# Slowly-varying functions near zero

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.

• 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 $$x_n < x \leqslant x_{n+1}$$.