Proposition 4

Let $p$ and $q$ be two (H) functions which are strictly positive on an interval $[x_0, +\infty[$.

1° If $q \ll p/p'$ one has $p(x + q(x)) \sim p(x)$. 2° If both $q \ll p/p'$ and $q(x) \ll x$ then $p(x - q(x)) \sim p(x)$.

The two parts of this proposition are clear if $p \sim k$ (constant $\neq 0$); one may therefore suppose that $p(x) \ll 1$ (otherwise one applies the argument to $1/p$). One then deduces $p'(x) \ll 1$.

1° One can write $p(x + q(x)) = p(x) + q(x)p'(x + \theta q(x))$ with $0 \leq \theta \leq 1$ (I, p. 14, corollary). Since $|p'(x)|$ tends to 0 as $x$ tends to $+\infty$, and is equal to an (H) function on a neighbourhood of $+\infty$, it is decreasing on an interval $[x_1, +\infty[$, so, for $x \geq x_1$, one has $|p'(x + \theta q(x))| \leq |p'(x)|$; since $qp' \ll p$ one has $p(x + q(x)) \sim p(x)$.

2° The condition $q(x) \ll x$ ensures that $x - q(x)$ tends to $+\infty$ with $x$. Again one has $p(x - q(x)) = p(x) - q(x)p'(x - \theta p(x))$ with $0 \leq \theta \leq 1$. The same argument as in the first part of the proof shows that, for $x$ sufficiently large, $|p'(x - \theta q(x))| \leq |p'(x - q(x))|$. It reduces to showing that $q(x) \frac{p'(x - q(x))}{p(x - q(x))}$ tends to 0 as $x$ tends to $+\infty$. The proposition is true if $p'/p \gg 1$ since then $|p'/p|$ is an (H) function, increasing for $x$ large enough; thus $q(x) \frac{|p'(x - q(x))|}{|p(x - q(x))|} \leq q(x) \frac{|p'(x)|}{|p(x)|}$, and $qp' \ll p$ by hypothesis. It is also true if $p'/p \sim k$ ($k$ constant $\neq 0$), for then

since $x - q(x)$ tends to $+\infty$. It only remains to examine the case where $p'/p \ll 1$. First suppose that $p(x)$ is of finite order relative to $x$, and so (V, p. 232, prop. 7) $p'(x)/p(x) \ll 1/x$. Then one has

so and one sees in this case that the proposition is true subject only to the hypothesis that $q(x) \ll x$. Finally we examine the case where $1/x \ll p'(x)/p(x) \ll 1$; the function $r = p'/p$ is then of finite order relative to $x$; since, by the preceding remark, prop. 4 of V, p. 255, is applicable to such a function, one has $p'(x-q(x))/p(x-q(x)) \sim p'(x)/p(x)$, and the hypothesis $qp' \ll p$ allows us to complete the proof.