# Convergence Theorems of Common Elements for Pseudocontractive Mappings and Monotone Mappings

Authors: Jae Ug Jeong
Journal: Abstract and Applied Analysis
 (2015)

---

## Abstract

An algorithm for treating pseudocontractive mappings and monotone mappings is proposed. Convergence analysis of algorithm is investigated in the framework of Hilbert spaces.


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## Body

## 1. Introduction

The motivation for common element problem is mainly due to its possible applications to mathematical modeling of concrete complex problems. The common element problems include mini-max problems, complementarily problems, equilibrium problems, common fixed point problems, and variational inequalities as special cases; see [1–7] and the references therein. It is well-known that the convex feasibility problem is a special case of the common zero (fixed) points of nonlinear mappings. And many important problems have reformulations which require finding zero points, for instance, evolution equations, complementarily problems, mini-max problems, and variational inequalities and optimization. For studying zero points of monotone mappings, the most well-known algorithm is the proximal point algorithm; see [8, 9] and the references therein. Regularization methods recently have been investigated for treating zero points of monotone mappings; see [2, 5, 6, 9] and references therein.In 2010, Takahashi et al. [6] studied zero point problems of the sum of two monotone mappings and fixed point problems of a nonexpansive mapping based on the following iterative algorithm:(1)
x
1
=
x
∈
C
,
y
n
=
α
n
x
+
(
1
-
α
n
)
J
r
n
(
I
-
r
n
A
)
x
n
,
x
n
+
1
=
β
n
x
n
+
(
1
-
β
n
)
T
y
n
,
∀
n
≥
1
,where C is a nonempty closed convex subset of a real Hilbert space H, {
α
n
} and {
β
n
} are real number sequences in (
0,1
), {
r
n
} is a positive sequence, T
:
C
→
C is a nonexpansive mapping, A
:
C
→
H is an inverse strongly monotone mapping, B
:
H
→
2
H is a maximal monotone mapping, and J
r
n
=
(
I
+
r
n
B
)
-
1, where I is the identity mapping. They proved that the sequence {
x
n
} generated in (1) converges strongly to some z
∈
F
(
T
)
∩
(
A
+
B
)
-
1
(
0
) provided that the control sequences satisfy some restrictions, where F
(
T
) is the set fixed points of T.In 2014, Shahzad and Zegeye [5] considered an iterative method for a common point of fixed points of Lipschitzian pseudocontractive mappings and zeros of sum of two monotone mappings based on the projection method in a real Hilbert space. To be more precise, they investigated the following algorithm:(2)
x
0
∈
C
,
y
n
=
(
1
-
β
n
)
x
n
+
β
n
T
x
n
,
x
n
+
1
=
P
C
1
-
α
n
θ
n
x
n
+
δ
n
T
y
n
+
γ
n
J
r
n
I
-
λ
n
A
x
n
,where C is a nonempty closed convex subset of a real Hilbert space H, {
α
n
}, {
β
n
}, {
θ
n
}, {
δ
n
}, and {
γ
n
} are real number sequences in (
0,1
), {
r
n
} is a positive sequence, T
:
C
→
C is a Lipschitzian pseudocontractive mapping, A
:
C
→
H is an inverse strongly monotone mapping, B
:
H
→
2
H is a maximal monotone mapping, and J
r
n
=
(
I
+
r
n
B
)
-
1. They proved that the sequence {
x
n
} generated in (2) converges strongly to the minimum-norm point x
∈
F
(
T
)
∩
(
A
+
B
)
-
1
(
0
) provided that the control sequences satisfy some restrictions.In this paper, we are concerned with the problem of finding a common element in the intersectionF
(
T
1
)
∩
F
(
T
2
)
∩
(
A
+
B
)
-
1
(
0
), where F
(
T
i
) denotes the fixed point set of the pseudocontractive mapping T
i, i
=
1,2, and (
A
+
B
)
-
1
(
0
) denotes the zero point set of the sum of an inverse strongly monotone mapping A and a maximal monotone mapping B. Applications to a common element of the set of common fixed points of Lipschitzian pseudocontractive mappings and solutions of variational inequality for α-inverse strongly monotone mappings are included. Our theorems improve and extend those announced by Shahzad and Zegeye [5], Takahashi et al. [6], and other authors with the related interest.


