Let $f: G \to \mathbb{C}$ be holomorphic, where $G$ is a domain subset
of $\mathbb{C}$. Let $|z-a| \leq R \subset G$. Show that if $|f(z)| \leq M$ for all $z \in |z-a|=R$ then for any $z_1,z_2 \in \{|z-a| \leq \frac{R}{2} \}$ then $|f(z_1)-f(z_2)| \leq \frac{4M}{R} |z_1-z_2|$
By Cauchy formula we get $f(z_i)= \frac{1}{2\pi i} \int_{|z-a| = \frac{R}{2}} \frac{f(z)}{z-z_i} dz$ for $i=1,2$. Then by ML-estimate $|f(z_1)-f(z_2)|=\frac{1}{2\pi} |\int_{|z-a| = \frac{R}{2}} f(z) \frac{z_1-z_2}{(z-z_1)(z-z_2)}| dz \leq \frac{R}{2}M\frac{|z_1-z_2|}{R^2}$.
It is not the same result in the question, I wonder if I have any mistake?
Best Answer
You should apply Cauchy's formula to the circle with radius $R$ around $a$: $$ |f(z_1)-f(z_2)|=\frac{1}{2\pi} \left|\int_{|z-a| = R} f(z) \frac{z_1-z_2}{(z-z_1)(z-z_2)}\right| dz $$ Then use that $|z-z_k| > \frac R2$ for $k=1, 2$.