# New Mixed Equilibrium Problems and Iterative Algorithms for Fixed Point Problems in Banach Spaces

Authors: Minjiang Chen; Jianmin Song; Shenghua Wang
Journal: Journal of Applied Mathematics
 (2014)

---

## Abstract

We first introduce a new mixed equilibrium problem with a relaxed monotone mapping in Banach spaces and prove the existence of solutions of the equilibrium problem. Then we introduce a new iterative algorithm for finding a common element of the set of solutions of the equilibrium problem and the set of fixed points of a quasi-ϕ-nonexpansive mapping and prove some strong convergence theorems of the iteration. Our results extend and improve the corresponding ones given by Wang et al., Takahashi and Zembayashi, and some others.


---

## Body

## 1. Introduction

LetE be a Banach space with the dual space E
* and let K be a nonempty closed convex subset of E. Let Θ be a bifunction from K
×
K to ℝ, where ℝ denotes the set of numbers. The equilibrium problem is to find x
*
∈
K such that
(1)
Θ
(
x
*
,
y
)
≥
0
,
∀
y
∈
K
.The set of solutions of the above equilibrium problem is denoted byEP
(
Θ
).In order to solve the equilibrium problem, the bifunctionΘ is usually to be assumed that following conditions are satisfied:(A1)
Θ
(
x
,
x
)
=
0 for all x
∈
K;
(A2)
Θ is monotone; that is, Θ
(
x
,
y
)
+
Θ
(
y
,
x
)
≤
0 for all x
,
y
∈
K;
(A3)
for allx
,
y
,
z
∈
K,
(2)
limsup
t
↓
0
Θ
(
t
z
+
(
1
-
t
)
x
,
y
)
≤
Θ
(
x
,
y
)
;
(A4)
for allx
∈
K, Θ
(
x
,
·
) is convex and lower semicontinuous.Recently, Takahashi and Zembayashi [1] extended the equilibrium problems and fixed point problems from Hilbert spaces to Banach spaces. More precisely, they gave the following iterative scheme:
(3)
x
0
=
x
∈
K
,
y
n
=
J
-
1
(
α
n
J
x
n
+
(
1
-
α
n
)
J
S
x
n
)
,
u
n
∈
K
such
that
Θ
(
u
n
,
y
)
+
1
r
n
〈
y
-
u
n
,
J
u
n
-
J
y
n
〉
≥
0
,
∀
y
∈
K
,
H
n
=
{
z
∈
K
:
ϕ
(
z
,
u
n
)
≤
ϕ
(
z
,
x
n
)
}
,
W
n
=
{
z
∈
K
:
〈
x
n
-
z
,
J
x
-
J
x
n
〉
≥
0
}
,
x
n
+
1
=
Π
H
n
∩
W
n
x
,
∀
n
≥
0
,
where S is a relatively nonexpansive mapping from K into itself such that F
(
S
)
∩
EP
(
Θ
)
≠
∅, J is the duality mapping on E, and {
α
n
}
⊂
[
0,1
] satisfies liminf
n
→
∞
α
n
(
1
-
α
n
)
>
0 and {
r
n
}
⊂
[
a
,
∞
) for some a
>
0. They proved that the sequence {
x
n
} generated by (3) converges strongly to Π
F
(
S
)
∩
EP
(
Θ
)
x, where Π
F
(
S
)
∩
EP
(
Θ
) is the generalized projection of E onto F
(
S
)
∩
EP
(
Θ
).Very recently, Qin et al. [2] introduced the following hybrid algorithm to solve the equilibrium problems and fixed point problems for quasi-ϕ-nonexpansive mappings in a uniformly convex and uniformly smooth Banach space E with a nonempty closed convex subset K of E:
(4)
x
0
∈
E
chosen
arbitrarily
,
C
1
=
K
,
x
1
=
Π
C
1
x
0
,
y
n
=
J
-
1
(
α
n
J
x
n
+
β
n
J
T
x
n
+
γ
n
J
S
x
n
)
,
u
n
∈
K
such
that
Θ
(
u
n
,
y
)
+
1
r
n
〈
y
-
u
n
,
J
u
n
-
J
y
n
〉
≥
0
,
∀
y
∈
K
,
C
n
+
1
=
{
z
∈
C
n
:
ϕ
(
z
,
u
n
)
≤
ϕ
(
z
,
x
n
)
}
,
x
n
+
1
=
Π
C
n
+
1
x
0
,
∀
n
≥
0
,
where T and S are two quasi-ϕ-nonexpansive mappings from K into itself such that F
=
F
(
T
)
∩
F
(
S
)
∩
EP
(
Θ
)
≠
∅, {
α
n
}, {
β
n
}, and {
γ
n
} are three sequences in [
0,1
] satisfying some certain conditions. They proved that the sequence {
x
n
} generated by (4) converges strongly to Π
F
x
0, where Π
F is the generalized projection of E onto F.In [3], Fang and Huang introduced a concept called a relaxed η-α-monotone mapping.A mappingA
:
K
→
E
* is said to be relaxed  η
-
α
-monotone  if there exist a mapping η
:
K
×
K
→
E and a function α
:
E
→
ℝ with α
(
t
z
)
=
t
p
α
(
z
) for all t
>
0 and z
∈
E, where p
>
1 is a constant, such that
(5)
〈
A
x
-
A
y
,
η
(
x
,
y
)
〉
≥
α
(
x
-
y
)
,
∀
x
,
y
∈
K
.Especially, ifη
(
x
,
y
)
=
x
-
y for all x
,
y
∈
K and α
(
z
)
=
κ
∥
z
∥
p, where p
>
1 and κ
>
1 are two constants, then A is said to be p
-monotone (see, e.g., [4–6]). They proved that, under some suitable assumptions, the following variational inequality is solvable: find x
∈
K such that
(6)
〈
A
x
,
η
(
y
,
x
)
〉
+
f
(
y
)
-
f
(
x
)
≥
0
,
∀
y
∈
K
,
where f is a function from K to ℝ
∪
{
∞
}. They also proved that the variational inequality (6) is equivalent to the following: find x
∈
K such that
(7)
〈
A
y
,
η
(
y
,
x
)
〉
+
f
(
y
)
-
f
(
x
)
≥
α
(
y
-
x
)
,
∀
y
∈
K
.In this paper, let us denote the set of solutions of the variational inequality (6) or (7) by Ω.It is worthy to notice that it is possible thatΩ is a singleton set, includes the finite elements or infinite elements. To show this, we see the following example.Example 1.
