Schwarz lemma

Statement

Let $D$ denote the open unit disc. Any holomorphic map $f:D\to D$ with $f(0)=0$ satisfies:

$|f(z)|\leq |z|\ \forall \ z\in D$

and:

$|f'(0)|\leq 1$

Moreover, if there is any point $z\neq 0$ such that $|f(z)|=|z|$ or if $|f'(0)|=1$, then $f$ is a rotation about zero, i.e. there exists $\alpha \in \mathbb {C}$ with $|\alpha |=1$, such that:

$f(z)=\alpha z\ \forall \ z\in D$

Facts used

• Maximum modulus principle

Applications

• Schwarz-Pick lemma

Proof

Consider the function:

$g(z):={\frac {f(z)}{z}}(z\neq 0),\qquad ,f'(0),(z=0)$

Clearly, $g$ is a holomorphic function on $D$.

Now, for any $r\in (0,1)$, we have:

$\max _{|z|=r}|g(z)|\leq {\frac {1}{r}}$

Thus, by the maximum modulus principle, we get:

$\max _{|z|\leq r}|g(z)|\leq {\frac {1}{r}}$

Taking the limit as $r\to 1$, we get:

$\max _{z\in D}|g(z)|\leq 1$

which yields that $|f(z)|\leq |z|$ for all $z$ and $|f'(0)|=1$. Moreover, if $|g(z)|=1$ for any $z\in D$, then the maximum modulus principle forces $g$ to be a constant function with modulus $1$.