While working out some thermodynamics, I came across the following problem, which may be viewed as a functional inequality:
Prove/disprove the existence of a differentiable function $f$ such that $$\Phi(x,y) = (x-y)\, (f'(x)-g'(y)) \geq 0, \qquad \forall\; (x, y)\in \Bbb R^2$$ for a given differentiable function $g$.
In the case where $g$ is convex, then the choice $f = g$ ends the proof.
Question: What happens if $g$ is not convex, e.g., if $g$ has an inflection point at $y=0$?
Attempt: Let us consider an example. In the case $g(y) = \frac13 y^3$, we look for $f$ such that \begin{aligned} \Phi(x,y) &= (x-y)\, (f'(x) - y^2) \\ &= x f'(x) - f'(x) y - x y^2 + y^3 \geq 0 \, . \end{aligned} By expanding $\Phi$ as a polynomial function of $y$, one observes that $\Phi$ becomes negative as $y \to {-\infty}$ for fixed $x$. Hence, there is no such function $f$. $\square$
The same conclusion is obtained for any function $g$ such that $y g'(y)$ decreases faster than identity as $y\to -\infty$. Nevertheless, this argument does not work for $g(y) = \tanh y$ for instance.
Can we find an appropriate $f$ in some particular cases? Or is it always impossible as soon as $g$ is nonconvex?