IMO 1983 SL 11

Problem

(FIN 2′) Let f : [0, 1] \toR be continuous and satisfy: bf(2x) = f(x), 0 \leqx \leq1/2; f(x) = b + (1 −b)f(2x −1), 1/2 \leqx \leq1, where b = 1+c 2+c, c > 0. Show that 0 < f(x)−x < c for every x, 0 < x < 1.

Solution

First suppose that the binary representation of x is finite: x = 0, a1a2 . . . an = n j=1 aj2−j, ai \in{0, 1}. We shall prove by induction on n that f(x) = n  j=1 b0 . . . bj−1aj, where bk = . −b if ak = 0, 1 −b if ak = 1. (Here a0 = 0.) Indeed, by the recursion formula, a1 = 0 ⇒f(x) = bf(n−1 j=1 aj+12−j) = b n−1 j=1 b1 . . . bjaj+1 hence f(x) = n−1 j=0 b0 . . . bjaj+1 as b0 = b1 = b; a1 = 1 ⇒f(x) = b + (1 −b)f(n−1 j=1 aj+12−j) = n−1 j=0 b0 . . . bjaj+1, as b0 = b, b1 = 1 −b. Clearly, f(0) = 0, f(1) = 1, f(1/2) = b > 1/2. Assume x = n j=0 aj2−j, and for k \geq2, v = x + 2−n−k+1, u = x + 2−n−k = (v + x)/2. Then f(v) = f(x) + b0 . . . bnbk−2 and f(u) = f(x) + b0 . . . bnbk−1 > (f(v) + f(x))/2. This means that the point (u, f(u)) lies above the line joining (x, f(x)) and (v, f(v)). By induction, every (x, f(x)), where x has a finite binary expansion, lies above the line joining (0, 0) and (1/2, b) if 0 < x < 1/2, or above the line joining (1/2, b) and (1, 1) if 1/2 < x < 1. It follows immediately that f(x) > x. For the second inequality, observe that f(x) −x = \infty  j=1 (b0 . . . bj−1 −2−j)aj < \infty  j=1 (bj −2−j)aj < \infty  j=1 (bj −2−j) = b 1 −b −1 = c. By continuity, these inequalities also hold for x with infinite binary rep- resentations.