Poisson Summation Formula

Theorem
Let $f : \R \to \C$ be a Schwarz function and $\hat f$ it's Fourier transform.

Then:


 * $\displaystyle \sum_{n \mathop \in \Z} f(n) = \sum_{m \mathop \in \Z} \hat f(m)$

Proof
The result has a simple proof using Fourier series.

Let:


 * $\displaystyle F(x) = \sum_{n \mathop \in \Z} f(x + n)$

Then $F(x)$ is $1$-periodic, and has Fourier coefficients:

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


 * $\displaystyle F(x) = \sum_{k \mathop \in \Z}\hat f(k)e^{ikx}$

Choosing $x = 0$ in this formula we have


 * $\displaystyle \sum_{n \mathop \in \Z} f(n) = \sum_{k \mathop \in \Z}\hat f(k)$

as required.