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Consider a Banach $\mathbb{R}$-space E and an element $u\in E''$ such that : for all sequence $\phi_n\in E'$ which $\sigma(E',E)$-converges to $\phi \in E'$, one has $\lim u(\phi_n)=u(\phi)$. Is it true that there is some $x\in E$ such that $u(f)=f(x)$ for all $f\in E'$ ?

remark : it is not clear for me that $u:E'\rightarrow \mathbb{R}$ is $\sigma(E',E)$-continuous

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    $\begingroup$ looks like a homework problem $\endgroup$
    – Nik Weaver
    Commented Jan 9, 2013 at 10:45
  • $\begingroup$ Really, Nik? It does not to me, but I might assign it the next time I teach a course in Banach space theory. @Terry: The answer is yes if E is separable but no in general; e.g. if $E=\ell_\infty$, because $\ell_\infty$ is a Shur space (that is, every weak$^*$ convergent sequence in $\ell_\infty^*$ is weakly convergent). $\endgroup$
    – Bill Johnson
    Commented Jan 9, 2013 at 14:38
  • $\begingroup$ Is Shur space = Schur space? $\endgroup$
    – Jochen Wengenroth
    Commented Jan 9, 2013 at 15:00
  • $\begingroup$ @Bill: based on the way it's phrased and the fact that there are easy counterexamples (I was thinking of the continuous functions on the extended long line) $\endgroup$
    – Nik Weaver
    Commented Jan 9, 2013 at 17:23
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    $\begingroup$ Voting to close to keep MO from resurrecting it. $\endgroup$
    – Bill Johnson
    Commented Feb 20, 2013 at 23:48

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