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RESEARC H Open Access
Strong convergence theorems for equilibrium
problems and fixed point problems: A new
iterative method, some comments and
applications
Zhenhua He
1
and Wei-Shih Du
2*
* Correspondence: wsdu@nknucc.
nknu.edu.tw
2
Department of Mathematics,
National Kaohsiung Normal
University, Kaohsiung 824, Taiwan
Full list of author information is
available at the end of the article
Abstract
In this paper, we introduce a new approach method to find a common element in
the intersection of the set of the solutions of a finite family of equilibrium problems
and the set of fixed points of a nonexpansive mapping in a real Hilbert space. Under
appropriate conditions, some strong convergence theorems are established. The
results obtained in this paper are new, and a few examples illustrating these results
are given. Finally, we point out that some ‘so-called’ mixed equilibrium problems and
generalized equilibrium problems in the literature are still usual equilibrium
problems.
2010 Mathematics Subject Classification: 47H09; 47H10, 47J25.
Keywords: strong convergence, iterative method, equilibrium problem, fixed point
problem
1 Introduction and preliminaries
Throughout this paper, we assume that H is a real Hilbert space with zero vector θ,

whose inner product and norm are denoted by 〈·, ·〉 and || · ||, respect ively. The sym-
bols N and ℝ are used to denote the sets of positive integers and real numbers, respec-
tively. Let K be a nonempty closed convex subset of H and T : K ® H be a mapping.
In this paper, the set of fixed points of T is denoted by F(T). We use symbols ® and
⇀ to denote strong and weak convergence, respectively.
For each point x Î H, there exists a unique nearest point in K, denoted by P
K
x, such
that

x − P
K
x

≤

x −
y 
, ∀
y
∈ K
.
The mapping P
K
is called the metric projection from H onto K. It is well known that
P
K
satisfies
x −
y

, P
K
x − P
K
y
≥P
K
x − P
K
y

2
for every x, y Î H. Moreover, P
K
x is characterized by the properties: for x Î H, and z
Î K,
He and Du Fixed Point Theory and Applications 2011, 2011:33
/>© 2011 He and Du; licensee S pringer. This i s an Open Access article distribu ted under the terms of the Creative Common s Attribution
License (http://creativecommons .org/license s/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium,
provided the original work is properly cited.
z
= P
K
(
x
)
⇔x − z, z − y≥0, ∀ y ∈ K
.
Let f be a bi-function from K × K into ℝ. The classical equilibrium problem is to find
x Î K such that

f
(
x, y
)
≥ 0, ∀ y ∈ K
.
(1:1)
Let EP(f) denote the set of all solutions of the problem (1.1). Since several probl ems
in phys ics, optimization, and economics r educe to find a solution of (1.1) (see, e.g.,
[1,2]), some authors had proposed some methods to find the solution of e quilibrium
problem (1.1); for instance, see [1-4]. We know that a mapping S is said to be nonex-
pansive mapping if for all x, y Î K,||Sx - Sy|| ≤ ||x - y||. Recently, some authors used
iterative method including composite iterative , CQ iterative, viscosity iterative etc. to
find a common element in the intersection of EP(f) and F(S); see, e.g., [5-11].
Let I be an index set. For each i Î I,letf
i
be a bi-function from K × K into ℝ.The
system of equilibrium problem is to find x Î K such that
f
i
(
x, y
)
≥ 0, ∀ y ∈ K and ∀i ∈ I
.
(1:2)
We know that

i
∈

I
EP( f
i
)
is the set of all solutions of the system of equilibrium pro-
blem (1.2).
For each i Î I,iff
i
(x, y)=〈A
i
x, y - x〉, where A
i
: K ® K is a nonlinear operator, then
the problem (1.2) becomes the following system of variational inequality problem:
Find an element x ∈ K such that A
i
x,
y
− x≥0, ∀
y
∈ K
.
(1:3)
It is obvious that the problem (1.3) is a special case of the problem (1.2).
The following Lemmas are crucial to our main results.
Lemma 1.1 (Demicloseness principle [12]) LetHbearealHilbertspaceandKa
closed convex subset of H. S : K ® H is a nonexpansive mapping. Then the mapping I -
S is demiclosed on K, where I is the identity mapping, i.e., x
n
⇀ x in K and (I - S)x

n
®
y implies that × Î K and (I - S)x = y.
Lemma 1.2 [13] Let {x
n
}and {y
n
} be bounded sequences in a Banach space E and let
{b
n
} be a sequence in [0,1] with 0 < lim inf
n®∞
b
n
≤ lim sup
n®∞
b
n
<1.Suppose x
n+1
=
b
n
y
n
+(1-b
n
)x
n
for al l integers n ≥ 0 and lim sup

n®∞
(||y
n+1
- y
n
||-||x
n+1
- x
n
||) ≤ 0,
then lim
n®∞
||y
n
- x
n
|| = 0.
Lemma 1.3 [5] Let H be a real Hilbert space. Then the following hold.
(a) ||x + y||
2
≤ ||y||
2
+2〈x, x + y〉 for all x, y Î H;
(b) ||ax +(1-a)y||
2
= a||x||
2
+(1-a)||y||
2
- a(1 - a)||x - y||

2
for all x, y Î H
and a Î ℝ;
(c) ||x - y||
2
=||x||
2
+||y||
2
-2〈x, y 〉 for all x, y Î H.
Lemma 1.4. [14] Let {a
n
} be a sequence of nonnegative real numbers satisfying the
following relation:
a
n+1
≤
(
1 − λ
n
)
a
n
+ γ
n
, n ≥ 0
.
If
He and Du Fixed Point Theory and Applications 2011, 2011:33
/>Page 2 of 15

