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Iterative methods for hierarchical common fixed point problems and variationalinequalities

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Abstract

The purpose of this paper is to deal with the problem of finding hierarchically acommon fixed point of a sequence of nearly nonexpansive self-mappings defined ona closed convex subset of a real Hilbert space which is also a solution of someparticular variational inequality problem. We introduce two explicit iterativeschemes and establish strong convergence results for sequences generatediteratively by the explicit schemes under suitable conditions. Our strongconvergence results include the previous results as special cases, and can beviewed as an improvement and refinement of several corresponding known resultsfor hierarchical variational inequality problems.

MSC: 47J20, 49J40.

1 Introduction

Variational inequality problems were initially studied by Stampacchia [1] in 1964. Since then, many kinds of variational inequalities have beenextended and generalized in several directions using novel and innovative techniques(see [25] and the references therein).

The classical variational inequality problem or the Stampacchia variationalinequality problem in a Hilbert space is defined as follows:

Let C be a nonempty closed convex subset of a real Hilbert space H,and let T:CH be a nonlinear mapping. Then the classicalvariational inequality problem is a problem of finding x C such that

T x , y x 0for all yC.
(1.1)

Problem (1.1) is denoted by VI(C,T). A point x C is a solution of VI(C,T) if and only if x is a fixed point of P C (IλT), where λ>0 is a constant, I is the identity mappingfrom C into itself, and P C is the metric projection from H onto aclosed convex subset C of H. The set of solutions of (1.1) isdenoted by Ω(C,T), that is,

Ω(C,T)= { x C : T x , y x 0  for all  y C } .

The variational inequality VI(C,T) is called a monotone variational inequality if theoperator T is a monotone operator. The problem of kind (1.1) is connectedwith the convex minimization problem, complementarity problem, saddle point problem,fixed point problem, Nash equilibrium problem, the problem of finding a pointxH satisfying Tx=0 and so on, and it has several applications indifferent branches of natural sciences, social sciences, management and engineering(see [611] and the references therein). In this context, we discuss the variationalinequality problem over the set of fixed points of a mapping which is known as ahierarchical variational inequality problem or a hierarchical fixed point problem.

The hierarchical variational inequality problem over the set of fixed points of anonexpansive mapping is defined as follows:

Let C be a nonempty closed convex subset of a real Hilbert space H,and let T,S:CC be two nonexpansive mappings. Then the hierarchicalvariational inequality problem is given as follows:

Find  x F(S) such that  ( I T ) x , y x 0 for all yF(S).
(1.2)

Problem (1.2) is denoted by VI F ( S ) (C,T). The set of solutions of (1.2) is denoted by Ω F ( S ) (C,T), that is,

Ω F ( S ) (C,T)= { x F ( S ) : ( I T ) x , y x 0  for all  y F ( S ) } .

It is easy to observe that VI F ( S ) (C,T) is equivalent to the fixed point problem x = P F ( S ) T x , that is, x is a fixed point of the nonexpansive mapping P F ( S ) T. Of course, if T=I, then the set of solutions of VI F ( S ) (C,T) is Ω F ( S ) (C,T)=F(S).

Firstly, Moudafi and Maingé [12] introduced an implicit iterative algorithm to solve problem (1.2) andproved weak and strong convergence results, and after that, many iterative methodshave been developed for solving hierarchical problem (1.2) by several authors (see,e.g., [1317]).

Very recently, Gu et al.[18] motivated and inspired by the results of Marino and Xu [15], Yao et al.[17] introduced and studied two iterative schemes for solving hierarchicalvariational inequality problem and proved the corresponding strong convergenceresults for the generated sequences in the context of a countable family ofnonexpansive mappings under suitable conditions on parameters.

In this paper, inspired by Gu et al.[18] and Sahu et al.[19], we introduce two explicit iterative schemes which generate sequences viaiterative algorithms. We prove that the generated sequences converge strongly to theunique solutions of particular variational inequality problems defined over the setof common fixed points of a sequence of nearly nonexpansive mappings. Our results,in one sense, extend the results of Gu et al.[18] to the sequence of nonexpansive mappings and, in another sense, to thesequence of nearly nonexpansive mappings, which is a wider class of sequence ofnonexpansive mappings. Our results also generalize the results of Cianciaruso etal.[13], Yao et al.[17], Moudafi [20], Xu [21] and many other related works.

