# Strong and weak convergence theorems for common zeros of finite accretive mappings

## Abstract

In this paper, we present a new iterative scheme to solve the problems of finding common zeros of finite m-accretive mappings in a real Hilbert space. Some strong and weak convergence theorems are established under different assumptions, which extends the corresponding works given by some authors.

MSC:47H05, 47H09.

## 1 Introduction and preliminaries

Let H be a real Hilbert space with the inner product $〈⋅,⋅〉$ and norm $∥⋅∥$, respectively. Then for $∀x,y∈H$, and $λ∈[0,1]$,

$∥ λ x + ( 1 − λ ) y ∥ 2 =λ ∥ x ∥ 2 +(1−λ) ∥ y ∥ 2 −λ(1−λ) ∥ x − y ∥ 2 .$
(1.1)

We write $x n ⇀x$ to indicate that the sequence ${ x n }$ converges weakly to x, and $x n →x$ implies that ${ x n }$ converges strongly to x.

Let C be a closed and convex subset of H. Then, for every point $x∈H$, there exists a unique nearest point in C, denoted by $P C x$, such that $∥x− P C x∥≤∥x−y∥$ for all $y∈C$. $P C$ is called the metric projection of H onto C. It is well known that $P C :H→C$ is characterized by the properties:

1. (i)

$〈x− P C x, P C x−y〉≥0$, for all $y∈C$ and $x∈H$;

2. (ii)

For every $x,y∈H$, $∥ P C x − P C y ∥ 2 ≤〈x−y, P C x− P C y〉$;

3. (iii)

$∥ P C x− P C y∥≤∥x−y∥$, for every $x,y∈H$;

4. (iv)

$x n ⇀ x 0$ and $P C x n → y 0$ imply that $P C x 0 = y 0$.

A mapping $f:C→C$ is called a contraction if there exists a constant $k∈(0,1)$ such that $∥f(x)−f(y)∥≤k∥x−y∥$, for $∀x,y∈C$. We use $∑ C$ to denote the collection of mappings f verifying the above inequality. That is, .

A mapping $T:C→C$ is said to be nonexpansive if $∥Tx−Ty∥≤∥x−y∥$, for $∀x,y∈C$. We use $F(T)$ to denote the fixed point set of T, that is, $F(T):={x∈C:Tx=x}$.

A mapping $A:H⊃D(A)→R(A)⊂H$ is called accretive if $〈x−y,Ax−Ay〉≥0$, for $∀x,y∈D(A)$ and it is called m-accretive if $R(I+λA)=H$, for $∀λ>0$. An m-accretive mapping A is demi-closed, that is, if ${ x n }⊂D(A)$ such that $x n ⇀x$ and $A x n →y$, then $x∈D(A)$ and $y=Ax$. Let $A − 1 0$ denote the set of zeros of A, that is, $A − 1 0:={x∈D(A):Ax=0}$. We denote by $J r A$ (for $r>0$) the resolvent of A, that is, $J r A := ( I + r A ) − 1$. Then $J r A$ is nonexpansive and $F( J r A )= A − 1 0$.

Interest in accretive mappings, which is an important class of nonlinear operators, stems mainly from their firm connection with equations of evolution. It is well known that many physically significant problems can be modeled by initial value problems of the form

$x ′ (t)+Ax(t)=0,x(0)= x 0 ,$
(1.2)

where A is an accretive mapping. Typical examples where such evolution equations occur can be found in the heat, wave or Schrödinger equations. If $x(t)$ is dependent on t, then (1.2) is reduced to

$Au=0,$
(1.3)

whose solutions correspond to the equilibrium of the system (1.2). Consequently, within the past 40 years or so, considerable research efforts have been devoted to methods for finding approximate solutions of (1.3). An early fundamental result in the theory of accretive operators, due to Browder [1]. One classical method for studying the problem $0∈Ax$, where A is an m-accretive mapping, is the following so-called proximal method (cf. [2]):

$x 0 ∈H, x n + 1 ≈ J r n x n ,n≥0,$
(1.4)

where $J r n := ( I + r n A ) − 1$. It was shown that the sequence generated by (1.4) converges weakly or strongly to a zero point of A under some conditions.

Recall that the following normal Mann iterative scheme to approximate the fixed point of a nonexpansive mapping $T:C→C$ was introduced by Mann [3]:

$x 0 ∈C, x n + 1 =(1− α n ) x n + α n T x n ,n≥0.$
(1.5)

It was proved that under some conditions, the sequence ${ x n }$ produced by (1.5) converges weakly to a point in $F(T)$.

Later, many mathematicians tried to combine the ideas of proximal method and Mann iterative method to approximate the zeros of m-accretive mappings; see, e.g. [411] and references therein.

Especially, in 2007, Qin and Su [4] presented the following iterative scheme:

${ x 1 ∈ C , y n = β n x n + ( 1 − β n ) J r n x n , x n + 1 = α n u + ( 1 − α n ) y n ,$
(1.6)

where $J r n = ( I + r n A ) − 1$. They showed that ${ x n }$ generated by the above scheme converges strongly to a zero of A.

Based on iterative schemes (1.4) and (1.5), Zegeye and Shahzad extended their discussion to the case of finite m-accretive mappings. They presented in [12] the following iterative scheme:

$x 0 ∈C, x n + 1 = α n u+(1− α n ) S r x n ,n≥0,$
(1.7)

where $S r = a 0 I+ a 1 J A 1 + a 2 J A 2 +⋯+ a l J A l$ with $J A i = ( I + A i ) − 1$ and $∑ i = 0 l a i =1$. If $⋂ i = 1 l A i − 1 (0)≠∅$, they proved that ${ x n }$ generated by (1.7) converges strongly to the common zeros of $A i$ ($i=1,2,…,l$) under some conditions.

Later, their work was extended to the following one presented by Hu and Liu in [13]:

$x 0 ∈C, x n + 1 = α n u+ β n x n + ϑ n S r n x n ,n≥0,$
(1.8)

where $S r n = a 0 I+ a 1 J r n A 1 + a 2 J r n A 2 +⋯+ a l J r n A l$ with $J r n A i = ( I + r n A i ) − 1$ and $∑ i = 0 l a i =1$. ${ α n },{ β n },{ ϑ n }⊂(0,1)$ and $α n + β n + ϑ n =1$. If $⋂ i = 1 l A i − 1 (0)≠∅$, they proved that ${ x n }$ converges strongly to the common zeros of $A i$ ($i=1,2,…,l$) under some conditions.

In this paper, based on the work of (1.6), (1.7), and (1.8), we present the following iterative scheme:

${ x 1 ∈ C , y n = β n f ( x n ) + ( 1 − β n ) S r n A m A m − 1 ⋯ A 1 x n , u n = ϑ n f ( y n ) + ( 1 − ϑ n ) W r n y n , x n + 1 = α n f ( u n ) + ( 1 − α n ) u n ,$
(A)

where $S r n A m A m − 1 ⋯ A 1 := J r n A m J r n A m − 1 ⋯ J r n A 1$, $W r n = a 0 I+ a 1 J r n B 1 + a 2 J r n B 2 +⋯+ a l J r n B l$, $J r n A i = ( I + r n A i ) − 1$ and $J r n B j = ( I + r n B j ) − 1$, for $i=1,2,…,m$; $j=1,2,…,l$. $∑ k = 0 l a k =1$, $f:C→C$ is a contraction, both ${ A i } i = 1 m$ and ${ B j } j = 1 l$ are finite families of m-accretive mappings. More details of iterative scheme (A) will be presented in Section 2. We shall prove a weak convergent theorem and a strong convergent theorem under different assumptions on ${ α n }$, ${ β n }$, ${ ϑ n }$, and ${ r n }$, respectively.

