Prove that a Banach space cannot be reflexive if some strict closed subspace of its dual space separates its points

Question

Let $X$ be a Banach space and let $Z$ be a closed subspace of $X^*$ such that $Z\neq X^*$. Suppose $Z$ separates the points in $X$, that is, if $x \in X$ and $x^*(x) = 0$ $\forall x^* \in Z$ then $x = 0$. Prove that $X$ is not reflexive, i.e. that the canonical injection $J_X$: $X$ $\rightarrow$ $X^{**}$ is not surjective.

How can I do that?

Answer 1

Since $Z$ separates points you have that the pre-annihilator $$Z_\bot :=\{x \in X \colon z(x)=0, \forall z \in Z \} = \{0\}.$$ Use HB to show that the annihilator of pre-annihilator, $$(Z_\bot)^\bot :=\{ x^* \in X^* \colon x^*(Z_\bot) =\{ 0\} \}$$ is equal to the weak-star closure of $Z$ i.e. $$ (Z_\bot)^\bot = \overline{Z}^{w^*}$$ and thus $$\overline{Z}^{w^*} = \{0\}^\bot = X^*.$$ Suppose that $X$ is reflexive so that $X^*$ is reflexive. Then $$X^*=\overline{Z}^{w^*} = \overline {Z}^w = \overline Z = Z$$ (by Mazur theorem; $Z$ is convex). However $X^* \neq Z$.

Answer 2

Let $u\in X^*\setminus Z$. By Hahn-Banach, $$\exists v\in X^{**}\quad v(Z)=\{0\}\text{ and }v(u)=1.$$ Such a $v$ cannot belong to the range of $J_X$, because $v=J_X(x)$ would imply $$Z(x)=\{0\}\text{ and }u(x)=1,$$ thereby contradicting the hypothesis that $Z$ separates the points.