IMO 2003 SL A2

Origin: AUS | Category: Algebra

Problem

Find all nondecreasing functions f : R \toR such that (i) f(0) = 0, f(1) = 1; (ii) f(a) + f(b) = f(a)f(b) + f(a + b −ab) for all real numbers a, b such that a < 1 < b.

Solution

We can rewrite (ii) as −(f(a) −1)(f(b) −1) = f(−(a −1)(b −1) + 1) −1. So putting g(x) = f(x+1)−1, this equation becomes −g(a−1)g(b−1) = g(−(a −1)(b −1)) for a < 1 < b. Hence −g(x)g(y) = g(−xy) for x < 0 < y, and g is nondecreasing with g(−1) = −1, g(0) = 0. (1) Conversely, if g satisfies (1), than f is a solution of our problem. Setting y = 1 in (1) gives −g(−x)g(1) = g(x) for each x > 0, and therefore (1) reduces to g(1)g(yz) = g(y)g(z) for all y, z > 0. We have two cases: (i) g(1) = 0. By (1) we have g(z) = 0 for all z > 0. Then any nonde- creasing function g : R \toR with g(−1) = −1 and g(z) = 0 for z \geq0 satisfies (1) and gives a solution: f is nondecreasing, f(0) = 0 and f(x) = 1 for every x \geq1 (ii) g(1) ̸= 0. Then the function h(x) = g(x) g(1) is nondecreasing and satisfies h(0) = 0, h(1) = 1, and h(xy) = h(x)h(y). Fix a > 0, and let h(a) = b = ak for some k \inR. It follows by induction that h(aq) = h(a)q =

(aq)k for every rational number q. But h is nondecreasing, so k \geq0, and since the set {aq | q \inQ} is dense in R+, we conclude that h(x) = xk for every x > 0. Finally, putting g(1) = c, we obtain g(x) = cxk for all x > 0. Then g(−x) = −xk for all x > 0. This g obviously satisfies (1). Hence f(x) = ⎧ ⎨ ⎩ c(x −1)k, if x > 1; 1, if x = 1; 1 −(1 −x)k, if x < 1, where c > 0 and k \geq0.