I've been set this problem recently and I'm having a lot of trouble with it. Any help would be much appreciated!
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a function with continuous derivatives of all orders and suppose that, for some $x_{0}\in \mathbb{R}$ the derivative $f'(x_{0})$ is non-zero. Write $f(x_{0})=y_{0}$.
(a) Show there exists an open interval D containing $x_{0}$ such that $f'(x)\neq 0$ for all $x\in D$.
(b) Define $F_{y_{0}}:D\rightarrow \mathbb{R}$ by $F_{y_{0}}(x)=x-\frac{f(x)-y_{0}}{f'(x)}$. Show that $F_{y_{0}}$ is a Lipschitz function with Lipschitz constant less than 1.
(c) Using the Contraction Mapping Principle, show that there is a unique function $g:f(D)\rightarrow D$ such that $f\circ g$ and $g\circ f$ are the identity maps on $D$ and $f(D)$ respectively.
(d) Show that g is continuous. Hint: Write $\mid g(y)-g(y_{0})\mid = \mid F_{y}(g(y))- F_{y_{0}}(g(y_{0})\mid \leq \mid F_{y}(g(y))- F_{y}(g(y_{0})) \mid + \mid F_{y}(g(y_{0})) - F_{y_{0}}(g(y_{0}))\mid$
(e) Show that g is differentiable.
Sorry for the lengthy question! I have managed to do parts (a) and (b), but I'm currently stuck on part (c). I have managed to derive that the unique fixed point of $F_{y_{0}}$ is $x_{0}$ , but I am unsure on how to find an inverse function for $f$. I tried defining new functions $F_{y}(x)=x-\frac{f(x)- y}{f'(x)}$ for $x\in D$, which also have fixed points but I'm unsure on how to proceed. Once again, apologies for the lengthy question!
The function $f: D \rightarrow f(D)$ is surjective. To prove that it has an inverse $g: f(D) \rightarrow D$ it suffices to show that $f$ is injective. Now, if $f(x) = y = f(x')$ for $x,x' \in D$, then $F_y$ has the two fixed points $x$ and $x'$. As you already proved that $F_y$ has a unique fixed point, you can conclude that $x = x'$ and so you get $g: f(D) \rightarrow D$ with the desired properties. Note that you could also define $g(y)$ to be the unique fixed point of $F_y$.