Poisson summation formula

Remarks

The proof periodises $f$: the function $F(x) = \sum_{n \in \Z} f(x + n)$ is smooth and has period $1$. Its Fourier coefficients are $\int_0^1 F(x) e^{-2\pi i k x} \, dx = \widehat{f}(k)$, so its Fourier series $\sum_k \widehat{f}(k) e^{2\pi i k x}$ converges to $F(x)$, and $x = 0$ gives the formula. Example: for $t > 0$ and $f(x) = e^{-\pi t x^2}$ one has $\widehat{f}(\xi) = t^{-1/2} e^{-\pi \xi^2 / t}$, and the formula gives $\sum_{n} e^{-\pi t n^2} = t^{-1/2} \sum_{k} e^{-\pi k^2/t}$, the transformation law of the Jacobi theta function. The Schwartz hypothesis can be weakened, but some decay and regularity of $f$ and $\widehat{f}$ is needed for both sides to converge to the same value. Let $G$ be a locally compact abelian group and $H$ a discrete subgroup with $G/H$ compact. A similar formula then holds, with the sum on the right over the characters of $G$ trivial on $H$. It needs suitable conditions on $f$ and a suitable normalisation of the Haar measures.

Going deeper

For a student

The formula is the basic link between a lattice and its dual lattice; in $\R^n$ it reads $\sum_{v \in L} f(v) = \frac{1}{\operatorname{vol}(\R^n/L)} \sum_{w \in L^*} \widehat{f}(w)$ for a lattice $L$ with dual lattice $L^*$. Applied to Gaussians it gives the functional equation of theta functions, which leads to the functional equation of the Riemann zeta function. In signal processing it underlies the sampling theorem, and in analytic number theory it is used to count lattice points.