Suppose $E$ is a Banach space, and $K\subseteq E^*$ is convex, and is closed and bounded with respect to weak-* topology. Is it true that $K$ is compact?

If $E$ is reflexive, then this is the case, since weak boundedness implies norm boundedness, and it easily follows from Banach-Alaoglu theorem that $K$ is compact (even without convexity). I wonder what happens if we drop the reflexive condition.

Any ideas? Thanks!

As user1952009 pointed out, the preceding argument also works for non-reflexive Banach spaces, see Given a Banach space $X$, are weak$^*$ bounded subsets of the dual space $X '$ also strongly bounded (with respect to the usual norm in $X '$)?.

  • $\begingroup$ what does boiunded for the weak-* topology means exactly ? it is not what appears in en.wikipedia.org/wiki/Uniform_boundedness_principle ? $\endgroup$ – reuns Mar 23 '16 at 22:00
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    $\begingroup$ @user1952009 Thanks, you are right. It works. I remembered the general theorem that weak boundedness is equivalent to original boundedness for locally convex spaces, and if $E$ is reflexive, weak-* topology coincides with weak topology. $\endgroup$ – Yai0Phah Mar 23 '16 at 22:12
  • $\begingroup$ The subset $A \subset E$ is weak* bounded means that for every weak*-open set $U \subset E$ there is a positive constant $\lambda$ such that $A \subset \lambda U$ $\endgroup$ – Daron Mar 23 '16 at 23:12

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