Showing $\lim_{n\to\infty}\int_{\mathbb{R}} f(x)f(x+n) dx=0$

Hint: There exists a continuous function $g$ with compact support such that $ \|f-g\|_2 <\epsilon$. $\int g(x)g(x+n)=0$ for $n$ sufficiently large and $|\int f(x)f(x+n)-\int g(x)g(x+n)| <\epsilon ( \|f\|_2+ \|g\|_2)$ by Holder's inequlaity.

A simpler proof: (continuity of $g$ is not required to we can simplify the proof as follows:)

Let $g(x)=f(x)$ for $|x| \leq N$ and $0$ for $|x| >N$. Observe that $ g(x)g(x+n)=0$ for all $x$ if $n >2N$ and that $\int |f-g|^{2} \to 0$ as $N \to \infty$. . Hence $\int g(x)g(x+n)dx=0$ fo $n >2N$. Now $$\int f(x)f(x+n) -\int g(x)g(x+n)$$ $$=\int f(x)f(x+n) -\int g(x)f(x+n)+\int g(x)f(x+n) -\int g(x)g(x+n).$$ Use Holder's inequality to show that this quantity tends to $0$ as $N \to \infty$.