# Iterative Methods for the Sum of Two Monotone Operators

Authors: Yeong-Cheng Liou
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We introduce an iterative for finding the zeros point of the sum of two monotone operators. We prove that the suggested method converges strongly to the zeros point of the sum of two monotone operators.


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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” quasi-variational inclusion problem is to find a u∈2H such that(1.1)0∈Ax+Bx.
The set of solutions of (1.1) 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 (1.1) includes many problems as special cases.(1)
IfB=∂ϕ:H→2H, where ϕ:H→R∪+∞ is a proper convex lower semicontinuous function and ∂ϕ is the subdif and if onlyerential of ϕ, then the variational inclusion problem (1.1) is equivalent to find u∈H such that
(1.2)〈Au,y-u〉+ϕ(y)-ϕ(u)≥0,∀y∈H,
which is called the mixed quasi-variational 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,
(1.3)δC={0,x∈C,+∞,x∉C,
then the variational inclusion problem (1.1) is equivalent to find u∈C such that
(1.4)〈Au,v-u〉≥0,∀v∈C.This problem is called Hartman-Stampacchia variational inequality (see, e.g., [6]).Recently, Zhang et al. [7] introduced a new iterative scheme for finding a common element of the set of solutions to the inclusion problem, and the set of fixed points of nonexpansive mappings in Hilbert spaces. Peng et al. [8] introduced another iterative scheme by the viscosity approximate method for finding a common element of the set of solutions of a variational inclusion with set-valued maximal monotone mapping and inverse strongly monotone mappings, the set of solutions of an equilibrium problem, and the set of fixed points of a nonexpansive mapping. For some related works, please see [9–27] and the references therein.Inspired and motivated by the works in the literature, in this paper, we introduce an iterative for solving the problem (1.1). We prove that the suggested method converges strongly to the zeros point of the sum of two monotone operators A+B.


## 2. Preliminaries

LetH be a real Hilbert space with inner product 〈·,·〉 and norm ∥·∥, respectively. Let C be a nonempty closed convex subset of H. Recall that a mapping A:C→H is said to be α-inverse strongly-monotone if and if only (2.1)〈Ax-Ay,x-y〉≥α‖Ax-Ay‖2
for some α>0 and for all x,y∈C. It is known that if A is α-inverse strongly monotone, then (2.2)‖Ax-Ay‖≤1α‖x-y‖
for all x,y∈C.LetB be a mapping of H into 2H. The effective domain of B is denoted by dom(B), that is,(2.3)dom(B)={x∈H:Bx≠∅}.
A multivalued mapping B is said to be a monotone operator on H if and if only (2.4)〈x-y,u-v〉≥0
for all x,y∈dom(B), u∈Bx, and v∈By. A monotone operator B on H is said to be maximal if and if only its graph is not strictly contained in the graph of any other monotone operator on H. Let B be a maximal monotone operator on H and let B-10={x∈H:0∈Bx}.For a maximal monotone operatorB on H and λ>0, we may define a single-valued operator: (2.5)JλB=(I+λB)-1:H→dom(B),
which is called the resolvent of B for λ. It is known that the resolvent JλB is firmly nonexpansive, that is, (2.6)‖JλBx-JλBy‖2≤〈JλBx-JλBy,x-y〉
for all x,y∈C and B-10=F(JλB) for all λ>0.The following resolvent identity is well known: forλ>0 and μ>0, there holds the following identity:(2.7)JλBx=JμB(μλx+(1-μλ)JλBx),x∈H.We use the following notation:(i)
xn⇀x stands for the weak convergence of (xn) to x;(ii)
xn→x stands for the strong convergence of (xn) to x.We need the following lemmas for the next section.Lemma 2.1 (see [28]).
LetC be a nonempty closed convex subset of a real Hilbert space H. Let the mapping A:C→H be α-inverse strongly monotone and let λ>0 be a constant. Then, one has
(2.8)‖(I-λA)x-(I-λA)y‖2≤‖x-y‖2+λ(λ-2α)‖Ax-Ay‖2,∀x,y∈C.
In particular, if 0≤λ≤2α, then I-λA is nonexpansive.Lemma 2.2 (see [29]).
Let{xn} and {yn} be bounded sequences in a Banach space X and let {βn} be a sequence in [0,1] with
(2.9)0<liminf⁡n→∞βn≤limsup⁡n→∞βn<1.
Suppose that
(2.10)xn+1=(1-βn)yn+βnxn
for all n≥0 and
(2.11)limsup⁡n→∞(‖yn+1-yn‖-‖xn+1-xn‖)≤0.
Then, limn→∞∥yn-xn∥=0.Lemma 2.3 (see [30]).
Assume that{an} is a sequence of nonnegative real numbers such that
(2.12)an+1≤(1-γn)an+δnγn,
where {γn} is a sequence in (0,1) and {δn} is a sequence such that  (1)
∑n=1∞γn=∞;(2)
limsupn→∞δn≤0 or ∑n=1∞|δnγn|<∞.Then limn→∞an=0.


