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

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

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

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

• If $x=-y$, what is $g(f(x)-f(y))$ supposed to be? It is allowed to be any quantity? – Carl Schildkraut Feb 14 at 3:56
• Well, $g$ is an even function -- at least on the values we care about, which is $R -R$, where $R$ is the range of $f$. – JonathanZ supports MonicaC 2 days ago
• @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 $f(x)-f(-x)$ – High GPA 2 days ago