The split variational inequality problem and its algorithm iteration

Available online at www.isr-publications.com/jnsa J. Nonlinear Sci. Appl., 10 (017), 649 661 Research Article Journal Homepage: www.tjnsa.com - www.isr-publications.com/jnsa The split variational inequality problem and its algorithm iteration Yonghong Yao a, Xiaoxue Zheng a, Limin Leng a, Shin Min Kang b,c, a Department of Mathematics, Tianjin Polytechnic University, Tianjin 300387, China. b Center for General Education, China Medical University, Taichung 4040, Taiwan. c Department of Mathematics and the RINS, Gyeongsang National University, Jinju 588, Korea. Communicated by X. Qin Abstract The split variational inequality problem under a nonlinear transformation has been considered. An iterative algorithm is presented to solve this split problem. Strong convergence results are obtained. c 017 All rights reserved. Keywords: Split variational inequality, iterative algorithm, nonlinear transformation. 010 MSC: 49J53, 49M37, 65K10, 90C5. 1. Introduction Let H be a real Hilbert space with its inner product, and norm. Let C H be a nonempty closed convex set. Let A : C H and ψ : C C be three nonlinear operators. Now we consider the following variational inequality of finding u C such that Au, ψ(v) ψ(u) 0, v C. (1.1) Denote the set of the solutions of (1.1) by VI(A, ψ, C). If ψ I, then (1.1) reduces to the variational inequality of finding u C such that Au, v u 0, v C. (1.) Denote the set of the solutions of (1.) by VI(A, C). Variational inequalities have played a critical and significant part in the study of several unrelated problems arising in physics, finance, economics, network analysis, elasticity, optimization, water resources, medical images and structural analysis. For some related work, please refer to Ceng et al. [5, 6], Cianciaruso et al. [1], Glowinski [13], Iusem [16], Korpelevič [17], Noor [1], Qin and Cho [19], Qin and Yao [0], Yao et al. [7, 9, 31, 3], Zegeye et al. [34], and Zhang et al. [35]. Corresponding author Email addresses: yaoyonghong@aliyun.com (Yonghong Yao), zhengxiaoxue1991@aliyun.com (Xiaoxue Zheng), lenglimin@aliyun.com (Limin Leng), smkang@gnu.ac.kr (Shin Min Kang) doi:10.436/jnsa.010.05.31 Received 016-1-6

Y.-H. Yao, X.-X. Zheng, L.-M. Leng, S.-M. Kang, J. Nonlinear Sci. Appl., 10 (017), 649 661 650 Recently, the split feasibility problem has been studied extensively, see [ 4, 7 9, 4 6, 8, 33]. The split feasibility problem is formulated as finding ˆv such that ˆv C and ψ(ˆv) Q, (1.3) where C and Q are two closed convex subsets of two Hilbert spaces H 1 and H, respectively, and ψ : H 1 H is a bounded linear operator. Such problems arise in the field of intensity-modulated radiation therapy. Special cases. (i) If C = Fix(S) and Q = Fix(T), then (1.3) reduces to the split common fixed point problem which was first introduced by Censor and Segal [11]. The reader can refer to He and Du [14, 15] and Yao et al. [30]. (ii) If C = VI(A, D) and Q = VI(B, E), then (1.3) reduces to the split variational inequality problem which was studied in [10]. In the present manuscript, we focus on the following split variational inequality problem of finding a point ˆv such that ˆv VI(A, ψ, C) and ψ(ˆv) VI(B, C). (1.4) Remark 1.1. In the existing literature, the split problem that requires to find a point of an operator in one space whose image under a linear transformation is a point of another operator in the image space. However, in (1.4), the transformation ψ is nonlinear. In order to solve (1.4), we introduce a new iterative algorithm. Under some mild assumptions, we show the strong convergence of the presented algorithm.. Preliminaries Let C be a nonempty closed convex subset of a real Hilbert space H. Definition.1. An operator ϑ : C C is said to be L-Lipschitz continuous if there exists a constant L > 0 such that ϑ(u) ϑ(u ) L u u for all u, u C. Definition.. An operator A : C H is said to be (1) monotone if Au Au, u u 0, u, u C; () strongly monotone if there exists a constant δ > 0 such that Au Au, u u δ u u, u, u C; (3) inverse strongly monotone if there exists η > 0 such that Let ψ : C C be a nonlinear operator. Au Au, u u η Au Au, u, u C. Definition.3. An operator A : C H is said to be η-inverse strongly ψ-monotone iff for all u, u C and for some η > 0. Au Au, ψ(u) ψ(u ) η Au Au

Y.-H. Yao, X.-X. Zheng, L.-M. Leng, S.-M. Kang, J. Nonlinear Sci. Appl., 10 (017), 649 661 651 Let A : H H be a set-valued operator. The effective domain of A is denoted by dom(a), that is, dom(a) = {x H : Ax }. Definition.4. A set-valued operator A is said to be monotone on H iff for all x, y dom(a), u Ax and v Ay. x y, u v 0 A monotone operator A on H is said to be maximal iff its graph is not strictly contained in the graph of any other monotone operator on H. For every point u H, there exists a unique nearest point in C, denoted by proj C [u] such that u proj C [u] u u for all u C. The mapping proj C is called the metric projection of H onto C. It is well-known that proj C is a typical firmly nonexpansive mapping and is characterized by the following property u proj C [u], u proj C [u] 0, u H, u C. (.1) Lemma.5 ([]). Let C be a nonempty closed convex subset of a real Hilbert space H and A : C H be a ξ-inverse strongly monotone mapping. Then, where δ > 0 is a constant. (I δa)u (I δa)u x y + δ(δ ξ) Au Au, u, u C, (.) Lemma.6 ([3]). Let {ϖ n } [0, ), {µ n } (0, 1), and {ρ n } be three sequences such that Assume the following restrictions are satisfied (i) n=1 µ n = ; (ii) lim n ρ n µ n 0 or n=1 ρ n <. Then lim n ϖ n = 0. ϖ n+1 (1 µ n )ϖ n + ρ n, n 1. Lemma.7 ([18]). Let {w n } be a sequence of real numbers. Assume {w n } does not decrease at infinity, that is, there exists at least a subsequence {w nk } of {w n } such that w nk w nk +1 for all k 0. For every n N 0, define an integer sequence {τ(n)} as τ(n) = max{i n : w ni < w ni +1}. Then τ(n) as n and for all n N 0 max{w τ(n), w n } w τ(n)+1. 3. Main results In this section, we study the split variational inequality problem and the convergence analysis of its iterative algorithm. Let H be a real Hilbert space and C H be a nonempty closed convex set. Let ψ : C C be a weakly continuous and δ-strongly monotone mapping such that its range R(ψ) = C. Let A : C H be an η-inverse strongly ψ-monotone mapping. Let B : C H be a ξ-inverse strongly monotone mapping.

