If $f,g\in L^2(\mathbb{R})$ then $\lim_{n\to \infty}\int_{\mathbb{R}}f(x)g(x+n)=0$

Question

I don't know if this is true or false. I tried counterexamples but couldn't find any so then I tried to prove this with some theorems regarding convolutions because it looks a little bit like a convolution but couldn't find any that I could use. Can someone give me a hint?

Answer 1

Let $f,g$ be square integrable functions (not equivalent classes of functions). Let $S_{f}=\{x\in\mathbb{R}\mid f(x)\neq0\}$ and $S_{g}=\{x\in\mathbb{R}\mid g(x)\neq0\}$ be the supports of $f$ and $g$ respectively.

Case 1. $S_{f},S_{g}$ are bounded subsets of $\mathbb{R}$: Note that for sufficiently large $n$, $f(x)g(x+n)=0$ for all $x$, so $\int f(x)g(x+n)dx=0$.

Case 2. Let $f_{n}(x)=\begin{cases} f(x), & \mbox{if }x\in[-n,n]\\ 0, & \mbox{otherwise} \end{cases}$ and $g_{n}=\begin{cases} g(x), & \mbox{if }x\in[-n,n]\\ 0, & \mbox{otherwise} \end{cases},$ then $f_{n},g_{n}\in L^{2}$ with bounded supports. Note that $|f_{n}-f|^{2}\leq|f|^{2}$ and $|f_{n}-f|^{2}\rightarrow0$, so by Dominated Convergence Theorem, $||f_{n}-f||_{2}\rightarrow0$. Similarly, $||g_{n}-g||_{2}\rightarrow0$. Choose $M>0$ such that $||f||_{2}\leq M$ and $||g||_{2}\leq M$. Let $\varepsilon>0$ be given. Choose $N$ such that $||f_{N}-f||_{2}<\frac{\varepsilon}{M}$ and $||g_{N}-g||_{2}<\frac{\varepsilon}{M}$. By Case 1, choose $K$ such that $\int f_{N}(x)g_{N}(x+k)dx=0$ whenever $k\geq K$. For any $k\geq K$, we have that \begin{eqnarray*} & & \int f(x)g(x+k)dx\\ & = & \int\left[f(x)g(x+k)-f_{N}(x)g(x+k)\right]dx+\int\left[f_{N}(x)g(x+k)-f_{N}(x)g_{N}(x+k)\right]dx. \end{eqnarray*} By Cauchy-Schwarz inequality, \begin{eqnarray*} & & \left|\int\left[f(x)g(x+k)-f_{N}(x)g(x+k)\right]dx\right|\\ & \leq & ||f-f_{N}||_{2}||g(\cdot+k)||_{2}\\ & = & ||f-f_{N}||_{2}||g||_{2}\\ & \leq & \varepsilon. \end{eqnarray*} By Cauchy-Schwarz inequality again, \begin{eqnarray*} & & \left|\int\left[f_{N}(x)g(x+k)-f_{N}(x)g_{N}(x+k)\right]dx\right|\\ & \leq & ||f_{N}||_{2}\cdot||g(\cdot+k)-g_{N}(\cdot+k)||_{2}\\ & \leq & ||f||_{2}\cdot||g-g_{N}||_{2}\\ & \leq & \varepsilon. \end{eqnarray*} This shows that $\lim_{k\rightarrow\infty}\int f(x)g(x+k)dx=0.$

Answer 2

The basics of the Fourier transform, $F,$ on $L^2(\mathbb R)$ lead to an nice solution: We have

$$ \int_{\mathbb R} f(x)g(x+n)\,dx = \int_{\mathbb R} F(f)(t)F(g(x+n))(t)\,dt$$ $$ = \int_{\mathbb R} F(f)(t)\cdot F(g)(t)e^{int}\,dt.$$

Now $F(f),F(g)\in L^2,$ so $F(f)\cdot F(g)\in L^1.$ Hence by the Riemann Lebesgue lemma, the last integral $\to 0$ as $n\to \infty,$ and we're done.