Example

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

Topological Vector Spaces, II, p. 16, § 2. Convex sets

If f is an affine linear function on E, then f and $f^2$ are convex functions on E; this is obvious for f since

f(\lambda x + (1 - \lambda) y) = \lambda f(x) + (1 - \lambda) f(y);

on the other hand, if

\alpha = f(x), \beta = f(y)

, then;

\lambda \alpha^2 + (1 - \lambda) \beta^2 - (\lambda \alpha + (1 - \lambda) \beta)^2 = \lambda(1 - \lambda) (\alpha - \beta)^2 \geq 0

for $0 < \lambda < 1$ ; further, the restriction of $f^2$ to an affine line $D \subset E$ is strictly convex if $f|D$ is not a constant.

A real-valued function $f$ , defined over $X$ , is concave (resp. strictly concave) if — $f$ is convex (resp. strictly convex). That is to say, for every two distinct points $x, y$ of $X$ and every number $\lambda$ , such that $0 < \lambda < 1$ , we have

f(\lambda x + (1 - \lambda) y) \geqslant \lambda f(x) + (1 - \lambda) f(y)

(resp.

f(\lambda x + (1 - \lambda) y) > \lambda f(x) + (1 - \lambda) f(y)).

A mapping of $X$ in $\mathbf{R}$ is affine if it is both convex and concave (cf. II, p. 78, exerc. 11).

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

Cited by

Nothing.

Cites

II, p. 78