Bourbaki

Proposition 2

Tag 08GX, permanent. Cite it as [Bourbaki, Tag 08GX].

Functions of a Real Variable, chapter V, Appendix 0. Hardy Fields. (H) Functions

Let ff be a function belonging to a Hardy field K\mathscr{K}.

1 If ff is of infinite order relative to xx, then, for every integer n>0n > 0,

f(n)(x)∼(f′(x))n(f(x))n−1. f^{(n)}(x) \sim \frac{(f'(x))^n}{(f(x))^{n-1}}.
(2)

2° If ff is of finite order μ\mu relative to xx then, for every n>0n > 0,

f(n)(x)∼μ(μ−1)…(μ−n+1)f(x)xn f^{(n)}(x) \sim \mu(\mu - 1) \ldots (\mu - n + 1) \frac{f(x)}{x^n}
∼(μ−1)…(μ−n+1)μn−1(f′(x))n(f(x))n−1 \sim \frac{(\mu - 1) \ldots (\mu - n + 1)}{\mu^{n-1}} \frac{(f'(x))^n}{(f(x))^{n-1}}
(3) except if μ\mu is an integer ⩾0\geqslant 0 and n>μn > \mu.

1 If ff is of infinite order relative to xx one has log⁡∣f∣≫log⁡x\log |f| \gg \log x, so, since log⁡∣f∣\log |f| and log⁡x\log x are comparable to any order, f′/f≫1/xf'/f \gg 1/x. Put g=f′/fg = f'/f; since gg is equal to a function in K\mathscr{K} on a neighbourhood of +∞+\infty one deduces from 1/g≪x1/g \ll x, that g′/g2≪1g'/g^2 \ll 1, and so g′/g≪g=f′/fg'/g \ll g = f'/f, or again fg′≪gf′fg' \ll gf'. From the relation f′=fgf' = fg one deduces by differentiating that

f′′=fg′+gf′∼gf′ f'' = fg' + gf' \sim gf'
or again f′′/f′∼f′/ff''/f' \sim f'/f. The same argument, applied to f(n)f^{(n)} instead of ff, shows, by induction on nn, that f(n)/f(n−1)∼f′/ff^{(n)}/f^{(n-1)} \sim f'/f; whence relation (2).

2 If ff is of finite order μ\mu relative to xx, and if μ≠0\mu \neq 0, one has log⁡∣f∣∼μlog⁡x\log |f| \sim \mu \log x, whence, on differentiating, f′(x)∼μf(x)xf'(x) \sim \mu \frac{f(x)}{x}; one deduces that f′f' is of order μ−1\mu - 1 relative to xx, which allows one to apply the same argument by induction on nn so long as μ≠n\mu \neq n, whence formula (3) when μ\mu is not an integer ⩾0\geqslant 0 and <n< n.

When ff is of integral order p⩾0p \geqslant 0 relative to xx one can write f(x)=xpf1(x)f(x) = x^p f_1(x), where f1f_1 is of order 0 relative to xx. By prop. 2 one has

f(p)∼p!f1. f^{(p)} \sim p! f_1.
To evaluate the derivatives of order n>pn > p one may restrict oneself to the case p=0p = 0. Then one has log⁡∣f∣≪log⁡x\log |f| \ll \log x, whence f′(x)/f(x)≪1/xf'(x)/f(x) \ll 1/x, in other words xf′(x)≪f(x)x f'(x) \ll f(x); if ff is not equivalent to a constant k≠0k \neq 0 one has, on differentiating this relation (V, p. 232, prop. 7), xf′′(x)+f′(x)≪f′(x)x f''(x) + f'(x) \ll f'(x), which implies that xf′′(x)∼−f′(x)x f''(x) \sim -f'(x). Taking account of this formula, one sees by induction on nn that f(n)f^{(n)} is of order ⩽−n\leqslant -n relative to xx, and that
f(n)∼(−1)n+1(n−1)!f′(x)xn−1. f^{(n)} \sim (-1)^{n+1} (n-1)! \frac{f'(x)}{x^{n-1}}.
(4)

If ff is equivalent to a constant k≠0k \neq 0 one has f(x)=k+f2(x)f(x) = k + f_2(x) with f2≪1f_2 \ll 1, and it reduces to studying the derivatives of f2f_2.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Transcribed from the printed volume. Not translated.