Unit ball of $X^{**}$ is weakly compact!

Question

Is it true that the closed unit ball in $X^{**}$ is compact with respect to the weak topology on $X^{**}$, where $X$ is a Banach space? If so, how can we prove it?

"Closed unit ball of $Y^*$ is compact in weak topology for any Banach space". I don't think this is true — NewB, Feb 08 '19 at 15:32
@mathworker21 This is true for the weak$^*$-topology, not the weak topology. — Aweygan, Feb 08 '19 at 15:40

score 4 · Accepted Answer · answered Feb 08 '19 at 15:39

4

The unit ball of any Banach space $X$ is compact with respect to the weak topology if and only if $X$ is reflexive (a good exercise, which I recommend trying). Since a Banach space is reflexive if and only if $X^*$ is reflexive, we have

If $X$ is a Banach space, then the unit ball of $X^{**}$ is weakly compact if and only if $X$ is reflexive.

answered Feb 08 '19 at 15:39

Aweygan

23,883

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Thank you @Aweygan . Will certainly try that exercise. – NewB Feb 08 '19 at 15:41
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You're welcome, glad to help! – Aweygan Feb 08 '19 at 15:42

score 2 · Answer 2 · answered Feb 08 '19 at 15:40

It is not true in general. The unit ball in an arbitrary Banach space is weakly compact if and only this Banach space is reflexive (see here for references and a proof sketch). The second dual of a Banach space is not necessarily reflexive; in fact, the dual of a Banach space is reflexive if and only if the Banach space itself is reflexive (see here). Thus, the unit ball in $Y^{\ast\ast}$ is weakly compact if and only if $Y$ is reflexive.

Unit ball of $X^{**}$ is weakly compact!

2 Answers2