Proposition 12

Let g be a function in $\mathscr{H}(\mathfrak{F}, \mathbf{R})$ such that $\lim_{\mathfrak{F}} g = +\infty$; let f be a function in $\mathscr{H}(\mathfrak{F}, \mathbf{R})$.

a) For f to be of order $+\infty$ relative to g it is necessary and sufficient that $f \gg g^\alpha$ for every $\alpha \geq 0$.

b) For f to be of order $-\infty$ relative to g it is necessary and sufficient that $f \ll g^{-\alpha}$ for every $\alpha > 0$.

c) For f to be of finite order equal to $\rho$ relative to g it is necessary and sufficient that, for every $\varepsilon > 0$, one has $g^{\rho-\varepsilon} \ll f \ll g^{\rho+\varepsilon}$.

Let us prove c) for example. If the order of f relative to g is $\rho$ then for every $\varepsilon > 0$ there exists a set $M \in \mathfrak{F}$ such that, for every $x \in M$, we have

or $(g(x))^{\rho - \frac{\varepsilon}{2}} \leq |f(x)| \leq (g(x))^{\rho + \frac{\varepsilon}{2}}$; since $\lim_{\mathfrak{F}} g = +\infty$ one thus has $g^{\rho-\varepsilon} \ll f \ll g^{\rho+\varepsilon}$ for every $\varepsilon > 0$; the converse is immediate. The proofs of a) and b) are similar.

Note that if f is of finite order $\rho$ relative to g then $fg^{-\rho}$ is of order 0 relative to g, and conversely; if $f_1$ (resp. $f_2$) is of order $\rho_1$ (resp. $\rho_2$) relative to g, and if $\rho_1 + \rho_2$ is defined, then $f_1 f_2$ is of order $\rho_1 + \rho_2$ relative to g.