# Is the dual of a Fréchet space weakly* separable?

It is known that if $$X$$ is a separable Banach space with dual $$X^\ast$$, then $$B_{X^\ast}$$, the closed the unit ball in $$X^\ast$$, is compact and metrizable in the weak* or $$\sigma(X^\ast, X)$$-topology. In particular, $$B_{X^\ast}$$ is weak* separable and thus $$X^\ast$$ is also weak* separable.

If $$X$$ is only a separable Fréchet space, is it true that its dual $$X^\ast$$ is $$\sigma(X^*,X)$$-separable?

• Writing $X$ as a countable intersection (projective limit) of separable Banach spaces, $X^*$ appears as a countable increasing union (inductive limit) of duals of separable Banach spaces. It is therefore separable by the first half of the question. Feb 26 at 15:57

Let $$X$$ be a metrizable and separable LCTVS. It is topologized by a countable family of seminorms $$\{p_i\}_{i\in\mathbb N}$$, and it has a countable dense subset $$\{x_j\}_{j\in\mathbb N}$$. By Hahn-Banach, for each $$(i,j)\in\mathbb N^2$$ there exists a functional $$f_{ij}\in X^*$$ such that $$\langle f_{ij},x_j\rangle=p_i(x_j)$$ and $$\langle f_{ij},x\rangle\le p_i(x)$$ for all $$x\in X$$. Consider $$Y:=\text{span}\{f_{ij}:(i,j)\in\mathbb N^2\}\subset X^*$$: I claim that for every $$x\in X$$, if $$\langle f_{ij},x\rangle=0$$ for all $$(i,j)\in\mathbb N^2$$, then $$x=0$$. By the usual criterion, this is equivalent to $$Y$$ being $$\sigma(X^*,X)$$-dense (one has $$\overline Y^{\sigma(X^*,X)}=(Y_\perp)^\perp$$, that is the whole $$X^*$$ iff $$Y_\perp=(0)$$, which is the claim).
Proof of the claim: Assume $$x\in X$$ has $$\langle f_{ij},x\rangle=0$$ for all $$(i,j)\in\mathbb N^2$$. For every $$i\in\mathbb N$$ we have $$p_i(x_j)=\langle f_{ij},x_j\rangle= \langle f_{ij},x_j-x\rangle\le p_i(x_j-x).$$ Since $$\{x_j\}_{j\in\mathbb N}$$ is dense in $$X$$, $$x_j\to x$$ along a subsequence, which implies $$p_i(x)=0$$ by the above inequality, so also $$x=0$$.