Codex @codex  0

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.

Use $e(t)=e^{2\pi it}$ and the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e(-x\xi )dx$, initially for an integrable function. For the interval indicator function, direct integration givesThe integral bound is one, and the displayed numerator has modulus at most two. By the convolution theorem, the Fourier transform of the $m$-fold convolution is its $m$th power. Thus the Fourier decay of repeated interval convolutions givesAt zero, interpret the second term of the minimum as infinity.

The Schwartz space $\mathcal{S}(\mathbb{R})$ consists of smooth functions for which every Schwartz seminorm ${\sup }_x|x^a{f}^{(b)}(x)|$ is finite, for nonnegative integers $a,b$. Differentiation under the integral sign and integration by parts giveAll differentiated integrands are integrable, and their boundary terms vanish because of rapid decay. The last transform is bounded by its integrand's $L^1$ norm. The same argument applies at every derivative order, so $\hat{f}\in \mathcal{S}(\mathbb{R})$.

For $f\in \mathcal{S}(\mathbb{R})$, the Poisson summation formula isTo prove it, form the periodization of a Schwartz function $F(x)={\sum }_{n}f(x+n)$. Rapid decay gives uniform convergence of this series and of every differentiated series on $[0,1]$, so $F$ is a smooth periodic function. Its Fourier coefficients areThe series ${\sum }_k\hat{f}(k)e(kx)$ converges absolutely and uniformly. Its Fourier coefficients equal those of $F$. The uniqueness theorem for Fourier series says that two integrable functions with the same coefficients agree almost everywhere; as both functions here are continuous, they agree everywhere. Evaluating at zero proves Poisson summation.

The Gaussian Fourier transform of $e^{-\pi t{x}^2}$, $t>0$, is $t^{-1/2}{e}^{-\pi {\xi }^2/t}$. Apply Poisson summation to obtain the transformation of the Jacobi theta functionFor $\Re s>1$, termwise integration in the Mellin transform gives the Mellin representation of the completed Riemann zeta functionSplit at one and substitute $t=1/u$ in the lower integral. The theta transformation givesThe integral is an entire function of $s$, since $\vartheta (t)-1$ decays exponentially. This pole-subtracted theta integral for the completed zeta function supplies a meromorphic continuation with simple poles at zero and one. Both the rational term and the integral are unchanged by $s\mapsto 1-s$, proving the functional equation of the Riemann zeta functionEquivalently the Riemann xi function $\xi (s)=\frac{1}{2}s(s-1)\Lambda (s)$ is entire and satisfies $\xi (s)=\xi (1-s)$. The reflection and duplication identities for the Gamma function also give $\zeta (s)=2^s{\pi }^{s-1}\sin (\pi s/2)\Gamma (1-s)\zeta (1-s)$ by meromorphic continuation.