Open Access

The combination of the set of solutions of equilibrium problem for convergence theorem of the set of fixed points of strictly pseudo-contractive mappings and variational inequalities problem

Fixed Point Theory and Applications20132013:291

https://doi.org/10.1186/1687-1812-2013-291

Received: 12 June 2013

Accepted: 22 August 2013

Published: 9 November 2013

Abstract

For the purpose of this article, we introduce a new problem using the concept of equilibrium problem and prove the strong convergence theorem for finding a common element of the set of fixed points of an infinite family of κ i -strictly pseudo contractive mappings and of a finite family of the set of solutions of equilibrium problem and variational inequalities problem. Furthermore, we utilize our main theorem for the numerical example.

Keywords

strictly pseudo-contractive mapping S-mapping variational inequality the combination of equilibrium problem

1 Introduction

Let C be a nonempty closed convex subset of a real Hilbert space H with the inner product , and the norm . Let A be a strongly positive linear bounded operator on H if there is a constant γ ¯ > 0 with the property
A x , x γ ¯ x 2 , x H .

We now recall some well-known concepts and results as follows.

Definition 1.1 Let B : C H be a mapping. Then B is called
  1. (i)
    monotone if
    B x B y , x y 0 , x , y C ;
     
  2. (ii)
    υ-strongly monotone if there exists a positive real number υ such that
    B x B y , x y υ x y 2 , x , y C ;
     
  3. (iii)
    ξ-inverse-strongly monotone if there exists a positive real number ξ such that
    x y , B x B y ξ B x B y 2 , x , y C .
     
Definition 1.2 Let T : C C be a mapping. Then
  1. (i)

    an element x C is said to be a fixed point of T if T x = x and Fix ( T ) = { x C : T x = x } denotes the set of fixed points of T;

     
  2. (ii)
    a mapping T is called nonexpansive if
    T x T y x y , x , y C ;
     
  3. (iii)
    T is said to be κ-strictly pseudo-contractive if there exists a constant κ [ 0 , 1 ) such that
    T x T y 2 x y 2 + κ ( I T ) x ( I T ) y 2 , x , y C .
    (1.1)
     

Note that the class of κ-strict pseudo-contractions strictly includes the class of nonexpansive mappings.

Fixed point problems arise in many areas such as the vibration of masses attached to strings or nets (see the book by Cheng [1]) and a network bandwidth allocation problem [2] which is one of the central issues in modern communication networks. In applications to neural networks, fixed point theorems can be used to design a dynamic neural network in order to solve steady state solutions [3]. For general information on neural networks, see the books by Robert [4] or by Haykin [5].

Let G : C H . The variational inequality problem is to find a point u C such that
G u , v u 0
(1.2)

for all v C . The set of solutions of the variational inequality is denoted by VI ( C , G ) .

Variational inequalities were introduced and investigated by Stampacchia [6] in 1964. It is now well known that variational inequalities cover as diverse disciplines as partial differential equations, optimal control, optimization, mathematical programming, mechanics and finance; see [79].

Let F : C × C R be a bifunction. The equilibrium problem for F is to determine its equilibrium point, i.e., the set
EP ( F ) = { x C : F ( x , y ) 0 , y C } .
(1.3)

Equilibrium problems, which were introduced by [10] in 1994, have had a great impact and influence on the development of several branches of pure and applied sciences. Numerous problems in physics, optimization and economics are related to seeking some elements of EP ( F ) ; see [10, 11]. Many authors have studied an iterative scheme for the equilibrium problem; see, for example, [1114].

In 2005, Combettes and Hirstoaga [11] introduced some iterative schemes for finding the best approximation to the initial data when EP ( F ) is nonempty and proved the strong convergence theorem.

In 2007, Takahashi and Takahashi [14] proved the following theorem.

Theorem 1.1 Let C be a nonempty closed convex subset of H. Let F be a bifunction from C × C to satisfying
  1. (A1)

    F ( x , x ) = 0 for all x C ;

     
  2. (A2)

    F is monotone, i.e., F ( x , y ) + F ( y , x ) 0 for all x , y C ;

     
  3. (A3)
    For each x , y , z C ,
    lim t 0 + F ( t z + ( 1 t ) x , y ) F ( x , y ) ;
     
  4. (A4)

    For each x C , y F ( x , y ) is convex and lower semicontinuous;

     
and let S be a nonexpansive mapping of C into H such that F ( S ) EP ( F ) . Let f be a contraction of H into itself, and let { x n } and { u n } be sequences generated by x 1 H and
F ( u n , y ) + 1 r n y u n , u n x 0 , y C , x n + 1 = α n f ( x n ) + ( 1 α n ) S u n

for all n N , where { α n } [ 0 , 1 ] and { r n } [ 0 , 1 ] satisfy some control conditions. Then { x n } and { u n } converge strongly to z F ( S ) EP ( F ) , where z = P F ( S ) EP ( F ) f ( z ) .

For i = 1 , 2 , , N , let F i : C × C R be bifunctions and a i ( 0 , 1 ) with i = 1 N a i = 1 . Define the mapping i = 1 N a i F i : C × C R . The combination of equilibrium problem is to find x C such that
( i = 1 N a i F i ) ( x , y ) 0 , y C .
(1.4)
The set of solutions (1.4) is denoted by
EP ( i = 1 N a i F i ) = { x C : ( i = 1 N a i F i ) ( x , y ) 0 , y C } .

If F i = F , i = 1 , 2 , , N , then (1.4) reduces to (1.3).

Motivated by Theorem 1.1 and (1.4), we prove the strong convergence theorem for finding a common element of the set of fixed points of an infinite family of κ i -strictly pseudo-contractive mappings and a finite family of the set of solutions of equilibrium problem and variational inequalities problem.

2 Preliminaries

Let H be a real Hilbert space and C be a nonempty closed convex subset of H. We denote weak convergence and strong convergence by ‘’ and ‘→’, respectively. In a real Hilbert space H, it is well known that
α x ( 1 α ) y 2 = α x 2 + ( 1 α ) y 2 α ( 1 α ) x y 2

for all x , y H and α [ 0 , 1 ] .

Recall that the (nearest point) projection P C from H onto C assigns to each x H the unique point P C x C satisfying the property
x P C x = min y C x y .

The following lemmas are needed to prove the main theorem.

Lemma 2.1 [15]

For given z H and u C ,
u = P C z u z , v u 0 , v C .
It is well known that P C is a firmly nonexpansive mapping of H onto C and satisfies
P C x P C y 2 P C x P C y , x y , x , y H .

