$C^{\alpha}$ estimates for spectral fractional Laplacian Consider the following problem with $\Omega_{\epsilon}$ is bounded for each fixed $\epsilon>0$ and $s\in (0, 1)$. Let $u_{\epsilon}$ be a classical solution of
\begin{equation}
  \ \  \left\{\begin{aligned}
(-\Delta)^su_{\epsilon}&=f_{\epsilon}  &&\text{in } \Omega_{\epsilon}  \\
u_{\epsilon} & =g_{\epsilon} &&\text{ in } \partial \Omega_{\epsilon} 
  \end{aligned}
  \right.
\end{equation} 
Suppose $\Omega_{\epsilon}\to \mathbb R^N$ as ${\epsilon}\to 0.$ Let $u_{\epsilon}, f_{\epsilon}$ is uniformly bounded in every compact set. Is it possible to derive local uniform $C^{\alpha}$ estimates for $u_{\epsilon}$. Any reference is welcome. 
 A: (Too long for a comment.)
Not sure if this will work for you: you can "cheat" and switch to the fractional Laplacian in full space. I mean, if you extend your $u_\epsilon$ to all of $\mathbb{R}^N$ so that $u_\epsilon(x) = 0$ for $x \notin \Omega_\epsilon$, then $$(-\Delta)^s u_\epsilon = \tilde f_\epsilon$$ for some $\tilde f_\epsilon$, which is not much different from $f_\epsilon$ except near the boundary of $\Omega_\epsilon$. I mean, $$\tilde f_\epsilon = f_\epsilon + ((-\Delta)^s - (-\Delta_{\Omega_\epsilon})^s) u_\epsilon,$$ and the kernel of the non-local operator $(-\Delta)^s - (-\Delta_{\Omega_\epsilon})^s$ is given by $$K(x, y) = c_{N,s} \int_0^\infty (p_t(y - x) - p_t^{\Omega_\epsilon}(y - x)) t^{-1 - s} ds ,$$ where $p_t$ is the heat kernel for $\Delta$, while $p_t^\Omega$ is the Dirichlet heat kernel for $\Delta$ in $\Omega$. A simple probabilistic bound for $p_t(y - x) - p_t^\Omega(x, y)$ in terms of the hitting time of $\partial \Omega$ should now give a decent estimate for $K(x, y)$, and the result you need should follow from standard regularity theory for $(-\Delta)^s$. This ideas are inspired by a series of articles by Song and Vondraček, see, e.g., here.
By the way, how exactly is your "spectral fractional Laplacian" defined so that it includes non-zero boundary condition $g_\epsilon$? (In the above I assumed that $g_\epsilon = 0$).
