operator $G_s$ is of order $2$, so of the same order as the complex Laplacian $\Box$. Moreover notice that $G_s u$ only depends on the boundary values of $u$ and $\dbar u$ and their derivatives up to order $2s$, and that in general $G_s$ is not diagonal. An analysis of the analogue of the operator $G_s$ in the case of the de Rham complex, appears in [@FKP1].

[Proof of Theorem 3.1]{} Let $\phi\in C^\infty_{(0,q)} (\overline{\Omega})$ and $\psi\in C^\infty_{(0,q+1)} (\overline{\Omega})$. Using Green’s formula we have $$\begin{aligned}
\label{>>>}
\la \dbar\phi,\psi\ra_s
= & \la \phi,\dbar^* \psi\ra_s = \la \phi,\vartheta \psi\ra_s
    +\la\phi,\K\psi\ra_s \notag \\
= & \la \phi,\vartheta \psi\ra_s + \sum_{0\le|\alpha|\le s} \CP
    \sum_{KkI} \e{K}{kI}
\int_{b\Omega}D^\alpha \phi_I \overline{D^\alpha \psi_K}
\pd{\varrho}{\bar z_k} .\end{aligned}$$ Recall that the $(0,q+1)$ form $\psi$ belongs to $\dom \dbar^*$ if and only if there exists a constant $C_\psi >0$ such that $|\la \dbar \phi,\psi\ra_s|\le C_\psi \|\phi\|_s$ for all $\phi\in\dom\dbar$. Hence $\psi\in\dom\dbar^*$ if and only if the boundary terms in the calculation (\[&gt;&gt;&gt;\]) above can be bounded by $C_\psi \|\phi\|_s$. By the Sobolev trace theorem we can bound the terms of the form $$\int_{b\Omega} D^\alpha \phi_I \overline{D^\alpha \psi
_K}
\pd{\varrho}{\bar z_k}$$ when $|\alpha|\le s-1$. Thus it suffices to consider the sum $$\sum_{|\alpha|= 