Lemma 2.2 [16]

Each Hilbert space H satisfies Opial’s condition, i.e., for any sequence { x n } H with x n x , the inequality
lim inf n x n x < lim inf n x n y

holds for every y H with y x .

Lemma 2.3 [17]

Let { s n } be a sequence of nonnegative real numbers satisfying
s n + 1 ( 1 α n ) s n + δ n , n 0 ,
where α n is a sequence in ( 0 , 1 ) and { δ n } is a sequence such that
  1. (1)

    n = 1 α n = ,

     
  2. (2)

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

     

Then lim n s n = 0 .

Lemma 2.4 Let H be a real Hilbert space. Then
x + y 2 x 2 + 2 y , x + y ,

for all x , y H .

Lemma 2.5 [15]

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 u C . Then, for λ > 0 ,
u = P C ( I λ A ) u u VI ( C , A ) ,

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

Lemma 2.6 [18]

Let C be a nonempty closed convex subset of a real Hilbert space H, and let S : C C be a self-mapping of C. If S is a κ-strictly pseudo-contractive mapping, then S satisfies the Lipschitz condition
S x S y 1 + κ 1 κ x y , x , y C .
For solving the equilibrium problem for a bifunction F : C × C R , let us assume that F and C satisfy the following conditions:
  1. (A1)

    F ( x , x ) = 0 for all x C ;

     
  2. (A2)

    F is monotone, i.e., F ( x , y ) + F ( y , x ) 0 for all x , y C ;

     
  3. (A3)
    For each x , y , z C ,
    lim t 0 F ( t z + ( 1 t ) x , y ) F ( x , y ) ;
     
  4. (A4)

    For each x C , y F ( x , y ) is convex and lower semicontinuous.

     

By using the concept of equilibrium problem, we have Lemma 2.7.

Lemma 2.7 Let C be a nonempty closed convex subset of a real Hilbert space H. For i = 1 , 2 , , N , let F i : C × C R be bifunctions satisfying (A1)-(A4) with i = 1 N EP ( F i ) . Then
EP ( i = 1 N a i F i ) = i = 1 N EP ( F i ) ,

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

Proof It is easy to show that i = 1 N EP ( F i ) EP ( i = 1 N a i F i ) .

Let x 0 EP ( i = 1 N a i F i ) and x i = 1 N EP ( F i ) . Then we have
( i = 1 N a i F i ) ( x 0 , y ) 0 , y C
(2.1)
and
F k ( x , y ) 0 , for all  k = 1 , 2 , , N ,  and  y C .
(2.2)
From (2.2) and x 0 C , we have
F k ( x , x 0 ) 0 ,
(2.3)
for all k = 1 , 2 , , N . From (2.3) and (A2), we obtain
F k ( x 0 , x ) F k ( x , x 0 ) + F k ( x 0 , x ) 0 .
(2.4)
Since x C , it follows from (2.1) that
i = 1 N a i F i ( x 0 , x ) 0 .
(2.5)
Applying (2.5), for each k = 1 , 2 , , N , we obtain
0 i = 1 N a i F i ( x 0 , x ) = i = 1 k 1 a i F i ( x 0 , x ) + a k F k ( x 0 , x ) + i = k + 1 N a i F i ( x 0 , x ) .
(2.6)
From (2.3), (2.6) and (A2), it follows that
a k F k ( x 0 , x ) i = 1 k 1 a i F i ( x 0 , x ) i = k + 1 N a i F i ( x 0 , x ) i = 1 k 1 a i F i ( x , x 0 ) + i = k + 1 N a i F i ( x , x 0 ) 0 .
(2.7)
Inequalities (2.7) and (2.4) guarantee that
F k ( x 0 , x ) = 0 for every  k = 1 , 2 , , N .
(2.8)
By using (2.8) and (A1), deduce that
x 0 = x .
It implies that
x 0 i = 1 N EP ( F i ) .
Therefore,
EP ( i = 1 N a i F i ) i = 1 N EP ( F i ) .
Hence, we have
EP ( i = 1 N a i F i ) = i = 1 N EP ( F i ) .

 □

Example 2.8 Let be the set of real numbers, and let bifunctions F i : R × R R , i = 1 , 2 , 3 , be defined by
F 1 ( x , y ) = x 2 + y 2 , F 2 ( x , y ) = 2 x 2 + x y + y 2 , F 3 ( x , y ) = x 2 2 2 x y + 5 y 2 2 , x , y R .
It is easy to check that F i ( x , y ) satisfy (A1)-(A4) for every i = 1 , 2 , 3 and
i = 1 3 EP ( F i ) = { 0 } .
(2.9)
By choosing a 1 = 1 12 , a 2 = 2 3 and a 3 = 1 4 , we obtain
i = 1 3 a i F i ( x , y ) = 1 24 ( 37 x 2 + 4 x y + 33 y 2 ) .
(2.10)
From (2.10), we have
EP ( i = 1 3 a i F i ) = { 0 } .
(2.11)
From (2.9) and (2.11), we obtain
EP ( i = 1 3 a i F i ) = i = 1 3 EP ( F i ) = { 0 } .

Remark 2.9 By using Lemma 2.7, we can guarantee the result of Example 2.8.

Lemma 2.10 [10]

Let C be a nonempty closed convex subset of H, and let F be a bifunction of C × C into satisfying (A1)-(A4). Let r > 0 and x H . Then there exists z C such that
F ( z , y ) + 1 r y z , z x 0 , y C .

Lemma 2.11 [11]

Assume that F : C × C R satisfies (A1)-(A4). For r > 0 , define a mapping T r : H C as follows:
T r ( x ) = { z C : F ( z , y ) + 1 r y z , z x 0 , y C }
for all x H . Then the following hold:
  1. (i)

    T r is single-valued;

     
  2. (ii)
    T r is firmly nonexpansive, i.e., for any x , y H ,
    T r ( x ) T r ( y ) 2 T r ( x ) T r ( y ) , x y ;
     
  3. (iii)

    Fix ( T r ) = EP ( F ) ;

     
  4. (iv)

    EP ( F ) is closed and convex.

