TITLE: Associative mean
QUESTION [9 upvotes]: Can there be a function $m(a,b)$ that is both associative and a mean, i.e., 
$\min (a,b) \leq m(a,b) \leq \max (a,b)$?  The obvious solutions are $m(a,b) = \max(a,b)$ or $\min(a,b)$, but are there more?

REPLY [3 votes]: Let us, indeed, describe "monotone, continuous, associative means" on $[0,1]^2$. I'll just write $a*b$ instead of $m(a,b)$ to make long expressions easier to comprehend. Let $c=0*1,d=1*0$.  Then, for every $x\le c\le y$, we have
$c\le c*y=(0*1)*y=0*(1*y)\le 0*y\le 0*1=c$ and, thereby, $c=0*y\le x*y\le c*y=c$, i.e., $x\le c\le y$ implies $x*y=c$.
Let now $a$ be any number with $q=0*a<a$. Repeating the same argument on $[0,a]$ instead of $[0,1]$, we see that for every $x\le q\le y$, we have $x*y=q$. Thus the function $y\mapsto 0*y$ is identically $q$ on $(q,a)$. If we have $a'>a$ such that $q'=0*a'>q$ at all, then, by continuity we can find such $a'$ with $q'<a$. But then $0*y$ should be simultaneously $q$ and $q'$ on $(q',a)$, which is impossible. Thus, if $q=0*a<a$, then $q=0*a'$ for all $a'\ge a$, i.e., $y\mapsto 0*y=y$ up to a certain point $y_0$ and then constant $y_0$, i.e., $0*y=\min(y,y_0)$. But then this constant must be $0*1=c$. Similarly we can conclude that $x*1=\max(x,c)$. Thus, if $x\le y$, then
$$
x\le x*y\le x*1=\max(x,c), \min(y,c)= 0*y\le x*y\le y 
$$
But for $x\le y$ on one side of $c$ one of these double inequalities is a tight squeeze, so if $x\le y\le c$, then $x*y=y$, if $c\le x\le y$, then $x*y=x$, and we have already seen that if $x\le c\le y$, then $x*y=c$. Thus we have the full description of $x*y$ for $x\le y$. The case $x\ge y$ is treated in the same way using $d$.