8
$\begingroup$

Please help me to understand why the unit ball of a separable banach space is metrizable, when it is given the induced topology from the weak topology. Specifically, my problem is I think I'd be able to do it if I knew the duel was separable, but not with the current given information. (If I had this additional information, I'd use the infinite sum metric tricks that one uses all too often.)

The underlying field can be real or complex.

$\endgroup$
  • 1
    $\begingroup$ A weak* topology is defined on a dual. You mean: the unit ball of the dual is metrizable when $X$ is separable. By Banach-Alaoglu, it is a nice compact metric space. See here. $\endgroup$ – Julien Apr 5 '13 at 20:53
  • $\begingroup$ No I'm sorry. I meant that supposedly one can metrize the unit ball of the weak topology of the original space? $\endgroup$ – Jeff Apr 5 '13 at 21:41
  • $\begingroup$ Then you need to edit your question. And I think your assumption is most likely that the dual is separable. See here in this case. $\endgroup$ – Julien Apr 5 '13 at 22:10
  • $\begingroup$ If the dual space is separable then yup's answer (or Banach-Alaoglu) does the trick: the inclusion $J \colon X \to X^{\ast\ast}$ is a homeomorphism from $X$ with the weak topology to $X^{\ast\ast}$ with the weak*-topology and the unit ball in $X^{\ast\ast}$ is metrizable. $\endgroup$ – Martin Apr 5 '13 at 22:21
  • $\begingroup$ The unit ball in $\ell^1$ is not metrizable in its weak topology: the unit sphere is norm-closed and the weak closure of the unit sphere is the entire unit ball; yet the weak and norm-topologies have the same convergent sequences by the Schur property of $\ell^1$. $\endgroup$ – Martin Apr 5 '13 at 22:28
9
$\begingroup$

I assume you mean metrizability of the unit ball in the dual space. Since you don't ask about compactness, we can cheat and assume Alaoglu's theorem that the unit ball is compact in the weak*-topology. Since a continuous injective map from a compact space to a Hausdorff space is a homeomorphism onto its image, it suffices to find a continuous and injective map into some metrizable space.

Choose a dense sequence $\langle x_n : n \in \mathbb{N}\rangle$ in the unit sphere of $X$, this exists by separability of $X$. Let $B'$ be the unit ball in the dual space. By the definition of the weak*-topology we have a continuous map $f_n \colon B' \to D$ given by $f_n(\varphi) = \varphi(x_n)$ where $D = \{z \in \mathbb{C} : |z| \leq 1\}$ is the closed unit disc in $\mathbb{C}$. These assemble to a continuous map $f \colon B' \to D^{\mathbb{N}}$ given by $$f(\varphi) = \langle f_n(\varphi) : n \in \mathbb{N}\rangle = \langle \varphi(x_n) : n \in \mathbb{N}\rangle$$ by the definition of the product topology on $D^\mathbb{N}$. I claim that this map is injective. Indeed, if $\varphi_1 \neq \varphi_2$ then there is $x \in X$ such that $\varphi_1(x) \neq \varphi_2(x)$. Normalizing $x$ we can assume hat $\|x\| = 1$. By choosing $x_n$ close enough to $x$ we will have $\varphi_1(x_n) \neq \varphi_2(x_n)$. This implies that $f(\varphi_1) \neq f(\varphi_2)$ and we're done.

To be a little more explicit, the standard metric on $D^\mathbb{N}$ is $d(a,b) = \sum_{n=1}^\infty 2^{-n} |a_n - b_n|$ and since $f$ is injective and continuous, $\delta(\varphi_1,\varphi_2) = d(f(\varphi_1),f(\varphi_2))$ is a continuous metric on $B'$ with the weak*-topology and this shows metrizability of $B'$.

We assumed Alaoglu's theorem. If we don't do that there are two more things to do:

  1. show that convergence with respect to $\delta$ is equivalent to weak*-convergence.
  2. show that $B'$ is compact with respect to $\delta$.
$\endgroup$

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.