# regular and slowly varying functions examples $e^{\log(x)}$, $e^{\lfloor \log(x) \rfloor}$, $2 + \sin(x)$ and $e^{(\log(x))^\beta}$

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.

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?

• 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. May 29, 2019 at 20:37

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?)
• You can't really use $g_3(x)=\sin x$ since you'd have to divide by zero. May 29, 2019 at 20:40
• 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. May 29, 2019 at 20:55