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The Scientific World Journal logoLink to The Scientific World Journal
. 2014 Feb 20;2014:390956. doi: 10.1155/2014/390956

Strong Convergence Algorithm for Split Equilibrium Problems and Hierarchical Fixed Point Problems

Abdellah Bnouhachem 1,2,*
PMCID: PMC3950372  PMID: 24701164

Abstract

The purpose of this paper is to investigate the problem of finding the approximate element of the common set of solutions of a split equilibrium problem and a hierarchical fixed point problem in a real Hilbert space. We establish the strong convergence of the proposed method under some mild conditions. Several special cases are also discussed. Our main result extends and improves some well-known results in the literature.

1. Introduction

Let H be a real Hilbert space, whose inner product and norm are denoted by 〈·, ·〉 and ||·||. Let C be a nonempty closed convex subset of H. We introduce the following definitions which are useful in the following analysis.

Definition 1 —

The mapping T:  CH is said to be

  • (a)
    monotone, if
    TxTy,xy0,x,yC; (1)
  • (b)
    strongly monotone, if there exists α > 0 such that
    TxTy,xyα||xy||2,x,yC; (2)
  • (c)
    α-inverse strongly monotone, if there exists α > 0 such that
    TxTy,xyα||TxTy||2,x,yC; (3)
  • (d)
    nonexpansive, if
    ||TxTy||||xy||,x,yC; (4)
  • (e)
    k-Lipschitz continuous, if there exists a constant k > 0 such that
    ||TxTy||k||xy||,x,yC; (5)
  • (f)
    contraction on C, if there exists a constant 0 ≤ k < 1 such that
    ||TxTy||k||xy||,x,yC. (6)

It is easy to observe that every α-inverse strongly monotone T is monotone and Lipschitz continuous. It is well known that every nonexpansive operator T:  HH satisfies, for all (x, y) ∈ H × H, the inequality

(xT(x))(yT(y)),T(y)T(x)12||(T(x)x)(T(y)y)||2 (7)

and therefore, we get, for all (x, y) ∈ H × Fix⁡(T),

xT(x),yT(x)12||T(x)x||2. (8)

See, for example, [1, Theorem 1], and [2, Theorem 3].

The fixed point problem for the mapping T is to find xC such that

Tx=x. (9)

We denote by F(T) the set of solutions of (9). It is well known that F(T) is closed and convex and P F(T) is well defined (see [3]).

The equilibrium problem denoted by EP is to find xC such that

F(x,y)0,yC. (10)

The solution set of (10) is denoted by EP⁡(F). Numerous problems in physics, optimization, and economics reduce to finding a solution of (10); see [47]. In 1997, Combettes and Hirstoaga [8] introduced an iterative scheme of finding the best approximation to the initial data when EP⁡(F) is nonempty. In 2007, Plubtieng and Punpaeng [6] introduced an iterative method for finding the common element of the set F(T)∩EP⁡(F).

Recently, Censor et al. [9] introduced a new variational inequality problem which we call the split variational inequality problem (SVIP). Let H 1 and H 2 be two real Hilbert spaces. Given operators f:  H 1H 1 and g:  H 2H 2, a bounded linear operator A:  H 1H 2, and nonempty, closed, and convex subsets CH 1 and QH 2, the SVIP is formulated as follows: find a point x* ∈ C such that

f(x),xx0xC (11)

and such that

y=AxQsolves  g(y),yy0yQ. (12)

In [10], Moudafi introduced an iterative method which can be regarded as an extension of the method given by Censor et al. [9] for the following split monotone variational inclusions:

Find  xH1such  that  0f(x)+B1(x) (13)

and such that

y=AxH2solves  0g(y)+B2(y), (14)

where B i: H i → 2Hi is a set-valued mapping for i = 1,2. Later Byrne et al. [11] generalized and extended the work of Censor et al. [9] and Moudafi [10].

Very recently, Kazmi and Rizvi [12] studied the following pair of equilibrium problems called split equilibrium problem: let F 1:  C × CR and F 2: Q × QR be nonlinear bifunctions and let A: H 1H 2 be a bounded linear operator; then, the split equilibrium problem (SEP) is to find x* ∈ C such that

F1(x,x)0,xC, (15)

and such that

y=AxQsolves  F2(y,y)0,yQ. (16)

The solution set of SEP (15)-(16) is denoted by Λ = {pEP⁡(F 1) : ApEP⁡(F 2)}.

