Basis of $W^{1,p}_0\cap L^2$ using $(\lambda_i, v)_{H^s_0} = \mu_i(\lambda_i, v)_{L^2}$

Question

Let $p > 1$. Define $\lambda_i$ by the eigenfunctions of the problem $$(\lambda_i, v)_{H^s_0} = \mu_i(\lambda_i, v)_{L^2}\quad\text{for all $v \in H^s_0(\Omega)$},$$ where $s$ is chosen large enough so that $H^s_0(\Omega) \subset W^{1,p}_0(\Omega)$ continuously (this is what is done in page 3 (or 999) of this.)

Is there a nice way to see why the $\lambda_i \in H^s_0(\Omega)$ are smooth, and why they form a basis of $W^{1,p}_0 \cap L^2$ which is orthonormal in $L^2$?

I'm not even sure what the operator defining the eigenvalue problem is.

Edit: comments on this thread refer to the old version.

Answer 1

Define $A:H^s_0 \to (H^s_0)^*$ by $\langle Au, v \rangle = (u,v)_{H^s_0}$ and consider the equation $\langle Au, v \rangle = (f,v){L^2}.$

$u$ exists uniquely, and the solution map $T(f) = u$, $T:L^2 \to L^2$ is compact. Then one can apply the Hilbert-Schmidt theorem.