Open Access

A new relaxed extragradient-like algorithm for approaching common solutions of generalized mixed equilibrium problems, a more general system of variational inequalities and a fixed point problem

Fixed Point Theory and Applications20132013:126

https://doi.org/10.1186/1687-1812-2013-126

Received: 20 August 2012

Accepted: 29 April 2013

Published: 15 May 2013

Abstract

In this paper, we introduce a new iterative algorithm by the relaxed extragradient-like method for finding a common element of the set of solutions of generalized mixed equilibrium problems, the set of solutions of a more general system of variational inequalities for finite inverse strongly monotone mappings and the set of solutions of a fixed point problem of a strictly pseudocontractive mapping in a Hilbert space. Then we prove strong convergence of the scheme to a common element of the three above described sets.

Keywords

generalized mixed equilibrium problemsmore general system of variational inequalitiesfixed pointinverse strongly monotone mappingstrictly pseudocontractive mapping

1 Introduction

In this paper, we assume that H is a real Hilbert space with the inner product , and the induced norm and C is a nonempty closed convex subset of H. P C denotes the metric projection of H onto C and F ( T ) denotes the fixed points set of a mapping T. The sequence { x n } converges weakly to x which is denoted by x n x .

Let φ : C R be a real-valued function and let A : C H be a nonlinear mapping. Suppose that F : C × C R is a bifunction.

The generalized mixed equilibrium problem is to find x C (see, e.g., [16]) such that
F ( x , y ) + φ ( y ) φ ( x ) + A x , y x 0 , y C .
(1.1)

The set of solutions of (1.1) is denoted by GMEP ( F , φ , A ) .

If φ 0 , then problem (1.1) reduces to the generalized equilibrium problem, which is to find x C (see, e.g., [79]) such that
F ( x , y ) + A x , y x 0 , y C .
(1.2)

The set of solutions of (1.2) is denoted by GEP ( F , A ) .

If A 0 , then problem (1.1) reduces to the mixed equilibrium problem, which is to find x C (see, e.g., [1013]) such that
F ( x , y ) + φ ( y ) φ ( x ) 0 , y C .
(1.3)

The set of solutions of (1.3) is denoted by MEP ( F , φ ) .

If F 0 , then problem (1.1) reduces to the mixed variational inequality of Browder type, which is to find x C (see, e.g., [3, 14]) such that
φ ( y ) φ ( x ) + A x , y x 0 , y C .
(1.4)

The set of solutions of (1.4) is denoted by MVI ( C , φ , A ) .

If φ 0 , A 0 , then problem (1.1) reduces to the equilibrium problem, which is to find x C (see, e.g., [1517]) such that
F ( x , y ) 0 , y C .
(1.5)

The set of solutions of (1.5) is denoted by EP ( F ) .

If F 0 , φ 0 , then problem (1.1) reduces to the variational inequality, which is to find x C (see, e.g., [1829]) such that
A x , y x 0 , y C .
(1.6)

The set of solutions of (1.6) is denoted by VI ( C , A ) .

If F 0 , A 0 , then problem (1.1) reduces to the minimized problem, which is to find x C (see, e.g., [1828]) such that
φ ( y ) φ ( x ) 0 , y C .
(1.7)

The set of solutions of (1.7) is denoted by Argmin ( φ ) .

Let A , B : C H be two mappings. Ceng et al. [2] considered the following problem of finding ( x , y ) C × C such that
{ λ A y + x y , x x 0 , x C , μ B x + y x , x y 0 , x C ,
(1.8)

which is called a general system of variational inequalities where λ > 0 and μ > 0 are two constants. In particular, if A = B , x = y , then problem (1.8) reduces to the classical variational inequality problem (1.6).

In order to find the common element of the solutions of problem (1.8) and the set of fixed points of one nonexpansive mapping S, Ceng et al. [2] studied the following algorithm: fix u C , x 0 C , and
{ z n = T λ n ( Θ , φ ) ( x n λ n F x n ) , y n = T μ 1 G 1 [ T μ 2 G 2 ( z n μ 2 B 2 z n ) μ 1 B 1 T μ 2 G 2 ( z n μ 2 B 2 z n ) ] , x n + 1 = α n u + β n x n + γ n y n + δ n S y n , n 0 .
(1.9)

Under appropriate conditions, they obtained one strong convergence theorem.

Let C be a nonempty closed convex subset of a real Hilbert space H. Let { A i } i = 1 N : C H be a family of mappings. Cai and Bu [1] considered the following problem of finding ( x 1 , x 2 , , x N ) C × C × × C such that
{ λ N A N x N + x 1 x N , x x 1 0 , x C , λ N 1 A N 1 x N 1 + x N x N 1 , x x N 0 , x C , λ 2 A 2 x 2 + x 3 x 2 , x x 3 0 , x C , λ 1 A 1 x 1 + x 2 x 1 , x x 2 0 , x C .
(1.10)
And (1.10) can be rewritten as
{ x 1 ( I λ N A N ) x N , x x 1 0 , x C , x N ( I λ N 1 A N 1 ) x N 1 , x x N 0 , x C , x 3 ( I λ 2 A 2 ) x 2 , x x 3 0 , x C , x 2 ( I λ 1 A 1 ) x 1 , x x 2 0 , x C ,
(1.11)

which is called a more general system of variational inequalities in Hilbert spaces, where λ i > 0 for all i { 1 , 2 , , N } . The set of solutions to (1.10) is denoted by Ω. In particular, if N = 2 , A 1 = B , A 2 = A , λ 1 = μ , λ 2 = λ , x 1 = x , x 2 = y , then problem (1.10) reduces to problem (1.8).

In order to find a common element of the solutions of problem (1.10) and the common fixed points of a family of strictly pseudocontractive mappings, Cai and Bu [1] studied the following algorithm: pick any x 0 H , set C 1 = C , x 1 = P C 1 x 0 , and
{ u n = T r M , n ( F M , φ M ) ( I r M , n B M ) T r M 1 , n ( F M 1 , φ M 1 ) u n = × ( I r M 1 , n B M 1 ) T r 1 , n ( F 1 , φ 1 ) ( I r 1 , n B 1 ) x n , y n = P C ( I λ N A N ) P C ( I λ N 1 A N 1 ) P C ( I λ 2 A 2 ) P C ( I λ 1 A 1 ) u n , z n = α n y n + ( 1 α n ) S n y n , C n + 1 = { z C n : z n z x n z } , x n + 1 = P C n + 1 x 0 , n 1 .
(1.12)

Under suitable conditions, they also obtained one strong convergence theorem.

