Zero Theorems of Accretive Operators

Malaysian Mathematical Sciences Society

http://math.usm.my/bulletin

Zero Theorems of Accretive Operators

Yan Hao

School of Mathematics, Physics and Information Science, Zhejiang Ocean University, Zhoushan 316004, China

[email protected]

Abstract. In this paper, we introduce and analysis an iterative method for finding a common zero of twom-accretive operators. Under appropriate re- strictions imposed on the parameters, we obtain a convergence theorem in a real Banach space.

2010 Mathematics Subject Classification: 47H05, 47H09, 47H10

Keywords and phrases: Accretive operator, zero, nonexpansive mapping, fixed point.

1. Introduction and preliminaries

Throughout this paper, we always assume thatE is a real Banach space. Let E^∗ be the dual space of E. Let ϕ: [0,∞] :=R⁺ → R⁺ be a continuous and strictly increasing function such thatϕ(0) = 0 andϕ(t)→ ∞as t→ ∞. This function ϕis called a gauge function. The duality mappingJϕ:E→E^∗associated with a gauge functionϕis defined by

Jϕ(x) ={f^∗∈E^∗:hx, f^∗i=kxkϕ(kxk), kf^∗k=ϕ(kxk)}, ∀x∈E, where h·,·i denotes the generalized duality pairing. In the case that ϕ(t) = t,we writeJ forJϕ and callJ the normalized duality mapping.

Following Browder [2], we say that a Banach space E has a weakly continuous duality mapping if there exists a gauge ϕ for which the duality mapping Jϕ(x) is single-valued and weak-to-weak^∗ sequentially continuous (i.e., if{xn} is a sequence in E weakly convergent to a point x, then the sequence Jϕ(xn) converges weakly^∗ toJ_ϕx). It is known thatl^p has a weakly continuous duality mapping with a gauge functionϕ(t) =t^p−1 for all 1< p <∞.Set

Φ(t) = Z t

0

ϕ(τ)dτ, ∀t≥0, then

Jϕ(x) =∂Φ(kxk), ∀x∈E,

Communicated bySriwulan Adji.

Received:June 1, 2009;Revised: September 4, 2009.

where∂ denotes the sub-differential in the sense of convex analysis.

The norm ofE is said to be Gˆateaux differentiable (andE is said to be smooth) if

limt→0

kx+tyk − kxk t

exists for each x, y in its unit sphere U = {x ∈ E : kxk = 1}. It is said to be uniformly Fr´echet differentiable (andE is said to be uniformly smooth) if the limit is attained uniformly for (x, y)∈U×U.

A Banach spaceE is said to strictly convex if and only if kxk=kyk=k(1−λ)x+λyk

for allx, y∈E and 0< λ <1 implies thatx=y. E is said to uniformly convex if, for any∈(0,2], there exists δ >0 such that, for anyx, y∈U,

kx−yk ≥ implies

x+y 2

≤1−δ.

Let C be a nonempty, closed and convex subset of E. Recall that a mapping f :C→C is said to be α-contractive if there exists a constantα∈(0,1) such that

kf x−f yk ≤αkx−yk, ∀x, y∈C.

In this paper, we use ΠC to denote the collection of all contractive mappings on C. That is, ΠC ={f|f :C→C is a contractive mapping}. Recall that a mapping T :C→C is said to be nonexpansive if

kT x−T yk ≤ kx−yk, ∀x, y∈C.

In this paper, we use F(T) to denote the set of fixed points of T. The class of nonexpansive mapping is a kind of important nonlinear mapping which was studied by many authors; see, for example, [1–25].

One classical way to study nonexpansive mappings is to use contractions to ap- proximate a nonexpansive mapping [3, 18]. More precisely, taket∈(0,1) and define a contractionTt:C→C by

(1.1) T_tx=tu+ (1−t)T x, ∀x∈C,

where u∈ C is a fixed point. Banach’s contraction mapping principle guarantees thatTthas a unique fixed pointxtinC. That is,

(1.2) xt=tu+ (1−t)T xt.

