Let $X$ be a separable Banach space such that $X$ and its dual $X^*$ have Radon-Nikodym property. Let $C$ be a convex, closed and bounded subset of $X$.
Can we say that $C$ is weakly compact or weakly locally compact?
An idea please
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1 Answer
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The closed unit ball of a Banach space $X$ is weakly compact if and only if it is reflexive. So if $X$ is a non-reflexive space such that $X$ and its dual have RNP, take $C$ to be the unit ball to see that the answer to your first question is negative.
For instance, takes $X$ to be the classical James space; its dual is separable, hence has RNP; its bidual is also separable, hence has RNP; and $X$ is a closed subspace of $X^{**}$ hence also has RNP.
I suspect that the same counterexample also works for "locally weakly compact" but I have not checked the details.
answered Jun 3, 2020 at 3:32
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1$\begingroup$ Perhaps the reason for a conjecture like the OP is that the James space is relatively unknonwn. See mathoverflow.net/a/43987/454 $\endgroup$ Commented Jun 3, 2020 at 11:08
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$\begingroup$ @GeraldEdgar Indeed. I must confess that I spent quite a while scratching my head and thinking of "classical" spaces before eventually hittting on the idea "when can the bidual be separable without the space being reflexive" $\endgroup$ Commented Jun 3, 2020 at 16:21
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1$\begingroup$ If $C$ is closed and convex, it's weakly closed; so weak compactness and weak local compactness are the same. $\endgroup$ Commented Jun 3, 2020 at 17:27
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$\begingroup$ @DirkWerner Thanks: I suspected as much but was too tired when I wrote the original example to check that I wasn't making a silly oversight $\endgroup$ Commented Jun 3, 2020 at 18:09