The Split Hierarchical Monotone Variational Inclusions Problems and Fixed Point Problems for Nonexpansive Semigroup

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1 International Mathematical Forum, Vol. 11, 2016, no. 8, HIKARI Ltd, The Split Hierarchical Monotone Variational Inclusions Problems and Fixed Point Problems for Nonexpansive Semigroup Zhaoli Ma 1, Lin Wang 2 and Jinhong Zhao 1 1 School of Information Engineering The College of Arts and Sciences Yunnan Normal University Kunming, Yunnan, , P. R. China 2 College of Statistics and Mathematics Yunnan University of Finance and Economics Kunming, Yunnan, , P. R. China Copyright c 2016 Zhaoli Ma, Lin Wang and Jinhong Zhao. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract The purpose of this paper is to study a split hierarchical monotone variational inclusion problem which includes split variational inequality problems, split common fixed point problems, split monotone variational inclusion problems, split convex minimization problems, etc., as special cases, and fixed point problems for nonexpansive semigroup in the setting of Hilbert spaces. For solving this kind problems, some new iterative algorithms are proposed. Under suitable conditions, some weak and strong convergence theorems for the sequences generated by the proposed algorithms are proved. The results presented in the paper improve and extend some recent corresponding results. Mathematics Subject Classifications: 47H09, 47J25 Keywords: split hierarchical monotone variational inclusions, Fixed point problems, convergence, nonexpansive semigroup, Hilbert spaces

2 396 Zhaoli Ma, Lin Wang and Jinhong Zhao 1 Introduction Let H 1 and H 2 be real Hilbert spaces with inner product, and norm, C H 1 and Q H 2 be nonempty, closed, and convex sets, respectively. The mapping T : H 1 H 1 is said to be nonexpansive if for all x, y H 1 T x T y x y. (1.1) A one-parameter family F := {T (t) : t 0} is said to be a nonexpansive semigroup on H 1 if the following conditions are satisfied: (i) T (0)x = x, for all x H 1 ; (ii) T (s + t) = T (s)t (t), for all s, t 0 ; (iii) for each x H 1, the mapping t T (t)x is continuous; (iv) T (t)x T (t)y x y, for all x, y H 1 and t 0 We denote by F (F) the set of all common fixed points of F, that is, F (F) := {x E : T (t)x = x, 0 t < } = t 0 F (T (t)). (1.2) It is well known that F (F) is closed and convex (see [1]). Let A : H 1 H 2 be a bounded linear operator, and f : H 1 H 1 and g : H 2 H 2 be two given operators. Recently, Censor et al. [2] introduced the following split variational inequality problem (SVIP): and such that F ind x C such that f(x ), x x 0, for all x C, (1.3) y := Ax Q solves g(y ), y y 0, for all y Q, (1.4) Let Λ denote the solution set of the SVIP, that is, Λ = {x solves(1.3) : Ax solves (1.4)}. If the sets C and Q are the set of fixed points of the operators T : H 1 H 1 and S : H 2 H 2, respectively, then the SVIP is called a split hierarchical variational inequality problem (SHVIP). The split variational inequality problem (SVIP) contains the split feasibility problem (SFP) as a special case. For further details of the SFP, we refer to [3-9] and the references therein. If f and g are convex and differentiable, then the SVIP is equivalent to the following split minimization problem: minf(x), subject to x C, (1.5)

