Let $0<a<b<1$ be reals numbers and g(x)=\left\{\begin{array}{cc}x+1-a,&\mbox{ if } 0$n+1$ reals numbers $0<x_0<x_1<...<x_n<1$, for which $g^{[n]}(x_i)=x_i \ (0 \leq i \leq n)$. Prove that there exists a positive integer $N$, such that $g^{[N]}(x)=x$ for all $0<x<1$.
($g^{[n]}(x)= \underbrace{g(g(....(g(x))....))}_{\text{n times}}$)
Official solution [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=125&t=365714&p=2011659#p2011659]here