It is unclear, in general, what the behavior of x_t is as t→0, even if T has a fixed point. However, in the case of T having a fixed point, Browder [3] proved that if E is a Hilbert space, then x_t converges strongly to a fixed point of T. Reich [18]

extended Broweder’s result to the setting of Banach spaces and proved that ifE is a uniformly smooth Banach space, thenxt converges strongly to a fixed point ofT and the limit defines the (unique) sunny nonexpansive retraction fromContoF(T).

Xu [22] proved that Browder’s results still hold in reflexive Banach spaces which have a weakly continuous duality mapping; see [22] for more details.

Recall that if C and D are nonempty subsets of a Banach space E such that C is nonempty closed convex and D ⊂ C, then a map Q : C → D is called a retraction fromContoD providedQ(x) =xfor allx∈D. A retractionQ:C→D

is sunny provided Q(x+t(x−Q(x))) = Q(x) for all x ∈ C and t ≥ 0 whenever x+t(x−Q(x))∈C. A sunny nonexpansive retraction is a sunny retraction which is also nonexpansive. Sunny nonexpansive retractions are characterized as follows [9, 19]:

IfEis a smooth Banach space, thenQ:C→Dis a sunny nonexpansive retraction if and only if there holds the inequality

(1.3) hx−Qx, J(y−Qx)i ≤0, ∀x∈C, y∈D.

Reich [18] showed that ifEis uniformly smooth and ifDis the fixed point set of a nonexpansive mapping fromCinto itself, then there is a unique sunny nonexpansive retraction fromC ontoDand it can be constructed as follows.

Theorem 1.1. Let E be a uniformly smooth Banach space and letT :C→C be a nonexpansive mapping with a fixed point. For each fixed u∈C and every t∈(0,1), the unique fixed pointx_t∈C of the contractionC 3x7→tu+ (1−t)T x converges strongly as t→0 to a fixed point ofT. Define Q:C→D by Qu=s−lim_t→0x_t. ThenQis the unique sunny nonexpansive retract fromContoD; that is, Q satisfies the property:

hu−Qu, J(y−Qu)i ≤0, ∀u∈C, y∈D.

If E is a reflexive Banach space which has a weakly continuous duality map, then Q:C →D is a sunny nonexpansive retraction if and only if there holds the inequality

(1.4) hx−Qx, J_ϕ(y−Qx)i ≤0, ∀x∈C, y∈D.

In 2006, Xu [22] obtained an analogue of Theorem 1.1 in a reflexive Banach space.

To be more precise, he proved the following result.

Theorem 1.2. Let E be a reflexive Banach space and has a weakly continuous duality map J_ϕ(x) with gauge ϕ. Let C be closed convex subset of E and let T : C → C be a nonexpansive mapping. Fix u∈C and t ∈ (0,1). Let x_t ∈ C be the unique solution in C to the equation (1.2). Then T has a fixed point if and only if {x_t} remains bounded as t → 0⁺, and in this case, {x_t} converges as t → 0⁺ strongly to a fixed point of T.

Recall that a mappingAwith domainD(A) and rangeR(A) inEis accretive, if for eachxi∈D(A) andyi∈Axi(i= 1,2),there exists ajϕ(x2−x1)∈Jϕ(x2−x1) such that

hy2−y1, jϕ(x2−x1)i ≥0.

An accretive operatorAism-accretive ifR(I+rA) =Efor eachr >0. Through- out this article we always assume thatA ism-accretive and has a zero (i.e., the in- clusion 0∈A(z) is solvable). For eachr >0, we denote byJrthe resolvent ofA, i.e., Jr= (I+rA)⁻¹. Note that ifAism-accretive, thenJr:E→D(A) is nonexpansive and F(Jr) = F for all r > 0. We also denote byAr the Yosida approximation of A, i.e., Ar = ¹_r(I−Jr). It is known that Jr is a nonexpansive mapping from E to C:=D(A) which will be assumed convex.

