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Let $\alpha$ be real numbers and let $f\colon\mathbb{R}\to \mathbb{C}$ be a function in $L^2 (\mathbb{R})$ (actually smooth and compactly supported, but this doesn't seem to be relevant). I am interested in the following integral average:

$$\frac{1}{R}\int_{0}^{R}\int_{\mathbb{R}}\bar{f}\left(x\right)e^{2\pi i\alpha rx}f\left(x+r\right) dxdr$$

and in its convergence as $R \to \infty$. I have reason to believe that it converges to zero, but offhand I don't see why it converges at all. Does anyone have any ideas about that?

Thanks for any help.

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For me, it seems that the inner intergal is independent of the variable $r$, by the substitution $x+r\mapsto x$. That is, under the integral average its value is conserved. –  sos440 Sep 5 '12 at 17:42
    
Sorry, I got confused with a different expression. It should be okay now. –  Mark Sep 5 '12 at 18:04
    
Do the substitution $Rs=r$, then cut the integral $\int_0^{\epsilon}+\int_{\varepsilon}^1$ and look at the supports. –  Davide Giraudo Sep 5 '12 at 18:48

1 Answer 1

up vote 1 down vote accepted

Let $$I_R:=\frac{1}{R}\int_0^R\int_{\Bbb{R}}\bar f\left(x\right)e^{2\pi i\alpha rx}f\left(x+r\right) dxdr.$$ We use the substitution $Rs=r$ to get $$I_R=\int_0^1\int_{\Bbb R}\bar f(x)e^{2\pi i\alpha Rsx}f(x+Rs)dxds.$$ This gives $$I_R\leq \varepsilon |\operatorname{supp}f|\cdot\sup_{t\in \Bbb R}|f(t)|+\int_\varepsilon^1\int_{\Bbb R}\bar f(x)e^{2\pi i\alpha Rsx}f(x+Rs)dxds.$$ We assume that the support of $f$ is contained in $[-A,A]$. The support of $f(x)f(x+Rs)$ is contained in $[-A,A]\cap [-A-Rs,A-Rs]\subset [-A,A]\cap [-A-R,A-R\varepsilon]$. It's the emptyset when $A-R\varepsilon< -A$ i.e. $R>\frac{2A}{\varepsilon}$.

By a density argument, we can extend the result to the functions in $L^2(\Bbb R)$.

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Looks good to me. –  Mark Sep 6 '12 at 17:28

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