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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?

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  • $\begingroup$ "Closed unit ball of $Y^*$ is compact in weak topology for any Banach space". I don't think this is true $\endgroup$ – Operator Zebra Feb 8 at 15:32
  • $\begingroup$ @mathworker21 This is true for the weak$^*$-topology, not the weak topology. $\endgroup$ – Aweygan Feb 8 at 15:40
  • $\begingroup$ my apologies you guys $\endgroup$ – mathworker21 Feb 8 at 15:49
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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.

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  • $\begingroup$ Thank you @Aweygan . Will certainly try that exercise. $\endgroup$ – Operator Zebra Feb 8 at 15:41
  • $\begingroup$ You're welcome, glad to help! $\endgroup$ – Aweygan Feb 8 at 15:42
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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.

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