Let S be the set of functions f defined on reals in the closed interval [0, 1] with non-negative real values such that f(1) = 1 and f(x) + f(y) ≤ f(x + y) for all x, y such that x + y ≤ 1. What is the smallest k such that f(x) ≤ kx for all f in S and all x?
Solution
Answer: k = 2.
Consider the function f(x) = 0 for 0 ≤ x ≤ 1/2, 1 for 1/2 < x ≤ 1. If x + y ≤ 1, then at least one of x, y is ≤ 1/2, so at least one of f(x), f(y) is 0. But f is obviously increasing, so the other of f(x), f(y) is ≤ f(x + y). Thus f satisfies the conditions. But f(1/2 + ε) = 1, so k cannot be smaller than 2.
So now let f be any function satisfying the conditions. We wish to show that f(x) ≤ 2x. Putting y = 1-x, we get f(x) + f(y) <= f(1) = 1. But f(y) is non-negative, so f(x) ≤ 1. Put y = x ≤ 1/2. Then 2f(x) ≤ f(2x). A simple induction gives 2nf(x) ≤ f(2nx) for x ≤ 1/2n. Now take any x in [0, 1]. If x > 1/2, then 2x > 1, so f(x) < 2x. If x ≤ 1/2, choose n ≥ 1, so that 1/2n+1 < x ≤ 1/2n. Then 2nf(x) ≤ f(2nx) ≤ 1, so f(x) ≤ 1/2n ≤ 2x.
© John Scholes
jscholes@kalva.demon.co.uk
23 Aug 2002
Last corrected/updated 25 Nov 03