Order unit

For the ordering cone ${\displaystyle K\subseteq X}$ in the vector space ${\displaystyle X}$, the element ${\displaystyle e\in K}$ is an order unit (more precisely an ${\displaystyle K}$-order unit) if for every ${\displaystyle x\in X}$ there exists a ${\displaystyle \lambda _{x}>0}$ such that ${\displaystyle \lambda _{x}e-x\in K}$ (i.e. ${\displaystyle x\leq _{K}\lambda _{x}e}$).[3]

Let ${\displaystyle X=\mathbb {R} }$ be the real numbers and ${\displaystyle K=\mathbb {R} _{+}=\{x\in \mathbb {R} :x\geq 0\}}$, then the unit element ${\displaystyle 1}$ is an order unit.

Let ${\displaystyle X=\mathbb {R} ^{n}}$ and ${\displaystyle K=\mathbb {R} _{+}^{n}=\{x\in \mathbb {R}$ :\forall i=1,\ldots ,n:x_{i}\geq 0\}} , then the unit element ${\displaystyle {\vec {1}}=(1,\ldots ,1)}$ is an order unit.

Each interior point of the positive cone of an ordered TVS is an order unit.[2]

Each order unit of an ordered TVS is interior to the positive cone for the order topology.[2]

If (X, ≤) is a preordered vector space over the reals with order unit u, then the map ${\displaystyle p(x):=\inf \left\{t\in \mathbb {R} :x\leq tu\right\}}$ is a sublinear functional.[4]

Suppose (X, ≤) is an ordered vector space over the reals with order unit u whose order is Archimedean and let U = [-u, u]. Then the Minkowski functional p_U of U (defined by ${\displaystyle p_{U}(x):=\left\{r>0:x\in r[-u,u]\right\}}$) is a norm called the order unit norm. It satisfies p_U(u) = 1 and the closed unit ball determined by p_U is equal to [-u, u] (i.e. [-u, u] = \{ x \in X : p_U(x) ≤ 1 \}.[4]

• Narici, Lawrence; Beckenstein, Edward (2011). Topological Vector Spaces. Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press. ISBN 978-1584888666. OCLC 144216834.
• Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. 8 (Second ed.). New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.