KYUNGPOOK Math. J. 2019; 59(3): 445-464  
Strong Convergence Theorems for Common Points of a Finite Family of Accretive Operators
Jae Ug Jeong∗ and Soo Hwan Kim
Department of Mathematics, Dongeui University, Pusan 47340, South Korea
e-mail : jujeong@deu.ac.kr and sh-kim@deu.ac.kr
accretive operator, nonexpansive mapping, zero point, resolvent operator.
Received: March 18, 2016; Revised: June 25, 2019; Accepted: August 5, 2019; Published online: September 23, 2019.
© Kyungpook Mathematical Journal. All rights reserved.

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Abstract

In this paper, we propose a new iterative algorithm generated by a finite family of accretive operators in a q-uniformly smooth Banach space. We prove the strong convergence of the proposed iterative algorithm. The results presented in this paper are interesting extensions and improvements of known results of Qin et al. [Fixed Point Theory Appl. 2014(2014): 166], Kim and Xu [Nonlinear Anal. 61(2005), 51–60] and Benavides et al. [Math. Nachr. 248(2003), 62–71].

Introduction
roduction

In the real world, many important problems, have reformulations which require finding zero points of some nonlinear operators. This is true of such problems as inverse problems, transportation problems, evolution equations, complementarity problems, mini-max problems, variational inequalities and optimization problems (see [7, 10, 11, 15, 19, 27] and the references therein). In the past several decades, many authors have studied the existence and convergence of different schemes for finding zero points for maximal monotone operators (see [20, 23, 24] and the references therein).

Let E be a real Banach space with dual space E*, and C be a nonempty closed convex subset of E. Recall that a mapping T : CC is said to be nonexpansive if TxTyxyfor all x, yC. We denote by Fix(T) the set of fixed points of T, i.e., Fix(T) = {xC : Tx = x}. Let I denote the identity operator on E. An operator AE × E with domain D(A) = {zE : Azϕ} and range R(A) = ∪{Az : zD(A)} is said to be accretive if for t > 0 and x, yD(A), ‖xy‖ ≤ ‖xy + t(uv)‖ for every uAx, vAy. It follows from Kato [13] that A is accretive if and only if for x, yD(A), there exists jq(xy) such that 〈uv, jq(xy)〉 ≥ 0. An accretive operator is said to be m-accretive if R(I + rA) = E for all r > 0. In a real Hilbert space H, an operator A is m-accretive if and only if A is maximal monotone.

One of the major problems in the theory of accretive operators is as follows: Find a point zE such that 0Az,where A is an operator from E into E*. A point zE is called a zero point of A. The set of zero points of the operator A is denoted by A−1(0). If A is m-accretive, then the solution set A−1(0) is closed and convex. For an accretive operator A, we can define a nonexpansive single-valued mapping Jr: R(I + rA) → D(A) by Jr= (I + rA)−1 for each r > 0, which is called the resolvent of A.

A well-known method for solving zeros of maximal monotone operator is a proximal point algorithm. Let A be a maximal monotone operator in a Hilbert space H. The proximal point algorithm generates, for starting x1 = xH, a sequence {xn} in H by xn+1=Jλnxn,n1,where {λn} ⊂ (0, ∞) and Jλn = (I + λnA)−1. Rockafellar [24] proved that the sequence {xn} defined by (1.1) converges weakly to an element of A−1(0).

Recently, many authors have investigated the strong convergence of a proximal point algorithm. Strong convergence theorems for zero points of accretive operators were established in [12, 14, 22, 29] and the references therein.

In 2003, Benavides et al. [4] proposed one iterative scheme for approaching a zero of an m-accretive operator in a uniformly convex Banach space as follows: {x0C,xn+1=αnxn+(1αn)Jλnxn,n0.

Recently, Kim and Xu [16] presented the following iteration scheme in a uniformly smooth Banach space: {x0,uE,xn+1=αnu+(1αn)Jλnxn,n0.And they proved the sequence {xn} converges strongly to a zero of an m-accretive operator.

