Let $\Omega$ be a bounded subset of $\mathbb{R}^n$. We know that the Laplace operator \begin{align} \Delta \colon H_0^1(\Omega) \to L^2(\Omega) \end{align} admits an inverse operator \begin{align} A \colon L^2(\Omega) \to H_0^1(\Omega) \end{align} which is bounded and then (by Poincaré inequality and Rellich embedding theorem) can be viewed as a compact operator from $L^2(\Omega)$ to $ L^2(\Omega)$.
Now I have a basic problem: since $\Delta$ is unbounded, how can $A$ be a bounded operator? Wouldn't this contradict the open mapping theorem?
Basically, the problem is the norm of the spaces where you define the operators.
The Laplace operator $\Delta$ has domain $H^2_0 (\Omega)$. If one endows this Sobolev space with its norm, then $\Delta:H^2_0 (\Omega) \to L^2(\Omega)$ is bounded. So there is no contradiction with the open mapping thm.
On the other hand, note that we cannot apply the open mapping theorem to $A: L^2(\Omega) \to H^2_0(\Omega)$, if we consider the $L^2$ norm on $H^2(\Omega)$ (since it is not a Banach space).