Banach Spaces V: A Closer Look at the w- and the w -Topologies

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 φ φ(x) K, x X continuous. Equivalently, for a net (φ λ ) λ Λ and a functional φ in X w, one has φ λ φ, if and only if φ λ (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 : φ 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 (r n ) n=1 (0, ), with lim n r n =, 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 φ tφ X are w -homeomorphisms, and (ta) r = (A) r/t, r, t > 0. For the second statement, we fix r > 0, net (ψ λ ) λ Λ (A+φ) r, w such that ψ λ 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 ψ λ φ ψ λ + φ r + φ, it follows that (ψ λ φ) λ Λ is a net in (A) r+ φ, 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 (φ n ) 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 (φ n ) n=1 is automatically bounded, so there exists r > 0, such that φ n (A) r, n, and then, by the. w obvious implication φ n φ φ n 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 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 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 A, there exists a sequence (x n ) n=1 X, such that ( ) lim n x n = 0; ( ) for every φ A, there exists n N, such that φ(x n ) > 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 φ 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 F 0 = {0}, and let us construct recursively an whole sequence (F n ) n=0 of finite subsets of X, such that for each n we have: (a) F n (X ) 1/n, and (b) (A) n F 0 F n 1 =, 2

Assume F 0 = {0}, F 1,..., F N 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 F N (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 F N, we argue by contradiction. Assume (A) N+1 F 0 F N 1 G, 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 P fin ((X ) 1/N ) are non-empty, then by the Intersection Property, it follows that G P fin ((X ) 1/N ) K G is non-empty. In particular, if we pick an 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 φ N, so our particular φ in fact belongs to (A) N+1 F 0 F N 1 (X ) N = (A) N F 0 F N 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 F n, satisfying (a) and (b), we now form a sequnece (x n ) n=1, which lists the sets F 1, F 2,... in order, which means that we have a sequence of integers 0 = k 0 < k 1 <..., such that F j = {x n : k j 1 < n k j }, 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 φ. By (b), there exists j {1,..., N}, such that φ Fj, so there exists some x = x n F j, such that φ(x n ) > 1. Having proved Claim 3, let us fix (x n ) n=1 satisfying ( ) and ( ), and consider the operator T : X φ ( φ(x n ) ) n=1 N Using ( ), it is pretty clear that T is linear and has Range T c 0,K the Banach space of all sequences in K that converge to 0. Moreover, T is also continuous, since φ(x n ) φ x n φ, n. Since T is linear and A is convex 1, so is the set T (A) c 0,K. In c 0,K consider the set B = {b c 0,K : b < 1}. Of course, B is convex and open in c 0,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 : c 0,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 0,K ) is isometrically identified with l 1 K, so the functional η can be represented by a vector t = (t n ) n=1 l 1 K as η(y) = K. t n y n, y = (y n ) n=1 c 0,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 η(t φ) = t n φ(x n ) = n=1 φ(t n x n ). (3) The point is that, by condition ( ) we have since t n x n = t n x n t n, so n=1 t nx n <. Since X is a Banach space, by the Summability Test it follows that the series n=1 t nx n is convergent to some x X, so (3) becomes Going back to (2) we now have: n=1 η(t φ) = φ(x), φ X. 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 of 0, which is disjoint from A. : Re ψ < α} is clearly a w -neighborhood V Corollary 1. Given a Banach space X, and a linear subspace Z X, the following conditions are equivalent: (i) Z is w -closed in X ; (ii) for every r > 0, the ball (Z) r is w -closed in X ; (ii ) there exists r > 0, such that (Z) r is w -closed in X. Proof. By Theorem 1, all we need to prove is the implication (ii ) (ii). But this is trivial, using positive dilations. (See also Claim 1 from the proof above.) Another more interesting application of the Krein-Smulian Theorem concerns the transpose. Recall that, given topological spaces X and Y, and a linear continuous operator T : X Y, one defines its transpose as the operator T : Y φ φ T X. In general, T is continuous, when both X and Y are equipped with their respective w - topologies. When both X and Y are normed vector spaces, we know that T is also norm continuous. Although, the result below is a special case of the more general result that follows it (Theorem 2 below), we state it here separately, for technical purposes. Lemma 1. If X and Y are Banach spaces, and T L(X, Y) is injective, then the following conditions are equivalent: (i) T L(Y, X ) is invertible; (ii) T L(Y, X ) is bounded from below, that is, there exists D > 0, such that T φ D φ, φ Y. (4) 4

