# 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 $$a>0$$.

• slowly varying means?
– zhw.
Dec 30, 2019 at 0:33

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 $$a \neq 1$$ -- 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.