# Strong Convergence Theorems for Solutions of Equilibrium Problems and Common Fixed Points of a Finite Family of Asymptotically Nonextensive Nonself Mappings

Authors: Lijuan Zhang; Hui Tong; Ying Liu
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

An iterative algorithm for finding a common element of the set of common fixed points of a finite family of asymptotically nonextensive nonself mappings and the set of solutions for equilibrium problems is discussed. A strong convergence theorem of common element is established in a uniformly smooth and uniformly convex Banach space.


---

## Body

## 1. Introduction

LetE be a real Banach space with norm ∥
·
∥, let E
* denote the dual of E, and let 〈
x
,
f
〉 denote the value of f
∈
E
* at x
∈
E. Suppose that C is a nonempty, closed convex subset of E. Let f be a bifunction of C
×
C into R, where R is the set of real numbers. The equilibrium problem for f
:
C
×
C
→
R is to find x
∈
C such that
(1)
f
(
x
,
y
)
≥
0
,
∀
y
∈
C
.
The set of solutions of (1) is denoted by E
P
(
f
). Given a mapping T
:
C
→
E
*, let f
(
x
,
y
)
=
〈
T
x
,
y
-
x
〉 for all x
,
y
∈
C. Then, p
∈
E
P
(
f
) if and only if 〈
T
p
,
y
-
p
〉
≥
0 for all y
∈
C; that is, p is a solution of the variational inequality. Numerous problems in physics, optimization, and economics reduce to find a solution of (1). Some methods have been proposed to solve the equilibrium problems; see [1–5].LetJ be the normalized duality mapping from E into 2
E
* given by
(2)
J
x
=
{
f
∈
E
*
:
〈
x
,
f
〉
=
∥
x
∥
∥
f
∥
,
∥
x
∥
=
∥
f
∥
}
for all x
∈
E. It is well known that if E is uniformly smooth, then J is uniformly norm-to-norm continuous on each bounded subset of E. It is also well known that E is uniformly smooth if and only if E
* is uniformly convex.LetC be a nonempty closed convex subset of a Hilbert space H and let P
C
:
H
→
Cbe the metric projection of H onto C; then P
C is nonexpansive. This fact actually characterizes Hilbert spaces and consequently it is not available in more general Banach spaces. In this connection, Alber [6] recently introduced a generalized projection operator Π
C in a Banach space E which is an analogue of the metric projection in Hilbert spaces. Consider the functional defined by
(3)
ϕ
(
x
,
y
)
=
∥
x
∥
2
-
2
〈
x
,
J
y
〉
+
∥
y
∥
2
,
∀
x
,
y
∈
E
.
Observe that, in a Hilbert space H, (3) reduces to ϕ
(
x
,
y
)
=
∥
x
-
y
∥
2
,
x
,
y
∈
H. The generalized projection Π
C
:
E
→
C is a map that assigns to an arbitrary point x
∈
E the minimum point of the functional ϕ
(
x
,
y
); that is, Π
C
x
=
x
¯, where x
¯ is the solution to the minimization problem
(4)
ϕ
(
x
¯
,
x
)
=
min
⁡
y
∈
C
ϕ
(
y
,
x
)
,existence and uniqueness of the operator Π
C follows from the properties of the functional ϕ
(
x
,
y
) and strict monotonicity of the mapping J. In Hilbert spaces, Π
C
=
P
C. It is obvious from the definition of function ϕ that
(5)
(
∥
x
∥
-
∥
y
∥
)
2
≤
ϕ
(
y
,
x
)
≤
(
∥
x
∥
+
∥
y
∥
)
2
,
∀
x
,
y
∈
E
,
(6)
ϕ
(
x
,
y
)
=
ϕ
(
x
,
z
)
+
ϕ
(
z
,
y
)
+
2
〈
x
-
z
,
J
z
-
J
y
〉
,
hhhhhhhhhhhhhhhhhhhhi
∀
x
,
y
,
z
∈
E
.
(7)
ϕ
(
x
,
y
)
=
〈
x
,
J
x
-
J
y
〉
+
〈
y
-
x
,
J
y
〉
≤
∥
x
∥
∥
J
x
-
J
y
∥
+
∥
y
-
x
∥
∥
y
∥
,
∀
x
,
y
,
z
∈
E
.LetC be a nonempty subset of E and let T
:
C
→
E be a mapping. The set of fixed points of T is denoted by F
(
T
). T
:
C
→
E is called asymptotically nonextensive if and only if there exists a sequence {
k
n
}
⊂
[
1
,
∞
) with lim
⁡
n
→
∞
k
n
=
1, such that
(8)
ϕ
(
T
(
Π
C
T
)
n
-
1
x
,
T
(
Π
C
T
)
n
-
1
y
)
≤
k
n
ϕ
(
x
,
y
)
,
hhhhhhhhhhhhhhhhhh
∀
x
,
y
∈
C
,
n
≥
1
.
Asymptotically nonextensive mappings coincide with asymptotically nonexpansive mappings in Hilbert spaces.In [7], Chidume et al. studied the fixed point problem of an asymptotically nonextensive nonself mapping and obtained weak convergence theorem. Recently, in [8], liu introduced the following iterative scheme for approximating a common fixed point of two asymptotically nonextensive nonself mappings in a uniformly smooth and uniformly convex Banach space:
(9)
y
n
=
Π
C
(
J
-
1
(
β
n
J
x
n
+
(
1
-
β
n
)
J
S
(
Π
C
S
)
n
-
1
x
n
)
)
,
x
n
+
1
=
Π
C
(
J
-
1
(
α
n
J
x
n
+
(
1
-
α
n
)
J
T
(
Π
C
T
)
n
-
1
y
n
)
)
.
Liu obtained strong convergence theorem.Inspired and motivated by the facts above, the purpose of this paper is to prove a strong convergence theorem for finding a common element of the set of common fixed points of a finite family of asymptotically nonextensive nonself mappings and the set of solutions for equilibrium problems in a uniformly smooth and uniformly convex Banach space.


