Complex analysis review 5

Maximum modulus principle and Schwarz lemma

Average Vaule Properties

$$\begin{align*} f(z_0) &= \frac{1}{2\pi i}\int_{\partial D(z_0,r)}\frac{f(\xi)}{\xi-z_0}d\xi\\ &=\frac{1}{2\pi i}\int_0^{2\pi}\frac{f(z_0+re^{it})ire^{it}}{re^{it}}dt\\ &=\frac{1}{2\pi}\int_0^{2\pi}f(z_0+re^{it})dt. \end{align*}$$

Shows that $f(z_0)$ is equal to the integration average on the circle $\partial D(z_0,r)$.

Maximum modulus principle

Suppose that $f(z)$ is analytic on $U\subset \mathbb{C}$, and there is a $z_0\in U$ such that $|f(z_0)|\geq |f(z)|,\forall z\in U$, then $f(z)$ must be a constant function on $U$.

Multiple by a constant with modulus 1, such that $M = f(z_0)\geq0$, let

$$S = \{z\in U|f(z) = f(z_0\}.$$
Then $S\neq \emptyset$. Since $f$ is a continuous function on $U$, so $S$ is closed set. Now we want to prove that $S$ is also an open set, then since $U$ is a single connected domain, we concluded that $S=U$.

If $w\in S$, choose $r$, such that $D(w,r)\subset U$, and let $0<r'<r$, then

$$M = f(w) = |\frac{1}{2\pi}\int_0^{2\pi}f(w+r'e^{it})dt| \leq M.$$
So
$$f(w+r'e^{it}) = |f(w+r'e^{it})| = M$$
for any $t$ and $0<r'<r$, which means
$$D(w,r')\subset S.$$
From the maximum modulus principle, we have that if $f$ is analytic on $U\subset \mathbb{C}$, $U$ is a bounded domain, and $f$ is continuous on $\bar{U}$, then if $f$ is not identical to a constant function, $|f(z)|$ can only attains its maximum on the boundary $\partial U$.

Schwarz Lemma

Suppose that $f$ is an analytic function which maps $D=D(0,1)$ into $D$, and $f(0) = 0$, then

$$|f(z)|\leq |z|,\quad |f'(0)|\leq 1.$$
Moreover $|f(z)| =|z|,z\neq 0$ or $|f'(0)|=1$ holds if and only if $f(z) = e^{i\tau}z.\tau\in \mathbb{R}$.

Let

$$G(z) = \left\{ \begin{align*} &\frac{f(z)}{z}& if \quad z\neq 0\\ &f'(0)& if\quad z=0 \end{align*} \right.$$
Then $G(z)$ is analytic on $D$. Consider $\{z||z|\leq 1-\epsilon\}$, by maximum modulus principle, we have
$$|G(z)|\leq \frac{\max_{|z|=1-\epsilon}|f(z)|}{1-\epsilon}< \frac{1}{1-\epsilon}.$$
Let $\epsilon \rightarrow 0$, we have $|G(z)|\leq 1$ on $D$. Therefore, when $z\neq0$, $|f(z)|\leq z$ and when $z=0$, $|G(0)|=|f'(0)|\leq 1$. The case when equality holds are easy.

$Aut(D)$

Let $a\in D$,

$$\phi_a(z) = \frac{-z+a}{1-\bar{a}z}\in Aut(D)$$
And $\phi_a^{-1} = \phi_a$. Then mapping above is called the Mobius mapping.

Let $\tau\in\mathbb{R}$, and define the rotation mapping

$$\xi =\rho_\tau(z)= e^{i\tau}z$$

Theorem

If $f\in Aut(D)$, then there is $a\in D,\tau\in \mathbb{R}$, such that

$$f(z) = \phi_a\circ\rho_\tau(z).$$
Which means that the element of $Aut(D)$ is the component of Mobius tranforamtion and rotation transformation.

Let $b=f(0)$ then let

$$G = \phi_b\circ f$$
$G$ is also in $Aut(D)$, and $G(0)=\phi_b\circ f(0) = \phi_b(b)=0$, by Schwarz lemma, we have $|G'(0)|\leq 1$. And $G$ is invertible with $G^{-1}\in Aut(D),G^{-1}(0)=0$. By Schwarz lemma again,
$$|\frac{1}{G'(0)}| = |(G^{-1})'(0)|\leq 1.$$
Then $|G'(0)|=1$. Then
$$G(z) = e^{i\tau}z = \rho_\tau(z).$$
Which shows that
$$f = \phi_{-b}\circ \rho_\tau.$$

Schwarz-Pick lemma

Suppose that $f$ is an analytic function which maps $D=D(0,1)$ into $D$, and $z_1,z_2\in D$, $w_1 = f(z_1),w_2 =f(z_2)$, then

$$|\frac{w_1-w_2}{1-w_1\bar{w}_2}|\leq |\frac{z_1-z_2}{1-z_1\bar{z}_2}|,$$
and
$$\frac{|dw|}{1-|w|^2} \leq \frac{|dz|}{1-|z|^2}.$$
Construct
$$\phi(z) = \frac{z+z_1}{1+\bar{z}_1z},\quad \psi(z) = \frac{z-w_1}{1-\bar{w}_1z}$$
Then $\phi,\psi\in Aut(D)$.

And consider $\psi\circ f\circ \phi$, use Schwarz lemma, then the remaining are easy (let $z = \phi^{-1}(z_2)$).

In the above theorem, we actually define a measure called Poincare measure. Then if that $f$ is an analytic function which maps $D=D(0,1)$ into , the Poincare is nonincreasing.