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Suppose $f\geq 0$ be a measurable function such that

$$\int_Bf\leq \frac{m(B)}{1+m(B)}$$ for any ball.

Then show that $$\int_{\mathbb{R}^n}f(kx)g(x)\to 0$$

as $k\to \infty$ for any $g$ integrable in $\mathbb{R}^n$.

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    Do you mean $\int_B f(x)\ dx \le \frac{m(B)}{1+m(B)}$?2012-03-07
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    yeah, I corrected.2012-03-07
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    Sorry, f is positive.2012-03-07

1 Answers 1

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This is kind of a Riemann-Lebesgue Lemma, but I'm not sure if we can obtain a proof just from that.

Anyway, let us first note that if we prove the assertion for $g\geq0$ we are done, since $$ \left|\int_{\mathbb{R}^n}f(kx)g(x)\,dm\right|\leq\int_{\mathbb{R}^n}f(kx)|g(x)|\,dm. $$

The hypothesis on $f$ implies that $f$ is integrable and $\int f\,dm\leq1$. Indeed, let $B_n$ be the ball of radius $n$ centered at the origin. Then, by Monotone Convergence, $$ \int f\,dm=\lim_n\int f\; 1_{B_n}\,dm=\lim_n\int_{B_n} f\,dm\leq 1. $$

Knowing that $f$ is integrable, almost every point in $\mathbb{R}^n$ is a Lebesgue point, so $f\leq1$ a.e. by the inequality $$ \frac1{m(B)}\int_Bf\,dm\leq\frac1{1+m(B)}. $$

The integrability of $g$ implies that $$ \lim_{K\to\infty}\int_{g>K}g=0. $$ Given $\varepsilon>0$, we choose $K$ such that $\int_{g>K}g\,dm<\varepsilon$. Then, writing $g_K$ for the function $g\;1_{g\leq K}$, \begin{eqnarray} \int f(kx)g(x)\,dm&=&\int_{g\leq K}f(kx)g(x)\,dm +\int_{g>K}f(kx)g(x)\,dm\\ &\leq&\int f(kx)g_K(x)\,dm +\varepsilon\\ &=&\frac1{k^n}\int f(t) g_K(t/k)\,dm+\varepsilon\\ &\leq&\frac{K}{k^n}+\varepsilon. \end{eqnarray} Thus $$ \limsup_{k\to\infty}\int f(kx)g(x)\,dm\leq\varepsilon. $$ As $\varepsilon$ was arbitrary, we conclude that the $\limsup$ is zero, and so the limit exists and is zero: $$ \lim_{k\to\infty}\int f(kx)g(x)\,dm=0. $$

(thanks to Norbert and to Nick Strehlke for the ideas to shorten the proof)

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    This solution seems to me too complicated. Once you have proved that $\Vert f\Vert_\infty\leq 1$. You can note that $$\int_{\mathbb{R}^n}f(kx)g(x)d\mu(x)=k^{-n}\int_{\mathbb{R}^n}f(t)g(t/k)d\mu(t)$$ And apply dominated convergence theorem to the last integral to get $0$. Am I mistaken somewhere?2012-03-08
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    I think you are right. My proof just came that way as I started thinking about the problem, and it took me a while to notice that the hypothesis implies $\|f\|_\infty\leq1$, so by the time I was in position to notice your observation, I was already involved in the more complicated proof. Actually, I usually make an effort to try an avoid "simple functions' proofs", but I didn't immediately see the way here.2012-03-08
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    So, why don't you edit your solution to shorter one?2012-03-08
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    I started doing it, but then I noticed that is not obvious to me how to apply dominated convergence to your integral; what would the dominating function be?2012-03-08
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    I Little hard but there is a pretty nice answer! I will check the details. But I believe it is a correct answer.2012-03-08
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    Please do check the details. Also, I'm waiting for Norbert's answer; then I'll shorten the answer as per his suggestion.2012-03-08
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    Here's one way to use Norbert's observation to shorten the proof (though it doesn't use dominated convergence). Assuming $|g|$ is bounded by $M$, the integral that Norbert wrote down is no bigger than $k^{-n}M\|f\|_{L^1} \leq k^{-n} M$. That gives the result for bounded functions. In general, choose $g_\epsilon$ bounded with $\|g - g_\epsilon\|_{L^1} < \epsilon$, bound the integral of $f(kx)$ against $g(x)$ by $\|f\|_{L^\infty}\|g-g_\epsilon\|_{L^1} + \int f(kx)g_\epsilon(x)\,dx$, and use $\|f\|_{L^\infty} \leq 1$.2012-03-09
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    Thanks. It's edited now to include these ideas. I like it a lot better with no simple functions!2012-03-09
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    Nice answer, it proves overall that cooperating is more efficient than competing!2012-03-10