If the Fourier transform $F(k)$ of $f(x)$ vanishes outside of the interval $(-1,1)$ then, by virtue of Poisson summation, $$\sum_{n=-\infty}^\infty f(x+n)=\sum_{n=-\infty}^\infty F(n)e^{2\pi inx}=F(0)$$ independent of $x$.
Carlo Beenakker
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