Schwarz's Lemma/Corollary

Corollary to Schwarz's Lemma
Let $D$ be the unit disk centred at $0$.

Let $f : D \to \C$ be a holomorphic function.

If:


 * $\cmod {\map f \omega} = \cmod \omega$ for some $\omega \in D \setminus \set 0$

or:


 * $\cmod {\map {f'} 0} = 1$

then $\map f z = a z$ for all $z \in D$ for some $a \in \C$ with $\cmod a = 1$.

Proof
Let $g : D \to \C$ be a complex function with:


 * $\map g z = \begin{cases}\frac {\map f z} z & z \ne 0 \\ \map {f'} 0 & z = 0\end{cases}$

By Schwarz's Lemma: Lemma, $g$ is holomorphic on $D$.

By Schwarz's Lemma, we have:


 * $\cmod {\map g z} \le 1$

for all $z \in D$.

Suppose that:


 * $\cmod {\map f \omega} = \cmod \omega$ for some $\omega \in D \setminus \set 0$.

Then, for this $\omega$, we have:


 * $\cmod {\map g \omega} = 1$

Hence $g$ attains a maximum at $\omega \in D \setminus \set 0$.

As $D$ is a domain, by the Maximum Modulus Principle $g$ is therefore constant.

Suppose instead that:


 * $\cmod {\map {f'} 0} = 1$

Then:


 * $\cmod {\map g 0} = 1$

In this case, $g$ attains a maximum at $0 \in D$, again implying that $g$ is constant.

That is, there exists some $a \in \C$ such that:


 * $\map g z = a$

Note that we then have:


 * $\cmod {\map g 0} = \cmod a = 1$

For $z \in D \setminus \set 0$, we therefore have:


 * $\dfrac {\map f z} z = a$

That is:


 * $\map f z = a z$

As $\map f 0 = 0$, we have:


 * $\map f z = a z$

for all $z \in D$, for some $a \in \C$ with $\cmod a = 1$.