# Ngô Quốc Anh

## December 17, 2009

### Schwarz’s Lemma, Schwarz-Pick theorem, and some applications involving inequalities

Filed under: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 10:52

In mathematics, the Schwarz lemma, named after Hermann Amandus Schwarz, is a result in complex analysis about holomorphic functions defined on the open unit disk.

Schwarz’s Lemma: Let $D=\{z : |z|<1\}$ be the open unit disk in the complex plane $\mathbb C$. Let $f : D \to \overline D$ be a holomorphic function with $f(0)=0$. The Schwarz lemma states that under these circumstances $|f(z)| \leq |z|$ for all $z \in D$, and $|f'(0)| \leq 1$. Moreover, if the equality $|f(z)|=|z|$ holds for any $z \ne 0$, or $|f'(0)|=1$ then $f$ is a rotation, that is, $f(z)=az$ with $|a=1$.

This lemma is less celebrated than stronger theorems, such as the Riemann mapping theorem, which it helps to prove; however, it is one of the simplest results capturing the “rigidity” of holomorphic functions. No similar result exists for real functions, of course. To prove the lemma, one applies the maximum modulus principle to the function $\frac{f(z)}{z}$.

Proof: Let $g(z)=\frac{f(z)}{z}$. The function $g(z)$ is holomorphic in $D$ (excluding $0$) since $f(0)=0$ and $f$ is holomorphic. Let $D_r$ be a closed disc within $D$ with radius $r$. By the maximum modulus principle,

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

for all $z$ in $D_r$ and all $z_r$ on the boundary of $D_r$. As $r$ approaches $1$ we get $|g(z)| \leq 1$. Moreover, if there exists a $z_0$ in $D$ such that $g(z_0)=1$. Then, applying the maximum modulus principle to $g$, we obtain that $g$ is constant, hence $f(z)=kz$, where $k$ is constant and $|k|=1$. This is also the case if $|f'(0)|=1$.

A variant of the Schwarz lemma can be stated that is invariant under analytic automorphisms on the unit disk, i.e. bijective holomorphic mappings of the unit disc to itself. This variant is known as the Schwarz-Pick theorem (after Georg Pick):

Schwarz-Pick theorem: Let $f : D \to D$ be holomorphic. Then, for all $z_1, z_2 \in D$,

$\displaystyle\left|\frac{f(z_1)-f(z_2)}{1-\overline{f(z_1)}f(z_2)}\right| \le \frac{\left|z_1-z_2\right|}{\left|1-\overline{z_1}z_2\right|}$

and, for all $z \in D$

$\displaystyle\frac{\left|f'(z)\right|}{1-\left|f(z)\right|^2} \le \frac{1}{1-\left|z\right|^2}$.

The expression

$\displaystyle d(z_1,z_2)=\tanh^{-1}\left(\frac{\left|z_1-z_2\right|}{\left|1-\overline{z_1}z_2\right|}\right)$

is the distance of the points $z_1, z_2$ in the Poincaré metric, i.e. the metric in the Poincaré disc model for hyperbolic geometry in dimension two. The Schwarz-Pick theorem then essentially states that a holomorphic map of the unit disk into itself decreases the distance of points in the Poincaré metric. If equality holds throughout in one of the two inequalities above (which is equivalent to saying that the holomorphic map preserves the distance in the Poincaré metric) , then $f$ must be an analytic automorphism of the unit disc, given by a Möbius transformation mapping the unit disc to itself.

An analogous statement on the upper half-plane $\mathbb H$ can be made as follows:

Let $f: \mathbb H \to \mathbb H$ be holomorphic. Then, for all $z_1, z_2 \in \mathbb H$,

$\displaystyle\left|\frac{f(z_1)-f(z_2)}{\overline{f(z_1)}-f(z_2)}\right| \le \frac{\left|z_1-z_2\right|}{\left|\overline{z_1}-z_2\right|}$.

This is an easy consequence of the Schwarz-Pick theorem mentioned above: One just needs to remember that the Cayley transform

$\displaystyle W(z) = \frac{z-i}{z + i}$

maps the upper half-plane $\mathbb H$ conformally onto the unit disc $D$. Then, the map $W \circ f \circ W^{-1}$ is a holomorphic map from $D$ onto $D$. Using the Schwarz-Pick theorem on this map, and finally simplifying the results by using the formula for $W$, we get the desired result. Also, for all $z \in \mathbb H$,

$\displaystyle\frac{\left|f'(z)\right|}{\mbox{Im }f(z)} \le \frac{1}{\mbox{Im }(z)}$.

If equality holds for either the one or the other expressions, then $f$ must be a Möbius transformation with real coefficients. That is, if equality holds, then

$\displaystyle f(z)=\frac{az+b}{cz+d}$

with $a$, $b$, $c$, $d$ being real numbers, and $ad-bc>0$.

