I am revising linear algebra, and am a bit stuck on this problem.
So, if you define a natural isomorphism $f$ between $V$ and its double dual $V''$, you get $$f:V\rightarrow V''$$ $$v\mapsto E_v$$ where, $$E_v(f)=f(v),\forall v \in V'$$ I am then asked to show, that if $U$ is a subspace of $V$, under this natural isomorphism, $f|_U$ is a bijection between $U$ and $U^{00}$.
I am not sure how to do this, I see that if $u$ is mapped to its evaluation map, then the evaluation map on all the functions in $U^0$ is clearly 0, but from here, I do not know how to show the bijection.
Any help appreciated, thank you.
Here's some different approach:
For given vector space $V$ and its subspace $U$, $ dim U + dim U^0 = dim V$.
Also, by taking dual, $ dim U^0 + dim U^{00} = dim V^* $.
Since $dim V = dim V^*$, this results $dim U = dim U^{00}$.
By definition, $U$ is a subspace of $U^{00}$.
Hence $dim U = dim U^{00}$ implies $ U = U^{00}$.