How to prove this fact about derivatives?

Question

Could you help me prove this, please?

We say f:R --> R is uniformly differentiable if, for every eps > 0, there's delta > 0 such that |(f(y) - f(x))/( y - x) - f'(x)| < eps for every x and y such that 0 < | y - x | < delta.

Show that f is uniformly differentiable if, and only if, f' is uniformly continuous on R.

Thank you

Answer 1

Suppose that f is uniformly differentiable. Let ε>0, and choose δ so that ∀x∀y, 0<|y-x|<δ ⇒ |(f(y) - f(x))/(y-x) - f'(x)| < ε/2. Then let x and y be any two points such that |x-y| < δ. Then either x=y, in which case f'(x) = f'(y), and |f'(x) - f'(y)|=0<ε, or else x≠y, in which case we have that |(f(y) - f(x))/(y-x) - f'(x)| < ε/2 and also that |(f(x) - f(y))/(x-y) - f'(y)| < ε/2 (simply interchanging the roles of x and y). Then we have that |f'(x) - f'(y)| = |f'(x) - (f(y) - f(x))/(y-x) + (f(y) - f(x))/(y-x) - f'(y)| ≤ |f'(x) - (f(y) - f(x))/(y-x)| + |(f(y) - f(x))/(y-x) - f'(y)| < ε/2+ε/2 = ε. Thus in either case |x-y|<δ ⇒ |f'(x) - f'(y)|<ε, so f' is uniformly continuous.

Conversely, suppose that f' is uniformly continuous. Let ε>0, and choose δ so that ∀x∀y, |y-x|<δ ⇒ |f'(y) - f'(x)|<ε. Then let x and y be any two points such that 0<|y-x|<δ. By the mean value theorem, there exists a point c between x and y such that f'(c) = (f(x) - f(y))/(x-y). Since c is between x and y, |c-x| ≤ |y-x| < δ. So we have |(f(y) - f(x))/(y-x) - f'(x)| = |f'(c) - f'(x)| < ε. Therefore, f is uniformly differentiable.