     
Remark 2.12 From Lemma 2.7, it is easy to see that i = 1 N a i F i satisfies (A1)-(A4). By using Lemma 2.11, we obtain
Fix ( T r ) = EP ( i = 1 N a i F i ) = i = 1 N EP ( F i ) ,

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

Definition 2.1 [19]

Let C be a nonempty convex subset of a real Hilbert space. Let T i , i = 1 , 2 ,  , be 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 ] , α 1 j + α 2 j + α 3 j = 1 . For every n N , we define the mapping S n : C C as follows:
U n , n + 1 = I , U n , n = α 1 n T n U n , n + 1 + α 2 n U n , n + 1 + α 3 n I , U n , n 1 = α 1 n 1 T n 1 U n , n + α 2 n 1 U n , n + α 3 n 1 I , U n , k + 1 = α 1 k + 1 T k + 1 U n , k + 2 + α 2 k + 1 U n , k + 2 + α 3 k + 1 I , U n , k = α 1 k T k U n , k + 1 + α 2 k U n , k + 1 + α 3 k I , U n , 2 = α 1 2 T 2 U n , 3 + α 2 2 U n , 3 + α 3 2 I , S n = U n , 1 = α 1 1 T 1 U n , 2 + α 2 1 U n , 2 + α 3 1 I .

This mapping is called S-mapping generated by T n , , T 1 and α n , α n 1 , , α 1 .

Lemma 2.13 [19]

Let C be a nonempty closed convex subset of a real Hilbert space H. Let { T i } i = 1 be κ i -strictly pseudo-contractive mappings of C into itself with i = 1 Fix ( T i ) and κ = sup i N κ 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 2 j b < 1 and α 1 j , α 2 j , α 3 j ( κ , 1 ) for all j = 1 , 2 ,  . For every n N , let S n be an S-mapping generated by T n , T n 1 , , T 1 and α n , α n 1 , , α 1 . Then, for every x C and k N , lim n U n , k x exists.

For every k N and x C , Kangtunyakarn [19] defined the mapping U , k and S : C C as follows:
lim n U n , k x = U , k x
and
lim n S n x = lim n U n , 1 x = S x .

Such a mapping S is called S-mapping generated by T n , T n 1 , and α n , α n 1 ,  .

Remark 2.14 [19]

For every n N , S n is nonexpansive and lim n sup x D S n x S x = 0 for every bounded subset D of C.

Lemma 2.15 [19]

Let C be a nonempty closed convex subset of a real Hilbert space H. Let { T i } i = 1 be κ i -strictly pseudo-contractive mappings of C into itself with i = 1 Fix ( T i ) and κ = sup i N κ 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 2 j b < 1 and α 1 j , α 2 j , α 3 j ( κ , 1 ) for all j = 1 , 2 ,  . For every n N , let S n and S be S-mappings generated by T n , T n 1 , , T 1 and α n , α n 1 , , α 1 and T n , T n 1 , and α n , α n 1 ,  , respectively. Then Fix ( S ) = i = 1 Fix ( T i ) .

Lemma 2.16 Let C be a nonempty closed convex subset of a real Hilbert space H. For every i = 1 , 2 , , N , let A i be a strongly positive linear bounded operator on a Hilbert space H with coefficient γ i > 0 and γ ¯ = min i = 1 , 2 , , N γ i . Let { a i } i = 1 N ( 0 , 1 ) with i = 1 N a i = 1 . Then the following properties hold:
  1. (i)

    I ρ i = 1 N a i A i 1 ρ γ ¯ and I ρ i = 1 N a i A i is a nonexpansive mapping for every 0 < ρ < A i 1 ( i = 1 , 2 , , N ).

     
  2. (ii)

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

     
Proof To show (i), it is obvious that I ρ i = 1 N a i A i is a positive linear bounded operator on H, which yields that
I ρ i = 1 N a i A i = sup { | I ρ i = 1 N a i A i x , x | : x H , x = 1 } .
(2.12)
Since A i is a strongly positive operator for all i = 1 , 2 , , N , we get
i = 1 N a i A i x , x = i = 1 N a i A i x , x i = 1 N a i γ i x 2 i = 1 N a i γ ¯ x 2 = γ ¯ x 2 ,
(2.13)

which implies that i = 1 N a i A i is a γ ¯ -strongly positive operator.

Let x = 1 . Then, by using (2.13), we obtain
( I ρ i = 1 N a i A i ) x , x = x ρ i = 1 N a i A i x , x = x 2 ρ i = 1 N a i A i x , x ( 1 ρ γ ¯ ) x 2 = 1 ρ γ ¯ .
(2.14)
From (2.12) and (2.14), we have
I ρ i = 1 N a i A i 1 ρ γ ¯ .
(2.15)
Next, we show that I ρ i = 1 N a i A i is a nonexpansive mapping. Let x , y C . Then, using (2.15), we obtain
( I ρ i = 1 N a i A i ) x ( I ρ i = 1 N a i A i ) y = ( I ρ i = 1 N a i A i ) ( x y ) ( I ρ i = 1 N a i A i ) x y ( 1 ρ γ ¯ ) x y x y .

Hence, I ρ i = 1 N a i A i is a nonexpansive mapping.

To prove (ii), it is easy to see that
i = 1 N VI ( C , A i ) VI ( C , i = 1 N a i A i ) .
(2.16)
Let x 0 VI ( C , i = 1 N a i A i ) and x i = 1 N VI ( C , A i ) . Then we have
y x 0 , i = 1 N a i A i x 0 0 , y C .
(2.17)
From (2.16), we have x VI ( C , i = 1 N a i A i ) . It implies that
y x , i = 1 N a i A i x 0 , y C .
(2.18)
From (2.17), (2.18) and x , x 0 C , we obtain
x x 0 , i = 1 N a i A i x 0 0
(2.19)
and
x 0 x , i = 1 N a i A i x 0 .
(2.20)
By summing up (2.19) and (2.20), we have
0 x 0 x , i = 1 N a i A i x i = 1 N a i A i x 0 = i = 1 N a i x 0 x , A i x A i x 0 i = 1 N a i γ i x 0 x 2 i = 1 N a i γ ¯ x 0 x 2 = γ ¯ x 0 x 2 .

It implies that x 0 = x .

Then we can conclude that x 0 i = 1 N VI ( C , A i ) . Therefore
VI ( C , i = 1 N a i A i ) i = 1 N VI ( C , A i ) .
Hence, we have
VI ( C , i = 1 N a i A i ) = i = 1 N VI ( C , A i ) .

 □

3 Main result

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. For i = 1 , 2 , , N , let F i : C × C R be a bifunction satisfying (A1)-(A4), and let A i : C H be a strongly positive linear bounded operator on H with coefficient γ i > 0 and γ ¯ = min i = 1 , 2 , , N γ i . Let { T i } i = 1 be an infinite family of κ i -strictly pseudo-contractive mappings of C into itself, and let σ 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 , α 1 j + α 2 j η < 1 and α 1 j , α 2 j , α 3 j [ p , q ] ( κ , 1 ) for all j N . For every n N , let S n and S be the S-mappings generated by T n , T n 1 , , T 1 and σ n , σ n 1 , , σ 1 and T n , T n 1 , and σ n , σ n 1 ,  , respectively. Assume that F = i = 1 N EP ( F i ) i = 1 N VI ( C , A i ) i = 1 Fix ( T i ) . Let the sequences { x n } and { u n } be generated by x 1 , u C and
{ i = 1 N a i F i ( u n , y ) + 1 r n y u n , u n x n 0 , y C , x n + 1 = β n ( α n u + ( 1 α n ) S n x n ) + ( 1 β n ) P C ( I ρ n i = 1 N b i A i ) u n , n 1 ,
(3.1)
where { α n } , { β n } , { ρ n } ( 0 , 1 ) and 0 a i , b i 1 , for every i = 1 , 2 , , N , satisfy the following conditions:
  1. (i)

    n = 1 α n = and lim n α n = 0 ,

     
  2. (ii)

    0 < a β n b < 1 , n N ,

     
  3. (iii)

    lim n ρ n = 0 ,

     
  4. (iv)

    0 < c r n d < 1 , n N ,

     
  5. (v)

    i = 1 N a i = i = 1 N b i = 1 ,

     
  6. (vi)

    n = 1 | α n + 1 α n | < , n = 1 | β n + 1 β n | < , n = 1 | r n + 1 r n | < , n = 1 | ρ n + 1 ρ n | < and n = 1 α 1 n < .

     

Then the sequences { x n } and { u n } converge strongly to z 0 = P F u .

Proof Since ρ n 0 as n , without loss of generality, we may assume that ρ n < 1 A i , n N and i = 1 , 2 , , N .

The proof will be divided into five steps.

Step 1. We will show that { x n } is bounded.

Since i = 1 N a i F i satisfies (A1)-(A4) and
i = 1 N a i F i ( u n , y ) + 1 r n y u n , u n x n 0 , y C ,

by Lemma 2.11 and Remark 2.12, we have u n = T r n x n and Fix ( T r n ) = i = 1 N EP ( F i ) .

Let z F . Since z i = 1 N VI ( C , A i ) , by Lemma 2.5 and Lemma 2.16, we have
VI ( C , i = 1 N b i A i ) = F ( P C ( I ρ n i = 1 N b i A i ) ) .
From Lemma 2.16 and nonexpansiveness of T r n , we have
x n + 1 z = β n ( α n u + ( 1 α n ) S n x n ) + ( 1 β n ) P C ( I ρ n i = 1 N b i A i ) u n z = β n ( α n u + ( 1 α n ) S n x n z ) + ( 1 β n ) ( P C ( I ρ n i = 1 N b i A i ) u n z ) β n α n ( u z ) + ( 1 α n ) ( S n x n z ) + ( 1 β n ) P C ( I ρ n i = 1 N b i A i ) u n z β n ( α n u z + ( 1 α n ) x n z ) + ( 1 β n ) u n z = β n ( α n u z + ( 1 α n ) x n z ) + ( 1 β n ) T r n x n z β n ( α n u z + ( 1 α n ) x n z ) + ( 1 β n ) x n z max { u z , x 1 z } .

By induction on n, for some M > 0 , we have x n z M , n N . It implies that { x n } is bounded and so { u n } is a bounded sequence.

Step 2. We will show that lim n x n + 1 x n = 0 .

