Let $D$ be the unit disk.
I already proved this lemma:
Let $f$ be a real continuous harmonic function defined on $D$. Suppose that, after a rotation of the coordinates, $f(e^{it})-f(e^{-it})\geq 0$ on $[0,\pi]$ and there are $a,b \in R$ such that $0\leq a < b \leq \pi$ and for $t\in[a,b]$ we have $f(e^{it})-f(e^{-it})>0$. Than $f$ has no critical points in $D$.
Now I have to prove the Rado Kneser Choquet theorem:
Let $\Omega$ be a bounded convex domain with a Jordan curve $\Gamma$ as contour. If $\hat{f}$ is a continuous mapping from $\partial D$ to $\Gamma$ with the property that $\hat{f}(e^{it})$ runs once around $\Gamma$ if $e^{it}$ runs around the unit circle. Then the harmonic extension $f$ van $\hat{f}$ is an injective harmonic mapping from $D$ onto $\Omega$.
By 'harmonic extension', I mean the harmonic extension of the Poisson integral formula:
let $\hat{f}(e^{i\theta})$ be a complex function that is that is piecewise continuous and bounded for $\theta \in [0,2\pi]$. Then we call the function $f(z)=\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-|z|^2}{|e^{it}-z|^2}\hat{f}(e^{it})dt$ the harmonic extension of $\hat{f}$.
Now I'm trying to prove the Rado Kneser Choquet theorem, and this is what I already have:
First we prove that $f$ is local injetive in every point of $D$. Suppose $f=u+iv$ is not local injective in $z_{0} \in D$. Then we have $$\begin{vmatrix} \frac{\partial u}{\partial x}(z_{0}) & \frac{\partial v}{\partial x}(z_{0}) \\ \frac{\partial u}{\partial y}(z_{0}) & \frac{\partial v}{\partial y}(z_{0})\end{vmatrix} = 0. $$ Consequently,
$$\left\{ \begin{matrix} a\frac{\partial u}{\partial x}(z_0)+b\frac{\partial v}{\partial x}(z_0)=0 \\ a\frac{\partial u}{\partial y}(z_0)+b\frac{\partial v}{\partial y}(z_0)=0 \end{matrix} \right. $$
had a real solution $(a,b)\neq (0,0)$. We define $\psi:=au+bv$. then we have that $z_0$ is a critical point of $\psi$. But $\psi$ satisfies the conditions of the lemma:
$u$ and $v$ are harmonic so $\psi$ is harmonic.
$u$ and $v$ are real so $\psi$ is real too.
$\psi$ is continuous.That's because $f=u+iv$ is harmonic, so $u$ and $v$ are continuous.
$\psi(e^{it})-\psi(e^{-it}) \geq 0$ on $[0,\pi]$
$\psi(e^{it})-\psi(e^{-it})> 0$ on $[a,b]$
But I don't know how to prove 4. and 5. Can anyone help me please?