Sei nun
f
{\displaystyle f}
γ
:=
(
1
1
0
1
)
∈
Γ
(
1
)
f
(
z
)
=
f
(
γ
.
z
)
=
f
(
z
+
1
)
∀
z
∈
H
{\displaystyle \gamma :={\begin{pmatrix}1&1\\0&1\\\end{pmatrix}}\in \Gamma (1)\,f(z)=f(\gamma .z)=f(z+1)\,\forall \,z\in \mathbb {H} }
f
{\displaystyle f}
f
(
x
+
i
y
)
=
∑
n
=
−
∞
∞
a
n
(
y
)
e
2
π
i
n
x
{\displaystyle f(x+iy)=\sum _{n=-\infty }^{\infty }a_{n}(y)e^{2\pi inx}}
Wir beobachten außerdem :
f
{\displaystyle f}
a
0
(
y
)
=
0
∀
y
>
0
{\displaystyle a_{0}(y)=0\,\,\,\forall y>0}
0
=
∫
0
1
f
(
z
+
t
)
d
t
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0=\int _{0}^{1}f(z+t)dt}
=
∫
0
1
∑
n
=
−
∞
∞
a
n
(
y
)
e
2
π
i
n
(
x
+
t
)
d
t
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\int _{0}^{1}\sum _{n=-\infty }^{\infty }a_{n}(y)e^{2\pi in(x+t)}dt}
=
(
1
)
∑
n
=
−
∞
∞
a
n
(
y
)
e
2
π
i
n
x
∫
0
1
e
2
π
i
n
t
d
t
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overset {(1)}{=}}\sum _{n=-\infty }^{\infty }a_{n}(y)e^{2\pi inx}\int _{0}^{1}e^{2\pi int}dt}
=
a
0
(
y
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=a_{0}(y)}
wobei in (1) benutzt wurde, dass die Reihe
∑
n
=
−
∞
∞
a
n
(
y
)
e
2
π
i
n
(
x
+
t
)
d
t
{\displaystyle \sum _{n=-\infty }^{\infty }a_{n}(y)e^{2\pi in(x+t)}dt}
y
>
0
{\displaystyle y>0}
Lemma : Fourierkoeffizienten einer Maaßschen Wellenform
Bearbeiten
Sei
λ
∈
C
{\displaystyle \lambda \in \mathbb {C} }
Δ
{\displaystyle \Delta }
ν
∈
C
{\displaystyle \nu \in \mathbb {C} }
λ
=
1
4
−
ν
2
{\displaystyle \lambda ={\frac {1}{4}}-\nu ^{2}}
K
s
(
y
)
:=
1
2
∫
0
∞
e
−
y
(
t
+
t
−
1
)
/
2
t
s
d
t
t
,
s
∈
C
,
y
>
0
{\displaystyle K_{s}(y):={\frac {1}{2}}\int _{0}^{\infty }e^{-y(t+t^{-1})/2}t^{s}{\frac {dt}{t}}\,\,,s\in \mathbb {C} \,,\,y>0}
f
{\displaystyle f}
a
n
(
y
)
=
c
n
y
K
ν
(
2
π
|
r
|
y
)
,
c
n
∈
C
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a_{n}(y)=c_{n}{\sqrt {y}}K_{\nu }(2\pi |r|y)\,\,\,,\,\,\,c_{n}\in \mathbb {C} }
falls
n
≠
0
{\displaystyle n\neq 0}
n
=
0
{\displaystyle n=0}
a
0
(
y
)
=
c
0
y
1
2
−
ν
+
d
0
y
1
2
+
ν
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a_{0}(y)=c_{0}y^{{\frac {1}{2}}-\nu }+d_{0}y^{{\frac {1}{2}}+\nu }}
c
0
,
d
0
∈
C
{\displaystyle c_{0},\,d_{0}\in \mathbb {C} }
Beweis : Es gilt
Δ
(
f
)
=
(
1
4
−
ν
2
)
f
{\displaystyle \Delta (f)=({\frac {1}{4}}-\nu ^{2})f}
n
∈
Z
{\displaystyle n\in \mathbb {Z} }
a
n
=
∫
0
1
f
(
x
+
i
y
)
e
−
2
π
i
n
x
d
x
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a_{n}=\int _{0}^{1}f(x+iy)e^{-2\pi inx}dx}
Zusammen folgt für
n
∈
Z
{\displaystyle n\in \mathbb {Z} }
(
1
4
