Math 411: Honours Complex Variables - University of Alberta

30 CHAPTER 5. CAUCHY’S INTEGRAL THEOREM AND FORMULA
with
ℓ(∂∆n) = 1
2nℓ(∂∆) and �� ���
� ��
� �
f�
� ≤ 4n�
�
�
�
f�
�
for n ∈ N.
Let z0 ∈ � ∞
n=1 ∆n, and define
∂∆
∂∆n
r: D → C, z ↦→ f(z)−f(z0)−f ′ (z0)(z −z0),
|r(z)| � �
so that lim = 0 and (r −f) =
z→z0 |z −z0| γ γ [−f(z0)−f ′ (z0)(z −z0)]dz = 0 for each
closed curve γ in D (noting that the integrand has an antiderivative). Consequently,
��
�
�
�
� ��
� �
f�
� ≤ 4n�
�
�
�
r�
�
∂∆
∂∆n
for n ∈ N. Let ǫ > 0 and choose δ > 0 such that
� �
�
�
r(z) �
�
�z
−z0
� ≤
ǫ
[ℓ(∂∆)] 2
for all z ∈ D with |z −z0|< δ. Choose n ∈ N such that ∆n ⊂ Bδ(z0). For z ∈ ∆n,
this means that
|z −z0|≤ ℓ(∂∆n) = 1
2nℓ(∂∆). We thus obtain:
���� � � ��
� �
f�
� ≤ 4n�
�
�
�
r�
�
∂∆
∂∆n
≤ 4 n ℓ(∂∆n) sup
ζ∈∂∆n
≤ 2 n ℓ(∂∆) sup
ζ∈∂∆n
≤ ǫ.
As ǫ > 0 was arbitrary, this proves the claim.
|r(ζ)|= 2 n ℓ(∂∆) sup
ǫ
|ζ −z0|
[ℓ(∂∆)] 2
� �� �
≤ 1
2 nℓ(∂∆)
ζ∈∂∆n
|r(ζ)|
Definition. A set D ⊂ C is called star shaped if there exists z0 ∈ D such that
[z0,z] ⊂ D for each z ∈ D. The point z0 is called a center for D.
Theorem 5.2. Let D ⊂ C be open and star shaped with center z0, and let f: D → C
be continuous such that �
f(ζ)dζ = 0
∂∆
for each triangle ∆ ⊂ D with z0 as a vertex. Then f has an antiderivative.