Poisson Summation Formula

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

Let $\hat f$ be its Fourier transform.

Then:


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

Proof
Let:


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

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

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


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

Choosing $x = 0$ in this formula:


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

as required.