## 2. Preliminaries

LetH be a real Hilbert space with the inner product 〈
·
,
·
〉 and the norm ·. Let C be a nonempty closed convex subset of H and let P
C be the metric projection from H onto C. Let T
:
C
→
C be a mapping. In this paper, we use F
(
T
) to denote the fixed point set of T; that is, F
(
T
)
=
{
x
∈
C
:
x
=
T
x
}.Recall thatT is nonexpansive if (3)
T
x
-
T
y
≤
x
-
y
,
∀
x
,
y
∈
C
.
T is said to be a γ-strictly pseudocontractive mapping if there exists γ
∈
[
0,1
) such that (4)
T
x
-
T
y
2
≤
x
-
y
2
+
γ
I
-
T
x
-
I
-
T
y
2
,
∀
x
,
y
∈
C
.Note that the class of γ-strictly pseudocontractive mappings includes the class of nonexpansive mappings as a special case. T is said to be a pseudocontractive mapping if(5)
T
x
-
T
y
2
≤
x
-
y
2
+
I
-
T
x
-
I
-
T
y
2
,
∀
x
,
y
∈
C
.We note that inequalities (4) and (5) can be equivalently written as (6)
x
-
y
,
T
x
-
T
y
≤
x
-
y
2
-
γ
(
I
-
T
)
x
-
(
I
-
T
)
y
2for some γ
>
0 and (7)
x
-
y
,
T
x
-
T
y
≤
x
-
y
2
,
∀
x
,
y
∈
C
,respectively. Note that the class of γ-strictly pseudocontractive mappings is contained in the class of pseudocontractive mappings. We note that the inclusion is proper. We remark that T is a γ-strictly pseudocontractive mapping if and only if I
-
T is a γ-inverse strongly monotone mapping and T is a pseudocontractive mapping if and only if I
-
T is a monotone mapping.LetA
:
C
→
H be a mapping and A
-
1
0 stands for the zero point set of A; that is, A
-
1
0
=
{
x
∈
C
:
A
x
=
0
}. Recall that A is said to be monotone if (8)
〈
A
x
-
A
y
,
x
-
y
〉
≥
0
,
∀
x
,
y
∈
C
.
A is said to be α-inverse strongly monotone if there exists a constant α
>
0 such that (9)
A
x
-
A
y
,
x
-
y
≥
α
A
x
-
A
y
2
,
∀
x
,
y
∈
C
.It is not hard to see that α-inverse strongly monotone mappings are Lipschitz continuous with constant L
=
1
/
α; that is, A
x
-
A
y
≤
(
1
/
α
)
x
-
y for all x
,
y
∈
C.Recall that the classical variational inequality, denoted byV
I
(
C
,
A
), is to find u
∈
C such that (10)
A
u
,
v
-
u
≥
0
,
∀
v
∈
C
.A multivalued mappingB
:
H
→
2
H with the domain D
(
B
)
=
{
x
∈
H
:
B
x
≠
ϕ
} and the range R
(
B
)
=
{
B
x
:
x
∈
D
(
B
)
} is said to be monotone if, for x
1
∈
D
(
B
), x
2
∈
D
(
B
), y
1
∈
B
x
1, and y
2
∈
B
x
2, we have 〈
x
1
-
x
2
,
y
1
-
y
2
〉
≥
0. A monotone mapping B is said to be maximal if its graph G
(
B
)
=
{
(
x
,
y
)
:
y
∈
B
x
} is not properly contained in the graph of any other monotone mapping. Let B
:
H
→
2
H be a maximal monotone mapping. Then we can define, for each λ
>
0, a nonexpansive single-valued mapping J
λ
:
H
→
H by J
λ
=
(
I
+
λ
B
)
-
1. It is called the resolvent of B. We know that B
-
1
0
=
F
(
J
λ
) for all λ
>
0 and J
λ is firmly nonexpansive.Lemma 1.
LetH be a real Hilbert space. Then, for any given x
,
y
∈
H, the following inequality holds: (11)
x
+
y
2
≤
x
2
+
2
y
,
x
+
y
.Lemma 2 (see [10]).
LetC be a convex subset of a real Hilbert space H. Let x
∈
H. Then x
0
=
P
C
x if and only if (12)
z
-
x
0
,
x
-
x
0
≤
0
,
∀
z
∈
C
.Lemma 3 (see [2]).
LetC be a nonempty closed convex subset of a real Hilbert space H. Let A
:
C
→
H be a mapping and let B
:
H
→
2
H be a maximal monotone mapping. Then F
(
J
λ
(
I
-
λ
A
)
)
=
(
A
+
B
)
-
1
0.Lemma 4 (see [11]).
LetH be a Hilbert space. Let B
1
:
D
(
B
1
)
⊆
H
→
2
H and let B
2
:
D
(
B
2
)
⊆
H
→
2
H be maximal monotone mappings. Suppose that D
(
A
)
∩
i
n
t
(
D
(
B
)
)
≠
ϕ. Then B
1
+
B
2 is a maximal monotone mapping.Lemma 5 (see [4]).
Let{
a
n
} be a sequence of real numbers. Assume that there exists a subsequence {
n
i
} of {
n
} such that a
n
i
<
a
n
i
+
1 for all i
∈
N. Then there exists a nondecreasing sequence {
m
k
}
⊂
N such that m
k
→
∞ and the following properties are satisfied by all sufficiently large numbers k
∈
N: (13)
a
m
k
≤
a
m
k
+
1
,
a
k
≤
a
m
k
+
1
.Lemma 6 (see [12]).
LetH be a real Hilbert space. Then, for all x
i
∈
H and α
i
∈
[
0,1
] for i
=
1,2
,
…
,
n such that α
1
+
α
2
+
⋯
+
α
n
=
1, the following equality holds: (14)
α
0
x
0
+
α
1
x
1
+
⋯
+
α
n
x
n
2
=
∑
i
=
0
n
‍
α
i
x
i
2
-
∑
0
≤
i
,
j
≤
n
‍
α
i
,
α
j
x
i
-
x
j
2
.Lemma 7 (see [7]).
LetC be a nonempty closed convex subset of a real Hilbert Hilbert and let T
:
C
→
C be a continuous pseudocontractive mapping. Then(i)
F
(
T
) is a closed convex subset of C;
(ii)
(
I
-
T
) is demiclosed at zero; that is, if {
x
n
} is a sequence in C such that x
n
⇀
x and T
x
n
-
x
n
→
0 as n
→
∞, then x
=
T
x.Lemma 8 (see [13]).
Let{
a
n
} be a sequence of nonnegative real numbers satisfying the following relation: (15)
a
n
+
1
≤
(
1
-
α
n
)
a
n
+
α
n
δ
n
,
n
≥
n
0
,where {
α
n
}
⊂
(
0,1
) and {
δ
n
}
⊂
R satisfy the following conditions: l
i
m
n
→
∞
α
n
=
0, ∑
n
=
1
∞
‍
α
n
=
∞, and l
i
m
s
u
p
n
→
∞
δ
n
≤
0. Then l
i
m
n
→
∞
a
n
=
0.


