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The question is as follows:

Let $0<\alpha<1$. Let $f(z)$ be analytic on $\mathbb{D}$. If $|f(z)-f(w)|\leq C|z-w|^{\alpha}$ for any $z,w\in\mathbb{D}$ and a constant $C$, then there exists a constant $A=A(C)<\infty$ such that $|f'(z)|\leq A(1-|z|)^{\alpha-1}$.

My first attempt was to use Schwarz-Pick. But then I realized that the function might not be bounded. Then I found an analogous question: A Pick Lemma like problem. The solution uses Cauchy integral formula. I suppose the same method should be applied here, but I don't know how to proceed. Any hint will be appreciated.

kzkzkzz
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    Hints: 1. In the integral formula in the linked answer the numerator of the integrand can be replaced by $f(\zeta)-f(z)$ without changing the value of the integral. 2. The Lipschitz condition makes it possible to extend $f$ to a continuous function on the closed unit disk. – leoli1 Aug 09 '21 at 18:53
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    (the second hint in my previous comment is actually unnecessary, it shows that $f$ is bounded which isn't needed here) – leoli1 Aug 09 '21 at 19:11

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Based on the hint given by leoli1 in the comment, I can complete the answer: \begin{equation} |f'(z)|=|\frac{1}{2\pi i}\int_{|w|=r}\dfrac{f(w)}{(w-z)^2}dw| =|\frac{1}{2\pi i}\int_{|w|=r}\dfrac{f(w)-f(z)}{(w-z)^2}dw|\leq \frac{1}{2\pi}\int_{|w|=r}\dfrac{|f(z)-f(w)|}{|z-w|^2}dw \leq \frac{1}{2\pi}\int_{|w|=r}\dfrac{C|z-w|^{\alpha}}{|z-w|^2}dw=\frac{1}{2\pi}\int_{|w|=r}C|z-w|^{\alpha-2}dw\leq\frac{1}{2\pi}\int_{|w|=r}C'r^{\alpha-2}dw=\frac{1}{2\pi}C'r^{\alpha-1}. \end{equation} Letting $r\rightarrow 1-|z|$, we are done.

kzkzkzz
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