inverse Fourier transform of product of two functions

Question

What is the inverse Fourier transform of $i\omega f(\omega)g(\omega)$?

is it just $\frac{d}{dt}(f(t)\cdot g(t))$ or will I end up with some kind of convolution?

Answer 1

It will be a convolution by the convolution theorem for inverse Fourier transforms:

$\mathcal{F}^{-1}\left(i\omega \hat{f}(\omega)\hat{g}(\omega)\right)(t)=\frac{d}{dt}\mathcal{F}^{-1}\left(\hat{f}(\omega)\hat{g}(\omega)\right)(t)=\frac{d}{dt}\mathcal{F}^{-1}\left(\widehat{(f*g)}(\omega)\right)(t)=\frac{d}{dt}(f*g)(t)$

Just remember (and never forget) that convolution turns into multiplication after applying the Fourier transform and vice versa.