= Solution
Suppose initially that $f,g$ are absolutely integrable, so <Fubini's theorem> justifies exchanging the integrations. The transform of their <convolution> is
$$
\widetilde h(\omega)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(t)g(z-t)e^{-i\omega z}\,dt\,dz.
$$
Set $s=z-t$, so $z=s+t$ and $dz=ds$. The exponential separates:
$$
\widetilde h(\omega)=\left[\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\,dt\right]
\left[\int_{-\infty}^{\infty}g(s)e^{-i\omega s}\,ds\right].
$$
Therefore the <convolution theorem> is
$$
\boxed{\widetilde{f*g}(\omega)=\widetilde f(\omega)\widetilde g(\omega).}
$$
No extra $2\pi$ occurs in this direction with the specified normalization.
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