Split equality problem with equilibrium problem, variational inequality problem, and fixed point problem of nonexpansive semigroups

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1 Available online at J. Nonlinear Sci. Appl., 10 (2017), Research Article Journal Homepage: Split equality problem with equilibrium problem, variational inequality problem, and fixed point problem of nonexpansive semigroups Abdul Latif a,, Mohammad Eslamian b a Department of Mathematics, King Abdulaziz University, P. O. Box 80203, Jeddah 21589, Saudi Arabia. b Department of Mathematics, University of Science and Technology of Mazandaran, P. O. Box , Behshahr, Iran. Communicated by Z. Kadelburg Abstract In this paper, we present a new algorithm for the split equality problem for finding a common element of solution of equilibrium problem, solution of variational inequality problem for monotone and Lipschitz continuous operators, and common fixed point of nonexpansive semigroups. We establish strong convergence of the algorithm in an infinite dimensional Hilbert spaces. Our results improve and generalize some recent results in the literature. c 2017 All rights reserved. Keywords: Split equality problem, equilibrium problem, variational inequality, nonexpansive semigroups, fixed point MSC: 47J25, 47N10, 65J15, 90C Introduction Let C be a nonempty closed convex subset of H. Let Φ : C C R be a bifunction. In [6] it was shown that a broad class of problems in optimization, such as variational inequality, convex minimization, fixed point, and Nash equilibrium problems can be formulated as the equilibrium problem associated to the bifunction Φ and the set C: find x C such that Φ(x, y) 0, y C. A point x C solving this problem is said to be an equilibrium point. The set of solutions is denoted by EP(Φ). Numerous problems in physics, optimization, and economics are reduced to find a solution of the equilibrium problem; see [36]. Many techniques and algorithms have been devised to analyze the existence and approximation of a solution to equilibrium problems; see [16]. Let F : H H be a nonlinear operator. It is well-known that the variational inequality problem is to find u C such that Fu, v u 0, v C. (1.1) Corresponding author addresses: alatif@kau.edu.sa (Abdul Latif), mhmdeslamian@gmail.com (Mohammad Eslamian) doi: /jnsa Received

2 A. Latif, M. Eslamian, J. Nonlinear Sci. Appl., 10 (2017), We denote by VI(C, F) the solution set of (1.1). The theory of variational inequalities has played an important role in the study of a wide class of problems arising in pure and applied sciences including mechanics, optimization and optimal control, partial differential equation, operations research and engineering sciences. During the last decades this problem has been studied by many authors, (see [5, 23, 25, 37, 38]). We recall the following definition on F : H H. The operator F is called Lipschitz continuous on C H with constant L > 0 if F(x) F(y) L x y, x, y C; nonexpansive on C if monotone on C if F(x) F(y) x y, x, y C; F(x) F(y), x y 0, x, y C; inverse strongly monotone with constant β > 0, (β-ism) if F(x) F(y), x y β F(x) F(y) 2, x, y C. We note that every β-inverse strongly monotone operator is monotone and Lipschitz continuous. It is known that if F is β-inverse strongly monotone, and λ (0, 2β), then P C (I λf) is nonexpansive. It is worth noting that there exists a monotone Lipschitz continuous operator F such that P C (I λf) fails to be nonexpansive [15]. A family T := {T(s) : 0 s < } of mappings on C is called a nonexpansive semigroup if it satisfies the following conditions: (i) T(0)x = x for all x C; (ii) T(s + t) = T(s)T(t) for all s, t 0; (iii) T(s)x T(s)y x y for all x, y C and s 0; (iv) for all x C, s T(s)x is continuous. We use F(T) to denote the common fixed point set of the semigroup T, i.e., F(T) = {x C : T(s)x = x, s 0}. It is well-known that F(T) is closed and convex [7]. A nonexpansive semigroup T on C is said to be uniformly asymptotically regular (in short, u.a.r.) on C if for all h 0 and any bounded subset E of C, For each h 0, define σ t (x) = 1 t t 0 lim sup T(h)(T(t)x) T(t)x = 0. t x E T(s)x ds. Then lim sup T(h)(σ t (x)) σ t (x) = 0, t x E provides E be a closed bounded convex subset of C. It is known that the set {σ t (x) : t > 0} is a u.a.r. nonexpansive semigroup; see [14]. The other examples of u.a.r. operator semigroup can be found in [1]. It is well-known that the semigroup result has implications in partial differential equation theory, evolutionary equation theory, and fixed point theory. The nonexpansive semigroup is directly linked to solutions of differential equations and has been studied by several authors (see, for example, [29, 39] and the references therein). In the last years, many authors studied the problems of finding a common element of the set of fixed points of nonlinear operator and the set of solutions of an equilibrium problem (and the set of solutions of variational inequality problem) in the framework of Hilbert spaces, see, for instance, [2, 11, 18, 20, 24, 31] and the references therein. The motivation for studying such a problem is in its possible application to

