Schwarz lemma

Statement

Let $\mathbb{D} = \{z \in \mathbb{C} : |z| < 1\}$ be the open unit disc. Let $f : \mathbb{D} \to \mathbb{D}$ be holomorphic on $\mathbb{D}$ with $f(0) = 0$. Then $|f(z)| \le |z|$ for each $z \in \mathbb{D}$, and $|f'(0)| \le 1$. If $|f(z_0)| = |z_0|$ for some $z_0 \neq 0$ in $\mathbb{D}$, or if $|f'(0)| = 1$, then there is $a \in \mathbb{C}$ with $|a| = 1$ such that $f(z) = a z$ for each $z \in \mathbb{D}$.

Remarks

Proof idea: $g(z) = f(z)/z$, with $g(0) = f'(0)$, is holomorphic on $\mathbb{D}$. The maximum modulus principle on the discs $|z| \le r$ gives $|g| \le 1/r$ for each $r < 1$, so $|g| \le 1$. It follows that the bijective holomorphic maps $\mathbb{D} \to \mathbb{D}$ with $f(0) = 0$ are the rotations $z \mapsto az$ with $|a| = 1$. The Schwarz–Pick theorem is the version without the condition $f(0) = 0$.

Explanations

L1 human

A holomorphic map from the unit disc to itself that keeps $0$ fixed cannot move any point farther away from $0$. The only maps that keep the distance to $0$ of some point are the rotations.