If $X$ is separable space then $X^∗$ is separable in all topologies $\tau$ such that $(X^∗,\tau)^∗ =X$?

Question

Let $(X,\|.\|_{X})$ be a separable Banach space and the associated dual space is denoted by $X^*$. By $w^*$ we shall indicate the weak$-*$ topology on $X^*$.

Let $B_{X^∗}= \{x^∗ \in X^∗ : \|x^∗\|_{X^∗}\leq 1\}$. Since $X$ is separable, the set $B_{X^∗}$ furnished with the relative $w^∗-$topology is compact (by the Alaoglu theorem) and metrizable (see Theorem I.5.85). Note that $$ X^*=\bigcup_{n}{nB_{X^*}} $$ hence $X^∗_{w^∗}$ (the space $X^∗$ furnished with the $w^∗-$topology) is separable.

Can we say that : $X^∗$ is separable in all topologies $\tau$ such that $(X^∗,\tau)^∗ =X$?

Answer 1

Yes. Separability of a locally convex space is equivalent to the existence of a dense countable-dimensional subspace. All compatible topologies for a dual pair (i.e. those that give the same continuous linear functionals) agree on the closures of convex sets.