3 A. Latif, M. Eslamian, J. Nonlinear Sci. Appl., 10 (2017), mathematical models whose constraints can be expressed as fixed-point problems and/or equilibrium problem. This happens, in particular, in the practical problems as signal processing, network resource allocation, and image recovery; see, for instance, [24, 31]. Let C and Q be nonempty closed convex subsets of real Hilbert spaces H 1 and H 2, respectively. The split feasibility problem (SFP) is formulated as: to find x C and Ax Q, where A : H 1 H 2 is a bounded linear operator. In 1994, Censor and Elfving [11] first introduced the SFP in finite-dimensional Hilbert spaces for modeling inverse problems which arise from phase retrievals and in medical image reconstruction [8]. The split feasibility problem has received much attention due to its applications in signal processing and image reconstruction [9], with particular progress in intensitymodulated therapy [10]. Recently, Moudafi [33] introduced the following split equality problem. Let A : H 1 H 3, B : H 2 H 3 be two bounded linear operators, let C and Q be nonempty closed convex subsets of real Hilbert spaces H 1 and H 2, respectively. The split equality problem (SEP) is to find x C, y Q such that Ax = By, which allows asymmetric and partial relations between the variables x and y. The interest is to cover many situations, for instance in decomposition methods for PDEs, applications in game theory and in intensity-modulated radiation therapy (IMRT). In decision sciences, this allows consideration of agents that interplay only via some components of their decision variables (see [3, 4]). Since every nonempty closed convex subset of a Hilbert space can be regarded as a set of fixed points of a projection, Moudafi [34] introduced the following split equality fixed point problem (SEFP). Let A : H 1 H 3, B : H 2 H 3 be two bounded linear operators, let S : H 1 H 1 and T : H 2 H 2 be two nonlinear operators such that Fix(S) and Fix(T). The split equality fixed point problem (SEFP) is to find x Fix(S), y Fix(T) such that Ax = By. (1.2) If H 2 = H 3 and B = I, then the split equality fixed point problem (1.1) reduces to the split common fixed point problem (SCFP) originally introduced in Censor and Segal [13] which is to find x Fix(S) with Ax Fix(T). Algorithms for solving the SEP and SCFP received great attention (see [17, 19, 21, 22, 26 28, 30, 32, 35, 40, 42] and references therein). In [35], Moudafi and Al-Shemas introduced the following simultaneous iterative method to solve SEFP (1.2) { x n+1 = S(x n γ n A (Ax n By n )), y n+1 = T(y n + γ n A n 0, (1.3) (Ax n By n )), 2 for firmly quasi-nonexpansive operators S and T, where γ n (ɛ, λ A +λ B ɛ) where λ A and λ B stand for the spectral radius of A A and B B, respectively. We note that in the above algorithm the determination of the step-size {γ n } depends on the operator (matrix) norms A and B (or the largest eigenvalues of A A and B B). In order to implement the alternating algorithm (1.3) for solving SEFP (1.2), the computation (or, at least, estimation) of the operator norms of A and B, which is in general not an easy task in practice, will be required. To overcome this difficulty, Lopez et al. [30] introduced a method for estimating the step-sizes which does not require any knowledge of the operator norms for solving the split feasibility problems. Inspired by them, Zhao [42] and Dong et al. [17] introduced some new choice of the step-size sequence {γ n } for the simultaneous Mann iterative algorithm to solve SEFP. Now, we consider a new type of split equality problem with equilibrium problem, variational inequality problem and fixed point problem as follows: Let H 1, H 2, and H 3, be real Hilbert spaces, A : H 1 H 3 and B : H 2 H 3 be bounded linear operators, and let C and Q be two nonempty closed convex subsets of H 1 and H 2, respectively. Let

