Let $\Omega$ be bounded and smooth domain in $\mathbb{R}^n$, $s\in(0,1)$, $e_1\in \mathbb{H}^s(\Omega)$ the first eigenfunction of fractional laplacian $(-\Delta)^s$ with eigenvalue $\lambda_1>0$, in weak formulation, that is: $$ \frac{C(n,s)}{2}\int_{\mathbb{R}^n\times\mathbb{R}^n}\frac{(e_1(x)-e_1(y))(\phi(x)-\phi(y))}{|x-y|^{n+2s}}\,dx\,dy=\lambda_1\int_\Omega e_1(x)\phi(x)\,dx, \quad\forall\phi\in \mathbb{H}^s(\Omega). $$ I know that $e_1$ is continuous on whole $\mathbb{R}^n$. I want to prove that: $$ (-\Delta)^se_1(x)=\lambda_1e_1(x), \quad\forall x\in\Omega,$$ but i have no idea on how to proceed. Any help would be appreciated.

Here: $$ \mathbb{H}^s(\Omega)=\{u\in H^s(\mathbb{R}^n): u=0\,\, \text{ q.o. }\in \mathbb{R}^n\setminus\Omega\},$$ and: $$ (-\Delta)^su(x):=\frac{C(n,s)}{2}\int_{\mathbb{R}^n}\frac{2u(x)-u(x+y)-u(x-y)}{|y|^{n+2s}}\,dy,\quad\forall x\in\mathbb{R}^n, \forall u\in\mathcal{S}(\mathbb{R}^n). $$ Moreover, how i can define $(-\Delta)^s$ for less regular function?

The Cauchy process and the Steklov problemby Rodrigo Bañuelos and Tadeusz Kulczycki (JFA 2004, DOI:10.1016/j.jfa.2004.02.005) is a good source. $\endgroup$Ten equivalent definitions of the fractional Laplace operator, DOI:10.1515/fca-2017-0002. This is about definitions in all of $\mathbb R^n$, but still hopefully related. $\endgroup$2more comments