If the Fourier transform $F(k)$ of $f(x)$ vanishes outside of the interval $(-1,1)$ then, by virtue of <A HREF="https://en.wikipedia.org/wiki/Poisson_summation_formula">Poisson summation,</A>
$$\sum_{n=-\infty}^\infty f(x+n)=\sum_{n=-\infty}^\infty F(n)e^{2\pi inx}=F(0)$$ independent of $x$.