It is easy to verify the following lemma.
Proof.
Let
F
be a Douglas metric. It is easy to verify that
Q
/
s
is not constant; then by Lemma 12,
β
is closed. Thus
s
0
=
0
and
s
0
i
=
0
. By Lemma 11 and (37), we have
(39)
T
i

1
n
+
1
∂
T
m
∂
y
m
y
i
=
γ
j
k
i
(
x
)
y
j
y
k
,
where
γ
j
k
i
(
x
)
are some set of local scalar functions and
(40)
T
i
=
H
r
00
b
i
,
∂
T
m
∂
y
m
=
H
′
α

1
(
b
2

s
2
)
r
00
+
2
H
r
0
,
H
′
=

q
(
q

1
)
[
2
s
(
1

q
2
)
+
(
2

q
)
]
2
[
s
2
(
1

q
2
)
+
s
(
2

q
)
+
1
+
b
2
q
(
q

1
)
]
2
.
We rewrite (39) as a polynomial in
y
i
and
α
, which is linear in
α
. This gives
(41)
q
(
q

1
)
α
2
2
[
β
2
(
1

q
2
)
+
α
β
(
2

q
)
+
α
2
(
1
+
b
2
q
(
q

1
)
)
]
r
o
o
b
i
+
1
n
+
1
×
q
(
q

1
)
[
2
β
(
1

q
2
)
+
α
(
2

q
)
]
2
[
β
2
(
1

q
2
)
+
α
β
(
2

q
)
+
α
2
(
1
+
b
2
q
(
q

1
)
)
]
2
×
(
b
2
α
2

β
2
)
r
00
y
i

1
n
+
1
×
q
(
q

1
)
α
2
[
β
2
(
1

q
2
)
+
α
β
(
2

q
)
+
α
2
(
1
+
b
2
q
(
q

1
)
)
]
r
0
y
i
=
γ
j
k
i
(
x
)
y
j
y
k
.
Namely,
(42)
[
n
+
1
]
q
(
q

1
)
α
2
×
[
β
2
(
1

q
2
)
+
α
β
(
2

q
)
+
α
2
(
1
+
b
2
q
(
q

1
)
)
]
r
00
b
i
+
q
(
q

1
)
[
2
β
(
1

q
2
)
+
α
(
2

q
)
]
×
(
b
2
α
2

β
2
)
r
00
y
i

2
q
(
q

1
)
α
2
×
[
β
2
(
1

q
2
)
+
α
β
(
2

q
)
+
α
2
(
1
+
b
2
q
(
q

1
)
)
]
r
0
y
i
=
2
γ
j
k
i
(
x
)
y
j
y
k
[
n
+
1
]
[
β
2
(
1

q
2
)
+
α
β
(
2

q
)
=
2
γ
j
k
i
(
x
)
y
j
y
k
[
n
+
1
]
w
w
+
α
2
(
1
+
b
2
q
(
q

1
)
)
]
2
.
Contracting (42) with
b
i
yields
(43)
[
n
+
1
]
q
(
q

1
)
α
2
×
[
β
2
(
1

q
2
)
+
α
β
(
2

q
)
+
α
2
(
1
+
b
2
q
(
q

1
)
)
]
r
00
b
2
+
q
(
q

1
)
[
2
β
(
1

q
2
)
+
α
(
2

q
)
]
×
(
b
2
α
2

β
2
)
r
00
β

2
q
(
q

1
)
α
2
×
[
β
2
(
1

q
2
)
+
α
β
(
2

q
)
+
α
2
(
1
+
b
2
q
(
q

1
)
)
]
r
0
β

T
[
n
+
1
]
[
β
2
(
1

q
2
)
+
α
β
(
2

q
)

T
[
n
+
1
]
w
w
+
α
2
(
1
+
b
2
q
(
q

1
)
)
]
2
=
0
,
where
T
:
=
2
γ
j
k
i
(
x
)
y
j
y
k
b
i
.
In the following proof we consider the following four cases.
Case 1. Suppose
r
00
and
T
are divisible by
α
2
; then there are two scalar functions
λ
=
λ
(
x
)
and
τ
=
τ
(
x
)
such that
(44)
T
=
λ
α
2
,
r
00
=
τ
α
2
,
r
0
=
τ
β
.
Plugging (44) into (43), we get
(45)
A
1
α
6
+
A
2
α
5
β
+
A
3
α
4
β
2
+
A
4
α
3
β
3
+
A
5
α
2
β
4
=
0
,
where
(46)
A
1
=
[
n
+
1
]
q
(
q

