# A Hybrid Proximal Algorithm for the Sum of Monotone Operators with Multivalued Mappings

Authors: N. Kaewyong; L. Kittiratanawasin; C. Pukdeboon; K. Sitthithakerngkiet
Journal: Journal of Function Spaces
 (2018)

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## Abstract

We modify a hybrid method and a proximal point algorithm to iteratively find a zero point of the sum of two monotone operators and fixed point of nonspreading multivalued mappings in a Hilbert space by using the technique of forward-backward splitting method. The strong convergence theorem is established and the illustrative numerical example is presented on this work. The results of this paper extend and improve some well-known results in the literature.


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## Body

## 1. Introduction

In a Hilbert space, many authors have intensively studied the convergence of finding a zero point for monotone operators, that is, to find a pointx∈H such that(1)0∈Ax,where A is a monotone operator and the set of zero point of A is denoted by A-1(0). The first method for finding a zero point is introduced by Martinet [1] in 1970, it is well known as theproximal point algorithm (PPA) which generates a sequence(2)xn+1=JrnAxn,∀n∈N,where JrnA=(I+rnA)-1 is the resolvent operator of maximal monotone operator A, I is the identity mapping and {rn}⊂(0,∞) is a regularization sequence. It can be related to many kinds of important problems, such as convex minimization problems, equilibrium problems, and variational inequality problems. An iterative (2) is equivalent to(3)xn∈xn+1+rnAxn+1∀n∈N.It is known that JrnA can be reduced to(4)xn+1=argminϕy+12rnxn-y2,y∈H∀n∈N,if we let ϕ(x):H→R∪∞ be a proper convex and lower semicontinuous function.Later, Rockafellar [2] presented an inexact variant of the following method:(5)xn+1=JrnAxn+en,∀n∈N,where {en} is an error sequence. Rockafellar [2] proved that if en→0 quickly enough such that ∑n=1∞‍en<∞, liminfn→∞rn>0, and A-1(0)≠∅, then the sequence {xn} converges weakly to a solution of a zero point of A.In 1979, Lions and Mercier [3] presented the splitting algorithms to iteratively find zero point of the sum of two nonlinear operators. This algorithm is extended to solve the nonlinear equations seeking a solution of the following inclusion problem:(6)0∈A+B,where A and B are two monotone operators. The inclusion problem can be formulated to many important problems, such as a stationary solution of the initial value problem of the evolution equation [3], the minimization problem [4], which is widely used in image recovery, signal processing, and machine learning, equilibrium problems, and variational inequality problems; see [5]. A splitting method for solving the inclusion problem (6) intends an iterative method for which each iteration involves only with the individual operators A and B but not A+B. Lions and Mercier [3] introduced the nonlinear splitting iterative algorithms to solve the inclusion problem (6), generated by(7)xn+1=2JrA-I2JrB-Ixn,∀n∈N,(8)xn+1=JrA2JrB-I+I-JrBxn,∀n∈N,where JrA and JrB are the resolvent operators of monotone operators A and B, respectively, with r>0. The algorithm (7) is called the nonlinear Peaceman-Rachford splitting iterative algorithm. Since (2JrA-I)(2JrB-I) is merely nonexpansive operator then it fails, in general, to converge but the mean averages of xn can be weakly convergent; for more details see [6]. However, the algorithm, known as the nonlinear Douglas-Rachford splitting iterative algorithm (8), always converges in the weak topology to a point JrBx because the operator JrA(2JrB-I)+(I-JrB) is firmly nonexpansive.The extendedPPA is introduced by Manaka and Takahashi [7] to the case of sum of two monotone operators A and B by using the technique of forward-backward splitting method which generates a sequence {xn} defined by(9)x1∈C,xn+1=αnxn+1-αnTJrnAI-rnBxn,∀n≥1,where T is a nonexpansive mapping on a nonempty closed convex subset C of H, JrnA is the resolvent of a maximal monotone operator A with {rn} being a positive sequence, B is an inverse strongly monotone mapping, and {αn} is a sequence in (0,1). This algorithm shows that a sequence {xn} converges weakly to some point z∈Fix(T)∩(A+B)-1(0) provided that the control sequence satisfies some conditions.In 2014, Cho et al. [8] presented the strong convergence theorem for the solution set Fix(T)∩(A+B)-1(0) in a Hilbert space by using the following iterative scheme:(10)x1∈C,zn=αnfxn+1-αnxn,yn=JrnAzn-rnBzn+en,xn+1=βnxn+1-βnγnyn+1-γnTyn∀n∈N,where {αn}, {βn}, {γn} are sequences in (0, 1), {rn} is a positive sequence, T is a strictly pseudo-contractive mapping with k∈[0,1), and f is a contractive mapping.An iterative algorithm for finding an approximate solution of the sum of two monotone operators and fixed point of several type mappings has received a lot of attention more recently; for more details, see [9–11].On the other hand, Iemoto and Takahashi [12] study the approximation of common fixed points of a nonexpansive mapping T and a nonspreading mapping S (i.e., 2Sx-Sy2≤Sx-y2+x-Sy2) in a Hilbert space by using iterative scheme:(11)x1∈C,xn+1=1-αnxn+αnβnSxn+1-βnTxn,where {αn},{βn}⊂[0,1]. They proved that if Fix(S)∩Fix(T) is nonempty then the sequence generated by (11) converges weakly to common fixed point of S and T. For the extension of mappings, many authors have studied the convergence theorems of multivalued mappings (see [13–15]).In 2016, Suantai et al. [16] considered iterative schemes for solving split equilibrium problems and fixed point problems of nonspreading multivalued mappings in Hilbert spaces and proved that the modified Mann iteration converges weakly to a common solution of the considered problems.Inspired by [8, 16], in this paper, we present the convergence analysis on the set Fix(T)∩(A+B)-1(0), where T is a nonspreading multivalued mapping in a Hilbert space. The results of this paper extend and improve some well-known results in the literature. Furthermore, the illustrative numerical example is presented.


