$X^*$ with its weak*-topology is of the first category in itself

Question

Let $X$ be an infinite-dimensional Fréchet space. Prove that $X^*$,with its weak*-topology is of the first category in itself.

Answer 1

Here is a proof for a Banach space $X$. Clearly,

$$X^*=\bigcup_{n\ge1}nB^*(0;1),$$

where $B^*:=B^*(0;1)$ is the closed unit ball in $X^*$. It suffices to prove that $\textrm{int}_{w^*}B^*=\emptyset$.

Assuming the contrary one gets that $0\in \textrm{int}_{w^*}B^*$ since $\textrm{int}_{w^*}B^*$ is convex symmetric. Hence $\exists x_1,x_2,..,x_n\in X,\ \epsilon>0$ such that

$$V_{x_1,x_2,..,x_n;\epsilon}:=\{x^*\in X^*\mid |x^*(x_i)|<\epsilon,\ \forall i=\overline{1,n}\}\subset B^*.$$

From this inclusion we see that $\cap_{i=\overline{1,n}} \textrm{Ker}\ x_i$ is a bounded subspace so it must be equal to $\{0\}$.

Now for every $x\in X$ we have $x^*(x)=$ whenever $x^*(x_i)=0$, for every $i=\overline{1,n}$, that is, by the Kernel's Theorem, every $x$ is a linear combination of $\{x_1,x_2,..,x_n\}$, i.e., $X$ is finite dimensional, a contradiction.

You adapt it for a Frechet space and/or put all the necessary details.