1
)
[
1
+
b
2
q
(
q

1
)
]
b
2
τ

[
n
+
1
]
[
1
+
b
2
q
(
q

1
)
]
2
λ
,
A
2
=
[
n
+
2
]
q
(
q

1
)
(
2

q
)
b
2
τ

2
[
n
+
1
]
(
2

q
)
[
1
+
b
2
q
(
q

1
)
]
λ
,
A
3
=
[
n
+
2
]
q
(
q

1
)
(
1

q
2
)
b
2
τ

2
q
(
q

1
)
[
1
+
b
2
q
(
q

1
)
]
τ

[
n
+
1
]
(
2

q
)
2
λ

2
[
n
+
1
]
×
(
1

q
2
)
[
1
+
b
2
q
(
q

1
)
]
λ
,
A
4
=

3
q
(
q

1
)
(
2

q
)
τ

2
[
n
+
1
]
(
1

q
2
)
(
2

q
)
λ
,
A
5
=

4
q
(
q

1
)
(
1

q
2
)
τ

[
n
+
1
]
(
1

q
2
)
2
λ
.
Because
α
even
is polynomial in
y
i
, from (45), we obtain
(47)
A
1
α
6
+
A
3
α
4
β
2
+
A
5
α
2
β
4
=
0
,
A
2
α
5
β
+
A
4
α
3
β
3
=
0
.
By (47), we get
(48)
α
2
[
A
1
α
2
+
A
3
β
2
]
=

A
5
β
4
,
A
2
α
2
=

A
4
β
2
.
Because
β
4
is not divisible by
α
2
and
[
A
1
α
2
+
A
3
β
2
]
, we get
A
5
=
0
,
A
1
α
2
+
A
3
β
2
=
0
.
β
2
is not divisible by
α
2
; we get
(49)
A
1
=
A
2
=
A
3
=
A
4
=
0
.
By
A
4
=
A
5
=
0
, we get
τ
=
λ
=
0
. Thus
r
00
=
r
0
=
T
=
0
; that is,
β
is parallel with respect to
α
.
Case 2. Suppose that
r
00
is not divisible by
α
2
, but
T
is divisible by
α
2
; namely, there is a scalar function
λ
=
λ
(
x
)
such that
(50)
T
=
λ
α
2
.
Plugging (50) into (43) yields
(51)
B
1
α
6
+
B
2
β
α
5
+
(
B
3
r
00
+
B
4
β
r
0
)
α
4
+
(
B
5
β
r
00
+
B
6
β
2
r
0
+
B
7
β
3
)
α
3
+
(
B
8
β
2
r
00
+
B
9
β
3
r
0
+
B
10
β
4
)
α
2
+
B
11
β
3
r
00
α
+
B
12
β
4
r
00
=
0
,
where
(52)
B
1
=

