Timeline for Scaled Harnack inequality $\sup_{B_r} v \le c\,(1-r)^{-p}\, \inf_{B_r} v$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| S Sep 3, 2019 at 0:57 | history | suggested | Ali Taghavi |
I add a tag.
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| Sep 3, 2019 at 0:15 | review | Suggested edits | |||
| S Sep 3, 2019 at 0:57 | |||||
| Aug 31, 2019 at 9:41 | vote | accept | Riku | ||
| Aug 30, 2019 at 15:07 | comment | added | Connor Mooney | Sure, please see my answer below. | |
| Aug 30, 2019 at 15:06 | answer | added | Connor Mooney | timeline score: 3 | |
| Aug 30, 2019 at 9:52 | comment | added | Riku | @ConnorMooney How does the last inequality in your remark give that polynomial growth $(1-r)^{-p}$? Could you add more details on this in an answer, please? | |
| Aug 29, 2019 at 20:56 | comment | added | Connor Mooney | It follows from applying the usual Harnack inequality to a sequence of balls. For example, applying Harnack in $B_1$ gives $u|_{B_{1/2}} \leq Cu(0)$. Applying it in balls of radius $1/2$ centered at points on $\partial B_{1/2}$ gives $u|_{B_{3/4}} \leq C^2u(0)$. Continuing we get $u|_{B_{1-2^{-k}}} \leq C^ku(0)$, giving polynomial growth near the boundary with power $p \sim \log C / \log 2$. | |
| Aug 29, 2019 at 16:06 | history | edited | Riku | CC BY-SA 4.0 |
edited title
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| Aug 28, 2019 at 16:43 | history | asked | Riku | CC BY-SA 4.0 |