If $f:\mathbb R^+ \to \mathbb R^+$, and for all $y>0$, the following equation holds (i.e. it is slowly varying fuction.) $$\lim_{x\to +\infty} \frac{f(xy)}{f(x)}=1$$

How to prove the following? $$\lim_{x\to +\infty} x^\rho f(x) = \begin{cases}0 & \rho<0\\\infty & \rho>0\end{cases}$$

Notice that $f$ is not necessarily continuous. This problem seems simple but I can't solve it.


A nice way to solve this is use the Heine definition of limit that is: $$\lim_{x \to \infty }f(x)=L $$ iff all sequence of $x_n$ converging to $\infty$ the sequence of $f(x_n)$ convegerse to $L$. Now if $f$ is bounded then its easy. Let's assume $f$ is not bounded. $f$ will converge to $\infty$ else we will have a subsequence converging to a real limit. all the mulipieceis of that sequence will also converge to the same limit by: $$lim_{n \to \infty}\frac{f(x_n y)}{f(x_n)}=1$$ This means that every sequence will converge to this real limit because if $h_i$ is a sequence then $x_i \cdot \frac{h_{i'}}{x_{i'}}$ will converge to the same limit for all $i' \in \mathbb{N}$. This will lead to a contradiction so $f$ has to converge to $\infty$.
If $\rho >0$ it is easy. Now we will check what happens if $\rho <0$. We will have: $$lim_{n \to \infty}\frac{(x_n y)^{\rho}f(x_n y)}{x_n^{\rho}f(x_n)}=y^{\rho}<1$$ for $y> 1$ so by delmabart test: $$\sum_{n=0}^{\infty}(xy^{n})^{\rho}f(xy^{n})$$ converges for all $x>0$ and $y>1$ and we know that: $$\lim_{n \to \infty}(xy^n)^{\rho}f(xy^n)=0$$ It is left to show that this implies that for a general series we have: $$\lim_{n \to \infty}x_n^{\rho}f(x_n)=0$$

  • $\begingroup$ That's nice! And I found that using $g(x)=1/f(x)$ will yield the equation when $\rho>0$. $\endgroup$ – user482401 Mar 19 '18 at 15:56

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