# Projection Algorithms for Variational Inclusions

Authors: Youli Yu; Pei-Xia Yang; Khalida Inayat Noor
Journal: Journal of Applied Mathematics
 (2012)

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## Abstract

We present a projection algorithm for finding a solution of a variational inclusion problem in a real Hilbert space. Furthermore, we prove that the proposed iterative algorithm converges strongly to a solution of the variational inclusion problem which also solves some variational inequality.


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## Body

## 1. Introduction

LetH be a real Hilbert space. Let B:H→H be a single-valued nonlinear mapping and R:H→2H be a set-valued mapping. Now we concern the following variational inclusion, which is to find a point x∈H such that(1.1)θ∈B(x)+R(x),
where θ is the zero vector in H. The set of solutions of problem (1.1) is denoted by I(B,R). If H=Rm, then problem (1.1) becomes the generalized equation introduced by Robinson [1]. If B=0, then problem (1.1) becomes the inclusion problem introduced by Rockafellar [2]. It is known that (1.1) 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, game theory, and so forth. Also various types of variational inclusions problems have been extended and generalized. Recently, Zhang et al. [3] introduced a new iterative scheme for finding a common element of the set of solutions to the problem (1.1) and the set of fixed points of nonexpansive mappings in Hilbert spaces. Peng et al. [4] 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, see [5–28] and the references therein.Inspired and motivated by the works in the literature, in this paper, we present a projection algorithm for finding a solution of a variational inclusion problem in a real Hilbert space. Furthermore, we prove that the proposed iterative algorithm converges strongly to a solution of the variational inclusion problem which also solves some variational inequality.


## 2. Preliminaries

LetH be a real Hilbert space with inner product 〈·,·〉 and norm ∥·∥. Let C be a nonempty closed convex subset of H. Recall that a mapping B:C→C is said to be α-inverse strongly monotone if there exists a constant α>0 such that 〈Bx-By,x-y〉≥α∥Bx-By∥2,for all x,y∈C. A mapping A is strongly positive on H if there exists a constant μ>0 such that 〈Ax,x〉≥μ∥x∥2 for all x∈H.For anyx∈H, there exists a unique nearest point in C, denoted by PC(x), such that(2.1)‖x-PC(x)‖≤‖x-y‖,forally∈C.
Such a PC is called the metric projection of H onto C. We know that PC is nonexpansive. Further, for x∈H and x*∈C,(2.2)x*=PC(x)⟺〈x-x*,x*-y〉≥0forally∈C.A set-valued mappingT:H→2H is called monotone if, for all x,y∈H, f∈Tx and g∈Ty imply 〈x-y,f-g〉≥0. A monotone mapping T:H→2H is maximal if its graph G(T) is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping T is maximal if and only if, for (x,f)∈H×H, 〈x-y,f-g〉≥0 for every (y,g)∈G(T) implies f∈Tx.Let the set-valued mappingR:H→2H be maximal monotone. We define the resolvent operator JR,λ associated with R and λ as follows:(2.3)JR,λ=(I+λR)-1(x),x∈H,
where λ is a positive number. It is worth mentioning that the resolvent operator JR,λ is single-valued, nonexpansive, and 1-inverse strongly monotone, and that a solution of problem (1.1) is a fixed point of the operator JR,λ(I-λB) for all λ>0, see for instance [29].Lemma 2.1 (see [30]).
LetR:H→2H be a maximal monotone mapping and B:H→H be a Lipschitz-continuous mapping. Then the mapping (R+B):H→2H is maximal monotone.Lemma 2.2 (see [8]).
Let{xn} and {yn} be bounded sequences in a Banach space X and let {βn} be a sequence in [0,1] with 0<liminfn→∞βn≤limsupn→∞βn<1. Suppose xn+1=(1-βn)yn+βnxn for all integers n≥0 and limsupn→∞(∥yn+1-yn∥-∥xn+1-xn∥)≤0. Then, limn→∞∥yn-xn∥=0.Lemma 2.3 (see [31]).
Assume{an} is a sequence of nonnegative real numbers such that an+1≤(1-γn)an+δn, where {γn} is a sequence in (0,1) and {δn} is a sequence such that(1)
∑n=1∞γn=∞;(2)
limsupn→∞δn/γn≤0 or ∑n=1∞|δn|<∞.Then limn→∞an=0.


## 3. Main Result

In this section, we will prove our main result. First, we give some assumptions on the operators and the parameters. Subsequently, we introduce our iterative algorithm for finding solutions of the variational inclusion (1.1). Finally, we will show that the proposed algorithm has strong convergence.In the sequel, we will assume that(A1)
C is a nonempty closed convex subset of a real Hilbert space H;(A2)
A is a strongly positive bounded linear operator with coefficient 0<μ<1, R:H→2H is a maximal monotone mapping and B:C→C is an α-inverse strongly monotone mapping;(A3)
λ>0 is a constant satisfying λ<2α.Now we introduce the following iteration algorithm.Algorithm 3.1.
For givenx0∈C arbitrarily, compute the sequence {xn} as follows:
(3.1)xn+1=(1-βn)xn+βnPC[(I-αnA)JR,λ(I-λB)xn],n≥0,
where {αn} and {βn} are two real sequences in [0,1].Now we study the strong convergence of the algorithm (3.1)Theorem 3.2.
Suppose thatI(B,R)≠∅. Assume the following conditions are satisfied:(i)
limn→∞αn=0;(ii)
∑n=0∞αn=∞;(iii)
0<liminfn→∞βn≤limsupn→∞βn<1.Then the sequence {xn} generated by (3.1) converges strongly to x̃∈I(B,R) which solves the following variational inequality:
(3.2)〈Ax,y-x〉≥0,∀y∈I(B,R).Proof.
Takex*∈I(B,R). It is clear that
(3.3)JR,λ(x*-λBx*)=x*.
We divide our proofs into the following five steps:(1)
the sequence{xn} is bounded.(2)
∥xn+1-xn∥→0.(3)
∥Bxn-Bx*∥→0.(4)
limsupn→∞〈Ax̃,xn-x̃〉≥0 where x̃=PI(B,R)(I-A)(x̃).(5)
xn→x̃.Proof of (1.1).
SinceB is α-inverse strongly monotone, we have
(3.4)‖(I-λB)x-(I-λB)y‖2≤‖x-y‖2+λ(λ-2α)‖Bx-By‖2.
It is clear that if 0≤λ≤2α, then (I-λB) is nonexpansive. Set yn=JR,λ(xn-λBxn),n≥0. It follows that
(3.5)‖yn-x*‖=‖JR,λ(xn-λBxn)-JR,λ(x*-λBx*)‖≤‖(xn-λBxn)-(x*-λBx*)‖≤‖xn-x*‖.
Since A is linear bounded self-adjoint operator on H, then
(3.6)‖A‖=sup⁡{|〈Au,u〉|:u∈H,‖u‖=1}.
Observe that
(3.7)〈(I-αnA)u,u〉=1-αn〈Au,u〉≥1-αn‖A‖≥0,
that is to say I-αnA is positive. It follows that
(3.8)‖(I-αnA)‖=sup⁡{〈(I-αnA)u,u〉:u∈H,‖u‖=1}=sup⁡{1-αn〈Au,u〉:u∈H,‖u‖=1}≤1-αnμ.
From (3.1), we deduce that
(3.9)‖xn+1-x*‖=‖(1-βn)xn+βnPC[(I-αnA)yn]-x*‖≤(1-βn)‖xn-x*‖+βn‖[(I-αnA)(yn-x*)]-αnAx*‖≤(1-βn)‖xn-x*‖+(1-αnμ)βn‖yn-x*‖+αnβn‖Ax*‖≤(1-αnβnμ)‖xn-x*‖+αnβn‖Ax*‖≤max⁡{‖x0-x*‖,‖Ax*‖μ}.
Therefore, {xn} is bounded.Proof of (3.1).
Setzn=PC[(I-αnA)JR,λ(I-λB)xn] for all n≥0. Then, we have
(3.10)‖zn-zn-1‖=‖PC[(I-αnA)yn]-PC[(I-αn-1A)yn-1]‖≤‖[(I-αnA)yn]-[(I-αn-1A)yn-1]‖=‖(I-αnA)(yn-yn-1)+(αn-1-αn)Ayn-1‖≤‖(I-αnA)(yn-yn-1)‖+‖(αn-1-αn)Ayn-1‖≤(1-αnμ)‖yn-yn-1‖+|αn-αn-1|‖Ayn-1‖.
Note that
(3.11)‖yn-yn-1‖=‖JR,λ(xn-λBxn)-JR,λ(xn-1-λBxn-1)‖≤‖(xn-λBxn)-(xn-1-λBxn-1)‖≤‖xn-xn-1‖.
Substituting (3.11) into (3.10), we get
(3.12)‖zn-zn-1‖≤(1-αnμ)‖xn-xn-1‖+|αn-αn-1|‖Ayn-1‖.
Therefore,
(3.13)limsup⁡n→∞(‖zn-zn-1‖-‖xn-xn-1‖)≤0.
This together with Lemma 2.2 imply that
(3.14)lim⁡n→∞‖zn-xn‖=0.
Hence,
(3.15)lim⁡n→∞‖xn+1-xn‖=lim⁡n→∞βn‖zn-xn‖=0.Proof of (3.4).
From (3.4), we get
(3.16)‖yn-x*‖2=‖JR,λ(xn-λBxn)-JR,λ(x*-λBx*)‖2≤‖(xn-λBxn)-(x*-λBx*)‖2≤‖xn-x*‖2+λ(λ-2α)‖Bxn-Bx*‖2.
By (3.1), we obtain
(3.17)‖xn+1-x*‖2≤(1-βn)‖xn-x*‖2+βn‖PC[(I-αnA)yn]-x*‖2≤(1-βn)‖xn-x*‖2+βn‖(I-αnA)yn-x*‖2=(1-βn)‖xn-x*‖2+βn(‖yn-x*-αnAyn‖2)=(1-βn)‖xn-x*‖2+βn(‖yn-x*‖2-2αn〈yn-x*,Ayn〉+αn2‖Ayn‖2)≤(1-βn)‖xn-x*‖2+βn(‖yn-x*‖2+αn(2‖yn-x*‖‖Ayn‖+‖Ayn‖2))≤(1-βn)‖xn-x*‖2+βn(‖yn-x*‖2+αnM),
where M>0 is some constant satisfying supn{2∥yn-x*∥∥Ayn∥+∥Ayn∥2}≤M. From (3.16) and (3.17), we have
(3.18)‖xn+1-x*‖2≤‖xn-x*‖2+λ(λ-2α)βn‖Bxn-Bx*‖2+αnM.
Thus,
(3.19)λ(2α-λ)βn‖Bxn-Bx*‖2≤‖xn-x*‖2-‖xn+1-x*‖2+αnM≤(‖xn-x*‖+‖xn+1-x*‖)‖xn+1-xn‖+αnM,
which imply that
(3.20)lim⁡n→∞‖Bxn-Bx*‖=0.Proof of (3.10).
SinceJR,λ is 1-inverse strongly monotone, we have
(3.21)‖yn-x*‖2=‖JR,λ(xn-λBxn)-JR,λ(x*-λBx*)‖2≤〈xn-λBxn-(x*-λBx*),yn-x*〉=12(‖xn-λBxn-(x*-λBx*)‖2+‖yn-x*‖2-‖xn-λBxn-(x*-λBx*)-(yn-x*)‖2)≤12(‖xn-x*‖2+‖yn-x*‖2-‖xn-yn-λ(Bxn-Bx*)‖2)=12(‖xn-x*‖2+‖yn-x*‖2-‖xn-yn‖2+2λ〈Bxn-Bx*,xn-yn〉-λ2‖Bxn-Bx*‖2),
which implies that
(3.22)‖yn-x*‖2≤‖xn-x*‖2-‖xn-yn‖2+2λ‖Bxn-Bx*‖‖xn-yn‖.
Substitute (3.22) into (3.17) to get
(3.23)‖xn+1-x*‖2≤‖xn-x*‖2-βn‖xn-yn‖2+2λ‖Bxn-Bx*‖‖xn-yn‖+αnM.
Then we derive
(3.24)βn‖xn-yn‖2≤(‖xn-x*‖+‖xn+1-x*‖)‖xn+1-xn‖+2λ‖Bxn-Bx*‖‖xn-yn‖+αnM.
So, we have
(3.25)lim⁡n→∞‖xn-yn‖=0.
We note thatPI(B,R)(I-A) is a contraction. As a matter of fact,
(3.26)‖PI(B,R)(I-A)x-PI(B,R)(I-A)y‖≤‖(I-A)x-(I-A)y‖≤‖I-A‖‖x-y‖≤(1-μ)‖x-y‖,
for all x,y∈H. Hence PI(B,R)(I-A) has a unique fixed point, say x̃∈I(B,R). That is x̃=PI(B,R)(I-A)(x̃). This implies that 〈Ax̃,y-x̃〉≥0 for all y∈I(B,R). Next, we prove that
(3.27)limsup⁡n→∞〈Ax̃,xn-x̃〉≥0.
First, we note that there exists a subsequence{xni} of {xn} such that
(3.28)limsup⁡n→∞〈Ax̃,xn-x̃〉=lim⁡j→∞〈Ax̃,xnj-x̃〉.
Since {xnj} is bounded, there exists a subsequence {xnji} of {xnj} which converges weakly to w. Without loss of generality, we can assume that xnj⇀w.
Next, we show thatw∈I(B,R). In fact, since B is α-inverse strongly monotone, B is Lipschitz-continuous monotone mapping. It follows from Lemma 2.1 that R+B is maximal monotone. Let (v,g)∈G(R+B), that is, g-Bv∈R(v). Again since yni=JR,λ(xni-λBxn-i), we have xni-λBxni∈(I+λR)(yni), that is, (1/λ)(xni-yni-λBxni)∈R(yni). By virtue of the maximal monotonicity of R+B, we have
(3.29)〈v-yni,g-Bv-1λ(xni-yni-λBxni)〉≥0,
and so
(3.30)〈v-yni,g〉≥〈v-yni,Bv+1λ(xni-yni-λBxni)〉=〈v-yni,Bv-Byni+Byni-Bxni+1λ(xni-yni)〉≥〈v-yni,Byni-Bxni〉+〈v-yni,1λ(xni-yni)〉.
It follows from ∥xn-yn∥→0, ∥Bxn-Byn∥→0 and yni⇀w that
(3.31)lim⁡ni→∞〈v-yni,g〉=〈v-w,g〉≥0.
It follows from the maximal monotonicity of B+R that θ∈(R+B)(w), that is, w∈I(B,R). Therefore, w∈I(B,R). It follows that
(3.32)limsup⁡n→∞〈Ax̃,xn-x̃〉=lim⁡j→∞〈Ax̃,xnj-x̃〉=〈Ax̃,w-x̃〉≥0.Proof of (3.11).
First, we note thatzn=PC[(I-αnA)yn], then for all x∈C, we have 〈zn-(I-αnA)yn,zn-x〉≤0. Thus,
(3.33)‖zn-x̃‖2=〈zn-x̃,zn-x̃〉=〈zn-(I-αnA)yn+(I-αnA)yn-x̃,zn-x̃〉=〈zn-(I-αnA)yn,zn-x̃〉+〈(I-αnA)yn-x̃,zn-x̃〉≤〈(I-αnA)(yn-x̃)-αnAx̃,zn-x̃〉=〈(I-αnA)(yn-x̃),zn-x̃〉+αn〈-Ax̃,zn-x̃〉≤‖(I-αnA)(yn-x̃)‖‖zn-x̃‖+αn〈-Ax̃,zn-x̃〉≤(1-αnμ)2(‖xn-x̃‖2+‖zn-x̃‖2)+αn〈-Ax̃,zn-x̃〉,
that is,
(3.34)‖zn-x̃‖2≤(1-αnμ)‖xn-x̃‖2+αn1+αnμ〈-Ax̃,zn-x̃〉.
So,
(3.35)‖xn+1-x̃‖2≤(1-βn)‖xn-x*‖2+βn‖zn-x*‖2≤(1-βn)‖xn-x*‖2+βn(1-αnμ)‖xn-x̃‖2+αnβn1+αnμ〈-Ax̃,zn-x̃〉=(1-αnβnμ)‖xn-x*‖2+αnβn1+αnμ〈-Ax̃,zn-x̃〉=(1-δn)‖xn-x*‖2+δnσn,
where δn=αnβnμ and σn=(1/(1+αnμ)μ)〈-Ax̃,zn-x̃〉. It is easy to see that ∑n=1∞δn=∞ and limsupn→∞σn≤0. Hence, by Lemma 2.3, we conclude that the sequence {xn} converges strongly to x̃. This completes the proof.


## 4. Conclusion

The results proved in this paper may be extended for multivalued variational inclusions and related optimization problems.



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