Das Ziel dieses Artikels ist es, alle biholomorphen Abbildungen
D
→
D
{\displaystyle \mathbb {D} \to \mathbb {D} }
Sei
f
:
D
→
D
{\displaystyle f\colon \mathbb {D} \to \mathbb {D} }
z
0
∈
D
{\displaystyle z_{0}\in \mathbb {D} }
λ
∈
C
{\displaystyle \lambda \in \mathbb {C} }
|
λ
|
=
1
{\displaystyle |\lambda |=1}
f
(
z
)
=
λ
z
−
z
0
1
−
z
¯
0
z
{\displaystyle f(z)=\lambda {\frac {z-z_{0}}{1-{\bar {z}}_{0}z}}}
Der Einheitskreis
D
{\displaystyle \mathbb {D} }
und sein Bild unter
f
(
z
)
=
λ
z
−
z
0
1
−
z
¯
0
z
{\displaystyle f(z)=\lambda {\frac {z-z_{0}}{1-{\bar {z}}_{0}z}}}
z
0
=
2
5
(
1
+
i
)
{\displaystyle z_{0}={\frac {2}{5}}(1+i)}
λ
=
2
−
1
/
2
(
1
+
i
)
{\displaystyle \lambda =2^{-1/2}(1+i)}
Umgekehrt sind alle solchen Abbildungen Automorphismen von
D
{\displaystyle \mathbb {D} }
Bevor wir den Satz beweisen, wollen wir noch eine wichtige Folgerung notieren:
Sei
z
0
∈
D
{\displaystyle z_{0}\in \mathbb {D} }
D
{\displaystyle \mathbb {D} }
f
(
0
)
=
z
0
{\displaystyle f(0)=z_{0}}
f
′
(
0
)
>
0
{\displaystyle f'(0)>0}
Zunächst zur Eindeutigkeit: Sind
f
,
g
{\displaystyle f,g}
h
:=
f
−
1
∘
g
:
D
→
D
{\displaystyle h:=f^{-1}\circ g\colon \mathbb {D} \to \mathbb {D} }
h
(
0
)
=
0
{\displaystyle h(0)=0}
λ
{\displaystyle \lambda }
z
1
{\displaystyle z_{1}}
h
(
z
)
=
λ
z
−
z
1
1
−
z
¯
1
z
{\displaystyle h(z)=\lambda {\frac {z-z_{1}}{1-{\bar {z}}_{1}z}}}
0
=
h
(
0
)
=
λ
(
−
z
1
)
⟺
z
1
=
0
{\displaystyle 0=h(0)=\lambda (-z_{1})\iff z_{1}=0}
h
(
z
)
=
λ
z
{\displaystyle h(z)=\lambda z}
λ
=
h
′
(
0
)
=
(
f
−
1
∘
g
)
′
(
0
)
=
1
f
′
(
(
f
−
1
∘
g
)
(
0
)
)
⋅
g
′
(
0
)
=
g
′
(
0
)
f
′
(
0
)
>
0
{\displaystyle \lambda =h'(0)=(f^{-1}\circ g)'(0)={\frac {1}{f'{\big (}(f^{-1}\circ g)(0){\big )}}}\cdot g'(0)={\frac {g'(0)}{f'(0)}}>0}
λ
=
1
{\displaystyle \lambda =1}
h
=
i
d
{\displaystyle h=\mathrm {id} }
λ
=
1
{\displaystyle \lambda =1}
Zur Existenz: Definiere
g
:
D
→
D
{\displaystyle g\colon \mathbb {D} \to \mathbb {D} }
g
(
z
)
:=
λ
z
−
z
0
1
−
z
¯
0
z
{\displaystyle g(z):=\lambda {\frac {z-z_{0}}{1-{\bar {z}}_{0}z}}}
λ
{\displaystyle \lambda }
g
′
(
z
0
)
>
0
{\displaystyle g'(z_{0})>0}
g
{\displaystyle g}
g
(
z
0
)
=
0
{\displaystyle g(z_{0})=0}
f
:=
g
−
1
{\displaystyle f:=g^{-1}}
f
(
0
)
=
z
0
{\displaystyle f(0)=z_{0}}
f
′
(
0
)
=
1
/
g
′
(
z
0
)
>
0
{\displaystyle f'(0)=1/g'(z_{0})>0}
Als erstes zeigen wir, dass alle diese Abbildungen Automorphismen sind. Sei also
z
0
∈
D
{\displaystyle z_{0}\in \mathbb {D} }
