= Solution
Use the <Fourier transform> convention
$$
\widetilde f(\omega)=\int_{-\infty}^{\infty}e^{-i\omega t}f(t)\,dt,\qquad
f(t)=\frac1{2\pi}\int_{-\infty}^{\infty}e^{i\omega t}\widetilde f(\omega)\,d\omega.
$$
The <convolution> is $(f*g)(t)=\int_{-\infty}^{\infty}f(\tau)g(t-\tau)\,d\tau$. For integrable $f,g$, <Fubini's theorem> and the substitution $s=t-\tau$ give
$$
\begin{aligned}
\widetilde{f*g}(\omega)
&=\int_{\mathbb R}\int_{\mathbb R}e^{-i\omega t}f(\tau)g(t-\tau)\,d\tau\,dt\\
&=\left(\int_{\mathbb R}e^{-i\omega\tau}f(\tau)\,d\tau\right)\left(\int_{\mathbb R}e^{-i\omega s}g(s)\,ds\right).
\end{aligned}
$$
Hence the <convolution theorem> is
$$
\boxed{\widetilde{f*g}(\omega)=\widetilde f(\omega)\widetilde g(\omega).}
$$
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