# Regarding metrizability of weak/weak* topology and separability of Banach spaces.

Let $X$ be a Banach space, $X^*$ its dual, $\mathcal{B}$ the unit ball of $X$ and $\mathcal{B}^*$ the unit ball of $X^*$. The following result is well-known

Theorem. If $X$ is separable, then $\mathcal{B}^*$ endowed with the weak*-topology is metrizable.

A solution would be the following: Let $(x_n)_{n\in\mathbb{N}}$ be a dense subset of $\mathcal{B}$. Then $d(\phi_1,\phi_2)=\sum_{n=1}^\infty 2^{-n}|\phi_1(x_n)-\phi_2(x_n)|$ is a metric which induces the weak*-topology of $\mathcal{B}^*$.

I was wondering about the converse:

Question 1. Suppose that the weak*-topology of $\mathcal{B}^*$ is metrizable. Is $X$ (norm-)separable?

My thought was: By Banach-Alaoglu $\mathcal{B}^*$ is weak*-compact. If it is also metrizable, then it is (weak*-)separable, and it follows easily that $X^*=\bigcup_{n=1}^\infty\mathcal{B}^*$ is also weak*-separable. The problem is that this does not imply that $X$ is separable ($\ell_\infty$ is a counterexample - weak$^*$-separability of $l_\infty^*$.)

My second idea was to try to compare a metric which induces the weak*-topology of $\mathcal{B}^*$ with some metric as in the proof of the theorem above, but I did not succeed.

We can also consider all the "dual statements" and ask similar questions. More precisely, we know that the following holds:

Theorem. If $X^*$ is norm-separable, then the weak topology of $\mathcal{B}$ is metrizable.

We prove this by defining $d(x,y)=\sum_{n=1}^\infty 2^{-n}|\phi_n(x-y)|$ for a dense sequence $(\phi_n)_{n\in\mathbb{N}}\subseteq\mathcal{B}^*$, as expected. Then we ask the question:

Question 2. If the weak topology of $\mathcal{B}$ is metrizable, is $X^*$ necessarily norm-separable (so $X$ is norm-separable as well)?

An alternative approach to answer Question 2 would be the following: Suppose $d$ is a metric inducing the weak topology of $\mathcal{B}$. Changing $d$ by $\frac{d}{1+d}$ if necessary, assume that $d$ is bounded. Let's identify $X$ as a subspace of $X^{**}$ via the canonical mapping. Goldstine's Theorem implies that the only possible metric in $\mathcal{B}^{**}$ extending $d$ and inducing the weak*-topology has to be given by $$d'(\Lambda,\Gamma)=\inf\left\{\operatorname{diam}(U\cap X):U\text{ is weak*-open and }\Lambda,\Gamma\in U\right\}.$$ If Question 1 has a positive answer (and all the rest stated is true as well), then Question 1 implies that $X^*$ is separable.

One last comment: If $X$ is reflexive, then both questions have positive answers. For Question 1, we know (as above) that $X^*$ is weakly*-separable. But reflexivity of $X$ implies that the weak and weak*-topologies of $X^*$ are equal, so $X^*$ is weakly separable, hence separable, so $X$ must also be separable. A solution for question 2 follows similarly.

• Yes to both questions. The first is a common result. You can find a proof in Theorem 2.6.23 of Megginson's An Introduction to Banach Space Theory. The second result seems less common; but you can find a proof here (see the preceding page there for your first question). – David Mitra Aug 31 '14 at 8:53