$X^*$ 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 $x^*(x)=0$ 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.

Answer 2

Let $V_n=\{x\in X:d(0,x)<1/n\}$ and consider polars $$V_{n}^*=\{\Lambda\in X^*:|\Lambda(x)|\le 1(x\in V_n)\}$$ We have $X^*=\bigcup_{n=1}^\infty V^*_n$ since each $V_n^*$ is weak*-compact by Banach-Alaoglu. Hence it is enough to assume one of $V^*_n$ has non-empty interior ,

this implies $\exists\{x_1,\cdots,x_k\}\subset X$ for some $k\in\mathbb{N}$ such that $$\Lambda(x_i)=0~(\forall 1\le i\le k) \implies \Lambda=0~( \forall \Lambda\in X^*)~(1)$$ Since all weak*-continuous linear functional on $X^*$ have form $\Lambda\mapsto \Lambda(x)$ for some $x\in X$, $(1)$ implies that $X^*{}^*$ has finite dimension.

It is enough to prove that $M=\bigcap_{\Lambda\in X^*}\operatorname{Ker}(\Lambda)=\{0\}$ since when this is combined with $X^*{}^*$ has finite dimension implies $X$ has finite dimension.

Suppose that $\exists x\in M$ such that $x\neq 0$ , since $M$ is a closed subspace of a Frechet space $M$ is Frechet and for every continuous linear functional $f$ on a subspace of a locally convex space $Y$ there exists $\Lambda\in Y^*$ such that $\Lambda=f$ on that subspace , this implies $$M^*=\{0\}.$$ $M$ being Frechet , its topology is induced by countable family of semi-norms $(p_n)_{n\in\mathbb{N}}$ and $M$ is complete w.r.t this family.

If $p_n(x)\neq 0$ for some $x\neq 0$ in $M$ and some $n\in \mathbb{N}$ then there exists $K>0$ such that $1<Kp_n(x)$ hence if $f(tx)=t$ then $f$ extends to a non-constant continuous linear functional on $M$.

Hence $p_n(x)=0$ for all $x\in M$ and for all $n\in\mathbb{N}$.

Since $M$ is complete w.r.t this family of semi-norms every sequence in $M$ converges to $0$ this is impossible unless $M=\{0\}$.