# A New Iterative Scheme of Modified Mann Iteration in Banach Space

Authors: Jinzuo Chen; Dingping Wu; Caifen Zhang
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We introduce the modified iterations of Mann's type for nonexpansive mappings and asymptotically nonexpansive mappings to have the strong convergence in a uniformly convex Banach space. We study approximation of common fixed point of asymptotically nonexpansive mappings in Banach space by using a new iterative scheme. Applications to the accretive operators are also included.


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## Body

## 1. Introduction

LetE be a real Banach space, C a nonempty closed convex subset of E, and T:C→C a mapping. Recall that T is a nonexpansive mapping [1] if ∥Tx-Ty∥≤∥x-y∥ for all x,y∈C, and T is asymptotically nonexpansive [2] if there exists a sequence {kn} with kn≥1 for all n and lim⁡n→∞kn=1 such that ∥Tnx-Tny∥≤kn∥x-y∥ for all integers n≥1 and x,y∈C. A point x∈C is a fixed point of T provided Tx=x. Denote by Fix⁡(T) the set of fixed points of T; that is, Fix⁡(T)={x∈C:Tx=x}.Iterative methods are often used to solve the fixed point equationTx=x. One classical iteration process is introduced in 1953 by Mann [3] which is well known as Mann iteration process and is defined as follows:
(1)xn+1=αnxn+(1-αn)Txn,n≥0,
where the sequence {αn} is chosen in (0,1) and the initial guess x0∈C is arbitrarily chosen.There exists rich literature on the convergence of Mann iteration for different classes of operators considered on various spaces. Mann’s iteration method (1) has been proved to be a powerful method for solving nonlinear operator equations involving nonexpansive mapping, asymptotically nonexpansive mapping, and other kinds of nonlinear mapping; see [3–11] and the references therein.It is known that Mann’s iteration method (1) is in general not strongly convergent for nonexpansive mappings. So to get strong convergence, one has to modify the iteration method (1). In this regard, we will show the modified iteration in Section 3.Motivated and inspired by the research going on in these fields, we suggest and analyze now new modified Mann’s iteration for finding the common fixed point of the nonexpansive mappings and asymptotically nonexpansive mappings in Banach space. We propose the modified Mann’s iteration and consider the strong convergence of the approximate solutions for nonexpansive and asymptotically nonexpansive in Banach space.We suggest and analyze the following iterative:(2)x0∈Cchosenarbitrarily,yn=βnxn+(1-βn)Txn,zn=γnxn+(1-γn)Sxn,xn+1=αnyn+(1-αn)Rzn,n≥0,(3)x0∈Cchosenarbitrarily,yn=βnxn+(1-βn)Tnxn,zn=γnxn+(1-γn)Snxn,xn+1=αnyn+(1-αn)zn,n≥0,(4)x0∈Cchosenarbitrarily,yn=βnxn+(1-βn)J1,nxn,zn=γnxn+(1-γn)J2,nxn,xn+1=αnyn+(1-αn)J3,nzn,n≥0,
and if there exists two sequences {xn′} and {xn′′} generated by
(5)x0∈Cchosenarbitrarily,yn=βnxn′+(1-βn)Tnxn′,zn=γnxn′′+(1-γn)Snxn′′,xn+1′=αnzn+(1-αn)yn,xn+1′′=αnyn+(1-αn)zn,n≥0,(6)x0∈Cchosenarbitrarily,yn=βnxn′+(1-βn)J1,nxn′,zn=γnxn′′+(1-γn)J2,nxn′′,xn+1′=αnzn+(1-αn)yn,xn+1′′=αnyn+(1-αn)zn,n≥0.Our second modification of Mann’s iteration method (1) is adaption to (2) for finding a zero of an m-accretive operator A, for which we assume that the zero set A-1(0)≠∅. Our iterations process {xn} is given by (4), and sequences {xn′} and {xn′′} are as follows (6), where, for each r>0, Jr=(I+rA)-1 is the resolvent of A. We prove that not only {xn}n∞ defined by (4) but also {xn′} and {xn′′} generated by (6) converge strongly to a zero of A under certain assumptions in a uniformly Banach space.We writexn→x to indicate that the sequence {xn} converges strongly to x. Using F is to denote the set of common fixed point of the mappings T, S, and R, and using F¯ is to denote the set of common fixed points of the mappings T and S.


## 2. Preliminaries

This section collects some lemmas, which will be used in the proofs for the main results in the next section.Lemma 1 (see [8]).
Let{an}, {bn}, and {δn} be sequences of nonnegative real numbers satisfying the inequality
(7)an+1≤(1+δn)an+bn,n≥1.
If ∑n=1∞δn<∞ and ∑n=1∞bn<∞, then(1)
lim⁡n→∞an exists;(2)
lim⁡n→∞an=0 whenever liminf⁡n→∞an=0.Lemma 2 (see [12]).
Suppose thatE is a uniformly convex Banach space and 0<tn<1 for all n∈N. Let {xn} and {yn} be two sequences of E such that limsup⁡n→∞∥xn∥≤r, limsup⁡n→∞∥yn∥≤r and lim⁡n→∞∥tnxn+(1-tn)yn∥=r hold for some r≥0; then lim⁡n→∞∥xn-yn∥=0.Lemma 3 (see [10]).
A mappingT: C→C with a nonempty fixed point set F in C will be said to satisfy Condition (I).
If there is a nondecreasing functionf:[0,∞)→[0,∞) with f(0)=0,f(r)>0 for all r∈(0,∞) such that ∥x-Tx∥≥f(d(x,F)) for all x∈C, where  d(x,F)=inf⁡{∥x-p∥:p∈F}.Lemma 4 (see [13]).
Given a numberr>0, let E be a uniformly convex Banach space; then there exists a continuous strictly increasing function φ:[0,∞]→[0,∞] with φ(0)=0, such that
(8)∥λx+μy+γz∥2≤λ∥x∥2+μ∥y∥2+γ∥z∥2-λμφ(∥x-y∥),
for all λ,μ,γ∈[0,1] and x,y,z∈E such that ∥x∥≤r, ∥y∥≤r and ∥z∥≤r.Lemma 5 (see [14]).
Let{αn}, {βn} be sequences of nonnegative real numbers such that ∑n=1∞αn=∞. If ∑n=1∞αnβn<∞, then liminf⁡n→∞βn=0.Lemma 6 (see [15]).
Forλ>0, μ>0, and x∈E, the following identity holds
(9)Jλx=Jμ(μλ+(1-μλ)Jλx).


