Timeline for Global Poincaré type estimate
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 19, 2016 at 0:51 | comment | added | Ali | I think that the inequality also works for $-1<\delta <0$ but the proof will be probably harder! | |
| Jul 17, 2016 at 2:19 | comment | added | Willie Wong | Ah, I see what you mean. Then I don't have the answer handy. | |
| Jul 17, 2016 at 2:15 | comment | added | Ali | I don't think that is true. The constant coming out of your proof is $(\frac{\delta}{2})^2$ which is not strong enough to conclude the proof since $d+1 >d$ | |
| Jul 17, 2016 at 2:07 | comment | added | Willie Wong | Your "sharper" inequality is not sharper in any way. Note that $\int \langle x\rangle^{\delta - 3} |x_1|^2 u^2 \leq \|u\|^2_{L^2_{d-1}}$. What you want follows immediately from what has been proven. Note that in my answer the constant $C$ depends on $\delta$ and degenerates as $\delta \searrow 0$. | |
| Jul 16, 2016 at 15:35 | comment | added | Ali | Thank you @Willie Wong for the nice proof. I think a sharper version of the estimate holds for at least some $\delta$ near zero that I have added to the question. Do you have any insights on that? | |
| Jul 16, 2016 at 5:03 | history | edited | Willie Wong | CC BY-SA 3.0 |
Oops, scaled the wrong way.
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| Jul 16, 2016 at 4:58 | vote | accept | Ali | ||
| Jul 16, 2016 at 3:58 | history | edited | Willie Wong | CC BY-SA 3.0 |
added 383 characters in body
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| Jul 16, 2016 at 3:34 | history | edited | Willie Wong | CC BY-SA 3.0 |
deleted 241 characters in body
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| Jul 16, 2016 at 3:27 | history | answered | Willie Wong | CC BY-SA 3.0 |