Putting D = i = 1 N b i A i , from the definition of x n , we have
x n + 1 x n = β n ( α n u + ( 1 α n ) S n x n ) + ( 1 β n ) P C ( I ρ n D ) u n ( β n 1 ( α n 1 u + ( 1 α n 1 ) S n 1 x n 1 ) + ( 1 β n 1 ) P C ( I ρ n 1 D ) u n 1 ) = β n ( α n u + ( 1 α n ) S n x n ) β n ( α n 1 u + ( 1 α n 1 ) S n 1 x n 1 ) + β n ( α n 1 u + ( 1 α n 1 ) S n 1 x n 1 ) β n 1 ( α n 1 u + ( 1 α n 1 ) S n 1 x n 1 ) + ( 1 β n ) P C ( I ρ n D ) u n ( 1 β n ) P C ( I ρ n D ) u n 1 + ( 1 β n ) P C ( I ρ n D ) u n 1 ( 1 β n ) P C ( I ρ n 1 D ) u n 1 + ( 1 β n ) P C ( I ρ n 1 D ) u n 1 ( 1 β n 1 ) P C ( I ρ n 1 D ) u n 1 β n [ | α n α n 1 | u + ( 1 α n ) S n x n ( 1 α n ) S n x n 1 + ( 1 α n ) S n x n 1 ( 1 α n ) S n 1 x n 1 + ( 1 α n ) S n 1 x n 1 ( 1 α n 1 ) S n 1 x n 1 ] + | β n β n 1 | [ α n 1 u + ( 1 α n 1 ) S n 1 x n 1 ] + ( 1 β n ) P C ( I ρ n D ) u n P C ( I ρ n D ) u n 1 + ( 1 β n ) P C ( I ρ n D ) u n 1 P C ( I ρ n 1 D ) u n 1 + | β n β n 1 | P C ( I ρ n 1 D ) u n 1 β n [ | α n α n 1 | u + ( 1 α n ) S n x n S n x n 1 + ( 1 α n ) S n x n 1 S n 1 x n 1 + | α n α n 1 | S n 1 x n 1 ] + | β n β n 1 | [ α n 1 u + ( 1 α n 1 ) S n 1 x n 1 ] + ( 1 β n ) u n u n 1 + ( 1 β n ) ( I ρ n D ) u n 1 ( I ρ n 1 D ) u n 1 + | β n β n 1 | P C ( I ρ n 1 D ) u n 1 β n [ | α n α n 1 | u + ( 1 α n ) x n x n 1 + ( 1 α n ) S n x n 1 S n 1 x n 1 + | α n α n 1 | S n 1 x n 1 ] + | β n β n 1 | [ α n 1 u + ( 1 α n 1 ) S n 1 x n 1 ] + ( 1 β n ) u n u n 1 + ( 1 β n ) | ρ n ρ n 1 | D u n 1 + | β n β n 1 | P C ( I ρ n 1 D ) u n 1 .
(3.2)
By using the same method as in step 2 of Theorem 3.1 in [20], we have
S n x n 1 S n 1 x n 1 α 1 n 2 1 κ x n 1 z .
(3.3)
Since u n = T r n x n , by utilizing the definition of T r n , we obtain
i = 1 N a i F i ( T r n x n , y ) + 1 r n y T r n x n , T r n x n x n 0 , y C ,
(3.4)
and
i = 1 N a i F i ( T r n + 1 x n + 1 , y ) + 1 r n + 1 y T r n + 1 x n + 1 , T r n + 1 x n + 1 x n + 1 0 , y C .
(3.5)
From (3.4) and (3.5), it follows that
i = 1 N a i F i ( T r n x n , T r n + 1 x n + 1 ) + 1 r n T r n + 1 x n + 1 T r n x n , T r n x n x n 0 ,
(3.6)
and
i = 1 N a i F i ( T r n + 1 x n + 1 , T r n x n ) + 1 r n + 1 T r n x n T r n + 1 x n + 1 , T r n + 1 x n + 1 x n + 1 0 .
(3.7)
From (3.6) and (3.7) and the fact that i = 1 N a i F i satisfies (A2), we have
1 r n T r n + 1 x n + 1 T r n x n , T r n x n x n + 1 r n + 1 T r n x n T r n + 1 x n + 1 , T r n + 1 x n + 1 x n + 1 0 ,
which implies that
T r n x n T r n + 1 x n + 1 , T r n + 1 x n + 1 x n + 1 r n + 1 T r n x n x n r n 0 .
It follows that
T r n + 1 x n + 1 T r n x n , T r n x n T r n + 1 x n + 1 + T r n + 1 x n + 1 x n r n r n + 1 ( T r n + 1 x n + 1 x n + 1 ) 0 .
(3.8)
From (3.8), we obtain
T r n + 1 x n + 1 T r n x n 2 T r n + 1 x n + 1 T r n x n , T r n + 1 x n + 1 x n r n r n + 1 ( T r n + 1 x n + 1 x n + 1 ) = T r n + 1 x n + 1 T r n x n , x n + 1 x n + ( 1 r n r n + 1 ) ( T r n + 1 x n + 1 x n + 1 ) T r n + 1 x n + 1 T r n x n x n + 1 x n + ( 1 r n r n + 1 ) ( T r n + 1 x n + 1 x n + 1 ) T r n + 1 x n + 1 T r n x n [ x n + 1 x n + | 1 r n r n + 1 | T r n + 1 x n + 1 x n + 1 ] = T r n + 1 x n + 1 T r n x n [ x n + 1 x n + 1 r n + 1 | r n + 1 r n | T r n + 1 x n + 1 x n + 1 ] T r n + 1 x n + 1 T r n x n [ x n + 1 x n + 1 d | r n + 1 r n | T r n + 1 x n + 1 x n + 1 ] ,
which yields that
u n + 1 u n x n + 1 x n + 1 d | r n + 1 r n | u n + 1 x n + 1 .
(3.9)
From (3.9), we have
u n u n 1 x n x n 1 + 1 d | r n r n 1 | u n x n .
(3.10)
By substituting (3.3) and (3.10) into (3.2), we obtain
x n + 1 x n β n [ | α n α n 1 | u + ( 1 α n ) x n x n 1 + ( 1 α n ) α 1 n 2 1 κ x n 1 z + | α n α n 1 | S n 1 x n 1 ] + | β n β n 1 | [ α n 1 u + ( 1 α n 1 ) S n 1 x n 1 ] + ( 1 β n ) [ x n x n 1 + 1 d | r n r n 1 | u n x n ] + ( 1 β n ) | ρ n ρ n 1 | D u n 1 + | β n β n 1 | P C ( I ρ n 1 D ) u n 1 β n ( 1 α n ) x n x n 1 + ( 1 β n ) x n x n 1 + | α n α n 1 | u + α 1 n 2 1 κ x n 1 z + | α n α n 1 | S n 1 x n 1 + | β n β n 1 | [ u + S n 1 x n 1 ] + 1 d | r n r n 1 | u n x n + | ρ n ρ n 1 | D u n 1 + | β n β n 1 | P C ( I ρ n 1 D ) u n 1 ( 1 α n β n ) x n x n 1 + | α n α n 1 | K + α 1 n 2 1 κ K + | α n α n 1 | K + | β n β n 1 | 2 K + 1 d | r n r n 1 | K + | ρ n ρ n 1 | K + | β n β n 1 | K = ( 1 α n β n ) x n x n 1 + 2 K | α n α n 1 | + 2 K 1 κ α 1 n + 3 K | β n β n 1 | + K d | r n r n 1 | + K | ρ n ρ n 1 | ,
(3.11)
where K = max n N { u , x n z , S n x n , u n x n , D u n , P C ( I ρ n D ) u n } . From (3.11), conditions (i), (ii), (vi) and Lemma 2.3, we obtain
lim n x n + 1 x n = 0 .
(3.12)

Step 3. We will show that lim n u n x n = lim n P C ( I ρ n D ) x n x n = lim n S n x n x n = 0 , where D = n = 1 N b i A i .

