Question about the proof $X'$ reflexive $\Rightarrow X$ reflexive.

Question

I have a doubt in the proof I have been given of the fact:

For a Banach space $X$, if $X'$ is reflexive then $X$ is reflexive. This is proven by showing first theorem 1 and theorem 2, which I quote here:

Theorem 1: If a Banach space $X$ is reflexive then its dual space $X'$ is reflexive.

Theorem 2: A closed subspace of a reflexive Banach space is reflexive.

Proof:

Proof: Suppose $X'$ is reflexive. By Theorem 1 it follows that $X''$ is reflexive. If we consider the Canonical mapping $J : X \to X''$ it follows that $J(X)$ is a subspace of $X''$. Since $X \cong J(X)$ and $X$ is a Banach space, then $J(X)$ is a Banach space and hence closed. By Theorem 2 we can conclude that $J(X)$ is reflexive. But since $X \cong J(X)$ conclude that $X$ is reflexive.

The question is, Why is $J(X) \cong X$ ?, that is preciselly what I want to show, but I haven't been able to conclude it...

Answer 1

In this context, $\cong$ is to be interpreted as “isometrically isomorphic.” This means that there exists a map $f:X\to J(X)$ such that

• $f$ is linear;
• $f$ is bijective; and
• $f$ preserves norms: $\|f(x)\|_{J(x)}=\|x\|_X$ for each $x\in X$.

The natural candidate for such a function $f$ is the canonical mapping $J:X\to J(X)$ itself. By construction, it is surjective. It is not difficult to check that it is linear. In addition, one can use the Hahn–Banach theorem to establish that it preserves norms. This latter property actually implies also that $J$ is injective.