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2.3 the Theorem of Eberlein Smulian 2.3. THE THEOREM OF EBERLEIN SMULIAN 41 2.3 The Theorem of Eberlein Smulian For infinite dimensional Banach spaces the weak topology is not metrizable (see Exercise 1). Nevertheless compactness in the weak topology can be characterized by sequences. Theorem 2.3.1. [The Theorem of Eberlein- Smulian] Let X be a Banach space. For subset K the following are equivalent. σ(X,X∗) a) K is relatively σ(X, X∗) compact, i.e. K is compact. b) Every sequence in K contains a σ(X, X∗)-convergent subsequence. c) Every sequence in K has a σ(X, X∗)-accumulation point. We will need the following Lemma. Lemma 2.3.2. Let X be a Banach space and assume that there is a count- able set C = xn∗ : n N BX , so that C = 0 . We say that C is total { ∈ }⊂ ∗ ⊥ { } for X. Consider for x, y ∞ n d(x, y)= 2− x∗ ,x y . |# n − %| n=1 ! Then d is a metric on X, and for any σ(X, X∗)-compact set K, σ(X, X∗) coincides on K with the metric generated by d. Proof. It is clear that d is a metric on X. Assume that K X is weak ⊂ compact. By the Theorem of Banach Steinhaus 2.1.12 K is therefore norm bounded and we consider the identity I as map from the space (K,σ(X, X ) ∗ ∩ K) to (K, K)( being the topology generated by d). Then I is con- Td ∩ Td tinuous: Indeed, if (x : i I) is a net which is converging in σ(X, X ) to i ∈ ∗ some x K and if ε> 0 is arbitrary, we first use the boundedness of K to ∈ find n N so that ∈ ∞ j 1 n+1 2 − xj∗,xi x 2− sup x <ε/2, |# − %|≤ x K ( ( j=!n+1 ∈ and then we choose i I so that n 2j 1 x ,x x <ε/2 for all i I, 0 ∈ j=1 − |# j∗ i − %| ∈ with i i0,. It follows that d(xi,x) <ε. ≥ " Since images of compact sets under continuous functions are compact, and thus (by bijectivity of I) images of σ(X, X∗)-open sets in K under I are open in (K, ) it follows that I is a homeomorphism. T 42CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS Lemma 2.3.3. Assume that X is separable. Then there is a countable total set C X . ⊂ ∗ Proof. Let D X be dense, and choose by the Corollary 1.4.6 of the The- ⊂ orem of Hahn Banach for each element x D, an element y S so ∈ x∗ ∈ X∗ that y ,x = x . Put C = y : x D . If x X, x = 0, is arbi- # x∗ % ( ( { x∗ ∈ } ∈ * trary then there is a sequence (xk) D, so that limk xk = x, and thus ⊂ →∞ limk yx∗ ,x = x > 0. Thus there is a x∗ C so that x∗,x = 0, →∞# k % ( ( ∈ # %* which implies that C is total. Proof of Theorem 2.3.1. “(a) (b)” Assume that K is σ(X, X )-compact ⇒ ∗ (if necessary, pass to the closure) and let (x ) K be a sequence, and put n ⊂ X0 = span(xn : n N). X0 is a separable Banach space. By Proposition ∈ 2.1.5 the topology σ(X0,X0∗) coincides with the restriction of σ(X, X∗) to X . It follows therefore that K = K X is σ(X ,X )-compact. Since X 0 0 ∩ 0 0 0∗ 0 is separable, by Lemma 2.3.3 there exists a countable set C BX , so that ⊂ 0∗ C = 0 . ⊥ { } It follows therefore from Lemma 2.3.2 that (K ,σ(X ,X ) K ) is 0 0 0∗ ∩ 0 metrizable and thus (xn) has a convergent subsequence in K0. Again, using the fact that on X0 the weak topology coincides with the weak topology on X, we deduce our claim. “(b) (c)” clear. ⇒ “(c) (a)” Assume K X satisfies (c). We first observe that K is (norm) ⇒ ⊂ bounded. Indeed, for x∗ X∗, the set Ax = x∗,x : x K K is ∈ ∗ {# % ∈ }⊂ the continuous image of A (under x∗) and thus has the property that every sequence has an accumulation point in K. This implies that Ax∗ is bounded in K for all x∗ X∗, but this implies by the Banach Steinhaus Theorem ∈ 2.1.13 that A X must be bounded. ⊂ Let χ : X$ X∗∗ be the canonical embedding. By the Theorem of → σ(X∗∗,X∗) Alaoglu 2.1.8, it follows that χ(K) is σ(X∗∗,X∗)-compact. There- σ(X ,X ) fore it will be enough to show that χ(K) ∗∗ ∗ χ(x) (because this ⊂ would imply that every net (χ(x ):i I) χ(K) has a subnet which i ∈ ⊂ σ(χ(X),X∗)) converges to some element χ(x) χ(X)). σ(X∗∗,X∗) ∈ So fix