In complex analysis, the Schwarz lemma is one of the results for holomorphic functions from an open unit disc itself. However, it is one of the simplest results catching the rigidity of holomorphic functions. The theorem can be defined by taking an analytic function f(z). In this article, you will learn the statement and proof of the Schwarz Lemma, along with a related theorem and a solved example.

Click here to understand what an analytic function is.

Schwarz Lemma Statement

Suppose f(z) is an analytic function on the unit disc such that f(z) ≤ 1 for all z and f(0) = 0 then |f(z)| ≤ |z| and |f’(0)| ≤ 1.

Also, if |f(z)| = |z| for some z ≠ 0 or if |f’(0)| = 1, then f is a rotation about 0, that means f(z) = az, where a is a complex constant such that |a| = 1.

Schwarz Lemma Proof

The proof of the Schwarz lemma is a direct application of the maximum modulus principle on the function g(z) such that,

Then, g(z) is holomorphic on D as g : D → C is a complex function.

Let r be a real number such that 0 < r < 1.

If D_r = {z : |z| ≤ r} indicates the closed disc of radius r and origin is the centre.

By maximum modulus principle, we have for r < 1 and for any z ∈ D_r, there exists z_r on the boundary of the closed disc D_r, so

|g(z)| ≤ |g(z_r)|

As we know,

|g(z_r)| = |f(z_r)|/ |z_r| ≤ 1/r

As r approaches 1, we get |g(z)| ≤ 1 since |f(z_r)|/|z_r| ≤ 1.

Also, if |g(z)| ≤ 1, |f(z)/ z| ≤ 1

That means, |f(z)| ≤ |z|

Suppose |f(z)| = |z| for some non-zero z ∈ D, or |f'(0)| = 1 then |g(z)| = 1 at some point of D.

Thus, by the maximum modulus principle, g(z) must be a constant on D, i.e., g(z) = a such that |a| = 1.

Therefore, f(z) = az.

Hence proved.

Schwarz–Pick theorem

One of the Schwarz lemma variants is the Schwarz–Pick theorem, which illustrates the analytic automorphisms of the unit disc. That means bijective holomorphic mappings of the unit disc to itself.

Statement of Schwarz-Pick Lemma (Theorem):

Suppose f: D → D is a holomorphic function, then

Schwarz Lemma Solved Example

Question:

Suppose f is analytic on the closed unit disk, f(0) = 0, and |f(z)| ≤ |e^z| whenever |z| = 1. How big can f((1 + i)/2) be?

Solution:

We know that,

|f(z)| ≤ max{|e^z| : |z| = 1} ≤ max{|e^{cos θ}| : θ ∈ [0, 2π]} = e

By the Schwarz lemma, we have;

|f(z)| ≤ e|z| for any value of z in the closed unit disc.

Hence, |f((1 + i)/2)| ≤ e|(1 + i)/2| = e/√2.

Schwarz Lemma Probelms

1. Let f(z) be an analytic function in the upper half plane {z ∈ C : Im(z) > 0}. Suppose |f(z)| < 1 for all z in the domain of f, and f(i) = 0. Determine the largest possible value of |f(2i)|.
2. Does there exist an analytic function f : D → D with f(1/2) = 3/4 and f'(1/2) = 2/3?
3. When does the equality in Schwarz lemma hold?