# Positive Solutions for a Third Order Nonlinear Neutral Delay Difference Equation

Authors: Zeqing Liu; Xiaochen Wang; Shin Min Kang; Young Chel Kwun
Journal: Abstract and Applied Analysis
 (2015)

---

## Abstract

The existence, multiplicity, and properties of positive solutions for a third order nonlinear neutral delay difference equation are discussed. Six examples are given to illustrate the results presented in this paper.


---

## Body

## 1. Introduction and Preliminaries

Recently, some researchers used the Reccati transformation techniques, fixed point theorems, and iterative algorithms to study the oscillation, nonoscillation, asymptotic properties, and solvability for linear and nonlinear third order difference equations and systems; see, for example, [1–6] and the references therein. In particular, Saker [4], Andruch-Sobilo and Migda [1], and Grace and Hamedani [2] discussed the oscillation for the following third order difference equations:
(1)
Δ
3
x
(
n
)
+
p
(
n
)
x
(
n
+
1
)
=
0
,
n
≥
n
0
,
Δ
3
(
x
(
n
)
-
p
(
n
)
x
(
σ
(
n
)
)
)
±
q
(
n
)
x
(
τ
(
n
)
)
=
0
,
n
≥
n
0
,
Δ
3
(
x
(
n
)
-
x
(
n
-
τ
)
)
±
q
(
n
)
x
n
-
σ
3
sgn
x
(
n
-
σ
)
=
0
,
n
≥
0
.
Making use of the Schauder fixed point theorem, Banach fixed point theorem, and Mann iterative schemes, Yan and Liu [5] and Liu et al. [3], respectively, proved the existence of a bounded nonoscillatory solution for the third order difference equation:
(2)
Δ
3
x
(
n
)
+
f
n
,
x
n
,
x
n
-
τ
=
0
,
n
≥
n
0
and the existence of positive solutions and convergence of the Mann iterative schemes for the following third order nonlinear neutral delay difference equation:
(3)
Δ
3
x
n
+
b
n
x
n
-
τ
+
Δ
h
n
,
x
h
1
n
,
x
h
2
n
,
…
,
x
h
k
n
+
f
(
n
,
x
(
f
1
(
n
)
)
,
x
(
f
2
(
n
)
)
,
…
,
x
(
f
k
(
n
)
)
)
=
c
(
n
)
,
n
≥
n
0
.However, the following third order nonlinear neutral delay difference equation:(4)
Δ
3
x
n
+
b
n
x
n
-
τ
+
c
n
+
Δ
2
h
n
,
x
h
1
n
,
…
,
x
h
k
n
+
Δ
g
n
,
x
g
1
n
,
…
,
x
g
k
n
+
f
n
,
x
f
1
n
,
…
,
x
f
k
n
=
d
n
,
n
≥
n
0
,
where τ
,
k
,
n
0
∈
N, {
c
(
n
)
}
n
∈
N
n
0, {
b
(
n
)
}
n
∈
N
n
0, and {
d
(
n
)
}
n
∈
N
n
0 are real sequences, f
,
g
,
h
∈
C
(
N
×
R
k
,
R
) and f
l
,
g
l
,
h
l
:
N
n
0
→
N with
(5)
lim
n
→
∞
f
l
n
=
lim
n
→
∞
g
l
n
=
lim
n
→
∞
h
l
n
=
+
∞
,
l
∈
1,2
,
…
,
k
has not been studied. The purpose of this paper is to study solvability of (4). By utilizing the Krasnoselskii fixed point theorem, Schauder fixed point theorem and some new techniques, we establish the existence, multiplicity, and properties of positive solutions of (4). Six examples are constructed to illuminate our results.Throughout this paper, we assume thatΔ is the forward difference operator defined by Δ
x
(
n
)
=
x
(
n
+
1
)
-
x
(
n
), R
=
(
-
∞
,
+
∞
), R
+
=
[
0
,
+
∞
), and N
0 and N denote the sets of nonnegative integers and positive integers, respectively,
(6)
N
t
=
{
n
:
n
∈
N
0
with
n
≥
t
}
,
t
∈
N
0
,
α
=
inf
⁡
{
f
l
(
n
)
,
g
l
(
n
)
,
h
l
(
n
)
:
1
≤
l
≤
k
,
n
∈
N
n
0
}
,
β
=
min
⁡
n
0
-
τ
,
α
∈
N
.
Let l
β
∞ denote the Banach space of all real sequences x
=
{
x
(
n
)
}
n
∈
N
β with norm
(7)
x
=
sup
⁡
n
∈
N
β
x
n
n
<
+
∞
for
x
=
x
n
n
∈
N
β
∈
l
β
∞
.
Let A, B, A
*, B
* and c
* be positive constants, T
∈
N, {
c
(
n
)
}
n
∈
N
β, {
A
(
n
)
}
n
∈
N
β, and {
B
(
n
)
}
n
∈
N
β be real sequences with
(8)
B
n
=
B
+
c
n
n
>
A
n
=
A
-
c
n
n
,
n
∈
N
β
,
c
n
=
c
n
0
,
β
≤
n
≤
n
0
-
1
,
c
*
≥
sup
⁡
n
∈
N
β
c
n
n
,
A
*
=
A
-
c
*
,
B
*
=
B
+
c
*
.
Put
(9)
Ω
A
*
,
B
*
,
T
=
x
=
x
n
n
∈
N
β
∈
l
β
∞
:
A
(
T
)
≤
x
n
n
≤
B
T
,
x
n
n
β
≤
n
<
T
;
A
n
≤
x
n
n
≤
B
(
n
)
,
n
≥
T
.
It is easy to see that Ω
(
A
*
,
B
*
,
T
) is a bounded closed and convex subset of l
β
∞.By a solution of (4), we mean a sequence {
x
(
n
)
}
n
∈
N
β
∈
l
β
∞ with a positive integer T
≥
n
0
+
τ
+
α such that (4) holds for all n
≥
T.The following lemmas play important roles in this paper.Lemma 1 (see [6]).
A bounded and uniformly Cauchy subsetC
⊆
l
β
∞ is relatively compact.Lemma 2 (Krasnoselskii fixed point theorem).
LetX be a Banach space, D a bounded closed convex subset of X, and S
,
G
:
D
→
X mappings such that S
x
+
G
y
∈
D for every pair x
,
y
∈
D. If S is a contraction and G is completely continuous, then the equation
(10)
S
x
+
G
x
=
x
has a solution in D.Lemma 3 (Schauder fixed point theorem).
LetD be a nonempty closed convex subset of a Banach space X and T
:
D
→
D a continuous mapping such that T
(
D
) is a relatively compact subset of X. Then T has at least one fixed point in D.Lemma 4.
Letτ
,
n
∈
N and {
q
(
k
)
}
k
∈
N
⊂
R
+. Then(i)
∑
i
=
1
∞
‍
∑
t
=
n
+
i
τ
∞
‍
q
(
t
)
≤
∑
t
=
n
+
τ
∞
‍
(
t
/
τ
)
q
(
t
);
(ii)
∑
i
=
1
∞
‍
∑
s
=
n
+
i
τ
∞
‍
∑
t
=
s
∞
‍
q
(
t
)
≤
∑
t
=
n
+
τ
∞
‍
(
t
2
/
τ
)
q
(
t
);
(iii)
∑
p
=
1
∞
‍
∑
i
=
n
+
p
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
q
(
t
)
≤
∑
t
=
n
+
τ
∞
‍
(
t
3
/
τ
)
q
(
t
).Proof.
(i) Let[
t
] denote the largest integer number not exceeding t
∈
R
+. It is clear that
(11)
∑
i
=
1
∞
‍
∑
t
=
n
+
i
τ
∞
‍
q
t
=
∑
t
=
n
+
τ
∞
‍
1
+
t
-
n
-
τ
τ
q
t
≤
∑
t
=
n
+
τ
∞
‍
t
τ
q
t
.
(ii) It follows from (i) that(12)
∑
i
=
1
∞
∑
s
=
n
+
i
τ
∞
‍
∑
t
=
s
∞
‍
q
t
=
∑
i
=
1
∞
‍
∑
t
=
n
+
i
τ
∞
‍
1
+
t
-
n
-
i
τ
q
t
≤
∑
i
=
1
∞
‍
∑
t
=
n
+
i
τ
∞
‍
t
q
t
≤
∑
t
=
n
+
τ
∞
‍
t
2
τ
q
t
.
(iii) It follows from (ii) that(13)
∑
p
=
1
∞
‍
∑
i
=
n
+
p
τ
∞
‍
∑
s
=
i
∞
‍
‍
∑
t
=
s
∞
‍
q
t
=
∑
p
=
1
∞
‍
∑
i
=
n
+
p
τ
∞
‍
∑
t
=
i
∞
‍
t
-
i
+
1
q
t
≤
∑
p
=
1
∞
‍
∑
i
=
n
+
p
τ
∞
‍
∑
t
=
i
∞
‍
t
q
(
t
)
≤
∑
t
=
n
+
τ
∞
‍
t
3
τ
q
(
t
)
.
This completes the proof.


## 2. Existence of Positive Solutions

Now we discuss the existence, multiplicity, and properties of positive solutions of (4) under various conditions on the sequence {
b
(
n
)
}
n
∈
N
β
⊆
R.Theorem 5.
Assume that there exist a constantb
* and three nonnegative sequences {
F
n
}
n
∈
N
n
0, {
G
n
}
n
∈
N
n
0, and {
H
n
}
n
∈
N
n
0 satisfying
(14)
A
+
b
*
B
*
<
B
,
0
≤
b
(
n
)
≤
b
*
<
1
,
n
∈
N
n
0
,
(15)
f
n
,
u
1
,
u
2
,
…
,
u
k
≤
F
n
,
g
n
,
u
1
,
u
2
,
…
,
u
k
≤
G
n
,
h
n
,
u
1
,
u
2
,
…
,
u
k
≤
H
n
,
n
,
u
l
∈
N
n
0
×
R
+
∖
0
,
1
≤
l
≤
k
,
(16)
lim
n
→
∞
1
n
∑
i
=
n
∞
‍
max
⁡
H
i
,
∑
s
=
i
∞
‍
G
s
=
0
,
(17)
lim
n
→
∞
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
max
⁡
F
t
,
d
t
=
0
.
Then(i)
equation (4) possesses a positive solution x
=
{
x
(
n
)
}
n
∈
N
β
∈
l
β
∞ with
(18)
lim
n
→
∞
x
(
n
)
+
b
(
n
)
x
(
n
-
τ
)
+
c
(
n
)
n
∈
A
+
b
*
B
*
,
B
,
(19)
A
*
≤
liminf
⁡
n
→
∞
x
(
n
)
n
≤
limsup
⁡
n
→
∞
x
(
n
)
n
≤
B
*
;
(ii)
equation (4) possesses uncountably many positive solutions in l
β
∞.Proof.
(i) LetL
∈
(
A
+
b
*
B
*
,
B
). It follows from (16) and (17) that there exists T
≥
n
0
+
τ
+
α satisfying
(20)
1
T
∑
i
=
T
∞
‍
H
i
+
∑
s
=
i
∞
‍
G
s
+
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
<
min
⁡
{
L
-
A
-
b
*
B
*
,
B
-
L
}
.
Define two mappings S
L and G
L
:
Ω
(
A
*
,
B
*
,
T
)
→
l
β
∞ by
(21)
S
L
x
n
=
n
L
-
b
n
x
n
-
τ
-
c
n
,
n
≥
T
,
n
T
(
S
L
x
)
(
T
)
,
β
≤
n
<
T
,
(22)
(
G
L
x
)
(
n
)
=
∑
i
=
n
∞
‍
h
i
,
x
h
1
i
,
…
,
x
h
k
i
-
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
s
,
x
g
1
s
,
…
,
x
g
k
s
+
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
t
,
x
f
1
t
,
…
,
x
f
k
t
x
f
k
t
-
d
(
t
)
,
n
≥
T
,
n
T
G
L
x
T
,
β
≤
n
<
T
,
for any x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
).
Now we show that(23)
S
L
x
+
G
L
y
∈
Ω
(
A
*
,
B
*
,
T
)
,
x
,
y
∈
Ω
(
A
*
,
B
*
,
T
)
;
(24)
S
L
x
-
S
L
y
≤
b
*
x
-
y
,
x
,
y
∈
Ω
A
*
,
B
*
,
T
,
(25)
G
L
y
≤
B
,
y
∈
Ω
(
A
*
,
B
*
,
T
)
.
Using (14), (15), and (20)–(22), we get that for any x
=
{
x
(
n
)
}
n
∈
N
β
,
y
=
{
y
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
)
(26)
S
L
x
n
+
G
L
y
n
n
=
L
-
b
n
n
x
n
-
τ
-
c
n
n
+
1
n
∑
i
=
n
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
≤
L
+
c
n
n
+
1
n
∑
i
=
n
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
d
t
≤
L
+
c
n
n
+
1
T
∑
i
=
T
∞
‍
H
i
+
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
G
s
+
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
<
L
+
c
n
n
+
min
⁡
⁡
L
-
A
-
b
*
B
*
,
B
-
L
≤
B
+
|
c
(
n
)
|
n
=
B
(
n
)
,
n
≥
T
,
S
L
x
n
+
G
L
y
n
n
=
n
T
·
S
L
x
T
+
G
L
y
T
n
≤
B
T
,
β
≤
n
<
T
,
S
L
x
n
+
G
L
y
n
n
=
L
-
b
n
n
x
n
-
τ
-
c
n
n
+
1
n
∑
i
=
n
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
-
d
t
≥
L
-
n
-
τ
n
b
n
x
n
-
τ
n
-
τ
-
c
n
n
-
1
n
∑
i
=
n
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
-
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
d
t
≥
L
-
b
*
B
*
-
c
n
n
-
1
T
∑
i
=
T
∞
‍
H
i
-
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
G
s
-
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
>
L
-
b
*
B
*
-
c
n
n
-
min
⁡
⁡
L
-
A
-
b
*
B
*
,
B
-
L
≥
A
-
c
n
n
=
A
(
n
)
,
n
≥
T
,
S
L
x
n
+
G
L
y
n
n
=
n
T
·
(
S
L
x
)
(
T
)
+
(
G
L
y
)
(
T
)
n
≥
A
(
T
)
,
β
≤
n
<
T
,
S
L
x
n
-
S
L
y
n
n
=
b
n
n
-
τ
n
x
n
-
τ
-
y
n
-
τ
n
-
τ
≤
b
*
x
-
y
,
n
≥
T
,
S
L
x
n
-
S
L
y
n
n
=
n
T
S
L
x
T
-
S
L
y
T
n
≤
b
*
x
-
y
,
β
≤
n
<
T
,
G
L
y
n
n
=
1
n
∑
i
=
n
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
∑
i
=
n
∞
‍
t
,
y
f
1
t
,
…
,
y
f
k
t
-
d
t
≤
1
n
∑
i
=
n
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
d
t
≤
1
T
∑
i
=
T
∞
‍
H
i
+
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
G
s
+
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
<
min
⁡
L
-
A
-
b
*
B
*
,
B
-
L
≤
B
,
n
≥
T
,
G
L
y
n
n
=
n
T
G
L
y
T
n
≤
B
,
β
≤
n
<
T
,
which yield the fact that (23)–(25) hold.
