Let $f:[0, \infty) \rightarrow \mathbb{R}\_+$ be a strictly increasing function. I recently found out that if $f$ grows faster than any polynomial in the sense that $$ \limsup\_{x \rightarrow \infty} \frac{f(x)}{x^k} = \infty \; \; \; \forall \; k > 0 $$
then for any $\alpha \in (0,1)$ it must be the case that $$ \limsup\_{x \rightarrow \infty} \frac{f(x)}{f(\alpha x)} = \infty $$
I'm wondering if the converse is also true. That is, if a strictly increasing function $f$ satisfies $f=O(x^k)$ for some $k > 0 $, then does it necessarily follow that $$ \limsup\_{x \rightarrow \infty} \frac{f(x)}{f(\alpha x)} < \infty \; \; \; \; \; \forall \; \; \alpha \in (0,1) $$
I wasn't able to find a reference for this anywhere and my intuition tells me this should be true so a counterexample or reference to a proof would be greatly appreciated.



Equivalent characterization of polynomial order
The claim about the $\limsup$ does not follow from the $O(x^k)$ hypothesis.


Let $f(x)=\exp(k \exp(\lfloor\log(\log(x))\rfloor)) + x$.


Then $f(x)$ is $O(x^k)$, since $f(x) \le x^k + x$, with equality where $\log(\log(x))$ is an integer.


For $\alpha \in (0,1)$, let $n$ be an integer such that $\exp(\exp(n-1)) / \exp(\exp(n)) < \alpha/2$. Let $x=\exp(\exp(n))$. Then


$$\frac{f(x)}{f(\alpha x)} \ge \frac{\exp(k \exp(n))}{\exp(k \exp(n-1))} = \exp\Big(k (e-1) \exp(n-1)\Big)$$ 
which is an unbounded function of $n$.


