Timeline for weak closedness of the unit ball for a dual pair of Banach space
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 14, 2018 at 16:56 | vote | accept | Littlefield | ||
| Feb 14, 2018 at 14:01 | answer | added | Nik Weaver | timeline score: 10 | |
| Feb 14, 2018 at 13:04 | comment | added | Nik Weaver | Argh, I meant to delete that part! Give me a minute to think of a counterexample. | |
| Feb 14, 2018 at 12:48 | comment | added | Littlefield | Yes, by dual pairs I understand that $Y$ separates points of $X$. So, why is the answer yes? | |
| Feb 14, 2018 at 12:47 | comment | added | Nik Weaver | @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. | |
| Feb 14, 2018 at 11:59 | comment | added | Littlefield | 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. | |
| Feb 14, 2018 at 11:37 | comment | added | erz | Consider $X$ to be of dimension at least $2$ and $Y$ to be a one-dimensional subspace of $X^*$ with the norm of $X^*$. | |
| Feb 14, 2018 at 9:36 | review | First posts | |||
| Feb 14, 2018 at 9:38 | |||||
| Feb 14, 2018 at 9:35 | history | asked | Littlefield | CC BY-SA 3.0 |