3 Split hierarchical monotone variational inclusions problems 397 and such that y := Ax Q solves ming(y), subject to y Q, (1.6) For further details on the equivalence between a variational inequality and an optimization problem, see[10]. In 2011, Moudafi [11] introduced the following split monotone variational inclusion problem (SMVIP): and such that F ind x H 1 such that 0 f(x ) + B 1 (x ), (1.7) y := Ax H 2 solves 0 g(y ) + B 2 (y ), (1.8) Let Ψ denote the solution set of SMVIP, that is Ψ = {x solves (1.7) : Ax solves(1.8)} where B 1 : H 1 2 H 1 and B 2 : H 2 2 H 2 are multi-valued maximal monotone mappings, f : H 1 H 1 and g : H 2 H 2 be two given operators. If f 0 and g 0, then split monotone variational inclusion problem (SMVIP) reduces to the following split variational inclusion problem (SVIP): and such that F ind x H 1 such that 0 B 1 (x ), (1.9) y := Ax H 2 solves 0 B 2 (y ), (1.10) which constitutes a pair of variational inclusion problems connected with a bounded linear operator A in two different Hilbert spaces H 1 and H 2. The split monotone variational inclusion problem includes as special cases: the split common fixed point problem, the split variational inequality problem, the split zero problem, and the split feasibility problem, split monotone variational inclusion problems, split convex minimization problems and so on. This formalism is also at the core of the modeling of many inverse problems arising for phase retrieval and other real-world problems; for instance, in sensor networks in computerized tomography and data compression; see [12,13] and the references therein. To solve the SMVIP, Moudafi [11] introduced the following iterative method to solve split monotone variational inclusion problem and obtained some weak convergence theorems: Let > 0 and x 0 be the initial point, compute iterative sequence {x n } generated by the following scheme: x n+1 = U(x n + γa (T I)Ax n ), for all n N, (1.11)

4 398 Zhaoli Ma, Lin Wang and Jinhong Zhao where γ (0, 1 ) with L being the spectral radius of the operator L A A, U := J B 1 (I f) and T := J B 2 (I g) and J B 1 and J B 2 are the resolvents of B 1 and B 2, respectively. Recently, Ansari et al. [14] introduced the following split hierarchical monotone variational inclusion problem (SHMVIP): and such that F indx F ix(t ) such that 0 f(x ) + B 1 (x ), (1.12) y := Ax F ix(s) solves 0 g(y ) + B 2 (y ), (1.13) we denoted by Γ the solution set of (SHMVIP): that is Γ = {x solves (1.12) : Ax solves (1.13)} In 2015, Ansari et al. [14] modified iteration scheme (1.11) to the case of a split hierarchical monotone variational inclusion problem and the fixed point problem of a nonexpansive mapping. To be more precise, they proved the following weak convergence theorem. Theorem 1.1 [14] Let A : H 1 H 1 be a bounded linear operator, f : H 1 H 1 be an α 1 -inverse strongly monotone operator, T : H 1 H 1 be a strongly nonexpansive operator such that F (T ), g : H 2 H 2 be an α 2 -inverse strongly monotone operator, and S : H 2 H 2 be a nonexpansive operator such that F (S), and α = min{α 1, α 2 }. Consider the operator U := J B 1 (I f) and V := J B 2 (I g) with (0, 2α), and B 1 : H 1 H 1 and B 2 : H 2 H 2 are two maximal monotone set-valued mappings with nonempty values. Let {x n } be a sequence generated by: x n H 1 x n+1 = T U(x n + γa (SV I)Ax n, n 1, (1.14) where γ (0, 1 A 2 ), then the sequence {x n } converges weakly to an element x Γ, provided Γ. In this paper, Motivated and inspired by the recent research going on in the direction of split variational inclusion problems and split common fixed point problems, we introduce a new iterative scheme to approximate a common element of the set of solutions of a split hierarchical monotone variational inclusion and the set of common fixed points of one-parameter nonexpansive semigroups in the setting of two Hilbert spaces. Under some suitable conditions on parameters, some weak and strong convergence theorems for the sequences generated by the proposed algorithms are proved. The results presented in the paper improve and extend some recent corresponding results.

