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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?

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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.

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Yes. The cleanest proof (that is, the proof where actually justifying the picky details is easiest) is probably using Morera's Theorem.

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