x∗∗ χ(K) . Recursively we will choose for each k N, 0 ∈ ∈ xk K, finite sets A∗ SX , if k N0, so that ∈ k ⊂ ∗ ∈ 1 (2.3) x∗∗ χ(x ),x∗ < for all x∗ A∗, # 0 − k % k ∈ j 0 j<k # # ≤$ # # 2.3. THE THEOREM OF EBERLEIN SMULIAN 43 x∗∗ (2.4) x∗∗ span(x0∗∗,χ(xj), 1 j k) x∗∗ max x∗∗,x∗ ( (. ∀ ∈ ≤ ≤ ( (≥x A∗ |# %|≥ 2 ∗∈ k For k = 0 choose A = x , x S , with x (x ) x /2, then 0∗ { ∗} ∗ ∈ X ∗ | ∗ 0∗∗ |≥( 0∗∗( condition (2.4) is satisfied, while condition (2.3) is vacuous. Assuming that x1,x2, . xk 1 and A0∗,A1∗,...,Ak∗ 1 have been chosen for some k>1, we can first choose−x K so that (2.3)− is satisfied (since A is k ∈ j∗ finite for j =1, 2, . k 1), and then, since span(x ,χ(x ),j k) is a finite − 0∗∗ j ≤ dimensional space we can choose A S so that (2.4) holds. k∗ ⊂ X∗ By our assumption (c) the sequence (xk) has an σ(X, X∗)- accumulation point x0. By Proposition 2.1.7 it follows that x0 Y = span(xk : k N)(·( = σ(X,X∗) ∈ ∈ span(xk : k N) . ∈ We will show that x0∗∗ = χ(x0) (which will finish the proof ). First note that for any x∗ j Aj∗ ∈ ∈N x0∗∗ χ(x0),x%∗ lim inf x0∗∗ χ(xk),x∗ + x∗,xk x0 =0. # − % ≤ k # − % # − % →∞ # # &# # # #' Secondly# consider the# space Z =# span(x∗∗,χ(xk),k# #N)(·( X∗∗ #it follows 0 ∈ ⊂ from (2.4) that the set of restrictions of elements of k∞=1 Ak∗ to Y is total in Z and thus that % ∞ x∗∗ χ(x ) Z A∗ = 0 , 0 − 0 ∈ ∩ k { } ( k$=1 )⊥ which implies our claim. Exercises 1. Prove that if X is a separable Banach space (BX∗ ,σ(X∗,X)) is metriz- able. 2. For an infinite dimensional Banach space prove that (X,σ(X, X )) is ∗ ∗ not metrizable. Hint: Exercise 4 in Section 2.1 3. Prove that for two Banach spaces X and Y , the adjoint of a lin- ear bounded operator T : X Y is w continuous (i.e σ(Y ,Y) → ∗ ∗ − σ(X∗,X)continuous). 4. Show that %1 isometric to a subspace of C[0, 1]. 5. Show that % is not complemented in C[0, 1]. ∗ 1 44CHAPTER 2. WEAK TOPOLOGIES, REFLEXIVITY, ADJOINT OPERATORS 2.4 The Principle of Local Reflexivity In this section we we proof a result of J. Lindenstrauss and H. Rosenthal [LR] which states that for a Banach space X the finite dimensional subspaces of the bidual X∗∗ are in a certain similar to the finite dimensional subspaces of X. Theorem 2.4.1. [LR] [The Principle of Local Reflexivity] Let X be a Banach space and let F X and G X be finite dimensional ⊂ ∗∗ ⊂ ∗ subspaces of X∗∗ and X∗ respectively. Then, given ε> 0, there is a subspace E of X containing F X (we ∩ identify X with its image under the canonical embedding) with dim E = dim F and an isomorphism T : F E with T T 1 1+ε such that → ( (·( − (≤ (2.5) T (x)=x if x F X and ∈ ∩ (2.6) x∗,T(x∗∗) = x∗∗,x∗ if x∗ G, x∗∗ F. # % # % ∈ ∈ We need several Lemmas before we can prove Theorem 2.4.1. The first one is a corollary the Geometric Hahn-Banach Theorem Proposition 2.4.2. [Variation of Geometrical Version of the Theorem of Hahn Banach] Assume that X is a Banach space and C X is convex with C = ⊂ ◦ * ∅ and let x X C (so x could be in the boundary of C). Then there exists ∈ \ an x X so that ∗ ∈ ∗ 0 x∗,z < 1= x∗,x for all z C , 0# % # % ∈ and, if moreover C is absolutely convex (i.e. if ρx C for all x C and ∈ ∈ ρ K, with ρ 1), then ∈ | |≤ 0 x∗,z < 1= x∗,x for all z C . |# %| # % ∈ Lemma 2.4.3. Assume T : X Y is a bounded linear operator between → the Banach spaces X and Y and assume that T (X) is closed. Suppose that for some y Y there is an x X with x < 1, so ∈ ∗∗ ∈ ∗∗ ( ∗∗( that T (x )=y. Then there is an x X, with x < 1 so that T (x)=y. ∗∗ ∗∗ ∈ ( ( Proof. We first show that there is an x X so that T (x)=y. Assume this ∈ where not true, then we could find by the Hahn-Banach Theorem (Corollary 1.4.5) an element y Y so that y (z) = 0 for all z T (X) and y ,y =1 ∗ ∈ ∗ ∗ ∈ # ∗ % 2.4. THE PRINCIPLE OF LOCAL REFLEXIVITY 45 (T (X) is closed). But this yields T (y ),x = y ,T(x) = 0, for all x X, # ∗ ∗ % # ∗ % ∈ and, thus, T ∗(y∗) = 0.
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