How to choose function $\sum_{m\in \mathbb Z} (-1)^m f(x+m) f(x-m+n)=0$?

Question

Can we expect to choose a function $f:\mathbb R \to \mathbb R$ (nonzero compactly supported) so that

$\sum_{m\in \mathbb Z} (-1)^m f(x+m) f(x-m+n)=0$ for all $x\in \mathbb R$ and $n\in \mathbb Z$?

Answer 1

No.

Claim If $f$ has compact support and is non-trivial, then there exists some $x_*\in \mathbb{R}$ and $n_* \in \mathbb{Z}$ such that the sum $\sum_{m\in \mathbb{Z}} (-1)^m f(x_* + m) f(x_*-m + n_*) \neq 0$.

Proof: let $n_* = 0$.

Take $y_0 = \sup \{f \neq 0\} < \infty$. Then there exists $x_* \in [y_0 - 1/3, y_0]$ such that $f(x_*) \neq 0$ but $f(x_* + m) \equiv 0$ for every integer $m \geq 1$.

For this $x_*$ (and $n_* = 0$) we can evaluate the sum

$$ \sum_{m} (-1)^m f(x_* + m) f(x_* - m) = f(x_*)^2 > 0$$

since the summand vanishes for every $m \neq 0$.