Schwartz functions on the real line have explicit Fourier interpolation

You need to know: Set ${\mathbb R}$ of real numbers, even function $f:{\mathbb R} \to {\mathbb R}$ (such that $f(x)=f(-x)$ for all $x\in{\mathbb R}$ ), derivative, k-th derivative $f^{(k)}(x)$ , integration, set ${\mathbb C}$ of complex numbers, Fourier transform $\hat{f}(t)=\int_{-\infty}^{\infty} f(x) e^{-2\pi i t x} dx$ of an integrable function $f:{\mathbb R} \to {\mathbb R}$ , absolute convergence of infinite series.

Background: Function $f:{\mathbb R}\to{\mathbb R}$ is called a Schwartz function if there exist all derivatives $f^{(k)}(x)$ for all $k=1,2,3,\dots$ and for all $x\in{\mathbb R}$ , and, for every k and $\gamma\in{\mathbb R}$ , there is a constant $C(k,\gamma)$ such that $|x^\gamma f^{(k)}(x)| \leq C(k,\gamma), \, \forall x\in {\mathbb R}$ .

The Theorem: On 1st January 2017, Danylo Radchenko and Maryna Viazovska submitted to arxiv a paper in which they proved the existence of a collection of even Schwartz functions $a_n:{\mathbb R}\to{\mathbb R}$ with the property that for any even Schwartz function $f:{\mathbb R} \to {\mathbb R}$ and any $x\in{\mathbb R}$ we have $f(x)=\sum\limits_{n=0}^\infty a_n(x)f(\sqrt{n})+\sum\limits_{n=0}^\infty \hat{a_n}(x)\hat{f}(\sqrt{n})$ , where the right-hand side converges absolutely.

Short context: The classical Whittaker-Shannon interpolation formula states that if the Fourier transform $\hat{f}$ of function $f:{\mathbb R} \to {\mathbb R}$ is supported in $[-w/2,w/2]$ , then $f(x)=\sum\limits_{n=-\infty}^\infty f(n/w)\text{sinc}(wx-n)$ , where $\text{sinc}(x) = \sin(\pi x)/(\pi x)$ . The formula has numerous applications, in particular it allows to construct a “nice” continuous function which approximates a given sequence of real numbers. However, it does not work for functions whose Fourier transform has unbounded support. The Theorem provides a similar formula which works for arbitrary Schwartz functions. In particular, it implies that if $f:{\mathbb R} \to {\mathbb R}$ is an even Schwartz function such that $f(\sqrt{n})=\hat{f}(\sqrt{n})=0$ for $n=0,1,2,\dots$ , then $f(x)=0$ for all $x\in{\mathbb R}$ .