(i) l
n
Î [0,1],
∞

n=0
λ
n
=
∞
or, equivalently,

∞
n
=
0
(1 − λ
n
)=0
;
(ii)
lim sup
n→∞
γ
n
λ
n
≤
0
or

∞

n=0
|γ
n
| <
∞
,
then
lim
n
→∞
a
n
=
0
.
Lemma 1.5 [1] Let K be a nonempty closed convex subset of H and F be a bi-function
of K × K into ℝ satisfying the following conditions.
(A1) F(x, x)=0for all × Î K;
(A2) F is monotone, that is, F(x, y)+F(y, x) ≤ 0 for all x, y Î K;
(A3) for each x, y, z Î K,
lim
t
↓
0
F( tz +(1− t)x, y) ≤ F(x, y)
;
(A4) for each × Î K, y ® F (x, y) is convex and lower semi-continuous.Let r >0and ×
Î H. Then, there exists z Î K such that

F( z , y)+
1
r
y − z, z − x≥0, for all y ∈ K
.
Lemma 1.6 [3] Let K be a nonempty closed convex subset of H and let F be a bi-
function of K × K into R satisfying (A1) - (A4).Forr>0 and × Î H, define a mapping
T
r
: H ® K as follows:
T
r
(x)=

z ∈ K : F(z, y)+
1
r
y − z, z − x≥0, ∀ y ∈ K

for all × Î H. Then the following hold:
(i) T
r
is single-valued;
(ii) T
r
is firmly nonexpansive, that is, for any x, y Î H,
 T
r
x − T
r

y

2
≤T
r
x − T
r
y
, x −
y

;
(iii) F(T
r
)=EP (F);
(iv) EP(F) is closed and convex.
2 Main results and their applications
Let I = {1, 2, , k} be a finite index set, where k Î N. For each i Î I, let f
i
be a bi-func-
tions from K × K into ℝ satisfying the conditions (A1)-(A4). Denote
T
i
r
n
: H →
K
by
T
i

r
n
(x)=

z ∈ K : f
i
(z, y)+
1
r
n
y − z, z − x≥0, ∀ y ∈ K

.
For each (i, n) Î I×N, applying Lemmas 1.5 and 1.6,
T
i
r
n
is a firmly nonexpansive
single-valued mapping such that
F( T
i
r
n
)=EP(f
i
)
is closed and convex. For each i Î I,
let
u

i
n
= T
i
r
n
x
n
, n Î N.
First, let us consider the following example.
He and Du Fixed Point Theory and Applications 2011, 2011:33
/>Page 3 of 15
Example A Let f
i
:[-1,0]×[-1,0] ®ℝ be defined by f
i
(x, y) = (1+x
2i
)(x-y), i =1,2,3.
It is easy to see that for any i Î {1, 2, 3}, f
i
(x, y) satisfies the conditions (A1)-(A4) and

3
i
=1
EP( f
i
)={0
}

.LetSx = x
3
and
g
x =
1
2
x
, ∀ x Î [-1, 0] Then g is a
1
2
-contraction from
K into itself and S : K ® K is a nonexpansive mapping with


3
i=1
EP( f
i
))


F( S)={0
}
. Let l Î (0, 1), {r
n
} ⊂ [1, + ∞)and{a
n
} ⊂ (0,1) satisfy the
conditions (i) lim

n® ∞
a
n
=0,and(ii)

∞
n
=1
α
n
=+
∞
, or equivalently,

∞
n
=1
(
1 − α
n
)
=
0
; e.g., let
λ =
1
3
,{a
n
} ⊂ (0, 1) and {r

n
} ⊂ [1, + ∞) be given by
α
n
=

0, if n is even;
1
n
,ifn is odd.
and r
n
=

2, if n is even
;
2 −
1
n
,ifn is odd.
Define a sequence {x
n
}by
⎧
⎪
⎪
⎪
⎪
⎪
⎪

⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩
x
1
∈ [−1, 0],
u
i
n
= T
i
r
n
x
n
, i = 1,2,3,
x
n+1
= α
n
g(x
n
)+(1− α
n
)y

n
,
y
n
=(1− λ)x
n
+ λSz
n
,
z
n
=
u
1
n
+ u
2
n
+ u
3
n
3
, ∀n ∈
N
.
(2:1)
Then the sequences {x
n
} and
{u

i
n
}
, i = 1, 2, 3, defined by (2.1) all strongly converge to
0.
Proof
(a) By Lemmas 1.5 and 1.6, (2.1) is well defined.
(b) Let K = [-1, 0]. For each i Î {1, 2, 3}, define
L
i
(y, z, v, r)=(z − y)

(1 + z
2i
) −
1
r
(z − v)

∀y, z, v ∈ K, ∀r ≥ 1
.
We claim that for each v Î K and any i Î {1, 2, 3}, there exists a unique z =0Î K
such that
(
P
)
L
i
(
y, z, v, r

