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Convergence theorems for a system of equilibrium problems and fixed point problems of a strongly nonexpansive sequence

Abstract

The purpose of this paper is to prove a strong convergence theorem of an iterative scheme associated to a strongly nonexpansive sequence for finding a common element of the set of equilibrium problems and the set of fixed point problems of a pair of sequences of nonexpansive mappings where one of them is a strongly nonexpansive sequence. Moreover, in the last section, by using our main result, we obtain a strong convergence theorem of an iterative scheme associated to a strongly nonexpansive sequence for finding a common element of the set of a finite family of equilibrium problems and the set of fixed point problems of a pair of sequences of nonexpansive mappings where one of them is a strongly nonexpansive sequence in a Hilbert space, and we also give some examples to support our main result.

1 Introduction

Throughout this paper, we assume that H is a real Hilbert space with the inner product , and the norm . A mapping T of C into itself is called nonexpansive if TxTyxy for all x,yH. The set of fixed points of T is denoted by F(T), i.e., F(T)={xH:Tx=x}. It is known that F(T) is closed and convex if T is nonexpansive. Let P C be a metric projection of H onto C, i.e., for xH, P C x satisfies the property

x P C x= min y C xy.

We use and to denote weak and strong convergence, respectively. Let { T n } be a sequence of mappings of C into H. The set of common fixed points of { T n } is denoted by F({ T n })= n = 1 F( T n ). Recall the main concepts as follows:

  1. (1)

    A sequence { z n } in C is said to be an approximate fixed point sequence of { T n } if z n T n z n 0. The set of all bounded approximate fixed point sequences of { T n } is denoted by F ˜ ({ T n }); see [1]. It is clear that if { T n } has a common fixed point, then F ˜ ({ T n }) is nonempty.

  2. (2)

    A sequence { T n } is said to be a strongly nonexpansive sequence if each T n is nonexpansive and

    x n y n ( T n x n T n y n )0,

    whenever { x n } and { y n } are sequences in C such that { x n y n } is bounded and x n y n T n x n T n y n 0.

  3. (3)

    A sequence { T n } having a common fixed point is said to satisfy the condition (Z) if every weak cluster point of { x n } is a common fixed point whenever { x n } F ˜ ({ T n }).

  4. (4)

    A sequence { T n } of nonexpansive mappings of C into H is said to satisfy the condition (R) if

    lim n sup y D T n + 1 y T n y=0

    for every nonempty bounded subset D of C; see [2].

Example 1.1 Let be a set of real numbers. For every nN, the mapping T n :RR is defined by T n x= 1 n x for all xR.

Then { T n } is a nonexpansive sequence, but it is not a strongly nonexpansive sequence.

Example 1.2 For every nN, the mapping T n :[0,1][0,1] is defined by T n x=(1 1 n )x for all x[0,1].

Then { T n } is a strongly nonexpansive sequence.

Solution It is easy to see that T n is a nonexpansive mapping for all nN.

Let { x n } and { y n } be sequences in [0,1] with { x n y n } being bounded and | x n y n || T n x n T n y n |0 as n.

Since x n y n ( T n x n T n y n )= 1 n ( x n y n ), for all nN, then we have

x n y n ( T n x n T n y n )0as n.

Then { T n } is a strongly nonexpansive sequence.

Let G:C×CR be a bifunction. The equilibrium problem for G is to determine its equilibrium points, i.e., the set

EP(G)= { x G : G ( x , y ) 0 , y C } .

It is a unified model of several problems, namely, variational inequality problem, complementary problem, saddle point problem, optimization problem, fixed point problem, etc.; see [35]. Several iterative methods have been proposed to solve the equilibrium problem; see, for instance, [68]. In 2005, Combettes and Hirstoaga [4] introduced some iterative schemes of finding the best approximation to the initial data when EP(G) is nonempty and proved a strong convergence theorem.

Also in [4], Combettes and Hiratoaga, following [3], defined S r :HC by

S r (x)= { z C : G ( z , y ) + 1 r y z , z x 0 , y C } .

They proved that under suitable hypotheses S r is single-valued and firmly nonexpansive with F( S r )=EP(G).

In 2007, Takahashi and Takahashi [9] proved the following theorem.

Theorem 1.3 Let C be a nonempty closed convex subset of H. Let G be a bifunction from C×C to satisfying

  1. (A1)

    G(x,x)=0, xC;

  2. (A2)

    G is monotone, i.e., G(x,y)+G(y,x)0, x,yC;

  3. (A3)

    x,y,zC, lim t 0 + G(tz+(1t)x,y)G(x,y);

  4. (A4)

    xC, yG(x,y) is convex and lower semicontinuous;

and let S be a nonexpansive mapping of C into H such that F(S)EP(G). Let f be a contraction of H into itself, and let { x n } and { u n } be sequences generated by x 1 H and

G ( u n , y ) + 1 r n y u n , u n x n 0 , y C , x n + 1 = α n f ( x n ) + ( 1 α n ) S u n

for all nN, where { α n }[0,1] and { r n }(0,1) satisfy

  1. (C1)

    lim n α n =0;

  2. (C2)

    n = 1 α n =;

  3. (C3)

    n = 1 | α n + 1 α n |<;

and lim inf n r n >0 and n = 1 | r n + 1 r n |<.

Then { x n } and { u n } converge strongly to zF(S)EP(G), where z= P F ( S ) EP ( G ) f(z).

Very recently, in 2011, Aoyama and Kimura [10] proved a strong convergence theorem of the iterative scheme of { x n } associated to a strongly nonexpansive sequence as follows.

Theorem 1.4 Let H be a Hilbert space, let C be a nonempty closed convex subset of H, and let { S n } and { T n } be sequences of nonexpansive self-mappings of C. Suppose that F=F({ S n })F({ T n }) is nonempty, both { S n } and { T n } satisfy the conditions (R) and (Z), and { S n } or { T n } is a strongly nonexpansive sequence. Let { α n } and { β n } be sequences in [0, 1] such that

lim n α n =0, n = 1 α n = and 0< lim inf n β n lim sup n β n <1.

Let x,uC and let { x n } be a sequence in C defined by x 1 =xC and

x n + 1 = β n x n +(1 β n ) S n ( α n u + ( 1 α n ) T n x n )

for all nN. Then { x n } converges strongly to P F u.

