# Are bounded subsets of a locally convex space relatively weakly compact in the bidual?

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.
• 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? – serenus Jul 8 at 13:32
• 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 $X$?). – serenus Jul 8 at 13:42
• 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. – Jochen Jul 8 at 13:52