Schwarz lemma

Statement

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

$|f(z)|\le \left|z\right|\mathrm{\forall }z\in D$

and:

$|f^{\prime }(0)|\le 1$

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

$f(z)=\alpha z\mathrm{\forall }z\in D$

Facts used

• Maximum modulus principle

Applications

• Schwarz-Pick lemma

Proof

Consider the function:

$g(z):=\frac{f(z)}{z}(z\ne 0),,f^{\prime }(0),(z=0)$

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

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

${max}_{\left|z\right|=r}|g(z)|\le \frac{1}{r}$

Thus, by the maximum modulus principle, we get:

${max}_{\left|z\right|\le r}|g(z)|\le \frac{1}{r}$

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

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

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