# Weak and Strong Convergence Theorems for Zeroes of Accretive Operators in Banach Spaces

Authors: Yanlai Song; Luchuan Ceng
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

The purpose of this paper is to present two new forward-backward splitting schemes with relaxations and errors for finding a common element of the set of solutions to the variational inclusion problem with two accretive operators and the set of fixed points of nonexpansive mappings in infinite-dimensional Banach spaces. Under mild conditions, some weak and strong convergence theorems for approximating this common elements are proved. The methods in the paper are novel and different from those in the early and recent literature. Our results can be viewed as the improvement, supplementation, development, and extension of the corresponding results in the very recent literature.


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## Body

## 1. Introduction

LetC be a nonempty closed convex subset of a real Hilbert space H. Let A:C→H be a single-valued nonlinear mapping and let B:H→2H be a multivalued mapping. The so-called quasivariational inclusion problem is to find a z∈H such that
(1)0∈(A+B)z.
The set of solutions of (1) is denoted by (A+B)-1(0). A number of problems arising in structural analysis, mechanics, and economics can be studied in the framework of this kind of variational inclusions; see, for instance, [1–4]. The problem (1) includes many problems as special cases as follows.(1)
IfB=∂ϕ:H→2H, where ϕ:H→ℝ∪{+∞} is a proper convex lower semicontinuous function and ∂ϕ is the subdifferential of ϕ, then the variational inclusion problem (1) is equivalent to find u∈H such that
(2)〈Au,y-u〉+ϕ(y)-ϕ(u)≥0,∀y∈H,
which is called the mixed quasivariational inequality (see, Noor [5]).(2)
IfB=∂δC, where C is a nonempty closed convex subset of H and δC:H→[0,∞] is the indicator function of C, that is,
(3)δC={0,x∈C,+∞,x∉C,then the variational inclusion problem (1) is equivalent to find u∈C such that
(4)〈Au,v-u〉≥0,∀v∈C.
This problem is called Hartman-Stampacchia variational inequality (see, e.g., [6]).In [7], Zhang et al. investigated the problem of finding a common element of the set of solutions to the inclusion problem and the set of fixed points of nonexpansive mappings by considering the following iterative algorithm:
(5)yn=JM,λ(xn-λAxn),xn+1=αnx+(1-αn)Syn,
where A:H→H is an α-cocoercive mapping, M:H→2H is a maximal monotone mapping, S:H→H is a nonexpansive mapping, and {αn} is a sequence in [0,1]. Under mild conditions, they obtained a strong convergence theorem.In [8], Manaka and Takahashi introduced the following iteration:
(6)x1∈C,xn+1=αnxn+(1-αn)SJλn(I-λnA)xn,00000000000000000000000000n≥1,
where {αn} is a sequence in (0, 1), {λn} is a positive sequence, S:C→C is a nonexpansive mapping, A:C→H is an inversely strongly monotone mapping, B:D(B)⊂C→2H is a maximal monotone operator, and Jλn=(I+λnB)-1 is the resolvent of B. They showed that the sequence {xn} generated in (6) converges weakly to some z∈(A+B)-1(0)∩F(S) provided that the control sequence satisfies some restrictions.It is well known that, the quasivariational inclusion problem in the setting of Hilbert spaces has been extensively studied in the literature; see, for instance, [3–14]. However, there is little work in the existing literature on this problem in the setting of Banach spaces (though there were some work on finding a common zero of a finite family of accretive operators [15–17]). The main difficulties are due to the fact that the inner product structure of a Hilbert space fails to be true in a Banach space. To overcome these difficulties, López et al. [18] use the new technique to carry out certain initiative investigations on splitting methods for accretive operators in Banach spaces. They considered the following algorithms with errors in Banach spaces:
(7)xn+1=(1-αn)xn+αn(JrnB(xn-rn(Axn+an))+bn),(8)xn+1=αnu+(1-αn)(JrnB(xn-rn(Axn+an))+bn),
where u∈E,{an},{bn}⊂E, and JrnB=(I+rnB)-1 are the resolvent of B. Then they studied the weak and strong convergence of the algorithms (7) and (8), respectively.Motivated and inspired by Zhang et al. [7], Manaka and Takahashi [8], López et al. [18], and Cho et al. [19], the purpose of this paper is to introduce two iterative forward-backward splitting methods for finding a common element of the set of solutions of the variational inclusion problem (1) with m-accretive operators and inverse strongly accretive operators and the set of fixed points of nonexpansive mappings in the setting of Banach space. Under suitable conditions, some weak and strong convergence theorems for approximating to this common elements are proved. The results presented in the paper not only improve and extend the main result in Zhang et al. [7], but also replenish and extend the corresponding results in Manaka and Takahashi [8], López et al. [18], and Cho et al. [19].


