Prove that the Reisz map $\displaystyle \mathcal{R}:V^x\to V$ defined by $\displaystyle <\mathcal{R}(f),y>=f(y)$ for all y in V, is an isomorphism.

Let $\displaystyle \mu, \lambda\in F$

$\displaystyle \mathcal{R}(\lambda f+\mu g),y>=(\lambda f+\mu g)(y)=$

$\displaystyle \lambda f(y)+\mu g(y)=<\lambda\mathcal{R}(f),y>+<\mu\mathcal{R}(g), y>$

I am not sure how to show it is onto and one to one with the Reisz map.