In order to prove our main results, we need the following lemmas.

By using the properties of the metric projection and m-accretive mappings, we can easily prove the following two lemmas.

Lemma 1.1 For $∀x∈H$ and $∀y∈C$, $∥ P C x − y ∥ 2 + ∥ P C x − x ∥ 2 ≤ ∥ y − x ∥ 2$.

Lemma 1.2 For $∀y∈ A − 1 0$, $∀x∈H$ and $r>0$,

$∥ ( I + r A ) − 1 x − y ∥ 2 + ∥ ( I + r A ) − 1 x − x ∥ 2 ≤ ∥ y − x ∥ 2 .$

Lemma 1.3 ([14])

Let ${ a n }$ and ${ b n }$ be two sequences of nonnegative real numbers satisfying

$a n + 1 ≤ a n + b n ,∀n≥1.$

If $∑ n = 1 ∞ b n <+∞$, then $lim n → ∞ a n$ exists.

Lemma 1.4 ([15])

Let H be a real Hilbert space and A be an m-accretive mapping. For $λ,μ>0$ and $x∈H$, we have

$J λ A x= J μ A ( μ λ x + ( 1 − μ λ ) J λ A x ) ,$

where $J λ A = ( I + λ A ) − 1$ and $J μ A = ( I + μ A ) − 1$.

Lemma 1.5 ([16])

Let H be a real Hilbert space and C be a closed convex subset of H. Let $T:C→C$ be a nonexpansive mapping with $F(T)≠∅$, and $f∈ ∑ C$. Then $z t$, defined by

$z t =tf( z t )+(1−t)T z t , z t ∈C,$

converges strongly to a point in $F(T)$. If one defines $Q: ∑ C →F(T)$ by $Q(f):= lim t → 0 z t$, $f∈ ∑ C$, then $Q(f)$ solves the following variational inequality:

$〈 ( I − f ) Q ( f ) , Q ( f ) − p 〉 ≤0,∀p∈F(T).$

Lemma 1.6 ([17])

Let ${ a n }$, ${ b n }$, and ${ c n }$ be three sequences of nonnegative real numbers satisfying

$a n + 1 ≤(1− c n ) a n + b n c n ,∀n≥1,$

where ${ c n }⊂(0,1)$ such that

1. (i)

$c n →0$ and $∑ n = 1 ∞ c n =+∞$,

2. (ii)

either $lim sup n → ∞ b n ≤0$ or $∑ n = 1 ∞ | b n c n |<+∞$.

Then $lim n → ∞ a n =0$.

Lemma 1.7 In a Hilbert space H, we can easily get the following inequality:

$∥ x + y ∥ 2 ≤ ∥ x ∥ 2 +2〈y,x+y〉,∀x,y∈H.$

## 2 Weak and strong convergence theorems

Lemma 2.1 Let H be a real Hilbert space, C be a nonempty closed and convex subset of H and $A i , B j :C→C$ ($i=1,2,…,m$; $j=1,2,…,l$) be finitely many m-accretive mappings such that $D:=( ⋂ i = 1 m A i − 1 0)∩( ⋂ j = 1 l B j − 1 0)≠∅$. Suppose $S r A m A m − 1 ⋯ A 1 := J r A m J r A m − 1 ⋯ J r A 1$ and $W r := a 0 I+ a 1 J r B 1 + a 2 J r B 2 +⋯+ a l J r B l$, where $J r A i = ( I + r A i ) − 1$ ($i=1,2,…,m$), $J r B j = ( I + r B j ) − 1$ ($j=1,2,…,l$), $a k ∈(0,1)$, $k=0,1,…,l$, $∑ k = 0 l a k =1$, and $r>0$. Then $S r A m A m − 1 ⋯ A 1 :C→C$ and $W r :C→C$ are nonexpansive.

Lemma 2.1 can easily be obtained in view of the facts that $( I + r A i ) − 1$ and $( I + r B j ) − 1$ are nonexpansive, $i=1,2,…,m$; $j=1,2,…,l$.

Theorem 2.1 Let H, C, D, and $A i , B j :C→C$ ($i=1,2,…,m$; $j=1,2,…,l$) be the same as those in Lemma  2.1. Suppose that $D≠∅$. Let ${ x n }$ be generated by the iterative scheme (A). If ${ α n }$, ${ β n }$ and ${ ϑ n }$ are three sequences in $[0,1)$ such that $∑ n = 1 ∞ α n <+∞$, $∑ n = 1 ∞ β n <+∞$, $∑ n = 1 ∞ ϑ n <+∞$, ${ r n }⊂(0,+∞)$ with $lim n → ∞ r n =+∞$ and $f:C→C$ is a contraction with contractive constant $k∈(0,1)$. Then ${ x n }$ converges weakly to a point $v 0 ∈D$ satisfying

$lim n → ∞ ∥ x n − v 0 ∥= min y ∈ D lim n → ∞ ∥ x n −y∥.$
(2.1)

Proof We split our proof into five steps.

Step 1. ${ x n }$, ${ u n }$ and ${ y n }$ are all bounded.

We can easily know that $⋂ i = 1 m A i − 1 0⊂F( S r n A m ⋯ A 1 )$, and $⋂ j = 1 l B j − 1 0⊂F( W r n )$. Then for $∀p∈D$, from Lemma 2.1, we have

$∥ S r n A m ⋯ A 1 x n − p ∥ = ∥ S r n A m ⋯ A 1 x n − S r n A m ⋯ A 1 p ∥ ≤∥ x n −p∥.$
(2.2)

Based on (2.2), we know that

$∥ y n − p ∥ ≤ β n ∥ f ( x n ) − p ∥ + ( 1 − β n ) ∥ S r n A m ⋯ A 1 x n − p ∥ ≤ [ 1 − β n ( 1 − k ) ] ∥ x n − p ∥ + β n ∥ f ( p ) − p ∥ .$
(2.3)

Then (2.3) and Lemma 2.1 imply that

$∥ u n − p ∥ ≤ ϑ n ∥ f ( y n ) − f ( p ) ∥ + ϑ n ∥ f ( p ) − p ∥ + ( 1 − ϑ n ) ∥ y n − p ∥ ≤ [ 1 − β n ( 1 − k ) ] [ 1 − ϑ n ( 1 − k ) ] ∥ x n − p ∥ + [ ϑ n + β n − ϑ n β n ( 1 − k ) ] ∥ f ( p ) − p ∥ .$
(2.4)

Using (2.4), we know that

$∥ x n + 1 − p ∥ ≤ α n ∥ f ( u n ) − f ( p ) ∥ + α n ∥ f ( p ) − p ∥ + ( 1 − α n ) ∥ u n − p ∥ ≤ [ 1 − β n ( 1 − k ) ] [ 1 − α n ( 1 − k ) ] [ 1 − ϑ n ( 1 − k ) ] ∥ x n − p ∥ + { [ 1 − α n ( 1 − k ) ] [ ϑ n + β n − ϑ n β n ( 1 − k ) ] + α n } ∥ f ( p ) − p ∥ ≤ ∥ x n − p ∥ + ( ϑ n + β n + α n ) ∥ f ( p ) − p ∥ .$
(2.5)

Then Lemma 1.3 implies that $lim n → ∞ ∥ x n −p∥$ exists, which ensures that ${ x n }$ is bounded. Combining with the fact that f is a contraction and noticing (2.2), (2.3), and (2.4), we can easily know that ${f( x n )}$, ${ u n }$, ${ y n }$, ${f( u n )}$, ${f( y n )}$, ${ S r n A i ⋯ A 1 x n }$ ($i=1,2,…,m$), and ${ J r n B j x n }$ ($j=1,2,…,l$) are all bounded.

