# The Viscosity Approximation Forward-Backward Splitting Method for Zeros of the Sum of Monotone Operators

Authors: Oganeditse Aaron Boikanyo
Journal: Abstract and Applied Analysis
 (2016)

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## Abstract

We investigate the convergence analysis of the following general inexact algorithm for approximating a zero of the sum of a cocoercive operatorA and maximal monotone operators B with D(B)⊂H: xn+1=αnf(xn)+γnxn+δn(I+rnB)-1(I-rnA)xn+en, for n=1,2,…, for given x1 in a real Hilbert space H, where (αn), (γn), and (δn) are sequences in (0,1) with αn+γn+δn=1 for all n≥1, (en) denotes the error sequence, and f:H→H is a contraction. The algorithm is known to converge under the following assumptions on δn and en: (i) (δn) is bounded below away from 0 and above away from 1 and (ii) (en) is summable in norm. In this paper, we show that these conditions can further be relaxed to, respectively, the following: (i) (δn) is bounded below away from 0 and above away from 3/2 and (ii) (en) is square summable in norm; and we still obtain strong convergence results.


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## Body

## 1. Introduction

LetH be a real Hilbert space endowed with the inner product 〈·,·〉 and norm ·. Let us consider the problem (1)findapointx∈Hsuchthat0∈Ax+Bx,where A and B are maximal monotone operators. The literature on problem (1) exists (see [1–5] and the references therein). Note that two possibilities exist here: either A+B is maximal monotone or A+B is not maximal monotone. (For conditions that ensure that A+B is maximal monotone, we refer the reader to [6].) One of the iterative procedures used to solve problem (1), in the absence of the maximality of A+B, is by using splitting methods, which have received much attention in the recent past due to their applications in image recovery, signal processing, and machine learning. One of the popular iterative methods used for solving problem (1) is the forward-backward method introduced by Passty [7] in 1979 which defines a sequence (xn) by (2)xn+1=I+rnB-1I-rnAxn,forn=1,2,…,where (rn) is a sequence of positive numbers, A and B are maximal monotone operators with D(B)⊂D(A), and A is single valued. Since its inception, the splitting method (2) has received much attention from several authors including Tseng [8], Mercier [9], Gabay [10], and Chen [11]. The projected gradient method is in fact a special case of scheme (2). In this case, A is taken as the gradient of a function and B as the subdifferential of the indicator function of a closed and convex subset of a real Hilbert space.In the case whenT≔A+B is maximal monotone, the most popular iterative method for solving (1) is the proximal point algorithm (PPA) which was first introduced by Martinet [12] in 1970 and later developed by Rockafellar [13]. Due to the failure of Rockafellar’s PPA to converge strongly [14] for arbitrary maximal monotone operators, several authors [15–23] have presented modified versions of the PPA that always converge strongly. Recently, Yao and Noor [22] proposed the contraction proximal point algorithm (3)xn+1=αnu+γnxn+δnJrnBxn+en,forn=1,2,…,where B is a maximal monotone operator, (αn), (γn), and (δn) are sequences in (0,1) with αn+γn+δn=1 for all n≥1, JrnB≔(I+rnB)-1, (en) is a sequence of computational errors, u∈H is given, and rn>0. Algorithm (3) was used to find zeros of maximal monotone operators. The viscosity approximation method which is also used for finding zeros of maximal monotone operators was introduced by Takahashi [19] as a scheme that generates