# Approximating Iterations for Nonexpansive and Maximal Monotone Operators

Authors: Zhangsong Yao; Sun Young Cho; Shin Min Kang; Li-Jun Zhu
Journal: Abstract and Applied Analysis
 (2015)

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## Abstract

We present two algorithms for finding a zero of the sum of two monotone operators and a fixed point of a nonexpansive operator in Hilbert spaces. We show that these two algorithms converge strongly to the minimum norm common element of the zero of the sum of two monotone operators and the fixed point of a nonexpansive operator.


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## Body

## 1. Introduction

Throughout, we assume thatH is a real Hilbert space with inner product 〈·,·〉 and norm ∥·∥, respectively. Let C⊂H be a nonempty closed convex set.Definition 1.
An operatorS:C→C is said to benonexpansive if(1)Su-Sv≤u-vfor all u,v∈C.We denote byFix(S) the set of fixed points of S.Definition 2.
An operatorA:C→H is said to be ξ-inverse strong monotone if (2)Au-Av,u-v≥ξAu-Av2for some ξ>0 and for all u,v∈C.It is known that ifA is ξ-inverse strong monotone, then A is 1/ξ-lipschitz, that is, (3)Au-Av≤1ξu-v,for all u,v∈C. Furthermore,(4)I-δAu-I-δAv2≤u-v2+δδ-2ξAu-Av2,∀u,v∈C.In particular, if δ∈(0,2ξ), then I-δA is nonexpansive.LetB:H→2H be a set-valued operator. The effective domain of B is denoted by dom(B), that is, dom(B)={x∈H:Bx≠∅}.Definition 3.
A multivalued operatorB is said to be amonotone on H if and only if(5)x-y,u-v≥0 for all x,y∈dom(B), u∈Bx, and v∈By.A monotone operatorB on H is said to bemaximal if and only if its graph is not strictly contained in the graph of any other monotone operator on H. We denote by B-10 the set of the zero points of B, that is, B-10={x∈H:0∈Bx}.Forλ>0, we define a single-valued operator (6)JλB=I+λB-1:H⟶domB, which is called the resolvent of B for λ. It is known that the resolvent JλB is firmly nonexpansive, that is,(7)JλBu-JλBv2≤JλBu-JλBv,u-v,for all u,v∈C and B-10=Fix(JλB) for all λ>0.In the present paper, we consider the variational inclusion of finding a zerox∈H of the sum of two monotone operators A and B such that(8)0∈Ax+Bx,where A:H→H is a single-valued operator and B:H→2H is a set-valued operator. The set of solutions of problem (8) is denoted by (A+B)-1(0).Special Cases. (i) If H=Rm, then problem (8) becomes the generalized equation introduced by Robinson [1].(ii) IfA=0, then problem (8) becomes the inclusion problem introduced by Rockafellar [2].It is known that (8) provides a convenient framework for the unified study of optimal solutions in many optimization related areas including mathematical programming, complementarity, variational inequalities, optimal control, mathematical economics, equilibria, and game theory. Also various types of variational inclusions problems have been extended and generalized. For related work, please see [3–20].Zhang et al. [21] introduced the following iterative algorithm for finding a common element of the set of solutions to the problem (8) and the set of fixed points of a nonexpansive operator:(9)xn+1=αnx0+1-αnSJλBxn-λAxn,where S:C→C is a nonexpansive operator. Under some mild conditions, they prove that the sequence {xn} converges strongly to x∗∈Fix(S)∩(A+B)-1(0).Recently, Takahashi et al. [22] introduced another iterative algorithm for finding a zero of the sum of two monotone operators and a fixed point of a nonexpansive operator(10)xn+1=βnxn+1-βnSαnx0+1-αnJλnBxn-λnAxnfor all n≥0. Under some assumptions, they proved that the sequence {xn} converges strongly to a point of Fix(S)∩(A+B)-1(0).Motivated and inspired by (9) and (10), in the present paper, we suggest two algorithms(11)xt=JλB1-tSxt-λASxt,t∈0,1,(12)xn+1=βnxn+1-βnJλnB1-αnSxn-λnASxn,n≥0.It is obvious that (12) is very different from (9) and (10). Furthermore, we prove that both (11) and (12) converge strongly to the minimum norm element in Fix(S)∩(A+B)-10. It should be pointed out that we do not use the metric projection in (11) and (12).


