8
$\begingroup$

The Krein-Smulian Theorem states that for a convex set S, having weak*-closed intersections with closed balls implies being weak*-closed.

I would be really happy to have an example of a NOT weak-closed convex subset $S$ of a Banach space, such that the intersection with the closed balls are always weak-closed.

In this post, there are some counterexamples for the "convex" hypothesis. I would like to have counterexamples for the "weak*" hypothesis.

$\endgroup$
4
+100
$\begingroup$

This is true for the weak topology too and it is easy to prove.

Proof. By the geometric form of the Hahn-Banach theorem, for a convex subset of $X$ to be closed is the same as to be weakly closed so it is enough to show that $S$ is closed.

Suppose that $S\subset X$ is convex and $S\cap B$ is (weakly) closed for every closed ball $B$ in $X$. Let $(x_n)_{n=1}^\infty$ be a sequence in $S$ which converges to some $x\in X$. Then $(x_n)_{n=1}^\infty$ is bounded, so it is contained in some ball $B$. However $S\cap B$ is closed, so $x\in S\cap B$. Consequently, $S$ is (weakly) closed. $\square$

$\endgroup$
  • $\begingroup$ Thanks, Tomek! It seems Mazur's lemma is for sequences. But, is "sequentially weakly closed" the same as closed in all Banach spaces? $\endgroup$ – André Caldas May 12 '17 at 22:21
  • $\begingroup$ So, my conclusion is that the difficulty in proving Krein-Smulian rests in the fact that when you have a strongly closed convex set $S \subset X^*$ and a point $\phi \in X^*$, it is not as easy to find $x \in X$ separating $S$ and $\phi$, as it is easy to apply Hahn-Banach and get an $F \in X^{**}$. $\endgroup$ – André Caldas May 13 '17 at 10:26
  • $\begingroup$ This is so disappointing! :-P $\endgroup$ – André Caldas May 13 '17 at 10:30
  • $\begingroup$ I'd like to take this opportunity and suggest to professors teaching the subject and to authors writing about it, that they introduce the theorem by first showing how easy it is to prove for the strong and weak topologies... and then, arguing that it is not so simple in the weak* case because Hahn-Banach cannot be directly applied. Thank you very much, Tomek! :-) $\endgroup$ – André Caldas May 13 '17 at 10:34
  • $\begingroup$ I think it would be nice if you could edit your post in order to avoid talking about sequences. $\endgroup$ – André Caldas May 15 '17 at 0:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.