To show this, let z F . Since u n = T r n x n and T r n is a firmly nonexpansive mapping, then we obtain
z T r n x n 2 = T r n z T r n x n 2 T r n z T r n x n , z x n = 1 2 ( T r n x n z 2 + x n z 2 T r n x n x n 2 ) ,
which yields that
u n z 2 x n z 2 u n x n 2 .
(3.13)
By nonexpansiveness of P C ( I ρ n D ) , (3.13) and the definition of x n , we have
x n + 1 z 2 = β n ( α n u + ( 1 α n ) S n x n ) + ( 1 β n ) P C ( I ρ n D ) u n z 2 = β n ( α n ( u z ) + ( 1 α n ) ( S n x n z ) ) + ( 1 β n ) ( P C ( I ρ n D ) u n z ) 2 β n α n ( u z ) + ( 1 α n ) ( S n x n z ) 2 + ( 1 β n ) P C ( I ρ n D ) u n z 2 β n ( α n u z 2 + ( 1 α n ) S n x n z 2 ) + ( 1 β n ) u n z 2 β n ( α n u z 2 + ( 1 α n ) x n z 2 ) + ( 1 β n ) ( x n z 2 u n x n 2 ) α n u z 2 + x n z 2 ( 1 β n ) u n x n 2 ,
which implies that
( 1 β n ) u n x n 2 α n u z 2 + x n z 2 x n + 1 z 2 α n u z 2 + ( x n z + x n + 1 z ) x n + 1 x n .
(3.14)
By (3.12), (3.14), conditions (i) and (ii), we have
lim n u n x n = 0 .
(3.15)
Put w n = α n u + ( 1 α n ) S n x n . By the definition of x n and z F , we obtain
x n + 1 z 2 = β n w n + ( 1 β n ) P C ( I ρ n D ) u n z 2 = β n ( w n z ) + ( 1 β n ) ( P C ( I ρ n D ) u n z ) 2 = β n w n z 2 + ( 1 β n ) P C ( I ρ n D ) u n z 2 β n ( 1 β n ) w n P C ( I ρ n D ) u n 2 β n α n u + ( 1 α n ) S n x n z 2 + ( 1 β n ) u n z 2 β n ( 1 β n ) w n P C ( I ρ n D ) u n 2 β n ( α n u z 2 + ( 1 α n ) S n x n z 2 ) + ( 1 β n ) x n z 2 β n ( 1 β n ) w n P C ( I ρ n D ) u n 2 β n ( α n u z 2 + ( 1 α n ) x n z 2 ) + ( 1 β n ) x n z 2 β n ( 1 β n ) w n P C ( I ρ n D ) u n 2 α n u z 2 + x n z 2 β n ( 1 β n ) w n P C ( I ρ n D ) u n 2 ,
which yields that
β n ( 1 β n ) w n P C ( I ρ n D ) u n 2 α n u z 2 + x n z 2 x n + 1 z 2 α n u z 2 + ( x n z + x n + 1 z ) x n + 1 x n .
(3.16)
By (3.12) and conditions (i) and (ii), we have
lim n w n P C ( I ρ n D ) u n = 0 .
(3.17)
By the definition of x n , we obtain
x n + 1 P C ( I ρ n D ) u n = β n ( w n P C ( I ρ n D ) u n ) .
(3.18)
By (3.18) and the definition of x n , we have
x n P C ( I ρ n D ) x n x n x n + 1 + x n + 1 P C ( I ρ n D ) u n + P C ( I ρ n D ) u n P C ( I ρ n D ) x n x n x n + 1 + β n w n P C ( I ρ n D ) u n + u n x n .
From (3.12), (3.15) and (3.17), we have
lim n x n P C ( I ρ n D ) x n = 0 .
(3.19)
Since
x n P C ( I ρ n D ) u n x n P C ( I ρ n D ) x n + P C ( I ρ n D ) x n P C ( I ρ n D ) u n x n P C ( I ρ n D ) x n + x n u n ,
by using (3.15) and (3.19), we have
lim n x n P C ( I ρ n D ) u n = 0 .
(3.20)
By the definition of x n , we obtain
x n + 1 x n = β n ( α n u + ( 1 α n ) S n x n ) + ( 1 β n ) P C ( I ρ n D ) u n x n = β n ( α n u + ( 1 α n ) S n x n x n ) + ( 1 β n ) ( P C ( I ρ n D ) u n x n ) = β n ( α n ( u x n ) + ( 1 α n ) ( S n x n x n ) ) + ( 1 β n ) ( P C ( I ρ n D ) u n x n ) = α n β n ( u x n ) + β n ( 1 α n ) ( S n x n x n ) + ( 1 β n ) ( P C ( I ρ n D ) u n x n ) ,
from which it follows that
β n ( 1 α n ) S n x n x n α n β n u x n + x n + 1 x n + ( 1 β n ) P C ( I ρ n D ) u n x n .
From (3.12), (3.20), conditions (i) and (ii), we have
lim n S n x n x n = 0 .
(3.21)

Step 4. We will show that lim sup n u z , x n z 0 , where z = P F u .

To show this, choose a subsequence { x n k } of { x n } such that
lim sup n u z , x n z = lim k u z , x n k z .
(3.22)

Without loss of generality, we can assume that x n k ω as k , where ω C . From (3.15), we obtain u n k ω as k .

Assume that ω i = 1 N VI ( C , A i ) . Since i = 1 N VI ( C , A i ) = F ( P C ( I ρ n k D ) ) , we have ω P C ( I ρ n k D ) ω , where D = i = 1 N b i A i .

By nonexpansiveness of P C ( I ρ n k D ) , (3.19) and Opial’s condition, we obtain
lim inf k x n k ω < lim inf k x n k P C ( I ρ n k D ) ω lim inf k x n k P C ( I ρ n k D ) x n k + lim inf k P C ( I ρ n k D ) x n k P C ( I ρ n k D ) ω lim inf k x n k ω .
This is a contradiction. Then we have
ω i = 1 N VI ( C , A i ) .
(3.23)

Next, we will show that ω i = 1 Fix ( T i ) .

By Lemma 2.15, we have Fix ( S ) = i = 1 Fix ( T i ) . Assume that ω S ω . Using Opial’s condition, (3.21) and Remark 2.14, we obtain
lim inf k x n k ω < lim inf k x n k S ω lim inf k ( x n k S n k x n k + S n k x n k S n k ω + S n k ω S ω ) lim inf k x n k ω .
This is a contradiction. Then we have
ω Fix ( S ) = i = 1 Fix ( T i ) .
(3.24)

Next, we will show that ω i = 1 N EP ( F i ) .

Since
i = 1 N a i F i ( u n , y ) + 1 r n y u n , u n x n 0 , y C ,
and i = 1 N a i F i satisfies conditions (A1)-(A4), we obtain
1 r n y u n , u n x n i = 1 N a i F i ( y , u n ) , y C .
In particular, it follows that
y u n k , u n k x n k r n k i = 1 N a i F i ( y , u n k ) , y C .
(3.25)
From (3.15), (3.25) and (A4), we have
i = 1 N a i F i ( y , ω ) 0 , y C .
(3.26)
Put y t : = t y + ( 1 t ) ω , t ( 0 , 1 ] , we have y t C . By using (A1), (A4) and (3.26), we have
0 = i = 1 N a i F i ( y t , y t ) = i = 1 N a i F i ( y t , t y + ( 1 t ) ω ) t i = 1 N a i F i ( y t , y ) + ( 1 t ) i = 1 N a i F i ( y t , ω ) t i = 1 N a i F i ( y t , y ) .
It implies that
i = 1 N a i F i ( t y + ( 1 t ) ω , y ) 0 , t ( 0 , 1 ]  and  y C .
(3.27)
From (3.27), taking t 0 + and using (A3), we can conclude that
0 i = 1 N a i F i ( ω , y ) , y C .
Therefore, ω EP ( i = 1 N a i F i ) . By Lemma 2.7, we obtain EP ( i = 1 N a i F i ) = i = 1 N EP ( F i ) . It follows that
ω i = 1 N EP ( F i ) .
(3.28)

From (3.23), (3.24) and (3.28), we can deduce that ω F .

Since x n k ω and ω F , then, by Lemma 2.1, we can conclude that
lim sup n u z , x n z = lim k u z , x n k z = u z , ω z 0 .
(3.29)

Step 5. Finally, we will show that the sequence { x n } converges strongly to z = P F u .

