absolute value in a vector lattice
Let $V$ be a vector lattice over $\mathbb{R}$, and ${V}^{+}$ be its positive cone^{}. We define three functions from $V$ to ${V}^{+}$ as follows. For any $x\in V$,

•
${x}^{+}:=x\vee 0$,

•
${x}^{}:=(x)\vee 0$,

•
$x:=(x)\vee x$.
It is easy to see that these functions are welldefined. Below are some properties of the three functions:

1.
${x}^{+}={(x)}^{}$ and ${x}^{}={(x)}^{+}$.

2.
$x={x}^{+}{x}^{}$, since ${x}^{+}{x}^{}=(x\vee 0)(x)\vee 0=(x\vee 0)+(x\wedge 0)=x+0=x$.

3.
$x={x}^{+}+{x}^{}$, since ${x}^{+}+{x}^{}=x+2{x}^{}=x+(2x)\vee 0=(x2x)\vee (x+0)=x$.

4.
If $0\le x$, then ${x}^{+}=x$, ${x}^{}=0$ and $x=x$. Also, $x\le 0$ implies ${x}^{+}=0$, ${x}^{}=x$ and $x=x$.

5.
$x=0$ iff $x=0$. The “only if” part is obvious. For the “if” part, if $x=0$, then $(x)\vee x=0$, so $x\le 0$ and $x\le 0$. But then $0\le x$, so $x=0$.

6.
$rx=rx$ for any $r\in \mathbb{R}$. If $0\le r$, then $rx=(rx)\vee (rx)=r\left((x)\vee x\right)=rx=rx$. On the other hand, if $r\le 0$, then $rx=(rx)\vee (rx)=(r)\left(x\vee (x)\right)=rx=rx$.

7.
$x+y=x+y\vee xy$, since
$$LHS=(x)\vee x+(y)\vee y=(xy)\vee (x+y)\vee (xy)\vee (x+y)=RHS.$$ 
8.
(triangle inequality). $x+y\le x+y$, since $x+y\le x+y\vee xy=x+y$.
Properties 5, 6, and 8 satisfy the axioms of an absolute value^{}, and therefore $x$ is called the absolute value of $x$. However, it is not the “norm” of a vector in the traditional sense, since it is not a realvalued function.
Title  absolute value in a vector lattice 

Canonical name  AbsoluteValueInAVectorLattice 
Date of creation  20130322 17:03:16 
Last modified on  20130322 17:03:16 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  10 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 46A40 
Classification  msc 06F20 
Defines  absolute value 