Let S: CH be a nonexpansive mapping. The following problem is called a hierarchical fixed point problem: find xF(T) such that

xSx,yx0,yF(T). (17)

It is known that the hierarchical fixed point problem (17) links with some monotone variational inequalities and convex programming problems; see [13, 14]. Various methods [1520] have been proposed to solve the hierarchical fixed point problem. In 2010, Yao et al. [14] introduced the following strong convergence iterative algorithm to solve the problem (17):

yn=βnSxn+(1βn)xn,xn+1=PC[αnf(xn)+(1αn)Tyn],n0, (18)

where f:  CH is a contraction mapping and {α n} and {β n} are two sequences in (0,1). Under some certain restrictions on parameters, Yao et al. proved that the sequence {x n} generated by (18) converges strongly to zF(T), which is the unique solution of the following variational inequality:

(If)z,yz0,yF(T). (19)

In 2011, Ceng et al. [21] investigated the following iterative method:

xn+1=PC[αnρU(xn)+(IαnμF)(T(yn))],n0, (20)

where U is a Lipschitzian mapping and F is a Lipschitzian and strongly monotone mapping. They proved that under some approximate assumptions on the operators and parameters, the sequence {x n} generated by (20) converges strongly to the unique solution of the variational inequality

ρU(z)μF(z),xz0,x  Fix(T). (21)

In the present paper, inspired by the above cited works and by the recent works going in this direction, we give an iterative method for finding the approximate element of the common set of solutions of (15)-(16) and (17) in real Hilbert space. Strong convergence of the iterative algorithm is obtained in the framework of Hilbert space. We would like to mention that our proposed method is quite general and flexible and includes many known results for solving split equilibrium problems and hierarchical fixed point problems; see, for example, [13, 14, 1719, 2123] and relevant references cited therein.

2. Preliminaries

In this section, we recall some basic definitions and properties, which will be frequently used in our later analysis. Some useful results proved already in the literature are also summarized. The first lemma provides some basic properties of projection onto C.

Lemma 2 —

Let P C denote the projection of H onto C. Then, one has the following inequalities:

zPC[z],PC[z]v0,zH,vC;uv,PC[u]PC[v]||PC[u]PC[v]||2,u,vH;||PC[u]PC[v]||||uv||,u,vH;||uPC[z]||2||zu||2||zPC[z]||2,zH,uC. (22)

Assumption 3 (see [24]) —

Let F:  C × C → ℝ be a bifunction satisfying the following assumptions:

  • (i)

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

  • (ii)

    F is monotone; that is, F(x, y) + F(y, x) ≤ 0, for all x, yC;

  • (iii)

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

  • (iv)

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

  • (v)
    for fixed r > 0 and zC, there exists a bounded subset K of H 1 and xCK such that
    F(y,x)+1ryx,xz0,yCK. (23)

Lemma 4 (see [8]) —

Assume that F 1: C × C → ℝ satisfies Assumption 3. For r > 0 and for all xH 1, define a mapping T r F1: H 1C as follows:

TrF1(x)={zC:F1(z,y)+1ryz,zx0,yC}. (24)

Then the following hold:

  • (i)

    T r F1 is nonempty and single-valued;

  • (ii)
    T r F1 is firmly nonexpansive; that is,
    ||TrF1(x)TrF1(y)||2TrF1(x)TrF1(y),xy,x,yH1; (25)
  • (iii)

    F(T r F1) = EP(F 1);

  • (iv)

    EP(F 1) is closed and convex.

Assume that F 2:  Q × Q → ℝ satisfies Assumption 3. For s > 0 and for all uH 2, define a mapping T s F2:  H 2Q as follows:

TsF2(u)={vQ:F2(v,w)+1swv,vu0,wQ}. (26)

Then T s F2 satisfies conditions (i)–(iv) of Lemma 4. Consider F(T s F2) = EP⁡(F 2, Q), where EP⁡(F 2, Q) is the solution set of the following equilibrium problem:

find  yQsuch  that  F2(y,y)0,yQ. (27)

Lemma 5 (see [25]) —

Assume that F 1:  C × C → ℝ satisfies Assumption 3, and let T r F1 be defined as in Lemma 4. Let x, yH 1 and r 1, r 2 > 0. Then

||Tr2F1(y)Tr1F1(x)||||yx||+|r2r1r2|||Tr2F1(y)y||. (28)

Lemma 6 (see [26]) —

Let C be a nonempty closed convex subset of a real Hilbert space H. If  T:  CC is a nonexpansive mapping with Fix⁡(T) ≠ , then the mapping IT is demiclosed at 0; that is, if {x n} is a sequence in C weakly converging to x and if {(IT)x n} converges strongly to 0, then (IT)x = 0.

