Let $f:\mathbb{N} \to\mathbb{N}$ be a monotonous numbertheoretic function, and let $g:\mathbb{N} \to\mathbb{N}$ be such that
$$\forall m \exists n f(n)\leq g(n).$$
Is $f$ always primitive recursive in $g$, i.e., primitive recursively definable given $g$? My current intuition is that the answer is "No", since the gaps between values where $g$ catches up with $f$ may be much larger than what a fixed primitive recursion can deal with, but I don't know how to show this.
Rosza Péter long since gave an example of a computable function $p$ which is not primitive recursive but only takes the values 0 and 1.
Define $f$ by putting $f(n+ 1) = f(n) + 1$ if $p(n) = 0$ and $f(n) +2$ otherwise. $f$ is computable, monotone, not primitive recursive but dominated by e.g. $g = 3n$. And $f$ isn't primitive recursively definable given $g$.
Does that answer the intended question?
For Péter's function, or a version thereof, see my Intro to Gödel's Theorems, 2nd ed., p. 293, footnote. (Book available to download at https://logicmatters.net.)