Convolution theorem ($\widehat{f\ast g}=\hat{f}\hat{g}$)

For Lebesgue integrable functions $f,g$ on $\mathbb{R}$, use the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-2\pi ix\xi }dx$. ThenThus 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 ^[ref].

For the indicator function $q=1_{[0,1]}$,with $\mathrm{sinc}(0)=1$. Therefore the triangular function $q\ast q$ has Fourier transform $e^{-2\pi i\xi }{\mathrm{sinc}}^2(\xi )$. 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 $\Vert f{\Vert }_1\Vert g{\Vert }_1<\infty$. Set $u=x-y$ in the inner integral: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.

For Schwartz functions $f,g$ and the Fourier transform convention $\hat{f}(\xi )=\int f(x)e^{-ix\xi }dx$, the displayed formula exchanges a pointwise product for a convolution. Substitute Fourier inversion for $g$ into the Fourier transform of $fg$. Fubini's theorem, justified by $\Vert f{\Vert }_1\Vert \hat{g}{\Vert }_1<\infty$, givesUnlike the convolution theorem $\widehat{f\ast g}=\hat{f}\hat{g}$, this version cannot be asserted for arbitrary pairs in $L^1$ without additional hypotheses, since their pointwise product need not be an integrable function.

For $a>0$ and $x\ge 0$, split the convolution integral at $y=0$ and $y=x$. The three contributions are $e^{-ax}/(2a)$, $x{e}^{-ax}$ and $e^{-ax}/(2a)$. Evenness gives the displayed answer for all real $x$.

With Fourier transform convention $\tilde{f}(k)=\int f(x)e^{-ikx}dx$, one has $\tilde{f}(k)=2a/(a^2+k^2)$. The convolution theorem and Fourier inversion theorem therefore yieldThe value at $x=0$ is $\pi /(2a^3)$. Multiplying $f$ by $a/2$ instead gives a normalized Laplace distribution; its self-convolution is the density $a(1+a|x|)e^{-a|x|}/4$.

For $a,b>0$, the convolution of $e^{-a|x|}$ with the indicator of $[-b,b]$ is $2e^{-a|x|}\sinh (ab)/a$ outside the interval and $2[1-e^{-ab}\cosh (ax)]/a$ inside it. With Fourier convention $\tilde{f}(k)=\int f(x)e^{-ikx}dx$, their transforms are $2a/(a^2+k^2)$ and $2\sin (bk)/k$. Fourier inversion therefore evaluates ${\int }_{\mathbb{R}}\sin (bk)e^{ikx}/[k(a^2+k^2)]dk$ as $\pi h(x)/(2a)$.

For $F(x)=e^{-x}{1}_{x\ge 0}$, induction in the convolution integral givesIts Fourier transform in the angular-frequency convention is $(1+ik{)}^{-n}$.