Kim and Xu [12] studiedm-accretive operators by considering the following iter- ative algorithm

(1.5) xn+1=αnu+ (1−αn)Jr_nxn, n≥0,

where{αn}is a sequence in (0,1),u∈Cis a fixed point andJ_rn= (I+r_nA)⁻¹. They proved that the sequence{x_n}generated by the above iterative algorithm converges strongly to a zero point ofAin the framework of uniformly smooth Banach spaces.

Recently, Qin and Su [15] studied so-called modified Mann iterations (1.6)

(yn =βnxn+ (1−βn)Jr_nxn, x_n+1=α_nu+ (1−α_n)y_n,

where u ∈ C is a fixed point, {αn} and {βn} are sequences in (0,1) and Jr_n = (I+rnA)⁻¹. They obtained that the sequence generated by the above iterative algorithm converges strongly to a zero point ofAassume thatEis uniformly smooth.

They also proved that the conclusion still holds provided thatEis a reflexive Banach space which has a weak continuous duality map.

Viscosity approximation method, which was first introduced by Moudafi [14], for the problem of finding fixed points of nonexpansive mapping has been studied by many authors.

For a real number t ∈ (0,1) and a contractive mapping f ∈ Π_C, we define a mappingT_tx=tf(x) + (1−t)T xfor allx∈C.It is obviously thatT_tis a contractive mapping onC. In fact, for anyx, y∈C, we obtain

kTtx−Ttyk=kt(f(x)−f(y)) + (1−t)(T x−T y)k

≤αtkx−yk+ (1−t)kT x−T yk

≤αtkx−yk+ (1−t)kx−yk

= (1−t(1−α))kx−yk.

Letx_tbe the unique fixed point ofT_t. That is,x_tis the unique solution of the fixed point equation:

(1.7) xt=tf(xt) + (1−t)T xt.

Theorem 1.3. [5, 7]LetEbe a reflexive Banach space which has a weakly continu- ous duality mappingJ_ϕ(x). LetC be closed convex subset ofE andT :C→C be a nonexpansive mapping. Letf :C→Cbe a contractive mapping withF(f)6=∅. For anyt∈(0,1), let {x_t} be defined by(1.7), whereT is a nonexpansive mapping. De- fine a mappingQ: Π_C→F(T)by Q(f) := lim_t→0x_t, f ∈Π_C. Then then mapping Qis the sunny nonexpansive retraction fromΠ_C ontoF(T).

In [5], Chen and Zhu also considered the following iterative methods:

(1.8) x_n+1=α_nf(x_n) + (1−α_n)J_rnx_n, n≥0,

where {αn} is a sequence in (0,1), f : C → C is an α-contraction and Jr_n = (I+rnA)⁻¹. They proved that the sequence{xn} generated by the above iterative algorithm converges strongly to a zero point of A in a real Banach space which includes the corresponding results in Xu [22].

Recently, Chenet al. [6] further studied the following iterative method:

(1.9)

(yn =βnxn+ (1−βn)Jr_nxn, x_n+1=α_nf(x_n) + (1−α_n)y_n,

where f : C →C is an α-contraction,{αn} and {βn} are sequences in (0,1) and Jr_n= (I+rnA)⁻¹. They also obtained a zero theorem of the operatorA, see [6] for more details.

In this paper, we consider a pair of m-accretive operators instead of a single operator which was studied by Chen and Zhu [5], Chen et al. [6], Cho and Qin [7], Kim and Xu [12], Qin and Su [15] and Xu [22] in a reflexive Banach space which admits a weak continuous duality map. Strong convergence theorems are established. To prove the main result, we need the following results.

The first part of the next lemma is an immediate consequence of the subdifferential inequality and the proof of the second part can be found in [13].

Lemma 1.1. Assume that a Banach spaceE has a weakly continuous duality map- ping J_ϕ with a gaugeϕ.

(i) For allx, y∈E,the following inequality holds:

Φ(kx+yk)≤Φ(kxk) +hy, Jϕ(x+y)i.

