Seien
z
∈
G
T
(
A
)
{\textstyle z\in {\mathcal {G}}_{\mathcal {T}}(A)}
(
A
,
‖
⋅
‖
A
)
,
(
B
,
‖
|
⋅
|
‖
A
~
)
∈
T
e
(
K
)
{\textstyle \left(A,\left\|\cdot \right\|_{\mathcal {A}}\right),\left(B,\left\|\!\left|\cdot \right|\!\right\|_{\widetilde {\mathcal {A}}}\right)\in {\mathcal {T}}_{e}(\mathbb {K} )}
B
{\textstyle B}
T
{\textstyle {\mathcal {T}}}
A
{\textstyle A}
b
∈
B
{\textstyle b\in B}
z
{\textstyle z}
unital positiv .
Bearbeiten
Man betrachtet das auf der Algebraerweiterung
B
{\textstyle B}
‖
|
⋅
|
‖
A
~
{\displaystyle \left\|\!\left|\cdot \right|\!\right\|_{\widetilde {\mathcal {A}}}}
‖
|
⋅
|
‖
A
~
{\textstyle \left\|\!\left|\cdot \right|\!\right\|_{\widetilde {\mathcal {A}}}}
A
{\displaystyle A}
‖
⋅
‖
A
{\displaystyle \left\|\cdot \right\|_{\mathcal {A}}}
äquivalent ist.
Bearbeiten
Das von
B
{\displaystyle B}
A
{\displaystyle A}
‖
⋅
‖
A
~
{\textstyle \left\|\cdot \right\|_{\widetilde {\mathcal {A}}}}
τ
:
A
→
A
′
⊂
B
{\displaystyle \tau :A\to A'\subset B}
Algebraerweiterung ) d.h.
‖
⋅
‖
α
~
:
A
⟶
K
,
x
⟼
‖
|
τ
(
x
)
|
‖
α
~
mit
α
~
∈
A
~
{\displaystyle {\begin{array}{rccl}\left\|\cdot \right\|_{\widetilde {\alpha }}:&A&\longrightarrow &\mathbb {K} ,\\&x&\longmapsto &\left\|\!\left|\tau (x)\right|\!\right\|_{\widetilde {\alpha }}{\mbox{ mit }}{\widetilde {\alpha }}\in {\widetilde {\mathcal {A}}}\\\end{array}}}
Bearbeiten
Da
τ
:
A
→
A
′
⊂
B
{\displaystyle \tau :A\to A'\subset B}
A
′
{\displaystyle A'}
A
{\displaystyle A}
τ
{\displaystyle \tau }
τ
−
1
{\displaystyle \tau ^{-1}}
Äquivalenz der Gaugefunktionalsysteme
‖
⋅
‖
A
~
{\textstyle \left\|\cdot \right\|_{\widetilde {\mathcal {A}}}}
‖
⋅
‖
A
{\textstyle \left\|\cdot \right\|_{\mathcal {A}}}
Beweis 4 - Definition von Stetigkeitssequenzen
Bearbeiten
Nun definiert man damit folgendes System
‖
⋅
‖
A
~
×
N
0
{\textstyle \left\|\cdot \right\|_{{\widetilde {\mathcal {A}}}\times \mathbb {N} _{0}}}
‖
⋅
‖
(
α
~
,
k
)
:
A
⟶
K
,
x
⟼
‖
|
z
k
⋅
x
|
‖
α
~
mit
α
~
∈
A
~
.
