# properties of certain monotone functions

In the definitions of some partially ordered algebraic systems such as po-groups and po-rings, the multiplication is set to be compatible with the partial ordering on the universe in the following sense:

 $ab\leq ac\quad\mbox{iff}\quad b\leq c\qquad\mbox{ and }\qquad ab\leq cb\quad% \mbox{iff}\quad a\leq c$

This is no coincidence. In fact, these “definitions” are actually consequences of properties concerning monotone functions satisfying certain algebraic rules, which is the focus of this entry.

Recall that an $n$-ary function $f$ on a set $A$ is said to be monotone if it is monotone in each of its variables. In other words, for every $i=1,2,\ldots,n$, the function $f(a_{1},\ldots,a_{i-1},x,a_{i+1},\ldots,a_{n})$ is monotone in $x$, where each of the $a_{j}$ is a fixed but arbitrary element of $A$. We use the notation $\uparrow,\downarrow,\updownarrow$ to denote the monotonicity of each variable in $f$. For example, $(\uparrow,\uparrow,\uparrow)$ denotes a ternary isotone function, whereas $(\downarrow,\updownarrow)$ denotes a binary function which is antitone with respect to its first variable, and both isotone/antitone with respect to the second.

###### Proposition 1.

Let $f$ be an $n$-ary commutative monotone operation on a set $A$. Then $f$ is either isotone or antitone.

###### Proof.

Suppose $f$ is isotone (or antitone) in its first variable. Since $f(x,a_{1},\ldots,a_{n-1})=f(a_{1},x,\ldots,a_{n-1})=\cdots=f(a_{1},\ldots,a_{n% -1},x)$, $f$ is isotone (or antitone) in each of its remaining variables. ∎

###### Proposition 2.

Let $f$ be an $n$-ary monotone operation on a set $A$ with an identity element $e\in A$. In other words, $f(x,e,\ldots,e)=f(e,x,\ldots,e)=\cdots=f(e,e,\ldots,x)=x$. Then $f$ is either strictly isotone or strictly antitone.

###### Proof.

The proof is the same as the one before. Furthermore, if $f$ is isotone and $a, then $f(a,e,\ldots,e)=a, so the strict ordering is preserved. The same holds true if $f$ is antitone. ∎

###### Proposition 3.

Let $f$ be a binary monotone operation on a set $A$ such that it is isotone (antitone) with respect to its first variable. Suppose $g$ is a unary operation on $A$ such that $f(x,g(x))$ is a fixed element of $A$. Then $g$ is antitone (isotone).

###### Proposition 4.

Let $f$ be an $n$-ary associative monotone operation on a set $A$. Then

• $f$ is isotone if $n$ is even

• $f$ is either isotone, or is $(\underbrace{\uparrow,\ldots,\uparrow}_{m},\downarrow,\underbrace{\uparrow,% \ldots,\uparrow}_{m})$, if $n$ is odd, say $n=2m+1$.

###### Proof.

Suppose first that $n=2m$, $i\leq m$, and $g(x)=f(a_{1},\ldots,a_{i-1},x,\ldots,a_{m+1},\ldots,a_{2m})$ is antitone. Then $g(g(x))$ is isotone. By the associativity of $f$, $g(g(x))$ is

 $\displaystyle f(a_{1},\ldots,a_{i-1},f(a_{1},\ldots,a_{i-1},x,a_{i+1},\ldots,a% _{m},\ldots,a_{2m}),\ldots,a_{m+1},\ldots,a_{2m})$ $\displaystyle=$ $\displaystyle f(a_{1},\ldots,a_{i-1},a_{1},\ldots,a_{i-1},x,a_{i+1},\ldots,f(a% _{2m-i+1},\ldots,a_{2m},a_{i+1},\ldots,a_{m+1},\ldots,a_{2m})).$

In the second expression, the position of $x$ is $2i-2\leq 2m-1<2m$, therefore implying that $g(g(x))$ is antitone, which is a contradiction! Therefore, $g(x)$ is isotone. Now, if $i>m$, and $h(x)=f(b_{1},\ldots,b_{m},\ldots,b_{i-1},x,b_{i+1},\ldots,b_{2m})$ is antitone, then $h(h(x))$ is isotone. But

 $\displaystyle f(b_{1},\ldots,b_{m},\ldots,b_{i-1},f(b_{1},\ldots,b_{m},\ldots,% b_{i-1},x,b_{i+1},\ldots,b_{2m}),b_{i+1},\ldots,b_{2m})$ $\displaystyle=$ $\displaystyle f(f(b_{1},\ldots,b_{m},\ldots,b_{i-1},b_{1},\ldots,b_{2m-i+1}),% \ldots,b_{i-1},x,b_{i+1},\ldots,b_{2m},b_{i+1},\ldots,b_{2m})),$

and the position of $x$ is the second expression is $(i-1)-(2m-i+1)+2=2i-2m>1$, therefore implying that $h(h(x))$ is antitone, again a contradiction. As a result, $f$ is isotone for all $i=1,\ldots,n$.

The argument above also works when $n$ is odd, say $n=2m+1$ and $i\neq m+1$. Finally, since $f$ is monotone, it is monotone with respect to the $i$-th variable when $i=m+1$, so $f$ is one of the following three forms:

 $(\underbrace{\uparrow,\ldots,\uparrow}_{2m+1}),\qquad(\underbrace{\uparrow,% \ldots,\uparrow}_{m},\updownarrow,\underbrace{\uparrow,\ldots,\uparrow}_{m}),% \qquad(\underbrace{\uparrow,\ldots,\uparrow}_{m},\downarrow,\underbrace{% \uparrow,\ldots,\uparrow}_{m}),$

the first two of which imply that $f$ is isotone. ∎

An example of an associative function that is, say $(\uparrow,\downarrow,\uparrow)$, is given by

 $f:\mathbb{Z}^{3}\to\mathbb{Z}\qquad\mbox{where}\qquad f(x,y,z)=x-y+z.$

$f$ is associative since $f(f(r,s,t),u,v)=f(r,f(s,t,u),v)=f(r,s,f(t,u,v))=r-s+t-u+v$.

Title properties of certain monotone functions PropertiesOfCertainMonotoneFunctions 2013-03-22 19:03:31 2013-03-22 19:03:31 CWoo (3771) CWoo (3771) 9 CWoo (3771) Result msc 06F99 msc 08C99 msc 08A99