−
ν
2
)
a
n
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,({\frac {1}{4}}-\nu ^{2})a_{n}}
=
∫
0
1
(
1
4
−
ν
2
)
f
(
x
+
i
y
)
e
−
2
π
i
n
x
d
x
{\displaystyle =\int _{0}^{1}({\frac {1}{4}}-\nu ^{2})f(x+iy)e^{-2\pi inx}dx}
=
∫
0
1
(
Δ
f
)
(
x
+
i
y
)
e
−
2
π
i
n
x
d
x
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\int _{0}^{1}(\Delta f)(x+iy)e^{-2\pi inx}dx}
=
−
y
2
(
∫
0
1
∂
2
f
∂
x
2
(
x
+
i
y
)
e
−
2
π
i
n
x
d
x
+
∫
0
1
∂
2
f
∂
y
2
(
x
+
i
y
)
e
−
2
π
i
n
x
d
x
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-y^{2}\left(\int _{0}^{1}{\frac {\partial ^{2}f}{\partial x^{2}}}(x+iy)e^{-2\pi inx}dx+\int _{0}^{1}{\frac {\partial ^{2}f}{\partial y^{2}}}(x+iy)e^{-2\pi inx}dx\right)}
=
(
1
)
−
y
2
(
2
π
i
n
)
2
a
n
(
y
)
−
y
2
∂
2
∂
y
2
∫
0
1
f
(
x
+
i
y
)
e
−
2
π
i
n
x
d
x
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\overset {(1)}{=}}-y^{2}(2\pi in)^{2}a_{n}(y)-y^{2}{\frac {\partial ^{2}}{\partial y^{2}}}\int _{0}^{1}f(x+iy)e^{-2\pi inx}dx}
=
−
y
2
(
2
π
i
n
)
2
a
n
(
y
)
−
y
2
∂
2
∂
y
2
a
n
(
y
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-y^{2}(2\pi in)^{2}a_{n}(y)-y^{2}{\frac {\partial ^{2}}{\partial y^{2}}}a_{n}(y)}
=
4
π
2
n
2
y
2
a
n
(
y
)
−
y
2
∂
2
∂
y
2
a
n
(
y
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=4\pi ^{2}n^{2}y^{2}a_{n}(y)-y^{2}{\frac {\partial ^{2}}{\partial y^{2}}}a_{n}(y)}
In (1) wurde für den ersten Summanden benutzt, dass der n-te Fourierkoeffizient von
∂
2
f
∂
x
2
{\displaystyle {\frac {\partial ^{2}f}{\partial x^{2}}}}
(
2
π
i
n
)
2
a
n
(
y
)
{\displaystyle (2\pi in)^{2}a_{n}(y)}
y
2
∂
2
∂
y
2
a
n
(
y
)
+
(
1
4
−
ν
2
−
4
π
n
2
y
2
)
a
n
(
y
)
=
0
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y^{2}{\frac {\partial ^{2}}{\partial y^{2}}}a_{n}(y)+({\frac {1}{4}}-\nu ^{2}-4\pi n^{2}y^{2})a_{n}(y)=0}
Für
n
=
0
{\displaystyle n=0}
f
{\displaystyle f}
c
0
,
d
0
∈
C
{\displaystyle c_{0},d_{0}\in \mathbb {C} }
f
(
y
)
=
c
0
y
1
4
−
ν
+
d
0
y
1
4
+
ν
{\displaystyle f(y)=c_{0}y^{{\frac {1}{4}}-\nu }+d_{0}y^{{\frac {1}{4}}+\nu }}
Für
n
≠
0
{\displaystyle n\neq 0}
f
{\displaystyle f}
f
(
y
)
=
c
n
y
K
v
(
2
π
|
n
|
y
)
+
d
n
y
I
v
(
2
π
|
n
|
y
)
{\displaystyle f(y)=c_{n}{\sqrt {y}}K_{v}(2\pi |n|y)+d_{n}{\sqrt {y}}I_{v}(2\pi |n|y)}
c
n
,
d
n
∈
C
{\displaystyle c_{n},d_{n}\in \mathbb {C} }
K
v
(
s
)
{\displaystyle K_{v}(s)}
I
v
(
s
)
{\displaystyle I_{v}(s)}
f
{\displaystyle f\,\,\,}
a
n
(
y
)
=
c
n
y
K
v
(
2
π
|
n
|
y
)
{\displaystyle a_{n}(y)=c_{n}{\sqrt {y}}K_{v}(2\pi |n|y)}
d
n
=
0
{\displaystyle d_{n}=0}
c
n
∈
C
.
◻
{\displaystyle c_{n}\in \mathbb {C} .\,\,\,\square }
Satz : L-Funktion einer Maaßschen Wellenform
Bearbeiten
Sei
f
(
x
+
i
y
)
=
∑
n
≠
0
c
n
y
K
ν
(
2
π
|
n
|
y
)
e
2
π
i
n
x
{\displaystyle f(x+iy)=\sum _{n\neq 0}c_{n}{\sqrt {y}}K_{\nu }(2\pi |n|y)e^{2\pi inx}}
f
{\displaystyle f}
L
(
s
,
f
)
=
∑
n
=
1
∞
c
n
n
−
s
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,L(s,f)=\sum _{n=1}^{\infty }c_{n}n^{-s}}
Dann konvergiert die Reihe
L
(
s
,
f
)
{\displaystyle L(s,f)}
R
e
(
s
)
>
3
2
{\displaystyle Re(s)>{\frac {3}{2}}}
C
{\displaystyle \mathbb {C} }
Ist f gerade oder ungerade so definiert man
Λ
(
s
,
f
)
:=
π
−
s
Γ
(
s
+
ϵ
+
ν
2
)
Γ
(
s
+
ϵ
−
ν
2
)
L
(
s
,
f
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Lambda (s,f):=\pi ^{-s}\Gamma ({\frac {s+\epsilon +\nu }{2}})\Gamma ({\frac {s+\epsilon -\nu }{2}})L(s,f)}
wobei
ϵ
=
0
{\displaystyle \epsilon =0}
f
{\displaystyle f}
ϵ
=
−
1
{\displaystyle \epsilon =-1}
f
{\displaystyle f}
Λ
{\displaystyle \Lambda }
Λ
(
s
,
f
)
=
(
−
1
)
ϵ
Λ
(
1
−
s
,
f
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Lambda (s,f)=(-1)^{\epsilon }\Lambda (1-s,f)}
Beweis:
Sei f eine Maaß-Spitzenform.
Zuerst machen wir uns klar wie schnell die Fourierkoeffizienten von f wachsen.
Behauptung: Es gilt
c
n
=
O
(
n
1
2
)
{\displaystyle c_{n}={\mathcal {O}}(n^{\frac {1}{2}})}
Beweis : Da f eine Maas-Spitzenform ist, existieren
C
,
N
>
0
{\displaystyle C,N>0}
y
>
1
{\displaystyle y>1}
|
f
(
x
+
i
y
)
|
≤
C
y
N
{\displaystyle |f(x+iy)|\leq Cy^{N}}
y
<
1
2
{\displaystyle y<{\frac {1}{2}}}
ω
∈
D
{\displaystyle \omega \in D}
z
=
x
+
i
y
{\displaystyle z=x+iy}
Γ
(
1
)
{\displaystyle \Gamma (1)}
I
m
(
ω
)
≤
1
y
{\displaystyle Im(\omega )\leq {\frac {1}{y}}}
Γ
(
1
)
{\displaystyle \Gamma (1)}
y
<
1
2
{\displaystyle y<{\frac {1}{2}}}
|
f
(
x
+
i
y
)
|
≤
C
y
−
N
{\displaystyle |f(x+iy)|\leq Cy^{-N}}
y
<
1
2
{\displaystyle y<{\frac {1}{2}}}
|
c
n
|
y
|
K
ν
(
2
π
|
n
|
y
)
≤
∫
0
1
|
f
(
x
+
i
y
)
|
d
x
≤
C
y
−
N
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|c_{n}|{\sqrt {y}}|K_{\nu }(2\pi |n|y)\leq \int _{0}^{1}|f(x+iy)|dx\leq Cy^{-N}}
Für
n
∈
Z
,
|
n
|
>
2
{\displaystyle n\in \mathbb {Z} \,,|n|>2}
y
=
1
|
n
|
{\displaystyle y={\frac {1}{|n|}}}
|
c
n
|
≤
C
r
N
+
1
2
|
K
ν
(
2
π
)
|
−
1
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|c_{n}|\leq Cr^{N+{\frac {1}{2}}}|K_{\nu }(2\pi )|^{-1}}
Damit finden wir eine Konstante
D
≥
C
{\displaystyle D\geq C}
n
≠
0
{\displaystyle n\neq 0}
|
c
n
|
≤
D
r
N
+
1
2
|
K
ν
(
2
π
)
|
−
1
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|c_{n}|\leq Dr^{N+{\frac {1}{2}}}|K_{\nu }(2\pi )|^{-1}}
Nun fällt die K-Besselfunktion aber exponentiell schnell und f ist eine Maas-Spitzenform. Zusammen folgt, dass f auf
D
{\displaystyle D}
H
{\displaystyle \mathbb {H} }
N
=
0
{\displaystyle N=0}
|
c
n
|
≤
K
n
1
2
{\displaystyle |c_{n}|\leq Kn^{\frac {1}{2}}}
K
<
∞
{\displaystyle K<\infty }
c
n
=
O
(
n
1
2
)
.
