The Schwarz Lemma — Mathematics Knowledge System

Overview

The Schwarz lemma is a rigidity statement about holomorphic self-maps of the unit disk that fix the origin: if f is holomorphic on the open unit disk D, maps D into D, and f(0) = 0, then |f(z)| ≤ |z| for every z in D, and also |f'(0)| ≤ 1. Moreover, if equality holds anywhere in either inequality (short of the trivial case), f must be a rotation, f(z) = e^(iθ)z for some real θ. The proof is a slick one-line application of the maximum modulus principle to the auxiliary function g(z) = f(z)/z (made removable-singularity-free at 0 by g(0) = f'(0)). Despite its short proof, the Schwarz lemma is enormously consequential: it is the seed of the Schwarz–Pick lemma (which extends it to maps not fixing the origin, producing the hyperbolic metric on the disk), and it underlies rigidity and uniqueness results throughout geometric function theory, including the classification of automorphisms of the disk.

Intuition

Because f(0)=0, the power series of f has no constant term, so f(z) = a₁z + a₂z² + ⋯ = z·(a₁ + a₂z + ⋯); the function g(z) = f(z)/z is exactly that bracketed holomorphic function, and its value at 0 is a₁ = f'(0), so g extends holomorphically across the removable singularity at the origin. Now apply the maximum modulus principle to g on a disk of radius r < 1: since |f| < 1 on D, on the circle |z|=r we have |g(z)| = |f(z)|/r < 1/r, so the maximum of |g| on |z|≤r is at most 1/r. Letting r → 1⁻ (since f is defined on the full open disk), the bound 1/r → 1, so |g(z)| ≤ 1 everywhere on D. Translating back, |f(z)/z| ≤ 1, i.e., |f(z)| ≤ |z|; and |g(0)|=|f'(0)| ≤ 1 too. Equality anywhere forces |g| to attain an interior maximum, which by the maximum modulus principle means g is a constant of modulus 1 — a pure rotation.

Formal Definition

Definition

Let f : D → D be holomorphic on the open unit disk D = {z : |z| < 1}, with f(0) = 0:

|f(z)| \le |z| \quad \text{for all } z \in D

Schwarz lemma, pointwise bound

|f'(0)| \le 1

Schwarz lemma, derivative bound

\text{Equality in either bound at some } z_0 \ne 0 \text{ (or } |f'(0)|=1\text{)} \;\Longrightarrow\; f(z) = e^{i\theta}z \text{ for some } \theta \in \mathbb{R}

Rigidity / equality case

g(z) = \begin{cases} f(z)/z & z \ne 0 \\ f'(0) & z = 0 \end{cases}, \qquad g \text{ holomorphic on } D, \quad |g(z)| \le 1

Auxiliary function used in the proof

Worked Examples

f is holomorphic on all of ℂ (a polynomial), f(0)=0² =0, and for |z|<1, |f(z)|=|z|²<1, so f maps D into D.
$f(0)=0, \qquad |z|<1 \Rightarrow |f(z)|=|z|^2<1$
Check the pointwise bound at z=1/2: f(1/2) = 1/4.
$|f(1/2)| = \left|\left(\tfrac12\right)^2\right| = \frac14 \le \frac12 = |1/2|$
Check the derivative bound: f'(z)=2z, so f'(0)=0.
$|f'(0)| = 0 \le 1$

Answer: f(z)=z² satisfies both Schwarz lemma bounds strictly (|1/4| ≤ |1/2| and |f'(0)|=0≤1), consistent with f not being a rotation.

Practice Problems

Difficulty 5/10

Does f(z) = z/2 (holomorphic on D, f(0)=0, maps D into D since |z/2|<1/2<1) satisfy the Schwarz lemma bound |f(z)| ≤ |z|? Check at z = 0.8.

Difficulty 6/10

What is |f'(0)| for f(z) = z/2, and does it satisfy the Schwarz lemma's derivative bound?

Difficulty 8/10

Suppose f : D → D is holomorphic with f(0) = 0 and |f'(0)| = 1. Show f must be a rotation f(z) = e^{iθ}z, and explain why this rules out, e.g., f(z) = z + z²/2 on all of D even though its derivative at 0 is 1.

Quiz

Summary

If f : D → D is holomorphic with f(0)=0, then |f(z)| ≤ |z| for all z ∈ D and |f'(0)| ≤ 1.
The proof applies the maximum modulus principle to g(z)=f(z)/z, which extends holomorphically to D with g(0)=f'(0) and |g|≤1.
Equality in either bound at a single nonzero point forces f to be a rotation, f(z) = e^{iθ}z.
The Schwarz lemma is the seed of the Schwarz–Pick lemma and the hyperbolic metric on the disk, and underlies rigidity results about disk automorphisms.

References

WebsiteWikipedia — Schwarz lemma