## 2. Preliminaries

Throughout this paper, we denote byE and E* a real Banach space and the dual space of E, respectively. Let C be a subset of E and T be a mapping on C. We use F(T) to denote the set of fixed points of T. Let q>1 be a real number. The generalized duality mapping Jq:E→2E* is defined by
(9)Jq(x)={x*∈E*:〈x,x*〉=∥x∥q,∥x*∥=∥x∥q-1}
for all x∈E, where 〈·,·〉 denotes the generalized duality pairing between E and E*. In particular, J=J2 is called the normalized duality mapping and Jq(x)=∥x∥q-2J2(x) for x≠0. If E is a Hilbert space, then J=I, where I is the identity mapping. It is well known that if E is smooth, then Jq is single-valued, which is denoted by jq.The norm of a Banach spaceE is said to be Gâteaux differentiable if the limit
(10)lim⁡t→0∥x+ty∥-∥x∥t
exists for all x,y on the unit sphere S(E)={x∈E:∥x∥=1}. If, for each y∈S(E), the limit (10) is uniformly attained for x∈S(E), then the norm of E is said to be uniformly Gâteaux differentiable. The norm of E is said to be Fréchet differentiable if, for each x∈S(E), the limit (10) is attained uniformly for y∈S(E).LetρE:[0,1)→[0,1) be the modulus of smoothness of E defined by
(11)ρE(t)=sup⁡{12(∥x+y∥+∥x-y∥)0000000-1:x∈S(E),∥y∥≤t12}.A Banach spaceE is said to be uniformly smooth if ρE(t)/t→0 as t→0. Let q>1. A Banach space E is said to be q-uniformly smooth, if there exists a fixed constant c>0 such that ρE(t)≤ctq. It is well known that E is uniformly smooth if and only if the norm of E is uniformly Fréchet differentiable. If E is q-uniformly smooth, then q≤2 and E is uniformly smooth, and hence the norm of E is uniformly Fréchet differentiable, in particular, the norm of E is Fréchet differentiable. Typical examples of both uniformly convex and uniformly smooth Banach spaces are Lp, where p>1. More precisely, Lp is min⁡{p,2}-uniformly smooth for every p>1.A Banach spaceE is said to be uniformly convex if, for any ε∈(0,2], there exists δ>0 such that, for any x,y∈S(E), ∥x-y∥≥ε implies ∥(x+y)/2∥≤1-δ. It is known that a uniformly convex Banach space is reflexive and strictly convex.A Banach space is said to have the Kadec-Klee property if for every sequence{xn} in E, xn⇀x and ∥xn∥→∥x∥ together imply ∥xn-x∥→0.A Banach spaceE is said to satisfy Opial’s condition if for any sequence {xn} in E the condition that {xn} converges weakly to x∈E implies that the inequality
(12)liminf⁡n→∞∥xn-x∥≤liminf⁡n→∞∥xn-y∥
holds for every y∈E with y≠x.Definition 1.
A mappingT:C→E is said to be (1)
nonexpansive if(13)∥Tx-Ty∥≤∥x-y∥∀x,y∈C;(2)
r-contractive if for all x,y∈C, there exists r∈(0,1) such that
(14)∥Tx-Ty∥≤r∥x-y∥;(3)
accretive if for allx,y∈C, there exists jq(x-y)∈Jq(x-y) such that
(15)〈Tx-Ty,jq(x-y)〉≥0;(4)
η-strongly accretive if for all x,y∈C, there exists η>0 and jq(x-y)∈Jq(x-y) such that
(16)〈Tx-Ty,jq(x-y)〉≥η∥x-y∥q;(5)
μ-inverse strongly accretive if for all x,y∈C, there exists μ>0 and jq(x-y)∈Jq(x-y) such that
(17)〈Tx-Ty,jq(x-y)〉≥μ∥Tx-Ty∥q.Definition 2.
A set-valued mappingT:D(T)⊆E→2E is said to be(1)
accretive if for anyx,y∈D(T), there exists j(x-y)∈J(x-y), such that for all u∈T(x) and v∈T(y),
(18)〈u-v,j(x-y)〉≥0;(2)
m-accretive if T is accretive and (I+ρT)(D(T))=E for every (equivalently, for some) ρ>0, where I is the identity mapping.
LetM:D(M)→2E be m-accretive. The mapping JρM:E→D(M), defined by
(19)JρM(u)=(I+ρM)-1(u),∀u∈E,
is called the resolvent operator associated with M, where ρ is any positive number and I is the identity mapping. It is well known that JρM is single-valued and nonexpansive.In order to prove our main results, we need the following lemmas.Lemma 3 (see [20]).
LetE be a Banach space and Jq be a generalize duality mapping. Then for any given x,y∈E, the following inequality holds:
(20)∥x+y∥q≤∥x∥q+q〈y,jq(x+y)〉,00000000000jq(x+y)∈Jq(x+y).
In particular, we have, for any given x,y∈E,
(21)∥x+y∥2≤∥x∥2+2〈y,j(x+y)〉,j(x+y)∈J(x+y).Lemma 4 (see [21]).
Let{αn} be a sequence of nonnegative numbers satisfying the property:
(22)αn+1≤(1-γn)αn+bn+γncn,n∈ℕ,
where {γn}, {bn}, and {cn} satisfy the restrictions:(i)
∑n=1∞γn=∞,(ii)
bn≥0, ∑n=1∞bn<∞,(iii)
limsup⁡n→∞⁡cn≤0.Then, lim⁡n→∞αn=0.Lemma 5 (see [22]).
