In this section, first we give the following two lemmas (see the detailed proof in [3]).
Proof.
We first give the proof of the second inequality in (16).
Define
(19)
G
(
x
)
=
1
2
∫
A
1
(
x
)
T

A
1
(
x
)
[
α
k
(
t
)

z
t
(
x
,
t
)

2
h
h
h
h
h
h
h
h
h
h
h
+
β
k
(
x
,
t
)

z
x
(
x
,
t
)

2
]
d
t
.
Note that
(20)
G
(
0
)
=
1
2
∫
A
1
(
x
)
T

A
1
(
x
)
β
k
(
0
,
t
)

z
x
(
0
,
t
)

2
d
t
.
The derivative of the functional of
G
is
(21)
G
′
(
x
)
=
∫
A
1
(
x
)
T

A
1
(
x
)
[
1
2
α
k
(
t
)
z
t
(
x
,
t
)
z
t
,
x
(
x
,
t
)
hhhhhhhhhh
+
β
k
(
x
,
t
)
z
x
(
x
,
t
)
z
x
x
(
x
,
t
)
hhhhhhhhhh
+
1
2
β
k
,
x
(
x
,
t
)

z
x
(
x
,
t
)

2
]
d
t
+
(

1
2
A
1
′
(
x
)
)
×
∑
t
=
T

A
1
(
x
)
,
A
1
(
x
)
[
α
k
(
t
)

z
t
(
x
,
t
)

2
+
β
k
(
x
,
t
)

z
x
(
x
,
t
)

2
]
,
where
(22)
∫
A
1
(
x
)
T

A
1
(
x
)
α
k
(
t
)
z
t
(
x
,
t
)
z
t
,
x
(
x
,
t
)
d
t
=
[
α
k
(
t
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)
]

A
1
(
x
)
T

A
1
(
x
)

∫
A
1
(
x
)
T

A
1
(
x
)
[
α
k
,
t
(
t
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)
hhhhhhhhhh
+
α
k
(
t
)
z
t
t
(
x
,
t
)
z
x
(
x
,
t
)
]
d
t
.
By (12), it follows that
(23)
α
k
(
t
)
z
t
t
(
x
,
t
)
=
[
β
k
(
x
,
t
)
z
x
]
x
(
x
,
t
)

γ
k
(
x
)
z
t
x
(
x
,
t
)
=
β
k
,
x
(
x
,
t
)
z
x
(
x
,
t
)
+
β
k
(
x
,
t
)
z
x
x
(
x
,
t
)

γ
k
(
x
)
z
t
x
(
x
,
t
)
,
from which and using integrating by parts, we have that
(24)
∫
A
1
(
x
)
T

A
1
(
x
)
α
k
(
t
)
z
t
(
x
,
t
)
z
t
,
x
(
x
,
t
)
d
t
=
[
α
k
(
t
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)
]

A
1
(
x
)
T

A
1
(
x
)

∫
A
1
(
x
)
T

A
1
(
x
)
[
α
k
,
t
(
t
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)
hhhhhhhhhhhh
+
β
k
,
x
(
x
,
t
)

z
x
(
x
,
t
)

2
hhhhhhhhhhhh
+
β
k
(
x
,
t
)
z
x
x
(
x
,
t
)
z
x
(
x
,
t
)
hhhhhhhhhhhh

γ
k
(
x
)
z
t
x
(
x
,
t
)
z
x
(
x
,
t
)
]
d
t
=
[
α
k
(
t
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)
]

A
1
(
x
)
T

A
1
(
x
)

∫
A
1
(
x
)
T

A
1
(
x
)
[
α
k
,
t
(
t
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)
hhhhhhhhhhhh
+
β
k
,
x
(
x
,
t
)

z
x
(
x
,
t
)

2
hhhhhhhhhhhh
+
β
k
(
x
,
t
)
z
x
x
(
x
,
t
)
z
x
(
x
,
t
)
]
d
t
+
[
1
2
γ
k
(
x
)

z
x
(
x
,
t
)

2
]

A
1
(
x
)
T

A
1
(
x
)
=
[
+
1
2
γ
k
(
x
)

z
x
(
x
,
t
)

2
α
k
(
t
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)
hhhhh
+
1
2
γ
k
(
x
)

z
x
(
x
,
t
)

2
]

A
1
(
x
)
T

A
1
(
x
)

∫
A
1
(
x
)
T

A
1
(
x
)
[
α
k
,
t
(
t
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)
hhhhhhhhhhhh
+
β
k
,
x
(
x
,
t
)

z
x
(
x
,
t
)