Proof. The implication (i) (ii) is trivial, by the Inverse Mapping Principle. (If T is invertible, then its inverse T 1 : X Y is also norm-continuous. See BS I.) (ii) (i). Assume T is bounded from below, and let D > 0 be a constant satisfying (4). Consider the linear subspace Z = Range T X. Claim: Z is w -closed in X. Denote, for simplicity Range T by Z. According to Corollary 1, it suffices to show that the ball (Z) D is w -closed in X w. Start with a net (ψ λ ) λ Λ (Z) D, such that ψ λ ψ, for some ψ X, and let us prove that ψ (Z) D. As discussed before (use the fact that (X ) D is w -closed), it suffices to show that ψ Z. Since ψ λ Z = Range T, we know that there exists a unique φ λ Y such that ψ λ = T φ λ. (The uniqueness is due to the injectivity of T, which is automatic from (4).) Using (4) we also know that D ψ λ = T φ λ D φ λ, so the net (φ λ ) λ Λ satisfies φ λ 1, λ. Since (φ λ ) λ Λ is now a net in (Y ) 1, using Alaoglu s Theorem, it has a sub-net (φ η(σ) ) σ Σ implemented by a directed map η : Σ Λ which is w -convergent in Y to some φ Y. Since T is continuous with respect to the w -topologies, it follows that T w φ η(σ) T φ, in X. On the one hand, this means that (ψ η(σ) ) σ Σ which is a sub-net of (ψ λ ) λ Λ is w -convergent to T φ. On the other hand, w since ψ λ ψ, this forces ψ = T φ, so ψ indeed belongs to Range T = Z. Having proved the above Claim, we now see that, in order to prove that T is invertible, it suffices to prove that Z is w -dense in X. Using the dual pairing between X and X, this is equivalent (see DT I, Exercise 5-7) to the equality where the annihilator space of Z is constructed as Z ann = {0}, (5) Z ann = {x X : ψ(x) = 0, ψ Z}. But the equality (5) is pretty, for if we start with some x Z ann, then for every φ Y, the functional T φ = φ T belongs to Z, and this implies φ(t x) = 0, φ Y, which in turn forces T x = 0, and the injectivity of T forces x = 0. Theorem 2. Suppose X and Y are Banach spaces. For an opeartor T L(X, Y), the following conditions are equivalent: (i) T is invertible; (ii) T L(Y, X ) is invertible Proof. The implication (i) (ii) is trivial, because if T is invertible, then T is invertible, with inverse: (T 1 ). (ii) (i). Assume T is invertible. Using the implication (i) (ii), it follows that the double transpose T L(X, Y ) is invertible, hence bounded from below. Of course, if we use the bidual isometric embeddings Θ X : X X and Θ Y : Y Y, we have the identity Θ Y T = T Θ X. 5

Since both Θ X and Θ Y are isometric, the above identity implies the fact that T is bounded from below. In particular (see BS II) it follows that T has closed range, so in order to prove that T is invertible, it suffices to show that the subspace M = Range is norm-dense in X. Exactly as in the proof of Lemma 1, using the dual pairing between Y and Y, this is equivalent (see again DT I, Exercise 5-7) to the equality where the annihilator space of M is constructed as M ann = {0}, (6) M ann = {φ Y : φ(y) = 0, y M}. But the equality (6) is pretty, for if we start with some φ M ann, then for every x X, the element y = T x belongs to M, and this implies φ(t x) = 0, x X, which in turn forces T φ = 0, and the injectivity of T forces φ = 0. B. The Eberlein-Smulian Theorem In this sub-section we discuss various notions of (relative) compactness, which ultimately will be proved to be equivalent in the case of the weak topology on a normed vector space. Recall that, given a locally convex topological vector space (X, T), the w T -topology on X is defined to be the weakest topology that makes all functionals φ (X, T) continuous. w Equivalently, for a net (x λ ) λ Λ and a vector x in X, one has x T λ x, if and only if φ(x λ ) φ(x), φ (X, T). Recall also that, for a convex subset C X the T-closure C T and the w-closure C w T coincide. In this sub-section we deal exclusively with normed vector spaces X, with T being the norm topology, and we agree to remove T from the notation. Notations. Suppose X is a normed vector space. Denote by Θ : X X the standard bidual embedding, where for every x X, the linear continuous map Θx X is defined by: (Θx)(φ) = φ(x), φ X. We know that Θ is linear and isometric. If we consider the bidual Banach space X = (X ), then besides the norm topology, X carries its weak dual topology. In order to avoid any future complications, we denote this topology by w w, so by definition η λ η (in X ) means η λ (φ) η(φ). Remark 1. The linear operator Θ is also continuous, as a map Θ : (X, w) (X, w ). Furthermore, Θ is in fact a homeomorphism, when regarded as a map Θ : (X, w) (Θ(X ), w ). (7) (On the range Θ(X ), the topology we consider is the induced w -topology.) The fact that (7) is a homeomorphism is trivial, since by definition, one has the equivalence: x λ w x (in X ) Θx λ w Θx (in X ). (8) 6

