= Convolution theorem
{title2=$\widehat{f*g}=\widehat f\,\widehat g$}
{wiki}
For <Lebesgue integrable functions> $f,g$ on $\mathbb R$, use the <Fourier transform> convention $\widehat f(\xi)=\int_{\mathbb R}f(x)e^{-2\pi ix\xi}\,dx$. Then
$$
\widehat{f*g}(\xi)=\widehat f(\xi)\widehat g(\xi).
$$
Thus the <integral> combining all translated products becomes an ordinary multiplication at each <frequency>. Different <Fourier transform> normalizations can place a constant in this identity; fixing the convention prevents that ambiguity. Its use in filtering is illustrated by https://see.stanford.edu/materials/lsoftaee261/transcripts/TheFourierTransformAndItsApplications-Lecture09.pdf[Brad Osgood, Stanford EE261, Lecture 9, convolution and Fourier-domain filtering]{ref}.
For the <indicator function> $q=\mathbf1_{[0,1]}$,
$$
\widehat q(\xi)=e^{-\pi i\xi}\operatorname{sinc}(\xi),\qquad \operatorname{sinc}(\xi)=\frac{\sin(\pi\xi)}{\pi\xi},
$$
with $\operatorname{sinc}(0)=1$. Therefore the <triangular function> $q*q$ has <Fourier transform> $e^{-2\pi i\xi}\operatorname{sinc}^2(\xi)$. The extra factor of the <sinc function> improves the large-frequency bound from a constant times $1/|\xi|$ to one times $1/|\xi|^2$, matching the smoother output in the overlap picture. <Fourier decay of repeated interval convolutions> extends this calculation to further repetitions.
For a proof outline, <Fubini's theorem> permits exchanging the two <integrals> because the integral of $|f(y)g(x-y)|$ over $(x,y)$ is $\|f\|_1\|g\|_1<\infty$. Set $u=x-y$ in the inner <integral>:
$$
\begin{aligned}
\widehat{f*g}(\xi)&=\int_{\mathbb R}\!\int_{\mathbb R}f(y)g(x-y)e^{-2\pi ix\xi}\,dy\,dx\\
&=\int_{\mathbb R}\!\int_{\mathbb R}f(y)g(u)e^{-2\pi i(y+u)\xi}\,du\,dy\\
&=\widehat f(\xi)\widehat g(\xi).
\end{aligned}
$$
The factorization of the <complex exponential> is the mechanism behind the simplification. The <discrete convolution> version similarly turns coefficient sums into products of <generating functions>.
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