Timeline for Property about the fractional Laplacian
Current License: CC BY-SA 4.0
18 events
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| Apr 25, 2022 at 22:07 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Dec 26, 2021 at 22:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 28, 2021 at 21:00 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 29, 2021 at 20:07 | answer | added | Guilherme | timeline score: 1 | |
| Jul 29, 2021 at 20:01 | comment | added | Guilherme | @MateuszKwaśnicki in the reference mentioned, is called Parseval's identity or Plancherel's identity. But anyway, thank you very much. | |
| Jul 29, 2021 at 19:54 | comment | added | Mateusz Kwaśnicki | Yes, that is what I meant. (Sorry, I somehow keep saying Plancherel's theorem where I should say Parseval's identity.) | |
| Jul 29, 2021 at 16:40 | comment | added | Guilherme | @MateuszKwaśnicki I used the Parseval's identity contained in page $100$ of Iorio Jr, R., Iorio, V., Fourier Analysis and Partial Differential Equations, Cambridge Studies in Advanced Mathematics 70, Cambridge University Press, Cambridge, 2001. And the fact that $(-\Delta)^s$ can be seen as a pseudo-differential operator of the form $\widehat{(-\Delta)^sg}(n)=|\xi|^{2s}\widehat{g}(n)$ | |
| Jul 29, 2021 at 16:38 | comment | added | Guilherme | @MateuszKwaśnicki is correct \begin{eqnarray*} \int_{-\pi}^{\pi} |(-\Delta)^{s/2}f(x)|^2\; dx = 2\pi \sum_{n \in \mathbb{Z}} \left| |n|^{s}\widehat{f}(n) \right|^2 &=&2\pi \sum_{n \in \mathbb{Z}} |n|^{2s} \, \left|\widehat{f}(n) \right|^2 \\ &=& 2\pi \sum_{n \in \mathbb{Z}} |n|^{2s} \, \widehat{f}(n) \, \overline{\widehat{f}(n) } \\ & = & 2\pi \sum_{n \in \mathbb{Z}} (\widehat{-\Delta)^{s}} f (n) \, \overline{\widehat{f}(n) } \\ & = & \int_{-\pi}^{\pi} \overline{f(x)} \, ({-\Delta)^{s}} f (x) \; dx? \end{eqnarray*} | |
| Jul 29, 2021 at 8:41 | comment | added | Mateusz Kwaśnicki | Apply Plancherel's theorem. Or use the fact that $(-\Delta)^{s/2}$ is self-adjoint. | |
| Jul 28, 2021 at 23:21 | comment | added | Guilherme | @MateuszKwaśnicki Do you know any references to this result? More precisely, for $\int_{-\pi}^{\pi} \overline{f(x)}(-\Delta)^sf(x)\; dx= \int_{-\pi}^{\pi} |(-\Delta)^{s/2}f(x)|^2$? | |
| Jul 28, 2021 at 23:21 | history | undeleted | Guilherme | ||
| May 24, 2021 at 0:13 | history | deleted | Guilherme | via Vote | |
| May 22, 2021 at 18:02 | comment | added | Mateusz Kwaśnicki | Well, $(\sum a_n)(\sum b_n)$ is typically not equal to $\sum a_nb_n$... The integrals are, of course, equal, as long as $f$ is in the $L^2$ domain of $(-\Delta)^s$. | |
| May 22, 2021 at 12:29 | comment | added | Guilherme | @MateuszKwaśnicki and $\int_{-\pi}^{\pi} \overline{f(x)}(-\Delta)^sf(x)\; dx= \int_{-\pi}^{\pi} |(-\Delta)^{s/2}f(x)|^2$ holds? | |
| May 22, 2021 at 11:15 | comment | added | Guilherme | @MateuszKwaśnicki Why? | |
| May 22, 2021 at 8:11 | comment | added | Mateusz Kwaśnicki | This is not even true for $s = 1$. The error seems to lie in the second equality of your formula (2). | |
| May 21, 2021 at 19:59 | history | edited | Guilherme | CC BY-SA 4.0 |
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| May 21, 2021 at 19:52 | history | asked | Guilherme | CC BY-SA 4.0 |