In particular, for a subset A X, one has the equivalence A is w-compact (in X ) Θ(A) is w -compact (in X ). (9) The Topology terminology we are going to use in this sub-section is as follows. Definitions. Suppose X is a topological Hausdorff space and A X is some subset. A. We say that A cluster-compact, if every infinite subset S A admits a cluster point in A. (Recall that a point x is said to be an cluster point for S, if for every neighborhood V of x, the intersection V (S {x}) is non-empty. This condition is equivalent to the condition that V S is infinite.) B. We say that A sequentially compact, if every sequence in A has a subsequence which is convergent to some point in A. A. We say that A relatively cluster-compact in X, if every infinite subset S A admits a cluster point (in X). B. We say that A relatively sequentially compact in X, if every sequence in A has a subsequence which is convergent to some point in X. C. We say that A is relatively compact in X, if its closure A is compact in X. Comments. We know from Topology that the condition that compactness for A can be characterized in terms of nets by the following statement: ( ) The set A is compact in X, if an only if every net in A has a sub-net, which is convergent to some point in A. Using net convergence terminology, the cluster-compactness and relative cluster-compactness can be equivalently described by the following statements. (a) The set A is cluster-compact in X, if an only if every sequence in A has a sub-net, which is convergent to some point in A. (a ) The set A is relatively cluster-compact in X, if an only if every sequence in A has a sub-net, which is convergent to some point in X. Because of these characterizations, the phrase cluster-compact is sometimes replaced by countably compact. What is missing from this discussion is a relative version of ( ). One might be tempted to consider modified version of statement ( ), by introducing the following terminology: we say that A X is relatively net-compact in X, if every net in A has a sub-net, which is convergent to some point in X. One could then guess that relative net-compactness is equivalent to relative compactness. Unfortunately, this conjectured equivalence is not true in general. Fortunately, in what follows, we will not be concerned with this distinction. (As it turns out, for what we are interested in here, we are not going to use relative net-compactness at all, although Exercise 4 below suggests that we should not worry about this issue, in the setting of this sub-section.) Exercises 1-2. Suppose X is a topological Hausdorff space, and A X. 7

1. Prove statements (a) and (a ) above. 2. Prove the following implications (i) If A is compact in X, then A is cluster-compact in X; (ii) If A is sequentially compact in X, then A is cluster-compact in X; (i ) If A is relatively compact in X, then A is relatively cluster-compact in X; (ii ) If A is relatively sequentially compact in X, then A is relatively cluster-compact in X; The next two exercises are optional. Exercise 3. Prove that: A relatively compact A relatively net-compact. Exercise 4* Prove that, if X is either 2 a topological vector space, or a metric space, then for a set A X, one has the equivalence: A is relatively compact A is relatively net-compact. Comment. The implication A compact A sequentially compact is false in general. (See Exercise?? from BS II.) Remark 2. If X is metrizable, then for a subset A X, one has the equivalences (i) A compact A cluster-compact A sequentially compact (ii) A relatively compact A relatively cluster-compact A relatively sequentially compact In preparation for the main result in this sub-section (Theorem 3 below), we first prove the following elementary technical result. Lemma 2. Let Y and Z be normed vector spaces, and let V L(Y, Z) be a finite dimensional linear subspace. For every constant C > 1, there exists a finite subset F (Y) 1, such that: V C max y F V y, V V. Proof. Consider the unit sphere Ω = {V V : V = 1}. Since M is finite dimensional, Ω is compact, in the norm topology. For every F P fin ((Y) 1 ), let f F : Ω [0, ) be the function defined by f F (V ) = max y F V y, V Ω. If V 1, V 2 Ω and y F P fin ((Y) 1 ), then V 1 y V 2 y + (V 1 V 2 )y f F (V 2 ) + V 1 V 2 y f F (V 2 ) + V 1 V 2, so taking maximum (over y F) yields f F (V 1 ) f F (V 2 ) + V 1 V 2. By symmetry, we also have f F (V 2 ) f F (V 1 ) + V 1 V 2, so we obtain the inequality f F (V 1 ) f F (V 2 ) V 1 V 2, V 1, V 2 Ω, F P fin ((Y) 1 ). This proves that, for each F P fin ((Y) 1 ), the function f F : Ω [0, ) is continuous. Since, for F, G P fin ((Y) 1 ), we obviously have the implication: f F f G, it follows that the net (f F ) F Pfin ((Y) 1 ) is increasing. Finally, by the definition of the operator norm, it is pretty clear 2 In fact the same conclusion holds, if X is a uniform space. 8