)
≥ 0 ∀y ∈ K, ∀ r ≥
1
or, equivalently,
(1+z
2i
)(z−y)+
1
r
y− z, z− v =(1+z
2i
)(z−y)+
1
r
(y− z)(z− v) ≥ 0 ∀y ∈ K, ∀r ≥ 1
.
Obviously, z = 0 is a solution of the problem
(
P
)
. On the other hand, there does not
exist z Î [-1, 0) such that z-y≤ 0 and
(1 + z
2i
) −
1
r
(z − v) ≤
0
.Soz = 0 is the uniq ue

solution of the problem
(
P
)
.
(c) We notice that (2.1) is equivalent with (2.2), where
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
x
1
∈ [−1, 0],
f
i
(u
i

n
, y)+
i
r
n
y − u
i
n
, u
i
n
− x
n
≥0, ∀ y ∈ K, ∀i =1,2,3
,
x
n+1
= α
n
g(x
n
)+(1− α
n
)y
n
,
y
n
=(1− λ)x
n

+ λSz
n
,
z
n
=
u
1
n
+ u
2
n
+ u
3
n
3
, n ∈ N.
(2:2)
He and Du Fixed Point Theory and Applications 2011, 2011:33
/>Page 4 of 15
It is easy to see that {x
n
} ⊂ [-1, 0], so, by (b),
u
1
n
= u
2
n
= u

3
n
=
0
for all n Î N. We need
to prove x
n
® 0asn ® ∞. Since z
n
= 0 for all n Î N, we have y
n
=(1-l)x
n
and
x
n+1
= α
n
g(x
n
)+(1−α
n
)y
n
=
1
2
α
n
x

n
+(1−α
n
)(1−λ)x
n
=

1 −
1
2
α
n

− (1 − α
n
)λ

x
n
(2:3)
for all n Î N. For any n Î N, from (2.3), we have
|
x
n+1
| =

1 −
1
2
α

n

− (1 − α
n
)λ

|
x
n
|
≤

1 −
1
2
α
n

|
x
n
|
.
(2:4)
Hence {|x
n
|} is a strictly deceasing sequence and |x
n
| ≥ 0foralln Î N.So
lim

n
→∞
|
x
n
|
exists.
On the other hand, for any n, m Î N with n>m, using (2.4), we obtain
|
x
n+1
|≤

1 −
1
2
α
n

|
x
n
|
≤

1 −
1
2
α
n


1 −
1
2
α
n−1

|
x
n−1
|
≤···≤
n

j
=m

1 −
1
2
α
j

|
x
m
|
,
which implies
lim sup

n
→∞
|
x
n
|
≤ 0 ≤ lim inf
n→∞
|
x
n
|
. Therefore {x
n
} strongly converges to 0.
□
In this paper, motivated by the preceding Example A, we introduce a new iterative
algorithm for the problem of finding a common element in the set of sol utions to the
system of equilibrium problem and the set of fixed points of a nonexpansive mapping.
The following new strong convergence theorem is established in the framework of a
real Hilbert space H.
Theorem 2.1 Let K be a nonempty closed c onvex subset o f a real Hilbert space H
and I = {1, 2, , k} be a finit e inde x set. For e ach i Î I, let f
i
be a bi-function from K ×
Kintoℝ satisfying (A1)-(A4). Let S : K ® K be a nonexpansive mapping with

=



k
i=1
EP( f
i
)


F( S) =
∅
.Letl, r Î (0, 1) and g : K ® Kisar-contraction. Let
{x
n
} be a sequence generated in the following manner:
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩
x
1

∈ K,
u
i
n
= T
i
r
n
x
n
, ∀i ∈ I.
x
n+1
= α
n
g(x
n
)+(1− α
n
)y
n
,
y
n
=(1− λ)x
n
+ λSz
n
,
z

n
=
u
1
n
+ ···+ u
k
n
k
, ∀n ∈ N.
(D
H
)
If the above control coefficient sequences {a
n
} ⊂ (0, 1) and {r
n
} ⊂ (0, +∞) satisfy the
following restrictions:
(D1)
lim
n
→∞
α
n
=
0
,
∞


n
=1
α
n
=+
∞
and
lim
n
→∞
|α
n+1
− α
n
| =
0
;
(D2)
lim inf
n
→
∞
r
n
>
0
and
lim
n
→∞

|r
n+1
− r
n
| =
0
.
then the sequences {x
n
} and
{u
i
n
}
, for all i Î I, converge strongly to an element c = P
Ω
g
(c) Î Ω. The following conclusion is immediately drawn from Theorem 2.1.
He and Du Fixed Point Theory and Applications 2011, 2011:33
/>Page 5 of 15
Corollary 2.1 LetKbeanonemptyclosedconvexsubsetofarealHilbertspaceH.
Let f be a bi-function from K × K in to ℝ satisfying (A1)-(A4) and S : K ® Kbeanon-
expansive mapping with Ω = EP(f) ∩F(S) ≠ ∅. Let l, r Î (0,1) and g : K ® Kisar-
contraction. Let {x
n
} be a sequence generated in the following manner:
⎧
⎪
⎪
⎨

⎪
⎪
⎩
x
1
∈ K,
u
n
= T
r
n
x
n
,
x
n+1
= α
n
g(x
n
)+(1− α
n
)y
n
,
y
n
=(1− λ)x
n
+ λSu

n
, ∀n ∈ N
.
If the above control coefficient sequences {a
n
} ⊂ (0, 1) and {r
n
} ⊂ (0, +∞) satisfy all
the restrictions in Theorem 2.1, then the sequences {x
n
} and {u
n
} converge strongly to an
element c = P
Ω
g(c) Î Ω, respectively.
If f
i
(x, y) ≡ 0 for all (x, y) Î K × K in Theorem 2.1 and all i Î I, then, from the algo-
rithm (D
H
), we obtain
u
i
n
≡ P
K
(x
n
)

