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

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X is a frechet space if X is a locally convex F-space. – Ali Qurbani Jun 4 '12 at 16:51
X is an F-space if its topology τ is induced by a complete invariant metric d. – Ali Qurbani Jun 4 '12 at 16:53
Did you manage to do that when $X$ is a normed space? – Davide Giraudo Jun 4 '12 at 17:11
No,X is only topological vector space – Ali Qurbani Jun 4 '12 at 17:31
I know, but it's possible that some ideas use in the normed space case could help. For example, you can look at the ball of $X^*$ centered at $0$ with radius $n$ (when $X$ is normed). – Davide Giraudo Jun 4 '12 at 18:05
up vote 3 down vote accepted

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


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.

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