## 3. Main Results

Theorem 9.
LetC be a nonempty closed convex subset of a real Hilbert space H. Let T
1
,
T
2
:
C
→
C be Lipschitzian pseudocontractive mappings with Lipschitz constants L
1 and L
2, respectively. Let A
:
C
→
H be an α-inverse strongly monotone mapping and let B be a maximal monotone mapping such that the domain of B is subset of C. Assume that F
=
F
(
T
1
)
∩
F
(
T
2
)
∩
(
A
+
B
)
-
1
0
≠
ϕ. Let J
λ
n
=
(
I
+
λ
n
B
)
-
1, where {
λ
n
} is a positive real number sequence. Given x
1
,
u
∈
C, let {
x
n
} be the sequence generated by the following algorithm:(16)
z
n
=
(
1
-
c
n
)
x
n
+
c
n
T
2
x
n
,
y
n
=
(
1
-
β
n
)
x
n
+
β
n
T
1
x
n
,
x
n
+
1
=
P
C
α
n
u
+
1
-
α
n
f
f
f
×
θ
n
x
n
+
δ
n
T
1
y
n
+
γ
n
T
2
z
n
ξ
n
J
λ
n
(
I
-
λ
n
A
)
x
n
f
f
f
+
ξ
n
J
λ
n
(
I
-
λ
n
A
)
x
n
.Assume that the sequences {
α
n
}, {
β
n
}, {
c
n
}, {
θ
n
}, {
δ
n
}, {
γ
n
}, {
ξ
n
}, and {
λ
n
} satisfy the following restrictions:(a)
0
<
a
<
λ
n
<
b
<
2
α;
(b)
0
<
c
≤
θ
n
,
δ
n
,
γ
n
,
ξ
n
≤
d
<
1 and θ
n
+
δ
n
+
γ
n
+
ξ
n
=
1;
(c)
0
<
α
n
<
e
<
1, l
i
m
n
→
∞
α
n
=
0 and ∑
n
=
1
∞
‍
α
n
=
∞;
(d)
δ
n
+
γ
n
+
ξ
n
≤
β
n, c
n
≤
β
<
1
/
(
1
+
L
2
+
1
), for all n
≥
1,for some real numbers a
,
b
,
c
,
d
,
e
,
β
>
0, where L
=
m
a
x
{
L
1
,
L
2
}. Then {
x
n
} converges strongly to some point x
¯, where x
¯
=
P
F
u.Proof.
First, we show thatI
-
λ
n
A is nonexpansive.
Indeed, we have(17)
(
I
-
λ
n
A
)
x
-
(
I
-
λ
n
A
)
y
2
=
x
-
y
2
-
2
λ
n
〈
x
-
y
,
A
x
-
A
y
〉
+
λ
n
2
A
x
-
A
y
2
≤
x
-
y
2
-
λ
n
(
2
α
-
λ
n
)
A
x
-
A
y
2
.It follows from restriction (a) that I
-
λ
n
A is nonexpansive.
Letp
∈
F. It follows from (5), (16), and Lemmas 3 and 6 that(18)
x
n
+
1
-
p
2
=
P
C
α
n
u
+
1
-
α
n
f
f
×
θ
n
x
n
+
δ
n
T
1
y
n
+
γ
n
T
2
z
n
ξ
n
J
λ
n
I
-
λ
n
A
x
n
f
f
+
ξ
n
J
λ
n
I
-
λ
n
A
x
n
-
p
2
≤
α
n
u
+
(
1
-
α
n
)
+
ξ
n
J
λ
n
(
I
-
λ
n
A
)
x
n
×
θ
n
x
n
+
δ
n
T
1
y
n
+
γ
n
T
2
z
n
ξ
n
J
λ
n
(
I
-
λ
n
A
)
x
n
+
ξ
n
J
λ
n
(
I
-
λ
n
A
)
x
n
-
p
2
≤
α
n
u
-
p
2
+
(
1
-
α
n
)
×
θ
n
(
x
n
-
p
)
+
δ
n
(
T
1
y
n
-
p
)
J
λ
n
I
-
λ
n
A
x
n
-
p
+
γ
n
(
T
2
z
n
-
p
)
+
ξ
n
J
λ
n
I
-
λ
n
A
x
n
-
p
2
≤
α
n
u
-
p
2
+
(
1
-
α
n
)
×
θ
n
x
n
-
p
2
+
δ
n
T
1
y
n
-
p
2
J
λ
n
(
I
-
λ
n
A
)
x
n
-
p
2
+
γ
n
T
2
z
n
-
p
2
+
ξ
n
J
λ
n
(
I
-
λ
n
A
)
x
n
-
p
2
-
(
1
-
α
n
)
θ
n
δ
n
T
1
y
n
-
x
n
2
-
(
1
-
α
n
)
θ
n
γ
n
T
2
z
n
-
x
n
2
-
(
1
-
α
n
)
θ
n
ξ
n
J
λ
n
(
I
-
λ
n
A
)
x
n
-
x
n
2
≤
α
n
u
-
p
2
+
(
1
-
α
n
)
(
θ
n
+
ξ
n
)
x
n
-
p
2
+
(
1
-
α
n
)
δ
n
[
y
n
-
p
2
+
y
n
-
T
1
y
n
2
]
+
(
1
-
