Show that if $f(z)$ is analytic for $|z| \leq R$, $f(0)=0$ and $|f(z)| \leq M$ then $|f(z)| \leq \frac{M|z|}{R}$.

Question

Show that if $f(z)$ is analytic for $|z| \leq R$, $f(0)=0$ and $|f(z)| \leq M$ then $|f(z)| \leq \frac{M|z|}{R}$. I have no clue how to prove this. I tried to use the Cauchy's Inequality, but didn't get far. Please help.

Answer 1

Set $g(z):\mathbb{D}\to\mathbb{D}$ given by $g(z)=f(Rz)/M$. It is evident that $g$ is a holomorphic function as a composition of such and indeed it is $|g(z)|\leq1$, which is in fact $|g(z)|<1$, due to the maximum modulus principle. We also have $g(0)=f(0)=0$, so by Schwarz's lemma it is $|g(z)|\leq |z|$ for all $z\in\mathbb{D}$, thus $|f(Rz)|\leq M|z|$ for all $z\in\mathbb{D}$. Setting $Rz=w$ we get $|f(w)|\leq M|w|/R$ for all $w\in D(0,R)$ which is what we want to prove.