## 2. Lemmas

In this section, we collect several useful lemmas for our next section.First, the following resolvent equality is well known.Lemma 4.
Forλ>0 and λ†>0, one has(13)JλBu=Jλ†Bλ†λu+1-λ†λJλBu,∀u∈H.Lemma 5 (see [23]).
LetC⊂H be a closed convex set. Let S:C→C be a nonexpansive operator. Then Fix(S) is a closed convex subset of C and the operator I-S is demiclosed at 0.Lemma 6 (see [24]).
LetX be a Banach space. Let {un}⊂X and {vn}⊂X be two bounded sequences. Let the sequence {ζn}⊂(0,1) satisfy 0<lim_n→∞ζn≤lim¯n→∞ζn<1. Suppose un+1=(1-ζn)vn+ζnun for all n≥0 and lim¯n→∞(∥vn+1-vn∥-∥un+1-un∥)≤0. Then limn→∞∥un-vn∥=0.Lemma 7 (see [25]).
Let{σn}⊂[0,∞), {γn}⊂(0,1), and {δn}⊂R be three sequences satisfying (14)σn+1≤1-γnσn+δnγn. If ∑n=1∞‍γn=∞ and lim¯n→∞δn≤0 (or ∑n=1∞‍δnγn<∞), then limn→∞σn=0.


## 3. Strong Convergence Results

LetC⊂H be a nonempty closed convex set. Let A:C→H be a ϱ-inverse strong monotone operator. Let B be a maximal monotone operator on H such that dom(B)⊂C. Let S:C→C be a nonexpansive operator.Pick up a constantτ∈(0,2ϱ). For any t∈(0,2ϱ-τ/2ϱ), we define an operator (15)ψx=JτB1-tS-τASx,for all x∈C.SinceJτB, S, and I-τA/(1-t) are nonexpansive, we have(16)ψx-ψy=JτB1-tI-τ1-tASxmk1-tI-τ1-tASx-JτB1-tI-τ1-tASy≤1-tI-τ1-tASxmmmmkk-I-τ1-tASy≤1-tx-y,for any x,y∈C. Hence ψ is a contraction on C. We use xt to denote the unique fixed point of ψ in C. Thus, {xt} satisfies the fixed point equation(17)xt=JτB1-tSxt-τASxt.Next, we give the convergence analysis of (17).Theorem 8.
Assume thatFix(S)∩(A+B)-10≠∅. Then {xt} defined by (17) converges strongly, as t→0+, to the minimum norm element in Fix(S)∩(A+B)-1(0).Proof.
Choose anyz∈Fix(S)∩(A+B)-1(0). It is obvious that z=Sz=JτB(z-τAz) for all τ>0. So, we have(18)z=Sz=JτBz-τAz=JτBtz+1-tI-τ1-tASzfor all t∈(0,1).
From (17), we have(19)xt-z=JτB1-tI-τ1-tASxt-z=JτB1-tSxt-τ1-tASxtmm-JτBtz+1-tSz-τ1-tASz≤1-tSxt-τ1-tASxtmm-tz+1-tSz-τ1-tASz=1-tSxt-τ1-tASxtmkmmmm-Sz-τ1-tASz-tz≤1-tI-τ1-tASxt-I-τ1-tASz+tz≤1-txt-z+tz.Hence, we get(20)xt-z≤z.Thus, {xt} is bounded.