By nonexpansiveness of S n and P C ( I ρ n D ) , we have
x n + 1 z 2 = β n ( α n u + ( 1 α n ) S n x n ) + ( 1 β n ) P C ( I ρ n D ) u n z 2 = β n ( α n ( u z ) + ( 1 α n ) ( S n x n z ) ) + ( 1 β n ) ( P C ( I ρ n D ) u n z ) 2 = α n β n ( u z ) + β n ( 1 α n ) ( S n x n z ) + ( 1 β n ) ( P C ( I ρ n D ) u n z ) 2 β n ( 1 α n ) ( S n x n z ) + ( 1 β n ) ( P C ( I ρ n D ) u n z ) 2 + 2 α n β n u z , x n + 1 z ( β n ( 1 α n ) x n z + ( 1 β n ) u n z ) 2 + 2 α n β n u z , x n + 1 z ( β n ( 1 α n ) x n z + ( 1 β n ) x n z ) 2 + 2 α n β n u z , x n + 1 z = ( 1 α n β n ) 2 x n z 2 + 2 α n β n u z , x n + 1 z ( 1 α n β n ) x n z 2 + 2 α n β n u z , x n + 1 z .

From (3.29), conditions (i), (ii) and Lemma 2.3, we can conclude that { x n } converges strongly to z = P F u . By (3.15), we have { u n } converges strongly to z = P F u . This completes the proof. □

4 Application

In this section, we apply our main theorem to prove strong convergence theorems involving equilibrium problem, variational inequality problem and fixed point problem.

Theorem 4.1 Let C be a nonempty closed convex subset of a real Hilbert space H. For i = 1 , 2 , , N , let F : C × C R be a bifunction satisfying (A1)-(A4), and let A : C H be a strongly positive linear bounded operator on H with coefficient γ > 0 . Let { T i } i = 1 be an infinite family of κ i -strictly pseudo-contractive mappings of C into itself, and let σ 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 , α 1 j + α 2 j η < 1 and α 1 j , α 2 j , α 3 j [ p , q ] ( κ , 1 ) for all j N . For every n N , let S n and S be the S-mappings generated by T n , T n 1 , , T 1 and σ n , σ n 1 , , σ 1 and T n , T n 1 , and σ n , σ n 1 ,  , respectively. Assume that F = EP ( F ) VI ( C , A ) i = 1 Fix ( T i ) . Let the sequences { x n } and { u n } be generated by x 1 , u C and
{ F ( u n , y ) + 1 r n y u n , u n x n 0 , y C , x n + 1 = β n ( α n u + ( 1 α n ) S n x n ) + ( 1 β n ) P C ( I ρ n A ) u n , n 1 ,
(4.1)
where { α n } , { β n } , { ρ n } ( 0 , 1 ) and 0 a i , b i 1 , for every i = 1 , 2 , , N , satisfy the following conditions:
  1. (i)

    n = 1 α n = and lim n α n = 0 ,

     
  2. (ii)

    0 < a β n b < 1 , n N ,

     
  3. (iii)

    lim n ρ n = 0 ,

     
  4. (iv)

    0 < c r n d < 1 , n N ,

     
  5. (v)

    n = 1 | α n + 1 α n | < , n = 1 | β n + 1 β n | < , n = 1 | r n + 1 r n | < , n = 1 | ρ n + 1 ρ n | < and n = 1 α 1 n < .

     

Then the sequences { x n } and { u n } converge strongly to z 0 = P F u .

Proof Take F = F i and A = A i , i = 1 , 2 , , N . By Theorem 3.1, we obtain the desired conclusion. □

Theorem 4.2 Let C be a nonempty closed convex subset of a real Hilbert space H. For every i = 1 , 2 , , N , let A i : C H be a strongly positive linear bounded operator on H with coefficient γ i > 0 and γ ¯ = min γ i . Let { T i } i = 1 be an infinite family of κ i -strictly pseudo-contractive mappings of C into itself, and let σ 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 , α 1 j + α 2 j η < 1 and α 1 j , α 2 j , α 3 j [ p , q ] ( κ , 1 ) for all j N . For every n N , let S n and S be the S-mappings generated by T n , T n 1 , , T 1 and σ n , σ n 1 , , σ 1 and T n , T n 1 , and σ n , σ n 1 ,  , respectively. Assume that F = i = 1 N VI ( C , A i ) i = 1 Fix ( T i ) . Let the sequences { x n } and { u n } be generated by x 1 , u C and
x n + 1 = β n ( α n u + ( 1 α n ) S n x n ) + ( 1 β n ) P C ( I ρ n i = 1 N b i A i ) x n , n 1 ,
(4.2)
where { α n } , { β n } , { ρ n } ( 0 , 1 ) satisfy the following conditions:
  1. (i)

    n = 1 α n = and lim n α n = 0 ,

     
  2. (ii)

    0 < a β n b < 1 , n N ,

     
  3. (iii)

    lim n ρ n = 0 ,

     
  4. (iv)

    0 < c r n d < 1 , n N ,

     
  5. (v)

    i = 1 N b i = 1 ,

     
  6. (vi)

    n = 1 | α n + 1 α n | < , n = 1 | β n + 1 β n | < , n = 1 | r n + 1 r n | < , n = 1 | ρ n + 1 ρ n | < and n = 1 α 1 n < .

     

Then the sequences { x n } and { u n } converge strongly to z 0 = P F u .

Proof Put F i = 0 , i = 1 , 2 , , N . Then we have u n = P C x n = x n , n N . Therefore the conclusion of Theorem 4.2 can be obtained from Theorem 3.1. □

Theorem 4.3 Let C be a nonempty closed convex subset of a real Hilbert space H. For i = 1 , 2 , , N , let F i : C × C R be a bifunction satisfying (A1)-(A4). Let { T i } i = 1 be an infinite family of nonexpansive mappings of C into itself, and let σ 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 , α 1 j + α 2 j η < 1 and α 1 j , α 2 j , α 3 j [ p , q ] ( 0 , 1 ) for all j N . For every n N , let S n and S be the S-mappings generated by T n , T n 1 , , T 1 and σ n , σ n 1 , , σ 1 and T n , T n 1 , and σ n , σ n 1 ,  , respectively. Assume that F = i = 1 N EP ( F i ) i = 1 Fix ( T i ) . Let the sequences { x n } and { u n } be generated by x 1 , u C and
{ i = 1 N a i F i ( u n , y ) + 1 r n y u n , u n x n 0 , y C , x n + 1 = β n ( α n u + ( 1 α n ) S n x n ) + ( 1 β n ) P C u n , n 1 ,
(4.3)
where { α n } , { β n } , { ρ n } ( 0 , 1 ) and 0 a i 1 , for every i = 1 , 2 , , N , satisfy the following conditions:
  1. (i)

    n = 1 α n = and lim n α n = 0 ,

     
  2. (ii)

    0 < a β n b < 1 , n N ,

     
  3. (iii)

    lim n ρ n = 0 ,

     
  4. (iv)

    0 < c r n d < 1 , n N ,

     
  5. (v)

    i = 1 N a i = 1 ,

     
  6. (vi)

    n = 1 | α n + 1 α n | < , n = 1 | β n + 1 β n | < , n = 1 | r n + 1 r n | < , n = 1 | ρ n + 1 ρ n | < and n = 1 α 1 n < .

