The weak and weak$^*$ topologies of the dual of a reflexive Banach space coincide

Question

Let $(E, |\cdot|)$ be a normed vector space. Let $E^*, E^{**}$ be the dual and bidual of $E$ respectively. Let $\tau := \sigma(E^*, E^{**})$ and $\tau' := \sigma(E^{*}, E)$ be the weak and weak$^*$ topologies of $E^*$ respectively. If $E \cong E^{**}$, i.e., $E$ is isometrically isomorphic to $E^{**}$, then $\tau = \tau'$. This would be the case if $E$ is a reflexive Banach space or a Hilbert space in particular.

Could you confirm if my understanding is correct?

I think that the weak and weak star topologies coincide if and only if $E$ is reflexive, which (the latter) is not always the case when $E$ is isometrically isomorphic to $E^{**}$ — Evangelopoulos Foivos, Dec 02 '22 at 15:37
@EvangelopoulosPhoevos You are right! I have found a related question here. — Akira, Dec 02 '22 at 15:39

score 3 · Accepted Answer · answered Dec 03 '22 at 15:10

Reflexivity is the isomorphism of $E$ with $E^{**}$ via the specific map $x\longmapsto \hat x$, where $\hat x(\varphi)=\varphi(x)$. It is not enough for reflexivity to have $E\simeq E^{**}$ isometrically. This was established by R.C. James in PNAS 37 (1950), 174-178.

The weak and weak$^*$ topologies of the dual of a reflexive Banach space coincide

1 Answers1