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Superimposed algorithms for the split equilibrium problems and fixed point problems

Abstract

The split common problems for finding the equilibrium points and fixed points have been studied. A parallel superimposed algorithm is introduced to solve this split common problem. Strong convergence theorems are shown with some analysis techniques.

MSC:49J30, 47H09, 65K10.

1 Introduction

In the present article, our main purpose is to study the split problem. First, we recall some relevant background in the literature.

Problem 1: the split feasibility problem

Let C and Q be two nonempty closed convex subsets of Hilbert spaces H 1 and H 2 , respectively, and let A: H 1 H 2 be a bounded linear operator. The problem of finding a point x such that

x CandA x Q
(1.1)

is called the split feasibility; it was first introduced by Censor and Elfving [1] in finite dimensional Hilbert spaces. Such problems arise in the field of intensity-modulated radiation therapy when one attempts to describe physical dose constraints and equivalent uniform dose constraints within a single model. When C R N and Q R M are a single pair of sets, Censor and Elfving [1] introduced the simultaneous multi-projections algorithm:

x n + 1 = A 1 ( λ 1 I + λ 2 A A ) 1 ( λ 1 v 1 n + 1 + λ 2 A A v 2 n + 1 ) ,n0,
(1.2)

where λ 1 >0, λ 2 >0, λ 1 + λ 2 =1, v i n + 1 = P C i f i ( b n ) (i=1,2), and b n is the solution of the equation

i = 1 2 λ i f i ( b n )= i = 1 2 f i ( v i n ) .

Note that the simultaneous multi-projections algorithm (1.2) involve a matrix inversion A 1 at each iterative step. This is very time-consuming, particularly if the dimensions are large. In order to solve this problem, Byrne [2] derived a new algorithm, called the CQ-algorithm:

x n + 1 = P C ( x n τ A ( I P Q ) A x n ) ,n0,

where τ(0, 2 L ) with L being the largest eigenvalue of the matrix A A, I is the unit matrix or operator, and P C and P Q denote the orthogonal projections onto C and Q, respectively. The CQ-algorithm and its variant forms have now been studied for the split feasibility problem; see, for instance [313].

Problem 2: the split common fixed point problem

If every closed convex subset of a Hilbert space is the fixed point set of its associating projection, then the split feasibility problem becomes a special case of the split common fixed point problem of finding a point x with the property:

x Fix(U)andA x Fix(T).
(1.3)

This problem was first introduced by Censor and Segal [14] who invented an algorithm which generates a sequence { x n } according to the iterative procedure:

x n + 1 =U ( x n γ A ( I T ) A x n ) ,nN.
(1.4)

Moudafi [15] extended (1.4) to the following relaxed algorithm:

x n + 1 = U α n ( x n + γ A ( T β I ) A x n ) ,nN,

where β(0,1), α n (0,1) are relaxation parameters. Consequently, Wang and Xu [16] considered a general cyclic algorithm. Very recently, the split problem has also been extended to solve other problems, such as the split monotone variational inclusions and the split variational inequalities, please refer to [15, 1722] and [2325].

Problem 3: the equilibrium problem

Consider the following equilibrium problem: Finding x C such that

F ( x , x ) 0,xC,
(1.5)

where F:C×CR is a bifunction. We will denote by EP(F) the set of solutions of (1.5).

The equilibrium problems, in its various forms, found application in optimization problems, fixed point problems, and convex minimization problems; in other words, equilibrium problems are a unified model for problems arising in physics, engineering, economics, and so on (see [2629]).

Motivated by the split common fixed point problem and the equilibrium problem, He and Du [30] presented the following split equilibrium problem and fixed point problem:

Find a point  x Fix(T)EP(F) such that A x Fix(S)EP(G),
(1.6)

where Fix(S) and Fix(T) are the sets of fixed points of two nonlinear mappings S and T, respectively, EP(F) and EP(G) are the solution sets of two equilibrium problems with bifunctions F and G, respectively, and A is a bounded linear mapping. Denote the solution set of (1.6) by

Γ= { x Fix ( T ) EP ( F ) : A x Fix ( S ) EP ( G ) } .

Special cases

  1. 1.

    If F=0 and G=0, then (1.6) is reduced to the following split common fixed point problem, which has been considered by many authors, for example, [14, 15, 17] and [21]:

    Find a point  x Fix(T) such that A x Fix(S).
    (1.7)
  2. 2.

    If S= P Q and T= P C , then (1.7) is reduced to the split feasibility problem (1.1).

  3. 3.

    If S and T are all identity operators, then (1.6) is reduced to the split equilibrium problem which has been considered in [18]:

    Find a point  x EP(F) such that A x EP(G).

Based on the work in this direction, in this paper we will develop new algorithms to solve the split equilibrium problem and the fixed point problem (1.6). We first introduce a parallel superimposed algorithm. Consequently, strong convergence theorems are shown with some analysis techniques.

2 Preliminaries

Let H be a real Hilbert space with inner product , and norm , respectively. Let C be a nonempty closed convex subset of H.

Definition 2.1 A mapping T:CC is called nonexpansive if

TxTyxy

for all x,yC.

We will use Fix(T) to denote the set of fixed points of T, that is, Fix(T)={xC:x=Tx}.

Definition 2.2 A mapping f:CC is called contractive if

f ( x ) f ( y ) ρxy

for all x,yC and for some constant ρ(0,1). In this case, we call f is a ρ-contraction.

Definition 2.3 A linear bounded operator B:HH is called strongly positive if there exists a constant γ>0 such that

Bx,xγ x 2

for all x,yH.

Definition 2.4 We call P C :HC the metric projection if for each xH

x P C ( x ) =inf { x y : y C } .

It is well known that the metric projection P C :HC is characterized by

x P C ( x ) , y P C ( x ) 0

for all xH, yC. From this, we can deduce that P C is firmly nonexpansive, that is,

P C ( x ) P C ( y ) 2 x y , P C ( x ) P C ( y )
(2.1)

for all x,yH. Hence P C is also nonexpansive.

It is well known that in a real Hilbert space H, the following two equalities hold:

t x + ( 1 t ) y 2 =t x 2 +(1t) y 2 t(1t) x y 2
(2.2)

for all x,yH and t[0,1], and

x + y 2 = x 2 +2x,y+ y 2
(2.3)

for all x,yH. It follows that

x + y 2 x 2 +2y,x+y
(2.4)

for all x,yH.

Throughout this paper, we assume that a bifunction F:C×CR satisfies the following conditions:

  1. (H1)

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

  2. (H2)

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

  3. (H3)

    for each x,y,zC, lim t 0 F(tz+(1t)x,y)F(x,y);

  4. (H4)

    for each xC, yF(x,y) is convex and lower semicontinuous.

Lemma 2.5 ([31])

Let C be a nonempty closed convex subset of a real Hilbert space H. Let F:C×CR be a bifunction which satisfies conditions (H1)-(H4). Let r>0 and xC. Then there exists zC such that

F(z,y)+ 1 r yz,zx0,yC.

Further, if U r F (x)={zC:F(z,y)+ 1 r yz,zx0,yC}, then the following hold:

  1. (i)

    U r F is single-valued and U r F is firmly nonexpansive, i.e., for any x,yH, U r F x U r F y 2 U r F x U r F y,xy;

  2. (ii)

    EP(F) is closed and convex and EP(F)=Fix( U r F ).

Lemma 2.6 ([32])

Let the mapping U r F be defined as in Lemma  2.5. Then, for r,s>0 and x,yH,

U r F ( x ) U s F ( y ) xy+ | s r | s U s F ( y ) y .

Lemma 2.7 ([33])

Let { x n } and { y n } be two 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 =(1 β n ) y n + β n x n

for all n0 and

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

Then lim n y n x n =0.

Lemma 2.8 ([34])

Let C be a closed convex subset of a real Hilbert space H and let S:CC be a nonexpansive mapping. Then the mapping IS is demiclosed. That is, if { x n } is a sequence in C such that x n x weakly and (IS) x n y strongly, then (IS) x =y.

Lemma 2.9 ([35])

Assume that { a n } is a sequence of nonnegative real numbers such that

a n + 1 (1 γ n ) a n + δ n ,nN,

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 a n =0.

3 Main results

In this section, we introduce our algorithm and prove our main results.

Let H 1 and H 2 be two real Hilbert spaces and let C and D be two nonempty closed convex subsets of H 1 and H 2 , respectively. Let A: H 1 H 2 be a bounded linear operator with its adjoint A , B be a strongly positive bounded linear operator on H 1 with coefficient γ>0. Let f:CC be a ρ-contraction and F:C×CR and G:D×DR be two bifunctions satisfying the conditions (H1)-(H4). Let S:DD and T:CC be two nonexpansive mappings.

Algorithm 3.1 Taking x 0 H 1 arbitrarily, we define a sequence { x n } by the following:

x n + 1 = α n σf( x n )+ β n x n + ( ( 1 β n ) I α n B ) T U λ n F ( x n + δ A ( S U γ n G I ) A x n )
(3.1)

for all nN, where { λ n } and { γ n } are two real number sequences in (0,), δ(0, 1 A 2 ) and σ>0 are two constants and { α n } and { β n } are two real number sequences in (0,1).

Theorem 3.2 Suppose Γ and suppose the following conditions hold:

  1. (C1):

    lim n α n =0 and n = 1 α n =;

  2. (C2):

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

  3. (C3):

    lim inf n λ n >0 and lim n λ n + 1 λ n =1;

  4. (C4):

    lim inf n γ n >0 and lim n γ n + 1 γ n =1;

  5. (C5):

    σρ<γ.

Then the sequence { x n } generated by algorithm (3.1) converges strongly to p= Proj Γ (σf+IB)p, which solves the following VI:

( σ f B ) x , y x 0,yΓ.
(3.2)

Proof First, we know that the solution of (3.2) is unique. We denote the unique solution by p. That is, p= Proj Γ (σf+IB)p. Then we have pFix(T)EP(F) and ApFix(S)EP(G). Set z n = U γ n G A x n , y n = x n +δ A (S U γ n G I)A x n and u n = U λ n F ( x n +δ A (S U γ n G I)A x n ) for all nN. Then u n = U λ n F y n . From Lemma 2.5, we know that U λ n F and U γ n G are firmly nonexpansive. By these facts, we have the following conclusions:

z n Ap= U γ n G A x n A p A x n Ap,
(3.3)
u n p= U λ n F y n p y n p
(3.4)

and

S U γ n G A x n A p 2 U γ n G A x n A p 2 A x n A p 2 U γ n G A x n A x n 2 .
(3.5)

Applying Lemma 2.6, we deduce

u n + 1 u n = U λ n + 1 F y n + 1 U λ n F y n y n + 1 y n + | λ n + 1 λ n λ n + 1 | u n + 1 y n + 1
(3.6)

and

z n + 1 z n = U γ n + 1 G A x n + 1 U γ n G A x n A x n + 1 A x n + | γ n + 1 γ n γ n + 1 | z n + 1 A x n .
(3.7)

From (3.1), we have

x n + 1 p = α n ( σ f ( x n ) B p ) + β n ( x n p ) + ( ( 1 β n ) I α n B ) ( T u n p ) α n σ f ( x n ) f ( p ) + α n σ f ( p ) B p + β n x n p + ( 1 β n α n γ ) u n p .
(3.8)

Using (2.3), we get

y n p 2 = x n p + δ A ( S z n A x n ) 2 = x n p 2 + δ 2 A ( S z n A x n ) 2 + 2 δ x n p , A ( S z n A x n ) .
(3.9)

Since A is a linear operator with its adjoint A , we have

x n p , A ( S z n A x n ) = A ( x n p ) , S z n A x n = A x n A p + S z n A x n ( S z n A x n ) , S z n A x n = S z n A p , S z n A x n z n A x n 2 .
(3.10)

Again from (2.3), we obtain

S z n Ap,S z n A x n = 1 2 ( S z n A p 2 + S z n A x n 2 A x n A p 2 ) .
(3.11)

