Timeline for Convolution of $\mathscr{F}\{ \log \}(x) * \mu$ with compactly supported measure $\mu$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 28 at 16:58 | comment | added | Christian Remling | Yes, that would work if you restrict to $|t|<a$. | |
| Mar 28 at 16:30 | comment | added | Grandes Jorasses | @ChristianRemling thank you. Then the only option would be to truncate also from above and use the boundedness of $\hat{\mu}$, right? | |
| Mar 28 at 16:13 | comment | added | Christian Remling | No, not in general because $\psi$ gets large for large $|t|$ and $\widehat{\mu}$ is not guaranteed to have any decay, so we cannot automatically conclude that $\widehat{\mu}\psi\in L^2(|t|>a)$. | |
| Mar 28 at 16:03 | comment | added | Grandes Jorasses | @ChristianRemling Thank you very much for your answer! One more question: if I truncate the domain of $\psi(t)$ near zero and consider $\hat{\mu}(t)\psi(t) 1_{|t|>a}$ for some $a>0$, is the resulting distribution in $L^2$? | |
| Mar 28 at 14:12 | comment | added | Christian Remling | A bit more informally, you can also just observe that the product $\widehat{\mu}(t)\psi (t)$ is still tempered since $\widehat{\mu}$ is bounded. | |
| Mar 28 at 14:11 | comment | added | Christian Remling | See Theorem 4.3 here: mat.univie.ac.at/~stein/lehre/SoSem09/distrvo.pdf | |
| Mar 28 at 13:13 | history | edited | Grandes Jorasses | CC BY-SA 4.0 |
added 17 characters in body
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| Mar 28 at 12:02 | history | asked | Grandes Jorasses | CC BY-SA 4.0 |