Writing $f(x,y)$ as $\Phi(g(x) + h(y))$

Solution 1:

The statement is not true.

To make the counterexample simpler, I will take $f:[-1,1]^2 \rightarrow \mathbb{R}$. Let f(x,y)=xy. So we have

---0+++
---0+++
0000000
+++0---
+++0---

Consider the image of xy=0 in g(x)+h(y). Since it's a connected, compact set, the image is connected and compact, so an interval [m,n], with $\Phi([m,n])=0$. We may assume that $\Phi(n+\epsilon)>0$ and $\Phi(m-\epsilon)<0$. Let A=g(1), a=g(-1), B=h(1), b=h(-1).

Then we have $A+B>n, \; a+b>n, \; A+b<m, \; a+B<m$, which leads to a contradiction.

Solution 2:

Suppose $f$ is such that $f(x,0)$ and $f(0,y)$ are monotonic in $x$ and $y$ respectively over $[0,1]$. Since $f(x,0) = \Phi(g(x) + h(0))$, it must be the case that $g$ is monotonic over $[0,1]$. Similarly, $h$ is monotonic over $[0,1]$. So $g(x) + h(y)$ is the sum of monotonic functions. It's not hard to show that for any $(x,y)$ in the interior of $[0,1]^2$, there must be a point on the boundary with the same value of $g(x) + h(y)$, and therefore the same value of $\Phi(g(x) + h(y))$. Since it's easy to construct a continuous $f$ which satisfies the first assumption yet takes a larger range of values in the interior than it does on the boundary (for any $f$ that doesn't, try adding $M x(1-x)y(1-y)$ for large enough $M$), this contradicts the conjecture.