We may let $M 1$ = max{$sup{∥ x n ∥:n≥1}$, $sup{∥ y n ∥:n≥1}$, $sup{∥ u n ∥:n≥1}$, $sup{∥f( x n )∥:n≥1}$, $sup{∥f( y n )∥:n≥1}$, $sup{∥f( u n )∥:n≥1}$, $sup{∥ S r n A i ⋯ A 1 x n ∥:n≥1,i=1,2,…,m}$, $sup{∥ J r n B j x n ∥:n≥1,j=1,2,…,l}$}.

Step 2. $lim n → ∞ ∥ P D x n − x n ∥$ exists.

In fact, it follows from the property of $P D$ that

$∥ P D x n + 1 − x n + 1 ∥≤∥ P D x n − x n + 1 ∥.$
(2.6)

In view of Lemma 1.1, we know that for $∀v∈D$,

$∥ v − P D x n ∥ 2 ≤ ∥ v − x n ∥ 2 − ∥ P D x n − x n ∥ 2 ≤ ∥ x n − v ∥ 2 ,$
(2.7)

which implies that ${ P D x n }$ is bounded since ${ x n }$ is bounded from step 1. Then ${f( P D x n )}$ is also bounded.

Let $M 2 =max{sup{∥ P D x n ∥:n≥1},sup{∥f( P D x n )∥:n≥1}}$.

Noticing (2.5) and (2.6), we have

$∥ x n + 1 − P D x n + 1 ∥ ≤ ∥ x n − P D x n ∥ + ( ϑ n + β n + α n ) ∥ f ( P D x n ) − P D x n ∥ ≤ ∥ x n − P D x n ∥ + 2 ( ϑ n + β n + α n ) M 2 .$

Therefore, in view of Lemma 1.3, $lim n → ∞ ∥ P D x n − x n ∥$ exists.

Step 3. $P D x n → v 0$, where $v 0 ∈D$ satisfies (2.1), as $n→∞$.

We first claim that there exists a unique element $v 0 ∈D$ such that

$lim n → ∞ ∥ x n − v 0 ∥= min y ∈ D lim n → ∞ ∥ x n −y∥.$

In fact, if we let $h(y)= lim n → ∞ ∥ x n −y∥$, $∀y∈D$. Then we can easily find that $h(y):D→ R +$ is proper, strictly convex and lower-semi-continuous and $h(y)→+∞$ as $∥y∥→+∞$. This ensures that there exists a unique element $v 0 ∈D$ such that $h( v 0 )= min y ∈ D h(y)$.

From (2.7), we know that

$lim n → ∞ ∥ v 0 − P D x n ∥ 2 ≤ lim n → ∞ ( ∥ v 0 − x n ∥ 2 − ∥ P D x n − x n ∥ 2 ) = h 2 ( v 0 )− lim n → ∞ ∥ P D x n − x n ∥ 2 ≤0.$

Therefore, $P D x n → v 0$, as $n→∞$.

Step 4. $ω( x n )⊂D$, where $ω( x n )$ denotes the set consisting all of the weak limit points of ${ x n }$.

Since ${ x n }$ is bounded, then there exists a subsequence of ${ x n }$, for simplicity, we still denote it by ${ x n }$, such that $x n ⇀x$, as $n→∞$.

Since $∥⋅∥$ is convex, by using Lemma 1.2 and noticing (2.3), we have, for $∀p∈D$,

$∥ y n − p ∥ 2 ≤ β n ∥ f ( x n ) − p ∥ 2 + ( 1 − β n ) ∥ S r n A m ⋯ A 1 x n − p ∥ 2 ≤ β n ∥ f ( x n ) − p ∥ 2 + ( 1 − β n ) [ ∥ S r n A m − 1 ⋯ A 1 x n − p ∥ 2 − ∥ S r n A m ⋯ A 1 x n − S r n A m − 1 ⋯ A 1 x n ∥ 2 ] ≤ β n k ∥ x n − p ∥ 2 + ( 1 − β n ) [ ∥ x n − p ∥ 2 − ∥ S r n A m ⋯ A 1 x n − S r n A m − 1 ⋯ A 1 x n ∥ 2 ] + β n ∥ f ( p ) − p ∥ 2 + 2 β n k ∥ x n − p ∥ ∥ f ( p ) − p ∥ ≤ ∥ x n − p ∥ 2 − ( 1 − β n ) ∥ S r n A m ⋯ A 1 x n − S r n A m − 1 ⋯ A 1 x n ∥ 2 + β n ∥ f ( p ) − p ∥ 2 + 2 β n k ∥ x n − p ∥ ∥ f ( p ) − p ∥ .$
(2.8)

Then using (2.8), we have

$∥ u n − p ∥ 2 ≤ ϑ n ∥ f ( y n ) − p ∥ 2 + ( 1 − ϑ n ) ∥ W r n y n − W r n p ∥ 2 ≤ [ 1 − ϑ n ( 1 − k ) ] ∥ y n − p ∥ 2 + 2 ϑ n k ∥ y n − p ∥ ∥ f ( p ) − p ∥ + ϑ n ∥ f ( p ) − p ∥ 2 ≤ ∥ x n − p ∥ 2 − ( 1 − β n ) ∥ S r n A m ⋯ A 1 x n − S r n A m − 1 ⋯ A 1 x n ∥ 2 + ( ϑ n + β n ) ∥ f ( p ) − p ∥ 2 + 2 k ( β n ∥ x n − p ∥ + ϑ n ∥ y n − p ∥ ) ∥ f ( p ) − p ∥ ,$
(2.9)

which implies that

$∥ x n + 1 − p ∥ 2 ≤ [ 1 − α n ( 1 − k ) ] ∥ u n − p ∥ 2 + 2 α n k ∥ u n − p ∥ ∥ f ( p ) − p ∥ + α n ∥ f ( p ) − p ∥ 2 ≤ ∥ x n − p ∥ 2 − ( 1 − β n ) ∥ S r n A m ⋯ A 1 x n − S r n A m − 1 ⋯ A 1 x n ∥ 2 + ( α n + β n + ϑ n ) ∥ f ( p ) − p ∥ 2 + 2 k ( α n ∥ u n − p ∥ + β n ∥ x n − p ∥ + ϑ n ∥ y n − p ∥ ) ∥ f ( p ) − p ∥ .$
(2.10)

Thus

$0 ≤ ( 1 − β n ) ∥ S r n A m ⋯ A 1 x n − S r n A m − 1 ⋯ A 1 x n ∥ 2 ≤ ∥ x n − p ∥ 2 − ∥ x n + 1 − p ∥ 2 + ( α n + β n + ϑ n ) ∥ f ( p ) − p ∥ 2 + 2 k ( α n ∥ u n − p ∥ + β n ∥ x n − p ∥ + ϑ n ∥ y n − p ∥ ) ∥ f ( p ) − p ∥ .$
(2.11)

Since from the proof of step 1, we know that $lim n → ∞ ∥ x n −p∥$ exists, then $S r n A m ⋯ A 1 x n − S r n A m − 1 ⋯ A 1 x n →0$, as $n→∞$.

