Prove that there does not exist 1D real functions $f,g$ such that $h(x,y)=g(f(x)-f(y))$.
The problem seems really really easy because it is obvious that $x+y\neq f(x)-f(y)$. But I only have a really complicated approach outlined below, which is not necessarily correct. Do you have a simple proof?
By contradiction. suppose that $h=g(f(x)-f(y))$.
If we further assume that $h(a,b)\geq h(\alpha,\beta)$ and $h(b,c)\geq h(\beta,\gamma)$,
we must have: $h(a,b)\geq h(\alpha,\gamma)$.
Let $a=1$, $b=2$, $c=3$;
$\alpha=95.95$, $\beta=100$, $\gamma=103.44$.
We have: $h(\alpha,\gamma)=2.10$, which is less than $h(a,c)=4$
$h(a,b)=1/3$, $h(b,c)=1/5$, which are less than $h(\alpha,\beta)=0.339$, $h(\beta,\gamma)=0.2001$
Here is another approach but assuming the differentiability of $f$ and $g$. Assume $h=g(f(x)-f(y))$
The partial derivatives: $h_x=g'f_x$