Proposition 9

Let $f$ be a real finite function which is upper semi-continuous on an open interval $I \subset \mathbf{R}$. For $f$ to be convex on $I$ it is necessary and sufficient that for every point $M_x$ of the graph $G$ of $f$ every line locally above $G$ at this point should be locally on $G$ (at the point $M_x$).

The condition is necessary: indeed, if $f$ is convex on $I$ then at every point $M_a$ of the graph $G$ of $f$ there exists a support line $\Delta$ to $G$: now $\Delta$ is below $G$, so a fortiori locally below $G$ (I, p. 29); if a line $D$ is locally above $G$ at the point $M_a$ it is locally above $\Delta$, so must coincide with $\Delta$, and consequently is locally on $G$ at the point $M_a$.

The condition is sufficient. Indeed, suppose it is satisfied, and suppose that $f$ is not convex on $I$; then there are two points $a, b$ of $I$ ($a < b$) such that there are points $M_x$ of $G$ strictly above the segment $M_aM_b$ (fig. 4). In other words, the function $g(x) = f(x) - f(a) - \frac{f(b) - f(a)}{b - a}(x - a)$ takes values $> 0$ on $[a, b]$; since $g$ is finite and upper semi-continuous on this compact interval its least upper bound $k$ on $[a, b]$ is finite and $> 0$, and the set $g^{-1}(k)$ is closed and not empty (\emph{Gen. Top.}, IV, p. 361, th. 3 and prop. 1). Let $c$ be the greatest lower bound of $g^{-1}(k)$; we have $a < c < b$. and at the point $M_c$ the line D with equation $y = f(c) + \frac{f(b) - f(a)}{b - a}(x - c)$ lies locally above G; but it cannot be locally on G at this point, since, for $a < x < c$, one has $g(x) < k$, which signifies that $M_x$ is strictly below D. This has led us to a contradiction, which establishes the proposition.