LetE
=
ℝ
×
ℝ and K
=
[
0,1
]
×
[
0,1
]. Define the mappings A
:
K
→
E
* by A
(
x
1
,
x
2
)
=
(
x
1
,
x
2
) for all (
x
1
,
x
2
)
∈
K, α
:
E
→
ℝ by α
(
(
x
1
,
x
2
)
)
=
3
x
1
2
+
3
x
2
2 for all (
x
1
,
x
2
)
∈
E, and η
:
K
×
K
→
E by η
(
(
x
1
,
x
2
)
,
(
y
1
,
y
2
)
)
=
(
4
(
x
1
-
y
1
)
,
4
(
x
2
-
y
2
)
) for all (
x
1
,
x
2
)
×
(
y
1
,
y
2
)
∈
K
×
K. Then the mapping A is a relaxed η-α-monotone mapping. In fact, for all x
=
(
x
1
,
x
2
), y
=
(
y
1
,
y
2
)
∈
K, we have
(8)
〈
A
x
-
A
y
,
η
(
x
,
y
)
〉
=
〈
(
x
1
-
y
1
,
x
2
-
y
2
)
,
(
4
(
x
1
-
y
1
)
,
4
(
x
2
-
y
2
)
)
〉
=
16
[
(
x
1
-
y
1
)
2
+
(
x
2
-
y
2
)
2
]
≥
3
[
(
x
1
-
y
1
)
2
+
(
x
2
-
y
2
)
2
]
=
α
(
x
-
y
)
.
Hence,A is a relaxed η-α-monotone mapping. Now by defining the different function f we find the solution set Ω of (6). First we compute, for x
=
(
x
1
,
x
2
), y
=
(
y
1
,
y
2
)
∈
K,
(9)
〈
A
x
,
η
(
y
,
x
)
〉
+
f
(
y
)
-
f
(
x
)
=
〈
A
(
x
1
,
x
2
)
,
η
(
(
y
1
,
y
2
)
,
(
x
1
,
x
2
)
)
〉
+
f
(
y
)
-
f
(
x
)
=
〈
(
x
1
,
x
2
)
,
(
4
(
y
1
-
x
1
)
,
4
(
y
2
-
x
2
)
)
〉
+
f
(
y
)
-
f
(
x
)
=
4
x
1
(
y
1
-
x
1
)
+
4
x
2
(
y
2
-
x
2
)
+
f
(
y
)
-
f
(
x
)
.
(i)
Let the functionf
(
x
)
=
x
1
+
x
2 for all x
=
(
x
1
,
x
2
)
∈
K. It is easy to see that the solution set Ω
=
{
(
0,0
)
}. That is, only when x
=
(
0,0
), (6) holds for all y
∈
K.
(ii)
Let the functionf
(
1,1
)
=
-
20 and f
(
x
1
,
x
2
)
=
x
1
+
x
2 for all K
∖
{
(
1,1
)
}. By (9), it is not hard to see that the solution set Ω
=
{
(
0,0
)
,
(
1,1
)
}.
(iii)
Let the functionf
(
x
1
,
x
2
)
=
-
20 for all (
x
1
,
x
2
)
=
(
1
/
n
,
1
/
n
)
∈
K where n
∈
ℕ and f
(
x
1
,
x
2
)
=
x
1
+
x
2 for all K
∖
{
(
1
/
n
,
1
/
n
)
:
n
∈
ℕ
}. Then by (9) we see that Ω
=
{
(
0,0
)
,
(
1
/
n
,
1
/
n
)
:
n
∈
ℕ
}. In this case, Ω includes the unfinite elements.Recently, Wang et al. [7] proposed the following equilibrium problem in a Hilbert space H with a nonempty closed convex bounded subset K: find x
∈
K such that
(10)
Θ
(
x
,
y
)
+
〈
T
z
,
η
(
y
,
z
)
〉
≥
0
,
∀
y
∈
K
,
where T
:
K
→
H is a relaxed η-α-monotone mapping. They introduced a new algorithm for solving the equilibrium problem in Hilbert spaces.In [3], Fang and Huang did not give the algorithm to solve the variational inequality (6) or (7). In this paper, motivated and inspired by the above results, we introduced a new mixed equilibrium problem with the relaxed monotone mapping and prove the existence of solutions of the mixed equilibrium problem. Then we propose an iterative scheme to find the common element of the set of solutions of the mixed equilibrium problem and the set of fixed points of a quasi-ϕ-nonexpansive mapping in Banach spaces. In particular, the variational inequality (6) or (7) may be solved by the algorithm proposed in this paper. Our results extend and improve the corresponding ones given by Takahashi and Zembayashi [1], Fang and Huang [3], and Wang et al. [7].


## 2. Preliminaries

A Banach spaceE is said to be strictly convex if ∥
(
x
+
y
)
/
2
∥
<
1 for all x
,
y
∈
E with ∥
x
∥
=
∥
y
∥
=
1 and x
≠
y. It is said to be uniformly convex if lim
n
→
∞
∥
x
n
-
y
n
∥
=
0 for any two sequences {
x
n
} and {
y
n
} in E such that ∥
x
n
∥
=
∥
y
n
∥
=
1 and lim
n
→
∞
∥
(
x
n
+
y
n
)
/
2
∥
=
1.LetU
=
{
x
∈
E
:
∥
x
∥
=
1
} be the unit sphere of E. A Banach space E is said to be smooth provided
(11)
lim
t
→
0
∥
x
+
t
y
∥
-
∥
x
∥
t
exists for each x
,
y
∈
U. It is said to be uniformly smooth if the limit is attained uniformly for x
,
y
∈
E.LetE
* be the topological dual space of E and J be the normalized duality mapping from E into 2
E
* given by
(12)
J
x
=
{
x
*
∈
E
*
:
〈
x
,
x
*
〉
=
∥
x
∥
∥
x
*
∥
,
∥
x
*
∥
=
∥
x
∥
}
,
∀
x
∈
E
.It is well known that ifE is uniformly convex, then J is uniformly continuous on bounded subsets of E and if E is uniformly smooth, then E
* is uniformly convex.Recently, Alber [8] introduced a generalized projection operator Π
K in a Banach space E with a nonempty closed convex subset K, which is an analogue of the metric projection in Hilbert spaces:letϕ
:
E
×
E
→
ℝ be a function defined by
(13)
ϕ
(
x
,
y
)
=
∥
x
∥
2
-
2
〈
x
,
J
y
〉
+
∥
y
∥
2
,
∀
x
,
y
∈
E
.Thegeneralized projection  Π
K
:
E
→
K is a mapping that assigns to an arbitrary point x
∈
E the minimum point of the functional ϕ
(
x
,
y
); that is, Π
K
x
=
x
^, where x
^ is a solution to the minimization problem
(14)
ϕ
(
x
^
,
x
)
=
inf
y
∈
K
ϕ
(
y
,
x
)
.The existence and uniqueness ofΠ
K follows from the properties of the function ϕ
(
x
,
y
) and the strict monotonicity of the mapping J (see, e.g., [8–12]).LetE be a smooth, strictly convex, and reflexive Banach space E. Then, for all x
,
y
∈
E, we have the following [8, 9, 11]:
(15)
(
∥
y
∥
-
∥
x
∥
)
2
≤
ϕ
(
y
,
x
)
≤
(
∥
y
∥
+
∥
x
∥
)
2
,
ϕ
(
x
,
y
)
=
0
⟺
x
=
y
.LetT be a mapping from K into itself. A point p
∈
K is said to be an asymptotic fixed point of T [13] if K contains a sequence {
x
n
} which converges weakly to p such that
(16)
lim
n
→
∞
∥
x
n
-
T
x
n
∥
=
0
.The set of asymptotic fixed points ofT is denoted by F
^
(
T
). The mapping T is said to be relatively nonexpansive [14–16] if
(17)
F
^
(
T
)
=
F
(
T
)
,
ϕ
(
p
,
T
x
)
≤
ϕ
(
p
,
x
)
,
∀
x
∈
K
,
p
∈
F
(
T
)
.We recall that a mappingT
:
K
→
K is said to be nonexpansive if ∥
T
x
-
T
y
∥
≤
∥
x
-
y
∥ for all x
,
y
∈
K, T is said to be ϕ
-nonexpansive if ϕ
(
T
x
,
T
y
)
≤
ϕ
(
x
,
y
) for all x
,
y
∈
K, and T is said to be quasi-
ϕ
-nonexpansive if
(18)
F
(
T
)
≠
∅
,
ϕ
(
p
,
T
x
)
≤
ϕ
(
p
,
x
)
,
∀
x
∈
K
,
p
∈
F
(
T
)
.Notice that the class of quasi-ϕ-nonexpansive mappings is more general than the class of relatively nonexpansive mappings [14–16] which requires the strong restriction that F
^
(
T
)
=
F
(
T
). Some examples of quasi-ϕ-nonexpansive mappings may be found in [2].A mappingS
:
E
→
E is said to be closed if for any sequence {
x
n
}
⊂
E such that lim
n
→
∞
x
n
=
x
0 and lim
n
→
∞
S
x
n
=
y
0, then S
x
0
=
y
0.The following lemmas are needed for the proof of our main results in next section.Lemma 2 (see [10]).
