I'm trying to understand the general interaction/duality between Fredholm and compact operators and I ran into the following:
Let $L$ be the Laplacian or some elliptic operator on the Sobolev space $H=H^2(\Omega)\cap H_0^1(\Omega)$. Then for some positive $\mu_0$, $L+\mu_0$ is invertible with compact inverse, call it $K$. That is $KL+\mu_0K=\text{id}_{H}$ and $LK+\mu_0K=\text{id}_{L^2(\Omega)}$.
Now this would imply that $K$ is invertible modulo a compact operator (a scalar multiple of itself), so that $K$ is Fredholm. But surely a strictly singular operator cannot be Fredholm. Where is the error?
SOLVED: due to Rellich-Kondrakov, the inverse $K$ is compact from $L^2(\Omega)$ to $L^2(\Omega)$, but not from $L^2(\Omega)$ to $H$. The compact $K$ is the inverse of the unbounded Laplacian, which is obviously not Fredholm.