Next we show thatG
L is completely continuous. Let y
w
=
{
y
w
(
n
)
}
n
∈
N
β and y
=
{
y
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
) with
(27)
lim
w
→
∞
y
w
=
y
.
Using (16), (17), (27), and the continuity of f, g, and h, we know that for given ε
>
0, there exist T
1, T
2, T
3 and T
4
∈
N with T
4
>
T
3
>
T
2
>
T
1
>
T satisfying
(28)
1
T
max
⁡
⁡
∑
i
=
T
1
+
1
∞
‍
H
i
+
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
G
s
+
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
,
∑
i
=
T
T
1
‍
∑
s
=
T
2
+
1
∞
‍
G
s
+
∑
i
=
T
T
1
‍
∑
s
=
T
2
+
1
∞
‍
∑
t
=
s
∞
‍
F
t
+
∑
i
=
T
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
T
3
+
1
∞
‍
F
t
<
ε
16
,
(29)
1
T
max
⁡
⁡
∑
i
=
T
T
1
‍
h
i
,
y
w
h
1
i
,
…
,
y
w
h
k
i
-
h
(
i
,
y
(
h
1
(
i
)
)
,
…
,
y
(
h
k
(
i
)
)
)
,
∑
i
=
T
T
1
‍
∑
s
=
i
T
2
‍
g
(
s
,
y
w
(
g
1
(
s
)
)
,
…
,
y
w
(
g
k
(
s
)
)
)
-
g
(
s
,
y
(
g
1
(
s
)
)
,
…
,
y
(
g
k
(
s
)
)
)
,
∑
i
=
T
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
s
T
3
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
∑
i
=
T
T
1
‍
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
<
ε
16
,
w
≥
T
4
.
Combining (15), (22), (28), and (29), we infer that
(30)
G
L
y
w
-
G
L
y
=
sup
⁡
⁡
G
L
y
w
n
-
G
L
y
n
n
:
n
∈
N
β
=
max
⁡
⁡
sup
⁡
⁡
n
T
·
G
L
y
w
T
-
G
L
y
T
n
:
β
≤
n
<
T
,
sup
⁡
⁡
G
L
y
w
n
-
G
L
y
n
n
:
n
∈
N
T
=
sup
⁡
⁡
G
L
y
w
n
-
G
L
y
n
n
:
n
∈
N
T
≤
1
T
∑
i
=
T
∞
‍
h
i
,
y
w
h
1
i
,
…
,
y
w
h
k
i
-
h
i
,
y
h
1
i
,
…
,
y
h
k
i
+
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
w
g
1
s
,
…
,
y
w
g
k
s
-
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
=
1
T
∑
i
=
T
T
1
‍
h
i
,
y
w
h
1
i
,
…
,
y
w
h
k
i
i
,
y
h
1
i
,
…
,
y
h
k
i
-
h
i
,
y
h
1
i
,
…
,
y
h
k
i
+
1
T
∑
i
=
T
1
+
1
∞
‍
h
i
,
y
w
h
1
i
,
…
,
y
w
h
k
i
-
h
i
,
y
h
1
i
,
…
,
y
h
k
i
+
1
T
∑
i
=
T
T
1
‍
∑
s
=
i
T
2
‍
g
s
,
y
w
g
1
s
,
…
,
y
w
g
k
s
-
g
s
,
y
g
1
s
,
y
g
2
s
,
…
,
y
g
k
s
+
1
T
∑
i
=
T
T
1
‍
∑
s
=
T
2
+
1
∞
‍
g
s
,
y
w
g
1
s
,
…
,
y
w
g
k
s
-
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
T
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
w
g
1
s
,
…
,
y
w
g
k
s
-
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
T
∑
i
=
T
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
s
T
3
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
1
T
∑
i
=
T
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
T
3
+
1
∞
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
1
T
∑
i
=
T
T
1
‍
∑
s
=
T
2
+
1
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
1
T
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
<
ε
16
+
2
T
∑
i
=
T
1
+
1
∞
‍
H
i
+
ε
16
+
2
T
∑
i
=
T
T
1
‍
∑
s
=
T
2
+
1
∞
‍
G
s
+
2
T
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
G
s
+
ε
16
+
2
T
∑
i
=
T
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
T
3
+
1
∞
‍
F
t
+
2
T
∑
i
=
T
T
1
‍
∑
s
=
T
2
+
1
∞
‍
∑
t
=
s
∞
‍
F
t
+
2
T
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
<
ε
,
w
≥
T
4
,
which implies that G
L is continuous in Ω
(
A
*
,
B
*
,
T
).
It follows from (15), (22), and (28) that for any x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
) and t
1
,
t
2
≥
T
4
(31)
G
L
y
t
2
t
2
-
G
L
y
t
1
t
1
=
1
t
2
∑
i
=
t
2
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
t
1
∑
i
=
t
1
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
t
2
∑
i
=
t
2
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
t
1
∑
i
=
t
1
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
t
2
∑
i
=
t
2
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
-
d
t
-
1
t
1
∑
i
=
t
1
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
-
d
t
≤
2
T
4
∑
i
=
T
4
∞
‍
H
i
+
∑
i
=
T
4
∞
‍
∑
s
=
i
∞
‍
G
s
+
∑
i
=
T
4
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
<
ε
,
which means that G
L
(
Ω
(
A
*
,
B
*
,
T
)
) is uniformly Cauchy, which together with (25) and Lemma 1 yields that G
L
(
Ω
(
A
*
,
B
*
,
T
)
) is relatively compact. Hence G
L is completely continuous in Ω
(
A
*
,
B
*
,
T
). Thus (14), (24), and Lemma 2 ensure that the mapping S
L
+
G
L has a fixed point x
=
{
x
(
n
)
}
n
∈
N
β, which together with (21) and (22) implies that
(32)
x
n
=
n
L
-
b
n
x
n
-
τ
-
c
n
+
∑
i
=
n
∞
‍
h
i
,
x
h
1
i
,
…
,
x
h
k
i
-
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
s
,
x
g
1
s
,
…
,
x
g
k
s
+
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
,
n
≥
T
;
which gives that
(33)
Δ
x
n
+
b
n
x
n
-
τ
+
c
n
=
L
-
h
n
,
x
h
1
n
,
…
,
x
h
k
n
+
∑
s
=
n
∞
‍
g
s
,
x
g
1
s
,
…
,
x
g
k
s
-
∑
s
=
n
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
,
n
≥
T
+
τ
,
Δ
2
x
n
+
b
n
x
n
-
τ
+
c
n
=
-
Δ
h
n
,
x
h
1
n
,
…
,
x
h
k
n
-
g
n
,
x
g
1
n
,
…
,
x
g
k
n
+
∑
t
=
n
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
,
n
≥
T
+
τ
,
Δ
3
x
n
+
b
n
x
n
-
τ
+
c
n
=
-
Δ
2
h
n
,
x
h
1
n
,
…
,
x
h
k
n
-
Δ
g
n
,
x
g
1
n
,
…
,
x
g
k
n
-
f
n
,
x
f
1
n
,
…
,
x
f
k
n
-
d
n
,
n
≥
T
+
τ
,
that is, x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
) is a positive solution of (4). It follows from (15)–(17), and (32) that
(34)
x
n
+
b
n
x
n
-
τ
+
c
n
n
-
L
=
1
n
∑
i
=
n
∞
‍
h
i
,
x
h
1
i
,
…
,
x
h
k
i
-
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
s
,
x
g
1
s
,
…
,
x
g
k
s
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
≤
1
n
∑
i
=
n
∞
‍
H
i
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
G
s
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
⟶
0
as
n
⟶
∞
,
A
*
=
A
-
c
*
≤
A
-
c
n
n
=
A
n
≤
x
n
n
≤
B
n
=
B
+
c
n
n
≤
B
+
c
*
=
B
*
,
n
≥
T
,
which yield that
(35)
lim
n
→
∞
x
(
n
)
+
b
(
n
)
x
(
n
-
τ
)
+
c
(
n
)
n
=
L
∈
A
+
b
*
B
*
,
B
,
A
*
≤
liminf
⁡
n
→
∞
x
n
n
≤
limsup
⁡
n
→
∞
x
n
n
≤
B
*
;
that is, (18) and (19) hold.
(ii) LetL
1
,
L
2
∈
(
A
+
b
*
B
*
,
B
) and L
1
≠
L
2. As in the proof of (i), we infer that for each l
∈
{
1,2
}, there exist a constant T
l
≥
n
0
+
τ
+
α and two mappings S
L
l and G
L
l
:
Ω
(
A
*
,
B
*
,
T
l
)
→
l
β
∞ satisfying (20)–(22), where T
,
L
,
S
L and G
L are replaced by T
l
,
L
l
,
S
L
l and G
L
l, respectively, and S
L
l
+
G
L
l possesses a fixed point x
l
=
{
x
l
(
n
)
}
n
∈
N
β, which is a positive solution of (4); that is,
(36)
x
l
n
=
n
L
l
-
b
n
x
l
n
-
τ
-
c
n
+
∑
i
=
n
∞
‍
h
i
,
x
l
h
1
i
,
…
,
x
l
h
k
i
-
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
s
,
x
l
g
1
s
,
…
,
x
l
g
k
s
+
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
l
f
1
t
,
…
,
x
l
f
k
t
-
d
t
,
n
≥
T
l
.
Note that (16) and (17) imply that there exists T
*
>
max
⁡
{
T
1
,
T
2
} satisfying
(37)
1
T
*
∑
i
=
T
*
∞
‍
H
i
+
∑
i
=
T
*
∞
‍
∑
s
=
i
∞
‍
G
s
+
∑
i
=
T
*
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
<
|
L
1
-
L
2
|
4
,
which together with (15) and (36) means that for any n
≥
T
*
(38)
x
1
n
n
-
x
2
n
n
=
L
1
-
L
2
-
b
n
n
(
x
1
(
n
-
τ
)
-
x
2
(
n
-
τ
)
)
+
1
n
∑
i
=
n
∞
‍
h
(
i
,
x
1
(
h
1
(
i
)
)
,
…
,
x
1
(
h
k
(
i
)
)
)
-
h
(
i
,
x
2
(
h
1
(
i
)
)
,
…
,
x
2
(
h
k
(
i
)
)
)
-
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
(
s
,
x
1
(
g
1
(
s
)
)
,
…
,
x
1
(
g
k
(
s
)
)
)
-
g
(
s
,
x
2
(
g
1
(
s
)
)
,
…
,
x
2
(
g
k
(
s
)
)
)
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
1
f
1
t
,
…
,
x
1
f
k
t
b
n
n
t
,
x
1
f
1
t
,
…
,
x
1
f
k
t
-
f
(
t
,
x
2
(
f
1
(
t
)
)
,
…
,
x
2
(
f
k
(
t
)
)
)
≥
L
1
-
L
2
-
n
-
τ
b
n
n
x
1
n
-
τ
-
x
2
n
-
τ
n
-
τ
-
1
T
*
∑
i
=
T
*
∞
‍
h
(
i
,
x
1
(
h
1
(
i
)
)
,
…
,
x
1
(
h
k
(
i
)
)
)
-
h
(
i
,
x
2
(
h
1
(
i
)
)
,
…
,
x
2
(
h
k
(
i
)
)
)
-
1
T
*
∑
i
=
T
*
∞
‍
∑
s
=
i
∞
‍
g
(
s
,
x
1
(
g
1
(
s
)
)
,
…
,
x
1
(
g
k
(
s
)
)
)
-
g
(
s
,
x
2
(
g
1
(
s
)
)
,
…
,
x
2
(
g
k
(
s
)
)
)
-
1
T
*
∑
i
=
T
*
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
(
t
,
x
1
(
f
1
(
t
)
)
,
…
,
x
1
(
f
k
(
t
)
)
)
-
f
(
t
,
x
2
(
f
1
(
t
)
)
,
…
,
x
2
(
f
k
(
t
)
)
)
≥
L
1
-
L
2
-
b
*
x
1
-
x
2
-
2
T
*
∑
i
=
T
*
∞
‍
H
i
+
∑
i
=
T
*
∞
‍
∑
s
=
i
∞
‍
G
s
+
∑
i
=
T
*
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
>
L
1
-
L
2
2
-
b
*
x
1
-
x
2
,
which implies that
(39)
x
1
-
x
2
>
L
1
-
L
2
2
(
1
+
b
*
)
>
0
,
which yields that x
1
≠
x
2. Therefore (4) possesses uncountably many positive solutions in l
β
∞. This completes the proof.Theorem 6.
Assume that there exist constantsb
* and b
* and three nonnegative sequences {
F
n
}
n
∈
N
n
0, {
G
n
}
n
∈
N
n
0, and {
H
n
}
n
∈
N
n
0 satisfying (15)–(17) and
(40)
b
*
A
+
B
*
+
c
*
b
*
b
*
<
b
*
B
+
b
*
A
*
b
*
-
c
*
,
1
<
b
*
≤
b
n
≤
b
*
,
n
∈
N
n
0
.
Then(i)
equation (4) possesses a positive solution x
=
{
x
(
n
)
}
n
∈
N
β
∈
l
β
∞ with (19) and
(41)
lim
n
→
∞
x
n
+
b
n
x
n
-
τ
+
c
n
n
∈
b
*
A
+
B
*
+
c
*
b
*
b
*
,
b
*
B
+
b
*
A
*
b
*
-
c
*
;
(ii)
equation (4) possesses uncountably many positive solutions in l
β
∞.Proof.
(i) PutL
∈
(
b
*
A
+
(
B
*
+
c
*
)
(
b
*
/
b
*
)
,
b
*
B
+
b
*
A
*
/
b
*
-
c
*
). Observe that
(42)
lim
n
→
∞
b
*
A
+
1
+
τ
n
B
*
+
c
*
b
*
b
*
=
b
*
A
+
B
*
+
c
*
b
*
b
*
<
L
<
b
*
B
+
b
*
A
*
b
*
-
c
*
=
lim
n
→
∞
b
*
B
+
b
*
A
*
b
*
-
c
*
1
+
τ
n
,
which yields that there exists N
∈
N satisfying
(43)
b
*
A
+
B
*
+
c
*
b
*
b
*
<
b
*
A
+
1
+
τ
N
B
*
+
c
*
b
*
b
*
<
L
<
b
*
B
+
b
*
A
*
b
*
-
c
*
1
+
τ
N
<
b
*
B
+
b
*
A
*
b
*
-
c
*
.
