Clearly we need only consider the case when the indices are $1, 2, ..., p$ and $\lambda_i > 0$ for each $i$; the proposition is trivial if $p = 1$; we prove the result by induction on $p$. Put $\mu = \sum_{i=1}^{p-1} \lambda_i > 0$, and $y = \sum_{i=1}^{p-1} \frac{\lambda_i}{\mu} x_i$; the induction hypothesis implies that $y \in A$. Now as $\lambda_p = 1 - \mu$ and $\sum_{i=1}^p \lambda_i x_i = \mu y + (1 - \mu) x_p$, its follows from def. 1 that $\sum_{i=1}^p \lambda_i x_i$ belongs to $A$.

The image under $f$ of the closed segment with end points $x, y$ is the closed segment with end points $f(x), f(y)$, hence the first statement. We deduce that the inverse image of a closed segment of $F$ under $f$ contains each closed segment whose end points belong to it; the second statement of prop. 2 follows.

In particular the image of a convex set under a homothety or a translation is a convex set.

For these are the inverse images under $g$ of intervals of $\mathbf{R}$ and thus are convex.

With the notations of prop. 3 the points of a subset $M$ of an affine space are on the same side (resp. strictly on the same side) of the hyperplane $H$ if $M$ is contained in one of the half-spaces defined by $g(x) \geq 0, g(x) \leq 0$ (resp. $g(x) > 0$ or $g(x) < 0$).

Clearly the condition is necessary. Conversely suppose that it is satisfied and let $g(x) = 0$, be an equation defining H ($g$ is an affine linear mapping of E in $\mathbf{R}$). The set $g(A)$ is convex in $\mathbf{R}$, therefore it is an interval, and $0 \notin g(A)$. Hence $g(x)$ is of fixed sign for all $x \in A$.

The proposition follows immediately from def. 1 of II, p. 7.

In fact, each projection $\mathrm{pr}_i$ is a linear mapping and we have $A_i = \mathrm{pr}_i A$ and $A = \bigcap_{i \in I} \mathrm{pr}_i^{-1}(A_i)$; the proposition follows from props. 2 and 5 above.

For it is the image under an affine linear mapping of a rectangular parallelepiped, and this last is convex by prop. 6.

For $\alpha A + \beta B$ is the image of the convex subset $A \times B$ of $E \times E$ under the linear mapping $(x, y) \mapsto \alpha x + \beta y$ of $E \times E$ in $E$.

Denote the set of these linear combinations by C, clearly C is contained in every convex set which contains all the $A_i$ (II, p. 8, prop. 1); on the other hand $A_i \subset C$ for every $i$. All that remains to be proved is that C is convex. Let $x = \sum \lambda_i x_i$, $y = \sum \mu_i y_i$ be two points of C and $\alpha$ be a number such that $0 < \alpha < 1$, write γ_i = αλ_i + (1 - α) μ_i for every i ∈ I, and let J be the set (finite) of the indices of I for which γ_i ≠ 0. We can write αx + (1 - α)y = ∑_{i∈J} γ_i z_i, where

belongs to A_i for all i ∈ J; but $\sum_{i\in J} \gamma_i = \alpha \sum_{i\in I} \lambda_i + (1 - \alpha) \sum_{i\in I} \mu_i = 1$, and we see that $ax + (1 - \alpha) y \in C$. The proposition is proved.

The dimension of the affine linear variety (A, II, § 9.3) generated by the convex set A is called the dimension of A.

Let E be a vector space. The convex envelope C, of the balanced envelope of a set A in E is called the balanced convex envelope (or the symmetric convex envelope) of A; clearly it is the smallest symmetric convex set that contains A; it is also the convex envelope of $A \cup (-A)$, since every point of the balanced envelope of A belongs to a segment with extremities a and −a where $a \in A$. The set C coincides with the set of linear combinations $\sum_i \lambda_i x_i$ where $x_i \in A$ and $\sum_i |\lambda_i| \leq 1$; for it is clear that this set of points is convex and contains A and −A; it is sufficient to prove that it is contained in C, and for this we need consider only those linear combinations for which $\mu = \sum_i |\lambda_i| > 0$; we can then write $\sum_i \lambda_i x_i = \mu \cdot \sum_i \alpha_i y_i$ with $\alpha_i = \lambda_i/\mu$ and $y_i = x_i$, if $\lambda_i \geq 0$; and $\alpha_i = -\lambda_i/\mu$; $y_i = -x_i$ if $\lambda_i < 0$; clearly $\sum_i \alpha_i = 1$, and our assertion is proved.

There is a similar statement for linear mappings and balanced convex envelopes.

