Poisson Summation Formula

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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:

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


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.

$\blacksquare$


Source of Name

This entry was named for Siméon-Denis Poisson.


Sources