Exercise 12 solution: We need to show ρ(f) = p/q implies F has a p/q-periodic point. Suppose F has no p/q-periodic point. Then for all x ∈ R, F^{q}(x) ≠ x + p. By the Intermediate Value Theorem applied to the continuous function F^{q}(x) − x, either F^{q}(x) > x + p for all x ∈ R, or F^{q}(x) < x + p for all x ∈ R. Suppose F^{q}(x) > x + p for all x ∈ R. As in the solution of Exercise 11, when considering F^{q}(x)−x it suffices to take x ∈ [0, 1]. By the Extreme Value Theorem F^{q}(x) − x assumes a minimum value m on [0, 1]. By assumption, p < m. Let ϵ = m − p, so that F^{q}(x) ≥ x + p + ϵ for all x ∈ R. Since F^{q} is increasing,

F^{2q}(x) ≥ F^{q}(x + p + ϵ) ≥ (x + p + ϵ) + p + ϵ = x + 2(p + ϵ),

and continuing, F^{kq}(x) ≥ x + k(p + ϵ) for k a positive integer. Hence

ρ(f) = lim_{k→∞} \frac{F^{kq}(x) − x}{kq} ≥ \frac{p + ϵ}{q} > \frac{p}{q},

a contradiction.

Similarly, if F^{q}(x)−x < p for all x ∈ R, there is an ϵ > 0 such that F^{q}(x) ≤ x+p−ϵ for all x ∈ R. Then F^{kq}(x) ≤ x + k(p − ϵ), implying ρ(f) ≤ (p − ϵ)/q < p/q.

We have shown that if ρ(f) = p/q then F has a p/q-periodic point.