LetE be a uniformly convex and smooth Banach space and let {
x
n
}, {
y
n
} be two sequences of E. If ϕ
(
x
n
,
y
n
)
→
0 and either {
x
n
} or {
y
n
} is bounded, then x
n
-
y
n
→
0.Lemma 3 (see [8]).
LetK be a nonempty closed convex subset of a smooth Banach space E and x
∈
E. Then x
0
=
Π
K
x if and only if
(19)
〈
x
0
-
y
,
J
x
-
J
x
0
〉
≥
0
,
∀
y
∈
K
.Lemma 4 (see [8]).
LetE be a reflexive, strictly convex, and smooth Banach space, let K be a nonempty closed convex subset of E and x
∈
E. Then
(20)
ϕ
(
y
,
Π
K
x
)
+
ϕ
(
Π
K
x
,
x
)
≤
ϕ
(
y
,
x
)
,
∀
y
∈
K
.The following lemma is contained implicitly in Step 1 of Theorem 3.1 of [17].Lemma 5 (see [17]).
LetE be a uniformly convex and smooth Banach space, let K be a nonempty closed convex subset of E, and let T be a closed quasi-ϕ-nonexpansive mapping from K into itself. Then F
(
T
) is a closed convex subset of K.Lemma 6 (see [18]).
LetE be a uniformly convex Banach space and let B
r
(
0
) be a closed ball of E. Then there exists a continuous strictly increasing convex function g
:
[
0
,
∞
)
→
[
0
,
∞
) with g
(
0
)
=
0 such that
(21)
∥
λ
x
+
μ
y
+
γ
z
∥
2
≤
λ
∥
x
∥
2
+
μ
∥
y
∥
2
+
γ
∥
z
∥
2
-
λ
μ
g
(
∥
x
-
y
∥
)
for all x
,
y
,
z
∈
B
r
(
0
) and λ
,
μ
,
γ
∈
[
0,1
] with λ
+
μ
+
γ
=
1.


## 3. Main Results

For our main results, we introduce some definitions and lemmas as follows.Definition 7 (see [3]).
LetE be a Banach space with the dual space E
* and let K be a nonempty subset of E. Let A
:
K
→
E
* and η
:
K
×
K
→
E be two mappings. The mapping A
:
K
→
E
* is said to be η
-hemicontinuous if, for any fixed x
,
y
∈
K, the function f
:
[
0,1
]
→
(
-
∞
,
∞
) defined by f
(
t
)
=
〈
A
(
(
1
-
t
)
x
+
t
y
)
,
η
(
x
,
y
)
〉 is continuous at 0
+.Definition 8.
LetE be a Banach space with the dual space E
* and let K be a nonempty subset of E. A mapping F
:
K
→
2
E is called a KKM mapping if, for any {
x
1
,
…
,
x
n
}
⊂
K,
(22)
co
¯
{
x
1
,
…
,
x
n
}
⊂
⋃
i
=
1
n
F
(
x
i
)
,
where 2
E denotes the family of all the nonempty subsets of E.Lemma 9 (see [19]).
LetM be a nonempty subset of a Hausdorff topological vector space X and let F
:
M
→
2
X be a KKM mapping. If F
(
x
) is closed in X for all x
∈
X in K and compact for some x
∈
K, then ∩
x
∈
M
F
(
x
)
≠
∅.Lemma 10.
Let
E
be a reflexive Banach space with the dual space
E
*
and let
K
be a nonempty closed convex subset of
E
. Let
A
:
K
→
E
*
be an
η
-hemicontinuous and relaxed
η
-
α
-monotone mapping. Let
Θ
be a bifunction from
K
×
K
to
ℝ
satisfying
(
A
1
)
and
(
A
4
)
and let
f
be a proper convex function from
K
×
K
to
ℝ
∪
{
+
∞
}
. Let
r
>
0
and
z
∈
E
. Assume that
(i)
η
(
x
,
x
)
=
0 for all x
∈
K;
(ii)
for any fixedu
,
v
∈
K, the mapping x
↦
〈
A
v
,
η
(
x
,
u
)
〉 is convex.
Then the following problems (23) and (24) are equivalent:(I)
findx
∈
K such that
(23)
Θ
(
x
,
y
)
+
〈
A
x
,
η
(
y
,
x
)
〉
+
f
(
y
)
-
f
(
x
)
+
1
r
〈
y
-
x
,
J
x
-
J
z
〉
≥
0
,
∀
y
∈
K
;
(II)
findx
∈
K such that
(24)
Θ
(
x
,
y
)
+
〈
A
y
,
η
(
y
,
x
)
〉
+
f
(
y
)
-
f
(
x
)
+
1
r
〈
y
-
x
,
J
x
-
J
z
〉
≥
α
(
y
-
x
)
,
∀
y
∈
K
.Proof.
Letx
∈
K be a solution of the problem (23). Since A is relaxed η-α monotone, we have
(25)
Θ
(
x
,
y
)
+
〈
A
y
,
η
(
y
,
x
)
〉
+
f
(
y
)
-
f
(
x
)
+
1
r
〈
y
-
x
,
J
x
-
J
z
〉
≥
Θ
(
x
,
y
)
+
α
(
y
-
x
)
+
f
(
y
)
-
f
(
x
)
+
1
r
〈
y
-
x
,
J
x
-
J
z
〉
+
〈
A
x
,
η
(
y
,
x
)
〉
≥
α
(
y
-
x
)
,
∀
y
∈
K
.
Thus, x
∈
K is a solution of the problem (24).
