I have already shown that if $X$ is reflexive then $X^*$ is reflexive, but I need some help on the other direction. The canonical mapping is defined by $$ J : X \to X^{**}, \ J(x) (f) = f(x)$$  For the other direction suppose that $X^*$ is but $X$ is not reflexive. It follows that the canonical mapping $J : X \to X^{**}$ is not a surjection, so there must exist some element $\lambda \in X^{**}$ such that $\lambda \not \in J(X)$. Since $J$ is an injective linear isometry it follows that the set $J(X)$ is closed, and hence $$ \operatorname{dist} (J(X), \lambda) &gt; 0$$ It is also clear that the image of $X$ is a subspace of $X^{**}$. From a corollary of the Hahn-Banach theorem it follows that there is a non-zero $\Lambda \in X^{***}$ such that $$ \forall y \in J(X), \ \Lambda (y) = 0$$ Now since $\Lambda \in X^{***}$ and $X^*$ is reflexive it follows that the canonical map $J^* : X^{*} \to X^{***}$ is a surjection and hence a bijection. This implies that there exists $\alpha \in X^*$ such that $$ J^* (\alpha) = \Lambda$$ and in particular $$ \forall y \in J(X), \ J^* (\alpha) (y) = 0 \implies \forall y \in J(X), \  y(\alpha) = 0$$ Now if I am able to show that $\alpha = 0_{X^*}$ I am done since this implies that $$ \Lambda = J^*(0_{X^*}) = 0_{X^{***}}$$ which is a contradiction since $\Lambda$ is non-zero.   I am unsure how I would show this though. I suppose I could construct some $y \in J(X)$ which would do it. Any ideas?  Edit : I think I have it. We have $$ \forall y \in J(X), \ y (\alpha) = 0 \iff \forall x \in X, \ J(x) (\alpha) = 0 \iff \forall x \in X, \ \alpha(x) = 0$$ So $\alpha = 0_{X^*}$ and the theorem follows.
How to show that a Banach space $X$ is reflexive if and only if its dual $X'$ is reflexive?
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