A function $f$ is called regular varying with level $\alpha$ if for any $\lambda > 0$ it holds

$\lim_{x \rightarrow \infty} \frac{f(\lambda x)}{f(x)} = \lambda ^\alpha$.

A regular varying function with level $\alpha = 0$ is called slowly varying.

Given is following task:

Which of the following functions are regular varying?

$f_1(x) = e^{\log(x)}$

$f_2(x) = e^{\lfloor \log(x) \rfloor}$

Which of the following functions are slowly varying?

$f_3(x) = 2 + \sin(x)$

$f_4(x) = e^{(\log(x))^\beta}$, $\beta \in \mathbb R$

The first function is easy since I am interested in the limes for $x \rightarrow \infty$, I can use that for $x>0$ it holds that $f_1(x) = e^{\log(x)} = x$ and thus $\lim_{x \rightarrow \infty} \frac{f_1(\lambda x)}{f_1(x)} = \frac{\lambda x}{x} = \lambda$, so $f_1$ is regular varying with level $\alpha = 1$.

Do you have some hints for the other functions?

  • 2
    $\begingroup$ Taking $\lambda = e^{1/2}$ for $f_2$ then when $x=e^{n}$ with $n$ an integer, then you have $f_2(\lambda x)=f_2(x)$ so you'd need $\alpha=1.$ But when $x=e^{n+1/2}$ you have $f_2(x)=x,f_2(\lambda x)=ex$ so you have $\limsup f_2(\lambda x)/f_2(x)=\lambda^2$ and $\liminf f_2(\lambda x)/f_2(x) = 1.$ This means the limit can't exist. $\endgroup$ May 29, 2019 at 20:37

1 Answer 1


For $f_{2}(x)$, I would use the estimate $$ \log(x) - 1 \leq \lfloor \log(x) \rfloor \leq \log(x), $$ so $$ e^{-1} x \leq f_{2}(x) \leq x, $$ and $$ \log(\lambda) + \log(x) - 1 \leq \lfloor \log(x) \rfloor \leq \log(\lambda) + \log(x), $$ so $$ e^{-1} \lambda x \leq f_{2}(\lambda x) \leq \lambda x. $$

As for $f_3(x) = 2 + \sin(x)$, let $$\lambda = {1 \over 2}.$$ What is the behavior of $$ {f_{2}(\lambda x) \over f_{2}(x)}? $$ (Specifically, what are the minima and maxima?)

  • 2
    $\begingroup$ You can't really use $g_3(x)=\sin x$ since you'd have to divide by zero. $\endgroup$ May 29, 2019 at 20:40
  • $\begingroup$ I dont know how to use the inequalities, the deliver me $\frac{f_2(\lambda x)}{f_2(x)} \leq \frac{\lambda x}{f_2(x)} \leq \frac{\lambda x}{e^{-1} x} = e \lambda$ and $\frac{f_2(\lambda x)}{f_2(x)} \geq \frac{e^{-1} \lambda x}{f_2(x)} \geq \frac{e^{-1} \lambda x}{x} = e^{-1} \lambda$. The limit for $x \rightarrow \infty$ would have no effect since the inequalities are independent from $x$. I don't see where this is proves or disproves varying function. $\endgroup$ May 29, 2019 at 20:55
  • $\begingroup$ @ThomasAndrews, thanks. Edited. $\endgroup$
    – avs
    May 29, 2019 at 21:13

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