sequences (xn) by (4)xn+1=αnfxn+1-αnJrnBxn,forn=1,2,…,where f is a contraction from a closed and convex subset C of a reflexive Banach space X, B is an accretive operator (hence monotone if X is a Hilbert space) with the range condition D(B)¯⊂C⊂∩r>0R(I+rB), (αn) is a sequence in (0,1), JrnB≔(I+rnB)-1, and rn>0. Under appropriate conditions, strong convergence of scheme (4) is obtained.Recently, the forward-backward splitting method (2), the contraction proximal point algorithm (3), and the viscosity approximation method (4) were combined to obtain the algorithm(5)xn+1=αnfxn+γnxn+δnJrnBI-rnAxn+en,forn=1,2,…,where A:D(A)=H→H is a β-cocoercive operator, B:D(B)⊂H→H is a maximal monotone operator, (αn), (γn), and (δn) are sequences in (0,1) with αn+γn+δn=1 for all n≥1, JrnB≔(I+rnB)-1, (en) is a sequence of computational errors, u∈H is given, f:H→H is a contraction, and rn>0. They were able to prove that the sequence generated by (5) converges strongly to the unique fixed point of the operator PSf, where S≔(A+B)-10. Note that when en=0 for all n≥1, then it suffices to derive strong convergence of (5) when f:C→C, where C is a closed and convex subset of H, and B:D(B)⊂C→H. Note that, in the case when en≠0, if xn∈C for a given n≥1, the (n+1)th iterate xn+1 may fail to be in C. That is, algorithm (5) is not well defined if f:C→C.It is worthy of note that Noor’s algorithm (3) excludes the important case δn∈[1,2), the overrelaxed case. However, the overrelaxed factor may indeed speed up the rate of the algorithm (see [24]). That is why Wang and Cui [25] investigated the convergence of sequences generated by (6)xn+1=αnu+γnxn+δnJrnBxn+en,forn=1,2,…,for (αn)⊂(0,1), (γn)⊂(-1,1), and (δn)⊂(0,2) with αn+γn+δn=1 for all n≥1. A natural question thus arises: is it possible to relax the conditions on (γn) and (δn) used in algorithm (5) further? Our purpose in this paper is to affirmatively answer this question. Furthermore, we prove a strong convergence result associated with algorithm (7)xn+1=αnfxn+γnxn+δnJrnBI-rnAxn+en,forn=1,2,…,for (αn)⊂(0,1), (γn)⊂(-3/2,1), and (δn)⊂(0,3/2) with αn+γn+δn=1 for all n≥1, for the case when the sequence of error terms is square summable in norm. Our main result improves and refines similar results in the literature by using fewer conditions to derive strong convergence of (7). In addition, our results generalize many results in the literature such as [25, Theorem 1], [26, Theorem 1], and [27, Theorem 1].


## 2. Preliminary Results

LetH be a real Hilbert space endowed with the inner product 〈·,·〉 and norm ·. A map T:H→H is called a Lipschitz mapping if there exists L≥0 such that T(x)-T(y)≤Lx-y for all x,y∈H. The number L associated with T is called a Lipschitz constant. If L<1, we say that T is a contraction, and T is called nonexpansive if L=1. The set of fixed points of T is given by F(T)={x∈C:T(x)=x}. It is well known that if T is nonexpansive, then I-T is demiclosed at zero and F(T) is closed and convex; see [28]. Given an operator T:H→H, we say that I-T is demiclosed at zero if, for any sequence (xn) in H, the following implication holds: (8)xn⇀xI-Txn⟶0⟹x∈FixT.Here, xn⇀x means that (xn) converges weakly to x and xn→x is used to indicate that (xn) converges strongly to x. These notations will be used in the sequel.For a nonempty closed and convex subsetF of H, the metric projection (nearest point mapping) PF:H→F is defined as follows: given x∈H, PFx is the unique point in F having the property (9)x-PFx=infy∈F⁡x-y.Note that the projection operator is firmly nonexpansive and has the following characterization which will be