1
$\begingroup$

If it is given that $X$ is a reflexive Banach space. Let $K \subset X$ be a norm closed and norm bounded convex set. I want to show that $K$ is weakly compact. I have the following idea but I am not sure about how to prove one thing.

Idea for proof: $K$ is bounded therefore there exists some closed ball $B_{r}$(of radius $r$) such that $K \subset B_{r}$. We know that closed ball $B_{r} = rB_{1}$, therefore the closed unit ball $B_{1} = \frac{1}{r}B_{r}$. By Kakutani's Theorem $B_{1}$ is weakly compact. I'm not sure then if we can state that since $B_{1}$ is weakly compact it follows that $B_{r}$ is weakly compact. If it does then since $K$ is weakly closed(since it is norm closed) it follows that $K$ is also weakly compact.

So basically I want to know if it follows that if $B_{1}$ is weakly closed then any closed ball $B_{r}$ is weakly compact? If so, could I get a hint of how to show this? If not are there any suggestions regarding how to prove that $K \subset X$ is weakly compact?

$\endgroup$
  • 1
    $\begingroup$ Let $\mathcal{O}=\{O_{\alpha}\,\mid\,\alpha\in\mathcal{A}\}$ be an open cover of $B_r$. Then $\frac{1}{r}\mathcal{O}$ is an open cover of $B_1$ and hence, there is a finite subcover, $\frac{1}{r}\mathcal{O}'=\{\frac{1}{r}O_{\alpha_k}\,\mid\,k\in[1,n]\cap\mathbb{N}\}$. Since $B_1\subset\frac{1}{r}\bigcup\limits_{k=1}^n O_{\alpha_k}$, $B_r\subset\bigcup\limits_{k=1}^nO_{\alpha_k}$ and hence $B_r$ is compact. $\endgroup$ – Alex Schiff Jun 17 '14 at 17:25
  • 1
    $\begingroup$ A fancier fact: in any topological vector space, scaling is a homeomorphism, so it maps compact sets to compact sets. $\endgroup$ – Nate Eldredge Jun 17 '14 at 17:29
  • $\begingroup$ @NateEldredge yeah, I should have thought of that. Oh well. $\endgroup$ – Alex Schiff Jun 17 '14 at 17:31

Your Answer

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

Browse other questions tagged or ask your own question.