Suppose that $X$ is a Banach space and $F$ is a closed subspace of $X^{**}$. Consider $K(F)$ to be linear subspace of $X^{**}$ consisting of weak*-limits of w*-convergent sequences from $F$. Is it true that

$$K(K(F)) = K(F)? $$

I couldn't find any immediate counterexamples to this claim.

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    $\begingroup$ Why are you considering $F \subset X^{**}$ and not $F \subset X^{*}$? Typically, weak* convergence would be defined on the dual space of $X$, so is there a particular reason that you consider it on $X^{**}$? I mean, you could otherwise replace $X$ by $Y := X^*$ and all your statements would be down "one level", requiring one star less everywhere, i.e. $F \subset Y^{*}$, $K(F) \subset Y^{*}$.. $\endgroup$ – Andre Jul 27 '17 at 12:29
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    $\begingroup$ For just topological spaces, see math.stackexchange.com/questions/458364/… ... $\endgroup$ – GEdgar Jul 27 '17 at 13:08
  • $\begingroup$ also math.stackexchange.com/questions/1799452/… $\endgroup$ – GEdgar Jul 27 '17 at 13:12
  • $\begingroup$ So the hint would be: take a good counterexample in toplogical spaces, and convert it to a counterexample for weak* topology in some Banach space. $\endgroup$ – GEdgar Jul 27 '17 at 13:13
  • $\begingroup$ @GEdgar, this need not be easy as we deal with linear subspaces, rather than arbitrary sets. $\endgroup$ – user512365 Jul 27 '17 at 13:18

The answer is negative as proved by Mazurkiewicz:

S. Mazurkiewicz, Sur la dérivée faible d'un ensemble de fonctionnelles linéaires, Studia Math. 2 (1930), 68–71.


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