# How to prove the part of "precisely"? From Complex Analysis (By E. Stein) Page65 Exe.7

Question: Suppose $$f:D→C$$ is holomorphic. Show that the diameter $$d=\sup_{z,w\in D}|f(z) − f(w)|$$ of the image of f satisfies $$2|f'(0)|\leq d$$. Moreover, it can be shown that equality holds precisely when $$f$$ is linear, $$f(z)=a_0 + a_1z$$.

This problem is quoted directily from the book. The first part of the question (the inequality) is not difficult once the hint was given. However, I have been thinking about the second part (precisely) for ages.

It seems like Exe.9 in the book may have contributions to this question or just say the "Schwarz Lemma" related to conformal mappings. But what I am struggling with is I cannot restrict the image of $$f$$ to get a $$D \rightarrow D$$ map. Therefore I cannot meet the assumptions of "Schwarz Lemma".

How can this proposition be proved?

Exe.9 (Conclusion):

Let $$Ω$$ be a bounded open subset of $$C$$, and $$ϕ : Ω → Ω$$ a holomorphic function. If there exists a point $$z_0 ∈ Ω$$ such that $$ϕ(z_0) = z_0$$ and $$ϕ'(z_0) = 1$$ then $$ϕ$$ is linear.

Schwarz Lemma:

Let $$f : D → D$$ be holomorphic with $$f(0) = 0$$. Then

$$(i) |f(z)| ≤ |z|$$ for all $$z ∈ D$$.

$$(ii)$$ If for some $$z_0\neq 0$$ we have $$|f(z_0)| = |z_0|$$, then $$f$$ is a rotation.

$$(iii) |f'(0)| ≤ 1$$, and if equality holds, then $$f$$ is a rotation.

Personal solution to the first part:

Let $$C_r$$ denotes the circle of $$|\xi|=r$$.

Note that, by Cauchy formula, $$f'(0)=\frac{1}{2\pi i}\int_{C_r} \frac{f(\xi)}{\xi^2} d\xi$$=$$-\frac{1}{2\pi i}\int_{C_r} \frac{f(-\xi)}{\xi^2} d\xi$$.

Hence, $$2f'(0)=\frac{1}{2\pi i}\int_{C_r} \frac{f(\xi)-f(-\xi)}{\xi^2} d\xi$$.

Therefore, $$2|f'(0)|=|\frac{1}{2\pi}\int_{C_r} \frac{f(\xi)-f(-\xi)}{i \xi^2} d\xi| \leq r\sup_{\xi \in C_r} |\frac{f(\xi)-f(-\xi)}{i \xi^2}| \leq \frac{d}{r}$$

Since $$\forall r \in (0,1)$$ the inequality holds, $$2|f'(0)|\leq d$$.

QED

• Not all of us have the book, so it would be nice if you show how you did the first part. Apr 27 '19 at 11:58
• Thanks! I have added my solution to part one and the other also gave another approach to part one.
– 高廷宇
Apr 27 '19 at 13:25

First ,in the proof of the inequality, we use the fact $$|f(z)-f(-z)|\leq\sup_{z,w\in\mathbb{D}}|f(z)-f(w)|=C$$ Where $$z\in\mathbb{D}$$.
Notice that $$F(z)=f(z)-f(-z)$$,thus $$F(0)=0$$ , $$F'(0)=2f'(0)=d$$ . Actually , the condition in the ex 9 ,can be extended to the $$\phi'(z_0)=c>0$$, cause in the proof what $$\phi'$$ is just doesn't matter. So , by ex9, the reason is clear. Here is an answer of ex9 • You have flagged this question to be deleted. But you can delete your own answer --- click on the delete button at the bottom of your post. Apr 27 '19 at 22:45