Properties of slowly-varying functions at two large points

Question

Consider a slowly-varying function $L:(1,\infty) \mapsto (0,\infty)$, i.e. a function such that $L(cx)/L(x)\to1$ as $x\to\infty$ for all $c>0$. Assume that $\lim_{x \to \infty}L(x)=0$.

Question: is it true that, necessarily, for any sequences $x_n \to \infty$ and $c_n \to \infty$, $L(x_nc_n)/L(x_n)=O(1)$?

Comments: To prove this is true, I was trying to use the following well known representation for slowly varying functions $$ L(x)=c(x) \exp\left( \int_{x_0}^x \frac{\epsilon(s)}{s} ds \right) $$ where $c:(1, \infty) \mapsto (0, \infty)$ satisfies $\lim_{x \to \infty}|c(x)| \in(0,\infty)$ and $\epsilon:(1,\infty) \mapsto \mathbb{R} $ is a continous bounded function on $[x_0,\infty)$, for some $x_0>0$. Unfortunately, this representation does not guarantee that, e.g., $L(x)$ is asymptotically equivalent to a nonincreasing function, which would have guaranteed $L(x_nc_n)/L(x_n) \leq 1$ (see, e.g., Remark B.1.11 in de Haan and Ferreira (2006) Extreme Value Theory: An Introduction). So I got stuck, I'm not sure whether I'm missing anything stupid.

Answer 1

(I will use a definition of a slowly varying function from Wikipedia)

No, of course not. Here is an example:

Let us first of all for convenience switch to the additive formulation by considering $g(t) = \log f(\log t)$, then $|g(t+a)-g(t)| < \varepsilon$ for all $t$ big enough depending on $a$ and $\varepsilon$, we have $g(x)\to -\infty, x\to +\infty$ and we want $g(x_n+c_n) - g(x_n)\to +\infty$ for some sequences $x_n, c_n \to +\infty$.

For example, define $g(2^{2n}) = -2\log(n)$, $g(2^{2n+1}) = -\log(n)$, and between those values interpolate linearly, the derivative tends to zero, so the function is slowly varying, and $g$ clearly tends to $-\infty$. And yet, for $x_n = 2^{2n}, c_n = 2^{2n+1}-2^{2n}$ we get what we want.

In terms of your "representation for slowly varying functions", this essentially boils down to the fact that the integral $\int_1^{+\infty} \frac{1}{s}ds$ diverges, and it is a nice exercise to see where this was (in spirit) used in my answer.