In this paper, motivated and inspired by the above facts, we study a new iterative algorithm by the relaxed extragradient-like method for finding a common element of the set of solutions of generalized mixed equilibrium problems, the set of solutions of a more general system of variational inequalities for finite inverse strongly monotone mappings and the set of solutions of a fixed point problem of a strictly pseudocontractive mapping in a Hilbert space. Then we prove strong convergence of the scheme to a common element of the three above described sets.

2 Preliminaries

For solving the equilibrium problem, let us assume that the bifunction F satisfies 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 any x , y C ;

     
  3. (A3)
    F is weakly upper semicontinuous, i.e., for each x , y , z C ,
    lim sup t 0 F ( x + t ( z x ) , y ) F ( x , y ) ;
     
  4. (A4)

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

     
  5. (B1)
    For each x H and r > 0 , there exists a bounded subset D x C and y x C such that for any z C D x ,
    F ( z , y x ) + φ ( y x ) φ ( z ) + 1 r y x z , z x < 0 ;
     
  6. (B2)

    C is a bounded set.

     
Let H be a real Hilbert space. It is well known that
x + y 2 = x 2 + y 2 + 2 x , y
(2.1)
and
x 2 y 2 x y ( x + y )
(2.2)

for all x , y H .

Definition 2.1 Let C be a nonempty closed convex subset of a real Hilbert space H.
  1. (1)
    A mapping T : C C is said to be nonexpansive if
    T x T y x y , x , y C ;
     
  2. (2)
    A mapping T : C H is said to be L-Lipschitzian if there exists L > 0 such that
    T x T y L x y , x , y C ;
     
  3. (3)
    A mapping T : C C is said to be k-strictly pseudocontractive if there exists a constant k [ 0 , 1 ) such that
    T x T y 2 x y 2 + k ( I T ) x ( I T ) y 2 , x , y C .
    (2.3)

    It is obvious that k = 0 , then the mapping T is nonexpansive;

     
  4. (4)
    A mapping T : C H is said to be monotone if
    T x T y , x y 0 , x , y C ;
     
  5. (5)
    A mapping T : C H is said to be α-inverse-strongly monotone if there exists a positive real number α such that
    T x T y , x y α T x T y 2 , x , y C .
     

It is obvious that any α-inverse-strongly monotone mapping T is monotone and 1 α -Lipschitz continuous.

Definition 2.2 P C : H C is called a metric projection if for every point x H , there exists a unique nearest point in C, denoted by P C x , such that
x P C x x y , y C .

In order to prove our main results in the next section, we recall some lemmas.

Lemma 2.1 [30]

Let C be a nonempty closed convex subset of H and let T : C C be a k-strictly pseudocontractive mapping, then the following results hold:
  1. (1)
    equation (2.3) is equivalent to
    T x T y , x y x y 2 1 k 2 ( I T ) x ( I T ) y 2 , x , y C ;
    (2.4)
     
  2. (2)
    T is Lipschitz continuous with a constant 1 + k 1 k , i.e.,
    T x T y 1 + k 1 k x y , x , y C ;
    (2.5)
     
  3. (3)
    (Demi-closed principle) I T is demi-closed on C, that is,
    if x n x C and ( I T ) x n 0 , then x = T x .
     

Lemma 2.2 [1]

Let C be a nonempty closed convex subset of H and let T : C H be an α-inverse-strongly monotone mapping, then for all x , y C and λ > 0 , we have
( I λ T ) x ( I λ T ) y 2 = ( x y ) λ ( T x T y ) 2 = x y 2 2 λ T x T y , x y + λ 2 T x T y 2 x y 2 + λ ( λ 2 α ) T x T y 2 .

So, if 0 < λ 2 α , then I λ T is a nonexpansive mapping from C to H.

Lemma 2.3 Let C be a nonempty closed convex subset of H and let P C : H C be a metric projection, then
  1. (1)

    P C x P C y 2 x y , P C x P C y , x , y H ;

     
  2. (2)

    moreover, P C is a nonexpansive mapping, i.e., P C x P C y x y , x , y H ;

     
  3. (3)

    x P C x , y P C x 0 , x H , y C ;

     
  4. (4)

    x y 2 x P C x 2 + y P C x 2 , x H , y C .

     

Lemma 2.4 [4]

Let C be a nonempty closed convex subset of H. Assume that F : C × C R satisfies (A1)-(A4) and let φ : C R be a lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For r > 0 and x H , define a mapping T r ( F , φ ) : H C as follows:
T r ( F , φ ) ( x ) = { z C : F ( z , y ) + φ ( y ) φ ( z ) + 1 r y z , z x 0 , y C }
for all x H . Then the following hold:
  1. (1)

    for each x H , T r ( F , φ ) ( x ) and T r ( F , φ ) is single-valued;

     
  2. (2)
    T r ( F , φ ) is firmly nonexpansive, that is, for any x , y H ,
    T r ( F , φ ) ( x ) T r ( F , φ ) ( y ) 2 T r ( F , φ ) ( x ) T r ( F , φ ) ( y ) , x y ;
     
  3. (3)

    F ( T r ( F , φ ) ) = MEP ( F , φ ) ;

     
  4. (4)

    MEP ( F , φ ) is closed and convex.

