Timeline for Controling mixed derivatives
Current License: CC BY-SA 3.0
12 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Sep 11, 2014 at 11:37 | vote | accept | AlexR | ||
| Aug 22, 2014 at 8:28 | answer | added | Willie Wong | timeline score: 1 | |
| Aug 22, 2014 at 7:56 | history | edited | AlexR | CC BY-SA 3.0 |
added 337 characters in body
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| Aug 22, 2014 at 7:21 | history | edited | AlexR | CC BY-SA 3.0 |
added 51 characters in body
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| Aug 22, 2014 at 7:21 | comment | added | AlexR | @JoonasIlmavirta Actually, I forgot to mention that $\chi(\mathbb R) = \psi(\mathbb R) = [0,1]$. Editing that in. Again, thanks for looking at the problem :) | |
| Aug 21, 2014 at 16:29 | comment | added | Joonas Ilmavirta | For $N=0$ and $z\in\mathbb R$ your statement reads $\int_0^1t^z|1-\psi(t\tau)||\chi(t)|dt\leq C_0\int_0^1t^z|\chi(t)|dt$. You can make $\psi|_{[4/3,5/3]}$ arbitrarily large and choose $\tau$ large to violate this for any $C_0$. Are there perhaps more assumptions (or errors in what I wrote)? | |
| Aug 21, 2014 at 16:14 | history | edited | AlexR | CC BY-SA 3.0 |
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| Aug 21, 2014 at 16:13 | comment | added | AlexR | @JoonasIlmavirta Actually not, thanks for noticing :) | |
| Aug 21, 2014 at 14:45 | comment | added | Joonas Ilmavirta | Are you sure your parentheses are as they should? The LHS has $t^z(1-\psi\chi)$ but the RHS suggests that you might want $t^z(1-\psi)\chi$. | |
| Aug 21, 2014 at 14:05 | review | First posts | |||
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| Aug 21, 2014 at 14:04 | history | asked | AlexR | CC BY-SA 3.0 |