Timeline for Limit of $e^{-t(-\Delta)^\alpha}$ when $\alpha \to 1$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| May 26, 2021 at 19:52 | vote | accept | Migalobe | ||
| May 22, 2021 at 22:00 | comment | added | Mateusz Kwaśnicki | For the full space, and a global estimate, see my answer below. Alternative way would be to use Bochner's subordination formula — this might also work for fractional powers of other operators, in particular for the fractional Dirichlet Laplacian. If the domain is bounded, estimates of the least eigenvalue (or for the spectral gap in the Neumann case) enter into the game, as Nate Eldredge suggested. The Dirichlet fractional Laplacian seems to be least accessible, but also here some $\alpha$-continuity results should be available — I would have to search references. | |
| May 22, 2021 at 21:54 | answer | added | Mateusz Kwaśnicki | timeline score: 1 | |
| May 22, 2021 at 11:36 | history | edited | Migalobe | CC BY-SA 4.0 |
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| May 22, 2021 at 11:34 | comment | added | Migalobe | Thanks for your comments. @Mateusz I meant the fractional power in whole $\mathbb R^d$. I would also be interested to the techniques used in other cases. | |
| May 22, 2021 at 8:14 | comment | added | Mateusz Kwaśnicki | What do you mean by $(-\Delta)^\alpha$: the fractional power of $\Delta$ in full space, its restriction to $D$ with (say) zero exterior condition ("Dirichlet fractional Laplacian"), or the fractional power of (say) DIrichlet Laplacian in $D$ ("fractional Dirichlet Laplacian" or "spectral fractional Laplacian")? In each case some estimates are available, I think, but of course they are of different type. | |
| May 21, 2021 at 19:04 | comment | added | Nate Eldredge | For starters, the spectral theorem tells you that up to unitary transformation, you are working with $e^{-th^\alpha}$ for some nonnegative measurable function $h$. Then dominated convergence should get you convergence in $L^2$. If you know the spectral gap of $\Delta$ on your domain, then $h$ is bounded away from 0 and this may tell you more. | |
| May 21, 2021 at 18:37 | history | asked | Migalobe | CC BY-SA 4.0 |