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Question: Assume $(x_n)$ is a weakly convergent sequence in an infinite dimensional Banach space $X$. Show that $\overline{\operatorname{conv}}\{x_n\}$ does not have any interior point.

The hint given tells me to assume $x_n \to 0$ $(weak)$, and that $K = \overline{\operatorname{conv}}\{x_n\}$ contains $0$ in its interior. But I don't understand how to go from the general case to this specific case.

I'm mainly interested in this hint rather than the question itself.

Thank you.

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    $\begingroup$ Adding a constant to the sequence translates the weak limit and the closed convex hull by the same constant. So we can assume $x_n \rightharpoonup 0$. Suppose then that $K$ has an interior point $y$. Then $-y$ is an interior point of $\overline{\operatorname{conv}}\{ -x_n\}$. Mix the sequences $(x_n)$ and $(-x_n)$. $\endgroup$ – Daniel Fischer Oct 18 '16 at 22:17

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