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For $x_+ \in (0,\infty)$ let $f\colon(0,x_+] \to (0,\infty)$ be a continous differentiable function with $f(x) > 0$ and $f'(x) < 0$ for all $x \in (0,x_+]$. Moreover, we assume that $$\lim_{x \to 0} f(x) = \infty$$ 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 $x,y \in (0,x_+]$.

From my intuition this is not valid, however i am not able to derive a counter example; that function needs to be really steep in direction $x \to 0$.

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  • $\begingroup$ What about $f(x) = \frac{a}{x^p}$ for $a, p > 0$? You can take $\beta = a$. $\endgroup$
    – Michael Albanese
    Commented Jun 24, 2022 at 13:57
  • $\begingroup$ Why write both $(0,x_+]$ and $]0,x_+]$ in the same post? $\endgroup$
    – Gerald Edgar
    Commented Jun 24, 2022 at 15:42
  • $\begingroup$ @GeraldEdgar by accident, this has been fixed now. $\endgroup$
    – maximilian43
    Commented Jun 24, 2022 at 16:00

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A counterexample is given by $$f(x)=e^{1/x}.$$

Indeed, then $f(x)f(x)/f(xx)\to0$ as $x\downarrow0$, so that, for any real $x_+$, there is no real $\beta>0$ such that $f(x)f(y)\ge\beta f(xy)$ for all $x,y$ in $(0,x_+]$

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  • $\begingroup$ This was exactly what i was looking for, i was playing around with some exponential functions but couldn't find the right arguments, thank you very much! $\endgroup$
    – maximilian43
    Commented Jun 24, 2022 at 14:09
  • $\begingroup$ @maximilian43 : You are welcome. $\endgroup$
    – Iosif Pinelis
    Commented Jun 24, 2022 at 14:16

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