Timeline for To show a function is zero, assuming some integral conditions on its Fourier transform
Current License: CC BY-SA 4.0
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| Sep 8, 2022 at 13:47 | comment | added | Christian Remling | Alternatively, $\lim_{y\to 0+}\int e^{-y|t|}e^{ixt}\widehat{f}(t) = f(x)$ (up to a factor perhaps), for example because multiplying $\widehat{f}$ with $e^{-y|t|}$ is the same as convolving $f$ with the Poisson kernel. | |
| Sep 8, 2022 at 13:16 | comment | added | Giorgio Metafune | Usually for the Laplace transform one assumes that $g$ is exponentially bounded, that is $|g(t)| \leq Me^{Ct}$ so that the transform is defined for $Re\, y >C$. In this case $g=\hat f$ is even bounded. If you want to keep integrability assumptions, you may apply the argument to $e^{-t} \hat f(t)$. | |
| Sep 8, 2022 at 11:48 | comment | added | user483450 | @GiorgioMetafune To apply your argument we need to assume that $\widehat{f}$ is also integrable on $[0,\infty)$. But is the result true if $\widehat{f}$ is not integrable on $[0,\infty)$? | |
| Sep 8, 2022 at 11:34 | comment | added | Giorgio Metafune | If $x=0$ in (i), then the Laplace transform of $\hat f$ is 0 for $y>0$ and then $\hat f=0$ for $t>0$. | |
| Sep 8, 2022 at 10:16 | history | edited | user483450 | CC BY-SA 4.0 |
added 4 characters in body
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| Sep 8, 2022 at 8:40 | history | asked | user483450 | CC BY-SA 4.0 |