In the proof of a proposition on the exchangeability of complex differentiation and integration with regards to parameter integrals I am stuck at a step involving an upper bound of the difference quotient involving the Cauchy Integral Formula.
$\textbf{Proposition}$ Let $(X,\mathcal{A}$, $\mu$) be a measure space, $\Omega \subseteq \mathbb{C}$ an open set and $f: X \times \Omega \to \mathbb{C}$.
For $z \in \mathbb{C}$ define $F(z) = \int_X f(x,z) d\mu(x)$.
Furthermore assume that
- $\forall z \in \Omega$ the mapping $x \to f(x,z)$ is measurable,
- there exists a null set $N \in \mathcal{A}$ such that the mapping $z \to f(x,z)$ is holomorphic on $\Omega$ $\forall x \in N^c$,
- there exists an integrable function $g(x)$ such that $|f(x,z)| \leq g(x)$ $\forall x \in N^c$ $\forall z \in \Omega$.
Then $f$ is holomorphic on $\Omega$ and for $m \in \mathbb{N}_0$ we have $$ \left(\frac{d}{dz}\right)^m F(z) = \int_X \left(\frac{\partial}{\partial z}\right)^m f(x,z) d\mu(x).$$
An induction follows and in the step from $m$ to $m+1$ the following argument is used.
Let $(z_n)_{n\geq 1}$ be a sequence convergent to $z_0$, $z_n \in U_{2r}(z_0)$. By the Cauchy Integral Formula $$\frac{f^{(m)}(x, z_n)-f^{(m)}(x,z_0)}{z_n-z_0} = \frac{m!}{2\pi i (z_n-z_0)} \int_{\mathcal{C}} \frac{f(x,\zeta)}{(\zeta-z_n)^{m+1}}-\frac{f(x,\zeta)}{(\zeta-z_0)^{m+1}} d\zeta$$ with $\mathcal{C} = \{\zeta : |\zeta-z_0|=2r\}$.
And then the following upper bound is given: $$\left|\frac{f^{(m)}(x, z_n)-f^{(m)}(x,z_0)}{z_n-z_0}\right| \leq \frac{m!g(x)}{2\pi |z_n-z_0|} \int_{\mathcal{C}} \frac{|z_n-z_0|(\sum_{j=0}^m |\zeta-z_n|^j|\zeta-z_0|^{m-j})}{|\zeta-z_n|^{m+1}|\zeta-z_0|^{m+1}}d|\zeta|$$
Any hint to how this last bound is derived would be very much appreciated.