By (4) and (19), we derive(21)xt-z2≤1-tSxt-τ1-tASxtmkmmmmk-Sz-τ1-tASz+t-z2≤1-tSxt-τ1-tASxtmmmmmk-Sz-τ1-tASz2+tz2=1-tSxt-Sz-τ1-tASxt-ASz2+tz2=1-tτ21-t2Sxt-Sz2-2τ1-tmmmmkm×ASxt-ASz,Sxt-Szmmmmkm+τ21-t2ASxt-ASz2+tz2≤1-tτ21-t2Sxt-Sz2-2ϱτ1-tASxt-ASz2mmmmmk+τ21-t2ASxt-ASz2+tz2=1-tSxt-Sz2+τ1-t2τ-21-tϱmmmmmkτ1-t2×ASxt-ASz2+tz2≤1-txt-z2+τ1-tτ-21-tϱ×ASxt-ASz2+tz2.So,(22)τ1-t21-tϱ-τASxt-Az2≤tz2-txt-z2⟶0.Since 2(1-t)ϱ-τ>0 for all t∈(0,1-τ/2ϱ), we obtain(23)limt→0+ASxt-Az=0.Using the firm nonexpansivity of JτB, we have(24)xt-z2=JτB1-tSxt-τASxt-z2=JτB1-tSxt-τASxt-JτBz-τAz2≤1-tSxt-τASxt-z-τAz,xt-z=121-tSxt-τASxt-z-τAz2mmk+xt-z2mmk-1-tSxt-τASxt-τAz-xt2.Note that(25)1-tSxt-τASxt-z-τAz2=1-tSxt-τ1-tASxtmmmmmmm-Sz-τ1-tASz+t-z2≤1-tSxt-τ1-tASxtmmmmmm-Sz-τ1-tASz2+tz2≤1-txt-z2+tz2.Thus,(26)xt-z2≤121-txt-z2+tz2+xt-z2mmk-1-tSxt-τASxt-Az-xt2.It follows that(27)xt-z2≤1-txt-z2+tz2-1-tSxt-xt-τASxt-Az2=1-txt-z2+tz2-1-tSxt-xt2+2τ1-tSxt-xt,ASxt-Az-τ2ASxt-Az2≤1-txt-z2+tz2-1-tSxt-xt2+2τ1-tSxt-xtASxt-Az.Hence,(28)1-tSxt-xt2≤tz2+2τ1-tSxt-xtASxt-Az.This together with (23) implies that (29)limt→0+1-tSxt-xt=0.So,(30)limt→0+xt-Sxt=0.By (19), we have(31)xt-z2≤1-tSxt-τ1-tASxtmmmmmmi-z-τ1-tAz+t-z2=1-t2Sxt-τ1-tASxtmmmmmm-z-τ1-tAz2+2t1-t-z,Sxt-τ1-tASxtmmmmmmkim-z-τ1-tAz+t2z2≤1-t2xt-z2+2t1-t×-z,Sxt-τ1-tASxt-ASz-z+t2z2.It follows that(32)xt-z2≤-z,Sxt-τ1-tASxt-Az-z+t2z2+xt-z2+tzSxt-τ1-tASxt-Az-z≤-z,Sxt-z+t+ASxt-AzM,where M is some constant such that(33)supt∈0,2ϱ-τ/2ϱ12z2+xt-z2,mmmmmmmzSxt-τ1-tASxt-Az-z≤M.Now we show that {xt} is relatively norm-compact as t→0+. Assume {tn}⊂(0,2ϱ-τ/2ϱ) such that tn→0+ as n→∞. Put xn:=xtn. From (32), we have(34)xn-z2≤-z,Sxn-z+tn+ASxn-AzM.Since {xn} is bounded, without loss of generality, we may assume that xnj⇀x~∈C. Hence, xnj-τ/1-tnj(ASxnj-Az)⇀x~ because of ∥ASxn-Az∥→0 by (23). From (30), we have(35)limn→∞xn-Sxn=0.By Lemma 5 and (35), we deduce x~∈Fix(S).
