# Constant “periodization” of a function

Let $$w$$ be a rapidly decaying function on $$\mathbb{R}$$ such that $$\sum_{n \in \mathbb{Z}} w(x+n) = 0$$ for all $$x \in \mathbb{R}$$. Does that imply that $$w$$ is identically zero? What if we assume that $$w$$ is continuous?

• The sum over the integers is a doubly-infinite series. How are you defining its value? Something like $\lim_{m\to\infty}\sum_{-m}^m$? – Gerry Myerson Oct 15 '18 at 6:42
• Jupp, for example. But since the function is rapidly decaying, you can choose any enumeration of the integers and end up with the same value. – Matthias Ludewig Oct 15 '18 at 8:53
• Indeed if you assume say that $w$ is also smooth this periodization equals $\sum \hat w(n) e(n x)$, so the condition is equivalent to $\hat w(n) = 0$ at any integer $n$ (so the Fourier transform of any smooth rapidly decreasing function that is zero at integers gives a counter-example and these are essentially all of them). – Rodrigo Oct 15 '18 at 10:24

No, the Haar wavelet is a counter example (i.e. $$w(x) = \chi_{[0,1[}(x) - \chi_{[1,2[}(x)$$). Decay is arbitrarily fast and mollifying by convolution gives a smooth counterexample.