Let \(f\) be differentiable on \([a,b]\). Say that \(f\) is uniformly differentiable if for every \(\varepsilon>0\) there exists \(\delta>0\) such that \[ \left|\frac{f(x)-f(y)}{x-y}-f'(y)\right|<\varepsilon \] whenever \(x,y\in[a,b]\), \(x\ne y\), and \(|x-y|<\delta\).
Prove that \(f\) is uniformly differentiable on \([a,b]\) if and only if \(f'\) is continuous on \([a,b]\).
Solution
For distinct \(x,y\in[a,b]\), write \[ Q(x,y)=\frac{f(x)-f(y)}{x-y}. \] Then \[ Q(y,x)=Q(x,y). \]
1. If \(f'\) is continuous on \([a,b]\), then \(f'\) is uniformly continuouson \([a,b]\).
The interval \([a,b]\) is compact. Hence the Heine–Cantor theorem applied to the continuous function \(f'\) shows that \(f'\) is uniformly continuous.
2. If \(f'\) is continuous on \([a,b]\), then \(f\) is uniformlydifferentiable on \([a,b]\).
Let \(\varepsilon>0\). By step 1, choose \(\delta>0\) such that \[ \abs{u-v}<\delta \quad\Longrightarrow\quad \abs{f'(u)-f'(v)}<\varepsilon \] for all \(u,v\in[a,b]\).
Take distinct \(x,y\in[a,b]\) with \[ \abs{x-y}<\delta. \] The mean value theorem applied to the interval with endpoints \(x\) and \(y\) gives a point \(c\) strictly between them such that \[ Q(x,y)=f'(c). \] Since \(c\) lies between \(x\) and \(y\), \[ \abs{c-y}<\abs{x-y}<\delta. \] Therefore \[ \abs{Q(x,y)-f'(y)} = \abs{f'(c)-f'(y)} < \varepsilon. \] The same \(\delta\) works for every such pair \(x,y\), so \(f\) is uniformly differentiable.
3. If \(f\) is uniformly differentiable on \([a,b]\), then \(f'\) is uniformlycontinuous on \([a,b]\).
Let \(\varepsilon>0\). By uniform differentiability, choose \(\delta>0\) such that for distinct \(x,y\in[a,b]\) with \(\abs{x-y}<\delta\), \[ \abs{Q(x,y)-f'(y)}<\frac{\varepsilon}{2}. \] Apply the same estimate to the ordered pair \((y,x)\). Since \[ Q(y,x)=Q(x,y), \] we also have \[ \abs{Q(x,y)-f'(x)}<\frac{\varepsilon}{2}. \] Hence \[ \begin{aligned} \abs{f'(x)-f'(y)} &\leq \abs{f'(x)-Q(x,y)} + \abs{Q(x,y)-f'(y)}\\ &< \varepsilon. \end{aligned} \] If \(x=y\), the same conclusion is immediate. Hence the estimate holds for all \(x,y\in[a,b]\) with \(\abs{x-y}<\delta\). Thus \(f'\) is uniformly continuous on \([a,b]\).
4. If \(f\) is uniformly differentiable on \([a,b]\), then \(f'\) iscontinuous on \([a,b]\).
Every uniformly continuous function is continuous. Apply this to \(f'\) using step 3.
5. Therefore\[ \boxed{ f\text{ is uniformly differentiable on }[a,b] \quad\Longleftrightarrow\quad f'\text{ is continuous on }[a,b] }. \]
Step 2 proves the reverse implication, and step 4 proves the forward implication.
6. Q.E.D.
Step 5 is the required equivalence.