used in this paper: for any x∈H, (10)x-PFx,z-PFx≤0,∀z∈F.Recall that an operator T:H→H is said to be firmly nonexpansive if for every x,y∈H(11)Tx-Ty2≤x-y2-I-Tx-I-Ty2.It is clear that every firmly nonexpansive map is nonexpansive. A single-valued operator A:D(A)⊂H→H is called β-inverse strongly monotone (β-cocoercive) for a positive number β if (12)Ax-Ay,x-y≥βAx-Ay2∀x,y∈DA.It is known that, for a β-inverse strongly monotone mapping A, the map T≔I-rA is nonexpansive for all r∈(0,2β). A (possibly set-valued) nonlinear operator A:D(A)⊂H→H is said to be monotone if (13)x′-y′,x-y≥0∀x,y∈DA,x′∈Ax,y′∈Ay.In other words, A is monotone if its graph, G(A)={(x,y)∈H×H:x∈D(A),y∈Ax}, is a monotone subset of H×H. A monotone operator A is called maximal monotone if it is monotone and its graph is not properly contained in the graph of any other monotone operator. Note that a β-cocoercive operator is monotone. We know that A is maximal monotone if and only if the range of I+A is equal to H (i.e., R(I+A)=H). If A is maximal monotone and μ is a positive number, then the resolvent of A is a single-valued and firmly nonexpansive operator JμA:H→H defined by JμA(x)=(I+μA)-1(x). We note that JμA is everywhere defined on H. For more information, refer to [29].Finally, we recall some elementary inequalities in real Hilbert spaces. For everyx,y∈H, the inequality (14)x+y2≤x2+2y,x+yholds. If α,β are any real numbers in (0,1) with α+β=1, then, for any x,y∈H, (15)αx+βy2=αx2+βy2-αβx-y2.Moreover, for any a,b∈H, the inequality (16)2ab≤12a2+2b2can also be proved easily.Now, we establish two lemmas that will enable us to prove our main result.Lemma 1.
LetA be a β-cocoercive operator and let r be a positive real number satisfying r≤β. Then, T≔I-rA is firmly nonexpansive.Proof.
We have to show that(17)Tx-Ty2≤x-y2-I-Tx-I-Ty2=x-y2-r2Ax-Ay2.Using the definition of T, we have (18)Tx-Ty2=x-y-rAx-Ay2=x-y2+r2Ax-Ay2-2rAx-Ay,x-y≤x-y2+r2Ax-Ay2-2rβAx-Ay2,where the inequality follows from the fact that A is β-cocoercive. Since r≤β, it follows that -rβ≤-r2 and inequality (17) follows at once.Lemma 2.
LetT and S be firmly nonexpansive mappings. Then, (19)TSx-TSy2≤x-y2-12I-TSx-I-TSy2.In particular, TS is nonexpansive.Proof.
From the firmly nonexpansive property ofT, we get (20)TSx-TSy2≤Sx-Sy2-I-TSx-I-TSy2≤x-y2-I-Sx-I-Sy2-I-TSx-I-TSy-I-Sx-I-Sy2=x-y2-I-TSx-I-TSy2-2I-Sx-I-Sy2+2I-TSx-I-TSy,I-Sx-I-Sy,where the second inequality follows from the fact that S is firmly nonexpansive. Using (16) with a=(I-TS)x-(I-TS)y and b=(I-S)x-(I-S)y, we get (21)TSx-TSy2≤x-y2-I-TSx-I-TSy2-2I-Sx-I-Sy2+12I-TSx-I-TSy2+2I-Sx-I-Sy2=x-y2-12I-TSx-I-TSy2.The proof is complete.Remark 3.
LetB be maximal monotone and let A be a β-cocoercive mapping with β>0. Taking U=JrB and V=(I-rA) for some r≤β, both U and V are firmly nonexpansive (see Lemma 1). In addition, if S≔Fix(JrB(I-rA))≠∅, then, for any z∈S and x∈D(A), we obtain from Lemma 2 that (22)JrBI-rAx-z2≤x-z2-12x-JrBI-rAx2.We next recall some lemmas that will be useful in proving our main results. In the next two lemmas, it is assumed thatA is a β-cocoercive mapping with β>0 and B is maximal monotone.Lemma 4.
Ifλ and μ are any two positive real numbers, then (23)JλBI-λAx-JμBI-μAx≤1-λμx-JμBI-μAxholds for any x∈H.The proof of Lemma4 can be reproduced easily.Lemma 5 (see López et al. [3]).
Ifλ and μ are two positive real numbers such that μ≥λ, then (24)x-JλBI-λAx≤2x-JμBI-μAxholds for any x∈H.The next two lemmas are important in showing that under suitable assumptions our sequence(xn) generated by (7) is bounded. The proofs can be reproduced by following some ideas of [26].Lemma 6.