4 A. Latif, M. Eslamian, J. Nonlinear Sci. Appl., 10 (2017), T := {T(t) : t 0} and S := {S(t) : t 0} be two u.a.r. nonexpansive semigroups on H 1 and H 2, respectively. Let, F : H 1 H 1 be a monotone and L-Lipschitz continuous operator on C and G : H 2 H 2 be a monotone and K-Lipschitz continuous operator on Q and that Φ : C C R and Ψ : Q Q R be functions. We find a point x Fix(T) EP(Φ) VI(C, F) and y Fix(S) EP(Φ) VI(Q, G) such that Ax = By. In this paper, using the extragradient method introduced by Korpelevich [26], we present a new algorithm for the split equality problem for finding a common element of solution of equilibrium problem, solution of variational inequality problem for monotone and Lipschitz continuous operators, and common fixed point of nonexpansive semigroups. Our algorithm does not require any knowledge of the operator norms. We establish strong convergence of the algorithm in an infinite dimensional Hilbert space. Our results improve and generalize the result of Moudafi [34, 35], Censor et al. [12], Zhao [42], and many others. 2. Preliminaries In the sequel, we use the notions and for weak convergence and strong convergence, respectively. We need the following lemmas to prove our main result. Lemma 2.1 ([40]). Assume that {a n } is a sequence of nonnegative real numbers such that a n+1 (1 ϑ n )a n + ϑ n δ n, n 0, where {ϑ n } is a sequence in (0, 1) and {δ n } is a sequence in R such that (i) n=1 ϑ n = ; (ii) lim sup δ n 0 or n=1 ϑ nδ n <. Then lim a n = 0. Lemma 2.2 ([32]). Let {t n } be a sequence of real numbers such that there exists a subsequence {n i } of {n} such that t ni < t ni +1 for all i N. Then there exists a nondecreasing sequence {τ(n)} N such that τ(n) and the following properties are satisfied by all (sufficiently large) numbers n N: In fact t τ(n) t τ(n)+1, t n t τ(n)+1. τ(n) = max{k n : t k < t k+1 }. Lemma 2.3 ([41]). For each x 1,, x m H and α 1,, α m [0, 1] with m i=1 α i = 1, the following equality holds m α 1 x α m x m 2 = α i x i 2 α i α j x i x j 2. i=1 1 i<j m Given a nonempty closed convex set C H, the mapping that assigns every point x H to its unique nearest point in C is called the metric projection onto C and is denoted by P C, i.e., P C C and x P C x = inf y C x y. The metric projection P C is characterized by the fact that P C (x) C and y P C (x), x P C (x) 0, x H, y C. Lemma 2.4 ([2]). Let F : H H be a monotone and L-Lipschitz operator on C and λ be a positive number. Let u n = P C (x n F(x n )) and v n = P C (x n λf(u n )). Then for all x VI(C, F) we have v n x 2 x n x 2 (1 λ L) u n x n 2 (1 λ L) u n v n 2.

5 A. Latif, M. Eslamian, J. Nonlinear Sci. Appl., 10 (2017), For solving the equilibrium problem, let us assume that the bifunction Φ satisfies the following conditions: (A1) Φ(x, x) = 0 for all x C; (A2) Φ is monotone, i.e., Φ(x, y) + Φ(y, x) 0 for any x, y C; (A3) for each x, y, z C lim sup Φ(tz + (1 t)x, y) Φ(x, y); t 0 + (A4) for each x C, y Φ(x, y) is convex and lower semi-continuous. We know the following lemma which appeared implicitly in Blum et al. [6] and Combettes et al. [16]. Lemma 2.5 ([6, 16]). Let C be a nonempty closed convex subset of H and let Φ be a bifunction of C C into R satisfying (A1)-(A4). Let r > 0 and x H. Then, there exists z C such that Φ(z, y) + 1 y z, z x 0, y C. r Further, if U Φ r x = {z C : Φ(z, y) + 1 y z, z x 0, r y C}, then, the followings hold (i) U Φ r is single-valued; (ii) U Φ r is firmly nonexpansive, i.e., for any x, y H, (iii) Fix(U Φ r ) = EP(Φ); (iv) EP(Φ) is closed and convex. 3. Split equality problem U Φ r x U Φ r y 2 U Φ r x U Φ r y, x y ; In this section, by combining Halpern s iterative method and the extragradient method [26], we present our algorithm for solving the split equality problem with equilibrium problem, variational inequality problem, and fixed point problem in Hilbert spaces. Theorem 3.1. Let H 1, H 2, and H 3, be real Hilbert spaces, A : H 1 H 3 and B : H 2 H 3 be bounded linear operators and let C and Q be two nonempty closed convex subsets of H 1 and H 2, respectively. Let T := {T(t) : t 0} and S := {S(t) : t 0} be two u.a.r. nonexpansive semigroups on H 1 and H 2, respectively. Let, F : H 1 H 1 be a monotone and L- Lipschitz continuous operator on C and G : H 2 H 2 be a monotone and K- Lipschitz continuous operator on Q and that Φ : C C R and Ψ : Q Q R be functions satisfying conditions (A1)-(A4). Suppose Ω = {x Fix(T) VI(C, F) EP(Φ), y Fix(S) VI(Q, G) EP(Ψ) : Ax = By}. Let {x n } and {y n } be sequences generated by x 0, ϑ H 1, y 0, ζ H 2, and by z n = x n γ n A (Ax n By n ), η n = U Φ κ n,1 z n, u n = P C (η n λ n F(η n )), v n = P C (η n λ n F(u n )), x n+1 = α n ϑ + β n v n + δ n T(r n )v n w n = y n + γ n B (Ax n By n ), σ n = U Ψ κ n,2 w n, s n = P Q (σ n ρ n G(σ n )), t n = P Q (σ n ρ n G(s n )), y n+1 = α n ζ + β n t n + δ n S(ι n )t n, n 0, (3.1)

