The exercise is as follows

Exercise: Let $f : \overline{B(0,R)} \to \mathbb{C}$ be a holomorphism with $|f(z)| \leq M$, for some $M > 0$. Show that

$$\left| \frac{ f(z) - f(0)}{M^2 - \overline{f(0)}f(z)} \right| \leq \frac{|z|}{MR}.$$

My attempt: I defined the holomorphic function $g: B(0,1) \to \mathbb{C}$

$$ g(z) = M \frac{ f(Rz) - f(0)}{M^2 - \overline{f(0)}f(Rz)} $$

and the result would follow from Schwarz's Lemma if

  • $g(B(0,1)) \subset B(0,1)$
  • $g(0)=0$

The latter is easily seen, however I couldn't show the former.



2 Answers 2


It is easy to show that $g(0)=0.$

Note that $g(z)=\frac{\frac{f(Rz)}{M}-\frac{f(0)}{M}}{1-\frac{\overline{f(0)}}{M}\frac{f(Rz)}{M}}$, where $|\frac{f(Rz)}{M}|<1$ and $|\frac{f(0)}{M}|<1$.

Let $\frac{f(Rz)}{M}=a+bi$, and $\frac{f(0)}{M}=c+di$, then: $$g(z)=\frac{a+bi-c-di}{1-(a-bi)(c+di)}$$ and $$|g(z)|^2=\frac{(a-c)^2+(b-d)^2}{(1-ac-bd)^2+(cb-ad)^2}.$$ Since $(1-a^2-b^2)(1-c^2-d^2)>0$ (because of absolute values of $f/M<1$), doing simple calculation you can show that numerator of $|g(z)|^2$ is less then denomiantor, and so $|g(z)|^2<1$, and $g(B(0,1))$ belongs to $B(0,1).$

  • $\begingroup$ Hello! Thank you for your answer. It wasn't exactly what I was looking for, but I found a way using your second line. Give me few minutes and I will write answer. $\endgroup$
    – B. Rivas
    Feb 14, 2016 at 19:21

There is a theorem which could gives us some help here. It is in many books, for example, Rudin, Real and Complex Analysis, page 254.

12.4 Theorem: For any $\alpha \in B(0,1)$, define

$$\varphi_\alpha(z) = \frac{z - \alpha}{1-\overline{\alpha}z}.$$

For fixed $\alpha \in B(0,1)$ the function $\varphi_\alpha$ is a one-to-one holomorphism which maps $B(0,1)$ onto $B(0,1)$, $ \partial B(0,1)$ onto $\partial B(0,1)$ and $\alpha$ to $0$. The inverse of $\varphi_\alpha$ is $\varphi_{-\alpha}$.

Sketch of the proof $\varphi_\alpha$ is holomorphic in the whole plane, except for a pole at $\frac{1}{\overline{\alpha}}$ which lies outside of $\overline{B(0,1)}$. Straightforward substitution shows that $$\varphi_{-\alpha}(\varphi_\alpha(z))=z$$ Moreover, for $z \in \partial B(0,1)$ we have $$| \varphi_\alpha(z) | = \frac{|z|}{|\overline{z}|}=1.$$ Hence by the Modulus Maximum Principle $|\varphi_\alpha(z)| < 1 $. The inverse gives us the surjective part.

Solution for the exercise: Using the manipulation given by @Jane,

$$g(z) = \frac{ \frac{f(Rz)}{M} - \frac{f(0)}{M} }{ 1 - \frac{\overline{f(0)}}{M}\frac{f(Rz)}{M} }$$

we shall define $\alpha = \frac{f(0)}{M} \in B(0,1)$ and $h:B(0,1) \to B(0,1)$ with $h(z)=\frac{f(Rz)}{M}$.

Then by the theorem above,

$$g(z)= \varphi_\alpha ( h(z) ) $$

maps B(0,1) into B(0,1) and the results follows from Schwarz Lemma as already stated.


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