Timeline for Spectral theorem for unbounded self-adjoint operators on REAL Hilbert spaces
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 15, 2020 at 4:39 | answer | added | Jochen Glueck | timeline score: 2 | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Sep 21, 2014 at 14:37 | answer | added | TrialAndError | timeline score: 1 | |
| Mar 6, 2014 at 10:13 | answer | added | Jochen Wengenroth | timeline score: 6 | |
| Jan 18, 2014 at 8:47 | answer | added | alpha | timeline score: 1 | |
| Jan 17, 2014 at 13:16 | comment | added | Simon Henry | Yes because, as you mentioned the spectrum is real and if $U$ is a subset of $\mathbb{R}$ then $E_{x,y}(U)$ is defined as $\langle x ,P y \rangle$ where $P$ is the spectral projection of $T$ corresponding to $U$, and as far as i know you don't need complex number to define them. | |
| Jan 17, 2014 at 12:43 | comment | added | Jochen Wengenroth | Marc Palm: Thanks for sharing my "skepticism". 7891user: In the proof of his lemma 4 Leinfelder uses a kind of Cayley transform and writes "here we make use of the fact that $H$ is a complex Hilbert space". Simon Henry: If $\langle Tx,y\rangle = \int t dE_{x,y}(t)$ is real does this immediately imply that $E_{x,y}$ is a real measure? Pietro Meyer: As I mentioned in the question the unitary operator does not automatically map the "real part" of $H$ to real valued functions. How can one deal with this? Anyway, thank you all for thinking about the question. | |
| Jan 17, 2014 at 12:02 | comment | added | Marc Palm | I am not so sure whether things work out well: link.springer.com/article/10.1007%2FBF01364454#page-1. | |
| Jan 17, 2014 at 10:52 | comment | added | Pietro Majer | Although the complexification is not the unique way, it is a general and natural tool, so I'd suggest to consider it fixing all details. | |
| Jan 17, 2014 at 10:39 | comment | added | 7891user | Try the brilliant proof by Leinfelder in his article "A geometric proof of the spectral theorem for unbounded self-adjoint operators" in Math. Ann. vol. 242 which does not use the complexity of the Hilbert space. | |
| Jan 17, 2014 at 10:30 | comment | added | Simon Henry | If you use the version of the spectral theorem where the operator is written as an integral of its spectral measure then there is no longer any problems with comparing the initial Hilbert space and the space of real valued functions. | |
| Jan 17, 2014 at 10:16 | history | asked | Jochen Wengenroth | CC BY-SA 3.0 |