Fastest decay of Fourier transform of function of (one-sided or two-sided) exponential (or faster) decay

Question

Let $f:\mathbb{R}\to \mathbb{R}$ be a function in $L^2$ satisfying $|f(x)|\ll e^{-a_1 x}$, $a_1>0$, for $x\to \infty$. (Variant: assume as well that $|f(x)|\ll e^{a_2 x}$, $a_2>0$, for $x\to -\infty$.) What is the fastest $\widehat{f}(t)$ can decay (in both directions)?

EDIT: I should emphasize I would be very interested in a simple, concrete example that shows that the bound below (which is correct, as has been confirmed) is tight.

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I can show that $\widehat{f}(t)$ cannot decay as fast as $e^{-e^{r |t|}}$ (even in one direction) if $r\cdot \max(a_1 a_2) >\pi^2$. (A one-sided assumption is probably enough, though I haven't checked. ) Can that decay be reached for some $r>0$? (Perhaps even for $r = \pi^2/\max(a_2,a_2)$?)

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UPDATE: what if $|f(x)|$ decreases faster than any $e^{-A x}$, $A$ arbitrary, as $x\to \infty$? How fast can $\widehat{f}(t)$ decay then (in both directions)?

Answer 1

The simplest result can be stated when both your conditions hold. If they are satisfied, your Fourier transform is analytic and bounded from above in the strip $-a_2<\mathrm{Im}\, s<a_1$, and bounded from above in any smaller strip. Such function has a limit rate of decay $\exp\left(-e^{\pi|s|/a}\right)$, where $a=a_1+a_2$ is the width of your strip, in the sense that if $|\hat{f}(s)|\leq \exp\left(-e^{(\pi/a+\epsilon)|s|}\right)$ when $s\to +\infty$ or if $s\to-\infty$, then $f\equiv 0$. This is a simple consequence of the Phragmen-Lindelof Principle, see for example,

B. Levin, Lectures on entire functions, AMS 1996.

With a bit more labor (using Poisson integral for a strip) one can prove that this remains true when $a_1=0$ or $a_2=0$, that is when we have one-sided decay condition on $f$. More precisely, we have a bounded analytic $\hat{f}$ function in the strip $0<\mathrm{Im}\, s<a$, say continuous at the boundary, and we want to know how fast it can decrease on the real line. Let $u$ be the least harmonic majorant of $\log|\hat{f}|$. Then, as a harmonic function bounded from above, it must be equal to the Poisson integral of its boundary values. In particular, this Poisson integral must converge. This gives the maximum possible rate of decrease (of $u(x)$ to $-\infty$). Since Poisson kernel for the strip of width $a$ is $\mathrm{Im}\coth(\pi z/(2a))$ we obtain the result.

Function $\hat{f}(s)=\exp(-\cosh \pi s/a)$ shows that the result is best possible. This function is in $L^2$ so $f$ is also in $L^2$. Correct exponential decrease of $f$ is easy to prove directly.

Edit. I found a reference, which also contains an example that shows that it is best possible, even in the class of probability densities ($f\geq 0$):

Ju. Linnik and I. Ostrovskii, Decomposition of random variables and vectors, AMS 1977,

Chap. II, section 4, at the very end of this section. Their example has an additional feature that $f\geq 0$, and for this reason it requires some labor.