Let $X$ be a Banach space with norm $|x|_X$. (For example $X=\mathbb R$)
We assume that a function $F: X \rightarrow X$ is Lipschitz continuous but not a contraction map, hence:
$$ | F(x) - F(y) |_X < \lambda |x-y|_X $$ where $\lambda >1$.
Furthermore for each $x \in X$: $F^n(x)=F^{n-1}(F(x))$ converges to a $x^*(x)$, which is a fixed point of $F$.
Example
This seems to be possible for example:
$F(x)=e^{-x}$ on $X=[-2,\infty)$ is Lipschitz but not a contraction.
But $F^2(x)=e^{-e^{-x}}$ is a contraction and hence the convergence of $F^n$ is given.
Question:
Is (for the "everywhere convergence to a fixed point") necessarily that for some $k\in \mathbb N$, $F^k$ is a contraction map?
Do you have an example where this is not the case?
Let $X=\mathbb{R}$ and define for $x\ge0$ $$ f(x)=\begin{cases} x/2 & \text{ if }0\le x<4,\\ 2\,x-6 & \text{ if }4\le x<5,\\ x-1 & \text{ if } x\ge5, \end{cases}$$ and $f(x)=-f(-x)$ if $x<0$. Then $f$ is Lipschitz but not a contraction. We also have $f(0)=0$, $f(x)<x$ if $x>0$ and $f(x)>x$ if $x<0$. It is easy to prove that for any $x_0\in X$ the sequence of iterates $x_{n+1}=f(x_n)$ converges to the unique fixed point $0$. But for any $n\in\mathbb{N}$ the function $f^n$ ($f$ composed with itself $n$ times) is not a contraction, since in the graph of $f^n$ ($f$ composed with itself $n$ times) there will always be a segment with slope $2$. Here is the graph of $x$, $f$, $f^2$ and $f^3$.