## 2. Preliminaries

LetH be a real Hilbert space with inner product 〈·,·〉 and norm ·, and let C be a nonempty closed convex subset of H. For any x,y∈H and λ∈[0,1], we see that(12)x+y2≤x2+2y,x+y,λx+1-λy2=λx2+1-λy2-λ1-λx-y2.An operatorT:H→H is called a nonexpansive mapping if(13)Tx-Ty2≤x-y2,∀x,y∈H,and is called a firmly nonexpansive mapping if(14)Tx-Ty2≤Tx-Ty,x-y,∀x,y∈H.Clearly, the above inequality is equivalent to(15)Tx-Ty2≤x-y2-I-Tx-I-Ty2∀x,y∈H,where I is identity mapping. For any point x∈H, there exists a unique nearest point of C, denoted by PCx, such that(16)x-PCx≤x-y∀y∈C.The operator PC denotes the metric projection from H onto C. It is known that PC is a firmly nonexpansive mapping; that is,(17)PCx-PCy2≤PCx-PCy,x-y,∀x,y∈H.Furthermore, for any x∈H and z∈C, we note that z=PCx if and only if(18)x-z,z-y≥0∀y∈C.Any subsetC of a Hilbert space H is said to be proximinal if, for all x∈H, there exists y∈C such that(19)x-y=dx,C=inf⁡x-z:z∈C.In this paper, we denote the setsCB(C),K(C), and P(C) are the families of nonempty closed bounded subsets, nonempty compact subsets, and nonempty proximal subsets of C, respectively. TheHausdorff metric on CB(C) is defined by(20)HA,B=maxsupx∈Adx,B,supy∈Bdy,A,∀A,B∈CBC,where d(x,B)=infb∈Bx-b. Let T:C→CB(C) be a multivalue mapping, an element p∈C is called afixed point of  T if p∈Tp and we denote the fixed point set of a multivalue operator T by Fix(T). A multivalued mapping T:C→K(C) is said to benonexpansive if H(Tx,Ty)≤x-yforallx,y∈C and said to bequasi-nonexpansive if H(Tx,Tp)≤x-pforallx∈Candp∈Fix(T). In this paper, we focus on ak-nonspreading multivalued mapping T that satisfies, for all x,y∈C,(21)HTx,Ty2≤kdTx,y2+dx,Ty2,for some k>0.Condition (I). Let H be a Hilbert space and C be a subset of H. A multivalued mapping T:C→CB(C) is said to satisfyCondition (I) if x-p=d(x,Tp) for all x∈H and p∈Fix(T).Remark 1.
It is easy to see thatT satisfies Condition (I) if and only if Tp=p for all p∈Fix(T). We know that if T is nonexpansive, then T is quasi-nonexpansive. Clearly, if T is a 1/2-nonspreading and Fix(T)≠0, then T is quasi-nonexpansive. Example in [16] shows that T is a 1/2-nonspreading multivalued mapping which is not nonexpansive.A mappingB:C→H is called α-inverse strongly monotone, if there exists α>0 such that(22)x-y,Bx-By≥αBx-By2,for all x,y∈C. We see that if B is α-inverse strongly monotone, then 〈x-y,Bx-By〉≥0 and Bx-By≥1/αx-y for all x,y∈C. Moreover, for any constant r>0, it is easy to see that(23)I-rBx-I-rBy2≤x-y2-r2α-rBx-By2,where I is identity mapping. In particular, if r∈(0,2α), then (I-rB) is a nonexpansive mapping. For more example of inverse-strongly monotone mappings, see [17, 18].LetA be a mapping of H into 2H; the effective domain of A is denoted by dom(A); that is, dom(A)=x∈H:Ax≠0. A multivalued mapping A is said to be a monotone operator on H if (24)x-y,u-v≥0for all x,y∈dom(A), u∈Ax, and v∈Ay. A monotone operator A on H is said to be maximal if its graph is not properly contained in the graph of any other monotone operator on H. For maximal monotone operator A on H and r>0, we may define a single-valued operator JrA:H→dom(A) by JrA=(I+rA)-1, which is called the resolvent of A for r>0. If we let B:C→H be a single value operator and let A be a maximal monotone operator in H with D(A)⊃C and D(B)⊃C, then, using the concept by [19], for r>0,(25)FixJrAI-rB=A+B-10.Lemma 2 (see [16]).
LetC be a nonempty closed convex subset of a Hilbert space H and T:C→K(C) be a k-nonspreading multivalued mapping with k∈(0,1/2]. Let {xn} be a sequence in C such that xn⇀p and limn→∞xn-yn=0 for some yn∈Txn. Then p∈Tp.


