Timeline for Is the second weak derivative a self-adjoint operator?
Current License: CC BY-SA 4.0
17 events
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| Aug 24, 2021 at 23:21 | vote | accept | Maximilian Janisch | ||
| S Aug 24, 2021 at 21:03 | history | bounty ended | CommunityBot | ||
| S Aug 24, 2021 at 21:03 | history | notice removed | CommunityBot | ||
| Aug 24, 2021 at 7:22 | answer | added | phantomias | timeline score: 1 | |
| S Aug 16, 2021 at 19:05 | history | bounty started | Maximilian Janisch | ||
| S Aug 16, 2021 at 19:05 | history | notice added | Maximilian Janisch | Authoritative reference needed | |
| Aug 16, 2021 at 7:42 | history | edited | Maximilian Janisch | CC BY-SA 4.0 |
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| Aug 15, 2021 at 9:53 | comment | added | bathalf15320 | The F.T. is a Hilbert space isomorphism on $L^2$ of the line. When I learned Sobolev spaces, they were defined as the inverse images of weighted $L^2$-spaces under the F.T. These are, in turn. the domains of definition of multiplication operators and it is a well-known standard fact that the latter are self-adjoint. This method , of course, works in a much more general setting than that of your question. | |
| Aug 15, 2021 at 9:33 | history | edited | Maximilian Janisch | CC BY-SA 4.0 |
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| Aug 15, 2021 at 9:25 | comment | added | Maximilian Janisch | @NateEldredge Now updated :) | |
| Aug 15, 2021 at 9:25 | history | edited | Maximilian Janisch | CC BY-SA 4.0 |
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| Aug 14, 2021 at 20:26 | comment | added | Nate Eldredge | For (1), use Cauchy-Schwarz instead to estimate $\int \psi'' \phi_n - \int \psi'' \tilde{\psi} = \int \psi'' (\phi_n - \tilde{\psi})$. Notice that $\phi_n \to \tilde{\psi}$ in $L^2$. For the other side, proceed similarly and note that $\phi_n'' \to \tilde{\psi}''$ in $L^2$ by virtue of $W^{2,2}$ convergence. DCT isn't needed here. | |
| Aug 14, 2021 at 17:00 | history | edited | Maximilian Janisch | CC BY-SA 4.0 |
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| Aug 14, 2021 at 15:59 | comment | added | Jochen Glueck | @ChristianRemling: "'well known' (to those who know them well)" Nominated for the Mathematical Aphorism of the Week Award! :-) | |
| Aug 14, 2021 at 15:36 | comment | added | Christian Remling | Or you could observe that the deficiency indices of $T$ are zero. All these arguments are classical and "well known" (to those who know them well), so a walk to the library should really help you here. | |
| Aug 14, 2021 at 15:33 | comment | added | Christian Remling | Yes, this is true. It can be shown directly by methods similar to those discussed in my lecture notes here, example 11.1: math.ou.edu/~cremling/teaching/lecturenotes/fa-new/ln11.pdf In this particular case though, it would be much quicker to note that $T$ is multiplication by $-k^2$ on its natural domain after taking Fourier transforms. | |
| Aug 14, 2021 at 15:19 | history | asked | Maximilian Janisch | CC BY-SA 4.0 |