|
λ
|
=
1
{\displaystyle |\lambda |=1}
f
(
z
)
=
λ
z
−
z
0
1
−
z
¯
0
z
{\displaystyle f(z)=\lambda {\frac {z-z_{0}}{1-{\bar {z}}_{0}z}}}
f
{\displaystyle f}
|
f
(
z
)
|
=
|
z
−
z
0
|
|
1
−
z
¯
0
z
|
=
|
z
|
|
1
−
z
¯
z
0
|
|
1
−
z
¯
0
z
|
=
|
z
|
=
1
,
|
z
|
=
1
{\displaystyle |f(z)|={\frac {|z-z_{0}|}{|1-{\bar {z}}_{0}z|}}={\frac {|z||1-{\bar {z}}z_{0}|}{|1-{\bar {z}}_{0}z|}}=|z|=1,\quad |z|=1}
und
f
(
z
0
)
=
0
{\displaystyle f(z_{0})=0}
f
(
D
)
⊆
D
{\displaystyle f(\mathbb {D} )\subseteq \mathbb {D} }
f
{\displaystyle f}
f
{\displaystyle f}
f
(
z
)
=
w
⟺
λ
z
−
z
0
1
−
z
z
¯
0
=
w
⟺
z
−
z
0
=
λ
¯
w
(
1
−
z
z
¯
0
)
⟺
z
(
1
+
λ
¯
w
z
¯
0
)
=
λ
¯
w
+
z
0
⟺
z
=
λ
¯
w
+
z
0
1
+
λ
¯
w
z
¯
0
⟺
z
=
λ
¯
w
−
(
−
λ
z
0
)
1
−
(
−
λ
z
0
)
¯
w
{\displaystyle {\begin{array}{rl}f(z)&=w\\\iff \displaystyle \lambda {\frac {z-z_{0}}{1-z{\bar {z}}_{0}}}&=w\\\iff z-z_{0}&={\bar {\lambda }}w(1-z{\bar {z}}_{0})\\\iff z(1+{\bar {\lambda }}w{\bar {z}}_{0})&={\bar {\lambda }}w+z_{0}\\\iff z&=\displaystyle {\frac {{\bar {\lambda }}w+z_{0}}{1+{\bar {\lambda }}w{\bar {z}}_{0}}}\\\iff z&={\bar {\lambda }}\displaystyle {\frac {w-{(-\lambda z_{0})}}{1-{\overline {(-\lambda z_{0})}}w}}\end{array}}}
Also ist
f
−
1
{\displaystyle f^{-1}}
Zum Beweis der Tatsache, dass jeder Automorphismus von der behaupteten Form ist, betrachten wir zunächst den Spezialfall
f
(
0
)
=
0
{\displaystyle f(0)=0}
Lemma von Schwarz zunächst
|
f
(
z
)
|
≤
|
z
|
{\displaystyle |f(z)|\leq |z|}
z
∈
D
{\displaystyle z\in \mathbb {D} }
f
−
1
{\displaystyle f^{-1}}
|
z
|
≤
|
f
(
z
)
|
{\displaystyle |z|\leq |f(z)|}
|
f
(
z
)
|
=
|
z
|
{\displaystyle |f(z)|=|z|}
z
∈
D
{\displaystyle z\in \mathbb {D} }
f
{\displaystyle f}
f
(
z
)
=
λ
z
{\displaystyle f(z)=\lambda z}
|
λ
|
=
1
{\displaystyle |\lambda |=1}
Sei nun
f
(
0
)
=:
z
1
{\displaystyle f(0)=:z_{1}}
g
(
z
)
:=
z
−
z
1
1
−
z
¯
1
z
{\displaystyle g(z):={\frac {z-z_{1}}{1-{\bar {z}}_{1}z}}}
g
{\displaystyle g}
h
:=
g
∘
f
{\displaystyle h:=g\circ f}
D
{\displaystyle \mathbb {D} }
h
(
0
)
=
0
{\displaystyle h(0)=0}
h
(
z
)
=
λ
z
{\displaystyle h(z)=\lambda z}
|
λ
|
=
1
{\displaystyle |\lambda |=1}
f
(
z
)
=
g
−
1
(
λ
z
)
=
λ
z
+
z
1
1
+
z
¯
1
λ
z
=
λ
z
+
λ
¯
z
1
1
+
λ
¯
z
1
¯
z
{\displaystyle f(z)=g^{-1}(\lambda z)={\frac {\lambda z+z_{1}}{1+{\bar {z}}_{1}\lambda z}}=\lambda {\frac {z+{\bar {\lambda }}z_{1}}{1+{\overline {{\bar {\lambda }}z_{1}}}z}}}
mit
z
0
:=
−
λ
¯
z
1
{\displaystyle z_{0}:=-{\bar {\lambda }}z_{1}}