2 Preliminaries

Let C be a nonempty subset of a real Hilbert space H with the innerproduct , and the norm , respectively. A mapping T:CH is called

  1. (1)

    monotone if

    TxTy,xy0for all x,yC,
  2. (2)

    η-strongly monotone if there exists a positive real number η such that

    TxTy,xyη x y 2 for all x,yC,
  3. (3)

    k-Lipschitzian if there exists a constant k>0 such that

    TxTykxyfor all x,yC,
  4. (4)

    ρ-contraction if there exists a constant ρ(0,1) such that

    TxTyρxyfor all x,yC,
  5. (5)

    nonexpansive if

    TxTyxyfor all x,yC,
  6. (6)

    nearly nonexpansive [22, 23] with respect to a fixed sequence { a n } in [0,) with a n 0 if

    T n x T n y xy+ a n for all x,yC and nN.

Throughout this paper, we denote by I the identity mapping of H.Also, we denote by → and the strong convergence and weak convergence,respectively. The symbol stands for the set of all natural numbers and ω w ({ x n }) denotes the set of all weak subsequential limits of{ x n }.

Let C be a nonempty closed convex subset of H. Then, for anyxH, there exists a unique nearest point in C,denoted by P C (x), such that

x P C ( x ) =infxyfor all yC.

The mapping P C is called the metric projection fromH onto C.

It is observed that P C is a nonexpansive and monotone mapping fromH onto C (see Agarwal et al.[22] for other properties of projection operators).

Let C be a nonempty subset of a real Hilbert space H, and let S 1 , S 2 :CH be two mappings. We denote by B(C) the collection of all bounded subsets of C.The deviation between S 1 and S 2 on BB(C)[24], denoted by S 1 S 2 , B , is defined by

S 1 S 2 , B =sup { S 1 ( x ) S 2 ( x ) : x B } .

In what follows, we shall make use of the following lemmas and proposition.

Lemma 2.1 ([11])

Let C be a nonempty closed convex subset of a real Hilbert space H. IfxHandyC, theny= P C (x)if and only if the following inequality holds:

xy,zy0for allzC.

Lemma 2.2 ([25])

Letf:CHbe a λ-contraction mapping andT:CCbe nonexpansive. Then the following hold:

  1. (a)

    The mapping If is (1λ)-strongly monotone, i.e.,

    x y , ( I f ) x ( I f ) y (1λ) x y 2 for allx,yC.
  2. (b)

    The mapping IT is monotone, i.e.,

    x y , ( I T ) x ( I T ) y 0for allx,yC.

Lemma 2.3 ([22])

Let T be a nonexpansive self-mapping of a nonempty closed convex subset C of a real Hilbert space H. ThenITis demiclosed at zero, i.e., if{ x n }is a sequence in C weakly converging to somexCand the sequence{(IT) x n }strongly converges to 0, thenxF(T).

Lemma 2.4 ([26])

Let { t n } and { d n } be the sequences of nonnegative real numbers such that

t n + 1 (1 b n ) t n + c n + d n for allnN,

where{ b n }is a real number sequence in(0,1)and{ c n }is a real number sequence. Assume that the following conditionshold:

  1. (i)

    n = 1 d n <;

  2. (ii)

    n = 1 b n = and lim sup n c n b n 0.

Then lim n t n =0.

Lemma 2.5 ([24, 27])

Let C be a nonempty closed convex subset of a real Hilbert space H and t i >0 (i=1,2,3,,N) such that i = 1 N t i =1. Let T 1 , T 2 , T 3 ,, T N :CCbe nonexpansive mappings with i = 1 N F( T i )and letT= i = 1 N t i T i . Then T is nonexpansive from C into itself andF(T)= i = 1 N F( T i ).

