Proposition 1

Let E be an ordered vector space, L a mapping of E into $\mathbf{R}$ such that $L(x + y) = L(x) + L(y)$ and such that the relation $x \geq 0$ implies $L(x) \geq 0$; then $L(\lambda x) = \lambda L(x)$ for every scalar $\lambda$ and every $x \geq 0$.

Since $L(-x) = -L(x)$ ($L$ being a representation of the additive group E in $\mathbf{R}$), we can restrict ourselves to the case that $\lambda \geq 0$. For every integer $n \geq 0$, we have $L(nx) = nL(x)$, whence $L((1/n)x) = (1/n)L(x)$ and consequently $L(rx) = rL(x)$ for every rational number $r \geq 0$. On the other hand, $L$ is increasing in E; if $r$ and $r'$ are rational numbers such that $r \leq \lambda \leq r'$, it follows that $rL(x) \leq L(\lambda x) \leq r'L(x)$; since $rL(x)$ and $r'L(x)$ differ from $\lambda L(x)$ as little as we like, we have $L(\lambda x) = \lambda L(x)$.