Going back to (2.8) again, we know that

$∥ y n − p ∥ 2 ≤ β n ∥ f ( x n ) − p ∥ 2 + ( 1 − β n ) ∥ S r n A m − 1 ⋯ A 1 x n − p ∥ 2 ≤ β n ∥ f ( x n ) − p ∥ 2 + ( 1 − β n ) [ ∥ S r n A m − 2 ⋯ A 1 x n − p ∥ 2 − ∥ S r n A m − 1 ⋯ A 1 x n − S r n A m − 2 ⋯ A 1 x n ∥ 2 ] ≤ β n k ∥ x n − p ∥ 2 + ( 1 − β n ) [ ∥ x n − p ∥ 2 − ∥ S r n A m − 1 ⋯ A 1 x n − S r n A m − 2 ⋯ A 1 x n ∥ 2 ] + β n ∥ f ( p ) − p ∥ 2 + 2 β n k ∥ x n − p ∥ ∥ f ( p ) − p ∥ ≤ ∥ x n − p ∥ 2 − ( 1 − β n ) ∥ S r n A m − 1 ⋯ A 1 x n − S r n A m − 2 ⋯ A 1 x n ∥ 2 + β n ∥ f ( p ) − p ∥ 2 + 2 β n k ∥ x n − p ∥ ∥ f ( p ) − p ∥ .$
(2.12)

Then using (2.12), repeating the processes of (2.9)-(2.11), we know that

By using the inductive method, we have the following results:

$S r n A m − 2 ⋯ A 1 x n − S r n A m − 3 ⋯ A 1 x n → 0 , ⋯ ( I + r n A 1 ) − 1 x n − x n → 0 ,$

as $n→∞$. Therefore, $( I + r n A 1 ) − 1 x n ⇀x$, …, $S r n A m A m − 1 ⋯ A 1 x n = ( I + r n A m ) − 1 ⋯ ( I + r n A 1 ) − 1 x n ⇀x$, as $n→∞$.

Let $v n , 1 = ( I + r n A 1 ) − 1 x n$, then $A 1 v n , 1 = x n − v n , 1 r n →0$, since $r n →+∞$ and both ${ x n }$ and ${ v n , 1 }$ are bounded. This ensures that $x∈ A 1 − 1 0$.

Let $v n , 2 = ( I + r n A 2 ) − 1 ( I + r n A 1 ) − 1 x n = ( I + r n A 2 ) − 1 v n , 1$, then $A 2 v n , 2 = v n , 1 − v n , 2 r n →0$, which implies that $x∈ A 2 − 1 0$.

By induction, let $v n , m = ( I + r n A m ) − 1 ⋯ ( I + r n A 1 ) − 1 x n = ( I + r n A m ) − 1 v n , m − 1$, then $A m v n , m = v n , m − 1 − v n , m r n →0$, which implies that $x∈ A m − 1 0$. Thus $x∈ ⋂ i = 1 m A i − 1 0$.

Next, we shall show that $x∈ ⋂ j = 1 l B j − 1 0$.

From step 1, we may assume that there exists $M 3 >0$ such that $2∥ x n −p∥∥f(p)−p∥+ ∥ f ( p ) − p ∥ 2 ≤ M 3$, $2∥ y n −p∥∥f(p)−p∥+ ∥ f ( p ) − p ∥ 2 ≤ M 3$ and $2∥ u n −p∥∥f(p)−p∥+ ∥ f ( p ) − p ∥ 2 ≤ M 3$.

Now, computing the following, $∀p∈D$:

$∥ y n − p ∥ 2 ≤ [ 1 − β n ( 1 − k ) ] ∥ x n − p ∥ 2 + β n ∥ f ( p ) − p ∥ 2 + 2 β n k ∥ x n − p ∥ ∥ f ( p ) − p ∥ ≤ [ 1 − β n ( 1 − k ) ] ∥ x n − p ∥ 2 + β n M 3 .$
(2.13)

By using Lemma 1.2,

$∥ u n − p ∥ 2 ≤ k ϑ n ∥ y n − p ∥ 2 + 2 ϑ n k ∥ f ( p ) − p ∥ ∥ y n − p ∥ + ϑ n ∥ f ( p ) − p ∥ 2 + ( 1 − ϑ n ) ( a 0 ∥ y n − p ∥ 2 + ∑ j = 1 l a j ∥ ( I + r n B j ) − 1 y n − p ∥ 2 ) ≤ k ϑ n ∥ y n − p ∥ 2 + 2 ϑ n k ∥ f ( p ) − p ∥ ∥ y n − p ∥ + ϑ n ∥ f ( p ) − p ∥ 2 + ( 1 − ϑ n ) [ a 0 ∥ y n − p ∥ 2 + ∑ j = 1 l a j ( ∥ y n − p ∥ 2 − ∥ ( I + r n B j ) − 1 y n − y n ∥ 2 ) ] = [ 1 − ϑ n ( 1 − k ) ] ∥ y n − p ∥ 2 + 2 ϑ n k ∥ y n − p ∥ ∥ f ( p ) − p ∥ + ϑ n ∥ f ( p ) − p ∥ 2 − ( 1 − ϑ n ) ∑ j = 1 l a j ∥ ( I + r n B j ) − 1 y n − y n ∥ 2 ≤ ∥ y n − p ∥ 2 − ( 1 − ϑ n ) ∑ j = 1 l a j ∥ ( I + r n B j ) − 1 y n − y n ∥ 2 + ϑ n M 3 .$
(2.14)

Then (2.13) and (2.14) imply that

$∥ x n + 1 − p ∥ 2 ≤ [ 1 − α n ( 1 − k ) ] ∥ u n − p ∥ 2 + 2 α n k ∥ u n − p ∥ ∥ f ( p ) − p ∥ + α n ∥ f ( p ) − p ∥ 2 ≤ [ 1 − α n ( 1 − k ) ] ∥ u n − p ∥ 2 + α n M 3 ≤ [ 1 − α n ( 1 − k ) ] [ ∥ y n − p ∥ 2 − ( 1 − ϑ n ) ∑ j = 1 l a j ∥ ( I + r n B j ) − 1 y n − y n ∥ 2 + ϑ n M 3 ] + α n M 3 ≤ [ 1 − α n ( 1 − k ) ] [ 1 − β n ( 1 − k ) ] ∥ x n − p ∥ 2 + [ 1 − α n ( 1 − k ) ] M 3 ( β n + ϑ n ) + α n M 3 − [ 1 − α n ( 1 − k ) ] ( 1 − ϑ n ) ∑ j = 1 l a j ∥ ( I + r n B j ) − 1 y n − y n ∥ 2 .$
(2.15)

From step 1, we know that $lim n → ∞ ∥ x n −p∥$ exists, then (2.15) implies that

(2.16)

From the iterative scheme (A), $β n →0$, and the results of step 1, we know that

Then $y n ⇀x$, since $S r n A m A m − 1 ⋯ A 1 x n ⇀x$, as $n→∞$.

Thus from (2.16), we have $( I + r n B j ) − 1 y n ⇀x$, imitating the proof of $x∈ ⋂ i = 1 m A i − 1 0$, we can see that $x∈ ⋂ j = 1 l B j − 1 0$, and then $x∈D$.

Step 5. $x n ⇀ v 0 = lim n → ∞ P D x n$.

In fact, for $∀y∈D$,

$〈 P D x n −y, P D x n − x n 〉≤0.$
(2.17)

From step 3, we know that $P D x n → v 0$, as $n→∞$. Let ${ x n i }$ be a subsequence of ${ x n }$ which is weakly convergent to $x 0$. Then $x 0 ∈D$ from step 4. Taking the limits on both sides of (2.17), we know that $〈 v 0 −y, v 0 − x 0 〉≤0$.

