# Averaging and fractional Laplacian

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

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

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 $$|a A - b B| \leqslant |a| |A - B| + |B| |a - b| .$$