Lemma 7 (see [21]) —

Let U:  CH be τ-Lipschitzian mapping and let F:  CH be a k-Lipschitzian and η-strongly monotone mapping; then for 0 ≤ ρτ < μη, μFρU is μηρτ-strongly monotone; that is,

(μFρU)x(μFρU)y,xy(μηρτ)||xy||2,x,yC. (29)

Lemma 8 (see [27]) —

Suppose that λ ∈ (0,1) and μ > 0. Let F:  CH be an k-Lipschitzian and η-strongly monotone operator. In association with nonexpansive mapping T:  CC, define the mapping T λ:  CH by

Tλx=TxλμFT(x),xC. (30)

Then T λ is a contraction provided that μ < (2η/k 2); that is,

||TλxTλy||(1λν)||xy||,x,yC, (31)

where ν=1-1-μ(2η-μL2).

Lemma 9 (see [28]) —

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

an+1(1γn)an+δn, (32)

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

  1. n=1 γ n = ;

  2. limsup⁡n δ n/γ n ≤ 0 or ∑n=1 |δ n| < .

Then lim⁡n a n = 0.

Lemma 10 (see [29]) —

Let C be a closed convex subset of H. Let {x n} be a bounded sequence in H. Assume that

  1. the weak w-limit set w w(x n) ⊂ C where w w(x n) = {x:  x nix};

  2. for each zC, lim⁡n||x nz|| exists.

Then {x n} is weakly convergent to a point in C.

3. The Proposed Method and Some Properties

In this section, we suggest and analyze our method and we prove a strong convergence theorem for finding the common solutions of the split equilibrium problem (15)-(16) and the hierarchical fixed point problem (17).

Let H 1 and H 2 be two real Hilbert spaces and let CH 1 and QH 2 be nonempty closed convex subsets of Hilbert spaces H 1 and H 2, respectively. Let A:  H 1H 2 be a bounded linear operator. Assume that F 1: C × C → ℝ and F 2: Q × Q → ℝ are the bifunctions satisfying Assumption 3 and F 2 is upper semicontinuous in first argument. Let S, T:  CC be a nonexpansive mapping such that Λ∩F(T) ≠ . Let F:  CC be an k-Lipschitzian mapping and η-strongly monotone and let U:  CC be τ-Lipschitzian mapping. Now we introduce the proposed method as follows.

Algorithm 11 —

For a given x 0C arbitrarily, let the iterative sequences {u n}, {x n}, and {y n} be generated by

un=TrnF1(xn+γA(TrnF2I)Axn);yn=βnSxn+(1βn)un;xn+1=PC[αnρU(xn)+(IαnμF)(T(yn))],n0, (33)

where {r n}⊂(0,2ς) and γ ∈ (0,1/L), L is the spectral radius of the operator A*A, and A* is the adjoint of A. Suppose that the parameters satisfy 0 < μ < (2η/k 2), 0 ≤ ρτ < ν, where ν=1-1-μ(2η-μk2). And {α n} and {β n} are sequences in (0,1) satisfying the following conditions:

  1. lim⁡n α n = 0 and ∑n=1 α n = ;

  2. lim⁡n(β n/α n) = 0;

  3. n=1 | α n−1α n | < and ∑n=1 |β n−1β n| < ;

  4. liminf⁡n r n < limsup⁡n r n < 2ς and ∑n=1 |r n−1r n| < .

Remark 12 —

Our method can be viewed as extension and improvement for some well-known results as follows

  1. The proposed method is an extension and improvement of the method of Wang and Xu [23] for finding the approximate element of the common set of solutions of a split equilibrium problem and a hierarchical fixed point problem in a real Hilbert space.

  2. If the Lipschitzian mapping U = f, F = I, ρ = μ = 1, we obtain an extension and improvement of the method of Yao et al. [14] for finding the approximate element of the common set of solutions of a split equilibrium problem and a hierarchical fixed point problem in a real Hilbert space.

  3. The contractive mapping f with a coefficient α ∈ [0,1) in other papers (see [14, 19, 22, 27]) is extended to the cases of the Lipschitzian mapping U with a coefficient constant γ ∈ [0, ).

This shows that Algorithm 11 is quite general and unifying.