## 3. Convergence to a Common Fixed Point of Nonexpansive Mappings

In this part, we prove our main theorem for finding a common fixed point of nonexpansive mappings in Banach space.Theorem 7.
LetC be a nonempty closed convex subset of a uniformly convex Banach space E, and let T, S, and R be three nonexpansive commuting mappings of C satisfying Condition (I) and F≠∅. Given that {αn}, {βn}, and {γn} are sequences in (0,1) such that ∑‍αn<∞, ∑‍γnβn=∞, ∑‍(1-γn)<∞ for all n≥1.
Define a sequence{xn}n=0∞ in C by algorithm (2); then {xn}n=0∞ strongly converges to a common fixed point of T, S, and R.Proof.
First, we observe that{xn} is bounded; if we take an arbitrary fixed point q of F, noting that ∥yn-q∥≤∥xn-q∥ and ∥zn-q∥≤∥xn-q∥, we have
(10)∥xn+1-q∥=∥αnyn+(1-αn)Rzn-q∥≤αn∥yn-Rzn∥+∥Rzn-q∥≤(1+2αn)∥xn-q∥.
By Lemma 1 and ∑‍αn<∞, lim⁡n→∞∥xn-q∥ exists. Denote
(11)lim⁡n→∞∥xn-q∥=c.
Hence, {xn} is bounded, so are {yn} and {zn}. Now
(12)∥xn+1-q∥=∥αnyn+(1-αn)Rzn-q∥≤αn∥yn-Rzn∥+∥zn-q∥.
Since ∥zn-q∥≤∥xn-q∥, this implies that
(13)lim⁡n→∞∥zn-q∥=lim⁡n→∞∥xn-q∥=c.
Moreover, ∥Sxn-q∥≤∥xn-q∥ implies that
(14)limsup⁡n→∞∥Sxn-q∥≤c.
Thus,
(15)c=lim⁡n→∞∥zn-q∥=lim⁡n→∞∥γnxn+(1-γn)Sxn-q∥=lim⁡n→∞∥γn(xn-q)+(1-γn)(Sxn-q)∥,
given by Lemma 2 that
(16)lim⁡n→∞∥Sxn-xn∥=0.
By (10) and ∑‍αn<∞, then we have
(17)∥xn+m-q∥≤(1+2αn+m-1)∥xn+m-1-q∥≤e2αn+m-1∥xn+m-1-q∥≤⋯≤e2∑i=nn+m-1αi∥xn-q∥.
That is,
(18)∥xn+m-q∥≤M∥xn-q∥,
where M=e2∑i=nn+m-1αi for all m,n∈ℕ, for all q∈F and for M>0.
Next, we prove that{xn}n=0∞ is a Cauchy sequence.
Sinceq∈F arbitrarily and lim⁡n→∞∥xn-q∥ exists, consequently, d(xn,F) exists by Lemma 3. From Lemma 3 and (16), we get
(19)lim⁡n→∞f(d(xn,F))≤lim⁡n→∞∥xn-Sxn∥=0.
Since f:[0,∞)→[0,∞) is a nondecreasing function satisfying f(0)=0, f(r)>0 for all r∈(0,∞), therefore, we have lim⁡n→∞d(xn,F)=0.
Letε>0; since lim⁡n→∞d(xn,F)=0, therefore, there exists a constant n0 such that, for all n≥n0, we have d(xn,F)≤ε/2M. There must exist p1∈F, such that
(20)d(xn,p1)≤ε2M.
From (18), it can be obtained that, when n≥n0 and m∈ℕ,
(21)∥xn+m-xn∥≤∥xn+m-p1∥+∥xn-p1∥≤2M∥xn-p1∥≤2M·ε2M=ε.
This implies that{xn}n=0∞ is a Cauchy sequence in a closed convex subset C of a Banach space E. Thus, it must converge to a point in C; let lim⁡n→∞xn=p.
