Let R be the reals. f : R → R is twice differentiable, f '' is continuous and f(0) = 0. Define g : R → R by g(x) = f(x)/x for x ≠ 0, g(0) = f '(0). Show that g is differentiable and that g' is continuous.
Solution
A routine application of the mean value theorem.
The only issue is x = 0, because elsewhere we have simply g'(x) = f '(x)/x - f(x)/x2.
Given x, we can find ξ such that f(x) = f(0) + x f '(0) + 1/2 x2 f ''(ξ) = x f '(0) + 1/2 x2 f ''(ξ). Hence limx→0 (g(x) - f '(0) )/x = limx→0 1/2 f ''(ξ) = 1/2 f ''(0). So g'(0) exists.
Now limx→0 g'(x) = limx→0 (x f '(x) - f(x))/x2. But f '(x) = f '(0) + x f ''(ζ), and f(x) = x f '(0) + 1/2 x2 f ''(ξ) with ξ, ζ ∈ (0, x). So limx→0 g'(x) = limx→0 f ''(ζ) - 1/2 f ''(ξ) = 1/2 f ''(0) = g'(0). So g' is continuous at 0.
© John Scholes
jscholes@kalva.demon.co.uk
26 Nov 1999