Banach Space is Reflexive iff Normed Dual is Reflexive

Theorem
Let $\Bbb F \in \set {\R, \C}$.

Let $X$ be a Banach space over $\Bbb F$.

Let $X^\ast$ be the normed dual space of $X$.

Then:


 * $X$ is reflexive $X^\ast$ is reflexive.

Proof
Let $X^{\ast \ast}$ be the second normed dual of $X$.

Let $X^{\ast \ast \ast}$ be the normed dual of $X^{\ast \ast}$.

Necessary Condition
Suppose that $X$ is reflexive.

We want to show that $X^\ast$ is reflexive.

That is, for each $\Phi \in X^{\ast \ast \ast}$ we aim to find $\phi \in X^\ast$ such that:


 * $\map \Phi F = \map {\phi^\wedge} F = \map {\map {J_{X^\ast} } f} F$ for each $F \in X^{\ast \ast}$

where $J_{X^\ast}$ is the evaluation linear transformation $X^\ast \to X^{\ast \ast \ast}$.

Since $X$ is reflexive, for each $F \in X^{\ast \ast}$ there exists $x \in X$ such that:


 * $F = x^\wedge = \map {J_X} x$

where $J_X$ is the evaluation linear transformation $X \to X^{\ast \ast}$.

So, it suffices to find $\phi$ such that for each $x \in X$ we have:

for each $x \in X$.

We show that this actually defines an element $\phi \in X^\ast$.