In particular, for all x, y∈E,

kx+yk²≤ kxk²+ 2hy, J(x+y)i.

(ii) Assume that a sequence{xn}inE converges weakly to a pointx∈E. Then the following identity holds:

lim sup

n→∞

Φ(kx_n−yk) = lim sup

n→∞

Φ(kx_n−xk) + Φ(ky−xk), ∀x, y∈E.

Lemma 1.2. [4]Let C be a closed convex subset of a strictly convex Banach space E. LetT₁ andT₂be two nonexpansive mappings on C. Suppose that F(T₁)∩F(T₂) is nonempty. Then a mappingT on C defined by

T x=λT1x+ (1−λ)T2x, ∀x∈C is well defined, nonexpansive and F(T) =F(T1)∩F(T2)holds.

Lemma 1.3. [23]Assume that{αn}is a sequence of nonnegative real numbers such that

αn+1≤(1−γn)αn+δn, ∀n≥1,

where{γn}is a sequence in(0,1) and{δn} is a sequence such that (i) P∞

n=1γn =∞;

(ii) lim sup_n→∞^δ_γⁿ

n ≤0.

Thenlim_n→∞αn= 0.

Lemma 1.4. [13] Let E be a Banach space satisfying a weakly continuous duality map, C a nonempty closed convex subset of E and T : C → C a nonexpansive mapping with a fixed point. Then I−T is demi-closed at zero, i.e., if {xn} is a sequence inC which converges weakly toxand if the sequence{(I−T)xn}converges strongly to zero, thenx=T x.

2. Main results

Theorem 2.1. Let E be a strictly convex and reflexive Banach space which has a weakly continuous duality map J_ϕ and f ∈ Π_C. Let A and B be m-accretive operators in E such thatC :=D(A)∩D(B) is convex. Let{β_n} be a real number sequences in(0,1). Let {x_n} be a sequence generated by the following manner:





 x₀∈C,

yn=βnJrxn+ (1−βn)Jsxn,

x_n+1=α_nf(x_n) + (1−α_n)y_n, n≥0,

wherer, s >0,Jr= (I+rA)⁻¹andJs= (I+sB)⁻¹. Assume thatA⁻¹(0)∩B⁻¹(0)6=

∅. If the above control sequences satisfy the following restrictions:

(a) limn→∞αn= 0,P∞

n=1αn =∞ andP∞

n=1|αn+1−αn|<∞;

(b) limn→∞βn=β∈(0,1) andP∞

n=1|βn+1−βn|<∞,

then the sequence {xn} converges strongly to some pointQ(f)∈A⁻¹(0)∩B⁻¹(0), whereQis the sunny nonexpansive retraction Q: Π_C→A⁻¹(0)∩B⁻¹(0).

Proof. First, we prove that{x_n}is bounded. For anyp∈A⁻¹(0)∩B⁻¹(0), we see that

kyn−pk=kβnJrxn+ (1−βn)Jsxn−pk

=kβn(Jrxn−Jrp) + (1−βn)(Jsxn−Jsp)k

≤β_nkJrx_n−J_rpk+ (1−β_n)kJsx_n−J_spk

≤βnkxn−pk+ (1−βn)kxn−pk

=kxn−pk.

It follows that

kx_n+1−pk=kα_nf(x_n) + (1−α_n)y_n−pk

=kαn[f(xn)−f(p)] +αn[f(p)−p] + (1−αn)(yn−p)k

≤αnkf(xn)−f(p)k+αnkf(p)−pk+ (1−αn)kyn−p)k (2.1)

≤α_nαkx_n−pk+α_nkf(p)−pk+ (1−α_n)kx_n−p)k

= [1−αn(1−α)]kxn−pk+αnkf(p)−pk.

Putting B = max{kx0−pk,^kf(p)−pk_1−α }, we show that kxn−pk ≤B for all n≥ 0.

