Timeline for Minimum of two plurisubharmonic functions
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Nov 11, 2020 at 12:54 | vote | accept | Luka Thaler | ||
| Sep 23, 2014 at 12:43 | comment | added | ACL | In the real case, take $D=\mathopen]-1;1\mathclose[$, $u(x)=x$ and $v(x)=-x$. One has $\Phi(x)=-|x|$, $\Phi|_{\partial D}=-1$ hence $\Phi(x)>\sup_{\partial D}\Phi$ for every $x\in D$. | |
| Sep 22, 2014 at 18:54 | answer | added | Alexandre Eremenko | timeline score: 2 | |
| Sep 22, 2014 at 18:08 | history | edited | Luka Thaler | CC BY-SA 3.0 |
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| Sep 22, 2014 at 18:00 | comment | added | Luka Thaler | Yes thank you I am familiar with that. I've just seen that I've asked the same question twice. The edited version can be seen above. Sorry for inconvenience. | |
| Sep 22, 2014 at 17:52 | history | edited | Luka Thaler | CC BY-SA 3.0 |
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| Sep 22, 2014 at 16:26 | comment | added | ACL | It suffices that min {u,v} = u everywhere. More seriously (it is easier to visualize the real analogue of convex functions), if $u$ and $v$ are distinct linear functions on the real line and $x\in\mathbf R$ is such that $u(x)=v(x)$, then $\min\{u,v\}$ is concave and is not convex. | |
| Sep 22, 2014 at 15:34 | history | asked | Luka Thaler | CC BY-SA 3.0 |