Let C be a nonempty subset of a real Hilbert space H. LetT:= { T n } n = 1 be a sequence of mappings from C intoitself. We denote by F(T) the set of common fixed points of the sequence, that is,F(T)= n = 1 F( T n ). Fix a sequence { a n } in [0,) with a n 0, and let { T n } be a sequence of mappings from C intoH. Then the sequence { T n } is called a sequence of nearly nonexpansivemappings[19] with respect to a sequence { a n } if

T n x T n yxy+ a n for all x,yC and nN.

One can observe that the sequence of nonexpansive mappings is essentially a sequenceof nearly nonexpansive mappings.

We now introduce the following:

Let C be a nonempty closed convex subset of a Hilbert space H. LetT= { T n } n = 1 be a sequence of nearly nonexpansive mappings fromC into itself with sequence { a n } such that n = 1 F( T n ). Let T:CC be a mapping such that Tx= lim n T n x for all xC with F(T)= n = 1 F( T n ). Then { T n } is said to satisfy condition (G) iffor each sequence { x n } in C with x n w and x n T i x n 0 for all iN imply wF(T).

Proposition 2.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let{ T n }be a sequence of nonexpansive mappings from C into itself. Then{ T n }satisfies condition (G).

Proposition 2.2 Let C be a nonempty closed convex subset of a real Hilbert space H. LetT= { T n } n = 1 be a sequence of nearly nonexpansive mappings from C into itself with sequence{ a n }such that n = 1 F( T n ). Then

T n x x 2 2x T n x,x x ˜ + ( a n + 2 x x ˜ ) a n

for allxCand x ˜ n = 1 F( T n ).

Proof Let xC and x ˜ n = 1 F( T n ). Then

T n x T n x ˜ 2 ( x x ˜ + a n ) 2 = x x ˜ 2 + ( a n + 2 x x ˜ ) a n .

Since x ˜ n = 1 F( T n ), we have

x x ˜ 2 + ( a n + 2 x x ˜ ) a n T n x x ˜ 2 = ( T n x x ) + ( x x ˜ ) 2 = T n x x 2 + x x ˜ 2 + 2 T n x x , x x ˜ .

Therefore,

T n x x 2 2x T n x,x x ˜ + ( a n + 2 x x ˜ ) a n .

 □

Proposition 2.3 Let C be a nonempty closed convex subset of a real Hilbert space H. LetT= { T n } n = 1 be a sequence of nearly nonexpansive mappings from C into itself with sequence{ a n }. Then

( I T n ) x ( I T n ) y , x y a n xyfor allx,yCandnN.

Proof Let x,yC. Then

( I T n ) x ( I T n ) y , x y = x y 2 T n x T n y , x y x y 2 T n x T n y x y x y 2 ( x y + a n ) x y = a n x y .

 □

3 Main result

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Letf:CHbe a λ-contraction, and let{ S n }be a sequence of nonexpansive mappings from C into itself. Let S be a nonexpansive mapping from C into itself such that lim n S n x=Sxfor allxC. LetT= { T n } n = 1 be a sequence of uniformly continuous nearly nonexpansive mappings from C into itself with sequence{ a n }such thatF(T). Let T be a mapping from C into itself defined byTx= lim n T n xfor allxC. Suppose thatF(T)= n = 1 F( T n ). For arbitrary x 1 C, consider the sequence{ x n }generated by the following iterative process:

{ x 1 C , y n = ( 1 β n ) x n + β n S n x n , x n + 1 = P C [ α n f ( x n ) + i = 1 n ( α i 1 α i ) T i y n ]
(3.1)

for allnN, where α 0 =1, { α n }is a strictly decreasing sequence in(0,1)and{ β n }is a sequence in(0,1)satisfying the conditions:

  1. (i)

    lim n α n =0, n = 1 α n =;

  2. (ii)

    n = 1 ( α n 1 α n )<, n = 1 | β n 1 β n |< and lim n β n α n =τ(0,);

  3. (iii)

    lim n ( ( α n 1 α n ) + | β n β n 1 | ) α n β n =0;

  4. (iv)

    there exists a constant N>0 such that 1 α n | 1 β n 1 β n 1 |N;

  5. (v)

    lim n γ n β n =0, where γ n := i = 1 n ( α i 1 α i ) a i , and either n = 1 S n + 1 S n , B < or lim n S n + 1 S n , B α n =0 for each BB(C).

