Proposition 7

Let $f$ be a finite real function defined on an interval $I \subset \mathbf{R}$. For $f$ to be convex on $I$ it is necessary and sufficient that for every pair of numbers $a, b$ of $I$ such that $a < b$, and for every real number $\mu$, the function $f(x) + \mu x$ attains its supremum on $[a, b]$ at one of the points $a, b$.

The condition is necessary; indeed, since $\mu x$ is convex on $\mathbf{R}$, the function $f(x) + \mu x$ is convex on $I$; one can therefore restrict oneself to the case $\mu = 0$. Then, for

one has

The condition is sufficient. Let us take $\mu = -\frac{f(b) - f(a)}{b - a}$ and let $g(x) = f(x) + \mu x$; one has $g(a) = g(b)$ and therefore $g(x) \leq g(a)$ for all $x \in [a, b]$, and one can check immediately that this inequality is equivalent to the inequality (1) where one replaces $z$ by $a$ and $x'$ by $b$.