Timeline for Fourier transform of a bounded function
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| May 9, 2013 at 16:24 | comment | added | BS. | No, because modifying the signum function, say on [-1,1], to render it continuous (or even smooth), would just add an $L^1$ function to it, hence a $C_0$ function to its Fourier transform. On the other hand continuous [positive definite][1] functions on $\mathbb{R}$ are exactly the Fourier transforms of finite positive measures, by Bochner's theorem. [1]: en.wikipedia.org/wiki/Positive-definite_function | |
| May 9, 2013 at 16:04 | comment | added | Matthias Ludewig | I suppose, continuity of the function does not help either? | |
| May 9, 2013 at 15:38 | vote | accept | Matthias Ludewig | ||
| May 9, 2013 at 14:36 | history | answered | BS. | CC BY-SA 3.0 |