Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 53 / 1 / c / Solution

For $1<s\le \sqrt{2}$, use centered tent-map period-doubling renormalization. LetThe image $T_s(J_s)=[1+s-s^2,1]$ lies to the right of $J_s$, with disjoint interiors, because $1+s-s^2\ge s-1$. It lies in the positive branch. Consequently,and its range $[-c,(s^2-1)c]$ is contained in $J_s$. The orientation-reversing affine map $h_s$ gives the exact topological conjugacyThus renormalization squares the slope and halves the return period. Since $s>1$, there is a least $n\ge 0$ for which $s^{2^n}>\sqrt{2}$. Minimality implies $s^{2^n}\le 2$, and all previous slopes are in the renormalizable range. Successive affine maps therefore conjugate the restriction of $T_s^{2^n}$ on a nested central interval to $T_{s^{2^n}}$.

By the preceding part, an iterate $T_{s^{2^n}}^m$ has a horseshoe for an interval map. Pull its two branch intervals and its range interval back through the composed conjugacy. They give a horseshoe for $T_s^{2^{n}m}$. ThereforeAt the endpoint $s=\sqrt{2}$, one renormalization gives the full slope-two tent map, so the argument includes that endpoint without assuming the strict inequality from part (a).