     

Lemma 2.5 [3]

Let C be a nonempty closed convex subset of H. Assume that F : C × C R satisfies (A1)-(A4), B : C H is a continuous monotone mapping and let φ : C R be a lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For r > 0 and x H , define a mapping K r ( F , φ ) : H C as follows:
K r ( F , φ ) ( x ) = { z C : F ( z , y ) + φ ( y ) φ ( z ) + B x , y z + 1 r y z , z x 0 , y C }
for all x H . Then the following hold:
  1. (1)

    for each x H , K r ( F , φ ) ( x ) and K r ( F , φ ) is single-valued;

     
  2. (2)
    K r ( F , φ ) is firmly nonexpansive, that is, for any x , y H ,
    K r ( F , φ ) ( x ) K r ( F , φ ) ( y ) 2 K r ( F , φ ) ( x ) K r ( F , φ ) ( y ) , x y ;
     
  3. (3)

    F ( K r ( F , φ ) ) = GMEP ( F , φ , B ) ;

     
  4. (4)

    GMEP ( F , φ , B ) is closed and convex.

     
Lemma 2.6 Let C be a nonempty closed convex subset of H. Let { F k } k = 1 M be a family of bifunctions from C × C into satisfying (A1)-(A4), let { φ k } k = 1 M be a family of lower semicontinuous functions from C into , and let { B k } k = 1 M be a family of β k -inverse-strongly monotone mappings from C into H. For F k and φ k , k = 1 , 2 , , M , assume that either (B1) or (B2) holds. Let T : C H be a mapping defined by
T ( x ) = T r M ( F M , φ M ) ( I r M B M ) T r M 1 ( F M 1 , φ M 1 ) ( I r M 1 B M 1 ) T r 1 ( F 1 , φ 1 ) ( I r 1 B 1 ) x , x C .
Putting Θ 0 = I , where I is an identity mapping,
Θ k = T r k ( F k , φ k ) ( I r k B k ) T r k 1 ( F k 1 , φ k 1 ) ( I r k 1 B k 1 ) T r 1 ( F 1 , φ 1 ) ( I r 1 B 1 ) , k = 1 , 2 , , M .
If x k = 1 M GMEP ( F k , φ k , B k ) and 0 < r k 2 β k , k = 1 , 2 , , M , then
  1. (1)

    Θ k x = x , k = 1 , 2 , , M ;

     
  2. (2)

    T is nonexpansive.

     
Proof (1) Since { B k } k = 1 M is a family of β k -inverse-strongly monotone mappings from C into H, so they are continuous monotone mappings. Observe that
T r k ( F k , φ k ) ( I r k B k ) x = { z C : F k ( z , y ) + φ k ( y ) φ k ( z ) + 1 r k y z , z ( I r k B k ) x 0 , y C } = { z C : F k ( z , y ) + φ k ( y ) φ k ( z ) + B k x , y z + 1 r k y z , z x 0 , y C } = K r k ( F k , φ k ) ( x ) .
By Lemma 2.5, we know that if x k = 1 M GMEP ( F k , φ k , B k ) then x is the fixed point of the mapping K r k ( F k , φ k ) , k = 1 , 2 , , M , so we have
x = K r k ( F k , φ k ) ( x ) = T r k ( F k , φ k ) ( I r k B k ) x ,
(2.6)
which implies that x is a fixed point of the mapping T r k ( F k , φ k ) ( I r k B k ) . Therefore we get
Θ k x = x , k = 1 , 2 , , M .
(2) Since T r ( F , φ ) is firmly nonexpansive, then it is obvious that T r ( F , φ ) is nonexpansive. And from Lemma 2.2, we have
T ( x ) T ( y ) = Θ M x Θ M y = T r M ( F M , φ M ) ( I r M B M ) Θ M 1 x T r M ( F M , φ M ) ( I r M B M ) Θ M 1 y ( I r M B M ) Θ M 1 x ( I r M B M ) Θ M 1 y Θ M 1 x Θ M 1 y Θ 0 x Θ 0 y = x y ,

which implies T is nonexpansive. □

Lemma 2.7 [1]

Let C be a nonempty closed convex subset of H. Let A i be α i -inverse-strongly monotone from C into H, respectively, where i { 1 , 2 , , N } . Let G : C C be a mapping defined by
G ( x ) = P C ( I λ N A N ) P C ( I λ N 1 A N 1 ) P C ( I λ 2 A 2 ) P C ( I λ 1 A 1 ) x , x C .

If 0 < λ i 2 α i , i = 1 , 2 , , N , then G is nonexpansive.

Proof Put Ω i = P C ( I λ i A i ) P C ( I λ i 1 A i 1 ) P C ( I λ 1 A 1 ) , i = 1 , 2 , , N , and Ω 0 = I , where I is an identity mapping. Since P C is nonexpansive and from Lemma 2.2, we have
G ( x ) G ( y ) = Ω N x Ω N y = P C ( I λ N A N ) Ω N 1 x P C ( I λ N A N ) Ω N 1 y ( I λ N A N ) Ω N 1 x ( I λ N A N ) Ω N 1 y Ω N 1 x Ω N 1 y Ω 0 x Ω 0 y x y ,

which implies G is nonexpansive. □

Lemma 2.8 Let C be a nonempty closed convex subset of H. Let A i : C H be a nonlinear mapping, where i = 1 , 2 , , N . For given x i C , i = 1 , 2 , , N , ( x 1 , x 2 , , x N ) is a solution of problem (1.10) if and only if
x 1 = P C ( I λ N A N ) x N , x i = P C ( I λ i 1 A i 1 ) x i 1 , i = 2 , 3 , , N ,
(2.7)
that is,
x 1 = P C ( I λ N A N ) P C ( I λ N 1 A N 1 ) P C ( I λ 2 A 2 ) P C ( I λ 1 A 1 ) x 1 .

Proof () From Lemma 2.3(3), it is obvious that (2.7) is the solution of problem (1.10).

() Since
λ N A N x N + x 1 x N , x x 1 0 , x C x 1 ( I λ N A N ) x N , x x 1 0 , x C ( I λ N A N ) x N x 1 , ( I λ N A N ) x N x 1 ( I λ N A N ) x N + x 0 , x C ( I λ N A N ) x N x 1 2 ( I λ N A N ) x N x 1 , ( I λ N A N ) x N x , x C ( I λ N A N ) x N x 1 ( I λ N A N ) x N x , x C x 1 = P C ( I λ N A N ) x N .
Similarly, we get
x i = P C ( I λ i 1 A i 1 ) x i 1 , i = 2 , 3 , , N .
Therefore we have
x 1 = P C ( I λ N A N ) P C ( I λ N 1 A N 1 ) P C ( I λ 2 A 2 ) P C ( I λ 1 A 1 ) x 1 ,

which completes the proof. □

From Lemma 2.8, we know that x 1 = G ( x 1 ) , that is, x 1 is a fixed point of the mapping G, where G is defined by Lemma 2.7. Moreover, if we find the fixed point x 1 , it is easy to solve the other points by (2.7).