For allϵ>0, as lim⁡n→∞xn=p, thus, there exists a number n1 such that, when n2≥n1,
(22)∥xn2-p∥≤ϵ4.
In fact,lim⁡n→∞d(xn,F)=0 implies that using number n2 above, when n≥n2, we have d(xn,F)≤ϵ/8. In particular, d(xn2,F)≤ϵ/8. Thus, there must exist p-∈F, such that
(23)∥xn2-p-∥=d(xn2,p-)≤ϵ8.
From (22) and (23), we get
(24)∥Sp-p∥=∥Sp-p-+Sxn2-p-+p--xn2+xn2-p+p--Sxn2∥≤∥Sp-p-∥+∥xn2-p-∥+∥xn2-p∥+2∥Sxn2-p-∥≤∥p-p-∥+3∥xn2-p-∥+∥xn2-p∥≤4∥xn2-p-∥+2∥xn2-p∥≤4ϵ8+2ϵ4=ϵ.
As ϵ is an arbitrary positive number, thus, Sp=p.
Let(25)un+1=γnxn+βnTxn+(1-βn-γn)Txn.
Then we have
(26)∥un+1-q∥2=∥γnxn+βnTxn+(1-βn-γn)Txn-q∥2≤γn∥xn-q∥2+βn∥Txn-q∥2+(1-βn-γn)∥Txn-q∥2-γnβnφ∥xn-Txn∥≤∥xn-q∥2-γnβnφ∥xn-Txn∥;
hence,
(27)γnβnφ∥xn-Txn∥≤∥xn-q∥2-∥un+1-q∥2,
for q∈F. Summing from n=1 to ∞, we have
(28)∑n=1∞‍γnβnφ∥xn-Txn∥≤∑n=1∞‍(∥xn-q∥2-∥un+1-q∥2)≤2K∑n=1∞‍∥un+1-xn∥≤4K2∑n=1∞‍(1-γn),
where K=sup⁡n∈ℕ{∥xn-q∥}; since ∑n=1∞(1-γn)<∞, we get
(29)∑n=1∞‍γnβnφ∥xn-Txn∥≤4K2∑n=1∞‍(1-γn)<∞.
Since ∑n=1∞γnβn=∞, from Lemma 5, we get liminf⁡n→∞φ∥Txn-xn∥=0. Hence,
(30)liminf⁡n→∞∥Txn-xn∥=0.
Since T and R are nonexpansive mappings, we have
(31)∥Txn+1-xn+1∥=∥Txn+1-RTxn+RTxn+αnRTxn-αnRTxn-αnyn-(1-αn)Rzn∥≤∥Txn+1-RTxn∥+(1-αn)∥RTxn-Rzn∥+αn∥RTxn-yn∥≤∥xn+1-Rxn∥+(1-αn)∥Txn-zn∥+αn∥RTxn-yn∥≤αn∥yn-Rxn∥+2(1-αn)∥zn-xn∥+(1-αn)∥Txn-xn∥+αn∥Txn-yn∥+αn∥Rxn-xn∥≤αn∥yn-xn∥+2αn∥Rxn-xn∥+2(1-αn)(1-γn)∥Sxn-xn∥+(1-αn)∥Txn-xn∥+αnβn∥Txn-xn∥≤∥Txn-xn∥+2αn∥Rxn-xn∥+2(1-αn)(1-γn)∥Sxn-xn∥.
Since ∑‍αn<∞, ∑n=1∞(1-γn)<∞, it follows from Lemma 1 that lim⁡n→∞∥Txn-xn∥ exists. Therefore, from (30), we get
(32)lim⁡n→∞∥Txn-xn∥=0.
Let
(33)vn+1=γnxn+βnRxn+(1-βn-γn)Rxn.
Using the same argument we can get
(34)liminf⁡n→∞∥Rxn-xn∥=0.
Since R is a nonexpansive mapping, we have
(35)∥Rxn+1-xn+1∥=∥Rxn+1-Rxn+Rxn+αnRxn-αnRxn-αnyn-(1-αn)Rzn∥≤∥xn+1-xn∥+(1-αn)∥xn-zn∥+αn∥Rxn-xn∥+αn∥yn-xn∥≤αn∥Rxn-xn∥+2αn∥yn-xn∥+(1-αn)∥Rzn-xn∥+(1-αn)∥zn-xn∥≤∥Rxn-xn∥+2αn∥yn-xn∥+2(1-αn)∥zn-xn∥=∥Rxn-xn∥+2αn(1-βn)∥Txn-xn∥+2(1-αn)(1-γn)∥Sxn-xn∥.
Since ∑‍αn<∞, ∑‍(1-γn)<∞, it follows from Lemma 1 that lim⁡n→∞∥Rxn-xn∥ exists. Therefore, from (34), we get
(36)lim⁡n→∞∥Rxn-xn∥=0.
Then using the same argument, we can show that{xn}n=0∞ converges strongly to a common fixed point of T, S, and R.


