Neumann spectral fractional Laplacian in bounded domains with Lipschitz boundary
It is not difficult to show that the Poisson problem for the Neumann spectral fraction Laplacian
\[(-\Delta_N)_{SFL}^s u = f, \qquad \text{in } \Omega.\]preserves $L^2$ and $H^1$ regularity in bounded domains $\Omega$ such that $\partial \Omega$ is Lipschitz. This is useful, in particular, for numerical analysis where the domain is a polygon.
It is known (see [1] and there references therein) that Lipschitz domains have the $H^1$ extension property, meaning that there is an operator $E: H^1 (\Omega) \to H^1 (\mathbb R^n)$ such that $E(u) = u$ in $\Omega$. Hence, it is not difficult to show that the resolvent of the classical Neumann problem
\[\begin{cases} -\Delta u = f & \text{in } \Omega, \\ \frac{\partial u}{\partial n} = 0 & \text{on } \partial \Omega \end{cases}\]is an operator $(-\Delta_N)^{-1}: L^2 (\Omega) \to L^2 (\Omega)$ which is compact. Hence it admits an eigen-decomposition, which we take $L^2$ orthonormal. We have $\lambda_1 = 0$ and $\varphi_1$ constant.
With this $(-\Delta_N)^{\pm s}$ is defined as follows. If $\int_\Omega f = 0$, using the spectral decomposition the solution of with $\int_\Omega u = 0$ you get
\[u =(-\Delta_N)^{- s}f =\sum_{i>1}\lambda_i^{- s} \langle f,\varphi_i \rangle \varphi_i.\]In other words, $\langle u, \varphi_1\rangle = 0$, and for $i > 1$ we have $\langle u, \varphi_i \rangle = \lambda_i^{-s} \langle f,\varphi_i\rangle$. In this setting, you can observe
\[\begin{aligned} ||v||_{L^2}^2 &= \sum _{i \ge 1} \langle v, \varphi_i \rangle^2 \\ ||D v||_{L^2}^2 &= \langle v, -\Delta v \rangle = \sum _{i > 1} \lambda_i \langle v, \varphi_i \rangle^2 \end{aligned}\]It is therefore easy to compute
\[||D^k v||^2_{L^2} \le \lambda_2^{-s} ||D^k f||^2 _{L^2}, \qquad \text{for } k = 0,1 .\]To recover improvements of regularity, for example into “$H^{1+s}$ spaces”, one has to be careful. There are several different definitions (Sobolev-Slobodeckij, Hardy, … ). These can cause difficulties with the regularity of $\Omega$.
We point to [2].