From (3.5), (3.10), and (3.11), we have

x n p , A ( S z n A x n ) = 1 2 ( S z n A p 2 + S z n A x n 2 A x n A p 2 ) S z n A x n 2 1 2 ( A x n A p 2 z n A x n 2 + S z n A x n 2 A x n A p 2 ) S z n A x n 2 = 1 2 z n A x n 2 1 2 S z n A x n 2 .
(3.12)

Substituting (3.12) into (3.9) to deduce

y n p 2 x n p 2 + δ 2 A 2 S z n A x n 2 + 2 δ ( 1 2 z n A x n 2 1 2 S z n A x n 2 ) = x n p 2 + ( δ 2 A 2 δ ) S z n A x n 2 δ z n A x n 2 x n p 2 .

It follows that

y n p x n p.

Thus, from (3.8), we get

x n + 1 p α n σ ρ x n p + α n σ f ( p ) B p + β n x n p + ( 1 β n α n γ ) x n p = [ 1 ( γ σ ρ ) α n ] x n p + α n σ f ( p ) B p max { x n p , σ f ( p ) B p γ σ ρ } .

The boundedness of the sequence { x n } follows.

Next, we estimate u n + 1 u n . Observe that

y n + 1 y n 2 = x n + 1 x n + δ [ A ( S z n + 1 A x n + 1 ) A ( S z n A x n ) ] 2 = x n + 1 x n 2 + δ 2 A ( S z n + 1 A x n + 1 ) A ( S z n A x n ) 2 + 2 δ x n + 1 x n , A [ ( S z n + 1 A x n + 1 ) ( S z n A x n ) ] x n + 1 x n 2 + δ 2 A 2 S z n + 1 S z n ( A x n + 1 A x n ) 2 + 2 δ A x n + 1 A x n , S z n + 1 S z n ( A x n + 1 A x n ) = x n + 1 x n 2 + δ 2 A 2 S z n + 1 S z n ( A x n + 1 A x n ) 2 + 2 δ S z n + 1 S z n , S z n + 1 S z n ( A x n + 1 A x n ) 2 δ S z n + 1 S z n ( A x n + 1 A x n ) 2 = x n + 1 x n 2 + δ 2 A 2 S z n + 1 S z n ( A x n + 1 A x n ) 2 + δ ( S z n + 1 S z n 2 + S z n + 1 S z n ( A x n + 1 A x n ) 2 A x n + 1 A x n 2 ) 2 δ S z n + 1 S z n ( A x n + 1 A x n ) 2 = x n + 1 x n 2 + ( δ 2 A 2 δ ) S z n + 1 S z n ( A x n + 1 A x n ) 2 + δ ( S z n + 1 S z n 2 A x n + 1 A x n 2 ) x n + 1 x n 2 + ( δ 2 A 2 δ ) S z n + 1 S z n ( A x n + 1 A x n ) 2 + δ ( z n + 1 z n 2 A x n + 1 A x n 2 ) .
(3.13)

Since δ(0, 1 A 2 ), we derive by virtue of (3.7) and (3.13) that

y n + 1 y n 2 x n + 1 x n 2 + δ | γ n + 1 γ n γ n + 1 | ( z n + 1 z n + A x n + 1 A x n ) .
(3.14)

According to (3.6) and (3.14), we have

u n + 1 u n 2 = U λ n + 1 F y n + 1 U λ n F y n 2 ( y n + 1 y n + | λ n + 1 λ n λ n + 1 | u n + 1 y n + 1 ) 2 y n + 1 y n 2 + | λ n + 1 λ n λ n + 1 | ( 2 y n + 1 y n u n + 1 y n + 1 + | λ n + 1 λ n λ n + 1 | u n + 1 y n + 1 2 ) x n + 1 x n 2 + | λ n + 1 λ n λ n + 1 | ( 2 y n + 1 y n u n + 1 y n + 1 + | λ n + 1 λ n λ n + 1 | u n + 1 y n + 1 2 ) + δ | γ n + 1 γ n γ n + 1 | ( z n + 1 z n + A x n + 1 A x n ) x n + 1 x n 2 + ( | λ n + 1 λ n λ n + 1 | + δ | γ n + 1 γ n γ n + 1 | ) M ,
(3.15)

where M>0 is a constant such that

sup n { 2 y n + 1 y n u n + 1 y n + 1 + | λ n + 1 λ n λ n + 1 | u n + 1 y n + 1 2 + δ ( z n + 1 z n + A x n + 1 A x n ) } M .