Conversely, letx
∈
K be a solution of the problem (24) and let y
∈
K be any point with f
(
y
)
<
∞. From (24), it follows that f
(
x
)
<
∞. Letting
(26)
y
t
=
(
1
-
t
)
x
+
t
y
,
∀
t
∈
(
0,1
)
,
we have y
t
∈
K. Since x
∈
K is a solution of the problem (24), it follows that
(27)
Θ
(
x
,
y
t
)
+
〈
A
y
t
,
η
(
y
t
,
x
)
〉
+
f
(
y
t
)
-
f
(
x
)
+
1
r
〈
y
t
-
x
,
J
x
-
J
z
〉
≥
α
(
y
t
-
x
)
=
t
p
α
(
y
-
x
)
.
The convexity off and conditions (i), (ii), (A1), and (A4) imply that
(28)
f
(
y
t
)
-
f
(
x
)
=
f
(
(
1
-
t
)
x
+
t
y
)
-
f
(
x
)
≤
t
(
f
(
y
)
-
f
(
x
)
)
,
〈
A
y
t
,
η
(
y
t
,
x
)
〉
≤
(
1
-
t
)
〈
A
y
t
,
η
(
x
,
x
)
〉
+
t
〈
A
y
t
,
η
(
y
,
x
)
〉
=
t
〈
A
(
x
+
t
(
y
-
x
)
)
,
η
(
y
,
x
)
〉
,
Θ
(
x
,
y
t
)
≤
(
1
-
t
)
Θ
(
x
,
x
)
+
t
Θ
(
x
,
y
)
=
t
Θ
(
x
,
y
)
.
Thus, it follows from (27)-(28) that
(29)
Θ
(
x
,
y
)
+
〈
A
(
x
+
t
(
y
-
x
)
)
,
η
(
y
,
x
)
〉
+
f
(
y
)
-
f
(
x
)
+
1
r
〈
y
-
x
,
J
x
-
J
z
〉
≥
t
p
-
1
α
(
y
-
x
)
,
∀
y
∈
K
.
SinceA is η-hemicontinuous and p
>
1, letting t
→
0 in (29), we get
(30)
Θ
(
x
,
y
)
+
〈
A
x
,
η
(
y
,
x
)
〉
+
f
(
y
)
-
f
(
x
)
+
1
r
〈
y
-
x
,
J
x
-
J
z
〉
≥
0
for all y
∈
K with f
(
y
)
<
∞. When f
(
y
)
=
+
∞, the relation
(31)
Θ
(
x
,
y
)
+
〈
A
x
,
η
(
y
,
x
)
〉
+
f
(
y
)
-
f
(
x
)
+
1
r
〈
y
-
x
,
J
x
-
J
z
〉
≥
0
is trivial. Therefore, x
∈
K is also a solution of the problem (23). This completes the proof.Lemma 11.
LetE be a reflexive Banach space with the dual space E
* and let K be a nonempty closed convex bounded subset of E. Let A
:
K
→
E
* be an η-hemicontinuous and relaxed η-α-monotone mapping, let Θ be a bifunction from K
×
K to ℝ satisfying (
A
1
) and (
A
4
), and let f be a proper convex function from K
×
K to ℝ
∪
{
+
∞
}. Let r
>
0 and z
∈
E. Assume that (i)
η
(
x
,
y
)
+
η
(
y
,
x
)
=
0 for all x
,
y
∈
K;
(ii)
for any fixedu
,
v
∈
K, the mapping x
↦
〈
A
v
,
η
(
x
,
u
)
〉 is convex and lower semicontinuous;
(iii)
α
:
E
→
ℝ is weakly lower semicontinuous; that is, for any net {
x
β
}, {
x
β
} converges to x in σ
(
E
,
E
*
) implying that α
(
x
)
≤
liminf
α
(
x
β
). Then the problem (23) is solvable.Proof.
Define two set-valued mappingsF
,
G
:
K
→
2
E as follows:
(32)
F
(
y
)
=
{
+
1
r
〈
y
-
x
,
J
x
-
J
z
〉
≥
0
x
∈
K
:
Θ
(
x
,
y
)
+
〈
A
x
,
η
(
y
,
x
)
〉
+
f
(
y
)
-
f
(
x
)
+
1
r
〈
y
-
x
,
J
x
-
J
z
〉
≥
0
}
,
∀
y
∈
K
,
G
(
y
)
=
{
+
1
r
〈
y
-
x
,
J
x
-
J
z
〉
≥
α
(
y
-
x
)
x
∈
K
:
Θ
(
x
,
y
)
+
〈
A
y
,
η
(
y
,
x
)
〉
+
f
(
y
)
-
f
(
x
)
+
1
r
〈
y
-
x
,
J
x
-
J
z
〉
≥
α
(
y
-
x
)
}
,
∀
y
∈
K
,
respectively.
Now, we claim thatF is a KKM mapping. If F is not a KKM mapping, then there exist {
y
1
,
…
,
y
n
}
⊂
K and t
i
>
0, i
=
1
,
…
,
n, such that
(33)
∑
i
=
1
n
t
i
=
1
,
y
=
∑
i
=
1
n
t
i
y
i
∉
⋃
i
=
1
n
F
(
y
i
)
.
By the definition ofF, we have
(34)
Θ
(
y
,
y
i
)
+
〈
A
y
,
η
(
y
i
,
y
)
〉
+
f
(
y
i
)
-
f
(
y
)
+
1
r
〈
y
i
-
y
,
J
y
-
J
z
〉
<
0
,
∀
i
=
1
,
…
,
n
.
It follows that(35)
0
=
Θ
(
y
,
y
)
=
Θ
(
y
,
∑
i
=
1
n
t
i
y
i
)
+
〈
A
y
,
η
(
∑
i
=
1
n
t
i
y
i
,
y
)
〉
≤
∑
i
=
1
n
t
i
Θ
(
y
,
y
i
)
+
∑
i
=
1
n
t
i
〈
A
y
,
η
(
y
i
,
y
)
〉
<
∑
i
=
1
n
t
i
(
f
(
y
)
-
f
(
y
i
)
+
1
r
〈
y
-
y
i
,
J
y
-
J
z
〉
)
=
f
(
y
)
-
∑
i
=
1
n
t
i
f
(
y
i
)
≤
f
(
y
)
-
f
(
y
)
=
0
,
which is a contradiction. This implies that F is a KKM mapping.
Now, we prove that(36)
F
(
y
)
⊂
G
(
y
)
,
∀
y
∈
K
.
For any giveny
∈
K, letting x
∈
F
(
y
), then
(37)
Θ
(
x
,
y
)
+
〈
A
x
,
η
(
y
,
x
)
〉
+
f
(
y
)
-
f
(
x
)
+
1
r
〈
y
-
x
,
J
x
-
J
z
〉
≥
0
.
SinceA is relaxed η-α-monotone, we have
(38)
Θ
(
x
,
y
)
+
〈
A
y
,
η
(
y
,
x
)
〉
+
f
(
y
)
-
f
(
x
)
+
1
r
〈
y
-
x
,
J
x
-
J
z
〉
≥
Θ
(
x
,
y
)
+
〈
A
x
,
η
(
y
,
x
)
〉
+
α
(
y
-
x
)
+
f
(
y
)
-
f
(
x
)
+
1
r
〈
y
-
x
,
J
x
-
J
z
〉
≥
α
(
y
-
x
)
,
which implies that x
∈
G
(
y
) and so
(39)
F
(
y
)
⊂
G
(
y
)
,
∀
y
∈
K
.
