Suppose that $\langle X,Y\rangle$ is a dual pair of Banach spaces satisfying $|\langle x,y\rangle|\leq \Vert x\Vert\Vert y\Vert$ for all $x\in X$, $y\in Y$.
Is it true that the unit ball of $X$ is $\sigma(X,Y)$-closed?
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1$\begingroup$ Consider $X$ to be of dimension at least $2$ and $Y$ to be a one-dimensional subspace of $X^*$ with the norm of $X^*$. $\endgroup$– erzCommented Feb 14, 2018 at 11:37
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$\begingroup$ This is not enough, as far as I see. For instance, consider $X:=(R^2,|\cdot|_2$. If $\langle X,Y\rangle$ is a dual pair with $Y$ one dimensional, then $Y$ is spanned by a vector $(x_1,x_2)$ with $x_i\neq 0$. Then that weak convergence of sequences in $B_X$ is equivalent to the strong convergence. $\endgroup$– LittlefieldCommented Feb 14, 2018 at 11:59
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1$\begingroup$ @Littlefield: think harder about the fact that the $\sigma(X,Y)$ topology may be non-Hausdorff. Or did you want to assume that $Y$ separates points of $X$? Then the answer is yes. $\endgroup$– Nik WeaverCommented Feb 14, 2018 at 12:47
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$\begingroup$ Yes, by dual pairs I understand that $Y$ separates points of $X$. So, why is the answer yes? $\endgroup$– LittlefieldCommented Feb 14, 2018 at 12:48
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$\begingroup$ Argh, I meant to delete that part! Give me a minute to think of a counterexample. $\endgroup$– Nik WeaverCommented Feb 14, 2018 at 13:04
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Okay, even assuming that $Y$ separates points the answer is still no. Take $X = l^1$ and let $Y$ be the set of elements $(a_n)$ of $l^\infty$ which satisfy $\lim a_n = 2a_1$. It's easy to see that $Y$ separates the points of $X$, but $e_n \to 2e_1$ weakly where $(e_n)$ is the standard basis of $l^1$. So $2e_1$ is in the weak closure of the unit ball.
answered Feb 14, 2018 at 14:01