Lemma 13 —

Let x* ∈ Λ∩F(T). Then {x n},  {u n}, and {y n} are bounded.

Proof —

Let x* ∈ Λ∩F(T); we have x* = T rn F1(x*) and Ax* = T rn F2(Ax*). Then

||unx||2=||TrnF1(xn+γA(TrnF2I)Axn)x||2=||TrnF1(xn+γA(TrnF2I)Axn)TrnF1(x)||2||xn+γA(TrnF2I)Axnx||2=||xnx||2+γ2||A(TrnF2I)Axn||2+2γxnx,A(TrnF2I)Axn=||xnx||2+γ2(TrnF2I)Axn,AA(TrnF2I)Axn+2γxnx,A(TrnF2I)Axn. (34)

From the definition of L, it follows that

γ2(TrnF2I)Axn,AA(TrnF2I)AxnLγ2(TrnF2I)Axn,(TrnF2I)Axn=Lγ2||(TrnF2I)Axn||2. (35)

It follows from (8) that

2γxnx,A(TrnF2I)Axn=2γA(xnx),(TrnF2I)Axn=2γA(xnx)+(TrnF2I)Axn(TrnF2I)Axn,(TrnF2I)Axn=2γ(TrnF2AxnAx,(TrnF2I)Axn||(TrnF2I)Axn||2)2γ(12||(TrnF2I)Axn||2||(TrnF2I)Axn||2)=γ||(TrnF2I)Axn||2. (36)

Applying (36) and (35) to (34) and from the definition of γ, we get

||unx||2||xnx||2+γ(Lγ1)||(TrnF2I)Axn||2||xnx||2. (37)

Denote V n = α n ρU(x n)+(Iα n μF)(T(y n)). Next, we prove that the sequence {x n} is bounded; without loss of generality we can assume that β nα n for all n ≥ 1. From (33), we have

||xn+1x||=||PC[Vn]PC[x]||||αnρU(xn)+(IαnμF)(T(yn))x||αn||ρU(xn)μF(x)||+||(IαnμF)(T(yn))(IαnμF)T(x)||=αn||ρU(xn)ρU(x)+(ρUμF)(x)||+||(IαnμF)(T(yn))(IαnμF)T(x)||αnρτ||xnx||+αn||(ρUμF)(x)||+(1αnν)||ynx||αnρτ||xnx||+αn||(ρUμF)(x)||+(1αnν)||βnSxn+(1βn)unx||αnρτ||xnx||+αn||(ρUμF)(x)||+(1αnν)(βn||SxnSx||+βn||Sxx||+(1βn)||unx||)αnρτ||xnx||+αn||(ρUμF)(x)||+(1αnν)(βn||SxnSx||+βn||Sxx||+(1βn)||xnx||)αnρτ||xnx||+αn||(ρUμF)(x)||+(1αnν)(βn||xnx||+βn||Sxx||+(1βn)||xnx||)=(1αn(νρτ))||xnx||+αn||(ρUμF)(x)||+(1αnν)βn||Sxx||(1αn(νρτ))||xnx||+αn||(ρUμF)(x)||+βn||Sxx||(1αn(νρτ))||xnx||+αn(||(ρUμF)(x)||+||Sxx||)=(1αn(νρτ))||xnx||+αn(νρτ)νρτ(||(ρUμF)x||+||Sxx||)  max{||xnx||,1νρτ×(||(ρUμF)(x)||+||Sxx||)}, (38)

where the third inequality follows from Lemma 8.

By induction on n, we obtain ||x nx*|| ≤ max⁡{||x 0x*||, (1/(1 − ρ))(||(ρUμF)x*|| + ||Sx* − x*||)}, for n ≥ 0 and x 0C. Hence {x n} is bounded and, consequently, we deduce that {u n}, {y n}, {S(x n)}, {T(x n)}, {F(T(y n))}, and {U(x n)} are bounded.

Lemma 14 —

Let x* ∈ Λ∩F(T) and {x n} the sequence generated by the Algorithm 11. Then one has

  1. lim⁡n||x n+1x n|| = 0;

  2. the weak w-limit set w w(x n) ⊂ F(T), (w w(x n) = {x : x nix}).

