[Math] A example on Fourier tranform of a continous compactly supported function

Question

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
$$|f^{ft}(t)|=O(e^{-k\left|t\right|^{\gamma}}),\, \left|(f^{ft})'(t)\right|=O(e^{-k\left|t\right|^{\gamma}}).$$
So far I have not found such a function.
Can someone help me? Thank so much.

Answer 1

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 $|g(x)|\leq Ae^{-k|x|^\gamma}$ for $k>0$ and $\gamma>1$. Denote $I_m=[-m,m]$.

By Benedick's theorem there is a constant $C>0$ so that $$ \|f\|_{L^2(\mathbb R)} \leq Ce^{2Cm|S|}\|g\|_{L^2(\mathbb R\setminus I_m)}. $$ If $m\geq\gamma^{-1/(\gamma-1)}$, then $x\geq m$ implies $x^\gamma-x\geq m^\gamma-m$. Therefore $$ \|g\|_{L^2(\mathbb R\setminus I_m)}^2 = \sum_\pm\int_m^\infty|g(\pm x)|^2dx \leq \sum_\pm\int_m^\infty A^2e^{-2kx^\gamma} dx \leq 2A^2e^{-2km^\gamma}\int_m^\infty e^{-2k(x-m)} dx = 2A^2e^{-2km^\gamma}(2k)^{-1}. $$ Inserting this to the previous estimate gives $$ \|f\|_{L^2(\mathbb R)} \leq Ck^{-1/2}Ae^{2Cm|S|-km^\gamma}. $$ Since $\gamma>1$, taking the limit $m\to\infty$ gives $\|f\|_{L^2(\mathbb R)}=0$ so $f=g=0$.