Exercise 11 solution:

(a) Suppose there exist integers p, q and a real x satisfying F^{q}(x) = x + p. then

lim_{m→∞} \frac{F^{mq}(x) − x}{mq} = lim_{m→∞} \frac{x + mp − x}{mq} = \frac{p}{q}.

Let 0 ≤ r < q be an integer, and let y = k + ⟨y⟩, k an integer. Then F (y) − y = F (k + ⟨y⟩) − (k + ⟨y⟩) = F (⟨y⟩) − ⟨y⟩, so when considering F (y) − y it suffices to take y ∈ [0, 1]. Since F^{r} is continuous, it is bounded on [0, 1]. Hence there exists a constant M such that |F^{r}(y) − y| ≤ M for all y ∈ R and all 0 ≤ r < q.

Let n ∈ Z, and write n = mq + r, 0 ≤ r < q. Pick the constant M as above. Then

\frac{F^{n}(x) − x}{n} = \frac{F^{mq+r}(x) − x}{mq + r} = \frac{F^{mq+r}(x) − F^{mq}(x)}{mq + r} + \frac{F^{mq}(x) − x}{mq + r}

= \frac{mq}{mq + r} · \frac{F^{mq}(x) − x}{mq} + \frac{F^{r}(F^{mq}(x)) − F^{mq}(x)}{mq + r}.

Note that

|\frac{F^{r}(F^{mq}(x)) − F^{mq}(x)}{mq + r}| ≤ \frac{M}{n} ,

and hence approaches 0 as n → ∞. Since 0 ≤ r < q, mq/(mq + r) → 1 as n → ∞. Thus

lim_{n→∞} \frac{F^{n}(x) − x}{n} = lim_{m→∞} \frac{F^{mq}(x) − x}{mq} = \frac{p}{q}.

We have shown that if F has a p/q-periodic point then the rotation number is the rational p/q.

(b) Suppose F has no p/q-periodic points. Fix n > 0. Then for all x ∈ R, and for all integers k, F^{n}(x) − x ≠ k. So by the Intermediate Value Theorem, there is an integer k_{n} such that k_{n} < F^{n}(x) − x < k_{n} + 1 for all x ∈ R. Applying this inequality with x-values of 0, F^{n}(0), F^{2n}(0), …, we have

k_{n} < F^{n}(0) < k_{n} + 1
k_{n} < F^{2n}(0) − F^{n}(0) < k_{n} + 1

⋮

k_{n} < F^{mn}(0) − F^{(m−1)n}(0) < k_{n} + 1.

Adding these inequalities yields mk_{n} < F^{mn}(0) < m(k_{n} + 1), or

\frac{k_{n}}{n} < \frac{F^{mn}(0)}{mn} < \frac{k_{n}}{n} + \frac{1}{n}. Also, \frac{k_{n}}{n} < \frac{F^{n}(0)}{n} < \frac{k_{n}}{n} + \frac{1}{n},

so that

|\frac{F^{mn}(0)}{mn} − \frac{F^{n}(0)}{n}| < \frac{1}{n}.

Interchanging the roles of n and m, we also have

|\frac{F^{mn}(0)}{mn} − \frac{F^{m}(0)}{m}| < \frac{1}{m}.

Combining these two inequalities yields

|\frac{F^{n}(0)}{n} − \frac{F^{m}(0)}{m}| < \frac{1}{n} + \frac{1}{m}.

Since n and m were arbitrary, we have that \{\frac{F^{n}(0)−0}{n}\} is a Cauchy sequence. Since R is complete this sequence must converge. Finally, since ρ(f) is independent of x, the rotation number exists for all x ∈ R.