Y.-H. Yao, X.-X. Zheng, L.-M. Leng, S.-M. Kang, J. Nonlinear Sci. Appl., 10 (017), 649 661 65 Remark 3.1. For u, v C, since ψ is δ-strongly monotone, we get Then, δ u v ψ(u) ψ(v), u v ψ(u) ψ(v) u v. ψ(u) ψ(v) δ u v, u, v C. (3.1) In view of the assumptions on the operator ψ, we know ψ 1 is single-valued and Lipschitz continuous. Problem 3.. The split variational inequality problem is to find û such that û VI(A, ψ, C) and ψ(û) VI(B, C). (3.) Denote the set of solutions of (3.) by Λ, i.e., Λ = VI(A, ψ, C) ψ 1 (VI(B, C)). Throughout, we assume Λ. Algorithm 3.3. For given initial value x 0 C, define the sequence {x n } by the following form u n = proj C [ψ(x n ) ζ n Ax n ], v n = proj C [η n µϑ(x n ) + (1 η n )(u n δ n Bu n )], ψ(x n+1 ) = ξ n ψ(x n ) + (1 ξ n )v n, n 0, (3.3) where ϑ : C H is an L-Lipschitz continuous mapping, {η n }, {ξ n }, and {δ n } are three real number sequences in [0, 1], {ζ n } is a real number sequence in (0, ), and 0 < µ < δ/l is a constant. Remark 3.4. The solution of the variational inequality of finding x Λ such that µϑ(x ) ψ(x ), ψ(x) ψ(x ) 0, x Λ (3.4) is unique. As a matter of fact, assume that both x and x solve (3.4). Then, µϑ(x ) ψ(x ), ψ( x) ψ(x ) 0 and µϑ( x) ψ( x), ψ(x ) ψ( x) 0. Adding up the above two inequalities, we deduce It follows that µϑ( x) ψ( x) µϑ(x ) + ψ(x ), ψ(x ) ψ( x) 0. ψ(x ) ψ( x) µ ϑ(x ) ϑ( x), ψ(x ) ψ( x) µ ϑ(x ) ϑ( x) ψ(x ) ψ( x). This together with (3.1) implies that δ x x ψ(x ) ψ( x) µ ϑ(x ) ϑ( x) µl x x. We get x = x immediately because of µl < δ. So, the variational inequality (3.4) has a unique solution denoted by û. Proposition 3.5. Let C be a nonempty closed convex subset of a real Hilbert space H. Let A : C H be an η-inverse strongly ψ-monotone mapping. Then, Proof. In fact, (ψ(x) ζax) (ψ(y) ζay) ψ(x) ψ(y) + ζ(ζ η) Ax Ay, x, y C. (3.5) (ψ(x) ζax) (ψ(y) ζay) = ψ(x) ψ(y) ζ Ax Ay, ψ(x) ψ(y) + ζ Ax Ay ψ(x) ψ(y) ζη Ax Ay + ζ Ax Ay ψ(x) ψ(y) + ζ(ζ η) Ax Ay.

Y.-H. Yao, X.-X. Zheng, L.-M. Leng, S.-M. Kang, J. Nonlinear Sci. Appl., 10 (017), 649 661 653 Theorem 3.6. Assume the following conditions are satisfied: (C1): lim n η n = 0 and n η n = ; (C): 0 < lim n ξ n lim n ξ n < 1 and δ (Lµ, η); (C3): 0 < lim n ζ n lim n ζ n < η and 0 < lim n δ n lim n δ n < ξ. Then the sequence {x n } generated by (3.3) converges strongly to û Λ which solves the variational inequality (3.4). Proof. Note that û VI(A, ψ, C) and ψ(û) VI(B, C). Using (.1), we deduce ψ(û) = proj C [ψ(û) ζ n Aû] = proj C [ψ(û) δ n Bψ(û)] for all n 0. First, from (.), we have (u n δ n Bu n ) (ψ(û) δ n Bψ(û)) u n ψ(û) + δ n (δ n ξ) Bu n Bψ(û) u n ψ(û). By (3.5), we get u n ψ(û) = proj C [ψ(x n ) ζ n Ax n ] proj C [ψ(û) ζ n Aû] (ψ(x n ) ζ n Ax n ) (ψ(û) ζ n Aû) ψ(x n ) ψ(û) + ζ n (ζ n η) Ax n Aû ψ(x n ) ψ(û) and ψ(x n+1 ) ζ n+1 Ax n+1 (ψ(x n ) ζ n+1 Ax n ) ψ(x n+1 ) ψ(x n ) + ζ n+1 (ζ n+1 η) Ax n+1 Ax n. From (3.1), (3.3), (3.6), and (3.7), we have v n ψ(û) = proj C [η n µϑ(x n ) + (1 η n )(u n δ n Bu n )] proj C [ψ(û) δ n Bψ(û)] η n (µϑ(x n ) ψ(û) + δ n Bψ(û)) + (1 η n )((u n δ n Bu n ) (ψ(û) δ n Bψ(û))) η n µϑ(x n ) µϑ(û) + η n µϑ(û) ψ(û) + δ n Bψ(û) + (1 η n ) (u n δ n Bu n ) (ψ(û) δ n Bψ(û)) η n µl x n û + η n µϑ(û) ψ(û) + δ n Bψ(û) + (1 η n ) u n ψ(û) η n µl/δ ψ(x n ) ψ(û) + η n µϑ(û) ψ(û) + δ n Bψ(û) + (1 η n ) ψ(x n ) ψ(û) = [1 (1 µl/δ)η n ] ψ(x n ) ψ(û) + η n µϑ(û) ψ(û) + δ n Bψ(û) [1 (1 µl/δ)η n ] ψ(x n ) ψ(û) + η n ( µϑ(û) ψ(û) + ξ Bψ(û) ). (3.6) (3.7) (3.8) By combination of (3.6), (3.7), and (3.8), we obtain v n ψ(û) η n (µϑ(x n ) ψ(û) + δ n Bψ(û)) + (1 η n )((u n δ n Bu n ) (ψ(û) δ n Bψ(û))) η n µϑ(x n ) ψ(û) + δ n Bψ(û) + (1 η n ) (u n δ n Bu n ) (ψ(û) δ n Bψ(û)) η n µϑ(x n ) ψ(û) + δ n Bψ(û) + (1 η n )[ u n ψ(û) + δ n (δ n ξ) Bu n Bψ(û) ] η n µϑ(x n ) ψ(û) + δ n Bψ(û) + (1 η n )[ ψ(x n ) ψ(û) + δ n (δ n ξ) Bu n Bψ(û) + ζ n (ζ n η) Ax n Aû ] η n µϑ(x n ) ψ(û) + δ n Bψ(û) + (1 η n ) ψ(x n ) ψ(û). (3.9) According to (3.3) and (3.8), we have ψ(x n+1 ) ψ(û) ξ n ψ(x n ) ψ(û) + (1 ξ n ) v n ψ(û)

Y.-H. Yao, X.-X. Zheng, L.-M. Leng, S.-M. Kang, J. Nonlinear Sci. Appl., 10 (017), 649 661 654 ξ n ψ(x n ) ψ(û) + (1 ξ n )[1 (1 µl/δ)η n ] ψ(x n ) ψ(û) + (1 ξ n )η n ( µϑ(û) ψ(û) + ξ Bψ(û) ) = [1 (1 µl/δ)(1 ξ n )η n ] ψ(x n ) ψ(û) µϑ(û) ψ(û) + ξ Bψ(û) + (1 µl/δ)(1 ξ n )η n. 1 µl/δ By induction { ψ(x n ) ψ(û) max ψ(x 0 ) ψ(û), } µϑ(û) ψ(û) + ξ Bψ(û). 1 µl/δ It follows that x n û 1 δ ψ(x n) ψ(û) 1 δ max { ψ(x 0 ) ψ(û), } µϑ(û) ψ(û) + ξ Bψ(û). 1 µl/δ Hence, {ψ(x n )}, {x n }, {u n }, and {v n } are all bounded. From (3.3), we have ψ(x n+1 ) ψ(x n ) = (1 ξ n )(v n ψ(x n )). (3.10) Thus, Observe that ψ(x n+1 ) ψ(x n ), ψ(x n ) ψ(û) = (1 ξ n ) v n ψ(x n ), ψ(x n ) ψ(û). (3.11) ψ(x n+1 ) ψ(x n ), ψ(x n ) ψ(û) = ψ(x n+1 ) ψ(û) ψ(x n ) ψ(û) ψ(x n+1 ) ψ(x n ) (3.1) and v n ψ(x n ), ψ(x n ) ψ(û) = v n ψ(û) ψ(x n ) ψ(û) v n ψ(x n ). (3.13) By virtue of (3.11), (3.1), and (3.13), we deduce ψ(x n+1 ) ψ(û) ψ(x n ) ψ(û) ψ(x n+1 ) ψ(x n ) = (1 ξ n )[ v n ψ(û) ψ(x n ) ψ(û) v n ψ(x n ) ]. (3.14) Combining (3.9), (3.10) with (3.14), we have ψ(x n+1 ) ψ(û) ψ(x n ) ψ(û) = (1 ξ n )[ v n ψ(û) ψ(x n ) ψ(û) v n ψ(x n ) ] + (1 ξ n ) v n ψ(x n ) = (1 ξ n )[ v n ψ(û) ψ(x n ) ψ(û) ] ξ n (1 ξ n ) v n ψ(x n ) (1 ξ n )[η n µϑ(x n ) ψ(û) + δ n Bψ(û) + (1 η n ) u n ψ(û) ψ(x n ) ψ(û) ] ξ n (1 ξ n ) v n ψ(x n ). (3.15) Returning to (3.8), we get v n ψ(û) [1 (1 µl/δ)η n ] ψ(x n ) ψ(û) + (1 µl/δ)η n ( µϑ(û) ψ(û) + ξ Bψ(û) (1 µl/δ) ). (3.16) Next, we consider two possible cases. Case 1. Assume there exists some integer m > 0 such that { ψ(x n ) ψ(û) } is decreasing for all n m.

