Hardy Fields. (H) Functions

(1)

then there exists an interval $[x_1, +\infty[$ on which $y$ does not change sign.

Indeed, let us put $z(x) = y(x) \exp \left( - \int_{x_0}^x a(t) \, dt \right)$ (cf. IV, p. 183); then, by (1), $z'(x) = b(x) \exp \left( - \int_{x_0}^x a(t) \, dt \right)$. If $b(x) \geqslant 0$ for $x \geqslant x_0$ then $z$ is increasing on this interval, so either is $< 0$ on all this interval, or is zero on an interval $[x_1, +\infty[$, or else is $> 0$ on an interval $[x_1, +\infty[$; since $y$ has the same sign as $z$ the proposition is proved in this case. The argument is similar if $b(x) \leqslant 0$ for $x \geqslant x_0$.

The proposition is trivial if $p(u)$ has coefficients which are identically zero on $[x_0, +\infty[$, or if $p(u)$ is of degree 0 in $u$, since any function in $\mathfrak{K}$ is of constant sign on an interval $[x_1, +\infty[$. Suppose that $p(u)$ is of degree $n > 0$; the leading coefficient $c$ of $p(u)$ is then $\neq 0$ on an interval $[\alpha, +\infty[$; one can thus write $p(u) = c(u^n + c_1 u^{n-1} + \cdots + c_n)$ where $c, c_1, c_2, \ldots, c_n$ are functions belonging to $\mathfrak{K}$ and differentiable on $[\alpha, +\infty[$; so it is enough to prove the lemma for $c = 1$. We argue by induction on $n$; one has

where $q(u)$ is a polynomial of degree $\leq n - 1$, with coefficients in $\mathfrak{K}$. By hypothesis the functions $na(x)$ and $q(y(x))$ do not change sign on an interval $[ \beta, +\infty[$; the lemma is thus a consequence of lemma 1.

Indeed, by lemma 2, there exists an interval $[ x_1, +\infty[$ on which $r(y)$ is defined, continuous, and does not change sign, from which it follows immediately that $\mathfrak{K}(y)/\mathbf{R}_\infty$ is a field; moreover, since

(where $r'(y) = (p'(y)q(y) - p(y)q'(y))/(q(y))^2$ is defined by hypothesis on a neighbourhood of $+\infty$), the derivative of every function in $\mathfrak{K}(y)$ belongs to $\mathfrak{K}(y)$, which proves that $\mathfrak{K}(y)$ satisfies the conditions of def. 1 of V, p. 247.

It is clear that $\mathfrak{K}(y)/\mathbf{R}_\infty$ is obtained by the algebraic adjunction to $\mathfrak{K}/\mathbf{R}_\infty$ of the class of $y$ modulo $\mathbf{R}_\infty$. Again one says that $\mathfrak{K}(y)$ is obtained by adjoining $y$ to $\mathfrak{K}$.

Indeed, $(\log|y|)' = y'/y$ is equal to a function in $\mathfrak{K}$ on an interval $[ x_0, +\infty[$.

COROLLARY 2 If $y$ is any function in $\mathfrak{K}$, then $\mathfrak{K}(e^y)$ is a Hardy field.

Indeed, $(e^y)' = e^y y'$, and $y'$ is equal to a function in $\mathfrak{K}$ on an interval $[ x_0, +\infty[$.

COROLLARY 3 If $\mathfrak{K}$ contains the real constants, and if $y$ is a function in $\mathfrak{K}$, not identically zero on a neighbourhood of $+\infty$, then $\mathfrak{K}(|y|^{\alpha})$ is a Hardy field for every real number $\alpha$.

Indeed, $\frac{d}{dx}(|y|^{\alpha}) = |y|^{\alpha} (\alpha y'/y)$, and $\alpha y'/y$ is equal to a function in $\mathfrak{K}$ on an interval $[ x_0, +\infty[$.

Finally, we note that if $y$ is the primitive of any function in $\mathfrak{K}$ then $\mathfrak{K}(y)$ is again a Hardy field.

