I was reading Kreyszig's proof that every Hilbert space is reflexive. I understand everything except one thing, which seems simple.
Defining $A:H'\to H$ by $Af=z$, where $z$ is given by the Riesz representation $f(x)=\langle x,z\rangle$, how can we conclude that $A$ is bijective? I know that injectivity comes from the Riesz Representation Theorem, but I can't see how to show that for each $z\in H$ there exists $f\in H'$ such that $Af=z$.
Thanks in advance!
Best Answer
In the comments, you were given an answer to your question. Allow me to give another perspective to prove this result. The proof contains these ingredients, but does not mention them explicitly.
By the Riesz representation theorem, you are given a surjective antilinear isometric map $$H \to H^*$$
Similarly, there is a surjective antilinear isometric map $$H^* \to H^{**}.$$
We can then consider the composition $$H \to H^{**}$$ which is surjective, linear and isometric.
Show that this composition is the canonical map of $H$ into its bidual!