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Any real function can be expressed as a function of the sum of two monotonic real functions?

More precisely, for real function p(x, y), there exist continuous real functions P(x), h(x,y), g(x) such that:


Where $P(x), h(x,y), g(x)$ are arbitrary satisfying $\frac {d(g(x))}{dx}>0$, $\frac {\partial h(x,y)}{\partial y}>0$

This is equivalent to mine another question “Solving Functional Equation”. By letting $h(x,y)=ln(w(x,y)), g(x)=-ln(u(x)), p(x,y)=\frac {u(f(x,y))}{u(x)}$, we have:

$p(x,y)=\frac {u(f(x,y))}{u(x)}=F[ln(w(x,y))-ln(u(x))]=F(ln \frac {w(x,y)}{u(x)})=\Psi (\frac {w(x,y)}{u(x)})$

Thank you very much!

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You seem to be missing a continuity hypothesis on $p$. – Joel David Hamkins Dec 3 '10 at 11:57
I'm voting to close as the question is too localised. But a quick counterexample: Let $p(0,y)$ be some function such that $p(0,y) = 0$. Since $h$ is continuous and monotonic in $y$, there exists some $z_0 \in \mathbb{R}$ and a small $\delta$ such that $P(z) = 0$ whenever $|z-z_0| = \delta$, and such that there is some $(0,y_0)$ such that $h(0,y_0) + g(0) = z_0$. Now using continuity of $g, h$, the function $P(h(x,y) + g(x))$ must vanish on a neighborhood of $(0,y_0)$. So a function like $p(x,y) = xy$ cannot be written in the form you want. – Willie Wong Dec 3 '10 at 12:00
By setting $P(x)=e^x, g(x)=ln(x), h(x,y)=ln(y)$, we have: $p(x,y)=xy=e^{ln(xy)}=e^{ln(x)+ln(y)}=P(g(x)+h(x,y))$ Where: $\frac {\partial lnx}{\partial x}=\frac {1}{x}>0$, $\frac {\partial lny}{\partial y}=\frac {1}{y}>0$ for $x>0, y>0$. – Wang Tao Dec 4 '10 at 2:40

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