If the dual unit ball of a normed space $X$ is metrizable in the weak-$*$ topology then $X$ is separable

Question

Let $X$ be a normed space and $(B_{X^*},w^*)$ be the unit ball of the dual space $X^*$ endowed with the weak-$*$ topology.

Here is a proof a the fact that if $(B_{X^*},w^*)$ is metrizable then $X$ is separable :

Set $K :=(B_{X^*},w^*)$. Since $K$ is metrizable we have that $C(K)$ - the space of continuous functions over $K$ - is separable (proved as a lemma). Consider the function $\Lambda : X \rightarrow C(K)$ defined by $x \mapsto \widehat{x}\big|_K$ where $\widehat{x}\big|_K(x^*) = x^*(x)$ for all $x \in X$. This function is well-defined and $$ \big\|\widehat{x}\big|_K\big\|_\infty = \sup\big\{|x^*(x)| : x^* \in > B_{X^*}\big\} = \|x\|$$ by Hahn-Banach. Hence $X$ is separable.

I don't get why it should be obvious that $X$ is separable. Is the function $\Lambda$ onto ?

Answer 1

The subset $\Lambda(X)$ is a subset of a separable metric space, hence there exists a sequence $\{\Lambda(x_l),l\in\mathbb N\} $ where $x_l\in X$, which is dense in $\Lambda(X)$. Using the fact that $\Lambda$ is an isometry, we get that the sequence $\{x_l,l\in\mathbb N\} $ is dense in $X$.