Closed subspace of a reflexive Banach space is reflexive

Question

I'm studying Conway's functional Analysis by myself. In page 132 of his book, for showing every Closed subspace M of a reflexive Banach space X is reflexive, he says $\sigma(X,X^*)_{|_{M}}=\sigma(M,M^*)$. But I can not understand how it is. Please regard me. Thanks in advance

Answer 1

You can use the criterion that reflexivity is equivalent to the unit ball being compact in the weak topology. Then the intersection of the unit ball with the closed subspace (which is still closed in the weak topology) is also compact, hence $E\subseteq X$ is reflexive.

Edit: If you want to do it directly, you can use the Hahn Banach theorem to do it directly, since a closed subspace is characterized by the linear functionals which vanish on it. Then if $E\subseteq X$ is a closed subspace, we see

$$E^*\cong Hom_{\Bbb F}(X,\Bbb F)/\{f\in X^* : f(E)=0\}$$

where here $\Bbb F$ is your field, either $\Bbb R$ or $\Bbb C$. Then dualizing this gives rise to the isomorphism since the map

$$\begin{cases}X\to X^{**}=Hom_{\Bbb F}(Hom_{\Bbb F}(X,\Bbb F), \Bbb F) \\ x\mapsto E_x\end{cases}$$

is surjective you conclude the same holds passing through the quotient.