that, for every T Ω, one has the equality lim F f F (T ) = T = 1. (On the one hand, it is clear that f F (V ) V = 1. On the other hand, for every V Ω and every ε > 0, there exists y V,ε (Y) 1 such that V y V,ε V ε = 1 ε, so for every F P fin ((Y) 1 ) satisfying F {y V,ε }, we have 1 f F (V ) V y V,ε 1 ε.) This argument proves that the increasing net (f F ) F Pfin ((Y) 1 ) C R (Ω) converges point-wise to the constant (continuous) function 1. By Dini s Theorem (see BS III), the same net will converge uniformly on Ω. In particular, for every ε > 0, there exists F ε P fin ((Y) 1 ), such that: f Fε (V ) 1 ε, V Ω. Of course, this yields 1 (1 ε) 1 max y Fε V y, V Ω, so by homogeneity (every V V can be written as V = V V 1, with V 1 Ω), we also get V (1 ε) 1 max y F ε V y, V V, and the Lemma follows, by choosing ε > 0 small enough, so that (1 ε) 1 C. With these preparations we are now in position to prove the following fundamental result. Theorem 3 (Day-Whitley). Let X be a normed vector space, let A be a non-empty relatively cluster-compact subset in (X, w), and let Θ : X X be the bidual embedding. For any η Θ(A) w, there exist x X and a sequence (a n ) n=1 A, such that Θx = η and a n w x. Proof. The key step is contained in the following Claim 1: There exist an increasing sequence (F n ) n=1 of finite subsets in (X ) 1 and a sequence (a n ) n=1 A, such that (a) µ 2 max φ Fn µ(φ), µ span{η, Θa 1,..., Θa n }, (b) (η Θan ) span Fn 1 1/n, for every n N. Start off by choosing some φ 0 (X ) 1 such that η 2 η(φ 0 ), and put F 0 = {φ 0 }. The two sequences are constructed recursively as follows. Assume the elements a 1,..., a N 1 A and the finite sets F 0 F 1 F N 1 (X ) 1 have been constructed, so that (a) and holds for 0 n N 1 and (b) holds for 1 n N 1, and let us indicate how the next finite set F N satisfying F N 1 F N (X ) 1 and the next element a N A can be constructed, so that (a) and (b) hold for n = N and (b). (At start we have N = 1. We have nothing to verify in condition (b). Condition (a) is satisfied, by our choice of φ 0.) First of all, if we work on the finite dimensional dual space L = [span F N 1 ], any two norms on it are equivalent. In particular, there exists some constant M > 0, such that µ M max φ FN 1 µ(φ), µ L. Of course if we start with arbitrary functionals ν X and we restrict them to span F N 1 we obtain functionals in L, so we now have: ν span FN 1 M max φ F N 1 ν(φ), ν X. (10) 9

We know that η belongs to the w -closure of Θ(A), so for every finite subset G X and every ε > 0, there exists x = x G,ε A, such that max φ G (η Θx)(φ) < ε. Choosing then ε such that ε 1 MN, and letting G = F N 1, there exists a A, such that max (η Θa)(φ) < 1 φ F N 1 MN, so by (10) we have (η Θa) span FN 1 1/N, so our next element a N = a clearly satisfies condition (b) with n = N. In order to construct the finite set F N we argue as follows. Consider the finite dimensional space V N = span{η, Θa 1,..., Θa N } X, and apply Lemma 2 with Y = X, Z = K, and C = 2, to obtain a finite subset F (X ) 1, such that µ 2 max µ(φ), µ V. φ F The desired set F N can now be defined as F N 1 sup F. For the remainder of the proof, we fix two sequences (a n ) n=1 A and (F n ) n=1 as in Claim 1. Denote the union n=1 F n by B. As discussed at the beginning of this sub-section, the norm closure and the weak closure of span{a n : n N } coincide, i.e. span{a m : n N} = span{a m : n N} w. (11) Denote this closure by E, and consider the linear subspace M = span(θ(e) {η}) = Θ(E) + Kη X. Claim 2: If we define, for every µ X, the quantity µ B = sup φ B µ(φ), then. B is a norm on M, which is equivalent to the dual norm from X, namely one has: 1 µ µ 2 B µ, (12) for every µ M. It is pretty obvious that. B is a seminorm on X. Of course, the second inequality in (12) holds for all µ X. To prove the first inequality, we fix µ M and ε > 0. Write µ = Θ(x) + αη, with x E and α K. By the definition of E, we know that there exists a span{a n : n N}, such that x a ε. Since a span{a 1,..., a n }, for some n, the functional ν = Θ(a) + αη belongs to span{η, Θa 1,..., Θa n }, so by property (a) it follows that ν 2 max ν(φ) 2 ν B. (13) φ F n Using the first inequality in (12), and the fact that Θ is isometric, we know that: ν µ B ν µ = (Θa + αη) (Θx + αη) = Θ(a x) = a x ε. These estimates yield the inequalities µ ε ν and ν B µ B + ε (here we use the fact that. B is a seminomrm), so if we go back to (13), we obtain µ ε ν 2 ν B 2 µ B + 2ε. 10

