absolute value in a vector lattice
AbsoluteValueInAVectorLattice
2007-05-07 15:50:34
2008-06-21 15:34:42
Definition
vector lattice
absolute value
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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 $x\in V$,
\begin{itemize}
\item $x^+:=x\vee 0$,
\item $x^-:=(-x)\vee 0$,
\item $|x|:=(-x)\vee x$.
\end{itemize}
It is easy to see that these functions are well-defined. Below are some properties of the three functions:
\begin{enumerate}
\item $x^+=(-x)^-$ and $x^-=(-x)^+$.
\item $x=x^+-x^-$, since $x^+-x^-=(x\vee 0)-(-x)\vee 0=(x\vee 0)+(x\wedge 0)=x+0=x$.
\item $|x|=x^++x^-$, since $x^++x^-=x+2x^-=x+(-2x)\vee 0=(x-2x)\vee (x+0)=|x|$.
\item If $0\le x$, then $x^+=x$, $x^-=0$ and $|x|=x$. Also, $x\le 0$ implies $x^+=0$, $x^-=-x$ and $|x|=-x$.
\item $|x|=0$ 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 $x=0$.
\item $|rx|=|r||x|$ for any $r\in \mathbb{R}$. If $0\le r$, then $|rx|=(-rx)\vee (rx)=r\big((-x)\vee x\big)=r|x|=|r||x|$. On the other hand, if $r\le 0$, then $|rx|=(-rx)\vee (rx)=(-r)\big(x\vee (-x)\big)=-r|x|=|r||x|$.
\item $|x|+|y|=|x+y|\vee |x-y|$, since $$LHS=(-x)\vee x+(-y)\vee y=(-x-y)\vee (-x+y)\vee (x-y)\vee (x+y)=RHS.$$
\item (triangle inequality). $|x+y|\le |x|+|y|$, since $|x+y|\le |x+y|\vee |x-y|=|x|+|y|$.
\end{enumerate}
Properties 5, 6, and 8 satisfy the axioms of an absolute value, and therefore $|x|$ is called the \emph{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.