Assume \(f : [0,1] \to \mathbb{R}\) is uniformly continuous, increasing and convex. Prove \(f\) is differentiable almost everywhere and
\[f(1) - f(0) = \int_0^1 f'(x) \, dx.\]
Solution
1. Prove differentiability almost everywhere.
Since \(f\) is increasing on \([0,1]\), Lebesgue’s theorem on monotone functions implies that \(f\) is differentiable almost everywhere on \((0,1)\). Moreover, \[ f'(x)\ge0 \] at every point where the derivative exists.
2. Use convexity to obtain absolute continuity on interior compact intervals.
Fix \[ 0<a<b<1. \] Choose \(a_0,b_0\) with \[ 0<a_0<a<b<b_0<1. \] For any \(a\le x<y\le b\), convexity implies monotonicity of secant slopes, so \[ \frac{f(a)-f(a_0)}{a-a_0} \le \frac{f(y)-f(x)}{y-x} \le \frac{f(b_0)-f(b)}{b_0-b}. \] Hence the secant slopes on \([a,b]\) are uniformly bounded. Therefore \(f\) is Lipschitz on \([a,b]\), and hence absolutely continuous there.
The Fundamental Theorem of Calculus for absolutely continuous functions gives \[ f(b)-f(a)=\int_a^b f'(x)\,dx. \]
3. Let the interior interval expand to \([0,1]\).
Take \[ a_n=1/n, \qquad b_n=1-1/n \] for \(n\) large. Since \(f\) is uniformly continuous on \([0,1]\), \[ f(a_n)\to f(0), \qquad f(b_n)\to f(1). \] By Step 2, \[ f(b_n)-f(a_n)=\int_{a_n}^{b_n}f'(x)\,dx. \] Because \(f'\ge0\) almost everywhere and the intervals \([a_n,b_n]\) increase to \((0,1)\), the Monotone Convergence Theorem yields \[ \int_0^1 f'(x)\,dx =\lim_{n\to\infty}\int_{a_n}^{b_n}f'(x)\,dx. \] Therefore \[ \boxed{ f(1)-f(0)=\int_0^1 f'(x)\,dx.} \]