Duality argument for elliptic regularity

Question

M. Dauge proved in [1] the regularity property "$\Delta u \in (W^1_{p'})^*$ $\Rightarrow$ $u \in W^1_p$" for Dirichlet and Neumann problem in domains with piecewise smooth boundaries, for $p>3$. (See Corollary 3.10.) Then the author states (Remark 3.11):

By a duality argument it is easy to deduce from previous statement that the Laplace operator [between $W^1_p$ and $(W^1_{p'})^*$] is an isomorphism [...] when $3/2-\epsilon <p<3+\epsilon$ for a $\epsilon > 0$.

What is “a duality argument” and which theorem should be used in this case?

[1] Dauge, M. Neumann and mixed problems on curvilinear polyhedra. Integral Equations Operator Theory 15 (1992), no. 2, 227–261.

Answer 1

The property "$\Delta u \in (W^1_{p'})^* \implies u \in W^1_p$" implies that the weak Laplacian $\Delta \colon W^1_p \to (W^1_{p'})^*$ has a continuous linear inverse, when $p>2$.

This follows from the definition of $\Delta$ in $(W^1_{p'})^*$ by restriction from $(W^1_2)^*$: Note that $(W^1_{p'})^* \subset (W^1_2)^*$. The domain $D_p$ of $\Delta$ in $(W^1_{p'})^*$ consists of exactly those $u \in W^1_2 \cap (W^1_{p'})^*$ for which $\Delta u \in (W^1_{p'})^*$. Since $\Delta \colon W^1_2 \to (W^1_2)^*$ is an isomorphism (Lax Milgram lemma), $\Delta \colon D_p \to (W^1_{p'})^*$ is certainly still bijective. The elliptic regularity question is now whether $D_p$ in fact coincides with $W^1_p$. (In fact, if we equip $D_p$ with the graph norm induced by $\Delta$, then the Laplacian also admits a continuous inverse in this setting by the open mapping theorem.) This is exactly what is answered in the affirmative by "$\Delta u \in (W^1_{p'})^* \implies u \in W^1_p$".

Accordingly, if $\Delta \colon W^1_p \to (W^1_{p'})^*$ has a continuous linear inverse, then its adjoint operator $\Delta^* \colon W^1_{p'} \to (W^1_p)^*$ also has a continuous linear inverse. But this adjoint operator is exactly what is meant by the Laplacian $\Delta$ on $(W^1_p)^*$, which is a definition by duality.