Find all real-valued functions f on the reals such that (1) f(1) = 1, (2) f(-1) = -1, (3) f(x) ≤ f(0) for 0 < x < 1, (4) f(x + y) ≥ f(x) + f(y) for all x, y, (5) f(x + y) ≤ f(x) + f(y) + 1 for all x, y.
Solution
Answer: f(x) = [x].
f(x+1) >= f(x) + f(1) = f(x) + 1 by (4) and (1). But f(x) ≥ f(x+1) + f(-1) = f(x+1) - 1 by (4) and (2). Hence f(x+1) = f(x) + 1.
In particular, 1 = f(1) = f(0+1) = f(0) + 1, so f(0) = 0. Hence, by (3), f(x) ≤ 0 for 0 < x < 1. But, by (5), 1 = f(1) = f(x + 1-x) ≤ f(x) + f(1-x) + 1, so f(x) + f(1-x) ≥ 0. But if 0 < x < 1, then also 0 < 1-x < 1, so f(x) = f(1-x) = 0.
Thus we have established that f(x) = 0 for 0 ≤ x < 1, and f(x+1) = f(x) + 1. It follows that f(x) = [x] for all x.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002