a function that is analytic which satisfies a Lipschitz condition

Question

Let $f$ be a function that is analytic in an open set U and satisfies a Lipschitz condition of order $\alpha$ ($0< \alpha \leq 1),$ i.e., $|f(z_2)-f(z_1)|< m|z_1-z_2|^{\alpha}$ for $z_1, z_2 \in U,$ $m$ a constant. If the open disk $D(z_0,r )\subset U$ then $|f'(z_0)|\leq mr^{\alpha-1}. $

I haven't worked much with "Lipschitz condition of order $\alpha$"so I do not understand the question very much. Maybe using Cauchy's estimatives...

Thanks

Answer 1

\begin{align*} |f'(z_{0})|&=\left|\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)-f(z_{0})}{(z-z_{0})^{2}}dz+\frac{1}{2\pi i}\int_{\gamma}\frac{f(z_{0})}{(z-z_{0})^{2}}dz\right|\\ &\leq\frac{1}{2\pi}\int_{0}^{1}\frac{|f(z_{0}+re^{2\pi it})-f(z_{0})|}{r^{2}}2\pi rdt\\ &\leq m\int_{0}^{1}\frac{r^{\alpha}}{r}dt\\ &=mr^{\alpha-1}. \end{align*}