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Suppose $f,g\in L^2(\mathbb{R})$, how to show that $$\lim_{n\rightarrow \infty}\int_\mathbb{R}f(x)g(x+n)=0$$

If functions in $L^p(\mathbb{R})$ vanish outside of a set of finite measure, then we can assume $E=\{x|f(x)\neq 0\}$ and $F=\{x|g(x)\neq 0\}$ and $E$ and $F$ are measurable. And we defien $F_n=\{x|g(x+n)\neq 0\}$.

So can we find $n$ such that $m(F_n\cap E)\leq1/n$?

I have another idea that is to use Simple Approximation Theorem to prove.

Here is the Simple Approximation Theorem: enter image description here

update

In general it is not true that functions in $L^p(\mathbb{R})$ vanish outside of a set of finite measure(counterexample see @Ant 's comment). However it is true that simple functions in $L^p(\mathbb{R})$ vanish outside of a set of finite measure.

DuFong
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  • $f(x) = 1/x^2$ never vanishes but is in $L^2((1, \infty))$. One can readily modify it to make it $L^2(\mathbb R)$. They tend to $0$ as $x \to \infty$ though – Ant May 07 '16 at 18:20
  • How could it be that $f$ and $g$ vanish outside a set of finite measure, yet $m(\mathbb R\setminus E)=m(\mathbb R\setminus F)=0$ - meaning they vanish almost nowhere? (And, do you know any explicit examples of $L^2$ functions?) – Milo Brandt May 07 '16 at 18:21
  • What is your "Simple Approximation theorem"? – HarshCurious May 07 '16 at 18:21
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    @Ant That's not true, at least if you mean in some standard sense of "tending to zero". $\sum_{i=0}^{\infty}\chi_{[i,i+2^{-i}]}$ is an $L^2$ function (where $\chi_I$ is the indicator function of $I$) which does not tend to $0$ as $x$ goes to infinity. – Milo Brandt May 07 '16 at 18:23
  • @Ant, $f(x)=1/x^2$ is not in $L^2(\mathbb{R})$. – DuFong May 07 '16 at 18:23
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    @Benjamin That's why I said $(1, \infty)$. But you can modify it to avoid problems in $0$, for example a function equal to $1/x^2$ outside $(-1, 1)$ and equal to $1$ in $(-1, 1)$. This is $L^2(\mathbb R)$ – Ant May 07 '16 at 18:26
  • @MiloBrandt You're right, that's something I always forget, even though I had seen counter examples before. Is there another sense of "tending to zero" such that it works? :-) – Ant May 07 '16 at 18:28
  • @MiloBrandt $m(\mathbb R\setminus E)=m(\mathbb R\setminus F)=0$ cannot be true. I deleted it. – DuFong May 07 '16 at 18:34
  • @Ant yes: if the limit exists, then it is equal to zero. –  May 07 '16 at 18:34

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Hints: Supppose first $f$ vanishes outside $[-a,a].$ Show $\int_{-a}^a f(x) g(n+x) \, dx\to 0$ by applying Cauchy-Schwartz and the fact that $\int_{-a}^a |g(n+x)|^2 \, dx\to 0.$ For the general $f\in L^2,$ approximate $f$ by $f\chi_{[-a,a]}$ for large $a.$

zhw.
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  • You mean $\lim_{n\rightarrow \infty }\lim_{a\rightarrow \infty}\int_{-a}^{a}f(x)g(x+n)dx=\lim_{a\rightarrow \infty }\lim_{n\rightarrow \infty}\int_{-a}^{a}f(x)g(x+n)dx$, right? But can we change the order of the limit? – DuFong May 07 '16 at 19:27
  • Here someone talks the sufficient condition for change order of limits: http://math.stackexchange.com/questions/15240/when-can-you-switch-the-order-of-limits – DuFong May 07 '16 at 19:29
  • Yes, it turns out we can change the order of limits here. But I think the proof is easier - and you'll learn more - by not scouring the internet for a general change of limit result that applies. Rather, just do it. – zhw. May 08 '16 at 19:23