Show that there is no real-valued function f on the reals such that ( f(x) + f(y) )/2 ≥ f( (x+y)/2 ) + |x - y| for all x, y.
Solution
Put x = a + b, y = a - b with b > 0. Then we have f(a) ≤ 1/2 f(a+b) + 1/2 f(a-b) - 2b. Also f(a + b/2) ≤ 1/2 f(a) + 1/2 f(a+b) - b, f(a - b/2) ≤ 1/2 f(a-b) + 1/2 f(a) - b, and f(a) ≤ 1/2 f(a - b/2) + 1/2 f(a + b/2) - b ≤ 1/4 f(a-b) + 1/2 f(a) + 1/4 f(a+b) - 2b. Hence f(a) ≤ 1/2 f(a-b) + 1/2 f(a+b) - 4b. But a and b are arbitrary (apart from b > 0) so this argument can now be repeated to show that f(a) ≤ 1/2 f(a-b) + 1/2 f(a+b) + 2nb for any positive integer n. Contradiction.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002