Proof —

Since u n = T rn F1(x n + γA*(T rn F2I)Ax n) and u n−1 = T rn−1 F1(x n−1 + γA*(T rn−1 F2I)Ax n−1)  it follows from Lemma 5 that

||unun1||||xnxn1+γ(A(TrnF2I)AxnA(Trn1F2I)Axn1)||+|1rn1rn|||TrnF1(xn+γA(TrnF2I)Axn)(xn+γA(TrnF2I)Axn)||||xnxn1γAA(xnxn1)||+γ||A||||TrnF2AxnTrn1F2Axn1||+|1rn1rn|||TrnF1(xn+γA(TrnF2I)Axn)(xn+γA(TrnF2I)Axn)||(||xnxn1||22γ||A(xnxn1)||2+γ2||A||4||xnxn1||2)1/2+γ||A||(||A(xnxn1)||+|1rn1rn|||TrnF2AxnAxn||)+|1rn1rn|||TrnF1(xn+γA(TrnF2I)Axn)(xn+γA(TrnF2I)Axn)||(12γ||A||2+γ2||A||4)1/2||xnxn1||+γ||A||2||xnxn1||+γ||A|||1rn1rn|||TrnF2AxnAxn||+|1rn1rn|||TrnF1(xn+γA(TrnF2I)Axn)(xn+γA(TrnF2I)Axn)||=(1γ||A||2)||xnxn1||+γ||A||2||xnxn1||+γ||A|||1rn1rn|||TrnF2AxnAxn||+|1rn1rn|||TrnF1(xn+γA(TrnF2I)Axn)(xn+γA(TrnF2I)Axn)||=||xnxn1||+γ||A|||1rn1rn|||TrnF2AxnAxn||+|1rn1rn|||TrnF1(xn+γA(TrnF2I)Axn)(xn+γA(TrnF2I)Axn)||=||xnxn1||+|rnrn1rn|(γ||A||σn+χn), (39)

where σ n : = ||T rn F2 Ax nAx n|| and χ n : = ||T rn F1(x n + γA*(T rn F2I)Ax n) − (x n + γA*(T rn F2I)Ax n)||. Without loss of generality, let us assume that there exists a real number μ such that r n > μ > 0, for all positive integers n. Then we get

||un1un||||xn1xn||+1μ|rn1rn|(γ||A||σn+χn). (40)

From (33) and the above inequality, we get

||ynyn1||=||βnSxn+(1βn)un(βn1Sxn1+(1βn1)un1)||=||βn(SxnSxn1)+(βnβn1)Sxn1+(1βn)(unun1)+(βn1βn)un1||βn||xnxn1||+(1βn)||unun1||+|βnβn1|(||Sxn1||+||un1||)βn||xnxn1||+(1βn)×{||xn1xn||+1μ|rn1rn|(γ||A||σn+χn)}+|βnβn1|(||Sxn1||+||un1||)||xnxn1||+1μ|rn1rn|  (γ||A||σn+χn)+|βnβn1|(||Sxn1||+||un1||). (41)

Next, we estimate

||xn+1xn||=||PC[Vn]PC[Vn1]||||αnρ(U(xn)U(xn1))+(αnαn1)ρU(xn1)+(IαnμF)(T(yn))(IαnμF)T(yn1)+(IαnμF)(T(yn1))(Iαn1μF)(T(yn1))||αnρτ||xnxn1||+(1αnν)||ynyn1||+|αnαn1|(||ρU(xn1)||+||μF(T(yn1))||), (42)

where the second inequality follows from Lemma 8. From (41) and (42), we have

||xn+1xn||αnρτ||xnxn1||+(1αnν)×{||xnxn1||+1μ|rn1rn|(γ||A||σn+χn)+|βnβn1|(||Sxn1||+||un1||)}+|αnαn1|(||ρU(xn1)||+||μF(T(yn1))||)(1(νρτ)αn)||xnxn1||+1μ|rn1rn|(γ||A||σn+χn)+|βnβn1|(||Sxn1||+||un1||)+|αnαn1|(||ρU(xn1)||+||μF(T(yn1))||)(1(νρτ)αn)||xnxn1||+M(1μ|rn1rn|+|βnβn1|+|αnαn1|), (43)

where

M=  max{supn1(γ||A||σn+χn),supn1(||Sxn1||+||un1||),supn1(||ρU(xn1)||+||μF(T(yn1))||)}. (44)

It follows from conditions (a)–(d) of Algorithm 11 and Lemma 9 that

limn||xn+1xn||=0. (45)

