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Is the proof given of $\text{Proposition (1)}$ vaild, and also is there an alternate way to approach this problem without relying on Leibniz Rule ?

$\text{Proposition (1)}$

Suppose that $U \subset \mathbb{C}$ is an open set. Let $F \in C^{0}(U).$ Suppose that for every $\overline D(z,r) \subset U$ and $\gamma$ the curve surrounding this disc (with counter-clockwise orientation) and all $w \in D(z,r)$ it holds that in $(1.1)$

$(1.1)$

$$F(w) = \frac{1}{2 \pi i} \oint_{\gamma} \frac{F(\zeta)}{\zeta - w}d \zeta$$

Prove that $F$ is holomorphic

In order to prove $(1.1)$, one needs the sufficient criterion for a function to be considered holomorphic this is carefully constructed in $(2.2)$

$\text{Lemma (2)}$

$\text{Definition}\, \, (2.2)$

A continuously differentiable $(C^{k})$ function $f: U \rightarrow \mathbb{C}$ defined on an open subset $U$ of $\mathbb{C}$ is said to be holomorphic if

$$\partial_{\overline z}f = 0$$

at every point of $U.$

In view of $(2.2)$ it becomes necessary to formulate the following claim in $(3.3)$

$(3.3)$

$$ \partial_{\overline z} F(w) = 0 $$

Utilizing, Differentiation Under the Integral Sign it's trivial to see that

$$\frac{1}{2 \pi i} \bigg( \partial_{\overline z} F\oint_{\gamma} \frac{F(\zeta)}{\zeta - w}d \zeta \bigg) = \frac{1}{2 \pi i} \oint_{\gamma} \partial_{\overline z} F \big(\frac{F(\zeta)}{\zeta - w} \big )d \zeta = 0$$

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    $\begingroup$ Isn't in definition 2.2 the condition $\partial_{\bar{z}}f=0$.? $\endgroup$ – Nosrati Jul 7 '18 at 20:56
  • $\begingroup$ Sorry about the typo I'll fix that $\endgroup$ – Zophikel Jul 7 '18 at 21:03
  • $\begingroup$ The last line needs $\partial_{\bar z}$ a few places, too. $\endgroup$ – Ted Shifrin Jul 7 '18 at 22:10
  • $\begingroup$ I Will address the typo thanks for noticing @TedShifrin $\endgroup$ – Zophikel Jul 8 '18 at 13:39
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HINT for an alternative approach: Use Morera's Theorem (and change the order of integration).

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