Let $U$ be a disc (of possibly infinite radius) in $\mathbb{C}$. Then $U$ has the following property:
(P) For every $g \in C^{\infty}_c(U)$, there is an $f \in C^{\infty}_c(U)$ such that $\frac{\partial}{\partial \bar z}f = g$.
Does the argument (A) below establish the property (Q) of $U$?
(Q) For every $g \in C^{\infty}(U)$, there is an $f \in C^{\infty}(U)$ such that $\frac{\partial}{\partial \bar z}f = g$.
(A): Let $\mathcal{U} = \lbrace U_i \rbrace_{I}$ be a locally finite open cover of $U$ such that each $U_i$ has compact closure. Let $\lbrace \phi_i \rbrace$ be a $C^{\infty}$ partition of unity on $U$ such that $\text{supp}(\phi_i) \subset U_i$ for each $i \in I$. Denote $g_i:=\phi_i g$.
By (P), there are $f_i \in C^{\infty}_c(U)$ such that $\frac{\partial}{\partial \bar z}f_i = g_i$ on $U$.
It is meaningful to write $g = \sum_i g_i$ on $U$, since $\mathcal{U}$ is locally finite. Set $f := \sum_i f_i$. We confirm the equation in (Q) at a point $x \in U$. Note that $x$ has a neighborhood $V$ such that, for a fixed subset $I_x \subset I$, we have:
- $f(y) = \sum_{i \in I_x} f_i(y)$ for all $y \in V$
- $\sum_{i \in I_x} g_i(x) = g(x)$
Thus:
$$ \frac{\partial f}{\partial \bar z}(x) = \sum_{i \in I_x} \frac{\partial f_i}{\partial \bar z}(x) = \sum_{i \in I_x} g_i(x) = g(x)$$
The question is, once again: Does the argument (A) establish the property (Q) of $U$?