9
$\begingroup$

Let $V$ be a Banach space. Show that the closed unit ball in $V$ is also closed in the weak topology.


I know this is a consequence of the statement any closed convex subset in $V$ is closed in the weak topology, which the proof used the geometric Hahn-Banach theorem. My question is: does this problem have an elementary proof without using Hahn-Banach? Any help is appreciated.

$\endgroup$
11
  • $\begingroup$ +1 Seems kind of hard, since the claim comes down to showing existence of certain functional(s), which comes down to Hahn-Banach. $\endgroup$
    – Michael
    Mar 16, 2014 at 10:36
  • $\begingroup$ In dual Banach spaces the unit ball is even weak*-closed (without Hah-Banach) and hence also weakly closed. However, this does not help for the general case. $\endgroup$
    – Jochen
    Mar 16, 2014 at 17:19
  • 2
    $\begingroup$ @Jochen I am not sure if quoting Banach-Alaoglu instead of Hahn-Banach makes it "elementary" as the OP intended. $\endgroup$
    – Michael
    Mar 16, 2014 at 19:30
  • $\begingroup$ @Michael You do not need Banach-Alaoglu: $B_{X^*}=\bigcap_{x\in B_X} \lbrace f\in X^*: |f(x)|\le 1 \rbrace$ is weak*-closed. $\endgroup$
    – Jochen
    Mar 17, 2014 at 7:26
  • $\begingroup$ I wanted to use a relation of this type $B_V =\bigcap_{f \in V*,\|f\| \leq 1} \{x \in V \mid |\langle f,x \rangle | \leq 1 \}$ But Hanh - Banach is needed to say at least that such intersection is not empty. In fact, I think that if you want to speak about functionals you need HB, I don't know NON-equivalent results which grant the non triviality of $V^*$. $\endgroup$
    – Riccardo
    Mar 17, 2014 at 8:33

2 Answers 2

3
$\begingroup$

At first I wanted to follow the reasoning showed in "Brezis - Functional Analysis Sobolev Spaces and Partial Differential Equations" at pages 59-60, riassumed in this identity $$B_V =\bigcap_{f \in V^*,\|f\| \leq 1} \{x \in V \mid |\langle f,x \rangle | \leq 1 \}$$ (two words about it: it relies on the characterization of open/closed set in the weak topology and property of the operatorial norm and -obviously- some corollaries of HB, existence of $f_{x_0}$ such that $f(x_0)=\|x_0\|$ and $\|f_{x_0}\|=1$) and so it would be not suitable for the question here.

But there is a more deep use of HB here. An immediate corollary of HB (adapted to these hypothesis) is

Corollary. Let $V$ a non trivial Banach Space, then $V^*$ is not trivial (it contains other elements than the zero map)

Without HB there is nothing (modulo equivalent forms of it) (as far as I know at least) that grants us the non triviality of $V^*$, in the infinite dimensional case obviously. Without assuming HB the first intersection is non empty but can't be refined more than $$\bigcap_{f \in V^*,\|f\| \leq 1} \{x \in V \mid |\langle f,x \rangle | \leq 1 \} = V$$ because we don't have (we can't write down explicitly) any functional other than the zero map.

In this (rather pathological) case, the weak topology could be the trivial one, the only open set is $V$, and so $B_v$ is not closed, and this observation should show the deep problem in not assuming HB.

$\endgroup$
1
  • $\begingroup$ Thank you so much for organizing out all this! I also think HB is the only thing that can guarantee the non-triviality of the functional space. Your answer teaches me that we not only need HB for this problem but also for everything else involving dual spaces. $\endgroup$
    – user112564
    Mar 19, 2014 at 5:25
3
$\begingroup$

If $x_n \to x$ weakly then we have that $\lambda x_n \to \lambda x$ for all $\lambda \in V^*$and $|\lambda x_n| \leq \|\lambda\| \|x_n\|$. Dividing both sides by $\|\lambda\|$ gives $$ \frac{|\lambda x_n|}{\|\lambda \|} \leq \|x_n\|.$$ Taking $n \to \infty$ and substituting in $\|x\| = \sup_{\lambda \in V^*} \frac{|\lambda x|}{\|\lambda \|}$ gives $$\|x\| \leq \liminf_{n \to \infty} \|x_n\|.$$Then any limit $x$ of $x_n$ with $\|x_n \| \leq 1$ for all $n$ will necessarily have $\|x\| \leq 1$.

$\endgroup$
4
  • $\begingroup$ In the first sentence you mean: for all $\lambda \in V'$, right? $\endgroup$ Mar 17, 2014 at 7:06
  • 1
    $\begingroup$ I suspect that $||x||=\sup \{\lambda x \colon \lambda\in V', ||\lambda||=1\}$ uses Hahn-Banach. $\endgroup$ Mar 17, 2014 at 7:08
  • $\begingroup$ I suspect that's the case, oh well. $\endgroup$ Mar 17, 2014 at 7:24
  • $\begingroup$ @bringingdownthegauss We can't use weakly convergent sequences here, we have to use nets. But the argument stays the same. Otherwise, this won't prove that $B_V$ is weakly closed, but only that it's weakly sequentially closed. $\endgroup$
    – Jakobian
    Sep 26, 2022 at 14:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .