Available online at www.tjnsa.com J. Nonlinear Sci. Appl. 9 (2016), 61–74 Research Article Algorithms for the variational inequalities and fixed point problems Yaqiang Liua , Zhangsong Yaob,∗, Yeong-Cheng Liouc , Li-Jun Zhud a School of Management, Tianjin Polytechnic University, Tianjin 300387, China. b School of Information Engineering, Nanjing Xiaozhuang University, Nanjing 211171, China. c Department of Information Management, Cheng Shiu University, Kaohsiung 833, Taiwan and Center for General Education, Kaohsiung Medical University, Kaohsiung 807, Taiwan. d School of Mathematics and Information Science, Beifang University of Nationalities, Yinchuan 750021, China. Communicated by Yeol Je Cho Abstract A system of variational inequality and fixed point problems is considered. Two algorithms have been constructed. Our algorithms can find the minimum norm solution of this system of variational inequality c and fixed point problems. 2016 All rights reserved. Keywords: Variational inequality, monotone mapping, nonexpansive mapping, fixed point, minimum norm. 2010 MSC: 47H05, 47H10, 47H17. 1. Introduction Variational inequality problems were initially studied by Stampacchia [17] in 1964. Variational inequalities have applications in diverse disciplines such as physical, optimal control, optimization, mathematical programming, mechanics and finance, see [12], [16], [17], [29] and the references therein. The main purpose of this paper is devoted to find the minimum norm solution of some system of variational inequality and fixed point problems. Let H be a real Hilbert space with inner product h·, ·i and norm k · k, respectively. Let C be a nonempty closed convex subset of H. ∗ Corresponding author Email addresses: yani3115791@126.com (Yaqiang Liu), yaozhsong@163.com (Zhangsong Yao), simplex_liou@hotmail.com (Yeong-Cheng Liou), zhulijun1995@sohu.com (Li-Jun Zhu) Received 2015-05-12 Y. Liu, Z. Yao, Y. C. Liou, L. J. Zhu, J. Nonlinear Sci. Appl. 9 (2016), 61–74 62 Definition 1.1. A mapping F : C → H is called ζ-inverse strongly monotone if there exists a real number ζ > 0 such that hFx − Fy, x − yi ≥ ζkFx − Fyk2 , ∀x, y ∈ C. Definition 1.2. A mapping R : C → H is called κ-contraction, if there exists a constant κ ∈ [0, 1) such that kR(x) − R(y)k ≤ κkx − yk for all x, y ∈ C. Definition 1.3. A mapping N : C → C is said to be nonexpansive if kN x − N yk ≤ kx − yk, ∀x, y ∈ C. We use F ix(N ) to denote the set of fixed points of N . Definition 1.4. We call P rojC : H → C is the metric projection if P rojC : H → C assigns to each point x ∈ C the unique point P rojC x ∈ C satisfying the property kx − P rojC xk = inf kx − yk =: d(x, C). y∈C Let F : C → H be a nonlinear mapping. Recall that the classical variational inequality is to find x∗ ∈ C such that hFx∗ , x − x∗ i ≥ 0 for all x ∈ C. (1.1) The set of solutions of the variational inequality (1.1) is denoted by V I(F, C). The variational inequality problem has been extensively studied in the literature. Related works, please see, e.g. [1]-[11], [13], [15], [19]-[28], [30]-[34] and the references therein. For finding an element of F ix(N ) ∩ V I(F, C), Takahashi and Toyoda [19] introduced the following iterative scheme: xn+1 = ζn xn + (1 − ζn )N P rojC (xn − ηn Fxn ), n ≥ 0, (1.2) where P rojC is the metric projection of H onto C, {ζn } is a sequence in (0, 1), and {ηn } is a sequence in (0, 2ζ). Takahashi and Toyoda showed that the sequence {xn } converges weakly to some z ∈ F ix(N ) ∩ V I(F, C). Consequently, Nadezhkina and Takahashi [11] and Zeng and Yao [34] proposed some so-called extragradient methods for finding a common element of the set of fixed points of a nonexpansive mapping and the set of solutions of a variational inequality problem. Recently, Ceng, Wang and Yao [3] considered a general system of variational inequality of finding x∗ ∈ C such that ( hηFy ∗ + x∗ − y ∗ , x − x∗ i ≥ 0, ∀x ∈ C, (SV I) hξGx∗ + y ∗ − x∗ , x − y ∗ i ≥ 0, ∀x ∈ C, where F, G : C → H are two nonlinear mappings, y ∗ = P rojC (x∗ −ξGx∗ ), η > 0 and ξ > 0 are two constants. The solutions set of SVI is denoted by Ω. If take F = G, then SVI reduces to finding x∗ ∈ C such that ( hηFy ∗ + x∗ − y ∗ , x − x∗ i ≥ 0, ∀x ∈ C, hξFx∗ + y ∗ − x∗ , x − y ∗ i ≥ 0, ∀x ∈ C, which is introduced by Verma [20] (see also Verma [21]). Further, if we add up the requirement that x∗ = y ∗ , then SVI reduces to the classical variational inequality problem (1.1). For finding an element of F ix(N ) ∩ Ω, Ceng, Wang and Yao [3] introduced the following relaxed extragradient method: ( n y = P rojC (xn − ξGxn ), (1.3) xn+1 = ζn u + βn xn + γn N P rojC (y n − ηFy n ), n ≥ 0. Y. Liu, Z. Yao, Y. C. Liou, L. J. Zhu, J. Nonlinear Sci. Appl. 9 (2016), 61–74 63 They proved the strong convergence of the above method to some element in F ix(N ) ∩ Ω. On the other hand, in many problems, it is needed to find a solution with minimum norm. A typical example is the least-squares solution to the constrained linear inverse problem, see [14]. It is our purpose in this paper that we construct two methods, one implicit and one explicit, to find the minimum norm element in F ix(N ) ∩ Ω; namely, the unique solution x∗ to the quadratic minimization problem: x∗ = arg kxk2 . min x∈F ix(N )∩Ω We obtain two strong convergence theorems. 