Therefore,

u n + 1 u n x n + 1 x n + ( | λ n + 1 λ n λ n + 1 | + δ | γ n + 1 γ n γ n + 1 | ) M .
(3.16)

From (3.1), we write x n + 1 = β n x n +(1 β n ) w n where w n =T u n + α n 1 β n (σf( x n )BT u n ) for all nN. Then we have

w n + 1 w n = T u n + 1 T u n + α n + 1 1 β n + 1 ( σ f ( x n + 1 ) B T u n + 1 ) α n 1 β n ( σ f ( x n ) B T u n ) T u n + 1 T u n + α n + 1 1 β n + 1 σ f ( x n + 1 ) B T u n + 1 + α n 1 β n σ f ( x n ) B T u n u n + 1 u n + α n + 1 1 β n + 1 σ f ( x n + 1 ) B T u n + 1 + α n 1 β n σ f ( x n ) B T u n x n + 1 x n + ( | λ n + 1 λ n λ n + 1 | + δ | γ n + 1 γ n γ n + 1 | ) M + α n + 1 1 β n + 1 σ f ( x n + 1 ) B T u n + 1 + α n 1 β n σ f ( x n ) B T u n .

Noting the condition (C1) and the boundedness of the sequences { u n + 1 }, { y n + 1 }, { z n + 1 }, {A x n }, {f( x n )}, and {BT u n }, we have

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

By Lemma 2.7, we deduce

lim n x n w n =0.

Hence,

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

Since x n + 1 x n = α n (σf( x n )BT u n )+(1 β n )(T u n x n ), we obtain

T u n x n 1 β n { α n σ f ( x n ) B T u n + x n + 1 x n } .

Thus,

lim n x n T u n =0.
(3.18)

Using the firmly nonexpansiveness of U λ n F , we have

u n p 2 = U λ n F y n p 2 y n p 2 U λ n F y n y n 2 = y n p 2 u n y n 2 = y n p 2 u n x n δ A ( S U γ n G I ) A x n 2 = y n p 2 u n x n 2 δ 2 A ( S U γ n G I ) A x n 2 + 2 δ u n x n , A ( S U γ n G I ) A x n .
(3.19)

Applying (2.4) to (3.1) to deduce

x n + 1 p 2 = α n ( σ f ( x n ) B p ) + β n ( x n T u n ) + ( I α n B ) ( T u n p ) 2 ( I α n B ) ( T u n p ) + β n ( x n T u n ) 2 + 2 α n σ f ( x n ) B p , x n + 1 p [ I α n B T u n p + β n x n T u n ] 2 + 2 α n σ f ( x n ) B p x n + 1 p [ ( 1 α n γ ) u n p + β n x n T u n ] 2 + 2 α n σ f ( x n ) B p x n + 1 p = ( 1 α n γ ) 2 u n p 2 + β n 2 x n T u n 2 + 2 ( 1 α n γ ) β n u n p x n T u n + 2 α n σ f ( x n ) B p x n + 1 p .
(3.20)

It follows from (3.19) that

x n + 1 p 2 x n p 2 u n y n 2 + β n 2 x n T u n 2 + 2 ( 1 α n γ ) β n u n p x n T u n + 2 α n σ f ( x n ) B p x n + 1 p .
(3.21)

Then

u n y n 2 x n p 2 x n + 1 p 2 + β n 2 x n T u n 2 + 2 ( 1 α n γ ) β n u n p x n T u n + 2 α n σ f ( x n ) B p x n + 1 p ( x n p + x n + 1 p ) x n + 1 x n + β n 2 x n T u n 2 + 2 ( 1 α n γ ) β n u n p x n T u n + 2 α n σ f ( x n ) B p x n + 1 p .

