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So, there are no general expressions for large-frequency asymptotics of Fourier transforms, but I am interested in any techniques that would allow to establish upper bounds on the rate of decrease of Fourier transforms (I am talking about cases when the decrease is faster than any inverse power of the frequency).

For example, one may have a function $f(t)$ and its Fourier transform $f(\omega)$ that cannot be evaluated explicitly, $f(\omega)=\frac{1}{2\pi}\int dt e^{i\omega t} f(t)$. It is known that if $f(t)$ is infinitely differentiable, then $f(\omega)$ decreases faster than $O(1/\omega^n)$ for any $n$ at large $\omega$. How could one go about proving more refined bounds on the asymptotic decrease? For example, what usable conditions could one set up for $f(t)$ to ensure that $f(\omega)$ decreases faster than $O(e^{-a\omega^2})$ for some $a$ at large $\omega$?

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I asked a question similar to this, got a nice answer, perhaps of use to you.… – Alice Sep 22 '11 at 17:41
Thank you, Alice, it does seem relevant and I have located the portion of Reed&Simon vol.2 dealing with the question. However, what is discussed there is $O(e^{-a\omega})$ decrease, whereas I am specifically interested in proving $O(e^{-a\omega^2})$ decrease (for a specific function that has come up in my research). Any ideas? – eoe Sep 23 '11 at 4:31
To reduce ambiguity, it might be better to use a different symbol to indicate the Fourier transform of a function, e.g., $Ff$ or $\hat{f}$. At any rate, one sufficient condition is that $f$ is a finite sum of functions of the form $P_b(t)e^{-bt^2}$ for $b$ a positive real number and $P_b$ a polynomial. You can extend this to allow an infinite sum under suitably restrictive growth conditions (but I do not know an example off-hand). – S. Carnahan Sep 26 '11 at 9:18

Have a look at Generalized Functions, vol. 2, by Gel'fand and Shilov. They discuss what is nowadays called Gelfand--Shilov spaces. These are subspaces of the Schwartz space such that a function $f$ and its Fourier transform $\hat f$ decrease like $|f(x)| \leq A \exp(- c |x|^{1/a})$ and $|\hat f(\xi)| \leq A \exp(- d |\xi|^{1/b})$, $a+b \geq 1$.

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