0

Let $E$ be a banach space. I have already shown that if $E$ is reflexive then $E^{*}$ is reflexive. Now I want to show that

if the dual space $E^{*}$ is reflexive, then $E$ is reflexive.

If $E^{*}$ is reflexive then, as a corolary, $E^{**}$ is reflexive. Therefore $E^{**}=E^{****}$.

Then, as the canonical injection $J$ is an isometry, $J(E)$ is closed in $E^{**}$, so it is also reflexive, so $J(E)=J(E)^{**} \subset E^{****}$.

How does one conclude from here that $E=E^{**}$?

D1X
  • 2,342

1 Answers1

2

Since $E^*$ is reflexive you got that $E^{**}$ is reflexive (you showed that). Since $E$ is isometric isomorphic to a closed, complete subset of $E^{**}$ you know that $E$ is also reflexive because every closed subspace of a reflexive space is also reflexive.

Hope that it helps you :)

Yaddle
  • 5,246