Bourbaki

Proposition 20

Tag 07CE, permanent. Cite it as [Bourbaki, Tag 07CE].

Topological Vector Spaces, II, p. 17, § 2. Convex sets

Let ff be a convex function, defined over a convex set XX of the affine space E. Then for every family (xi)1≤i≤p(x_i)_{1 \leq i \leq p} of pp points of X and every family (λi)1≤i≤p(\lambda_i)_{1 \leq i \leq p} of pp real numbers, all ≥0\geq 0, such that ∑i=1pλi=1\sum_{i=1}^p \lambda_i = 1, we have:

(3)

f(∑i=1pλixi)≤∑i=1pλif(xi). f\left( \sum_{i=1}^p \lambda_i x_i \right) \leq \sum_{i=1}^p \lambda_i f(x_i).

If ff is strictly convex and if λi>0\lambda_i > 0 for all ii, then

(4)

f(∑i=1pλixi)<∑i=1pλif(xi), f\left( \sum_{i=1}^p \lambda_i x_i \right) < \sum_{i=1}^p \lambda_i f(x_i),
unless all the xix_i are equal.

The inequality (3) follows from II, prop. 19 above and II, p. 8, prop. 1. Suppose that the xix_i are not all equal (which implies p≥2p \geq 2) and that the λi\lambda_i are all >0> 0; then the point z=∑i=1pλixiz = \sum_{i=1}^p \lambda_i x_i differs from at least one xix_i. Suppose for definiteness that z≠x1z \neq x_1, write z=λ1x1+(1−λ1)y1z = \lambda_1 x_1 + (1 - \lambda_1) y_1 where y1=∑i=2pλi1−λ1xiy_1 = \sum_{i=2}^p \frac{\lambda_i}{1 - \lambda_1} x_i. Then y1≠x1y_1 \neq x_1 and, as 0<λ1<10 < \lambda_1 < 1, we have, by hypothesis,

f(z)<λ1f(x1)+(1−λ1)f(y1). f(z) < \lambda_1 f(x_1) + (1 - \lambda_1) f(y_1).
But by (3) f(y1)≤∑i=2pλi1−λ1f(xi)f(y_1) \leq \sum_{i=2}^p \frac{\lambda_i}{1 - \lambda_1} f(x_i), and the inequality (4) follows.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

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