## 2. Preliminaries

LetE be a real Banach space. When {
x
n
} is a sequence in E, we denote strong convergence of {
x
n
} to x
∈
E by x
n
→
x and weak convergence by x
n
⇀
x. E is said to have the Kadec-Klee property if and only if for a sequence {
x
n
} of E satisfying that x
n
⇀
x
∈
E and ∥
x
n
∥
→
∥
x
∥, then x
n
→
x. It is known that if E is uniformly convex, then E has the Kadec-Klee property.A mappingT
:
C
→
C is said to be closed; if for any sequence {
x
n
}
⊂
C with x
n
→
x and T
x
n
→
y, then T
x
=
y.Lemma 1.
LetE be a uniformly smooth and strictly convex Banach space which enjoys the Kadec-Klee property, let C be a nonempty, closed, and convex subset of E, and let T
:
C
→
E be an asymptotically nonextensive nonself mapping with a sequence {
k
n
}
⊂
[
1
,
∞
) such that T is closed. Then F
(
T
) is closed and convex.Proof.
Takex
,
y
∈
F
(
T
),  t
∈
(
0,1
). Put z
:
=
t
x
+
(
1
-
t
)
y. Using the same argument presented in the proof of [9, Theorem 2.1, page 854-855], we can obtain that lim
⁡
n
→
∞
T
(
Π
C
T
)
n
-
1
z
=
z. By the continuity of Π
C, we have
(10)
lim
⁡
n
→
∞
(
Π
C
T
)
n
z
=
z
.
Therefore,
(11)
lim
⁡
n
→
∞
(
(
Π
C
T
)
n
z
-
T
(
Π
C
T
)
n
z
)
=
0
.
By (10), (11) and the closedness of T, we have z
∈
F
(
T
) which implies that F
(
T
) is convex.
Letx
n
∈
F
(
T
) and x
n
→
q; then, we have x
n
-
T
x
n
→
0. It follows from the closedness of  T that q
∈
F
(
T
). This implies that F
(
T
) is closed.Lemma 2 (see [6]).
LetE be a reflexive, strictly convex, and smooth Banach space; let C be a nonempty, closed, and convex subset of E. Then the following conclusions hold:(1)
ϕ
(
y
,
Π
C
x
)
+
ϕ
(
Π
C
x
,
x
)
≤
ϕ
(
y
,
x
)
,
f
o
r
a
l
l
y
∈
C, and x
∈
E;
(2)
ifx
∈
E and z
∈
C, then z
=
Π
C
x if and only if 〈
z
-
y
,
J
x
-
J
z
〉
≥
0
,
f
o
r
a
l
l
y
∈
C;
(3)
forx
,
y
∈
E
,
ϕ
(
y
,
x
)
=
0 if and only if x
=
y.Lemma 3 (see [10]).
LetE be a uniformly convex and smooth Banach space and let {
y
n
}
,
{
z
n
} be two sequences of E. If ϕ
(
y
n
,
z
n
)
→
0 and either {
y
n
} or {
z
n
} is bounded, then y
n
-
z
n
→
0.Lemma 4 (see [11]).
LetE be a smooth and uniformly convex Banach space and let r
>
0. Then there exists a strictly increasing, continuous, and convex function g
:
[
0,2
r
]
→
R such that g
(
0
)
=
0 and
(12)
∥
t
x
+

