Proof of Proposition <xref reftype="statement" rid="prop2.1">6</xref>.
Let
p
∈
M
, and we choose a local orthonormal frame field of the metric
G
around
p
. Then,
(65)
Φ
=
∑
(
ρ
,
j
)
2
ρ
2
,
Φ
,
i
=
2
∑
ρ
,
j
ρ
,
j
i
ρ
2

2
ρ
,
i
∑
(
ρ
,
j
)
2
ρ
3
,
Δ
Φ
=
2
∑
(
ρ
,
j
i
)
2
ρ
2
+
2
∑
ρ
,
j
ρ
,
j
i
i
ρ
2

8
∑
ρ
,
j
ρ
,
i
ρ
,
j
i
ρ
3
+
(
n
+
6
)
Φ
2

2
τ
Φ
,
where we used (20). In the case
Φ
(
p
)
=
0
, it is easy to get, at
p
,
(66)
Δ
Φ
≥
2
∑
(
ρ
,
i
j
)
2
ρ
2
.
Now, we assume that
Φ
(
p
)
≠
0
. Choose a local orthonormal frame field of the metric
G
around
p
such that
ρ
,
1
(
p
)
=
∥
∇
ρ
∥
(
p
)
>
0
,
ρ
,
i
(
p
)
=
0
, for all
i
>
1
. Then,
(67)
Δ
Φ
=
2
(
1

δ
+
δ
)
∑
(
ρ
,
i
j
)
2
ρ
2
+
2
∑
ρ
,
j
ρ
,
j
i
i
ρ
2

8
(
ρ
,
1
)
2
ρ
,
11
ρ
3
+
(
n
+
6
)
Φ
2

2
τ
Φ
,
where
1
>
δ
>
0
is a constant to be determined later. Applying (20), we obtain
(68)
2
∑
(
ρ
,
i
j
)
2
ρ
2
≥
2
n
n

1
(
ρ
,
11
)
2
ρ
2
+
4
∑
i
>
1
(
ρ
,
1
i
)
2
ρ
2
+
2
n
n

1
(
ρ
,
1
)
2
ρ
,
11
ρ
3
+
n
2
2
(
n

1
)
Φ
2

4
n

1
ρ
,
11
ρ
τ
+
2
n

1
τ
2

2
n
n

1
Φ
τ
.
An application of the Ricci identity shows that
(69)
2
ρ
2
∑
ρ
,
j
ρ
,
j
i
i
=

2
n
(
ρ
,
1
)
2
ρ
,
11
ρ
3
+
n
(
ρ
,
1
)
4
ρ
4
+
2
R
11
(
ρ
,
1
)
2
ρ
2
+
2
ρ
,
1
ρ
2
(
ρ
τ
)
,
1
.
Substituting (68) and (69) into (67), we obtain
(70)
Δ
Φ
≥
2
δ
∑
(
ρ
,
i
j
)
2
ρ
2
+
(

2
n

8
+
2
n
(
1

δ
)
n

1
)
×
(
ρ
,
1
)
2
ρ
,
11
ρ
3
+
2
R
11
(
ρ
,
1
)
2
ρ
2
+
(
n
2
(
1

δ
)
2
(
n

1
)
+
2
(
n
+
3
)
)
(
ρ
,
1
)
4
ρ
4
+
2
ρ
,
1
ρ
2
(
ρ
τ
)
,
1

4
n

2

2
n
δ
n

1
Φ
τ
+
(
1

δ
)
×
(
2
n
n

1
(
ρ
,
11
)
2
ρ
2
+
4
∑
i
>
1
(
ρ
,
1
i
)
2
ρ
2
h
h
h
h
2
n
n

1
(
ρ
,
11
)
2
ρ
2
+
4
∑
i
>
1
(
ρ
,
1
i
)
2
ρ
2

4
n

1
ρ
,
11
ρ
τ
+
2
n

1
τ
2
)
.
Note that
(71)
(
ρ
,
11
)
2
ρ
2
=
1
4
∑
(
Φ
,
i
)
2
Φ

∑
i
>
1
(
ρ
,
1
i
)
2
ρ
2
+
2
(
ρ
,
1
)
2
ρ
,
11
ρ
3

(
ρ
,
1
)
4
ρ
4
.
Then, (70) and (71) together give us
(72)
Δ
Φ
≥
2
δ
∑
(
ρ
,
i
j
)
2
ρ
2
+
n
(
1

δ
)
2
(
n

1
)
∑
(
Φ
,
i
)
2
Φ
+
(
6
n
(
1

δ
)
n

1

2
(
n
+
4
)
)
(
ρ
,
1
)
2
ρ
,
11
ρ
3
+
2
R
11
(
ρ
,
1
)
2
ρ
2
+
[
(
n
2

4
n
)
(
1

δ
)
2
(
n

1
)
+
2
(
n
+
3
)
]
(
ρ
,
1
)
4
ρ
4
+
1

δ
n

1
(
2
τ
2

4
ρ
,
11
ρ
τ
)