## 4. Convergence to a Common Fixed Point of Asymptotically Nonexpansive Mappings

### 4.1. There Exists One Sequence{xn}

In this part, we prove our main theorem for finding a common fixed point of asymptotically nonexpansive mappings in Banach space in the case of one sequence.Theorem 8.
LetC be a nonempty closed convex subset of a uniformly convex Banach space E, and let T and S be two asymptotically nonexpansive mappings of C satisfying Condition (I) and F¯≠∅. Given {αn}, {βn}, and {γn} are sequences in (0,1) and {kn} with kn≥1 such that ∑‍αn<∞, ∑‍(1-βn)<∞ and ∑‍(kn-1)<∞ for all n≥1.
Define a sequence{xn}n=0∞ in C by algorithm (3); then {xn}n=0∞ strongly converges to a common fixed point of T and S.Proof.
First, we observe that{xn} is bounded; if we take an arbitrary fixed point q of F¯, noting that ∥yn-q∥≤kn∥xn-q∥ and ∥zn-q∥≤kn∥xn-q∥, we have
(37)∥xn+1-q∥=∥αnyn+(1-αn)zn-q∥≤αn∥yn-q∥+(1-αn)∥Zn-q∥≤(1+kn-1)∥xn-q∥.
By Lemma 1 and ∑‍(kn-1)<∞, thus, lim⁡n→∞∥xn-q∥ exists. Denote lim⁡n→∞∥xn-q∥=c, and put k∞=sup⁡{kn:n≥1}<∞. Hence, {xn} is bounded, so are {yn} and {zn}. Now
(38)∥xn+1-q∥=∥αnyn+(1-αn)zn-q∥≤αn∥yn-zn∥+∥zn-q∥.
Since ∥zn-q∥≤kn∥xn-q∥, this implies that
(39)lim⁡n→∞∥zn-q∥=lim⁡n→∞∥xn-q∥=c.
Moreover, ∥Snxn-q∥≤kn∥xn-q∥ implies that
(40)limsup⁡n→∞∥Snxn-q∥≤c.
Thus,
(41)c=lim⁡n→∞∥zn-q∥=lim⁡n→∞∥γnxn+(1-γn)Snxn-q∥=lim⁡n→∞∥γn(xn-q)+(1-γn)(Snxn-q)∥,
given by Lemma 2 that
(42)lim⁡n→∞∥Snxn-xn∥=0.
Now,
(43)∥zn-xn∥=∥γnxn+(1-γn)Snxn-xn∥≤(1-γn)∥(Snxn-xn)∥.
Hence, by (42),
(44)lim⁡n→∞∥zn-xn∥=0.
Also note that
(45)∥xn+1-xn∥=∥αnyn+(1-αn)zn-xn∥≤αn∥yn-xn∥+(1-αn)∥zn-xn∥,
so that condition ∑‍αn<∞ and (44) give
(46)lim⁡n→∞∥xn+1-xn∥=0.
Next, we show lim⁡n→∞∥xn+1-Sxn∥=0.
We have(47)∥xn+1-Sxn+1∥≤∥xn+1-Sn+1xn+1∥+∥Sn+1xn+1-Sn+1xn∥+∥Sn+1xn-Sxn+1∥≤∥xn+1-Sn+1xn+1∥+k∞∥xn+1-xn∥+k∞∥Snxn-xn+1∥≤∥xn+1-Sn+1xn+1∥+2k∞∥xn+1-xn∥+k∞∥Snxn-xn∥.
Hence, by (42) and (46), we get
(48)lim⁡n→∞∥xn-Sxn∥=0.
By (37) we assume that kn=1+rn, so ∑‍rn<∞ for ∑‍(kn-1)<∞, and now,
(49)∥xn+m-q∥≤(1+rn+m-1)∥xn+m-1-q∥≤ern+m-1∥xn+m-1-q∥≤⋯≤e∑i=nn+m-1ri∥xn-q∥.
That is,
(50)∥xn+m-q∥≤M∥xn-q∥,
where M=e∑i=nn+m-1ri, for all m,n∈ℕ, for all q∈F¯ and for M>0.
Next, we prove that{xn}n=0∞ is a Cauchy sequence.
As proved in Theorem7, it is easy to see that {xn}n=0∞ is a Cauchy sequence in a closed convex subset C of a Banach space E. Thus, it must converge to a point in C; let lim⁡n→∞xn=p.
For allϵ>0, as lim⁡n→∞xn=p, thus, there exists a number n1 such that, when n2≥n1,
(51)∥xn2-p∥≤ϵ2+2k∞.
In fact,lim⁡n→∞d(xn,F¯)=0 implies that using number n2 above, when n≥n2, we get d(xn,F¯)≤ϵ/(2+6k∞). In particular, d(xn2,F¯)≤ϵ/(2+6k∞). Thus, there must exist p-∈F¯, such that
(52)∥xn2-p-∥=d(xn2,p-)=ϵ2+6k∞.
From (51) and (52), using the same argument in Theorem 7, we get Sp=p.
Now, we return to provelim⁡n→∞∥xn-Txn∥=0.
It is easy to see thatlim⁡n→∞∥yn-xn∥=0, and then combined with (44), we have lim⁡n→∞∥zn-yn∥=0; thus, lim⁡n→∞∥zn-q∥=lim⁡n→∞∥yn-q∥, and then we get
(53)lim⁡n→∞∥yn-q∥=lim⁡n→∞∥xn-q∥=c.
Moreover, ∥Tnxn-q∥≤kn∥xn-q∥ implies that
(54)limsup⁡n→∞∥Tnxn-q∥≤c;
thus,
(55)lim⁡n→∞∥xn-Tnxn∥=0.
Therefore, we have
(56)lim⁡n→∞∥xn-Txn∥=0.
Thus, we can show that {xn}n=0∞ converges strongly to a common fixed point of T and S immediately.


