Definition 1

*Given a subset $\mathfrak{K}$ of $\mathscr{H}(\mathfrak{F}, \mathbf{R})$ one says that $\mathfrak{K}/R_\infty$ (the canonical image of $\mathfrak{K}$ in $\mathscr{H}(\mathfrak{F}, \mathbf{R})/R_\infty$) is a Hardy field, if $\mathfrak{K}$ satisfies the following conditions:

1. $\mathfrak{K}/R_\infty$ is a subfield of the ring $\mathscr{H}(\mathfrak{F}, \mathbf{R})/R_\infty$. 2. Every function in $\mathfrak{K}$ is continuous and differentiable on an interval $[a, +\infty[$ (depending on the function), and the class with respect to $R_\infty$ of its derivative belongs to $\mathfrak{K}/R_\infty$.

The hypothesis that $\mathfrak{K}/R_\infty$ is a field is equivalent to the following conditions: if $f \in \mathfrak{K}$ and $g \in \mathfrak{K}$ then $f + g$ and $fg$ are equal to functions in $\mathfrak{K}$ on some set in $\mathfrak{F}$; further, if $f$ is not identically zero on a set in $\mathfrak{F}$ then there exists a set $M$ in $\mathfrak{F}$ on which $f$ does not vanish, and $1/f$ is equal to a function from $\mathfrak{K}$ on $M$; by condition 2' one can always assume that $M$ is taken so that $f$ is continuous on $M$, and consequently has constant sign on this interval.

By an abuse of language, if $\mathfrak{K}$ is such that $\mathfrak{K}/R_\infty$ is a Hardy field we shall say in what follows that $\mathfrak{K}$ is itself a Hardy field.