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Let $X$ be a Banach space, $S$ a subset of $X$. What is the closure of $S$ with respect to the weak topology?

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It is often very difficult to calculate. A point $y\in X$ is in the weak closure if you can not enclose it in a weak neighborhood disjoint from $S$, i.e. if for every $\epsilon>0$ and linear functionals $\ell_1,\ldots,\ell_k\in X'$ the set $N=\{ x\in X : |\ell_i(y-x)|<\epsilon, i=1,\ldots,k \}$ intersects $S$. Note that $k$ must be finite.

In finite dimension these neighborhoods generate the usual Banach topology so there is no difference taking norm or weak closure. In infinite dimension this is no longer so. Every such neighborhood is unbounded and contains infinite dimensional affine sub-spaces (related to the intersection of $\ker \ell_i$). Take the unit sphere $S=\{x\in X: \|x\|=1\}$. Its weak closure is the closed unit ball $B=\{x\in X: \|x\|\leq 1\}$ because any weak neighborhood of an interior point intersects the boundary. An exterior point may be separated through Hahn-Banach so you don't get more. If $S$ is the finite union of disjoint spheres then the weak closure is the union of the corresponding closed balls. If you have a concrete case to consider you might want to post it?

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  • $\begingroup$ Thank you for your answer! I have to show that all the point of X such that there is a sequence of point of S converging weakly to, belong to the closure of S in the weak topology. $\endgroup$ Jul 26, 2016 at 10:16
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    $\begingroup$ You are welcome. You have probably already deduced this, but anyway, if $N$ is a neighborhood as above, and $x_n\rightarrow y$ then $\ell_i(y-x_n)$ goes to zero for each $i$ so $S$ intersects $N$. In fact, the weak closure contains the strong closure (and whence the points that you describe) $\endgroup$
    – H. H. Rugh
    Jul 26, 2016 at 10:36
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    $\begingroup$ now is really clear, thank you. $\endgroup$ Jul 26, 2016 at 10:52

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