Is a nondecreasing function bounded by a slowly varying function also slowly varying? I have a seemingly simple question that I haven't been able to resolve: > **Question.** Suppose that $f:(0,\infty)\to(0,\infty)$ is nondecreasing. Suppose that $f\leq L$ for a slowly varying function $L$. Does this imply that $f$ is also slowly varying? For example, If we have some arbitrary function $f$ such that $0\leq f\leq L$ where $L$ is slowly varying, then obviously this doesn't imply that $g$ is also slowly varying. To see this, we can take $f(x)=\sin(x)+2$. However, if we also assume that $f$ is nondecreasing, the question strikes me as nontrivial. For instance, a bounded nondecreasing function is convergent, hence slowly varying, and so the claim is at least true in this limited context. * * * **Edit.** I mean slowly varying in the standard sense of the term, i.e., $L$ is slowly varying if and only if $$\lim_{x\to\infty}\frac{L(ax)}{L(x)}=1$$ for every $a>0$.

No. Take $L(x) = \log(x + 1)$, and let $$f(x) = \exp(\lfloor \log \log (x + 1) \rfloor)$$ where $\lfloor \cdot \rfloor$ denotes the floor function. Then $L$ is slowly varying and $f \leq L$, but for $f$ the required limit does not exist for any $a \
eq 1$ \-- in fact for all $a > 1$ we have $\limsup_{x \to \infty} f(ax)/f(x) = e$, since $f$ scales by a factor of $e$ at the jump discontinuity at each point of the form $x = \exp(\exp(n)) - 1$.

If you would want $f$ to be continuous, then we can just remove each of the discontinuities at these points, replacing the jumps with sufficiently steep linear decreases.