& \le c(\|f_1\|_s^2 +\|f_2\|_s^2)\\
& =c\|f\|_s^2 . \qquad \qquad \qquad \qquad \qed\end{aligned}$$

We let $N_s$ be the operator on ${W^s_{(0,q)}(\Omega)}$ defined by $$\label{box}
(\dbar\dbar^* +\dbar^*\dbar)N_s f=f$$ for all $f\in{W^s_{(0,q)}(\Omega)}$. \[Notice that the harmonic space for the operator on the left side of (\[box\]) is just the zero space—by the preceding arguments. Therefore this last condition uniquely defines $N_s$.\] We call $N_s$ the [*Neumann operator*]{} for the $\dbars$-Neumann problem. Thus we have proved that $N_s$ is a bounded operator from ${W^s_{(0,q)}(\Omega)}$ into itself, for $0<q\le n$.

[ *We want to stress the fact that the results of the present section are [*independent*]{} of the particular choice of the Sobolev inner product. In fact, the same arguments work for any $s\ge0$, not necessarily integral, and any choice of an equivalent norm in ${W^s_{(0,q)}(\Omega)}$. The form of the inner product (\[Sobolev\]) will only become relevant in the next section since we seek explicit formulas for $\dbar^*$ and its domain.* ]{}

The Hilbert space adjoint $\dbar^*$ of $\dbar$
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In this section we wish to make the $\dbars$-Neumann problem more explicit by calculating the domain of $\dbar^*$, and the operator $\dbar^*$ itself by showing how it actually operates on the $(0,q)$-forms in its domain. 