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

Could you prove or disprove the following statement?

Let $f\colon[0,1]^2\rightarrow \mathbb R$ be a continuous function. Then there are continuous functions $g,\ h\colon [0,1]\rightarrow \mathbb R$ and $\Phi\colon \mathbb R \to \mathbb R$ such that $$f(x,y) = \Phi(g(x) + h(y)).$$

(This problem popped up in my mind while I was thinking about this related one on MO. I couldn't find an easy proof or a disproof. This version is much weaker than the one asked at MO, since $g$ and $h$ do depend on $f$ here.)

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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

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0000000
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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.

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I'm a little bit confused here. Are you sure you want $\Phi(m + \epsilon) > 0$ and $\Phi(n - \epsilon) < 0$? Don't you mean $\Phi(m - \epsilon) > 0$ and $\Phi(n + \epsilon) < 0$ or another variant? Probably I'm just being stupid and missing something... –  Adrián Barquero Nov 29 '10 at 3:45
@Adrian: No, you're completely right! Thanks for the correction; I was too hasty and mixed up m and n. It's fixed now. –  Jonas Kibelbek Nov 29 '10 at 4:48

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.

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Consider $f(x,y) = x+y$. It satisfies your assumptions, but doesn't lead to a counterexample. –  AgCl Nov 29 '10 at 3:07
@AgCl: You're right, I shouldn't have said "Choose any $f$". I'll edit. –  Rahul Nov 29 '10 at 3:14