Integrating the fractional Laplacian over $\Omega$ Let $\Omega$ be a bounded regular open subset of $\mathbb{R}^N$, $N\geq 1$.
I want to know if this statement is true : for $u\geq 0$ in $\Omega$, and $u=0$ in $\mathbb{R}^N\setminus \Omega$,
$$
\int_\Omega (-\Delta)^s u(x) dx\geq 0.
$$
I reasoned by the fact that, for $s\in (0,1)$,
$$
\int_\Omega (-\Delta)^s u(x) dx=\int_\Omega P.V.\int_{\mathbb{R}^N} \frac{u(x)-u(y)}{|x-y|^{N+2s}} dy dx \geq \int_\Omega P.V.\int_{\mathbb{R}^N\setminus \Omega} \frac{u(x)-u(y)}{|x-y|^{N+2s}} dy dx = \int_\Omega P.V.\int_{\mathbb{R}^N\setminus \Omega} \frac{u(x)}{|x-y|^{N+2s}} dy dx\geq 0, 
$$
as  $u=0$ in $\mathbb{R}^N\setminus \Omega$.
Am I correct?
Thank you in advance for your help.
 A: Yes, the result is true and your proof is correct, but the first inequality you have is actually an equality. Let $u\in C^2(\Omega)\cap L^\infty(\mathbb R^n)$. You write $$\int_\Omega \int_{\mathbb R^n} \frac{u(x)-u(y)}{\vert x - y \vert^{n+2s} } \, dy dx \geqslant \int_\Omega \int_{\mathbb R^n \setminus\Omega} \frac{u(x)-u(y)}{\vert x - y \vert^{n+2s} } \, dy dx $$ which is equivalent to the inequality $$\int_\Omega \int_{\Omega} \frac{u(x)-u(y)}{\vert x - y \vert^{n+2s} } \, dy dx \geqslant 0. $$ This is true, but actually $$\int_\Omega \int_{\Omega} \frac{u(x)-u(y)}{\vert x - y \vert^{n+2s} } \, dy dx = -\int_\Omega \int_{\Omega} \frac{u(y)-u(x)}{\vert y - x \vert^{n+2s} } \, dy dx = -\int_\Omega \int_{\Omega} \frac{u(\tilde x)-u(\tilde y)}{\vert \tilde x - \tilde y \vert^{n+2s} } \, d\tilde y d\tilde x$$ by making the change of coordinates $\tilde x =y$ and $\tilde y =x$, so actually $$ \int_\Omega \int_{\Omega} \frac{u(x)-u(y)}{\vert x - y \vert^{n+2s} } \, dy dx =0.$$ Thus, $$\int_\Omega (-\Delta)^s u(x) \, dx = \int_\Omega \int_{\mathbb R^n \setminus \Omega} \frac{u(x)-u(y)}{\vert x - y \vert^{n+2s}}\,dydx, \tag{$\ast$}$$ so for $u=0$ in $\mathbb R^n\setminus \Omega$ and $u \geqslant 0$ in $\Omega$, the result follows exactly as you said.

As a fun side note, often we define $$ \mathcal N_s u(x) = \int_\Omega \frac{u(x)-u(y)}{\vert x - y \vert^{n+2s}}\, dy$$ which is a sort of fractional Neumann derivative. Then $(\ast)$ can be rewritten as $$\int_\Omega (-\Delta)^s u(x) \, dx = -\int_{\mathbb R^n \setminus \Omega} \mathcal N_su(x)\, dx $$ which is a nonlocal analogue of the identity $$\int_\Omega \Delta u \, dx = \int_{\partial \Omega} \partial_\nu u \, \mathcal H^{n-1}_x, $$ see Nonlocal problems with Neumann boundary conditions by Dipierro, Ros-Oton, and Valdinoci.
