I have a continuous function $\mu(x,y)=G(h\_1(x+y), h\_2(|x-y|))$ such that $\mu(x,x)=0$ and $\mu(x,y)+\mu(y,z)=\mu(x,z)$ for $x\le y\le z$. I want to show that the only possible case is $\mu(x,y)=c\cdot (x+y)|x-y|$ for some constant $c$.


If I assume that I have a polynomial, no problem, since I have elementary symmetric polynomial as the basis of symmetric polynomials. My question: is it possible for $\mu(x,y)$ to not be a polynomial? 



Symmetric function
If $g(s,t) = a(s+t) - a(s-t)$ for an arbitrary continuous function $a$, $u(x,y) = g(x+y, y-x) = a(2y) - a(2x)$ satisfies $u(x,x) = 0$ and $u(x,y) + u(y,z) = u(x,z)$ for all $x,y,z$.  

Then $\mu(x,y) = g(x+y,|x-y|)$ agrees with $u(x,y)$ when $x \le y$, and so
$\mu(x,x) = u(x,x) = 0$ and $\mu(x,y) + \mu(y,z) = u(x,y) + u(y,z) = u(x,z) = \mu(x,z)$ for $x \le y \le z$. There are lots more solutions than $c (x+y)|x-y|$.


