v=g$, and we have the estimate $$\| v\|_s \le c_s \|g\|_s .$$ We can apply the same argument to the $\dbar^*$-equation, i.e. given any $f$ with $\dbar^*_{(0,q)}f=0$, we can find $u$ orthogonal to $\ker\dbar^*_{(0,q+1)}$ such that $\dbar^*_{(0,q+1)} u=f$, with the estimate $$\| u\|_s \le c_s \|f\|_s .$$ We shall call such solutions $u$ and $v$ the $s$-[*canonical*]{} solution to the $\dbar$ and $\dbar^*$ equation, respectively.

We now establish the solvability of the $\dbars$-Neumann problem. We shall suppress the subscripts on the operators $\dbar$ and $\dbar^*$ (used to denote the space of forms that is being acted upon), since this will be clear from context. Let $f\in{W^s_{(0,q)}(\Omega)}$. Then $f$ can be uniquely written as $f=f_1+f_2$ with $f_1 \in \rg\dbar$ and $f_2\in\rg \dbar^*$. Let $g_1,g_2$ be the canonical solution of $\dbar g_1=f_1$, and $\dbar^* g_2=f_2$, respectively. Since $g_1\perp \ker \dbar$ we have that $g_1\in\rg\dbar^*$, and therefore $\dbar^* g_1=0$. Analogously, $g_2\in\rg\dbar$ and $\dbar g_2=0$.

Thus we can canonically select $u_1,u_2$ such that $\dbar^* u_1=g_1$ and $\dbar u_2 =g_2$. Setting $u=u_1+u_2$ we obtain that $$(\dbar\dbar^* +\dbar^* \dbar)u=f,$$ and the desired estimate follows from the corresponding ones for $\dbar$ and $\dbar^*$: $$\begin{aligned}
\|
u\|_s^2 
& = \|u_1\|_s^2 +\|u_2\|_s^2 \\
& \le c( \|g_1\|_s^2 +\|g_2\|_s^2 )\\