Lemma 2.9 [31]

Let { x n } and { y 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 ) y n for all integers n 0 and lim sup n ( y n + 1 y n x n + 1 x n ) 0 . Then lim n y n x n = 0 .

Lemma 2.10 [30]

Let T : C C be a nonexpansive mapping with F ( T ) . If x n x and ( I T ) x n y , then ( I T ) x = y .

Lemma 2.11 [30]

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

    n = 1 γ n = ;

     
  2. (2)

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

     

Then lim n α n = 0 .

3 Main results

In this section, we state and verify our main results. We have the following theorem.

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let { F k } k = 1 M be a family of bifunctions from C × C into satisfying (A1)-(A4), let { φ k } k = 1 M : C R be a family of lower semicontinuous and convex functions and let { B k } k = 1 M be a family of β k -inverse-strongly monotone mappings from C into H. Let A i be α i -inverse-strongly monotone from C into H, respectively, where i { 1 , 2 , , N } . Let S be a δ-strict pseudocontractive mapping from C into itself such that F = [ k = 1 M GMEP ( F k , φ k , B k ) ] F ( G ) F ( S ) , where G is defined by Lemma 2.7. For F k and φ k , k = 1 , 2 , , M , assume that either (B1) or (B2) holds. Pick any x 0 C , let { x n } C be a sequence generated by
{ z n = T r M , n ( F M , φ M ) ( I r M , n B M ) T r M 1 , n ( F M 1 , φ M 1 ) z n = × ( I r M 1 , n B M 1 ) T r 1 , n ( F 1 , φ 1 ) ( I r 1 , n B 1 ) x n , y n = P C ( I λ N A N ) P C ( I λ N 1 A N 1 ) P C ( I λ 2 A 2 ) P C ( I λ 1 A 1 ) z n , x n + 1 = a n x 0 + b n x n + c n y n + d n S y n , n 0 ,
(3.1)
where λ i ( 0 , 2 α i ) , i = 1 , 2 , , N , δ ( 0 , 1 ) . { a n } , { b n } , { c n } , { d n } [ 0 , 1 ] satisfy the following conditions:
  1. (i)

    a n + b n + c n + d n = 1 and ( c n + d n ) δ c n for all n 0 ;

     
  2. (ii)

    lim n a n = 0 and n = 0 a n = ;

     
  3. (iii)

    0 < lim inf n b n lim sup n b n < 1 and lim inf n d n > 0 ;

     
  4. (iv)

    lim n ( c n + 1 1 b n + 1 c n 1 b n ) = 0 ;

     
  5. (v)

    0 < lim inf n r k , n lim sup n r k , n < 2 β k , k = 1 , 2 , , M .

     

Then { x n } C converges strongly to P F x 0 .

Proof Putting
Θ n k = T r k , n ( F k , φ k ) ( I r k , n B k ) T r k 1 , n ( F k 1 , φ k 1 ) ( I r k 1 , n B k 1 ) T r 1 , n ( F 1 , φ 1 ) ( I r 1 , n B 1 ) , k { 1 , , M } , n N ,
and
Ω i = P C ( I λ i A i ) P C ( I λ i 1 A i 1 ) P C ( I λ 2 A 2 ) P C ( I λ 1 A 1 ) , i { 1 , 2 , , N } ,

Θ n 0 = Ω 0 = I , where I is the identity mapping on H. Then we have that z n = Θ n M x n and y n = Ω N z n . From Lemma 2.6 and Lemma 2.7, it can be seen easily that Θ n k and Ω i are nonexpansive, where k { 1 , 2 , , M } , i { 1 , 2 , , N } . We divide the proof into six steps.

Step 1. Firstly, we show that { x n } is bounded.

Indeed, take p F arbitrarily. Since p = Θ n k p = S p , k { 1 , 2 , , M } , n N . By Lemma 2.6, we have
z n p = Θ n M x n Θ n M p x n p .
(3.2)
It follows from Lemma 2.7 and (3.2) that
y n p = Ω N z n Ω N p z n p x n p .
(3.3)
Furthermore, from (3.1), we have
x n + 1 p = a n x 0 + b n x n + c n y n + d n S y n p = a n ( x 0 p ) + b n ( x n p ) + c n ( y n p ) + d n ( S y n p ) a n x 0 p + b n x n p + c n ( y n p ) + d n ( S y n p ) .
(3.4)
Since ( c n + d n ) δ c n , (2.3) and (2.4), we have
c n ( y n p ) + d n ( S y n p ) 2 = c n 2 y n p 2 + d n 2 S y n p 2 + 2 c n d n S y n p , y n p c n 2 y n p 2 + d n 2 [ y n p 2 + δ y n S y n 2 ] + 2 c n d n [ y n p 2 1 δ 2 y n S y n 2 ] = ( c n + d n ) 2 y n p 2 + [ d n 2 δ ( 1 δ ) c n d n ] y n S y n 2 = ( c n + d n ) 2 y n p 2 + d n [ ( c n + d n ) δ c n ] y n S y n 2 ( c n + d n ) 2 y n p 2 ,
which implies that
c n ( y n p ) + d n ( S y n p ) ( c n + d n ) y n p .
(3.5)
From (3.2)-(3.5) it follows that
x n + 1 p a n x 0 p + b n x n p + c n ( y n p ) + d n ( S y n p ) a n x 0 p + b n x n p + ( c n + d n ) y n p a n x 0 p + b n x n p + ( c n + d n ) x n p = a n x 0 p + ( 1 a n ) x n p .
So, we have
x n + 1 p max { x 0 p , x n p } , n 0 .
By induction, we obtain that
x n p x 0 p , n 0 .

Hence, { x n } is bounded. Consequently, we deduce immediately that { z n } , { y n } , { S y n } are bounded.

