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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.2012-06-04
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    X is an F-space if its topology τ is induced by a complete invariant metric d.2012-06-04
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    Did you manage to do that when $X$ is a normed space?2012-06-04
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    No,X is only topological vector space2012-06-04
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    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).2012-06-04
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    give the open mapping theorem a thought. on a banach space, since the inclusion $X^* \rightarrow X^*$ is continuous where the former has the norm topology, then by the open mapping, if the image is 2nd category the map will be invertible & the topologies equivalent.2012-06-04

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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.

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    What do you mean by $\cap Ker x_i$? how should I see them as applications?2017-06-19
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    @Irene $X\subset X^{**}$ so every $x$ can be seen as a linear functional on $X^*$.2017-06-20