As in the proof of Theorem  2.2 in [3], also, we can obtain
(40)
⋂
y
∈
K
F
(
y
)
=
⋂
y
∈
K
G
(
y
)
≠
∅
.
Hence, there existsx
∈
K such that
(41)
Θ
(
x
,
y
)
+
〈
A
x
,
η
(
y
,
x
)
〉
+
f
(
y
)
-
f
(
x
)
+
1
r
〈
y
-
x
,
J
x
-
J
z
〉
≥
0
,
∀
y
∈
K
.
This completes the proof.Next, let us denote the set of solutions of the following mixed equilibrium problem byEP
(
Θ
,
A
): find x
∈
K such that
(42)
Θ
(
x
,
y
)
+
〈
A
x
,
η
(
y
,
x
)
〉
+
f
(
y
)
-
f
(
x
)
≥
0
,
∀
y
∈
K
.It is easy to see that ifΘ
(
x
,
y
)
≡
0 for all x
,
y
∈
K, then the above mixed equilibrium problem reduces to the variational inequality problem (6) or (7).Lemma 12.
LetE be a uniformly smooth, strictly convex Banach space with the dual space E
* and let K be a nonempty closed convex bounded subset of E. Let A
:
K
→
E
* be an η-hemicontinuous and relaxed η-α-monotone mapping, let Θ be a bifunction from K
×
K to ℝ satisfying (A1), (A2), and (A4), and let f be a proper convex function from K
×
K to ℝ
∪
{
+
∞
}. Let r
>
0 and define a mapping T
r
:
E
→
K as follows:
(43)
T
r
(
x
)
=
{
+
1
r
〈
y
-
z
,
J
z
-
J
x
〉
≥
0
f
o
r
a
l
l
y
∈
K
z
∈
K
:
Θ
(
z
,
y
)
+
〈
A
z
,
η
(
y
,
z
)
〉
+
f
(
y
)
-
f
(
z
)
+
1
r
〈
y
-
z
,
J
z
-
J
x
〉
≥
0
,
∀
y
∈
K
}
for all x
∈
E. Assume that
(i)η
(
x
,
y
)
+
η
(
y
,
x
)
=
0, for all x
,
y
∈
K;
(ii) for any fixedu
,
v
∈
K, the mapping x
↦
〈
A
v
,
η
(
x
,
u
)
〉 is convex and lower semicontinuous;
(iii)α
:
E
→
ℝ is weakly lower semicontinuous; that is, for any net {
x
β
}, x
β converges to x in σ
(
E
,
E
*
) implying that α
(
x
)
≤
liminf
α
(
x
β
);
(iv) for anyx
,
y
∈
K, α
(
x
-
y
)
+
α
(
y
-
x
)
≥
0;
(v)〈
A
(
t
z
1
+
(
1
-
t
)
z
2
)
,
η
(
y
,
t
z
1
+
(
1
-
t
)
z
2
)
〉
≥
t
〈
A
z
1
,
η
(
y
,
z
1
)
〉
+
(
1
-
t
)
〈
A
z
2
,
η
(
y
,
z
2
)
〉, for any z
1
,
z
2
,
y
∈
K and t
∈
[
0,1
].
Then the following hold:(1)
T
r is single-valued;
(2)
T
r is a firmly nonexpansive-type mapping; that is, for all x
,
y
∈
E,
(44)
〈
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
)
=
EP
(
Θ
,
A
);
(4)
T
r is quasi-ϕ-nonexpansive satisfying ϕ
(
w
,
T
r
x
)
+
ϕ
(
T
r
x
,
x
)
≤
ϕ
(
w
,
x
) for all w
∈
F
(
T
r
) and x
∈
E;
(5)
EP
(
Θ
,
A
) is closed and convex.Proof.
We claim thatT
r is single-valued. Indeed, for any x
∈
C and r
>
0, let z
1
,
z
2
∈
T
r
x. Then we have
(45)
Θ
(
z
1
,
z
2
)
+
〈
A
z
1
,
η
(
z
2
,
z
1
)
〉
+
f
(
z
2
)
-
f
(
z
1
)
+
1
r
〈
z
2
-
z
1
,
J
z
1
-
J
x
〉
≥
0
,
Θ
(
z
2
,
z
1
)
+
〈
A
z
2
,
η
(
z
1
,
z
2
)
〉
+
f
(
z
1
)
-
f
(
z
2
)
+
1
r
〈
z
1
-
z
2
,
J
z
2
-
J
x
〉
≥
0
.
Adding the two inequalities, it follows from (i) that(46)
Θ
(
z
1
,
z
2
)
+
Θ
(
z
2
,
z
1
)
+
〈
A
z
1
-
A
z
2
,
η
(
z
2
,
z
1
)
〉
+
1
r
〈
z
1
-
z
2
,
J
z
2
-
J
z
1
〉
≥
0
.
Thus, from (A2), we have(47)
〈
A
z
1
-
A
z
2
,
η
(
z
2
,
z
1
)
〉
+
1
r
〈
z
1
-
z
2
,
J
z
2
-
J
z
1
〉
≥
0
;
that is,
(48)
1
r
〈
z
1
-
z
2
,
J
z
2
-
J
z
1
〉
≥
〈
A
z
2
-
A
z
1
,
η
(
z
2
,
z
1
)
〉
.
SinceT is relaxed η-α-monotone and r
>
0, one has
(49)
〈
z
1
-
z
2
,
J
z
2
-
J
z
1
〉
≥
r
α
(
z
2
-
z
1
)
.
In (48), exchanging z
1 to z
2, we get
(50)
1
r
〈
z
2
-
z
1
,
J
z
1
-
J
z
2
〉
≥
〈
A
z
1
-
A
z
2
,
η
(
z
1
,
z
2
)
〉
≥
α
(
z
1
-
z
2
)
;
that is,
(51)
〈
z
1
-
z
2
,
J
z
2
-
J
z
1
〉
≥
r
α
(
z
1
-
z
2
)
.
Thus, it follows from (49), (51), and (iv) that
(52)
〈
z
1
-
z
2
,
J
z
2
-
J
z
1
〉
≥
0
.
Since E is strictly convex, we have z
1
=
z
2.
Next, we claim thatT
r is a firmly nonexpansive-type mapping. Indeed, for any x
,
y
∈
K, we have
(53)
Θ
(
T
r
x
,
T
r
y
)
+
〈
A
T
r
x
,
η
(
T
r
y
,
T
r
x
)
〉
+
f
(
T
r
y
)
-
f
(
T
r
x
)
+
1
r
〈
T
r
y
-
T
r
x
,
J
T
r
x
-
J
x
〉
≥
0
,
Θ
(
T
r
y
,
T
r
x
)
+
〈
A
T
r
y
,
η
(
T
r
x
,
T
r
y
)
〉
+
f
(
T
r
x
)
-
f
(
T
r
y
)
+
1
r
〈
T
r
x
-
T
r
y
,
J
T
r
y
-
J
y
〉
≥
0
.