[
n
+
1
]
[
1
+
b
2
q
(
q

1
)
]
2
λ
,
B
2
=

2
[
n
+
1
]
(
2

q
)
[
1
+
b
2
q
(
q

1
)
]
λ
,
B
3
=
[
n
+
1
]
q
(
q

1
)
[
1
+
b
2
q
(
q

1
)
]
b
2
,
B
4
=

2
q
(
q

1
)
[
1
+
b
2
q
(
q

1
)
]
,
B
5
=
[
n
+
2
]
q
(
q

1
)
(
2

q
)
b
2
,
B
6
=

2
q
(
q

1
)
(
2

q
)
,
B
7
=

2
[
n
+
1
]
(
2

q
)
(
1

q
2
)
λ
,
B
8
=
[
n
+
3
]
q
(
q

1
)
(
1

q
2
)
b
2
,
B
9
=

2
q
(
q

1
)
(
1

q
2
)
,
B
10
=

[
n
+
1
]
(
1

q
2
)
2
λ
,
B
11
=

q
(
q

1
)
(
2

q
)
,
B
12
=

q
(
q

1
)
(
1

q
2
)
.
Because
α
even
is polynomial in
y
i
, from (51), we obtain
(53)
B
1
α
6
+
(
B
3
r
00
+
B
4
β
r
0
)
α
4
wwii
+
(
B
8
β
2
r
00
+
B
9
β
3
r
0
+
B
10
β
4
)
α
2
wwii
+
B
12
β
4
r
00
=
0
,
(54)
B
2
α
5
β
+
(
B
5
β
r
00
+
B
6
β
2
r
0
+
B
7
β
3
)
α
3
+
B
11
β
3
r
00
α
=
0
.
From (54), we have
(55)
α
2
[
B
2
α
2
+
(
B
5
r
00
+
B
6
β
r
0
+
B
7
β
2
)
]
=

B
11
β
2
r
00
.
Because
β
2
is not divisible by
α
2
,
r
00
must be divisible by
α
2
. This contradicts our assumption. Hence, this case is impossible.
Case 3. Suppose that
T
is not divisible by
α
2
, but
r
00
is divisible by
α
2
; namely, there is a scalar function
τ
=
τ
(
x
)
such that
(56)
r
00
=
τ
α
2
,
r
0
=
τ
β
.
Plugging (56) into (43) yields
(57)
C
1
α
6
+
C
2
β
α
5
+
(
C
3
β
2
+
C
4
T
)
α
4
wwii
+
(
C
5
β
3
+
C
6
β
T
)
α
3
+
(
C
7
β
4
+
C
8
β
2
T
)
α
2
wwii
+
C
9
β
3
T
α
+
C
10
β
4
T
=
0
,
where
(58)
C
1
=
[
n
+
1
]
q
(
q

1
)
[
1
+
b
2
q
(
q

1
)
]
b
2
τ
,
C
2
=
[
n
+
2
]
q
(
q

1
)
(
2

q
)
b
2
τ
,
C
3
=
[
n
+
3
]
q
(
q

1
)
(
1

q
2
)
b
2
τ

2
q
(
q

1
)
[
1
+
b
2
q
(
q

1
)
]
τ
,
C
4
=
[
n
+
1
]
[
1
+
b
2
q
(
q

1
)
]
2
,
C
5
=

3
q
(
q

1
)
(
2

q
)
τ
,
C
6
=

2
[
n
+
1
]
(
2

q
)
[
1
+
b
2
q
(
q

1
)
]
,
C
7
=

4
q
(
q

1
)
(
1

q
2
)
τ
,
C
8
=

2
[
n
+
1
]
(
1

q
2
)
[
1
+
b
2
q
(
q

1
)
]
,
C
9
=

2
[
n
+
1
]
(
2

q
)
(
1

q
2
)
,
C
10
=
[
n
+
1
]
(
1

q
2
)
2
.
Because
α
even
is polynomial in
y
i
, from (57), we obtain
(59)
C
1
α
6
+
(
C
3
β
2
+
C
4
T
)
α
4
+
(
C
7
β
4
+
C
8
β
2
T
)
α
2
wwii
+
C
10
β
4
T
=
0
,
(60)
C
2
β
α
5
+
(
C
5
β
3
+
C
6
β
T
)
α
3
+
C
9
β
3
T
α
=
0
.
From (60), we have
(61)
α
2
[
C
2
α
2
+
(
C
5
β
2
+
C
6
T
)
]
=

C
9
β
2
T
.
Because
β
2
is not divisible by
α
2
, then
T
must be divisible by
α
2
. This contradicts our assumption. Hence, this case is impossible, too.
Case 4. Suppose that
T
and
r
00
are not divisible by
α
2
. From (43), we have
(62)
(
D
1
r
00
+
D
2
r
0
β
+
D
3
T
)
α
4
+
(
D
4
r
00
β
+
D
5
r
0
β
2
+
D
6
β
T
)
α
3
+
(
D
7
r
00
β
2
+
D
8
r
0
β
3
+
D
9
β
2
T
)
α
2
+
(
D
10
r
00
β
3
+
D
11
β
3
T
)
α
+
(
D
12
r
00
β
4
+
D
13
β
4
T
)
=
0
,
where
(63)
D
1
=
[
n
+
1
]
q
(
q

