Biholomorphic function with bounded derivative Let $U$ be an open subsets of $\mathbb{C}$ and $\psi:U\rightarrow U$ a biholomorphic function (a holomorphic function with holomorphic inverse). I want to prove the following statement:
For every $\epsilon >0$ there exists $\delta>0$ such that if $\frac{1}{1+\delta}<|\psi'(z)|<1+\delta$ for every $z\in U$ then $|\psi''(z)|<\epsilon$ for every $z\in U$. 
This means that if $\psi$ is biholomorphic and the module of its derivative is everywhere close to 1 then $\psi$ is "almost" the identity.
I've thought about it but I wasn't able to find anything that works. Using $\psi(\psi^{-1})=Id$ I just obtained that if $\frac{1}{1+\delta}<|\psi'(z)|<1+\delta$ then also $\frac{1}{1+\delta}<|(\psi^{-1})'(z)|<1+\delta$ and that $\frac{1}{(1+\delta)^3}<|\frac{\psi''(\psi^{-1})}{(\psi^{-1})''}|<(1+\delta)^3$ but this isn't enough to conclude.
Can anyone help me? Thank you very much!
 A: For the case of $U =\mathbb D,$ the open unit disc: If $f:\mathbb D\to \mathbb D$ is biholomorphic, then 
$$f(z) = c \frac{a-z}{1-\bar a z}$$
for some $a \in \mathbb D$ and $|c|=1.$ We can ignore $c.$ It follows that
$$f'(z) = \frac{|a|^2-1}{(1-\bar a z)^2},\,\, f''(z) = \frac{2\bar a (|a|^2-1)}{(1-\bar a z)^3}.$$
Now if $1-\delta < |f'(z)| < 1+ \delta$ for $z\in \mathbb D,$ then $|f'(0)| = 1-|a|^2$ satisfies the same inequality. Hence $|a| <\sqrt {\delta}.$ Thus
$$|f''(z)| = \frac{2|\bar a (|a|^2-1)|}{|1-\bar a z|^3}  \le \frac{2\sqrt {\delta}}{(1-|a|)^3} \le \frac{2\sqrt {\delta}}{(1-\delta)^3}$$
for $z\in \mathbb D.$ This gives the result.
A: *

*Assume that $U$ is the unit disk.
Let $f(z) = \log \psi'(z)$ that is holomorphic on $U$. 
Since $Re(f(z)) = \log |\psi'(z)|$ you have that $|Re(f(z))| < C_0 \delta$, and by the Schwarz integral formula, for $|z| < 1-2\epsilon$ :
$$2 \pi f(z) = \frac{1}{i} \int_{|s| = 1-\epsilon} \frac{s+ z}{s - z} \text{Re}(f(s)) \, \frac{ds}{s}+ 2\pi i\,\text{Im}(f(0))$$
So that $2 \pi |f(z)-i\, Im(f(0))| < 2\pi (1-\epsilon)\left(\frac{(1-\epsilon)+(1-2\epsilon)}{(1-\epsilon)-(1-2\epsilon)}C_0\delta\frac{1-\epsilon}{1-\epsilon}\right)< C_1 \delta \epsilon^{-1}$.
Hence, with the Cauchy integral formula $$\left|\frac{\psi''(z)}{\psi'(z)}\right| = |f'(z)| = \left|\int_{|s-z|=r} \frac{f(s)-i \,\text{Im}(f(0))}{(s-z)^2}ds\right| < C_2 \delta  \epsilon^{-2}$$ and $|\psi''(z)| < C_3 \delta \epsilon^{-2}$. 


*

*Overall you have $$|\psi''(z)| < C \frac{\delta}{ \min(1,d(z,\partial U))^2}$$ where $d(z,\partial U)$ is the distance to the boundary of $U$, and by looking at $\psi_{\mid V}$ where $V$ is an open disk $\subset U$, this will stay true for any $U$.

*Note that for a general holomorphic function $U \to U$, you can't do better than $|\psi''(z)| < C \frac{\delta}{ \min(1,d(z,\partial U))^{1-\epsilon}}$ : take $U$ the unit disk and $\psi(z) = \alpha z+\delta \, (1-z)^{1+\epsilon}, \psi'(z) = \alpha-\delta \, (1-z)^{\epsilon}(1+\epsilon),\psi''(z) = \delta \, (1-z)^{-1+\epsilon}(1+\epsilon)\epsilon$ is unbounded as $z \to 1$.

*So the question is how to use that $\psi$ is bi-holomorphic $U \to U$, and if it changes something.
