let $f,g\in L^2(\mathbb R)$ and note the sequence $x_n=|\int_\mathbb R f(x+n)g(x)dx|$. Does $x_n$ converge as a sequence in $\mathbb R$?
My guess would be yes, and that the limit is $0$, because in the edges $f$ should "tend" to zero, otherwise it would be in $L^2$. But I don't quite know.
Assume $f,g\in L^{2}(\mathbb{R})$. Let $\epsilon > 0$ be given, and write $$ \int_{-\infty}^{\infty}f(x+n)g(x)dx=\int_{-n-R}^{-n+R}f(x+n)g(x)dx+\int_{|x+n| \ge R}|f(x+n)g(x)dx. $$ Therefore, the above is bounded by $$ \|f\|_{L^{2}}\left(\int_{-\infty}^{-n+R}|g(x)|^{2}dx\right)^{1/2} + \left(\int_{|x| \ge R}|f(x)|^{2}dx\right)^{1/2}\|g\|_{L^{2}}. $$ Let $\epsilon > 0$ be given. The second term above tends to $0$ as $R\rightarrow\infty$ and, hence, there exists $R$ large enough that the second term above is bounded by $\epsilon/2$. Then, for this fixed $R > 0$, the first term above tends to $0$ as $n\rightarrow\infty$. So there exists a positive integer $N$ such that the first term is bounded by $\epsilon/2$ whenever $n \ge N$. Finally, $$ \left|\int_{-\infty}^{\infty}f(x+n)g(x)dx\right| < \epsilon \mbox{ whenever } n \ge N. $$
