IMC 1998 · Day 1 · P3

a) Fix $x = x_0 \in (0,1)$. If we denote $x_n = f_n(x_0)$, $n = 1, 2, \dots$ it is easy to see that $x_1 \in (0, 1/2]$, $x_1 \le f(x_1) \le 1/2$ and $x_n \le f(x_n) \le 1/2$ (by induction). Then $(x_n)_n$ is a bounded non-decreasing sequence and, since $x_{n+1} = 2 x_n (1 - x_n)$, the limit $l = \lim_{n \to \infty} x_n$ satisfies $l = 2l(1-l)$, which implies $l = 1/2$. Now the monotone convergence theorem implies that $$\lim_{n \to \infty} \int_0^1 f_n(x)\,dx = 1/2.$$

b) We prove by induction that $$\tag{1} f_n(x) = \frac{1}{2} - 2^{2^n - 1} \left( x - \frac{1}{2} \right)^{2^n}$$ holds for $n = 1, 2, \dots$. For $n = 1$ this is true, since $f(x) = 2x(1-x) = \frac{1}{2} - 2(x - \frac{1}{2})^2$. If (1) holds for some $n = k$, then we have $$\begin{align*} f_{k+1}(x) &= f_k(f(x)) = \frac{1}{2} - 2^{2^k - 1} \left( \left( \tfrac{1}{2} - 2 (x - \tfrac{1}{2})^2 \right) - \tfrac{1}{2} \right)^{2^k} \\ &= \frac{1}{2} - 2^{2^k - 1} \left( -2 (x - \tfrac{1}{2})^2 \right)^{2^k} = \frac{1}{2} - 2^{2^{k+1} - 1} \left( x - \tfrac{1}{2} \right)^{2^{k+1}} \end{align*}$$ which is (2) for $n = k+1$.

Using (1) we can compute the integral, $$\int_0^1 f_n(x)\,dx = \left[ \frac{x}{2} - \frac{2^{2^n - 1}}{2^n + 1} \left( x - \frac{1}{2} \right)^{2^n + 1} \right]_{x=0}^{1} = \frac{1}{2} - \frac{1}{2(2^n + 1)}.$$