Let $F\in\mathcal D$, $\phi=F-id$, show that $\left\lvert \cfrac{F^n(0)}{n} - F(0)\right\rvert\le\cfrac{n-1}{n} ,\forall\, n\in\mathbb N$
Note: We define $\mathcal D$ as the set of increasing homomorphisms (under $\mathbb R$) s.t $F=\phi +id$, where $\phi:\mathbb R\to\mathbb R$ is a continuous $1-$periodic function.
Note: $f^n:=f\circ f\circ\dots\circ f,$ $n$ times.
I showed that if $\rho(F) :=\lim_{n\to\infty}\cfrac{F^n(0)}{n}$, then
$$\min\{F^n(x)-x:x\in\mathbb R\}\le n\rho(F)\le \max\{F^n(x)-x:x\in\mathbb R\}$$
but i'm note sure if that's useful here, any hints?
Thank you!
For all $x \in [0,1]$, we see that : $$\phi(x)+x- \phi(0)= F(x)-F(0) \ge 0$$ so $\phi(x) -\phi(0) \ge -1 \quad \forall x \in [0,1]$, but we have in addition that $\phi$ is 1-periodic, hence $$\phi(x)-\phi(0) \ge -1 \quad \forall x$$ Similarly, for all $x \in [-1,0]$, we can imply a similar result, then : $$\phi(x) -\phi(0 ) \le 1 \quad \forall x$$ In other words, $$ | \phi(x) -\phi(0)| \le 1$$ for all $x$.
So for all $n$ we have:
$$| F^{n+1}(0)-(n+1)F(0)| \le | F^{n}(0)- nF(0)| +| \phi( F^{n}(0))-\underbrace{F(0)}_{= \phi(0)}| \le | F^{n}(0)- nF(0)| +1$$
Hence forth the conclusion.