5 Split hierarchical monotone variational inclusions problems Preliminaries We now recall some definitions and elementary facts which will be used in the proofs of our main results. A multi-valued mapping B : H 1 2 H 1 is called monotone if, for all x, y H 1, u Bx and v By such that x y, u v 0. (2.1) A monotone mapping B is maximal if the Graph (B) is not properly contained in the graph of any other monotone mapping. It is well known that a monotone mapping B is maximal if and only if for (x, u) H 1 H 1, x y, u v 0 for every (y, v) Graph(B) implies that u Bx. Let B : H 1 2 H 1 be a multi-valued monotone mapping. Then the resolvent mapping J B : H 1 H 1 associated with B is defined by J B (x) := (I + B) 1 (x), x H 1, (2.2) for some > 0, where I stands for the identity operator on H 1. Note that for all > 0 the resolvent operator J B is single-valued, nonexpansive, and firmly nonexpansive. see[14] Let B 1 : H 1 2 H 1 and B 2 : H 2 2 H 2 be set-valued mappings with nonempty values, and let f : H 1 H 1 and g : H 2 H 2 be mappings. Let T : H 1 H 1 and S : H 2 H 2 be operators such that F ix(t ) and F ix(s). Let U := J B 1 (I f) and V := J B 2 (I g), Ansari et al. [14] proved that (1.12) and (1.13) can be rewritten as and such that F ind x F ix(t ) such that x J B 1 (I f), (2.3) y := Ax F ix(s) solves 0 y J B 2 (I g) (2.4) Definition 2.1 An operator T : H H is said to be : (i) strongly nonexpansive [15,16,17] if T is nonexpansive and lim (x n y n ) (T x n T y n ) = 0, (2.5) n whenever {x n } and {y n } are bounded sequences in H and lim ( x n y n T x n T y n ) = 0; (2.6) n (ii) averaged nonexpansive [18] if it can be written as T = (1 α)i + αs, (2.7)

6 400 Zhaoli Ma, Lin Wang and Jinhong Zhao where α (0, 1), I is the identity operator on H, and S : H H is a nonexpansive mapping; (iii) firmly nonexpansive if T x T y 2 x y, T x T y, for all x, y H; (iv) α-inverse strongly monotone (α-ism) if there exists a constant α > 0 such that T x T y, x y α T x T y 2, for all x, y H. Proposition 2.2 [18] Let T : H H be an operator (i) If T is ν-ism, then for γ > 0, γt is ν -ism; γ (ii) T is averaged if and only if the complement I T is ν-ism for some ν > 1. 2 Indeed, for α (0, 1), T is α-averaged if and only if I T is 1 -ism; 2α (iii) The composite of finitely many averaged mappings is averaged. Let ϕ : H H be a given single-valued α-inverse strongly monotone operator and (0, 2α). Then (I ϕ) is averaged. Indeed, since ϕ is α-inverse strongly monotone, ϕ is α -ism. Thus, I ϕ is averaged as α > 1 2. Since J B is firmly nonexpansive, and therefore, it is averaged. It is well known that the composition of averaged mapping is averaged, thus J B (I ϕ) is averaged. Since every averaged mapping is strongly nonexpansive [17], it follows that J B (I ϕ) is also strongly nonexpansive. Lemma 2.3 [14] Let φ : H H be a given single-valued operator, B : H 2 H be maximal monotone set-valued mapping. Then 0 φ(x ) + B(x ) x F ix(j B (I φ))x. (2.8) Let T : C C be a mapping with F (T ). Then T is said to be demiclosed at zero if for any {x n } C with x n x and x n T x n 0, x = T x. A mapping T : C C is said to be semi compact, if for any sequence {x n } in C such that x n T x n 0, (n ), there exists subsequence {x nj } of {x n } such that {x xj } converges strongly to x C. A Banach space E is said to satisfy Opial property if for any sequence {x n } in E, x n x, for any y E with y x, we have lim inf n x n x < lim inf n x n y. (2.9) It is well known that every Hilbert space satisfies Opial s condition. Lemma 2.4 ([19]) Let C be a nonempty a nonempty, closed, and convex subset of a real Hilbert space H and T : C C be a nonexpansive operator with F ix(t ), If the sequence {x n } C converges weakly to x and the sequence {(I T )x n } converges strongly to y, then (I T )x = y. In particular, if y = 0, then x F ix(t ).

