# Symmetry of fractional laplacian

Let $$\Omega\subset\mathbb{R}^n$$, let $$s\in [1/2,1)$$, let $$u\in C^{1,2s-1+\epsilon}(\Omega)$$ such that: $$u=0$$ on $$\mathbb{R}^n\setminus\Omega$$, and: $$u\in C^{0,s}(\mathbb{R}^n)$$, is true that: $$\int_{\mathbb{R}^n}\phi(-\Delta)^su\,dx=\int_{\mathbb{R}^n}u(-\Delta)^s\phi,\quad\forall\phi\in C^\infty_c(\mathbb{R}^n)?$$ I kwon only that: $$\int_{\mathbb{R}^n}\phi(-\Delta)^sf\,dx=\int_{\mathbb{R}^n}f(-\Delta)^s\phi\,dx,\quad\forall f,\phi\in \mathcal{S}(\mathbb{R}^n).$$ I have no idea on how to go on, any help is appreciated.

• The proof follows by approximation, but the correct setting is $u \in H^{2s}_0(\Omega)$. Whether your condition on $u$ implies that $u$ is in the Sobolev space $H^{2s}_0(\Omega)$ may depend on the exact definition of $C^{k,\alpha}(\Omega)$, as pointed out in many of your previous questions. Commented Nov 17, 2020 at 9:13
• Can you give the details of this approximation in $H^{2s}(\Omega)$, please? I do not know any approximation results in $H^{2s}(\Omega)$.
– inoc
Commented Nov 17, 2020 at 9:20
• @MateuszKwaśnicki: Moreover, for $u\in C^{1,2s-1+\epsilon}(\mathbb{R}^n)$ i mean that: $u$ is $(2s-1+\epsilon)$-Hölder continuous and $\nabla u$ is $(2s-1+\epsilon)$-Hölder continuous. This hypothesis implies that: $u\in H^{2s}(\Omega)$?
– inoc
Commented Nov 17, 2020 at 9:35
• Any $H^{2s}(\mathbb R^n)$ function $u$ can be approximated by a sequence $u_n$ of Schwartz functions in the norm of $H^{2s}(\mathbb R^n)$ (so that $u_n$ converges to $u$ and $(-\Delta)^s u_n$ converges to $(-\Delta)^s u$, both limits in $L^2(\mathbb R^n)$). This can be found in most textbooks on Sobolev spaces, I believe. I am sure it is proved in Samko's Hypersingular integrals and Their Applications. Commented Nov 17, 2020 at 10:07
• Regarding your other comment: I already explained that some authors use $C^{k,\alpha}(\Omega)$ for the class of functions with $k$-th partial derivatives Hölder continuous in any compact subset of $\Omega$, while others require them to by Hölder continuous in all of $\Omega$. In the latter case, $u$ extends to a function in $C^{k,\alpha}(\overline \Omega)$, so with this definition the assumptions in your questions would be odd (if $u \in C^{1,\alpha}(\Omega)$, then automatically $y \in C^{1,\alpha}(\mathbb R^n)$). Commented Nov 17, 2020 at 10:12

If $$x, y \in \Omega$$, then $$|u(y) - u(x) - \nabla u(x) \cdot (y - x)| \leqslant C |y - x|^{2 s + \epsilon} ,$$ and so the integral $$\iint_{\Omega \times \Omega} \frac{u(y) - u(x) - \nabla u(x) \cdot (y - x)}{|y - x|^{n + 2 s}} \, \phi(x) dx dy$$ converges absolutely.

Denote $$d(x) = \operatorname{dist}(x, \partial \Omega)$$. If $$x \in \Omega$$, $$y \in \Omega^c$$, then $$|u(x)| \leqslant C d(x)$$ (because $$\nabla f$$ is bounded) and $$u(y) = 0$$. Thus, $$|u(y) - u(x) - \nabla u(x) \cdot (y - x)| \leqslant |u(x)| + |\nabla u(x)| \, |y - x| \leqslant C d(x) + C |y - x| .$$ Furthermore, $$\int_{\Omega^c} \frac{1}{|y - x|^{n + 2 s}} \, dy \leqslant \frac{1}{(d(x))^{2s}}$$ and $$\int_{\Omega^c} \frac{|y - x|}{|y - x|^{n + 2 s}} \, dy \leqslant \frac{1}{(d(x))^{2s - 1}} \, .$$ Finally, $$1 / (d(x))^{2s - 1}$$ is integrable. It follows that the integral $$\iint_{\Omega \times \Omega^c} \frac{u(y) - u(x) - \nabla u(x) \cdot (y - x)}{|y - x|^{n + 2 s}} \, \phi(x) dx dy$$ converges absolutely, too.

Similarly, if $$x \in \Omega^c$$ and $$y \in \Omega$$, we find that $$|u(y) - u(x) - \nabla u(x) \cdot (y - x)| \leqslant |u(y)| \leqslant C d(y) ,$$ and since $$\int_\Omega \frac{1}{|y - x|^{n + 2 s}} \, dy \leqslant \min \biggl\{ \frac{1}{(d(x))^{2s}} , \frac{C |\Omega|}{|x|^{n + 2 s}} \biggr\} ,$$ we have absolute convergence of $$\iint_{\Omega^c \times \Omega} \frac{u(y) - u(x) - \nabla u(x) \cdot (y - x)}{|y - x|^{n + 2 s}} \, \phi(x) dx dy .$$

Finally, the integral over $$\Omega^c \times \Omega^c$$ is identically zero.

We conclude that the integral $$\iint_{\mathbb R^n \times \mathbb R^n} \frac{u(y) - u(x) - \nabla u(x) \cdot (y - x)}{|y - x|^{n + 2 s}} \, \phi(x) dx dy$$ converges absolutely. Now the usual argument applies: \begin{aligned} \int_\Omega (-\Delta)^s u(x) \phi(x) dx & = \iint_{\mathbb R^n \times \mathbb R^n} \frac{u(y) - u(x) - \nabla u(x) \cdot (y - x)}{|y - x|^{n + 2 s}} \, \phi(x) dx dy \\ & = \lim_{\delta \to 0^+} \iint_{|x - y| > \delta} \frac{u(y) - u(x) - \nabla u(x) \cdot (y - x)}{|y - x|^{n + 2 s}} \, \phi(x) dx dy \\ & = \lim_{\delta \to 0^+} \iint_{|x - y| > \delta} \frac{u(y) - u(x)}{|y - x|^{n + 2 s}} \, \phi(x) dx dy \\ & = \lim_{\delta \to 0^+} \iint_{|x - y| > \delta} \frac{\phi(y) - \phi(x)}{|y - x|^{n + 2 s}} \, u(x) dx dy \\ & = \lim_{\delta \to 0^+} \iint_{|x - y| > \delta} \frac{\phi(y) - \phi(x) - \nabla \phi(x) (y - x)}{|y - x|^{n + 2 s}} \, u(x) dx dy \\ & = \int_\Omega (-\Delta)^s \phi(x) u(x) dx . \end{aligned} (Here the second equality follows by dominated convergence, the fourth one by Fubini, and the sixth one again by dominated convergence.)

• Regarding the integral expression for $(-\Delta)^s u(x)$: isn't this the very definition of $(-\Delta)^s u(x)$? The term $|y - x|^{-n - 2s} \nabla u(x) \cdot (y - x)$ disappears because it has zero integral over $\mathbb R^n \setminus B(x, r)$ for every $r > 0$. Commented Nov 24, 2020 at 11:58
• (The above comment answers questions raised in other comments that now seem deleted.) Commented Nov 24, 2020 at 11:59