Let(sn) be a sequence of nonnegative real numbers satisfying (25)sn+1≤1-αn1+ϵnsn+αnK+ξnfor some K>0, where (αn)⊂(0,1) and (ϵn),(ξn)∈l1 are sequences of positive real numbers. Then, (sn) is bounded.Lemma 7.
Let(sn) be a sequence of nonnegative real numbers satisfying (26)sn+1≤1-αn1+ϵnsn+αnK1+ϵnK2,where K1,K2 are positive constants, (αn) is a sequence in (0,1), and ϵn∈[0,∞) with ϵn(3-2αn)≤αn for all n≥0. Then, (sn) is bounded.The last two lemmas will be vital in deducing strong convergence of the sequence generated by (7).Lemma 8 (see Xu [20]).
Let(sn) be a sequence of nonnegative real numbers satisfying (27)sn+1≤1-ansn+anbn+cn,n≥0,where (an), (bn), and (cn) satisfy the following conditions: (i) (an)⊂[0,1], with ∑n=0∞‍an=∞, (ii) cn≥0 for all n≥0 with ∑n=0∞‍cn<∞, and (iii) limsupn→∞bn≤0. Then, limn→∞sn=0.Lemma 9 (see Maingé [30]).
Let(sk) be a sequence of real numbers that does not decrease at infinity, in the sense that there exists a subsequence (skj) of (sk) such that skj<skj+1 for all j≥0. Define an integer sequence (mk)k≥k0 as (28)mk=max⁡k0≤l≤k:sl<sl+1.Then, mk→∞ as k→∞ and for all k≥k0(29)max⁡smk,sk≤smk+1.


## 3. Main Results

For anyr∈(0,β], the map JrB(I-rA), being the composition of nonexpansive maps, is nonexpansive. Since the fixed point set (if it is not empty) of a nonexpansive map is closed and convex, it follows that if S≔(A+B)-1(0)≠∅, then S is closed and convex. Therefore, the map PS is well defined and nonexpansive. Also, for any closed and convex set C, the map PSf:C→S is a κ-contraction whenever f:C→C is a κ-contraction. This information will be used in this section.Theorem 10.
LetH be a real Hilbert space and let f:H→H be a κ-contraction with κ<1/2. Assume that A:D(A)=H→H is a β-cocoercive operator and B:D(B)⊂H→H is a maximal monotone operator with S≔(A+B)-1(0)≠∅. For x0∈H, let (xn) be a sequence generated by (7) with αn∈(0,1), γn∈-3/2,1, and δn∈0,3/2 satisfying αn+γn+δn=1, rn∈(0,β], and (en) is a sequence of errors in H. Then, (xn) converges strongly to the unique fixed point z of PSf, provided that(i)
limn→∞αn=0 and ∑n=0∞‍αn=∞,(ii)
liminfn→∞rn>0,(iii)
0<liminfn→∞δn≤limsupn→∞δn<3/2,(iv)
en≤ηnxn-JrnB(I-rnA)xn with ∑n=0∞‍ηn2<∞.Proof.