Then the sequence{ x n }converges strongly to a point x n = 1 F( T n ), which is the unique solution of the followingvariational inequality:

1 τ ( I f ) x + ( I S ) x , x x 0for allx n = 1 F( T n ).
(3.2)

Proof It was proved in [17] that variational inequality problem (3.2) has the unique solution. Letp n = 1 F( T n ). We now break the proof into the following steps.

Step 1. { x n } is bounded.

From (3.1), we have

y n p = ( 1 β n ) x n + β n S n x n p ( 1 β n ) x n p + β n ( S n x n S n p + S n p p ) x n p + β n S n p p .

It follows that

x n + 1 p = P C [ α n f ( x n ) + i = 1 n ( α i 1 α i ) T i y n ] P C ( p ) α n f ( x n ) + i = 1 n ( α i 1 α i ) T i y n p = α n ( f ( x n ) p ) + i = 1 n ( α i 1 α i ) ( T i y n p ) α n ( f ( x n ) f ( p ) + f ( p ) p ) + i = 1 n ( α i 1 α i ) T i y n T i p α n λ x n p + α n f ( p ) p + i = 1 n ( α i 1 α i ) ( y n p + a i ) α n λ x n p + α n f ( p ) p + ( 1 α n ) ( x n p + β n S n p p ) + γ n [ 1 α n ( 1 λ ) ] x n p + α n f ( p ) p + β n S n p p + γ n .

Note that lim n β n α n =τ(0,) and lim n γ n β n =0, so there exists a constant K>0 such that

α n f ( p ) p + β n S n p p + γ n α n Kfor all nN.

Thus, we have

x n + 1 p ( 1 α n ( 1 λ ) ) x n p + α n K max { x n p , K 1 λ } for all  n N .

Hence { x n } is bounded. So, {f( x n )}, { y n }, { T i x n } and { T i y n } are bounded.

Step 2. x n + 1 x n 0 as n.

Set u n := α n f( x n )+ i = 1 n ( α i 1 α i ) T i y n for all nN. Set M:= sup n > 1 {f( x n 1 )+ T n y n 1 + S n 1 x n 1 + x n 1 }. From (3.1) we have

x n + 1 x n = P C ( u n ) P C ( u n 1 ) u n u n 1 = α n ( f ( x n ) f ( x n 1 ) ) + ( α n α n 1 ) f ( x n 1 ) + i = 1 n ( α i 1 α i ) ( T i y n T i y n 1 ) + ( α n 1 α n ) T n y n 1 α n f ( x n ) f ( x n 1 ) + i = 1 n ( α i 1 α i ) ( y n y n 1 + a i ) + ( α n 1 α n ) ( f ( x n 1 ) + T n y n 1 ) α n λ x n x n 1 + i = 1 n ( α i 1 α i ) y n y n 1 + ( α n 1 α n ) ( f ( x n 1 ) + T n y n 1 ) + γ n α n λ x n x n 1 + ( 1 α n ) y n y n 1 + ( α n 1 α n ) M + γ n .
(3.3)

Set B:={ x n }. Now, from (3.1) we have

y n y n 1 = ( ( 1 β n ) x n + β n S n x n ) ( ( 1 β n 1 ) x n 1 + β n 1 S n 1 x n 1 ) = ( 1 β n ) ( x n x n 1 ) + ( β n 1 β n ) x n 1 + β n ( S n x n S n x n 1 ) + β n ( S n x n 1 S n 1 x n 1 ) + ( β n β n 1 ) S n 1 x n 1 ( 1 β n ) ( x n x n 1 ) + ( β n 1 β n ) x n 1 + β n ( S n x n S n x n 1 ) + ( β n β n 1 ) S n 1 x n 1 + β n S n S n 1 , B x n x n 1 + | β n β n 1 | M + β n S n S n 1 , B .
(3.4)