This together with (C1), (3.17), and (3.18) implies that

lim n u n y n =0.
(3.22)

From (3.20), we have

x n + 1 p 2 ( 1 α n γ ) 2 u n p 2 + β n 2 x n T u n 2 + 2 ( 1 α n γ ) β n u n p x n T u n + 2 α n σ f ( x n ) B p x n + 1 p y n p 2 + β n 2 x n T u n 2 + 2 ( 1 α n γ ) β n u n p x n T u n + 2 α n σ f ( x n ) B p x n + 1 p x n p 2 + ( δ 2 A 2 δ ) S z n A x n 2 δ z n A x n 2 + β n 2 x n T u n 2 + 2 ( 1 α n γ ) β n u n p x n T u n + 2 α n σ f ( x n ) B p x n + 1 p .

Hence,

( δ δ 2 A 2 ) S z n A x n 2 + δ z n A x n 2 x n p 2 x n + 1 p 2 + β n 2 x n T u n 2 + 2 ( 1 α n γ ) β n u n p x n T u n + 2 α n σ f ( x n ) B p x n + 1 p ( x n p + x n + 1 p ) x n + 1 x n + β n 2 x n T u n 2 + 2 ( 1 α n γ ) β n u n p x n T u n + 2 α n σ f ( x n ) B p x n + 1 p ,

which implies that

lim n S z n A x n = lim n z n A x n =0.

So,

lim n S z n z n =0.
(3.23)

Note that

y n x n = δ A ( S U γ n G I ) A x n δ A S z n A x n .

Therefore,

lim n x n y n =0.
(3.24)

From (3.18), (3.22), and (3.24), we get

lim n x n T x n =0.
(3.25)

Now, we show that lim sup n (σfB)p, x n p0. Choose a subsequence { x n i } of { x n } such that

lim sup n ( σ f B ) p , x n p = lim i ( σ f B ) p , x n i p .
(3.26)

Since the sequence { x n i } is bounded, we can choose a subsequence { x n i j } of { x n i } such that x n i j z. For the sake of convenience, we assume (without loss of generality) that x n i z. Consequently, we derive from the above conclusions that

y n i z, u n i z,A x n i zand z n i Az.
(3.27)

By the demi-closed principle of the nonexpansive mappings S and T (see Lemma 2.8), we deduce zFix(T) and AzFix(S) (according to (3.25) and (3.23), respectively).

Next, we show that zEP(F). Since u n = U λ n F y n , we have

F( u n ,y)+ 1 λ n y u n , u n y n 0,yC.
(3.28)

It follows from the monotonicity of F that

1 λ n y u n , u n y n F(y, u n ),
(3.29)

and hence

y u n i , u n i y n i λ n i F(y, u n i ).
(3.30)

Since u n y n 0, u n i z, and lim inf n λ n >0, we obtain u n i y n i λ n i 0. It follows that 0F(y,z). For t with 0<t1 and yC, let y t =ty+(1t)zC. It follows that F( y t ,z)0. So,

0=F( y t , y t )tF( y t ,y)+(1t)F( y t ,z)tF( y t ,y).
(3.31)

Therefore, 0F( y t ,y). Thus 0F(z,y). This implies that zEP(F). Similarly, we can prove that AzEP(G). To this end, we deduce zFix(T)EP(F) and AzFix(S)EP(G). That is to say, zΓ. Therefore,

lim sup n ( σ f B ) p , x n p = lim i ( σ f B ) p , x n i p = lim i ( σ f B ) p , z p 0 .
(3.32)

Finally, we prove x n p. From (3.1), we have

x n + 1 p 2 = α n ( σ f ( x n ) B p ) + β n ( x n p ) + ( ( 1 β n ) I α n B ) ( T u n p ) , x n + 1 p = α n σ f ( x n ) B p , x n + 1 p + β n x n p , x n + 1 p + ( ( 1 β n ) I α n B ) ( T u n p ) , x n + 1 p α n σ f ( x n ) f ( p ) , x n + 1 p + α n σ f ( p ) B p , x n + 1 p + β n x n p x n + 1 p + ( 1 β n α n γ ) T u n p x n + 1 p [ 1 ( γ σ ρ ) α n ] x n p x n + 1 p + α n σ f ( p ) B p , x n + 1 p 1 ( γ σ ρ ) α n 2 x n p 2 + 1 2 x n + 1 p 2 + α n σ f ( p ) B p , x n + 1 p .
(3.33)

