Exercise 13 solution: First note that if G : R → R is continuous and lim_{n→∞} G^{n}(x) = x_{0}, then

x_{0} = lim_{n→∞} G^{n}(x) = lim_{n→∞} G^{n+1}(x) = lim_{n→∞} G(G^{n}(x)) = G(x_{0}),

so that x_{0} is a fixed point of G.

Now let x ∈ R. If G(x) = x then lim_{n→∞} G^{n}(x) = x. Suppose G(x) ≠ x.

(i) If G(x) < x then, since G is strictly increasing, G^{2}(x) < G(x), G^{3}(x) < G^{2}(x), and continuing, G^{n+1}(x) < G^{n}(x) for all n ≥ 0. Applying the Monotone Convergence Theorem to the decreasing sequence {G^{n}(x)}, either G^{n}(x) → −∞ as n → ∞, or G^{n}(x) → x_{0} as n → ∞. In the latter case x_{0} is a fixed point of G as shown above.

(ii) If G(x) > x, then the sequence {G^{n}(x)} is increasing. Hence either G^{n}(x) → ∞ as n → ∞, or G^{n}(x) → x_{0} as n → ∞, and we are done.