## 3. Main Results

Theorem 3.
LetH be a real Hilbert space and C be a nonempty closed convex subset of Hilbert spaces H. Let A:D(A)⊆C→2H be a maximal monotone operator, B:C→H be an α-inverse strongly monotone operator, and T:C→K(C) be a 1/2-nonspreading multivalued mapping. Assume that Θ=Fix(T)∩(A+B)-1(0)≠∅ and {xn} is a sequence defined by(26)x0∈Carbitrarily,un=JrnAI-rnBxn,yn∈anxn+1-anTun,Dn+1=x∈Dn:yn-x2≤xn-x2,xn+1=PDn+1x0,∀n≥0,D0=C⊂Hfor all n≥0, where {an}⊂(0,1) and {rn} is a real number sequence in (0,2α). Suppose that the following conditions hold:(a)
limn→∞an=0, and ∑n=1∞‍an=∞,(b)
limn→∞rn=r and r∈(0,α).
Then, the sequence{xn} converges strongly to a point x∗∈Θ.Proof.
First, we will show thatΘ⊂Dn,∀n≥0 by using by the mathematical induction. Clearly, Θ⊂C=D0 and assume that Θ⊂Dn for some n≥0. Let p∈Θ be fixed. So, we can obtain that p∈Tp and p=JrnA(I-rnB)p and since JrnA and (I-rnB) are nonexpansive mappings, we have(27)un-p2=JrnAI-rnBxn-JrnAI-rnBp2≤I-rnBxn-I-rnBp2≤xn-p2.Since yn∈anxn+(1-an)Tun, there is zn∈Tun such that yn=anxn+(1-an)zn and then we get(28)yn-p2=anxn+1-anzn-p2=anxn-p2+1-anzn-p2-an1-anxn-zn2.From (27), (28), and Condition (I), it follows that(29)yn-p2≤anxn-p2+1-anzn-p2=anxn-p2+1-andzn,Tp2≤anxn-p2+1-anHTun,Tp2≤anxn-p2+1-anun-p2≤anxn-p2+1-anxn-p2=xn-p2.That is, p∈Dn+1 and so Θ⊂Dn+1. Therefore, Θ⊂Dn for all n≥0.
By the assumptions, we can conclude thatDn is nonempty closed convex subset of H and then Θ⊂Dn+1⊂Dn,∀n≥0. For fixed p∈Θ and from xn+1=PDn+1x0, we obtain that(30)xn+1-x0=PDn+1x0-x0≤p-x0.This implies that the sequence {xn} is bounded. Since xn=PDnx0 and Dn+1⊂Dn,∀n≥0, by the properties of the metric projection, we have(31)x0-xm,xm-xn≥0,for any n≥m≥0. Next, we want to show that {xn} is a Cauchy sequence. We compute(32)xn-x02=xn-x0,xn-x0=xn-x0,xn-xn+1+xn+1-x0=xn-x0,xn-xn+1+xn-x0,xn+1-x0.This implies that(33)x0-xn,xn-xn+1=-xn-x02+xn-x0,xn+1-x0=-xn-x02+xn-x0xn+1-x0.By (31) and (33), we get that(34)0≤-xn-x02+xn-x0xn+1-x0.Therefore,(35)xn-x0≤xn+1-x0,∀n≥0.We have from (30) and (35) that limn→∞xn-x0 exists. For any n≥m≥0, by using (31) again, we get(36)xn-x02=xn-xm+xm-x02=xn-xm2+xm-x02+2xn-xm,xm-x0≥xn-xm2+xm-x02.Consequently, we obtain that(37)xn-xm2≤xn-x02-xm-x02.Hence, as m→∞ and n→∞, we have(38)xn-xm⟶0.Therefore the sequence {xn} is a Cauchy sequence. Without loss of generality, we can assume that(39)xn⟶x∗asn⟶∞.
Next, we will prove thatlimn→∞zn-un=0 for some zn∈Tun by dividing the proof into 4 steps.
Step I. We will prove that limn→∞yn-xn=0.
Sincexn+1=PDn+1x0∈Dn+1⊂Dn, from (26), we have(40)yn-xn+12≤xn-xn+12and we obtain(41)yn-xn+1≤xn-xn+1.By (38), we conclude that(42)limn→∞yn-xn+1=limn→∞xn-xn+1=0.Consider(43)yn-xn≤yn-xn+1+xn+1-xn.Then, by (38) and (42), we obtain that (44)limn→∞yn-xn=0.