Y.-H. Yao, X.-X. Zheng, L.-M. Leng, S.-M. Kang, J. Nonlinear Sci. Appl., 10 (017), 649 661 655 In this case, we know that lim n ψ(x n ) ψ(û) exists. From (3.15) and (3.16), we have ξ n (1 ξ n ) v n ψ(x n ) ψ(x n ) ψ(û) ψ(x n+1 ) ψ(û) + (1 ξ n )[ v n ψ(û) ψ(x n ) ψ(û) ] ψ(x n ) ψ(û) ψ(x n+1 ) ψ(û) + (1 µl/δ)η n ( µϑ(û) ψ(û) + ξ Bψ(û) (1 µl/δ) 0. ) This together with (C) implies that Furthermore, it follows from (3.10) that lim v n ψ(x n ) = 0. n By the convexity of the norm and (3.9), we have lim n ψ(x n+1) ψ(x n ) = 0. (3.17) ψ(x n+1 ) ψ(û) = ξ n (ψ(x n ) ψ(û)) + (1 ξ n )(v n ψ(û)) ξ n ψ(x n ) ψ(û) + (1 ξ n ) v n ψ(û) ξ n ψ(x n ) ψ(û) + η n (1 ξ n ) µϑ(x n ) ψ(û) + δ n Bψ(û) + (1 η n )(1 ξ n ) ψ(x n ) ψ(û) + (1 η n )(1 ξ n )δ n (δ n ξ) Bu n Bψ(û) + (1 η n )(1 ξ n )ζ n (ζ n η) Ax n Aû ψ(x n ) ψ(û) + η n (1 ξ n ) µϑ(x n ) ψ(û) + δ n Bψ(û) + (1 η n )(1 ξ n )δ n (δ n ξ) Bu n Bψ(û) + (1 η n )(1 ξ n )ζ n (ζ n η) Ax n Aû. (3.18) Thus, (1 η n )(1 ξ n )δ n (ξ δ n ) Bu n Bψ(û) + (1 η n )(1 ξ n )ζ n (η ζ n ) Ax n Aû ψ(x n ) ψ(û) ψ(x n+1 ) ψ(û) + η n (1 ξ n ) µϑ(x n ) ψ(û) + δ n Bψ(û) ( ψ(x n ) ψ(û) + ψ(x n+1 ) ψ(û) ) ψ(x n+1 ) ψ(x n ) + η n (1 ξ n ) µϑ(x n ) ψ(û) + δ n Bψ(û) 0 (by (C1) and (3.17)). Since lim inf n (1 ξ n )(1 η n )ζ n (η ζ n ) > 0 and lim inf n (1 η n )(1 ξ n )δ n (ξ δ n ) > 0, we obtain lim n Ax n Aû = 0 and lim n Bu n Bψ(û) = 0. (3.19) Set z n = ψ(x n ) ζ n Ax n (ψ(û) ζ n Aû) for all n. Using the property (.1) of projection, we get u n ψ(û) = proj C [ψ(x n ) ζ n Ax n ] proj C [ψ(û) ζ n Aû] z n, u n ψ(û) = 1 { zn + u n ψ(û) z n u n + ψ(û) } 1 { ψ(xn ) ψ(û) + u n ψ(û) ψ(x n ) u n ζ n (Ax n Aû) }

Y.-H. Yao, X.-X. Zheng, L.-M. Leng, S.-M. Kang, J. Nonlinear Sci. Appl., 10 (017), 649 661 656 = 1 { ψ(xn ) ψ(û) + u n ψ(û) ψ(x n ) u n ζ n Ax n Aû + ζ n ψ(x n ) u n, Ax n Aû }. It follows that u n ψ(û) ψ(x n ) ψ(û) ψ(x n ) u n + ζ n ψ(x n ) u n Ax n Aû. (3.0) Set z n = u n δ n Bu n (ψ(û) δ n Bψ(û)) for all n. We also have v n ψ(û) = proj C [η n µϑ(x n ) + (1 η n )(u n δ n Bu n )] proj C [ψ(û) δ n Bψ(û)] η n (µϑ(x n ) ψ(û) + δ n Bψ(û)) + (1 η n )z n, v n ψ(û) = 1 { ηn (µϑ(x n ) ψ(û) + δ n Bψ(û)) + (1 