Let \(f\) be differentiable on \([a, b]\). Say that \(f\) is uniformly differentiable iff
\[ \forall \varepsilon > 0,\, \exists \delta > 0 \text{ such that } \quad \abs{x-y} < \delta \implies \abs{ {f(x) - f(y) \over x-y} - f'(y)} < \eps .\]
Prove that \(f\) is uniformly differentiable on \([a, b] \iff f'\) is continuous on \([a, b]\).
Solution
\(\implies\): Fix \(\eps>0\) and choose \(\delta = \delta(\eps)\) to get a bound corresponding to \(\eps/2\), then for all \(x,y\) with \(\abs{x-y} < \delta\) on \([a, b]\), we have \[ \abs{f'(x) - f'(y) } \leq \abs{f'(x) - {f(x) - f(y) \over x- y} } + \abs{ {f(x) - f(y) \over x-y} - f'(y)} < \eps .\] This uses uniformity to bound by \(\eps/2\) the terms involving \(f'(x)\) and \(f'(y)\) respectively. So \(f'\) is in fact uniformly continuous on \([a, b]\).
\(\impliedby\): Let \(\eps> 0\) and \(x,y\in [a, b]\) be arbitrary. Then by the MVT, we can a \(\xi\in [x, y]\) with \(f'(\xi)(x-y) = f(x) - f(y)\). Then use continuity of \(f'\) to choose \(\delta = \delta(\eps, x, y)\) so that \(\abs{x-y} < \delta \implies \abs{f(x) - f(y)} < \eps\), and note that \(\abs{x-\xi} \leq \abs{x-y} < \delta\), so \[ \abs{ {f(x) - f(y) \over x-y } - f'(y) } = \abs{ f'(\xi) - f'(y)} < \eps .\]