Step II. We will show that limn→∞Bxn-Bp=0.
Note that(45)yn-p2=anxn+1-anTJrnAI-rnBxn-p2=anxn-p2+1-anTJrnAI-rnBxn-p2-an1-anxn-TJrnAI-rnBxn2≤anxn-p2+1-anTJrnAI-rnBxn-p2=anxn-p2+1-anTun-p2=anxn-p2+1-anzn-p2for  somezn∈Tun=anxn-p2+1-andzn,Tp2≤anxn-p2+1-anHTun,Tp2≤anxn-p2+1-anun-p2=anxn-p2+1-anJrnAI-rnBxn-JrnAI-rnBp2.It follows that(46)yn-p2≤anxn-p2+1-anI-rnBxn-I-rnBp2≤anxn-p2+1-anxn-p2-2rn2α-rnBxn-Bp=anxn-p2+1-anxn-p2-2rn2α-rn1-anBxn-Bp2.This implies that(47)2rn2α-rn1-anBxn-Bp2≤xn-p2-yn-p2.By using (44), rn∈(0,2α), and an∈(0,1), then we conclude that(48)limn→∞Bxn-Bp=0.
Step III. We will show that limn→∞JrnA(I-rnB)xn-xn=limn→∞un-xn=0.
Note that(49)JrnAI-rnBxn-p2=JrnAI-rnBxn-JrnAI-rnBp2≤I-rnBxn-I-rnBp,JrnAI-rnBxn-JrnAI-rnBp=JrnAI-rnBxn-p,I-rnBxn-I-rnBp.Consider(50)JrnAI-rnBxn-p-I-rnBxn-I-rnBp2=JrnAI-rnBxn-p2-2JrnAI-rnBxn-p,I-rnBxn-I-rnBp+I-rnBxn-I-rnBp2.Then, we get(51)JrnAI-rnBxn-p,I-rnBxn-I-rnBp=12JrnAI-rnBxn-p2+I-rnBxn-I-rnBp2-JrnAI-rnBxn-p-I-rnBxn-I-rnBp2.From (49) and (51), we obtain that(52)JrnAI-rnBxn-p2≤12JrnAI-rnBxn-p2+I-rnBxn-I-rnBp2-JrnAI-rnBxn-p-I-rnBxn-I-rnBp2=12JrnAI-rnBxn-p2+I-rnBxn-I-rnBp2-JrnAI-rnBxn-xn-rnBp-Bxn2≤12JrnAI-rnBxn-p2+xn-p2-JrnAI-rnBxn-xn2+2rnJrnAI-rnBxn-xnBxn-Bp-rn2Bp-Bxn2.Form (52), we have(53)JrnAI-rnBxn-p2≤xn-p2-JrnAI-rnBxn-xn2-rn2Bxn-Bp2+2rnJrnAI-rnBxn-xnBxn-Bp.Then, from (45) and (53), we obtain that(54)yn-p2≤anxn-p2+1-anJrnAI-rnBxn-p2≤anxn-p2+1-anxn-p2-JrnAI-rnBxn-xn2+2rnJrnAI-rnBxn-xnBxn-Bp-rn2Bxn-Bp2=xn-p2-1-anJrnAI-rnBxn-xn2+1-an2rnJrnAI-rnBxn-xnBxn-Bp+rn2Bxn-Bp2.It follows that(55)1-anJrnAI-rnBxn-xn2≤xn-p2-yn-p2+1-an2rnJrnAI-rnBxn-xnBxn-Bp-rn2Bxn-Bp2.Form (55), in view of conditions (b), and (44), we conclude that(56)limn→∞JrnAI-rnBxn-xn=limn→∞un-xn=0.
Step IV. We will show that limn→∞un-zn=0, for some zn∈Tun.
By using Condition (I) and (28), we obtain that(57)yn-p2=anxn-p+1-anzn-p2for  somezn∈Tun=anxn-p2+1-anzn-p2-an1-anxn-zn2=anxn-p2+1-andzn,Tp2-a1-anxn-zn2≤anxn-p2+1-anHTun,Tp2-an1-anxn-zn2≤anxn-p2+1-anun-p2-an1-anxn-zn2≤xn-p2-an1-anxn-zn2.This implies that(58)an1-anxn-zn2≤xn-p2-yn-p2.Form (44), we conclude that(59)limn→∞xn-zn=0.Consider(60)zn-un≤zn-xn+xn-un.Therefore, we get from (56) and (59)(61)limn→∞zn-un=0.
Finally, We will prove thatx∗∈Fix(T)∩(A+B)-1(0). Since xn→x∗ and by (56), we get that un→x∗ also. Since zn∈Tun and limn→∞un-zn=0, by using Lemma 2, we have x∗∈Fix(T).
Consider(62)JrnAI-rnBx∗-x∗≤JrnAI-rnBxn-JrnAI-rnBx∗+JrnAI-rnBxn-xn+xn-x∗≤2xn-x∗+JrnAI-rnBxn-xn.Since xn→x∗ and by (56), we get that(63)limn→∞JrnAI-rnBx∗-x∗=0.That is, JrA(I-rB)x∗=x∗ as n→∞, which implies that x∗∈(A+B)-1(0). Therefore, we conclude that x∗∈Θ which completes the proof.