### 4.2. There Exist Two Sequences{xn′} and {xn′′}

In this part, we prove our main theorem for finding a common fixed point of asymptotically nonexpansive mappings in Banach space in the case of two sequences.Theorem 9.
LetC be a nonempty bounded closed convex subset of a uniformly convex Banach space E, and let T and S be two asymptotically nonexpansive mappings of C satisfying Condition (I) and F¯≠∅. Given {αn}, {βn}, and {γn} are sequences in (0,1) and {kn} with 1≤kn<∞ such that ∑‍αn<∞, ∑‍βn<∞, βn>γn and ∑‍(kn-1)<∞ for all n≥1.
Define two sequences{xn′}n=0∞ and {xn′′}n=0∞ in C by the algorithm (5); then {xn′}n=0∞ and {xn′′}n=0∞ strongly converge to a common fixed point of T and S.Proof.
By the boundedness ofC, we obverse that both {xn′} and {xn′′} are bounded; if we take an arbitrary fixed point q of F¯, noting that ∥yn-q∥≤kn∥xn′-q∥ and ∥zn-q∥≤kn∥xn′′-q∥, we have
(57)∥xn+1′-q∥=∥αnzn+(1-αn)yn-q∥≤αn∥zn-yn∥+∥yn-q∥=(1+αnkn+kn-1)∥xn′-q∥+αnkn∥xn′′-q∥.
By Lemma 1 and ∑‍αn<∞, ∑‍(kn-1)<∞, thus, lim⁡n→∞∥xn′-q∥ exists. Denote
(58)lim⁡n→∞∥xn′-q∥=c′.
Similarly, we have
(59)lim⁡n→∞∥xn′′-q∥=c′′.
Since both {xn′} and {xn′′} are bounded, put k∞=sup⁡{kn:n≥1}<∞. We get that {yn} and {zn} are bounded. Now
(60)∥xn+1′-q∥=∥αnzn+(1-αn)yn-q∥≤αn∥zn-yn∥+∥yn-q∥.
Since ∥yn-q∥≤kn∥xn′-q∥, this implies that
(61)lim⁡n→∞∥yn-q∥=lim⁡n→∞∥xn′-q∥=c′.
Moreover, ∥Tnxn′-q∥≤kn∥xn′-q∥ implies that
(62)limsup⁡n→∞∥Tnxn′-q∥≤c′.
Thus,
(63)c′=lim⁡n→∞∥yn-q∥=lim⁡n→∞∥βnxn′+(1-βn)Tnxn′-q∥=lim⁡n→∞∥βn(xn′-q)+(1-βn)(Tnxn′-q)∥,
given by Lemma 2 that
(64)lim⁡n→∞∥Tnxn′-xn′∥=0.
Also note that
(65)∥xn+1′-xn′∥=∥αnzn+(1-αn)yn-xn′∥≤αn∥zn-yn∥+(1-βn)∥(Tnxn′-xn′)∥,
so that condition ∑‍αn<∞ and (64) give
(66)lim⁡n→∞∥xn+1′-xn′∥=0.
We have
(67)∥xn+1′-Txn+1′∥≤∥xn+1′-Tn+1xn+1′∥+∥Tn+1xn+1′-Tn+1xn′∥+∥Tn+1xn′-Txn+1′∥≤∥xn+1′-Tn+1xn+1′∥+k∞∥xn+1′-xn′∥+k∞∥Tnxn′-xn+1′∥≤∥xn+1′-Tn+1xn+1′∥+2k∞∥xn+1′-xn′∥+k∞∥Tnxn′-xn′∥.
Hence, by (64) and (66), we get
(68)lim⁡n→∞∥xn′-Txn′∥=0,
and then we assume that kn=1+rn, so ∑‍rn<∞ for ∑‍(kn-1)<∞; now by (57), we obtain that
(69)∥xn+m′-q∥≤(1+αn+m-1)(1+rn+m-1)∥xn+m-1′-q∥+sn+m-1≤eαn+m-1ern+m-1∥xn+m-1′-q∥+sn+m-1≤…≤e∑i=nn+m-1αie∑i=nn+m-1ri∥xn′-q∥+e∑i=nn+m-1αie∑i=nn+m-1ri∑i=nn+m-1‍si,
where si=αiki∥xn′′-q∥.
By∑‍αn<∞ and the convergence of {rn}, that is,
(70)∥xn+m′-q∥≤M(∥xn′-q∥+∑i=n∞‍si),
where M=e∑i=nn+m-1αie∑i=nn+m-1ri, si=αiki∥xn′′-q∥, for all m,n∈ℕ, q∈F¯ and for M>0.
Next, we prove that{xn′}n=0∞ is a Cauchy sequence.
Sinceq∈F¯ arbitrarily and lim⁡n→∞∥xn′-q∥ exists, consequently, d(xn′,F¯) exists by Lemma 3. From Lemma 3 and (68), we get
(71)lim⁡n→∞f(d(xn′,F¯))≤lim⁡n→∞∥xn′-Txn′∥=0.
Since f:[0,∞)→[0,∞) is a nondecreasing function satisfying f(0)=0, f(r)>0 for all r∈(0,∞), therefore, we have
(72)lim⁡n→∞d(xn′,F¯)=0.
Let ε>0, since lim⁡n→∞d(xn′,F¯)=0 and ∑i=1∞si<∞, therefore, there exists a constant n0 such that, for all n≥n0, we have
(73)d(xn′,F¯)≤ε3M,∑j=n0∞‍sj≤ε6M,
in particular,
(74)d(xn0′,F¯)≤ε3M.
There must exist p1∈F¯ such that
(75)d(xn0′,p1)≤ε3M.
From (70), it can be obtained that, when n≥n0,
(76)∥xn+m′-xn′∥≤∥xn+m′-p1∥+∥xn′-p1∥≤2M(∥xn0′-p1∥+∑j=n0n0+m-1‍sj)≤2M(ε3M+ε6M)=ε.
This implies that {xn′}n=0∞ is a Cauchy sequence in a closed convex subset C of a Banach space E. Thus, it must converge to a point in C; let lim⁡n→∞xn′=x′.
For allϵ>0, on lines similar to Theorem 8, from lim⁡n→∞xn′=x′, it can be proved that ∥Tx′-x′∥≤ϵ. As ϵ is an arbitrary positive number, thus, Tx′=x′. Similarly, we have lim⁡n→∞∥Sxn′′-xn′′∥=0, and then Sx′′=x′′ (xn′′→x′′ as n→∞).
Finally, we provex′=x′′.
Letwn+1=αnTnxn′+(1-αn)Snxn′, and put ∥xn′′-q∥≥∥xn′-q∥, for all n≥1. Then,
(77)∥wn+1-q∥≤αnkn∥xn′-q∥+(1-αn)kn∥xn′′-q∥≤knmax⁡⁡{∥xn′-q∥,∥xn′′-q∥}≤kn∥xn′′-q∥.
We have lim⁡n→∞∥wn+1-q∥≤lim⁡n→∞∥xn′′-q∥. Now,
(78)∥xn+1′′-q∥=∥αnyn+(1-αn)zn-q∥≤βn∥αnxn′+(1-αn)xn′′∥+∥αnTnxn′+(1-αn)Snxn′′-q∥=βn∥αnxn′+(1-αn)xn′′∥+∥wn+1-q∥.
Since ∑‍βn<∞ and the boundedness of {xn′} and {xn′′}, we get
(79)lim⁡n→∞∥wn+1-q∥=c′′.
Moreover, ∥Tnxn′-q∥≤kn∥xn′-q∥ and ∥Snxn′′-q∥≤kn∥xn′′-q∥ imply that
(80)limsup⁡n→∞∥Tnxn′-q∥≤c′′,limsup⁡n→∞∥Snxn′′-q∥≤c′′.
Thus,
(81)c′′=lim⁡n→∞∥wn+1-q∥=∥αnTnxn′+(1-αn)Snxn′′-q∥=∥αn(Tnxn′-q)+(1-αn)(Snxn′′-q)∥,
given by Lemma 2 that
(82)lim⁡n→∞∥Snxn′′-Tnxn′∥=0.
So
(83)lim⁡n→∞∥xn+1′′-xn+1′′∥=lim⁡n→∞|2αn-1|∥zn-yn∥=lim⁡n→∞|2αn-1|∥γn(xn′′-Snxn′′)-βn(xn′-Tnxn′)+(Snxn′′-Tnxn′)∥,
so we obtain that lim⁡n→∞∥xn+1′-xn+1′′∥=0 for (64) and (82), and it means x′=x′′. This completes the proof.


