Let f be a convex function, defined over a convex set X of the affine space E. Then for every family (xi)1≤i≤p of p points of X and every family (λi)1≤i≤p of p real numbers, all ≥0, such that ∑i=1pλi=1, we have:
(3)
f(i=1∑pλixi)≤i=1∑pλif(xi).
If f is strictly convex and if λi>0 for all i, then
(4)
f(i=1∑pλixi)<i=1∑pλif(xi), unless all the
xi are equal.
The inequality (3) follows from II, prop. 19 above and II, p. 8, prop. 1. Suppose that the xi are not all equal (which implies p≥2) and that the λi are all >0; then the point z=∑i=1pλixi differs from at least one xi. Suppose for definiteness that z=x1, write z=λ1x1+(1−λ1)y1 where y1=∑i=2p1−λ1λixi. Then y1=x1 and, as 0<λ1<1, we have, by hypothesis,
f(z)<λ1f(x1)+(1−λ1)f(y1). But by (3)
f(y1)≤∑i=2p1−λ1λif(xi), and the inequality (4) follows.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.