Adding the above two inequalities, by (i) and (A2), we get(54)
〈
A
T
r
x
-
A
T
r
y
,
η
(
T
r
y
,
T
r
x
)
〉
+
1
r
〈
T
r
y
-
T
r
x
,
J
T
r
x
-
J
T
r
y
-
J
x
+
J
y
〉
≥
0
;
that is,
(55)
1
r
〈
T
r
y
-
T
r
x
,
J
T
r
x
-
J
T
r
y
-
J
x
+
J
y
〉
≥
〈
A
T
r
y
-
A
T
r
x
,
η
(
T
r
y
,
T
r
x
)
〉
.
Repeating the process from (48) to (52), we can obtain
(56)
〈
T
r
y
-
T
r
x
,
J
T
r
x
-
J
T
r
y
-
J
x
+
J
y
〉
≥
0
.
Therefore, we have(57)
〈
T
r
x
-
T
r
y
,
J
T
r
x
-
J
T
r
y
〉
≤
〈
T
r
x
-
T
r
y
,
J
x
-
J
y
〉
,
which shows that T
r is firmly nonexpansive.
Next, we claim thatF
(
T
r
)
=
EP
(
Θ
,
A
). Indeed, we have the following:
(58)
u
∈
F
(
T
r
)
⟺
u
=
T
r
u
⟺
Θ
(
u
,
y
)
+
〈
A
u
,
η
(
y
,
u
)
〉
+
f
(
y
)
-
f
(
u
)
+
1
r
〈
y
-
u
,
J
u
-
J
u
〉
≥
0
,
∀
y
∈
K
⟺
Θ
(
u
,
y
)
+
〈
A
u
,
η
(
y
,
u
)
〉
+
f
(
y
)
-
f
(
u
)
≥
0
,
∀
y
∈
K
⟺
u
∈
EP
(
Θ
,
A
)
.
Finally, we prove (4) and (5). Indeed, we can obtain (4) directly from Lemma 2.8 and Lemma 2.9 of [1]. Moreover, since T
r is quasi-ϕ-nonexpansive, one sees that T
r is closed and convex from Lemma 5. Therefore, EP
(
Θ
,
A
) is also closed and convex. This completes the proof.Remark 13.
In (42), if A
(
x
)
≡
0 for all x
∈
K, then the equilibrium problem (42) is reduced to the equilibrium problem in [20] and if η
(
y
,
x
)
=
y
-
x for all x
,
y
∈
K, then then the equilibrium problem (42) is reduced to the equilibrium problem in [21].In the following, we give the main result of this paper.Theorem 14.
LetK be a nonempty closed convex bounded subset of a uniformly convex, uniformly smooth Banach space E with the dual space E
*. Let A
:
K
→
E
* be an η-hemicontinuous and relaxed η-α monotone mapping, let Θ be a bifunction from K
×
K to ℝ satisfying (A1)–(A4), and let f be a proper convex and lower semicontinuous function from K
×
K to ℝ
∪
{
+
∞
}. Let S
:
E
→
E be a closed quasi-ϕ-nonexpansive mapping such that F
=
F
(
S
)
∩
EP
(
Θ
,
A
)
≠
∅. Assume that the conditions (i)–(v), Lemma 12, and the following condition hold:
(vi) for allx
,
y
,
z
,
w
∈
K,
(59)
limsup
t
↓
0
〈
A
z
,
η
(
x
,
t
y
+
(
1
-
t
)
w
)
〉
≤
〈
A
z
,
η
(
x
,
w
)
〉
.
Let{
x
n
} be a sequence in K generated by the following manner:
(60)
x
1
=
x
∈
K
c
h
o
s
e
n
a
r
b
i
t
r
a
r
i
l
y
,
y
n
=
J
-
1
(
α
n
J
x
n
+
(
1
-
α
n
)
J
S
x
n
)
,
u
n
∈
K
s
u
c
h
t
h
a
t
Θ
(
u
n
,
y
)
+
〈
A
u
n
,
η
(
y
,
u
n
)
〉
+
f
(
y
)
-
f
(
u
n
)
+
1
r
n
〈
y
-
u
n
,
J
u
n
-
J
y
n
〉
≥
0
,
∀
y
∈
K
,
C
n
=
{
z
∈
K
:
ϕ
(
z
,
u
n
)
≤
ϕ
(
z
,
x
n
)
}
,
D
n
=
⋂
i
=
1
n
C
i
,
x
n
+
1
=
Π
D
n
x
,
∀
n
≥
1
,
where J is the duality mapping on E, {
α
n
}
⊂
(
0,1
) with 0
<
liminf
n
→
∞
α
n
(
1
-
α
n
), and {
r
n
}
⊂
(
r
,
∞
) with r
>
0. Then the sequence {
x
n
} generated by (52) converges strongly to Π
F
x.Proof.
First, since the inequalityϕ
(
z
,
u
n
)
≤
ϕ
(
z
,
x
n
) is equivalent to the following:
(61)
2
(
〈
z
,
J
x
n
〉
-
〈
z
,
J
u
n
〉
)
≤
∥
x
n
∥
2
-
∥
u
n
∥
2
,
C
n is closed and convex for all n
≥
1. It follows from the definition of D
n that D
n is also closed and convex for all n
≥
1.
On the other hand, since everyT
r
n is quasi-ϕ-nonexpansive, according to Lemma 12 (4), we have, for any w
∈
F,
(62)
ϕ
(
w
,
u
n
)
=
ϕ
(
w
,
T
r
n
y
n
)
≤
ϕ
(
w
,
y
n
)
=
ϕ
(
w
,
J
-
1
(
α
n
J
x
n
+
(
1
-
α
n
)
J
S
x
n
)
)
=
∥
w
∥
2
-
2
α
n
〈
w
J
x
n
〉
-
2
(
1
-
α
n
)
〈
w
,
J
S
x
n
〉
+
∥
α
n
J
x
n
+
(
1
-
α
n
)
J
S
x
n
∥
2
≤
∥
w
∥
2
-
2
α
n
〈
w
J
x
n
〉
-
2
(
1
-
α
n
)
〈
w
,
J
S
x
n
〉
+
α
n
∥
x
n
∥
2
+
(
1
-
α
n
)
∥
J
S
x
n
∥
2
=
α
n
ϕ
(
w
,
x
n
)
+
(
1
-
α
n
)
ϕ
(
w
,
S
x
n
)
≤
α
n
ϕ
(
w
,
x
n
)
+
(
1
-
α
n
)
ϕ
(
w
,
x
n
)
=
ϕ
(
w
,
x
n
)
,
which shows that w
∈
C
n for all n
≥
1. This implies that F
⊂
C
n for all n
≥
1. The definition of D
n shows that F
⊂
D
n for all n
≥
1. Hence, the sequence {
x
n
} generated by the algorithm (60) is well defined.
Note thatD
n
⊂
D
n
-
1 for all n
≥
1. Hence, by x
n
+
1
=
Π
D
n
x
∈
D
n
⊂
D
n
-
1 and x
n
=
Π
D
n
-
1
x, we have
(63)
ϕ
(
x
n
,
x
)
≤
ϕ
(
x
n
+
1
,
x
)
,
∀
n
≥
1
.
Therefore,{
ϕ
(
x
n
,
x
)
} is nondecreasing. Since {
x
n
}
⊂
K, {
ϕ
(
x
n
,
x
)
} is bounded. It follows that the limit of {
ϕ
(
x
n
,
x
)
} exists.
