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

Question

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.

Answer 1

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 $N \to \infty$.

Answer 2

In your attempt, try $\int \Big(f(x)\mathbb{1}_{[-N,N]}(x)\Big)\Big(f(x+n)\mathbb{1}_{[-N,N]}(x+n)\Big)\,dx$

Notice that for all $n$ large enough, $\mathbb{1}_{[-N,N]}\mathbb{1}_{[-N-n,N-n]}=0$

This in a way is similar to the suggestion by Prof. Rama Murthy since

$\|f\mathbb{1}_N-f\|_2\xrightarrow{N\rightarrow\infty}0$

Thus setting $f_N=f\mathbb{1}_N$ and $\tau_{-n}f=f(\cdot+n)$ for all $f\in L_2$, and using translation invariance of Lebesgue's measure ($\|\tau_{-n}f\|_2=\|f\|_2$ for all $n$), $\|f_N\|_2\leq\|f\|_2$ for ll $N$, and Cauchy-Schwartz inequality, we obtain

\begin{aligned} \|f\,\tau_{-n}f\|_2&\leq\|f\,\tau_{-n}f-f_N\,\tau_{-n}f_N\|_2+\|f_N\,\tau_{-n}f_N\|_2\\ &\leq\|f-f_N\|_2\|\tau_{-n}f\|_2+ \|f_N\|\|\tau_{-n}f-\tau_{-n}f_N\|_2+\|f_N\,\tau_{-n}f_N\|_2\\ &\leq2\|f-f_N\|_2\|f\|_2 + \|f_N\,\tau_{-n}f_N\|_2\ \end{aligned}

The rest should be straight forward.

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For similar proof, if you are aware of the fact that $\mathcal{C}_{00}(\mathbb{R})$ is dense in $L_2$, you may show the result holds for continuous functions of compact support.

Answer 3

Pick $\epsilon\gt0$. Find $n$ so that $$ \int_{|x|\ge n/2}f(x)^2\,\mathrm{d}x\le\frac{\epsilon^2}{4\|f\|_2^2}\tag1 $$ Then $$ \begin{align} &\int_{-\infty}^\infty|f(x)|\,|f(x+n)|\,\mathrm{d}x\\ &=\int_{-\infty}^{-n/2}\color{#090}{|f(x)|}\,\color{#E80}{|f(x+n)|}\,\mathrm{d}x+\int_{-n/2}^\infty\color{#C00}{|f(x)|}\,\color{#00F}{|f(x+n)|}\,\mathrm{d}x\tag2\\ &\le\color{#090}{\left(\int_{-\infty}^{-n/2}f(x)^2\,\mathrm{d}x\right)^{1/2}}\color{#E80}{\left(\int_{-\infty}^{-n/2}f(x+n)^2\,\mathrm{d}x\right)^{1/2}}\\ &+\color{#C00}{\left(\int_{-n/2}^\infty f(x)^2\,\mathrm{d}x\right)^{1/2}}\color{#00F}{\left(\int_{-n/2}^\infty f(x+n)^2\,\mathrm{d}x\right)^{1/2}}\tag3\\ &=\color{#090}{\left(\int_{-\infty}^{-n/2}f(x)^2\,\mathrm{d}x\right)^{1/2}}\color{#E80}{\left(\int_{-\infty}^{n/2}f(x)^2\,\mathrm{d}x\right)^{1/2}}\\ &+\color{#C00}{\left(\int_{-n/2}^\infty f(x)^2\,\mathrm{d}x\right)^{1/2}}\color{#00F}{\left(\int_{n/2}^\infty f(x)^2\,\mathrm{d}x\right)^{1/2}}\tag4\\ &\le\color{#090}{\frac\epsilon{2\|f\|_2}}\color{#E80}{\|f\|_2}+\color{#C00}{\|f\|_2}\color{#00F}{\frac\epsilon{2\|f\|_2}}\tag5\\[6pt] &=\epsilon\tag6 \end{align} $$ Explanation:
$(2)$: break the domain of integration at $-n/2$
$(3)$: Cauchy-Schwarz (twice)
$(4)$: substitute $x\mapsto x-n$ in the orange and blue integrals
$(5)$: apply $(1)$ to the green and blue integrals
$(6)$: simplify