Find the Fourier transform of the absolute value of a function, given its Fourier transform

Question

It is possible to find the following Fourier transform? $ \widehat{\vert f\vert}(\xi)$, provided $ f $ has Fourier transform. I have not found any information about this statement.

Answer 1

Generally, nonlinear transforms do not play well with Fourier transforms. One exception is $|f|^2$: writing it out as $f\bar f$ and observing that $$ \widehat{ \bar f}(\xi) = \overline{\hat f(-\xi)} $$ we get that $\widehat{|f|^2} $ is the convolution of $\hat f$ with $\overline{\hat f(-\xi)}$.

But without the square it's much worse. Consider that when $f$ is smooth (and thus $\hat f$ decays quickly), the absolute value $|f|$ is generally not differentiable (and so its Fourier transform will not decay so quickly anymore). The point being, $|f|$ and $f$ can look very differently on the Fourier transform side.

Since $|f|=f\operatorname{sign}f$, the question becomes: will you be able to find the Fourier transform of $\operatorname{sign} f$, and to convolve it with $\hat f$? Chances are that the answer is no.