Timeline for Regularity of Fourier transforms of $L^p$ functions for $2<p\le\infty$
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 10, 2018 at 11:48 | vote | accept | Dominic Wynter | ||
| Jan 10, 2018 at 3:54 | history | edited | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |
Added explanation
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| Jan 10, 2018 at 3:43 | history | edited | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |
obsolete passage removed
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| Jan 10, 2018 at 3:34 | history | edited | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |
small rephrasing
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| Jan 10, 2018 at 3:24 | history | edited | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |
small rephrasing, added remark
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| Jan 10, 2018 at 3:18 | history | edited | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |
small rephrasing
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| Jan 10, 2018 at 3:12 | comment | added | Pedro Lauridsen Ribeiro | I've added the missing discussion on the title's question. Moreover, any $m\in\mathscr{S}′$ yields $T_m$ as a map from $\mathscr{S}$ into $\mathscr{S}′$, since it's just convolution with $\check{m}$ which also belongs to $\mathscr{S}′$. Since $L^\infty\subset\mathscr{S}'$, any Fourier multiplier fits into this picture. In other words, it makes sense to ask whether $T_m$ maps $\mathscr{S}$ into $L^p$ for a given $m\in L^\infty$ and, if that's the case, whether it's bounded in the $L^p$ norm. | |
| Jan 10, 2018 at 2:57 | history | edited | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |
Grammar fix, incomplete sentence amended
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| Jan 10, 2018 at 2:39 | history | edited | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |
Added previously missing discussion on title's question
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| Jan 10, 2018 at 2:34 | comment | added | Dominic Wynter | Actually, I’ve found a potential issue in your answer — $m$ need not be smooth, according to the Mikhlin Multiplier Theorem. Therefore an upper bound on the order is in fact required to do what you’re suggesting. | |
| Jan 10, 2018 at 2:26 | comment | added | Willie Wong | @MonstrousMoonshine: I agree, but you did (unfortunately) use the Mikhlin-Hormander theorem as a motivating example. It may pay to edit the question so that the motivation is somewhat de-emphasized. | |
| Jan 10, 2018 at 0:27 | comment | added | Dominic Wynter | That’s a bit besides the point of my question though. I was wondering specifically about just how “rough” the Fourier transform of a bounded function can get. | |
| Jan 9, 2018 at 20:36 | history | edited | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |
Added explanation
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| Jan 9, 2018 at 20:25 | history | answered | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |