Poisson Summation Formula

Given $f\in \mathcal{S}(\mathbb{R})$, we construct a function on the circle with $F_1(x)={\sum }_{n=-\infty }^{\infty }f(x+n)$

Since $f$ is rapidly decreasing, this converges absolutely and uniformly on every compact subset of $\mathbb{R}$, so $F_1$ is continuous

$F_1$ is periodic with period $1$, so we call it the periodization of $f$

Another way we could make $f$ periodic is with the Fourier transform evaluated at discrete points, say $F_2(x)={\sum }_{n=-\infty }^{\infty }\hat{f}(n)e^{2\pi inx}$

Again, this converges absolutely and uniformly, and is also of period $1$

Theorem: ${\sum }_{n=-\infty }^{\infty }f(x+n)={\sum }_{n=-\infty }^{\infty }\hat{f}(n)e^{2\pi inx}$

We prove this by showing the Fourier coefficients are the same (which is sufficient since they are continuous),
${\int }_0^1\left({\sum }_{n=-\infty }^{\infty }f(x+n)\right)e^{-2\pi imx}dx={\sum }_{n=-\infty }^{\infty }{\int }_0^{1}f(x+n)e^{-2\pi imx}dx$
$={\sum }_{n=-\infty }^{\infty }{\int }_n^{n+1}f(y)e^{-2\pi imy}dy$
$={\int }_{-\infty }^{\infty }f(y)e^{-2\pi imy}dy=\hat{f}(m)$

${\int }_0^1\left({\sum }_{n=-\infty }^{\infty }\hat{f}(n)e^{2\pi inx}\right)e^{-2\pi imx}dx={\sum }_{n=-\infty }^{\infty }\hat{f}(n){\int }_0^1{e}^{2\pi i(n-m)x}dx=\hat{f}(m)$

This formula has interesting applications to theta and zeta functions, heat kernels, and Poisson kernels