Let $X$ be an infinite-dimensional Fréchet space. Prove that $X^*$,with its weak*-topology is of the first category in itself.
$X^*$ with its weak*-topology is of the first category in itself
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0give 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
1 Answers
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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0@Irene $X\subset X^{**}$ so every $x$ can be seen as a linear functional on $X^*$. – 2017-06-20