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Help me to prove this please:

Let $X$ be a separable Banach space and $\hat{x}\in X^{**}$. If for all converging sequence $f_{n}\to_{\sigma(X^{*},X)} f$ in $X^{*}$ we have $\hat{x}(f_{n})\to\hat{x}(f)$, then $\hat{x}\in X$.

A.Γ.
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Veridian Dynamics
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  • Look at this answer – A.Γ. May 27 '17 at 20:48
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    @A.Γ. But the hypotheses only say that $\hat{x}$ is $\sigma(X^{\ast},X)$-sequentially continuous. Since $\sigma(X^{\ast},X)$ isn't metrisable when $X$ is an infinite-dimensional Banach space, deducing the $\sigma(X^{\ast},X)$-continuity of $\hat{x}$ needs to be done. – Daniel Fischer May 27 '17 at 20:55
  • @DanielFischer Then I misinterpret the question and took it granted from the separability of $X$. Perhaps it is sufficient that equicontinuous subsets of $X^*$ are metrisable. – A.Γ. May 27 '17 at 21:42
  • I've tracked the result down to what I believe to be the original one. Theorem [Banach, Théorie des Opérations Linéaires, Subwencji Funduszu Kultury Narodowej, 1932, page 124] Let $X$ be a separable Banach space. A subspace $Z$ of $X^$ is weak- closed if and only if it is weak-* sequentially closed. – A.Γ. May 28 '17 at 08:52
  • @A.Γ. that's precisely what i wanted. Thanks! – Veridian Dynamics May 28 '17 at 20:15
  • @Basti You can also look at Krein-Smulian theorem and corollaries. – A.Γ. May 28 '17 at 21:48
  • @Basti .. or even better here – A.Γ. May 28 '17 at 21:55

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Let $r>0,\ B^{*}:=\{f\in X^{*}:\|f\|\leq r\}$. By the Krein-Smulian theorem, $S:=\hat{x}^{-1}(\{0\})$ is weak-closed if we can prove that $S\cap B^{*}$ is weak-closed. Let $x^{*}$ be in the weak-closure of $S\cap B$. Then $x^{*}\in B^{*}$, but since $X$ is separable, $B^{*}$ is metrizable in the weak-topology and therefore (using the hypithesis on $\hat{x}$) $<\hat{x},x^{*}>=0$. So, $S$ is closed and $\hat{x}$ continuous.

Veridian Dynamics
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  • To introduce $x^$ is not really necessary. $S$ weak closed iff $S\cap B^$ is weak closed, but $S\cap B^$ is metrisable, hence, $S\cap B^$ weak* closed iff $S\cap B^$ weak sequentially closed iff $S$ is weak* sequentially closed. The latter is assumed. The last iff is easy due to the fact that weak* convergent sequences are bounded. – A.Γ. May 31 '17 at 08:10