= Solution
For <integrable functions> $f,g$, <Fubini's theorem> justifies exchanging the two <integrals> in the <Fourier transform> of the <convolution>. Set $s=x-t$ to obtain
$$
\begin{aligned}
\widetilde h(k)&=\int_{\mathbb R}\int_{\mathbb R}f(t)g(x-t)e^{-ikx}\,dt\,dx\\
&=\int_{\mathbb R}\int_{\mathbb R}f(t)g(s)e^{-ik(s+t)}\,ds\,dt\\
&=\left[\int_{\mathbb R}f(t)e^{-ikt}dt\right]\left[\int_{\mathbb R}g(s)e^{-iks}ds\right].
\end{aligned}
$$
This proves the <convolution theorem> with the stated normalization:
$$
\boxed{\widetilde h(k)=\widetilde f(k)\widetilde g(k).}
$$
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