Let{an}, {bn}, and {δn} be sequences of nonnegative real numbers satisfying the inequality
(23)an+1≤(1+δn)an+bn,∀n=1,2,….
If ∑n=0∞δn<∞ and ∑n=0∞bn<∞, then lim⁡n→∞an exists. In particular, lim⁡n→∞an=0 whenever there exists a subsequence {ank} in {an} which strongly converges to zero.Lemma 6 (see [20]).
Let1<p<∞, q∈(1,2], r>0 be given.(i)
IfE is uniformly convex, then there exists a continuous, strictly increasing, and convex function φ:[0,∞)→[0,∞) with φ(0)=0 such that
(24)∥λx+(1-λ)y∥p≤λ∥x∥p+λ∥y∥p-Wp(λ)φ(∥x-y∥),000x,y∈Br,0≤λ≤1,
where Wp(λ)=λp(1-λ)+(1-λ)pλ, Br={z∈E:∥z∥≤r}.(ii)
IfE be a real q-uniformly smooth Banach space, then there exists a constant Cq>0 such that
(25)∥x+y∥q≤∥x∥q+q〈y,Jq(x)〉+Cq∥y∥q,∀x,y∈E.Lemma 7 (see [23]).
LetE be a uniformly convex Banach space, C a closed convex subset of E, and T:C→E a nonexpansive mapping. Then, I-T is demiclosed at zero.Lemma 8 (see [24]).
IfE is a uniformly convex Banach space and C is a closed convex bounded subset of E, there is a continuous strictly increasing function g:[0,∞)→[0,∞) with g(0)=0 such that
(26)g(∥S(tx+(1-t)y)-(αSx+(1-α)Sy)∥)≤∥x-y∥-∥Sx-Sy∥
for all x,y∈C,t∈[0,1] and nonexpansive mappings S:C→E.Lemma 9 (see [25]).
LetE be a real reflexive Banach space such that its dual E* has the Kadec-Klee property. Let {xn} be a bounded sequence in E and x*,y*∈ωw(xn); here ωw(xn) denotes the weak w-limit set of {xn}. Suppose lim⁡n→∞∥txn+(1-t)x*-y*∥ exists for all t∈[0,1]. Then x*=y*.Lemma 10.
LetC be a nonempty closed convex subset of a real q-uniformly smooth Banach space E. Let the mapping A:C→E be a α-inverse strongly accretive operator. Then the following inequality holds:
(27)∥(I-λA)x-(I-λA)y∥q≤∥x-y∥q-λ(qα-Cqλq-1)∥Ax-Ay∥q.
In particular, if 0<λ≤(qα/Cq)1/(q-1), then ∥I-λA∥ is nonexpansive.Proof.
Indeed, for allx,y∈C, it follows from Lemma 6 that
(28)∥(I-λA)x-(I-λA)y∥q=∥(x-y)-λ(Ax-Ay)∥q≤∥x-y∥q-qλ〈Ax-Ay,jq(x-y)〉+Cqλq∥Ax-Ay∥q≤∥x-y∥q-qαλ∥Ax-Ay∥q+Cqλq∥Ax-Ay∥q≤∥x-y∥q-λ(qα-Cqλq-1)∥Ax-Ay∥q.
It is clear that if 0<λ≤(qα/Cq)1/(q-1), then I-λA is nonexpansive. This completes the proof.Lemma 11.
Assume thatC is a nonempty closed subset of a real uniformly convex and q-uniformly smooth Banach space E. Suppose that A is a single-valued and α-inverse strongly accretive operator for some α>0 and B is an m-accretive operator in E, with C⊂D(A) and C⊂D(B). Moreover, denote Jr by
(29)Jr≡JrB=(I+rB)-1
and Tr by
(30)Tr=Jr(I-rA)=(I+rB)-1(I-rA).
Then, it is holds for all r>0 that F(Tr)=(A+B)-1(0).Proof.
From the definition ofTr, it follows that
(31)x=Trx⟺x=(I+rB)-1(I-rA)x⟺(I-rA)x∈(I+rB)x⟺0∈(A+B)x.
This lemma alludes to the fact that in order to solve the inclusion problem (1), it suffices to find a fixed point of Tr. Since Tr is already split, an iterative algorithm for Tr corresponds to a splitting algorithm for (1). However, to guarantee convergence (weak or strong) of an iterative algorithm for Tr, we need good metric properties of Tr such as nonexpansivity. To this end, some relate geometric conditions on the underlying space E are very necessary (see Lemmas 12 and 13 below).Lemma 12 (see [18]).