It follows from (16) and (17) that there exist θ
∈
(
0,1
) and T
>
max
⁡
{
n
0
+
τ
+
α
,
N
} satisfying
(44)
θ
=
1
b
*
1
+
τ
T
,
(45)
1
T
∑
i
=
T
+
τ
∞
‍
H
i
+
∑
s
=
i
∞
‍
G
s
+
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
<
min
⁡
b
*
B
+
b
*
A
*
b
*
-
c
*
1
+
τ
N
-
L
,
b
*
L
b
*
b
*
L
b
*
-
b
*
A
-
1
+
τ
N
B
*
+
c
*
<
min
⁡
b
*
B
+
b
*
A
*
b
*
-
c
*
1
+
τ
T
-
L
,
b
*
L
b
*
b
*
L
b
*
-
b
*
A
-
1
+
τ
T
B
*
+
c
*
.
Define two mappings S
L and G
L
:
Ω
(
A
*
,
B
*
,
T
)
→
l
β
∞ by
(46)
S
L
x
n
=
n
L
b
n
+
τ
-
x
n
+
τ
b
n
+
τ
-
c
n
+
τ
b
n
+
τ
,
n
≥
T
,
n
T
S
L
x
T
,
β
≤
n
<
T
,
(47)
G
L
x
n
=
1
b
(
n
+
τ
)
∑
i
=
n
+
τ
∞
‍
h
(
i
,
x
(
h
1
(
i
)
)
,
…
,
x
(
h
k
(
i
)
)
)
-
1
b
(
n
+
τ
)
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
g
(
s
,
x
(
g
1
(
s
)
)
,
…
,
x
(
g
k
(
s
)
)
)
+
1
b
(
n
+
τ
)
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
(
t
,
x
(
f
1
(
t
)
)
,
…
,
x
(
f
k
(
t
)
)
)
-
d
(
t
)
,
n
≥
T
,
n
T
(
G
L
x
)
(
T
)
,
β
≤
n
<
T
for any x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
).
Now we show that (23), (48), and (49) below hold
(48)
S
L
x
-
S
L
y
≤
θ
x
-
y
,
x
,
y
∈
Ω
(
A
*
,
B
*
,
T
)
;
(49)
G
L
y
≤
B
+
A
*
b
*
,
y
∈
Ω
(
A
*
,
B
*
,
T
)
.
Using (15) and (44)–(47), we get that for any x
=
{
x
(
n
)
}
n
∈
N
β
,
y
=
{
y
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
)
(50)
S
L
x
n
+
G
L
y
n
n
=
L
b
n
+
τ
-
x
n
+
τ
n
b
n
+
τ
-
c
n
+
τ
n
b
n
+
τ
+
1
n
b
n
+
τ
∑
i
=
n
+
τ
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
n
b
n
+
τ
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
-
d
t
≤
L
b
*
-
x
n
+
τ
n
+
τ
b
n
+
τ
+
n
+
τ
n
b
n
+
τ
·
c
n
+
τ
n
+
τ
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
d
t
≤
L
b
*
-
A
*
b
*
+
c
*
b
*
1
+
τ
T
+
1
T
b
*
∑
i
=
T
+
τ
∞
‍
H
i
+
1
T
b
*
∑
i
=
T
+
τ
∞
‍
∑
s
=
i
∞
‍
G
s
+
1
T
b
*
∑
i
=
T
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
<
L
b
*
-
A
*
b
*
+
c
*
b
*
1
+
τ
T
+
1
b
*
min
⁡
⁡
b
*
B
+
b
*
A
*
b
*
-
c
*
1
+
τ
T
-
L
,
b
*
b
*
L
-
b
*
A
-
1
+
τ
T
B
*
+
c
*
≤
B
≤
B
(
n
)
,
n
≥
T
,
S
L
x
n
+
G
L
y
n
n
=
n
T
·
S
L
x
T
+
G
L
y
T
n
≤
B
(
T
)
,
β
≤
n
<
T
,
S
L
x
n
+
G
L
y
n
n
=
L
b
n
+
τ
-
x
n
+
τ
n
b
n
+
τ
-
c
n
+
τ
n
b
n
+
τ
+
1
n
b
n
+
τ
∑
i
=
n
+
τ
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
n
b
n
+
τ
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
-
d
t
≥
L
b
*
-
n
+
τ
n
b
n
+
τ
·
x
n
+
τ
n
+
τ
-
n
+
τ
n
b
n
+
τ
·
c
n
+
τ
n
+
τ
-
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
n
b
n
+
τ
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
-
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
d
t
≥
L
b
*
-
1
b
*
(
1
+
τ
T
)
B
*
-
c
*
b
*
(
1
+
τ
T
)
-
1
T
b
*
∑
i
=
T
+
τ
∞
‍
H
i
-
1
T
b
*
∑
i
=
T
+
τ
∞
‍
∑
s
=
i
∞
‍
G
s
-
1
T
b
*
·
∑
i
=
T
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
>
L
b
*
-
1
b
*
1
+
τ
T
B
*
+
c
*
-
1
b
*
min
⁡
b
*
B
+
b
*
A
*
b
*
-
c
*
(
1
+
τ
T
)
-
L
,
b
*
b
*
L
-
b
*
A
-
(
1
+
τ
T
)
(
B
*
+
c
*
)
≥
A
≥
A
(
n
)
,
n
≥
T
,
S
L
x
n
+
G
L
y
n
n
=
n
T
·
S
L
x
T
+
G
L
y
T
n
≥
A
(
T
)
,
β
≤
n
<
T
,
S
L
x
n
-
S
L
y
n
n
=
n
+
τ
n
b
n
+
τ
x
n
+
τ
-
y
n
+
τ
n
+
τ
≤
θ
x
-
y
,
n
≥
T
,
S
L
x
n
-
S
L
y
n
n
=
n
T
S
L
x
T
-
S
L
y
T
n
≤
θ
x
-
y
,
β
≤
n
<
T
,
G
L
y
n
n
=
1
n
b
n
+
τ
∑
i
=
n
+
τ
∞
‍
h
(
i
,
y
(
h
1
(
i
)
)
,
…
,
y
(
h
k
(
i
)
)
)
-
1
n
b
n
+
τ
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
[
f
(
t
,
y
(
f
1
(
t
)
)
,
…
,
y
(
f
k
(
t
)
)
)
-
d
(
t
)
]
≤
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
d
t
≤
1
T
b
*
∑
i
=
T
+
τ
∞
‍
H
i
+
1
T
b
*
∑
i
=
T
+
τ
∞
‍
∑
s
=
i
∞
‍
G
s
+
1
T
b
*
∑
i
=
T
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
<
1
b
*
min
⁡
⁡
b
*
B
+
b
*
A
*
b
*
-
c
*
(
1
+
τ
T
)
-
L
,
b
*
b
*
L
-
b
*
A
-
(
1
+
τ
T
)
(
B
*
+
c
*
)
≤
B
+
A
*
b
*
,
n
≥
T
,
G
L
y
n
n
=
n
T
G
L
y
T
n
≤
B
+
A
*
b
*
,
β
≤
n
<
T
,
which yield (23), (48), and (49).
Next we show thatG
L is completely continuous. Let y
w
=
{
y
w
(
n
)
}
n
∈
N
β and y
=
{
y
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
) with (27). Using (16), (17), (27), (47), and the continuity of f, g, and h, we know that for given ε
>
0, there exist T
1, T
2, T
3, and T
4
∈
N with T
4
>
T
3
>
T
2
>
T
1
>
T
+
τ satisfying
(51)
1
T
b
*
max
⁡
∑
i
=
T
1
+
1
∞
‍
H
i
+
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
G
s
+
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
,
∑
i
=
T
+
τ
T
1
‍
∑
s
=
T
2
+
1
∞
‍
G
s
⁡
+
∑
i
=
T
+
τ
T
1
‍
∑
s
=
T
2
+
1
∞
‍
∑
t
=
s
∞
‍
F
t
+
∑
i
=
T
+
τ
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
T
3
+
1
∞
‍
F
t
<
ε
16
,
(52)
1
T
b
*
max
⁡
∑
i
=
T
+
τ
T
1
‍
h
i
,
y
w
h
1
i
,
…
,
y
w
h
k
i
-
h
i
,
y
h
1
i
,
…
,
y
h
k
i
,
∑
i
=
T
+
τ
T
1
‍
∑
s
=
i
T
2
‍
g
s
,
y
w
g
1
s
,
…
,
y
w
g
k
s
-
g
s
,
y
g
1
s
,
…
,
y
g
k
s
,
∑
i
=
T
+
τ
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
s
T
3
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
∑
i
=
T
+
τ
T
1
‍
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
<
ε
16
,
w
≥
T
4
.
Combining (15), (47), (51), and (52), we infer that
(53)
G
L
y
w
-
G
L
y
=
sup
⁡
⁡
G
L
y
w
n
-
G
L
y
n
n
:
n
∈
N
β
=
max
⁡
⁡
sup
⁡
⁡
n
T
·
G
L
y
w
T
-
G
L
y
T
n
:
β
≤
n
<
T
,
sup
⁡
⁡
G
L
y
w
n
-
G
L
y
n
n
:
n
∈
N
T
≤
sup
⁡
⁡
1
n
b
n
+
τ
∑
i
=
n
+
τ
∞
‍
·
∑
i
=
n
+
τ
∞
‍
h
i
,
y
w
h
1
i
,
…
,
y
w
h
k
i
-
h
i
,
y
h
1
i
,
…
,
y
h
k
i
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
w
g
1
s
,
…
,
y
w
g
k
s
-
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
:
1
n
b
n
+
τ
∑
i
=
n
+
τ
∞
‍
n
∈
N
T
≤
1
T
b
*
∑
i
=
T
+
τ
T
1
‍
h
i
,
y
w
h
1
i
,
…
,
y
w
h
k
i
-
h
i
,
y
h
1
i
,
…
,
y
h
k
i
+
1
T
b
*
∑
i
=
T
1
+
1
∞
‍
h
i
,
y
w
h
1
i
,
…
,
y
w
h
k
i
-
h
i
,
y
h
1
i
,
…
,
y
h
k
i
+
1
T
b
*
∑
i
=
T
+
τ
T
1
‍
∑
s
=
i
T
2
‍
g
(
s
,
y
w
(
g
1
(
s
)
)
,
…
,
y
w
(
g
k
(
s
)
)
)
-
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
T
b
*
∑
i
=
T
+
τ
T
1
‍
∑
s
=
T
2
+
1
∞
‍
g
s
,
y
w
g
1
s
,
…
,
y
w
g
k
s
-
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
T
b
*
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
w
g
1
s
,
…
,
y
w
g
k
s
-
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
T
b
*
∑
i
=
T
+
τ
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
s
T
3
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
1
T
b
*
∑
i
=
T
+
τ
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
T
3
+
1
∞
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
1
T
b
*
∑
i
=
T
+
τ
T
1
‍
∑
s
=
T
2
+
1
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
1
T
b
*
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
<
ε
16
+
2
T
b
*
∑
i
=
T
1
+
1
∞
‍
H
i
+
ε
16
+
2
T
b
*
∑
i
=
T
+
τ
T
1
‍
∑
s
=
T
2
+
1
∞
‍
G
s
+
2
T
b
*
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
G
s
+
ε
16
+
2
T
b
*
∑
i
=
T
+
τ
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
T
3
+
1
∞
‍
F
t
+
2
T
b
*
∑
i
=
T
+
τ
T
1
‍
∑
s
=
T
2
+
1
∞
‍
∑
t
=
s
∞
‍
F
t
+
2
T
b
*
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
<
ε
,
w
≥
T
4
,
which implies that G
L is continuous in Ω
(
A
*
,
B
*
,
T
).
It follows from (47) and (51) that for any x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
) and t
1
,
t
2
≥
T
4
(54)
G
L
y
t
2
t
2
-
G
L
y
t
1
t
1
=
1
t
2
b
n
+
τ
∑
i
=
t
1
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
-
d
t
·
∑
i
=
t
2
+
τ
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
t
1
b
n
+
τ
·
∑
i
=
t
1
+
τ
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
t
2
b
n
+
τ
·
∑
i
=
t
2
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
t
1
b
n
+
τ
·
∑
i
=
t
1
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
t
2
b
n
+
τ
·
∑
i
=
t
2
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
-
d
t
-
1
t
1
b
n
+
τ
·
∑
i
=
t
1
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
-
d
t
≤
2
T
4
b
*
·
∑
i
=
T
4
+
τ
∞
‍
H
i
+
∑
i
=
T
4
+
τ
∞
‍
∑
s
=
i
∞
‍
G
s
+
∑
i
=
T
4
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
<
ε
,
which means that G
L
(
Ω
(
A
*
,
B
*
,
T
)
) is uniformly Cauchy, which together with (49) and Lemma 1 yields that G
L
(
Ω
(
A
*
,
B
*
,
T
)
) is relatively compact. That is, G
L is completely continuous in Ω
(
A
*
,
B
*
,
T
). Thus (44), (48), and Lemma 2 ensure that the mapping S
L
+
G
L has a fixed point x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
), which together with (46) and (47) yields that
(55)
x
n
=
n
L
b
n
+
τ
-
x
n
+
τ
b
n
+
τ
-
c
n
+
τ
b
n
+
τ
+
1
b
n
+
τ
∑
i
=
n
+
τ
∞
‍
h
i
,
x
h
1
i
,
…
,
x
h
k
i
-
1
b
n
+
τ
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
x
g
1
s
,
…
,
x
g
k
s
+
1
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
,
n
≥
T
,
which means that
(56)
x
n
+
τ
+
b
n
+
τ
x
n
+
c
n
+
τ
=
n
L
+
∑
i
=
n
+
τ
∞
‍
h
i
,
x
h
1
i
,
…
,
x
h
k
i
-
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
x
g
1
s
,
…
,
x
g
k
s
+
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
,
n
≥
T
.
It follows from (56) that
(57)
Δ
x
n
+
b
n
x
n
-
τ
+
c
n
=
L
-
h
n
,
x
h
1
n
,
…
,
x
h
k
n
+
∑
s
=
n
∞
‍
g
s
,
x
g
1
s
,
x
g
2
s
,
…
,
x
g
k
s
-
∑
s
=
n
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
,
n
≥
T
+
τ
;
Δ
3
x
n
+
b
n
x
n
-
τ
+
c
n
=
-
Δ
2
h
n
,
x
h
1
n
,
…
,
x
h
k
n
-
Δ
g
n
,
x
g
1
n
,
…
,
x
g
k
n
-
f
n
,
x
f
1
n
,
…
,
x
f
k
n
-
d
n
,
n
≥
T
+
τ
,
that is, x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
) is a positive solution of (4). By means of (15)–(17) and (56), we deduce that
(58)
x
n
+
b
n
x
n
-
τ
+
c
n
n
-
L
=
-
τ
n
+
1
n
∑
i
=
n
∞
‍
h
i
,
x
h
1
i
,
…
,
x
h
k
i
-
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
s
,
x
g
1
s
,
…
,
x
g
k
s
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
∑
i
=
n
∞
‍
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
(
t
)
≤
τ
n
+
1
n
∑
i
=
n
∞
‍
H
i
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
G
s
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
⟶
0
as
n
⟶
∞
,
which ensures that (41) holds. The proof of (19) is similar to that of Theorem 5 and is omitted.
