5
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$$h(x,y)=\frac{(x-y)^3}{x+y}$$

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?


My try:

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: $h(a,b)\geq h(\alpha,\gamma)$.

However, since

$$h(x,y)=\frac{(x-y)^3}{x+y}$$

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: $h_x=g'f_x$

$h_y=g'f_y$

$h_x/h_y=f_x/f_y$

Therefore

$$h_x/h_y=\frac{x+2y}{-2x-y}=f_x/f_y$$

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  • 1
    $\begingroup$ If $x=-y$, what is $g(f(x)-f(y))$ supposed to be? It is allowed to be any quantity? $\endgroup$ – Carl Schildkraut Feb 14 at 3:56
  • $\begingroup$ Well, $g$ is an even function -- at least on the values we care about, which is $R -R$, where $R$ is the range of $f$. $\endgroup$ – JonathanZ supports MonicaC 2 days ago
  • 1
    $\begingroup$ @CarlSchildkraut $h$ is undefined at $(x,-x)$ (i.e. the domain of $h$ does not include any points such that $x=-y$). $g$ is also undefined at the point $f(x)-f(-x)$ $\endgroup$ – High GPA 2 days ago

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