α
n
)
γ
n
[
z
n
-
p
2
+
z
n
-
T
2
z
n
2
]
-
(
1
-
α
n
)
θ
n
δ
n
T
1
y
n
-
x
n
2
-
(
1
-
α
n
)
θ
n
γ
n
T
2
z
n
-
x
n
2
-
(
1
-
α
n
)
θ
n
ξ
n
J
λ
n
(
I
-
λ
n
A
)
x
n
-
x
n
2
.It follows from (5), (16), and Lemma 6 that(19)
z
n
-
p
2
=
(
1
-
c
n
)
(
x
n
-
p
)
+
c
n
(
T
2
x
n
-
p
)
2
=
(
1
-
c
n
)
x
n
-
p
2
+
c
n
T
2
x
n
-
p
2
-
c
n
(
1
-
c
n
)
x
n
-
T
2
x
n
2
≤
(
1
-
c
n
)
x
n
-
p
2
+
c
n
x
n
-
p
+
x
n
-
T
2
x
n
2
-
c
n
(
1
-
c
n
)
x
n
-
T
2
x
n
2
=
x
n
-
p
2
+
c
n
2
x
n
-
T
2
x
n
2
,
y
n
-
p
2
=
(
1
-
β
n
)
(
x
n
-
p
)
+
β
n
(
T
1
x
n
-
p
)
2
≤
x
n
-
p
2
+
β
n
2
x
n
-
T
1
x
n
2
.Similarly, we have that (20)
y
n
-
T
1
y
n
2
=
(
1
-
β
n
)
(
x
n
-
T
1
y
n
)
+
β
n
(
T
1
x
n
-
T
1
y
n
)
2
=
(
1
-
β
n
)
x
n
-
T
1
y
n
2
+
β
n
T
1
x
n
-
T
1
y
n
2
-
β
n
(
1
-
β
n
)
x
n
-
T
1
x
n
2
≤
(
1
-
β
n
)
x
n
-
T
1
y
n
2
+
β
n
L
2
x
n
-
y
n
2
-
β
n
(
1
-
β
n
)
x
n
-
T
1
x
n
2
=
(
1
-
β
n
)
x
n
-
T
1
y
n
2
-
β
n
(
1
-
β
n
2
L
2
-
β
n
)
x
n
-
T
1
x
n
2
.
z
n
-
T
2
z
n
2
=
(
1
-
c
n
)
(
x
n
-
T
2
z
n
)
+
c
n
(
T
2
x
n
-
T
2
z
n
)
2
≤
(
1
-
c
n
)
x
n
-
T
2
z
n
2
-
c
n
(
1
-
c
n
2
L
2
-
c
n
)
x
n
-
T
2
x
n
2
.Substituting (19) and (20) into (18), we obtain that(21)
x
n
+
1
-
p
2
≤
α
n
u
-
p
2
+
(
1
-
α
n
)
(
θ
n
+
ξ
n
)
x
n
-
p
2
+
1
-
α
n
δ
n
x
n
-
p
2
+
β
n
2
x
n
-
T
1
x
n
2
+
(
1
-
α
n
)
δ
n
(
1
-
β
n
)
x
n
-
T
1
y
n
2
-
β
n
(
1
-
β
n
2
L
2
-
β
n
)
x
n
-
T
1
x
n
2
+
1
-
α
n
γ
n
x
n
-
p
2
+
c
n
2
x
n
-
T
2
x
n
2
+
(
1
-
α
n
)
γ
n
(
1
-
c
n
)
x
n
-
T
2
z
n
2
-
c
n
(
1
-
c
n
2
L
2
-
c
n
)
x
n
-
T
2
x
n
2
-
(
1
-
α
n
)
θ
n
δ
n
T
1
y
n
-
x
n
2
-
(
1
-
α
n
)
θ
n
γ
n
T
2
z
n
-
x
n
2
-
(
1
-
α
n
)
θ
n
ξ
n
J
λ
n
(
I
-
λ
n
A
)
x
n
-
x
n
2
=
α
n
u
-
p
2
+
(
1
-
α
n
)
x
n
-
p
2
-
(
1
-
α
n
)
δ
n
β
n
(
1
-
2
β
n
-
β
n
2
L
2
)
x
n
-
T
1
x
n
2
-
(
1
-
α
n
)
γ
n
c
n
(
1
-
2
c
n
-
c
n
2
L
2
)
x
n
-
T
2
x
n
2
+
(
1
-
α
n
)
δ
n
(
δ
n
+
γ
n
+
ξ
n
-
β
n
)
T
1
y
n
-
x
n
2
+
(
1
-
α
n
)
γ
n
(
δ
n
+
γ
n
+
ξ
n
-
c
n
)
T
2
z
n
-
x
n
2
-
(
1
-
α
n
)
θ
n
ξ
n
J
λ
n
(
I
-
λ
n
A
)
x
n
-
x
n
2
.In view of restriction (d), we find that (22)
1
-
2
β
n
-
β
n
2
L
2
≥
1
-
2
β
-
β
2
L
2
>
0
,
1
-
2
c
n
-
c
n
2
L
2
≥
1
-
2
β
-
β
2
L
2
>
0
,
δ
n
+
γ
n
+
ξ
n
-
β
n
≤
0
,
δ
n
+
γ
n
+
ξ
n
-
c
n
≤
0
,for all n
≥
1. It follows from (21) and (22) that (23)
x
n
+
1
-
p
2
≤
α
n
u
-
p
2
+
(
1
-
α
n
)
x
n
-
p
2
.Putting M
=
m
a
x
{
u
-
p
2
,
x
1
-
p
2
}, we find that x
n
-
p
2
≤
M for all n
≥
1.
Indeed, it is clear thatx
2
-
p
2
≤
M. Suppose that x
m
-
p
≤
M for some positive integer m. It follows that (24)
x
m
+
1
-
p
2