## 4.1. There Exists One Sequence{xn}

In this part, we prove our main theorem for finding a common fixed point of asymptotically nonexpansive mappings in Banach space in the case of one sequence.Theorem 8.
LetC be a nonempty closed convex subset of a uniformly convex Banach space E, and let T and S be two asymptotically nonexpansive mappings of C satisfying Condition (I) and F¯≠∅. Given {αn}, {βn}, and {γn} are sequences in (0,1) and {kn} with kn≥1 such that ∑‍αn<∞, ∑‍(1-βn)<∞ and ∑‍(kn-1)<∞ for all n≥1.
Define a sequence{xn}n=0∞ in C by algorithm (3); then {xn}n=0∞ strongly converges to a common fixed point of T and S.Proof.
First, we observe that{xn} is bounded; if we take an arbitrary fixed point q of F¯, noting that ∥yn-q∥≤kn∥xn-q∥ and ∥zn-q∥≤kn∥xn-q∥, we have
(37)∥xn+1-q∥=∥αnyn+(1-αn)zn-q∥≤αn∥yn-q∥+(1-αn)∥Zn-q∥≤(1+kn-1)∥xn-q∥.
By Lemma 1 and ∑‍(kn-1)<∞, thus, lim⁡n→∞∥xn-q∥ exists. Denote lim⁡n→∞∥xn-q∥=c, and put k∞=sup⁡{kn:n≥1}<∞. Hence, {xn} is bounded, so are {yn} and {zn}. Now
(38)∥xn+1-q∥=∥αnyn+(1-αn)zn-q∥≤αn∥yn-zn∥+∥zn-q∥.
Since ∥zn-q∥≤kn∥xn-q∥, this implies that
(39)lim⁡n→∞∥zn-q∥=lim⁡n→∞∥xn-q∥=c.
Moreover, ∥Snxn-q∥≤kn∥xn-q∥ implies that
(40)limsup⁡n→∞∥Snxn-q∥≤c.
Thus,
(41)c=lim⁡n→∞∥zn-q∥=lim⁡n→∞∥γnxn+(1-γn)Snxn-q∥=lim⁡n→∞∥γn(xn-q)+(1-γn)(Snxn-q)∥,
given by Lemma 2 that
(42)lim⁡n→∞∥Snxn-xn∥=0.
Now,
(43)∥zn-xn∥=∥γnxn+(1-γn)Snxn-xn∥≤(1-γn)∥(Snxn-xn)∥.
Hence, by (42),
(44)lim⁡n→∞∥zn-xn∥=0.
Also note that
(45)∥xn+1-xn∥=∥αnyn+(1-αn)zn-xn∥≤αn∥yn-xn∥+(1-αn)∥zn-xn∥,
so that condition ∑‍αn<∞ and (44) give
(46)lim⁡n→∞∥xn+1-xn∥=0.
Next, we show lim⁡n→∞∥xn+1-Sxn∥=0.
We have(47)∥xn+1-Sxn+1∥≤∥xn+1-Sn+1xn+1∥+∥Sn+1xn+1-Sn+1xn∥+∥Sn+1xn-Sxn+1∥≤∥xn+1-Sn+1xn+1∥+k∞∥xn+1-xn∥+k∞∥Snxn-xn+1∥≤∥xn+1-Sn+1xn+1∥+2k∞∥xn+1-xn∥+k∞∥Snxn-xn∥.
Hence, by (42) and (46), we get
(48)lim⁡n→∞∥xn-Sxn∥=0.
By (37) we assume that kn=1+rn, so ∑‍rn<∞ for ∑‍(kn-1)<∞, and now,
(49)∥xn+m-q∥≤(1+rn+m-1)∥xn+m-1-q∥≤ern+m-1∥xn+m-1-q∥≤⋯≤e∑i=nn+m-1ri∥xn-q∥.
That is,
(50)∥xn+m-q∥≤M∥xn-q∥,
where M=e∑i=nn+m-1ri, for all m,n∈ℕ, for all q∈F¯ and for M>0.
Next, we prove that{xn}n=0∞ is a Cauchy sequence.
As proved in Theorem7, it is easy to see that {xn}n=0∞ is a Cauchy sequence in a closed convex subset C of a Banach space E. Thus, it must converge to a point in C; let lim⁡n→∞xn=p.
For allϵ>0, as lim⁡n→∞xn=p, thus, there exists a number n1 such that, when n2≥n1,
(51)∥xn2-p∥≤ϵ2+2k∞.
In fact,lim⁡n→∞d(xn,F¯)=0 implies that using number n2 above, when n≥n2, we get d(xn,F¯)≤ϵ/(2+6k∞). In particular, d(xn2,F¯)≤ϵ/(2+6k∞). Thus, there must exist p-∈F¯, such that
(52)∥xn2-p-∥=d(xn2,p-)=ϵ2+6k∞.
From (51) and (52), using the same argument in Theorem 7, we get Sp=p.
Now, we return to provelim⁡n→∞∥xn-Txn∥=0.
It is easy to see thatlim⁡n→∞∥yn-xn∥=0, and then combined with (44), we have lim⁡n→∞∥zn-yn∥=0; thus, lim⁡n→∞∥zn-q∥=lim⁡n→∞∥yn-q∥, and then we get
(53)lim⁡n→∞∥yn-q∥=lim⁡n→∞∥xn-q∥=c.
Moreover, ∥Tnxn-q∥≤kn∥xn-q∥ implies that
(54)limsup⁡n→∞∥Tnxn-q∥≤c;
thus,
(55)lim⁡n→∞∥xn-Tnxn∥=0.
Therefore, we have
(56)lim⁡n→∞∥xn-Txn∥=0.
Thus, we can show that {xn}n=0∞ converges strongly to a common fixed point of T and S immediately.


