Timeline for Generalization to the spectral Theorem of self-adjoint operators
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Jan 2, 2018 at 15:45 | history | edited | Schüler | CC BY-SA 3.0 |
improved formatting
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| Dec 2, 2017 at 8:18 | history | edited | Schüler | CC BY-SA 3.0 |
improved formating
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| Dec 1, 2017 at 18:40 | comment | added | LSpice | Your question doesn't quite make it clear, but you want a single $U$ that works for both $S_k$, right? (Otherwise I guess that @JochenGlueck's comment gives it immediately, modulo the $\sigma$-finiteness issue.) | |
| Dec 1, 2017 at 15:39 | vote | accept | Schüler | ||
| Dec 1, 2017 at 14:48 | answer | added | Matthew Daws | timeline score: 9 | |
| Dec 1, 2017 at 13:52 | history | edited | Schüler | CC BY-SA 3.0 |
improved formating
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| Dec 1, 2017 at 13:28 | comment | added | Matthew Daws | The answer will be negative if (and only if) you can find $S_1,S_2$ commuting, with $MS_i$ self-adjoint for both $i$, but with $MS_1$ not commuting with $MS_2$. | |
| Dec 1, 2017 at 12:59 | history | edited | Schüler | CC BY-SA 3.0 |
improved formating
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| Dec 1, 2017 at 12:55 | comment | added | Schüler | I have edited my question. | |
| Dec 1, 2017 at 12:53 | history | edited | Schüler | CC BY-SA 3.0 |
improved formating
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| Dec 1, 2017 at 12:52 | history | undeleted | Schüler | ||
| Dec 1, 2017 at 10:33 | history | deleted | Schüler | via Vote | |
| Dec 1, 2017 at 10:26 | comment | added | Jochen Glueck | Well, that depends. If $X$ can be chosen to be $\sigma$-finite, then $X$ can also be chosen to be finite by a simple change-of-density argument. However, I doubt that $X$ can always be chosen to be $\sigma$-finite in the spectral theorem (in particular, if $F$ is not separable). To be sure, I suggest to look it up in a textbook or a monograph where the spectral theorem is treated. | |
| Dec 1, 2017 at 10:16 | comment | added | Schüler | Thank you. Do you mean that $\mu(X)$ can be equal to $\infty$? | |
| Dec 1, 2017 at 10:11 | comment | added | Jochen Glueck | Your assumptions imply that $(MS)^* = S^*M = MS$, so $MS$ is self-adjoint, which shows that your claim is true, right? (I'm not quite convinced, however, that $(X,\mu)$ can be chosen as a ($\sigma$-)finite space, in particular if $F$ is not assumed to be separable.) | |
| Dec 1, 2017 at 9:07 | history | asked | Schüler | CC BY-SA 3.0 |