For x 1 ,u,vC, let { u n }, { v n }, { y n } and { x n } be the sequences defined by

{ F 1 ( u n , u ) + 1 r n u u n , u n x n 0 , F 2 ( v n , v ) + 1 s n v v n , v n x n 0 , y n = δ n u n + ( 1 δ n ) v n , x n + 1 = β n x n + ( 1 β n ) S n ( α n f ( T n y n ) + ( 1 α n ) T n y n ) , n 1 ,
(1.1)

where f:CC is a contractive mapping with α(0, 1 2 ) and { S n }, { T n } are sequences of nonexpansive mappings, one of them is a strongly nonexpansive sequence.

In this paper, inspired and motivated by [10] and [9], we prove that a strong convergence theorem of the iterative scheme { x n } defined by (1.1) converges strongly to z= P F f(z), where F=EP( F 1 )EP( F 2 )F({ S n })F({ T n }), under the conditions (R) and (Z) and suitable conditions of { r n }, { s n }, { α n }, { β n } and { δ n }.

2 Preliminaries

In this section, we need the following lemmas to prove our main result in the next section.

Lemma 2.1 (See [11])

Given xH and yC. Then P C x=y if and only if the following inequality holds:

xy,yz0,zC.

Lemma 2.2 (See [12])

Let { s n } be a sequence of nonnegative real numbers satisfying

s n + 1 =(1 α n ) s n + α n β n ,n0,

where { α n }, { β n } satisfy the conditions

  1. (1)

    { α n }[0,1], n = 1 α n =;

  2. (2)

    lim sup n β n 0 or n = 1 | α n β n |<.

Then lim n s n =0.

Lemma 2.3 (See [13])

Let { x n } and { z n } be bounded sequences in a Banach space X, and let { β n } be a sequence in [0,1] with 0< lim inf n β n lim sup n β n <1. Suppose that

x n + 1 = β n x n +(1 β n ) z n

for all integers n0 and

lim sup n ( z n + 1 z n x n + 1 x n ) 0.

Then lim n x n z n =0.

Lemma 2.4 (See [14])

Let C be a closed convex subset of a strictly convex Banach space E. Let { T n :nN} be a sequence of nonexpansive mappings on C. Suppose that n = 1 F( T n ) is nonempty. Let { λ n } be a sequence of positive numbers with n = 1 λ n =1. Then a mapping S on C defined by

S(x)= n = 1 λ n T n x

for all xC is well defined, nonexpansive and F(S)= n = 1 F( T n ) holds.

Lemma 2.5 (See [4])

Let C be a nonempty closed convex subset of a Hilbert space H, and let G:C×CR satisfy

  1. (A1)

    G(x,x)=0, xC;

  2. (A2)

    G is monotone, i.e., G(x,y)+G(y,x)0, x,yC;

  3. (A3)

    x,y,zC, lim t 0 + G(tz+(1t)x,y)G(x,y);

  4. (A4)

    xC, yG(x,y) is convex and lower semicontinuous.

For xH and r>0, define a mapping S r :HC as follows:

S r (x)= { z C : G ( z , y ) + 1 r y z , z x 0 , y C } .

Then S r is well defined and the following hold:

  1. (1)

    S r is single-valued;

  2. (2)

    S r is firmly nonexpansive, i.e., S r ( x ) S r ( y ) 2 S r (x) S r (y),xy, x,yH;

  3. (3)

    F( S r )=EP(G);

  4. (4)

    EP(G) is closed and convex.

Lemma 2.6 (See [11]) (Demiclosedness principle)

Assume that T is a nonexpansive self-mapping of a closed convex subset C of a Hilbert space H. If T has a fixed point, then IT is demiclosed. That is, whenever { x n } is a sequence in C weakly converging to some xC and the sequence {(IT) x n } converges strongly to some y, it follows that (IT)x=y. Here, I is the identity mapping of H.

Lemma 2.7 Let H be a real Hilbert space. Then, for all x,yH,

x + y 2 x 2 +2y,x+y.

Lemma 2.8 (See [10])

Let H be a Hilbert space, let C be a nonempty subset of H, and let { S n } and { T n } be the sequences of nonexpansive self-mappings of C. Suppose that { S n } and { T n } satisfy the condition (R) and that { T n y:nN,yD} is bounded for any bounded subset D of C. Then { S n T n } satisfies the condition (R).

Lemma 2.9 (See [1])

Let H be a Hilbert space, let C be a nonempty subset of H, and let { S n } and { T n } be the sequences of nonexpansive self-mappings of C. Suppose that { S n } or { T n } is a strongly nonexpansive sequence and that F ˜ ({ S n }) F ˜ ({ T n }) is nonempty. Then F ˜ ({ S n }) F ˜ ({ T n })= F ˜ ({ S n T n }).

3 Main result

Theorem 3.1 Let H be a Hilbert space, let C be a nonempty closed convex subset of H. Let F 1 and F 2 be two bifunctions from C×C into satisfying (A1)-(A4), respectively, and let { S n } and { T n } be sequences of nonexpansive self-mappings of C with F=EP( F 1 )EP( F 2 )F({ S n })F({ T n }). Let { T n } or { S n } be a sequence of strongly nonexpansive mappings, and let f:CC be a contractive mapping with α(0, 1 2 ). Let { x n }, { u n }, { v n } be sequences generated by x 1 ,u,vC and

{ F 1 ( u n , u ) + 1 r n u u n , u n x n 0 , F 2 ( v n , v ) + 1 s n v v n , v n x n 0 , y n = δ n u n + ( 1 δ n ) v n , x n + 1 = β n x n + ( 1 β n ) S n ( α n f ( T n y n ) + ( 1 α n ) T n y n ) , n 1 ,
(3.1)

where { α n },{ β n }[0,1], { r n },{ s n }(a,b)[0,1]. Assume that the following conditions hold:

  1. (i)

    lim n α n =0 and n = 1 α n =;

  2. (ii)

    0< lim inf n β n lim sup n β n <1;

  3. (iii)

    n = 0 | r n + 1 r n |, n = 0 | s n + 1 s n |<;

  4. (iv)

    lim n δ n =δ(0,1);

  5. (v)

    { S n } and { T n } satisfy the conditions R and Z.

Then the sequences { x n }, { u n }, { v n }, { y n } converge strongly to z= P F f(z).

Proof Let vF. From the definition of x n , we have

x n + 1 v = β n ( x n v ) + ( 1 β n ) ( S n ( α n f ( T n y n ) + ( 1 α n ) T n y n ) v ) β n x n v + ( 1 β n ) α n f ( T n y n ) + ( 1 α n ) T n y n v β n x n v + ( 1 β n ) ( α n f ( T n y n ) v + ( 1 α n ) T n y n v ) β n x n v + ( 1 β n ) ( α n f ( T n y n ) f ( v ) + α n f ( v ) v + ( 1 α n ) T n y n v ) β n x n v + ( 1 β n ) ( α n α y n v + α n f ( v ) v + ( 1 α n ) y n v ) = β n x n v + ( 1 β n ) ( α n f ( v ) v + ( 1 α n ( 1 α ) ) y n v ) .
(3.2)

From Lemma 2.5 and (3.1), we have EP( F 1 )=F( S r n ), EP( F 2 )=F( S s n ), S r n x n = u n and S s n x n = v n . By vF and the nonexpansiveness of S r n and S s n , we have

y n v = δ n ( u n v ) + ( 1 δ n ) ( v n v ) δ n u n v + ( 1 δ n ) v n v = δ n S r n x n v + ( 1 δ n ) S s n x n v x n v .
(3.3)

Substituting (3.3) into (3.2), we have

x n + 1 v β n x n v + ( 1 β n ) ( α n f ( v ) v + ( 1 α n ( 1 α ) ) y n v ) β n x n v + ( 1 β n ) ( α n f ( v ) v + ( 1 α n ( 1 α ) ) x n v ) = β n x n v + ( 1 β n ) α n f ( v ) v + ( 1 β n ) ( 1 α n ( 1 α ) ) x n v = ( 1 β n ) α n f ( v ) v + ( 1 α n ( 1 β n ) ( 1 α ) ) x n v max { x n v , f ( v ) v 1 α } .

By induction we can conclude that { x n } is bounded and so are { u n }, { v n }, { y n }. Next, we show that F ˜ ({ S n A n })= F ˜ ({ S n }) and F ˜ ({ A n T n })= F ˜ ({ T n }), where A n = α n f+(1 α n )I.

Let { z n } be a bounded sequence in C. From the nonexpansiveness of S n , we have

S n A n z n S n z n A n z n z n = α n f ( z n ) z n .
(3.4)

From (3.4) and α n 0 as n, we have

lim n S n A n z n S n z n =0.
(3.5)

Let { z n } F ˜ ({ S n A n }), then we have

z n S n z n z n S n A n z n + S n A n z n S n z n .

From (3.5), we have

lim n z n S n z n =0,

which implies that { z n } F ˜ ({ S n }). It follows that

F ˜ ( { S n A n } ) F ˜ ( { S n } ) .
(3.6)

Let { z n } F ˜ ({ S n }), then we have

z n S n A n z n z n S n z n + S n z n S n A n z n .

From (3.5), we have

lim n z n S n A n z n =0,

which implies that { z n } F ˜ ({ S n A n }). It follows that

F ˜ ( { S n } ) F ˜ ( { S n A n } ) .
(3.7)

From (3.6) and (3.7), we have

F ˜ ( { S n } ) = F ˜ ( { S n A n } ) .
(3.8)

Let { z n } be a bounded sequence in C, then we have { T n z n } is bounded and so is {f( T n z n )}. Since

A n T n z n T n z n = α n f ( T n z n ) T n z n

and α n 0 as n, we have

lim n A n T n z n T n z n =0.
(3.9)

Let { z n } F ˜ ({ A n T n }), then we have

z n T n z n z n A n T n z n + A n T n z n T n z n .

From (3.9), we have

lim n z n T n z n =0,

which implies that

{ z n } F ˜ ( { T n } ) .

It follows that

F ˜ ( { A n T n } ) F ˜ ( { T n } ) .
(3.10)

Let { z n } F ˜ ({ T n }), then we have

z n A n T n z n z n T n z n + T n z n A n T n z n .

From (3.9), we have

lim n z n A n T n z n =0,

which implies that

{ z n } F ˜ ( { A n T n } ) .

It follows that

F ˜ ( { T n } ) F ˜ ( { A n T n } ) .
(3.11)

From (3.10) and (3.11), we have

F ˜ ( { T n } ) = F ˜ ( { A n T n } ) .
(3.12)

Next, we show that

F ˜ ( { S n A n T n } ) = F ˜ ( { S n } ) F ˜ ( { T n } ) .

Since is nonempty, from (3.8), (3.12), we have

F ˜ ( { S n A n } ) F ˜ ( { T n } ) = F ˜ ( { S n } ) F ˜ ( { T n } )
(3.13)

and

F ˜ ( { S n } ) F ˜ ( { A n T n } ) = F ˜ ( { S n } ) F ˜ ( { T n } ) .
(3.14)

Suppose that { S n } is a strongly nonexpansive sequence. From (3.14) and Lemma 2.9, we have

F ˜ ( { S n A n T n } ) = F ˜ ( { S n } ) F ˜ ( { A n T n } ) = F ˜ ( { S n } ) F ˜ ( { T n } ) .
(3.15)

On the other hand, suppose that { T n } is a strongly nonexpansive sequence. From (3.13) and Lemma 2.9, we have

F ˜ ( { S n A n T n } ) = F ˜ ( { S n A n } ) F ˜ ( { T n } ) = F ˜ ( { S n } ) F ˜ ( { T n } ) .
(3.16)

From (3.16) and (3.15), we have F ˜ ({ S n A n T n })= F ˜ ({ S n }) F ˜ ({ T n }). Next, we show that { A n } and { S n A n T n } satisfy the condition (R). It is easy to see that A n is a nonexpansive mapping for every nN and that { A n y:nN,yD} is bounded, where D is a bounded subset of C. Let yD, then we have

A n + 1 y A n y = α n + 1 f ( y ) + ( 1 α n + 1 ) y α n f ( y ) ( 1 α n ) y | α n + 1 α n | f ( y ) + | α n + 1 α n | y .

From the condition (i), we have

lim n sup y D A n + 1 y A n y=0.

It follows that { A n } satisfies the condition (R). From Lemma 2.8, we have that { S n A n } satisfies the condition (R). From the nonexpansiveness of T n and F, we have { T n y:nN,yD} is bounded for any bounded subset D of C. From Lemma 2.8, we have that { S n A n T n } satisfies the condition (R).

Next, we show that

lim n x n + 1 x n =0.
(3.17)

Put

x n + 1 = β n x n +(1 β n ) w n ,
(3.18)

where w n = S n ( α n f( T n y n )+(1 α n ) T n y n ). From the definition of w n , we have

w n + 1 w n = S n + 1 A n + 1 T n + 1 y n + 1 S n A n T n y n S n + 1 A n + 1 T n + 1 y n + 1 S n A n T n y n + 1 + S n A n T n y n + 1 S n A n T n y n sup y D S n + 1 A n + 1 T n + 1 y S n A n T n y + y n + 1 y n ,
(3.19)

where D is a bounded subset of C. Besides, we have

y n + 1 y n = δ n + 1 u n + 1 + ( 1 δ n + 1 ) v n + 1 δ n u n ( 1 δ n ) v n = δ n + 1 u n + 1 δ n + 1 u n + δ n + 1 u n ( 1 δ n + 1 ) v n + ( 1 δ n + 1 ) v n + ( 1 δ n + 1 ) v n + 1 δ n u n ( 1 δ n ) v n = δ n + 1 ( u n + 1 u n ) + ( δ n + 1 δ n ) u n + ( 1 δ n + 1 ) ( v n + 1 v n ) + ( δ n δ n + 1 ) v n δ n + 1 u n + 1 u n + | δ n + 1 δ n | u n + ( 1 δ n + 1 ) v n + 1 v n + | δ n δ n + 1 | v n .
(3.20)

From (3.1) and Lemma 2.5, we have u n = S r n x n . This implies that

F 1 ( u n ,u)+ 1 r n u u n , u n x n 0for all uC
(3.21)

and

F 1 ( u n + 1 ,u)+ 1 r n + 1 u u n + 1 , u n + 1 x n + 1 0for all uC.
(3.22)

Putting u= u n + 1 in (3.21) and u= u n in (3.22), we have

F 1 ( u n , u n + 1 )+ 1 r n u n + 1 u n , u n x n 0
(3.23)

and

F 1 ( u n + 1 , u n )+ 1 r n + 1 u n u n + 1 , u n + 1 x n + 1 0.
(3.24)

Summing up the last two inequalities and using (A2), we obtain

u n + 1 u n , u n x n r n u n + 1 x n + 1 r n + 1 0.

This implies that

u n + 1 u n , u n u n + 1 + u n + 1 x n r n r n + 1 ( u n + 1 x n + 1 ) 0.

Hence,

u n + 1 u n 2 u n + 1 u n , u n + 1 x n r n r n + 1 ( u n + 1 x n + 1 ) = u n + 1 u n , u n + 1 x n + 1 + x n + 1 x n r n r n + 1 ( u n + 1 x n + 1 ) = u n + 1 u n , x n + 1 x n + ( 1 r n r n + 1 ) ( u n + 1 x n + 1 ) u n + 1 u n ( x n + 1 x n + 1 r n + 1 | r n + 1 r n | u n + 1 x n + 1 ) u n + 1 u n ( x n + 1 x n + 1 a | r n + 1 r n | u n + 1 x n + 1 ) .

Then we have

u n + 1 u n x n + 1 x n + 1 a | r n + 1 r n | u n + 1 x n + 1 .
(3.25)

From (3.1) and Lemma 2.5, we have v n = S s n x n . This implies that

F 2 ( v n ,v)+ 1 s n v v n , v n x n 0for all vC.

By using the same method as (3.25), we have

v n + 1 v n x n + 1 x n + 1 a | s n + 1 s n | v n + 1 x n + 1 .
(3.26)

Substituting (3.25) and (3.26) into (3.20), we have

y n + 1 y n δ n + 1 u n + 1 u n + | δ n + 1 δ n | u n + ( 1 δ n + 1 ) v n + 1 v n + | δ n δ n + 1 | v n δ n + 1 ( x n + 1 x n + 1 a | r n + 1 r n | u n + 1 x n + 1 ) + ( 1 δ n + 1 ) ( x n + 1 x n + 1 a | s n + 1 s n | v n + 1 x n + 1 ) + 2 M | δ n δ n + 1 | x n + 1 x n + 1 a | r n + 1 r n | u n + 1 x n + 1 + 1 a | s n + 1 s n | v n + 1 x n + 1 + 2 M | δ n δ n + 1 | ,
(3.27)

where M= sup n N { u n , v n }. Substituting (3.27) into (3.19), we have

w n + 1 w n sup y D S n + 1 A n + 1 T n + 1 y S n A n T n y + y n + 1 y n sup y D S n + 1 A n + 1 T n + 1 y S n A n T n y + x n + 1 x n + 1 a | r n + 1 r n | u n + 1 x n + 1 + 1 a | s n + 1 s n | v n + 1 x n + 1 + 2 M | δ n δ n + 1 | .
(3.28)

From (3.28), the conditions (iii), (iv) and { S n A n T n } satisfying the condition (R), we have

lim sup n ( w n + 1 w n x n + 1 x n ) 0.
(3.29)

From Lemma 2.3 and the definition of x n , we have

lim n x n w n =0.
(3.30)

From the definition of x n , we have

x n + 1 x n =(1 β n )( w n x n ).
(3.31)

From (3.30), (3.31) and the condition (ii), we have

lim n x n + 1 x n =0.

Next, we show that

lim n y n x n =0.

From the definition of y n , we have

y n x n δ n u n x n +(1 δ n ) v n x n .
(3.32)

Next, we show that

lim n u n x n = lim n v n x n =0.

Let vF. From the definition of x n , we have

x n + 1 v 2 β n x n v 2 + ( 1 β n ) S n ( α n f ( T n y n ) + ( 1 α n ) T n y n ) v 2 β n x n v 2 + ( 1 β n ) α n ( f ( T n y n ) v ) + ( 1 α n ) ( T n y n v ) 2 β n x n v 2 + ( 1 β n ) ( α n f ( T n y n ) v 2 + ( 1 α n ) T n y n v 2 ) β n x n v 2 + ( 1 β n ) ( α n f ( T n y n ) v 2 + ( 1 α n ) y n v 2 ) β n x n v 2 + ( 1 β n ) ( α n f ( T n y n ) v 2 + ( 1 α n ) ( δ n u n v 2 + ( 1 δ n ) v n v 2 ) ) .
(3.33)

From the firm nonexpansiveness of S r n and u n = S r n x n , we have

u n v 2 = S r n x n S r n v 2 u n v , x n v = 1 2 ( u n v 2 + x n v 2 u n x n 2 ) .

It implies that

u n v 2 x n v 2 u n x n 2 .
(3.34)

Since S s n is a firmly nonexpansive mapping and v n = S s n x n , by using the same method as (3.34), we have

v n v 2 x n v 2 v n x n 2 .
(3.35)

Substituting (3.34), (3.35) into (3.33), we have

x n + 1 v 2 β n x n v 2 + ( 1 β n ) ( α n f ( T n y n ) v 2 + ( 1 α n ) ( δ n u n v 2 + ( 1 δ n ) v n v 2 ) ) β n x n v 2 + ( 1 β n ) ( α n f ( T n y n ) v 2 + ( 1 α n ) ( δ n ( x n v 2 u n x n 2 ) + ( 1 δ n ) ( x n v 2 v n x n 2 ) ) ) = β n x n v 2 + ( 1 β n ) ( α n f ( T n y n ) v 2 + ( 1 α n ) ( δ n x n v 2 δ n u n x n 2 + ( 1 δ n ) x n v 2 ( 1 δ n ) v n x n 2 ) ) = β n x n v 2 + ( 1 β n ) ( α n f ( T n y n ) v 2 + ( 1 α n ) ( x n v 2 δ n u n x n 2 ( 1 δ n ) v n x n 2 ) ) = β n x n v 2 + ( 1 β n ) α n f ( T n y n ) v 2 + ( 1 α n ) ( 1 β n ) ( x n v 2 δ n u n x n 2 ( 1 δ n ) v n x n 2 ) = β n x n v 2 + ( 1 β n ) α n f ( T n y n ) v 2 + ( 1 α n ) ( 1 β n ) x n v 2 δ n ( 1 α n ) ( 1 β n ) u n x n 2 ( 1 δ n ) ( 1 α n ) ( 1 β n ) v n x n 2 x n v 2 + α n f ( T n y n ) v 2 δ n ( 1 α n ) ( 1 β n ) u n x n 2 ( 1 δ n ) ( 1 α n ) ( 1 β n ) v n x n 2 .
(3.36)

From (3.36), we have

δ n ( 1 α n ) ( 1 β n ) u n x n 2 x n v 2 x n + 1 v 2 + α n f ( T n y n ) v 2 ( 1 δ n ) ( 1 α n ) ( 1 β n ) v n x n 2 ( x n v + x n + 1 v ) x n + 1 x n + α n f ( T n y n ) v 2 ( 1 δ n ) ( 1 α n ) ( 1 β n ) v n x n 2 ( x n v + x n + 1 v ) x n + 1 x n + α n f ( T n y n ) v 2 .

From the conditions (i), (ii), (iv) and (3.17), we have

lim n u n x n =0.
(3.37)

By using the method as (3.37), we have

lim n v n x n =0.
(3.38)

From (3.32), (3.37) and (3.38), we have

lim n y n x n =0.
(3.39)

Next, we show that

{ y n } F ˜ ( { S n } ) F ˜ ( { T n } ) .
(3.40)

Since

S n A n T n y n y n S n A n T n y n x n + x n y n = w n x n + x n y n ,

from (3.30) and (3.39), we have

lim n S n A n T n y n y n =0.

Since { y n } is bounded, we have

{ y n } F ˜ ( { S n A n T n } ) .
(3.41)

Since F ˜ ({ S n A n T n })= F ˜ ({ S n }) F ˜ ({ T n }) and (3.41), we have (3.40).

Next, we show that

lim n S n m n m n =0,

where m n = α n f( T n y n )+(1 α n ) T n y n . From the definition of m n , we have

S n m n m n S n m n x n + m n x n = S n m n x n + α n ( f ( T n y n ) x n ) + ( 1 α n ) ( T n y n x n ) w n x n + α n f ( T n y n ) x n + ( 1 α n ) T n y n x n w n x n + α n f ( T n y n ) x n + T n y n y n + y n x n .

From (3.39), (3.40), (3.30) and the condition (i), we have

lim n S n m n m n =0.

Next, we show that

lim sup n f ( z ) z , m n z 0,

where z= P F f(z). Since { y n } is bounded, there exists a subsequence { y n i } of { y n } converging weakly to v, that is, y n i v as i. From (3.40), { S n } and { T n } satisfying the condition (Z), we have vF({ S n })F({ T n }).

Define the mapping Q:CC by

Q(x)=δ S r n x+(1δ) S s n xfor all xC,

where lim n δ n =δ(0,1). From the nonexpansiveness of S r n , S s n and Lemma 2.4, we have

F(Q)=F( S r n )F( S s n )=EP( F 1 )EP( F 2 ).

From the definitions of y n and Q, we have

x n Q x n x n y n + y n Q x n x n y n + δ n u n + ( 1 δ n ) v n δ S r n x n ( 1 δ ) S s n x n x n y n + | δ n δ | u n + | δ n δ | v n .
(3.42)

From (3.39), (3.42) and the condition (iv), we have

lim n x n Q x n =0.
(3.43)

From (3.39) and y n i v as i, we have x n i v as i. By (3.43), x n i v as i and Lemma 2.6, we have

vF(Q)=EP( F 1 )EP( F 2 ).

Hence,

vEP( F 1 )EP( F 2 )F ( { S n } ) F ( { T n } ) =F.
(3.44)

By (3.40), (3.44) and the condition (i), we have

lim sup n f ( z ) z , m n z = lim sup n ( α n f ( z ) z , f ( T n y n ) T n y n + f ( z ) z , T n y n z ) = lim i ( α n i f ( z ) z , f ( T n i y n i ) T n i y n i + f ( z ) z , T n i y n i z ) = lim i ( α n i f ( z ) z , f ( T n i y n i ) T n i y n i + f ( z ) z , T n i y n i y n i + f ( z ) z , y n i z ) = f ( z ) z , v z 0 .

Finally, we show that the sequence { x n } converges strongly to z= P F f(z). From the definition of { x n }, we have

x n + 1 z 2 = β n ( x n z ) + ( 1 β n ) ( S n m n z ) 2 β n x n z 2 + ( 1 β n ) S n m n z 2 β n x n z 2 + ( 1 β n ) m n z 2 .
(3.45)

Since m n = α n f( T n y n )+(1 α n ) T n y n , we have

m n z 2 = α n ( f ( T n y n ) z ) + ( 1 α n ) ( T n y n z ) 2 ( 1 α n ) 2 T n y n z 2 + 2 α n f ( T n y n ) z , m n z ( 1 α n ) x n z 2 + 2 α n f ( T n y n ) f ( z ) , m n z + 2 α n f ( z ) z , m n z ( 1 α n ) x n z 2 + 2 α n α x n z m n z + 2 α n f ( z ) z , m n z ( 1 α n ) x n z 2 + α n α ( x n z 2 + m n z 2 ) + 2 α n f ( z ) z , m n z = ( 1 α n ) x n z 2 + α n α x n z 2 + α n α m n z 2 + 2 α n f ( z ) z , m n z = ( 1 α n ( 1 α ) ) x n z 2 + α n α m n z 2 + 2 α n f ( z ) z , m n z .

This implies that

m n z 2 1 α n ( 1 α ) 1 α n α x n z 2 + 2 α n 1 α n α f ( z ) z , m n z = 1 α n α + α n α α n ( 1 α ) 1 α n α x n z 2 + 2 α n 1 α n α f ( z ) z , m n z = ( 1 α n ( 1 2 α ) 1 α n α ) x n z 2 + 2 α n 1 α n α f ( z ) z , m n z .
(3.46)

Substituting (3.46) into (3.45), we have

x n + 1 z 2 β n x n z 2 + ( 1 β n ) m n z 2 β n x n z 2 + ( 1 β n ) ( ( 1 α n ( 1 2 α ) 1 α n α ) x n z 2 + 2 α n 1 α n α f ( z ) z , m n z ) β n x n z 2 + ( 1 β n ) ( 1 α n ( 1 2 α ) 1 α n α ) x n z 2 + 2 α n ( 1 β n ) 1 α n α f ( z ) z , m n z = β n x n z 2 + ( ( 1 β n ) α n ( 1 2 α ) ( 1 β n ) 1 α n α ) x n z 2 + 2 α n ( 1 β n ) 1 α n α f ( z ) z , m n z = ( 1 α n ( 1 2 α ) ( 1 β n ) 1 α n α ) x n z 2 + α n ( 1 β n ) ( 1 2 α ) 1 α n α 2 f ( z ) z , m n z ( 1 2 α ) .
(3.47)

Applying (3.47), the conditions (i), (ii) and Lemma 2.2, we have { x n } converges strongly to z= P F f(z). From (3.39), (3.37) and (3.38), it is easy to see that { y n }, { u n }, { v n } converge strongly to z= P F f(z). This completes the proof. □

4 Applications

In this section, we give three examples for a strongly nonexpansive sequence and prove a strong convergence theorem associated to the variational inequality problem.

Before we give three examples, we need the following definition and lemmas.

Definition 4.1 Let C be a nonempty closed convex subset of a real Hilbert space H. A mapping A:CH is called an α-inverse strongly monotone mapping if there exists an α>0 such that

xy,axAyα A x A y 2

for all x,yC.

A mapping A:CH is called α-strongly monotone if there exists α>0 such that

xy,axAyα x y 2

for all x,yC.

A mapping T:CC is called a κ-strictly pseudo-contractive mapping if there is κ[0,1) such that

T x T y 2 x y 2 +κ ( I T ) x ( I T ) y 2
(4.1)

for all x,yC.

Then (4.1) is equivalent to

x y , ( I T ) x ( I T ) y 1 κ 2 ( I T ) x ( I T ) y 2

for all x,yC.

The set of solutions of the variational inequality problem of the mapping A:CH is denoted by VI(C,A), that is,

VI(C,A)= { x C : y x , A x 0 , y C } .

Let A,B:CH be two mappings. In 2013, Kangtunyakarn [15] modified VI(C,A) as follows:

VI ( C , a A + ( 1 a ) B ) = { x C : y x , ( a A + ( 1 a ) B ) x 0 , y C  and  a ( 0 , 1 ) } .

Remark 4.1 If T:CC is a κ-strictly pseudo-contractive mapping with F(T), then (IT) is a 1 κ 2 -inverse strongly monotone mapping and F(T)=VI(C,IT).

Lemma 4.2 (See [16])

Let H be a Hilbert space, let C be a nonempty closed convex subset of H, and let A be a mapping of C into H. Let uC. Then, for λ>0,

u= P C (IλA)uuVI(C,A),

where P C is the metric projection of H onto C.

Lemma 4.3 (See [15])

Let C be a nonempty closed convex subset of a real Hilbert space H, and let A,B:CH be α and β-inverse strongly monotone mappings, respectively, with α,β>0 and VI(C,A)VI(C,B). Then

VI ( C , a A + ( 1 a ) B ) =VI(C,A)VI(C,B),a(0,1).
(4.2)

Furthermore, if 0<γ<2η, where η=min{α,β}, we have Iγ(aA+(1a)B) is a nonexpansive mapping.

Example 4.4 Let T:CC be a κ-strictly pseudo-contractive mapping with F(T). Let { λ n } be a sequence of positive real numbers such that

0< inf n N λ n sup n N λ n <1κand lim n ( λ n + 1 λ n )=0,

and let { T n } be a sequence of mappings defined by T n = P C (I λ n (IT)). Then { T n } is a strongly nonexpansive sequence satisfying the conditions (R) and (Z).

Proof Since T is a κ-strictly pseudo-contractive mapping, then IT is 1 κ 2 -inverse strongly monotone. From Example 4.3 in [10], we have { T n } is a strongly nonexpansive sequence satisfying the conditions (R) and (Z). □

Example 4.5 Let A,B:CH be α,β-inverse strongly monotone mappings, respectively, with γ ¯ =min{α,β} and VI(C,A)VI(C,B). Let { λ n } be a sequence of positive real numbers such that

0< inf n N λ n sup n N λ n <2 γ ¯ and lim n ( λ n + 1 λ n )=0,

and let { T n } be a sequence of mappings defined by T n = P C (I λ n D), where D=aA+(1a)B for all a(0,1). Then { T n } is a strongly nonexpansive sequence satisfying the conditions (R) and (Z).

Proof Let x,yC, then we have

x y , D x D y = x y , ( a A + ( 1 a ) B ) x ( a A + ( 1 a ) B ) y a x y , A x A y + ( 1 a ) x y , B x B y a α A x A y 2 + ( 1 a ) β B x B y 2 γ ¯ ( a A x + ( 1 a ) B x a A y ( 1 a ) B y 2 ) γ ¯ D x D y 2 .

Then D is a γ ¯ -inverse strongly monotone mapping. From Example 4.3 in [10], we have that { T n } is a strongly nonexpansive sequence satisfying the conditions (R) and (Z). □

Example 4.6 Let A:CH be an α-strongly monotone and L-Lipschitzian mapping with VI(C,A). Let { λ n } be a sequence of positive real numbers such that

0< inf n N λ n sup n N λ n < 2 α L 2 and lim n ( λ n + 1 λ n )=0,

and let { T n } be a sequence of mappings defined by T n = P C (I λ n A). Then { T n } is a strongly nonexpansive sequence satisfying the conditions (R) and (Z).

Proof Let x,yC, then we have

x y , A x A y α x y 2 α L 2 A x A y 2 .

Then A is an α L 2 -inverse strongly monotone mapping. From Example 4.3 in [10], we have that { T n } is a strongly nonexpansive sequence satisfying the conditions (R) and (Z). □

Example 4.7 (See [10])

Let { R n } be a sequence of nonexpansive mappings of C into itself having a common fixed point, and let { μ n } be a sequence in [0,1]. For each nN, a W-mapping [17] T n generated by R n , R n 1 ,, R 1 and μ n , μ n 1 ,, μ 1 is defined as follows:

U n , n = μ n R n + ( 1 μ n ) I , U n , n 1 = μ n 1 R n 1 U n , n + ( 1 μ n 1 ) I , U n , n 2 = μ n 2 R n 2 U n , n 1 + ( 1 μ n 2 ) I , U n , k = μ k R k U n , k + 1 + ( 1 μ k ) I , U n , 2 = μ 2 R 2 U n , 3 + ( 1 μ 2 ) I , T n = U n , 1 = μ 1 R 1 U n , 2 + ( 1 μ 1 ) I .

If 0< μ 1 1 and 0< μ n b, for all n2 and 0<b<1, then { T n } satisfies the conditions (R) and (Z).

By using our main result and these three examples, we obtain the following results.

Theorem 4.8 Let H be a Hilbert space, let C be a nonempty closed convex subset of H. Let F 1 and F 2 be two bifunctions from C×C into satisfying (A1)-(A4), respectively. Let T:CC be a κ-strictly pseudo-contractive mapping with F(T). Let { λ n } be a sequence of positive real numbers such that

0< inf n N λ n sup n N λ n <1κand lim n ( λ n + 1 λ n )=0,

and let { T n } be a sequence of mappings defined by T n = P C (I λ n (IT)). Let { R n } be a sequence of nonexpansive mappings of C into itself having a common fixed point, and let { μ n } be a sequence in [0,1]. For each nN, W n is a W-mapping generated by R n , R n 1 ,, R 1 and μ n , μ n 1 ,, μ 1 . Assume that F=EP( F 1 )EP( F 2 )F({ R n })F(T). Let f:CC be a contractive mapping with α(0, 1 2 ). Let { x n }, { u n }, { v n } be sequences generated by x 1 ,u,vC and

{ F 1 ( u n , u ) + 1 r n u u n , u n x n 0 , F 2 ( v n , v ) + 1 s n v v n , v n x n 0 , y n = δ n u n + ( 1 δ n ) v n , x n + 1 = β n x n + ( 1 β n ) W n ( α n f ( T n y n ) + ( 1 α n ) T n y n ) , n 1 ,
(4.3)

where { α n },{ β n }[0,1], { r n },{ s n }(a,b)[0,1]. Assume that the following conditions hold:

  1. (i)

    lim n α n =0 and n = 1 α n =;

  2. (ii)

    0< lim inf n β n lim sup n β n <1;

  3. (iii)

    n = 0 | r n + 1 r n |, n = 0 | s n + 1 s n |<;

  4. (iv)

    lim n δ n =δ(0,1).

Then the sequences { x n }, { u n }, { v n }, { y n } converge strongly to z= P F f(z).

Proof From Example 4.4, we have { T n } is a strongly nonexpansive sequence satisfying the conditions (R) and (Z). From Lemma 4.2, we have F( T n )=F( P C (I λ n (IT)))=VI(C,IT)=F(T) for all nN. It implies that F({ T n })=F(T). From [18], we have F({ W n })=F({ R n }). It follows that F=EP( F 1 )EP( F 2 )F({ W n })F({ T n }). From Example 4.7, we have { W n } is a nonexpansive sequence satisfying the conditions (R) and (Z). By Theorem 3.1, we can conclude the desired result. □

Theorem 4.9 Let H be a Hilbert space, let C be a nonempty closed convex subset of H. Let F 1 and F 2 be two bifunctions from C×C into satisfying (A1)-(A4), respectively. Let A,B:CH be α,β-inverse strongly monotone mappings, respectively, with γ ¯ =min{α,β} and VI(C,A)VI(C,B). Let { λ n } be a sequence of positive real numbers such that

0< inf n N λ n sup n N λ n <2 γ ¯ and lim n ( λ n + 1 λ n )=0,

and let { T n } be a sequence of mappings defined by T n = P C (I λ n D), where D=aA+(1a)B for all a(0,1). Let { R n } be a sequence of nonexpansive mappings of C into itself having a common fixed point, and let { μ n } be a sequence in [0,1]. For each nN, W n is a W-mapping generated by R n , R n 1 ,, R 1 and μ n , μ n 1 ,, μ 1 . Assume that F=EP( F 1 )EP( F 2 )F({ R n })VI(C,A)VI(C,B). Let f:CC be a contractive mapping with α(0, 1 2 ). Let { x n }, { u n }, { v n } be sequences generated by x 1 ,u,vC and

{ F 1 ( u n , u ) + 1 r n u u n , u n x n 0 , F 2 ( v n , v ) + 1 s n v v n , v n x n 0 , y n = δ n u n + ( 1 δ n ) v n , x n + 1 = β n x n + ( 1 β n ) W n ( α n f ( T n y n ) + ( 1 α n ) T n y n ) , n 1 ,
(4.4)

where { α n },{ β n }[0,1], { r n },{ s n }(a,b)[0,1]. Assume that the following conditions hold:

  1. (i)

    lim n α n =0 and n = 1 α n =;

  2. (ii)

    0< lim inf n β n lim sup n β n <1;

  3. (iii)

    n = 0 | r n + 1 r n |, n = 0 | s n + 1 s n |<;

  4. (iv)

    lim n δ n =δ(0,1).

Then the sequences { x n }, { u n }, { v n }, { y n } converge strongly to z= P F f(z).

Proof From Example 4.5, we have { T n } is a strongly nonexpansive sequence satisfying the conditions (R) and (Z). From Lemmas 4.2 and 4.3, we have F( T n )=F( P C (I λ n D))=VI(C,D)=VI(C,A)VI(C,B) for all nN. It implies that F({ T n })=VI(C,A)VI(C,B). From [18], we have F({ W n })=F({ R n }). It follows that F=EP( F 1 )EP( F 2 )F({ W n })F({ T n }). From Example 4.7, we have { W n } is a nonexpansive sequence satisfying the conditions (R) and (Z). By Theorem 3.1, we can conclude the desired result. □

Theorem 4.10 Let H be a Hilbert space, let C be a nonempty closed convex subset of H. Let F 1 and F 2 be two bifunctions from C×C into satisfying (A1)-(A4), respectively. Let A:CH be an α-strongly monotone and L-Lipschitzian mapping with VI(C,A). Let { λ n } be a sequence of positive real numbers such that

0< inf n N λ n sup n N λ n < 2 α L 2 and lim n ( λ n + 1 λ n )=0,

and let { T n } be a sequence of mappings defined by T n = P C (I λ n A). Let { R n } be a sequence of nonexpansive mappings of C into itself having a common fixed point, and let { μ n } be a sequence in [0,1]. For each nN, W n is a W-mapping generated by R n , R n 1 ,, R 1 and μ n , μ n 1 ,, μ 1 . Assume that F=EP( F 1 )EP( F 2 )F({ R n })VI(C,A). Let f:CC be a contractive mapping with α(0, 1 2 ). Let { x n }, { u n }, { v n } be sequences generated by x 1 ,u,vC and

{ F 1 ( u n , u ) + 1 r n u u n , u n x n 0 , F 2 ( v n , v ) + 1 s n v v n , v n x n 0 , y n = δ n u n + ( 1 δ n ) v n , x n + 1 = β n x n + ( 1 β n ) W n ( α n f ( T n y n ) + ( 1 α n ) T n y n ) , n 1 ,
(4.5)

where { α n },{ β n }[0,1], { r n },{ s n }(a,b)[0,1]. Assume that the following conditions hold:

  1. (i)

    lim n α n =0 and n = 1 α n =;

  2. (ii)

    0< lim inf n β n lim sup n β n <1;

  3. (iii)

    n = 0 | r n + 1 r n |, n = 0 | s n + 1 s n |<;

  4. (iv)

    lim n δ n =δ(0,1).

Then the sequences { x n }, { u n }, { v n }, { y n } converge strongly to z= P F f(z).

Proof From Example 4.6, we have { T n } is a strongly nonexpansive sequence satisfying the conditions (R) and (Z). From Lemma 4.2, we have F( T n )=F( P C (I λ n A))=VI(C,A) for all nN. It implies that F({ T n })=VI(C,A). From [18], we have F({ W n })=F({ R n }). It follows that F=EP( F 1 )EP( F 2 )F({ W n })F({ T n }). From Example 4.7, we have { W n } is a nonexpansive sequence satisfying the conditions (R) and (Z). By Theorem 3.1, we can conclude the desired result. □

Theorem 4.11 Let H be a Hilbert space, let C be a nonempty closed convex subset of H. Let F 1 be a bifunction from C×C into satisfying (A1)-(A4), and let { S n } and { T n } be sequences of nonexpansive self-mappings of C with F=EP( F 1 )F({ S n })F({ T n }). Let { T n } or { S n } be a sequence of strongly nonexpansive mappings, and let f:CC be a contractive mapping with α(0, 1 2 ). Let { x n }, { u n } be sequences generated by x 1 ,uC and

{ F 1 ( u n , u ) + 1 r n u u n , u n x n 0 , x n + 1 = β n x n + ( 1 β n ) S n ( α n f ( T n u n ) + ( 1 α n ) T n u n ) , n 1 ,
(4.6)

where { α n },{ β n }[0,1], { r n },{ s n }(a,b)[0,1]. Assume that the following conditions hold:

  1. (i)

    lim n α n =0 and n = 1 α n =;

  2. (ii)

    0< lim inf n β n lim sup n β n <1;

  3. (iii)

    n = 0 | r n + 1 r n |<;

  4. (iv)

    { S n } and { T n } satisfy the conditions R and Z.

Then the sequences { x n }, { u n } converge strongly to z= P F f(z).

Proof Put F 1 F 2 , s n = r n and u n = v n . From Theorem 3.1, we can conclude the desired conclusion. □

The following result can be obtained from Theorem 3.1. We, therefore, omit the proof.

Theorem 4.12 Let H be a Hilbert space, let C be a nonempty closed convex subset of H. Let F i be bifunctions from C×C into , for every i=1,2,,N, satisfying (A1)-(A4), and let { S n } and { T n } be sequences of nonexpansive self-mappings of C with F= i = 1 N EP( F i )F({ S n })F({ T n }). Let { T n } or { S n } be a sequence of strongly nonexpansive mappings, and let f:CC be a contractive mapping with α(0, 1 2 ). Let { x n }, { u n }, { v n } be sequences generated by x 1 , u i C, for every i1,2,,N, and

{ F i ( u n i , u i ) + 1 r n i u u n i , u n i x n 0 , y n = i = 1 N δ n i u n i , x n + 1 = β n x n + ( 1 β n ) S n ( α n f ( T n y n ) + ( 1 α n ) T n y n ) , n 1 ,
(4.7)

where { α n },{ β n }[0,1], { r n },{ s n }(a,b)[0,1]. Assume that the following conditions hold:

  1. (i)

    lim n α n =0 and n = 1 α n =;

  2. (ii)

    0< lim inf n β n lim sup n β n <1;

  3. (iii)

    n = 0 | r n + 1 i r n i |<, i=1,2,,N;

  4. (iv)

    i = 1 N δ n i =1;

  5. (v)

    lim n δ n i = δ i (0,1), i=1,2,,N;

  6. (vi)

    { S n } and { T n } satisfy the conditions R and Z.

Then the sequences { x n }, { y n } and { u n i }, for every i=1,2,,N, converge strongly to z= P F f(z).

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Acknowledgements

This research was supported by the Research Administration Division of King Mongkut’s Institute of Technology Ladkrabang.

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Kangtunyakarn, A. Convergence theorems for a system of equilibrium problems and fixed point problems of a strongly nonexpansive sequence. Fixed Point Theory Appl 2013, 193 (2013). https://doi.org/10.1186/1687-1812-2013-193

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