Since the inequality µ ε 2 µ B + 2ε holds for all ε > 0, we must have µ 2 µ B, thus proving the desired first inequality from (12). Claim 3: lim n φ(a n ) = η(φ), φ B. Indeed, if we fix some φ B, and we choose k, such that φ F k, for every n > k, we have F k F n 1, so using condition (b) it follows that: η(φ) φ(a n ) = (η Θa n )(φ) (η Θan ) span Fn 1 φ 1/n, n > k, and the Claim follows. The next step is the most important one. Claim 4: Whenever (a σ(λ) ) λ Λ is a sub-net of (a n ) n=1 implemented by a directed map σ : Λ N which is w-convergent to some x X, it follows that Θx = η. Since a σ(λ) w x, it follows that x belongs to the w-closure E = span{a n : n N} w. Using Claim 2, it follows that η Θx 2 η Θx B. (14) Fix for the moment some φ B. On the one hand, since the net ( φ(a σ(λ) ) ) is a clearly a λ Λ sub-net of the sequence ( φ(a n ) ), which by Claim 3 converges to η(φ), we also have n=1 lim λ φ(a σ(λ) ) = η(φ). (15) On the other hand, since a w σ(λ) x, it follows that lim λ φ(a σ(λ) ) = φ(x), so by (15) we now have η(φ) = φ(x), which can equivalently be written as: (η Θx)(φ) = 0. Since this equality holds for arbitrary φ B, it forces η Θx B = 0, which by (14) forces η Θx = 0, i.e. η = Θx. To finish the proof of the Theorem, we argue as follows. Since A is assumed to be relatively cluster-compact in (X, w), there exists some x X, a directed set Λ, and a directed map σ : Λ N, such that the sub-net (a σ(λ) ) λ Λ is convergent to x in the weak topology. By Claim 4, we know that Θx = η. We wish to prove that in fact a w n x 0. Argue by contradiction, assuming the existence of some w-neighborhood W of x, and of sequence of integers 1 k(1) < k(2) <..., such that a k(n) W, n. Again, using the relative cluster-compactness, there exists a directed set Λ, a directed map σ : Λ N, and some x X, so that the sub-net (a k(σ (λ ))) λ Λ is w-convergent to x. Since this is a sub-net of a sub-sequence of (a n ) n=1, it is a sub-net of (a n ) n=1, which by Claim 4 forces Θx = η = Θx. By the injectivity of Θ we must have x = x, which is impossible, since a k(σ (λ )) W, λ Λ. Corollary 2. If X is a normed vector space, and A is relatively cluster-compact in (X, w), then for every a A w, there exists a sequence (a n ) n=1 A, such that a w n a. Proof. We know that there exists a net (b λ ) λ Λ A, with b λ w a. Since Θ : (X, w) (X, w w ) is continuous, it follows that Θb λ Θa, so Θa belongs to Θ(A) w. By Theorem 3, applied to η = Θa, there exists x X and a sequence (a n ) n=1 A, such that Θx = η = Θa, and a n w x. Of course, by the injectivity of Θ it follows that x = a, and we are done. 11

Theorem 4 (Eberlein-Smulian relative version). Given a normed vector space X and a non-empty subset A X, the following are equivalent: (i) A is relatively compact in (X, w); (ii) A is relatively cluster-compact in (X, w); (iii) A is relatively sequentially compact in (X, w). Proof. As discussed previously (see Exercise 2), we only need to prove the implications already know that the implications (i) (ii) (iii). (ii) (i). Assume A is relatively cluster-compact, and let us prove that A is relatively compact in (X, w). We first notice that, for every φ X, the set φ(a) K is clearly relatively cluster-compact in K. By Remark 2, it follows that φ(a) is relatively compact in K. In particular, φ(a) K is bounded, for every φ X. By the Uniform Boundedness Principle (see Exercise?? from BS II), it follows that A is bounded in X, and the same will be true for the set Θ(A) X. In particular, there exists r > 0, such that Θ(A) (X ) r, and since (X ) r is compact in (X, w ) by Alaoglu s Theorem it follows that the closure C = Θ(A) w (X ) r is also compact in (X, w ). By Theorem 3, however, we know that C Θ(X ), so one can write C = Θ(B), for some B X. It is trivial that we have the inclusion B A, since Θ is injective. Finally, by Remark 1, we know that B is compact in (X, w), and therefore A which is contained in B is indeed relatively compact in (X, w). (ii) (iii). Assume A is relatively cluster-compact in (X, w), and (x n ) n=1 is a sequence in A, and let us indicate how a w-convergent subsequence can be constructed. Of course, if there exist an infinite set M N, such that x m = x n, m, n M, we are done. Assume no such M exists, so passing to a subsequence we can also assume that m n x m x n. (16) Consider then the infinite subset S = {x n : n N} A, which is again a relatively clustercompact in (X, w). Pick then some x X, which is a cluster point for S in (X, w). Since x is also a cluster point for S {x}, by removing, if necessary, one term from the sequence (x n ) n, we can assume that x n x, n N. (17) Since x S w, by Corollary 2 there exists a sequence (a m ) m=1 S, such that a w m x. If there existed an infinite subset M N, such that a m = a n, m, n M, then (a m ) m=1 would have a constant subsequence, which is impossible by (17). So passing to a subsequence, we can also assume that m n a m a n. (18) Of course, by the construction of S, there exists a sequence of (k m ) m=1 N, such that a m = x km, m. By (18) and (16) it is clear that m n k m k n, so clearly the sequence (k m ) m=1 N possesses a strictly increasing subsequence (k rn ) n=1. It is then clear that the sequence a rn = x krn is a subsequence of both (a m ) m and (x n ) n, which is w-convergent to x. 12

