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

More precisely, for real function '$p(x, y): \Re^2 \to \Re^1$', there exist $C^1$ continuous real functions '$P(x), h(x,y), g(x)$' such that:

$p(x,y)=P(h(x,y)+g(x))$

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!

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:

$p(x,y)=P(h(x,y)+g(x))$

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!