R is the reals. [a, b] is an interval with b ≥ a + 2. f : [a, b] → R is twice differentiable and |f(x)| ≤ 1 and |f ''(x)| ≤ 1. Show that |f '(x)| ≤ 2.
Solution
Take the interval to be [-1, 1]. Taylor's formula gives f(1) = f(x) + (1 - x) f '(x) + (1 - x)2f ''(h)/2, f(-1) = f(x) + (-1 - x) f '(x) + (-1 - x)2f ''(k)/2 for some h, k in the interval (note that we are expanding about x).
Subtracting, f(1) - f(-1) = 2 f '(x) + (1 - x)2f ''(h)/2 - (1 + x)2f ''(k)/2. Hence 2 |f '(x)| ≤ |f(1)| + |f(-1)| + (1 - x)2|f ''(h)|/2 + (1 + x)2|f ''(k)|/2 ≤ 2 + (1 - x)2/2 + (1 + x)2/2 = 3 + x2 ≤ 4. So |f '(x)| ≤ 2.
© John Scholes
jscholes@kalva.demon.co.uk
5 Feb 2002