(
1
-
t
)
y
∥
2

≤

t
∥
x
∥
2
+
(
1
-
t
)
∥
y
∥
2
-
t
(
1
-
t
)
g
(
∥
x
-
y
∥
)
for all x
,
y
∈
B
r and t
∈
[
0,1
], where B
r
=
{
z
∈
E
:
∥
z
∥
≤
r
}.For solving the equilibrium problem, let us assume that a bifunctionf
:
C
×
C
→
R satisfies the following conditions:(A1)f
(
x
,
x
)
=
0 for all x
∈
C;
(A2)f is monotone; that is, f
(
x
,
y
)
+
f
(
y
,
x
)
≤
0 for all x
,
y
∈
C;
(A3) for eachx
,
y
,
z
∈
C,
(13)
lim
⁡
t
→
0
f
(
t
z
+
(
1
-
t
)
x
,
y
)
≤
f
(
x
,
y
)
;
(A4) for eachx
∈
C, y
↦
f
(
x
,
y
) is convex and lower semicontinuous.Lemma 5 (see [12]).
LetC be a closed convex subset of a smooth, strictly convex, and reflexive Banach space E, let f be a bifunction from C
×
C to R satisfying (A1)–(A4), and for r
>
0 and x
∈
E, define a mapping T
r
:
E
→
C as follows:
(14)
T
r
x
=
{
z
∈
C
:
f
(
z
,
y
)
+
1
r
〈
y
-
z
,
J
z
-
J
x
〉
≥
0
,
∀
y
∈
C
}
.
Then the following conclusions hold:(1)
T
r is single-valued;
(2)
T
r is firmly nonexpansive; that is, for any x
,
y
∈
E,
(15)
〈
T
r
x
-
T
r
y
,
J
T
r
x
-
J
T
r
y
〉
≤
〈
T
r
x
-
T
r
y
,
J
x
-
J
y
〉
;
(3)
F
(
T
r
)
=
E
P
(
f
);
(4)
E
P
(
f
) is closed and convex;
(5)
ϕ
(
q
,
T
r
x
)
+
ϕ
(
T
r
x
,
x
)
≤
ϕ
(
q
,
x
)
,
f
o
r
a
l
l
q
∈
F
(
T
r
).


