Timeline for When does the adjoint operator map closed convex subsets to closed convex subset?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 26, 2012 at 11:29 | vote | accept | Andy Teich | ||
| Jun 13, 2012 at 13:37 | comment | added | Nik Weaver | Since the question is changing, maybe you'd better edit your original post to clarify what it is you really want. | |
| Jun 13, 2012 at 13:26 | comment | added | Andy Teich | So it is sufficient to assume that $C$ is weak*-closed and $M$ is weak*-closed. Is it also necessary to have that $C$ is weak*-closed?Can we give conditions on $C$, $M$ or $T'$ to have that for any $C\subseteq M$ the image $T'(C)$ is weak*-closed in $X'$? | |
| Jun 13, 2012 at 12:51 | comment | added | Nik Weaver | Weak* compact (which is the same as weak* closed, if it's bounded). | |
| Jun 13, 2012 at 12:45 | comment | added | Andy Teich | If I understand you correctly, it also would suffice to assume that $M$ itself is weak*-closed...? | |
| Jun 13, 2012 at 12:14 | history | answered | Nik Weaver | CC BY-SA 3.0 |