## 4.2. There Exist Two Sequences{xn′} and {xn′′}

In this part, we prove our main theorem for finding a common fixed point of asymptotically nonexpansive mappings in Banach space in the case of two sequences.Theorem 9.
LetC be a nonempty bounded closed convex subset of a uniformly convex Banach space E, and let T and S be two asymptotically nonexpansive mappings of C satisfying Condition (I) and F¯≠∅. Given {αn}, {βn}, and {γn} are sequences in (0,1) and {kn} with 1≤kn<∞ such that ∑‍αn<∞, ∑‍βn<∞, βn>γn and ∑‍(kn-1)<∞ for all n≥1.
Define two sequences{xn′}n=0∞ and {xn′′}n=0∞ in C by the algorithm (5); then {xn′}n=0∞ and {xn′′}n=0∞ strongly converge to a common fixed point of T and S.Proof.
By the boundedness ofC, we obverse that both {xn′} and {xn′′} are bounded; if we take an arbitrary fixed point q of F¯, noting that ∥yn-q∥≤kn∥xn′-q∥ and ∥zn-q∥≤kn∥xn′′-q∥, we have
(57)∥xn+1′-q∥=∥αnzn+(1-αn)yn-q∥≤αn∥zn-yn∥+∥yn-q∥=(1+αnkn+kn-1)∥xn′-q∥+αnkn∥xn′′-q∥.
By Lemma 1 and ∑‍αn<∞, ∑‍(kn-1)<∞, thus, lim⁡n→∞∥xn′-q∥ exists. Denote
(58)lim⁡n→∞∥xn′-q∥=c′.
Similarly, we have
(59)lim⁡n→∞∥xn′′-q∥=c′′.
Since both {xn′} and {xn′′} are bounded, put k∞=sup⁡{kn:n≥1}<∞. We get that {yn} and {zn} are bounded. Now
(60)∥xn+1′-q∥=∥αnzn+(1-αn)yn-q∥≤αn∥zn-yn∥+∥yn-q∥.
Since ∥yn-q∥≤kn∥xn′-q∥, this implies that
(61)lim⁡n→∞∥yn-q∥=lim⁡n→∞∥xn′-q∥=c′.
Moreover, ∥Tnxn′-q∥≤kn∥xn′-q∥ implies that
(62)limsup⁡n→∞∥Tnxn′-q∥≤c′.
Thus,
(63)c′=lim⁡n→∞∥yn-q∥=lim⁡n→∞∥βnxn′+(1-βn)Tnxn′-q∥=lim⁡n→∞∥βn(xn′-q)+(1-βn)(Tnxn′-q)∥,
given by Lemma 2 that
(64)lim⁡n→∞∥Tnxn′-xn′∥=0.
Also note that
(65)∥xn+1′-xn′∥=∥αnzn+(1-αn)yn-xn′∥≤αn∥zn-yn∥+(1-βn)∥(Tnxn′-xn′)∥,
so that condition ∑‍αn<∞ and (64) give
(66)lim⁡n→∞∥xn+1′-xn′∥=0.
We have
(67)∥xn+1′-Txn+1′∥≤∥xn+1′-Tn+1xn+1′∥+∥Tn+1xn+1′-Tn+1xn′∥+∥Tn+1xn′-Txn+1′∥≤∥xn+1′-Tn+1xn+1′∥+k∞∥xn+1′-xn′∥+k∞∥Tnxn′-xn+1′∥≤∥xn+1′-Tn+1xn+1′∥+2k∞∥xn+1′-xn′∥+k∞∥Tnxn′-xn′∥.
Hence, by (64) and (66), we get
(68)lim⁡n→∞∥xn′-Txn′∥=0,
and then we assume that kn=1+rn, so ∑‍rn<∞ for ∑‍(kn-1)<∞; now by (57), we obtain that
(69)∥xn+m′-q∥≤(1+αn+m-1)(1+rn+m-1)∥xn+m-1′-q∥+sn+m-1≤eαn+m-1ern+m-1∥xn+m-1′-q∥+sn+m-1≤…≤e∑i=nn+m-1αie∑i=nn+m-1ri∥xn′-q∥+e∑i=nn+m-1αie∑i=nn+m-1ri∑i=nn+m-1‍si,
where si=αiki∥xn′′-q∥.
By∑‍αn<∞ and the convergence of {rn}, that is,
(70)∥xn+m′-q∥≤M(∥xn′-q∥+∑i=n∞‍si),
where M=e∑i=nn+m-1αie∑i=nn+m-1ri, si=αiki∥xn′′-q∥, for all m,n∈ℕ, q∈F¯ and for M>0.
Next, we prove that{xn′}n=0∞ is a Cauchy sequence.
Sinceq∈F¯ arbitrarily and lim⁡n→∞∥xn′-q∥ exists, consequently, d(xn′,F¯) exists by Lemma 3. From Lemma 3 and (68), we get
(71)lim⁡n→∞f(d(xn′,F¯))≤lim⁡n→∞∥xn′-Txn′∥=0.
Since f:[0,∞)→[0,∞) is a nondecreasing function satisfying f(0)=0, f(r)>0 for all r∈(0,∞), therefore, we have
(72)lim⁡n→∞d(xn′,F¯)=0.
Let ε>0, since lim⁡n→∞d(xn′,F¯)=0 and ∑i=1∞si<∞, therefore, there exists a constant n0 such that, for all n≥n0, we have
(73)d(xn′,F¯)≤ε3M,∑j=n0∞‍sj≤ε6M,
in particular,
(74)d(xn0′,F¯)≤ε3M.
There must exist p1∈F¯ such that
(75)d(xn0′,p1)≤ε3M.
From (70), it can be obtained that, when n≥n0,
(76)∥xn+m′-xn′∥≤∥xn+m′-p1∥+∥xn′-p1∥≤2M(∥xn0′-p1∥+∑j=n0n0+m-1‍sj)≤2M(ε3M+ε6M)=ε.
This implies that {xn′}n=0∞ is a Cauchy sequence in a closed convex subset C of a Banach space E. Thus, it must converge to a point in C; let lim⁡n→∞xn′=x′.
For allϵ>0, on lines similar to Theorem 8, from lim⁡n→∞xn′=x′, it can be proved that ∥Tx′-x′∥≤ϵ. As ϵ is an arbitrary positive number, thus, Tx′=x′. Similarly, we have lim⁡n→∞∥Sxn′′-xn′′∥=0, and then Sx′′=x′′ (xn′′→x′′ as n→∞).
Finally, we provex′=x′′.
Letwn+1=αnTnxn′+(1-αn)Snxn′, and put ∥xn′′-q∥≥∥xn′-q∥, for all n≥1. Then,
(77)∥wn+1-q∥≤αnkn∥xn′-q∥+(1-αn)kn∥xn′′-q∥≤knmax⁡⁡{∥xn′-q∥,∥xn′′-q∥}≤kn∥xn′′-q∥.
We have lim⁡n→∞∥wn+1-q∥≤lim⁡n→∞∥xn′′-q∥. Now,
(78)∥xn+1′′-q∥=∥αnyn+(1-αn)zn-q∥≤βn∥αnxn′+(1-αn)xn′′∥+∥αnTnxn′+(1-αn)Snxn′′-q∥=βn∥αnxn′+(1-αn)xn′′∥+∥wn+1-q∥.
Since ∑‍βn<∞ and the boundedness of {xn′} and {xn′′}, we get
(79)lim⁡n→∞∥wn+1-q∥=c′′.
Moreover, ∥Tnxn′-q∥≤kn∥xn′-q∥ and ∥Snxn′′-q∥≤kn∥xn′′-q∥ imply that
(80)limsup⁡n→∞∥Tnxn′-q∥≤c′′,limsup⁡n→∞∥Snxn′′-q∥≤c′′.
Thus,
(81)c′′=lim⁡n→∞∥wn+1-q∥=∥αnTnxn′+(1-αn)Snxn′′-q∥=∥αn(Tnxn′-q)+(1-αn)(Snxn′′-q)∥,
given by Lemma 2 that
(82)lim⁡n→∞∥Snxn′′-Tnxn′∥=0.
So
(83)lim⁡n→∞∥xn+1′′-xn+1′′∥=lim⁡n→∞|2αn-1|∥zn-yn∥=lim⁡n→∞|2αn-1|∥γn(xn′′-Snxn′′)-βn(xn′-Tnxn′)+(Snxn′′-Tnxn′)∥,
so we obtain that lim⁡n→∞∥xn+1′-xn+1′′∥=0 for (64) and (82), and it means x′=x′′. This completes the proof.