η n )z n + v n ψ(û) η n (µϑ(x n ) ψ(û) + δ n Bψ(û)) + (1 η n )z n v n + ψ(û) } 1 { ηn µϑ(x n ) ψ(û) + δ n Bψ(û) + (1 η n ) u n ψ(û) + v n ψ(û) η n (µϑ(x n ) ψ(û) + δ n Bψ(û) z n) + u n v n δ n (Bu n Bψ(û)) } = 1 { ηn µϑ(x n ) ψ(û) + δ n Bψ(û) + (1 η n ) u n ψ(û) + v n ψ(û) u n v n η n µϑ(x n ) ψ(û) + δ n Bψ(û) z n + δ n η n Bu n Bψ(û), µϑ(x n ) ψ(û) + δ n Bψ(û) z n δ n Bu n Bψ(û) + δ n u n v n, Bu n Bψ(û) η n u n v n, µϑ(x n ) ψ(û) + δ n Bψ(û) z n }. It follows that v n ψ(û) η n µϑ(x n ) ψ(û) + δ n Bψ(û) + (1 η n ) u n ψ(û) + δ n η n Bu n Bψ(û) µϑ(x n ) ψ(û) + δ n Bψ(û) z n u n v n + δ n u n v n Bu n Bψ(û) + η n u n v n µϑ(x n ) ψ(û) + δ n Bψ(û) z n. (3.1) In the light of (3.18), (3.0), and (3.1), we derive ψ(x n+1 ) ψ(û) ξ n ψ(x n ) ψ(û) + (1 ξ n ) v n ψ(û) ψ(x n ) ψ(û) + η n µϑ(x n ) ψ(û) + δ n Bψ(û) (1 η n ) ψ(x n ) u n (1 ξ n ) u n v n + ζ n ψ(x n ) u n Ax n Aû + δ n u n v n Bu n Bψ(û) + δ n η n Bu n Bψ(û) µϑ(x n ) ψ(û) + δ n Bψ(û) z n + η n u n v n µϑ(x n ) ψ(û) + δ n Bψ(û) z n. Then, (1 η n ) ψ(x n ) u n + (1 ξ n ) u n v n ( ψ(x n ) ψ(û) + ψ(x n+1 ) ψ(û) ) ψ(x n+1 ) ψ(x n ) + ζ n ψ(x n ) u n Ax n Aû + δ n u n v n Bu n Bψ(û) + δ n η n Bu n Bψ(û) µϑ(x n ) ψ(û) + δ n Bψ(û) z n

Y.-H. Yao, X.-X. Zheng, L.-M. Leng, S.-M. Kang, J. Nonlinear Sci. Appl., 10 (017), 649 661 657 Therefore, + η n u n v n µϑ(x n ) ψ(û) + δ n Bψ(û) z n + η n µϑ(x n ) ψ(û) + δ n Bψ(û) 0 (by (C1), (3.17), and (3.19)). lim n ψ(x n ) u n = 0 and lim n u n v n = 0. (3.) Now, we prove lim n µϑ(û) ψ(û), v n ψ(û) 0. Choose a subsequence {v ni } of {v n } such that lim n µϑ(û) ψ(û), v n ψ(û) = lim i µϑ(û) ψ(û), v ni ψ(û). (3.3) Since {x ni } is bounded, there exists a subsequence {x nij } of {x ni } which converges weakly to some point z C. Without loss of generality, we may assume that x ni z. This implies that ψ(x ni ) ψ(z) due to the weak continuity of ψ. Thus, v ni ψ(z) by (3.). Next, we show z VI(A, ψ, C). Set { Au + N C (u), u C, Ru =, u C. By [35], we know that R is maximal ψ-monotone. Let (u, w) G(R). Since w Bu N C (u) and x n C, we have ψ(u) ψ(x n ), w Au 0. Noting that u n = proj C [ψ(x n ) ζ n Ax n ], we get