◻
{\displaystyle c_{n}={\mathcal {O}}(n^{\frac {1}{2}})\,.\,\,\square }
K
ν
{\displaystyle K_{\nu }}
Behauptung: Für
R
e
(
s
)
>
|
R
e
(
ν
)
|
{\displaystyle Re(s)>|Re(\nu )|}
∫
0
∞
K
ν
(
y
)
y
s
d
y
y
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int _{0}^{\infty }K_{\nu }(y)y^{s}{\frac {dy}{y}}}
absolut und es gilt
∫
0
∞
K
ν
(
y
)
y
s
d
y
y
=
2
s
−
2
Γ
(
s
+
ν
2
)
Γ
(
s
−
ν
2
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int _{0}^{\infty }K_{\nu }(y)y^{s}{\frac {dy}{y}}=2^{s-2}\Gamma ({\frac {s+\nu }{2}})\Gamma ({\frac {s-\nu }{2}})}
Beweis: Nach Definition gilt
∫
0
∞
K
ν
(
y
)
y
s
d
y
y
=
1
2
∫
0
∞
∫
0
∞
e
−
y
(
t
+
t
−
1
)
/
2
t
s
d
t
t
d
y
y
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int _{0}^{\infty }K_{\nu }(y)y^{s}{\frac {dy}{y}}={\frac {1}{2}}\int _{0}^{\infty }\int _{0}^{\infty }e^{-y(t+t^{-1})/2}t^{s}{\frac {dt}{t}}{\frac {dy}{y}}}
Wir wenden nun die Transformationsformel auf den Diffeomorphismus
Φ
:
(
0
,
∞
)
×
(
0
,
∞
)
→
(
0
,
∞
)
×
(
0
,
∞
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Phi :(0,\infty )\times (0,\infty )\to (0,\infty )\times (0,\infty )}
Φ
(
t
,
y
)
(
1
2
t
−
1
y
,
1
2
t
y
)
=
(
u
,
v
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Phi (t,y)({\frac {1}{2}}t^{-1}y,{\frac {1}{2}}ty)=(u,v)}
an. Wir erhalten
y
=
2
u
v
{\displaystyle y=2{\sqrt {uv}}}
t
=
v
u
{\displaystyle t={\sqrt {\frac {v}{u}}}}
D
Φ
(
t
,
y
)
=
1
2
(
−
y
t
2
1
t
y
t
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,D\Phi (t,y)={\frac {1}{2}}{\begin{pmatrix}-{\frac {y}{t^{2}}}&{\frac {1}{t}}\\y&t\\\end{pmatrix}}}
mit Determinante
d
e
t
(
D
Φ
)
(
t
,
y
)
=
−
y
2
t
{\displaystyle det(D\Phi )(t,y)=-{\frac {y}{2t}}}
1
2
2
s
−
1
∫
0
∞
∫
0
∞
e
−
u
−
v
v
(
s
+
ν
)
/
2
u
(
s
−
v
)
/
2
d
v
v
d
u
u
=
2
s
−
2
Γ
(
s
+
ν
2
)
Γ
(
s
−
ν
2
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\frac {1}{2}}2^{s-1}\int _{0}^{\infty }\int _{0}^{\infty }e^{-u-v}v^{(s+\nu )/2}u^{(s-v)/2}{\frac {dv}{v}}{\frac {du}{u}}=2^{s-2}\Gamma ({\frac {s+\nu }{2}})\Gamma ({\frac {s-\nu }{2}})}
und dieses konvergiert absolut für
R
e
(
s
)
>
|
R
e
(
ν
)
|
◻
{\displaystyle Re(s)>|Re(\nu )|\,\,\,\square }
Nun zum Beweis des Satzes. Ist f gerade oder ungerade folgt aus der Eindeutigkeit der Fourierkoeffizienten
a
n
=
(
−
1
)
ϵ
{\displaystyle a_{n}=(-1)^{\epsilon }}
n
∈
N
{\displaystyle n\in \mathbb {N} }
