# Vanishing derivative at infinity implies slowly varying

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)

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

## 2 Answers

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.

• Do you mean mean value theorem?
– user515599
Jun 17, 2020 at 16:19
• Of course. My bad. Jun 17, 2020 at 16:21
• Your A appeared as I was typing mine. I am an extremely slow typist........ +1 Jun 17, 2020 at 16:29

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 $$y_x\ge \min (\log x,\,\log \lambda x).$$