order bound dual

In mathematics, specifically in order theory and functional analysis, the order bound dual of an ordered vector space $X$ is the set of all linear functionals on $X$ that map order intervals, which are sets of the form $$, b:= \, to bounded sets.
The order bound dual of $X$ is denoted by $X^.$ This space plays an important role in the theory of ordered topological vector spaces.

Canonical ordering

An element $g$ of the order bound dual of $X$ is called positive if $x \geq 0$ implies $\operatorname(f(x)) \geq 0.$
The positive elements of the order bound dual form a cone that induces an ordering on $X^$ called the .
If $X$ is an ordered vector space whose positive cone $C$ is generating (meaning $X = C - C$) then the order bound dual with the canonical ordering is an ordered vector space.

Properties

The order bound dual of an ordered vector spaces contains its order dual.
If the positive cone of an ordered vector space $X$ is generating and if for all positive $x$ and $x$ we have $$, x+ , y= , x + y then the order dual is equal to the order bound dual, which is an order complete vector lattice under its canonical ordering.

Suppose $X$ is a vector lattice and $f$ and $g$ are order bounded linear forms on $X.$
Then for all $x \in X,$
# $\sup(f, g)(, x, ) = \sup \$
# $\inf(f, g)(, x, ) = \inf \$
# $, f, (, x, ) = \sup \$
# $, f(x), \leq , f, (, x, )$
# if $f \geq 0$ and $g \geq 0$ then $f$ and $g$ are lattice disjoint if and only if for each $x \geq 0$ and real $r > 0,$ there exists a decomposition $x = a + b$ with $a \geq 0, b \geq 0, \text f(a) + g(b) \leq r.$

See also

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References

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{{Functional analysis

Functional analysis