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
$$\frac{1}{2 \pi i} \bigg( \partial_{\overline z} F\oint_{\gamma} \frac{F(\zeta)}{\zeta – w}d \zeta \bigg) = \frac{1}{2 \pi i} \oint_{\gamma} \partial_{\overline z} F \big(\frac{F(\zeta)}{\zeta – w} \big )d \zeta = 0$$
Best Answer
HINT for an alternative approach: Use Morera's Theorem (and change the order of integration).