Timeline for Superharmonic extension
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Feb 22, 2021 at 20:03 | history | bounty ended | CommunityBot | ||
| S Feb 22, 2021 at 20:03 | history | notice removed | CommunityBot | ||
| S Feb 14, 2021 at 19:01 | history | bounty started | M. Rahmat | ||
| S Feb 14, 2021 at 19:01 | history | notice added | M. Rahmat | Authoritative reference needed | |
| Feb 14, 2021 at 18:06 | comment | added | M. Rahmat | @AlexandreEremenko. In dimension 2, by using the Kelvin transformation and sending $y_0$ to infinity, we obtain a subharmonic function outside $y_0$. But I don't see why the restriction of this function to $K$ is identical to $u$ (?) | |
| Feb 12, 2021 at 11:21 | comment | added | Alexandre Eremenko | In dimension 2, just send this $y_0$ to infinity, and use the known result. In other dimensions you may repeat the arguments from the proof of the result for $y_0=\infty$. | |
| Feb 12, 2021 at 4:50 | comment | added | M. Rahmat | Thanks. Yes, $y_0$ is an isolated point of $K$ indeed. But in this case how to construct the function? | |
| Feb 12, 2021 at 2:02 | comment | added | Alexandre Eremenko | It depends on what point $y_0$ you allow to use. If $y_0$ is an accumulation point of $K$ and $\lim_{y\to y_0}u(y)$ is finite then such extension is not possible. If $y_0$ is an isolated point of $K$ then it is possible. | |
| Feb 12, 2021 at 0:52 | history | asked | M. Rahmat | CC BY-SA 4.0 |