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:

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.