≤
α
m
u
-
p
2
+
(
1
-
α
m
)
x
m
-
p
2
≤
α
m
M
+
(
1
-
α
m
)
M
=
M
.This finds that {
x
n
} is bounded and hence {
y
n
} and {
z
n
} are bounded.
Letw
n
=
θ
n
x
n
+
δ
n
T
1
y
n
+
γ
n
T
2
z
n
+
ξ
n
J
λ
n
(
I
-
λ
n
A
)
x
n. Then we see that x
n
+
1
=
P
C
(
α
n
u
+
(
1
-
α
n
)
w
n
). Put x
¯
=
P
F
u. Using (16), (19), and (20) and Lemmas 1 and 6, we find that (25)
x
n
+
1
-
x
¯
2
≤
α
n
(
u
-
x
¯
)
+
(
1
-
α
n
)
(
w
n
-
x
¯
)
2
≤
(
1
-
α
n
)
w
n
-
x
¯
2
+
2
α
n
u
-
x
¯
,
α
n
u
-
x
¯
+
1
-
α
n
w
n
-
x
¯
≤
(
1
-
α
n
)
θ
n
x
n
-
x
¯
2
+
(
1
-
α
n
)
δ
n
T
1
y
n
-
x
¯
2
+
(
1
-
α
n
)
γ
n
T
2
z
n
-
x
¯
2
+
(
1
-
α
n
)
ξ
n
J
λ
n
(
I
-
λ
n
A
)
x
n
-
x
¯
2
-
(
1
-
α
n
)
θ
n
δ
n
T
1
y
n
-
x
n
2
-
(
1
-
α
n
)
θ
n
γ
n
T
2
z
n
-
x
n
2
-
(
1
-
α
n
)
θ
n
ξ
n
J
λ
n
(
I
-
λ
n
A
)
x
n
-
x
n
2
+
2
α
n
2
u
-
x
¯
2
+
2
α
n
1
-
α
n
u
-
x
¯
,
w
n
-
x
¯
≤
(
1
-
α
n
)
θ
n
x
n
-
x
¯
2
+
1
-
α
n
δ
n
y
n
-
x
¯
2
+
y
n
-
T
1
y
n
2
+
1
-
α
n
γ
n
z
n
-
x
¯
2
+
z
n
-
T
2
z
n
2
+
(
1
-
α
n
)
ξ
n
x
n
-
x
¯
2
-
(
1
-
α
n
)
θ
n
δ
n
T
1
y
n
-
x
n
2
-
(
1
-
α
n
)
θ
n
γ
n
T
2
z
n
-
x
n
2
-
(
1
-
α
n
)
θ
n
ξ
n
J
λ
n
(
I
-
λ
n
A
)
x
n
-
x
n
2
+
2
α
n
2
u
-
x
¯
2
+
2
α
n
1
-
α
n
u
-
x
¯
,
w
n
-
x
¯
≤
(
1
-
α
n
)
(
θ
n
+
ξ
n
)
x
n
-
x
¯
2
+
1
-
α
n
δ
n
x
n
-
x
¯
2
+
β
n
2
x
n
-
T
1
x
n
2
+
(
1
-
α
n
)
δ
n
(
1
-
β
n
)
x
n
-
T
1
y
n
2
-
β
n
(
1
-
β
n
2
L
2
-
β
n
)
x
n
-
T
1
x
n
2
+
(
1
-
α
n
)
γ
n
[
x
n
-
x
¯
2
+
c
n
2
x
n
-
T
2
x
n
2
]
+
(
1
-
α
n
)
γ
n
(
1
-
c
n
)
x
n
-
T
2
z
n
2
-
c
n
(
1
-
c
n
2
L
2
-
c
n
)
x
n
-
T
2
x
n
2
-
(
1
-
α
n
)
θ
n
δ
n
T
1
y
n
-
x
n
2
-
(
1
-
α
n
)
θ
n
γ
n
T
2
z
n
-
x
n
2
-
(
1
-
α
n
)
θ
n
ξ
n
J
λ
n
(
I
-
λ
n
A
)
x
n
-
x
n
2
+
2
α
n
2
u
-
x
¯
2
+
2
α
n
1
-
α
n
u
-
x
¯
,
w
n
-
x
¯
,which implies from (22) that(26)
x
n
+
1
-
x
¯
2
≤
(
1
-
α
n
)
x
n
-
x
¯
2
-
(
1
-
α
n
)
δ
n
β
n
(
1
-
2
β
n
-
β
n
2
L
2
)
x
n
-
T
1
x
n
2
+
(
1
-
α
n
)
δ
n
(
δ
n
+
ξ
n
+
γ
n
-
β
n
)
T
1
y
n
-
x
n
2
-
(
1
-
α
n
)
γ
n
c
n
(
1
-
2
c
n
-
c
n
2
L
2
)
x
n
-
T
2
x
n
2
+
(
1
-
α
n
)
γ
n
(
δ
n
+
ξ
n
+
γ
n
-
c
n
)
T
2
z
n
-
x
n
-
(
1
-
α
n
)
θ
n
ξ
n
J
λ
n
(
I
-
λ
n
A
)
x
n
-
x
n
2
+
2
α
n
2
u
-
x
¯
2
+
2
α
n
1
-
α
n
u
-
x
¯
,
w
n
-
x
¯
≤
(
1
-
α
n
)
x
n
-
x
¯
2
+
2
α
n
2
u
-
x
¯
2
+
2
α
n
(
1
-
α
n
)
〈
u
-
x
¯
,
w
n
-
x
¯
〉
.
Now we consider two cases.
Case 1. Suppose that there exists n
0
∈
N such that {
x
n
-
x
¯
} is decreasing for all n
≥
n
0. Then we get that {
x
n
-
x
¯
} is convergent. It follows from (22) and (26) that(27)
x
n
-
T
1
x
n
⟶
0
,
x
n
-
T
2
x
n
⟶
0
,
x
n
-
J
λ
n
(
I
-
λ
n
A
)
x
