Let a periodic orbit of $f$ be a set of points $P$ such that $\forall p\in P:\exists n\in\Bbb N_{>0}:f^n(p)\in P$ where $f^n$ indicates composition.
Must a continuous function on an open interval with no periodic orbits, other than a single fixed point (i.e. satisfying $f(p)=p$), necessarily contract to $p$ on iteration?
Let $f:(l,r)\to(l,r)$ be a continuous function on an open interval of $\Bbb R$, having precisely one fixed point $p$ satisfying $l<p<r$.
Then we need to show that $\lim_{n\to\infty}f^n(x)=p$ or prove otherwise.
contract on iteration to $p$ as its limit?
For what types of $f$ will it converge in finitely many iterations for all $x\in(l,r)$?
I'm aware of Sharkovskii's theorem which is related but doesn't seem to help. I think something much more elementary is needed.