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I am currently studying complex analysis, and I came across an exercice that I could only partially solve, here it is :

Let $\Omega$ be a connected bounded set of $\mathbb{C}$, let $f : \Omega \rightarrow \Omega$ and holomorphic function and $a\in \Omega$ a fixed point of $f$ (i.e. $f(a) = a$). We define $f_n = (f\circ f \circ f ...)$ n times. Then prove that :

(A) $|f'(a)| \leq 1$

(B) $|f(a)| = 1 \iff f$ is bijective ;

(C) if $|f'(a)| < 1$, then $(f_n)$ converges uniformly on all compacts of $\Omega$ to the constant function equal to $a$.

So the first part of the exercice is to show that these properties hold in the case where $\Omega$ is simply connected. What I did was the following.

(A) Consider $f_n$. Using the Cauchy integral formula, we can show that $|f_n'(a)|$ must be bounded in modulus. But since $f_n'(a)= (f'(a))^n$ then $|(f'(a))^n|$ must be bounded and so necessarily $|f'(a)| \leq 1 $

(B) If $f$ is bijective then $f^{-1}$ satisfies all the condition for (A) to hold, and then, we have that $1 \leq |f'(a)| \leq 1$ as requested. If $|f'(a)| = 1$, I don't really know how to proceed... All I know is that $f'_n(a)$ will converge to $1$... But then how should I apply it ?? Or should I even use it ?

(C) Here, apart from the fact that I know that the function to which $f_n$ converges, let it be $g(z)$, has $g'(a)=0$, I don't know how to show that the derivative is actually $0$ everywhere. I think there should be an argument using the uniform convergence of the functions $f_n$, but that part too, I don't have a clue how to show it...

There is also a next part to the exercice, where these properties are to be shown in the general case. However I am primarily interested in understanding them for the simpler case of a simply connected set.

Domates
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Frotaur
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  • For (B) please look here: http://math.stackexchange.com/questions/774775/holomorphic-function-is-bijective-if-neutral-fixed-point –  Mar 22 '16 at 13:34
  • For a simply connected domain, the intention is that you conjugate $f$ with a biholomorphic map $T \colon \Omega \to \mathbb{D}$ that maps the fixed point to $0$. Then A and B follow immediately from the Schwarz lemma. In general, you argue with the normality of the family ${ f_n : n \in \mathbb{N}}$. For C (regardless of whether you look at a simply connected domain or not), by the continuity of $f'$ there is a closed disk $\overline{D_r(a)} = { z : \lvert z-a\rvert \leqslant r}$ such that $\lvert f'(z)\rvert < 1$ on that disk. The Banach fixed point theorem and normality finish it. – Daniel Fischer Mar 22 '16 at 16:59

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