Let us denoteρn≔δn(1-αn)-1 and yn≔(1-ρn)xn+ρnJrnB(I-rnA)xn. Then, from (7), xn+1=αnf(xn)+(1-αn)yn+en. In order to prove that (xn) is bounded, we note that, from the condition αn→0 as n→∞, one may assume without loss of generality that αn<1/2 for all n. If z is the unique fixed point of PSf, then z∈S and from (15) we have (30)xn+1-z2=αnfxn-z+en+1-αnyn-z+en2=αnfxn-z+en2+1-αnyn-z+en2-αn1-αnfxn-yn2≤αnfxn-z+en2+1-αnyn-z+en2.To estimate yn-z+en2, let us first observe that (31)yn-z2=xn-z+ρnJrnBI-rnAxn-xn2=xn-z2+ρn2JrnBI-rnAxn-xn2-ρn2xn-z,xn-JrnBI-rnAxn=xn-z2+ρn2JrnBI-rnAxn-xn2-ρnxn-z2+xn-JrnBI-rnAxn2-JrnBI-rnAxn-z2=1-ρnxn-z2+ρnJrnBI-rnAxn-z2-ρn1-ρnxn-JrnBI-rnAxn2.Making use of inequality (22), we obtain (32)yn-z2≤xn-z2-ρn32-ρnxn-JrnBI-rnAxn2.Therefore, it follows that (33)yn-z+en2=yn-z2+en2+2yn-z,en≤yn-z2+en2+ρn3-2ρnen24ηn2+4ηn2ρn3-2ρnyn-z2=1+4ηn2ρn3-2ρnyn-z2+ρn3-2ρnen24ηn2.Note that condition (iii) of the theorem is equivalent to 0<liminfn→∞ρn≤limsupn→∞ρn<3/2. Then, there exist positive real numbers ρ1 and ρ2 such that ρ1≤ρn≤ρ2<3/2. From this condition, we derive 4/ρn(3-2ρn)≤C for some positive constant C. Denote ϵn≔ηn2C. Since (ηn2)∈l1, it follows that (ϵn)∈l1 as well. Therefore, from (32) and condition (iv) of the theorem, we have (34)yn-z+en2≤1+ϵnxn-z2-ρn3-2ρn4xn-JrnBI-rnAxn2.On the other hand, by similar arguments as above, we have (35)fxn-z+en2=fxn-z2+en2+2fxn-z,en≤fxn-z2+en2+ρn3-2ρnen24ηn2+4ηn2ρn3-2ρnfxn-z2=1+4ηn2ρn3-2ρnfxn-z2+ρn3-2ρnen24ηn2,where the inequality follows from (16). Moreover, using the property that f is a κ-contraction, we get (36)fxn-z2=fxn-fz2+fz-z2+2fxn-fz,fz-z≤2fxn-fz2+2fz-z2≤2κxn-z2+2fz-z2.Since en≤ηnxn-JrnB(I-rnA)xn, it follows that (37)fxn-z+en2≤1+ϵn2κxn-z2+2fz-z2+1+ϵnρn3-2ρn4xn-JrnBI-rnAxn2.Combining this last inequality with (34), it follows from (30) that(38)xn+1-z2≤1-αn1-2κ1+ϵnxn-z2+αnK+ϵnK-1-2αn1+ϵnρn3-2ρn4xn-JrnBI-rnAxn2,where K=2f(z)-z2. Since αn<1/2 for very large n, by our assumption, the inequality above reduces to (39)xn+1-z2≤1-αn1-2κ1+ϵnxn-z2+αnK+ϵnK.Applying Lemma 6 with sn≔xn-z2 and ξn≔ϵnK, we derive that the sequence (xn-z) is bounded. Therefore, (xn) is bounded.
We next show that(40)xn+1-z2≤1-αn1-2κxn-z2+αndn+ϵnM-L1-αnxn-JrnBI-rnAxn2holds for some positive constants L and M, where dn denotes (41)dn≔2fz-z+en,xn+1-z+αn+enM.To this end, we first note that the condition ρn<3/2 and (32) imply that (42)yn-z≤xn-z.Therefore, from the equality xn+1=αnf(xn)+(1-αn)yn+en and using the fact that f is a κ-contraction, we obtain (43)xn+1-z≤αnfxn-z+1-αnyn-z+en≤αnfxn-fz+fz-z+1-αnxn-z+en≤αnκxn-z+fz-z+1-αnxn-z+en=1-αn1-κxn-z+αnfz-z+en≤xn-z+αnfz-z+en.On the other hand, it follows from (14) that (44)xn+1-z2=1-αnyn-z+en+αnfxn-z+en2≤1-αnyn-z+en2+2αnfxn-z+en,xn+1-z=1-αnyn-z+en2+2αnfz-z+en,xn+1-z+2αnfxn-fz,xn+1-z≤1-αnyn-z+en2+2αnfz-z+en,xn+1-z+2αnfxn-fzxn+1-z.Again, using the fact that f is a κ-contraction with κ<1/2, we have (45)xn+1-z2≤1-αnyn-z+en2+2αnfz-z+en,xn+1-z+2αnκxn-zxn+1-z≤1-αnyn-z+en2+2αnfz-z+en,xn+1-z+2αnκxn-zxn-z+αnfz-z+en≤1-αnyn-z+en2+2αnfz-z+en,xn+1-z+2αnκxn-z2+αnαn+enM,where M=(1+f(z)-z)supnxn-z. Combining this inequality with (34), we obtain (46)xn+1-z2≤1-αn1-2κ1+ϵnxn-z2+αnαn+enM-1-αn1+ϵnρn3-2ρn4xn-JrnBI-rnAxn2+2αnfz-z+en,xn+1-z.Using condition (iii) of the theorem and the boundedness of (xn), we readily get (40).