(ii) LetL
1
,
L
2
∈
(
b
*
A
+
(
B
*
+
c
*
)
(
b
*
/
b
*
)
,
b
*
B
+
b
*
A
*
/
b
*
-
c
*
) and L
1
≠
L
2. As in the proof of (i), we deduce that, for each l
∈
{
1,2
}, there exist θ
l
∈
(
0,1
), T
l
≥
n
0
+
τ
+
α, and two mappings S
L
l and G
L
l
:
Ω
(
A
*
,
B
*
,
T
l
)
→
l
β
∞ satisfying (43)–(47), where θ, T, L, S
L, and G
L are replaced by θ
l, T
l, L
l, S
L
l, and G
L
l, respectively, and S
L
l
+
G
L
l possesses a fixed point x
l
=
{
x
l
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
l
), which is a positive solution of (4); that is,
(59)
x
l
n
=
n
L
b
n
+
τ
-
x
l
n
+
τ
b
n
+
τ
-
c
n
+
τ
b
n
+
τ
+
1
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
h
i
,
x
l
h
1
i
,
…
,
x
l
h
k
i
-
1
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
x
l
g
1
s
,
…
,
x
l
g
k
s
+
1
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
l
f
1
t
,
…
,
x
l
f
k
t
t
,
x
l
f
1
t
,
…
,
x
l
f
k
t
-
d
t
,
n
≥
T
l
.
Note that (16) and (17) imply that there exists T
*
>
max
⁡
{
T
1
,
T
2
} satisfying
(60)
1
T
*
b
*
∑
i
=
T
*
+
τ
∞
‍
H
i
+
∑
i
=
T
*
+
τ
∞
‍
∑
s
=
i
∞
‍
G
s
+
∑
i
=
T
*
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
<
L
1
-
L
2
4
b
*
.
In view of (15), (59), and (60), we infer that for any n
≥
T
*
(61)
x
1
n
n
-
x
2
n
n
=
L
1
-
L
2
b
n
+
τ
-
x
1
n
+
τ
-
x
2
n
+
τ
n
b
n
+
τ
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
h
i
,
x
1
h
1
i
,
…
,
x
1
h
k
i
-
h
i
,
x
2
h
1
i
,
…
,
x
2
h
k
i
-
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
g
(
s
,
x
1
(
g
1
(
s
)
)
,
…
,
x
1
(
g
k
(
s
)
)
)
-
g
s
,
x
2
g
1
s
,
…
,
x
2
g
k
s
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
(
t
,
x
1
(
f
1
(
t
)
)
,
…
,
x
1
(
f
k
(
t
)
)
)
x
1
n
+
τ
-
x
2
n
+
τ
n
b
n
+
τ
-
f
t
,
x
2
f
1
t
,
…
,
x
2
f
k
t
≥
L
1
-
L
2
b
*
-
n
+
τ
n
b
n
+
τ
x
1
n
+
τ
-
x
2
n
+
τ
n
+
τ
-
1
T
*
b
*
∑
i
=
T
*
+
τ
∞
‍
h
i
,
x
1
h
1
i
,
…
,
x
1
h
k
i
-
h
i
,
x
2
h
1
i
,
…
,
x
2
h
k
i
-
1
T
*
b
*
∑
i
=
T
*
+
τ
∞
‍
∑
s
=
i
∞
‍
g
(
s
,
x
1
(
g
1
(
s
)
)
,
…
,
x
1
(
g
k
(
s
)
)
)
-
g
s
,
x
2
g
1
s
,
…
,
x
2
g
k
s
-
1
T
*
b
*
∑
i
=
T
*
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
(
t
,
x
1
(
f
1
(
t
)
)
,
…
,
x
1
(
f
k
(
t
)
)
)
-
f
t
,
x
2
f
1
t
,
…
,
x
2
f
k
t
≥
L
1
-
L
2
b
*
-
1
b
*
1
+
τ
T
*
x
1
-
x
2
-
2
T
*
b
*
·
∑
i
=
T
*
+
τ
∞
‍
H
i
+
∑
i
=
T
*
+
τ
∞
‍
∑
s
=
i
∞
‍
G
s
+
∑
i
=
T
*
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
>
L
1
-
L
2
2
b
*
-
1
b
*
(
1
+
τ
T
*
)
x
1
-
x
2
,
which implies that
(62)
x
1
-
x
2
>
L
1
-
L
2
2
b
*
(
1
+
(
1
/
b
*
)
(
1
+
τ
/
T
*
)
)
>
0
,
which yields that x
1
≠
x
2. That is, (4) possesses uncountably many positive solutions in l
β
∞. This completes the proof.Theorem 7.
Assume that there exist constantsb
* and b
* and three nonnegative sequences {
F
n
}
n
∈
N
n
0, {
G
n
}
n
∈
N
n
0, and {
H
n
}
n
∈
N
n
0 satisfying (15)–(17) and
(63)
A
<
B
+
b
*
B
*
,
-
1
<
b
*
≤
b
(
n
)
≤
b
*
≤
0
,
n
∈
N
n
0
.
Then(i)
equation (4) possesses a positive solution x
=
{
x
(
n
)
}
n
∈
N
β
∈
l
β
∞ with (19) and
(64)
lim
n
→
∞
x
(
n
)
+
b
(
n
)
x
(
n
-
τ
)
+
c
(
n
)
n
∈
(
A
,
B
+
b
*
B
*
)
;
(ii)
equation (4) possesses uncountably many positive solutions in l
β
∞.Proof.
(i) LetL
∈
(
A
,
B
+
b
*
B
*
). It follows from (16) and (17) that there exists T
≥
n
0
+
τ
+
α satisfying
(65)
1
T
∑
i
=
T
∞
‍
H
i
+
∑
s
=
i
∞
‍
G
s
+
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
<
min
⁡
{
L
-
A
,
B
+
b
*
B
*
-
L
}
.
Define two mappings S
L and G
L
:
Ω
(
A
*
,
B
*
,
T
)
→
l
β
∞ by (21) and (22).
Now we show that (23), (66) below hold:
(66)
S
L
x
-
S
L
y
≤
b
*
x
-
y
,
x
,
y
∈
Ω
(
A
*
,
B
*
,
T
)
;
G
L
y
≤
B
,
y
∈
Ω
(
A
*
,
B
*
,
T
)
.
Using (15), (21), (22), (63), and (65), we get that for any x
=
{
x
(
n
)
}
n
∈
N
β
,
y
=
{
y
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
),
(67)
S
L
x
n
+
G
L
y
n
n
=
L
-
b
n
n
x
n
-
τ
-
c
n
n
+
1
n
∑
i
=
n
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
-
d
t
≤
L
-
n
-
τ
b
n
n
·
x
n
-
τ
n
-
τ
+
c
n
n
+
1
n
∑
i
=
n
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
d
t
≤
L
-
b
*
B
*
+
c
n
n
+
1
T
∑
i
=
T
∞
‍
H
i
+
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
G
s
+
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
<
L
-
b
*
B
*
+
c
n
n
+
min
⁡
⁡
L
-
A
,
B
+
b
*
B
*
-
L
≤
B
+
c
n
n
=
B
(
n
)
,
n
≥
T
,
S
L
x
n
+
G
L
y
n
n
=
n
T
·
S
L
x
T
+
G
L
y
T
n
S
L
x
n
+
G
L
y
n
n
≤
B
(
T
)
,
β
≤
n
<
T
,
S
L
x
n
+
G
L
y
n
n
=
L
-
b
n
n
x
n
-
τ
-
c
n
n
+
1
n
∑
i
=
n
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
-
d
t
≥
L
-
c
n
n
-
1
n
∑
i
=
n
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
-
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
d
t
≥
L
-
c
n
n
-
1
T
∑
i
=
T
∞
‍
H
i
-
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
G
s
-
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
>
L
-
c
n
n
-
min
⁡
⁡
{
L
-
A
,
B
+
b
*
B
*
-
L
}
≥
A
-
c
n
n
=
A
(
n
)
,
n
≥
T
,
S
L
x
n
+
G
L
y
n
n
=
n
T
·
S
L
x
T
+
G
L
y
T
n
≥
A
(
T
)
,
β
≤
n
<
T
,
S
L
x
n
-
S
L
y
n
n
=
b
n
n
-
τ
n
x
n
-
τ
-
y
n
-
τ
n
-
τ
≤
b
*
x
-
y
,
n
≥
T
,
S
L
x
n
-
S
L
y
n
n
=
n
T
S
L
x
T
-
S
L
y
T
n
≤
b
*
x
-
y
,
β
≤
n
<
T
,
G
L
y
n
n
=
1
n
∑
i
=
n
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
[
f
(
t
,
y
(
f
1
(
t
)
)
,
…
,
y
(
f
k
(
t
)
)
)
-
d
(
t
)
]
≤
1
n
∑
i
=
n
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
d
t
≤
1
T
∑
i
=
T
∞
‍
H
i
+
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
G
s
+
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
<
min
⁡
{
L
-
A
,
B
+
b
*
B
*
-
L
}
≤
B
,
n
≥
T
,
G
L
y
n
n
=
n
T
G
L
y
T
n
≤
B
,
β
≤
n
<
T
,
which yield (21) and (66). The rest of the proof is similar to that of Theorem 5. This completes the proof.Theorem 8.
Assume that there exist constantsb
* and b
* and three nonnegative sequences {
F
n
}
n
∈
N
n
0, {
G
n
}
n
∈
N
n
0, and {
H
n
}
n
∈
N
n
0 satisfying (15)–(17) and
(68)
b
*
B
+
B
*
+
c
*
<
b
*
A
+
A
*
-
b
*
c
*
b
*
<
0
,
b
*
≤
b
n
≤
b
*
<
-
1
,
n
∈
N
n
0
.
Then(i)
equation (4) possesses a positive solution x
=
{
x
(
n
)
}
n
∈
N
β
∈
l
β
∞ with (19) and
(69)
lim
n
→
∞
x
n
+
b
n
x
n
-
τ
+
c
n
n
∈
(
b
*
B
+
B
*
+
c
*
,
b
*
A
+
A
*
-
b
*
c
*
b
*
)
;
(ii)
equation (4) possesses uncountably many positive solutions in l
β
∞.Proof.
(i) LetL
∈
(
b
*
B
+
(
B
*
+
c
*
)
,
b
*
A
+
A
*
-
b
*
c
*
/
b
*
). Notice that
(70)
lim
n
→
∞
b
*
B
+
B
*
+
c
*
1
+
τ
n
=
b
*
B
+
B
*
+
c
*
<
L
<
b
*
A
+
A
*
-
b
*
c
*
b
*
=
lim
n
→
∞
[
b
*
A
+
A
*
-
b
*
c
*
b
*
(
1
+
τ
n
)
]
,
which means that there exists N
∈
N satisfying
(71)
b
*
B
+
B
*
+
c
*
<
b
*
B
+
B
*
+
c
*
1
+
τ
N
<
L
<
b
*
A
+
A
*
-
b
*
c
*
b
*
1
+
τ
N
<
b
*
A
+
A
*
-
b
*
c
*
b
*
.
It follows from (16) and (17) that there exist θ
∈
(
0,1
) and T
>
max
⁡
{
N
,
n
0
+
τ
+
α
} satisfying
(72)
θ
=
1
b
*
(
1
+
τ
T
)
,
1
T
∑
i
=
T
+
τ
∞
‍
H
i
+
∑
s
=
i
∞
‍
G
s
+
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
<
min
⁡
⁡
L
-
b
*
B
-
1
+
τ
N
B
*
+
c
*
,
b
*
A
+
b
*
A
*
b
*
-
c
*
(
1
+
τ
N
)
-
b
*
b
*
L
<
min
⁡
⁡
L
-
b
*
B
-
(
1
+
τ
T
)
(
B
*
+
c
*
)
,
b
*
A
+
b
*
A
*
b
*
-
c
*
(
1
+
τ
T
)
-
b
*
b
*
L
.
Define two mappings S
L and G
L
:
Ω
(
A
*
,
B
*
,
T
)
→
l
β
∞ by (46) and (47).
Now we show that (23), (25), and (48) hold. Using (15), (46), (47), (68), and (72), we get that for any x
=
{
x
(
n
)
}
n
∈
N
β
,
y
=
{
y
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
)
(73)
S
L
x
n
+
G
L
y
n
n
=
L
b
n
+
τ
-
x
n
+
τ
n
b
n
+
τ
-
c
n
+
τ
n
b
n
+
τ
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
-
d
t
≤
L
b
*
-
n
+
τ
n
b
n
+
τ
·
x
n
+
τ
n
+
τ
-
n
+
τ
n
b
n
+
τ
·
c
n
+
τ
n
+
τ
-
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
-
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
d
t
≤
L
b
*
-
1
b
*
1
+
τ
T
B
*
+
c
*
b
*
1
+
τ
T
-
1
T
b
*
∑
i
=
T
+
τ
∞
‍
H
i
-
1
T
b
*
∑
i
=
T
+
τ
∞
∑
s
=
i
∞
‍
G
s
-
1
T
b
*
∑
i
=
T
+
τ
∞
∑
s
=
i
∞
∑
t
=
s
∞
‍
F
t
+
d
t
<
L
b
*
-
1
b
*
1
+
τ
T
B
*
+
c
*
-
1
b
*
min
⁡
⁡
L
-
b
*
B
-
1
+
τ
T
B
*
+
c
*
,
1
+
τ
T
-
b
*
b
*
b
*
A
+
b
*
A
*
b
*
-
c
*
1
+
τ
T
-
b
*
b
*
L
≤
B
≤
B
(
n
)
,
n
≥
T
,
S
L
x
n
+
G
L
y
n
n
=
n
T
·
S
L
x
T
+
G
L
y
T
n
≤
B
(
T
)
,
β
≤
n
<
T
,
S
L
x
n
+
G
L
y
n
n
=
L
b
n
+
τ
-
x
n
+
τ
n
b
n
+
τ
-
c
n
+
τ
n
b
n
+
τ
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
-
d
t
≥
L
b
*
-
n
+
τ
n
b
n
+
τ
·
x
n
+
τ
n
+
τ
+
n
+
τ
n
b
n
+
τ
·
c
n
+
τ
n
+
τ
+
1
n
b
n
+
τ
∑
i
=
n
+
τ
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
d
t
≥
L
b
*
-
A
*
b
*
+
c
*
b
*
(
1
+
τ
T
)
+
1
T
b
*
∑
i
=
T
+
τ
∞
‍
H
i
+
1
T
b
*
∑
i
=
T
+
τ
∞
‍
∑
s
=
i
∞
‍
G
s
+
1
T
b
*
∑
i
=
T
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
≥
L
b
*
-
A
*
b
*
+
c
*
b
*
1
+
τ
T
+
1
b
*
min
⁡
⁡
L
-
b
*
B
-
(
1
+
τ
T
)
(
B
*
+
c
*
)
,
b
*
A
+
b
*
A
*
b
*
-
c
*
(
1
+
τ
T
)
-
b
*
b
*
L
≥
A
≥
A
(
n
)
,
n
≥
T
,
S
L
x
n
+
G
L
y
n
n
=
n
T
·
S
L
x
T
+
G
L
y
T
n
≥
A
(
T
)
,
β
≤
n
<
T
,
S
L
x
n
-
S
L
y
n
n
=
n
+
τ
n
b
n
+
τ
x
n
+
τ
-
y
n
+
τ
n
+
τ
≤
θ
x
-
y
,
n
≥
T
,
S
L
x
n
-
S
L
y
n
n
=
n
T
S
L
x
T
-
S
L
y
T
n
≤
θ
x
-
y
,
β
≤
n
<
T
,
G
L
y
n
n
=
1
n
b
n
+
τ
∑
i
=
n
+
τ
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
n
b
n
+
τ
∑
i
=
n
+
τ
∞
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
-
d
t
≤
1
n
b
n
+
τ
∑
i
=
n
+
τ
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
+
1
n
b
n
+
τ
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
d
t
≤
1
T
b
*
∑
i
=
T
+
τ
∞
‍
H
i
+
1
T
b
*
∑
i
=
T
+
τ
∞
‍
∑
s
=
i
∞
‍
G
s
+
1
T
b
*
·
∑
i
=
T
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
≤
-
1
b
*
min
⁡
⁡
L
-
b
*
B
-
1
+
τ
T
B
*
+
c
*
,
b
*
b
*
b
*
A
+
b
*
A
*
b
*
-
c
*
1
+
τ
T
-
b
*
b
*
L
≤
B
,
n
≥
T
,
G
L
y
n
n
=
n
T
G
L
y
T
n
≤
B
,
β
≤
n
<
T
,
which yield (23), (25), and (48).