Letting $y= x 0$, we have $x 0 = v 0$.

Supposing ${ x n j }$ is another subsequence of ${ x n }$ such that $x n j ⇀ x 1$ as $j→∞$. Then repeating the above proof, we have $x 1 = v 0$. Since all of the weakly convergent subsequences of ${ x n }$ converge to the same element $v 0$, then the whole sequence ${ x n }$ converges weakly to $v 0$.

This completes the proof. □

Remark 2.1 To prove the strong convergence result in Theorem 2.2, we need to prove the following two lemmas first and some new proof techniques can be seen.

Lemma 2.2 Let $H,C,D, A i , B j :C→C$ ($i=1,2,…,m$; $j=1,2,…,l$), $S r A m A m − 1 ⋯ A 1$ and $W r$ be the same as those in Lemma  2.1. Suppose that $D≠∅$. Then $F( S r A m A m − 1 ⋯ A 1 )= ⋂ i = 1 m A i − 1 0$ and $F( W r )= ⋂ j = 1 l B j − 1 0$, for $∀r>0$.

Proof It is easy to check $⋂ i = 1 m A i − 1 0⊂F( S r A m A m − 1 ⋯ A 1 )$ and $⋂ j = 1 l B j − 1 0⊂F( W r )$, for $∀r>0$.

Next, we shall show that $F( W r )⊂ ⋂ j = 1 l B j − 1 0$.

For $∀p∈F( W r )$, $∀q∈ ⋂ j = 1 l B j − 1 0$. Since $⋂ j = 1 l B j − 1 0⊂F( W r )$, then $q= W r q$. Thus

$∥ q − p ∥ = ∥ a 0 ( q − p ) + a 1 ( J r B 1 q − J r B 1 p ) + ⋯ + a l ( J r B l q − J r B l p ) ∥ ≤ a 0 ∥ q − p ∥ + a 1 ∥ J r B 1 q − J r B 1 p ∥ + ⋯ + a l ∥ J r B l q − J r B l p ∥ ≤ ∥ q − p ∥ .$

Then $a 0 (∥q−p∥−∥q−p∥)+ a 1 (∥q−p∥−∥ J r B 1 q− J r B 1 p∥)+⋯+ a l (∥q−p∥−∥ J r B l q− J r B l p∥)=0$. Since $∥q−p∥−∥ J r B j q− J r B j p∥≥0$, $j=1,2,…,l$, then $∥q−p∥−∥ J r B j q− J r B j p∥=0$, $j=1,2,…,l$. That is,

$∥q−p∥= ∥ J r B j q − J r B j p ∥ = ∥ q − J r B j p ∥ ,j=1,2,…,l.$
(2.18)

By using Lemma 1.2 and (2.18), we know that $∥ p − J r B j p ∥ 2 ≤ ∥ q − p ∥ 2 − ∥ q − J r B j p ∥ 2 =0$, $j=1,2,…,l$. Thus $p= J r B j p$, which implies that $p∈ B j − 1 0$, $j=1,2,…,l$. Then $F( W r )⊂ ⋂ j = 1 l B j − 1 0$, for $r>0$.

Finally, we shall show that $F( S r A m A m − 1 ⋯ A 1 )⊂ ⋂ i = 1 m A i − 1 0$.

For $∀p∈F( S r A m A m − 1 ⋯ A 1 )$, then $p= S r A m A m − 1 ⋯ A 1 p$. Let $q∈ ⋂ i = 1 m A i − 1 0$, then $q= S r A m A m − 1 ⋯ A 1 q$, since $⋂ i = 1 m A i − 1 0⊂F( S r A m A m − 1 ⋯ A 1 )$. Therefore,

$∥ q − p ∥ = ∥ S r A m A m − 1 ⋯ A 1 q − S r A m A m − 1 ⋯ A 1 p ∥ ≤ ∥ S r A m − 1 A m − 2 ⋯ A 1 q − S r A m − 1 A m − 2 ⋯ A 1 p ∥ ≤ ∥ S r A m − 2 A m − 3 ⋯ A 1 q − S r A m − 2 A m − 3 ⋯ A 1 p ∥ ≤ ⋯ ≤ ∥ ( I + r A 1 ) − 1 q − ( I + r A 1 ) − 1 p ∥ ≤ ∥ q − p ∥ .$
(2.19)

From (2.19), we know that

$∥ q − ( I + r A 1 ) − 1 p ∥ =∥q−p∥.$
(2.20)

Noticing that (2.20) and (2.18) have the same form, then repeating the proof of $p= J r B j p$, we know that $p= ( I + r A 1 ) − 1 p$ and then $p∈ A 1 − 1 0$.

Since $p∈ A 1 − 1 0$, using (2.19) again, we know that

$∥ q − p ∥ = ∥ ( I + r A 2 ) − 1 ( I + r A 1 ) − 1 q − ( I + r A 2 ) − 1 ( I + r A 1 ) − 1 p ∥ = ∥ q − ( I + r A 2 ) − 1 p ∥ .$
(2.21)

Repeating the above proof again, $p∈ A 2 − 1 0$.

By induction, we have $p∈ A m − 1 0$. Therefore, $F( S r A m A m − 1 ⋯ A 1 )⊂ ⋂ i = 1 m A i − 1 0$.

This completes the proof. □

Lemma 2.3 Let $H,C,D, A i , B j :C→C$ ($i=1,2,…,m$; $j=1,2,…,l$), $S r A m A m − 1 ⋯ A 1$ and $W r$ be the same as those in Lemma  2.1. Suppose that $D≠∅$. Then $W r S r A m A m − 1 ⋯ A 1 :C→C$ is nonexpansive and $F( W r S r A m A m − 1 ⋯ A 1 )=D$, for $∀r>0$.

Proof It is easy to check that $W r S r A m A m − 1 ⋯ A 1 :C→C$ is nonexpansive. We are left to show that $F( W r S r A m A m − 1 ⋯ A 1 )=D$.

$∀p∈D$, then, from Lemma 2.2, $p= S r A m A m − 1 ⋯ A 1 p$ and $p= W r p$. Thus $p= W r S r A m A m − 1 ⋯ A 1 p$, which implies that $D⊂F( W r S r A m A m − 1 ⋯ A 1 )$.

On the other hand, let $p∈F( W r S r A m A m − 1 ⋯ A 1 )$, then $p= W r S r A m A m − 1 ⋯ A 1 p$. Let $q∈D$, then $q= W r S r A m A m − 1 ⋯ A 1 q$, since $D⊂F( W r S r A m A m − 1 ⋯ A 1 )$. Then Lemma 2.1 ensures that

$∥ p − q ∥ ≤ ∥ S r A m A m − 1 ⋯ A 1 p − S r A m A m − 1 ⋯ A 1 q ∥ ≤ ∥ S r A m − 1 ⋯ A 1 p − S r A m − 1 ⋯ A 1 q ∥ ≤ ⋯ ≤ ∥ J r A 1 p − J r A 1 q ∥ ≤ ∥ p − q ∥ ,$

which implies that

$∥ J r A 1 p − q ∥ = ∥ S r A 2 A 1 p − q ∥ =⋯= ∥ S r A m A m − 1 ⋯ A 1 p − q ∥ =∥p−q∥.$

Using the same method as that in Lemma 2.2, $p∈ ⋂ i = 1 m A i − 1 0$. Thus $p= S r A m A m − 1 ⋯ A 1 p$. Since $p= W r S r A m A m − 1 ⋯ A 1 p$, then $p= W r p$, which implies that $p∈ ⋂ j = 1 l B j − 1 0$ from Lemma 2.2. Therefore, $F( W r S r A m A m − 1 ⋯ A 1 )⊂D$.

