Let $f:\mathbb{R}^+\to\mathbb{R}^+$ be such that $f(0)=0$, $f'(x)>0$ and $f''(x)>0$ (increasing and convex).
For $K\geq 0$ let
\begin{align}
g(x) = \frac{xf'(x)}{K+f(x)}.
\end{align}
Is the function $g$ increasing? It is increasing in standard cases like $f(x)=x^{\alpha}$ for $\alpha\geq 1$, and $f(x)=e^{x}-1$. Are there obvious counterexamples?
If $K=0$, then $g'(x)>0$ is equivalent to
\begin{align}
x\frac{f''(x)}{f'(x)} \geq x\frac{f(x)}{f'(x)} – 1.
\end{align}
Is this an interpretable condition? I have seen $x\frac{f''(x)}{f'(x)}$ referred to as the curvature of a function.
Best Answer
A counterexample for $K=0$ is the function $$ f(x) = e^{-x} + x - 1 $$ which satisfies all the given conditions on $(0, \infty)$. But $$ g(x) = \frac{x f'(x)}{f(x)}= \frac{x(1-e^{-x})}{e^{-x}+x-1} $$ satisfies $$ g(1) = e-1 > 1 \\ \lim_{x \to \infty} g(x) = 1 $$ so that $g$ can not be increasing everywhere.
A counterexample for arbitrary $K > 0$ is $$ f(x) = 2K(e^{-2Kx} + 2Kx - 1) \, . $$ Then $$ g(x) = \frac{x f'(x)}{K+f(x)}= \frac{4Kx(1-e^{-2Kx})}{2(e^{-2Kx}+2Kx)-1} = h(2Kx) $$ with $$ h(x) = \frac{2x(1-e^{-x})}{2(e^{-x}+x)-1} \, . $$ But $$ h(4) = \frac{8-8e^{-4}}{7+2e^{-4}} = 1 + \frac{e^4-10}{7e^4 + 2}> 1 $$ and $\lim_{x \to \infty} h(x) = 1$, so that $h$ (and therefore $g$) can not be increasing everywhere.