In 2014, Qin et al. [21] studied the problem of modifying the proximal point algorithm to have strong convergence for the sum of two accretive operators in a real q-uniformly smooth Banach space. To be more precise, they considered the following iterative process: {x0C,xn+1=αnf(xn)+βnJλn(xnλnAxn+en)+γnfn,n0.They proved that the sequence {xn} generated in the above iterative process converges strongly to x = Proj(A+B)−1(0)f(x).

Inspired and motivated by the corresponding convergence results of (1.2)(1.4), we are concerned with the problem of finding a common zero point of a family of accretive operators based on the proximal point algorithm. A strong convergence theorem is established in a real q-uniformly smooth Banach space.The results obtained in this paper improve and extend the results of Qin et al. [21], Kim and Xu [16], Benavides et al. [4] and some other results in this direction.

Preliminaries
iminaries

Let C be a subset of a real Banach space E. Let E* be a dual space of E and q > 1 be a real number. We recall that the generalized duality mapping Jq: E → 2E* is defined by Jq(x)={x*E*:x,x*=xq,x*=xq1},xE.In particular, J = J2 is called a normalized duality mapping and Jq(x) = ‖xq−2J2(x) for x ≠ 0. We know that Jqis single-valued if E is smooth, which is denoted by jq. If E is a Hilbert space, then J = I, the identity mapping.

Let S(E) = {xE : ‖x‖ = 1}. A Banach space E is called uniformly smooth if ρE(t)t0 as t → 0, where ρE: [0, ∞) → [0, ∞) is the modulus of smoothness of E which is defined by ρE(t)=sup{12(x+y+xy)1:xS(E),yt}.A Banach space E is said to be q-uniformly smooth if there exists a constant c > 0 such that ρE(t) ≤ ctq. If E is q-uniformly smooth, then q ≤ 2 and E is uniformly smooth. It is shown in [8] that there is no Banach space which is q-uniformly smooth with q > 2. Hilbert spaces, Lp(or lp) spaces and Sobolev space Wmp, where p ≥ 2, are 2-uniformly smooth.

Definition 2.1

A mapping T : CE is said to be

  • η-strongly accretive if for all x, yC, there exists η > 0 and jq(xy) ∈ Jq(xy) such that TxTy,jq(xy)ηxyq;

  • µ-inverse strongly accretive if for all x, yC, there exists µ > 0 and jq(xy) ∈ Jq(xy) such that TxTy,jq(xy)μTxTyq.

  • L-Lipschitzian if for all x, yC, there exists L > 0 such that TxTyLxy.

  • κ-contractive if for all x, yC, there exists a constant κ ∈ (0, 1) such that TxTyκxy.Let D be a nonempty subset of C. Let Q : CD be a mapping. Then Q is said to be sunny if Q(tx + (1 − t)Qx) = Qx whenever tx + (1 − t)QxC for xC and t ≥ 0. Q is called a retraction if Qx = x for all xD. Next we give the following lemmas which play an important role in this article.

Lemma 2.1

([18]) Let q > 1. Then the following inequality holds:ab1qaq+q1qbqq1for arbitrary positive real numbers a, b.

Lemma 2.2

Let q > 1. For any two nonnegative real numbers a and b, we have(a+b)q2q(aq+bq).

Lemma 2.3

([28]) Let E be a real q-uniformly smooth Banach space. Then the following inequality holds:x+yqxq+qy,jq(x)+Kqyq,x,yE,where Kq is some fixed positive constant.

Lemma 2.4

([3]) Let E be a Banach space and A be an m-accretive operator. For λ > 0, µ > 0 and xE, we haveJλx=Jμ(μλx+(1μλ)Jλx),where Jλ= (I + λA)−1and Jµ= (I + µA)−1.

Lemma 2.5

([25]) Let {xn} and {yn} be bounded sequences in a Banach space E and {βn} be a sequence in (0, 1) with 0 < lim infn→∞βn≤ lim supn→∞βn < 1. Suppose that xn+1 = (1 − βn)yn+ βnxn for every n ≥ 1 andlimsupn(yn+1ynxn+1xn)0.Then limn→∞ynxn‖ = 0.

Lemma 2.6

([20]) Let C be a nonempty closed convex subset of a real uniformly smooth Banach space E. Let f : CC be a contractive mapping and let T : CC be a nonexpansive mapping. For each t ∈ (0, 1), let xt be the unique solution of the equation x = tf(x) + (1 − t)Tx. Then {xt} converges strongly to a fixed point x̄ = QFix(T)f().