     

Then the sequences { x n } and { u n } converge strongly to z 0 = P F u .

Proof Put A i = 0 , i = 1 , 2 , , N . Let κ i = 0 , then T i is a nonexpansive mapping for every i = 1 , 2 ,  . The result of Theorem 4.3 can be obtained by Theorem 3.1. □

5 Example and numerical results

In this section, an example is given to support Theorem 3.1.

Example 5.1 Let be the set of real numbers, and let the mapping A i : R R be defined by A i x = i x 2 , x R and b i = 7 8 i + 1 N 8 N for every i = 1 , 2 , , N . For n N , let the mapping T n : R R be defined by
T n x = 7 n 7 n + 1 x , x R ,
and let F : R R be defined by
F i ( x , y ) = i ( 3 x 2 + x y + 2 y 2 ) , x , y R .
Furthermore, let a i = 2 3 i + 1 N 3 N for every i = 1 , 2 , , N . Then we have
i = 1 N a i F i ( x , y ) = i = 1 N ( 2 3 i + 1 N 3 N ) i ( 3 x 2 + x y + 2 y 2 ) = E ( 3 x 2 + x y + 2 y 2 ) ,
where E = i = 1 N ( 2 3 i + 1 N 3 N ) i . It is easy to check that i = 1 N a i F i satisfies all the conditions of Theorem 3.1 and EP ( i = 1 N a i F i ) = i = 1 N EP ( F i ) = { 0 } . Let { x n } and { u n } be the sequences generated by (3.1). By the definition of F, we have
0 i = 1 N a i F i ( u n , y ) + 1 r n y u n , u n x n = E ( 3 u n 2 + u n y + 2 y 2 ) + 1 r n ( y u n ) ( u n x n ) = E ( 3 u n 2 + u n y + 2 y 2 ) + 1 r n ( y u n y x n u n 2 + u n x n ) 0 E r n ( 3 u n 2 + u n y + 2 y 2 ) + ( y u n y x n u n 2 + u n x n ) = 3 E u n 2 r n + E u n r n y + 2 E y 2 r n + y u n y x n u n 2 + u n x n = 2 E r n y 2 + ( E u n r n + u n x n ) y + u n x n u n 2 3 E u n 2 r n .
Let G ( y ) = 2 E r n y 2 + ( u n ( E r n + 1 ) x n ) y + u n x n u n 2 3 E u n 2 r n . G ( y ) is a quadratic function of y with coefficient a = 2 E r n , b = u n ( E r n + 1 ) x n , and c = u n x n u n 2 3 E u n 2 r n . Determine the discriminant Δ of G as follows:
Δ = b 2 4 a c = ( u n ( E r n + 1 ) x n ) 2 4 ( 2 E r n ) ( u n x n u n 2 3 E u n 2 r n ) = u n 2 ( E r n + 1 ) 2 2 x n u n ( E r n + 1 ) + x n 2 8 E r n u n x n + 24 E 2 u n 2 r n 2 + 8 E u n 2 r n = E 2 u n 2 r n 2 + 2 E r n u n 2 + u n 2 2 E x n u n r n 2 x n u n + x n 2 8 E r n u n x n + 24 E 2 u n 2 r n 2 + 8 E u n 2 r n = u n 2 + 10 E r n u n 2 + 25 E 2 u n 2 r n 2 2 x n u n 10 E x n u n r n + x n 2 = ( u n + 5 E u n r n ) 2 2 x n ( u n + 5 E u n r n ) + x n 2 = ( u n + 5 E u n r n x n ) 2 .
We know that G ( y ) 0 , y R . If it has at most one solution in , then Δ 0 , so we obtain
u n = x n 1 + 5 i = 1 N ( 2 3 i + 1 N 3 N ) i r n .
(5.1)
For every n N , x , y R and T n x = 7 n 7 n + 1 x , then we have T n is a nonexpansive mapping. It implies that T is 0-strictly pseudo-contractive for every n N . For every j = 1 , 2 ,  , let α 1 j = 3 5 j 2 + 3 , α 2 j = 3 j 2 5 j 2 + 3 , α 3 j = 2 j 2 5 j 2 + 3 . Then σ j = ( 3 5 j 2 + 3 , 3 j 2 5 j 2 + 3 , 2 j 2 5 j 2 + 3 ) for all j = 1 , 2 ,  . Since S n is an S-mapping generated by T n , T n 1 , , T 1 and σ n , σ n 1 , , σ 1 , we obtain
U n , n + 1 x n = x n , U n , n x n = ( 3 5 n 2 + 3 ( 7 n 7 n + 1 ) U n , n + 1 + ( 3 n 2 5 n 2 + 3 ) U n , n + 1 + 2 n 2 5 n 2 + 3 ) x n , U n , n 1 x n = ( 3 5 ( n 1 ) 2 + 3 ( 7 ( n 1 ) 7 ( n 1 ) + 1 ) U n , n + ( 3 ( n 1 ) 2 5 ( n 1 ) 2 + 3 ) U n , n U n , n 1 x n = + 2 ( n 1 ) 2 5 ( n 1 ) 2 + 3 ) x n , U n , k + 1 x n = ( 3 5 ( k + 1 ) 2 + 3 ( 7 ( k + 1 ) 7 ( k + 1 ) + 1 ) U n , k + 2 + ( 3 ( k + 1 ) 2 5 ( k + 1 ) 2 + 3 ) U n , k + 1 U n , k + 1 x n = + 2 ( k + 1 ) 2 5 ( k + 1 ) 2 + 3 ) x n , U n , k x n = ( 3 5 k 2 + 3 ( 7 k 7 k + 1 ) U n , k + 1 + ( 3 k 2 5 k 2 + 3 ) U n , k + 2 k 2 5 k 2 + 3 ) x n , U n , 2 x n = ( 3 23