The Cauchy Integral Formula for a disk is stated as follows:
Let $f$: D $\to \mathbb C$ and $ z_0\in D$ If $f$ is analytic, then for every $ r\gt0$ with $\overline{B_r(z_0)} \subset D$ we have: $$f(z)=\frac{1}{2\pi i}\int_{\partial B_r(z_0)}\frac{f(w)}{w-z}dw$$
Is the converse true? That is, are all functions that satisfy this integral equation, automatically analytic? Or is there some function that satisfies this relation, but is not analytic?
I believe the answer is yes, as you can move the $\frac{\partial}{\partial \overline{z}}$ operator through the integral sign, and since $\frac{\partial}{\partial \overline{z}}(\frac{f(w)}{w-z})=0$, we have $\frac{\partial}{\partial \overline{z}}f(z)=0$, i.e $f$ is holomorphic. Another argument is to expand $\frac{f(w)}{w-z}$ as a power series (I forget how it goes exactly, but it is in every complex variable book) and this shows that $f(z)$ is analytic. And this is precisely the argument that shows analytic and holomorphic are the same thing.
Yes. The cleanest proof (that is, the proof where actually justifying the picky details is easiest) is probably using Morera's Theorem.