If C contains a line through 0 then clearly $C'$ is not convex. Suppose now that C is proper and let x, y be two points of $C'$. The closed segment with end points x, y is contained in C; if it contains 0 then $\lambda x + (1 - \lambda) y = 0$ for some $\lambda$ with $0 < \lambda < 1$, therefore $x = \mu y$ with $\mu < 0$. Thus C contains the line through 0 and x, contrary to hypothesis.

For the condition $\lambda C \subset C$ for all $\lambda > 0$ characterises the cones. If C is convex we have $C + C = \frac{1}{2}C + \frac{1}{2}C = C$ (II, p. 8, Remark). Conversely, if the cone C is such that $C + C \subset C$, then for $0 < \lambda < 1$, we have $\lambda C + (1 - \lambda) C = C + C \subset C$, which shows that C is convex.

For, if $V = C - C$, then V is non empty, we have $\lambda V = V$ for all $\lambda \neq 0$, and $V + V = C + C - (C + C) \subset C - C = V$, which shows that V is a vector subspace. Finally every vector subspace that contains C also contains V.

For, if $W = C \cap (-C)$, then W is non-empty and $\lambda W = W$ for all $\lambda \neq 0$, also

which shows that W is a vector subspace. Clearly every vector subspace contained in C is also contained in W.

Obviously, if $f$ is a linear mapping of E in a vector space F, then $f(C)$, the image of a convex cone C in E, is a convex cone in F. Every intersection of convex cones (with vertex 0) in E is a convex cone. For every subset A of E the intersection of convex cones containing A (these exist, E itself is one such cone) is the smallest convex cone that contains A; it is called the convex cone generated by A.

In fact, it is obvious that C, the set of such points, is a convex cone containing the union of the $C_i$, and that it is contained in any convex cone which contains this union.

It is sufficient to see that, if a convex cone contains a point $x \neq 0$ of A then it also contains the half-line $C_x$ of the points $\lambda x$ where $\lambda$ varies in the set of positive numbers and that $C_x$ is a convex cone.

The set C is clearly a cone; it is sufficient to show that C is convex. Let $\lambda x, \mu y$ be two points of C ($\lambda > 0, \mu > 0, x \in A, y \in A$). Let $\alpha, \beta$ be two numbers $> 0$ such that $\alpha + \beta = 1$. Then $\alpha \lambda x + \beta \mu y = (\alpha \lambda + \beta \mu) z$, with $z \in A$, and $\alpha \lambda + \beta \mu > 0$; hence $\alpha \lambda x + \beta \mu y \in C$.

(ii) Conversely, if P is a pointed convex cone in E, then the relation $y - x \in P$ is a preorder relation on E, and the preorder structure that it defines is the only one that is compatible with the vector space structure of E and for which P is the set of elements $\geqslant 0$.

(iii) The relation $y - x \in P$, with P a pointed convex cone, is an order relation on E if and only if P is a proper cone.

(i) Axioms (EO_I) and (EO_{II}) imply $P + P \subset P$ and $\lambda P \subset P$ for all $\lambda > 0$. As $0 \in P$, it follows that P is a pointed convex cone (II, p. 11, prop. 10).

(ii) Conversely, if P is a pointed convex cone, the relation $P + P \subset P$ implies that the relation $y - x \in P$ is a preorder compatible with the additive group structure of E (A, VI, p. 3, prop. 3); clearly writing it $x \leqslant y$, the set P is identical with the set of $x \geqslant 0$; further the relation $\lambda P \subset P$ for all $\lambda \geqslant 0$ shows that axiom (EO_{II}) is satisfied.

(iii) To say that P is proper means that $P \cap (-P) = \{0\}$ (II, p. 11, cor. 2), hence that $y - x \in P$ is an order relation.

For, let A be a convex set; the mapping $(x, y) \mapsto \lambda x + (1 - \lambda) y$, where $0 < \lambda < 1$, is continuous in $E \times E$ and maps $A \times A$ in A; thus (GT, I, § 2.1, th. 1) it maps $\overline{A} \times \overline{A}$ in $\overline{A}$, which shows that $\overline{A}$ is convex. Similarly, if C is a convex cone with vertex 0 then $\overline{C} + \overline{C} \subset \overline{C}$ and $\lambda \overline{C} \subset \overline{C}$ for all $\lambda > 0$.

