Proposition 1

— Let E be a preordered vector space and V be a vector subspace of E such that every element of E is bounded above by an element of V. Given a linear form f on V that is positive for the preordered vector space structure of V (induced by that of E) there exists a non-empty set $S_f$ of positive linear forms on E, each being an extension of f. If $h \in S_f$ then the values $h(a)$ for $a \in E$ lie in the interval $[\alpha', \alpha'']$, where

I. Special case.

Suppose firstly that $E = V + \mathbf{R}a$. Since the proposition is trivial if $a \in V$, we confine ourselves to the case $a \notin V$. The conditions on V imply that the set $A''$ of $y \in V$ such that $a \leq y$ is not empty; similarly the set $A'$ of $z \in V$ such that $-z \geq -a$ (\emph{i.e.} $z \leq a$) is not empty. For $y \in A''$ and $z \in A'$, we have $z \leq a \leq y$, and thus by hypothesis $f(z) \leq f(y)$. Thus $\alpha', \alpha''$ are finite and $\alpha' \leq \alpha''$. Any linear form $f_1$ on E that extends $f$ is completely determined by $f_1(a)$ and for all $\lambda \in \mathbf{R}$ and all $x \in V$, we have

Thus $f_1$ is positive if and only if the relations

imply

As $f(\mu x) = \mu f(x)$ and the relations $x \geq 0$ and $\mu x \geq 0$ are equivalent for $\mu > 0$, it is sufficient to show that (2) implies (3) in the particular cases $\lambda = 0, \lambda = 1$ and $\lambda = -1$. For $\lambda = 0$, the fact that (2) implies (3) follows from the hypothesis that $f$ is positive. For $\lambda = 1$, to say that (2) implies (3) means that for $-x \in A'$, we have $f_1(a) \geq f(-x)$, \emph{i.e.} $f_1(a) \geq \alpha'$; for $\lambda = -1$, (2) implies (3), means that for $x \in A''$, we have $f(x) \geq f_1(a)$, \emph{i.e.} $f_1(a) \leq \alpha''$. The proposition is therefore proved in this case.

II. General case.

Let $\mathfrak{F}$ be the set of pairs $(W, g)$ where $W$ is a vector subspace of $E$ containing $V$ and $g$ is a positive linear form on $W$ which is an extension of $f$. We order $\mathfrak{F}$ putting $(W, g) \leq (W', g')$ if $W \subset W'$ and if $g'$ is an extension of $g$. Clearly $\mathfrak{F}$ is inductive and by th. 2 of S, III, § 2.4, there is a maximal element $(W_0, g_0)$. Suppose $W_0 \neq E$. Then there exists a vector $b \notin W_0$, and, if $W_1 = W_0 + \mathbf{R}b$, the special case above shows that there exists a positive linear form on $W_1$ which is an extension of $g_0$; this contradicts the hypothesis that $(W_0, g_0)$ is maximal. Thus $W_0 = E$, and the first part of the proposition is proved. When $a \in V$, the second assertion is obviously true with $\alpha' = \alpha'' = f(a)$; if, on the contrary, $a \notin V$ and one puts $V_1 = V + \mathbf{R}a$, the second assertion follows from the special case I of the proof.