Let $r \in(0,1)$, $a$ is real.
$$\left | (1-z)^{a}-(1-w)^{a}\right |\leq C \left | z-w\right |$$
for all $\left | z\right |\leq r$ and $\left | w\right |\leq r$.
Here is my idea:
We can see that $(1-z)^{a}$is an analytic function($\left | z\right |\leq r$). So we have $f(z)=\sum_{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}z^{n}$,and $f^{(n)}(0)=(-1)^{n}a(a-1)\cdots (a-n+1)$. So we have$$(1-z)^{a}-(1-w)^{a}=\sum_{n=0}^{\infty }\frac{(-1)^{n}a(a-1)\cdots (a-n+1)}{n!}(z^{n}-w^{n})=\sum_{n=0}^{\infty }\frac{(-1)^{n}a(a-1)\cdots (a-n+1)}{n!}(z-w)(z^{n-1}+\cdots w^{n-1}).$$ And then we have$$|(1-z)^{a}-(1-w)^{a}|\leq \sum_{n=0}^{\infty }\frac{(-1)^{n}a(a-1)\cdots (a-n+1)}{n!}|z-w|nr^{n}.$$ But how to continue? How can we prove this series is convergent?
Some thoughts (Written in 2023/06/15)
We have \begin{align*} \left|\frac{(1 - z)^a - (1 - w)^a}{z - w}\right| &= \left|\frac{1}{z - w}\int_{w}^z a(1 - x)^{a - 1}\,\mathrm{d} x\right|\\ &= \left|\int_0^1 a\Big(1 - (w + t(z - w))\Big)^{a - 1}\,\mathrm{d} t\right|\\ &\le |a|\int_0^1 \left|1 - (w + t(z - w))\right|^{a-1}\,\mathrm{d} t\\ &\le |a|\max((1+r)^{a-1}, (1 - r)^{a-1}) \end{align*} where we use $|1 - (w + t(z - w))| \le 1 + |w + t(z - w)| \le 1 + r$ and $|1 - (w + t(z - w))| \ge 1 - |w + t(z-w)| \ge 1 - r$.