Order unit explained

An order unit is an element of an ordered vector space which can be used to bound all elements from above.[1] In this way (as seen in the first example below) the order unit generalizes the unit element in the reals.

According to H. H. Schaefer, "most of the ordered vector spaces occurring in analysis do not have order units."

Definition

For the ordering cone

K\subseteqX

in the vector space

X

, the element

e\inK

is an order unit (more precisely a

K

-order unit) if for every

x\inX

there exists a

λx>0

such that

λxe-x\inK

(that is,

x\leqKλxe

).[2]

Equivalent definition

The order units of an ordering cone

K\subseteqX

are those elements in the algebraic interior of

K;

that is, given by

\operatorname{core}(K).

Examples

Let

X=\R

be the real numbers and

K=\R+=\{x\in\R:x\geq0\},

then the unit element

1

is an .

Let

X=\Rn

and

K=

n
\R
+

=\left\{xi\in\R:foralli=1,\ldots,n:xi\geq0\right\},

then the unit element

\vec{1}=(1,\ldots,1)

is an .

Each interior point of the positive cone of an ordered topological vector space is an order unit.

Properties

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

If

(X,\leq)

is a preordered vector space over the reals with order unit

u,

then the map

p(x):=inf\{t\in\R:x\leqtu\}

is a sublinear functional.

Order unit norm

Suppose

(X,\leq)

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

pU

of

U,

defined by

pU(x):=inf\{r>0:x\inr[-u,u]\},

is a norm called the . It satisfies

pU(u)=1

and the closed unit ball determined by

pU

is equal to

[-u,u];

that is,

[-u,u]=\left\{x\inX:pU(x)\leq1\right\}.

Notes and References

  1. Book: Convex Cones. Fuchssteiner. Benno. Lusky. Wolfgang. 9780444862907. Elsevier. 1981.
  2. Book: Charalambos D. Aliprantis. Rabee Tourky. Cones and Duality. 9780821841464. American Mathematical Society. 2007.