Now, using (3.4) in (3.3), we obtain that

x n + 1 x n u n u n 1 α n λ x n x n 1 + ( 1 α n ) [ x n x n 1 + | β n β n 1 | M + β n S n S n 1 , B ] + ( α n 1 α n ) M + γ n ( 1 α n ( 1 λ ) ) x n x n 1 + M [ ( α n 1 α n ) + | β n β n 1 | ] + ( 1 α n ) β n S n S n 1 , B + γ n ( 1 α n ( 1 λ ) ) x n x n 1 + M [ ( α n 1 α n ) + | β n β n 1 | ] + β n S n S n 1 , B + γ n .
(3.5)

Thus, by using conditions (i), (v), n = 1 ( α n 1 α n )<, n = 1 | β n 1 β n |< and applying Lemma 2.4, we conclude that

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

Step 3. We claim lim n x n T i x n =0 for all iN.

Since T i x n C for all iN and i = 1 n ( α i 1 α i )+ α n =1, we get

i = 1 n ( α i 1 α i ) T i x n + α n zCfor all zC.

Noticing x n + 1 = P C ( u n ) and fixing z n = 1 F( T n ), from (3.1) we have

i = 1 n ( α i 1 α i ) ( x n T i x n ) = P C ( u n ) + ( 1 α n ) x n ( i = 1 n ( α i 1 α i ) T i x n + α n z ) + α n z x n + 1 = P C ( u n ) P C ( i = 1 n ( α i 1 α i ) T i x n + α n z ) + ( 1 α n ) ( x n x n + 1 ) + α n ( z x n + 1 ) .

Hence,

i = 1 n ( α i 1 α i ) x n T i x n , x n p = P C ( u n ) P C ( i = 1 n ( α i 1 α i ) T i x n + α n z ) , x n p + ( 1 α n ) ( x n x n + 1 ) , x n p + α n z x n + 1 , x n p u n i = 1 n ( α i 1 α i ) T i x n α n z x n p + ( 1 α n ) x n x n + 1 x n p + α n z x n + 1 x n p = α n ( f ( x n ) z ) + i = 1 n ( α i 1 α i ) ( T i y n T i x n ) x n p + ( 1 α n ) x n x n + 1 x n p + α n z x n + 1 x n p α n f ( x n ) z x n p + i = 1 n ( α i 1 α i ) ( y n x n + a i ) x n p + ( 1 α n ) x n x n + 1 x n p + α n z x n + 1 x n p α n f ( x n ) z x n p + ( 1 α n ) y n x n x n p + ( 1 α n ) x n x n + 1 x n p + α n z x n + 1 x n p + γ n x n p α n f ( x n ) z x n p + ( 1 α n ) β n S n x n x n x n p + ( 1 α n ) x n x n + 1 x n p + α n z x n + 1 x n p + γ n R ( 2 α n + β n ) M + ( 1 α n ) x n x n + 1 R + γ n R ,
(3.7)

where R is a positive constant such that x n pR for all nN and

M = sup n N { f ( x n ) z x n p , S n x n x n x n p , z x n + 1 x n p } .

Set ξ n := 1 2 ( a n +2 x n p) a n . From Proposition 2.2, using (3.7), we obtainthat

1 2 i = 1 n ( α i 1 α i ) x n T i x n 2 i = 1 n ( α i 1 α i ) x n T i x n , x n p + ξ n ( 2 α n + β n ) M + ( 1 α n ) x n x n + 1 R + γ n R .

Using (3.6), condition (i), lim n γ n β n =0 and lim n β n α n =τ(0,), we have

lim n i = 1 n ( α i 1 α i ) x n T i x n =0.

Since ( α i 1 α i ) x n T i x n i = 1 n ( α i 1 α i ) x n T i x n for all iN and { α n } is strictly decreasing, we have

lim n x n T i x n =0for all iN.

Step 4. y n T i y n 0 as n for all iN.

