My question is concerned with the existence of compactly supported functions whose its Fourier transform satisfies a given condition: For $\gamma\ge 1$, one can prove that there is no compactly supported function $f$ satisfying $\left {\hat f\left( t \right)} \right = O\left( {{e^{  k{{\left t \right}^\gamma }}}} \right)$ as $t\to +\infty$?, for some $k>0$. Here $\hat f$ denotes the usual Fourier transform of $f$. Now I am interested in the case of $0<\gamma<1$. For $\gamma=1/2$, I can give an example of a such function, for example the standard bump function $\varphi \left( x \right) = \exp \left\{ {  \frac{1}{{1  {x^2}}}} \right\}{\chi _{\left( {  1,1} \right)}}\left( x \right)$ having $\left {\hat \varphi\left( t \right)} \right = O\left( {{e^{  \sqrt {\left t \right} }}} \right)$ (see in http://en.wikipedia.org/wiki/Bump_function). My question is: Given $\gamma\in(0,1)$. Is there a compactly supported function $f$ such that $\left {\hat f\left( t \right)} \right = O\left( {{e^{  k{{\left t \right}^\gamma }}}} \right)$ as $t\to +\infty$?
3 Answers
As Denis Chaperon de Lauzières says, the precise answer to this question is given by the BeurlingMalliavin theorems. They give a very precise characterization of weights $w\geq 1$ for which there exists a Fourier transform $f$ of a function with bounded support, such that $f(x)w(x)$ is bounded. The necessary condition is that $$\int \frac{\log w(x)}{1+x^2}dx<\infty.$$ This is the famous "Logarithmic integral". To make it also sufficient one needs some regularity. For example, if $\log w$ is uniformly continuous then this is also sufficient. Another regularity condition is that $w$ is itself a functon of exponential type with $$\int\frac{\log^+w(x)}{1+x^2}dx<\infty.$$
References: BeurlingMalliavin, Acta math. 107 (1962) 291309, P. Koosis, Logarithmic Integral, P. Koosis, Lecons sur les theoremes de BeurlingMalliavin.
The oldest reference I know which constructs compactlysupported functions with as good as possible decay of the Fourier transform at infinity is A. Ingham, "A note on Fourier transforms", J. London Math. Soc. s1–9 (1934), 29–32.
See also Th. 1.3.5 in Hörmander's "The analysis of linear partial differential operators I: distribution theory and Fourier analysis", SpringerVerlag, 2003.
(I think more refined questions are addressed by the "Beurling  Malliavin multiplier theorem").
Yes there is. What you need is the PaleyWiener Theorem as in http://en.wikipedia.org/wiki/Paley–Wiener_theorem


$\begingroup$ The PaleyWiener theorem shows that if $\widehat{f}(\xi) = O(e^{k\xi})$ for some positive $k$, then $f$ extends to a holomorphic function in a horizontal strip containing $\mathbb{R}$, and so vanishes on an open set if and only if $f$ is uniformly zero. $\endgroup$ Oct 23, 2014 at 17:39

1$\begingroup$ So if $\widehat{f}(\xi) = O(e^{k\xi^{\gamma}})$ with $\gamma \geq 1$ and $k > 0$, then this same method works. But for $\gamma < 1$ this says nothing. $\endgroup$ Oct 23, 2014 at 17:40

2$\begingroup$ In general, what you are asking about are functions in GelfandShilov spaces. Perhaps there is something in the literature about these. I'd be extremely interested to know if it's true that if $\widehat{f} = O(e^{k\xi^{\gamma}})$ for all $\gamma < 1$, then $\widehat{f}$ must be analytic on $\mathbb{R}$. But I don't know the answer. $\endgroup$ Oct 23, 2014 at 17:43