Duality of Vector Spaces and Topological Vector Spaces

Question

Let $K$ be a field, and put $K \text{-vect}$ for the category of $K$-vector spaces. There is an adjunction $[-, K]_{K \text{-vect}} : K \text{-vect}^{op} \leftrightarrow K \text{-vect} : [-, K]_{K \text{-vect}}$ where $[-, K]_{K \text{-vect}}$ sends a vector space $V$ to $[V, K ]_{K \text{-vect}}$, the $K$ vector space of linear maps from $V$ to $K$.

Evidently, the vector space structure on $V^*$ alone is not enough to recover $V$. I wonder, though, about putting a topology on it. For each $a \in V$, there is a map $\hat{a} : V^* \rightarrow K$ sending $\phi$ to $\phi(a)$. Give $V^*$ the weakest topology such that $\hat{a}$ is continuous for each $a \in V$.

Question: form $[V^*, K]_{K \text{-topvect}}$, the vector space of continuous linear maps of topological vector spaces from $V^*$ to $K$. Is this vector space isomorphic to $V$?

If this turns out to be true, then we have a functor $F : K \text{-vect}^{op} \rightarrow K \text{-topvect}$ and a functor $G : K \text{-topvect} \rightarrow K \text{-vect}^{op}$, such that $G \circ F \cong 1_{K \text{-vect}^{op}}$

Answer 1

Yes, $V$ and $[V^*,K]_{\operatorname{K-topvect}}$ are indeed canonically isomorphic as vector spaces.

To see this we should first note, that the map $V\to [V^*,K]_{\operatorname{K-topvect}}$ is obviously injective. So it remains to prove that it is surjective. Suppose that a functional $f: V^*\to K$ is continuous, then it is bounded on a neighbourhood $U$ of zero in $V^*$. This neighbourhood can be chosen as a basic neighbourhood of zero in your topology, i. e. $$ U=\{g\in V^*: \forall i=1,...,n\quad |g(a_i)|\le 1\} $$ for some finite set of vectors $a_1,...,a_n\in V$. This implies that $f$ is bounded on the set of common zeroes of $\widehat{a_i}$, $$ \bigcap_{i=1}^n\operatorname{Ker}\widehat{a_i}=\{g\in V^*: \forall i=1,...,n\quad |g(a_i)|=0\} $$ and since this set is a vector subspace in $V^*$, $f$ vanishes on it: $$ f\Big|_{\bigcap_{i=1}^n\operatorname{Ker}\widehat{a_i}}=0 $$ This in its turn implies that $f$ is a linear combination of $\widehat{a_i}$: $$ f=\sum_{i=1}^n\lambda_i\cdot \widehat{a_i}, \qquad \lambda_i\in K. $$ And therefore $f$ is the image (under the mapping $a\mapsto\hat{a}$) of the vector $$ x=\sum_{i=1}^n\lambda_i\cdot a_i\in V. $$ (Another way to prove this is the reference to the Mackey-Arens theorem.)