Another proof of reflexive Hilbert spaces

Question

Here is the proof that every Hilbert space is refexive:

Let $\varphi\in\mathcal{H^{**}}$ be arbitrary. By Riesz, there is a unique $f_\varphi\in\mathcal{H^*}$ with

$\varphi(f)=\langle\,f,f_\varphi\rangle$ for all $f \in\mathcal{H^*} $.

Using the same notation and theorem, we have

$\hat{y}_{f_\varphi}(f)= f(y_{f_\varphi})=\langle\,y_{f_\varphi},y_f\rangle=\langle\,f,f_\varphi\rangle=\varphi(f)$

This implies $\hat{y}_{f_\varphi}=\varphi$, thus $\mathcal{H}$ reflexive.

I understood all the steps except for the last implication. Basically, we just showed that $2$ functionals from bi-dual space $\mathcal{H^{**}}$ are the same, why would it imply that $\mathcal{H}$ is reflexive? Any explanation would be highly appreciated!

Answer 1

Let $\Phi:\mathcal H \to \mathcal H^{**}$ denote the canonical injection, AKA the evaluation map (in the notation of the proof, $\Phi(x) = \hat x$)t. We want to prove that $\Phi$ is surjective. In other words: we want to prove that for any $\varphi \in \mathcal H^{**}$, there exists a $y \in \mathcal H$ such that $\Phi(y) = \varphi$.

So, begin with any $\varphi$. By the RRT, there exists a unique $f_{\varphi}$ such that for all $f \in \mathcal H^*$, $\varphi(f) = \langle f, f_{\varphi}\rangle$.

Note that this requires that requires an inner product on $\mathcal H^*$. Recall how such an inner product is defined: RRT says that there exists a $y_f$ for every $f \in \mathcal H^*$ such that for $y \in \mathcal H$, we have $f(y) = \langle y,y_f\rangle$. With this established, we define $$ \langle f,g \rangle := \langle y_f,y_g\rangle. $$

We claim that $\Phi(y_{f_{\varphi}}) = \varphi$ (that is, $y_{f_{\varphi}}$ is "the $y$ that we're looking for"). Indeed, we note that for any $f \in \mathcal H^*$, we have $$ [\Phi(y_{f_{\varphi}})](f) = f(y_{f_{\varphi}}) = \langle y_{f_\varphi},y_f \rangle = \langle f, f_{\varphi}\rangle = \varphi(f) $$

Answer 2

The point is that $\varphi \in \mathcal{H}^{**}$ was arbitrary, and your proof shows that it agrees with another element of $\mathcal{H}^{**}$ which has a particular form, thus showing that every element of $\mathcal{H}^{**}$ has that special form. To see why that is a proof of reflexivity, let's review the general situation:

Let $X$ be a Banach space. For any $y \in X$, we have a functional $\varphi_y \in X^{**}$ defined by $\varphi_y(f) = f(y)$. The map $y \mapsto \varphi_y$ is always an isometric linear embedding of $X$ into $X^{**}$, so the issue is to check that this map is surjective.

Proving surjectivity is exactly this: Given any $\varphi \in X^{**}$, find a $y \in X$ such that for all $f \in X^*$, $\varphi(f) = f(y)$. The proof you've given here is exactly finding such a $y$.