Step 2. Next, we prove that lim n x n + 1 x n = 0 .

Indeed, define x n + 1 = b n x n + ( 1 b n ) w n for all n 0 . It follows that
w n + 1 w n = x n + 2 b n + 1 x n + 1 1 b n + 1 x n + 1 b n x n 1 b n = a n + 1 x 0 + c n + 1 y n + 1 + d n + 1 S y n + 1 1 b n + 1 a n x 0 + c n y n + d n S y n 1 b n = a n + 1 x 0 1 b n + 1 a n x 0 1 b n + c n + 1 ( y n + 1 y n ) + d n + 1 ( S y n + 1 S y n ) 1 b n + 1 + ( c n + 1 1 b n + 1 c n 1 b n ) y n + ( d n + 1 1 b n + 1 d n 1 b n ) S y n .
(3.6)
Observe that
c n + 1 ( y n + 1 y n ) + d n + 1 ( S y n + 1 S y n ) 2 = c n + 1 2 y n + 1 y n 2 + d n + 1 2 S y n + 1 S y n 2 + 2 c n + 1 d n + 1 S y n + 1 S y n , y n + 1 y n c n + 1 2 y n + 1 y n 2 + d n + 1 2 [ y n + 1 y n 2 + δ ( y n + 1 S y n + 1 ) ( y n S y n ) 2 ] + 2 c n + 1 d n + 1 [ y n + 1 y n 2 1 δ 2 ( y n + 1 S y n + 1 ) ( y n S y n ) 2 ] = ( c n + 1 + d n + 1 ) 2 y n + 1 y n 2 + [ d n + 1 2 δ ( 1 δ ) c n + 1 d n + 1 ] ( y n + 1 S y n + 1 ) ( y n S y n ) 2 = ( c n + 1 + d n + 1 ) 2 y n + 1 y n 2 + d n + 1 [ ( c n + 1 + d n + 1 ) δ c n + 1 ] × ( y n + 1 S y n + 1 ) ( y n S y n ) 2 ( c n + 1 + d n + 1 ) 2 y n + 1 y n 2 ,
which implies that
c n + 1 ( y n + 1 y n ) + d n + 1 ( S y n + 1 S y n ) ( c n + 1 + d n + 1 ) y n + 1 y n .
(3.7)
Since
z n + 1 z n = Θ n M x n + 1 Θ n M x n x n + 1 x n ,
(3.8)
then we have
y n + 1 y n = Ω N z n + 1 Ω N z n z n + 1 z n x n + 1 x n .
(3.9)
Hence it follows from (3.6), (3.7), (3.9) and c n + 1 + d n + 1 1 b n + 1 = 1 a n + 1 b n + 1 1 b n + 1 < 1 that
w n + 1 w n ( a n + 1 1 b n + 1 + a n 1 b n ) x 0 + c n + 1 ( y n + 1 y n ) + d n + 1 ( S y n + 1 S y n ) 1 b n + 1 + | c n + 1 1 b n + 1 c n 1 b n | y n + | d n + 1 1 b n + 1 d n 1 b n | S y n ( a n + 1 1 b n + 1 + a n 1 b n ) x 0 + c n + 1 + d n + 1 1 b n + 1 x n + 1 x n + | c n + 1 1 b n + 1 c n 1 b n | y n + | a n + 1 + c n + 1 1 b n + 1 a n + c n 1 b n | S y n a n + 1 1 b n + 1 ( x 0 + S y n ) + a n 1 b n ( x 0 + S y n ) + x n + 1 x n + | c n + 1 1 b n + 1 c n 1 b n | ( y n + S y n ) .
(3.10)
Consequently, it follows from (3.10), conditions (ii), (iv) and { y n } , { S y n } are bounded that
lim sup n ( w n + 1 w n x n + 1 x n ) lim sup n { a n + 1 1 b n + 1 ( x 0 + S y n ) + a n 1 b n ( x 0 + S y n ) + | c n + 1 1 b n + 1 c n 1 b n | ( y n + S y n ) } = 0 .
Hence, by Lemma 2.9, we get lim n w n x n = 0 . Thus, from condition (iii), we have
lim n x n + 1 x n = lim n ( 1 b n ) w n x n = 0 .
(3.11)

Step 3. We show that lim n B k Θ n k 1 x n B k p = 0 , k = 1 , 2 , , M and lim n A i Ω i 1 z n A i Ω i 1 p = 0 , i = 1 , 2 , , N .