Next, we show that lim⁡n||u nx n|| = 0. Since x* ∈ Λ∩F(T) by using (34) and (37), we obtain

||xn+1x||2=PC[Vn]x,xn+1x=PC[Vn]Vn,PC[Vn]x+Vnx,xn+1xαn(ρU(xn)μF(x)+(IαnμF)(T(yn)))(IαnμF)(T(x)),xn+1x=αnρ(U(xn)U(x)),xn+1x+αnρU(x)μF(x),xn+1x+(IαnμF)(T(yn))(IαnμF)(T(x)),xn+1xαnρτ||xnx||||xn+1x||+αnρU(x)μF(x),xn+1x+(1αnν)||ynx||||xn+1x||αnρτ2(||xnx||2+||xn+1x||2)+αnρU(x)μF(x),xn+1x+(1αnν)2(||ynx||2+||xn+1x||2)(1αn(νρτ))2||xn+1x||2+αnρτ2||xnx||2+αnρU(x)μF(x),xn+1x+(1αnν)2(βn||Sxnx||2+(1βn)||unx||2)(1αn(νρτ))2||xn+1x||2+αnρτ2||xnx||2+αnρU(x)μF(x),xn+1x+(1αnν)βn2||Sxnx||2+(1αnν)(1βn)2×{||xnx||2+γ(Lγ1)||(TrnF2I)Axn||2}, (46)

where the last inequality follows from (37), which implies that

||xn+1x||2αnρτ1+αn(νρτ)||xnx||2+2αn1+αn(νρτ)ρU(x)μF(x),xn+1x+(1αnν)βn1+αn(νρτ)||Sxnx||2+(1αnν)(1βn)1+αn(νρτ)×{||xnx||2+γ(Lγ1)||(TrnF2I)Axn||2}αnρτ1+αn(νρτ)||xnx||2+2αn1+αn(νρτ)ρU(x)μF(x),xn+1x+||xnx||2+(1αnν)βn1+αn(νρτ)||Sxnx||2(1αnν)(1βn)γ(1Lγ)1+αn(νρτ)||(TrnF2I)Axn||2. (47)

Then from the above inequality, we get

(1αnν)(1βn)γ(1Lγ)1+αn(νρτ)||(TrnF2I)Axn||2αnρτ1+αn(νρτ)||xnx||2+2αn1+αn(νρτ)×ρU(x)μF(x),xn+1x+βn||Sxnx||2+||xnx||2||xn+1x||2αnρτ1+αn(νρτ)||xnx||2+2αn1+αn(νρτ)×ρU(x)μF(x),xn+1x+βn||Sxnx||2+(||xnx||+||xn+1x||)×||xn+1xn||. (48)

Since γ(1 − ) > 0, lim⁡n||x n+1x n|| = 0, α n → 0, and β n → 0, we obtain

limn||(TrnF2I)Axn||=0. (49)

Since T rn F1 is firmly nonexpansive, we have

||unx||2=||TrnF1(xn+γA(TrnF2I)Axn)TrnF1(x)||2unx,xn+γA(TrnF2I)Axnx=12{||unx||2+||xn+γA(TrnF2I)Axnx||2||unx[xn+γA(TrnF2I)Axnx]||2}=12{||unx||2+||xn+γA(TrnF2I)Axnx||2||unxnγA(TrnF2I)Axn||2}12{||unx||2+||xnx||2||unxnγA(TrnF2I)Axn||2}=12{||unx||2+||xnx||2[||unxn||2+γ2||A(TrnF2I)Axn||22γunxn,A(TrnF2I)Axn]}, (50)

where the last inequality follows from (34) and (37). Hence, we get

||unx||2||xnx||2||unxn||2+2γ||AunAxn||||(TrnF2I)Axn||. (51)

From (46) and the above inequality, we have

||xn+1x||2(1αn(νρτ))2||xn+1x||2+αnρτ2||xnx||2+αnρU(x)μF(x),xn+1x+(1αnν)2(βn||Sxnx||2+(1βn)||unx||2)(1αn(νρτ))2||xn+1x||2+αnρτ2||xnx||2+αnρU(x)μF(x),xn+1x+(1αnν)2×{βn||Sxnx||2+(1βn)×(||xnx||2||unxn||2+2γ||AunAxn||||(TrnF2I)Axn||)}, (52)

which implies that

||xn+1x||2αnρτ1+αn(νρτ)||xnx||2+2αn1+αn(νρτ)×ρU(x)μF(x),xn+1x+(1αnν)βn1+αn(νρτ)||Sxnx||2+(1αnν)(1βn)1+αn(νρτ)×{||xnx||2||unxn||2+2γ||AunAxn||||(TrnF2I)Axn||}αnρτ1+αn(νρτ)||xnx||2+2αn1+αn(νρτ)×ρU(x)μF(x),xn+1x+(1αnν)βn1+αn(νρτ)||Sxnx||2+||xnx||2+(1αnν)(1βn)1+αn(νρτ)×{||unxn||2+2γ||AunAxn||||(TrnF2I)Axn||}. (53)

