A convolution identity (revision of the question "Is this convolution identity known?")

Question

I have deleted the content of the original post.

The following exercise is inspired by answers to the questions this and this, but no knowledge of probability theory is required. The solution is straightforward assuming familiarity with the relevant theorems/properties. The proof is supposed to be informal.

Let $\hat \varphi (\omega)$ denote the Fourier transform of a function $\varphi$, and $*$ convolution. Express $\hat g(\omega)$ in terms of $\hat f(\omega)$ and a constant $a \in (0,1)$, so that $$ (f_2 * g)(t) = at(f * g)(t), $$ where $f_2 (x) = xf(x)$.

Answer 1

I may be messing up my usage of Fourier transforms and convolutions with distributions, but I think this can be solved with a few identities (namely how transforms work on products and convolutions, and how convolutions work with derivatives and Dirac deltas). Given the starting point $(xf) * g = a t (f * g)$, take the (unitary, ordinary frequency) transform to obtain

$(\hat{x}*\hat{f} ) \hat{g} = a\hat{x} * (\hat{f} \hat{g}) \implies (\delta'*\hat{f} ) \hat{g} = a\delta' * (\hat{f} \hat{g}) \implies \hat{f}' \hat{g} = a (\hat{f}\hat{g})'$

If you foil and rewrite you get the simple differential equation $a \hat{g} ' + (a-1) (\hat{f}' / \hat{f}) \hat{g} =0 $, which is easy to solve and comes out with $\hat{g} = \hat{f}^{1/a - 1}$ (I hope I did that all correctly!).

[EDIT - thanks to below comment.]

Answer 2

My answer is similar to chroma's answer, only that I get $$ (xf) * g = at(f * g) \Rightarrow \hat f'\hat g = a(\hat f \hat g)' $$ straight from the following standard property (combined, of course, with the convolution theorem): if $\psi = x \varphi$, then $\hat \psi = i \hat \varphi'$.