Proposition 19

Let $X$ be a convex set of the affine space $E$; and let $f$ be a real-valued function defined over $X$. Denote the set of points $(x, a) \in E \times \mathbf{R}$ for which $x \in X$ and $f(x) \leqslant a$ (resp. $x \in X$ and $f(x) < a$) by $F$ (resp. $F'$). Then the following conditions are equivalent:

a) The function $f$ is convex. b) The set $F$ in the affine space $E \times \mathbf{R}$ is convex. c) The set $F'$ in the affine space $E \times \mathbf{R}$ is convex.

We show that $a) \Rightarrow c)$. Let $(x, a)$ and $(y, b)$ be two points of $F'$ and $0 < \lambda < 1$, then $f(x) < a,\ f(y) < b$ and if $f$ is convex

which shows that the point $\lambda(x, a) + (1 - \lambda)(y, b)$ of $E \times \mathbf{R}$ belongs to $F'$. Thus $F'$ is convex.

Next we show that $c) \Rightarrow b)$. If $(x, a), (y, b)$ are two points of $F$ then for every $\varepsilon > 0$, $(x, a + \varepsilon)$ and $(y, b + \varepsilon)$ belong to $F'$ and, if $0 < \lambda < 1$, the same is true of $(\lambda x + (1 - \lambda) y, \lambda a + (1 - \lambda) b + \varepsilon)$; by the definition of $F$ this implies that $(\lambda x + (1 - \lambda) y, \lambda a + (1 - \lambda) b)$ belongs to $F$.

Finally $b) \Rightarrow a)$, for (with the above notation), if $(\lambda x + (1 - \lambda) y, \lambda a + (1 - \lambda) b)$ belongs to $F$ then

provided $a \geqslant f(x)$ and $b \geqslant f(y)$; hence (1) follows and $f$ is convex.