∫
0
∞
f
(
i
y
)
y
s
−
1
2
d
y
y
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int _{0}^{\infty }f(iy)y^{s-{\frac {1}{2}}}{\frac {dy}{y}}}
=
2
∫
0
∞
∑
n
=
1
∞
c
n
y
s
−
1
K
ν
(
2
π
n
y
)
d
y
{\displaystyle =2\int _{0}^{\infty }\sum _{n=1}^{\infty }c_{n}y^{s-1}K_{\nu }(2\pi ny)\,dy}
=
2
∑
n
=
1
∞
c
n
∫
0
∞
y
s
−
1
K
ν
(
2
π
n
y
)
d
y
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\sum _{n=1}^{\infty }c_{n}\int _{0}^{\infty }y^{s-1}K_{\nu }(2\pi ny)dy}
=
2
∑
n
=
1
∞
c
n
(
2
π
n
)
−
s
∫
0
∞
y
s
−
1
K
ν
(
y
)
d
y
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\sum _{n=1}^{\infty }c_{n}(2\pi n)^{-s}\int _{0}^{\infty }y^{s-1}K_{\nu }(y)dy}
=
2
(
2
π
)
−
s
(
∑
n
=
1
∞
c
n
n
−
s
)
2
s
−
2
Γ
(
s
+
ν
2
)
Γ
(
s
−
ν
2
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2(2\pi )^{-s}\left(\sum _{n=1}^{\infty }c_{n}n^{-s}\right)2^{s-2}\Gamma \left({\frac {s+\nu }{2}}\right)\Gamma \left({\frac {s-\nu }{2}}\right)}
=
1
2
π
s
Γ
(
s
+
ν
2
)
Γ
(
s
−
ν
2
)
L
(
s
,
f
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,={\frac {1}{2}}\pi ^{s}\Gamma \left({\frac {s+\nu }{2}}\right)\Gamma \left({\frac {s-\nu }{2}}\right)L(s,f)}
=
1
2
Λ
(
s
,
f
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,={\frac {1}{2}}\Lambda (s,f)}
Das vertauschen der Reihenfolge von Integral und Summe zeigt man zum Beispiel mit majorisierter Konvergenz, wobei man ausnutzt, dass für die K-Besselfunktion für
y
>
4
{\displaystyle y>4}
|
K
s
|
≤
e
−
y
2
K
R
e
(
s
)
(
2
)
{\displaystyle |K_{s}|\leq e^{-{\frac {y}{2}}}K_{Re(s)}(2)}
f
(
i
y
)
{\displaystyle f(iy)}
y
→
∞
{\displaystyle y\to \infty }
Wir definieren nun
Λ
1
(
s
,
f
)
:=
∫
1
∞
f
(
i
y
)
y
s
−
1
2
d
y
y
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Lambda _{1}(s,f):=\int _{1}^{\infty }f(iy)y^{s-{\frac {1}{2}}}{\frac {dy}{y}}}
Λ
2
(
s
,
f
)
:=
∫
0
1
f
(
i
y
)
y
s
−
1
2
d
y
y
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Lambda _{2}(s,f):=\int _{0}^{1}f(iy)y^{s-{\frac {1}{2}}}{\frac {dy}{y}}}
Damit gilt
Λ
=
2
(
Λ
1
+
Λ
2
)
{\displaystyle \Lambda =2(\Lambda _{1}+\Lambda _{2})}
f
(
i
y
)
{\displaystyle f(iy)}
y
→
∞
{\displaystyle y\to \infty }
Λ
1
(
s
,
f
)
{\displaystyle \Lambda _{1}(s,f)}
s
∈
C
{\displaystyle s\in \mathbb {C} }
Λ
1
{\displaystyle \Lambda _{1}}
f
{\displaystyle f}
Γ
(
1
)
{\displaystyle \Gamma (1)}
f
(
i
y
)
=
f
(
(
0
1
−
1
0
)
.
