Iterative algorithms for variational inclusions, mixed equilibrium and fixed point problems with...

Cent. Eur. J. Math. • 9(3) • 2011 • 640-656DOI: 10.2478/s11533-011-0021-3

Central European Journal of Mathematics

Iterative algorithms for variational inclusions,mixed equilibrium and fixed point problemswith application to optimization problems

Research Article

Yonghong Yao1∗, Yeol Je Cho2† , Yeong-Cheng Liou3‡

1 Department of Mathematics, Tianjin Polytechnic University, Tianjin, China

2 Department of Mathematics Education and RINS, Gyeongsang National University, Chinju, Korea

3 Department of Information Management, Cheng Shiu University, Kaohsiung, Taiwan

Received 4 January 2010; accepted 4 February 2011

Abstract: In this paper, we introduce an iterative algorithm for finding a common element of the set of solutions of a mixedequilibrium problem, the set of fixed points of a nonexpansive mapping, and the the set of solutions of a variationalinclusion in a real Hilbert space. Furthermore, we prove that the proposed iterative algorithm converges stronglyto a common element of the above three sets, which is a solution of a certain optimization problem related to astrongly positive bounded linear operator.

MSC: 49J30, 47H10, 47H17, 49M05

Keywords: Variational inclusion • Mixed equilibrium problem • Fixed point • Optimization problem • Nonexpansive mapping •Strong convergence© Versita Sp. z o.o.

1. Introduction

Let C be a nonempty closed convex subset of a real Hilbert space H. Let F : C → H be a nonlinear mapping, φ : C → Ra function, and Θ a bifunction of C×C into R.∗ E-mail: [email protected]† E-mail: [email protected]‡ E-mail: [email protected]

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UnauthenticatedDownload Date | 3/21/16 11:08 AM

Y. Yao, Y.J. Cho, Y.-C. Liou

Consider the following mixed equilibrium problem: Find x∗ ∈ C such thatΘ(x∗, y) + φ(y)− φ(x∗) + 〈Fx∗, y− x∗〉 ≥ 0, y ∈ C. (1)

If F = 0, then (1) is the following mixed equilibrium problem: Find x∗ ∈ C such thatΘ(x∗, y) + φ(y)− φ(x∗) ≥ 0, y ∈ C ; (2)

it was considered by Ceng and Yao [7]. If φ = 0, then (1) becomes the following equilibrium problem: Find x∗ ∈ C suchthat Θ(x∗, y) + 〈Fx∗, y− x∗〉 ≥ 0, y ∈ C, (3)studied by Takahashi and Takahashi [42]. For φ = 0 and F = 0, (1) is the following equilibrium problem: Find x∗ ∈ Csuch that Θ(x∗, y) ≥ 0, y ∈ C. (4)At last, if Θ(x, y) = 0 for all x, y ∈ C , then (1) is the following variational inequality problem: Find x∗ ∈ C such that

φ(y)− φ(x∗) + 〈Fx∗, y− x∗〉 ≥ 0, y ∈ C. (5)The sets of solutions of (1)–(5) are denoted by EP(1)–EP(5), respectively.Mixed equilibrium problems include fixed point problems, optimization problems, variational inequality problems, Nashequilibrium problems and equilibrium problems as special cases; see, for example, [5, 11, 12, 28, 49]. Some methods havebeen proposed to solve mixed equilibrium problems and equilibrium problems; see, for example, [7–10, 15, 20, 23, 28, 35,40, 41, 47–50].In 1997, Combettes and Hirstoaga [15] introduced an iterative method of finding the best approximation to the initialdata and proved a strong convergence theorem. Subsequently, Takahashi and Takahashi [41] introduced another iterativescheme for finding a common element of the set of solutions of the equilibrium problem (2) and the set of fixed pointpoints of a nonexpansive mapping. Furthermore, Yao et al. [47, 48] introduced some new iterative schemes for findinga common element of the set of solutions of the equilibrium problem (2) and the set of common fixed points of finitely(infinitely) nonexpansive mappings.Very recently, Ceng and Yao [7] considered a new iterative scheme for finding a common element of the set of solutionsof a mixed equilibrium problem and the set of common fixed points of finitely many nonexpansive mappings. Peng andYao [34] developed the CQ method and obtained some strong convergence results for finding a common element of theset of solutions of the mixed equilibrium problem (1), the set of solutions of a variational inequality and the set of fixedpoints of a nonexpansive mapping. Their results extend and improve the corresponding results in [15, 31, 41, 47].Recall that a mapping f : C → C is called a ρ-contraction, ρ ∈ [0, 1), if

‖f(x)− f(y)‖ ≤ ρ‖x − y‖, x, y ∈ C.

A mapping T : C → C is said to be nonexpansive if‖Tx − Ty‖ ≤ ‖x − y‖, x, y ∈ C.

Denote the set of fixed points of T by Fix (T ). A mapping B : C → C is said to be β-inverse strongly monotone, β > 0,if〈Bx − By, x − y〉 ≥ β‖Bx − By‖2, x, y ∈ C.

Recall that a mapping A is strongly positive on H if there exists a constant µ > 0 such that〈Ax, x〉 ≥ µ‖x‖2, x ∈ H.

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Iterative algorithms for variational inclusions, mixed equilibrium and fixed point problems with application to optimization problems

Let B : H → H be a single-valued nonlinear mapping and R : H → 2H be a set-valued mapping. We are concerned withthe following variational inclusion: Find a point x ∈ H such thatθ ∈ B(x) + R(x), (6)

where θ is the zero vector in H. The set of solutions of the problem (6) is denoted by I(B,R).Remark 1.1.(1) If R = ∂φ : H → 2H in (6), where φ : H → R is a proper convex lower semi-continuous function and ∂φ is thesub-differential of φ, then the variational inclusion (6) is equivalent to the following problem: Find x ∈ H such that

〈Bx, v − x〉+ φ(y)− φ(x) ≥ 0, v, y ∈ H,

which is called the mixed quasi-variational inequality in Noor [32].(2) M = ∂δC in (6), where C is a nonempty closed convex subset of H and δC : H → [0,∞] is the indicator functionof C , i.e.,δC (x) = 0 if x ∈ C,+∞ otherwise,then the variational inclusion (6) is equivalent to the following problem: Find x ∈ H such that〈Bx, v − x〉 ≥ 0, v ∈ H.

This is the Hartman–Stampacchia variational inequality.Remark 1.2.(1) If H = Rm, then the problem (6) is the generalized equation introduced by Robinson [36].(2) If B = 0, then the problem (6) is the inclusion problem introduced by Rockafellar [37].It is known that the problem (6) provides a convenient framework for the unified study of optimal solutions in manyoptimization related areas including mathematical programming, complementarity, variational inequalities, optimal con-trol, mathematical economics, equilibria and game theory, etc. Also, various types of variational inclusions problemshave been extended and generalized by many authors (for more details, see [1–3, 13, 17–19, 21, 22, 24, 30, 43, 44] andreferences therein).Recently, Kocourek et al. [27] and Zhang et al. [50] introduced a new iterative scheme for finding a common element ofthe set of solutions to the problem (6) and the set of fixed points of nonexpansive, nonspreading and hybrid mappings inHilbert spaces. Peng et al. [33] introduced another iterative scheme by the viscosity approximate method for finding acommon element of the set of solutions of a variational inclusion with set-valued maximal monotone mapping and inversestrongly monotone mappings, the set of solutions of an equilibrium problem and the set of fixed points of a nonexpansivemapping.On the other hand, the following optimization problem has been studied extensively by many authors:

minx∈Ω′

[ν2 〈Ax, x〉+ 12 ‖x − u‖2 − h(x)] ,

where Ω′ = ⋂∞n=1 Cn, C1, C2, . . . are closed convex subsets of H such that ⋂∞n=1 Cn 6= ∅, u ∈ H, ν ≥ 0 is a real number,

A is a strongly positive bounded linear operator on H and h is a potential function for γf (i.e., h′(x) = γf(x) for allx ∈ H). For this kind of optimization problems, see, for example, Bauschke and Borwein [4], Combettes [14], Deutschand Yamada [16], and Xu [45] when Ω′ = ⋂N

n=1 Cn and h(x) = 〈x, b〉 for any b ∈ H.In this paper we introduce an iterative algorithm for finding a common element of the set of solutions of a mixedequilibrium problem, the set of fixed points of a nonexpansive mapping, and the set of solutions of a variational inclusionin a real Hilbert space. Furthermore, we prove that the proposed iterative algorithm converges strongly to a commonelement of the above three sets, which is a solution of the optimization problem related to a strongly positive boundedlinear operator.642



2. Preliminaries

Let H be a real Hilbert space with an inner product 〈·, ·〉 and norm ‖ · ‖. Let C be a nonempty closed convex subset ofH. Then, for any x ∈ H, there exists a unique nearest point in C , denoted by PC (x), such that

‖x − PC (x)‖ ≤ ‖x − y‖, y ∈ C.

Such a PC is called the metric projection of H onto C . We know that PC is nonexpansive. Further, for all x ∈ H andx∗ ∈ C ,

x∗ = PC (x) ⇐⇒ 〈x − x∗, x∗ − y〉 ≥ 0, y ∈ C.

A set-valued mapping T : 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 monotonemapping. It is known that a monotone mapping T is maximal if, and only if, for all (x, f) ∈ H×H, 〈x − y, f − g〉 ≥ 0 forall (y, g) ∈ G(T ), implies f ∈ Tx.Let a set-valued mapping R : H → 2H be maximal monotone. We define the resolvent operator JR,λ associated with Rand λ as follows: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, nonexpansiveand 1-inverse strongly monotone (see, for example, [6]) and, further, a solution of the problem (6) is a fixed point of theoperator JR,λ(I − λB) for all λ > 0; see, for example, [29].Throughout this paper, we assume that a bifunction Θ: H×H → R and a convex function φ : H → R satisfy the followingconditions:(H1) Θ(x, x) = 0 for all x ∈ H;(H2) Θ is monotone, i.e., Θ(x, y) + Θ(y, x) ≤ 0 for all x, y ∈ H;(H3) for any y ∈ H, x 7→ Θ(x, y) is weakly upper semi-continuous;(H4) for any x ∈ H, y 7→ Θ(x, y) is convex and lower semi-continuous;(H5) for any x ∈ H and r > 0, there exist a bounded subset Dx ⊂ H and yx ∈ H such that, for any z ∈ H \Dx ,

Θ(z, yx ) + φ(yx )− φ(z) + 1r 〈yx − z, z − x〉 < 0.

Lemma 2.1 ([34]).Let H be a real Hilbert space. Let Θ: H×H → R be a bifunction and φ : H → R be a proper lower semicontinuous andconvex function. For any r > 0 and x ∈ H, define a mapping Sr : H → H as follows: for all x ∈ H,

Sr(x) = z ∈ H : Θ(z, y) + φ(y)− φ(z) + 1r 〈y− z, z − x〉 ≥ 0, y ∈ H .

Assume that the conditions (H1)–(H5) hold. Then we have the following:(1) for each x ∈ H, Sr(x) 6= ∅ and Sr is single-valued;

(2) Sr is firmly nonexpansive, i.e., for any x, y ∈ H, ‖Srx − Sry‖2 ≤ 〈Srx − Sry, x − y〉;(3) Fix (Sr) = EP(1);(4) EP(1) is closed and convex.

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Lemma 2.2 ([39]).Let xnn and znn be bounded sequences in a Banach space X and βnn be a sequence in [0, 1] with

0 < lim infn→∞

βn ≤ lim supn→∞

βn < 1.Suppose that xn+1 = (1− βn)zn + βnxn for all n ≥ 0 and

lim supn→∞

(‖zn+1 − zn‖ − ‖xn+1 − xn‖) ≤ 0.

Then limn→∞‖zn − xn‖ = 0.

Let Tn∞n=1 be an infinite family of nonexpansive mappings Tn : H → H and λ1, λ2, . . . be real numbers such that0 ≤ λi ≤ 1 for all i ≥ 1. For any n ≥ 1, define a mapping Wn of C into H as follows:Un,n+1 = I,Un,n = λnTnUn,n+1 + (1− λn)I,

Un,n−1 = λn−1Tn−1Un,n + (1− λn−1)I,. . .

Un,k = λkTkUn,k+1 + (1− λk )I,Un,k−1 = λk−1Tk−1Un,k + (1− λk−1)I,

. . .Un,2 = λ2T2Un,3 + (1− λ2)I,Wn = Un,1 = λ1T1Un,2 + (1− λ1)I.

(7)

Such a Wn is usually called the W -mapping generated by Tn, Tn−1, . . . , T1 and λn, λn−1, . . . , λ1.We need the following lemmas for our main results.Lemma 2.3 ([38, 48]).Let H be a real Hilbert space. Let Tn∞n=1 be an infinite family of nonexpansive mappings Tn : H → H such that⋂∞n=1 Fix (Tn) 6= ∅. Let λ1, λ2, . . . be real numbers such that 0 < λi ≤ b < 1 for any i ∈ N. Then we have the following:(1) for all x ∈ H and k ∈ N, the limit lim

n→∞Un,kx exists;

(2) Fix (W ) = ⋂∞n=1 Fix (Tn) where Wx = limn→∞

Wnx = limn→∞

Un,1x for all x ∈ C ;

(3) for any bounded sequence xnn in H, limn→∞‖Wxn −Wnxn‖ = 0.

Lemma 2.4 ([6]).Let R : H → 2H be a maximal monotone mapping and B : H → H be a Lipschitz and continuous monotone mapping.Then the mapping R + B : H → 2H is maximal monotone.

Lemma 2.5 ([25, 26]).Let C be a nonempty closed convex subset of a real Hilbert space H and g : C → R ∪ ∞ be a proper lower-semicontinuous differentiable convex function. If x∗ is a solution to the minimization problem

g(x∗) = infx∈C

g(x),644



then〈g′(x), x − x∗〉 ≥ 0, x ∈ C.

In particular, if x∗ solves the optimization problem

minx∈C

[ν2 〈Ax, x〉+ 12 ‖x − u‖2 − h(x)] ,

then ⟨u+ (γf − (I + νA)) x∗, x − x∗⟩ ≤ 0.

Lemma 2.6 ([46]).Assume that ann is a sequence of nonnegative real numbers such that an+1 ≤ (1 − γn)an + δn for all n ≥ 1, whereγnn is a sequence in (0, 1) and δnn is a sequence such that

(i) ∞∑n=1 γn =∞,

(ii) lim supn→∞

δnγn ≤ 0 or

∞∑n=1 |δn| <∞.

Then limn→∞

an = 0.

3. Main results

In this section, we first give some assumptions on the relevant operators and parameters. Subsequently, we introduceour iterative algorithm for finding a common element of the set of solutions of a mixed equilibrium problem, the set offixed points of a nonexpansive mapping, and the set of solutions of a variational inclusion. Finally, we show that theproposed algorithm is strongly convergent.Let H be a real Hilbert space. Let φ : H → R be a lower semicontinuous and convex function and Θ: H×H → R bea bifunction satisfying the conditions (H1)–(H5). Let A be a strongly positive bounded linear operator with coefficientµ > 0 and R : H → 2H be a maximal monotone mapping. Let the mappings F,B : H → H be α- and β-inverse stronglymonotone, respectively. Let f : H → H be a ρ-contraction and Tn∞n=1 be an infinite family of nonexpansive mappingsTn : H → H. Let r > 0, γ > 0, and λ > 0 be three constants such that r < 2α , λ < 2β, and 0 < γ < (1+ν)µ

ρ .Now we introduce an iteration algorithm.Algorithm 3.1.For any fixed u ∈ H and x0 ∈ H, compute the sequences xnn and unn as follows: for any y ∈ H and n ≥ 1,

Θ(un, y) + φ(y)− φ(un) + 1r⟨y− un, un − (xn − rFxn)⟩ ≥ 0,

xn+1 = αn(u+ γf(xn)) + βnxn + [(1− βn)I − αn(I + νA)]Wn JR,λ(un − λBun), (8)where αnn, βnn are two real sequences in [0, 1] and Wn is the W -mapping defined by (7).Now, we study the strong convergence of the hybrid iterative algorithm (8) as follows.Theorem 3.2.Suppose that Ω = ⋂∞n=1 Fix (Tn) ∩ EP(1) ∩ I(B,R) 6= ∅ and the following conditions hold:

(i) limn→∞

αn = 0 and∞∑n=0 αn =∞;

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(ii) 0 < lim infn→∞

βn ≤ lim supn→∞

βn < 1.

Then the sequence xnn generated by (8) converges strongly to a point x∗ ∈ Ω which solves the following optimizationproblem: min

x∈Ω[ν2 〈Ax, x〉+ 12 ‖x − u‖2 − h(x)] , (OP)

where h is a potential function for γf .

Proof. Let p be an element of Ω, that is,Sr(p− rFp) = JR,λ(p− λBp) = Tn(p) = p, n ≥ 0.

We divide our proof into the following steps:(a) the sequences xnn and unn are bounded;(b) ‖xn+1 − xn‖ → 0 as n→∞;(c) ‖Fxn − Fp‖ → 0 and ‖Bun − Bp‖ → 0 as n→∞;(d) ‖xn −Wxn‖ → 0 as n→∞;(e) lim sup

n→∞

⟨u+ γf(x∗)− (I + νA) x∗, xn − x∗⟩ ≤ 0, where x∗ is a solution of (OP);

(f) xn → x∗ as n→∞.(a) From conditions (i) and (ii), we may assume, without loss of generality, that αn ≤ (1− βn)(1 + ν‖A‖)−1 for all n ≥ 1.Since A is a linear bounded self-adjoint operator on H, we have ‖A‖ = sup|〈Ax, x〉| : x ∈ H, ‖x‖ = 1. Observe that

⟨((1− βn)I − αn(I + νA))x, x⟩ = 1− βn − αn − αnν〈Ax, x〉 ≥ 1− βn − αn − αnν‖A‖ ≥ 0,that is, (1− βn)I − αn(I + νA) is positive. It follows that

∥∥(1− βn)I − αn(I + νA)∥∥ = sup⟨((1− βn)I − αn(I + νA))x, x⟩ : x ∈ H, ‖x‖ = 1= sup1− βn − αn − αnν〈Ax, x〉 : x ∈ H, ‖x‖ = 1 ≤ 1− βn − αn(1 + νµ).Since F is α-inverse strongly monotone and B is β-inverse strongly monotone, we have

∥∥(I − rF )x − (I − rF )y∥∥2 ≤ ‖x − y‖2 + r(r − 2α)‖Fx − Fy‖2and ∥∥(I − λB)x − (I − λB)y∥∥2 ≤ ‖x − y‖2 + λ(λ− 2β)‖Bx − By‖2. (9)It is clear that, if 0 ≤ r ≤ 2α and 0 ≤ λ ≤ 2β, then (I− rF ) and (I− λB) are all nonexpansive. Set yn = JR,λ(un− λBun)for all n ≥ 0. It follows that

‖yn − p‖ = ∥∥JR,λ(un − λBun)− JR,λ(p− λBp)∥∥ ≤ ∥∥(un − λBun)− (p− λBp)∥∥ ≤ ‖un − p‖.By Lemma 2.1, we have un = Sr(xn − rFxn) for all n ≥ 0. Then,

‖un − p‖2 = ∥∥Sr(xn − rFxn)− Sr(p− rFp)∥∥2 ≤ ∥∥xn − rFxn − (p− rFp)∥∥2≤ ‖xn − p‖2 + r(r − 2α)‖Fxn − Fp‖2 ≤ ‖xn − p‖2 (10)

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and so‖yn − p‖ ≤ ‖xn − p‖. (11)From (8) and (11), we deduce that

‖xn+1 − p‖ = ∥∥αn(u+ γf(xn)) + βnxn + [(1− βn)I − αn(I + νA)]Wnyn − p∥∥= ∥∥αnu+ αn

(γf(xn)− (I + νA)p)+ βn(xn − p) + [(1− βn)I − αn(I + νA)](Wnyn − p)∥∥

≤(1− βn − αn(1 + νµ))‖yn − p‖+ βn‖xn − p‖+ αn‖u‖+ αn

∥∥γf(xn)− (I + νA)p∥∥≤(1− αn(1 + νµ))‖xn − p‖+ αn‖u‖+ αnγ‖f(xn)− f(p)‖+ αn

∥∥γf(p)− (I + νA)p∥∥≤(1− αn(1 + νµ))‖xn − p‖+ αn‖u‖+ αnγρ‖xn − p‖+ αn

∥∥γf(p)− (I + νA)p∥∥= [1− ((1 + νµ)− γρ) αn]‖xn − p‖+ αn(∥∥γf(p)− (I + νA)p∥∥+ ‖u‖)

≤ max‖x0 − p‖,∥∥γf(p)− (I + νA)p∥∥+ ‖u‖(1 + νµ)− γρ

.

Therefore, xnn is bounded which implies that unn, ynn, Wnynn, AWnynn, and f(xn)n are all bounded.(b) Define xn+1 = βnxn + (1− βn)vn for all n ≥ 0. From the definition of vn, it follows thatvn+1 − vn = xn+2 − βn+1xn+11− βn+1 − xn+1 − βnxn1− βn= αn+1(u+ γf(xn+1)) + ((1− βn+1)I − αn+1(I + νA))Wn+1yn+11− βn+1

− αn(u+ γf(xn)) + ((1− βn)I − αn(I + νA))Wnyn1− βn= αn+11− βn+1 (u+ γf(xn+1))− αn1− βn (u+ γf(xn)) +Wn+1yn+1 −Wnyn

+ αn1− βn (I + νA)Wnyn −αn+11− βn+1 (I + νA)Wn+1yn+1

= αn+11− βn+1[u+ γf(xn+1)− (I + νA)Wn+1yn+1]+ αn1− βn [(I + νA)Wnyn − u− γf(xn)]

+Wn+1yn+1 −Wn+1yn +Wn+1yn −Wnyn.

It follows that‖vn+1 − vn‖ − ‖xn+1 − xn‖ ≤ αn+11− βn+1

(‖u‖+ ‖γf(xn+1)‖+ ∥∥(I + νA)Wn+1yn+1∥∥)

+ αn1− βn (∥∥(I + νA)Wnyn∥∥+ ‖u‖+ ‖γf(xn)‖)+ ‖Wn+1yn+1 −Wn+1yn‖+ ‖Wn+1yn −Wnyn‖ − ‖xn+1 − xn‖

≤ αn+11− βn+1(‖u‖+ ‖γf(xn+1)‖+ ∥∥(I + νA)Wn+1yn+1∥∥)+ αn1− βn (∥∥(I + νA)Wnyn

∥∥+ ‖u‖+ ‖γf(xn)‖)+ ‖Wn+1yn −Wnyn‖+ ‖yn+1 − yn‖ − ‖xn+1 − xn‖.

(12)

Since Ti and Un,i are nonexpansive, it follows from (7) that‖Wn+1yn −Wnyn‖ = ‖λ1T1Un+1,2yn − λ1T1Un,2yn‖

≤ λ1‖Un+1,2yn − Un,2yn‖ = λ1‖λ2T2Un+1,3yn − λ2T2Un,3yn‖≤ λ1λ2‖Un+1,3yn − Un,3yn‖ ≤ . . . ≤ λ1λ2 · · · λn‖Un+1,n+1yn − Un,n+1yn‖ ≤ M

n∏i=1 λi,

(13)

where M > 0 is a constant such thatsup‖Un+1,n+1yn − Un,n+1yn‖ : n ≥ 0 ≤ M.

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Note that‖yn+1 − yn‖ = ∥∥JR,λ(un+1 − λBun+1)− JR,λ(un − λBun)∥∥ ≤ ∥∥(un+1 − λBun+1)− (un − λBun)∥∥

≤ ‖un+1 − un‖ = ∥∥Sr(xn+1 − rFxn+1)− Sr(xn − rFxn)∥∥≤∥∥(xn+1 − rFxn+1)− (xn − rFxn)∥∥ ≤ ‖xn+1 − xn‖.

(14)Substituting (13) and (14) into (12), we get

‖vn+1 − vn‖ − ‖xn+1 − xn‖ ≤ αn+11− βn+1(‖u‖+ ‖γf(xn+1)‖+ ∥∥(I + νA)Wn+1yn+1∥∥)

+ αn1− βn (∥∥(I + νA)Wnyn∥∥+ ‖u‖+ ‖γf(xn)‖)+M

n∏i=1 λi,

what implies that lim supn→∞

(‖vn+1 − vn‖ − ‖xn+1 − xn‖) ≤ 0.

Hence, by Lemma 2.2, we have limn→∞‖vn − xn‖ = 0 and, consequently,limn→∞‖xn+1 − xn‖ = lim

n→∞(1− βn) ‖vn − xn‖ = 0. (15)

(c) We can write (8) asxn+1 = αn

(u+ γf(xn)− (I + νA)Wnyn

)+ βn (xn −Wnyn) +Wnyn.

It follows that‖xn −Wnyn‖ ≤ ‖xn − xn+1‖+ ‖xn+1 −Wnyn‖

≤ ‖xn − xn+1‖+ αn∥∥u+ γf(xn)− (I + νA)Wnyn

∥∥+ βn‖xn −Wnyn‖.

That is,‖xn −Wnyn‖ ≤

11− βn ‖xn − xn+1‖+ αn1− βn ∥∥u+ γf(xn)− (I + νA)Wnyn∥∥.

This, together with αn → 0 and (15), implies thatlimn→∞‖xn −Wnyn‖ = 0. (16)

From (9) and (10) follows‖yn − p‖2 = ∥∥JR,λ(un − λBun)− JR,λ(p− λBp)∥∥2 ≤ ∥∥(un − λBun)− (p− λBp)∥∥2

≤ ‖un − p‖2 + λ(λ− 2β) ‖Bun − Bp‖2≤ ‖xn − p‖2 + r(r − 2α) ‖Fxn − Fp‖2 + λ(λ− 2β) ‖Bun − Bp‖2. (17)

Thus, by (8), we obtain‖xn+1 − p‖2 = ∥∥αn(u+ γf(xn)− (I + νA)p)+ βn(xn −Wnyn) + (I − αn(I + νA)) (Wnyn − p)∥∥2

≤∥∥(I − αn(I + νA))(Wnyn − p) + βn(xn −Wnyn)∥∥2 + 2αn⟨u+ γf(xn)− (I + νA)p, xn+1 − p⟩

≤[∥∥I − αn(I + νA)∥∥ ‖yn − p‖+ βn‖xn −Wnyn‖

]2 + 2αn∥∥u+ γf(xn)− (I + νA)p∥∥ ‖xn+1 − p‖≤[(1− αn (1 + νµ))‖yn − p‖+ βn‖xn −Wnyn‖

]2 + 2αn∥∥u+ γf(xn)− (I + νA)p∥∥ ‖xn+1 − p‖= (1− αn(1 + νµ))2‖yn − p‖2 + β2n‖xn −Wnyn‖2 + 2(1− αn(1 + νµ))βn‖yn − p‖ ‖xn −Wnyn‖+ 2αn∥∥u+ γf(xn)− (I + νA)p∥∥ ‖xn+1 − p‖.

(18)

648



From (17) and (18), it follows that‖xn+1 − p‖2 ≤ ‖xn − p‖2 + r(r − 2α)‖Fxn − Fp‖2 + λ(λ− 2β) ‖Bun − Bp‖2 + β2

n‖xn −Wnyn‖2+ 2(1− αn(1 + νµ))βn‖yn − p‖ ‖xn −Wnyn‖+ 2αn∥∥u+ γf(xn)− (I + νA)p∥∥ ‖xn+1 − p‖,and so

r(2α − r)‖Fxn − Fp‖2 + λ(2β − λ)‖Bun − Bp‖2 ≤ ‖xn − p‖2 − ‖xn+1 − p‖2 + β2n‖xn −Wnyn‖2+ 2(1− αn(1 + νµ))βn‖yn − p‖ ‖xn −Wnyn‖+ 2αn∥∥u+ γf(xn)− (I + νA)p∥∥ ‖xn+1 − p‖

≤(‖xn − p‖+ ‖xn+1 − p‖)‖xn+1 − xn‖+ β2

n‖xn −Wnyn‖2+ 2(1− αn(1 + νµ))βn‖yn − p‖ ‖xn −Wnyn‖+ 2αn∥∥u+ γf(xn)− (I + νA)p∥∥ ‖xn+1 − p‖.(19)

Therefore, by (15), (16), and (19), we deduce thatlimn→∞‖Fxn − Fp‖ = 0, lim

n→∞‖Bun − Bp‖ = 0.

(d) Since Sr is firmly nonexpansive, we have‖un − p‖2 = ∥∥Sr(xn − rFxn)− Sr(p− rFp)∥∥2 ≤ ⟨xn − rFxn − (p− rFp), un − p⟩

= 12(∥∥xn − rFxn − (p− rFp)∥∥2 + ‖un − p‖2 − ∥∥xn − rFxn − (p− rFp)− (un − p)∥∥2)≤ 12(‖xn − p‖2 + ‖un − p‖2 − ∥∥xn − un − r(Fxn − Fp)∥∥2)= 12(‖xn − p‖2 + ‖un − p‖2 − ‖xn − un‖2 + 2r 〈Fxn − Fp, xn − un〉 − r2‖Fxn − Fp‖2),

which implies that‖un − p‖2 ≤ ‖xn − p‖2 − ‖xn − un‖2 + 2r ‖Fxn − Fp‖ ‖xn − un‖. (20)

Since JR,λ is 1-inverse strongly monotone, we have‖yn − p‖2 = ∥∥JR,λ(un − λBun)− JR,λ(p− λBp)∥∥2 ≤ ⟨un − λBun − (p− λBp), yn − p⟩

= 12(‖un − λBun − (p− λBp)‖2 + ‖yn − p‖2 − ∥∥un − λBun − (p− λBp)− (yn − p)∥∥2)≤ 12(‖un − p‖2 + ‖yn − p‖2 − ∥∥un − yn − λ (Bun − Bp)∥∥2)= 12(‖un − p‖2 + ‖yn − p‖2 − ‖un − yn‖2 + 2λ 〈Bun − Bp, un − yn〉 − λ2‖Bun − Bp‖2).

This implies that‖yn − p‖2 ≤ ‖un − p‖2 − ‖un − yn‖2 + 2λ ‖Bun − Bp‖ ‖un − yn‖. (21)

Thus, by (20) and (21), we obtain‖yn − p‖2 ≤ ‖xn − p‖2 − ‖xn − un‖2 + 2r ‖Fxn − Fp‖ ‖xn − un‖ − ‖un − yn‖2 + 2λ ‖Bun − Bp‖ ‖un − yn‖. (22)

649



Substituting (22) into (18), we have‖xn+1 − p‖2 ≤ (1− αn(1 + νµ))2‖yn − p‖2 + β2

n‖xn −Wnyn‖2+ 2(1− αn(1 + νµ))βn‖yn − p‖ ‖xn −Wnyn‖+ 2αn∥∥u+ γf(xn)− (I + νA)p∥∥ ‖xn+1 − p‖≤(1− αn(1 + νµ))2 ‖xn − p‖2 − ‖xn − un‖2 + 2r ‖Fxn − Fp‖ ‖xn − un‖

− ‖un − yn‖2 + 2λ ‖Bun − Bp‖ ‖un − yn‖+ 2(1− αn(1 + νµ))βn‖yn − p‖ ‖xn −Wnyn‖+ 2αn∥∥u+ γf(xn)− (I + νA)p∥∥ ‖xn+1 − p‖+ β2n ‖xn −Wnyn‖2

and hence(1− αn(1 + νµ))2(‖xn − un‖2 + ‖un − yn‖2)≤(‖xn − p‖+ ‖xn+1 − p‖) ‖xn+1 − xn‖+ 2r ‖Fxn − Fp‖ ‖xn − un‖+ 2λ ‖Bun − Bp‖ ‖un − yn‖+ 2 (1− αn(1 + νµ))βn ‖yn − p‖ ‖xn −Wnyn‖+ 2αn ∥∥u+ γf(xn)− (I + νA)p∥∥ ‖xn+1 − p‖+ β2

n‖xn −Wnyn‖2.Thus, we have lim

n→∞‖xn − un‖ = 0, lim

n→∞‖un − yn‖ = 0.

With the observation that‖Wnyn − yn‖ ≤ ‖Wnyn − xn‖+ ‖xn − un‖+ ‖un − yn‖,we have

‖Wnyn − yn‖ → 0, n→∞.

Note that ‖Wyn−yn‖ ≤ ‖Wyn−Wnyn‖+‖Wnyn−yn‖. From this and Lemma 2.3, it follows that limn→∞‖Wyn−yn‖ = 0.Thus, lim

n→∞‖xn −Wxn‖ = 0.

(e) Next, we show that lim supn→∞

⟨u+ γf(x∗)− (I + νA) x∗, xn − x∗⟩ ≤ 0,

where x∗ is a solution of (OP). First, note that there exists a subsequence xnj j of xnn such thatlim supn→∞

⟨u+ (γf − (I + νA)) x∗, xn − x∗⟩ = lim

j→∞

⟨u+ (γf − (I + νA)) x∗, xnj − x∗⟩.

Since xnj j is bounded, there exists a subsequence xnjii of xnj j which converges weakly to w. Without loss ofgenerality, we assume that xnj j converges weakly to the point w. Since ‖Wxn − xn‖ → 0, it follows from the demi-closed principle of nonexpansive mappings that w ∈ Fix (W ).Now, we show that w ∈ EP(1). By un = Sr(xn − rFxn), it follows thatΘ(un, y) + φ(y)− φ(un) + 1

r⟨y− un, un − (xn − rFxn)⟩ ≥ 0, y ∈ H.

From (H2), we haveφ(y)− φ(un) + 1

r⟨y− un, un − (xn − rFxn)⟩ ≥ Θ(y, un), y ∈ H,

and henceφ(y)− φ(uni ) +⟨y− uni , uni − (xni − rFxni )

r

⟩≥ Θ(y, uni ), y ∈ H. (23)

650



For any t ∈ (0, 1] and y ∈ H, let yt = ty+ (1− t)w. From (23), we have〈yt − uni , Fyt〉 ≥ 〈yt − uni , Fyt〉 − φ(yt) + φ(uni )−⟨yt − uni , uni − (xni − rFxni )

r

⟩+ Θ(yt , uni )= ⟨yt − uni , Fyt − Funi⟩+ ⟨yt − uni , Funi − Fxni⟩− φ(yt) + φ(uni )−⟨yt − uni ,

uni − xnir

⟩+ Θ(yt , uni ).Since ‖uni − xni‖ → 0, we have ‖Funi − Fxni‖ → 0. Further, from the inverse strong monotonicity of F , we have〈yt −uni , Fyt −Funi〉 ≥ 0 and so, from (H4) and the weak lower semi-continuity of φ, uni−xni

r → 0 and uni → w weakly,〈yt − w, Fyt〉 ≥ −φ(yt) + φ(w) + Θ(yt , w). (24)

Thus, from (H1), (H4) and (24), we also have0 = Θ(yt , yt) + φ(yt)− φ(yt) ≤ tΘ(yt , y) + (1− t) Θ(yt , w) + tφ(y) + (1− t)φ(w)− φ(yt)= t

[Θ(yt , y) + φ(y)− φ(yt)]+ (1− t)[Θ(yt , w) + φ(w)− φ(yt)]≤ t[Θ(yt , y) + φ(y)− φ(yt)]+ (1− t)〈yt − w, Fyt〉 = t

[Θ(yt , y) + φ(y)− φ(yt)]+ (1− t)t〈y− w, Fyt〉and hence 0 ≤ Θ(yt , y) + φ(y)− φ(yt) + (1− t)〈y− w, Fyt〉.Letting t → 0, we have, for any y ∈ H,

Θ(w, y) + φ(y)− φ(w) + 〈y− w, Fw〉 ≥ 0,which implies that w ∈ EP(1).Next, we show that w ∈ I(B,R)). In fact, since B is β-inverse strongly monotone, B is Lipschitz continuous monotonemapping. It follows from Lemma 2.4 that R +B is maximal monotone. Let (v, g) ∈ G(R +B), i.e., g−Bv ∈ R(v). Again,since yni = JR,λ(uni − λBun−i), we have uni − λuni ∈ (I + λR)(yni ), i.e., 1

λ (uni − yni − λBuni ) ∈ R(yni ). By virtue of themaximal monotonicity of R + B, we have⟨v − yni , g− Bv −

1λ (uni − yni − λBuni )⟩ ≥ 0

and so〈v − yni , g〉 ≥

⟨v − yni , Bv + 1

λ (uni − yni − λBuni )⟩ = ⟨v − yni , Bv − Byni + Byni − Buni + 1λ (uni − yni )⟩

≥ 〈v − yni , Byni − Buni〉+⟨v − yni , 1λ (uni − yni )⟩ .It follows from ‖un − yn‖ → 0, ‖Bun − Byn‖ → 0 and yni → w weakly that

limni→∞〈v − yni , g〉 = 〈v − w, g〉 ≥ 0.

Since B + R is maximal monotone, we have θ ∈ (R + B)(w), i.e., w ∈ I(B,R). Therefore, it follows that w ∈ Ω andlim supn→∞

⟨u+ (γf − (I + νA)) x∗, xn − x∗⟩ = lim

j→∞

⟨u+ (γf − (I + νA)) x∗, xnj − x∗⟩

= ⟨u+ (γf − (I + νA)) x∗, w − x∗⟩ ≤ 0.651



(f) From (8), we have‖xn+1 − x∗‖2 = ∥∥αn(u+ γf(xn)− (I + νA)x∗)+ βn(xn − x∗) + ((1− βn)I − αn(I + νA)) (Wnyn − x∗)∥∥2

≤∥∥βn(xn − x∗) + ((1− βn)I − αn(I + νA))(Wnyn − x∗)∥∥2 + 2αn⟨u+ γf(xn)− (I + νA) x∗, xn+1 − x∗⟩

≤[∥∥((1− βn)I − αn(I + νA))(Wnyn − x∗)∥∥+ ‖βn(xn − x∗)‖]2 + 2αnγ ⟨f(xn)− f(x∗), xn+1 − x∗⟩+ 2αn ⟨u+ γf(x∗)− (I + νA) x∗, xn+1 − x∗⟩

≤[(1− βn − αn(1 + ν) µ) ‖yn − x∗‖+ βn‖xn − x∗‖

]2 + 2αnγρ ‖xn − x∗‖ ‖xn+1 − x∗‖+ 2αn ⟨u+ γf(x∗)− (I + νA) x∗, xn+1 − x∗⟩≤(1− αn(1 + ν) µ)2‖xn − x∗‖2 + αnγρ

‖xn − x∗‖2 + ‖xn+1 − x∗‖2+ 2αn ⟨u+ γf(x∗)− (I + νA) x∗, xn+1 − x∗⟩,

that is,‖xn+1 − x∗‖2 ≤ 1− 2αn(1 + ν)µ + α2

n(1 + ν)2µ2 + αnγρ1− αnγρ ‖xn − x∗‖2 + 2αn1− αnγρ ⟨u+ γf(x∗)− (I + νA) x∗, xn+1 − x∗⟩= [1− 2 ((1 + ν)µ − γρ) αn1− αnγρ

]‖xn − x∗‖2 + (

αn(1 + ν)µ)21− αnγρ ‖xn − x∗‖2

+ 2αn1− αnγρ ⟨u+ γf(x∗)− (I + νA) x∗, xn+1 − x∗⟩≤[1− 2 ((1 + ν)µ − γρ) αn1− αnγρ

]‖xn − x∗‖2

+ 2 ((1 + ν)µ − γρ) αn1− αnγρ

αn(1 + ν)µ2M12 ((1 + ν)µ − γρ) + 1(1 + ν)µ − γρ ⟨u+ γf(x∗)− (I + νA) x∗, xn+1 − x∗⟩= (1− δn) ‖xn − x∗‖2 + δnσn,

whereM1 = sup‖xn − x∗‖2 : n ≥ 1, δn = 2 ((1 + ν)µ − γρ) αn1− αnγρ ,

σn = αn(1 + ν)µ2M12 ((1 + ν)µ − γρ) + 1(1 + ν)µ − γρ ⟨u+ γf(x∗)− (I + νA) x∗, xn+1 − x∗⟩.It is easy to see that ∑∞

n=1 δn =∞ and lim supn→∞ σn ≤ 0. Hence, by Lemma 2.6, we conclude that the sequence xnnconverges strongly to x∗. This completes the proof.From Theorem 3.2, we can easily deduce the following.Corollary 3.3.For any x0 ∈ H, let the sequences xnn and unn in H be generated iteratively as follows: for any y ∈ H and n ≥ 1,Θ(un, y) + φ(y)− φ(un) + 1

r⟨y− un, un − (xn − rFxn)⟩ ≥ 0,

xn+1 = αn(u+ γf(xn)) + βnxn + (1− αn − βn)WnJR,λ(un − λBun),where αnn, βnn are two real sequences in [0, 1] and Wn is the W -mapping defined by (7). Suppose that Ω =⋂∞n=1 Fix (Tn) ∩ EP(1) ∩ I(B,R) 6= ∅ and conditions (i)–(ii) of Theorem 3.2 hold. Then the sequence xnn converges

strongly to a point x∗ ∈ Ω, which solves the following variational inequality:⟨u+ γf(x∗)− x∗, y− x⟩ ≤ 0, y ∈ Ω.

652



Corollary 3.4.For any x0 ∈ H, let the sequences xnn and unn in H be generated iteratively as follows: for any y ∈ H and n ≥ 1,Θ(un, y) + φ(y)− φ(un) + 1

r 〈y− un, un − xn〉 ≥ 0,xn+1 = αnγf(xn) + βnxn + [(1− βn)I − αnA]WnJR,λ(un − λBun),

where αnn, βnn are two real sequences in [0, 1] and Wn is the W -mapping defined by (7). Suppose that Ω =⋂∞n=1 Fix (Tn) ∩ EP(2) ∩ I(B,R) 6= ∅ and conditions (i)–(ii) of Theorem 3.2 hold. Then the sequence xnn converges

strongly to a point x∗ ∈ Ω, where x∗ = PΩ(γf(x∗) + (I − A) x∗) is a solution of the following variational inequality:⟨γf(x)− Ax, y− x⟩ ≤ 0, y ∈ Ω. (25)

Corollary 3.5.For any x0 ∈ H, let the sequences xnn and unn in H be generated iteratively as follows: for any y ∈ H and n ≥ 1,Θ(un, y) + φ(y)− φ(un) + 1

r⟨y− un, un − (xn − rFxn)⟩ ≥ 0,

xn+1 = αnγf(xn) + βnxn + [(1− βn)I − αnA]JR,λ(un − λBun),where αnn and βnn are two real sequences in [0, 1]. Suppose that Ω = EP(1)∩ I(B,R) 6= ∅ and conditions (i)–(ii) ofTheorem 3.2 hold. Then the sequence xnn converges strongly to a point x∗ ∈ Ω, where x∗ = PΩ(γf(x∗) + (I − A) x∗)is a solution of the variational inequality (25).Corollary 3.6.For any x0 ∈ H, let the sequences xnn and unn in H be generated iteratively as follows: for any y ∈ H and n ≥ 1,Θ(un, y) + 1

r⟨y− un, un − (xn − rFxn)⟩ ≥ 0,

xn+1 = αnγf(xn) + βnxn + [(1− βn)I − αnA]WnJR,λ(un − λBun),where αnn, βnn are two real sequences in [0, 1] and Wn is the W -mapping defined by (7). Suppose that Ω =⋂∞n=1 Fix (Tn) ∩ EP(3) ∩ I(B,R) 6= ∅ and conditions (i)–(ii) of Theorem 3.2 hold. Then the sequence xnn converges

strongly to a point x∗ ∈ Ω, where x∗ = PΩ(γf(x∗) + (I − A) x∗) is a solution of the variational inequality (25).Corollary 3.7.For any x0 ∈ H, compute the sequences xnn and unn in H as follows: for any y ∈ H and n ≥ 1,Θ(un, y) + 1

r⟨y− un, un − (xn − rFxn)⟩ ≥ 0,

xn+1 = αnγf(xn) + βnxn + [(1− βn)I − αnA]JR,λ(un − λBun),where αnn and βnn are two real sequences in [0, 1]. Suppose that Ω = EP(3) ∩ I(B,R) 6= ∅ and conditions (i)–(ii)of Theorem 3.2 hold. Then the sequence xnn converges strongly to a point x∗ ∈ Ω, where x∗ = PΩ(γf(x∗) + (I − A)x∗)is a solution of the variational inequality (25).Corollary 3.8.For any x0 ∈ H, compute the sequences xnn and unn in H as follows: for any y ∈ H and n ≥ 1,Θ(un, y) + 1

r⟨y− un, un − (xn − rFxn)⟩ ≥ 0,

xn+1 = αnf(xn) + βnxn + (1− αn − βn)JR,λ(un − λBun),where αnn and βnn are two real sequences in [0, 1]. Suppose that Ω = EP(3) ∩ I(B,R) 6= ∅ and conditions (i)–(ii)of Theorem 3.2 hold. Then the sequence xnn converges strongly to a point x∗ ∈ Ω, where x∗ = PΩf(x∗) is a solutionof the following variational inequality:

〈f(x)− x, y− x〉 ≤ 0, y ∈ Ω.653



Acknowledgements

The first author was supported in part by Colleges and Universities Science and Technology Development Foundation(20091003) of Tianjin and NSFC 11071279. The second author was supported by the Korea Research FoundationGrant funded by the Korean Government (KRF-2008-313-C00050). The third author was supported in part by NSC99-2221-E-230-006.

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