Proposition 5

Let g be an (H) function not equivalent to a constant $\neq 0$, and such that $g(x) \ll x$, and let $u(x)$ be the inverse function of $x - g(x)$, defined on a neighbourhood of $+\infty$. Let $(u_n)$ be the sequence of functions defined, by recursion on $n$, by the conditions $u_0(x) = x$, $u_n(x) = x + g(u_{n-1}(x))$ for $n \geqslant 1$; then $u_n \gg 1$ and

(11)

Let us put $y = u(x)$, $y_n = u_n(x)$; then $x = y - g(y)$, $y_0 = x$ and $y_n = x + g(y_{n+1})$. One first deduces that $x/y = 1 - \frac{g(y)}{y}$; since $y$ tends to $+\infty$ with $x$, the hypothesis $g(x) \ll x$ shows that $y = u(x) \sim x = y_0$; further,

where $z$ belongs to the interval with endpoints $x, y$; when $x$ tends to $+\infty$ so does $z$, and since $g(x) \ll x$, $g' \ll 1$, thus $g'(z)$ tends to 0, and consequently

whence

(12)

In the second place we show, by recursion on $n$, that when $x$ tends to $+\infty$ one has $u_n \gg 1$, and

(13)

Indeed, $y - y_n = g(y) - g(y_{n-1}) = (y - y_{n-1})g'(z_{n-1})$, where $z_{n-1}$ belongs to the interval with endpoints $y$ and $y_{n-1}$; by the induction hypothesis $z_{n-1}$ tends to $+\infty$ with $x$, and so $g'(z_{n-1})$ tends to 0, which proves (13). One deduces from this relation and from (12) that $u(x) - u_n(x) \ll u(x) - x \sim g(x) \ll x \sim u(x)$, whence $u_n(x) \sim u(x)$ and consequently $u_n \gg 1$. Finally, the relation $u(x) - u_n(x) \ll u(x) - x$ can also be written $(u(x) - x) - (u_n(x) - x) \ll u(x) - x$, whence

(14)

To prove (11), first remark that if $t(x)$ is a function such that $t(x) - x \sim g(x)$, then one has $g'(t(x)) \sim g'(x)$. Indeed, for any $\varepsilon > 0$, for $x$ sufficiently large $g'$ is monotone, so $g'(t(x))$ lies between $g'(x + (1 + \varepsilon)g(x))$ and $g'(x + (1 - \varepsilon)g(x))$. Prop. 4 of V, p. 255, shows that $g'(t(x)) \sim g'(x)$, provided one has established the relation $g \ll g'/g''$. Now, if $g$ is of infinite order relative to $x$ one has (V, p. 251, prop. 2) $g''/g' \sim g'/g$, and, since $g' \ll 1, g \ll g/g' \sim g'/g''$; if $g$ is of finite order $\mu$ relative to $x$ one necessarily has $\mu \leq 1$; if $\mu < 1$, since $g$ is not equivalent to a constant $\neq 0$, formulae (3) and (4) (V, p. 251) show that $g''/g' \sim k/x$ ($k$ constant $\neq 0$), whence again $g \ll g'/g''$; finally, if $\mu = 1$ then $g'$ is of order 0 relative to $x$, so $g''/g' \ll 1/x$, and so again $g \ll g'/g''$.

This being so, since $z_{n-1}$ lies between $y$ and $y_n$, it follows from (14) that $z_{n-1} - x \sim g(x)$, whence $g'(z_{n-1}) \sim g'(x)$ by the above; thus

whence we obtain (11) by induction on $n$.