Timeline for Weighted Sobolev norm in terms of Spherical harmonics coefficients
Current License: CC BY-SA 4.0
11 events
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| Sep 28, 2021 at 17:52 | comment | added | Willie Wong | Your Sobolev norm includes purely angular derivatives: when those hit, if your $h$ is in $H^s$ and not $H^{s+1}$, and $k > s$, the integral over $\mathbb{S}^2$ will converge for no $r$. | |
| Sep 28, 2021 at 15:31 | comment | added | Laithy | Thank you so much @WillieWong for always helping me. I have tried to compute what $L_{lm}$ are but I keep getting stuck. I will try again today. Only if you have time, please provide more details or at least a guess at what the $L_{lm}$ are. I forgot to say that $f_l$ is analytic and so its derivatives satisfy the appropriate assumptions. My initial guess is that $k$ can be chosen to be bigger than $s$. But you are saying thats not correct. So I am a bit confused. I will think about it again. | |
| Sep 28, 2021 at 14:28 | comment | added | Willie Wong | For Question #2: If you only assume decay of $f_l$ and not its higher derivatives, then generally your function will not be in $H^k_\delta$ for any $k > 0$. With appropriate assumptions of decay of higher derivatives up to order $s'$, then the best $k$ you can choose is $\min(s,s')$, and the best $\delta$ is a matter of explicit computation based on the assumed decay rates. | |
| Sep 28, 2021 at 14:23 | comment | added | Willie Wong | Question #1 seems to be just a matter of explicit computation, no? The integrals over $\mathbb{S}^2$ you can just use the orthogonality properties of $Y_{lm}$, and you will get a family of linear differential operators on $[1,\infty)$ (let's call them $L_{lm}$) such that the Sobolev norm looks something like $\sum \| L_{lm} c_{lm} \|^2_{L^2}$. | |
| Sep 28, 2021 at 3:23 | history | edited | Laithy | CC BY-SA 4.0 |
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| Sep 21, 2021 at 5:17 | comment | added | Laithy | ahh i didn't word the question properly. $f$ is a fixed function of $r$ only. I rewrote the question. | |
| Sep 21, 2021 at 5:16 | history | edited | Laithy | CC BY-SA 4.0 |
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| Sep 21, 2021 at 5:07 | history | edited | Laithy | CC BY-SA 4.0 |
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| Sep 21, 2021 at 3:22 | comment | added | user378654 | Is $f$ fixed and you do not care about the dependence on it? If no, (1) seems false: the weighted norm also measures derivatives of $f$. | |
| Sep 21, 2021 at 2:56 | history | edited | Laithy | CC BY-SA 4.0 |
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| Sep 20, 2021 at 21:13 | history | asked | Laithy | CC BY-SA 4.0 |