Proposition 5

— Let $L$ be a positive linear form on a Riesz space $E$. In order that a positive linear form $M$ on $E$ belong to the band generated by $L$ in $\Omega$, it is necessary and sufficient that, for every $x \geq 0$ in $E$ and every number $\varepsilon > 0$, there exist a number $\delta > 0$ such that the relations $0 \leq y \leq x$ and $L(y) \leq \delta$ imply $M(y) \leq \varepsilon$.

Let us first show that the condition is necessary. If $M \geq 0$ belongs to the band generated by $L$ in $\Omega$, then (§ 1, No. 5, Cor. of Prop. 6)

If one sets

$U_n$ is thus a positive linear form on $E$ and $\inf_n U_n = 0$ in $\Omega$; consequently (Lemma) $U_n(x)$ tends to 0 as $n$ tends to infinity, and there exists an $n$ such that $U_n(x) \leq \varepsilon/2$. Fixing such an $n$, we have $U_n(y) \leq \varepsilon/2$ for all $y$ such that $0 \leq y \leq x$, thus the relation $0 \leq y \leq x$ implies

if $y$ is such that $L(y) \leq \varepsilon/2n$ it follows that $M(y) \leq \varepsilon$, which establishes our assertion.

Let us now show that the condition is sufficient. For every positive linear form $M$ on $E$, one can write $M = U + V$, where $U$ belongs to the band generated by $L$ in $\Omega$ and where $V$ is alien to $L$, $U$ and $V$ being positive (§ 1, No. 5, Th. 1). If $M$ satisfies the condition of the statement then so does $V = M - U$, since $0 \leq V \leq M$. From this we will deduce that $V = 0$. For every $x \geq 0$ in $E$ and every number $\eta > 0$, there exist two elements $y \geq 0, z \geq 0$ of $E$ such that $x = y + z$ and $L(y) + V(z) \leq \eta$ (Prop. 4); given an arbitrary number $\varepsilon > 0$, choose $\eta \leq \varepsilon$ so that the relations $0 \leq u \leq x$ and $L(u) \leq \eta$ imply $V(u) \leq \varepsilon$; with $y$ and $z$ then determined as above, we have $L(y) \leq \eta$, therefore $V(y) \leq \varepsilon$ and so

since $\varepsilon$ is arbitrary, we have $V(x) = 0$ for every $x \geq 0$, that is, $V = 0$.