Assume thatC is a nonempty closed subset of a real uniformly convex and q-uniformly smooth Banach space E. Suppose that A is a single-valued and α-inverse strongly accretive operator for some α>0 and B is an m-accretive operator in E, with C⊂D(A) and C⊂D(B). Then, the following relations hold.(i)
Given0<s≤r and x∈E,
(32)∥Tsx-Trx∥≤|1-sr|∥x-Trx∥,∥x-Tsx∥≤2∥x-Trx∥.(ii)
Givens>0, there exists a continuous, strictly increasing, and convex function ϕq:[0,∞)→[0,∞) with ϕq(0)=0 such that for all x,y∈Bs,
(33)∥Trx-Try∥q≤∥x-y∥q-r(αq-rq-1Cq)∥Ax-Ay∥q-ϕq(∥(I-Jr)(I-rA)x-(I-Jr)0000000×(I-rA)y∥).Lemma 13.
Assume thatC is a nonempty closed subset of a real uniformly convex and q-uniformly smooth Banach space E. Suppose that S is a nonexpansive mapping and A is a single-valued and α-inverse strongly accretive operator for some α>0 and B is a m-accretive operator in E, with C⊂D(A) and C⊂D(B). Assume 0<r<(qα/Cq)1/(q-1). Then F(STr)=F(S)∩F(Tr).Proof.
Suppose thatx1∈F(STr), it is sufficient to show that x1∈F(S)∩F(Tr). Indeed, for x2∈F(S)∩F(Tr), we have by Lemma 12 that
(34)∥x1-x2∥q=∥STrx1-STrx2∥q≤∥Trx1-Trx2∥q≤∥x1-x2∥q-r(αq-rq-1Cq)×∥Ax1-Ax2∥q-ϕq(∥(I-Jr)(I-rA)x10000000-(I-Jr)(I-rA)x2∥.
The property of ϕ and the condition 0<r<(qα/Cq)1/(q-1) together imply that
(35)∥Ax1-Ax2∥=∥(I-Jr)(I-rA)x1-(I-Jr)(I-rA)x2∥=0.
It turns out that
(36)∥x1-Trx1-x2+Trx2∥=0,
which imply
(37)Trx1=x1.
Noticing the assumption of x1=STrx1, we can deduce x1=Sx1. This means that x1∈F(S)∩F(Tr).
Next we give a weak convergence theorem in a Banach spaceE.


## 3. Main Results

Theorem 14.
LetE be a uniformly convex and q-uniformly smooth Banach space. Let A:E→E be α-inverse strongly accretive, let B:E→2E be m-accretive, and let S:E→E be nonexpansive. Assume that F(S)∩(A+B)-1(0)≠∅. We define a sequence {xn} by the perturbed iterative scheme:
(38)xn+1=(1-αn)xn+αnS(Jrn(xn-rn(Axn+an))+bn),000000000000000000000000000∀n≥1,
where Jrn=(I+rnB)-1, {an},{bn}⊂E, {αn}⊂(0,1], and {rn}⊂(0,+∞). Assume that(i)
∑n=1∞∥an∥<∞ and ∑n=1∞∥bn∥<∞;(ii)
0<liminf⁡n→∞⁡αn≤limsup⁡n→∞⁡αn<1;(iii)
0<liminf⁡n→∞⁡rn≤limsup⁡n→∞⁡rn<(qα/Cq)1/(q-1).Then {xn} converges weakly to some point z^∈F(S)∩(A+B)-1(0).Proof.
We divide the proof into several steps.
First we prove thatlim⁡n→∞∥xn-z∥ exist for any point z∈F(S)∩(A+B)-1(0).
PuttingTn=Jrn(I-rnA)=(I+rnB)-1(I-rnA), one has
(39)S(Jrn(xn-rn(Axn+an))+bn)=STnxn+gn,
where
(40)gn=S(Jrn(xn-rn(Axn+an))+bn)-STnxn.
Then the iterative formula (38) turns into the form
(41)xn+1=(1-αn)xn+αn(STnxn+gn).
Thus, by virtue of nonexpansivity of S and Jrn, it follows that
(42)∥gn∥=∥S(Jrn(xn-rn(Axn+an))+bn)-STnxn∥≤∥Jrn(xn-rn(Axn+an))-Tnxn∥+∥bn∥≤rn∥an∥+∥bn∥.
It follows from (42) and the condition (i) that
(43)∑n=1∞‍∥gn∥<∞.
Since z∈F(S)∩(A+B)-1(0), we can deduce z∈F(S)∩F(Tn) due to Lemma 13. And Tn is nonexpansive due to Lemma 12 and condition (iii). Therefore, we can get from (41) that
(44)∥xn+1-z∥=(1-αn)∥xn-z∥+αn∥STnxn+gn-z∥≤(1-αn)∥xn-z∥+αn∥STnxn-z∥+αn∥gn∥≤(1-αn)∥xn-z∥+αn∥xn-z∥+αn∥gn∥≤∥xn-z∥+αn∥gn∥.
In view of (43), (44), and Lemma 5, we can get that lim⁡n→∞∥xn-z∥ exists. Therefor {xn} is bounded.
Next, we showlim⁡n→∞∥Tnxn-xn∥=0.