Next we show thatG
L is completely continuous. Let y
w
=
{
y
w
(
n
)
}
n
∈
N
β and y
=
{
y
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
) with (27). Using (16), (17), (27), and the continuity of f, g, and h, we know that for given ε
>
0, there exist T
1, T
2, T
3, and T
4
∈
N with T
4
>
T
3
>
T
2
>
T
1
>
T
+
τ satisfying
(74)
1
T
b
*
max
⁡
⁡
∑
i
=
T
1
+
1
∞
‍
H
i
+
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
G
s
+
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
,
∑
i
=
T
+
τ
T
1
‍
∑
s
=
T
2
+
1
∞
‍
G
s
+
∑
i
=
T
+
τ
T
1
‍
∑
s
=
T
2
+
1
∞
‍
∑
t
=
s
∞
‍
F
t
+
∑
i
=
T
+
τ
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
T
3
+
1
∞
‍
F
t
<
ε
16
,
1
T
b
*
max
⁡
⁡
∑
i
=
T
+
τ
T
1
‍
h
i
,
y
w
h
1
i
,
…
,
y
w
h
k
i
-
h
i
,
y
h
1
i
,
…
,
y
h
k
i
,
∑
i
=
T
+
τ
T
1
∑
s
=
i
T
2
‍
g
s
,
y
w
g
1
s
,
…
,
y
w
g
k
s
-
g
s
,
y
g
1
s
,
…
,
y
g
k
s
,
∑
i
=
T
+
τ
T
1
∑
s
=
i
T
2
∑
t
=
s
T
3
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
∑
i
=
T
+
τ
T
1
‍
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
<
ε
16
,
w
≥
T
4
.
Combining (15), (47), and (74), we infer that
(75)
G
L
y
w
-
G
L
y
=
sup
⁡
⁡
G
L
y
w
n
-
G
L
y
n
n
:
n
∈
N
β
=
max
⁡
⁡
sup
⁡
⁡
n
T
·
G
L
y
w
T
-
G
L
y
T
n
:
β
≤
n
<
T
,
sup
⁡
⁡
G
L
y
w
n
-
G
L
y
n
n
:
n
∈
N
T
≤
sup
⁡
⁡
1
n
b
n
+
τ
∑
i
=
n
+
τ
∞
‍
·
∑
i
=
n
+
τ
∞
‍
h
i
,
y
w
h
1
i
,
…
,
y
w
h
k
i
-
h
i
,
y
h
1
i
,
…
,
y
h
k
i
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
∑
s
=
i
∞
‍
g
s
,
y
w
g
1
s
,
…
,
y
w
g
k
s
-
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
n
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
:
∑
i
=
n
+
τ
∞
‍
n
∈
N
T
≤
1
T
b
*
∑
i
=
T
+
τ
T
1
‍
h
i
,
y
w
h
1
i
,
…
,
y
w
h
k
i
-
h
i
,
y
h
1
i
,
…
,
y
h
k
i
+
1
T
b
*
∑
i
=
T
1
+
1
∞
‍
h
i
,
y
w
h
1
i
,
…
,
y
w
h
k
i
-
h
i
,
y
h
1
i
,
…
,
y
h
k
i
+
1
T
b
*
·
∑
i
=
T
+
τ
T
1
‍
∑
s
=
i
T
2
‍
g
s
,
y
w
g
1
s
,
…
,
y
w
g
k
s
-
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
T
b
*
·
∑
i
=
T
+
τ
T
1
∑
s
=
T
2
+
1
∞
‍
g
s
,
y
w
g
1
s
,
…
,
y
w
g
k
s
-
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
T
b
*
·
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
w
g
1
s
,
…
,
y
w
g
k
s
-
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
T
b
*
·
∑
i
=
T
+
τ
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
s
T
3
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
1
T
b
*
·
∑
i
=
T
+
τ
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
T
3
+
1
∞
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
1
T
b
*
·
∑
i
=
T
+
τ
T
1
‍
∑
s
=
T
2
+
1
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
+
1
T
b
*
·
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
w
f
1
t
,
…
,
y
w
f
k
t
-
f
t
,
y
f
1
t
,
…
,
y
f
k
t
<
ε
16
+
2
T
b
*
∑
i
=
T
1
+
1
∞
‍
H
i
+
ε
16
+
2
T
b
*
∑
i
=
T
+
τ
T
1
∑
s
=
T
2
+
1
∞
‍
G
s
+
2
T
b
*
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
G
s
+
ε
16
+
2
T
b
*
∑
i
=
T
+
τ
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
T
3
+
1
∞
‍
F
t
+
2
T
b
*
∑
i
=
T
+
τ
T
1
‍
∑
s
=
T
2
+
1
∞
‍
∑
t
=
s
∞
‍
F
t
+
2
T
b
*
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
<
ε
,
w
≥
T
4
,
which implies that G
L is continuous in Ω
(
A
*
,
B
*
,
T
).
It follows from (47) and (68) that for any x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
) and t
1
,
t
2
≥
T
4
(76)
G
L
y
t
2
t
2
-
G
L
y
t
1
t
1
=
1
t
2
b
n
+
τ
∑
i
=
t
2
+
τ
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
t
1
b
n
+
τ
∑
i
=
t
1
+
τ
∞
‍
h
i
,
y
h
1
i
,
…
,
y
h
k
i
-
1
t
2
b
n
+
τ
∑
i
=
t
2
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
t
1
b
n
+
τ
∑
i
=
t
1
+
τ
∞
‍
∑
s
=
i
∞
‍
g
s
,
y
g
1
s
,
…
,
y
g
k
s
+
1
t
2
b
n
+
τ
·
∑
i
=
t
2
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
-
d
t
-
1
t
1
b
n
+
τ
·
∑
i
=
t
1
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
y
f
1
t
,
…
,
y
f
k
t
-
d
t
≤
2
T
4
b
*
·
∑
i
=
T
4
+
τ
∞
‍
H
i
+
∑
i
=
T
4
+
τ
∞
‍
∑
s
=
i
∞
‍
G
s
+
∑
i
=
T
4
+
τ
∞
∑
s
=
i
∞
∑
t
=
s
∞
‍
F
t
+
d
t
<
ε
,
which means that G
L
(
Ω
(
A
*
,
B
*
,
T
)
) is uniformly Cauchy, which together with (25) and Lemma 1 yields that G
L
(
Ω
(
A
*
,
B
*
,
T
)
) is relatively compact. That is, G
L is completely continuous in Ω
(
A
*
,
B
*
,
T
). Thus (25), (48), and Lemma 2 ensure that the mapping S
L
+
G
L has a fixed point x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
), which together with (46) and (47) implies that
(77)
x
n
=
n
L
b
n
+
τ
-
x
n
+
τ
b
n
+
τ
-
c
n
+
τ
b
n
+
τ
+
1
b
n
+
τ
·
∑
i
=
n
+
τ
∞
‍
h
i
,
x
h
1
i
,
…
,
x
h
k
i
-
1
b
n
+
τ
·
∑
i
=
n
+
τ
∞
∑
s
=
i
∞
‍
g
s
,
x
g
1
s
,
…
,
x
g
k
s
+
1
b
n
+
τ
·
∑
i
=
n
+
τ
∞
∑
s
=
i
∞
∑
t
=
s
∞
‍
[
f
(
t
,
x
(
f
1
(
t
)
)
,
…
,
x
(
f
k
(
t
)
)
)
-
d
(
t
)
]
,
n
≥
T
,
That is, x
=
{
x
(
n
)
}
n
∈
N
β is a positive solution of (4). The rest of the proof is similar to that of Theorem 6 and is omitted. This completes the proof.Theorem 9.
Assume that there exist three nonnegative sequences{
F
n
}
n
∈
N
n
0, {
G
n
}
n
∈
N
n
0, and {
H
n
}
n
∈
N
n
0 satisfying (15),
(78)
lim
n
→
∞
1
n
max
⁡
∑
t
=
n
∞
‍
t
H
t
,
∑
t
=
n
∞
‍
t
2
G
t
=
0
,
(79)
lim
n
→
∞
1
n
∑
t
=
n
∞
‍
t
3
max
⁡
{
F
t
,
d
t
}
=
0
,
(80)
lim
n
→
∞
1
n
∑
i
=
1
∞
‍
c
n
+
i
τ
=
0
,
(81)
b
(
n
)
=
-
1
,
n
∈
N
n
0
.
Then(i)
equation (4) possesses a positive solution x
=
{
x
(
n
)
}
n
∈
N
β
∈
l
β
∞ with (19) and
(82)
lim
n
→
∞
x
(
n
)
-
x
(
n
-
τ
)
+
c
(
n
)
n
=
0
;
(ii)
equation (4) possesses uncountably many positive solutions in l
β
∞.Proof.
(i) LetL
∈
(
A
,
B
). It follows from (78)–(80) that there exists T
≥
n
0
+
τ
+
α satisfying
(83)
1
n
∑
i
=
1
∞
‍
c
n
+
i
τ
<
1
2
min
⁡
B
-
L
,
L
-
A
,
n
∈
N
T
,
(84)
1
T
∑
t
=
T
∞
‍
t
H
t
+
1
T
∑
t
=
T
∞
‍
t
2
G
t
+
1
T
∑
t
=
T
∞
‍
t
3
F
t
+
d
t
<
1
2
min
⁡
{
B
-
L
,
L
-
A
}
.
Define a mapping S
L
:
Ω
(
A
*
,
B
*
,
T
)
→
l
β
∞ by
(85)
S
L
x
n
=
n
L
+
∑
i
=
1
∞
‍
c
n
+
i
τ
-
∑
i
=
1
∞
‍
∑
t
=
n
+
i
τ
∞
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
∑
i
=
1
∞
‍
∑
s
=
n
+
i
τ
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
-
∑
p
=
1
∞
‍
∑
i
=
n
+
p
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
,
n
≥
T
,
n
T
S
L
x
T
,
β
≤
n
<
T
,
for any x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
).
Now we show that(86)
S
L
x
∈
Ω
(
A
*
,
B
*
,
T
)
,
x
∈
Ω
(
A
*
,
B
*
,
T
)
;
(87)
S
L
x
≤
B
,
x
∈
Ω
(
A
*
,
B
*
,
T
)
.
It follows from (15), (83)–(85), and Lemma 4 that for any x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
)
(88)
S
L
x
n
n
-
L
=
1
n
∑
i
=
1
∞
‍
c
(
n
+
i
τ
)
-
1
n
∑
i
=
1
∞
‍
∑
t
=
n
+
i
τ
∞
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
1
n
∑
i
=
1
∞
‍
∑
s
=
n
+
i
τ
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
-
1
n
∑
p
=
1
∞
∑
i
=
n
+
p
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
∑
i
=
1
∞
‍
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
(
t
)
≤
1
n
∑
i
=
1
∞
‍
c
n
+
i
τ
+
1
n
∑
i
=
1
∞
‍
∑
t
=
n
+
i
τ
∞
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
1
n
∑
i
=
1
∞
‍
∑
s
=
n
+
i
τ
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
n
∑
p
=
1
∞
‍
∑
i
=
n
+
p
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
f
t
,
x
f
1
t
,
…
,
x
f
k
t
+
d
t
<
1
2
min
⁡
⁡
B
-
L
,
L
-
A
+
1
T
∑
t
=
T
∞
‍
t
H
t
+
1
T
∑
t
=
T
∞
‍
t
2
G
t
+
1
T
∑
t
=
T
∞
‍
t
3
F
t
+
d
t
<
min
⁡
{
B
-
L
,
L
-
A
}
,
n
≥
T
,
S
L
x
n
n
-
L
=
n
T
·
S
L
x
T
n
-
L
S
L
x
n
n
-
L
<
min
⁡
{
B
-
L
,
L
-
A
}
,
β
≤
n
<
T
,
which yields that
(89)
A
n
≤
A
≤
L
-
min
⁡
⁡
B
-
L
,
L
-
A
<
S
L
x
n
n
<
L
+
min
⁡
{
B
-
L
,
L
-
A
}
≤
B
≤
B
(
n
)
,
n
∈
N
β
;
that is, (86) and (87) hold.
Next we show thatS
L is continuous and S
L
(
Ω
(
A
*
,
B
*
,
T
)
) is uniformly Cauchy. Let x
w
=
{
x
w
(
n
)
}
n
∈
N
β and x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
) with
(90)
lim
w
→
∞
x
w
=
x
.
Using (15), (78), and (80) the continuity of f, g, and h, we know that for given ε
>
0, there exist T
2
>
T
1
>
T satisfying
(91)
1
n
∑
i
=
1
∞
‍
c
n
+
i
τ
<
ε
16
,
∀
n
∈
N
T
1
,
1
T
max
⁡
⁡
∑
t
=
T
T
1
‍
t
h
t
,
x
w
h
1
t
,
…
,
x
w
h
k
t
-
h
t
,
x
h
1
t
,
…
,
x
h
k
t
,
∑
t
=
T
T
1
‍
t
2
g
t
,
x
w
g
1
t
,
…
,
x
w
g
k
t
-
g
t
,
x
g
1
t
,
…
,
x
g
k
t
,
∑
t
=
T
T
1
‍
t
3
f
t
,
x
w
f
1
t
,
…
,
x
w
f
k
t
∑
t
=
T
T
1
‍
-
f
t
,
x
f
1
t
,
…
,
x
f
k
t
<
ε
16
,
w
≥
T
2
,
1
T
∑
t
=
T
1
+
1
∞
‍
t
H
t
+
∑
t
=
T
1
+
1
∞
‍
t
2
G
t
+
∑
t
=
T
1
+
1
∞
‍
t
3
F
t
+
d
t
<
ε
16
.
