4
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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: $$\lim_{n \to \infty}x_n^{\rho}f(x_n)=0$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.