Hence

(1αnν)(1βn)1+αn(νρτ)||unxn||2αnρτ1+αn(νρτ)||xnx||2+2αn1+αn(νρτ)ρU(x)μF(x),xn+1x+(1αnν)βn1+αn(νρτ)||Sxnx||2+2(1αnν)(1βn)γ1+αn(νρτ)||AunAxn||||(TrnF2I)Axn||+||xnx||2||xn+1x||2αnρτ1+αn(νρτ)||xnx||2+2αn1+αn(νρτ)ρU(x)μF(x),xn+1x+(1αnν)βn1+αn(νρτ)||Sxnx||2+2(1αnν)(1βn)γ1+αn(νρτ)||AunAxn||||(TrnF2I)Axn||+(||xnx||+||xn+1x||)||xn+1xn||. (54)

Since lim⁡n||x n+1x n|| = 0, α n → 0, β n → 0, and lim⁡n||(T rn F2I)Ax n|| = 0, we obtain

limn||unxn||=0. (55)

Now, let z ∈ Λ∩F(T); since T(x n) ∈ C, we have

||xnT(xn)||||xnxn+1||+||xn+1T(xn)||=||xnxn+1||+||PC[Vn]PC[T(xn)]||||xnxn+1||+||αn(ρU(xn)μF(T(yn))+T(yn)T(xn))||||xnxn+1||+αn||ρU(xn)μF(T(yn))||+||ynxn||||xnxn+1||+αn||ρU(xn)μF(T(yn))||+||βnSxn+(1βn)unxn||||xnxn+1||+αn||ρU(xn)μF(T(yn))||+βn||Sxnxn||+(1βn)||unxn||. (56)

Since lim⁡n||x n+1x n|| = 0,  α n → 0,  β n → 0,  ||ρU(x n) − μF(T(y n))||, and ||Sx nx n|| are bounded and lim⁡n||x nu n|| = 0, we obtain

limn||xnT(xn)||=0. (57)

Since {x n} is bounded, without loss of generality, we can assume that x nx* ∈ C. It follows from Lemma 6 that x* ∈ F(T). Therefore w w(x n) ⊂ F(T).

Theorem 15 —

The sequence {x n} generated by Algorithm 11 converges strongly to z, which is the unique solution of the variational inequality

ρU(z)μF(z),xz0,xΛF(T). (58)

Proof —

Since {x n} is bounded x nw and from Lemma 14, we have wF(T). Next, we show that wEP⁡(F 1). Since u n = T rn F1(x n + γA*(T rn F2I)Ax n), we have

F1(un,y)+1rnyun,unxn1rnyun,γA(TrnF2I)Axn0,yC. (59)

It follows from monotonicity of F 1 that

1rnyun,γA(TrnF2I)Axn+1rnyun,unxnF1(y,un),yC, (60)
1rnkyunk,γA(TrnkF2I)Axnk+yunk,unkxnkrnkF1(y,unk),yC. (61)

Since lim⁡n||u nx n|| = 0, lim⁡n||(T rn F2I)Ax n|| = 0, and x nw, it easy to observe that u nkw. It follows by Assumption 3(iv) that F 1(y, w) ≤ 0, for all yC.

For any 0 < t ≤ 1 and yC, let y t = ty + (1 − t)w; we have y tC. Then, from Assumptions 3((i) and (iv)), we have

0=F1(yt,yt)tF1(yt,y)+(1t)F1(yt,w)tF1(yt,y). (62)

Therefore F 1(y t, y) ≥ 0. From Assumption 3(iii), we have F 1(w, y) ≥ 0, which implies that wEP⁡(F 1).

Next, we show that AwEP⁡(F 2). Since {x n} is bounded and x nw, there exists a subsequence {x nk} of  {x n} such that x nkw and since A is a bounded linear operator, Ax nkAw. Now set v nk = Ax nkT rnk F2 Ax nk. It follows from (49) that lim⁡k v nk = 0 and Ax nkv nk = T rnk F2 Ax nk. Therefore from the definition of T rnk F2, we have

F2(Axnkvnk,y)+1rnky(Axnkvnk),(Axnkvnk)Axnk0,yC.   (63)

Since F 2 is upper semicontinuous in first argument, taking lim sup to above inequality as k and using Assumption 3(iv), we obtain

F2(Aw,y)0yC, (64)

which implies that AwEP⁡(F 2) and hence w ∈ Λ.

