# 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.

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 $$|h(z)|\leq|\varphi(z)|$$.

• Are you aware that $T(w) = \frac{w-f(z_0)}{1-\overline{f(z_0)}w}$ maps the unit disk onto itself? Commented Jun 17 at 6:47

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