I want to show that all Hilbert spaces are reflexive. I have found the following proof on StackExchange:
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?
Best Answer
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 $T = S_2\circ S_1$