Theorem 5 (Eberlein-Smulian absolute version). Given a normed vector space X and a non-empty subset A X, the following are equivalent: (i) A is compact in (X, w); (ii) A is cluster-compact in (X, w); (iii) A is sequentially compact in (X, w). Proof. As before, we only need the implications (i) (ii) (iii). Note that (i) is equivalent to A is closed and relatively compact in (X, w), and (iii) is equivalent to: A is closed and relatively sequentially compact in (X, w). Clearly, cluster-compact relatively cluster-compact, so by Theorem 4, it suffices to prove the implication cluster-compact closed. (Of course, everything here refers to the weak topology.) Assume A is cluster-compact in (X, w), and let us prove that A is w-closed. Start with some point x A w, and let us prove that x belongs to A. On the one hand, by Corollary 2, there exists a sequence (a n ) n=1 A, such that a w n x. On the other hand, by clustercompactness there exists a sub-net (a σ(λ) ) λ Λ of (a n ) n=1, which is w-convergent to some point a A. Of course, since a w n x, all the subnets w-converge to x, so we have a = x, which means that x A. Corollary 3. Suppose X is a normed vector space. For a subset A X, the following are equivalent: (i) A is compact in (X, w); (ii) for every separable norm-closed linear subspace Y X, the set A Y is compact in (Y, w). Proof. We know that every norm-closed linear subspace in X is also w-closed. (i) (ii). This implication is pretty obvious (and holds for arbitrary Y s), since for any norm-closed linear subspace Y X, the w-topology of Y coincides with the induced w-topology from X. This is due to the fact that every linear norm-continuous functional φ : Y K can be extended to a linear norm-continuous functional on X, using the Hahn- Banach Theorem. Therefore, if Y is norm-closed, hence w-closed in X, and A is w-compact in X, then A Y is w-compact in X, hence also compact in (Y, w). (ii) (i). Assume condition (ii) holds, and let us prove that A is compact in (X, w). By Theorem 5 it suffices to show that A is sequentially compact. Start with some sequence (a n ) n=1 A, and let us indicate how one can construct a subsequence, which is w-convergent to some point in A. Consider the linear subspace Y = span{a n : n N}, which is clearly separable. Since A Y is compact in (Y, w), and a n A Y, again by Theorem 5, there exists a subsequence (a kn ) n=1, which converges in (Y, w) to some a A. Since the w-topology on Y is the induced w-topology from X, it follows that a kn a in (X, w) as well, and we are done. Exercise 6. Prove the relative version of Corollary 3, namely the fact that the following two conditions are equivalent as well: 13

(i) A is relatively compact in (X, w); (ii) for every separable norm-closed linear subspace Y X, the set A Y is relatively compact in (Y, w). C. Applications One important application of the Eberlein-Smulian Theorem concerns the Reflexivity Problem: Given a Banach space X, decide when the bidual embedding Θ : X X is surjective. Since Θ is already an isometric linear map, the surjectivity of Θ will imply that one will have an isometric linear isomorphism Θ : X X. Definition. A normed vector space X, for which Θ is surjective, is called reflexive. As noted above, this condition implies that Θ : X X is an isometric linear isomorphism, so (since duals are Banach) X is necessarily a Banach space. Exercise 6. Let X be a normed vector space. (i) Prove that a linear functional Σ : X K is w -continuous, if and only if there exists φ X, such that Σ(η) = η(φ), η X. (ii) Define, for every set A X its annihilator in X (see also DT I) as A ann = {φ X : η(φ) = 0, η A}. Prove that, given A X and a w -closed linear subspace Z Z, the condition Z = span A w is equivalent to the equality A ann = Z ann. (iii) Using part (ii), prove that the linear subspace Θ(X ) is w -dense in X. Conclude that X is reflexive, if and only if Θ(X ) is w -closed in X. Exercise 7. Assume X is a normed vector space and η X. (i) Prove that η belongs to Θ(X ), if and only if η : X K is w -continuous. (ii) Prove that, if X is a Banach space, then the above condition is also equivalent to the condition that η is w -continuous on the unit ball (X ) 1. (Hint: For (ii) consider the linear subspace Z = Ker η X. Argue that, if the map η (X ) 1 w -continuous on the unit ball (X ) 1, then (Z) 1 is w -closed in (X ) 1, hence also w -closed in X. Use Corollary 1 to conclude that Z is w -closed in X, so see TVS?? φ is indeed w -continuous.) Reflexivity admits the following characterization in terms of the w-topology. Proposition 1 (Local Reflexivity Criterion). For a normed vector space X, the following are equivalent: (i) X is reflexive; (i ) every separable norm-closed linear subspace Y X is reflexive; 14

