Root of a function? (Proof with Banach theorem) Given is a function $f$: $\left [ -1,1 \right ]\rightarrow \mathbb{R}$, which is continuous and differentiable. The function $g$: $\left [ -1,1 \right ]\rightarrow \left [ -2,2 \right ]$ is a bijective approximation of $f$, such that $\left | g'(x)- f'(x) \right |\leq \frac{1}{2}inf_{y\in \left [ -1,1 \right ]}g'(y)$ and $\left | f(x)-g(x) \right |< 1$ for all $x\in \left [ -1,1 \right ]$.
We are looking for a root of the function $f$ in the interval $\left [ -1,1 \right ]$. Choose an $x_{1}:=0$ and consider the iteration formula $x_{n+1}:=g^{-1}(g(x_{n}) - f(x_{n}))$.
Show by using the Banach Fixed point theorem, that the sequence $\left \{ x_{n} \right \}$ converges to a point $x_{\infty }$, which is a root of the function $f$.
Reminder: (Banach fixed point theorem) Let $D\subset \mathbb{R}^{n}$ closed and $g:D\rightarrow D$ is a contraction. The $g$ has a fixed point $x_{\infty }\in D$ and the sequence $x_{0}\in D$, $x_{n+1}=g(x_{n})$ converges to $x_{\infty }$.
So, i guess the idea is to show that $f$ is a contraction on the interval $\left [ -1,1 \right ]$, this means that i have to show that there exist a $k\in (0,1)$ such that $\left | f(x)-f(y) \right |\leq k\left | x-y \right |$ for all $x,y\in \left [ -1,1 \right ]$. Maybe this is not enough for the proof...how can i use the additional information about the approximaion function $g$? 
Can anyone give me a hint, please? I would be happy to read any comments and remarks. Thank you in advance!

Define $D: [-1,1] \rightarrow \mathbb{R} \ $ by $ \ D(x) := g^{-1}(g(x) - f(x))$. Note that a fixed point of $D$ is what we are after. So we must show $D$ is a contraction. 
Note that for a function $h$ with suitable properties $(h^{-1})'(x) = \frac{1}{h'(h^{-1}(x))}.$
Note that by the Mean Value Theorem 
$$ \forall x,y \in [-1,1] \ \ \ \frac{|D(x) - D(y)|}{|x-y|} \leq \sup_{x \in [-1,1]}|D'(x)|,$$
so it suffices to bound the latter by a number in $(0,1)$. 
But 
\begin{aligned}
|D'(x)| &= \frac{|g'(x) - f'(x)|}{|g'(g^{-1}(g(x) - f(x))|} \\ 
&\leq \frac{1}{2}\frac{\inf_{[-1,1]}|g'(x)|}{|g'(g^{-1}(g(x) - f(x)))|}\\
&\leq \frac{1}{2} \frac{\inf_{[-1,1]}|g'(x)|}{\inf_{[-1,1]}|g'(g^{-1}(x))|}\\
&\leq \frac{1}{2} \frac{\inf_{[-1,1]}|g'(x)|}{\inf_{[-2,2]}|g'(g^{-1}(x))|}\\
&= \frac{1}{2} \frac{\inf_{[-1,1]}|g'(x)|}{\inf_{[-1,1]}|g'(x)|} = \frac{1}{2} < 1,\\
\end{aligned}
and we are done. 
NB $\quad$ In the second in inequality I use the fact that $\ \sup_{x \in [-1,1]}|(f(x) - g(x)| \leq 1$.
In the penultimate equality I use the fact that $g^{-1}$ is a bijection from $[-2,2]$ to $[-1,1]$.
I hope this helps,
Frank.
