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The question is quite elementary but nonetheless no proof or counter example comes to mind immediately.

Suppose that $X$ is a Banach space and $\{x_n\}$ is a sequence in $X$ such that $(x_n,y)$ converges for every $y$ in the dual of $X$. Does this imply that $x_n$ converges weakly to some element $x\in X$. Under some form of sequential compactness for example if $X$ is reflexive and separable then this is quite easy to see. Does it hold in general ?

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    $\begingroup$ The limit should define an element in the bidual $X^{**}$. In this language what you ask is whether that element is in (the canonical image of) $X\subset X^{**}$. I think in general there's no reason to expect that. In fact if I'm not mistaken this is related to being weakly sequentially complete. $\endgroup$
    – Teri
    Commented Sep 18 at 9:02
  • $\begingroup$ Thanks, I was not aware of this concept. Actually it is exactly the definition of weak sequentially completeness. $\endgroup$
    – an_ordinary_mathematician
    Commented Sep 18 at 9:36
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    $\begingroup$ $(1,0.0.\dots)$, $(1,1,0,0,\dots$, $(1,1,1,0,0,\dots$ and so on in $c_0$. $\endgroup$
    – terceira
    Commented Sep 18 at 12:11
  • $\begingroup$ Another example: $C[0,2]$ and $f_n(x)=\min(x^n,1).$ $\endgroup$
    – Ryszard Szwarc
    Commented Oct 5 at 17:43

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