It follows from Lemma 2.6 that
z n p 2 = Θ n M x n Θ n M p 2 Θ n k x n Θ n k p 2 = T r k , n ( F k , φ k ) ( I r k , n B k ) Θ n k 1 x n T r k , n ( F k , φ k ) ( I r k , n B k ) Θ n k 1 p 2 ( I r k , n B k ) Θ n k 1 x n ( I r k , n B k ) Θ n k 1 p 2 Θ n k 1 x n p 2 + r k , n ( r k , n 2 β k ) B k Θ n k 1 x n B k p 2 x n p 2 + r k , n ( r k , n 2 β k ) B k Θ n k 1 x n B k p 2 .
(3.12)
By Lemma 2.3 and Lemma 2.2, we have
y n p 2 = P C ( I λ N A N ) Ω N 1 z n P C ( I λ N A N ) Ω N 1 p 2 ( I λ N A N ) Ω N 1 z n ( I λ N A N ) Ω N 1 p 2 Ω N 1 z n Ω N 1 p 2 + λ N ( λ N 2 α N ) A N Ω N 1 z n A N Ω N 1 p 2 .
By induction, we get
y n p 2 z n p 2 + i = 1 N λ i ( λ i 2 α i ) A i Ω i 1 z n A i Ω i 1 p 2 .
(3.13)
From condition (i) and (3.5), we get
x n + 1 p 2 = a n ( x 0 p ) + b n ( x n p ) + c n ( y n p ) + d n ( S y n p ) , x n + 1 p = a n x 0 p , x n + 1 p + b n x n p , x n + 1 p + c n ( y n p ) + d n ( S y n p ) , x n + 1 p a n x 0 p , x n + 1 p + b n x n p x n + 1 p + c n ( y n p ) + d n ( S y n p ) x n + 1 p a n x 0 p , x n + 1 p + b n x n p x n + 1 p + ( c n + d n ) y n p x n + 1 p a n x 0 p , x n + 1 p + b n 2 ( x n p 2 + x n + 1 p 2 ) + c n + d n 2 ( y n p 2 + x n + 1 p 2 ) ,
that is,
x n + 1 p 2 2 a n 1 + a n x 0 p , x n + 1 p + b n 1 + a n x n p 2 + c n + d n 1 + a n y n p 2 .
(3.14)
So, in terms of (3.12) and (3.13), we have
x n + 1 p 2 2 a n 1 + a n x 0 p x n + 1 p + b n 1 + a n x n p 2 + c n + d n 1 + a n { z n p 2 + i = 1 N λ i ( λ i 2 α i ) A i Ω i 1 z n A i Ω i 1 p 2 } 2 a n 1 + a n x 0 p x n + 1 p + b n 1 + a n x n p 2 + c n + d n 1 + a n { x n p 2 + r k , n ( r k , n 2 β k ) B k Θ n k 1 x n B k p 2 + i = 1 N λ i ( λ i 2 α i ) A i Ω i 1 z n A i Ω i 1 p 2 } 2 a n 1 + a n x 0 p x n + 1 p + 1 a n 1 + a n x n p 2 + c n + d n 1 + a n { r k , n ( r k , n 2 β k ) B k Θ n k 1 x n B k p 2 + i = 1 N λ i ( λ i 2 α i ) A i Ω i 1 z n A i Ω i 1 p 2 } .
Therefore,
r k , n ( 2 β k r k , n ) B k Θ n k 1 x n B k p 2 + i = 1 N λ i ( 2 α i λ i ) A i Ω i 1 z n A i Ω i 1 p 2 2 a n c n + d n x 0 p x n + 1 p + 1 a n c n + d n ( x n p 2 x n + 1 p 2 ) 2 a n c n + d n x 0 p x n + 1 p + 1 a n c n + d n x n + 1 x n ( x n + 1 p + x n p ) .
Since lim n a n = 0 , 0 < lim inf n r k , n lim sup n r k , n < 2 β k , k = 1 , 2 , , M , λ i ( 0 , 2 α i ) , i = 1 , 2 , , N , δ ( 0 , 1 ) , lim inf n ( c n + d n ) > 0 and { x n } is bounded, we have
lim n B k Θ n k 1 x n B k p = 0 , k = 1 , 2 , , M ,
(3.15)
and
lim n A i Ω i 1 z n A i Ω i 1 p = 0 , i = 1 , 2 , , N .
(3.16)

Step 4. We prove that lim n S y n y n = 0 .

