SEQUENTIAL PROPERTIES OF THE WEAK TOPOLOGY IN A ...

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4 OLAV NYGAARD, AGDER UNIVERSITY COLLEGEE ′ 3 = {x ∗ n(3)+1 , x∗ n(3)+2 , ...x∗ n(4)W 4 ={y ∗∗ ∈ X ∗∗ : |(y ∗∗ − x ∗∗ )(x ∗ i )| < 1 }3 , 1 ≤ i ≤ n(4) .a 4 ∈ W 4 ∩ Aand go on with E 4 to produce W 5 and thus a 5 and so on.Step 2: Some consequences of the contruction.Ok, we have the sequence (a i ). What properties does it have? We know, sinceA is relatively weakly sequentially compact that it has a subsequence (a nk ) with aweak limit x and, by Mazur’s theorem x ∈ [a i ]. But then,x ∗∗ − x ∈ [x ∗∗ , x ∗∗ − a 1 , x ∗∗ − a 2 , x ∗∗ − a 3 ] def= Y.The way Y is built gives us a possibility to control norms there, we have for anyy ∗∗ ∈ Y that‖y ∗∗ ‖≤ sup |y ∗∗ (x ∗ m)|.m2So more than half the norm is found along the sequence x ∗ m. But this is also thecase if we weak-star close Y , and soStep 3: x ∗∗ = x.‖x ∗∗ − x‖ ≤ 2 sup |(x ∗∗ − x)(x ∗ m)|.mThis follows as soon as we can show that for any ɛ > 0 and any m, |(x ∗∗ −x)(x ∗ m)| < ɛ/2. So, let m be arbitrary. We know that x is the weak limit of (a nk ),so x ∗ m(a nk − x) → 0. Thus, we can find a natural number p(m) such that n k > passures |x ∗ m(a nk − x)| < ɛ/4. Now, if necessary, increase p such that n k > p impliesthat also |x ∗∗ − a nk | < ɛ/4. By the triangle inequalityand we are done.|(x ∗∗ − x)(x ∗ m)| ≤ |(x ∗∗ − a nk )(x ∗ m)| + |x ∗ m(x ∗∗ − x)|,A remark is that we have only used very elementary means in a clever way.We needed only the Banach-Steinhaus theorem, Mazur’s theorem and Alaoglu’stheorem from functional analysis, together with basics from topology. Moreover,Observation 1 was used just to avoid stating that A is bounded. Thus, Eberlein’stheorem goes through for bounded sets in any normed space X.If one thought of proving Eberlein’s theorem with pointwise topologies on Γalong the same lines, one quickly comes into many problems. But if Γ has theproperty that bounded Γ-convergent sequences are weakly convergent, then Eberlein’stheorem holds for bounded sets.Definition 2.5. A set Γ ⊂ X ∗ is called a Rainwater set if every bounded Γ-convergent sequence is weakly convergent.We obtain the following theorem:Theorem 2.6. (Generalized Eberlein’s theorem) Let Γ be a Rainwater set in X ∗and assume that A is a bounded set in X. Then, if A is relatively sequentiallyΓ-compact, it is relatively weakly compact.□
SEQUENTIAL PROPERTIES OF THE WEAK TOPOLOGY IN A BANACH SPACE. 5We will meet Rainwater sets later on, we just mention now that the extremepoints of B X ∗ provides a such set.3. Smulian’s theoremSuppose A ⊂ X is relatively weakly compact. Then it is weakly bounded, hencebounded. Every sequence in A must now have a weak limit point. We will showthat this limit point in fact is the weak limit of a sequence in A, and this is theessence of Smulian’s theorem. We follow the very wise leadership of Dunford andSchwartz. Recall that a set Γ ⊂ X ∗ is called total if its span is weak-star dense inX ∗ .Theorem 3.1. (Smulian 1940) Suppose a subset A of a Banach space X is relativelyweakly compact. Then it is relatively weakly sequentially compact.Proof. First we show that we may assume X is separable. To this end, suppose thetheorem has been proved in the separable case and let A ⊂ X be relatively weaklycompact. Let (a n ) be a sequence in A. Let Y = [a n ]. Then Y is a separable, closedsubspace. A ∩ Y is relatively weakly compact in Y . Now pick a subsequence (a nk )that converges weakly to some y ∈ Y . Then, since every x ∗ ∈ X ∗ is nothing butan extension of some y ∗ ∈ Y ∗ , (a nk ) converges weakly to y in X as well.So, the situation is reduced to studying a relatively weakly compact compactset A in a separable Banach space X. We take a sequence (a n ) in A, which weknow has a limit point x ∈ X and we try to find a subsequence of (a n ) that couldhave the possibility to converge weakly to x. So, first of all we need a subsequencethat has a tendency of converging weakly to something. For this, suppose H is acountable subset of S X ∗, say (x ∗ n). Look at the collectionx ∗ 1a 1 x ∗ 1a 2 x ∗ 1a 3 x ∗ 1a 4 · · ·x ∗ 2a 1 x ∗ 2a 2 x ∗ 2a 3 x ∗ 2a 4 · · ·x ∗ 3a 1 x ∗ 3a 2 x ∗ 3a 3 x ∗ 3a 4 · · ·....By the Bolzano-Weierstrass theorem, we can pick subsequences such that everyhorizontal line consist of convergent scalar sequences. Assume so has been done.Then take the diagonal sequence, and denote the corresponding subsequence of (a n )by (y n ). By construction lim m x ∗ y m exists for every x ∗ ∈ H, so some kind of weakconvergence seems to emerge.Now we will argue that there is a y 0 ∈ X such that lim m x ∗ y m = x ∗ y 0 for everyx ∗ ∈ H. We know (y n ) has a weak limit point y 0 . This means that every weakneighborhood at y 0 contains at least one (y n ). In particular all the sets{x ∈ X : |x ∗ (x − y 0 )| < ɛ}contains some y m . Since x ∗ y m converges for every x ∗ ∈ H, the limit has to be x ∗ y 0 .This can be done for every countable H. Let us now make H ”big” to get closerto weak convergence of (y n ). Let (x n ) be dense in S X and pick x ∗ n ⊂ S X ∗ such thatx ∗ nx n = 1. Let H = (x ∗ n). Then H is total in X ∗ . To see this, let x ∈ S X be suchthat x ∗ nx = 0 for every n and let x ∗ ∈ S X ∗. Let ɛ > 0 and find some x n such that‖x − x n ‖ < ɛ. Then, since ‖x ∗ ‖ = 1, |x ∗ x| < ɛ. Since ɛ was arbitrary, it followsfrom the Hahn-Banach separation theorem that H is total.
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