$X^*$ with $weak^*$-topology is first category in itself when $X$ is infinite dimensional Fréchet space

Question

I was reading Rudin's Functional Analysis and stuck at the problem in exercise 11 on page 87: Let $X$ be infinite-dimensional Fréchet space, then $X^*$ with its $weak^*$-topology is of the first category in itself. Here is my attempt.

Consider the a basic open set at origin corresponding to $\{x_1,...,x_n\}\subseteq X$ of $weak^*$-topology on $X^*$ given by $\{x^*\in X^* : |x^*(x_i)|<\epsilon, \forall i=1,2,..,n\}$. Now let $y\in X$ and consider a particular $weak^*$ nbd $\{x^*\in X^*: |x^*(y)|<1\}$ which contains a basic nbd i.e. for some $x_1,x_2,...,x_n\in X$ and $\epsilon>0$ we have $$\{x^*\in X^* : |x^*(x_i)|<\epsilon, \forall i=1,2,..,n\}\subseteq \{x^*\in X^*: |x^*(y)|<1\}.$$ Now this implies that if $z^*\in X^*$ and $|z^*(x_i)|=0, \forall x_i$ with $i=1,2,...,n$ then $tz^* \in \{x^*\in X^*: |x^*(y)|<1\}$ for each $t$ in scalar field $\Bbb K$ and this implies $|tz^*(y)|<1, \forall t\in \Bbb K\implies z^*=0$. Now for $x\in X$ defining $f_x:X^*\rightarrow \Bbb K$ by $f_x(\Lambda)=\Lambda(x), \forall \Lambda \in X^*$ we have $$\ker(f_y)\supseteq \bigcap_{i=1}^n \ker(f_{x_i})$$ so that $f_y$ can be written as linear combination of $f_{x_i},i=1,2,...,n$ and this implies $y$ can be written as linear combination of $x_1,x_2,...,x_n$.

Any suggestion or solution will help me. Thanks in advance.

Answer 1

Take a decreasing basic sequence of $0$-neighbourhoods $U_n$ in $X$ (i.e., every $0$-neighbourhood contains some $U_n$, such a sequence exists because $X$ is metrizable). The polars $U_n^\circ$ are weak$^*$-closed with $X^*=\bigcup_{n\in\mathbb N} U_n^\circ$. By your description of the basic weak$^*$ open sets in $X^*$ it is clear that the interior of each $U_n^\circ$ is empty. (A different argument would use Alaoglu's theorem: If $U_n^\circ$ had interior points it would be a compact $0$-neighbourhood in $X^*$, and any topological vector space with a compact $0$-neighbourhood is finite dimensional.)

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 $n$.

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

I tried to make it clear that I'm using every part of hypothesis. Hope it helps :)