Metrizability

Let $X$ be an infinite-dimensional Banach space. Then,

1. $(X,$wk $)$ is never metrizable.
2. $(X^*,$wk $^*)$ is never metrizable.
3. $($ball $X,$   relative    wk $)$ may or may not be metrizable.
4. $($ball $X^*,$   relative wk $^*)$ is metrizable iff $X$ is separable.

Proof. (of 1)
Fix $n \in \mathbb{N}$. Let $S_n = \{ x \in X : \Vert x\Vert = n \}$.
Claim: $0 \in$   wk cl $S_n$.
We build a net in $S_n$ indexed by the basic neighborhoods of 0 that converges weakly to 0.
Given $U_1$ a basic neighborhood of 0, there is $x_U \in S_n \cap U$. Check $x_U \rightarrow 0$ weakly.

Suppose there is a metric $d$ giving the weak topology on $X$. Since $0 \in$   wk cl $(S_n)$, there is $x_n \in S_n$ such that $d(0,x_n) < \frac1n$.
Then $x_n \rightarrow 0$ weakly and $\Vert x_n\Vert = n \rightarrow \infty$ as $n \rightarrow \infty$. This contradicts PUB.
$\qedsymbol$

The proof of 2 is similar to that of 1.

Proof. (of 3)
Example of nonmetrizable situation. Take $X = \ell^1$. By HW (Assignment 1, Q4), $x_n \rightarrow x$ weakly in $\ell^1$ if and only if $x_n \rightarrow x$ in the norm.
If the relative weak topology was metrizable, it would be the same, it would be the same as the norm topology.
$\qedsymbol$

Theorem 2.0.1   $($   ball $X^*,$   wk $^*)$ is metrizable iff $X$ is separable.

Proof. Assume $X$ is separable. Let $S$ be the countable dense set. For each $s \in S$, let

$$D_s := \{ \alpha \in \mathbb{F}: \vert\alpha\vert \leq 1 \}$$

Let $D = \displaystyle \prod_{s \in S} D_s$, $D$ is a compact metric space ($S$ is countable).
Define $\tau:$   ball $X^* \rightarrow D$ by $\tau(f)(s) = f(s)$. As in Alaoglu, if $f_\lambda \rightarrow f$ in the weak star topology, then $\tau(f_\lambda) \rightarrow \tau(f)$ in the product topology. So $\tau$ is continuous. As $\tau$ is 1-1 and ball $X^*$ is wk $^*$-compact, $\tau$ is a homeomorphism (A2.8).
Thus, $\tau($ball $X^*) \subset D$ is metrizable and since $\tau$ is a homeomorphism, ball $X^*$ is metrizable.

Assume $($ball $X^*,$   wk $^*)$ is metrizable. Choose a countable local base at 0, i.e., $U_n, n \in \mathbb{N}$ such that $U_n$ is an open set containing 0. For all $V$ open, $0 \in V$, there exists an $n$ such that $U_n \subset V$. So our topology is Hausdorff, $$\bigcap_{n \in \mathbb{N}} U_n = \{ 0 \}$$/.

For each $U_n$, there is a finite set $S_n \subset X$,

$$U_n := \{ f \in X^* : \vert f(x)\vert < 1, \forall x \in S_n \}$$

Let $S = \bigcup S_n$, a countable set. Claim: $X = \overline{span} S$ (which is true iff $S^\perp = \{ 0 \}$ )
If $f \in S^\perp$, then $f(s) = 0$ for all $s$. So, $f \in U_n$ for all $n$, so $f \in n U_n = \{ 0 \}$
Thus, the set of rational linear combinations of elements of $S$ is a countable dense set. $\qedsymbol$