Set
Δ
0
:=
D
{\displaystyle \Delta _{0}:=D}
(
Δ
n
)
n
≥
0
{\displaystyle (\Delta _{n})_{n\geq 0}}
1.
Δ
n
⊆
Δ
n
−
1
{\displaystyle \Delta _{n}\subseteq \Delta _{n-1}}
2.
L
(
∂
Δ
n
)
=
2
−
n
L
(
∂
D
)
{\displaystyle {\mathcal {L}}(\partial \Delta _{n})=2^{-n}{\mathcal {L}}(\partial D)}
L
{\displaystyle {\mathcal {L}}}
length of a curve
3.
|
∫
∂
D
f
(
z
)
,
d
z
|
≤
4
n
|
∫
∂
Δ
n
f
(
z
)
,
d
z
|
{\displaystyle \left|\int _{\partial D}f(z),dz\right|\leq 4^{n}\left|\int _{\partial \Delta _{n}}f(z),dz\right|}
For
n
≥
0
{\displaystyle n\geq 0}
Δ
n
{\displaystyle \Delta _{n}}
Δ
n
{\displaystyle \Delta _{n}}
Δ
n
+
1
i
{\displaystyle \Delta _{n+1}^{i}}
1
≤
i
≤
4
{\displaystyle 1\leq i\leq 4}
|
∫
∂
Δ
n
f
(
z
)
d
z
|
=
|
∑
i
=
1
4
∫
∂
Δ
n
+
1
i
f
(
z
)
d
z
|
≤
∑
i
=
1
4
|
∫
∂
Δ
n
+
1
i
f
(
z
)
d
z
|
≤
max
i
|
∫
∂
Δ
n
+
1
i
f
(
z
)
d
z
|
{\displaystyle {\begin{array}{rl}\displaystyle \left|\int _{\partial \Delta _{n}}f(z)\,dz\right|&=\displaystyle \left|\sum _{i=1}^{4}\int _{\partial \Delta _{n+1}^{i}}f(z)\,dz\right|\\&\leq \displaystyle \sum _{i=1}^{4}\left|\int _{\partial \Delta _{n+1}^{i}}f(z)\,dz\right|\\&\leq \displaystyle \max _{i}\left|\int _{\partial \Delta _{n+1}^{i}}f(z)\,dz\right|\end{array}}}
Choose
1
≤
i
≤
4
{\displaystyle 1\leq i\leq 4}
|
∫
∂
Δ
n
+
1
i
f
(
z
)
,
d
z
|
=
max
i
|
∫
∂
Δ
n
+
1
i
f
(
z
)
,
d
z
|
{\displaystyle \left|\int _{\partial \Delta _{n+1}^{i}}f(z),dz\right|=\max _{i}\left|\int _{\partial \Delta _{n+1}^{i}}f(z),dz\right|}
Δ
n
+
1
:=
Δ
n
+
1
i
{\displaystyle \Delta _{n+1}:=\Delta _{n+1}^{i}}
Δ
n
+
1
⊆
Δ
n
{\displaystyle \Delta _{n+1}\subseteq \Delta _{n}}
L
(
∂
Δ
n
+
1
)
=
1
2
L
(
∂
Δ
n
)
=
2
−
(
n
+
1
)
L
(
∂
D
)
{\displaystyle {\mathcal {L}}(\partial \Delta _{n+1})={\frac {1}{2}}{\mathcal {L}}(\partial \Delta _{n})=2^{-(n+1)}{\mathcal {L}}(\partial D)}
|
∫
∂
D
f
(
z
)
,
d
z
|
≤
4
n
|
∫
∂
Δ
n
f
(
z
)
,
d
z
|
≤
4
n
+
1
|
∫
∂
Δ
n
+
1
f
(
z
)
,
d
z
|
.
