# Does this problem solve with Schwarz Lemma? [closed]

Let $f:D\to D$ be analytic where $D=\{z:|z|<1\}$. show that $$\left|\dfrac{f(z)-f(w)}{1-\overline{f(z)}f(w)}\right|\leqslant \left|\dfrac{z-w}{1-\bar{z}w}\right|\quad w,z\in D$$ and $$\dfrac{|f'(z)|}{1-|f(z)|^2}\leqslant \dfrac{1}{1-|z|^2}\quad z\in D$$

• I think that I should use the function $\tau_a(z)=\dfrac{a-z}{1-\bar{a}z}$ But I don't know how can I use it? – Fair Feb 17 '18 at 18:27

As a corollary of Schwarz Lemma, the following lemma can easily be proven.

If $f(z)$ is analytic for $|z| < 1$, and $|f(z)| <1$ for $|z| <1$ and $f(0) = 0$, we have $|f'(0)| \leq 1$

Now fix $z_0 \in \mathbb{D}$ and Let $w_0 = f(z_0)$. Define $g(z)$ and $h(z)$ as follows

$$g(z) = \frac{z+z_0}{1+ \bar{z_0}z} \ \ \ \ \ \ h(w) = \frac{w - w_0}{1 - \bar{w_0}w}$$

Notice $(h \circ f \circ g) (0) = 0$ and $(h \circ f \circ g)$ satisfies the lemma above. Thus, we have $$(h \circ f \circ g)'(0) = |h'(w_0)f'(z_0)g'(0)| \leq 1$$

Calculating, we see that $$h'(w_0) = \frac{1}{1-|w_0|^2} \ \ \ \ \ \ \ g'(0) = \frac{1}{1-|z_0|^2}$$ Rewriting our inequality , we have $$|f'(z_0)| \leq \frac{1-|w_0|^2}{1-|z_0|^2}$$ Since $f(z_0) = w_0$, we have the second inequality.

I should add, the second inequality is a well known lemma called Pick's Lemma.