The Laplacian is an isomorphism between $W^{1,p}_0(\Omega)$ and $W^{-1,p}(\Omega)$

Question

Let $\Omega$ be a smooth open subset of $\mathbb{R}^n$. Is the Laplacian $\Delta:W^{1,p}_0(\Omega)\to W^{-1,p}(\Omega)$ an isomorphism? If so, is there a reference where I can find a proof of this result?

Answer 1

I am by no means an expert in this field, so if there is anything wrong with my answer or if I interpreted something incorrectly please let me know!

According to the paper An $L^p$-estimate for the gradient of solution of second order divergence equations by Meyers (1963), https://eudml.org/doc/83302, the (weak) Laplacian is an isomorphism between $W^{1,p}_0(\Omega)\to W^{-1,p}(\Omega)$ for certain $p$. The result can be found in Theorems 1 and 2 on Page 198 by taking $A=\mathbf{I}$, the identity operator. Simplified (by using the characterization of $W^{-1,p}(\Omega)$), the theorem states the following:

Let $n\geq 2$ and let $\Omega$ be a bounded connected open set in $\mathbb{R}^n$. Then there exists some $p_0 > 2$ such that for all $p\in[2,p_0)$ we have that if $f\in W^{-1,p}(\Omega)$ then there is a unique weak solution, $u$, to $\Delta u = f$ with $u\in W^{1,p}_0(\Omega)$ and there is a constant $C$ such that $$\|u\|_{W^{1,p}_0(\Omega)}\leq C\|f\|_{W^{-1,p}(\Omega)}.$$ Hence $\Delta^{-1}:W^{-1,p}(\Omega)\to W^{1,p}_0(\Omega)$ is well-defined, linear, and continuous. In particular, we deduce that $\Delta\circ \Delta^{-1} = \mathrm{id}_{W^{-1,p}(\Omega)}$ and $\Delta^{-1}\circ\Delta=\mathrm{id}_{W^{1,p}_0(\Omega)}$, with both maps continuous. That is to say that $\Delta:W^{1,p}_0(\Omega)\to W^{-1,p}(\Omega)$ is an isomorphism as desired.