Let $X$ be a Hausdorff locally convex topological vector space, and let $A$ be a bounded subset of $X$. Is it true that the weak closure $\overline{A}^{\sigma(X'', X''')}$ is compact in the bidual $(X'', \sigma(X'', X'''))$?


No. For a counterexample take the non-reflexive Banach space $X=c_0$ and let $A$ be its closed unit ball. Then $X$ and hence $A$ are closed in the normed bidual $X''$ and, since for convex sets in locally convex spaces the closure equals the weak closure, $A$ is closed in $(X'',\sigma(X'',X'''))$. If $A=\overline{A}^{\sigma(X'',X''')}$ were compact, the Krein-Milman theorem implies that it has extreme points which is not the case.

  • $\begingroup$ I think that in my post what is missing is the topology that the dual $X′$ is equipped with, isn't it? In your example, we assume that $X′$ is equipped with the norm topology, right? Or, do we really need a topology on the dual $X'$ in this case? $\endgroup$ – serenus Jul 8 at 13:32
  • $\begingroup$ And, for $X$ and $A$ as above, I think that the weak$^*$ closure $\overline{A}^{\sigma(X'', X')}$ is compact in $(X'',\sigma(X'', X'))$, which is the case when, in particular, $X$ is a reflexive Fréchet space (again, in this case should we put a topology on the dual $X'$, for example, the topology of uniform convergence on bounded subsets of $X$?). $\endgroup$ – serenus Jul 8 at 13:42
  • $\begingroup$ I think that the standard definition of the bidual is $X''=(X',\beta(X',X))'$, if $X$ is normed then this is indeed the norm topology of $X'$. If $X$ is reflexive then $\sigma(X'',X')$ is equal to the weak topology $\sigma(X,X')$ and then all bounded and weakly closed subsets of $X$ are weakly compact. $\endgroup$ – Jochen Jul 8 at 13:52

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