{\displaystyle \left\|\cdot \right\|_{({\widetilde {\alpha }},k)}:A\longrightarrow \mathbb {K} ,x\longmapsto \left\|\!\left|z^{k}\cdot x\right|\!\right\|_{\widetilde {\alpha }}{\mbox{ mit }}{\widetilde {\alpha }}\in {\widetilde {\mathcal {A}}}.}
Beweis 5 - Stetigkeit der Multiplikation
Bearbeiten
Für alle
α
~
∈
A
~
{\displaystyle {\widetilde {\alpha }}\in {\widetilde {\mathcal {A}}}}
β
~
∈
A
~
{\textstyle {\widetilde {\beta }}\in {\widetilde {\mathcal {A}}}}
x
∈
A
{\textstyle x\in A}
|
‖
x
⋅
y
‖
|
α
~
≤
|
‖
x
‖
|
β
~
⋅
|
‖
y
‖
|
β
~
{\displaystyle \left|\!\left\|x\cdot y\right\|\!\right|_{\widetilde {\alpha }}\leq \left|\!\left\|x\right\|\!\right|_{\widetilde {\beta }}\cdot \left|\!\left\|y\right\|\!\right|_{\widetilde {\beta }}}
Beweis 6 - Anwendung auf die Gaugefunktionale
Bearbeiten
Man erhält über die Stetigkeit der Multiplikation folgende Ungleichungen:
‖
x
‖
α
~
=
|
‖
x
‖
|
α
~
=
|
‖
z
k
⋅
b
k
⋅
x
‖
|
α
~
≤
|
‖
z
k
‖
|
β
~
⋅
|
‖
b
k
⋅
x
‖
|
β
~
≤
|
‖
z
k
‖
|
γ
⏟
D
k
(
α
~
)
⋅
‖
x
‖
(
α
~
,
k
)
=
D
k
(
α
~
)
⋅
‖
x
‖
(
α
~
,
k
)
{\displaystyle {\begin{array}{rcl}\left\|x\right\|_{\widetilde {\alpha }}&=&\left|\!\left\|x\right\|\!\right|_{\widetilde {\alpha }}=\left|\!\left\|z^{k}\cdot b^{k}\cdot x\right\|\!\right|_{\widetilde {\alpha }}\\&\leq &\left|\!\left\|z^{k}\right\|\!\right|_{\widetilde {\beta }}\cdot \left|\!\left\|b^{k}\cdot x\right\|\!\right|_{\widetilde {\beta }}\\&\leq &\underbrace {\left|\!\left\|z^{k}\right\|\!\right|_{\gamma }} _{D_{k}({\widetilde {\alpha }})}\cdot \left\|x\right\|_{({\widetilde {\alpha }},k)}=D_{k}({\widetilde {\alpha }})\cdot \left\|x\right\|_{({\widetilde {\alpha }},k)}\\\end{array}}}
Unter Verwendung der Unitalen Positivität der verwendeten Gaugfunktionalsysteme erhält man insbesondere:
0
<
‖
e
A
‖
α
~
≤
D
k
(
α
~
)
⋅
‖
z
k
⋅
e
A
‖
β
=
D
k
(
α
~
)
⏟
>
0
⋅
‖
z
k
‖
β
~
⏟
>
0
{\displaystyle 0<\left\|e_{A}\right\|_{\widetilde {\alpha }}\leq D_{k}({\widetilde {\alpha }})\cdot \left\|z^{k}\cdot e_{A}\right\|_{\beta }=\underbrace {D_{k}({\widetilde {\alpha }})} _{>0}\cdot \underbrace {\left\|z^{k}\right\|_{\widetilde {\beta }}} _{>0}}
Bearbeiten
Insgesamt gilt die folgende Abschätzung
durch die Äquivalenz der Gaugefunktionalsysteme auch für das gegebene Gaugefunktionalsystem
‖
⋅
‖
A
{\displaystyle \left\|\cdot \right\|_{\mathcal {A}}}
α
∈
A
{\displaystyle \alpha \in {\mathcal {A}}}
C
1
>
0
{\displaystyle C_{1}>0}
α
~
∈
A
~
{\displaystyle {\widetilde {\alpha }}\in {\widetilde {\mathcal {A}}}}
‖
⋅
‖
α
≤
C
1
⋅
‖
⋅
‖
α
~
{\displaystyle \left\|\cdot \right\|_{\alpha }\leq C_{1}\cdot \left\|\cdot \right\|_{\widetilde {\alpha }}}
Für diese
α
~
∈
A
~
{\displaystyle {\widetilde {\alpha }}\in {\widetilde {\mathcal {A}}}}
D
k
(
α
~
)
>
0
{\displaystyle D_{k}({\widetilde {\alpha }})>0}
β
~
∈
A
~
{\displaystyle {\widetilde {\beta }}\in {\widetilde {\mathcal {A}}}}
‖
x
‖
α
~
≤
D
k
(
α
~
)
⋅
‖
z
k
⋅
x
‖
β
~
{\displaystyle \left\|x\right\|_{\widetilde {\alpha }}\leq D_{k}({\widetilde {\alpha }})\cdot \left\|z^{k}\cdot x\right\|_{\widetilde {\beta }}}
