I'm trying to prove that Dual of a infinite-dimensional Fréchet space(locally convex and compatible with complete invariant metric) with its weak*-topology is of first category in itself.
This is a problem from rudin's functional analysis book (pg 87).
My approach : Let $X$ be the space mentioned above with metric $d$,
put $V_n=\{x\in X:d(0,x)<1/n\}$ and let $V_n^*=\{\Lambda\in X^*:|\Lambda(x)|\le 1 (\forall x\in V_n)\}$ be polar of $V_n$, this is clearly compact and it suffices to show each $V_n^*$ has empty interior in $X^*$
since, $X^*=\bigcup_{n=1}^\infty V_n^*$.
If some $V_n^*$ has non-empty interior then $F+V\subset V_n^*$ where $F\in V_n^*$ and $V=\{\Lambda:|\Lambda(x_i)|<\varepsilon (1\le i\le j)\}$ for some $\{x_1,\cdots ,x_j\}\subset X$ and $\varepsilon>0$,
this implies if $\Lambda\in X^*$ and $\Lambda(x_1)=\cdots=\Lambda(x_j)=0$ then $\Lambda=0$ which inturn implies $X$ has finite dimension.
My approach above didn't use local convexity nor completeness , $L^p$ for $0<p<1$ gives a counter example (it isn't locally convex).
Where did my approach went wrong ? How do I prove the statement in question?
The Banach-Alaoglu theorem holds for all topological vector spaces and every Hausdorff topological vector space which has a precompact non-empty open set is finite dimensional. You need however local convexity of the given metrizable space (completeness is not needed) to have the dual infinite dimensional. If the weak$^*$-dual is finite dimensional it is, of course, of second category in itself.