A Banach space is reflexive if and only if its dual is reflexive How to show that a Banach space $X$ is reflexive if and only if its dual $X'$ is reflexive?
 A: I'm assuming that you have two Theorems at your disposal, easily proven:
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.
Claim: Let $X$ be a Banach space. If $X'$ is reflexive then $X$ is reflexive.
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.
Moreover a consequence of this is that a Banach space is reflexive if and only if its dual space is reflexive.
A: It can be shown directly:
Let $J : X \to X^{**}$ and $J_*: X^* \to X^{***}$ be the canonical injections. Suppose by contradiction that $JX \subsetneq X^{**}$; using Hahn-Banach theorem, there exists $\zeta \in X^{***}$ such that $\zeta \neq 0$ and $\zeta \equiv 0$ on $JX$. 
Because $X^*$ is reflexive, there exists $\theta \in X^*$ such that $\zeta = J_*\theta$. For all $x \in X$:
$$0= \langle \zeta,Jx \rangle= \langle J_*\theta,Jx \rangle = \langle Jx,\theta \rangle= \langle \theta,x \rangle$$
You deduce that $\theta=0$ and therefore $\zeta=0$: a contradiction.
A: Really a Banach space $X$ is reflexive if and only if $X'$ is reflexive.

$$X\textrm{ is reflexive}\Longrightarrow X'\textrm{ is reflexive.}\tag{1}$$

Proof.
By Banach-Alaoglu-Bourbaki theorem the closed ball $B_{X'}$ is closed w.r.t. the weak-* topology $\sigma(X',X)$. By the reflexivity of $X$ we have $\sigma(X',X'')=\sigma(X',X).$ So $B_{X'}$ is closed w.r.t. the weak topology $\sigma(X',X),$ that is $X'$ is reflexive.$\square$

$$X'\textrm{ is reflexive}\Longrightarrow X\textrm{ is reflexive.}\tag{2}$$

Proof.
By hypothesis and by (1) we get that $X''$ is reflexive, and therefore even its closed vector subspace $J(X)$ is reflexive. But the canonical injection $J:X\to X''$ is an isometry so $X$ is reflexive.$\square$
A: Here's a different, more geometric approach that comes from Folland's book, exercise 5.24
Let $\widehat X$, $\widehat{X^*}$ be the natural images of $X$ and $X^*$ in $X^{**}$ and $X^{***}$.
Define $\widehat X^0 = \{F\in X^{***}: F(\widehat x) = 0  \text{ for all } \widehat x \in \widehat X\}$
1) It isn't hard to show that $\widehat{X^*} \bigcap \widehat X^0 = \{0\}$.
2) Furthermore, $\widehat{X^*} + \widehat X^0 = X^{***}$. To show this, let $f\in X^{***}$, and define $l \in X^*$ by $l(x) = f(\widehat x)$ for all $x\in X$.
Then $f(\phi) = \widehat l(\phi) + [f(\phi) - \widehat l(\phi)]$. 
Clearly $\widehat l \in \widehat{X^*}$, and we claim $f - \widehat l \in \widehat X^0$. Let $\widehat x \in \widehat X$. Then $f(\widehat x) - \widehat l ( \widehat x) = f(\widehat x) - \widehat x (l) = f(\widehat x) - l(x) = 0$
Now that 1) and 2) are verified, we prove the claim:
If $X$ is reflexive, then $\widehat X^0 = \{0\}$, and so $X^{***} = \widehat{X^*}$, so $X^*$ is reflexive.
If $X^*$ is reflexive, then $X^{***} = \widehat{X^*}$, so $\widehat X^0 = \{0\}$. Since $\widehat X$ is a closed subspace of $X^{**}$ (on assumption $X$ is Banach), if $\widehat X$ were a proper subspace of $X^{**}$, we would be able to use Hahn-Banach to construct an $F \in X^{***}$ such that $F$ is zero on $\widehat X$ and has ||F|| = 1. This, however, would contradict $\widehat X^0 = \{0\}$. So we conclude $\widehat X  = X^{**}$.
