§ 3. Some Applications of Cauchy’s Method
115
Indeed, formula (18) can also be written as
2πixf
(x) = −
sϕ(s)x
−s ds .
Passing from f (x) to xf
(x) is done by replacing ϕ(s) by −sϕ(s), a function
still satisfying conditions (i) and (ii) in the statement. Hence (20’) can again
be applied in this case provided that the residue a k of ϕ(s) at s = −k is
replaced by that of −sϕ(s), namely ka k since the point s = −k is a simple
pole of ϕ; so
xf
(x) ≈
ka k x
k ,
which gives (20”).
Iterating the argument, for all n, the derivative f
(n) (x) is, therefore, seen
to have an asymptotic expansion in the neighbourhood of 0, obtained by
differentiating term by term that of f n times. To deduce that f can be
extended to a C
∞ function on x ≥ 0, where this is a large inequality, it
remains to prove a rather easy general result:
Lemma 1. Let f be a function defined and infinitely differentiable on an
open interval 0 < x < b. f can be extended to an infinitely differentiable
function on the interval 0 ≤ x < b if and only if the following conditions
hold :
(a) f has an asymptotic expansion
f (x) ≈
k∈N
a k x
k , x −→ 0 ;
(b) for all n ∈ N, the derivative f
(n) (x) has an asymptotic expansion obtained by differentiating term by term that of f n times.
First of all, the relation f (x) = a 0 + a 1 x + o(x) shows both that f (x)
tends to a 0 as x approaches 0 and that if we define f (0) = a 0 , then the
function f thus extended has a derivative equal to a 1 at the origin. Since,
by (ii), f
(x) ≈ a 1 + 2a 2 x + . . . and so lim f
(x) = a 1 , the extension of f at
0 ≤ x < b is C
1 . Applying these arguments to f
instead of f , f
is seen to
have as extension a C
1 function, so that f is C
2 , etc.
A variation of lemma 2: suppose that
lim
x=0+
f
(n) (x) = f
(n) (0+)
exists for all n. Indeed, for 0 < x < x + h,
|f (x + h) − f (x) − f
(x)h| ≤ h. sup |f
(x + k) − f
(x)| ,
where the sup is extended to k ∈ [0, h]; since f
(0+) exists, this sup is ≤ r
for sufficiently small h, and so as x tends to 0,
115
Indeed, formula (18) can also be written as
2πixf
(x) = −
sϕ(s)x
−s ds .
Passing from f (x) to xf
(x) is done by replacing ϕ(s) by −sϕ(s), a function
still satisfying conditions (i) and (ii) in the statement. Hence (20’) can again
be applied in this case provided that the residue a k of ϕ(s) at s = −k is
replaced by that of −sϕ(s), namely ka k since the point s = −k is a simple
pole of ϕ; so
xf
(x) ≈
ka k x
k ,
which gives (20”).
Iterating the argument, for all n, the derivative f
(n) (x) is, therefore, seen
to have an asymptotic expansion in the neighbourhood of 0, obtained by
differentiating term by term that of f n times. To deduce that f can be
extended to a C
∞ function on x ≥ 0, where this is a large inequality, it
remains to prove a rather easy general result:
Lemma 1. Let f be a function defined and infinitely differentiable on an
open interval 0 < x < b. f can be extended to an infinitely differentiable
function on the interval 0 ≤ x < b if and only if the following conditions
hold :
(a) f has an asymptotic expansion
f (x) ≈
k∈N
a k x
k , x −→ 0 ;
(b) for all n ∈ N, the derivative f
(n) (x) has an asymptotic expansion obtained by differentiating term by term that of f n times.
First of all, the relation f (x) = a 0 + a 1 x + o(x) shows both that f (x)
tends to a 0 as x approaches 0 and that if we define f (0) = a 0 , then the
function f thus extended has a derivative equal to a 1 at the origin. Since,
by (ii), f
(x) ≈ a 1 + 2a 2 x + . . . and so lim f
(x) = a 1 , the extension of f at
0 ≤ x < b is C
1 . Applying these arguments to f
instead of f , f
is seen to
have as extension a C
1 function, so that f is C
2 , etc.
A variation of lemma 2: suppose that
lim
x=0+
f
(n) (x) = f
(n) (0+)
exists for all n. Indeed, for 0 < x < x + h,
|f (x + h) − f (x) − f
(x)h| ≤ h. sup |f
(x + k) − f
(x)| ,
where the sup is extended to k ∈ [0, h]; since f
(0+) exists, this sup is ≤ r
for sufficiently small h, and so as x tends to 0,