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 $s \in (0,1)$. I want to show that the operator $(-\Delta)^s : D((-\Delta)^s) \subset L^2(\Omega) \to L^2(\Omega)$ is closed. Here, $\Omega \subset \mathbb{R}^N$ is a bounded and smooth domain with $u = 0$ in $\mathbb{R}^N \backslash \Omega$.
Attempt:
Let $(u_n) \subset D((-\Delta)^s) $ with $u_n\to u$ in $L^2(\Omega)$ and $(-\Delta)^su_n \to v$ in $L^2(\Omega)$ . By Fatou's Lemma, I was able to show that $u \in D((-\Delta)^s)$. To show that $(-\Delta)^su = v$ a.e on $\Omega$, observe that since $u_n \to u$ in $L^2(\Omega)$, there is a subsequence $(u_{n_k})$ such that $u_{n_k} \to u$ a.e on $\Omega$ and $g \in L^2(\Omega)$ such that $|u_{n_k}| \leqslant g$ a.e. on $\Omega$. Thus
$$\frac{u_{n_k}(x) - u_{n_k}(y)}{|x - y|^{N + 2s}} \to\frac{u(x) - u(y)}{|x - y|^{N + 2s}} \ \mbox{a.e} \ on \ \Omega.$$ But, to use the Dominated Convergence Theorem and show that
$$\int_{\Omega} \frac{u_{n_k}(x) - u_{n_k}(y)}{|x - y|^{N + 2s}}dy \to \int_{\Omega} \frac{u(x) - u(y)}{|x - y|^{N + 2s}}dy,$$
we need to show that the above integrand is dominated by an integrable function. For example, we have
$$\frac{|u_{n_k}(x) - u_{n_k}(y)|}{|x - y|^{N + 2s}}\leqslant \frac{g(x) + g(y)}{|x - y|^{N + 2s}} \ \mbox{a.e} \ on \ \Omega.$$
But there is no guarantee that the function $\frac{g(x) + g(y)}{|x - y|^{N + 2s}} $ is integrable due to the singularity of $\frac{1}{|x - y|^{N + 2s}}$. Or is there?
