Exercise 23 solution:

(a) Let ω_{1} < ω_{2}. Then for all x ∈ R, F_{ω_{1}}(x) < F_{ω_{2}}(x). We claim F_{ω_{1}}^{n}(x) < F_{ω_{2}}^{n}(x) for all x ∈ R and all n ≥ 1. The n = 1 case is given above. Suppose F_{ω_{1}}^{n}(x) < F_{ω_{2}}^{n}(x) for all x ∈ R and some n > 1. Then F_{ω_{1}}^{n+1}(x) = F_{ω_{1}}(F_{ω_{1}}^{n}(x)) < F_{ω_{1}}(F_{ω_{2}}^{n}(x)) since F_{ω_{1}} is strictly increasing. And F_{ω_{1}}(F_{ω_{2}}^{n}(x)) < F_{ω_{2}}(F_{ω_{2}}^{n}(x)) = F_{ω_{2}}^{n+1}(x) by the inductive hypothesis.

(b) F_{0}(x) = x + \frac{ϵ}{2π} sin(2πx). Since F_{0}(0) = 0, F_{0}^{n}(0) = 0 for all n ≥ 1, and

ρ(0) = lim_{n→∞} \frac{F_{0}^{n}(0) − 0}{n} = 0.

We now show that for δ sufficiently small, F_{1−δ} has a 1/1-periodic point, i.e., F_{1−δ}(x) = x + 1 for some x ∈ R. Let δ < ϵ/2π, so that sin(2πx) = δ2π/ϵ has a solution x = x_{0}. Then F_{1−δ}(x_{0}) − x_{0} − 1 = −δ + \frac{ϵ}{2π} sin(2πx_{0}) = 0, so that x_{0} is a 1/1-periodic point for F_{1−δ}. Thus F_{1−δ} has a 1/1-periodic point for all 0 < δ < ϵ/2π.

(c) Let ϵ > 0 and ω_{0} ∈ [0, 1) be given. Pick n ∈ Z such that 2/n < ϵ. Pick k ∈ Z satisfying k − 1 < F_{ω_{0}}^{n}(0) < k + 1. Using the fact F_{ω_{0}}^{n} is strictly increasing (as in the solution of Exercise 12), m(k − 1) < F_{ω_{0}}^{mn}(0) < m(k + 1) for all m. Since F_{ω}^{n}(0) is a continuous function of ω, we can pick δ > 0 such that |ω − ω_{0}| < δ implies k − 1 < F_{ω}^{n}(0) < k + 1. Hence m(k − 1) < F_{ω}^{mn}(0) < m(k + 1) for all m, and we have |F_{ω}^{mn}(0) − F_{ω_{0}}^{mn}(0)| < 2m. Hence

|\frac{F_{ω}^{mn}(0) − 0}{mn} − \frac{F_{ω_{0}}^{mn}(0) − 0}{mn}| < \frac{2}{n},

yielding |ρ(ω) − ρ(ω_{0})| < ϵ. Thus ρ(ω) is continuous on [0, 1).