It is easy to see that the result holds forn = 0. We assume that the result holds for somen≥0. From (2.1), we can see thatkxn+1−pk ≤B. This shows that the sequence{xn} is bounded. Note that

yn−yn−1

=βnJrxn+ (1−βn)Jsxn−[β_n−1Jrx_n−1+ (1−β_n−1)Jsx_n−1]

=β_n(J_rx_n−J_rx_n−1) + (1−β_n)(J_sx_n−J_sx_n−1) + (J_rx_n−1−J_sx_n−1)(β_n−β_n−1).

This implies that

kyn−yn−1k ≤βnkJrxn−Jrxn−1k+ (1−βn)kJsxn−Jsxn−1k

+kJrx_n−1−J_sx_n−1k|βn−β_n−1|

≤βnkxn−x_n−1k+ (1−βn)kxn−x_n−1k +kJrx_n−1−Jsx_n−1k|βn−β_n−1|

≤ kxn−x_n−1k+|βn−β_n−1|M1, (2.2)

whereM₁is an appropriate constant such that M₁≥sup_n≥1{kJ_rx_n−1−J_sx_n−1k}.

On the other hand, we have

x_n+1−x_n= [α_nf(x_n) + (1−α_n)y_n]−[α_n−1f(x_n−1) + (1−α_n−1)y_n−1]

=αn[f(xn)−f(xn−1)] + (1−αn)(yn−yn−1) + [f(xn)−y_n−1](αn−α_n−1).

This gives that

kxn+1−xnk ≤αnkf(xn)−f(x_n−1)k+ (1−αn)kyn−y_n−1k +kf(x_n)−y_n−1k|α_n−α_n−1|

≤αnαkxn−xn−1k+ (1−αn)kyn−yn−1k+|αn−αn−1|M2, (2.3)

where M₂ is an appropriate constant such that M₂ ≥ sup_n≥1{kf(x_n)−y_n−1k}.

Substituting (2.2) into (2.3), we see that

(2.4) kxn+1−xnk ≤[1−αn(1−α)]kxn−x_n−1k+ (|βn−β_n−1|+|αn−α_n−1|)M3, where M3 is an appropriate constant such that M3 = max{M1, M2}. From the conditions (a), (b) and applying Lemma 1.3 to (2.4), we obtain that

(2.5) lim

n→∞kxn−xn+1k= 0.

Define an operatorW :C→C by

W x:=βJrx+ (1−β)Jsx, ∀x∈C,

where (0,1) 3β = lim_n→∞βn. From Lemma 1.2, we see that W is nonexpansive withF(W) =F(Jr)∩F(Js) =A⁻¹(0)∩B⁻¹(0).

Next, we prove thatkxn−W xnk →0 asn→ ∞.Note that

yn−W xn=βnJrxn+ (1−βn)Jsxn−[βJrxn+ (1−β)Jsxn]

= (β_n−β)J_rx_n+ (β−β_n)J_sx_n. It follows from the condition (b) that

(2.6) ky_n−W x_nk= 0.

On the other hand, we have

kx_n−W x_nk=kx_n−x_n+1+x_n+1−W x_nk

=kxn−xn+1+αnf(xn) + (1−αn)yn−W xnk

=kxn−xn+1+αn(f(xn)−W xn) + (1−αn)(yn−W xn)k

=kxn−xn+1k+αnkf(xn)−W xnk+ (1−αn)kyn−W xnk.

It follows from the condition (a), (2.5) and (2.6) that

(2.7) lim

n→∞kxn−W xnk= 0.

Next, we show that lim sup

n→∞

h(I−f)Q(f), Jϕ(Q(f)−xn)i ≤0, whereQ: ΠC → A⁻¹(0)∩B⁻¹(0) is the sunny nonexpansive retraction. To show it, we may choose a subsequence{xn_i}of{xn} such that

(2.8) lim sup

n→∞

h(I−f)Q(f), Jϕ(Q(f)−xn)i= lim

i→∞h(I−f)Q(f), Jϕ(Q(f)−xn_i)i.

