I'm working on the following problem:
Suppose $f:U\to U$ is analytic on a bounded connected domain $U\subset\mathbb C$. Show that if $0\in U$, $f(0)=0$ and $f'(0)=1$, then $f=\text{id}$ on all of $U$.
A hint I am given is to compute the first nonzero coefficient in the Taylor expansion of the function $f_n(z)-z$ at $0$, where $f_n=\underbrace{f\circ\cdots\circ f}_{n}$, but I'm not quite sure how to use this hint. Does anybody have any hints on how to solve this problem?
The crucial result one uses here is that if $0 \in U$ bounded region and $g:U \to U$ an analytic self-map of $U$ with Taylor series at the origin $\sum a_k(g)z^k$ then $|a_k(g)| \le C_{k,U}$ for some constant independent of $g$ (depending only on $k,U$)
Assume this and use the hint considering $f(z)=z+a_2z^2+...a_nz^n+..$ as in the hypothesis and let's prove that $a_n=0, n \ge 2$ by induction (this proves that $f(z)=z$ on an open set containing zero where the expansion is valid, hence by the identity theorem, $f(z)=z$ in the whole region $U$).
Notice that for $f_n=\underbrace{f\circ\cdots\circ f}_{n}$ we have $f_2(z)=f(z)+a_2f(z)^2+..=z+2a_2z^2+z^3g_2$ so immediately $f_n(z)=z+na_2+z^3g_n$, while if $U$ is contained in a ball centered at $0$ and of radius $M$, we have $|f_n(z)| \le M, z \in U$ as all $f_n$ are self-maps of $U$ hence $n|a_2|=|a_2(f_n)| \le C_2$ which forces $a_2=0$
Now assuming $a_2=..=a_{k-1}=0, k \ge 3$. let's show that $a_k=0$ and the method should be clear as now $f(z)=z+a_kz^k+z^{k+1}h_1$ implies $f_2(z)=f(z)+a_kf(z)^k+...=z+2a_kz^k+z^{k+1}h_2$ and hence $f_n(z)=z+na_kz^k+z^{k+1}h_n$, so again by the result above $n|a_k| \le C_k$ for all $n$ hence $a_k=0$ and we are done modulo said result
(which is an easy consequence of Cauchy theorem applied to a closed disc centered at zero of fixed radius $r$ contained in $U$ with boundary circle $C$, for which $|a_k(g)|=|g^{(k)}(0)/k!|=|\frac{1}{2\pi i}\oint_C \frac{g(w)dw}{(z-w)^{k+1}}| \le \frac{M}{2\pi r^k}=C_{k,U}$ as required, where as before the disc centered at zero and of radius $M$ conatins $U$, so any self-map of $U$, satisfies $|g(z)| \le M$))
We wish to use induction to show $f(z)=z+a_nz^n+O(z^{n+1})$ for all $n>1$. Given some $n$ for which this holds, use induction to show that $f_k(z)=z+ka_nz^n+O(z^{n+1})$, where $f_k$ represents $k$ compositions of $f$ with itself. Now use the Cauchy inequalities to show that $|f_k^{(n)}(0)|=n!k|a_n|$ is bounded by something in terms of $n$ but not in terms of $k$. Then we may take $k\to\infty$ to show that $a_n=0$. This holds for all $n\in\mathbb N$, which completes the argument.
