Timeline for Convolution vanishes on an interval
Current License: CC BY-SA 3.0
7 events
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Aug 17, 2015 at 14:39 | comment | added | Uchiha | Another question. In the argument it seems to require that $g$ is a tempered distribution. My feeling is that the statement might hold even for functions with exponential growth. Could you comment on this? Thank you. | |
Aug 14, 2015 at 16:55 | vote | accept | Uchiha | ||
Aug 14, 2015 at 15:19 | comment | added | Jean Duchon | $f$ itself is $\propto N'$, and $N'*g = N*(g')=(N*g)'$ | |
Aug 14, 2015 at 11:24 | comment | added | Uchiha | Thanks for the reply. I just realized: would the Fourier transform of $f\ast g$ be the Fourier transform $\hat{f}$ of $f$ (which is derivative of Gaussian) times the Fourier transform of $g$. So the differentiation seems to be applied on $\hat{f}$ instead of $\hat{g}$. How does the argument work then? | |
Aug 12, 2015 at 18:45 | vote | accept | Uchiha | ||
Aug 14, 2015 at 11:42 | |||||
Aug 12, 2015 at 15:23 | history | edited | Jean Duchon | CC BY-SA 3.0 |
deleted 10 characters in body
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Aug 12, 2015 at 11:17 | history | answered | Jean Duchon | CC BY-SA 3.0 |