Use Schwarz lemma to prove that $|\frac{f(z)-f(z_0)}{1-\overline{f(z_0)}f(z)}|\leq|\frac{z-z_0}{1-\overline{z_0}z}|$ with the following conditions.

Question

In $D:|z|<1$,$f(z)$ is analytic and $|f(z)|<1$.Suppose that $|z_0|<1$,prove that in region $D$,$$|\frac{f(z)-f(z_0)}{1-\overline{f(z_0)}f(z)}|\leq|\frac{z-z_0}{1-\overline{z_0}z}|$$

I want to use to the Schwarz lemma to prove it.Since $\omega=\varphi(z):=\frac{z-z_0}{1-\overline{z_0}z}$ is the univalent function of $C^\infty$, $\varphi^{-1}(0)=z_0$. Define $h(z):=\frac{f(z)-f(z_0)}{1-\overline{f(z_0)}f(z)}$,so $h(\varphi^{-1}(0))=0$. But I'm stuck with proving $|h(\varphi^{-1}(z))|<1$ in region $D$.

If it is right, $|h(\varphi^{-1}(z))|\leq|z|$ in region $D$ because of the Schwarz lemma. So $|h(z)|\leq|\varphi(z)|$.

Answer 1

It suffices to show $\varphi_{z_0}(z)=\frac{z-z_0}{1-\overline{z_0}z}$ send the unit disc to itself (in fact a biholomorphic map on the disc) as long as $|z_0|<1$, because then we have $$h\circ \varphi_{z_0}^{-1} = \varphi_{f(z_0)}\circ f \circ\varphi_{z_0}^{-1}$$

This is standard result for Mobius transformations, but not a hard one. Note that $\phi_{z_0}$ has only one singularity $z=\frac{1}{\overline{z_0}}$ that is outside the unit disc, therefore by maximum modulus principle,

$$\max_{|z|<1}\big|\frac{z-z_0}{1-\overline{z_0}z}\big|<\max_{|z|=1}\big|\frac{z-z_0}{1-\overline{z_0}z}\big|=\max_{|z|=1}\big|\frac{z-z_0}{\overline{z}(1-\overline{z_0}z)}\big|=\max_{|z|=1}\big|\frac{z-z_0}{\overline{z}-\overline{z_0}}\big|=1$$