Poisson summation formula

Step 1 (periodize). For a Schwartz function $f$, define $F(x) = \sum_{n=-\infty}^{\infty} f(x+n)$. Rapid decay of $f$ makes this sum converge absolutely and uniformly, and $F$ is a smooth function with period 1, so it has its own Fourier series on the circle.

Step 2 (compute the Fourier coefficients of the periodization). The $k$-th Fourier coefficient of $F$ is $c_k = \int_0^1 F(x) e^{-2\pi i k x}\,dx = \sum_{n} \int_0^1 f(x+n) e^{-2\pi i k x}\,dx$. Substituting $y = x+n$ in each term and using periodicity of $e^{-2\pi i k x}$ in $n$ reassembles the pieces into a single integral over all of $\mathbb{R}$: $c_k = \int_{-\infty}^{\infty} f(y) e^{-2\pi i k y}\,dy = \hat f(k)$.

Step 3 (evaluate at x=0). Since $F$ is smooth, its Fourier series converges to it pointwise, in particular at $x=0$: $F(0) = \sum_k c_k e^{0}$, i.e. $\sum_{n=-\infty}^{\infty} f(n) = \sum_{k=-\infty}^{\infty} \hat f(k)$. The left side is literally $F(0) = \sum_n f(n)$ by definition, which finishes the proof.