Let f be a function belonging to a Hardy field K.
1 If f is of infinite order relative to x, then, for every integer n>0,
f(n)(x)∼(f(x))n−1(f′(x))n. (2)
2° If f is of finite order μ relative to x then, for every n>0,
f(n)(x)∼μ(μ−1)…(μ−n+1)xnf(x) ∼μn−1(μ−1)…(μ−n+1)(f(x))n−1(f′(x))n (3) except if
μ is an integer
⩾0 and
n>μ.
1 If f is of infinite order relative to x one has log∣f∣≫logx, so, since log∣f∣ and logx are comparable to any order, f′/f≫1/x. Put g=f′/f; since g is equal to a function in K on a neighbourhood of +∞ one deduces from 1/g≪x, that g′/g2≪1, and so g′/g≪g=f′/f, or again fg′≪gf′. From the relation f′=fg one deduces by differentiating that
f′′=fg′+gf′∼gf′ or again
f′′/f′∼f′/f. The same argument, applied to
f(n) instead of
f, shows, by induction on
n, that
f(n)/f(n−1)∼f′/f; whence relation (2).
2 If f is of finite order μ relative to x, and if μ=0, one has log∣f∣∼μlogx, whence, on differentiating, f′(x)∼μxf(x); one deduces that f′ is of order μ−1 relative to x, which allows one to apply the same argument by induction on n so long as μ=n, whence formula (3) when μ is not an integer ⩾0 and <n.
When f is of integral order p⩾0 relative to x one can write f(x)=xpf1(x), where f1 is of order 0 relative to x. By prop. 2 one has
f(p)∼p!f1. To evaluate the derivatives of order
n>p one may restrict oneself to the case
p=0. Then one has
log∣f∣≪logx, whence
f′(x)/f(x)≪1/x, in other words
xf′(x)≪f(x); if
f is not equivalent to a constant
k=0 one has, on differentiating this relation (V, p. 232, prop. 7),
xf′′(x)+f′(x)≪f′(x), which implies that
xf′′(x)∼−f′(x). Taking account of this formula, one sees by induction on
n that
f(n) is of order
⩽−n relative to
x, and that
f(n)∼(−1)n+1(n−1)!xn−1f′(x). (4)
If f is equivalent to a constant k=0 one has f(x)=k+f2(x) with f2≪1, and it reduces to studying the derivatives of f2.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.