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.
Best Answer
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.)