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

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 $$\lim_{x \to 0}F(x) = -\infty$$.

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 $$0$$.