2. Preliminaries Let C be a nonempty closed convex subset of H. The following lemmas are useful for our main results. Lemma 2.1. Given x ∈ H and z ∈ C. (i) z = P rojC x if and only if there holds the relation: hx − z, y − zi ≤ 0 for all y ∈ C. (ii) z = P rojC x if and only if there holds the relation: kx − zk2 ≤ kx − yk2 − ky − zk2 for all y ∈ C. (iii) There holds the relation hP rojC x − P rojC y, x − yi ≥ kP rojC x − P rojC yk2 for all x, y ∈ H. Consequently, P rojC is nonexpansive and monotone. Lemma 2.2 ([3]). Let C be a nonempty closed convex subset of a real Hilbert space H. Let the mapping F : C → H be ζ-inverse strongly monotone. Then, we have k(I − ηF)x − (I − ηF)yk2 ≤ kx − yk2 + η(η − 2ζ)kFx − Fyk2 , ∀x, y ∈ C. In particular, if 0 ≤ η ≤ 2ζ, then I − ηF is nonexpansive. Lemma 2.3 ([3]). x∗ is a solution of SVI if and only if x∗ is a fixed point of the mapping U : C → C defined by U(x) = P rojC [P rojC (x − ξGx) − ηFP rojC (x − ξGx)], ∀x ∈ C, where y ∗ = P rojC (x∗ − ξGx∗ ). In particular, if the mappings F, G : C → H are ζ-inverse strongly monotone and δ-inverse strongly monotone, respectively, then the mapping U is a nonexpansive mapping provided η ∈ (0, 2ζ) and ξ ∈ (0, 2δ). Lemma 2.4 ([18]). Let {xn } and {y n } be bounded sequences in a Banach space X and let {δn } be a sequence in [0, 1] with 0 < lim inf δn ≤ lim sup δn < 1. Suppose xn+1 = (1 − δn )y n + δn xn for all integers n ≥ 0 and n→∞ n→∞ lim sup(ky n+1 − y n k − kxn+1 − xn k) ≤ 0. Then, lim ky n − xn k = 0. n→∞ n→∞ Lemma 2.5 ([10]). Let C be a closed convex subset of a real Hilbert space H and let N : C → C be a nonexpansive mapping. Then, the mapping I − N is demiclosed. That is, if {xn } is a sequence in C such that xn → x∗ weakly and (I − N )xn → y strongly, then (I − N )x∗ = y. Y. Liu, Z. Yao, Y. C. Liou, L. J. Zhu, J. Nonlinear Sci. Appl. 9 (2016), 61–74 64 Lemma 2.6 ([23]). Assume {an } is a sequence of nonnegative real numbers such that an+1 ≤ (1 − γn )an + δn γn , where {γn } is a sequence in (0, 1) and {δn } is a sequence such that P∞ (1) n=1 γn = ∞; P (2) lim supn→∞ δn ≤ 0 or ∞ n=1 |δn γn | < ∞. Then limn→∞ an = 0. 3. Main results In this section we will introduce two schemes for finding the unique point x∗ which solves the quadratic minimization kx∗ k2 = min kxk2 . (3.1) x∈F ix(N )∩Ω Let C be a nonempty closed convex subset of a real Hilbert space H. Let R : C → H be a κ-contraction. Let the mappings F, G : C → H be ζ-inverse strongly monotone and δ-inverse strongly monotone, respectively. Suppose η ∈ (0, 2ζ) and ξ ∈ (0, 2δ). Let N : C → C be a nonexpansive mapping. For each t ∈ (0, 1), we study the following mapping T t given by T t x = N P rojC [tR(x) + (1 − t)P rojC (I − ηF)P rojC (I − ξG)x], ∀x ∈ C. Since the mappings N , P rojC , I − ηF and I − ξG are nonexpansive, we can check easily that kT t x − T t yk ≤ [1 − (1 − κ)t]kx − yk which implies that T t is a contraction. Then there exists a unique fixed point xt of T t in C such that t t t z = P rojC (x − ξGx ), y t = P rojC (z t − ηFz t ), t t (3.2) t x = N P rojC [tR(x ) + (1 − t)y ]. In particular, if