Show that if the Riesz map is surjective on $H$, then $H$ is a Hilbert space

Question

Let $H$ be a vector space equipped with an inner product $(\cdot, \cdot)$ and $f:H\to H',\ f(x)=(\cdot,x)$ surjective. Now, why $H$ is a Hilbert space? The other direction is clear by Riesz' representation theorem but what about this?

Answer 1

The map $f$ is clearly injective, and an isometry. Suppose it is surjective : then $H$ is isometric to its dual space. For any normed vector space $X$ and Banach space $Y$, the space of continuous linear maps $$\mathrm{L}(X,Y)$$ equipped with the usual norm is automatically a Banach space, in particular for any normed vector space $X$, its topological dual $X'$ is a Banach space. Thus, $H$ is isometric to a Banach space, complete and so $H$ is a Hilbert space.