Hilbert Spaces are Reflexive I want to show that all Hilbert spaces are reflexive.  I have found the following proof on StackExchange:
Hilbert Space is reflexive
However, I do not understand it.  Essentially, we want to show that for all $g \in X^{**}$, (X is some Hilbert space) there exists a unique $x \in X$ such that $g(h) = h(x)$ for all $h \in X^*$.  Following the OP's logic, we should apply the Riesz-Fréchet Representation Theorem (RRT) twice:
Pick any $g \in X^{**}$.  Then, since $g$ is bounded on $X^*$, by RRT there exists a unique $f \in X^*$ such that $||g|| = ||f||$, and $g(h) = \langle h,f \rangle$ for all $h \in X^*$.  Now apply RRT to $f$ to get a unique $x \in X$ such that $||f|| = ||x||$ and $f(y) = \langle y,x \rangle$ for all $y \in X$.  It follows that:
$f(x) = \langle x,x\rangle = ||x||^2 = ||f||^2 = \langle f,f \rangle = g(f)$.
We have shown that for any $g \in X^{**}$ that there exist unique $x \in X$, $f \in X^*$ such that $f(x) = g(f)$.  This is not quite what we want.  We want this to hold for a general $h \in X^*$.
According to icurays1, we have basically defined a bijective mapping $T:X^{**} \to X$.  Why is T bijective, and why does this give us that $X$ is reflexive?
 A: Was this same question posted earlier? I have reposted the answer here in any case.
For any normed linear space, there is a natural map $T:X \to X^{\ast \ast}$ given by
$$
T(x)(f) = f(x) \quad\forall f\in X^{\ast}
$$
One can show that this is a linear isometry.
Reflexivity means that $T$ is surjective.
Now, in the case of a Hilbert space, there is a natural anti-linear isometry
$$
S_1 : H\to H^{\ast}
$$
given by $S_1(y)(x) = \langle x,y\rangle$. The Riesz Representation theorem says that this map is surjective, and hence an isomorphism.
In particular, $H^{\ast}$ is a Hilbert space, and so there is an anti-linear isomorphism
$$
S_2 : H^{\ast} \to H^{\ast\ast}
$$
Now what you need to check is that $T = S_2\circ S_1$
A: There are a couple of mistakes in your recreation of the proof. You have confused the roles of $h$ and $f$.
In the first application of $RRT$, you should have found $f$ such that 
$$ g(h) = \langle h, f \rangle $$
The version you posted makes no sense since the LHS doesn't depend on $h$, yet it is supposed to hold for all $h$.
In the second application of the theorem, you should find $x$ such that $h(x) = \langle h, f \rangle$ for all $h$ in $X^{*}$.
