# Poisson Summation Formula

## Theorem

Let $f: \R \to \C$ be a Schwarz function.

Let $\hat f$ be its Fourier transform.

Then:

$\ds \sum_{n \mathop \in \Z} \map f n = \sum_{m \mathop \in \Z} \map {\hat f} m$

## Proof

Let:

$\ds \map F x = \sum_{n \mathop \in \Z} \map f {x + n}$

Then $\map F x$ is $1$-periodic (because of absolute convergence), and has Fourier coefficients:

 $\ds \hat F_k$ $=$ $\ds \int_0^1 \sum_{n \mathop \in \Z} \map f {x + n} e^{-2\pi i k x} \rd x$ $\ds$ $=$ $\ds \sum_{n \mathop \in \Z} \int_0^1 \map f {x + n} e^{-2\pi i k x} \rd x$ because $f$ is Schwarz, so convergence is uniform $\ds$ $=$ $\ds \sum_{n \mathop \in \Z} \int_n^{n + 1} \map f x e^{-2\pi i k x} \rd x$ $\ds$ $=$ $\ds \int_\R \map f x e^{-2\pi i k x} \rd x$ $\ds$ $=$ $\ds \map {\hat f} k$ where $\hat f$ is the Fourier transform of $f$

Therefore by the definition of the Fourier series of $f$:

$\ds \map F x = \sum_{k \mathop \in \Z} \map {\hat f} k e^{i k x}$

Choosing $x = 0$ in this formula:

$\ds \sum_{n \mathop \in \Z} \map f n = \sum_{k \mathop \in \Z} \map {\hat f} k$

as required.

$\blacksquare$

## Source of Name

This entry was named for Siméon-Denis Poisson.