Is a Frechet space separable, if its dual is?

Question

It is known, and not too difficult to prove that if a dual of a normed space is separable, it is separable itself. Does the same hold for Frechet spaces?

(By the dual I mean the dual endowed with the strong topology.)

Answer 1

Every Fréchet space $X$ is an inverse limit of a sequence of Banach spaces since the topology of $X$ is induced by a countable family of seminorms $\|\cdot\|_n$ (thay may easily be arranged to be decreasing).

Suppose that the strong dual of $X$ is separable. Set $X_n = X / \{x\colon \|x\|_n = 0\}$. Each $X_n$ is a Banach space and $X$ admits a natural surjection onto each $X_n$ and so $X^*$ is an inductive limit of $X_n^*$s. In particular, each $X_n^*$ sits homeomorphically in $X^*$, which we assume is separable. Hence all $X_n$s have separable duals so they are separable themselves (here we use the fact that a metrisable subspace of a separable topological vector space is separable; see Corollary 4 here for a more general statement).

Being an inverse limit is nothing but being a certain subspace of the Cartesian product $\prod_{n=1}^\infty X_n$, which is metrisable and separable, hence so is every subspace thereof. Consequently, $X$ is separable.

Answer 2

I think so (no books at hand for a reference). All we need for the normed proof to go through, is some sort of Hahn-Banach theorem that gives us enough functionals to works with. And Fréchet spaces have enough functionals.