The solution of Cauchy-Riemann equation Can you give me an example that there is a $f \in C_0^{\infty}(\mathbb C)$, such that the equation $\bar \partial u=f$ has no  $C_0^{\infty}(\mathbb C)$ solution?
 A: If $f\in C^\infty_0(\mathbb{C})$, then $\displaystyle u:\zeta\mapsto \frac{1}{2i\pi}\int_{\mathbb{C}}\frac{f(z)}{z-\zeta}dz\wedge d\bar{z}$ is also in $C^\infty_0(\mathbb{C})$ and is a solution to $\bar{\partial}u=f$
A: With respect to the new bounty:
Here's a way to see this. Suppose $u$ is a compactly supported solution to $\bar{\partial}u=f$. Then we have, for large enough $R>0$
$$0=\int_{\partial D(0,R)} u(z)dz = 2i\int_{D(0,R)} \bar{\partial}u(z)\: d\bar{z}\wedge dz=2i\int_{D(0,R)} f(z)\:d\bar{z}\wedge dz$$
Taking $R\to\infty$, this implies that if a compactly supported solution to this equation exists, then $$\int_{\mathbb{C}}f(z)\:d\bar{z}\wedge dz=0$$Then, consider the standard $\mathscr{C}^{\infty}_0(\mathbb{C})$ bump function $\varphi$ such that $\varphi\geq0, \text{supp}(\varphi)\subset \mathbb{D}$, and $$\int_{\mathbb{C}} \varphi(z)\: d\bar{z}\wedge dz=1$$ (the construction of which is in most graduate real analysis texts). Then there exists no compactly supported $u$ such that $\bar{\partial}u=\varphi$.
Edit: Here's the explanation of the comment. Suppose $\int_{\mathbb{C}}z^nf(z)dA=0$ for all $n\geq0$. Let $D$ be a large enough ball centered at $0$ such that $\text{supp}(f)\subset D$. Consider the solution
$$u(z)=\frac{1}{\pi}\int_{D}\frac{f(\omega)}{z-\omega}dA(\omega)$$If $|z|>|\omega|$ (i.e. $z\not\in \bar{D}$). Then \begin{align*}u(z)&=\frac{1}{\pi}\int_{D}\frac{f(\omega)}{z-\omega}dA(\omega)=\frac{1}{\pi}\int_{D}\frac{f(\omega)}{z(1-\frac{\omega}{z})}dA(\omega)=\frac{1}{\pi}\int_{D}f(\omega)\sum_{n=0}^{\infty}\omega^nz^{-n-1}dA(\omega)\\&=\frac{1}{\pi}\sum_{n=0}^{\infty}\left(\int_{D}f(\omega)\omega^ndA(\omega)\right)z^{-n-1}=\frac{1}{\pi}\sum_{n=0}^{\infty}\left(\int_{\mathbb{C}}f(\omega)\omega^ndA(\omega)\right)z^{-n-1}\\&=0\end{align*}So $u=0$ outside of $D$, so $\text{supp}(u)\subset D$ and thus $u$ is compactly supported.
A: We take a $f\in C^\infty_0(\mathbb{C})$ such that supp$f$ is contained in the unit ball and $f \geq 0$ and $f=1$ on some smaller ball. 
If we have a compactly-supported solution for the equation, it has to be $\displaystyle u:\zeta\mapsto \frac{1}{2i\pi}\int_{\mathbb{C}}\frac{f(z)}{z-\zeta}dz\wedge d\bar{z}$ since $\mathop {\lim }\limits_{\left| \zeta  \right| \to \infty } u(\zeta ) = \infty $.
But if we examine the integral using polar coordinate more carefully, we will find Re$u(x,0)>0$ when $x$ is sufficiently large.
