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converges.
PROOF
Let
F
be a fundamental parallelogram for
L
and let
D
be the
length of the longer diagonal of
F
.Then
|z|≤D
for all
z ∈ F
.Let
ω
=
m
1
ω
1
+
m
2
ω
2
∈ L
with
|ω|≥
2
D
.If
x
1
,x
2
are real numbers with
m
i
≤ x
i
<
m
i
+1, then
ω
and
x
1
ω
1
+
x
2
ω
2
differ by an element of
F
,so
1
2
|
|
m
1
ω
1
+
m
2
ω
2
|≥|
x
1
ω
1
+
x
2
ω
2
|−
D
≥|
x
1
ω
1
+
x
2
ω
2
|−
m
1
ω
1
+
m
2
ω
2
|
,
since
|
m
1
ω
1
+
m
2
ω
2
|≥
2
D
. Therefore,
2
3
|x
1
ω
1
+
x
2
ω
2
|.
|m
1
ω
1
+
m
2
ω
2
|≥
Similarly,
|
x
1
ω
1
+
x
2
ω
2
|≥
D.
Comparing the sum to an integral yields
(3
/
2)
k
|x
1
ω
1
+
x
2
ω
2
|
1
|ω|
k
≤
(1
/
area of
F
)
k
dx
1
dx
2
.
|
ω
|≥
2
D
|
x
1
ω
1
+
x
2
ω
2
|≥
D
The change of variables defined by
u
+
iv
=
x
1
ω
1
+
x
2
ω
2
changes the integral
to
(
u
2
+
v
2
)
k/
2
du dv
=
C
2
π
∞
1
1
r
k
rdrdθ<∞,
C
θ
=0
r
=
D
|
u
+
iv
|≥
D
where
C
=(3
/
2)
k
/
(area of
F
). Therefore, the sum for
2
D
converges.
Since there are only finitely many
ω
with
|ω| <
2
D
,wehaveshownthatthe
sum converges.
|
ω
|≥
Lemma 9.4 and Equation 9.2 imply that the sum of the terms in Equa-
tion 9.1 with
|ω|≥
2
M
converges absolutely and uniformly for
z ∈ C
.Since
only finitely many terms have been omitted, we obtain (1). Since a uniform
limit of analytic functions is analytic,
℘
(
z
) is analytic for
z
∈ L
.If
z ∈ L
,
ω
)
2
then the sum of the terms for
ω
=
z
is analytic near
z
,sotheterm1
/
(
z
−
causes
℘
to have a double pole at
z
.Thisproves(2).
To prove (3), note that
ω
∈
L
if and only if
−
ω
∈
L
. Therefore, in the sum
for
℘
(
−
z
), we can take the sum over
−
ω
∈
L
. The terms of this sum are of
the form
1
(
−z
+
ω
)
2
−
1
(
−ω
)
2
1
(
z − ω
)
2
−
1
ω
2
.
=
Therefore the sum for
℘
(
−z
) equals the sum for
℘
(
z
).
The proof of (4) would be easy if we could ignore the terms 1
/ω
2
,since
changing
z
to
z
+
ω
would simply shift the summands. However, these terms
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