LetM1>0 be such that ∥xn∥<M1, for all n∈ℕ, and let s=q(M1+∥z∥)q-1. By (41), Lemmas 3, and 12, we have
(45)∥xn+1-z∥q=∥(1-αn)(xn-z)+αn(STnxn+gn-z)∥q≤∥(1-αn)(xn-z)+αn(STnxn-z)∥q+αnq〈gn,jq(xn+1-z)〉≤(1-αn)∥xn-z∥q+αn∥STnxn-z∥q+αnq∥gn∥∥xn+1-z∥q-1≤(1-αn)∥xn-z∥q+αn∥Tnxn-z∥q+αnq∥gn∥∥xn+1-z∥q-1≤∥xn-z∥q-αnrn(αq-rnq-1Cq)×∥Axn-Az∥q-αnϕq×(∥(I-Jrn)(I-rnA)xn000-(I-Jrn)(I-rnA)z∥)+αns∥gn∥≤∥xn-z∥q-αnrn(αq-rnq-1Cq)×∥Axn-Az∥q-αnϕq×(∥xn-rnAxn-Tnxn+rnAz∥)+αns∥gn∥.
Meanwhile, by the fact that ar-br≤rar-1(a-b), ∀r≥1, and (45), we can get that
(46)αnrn(αq-rnq-1Cq)∥Axn-Az∥q+αnϕq(∥xn-rnAxn-Tnxn+rnAz∥)≤∥xn-z∥q-∥xn+1-z∥q+αns∥gn∥≤q∥xn-z∥q-1(∥xn-z∥-∥xn+1-z∥)+αns∥gn∥.
Thanks to the conditions of (ii),(iii), and (43) and existence of lim⁡n→∞∥xn-z∥, one has
(47)lim⁡n→∞∥Axn-Az∥=lim⁡n→∞∥xn-rnAxn-Tnxn+rnAz∥=0.
Consequently,
(48)lim⁡n→∞∥Tnxn-xn∥=0.
Step  1. We prove lim⁡n→∞∥STnxn-Tnxn∥=0.
Noticing (43) and Lemma 6, we have
(49)∥xn+1-z∥2=∥(1-αn)(xn-z)+αn(STnxn+gn-z)∥2≤(1-αn)∥xn-z∥2+αn∥STnxn+gn-z∥2-W2(αn)φ(∥STnxn+gn-xn∥)≤(1-αn)∥xn-z∥2+αn(∥STnxn-z∥2+2∥STnxn-z∥∥gn∥0000000+∥gn∥2)-W2(αn)φ(∥STnxn+gn-xn∥)≤(1-αn)∥xn-z∥2+αn(∥xn-z∥2+2∥xn-z∥∥gn∥+∥gn∥2)-W2(αn)φ(∥STnxn+gn-xn∥)≤∥xn-z∥2+2αn∥xn-z∥∥gn∥+αn∥gn∥2-W2(αn)φ(∥STnxn+gn-xn∥)≤∥xn-z∥2+2∥xn-z∥∥gn∥+∥gn∥2-W2(αn)φ(∥STnxn+gn-xn∥),
which implies
(50)W2(αn)φ(∥STnxn+gn-xn∥)≤∥xn-z∥2-∥xn+1-z∥2+2∥xn-z∥∥gn∥+∥gn∥2,
where W2(αn)=αn(1-αn). From (50), assumptions (ii), (43), and existence of lim⁡n→∞∥xn-z∥, it turns out that
(51)lim⁡n→∞φ(∥STnxn+gn-xn∥)=0.
It follows from the property of φ that
(52)lim⁡n→∞∥STnxn+gn-xn∥=0.
Noticing (43), we have
(53)lim⁡n→∞∥STnxn-xn∥=0.
By (48) and (53), we get
(54)lim⁡n→∞∥STnxn-Tnxn∥=0.
Step  2. We prove ωw(xn)⊂F(S)∩(A+B)-1(0).
Sinceliminf⁡n→∞rn>0, there exists ε>0 such that rn≥ε for all n≥1. Then, by Lemma 12, we have
(55)lim⁡n→∞∥Tεxn-xn∥≤2lim⁡n→∞∥Tnxn-xn∥=0.
It follows from (48), (53), and (54) that
(56)∥STεxn-xn∥≤∥STεxn-STnxn∥+∥STnxn-xn∥≤∥Tεxn-Tnxn∥+∥STnxn-xn∥≤∥Tεxn-xn∥+∥xn-Tnxn∥+∥STnxn-xn∥⟶0.
By Lemmas 7 and 13, we get
(57)ωw(xn)⊂F(STε)=F(S)⋂‍(A+B)-1(0).
Finally, we show that {xn} converges weakly to a fixed point of z^∈F(S)∩(A+B)-1(0).
Indeed, it suffices to show thatωw(xn) consists of exactly only one point. To this end, we suppose that two different points z^ and z~ are in ωw(xn). Then there exist two different subsequences {ni} and {nj} such that xni⇀z^ and xnj⇀z~ as i→∞ and j→∞. Define Sn,m:D→E by
(58)Sn,m=Vn+m-1Vn+m-2⋯Vn,Vn=(1-αn)I+αnSTn.