n
⟶
0
,as n
→
∞. Also we obtain from (27) that (28)
y
n
-
x
n
=
β
n
x
n
-
T
1
x
n
⟶
0
,
z
n
-
x
n
=
c
n
x
n
-
T
2
x
n
⟶
0
,as n
→
∞. In view of the Lipschitz continuity of T
1
,
T
2 and (27) and (28), we find that(29)
T
1
y
n
-
x
n
≤
T
1
y
n
-
T
1
x
n
+
T
1
x
n
-
x
n
≤
L
y
n
-
x
n
+
T
1
x
n
-
x
n
⟶
0
a
s
n
⟶
∞
,
(30)
T
2
z
n
-
x
n
≤
T
2
z
n
-
T
2
x
n
+
T
2
x
n
-
x
n
≤
L
z
n
-
x
n
+
T
2
x
n
-
x
n
⟶
0
a
s
n
⟶
∞
.It follows from (27), (29), and (30) that(31)
w
n
-
x
n
≤
δ
n
T
1
y
n
-
x
n
+
γ
n
T
2
z
n
-
x
n
+
ξ
n
J
λ
n
(
I
-
λ
n
A
)
x
n
-
x
n
⟶
0
a
s
n
⟶
∞
.Since {
w
n
} is a bounded subset of H, we can choose a subsequence {
w
n
i
} of {
w
n
} such that w
n
i
⇀
w and (32)
l
i
m
s
u
p
n
→
∞
u
-
x
¯
,
w
n
-
x
¯
=
l
i
m
i
→
∞
u
-
x
¯
,
w
n
i
-
x
¯
.It follows from (31) that x
n
i
⇀
w. By (27) and Lemma 7, we obtain that w
∈
F
(
T
1
) and w
∈
F
(
T
2
).
Next, we show thatw
∈
(
A
+
B
)
-
1
0.
Notice that(33)
J
λ
n
(
I
-
λ
n
A
)
x
n
-
p
2
=
J
λ
n
(
I
-
λ
n
A
)
x
n
-
J
λ
n
(
I
-
λ
n
A
)
p
2
≤
(
I
-
λ
n
A
)
x
n
-
(
I
-
λ
n
A
)
p
2
=
x
n
-
p
2
-
2
λ
n
x
n
-
p
,
A
x
n
-
A
p
+
λ
n
2
A
x
n
-
A
p
2
≤
x
n
-
p
2
-
2
α
λ
n
A
x
n
-
A
p
2
+
λ
n
2
A
x
n
-
A
p
2
=
x
n
-
p
2
-
λ
n
(
2
α
-
λ
n
)
A
x
n
-
A
p
2
.It follows from (27) that(34)
λ
n
(
2
α
-
λ
n
)
A
x
n
-
A
p
2
≤
x
n
-
p
2
-
J
λ
n
(
I
-
λ
n
A
)
x
n
-
p
=
x
n
-
p
+
J
λ
n
I
-
λ
n
A
x
n
-
p
×
x
n
-
J
λ
n
(
I
-
λ
n
A
)
x
n
⟶
0
a
s
n
⟶
∞
.Hence we get (35)
l
i
m
n
→
∞
A
x
n
-
A
p
2
=
0
.Putting h
n
=
J
λ
n
(
I
-
λ
n
A
)
x
n, we find that (
(
x
n
i
-
h
n
i
)
/
λ
n
i
)
-
A
x
n
i
∈
B
h
n
i. Since B is monotone, we get that, for any (
u
,
v
)
∈
G
(
B
), (36)
h
n
i
-
u
,
x
n
i
-
h
n
i
λ
n
i
-
A
x
n
i
-
v
≥
0
,where G
(
B
)
=
{
(
x
,
w
)
∈
H
×
H
:
x
∈
D
(
B
)
,
w
∈
B
x
}. Since 〈
x
n
i
-
w
,
A
x
n
i
-
A
w
〉
≥
α
A
x
n
i
-
A
w
2, x
n
i
⇀
w, and A
x
n
i
→
A
p as i
→
∞, we have A
x
n
i
→
A
w. Thus, letting i
→
∞, we obtain from (27) and (36) that w
-
u
,
-
A
w
-
v
≥
0. This means -
A
w
∈
B
w, that is, 0
∈
(
A
+
B
)
w. Hence we get w
∈
(
A
+
B
)
-
1
0. This implies from Lemma 2 that(37)
l
i
m
s
u
p
n
→
∞
u
-
x
¯
,
w
n
-
x
¯
=
l
i
m
i
→
∞
u
-
x
¯
,
w
n
i
-
x
¯
=
u
-
x
¯
,
w
-
x
¯
≤
0
.
On the other hand, we have from (26) that(38)
x
n
+
1
-
x
¯
2
≤
(
1
-
α
n
)
x
n
-
x
¯
2
+
α
n
2
α
n
u
-
x
¯
2
α
n
u
-
x
¯
2
+
1
-
α
n
u
-
x
¯
,
w
n
-
x
¯
.From Lemma 8 and (37), we find that l
i
m
n
→
∞
x
n
-
x
¯
=
0.
Case 2. Suppose that there exists a subsequence {
n
i
} of {
n
} such that (39)
x
n
i
-
x
¯
<
x
n
i
+
1
-
x
¯
,for all i
∈
