= Solution
Use the <Fourier transform> convention
$$
\widetilde f(k)=\int_{\mathbb R}f(x)e^{-ikx}\,dx,\qquad
f(x)=\frac1{2\pi}\int_{\mathbb R}\widetilde f(k)e^{ikx}\,dk.
$$
Decay alone does not guarantee existence of the ordinary integrals, so impose the usual integrability and regularity assumptions. One sufficient version of the <Fourier inversion theorem> is that $f$ and $\widetilde f$ are integrable and $f$ is <continuous>, in which case the inversion integral recovers $f$. Under the standard piecewise-smooth Dirichlet assumptions, inversion instead gives the midpoint of the two one-sided limits at a jump, with a suitable improper integral interpretation.
For integrable $f,g$, define $(f*g)(x)=\int_{\mathbb R}f(x-y)g(y)\,dy$. The <convolution theorem> states
$$
\boxed{\widetilde{f*g}(k)=\widetilde f(k)\widetilde g(k).}
$$
To prove it, absolute integrability permits <Fubini's theorem>. Set $t=x-y$ in the inner integral:
$$
\begin{aligned}
\widetilde{f*g}(k)
&=\int_{\mathbb R}\int_{\mathbb R}f(x-y)g(y)e^{-ikx}\,dy\,dx\\
&=\int_{\mathbb R}g(y)e^{-iky}\,dy\int_{\mathbb R}f(t)e^{-ikt}\,dt.
\end{aligned}
$$
The absolute double integral is $\|f\|_1\|g\|_1$, justifying the interchange. The computations below assume $a,b>0$ and use the source PDF's factor $e^{ikx}$ in the final integral, omitted in the converted TeX.
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