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I was trying to solve the following problem:

Let $f:\mathbb D\rightarrow \mathbb D$ be holomorphic with $f(0)=0$ and $f(1/2)=0,$ where $\mathbb D=\{z:|z|<1\}.$ Then which of the following statements are correct?

(a)$|f'(1/2)|\leq 4/3,$

(b)$|f'(0)|\leq 1,$

(c)$|f'(1/2)|\leq 4/3$ and $|f'(0)|\leq 1,$

(d)$f(z)=z$ for $z\in \mathbb D.$

My Attempts: Clearly, here we can apply Schwarz Lemma.By this lemma,we can say that $|f(z)|\leq |z|$ for all $z \in\mathbb D$ and $|f'(0)|\leq 1$ and so i can say option $(b)$ is definitely correct.But i can not say anything about option$(a)$.Please help.Thanks in advance for your time.

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Use the Schwarz Lemma to check each statement. The Schwarz Lemma applies because $f:D \to D$ is holomorphic and $f(0) = 0$. – Conan Wong Dec 15 '12 at 17:12
@ConanWong Thanks a lot.I have edited my attempts following your advice.But still some more work to do. – user52976 Dec 15 '12 at 17:31

2 Answers 2

up vote 0 down vote accepted

Statement (b) follows from the Schwarz inequality. A similar trick can be applied for $z = \tfrac{1}{2}$:

$$g(z) = f(z)\frac{2-z}{2z-1}$$ is holomorphic on $\mathbb{D}$ and $|g(z)| \leq 1$. Also $g(\tfrac{1}{2}) = \tfrac{3}{4}f'(\tfrac{1}{2})$.

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Sir,pardon my ignorance but i could not understand the steps $|g(z)|\leq 1$ and also $g(1/2)=3/4f′(1/2).$ Can you explain a little bit more? – user52976 Dec 15 '12 at 17:58
As i see that $|g(z)|=|f(z)||(2-z)/(2z-1)|\leq |z||(2-z)/(2z-1)|$ that is $|g(z)|\leq |(2z-z^2)/(2z-1)|$ but i see that $|(2z-z^2)/(2z-1)|$ is greater than $1$ for $ z\in \mathbb D$. I am sure i am wrong but could not find out.... – user52976 Dec 15 '12 at 18:05
$1/z$ is also $>1$ on $\mathbb{D}$ but that does not harm the Scharz argument. It's the maximum modulus principle that saves the day. Same here. $z \mapsto (2z-1)/(2-z)$ is a Möbius transformation of the disc (hence bounded) that maps $\tfrac{1}{2}$ to $0$. – WimC Dec 15 '12 at 18:21

First, as @Conan Wong commented, use Schwarz lemma to say that $f(z)=z\cdot g(z)$ where $g:\mathbb{D}\to\mathbb{D}$ is holomorphic. Moreover, $g(1/2)=0$. Consider the map $\varphi_a=\frac{z-a}{1-\bar{a}z}$. This map is injective, satisfies $\varphi_a(\mathbb{D})=\mathbb{D}$ for any $a\in \mathbb{D}$ and $\varphi_a(a)=0$. Hence the map $\varphi(z)=\varphi_{1/2}(z)=\frac{2z-1}{2-z}$ preserves $\mathbb{D}$ and sends $1/2$ to $0$.
Look at $h(z)=g(\varphi^{-1}(z))$. We have $h(0)=0$ and $h:\mathbb{D}\to\mathbb{D}$ is holomorphic, so we can apply Schwarz lemma to $h$ once again, to get $h(z)=g(\varphi^{-1}(z))=zk(z)$ where $k:\mathbb{D}\to\mathbb{D}$ is holomorphic. So $g(z)=\varphi(z)k(\varphi(z))$. In conclusion, $f(z)=z\cdot\frac{2z-1}{2-z}\cdot K(z)$ where $K:\mathbb{D}\to\mathbb{D}$ is holomorphic. Can you continue?

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thanks a lot sir.I am trying to understand your arguments and will try to take it forward. – user52976 Dec 15 '12 at 17:42

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