N. By Lemma 5, there exists a nondecreasing sequence {
m
k
}
⊂
N such that m
k
→
∞ and (40)
x
m
k
-
x
¯
≤
x
m
k
+
1
-
x
¯
,
x
k
-
x
¯
≤
x
m
k
+
1
-
x
¯
,for all k
∈
N. From (22) and (26), we have x
m
k
-
T
1
x
m
k
→
0, x
m
k
-
T
2
x
m
k
→
0, and x
m
k
-
J
λ
m
k
(
I
-
λ
m
k
A
)
x
m
k
→
0 as k
→
∞. Thus, like in Case 1, we obtain w
m
k
-
x
m
k
→
0 and (41)
l
i
m
s
u
p
k
→
∞
u
-
x
¯
,
w
m
k
-
x
¯
≤
0
.From (26) and (40), we have(42)
α
m
k
x
m
k
-
x
¯
2
≤
x
m
k
-
x
¯
2
-
x
m
k
+
1
-
x
¯
2
+
2
α
m
k
α
m
k
u
-
x
¯
2
+
1
-
α
m
k
u
-
x
¯
,
w
m
k
-
x
¯
≤
2
α
m
k
α
m
k
u
-
x
¯
2
+
1
-
α
m
k
u
-
x
¯
,
w
m
k
-
x
¯
.Applying (41) and α
m
k
>
0, we have x
m
k
-
x
¯
→
0 as k
→
∞. It implies that x
m
k
+
1
-
x
¯
→
0 as k
→
∞. By (40), we have x
k
→
x
¯ as k
→
∞.
Therefore, from the above two cases, we can conclude that the sequence{
x
n
} converges strongly to x
¯
=
P
F
u. This completes the proof.From Lemma4, we have the following result.Corollary 10.
LetC be a nonempty closed convex subset of a real Hilbert space H such that i
n
t
(
C
)
≠
ϕ. Let T
1
,
T
2
:
C
→
C be Lipschitzian pseudocontractive mappings with Lipschitz constants L
1 and L
2, respectively. Let B
1
:
D
(
B
1
)
→
2
H and B
2
:
D
(
B
2
)
→
2
H be maximal monotone mappings such that D
(
B
1
)
∩
i
n
t
(
D
(
B
2
)
)
≠
ϕ. Assume that F
=
F
(
T
1
)
∩
F
(
T
2
)
∩
(
B
1
+
B
2
)
-
1
(
0
)
≠
ϕ. Let J
λ
n
=
(
I
+
λ
n
(
B
1
+
B
2
)
)
-
1, where {
λ
n
} is a positive real number sequence. Given x
1
,
u
∈
C, let {
x
n
} be the sequence generated by the following algorithm:(43)
z
n
=
(
1
-
c
n
)
x
n
+
c
n
T
2
x
n
,
y
n
=
(
1
-
β
n
)
x
n
+
β
n
T
1
x
n
,
x
n
+
1
=
P
C
α
n
u
+
(
1
-
α
n
)
θ
n
x
n
+
δ
n
T
1
y
n
+
γ
n
T
2
z
n
+
ξ
n
J
λ
n
x
n
×
θ
n
x
n
+
δ
n
T
1
y
n
+
γ
n
T
2
z
n
+
ξ
n
J
λ
n
x
n
,
∀
n
≥
1
.Assume that the sequences {
α
n
}, {
β
n
}, {
c
n
}, {
θ
n
}, {
δ
n
}, {
γ
n
}, {
ξ
n
}, and {
λ
n
} satisfy the following restrictions:(a)
0
<
a
<
λ
n
<
b
<
1;
(b)
0
<
c
≤
θ
n
,
δ
n
,
γ
n
,
ξ
n
≤
d
<
1, and θ
n
+
δ
n
+
γ
n
+
ξ
n
=
1;
(c)
0
<
α
n
<
e
<
1, l
i
m
n
→
∞
α
n
=
0, and ∑
n
=
1
∞
‍
α
n
=
∞;
(d)
δ
n
+
γ
n
+
ξ
n
≤
β
n
,
c
n
≤
β
<
1
/
(
1
+
L
2
+
1
), for all n
≥
1,for some real numbers a
,
b
,
c
,
d
,
e
>
0, where L
=
m
a
x
{
L
1
,
L
2
}. Then {
x
n
} converges strongly to some point x
¯, where x
¯
=
P
F
u.Remark 11.
IfT
1
=
T, T
2
=
I (the identity mapping), and u
=
0, then Theorem 9 reduces to Theorem 3.1 of Shahzad and Zegeye [6]. Thus, Theorem 9 covers Theorem 3.1 of Shahzad and Zegeye [6] as a special case.