Noticing that lim n β n α n =τ(0,) and using condition (i), we have β n 0 as n. Therefore, we obtain that

y n x n = β n S n x n x n 0as n.
(3.8)

So that for all iN, we have

y n T i x n y n x n + x n T i x n 0as n.
(3.9)

Since each T i is uniformly continuous, from (3.8) and (3.9), wehave

y n T i y n y n T i x n + T i x n T i y n 0as n

for all iN.

Step 5. lim n x n + 1 x n β n =0 and lim n u n u n 1 β n = u n u n 1 α n =0.

From (3.5) we obtain that

x n + 1 x n β n u n u n 1 β n ( 1 α n ( 1 λ ) ) x n x n 1 β n + M ( ( α n 1 α n ) β n + | β n β n 1 | β n ) + S n S n 1 , B + γ n β n = ( 1 α n ( 1 λ ) ) x n x n 1 β n 1 + ( 1 α n ( 1 λ ) ) x n x n 1 ( 1 β n 1 β n 1 ) + M ( ( α n 1 α n ) β n + | β n β n 1 | β n ) + S n S n 1 , B + γ n β n .
(3.10)

We observe that

( 1 α n ( 1 λ ) ) ( 1 β n 1 β n 1 ) α n 1 α n | 1 β n 1 β n 1 | α n N.

Set

μ n = α n ( 1 λ ) , φ n = α n N x n x n 1 + M ( ( α n 1 α n ) β n + | β n β n 1 | β n ) , ν n = S n S n 1 , B + γ n β n .

From (3.10) we obtain that

x n + 1 x n β n u n u n 1 β n ( 1 α n ( 1 λ ) ) x n x n 1 β n 1 + α n N x n x n 1 + M ( ( α n 1 α n ) β n + | β n β n 1 | β n ) + S n S n 1 , B + γ n β n ( 1 α n ( 1 λ ) ) u n 1 u n 2 β n 1 + α n N x n x n 1 + M ( ( α n 1 α n ) β n + | β n β n 1 | β n ) + S n S n 1 , B + γ n β n ( 1 μ n ) u n 1 u n 2 β n 1 + φ n + ν n .

Using conditions (i), (iii), (v) and applying Lemma 2.4, we have

lim n x n + 1 x n β n =0and lim n u n u n 1 β n = u n u n 1 α n =0.

Step 6. Set v n := x n x n + 1 ( 1 α n ) β n and u n := α n f( x n )+ i = 1 n ( α i 1 α i ) T i y n for all nN. Then, for any z n = 1 F( T n ), we can calculate v n , x n z.

From (3.1) we have

x n + 1 = P C ( u n ) u n + α n f( x n )+ i = 1 n ( α i 1 α i )( T i y n y n )+(1 α n ) y n ,

which gives that

x n x n + 1 = ( 1 α n ) x n + α n x n ( P C ( u n ) u n + α n f ( x n ) + i = 1 n ( α i 1 α i ) ( T i y n y n ) + ( 1 α n ) y n ) = ( 1 α n ) β n ( x n S n x n ) + ( u n P C ( u n ) ) + i = 1 n ( α i 1 α i ) ( y n T i y n ) + α n ( x n f ( x n ) ) .

Then we conclude that

x n x n + 1 ( 1 α n ) β n = x n S n x n + 1 ( 1 α n ) β n ( u n P C ( u n ) ) + 1 ( 1 α n ) β n i = 1 n ( α i 1 α i ) ( y n T i y n ) + α n ( 1 α n ) β n ( x n f ( x n ) ) .

