Why is the following statement valid? Note, $V$ is locally convex Hausdorff topological vector space over $\mathbb{C}$ and $V'$ is the space of all continuous linear maps from $V \to \mathbb{C}$.
$V$ is finite dimensional if and only if $V'$ with the weak topology is normable.
I believe that since $V$ is finite dimensional then $V'$ is finite dimensional. Thus all topologies on $V'$ are the same and $V'$ is normable. However, the other direction I am clueless as to the reasoning.
Clue: If $V’$ is infinite dimensional, then for every finite set $F\subset V’’$ of functionals $\bigcap \{\ker: f\in F\}$ should be a non-zero subspace which, therefore can not be contained in a ball $\{f\in V’:||f||<1\}$.