## 4. Applications

In this section, we will consider equilibrium problems and variational inequalities.LetC be a nonempty closed convex subset of a real Hilbert space H. Let F be a bifunction of C
×
C into R, where R denotes the set of real numbers. Recall the following equilibrium problem: find x
∈
C such that(44)
F
(
x
,
y
)
≥
0
,
∀
y
∈
C
.We use E
P
(
F
) to denote the solution set of the equilibrium problem. To study the equilibrium problems, we assume that F 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, (45)
l
i
m
s
u
p
t
↓
0
F
(
t
z
+
(
1
-
t
)
x
,
y
)
≤
F
(
x
,
y
)
;
(A4)
for eachx
∈
C, y
↦
F
(
x
,
y
) is convex and lower semicontinuousLemma 12 (see [1]).
LetC be a nonempty closed convex subset of a real Hilbert space and let F
:
C
×
C
→
R be a bifunction satisfying (A1)–(A4). Then, for any r
>
0 and x
∈
H, there exists z
∈
C such that (46)
F
(
z
,
y
)
+
1
r
y
-
z
,
z
-
x
≥
0
,
∀
y
∈
C
.Further, define (47)
T
r
x
=
z
∈
C
:
F
z
,
y
+
1
r
y
-
z
,
z
-
x
≥
0
,
∀
y
∈
C
,for all r
>
0 and x
∈
H. Then the following hold: (a)
T
r is single-valued;
(b)
T
r is firmly nonexpansive; that is, for any x
,
y
∈
H, (48)
T
r
x
-
T
r
y
2
≤
T
r
x
-
T
r
y
,
x
-
y
;
(c)
F
(
T
r
)
=
E
P
(
F
);
(d)
E
P
(
F
) is closed and convex.Lemma 13 (see [13]).
LetC be a nonempty closed convex subset of a real Hilbert space H, let F be a bifunction from C
×
C to R which satisfies (A1)–(A4), and let A
F be a multivalued mapping of H into itself defined by(49)
A
F
x
=
z
∈
H
:
F
x
,
y
≥
y
-
x
,
z
,
∀
y
∈
C
,
x
∈
C
,
ϕ
,
x
∉
C
.Then A
F is a maximal monotone mapping with the domain D
(
A
F
)
⊂
C, E
P
(
F
)
=
A
F
-
1
0, and (50)
T
r
(
x
)
=
I
+
r
A
F
-
1
x
,
∀
x
∈
H
,
r
>
0
,where T
r is defined as in (47).Now we consider an equilibrium problem. Using Lemmas12 and 13, the following result holds.Theorem 14.
LetC be a nonempty closed convex subset of a real Hilbert space H. Let T
1
,
T
2
:
C
→
C be Lipschitzian pseudocontractive mappings with Lipschitz constants L
1 and L
2, respectively. Assume that F
=
F
(
T
1
)
∩
F
(
T
2
)
∩
E
P
(
F
)
≠
ϕ. Given x
1
,
u
∈
C, let {
x
n
} be the sequence generated by the following algorithm:(51)
z
n
=
(
1
-
c
n
)
x
n
+
c
n
T
2
x
n
,
y
n
=
(
1
-
β
n
)
x
n
+
β
n
T
1
x
n
,
u
n
∈
C
s
u
c
h
t
h
a
t
F
u
n
,
v
+
1
r
n
v
-
u
n
,
u
n
-
x
n
≥
0
,
∀
v
∈
C
,
x
n
+
1
=
P
C
α
n
u
+
(
1
-
α
n
)
×
θ
n
x
n
+
δ
n
T
1
y
n
+
γ
n
T
2
z
n
+
ξ
n
u
n
,
∀
n
≥
1
.Assume that the sequences {
α
n
}, {
β
n
}, {
c
n
}, {
θ
n
}, {
δ
n
}, {
γ
n
}, {
ξ
n
}, and {
r
n
} satisfy the following restrictions:(a)
l
i
m
i
n
f
n
→
∞
r
n
>
0 and l
i
m
n
→
∞
r
n
+
1
-
r
n
=
0;
(b)
0
<
c
≤
θ
n
,
δ
n
,
γ
n, ξ
n
≤
d
<
1, and θ
n
+
δ
n
+
γ
n
+
ξ
n
=
1;
(c)
0
<
α
n
<
e
<
1, l
i
m
n
→
∞
α
n
=
0, and ∑
n
=
1
∞
‍
α
n
=
∞;
(d)
δ
n
+
γ
n
+
ξ
n
≤
β
n, c
n
<
β
<
1
/
(
1
+
L
2
+
1
), for all n
≥
1,for some real numbers c
,
d
,
e
>
0, where L
=
m
a
x
{
L
1
,
L
2
}. Then {
x
n
} converges strongly to some point x
¯, where x
¯
=
P
F
u.
Letf
:
H
→
(
-
∞
,
+
∞
] be a proper convex lower semicontinuous function. Then the subdifferential of ∂
f of f is defined as follows: (52)
∂
f
x
=
y
∈
H
:
f
z
≥
f
x
+
z
-
x
,
y
,
z
∈
H
,
∀
x
∈
H
.From Rockafellar [14], we find that ∂
f is maximal monotone. It is easy to verify that 0
∈
∂
f
(
x
) if and only if f
(
x
)
=
m
i
n
y
∈
H
f
(
y
). Let I
C be the indicator function of C; that is,(53)
I
C
x
=
0
,
x
∈
C
,
+
∞
,
x
∉
C
.Then I
C
:
H
→
(
-
∞
,
+
∞
] is a proper convex lower semicontinuous function and ∂
I
C is a maximal monotone mapping.Lemma 15 (see [6]).
LetC be a nonempty closed convex subset of a real Hilbert space H, let P
C be the metric projection from H onto C, and let ∂
I
C be the subdifferential of I
C, where I
C is the indicator function of C and let J
λ
=
(
I
+
λ
∂
I
C
)
-
1. Then (54)
y
=
J
λ
x
⟺
y
=
P
C
x
,
x
∈
H
,
y
∈
C
.Now we consider a variational inequality problem.Theorem 16.
LetC be a nonempty closed convex subset of a real Hilbert space H. Let T
1
,
T
2
:
C
→
C be Lipschitzian pseudocontractive mappings with Lipschitz constants L
1 and L
2, respectively. Let A
:
C
→
H be an α-inverse strongly monotone mapping. Assume that F
(
T
1
)
∩
F
(
T
2
)
∩
V
I
(
C
,
A
)
≠
ϕ. Given x
1
,
u
∈
C, let {
x
n
} be the sequence generated by the following algorithm:(55)
z
n
=
(
1
-
c
n
)
x
n
+
c
n
T
2
x
n
,
y
n
=
(
1
-
β
n
)
x
n
+
β
n
T
1
x
n
,
x
n
+
1
=
P
C
α
n
u
+
(
1
-
α
n
)
×
θ
n
x
n
+
δ
n
T
1
y
n
+
γ
n
T
2
z
n
+
ξ
n
P
C
(
I
-
λ
n
A
)
x
n
,
∀
n
≥
1
.Assume that the sequences {
α
n
}, {
β
n
}, {
c
n
}, {
θ
n
}, {
δ
n
}, {
γ
n
}, {
ξ
n
}, and {
λ
n
} satisfy the following restrictions:(a)
0
<
a
<
λ
n
<
b
<
2
α;
(b)
0
<
c
≤
θ
n
,
δ
n
,
γ
n
,
ξ
n
≤
d
<
1, and θ
n
+
δ
n
+
γ
n
+
ξ
n
=
1;
(c)
0
<
α
n
<
e
<
1, l
i
m
n
→
∞
α
n
=
0, and ∑
n
=
1
∞
‍
α
n
=
∞;
(d)
δ
n
+
γ
n
+
ξ
n
≤
β
n, c
n
<
β
<
1
/
(
1
+
L
2
+
1
), for all n
≥
1,for some real numbers a
,
b
,
c
,
d
,
e
>
0, where L
=
m
a
x
{
L
1
,
L
2
}. Then {
x
n
} converges strongly to some point x
¯, where x
¯
=
P
F
u.Proof.
PutB
=
∂
I
C in Theorem 9. Then we get that(56)
x
∈
A
+
∂
I
C
-
1
0
⟺
0
∈
A
x
+
∂
I
C
x
⟺
-
A
x
∈
∂
I
C
x
⟺
A
x
,
y
-
x
≥
0
⟺
x
∈
V
I
C
,
A
.From Lemma 15, we can conclude the desired conclusion immediately.



---

*Source: 383579-2015-07-27.xml*