we take R ≡ 0, then (3.2) reduces to t t t z = P rojC (x − ξGx ), y t = P rojC (z t − ηFz t ), t x = N P rojC [(1 − t)y t ]. (3.3) We next prove that the implicit methods (3.2) and (3.3) both converge. Theorem 3.1. Suppose Γ := F ix(N ) ∩ Ω 6= ∅. Then the net {xt } generated by the implicit method (3.2) converges in norm, as t → 0+ , to the unique solution x∗ of the following variational inequality x∗ ∈ Γ, h(I − R)x∗ , x − x∗ i ≥ 0, x ∈ Γ. (3.4) In particular, if we take R = 0, then the net {xt } defined by (3.3) converges in norm, as t → 0+ , to the minimum norm element in Γ, namely, the unique solution x∗ to the quadratic minimization problem: x∗ = arg min kxk2 . x∈Γ Proof. First, we prove that {xt } is bounded. Take u ∈ Γ. From Lemma 2.3, we have u = N u and u = P rojC [P rojC (u − ξGu) − ηFP rojC (u − ξGu)]. (3.5) Y. Liu, Z. Yao, Y. C. Liou, L. J. Zhu, J. Nonlinear Sci. Appl. 9 (2016), 61–74 65 Put v = P rojC (u − ξGu). Then u = P rojC (v − ηFv). From Lemma 2.2, we note that kz t − vk = kP rojC (xt − ξGxt ) − P rojC (u − ξGu)k ≤ kxt − uk, and ky t − uk = kP rojC (z t − ηFz t ) − P rojC (v − ηFv)k ≤ kz t − vk. It follows from (3.2) that kxt − uk = kN P rojC [tR(xt ) + (1 − t)y t ] − N P rojC uk ≤ kt(R(xt ) − u) + (1 − t)(y t − u)k ≤ tkR(xt ) − R(u)k + tkR(u) − uk + (1 − t)ky t − uk ≤ tκkxt − uk + tkR(u) − uk + (1 − t)kxt − uk = [1 − (1 − κ)t]kxt − uk + tkR(u) − uk, that is, kxt − uk ≤ kR(u) − uk . 1−κ Hence, {xt } is bounded and so are {y t }, {z t } and {R(xt )}. Now we can choose a constant M > 0 such that n sup 2kR(xt ) − ukky t − uk + kR(xt ) − uk2 , 2ξkxt − z t − (u − v)k, t o 2ηkz t − y t + (u − v)k, ky t − R(xt )k2 ≤ M. Since F is ζ-inverse strongly monotone and G is δ-inverse strongly monotone, we have from Lemma 2.2 that ky t − uk2 = k(I − ηF)z t − (I − ηF)vk2 ≤ kz t − vk2 + η(η − 2ζ)kFz t − Fvk2 , (3.6) and kz t − vk2 = k(I − ξG)xt − (I − ξG)uk2 ≤ kxt − uk2 + ξ(ξ − 2δ)kGxt − Guk2 . (3.7) Combining (3.6) with (3.7) to get ky t − uk2 = k(I − ηF)z t − (I − ηF)vk2 ≤ kxt − uk2 + η(η − 2ζ)kFz t − Fvk2 + ξ(ξ − 2δ)kGxt − Guk2 . (3.8) From (3.2) and (3.8), we have kxt − uk2 ≤ k(1 − t)(y t − u) + t(R(xt ) − u)k2 = (1 − t)2 ky t − uk2 + 2t(1 − t)hR(xt ) − u, y t − ui + t2 kR(xt ) − uk2 = ky t − uk2 + tM t 2 (3.9) t ≤ kx − uk + η(η − 2ζ)kFz − Fvk + ξ(ξ − 2δ)kGxt − Guk2 + tM, 2 Y. Liu, Z. Yao, Y. C. Liou, L. J. Zhu, J. Nonlinear Sci. Appl. 9 (2016), 61–74 66 that is, η(2ζ − η)kFz t − Fvk2 + ξ(2δ − ξ)kGxt − Guk2 ≤ tM → 0. Since η(2ζ − η) > 0 and ξ(2δ − ξ) > 0, we derive lim kFz t − Fvk = 0 and lim kGxt − Guk = 0. t→0 (3.10) t→0 From Lemma 2.1 and (3.2), we obtain kz t − vk2 = kP rojC (xt − ξGxt ) − P rojC (u − ξGu)k2 ≤ h(xt − ξGxt ) − (u − ξGu), z t − vi 1 t = k(x − ξGxt ) − (u − ξGu)k2 + kz t − vk2 2 − k(xt − u) − ξ(Gxt − Gu) − (z t − v)k2 1 