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

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 $f_2 (x) = xf(x)$.

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If you know the solution, why are you asking the question? – Nate Eldredge Nov 30 '10 at 16:14
Since it's interesting, probably not well-known (to say the least), motivated by some recent posts here, and since I found the solution only after the original question was posted (I formulated the original question too hastily). – Shai Covo Nov 30 '10 at 21:29

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.]

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Almost. the diff-eq you should end up with is $a \hat{g}' + (a-1) (\hat{f}' / \hat{f}) \hat{g}$ = 0. So you need to change the power raised in your final answer a little bit. – Willie Wong Nov 30 '10 at 16:40
That's what I get for using a bedside scrap of paper for scratchwork. :/ – chroma Nov 30 '10 at 16:55

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 $\hat \psi = i \hat \varphi'$.

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