I am cross-posting the question below, which I asked in Mathematics StackExchange a week ago and did not receive answers there. Thank you for your help!

It is well known that, if $x\mapsto f(x)$ is a slowly-varying function, then $$ \lim_{x\to\infty}\sup_{\lambda\in K}\left|\frac{f(\lambda x)}{f(x)}-1\right| = 0, $$ where $K\subset (0,\infty)$ is a compact set.

Of course, it is quite important that $K$ does not contain 0 in the above result. However, I am wondering if this uniform convergence (or, in fact, some other property of slowly-varying functions such as their explicit representation in terms of the exponential function, see below) can be exploited to prove something like the following:

Let $x\mapsto\lambda(x)$ be a function that goes to zero "sufficiently slowly" as $x\to\infty$ and let $x\mapsto f(x)$ be a slowly-varying function, then $$ \lim_{x\to\infty}\frac{f(\lambda(x) x)}{f(x)} = 1. $$

Can this be proven under some assumptions on $\lambda(\cdot)$?

One might be tempted to find some conditions on $\lambda(\cdot)$ via the representation $$ f(x) = \exp\left(\eta(x) + \int_c^{x}\frac{\varepsilon(t)}{t}\mathrm d t\right), $$ where $c>0$ is some arbitrary constant, $x\mapsto\eta(x)$ is measurable, bounded, and admits a limit as $x\to\infty$, and $t\to\varepsilon(t)$ is measurable, bounded, and converges to zero as $t\to\infty$. We get $$ \frac{f(\lambda(x)x)}{f(x)} = \exp(\eta(\lambda(x)x)-\eta(x))\times\exp\left(\int_{\lambda(x)x}^{x}\frac{\varepsilon(t)}{t}\mathrm d t\right). $$ If $\lambda(x)x\to\infty$ as $x\to\infty$, then the first term on the right-hand side converges to zero. However, bounding the second term on the right-hand side is difficult if one lacks any sort of estimate or bound on $\varepsilon(\cdot)$. It seems to me that the best one can do is exploit that $\varepsilon(t)\leq C$ for some $C>0$ and all $t>c$, and get \begin{align*} \exp\left(\int_{\lambda(x)x}^{x}\frac{\varepsilon(t)}{t}\mathrm d t\right) &\leq \exp\left(\int_{\lambda(x)x}^{x}\frac{C}{t}\mathrm d t\right)\\ % &=\exp\left(C(\log(x)-\log(\lambda(x)x)\right)\\ % &=\exp(-C\log(\lambda(x)), \end{align*} but this is insufficient since $\lambda(x)\to0$ as $x\to\infty$. So it seems that one has to exploit the fact that $\varepsilon(t)\to0$ as $t\to\infty$, but without more explicit estimates I would not know how.


1 Answer 1


Not sure if this answers the question.

  • For a given $f$ slowly varying at $\infty$, one can find $\lambda$ convergent to $0$ at $\infty$ such that $f(\lambda(x)) / f(x)$ converges to $1$ as $x \to \infty$. Indeed, for every $n = 1, 2, \ldots$ find $x_n$ such that $$\biggl|\frac{f(x/n)}{f(x)} - 1\biggr| < \frac{1}{n}$$ when $x > x_n$, and simply let $\lambda(x) = \tfrac{1}{n}$ when $x_n < x \leqslant x_{n+1}$.

  • For a given $\lambda > 0$ convergent to zero, one can find $f$ slowly varying at $\infty$ such that $f(\lambda(x)) / f(x)$ converges to $0$ (rather than to $1$) as $x \to \infty$. Indeed, for $n = 1, 2, \ldots$ find $x_n$ such that $\lambda(x) < \tfrac{1}{n^n}$ when $x > x_n$ (and assume $x_n > n x_{n-1}$ for simplicity). Now we define $f$ so that $f(x) = c_n \sqrt[n]x$ when $x_n < x \leqslant x_{n+1}$, where $c_n$ are chosen in such a way that $f$ is continuous. Then it is easy to see that $f$ is slowly varying, but $f(\lambda(x)) / f(x) \leqslant \tfrac{1}{n}$ when $x_n < x \leqslant x_{n+1}$.

De Hahn's classes of slowly varying functions may be what you are after.

  • $\begingroup$ It seems that your first bullet point was precisely what I was after. Thank you very much! $\endgroup$
    – Indigo
    Dec 15, 2020 at 15:09

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