Combining (15), (91), and Lemma 4, we infer that
(92)
S
L
x
w
-
S
L
x
=
sup
⁡
⁡
S
L
x
w
n
-
S
L
x
n
n
:
n
∈
N
β
=
max
⁡
⁡
sup
⁡
⁡
n
T
S
L
x
w
T
-
S
L
x
T
n
:
β
≤
n
<
T
,
sup
⁡
⁡
S
L
x
w
n
-
S
L
x
n
n
:
n
∈
N
T
≤
sup
⁡
⁡
1
n
∑
i
=
1
∞
‍
∑
t
=
n
+
i
τ
∞
‍
h
t
,
x
w
h
1
t
,
…
,
x
w
h
k
t
-
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
1
n
∑
i
=
1
∞
‍
∑
s
=
n
+
i
τ
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
w
g
1
t
,
…
,
x
w
g
k
t
-
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
n
∑
p
=
1
∞
∑
i
=
n
+
p
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
w
f
1
t
,
…
,
x
w
f
k
t
-
f
t
,
x
f
1
t
,
…
,
x
f
k
t
:
∑
i
=
1
∞
‍
n
∈
N
T
≤
1
T
∑
t
=
T
+
τ
∞
‍
t
h
t
,
x
w
h
1
t
,
…
,
x
w
h
k
t
-
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
1
T
∑
t
=
T
+
τ
∞
‍
t
2
g
t
,
x
w
g
1
t
,
…
,
x
w
g
k
t
-
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
T
∑
t
=
T
+
τ
∞
‍
t
3
f
t
,
x
w
f
1
t
,
…
,
x
w
f
k
t
-
f
t
,
x
f
1
t
,
…
,
x
f
k
t
≤
1
T
∑
t
=
T
T
1
‍
t
h
t
,
x
w
h
1
t
,
…
,
x
w
h
k
t
-
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
1
T
∑
t
=
T
1
+
1
∞
‍
t
h
t
,
x
w
h
1
t
,
…
,
x
w
h
k
t
-
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
1
T
∑
t
=
T
T
1
‍
t
2
g
t
,
x
w
g
1
t
,
…
,
x
w
g
k
t
-
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
T
∑
t
=
T
1
+
1
∞
‍
t
2
g
t
,
x
w
g
1
t
,
…
,
x
w
g
k
t
-
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
T
∑
t
=
T
T
1
‍
t
3
f
t
,
x
w
f
1
t
,
…
,
x
w
f
k
t
-
f
t
,
x
f
1
t
,
…
,
x
f
k
t
+
1
T
∑
t
=
T
1
+
1
∞
‍
t
3
f
t
,
x
w
f
1
t
,
…
,
x
w
f
k
t
-
f
t
,
x
f
1
t
,
…
,
x
f
k
t
<
ε
16
+
2
T
∑
t
=
T
1
+
1
∞
‍
t
H
t
+
ε
16
+
2
T
∑
t
=
T
1
+
1
∞
‍
t
2
G
t
+
ε
16
+
2
T
∑
i
=
T
1
+
1
∞
‍
t
3
F
t
<
ε
,
w
≥
T
2
,
which implies that S
L is continuous in Ω
(
A
*
,
B
*
,
T
). It follows from (80), (85), and Lemma 4 that for any x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
) and t
1
>
t
2
≥
T
2
(93)
S
L
x
t
1
t
1
-
S
L
x
t
2
t
2
=
1
t
1
∑
i
=
1
∞
‍
c
(
t
1
+
i
τ
)
-
1
t
2
∑
i
=
1
∞
‍
c
(
t
2
+
i
τ
)
-
1
t
1
∑
i
=
1
∞
‍
∑
t
=
t
1
+
i
τ
∞
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
1
t
2
∑
i
=
1
∞
‍
∑
t
=
t
2
+
i
τ
∞
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
1
t
1
∑
i
=
1
∞
‍
∑
s
=
t
1
+
i
τ
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
-
1
t
2
∑
i
=
1
∞
‍
∑
s
=
t
2
+
i
τ
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
-
1
t
1
∑
p
=
1
∞
‍
∑
i
=
t
1
+
p
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
(
t
)
+
1
t
2
∑
p
=
1
∞
‍
∑
i
=
t
2
+
p
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
∑
i
=
1
∞
‍
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
≤
1
t
1
∑
i
=
1
∞
‍
c
t
1
+
i
τ
+
1
t
2
∑
i
=
1
∞
‍
c
t
2
+
i
τ
+
1
t
1
∑
t
=
t
1
∞
‍
t
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
1
t
2
∑
t
=
t
2
∞
‍
t
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
1
t
1
∑
t
=
t
1
∞
‍
t
2
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
t
2
∑
t
=
t
2
∞
‍
t
2
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
t
1
∑
t
=
t
1
∞
‍
t
3
f
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
+
1
t
2
∑
t
=
t
2
∞
‍
t
3
f
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
<
2
ε
16
+
2
T
2
∑
t
=
T
2
∞
‍
t
H
t
+
2
T
2
∑
t
=
T
2
∞
‍
t
2
G
t
+
2
T
2
∑
t
=
T
2
∞
‍
t
3
[
F
t
+
d
t
]
<
ε
,
which means that S
L
(
Ω
(
A
*
,
B
*
,
T
)
) is uniformly Cauchy, which together with (87) and Lemma 1 yields that S
L
(
Ω
(
A
*
,
B
*
,
T
)
) is relatively compact. It follows from Lemma 3 that the mapping S
L has a fixed point x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
); that is,
(94)
x
n
=
n
L
+
∑
i
=
1
∞
‍
c
n
+
i
τ
-
∑
i
=
1
∞
‍
∑
t
=
n
+
i
τ
∞
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
∑
i
=
1
∞
‍
∑
s
=
n
+
i
τ
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
-
∑
p
=
1
∞
‍
∑
i
=
n
+
p
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
,
n
≥
T
,
which gives that
(95)
x
n
-
x
n
-
τ
=
τ
L
-
c
n
+
∑
t
=
n
∞
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
-
∑
s
=
n
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
,
n
≥
T
+
τ
.
It is easy to verify that (95) implies that
(96)
Δ
x
n
-
x
n
-
τ
+
c
n
=
-
h
n
,
x
h
1
n
,
…
,
x
h
k
n
+
∑
t
=
n
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
-
∑
s
=
n
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
,
n
≥
T
+
τ
,
Δ
2
x
n
-
x
n
-
τ
+
c
n
=
-
Δ
h
n
,
x
h
1
n
,
…
,
x
h
k
n
-
g
n
,
x
g
1
n
,
…
,
x
g
k
n
+
∑
t
=
n
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
,
n
≥
T
+
τ
,
which yields that
(97)
Δ
3
x
n
-
x
n
-
τ
+
c
n
=
-
Δ
2
h
n
,
x
h
1
n
,
…
,
x
h
k
n
-
Δ
g
n
,
x
g
1
n
,
…
,
x
g
k
n
-
f
n
,
x
f
1
n
,
…
,
x
f
k
n
-
d
n
,
n
≥
T
+
τ
,
which together with (81) gives that x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
) is a positive solution of (4). It follows from (78), (79), (95), and Lemma 4 that
(98)
x
n
-
x
n
-
τ
+
c
n
n
=
τ
L
n
+
1
n
∑
t
=
n
∞
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
-
1
n
∑
s
=
n
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
∑
t
=
n
∞
‍
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
≤
τ
L
n
+
1
n
∑
t
=
n
∞
‍
H
t
+
1
n
∑
s
=
n
∞
‍
∑
t
=
s
∞
‍
G
t
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
H
t
+
d
t
≤
τ
L
n
+
1
n
∑
t
=
n
∞
‍
H
t
+
1
n
∑
t
=
n
∞
‍
t
G
t
+
1
n
∑
t
=
n
∞
‍
t
2
H
t
+
d
t
⟶
0
as
n
⟶
∞
;
that is, (82) holds. The proof of (19) is similar to that of Theorem 5 and is omitted.
(ii) LetL
1
,
L
2
∈
(
A
,
B
) and L
1
≠
L
2. Similarly we conclude that for each l
∈
{
1,2
}, there exist a constant T
l
≥
n
1
+
τ
+
|
α
| and a mapping S
L
l
:
Ω
(
A
*
,
B
*
,
T
l
)
→
l
β
∞ satisfying (83)–(87), where T, L, and S
L are replaced by T
l, L
l, and S
L
l, respectively, and S
L
l possesses a fixed point x
l
=
{
x
l
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
l
), which is a positive solution of (4); that is,
(99)
x
l
n
=
n
L
l
+
∑
i
=
1
∞
‍
c
n
+
i
τ
-
∑
i
=
1
∞
‍
∑
t
=
n
+
i
τ
∞
‍
h
t
,
x
l
h
1
t
,
…
,
x
l
h
k
t
+
∑
i
=
1
∞
‍
∑
s
=
n
+
i
τ
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
l
g
1
t
,
…
,
x
l
g
k
t
-
∑
p
=
1
∞
‍
∑
i
=
n
+
p
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
l
f
1
t
,
…
,
x
l
f
k
t
t
,
x
l
f
1
t
,
…
,
x
l
f
k
t
-
d
(
t
)
,
n
≥
T
l
.
Note that (79) and (80) imply that there exists T
*
>
max
⁡
{
T
1
,
T
2
} satisfying
(100)
1
T
*
∑
t
=
T
*
∞
‍
t
H
t
+
∑
t
=
T
*
∞
‍
t
2
G
t
+
∑
t
=
T
*
∞
‍
t
3
F
t
<
L
1
-
L
2
4
.
In view of (15), (99), (100), and Lemma 4, we infer that for any n
≥
T
*
(101)
x
1
n
n
-
x
2
n
n
=
L
1
-
L
2
∑
i
=
1
∞
‍
-
1
n
∑
i
=
1
∞
‍
∑
t
=
n
+
i
τ
∞
‍
h
t
,
x
1
h
1
t
,
…
,
x
1
h
k
t
-
h
t
,
x
2
h
1
t
,
…
,
x
2
h
k
t
+
1
n
∑
i
=
1
∞
‍
∑
s
=
n
+
i
τ
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
1
g
1
t
,
…
,
x
1
g
k
t
-
g
t
,
x
2
g
1
t
,
…
,
x
2
g
k
t
-
1
n
∑
p
=
1
∞
‍
∑
i
=
n
+
p
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
1
f
1
t
,
…
,
x
1
f
k
t
∑
i
=
1
∞
‍
-
f
t
,
x
2
f
1
t
,
…
,
x
2
f
k
t
≥
L
1
-
L
2
-
1
T
*
∑
i
=
1
∞
‍
∑
t
=
T
*
+
i
τ
∞
‍
h
t
,
x
1
h
1
t
,
…
,
x
1
h
k
t
-
h
t
,
x
2
h
1
t
,
…
,
x
2
h
k
t
-
1
T
*
∑
i
=
1
∞
‍
∑
s
=
T
*
+
i
τ
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
1
g
1
t
,
…
,
x
1
g
k
t
-
g
t
,
x
2
g
1
t
,
…
,
x
2
g
k
t
-
1
T
*
∑
p
=
1
∞
‍
∑
i
=
T
*
+
p
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
1
f
1
t
,
…
,
x
1
f
k
t
-
f
t
,
x
2
f
1
t
,
x
2
f
2
t
,
…
,
t
,
x
1
f
1
t
,
…
,
x
1
f
k
t
x
2
f
k
t
≥
L
1
-
L
2
-
2
T
*
∑
t
=
T
*
∞
‍
t
H
t
+
∑
t
=
T
*
∞
‍
t
2
G
t
+
∑
t
=
T
*
∞
‍
t
3
F
t
>
L
1
-
L
2
2
>
0
,
which yields that x
1
≠
x
2. Thus (4) possesses uncountably many positive solutions in Ω
(
A
*
,
B
*
,
T
). This completes the proof.Theorem 10.
Assume that there exist three nonnegative sequences{
F
n
}
n
∈
N
n
0, {
G
n
}
n
∈
N
n
0, and {
H
n
}
n
∈
N
n
0 satisfying (15)–(17), (80), and
(102)
b
n
=
1
,
n
∈
N
n
0
.
Then(i)
equation (4) possesses uncountably many positive solutions x
=
{
x
(
n
)
}
n
∈
N
β
∈
l
β
∞ with (19) and
(103)
lim
n
→
∞
x
n
+
x
n
-
τ
+
c
n
n
∈
(
2
A
,
2
B
)
;
(ii)
equation (4) possesses uncountably many positive solutions in l
β
∞.Proof.
(i) LetL
∈
(
A
,
B
). It follows from (15)–(17) and (80) that there exists T
≥
n
0
+
τ
+
α satisfying (83) and
(104)
1
T
∑
t
=
T
∞
‍
H
t
+
1
T
∑
s
=
T
∞
‍
∑
t
=
s
∞
‍
G
t
+
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
<
1
2
min
⁡
{
L
-
A
,
B
-
L
}
.
Define a mapping S
L
:
Ω
(
A
*
,
B
*
,
T
)
→
l
β
∞ by
(105)
S
L
x
n
=
n
L
+
∑
i
=
1
∞
‍
-
1
i
c
(
n
+
i
τ
)
+
∑
s
=
1
∞
‍
∑
t
=
n
+
2
s
-
1
τ
n
+
2
s
τ
-
1
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
-
∑
i
=
1
∞
‍
∑
s
=
n
+
2
i
-
1
τ
n
+
2
i
τ
-
1
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
∑
p
=
1
∞
‍
∑
i
=
n
+
(
2
p
-
1
)
τ
n
+
2
p
τ
-
1
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
(
t
)
,
n
≥
T
,
n
T
S
L
x
T
,
β
≤
n
<
T
for any x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
).