4
n

2

2
n
δ
n

1
Φ
τ
+
2
ρ
,
1
ρ
2
(
ρ
τ
)
,
1
.
Using the Schwarz inequality gives
(73)
2
ρ
,
11
ρ
τ
≤
7
3
∑
(
ρ
,
i
j
)
2
ρ
2
+
3
7
τ
2
.
Using
(74)
(
ρ
,
1
)
2
ρ
,
11
ρ
3
=
1
2
Φ
,
i
ρ
,
i
ρ
+
Φ
2
,
and choosing
δ
=
7
/
(
3
n
+
4
)
, we get
(75)
Δ
Φ
≥
n
(
1

δ
)
2
(
n

1
)
∑
(
Φ
,
i
)
2
Φ
+
(
3
n
(
1

δ
)
n

1

(
n
+
4
)
)
×
∑
Φ
,
i
ρ
,
i
ρ
+
2
R
11
(
ρ
,
1
)
2
ρ
2
+
[
(
n
2
+
8
n
)
(
1

δ
)
2
(
n

1
)

2
]
Φ
2
+
8
(
1

δ
)
7
(
n

1
)
τ
2

4
n

2

2
n
δ
n

1
Φ
τ
+
2
ρ
,
1
ρ
2
(
ρ
τ
)
,
1
.
In the following, we will calculate the terms
R
11
(
(
ρ
,
1
)
2
/
ρ
2
)
and
(
ρ
,
1
/
ρ
2
)
(
ρ
τ
)
,
1
. Note that (17) is invariant under an affine transformation of coordinates that preserved the origin. So, we can choose the coordinates
x
1
,
x
2
,
…
,
x
n
such that
f
i
j
(
p
)
=
δ
i
j
and
∂
ρ
/
∂
x
1
=
∥
grad
ρ
∥
(
p
)
>
0
,
(
∂
ρ
/
∂
x
i
)
(
p
)
=
0
, for all
i
>
1
. From (19), we easily obtain
(76)
ρ
,
i
j
=
ρ
i
j
+
A
i
j
1
ρ
,
1
=
ρ
,
i
ρ
,
j
ρ

A
i
j
1
ρ
,
1
+
A
i
j
k
f
k
ρ
n
+
2
.
Thus, we get
(77)
Φ
,
i
=
2
ρ
,
1
ρ
,
1
i
ρ
2

2
ρ
,
i
(
ρ
,
1
)
2
ρ
3
=

2
A
i
11
(
ρ
,
1
)
2
ρ
2
+
2
ρ
,
1
f
k
A
k
i
1
(
n
+
2
)
ρ
,
(78)
∑
Φ
,
i
ρ
,
i
ρ
=

2
A
111
(
ρ
,
1
)
3
ρ
3
+
2
f
k
A
k
11
n
+
2
(
ρ
,
1
)
2
ρ
2
.
By the same method, as deriving (69), we have
(79)
∑
(
A
m
l
1
)
2
≥
n
n

1
∑
(
A
i
11
)
2

2
n

1
A
111
∑
A
i
i
1
+
1
n

1
(
∑
A
i
i
1
)
2
.
Note that
∑
A
i
i
1
=
(
(
n
+
2
)
/
2
)
(
ρ
1
/
ρ
)
. Therefore, by (14), (77), (78), and (79), we obtain
(80)
2
R
11
(
ρ
,
1
)
2
ρ
2
=
2
∑
(
A
k
j
1
)
2
(
ρ
,
1
)
2
ρ
2

(
n
+
2
)
A
111
(
ρ
,
1
)
3
ρ
3
≥
n
2
(
n

1
)
∑
(
Φ
,
i

2
(
ρ
,
1
f
k
A
k
i
1
/
(
n
+
2
)
ρ
)
)
2
Φ
+
(
n
+
2
)
(
n
+
1
)
2
(
n

1
)
∑
Φ
,
i
ρ
,
i
ρ

n
+
1
n

1
f
k
A
k
11
(
ρ
,
1
)
2
ρ
2
+
(
n
+
2
)
2
2
(
n

1
)
Φ
2
.
On the other hand, we have
(81)
2
ρ
,
1
ρ
2
(
ρ
τ
)
,
1
=
2
Φ
τ
+
1
n
+
2
∑
A
1
i
k
f
k
f
i
ρ
,
1
ρ
+
Φ
.
Then, inserting (80) and (81) into (75), we get
(82)
Δ
Φ
≥
2
n