1
)
[
1
+
b
2
q
(
q

1
)
]
b
2
,
D
2
=

2
q
(
q

1
)
[
1
+
b
2
q
(
q

1
)
]
,
D
3
=

[
n
+
1
]
[
1
+
b
2
q
(
q

1
)
]
2
,
D
4
=
[
n
+
2
]
q
(
q

1
)
(
2

q
)
b
2
,
D
5
=

2
q
(
q

1
)
(
2

q
)
,
D
6
=

2
[
n
+
1
]
[
1
+
b
2
q
(
q

1
)
]
,
D
7
=
[
n
+
3
]
q
(
q

1
)
(
1

q
2
)
,
D
8
=

2
q
(
q

1
)
(
1

q
2
)
,
D
9
=

[
n
+
1
]
[
2
(
1

q
2
)
[
1
+
b
2
q
(
q

1
)
]
+
(
2

q
)
2
]
,
D
10
=

q
(
q

1
)
(
2

q
)
,
D
11
=

2
[
n
+
1
]
(
2

q
)
(
1

q
2
)
,
D
12
=

2
q
(
q

1
)
(
1

q
2
)
,
D
13
=

[
n
+
1
]
(
1

q
2
)
2
.
Because
α
even
is polynomial in
y
i
, from (62), we get
(64)
(
D
1
r
00
+
D
2
r
0
β
+
D
3
T
)
α
4
+
(
D
7
r
00
β
2
+
D
8
r
0
β
3
+
D
9
β
2
T
)
α
2
+
(
D
12
r
00
β
4
+
D
13
β
4
T
)
=
0
,
(
D
4
r
00
β
+
D
5
r
0
β
2
+
D
6
β
T
)
α
3
+
(
D
10
r
00
β
3
+
D
11
β
3
T
)
α
=
0
.
We get
(65)
α
2
[
(
D
1
r
00
+
D
2
r
0
β
+
D
3
T
)
α
2
α
2
[
+
(
D
7
r
00
β
2
+
D
8
r
0
β
3
+
D
9
β
2
T
)
]
=

β
4
[
D
12
r
00
+
D
13
T
]
,
α
2
[
D
4
r
00
+
D
5
r
0
β
+
D
6
T
]
=

β
2
[
D
10
r
00
+
D
11
T
]
.
Because
β
4
and
β
2
are not divisible by
α
2
,
[
D
12
r
00
+
D
13
T
]
and
[
D
10
r
00
+
D
11
T
]
are divisible by
α
2
; hence there are two scalar functions
K
1
=
K
1
(
x
)
and
K
2
=
K
2
(
x
)
such that
(66)
D
12
r
00
+
D
13
T
=
K
1
α
2
,
D
10
r
00
+
D
11
T
=
K
2
α
2
.
Let
(67)
A
:
=
[
D
12
D
13
D
10
D
11
]
.
We have
det
A
=
3
[
n
+
1
]
q
(
q

1
)
(
2

q
)
(
1

q
2
)
2
≠
0
; then, we get
(68)
A

1
≔
[

2
3
q
(
q

1
)
(
1

q
2
)
1
3
q
(
q

1
)
(
2

q
)
1
3
[
n
+
1
]
(
1

q
2
)
2

2
3
[
n
+
1
]
(
2

q
)
(
1

q
2
)
]
.
From (66) and (68), we obtain
(69)
r
00
=
[

2
3
q
(
q

1
)
(
1

q
2
)
K
1
+
1
3
[
n
+
1
]
(
1

q
2
)
2
K
2
]
α
2
,
T
=
[
1
3
q
(
q

1
)
(
2

q
)
K
1
T
=
[
+

2
3
[
n
+
1
]
(
2

q
)
(
1

q
2
)
K
2
]
α
2
.
Thus
r
00
and
T
are divisible by
α
2
. This contradicts our assumption; hence, this case is impossible too. So only Case 1 is possible and in this case we proved that
r
00
=
0
; that is,
β
is parallel with respect to
α
.
Conversely, if
β
is parallel with respect to
α
, by Lemma 5, we get
G
i
=
G
α
i
. Because
α
is a Riemannian metric then
D
j
k
l
i
=
0
.