(ii) the unit ball (X ) 1 is compact in (X, w). Proof. Since for every norm-closed linear subspace Y X, one has the equality (Y) 1 = (X ) 1 Y, by Corollary 3 it follows that condition (ii) is equivalent to the condition: (ii ) for every separable norm-closed linear subspace Y X, the unit ball (Y) 1 is compact in (Y, w), therefore it suffices to prove the equivalence (i) (ii). Since Θ : (X, w) (Θ(X ), w ) is a homeomorphism, its restriction to the unit balls Θ 1 : ((X ) 1, w) ((Θ(X )) 1, w ) is also a homeomorphism. This means that condition (ii) is equivalent to the condition: ( ) (Θ(X )) 1 is compact in (X, w ). Since the w -topology on X is, of course the w -topology on the topological dual of the Banach space X, and Θ(X ) is a linear subspace, by Corollary 1, it follows that condition ( ) is equivalent to the condition ( ) Θ(X ) is closed in (X, w ). Finally, by Exercise 6, condition ( ) is equivalent to (i). Comments. Condition (i ) is referred to as local reflexivity. Note that, since the unit ball is always w-closed, condition (ii) can slightly be weakened, by requiring that (X ) 1 be relatively compact in (X, w). Reflexivity is obviously a topological condition. In particular: if X is reflexive, and Y is a Banach space, such that there exists a linear continuous isomorphism T : X Y, then Y is reflexive; if X is a reflexive complex Banach space, then X is also reflexive as a real Banach space; conversely, if X is a complex Banach space, which contains a real norm-closed subspace X 0, such that ix 0 X 0 = {0}, ix 0 + X 0, and X 0 is a real reflexive Banach space, then X is reflexive. Exercise 8. Prove the above three statements. Exercise 9. Assume X is reflexive, and Y is a norm-closed linear subspace. Show that both Y and X /Y are reflexive. Exercise 10. Let X be a Banach space. Prove that the following are equivalent: (i) X is reflexive; (ii) X is reflexive; (iii) X is reflexive. 15

Examples. Not all Banach spaces are reflexive. For instance c 0,K (S), l 1 K (S), and l K (S), provided, of course, that S is infinite. The Banach spaces l p K (S), 1 < p < are reflexive. All finite dimensional spaces are reflexive. The Banach spaces C K (Ω) with Ω compact Hausdorff space are not reflexive, unless Ω is a finite set (in which case C K (Ω) is finite dimensional). For more on the non-reflexivity of C K (Ω), see Exercises 11-13 below. Exercise 11-13. Let Ω be a compact Hausdorff space. 11. Prove that, for a sequence (f n ) n=1 and a function f in C(Ω), the following are equivalent: (i) f n w f; (ii) (f n ) n=1 is bounded, and f n (ω) f(ω), ω Ω. (Hint: Condition (i) is equivalent to: lim n Ω f ndµ = f dµ, for every positive Ω Radon measure.) 12. Assume for the moment that Ω is metrizable and infinite. Let ω 0 Ω be cluster point. (i) Construct a decreasing sequence of functions (f n ) n=1 C(Ω), such that f n (ω 0 ) = 1, n, and lim n f n (ω) = 0, ω ω 0. (ii) Show that no subsequence of (f n ) n=1 can converge in the w-topology to a continuous function f C(Ω). (iii) Conclude that C(Ω) is not reflexive. 13*. Prove that if Ω is arbitrary, but infinite, then C(Ω). (Hint: If Ω is infinite, C(Ω) is infinite dimensional. Pick a linearly independent countable set S C(Ω), and let A be the unital involutive subalgebra of C(Ω) generated by S, so that B is separable. By Exercises?? and?? from BS III, the norm closure A is isometrically isomorphic to C(Ω 0 ) for some infinite metrizable compact space Ω 0. By the preceding Exercise, A is not reflexive.) For the remainder of this sub-section we present another beautiful application (Theorem 6 below), concerning closed convex hulls. We begin with a technical result, which contains a special case of Theorem 6. Lemma 3. Suppose X is a separable Banach space, and Ω X is a w-compact subset. Consider the space 3 M K (Ω) of all K-valued Radon measures on Ω, equipped with the w - topology. (i) For every µ M K (Ω), there exists a unique point x µ X, such that 4 φ(x µ ) = Ω φ(x) dµ(x), φ X. 3 It is understood that Ω is equipped here with the induced w-topology, which turns it into a compact Hausdorff space. 4 Any φ X is w-continuous, so the restriction φ Ω belongs to C K (Ω). 16

