Duality of Vector Spaces and Topological Vector Spaces

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 $$G \circ F \cong 1_{K \text{-vect}^{op}}$$

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.)