Now, from(ϵn)∈l1, we can find ν>0 such that (47)ν≔limn→∞⁡∑k=1nϵk.Therefore, if we denote sn≔xn-z2+tnM, where (tn) is a nonnegative sequence that converges to zero and is defined by (48)tn=ν-∑k=1n-1ϵk,then (sn) converges to zero strongly if and only if (xn-z) does. In addition, we rewrite inequality (40) in the form (49)sn+1≤1-αn1-2κsn+αnbn-L1-αnxn-JrnBI-rnAxn2,where bn denotes (50)bn≔2fz-z+en,xn+1-z+αn+tn+enM.The next step is to show that (sn) converges strongly to zero. We achieve this by considering two possible cases on the sequence (sn).
Case
1. We assume that (sn) is eventually decreasing (i.e., there exists n0≥1 such that (sn) is decreasing for all n≥n0). Then, (sn) converges and rearranging terms in (49) we obtain (51)L1-αnxn-JrnBI-rnAxn2≤1-αn1-2κsn-sn+1+αnbn.Since (xn) is bounded, we pass to the limit in the above inequality to get (52)limn→∞⁡xn-JrnBI-rnAxn=0.Take a subsequence (xnk) of (xn) converging weakly to x^ such that (53)limsupn→∞⁡fz-z,xn-z=limk→∞⁡fz-z,xnk-z=fz-z,x^-z.Note that, from (7), we derive the inequality (54)xn+1-xn≤αnfxn-xn+δnxn-JrnBI-rnAxn+en,which together with (52), the boundedness of (xn), and en→0 as n→∞ implies that (55)limn→∞⁡xn+1-xn=0.The above limit implies that (56)limsupn→∞⁡fz-z+en,xn+1-z=fz-z,x^-z.If r>0 is the lower bound of (rn), then we have from Lemma 5(57)xn-JrBI-rAxn≤2xn-JrnBI-rnAxn,which together with (52) implies that (58)limn→∞⁡xn-JrBI-rAxn=0.Since S=Fix(JrnB(I-rnA))=Fix(JrB(I-rA)) for all n≥1 and JrB(I-rA) is nonexpansive, it follows that I-JrB(I-rA) is demiclosed at zero; see [29, page 20]. Therefore, from (58) and the property that I-JrB(I-rA) is demiclosed at zero, we conclude that x^∈S. Hence, from the characterization of projections, we conclude that (59)limsupn→∞⁡fz-z+en,xn+1-z≤0.By conditions (i) and (iv) of the theorem and the fact that tn→0 as n→∞, it follows that (60)limsupn→∞⁡bn≤0.Finally, we derive from inequality (49) that (61)sn+1≤1-αn1-2κsn+αnbn.The conclusion that sn→0 follows from Lemma 8.