Noticing that x n + 1 = P C ( u n ). For any z n = 1 F( T n ), we have

v n , x n z = 1 ( 1 α n ) β n u n P C ( u n ) , P C ( u n 1 ) z + x n S n x n , x n z + 1 ( 1 α n ) β n i = 1 n ( α i 1 α i ) y n T i y n , x n z + α n ( 1 α n ) β n ( I f ) x n , x n z .
(3.11)

By using Lemma 2.2, we obtain that

x n S n x n , x n z = ( I S n ) x n ( I S n ) z , x n z + ( I S n ) z , x n z ( I S n ) z , x n z ,
(3.12)
( I f ) x n , x n z = ( I f ) x n ( I f ) z , x n z + ( I f ) z , x n z ( 1 λ ) x n z 2 + ( I f ) z , x n z ,
(3.13)
y n T i y n , x n z = ( I T i ) y n ( I T i ) z , x n y n + ( I T i ) y n ( I T i ) z , y n z ( I T i ) y n ( I T i ) z , x n y n a i y n z = β n ( I T i ) y n , x n S n x n a i y n z for all  i N .
(3.14)

Also, from Lemma 2.1 we get that

u n P C ( u n ) , P C ( u n 1 ) z = u n P C ( u n ) , P C ( u n 1 ) P C ( u n ) + u n P C ( u n ) , P C ( u n ) z u n P C ( u n ) , P C ( u n 1 ) P C ( u n ) .
(3.15)

Substituting (3.12), (3.13), (3.14) and (3.15) into (3.11), we have

v n , x n z 1 ( 1 α n ) β n u n P C ( u n ) , P C ( u n 1 ) P C ( u n ) + ( I S n ) z , x n z + 1 ( 1 α n ) i = 1 n ( α i 1 α i ) [ ( I T i ) y n , x n S n x n a i y n z ] + α n ( 1 α n ) β n ( I f ) z , x n z + α n ( 1 λ ) ( 1 α n ) β n x n z 2 .
(3.16)

Step 7. x n x ˜ , where x ˜ is a strong cluster point of the sequence{ x n }.

From (3.16) we observe that

x n z 2 ( 1 α n ) β n α n ( 1 λ ) [ v n , x n z ( I S n ) z , x n z ] 1 1 λ ( I f ) z , x n z β n α n ( 1 λ ) i = 1 n ( α i 1 α i ) [ ( I T i ) y n , x n S n x n a i y n z ] + u n u n 1 α n ( 1 λ ) u n P C ( u n ) .

Noticing that v n = x n x n + 1 ( 1 α n ) β n 0, u n u n 1 α n 0, and for all iN, (I T i ) y n 0 as n. It is easy to see that a weak cluster point of{ x n } is also a strong cluster point. Note that thesequence { x n } is bounded, then there exists a subsequence{ x n k } of { x n } which converges to a point x C. We also have, for all iN, (I T i ) x n 0 as n. By using condition (G), we get that x F(T). Thus, for all z n = 1 F( T n ), we obtain that

( I f ) x n k , x n k z ( 1 α n k ) β n k α n k v n k , x n k z 1 α n k u n k P C ( u n k ) u n k 1 u n k ( 1 α n k ) β n k α n k ( I S n k ) z , x n k z β n k α n k i = 1 n k ( α i 1 α i ) ( I T i ) y n k , x n k S n k x n k + β n k γ n k α n k y n k z .

Now, letting k, we have

( I f ) x , x z τ ( I S ) z , x z for all z n = 1 F( T n ).

Now, since variational inequality problem (3.2) has the unique solution, then we getthat ω w ( x n )={ x ˜ }. Note that every weak cluster point of the sequence{ x n } is also a strong cluster point. Then we have lim n x n = x ˜ . □

Recently, Marino et al.[28] used a different approach to obtain the convergence of a more generaliterative method that involves an equilibrium problem. We now present the result ofGu et al. [[18], Theorem 3.5] as a corollary.

Corollary 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Letf:CHbe a λ-contraction. LetS:CCbe a nonexpansive mapping, and letT= { T n } n = 1 be a countable family of nonexpansive mappings from C into itself such thatF(T)= n = 1 F( T n ). For arbitrary x 1 C, consider the sequence{ x n }generated by the following iterative process:

{ x 1 C , y n = β n S x n + ( 1 β n ) x n , x n + 1 = P C [ α n f ( x n ) + i = 1 n ( α i 1 α i ) T i y n ]

for allnN, where α 0 =1, { α n }is a strictly decreasing sequence in(0,1)and{ β n }is a sequence in(0,1)satisfying conditions (i)-(iv) of Theorem  3.1. Then thesequence{ x n }converges strongly to a point x n = 1 F( T n ), which is the unique solution of the followingvariational inequality:

1 τ ( I f ) x + ( I S ) x , x x 0for allx n = 1 F( T n ).