Lemma 2.7

([2]) Let E be a real Banach space and let C be a nonempty closed and convex subset of E. Let A : CE be a single-valued operator and let M : E → 2Ebe an m-accretive operator. Then Fix(Jr(IrA)) = (A + M)−1(0), where Jr= (I + rM)−1.

Lemma 2.8

([17]) Let {an} be a sequence of nonnegative real numbers such that an+1 ≤ (1 − tn)an+ bn+ cn, where {cn} is a sequence of nonnegative real numbers, {tn} ⊂ (0, 1) and {bn} is a number sequence. Assume thatn=0tn=, limsupnbntn0andn=0cn<. Then limn→∞an= 0.

Main Results
n Results

Theorem 3.1

Let E be a real q-uniformly smooth Banach space with the constant Kq and C be a nonempty closed convex subset of E. Let N ≥ 1 be some positive integer. For each i = 1, 2, · · · , N, let Ai: CE be an ηi-strongly accretive and Li-Lipschitzian mapping. Let M : D(M)(⊆ C) → 2Ebe an m-accretive operator and f : CC be a κ-contraction. Assume that(i=1NλiAi+M)1(0)ϕ. Let {xn} be a sequence generated in the following iterative process:{x0C,xn+1=αnf(xn)+βnJrn((Irni=1NλiAi)xn+en)+γnxn+δngn,n0,where Jrn= (I + rnM)−1, {en} is a sequence in E, {gn} is a bounded sequence in E and {rn} is a positive real number sequence. Suppose that {αn}, {βn}, {γn}, {δn} and {λn} are sequences in [0, 1] satisfying the following restrictions:

  • αn+ βn+ γn+ δn= 1 andi=1Nλi=1;

  • limn→∞αn= 0 andn=1αn=;

  • 0 < lim infn→∞γn≤ lim supn→∞γn < 1;

  • n=0en<andn=0δn<;

  • rnµ > 0 for each n ≥ 0, rn[qη(2N1L)qKq]1q1andn=1|rnrn1|<,

where η = min{η1, η2, · · · , ηN} and L = max{L1, L2, · · · , LN}. Then the sequence {xn} converges strongly tox¯=Q(i=1NλiAi+M)1(0)f(x¯), which is the unique solution to the following variational inequality:f(x¯)x¯,jq(px¯)0,p(i=1NλiAi+M)1(0),whereQ(i=1NλiAi+M)1(0)is the unique sunny nonexpansive retraction of C onto(i=1NλiAi+M)1(0).

Proof

First, we show that i=1NλiAi is an η(2N1L)q-inverse strongly accretive mapping.

For all x, yC, from Lemma 2.2, we have (i=1NλiAi)x(i=1NλiAi)yq=(λ1A1xλ1A1y)+[(i=2NλiAi)x(i=2NλiAi)y]q2q(λ1A1xλ1A1yq+(i=2NλiAi)x(i=2NλiAi)yq)2(N1)qi=1NλiAixAiyq.Since Ai: CE is ηi-strongly accretive and Li-Lipschitzian continuous, it follows from (3.2) that (i=1NλiAi)x(i=1NλiAi)y,jq(xy)=i=1NλiAixAiy,jq(xy)i=1Nλiηixyqi=1NλiηiLiqAixAiyqi=1NλiηLqAixAiyqη(2N1L)q(i=1NλiAi)x(i=1NλiAi)yq,x,yC.This proves that i=1NλiAi is η(2N1L)q -inverse strongly accretive.

Next, we show that Irni=1NλiAi is a nonexpansive mapping.

In view of Lemma 2.3 and (3.3), we find that (Irni=1NλiAi)x(Irni=1NλiAi)yqxyqqrn(i=1NλiAi)x(i=1NλiAi)y,jq(xy)+Kqrnq(i=1NλiAi)x(i=1NλiAi)yqxyq(qη(2N1L)qKqrnq1)rn(i=1NλiAi)x(i=1NλiAi)yq.From condition (5), we find that Irni=1NλiAi is a nonexpansive mapping.