This completes the proof. □

Theorem 2.2 Suppose H, D, C, ${ A i } i = 1 m$, ${ B j } j = 1 l$ and f are the same as those in Theorem  2.1. Let ${ x n }$ be generated by the iterative scheme (A). If ${ α n }$, ${ β n }$ and ${ ϑ n }$ are three sequences in $(0,1)$ and ${ r n }⊂(0,+∞)$ satisfy the following conditions:

1. (i)

$∑ n = 1 ∞ | α n + 1 − α n |<+∞$, and $α n →0$, as $n→∞$;

2. (ii)

$∑ n = 1 ∞ β n =+∞$, $∑ n = 1 ∞ | β n + 1 − β n |<+∞$, and $β n →0$, as $n→∞$;

3. (iii)

$∑ n = 1 ∞ | ϑ n + 1 − ϑ n |<+∞$, and $ϑ n →0$, as $n→∞$;

4. (iv)

$∑ n = 1 ∞ | r n + 1 − r n |<+∞$, and $r n → r ∗ ≥ε>0$, as $n→∞$.

Then ${ x n }$ converges strongly to a point $p 0 ∈D$, which is the unique solution of the following variational inequality:

$〈 f ( p 0 ) − p 0 , p 0 − q 〉 ≥0,∀q∈D.$
(2.22)

Proof We shall split the proof into five steps:

Step 1. ${ x n }$ is bounded.

$∀ p ∈ D , ∥ y n − p ∥ ≤ [ 1 − β n ( 1 − k ) ] ∥ x n − p ∥ + β n ∥ f ( p ) − p ∥ , ∥ u n − p ∥ ≤ [ 1 − ϑ n ( 1 − k ) ] ∥ y n − p ∥ + ϑ n ∥ f ( p ) − p ∥ .$

Letting $δ n = α n + β n + ϑ n −( α n β n + α n ϑ n + β n ϑ n )(1−k)+ α n β n ϑ n ( 1 − k ) 2$. Then

$∥ x n + 1 − p ∥ ≤ [ 1 − α n ( 1 − k ) ] ∥ u n − p ∥ + α n ∥ f ( p ) − p ∥ ≤ [ 1 − α n ( 1 − k ) ] [ 1 − β n ( 1 − k ) ] [ 1 − ϑ n ( 1 − k ) ] ∥ x n − p ∥ + { [ 1 − α n ( 1 − k ) ] ϑ n + α n + [ 1 − α n ( 1 − k ) ] [ 1 − ϑ n ( 1 − k ) ] β n } ∥ f ( p ) − p ∥ = [ 1 − δ n ( 1 − k ) ] ∥ x n − p ∥ + δ n ∥ f ( p ) − p ∥ ≤ max { ∥ x n − p ∥ , 1 1 − k ∥ f ( p ) − p ∥ } , n ≥ 1 .$

By induction, $∥ x n −p∥≤max{∥ x 1 −p∥, 1 1 − k ∥f(p)−p∥}$, $n≥1$. Thus ${ x n }$ is bounded.

Step 2. $lim n → ∞ ∥ x n + 1 − x n ∥=0$ and $lim n → ∞ ∥ x n − u n ∥=0$.

In fact,

$∥ y n − y n − 1 ∥ ≤ | β n − β n − 1 | ∥ f ( x n ) − S r n A m ⋯ A 1 x n ∥ + β n − 1 ∥ f ( x n ) − f ( x n − 1 ) ∥ + ( 1 − β n − 1 ) ∥ S r n A m ⋯ A 1 x n − S r n − 1 A m ⋯ A 1 x n − 1 ∥ ≤ 2 M 1 | β n − β n − 1 | + β n − 1 k ∥ x n − x n − 1 ∥ + ( 1 − β n − 1 ) ∥ S r n A m ⋯ A 1 x n − S r n − 1 A m ⋯ A 1 x n − 1 ∥ .$
(2.23)

Next we discuss $∥ S r n A m ⋯ A 1 x n − S r n − 1 A m ⋯ A 1 x n − 1 ∥$.

If $r n − 1 ≤ r n$, then in view of Lemma 1.4,

$∥ J r n A 1 x n − J r n − 1 A 1 x n − 1 ∥ = ∥ J r n − 1 A 1 ( r n − 1 r n x n + ( 1 − r n − 1 r n ) J r n A 1 x n ) − J r n − 1 A 1 x n − 1 ∥ ≤ ∥ r n − 1 r n x n + ( 1 − r n − 1 r n ) J r n A 1 x n − x n − 1 ∥ ≤ r n − 1 r n ∥ x n − x n − 1 ∥ + ( 1 − r n − 1 r n ) ∥ J r n A 1 x n − x n − 1 ∥ ≤ ∥ x n − x n − 1 ∥ + r n − r n − 1 ε ∥ J r n A 1 x n − x n − 1 ∥ .$
(2.24)

For $∀p∈D$, let $M 4 = M 1 +∥p∥$, then

$∥ J r n A 1 x n − x n − 1 ∥ ≤ ∥ ( I + r n A 1 ) − 1 x n − p ∥ + ∥ p − x n − 1 ∥ ≤ ∥ x n − p ∥ + ∥ p − x n − 1 ∥ ≤ 2 M 4 .$
(2.25)

From (2.24) and (2.25), we know that

$∥ J r n A 1 x n − J r n − 1 A 1 x n − 1 ∥ ≤∥ x n − x n − 1 ∥+2 M 4 r n − r n − 1 ε .$
(2.26)

Notice that $S r n A 2 A 1 x n = J r n A 2 J r n A 1 x n$ and $S r n − 1 A 2 A 1 x n − 1 = J r n − 1 A 2 J r n − 1 A 1 x n − 1$; similar to (2.26), we have

$∥ S r n A 2 A 1 x n − S r n − 1 A 2 A 1 x n − 1 ∥ ≤ ∥ J r n A 1 x n − J r n − 1 A 1 x n − 1 ∥ +2 M 4 r n − r n − 1 ε .$
(2.27)

Following from (2.26) and (2.27), we have

$∥ S r n A 2 A 1 x n − S r n − 1 A 2 A 1 x n − 1 ∥ ≤∥ x n − x n − 1 ∥+2×2 M 4 r n − r n − 1 ε .$

Then by induction, we can get the following result:

$∥ S r n A m ⋯ A 1 x n − S r n − 1 A m ⋯ A 1 x n − 1 ∥ ≤∥ x n − x n − 1 ∥+2×m M 4 r n − r n − 1 ε .$
(2.28)

Putting (2.28) into (2.23), and letting $M 5 =max{ 2 × m M 4 ε ,2 M 1 }$,

$∥ y n − y n − 1 ∥ ≤ [ 1 − β n ( 1 − k ) ] ∥ x n − x n − 1 ∥ + 2 × m M 4 ε ( r n − r n − 1 ) + 2 M 1 | β n − β n − 1 | ≤ [ 1 − β n ( 1 − k ) ] ∥ x n − x n − 1 ∥ + M 5 [ ( r n − r n − 1 ) + | β n − β n − 1 | ] .$
(2.29)

If $r n ≤ r n − 1$, then imitating the above proof, we have

$∥ y n − y n − 1 ∥≤ [ 1 − β n ( 1 − k ) ] ∥ x n − x n − 1 ∥+ M 5 [ ( r n − 1 − r n ) + | β n − β n − 1 | ] .$
(2.30)