Thus we have

wΛF(T). (65)

Observe that the constants satisfy 0 ≤ ρτ < ν and

kηk2η212μη+μ2k212μη+μ2η21μ(2ημk2)1μημη11μ(2ημk2)μην. (66)

Therefore from Lemma 7, the operator μFρU is μηρτ strongly monotone, and we get the uniqueness of the solution of the variational inequality (58) and denote it by z ∈ Λ  ∩  F(T).

Next, we claim that lim⁡sup⁡nρU(z) − μF(z), x nz〉≤0. Since {x n} is bounded, there exists a subsequence {x nk} of {x n} such that

limsupnρU(z)μF(z),xnz=limsupkρU(z)μF(z),xnkz=ρU(z)μF(z),wz0. (67)

Next, we show that x nz. Consider

||xn+1z||2=PC[Vn]z,xn+1z=PC[Vn]Vn,PC[Vn]z+Vnz,xn+1zαn(ρU(xn)μF(z))+(IαnμF)(T(yn))(IαnμF)(T(z)),xn+1zαnρ(U(xn)U(z)),xn+1z+αnρU(z)μF(z),xn+1z+(IαnμF)(T(yn))(IαnμF)(T(z)),xn+1zαnρτ||xnz||||xn+1z||+αnρU(z)μF(z),xn+1z+(1αnν)||ynz||||xn+1z||αnρτ||xnz||||xn+1z||+αnρU(z)μF(z),xn+1z+(1αnν){βn||SxnSz||+βn||Szz||+(1βn)||unz||}||xn+1z||αnρτ||xnz||||xn+1z||+αnρU(z)μF(z),xn+1z+(1αnν){βn||xnz||+βn||Szz||+(1βn)||xnz||}||xn+1z||=(1αn(νρτ))||xnz||||xn+1z||+αnρU(z)μF(z),xn+1z+(1αnν)βn||Szz||||xn+1z||1αn(νρτ)2(||xnz||2+||xn+1z||2)+αnρU(z)μF(z),xn+1z+(1αnν)βn||Szz||||xn+1z|| (68)

which implies that

||xn+1z||21αn(νρτ)1+αn(νρτ)||xnz||2+2αn1+αn(νρτ)ρU(xn)μF(z),xn+1z+2(1αnν)βn1+αn(νρτ)||Szz||||xn+1z||(1αn(νρτ))||xnz||2+2αn(νρτ)1+αn(νρτ)×{1νρτρU(z)μF(z),xn+1z+(1αnν)βnαn(νρτ)||Szz||||xn+1z||}. (69)

Let γ n = α n(νρτ) and δ n = (2α n(νρτ)/(1 + α n(νρτ))){(1/(νρτ))〈ρU(z) − μF(z), x n+1z〉 + ((1 − α n ν)β n/α n(νρτ))||Szz||||x n+1z||}.

Since

n=1αn=,limsupn{1νρτρU(z)μF(z),xn+1z+(1αnν)βnαn(νρτ)||Szz||||xn+1z||}0. (70)

It follows that

n=1γn=,limsupnδnγn0. (71)

Thus all the conditions of Lemma 9 are satisfied. Hence we deduce that x nz. This completes the proof.

Remark 16 —

In hierarchical fixed point problem (17), if S = I − (ρUμF), then we can get the variational inequality (58). In (58), if U = 0, then we get the variational inequality 〈F(z), xz〉 ≥ 0, for all x ∈ Λ∩F(T), which is just the variational inequality studied by Suzuki [27] extending the common set of solutions of a system of variational inequalities, a split equilibrium problem, and a hierarchical fixed point problem.

4. Conclusions

In this paper, we suggest and analyze an iterative method for finding the approximate element of the common set of solutions of (15)-(16) and (17) in real Hilbert space, which can be viewed as a refinement and improvement of some existing methods for solving a split equilibrium problem and a hierarchical fixed point problem. Some existing methods (e.g., [13, 14, 1719, 2123]) can be viewed as special cases of Algorithm 11. Therefore, the new algorithm is expected to be widely applicable.

Acknowledgment

The author would like to thank Professor Omar Halli, Rector, Ibn Zohr University, for providing excellent research facilities.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

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