IMO 1972 SL 1

Origin: BUL

Problem

Let f and ϕ be real functions defined on the set R satisfying the functional equation f(x + y) + f(x −y) = 2ϕ(y)f(x), (1) for arbitrary real x, y (give examples of such functions). Prove that if f(x) is not identically 0 and |f(x)| \leq1 for all x, then |ϕ(x)| \leq1 for all x.

Solution

Suppose that f(x0) ̸= 0 and for a given y define the sequence xk by the formula xk+1 = . xk + y, if |f(xk + y)| \geq|f(xk −y)|; xk −y, otherwise. It follows from (1) that |f(xk+1)| \geq|ϕ(y)||f(xk)|; hence by induction, |f(xk)| \geq|ϕ(y)|k|f(x0)|. Since |f(xk)| \leq1 for all k, we obtain |ϕ(y)| \leq1. Second solution. Let M = sup f(x) \leq1, and xk any sequence, possibly constant, such that f(xk) \toM, k \to\infty. Then for all k, |ϕ(y)| = |f(xk + y) + f(xk −y)| 2|f(xk)| \leq 2M 2|f(xk)| \to1, k \to\infty.