Proposition 9

Let $f$ and $g$ be two functions in $\mathscr{H}(\mathfrak{F}, \mathbf{R})$, strictly positive on a set in $\mathfrak{F}$. For $f$ and $g$ to be comparable it is necessary and sufficient that for every number $t \geqslant 0$, with at most one exception, the function $f - tg$ has constant sign $^3$ on a set in $\mathfrak{F}$.

The condition is necessary. Indeed, if $f \ll g$ one has $f - tg \sim -tg$ except for $t = 0$, so $f - tg$ is strictly negative on a set in $\mathfrak{F}$, except for $t = 0$; if $f \gg g$ then $f - tg$ is strictly positive on a set in $\mathfrak{F}$ for all $t$; finally, if $f \sim kg$ ($k$ constant $> 0$), then $f - t g \sim (k - t)g$ except for $t = k$, so, except perhaps for $t = k$, $f - t g$ has the sign of $k - t$ on a set in $\mathfrak{F}$.

The condition is sufficient. Indeed, suppose that the ratio $f/g$ has two distinct cluster points $\alpha < \beta$ along $\mathfrak{F}$. For every number $t$ such that $\alpha < t < \beta$ there then exist in every set $X \in \mathfrak{F}$ two points $x_1, x_2$ such that $f(x_1)/g(x_1) < t$ and $f(x_2)/g(x_2) > t$; thus $f(x) - t g(x)$ does not have constant sign on $X$; we have arrived at a conclusion incompatible with the hypothesis. It follows that $f/g$ can have only one cluster value (finite or infinite) along the filter with base $\mathfrak{F}$, and consequently, (Gen. Top., I, p. 85, corollary) has this value as its limit along $\mathfrak{F}$.