Timeline for Proof verification for a theorem about a harmonic function on the unit disc [closed]
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 6, 2022 at 21:40 | comment | added | Maths Wizzard | Hi, what I am using from the first theorem is not the statement. I am using the proof. More specifically, I am using the fact that we proved that the maximum is always achieved on the boundary. Does that make moresense? | |
| Jul 6, 2022 at 21:26 | history | closed |
Alexandre Eremenko Yemon Choi Alexey Ustinov M.G. Neil Strickland |
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| Jul 6, 2022 at 11:49 | comment | added | username | The "but u=0 on the boundary" part doesn't hold. | |
| Jul 5, 2022 at 15:21 | comment | added | Maths Wizzard | Hi @username , apologies for the unclarity. I set $z_0$ to be the point of discontinuity on the unit disc. I set $(a,b)$ to be a point in the closed unit disc, with $z_0$ removed, such that $u(a,b)>0$. Is this clearer? Also, I are you sure that a similar argument to the above theorem doesn't hold? I am not saying it applies directly, but an identical proof works. | |
| Jul 5, 2022 at 9:07 | comment | added | username | At the beginning. Did you set $z_0=(a,b)$? The argument from the above Theorem does not apply directly, since precisely $u\neq0$ on the boundary, and isn't continuous. Where are the points of discontinuity in your discussion? | |
| Jul 5, 2022 at 6:46 | review | Close votes | |||
| Jul 6, 2022 at 21:26 | |||||
| Jul 5, 2022 at 6:33 | answer | added | Alexandre Eremenko | timeline score: 1 | |
| S Jul 5, 2022 at 0:11 | review | First questions | |||
| Jul 5, 2022 at 6:28 | |||||
| S Jul 5, 2022 at 0:11 | history | asked | Maths Wizzard | CC BY-SA 4.0 |