n
δ
2
(
n

1
)
∑
(
Φ
,
i
)
2
Φ

(
n
+
2
)
(
n

5
)
+
6
n
δ
2
(
n

1
)
×
∑
Φ
,
i
ρ
,
i
ρ
+
Φ
+
2
(
n
+
2
)
2

(
n
2
+
8
n
)
δ
2
(
n

1
)
Φ
2
+
8
(
1

δ
)
7
(
n

1
)
τ
2

2
n
(
1

δ
)
n

1
Φ
τ
+
1
n
+
2
×
∑
A
1
i
k
f
k
f
i
ρ
,
1
ρ

n
+
1
n

1
(
ρ
,
1
)
2
ρ
2
f
k
A
k
11

2
n
(
n

1
)
(
n
+
2
)
∑
Φ
,
i
f
k
A
k
i
1
Φ
+
2
n
(
n

1
)
(
n
+
2
)
2
×
∑
(
f
k
A
k
i
1
)
2
.
Using (77), we have
(83)
1
n
+
2
∑
A
1
i
k
f
k
f
i
ρ
,
1
ρ

n
+
1
n

1
(
ρ
,
1
)
2
ρ
2
f
k
A
k
11
=
1
2
f
i
Φ
,
i

2
n

1
(
ρ
,
1
)
2
ρ
2
f
k
A
k
11
.
One observes that the Schwarz inequality gives
(84)
2
n
(
n

1
)
(
n
+
2
)
∑
Φ
,
i
f
k
A
k
i
1
Φ
≤
9
n
8
(
n

1
)
∑
(
Φ
,
i
)
2
Φ
+
8
n
9
(
n

1
)
(
n
+
2
)
2
×
∑
(
f
k
A
k
i
1
)
2
,
2
n

1
(
ρ
,
1
)
2
ρ
2
f
k
A
k
11
≤
9
(
n
+
2
)
2
10
n
(
n

1
)
Φ
2
+
10
n
9
(
n

1
)
(
n
+
2
)
2
×
∑
(
f
k
A
k
i
1
)
2
,
2
n
Φ
τ
≤
τ
2
+
n
2
Φ
2
.
Note that by (17) we have
(85)
1
4
f
i
j
Φ
j
f
i
=
n
+
2
2
f
i
j
Φ
j
(
ln
ρ
)
i
+
1
4
Φ
j
x
j
=
n
+
2
2
f
i
j
Φ
j
(
ln
ρ
)
i
+
1
4
f
i
j
∂
Φ
∂
x
i
∂
u
∂
x
j
.
Then, inserting these estimates into (82) yields Proposition 6.
Proof of Corollary <xref reftype="statement" rid="coro2.3">8</xref>.
Now, we will calculate the term
(
ln
ρ
)
,
i
j
k
. In particular, if
f
satisfies PDE (4), choose the coordinate
(
x
1
,
x
2
,
…
,
x
n
)
such that
f
i
j
(
p
)
=
δ
i
j
; then we have
(86)
(
ln
ρ
)
,
i
j
k
=
1
n
+
2
(
A
i
j
k
+
A
i
j
k
,
p
f
,
p
)

(
ln
ρ
)
,
l
A
i
j
k
,
l
+
A
i
j
l
A
k
l
p
(
3
(
ln
ρ
)
,
p

2
n
+
2
f
,
p
)
.
Using (17), we have
(87)
3
(
ln
ρ
)
,
p

2
n
+
2
f
,
p
=

1
n
+
2
(
f
+
u
)
,
p
+
(
ln
ρ
)
,
p
.
By the Young inequality and the Schwarz inequality, we have
(88)
n
+
2
n
(
n

1
)
∑
A
i
j
k
A
i
j
l
A
k
l
h
(
ln
ρ
)
,
h
≤
1
2
J
2
+
16
n
2
(
n

1
)
2
(
n
+
2
)
4
Φ
2
,
1
n
(
n

1
)
∑
A
i
j
k
A
i
j
l
,
k
f
l
=
1
2
∑
J
,
l
f
,
l
=
1
4
〈
∇
J
,
∇
(
f
+
u
)
〉
+
n
+
2
2
〈
∇
J
,
∇
ln
ρ
〉
,
n
+
2
n
(
n

1
)
∑
A
i
j
k
A
i
j
l
,
k
(
ln
ρ
)
,
l
=
n
+
2
2
∑
J
,
i
(
ln
ρ
)
,
i
≤
1
n
(
n

1
)
∑
(
A
i
j
k
,
l
)
2
+
(
n
+
2
)
2
4
J
Φ
≤
1
n
(
n

1
)
∑
(
A
i
j
k
,
l
)
2
+
1
4
J
2
+
(
n
+
2
)
4
16
Φ
2
,
1
n
(
n

1
)
∑
A
i
j
k
A
j
i
l
A
k
l
p
(
f
+
u
)
,
p
≤
n
(
n

1
)
∥
∇
(
f
+
u
)
∥
J
3
/
2
.
Thus, by inserting (88) into Lemma 7, we obtain Corollary 8.