## 3. Main Results

Theorem 6.
LetC be a nonempty closed convex subset of a uniformly smooth and uniformly convex Banach space E. Let f be a bifunction from C
×
C to R satisfying (A1)–(A4), and let N be some positive integer. Let S
i
:
C
→
E be a closed asymptotically nonextensive nonself mapping with sequence {
k
n
,
i
}
⊂
[
1
,
∞
) such that ∑
n
=
1
∞
(
k
n
,
i
-
1
)
<
∞ for every 1
≤
i
≤
N. Suppose that Ω
=
⋂
i
=
1
N
‍
F
(
S
i
)
∩
E
P
(
f
) is nonempty and bounded. Let {
x
n
} be a sequence generated by the following manner:
(16)
x
0
∈
E
,
C
1
=
C
,
x
1
=
Π
C
1
x
0
,
y
n
=
J
-
1
(
α
n
,
0
J
x
n
+
∑
i
=
1
N
α
n
,
i
J
S
i
(
Π
C
S
i
)
n
-
1
x
n
)
,
u
n
=
T
r
n
y
n
,
C
n
+
1
=
{
z
∈
C
n
:
ϕ
(
z
,
u
n
)
≤
ϕ
(
z
,
x
n
)
+
θ
n
}
,
x
n
+
1
=
Π
C
n
+
1
x
0
,
where θ
n
=
(
k
n
-
1
)
sup
⁡
z
∈
Ω
ϕ
(
z
,
x
n
), k
n
=
max
⁡
{
k
n
,
i
}. {
α
n
,
i
} is a real number sequence in (
0,1
) for every 0
≤
i
≤
N, {
r
n
} is a real number sequence in [
a
,
∞
), where a is some positive real number. Assume that ∑
i
=
0
N
α
n
,
i
=
1 and liminf
⁡
n
→
∞
α
n
,
0
α
n
,
i
>
0 for every 1
≤
i
≤
N. Then {
x
n
} converges strongly to Π
Ω
x
0.Proof.
First, we show thatC
n is closed and convex. From the definitions of C
n, it is obvious C
n is closed. Moreover, since ϕ
(
z
,
u
n
)
≤
ϕ
(
z
,
x
n
)
+
θ
n is equivalent to 2
〈
z
,
J
x
n
-
J
u
n
〉
≤
∥
x
n
∥
2
-
∥
u
n
∥
2
+
θ
n, it follows that C
n is convex. From Lemmas 1 and 5, we have that Ω is closed and convex. Then {
x
n
} is well defined.
Next, we proveΩ
⊂
C
n for all n
≥
1. Ω
⊂
C
1
=
C is obvious. Suppose that Ω
⊂
C
n for some n
≥
2; for each z
∈
Ω, from Lemma 5, we have
(17)
ϕ
(
z
,
u
n
)
=
ϕ
(
z
,
T
r
n
y
n
)
≤
ϕ
(
z
,
y
n
)
=
∥
z
∥
2
-
2
〈
z
,
α
n
,
0
J
x
n
+
∑
i
=
1
N
α
n
,
i
J
S
i
(
Π
C
S
i
)
n
-
1
x
n
〉
+
∥
α
n
,
0
J
x
n
+
∑
i
=
1
N
α
n
,
i
J
S
i
(
Π
C
S
i
)
n
-
1
x
n
∥
2
≤
∥
z
∥
2
-
2
α
n
,
0
〈
z
,
J
x
n
〉
-
2
∑
i
=
1
N
α
n
,
i
〈
z
,
J
S
i
(
Π
C
S
i
)
n
-
1
x
n
〉
+
α
n
,
0
∥
x
n
∥
2
+
∑
i
=
1
N
α
n
,
i
∥
S
i
(
Π
C
S
i
)
n
-
1
x
n
∥
2
=
α
n
,
0
ϕ
(
z
,
x
n
)
+
∑
i
=
1
N
α
n
,
i
ϕ
(
z
,
S
i
(
Π
C
S
i
)
n
-
1
x
n
)
≤
α
n
,
0
ϕ
(
z
,
x
n
)
+
∑
i
=
1
N
α
n
,
i
k
n
,
i
ϕ
(
z
,
x
n
)
≤
α
n
,
0
ϕ
(
z
,
x
n
)
+
∑
i
=
1
N
α
n
,
i
k
n
ϕ
(
z
,
x
n
)
=
ϕ
(
z
,
x
n
)
+
(
1
-
α
n
,
0
)
(
k
n
-
1
)
ϕ
(
z
,
x
n
)
≤
ϕ
(
z
,
x
n
)
+
θ
n
.
This implies that z
∈
C
n
+
1, and so Ω
⊂
C
n
+
1. From x
n
=
Π
C
n
x
0, one sees
(18)
〈
x
n
-
u
,
J
x
0
-
J
x
n
〉
≥
0
,
∀
u
∈
C
n
.
Since Ω
⊂
C
n
+
1, we arrive at
(19)
〈
x
n
-
z
,
J
x
0
-
J
x
n
〉
≥
0
,
∀
z
∈
Ω
.
Next we show that the sequence{
x
n
} is bounded. From Lemma 2, we have
(20)
ϕ
(
x
n
,
x
0
)
=
ϕ
(
Π
C
n
x
0
,
x
0
)

