If $(^0E)^0=E$ for every closed subspace $E\subset X^*$ then $X$ is reflexive.

Question

Let $X$ be a normed vector space, $X^*$ be its dual and $X^{**}$ be its bidual.

For every $E \subset X^*$ we define the left-annihilator of $E$ as $$^0E=\{x \in X : f(x)=0 \ \forall f \in E\}.$$ And for every $F \subset X$ we define the annihilator of $F$ as $$F^0=\{f \in X^*: f(x)=0 \ \forall x \in F\}.$$ It is easy to see that $E \subset (^0E)^0$.

Suppose that for every closed subspace $E \subset X^*$ we have $$E = (^0E)^0.$$

I want to prove that it implies $X$ is reflexive.

So, we want to show that the natural isometric linear injection $J_X:X \to X^{**}$, $J_X(x)(\varphi)=\varphi(x)$, is surjective.

That is, given $\Psi \in X^{**}$ there is $x \in X$ such that for every $\varphi \in X^*$ we have $$\Psi(x)(\varphi)=\varphi(x).$$

These are some ideas I had but I didn't know how to use them:

Define $E:=null(\Psi) \subset X^*$. I think we should consider the case $E\neq X^*$ and then pick the $x\neq 0$ in $$\bigcap_{f \in E}null(f),$$ that is, $x \in (^0E)$.

I also know that we have a natural isometric isomorphism $\Phi: X^*/E \to (^0E)^*$, $\Phi([f])=f|_{(^0E)}$.

I appreciate any ideas or comments!

Answer 1

This should follow from the following facts:

• $(^0E)^0 = E$ if and only if $E$ is weak-$\star$ closed
• $\operatorname{null}(f)$ is closed if and only if $f$ is continuous
• $\operatorname{null}(f)$ is weak-$\star$ closed if and only if $f$ is weak-$\star$ continuous if and only if $f$ belongs to the image of $J_X$.