If we endow $[0,1[$ with the metric $$d(x,y) := \inf_{k\in\mathbb{Z}} |x-y+k|,$$ then $[0,1[$ is homeomorphic to $S^{1}$. For a fixed $N\in\mathbb{N}$ we consider the map $$f\colon [0,1[\rightarrow [0,1[, \qquad f(x):= Nx - \lfloor Nx\rfloor = Nx \mod 1.$$ Given an integer $n\in\mathbb{N}$ we can compare orbit segments of length $n$ with the metric $$d_{n}(x,y) := \max_{0\leq j\leq n-1}d(f^{j}(x),f^{j}(y)).$$ I want to prove that if $d_{n}(x,y)<1/N$, then $d_{n}(x,y) = N^{n-1}|x-y|$.
It is not hard to prove that $d_{n}(x,y)\leq N^{n-1}d(x,y)\leq N^{n-1}|x-y|$, even without the assumption that $d_{n}(x,y)<1/N$. Any suggestions for the other inequality (i.e. $d_{n}(x,y) \geq N^{n-1}|x-y|$) would be greatly appreciated. Thanks in advance!
The distance $d$ that you describe is the distance of two points $x$, $y$ if we imagine them being on $\mathcal{S}^1$. If $d(x,y)<1/N$ then it is easy to see that $d(f(x),f(y))=Nd(x,y)$ (if you imagine the points being complex numbers on the unit circle the function $f$ simply multiplies their arguments by $N$). Now the condition $$d_n(x,y)<1/N,$$ for some $n\in\mathbb{N}$, implies that $d(f^i(x),f^i(y))<1/N,$ for all $ i\leq n-1$. Hence $$d(f^i(x),f^i(y))=Nd\left(f^{i-1}(x),f^{i-1}(y)\right)=N^i d(x,y), \text{for all}\hspace{1mm} 1\leq i\leq n-1.$$ Obviously the biggest number among these is $$d\left(f^{n-1}(x),f^{n-1}(y)\right)=N^{n-1}d(x,y)=N^{n-1}|x-y|,$$ where the last equality holds because $d(x,y)$ is small. Thus $d_n(x,y)=N^{n-1}|x-y|$.