≤

ϕ
(
z
,
x
0
)
-
ϕ
(
z
,
x
n
)
≤
ϕ
(
z
,
x
0
)
,
for each z
∈
Ω
⊂
C
n and for all n
≥
1. Therefore, the sequence {
ϕ
(
x
n
,
x
0
)
} is bounded. It follows from (5) that the sequence {
x
n
} is also bounded. By the assumption, we have
(21)
lim
⁡
n
→
∞
θ
n
=
0
.
On the other hand, noticing that x
n
=
Π
C
n
x
0 and x
n
+
1
=
Π
C
n
+
1
x
0
∈
C
n
+
1
⊂
C
n, one has
(22)
ϕ
(
x
n
,
x
0
)
≤
ϕ
(
x
n
+
1
,
x
0
)
for all n
≥
1. Therefore, {
ϕ
(
x
n
,
x
0
)
} is nondecreasing. It follows that the limit of {
ϕ
(
x
n
,
x
0
)
} exists. By the definition of C
n, one has that C
m
⊂
C
n and x
m
=
Π
C
m
x
0
∈
C
n for any positive integer m
≥
n. It follows that
(23)
ϕ
(
x
m
,
x
n
)
=
ϕ
(
x
m
,
Π
C
n
x
0
)
≤
ϕ
(
x
m
,
x
0
)
-
ϕ
(
Π
C
n
x
0
,
x
0
)
=
ϕ
(
x
m
,
x
0
)
-
ϕ
(
x
n
,
x
0
)
.
Letting m
,
n
→
∞ in (23), we have ϕ
(
x
m
,
x
n
)
→
0. It follows from Lemma 3 that x
m
-
x
n
→
0 as m
,
n
→
∞. Hence, {
x
n
} is a Cauchy sequence. Since E is a Banach space and C is a closed and convex, one can assume that x
n
→
x
¯
∈
C as n
→
∞.
Next we show thatx
¯
∈
⋂
i
=
1
N
‍
F
(
S
i
). By taking m
=
1 in (23), we have that
(24)
lim
⁡
n
→
∞
ϕ
(
x
n
+
1
,
x
n
)
=
0
.
From Lemma 3, we have
(25)
lim
⁡
n
→
∞
∥
x
n
+
1
-
x
n
∥
=
0
.
Noticing that x
n
+
1
∈
C
n
+
1, we obtain
(26)
ϕ
(
x
n
+
1
,
u
n
)
≤
ϕ
(
x
n
+
1
,
x
n
)
+
θ
n
.
It follows from (21) and (24) that
(27)
lim
⁡
n
→
∞
ϕ
(
x
n
+
1
,
u
n
)
=
0
.
From Lemma 3, we have
(28)
lim
⁡
n
→
∞
∥
x
n
+
1
-
u
n
∥
=
0
.
Combining (25) with (28), we obtain that
(29)
lim
⁡
n
→
∞
∥
x
n
-
u
n
∥
=
0
.
It follows from x
n
→
x
¯ as n
→
∞ that u
n
→
x
¯
,asn
→
∞. Since J is uniformly norm-to-norm continuous on each bounded set, we have
(30)
lim
⁡
n
→
