If $u\in C_{0}^{\infty}(\mathbb{C})$ and $\frac{\partial u}{\partial\overline{z}}$ is a real valued nonnegative function, then $u\equiv 0$. The problem:
If $u\in C_{0}^{\infty}(\mathbb{C})$ and $\frac{\partial u}{\partial\overline{z}}$ is a real valued nonnegative function, then $u\equiv 0$.
Thoughts:
I believe this is done in the following fashion:
First, I am supposed to prove that if $u\in C_{0}^{\infty}(\mathbb{C})$ and $n\geq 0$ is an integer, then
$$\iint_{\mathbb{C}}\frac{\partial u}{\partial \overline{z}}\cdot z^n \,dz\wedge d \overline{z}=0.$$
Second, I am suppose to prove, conversely, that if $f\in C_{0}^{\infty}(\mathbb{C})$ and
$$\iint_{\mathbb{C}} f(z)\cdot z^n d z\wedge d \overline{z}=0$$
for all $n\geq 0$, then
$$u(w)=\frac{1}{2\pi i}\iint_{\mathbb{C}}\frac{f(z)}{z-w} d z\wedge d \overline{z}$$ has compact support.
Finally, I believe I am supposed to use this fact to conclude and resolve my problem. I am not sure how one might achieve any of the three steps.
 A: The first proposition follows from Stokes theorem or integration by parts as $$\int_{\mathbb{C}} \frac{\partial u}{\partial \overline{z}} z^n\,dz \wedge \,d\overline{z} = - \int_{\mathbb{C}} u \frac{\partial }{\partial \overline{z}} (z^n)\,dz \wedge \,d\overline{z} = 0.$$
The second one follows from observing that if $\text{supp} (f) \subset B_R(0)$ then for $|w| > R$ we have \begin{align*} u(w) = \int_{\mathbb{C}} \frac{f(z)}{z-w}\,dz \wedge \,d\overline{z} &= -\frac{1}{w} \int_{\mathbb{C}} \frac{f(z)}{1 - (z/w)}\,dz \wedge \,d\overline{z} \\&= -\frac{1}{w}\sum\limits_{n=0}^{\infty} \frac{1}{w^n}\int_{\mathbb{C}} f(z)z^n\,dz \wedge \,d\overline{z} = 0.\end{align*}
Combined these two propositions say that the solution to $\overline{\partial_z} u = f$ for $f \in C_c^{\infty} (\mathbb{C})$ has compact support iff $\displaystyle \int_{\mathbb{C}} f(z)z^n \,dz \wedge \,d\overline{z} = 0, \, \forall n \ge 0$.
Now, suppose $\overline{\partial_z} u = f \ge 0$ (real valued and positive) then $\partial_z \overline{u} = \overline{\overline{\partial_z} u} = \overline{f} = f = \overline{\partial}_z u$. Therefore, using integration by parts we have $$0 = \int_{\mathbb{C}} (\overline{\partial_z} u) z^n\,dz \wedge \,d\overline{z} = \int_{\mathbb{C}} (\partial_z \overline{u}) z^n\,dz \wedge \,d\overline{z} = -n\int_{\mathbb{C}} \overline{u} z^{n-1}\,dz \wedge \,d\overline{z},\, \forall n \ge 1.$$ That is if $v$ solves the $\overline{\partial}$-problem $$\overline{\partial_z} v = \overline{u}$$ then, $v$ is a smooth function with compact support by the equivalence established above. Furthermore, taking conjugate on both sides we note that $$\partial_z \overline{v} = u \implies \Delta \overline{v} = 4 \overline{\partial_z}\partial_z (\overline{v}) = 4\overline{\partial_z} u = 4f \ge 0.$$ That is $\overline{v}$ is a subharmonic function with compact support and hence must be identically $0$. This proves $u \equiv 0$.