Now we show that (86) and (87) hold. It follows from (15), (83), (104), and (105) that for any x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
)
(106)
S
L
x
n
n
-
L
=
1
n
∑
i
=
1
∞
‍
-
1
i
c
(
n
+
i
τ
)
+
1
n
∑
s
=
1
∞
‍
∑
t
=
n
+
(
2
s
-
1
)
τ
n
+
2
s
τ
-
1
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
-
1
n
∑
i
=
1
∞
‍
∑
s
=
n
+
(
2
i
-
1
)
τ
n
+
2
i
τ
-
1
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
n
∑
p
=
1
∞
‍
∑
i
=
n
+
(
2
p
-
1
)
τ
n
+
2
p
τ
-
1
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
∑
s
=
1
∞
‍
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
≤
1
n
∑
i
=
1
∞
‍
c
n
+
i
τ
+
1
n
∑
s
=
1
∞
‍
∑
t
=
n
+
(
2
s
-
1
)
τ
n
+
2
s
τ
-
1
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
1
n
∑
i
=
1
∞
‍
∑
s
=
n
+
(
2
i
-
1
)
τ
n
+
2
i
τ
-
1
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
n
∑
p
=
1
∞
‍
∑
i
=
n
+
(
2
p
-
1
)
τ
n
+
2
p
τ
-
1
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
f
t
,
x
f
1
t
,
…
,
x
f
k
t
f
t
,
x
f
1
t
,
…
,
x
f
k
t
x
f
k
t
+
d
t
<
1
2
min
⁡
⁡
L
-
A
,
B
-
L
+
1
T
∑
t
=
T
∞
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
1
T
∑
s
=
T
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
f
t
,
x
f
1
t
,
…
,
x
f
k
t
+
d
t
≤
1
2
min
⁡
⁡
L
-
A
,
B
-
L
+
1
T
∑
t
=
T
∞
‍
H
t
+
1
T
∑
s
=
T
∞
‍
∑
t
=
s
∞
‍
G
t
+
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
<
min
⁡
L
-
A
,
B
-
L
,
n
≥
T
,
S
L
x
n
n
-
L
=
n
T
·
S
L
x
T
n
-
L
<
min
⁡
{
L
-
A
,
B
-
L
}
,

β
≤
n
<
T
,
S
L
x
n
n
=
L
+
1
n
∑
i
=
1
∞
‍
-
1
i
c
(
n
+
i
τ
)
+
1
n
∑
s
=
1
∞
‍
∑
t
=
n
+
2
s
-
1
τ
n
+
2
s
τ
-
1
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
-
1
n
∑
i
=
1
∞
‍
∑
s
=
n
+
2
i
-
1
τ
n
+
2
i
τ
-
1
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
n
∑
p
=
1
∞
‍
∑
i
=
n
+
2
p
-
1
τ
n
+
2
p
τ
-
1
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
∑
t
=
s
∞
‍
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
≤
L
+
1
n
∑
i
=
1
∞
‍
c
n
+
i
τ
+
1
n
∑
s
=
1
∞
‍
∑
t
=
n
+
2
s
-
1
τ
n
+
2
s
τ
-
1
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
1
n
∑
i
=
1
∞
‍
∑
s
=
n
+
2
i
-
1
τ
n
+
2
i
τ
-
1
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
n
∑
p
=
1
∞
‍
∑
i
=
n
+
2
p
-
1
τ
n
+
2
p
τ
-
1
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
x
f
k
t
+
d
t
<
L
+
1
2
min
⁡
⁡
L
-
A
,
B
-
L
+
1
T
∑
t
=
T
∞
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
1
T
∑
s
=
T
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
f
t
,
x
f
1
t
,
…
,
x
f
k
t
+
d
t
≤
L
+
1
2
min
⁡
⁡
L
-
A
,
B
-
L
+
1
T
∑
t
=
T
∞
‍
H
t
+
1
T
∑
s
=
T
∞
‍
∑
t
=
s
∞
‍
G
t
+
1
T
∑
i
=
T
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
<
L
+
min
⁡
{
L
-
A
,
B
-
L
}
≤
B
,
n
≥
T
,
S
L
x
T
T
=
n
T
S
L
x
T
n
≤
B
,
β
≤
n
<
T
,
which yield that (86) and (87) hold.
Next we show thatS
L is continuous and S
L
(
Ω
(
A
*
,
B
*
,
T
)
) is uniformly Cauchy. Let x
w
=
{
x
w
(
n
)
}
n
∈
N
β and x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
) with (90). Using (16), (17), (80), and the continuity of f, g, and h, we know that for given ε
>
0, there exist T
1, T
2, T
3, and T
4
∈
N with T
4
>
T
3
>
T
2
>
T
1
>
T
+
τ satisfying
(107)
1
T
max
⁡
⁡
∑
t
=
T
+
τ
T
1
‍
h
t
,
x
w
h
1
t
,
…
,
x
w
h
k
t
-
h
t
,
x
h
1
t
,
…
,
x
h
k
t
,
∑
s
=
T
+
τ
T
1
‍
∑
t
=
s
T
2
‍
g
t
,
x
w
g
1
t
,
…
,
x
w
g
k
t
-
g
t
,
x
g
1
t
,
…
,
x
g
k
t
,
∑
i
=
T
+
τ
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
s
T
3
‍
f
t
,
x
w
f
1
t
,
…
,
x
w
f
k
t
∑
t
=
T
+
τ
T
1
‍
-
f
t
,
x
f
1
t
,
…
,
x
f
k
t
<
ε
16
,
w
≥
T
4
,
(108)
1
T
max
⁡
⁡
∑
t
=
T
1
+
1
∞
‍
H
t
+
∑
s
=
T
1
+
1
∞
‍
∑
t
=
s
∞
‍
G
t
+
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
,
∑
s
=
T
+
τ
T
1
‍
∑
t
=
T
2
+
1
∞
‍
G
t
+
∑
i
=
T
+
τ
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
T
3
+
1
∞
‍
F
t
+
∑
i
=
T
+
τ
T
1
‍
∑
s
=
T
2
+
1
∞
‍
∑
t
=
s
∞
‍
F
t
<
ε
16
,
(109)
1
n
∑
i
=
1
∞
‍
c
n
+
i
τ
<
ε
16
,
n
≥
T
4
.
Combining (15) and (105)–(108), we infer that
(110)
S
L
x
w
-
S
L
x
≤
1
n
∑
s
=
1
∞
‍
∑
t
=
n
+
(
2
s
-
1
)
τ
n
+
2
s
τ
-
1
‍
h
t
,
x
w
h
1
t
,
…
,
x
w
h
k
t
-
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
1
n
∑
i
=
1
∞
‍
∑
s
=
n
+
(
2
i
-
1
)
τ
n
+
2
i
τ
-
1
‍
∑
t
=
s
∞
‍
g
t
,
x
w
g
1
t
,
…
,
x
w
g
k
t
-
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
n
∑
p
=
1
∞
‍
∑
i
=
n
+
(
2
p
-
1
)
τ
n
+
2
p
τ
-
1
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
w
f
1
t
,
…
,
x
w
f
k
t
-
f
t
,
x
f
1
t
,
…
,
x
f
k
t
≤
1
T
∑
t
=
T
+
τ
T
1
‍
h
t
,
x
w
h
1
t
,
…
,
x
w
h
k
t
-
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
1
T
∑
t
=
T
1
+
1
∞
‍
h
t
,
x
w
h
1
t
,
…
,
x
w
h
k
t
-
h
t
,
x
h
1
t
,
…
,
x
h
k
t
+
1
T
∑
s
=
T
+
τ
T
1
‍
∑
t
=
s
T
2
‍
g
t
,
x
w
g
1
t
,
…
,
x
w
g
k
t
-
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
T
∑
s
=
T
+
τ
T
1
‍
∑
t
=
T
2
+
1
∞
‍
g
t
,
x
w
g
1
t
,
…
,
x
w
g
k
t
-
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
T
∑
s
=
T
1
+
1
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
w
g
1
t
,
…
,
x
w
g
k
t
-
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
T
∑
i
=
T
+
τ
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
s
T
3
‍
f
t
,
x
w
f
1
t
,
…
,
x
w
f
k
t
-
f
t
,
x
f
1
t
,
…
,
x
f
k
t
+
1
T
∑
i
=
T
+
τ
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
T
3
+
1
∞
‍
f
t
,
x
w
f
1
t
,
…
,
x
w
f
k
t
-
f
t
,
x
f
1
t
,
…
,
x
f
k
t
+
1
T
∑
i
=
T
+
τ
T
1
‍
∑
s
=
T
2
+
1
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
w
f
1
t
,
…
,
x
w
f
k
t
-
f
t
,
x
f
1
t
,
…
,
x
f
k
t
+
1
T
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
w
f
1
t
,
…
,
x
w
f
k
t
-
f
t
,
x
f
1
t
,
…
,
x
f
k
t
<
ε
16
+
2
T
∑
t
=
T
1
+
1
∞
‍
H
t
+
ε
16
+
2
T
∑
s
=
T
+
τ
T
1
‍
∑
t
=
T
2
+
1
∞
‍
G
t
+
2
T
∑
s
=
T
1
+
1
∞
‍
∑
t
=
s
∞
‍
G
t
+
ε
16
+
2
T
∑
i
=
T
+
τ
T
1
‍
∑
s
=
i
T
2
‍
∑
t
=
T
3
+
1
∞
‍
F
t
+
2
T
∑
i
=
T
+
τ
T
1
‍
∑
s
=
T
2
+
1
∞
‍
∑
t
=
s
∞
‍
F
t
+
2
T
∑
i
=
T
1
+
1
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
<
ε
,
w
≥
T
4
,
which implies that S
L is continuous in Ω
(
A
*
,
B
*
,
T
). It follows from (15), (108), and (109) that for any x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
) and t
1
,
t
2
≥
T
4
(111)
S
L
x
t
1
t
1
-
S
L
x
t
2
t
2
=
1
t
1
∑
i
=
1
∞
‍
-
1
i
c
(
t
1
+
i
τ
)
-
1
t
2
∑
i
=
1
∞
‍
-
1
i
c
(
t
2
+
i
τ
)
+
1
t
1
∑
s
=
1
∞
‍
∑
t
=
t
1
+
(
2
s
-
1
)
τ
t
1
+
2
s
τ
-
1
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
-
1
t
2
∑
s
=
1
∞
‍
∑
t
=
t
2
+
(
2
s
-
1
)
τ
t
2
+
2
s
τ
-
1
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
-
1
t
1
∑
i
=
1
∞
‍
∑
s
=
t
1
+
(
2
i
-
1
)
τ
t
1
+
2
i
τ
-
1
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
t
2
∑
i
=
1
∞
‍
∑
s
=
t
2
+
(
2
i
-
1
)
τ
t
2
+
2
i
τ
-
1
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
t
1
∑
p
=
1
∞
‍
∑
i
=
t
1
+
(
2
p
-
1
)
τ
t
1
+
2
p
τ
-
1
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
-
1
t
2
∑
p
=
1
∞
‍
∑
i
=
t
2
+
(
2
p
-
1
)
τ
t
2
+
2
p
τ
-
1
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
∑
i
=
1
∞
‍
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
(
t
)
≤
1
t
1
∑
i
=
1
∞
‍
c
t
1
+
i
τ
+
1
t
2
∑
i
=
1
∞
‍
c
t
2
+
i
τ
+
2
T
4
∑
t
=
T
4
+
τ
∞
‍
H
t
+
2
T
4
∑
s
=
T
4
+
τ
∞
‍
∑
t
=
s
∞
‍
G
t
+
2
T
4
∑
i
=
T
4
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
<
ε
,
which means that S
L
(
Ω
(
A
*
,
B
*
,
T
)
) is uniformly Cauchy, which together with (87) and Lemma 1 yields that S
L
(
Ω
(
A
*
,
B
*
,
T
)
) is relatively compact. It follows from Lemma 3 that the mapping S
L has a fixed point x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
); that is,
(112)
x
n
=
n
L
+
∑
i
=
1
∞
‍
-
1
i
c
n
+
i
τ
+
∑
s
=
1
∞
‍
∑
t
=
n
+
2
s
-
1
τ
n
+
2
s
τ
-
1
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
-
∑
i
=
1
∞
‍
∑
s
=
n
+
2
i
-
1
τ
n
+
2
i
τ
-
1
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
∑
p
=
1
∞
∑
i
=
n
+
2
p
-
1
τ
n
+
2
p
τ
-
1
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
,
n
≥
T
,
which gives that
(113)
x
n
+
x
n
-
τ
=
2
n
-
τ
L
-
c
n
+
∑
t
=
n
∞
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
-
∑
s
=
n
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
,
n
≥
T
+
τ
.
It follows from (113) that
(114)
Δ
x
n
+
x
n
-
τ
+
c
n
=
2
L
-
h
n
,
x
h
1
n
,
…
,
x
h
k
n
+
∑
t
=
n
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
-
∑
s
=
n
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
,
n
≥
T
+
τ
,
Δ
2
x
n
+
x
n
-
τ
+
c
n
=
-
Δ
h
n
,
x
h
1
n
,
…
,
x
h
k
n
-
g
n
,
x
g
1
n
,
…
,
x
g
k
n
+
∑
t
=
n
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
t
,
n
≥
T
+
τ
,
which yields that
(115)
Δ
3
x
n
+
x
n
-
τ
+
c
n
=
-
Δ
2
h
n
,
x
h
1
n
,
…
,
x
h
k
n
-
Δ
g
n
,
x
g
1
n
,
…
,
x
g
k
n
-
f
n
,
x
f
1
n
,
…
,
x
f
k
n
-
d
n
,
n
≥
T
+
τ
,
which together with (102) means that x
=
{
x
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
) is a positive solution of (4). In view of (15)–(17) and (113), we get that
(116)
x
n
+
x
n
-
τ
+
c
n
n
-
2
L
=
-
τ
L
n
+
1
n
∑
t
=
n
∞
‍
h
t
,
x
h
1
t
,
…
,
x
h
k
t
-
1
n
∑
s
=
n
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
g
1
t
,
…
,
x
g
k
t
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
f
1
t
,
…
,
x
f
k
t
∑
t
=
n
∞
‍
t
,
x
f
1
t
,
…
,
x
f
k
t
-
d
(
t
)
≤
τ
L
n
+
1
n
∑
t
=
n
∞
‍
H
t
+
1
n
∑
s
=
n
∞
‍
∑
t
=
s
∞
‍
G
t
+
1
n
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
+
d
t
⟶
0
as
n
⟶
∞
;
that is, (103) holds. Similar to the proof of Theorem 5, we deduce that (19) holds.
(ii) LetL
1
,
L
2
∈
(
A
,
B
) and L
1
≠
L
2. Similarly we obtain that for each l
∈
{
1,2
}, there exist a constant T
l
≥
n
0
+
τ
+
α and two mappings S
L
l
:
Ω
(
A
*
,
B
*
,
T
l
)
→
l
β
∞ satisfying (83), (104), and (105), where T, L, and S
L are replaced by T
l, L
l, and S
L
l, respectively, and S
L
l possesses a fixed point x
l
=
{
x
l
(
n
)
}
n
∈
N
β
∈
Ω
(
A
*
,
B
*
,
T
l
), which is a positive solution of (4); that is,
(117)
x
l
n
=
n
L
l
+
∑
i
=
1
∞
‍
-
1
i
c
(
n
+
i
τ
)
+
∑
i
=
1
∞
‍
∑
t
=
n
+
(
2
s
-
1
)
τ
n
+
2
s
τ
-
1
‍
h
t
,
x
l
h
1
t
,
…
,
x
l
h
k
t
-
∑
i
=
1
∞
‍
∑
s
=
n
+
(
2
i
-
1
)
τ
n
+
2
i
τ
-
1
‍
∑
t
=
s
∞
‍
g
t
,
x
l
g
1
t
,
…
,
x
l
g
k
t
+
∑
p
=
1
∞
‍
∑
i
=
n
+
(
2
p
-
1
)
τ
n
+
2
p
τ
-
1
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
l
f
1
t
,
…
,
x
l
f
k
t
t
,
x
l
f
1
t
,
…
,
x
l
f
k
t
-
d
(
t
)
,
n
≥
T
l
.
