I am trying to find a continuous compactly supported function $f$ such that the Fourier transform $f^{ft}$ and derivative $(f^{ft})'$ of the $f^{ft}$ decay faster than exponential rates, that is $$f^{ft}(t)=O(e^{k\leftt\right^{\gamma}}),\, \left(f^{ft})'(t)\right=O(e^{k\leftt\right^{\gamma}}).$$ So far I have not found such a function. Can someone help me? Thank so much.

$\begingroup$ Doesn't this run into problems in view of the following form of the uncertainty principle? en.wikipedia.org/wiki/Uncertainty_principle#Harmonic_analysis $\endgroup$ – Yemon Choi Aug 23 '14 at 12:42

$\begingroup$ Note than by the Schwartz's Paley–Wiener theorem such a function should be an entire function of finite order, i.e. growing not faster than $e^{a z}$ for some $a>0$. Roughly speaking it is a sum of exponents. So may be a related question would be: can such $f$ decay faster than exponent on a line? $\endgroup$ – Andrew Aug 23 '14 at 13:04
There are no such functions if $\gamma>1$. Even more is true:
Suppose $f\in L^2(\mathbb R)$ is supported in a set $S$ of finite measure and assume its Fourier transform $g=\hat f$ satisfies $g(x)\leq Ae^{kx^\gamma}$ for $k>0$ and $\gamma>1$. Denote $I_m=[m,m]$.
By Benedick's theorem there is a constant $C>0$ so that $$ \f\_{L^2(\mathbb R)} \leq Ce^{2CmS}\g\_{L^2(\mathbb R\setminus I_m)}. $$ If $m\geq\gamma^{1/(\gamma1)}$, then $x\geq m$ implies $x^\gammax\geq m^\gammam$. Therefore $$ \g\_{L^2(\mathbb R\setminus I_m)}^2 = \sum_\pm\int_m^\inftyg(\pm x)^2dx \leq \sum_\pm\int_m^\infty A^2e^{2kx^\gamma} dx \leq 2A^2e^{2km^\gamma}\int_m^\infty e^{2k(xm)} dx = 2A^2e^{2km^\gamma}(2k)^{1}. $$ Inserting this to the previous estimate gives $$ \f\_{L^2(\mathbb R)} \leq Ck^{1/2}Ae^{2CmSkm^\gamma}. $$ Since $\gamma>1$, taking the limit $m\to\infty$ gives $\f\_{L^2(\mathbb R)}=0$ so $f=g=0$.

1$\begingroup$ 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 $\gamma<1$. $\endgroup$ – makt Aug 23 '14 at 13:49

$\begingroup$ 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 $0<\gamma<1$ ? $\endgroup$ – Cao Aug 23 '14 at 13:56

$\begingroup$ @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?) $\endgroup$ – Joonas Ilmavirta Aug 23 '14 at 14:04

3$\begingroup$ 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 $\leftt\right\to\infty$ (\gamma=\frac{1}{2}). See in en.wikipedia.org/wiki/Bump_function $\endgroup$ – Cao Aug 23 '14 at 14:13