Proof of the Schwarz Lemma

The Schwarz lemma is a fundamental rigidity result constraining holomorphic self-maps of the unit disk that fix the origin. Despite its simple statement, it has far-reaching consequences.

Statement

Theorem7.8Schwarz lemma (restated)

Let $f: \mathbb{D} \to \mathbb{D}$ be holomorphic with $f(0) = 0$ . Then:

$|f(z)| \leq |z|$ for all $z \in \mathbb{D}$ .
$|f'(0)| \leq 1$ .
If equality holds in (1) for some $z \neq 0$ or in (2), then $f$ is a rotation: $f(z) = e^{i\theta}z$ .

Proof

Construction of the auxiliary function.

Define $g: \mathbb{D} \to \mathbb{C}$ by

$g(z) = \begin{cases} f(z)/z & \text{if } z \neq 0, \\ f'(0) & \text{if } z = 0. \end{cases}$

Since $f(0) = 0$ , the function $f(z)/z$ has a removable singularity at $z = 0$ (its Taylor series is $a_1 + a_2 z + a_3 z^2 + \cdots$ where $f(z) = a_1 z + a_2 z^2 + \cdots$ ). Setting $g(0) = a_1 = f'(0)$ makes $g$ holomorphic on all of $\mathbb{D}$ .

Applying the maximum modulus principle.

For $0 < r < 1$ , on the circle $|z| = r$ :

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

since $|f(z)| \leq 1$ ( $f$ maps into $\mathbb{D}$ ). By the maximum modulus principle, $|g(z)| \leq 1/r$ for all $|z| \leq r$ .

Taking the limit $r \to 1^-$ .

Since $|g(z)| \leq 1/r$ for every $r \in (|z|, 1)$ , letting $r \to 1^-$ gives $|g(z)| \leq 1$ for all $z \in \mathbb{D}$ .

This means $|f(z)/z| \leq 1$ , i.e., $|f(z)| \leq |z|$ for all $z \in \mathbb{D}$ . Also $|g(0)| = |f'(0)| \leq 1$ .

Rigidity (equality case).

If $|g(z_0)| = 1$ for some $z_0 \in \mathbb{D}$ (including $z_0 = 0$ ), then $|g|$ attains its maximum in $\mathbb{D}$ . By the maximum modulus principle, $g$ is constant: $g(z) = e^{i\theta}$ for some $\theta \in \mathbb{R}$ . Therefore $f(z) = e^{i\theta}z$ is a rotation. $\blacksquare$

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Corollaries

RemarkSchwarz-Pick lemma derivation

The Schwarz-Pick lemma follows by "conjugation." Given $f: \mathbb{D} \to \mathbb{D}$ holomorphic and $a \in \mathbb{D}$ , let $\varphi_a(z) = (z-a)/(1-\bar{a}z)$ be the Mobius automorphism swapping $0$ and $a$ . Apply the Schwarz lemma to

$g = \varphi_{f(a)} \circ f \circ \varphi_a^{-1}: \mathbb{D} \to \mathbb{D}, \quad g(0) = 0.$

Then $|g(z)| \leq |z|$ translates to $|\varphi_{f(a)}(f(\varphi_a^{-1}(z)))| \leq |z|$ , which gives the Schwarz-Pick inequality.

ExampleApplication: bounded derivative

If $f: \mathbb{D} \to \mathbb{D}$ is holomorphic, then for all $z \in \mathbb{D}$ :

$|f'(z)| \leq \frac{1 - |f(z)|^2}{1 - |z|^2}.$

In particular, $|f'(0)| \leq 1 - |f(0)|^2 \leq 1$ , with equality iff $f$ is a Mobius automorphism.

This estimate is sharp: the Mobius transformation $\varphi_a(z) = (z-a)/(1-\bar{a}z)$ satisfies $|\varphi_a'(z)| = (1-|a|^2)/|1-\bar{a}z|^2 = (1-|\varphi_a(z)|^2)/(1-|z|^2)$ .

RemarkGeometric interpretation

The Schwarz-Pick lemma states that holomorphic self-maps of $\mathbb{D}$ are contractions in the Poincare metric. The isometries are precisely the Mobius automorphisms $\text{Aut}(\mathbb{D})$ . This connects complex analysis to hyperbolic geometry in a profound way.