88
VIII – Cauchy Theory
outside the real axis and decomposes into the two functions F
+ and F
− on
the upper and lower half-planes H
+ and H
− . These functions being given by
the Fourier integrals occurring in (3), they extend by continuity to the closed
half-planes Im(z) ≥ 0 and Im(z) ≤ 0, provided these integrals converge for
real z, which assumes that f ∈ F
1 (R) since f has already been assumed to
be integrable. It is not in general possible to go from one to the other by
analytic extension since, as we will see, their limit values on the real axis
do not coincide. Nonetheless, it is necessary to observe that the left hand
side of (3) continues to be well-defined in a neighbourhood of z if z does not
belong to the support S of f , so that (3) defines a holomorphic function on
the connected open set H
+
∪ H
−
∪ (R − S) of C if the open subset R − S
of R is not empty; under these circumstances, the analytic extension across
R − S is possible, which does not prevent the function obtained to suffer a
discontinuities across S.
Set
u f (x, y) = F
+ (z) − F
− (¯ z) =
1
2πi
1
t − z
−
1
t − ¯
z
f (t)dt =
(11.4)
=
1
π
y
|t − z| 2 f (t)dt ,
where integration is as always over R when it is not specified. (2’) and (2”)
show that
54
u f (x, y) =
+∞
0
ˆ
f (u) exp(−2πuy)e(ux)du +
(11.5)
+
0
−∞
ˆ
f (u) exp(+2πuy)e(ux)du =
=
e(xu) exp(−2πy|u|) ˆ
f (u)du .
As y tends to +0, the function exp(−2πy|u|) converges uniformly to 1 on
every compact set by remaining ≤ 1. Therefore, the most elementary version
of the dominated convergence theorem shows that
lim
y→+0
F
+ (x + iy) − F
− (x − iy)
=
(11.6)
= lim
y→+0
u f =
e(xu) ˆ
f (u)du = f (x)
if f ∈ F
1 (R), which explains the impossibility of “ gluing back ” F
+ and F
−
into a single holomorphic function on C.
54 These computations have already been used in Chapter VII, § 6, n
◦ 30 to prove
Fourier’s inversion formula (Theorem 26).
VIII – Cauchy Theory
outside the real axis and decomposes into the two functions F
+ and F
− on
the upper and lower half-planes H
+ and H
− . These functions being given by
the Fourier integrals occurring in (3), they extend by continuity to the closed
half-planes Im(z) ≥ 0 and Im(z) ≤ 0, provided these integrals converge for
real z, which assumes that f ∈ F
1 (R) since f has already been assumed to
be integrable. It is not in general possible to go from one to the other by
analytic extension since, as we will see, their limit values on the real axis
do not coincide. Nonetheless, it is necessary to observe that the left hand
side of (3) continues to be well-defined in a neighbourhood of z if z does not
belong to the support S of f , so that (3) defines a holomorphic function on
the connected open set H
+
∪ H
−
∪ (R − S) of C if the open subset R − S
of R is not empty; under these circumstances, the analytic extension across
R − S is possible, which does not prevent the function obtained to suffer a
discontinuities across S.
Set
u f (x, y) = F
+ (z) − F
− (¯ z) =
1
2πi
1
t − z
−
1
t − ¯
z
f (t)dt =
(11.4)
=
1
π
y
|t − z| 2 f (t)dt ,
where integration is as always over R when it is not specified. (2’) and (2”)
show that
54
u f (x, y) =
+∞
0
ˆ
f (u) exp(−2πuy)e(ux)du +
(11.5)
+
0
−∞
ˆ
f (u) exp(+2πuy)e(ux)du =
=
e(xu) exp(−2πy|u|) ˆ
f (u)du .
As y tends to +0, the function exp(−2πy|u|) converges uniformly to 1 on
every compact set by remaining ≤ 1. Therefore, the most elementary version
of the dominated convergence theorem shows that
lim
y→+0
F
+ (x + iy) − F
− (x − iy)
=
(11.6)
= lim
y→+0
u f =
e(xu) ˆ
f (u)du = f (x)
if f ∈ F
1 (R), which explains the impossibility of “ gluing back ” F
+ and F
−
into a single holomorphic function on C.
54 These computations have already been used in Chapter VII, § 6, n
◦ 30 to prove
Fourier’s inversion formula (Theorem 26).