§ 3. Some Applications of Cauchy’s Method
87
in two different ways. The easiest proof consists in applying the (genuine)
Lebesgue-Fubini theorem; also with a few acrobatics, we could just consider
the version proved in Chapter V, n
◦ 33 for lsc functions (separate the real
and imaginary parts of the functions in question), but it is not worth it.
Choosing
51 g = ϕ, for any function f ∈ L
1 (R),
+∞
0
ˆ
f (u)u
s−1 e(zu)du = Γ (s)
R
f (t) [2πi(t − z)]
−s dt
(11.1)
for Re(s) > 0 and Im(z) > 0 since these conditions imply g ∈ L
1 (R). In any
event, the case s = 1 shows that for every f ∈ L
1 (R),
52
+∞
0
ˆ
f (u)e(zu)du =
1
2πi
R
f (t)
t − z
dt = F
+ (z) for Im(z) > 0 ,
(11.2’)
where F
+ is defined and holomorphic for Im(z) > 0. The similarity of this
result with Cauchy’s integral formula will not escape anyone, but it is misleading: F
+ (z) is not a holomorphic function reducing to f on R. If this formula
is applied to f (−t), a function whose Fourier transform is ˆ
f (−u), and if z is
replaced by −z, then similarly
0
−∞
ˆ
f (u)e(zu)du = −
1
2πi
f (t)
t − z
dt = −F
− (z) for , Im(z) < 0
(11.2”)
where F
− is defined and holomorphic for Im(z) < 0. This leads us to associate
to every integrable function f on R the function
53
F (z) =
1
2πi
f (t)dt
t − z
=
+∞
0
ˆ
f (u)e(zu)du if Im(z) > 0
−
0
−∞
ˆ
f (u)e(zu)du if Im(z) < 0
(11.3)
analogous to the Poisson transform P f introduced in Chapter V, § 5 in the
case of the unit disc [see in particular formula (21.7)]. It is holomorphic
51 The reader will probably observe that ϕ is a regulated function on R only if
Re(s) > 1 or s = 1 since the factor x
s−1
+
is not bounded in the neighbourhood of
0 for Re(s) < 1 or, for Re(s) = 1, s = 1, does not approach a limit as x −→ +0.
All these difficulties disappear in Lebesgue theory.
52 (2’) could be directly obtained by computing the Fourier transform of the function equal to e(ζy) for y > 0 and to 0 otherwise, and by applying the general
formula (*).
53 f (t)dt could be replaced by a measure dμ(t) with finite total mass, and it is
precisely by studying such functions that Stieltjes was led to define his integrals.
Example : write the set of rational numbers as a sequence (un) and choose the
measure μ given by
f (t)dμ(t) =
f (un)/n
2 for continuous f with compact
support. The behaviour of the corresponding function
F (z) =
1/n
2 (un − z)
in the neighbourhood of the real axis is not obvious.
87
in two different ways. The easiest proof consists in applying the (genuine)
Lebesgue-Fubini theorem; also with a few acrobatics, we could just consider
the version proved in Chapter V, n
◦ 33 for lsc functions (separate the real
and imaginary parts of the functions in question), but it is not worth it.
Choosing
51 g = ϕ, for any function f ∈ L
1 (R),
+∞
0
ˆ
f (u)u
s−1 e(zu)du = Γ (s)
R
f (t) [2πi(t − z)]
−s dt
(11.1)
for Re(s) > 0 and Im(z) > 0 since these conditions imply g ∈ L
1 (R). In any
event, the case s = 1 shows that for every f ∈ L
1 (R),
52
+∞
0
ˆ
f (u)e(zu)du =
1
2πi
R
f (t)
t − z
dt = F
+ (z) for Im(z) > 0 ,
(11.2’)
where F
+ is defined and holomorphic for Im(z) > 0. The similarity of this
result with Cauchy’s integral formula will not escape anyone, but it is misleading: F
+ (z) is not a holomorphic function reducing to f on R. If this formula
is applied to f (−t), a function whose Fourier transform is ˆ
f (−u), and if z is
replaced by −z, then similarly
0
−∞
ˆ
f (u)e(zu)du = −
1
2πi
f (t)
t − z
dt = −F
− (z) for , Im(z) < 0
(11.2”)
where F
− is defined and holomorphic for Im(z) < 0. This leads us to associate
to every integrable function f on R the function
53
F (z) =
1
2πi
f (t)dt
t − z
=
+∞
0
ˆ
f (u)e(zu)du if Im(z) > 0
−
0
−∞
ˆ
f (u)e(zu)du if Im(z) < 0
(11.3)
analogous to the Poisson transform P f introduced in Chapter V, § 5 in the
case of the unit disc [see in particular formula (21.7)]. It is holomorphic
51 The reader will probably observe that ϕ is a regulated function on R only if
Re(s) > 1 or s = 1 since the factor x
s−1
+
is not bounded in the neighbourhood of
0 for Re(s) < 1 or, for Re(s) = 1, s = 1, does not approach a limit as x −→ +0.
All these difficulties disappear in Lebesgue theory.
52 (2’) could be directly obtained by computing the Fourier transform of the function equal to e(ζy) for y > 0 and to 0 otherwise, and by applying the general
formula (*).
53 f (t)dt could be replaced by a measure dμ(t) with finite total mass, and it is
precisely by studying such functions that Stieltjes was led to define his integrals.
Example : write the set of rational numbers as a sequence (un) and choose the
measure μ given by
f (t)dμ(t) =
f (un)/n
2 for continuous f with compact
support. The behaviour of the corresponding function
F (z) =
1/n
2 (un − z)
in the neighbourhood of the real axis is not obvious.