Next, we show thatx~∈(A+B)-10. Let v∈Bu. Note that xn=JτB((1-tn)Sxn-τASxn) for all n. Then, we have(36)1-tnSxn-τASxn∈I+τBxn.So,(37)1-tnτSxn-ASxn-xnτ∈Bxn.Since B is monotone, we have, for (u,v)∈B,(38)tnγfxnτ+1-tnτSxn-ASxn-xnτ-v,xn-u≥0⟹1-tnSxn-τASxn-xn-τv,xn-u≥0⟹ASxn+v,xn-ummmmk≤1τSxn-xn,xn-u-tnτSxn,xn-u⟹ASx~+v,xn-ummmmk≤1τSxn-xn,xn-u-tnτSxn,xn-ummmmmk+ASx~-ASxn,xn-u⟹ASx~+v,xn-ummmmk≤1τSxn-xnxn-u+tnτSxnxn-ummmmmk+ASx~-ASxnxn-u.It follows that(39)ASx~+v,x~-u≤1τSxnj-xnjxnj-u+tnjτSxnjxnj-u+ASx~-ASxnjxnj-u+ASx~+v,x~-xnj.Since(40)xnj-x~,ASxnj-ASx~≥ϱASxnj-ASx~2,ASxnj→ASz, and xnj⇀x~, we have ASxnj→ASx~. We also observe that tn→0 and ∥Sxn-xn∥→0. Then, from (39), we derive(41)ASx~+v,x~-u≤0.That is, 〈-Ax~-v,x~-u〉≥0. Since B is maximal monotone, we have -Ax~∈Bx~. This shows that 0∈(A+B)x~. Hence, we have x~∈Fix(S)∩(A+B)-10. Therefore, we can substitute x~ for z in (34) to get(42)xn-x~2≤-x~,Sxn-x~+tn+ASxn-Ax~M.Consequently, the weak convergence of {xn} to x~ actually implies that xn→x~. This has proved the relative norm-compactness of the net {xt} as t→0+.
From (34), we get(43)x~-z2≤-z,x~-z,∀z∈FixS∩A+B-10.That is,(44)x~,x~-z≤0,∀z∈FixS∩A+B-10.It follows that(45)x~≤z,∀z∈FixS∩A+B-10.It is obvious that x~=projFix(S)∩(A+B)-10(0) by (44). This denotes that the entire net {xt} converges to x~. This completes the proof.Next, we present another algorithm.Algorithm 9.
For givenx0∈C, define a sequence {xn}⊂C iteratively by(46)xn+1=ςnxn+1-ςnJτnB1-ϱnSxn-τnASxn,∀n≥0,where {τn}⊂(0,2ϱ), {ϱn}⊂(0,1), and {ςn}⊂(0,1).Theorem 10.
Suppose thatFix(S)∩(A+B)-10≠∅. Assume that the following conditions are satisfied:(i)
limn→∞ϱn=0 and ∑n‍ϱn=∞;(ii)
0<lim_n→∞ςn≤lim¯n→∞ςn<1;(iii)
a(1-ϱn)≤τn≤b(1-ϱn), where [a,b]⊂(0,2ϱ) and limn→∞(τn+1-τn)=0.Then {xn} generated by (46) converges strongly to a point x~=projFix(S)∩(A+B)-1(0)(0) which is the minimum norm element in Fix(S)∩(A+B)-1(0).Proof.