## 3. Main Results

In this section, we will prove our main result.Theorem 3.1.
LetC be a nonempty closed and convex subset of a real Hilbert space H. Let A be an α-inverse strongly monotone mapping of C into H and let B be a maximal monotone operator on H, such that the domain of B is included in C. Let JλB=(I+λB)-1 be the resolvent of B for λ>0. Suppose that (A+B)-10≠∅. For u∈C and given x0∈C, let {xn}⊂C be a sequence generated by
(3.1)xn+1=βnxn+(1-βn)JλnB(αnu+(1-αn)(xn-λnAxn))
for all n≥0, where {λn}⊂(0,2α), {αn}⊂(0,1), and {βn}⊂(0,1) satisfy  (i)
limn→∞αn=0 and ∑nαn=∞;(ii)
0<liminfn→∞βn≤limsupn→∞βn<1;(iii)
a≤λn≤b where [a,b]⊂(0,2α) and limn→∞(λn+1-λn)=0.Then {xn} generated by (3.1) converges strongly to x̃=P(A+B)-10(u).Proof.
First, we choose anyz∈(A+B)-10. Note that
(3.2)z=JλnB(z-λn(1-αn)Az)=JλnB(αnz+(1-αn)(z-λnAz))
for all n≥0. Since JλB is nonexpansive for all λ>0, we have
(3.3)‖JλnB(αnu+(1-αn)(xn-λnAxn))-z‖2=‖JλnB(αnu+(1-αn)(xn-λnAxn))-JλnB(αnz+(1-αn)(z-λnAz))‖2≤‖(αnu+(1-αn)(xn-λnAxn))-(αnz+(1-αn)(z-λnAz))‖2=‖(1-αn)((xn-λnAxn)-(z-λnAz))+αn(u-z)‖2.
Since A is α-inverse strongly monotone, we get
(3.4)‖(1-αn)((xn-λnAxn)-(z-λnAz))+αn(u-z)‖2≤(1-αn)‖(xn-λnAxn)-(z-λnAz)‖2+αn‖u-z‖2=(1-αn)‖(xn-z)-λn(Axn-Az)‖2+αn‖u-z‖2=(1-αn)(‖xn-z‖2-2λn〈Axn-Az,xn-z〉+λn2‖Axn-Az‖2)+αn‖u-z‖2≤(1-αn)(‖xn-z‖2-2αλn‖Axn-Az‖2+λn2‖Axn-Az‖2)+αn‖u-z‖2=(1-αn)(‖xn-z‖2+λn(λn-2α)‖Axn-Az‖2)+αn‖u-z‖2.
By (3.3) and (3.4), we obtain
(3.5)‖JλnB(αnu+(1-αn)(xn-λnAxn))-z‖2≤(1-αn)(‖xn-z‖2+λn(λn-2α)‖Axn-Az‖2)+αn‖u-z‖2≤(1-αn)‖xn-z‖2+αn‖u-z‖2.
It follows from (3.1) and (3.5) that
(3.6)‖xn+1-z‖2=‖βn(xn-z)+(1-βn)(JλnB(αnu+(1-αn)(xn-λnAxn))-z)‖2≤βn‖xn-z‖2+(1-βn)‖JλnB(αnu+(1-αn)(xn-λnAxn))-z‖2≤βn‖xn-z‖2+(1-βn)((1-αn)‖xn-z‖2+αn‖u-z‖2)=[1-(1-βn)αn]‖xn-z‖2+(1-βn)αn‖u-z‖2≤max⁡{‖xn-z‖2,‖u-z‖2}.
