Poisson summation formula

From ClaudIO

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

Notation. Here $\hat{f}$ is the Fourier transform in the convention of this vault,

\hat{f} (k) = \int_{- \infty}^{+ \infty} f (x) e^{- 2 π i k x} d x,

and both sums are over integer lattice points. The formula holds for good functions — e.g. $f$ a Schwartz-class function (or mildly weaker: $f$ continuous, $\sum f (n)$ and $\sum \hat{f} (n)$ absolutely convergent).

Why. The integer Dirac comb $Ш (x) = \sum_{n \in Z} δ (x - n)$ is self-dual under this Fourier convention: $\hat{Ш} = Ш$ . So pairing $f$ with the comb equals pairing $\hat{f}$ with the same comb, by the duality identity $⟨ f, g ⟩ = ⟨ \hat{f}, \hat{g} ⟩$ :

\underset{\int f (x) Ш (x) d x = \sum_{n} f (n)}{\underset{⏟}{⟨ f, Ш ⟩}} = \underset{\sum_{k} \hat{f} (k)}{\underset{⏟}{⟨ \hat{f}, \hat{Ш} ⟩}}

Intuition: sampling $f$ at integer points and summing equals doing the same to its Fourier transform — because the integer lattice is its own Fourier-dual lattice.

Beware: not every comb is self-dual. In the general framework, the Fourier transform of the weighted comb $\sum_{n} a_{n} δ (x - n)$ is the periodic function $\sum_{n} a_{n} e^{- 2 π i n ω}$ — a Fourier series with the weights $a_{n}$ as coefficients, not another comb. Self-duality is the degenerate case $a_{n} \equiv 1$ : then that Fourier series $\sum_{n} e^{- 2 π i n ω}$ collapses (as a distribution) into the comb $\sum_{k} δ (ω - k)$ . This is why Poisson summation needs the unit comb: pairing $f$ against $Ш$ samples $f$ at integers with no coefficients.