Fixing p(i=1NλiAi+M)1(0), we find from Lemma 2.7 that x1p=α0f(x0)+β0Jr0((Ir0i=1NλiAi)x0+e0)+γ0x0+δ0g0pα0f(x0)p+β0Jr0((Ir0i=1NλiAi)x0+e0)p+γ0x0p+δ0g0pα0κx0p+α0f(p)p+β0x0p+β0e0+γ0x0p+δ0g0p[1α0(1κ)]x0p+α0f(p)p+e0+δ0g0p.Next, we prove that xnpM1+i=0n1ei+i=0n1δigip,where M1=max{x0p,f(p)p1κ}<.

In view of (3.4), we find that (3.5) holds for n = 1. We assume that the result holds for some m. Notice that xm+1p=αmf(xm)+βmJrm((Irmi=1NλiAi)xm+em)+γmxm+δmgmpαmf(xm)p+βmJrm((Irmi=1NλiAi)xm+em)p+γmxmp+δmgmpαmκxmp+αmf(p)p+βmxmp+βmem+γmxmp+δmgmp[1αm(1κ)]xmp+αm(1κ)f(p)p1κ+em+δmgmpM1+i=0mei+i=0mδigip.This shows that (3.5) holds.

In view of the restriction (4), we find that the sequence {xn} is bounded.

Put yn=(Irni=1NλiAi)xn+en. Note that yn+1yn=(Irn+1i=1NλiAi)xn+1+en+1(Irni=1NλiAi)xnen(Irn+1i=1NλiAi)xn+1(Irn+1i=1NλiAi)xn+(Irn+1i=1NλiAi)xn(Irni=1NλiAi)xn+en+1+enxn+1xn+|rn+1rn|(i=1NλiAi)xn+en+1+en.Put zn=xn+1γnxn1γn. Now we compute ‖zn+1zn‖. Note that zn+1zn=xn+2γn+1xn+11γn+1xn+1γnxn1γn=αn+11γn+1(f(xn+1)Jrn+1(yn+1))+Jrn+1(yn+1)+δn+11γn+1(gn+1Jrn+1(yn+1))αn1γn(f(xn)Jrn(yn))Jrn(yn)δn1γn(gnJrn(yn)).

This yields zn+1znαn+11γn+1f(xn+1)Jrn+1(yn+1)+Jrn+1(yn+1)Jrn(yn)+αn1γnf(xn)Jrn(yn)+δn+11γn+1gn+1Jrn+1(yn+1)+δn1γngnJrn(yn).

Next, we estimate ‖Jrn+1(yn+1) − Jrn(yn)‖.

