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My question is: given a Fourier transform $\hat f$ of a function $f$, is it possible to estimate its asymptotic behaviour without performing the inverse transform?

Let me give a concrete example.

Suppose, a spherically symmetric real-valued function $f(x)$ (with $x\equiv|{\bf x}|$ and ${\bf x}\in \mathbb R^3$) has a Fourier transform of the following form $$ \hat f(k) = \frac{P_m(k)}{Q_n(k)} $$ where $P_m(k)$ and $Q_n(k)$ are certain polynomials of $k=|{\bf k}|$, and the degrees $m$ and $n$, some integer numbers less than 10. In the general case, it is quite hard (if at all possible) to calculate the inverse Fourier transform. However, suppose, we need not the function $f(x)$ itself, but only wish to determine its asymptotic behaviuor for $x\rightarrow \infty$. Such situations typically occur in physics, for instance in field theory, where the asymptotics of a field $f(x)$ is oftentimes significantly more relevant than the exact expression for this field.

So, can we extract any information about asymptotic behaviour of $f(x)$ form the Fourier image $\hat f(k)$ itself, without performing the complicated (and sometimes even technically impossible) inverse transform?

Thank you!

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    $\begingroup$ if $\hat{f}(k)\propto k^p$ for small $k$, then $f(x)$ has for large $|x|$ the asymptotic $|x|^{-p-3}$. $\endgroup$ – Carlo Beenakker Mar 25 at 15:30
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    $\begingroup$ The general principle is that the asymptotics of the function depend on the smoothness of the FT and vice versa. $\endgroup$ – Christian Remling Mar 25 at 15:39
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    $\begingroup$ @CarloBeenakker: I don't think this works (for example, take $\widehat{f}=1/(1+|k|^2)$, then $p=0$). (You perhaps had the much less general evaluation $\widehat{|k|^{-\alpha}}=c|x|^{\alpha-d}$ in mind.) $\endgroup$ – Christian Remling Mar 25 at 15:45
  • $\begingroup$ Thank you all for your comments! If possible, please, hint where one could read about the relevant theorems. I have found a post on a similar topic math.stackexchange.com/questions/1794619/…, but nothing more or less rigorous has been indicated there. Any help with the literature will be highly appreciated! Thank you guys! $\endgroup$ – jonathan wolf Mar 25 at 15:56
  • $\begingroup$ @jonathanwolf: The way to make this precise is to just integrate by parts in the Fourier inversion formula. $\endgroup$ – Christian Remling Mar 25 at 18:32

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