{\displaystyle \left|\int _{\partial D}f(z),dz\right|\leq 4^{n}\left|\int _{\partial \Delta _{n}}f(z),dz\right|\leq 4^{n+1}\left|\int _{\partial \Delta _{n+1}}f(z),dz\right|.}
This ensures
Δ
n
+
1
{\displaystyle \Delta _{n+1}}
Since all
Δ
n
{\displaystyle \Delta _{n}}
⋂
n
≥
0
Δ
n
≠
∅
{\displaystyle \bigcap _{n\geq 0}\Delta _{n}\neq \emptyset }
z
0
∈
⋂
n
≥
0
Δ
n
{\displaystyle z_{0}\in \bigcap _{n\geq 0}\Delta _{n}}
f
{\displaystyle f}
z
0
{\displaystyle z_{0}}
V
{\displaystyle V}
z
0
{\displaystyle z_{0}}
A
:
V
→
C
{\displaystyle A\colon V\to \mathbb {C} }
A
(
z
0
)
=
0
{\displaystyle A(z_{0})=0}
f
(
z
)
=
f
(
z
0
)
+
(
z
−
z
0
)
f
′
(
z
0
)
+
A
(
z
)
(
z
−
z
0
)
,
z
∈
V
.
{\displaystyle f(z)=f(z_{0})+(z-z_{0})f'(z_{0})+A(z)(z-z_{0}),\qquad z\in V.}
Since the function
z
↦
f
(
z
0
)
+
(
z
−
z
0
)
f
′
(
z
0
)
{\displaystyle z\mapsto f(z_{0})+(z-z_{0})f'(z_{0})}
n
≥
0
{\displaystyle n\geq 0}
Δ
n
⊆
V
{\displaystyle \Delta _{n}\subseteq V}
∫
∂
Δ
n
f
(
z
)
,
d
z
=
∫
∂
Δ
n
f
(
z
0
)
+
(
z
−
z
0
)
f
′
(
z
0
)
+
A
(
z
)
(
z
−
z
0
)
,
d
z
=
∫
∂
Δ
n
A
(
z
)
(
z
−
z
0
)
,
d
z
.
{\displaystyle \int _{\partial \Delta _{n}}f(z),dz=\int _{\partial \Delta _{n}}f(z_{0})+(z-z_{0})f'(z_{0})+A(z)(z-z_{0}),dz=\int _{\partial \Delta _{n}}A(z)(z-z_{0}),dz.}
Thus, due to the continuity of
A
{\displaystyle A}
A
(
z
0
)
=
0
{\displaystyle A(z_{0})=0}
|
∫
∂
D
f
(
z
)
d
z
|
≤
4
n
|
∫
∂
Δ
n
f
(
z
)
d
z
|
=
4
n
|
∫
∂
Δ
n
A
(
z
)
(
z
−
z
0
)
d
z
|
≤
4
n
⋅
L
(
∂
Δ
n
)
max
z
∈
∂
Δ
n
|
z
−
z
0
|
|
A
(
z
)
|
≤
4
n
⋅
L
(
∂
Δ
n
)
2
max
z
∈
∂
Δ
n
|
A
(
z
)
|
=
L
(
∂
D
)
max
z
∈
∂
Δ
n
|
A
(
z
)
|
→
L
(
∂
D
)
|
A
(
z
0
)
|
=
0
,
n
→
∞
.
{\displaystyle {\begin{array}{rl}\displaystyle \left|\int _{\partial D}f(z)\,dz\right|&\leq \displaystyle 4^{n}\left|\int _{\partial \Delta _{n}}f(z)\,dz\right|\\&=\displaystyle 4^{n}\left|\int _{\partial \Delta _{n}}A(z)(z-z_{0})\,dz\right|\\&\leq \displaystyle 4^{n}\cdot {\mathcal {L}}(\partial \Delta _{n})\max _{z\in \partial \Delta _{n}}|z-z_{0}||A(z)|\\&\leq \displaystyle 4^{n}\cdot {\mathcal {L}}(\partial \Delta _{n})^{2}\max _{z\in \partial \Delta _{n}}|A(z)|\\&=\displaystyle {\mathcal {L}}(\partial D)\max _{z\in \partial \Delta _{n}}|A(z)|\\&\to \displaystyle {\mathcal {L}}(\partial D)|A(z_{0})|=0,\qquad n\to \infty .\end{array}}}
Notation in the Proof
edit
