Timeline for A question about Fourier transform of function of the type $Q(x)(1+P(x))^{z}$
Current License: CC BY-SA 3.0
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| S Nov 8, 2016 at 11:50 | history | bounty ended | CommunityBot | ||
| S Nov 8, 2016 at 11:50 | history | notice removed | CommunityBot | ||
| Nov 4, 2016 at 0:26 | history | edited | Tomas | CC BY-SA 3.0 |
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| Nov 1, 2016 at 2:21 | history | edited | Tomas | CC BY-SA 3.0 |
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| Oct 31, 2016 at 12:02 | history | edited | Tomas | CC BY-SA 3.0 |
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| S Oct 31, 2016 at 10:49 | history | bounty started | Tomas | ||
| S Oct 31, 2016 at 10:49 | history | notice added | Tomas | Authoritative reference needed | |
| Oct 31, 2016 at 10:48 | history | edited | Tomas | CC BY-SA 3.0 |
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| Jun 2, 2016 at 8:53 | comment | added | Tomas | @PieroD'Ancona Thanks for the comment. I understand that it will be singular at origin. I'm interested in the case that when the function has suitable decay near $\infty$ (better than the constant), whether it's Fourier transform has some control near the origin. For example, consider $f=(1+|x|^2)^{-a}$ ($0<2a<n$), then we have the desired properties described above. I want to generalize this to more general situations. | |
| May 31, 2016 at 10:39 | comment | added | Piero D'Ancona | What I'm saying is that you always have singularities. This is just the simplest case | |
| May 31, 2016 at 6:31 | history | edited | Tomas | CC BY-SA 3.0 |
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| May 31, 2016 at 6:30 | comment | added | Tomas | @PieroD'Ancona, Thanks, I'm more interested in the case that $P\ne 0$. | |
| May 31, 2016 at 6:28 | comment | added | Tomas | @WhatsUp, I did require $P\ge 0$. Fixed it. Thanks. | |
| May 30, 2016 at 15:41 | comment | added | Piero D'Ancona | P=0, f=1, $\hat f=\delta$ | |
| May 30, 2016 at 13:35 | history | edited | Tomas | CC BY-SA 3.0 |
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| May 30, 2016 at 13:18 | comment | added | WhatsUp | Do you require $P(x) \geq 0$ for all $x$? In general the function $f$ is not defined on some $x$. Think about $f(x) = (1 - x^2)^{-1}$. | |
| May 30, 2016 at 13:09 | history | asked | Tomas | CC BY-SA 3.0 |