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