Considering $(-\Delta)^s: H^{2s}(\Omega) \to L^2(\Omega)$, is it possible to show that this operator is closed (continuous)? For instance, taking a sequence $(u_n) \subset H^{2s}(\Omega)$ with $u_n \to u$ in $H^{2s}(\Omega)$, we need to show that $(-\Delta)^su_n \to (-\Delta)^su$ in $L^2(\Omega)$.
Attempt:
Consider $$\lim_{n \to +\infty}\int_\Omega |(-\Delta^s)u_n(x)|^2 dx = \lim_{n \to +\infty} \int_\Omega \Big(\int_\Omega \frac{u_n(x) - u_n(y)}{|x - y|^{N + 2s}} dy\Big)^2 dx.$$
So my idea would be to use some inequality (perhaps something similar to the Poincaré inequality) to obtain the norm of $u_n$ in $H^{2s}(\Omega)$ on the right side. But, I don't know how to get rid of this one on the right side of the expression above.