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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

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  • $\begingroup$ This has nothing to do with the fractional Laplacian: you can replace $\phi$ and $(-\Delta)^s u$ by any pair of (uniformly) continuous functions. $\endgroup$ Commented Sep 12, 2021 at 7:01
  • $\begingroup$ @MateuszKwaśnicki Is that so? It would be great! How do you prove it? $\endgroup$
    – Riku
    Commented Sep 12, 2021 at 8:46

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(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 $|x-y|\leqslant\epsilon$ 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 $$ |a A - b B| \leqslant |a| |A - B| + |B| |a - b| . $$

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

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