# Poisson summation formula

Let $f:\mathbb{R}\to\mathbb{R}$ be an integrable function and let

 $\hat{f}(\xi)=\int_{\mathbb{R}}e^{-2\pi i\xi x}f(x)dx,\quad\xi\in\mathbb{R}.$
 $\sum_{n\in\mathbb{Z}}f(n)=\sum_{n\in\mathbb{Z}}\hat{f}(n).$ (1)

whenever $f$ is such that both of the above infinite sums are absolutely convergent.

Equation (1) is useful because it establishes a correspondence between Fourier series  and Fourier integrals. To see the connection, let

 $g(x)=\sum_{n\in\mathbb{Z}}f(x+n),\quad x\in\mathbb{R},$

be the periodic function obtained by pseudo-averaging11This terminology is at best a metaphor. The operation in question is not a genuine mean, in the technical sense of that word. $f$ relative to $\mathbb{Z}$ acting as the discrete group of translations on $\mathbb{R}$. Since $f$ was assumed to be integrable, $g$ is defined almost everywhere, and is integrable over $[0,1]$ with

 $\|g\|_{L^{\!1}[0,1]}\leq\|f\|_{L^{\!1}(\mathbb{R})}.$

Since $f$ is integrable, we may interchange integration and summation to obtain

 $\hat{f}(k)=\sum_{n\in\mathbb{Z}}\int_{0}^{1}f(x+n)e^{-2\pi ikx}dx=\int_{0}^{1}% e^{-2\pi ikx}g(x)dx$

for every $k\in\mathbb{Z}$. In other words, the restriction of the Fourier transform of $f$ to the integers gives the Fourier coefficients of the averaged, periodic function $g$. Since we have assumed that the $\hat{f}(k)$ form an absolutely convergent series, we have that

 $g(x)=\sum_{k\in\mathbb{Z}}\hat{f}(k)e^{2\pi ikx}$

in the sense of uniform convergence  . Evaluating the above equation at $x=0$, we obtain the Poisson summation formula (1).

Title Poisson summation formula PoissonSummationFormula 2013-03-22 13:27:25 2013-03-22 13:27:25 rmilson (146) rmilson (146) 16 rmilson (146) Theorem msc 42A16 msc 42A38 Poisson summation