## 5. Application: Convergence to a Zero of Accretive Operators

LetE be a real Banach space. Recall that an operator (possibly multivalued) A with domain D(A) and range R(A) in E is said to be accretive if, for each xi∈D(A) and yi∈Axi(i=1,2), there exists a j(x1-x2)∈J(x1-x2) such that
(84)〈y1-y2,j(x1-x2)〉≥0,
where J is the normalized duality map from E to the dual space E* given by
(85)J(x)={x*∈E*:〈x,x*〉=∥x∥2=∥x*∥2},x∈E.
An accretive operator A is m-accretive if R(I+rA)=E for all r>0. Denote the zero set of A by
(86)F:=A-1(0)={z∈D(A):0∈Az}.For anm-accretive operator A with F≠∅ and C=D(A)¯ convex, the problem of finding a zero of A, that is, (87)findz∈Csuchthat0∈Az,
has extensively been investigated due to its applications in related problems such as minimization problems, variational inequality problems, and nonlinear evolution equations.It is known that theresolvent of A, defined by
(88)Jr=(I+rA)-1,
for r>0, is a nonexpansive mapping from E to C and it is straightforward to see that F coincides with the fixed point set of Jr for any r>0. Therefore, (87) is equivalent to the fixed point problem z=Jrz. Then an interesting approach to solving this problem is via iterative methods for nonexpansive mappings. We need the resolvent identity [15].Theorem 10.
LetE be a uniformly convex Banach space, and let A be an m-accretive operator in E such that Fix⁡(Jr1)⋂‍Fix⁡(Jr2)⋂‍Fix⁡(Jr3)=A-1(0)≠∅, Jri:E→E is nonexpansive commuting mappings for all ri>0 (i=1,2,3) satisfying Condition (I). Given {αn}, {βn}, and {γn} are sequences in (0,1), such that ∑‍αn<∞, ∑‍γnβn=∞, ∑‍(1-γn)<∞ and ri,n≥ε for some ε>0 for all n≥1.
Define a sequence{xn}n=0∞ by algorithm (4); then {xn}n=0∞ strongly converges to a zero of A.Proof.
Take any arbitraryq∈A-1(0); it follows from Lemma 1 that lim⁡n→∞∥xn-q∥ exists. From Lemma 2, it can be shown that lim⁡n→∞∥Jri,nxn-xn∥=0(i=1,2,3). Since Jri:E→E is nonexpansive for all ri>0, it follows from Lemma 6 that lim⁡n→∞∥Jrixn-xn∥=0. Therefore, all the conditions in Theorem 7 are satisfied. The conclusion of Theorem 10 can be obtained from Theorem 7 immediately.Theorem 11.
LetC be a nonempty bounded closed convex subset of a uniformly convex Banach space E, and let A be an m-accretive operator in C such that Fix⁡(Jr1)⋂‍Fix⁡(Jr2)=A-1(0)≠∅, Jri:C→C is nonexpansive for all ri>0 (i=1,2) satisfying Condition (I). Given {αn}, {βn}, and {γn} are sequences in (0,1) such that ∑‍αn<∞, ∑‍βn<∞, βn>γn and ri,n≥ε for some ε>0 for all n≥1.
Define two sequences{xn′}n=0∞ and {xn′′}n=0∞ in C by algorithm (6); then {xn′}n=0∞ and {xn′′}n=0∞ strongly converge to a zero of A.Proof.
Only a sketch of the proof is given here.
Take any arbitraryq∈Fix⁡(Jr1)⋂‍Fix⁡(Jr2); it follows from Lemma 1 that lim⁡n→∞∥xn′-q∥ and lim⁡n→∞∥xn′′-q∥ exist. From Lemma 2, it can be shown that lim⁡n→∞∥Jr1,nxn′-xn′∥=0 and lim⁡n→∞∥Jr2,nxn′′-xn′′∥=0. Since Jri:E→E is nonexpansive for all ri>0 (i=1,2), it follows from lemma 2.6 that lim⁡n→∞∥Jr1xn′-xn′∥=0 and lim⁡n→∞∥Jr2xn′′-xn′′∥=0. Therefore, all the conditions in Theorems 7, 8, and 9, are satisfied and using the same argument in those theorems, the conclusion of Theorem 11 can be obtained immediately.



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*Source: 264909-2014-02-11.xml*