Let \(f:[0,\infty)\to\mathbb R\) be continuous and differentiable, with \(f(0)=0\), and suppose \(f'\) is increasing on \([0,\infty)\). Prove that \[ g(x)= \begin{cases} f(x)/x,&x>0,\\ f'(0),&x=0 \end{cases} \] is increasing.
Solution
1. For every \(x>0\),\[ \frac{f(x)}{x}\leq f'(x). \]
Fix \(x>0\). By the mean value theorem applied to \(f\) on \([0,x]\), there is \(c\in(0,x)\) such that \[ f'(c) = \frac{f(x)-f(0)}{x-0} = \frac{f(x)}{x}. \] Since \(f'\) is increasing and \(c<x\), \[ f'(c)\leq f'(x). \] Substitution gives the claim.
2. For every \(x>0\),\[ g'(x)\geq0. \]
For \(x>0\), \[ \begin{aligned} g'(x) &= \frac{x f'(x)-f(x)}{x^2}\\ &= \frac{f'(x)-f(x)/x}{x}. \end{aligned} \] The numerator is nonnegative by step 1, and \(x>0\).
3. The function \(g\) is increasing on \((0,\infty)\).
By step 2, \(g'(x)\geq0\) throughout \((0,\infty)\). The standard monotonicity theorem for differentiable functions therefore gives that \(g\) is increasing there.
4. For every \(x>0\),\[ g(0)\leq g(x). \]
The point \(c\in(0,x)\) furnished by the mean value theorem in step 1 satisfies \[ g(x) = \frac{f(x)}x = f'(c). \] Since \(0<c\) and \(f'\) is increasing on \([0,\infty)\), \[ f'(0)\leq f'(c). \] By the definition of \(g(0)\), this is exactly \(g(0)\leq g(x)\).
5. Therefore\[ \boxed{g\text{ is increasing on }[0,\infty)}. \]
Step 3 compares any two positive arguments, and step 4 compares \(0\) with every positive argument.
6. Q.E.D.
Step 5 is the required conclusion.