Let $f:\mathbb R\to [0, \infty)$ be function. I want's to construct periodic function out of this function. Define $F(x)=\sum_{k\in \mathbb Z} f(x+k).$ Then clearly $F$ is a periodic function with period one on $\mathbb R,$ that is, $F(x+1)=F(x)$ for all $x\in \mathbb R.$ [ In fact, $F(x+1)= \sum_{k=-\infty}^{\infty} f(x+1+k)=\sum_{k'=k+1\in \mathbb Z}f(x+k')= F(x)$]
Define $G(x)=\sum_{0\neq k \in \mathbb Z}f(x+k).$
Question: Can we expect $G$ is perodic on $\mathbb R$ with period one, that is, $G(x+1)=G(x)$?
My attempt: Note that $G(x+1)=\sum_{k=-\infty}^{-1} G(x+1+k)+ \sum_{k=1}^{\infty} G(x+1+k)$. If we do change of variables on summation, $k'=k+1,$ we may obtain, $G(x+1)=\sum_{k'=-\infty}^{0}f(x+k')+\sum_{k'=2}^{\infty} f(x+k')$