# Fastest decay of Fourier Transform for Generalized Functions of compact support

What is the fastest decay possible for the Fourier transform of a generalized function with compact support and finite value at the origin? I know that regular functions cannot attain exponential decay, due to the Paley-Wiener Theorem and that a short paper by Ingham (1934) suggests that the fastest decay in that case is of the order $e^{-|t| \epsilon(|t|)}$, where $\epsilon(t)$ satisfies $\int_0^\infty \frac{\epsilon(t)}{t} dt < \infty.$

Optimally, it seems we may take $\epsilon(t) = (\lg t \cdot \lg\lg t\cdot\lg\lg\lg t\ldots)^{-1}$, but not $\epsilon(t) = 1/\lg t$. My question is two-fold:

1. Are there (regular, as in non-generalized) functions with compactly supported and bounded Fourier transform whose asymptotic decay is $e^{-|t|/\lg|t|}$, or faster? In other words, how tight is Ingham's construction based on products of sinc functions?
2. Is it possible to beat the Paley-Wiener theorem if we move to the space of generalized functions (such as Gelfand-Shilov spaces), while keeping the Fourier Transform compactly supported and bounded at the origin?

A positive answer to the above questions would have great implications for stability results in Quantum Mechanics, such as the stability of topological quantum order under weak perturbations. In particular, it would greatly improve bounds on the strength of arbitrary quasi-local perturbations to topological quantum systems, which preserve the topological nature of the system.

If I read this right, decay $\exp(-|x|/\log (2+|x|))$ is impossible for $L^2$ functions at least (see the Theorem in section 1.1 of the paper, which says that one can not even find an $L^2$ function with such decay and Fourier transform supported on a half-line; indeed, there is a necessary and sufficient condition for the latter).
• Thank you Denis. It seems that this is an important theorem in functional analysis. Moreover, it seems that your reference answers my question for a large variety of classes of functions, such as $L^2$ and $L^\infty$. Still, I am not sure if it applies to all generalized functions, though it seems that way. – Spiros M. Apr 9 '12 at 4:32