Again, we present the result of Yao et al. [[17], Theorem 3.2] as a corollary.

Corollary 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Letf:CHbe a λ-contraction. LetS:CCbe a nonexpansive mapping, and letT:CCbe a nonexpansive mapping such thatF(T). For arbitrary x 1 C, consider the sequence{ x n }generated by the following iterative process:

{ x 1 C , y n = ( 1 β n ) x n + β n S x n , x n + 1 = P C [ α n f ( x n ) + ( 1 α n ) T y n ]

for allnN, where{ α n }and{ β n }are two sequences in(0,1)satisfying conditions (i)-(iv) of Theorem  3.1. Then thesequence{ x n }converges strongly to a point x F(T), which is the unique solution of the followingvariational inequality:

1 τ ( I f ) x + ( I S ) x , x x 0for allxF(T).

Theorem 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Letf:CHbe a λ-contraction, and let{ S n }be a sequence of nonexpansive mappings from C into itself. Let S be a nonexpansive mapping from C into itself such that lim n S n x=Sxfor allxC. Let t 1 , t 2 , t 3 ,, t N >0such that i = 1 N t i =1. Let T 1 , T 2 , T 3 ,, T N :CCbe nonexpansive mappings such that i = 1 N F( T i ), and assume thatT= i = 1 N t i T i . For arbitrary x 1 C, consider the sequence{ x n }generated by the following iterative process:

{ x 1 C , y n = β n S n x n + ( 1 β n ) x n , x n + 1 = P C [ α n f ( x n ) + ( 1 α n ) i = 1 N t i T i y n ]

for allnN, where{ α n }and{ β n }are two sequences in(0,1)satisfying conditions (i)-(iv) of Theorem  3.1. Then thesequence{ x n }converges strongly to a point x i = 1 N F( T i ), which is the unique solution of the followingvariational inequality:

1 τ ( I f ) x + ( I S ) x , x x 0for allx i = 1 N F( T i ).

Proof The proof follows from Lemma 2.5 andCorollary 3.2. □

4 Numerical example

We present an example to show the effectiveness and convergence of the sequencegenerated by the considered iterative scheme.

Example 4.1 Let H=R and C=[0,1]. Let T be a self-mapping defined byTx=1x for all xC. Let f:CH be a contraction mapping defined byf(x)= 2 3 x 1 6 for all xC, and let { S n } be a sequence of nonexpansive mappings fromC into itself defined by S n x=x+ 1 2 n such that Sx= lim n S n x for all xC, where S is a nonexpansive mapping definedby Sx=x for all xC. Define sequences { α n } and { β n } in (0,1) by α n = β n = 1 ( n + 1 ) 1 / 2 . Without loss of generality, we may assume that a n = 1 n 3 / 2 for all nN. For each nN, define T n :CC by

T n x={ 1 x if  x [ 0 , 1 ) , a n if  x = 1 .

It was shown in [19] that T={ T n } is a sequence of nearly nonexpansive mappings fromC into itself such that F(T)={ 1 2 } and Tx= lim n T n x for all xC, where T is a nonexpansive mapping.

One can observe that all the assumptions of Theorem 3.1 are satisfied, and thesequence { x n } generated by (3.1) converges to a unique solution 1 2 of variational inequality (3.2) overF(T).

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The authors would like to thank the editor and referees for useful comments andsuggestions.

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Correspondence to Shin Min Kang.

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Sahu, D., Kang, S.M. & Sagar, V. Iterative methods for hierarchical common fixed point problems and variationalinequalities. Fixed Point Theory Appl 2013, 299 (2013) doi:10.1186/1687-1812-2013-299

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Keywords

  • hierarchical variational inequality
  • metric projection mapping
  • nonexpansive mapping
  • sequence of nearly nonexpansive mappings