Then xn can be written
(59)xn+m=Sn,mxn+cn,m,
where
(60)cn,m=Vn+m-1(Vn+m-20000000000×(⋯Vn+1(Vnxn+αngn)00000000000000+αn+1gn+1⋯Vn+m-2)0000000000+αn+m-2gn+m-2)+αn+m-1gn+m-1-Sn,mxn.
Thanks to the nonexpansivity of Vn, we have
(61)∥cn,m∥≤∑k=nn+m-1‍∥αkgk∥≤∑k=nn+m-1‍∥gk∥.
It follows from (43) that
(62)lim⁡m,n→∞∥cn,m∥⟶0.
Let
(63)fn(t)=∥txn+(1-t)z^-z~∥,dn,m=Sn,m(txn+(1-t)z^)-(tSn,mxn+(1-t)z^).
Apply Lemma 8 to the closed convex bounded subset K≔c-o({xn}∪{z^}) to obtain
(64)g(∥dn,m∥)≤∥xn-z^∥-∥Sn,mxn-Sn,mz^∥≤∥xn-z^∥-∥xn+m-z^-cn,m∥≤∥xn-z^∥-∥xn+m-z^∥+∥cn,m∥.
Since lim⁡n→∞∥xn-z^∥ exists, (62) and (64) together imply that
(65)lim⁡m,n→∞dn,m⟶0.
Furthermore, we have
(66)fn+m(t)=∥txn+m+(1-t)z^-z~∥≤∥dn,m∥+∥Sn,m(txn+(1-t)z^)-z~∥+t∥cm,n∥≤∥dn,m∥+∥txn+(1-t)z^-z~∥+t∥cm,n∥=∥dn,m∥+fn(t)+t∥cm,n∥.
After taking first limsup⁡m→∞ and then liminf⁡n→∞ in (66) and using (62) and (65), we get
(67)limsup⁡m→∞fm(t)≤liminf⁡n→∞fn(t)+lim⁡m,n→∞(∥dm,n∥+∥cn,m∥)=liminf⁡n→∞fn(t).
So that lim⁡n→∞∥txn+(1-t)z^-z~∥ exists for all t∈[0,1]. It follows from Lemma 9 that z^=z~. This completes the proof.Remark 15.
Compared with the known results in the literature, our results are very different from those in the following aspects.(i)
Theorem14 improves and extends Theorem 3.1 of Manaka and Takahashi [8] and Theorem 3 of Kamimura and Takahashi [9] from Hilbert spaces to uniformly convex and q-uniformly smooth Banach spaces.(ii)
Theorem14 also improves and extends Theorem 3.6 of López et al. [18] from the problem of finding an element of (A+B)-1(0) to the problem of finding an element of (A+B)-1(0)∩F(S).Theorem 16.
LetE be a uniformly convex and q-uniformly smooth Banach space. Let A:E→E be α-inverse strongly accretive, let B:E→2E be m-accretive, let f:E→E be r-contractive, and let S:E→E be nonexpansive. Assume that F(S)∩(A+B)-1(0)≠∅. We define a sequence {xn} by the perturbed iterative scheme:
(68)xn+1=αnf(xn)+(1-αn)×S(Jrn(xn-rn(Axn+an))+bn),0000000000000000000000000∀n≥1,
where Jrn=(I+rnB)-1. Assume that {αn}⊂[0,1], {rn}⊂(0,+∞), and {an},{bn}⊂E satisfying the following conditions:(i)
∑n=1∞∥an∥<∞ and ∑n=1∞∥bn∥<∞;(ii)
∑n=1∞αn=∞lim⁡n→∞⁡αn=0 and ∑n=1∞|αn+1-αn|<∞;(iii)
0<liminf⁡n→∞rn≤limsup⁡n→∞rn<(qα/Cq)1/(q-1) and ∑n=1∞|rn+1-rn|<∞.Then {xn} converges strongly to some point z∈F(S)∩(A+B)-1(0), which is the unique solution of the variational inequality 〈f(z)-z,j(p-z)〉≤0, ∀p∈F(S)∩(A+B)-1(0).Proof.
Let{yn} be a sequence generated by
(69)yn+1=αnf(yn)+(1-αn)STnyn,
where Tn:=Jrn(I-rnA). Hence to show the desired result, it suffices to prove that yn→z. Indeed, since Jrn and (I-rnA) are both nonexpansive under the condition of (iii), it follows that
(70)∥yn+1-xn+1∥≤∥αnf(yn)+(1-αn)S(Jrn(yn-rnAyn))000000-αnf(xn)-(1-αn)000000×S(Jrn(xn-rn(Axn+an))+bn)∥≤(1-αn)∥S(Jrn(yn-rnAyn))0000000000000-S(Jrn(xn-rn(Axn+an))+bn)(Jrn(yn-rnAyn))∥+αnr∥yn-xn∥≤(1-αn)∥Jrn(yn-rnAyn)-Jrn(xn-rn(Axn+an))∥+∥bn∥+αnr∥yn-xn∥≤[1-αn(1-r)]∥yn-xn∥+rn∥an∥+∥bn∥.