Bearbeiten
Dann gibt es wieder mit der Äquivalenz der Gaugefunktionalsysteme
β
~
∈
A
~
{\displaystyle {\widetilde {\beta }}\in {\widetilde {\mathcal {A}}}}
β
∈
A
{\displaystyle \beta \in {\mathcal {A}}}
C
2
>
0
{\displaystyle C_{2}>0}
‖
⋅
‖
α
~
≤
C
2
⋅
‖
⋅
‖
β
{\displaystyle \left\|\cdot \right\|_{\widetilde {\alpha }}\leq C_{2}\cdot \left\|\cdot \right\|_{\beta }}
Bearbeiten
Insgesamt erhält man für alle
α
∈
A
{\displaystyle \alpha \in {\mathcal {A}}}
β
∈
A
{\displaystyle \beta \in {\mathcal {A}}}
D
k
(
α
)
>
0
{\displaystyle D_{k}(\alpha )>0}
‖
⋅
‖
A
{\displaystyle \left\|\cdot \right\|_{\mathcal {A}}}
x
∈
A
{\displaystyle x\in A}
‖
x
‖
α
≤
C
1
⋅
‖
x
‖
α
~
≤
C
1
⋅
D
k
(
α
~
)
⋅
‖
z
k
⋅
x
‖
β
~
≤
C
1
⋅
D
k
(
α
~
)
⋅
C
2
⏟
D
k
(
α
)
⋅
‖
z
k
⋅
x
‖
β
{\displaystyle {\begin{array}{rcl}\left\|x\right\|_{\alpha }&\leq &C_{1}\cdot \left\|x\right\|_{\widetilde {\alpha }}\\&\leq &C_{1}\cdot D_{k}({\widetilde {\alpha }})\cdot \left\|z^{k}\cdot x\right\|_{\widetilde {\beta }}\\&\leq &\underbrace {C_{1}\cdot D_{k}({\widetilde {\alpha }})\cdot C_{2}} _{D_{k}(\alpha )}\cdot \left\|z^{k}\cdot x\right\|_{\beta }\\\end{array}}}
Die obige Ungleichung
‖
x
‖
α
≤
D
k
(
α
)
⋅
‖
z
k
⋅
x
‖
β
{\displaystyle \left\|x\right\|_{\alpha }\leq D_{k}(\alpha )\cdot \left\|z^{k}\cdot x\right\|_{\beta }}
‖
x
‖
α
=
0
{\displaystyle \|x\|_{\alpha }=0}
‖
x
‖
α
>
0
{\displaystyle \|x\|_{\alpha }>0}
1
D
k
(
α
)
≤
1
‖
x
‖
α
⏟
=
(
‖
x
‖
α
−
1
p
)
p
⋅
‖
z
k
⋅
x
‖
β
=
‖
z
k
⋅
x
‖
x
‖
α
1
p
‖
β
{\displaystyle {\frac {1}{D_{k}(\alpha )}}\leq \underbrace {\frac {1}{\left\|x\right\|_{\alpha }}} _{=\left(\left\|x\right\|_{\alpha }^{-{\frac {1}{p}}}\right)^{p}}\cdot \left\|z^{k}\cdot x\right\|_{\beta }=\left\|z^{k}\cdot {\frac {x}{\left\|x\right\|_{\alpha }^{\frac {1}{p}}}}\right\|_{\beta }}
Das Infimum wird auch bei
p
{\displaystyle p}
‖
x
‖
α
=
1
{\displaystyle \|x\|_{\alpha }=1}
‖
x
‖
x
‖
α
1
p
‖
α
=
(
1
‖
x
‖
α
1
p
)
p
⋅
‖
x
‖
α
=
‖
x
‖
α
‖
x
‖
α
=
1
{\displaystyle \left\|{\frac {x}{\left\|x\right\|_{\alpha }^{\frac {1}{p}}}}\right\|_{\alpha }=\left({\frac {1}{\left\|x\right\|_{\alpha }^{\frac {1}{p}}}}\right)^{p}\cdot \left\|x\right\|_{\alpha }={\frac {\left\|x\right\|_{\alpha }}{\left\|x\right\|_{\alpha }}}=1}
Für
z
∈
A
{\displaystyle z\in A}
K
{\displaystyle {\mathcal {K}}}
‖
x
‖
α
=
1
{\displaystyle \left\|x\right\|_{\alpha }=1}
α
∈
A
{\displaystyle \alpha \in {\mathcal {A}}}
β
∈
A
{\displaystyle \beta \in {\mathcal {A}}}
ε
k
>
0
{\displaystyle \varepsilon _{k}>0}
x
∈
A
{\displaystyle x\in A}
ε
k
=
inf
‖
x
‖
α
=
1
‖
z
k
⋅
x
‖
β
{\displaystyle \varepsilon _{k}=\inf _{\left\|x\right\|_{\alpha }=1}\left\|z^{k}\cdot x\right\|_{\beta }}
Bearbeiten
Negation der Aussage (13) liefert dann: Es gibt ein
α
∈
A
{\textstyle \alpha \in {\mathcal {A}}}
β
∈
A
{\textstyle \beta \in {\mathcal {A}}}
k
(
β
)
∈
N
{\textstyle k(\beta )\in \mathbb {N} }
inf
‖
x
‖
α
=
1
‖
z
k
(
β
)
⋅
x
‖
β
=
0
{\displaystyle \inf _{\left\|x\right\|_{\alpha }=1}\left\|z^{k(\beta )}\cdot x\right\|_{\beta }=0}
Damit folgt die Behauptung.
◻
{\displaystyle \Box }
