Timeline for An inequality for Fourier transform
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 6, 2014 at 22:31 | comment | added | Gian Maria Dall'Ara | It seems to me that the inequality follows from Plancherel inequality and the identity $\widehat{(-\Delta)^{-\alpha}f}=|x|^{-2\alpha}\widehat{f}$. | |
| Mar 6, 2014 at 21:06 | review | Close votes | |||
| Mar 7, 2014 at 0:31 | |||||
| Mar 6, 2014 at 20:50 | comment | added | András Bátkai | I decided to roll it back to preserve here. It is not obviuos at the first sight that it is not true. | |
| Mar 6, 2014 at 20:49 | history | rollback | András Bátkai |
Rollback to Revision 1
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| Mar 6, 2014 at 20:41 | review | Low quality posts | |||
| Mar 6, 2014 at 20:51 | |||||
| Mar 6, 2014 at 20:39 | history | edited | user47005 | CC BY-SA 3.0 |
deleted 393 characters in body
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| Feb 24, 2014 at 21:30 | comment | added | Asaf | If $f$ is very close to the delta function, the Fourier transform will be very close to constant, and the LHS might not be bounded (say for $\alpha>0.5$). | |
| Feb 24, 2014 at 21:04 | comment | added | user47005 | $f(x)=0$, for $x\in D^C$. | |
| Feb 24, 2014 at 20:53 | comment | added | András Bátkai | How do you define the Fourier transform if your $f$ is only defined in $\Omega$? Do you consider some kind of extension? | |
| Feb 24, 2014 at 19:13 | history | asked | user47005 | CC BY-SA 3.0 |