Consider the equation $\bar{\partial} f=g$ on the complex plane. We may assume $g$ is compactly supported, but we need the case that $g$ is only assumed to be continuous. Is there a solution to this equation? (I mean classical solution.) If yes, is it the solution given by Cauchy integral formula?


1. If $g$ is $C^1$, then we can find the answer from almost all textbooks.
2. I tried to use an approximation method, but failed to find the correct estimate.


It seems to me that the answer is yes and is widely known to experts.



Inhomogeneous Cauchy–Riemann equation on complex plane with continous right hand side
The integral operator
$$Ph(z)=-\frac{1}{\pi}\int\int h(\zeta)\left(\frac{1}{\zeta-z}-\frac{1}{z}\right)dxdy$$
acts on $L^p$, $p>2$, and the result satiafies $(Ph)\_{\overline{z}}=h$
in the sense of distributions. For continuous $h$, this equation may not
have a classical $C^1$ solution.


Edit. The following example was suggested by user @Fedja.
It is known that $(Ph)\_z=Th,$ where $T$ is the following operator:
$$Th(z)=\lim\_{\epsilon\to 0}\left(-\frac{1}{\pi}\int\int\_{|\zeta-z|>\epsilon}\frac{h(\zeta)}{(z-\zeta)^2}dxdy\right).$$
So taking
$f=Ph$, with
$$h(z)=\frac{z^2}{|z|^2\log|z|},$$
which is continuous,
we obtain $f\_z(0)=\infty$, so $f$ is not in $C^1$.


Reference for these two operators $P$ and $T$ is L. Ahlfors, Lectures on quasiconformal mappings, Chap. V, A.


