Showing $\lim_{n\to\infty}\int_{\mathbb{R}} f(x)f(x+n) dx=0$ https://ift.tt/eA8V8J

Problem

Let $f(x)\in L^2(-\infty,\infty)$. Prove that $$ \lim_{n\to\infty}\int_{-\infty}^\infty f(x)f(x+n) dx=0. $$

My attempt

Let $N\in\mathbb{N}$ and $f_N(x):=f(x)\chi_{[-N,N]}(x)$.

For any $n\in\mathbb{N}$, consider

$$ \begin{split} \int_{-\infty}^\infty f(x)f_N(x+n)dx&=\int_{-\infty}^\infty f(x)f(x+n)\chi_{[-N,N]}(x+n)dx\\ &=\int_{-\infty}^\infty f(x)f(x+n)\chi_{[-N-n,N-n]}(x)dx\\ &=\int_{-N-n}^{N-n}f(x)f(x+n)dx \end{split} $$

This is where I get stuck.

We know for a.e. $x\in\mathbb{R}$ that $\lim_{n\to\infty} f(x+n)=0$. Thus, if I were allowed to interchange limit and integral, I would get the desired conclusion. I tried Dominated Convergence Theorem, but could not find a way to dominate the integrand by an integrable function not depending on $n$.

Maybe there is a different approach. Any hints? I'd like to be able to say the absolute value of the above integral is less than any $\varepsilon>0$ for $n$ large enough, and then since $N$ is arbitrary, the conclusion follows.

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