## 4. Numerical Example and Convergence Analysis

In this section, we give the following numerical example to confirm the convergence of Theorem3 by using the algorithm (26).Example 1.
LetH=R and C=[0,4]. Define the mappings A:D(A)⊆C→2H, B:C→H, and T:C→K(C) by the following:(64)Ax≔2x+14,Bx≔12x-1,Tx≔0,x∈0,20,e2-x,x∈2,4.We see that the proposed mappings satisfy the assumptions in Theorem3. For each rn>0, we obtain that JrnA(I-rnB)(x)=x(2-r)/2+4r. It is easy to see that a point 0 is in the fixed point sets of JrnA(I-rnB) and T; that is, 0∈Fix(T)∩(A+B)-1(0).In Figure1, these initial points are randomly chosen from the set D0=C and we find optimal solution in 20 steps. This indicates that the sequence xn in algorithm (26) converges to the same point; that is, 0∈Θ as a solution of this example. In this experiment, Figure 1 indicates the behaviour of xn for algorithm (26) that converges to the same solution; that is, 0∈Fix(T)∩(A+B)-1(0) as a solution of this example. Moreover, the decreasing on alpha function decreases rate of convergence to the optimal solutions which is shown in Figure 2. Figure 3 shows that Dn+1⊂Dn⊂⋯⊂D2⊂D1⊂D0=C. This means that the iteration of Cn will squeeze the area until we obtain the approximated solution.Figure 1
The comparison of convergent rate from 3 initial points.Figure 2
The comparison of convergent rate from 3 different alpha parameters.Figure 3
The behaviors of the setDn.



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