Let B be the compact set in $\mathbf{R}^n$ defined by the points $(\lambda_1, \lambda_2, ..., \lambda_n)$ where $\lambda_i \geq 0$ ($1 \leq i \leq n$), and $\sum_{i=1}^n \lambda_i = 1$. Define a continuous mapping of $B \times \prod_{i=1}^n A_i \subset \mathbf{R}^n \times E^n$ in E by the formula $(\lambda_1, \lambda_2, ..., \lambda_n, x_1, x_2, ..., x_n) \mapsto \sum_{i=1}^n \lambda_i x_i$. The convex envelope C of $\bigcup_{i=1}^n A_i$ is the image of $B \times \prod_{i=1}^n A_i$ under this mapping; as $B \times \prod_{i=1}^n A_i$ is compact and E is Hausdorff, it follows that C is compact.

In fact, let j be the canonical mapping of E in its Hausdorff completion $\hat{E}$; if C is the convex envelope of A, then $j(C)$ is the convex envelope of the finite set $j(A)$ in $\hat{E}$, hence $j(C)$ is compact (cor. 1) and therefore C is precompact (GT, II, § 4.2).

For any point y of this segment, let f be the homothety of centre y and ratio $\lambda < 0$, which transforms $x_0$ into x. If V is an open neighbourhood of $x_0$ contained in A, then $f(V)$ is a neighbourhood of x and therefore contains a point $f(z) \in A$; now

hence $y - f(z) = \frac{\lambda}{\lambda - 1} (z - f(z))$, so that y is transformed into z by the homothety g, of centre $f(z)$ and ratio $\mu = \lambda / (\lambda - 1)$; since $0 < \mu < 1$, g transforms V into a neighbourhood of 0 contained in A. The proposition is proved.

It follows from prop. 16, that if $\overset{\circ}{A}$ is not empty, then it is a convex set and every point of $\overline{A}$ is a cluster point of $\overset{\circ}{A}$. Next we show that every interior point of $\overline{A}$ belongs to $\overset{\circ}{A}$. Let x be an interior point of $\overline{A}$ and suppose, for definiteness that $x = 0$. Let V be a symmetric neighbourhood of 0 that is contained in $\overline{A}$ and let $y \in \overset{\circ}{A} \cap V$; now $-y \in \overline{A}$, and therefore, by prop. 16, we see that $0 \in \overset{\circ}{A}$, if $y \neq 0$; this is obviously true if $y = 0$.

Since homotheties of ratio $> 0$ and centre 0 transform C into itself, they do the same for $\dot{C}$, thus $\dot{C}$ is a cone; the remainder of the corollary follows from cor. 1 and the obvious remark that if C is not empty then $\overline{C}$ contains the vertex of C.

Let H be a closed hyperplane in the topological vector space E over $\mathbf{R}$; it has an equation of the form $f(x) = \alpha$, where $f$ is a continuous linear form that is not identically zero in E (I, p. 13, th. 1). The closed half spaces defined respectively by $f(x) \leqslant \alpha$ and $f(x) \geqslant \alpha$ are therefore closed convex sets; their complements defined respectively by $f(x) > \alpha$ and $f(x) < \alpha$, are open convex sets. We say that these half-spaces are the closed (resp. open) half spaces determined by H.

In fact suppose that H contains the origin and that $f(x) = 0$ is an equation of H; suppose, for definiteness, that $f(x) \geqslant 0$ for all $x \in A$. The half space formed by the points y such that $f(y) > -1$ contains at least one interior point, and, by translation, the same is true of the half space of points y such that $f(y) > 0$; this shows that H is closed (I, p. 11, corollary). Then we know that $f$ is a strict morphism of E on $\mathbf{R}$ (I, p. 13, corollary), therefore $f(\dot{A})$ is an open set in $\mathbf{R}$. This set cannot contain 0 or it would contain numbers $< 0$ contrary to hypothesis; it is thus contained in the open interval $]0, +\infty[$. On the other hand, the half space of those y for which $f(y) \geqslant 0$ is closed and contains A, therefore it contains $\overline{A}$.

Apply prop. 17 to the case $A = P$ where H is the hyperplane with the equation $f(x) = 0$.

For, let $b = \lim_{x \in H} x$; for every $y \in H$, the set of $x \in H$ such that $x \geq y$ is a set of the section filter of H, therefore b is a cluster point of this set; but as the set $x \geq y$ is closed in E, we have $b \geq y$, thus b is an upper bound of H. On the other hand, if a is an upper bound of H, then H is contained in the closed set $x \leq a$; as b is a cluster point of H, we have $b \leq a$, which completes the proof (II, p. 72, exerc. 42).

a) The function $f$ is convex. b) The set $F$ in the affine space $E \times \mathbf{R}$ is convex. c) The set $F'$ in the affine space $E \times \mathbf{R}$ is convex.

