Alternate Version of Schwarz Lemma

Question

Let $f$ be analytic on $\vert z \vert \leq R$ with

$$\vert f(z) \vert \leq M$$

on $\vert z \vert = R$. Prove then that

$$\vert f(z)-f(0) \vert \leq \frac{2M\vert z \vert}{R}.$$

I know by the maximum principle that

$$\max_{\vert z \vert \leq R}\{f(z)\} \leq M$$

Since the maximum on the closure is the same as the maximum on the boundary, thus

$$\vert f(0) \vert \leq M$$

as $0 \in \overline{B_R(0)}$. But Then I get stuck.. any hints greatly appreciated.

Answer 1

You can proceed as in the proof of the “usual” Schwarz Lemma: $$ h(z) = \frac{f(z)-f(0)}{z} $$ has a removable singularity at $z=0$, and $|h(z)| \le \frac{2M}{R}$ for $|z| =R$. Now apply the maximum modulus principle to $h$.