By induction, we have
(3.7)‖xn+1-z‖≤max⁡{‖x0-z‖,‖u-z‖}.
Therefore, {xn} is bounded. We deduce immediately that {Axn} is also bounded. Set un=αnu+(1-αn)(xn-λnAxn) for all n. Then {un} and {JλnBun} are bounded.
Next, we estimate∥Jλn+1Bun+1-JλnBun∥. In fact, we have
(3.8)‖Jλn+1Bun+1-JλnBun‖=‖Jλn+1B(αn+1u+(1-αn+1)(xn+1-λn+1Axn+1))-JλnB(αnu+(1-αn)(xn-λnAxn))‖≤‖Jλn+1B(αn+1u+(1-αn+1)(xn+1-λn+1Axn+1))-Jλn+1B(αnu+(1-αn)(xn-λnAxn))‖+‖Jλn+1B(αnu+(1-αn)(xn-λnAxn))-JλnB(αnu+(1-αn)(xn-λnAxn))‖≤‖(αn+1u+(1-αn+1)(xn+1-λn+1Axn+1))-(αnu+(1-αn)(xn-λnAxn))‖+‖Jλn+1Bun-JλnBun‖≤‖(I-λn+1A)xn+1-(I-λn+1A)xn‖+|λn+1-λn|‖Axn‖+αn+1(‖u‖+‖xn+1‖+λn+1‖Axn+1‖)+αn(‖u‖+‖xn‖+λn‖Axn‖)+‖Jλn+1Bun-JλnBun‖.
SinceI-λn+1A is nonexpansive for λn+1∈(0,2α), we have ∥(I-λn+1A)xn+1-(I-λn+1A)xn∥≤∥xn+1-xn∥. By the resolvent identity (2.7), we have
(3.9)Jλn+1Bun=JλnB(λnλn+1un+(1-λnλn+1)Jλn+1Bun).
It follows that
(3.10)‖Jλn+1Bun-JλnBun‖=‖JλnB(λnλn+1un+(1-λnλn+1)Jλn+1Bun)-JλnBun‖≤‖(λnλn+1un+(1-λnλn+1)Jλn+1Bun)-un‖≤|λn+1-λn|λn+1‖un-Jλn+1Bun‖.
So,
(3.11)‖Jλn+1Bun+1-JλnBun‖≤‖xn+1-xn‖+|λn+1-λn|‖Axn‖+αn+1(‖u‖+‖xn+1‖+λn+1‖Axn+1‖)+αn(‖u‖+‖xn‖+λn‖Axn‖)+|λn+1-λn|λn+1‖un-Jλn+1Bun‖.
Thus,
(3.12)limsup⁡n→∞(‖Jλn+1Bun+1-JλnBun‖-‖xn+1-xn‖)≤0.
From Lemma 2.2, we get
(3.13)lim⁡n→∞‖JλnBun-xn‖=0.
Consequently, we obtain
(3.14)lim⁡n→∞‖xn+1-xn‖=lim⁡n→∞(1-βn)‖JλnBun-xn‖=0.
From (3.5) and (3.6), we have
(3.15)‖xn+1-z‖2≤βn‖xn-z‖2+(1-βn)‖JλnB(αnu+(1-αn)(xn-λnAxn))-z‖2≤(1-βn){(1-αn)(‖xn-z‖2+λn(λn-2α)‖Axn-Az‖2)+αn‖u-z‖2}+βn‖xn-z‖2=[1-(1-βn)αn]‖xn-z‖2+(1-βn)λn(λn-2α)‖Axn-Az‖2+(1-βn)αn‖u-z‖2≤‖xn-z‖2+(1-βn)λn(λn-2α)‖Axn-Az‖2+(1-βn)αn‖u-z‖2.
