Dolbeault cohomology and analytic regularity Let $M$ be a complex analytic $n$-manifold. The Dolbeault cohomology complex is defined using a quotient space of smooth differential forms. My question is : would it make a big difference if we were to use $C^n$ sections instead, or $C^2$, or even differential forms in Sobolev spaces ? 
(obviously with at least enough regularity to define $\overline{\partial}$)
EDIT
A bit more precision : if I were to consider the quotient space
$A/B$, where $A$ is the space of $C^1$ $(p,q)$-forms $\alpha$ for which $\overline{\partial} \alpha=0$ and $B$ is the set of $\overline{\partial}$ of $C^2 (p,q-1)$-forms, would I get the same dimension as $H^{p,q}$ ? What if I were to replace $C^2$ and $C^1$ by Sobolev spaces with distributional derivatives of order $2$ and $1$ in $L^p$ ?
 A: You have to define a complex of chains. If you are considering class $C^1$ forms then $\bar{\partial}$ does not have values in class $C^1$ forms, hence you can hardly define $\bar{\partial} \circ \bar{\partial}$, which is crucial because in cohomology you ask that $Im(\bar{\partial}) \subset Ker(\bar{\partial})$.
A: Let $F$ be a locally free sheaf on an $n$-dimensional complex manifold $X$.
The obvious is to consider the complex 
$$
O_X(F) \to A_X^{0,0,r}(F) \to A_X^{0,1,r-1}(F)\to \dots \to A_X^{0,n,r-n}(F) \to 0
$$
where $A_X^{p,q,s}(F)$ denotes the space of $(p,q)$ forms on $X$ with values in $F$ of class $C^s$, and all differentials are $\overline \partial$.
This does not work if $r$ is an integer, but does work if $r=k+\alpha$ for $k>n$ an integer and $0<\alpha<1$, i.e., if the appropriate derivatives are Holder of exponent $\alpha$.  This complex does then compute the cohomology of the locally free sheaf $F$. This is because in this Holder class the Dolbeault resolution is exact and fine.
But there is a loss of domain when convolving with the Cauchy kernel, and it is harder to make this work on open domains, however pseudo convex.   