(ii) The correspondence Σ : M K (Ω) µ x µ X is linear and continuous as a map Σ : (M K (Ω), w ) (X, w). (iii) One has the equality 5 Σ ( Prob(Ω) ) = conv Ω. (19) In particular, the closed convex hull conv Ω is compact in (X, w). Proof. (i). Start off by defining, for every µ M K (Ω), the map η µ : X φ φ(x) dµ(x) K. It is pretty clear that η µ : X K is linear and norm-continuous. Claim 1: η µ is w -continuous on the unit ball (X ) 1. Since X is separable, the unit ball ((X ) 1, w ) is metrizable (see BS I), so it suffices to prove the following implication (sequential continuity): w φ (in (X ) 1 ) φ n (x) dµ(x) φ(x) dµ(x). φ n Ω But this is pretty obvious from Lebesgue s Dominated Convergence Theorem. (Without any loss of generality, we can assume that µ M + (Ω), so µ is an honest i.e. positive measure.) Claim 2: η µ is w -continuous on the whole space X. To prove this, it suffices to show that the linear subs-space Z = Ker η µ is w -closed. By Corollary 1 it suffices to show that the unit ball (Z) 1 is w -closed, and this is trivial from Claim 1. Using Claim 2, statement (i) now follows immediately, since every w -continuous linear functional η : X K is given by a unique x X as η(φ) = φ(x), φ X. (ii). Linearity of Σ is obvious. For continuity we start with a net (µ λ ) λ Λ M K (Ω), such w that µ λ 0, which means that f(x) dµ λ 0, f C K (Ω), (20) and we show that x µλ Ω w 0, which is equivalent to: Ω Ω Ω φ(x) dµ λ 0, φ X. (21) It is trivial, however, that (21) follows from (20), since the restriction of any φ X to Ω is w-continuous. (iii). First of all, since Prob(Ω) is convex and w -compact in M K (Ω), by part (ii) it follows that the set Σ ( Prob(Ω) ) is convex and w-compact in X. Secondly, if we consider 6 5 For convex sets C X, their norm-closures and w-closures coincide: C = C w. (See LCVS V.) 6 See BS III, end of sub-section D. 17

the point-mass measures δ x Prob(Ω), then by Remark 6 from BS III, we know that Prob(Ω) = conv{δ x : x Ω} w, so by the linearity and continuity condition (ii), combined with the compactness of both Prob(Ω) and Σ ( Prob(Ω) ), it follows that Σ ( Prob(Ω) ) = conv{σ(δ x ) : x Ω} w. The desired equality (19) is now obvious, since clearly Σ(δ x ) = x, x Ω. Theorem 6 (Krein-Smulian) Let X be a Banach space. For a subset A X, the following are equivalent: (i) A is relatively compact in (X, w); (ii) conv A is relatively compact in (X, w). In particular, if A is compact in (X, w), then conv A w (= conv A) is compact in (X, w). Proof. The implication (ii) (i) is trivial, since conv A A. (i) (ii). Using Theorem 4 it suffices to show that conv A is relatively sequentially compact. Start with some sequence (a n ) n=1 conv A, and let us indicate how to construct a w-convergent subsequence. Using the definition of the convex hull, each a n is a convex combination of finitely many points in A, so if we collect all these elements (that contribute to all a n, n N), it follows that there exists a countable subset B A and a norm-closed separable subspace Y, such that a n conv B Y, n. Consider then the set Ω = B w Y, which is compact in (Y, w). According to Lemma 3, the set C = conv Ω w is compact in (Y, w). Since a n C, n, by Theorem 5 it follows that (a n ) n=1 indeed has a w-convergent subsequence. The second statement in the Theorem is trivial. Remark 3. One can show that Lemma 3 is valid for arbitrary Banach spaces X. Indeed, using the fact that every µ M K (Ω) is a linear combination of probability Radon measures, it suffices to prove statement (i) in the case when µ Prob(Ω). In this case, we can extend µ to a probability Radon measure µ on the compact space Ω = conv Ω, defined by µ(b) = µ(b Ω), B Bor( Ω), so we can simply take x µ to be the µ-barycenter of Ω, as explained in BS III, sub-section E. The other statements are proved exactly as we did in Lemma 3. 18