Lemma 5

Let $I\subset \overline{\mathbf{R}}$ be an interval not containing $+\infty$, and let $f$ be an increasing convex function from I into $[-\infty ,+\infty [$.

Let $n$ be a natural number, and let

be elements of I.

Let $\varrho_0\geqslant \varrho_1\geqslant \cdots \geqslant \varrho_n$ be positive real numbers. Suppose that

for every integer $j$ such that $0\leqslant j\leqslant n$. Then

Suppose first that $a_i\in \mathring{I}$, and hence $f(a_i)\in \mathbf{R}$, for every $i$. Let $i$ be such that $0\leqslant i\leqslant n$, and let $(\alpha_i, \beta_i)$ be real numbers such that the line with equation $y=\alpha_ix+\beta_i$ is a support line of the graph of $f$ at the point $(a_i, f(a_i))$ (FVR, I, p. 37). Consequently we have $f(a_i) =\alpha_ia_i+\beta_i$ and $f(b_i)\geqslant \alpha_ib_i+\beta_i$ since the graph of $f$ lies above the support line, whence

Moreover, if $i < n$, we have

(loc. cit. and FVR, I, p. 36, cor. 1); moreover $\alpha_i\geqslant 0$ since $f$ is increasing (FVR, I, p. 22, corollary), whence $\varrho_i\alpha_i\geqslant \varrho_{i+1}\alpha_{i+1}\geqslant 0$ for $0\leqslant i < n$.

Let $j$ be an integer such that $0\leqslant j\leqslant n$. Put

so that $A_j\geqslant 0$ by hypothesis (5). Applying inequality (7) and then lemma 4, we deduce that

Consider the general case, and argue by induction on $n$. If one of the $a_i$ does not belong to the interior of I, we necessarily have $a_0=$ sup I or $a_n=$ inf I.

Suppose first that $a_n$ = inf I. Then we have $a_n\leqslant b_n$ and therefore $\varrho_nf(a_n)\leqslant \varrho_nf(b_n)$. The induction hypothesis, applied to the families $(a_0, . . . , a_{n-1}), (b_0, . . . , b_{n-1})$ and $(\varrho_0, . . . , \varrho_{n-1})$, implies

whence the desired inequality, by adding $\varrho_nf(a_n)$. The case where $a_0=$ sup I is treated similarly.