Indeed, utilizing firmly nonexpansive of T r k ( F k , φ k ) and Lemma 2.2, we have
Θ n k x n Θ n k p 2 = T r k , n ( F k , φ k ) ( I r k , n B k ) Θ n k 1 x n T r k , n ( F k , φ k ) ( I r k , n B k ) p 2 ( I r k , n B k ) Θ n k 1 x n ( I r k , n B k ) p , Θ n k x n p = 1 2 ( ( I r k , n B k ) Θ n k 1 x n ( I r k , n B k ) p 2 + Θ n k x n p 2 ( I r k , n B k ) Θ n k 1 x n ( I r k , n B k ) p ( Θ n k x n p ) 2 ) 1 2 ( Θ n k 1 x n p 2 + Θ n k x n p 2 Θ n k 1 x n Θ n k x n r k , n ( B k Θ n k 1 x n B k p ) 2 ) ,
which implies
Θ n k x n p 2 Θ n k 1 x n p 2 Θ n k 1 x n Θ n k x n r k , n ( B k Θ n k 1 x n B k p ) 2 = Θ n k 1 x n p 2 Θ n k 1 x n Θ n k x n 2 r k , n 2 B k Θ n k 1 x n B k p 2 + 2 r k , n Θ n k 1 x n Θ n k x n , B k Θ n k 1 x n B k p Θ n k 1 x n p 2 Θ n k 1 x n Θ n k x n 2 + 2 r k , n Θ n k 1 x n Θ n k x n , B k Θ n k 1 x n B k p .
(3.17)
From (3.14), (3.3), Lemma 2.6 and (3.17), we have
x n + 1 p 2 2 a n 1 + a n x 0 p , x n + 1 p + b n 1 + a n x n p 2 + c n + d n 1 + a n z n p 2 2 a n 1 + a n x 0 p , x n + 1 p + b n 1 + a n x n p 2 + c n + d n 1 + a n Θ n k x n Θ n k p 2 a n 1 + a n x 0 p , x n + 1 p + b n 1 + a n x n p 2 + c n + d n 1 + a n [ Θ n k 1 x n p 2 Θ n k 1 x n Θ n k x n 2 + 2 r k , n Θ n k 1 x n Θ n k x n , B k Θ n k 1 x n B k p ] 2 a n 1 + a n x 0 p , x n + 1 p + b n 1 + a n x n p 2 + c n + d n 1 + a n [ x n p 2 Θ n k 1 x n Θ n k x n 2 + 2 r k , n Θ n k 1 x n Θ n k x n , B k Θ n k 1 x n B k p ] 2 a n 1 + a n x 0 p x n + 1 p + b n 1 + a n x n p 2 + c n + d n 1 + a n [ x n p 2 Θ n k 1 x n Θ n k x n 2 + 2 r k , n Θ n k 1 x n Θ n k x n B k Θ n k 1 x n B k p ] .
It follows that
c n + d n 1 + a n Θ n k 1 x n Θ n k x n 2 2 a n 1 + a n x 0 p x n + 1 p + 1 a n 1 + a n x n p 2 x n + 1 p 2 + 2 r k , n ( c n + d n ) 1 + a n Θ n k 1 x n Θ n k x n B k Θ n k 1 x n B k p 2 a n 1 + a n x 0 p x n + 1 p + x n p 2 x n + 1 p 2 + 2 r k , n ( c n + d n ) 1 + a n Θ n k 1 x n Θ n k x n B k Θ n k 1 x n B k p 2 a n x 0 p x n + 1 p + x n + 1 x n ( x n p + x n + 1 p ) + 2 r k , n ( c n + d n ) 1 + a n Θ n k 1 x n Θ n k x n B k Θ n k 1 x n B k p .
Since lim inf n c n + d n 1 + a n > 0 , a n 0 , x n + 1 x n 0 and B k Θ n k 1 x n B k p 0 , we conclude that
lim n Θ n k x n Θ n k 1 x n = 0 , k = 1 , 2 , , M .
(3.18)
Therefore we get
x n z n = Θ n 0 x n Θ n M x n Θ n 0 x n Θ n 1 x n + Θ n 1 x n Θ n 2 x n + + Θ n M 1 x n Θ n M x n 0 as  n .
(3.19)
From Lemma 2.3(1), we obtain
Ω N z n Ω N p 2 = P C ( I λ N A N ) Ω N 1 z n P C ( I λ N A N ) Ω N 1 p 2 ( I λ N A N ) Ω N 1 z n ( I λ N A N ) Ω N 1 p , Ω N z n Ω N p = 1 2 ( ( I λ N A N ) Ω N 1 z n ( I λ N A N ) Ω N 1 p 2 + Ω N z n Ω N p 2 ( I λ N A N ) Ω N 1 z n ( I λ N A N ) Ω N 1 p ( Ω N z n Ω N p ) 2 ) 1 2 ( Ω N 1 z n Ω N 1 p 2 + Ω N z n Ω N p 2 Ω N 1 z n Ω N z n + Ω N p Ω N 1 p λ N ( A N Ω N 1 z n A N Ω N 1 p ) 2 ) ,
which implies that
Ω N z n Ω N p 2 Ω N 1 z n Ω N 1 p 2 Ω N 1 z n Ω N z n + Ω N p Ω N 1 p λ N ( A N Ω N 1 z n A N Ω N 1 p ) 2 = Ω N 1 z n Ω N 1 p 2 Ω N 1 z n Ω N z n + Ω N p Ω N 1 p 2 λ N 2 A N Ω N 1 z n A N Ω N 1 p ) 2 + 2 λ N Ω N 1 z n Ω N z n + Ω N p Ω N 1 p , A N Ω N 1 z n A N Ω N 1 p Ω N 1 z n Ω N 1 p 2 Ω N 1 z n Ω N z n + Ω N p Ω N 1 p 2 + 2 λ N Ω N 1 z n Ω N z n + Ω N p Ω N 1 p A N Ω N 1 z n A N Ω N 1 p .
(3.20)
By induction, we have
Ω N z n Ω N p 2 z n p 2 i = 1 N Ω i 1 z n Ω i z n + Ω i p Ω i 1 p 2 + 2 i = 1 N λ i Ω i 1 z n Ω i z n + Ω i p Ω i 1 p A i Ω i 1 z n A i Ω i 1 p x n p 2 i = 1 N Ω i 1 z n Ω i z n + Ω i p Ω N 1 p 2 + 2 i = 1 N λ i Ω i 1 z n Ω i z n + Ω i p Ω i 1 p A i Ω i 1 z n A i Ω i 1 p ,
that is,
y n p 2 x n p 2 i = 1 N Ω i 1 z n Ω i z n + Ω i p Ω i 1 p 2 + 2 i = 1 N λ i Ω i 1 z n Ω i z n + Ω i p Ω i 1 p A i Ω i 1 z n A i Ω i 1 p .
(3.21)
From (3.14) and (3.21),
x n + 1 p 2 2 a n 1 + a n x 0 p , x n + 1 p + b n 1 + a n x n p 2 + c n + d n 1 + a n [ x n p 2 i = 1 N Ω i 1 z n Ω i z n + Ω i p Ω i 1 p 2 + 2 i = 1 N λ i Ω i 1 z n Ω i z n + Ω i p Ω i 1 p A i Ω i 1 z n A i Ω i 1 p ] 2 a n 1 + a n x 0 p x n + 1 p + b n 1 + a n x n p 2 + c n + d n 1 + a n [ x n p 2 i = 1 N Ω i 1 z n Ω i z n + Ω i p Ω i 1 p 2 + 2 i = 1 N λ i Ω i 1 z n Ω i z n + Ω i p Ω i 1 p A i Ω i 1 z n A i Ω i 1 p ] .
It follows that
c n + d n 1 + a n i = 1 N Ω i 1 z n Ω i z n + Ω i p Ω i 1 p 2 2 a n 1 + a n x 0 p x n + 1 p + 1 a n 1 + a n x n p 2 x n + 1 p 2 + 2 ( c n + d n ) 1 + a n i = 1 N λ i Ω i 1 z n Ω i z n + Ω i p Ω i 1 p A i Ω i 1 z n A i Ω i 1 p 2 a n 1 + a n x 0 p x n + 1 p + x n p 2 x n + 1 p 2 + 2 ( c n + d n ) 1 + a n i = 1 N λ i Ω i 1 z n Ω i z n + Ω i p Ω i 1 p A i Ω i 1 z n A i Ω i 1 p 2 a n 1 + a n x 0 p x n + 1 p + x n + 1 x n ( x n + 1 p + x n p ) + 2 ( c n + d n ) 1 + a n i = 1 N λ i Ω i 1 z n Ω i z n + Ω i p Ω i 1 p A i Ω i 1 z n A i Ω i 1 p 2 a n x 0 p x n + 1 p + x n + 1 x n ( x n + 1 p + x n p ) + 2 ( c n + d n ) 1 + a n i = 1 N λ i Ω i 1 z n Ω i z n + Ω i p Ω i 1 p A i Ω i 1 z n A i Ω i 1 p .
Since lim inf n c n + d n 1 + a n > 0 , a n 0 , x n + 1 x n 0 and A i Ω i 1 z n A i Ω i 1 p 0 , we conclude that
lim n i = 1 N Ω i 1 z n Ω i z n + Ω i p Ω i 1 p = 0 .
(3.22)
Therefore, we get
z n y n = Ω 0 z n Ω N z n i = 1 N Ω i 1 z n Ω i z n + Ω i p Ω i 1 p 0 as  n .
(3.23)
Thus from (3.19) and (3.23), we have
lim n x n y n = 0 .
(3.24)
Observe that
d n S y n x n = d n S y n d n x n = x n + 1 a n x 0 b n x n c n y n d n x n = x n + 1 x n + ( 1 b n d n ) x n c n y n a n x 0 = x n + 1 x n + a n ( x n x 0 ) + c n ( x n y n ) x n + 1 x n + a n x n x 0 + c n x n y n .
Since lim inf n d n > 0 , x n + 1 x n 0 , a n 0 and x n y n 0 , we have
lim n S y n x n = 0 .
(3.25)
From (3.24) and (3.25), we conclude that
lim n S y n y n = 0 .
(3.26)

Step 5. In this step, we prove that lim sup n x 0 x ¯ , x n x ¯ 0 , where x ¯ = P F x 0 .

