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Let $X$ be a complete Hausdorff locally convex topological vector space, let $X''=(X',\beta(X',X))'$ be its standard bidual. Let $\sigma(X,X')$ be the weak topology on $X$ and let $\sigma(X'',X')$ be the weak* topology on $X''$.

Let $A$ be an absolutely convex (=balanced+convex) closed subset of $X$ such that its $\sigma(X'',X')$-closure in $X''$, that is, $\overline{A}^{\sigma(X'',X')}$ is compact in $(X'',\sigma(X'',X'))$. Is it true that the weak closure $\overline{A}^{σ(X,X')}=A$ is compact in $(X,\sigma(X,X'))$?

I think that this should be true since $(X,\sigma(X,X'))$ is a topological subspace of $(X'',\sigma(X'',X'))$. But I am unable to verify this. Can any body give a clue, or a hint, or an answer?

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No. The unit ball of $c_0$ is weakly closed but not weakly compact. Its closure in the bidual is the unit ball of $\ell^\infty $ which by Alaoglu is weak$^\ast$ compact.

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  • $\begingroup$ Thanks for the answer. It seems that the space $c_0$ is a good source of counter-examples. $\endgroup$
    – serenus
    Sep 22, 2019 at 19:59
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    $\begingroup$ Any non-reflexive Banach space would work here. $\endgroup$
    – Jochen
    Sep 22, 2019 at 20:01

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