1
$\begingroup$

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?

Improve this question
$\endgroup$
3
  • $\begingroup$ What can you tell us about $h_1,h_2,G$? $\endgroup$
    – Alan
    Commented Jun 23, 2013 at 17:58
  • $\begingroup$ $G,h_1,h_2$ are continuous functions. $\endgroup$
    – Kito
    Commented Jun 23, 2013 at 18:03
  • $\begingroup$ There are many other examples, e.g. $\mu(x,y)=|x^3-y^3|$. $\endgroup$
    – Sergei Ivanov
    Commented Jun 23, 2013 at 18:35

1 Answer 1

Reset to default
3
$\begingroup$

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 $c (x+y)|x-y|$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Indeed that's true. So I guess I was on the wrong path. thank you. $\endgroup$
    – Kito
    Commented Jun 23, 2013 at 19:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .