Skip to main content

Convergence analysis for the equilibrium problems with numerical results

Abstract

In this paper, we propose an iterative scheme modified from the work of Ceng et al. (Nonlinear Anal. Hybrid Syst. 4:743-754, 2010) and Plubtieng and Punpaeng (J. Math. Anal. Appl. 336(1):455-469, 2007) to prove the strong convergence theorem for approximating a common element of the set of fixed points of nonspreading mappings and a finite family of the set of solutions of the equilibrium problem. Using this result, we obtain the strong convergence theorem for a finite family of nonspreading mappings and a finite family of the set of solutions of equilibrium problem. Moreover, in order to compare numerical results between the combination of the equilibrium problem and the classical equilibrium problem, some examples are given in one- and two-dimensional spaces of real numbers.

1 Introduction

Throughout this paper, let C be a nonempty closed convex subset of a real Hilbert space H with the inner product , and the norm . We denote weak convergence and strong convergence by the notations ‘’ and ‘→’, respectively. We use to denote the set of real numbers and Fix(T) to represent the set of fixed points of T, where T is a mapping from C into itself.

In 2008, Kohsaka and Takahashi [1] introduced the nonspreading mapping T in Hilbert space H as follows:

2 T u T v 2 T u v 2 + u T v 2 ,u,vC.
(1.1)

In 2009, it was shown by Iemoto and Takahashi [2] that (1.1) is equivalent to the following equation:

T u T v 2 u v 2 +2uTu,vTv,for all u,vC.

Many researchers proved the strong convergence theorem for a nonspreading mapping and some related mappings in Hilbert space; see for example [36].

Let B:CH. The variational inequality problem is to find a point uC satisfying the following inequality:

Bu,vu0,
(1.2)

for all vC. Moreover, VI(C,B) is used to denote the set of solutions of (1.2).

Let Φ:C×CR be a bifunction. The classical equilibrium problem for Φ is to find uC satisfying the following inequality:

Φ(u,v)0,vC.
(1.3)

We use EP(Φ) to represent the set of solution of (1.3).

Let the bifunction Φ satisfy the following conditions for solving the equilibrium problem.

(A1) Φ(u,u)=0 for all uC;

(A2) Φ is monotone, i.e., Φ(u,v)+Φ(v,u)0 for all u,vC;

(A3) for each u,v,wC,

lim t 0 + Φ ( t w + ( 1 t ) u , v ) Φ(u,v);

(A4) for each uC, vΦ(u,v) is convex and lower semicontinuous.

In 1994, Blum and Oettli [7] showed that the classical equilibrium problem (1.3) covers monotone inclusion problems, saddle point problems, variational inequality problems, minimization problems, Nash equilibria in noncooperative games, vector equilibrium problems, and certain fixed point problems.

Let Ψ= { Φ i } i = 1 , 2 , , N be a finite family of bifunctions from C×C to . The system of equilibrium problem for Ψ is to determine common equilibrium points for Ψ= { Φ i } i = 1 , 2 , , N , that is, the set

EP(Ψ)= { u C : Φ i ( u , v ) 0 , v C , i 1 , 2 , , N } .
(1.4)

The problem (1.4) extends (1.3) to a system of such problems covering various forms of feasibility problems [8]. Several iterative algorithms are proposed to solve the equilibrium problems and a finite family of equilibrium problems; see, for instance, [814].

Example 1.1 Let Ψ= { Φ i } i = 1 , 2 , , N be a finite family of bifunctions from C×C to , where the bifunctions Φ i are defined by

Φ i (u,v)=i(vu)(v+2u3),for every u,vR.

For each i=1,2,,N, it is obvious that the Φ i (x,y) satisfy (A1)-(A4). Then we obtain

EP(Ψ)= i = 1 N EP( Φ i )={1}.

In 2010, Peng et al. [15] proposed the following iterative algorithm for solving a family of infinite nonexpansive mappings and a finite family of equilibrium problems in Hilbert space:

{ z 1 = z H , u n = T β n F m T β n F m 1 T β n F 2 T β n F 1 z n , v n = P C ( I s n A ) u n , z n + 1 = α n γ f ( W n z n ) + ( I α n B ) W n P C ( I r n A ) v n , n N .
(1.5)

Under some appropriate conditions, they proved that { z n }, { v n }, and { u n } converge strongly to q= P Ω (γf+(IB))(q), where Ω= i = 1 Fix( S i )VI(C,A) k = 1 m EP( F k ) and f is a contractive mapping on H.

Over the past few years, many researchers have started working on the methods for finding a common solution of a finite family of equilibrium problems in Hilbert space; see, for instance, [1618].

In 2013, Suwannaut and Kangtunyakarn [12] introduced the combination of equilibrium problem which is to find uC such that

( i = 1 N a i Φ i ) (u,v)0,vC,
(1.6)

where Φ i :C×CR are bifunctions and a i (0,1) with i = 1 N a i =1, for every i=1,2,,N. The set of solutions (1.6) is denoted by EP( i = 1 N a i Φ i ).

If Φ i =Φ, for all i=1,2,,N, then the combination of equilibrium problem (1.6) reduces to the classical equilibrium problem (1.3).

Moreover, they obtain Lemma 2.10 as shown in the next section.

Example 1.2 For every i=1,2,3, let the bifunctions Φ i :R×RR, be given by

Φ 1 ( u , v ) = ( v u ) ( v + u 2 ) , Φ 2 ( u , v ) = ( v u ) ( 3 v + 5 u 8 ) , Φ 3 ( u , v ) = ( v u ) ( 9 v + 12 u 21 ) , u , v R .

For all i=1,2,3, it is obvious that the Φ i (u,v) satisfy (A1)-(A4). Let a 1 = 1 2 , a 2 = 1 9 and a 3 = 7 18 , thus we have

i = 1 3 a i Φ i (u,v)= 1 18 (vu)(78v+103u181).

This implies that

EP ( i = 1 3 a i Φ i ) = i = 1 3 EP( Φ i )={1}.

Remark 1.3 For all i=1,2,,N, let the mapping A i :CH be defined by Φ i (u,v)= A i u,vu for all u,vC. For each i=1,2,,N, if Φ i (u,v)= A i u,vu0 for all u,vC, and i=1,2,,N, then EP( Φ i )=VI(C, A i ). Hence we have

EP ( i = 1 N a i Φ i ) = i = 1 N EP( Φ i )= i = 1 N VI(C, A i ).

After we have studied research related to equilibrium problems, we obtain the following question.

Question Is it possible to prove strong convergence theorem for a finite family of equilibrium problem using different method from the result of Peng et al. [15], Piri [17] and references therein?

Inspired and motivated by the work of Iemoto and Takahashi [2], Suwannaut and Kangtunyakarn [12] and related research, we propose an iterative scheme modified from the work of Plubtieng and Punpaeng [19] and Ceng et al. [11] to prove the strong convergence theorem for approximating a common element of the set of fixed points of a nonspreading mapping and a finite family of the set of solutions of equilibrium problems using Lemma 2.10 and a different method from the work of Peng et al. [15] and Piri [17] and references therein. Moreover, some examples are given in order to compare the numerical results between the combination of the equilibrium problem and the classical equilibrium problem.

2 Preliminaries

We now recall the following definition and well-known lemmas.

Definition 2.1

  1. (i)

    A is strongly positive operator on H if there exists a constant β>0 such that

    Au,uβ u 2 ,uH.
  2. (ii)

    T is a nonexpansive mapping if

    TuTvuv,u,vC.
  3. (iii)

    For every uH, there is a unique nearest point P C u in C such that

    u P C uuv,vC.

Such an operator P C is called the metric projection of H onto C.

Lemma 2.1 ([20])

For a given wH and uC,

u= P C wuw,vu0,vC.

Furthermore, P C is a nonexpansive mapping.

Lemma 2.2 ([21])

Each Hilbert space H satisfies Opial’s condition, i.e., for any sequence { u n }H with u n u, the inequality

lim inf n u n u< lim inf n u n v

holds for every vH with vu.

Lemma 2.3 ([22])

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

u n + 1 (1 β n ) u n + η n ,n0,

where α n is a sequence in (0,1) and { η n } is a sequence such that

  1. (1)

    n = 1 β n =,

  2. (2)

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

Then lim n u n =0.

Lemma 2.4 ([4])

Let H be a real Hilbert space. Then the following results hold:

  1. (i)

    For all u,vH and t[0,1],

    t u + ( 1 t ) v 2 =t u 2 +(1t) v 2 t(1t) u v 2 ,
  2. (ii)

    u + v 2 u 2 +2v,u+v, for each u,vH.

Lemma 2.5 ([20])

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. Then, for α>0,

Fix ( P C ( I α A ) ) =VI(C,A),

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

Lemma 2.6 ([23])

Assume A is a strongly positive linear bounded operator on a Hilbert space H with coefficient β>0 and 0<δ< A 1 . Then IδA1δβ.

Lemma 2.7 ([2])

Let C be a nonempty closed convex subset of H. Then a mapping T:CC is nonspreading if and only if

T u T v 2 u v 2 +2uTu,vTv,for all u,vC.

Remark 2.8 If T is a nonexpansive mapping and uTu,vTv0, for every u,vC, then T is a nonspreading mapping.

Lemma 2.9 Let C be a nonempty closed convex subset of a real Hilbert space H and let T:CC be a nonspreading mapping with Fix(T). Then we have the following statements:

  1. (i)

    Fix(T)=VI(C,IT);

  2. (ii)

    for every uC and vFix(T),

    P C ( I λ ( I T ) ) u v uv,where λ(0,1).

Proof To prove (i), let x Fix(T). Then x =T x . Since

v x , ( I T ) x =0,vC,

we have x VI(C,IT), from which it follows that Fix(T)VI(C,IT).

Next, we show VI(C,IT)Fix(T).

Let x ˜ VI(C,IT). This implies that

v x ˜ , ( I T ) x ˜ 0,vC.
(2.1)

Let x Fix(T). Then, by Lemma 2.7, we obtain

T x ˜ T x 2 x ˜ x 2 +2 x ˜ T x ˜ , x T x = x ˜ x 2 .
(2.2)

Observe that

T x ˜ x 2 = x ˜ x ( I T ) x ˜ 2 = x ˜ x 2 2 x ˜ x , ( I T ) x ˜ + ( I T ) x ˜ 2 .
(2.3)

From (2.1), (2.2), and (2.3), we get

( I T ) x ˜ 2 2 x ˜ x , ( I T ) x ˜ 0,

which yields x ˜ Fix(T). Therefore VI(C,IT)Fix(T).

To prove (ii), let uC and vFix(T). Since T is a nonspreading mapping and we have Lemma 2.7, we get

T u T v 2 u v 2 +2uTu,vTv= u v 2 .
(2.4)

Thus we have

T u v 2 = u v ( I T ) u 2 = u v 2 2 u v , ( I T ) u + ( I T ) u 2 .
(2.5)

From (2.4) and (2.5), we obtain

( I T ) u 2 2 u v , ( I T ) u .
(2.6)

From (i) and Lemma 2.5, we have

vFix(T)=VI(C,IT)=Fix ( P C ( I λ ( I T ) ) ) .
(2.7)

By the nonexpansiveness of P C , (2.6), and (2.7), we get

P C ( I λ ( I T ) ) u v 2 = P C ( I λ ( I T ) ) u P C ( I λ ( I T ) ) v 2 ( I λ ( I T ) ) u ( I λ ( I T ) ) v 2 = ( u v ) λ ( ( I T ) u ( I T ) v ) 2 = ( u v ) λ ( I T ) u 2 = u v 2 2 λ u v , ( I T ) u + λ 2 ( I T ) u 2 u v 2 λ ( I T ) u 2 + λ 2 ( I T ) u 2 = u v 2 λ ( 1 λ ) ( I T ) u 2 u v 2 ,

which implies that P C (Iλ(IT))uvuv. □

Lemma 2.10 ([12])

Let C be a nonempty closed convex subset of a real Hilbert space H. For i=1,2,,N, let Φ i :C×CR be bifunctions satisfying (A1)-(A4) with i = 1 N EP( Φ i ). Then

EP ( i = 1 N a i Φ i ) = i = 1 N EP( Φ i ),

where a i (0,1) for every i=1,2,,N and i = 1 N a i =1.

Lemma 2.11 ([7])

Let C be a nonempty closed convex subset of H and let Φ be a bifunction of C×C into satisfying (A1)-(A4). Let t>0 and uH. Then there exists wC such that

Φ(w,v)+ 1 t vw,wu0,vC.

Lemma 2.12 ([8])

Assume that Φ:C×CR satisfies (A1)-(A4). For t>0, define a mapping S t :HC as follows:

S t (x)= { w C : Φ ( w , v ) + 1 t v w , w u 0 , v C } ,

for all uH. Then the following hold:

  1. (i)

    S t is single-valued;

  2. (ii)

    S t is firmly nonexpansive, i.e., for each u,vH,

    S t ( u ) S t ( v ) 2 S t ( u ) S t ( u ) , u v ;
  3. (iii)

    Fix( S t )=EP(Φ);

  4. (iv)

    EP(Φ) is closed and convex.

Remark 2.13 ([12])

From Lemma 2.10, it is easy to see that i = 1 N a i Φ i satisfies (A1)-(A4). By using Lemma 2.12, we obtain

Fix( S t )=EP ( i = 1 N a i Φ i ) = i = 1 N EP( Φ i ),

where a i (0,1), for each i=1,2,,N, and i = 1 N a i =1.

3 Strong convergence theorem

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let be an α-contractive mapping on H and let A be a strongly positive linear bounded operator on H with coefficient γ ¯ and 0<γ< γ ¯ α . Let T:CC be a nonspreading mapping. For every i=1,2,,N, let Φ i :C×CR be a bifunction satisfying (A1)-(A4) with Ω:=Fix(T) i = 1 N EP( Φ i ). Let { Z n }, { Y n }, and { V n } be sequences generated by Z 1 H and

{ i = 1 N a i Φ i ( V n , y ) + 1 φ n y V n , V n Z n 0 , y C , Y n = θ n P C Z n + ( 1 θ n ) V n , Z n + 1 = δ n γ F ( Z n ) + ( I δ n A ) P C ( I ψ n ( I T ) ) Y n , n N ,
(3.1)

where { δ n },{ θ n },{ φ n },{ ψ n }(0,1), 0< a i <1, for all i=1,2,,N. Suppose the conditions (i)-(vi) hold.

  1. (i)

    lim n δ n =0 and n = 1 δ n =;

  2. (ii)

    0<τ θ n υ<1, for some τ,υ>0;

  3. (iii)

    n = 1 ψ n <;

  4. (iv)

    0<ϵ φ n η<1, for some ϵ,η>0;

  5. (v)

    n = 1 N a i =1;

  6. (vi)

    n = 1 | δ n + 1 δ n |<, n = 1 | θ n + 1 θ n |<, n = 1 | ψ n + 1 ψ n |<, n = 1 | φ n + 1 φ n |<.

Then the sequences { Z n }, { Y n }, and { V n } converge strongly to q= P Ω (IA+γF)q.

Proof The proof of this theorem is divided into five steps.

Step 1. Claim that { Z n } is a bounded sequence.

Since δ n 0 as n, without loss of generality, we assume δ n < 1 A , for every nN. Since i = 1 N a i Φ i satisfies (A1)-(A4) and

i = 1 N a i Φ i ( V n ,y)+ 1 φ n y V n , V n Z n 0,yC,

by Lemma 2.12 and Remark 2.13, we have V n = T φ n Z n and Fix( T φ n )= i = 1 N EP( Φ i ).

From Lemma 2.5 and Lemma 2.9(i), we obtain

Fix(T)=Fix ( P C ( I ψ n ( I T ) ) ) .

Let zΩ. By the nonexpansiveness of P C and T φ n , we have

Y n z θ n P C Z n z+(1 θ n ) T φ n Z n z Z n z.
(3.2)

From Lemma 2.6, Lemma 2.9(ii), and (3.2), we obtain

Z n + 1 z δ n γ F ( Z n ) A z + I δ n A P C ( I ψ n ( I T ) ) Y n z δ n γ F ( Z n ) F ( z ) + δ n γ F ( z ) A z + ( 1 δ n γ ¯ ) Y n z δ n γ α Z n z + δ n γ F ( z ) A z + ( 1 δ n γ ¯ ) Z n z = ( 1 δ n ( γ ¯ γ α ) ) Z n z + δ n γ F ( z ) A z max { Z 1 z , γ F ( z ) A z γ ¯ γ α } .

By induction, we obtain Z n zmax{ Z 1 z, γ F ( z ) A z γ ¯ γ α }, nN. It shows that { Z n } is bounded and so are { V n } and { Y n }.

Step 2. Show that lim n Z n + 1 Z n =0.

By the definition of Z n and Lemma 2.6, we obtain

Z n + 1 Z n δ n γ F ( Z n ) F ( Z n 1 ) + γ | δ n δ n 1 | F ( Z n 1 ) + I δ n A P C ( I ψ n ( I T ) ) Y n P C ( I ψ n 1 ( I T ) ) Y n 1 + ( I δ n A ) P C ( I ψ n 1 ( I T ) ) Y n 1 ( I δ n 1 A ) P C ( I ψ n 1 ( I T ) ) Y n 1 δ n γ α Z n Z n 1 + γ | δ n δ n 1 | F ( Z n 1 ) + ( 1 δ n γ ¯ ) ( I ψ n ( I T ) ) Y n ( I ψ n 1 ( I T ) ) Y n 1 + | δ n δ n 1 | A P C ( I ψ n 1 ( I T ) ) Y n 1 δ n γ α Z n Z n 1 + γ | δ n δ n 1 | F ( Z n 1 ) + ( 1 δ n γ ¯ ) [ θ n Z n Z n 1 + | θ n θ n 1 | P C Z n 1 + ( 1 θ n ) V n V n 1 + | θ n θ n 1 | V n 1 + ψ n ( I T ) Y n ( I T ) Y n 1 + | ψ n ψ n 1 | ( I T ) Y n 1 ] + | δ n δ n 1 | A P C ( I ψ n 1 ( I T ) ) Y n 1 .
(3.3)

Using the same method as in [12] (Step 2 of Theorem 3.1), we have

V n V n 1 Z n Z n 1 + 1 ϵ | φ n φ n 1 | V n Z n .
(3.4)

Substitute (3.3) into (3.4) to get

Z n + 1 Z n δ n γ α Z n Z n 1 + γ | δ n δ n 1 | F ( Z n 1 ) + ( 1 δ n γ ¯ ) [ Z n Z n 1 + | θ n θ n 1 | P C Z n 1 + 1 θ n ϵ | φ n φ n 1 | V n Z n + | θ n θ n 1 | V n 1 + ψ n ( I T ) Y n ( I T ) Y n 1 + | ψ n ψ n 1 | ( I T ) Y n 1 ] + | δ n δ n 1 | A P C ( I ψ n 1 ( I T ) ) Y n 1 ( 1 δ n ( γ ¯ γ α ) ) Z n Z n 1 + ( 1 + γ ) | δ n δ n 1 | K + 2 | θ n θ n 1 | K + 1 ϵ | φ n φ n 1 | K + 2 ψ n K + | ψ n ψ n 1 | K ,
(3.5)

where K= max n N { V n ,F( Z n ), V n Z n , P C Z n ,(IT) Y n ,A P C (I ψ n (IT)) Y n }. From (3.5), the conditions (i), (iii), (v), and Lemma 2.3, we have

lim n Z n + 1 Z n =0.
(3.6)

Step 3. Prove that lim n V n Z n = lim n P C (I ψ n (IT)) Z n Z n =0.

To claim this, let zΩ. Since V n = T φ n Z n and T φ n is a firmly nonexpansive mapping, we have

z T φ n Z n 2 = T φ n z T φ n Z n 2 T φ n z T φ n Z n , z Z n = 1 2 ( T φ n Z n z 2 + Z n z 2 T φ n Z n Z n 2 ) ,

from which it follows that

V n z 2 Z n z 2 V n Z n 2 .
(3.7)

By the definition of Z n , Lemma 2.6, Lemma 2.9(ii), and (3.7), we get

Z n + 1 z 2 = δ n ( γ F ( Z n ) A P C ( I ψ n ( I T ) ) Y n ) + ( P C ( I ψ n ( I T ) ) Y n z ) 2 P C ( I ψ n ( I T ) ) Y n z 2 + 2 δ n γ F ( Z n ) A P C ( I ψ n ( I T ) ) Y n , Z n + 1 z Y n z 2 + 2 δ n γ F ( Z n ) A P C ( I ψ n ( I T ) ) Y n Z n + 1 z θ n P C Z n z 2 + ( 1 θ n ) V n z 2 + 2 δ n γ F ( Z n ) A P C ( I ψ n ( I T ) ) Y n Z n + 1 z θ n Z n z 2 + ( 1 θ n ) ( Z n z 2 V n Z n 2 ) + 2 δ n γ F ( Z n ) A P C ( I ψ n ( I T ) ) Y n Z n + 1 z = Z n z 2 ( 1 θ n ) V n Z n 2 + 2 δ n γ F ( Z n ) A P C ( I ψ n ( I T ) ) Y n Z n + 1 z ,

which implies that

( 1 θ n ) V n Z n 2 ( Z n z + Z n + 1 z ) Z n + 1 Z n + 2 δ n γ F ( Z n ) A P C ( I ψ n ( I T ) ) Y n Z n + 1 z .

From (3.6), the conditions (i) and (ii), this yields

lim n V n Z n =0.
(3.8)

By Lemma 2.6 and Lemma 2.9(ii), we get

Z n + 1 z 2 P C ( I ψ n ( I T ) ) Y n z 2 + 2 δ n γ F ( Z n ) A P C ( I ψ n ( I T ) ) Y n , Z n + 1 z Y n z 2 + 2 δ n γ F ( Z n ) A P C ( I ψ n ( I T ) ) Y n Z n + 1 z = θ n P C Z n z 2 + ( 1 θ n ) V n z 2 θ n ( 1 θ n ) P C Z n V n 2 + 2 δ n γ F ( Z n ) A P C ( I ψ n ( I T ) ) Y n Z n + 1 z Z n z 2 θ n ( 1 θ n ) P C Z n V n 2 + 2 δ n γ F ( Z n ) A P C ( I ψ n ( I T ) ) Y n Z n + 1 z ,

from which it follows that

θ n ( 1 θ n ) P C Z n V n 2 ( Z n z + Z n + 1 z ) Z n + 1 Z n + 2 δ n γ F ( Z n ) A P C ( I ψ n ( I T ) ) Y n Z n + 1 z .

From (3.6), the conditions (i) and (ii), this implies that

lim n P C Z n V n =0.
(3.9)

Since

P C Z n Z n P C Z n V n + V n Z n ,

using (3.8) and (3.9), we have

lim n P C Z n Z n =0.
(3.10)

Since

Y n Z n θ n P C Z n Z n +(1 θ n ) V n Z n ,

by (3.8) and (3.10), thus we obtain

lim n Y n Z n =0.
(3.11)

Observe that

Z n P C ( I ψ n ( I T ) ) Y n Z n Z n + 1 + Z n + 1 P C ( I ψ n ( I T ) ) Y n = Z n Z n + 1 + δ n γ F ( Z n ) A P C ( I ψ n ( I T ) ) Y n ,

which implies by (3.6) and the condition (i) that

lim n Z n P C ( I ψ n ( I T ) ) Y n =0.
(3.12)

Since

Z n P C ( I ψ n ( I T ) ) Z n Z n P C ( I ψ n ( I T ) ) Y n + P C ( I ψ n ( I T ) ) Y n P C ( I ψ n ( I T ) ) Z n Z n P C ( I ψ n ( I T ) ) Y n + ( I ψ n ( I T ) ) Y n ( I ψ n ( I T ) ) Z n Z n P C ( I ψ n ( I T ) ) Y n + Y n Z n + ψ n ( I T ) Y n ( I T ) Z n ,

by (3.11), (3.12), and the condition (iii), we obtain

lim n Z n P C ( I ψ n ( I T ) ) Z n =0.
(3.13)

Step 4. Show that lim sup n γF(q)Aq, Z n q0, where q= P Ω (IA+γF)q.

First, take a subsequence { Z n k } of { Z n } such that

lim sup n γ F ( q ) A q , Z n q = lim k γ F ( q ) A q , Z n k q .

Since { Z n } is bounded, we can assume that Z n k ω as k. By (3.8), it follows that U n k ω as k.

Assume ωFix(T). Since Fix(T)=Fix( P C (I ψ n k (IT))), we have ω P C (I ψ n k (IT))ω. By the nonexpansiveness of P C , the condition (iii), (3.13), and Opial’s condition, we get

lim inf k Z n k ω < lim inf k Z n k P C ( I ψ n k ( I T ) ) ω lim inf k ( Z n k P C ( I ψ n k ( I T ) ) Z n k + P C ( I ψ n k ( I T ) ) Z n k P C ( I ψ n k ( I T ) ) ω ) lim inf k ( Z n k P C ( I ψ n k ( I T ) ) Z n k + ( I ψ n k ( I T ) ) Z n k ( I ψ n k ( I T ) ) ω ) lim inf k ( Z n k P C ( I ψ n k ( I T ) ) Z n k + Z n k ω + ψ n k ( I T ) Z n k ( I T ) ω ) lim inf k Z n k ω .

This is a contradiction. Then we have

ωFix(T).
(3.14)

By continuing the same argument as in [12] (Step 4 of Theorem 3.1), we obtain

ω i = 1 N EP( Φ i ).
(3.15)

From (3.14) and (3.15), we get ωΩ. Since Z n k ω as k, by Lemma 2.1 we can conclude that

lim sup n γ F ( q ) A q , Z n q = lim k γ F ( q ) A q , Z n k q = γ F ( q ) A q , ω q 0 .
(3.16)

Step 5. Finally, claim that the sequence { Z n } converges strongly to q= P Ω (IA+γF)q.

By Lemma 2.4, Lemma 2.6, and Lemma 2.9(ii), we obtain

Z n + 1 q 2 = δ n ( γ F ( Z n ) A q ) + ( I δ n A ) ( P C ( I ψ n ( I T ) ) Y n q ) 2 ( I δ n A ) ( P C ( I ψ n ( I T ) ) Y n q ) 2 + 2 δ n γ F ( Z n ) A q , Z n + 1 q ( 1 δ n γ ¯ ) 2 Y n q 2 + 2 δ n γ F ( Z n ) F ( q ) Z n + 1 q + 2 δ n γ F ( q ) A q , Z n + 1 q ( 1 δ n γ ¯ ) 2 ( θ n P C Z n q 2 + ( 1 θ n ) V n q 2 ) + 2 δ n γ α Z n q Z n + 1 q + 2 δ n γ F ( q ) A q , Z n + 1 q ( 1 δ n γ ¯ ) 2 Z n q 2 + δ n γ α ( Z n q 2 + Z n + 1 q 2 ) + 2 δ n γ F ( q ) A q , Z n + 1 q ,

which implies that

Z n + 1 q 2 ( 1 δ n γ ¯ ) 2 + δ n γ α 1 δ n γ α Z n q 2 + 2 δ n 1 δ n γ α γ F ( q ) A q , Z n + 1 q = ( 1 2 δ n ( γ ¯ γ α ) 1 δ n γ α ) Z n q 2 + 2 δ n ( γ ¯ γ α ) 1 δ n γ α ( δ n γ ¯ 2 2 ( γ ¯ γ α ) Z n q 2 + 1 γ ¯ γ α γ F ( q ) A q , Z n + 1 q ) .

From (3.16), the condition (i), and Lemma 2.3, we can conclude that { Z n } converges strongly to q= P Ω (IA+γF)q. By (3.8) and (3.11), we see that { V n } and { Y n } converge strongly to q= P Ω (IA+γF)q. This completes the proof. □

The following corollaries are direct results from Theorem 3.1.

Corollary 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let be an α-contractive mapping on H and let A be a strongly positive linear bounded operator on H with coefficient γ ¯ and 0<γ< γ ¯ α . Let T:CC be a nonspreading mapping. Let Φ:C×CR be a bifunction satisfying (A1)-(A4) with Ω:=Fix(T)EP(Φ). Let { Z n }, { Y n }, and { V n } be sequences generated by Z 1 H and

{ Φ ( V n , y ) + 1 φ n y V n , V n Z n 0 , y C , Y n = θ n P C Z n + ( 1 θ n ) V n , Z n + 1 = δ n γ F ( Z n ) + ( I δ n A ) P C ( I ψ n ( I T ) ) Y n , n N ,
(3.17)

where { δ n },{ θ n },{ φ n },{ ψ n }(0,1). Suppose the conditions (i)-(vi) hold.

  1. (i)

    lim n δ n =0 and n = 1 δ n =;

  2. (ii)

    0<τ θ n υ<1, for some τ,υ>0;

  3. (iii)

    n = 1 ψ n <;

  4. (iv)

    0<ϵ φ n η<1, for some ϵ,η>0;

  5. (v)

    n = 1 | δ n + 1 δ n |<, n = 1 | θ n + 1 θ n |<, n = 1 | ψ n + 1 ψ n |<, n = 1 | φ n + 1 φ n |<.

Then the sequences { Z n }, { Y n }, and { V n } converge strongly to q= P Ω (IA+γF)q.

Proof Put Φ= Φ i , for all i=1,2,,N. Using Theorem 3.1, the desired result is obtained. □

In 2007, Plubtieng and Punpaeng [19] introduced the general iterative method for an equilibrium problem and a nonexpansive mapping in Hilbert spaces. Let S be a nonexpansive mapping on H with Fix(S)EP(F). With an initial value z 1 H, the sequences { z n } and { v n } are generated by

{ F ( v n , y ) + 1 φ n y v n , v n z n 0 , y H , z n + 1 = α n γ f ( z n ) + ( I α n A ) S v n , n N ,
(3.18)

where { r n }(0,) and α n [0,1] satisfy some appropriate conditions. Then { z n } and { v n } converge strongly to a point z, where z= P Fix ( S ) EP ( F ) (IA+γf)(z).

Later, in 2010, Ceng et al. [11] studied the iterative scheme for equilibrium problem and an infinite family of nonexpansive mappings. Let 0<γα< γ ˜ . Let { α n } and { γ n } be sequences in (0,1). Starting with z 1 H, the sequences { z n } and { u n } are generated by the following iterative scheme:

{ F ( u n , y ) + 1 r n y u n , u n z n 0 , y H , v n = ( 1 γ n ) z n + γ n W n u n , z n + 1 = α n γ f ( v n ) + ( I α n A ) W n v n ,
(3.19)

where W n is a W-mapping generated by an infinite family of nonexpansive mappings and infinite real numbers. Then, under some suitable conditions, the sequences { z n } and { u n } converge strongly to z = P n = 1 F ( T n ) EP ( ϕ ) f ˜ ( z ), where f ˜ =Iθ(Aγf).

Remark 3.3 For Corollary 3.2, we prove the strong convergence theorem for equilibrium problem and a nonspreading mapping. Motivated by the results of Ceng et al. [11] and Plubtieng and Punpaeng [19], we consider the following statements, different from this work.

  1. (i)

    We investigate the iterative algorithm for a nonspreading mapping instead of using a nonexpansive mapping.

  2. (ii)

    We study the general iterative method by using the sequence Y n = θ n P C Z n +(1 θ n ) V n .

Corollary 3.4 Let C be a nonempty closed convex subset of a real Hilbert space H. Let be an α-contractive mapping on H and let A:HH be a strongly positive linear bounded operator with coefficient γ ¯ and 0<γ< γ ¯ α . Let T:CC be a nonspreading mapping with Fix(T). Let { Z n } be the sequence generated by Z 1 H and

Z n + 1 = δ n γF( Z n )+(I δ n A) P C ( I ψ n ( I T ) ) P C Z n ,nN,
(3.20)

where { δ n },{ ψ n }(0,1). Suppose the conditions (i)-(vi) hold.

  1. (i)

    lim n δ n =0 and n = 1 δ n =;

  2. (ii)

    n = 1 ψ n <;

  3. (iii)

    n = 1 | δ n + 1 δ n |<, n = 1 | ψ n + 1 ψ n |<.

Then the sequence { Z n } converges strongly to q= P Fix ( T ) (IA+γF)q.

Proof Take Φ i =0, for every i=1,2,,N. Then we have V n = P C Z n , for every nN. The result of Corollary 3.4 can be obtained by Theorem 3.1. □

4 Applications

By means of our main result, we obtain the strong convergence theorem for a finite family of nonspreading mappings and a finite family of equilibrium problems in the setting of Hilbert space. To prove this, the following definitions, remarks, and lemmas are needed.

Definition 4.1 A mapping A is quasi-nonexpansive if

Txpxp,for every xC and pFix(T).

Remark 4.1 If T:CC is nonspreading with Fix(T), then T is quasi-nonexpansive.

Example 4.2 Let an inner product ,: R 2 × R 2 R be defined by u,v=uv= u 1 v 1 + u 2 v 2 and a usual norm : R 2 R be given by u= u 1 2 + u 2 2 , for all u=( u 1 , u 2 ),v=( v 1 , v 2 ) R 2 . Let I=[1,100] and let T: I 2 I 2 be defined by

Tu= ( u 1 + 1 2 , 5 u 2 + 1 6 ) ,for all u=( u 1 , u 2 ) I 2 .

First, we show that T is a nonspreading mapping.

For every u,v I 2 , we obtain

T u T v 2 = ( u 1 + 1 2 , 5 u 2 + 1 6 ) ( v 1 + 1 2 , 5 v 2 + 1 6 ) 2 = ( 1 2 ( u 1 v 1 ) , 5 6 ( u 2 v 2 ) ) 2 = 1 4 ( u 1 v 1 ) 2 + 25 36 ( u 2 v 2 ) 2

and

2 u T u , v T v = 2 ( u 1 , u 2 ) ( u 1 + 1 2 , 5 u 2 + 1 6 ) , ( v 1 , v 2 ) ( v 1 + 1 2 , 5 v 2 + 1 6 ) = 2 ( u 1 1 2 , u 2 1 6 ) , ( v 1 1 2 , v 2 1 6 ) = 2 ( u 1 1 2 , u 2 1 6 ) ( v 1 1 2 , v 2 1 6 ) = 2 [ ( u 1 1 2 ) ( v 1 1 2 ) + ( u 2 1 6 ) ( v 2 1 6 ) ] = ( u 1 1 ) ( v 1 1 ) 2 + ( u 2 1 ) ( v 2 1 ) 18 0 .

This yields

u v 2 + 2 u T u , v T v u v 2 = ( u 1 v 1 , u 2 v 2 ) 2 = ( u 1 v 1 ) 2 + ( u 2 v 2 ) 2 > 1 4 ( u 1 v 1 ) 2 + 25 36 ( u 2 v 2 ) 2 = T u T v 2 .
(4.1)

Then T is a nonspreading mapping and we observe that Fix(T)={1}, where 1=(1,1). For every uI×I and 1Fix(T), from (4.1), we have

T u T 1 2 u 1 2 + 2 u T u , 1 T 1 = u 1 2 .

Therefore T is a quasi-nonexpansive mapping.

The following example shows that the converse of Remark 4.1 does not hold.

Example 4.3 Let I=[0,2] and let T: I 2 I 2 be defined by

Tu={ ( u 1 + 2 2 , u 2 + 2 2 ) if  u ( 1 , 2 ] × ( 1 , 2 ] , ( u 1 2 , u 2 2 ) if  u [ 0 , 1 ] × [ 0 , 1 ] .

First, show that T is quasi-nonexpansive for all u I 2 .

Observe that Fix(T)={2} if x(1,2]×(1,2] and Fix(T)={0} if u[0,1]×[0,1], where 2=(2,2) and 0=(0,0).

For any u(1,2]×(1,2], we have

( u 1 + 2 2 , u 2 + 2 2 ) ( 2 , 2 ) = ( u 1 2 2 , u 2 2 2 ) = 1 2 ( u 1 2 , u 2 2 ) = 1 2 ( u 1 , u 2 ) ( 2 , 2 ) < u 2 .

For every u[0,1]×[0,1], we obtain

( u 1 2 , u 2 2 ) ( 0 , 0 ) = ( u 1 2 , u 2 2 ) = 1 2 ( u 1 , u 2 ) < u .

Therefore T is a quasi-nonexpansive for all u I 2 .

Choose u=( 3 2 , 3 2 ) and v=( 1 2 , 1 2 ), we have

T ( 3 2 , 3 2 ) T ( 1 2 , 1 2 ) 2 = ( 7 4 , 7 4 ) ( 1 4 , 1 4 ) 2 = ( 6 4 , 6 4 ) 2 = ( 6 4 ) 2 + ( 6 4 ) 2 = 9 2 .

Thus we get

u v 2 + 2 u T u , v T v = ( 3 2 , 3 2 ) ( 1 2 , 1 2 ) 2 + 2 ( 3 2 , 3 2 ) T ( ( 3 2 , 3 2 ) ) , ( 1 2 , 1 2 ) T ( 1 2 , 1 2 ) = ( 1 , 1 ) 2 + 2 ( 3 2 , 3 2 ) ( 7 4 , 7 4 ) , ( 1 2 , 1 2 ) ( 1 4 , 1 4 ) = 2 + 2 ( 3 2 7 4 , 3 2 7 4 ) ( 1 2 1 4 , 1 2 1 4 ) = 2 + 2 ( 1 4 , 1 4 ) ( 1 4 , 1 4 ) = 2 2 8 = 14 8 .

Hence we have

T u T v 2 > u v 2 +2uTu,vTv.

By changing T from a nonspreading mapping to a quasi-nonexpansive mapping with Fix(T), we obtain the same result as shown in Lemma 2.9.

Remark 4.4 Let C be a nonempty closed convex subset of a real Hilbert space H and let T:CC be a quasi-nonexpansive mapping with Fix(T). Then we have the following statement:

  1. (i)

    Fix(T)=VI(C,IT);

  2. (ii)

    for every uC and vFix(T),

    P C ( I λ ( I T ) ) u v uv,where λ(0,1).

Definition 4.2 ([24])

Let C be a nonempty convex subset of a real Banach space. Let { T i } i = 1 N be a finite family of (nonexpansive) mappings of C into itself. For each j=1,2, , let α j =( α 1 j , α 2 j , α 3 j )I×I×I where I=[0,1] and α 1 j + α 2 j + α 3 j =1. Define the mapping S:CC as follows:

U 0 = I , U 1 = α 1 1 T 1 U 0 + α 2 1 U 0 + α 3 1 I , U 2 = α 1 2 T 2 U 1 + α 2 2 U 1 + α 3 2 I , U 3 = α 1 3 T 3 U 2 + α 2 3 U 2 + α 3 3 I , U N 1 = α 1 N 1 T N 1 U N 2 + α 2 N 1 U N 2 + α 3 N 1 I , S = U N = α 1 N T N U N 1 + α 2 N U N 1 + α 3 N I .

This mapping is called the S-mapping generated by T 1 , T 2 ,, T N and α 1 , α 2 ,, α N .

Lemma 4.5 ([25])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let { T i } i = 1 N be a finite family of nonspreading mappings of C into itself with i = 1 N Fix( T i ) and let α j =( α 1 j , α 2 j , α 3 j )I×I×I where I=[0,1], α 1 j + α 2 j + α 3 j =1, α 1 j , α 3 j (0,1) for all j=1,2,,N1 and α 1 N (0,1], α 3 N [0,1), α 2 j (0,1) for all j=1,2,,N. Let S be the S-mapping generated by T 1 , T 2 ,, T N and α 1 , α 2 ,, α N . Then Fix(S)= i = 1 N Fix( T i ) and S is a quasi-nonexpansive mapping.

Theorem 4.6 Let C be a nonempty closed convex subset of a real Hilbert space H. Let F:CC be an α-contractive mapping, let A:CC be a strongly positive linear bounded operator with coefficient γ ¯ and 0<γ< γ ¯ α . For i=1,2,, N ¯ , let Φ i :C×CR be a bifunction satisfying (A1)-(A4). Let T i :CC, for i=1,2,,N be a finite family of nonspreading mappings with Ω:= i = 1 N Fix( T i ) i = 1 N ¯ EP( Φ i ). Let ρ j =( α 1 j , α 2 j , α 3 j )I×I×I, j=1,2,,N, where I=[0,1], α 1 j + α 2 j + α 3 j =1, α 1 j , α 3 j (0,1) for all j=1,2,,N1 and α 1 N (0,1], α 3 N [0,1), α 2 j (0,1) for all j=1,2,,N, and let S be the S-mapping generated by T 1 , T 2 ,, T N and ρ 1 , ρ 2 ,, ρ N . Let { Z n }, { Y n }, and { V n } be sequences generated by Z 1 H and

{ i = 1 N ¯ a i Φ i ( V n , y ) + 1 φ n y V n , V n Z n 0 , y C , Y n = θ n P C Z n + ( 1 θ n ) V n , Z n + 1 = δ n γ F ( Z n ) + ( I δ n A ) P C ( I ψ n ( I S ) ) Y n , n N ,
(4.2)

where { δ n },{ θ n },{ φ n },{ ψ n }(0,1), 0< a i <1, for all i=1,2,, N ¯ . Suppose the conditions (i)-(vi) hold.

  1. (i)

    lim n δ n =0 and n = 1 δ n =;

  2. (ii)

    0<τ θ n υ<1, for some τ,υ>0;

  3. (iii)

    n = 1 ψ n <;

  4. (iv)

    0<ϵ φ n η<1, for some ϵ,η>0;

  5. (v)

    n = 1 N ¯ a i =1;

  6. (vi)

    n = 1 | δ n + 1 δ n |<, n = 1 | θ n + 1 θ n |<, n = 1 | ψ n + 1 ψ n |<, n = 1 | φ n + 1 φ n |<.

Then the sequences { Z n }, { Y n }, and { V n } converge strongly to q= P Ω (IA+γF)q.

Proof Using Remark 4.4, Lemma 4.5, and the same method as in Theorem 3.1, we have the desired conclusion. □

Remark 4.7 Theorem 4.6 can be considered as an improvement of Theorem 3.1 in the result of Tian and Jin [26] in the sense that some conditions are not assumed.

  1. (i)

    T ω :=(1ω)I+ωT, ω(0, 1 2 ),

  2. (ii)

    T is demi-closed on H,

where T is a quasi-nonexpansive mapping on H.

5 Examples for equilibrium problems and numerical results

In this section, the numerical examples are given for supporting Theorem 3.1. Using these examples, we see that our iteration for the combination of equilibrium problem converges faster than our iteration for the classical equilibrium problem.

Example 5.1 Let the mappings A:RR, F:RR, be defined by

A x = x 2 , F x = x 4 , for all  x R .

For every i=1,2,,N, let Φ i :[1,100]×[1,100]R and T:[1,100][1,100] be defined by

T x = 2 x + 5 7 , Φ i ( x , y ) = i ( y x ) ( y + 2 x 3 ) , for all  x , y [ 1 , 100 ] .

Put a i = 4 5 i + 1 N 5 N , for every i=1,2,,N. Let γ= 1 3 , δ n = 1 3 n , θ n = n 2 n + 3 , φ n = 2 n 3 n + 2 , and ψ n = 1 n 2 for every nN. Let the initial values be defined as in the following cases:

  1. (i)

    Z 1 =50, N=1, and n=10,

  2. (ii)

    Z 1 =50 and n=N=10.

Then, for both cases, the sequences { Z n }, { Y n }, and { V n } converge strongly to 1.

Solution. It is obvious that T is a nonspreading mapping and Fix(T)={1}.

Since a i = 4 5 i + 1 N 5 N , we obtain

i = 1 N a i Φ i (x,y)= i = 1 N ( 4 5 i + 1 N 5 N ) i(yx)(y+2x3)=μ(yx)(y+2x3),

where μ= i = 1 N ( 4 5 i + 1 N 5 N )i. It is clear that i = 1 N a i Φ i satisfies all conditions in Theorem 3.1 and EP( i = 1 N a i Φ i )= i = 1 N EP( Φ i )={1}. Then we have

Fix(T) i = 1 N EP( Φ i )={1}.

Observe that

0 i = 1 N a i Φ i ( V n , y ) + 1 φ n y V n , V n Z n 0 = μ ( y V n ) ( y + 2 V n 3 ) + 1 φ n ( y V n ) ( V n Z n ) 0 μ φ n ( y V n ) ( y + 2 V n 3 ) + ( y V n ) ( V n Z n ) 0 = μ φ n y 2 + ( μ V n φ n + V n Z n 3 μ φ n ) y 0 + 3 μ φ n V n V n 2 2 μ φ n V n 2 + V n Z n .
(5.1)

Let G(y)=μ φ n y 2 +(μ V n φ n + V n Z n 3μ φ n )y+3μ φ n V n V n 2 2μ φ n V n 2 + V n Z n . G(y) is a quadratic function of y with coefficients a=μ φ n , b=μ V n φ n + V n Z n 3μ φ n , and c=3μ φ n V n V n 2 2μ φ n V n 2 + V n Z n . Determine the discriminant Δ of G as follows:

Δ = b 2 4 a c = ( μ V n φ n + V n Z n 3 μ φ n ) 2 4 ( μ φ n ) ( 3 μ φ n V n V n 2 2 μ φ n V n 2 + V n Z n ) = 9 μ 2 φ n 2 6 μ φ n V n 18 μ 2 φ n 2 V n + V n 2 + 6 μ φ n V n 2 + 9 μ 2 φ n 2 V n 2 + 6 μ φ n Z n 2 V n Z n 6 μ φ n V n Z n + Z n 2 = ( V n 3 μ φ n + 3 μ φ n V n Z n ) 2 .

From (5.1), we have G(y)0, for every yR. If G(y) has at most one solution in , thus we have Δ0. This implies that

V n = Z n + 3 μ φ n 1 + 3 μ φ n ,
(5.2)

where μ= i = 1 N ( 4 5 i + 1 N 5 N )i. Put δ n = 1 3 n , θ n = n 2 n + 3 , φ n = 2 n 3 n + 2 , ψ n = 1 n 2 , nN. It is clear to see that the sequences { δ n }, { θ n }, { φ n }, and { ψ n } satisfy all conditions in Theorem 3.1. For every nN, from (5.2), we rewrite (3.1) as follows:

{ Y n = n 2 n + 3 P [ 1 , 100 ] Z n + ( 1 n 2 n + 3 ) 1 1 + 3 μ 2 n 3 n + 2 ( Z n + 3 μ 2 n 3 n + 2 ) , Z n + 1 = 1 36 n Z n + ( I 1 3 n A ) P [ 1 , 100 ] ( I 1 n 2 ( I T ) ) Y n , n N .
(5.3)

From Theorem 3.1, we can conclude that the sequences { Z n }, { Y n }, and { V n } generated by (5.3) converge strongly to 1.

For case (i), with N=1, we have μ=1. Then (5.2) becomes

V n = Z n + 3 φ n 1 + 3 φ n .
(5.4)

Then we have

{ Y n = n 2 n + 3 P [ 1 , 100 ] Z n + ( 1 n 2 n + 3 ) 1 1 + 3 2 n 3 n + 2 ( Z n + 3 2 n 3 n + 2 ) , Z n + 1 = 1 36 n Z n + ( I 1 3 n A ) P [ 1 , 100 ] ( I 1 n 2 ( I T ) ) Y n , n N .
(5.5)

From Corollary 3.2, we can conclude that the sequences { Z n }, { Y n }, and { V n } generated by (5.5) converge strongly to 1.

Table 1 and Figure 1 show the values of sequences { Z n }, { Y n }, and { V n } in two cases.

Figure 1
figure 1

The convergence of { V n } , { Y n } , and { Z n } with initial value Z 1 =50 .

Table 1 The values of { V n } , { Y n } , and { Z n } with Z 1 =50

Remark 5.2

  1. (i)

    From Table 1 and Figure 1, the sequences { Z n }, { Y n }, and { V n } converge to 1, where {1}=Fix(T) i = 1 N EP( Φ i ).

  2. (ii)

    For case (i), Corollary 3.2 guarantees the convergence of { Z n }, { Y n }, and { V n }.

  3. (iii)

    For case (ii), the convergence of { Z n }, { Y n }, and { V n } can be guaranteed by Theorem 3.1.

  4. (iv)

    The iteration (5.3) for the combination of equilibrium problem converges faster than the iteration (5.5) for the classical equilibrium problem.

Finally, we give the numerical example for our main theorem in two-dimensional space of real numbers.

Example 5.3 Let R 2 be the two-dimensional space of real numbers with an inner product ,: R 2 × R 2 R defined by u,v=uv= u 1 v 1 + u 2 v 2 and a usual norm : R 2 R given by u= u 1 2 + u 2 2 , for all u=( u 1 , u 2 ),v=( v 1 , v 2 ) R 2 . Let the mappings A: R 2 R 2 , F: R 2 R 2 be defined by

A u = ( u 1 2 , u 2 2 ) , F u = ( u 1 4 , u 2 4 ) , for all  u = ( u 1 , u 2 ) R 2 .

For every i=1,2,,N and I=[0,1], let Φ i : I 2 × I 2 R and T: I 2 I 2 be defined by

T u = ( u 1 + 1 2 , 5 u 2 + 1 6 ) , Φ i ( u , v ) = i ( v u ) ( v + 6 u 7 ) , for all  u = ( u 1 , u 2 ) , v = ( v 1 , v 2 ) I 2 ,

where 7=(7,7). Let γ= 1 3 , δ n = 1 3 n , θ n = n 2 n + 3 , φ n = 2 n 3 n + 2 , and ψ n = 1 n 2 for every nN.

It is clear that T is a nonspreading mapping and Fix(T)={1}, where 1=(1,1).

Put a i = 8 9 i + 1 N 9 N , for every i=1,2,,N. It is obvious that i = 1 N a i Φ i satisfies all conditions in Theorem 3.1 and EP( i = 1 N a i Φ i )= i = 1 N EP( Φ i )={1}, where 1=(1,1). Then we have

Fix(T) i = 1 N EP( Φ i )={1}.

Then, by Theorem 3.1, the sequences Z n =( Z n 1 , Z n 2 ), Y n =( Y n 1 , Y n 2 ), and V n =( V n 1 , V n 2 ) converge strongly to {1}.

Remark 5.4 From Example 5.3, putting ρ= i = 1 N ( 8 9 i + 1 N 9 N )i, we obtain

0 i = 1 N a i Φ i ( V n , y ) + 1 φ n y V n , V n Z n = ρ ( y V n ) ( y + 6 V n ( 7 , 7 ) ) + 1 φ n ( y V n ) ( V n Z n ) = ρ ( y 1 V n 1 , y 2 V n 2 ) ( y 1 + 6 V n 1 7 , y 2 + 6 V n 2 7 ) + 1 φ n ( y 1 V n 1 , y 2 V n 2 ) ( V n 1 Z n 1 , V n 2 Z n 2 ) = ρ ( ( y 1 V n 1 ) ( y 1 + 6 V n 1 7 ) + ( y 2 V n 2 ) ( y 2 + 6 V n 2 7 ) ) + 1 φ n ( ( y 1 V n 1 ) ( V n 1 Z n 1 ) + ( y 2 V n 2 ) ( V n 2 Z n 2 ) ) = ( ρ ( y 1 V n 1 ) ( y 1 + 6 V n 1 7 ) + 1 φ n ( y 1 V n 1 ) ( V n 1 Z n 1 ) ) + ( ρ ( y 2 V n 2 ) ( y 2 + 6 V n 2 7 ) + 1 φ n ( y 2 V n 2 ) ( V n 2 Z n 2 ) ) 0 ( ρ φ n ( y 1 V n 1 ) ( y 1 + 6 V n 1 7 ) + ( y 1 V n 1 ) (