Timeline for Poincaré inequality on annular regions
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 24, 2018 at 17:24 | vote | accept | Adi | ||
| Nov 29, 2017 at 5:58 | vote | accept | Adi | ||
| Nov 29, 2017 at 5:58 | |||||
| Nov 28, 2017 at 5:40 | review | Low quality posts | |||
| Nov 28, 2017 at 9:00 | |||||
| Nov 28, 2017 at 5:33 | history | edited | Fan Zheng | CC BY-SA 3.0 |
added 19 characters in body
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| Nov 28, 2017 at 5:33 | comment | added | Fan Zheng | It's me that is missing this condition, but there is an easy fix: let $f=x_1^+:=\max(x_1,0)$. | |
| Nov 28, 2017 at 2:56 | comment | added | Adi | Quick question, the measure condition is not satisfied for this function. It's only zero on the hyperplane $x_1=0$. Am i missing something from your answer? | |
| Nov 27, 2017 at 2:03 | comment | added | Fan Zheng | Let $r$ be very large and $t=r+1$. Let $f(x_1,\dots,x_n)=x_1$. Then the average of $f$ is $r$ while the average of $\nabla f$ is 1 (no matter which $L^p$ norm you take). This is enough to falsify your proposed inequality. | |
| Nov 23, 2017 at 11:05 | comment | added | Adi | Could you please elaborate a little? I expect there is a problem as $t \rightarrow r$, but I am trying to understand what subclass of functions admits such an inequality. | |
| Nov 22, 2017 at 15:31 | review | Low quality posts | |||
| Nov 22, 2017 at 15:46 | |||||
| Nov 22, 2017 at 15:15 | history | answered | Fan Zheng | CC BY-SA 3.0 |