In view of Lemma 2.4 and (3.6), we find that Jrn+1(yn+1)Jrn(yn)=Jrn[rnrn+1yn+1+(1rnrn+1)Jrn+1(yn+1)]Jrn(yn)rnrn+1(yn+1yn)+(1rnrn+1)(Jrn+1(yn+1)yn)yn+1yn+|rn+1rn|rn+1Jrn+1yn+1yn+1xn+1xn+|rn+1rn|((i=1NλiAi)xn+Jrn+1yn+1yn+1rn+1)+en+1+enxn+1xn+|rn+1rn|M2+en+1+en,where M2 is an appropriate constant such that M2supn0{(i=1NλiAi)xn+Jrn+1yn+1yn+1μ}. Substituting (3.8) into (3.7), we find that zn+1znαn1γn+1f(xn+1)Jrn+1(yn+1)+xn+1xn+|rn+1rn|M2+en+1+en+αn1γnf(xn)Jrn(yn)+δn+11γn+1gn+1Jrn+1(yn+1)+δn1γngnJrn(yn).So, we get zn+1znxn+1xnαn+11γn+1f(xn+1)Jrn+1(yn+1)+|rn+1rn|M2+en+1+en+αn1γnf(xn)Jrn(yn)+δn+11γn+1gn+1Jrn+1(yn+1)+δn1γngnJrn(yn).In view of the restrictions (2), (3), (4) and (5), we find that limsupn(zn+1znxn+1xn)0.It follows from Lemma 2.5 that limn→∞znxn‖ = 0. From the restriction (3), we obtain limnxn+1xn=0.Notice that xnJrnynxnxn+1+xn+1Jrnynxnxn+1+αnf(xn)Jrnyn+γnxnJrnyn+δngnJrnyn.It follows that (1γn)xnJrnynxnxn+1+αnf(xn)Jrnyn+δngnJrnyn.In view of the restrictions (2), (3), (4) and (3.9), we find that limnxnJrnyn=0.Notice that xnJrn(Irni=1NλiAi)xnxnJrnyn+JrnynJrn(Irni=1NλiAi)xnxnJrnyn+en.Since n=0en<, we see from (3.10) that limnxnJrn(Irni=1NλiAi)xn=0.Since M is an accretive mapping, we have M(Jr(Iri=1NλiAi)xn)M(Jrn(Irni=1NλiAi)xn),jq(Jr(Iri=1NλiAi)xnJrn(Irni=1NλiAi)xn)0.From M=Jr1Ir=Jrn1Irn, we obtain (1r1rn)xn+(1rn1r)Jrn(Irni=1NλiAi)xn,jq(Jr(Iri=1NλiAi)xnJrn(Irni=1NλiAi)xn)+1rJrn(Irni=1NλiAi)xnJr(Iri=1NλiAi)xn,jq(Jr(Iri=1NλiAi)xnJrn(Irni=1NλiAi)xn)0.So, we get Jr(Iri=1NλiAi)xnJrn(Irni=1NλiAi)xn,jq(Jr(Iri=1NλiAi)xnJrn(Irni=1NλiAi)xn)rnrrnxnJrn(Irni=1NλiAi)xn,jq(Jr(Iri=1NλiAi)xnJrn(Irni=1NλiAi)xn).It follows that Jr(Iri=1NλiAi)xnJrn(Irni=1NλiAi)xnqrnrrnxnJrn(Irni=1NλiAi)xn,jq(Jr(Iri=1NλiAi)xnJrn(Irni=1NλiAi)xn)xnJrn(Irni=1NλiAi)xnJr(Iri=1NλiAi)xnJrn(Irni=1NλiAi)xnq1.Hence we get from (3.11) that Jr(Iri=1NλiAi)xnJrn(Irni=1NλiAi)xnxnJrn(Irni=1NλiAi)xn0asn.This implies that limnJr(Iri=1NλiAi)xnxn=0.Since Jr(Iri=1NλiAi) is nonexpansive and f is contractive, we find that the mapping tf+(1t)Jr(Iri=1NλiAi) is contractive, where t ∈ (0, 1). Let ztbe the unique fixed point of the tf+(1t)Jr(Iri=1NλiAi), that is, zt=tf(zt)+(1t)Jr(Iri=1NλiAi)ztfor all t ∈ (0, 1). Put = limt→0zt. Then we have from Lemma 2.6 that x¯=Q(i=1NλiAi+M)1(0)f(x¯), where Q(i=1NλiAi+M)1(0) is the unique sunny non-expansive retraction from C onto (i=1NλiAi+M)1(0).

Next, we claim that lim supn→∞f() − , jq(xn)〉 ≤ 0.

Since ztxn=(1t)(Jr(Iri=1NλiAi)ztxn)+t(f(zt)xn),we see that for any t ∈ (0, 1), ztxnq=(1t)Jr(Iri=1NλiAi)ztxn,jq(ztxn)+tf(zt)xn,jq(ztxn)=(1t)(Jr(Iri=1NλiAi)ztJr(Iri=1NλiAi)xn,jq(ztxn)+Jr(Iri=1NλiAi)xnxn,jq(ztxn))+tf(zt)zt,jq(ztxn)+tztxn,jq(ztxn)(1t)(ztxnq+Jr(Iri=1NλiAi)xnxnztxnq1)+tf(zt)zt,jq(ztxn)+tztxnqztxnq+Jr(Iri=1NλiAi)xnxnztxnq1+tf(zt))zt,jq(ztxn).It follows that ztf(zt),jq(ztxn)1tJr(Iri=1NλiAi)xnxnztxnq1.By virtue of (3.12), we find that limnsupztf(zt),jq(ztxn)0.Since zt as t → 0 and the duality mapping jqis single-valued and strong-weak* uniformly continuous on bounded subsets of E, we see that |f(x¯)x¯,jq(xnx¯)ztf(zt),jq(ztxn)||f(x¯)x¯,jq(xnx¯)jq(xnzt)|+|f(x¯)x¯+ztf(zt),jq(xnzt)||f(x¯)x¯,jq(xnx¯)jq(xnzt)|+f(x¯)x¯+ztf(zt)xnztq10ast0.Hence, for any ε > 0, there exists δ > 0 such that for t ∈ (0, δ), the following inequality holds: f(x¯)x¯,jq(xnx¯)ztf(zt),jq(ztxn)+ε.This implies that limnsupf(x¯),x¯,jq(xnx¯)limnsupztf(zt),jq(ztxn)+ε.Using (3.13), we see that limnsupf(x¯)x¯,jq(xnx¯)0.

