Let $u,\phi:\mathbb R \to \mathbb R$ be smooth functions and $\Omega_\epsilon$ be a bounded domain in $\mathbb R$ with diameter $\epsilon>0$ (consider for exaple the ball $B_{\epsilon/2}(0)$). Is it true that $$\frac{1}{\Omega_\epsilon}\int_{\Omega_\epsilon} \phi (\Delta)^s u dx  \left( \frac{1}{\Omega_\epsilon}\int_{\Omega_\epsilon} (\Delta)^s u dx\right)\left( \frac{1}{\Omega_\epsilon}\int_{\Omega_\epsilon} \phi dx\right) \to 0 $$ as $\epsilon \to 0$? Here $(\Delta)^s$ denotes the fractional Laplacian operator

$\begingroup$ This has nothing to do with the fractional Laplacian: you can replace $\phi$ and $(\Delta)^s u$ by any pair of (uniformly) continuous functions. $\endgroup$– Mateusz KwaśnickiCommented Sep 12, 2021 at 7:01

$\begingroup$ @MateuszKwaśnicki Is that so? It would be great! How do you prove it? $\endgroup$– RikuCommented Sep 12, 2021 at 8:46
1 Answer
(I believe this may be too basic for this site, but too long for a comment.)
All that we need to assume is that $\phi$ and $(\Delta)^s u$ are uniformly continuous and bounded. (Continuity suffices if we additionally know that $\Omega_\epsilon$ are all contained in a bounded region.)
If $f(x)f(y)\leqslant\delta$ whenever $xy\leqslant\epsilon$ and the diameter of $\Omega$ is no greater than $\epsilon$, then — for an arbitrary $x_0 \in \Omega$ — we have $$\biggl \frac{1}{\Omega} \int_{\Omega} f(x) dx  f(x_0)\biggr\leqslant \delta.$$ Applying this three times, with $f = \phi$, $f = (\Delta)^s u$ and $f = \phi (\Delta)^s u$, we find that $$\begin{aligned} & \biggl \frac{1}{\Omega_\epsilon} \int_{\Omega_\epsilon} \phi (\Delta)^s u  \frac{1}{\Omega_\epsilon} \int_{\Omega_\epsilon} \phi \times \frac{1}{\Omega_\epsilon} \int_{\Omega_\epsilon} (\Delta)^s u \biggr \\ & \qquad \leqslant \biggl \frac{1}{\Omega_\epsilon} \int_{\Omega_\epsilon} \phi (\Delta)^s u  \phi(x_\epsilon) (\Delta)^s u(x_\epsilon) \biggr \\ & \qquad \qquad + \biggl \frac{1}{\Omega_\epsilon} \int_{\Omega_\epsilon} \phi \times \frac{1}{\Omega_\epsilon} \int_{\Omega_\epsilon} (\Delta)^s u  \phi(x_\epsilon) (\Delta)^s u(x_\epsilon) \biggr \\ & \qquad \leqslant 2 M \delta + 2 M \delta = 4 M \delta,\end{aligned}$$ Here $M$ is an upper bound for $\phi$ and $(\Delta)^s u$, $x_\epsilon$ is an arbitrary point of $\Omega_\epsilon$, and, for a given $\delta > 0$, $\epsilon$ is small enough as in the definition of uniform continuity of $\phi$ and $(\Delta)^s u$. In the last inequality we used twise the standard estimate $$ a A  b B \leqslant a A  B + B a  b . $$

$\begingroup$ Thanks! I see now that it's indeed basic, but still very helpful to me. :) $\endgroup$– RikuCommented Sep 13, 2021 at 9:14