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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.

user512365
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    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. $F \subset Y^{}$, $K(F) \subset Y^{*}$..

     – Andre Jul 27 '17 at 12:29
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    For just topological spaces, see https://math.stackexchange.com/questions/458364/why-is-the-sequential-closure-not-sequentially-closed ... – GEdgar Jul 27 '17 at 13:08
  • also https://math.stackexchange.com/questions/1799452/taking-two-times-the-sequential-closure-of-the-set-of-continuous-functions-in-th – GEdgar Jul 27 '17 at 13:12
  • 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. – GEdgar Jul 27 '17 at 13:13
  • @GEdgar, this need not be easy as we deal with linear subspaces, rather than arbitrary sets. – user512365 Jul 27 '17 at 13:18

1 Answers1

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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.

user512365
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