= Solution
Assume initially that $f,g$ are integrable, so <Fubini's theorem> justifies interchanging the integrals. Transform the <convolution> and put $u=x-z$:
$$
\begin{aligned}
\widetilde h(k)&=\int_{\mathbb R}\!\int_{\mathbb R}f(z)g(x-z)e^{-ikx}\,dz\,dx\\
&=\int_{\mathbb R}f(z)e^{-ikz}\,dz\int_{\mathbb R}g(u)e^{-iku}\,du\\
&=\boxed{\widetilde f(k)\widetilde g(k)}.
\end{aligned}
$$
This is the <convolution theorem>. No factor $2\pi$ appears in this direction because the forward transform in the question is unnormalized.
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