7 Split hierarchical monotone variational inclusions problems 401 In a real Hilbert space H, it is also well known that tx+(1 t)y 2 = t x 2 +(1 t) y 2 t(1 t) x y 2, t [0, 1], for all x, y H. (2.10) and 2 x, y = x 2 + y 2 x y 2, for all x, y H. (2.11) 3 Main Results Theorem 3.1 Let H 1 and H 2 be real Hilbert spaces. Let A : H 1 H 1 be a bounded linear operator and A be the adjoint of A, f : H 1 H 1 be a β 1 -inverse strongly monotone operator, {T (t) : t 0} : H 1 H 1 and {S(t) : t 0} : H 2 H 2 be uniformly asymptotically regular nonexpansive semigroups such that t 0 F (T (t)) and F (S(t)), g : H 2 H 2 be a β 2 - inverse strongly monotone operator, and B 1 : H 1 2 H 1 and B 2 : H 2 2 H 2 be two maximal monotone set-valued mappings with nonempty values. Let {x n } be a sequence generated by: x 1 H 1 { yn = x n + γa (S(t n )V I)Ax n, n 1, (3.1) x n+1 = (1 α n )y n + α n T (t n )U(y n ), where the operator U := J B 1 (I f) and V := J B 2 (I g) with (0, 2α), {t n } is sequence of real numbers, {α n }, {β i }(i = 1, 2) and γ satisfy the following conditions: (1) t n > 0 and lim n t n = ; (2) β = min{β 1, β 2 } and γ (0, 1 A 2 ). (3) lim inf n α n (1 α n ) > 0 (I) If Γ, then the sequence {x n } converges weakly to an element x Γ. (II) In addition, if Γ, U is semi-compact and there exists at least one T (t) {T (t) : t 0} is semi-compact, then {x n } converges strongly to an element x Γ. Proof. Now we prove the conclusion (I). We shall divide the proof into four steps. Step 1. We first show that the limit lim n x n p exists for any p Γ. For any given p Γ, then y n p 2 = x n + γa (S(t n )V I)Ax n p 2 = x n p 2 + γ 2 A (S(t n )V I)Ax n 2 + 2γ x n p, A (S(t n )V I)Ax n x n p 2 + γ 2 A 2 (S(t n )V I)Ax n 2 + 2γ x n p, A (S(t n )V I)Ax n, (3.2)

8 402 Zhaoli Ma, Lin Wang and Jinhong Zhao where, x n p,a (S(t n )V I)Ax n = Ax n Ap, (S(t n )V I)Ax n = (S(t n )V I)Ax n Ap + (S(t n )V I)Ax n + Ax n, (S(t n )V I)Ax n = S(t n )V Ax n Ap, (S(t n )V I)Ax n (S(t n )V I)Ax n 2 = 1 2 S(t n)v Ax n Ap (S(t n)v Ax n Ax n Ax n Ap 2 (S(t n )V I)Ax n V Ax n Ap (S(t n)v Ax n Ax n Ax n Ap 2 (S(t n )V I)Ax n Ax n Ap (S(t n)v Ax n Ax n Ax n Ap 2 (S(t n )V I)Ax n 2 = 1 2 (S(t n)v I)Ax n 2. It follows from (3.2), (3.3) that (3.3) y n p 2 x n p 2 + γ 2 A 2 (S(t n )V I)Ax n 2 γ (S(t n )V I)Ax n 2 = x n p 2 γ(1 γ A 2 ) (S(t n )V I)Ax n 2. Since 0 < γ < 1 A 2, so, 0 < γ A 2 < 1, and from (3.4), we have It from (3.1) and (3.4) that (3.4) y n p 2 x n p 2. (3.5) x n+1 p 2 = (1 α n )(y n p) + α n (T (t n )Uy n p) 2 = (1 α n ) y n p 2 + α n T (t n )Uy n p 2 α n (1 α n ) y n T (t n )Uy n 2 y n p 2 α n (1 α n ) y n T (t n )Uy n 2 x n p 2 γ(1 γ A 2 ) (S(t n )V I)Ax n 2 α n (1 α n ) y n T (t n )Uy n 2. (3.6) This implies that lim n x n p exists. So, we obtian that {x n } is bounded. Step 2. We prove that lim n y n T (t)uy n = 0, lim n (S(t)V I)Ax n = 0 and lim n V Ax n Ax n = 0. It follows from (3.6) that x n+1 p 2 x n p 2 γ(1 γ A 2 ) (S(t n )V I)Ax n 2 α n (1 α n ) y n T (t n )Uy n 2. (3.7)