Indeed, if $f$ belongs to a Hardy field $\mathscr{K}$, then for every integer $n > 0$ there exists an interval $[x_0, +\infty[$ on which $f$ is $n$ times differentiable, its $n^{th}$ derivative being equal to a function in $\mathscr{K}$ on this interval. It is therefore enough to show that any two functions $f, g$ of $\mathscr{K}$ are comparable. This is evident if one of the functions is identically zero on a neighbourhood of $+\infty$; one may therefore restrict oneself to the case where they are both strictly positive on a neighbourhood of $+\infty$. But then, for every real number $t$, $f - tg$ is equal to a function in $\mathscr{K}$ on a neighbourhood of $+\infty$, so has constant sign on a neighbourhood of $+\infty$, which proves the proposition (V, p. 217, prop. 9).

One deduces immediately from this proposition that, if a Hardy field $\mathscr{K}$ contains the real constants (as we shall always assume in what follows), and if $f$ and $g$ are any two functions in $\mathscr{K}$ then any two of the functions $e^f, e^g, \log |f|, \log |g|, |f|^{\alpha}, |g|^{\alpha}$ ($\alpha$ an arbitrary real), $\int_a f, \int_a g$ ($a$ being any real number in an interval $[x_0, +\infty[$ where $f$ and $g$ are regulated) are comparable (when they are defined); indeed, any two of these functions belong to a given Hardy field obtained by adjoining them successively to $\mathscr{K}$.

Similarly, every function $f(x)$ in a Hardy field $\mathscr{K}$ is comparable to $x$, since $x$ and $f(x)$ belong to the Hardy field obtained by adjoining $x$ to $\mathscr{K}$. One thus concludes (in particular) that $f$ is comparable to any order to every power $x^{\alpha}$, as well as to $\log x$ and to $e^x$.

One also sees that if $f$ and $g$ belong to the same Hardy field $\mathscr{K}$, if $g(x) > 0$ on an interval $[x_0, +\infty[$, and if $g(x)$ tends to 0 or to $+\infty$ as $x$ tends to $+\infty$, then the order of $f$ relative to $g$ (V, p. 219) is always defined.

Prop. 8 of V, p. 233, is therefore applicable to every function $f$ in a Hardy field, and proves that:

1) if $f$ is of order $+\infty$ relative to $x$ then $\int_a^x f(t)\,dt \sim (f(x))^2/f'(x)$.

2) if $f$ is of order $\mu > -1$ relative to $x$ then $\int_a^x f(t)\,dt \sim \frac{1}{\mu+1} x f(x)$.

3) if $f$ is of order $\mu < -1$ relative to $x$ then $\int_a^{+\infty} f(t)\,dt \sim -\frac{1}{\mu+1} x f(x)$.

4) if $f$ is of order $-\infty$ relative to $x$ then $\int_a^{+\infty} f(t)\,dt \sim -(f(x))^2/f'(x)$.

Further, we have the following proposition:

1 If $f$ is of infinite order relative to $x$, then, for every integer $n > 0$,

(2)

2° If $f$ is of finite order $\mu$ relative to $x$ then, for every $n > 0$,

(3) except if $\mu$ is an integer $\geqslant 0$ and $n > \mu$.

1 If $f$ is of infinite order relative to $x$ one has $\log |f| \gg \log x$, so, since $\log |f|$ and $\log x$ are comparable to any order, $f'/f \gg 1/x$. Put $g = f'/f$; since $g$ is equal to a function in $\mathscr{K}$ on a neighbourhood of $+\infty$ one deduces from $1/g \ll x$, that $g'/g^2 \ll 1$, and so $g'/g \ll g = f'/f$, or again $fg' \ll gf'$. From the relation $f' = fg$ one deduces by differentiating that

or again $f''/f' \sim f'/f$. The same argument, applied to $f^{(n)}$ instead of $f$, shows, by induction on $n$, that $f^{(n)}/f^{(n-1)} \sim f'/f$; whence relation (2).