Finally, we prove that xn as n → ∞.

By virtue of (3.1), Lemma 2.1 and 2.7, we obtain that xn+1x¯q=αnf(xn)+βnJrn((Irni=1NλiAi)xn+en)+γnxn+δngnx¯qαnf(xn)f(x¯),jq(xn+1x¯)+αnf(x¯)x¯,jq(xn+1x¯)+βnJrn((Irni=1NλiAi)xn+en)x¯xn+1x¯q1+γnxnx¯xn+1x¯q1+δngnx¯xn+1x¯q1αnκxnx¯xn+1x¯q1+αnf(x¯)x¯,jq(xn+1x¯)+βnJrn((Irni=1NλiAi)xn+en)Jrn(Irni=1NλiAi)x¯xn+1x¯q1+γnxnx¯xn+1x¯q1+δngnx¯xn+1x¯q1αnκxnx¯xn+1x¯q1+αnf(x¯)x¯,jq(xn+1x¯)+βn(xnx¯+en)xn+1x¯q1+γnxnx¯xn+1x¯q1+δngnx¯xn+1x¯q1(1αn(1κ))xnx¯xn+1x¯q1+αnf(x¯)x¯,jq(xn+1x¯)+enxn+1x¯q1+δngnx¯xn+1x¯q1(1αn(1κ))(1qxnx¯q+q1qxn+1x¯q)+αnf(x¯)x¯,jq(xn+1x¯)+enxn+1x¯q1+δngnx¯xn+1x¯q1.It follows that xn+qx¯q(1αn(1κ))xnx¯q+qαnf(x¯)x¯,jq(xn+1x¯)+νn,where νn= q(‖en‖‖xn+1q−1 + δngn‖|xn+1q−1). In view of the restriction (4), we find that n=0νn<. Using the restriction (2), (3.14) and Lemma 2.8, we see that {xn} converges strongly to . This completes the proof.

Remark 3.1

  • Theorem 3.1 is still valid in the framework of the spaces which are Hilbert spaces and Lp-spaces, where p ≥ 2.

  • Theorem 3.1 improves and extends the main result of Qin et al. [21], Kim and Xu [16] and Benavides et al. [4] in the following aspects:

    • From a single inverse strongly accretive mapping to the class of strongly accretive and Lipschitzian mappings:

    • From the problem of finding an element of (A + M)−1(0) or A−1(0) to problem of finding an element of (i=1NλiAi+M)1(0):

    • From a forward-backward splitting algorithm with computational errors to a generalized forward-backward splitting algorithm algorithm with computational errors.

From Theorem 3.1, we obtain the following results:

Corollary 3.1

Let E be a real q-uniformly smooth Banach space with constant Kq and C be a nonempty closed convex subset of E. Let A : CE be an η-strongly accretive and L-Lipschitzian mapping. Let M : D(M)(⊆ C) → 2Ebe an m-accretive operator and f : CC be a κ-contraction. Assume that (A + M)−1(0) ≠ ϕ. Let {xn} be a sequence generated in the following iterative process:{x0C,xn+1=αnf(xn)+βnJrn(IrnA)xn+en)+γnxn+δngn,n0,where Jrn = (I + rnM)−1, {en} is a sequence in E, {gn} is a bounded sequence in E and {rn} is a positive real number sequence. Suppose that {αn}, {βn}, {γn} and {δn} are sequences in [0, 1] satisfying the following restrictions:

  • αn+ βn+ γn+ δn= 1;

  • limn→∞αn= 0 andn=0αn=;

  • 0 < lim infn→∞γn≤ lim supn→∞γn < 1;

  • n=0en<andn=0δn<;

  • rn > µ > 0 for each n ≥ 0, rn[qηLqKq]1q1andn=1|rnrn1|<.