It follows that

x n + 1 p 2 [ 1 ( γ σ ρ ) α n ] x n p 2 +2 α n σ f ( p ) B p , x n + 1 p .
(3.34)

Applying Lemma 2.9 and (3.32) to (3.34), we deduce x n p. The proof is completed. □

Algorithm 3.3 Taking x 0 H 1 arbitrarily, we define a sequence { x n } by the following:

x n + 1 = α n σf( x n )+ β n x n + ( ( 1 β n ) I α n B ) T ( x n + δ A ( S I ) A x n )
(3.35)

for all nN, where δ(0, 1 A 2 ) and σ>0 are two constants and { α n } and { β n } are two real number sequences in (0,1).

Corollary 3.4 Suppose Γ 1 ={xFix(T):AxFix(S)} and suppose the following conditions hold:

  1. (C1):

    lim n α n =0 and n = 1 α n =;

  2. (C2):

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

  3. (C3):

    σρ<γ.

Then the sequence { x n } generated by algorithm (3.35) converges strongly to p= Proj Γ 1 (σf+IB)p, which solves the following VI:

( σ f B ) x , y x 0,y Γ 1 .

Algorithm 3.5 Taking x 0 H 1 arbitrarily, we define a sequence { x n } by the following:

x n + 1 = α n σf( x n )+ β n x n + ( ( 1 β n ) I α n B ) U λ n F ( x n + δ A ( U γ n G I ) A x n )
(3.36)

for all nN, where { λ n } and { γ n } are two real number sequences in (0,), δ(0, 1 A 2 ) and σ>0 are two constants and { α n } and { β n } are two real number sequences in (0,1).

Corollary 3.6 Suppose Γ 2 ={xEP(F):AxEP(G)} and suppose the following conditions hold:

  1. (C1):

    lim n α n =0 and n = 1 α n =;

  2. (C2):

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

  3. (C3):

    lim inf n λ n >0 and lim n λ n + 1 λ n =1;

  4. (C4):

    lim inf n γ n >0 and lim n γ n + 1 γ n =1;

  5. (C5):

    σρ<γ.

Then the sequence { x n } generated by algorithm (3.36) converges strongly to p= Proj Γ 2 (σf+IB)p, which solves the following VI:

( σ f B ) x , y x 0,y Γ 2 .

Algorithm 3.7 Taking x 0 H 1 arbitrarily, we define a sequence { x n } by the following:

x n + 1 = α n σf( x n )+ β n x n + ( ( 1 β n ) I α n B ) P C ( x n + δ A ( P Q I ) A x n )
(3.37)

for all nN, where δ(0, 1 A 2 ) and σ>0 are two constants and { α n } and { β n } are two real number sequences in (0,1).

Corollary 3.8 Suppose Γ 3 ={xC:AxQ} and suppose the following conditions hold:

  1. (C1):

    lim n α n =0 and n = 1 α n =;

  2. (C2):

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

  3. (C3):

    σρ<γ.

Then the sequence { x n } generated by algorithm (3.37) converges strongly to p= Proj Γ 3 (σf+IB)p, which solves the following VI:

( σ f B ) x , y x 0,y Γ 3 .

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Acknowledgements

Li-Jun Zhu was supported in part by NSFC 61362033 and NZ13087. Yeong-Cheng Liou was supported in part by NSC 101-2628-E-230-001-MY3 and NSC 101-2622-E-230-005-CC3.

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Correspondence to Yeong-Cheng Liou.

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All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

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Zhu, L., Yao, Z., Liou, Y. et al. Superimposed algorithms for the split equilibrium problems and fixed point problems. J Inequal Appl 2014, 380 (2014). https://doi.org/10.1186/1029-242X-2014-380

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Keywords

  • split equilibrium problem
  • split fixed point
  • nonexpansive mapping
  • parallel superimposed algorithms