, ∀ i Î I. So we have the following result .
Corollary 2.2 LetKbeanonemptyclosedconvexsubsetofarealHilbertspaceH.
Let S : K ® K be a nonexpansive mapping with F(S) ≠ ∅. Let l, r Î (0, 1) and g : K
® Kisar-contraction. Let {x
n
} be a sequence generated in the following manner:
⎧
⎨
⎩
x
1
∈ K,
x
n+1
= α
n
g(x
n
)+(1− α
n
)y
n
,
y
n
=(1− λ)x
n
+ λSP
K
(x

n
), ∀n ∈ N
.
If the above control coefficient sequences {a
n
} ⊂ (0, 1) satisfy
lim
n
→∞
α
n
=
0
,
lim
n
→∞
|α
n+1
− α
n
| =
0
and
lim
n
→∞
|α
n+1
− α

n
| =
0
, then the sequences {x
n
} converge strongly
to an element c = P
Ω
g(c) Î F (S).
As some interesting and important applications of Theorem 2.1 for optimization pro-
blems and fixed point problems, we have the following.
Application (I) of Theorem 2.1 We will give an iterative algorithm for the following
optimization problem with a nonempty common solution set:
min
x
∈
K
h
i
(x), i ∈{1, 2, , k},(OP
)
where h
i
(x), i Î {1, 2, , k}, are convex and l ower semi-continuous funct ions defined
on a closed convex subset K of a Hilbert space H (for example, h
i
(x)=x
i
, x Î K := [0,
1], i Î {1, 2, , k}).

If we put f
i
(x, y)=h
i
(y)-h
i
(x), i Î {1, 2, , k}, then

k
i
=1
EP( f
i
)
is the common solu-
tion set of the problem (OP), where

k
i
=1
EP( f
i
)
denote the common solution set of the
following equilibrium:
F
ind x ∈ K such that f
i
(
x, y

)
≥ 0, ∀ y ∈ K and ∀ i ∈{1, 2, , k}
.
For i Î {1, 2, , k}, it is obvious that the f
i
(x, y) satisfies the conditions (A1)-(A4). Let
S = I (identity mapping), then from (D
H
), we have the following algorithm
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩
h
i
(y) − h
i
(u
i
n
)+

1
r
n
y − u
i
n
, u
i
n
− x
n
≥0, ∀ y ∈ K and ∀ i ∈{1, 2, , k}
,
x
n+1
= α
n
g(x
n
)+(1− α
n
)y
n
,
y
n
=(1− λ)x
n
+ λz
n

,
z
n
=
u
1
n
+ ···+ u
k
n
k
, n ≥ 1.
(2:5)
He and Du Fixed Point Theory and Applications 2011, 2011:33
/>Page 6 of 15
where x
1
Î K, l Î (0, 1), g : K ® K is a r-contraction. From Theorem 2.1, we kno w
that {x
n
}and
{u
i
n
}
, i Î{1,2, , k}, generated by (2.5), strongly converge to an element of

k
i
=1

EP( f
i
)
if the coefficients {a
n
} and {r
n
} satisfy the conditions of Theorem 2.1.
Application ( II) of Theorem 2.1 Let H, K, I, l, r, g bethesameasTheorem2.1.Let
A
1
, A
2
, , A
k
: K ® K be k nonlinear mappings with
k

i
=1
F( A
i
) =
∅
.Foranyi Î I,putf
i
(x, y)=〈x-A
i
x, y-x〉, ∀ x, y Î K.Since


k
i
=1
EP( f
i
)=

k
i
=1
F( A
i
)
,wehave
k

i
=1
EP( f
i
) =
∅
.LetS = I (identity mapping) in the algorithm (D
H
). Then the sequences
{ x
n
}and
{u
i

n
}
, defined by the algorithm (D
H
), converge str ongly to a common fixed
point of {A
1
, A
2
, , A
k
}, respectively.
The following result is important in this paper.
Lemma 2.1 Let H be a real Hilbert space. Then for any x
1
, x
2
, x
k
Î H and a
1
, a
2
, ,
a
k
Î [0,1] with

k
i

=1
a
i
=
1
, k Î N, we have





k

i=1
a
i
x
i





2
=
k

i=1
a
i

 x
i

2
−
k−1

i=1
k

j
=i+1
a
i
a
j
 x
i
− x
j

2
.
(2:6)
Proof It is obvious that (2.6) is true if a
j
=1forsomej, so it suffices to show that
(2.6) i s true for a
j
≠ 1forallj. The proof is by mathematic induction on k.Clearly,

(2.6) is true for k =1.Letx
1
, x
2
Î H and a
1
, a
2
Î [0,1] with a
1
+ a
2
= 1. By Lemma
1.3, we obtain

a
1
x
1
+ a
2
x
2

2
= a
1

x
1


2
+ a
2

x
2

2
− a
1
a
2

x
1
− x
2

2
,
which means that (2.6) hold for k = 2. Suppose that (2.6) is true for k = l Î N.Let
x
1
, x
2
, , x
l
, x
l+1