Since E is reflexive, we may further assume that x_ni * x¯ for some ¯x ∈ C. From Lemma 1.4, we see that ¯x∈F(W) =A⁻¹(0)∩B⁻¹(0). Hence, we arrive at

lim sup

n→∞

h(I−f)Q(f), J_ϕ(Q(f)−x_n)i=h(I−f)Q(f), J_ϕ(Q(f)−x)i ≤¯ 0.

It follows from (2.5) that

(2.9) lim sup

n→∞

h(I−f)Q(f), Jϕ(Q(f)−xn+1)i ≤0.

Finally, we show thatxn→Q(f) asn→ ∞.Notice that

Φ(ky_n−Q(f)k) = Φ(kβ_n(J_rx_n−Q(f)) + (1−β_n)(J_sx_n−Q(f))k)≤Φ(kx_n−Q(f)k).

It follows from Lemma 1.1 that

Φ(kxn+1−Q(f)k) = Φ(kαn(f(x_n)−f(Q(f))) +α_n(f(Q(f))−Q(f)) + (1−αn)(yn−Q(f))k)

≤Φ(αnkf(xn)−f(Q(f))k+ (1−αn)kyn−Q(f)k) +α_nhf(Q(f))−Q(f), J_ϕ(x_n+1−Q(f))i

≤Φ(αnαkxn−Q(f)k+ (1−αn)kxn−Q(f)k) +αnhf(Q(f))−Q(f), Jϕ(xn+1−Q(f))i

≤Φ((1−α_n(1−α))kx_n−Q(f)k)

+αnhf(Q(f))−Q(f), Jϕ(xn+1−Q(f))i

≤(1−αn(1−α))Φ(kxn−Q(f)k)

+α_nhf(Q(f))−Q(f), J_ϕ(x_n+1−Q(f))i.

From Lemma 1.3, we see that Φ(kx_n+1−Q(f)k)→0 asn→ ∞. That is, kxn−Q(f)k →0 as n→ ∞.

This completes the proof.

IfJs=I, then identity mapping, then we have the following results from Theorem 2.1.

Corollary 2.1. Let E be a strictly convex and reflexive Banach space which has a weakly continuous duality map Jϕ and f ∈ΠC. Let Abe a m-accretive operator in

E such that C :=D(A) is convex. Let {βn} be a real number sequences in (0,1).

Let {xn} be a sequence generated by the following manner:





 x0∈C,

yn=βnJrxn+ (1−βn)xn,

xn+1=αnf(xn) + (1−αn)yn, n≥0,

where r >0 and J_r = (I+rA)⁻¹. Assume that A⁻¹(0)6=∅. If the above control sequences satisfy the following restrictions:

(a) lim_n→∞αn= 0,P∞

n=1αn =∞ andP∞

n=1|αn+1−αn|<∞;

(b) lim_n→∞βn=β∈(0,1) andP∞

n=1|βn+1−βn|<∞,

then the sequence{xn}converges strongly to some pointQ(f)∈A⁻¹(0), where Qis the sunny nonexpansive retraction Q: Π_C→A⁻¹(0).

IfA=B andr=sin Theorem 2.1, we have the following result immediately.

Corollary 2.2. Let E be a strictly convex and reflexive Banach space which has a weakly continuous duality map Jϕ and f ∈ΠC. Let Abe a m-accretive operator in E such thatC:=D(A)is convex. Let{xn}be a sequence generated by the following manner:

x₀∈C, x_n+1=α_nf(x_n) + (1−α_n)J_rx_n, n≥0,

where r >0 and Jr = (I+rA)⁻¹. Assume that A⁻¹(0)6=∅. If the above control sequences satisfy the following restrictions:

(a) lim_n→∞αn= 0 andP∞

n=1αn=∞;

(b) P∞

n=1|αn+1−αn|<∞,

then the sequence{xn}converges strongly to some pointQ(f)∈A⁻¹(0), where Qis the sunny nonexpansive retraction Q: ΠC→A⁻¹(0).

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