Exercise 10 solution:

(a) Let k be an integer. We have

lim_{n→∞} \frac{F^{n}(x + k) − (x + k)}{n} = lim_{n→∞} \frac{F^{n}(x) + k − x − k}{n} = lim_{n→∞} \frac{F^{n}(x) − x}{n}.

Thus, to show ρ(f) is the same for x_{1} ≠ x_{2}, it suffices to consider x_{1}, x_{2} ∈ [0, 1), so that x_{2} − 1 ≤ x_{1} ≤ x_{2} + 1.

(b) Recalling that F is strictly increasing on R, we have

lim_{n→∞} \frac{F^{n}(x_{2}) − x_{2}}{n} = lim_{n→∞} \frac{F^{n}(x_{2}) − 1 − x_{2} − 1}{n} = lim_{n→∞} \frac{F^{n}(x_{2} − 1) − (x_{2} + 1)}{n}

≤ lim_{n→∞} \frac{F^{n}(x_{1}) − x_{1}}{n} = lim_{n→∞} \frac{F^{n}(x_{1}) − 1 − x_{1} − 1}{n}

= lim_{n→∞} \frac{F^{n}(x_{1} − 1) − (x_{1} + 1)}{n} ≤ lim_{n→∞} \frac{F^{n}(x_{2}) − x_{2}}{n} .

Hence ρ(f) is independent of x.