Letz∈Fix(S)∩(A+B)-1(0). We have z=JτnB(z-τnAz)=JτnB(ϱnz+(1-ϱn)(z-τnAz/(1-ϱn))) for all n≥0. Since JτnB, S, and I-τnA/(1-ϱn) are nonexpansive, we have(47)JτnB1-ϱnSxn-τnASxn-z=JτnB1-ϱnSxn-τn1-ϱnASxnmk-JτnBϱnz+1-ϱnz-τn1-ϱnAz≤1-ϱnSxn-τn1-ϱnASxnmk-ϱnz+1-ϱnz-τn1-ϱnAz=1-ϱnSxn-τn1-ϱnASxnmmmmmmm-z-τn1-ϱnAz+ϱn-z≤1-ϱnxn-z+ϱnz.Thus,(48)xn+1-z≤ςnxn-z+1-ςn1-ϱnxn-z+1-ςnϱnz=1-ϱn1-ςnxn-z+1-ςnϱnz.By induction, we have(49)xn+1-z≤maxx0-z,z.Therefore, {xn} is bounded.
From (4) and (47), we derive(50)-z-τn1-ρnAz1-ϱnSxn-τn1-ϱnASxn-z-τn1-ϱnAzm-z-τn1-ϱnAzm+ϱn-z2≤1-ϱnSxn-τn1-ϱnASxn-z-τn1-ϱnAz2+ϱnz2=1-ϱnSxn-z-τn1-ϱnASxn-Az2+ϱnz2=1-ϱnτn21-ρn2Sxn-z2-2τn1-ϱnASxn-Az,Sxn-zmmmmmkmm+τn21-ϱn2ASxn-Az2+ϱnz2≤1-ϱnτn21-ϱn2xn-z2-2ϱτn1-ϱnASxn-Az2mmkmmkmkm+τn21-ϱn2ASxn-Az2+ϱnz2=1-ϱnxn-z2+τn1-ϱn2τn-21-ϱnϱmmmkmkmmτn1-ρn2×ASxn-Az2+ϱnz2.Set un=(1-ϱn)Sxn-τnASxn for all n≥0. Since τn-2(1-ϱn)ϱ≤0 for all n≥0, we obtain(51)JτnBun-z2mk≤1-ϱnxn-z2+τn1-ϱn2τn-21-ϱnϱmmmmkmmmτn1-ρn2×ASxn-Az2+ϱnz2.From (46), we have(52)xn+1-z2=ςnxn-z+1-ςnJτnBun-z2≤ςnxn-z2+1-ςnJτnBun-z2.Set yn=JτnB((1-ϱn)Sxn-τnASxn) for all n≥0. Then xn+1=ςnxn+(1-ςn)yn for all n≥0. Next, we estimate ∥xn+1-xn∥. In fact, we have(53)yn+1-yn=Jτn+1Bun+1-JτnBun≤Jτn+1Bun+1-Jτn+1Bun+Jτn+1Bun-JτnBun≤1-ϱn+1Sxn+1-τn+1ASxn+1mk-1-ϱnSxn-τnASxnmk+Jτn+1Bun-JτnBun=I-τn+1ASxn+1-I-τn+1ASxnmk+τn-τn+1ASxn+ϱnSxn-ϱn+1Sxn+1mk+Jτn+1Bun-JτnBun≤I-τn+1ASxn+1-I-τn+1ASxnmk+τn+1-τnASxn+ϱnSxnmk+ϱn+1Sxn+1+Jτn+1Bun-JτnBun.Since I-τn+1A is nonexpansive for τn+1∈(0,2ϱ), we have(54)I-τn+1ASxn+1-I-τn+1ASxn≤Sxn+1-Sxn≤xn+1-xn.From (13), we have (55)Jτn+1Bun=JτnBτnτn+1un+1-τnτn+1Jτn+1Bun.It follows that(56)Jτn+1Bun-JτnBun=JτnBτnτn+1un+1-τnτn+1Jτn+1Bun-JτnBun≤τnτn+1un+1-τnτn+1Jτn+1Bun-un≤τn+1-τnτn+1un-Jτn+1Bun.So,(57)yn+1-yn≤xn+1-xn+τn+1-τnASxn+ϱnSxn+ϱn+1Sxn+1+τn+1-τnτn+1un-Jτn+1Bun.Then,(58)yn+1-yn-xn+1-xn≤τn+1-τnASxn+ϱnSxn+ϱn+1Sxn+1+τn+1-τnτn+1un-Jτn+1Bun.Since ϱn→0, τn+1-τn→0 and