Indeed, take a subsequence { x n i } of { x n } such that
lim sup n x 0 x ¯ , x n x ¯ = lim n x 0 x ¯ , x n i x ¯ .
Since { x n } is bounded, there exists a subsequence of { x n } which converges weakly to x . Without loss of generality, we may assume that x n i x . From (3.18) and (3.24), we have Θ n i k x n i x , y n i x , where k { 1 , 2 , , M } . From (3.26) and Lemma 2.1, we have x = S x that is x F ( S ) . Utilizing Lemma 2.7, we known that G is nonexpansive. And from (3.23), we obtain
y n i G y n i = G z n i G y n i z n i y n i 0 as  i .

According to Lemma 2.10, we obtain ( I G ) x = 0 , that is, x F ( G ) .

Next we prove that x k = 1 M GMEP ( F k , φ k , B k ) . Since
Θ n k x n = T r k , n ( F k , φ k ) ( I r k , n B k ) Θ n k 1 x n , n 1 , k { 1 , 2 , , M } .
For all y C , we have
F k ( Θ n k x n , y ) + φ k ( y ) φ k ( Θ n k x n ) + B k Θ n k 1 x n , y Θ n k x n + 1 r k , n y Θ n k x n , Θ n k x n Θ n k 1 x n 0 .
By (A2), we have
φ k ( y ) φ k ( Θ n k x n ) + B k Θ n k 1 x n , y Θ n k x n + 1 r k , n y Θ n k x n , Θ n k x n Θ n k 1 x n F k ( y , Θ n k x n ) .
Replacing n by n i in the above inequality, we have
φ k ( y ) φ k ( Θ n i k x n i ) + B k Θ n i k 1 x n i , y Θ n i k x n i + 1 r k , n i y Θ n i k x n i , Θ n i k x n i Θ n i k 1 x n i F k ( y , Θ n i k x n i ) .
Let z t = t y + ( 1 t ) x for all t ( 0 , 1 ] and y C . This implies that z t C . Then we have
z t Θ n i k x n i , B k z t z t Θ n i k x n i , B k z t + φ k ( Θ n i k x n i ) φ k ( z t ) B k Θ n i k 1 x n i , z t Θ n i k x n i 1 r k , n i z t Θ n i k x n i , Θ n i k x n i Θ n i k 1 x n i + F k ( z t , Θ n i k ) = φ k ( Θ n i k x n i ) φ k ( z t ) + z t Θ n i k x n i , B k z t B k Θ n i k x n i + z t Θ n i k x n i , B k Θ n i k x n i B k Θ n i k 1 x n i z t Θ n i k x n i , Θ n i k x n i Θ n i k 1 x n i r k , n i + F k ( z t , Θ n i k x n i ) .
By (3.18) we have B k Θ n i k x n i B k Θ n i k 1 x n i 0 as i . Furthermore, by the monotonicity of B k , we obtain z t Θ n i k x n i , B k z t B k Θ n i k x n i 0 . Then from (A4), the lower semicontinuity of φ and
Θ n i k x n i Θ n i k 1 x n i r k , n i 0 , Θ n i k x n i x ,
we obtain that
z t x , B k z t φ k ( x ) φ k ( z t ) + F k ( z t , x ) .
(3.27)
Using (A1), (A4) and (3.27), we have
0 = F k ( z t , z t ) + φ k ( z t ) φ k ( z t ) t F k ( z t , y ) + ( 1 t ) F k ( z t , x ) + t φ k ( y ) + ( 1 t ) φ k ( x ) t φ k ( z t ) ( 1 t ) φ k ( z t ) = t [ F k ( z t , y ) + φ k ( y ) φ k ( z t ) ] + ( 1 t ) [ F k ( z t , x ) + φ k ( x ) φ k ( z t ) ] t [ F k ( z t , y ) + φ k ( y ) φ k ( z t ) ] + ( 1 t ) z t x , B k z t = t [ F k ( z t , y ) + φ k ( y ) φ k ( z t ) ] + ( 1 t ) t y x , B k z t ,
and hence
0 F k ( z t , y ) + φ k ( y ) φ k ( z t ) + ( 1 t ) y x , B k z t .
Letting t 0 , we have, for each y C ,
0 F k ( x , y ) + φ k ( y ) φ k ( x ) + y x , B k x .
This implies that x GMEP ( F k , φ k , B k ) . Hence x k = 1 M GMEP ( F k , φ k , B k ) . Therefore,
x F = [ k = 1 M GMEP ( F k , φ k , B k ) ] F ( G ) F ( S ) .
This together with the property of metric projection implies that
lim sup n x 0 x ¯ , x n x ¯ = lim n x 0 x ¯ , x n i x ¯ = x 0 x ¯ , x x ¯ 0 .
(3.28)

Step 6. Finally, we can easily show that x n x ¯ as n .

Indeed, from (3.14) and (3.3), we have
x n + 1 x ¯ 2 2 a n 1 + a n x 0 x ¯ , x n + 1 x ¯ + b n 1 + a n x n x ¯ 2 + c n + d n 1 + a n x n x ¯ 2 = 2 a n 1 + a n x 0 x ¯ , x n + 1 x ¯ + ( 1 2 a n 1 + a n ) x n x ¯ 2 .

It is clear that n = 1 2 a n 1 + a n =