Î H and a
1
, a
2
, , a
l
, a
l+1
Î [0, 1) with

l+1
i
=1
a
i
=
1
.Let
y =

l+1
i=2
a
i
1−a
1
x
i
.
Then applying the induction hypothesis we have






l+1

i=1
a
i
x
i





2
= a
1
x
1
+(1− a
1
)y
2
= a
1
 x
1


2
+(1− a
1
)  y
2
− a
1
(1 − a
1
)  x
1
− y
2
=
l+1

i=1
a
i
 x
i

2
−
1
1 − a
1
l


i=2
l+1

j=i+1
a
i
a
j
 x
i
− x
j

2
− a
1
(1 − a
1
)





l+1

i=2
a
i
1 − a

1
(
x
i
− x
1
)





2
=
l+1

i=1
a
i
 x
i

2
−
1
1 − a
1
l

i=2

l+1

j=i+1
a
i
a
j
 x
i
− x
j

2
− a
1
(1 − a
1
)
l+1

i=2
a
i
1 − a
1
 x
1
− x
i


2
+ a
1
(1 − a
1
)
l

i=2
l+1

j=i+1
a
i
1 − a
1
a
j
1 − a
1
 x
i
− x
j

2
=
l+1

i=1

a
i
 x
i

2
−
1
1 − a
1
l

i=2
l+1

j=i+1
a
i
a
j
 x
i
− x
j

2
−
l+1

i=2

a
1
a
i
 x
1
− x
i

2
+
a
1
1 − a
1
l

i=2
l+1

j=i+1
a
i
a
j
 x
i
− x
j


2
=
l+1

i=1
a
i
 x
i

2
−
l+1

i=2
a
1
a
i
 x
1
− x
i

2
−
l

i=2
l+1


j=i+1
a
i
a
j
 x
i
− x
j

2
=
l+1

i=1
a
i
 x
i

2
−
l

i=1
l+1

j
=i+1

a
i
a
j
 x
i
− x
j

2
.
He and Du Fixed Point Theory and Applications 2011, 2011:33
/>Page 7 of 15
Hence, the equality (2.6) is also true for k = l + 1. This completes the induction. □
3 Proof of Theorem 2.1
We will proceed with the following steps.
Step 1: There exists a unique c Î Ω ⊂ H such that P
Ω
g(c)=c.
Since P
Ω
g is a r-contraction on H, Banach contraction principle ensures that there
exists a unique c Î H such that c = P
Ω
g(c) Î Ω.
Step 2: We prove that the sequences {x
n
}, {y
n
}, {z

n
} and
{u
i
n
}
, ∀i Î I, are all bounded.
First, we notice that (D
H
) is equivalent with (Z
H
), where
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪

⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
x
1
∈ K
f
1
(u
1
n
, y)+

1
r
n
y − u
1
n
, u
1
n
− x
n
≥0, ∀ y ∈ K,
f
2
(u
2
n
, y)+
1
r
n
y − u
2
n
, u
2
n
− x
n
≥0, ∀ y ∈ K,

.
.
.
f
k
(u
k
n
, y)+
1
r
n
y − u
k
n
, u
k
n
− x
n
≥0, ∀ y ∈ K,
x
n+1
= α
n
g(x
n
)+(1− α
n
)y

n
,
y
n
=(1− λ)x
n
+ λSz
n
,
z
n
=
u
1
n
+ ···+ u
k
n
k
, n ∈ N.
(Z
H
)
For each i Î I, we have
||u
i
n
− c|| = ||T
i
r

n
x
n
− T
i
r
n
c|| ≤ ||x
n
− c||, ∀ n ∈ N
.
(3:1)
For any n Î N, from (Z
H
) we have

z
n
− c

≤

x
n
− c

and
 y
n
− c


≤

x
n
− c
 .
(3:2)
Since g is a r-contraction, it follows from (3.2) that
 x
n+1
− c ≤α
n


g(x
n
) − c


+(1− α
n
)


y
n
− c



≤ α
n


g(x
n
) − g(c)


+ α
n


g(c) − c


+(1− α
n
)


y
n
− c


≤ α
n
ρ


x
n
− c

+ α
n


g(c) − c


+(1− α
n
)

x
n
− c

=

1 − α
n
(1 − ρ)


x
n
− c


+ α
n
(1 − ρ)


g(c) − c


1 − ρ
≤ max

 x
n
− c ,
 g(c) − c 
1 − ρ

,forn ∈ N
.
By induction, we obtain
 x
n
− c ≤max

 x
1
− c ,
 g(c) − c 
1 − ρ


for all n ∈ N
,
which shows that {x
n
} is bounded. Also, we know that {y
n
}, {z
n
}and
{
u
i
n
}
, ∀i Î I,are
all
bounded.
Step 3: We prove lim
n®∞
||x
n+1
- x
n
|| = 0.
He and Du Fixed Point Theory and Applications 2011, 2011:33
/>Page 8 of 15
For each i Î I, since
u
i
n

−
1
,
u
i
n
∈
K
, from (Z
H
), we have
f
i
(u
i
n
, u
i
n−1
)+
1
r
n
u
i
n−1
− u
i
n
, u

i
n
− x
n
≥0
,
(3:3)
and
f
i
(u
i
n−1
, u
i
n
)+
1
r
n
−1
u
i
n
− u
i
n−1
, u
i
n−1

− x
n−1
≥0
.
(3:4)
By (3.3) and (3.4) and (A2),
0 ≤ r
n

f
i
(u
i
n
, u
i
n−1
)+f
i
(u
i
n−1
, u
i
n
)

+ u
i
n−1

− u
i
n
, u
i
n
− x
n
−
r
n
r
n−1
(u
i
n−1
− x
n−1
)