2 If $f$ is of finite order $\mu$ relative to $x$, and if $\mu \neq 0$, one has $\log |f| \sim \mu \log x$, whence, on differentiating, $f'(x) \sim \mu \frac{f(x)}{x}$; one deduces that $f'$ is of order $\mu - 1$ relative to $x$, which allows one to apply the same argument by induction on $n$ so long as $\mu \neq n$, whence formula (3) when $\mu$ is not an integer $\geqslant 0$ and $< n$.

When $f$ is of integral order $p \geqslant 0$ relative to $x$ one can write $f(x) = x^p f_1(x)$, where $f_1$ is of order 0 relative to $x$. By prop. 2 one has

To evaluate the derivatives of order $n > p$ one may restrict oneself to the case $p = 0$. Then one has $\log |f| \ll \log x$, whence $f'(x)/f(x) \ll 1/x$, in other words $x f'(x) \ll f(x)$; if $f$ is not equivalent to a constant $k \neq 0$ one has, on differentiating this relation (V, p. 232, prop. 7), $x f''(x) + f'(x) \ll f'(x)$, which implies that $x f''(x) \sim -f'(x)$. Taking account of this formula, one sees by induction on $n$ that $f^{(n)}$ is of order $\leqslant -n$ relative to $x$, and that (4)

If $f$ is equivalent to a constant $k \neq 0$ one has $f(x) = k + f_2(x)$ with $f_2 \ll 1$, and it reduces to studying the derivatives of $f_2$.

Denote by $\mathscr{K}$ the set of functions $f \in \mathscr{H}(\mathfrak{F}, \mathbf{R})$ having the following properties: for each function $f \in \mathscr{K}$ there is a finite number of Hardy fields $\mathscr{K}_1, \mathscr{K}_2, \ldots, \mathscr{K}_n$ (the number $n$ and the fields $\mathscr{K}_i$ depending on $f$) such that $f \in \mathscr{K}_n$ and, for $0 \leq i \leq n-1$, one has $\mathscr{K}_{i+1} = \mathscr{K}_i(u_{i+1})$ where $u_{i+1}$ is equal either to $e^{z_i}$ or to $\log |z_i|$ with $z_i$ belonging to $\mathscr{K}_i$ and not vanishing identically on a neighbourhood of $+\infty$. One says that $u_1, u_2, \ldots, u_n$ form a definition sequence for the field $\mathscr{K}_n$ and of the function $f$; the same function can naturally admit several definition sequences.

By def. 1 of V, p. 247, every function $f \in \mathscr{K}$, not identically zero on a neighbourhood of $+\infty$, has constant sign and is differentiable on an interval $[x_0, +\infty[$; if $f \in \mathscr{K}_n$, then $1/f$ and $f'$ are equal to functions in $\mathscr{K}_n$, thus to functions in $\mathscr{K}$, on a neighbourhood of $+\infty$. To see that $\mathscr{K}$ is a Hardy field it is enough to prove that if $f$ and $g$ are two functions in $\mathscr{K}$ then $f - g$ and $fg$ are equal to functions in $\mathscr{K}$ on a neighbourhood of $+\infty$. Now let $u_1, u_2, \ldots, u_m$ be a definition sequence for $f$, and $v_1, v_2, \ldots, v_n$ a definition sequence for $g$. The sequence $u_1, u_2, \ldots, u_m, v_1, v_2, \ldots, v_n$ obtained by concatenating the sequences $(u_i)$ and $(v_i)$ is again a definition sequence of a Hardy field $\mathscr{K}_{m+n}$ and this field contains $f$ and $g$, so $f - g$ and $fg$ are equal to functions in $\mathscr{K}_{m+n}$ on a neighbourhood of $+\infty$.

One says that the Hardy field $\mathscr{K}$ defined in the proof of prop. 3 is the (H) extension of the Hardy field $\mathscr{K}_0$.

If $\mathscr{K}'$ is another Hardy field possessing the properties stated in prop. 3, it follows from the construction of $\mathscr{K}$ that $\mathscr{K}/\mathbf{R}_\sim$ is contained in $\mathscr{K}'/\mathbf{R}_\sim$. By abuse of language, one says that the (H) extension of the Hardy field $\mathscr{K}_0$ is the smallest Hardy field $\mathscr{K}$ having these properties.

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.