By virtue of Lemma 4 and (70), one has lim⁡n→∞∥yn-xn∥=0.
We first prove that the sequences{yn} is bounded.
Thanks to (69) and Lemma 13, we have
(71)∥yn+1-z∥=∥αnf(yn)+(1-αn)STnyn-z∥≤αn∥f(yn)-z∥+(1-αn)∥STnyn-z∥≤αn∥f(yn)-f(z)∥+αn∥f(z)-z∥+(1-αn)∥STnyn-z∥≤αnr∥yn-z∥+αn∥f(z)-z∥+(1-αn)∥yn-z∥=[1-αn(1-r)]∥yn-z∥+αn∥f(z)-z∥≤max⁡⁡{∥f(z)-z∥1-r,∥yn-z∥}.
By induction, we have
(72)∥yn-z∥≤max⁡{∥f(z)-z∥1-r,∥y1-z∥},∀n≥1.
Hence, {yn} is bounded, so are {f(yn)} and {Tn(yn)}.
Next we prove that(73)lim⁡n→∞∥yn+1-yn∥⟶0.
Putting zn=Tnyn, it follows from Lemma 12 that
(74)∥zn+1-zn∥=∥Tn+1yn+1-Tnyn∥≤∥Tn+1yn+1-Tnyn+1∥+∥Tnyn+1-Tnyn∥≤|1-rαnrβn|∥yn+1-Jrβn(1-rβnA)yn+1∥+∥yn+1-yn∥≤|rβn-rαn|∥yn+1-Jrβn(1-rβnA)yn+1∥rβn+∥yn+1-yn∥≤|rn+1-rn|M2+∥yn+1-yn∥,
where M2>sup⁡n≥1{∥yn+1-Jrβn(1-rβnA)yn+1∥/rβn}, rαn=min⁡{rn+1,rn}, and rβn=max⁡{rn+1,rn}. Hence from (69) and (74) we have
(75)∥yn+1-yn∥=∥αnf(yn)+(1-αn)Szn-αn-1f(yn-1)0000+(1-αn-1)Szn-1∥=∥(αn-αn-1)(f(yn-1)-Szn-1)0000+(1-αn)(Szn-Szn-1)∥+αn∥f(yn)-f(yn-1)∥≤|αn-αn-1|∥f(yn-1)-Szn-1∥+(1-αn)∥Szn-Szn-1∥+αnr∥yn-yn-1∥≤|αn-αn-1|M3+(1-αn)∥zn-zn-1∥+αnr∥yn-yn-1∥≤|αn-αn-1|M3+|rn-rn-1|M2+[1-αn(1-r)]∥yn-yn-1∥,
where M3>sup⁡n≥1{∥f(yn)-Szn∥}. It follows from Lemma 4, (ii), and (iii) that ∥yn+1-yn∥→0.
Again from Lemmas3 and 12, we obtain
(76)∥yn+1-z∥q=∥αn(f(yn)-z)0000+(1-αn)(STnyn-z)∥q≤(1-αn)∥Tnyn-z∥q+qαn×〈f(yn)-z,jq(yn+1-z)〉.≤∥Tnyn-z∥q+qαnM4≤∥yn-z∥q-rn(αq-rnq-1Cq)∥Ayn-Az∥q-ϕq(∥yn-rnAyn-Tnyn+rnAz∥)+qαnM4,
where M4>sup⁡n≥1{〈f(yn)-z,jq(yn+1-z)〉}. It follows immediately from (76) that
(77)lim⁡n→∞∥Ayn-Az∥=lim⁡n→∞∥yn-rnAyn-Tnyn+rnAz∥=0.
Hence we obtain that
(78)lim⁡n→∞∥Tnyn-yn∥=0.
By condition (iii), there exists ε>0 such that rn≥ε for all n≥1. Then, by Lemma 12, we get
(79)lim⁡n→∞∥Tεyn-yn∥≤lim⁡n→∞2∥Tnyn-yn∥=0.
Step  1. We show lim⁡n→∞∥STεyn-yn∥=0.
From (73), (78), (79), and (ii), we have
(80)∥STεyn-yn∥≤∥STεyn-STnyn∥+∥STnyn-yn∥≤∥Tεyn-Tnyn∥+∥STnyn-yn∥≤∥Tεyn-yn∥+∥yn-Tnyn∥+∥STnyn-yn+1∥+∥yn+1-yn∥≤∥Tεyn-yn∥+∥yn-Tnyn∥+αn∥f(yn)-STnyn∥+∥yn+1-yn∥⟶0.
Lemmas 7 and 13 imply that
(81)ωw(yn)⊂F(STε)=F(S)⋂‍F(Tε).
Next we prove that
(82)limsup⁡n→∞〈f(z)-z,jq(yn-z)〉≤0.
Equivalently (should ∥yn-z∥≠0), we need to prove that
(83)limsup⁡n→∞〈f(z)-z,j(yn-z)〉≤0.
To this end, let zt satisfy zt=tf(zt)+(1-t)STεzt. By Xu’s Theorem 4.1 [26], we get zt→z∈F(STε) as t→0, which z is the unique solution of the variational inequality:
(84)〈f(z)-z,j(x-z)〉≤0,∀x∈F(STε).
