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:

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


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.


Sources

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