$\frac{(x-y)^3}{x+y}\neq g(f(x)-f(y))$?

Question

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

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?

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

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 $h(\alpha,\beta)=0.339$, $h(\beta,\gamma)=0.2001$

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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

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

Answer 1

I can't follow your attempts.^{Fn. 1}  Here's a correct solution, although I can't exactly call it simple.  This is a surprisingly hard problem!

Generalities

Note that $$f(x)-f(y)=(f(x)-f(0))-(f(y)-f(0))$$ Thus we may take $f(0)=0$ without loss of generality (replace $f$ with $x\mapsto f(x)-f(0)$).

$f$ has constant sign

Unless $x=0$, \begin{equation} g(f(x))=h(x,0)=x^2\geq0 \end{equation} Likewise, $$g(-f(y))=h(0,y)=-y^2\leq0$$ So $\mathrm{rng}(f)\cap\mathrm{rng}(-f)\subseteq\{0\}$, i.e. $f$ is either always nonnegative or always nonpositive.

Inverting $g$

Equation (1) also has another, more surprising, property: $g\circ f$, considered as a map $\mathbb{R}^+\to\mathbb{R}^+$, has an inverse.  To see this, note that $g(f(\sqrt{x}))=x$; likewise $\sqrt{g(f(x))}=x$.  Thus $g$ is injective and surjective: that is, invertible.

Likewise $g(-f(x))=-x^2$.  Thus $g$ is invertible on $\mathbb{R}^-$ too; the only place it might not be invertible is $0$.  But that last case is easily taken care of: for any $x\neq0$, we have $g(0)=h(x,x)=0$; whereas $g(f(x))=h(x,0)\neq0$.  Thus $g(0)=0$ is the only root of $g$, and $g$ is invertible on all of $\mathbb{R}=\mathbb{R}^+\sqcup\{0\}\sqcup\mathbb{R}^-$.

A recursion for $f$

Now, $h$ is homogenous of degree $2$: $h(a,1)x^2=h(ax,x)$  Thus $$g(f(a)-f(1))=g(f(ax)-f(x))$$ for any $x\neq0$ and $a\neq1$  Since $g$ can be inverted, $$f(a)-f(1)=f(ax)-f(x)$$ again for any $x\neq0$ and $a\neq1$  In particular, \begin{equation} f(2)-f(1)=f(2x)-f(x) \end{equation} Let $C=f(2)-f(1)$ and fix some $x_0>0$; then $f(2^kx_0)=kC+f(x_0)$ by iterating (2).

A contradiction at last

Now restrict to only positive values of $x$; by our remarks above, this forces $f$ to be all of one sign (either positive or negative).  But iterated (2) conflicts with our conclusion that $f$ is all of one sign.  If $f(x_0)$ is positive, then there is some large $k$ such that $f(2^{-k}x_0)=f(x_0)-kC<0$; if $f(x_0)$ is negative, then $f(2^kx_0)=kC+f(x_0)>0$.  Thus the proposed decomposition is impossible.

Footnote

^{Fn. 1} Where does your proposed monotonicity formulas for $h$ come from?  What should I conclude from the ratio of partial derivatives?