Note that (16) and (17) imply that there exists T
*
>
max
⁡
{
T
1
,
T
2
} satisfying
(118)
1
T
*
∑
t
=
T
*
+
τ
∞
‍
H
t
+
∑
s
=
T
*
+
τ
∞
‍
∑
t
=
s
∞
‍
G
t
+
∑
i
=
T
*
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
<
L
1
-
L
2
4
.
In view of (15), (117), and (118), we infer that for any n
≥
T
*
(119)
x
1
n
n
-
x
2
n
n
=
L
1
-
L
2
∑
s
=
1
∞
‍
∑
t
=
n
+
2
s
-
1
τ
n
+
2
s
τ
-
1
‍
+
1
n
∑
s
=
1
∞
‍
∑
t
=
n
+
(
2
s
-
1
)
τ
n
+
2
s
τ
-
1
‍
h
t
,
x
1
h
1
t
,
…
,
x
1
h
k
t
-
h
i
,
x
2
h
1
t
,
…
,
x
2
h
k
t
-
1
n
∑
i
=
1
∞
‍
∑
s
=
n
+
(
2
i
-
1
)
τ
n
+
2
i
τ
-
1
‍
∑
t
=
s
∞
‍
g
t
,
x
1
g
1
t
,
…
,
x
1
g
k
t
-
g
t
,
x
2
g
1
t
,
…
,
x
2
g
k
t
+
1
n
∑
p
=
1
∞
‍
∑
i
=
n
+
2
p
-
1
τ
n
+
2
p
τ
-
1
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
1
f
1
t
,
x
1
f
2
t
,
…
,
x
1
f
k
t
-
f
t
,
x
2
f
1
t
,
…
,
∑
s
=
1
∞
‍
∑
t
=
n
+
2
s
-
1
τ
n
+
2
s
τ
-
1
‍
x
2
f
k
t
≥
L
1
-
L
2
-
1
T
*
∑
t
=
T
*
+
τ
∞
‍
h
t
,
x
1
h
1
t
,
…
,
x
1
h
k
t
-
h
t
,
x
2
h
1
t
,
…
,
x
2
h
k
t
-
1
T
*
∑
s
=
T
*
+
τ
∞
‍
∑
t
=
s
∞
‍
g
t
,
x
1
g
1
t
,
…
,
x
1
g
k
t
-
g
t
,
x
2
g
1
t
,
…
,
x
2
g
k
t
-
1
T
*
∑
i
=
T
*
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
f
t
,
x
1
f
1
t
,
…
,
x
1
f
k
t
-
f
t
,
x
2
f
1
t
,
…
,
x
2
f
k
t
≥
L
1
-
L
2
-
2
T
*
∑
t
=
T
*
+
τ
∞
‍
H
t
+
∑
s
=
T
*
+
τ
∞
‍
∑
t
=
s
∞
‍
G
t
+
∑
i
=
T
*
+
τ
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
F
t
>
L
1
-
L
2
2
>
0
,
which yields that x
1
≠
x
2. Therefore (4) possesses uncountably many positive solutions in l
β
∞. This completes the proof.


## 3. Illustrative Examples

Now we suggest six examples to explain the results presented in Section2. Notice that none of the known results can be applied to these examples.Example 1.
Consider the third order nonlinear neutral delay difference equation:(120)
Δ
3
x
n
+
n
-
2
2
n
x
n
-
τ
+
-
1
n
n
+
1
n
+
Δ
2
1
n
2
+
x
n
-
1
+
Δ
1
n
3
+
2
x
2
n
2
-
n
+
1
n
4
+
x
4
n
-
2
=
-
1
n
n
6
+
n
4
-
1
,
n
≥
3
,
where τ
∈
N
∖
{
3
} is fixed. Let n
0
=
3, k
=
1, β
=
min
⁡
{
|
3
-
τ
|
,
1
}
=
1, A
=
3, B
=
12, b
*
=
1
/
2, c
*
=
3, B
*
=
14, A
*
=
1, and
(121)
b
n
=
n
-
2
2
n
,
c
n
=
-
1
n
n
+
1
n
,
f
n
,
u
=
1
n
4
+
u
4
,
g
n
,
u
=
1
n
3
+
2
u
2
,
h
n
,
u
=
1
n
2
+
u
,
d
n
=
-
1
n
n
6
+
n
4
+
1
,
h
1
n
=
n
-
1
,
g
1
n
=
n
2
-
n
,
f
1
n
=
n
-
2
,
F
n
=
1
n
4
,
G
n
=
1
n
3
,
H
n
=
1
n
2
,
∀
n
,
u
∈
N
n
0
×
R
.
Note that for any p
>
2 and q
>
3
(122)
0
≤
1
n
max
⁡
∑
i
=
n
∞
‍
∑
t
=
i
∞
‍
1
t
p
,
∑
i
=
n
∞
‍
∑
s
=
i
∞
‍
∑
t
=
s
∞
‍
1
t
q
≤
1
n
max
⁡
∑
t
=
n
∞
‍
1
t
p
-
1
,
∑
t
=
n
∞
‍
1
t
q
-
2
⟶
0
as
n
⟶
∞
.
It is easy to see that (14)–(17) are satisfied. It follows from Theorem 5 that (120) possesses a positive solution x
=
{
x
(
n
)
}
n
∈
N
β
∈
l
β
∞ satisfying (18) and (19). Moreover, (120) possesses uncountably many positive solutions in l
β
∞.Example 2.
Consider the third order nonlinear neutral delay difference equation:(123)
Δ
3
x
n
+
5
+
1
2
n
x
n
-
τ
+
2
+
1
2
n
+
Δ
2
1
n
3
+
n
+
1
x
6
2
n
-
3
+
Δ
2
2
n
4
+
x
n
+
5
+
2
+
sin
⁡
n
3
x
n
2
-
n
n
6
+
x
2
n
2
-
n
=
n
2
-
1
n
6
+
n
3
+
2
,
n
≥
2
,
where τ
∈
N
∖
{
2
} is fixed. Let n
0
=
2, k
=
1, β
=
min
⁡
{
|
2
-
τ
|
,
1
}
=
1, A
=
5, B
=
200, b
*
=
6, b
*
=
5, c
*
=
4, B
*
=
204, A
*
=
1, and
(124)
b
n
=
5
+
1
2
n
,
c
n
=
2
+
1
2
n
,
f
n
,
u
=
sin
⁡
n
3
u
n
6
+
u
2
,
g
n
,
u
=
2
2
n
4
+
u
+
2
,
h
n
,
u
=
1
n
3
+
n
+
1
u
6
,
d
n
=
n
2
-
1
n
6
+
n
3
+
2
,
f
1
n
=
n
2
-
n
,
g
1
n
=
n
+
5
,
h
1
n
=
2
n
-
3
,
F
n
=
1
n
6
,
G
n
=
1
n
4
,
H
n
=
1
n
3
,
n
,
u
∈
N
n
0
×
R
.
It follows from (122) that (15)–(17) and (40) hold. Thus Theorem 6 ensures that (123) possesses a positive solution x
=
{
x
(
n
)
}
n
∈
N
β
∈
l
β
∞ satisfying (19) and (41). Moreover, (123) possesses uncountably many positive solutions in l
β
∞.Example 3.
Consider the third order nonlinear neutral delay difference equation:(125)
Δ
3
x
n
+
2
-
n
2
n
x
n
-
τ
+
64
64
-
15
n
+
Δ
2
4
cos
⁡
x
n
2
-
2
2
n
3
+
x
2
n
-
1
·
Δ
1
n
6
+
x
n
2
+
n
x
4
n
2
-
2
n
+
sin
x
2
2
n
-
7
n
4
+
x
4
3
n
-
8
=
n
+
1
-
ln
⁡
n
n
8
+
n
5
+
3
,
n
≥
4
,
where τ
∈
N
∖
{
4
} is fixed. Let n
0
=
4, k
=
2, β
=
min
⁡
{
|
4
-
τ
|
,
1
}
=
1, A
=
30, B
=
300, b
*
=
-
1
/
4, b
*
=
-
1
/
2, c
*
=
20, A
*
=
10, B
*
=
320, and
(126)
b
n
=
2
-
n
2
n
,
c
n
=
64
64
-
15
n
,
f
n
,
u
,
v
=
sin
u
2
n
4
+
v
4
,
g
n
,
u
,
v
=
1
n
6
+
u
v
4
,
h
n
,
u
,
v
=
4
cos
⁡
v
2
n
3
+
u
,
d
n
=
n
+
1
-
ln
⁡
n
n
8
+
n
5
+
3
,
f
1
n
=
2
n
-
7
,
f
2
n
=
3
n
-
8
,
g
1
n
=
n
2
+
n
,
g
2
n
=
n
2
-
2
n
,
h
1
n
=
2
n
-
1
,
h
2
n
=
n
2
-
2
,
F
n
=
1
n
4
,
G
n
=
1
n
6
,
H
n
=
2
n
3
,
n
,
u
,
v
∈
N
n
0
×
R
2
.
It follows from (122) that (15)–(17) and (63) hold. Thus Theorem 7 ensures that (125) possesses a positive solution x
=
{
x
(
n
)
}
n
∈
N
β
∈
l
β
∞ satisfying (19) and (64). Moreover, (125) possesses uncountably many positive solutions in l
β
∞.Example 4.
Consider the third order nonlinear neutral delay difference equation(127)
Δ
3
x
n
+
1
-
10
n
2
-
10
n
n
2
+
n
x
n
-
τ
+
2
n
+
2
n
2
+
Δ
2
sin
x
2
n
2
-
1
n
4
+
2
x
n
2
+
2
n
+
Δ
1
n
7
+
x
n
+
10
3
+
x
2
(
5
n
-
4
)
+
cos
⁡
2
x
n
2
+
3
n
5
+
x
2
4
n
2
-
1
=
1
n
6
+
2
n
3
+
8
,
n
≥
1
,
where τ
∈
N
∖
{
1
}
=
1 is fixed. Let n
0
=
1, k
=
2, β
=
min
⁡
{
|
1
-
τ
|
,
1
}, A
=
10, B
=
200, b
*
=
-
4, b
*
=
-
5, c
*
=
5, A
*
=
5, B
*
=
205, and
(128)
b
n
=
1
-
10
n
2
-
10
n
n
2
+
n
,
c
n
=
2
n
+
2
n
2
,
f
n
,
u
,
v
=
cos
⁡
2
u
n
5
+
v
2
,
g
n
,
u
,
v
=
3
n
7
+
u
3
+
v
2
,
h
n
,
u
,
v
=
sin
v
n
4
+
2
u
,
d
n
=
1
n
6
+
2
n
3
+
8
,
f
1
n
=
n
2
+
3
,
f
2
n
=
4
n
2
,
g
1
n
=
n
+
10
,
g
2
n
=
5
n
-
4
,
h
1
n
=
n
2
+
2
n
,
h
2
n
=
2
n
2
-
1
,
F
n
=
1
n
5
,
G
n
=
1
n
7
,
H
n
=
1
n
4
,
n
,
u
,
v
∈
N
n
0
×
R
2
.
It follows from (122) that (15)–(17) and (68) hold. Thus Theorem 8 ensures that (127) possesses a positive solution x
=
{
x
(
n
)
}
n
∈
N
β
∈
l
β
∞ satisfying (19) and (69). Moreover, (127) possesses uncountably many positive solutions in l
β
∞.Example 5.
Consider the third order nonlinear neutral delay difference equation:(129)
Δ
3
x
n
-
x
n
-
τ
+
n
+
1
n
3
+
Δ
2
1
n
3
+
x
2
5
n
-
4
+
Δ
1
n
6
+
2
x
8
n
2
-
n
+
1
+
sin
x
3
(
3
n
-
2
)
n
8
+
3
=
-
1
n
(
n
+
1
)
/
2
n
10
+
n
2
+
3
,
n
≥
1
,
where τ
∈
N
∖
{
1
} is fixed. Let n
0
=
1, k
=
1, β
=
1, A
=
3, B
=
5, c
*
=
2, A
*
=
1, B
*
=
7, and
(130)
b
n
=
-
1
,
c
n
=
n
+
1
n
3
,
f
n
,
u
=
sin
u
3
n
8
+
3
,
g
n
,
u
=
1
n
6
+
2
u
8
,
h
n
,
u
=
1
n
3
+
u
2
,
d
n
=
-
1
n
n
+
1
/
2
n
10
+
n
2
+
3
,
f
1
n
=
3
n
-
2
,
g
1
n
=
n
2
-
n
+
1
,
h
1
n
=
5
n
-
4
,
F
n
=
1
n
8
,
G
n
=
1
n
6
,
H
n
=
1
n
3
,
∀
n
,
u
∈
N
n
0
×
R
+
∖
0
.
It follows from (122) that (15) and (78)–(81) are satisfied. Thus Theorem 9 ensures that (129) possesses a positive solution x
=
{
x
(
n
)
}
n
∈
N
β
∈
l
β
∞ satisfying (19) and (82). Moreover, (129) possesses uncountably many positive solutions in l
β
∞.Example 6.
Consider the third order nonlinear neutral delay difference equation:(131)
Δ
3
x
n
+
x
n
-
τ
+
2
n
+
4
n
3
+
Δ
2
1
n
4
+
2
x
2
3
n
-
4
+
Δ
1
n
8
+
x
3
n
-
2
+
sin
[
5
x
(
n
2
-
3
)
]
n
5
+
8
=
1
n
8
+
n
5
+
5
,
n
≥
2
,
where τ
∈
N
∖
{
2
} is fixed. Let n
0
=
2, k
=
1, β
=
1, A
=
100, B
=
101, c
*
=
1, A
*
=
99, B
*
=
102, and
(132)
b
n
=
1
,
c
n
=
2
n
+
4
n
3
,
f
n
,
u
=
sin
⁡
5
u
n
5
+
8
,
g
n
,
u
=
1
n
8
+
u
3
,
h
n
,
u
=
1
n
4
+
2
u
2
,
d
n
=
1
n
8
+
n
5
+
5
,
f
1
n
=
n
2
-
3
,
g
1
n
=
n
-
2
,
h
1
n
=
3
n
-
4
,
F
n
=
1
n
5
,
G
n
=
1
n
8
,
H
n
=
1
n
4
,
n
,
u
∈
N
n
0
×
R
.
It follows from (122) that (15)–(17), (80), and (100) hold. Thus Theorem 10 ensures that (131) possesses a positive solution x
=
{
x
(
n
)
}
n
∈
N
β
∈
l
β
∞ satisfying (19) and (103). Moreover, (131) possesses uncountably many positive solutions in l
β
∞.



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*Source: 658651-2015-03-02.xml*