Let $\phi:\mathbb{R^n}\to\mathbb{R}$ be a Lebesgue-measurable function, with the property that for every $n$-dimensional cube $Q$ in $\mathbb{R^n}$, we have $$ \left|\int_{Q}\phi(x)dx \right|\leq\frac{M m(Q)}{1+m(Q)} $$

for a constant $M$, where $m$ denotes the Lebesgue measure. Prove that if $f \in L^{1}(\mathbb{R}^n)$, $$ \lim_{k\to \infty}\int_{\mathbb{R^n}}\phi(kx)f(x)dx = 0 $$

From the first inequality, if you assume that $\int_{\mathbb{R^n}}=\lim_{m(Q)\to \infty}\int_{Q}$ (is this even true?), you can do: $$ \left|\int_{\mathbb{R}^n}\phi(x)dx\right|=\lim_{m(Q)\to \infty}\left|\int_{Q}\phi(x)dx\right|\leq \lim_{m(Q)\to \infty}\frac{M m(Q)}{1+m(Q)}=M $$

Maybe one can use this and apply Hölder's inequality on that product, with some sort of variable change, but I couldn't make it work. Does anyone have any suggestions?

  • 2
    $\begingroup$ You might start by assuming $f$ is smooth and compactly supported and then use a density argument. This helps for a number of reasons, one of which is that you don't know whether $\phi$ is in any $L^p$ space on all of $\mathbb{R}^n$. $\endgroup$ – Ian May 25 '15 at 21:49
  • $\begingroup$ @Ian: For the density argument to work, to need to know $\phi \in L^\infty$. But this follows from the assumed inequality using Lebesgues differentiation theorem. $\endgroup$ – PhoemueX May 26 '15 at 9:53

First off, $\phi \in L^{1}_{loc}(\mathbb{R}^n)$. Then, from Lebesgue's differentiation theorem, for a.e. $x\in \mathbb{R}^{n}$ $$ \left|\phi(x)\right|=\lim_{r\to 0}\left|\frac{1}{m(B(x,r))}\int_{B(x,r)}\phi(y)dy\right| \leq M, $$ which means that $\phi \in L^{\infty}$, with $\|\phi\|_{\infty}\leq M$. Given $f\in L^1(\mathbb{R}^n)$ and $\varepsilon>0$, take $\tilde{f}\in C^{\infty}_c(\mathbb{R}^n)$ to be smooth and compactly supported ($\text{supp } \tilde{f} \subset K$), with $\|f-\tilde{f}\|_1 < \varepsilon$. Then, for every $k\geq 0$ $$ \left| \int_{\mathbb{R}^n}\phi(kx)f(x)dx \right| \leq \left|\int_{\mathbb{R}^n} \phi(kx)\tilde{f}(x)dx\right| + \int_{\mathbb{R}^n}\left|\phi(kx)\right|\left|f(x)-\tilde{f}(x) \right|dx \leq \\ \leq \left|\int_{\mathbb{R}^n} \phi(kx)\tilde{f}(x)dx\right| + M\varepsilon $$

However, for $\tilde{f}$, let $Q$ be a cube containing $K$, $$ \left|\int_{\mathbb{R}^n} \phi(kx)\tilde{f}(x)dx\right| = \left|\int_{K} \phi(kx)\tilde{f}(x)dx\right| \leq \|\tilde{f}\|_{\infty}\left|\int_{Q}\phi(kx)dx \right| = \\ =\|\tilde{f}\|_{\infty}\left|\frac{1}{k^n}\int_{\tilde{Q}}\phi(y)dy \right| \leq \frac{\|\tilde{f}\|_{\infty}M m(\tilde{Q})}{k^{n}(1+m(\tilde{Q}))} \leq \frac{\|\tilde{f}\|_{\infty}M}{k^n} \to 0 $$

as $k\to \infty$. This concludes the proof.


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