Then the sequence {xn} converges strongly to x̄ = Q(A+M)−1(0)f(), which is the unique solution to the following variational inequality:f(x¯)x¯,jq(px¯)0,p(A+M)1(0).

Proof

Taking A1 = A2 = · · · = An= A in Theorem 3.1, then i=1NλiAi=A. So, we can get the desired conclusion easily.

Corollary 3.2

Let E be a real q-uniformly smooth Banach space with the constant Kq and C be a nonempty closed convex subset of E. Let A : CE be an η-strongly accretive and L-Lipschitzian mapping. Let M : D(M)(⊆ C) → 2Ebe an m-accretive operator. Assume that (A + M)−1(0) ≠ ϕ. Let {xn} be a sequence generated in the following iterative process:{x0C,xn+1=αnu+βnJrn((IrnA)xn+en)+γnxn,n0,where u is a fixed element in C, Jrn = (I + rnM)−1, {en} is a sequence in E and {rn} is a positive real number sequence. Suppose that {αn}, {βn} and {γn} are sequences in [0, 1] satisfying the following restrictions:

  • αn+ βn+ γn= 1;

  • limn→∞αn= 0 andn=0αn=;

  • 0 < lim infn→∞γn≤ lim supn→∞γn < 1;

  • n=0en<;

  • rn > µ > 0 for each n ≥ 0, rn[qηLqKq]1q1andn=1|rnrn1|<.

Then the sequence {xn} converges strongly to x̄ = Q(A+M)−1(0)u, which is the unique solution to the following variational inequality:ux¯,jq(px¯)0,p(A+M)1(0).

Proof

Taking the mapping f maps any element in C into a fixed element u and δn= 0 in Theorem 3.1, we can get the desired conclusion easily.

Applications
lications

In this section, we consider some applications of Theorem 3.1 in the framework of Hilbert spaces.

  • (I) Application to Theorem 3.1 for k-strict pseudocontractive mappings.

Let C be a nonempty closed convex subset of a Hilbert space H.

Definition 4.1

([6]) A mapping T : CC is said to be a ν-strict pseudocontractive mapping if there exists ν ∈ [0, 1) such that TxTy2xy2+ν(IT)x(IT)y2,x,yC.

Let T : CC be ν-strict pseudocontractive. Define a mapping A = IT : CC. It is easy to see that A is a 1ν2-inverse strongly monotone mapping and Fix(T) = A−1(0).

Lemma 4.1

([1]) Let C be a nonempty closed convex subset of a Hilbert space H. Given an integer N ≥ 1, assume that{Ti}i=1N : HH is a finite family of νi-strict pseudocontractive mappings. Suppose that{λi}i=1Nis a positive real sequence such thati=1Nλi=1. Theni=1NλiTiis a ν-strict pseudocontractive mapping with ν = max{νi: 1 ≤ iN} andFix(i=1NλiTi)=i=1NFix(Ti).

Theorem 4.1

Let C be a closed convex subset of a real Hilbert space H. For each i = 1, 2, · · · , N, Ti: CC is νi-strict pseudocontractive mapping. Let f : CC be a κ-contraction. Assume thati=1NFix(Ti)ϕ. Let {xn} be a sequence generated in the following iterative process:{x0C,xn+1=αnf(xn)+βn[Irn(Ii=1NλiTi)]xn+γnxn+δngn,n0,where {gn} is a bounded sequence in E and {rn} is a positive real number sequence. Suppose that {αn}, {βn}, {γn}, {δn} and {λn} are sequences in [0, 1] satisfying the following restrictions:

  • αn+ βn+ γn+ δn= 1 andi=1Nλi=1;

  • limn→∞αn= 0 andi=0αn=;

  • 0 < lim infn→∞γn≤ lim supn→∞γn < 1;

  • n=0δn<;

  • rn > µ > 0 for each n ≥ 0, rn≤ 1 − ν andn=1|rnrn1|<, where ν = max{νi: 1 ≤ iN}.

