§ 2. Cauchy’s Integral Formulas
57
(a) g(ζ, z) is a continuous function of (ζ, z) ∈ G × G,
(b) h(z) =
μ
g(ζ, z)dζ is a holomorphic function on G,
(c) it can be analytically extended to all of C,
(d) the entire function thus obtained approaches 0 at infinity.
Liouville’s theorem (Chap. VII, § 4, n
◦ 18) will then show that h(z) = 0
everywhere, proving (1’) and the implication (ii) =⇒ (iii).
(a) g is clearly a continuous function of the couple (ζ, z) on the subset of
G × G defined by the relation ζ = z. Continuity at each point (a, a) of the
“ diagonal ” of the Cartesian product G × G is less obvious.
To simplify the notation, suppose that a = 0. The Taylor series f (z) =
c n z
n de f at a = 0 converges and represents f on a disc of radius R > 0.
Let D be a disc |z| < r with r < R; hence
f (ζ) − f (z) =
cn
ζ
n − z
n
= (ζ − z)
n≥1
cn
ζ
n−1 + ζ
n−2 z + . . . + z
n−1
in D × D. As g(0, 0) = f
(0) = c 1 , it follows that
|g(ζ, z) − g(0, 0)| =
n≥2
c n
ζ
n−1 + ζ
n−2 z + . . . + z
n−1
≤
≤
n≥2
n |c n | r
n−1 .
This is a convergent series in r, without a constant term. As r approaches 0,
so does its sum, proving the continuity of g.
(b) It is then possible to define
h(z) =
g(ζ, z)dζ =
g [μ(t), z] μ
(t)dt ,
(6.3)
where integration is along the given path μ : I −→ G. As g [μ(t), z] is a
continuous function of (t, z) on I × I by (a), and for given t, is holomorphic
in z, the result is holomorphic in z by Theorem 9 stated below.
(c) Ext(μ) contains the exterior of a disc, its complement
K = Int(μ) ∪ Supp(μ)
is compact, and by assumption (ii), contained in G. Since G is open, and
hence distinct from K, the open set U = G ∩ Ext(μ) is not empty. So for
z ∈ U , z /
∈ Supp(μ), and writing
h(z) =
μ
f (ζ)
ζ − z
dζ − f (z)
μ
d(ζ)
ζ − z
=
(6.4)
=
μ
f (ζ)
ζ − z
dζ − 2πif (z) Ind μ (z) =
μ
f (ζ)
ζ − z
dζ
57
(a) g(ζ, z) is a continuous function of (ζ, z) ∈ G × G,
(b) h(z) =
μ
g(ζ, z)dζ is a holomorphic function on G,
(c) it can be analytically extended to all of C,
(d) the entire function thus obtained approaches 0 at infinity.
Liouville’s theorem (Chap. VII, § 4, n
◦ 18) will then show that h(z) = 0
everywhere, proving (1’) and the implication (ii) =⇒ (iii).
(a) g is clearly a continuous function of the couple (ζ, z) on the subset of
G × G defined by the relation ζ = z. Continuity at each point (a, a) of the
“ diagonal ” of the Cartesian product G × G is less obvious.
To simplify the notation, suppose that a = 0. The Taylor series f (z) =
c n z
n de f at a = 0 converges and represents f on a disc of radius R > 0.
Let D be a disc |z| < r with r < R; hence
f (ζ) − f (z) =
cn
ζ
n − z
n
= (ζ − z)
n≥1
cn
ζ
n−1 + ζ
n−2 z + . . . + z
n−1
in D × D. As g(0, 0) = f
(0) = c 1 , it follows that
|g(ζ, z) − g(0, 0)| =
n≥2
c n
ζ
n−1 + ζ
n−2 z + . . . + z
n−1
≤
≤
n≥2
n |c n | r
n−1 .
This is a convergent series in r, without a constant term. As r approaches 0,
so does its sum, proving the continuity of g.
(b) It is then possible to define
h(z) =
g(ζ, z)dζ =
g [μ(t), z] μ
(t)dt ,
(6.3)
where integration is along the given path μ : I −→ G. As g [μ(t), z] is a
continuous function of (t, z) on I × I by (a), and for given t, is holomorphic
in z, the result is holomorphic in z by Theorem 9 stated below.
(c) Ext(μ) contains the exterior of a disc, its complement
K = Int(μ) ∪ Supp(μ)
is compact, and by assumption (ii), contained in G. Since G is open, and
hence distinct from K, the open set U = G ∩ Ext(μ) is not empty. So for
z ∈ U , z /
∈ Supp(μ), and writing
h(z) =
μ
f (ζ)
ζ − z
dζ − f (z)
μ
d(ζ)
ζ − z
=
(6.4)
=
μ
f (ζ)
ζ − z
dζ − 2πif (z) Ind μ (z) =
μ
f (ζ)
ζ − z
dζ