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The functions $$f(x)=e^{(\ln x)^{1/3} \cos((\ln x)^{1/3}) } \quad g(x)=e^{\sqrt{\ln x} \cos((\ln x)^{1/3}) }$$

are oscillating but slowly varying at infinity, that is for all $\lambda >0$, we have $$\lim_{x\to \infty} \frac{f(\lambda x)}{{f(x)}}=\lim_{x\to \infty} \frac{g(\lambda x)}{{g(x)}}=1$$

this leads me to ask the following question:

Suppose $h$ is differentiable hence continuous (or maybe uniformly?) and positive on $[A,\infty)$ with $h'(x)\to 0$ as $x\to \infty$, then $$f(x)=e^{h(\log(x))}$$ is slowly varying at infinity.

For the concrete functions I had, namely $f$ and $g$, I simply used L'Hopital to show the exponent goes to zero, not sure if this would hold for arbitrary $h$.

(The reason probability theory is tagged is because this is from extreme value theory)

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  • $\begingroup$ it is in the question, the limit is 1 for all $\lambda >0$ $\endgroup$
    – user515599
    Jun 17, 2020 at 15:33
  • $\begingroup$ Ah my bad, I see :) $\endgroup$ Jun 17, 2020 at 15:36
  • $\begingroup$ For $h'(x)\to 0$ to make sense, you should start by assuming that $h$ is differentiable, which implies continuity, of course. $\endgroup$ Jun 17, 2020 at 15:52
  • $\begingroup$ Sure why not, looking for any meaningful statement $\endgroup$
    – user515599
    Jun 17, 2020 at 15:55

2 Answers 2

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Assume $h$ is differentiable on an interval $[A,\infty)$ and $h'(x)\to 0$ as $x\to \infty$. Then by the Mean Value Theorem, $$\frac{f(\lambda x)}{f(x)} = e^{h(\log \lambda + \log x)-h(\log x)} = e^{\log(\lambda) h'(\xi(x))}, $$ where $\log x\le \xi(x)\le \log x + \log\lambda$. So, $h'(\xi(x))\to 0$ as $x\to \infty$, which shows that your conjecture is correct. You don't need to assume $h$ is positive.

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  • $\begingroup$ Do you mean mean value theorem? $\endgroup$
    – user515599
    Jun 17, 2020 at 16:19
  • $\begingroup$ Of course. My bad. $\endgroup$ Jun 17, 2020 at 16:21
  • $\begingroup$ Your A appeared as I was typing mine. I am an extremely slow typist........ +1 $\endgroup$ Jun 17, 2020 at 16:29
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There exists $y_x$ in the closed interval with end-point(s) $\log x,\,\log \lambda x$ such that $$h(\log \lambda x)-h(\log x)=h'(y_x)(\log [\lambda x]-\log x)=h'(y_x)\log \lambda.$$ Therefore $$\frac {\exp (h(\log \lambda x))}{\exp (h(\log x))}=\frac {\exp (h(\log x)+h'(y_x)\log \lambda)}{\exp (h(\log x))}=$$ $$=\exp (h'(y_x)\log \lambda)$$ which $\to 1$ as $x\to \infty$ because $y_x\ge \min (\log x,\,\log \lambda x).$

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