i
y
)
=
f
(
1
−
i
y
)
=
f
(
i
1
y
)
{\displaystyle f(iy)=f\left({\begin{pmatrix}0&1\\-1&0\\\end{pmatrix}}.iy\right)=f\left({\frac {1}{-iy}}\right)=f\left(i{\frac {1}{y}}\right)}
Wir erhalten nun
Λ
2
(
s
,
f
)
:=
∫
0
1
f
(
i
y
)
y
s
−
1
2
d
y
y
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Lambda _{2}(s,f):=\int _{0}^{1}f(iy)y^{s-{\frac {1}{2}}}{\frac {dy}{y}}}
=
∫
1
∞
f
(
i
1
y
)
y
−
s
+
1
2
d
y
y
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\int _{1}^{\infty }f(i{\frac {1}{y}})y^{-s+{\frac {1}{2}}}{\frac {dy}{y}}}
=
∫
1
∞
f
(
i
y
)
y
(
1
−
s
)
−
1
2
d
y
y
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\int _{1}^{\infty }f(iy)y^{(1-s)-{\frac {1}{2}}}{\frac {dy}{y}}}
=
Λ
1
(
1
−
s
,
f
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\Lambda _{1}(1-s,f)}
Damit ist auch
Λ
2
{\displaystyle \Lambda _{2}}
Λ
{\displaystyle \Lambda }
L
(
s
,
f
)
{\displaystyle L\left(s,f\right)}
C
{\displaystyle \mathbb {C} }
Λ
{\displaystyle \Lambda }
Λ
(
s
,
f
)
=
2
(
Λ
1
(
s
,
f
)
+
Λ
2
(
s
,
f
)
=
2
(
Λ
1
(
s
,
f
)
+
Λ
1
(
1
−
s
,
f
)
)
=
Λ
(
1
−
s
,
f
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Lambda (s,f)=2(\Lambda _{1}(s,f)+\Lambda _{2}(s,f)=2(\Lambda _{1}(s,f)+\Lambda _{1}(1-s,f))=\Lambda (1-s,f)}
Wenn f ungerade ist, definiert man
g
(
z
)
=
1
4
π
i
∂
f
∂
x
(
z
)
=
∑
1
∞
c
n
n
y
K
ν
(
2
π
n
y
)
c
o
s
(
2
π
n
x
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,g(z)={\frac {1}{4\pi i}}{\frac {\partial f}{\partial x}}(z)=\sum _{1}^{\infty }c_{n}n{\sqrt {y}}K_{\nu }(2\pi ny)cos(2\pi nx)}
Dann rechnet man analog zu oben
∫
0
∞
g
(
i
y
)
y
s
+
1
2
d
y
y
=
Λ
(
s
,
f
)
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int _{0}^{\infty }g(iy)y^{s+{\frac {1}{2}}}{\frac {dy}{y}}=\Lambda (s,f)}
indem man wieder benutzt, dass die K-Besselfunktion exponential fällt. Wir definieren wieder
Λ
1
(
s
,
f
)
:=
∫
1
∞
g
(
i
y
)
y
s
+
1
2
d
y
y
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Lambda _{1}(s,f):=\int _{1}^{\infty }g(iy)y^{s+{\frac {1}{2}}}{\frac {dy}{y}}}
Λ
2
(
s
,
f
)
:=
∫
0
1
g
(
i
y
)
y
s
+
1
2
d
y
y
{\displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Lambda _{2}(s,f):=\int _{0}^{1}g(iy)y^{s+{\frac {1}{2}}}{\frac {dy}{y}}}
Auch
g
(
i
y
)
{\displaystyle g(iy)}
y
→
∞
{\displaystyle y\to \infty }
Λ
1
{\displaystyle \Lambda _{1}}
g
(
i
y
)
=
−
1
y
2
g
(
i
y
)
{\displaystyle g(iy)=-{\frac {1}{y^{2}}}g\left({\frac {i}{y}}\right)}
Λ
2
(
s
,
f
)
=
−
Λ
1
(
1
−
s
,
f
)
{\displaystyle \Lambda _{2}(s,f)=-\Lambda _{1}(1-s,f)}
Λ
{\displaystyle \Lambda }
◻
{\displaystyle \,\,\,\square }