lim_n→∞τn>0, we obtain (59)limsupn→∞yn+1-yn-xn+1-xn≤0.By Lemma 6, we get(60)limn→∞yn-xn=0.Consequently, we obtain(61)limn→∞xn+1-xn=limn→∞1-ςnyn-xn=0.From (51) and (52), we have(62)xn+1-z2≤ςnxn-z2+1-ςnJτnBun-z2≤1-ςn1-ϱn×xn-z2+τn1-ϱn2τn-21-ϱnϱmmmτn1-ρn2×ASxn-Az2+1-ςnϱnz2+ςnxn-z2=1-1-ςnϱnxn-z2+1-ςnτn1-ϱnτn-21-ϱnϱASxn-Az2+1-ςnϱnz2≤xn-z2+1-ςnτn1-ϱnτn-21-ϱnϱ×ASxn-Az2+1-ςnϱnz2.Then, we obtain(63)1-ςnτn1-ϱn21-ϱnϱ-τnASxn-Az2≤xn-z2-xn+1-z2+1-ςnϱnz2≤xn-z-xn+1-zxn+1-xn+1-ςnϱnz2.Since limn→∞ϱn=0, limn→∞∥xn+1-xn∥=0, and lim_n→∞(1-ςn)τn/1-ϱn(2(1-ϱn)ϱ-τn)>0, we have(64)limn→∞ASxn-Az=0.Next, we show ∥xn-Sxn∥→0. By using the firm nonexpansivity of JτnB, we have(65)JτnBun-z2m=JτnB1-ϱnSxn-τnASxn-JτnBz-τnAz2m≤1-ϱnSxn-τnASxn-z-τnAz,JτnBun-zm=12JτnB21-ϱnSxn-τnASxn-z-τnAz2mmmm+JτnBun-z2mmmm-1-ϱnSxn-τnASxn-Az-JτnBun2.Observe that(66)1-ϱnSxn-τnASxn-z-τnAz2=1-ϱnSxn-τn1-ϱnASxnmmmmmmmmi-z-τn1-ϱnAz+ϱn-z2≤1-ϱnSxn-τn1-ϱnASxnmmmmmmim-z-τn1-ϱnAz2+ϱnz2≤1-ϱnxn-z2+ϱnz2.Hence,(67)JτnBun-z2≤121-ϱnxn-z2+ϱnz2+JτnBun-z2mkm-1-ϱnSxn-JτnBun-τnASxn-Az2.It follows that(68)JτnBun-z2≤1-ϱnxn-z2+ϱnz2-1-ϱnSxn-JτnBun-τnASxn-Az2=1-ϱnxn-z2+ϱnz2mk-1-ϱnSxn-JτnBun2mk+2τn1-ϱnSxn-JτnBun,ASxn-Azmk-τn2ASxn-Az2≤1-ϱnxn-z2+ϱnz2-1-ϱnSxn-JτnBun2+2τn1-ϱnSxn-JτnBunASxn-Az.This together with (52) implies that(69)xn+1-z2≤ςnxn-z2+1-ςn1-ϱnxn-z2+1-ςnϱnz2-1-ςn1-ϱnSxn-JτnBun2+2τn1-ςn1-ϱnSxn-JτnBun×ASxn-Az=1-1-ςnϱnxn-z2+1-ςnϱnz2-1-ςn1-ϱnSxn-JτnBun2+2τn1-ςn1-ϱnSxn-JτnBun×ASxn-Az.Hence,(70)1-ςn1-ϱnSxn-JτnBun2≤xn-z2-xn+1-z2-1-ςnϱnxn-z2+1-ςnϱnz2+2τn1-ςn×1-ϱnSxn-JτnBunASxn-Az≤xn-z+xn+1-zxn+1-xnmmk+1-ςnϱnz2+2τn1-ςnmmk×1-ϱnSxn-JτnBunASxn-Az.Since lim¯n→∞ςn<1, ∥xn+1-xn∥→0, ϱn→0, and ∥ASxn-Az∥→0 (by (60)), we deduce (71)limn→∞1-ϱnSxn-JτnBun=0.This indicates that(72)limn→∞Sxn-JτnBun=0.Combining (60) and (72), we get(73)limn→∞xn-Sxn=0.Put x~=limt→0+xt=projFix(S)∩(A+B)-1(0)(0), where xt is the net defined by (17). We will finally show that xn→x~.
