Timeline for Resolvent of Laplacian
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| S May 18, 2013 at 9:33 | vote | accept | supersnail | ||
| May 18, 2013 at 9:33 | vote | accept | supersnail | ||
| S May 18, 2013 at 9:33 | |||||
| May 18, 2013 at 9:32 | vote | accept | supersnail | ||
| May 18, 2013 at 9:33 | |||||
| Apr 4, 2013 at 9:17 | answer | added | Liviu Nicolaescu | timeline score: 7 | |
| Apr 4, 2013 at 9:01 | comment | added | user80744 |
A variant of Nik's argument in his answer is to estimate the norm of the resolvent of $A=-\Delta$ by estimating the spectral integral $(A-z)^{-1}=\int_{\mathbb{R}\setminus\sigma} (t-z)^{-1}\,d E(t)$, when $z$ is not in the spectrum $\sigma$. For unit vectors $u,v$ the total variation of the scalar measure $d(u|E(t)v)$ is at most unity.
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| Apr 4, 2013 at 8:16 | comment | added | Marc Palm | Ah okay, that is fairly obvious. The norm is the supremum of the function involved. Thanks for the explanation. | |
| Apr 4, 2013 at 7:45 | comment | added | Nik Weaver | However, I thought it was simpler to take $z < 0$ since actually ${\rm spec}(-\Delta) \subseteq [0,\infty)$. | |
| Apr 4, 2013 at 7:44 | comment | added | Nik Weaver | @Marc: I think Sönke's estimate is right. As I mentioned, $\|R(z)\| = 1/{\rm dist}(z, {\rm spec}(-\Delta))$. The distance from $z$ to ${\rm spec}(-\Delta)$ is always at least $|{\rm Im}\, z|$ since ${\rm spec}(-\Delta) \subseteq {\bf R}$. | |
| Apr 4, 2013 at 7:12 | comment | added | Marc Palm | Just to be sure that I am not confusing the OP. I was only saying that the resolvent will not always be compact. Nevertheless, you get your estimate as a chep consequence of functional calculus as Nik Weaver explains. @Soenke Hansen: I would believe that your estimate needs at least some constant, or not? | |
| Apr 4, 2013 at 7:02 | comment | added | user80744 |
Assuming a self-adjoint realization of the Laplacian: Yes. The resolvent $R$ of a self-adjoint operator always satisfies the estimate $\|R(z)\|\leq |\mathrm{Im} z|^{-1}$ if $z$ is not real.
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| Apr 4, 2013 at 7:01 | comment | added | Marc Palm | The resolvent of the Laplacian is compact if $(M,g)$ is compact. In general, the spectrum of the Laplacian will not be discrete. Consider the classical example $-y^2( \partial^2_x + \partial^2_y)$ on $SL_2(\mathbb{Z}) \backslash \mathbb{H}$ for example. The spectrum has a continuous part. | |
| Apr 4, 2013 at 6:30 | answer | added | Nik Weaver | timeline score: 9 | |
| Apr 4, 2013 at 5:31 | history | asked | supersnail | CC BY-SA 3.0 |