Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 32E / b / iii

The map $F$ itself has a horseshoe only for $\mu =2$. Indeed, its Lipschitz constant is $\mu$. If two intervals $J_0,J_1$ form a horseshoe and their convex hull has length $L$, thenwhich forces $\mu \ge 2$. At $\mu =2$, take $J=(-1,1)$, $K_0=(-1,0)$ and $K_1=(0,1)$; both $K_i$ map onto $J$.

The Lipschitz constant of $F^2$ is ${\mu }^2$, so the same length argument makes $\mu \ge \sqrt{2}$ necessary for an $F^2$ horseshoe. It is sufficient. PutFor $\mu \ge \sqrt{2}$, one has $a<1/\mu \le b\le 1$. Withthe two relevant monotone branches satisfyThus $K_0,K_1\subset J$ are disjoint and $F^2(K_0)=F^2(K_1)=J$. Therefore