2
hhhhhhhhhhhh
+
β
k
(
x
,
t
)
z
x
x
(
x
,
t
)
z
x
(
x
,
t
)
]
d
t
.
We conclude, using (24), that
(25)
G
′
(
x
)
=

∫
A
1
(
x
)
T

A
1
(
x
)
[
1
2
α
k
,
t
(
t
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)
hhhhhhhhhhh
+
1
2
β
k
,
x
(
x
,
t
)

z
x
(
x
,
t
)

2
]
d
t
+
[
1
2
α
k
(
t
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)
hhhh
+
1
2
γ
k
(
x
)

z
x
(
x
,
t
)

2
]

A
1
(
x
)
T

A
1
(
x
)
+
(

1
2
A
1
′
(
x
)
)
×
∑
t
=
T

A
1
(
x
)
,
A
1
(
x
)
[
α
k
(
t
)

z
t
(
x
,
t
)

2
+
β
k
(
x
,
t
)

z
x
(
x
,
t
)

2
]
.
We will choose
A
1
(
x
)
later which satisfies
(26)
[
α
k
(
t
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)
+
1
2
γ
k
(
x
)

z
x
(
x
,
t
)

2
]

A
1
(
x
)
T

A
1
(
x
)
+
(

1
2
A
1
′
(
x
)
)
×
∑
t
=
T

A
1
(
x
)
,
A
1
(
x
)
[
α
k
(
t
)

z
t
(
x
,
t
)

2
+
β
k
(
x
,
t
)

z
x
(
x
,
t
)

2
]
≤
0
.
From (25) and (26), for
ε
>
0
, it concludes that
(27)
G
′
(
x
)
≤

∫
A
1
(
x
)
T

A
1
(
x
)
[
1
2
α
k
,
t
(
t
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)
hhhhhhhhhhh
+
1
2
β
k
,
x
(
x
,
t
)

z
x
(
x
,
t
)

2
]
d
t
=
∫
A
1
(
x
)
T

A
1
(
x
)
[
k
2
x
α
k
(
t
)

k
z
t
(
x
,
t
)
z
x
(
x
,
t
)
hhhhhhhhhh
+
k
2
x
α
k
(
t
)

z
x
(
x
,
t
)

2
]
d
t
≤

∫
A
1
(
x
)
T

A
1
(
x
)
α
k
(
t
)
z
t
(
x
,
t
)
k
α
k
(
t
)
z
x
(
x
,
t
)
d
t

+

∫
A
1
(
x
)
T

A
1
(
x
)
k
2
x
1

k
2
x
2
β
k
(
x
,
t
)

z
x
(
x
,
t
)

2
d
t

≤
1
2
ε
∫
A
1
(
x
)
T

A
1
(
x
)
α
k
(
t
)

z
t
(
x
,
t
)

2
d
t
+
ε
2
∫
A
1
(
x
)
T

A
1
(
x
)
k
2
1

k
2
x
2
β
k
(
x
,
t
)

z
x
(
x
,
t
)

2
d
t
+
∫
A
1
(
x
)
T

A
1
(
x
)
k
2
x
1

k
2
x
2
β
k
(
x
,
t
)

z
x
(
x
,
t
)

2
d
t
≤
1
2
1
ε
∫
A
1
(
x
)
T

A
1
(
x
)
α
k
(
t
)

z
t
(
x
,
t
)

2
d
t
+
1
2
k
2
(
ε
+
2
)
1

k
2
∫
A
1
(
x
)
T

A
1
(
x
)
β
k
(
x
,
t
)

z
x
(
x
,
t
)

2
d
t
.
Take
ε
=
(
1

k
)
/
k
; then it is easy to check that
(28)
1
ε
=
k
2
(
ε
+
2
)
1

k
2
.
Hence
(29)
G
′
(
x
)
≤
k
1

k
G
(
x
)
x
∈
(
0,1
)
.
By Gronwall inequality, there exists
C
>
0
such that
(30)
G
(
x
)
≤
C
G
(
0
)
.
Integrating (30) in
[
0,1
]
, we have
(31)
∫
0
1
G
(
x
)
d
x
≤
C
G
(
0
)
.
By (17), we deduce that
(32)
[
T

2
A
1
(
1
)
]
E
0
=
∫
A
1
(
1
)
T

A
1
(
1
)
E
0
d
t
≤
(
1
+
k
T
)
∫
A
1
(
1
)
T

A
1
(
1
)
E
(
t
)
d
t
=
(
1
+
k
T
)
1
2
∫
A
1
(
1
)
T

A
1
(
1
)
∫
0
1
[
α
k
(
t
)

z
t
(
x
,
t
)