≤u
i
n−1
− u
i
n
, u
i
n
− x

n
−
r
n
r
n
−1
(u
i
n−1
− x
n−1
),
which implies
u
i
n−1
− u
i
n
, u
i
n−1
− u
i
n
+ x
n
− x
n−1

+ x
n−1
− u
i
n−1
+
r
n
r
n
−1
(u
i
n−1
− x
n−1
)≤0
.
(3:5)
It follows from (3.5) that
 u
i
n
− u
i
n−1
≤x
n
− x
n−1

 +




r
n
− r
n−1
r
n−1




 x
n−1
− u
i
n−1
 for all n ∈ N
.
(3:6)
Let
M :=
1
k

k
i=1

 x
n−1
− u
i
n−1
<
∞
.Foranyn Î N,since
z
n
=
1
k
(u
1
n
+ ···+ u
k
n
)
,by
(3.6), we have
 z
n
− z
n−1
≤
1
k
k


i
=1
 u
i
n
− u
i
n−1
≤x
n
− x
n−1
 +M




r
n
− r
n−1
r
n−1




.
(3:7)

Set
v
n
=
x
n+1
− (1 − β
n
)x
n
β
n
,
(3:8)
where b
n
=1-(1-l)(1 - a
n
), n Î N. Then for each n Î N,
x
n+1
− x
n
= β
n
(
v
n
− x
n

)
(3:9)
and
v
n
=
α
n
g(x
n
)+λ(1 − α
n
)Sz
n
β
n
.
(3:10)
For any n Î N, since
v
n+1
− v
n
=
α
n+1
g(x
n+1
)
β

n+1
−
α
n
g(x
n
)
β
n
−
λ(1 − α
n
)Sz
n
β
n
+
λ(1 − α
n+1
)Sz
n+1
β
n+1
=
α
n+1
g(x
n+1
)
β

n+1
−
α
n
g(x
n
)
β
n
−
λ(1 − α
n
)(Sz
n
− Sz
n+1
)
β
n
− λ(
1 − α
n
β
n
−
1 − α
n+1
β
n+1
)Sz

n+1
,
He and Du Fixed Point Theory and Applications 2011, 2011:33
/>Page 9 of 15
by (3.7), it follows that
 v
n+1
− v
n
−x
n+1
− x
n
≤
α
n+1
 g
(
x
n+1
)

β
n+1
+
α
n
 g
(
x

n
)

β
n
+
λ
(
1 − α
n
)
 z
n
− z
n+1

β
n
+




1 − α
n
β
n
−
1 − α
n+1

β
n+1




 Sz
n+1
−x
n+1
− x
n

≤
α
n+1
 g(x
n+1
) 
β
n+1
+
α
n
 g(x
n
) 
β
n
+


λ(1 − α
n
)
β
n
− 1

 x
n+1
− x
n

+
M
β
n




r
n+1
− r
n
r
n





+




1 − α
n
β
n
−
1 − α
n+1
β
n+1




 Sz
n+1
 .
From this and (D1), (D2), we get
lim sup
n
→∞
{ v
n+1
− v
n

−x
n+1
− x
n
} ≤ 0
.
(3:11)
By Lemma 1.2 and (3.11),
lim
n
→
∞
 v
n
− x
n
 =0
.
(3:12)
Owing to (3.9) and (3.12), we obtain
lim
n
→
∞
 x
n+1
− x
n
 =0
.

(3:13)
Step 4: We show
lim
n→∞
 Su
i
n
− u
i
n
 =
0
.
By (3.6), (3.13) and (D2), we have
lim
n
→∞
 u
i
n+1
− u
i
n
 =0, ∀i ∈ I
.
From (Z
H
), we get
lim
n

→
∞
 x
n+1
− y
n
 = lim
n
→
∞
α
n
 g(x
n
) − y
n
 =0
.
(3:14)
Since ||x
n
- y
n
|| ≤ ||x
n
- x
n+1
|| + ||x
n+1
- y

n
||, by (3.13) and (3.14),
lim
n
→
∞
 y
n
− x
n
 =0
,
which implies that
lim
n→∞
 Sz
n
− x
n
 = lim
n→∞
1
λ
 y
n
− x
n
 =0
.
By Lemma 1.6,

 u
i
n
−c
2
=  T
i
r
n
x
n
−T
i
r
n
c
2
≤T
i
r
n
x
n
−T
i
r
n
c, x
n
−c =

1
2

 u
i
n
− c
2
+  x
n
− c
2
−u
i
n
− x
n

2

,
which yields that
 u
i
n
− c
2
≤x
n
− c

2
−u
i
n
− x
n

2
.
(3:15)
From (3.15) and Lemma 2.1,
 z
n
− c
2
=





k

i=1
1
k

u
i
n

− c






2
≤
1
k
k

i=1
 u
i
n
− c
2
≤x
n
− c
2
−
1
k
k

i=1
 u

i
n
− x
n

2
.
He and Du Fixed Point Theory and Applications 2011, 2011:33
/>Page 10 of 15
Since
 x
n+1
− c
2
≤ α
n
 g(x
n
) − c
2
+(1− α
n
)  y
n
− c
2
≤ α
n
 x
n

