Proposition 1

Let $f$ be a finite real function, convex (resp. strictly convex) on an interval $I \subset \mathbf{R}$. For every family $(x_i)_{1 \leq i \leq p}$ of $p \geq 2$ distinct points of $I$, and every family $(\lambda_i)_{1 \leq i \leq p}$ of $p$ real numbers such that $0 < \lambda_i < 1$ and $\sum_{i=1}^p \lambda_i = 1$, we have

(resp.

Since the proposition (for convex functions) reduces to the inequality (1) for $p = 2$ we argue by induction for $p > 2$. The number $\mu = \sum_{i=1}^{p-1} \lambda_i$ is $> 0$; it is immediate that if $a$ and $b$ are the smallest and largest of the $x_i$ then $a \leq \frac{\sum_{i=1}^{p-1} \lambda_i x_i}{\sum_{i=1}^{p-1} \lambda_i} \leq b$; in other words, the point $x = \frac{1}{\mu} \sum_{i=1}^{p-1} \lambda_i x_i$ belongs to $I$, and the induction hypothesis implies that $\mu f(x) \leq \sum_{i=1}^{p-1} \lambda_i f(x_i)$; moreover we have, from (1), that

One argues in the same way for strictly convex functions, starting from the inequality (2).

We say that a finite real function $f$ is concave (resp. strictly concave) on $I$ if $-f$ is convex (resp. strictly convex) on $I$. It comes to the same to say that for every pair $(x, x')$ of distinct points of $I$ and every $\lambda$ such that $0 < \lambda < 1$ one has

(resp. $f(\lambda x + (1 - \lambda)x') > \lambda f(x) + (1 - \lambda)f(x')$).