Case
2. The sequence (sk) is not eventually decreasing; that is, there is a subsequence (skj) of (sk) such that skj<skj+1 for all j≥0. In this case, we define an integer sequence (mk)k≥k0 as in Lemma 9. Note that the subsequence (smk) satisfies the condition smk≤smk+1 for all k≥n0. It then follows from (49) that (62)αmk1-2κsmk+1+L1-αmkxmk-JrmkBI-rmkAxmk2≤αmkbmk.Since 2κ<1, αmk→0 as k→∞, and (bmk) is bounded, we conclude that (63)limk→∞⁡xmk-JrmkBI-rmkAxmk=0.Using similar arguments as in Case 1, we conclude that (64)limsupk→∞⁡bmk≤0.In view of (62), we have (65)1-2κsmk+1≤bmk,which implies that smk+1→0 as k→∞. Since sk≤smk+1 for all k≥k0 (see (29)), we also have sk→0 as k→∞. Hence, xk→z as k→∞, and the proof is complete.Remark 11.
Observe that if(xn) is bounded, then condition (iv) of Theorem 10 implies that ∑n=0∞‍en2<∞. The latter condition is weaker than the condition ∑n=0∞‍en<∞ used in the existing literature. In addition, we did not require the conditions βn≥1/2, (rn)⊂(0,1), and limsupn→∞rn<1. Therefore, Theorem 10 is an improvement of the aforementioned theorem.Remark 12.
Theorem10 also improves and extends many results that exist in the literature, such as [22, Theorem 2.2], [27, Theorem 1], and [26, Theorem 1]. To see this, it is enough to take A=0 and f(x)=u for all x∈H, where u∈H is a given fixed vector.Theorem 13.
LetC be a nonempty, closed, and convex subset of H and let f:C→C be a κ-contraction with κ<1/2. Assume that A:D(A)=C→H is a β-cocoercive operator and B:D(B)⊂C→H is a maximal monotone operator with S≔(A+B)-1(0)≠∅. For x0∈C and en=0 for all n≥1, let (xn) be a sequence generated by (7) with αn,γn,δn∈(0,1) satisfying αn+γn+δn=1, and rn∈(0,β]. Then, (xn) converges strongly to the unique fixed point z of PSf, provided that (i)
limn→∞αn=0 and ∑n=0∞‍αn=∞,(ii)
liminfn→∞rn>0,(iii)
0<liminfn→∞δn.Proof.
In the proof of Theorem10, we used the condition limsupn→∞δn<3/2 repeatedly to conclude that (3-2ρn)>0 for all n≥0, where ρn=δn(1-αn)-1. So if δn∈(0,1), then automatically (3-2ρn)>0 for all n≥0.Remark 14.
We have dropped the conditionsβn≥1/2, rn∈(0,1), and limsupn→∞rn<1 used in the literature to derive strong convergence of the sequence generated by (7) under the conditions of Theorem 13. Therefore, Theorem 13 is an improvement of the aforementioned results.Theorem 15.
LetH be a real Hilbert space and let f:H→H be a κ-contraction with κ<1/2. Assume that A:D(A)=H→H is a β-cocoercive operator and B:D(B)⊂H→H is a maximal monotone operator with S≔(A+B)-1(0)≠∅. For x0∈H, let (xn) be a sequence generated by (7) with αn∈(0,1), γn∈-3/2,1, and δn∈0,3/2 satisfying αn+γn+δn=1, rn∈(0,β], and (en) is a sequence of errors in H. Then, (xn) converges strongly to the unique fixed point z of PSf, provided that (i)
limn→∞αn=0 and ∑n=0∞‍αn=∞,(ii)
liminfn→∞rn>0,(iii)
0<liminfn→∞δn≤limsupn→∞δn<3/2,(iv)
en≤ηnxn-JrnB(I-rnA)xn with limn→∞ηn2/αn=0.Proof.
Similar to the proof of Theorem10, we assume without loss of generality that αn<1/2 for all n. Following similar steps as in the proof of Theorem 10, we derive (66)xn+1-z2≤1-αn1-2κ1+ϵnxn-z2+αnK+ϵnK-1-2αn1+ϵnρn3-2ρn4xn-JrnBI-rnAxn2,where z is the unique fixed point of PSf, K≔2f(z)-z2, and ϵn≔Cηn2 for some positive constant C. Since αn≤1/2 for all n, by our assumption, the inequality above reduces to (67)xn+1-z2≤1-αn1-2κ1+ϵnxn-z2+αnK+ϵnK.It is clear from the condition ηn2/αn→0 as n→0 that ϵn/αn→0 as n→0, and we may therefore assume without loss of generality that ϵn(3-2αn)≤αn for all n≥0. The conclusion that (xn) is bounded follows on applying Lemma 7 with sn≔xn-z2.
