# 施瓦茨引理

$\Delta = \{z: | z | < 1\}$複平面中的开圆盘，$f:\Delta\to\Delta$是全纯函数，并有f(0)=0。那么

$| f(z) | \le | z |$

$| f(z) |=| z |\,$

$| f'(0) |=1\,$

## 证明

$g(z) = \frac{f(z)}{z}.\,$

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

## 施瓦茨—皮克定理

$f:\Delta\to\Delta$ 全纯。那么，对所有$z_1,z_2\in \Delta$

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

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

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

$f:\mathbb{H}\to\mathbb{H}$全纯。那么，对所有$z_1,z_2\in \mathbb{H}$

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

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

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

## 参考

• Jurgen Jost, Compact Riemann Surfaces (2002), Springer-Verlag, New York. ISBN 3-540-43299-X (See Section 2.3)