− c
2
+2α
n
L +(1− α
n
)  y
n
− c
2
≤ [1 − λ
(
1 − α
n
)
]  x
n
− c
2
+2α
n
L + λ
(
1 − α
n
)
 z
n
− c
2

where
L =max{2  g
(
c
)
− c  x
n
− c ,  g
(
c
)
− c
2
} < ∞
,
We have
1 − α
n
k
λ
k

i
=1
 u
i
n
−x
n


2
≤x
n
−c
2
−x
n+1
−c
2
+2α
n
L ≤ ( x
n
−c  +  x
n+1
−c )  x
n
−x
n+1
 +2α
n
L
.
(3:16)
Letting n ® ∞ in the inequality (3.16), we obtain
lim
n
→∞
 u
i

n
− x
n
 =0, ∀i ∈ I
.
(3:17)
Furthermore, it is easy to prove that
lim
n
→
∞
 z
n
− x
n
 = lim
n
→
∞
 u
i
n
− z
n
 =0 ∀i ∈ I
.
For any i Î I, since
 Su
i
n

− u
i
n
≤Su
i
n
− Sz
n
 +  Sz
n
− x
n
 +  x
n
− u
i
n

,
it implies
lim
n
→∞
 Su
i
n
− u
i
n
 =0

.
(3:18)
Step 5: Prove lim sup
n®∞
〈g(c)-q, x
n
- c〉 ≤ 0.
Take a subsequence
{x
n

}
of {x
n
} such that
lim sup
n
→∞
g(c) − c, x
n
− c = lim
→∞
g(c) − c, x
n

− c
.
(3:19)
Since
{

x
n

}
is bounded, there exists a subsequence of
{
x
n

}
which is still denoted by
{x
n

}
such that
x
n


z
as ℓ ® ∞. Notice that for each i Î I,
lim

→∞
 u
i
n

− x

n

 =
0
by
(3.17), so we also have
u
i
n


z
as ℓ ® ∞, ∀ i Î I.
We want to show z Î Ω.First,weshowthatz Î F(S). In fact, since
lim

→
∞
 (I − S)u
i
n

 = lim

→
∞
 Su
i
n


− u
i
n

 =
0
and
u
i
n


z
as ℓ ® ∞, by Lemma 1.1, we
have (I - S)z = θ or, equivalently, z Î F(S).
For each i Î I,since
f
i
(u
i
n

, y)+
1
r
n

y − u
i
n


, u
i
n

− x
n

≥
0
, ∀ y Î K,itfollowsfrom
(A2) that
1
r
n

y − u
i
n

, u
i
n

− x
n

≥f
i
(y, u

i
n

)+f
i
(u
i
n

, y)+
1
r
n

y − u
i
n

, u
i
n

− x
n

≥f
i
(y, u
i
n


)
,
and hence
y − u
i
n

,
u
i
n

− x
n

r
n

≥f
i
(y, u
i
n

), ∀y ∈ K
.
He and Du Fixed Point Theory and Applications 2011, 2011:33
/>Page 11 of 15
Applying (3.17) and (A4),

f
i
(
y, z
)
≤ 0, ∀y ∈ K
.
(3:20)
Let y Î K be given. Put y
t
= ty +(1-t)z, t Î (0, 1). Then y
t
Î K and f
i
(y
t
, z) ≤ 0for
all i Î I. By (A1) and (A4), we get
0=f
i
(
y
t
, y
t
)
≤ tf
i
(
y

t
, y
)
+
(
1 − t
)
f
i
(
y
t
, z
)
≤ tf
i
(
y
t
, y
)
∀i ∈ I
.
For any i Î I, by (A3), we have
f
i
(z, y) ≥ lim
t
↓
0

f
i
(ty +(1− t)z, y) = lim
t
↓
0
f
i
(y
t
, y) ≥ 0.
(3:21)
Hence, from (3.21),
z
∈

k
i
=1
EP( f
i
)
. Therefore, we proved
z
∈  =(

k
i
=1
EP( f

i
))

F( S
)
. On the other hand, by (3.19), we obtain
lim sup
n
→∞
g(c) − c, x
n
− c = g(c) − c, z − c≤0
.
(3:22)
Ste p 6: Finally, we pro ve {x
n
}and
{u
i
n
}
,foralli Î I, converge strongly to c = P
Ω
g(c)
Î Ω.
From (Z
H
) and (a) of Lemma 1.3, we have
||x
n+1

− c||
2
≤ (1 − α
n
)
2
||y
n
− c||
2
+2α
n
g(x
n
) − g(c)+g(c) − c, x
n+1
− c
≤ (1 − α
n
)
2
||x
n
− c||
2
+2α
n
ρ||x
n
− c|| ||x

n +1
− c|| +2α
n
g(c) − c, x
n +1
− c

≤ (1 − 2α
n
+ α
2
n
)||x
n
− c||
2
+2α
n
ρ||x
n
− c|| ||x
n
− x
n +1
|| +2α
n
ρ||x
n
− c||
2

+2α
n
g(c) − c, x
n+1
− c
=(1− 2(1 − ρ)α
n
)||x
n
− c||
2
+ α
2
n
||x
n
− c||
2
+2α
n
ρ||x
n
− c|| ||x
n
− x
n +1
||
+2α
n
g(c) − c, x

n+1
− c
=(1− 2(1 − ρ)α
n
)||x
n
− c||
2
+ α
2
n
||x
n
− c||
2
+2α
n
ρ||x
n
− c|| ||x
n
− x
n+1
||
+2α
n
g
(
c
)

− c, x
n+1
− c.
(3:23)
For any n Î ℤ, let
a
n
= ||x
n
− c||
2
,
b
n
= α
n
||x
n
− c||
2
+2ρ||x
n
− c|| ||x
n
− x
n+1
|| +2g(c) − c, x
n+1
− c
,