Again, similar to the proof of Theorem10, we obtain (68)xn+1-z2≤1-αn1-2κxn-z2+αnb^n-L1-αnxn-JrnBI-rnAxn2,for some positive constant L, where (b^n) is the sequence denoted by (69)b^n≔2fz-z+en,xn+1-z+αn+en+ϵnαnMwith an appropriate constant M>0. We will show that sn≔xn-z2→0 as n→∞ by considering the following two cases on the sequence (sn).
Case
1. Assume that (sn) is eventually decreasing (i.e., there exists n0≥1 such that (sn) is decreasing for all n≥n0). Then, (sn) converges and rearranging terms in (68) yields (70)L1-αnxn-JrnBI-rnAxn2+αn1-2κsn≤sn-sn+1+αnb^n,which implies that (71)limn→∞⁡xn-JrnBI-rnAxn=0.From this limit and Lemma 5, we obtain (72)limn→∞⁡xn-JrBI-rAxn=0,where r>0 is the lower bound of (rn). Then, following similar steps as in the proof of Theorem 10, we obtain (73)limsupn→∞⁡fz-z,xn+1-z≤0.As a consequence of ηn2/αn→0 and αn→0 as n→∞, we derive en→0 as n→∞ and also (74)limsupn→∞⁡b^n≤0.Therefore, from (68), we obtain (75)xn+1-z2≤1-αn1-2κxn-z2+αnb^n,which together with the last limit and Lemma 8 implies that xn-z→0 as n→∞.
Case
2. The sequence (sk) is not eventually decreasing; that is, there is a subsequence (skj) of (sk) such that skj<skj+1 for all j≥0. Again in this case we define an integer sequence (mk)k≥k0 as in Lemma 9. Note that the subsequence (smk) satisfies the condition smk≤smk+1 for all k≥k0. It then follows from (68) that (76)xmk+1-z2≤1-αmk1-2κxmk+1-z2+αmkb^mk-L1-αmkxmk-JrmkBI-rmkAxmk2,for some positive constant L, where (b^mk) is the sequence denoted by (77)b^mk≔2fz-z+emk,xmk+1-z+αmk+emk+ϵmkαmkMwith an appropriate constant M>0. Note that, from ηmk2/αmk→0 and αmk→0 as k→∞, we derive emk→0 as k→∞ and (78)limsupk→∞⁡b^mk=limsupk→∞⁡fz-z+emk,xmk+1-z.Rearranging terms in (76), we obtain (79)L1-αmkxmk-JrmkBI-rmkAxmk2+αmk1-2κsmk+1≤αmkb^mk,for smk+1=xmk+1-z2. Taking the limit as k→∞ and noting that (b^mk) is bounded, we arrive at (80)limk→∞⁡xmk-JrmkBI-rmkAxmk=0.Similar to the proof of Theorem 10, we deduce that (81)limsupk→∞⁡b^mk=limsupk→∞⁡fz-z+emk,xmk+1-z≤0.From (79), we derive (82)1-2κsmk+1≤b^mk,which together with the fact that 2κ<1 implies that smk+1→0 as k→∞. Since sk≤smk+1 for all k≥n0, it follows that sk→0 as k→∞. This completes the proof of the theorem.Remark 16.
If we takeA=0 and f(x)=u for all x∈H in Theorem 15, where u∈H is a given fixed vector, then we get an improvement and extension of [26, Theorem 2]. Strong convergence results that extend [22, Theorem 2.2] and [27, Theorem 1] and many other results associated with the contraction proximal point algorithm under the accuracy criterion given in condition (iv) of Theorem 15 can also be obtained.



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*Source: 2371857-2016-03-27.xml*