∞
∥
J
x
n
-
J
u
n
∥
=
0
.
On the other hand, we have
(31)
ϕ
(
z
,
x
n
)
-
ϕ
(
z
,
u
n
)
=
∥
x
n
∥
2
-
∥
u
n
∥
2
-
2
〈
z
,
J
x
n
-
J
u
n
〉
≤
∥
x
n
-
u
n
∥
(
∥
x
n
∥
+
∥
u
n
∥
)
+
2
∥
z
∥
∥
J
x
n
-
J
u
n
∥
.
We obtain that
(32)
lim
⁡
n
→
∞
(
ϕ
(
z
,
x
n
)
-
ϕ
(
z
,
u
n
)
)
=
0
.
Since E is a uniformly smooth Banach space, we know that E
* is a uniformly convex Banach space. From Lemma 4, we find that
(33)
ϕ
(
z
,
u
n
)
=
ϕ
(
z
,
T
r
n
y
n
)
≤
ϕ
(
z
,
y
n
)
=
∥
z
∥
2
-
2
〈
z
,
α
n
,
0
J
x
n
+
∑
i
=
1
N
α
n
,
i
J
S
i
(
Π
C
S
i
)
n
-
1
x
n
〉
+
∥
α
n
,
0
J
x
n
+
∑
i
=
1
N
α
n
,
i
J
S
i
(
Π
C
S
i
)
n
-
1
x
n
∥
2
≤
∥
z
∥
2
-
2
α
n
,
0
〈
z
,
J
x
n
〉
-
2
∑
i
=
1
N
α
n
,
i
〈
z
,
J
S
i
(
Π
C
S
i
)
n
-
1
x
n
〉
+
α
n
,
0
∥
x
n
∥
2
+
∑
i
=
1
N
α
n
,
i
∥
S
i
(
Π
C
S
i
)
n
-
1
x
n
∥
2
-
α
n
,
0
α
n
,
1
g
(
∥
J
x
n
-
J
S
1
(
Π
C
S
1
)
n
-
1
x
n
∥
)
=
α
n
,
0
ϕ
(
z
,
x
n
)
+
∑
i
=
1
N
α
n
,
i
ϕ
(
z
,
S
i
(
Π
C
S
i
)
n
-
1
x
n
)
-
α
n
,
0
α
n
,
1
g
(
∥
J
x
n
-
J
S
1
(
Π
C
S
1
)
n
-
1
x
n
∥
)
≤
α
n
,
0
ϕ
(
z
,
x
n
)
+
∑
i
=
1
N
α
n
,
i
k
n
,
i
ϕ
(
z
,
x
n
)
-
α
n
,
0
α
n
,
1
g
(
∥
J
x
n
-
J
S
1
(
Π
C
S
1
)
n
-
1
x
n
∥
)
≤
α
n
,
0
ϕ
(
z
,
x
n
)
+
∑
i
=
1
N
α
n
,
i
k
n
ϕ
(
z
,
x
n
)
-
α
n
,
0
α
n
,
1
g
(
∥
J
x
n
-
J
S
1
(
Π
C
S
1
)
n
-
1
x
n
∥
)
=
ϕ
(
z
,
x
n
)
+
(
1
-
α
n
,
0
)
(
k
n
-
1
)
ϕ
(
z
,
x
n
)
-
α
n
,
0
α
n
,
1
g
(
∥
J
x
n
-
J
S
1
(
Π
C
S
1
)
n
-
1
x
n
∥
)
≤
ϕ
(
z
,
x
n
)
+
θ
n
-
α
n
,
0
α
n
,
1
g
(
∥
J
x
n
-
J
S
1
(
Π
C
S
1
)
n
-
1
x
n
∥
)
.
Therefore we have
(34)
α
n
,
0
α
n
,
1
g
(
∥
J
x
n
-
J
S
1
(
Π
C
S
1
)
n
-
1
x
n
∥
)
≤
ϕ
(
z
,
x
n
)
-
ϕ
(
z
,
u
n
)
+
θ
n
.
From liminf
⁡
n
→
α
n
,
0
α
n
,
1
>
0 and (21), (32), we have
(35)
lim
⁡
n
→
∞
g
(
∥
J
x
n
-
J
S
1
(
Π