Proof: The proof of the Schwarz-Pick theorem follows from Schwarz’s lemma and the fact that a Möbius transformation of the form

$\displaystyle\frac{z-z_0}{\overline{z_0}z-1}$ where $|z_0|<1$

maps the unit circle to itself. Fix $z_1$ and define the Möbius transformations

$\displaystyle M(z)=\frac{z_1-z}{1-\overline{z_1}z}$ and $\displaystyle\phi(z)=\frac{f(z_1)-z}{1-\overline{f(z_1)}z}$.

Since $M(z_1)=0$ and the Möbius transformation is invertible, the composition $\varphi(f(M^{-1}(z)))$ maps $0$ to $0$ and the unit disk is mapped into itself. Thus we can apply Schwarz’s lemma, which is to say

$\displaystyle |\phi(f(M^{-1}(z)))|=\left|\frac{f(z_1)-f(M^{-1}(z))}{1-\overline{f(z_1)}f(M^{-1}(z))}\right| \leq |z|$.

Now calling $z_2=M^{-1}(z)$ (which will still be in the unit disk) yields the desired conclusion

$\displaystyle\left|\frac{f(z_1)-f(z_2)}{1-\overline{f(z_1)}f(z_2)}\right| \le \left|\frac{z_1-z_2}{1-\overline{z_1}z_2}\right|$.

To prove the second part of the theorem, we just let $z_2$ tend to $z_1$.

Application 1 (QE Berkeley Spring 1991). Let the function $f$ be analytic in the unit disc, with $|f(z)| \leq 1$ and $f(0)=0$. Assume that there is a number $r \in (0,1)$ such that $f(r)=f(-r)=0$. Prove that

$\displaystyle\left| {f\left( z \right)} \right| \leqslant \left| z \right|\left| {\frac{{{z^2} - {r^2}}} {{1 - {r^2}{z^2}}}} \right|$.

Solution. Schwartz’s lemma implies that the function $f_1(z)=\frac{f(z)}{z}$ satisfies $|f_1(z)| \leq 1$. The linear fractional map $z \mapsto \frac{z-r}{1-rz}$ sends the unit disc onto itself. Applying Schwartz’s lemma to the function

$\displaystyle f_2(z)=f_1\left( \frac{z-r}{1-rz} \right)$

we conclude that the function

$\displaystyle f_3(z)=\frac{f_1(z)}{\left(\frac{z-r}{1-rz}\right)}$

satisfies $|f_3(z)| \leq 1$. Similarly, the map $z \mapsto \frac{z+r}{1+rz}$ sends the unit disc onto itself, and Schwartz’s lemma applied to the function

$\displaystyle f_4(z)=\frac{f_3(z)}{\left( \frac{z+r}{1+rz} \right)}$

implies that the function $f_5(z) =f_3\left( \frac{z+r}{1+rz} \right)$ satisfies $|f_5(z)| \leq 1$. All together, then

$\displaystyle\left| {f\left( z \right)} \right| \leqslant \left| z \right|\left| {\frac{{z - r}} {{1 - rz}}} \right|\left| {\frac{{z + r}} {{1 + rz}}} \right|\left| {{f_5}\left( z \right)} \right| \leqslant \left| z \right|\left| {\frac{{z - r}} {{1 - rz}}} \right|\left| {\frac{{z + r}} {{1 + rz}}} \right|$.

which is the desired inequality.

Application 2 (QE NUS Spring 2009). Suppose $f$ is analytic in $D := \{ z \in \mathbb C : |z|<1\}$ with $|f(z)|<1$. Show that

$\displaystyle\frac{{\left| {f\left( 0 \right)} \right| - \left| z \right|}} {{1 - \left| {f\left( 0 \right)} \right|\left| z \right|}} \leqslant \left| {f\left( z \right)} \right| \leqslant \frac{{\left| {f\left( 0 \right)} \right| + \left| z \right|}} {{1 + \left| {f\left( 0 \right)} \right|\left| z \right|}}$

for all $z \in D$.

Application 3 (QE NUS Fall 2009). Is there an analytic function $f$ on $\Delta$ (unit disk in the complex plane with center $0$) such that $|f(z)|<1$ for $|z|<1$ with $f(0)=\frac{1}{2}$ and $f'(0)=\frac{3}{4}$? If so, find such an $f$. Is it unique?

1. The problems from NUS QE are from GTM by Conway.

Comment by roticv — September 23, 2010 @ 14:53

• Hi, thanks so much, btw, which problem from NUS you are mentioning? I do think the first one is standard but the second one may be new.

Comment by Ngô Quốc Anh — September 23, 2010 @ 16:16

2. Both questions and it’s given in the same order as in the book. The book has a hint to Spring 2009.

Comment by roticv — September 27, 2010 @ 1:01

• Thanks. I will check it.

Comment by Ngô Quốc Anh — September 27, 2010 @ 1:03

• Thanks roticv, I got it. They are on pages 132 and 133 respectively of the first book.

Comment by Ngô Quốc Anh — September 27, 2010 @ 1:07

3. can anybody please send me the answer for the “Application 2 (QE NUS Spring 2009) and Application 3 (QE NUS Fall 2009)” displayed above.