Letm
,
n
∈
ℕ with m
>
n. Since x
m
=
Π
D
m
-
1
x
∈
D
m
-
1
⊂
D
n
-
1, from Lemma 4 it follows that
(64)
ϕ
(
x
m
,
x
n
)
=
ϕ
(
x
m
,
Π
D
n
-
1
x
)
≤
ϕ
(
x
m
,
x
)
-
ϕ
(
Π
D
n
-
1
x
,
x
)
=
ϕ
(
x
m
,
x
)
-
ϕ
(
x
n
,
x
)
.
Lettingn
→
∞ in (64), it follows that ϕ
(
x
m
,
x
n
)
→
0. It follows from Lemma 2 that ∥
x
m
-
x
n
∥
→
0 as m
,
n
→
∞. Hence {
x
n
} is a Cauchy sequence. There exists a point x
*
∈
K such that x
n
→
x
* as n
→
∞.
Next, we show thatx
*
∈
F
(
S
). By taking p
=
1 in (64), one arrives at
(65)
lim
n
→
∞
ϕ
(
x
n
+
1
,
x
n
)
=
0
.
From Lemma2, it follows that
(66)
lim
n
→
∞
∥
x
n
+
1
-
x
n
∥
=
0
.
Noticing thatx
n
+
1
∈
D
n
⊂
C
n for all n
≥
1, we obtain
(67)
ϕ
(
x
n
+
1
,
u
n
)
≤
ϕ
(
x
n
+
1
,
x
n
)
.
Thus, it follows from (65) that
(68)
lim
n
→
∞
ϕ
(
x
n
+
1
,
u
n
)
=
0
.
It follows from Lemma2 that
(69)
lim
n
→
∞
∥
x
n
+
1
-
u
n
∥
=
0
.
Combining (66) with (69), one has
(70)
lim
n
→
∞
∥
x
n
-
u
n
∥
=
0
.
Sincex
n
→
x
* as n
→
∞, we have
(71)
u
n
⟶
x
*
(
n
⟶
∞
)
.
On the other hand, sinceJ is uniformly norm-to-norm continuous on bounded sets, one has
(72)
lim
n
→
∞
∥
J
x
n
-
J
u
n
∥
=
0
.
SinceE is a uniformly smooth Banach space, one knows that E
* is a uniformly convex Banach space. Let r
=
sup
n
≥
1
{
∥
x
n
∥
,
∥
S
x
n
∥
}. It follows from Lemma 6 that, for any w
∈
F,
(73)
ϕ
(
w
,
u
n
)
=
ϕ
(
w
,
T
r
n
y
n
)
≤
ϕ
(
w
,
y
n
)
=
ϕ
(
w
,
J
-
1
(
α
n
J
x
n
+
(
1
-
α
n
)
J
S
x
n
)
)
=
∥
w
∥
2
-
2
α
n
〈
w
,
J
x
n
〉
-
2
(
1
-
α
n
)
〈
w
,
J
S
x
n
〉
+
∥
α
n
J
x
n
+
(
1
-
α
n
)
J
S
x
n
∥
2
≤
∥
w
∥
2
-
2
α
n
〈
w
,
J
x
n
〉
-
2
(
1
-
α
n
)
〈
w
,
J
S
x
n
〉
+
α
n
∥
J
x
n
∥
2
+
(
1
-
α
n
)
∥
J
S
x
n
∥
2
-
α
n
(
1
-
α
n
)
g
(
∥
J
S
x
n
-
J
x
n
∥
)
=
α
n
ϕ
(
w
,
x
n
)
+
(
1
-
α
n
)
ϕ
(
w
,
S
x
n
)
-
α
n
(
1
-
α
n
)
g
(
∥
J
S
x
n
-
J
x
n
∥
)
≤
ϕ
(
w
,
x
n
)
-
α
n
(
1
-
α
n
)
g
(
∥
J
S
x
n
-
J
x
n
∥
)
.
It follows that(74)
α
n
(
1
-
α
n
)
g
(
∥
J
S
x
n
-
J
x
n
∥
)
≤
ϕ
(
w
,
x
n
)
-
ϕ
(
w
,
u
n
)
.
On the other hand, one has(75)
ϕ
(
w
,
x
n
)
-
ϕ
(
w
,
u
n
)
=
∥
x
n
∥
2
-
∥
u
n
∥
2
-
2
〈
w
,
J
x
n
-
J
u
n
〉
≤
∥
x
n
-
u
n
∥
(
∥
x
n
∥
+
∥
u
n
∥
)
+
2
∥
w
∥
∥
J
x
n
-
J
u
n
∥
.
Thus, it follows from (70) and (72) that
(76)
ϕ
(
w
,
x
n
)
-
ϕ
(
w
,
u
n
)
⟶
0
as
n
⟶
∞
.
Noting that0
<
liminf
n
→
∞
α
n
(
1
-
α
n
) and using (74) and (76), we get
(77)
g
(
∥
J
x
n
-
J
S
x
n
∥
)
⟶
0
(
n
⟶
∞
)
.
It follows from the continuousness ofg that
(78)
∥
J
x
n
-
J
S
x
n
∥
⟶
0
(
n
⟶
∞
)
.
SinceJ
-
1 is also uniformly norm-to-norm continuous on bounded sets, one gets
(79)
lim
n
→
∞
∥
x
n
-
S
x
n
∥
=
0
.
From the closeness of S, we can conclude that x
*
∈
F
(
S
).
Next, we show thatx
*
∈
EP
(
Θ
,
A
). From (62), we arrived at
(80)
ϕ
(
w
,
y
n
)
≤
ϕ
(
w
,
x
n
)
.
Fromu
n
=
T
r
n
y
n and Lemma 12 (4), one has
(81)
ϕ
(
u
n
,
y
n
)
=
ϕ
(
T
r
n
y
n
,
y
n
)
≤
ϕ
(
w
,
y
n
)
-
ϕ
(
w
,
T
r
n
y
n
)
≤
ϕ
(
w
,
x
n
)
-
ϕ
(
w
,
T
r
n
y
n
)
=
ϕ
(
w
,
x
n
)
-
ϕ
(
w
,
u
n
)
.
Thus, it follows from (76) that
(82)
ϕ
(
u
n
,
y
n
)
⟶
0
(
n
⟶
∞
)
.
From Lemma2, we get
(83)
∥
u
n
-
y
n
∥
⟶
0
(
n
⟶
∞
)
.
Noting thatu
n
=
T
r
n
y
n, one obtains
(84)
Θ
(
u
n
,
y
)
+
〈
A
u
n
,
η
(
y
,
u
n
)
〉
+
f
(
y
)
-
f
(
u
n
)
+
1
r
n
〈
y
-
u
n
,
J
u
n
-
J
y
n
〉
≥
0
,
∀
y
∈
K
.
From (A4) and (i), it follows that(85)
1
r
n
∥
y
-
u
n
∥
∥
J
u
n
-
J
y
n
∥
≥
〈
A
u
n
,
η
(
u
n
,
y
)
〉
+
f
(
u
n
)
-
f
(
y
)
-
Θ
(
u
n
,
y
)
≥
〈
A
u
n
,
η
(
u
n
,
y
)
〉
+
f
(
u
n
)
-
f
(
y
)
+
Θ
(
y
,
u
n
)
,
∀
y
∈
K
.
