Timeline for Fourier transform of Analytic Functions
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| May 27, 2012 at 8:52 | answer | added | Bazin | timeline score: 5 | |
| May 9, 2010 at 12:14 | vote | accept | jonalm | ||
| May 6, 2010 at 19:08 | history | edited | jonalm | CC BY-SA 2.5 |
specified the problem, added tag
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| May 6, 2010 at 15:47 | answer | added | Vectornaut | timeline score: 8 | |
| May 6, 2010 at 15:14 | answer | added | Nate Eldredge | timeline score: 14 | |
| May 6, 2010 at 14:52 | comment | added | Emerton | Dear Harry, One can tell from the question that the OP is willing to entertain constructions based on delta functions and other distributions. This should already be enough to show that your comment is off-base. | |
| May 6, 2010 at 14:31 | answer | added | fedja | timeline score: 21 | |
| May 6, 2010 at 13:17 | comment | added | Mark Meckes | Harry, I think Andrew is alluding to the fact that the Fourier transform is an automorphism of the space of tempered distributions (just as it is on L^2). So Fourier inversion applies for that whole space. This includes plenty of functions that are not in L^2, not to mention plenty of things that aren't functions. | |
| May 6, 2010 at 12:56 | comment | added | Harry Gindi | Fourier inversion, not the Fourier Transform. He's asking for conditions on the transform for the inverse to be analytic. Also, the space of tempered distributions is isomorphic to the topological dual of the Schwartz space. | |
| May 6, 2010 at 10:17 | comment | added | Andrew Stacey | Not true, Harry. You can take the Fourier transform of quite a lot of things. Not quite arbitrary distributions, it's true, but tempered distributions are okay. | |
| May 6, 2010 at 10:16 | comment | added | jonalm | But I guess there exist functions which are not Schwartz, but has a well defined Fourier transform (?). In either case, is there a general way to express a Schwartz function. Like a series expansion? | |
| May 6, 2010 at 10:10 | comment | added | Harry Gindi | Fourier inversion doesn't make sense in general. You need the function to be Schwartz (or at least L^2). | |
| May 6, 2010 at 9:04 | history | edited | jonalm | CC BY-SA 2.5 |
fixed grammar
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| May 6, 2010 at 8:57 | history | asked | jonalm | CC BY-SA 2.5 |