It follows that
(3.16)(1-βn)λn(2α-λn)‖Axn-Az‖2≤‖xn-z‖2-‖xn+1-z‖2+(1-βn)αn‖u-z‖2≤(‖xn-z‖-‖xn+1-z‖)‖xn+1-xn‖+(1-βn)αn‖u-z‖2.
Since limn→∞αn=0, limn→∞∥xn+1-xn∥=0, and liminfn→∞(1-βn)λn(2α-λn)>0, we have
(3.17)lim⁡n→∞‖Axn-Az‖=0.
Put x̃=P(A+B)-10(u). Set vn=xn-(λn/(1-αn))(Axn-Ax̃) for all n. Take z=x̃ in (3.17) to get ∥Axn-Ax̃∥→0. First, we prove limsupn→∞〈u-x̃,vn-x̃〉≤0. We take a subsequence {vni} of {vn} such that
(3.18)limsup⁡n→∞〈u-x̃,vn-x̃〉=lim⁡i→∞〈u-x̃,vni-x̃〉.
It is clear that {vni} is bounded due to the boundedness of {xn} and ∥Axn-Ax̃∥→0. Then, there exists a subsequence {vnij} of {vni} which converges weakly to some point w∈C. Hence, {xnij} also converges  weakly to w because of ∥vnij-xnij∥→0. By the similar argument as that in [31], we can show that w∈(A+B)-10. This implies that
(3.19)limsup⁡n→∞〈u-x̃,vn-x̃〉=lim⁡j→∞〈u-x̃,vnij-x̃〉=〈u-x̃,w-x̃〉.
Note that x̃=P(A+B)-10(u). Then, 〈u-x̃,w-x̃〉≤0,w∈(A+B)-10. Therefore,
(3.20)limsup⁡n→∞〈u-x̃,vn-x̃〉≤0.
Finally, we prove that xn→x̃. From (3.1), we have
(3.21)‖xn+1-x̃‖2≤βn‖xn-x̃‖2+(1-βn)‖JλnBun-x̃‖2=βn‖xn-x̃‖2+(1-βn)‖JλnBun-JλnB(x̃-(1-αn)λnAx̃)‖2≤βn‖xn-x̃‖2+(1-βn)‖un-(x̃-(1-αn)λnAx̃)‖2=βn‖xn-x̃‖2+(1-βn)‖αnu+(1-αn)(xn-λnAxn)-(x̃-(1-αn)λnAx̃)‖2=βn‖xn-x̃‖2+(1-βn)‖(1-αn)((xn-λnAxn)-(x̃-λnAx̃))+αn(u-x̃)‖2=βn‖xn-x̃‖2+(1-βn)×((1-αn)2‖(xn-λnAxn)-(x̃-λnAx̃)‖2+2αn(1-αn)〈u-x̃,(xn-λnAxn)-(x̃-λnAx̃)〉+αn2‖u-x̃‖2)≤βn‖xn-x̃‖2+(1-βn)×((1-αn)‖xn-x̃‖2+2αn(1-αn)〈u-x̃,xn-λn(Axn-Ax̃)-x̃〉+αn2‖u-x̃‖2)≤[1-(1-βn)αn]‖xn-x̃‖2+(1-βn)αn{2(1-αn)〈u-x̃,vn-x̃〉+αn‖u-x̃‖2}.
It is clear that ∑n(1-βn)αn=∞ and limsupn→∞(2(1-αn)〈u-x̃,vn-x̃〉+αn∥u-x̃∥2)≤0. We can therefore apply Lemma 2.3 to conclude that xn→x̃. This completes the proof.


