Positive, monotone decreasing function, with derivative limit in 0 equal to ∞ submultiplicative up to an factor?

Question

Related to this question.

For $x_+ \in (0,\infty)$, $a \in \mathbb{R}$ let $F\colon[0,x_+] \to [a,\infty)$ be a twice continuous differentiable (in $(0,x_+)$) function with $f := F'$, $f(x) > 0$, and $f'(x) < 0$ for all $x \in (0,x_+]$. Moreover, we assume that $$\lim_{x \to 0} f(x) = \infty$$ and $F(0) = a$ holds.

The question: Does this implies that there exists a $\beta \in (0,\infty)$ such that $f(x)f(y) \ge \beta f(xy)$ for all $x,y \in (0,x_+]$.

This special version came in my mind after i analyzed that the counterexample here relies on the fact that the function $f$ is so steep that the primitive integral $F$ has limit $\lim_{x \to 0}F(x) = -\infty$.

Answer 1

Define $f$ on $]0,e^{-2}]$ by $$f(x) = \frac{1}{-\sqrt{x}\ln(x)}.$$ Then $f$ is positive, decreasing since $$\frac{\mathrm{d}}{\mathrm{d}x}\big(-\sqrt{x}\ln(x)\big) = \frac{-\ln(x)-2}{2\sqrt{x}} \ge 0 \text{ for all } x \in (0,e^{-2}].$$ Moreover $f(x) \to +\infty$ as $x \to 0$. But since $f(x) = o(x^{-2/3})$ as $x \to 0$, we can define $F$ on $[0,e^{-2}]$ by $$F(x) = \int_0^x f(t) \mathrm{d}t.$$ The assumptions hold with $a=0$. Yet, for all $x \in (0,e^{-2}]$, $$\frac{f(x)^2}{f(x^2)} = \frac{-x \ln(x^2)}{x \ln^2(x)} = \frac{-2}{\ln(x)} \to 0 \textrm{ as } x \to 0,$$ so the ratio $f(x)f(y)/f(xy)$ is not bounded away from $0$.