t ≤ kx − uk2 + kz t − vk2 − k(xt − z t ) − ξ(Gxt − Gu) − (u − v)k2 2 1 t kx − uk2 + kz t − vk2 − kxt − z t − (u − v)k2 = 2 + 2ξhxt − z t − (u − v), Gxt − Gui − ξ 2 kGxt − Guk2 , and ky t − uk = kP rojC (z t − ηFz t ) − P rojC (v − ηFv)k2 ≤ hz t − ηFz t − (v − ηFv), y t − ui 1 t = kz − ηFz t − (v − ηFv)k2 + ky t − uk2 2 − kz t − ηFz t − (v − ηFv) − (y t − u)k2 1 t kz − vk2 + ky t − uk2 − kz t − y t + (u − v)k2 ≤ 2 + 2ηhFz t − Fv, z t − y t + (u − v)i − η 2 kFz t − Fvk2 1 t ≤ kx − uk2 + ky t − uk2 − kz t − y t + (u − v)k2 2 + 2ηhFz t − Fv, z t − y t + (u − v)i . Thus, we have kz t − vk2 ≤ kxt − uk2 − kxt − z t − (u − v)k2 + M kGxt − Guk, (3.11) ky t − uk2 ≤ kxt − uk2 − kz t − y t + (u − v)k2 + M kFz t − Fvk. (3.12) and By (3.9) and (3.11), we have kxt − uk2 ≤ ky t − vk2 + tM ≤ kz t − vk2 + tM ≤ kxt − uk2 − kxt − z t − (u − v)k2 + (kGxt − Guk + t)M. Y. Liu, Z. Yao, Y. C. Liou, L. J. Zhu, J. Nonlinear Sci. Appl. 9 (2016), 61–74 67 It follows that kxt − z t − (u − v)k2 ≤ (kGxt − Guk + t)M. Since kGxt − Guk → 0, we deduce that lim kxt − z t − (u − v)k = 0. t→0 (3.13) From (3.9) and (3.12), we have kxt − uk2 ≤ ky t − uk2 + tM ≤ kxt − uk2 − kz t − y t + (u − v)k2 + (kFz t − Fvk + t)M. It follows that kz t − y t + (u − v)k2 ≤ (kFz n − Fvk + t)M, which implies that lim kz t − y t + (u − v)k = 0. t→0 (3.14) Thus, combining (3.13) with (3.14), we deduce that lim kxt − y t k = 0. t→0 (3.15) We note that kxt − N y t k = kN P rojC [tR(xt ) + (1 − t)y t ] − N P rojC y t k ≤ tM → 0. Hence, kN y t − y t k ≤ kN y t − xt k + kxt − y t k → 0. Therefore, kxt − N xt k → 0. (3.16) At the same time, from (3.2) and Lemma 2.3, we have kxt − U(xt )k = kN P rojC [tR(xt ) + (1 − t)U(xt )] − N P rojC [U(xt )]k ≤ tM → 0. (3.17) Next we show that {xt } is relatively norm compact as t → 0. Let {tn } ⊂ (0, 1) be a sequence such that n n tn → 0 as n → ∞. Put xn := xt and y n := y t . From (3.15)-(3.17), we have kxn − y n k → 0, kxn − N xn k → 0 and kxn − U(xn )k → 0. From (3.2), we get kxt − uk2 = kN P rojC [tR(xt ) + (1 − t)y t ] − N uk2 ≤ ky t − u − ty t + tR(xt )k2 = ky t − uk2 − 2thy t , y t − ui + 2thR(xt ), y t − ui + t2 ky t − R(xt )k2 = ky t − uk2 − 2thy t − u, y t − ui − 2thu, y t − ui + 2thR(xt ) − R(u), y t − ui + 2thR(u), y t − ui + t2 ky t − R(xt )k2 ≤ [1 − 2(1 − κ)t]kxt − uk2 + 2thR(u) − u, y t − ui + t2 ky t − R(xt )k2 . (3.18) Y. Liu, Z. Yao, Y. C. Liou, L. J. Zhu, J. Nonlinear Sci. Appl. 9 (2016), 61–74 68 It follows that t 1 hu − R(u), u − y t i + ky t − R(xt )k2 1−κ 2(1 − κ) t 1 hu − R(u), u − y t i + M. ≤ 1−κ 2(1 − κ) kxt − uk2 ≤ In particular, kxn − uk2 ≤ 1 tn hu − R(u), u − y n i + M, 1−κ 2(1 − κ) u ∈ Γ. (3.19) By the boundedness of {xn }, without loss of generality, we assume that {xn } converges weakly to a point x∗ ∈ C. It is clear that y n → x∗ weakly. From (3.18) we can use Lemma 2.5 to get x∗ ∈ Γ. We substitute x∗ for u in (3.19) to get kxn − x∗ k2 ≤ 1 tn hx∗ − R(x∗ ), x∗ − y n i + M. 1−κ 2(1 − κ) So the weak convergence of {y n } to x∗ implies that xn → x∗ strongly. We prove the relative norm compactness of the net {xt } as t → 0. In (3.19), we take the limit as n → ∞ to get kx∗ − uk2 ≤ 1 hu − R(u), u − x∗ i, 1−κ u ∈ Γ. (3.20) Which implies that x∗ solves the following variational inequality x∗ ∈ Γ, h(I − R)u, u − x∗ i ≥ 0, u ∈ Γ. It equals to its dual variational inequality x∗ ∈ Γ, h(I − R)x∗ , u − x∗ i ≥ 0, u ∈ Γ. Therefore, x∗ = (P rojΓ R)x∗ . This shows that x∗ is the unique fixed point in Γ of the contraction P rojΓ R. This is sufficient to conclude that the entire net {xt } converges in norm to x∗ as t → 0. Setting R = 0, then (3.20) is reduced to kx∗ − uk2 ≤ hu, u − x∗ i, u ∈ Γ. Equivalently, kx∗ k2 ≤ hx∗ , ui, u ∈ Γ. This implies that kx∗ k ≤ kuk, u ∈ Γ. Therefore, x∗ is the minimum norm element in Γ. This completes the proof. Below we introduce an explicit scheme for finding the minimum-norm element in Γ. Theorem 3.2. Suppose Γ := F ix(N ) ∩ Ω 6= ∅. For given x0 ∈ C arbitrarily, let the sequences {xn }, {y n } and {z n } be generated iteratively by n n n z = PC (x − ξGx ), y n = PC (z n − ηFz n ), (3.21) n+1 n n n x = δn x + (1 − δn )N P rojC [ζn R(x ) + (1 − ζn )y ], n ≥ 0, where {ζn } and {δn } are two sequences in [0, 1] satisfying the following conditions: Y. Liu, Z. Yao, Y. C. Liou, L. J. Zhu, J. Nonlinear Sci. Appl. 9 (2016), 61–74 ∞ P (i) lim ζn = 0 and n→∞ 69 ζn = ∞; n=0 (ii) 0 < lim inf δn ≤ lim sup δn < 1. n→∞ n→∞ Then the sequence {xn } converges strongly to x∗ which is the unique solution of variational inequality (3.4). In particular, if R = 0, then x∗ is the minimum norm element in Γ. Proof. First, we prove that the sequences {xn }, {y n } and {z n } are bounded. Let v = P rojC (u − ξGu) and u = P rojC (v − ηFv). From (3.21), we get ky n − uk = kP rojC (z n − ηFz n ) − P rojC (v − ηFv)k ≤ kz n − vk = kP rojC (xn − ξGxn ) − P rojC (u − ξGu)k ≤ kxn − uk, and kxn+1 − uk = kδn (xn − u) + (1 − δn )(N P rojC [ζn R(xn ) + (1 − ζn )y n ] − u)k ≤ δn kxn − uk + (1 − δn )kζn (R(xn ) − u) + (1 − ζn )(y n − u)k ≤ δn kxn − uk + (1 − δn )[ζn kR(xn ) − R(u)k + ζn kR(u) − uk + (1 − ζn )ky n − uk] ≤ δn kxn − uk + (1 − δn )[ζn κkxn − uk + ζn kR(u) − uk + (1 − ζn )kxn − uk] = [1 − (1 − κ)(1 − δn )ζn ]kxn − uk + ζn (1 − δn )kR(u) − uk kR(u) − uk ≤ max{kxn − uk, }. 1−κ By induction, we obtain, for all n ≥ 0, kR(u) − uk kx − uk ≤ max kx0 − uk, . 