ψ(u) u n, u n (ψ(x n ) ζ n Ax n ) 0. It follows that Then, ψ(u) u n, u n ψ(x n ) ζ n + Ax n 0. ψ(u) ψ(x ni ), w ψ(u) ψ(x ni ), Av ψ(u) ψ(x ni ), Au ψ(u) u ni, u n i ψ(x ni ) ψ(u) u ni, Ax ni ζ ni = ψ(u) ψ(x ni ), Au Ax ni + ψ(u) ψ(x ni ), Ax ni ψ(u) u ni, u n i ψ(x ni ) ψ(u) u ni, Ax ni ζ ni ψ(u) u ni, u n i ψ(x ni ) ψ(x ni ) u ni, Ax ni. ζ ni (3.4) Since ψ(x ni ) u ni 0 and ψ(x ni ) ψ(z), we deduce that ψ(u) ψ(z), w 0 by taking i in (3.4). Thus, z R 1 0 by the maximal ψ-monotonicity of R. Hence, z VI(A, ψ, C). Next, we need to prove ψ(z) VI(B, C). Set { R Bv + N C (v), v C, v =, v C. By [1], we know that R is maximal monotone. Let (v, w) G(R ). Since w Bv N C (v) and v n C, we have v v n, w Bv 0. Noting that v n = proj C [η n µϑ(x n ) + (1 η n )(u n δ n Bu n )], we get v v n, v n [η n µϑ(x n ) + (1 η n )(u n δ n Bu n )] 0. It follows that v v n, v n u n + Bu n η n (µϑ(x n ) u n + δ n Bu n ) 0. δ n δ n

Y.-H. Yao, X.-X. Zheng, L.-M. Leng, S.-M. Kang, J. Nonlinear Sci. Appl., 10 (017), 649 661 658 Then, v v ni, w v v ni, Bv v v ni, Bv v v ni, v n i u ni v v ni, Bu ni δ ni + η n i v v ni, µϑ(x ni ) u ni + δ ni Bu ni δ ni = v v ni, Bv Bv ni + v v ni, Bv ni v v ni, v n i u ni δ ni v v ni, Bu ni + η n i v v ni, µϑ(x ni ) u ni + δ ni Bu ni δ ni v v ni, v n i u ni v v ni, Bu ni Bv ni δ ni + η n i δ ni v v ni, µϑ(x ni ) u ni + δ ni Bu ni. (3.5) Since v ni u ni 0 and v ni ψ(z), we deduce that v ψ(z), w 0 by taking i in (3.5). Thus, ψ(z) R 1 0 by the maximal monotonicity of R. Hence, ψ(z) VI(B, C). Therefore, z Λ. From (3.3), we obtain lim n µϑ(û) ψ(û), v n ψ(û) = lim i µϑ(û) ψ(û), ψ(x ni ) ψ(û) = µϑ(û) ψ(û), ψ(z) ψ(û) 0. (3.6) Note that v n ψ(û) = proj C [η n µϑ(x n ) + (1 η n )(u n δ n Bu n )] proj C [ψ(û) (1 η n )δ n Bψ(û)] It follows that Therefore, η n (µϑ(x n ) ψ(û)) + (1 η n )z n, v n ψ(û) η n µ ϑ(x n ) ϑ(û), v n ψ(û) + η n µϑ(û) ψ(û), v n ψ(û) + (1 η n ) u n δ n Bu n (ψ(û) δ n Bψ(û)) v n ψ(û) η n Lµ x n û v n ψ(û) + η n µϑ(û) ψ(û), v n ψ(û) + (1 η n ) u n ψ(û) v n ψ(û) η n (µl/δ) ψ(x n ) ψ(û) v n ψ(û) + η n µϑ(û) ψ(û), v n ψ(û) + (1 η n ) ψ(x n ) ψ(û) v n ψ(û) = [1 (1 Lµ/δ)η n ] ψ(x n ) ψ(û) v n