9 Split hierarchical monotone variational inclusions problems 403 From (3.7), we have γ(1 γ A 2 ) (S(t n )V I)Ax n 2 +α n (1 α n ) y n T (t n )Uy n 2 x n p 2 x n+1 p 2. (3.8) This implies that lim n y n T (t n )Uy n = 0. (3.9) and lim n (S(t n )V I)Ax n = 0, (3.10) Since {S(t)} is uniformly asymptotically regular nonexpansive semigroup, C is any bounded subset of H 1 containing {y n } and lim n t n =, for all t 0, then lim n T (t)(t (t n )Uy n ) T (t n )y n lim sup T (t)(t (t n )Ux) T (t n )x 0. n x C Since {T (t)x} is continuous on t for all x H 1, and (3.11) y n T (t)uy n y n T (t n )Uy n + T (t n )Uy n T (t)t (t n )Uy n + T (t)t (t n )Uy n T (t)uy n, (3.12) it follows from (3.6) and (3.12) that Similarly, Since, So, by (3.10), we obtain In addition, since lim n y n T (t)uy n = 0. (3.13) lim n (S(t)V I)Ax n = 0. (3.14) y n x n = x n + γa (St n )V I)Ax n x n = γ A (St n )V I)Ax n. (3.15) lim n y n x n = 0. (3.16) S(t n )V Ax n Ap Ax n Ap S(t n )V Ax n Ax n. (3.17) Taking the limit on both sides of the above inequality and by using (3.10), we have lim n S(t n )V Ax n Ap Ax n Ap = 0. (3.18) Therefore, lim n ( S(t n )V Ax n Ap Ax n Ap ) = 0. (3.19)

10 404 Zhaoli Ma, Lin Wang and Jinhong Zhao Since S and V := J B 2 (I g) are nonexpansive, from (3.19) that, So, S(t n )V Ax n Ap V Ax n Ap Ax n Ap. (3.20) S(t n )V Ax n Ap Ax n Ap V Ax n Ap Ax n Ap 0. (3.21) Therefore, lim n ( S(t n )V Ax n Ap Ax n Ap ) lim n ( V Ax n Ap Ax n Ap ) 0. (3.22) It follows from (3.19) and (3.22) that lim n ( V Ax n Ap Ax n Ap ) = 0. (3.23) Since V is averaged nonexpansive and every averaged nonexpansive map is strongly nonexpansive, so, V is strongly nonexpansive. By the definition of strong nonexpansiveness of V and the boundedness of {Ap} and {Ax n }, we have lim n V Ax n Ax n = 0. (3.24) Step 3. We prove that lim n Uy n y n = 0 and lim n Uy n T (t)uy n = 0 Since, It follows from (3.9) that So, y n p T (t n )Uy n p y n T (t n )Uy n. (3.25) lim n y n p T Uy n p = 0. (3.26) lim n ( y n p T Uy n p ) = 0. (3.27) By the nonexpansiveness of T (t) and U, we have and T (t n )Uy n p Uy n p y n p. (3.28) T (t n )Uy n p y n p Uy n p y n p 0. (3.29) By (3.27) and (3.29), we get lim n ( Uy n p y n p ) = 0. (3.30)

