . The result below should be understood as some sort of converse to Alaoglu’s Theorem. Theorem 1 (Krein-Smulian). Given a Banach space X , and a convex set A ⊂ X ∗, the following conditions are equivalent: (i) A is w∗-closed in X ∗; ∗ ∗ ∗ (ii) for every r > 0, the set (A)r = A ∩ (X )r is w -closed in X ; ∞ (ii’) there exists a sequence (rn)n=1 ⊂ (0, ∞), with limn→∞ rn = ∞, such that (A)rn is w∗-closed in X ∗, for every n. ∗ ∗ Proof. Let us observe that, if we define the set R = {r > 0 : (A)r is w -closed in X }, then ∗ r ∈ R ⇒ (0, r) ⊂ R, Indeed, if (A)r is w -closed, and 0 < s < r, then the trivial equality ∗ ∗ ∗ ∗ (A)s = (A)r ∩ (X )s, combined with the fact that (X )s is w -compact (hence w -closed) ∗ ∗ in X , forces (A)s to be w -closed. The above mentioned feature of the set R clearly shows that conditions (ii) and (ii’) are equivalent. ∗ Obviously, since one has the inclusions (A)r ⊂ (X )r, in conditions (ii) and (ii’), the phrase “w∗-closed” can be replaced with “w∗-compact.” ∗ ∗ The implication (i) ⇒ (ii) is trivial, again because (X )r is w -compact. (For this implication, the assumption that A is convex is not needed. Likewise, X can just be a normed vector space.) 1 The hard implication (ii) ⇒ (i) will be proved in several steps. Claim 1: Property (ii) is invariant under translations and positive dilations, that is, if A satisfies (ii) then: • tA satisfies (ii), for every t > 0; •A + φ satisfies (ii), for every φ ∈ X ∗; The first statement is obvious, since dilations X ∗ 3 φ 7−→ tφ ∈ X ∗ are w∗-homeomorphisms, and (tA)r = (A)r/t, ∀ r, t > 0. For the second statement, we fix r > 0, net (ψλ)λ∈Λ ⊂ (A+φ)r, ∗ w ∗ such that ψλ −→ ψ, for some ψ ∈ X , and we show that ψ ∈ (A + φ)r. On the one hand, ∗ ∗ ∗ ∗ since (X )r is w -compact (hence w -closed), it is clear that ψ ∈ (X )r, so it suffices to show that ψ ∈ A + φ. On the other hand, since kψλ − φk ≤ kψλk + kφk ≤ r + kφk, it follows that ∗ (ψλ − φ)λ∈Λ is a net in (A)r+kφk, which is w -convergent to ψ − φ, so using condition (ii) on A implies that ψ − φ belongs to A, thus ψ ∈ A + φ. Claim 2: If A satisfies condition (ii), then A is closed in the norm topology. ∞ We must prove that, whenever (φn)n=1 ⊂ A is a sequence which converges in norm ∗ ∞ to some φ ∈ X , it follows that φ ∈ A. But this is quite obvious, because (φn)n=1 is automatically bounded, so there exists r > 0, such that φn ∈ (A)r, ∀ n, and then, by the k . k w∗ obvious implication φn −−→ φ ⇒ φn −→ φ, and by condition (ii) it follows that φ ∈ (A)r. After these preparations, we now proceed with the proof. What we must show is the following implication: if φ ∈ X ∗ r A, there exists a w∗-neighborhood V of φ, such that A ∩ V = ∅. Using Claim 1, we can assume that φ = 0, so now we assume 0 6∈ A, and we must produce a w∗-neighborhood V of 0 in X ∗, which is disjoint from A. The most important step in the proof is contained in the following. ∞ Claim 3: Assuming 0 6∈ A, there exists a sequence (xn)n=1 ⊂ X , such that (∗) limn→∞ kxnk = 0; (∗∗) for every φ ∈ A, there exists n ∈ N, such that |φ(xn)| > 1; As we shall see shortly, the condition (∗∗) has something to do with absolute polar sets. Recall (see DT I) that, for every set M ⊂ X , its absolute polar in X ∗ is the set ∗ M = {φ ∈ X : |φ(x)| ≤ 1, ∀ x ∈ M}. Since A is norm closed, there exists ρ > 0, such that kφk ≥ ρ, ∀ φ ∈ A. By Claim 1, we can assume that ρ > 1. (Otherwise use suitable dilations in both X and X ∗.) In particular, we can assume that (A)1 = ∅. Put F0 = {0}, and let us construct recursively an whole ∞ sequence (Fn)n=0 of finite subsets of X , such that for each n we have: (a) Fn ⊂ (X )1/n, and (b) (A)n ∩ F0 ∪ · · · ∪ Fn−1 = ∅, 2 Assume F0 = {0}, F1,..., FN−1 have been constructed, so that (a) holds for 1 ≤ n ≤ N−1, and (b) holds for 1 ≤ n ≤ N, and let us indicate how FN ⊂ (X )1/N can be constructed, such that (b) holds for n = N + 1. (Since (A)1 = ∅, there is nothing to check at the start N = 1.) To construct FN , we argue by contradiction. Assume (A)N+1 ∩ F0 ∩ · · · ∩ FN−1 ∩ G 6= ∅, ∗ for all finite subsets G ⊂ (X )1/N . Since absolute polars M of sets M ⊂ X are w -closed, it ∗ ∗ follows that the set K = (A)N+1 ∩F0 ∪· · ·∩FN−1 ⊂ (X )N+1 is w -compact, so if we assume that all sets of the form K ∩ G , G ∈ Pfin((X )1/N ) are non-empty, then by the Intersection Property, it follows that T K ∩ G is non-empty. In particular, if we pick an G∈Pfin((X )1/N ) element φ in this intersection, then φ ∈ (A)N+1 ∩ F0 ∩ · · · ∩ FN−1, and: |φ(x)| ≤ 1, ∀ x ∈ (X )1/N . (1) It is trivial that any functional φ ∈ X ∗, that satisfies (1), must have norm kφk ≤ N, so our particular φ in fact belongs to ∗ (A)N+1 ∩ F0 ∩ · · · ∩ FN−1 ∩ (X )N = (A)N ∩ F0 ∩ · · · ∩ FN−1, which is impossible, since (b) holds for n = N, so the set on the right-hand side is empty. Having constructed the sequence of finite sets Fn, satisfying (a) and (b), we now form ∞ a sequnece (xn)n=1, which lists the sets F1, F2,... in order, which means that we have a sequence of integers 0 = k0 < k1 < . , such that Fj = {xn : kj−1 < n ≤ kj}, ∀ j ∈ N. Of course, by (a), condition (∗) is trivially satisfied. To check condition (∗∗) we start with some φ ∈ A and we choose N ∈ N, such that N + 1 ≥ kφk. By (b), there exists j ∈ {1,...,N}, such that φ 6∈ Fj , so there exists some x = xn ∈ Fj, such that |φ(xn)| > 1. ∞ Having proved Claim 3, let us fix (xn)n=1 satisfying (∗) and (∗∗), and consider the operator ∗ ∞ Y T : X 3 φ 7−→ φ(xn) n=1 ∈ K. N Using (∗), it is pretty clear that T is linear and has Range T ⊂ c0,K – the Banach space of all sequences in K that converge to 0. Moreover, T is also continuous, since |φ(xn)| ≤ 1 kφk · kxnk ≤ kφk, ∀ n. Since T is linear and A is convex , so is the set T (A) ⊂ c0,K. In c0,K consider the set B = {b ∈ c0,K : kbk∞ < 1}. Of course, B is convex and open in c0,K in the norm topology. Moreover, by (∗∗) it follows that B is disjoint from T (A), so using the Hahn-Banach Separation Theorem, there exists a linear continuous functional η : c0,K → K, and some α ∈ R, such that Re η(a) ≥ α > Re η(b), ∀ a ∈ T (A), b ∈ B. (2) We know that the dual Banach space (c )∗ is isometrically identified with `1 , so the func- 0,K K tional η can be represented by a vector t = (t )∞ ∈ `1 as n n=1 K ∞ X ∞ η(y) = tnyn, ∀ y = (yn)n=1 ∈ c0,K. n=1 1 It is only here that we use the assumption that A is convex. 3 In particular, for any φ ∈ X ∗ we have ∞ ∞ X X η(T φ) = tnφ(xn) = φ(tnxn). (3) n=1 n=1 P∞ The point is that, by condition (∗) we have since ktnxnk = |tn|·kxnk ≤ |tn|, so n=1 ktnxnk < P∞ ∞. Since X is a Banach space, by the Summability Test it follows that the series n=1 tnxn is convergent to some x ∈ X , so (3) becomes η(T φ) = φ(x), ∀ φ ∈ X ∗. Going back to (2) we now have: Re φ(x) ≥ α > Re η(b), ∀ φ ∈ A, b ∈ B. Of course, using b = 0 in the above inequality yields Re φ(x) ≥ α > 0, ∀ φ ∈ A. With x and α as above, the set V = {ψ ∈ X ∗ : Re ψ < α} is clearly a w∗-neighborhood V of 0, which is disjoint from A.