TITLE: Inhomogeneous Cauchy–Riemann equation on complex plane with continous right hand side
QUESTION [5 upvotes]: 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?

If $g$ is $C^1$, then we can find the answer from almost all textbooks.

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.

REPLY [5 votes]: 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.