Using subdifferential inequality, we deduce that
(85)∥zt-yn∥2=t〈f(zt)-yn,j(zt-yn)〉+(1-t)〈STεzt-yn,j(zt-yn)〉=t〈f(zt)-zt,j(zt-yn)〉+t〈zt-yn,j(zt-yn)〉+(1-t)〈STεzt-STεyn,j(zt-yn)〉+(1-t)〈STεyn-yn,j(zt-yn)〉≤t〈f(zt)-zt,j(zt-yn)〉+t∥zt-yn∥2+(1-t)∥zt-yn∥2+(1-t)×∥STεyn-yn∥∥zt-yn∥≤t〈f(zt)-zt,j(zt-yn)〉+∥zt-yn∥2+∥STεyn-yn∥∥zt-yn∥,
which implies that
(86)〈f(zt)-zt,j(yn-zt)〉≤∥STεyn-yn∥t∥zt-yn∥.
Using (80), taking the upper limit as n→∞ firstly and then as t→0 in (86), we have
(87)limsup⁡t→0limsup⁡n→∞〈f(zt)-zt,j(yn-zt)〉≤0.
Since E is a uniformly smooth Banach space, we have the duality mapping j is norm-to-norm uniformly on any bounded subset of E, which ensures that the limits limsup⁡t→0 and limsup⁡n→∞ are interchangeable, and we have
(88)limsup⁡n→∞〈f(z)-z,j(yn-z)〉≤0.
Finally, we show ∥yn-z∥→0.
By Lemma3 and the fact that ab≤(1/q)aq+((q-1)/q)bq/(q-1), we get
(89)∥yn+1-z∥q=∥αnf(yn)+(1-αn)STnyn-z∥q=〈αnf(yn)+(1-αn)STnyn0000-z,jq(yn+1-z)〉=αn〈f(yn)-f(z),jq(yn+1-z)〉+αn〈f(z)-z,jq(yn+1-z)〉+(1-αn)〈STnyn-z,jq(yn+1-z)〉≤αn∥f(yn)-f(z)∥∥yn+1-z∥q-1+αn〈f(z)-z,jq(yn+1-z)〉+(1-αn)∥yn-z∥∥yn+1-z∥q-1≤αnr∥yn-z∥∥yn+1-z∥q-1+αn〈f(z)-z,jq(yn+1-z)〉+(1-αn)×∥yn-z∥∥yn+1-z∥q-1≤[1-αn(1-r)]∥yn-z∥∥yn+1-z∥q-1+αn〈f(z)-z,jq(yn+1-z)〉≤[1-αn(1-r)]1q∥yn-z∥q+q-1q×∥yn+1-z∥q+αn〈f(z)-z,jq(yn+1-z)〉,
which implies that
(90)∥yn+1-z∥q≤[1-αn(1-r)]∥yn-z∥q+qαn〈f(z)-z,jq(yn+1-z)〉.
Apply Lemma 4 to (90) to conclude yn→z as n→∞. This completes the proof.Remark 17.
Theorem16 improves and extends Theorem 3.7 of López et al. [18] in the following ways: (i)
from the problem of finding an element of(A+B)-1(0) to the problem of finding an element of (A+B)-1(0)∩F(S),(ii)
from a fixed elementu in E to a contractive mapping f.Remark 18.
Theorem16 improves and extends Theorem 2.1 of Zhang et al. [7] in the following ways: (i)
from Hilbert spaces to uniformly convex andq-uniformly smooth Banach spaces,(ii)
from a fixed elementu in E to a contractive mapping f,(iii)
from a fixed positive numberλ to a positive sequence {rn}.As a direct consequence of Theorem16, we obtain the following result.Corollary 19.
LetH be a Hilbert space. Let A:H→H be α-inverse strongly monotone, let B:H→2H be maximal monotone, let f:E→E be r-contractive, and let S:H→H be nonexpansive. Assume that F(S)∩(A+B)-1(0)≠∅. We define a sequence {xn} by the perturbed iterative scheme:
(91)xn+1=αnf(xn)+(1-αn)×S(Jrn(xn-rn(Axn+an))+bn),∀n≥1,
where Jrn=(I+rnB)-1. Assume that {αn}⊂[0,1], {rn}⊂(0,+∞), and {an},{bn}⊂H satisfying the following conditions:(i)
∑n=1∞∥an∥<∞ and ∑n=1∞∥bn∥<∞;(ii)
∑n=1∞αn=∞lim⁡n→∞⁡αn=0 and ∑n=1∞|αn+1-αn|<∞;(iii)
0<liminf⁡n→∞⁡rn≤limsup⁡n→∞⁡rn<2α and ∑n=1∞|rn+1-rn|<∞.Then {xn} converges strongly to some point z∈F(S)∩(A+B)-1(0), which is the unique solution of the variational inequality 〈f(z)-z,p-z〉≤0, ∀p∈F(S)∩(A+B)-1(0).



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*Source: 943753-2014-03-13.xml*