Combining (2.29) and (2.30),

$∥ y n − y n − 1 ∥≤ [ 1 − β n ( 1 − k ) ] ∥ x n − x n − 1 ∥+ M 5 ( | r n − 1 − r n | + | β n − β n − 1 | ) .$
(2.31)

Similar to the discussion of (2.24), we have

$∥ W r n y n − W r n − 1 y n − 1 ∥ ≤ a 0 ∥ y n − y n − 1 ∥ + ∑ j = 1 l a j ∥ J r n B j y n − J r n − 1 B j y n − 1 ∥ ≤ a 0 ∥ y n − y n − 1 ∥ + ∑ j = 1 l a j ( ∥ y n − y n − 1 ∥ + | r n − r n − 1 | ε ∥ J r n B j y n − y n − 1 ∥ ) ≤ ∥ y n − y n − 1 ∥ + 2 M 1 | r n − r n − 1 | ε .$
(2.32)

Using (2.32), then

$∥ u n − u n − 1 ∥ ≤ ϑ n k ∥ y n − y n − 1 ∥ + | ϑ n − ϑ n − 1 | ( ∥ f ( y n − 1 ) ∥ + ∥ W r n − 1 y n − 1 ∥ ) + ( 1 − ϑ n ) ∥ W r n y n − W r n − 1 y n − 1 ∥ ≤ [ 1 − ϑ n ( 1 − k ) ] ∥ y n − y n − 1 ∥ + 2 M 1 | ϑ n − ϑ n − 1 | + 2 M 1 ε | r n − r n − 1 | .$
(2.33)

Based on (2.31) and (2.33), and letting $M 6 = M 5 + 2 M 1 ε$, we have

$∥ x n + 1 − x n ∥ ≤ α n ∥ f ( u n ) − f ( u n − 1 ) ∥ + | α n − α n − 1 | ∥ f ( u n − 1 ) ∥ + ( 1 − α n ) ∥ u n − u n − 1 ∥ + | α n − α n − 1 | ∥ u n − 1 ∥ ≤ [ 1 − α n ( 1 − k ) ] ∥ u n − u n − 1 ∥ + 2 M 1 | α n − α n − 1 | ≤ [ 1 − α n ( 1 − k ) ] [ 1 − ϑ n ( 1 − k ) ] ∥ y n − y n − 1 ∥ + 2 M 1 ( | ϑ n − ϑ n − 1 | + | α n − α n − 1 | ) + 2 M 1 ε | r n − r n − 1 | ≤ [ 1 − β n ( 1 − k ) ] ∥ x n − x n − 1 ∥ + M 6 ( | r n − r n − 1 | + | β n − β n − 1 | + | α n − α n − 1 | + | ϑ n − ϑ n − 1 | ) .$

In view of Lemma 1.6, we know that $∥ x n + 1 − x n ∥→0$, as $n→∞$. Combining with the fact that $∥ x n + 1 − u n ∥= α n ∥f( u n )− u n ∥→0$, we can easily know that $∥ x n − u n ∥≤∥ x n + 1 − x n ∥+∥ x n + 1 − u n ∥→0$, as $n→∞$.

Step 3. $∥ W r u n − u n ∥→0$, and $∥ S r A m A m − 1 ⋯ A 1 u n − u n ∥→0$, as $n→∞$. In view of Lemma 1.4 again, we know that

$∥ S r n A 1 x n − S r A 1 x n ∥ = ∥ J r A 1 ( r r n x n + ( 1 − r r n ) J r n A 1 x n ) − J r A 1 x n ∥ ≤ | 1 − r r n | ∥ J r n A 1 x n − x n ∥ ≤ 2 M 1 | 1 − r r n | ,$

and then

$∥ S r n A 2 A 1 x n − S r A 2 A 1 x n ∥ ≤ r r n ∥ J r n A 1 x n − J r A 1 x n ∥ + | 1 − r r n | ∥ S r n A 2 A 1 x n − J r A 1 x n ∥ ≤ 2 M 1 | 1 − r r n | ( r r n + 1 ) .$

By induction,

$∥ S r n A m ⋯ A 1 x n − S r A m ⋯ A 1 x n ∥ ≤2 M 1 | 1 − r r n | [ ( r r n ) m − 1 + ⋯ + r r n + 1 ] →0,$
(2.34)

as $n→∞$, since $r n → r ∗$.

$∀p∈D$, continuing the computation of (2.15), we have

$0 ≤ [ 1 − α n ( 1 − k ) ] ( 1 − ϑ n ) ∑ j = 1 l a j ∥ ( I + r n B j ) − 1 y n − y n ∥ 2 ≤ ∥ x n − p ∥ 2 − ∥ x n + 1 − p ∥ 2 + M 3 ( α n + β n + ϑ n ) .$

From step 2, we know that $∥ x n − x n + 1 ∥→0$, then $∥ ( I + r n B j ) − 1 y n − y n ∥→0$, $j=1,2,…,l$, which implies that

(2.35)

Noticing that $∥ u n − W r n y n ∥= ϑ n ∥f( y n )− W r n y n ∥→0$, and $∥ y n − S r n A m ⋯ A 1 x n ∥= β n ∥f( x n )− S r n A m ⋯ A 1 x n ∥→0$, as $n→∞$.

Combining with the facts of (2.34), (2.35), and step 2, we know that

Using Lemma 1.4 again, then

$∥ W r n y n − W r y n ∥≤ ∑ j = 1 l a j ∥ J r n B j y n − J r B j y n ∥ ≤2 M 1 (1− a 0 ) | 1 − r r n | →0.$

Since $∥ W r u n − W r y n ∥≤∥ u n − y n ∥≤ ϑ n ∥f( y n )− y n ∥+(1− ϑ n )∥ W r n y n − y n ∥→0$, then $∥ u n − W r u n ∥≤∥ u n − W r n y n ∥+∥ W r n y n − W r y n ∥+∥ W r y n − W r u n ∥→0$, as $n→∞$.

Step 4. $lim sup n → ∞ 〈f( p 0 )− p 0 , u n − p 0 〉≤0$, $lim sup n → ∞ 〈f( p 0 )− p 0 , x n + 1 − p 0 〉≤0$, $lim sup n → ∞ 〈f( p 0 )− p 0 , y n − p 0 〉≤0$, where $p 0$ satisfies (2.22).

Using Lemmas 1.5 and 2.3, we know that if we let $z t =tf( z t )+(1−t) W r S r A m A m − 1 ⋯ A 1 z t$, $r>0$ and $t∈(0,1)$, then $z t → p 0 ∈F( W r S r A m A m − 1 ⋯ A 1 )=D$, as $t→ 0 +$. And, $p 0$ satisfies (2.22).

From step 3, we may choose $t n ∈(0,1)$ such that $t n →0$, $∥ S r A m ⋯ A 1 u n − u n ∥ t n →0$, and $∥ W r u n − u n ∥ t n →0$, as $n→∞$.

