Timeline for A example on Fourier tranform of a continous compactly supported function
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Aug 23, 2014 at 14:13 | comment | added | Cao | Thank you. In case of $0<\gamma<1$, I have just found a example on a continous compactly supported function that its Fourier transform decays faster than exponential rate. This function is the standard bump function, $\Phi \left( x \right) = {e^{ - \frac{1}{{1 - {x^2}}}}}{\chi _{\left( { - 1,1} \right)}}\left( x \right)$ having $\left| {{\Phi ^{\operatorname{ft} }}\left( t \right)} \right| = O\left( {{{\left| t \right|}^{ - \frac{3}{4}}}{e^{ - \sqrt {\left| t \right|} }}} \right)$ as $\left|t\right|\to\infty$ (\gamma=\frac{1}{2}). See in en.wikipedia.org/wiki/Bump_function | |
| Aug 23, 2014 at 14:07 | history | edited | Joonas Ilmavirta | CC BY-SA 3.0 |
added 34 characters in body
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| Aug 23, 2014 at 14:04 | comment | added | Joonas Ilmavirta | @Cao, I don't know what happens if $0<\gamma<1$. My approach only really works when $\gamma>1$, but on the other hand I only need to assume that $f$ has a support of finite measure and I need no bound on $g'$. It would be very convenient if there were bounds on $g$ and $g'$ in the whole complex plane instead of only on the real axis. (Also: Welcome to MO! Have you taken the intro tour that you can find under the help button yet?) | |
| Aug 23, 2014 at 13:56 | comment | added | Cao | Your proof is very nice, and I understand well. Thank you so much. In the proof, you used the assumption $\gamma>1$. What happen if $0<\gamma<1$ ? | |
| Aug 23, 2014 at 13:49 | comment | added | asv | I think there is no such function for $\gamma\geq 1$, not only for $\gamma>1$ as it is proven above. Indeed the inverse Fourier transform of $g$ defines an entire function for $\gamma>1$, and complex analytic function in the strip $|Im(x)|<k$ for $\gamma=1$. In either case $f$ must be real analytic on $\mathbb{R}$, and hence cannot have compact support. However my argument does not work for $\gamma<1$. | |
| Aug 23, 2014 at 13:25 | history | answered | Joonas Ilmavirta | CC BY-SA 3.0 |