Theorem 14.1.10.label Let $E$ be a Banach space over $K \in \RC$, then the following are equivalent:
- (1)
$E$ is reflexive. That is, $E = E^{**}$.
- (2)
$\ol{B_E(0, 1)}^{E}$ is weakly compact.
- (3)
The weak and weak* topologies on $E^{*}$ coincide.
- (4)
$E^{*}$ is reflexive.
Proof. (1) $\Rightarrow$ (2): By the Banach-Alaoglu Theorem.
(2) $\Rightarrow$ (1): By Goldstine’s Theorem and Proposition 5.16.3,
\[\ol{B_{E^{**}}(0, 1)}^{E^{**}}= \ol{B_E(0, 1)}^{\sigma(E^{**}, E^*)}= \ol{B_E(0, 1)}^{E}\]
(1) $\Leftrightarrow$ (3): By Lemma 18.1.4.
(4) $\Rightarrow$ (1): By (1) $\Rightarrow$ (4), $E^{**}$ is reflexive. Since $\ol{B_{E}(0, 1)}^{E}$ is norm closed in $E^{**}$, it is weakly closed in $E^{**}$, and hence weak*-closed in $E^{**}$. By Goldstine’s Theorem, $\ol{B_E(0, 1)}^{E} = \ol{B_{E^{**}}(0, 1)}^{E^{**}}$.$\square$
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