2
d
t
hhhhhhhhhhhhhhhhhhhhh
+
β
k
(
x
,
t
)

z
x
(
x
,
t
)

2
]
d
t
d
x
.
Now choose that
A
1
(
x
)
also satisfies
(33)
A
1
′
(
x
)
>
0
,
A
1
(
1
)
≥
A
1
(
x
)
≥
A
1
(
0
)
≥
0
.
Then from (19), (31), and (32), it follows that
(34)
[
T

2
A
1
(
1
)
]
E
0
≤
(
1
+
k
T
)
1
2
∫
A
1
(
x
)
T

A
1
(
x
)
∫
0
1
[
α
k
(
t
)

z
t
(
x
,
t
)

2
d
t
hhhhhhhhhhhhhhhhhhhh
+
β
k
(
x
,
t
)

z
x
(
x
,
t
)

2
]
d
t
d
x
≤
(
1
+
k
T
)
∫
0
1
G
(
x
)
d
x
≤
C
(
1
+
k
T
)
G
(
0
)
,
from which and from (20), we have that
(35)
G
(
0
)
=
1
2
∫
A
1
(
x
)
T

A
1
(
x
)
β
k
(
0
,
t
)

z
x
(
0
,
t
)

2
d
t
≥
1
C
(
1
+
k
T
)
[
T

2
A
1
(
1
)
]
E
0
.
Let
(36)
T
k
*
≜
2
A
1
(
1
)
.
When
T
>
T
k
*
, by (35), (16) follows.
In the following, when
T
>
T
k
*
, we choose
A
1
(
x
)
which satisfies (33) and (26). In (26), for
0
<
ε
≤
1
, we have that
(37)

α
k
(
t
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)

=

α
k
(
t
)
z
t
(
x
,
t
)
α
k
(
t
)
z
x
(
x
,
t
)

=

α
k
(
t
)
A
1
′
(
x
)
z
t
(
x
,
t
)
α
k
(
t
)
A
1
′
(
x
)
z
x
(
x
,
t
)

≤
ε
2
α
k
(
t
)
A
1
′
(
x
)

z
t
(
x
,
t
)

2
+
1
2
ε
α
k
(
t
)
A
1
′
(
x
)

z
x
(
x
,
t
)

2
=
A
1
′
(
x
)
2
[
ε
α
k
(
t
)

z
t
(
x
,
t
)

2
]
+
A
1
′
(
x
)
2
[
1
ε
α
k
(
t
)
(
A
1
′
(
x
)
)
2

z
x
(
x
,
t
)

2
]
,

1
2
γ
k
(
x
)

z
x
(
x
,
t
)

2

≤
1
2
A
1
′
(
x
)

γ
k
(
x
)

A
1
′
(
x
)

z
x
(
x
,
t
)

2
.
Assume that
(38)
[
α
k
(
t
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)
+
1
2
γ
k
(
x
)

z
x
(
x
,
t
)

2
]

A
1
(
x
)
T

A
1
(
x
)
+
(

1
2
A
1
′
(
x
)
)
×
∑
t
=
T

A
1
(
x
)
,
A
1
(
x
)
[
α
k
(
t
)

z
t
(
x
,
t
)

2
+
β
k
(
x
,
t
)

z
x
(
x
,
t
)

2
]
≤
1
2
A
1
′
(
x
)
×
∑
t
=
T

A
1
(
x
)
,
A
1
(
x
)
{
(
ε

1
)
α
k
(
t
)

z
t
(
x
,
t
)

2
[
1
ε
α
k
(
t
)
(
A
1
′
(
x
)
)
2
+

γ
k
(
x
)

A
1
′
(
x
)

β
k
(
x
,
t
)
]
+
[
1
ε
α
k
(
t
)
(
A
1
′
(
x
)
)
2
+

γ
k
(
x
)

A
1
′
(
x
)

β
k
(
x
,
t
)
]
×
[
1
ε
α
k
(
t
)
(
A
1
′
(
x
)
)
2
+

γ
k
(
x
)

A
1
′
(
x
)

β
k
(
x
,
t
)
]

z
x
(
x
,
t
)

2
}
≤
0
.
We must take
0
<
ε
≤
1
and
A
1
(
x
)
satisfies
(39)
1
ε
α
k
(
t
)
(
A
1
′
(
x
)
)
2
+

γ
k
(
x
)

A
1
′
(
x
)