C
S
1
)
n
-
1
x
n
∥
)
=
0
.
Therefore, from the property of g we have
(36)
lim
⁡
n
→
∞
∥
J
x
n
-
J
S
1
(
Π
C
S
1
)
n
-
1
x
n
∥
=
0
.
Since J
-
1 is also uniformly norm-to-norm continuous on each bounded set, we have
(37)
lim
⁡
n
→
∞
∥
x
n
-
S
1
(
Π
C
S
1
)
n
-
1
x
n
∥
=
0
.
Using (7), (34), and (36), we have
(38)
lim
⁡
n
→
∞
ϕ
(
x
n
,
S
1
(
Π
C
S
1
)
n
-
1
x
n
)
=
0
.
By (6), we obtain
(39)
ϕ
(
x
n
,
S
1
x
n
)
=
ϕ
(
x
n
,
x
n
+
1
)
+
ϕ
(
x
n
+
1
,
S
1
x
n
)
+
2
〈
x
n
-
x
n
+
1
,
J
x
n
+
1
-
J
S
1
x
n
〉
=
ϕ
(
x
n
,
x
n
+
1
)
+
ϕ
(
x
n
+
1
,
S
1
(
Π
C
S
1
)
n
x
n
+
1
)
+
ϕ
(
S
1
(
Π
C
S
1
)
n
x
n
+
1
,
S
1
(
Π
C
S
1
)
n
x
n
)
+
ϕ
(
S
1
(
Π
C
S
1
)
n
x
n
,
S
1
x
n
)
+
2
〈
S
1
(
Π
C
S
1
)
n
x
n
+
1
-
S
1
(
Π
C
S
1
)
n
x
n
,
J
S
1
(
Π
C
S
1
)
n
x
n
-
J
S
1
x
n
〉
+
2
〈
x
n
+
1
-
S
1
(
Π
C
S
1
)
n
x
n
+
1
,
J
S
1
(
Π
C
S
1
)
n
x
n
+
1
-
J
S
1
x
n
〉
+
2
〈
x
n
-
x
n
+
1
,
J
x
n
+
1
-
J
S
1
x
n
〉
.
Since ϕ
(
x
n
,
(
Π
C
S
1
)
n
x
n
)
≤
ϕ
(
x
n
,
S
1
(
Π
C
S
1
)
n
-
1
x
n
), from (38), we have lim
⁡
n
→
∞
ϕ
(
x
n
,
(
Π
C
S
1
)
n
x
n
)
=
0. Since ϕ
(
S
1
(
Π
C
S
1
)
n
x
n
,
S
1
x
n
)
≤
k
1
ϕ
(
(
Π
C
S
1
)
n
x
n
,
x
n
), then
(40)
lim
⁡
n
→
∞
ϕ
(
S
1
(
Π
C
S
1
)
n
x
n
,
S
1
x
n
)
=
0
.
Applying (24), (38), (40), the definition of S
1, and Lemma 3 to (39), we obtain that
(41)
lim
⁡
n
→
∞
ϕ
(
x
n
,
S
1
x
n
)
=
0
.
From Lemma 3, we obtain that
(42)
lim
⁡
n
→
∞
∥
x
n
-
S
1
x
n
∥
=
0
.
In the same way, we can obtain
(43)
lim
⁡
n
→
∞
∥
x
n
-
S
i
x
n
∥
=
0
,
2
≤
i
≤
N
.
From the closedness of S
i
,
1
≤
i
≤
N, we have x
¯
∈
⋂
i
=
1
N
‍
F
(
S
i
).
Next, we showx
¯
∈
E
P
(
f
). From Lemma 5, we have
(44)
ϕ
(
u
n
,
y
n
)
=
ϕ
(
T
r
n
y
n
,
y
n
)
≤
ϕ
(
z
,
y
n
)
-
ϕ
(
z
,
T
r
n
y
n
