IMO 1985 SL 17

Problem

6a.(SWE 3)IMO6 The sequence f1, f2, . . . , fn, . . . of functions is defined for x > 0 recursively by f1(x) = x, fn+1(x) = fn(x)  fn(x) + 1 n  . Prove that there exists one and only one positive number a such that 0 < fn(a) < fn+1(a) < 1 for all integers n \geq1.

Solution

The statement of the problem is equivalent to the statement that there is one and only one a such that 1 −1/n < fn(a) < 1 for all n. We note that each fn is a polynomial with positive coefficients, and therefore increasing and convex in R+. Define xn and yn by fn(xn) = 1 −1/n and fn(yn) = 1. Since fn+1(xn) =  1 −1 n 2 +  1 −1 n  1 n = 1 −1 n and fn+1(yn) = 1+1/n, it follows that xn < xn+1 < yn+1 < yn. Moreover, the convexity of fn together with the fact that fn(x) > x for all x > 0 implies that yn −xn < fn(yn) −fn(xn) = 1/n. Therefore the sequences have a common limit a, which is the only number lying between xn and yn for all n. By the definition of xn and yn, the statement immediately follows.