Poisson Summation

Video. Poisson Summation
Theorem (Poisson Summation Formula)
Suppose that \(f\) is a Schwartz function on the real line. Then
\[ \sum_{n=-\infty}^\infty f(n) = \sum_{n=-\infty}^{\infty} \widehat{f}(n). \]
Note
There are higher-dimensional generalizations of the Poisson summation formula and generalizations to cases weaker than Schwartz functions. The key ideas are the same as the ones used below.

1. Key Observations Leading to Poisson Summation

Lemma
Suppose that \(f\) is a Schwartz function on the real line. Then the limit
\[ \sum_{m = -\infty}^\infty f(x+m) \]
converges uniformly on compact sets to a \(C^\infty\) function which is periodic of period 1. In particular it is the sum of its own Fourier series at every point.
Proof
Meta (Main Idea)
Use the Weierstrass M-test combined with the decay inequality for Schwartz functions. To establish smoothness, use the criteria for term-by-term differentiability (namely that the term-by-term derivative converges uniformly as well).
Lemma
Let \(f\) be as above and let
\[ F(x) := \sum_{m = -\infty}^\infty f(x+m). \]
Then for each integer \(n\),
\[ \widehat{F}(n) = \widehat{f}(n), \]
where the \(\widehat{\, \cdot\, }\) denotes Fourier coefficients on the left-hand side and the Fourier transform on the right-hand side.
Proof
Meta (Main Idea)
Write out the integral for the Fourier coefficients and use uniform convergence of the series to commute integration with summation.
Proof (Proof of Poisson Summation Formula)
Let
\[ F(x) := \sum_{m=-\infty}^\infty f(x+m). \]
Because \(F\) is differentiable at \(x=0\), it is the sum of its Fourier series there, i.e.,
\[{}\sum_{m =-\infty}^\infty f(m) = F(0){}\]
\[{}= \sum_{n=-\infty}^\infty \widehat{F}(n){}\]
\[{}= \sum_{n=-\infty}^\infty \widehat{f}(n).{}\]