Timeline for Extremal point and probability
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 27, 2012 at 7:38 | comment | added | Jochen Wengenroth | Okay, one one does not need induction. However, the inductive argument does not need that $C$ is closed. For example, it proves that $\int f d\mu$ always belongs to $C$. | |
| Nov 26, 2012 at 21:54 | comment | added | Pietro Majer | Actually we do not need induction: since $x$ is an extremal point there is a linear functional $u$ such that $u(y) < u(x):=t$ for all $y\in C\setminus\{x\}$ (this is a characterization); with the same argument one immediately has $f(y)\in C\cap\{u=t \}=\{x\}$ a.e. that is $f(y)=t$ a.e. The same argument works with any compact $C$ in a LCTVS, and w.r.to the Pettis integral. | |
| Nov 26, 2012 at 21:18 | vote | accept | Vincent Beck | ||
| Nov 26, 2012 at 21:18 | |||||
| Nov 26, 2012 at 21:10 | vote | accept | Vincent Beck | ||
| Nov 26, 2012 at 21:13 | |||||
| Nov 26, 2012 at 17:53 | vote | accept | Vincent Beck | ||
| Nov 26, 2012 at 17:53 | |||||
| Nov 26, 2012 at 17:39 | comment | added | Alexandre Eremenko | Yes, if $C$ is compact. See the end of my solution. Bauer's theorem applies in any locally convex space. | |
| Nov 26, 2012 at 15:23 | comment | added | Nate Eldredge | Is this fact still true in infinite-dimensional spaces? Say, a separable Frechet space. | |
| Nov 26, 2012 at 15:02 | history | answered | Jochen Wengenroth | CC BY-SA 3.0 |