Timeline for Is a sign-preserving operator on $L^2$ a multiplication?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 29, 2019 at 11:14 | history | edited | Mizar | CC BY-SA 4.0 |
added proof that v is bounded
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| Jul 29, 2019 at 11:08 | comment | added | Mizar | Ah yes, $T1$ is not bounded a priori, but you discover it along the proof. I will edit my answer slightly. To deduce the $\sigma$-finite case, say $\Omega=\bigsqcup E_i$ (disjoint union) with $\mu(E_i)<\infty$ and declare $v$ to be $T1_{E_i}$ on $E_i$ (tell me if you have trouble seeing how to conclude from here). | |
| Jul 29, 2019 at 11:00 | vote | accept | daw | ||
| Jul 29, 2019 at 11:00 | comment | added | daw | Interestingly, the assumption on $T$ already implies $T1\in L^\infty(\mu)$. | |
| Jul 29, 2019 at 10:59 | comment | added | daw | How would one define $v$ in the non-finite case? Then $1\not\in L^2$. | |
| Jul 29, 2019 at 10:49 | comment | added | daw | Thanks for the nice answer! I was missing the trick $Tu = u\cdot T1$. | |
| Jul 29, 2019 at 10:49 | vote | accept | daw | ||
| Jul 29, 2019 at 10:52 | |||||
| Jul 29, 2019 at 10:01 | history | edited | Mizar | CC BY-SA 4.0 |
added proof in $\sigma$-finite case
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| Jul 29, 2019 at 9:33 | history | edited | Mizar | CC BY-SA 4.0 |
added 397 characters in body; added 1 character in body; edited body
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| Jul 29, 2019 at 9:22 | history | answered | Mizar | CC BY-SA 4.0 |