Using Lemma 1.7,

$∥ z t n − u n ∥ 2 ≤ ( 1 − t n ) 2 ∥ W r S r A m ⋯ A 1 z t n − u n ∥ 2 + 2 t n 〈 f ( z t n ) − u n , z t n − u n 〉 ≤ ( 1 − t n ) 2 [ ∥ z t n − u n ∥ + ∥ u n − S r A m ⋯ A 1 u n ∥ + ∥ u n − W r u n ∥ ] 2 + 2 t n 〈 f ( z t n ) − z t n , z t n − u n 〉 + 2 t n ∥ z t n − u n ∥ 2 .$

Then

$〈 f ( z t n ) − z t n , u n − z t n 〉 ≤ t n 2 ∥ z t n − u n ∥ 2 + ( 1 − t n ) 2 t n ∥ z t n − u n ∥ ( ∥ S r A m ⋯ A 1 u n − u n ∥ + ∥ u n − W r u n ∥ ) + ( 1 − t n ) 2 2 t n ( ∥ S r A m ⋯ A 1 u n − u n ∥ + ∥ W r u n − u n ∥ ) 2 .$
(2.36)

Since ${ S r A m ⋯ A 1 u n }$, ${ W r u n }$, ${ x n }$, ${ u n }$ and ${ z t n }$ are all bounded, and $∥ S r A m ⋯ A 1 u n − u n ∥ t n →0$, and $∥ W r u n − u n ∥ t n →0$, from (2.36), $lim sup n → ∞ 〈f( z t n )− z t n , u n − z t n 〉≤0$.

Recalling that $z t n → p 0$, then $〈 z t n − p 0 , u n − z t n 〉→0$. Thus $lim sup n → ∞ 〈f( z t n )− p 0 , u n − z t n 〉≤0$. Since $〈f( z t n )− p 0 , u n − p 0 〉=〈f( z t n )− p 0 , u n − z t n 〉+〈f( z t n )− p 0 , z t n − p 0 〉$, then $lim sup n → ∞ 〈f( p 0 )− p 0 , u n − p 0 〉≤0$. Then from step 2, $lim sup n → ∞ 〈f( p 0 )− p 0 , x n + 1 − p 0 〉≤0$.

Noticing that

$〈 f ( p 0 ) − p 0 , y n − p 0 〉 = 〈 f ( p 0 ) − p 0 , y n − W r n y n 〉 + 〈 f ( p 0 ) − p 0 , W r n y n − u n 〉 + 〈 f ( p 0 ) − p 0 , u n − x n + 1 〉 + 〈 f ( p 0 ) − p 0 , x n + 1 − p 0 〉 ,$

and using (2.35), iterative scheme (A) and the result of step 2, we have $lim sup n → ∞ 〈f( p 0 )− p 0 , y n − p 0 〉≤0$.

Step 5. $x n → p 0$, which satisfies (2.22), as $n→∞$.

Using Lemma 1.7, we know that

$∥ y n − p 0 ∥ 2 ≤ [ 1 − β n ( 1 − k ) ] ∥ x n − p 0 ∥ 2 +2 β n 〈 f ( p 0 ) − p 0 , y n − p 0 〉 .$
(2.37)

We have

$∥ u n − p 0 ∥ 2 ≤ [ 1 − ϑ n ( 1 − k ) ] ∥ y n − p 0 ∥ 2 +2 ϑ n 〈 f ( p 0 ) − p 0 , u n − p 0 〉 .$
(2.38)

Letting $M 7 =max{ ( M 1 + ∥ p 0 ∥ ) 2 ,2( M 1 +∥ p 0 ∥)(∥f( p 0 )∥+∥ p 0 ∥)}$ and using (2.37) and (2.38), we have

$∥ x n + 1 − p 0 ∥ 2 ≤ [ 1 − α n ( 1 − k ) ] ∥ u n − p 0 ∥ 2 + 2 α n 〈 f ( p 0 ) − p 0 , x n + 1 − p 0 〉 ≤ [ 1 − α n ( 1 − k ) ] [ 1 − β n ( 1 − k ) ] [ 1 − ϑ n ( 1 − k ) ] ∥ x n − p 0 ∥ 2 + 2 [ 1 − α n ( 1 − k ) ] [ 1 − ϑ n ( 1 − k ) ] β n 〈 f ( p 0 ) − p 0 , y n − p 0 〉 + 2 [ 1 − α n ( 1 − k ) ] ϑ n 〈 f ( p 0 ) − p 0 , u n − p 0 〉 + 2 α n 〈 f ( p 0 ) − p 0 , x n + 1 − p 0 〉 ≤ [ 1 − ( 1 − k ) ( α n + β n + ϑ n ) ] ∥ x n − p 0 ∥ 2 + M 7 ( 1 − k ) 2 ( α n β n + β n ϑ n + α n ϑ n ) + 2 α n ϑ n ( 1 − k ) 〈 p 0 − f ( p 0 ) , u n − p 0 〉 + 2 ( α n β n + β n ϑ n ) ( 1 − k ) 〈 p 0 − f ( p 0 ) , y n − p 0 〉 + 2 α n β n ϑ n ( 1 − k ) 2 〈 f ( p 0 ) − p 0 , y n − p 0 〉 + 2 α n 〈 f ( p 0 ) − p 0 , x n + 1 − p 0 〉 + 2 β n 〈 f ( p 0 ) − p 0 , y n − p 0 〉 + 2 ϑ n 〈 f ( p 0 ) − p 0 , u n − p 0 〉 ≤ [ 1 − ( 1 − k ) ( α n + β n + ϑ n ) ] ∥ x n − p 0 ∥ 2 + M 7 ( 1 − k ) 2 ( α n β n + β n ϑ n + α n ϑ n ) + M 7 ( 1 − k ) ( α n β n + β n ϑ n + α n ϑ n ) + 2 M 7 α n β n ϑ n ( 1 − k ) 2 + 2 α n 〈 f ( p 0 ) − p 0 , x n + 1 − p 0 〉 + 2 β n 〈 f ( p 0 ) − p 0 , y n − p 0 〉 + 2 ϑ n 〈 f ( p 0 ) − p 0 , u n − p 0 〉 .$
(2.39)

Let $c n =( α n + β n + ϑ n )(1−k)$, then $c n →0$ and $∑ n = 1 ∞ c n =+∞$.

Let $b n$ = $M 7 [ ( 2 − k ) ( α n β n + β n ϑ n + α n ϑ n ) α n + β n + ϑ n + 2 ( 1 − k ) α n β n ϑ n α n + β n + ϑ n ]$ + $2 α n ( α n + β n + ϑ n ) ( 1 − k ) 〈f( p 0 )− p 0 , x n + 1 − p 0 〉$ + $2 ϑ n ( α n + β n + ϑ n ) ( 1 − k ) 〈f( p 0 )− p 0 , u n − p 0 〉$ + $2 β n ( α n + β n + ϑ n ) ( 1 − k ) 〈f( p 0 )− p 0 , y n − p 0 〉$.

Notice that $lim n → ∞ α n β n + β n ϑ n + α n ϑ n α n + β n + ϑ n =0$, $lim n → ∞ α n β n ϑ n α n + β n + ϑ n =0$ and from the results in step 4, we have $lim sup n → + ∞ b n ≤0$.

Using Lemma 1.6, $x n → p 0$, which satisfies (2.22), as $n→∞$.

This completes the proof. □

Remark 2.2 The iterative construction in this paper generalizes and extends some corresponding ones in [2, 4, 12, 13], etc., in Hilbert spaces.

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## Acknowledgements

This work was supported by the National Natural Science Foundation of China (No. 11071053), Natural Science Foundation of Hebei (No. A2014207010), Key Project of Science and Research of Hebei Education Department (ZH2012080) and Key Project of Science and Research of Hebei University of Economics and Business (2013KYZ01).

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Wei, L., Tan, R. Strong and weak convergence theorems for common zeros of finite accretive mappings. Fixed Point Theory Appl 2014, 77 (2014). https://doi.org/10.1186/1687-1812-2014-77