We show that $a) \Rightarrow c)$. Let $(x, a)$ and $(y, b)$ be two points of $F'$ and $0 < \lambda < 1$, then $f(x) < a,\ f(y) < b$ and if $f$ is convex

which shows that the point $\lambda(x, a) + (1 - \lambda)(y, b)$ of $E \times \mathbf{R}$ belongs to $F'$. Thus $F'$ is convex.

Next we show that $c) \Rightarrow b)$. If $(x, a), (y, b)$ are two points of $F$ then for every $\varepsilon > 0$, $(x, a + \varepsilon)$ and $(y, b + \varepsilon)$ belong to $F'$ and, if $0 < \lambda < 1$, the same is true of $(\lambda x + (1 - \lambda) y, \lambda a + (1 - \lambda) b + \varepsilon)$; by the definition of $F$ this implies that $(\lambda x + (1 - \lambda) y, \lambda a + (1 - \lambda) b)$ belongs to $F$.

Finally $b) \Rightarrow a)$, for (with the above notation), if $(\lambda x + (1 - \lambda) y, \lambda a + (1 - \lambda) b)$ belongs to $F$ then

provided $a \geqslant f(x)$ and $b \geqslant f(y)$; hence (1) follows and $f$ is convex.

In fact, it is the projection on $E$ of the intersection of $F$ (resp. $F'$) and the hyperplane $E \times \{\alpha\}$ in $E \times \mathbf{R}$.

(3)

If $f$ is strictly convex and if $\lambda_i > 0$ for all $i$, then

(4)

unless all the $x_i$ are equal.

The inequality (3) follows from II, prop. 19 above and II, p. 8, prop. 1. Suppose that the $x_i$ are not all equal (which implies $p \geq 2$) and that the $\lambda_i$ are all $> 0$; then the point $z = \sum_{i=1}^p \lambda_i x_i$ differs from at least one $x_i$. Suppose for definiteness that $z \neq x_1$, write $z = \lambda_1 x_1 + (1 - \lambda_1) y_1$ where $y_1 = \sum_{i=2}^p \frac{\lambda_i}{1 - \lambda_1} x_i$. Then $y_1 \neq x_1$ and, as $0 < \lambda_1 < 1$, we have, by hypothesis,

But by (3) $f(y_1) \leq \sum_{i=2}^p \frac{\lambda_i}{1 - \lambda_1} f(x_i)$, and the inequality (4) follows.

The condition is obviously necessary, we prove that it is sufficient. Let $x_0 \in X$ be a point such that $f$ is bounded above in a neighbourhood V of $x_0$; we show firstly that $f$ is continuous at $x_0$. By translation, we can restrict ourselves to the case when $x_0 = 0$ and $f(x_0) = 0$; moreover we can suppose that the neighbourhood V is balanced (I, p. 7, prop. 4). Suppose that $f(x) \leq a$ in V; for every $\varepsilon, 0 < \varepsilon < 1$, we observe that if $x \in \varepsilon V$, then $x/\varepsilon \in V$ and $-x/\varepsilon \in V$. Applying inequality (1) of II, p. 16 to the points $x/\varepsilon$ and 0 and to the number $\lambda = \varepsilon$, we see that $f(x) \leq \varepsilon f(x/\varepsilon) \leq \varepsilon a$; applying it to points $x$ and $-x/\varepsilon$ and the number $\lambda = 1/(1 + \varepsilon)$, gives $f(x) \geq -\varepsilon f(-x/\varepsilon) \geq -\varepsilon a$. Thus $f(x)$ is arbitrarily small in $\varepsilon V$, if $\varepsilon$ is sufficiently small, and $f$ is continuous at $x = 0$.

Now let $y$ be some point of X; since X is open, there is a number $\rho > 1$ such that $z = \rho y$ belongs to X. Let $g$ be the homothety $x \mapsto \lambda x + (1 - \lambda)z$ of centre $z$ and ratio $\lambda = 1 - \frac{1}{\rho}$, which transforms 0 into $y$; for every point $g(x) \in g(V)$, we have from (1)

Thus $f$ is bounded above in a neighbourhood of $y$ and hence, by the first part, is continuous at $y$. The proposition is proved.

We can suppose that X is not empty. Then there exist, in X, $n + 1$ affinely independent points $a_i$ ($0 \leq i \leq n$) and the convex envelope of these points, S, contains the open non-empty set formed of the points $\sum_{i=0}^n \lambda_i a_i$ with $0 < \lambda_i < 1$ for all $i$ and $\sum_{i=0}^n \lambda_i = 1$. By II, p. 17, prop. 20, $f$ is bounded above in S and therefore is continuous.

In a topological vector space of infinite dimensions there exist, in general, linear non-continuous forms (II, p. 80, exerc. 25) and thus convex functions that are not continuous at any point.