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# Iterative methods for hierarchical common fixed point problems and variationalinequalities

## 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  in 1964. Since then, many kinds of variational inequalities have beenextended and generalized in several directions using novel and innovative techniques(see  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 $T:C→H$ be a nonlinear mapping. Then the classicalvariational inequality problem is a problem of finding $x ∗ ∈C$ such that

(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 $P C (I−λT)$, 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,

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 point$x∈H$ satisfying $Tx=0$ and so on, and it has several applications indifferent branches of natural sciences, social sciences, management and engineering(see  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 $T,S:C→C$ be two nonexpansive mappings. Then the hierarchicalvariational inequality problem is given as follows:

(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,

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  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., ).

Very recently, Gu et al. motivated and inspired by the results of Marino and Xu , Yao et al. 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. and Sahu et al., 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. 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., Yao et al., Moudafi , Xu  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 $T:C→H$ is called

1. (1)

monotone if

2. (2)

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

3. (3)

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

4. (4)

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

5. (5)

nonexpansive if

6. (6)

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

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 any$x∈H$, there exists a unique nearest point in C,denoted by $P C (x)$, such that

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. for other properties of projection operators).

Let C be a nonempty subset of a real Hilbert space H, and let$S 1 , S 2 :C→H$ 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 $B∈B(C)$, 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 ()

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

$〈x−y,z−y〉≤0for allz∈C.$

Lemma 2.2 ()

Let$f:C→H$be a λ-contraction mapping and$T:C→C$be nonexpansive. Then the following hold:

1. (a)

The mapping $I−f$ is $(1−λ)$-strongly monotone, i.e.,

$〈 x − y , ( I − f ) x − ( I − f ) y 〉 ≥(1−λ) ∥ x − y ∥ 2 for allx,y∈C.$
2. (b)

The mapping $I−T$ is monotone, i.e.,

$〈 x − y , ( I − T ) x − ( I − T ) y 〉 ≥0for allx,y∈C.$

Lemma 2.3 ()

Let T be a nonexpansive self-mapping of a nonempty closed convex subset C of a real Hilbert space H. Then$I−T$is demiclosed at zero, i.e., if${ x n }$is a sequence in C weakly converging to some$x∈C$and the sequence${(I−T) x n }$strongly converges to 0, then$x∈F(T)$.

Lemma 2.4 ()

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

$t n + 1 ≤(1− b n ) t n + c n + d n for alln∈N,$

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$T 1 , T 2 , T 3 ,…, T N :C→C$be nonexpansive mappings with$⋂ i = 1 N F( T i )≠∅$and let$T= ∑ i = 1 N t i T i$. Then T is nonexpansive from C into itself and$F(T)= ⋂ i = 1 N F( T i )$.

Let C be a nonempty subset of a real Hilbert space H. Let$T:= { T n } n = 1 ∞$ 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 with respect to a sequence ${ a n }$ if

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. Let$T= { T n } n = 1 ∞$ be a sequence of nearly nonexpansive mappings fromC into itself with sequence ${ a n }$ such that $⋂ n = 1 ∞ F( T n )≠∅$. Let $T:C→C$ be a mapping such that $Tx= lim n → ∞ T n x$ for all $x∈C$ 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 $i∈N$ imply $w∈F(T)$.

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. Let$T= { T n } n = 1 ∞$be a sequence of nearly nonexpansive mappings from C into itself with sequence${ a n }$such that$⋂ n = 1 ∞ F( T n )≠∅$. Then

$∥ T n x − x ∥ 2 ≤2〈x− T n x,x− x ˜ 〉+ ( a n + 2 ∥ x − x ˜ ∥ ) a n$

for all$x∈C$and$x ˜ ∈ ⋂ n = 1 ∞ F( T n )$.

Proof Let $x∈C$ 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,

$∥ T n x − x ∥ 2 ≤2〈x− T n x,x− x ˜ 〉+ ( a n + 2 ∥ x − x ˜ ∥ ) a n .$

□

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

$〈 ( I − T n ) x − ( I − T n ) y , x − y 〉 ≥− a n ∥x−y∥for allx,y∈Candn∈N.$

Proof Let $x,y∈C$. 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. Let$f:C→H$be 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$lim n → ∞ S n x=Sx$for all$x∈C$. Let$T= { T n } n = 1 ∞$be a sequence of uniformly continuous nearly nonexpansive mappings from C into itself with sequence${ a n }$such that$F(T)≠∅$. Let T be a mapping from C into itself defined by$Tx= lim n → ∞ T n x$for all$x∈C$. Suppose that$F(T)= ⋂ n = 1 ∞ F( T n )$. 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 all$n∈N$, 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 $B∈B(C)$.

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  that variational inequality problem (3.2) has the unique solution. Let$p∈ ⋂ n = 1 ∞ F( T n )$. 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

Thus, we have

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 $n∈N$. 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 $i∈N$.

Since $T i x n ∈C$ for all $i∈N$ and $∑ i = 1 n ( α i − 1 − α i )+ α n =1$, we get

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 $∥ x n −p∥≤R$ for all $n∈N$ 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 $i∈N$ and ${ α n }$ is strictly decreasing, we have

Step 4. $∥ y n − T i y n ∥→0$ as $n→∞$ for all $i∈N$.

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

(3.8)

So that for all $i∈N$, we have

(3.9)

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

for all $i∈N$.

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 $n∈N$. 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)
(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 $i∈N$, $(I− T i ) y n →0$ 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 $i∈N$, $(I− T i ) x n →0$ 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

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. 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. [, Theorem 3.5] as a corollary.

Corollary 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H. Let$f:C→H$be a λ-contraction. Let$S:C→C$be a nonexpansive mapping, and let$T= { T n } n = 1 ∞$be a countable family of nonexpansive mappings from C into itself such that$F(T)= ⋂ n = 1 ∞ F( T n )≠∅$. 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 all$n∈N$, 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. [, Theorem 3.2] as a corollary.

Corollary 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let$f:C→H$be a λ-contraction. Let$S:C→C$be a nonexpansive mapping, and let$T:C→C$be a nonexpansive mapping such that$F(T)≠∅$. 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 all$n∈N$, 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:

$〈 1 τ ( I − f ) x ∗ + ( I − S ) x ∗ , x − x ∗ 〉 ≥0for allx∈F(T).$

Theorem 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let$f:C→H$be 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$lim n → ∞ S n x=Sx$for all$x∈C$. Let$t 1 , t 2 , t 3 ,…, t N >0$such that$∑ i = 1 N t i =1$. Let$T 1 , T 2 , T 3 ,…, T N :C→C$be nonexpansive mappings such that$⋂ i = 1 N F( T i )≠∅$, and assume that$T= ∑ i = 1 N t i T i$. 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 all$n∈N$, 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 by$Tx=1−x$ for all $x∈C$. Let $f:C→H$ be a contraction mapping defined by$f(x)= 2 3 x− 1 6$ for all $x∈C$, 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 $x∈C$, where S is a nonexpansive mapping definedby $Sx=x$ for all $x∈C$. 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 $n∈N$. For each $n∈N$, define $T n :C→C$ by

It was shown in  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 $x∈C$, 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) over$F(T)$.

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## Acknowledgements

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). https://doi.org/10.1186/1687-1812-2013-299 