β
k
(
x
,
t
)
≤
0
.
From (39), we derive
(40)
1
ε
(
A
1
′
(
x
)
)
2
+
2
k
x
A
1
′
(
x
)
α
k
(
t
)
+
k
2
x
2
α
k
2
(
t
)
≤
1
α
k
2
(
t
)
.
Let
ε
=
1
. Then it follows that
(41)
[
1
A
1
′
(
x
)
+
k
x
α
k
(
t
)
]
2
≤
1
α
k
2
(
t
)
;
that is,
(42)
1
A
1
′
(
x
)
+
x
α
k
(
t
)
≤
1
α
k
(
t
)
.
By (42), we deduce that
T
>
T
k
*
,
(43)
A
1
′
(
x
)
≥
α
k
(
t
)
1

k
x
≥
1
+
k
T
k
*
1

k
x
.
Integrating into
(
0
,
x
)
, we have
(44)
A
1
(
x
)
≥
1
+
k
T
k
*
k
[

ln
(
1

k
x
)
]
.
Hence, we choose
(45)
A
1
(
x
)
=
1
+
k
T
k
*
k
[

ln
(
1

k
x
)
]
,
which satisfies (33) and (26). It follows that
(46)
A
1
(
1
)
=
1
+
k
T
k
*
k
[

ln
(
1

k
)
]
.
From the definition of
T
k
*
(see (36)) and (46), we deduce that
k
∈
(
0,1

1
/
e
)
,
(47)
T
k
*
=

2
ln
(
1

k
)
k
[
1
+
2
ln
(
1

k
)
]
=
2
A
1
(
1
)
.
In the following, we give the proof of the second inequality in (16).
We choose
q
(
x
)
=
x

1
for
x
∈
[
0,1
]
in (18). Noting that
α
k
′
(
t
)
=
k
,
β
k
,
x
(
x
,
t
)
=

2
k
2
x
/
(
1
+
k
t
)
, and
γ
k
(
x
)
=

2
k
x
, it follows that
(48)
1
2
∫
0
T
β
k
(
0
,
t
)

z
x
(
0
,
t
)

2
d
t
=
∫
0
T
E
(
t
)
d
t

∫
0
T
∫
0
1
k
(
x

1
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)
d
x
d
t
+
∫
0
T
∫
0
1
k
2
x
(
x

1
)
1
+
k
t

z
x
(
x
,
t
)

2
d
x
d
t
+
∫
0
1
[

z
x
(
x
,
t
)

2
α
k
(
t
)
(
x

1
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)

k
x
(
x

1
)

z
x
(
x
,
t
)

2
]
d
x

0
T
.
Next, we estimate every term in the right side of (48). Notice that
1
≤
α
k
(
t
)
≤
1
+
k
T
and
0
<
(
1

k
2
)
/
(
1
+
k
T
)
≤
β
k
(
x
,
t
)
≤
1
for any
(
x
,
t
)
∈
Q
. By (17), we have
(49)
∫
0
T
E
(
t
)
d
t

∫
0
T
∫
0
1
k
(
x

1
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)
d
x
d
t
+
∫
0
T
∫
0
1
k
2
x
(
x

1
)
1
+
k
t

z
x
(
x
,
t
)

2
d
x
d
t
≤
∫
0
T
E
(
t
)
d
t
+
C
∫
0
T
∫
0
1
[

z
t
(
x
,
t
)

2
+

z
x
(
x
,
t
)

2
]
d
x
d
t
≤
∫
0
T
E
(
t
)
d
t
+
C
∫
0
T
∫
0
1
[
α
t
(
t
)

z
t
(
x
,
t
)

2
+
β
k
(
x
,
t
)

z
x
(
x
,
t
)

2
]
d
x
d
t
≤
C
∫
0
T
E
(
t
)
d
t
≤
C
E
0
.
On the other hand, for each
t
∈
[
0
,
T
]
, it holds that
(50)

∫
0
1
[

z
x
(
x
,
t
)

2
α
k
(
t
)
(
x

1
)
z
t
(
x
,
t
)
z
x
(
x
,
t
)

k
x
(
x

1
)

z
x
(
x
,
t
)

2
]
d
x
∫
0
1

≤
C
∫
0
T
E
(
t
)
d
t
≤
C
E
0
.
Therefore, by (48)–(50), we have
(51)
1
2
∫
0
T
β
k
(
0
,
t
)

z
x
(
0
,
t
)

2
d
t
≤
C
E
0
≤
C
(

z
0

H
0
1
(
0,1
)
2
+

z
1

L
2
(
0,1
)
2
)
.