Then the sequence {xn} converges strongly to x̄, wherex¯i=1NFix(Ti)is the unique solution to the following variational inequalityf(x¯)x¯,px¯0,pi=1NFix(Ti).

Proof

From Lemma 4.1, we know that i=1NλiTi is a ν-strict pseudocontractive mapping with ν = max{νi: 1 ≤ iN} and Fix(i=1NλiTi)=i=1NFix(Ti). It follows that Ii=1NλiTi is a 1ν2-inverse strongly monotone mapping. The conclusion of Theorem 4.1 can be obtained from Theorem 3.1 immediately.

  • (II) Application to equilibrium problems.

Let F : C × C → ℝ be a bifunction, where ℝ is a set of real numbers. The equilibrium problem for the function F is to find a point xC such that F(x,y)0,yC.The set of solutions of (4.2) is denoted by EP(F). For solving the equilibrium problem, we assume that F satisfies the following conditions (see [5]):

  • (A1) F (x, y) = 0 for all xC;

  • (A2) F is monotone, i.e., F (x, y) + F (y, x) ≤ 0 for all x, yC;

  • (A3) F is upper-hemicontinuous, i.e., for each x, y, zC, lim supt0+F(tz+(1t)x,y)F(x,y);

  • (A4) F (x, ·) is convex and weakly lower semicontinuous for each xC.

Lemma 4.2

([5]) Let C be a nonempty closed convex subset of H and let F be a bifunction of C × C intosatisfying (A1)–(A4). Let r > 0 and xH. Then there exists zC such thatF(z,y)+1ryz,zx0,yC.

Lemma 4.3.([9])

Assume that F : C × C → ℝ satisfies (A1)–(A4). For r > 0 and xH, define a mapping Tr: HC as follows:Tr(x)={zC:F(z,y)+1ryz,zx0,yC}for all xH. Then the following hold:

  • Tr is single-valued;

  • Tr is firmly nonexpansive, i.e., for all x, yH,TrxTry2TrxTry,xy;

  • Fix(Tr) = EP(F), ∀r > 0;

  • EP(F) is a closed and convex set.

Lemma 4.4

([26]) Let C be a nonempty closed convex subset of H and let F be a bifunction of C ×C intosatisfying (A1)–(A4). Let AF be a multi-valued mapping of H into itself defined byAFx={{zH:F(x,y)yx,z,yC},xC,ϕ,xC.ThenEP(F)=AF1(0)and AF is a maximal monotone operator with D(AF) ⊆ C. Further, for any xH and r > 0, Tr coincides with the resolvent of AF, i.e., Trx = (I + rAF)−1x.

Theorem 4.2

Let C be a nonempty closed convex subset of a real Hilbert space H. Let F : C × C → ℝ be a bifunction satisfying (A1)–(A4) and f : CC be a κ-contraction. AssumeAF1(0)ϕ. Let {xn} be a sequence generated in the following iterative process:{x0C,xn+1=αnf(xn)+βnTrn(xn+en)+γnxn+δngn,n0,where {en} is a sequence in H, {gn} is a bounded sequence in H. Suppose that {αn}, {βn}, {γn} and {δn} are sequences in [0, 1] satisfying the following restrictions:

  • αn+ βn+ γn+ δn= 1;

  • limn→∞αn= 0 andn=1αn=;

  • 0 < lim infn→∞γn≤ lim supn→∞γn < 1;

  • n=0en<andn=0δn<;

  • rn > µ > 0 for each n ≥ 0 andn=1rnrn1<.

Then the sequence {xn} converges strongly tox¯=PAF1(0)f(x¯), which is the unique solution to the following variational inequality:f(x¯)x¯,px¯0,pAF1(0).

Proof

Taking Ai= 0 for i = 1, 2, · · · , N and Jrn = Trn, iterative scheme (4.3) reduces to (3.1) in a Hilbert space and the desired conclusion follows immediately from Lemma 4.4 and Theorem 3.1. This completes the proof.

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