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 $\mathbb{R}^n$.

  • 2
    $\begingroup$ Do you mean $\int_B f(x)\ dx \le \frac{m(B)}{1+m(B)}$? $\endgroup$ – Robert Israel Mar 7 '12 at 21:14
  • $\begingroup$ yeah, I corrected. $\endgroup$ – checkmath Mar 7 '12 at 21:25
  • $\begingroup$ Sorry, f is positive. $\endgroup$ – checkmath Mar 7 '12 at 22:21

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)

  • $\begingroup$ 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? $\endgroup$ – Norbert Mar 8 '12 at 21:31
  • $\begingroup$ 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. $\endgroup$ – Martin Argerami Mar 8 '12 at 21:40
  • $\begingroup$ So, why don't you edit your solution to shorter one? $\endgroup$ – Norbert Mar 8 '12 at 22:00
  • 1
    $\begingroup$ 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 $\|f\|_{L^\infty} \leq 1$. $\endgroup$ – Nick Strehlke Mar 9 '12 at 2:24
  • 2
    $\begingroup$ Nice answer, it proves overall that cooperating is more efficient than competing! $\endgroup$ – checkmath Mar 10 '12 at 17:35

Your Answer

