Metrizability of the unit ball $B_{X^*}$.

Question

I am trying to prove the assertion:

If $X$ is a separable normed space, then the unit ball in $X^*$ with the weak* topology, $(B_{X^*},\sigma(X^*,X))$, is metrizable.

Firstly, I took $D=(x_n)\subset X$, $\overline{D}=X$. If $\sigma(X^*,D)$ is the topology in $X^*$ with countable local basis for $0$:

$$U_n= \{x^* \in X^*: \sup\limits_{1\leq j\leq n}|x^*(x_j)|< \varepsilon_n \}$$

How can I prove that $(X^*, \sigma(X^*,D))$ is a Hausdorff TVS? If I could prove this, than I should conclude that $(X^*, \sigma(X^*,D))$ is metrizable.

Then I have another problem: How can I use Alaoglu's Theorem to show that $\sigma(X^*,D)=\sigma(X^*,X)$ in $B_{X^*}$ ?

Answer 1

Let $D=(x_n)$ dense in $X$ and let $\sigma(X^*,D)$ the locally convex topology in $X^*$ with countable local basis at $0\in X^*$ given by: $$U_n=\{x^*\in X^*: \sup_{1\leq j\leq n}|x^*(x_j)|< 1/n \}$$ Then $(X^*,\sigma(X^*,D))$ is a Hausdorff TVS. In fact, if $x^*\neq 0$, as $x^*$ is continuous, there is $j_0$ such that $x^*(x_{j_0})\neq 0$, so we have $|x^*(x_{j_0})| \geq 1/n_0$. We have:

$$\forall m \geq n_0 \quad x^* \notin U_m $$

Now it is easy to see that $(x^* + U_{2n_0})$ is an open neighborhood of $x^*$ and $U_{2n_0} \cap (x^* + U_{2n_0})=\emptyset$. Hence, $(X^*,\sigma(X^*,D))$ is a Hausdorff TVS, so it is metrizable.

Clearly, $\sigma(X^*,D) \leq \sigma(X^*,X)$ and by Alaoglu's theorem, $B_{X^*}$ is $\sigma(X^*,X)$-compact. Then

$$ (B_{X^*},\sigma(X^*,X)) \stackrel{id}{\longrightarrow} (B_{X^*},\sigma(X^*,D)) $$

is an homeomorphism and $\sigma(X^*,X)=\sigma(X^*,D)$ in $B_{X^*}$, therefore $(B_{X^*},\sigma(X^*,X))$ is metrizable.

Answer 2

I don't immediately see if your approach leads to a solution (in particular I think you need "normed" for this to be true).

This is the way I would do it: assume that $\|\phi\|\leq c$ for all $\phi\in K$. Let $D=(x_n)$ be dense in the unit ball of $X$. Now define $$ d(\phi,\psi)=\sum_{k=1}^\infty2^{-k}|\phi(x_k)-\psi(x_k)|. $$ It is straighforward to check that this is a distance.

Assume $d(\phi_n,0)\to0$. Fix $x\in X$, $\varepsilon>0$. Then there exists $k$ with $\|x-x_k\|<\varepsilon$. Then $$ |\phi_n(x)|\leq\|\phi_n\|\,\|x-x_k\|+|\phi_n(x_k)|\leq c\,\varepsilon+2^k\,d(\phi_n,0). $$ Then $$ \limsup_n|\phi_n(x)|\leq c\varepsilon. $$ As $\varepsilon$ was arbitrary, we get $\limsup_n|\phi_n(x)|=0$, i.e. $\phi_n(x)\to0$. So $\phi_n\to0$ in the $\sigma(X^*,X)$ topology.

Conversely, if $\phi_n\to0$ weak*, then $\phi_n(x)\to0$ for all $x\in X$. Fix $\varepsilon>0$. Let $k\in\mathbb N$ with $2^{-k}<\varepsilon/(2c)$. Then there exists $n_0$ such that $|\phi_n(x_j)|<\varepsilon/2$ for $j=1,\ldots,k$ for all $n\geq n_0$. Then, for such $n$, $$ d(\phi_n,0)=\sum_{j=1}^k2^{-j}|\phi_n(x_j)|+\sum_{j=k+1}^\infty2^{-j}|\phi_n(x_j)| \leq\frac\varepsilon2\,\sum_{j=1}^k2^{-j}+c\sum_{j=k+1}^\infty2^{-j}\leq\frac\varepsilon2+\frac\varepsilon2=\varepsilon. $$