Let $f,g$ two positive functions ($\ge 0$) with domain $[-1,1]$ .
We want to describe conditions of $f$ and $g$ that leads to the conclusion that $f$ can dominate $g$ according to the following definition:
Definition : $f$ dominates $g$ if it exists $k>0$ so that $kf(x) \ge g(x)$ for every $x \in [-1,1]$.
--- MY THOUGHTS: One necessary condition is that $f(x)=0 \rightarrow g(x)=0$ i.e. the points where $f$ is null shall be also points where $g$ is null. In fact for such a point $x_0$, $kf(x_0)$ will be equal to zero for every value of $k$ and therefore could never dominate $g$ if $g(x_0)>0$.
This is not enough though, because for example $f(x)=x^4$ cannot dominate $g(x)=x^2$ for any value of $k$. The order of the Taylor expansion at $0$ forces $kf(x)$ to go faster to zero than $g(x)$ for every value of $k$. So this suggests a statement.
--- STATEMENT: Let $f,g$ be positive and analytic and suppose that on the border they are strictly positive. If $\overline x$ is such that $f(\overline x)=0$ one should have that $g(\overline x)=0$. Further, let the infinitesimal order of $f$ at $\overline x$ be $2k$ (so that the first $2k-1$ derivatives are $0$) and for $g$ let it be $2m$. Than if $k \le m$, $f$ can majorize $g$ in the sense above.
I took even orders of expansion because if the order was odd locally it could have negative values, which is against the assumptions on $f$ and $g$.
Is my reasoning correct ? Or can it be improved to reach a correct statement ? Are there weaker conditions ?