Show that \(f'\) bounded implies \(f\) is uniformly continuous.
Solution
Let \(f\) be real-valued and differentiable on an interval, with \(\abs{f'}\le M\). For \(x,y\) in the interval, the mean value theorem gives \(\xi\) between \(x\) and \(y\) with \[ \abs{f(x) - f(y)} = \abs{f'(\xi)} \abs{x-y} \le M\abs{x-y} .\] Given \(\eps>0\), \(\delta=\eps/M\) (any \(\delta\) if \(M=0\)) satisfies \(\abs{f(x)-f(y)}<\eps\) whenever \(\abs{x-y}<\delta\), independently of \(x\) and \(y\).