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I feel like this must be a monumentally stupid question. Say $X$ is a Banach space, $S\subset X^*$, and $x^*$ is in the weak* closure of $S$. Must $x^*$ lie in the weak* closure of some norm-bounded subset of $S$?

(If $x^*$ is the weak* limit of a sequence of elements of $S$ then this is clear by Banach-Steinhaus. But a convergent net of scalars need not be bounded...)

David C. Ullrich
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    The question does not ask whether this subset may depend on $x^*$. – Crostul Mar 20 '16 at 23:52
  • If $B$ is a closed ball of $X^$ containing $x^$, then $B$ is weakly* closed and norm-bounded. Maybe you can consider $S \cap B$? – Crostul Mar 20 '16 at 23:56
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    Yes, I know that a closed ball is weak* closed - in fact it's weak* compact. I haven't seen how that helps. Your comment that I don't specify whether the subset may depend on $x^$ makes very little sense to me, sorry. $S$ and $x^$ are given. – David C. Ullrich Mar 21 '16 at 00:06
  • if I understood well the definitions, $X^$ is a normed vector space for the operator norm. for any $y \in X^$ : $|y|{X^*}= \max{\textstyle x \in X, |x|X \le 1} |y(x)|$, and it is a Banach space since $X$ is complete. the weak closure $\overline{S}$ of $S \subset X^$ is $s \in \bar{S}$ iff there is a sequence $(s_n) \in S$ such that for every $x \in X$ : $|s(x) - s_n(x)| \to 0$. but this doesn't ensure that $s \in X^$ i.e. that $|s|{X^*} < \infty$ ? – reuns Mar 21 '16 at 00:33
  • or maybe I didn't understand what was the weak* closure ? – reuns Mar 21 '16 at 00:41
  • @user1952009 I guess so. ;-) – Friedrich Philipp Mar 21 '16 at 00:42
  • @FriedrichPhilipp : very funny... so, what means $s \in \bar{S}$ the weak* closure of $S \in X^*$ ? – reuns Mar 21 '16 at 00:46
  • @user1952009 The biggest problem with your comment is the word "sequence". The weak* topology is not metrizable. – David C. Ullrich Mar 21 '16 at 00:47
  • a closure is always adding the limits of some "so said converging" sequences... ? – reuns Mar 21 '16 at 00:48
  • @user1952009 Sorry, but why don't you just google it? – Friedrich Philipp Mar 21 '16 at 00:48
  • @user1952009 No, that is simply not true. – David C. Ullrich Mar 21 '16 at 00:49
  • @user1952009 +1 for wishing us fun. ;-) – Friedrich Philipp Mar 21 '16 at 00:58
  • @user1952009 Sorry, but it sounds like you simply don't understand the difference between a topologicall space and a metric space, and that's a long story... – David C. Ullrich Mar 21 '16 at 01:08
  • @DavidC.Ullrich : when reading wikipedia, I have the feeling it says what I wrote... https://en.wikipedia.org/wiki/Weak_topology#Weak-.2A_convergence – reuns Mar 21 '16 at 01:32
  • @user1952009 This is getting a little tedious. It does not say what you wrote. That bit on Wikipedia refers to nets, not sequences. – David C. Ullrich Mar 21 '16 at 01:34

3 Answers3

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The Krein-Smulian theorem says that for a convex set $S$, having weak*-closed intersections with closed balls implies being weak*-closed. The fact that convexity is required here suggests that the answer to your question should be negative. And it is.

Let $X=\ell^2$. In the dual space, also $\ell^2$, consider the "infinite ellipsoid" $$ S = \left\{y : \sum_{n=1}^\infty y_n^2/n^2=1\right\} $$ Every line through the origin meets $S$; therefore, $0$ is in the weak closure of $S$.

On the other hand, $0$ is not in the weak closure of any intersection $S\cap B_R$ where $B_R=\{y:\|y\|\le R\}$. Indeed, consider the weak-open sets $$ U_N = \left\{y: \sum_{n=1}^N y_n^2 < \frac12\right\} $$ If $N$ is large enough, then for every $y\in U_N\cap B_R $ we have $$ \sum_{n=1}^\infty y_n^2/n^2 < \frac12+ \frac{R^2}{(N+1)^2} < 1 $$ hence $U_N\cap (B_R\cap S)= \varnothing$.

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    Excellent - my $S$ actually is convex, so Krein-Smulllyan is just what I needed. – David C. Ullrich Mar 21 '16 at 01:20
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    @DavidC.Ullrich Lol, why didn't you say this before... ;-) – Friedrich Philipp Mar 21 '16 at 01:21
  • @FriedrichPhilipp Sorry. Should have realized convexity could be relevant. Heh, I said it must be a stupid question, just a little stupider than I thought. – David C. Ullrich Mar 21 '16 at 01:31
  • @DavidC.Ullrich As 404's answer shows, it's not that stupid. And I just realized that you might not have seen that convexity plays a role. So, why should you have mentioned it. Maybe you did not even check your set for convexity before. I admit I had forgotten about Krein-Shmulyan, but I knew that weak closure and norm-closure of convex sets are the same. So, I thought $S$ should probably not be convex for a counterexample. However, have fun applying Krein-Shmulyan. :-) – Friedrich Philipp Mar 21 '16 at 01:37
  • @404: Can you please explain your sentence: "Every line through the origin meets $S$; therefore, $0$ is in the weak closure of $S$."? The unit circle in $\mathbb{R}^n$ also has this property, but $0$ is not in its weak(*) closure. – gerw Mar 21 '16 at 11:49
  • @gerw Every weak neighborhood of 0 in an infinite-dimensional Hilbert space contains a iine (since it contains a subspace of finite codimension). –  Mar 21 '16 at 11:52
  • @404: Of course! -- and $\mathbb{R}^n$ is not infinite dimensional ;) – gerw Mar 21 '16 at 11:59
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In addition to the answer by 404, I have two further counterexamples in $\ell^2$: \begin{align*} A &= \{ \sqrt{n} \, e_n : n \in \mathbb{N}\} \\ B &= \{ e_m + m \, e_n : m,n \in \mathbb{N}, 1 \le m < n \} \end{align*} Again, $0$ is in the weak* closure, but all intersections of $A$ or $B$ with bounded sets are discrete! (It is also interesting to note that this implies that there is no sequence in $A$ or $B$ converging weak*ly to $0$.)

gerw
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The reference to Krein-Smulian in one of the answers is actually a little misleading, and is related to the following warning in Conway's Functional Analysis, page 161:

page 161

In fact, you can find a Banach space $X$ and $S\subseteq X^\ast$ a subspace (so certainly convex) such that $S$ is weak$^*$-dense in $X^\ast$, but there is a vector $f_0\in X^\ast$ such that no bounded net in $S$ converges to $f_0$; equivalently, $f_0$ is not in the weak$^\ast$-closure of $S$ intersected with any bounded subset of $X^\ast$.

I do not know a canonical reference for a construction. However, it is fairly easy to adapt this Mathoverflow answer. I wrote up the details here (github).

Matthew Daws
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