λ
n
=2
(
1 − ρ
)
α
n
,
and
γ
n
= α
n
b
n
.
From (3.23), we have
a
n+1
≤
(
1 − λ
n
)
a
n
+ γ
n
, ∀n ∈ N

.
It is easy to verify that all conditions of Lemma 1.4 are satisfied. Hence, applying
Lemma 1.4, we obtain lim
n®∞
a
n
= 0 which implies
lim
n
→
∞
||x
n
− c|| =0
,
or equivalence , {x
n
} s trongly converges to c. By (3 .17), we can prove that for any i Î
I,
{u
i
n
}
strongly converges to c. The proof of Theorem 2.1 is completed. □
He and Du Fixed Point Theory and Applications 2011, 2011:33
/>Page 12 of 15
4 Fu rther remarks
Let K be a nonempty closed convex subset of H and f be a bi-function of K × K into ℝ.
Remark 4.1 Recently, some authors introduced the followin g mixed equilibrium pro-
blem (MEP, for short) (see [15-17] and references therein) and generalized equilibrium

problem (GEP, for short) (see [18-20] and references therein):
(a) Mixed equilibrium problem [15-17]:
Find an element x ∈ C such that f
(
x, y
)
+ ϕ
(
y
)
− ϕ
(
x
)
≥ 0, ∀y ∈ C.
(
MEP
)
where  : C ® ℝ is a real-valued function.
(b) Generalized equilibrium problem [18-20]:
Find an element x ∈ C such that f
(
x, y
)
+
(
Ax, y − x
)
≥ 0, ∀y ∈ C.
(

GEP
)
where A : C ® H is a nonlinear operator.
In [15-17], the authors gave some iterative methods for finding the solution of MEP
when the bi-function f(x, y) admits the conditions (A1)-(A4) and the real-valued func-
tion  satisfies the following condition:
(A5)  : C ® ℝ is a proper lower semi-continuous and convex function.
However, in this case, w e argue that the prob lem MEP is still the equilibrium pro-
blem (1.1). In fact, if we put f
1
(x, y)=f(x, y), f
2
(x, y)=(y)-(x)andF(x, y)=f
1
(x, y)
+ f
2
(x, y) for each (x, y) Î C × C,thenf
1
(x, y) satisfies the conditions (A1)-(A4), f
2
(x,
y) satisfies the condition (A5) and the function  must satisfy the conditions (A1)-
(A4). This shows that for each (x, y) Î C × C, F(x, y) satisfies the conditions (A1)-
(A4). So, when we study the solution of MEP, we only need to study the solution of
the equilibrium (1.1). This also shows that some “ so-called” mixed equilibrium pro-
blem studied in [15-17] is still the equilibrium problem (1.1).
Remark 4. 2 Let us recall some well-known definitions. A mapping T : C ® C is said
to be
(1) v-expansive if ||Tx - Ty|| ≥ v||x - y|| for all x, y Î C. In particular, if v = 1, then

T is called expansive.
(2) v-strongly monotone if there exists a constant v>0 such that
Tx − T
y
, x −
y
≥v||x −
y
||
2
, ∀x,
y
∈ C
.
Clearly, any v-strongly monotone mapping is v-expansive.
(3) u-inverse strongly monotone if there exists a constant u>0 such that
Tx − T
y
, x −
y
≥u||Tx − T
y
||
2
, ∀x,
y
∈ C
.
(4) L-Lipschitz continuous if ||Tx - Ty|| ≤ L||x - y|| for all x, y Î C. In particular, if
L = 1, then T is called nonexpansive.

He and Du Fixed Point Theory and Applications 2011, 2011:33
/>Page 13 of 15
It is easy to see that a u-inv erse strongly mono tone operator is
1
u
-Lipschitz
continuous.
For the problem GEP, if the nonlinear operator A : C ® H is a u-inverse strongly
monotone operator and the bi-function f(x, y) admits the conditions (A1)-(A4), we
argue that the pro blem GEP is still the problem (1.1) and so it is indeed not a gen eral-
ization. In fact, if A is a u-inverse strongly monotone operator from C into H, then A
is a continuous operator. So, we obtain easily that the function (x, y) ® <Ax, y - x〉,
∀x, y Î C, satisfies the co nditions (A1)-(A4). Hence, if we put F(x, y)=f (x, y )+〈Ax, y
- x〉 ≥ 0, then the problem GEP studied in [18-20] is still the problem (1.1).
5 Conclusion
The problem MEP studied in [15-17] and the problem GEP studied in [18-20] are still
the problem (1.1) studied in the literature [5-11,21-24] and others.
Acknowledgements
Zhenhua He was supported by the Natural Science Foundation of Yunnan Province (2010ZC152) and the Scientific
Research Foundation from Yunnan Province Education Committee (08Y0338); Wei-Shih Du was supported by the
National Science Council of the Republic of China.
Author details
1
Department of Mathematics, Honghe University, Yunnan, 661100, China
2
Department of Mathematics, National
Kaohsiung Normal University, Kaohsiung 824, Taiwan
Authors’ contributions
Both authors contributed equally and significantly in writing this paper. Both authors read and approved the final
manuscript.

Competing interests
The authors declare that they have no competing interests.
Received: 2 April 2011 Accepted: 12 August 2011 Published: 12 August 2011
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Cite this article as: He and Du: Strong convergence theorems for equilibrium problems and fixed point
problems: A new iterative method, some comments and applications. Fixed Point Theory and Applications 2011
2011:33.
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