11 Split hierarchical monotone variational inclusions problems 405 Since {y n } is a bounded sequence and {p} being a constant sequence, also bounded, U is averaged nonexpansive and every averaged nonexpansive map is strongly nonexpansive, by the strong nonexpansiveness of U, we have In addition, since lim n Uy n y n = 0. (3.31) Uy n T (t)uy n Uy n y n + y n T (t)uy n. (3.32) It follows from (3.13) and (3.32) that lim n Uy n T (t)uy n = 0. (3.33) Step 4. We prove that {x n } converges weakly to x Γ. Since {x n } is a bounded sequence, there exists a subsequence {x ni } of {x n } converging weakly to x. In addition, since A is a bounded linear operator, we know that {Ax ni } converges weakly to Ax. Further, Since S(t)V is nonespansive for all t 0 and V is nonexpansive, by Lemma 2.4, (3.14) and (3.24), we know that S(t)V Ax = Ax and V Ax = Ax, this means that Ax F (S(t)) solves Ax F (V ) = F (J B 2 (I g)). On the other hand, for all f H, f(y n ) f(x ) = f(y n ) + f(x n ) f(x n ) f(x ) f(y n ) f(x n ) + f(x n ) f(x ) f y n x n + f(x n ) f(x ). (3.34) Since x n x and from (3.16), we have lim n f(y n ) f(x ) = 0, thus, y n x, it follows from (3.31) that Uy n x, since U is demiclosed at zero, we have Ux = x. Since T (t) is nonespansive for all t 0 and by Lemma 2.4, (3.31) and (3.33), we have T x = x, that is, x F (T (t)) solves x F (U) = F (J B 1 (I f)). This means that x Γ. assume that there exists another subsequence {x nj } of {x n } such that {x nj } converges weakly to y H 1, using the same argument above, we also know that y Γ. Since each Hilbert space possesses Opial s condition, we have lim inf n i x n i x < lim inf n i x n i y = lim inf x n y n = lim inf x n j y n j < lim inf n j x n j x = lim inf n x n x = lim inf n i x n i x. (3.35)

12 406 Zhaoli Ma, Lin Wang and Jinhong Zhao This is a contradiction. Therefore {x n } converges weakly to x Γ. The proof of conclusion(i) is completed. Next, we prove the conclusion(ii). Since U is semi-compact and there exists at least one T (t) {T (t) : t 0} is semi-compact, and lim n Uy n y n = 0 and lim n Uy n T (t)uy n = 0 for all t 0, there exist subsequence {y nj } of {y n } such that {y nj } converges strongly to µ H 1. By using (3.16) again, we know that the subsequence {x nj } of {x n } converges strongly to µ. Due to {x n } converges weakly to x, we obtain µ = x. By lim n x n x exists and lim n x nj x = 0, we know that {x n } converges strongly to x Γ. This completes the proof of the conclusion(ii). Corollary 3.2 Let H 1 and H 2 be real Hilbert spaces. Let A : H 1 H 1 be a bounded linear operator and A be the adjoint of A, f : H 1 H 1 be an β 1 -inverse strongly monotone operator, T : H 1 H 1 be a nonexpansive mapping such that F (T ), g : H 2 H 2 be an β 2 -inverse strongly monotone operator, S : H 2 H 2 be nonexpansive mapping such that F (S), and B 1 : H 1 2 H 1 and B 2 : H 2 2 H 2 are two maximal monotone set-valued mappings with nonempty values. Let {x n } be a sequence generated by: x 1 H 1 { yn = x n + γa (SV I)Ax n, n 1, (3.36) x n+1 = (1 α n )y n + α n T U(y n ), where the operator U := J B 1 (I f) and V := J B 2 (I g) with (0, 2α), {α n }, {β i }(i = 1, 2) and γ satisfy the following conditions: (1) β = min{β 1, β 2 } and γ (0, 1 A 2 ). (2) lim inf n α n (1 α n ) > 0 (1)If Γ, then the sequence {x n } converges weakly to an element x Γ. (II)In addition, if Γ, U and T are semi-compact, then {x n } converges strongly to an element x Γ. Acknowledgements. This work was supported by the National Natural Science Foundation of China (Grant No ), and the Science Foundation of Education Department of Yunnan Province (Grant No. 2014C220Y). References [1] F. E. Browder, Convergence of approximants to fixed points of nonexpansive nonlinear mappings in Banach spaces, Arch. Ration. Mech. and Anal., 24 (1967),

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