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Case 2: When
J
K
is an infinite index set, suppose, on the contrary, that there
i
exist an infinite subset
K
J
K
and
ε
>
0
such that
i
i
0
. (2.3)
g
ε
,
k
K
k
0
i
Utilizing (2.1),(2.2),(H 2 ),(H 3 ) and (2.3), we have
v
,
x
y
-
g
,
P
=
k
k
k
g
,
v
μ
v
,
x
y
=
μ
α
g
(
y
),
d
k
k
2
k
k
1
k
k
k
1
k
k
k
v
k
μ
μ
α
g
,
d
=
μ
μ
α
[
g
,
s
+
g
,
ω
]
0
1
k
k
k
0
1
k
k
k
k
k
2
μ
μ
α
[
c
g
g
ω
]
(2.4)
0
1
k
2
k
k
k
2
μ
μ
α
[
c
g
γ
g
(
q
+
p
g
)]
0
1
k
2
k
k
k
k
2
μ
μ
(
c
γ
p
γ
q
ε
)
α
g
0
1
2
k
k
0
k
k
for
k
On the other hand, by Algorithm 2.1, (H 1 ), (H 3 ) and (2.3), we have
K
.
i
v
,
x
y
P
=
k
k
k
x
y
=
α
d
k
k
k
k
k
v
k
α
(
c
g
+
γ
(
q
+
p
g
))
(2.5)
k
1
k
k
k
α
(
c
+
γ
p
+
γ
q
ε
)
g
k
1
k
k
0
k
for
Note that for every
k
K
.
i
k
K
,
there exists an integer k
such that
k
=
l
( k
)
i
1
,
i
that is,
k
+
1
=
l
(
k
)
(
i
1
1
K
.
i
1
Therefore by (1.6), for
k
K
,
we have
i
1
f
f
-
σλ
g
T
k
P
σλ
2
P
T
k
B
P
l
(
k
)
k
+
1
k
k
2
k
k
k
1
-
σλ
g
T
k
P
σλ
s
P
T
B
P
k
k
k
k
k
k
k
2
(2.6)
1
δ
g
T
k
P
=
-
σλ
g
T
k
P
σλ
P
T
B
P
k
k
2
k
P
T
k
B
P
k
k
k
k
k
σ
(
2
δ
)
=
-
λ
g
T
k
P
,
k
k
2
which, together with (2.4) and
σ
>
0
,
δ
<
2
, implies that
σ
(
2
δ
)
σ
(
2
δ
)
2
f
f
-
λ
g
T
k
P
-
μ
μ
(
c
γ
p
γ
q
ε
)
α
λ
g
.
(2.7)
l
(
k
)
k
+
1
2
k
k
2
0
1
2
k
k
0
k
k
k
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