1−κ n Hence, {xn } is bounded. Consequently, we deduce that {y n } ,{z n }, {R(xn )}, {Fz n } and {Gxn } are all bounded. Let M > 0 is a constant such that n sup ky n k + kR(xn )k, 2ky n − R(xn )kky n − uk + ky n − R(xn )k2 , n (kxn − uk + kxn+1 − uk), 2ξkxn − z n − (u − v)k, 2ηkz n − y n + (u − v)k, o 2ηkFz n − Fvkkz n − y n + (u − v)k, 2ξkxn − z n − (u − v)kkGxn − Gukk ≤ M. Next we show lim ky n − N y n k = 0. n→∞ Define xn+1 = δn xn + (1 − δn )un for all n ≥ 0. It follows from (3.21) that kun+1 − un k = kN P rojC [ζn+1 R(xn+1 ) + (1 − ζn+1 )y n+1 ] − N P rojC [ζn R(xn ) + (1 − ζn )y n ]k ≤ kζn+1 R(xn+1 ) + (1 − ζn+1 )y n+1 − ζn R(xn ) − (1 − ζn )y n k ≤ ky n+1 − y n k + ζn+1 (ky n+1 k + kR(xn+1 )k) + ζn (ky n k + kR(xn )k) ≤ kP rojC (z n+1 − ηFz n+1 ) − P rojC (z n − ηFz n )k + M (ζn+1 + ζn ) ≤ kz n+1 − z n k + M (ζn+1 + ζn ) = kP rojC (xn+1 − ξGxn+1 ) − P rojC (xn − ξGxn )k + M (ζn+1 + ζn ) ≤ kxn+1 − xn k + M (ζn+1 + ζn ). Y. Liu, Z. Yao, Y. C. Liou, L. J. Zhu, J. Nonlinear Sci. Appl. 9 (2016), 61–74 70 This together with (i) imply that lim sup kun+1 − un k − kxn+1 − xn k ≤ 0. n→∞ Hence by Lemma 2.4, we get lim kun − xn k = 0. Consequently, n→∞ lim kxn+1 − xn k = lim (1 − δn )kun − xn k = 0. n→∞ n→∞ By the convexity of the norm k · k, we have kxn+1 − uk2 = kδn (xn − u) + (1 − δn )(un − u)k2 ≤ δn kxn − uk2 + (1 − δn )kun − uk2 ≤ δn kxn − uk2 + (1 − δn )ky n − u − ζn (y n − R(xn ))k2 ≤ δn kxn − uk2 + (1 − δn )ky n − uk2 + ζn M. (3.22) From Lemma 2.2 and (3.21), we have ky n − uk2 ≤ kz n − vk2 + η(η − 2ζ)kFz n − Fvk2 ≤ kxn − uk2 + ξ(ξ − 2δ)kGxn − Guk2 + η(η − 2ζ)kFz n − Fvk2 . (3.23) Substituting (3.23) into (3.22), we have kxn+1 − uk2 ≤ δn kxn − uk2 + (1 − δn )[kxn − uk2 + ξ(ξ − 2δ)kGxn − Guk2 + η(η − 2ζ)kFz n − Fvk2 ] + ζn M = kxn − uk2 + (1 − δn )ξ(ξ − 2δ)kGxn − Guk2 + (1 − δn )η(η − 2ζ)kFz n − Fvk2 + ζn M. Therefore, (1 − δn )η(2ζ − η)kFz n − Fvk2 + (1 − δn )ξ(2δ − ξ)kGxn − Guk2 ≤ kxn − uk − kxn+1 − uk + ζn M ≤ (kxn − uk + kxn+1 − uk)kxn − xn+1 k + ζn M ≤ (kxn − xn+1 k + ζn )M. Since lim inf (1 − δn )η(2ζ − η) > 0, lim inf (1 − δn )ξ(2δ − ξ) > 0, kxn − xn+1 k → 0 and ζn → 0, we derive n→∞ n→∞ lim kFz n − Fvk = 0 and lim kGxn − Guk = 0. n→∞ n→∞ From Lemma 2.1 and (3.21), we obtain kz n − vk2 = kPC (xn − ξGxn ) − PC (u − ξGu)k2 ≤ h(xn − ξGxn ) − (u − ξGu), z n − vi 1 n k(x − ξGxn ) − (u − ξGu)k2 + kz n − vk2 − k(xn − u) − ξ(Gxn − Gu) − (z n − v)k2 = 2 1 n kx − uk2 + kz n − vk2 − k(xn − z n ) − ξ(Gxn − Gu) − (u − v)k2 ≤ 2 1 n kx − uk2 + kz n − vk2 − kxn − z n − (u − v)k2 = 2 + 2ξhxn − z n − (u − v), Gxn − Gui − ξ 2 kGxn − Guk2 , Y. Liu, Z. Yao, Y. C. Liou, L. J. Zhu, J. Nonlinear Sci. Appl. 9 (2016), 61–74 71 and ky n − uk = kP rojC (z n − ηFz n ) − P rojC (v − ηFv)k2 ≤ hz n − ηFz n − (v − ηFv), y n − ui 1 n = kz − ηFz n − (v − ηFv)k2 + ky n − uk2 2 − kz n − ηFz n − (v − ηFv) − (y n − u)k2 1 n ≤ kz − vk2 + ky n − uk2 − kz n − y n + (u − v)k2 2 + 2ηhFz n − Fv, z n − y n + (u − v)i − η 2 kFz n − Fvk2 1 n ≤ kx − uk2 + ky n − uk2 − kz n − y n + (u − v)k2 2 + 2ηhFz n − Fv, z n − y n + (u − v)i . Thus, we deduce kz n − vk2 ≤ kxn − uk2 − kxn − z n − (u − v)k2 + 2ξkxn − z n − (u − v)kkGxn − Guk ≤ kxn − uk2 − kxn − z n − (u − v)k2 + M kGxn − Guk (3.24) ky n − uk2 ≤ kxn − uk2 − kz n − y n + (u − v)k2 + M kFz n − Fvk. (3.25) and By (3.22) and (3.24), we have kxn+1 − uk2 ≤ δn kxn − uk2 + (1 − δn )ky n − uk2 + ζn M ≤ δn kxn − uk2 + (1 − δn )kz n − vk2 + ζn M ≤ δn kxn − uk2 + (1 − δn )[kxn − uk2 − kxn − z n − (u − v)k2 + M kGxn − Guk] + ζn M ≤ kxn − uk2 − (1 − δn )kxn − z n − (u − v)k2 + (kGxn − Guk + ζn )M. It follows that (1 − δn )kxn − z n − (u − v)k2 ≤ (kxn+1 − xn k + kGxn − Guk + ζn )M. Since lim inf (1 − δn ) > 0, ζn → 0, kxn+1 − xn k → 0 and kGxn − Guk → 0, we deduce that n→∞ lim kxn − z n − (u − v)k = 0. n→∞ From (3.22) and (3.25), we have kxn+1 − uk2 ≤ δn kxn − uk2 + (1 − δn )ky n − uk2 + ζn M ≤ δn kxn − uk2 + (1 − δn )[kxn − uk2 − kz n − y n + (u − v)k2 + M kFz n − Fvk] + ζn M ≤ kxn − uk2 − (1 − δn )kz n − y n + (u − v)k2 + (kFz n − Fvk + ζn )M. It follows that (1 − δn )kz n − y n + (u − v)k2 ≤ (kxn+1 − xn k + kFz n − Fvk + ζn )M, (3.26) Y. Liu, Z. Yao, Y. C. Liou, L. J. Zhu, J. Nonlinear Sci. Appl. 9 (2016), 61–74 72 which implies that lim kz n − y n + (u − v)k = 0. (3.27) n→∞ Thus, from (3.26) and (3.27), we deduce that lim kxn − y n k = 0. n→∞ Hence, kN y n − un k = kN P rojC y n − N P rojC [ζn R(xn ) + (1 − ζn )y n ]k ≤ ζn M → 0. Therefore, kN y n − y n k ≤ kN y n − un k + kun − xn k + kxn − y n k → 0. Next we prove lim suphx∗ − R(x∗ ), x∗ − y n i ≤ 0, n→∞ where x∗ = PΓ R(x∗ ). Indeed, we can choose a subsequence {yni } of {y n } such that lim suphx∗ − R(x∗ ), x∗ − y n i = lim hx∗ − R(x∗ ), x∗ − yni i. i→∞ n→∞ Without loss of generality, we may further assume that yni → z weakly, then it is clear that z ∈ Γ. Therefore, lim suphx∗ − R(x∗ ), x∗ − y n i = lim hx∗ − R(x∗ ), x∗ − zi ≤ 0. i→∞ n→∞ From (3.21), we have kxn+1 − x∗ k2 ≤ δn kxn − x∗ k2 + (1 − δn )kζn (R(xn ) − x∗ ) + (1 − ζn )(y n − x∗ )k2 ≤ δn kxn − x∗ k2 + (1 − δn )[(1 − ζn )2 ky n − x∗ k2 + 2ζn (1 − ζn )hR(xn ) − x∗ , y n − x∗ i + ζn 2 kR(xn ) − x∗ k2 ] = δn kxn − x∗ k2 + (1 − δn )[(1 − ζn )2 ky n − x∗ k2 + 2ζn (1 − ζn )hR(xn ) − R(x)∗ , y n − x∗ i + 2ζn (1 − ζn )hR(x∗ ) − x∗ , y n − x∗ i + ζn 2 kR(xn ) − x∗ k2 ] ≤ [1 − 2(1 − κ)(1 − δn )ζn ]kxn − x∗ k2 + 2ζn (1 − ζn )(1 − δn )hR(x∗ ) − x∗ , y n − x∗ i + (1 − δn )ζn 2 M = (1 − γ n ))kxn − x∗ k2 + δ n γ n , where γ n = 2(1 − κ)(1 − δn )ζn and δ n / = (1−ζn ) ∗ 1−κ hR(x) − x∗ , y n − x∗ i + ζn M 2(1−κ) . It is clear that ∞ P γn = ∞ n=0 and lim sup δ ≤ 0. Hence, all conditions of Lemma 2.6 are satisfied. Therefore, we immediately deduce that n→∞ xn → x∗ . Finally, if we take R = 0, by the similar argument as that Theorem 3.1, we deduce immediately that x∗ is a solution of (3.5). 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