Consider the fractional Laplacian defined by
$$(-\Delta)^s u(x) = P.V. \int_{\mathbb{R}^N} \frac{u(x) - u(y)}{|x - y|^{N + 2s}}dy, \ s \in (0,1).$$ Also consider that
$$D((-\Delta)^s) = \{u \in H^s(\Omega); (-\Delta)^su \in L^2(\Omega)\},$$
for some $s \in (0,1)$. Here, $\Omega \subset \mathbb{R}^N$ is a bounded and smooth domain with $u = 0$ in $\mathbb{R}^N \backslash \Omega$ and $N >2s$. The space $H^s(\Omega)$ is a usual fractional Sobolev space equipped with the norm $\|u\|_{H^s(\Omega)} = \int_\Omega u^2(x)dx + \int_\Omega \int_\Omega \frac{|u(x) - u(y)|^2}{|x - y|^{N + 2s}}dydx$.
I need to show that the operator $(-\Delta)^s : D((-\Delta)^s) \subset L^2(\Omega) \to L^2(\Omega) $ is m-accretive, that is that is, $-(-\Delta)^s$ is dissipative and that there is $\lambda_0 > 0$ such that the range $R(\lambda_0I + (-\Delta)^s) = L^2(\Omega)$. I managed to prove the dissipative. That is, $$\left<-(-\Delta)^s)u, u\right>_{L^2(\Omega)} = -C\left<u, u\right>_{H^s(\Omega)} \leqslant 0,$$ for all $u \in D((-\Delta)^s)$ and some $C > 0$.
But I can't show that the second part. My idea is to build a coercive continuous bilinear form and use Lax-Milgram to prove the maximality of the operator. Namely,
$a: X \times X \to \mathbb{R}$, where $a(u,v) = \left<(-\Delta^s)u+ u, v\right>_{L^2(\Omega)}$.
However, I have a doubt in which Hilbert space $X$ this bilinear form would be defined. If this is not possible, I can also obtain an extension of this operator so that this new operator is m-accretive.
