Schwarz's Lemma

Theorem

Let $D$ be the unit disk centred at $0$.

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

Let $\map f 0 = 0$ and $\cmod {\map f z} \le 1$ for all $z \in D$.

Then $\cmod {\map {f'} 0} \le 1$, and $\cmod {\map f z} \le \cmod z$ for all $z \in D$.

Corollary

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

First a lemma:

Lemma

Let $D$ be the unit disk centred at $0$.

Let $g : D \to \C$ be a complex function such that:

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

Then $g$ is holomorphic on $D$.

$\Box$

Let $g : D \to \C$ be a complex function such that:

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

By the lemma, $g$ is holomorphic on $D$.

Let $r$ be a real number with $0 < r < 1$.

Let $C_r$ be the closed disk of radius $r$ centred at $0$.

Since $C_r$ is compact, by Continuous Function on Compact Space is Bounded, $g$ attains its maximum on $C_r$.

Let $\alpha \in C_r$ be a point where this maximum is attained.

If $g$ is constant, then we can pick $\alpha$ to be any point on $C_r$.

If $g$ is non-constant, then by the maximum modulus principle, $\alpha$ necessarily lies on the boundary of $C_r$.

That is, $\cmod \alpha = r$ if $g$ is non-constant.

If $g$ is constant, pick $\alpha$ such that $\cmod \alpha = r$ to avoid splitting into cases.

With that, for all $z \in C_r$ we have:

$\cmod {\map g z} \le \cmod {\map g \alpha}$

Since $0$ lies in the interior of $C_r$, we must have $\alpha \ne 0$, so:

\(\ds \cmod {\map g \alpha}\)

\(=\)

\(\ds \cmod {\frac {\map f \alpha} \alpha}\)

\(\ds \)

\(=\)

\(\ds \frac {\cmod {\map f \alpha} } {\cmod \alpha}\)

\(\ds \)

\(\le\)

\(\ds \frac 1 r\)

since $\cmod {\map f z} \le 1$ on $D$, and $\cmod \alpha = r$

So, for all $z \in C_r$ we have:

$\cmod {\map g z} \le \dfrac 1 r$

Aiming for a contradiction, suppose there exists $\omega \in D$ with:

$\cmod \omega < 1$

$\cmod {\map g \omega} = 1 + \varepsilon > 1$

Note that we then have:

$0 < \dfrac 1 {1 + \varepsilon} < 1$

Note that on $C_R$ with $R > \map \max {\dfrac 1 {1 + \varepsilon}, \cmod \omega}$, we should have:

$\cmod {\map g z} \le \dfrac 1 R < 1 + \varepsilon$

However, we have $\omega \in C_R$ and:

$\cmod {\map g \omega} = 1 + \varepsilon$

a contradiction.

So, it must be the case that:

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

for all $z \in D$.

We have:

$\cmod {\map {f'} 0} = \cmod {\map g 0} \le 1$

Note that:

$\cmod {\map f 0} = 0$

so certainly:

$\cmod {\map f z} \le \cmod z$

for $z = 0$.

Take $z \in D \setminus \set 0$, from:

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

we have:

$\cmod {\dfrac {\map f z} z} \le 1$

that is:

$\cmod {\map f z} \le \cmod z$

hence the result.

$\blacksquare$

Source of Name

This entry was named for Karl Hermann Amandus Schwarz.