SinceJ is uniformly norm-to-norm continuous on bounded sets, one has
(86)
lim
n
→
∞
∥
J
u
n
-
J
y
n
∥
=
0
.
Noticing thatr
n
≥
r
>
0 for all n
≥
1, it follows from (A4), (ii), (85), and (86) that
(87)
0
≥
〈
A
x
*
,
η
(
x
*
,
y
)
〉
+
f
(
x
*
)
-
f
(
y
)
+
Θ
(
y
,
x
*
)
,
∀
y
∈
K
.
For all0
<
t
≤
1 and y
∈
K, define y
t
=
t
y
+
(
1
-
t
)
x
*. Noticing that x
*
,
y
∈
K, one obtains y
t
∈
K, which yields that
(88)
0
≥
〈
A
x
*
,
η
(
x
*
,
y
t
)
〉
+
f
(
x
*
)
-
f
(
y
t
)
+
Θ
(
y
t
,
x
*
)
.
It follows from (A1), (A4), (i), (ii), the convexity off, and (88) that
(89)
0
=
Θ
(
y
t
,
y
t
)
+
〈
A
x
*
,
η
(
y
t
,
y
t
)
〉
+
f
(
y
t
)
-
f
(
y
t
)
≤
t
[
Θ
(
y
t
,
y
)
+
〈
A
x
*
,
η
(
y
,
y
t
)
〉
+
f
(
y
)
-
f
(
y
t
)
]
+
(
1
-
t
)
[
Θ
(
y
t
,
x
*
)
+
〈
A
x
*
,
η
(
p
,
y
t
)
〉
+
f
(
x
*
)
-
f
(
y
t
)
]
≤
t
[
Θ
(
y
t
,
y
)
+
〈
A
x
*
,
η
(
y
,
y
t
)
〉
+
f
(
y
)
-
f
(
y
t
)
]
;
that is,
(90)
Θ
(
y
t
,
y
)
+
〈
A
x
*
,
η
(
y
,
y
t
)
〉
+
f
(
y
)
-
f
(
y
t
)
≥
0
.
Lettingt
↓
0, it follows from (A3), (vi), and the lower semicontinuity of f that
(91)
Θ
(
x
*
,
y
)
+
〈
A
x
*
,
η
(
y
,
x
*
)
〉
+
f
(
y
)
-
f
(
x
*
)
≥
0
,
∀
y
∈
K
.
This implies that x
*
∈
EP
(
Θ
,
A
).
Finally, we provex
*
=
Π
F
x. From x
n
+
1
=
Π
D
n
x, we see that
(92)
〈
x
n
+
1
-
z
,
J
x
-
J
x
n
+
1
〉
≥
0
,
∀
z
∈
D
n
.
SinceF
⊂
D
n for each n
≥
1, we have
(93)
〈
x
n
+
1
-
w
,
J
x
-
J
x
n
+
1
〉
≥
0
,
∀
w
∈
F
.
By taking the limit in (93), one has
(94)
〈
x
*
-
w
,
J
x
-
J
x
*
〉
≥
0
,
∀
w
∈
F
.
At this point, in view of Lemma 3, one sees that x
*
=
Π
F
x. This completes the proof.IfΘ
≡
0 in Theorem 14, then we have the following result.Corollary 15.
LetK be a nonempty closed convex bounded subset of a uniformly convex, uniformly smooth Banach space E with the dual space E
*. Let A
:
K
→
E
* be an η-hemicontinuous and relaxed η-α-monotone mapping and let f be a proper convex and lower semicontinuous function from K
×
K to ℝ
∪
{
+
∞
}. Let S
:
E
→
E be a closed quasi-ϕ-nonexpansive mapping such that F
=
F
(
S
)
∩
Ω
≠
∅. Assume that the conditions (i)–(v), Lemma 12, and the following condition hold:
(vi) for allx
,
y
,
z
,
p
∈
K,
(95)
limsup
t
↓
0
〈
A
z
,
η
(
x
,
t
y
+
(
1
-
t
)
p
)
〉
≤
〈
A
z
,
η
(
x
,
p
)
〉
.
Let{
x
n
} be a sequence in K generated by the following manner:
(96)
x
1
=
x
∈
K
c
h
o
s
e
n
a
r
b
i
t
r
a
r
i
l
y
,
y
n
=
J
-
1
(
α
n
J
x
n
+
(
1
-
α
n
)
J
S
x
n
)
,
u
n
∈
K
s
u
c
h
t
h
a
t
〈
A
u
n
,
η
(
y
,
u
n
)
〉
+
f
(
y
)
-
f
(
u
n
)
≥
0
,
∀
y
∈
K
,
C
n
=
{
z
∈
K
:
ϕ
(
z
,
u
n
)
≤
ϕ
(
z
,
x
n
)
}
,
D
n
=
⋂
i
=
1
n
C
i
,
x
n
+
1
=
Π
D
n
x
,
∀
n
≥
1
,
where J is the duality mapping on E, {
α
n
}
⊂
(
0,1
) with 0
<
liminf
n
→
∞
α
n
(
1
-
α
n
), and {
r
n
}
⊂
(
r
,
∞
) with r
>
0. Then the sequence {
x
n
} generated by (96) converges strongly to Π
F
x.IfA
≡
0 and f
≡
0 in Theorem 14, then we have following.Corollary 16.
LetK be a nonempty closed convex subset of a uniformly convex, uniformly smooth Banach space E with the dual space E
*. Let Θ be a bifunction from K
×
K to ℝ satisfying (A1)–(A4) and let S
:
E
→
E be a closed quasi-ϕ-nonexpansive mapping such that F
=
F
(
S
)
∩
EP
(
Θ
)
≠
∅. Let {
x
n
} be a sequence generated by the following manner:
(97)
x
1
=
x
∈
E
c
h
o
s
e
n
a
r
b
i
t
r
a
r
i
l
y
,
y
n
=
J
-
1
(
α
n
J
x
n
+
(
1
-
α
n
)
J
S
x
n
)
,
u
n
∈
K
s
u
c
h
t
h
a
t
Θ
(
u
n
,
y
)
+
1
r
n
〈
y
-
u
n
,
J
u
n
-
J
y
n
〉
≥
0
,
∀
y
∈
K
,
C
n
=
{
z
∈
E
:
ϕ
(
z
,
u
n
)
≤
ϕ
(
z
,
x
n
)
}
,
D
n
=
⋂
i
=
1
n
C
i
,
x
n
+
1
=
Π
D
n
x
,
∀
n
≥
1
,
where J is the duality mapping on E, {
α
n
}
⊂
(
0,1
) with 0
<
liminf
n
→
∞
α
n
≤
limsup
n
→
∞
α
n
<
1 and {
r
n
}
⊂
(
r
,
∞
) with r
>
0. Then the sequence {
x
n
} generated by (97) converges strongly to Π
F
x.Remark 17.
Theorem14 improves the corresponding ones of Wang et al. [7] from Hilbert spaces to Banach spaces and those of Takahashi and Zembayashi [1] from relative nonexpansive mapping to quasi-ϕ-nonexpansive mapping.



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*Source: 193749-2014-01-09.xml*