Information Technology Reference
In-Depth Information
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
Search WWH ::
Custom Search