Setvn=xn-τn/1-ϱn(ASxn-Ax~) for all n. Take z=x~ in (64) to get ∥ASxn-Ax~∥→0. First, we prove lim¯n→∞〈-x~,Sxn-x~〉≤0. We take a subsequence {Sxni} of {Sxn} such that (74)lim¯n→∞⁡-x~,Sxn-x~=limi→∞⁡-x~,Sxni-x~.It is clear that {Sxni} is bounded due to the boundedness of {Sxn} and ∥ASxn-Ax~∥→0. Then, there exists a subsequence {Sxnij} of {Sxni} which converges weakly to some point w∈C. Hence, {xnij} and {ynij} also converge weakly to w because of ∥Sxnij-xnij∥→0 and ∥xnij-ynij∥→0. By the demiclosedness principle of the nonexpansive mapping (see Lemma 5) and (73), we deduce w∈Fix(S). Furthermore, by similar argument as that of Theorem 8, we can show that w is also in (A+B)-1(0). Hence, we have w∈Fix(S)∩(A+B)-1(0). This implies that(75)lim¯n→∞⁡-x~,Sxn-x~=limj→∞-x~,Sxnij-x~=-x~,w-x~.Note that x~=projFix(S)∩(A+B)-1(0)(0). Then, 〈-x~,w-x~〉≤0, w∈Fix(S)∩(A+B)-1(0). Therefore,(76)lim¯n→∞⁡-x~,Sxn-x~≤0.From (46), we have(77)
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.It is clear that ∑n‍2(1-ςn)ϱn=∞ and(78)limsupn→∞x~2τnx~ASxn-Ax~+1-ϱnmmmk×-x~,Sxn-x~+ϱnx~2+xn-x~2≤0.By Lemma 7, we conclude that xn→x~. This completes the proof.Corollary 11.
Suppose that(A+B)-1(0)≠∅. Let τ be a constant satisfying a≤τ≤b, where [a,b]⊂(0,2ϱ). For t∈(0,1-τ/(2ϱ)), let {xt}⊂C be a net generated by (79)xt=JτB1-txt-τAxt.Then the net {xt} converges strongly, as t→0+, to a point x~=proj(A+B)-1(0)(0) which is the minimum norm element in (A+B)-1(0).Corollary 12.
Suppose that(A+B)-1(0)≠∅. For given x0∈C, let {xn}⊂C be a sequence generated by (80)xn+1=ςnxn+1-ςnJτnB1-ϱnxn-τnAxnfor all n≥0, where {τn}⊂(0,2ϱ), {ϱn}⊂(0,1), and {ςn}⊂(0,1) satisfy(i)
limn→∞ϱn=0 and ∑n‍ϱn=∞;(ii)
0<lim_n→∞ςn≤lim¯n→∞ςn<1;(iii)
a(1-ϱn)≤τn≤b(1-ϱn), where [a,b]⊂(0,2ϱ) and limn→∞(τn+1-τn)=0.Then {xn} converges strongly to a point x~=proj(A+B)-1(0)(0) which is the minimum norm element in (A+B)-1(0).



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*Source: 451320-2015-08-27.xml*