Is $\sum_{m\in \mathbb Z} f(x-m)f(x-n) \in L^2(a,b)$ if $f\in L^2(\mathbb R)$?

Question

Let $f\in L^2(\mathbb R)$ and $0<a<b< \infty,$ put $A= \{x\in \mathbb R: a<|x|<b\}.$ Fix $n\in \mathbb Z.$ Define $$ F_n(x)=\sum_{m\in \mathbb Z} f(x-n-m)f(x-m)$$ We note that $F_n$ is a periodic function with period 1. And hence we cannot expect $F_n$ belongs to $L^2(\mathbb R).$

But can we expect $F_n\in L^2(A)$?

I'm also curios to know what happens of summation over $n$? Specifically, define $$G(x)= \sum_{n\in \mathbb Z} F_n(x).$$

Can we expect $G\in L^2(A)$?