ψ(û) + η n µϑ(û) ψ(û), v n ψ(û) 1 (1 Lµ/δ)η n ψ(x n ) ψ(û) + 1 v n ψ(û) + η n µϑ(û) ψ(û), v n ψ(û). v n ψ(û) [1 (1 Lµ/δ)η n ] ψ(x n ) ψ(û) + η n µϑ(û) ψ(û), v n ψ(û). ψ(x n+1 ) ψ(û) ξ n ψ(x n ) ψ(û) + (1 ξ n ) v n ψ(û) ξ n ψ(x n ) ψ(û) + (1 ξ n )[1 (1 µl/δ)η n ] ψ(x n ) ψ(û) + (1 ξ n )η n µϑ(û) ψ(û), v n ψ(û) = [1 (1 µl/δ)(1 ξ n )η n ] ψ(x n ) ψ(û) (3.7) + (1 ξ n )η n µϑ(û) ψ(û), v n ψ(û) = [1 (1 µl/δ)(1 ξ n )η n ] ψ(x n ) ψ(û)

Y.-H. Yao, X.-X. Zheng, L.-M. Leng, S.-M. Kang, J. Nonlinear Sci. Appl., 10 (017), 649 661 659 ( ) + (1 µl/δ)(1 ξ n )η n 1 µl/δ µϑ(û) ψ(û), v n ψ(û). We can apply Lemma.6 to (3.7) to conclude that ψ(x n ) ψ(û) and x n û. Case. Assume there exists an integer n 0 such that ψ(x n0 ) ψ(û) ψ(x n0 +1) ψ(û). At this case, we set ω n = { ψ(x n ) ψ(û) }. Then, we have ω n0 ω n0 +1. Define an integer sequence {τ n } for all n n 0 as follows: τ(n) = max{l N n 0 l n, ω l ω l+1 }. It is clear that τ(n) is a non-decreasing sequence satisfying and lim τ(n) = n ω τ(n) ω τ(n)+1, for all n n 0. By the similar argument as that of (3.6) and (3.7), we can prove that lim n µϑ(û) ψ(û), v τ(n) ψ(û) 0 (3.8) and ω τ(n)+1 [1 (1 µl/δ)(1 ξ τ(n))η τ(n) ]ω τ(n) ( ) + (1 µl/δ)(1 ξ τ(n) )η τ(n) 1 µl/δ µϑ(û) ψ(û), v τ(n) ψ(û). Since ω τ(n) ω τ(n)+1, we have from (3.9) that (3.9) ω τ(n) 1 µl/δ µϑ(û) ψ(û), v τ(n) ψ(û). (3.30) Combining (3.8) with (3.30), we have and hence From (3.8) and (3.9), we also obtain lim n ω τ(n) 0, lim ω τ(n) = 0. (3.31) n This together with (3.31) implies that Applying Lemma.7 we get lim n ω τ(n)+1 lim n ω τ(n). lim ω τ(n)+1 = 0. n 0 ω n max{ω τ(n), ω τ(n)+1 }. Therefore, ω n 0. That is, x n û. This completes the proof. Algorithm 3.7. For given initial value x 0 C, define the sequence {x n } by the following form u n = proj C [x n ζ n Ax n ], v n = proj C [η n µϑ(x n ) + (1 η n )(u n δ n Bu n )], x n+1 = ξ n x n + (1 ξ n )v n, n 0, (3.3) where ϑ : C H is an L-Lipschitz continuous mapping, {η n }, {ξ n }, and {δ n } are three real number sequences in [0, 1], {ζ n } is a real number sequence in (0, ), and 0 < µ < 1/L is a constant.

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