## 4. Applications

Next, we consider the problem for finding the minimum norm solution of a mathematical model related to equilibrium problems. LetC be a nonempty, closed, and convex subset of a Hilbert space and let G:C×C→R be a bifunction satisfying the following conditions:(E1)
G(x,x)=0 for all x∈C;(E2)
G is monotone, that is, G(x,y)+G(y,x)≤0 for all x,y∈C;(E3)
for allx,y,z∈C, limsupt↓0G(tz+(1-t)x,y)≤G(x,y);(E4)
for allx∈C, G(x,·) is convex and lower semicontinuous.Then, the mathematical model related to equilibrium problems (with respect toC) is to find x̃∈C such that(4.1)G(x̃,y)≥0
for all y∈C. The set of such solutions x̃ is denoted by EP(G). The following lemma appears implicitly in Blum and Oettli [32].Lemma 4.1.
LetC be a nonempty, closed, and convex subset of H and let G be a bifunction of C×C into R satisfying (E1)–(E4). Let r>0 and x∈H. Then, there exists z∈C such that
(4.2)G(z,y)+1r〈y-z,z-x〉≥0,∀y∈C.The following lemma was given by Combettes and Hirstoaga [33].Lemma 4.2.
Assume thatG:C×C→R satisfies (E1)–(E4). For r>0 and x∈H, define a mapping Tr:H→C as follows:
(4.3)Tr(x)={z∈C:G(z,y)+1r〈y-z,z-x〉≥0,∀y∈C}
for all x∈H. Then, the following holds:  (1)
Tr is single valued;(2)
Tr is a firmly nonexpansive mapping, that is, for all x,y∈H,
(4.4)‖Trx-Try‖2≤〈Trx-Try,x-y〉;(3)
F(Tr)=EP(G);(4)
EP(G) is closed and convex.We call suchTr the resolvent of G for r>0. Using Lemmas 4.1 and 4.2, we have the following lemma. See [34] for a more general result.Lemma 4.3.
LetH be a Hilbert space and let C be a nonempty, closed, and convex subset of H. Let G:C×C→R satisfy (E1)–(E4). Let AG be a multivalued mapping of H into itself defined by
(4.5)AGx={{z∈H:G(x,y)≥〈y-x,z〉,∀y∈C},∅,x∈C,x∉C.
Then, EP(G)=AG-1(0) and AG is a maximal monotone operator with dom(AG)⊂C. Further, for any x∈H and r>0, the resolvent Tr of G coincides with the resolvent of AG; that is,
(4.6)Trx=(I+rAG)-1x.Form Lemma4.3 and Theorems 3.1, we have the following result.Theorem 4.4.
LetC be a nonempty, closed, and convex subset of a real Hilbert space H. Let G be a bifunction from C×C→R satisfying (E1)–(E4) and let Tλ be the resolvent of G for λ>0. Suppose EP(G)≠∅. For u∈C and given x0∈C, let {xn}⊂C be a sequence generated by
(4.7)xn+1=βnxn+(1-βn)Tλn(αnu+(1-αn)xn)
for all n≥0, where {λn}⊂(0,∞), {αn}⊂(0,1), and {βn}⊂(0,1) satisfy  (i)
limn→∞αn=0 and ∑nαn=∞;(ii)
0<liminfn→∞βn≤limsupn→∞βn<1;(iii)
a≤λn≤b where [a,b]⊂(0,∞) and limn→∞(λn+1-λn)=0.Then {xn} converges strongly to a point x̃=PEP(G)(u).



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