)
≤
ϕ
(
z
,
y
n
)
-
ϕ
(
z
,
u
n
)
=
ϕ
(
z
,
x
n
)
+
θ
n
-
ϕ
(
z
,
u
n
)
.
It follows from (21) and (32) that lim
⁡
n
→
∞
ϕ
(
u
n
,
y
n
)
=
0. From Lemma 3, we see that
(45)
lim
⁡
n
→
∞
∥
y
n
-
u
n
∥
=
0
.
Since J is uniformly norm-to-norm continuous on each bounded set, we have
(46)
lim
⁡
n
→
∞
∥
J
y
n
-
J
u
n
∥
=
0
.
From r
n
≥
a, we have
(47)
lim
⁡
n
→
∞
∥
J
u
n
-
J
y
n
∥
r
n
=
0
.
By u
n
=
T
r
n
y
n, we have
(48)
f
(
u
n
,
y
)
+
1
r
n
〈
y
-
u
n
,
J
u
n
-
J
y
n
〉
≥
0
,
∀
y
∈
C
.
From (A2), we have
(49)
∥
y
-
u
n
∥
∥
J
u
n
-
J
y
n
∥
r
n
≥
1
r
n
〈
y
-
u
n
,
J
u
n
-
J
y
n
〉
≥
-
f
(
u
n
,
y
)
≥
f
(
y
,
u
n
)
,
∀
y
∈
C
.
Letting n
→
∞, we have from (A4), (47) and u
n
→
x
¯, as n
→
∞ that.
(50)
f
(
y
,
x
¯
)
≤
0
,
∀
y
∈
C
.
For 0
<
t
<
1 and y
∈
C, let y
t
=
t
y
+
(
1
-
t
)
x
¯. Since y
∈
C and x
¯
∈
C, we have y
t
∈
C and hence f
(
y
t
,
x
¯
)
≤
0. So, from (A1) and (A4) we have
(51)
0
=
f
(
y
t
,
y
t
)
≤
t
f
(
y
t
,
y
)
+
(
1
-
t
)
f
(
y
t
,
x
¯
)
≤
t
f
(
y
t
,
y
)
.
Dividing by t, we have
(52)
f
(
y
t
,
y
)
≥
0
,
∀
y
∈
C
.
Letting t
→
0, from (A3), we have f
(
x
¯
,
y
)
≥
0
,
for
all
y
∈
C. Therefore,  x
¯
∈
E
P
(
f
).
Finally, we showx
¯
=
Π
Ω
x
0. By taking limit in (19), we have
(53)
〈
x
¯
-
z
,
J
x
0
-
J
x
¯
〉
≥
0
,
∀
z
∈
Ω
.
At this point, in view of Lemma 2, we have that x
¯
=
Π
Ω
x
0. This completes the proof.Remark 7.
Theorem6 improves the main theorem in [8] in the following senses.(1)
Theorem6 generalizes this theorem from two asymptotically nonextensive operators to a finite family of asymptotically nonextensive operators.
(2)
Theorem6 removes the condition that S
i is completely continuous or semicompact.



---

*Source: 494632-2014-04-29.xml*