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Timeline for Convolution vanishes on an interval

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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
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Aug 12, 2015 at 11:17 history answered Jean Duchon CC BY-SA 3.0