\sum_{KkJ}\e{K}{kJ}\int_{b\Omega} D^\alpha
\phi_J \overline{D^\alpha \psi_K}\pd{\varrho}{\bar z_k} \, ;$$ so it follows that $$\label{bnry-eq-K}
\la \phi,\K\psi\ra_s = \sum_{0\le|\alpha|\le s} \CP
\sum_{KkJ}\e{K}{kJ}\int_{b\Omega} D^\alpha 
\phi_J \overline{D^\alpha \psi_K}\pd{\varrho}{\bar z_k} .$$ By choosing $\phi$ with compact support in $\Omega$ we find that $\K\psi$ satisfies $$\begin{aligned}
0 = & \la\phi,\K\psi\ra_s \\
= & \sum_{|J|=q}\sum_{0\le|\alpha|\le s} \CP
\int_\Omega D^\alpha \phi_J \overline{D^\alpha (\K\psi)_J} \\
= & \sum_{|J|=q}\sum_{0\le|\alpha|\le s} (-1)^{|\alpha|} \CP
    \int_\Omega \phi_J \overline{D^{2\alpha} (\K\psi)_J} .\end{aligned}$$ Since this holds for all $\phi\in C^\infty_{(0,q)} (\Omega)$ with compact support in $\Omega$, we see that $(\K\psi)_J$ must satisfy the equation $$0=\sum_{0\le|\alpha|\le s}(-1)^{|\alpha|}\CP D^{2\alpha}(\K\psi)_J 
=\sum_{j=0}^{s} (-\Delta)^j (\K\psi)_J \quad\text{on }\Omega$$ for all $J$, which is the equation on the interior of $\Omega$ that appears in (\[BV\]).

Now we move on to consider the boundary conditions that $\K\psi$ must satisfy. For $\phi\in C^\infty_{(0,q)}(\overline{\Omega})$, by repeatedly applying Green’s theorem to the left hand side of equation (\[bnry-eq-K\]), and recalling equation (\[iteration\]), we have $$\begin{aligned}
\la \phi,\K\psi\ra_s 
& = \sum_{|J|=q} \biggl ( \la \phi_J, (\K\psi)_J \ra_0
+\sum_{j=1}^{2n}\la D_j \phi_J ,D_j (\K\psi)_J \ra_{s-1} \biggr) \\
& = \sum_{|J|=q}\biggl ( \int_\Omega \phi_J \overline{(\K\psi)_J} 
+\sum_{i=1}^{2n} \int_\Omega D_i \phi_J \overline{D_i (\K\psi)_J} \\ 
& \qquad\qquad + \sum_{1\le|\beta|\le s-1} \CPb \sum_{i=1}^{2n} 
\int_\Omega D_i D^\beta \phi_J \overline{D_i 
D^\beta (\K\psi)_J} \biggr )  \\ 
& = 