Suppose that $(x_{n})$ is a contractive sequence in $\mathbb{R}$ and that $f:\mathbb{R}\to\mathbb{R}$ is a differentiable function with $x_{n+1}=f(x_{n})$ for all $n\in\mathbb{N}$.
Is there a example so that $|f'(x)|>1$ on the interval containing all $x_{n}$'s?
There seems to be a function that satisfies the above condition, but it does not come up well.
Give some advice! Thank you!
A sequence $\left({x_{n}}\right)_{n\geq 1}$ is contractive iff there exists a constant $c\in (0,1)$, such that, for all $n\geq 1$, $$|{x_{n+2}-x_{n+1}}|\leq c |{x_{n+1}-x_{n}}|.$$
If $x_{n+1}>x_n$, by the Mean Value Theorem, there exists $t\in (x_{n},x_{n+1})$ such that $$|{x_{n+2}-x_{n+1}}|=|f(x_{n+1})-f(x_{n})|=|f'(t)||{x_{n+1}-x_{n}}|\leq c |{x_{n+1}-x_{n}}|$$ which implies that $|f'(t)|\leq c<1$. So $|f'|$ can't be $>1$ everywhere in the interval containing all the contractive sequence.
On the other hand, $f'$ can be $>1$ at some points in that interval. Take $f(x)=x^2$ with $x_1=3/5>1/2$. Then $\left({x_{n}}\right)_{n\geq 1}$ is contractive with $c=24/25<1$, but $f'(x)>2x>1$ in $(1/2,x_1]\subset (0,x_1]$ (which contains the sequence $\left({x_{n}}\right)_{n\geq 1}$ since $x_n\downarrow 0$).
Another example: $f(x)=x+\sin(x)$ with $x_1\in (0,\pi/2)$. Then $x_n\uparrow\pi$ and it is eventually contractive, but $f'(x)=1+\cos(x)>1$ in $(0,\pi/2)$.