Timeline for About the proof of higher regularity boundary Harnack inequality
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| May 6, 2022 at 6:09 | vote | accept | Holden Lyu | ||
| Sep 10, 2021 at 3:55 | comment | added | Holden Lyu | @AlessandroDellaCorte Thank you anyway. It’s happy to discuss with you. | |
| Sep 9, 2021 at 15:30 | comment | added | Alessandro Della Corte | Yeah, you're right. | |
| Sep 9, 2021 at 15:10 | answer | added | Mateusz Kwaśnicki | timeline score: 4 | |
| Sep 9, 2021 at 13:11 | history | edited | Alessandro Della Corte | CC BY-SA 4.0 |
added 4 characters in body; edited tags
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| Sep 9, 2021 at 9:42 | comment | added | Holden Lyu | @AlessandroDellaCorte I don’t think so. You may read a paper of Junyan Zheng. Or, you can consider an example that $y=|x|^{1+\alpha}$, which fits the condition of g, but you can see that there is no interior ball at (0,0) | |
| Sep 9, 2021 at 9:35 | comment | added | Alessandro Della Corte | In the proof of Hopf Lemma, the smoothness of the boundary is used when claiming that there is a small ball contained in $\Omega$ whose closure is tangent to the boundary at $x$ and intersects the boundary only at $x$. But isn't this what the authors assume at the fourth line of Section 2? (although they should also have added that they keep the assumption $x_n>g(x')$ made in the fourth line of the Introduction). | |
| Sep 9, 2021 at 9:18 | comment | added | Holden Lyu | @AlessandroDellaCorte But, as the boundary is $C^{1, \alpha}$, we cannot get the interior ball condition. How to use Hopf lemma here? | |
| Sep 9, 2021 at 8:39 | comment | added | Holden Lyu | @AlessandroDellaCorte yes I do want to know how to get this condition. Did they claim it? I have no idea ever reading it… Could you please tell where they claimed it? | |
| Sep 9, 2021 at 8:34 | comment | added | Holden Lyu | @AlessandroDellaCorte Yes it is the paper I’m reading. The other conditions doesn’t seem that essential. Could you please tell that where the claim $u_{\nu}$ is bounded? It seems that I’ve missed the condition. | |
| Sep 9, 2021 at 8:34 | comment | added | Alessandro Della Corte | This is the paper, right? arxiv.org/pdf/1403.2588.pdf | |
| Sep 9, 2021 at 7:35 | history | asked | Holden Lyu | CC BY-SA 4.0 |