IMO 1977 LL FIN46

Origin: FIN

Problem

Let f be a strictly increasing function defined on the set of real numbers. For x real and t positive, set g(x, t) = f(x + t) −f(x) f(x) −f(x −t). Assume that the inequalities 2−1 < g(x, t) < 2 hold for all positive t if x = 0, and for all t \leq|x| otherwise. Show that 14−1 < g(x, t) < 14 for all real x and positive t.

Solution

We need to consider only the case t > |x|. There is no loss of generality in assuming x > 0. To obtain the estimate from below, set a1 = f  −x + t  −f(−(x + t)), a2 = f(0) −f  −x + t  , a3 = f x + t  −f(0), a4 = f(x + t) −f x + t  . Since −(x + t) < x −t and x < (x + t)/2, we have f(x) −f(x −t) \leq a1 + a2 + a3. Since 2−1 < aj+1/aj < 2, it follows that g(x, t) > a4 a1 + a2 + a3

a3/2 4a3 + 2a3 + a3 = 14−1. To obtain the estimate from above, set

b1 = f(0) −f  −x + t  , b2 = f x + t  −f(0), b3 = f 2(x + t)  −f x + t  , b4 = f(x + t) −f 2(x + t)  . If t < 2x, then x −t < −(x + t)/3 and therefore f(x) −f(x −t) \geqb1. If t \geq2x, then (x + t)/3 \leqx and therefore f(x) −f(x −t) \geqb2. Since 2−1 < bj+1/bj < 2, we get g(x, t) < b2 + b3 + b4 min{b1, b2} < b2 + 2b2 + 4b2 b2/2 = 14.