absolute value in a vector lattice

Let $V$ be a vector lattice over $ℝ$, 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 well-defined. 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 $x^{+}+x^{-}=x+2x^{-}=x+(-2x)\vee 0=(x-2x)\vee (x+0)=|x|$.

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

7. $|x|+|y|=|x+y|\vee |x-y|$, since

   $$LHS=(-x)\vee x+(-y)\vee y=(-x-y)\vee (-x+y)\vee (x-y)\vee (x+y)=RHS.$$

8. (triangle inequality). $|x+y|\le |x|+|y|$, since $|x+y|\le |x+y|\vee |x-y|=|x|+|y|$.

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 real-valued function.