6 A. Latif, M. Eslamian, J. Nonlinear Sci. Appl., 10 (2017), where the step-size γ n is chosen in such a way that γ n (ɛ, 2 Ax n By n 2 B (Ax n By n ) 2 + A ɛ), n Π, (Ax n By n ) 2 otherwise γ n = γ (γ being any nonnegative value), where the index set Π = {n : Ax n By n 0}. Let the sequences {r n }, {ι n }, {κ n,1 }, {κ n,2 }, {α n }, {β n }, {δ n }, {λ n }, and {ρ n } satisfy the following conditions: (i) α n + β n + δ n = 1, and lim inf n β n δ n > 0; (ii) lim r n = and lim ι n = ; (iii) λ n [a, b] (0, 1 L ) and ρ n [c, d] (0, 1 K ); (iv) lim α n = 0, n=0 α n = ; (v) lim inf n κ n,1 > 0 and lim inf n κ n,2 > 0. Then, the sequence {(x n, y n )} converges strongly to (x, y ) Ω. Proof. Take (x, y ) Ω. Since the operator U Φ κ n,1 is firmly nonexpansive (see Lemma 2.5), we have Also, we have and From Lemma 2.4 we deduce η n x 2 = U Φ κ n,1 z n x 2 z n x 2 z n η n 2. (3.2) σ n y 2 = U Ψ κ n,2 w n y 2 w n y 2 w n σ n 2. (3.3) v n x 2 η n x 2 (1 λ n L) η n u n 2 (1 λ n L) u n v n 2, (3.4) t n y 2 σ n y 2 (1 ρ n K) σ n s n 2 (1 ρ n K) t n s n 2. (3.5) Applying Lemma 2.3 and inequalities (3.2) and (3.4) we have x n+1 x 2 = α n ϑ + β n v n + δ n T(r n )v n x 2 α n ϑ x 2 + β n v n x 2 + δ n T(r n )v n x 2 β n δ n T(r n )v n v n 2 α n ϑ x 2 + β n v n x 2 + δ n v n x 2 β n δ n T(r n )v n v n 2 α n ϑ x 2 + (1 α n ) z n x 2 β n δ n T(r n )v n v n 2 (1 α n ) z n η n 2 (1 α n )(1 λ n L) η n u n 2 (1 α n )(1 λ n L) u n v n 2. Similarly, from inequalities (3.3) and (3.5) we have y n+1 y 2 = α n ζ + β n t n + δ n S(ι n )t n y 2 From algorithm (3.1) we have that α n ζ y 2 + β n t n y 2 + δ n S(ι n )t n y 2 β n δ n S(ι n )t n t n 2 α n ζ y 2 + (1 α n ) w n y 2 β n δ n S(ι n )t n t n 2 (1 α n ) w n σ n 2 (1 α n )(1 ρ n K) σ n s n 2 (1 α n )(1 ρ n K) t n s n 2. z n x 2 = x n γ n A (Ax n By n ) x 2 = x n x 2 + γ 2 n A (Ax n By n ) 2 2γ n x n x, A (Ax n By n ) = x n x 2 + γ 2 n A (Ax n By n ) 2 2γ n Ax n Ax, (Ax n By n ) = x n x 2 + γ 2 n A (Ax n By n ) 2 γ n Ax n Ax 2 γ n Ax n By n 2 + γ n By n Ax 2. (3.6) (3.7)

7 A. Latif, M. Eslamian, J. Nonlinear Sci. Appl., 10 (2017), By similar way we obtain that w n y 2 = y n + γ n B (Ax n By n ) y 2 = y n y 2 + γ 2 n B (Ax n By n ) 2 γ n By n By 2 γ n Ax n By n 2 + γ n Ax n By 2. By adding the two last inequalities and by taking into account the fact that Ax = By we obtain z n x 2 + w n y 2 = x n x 2 + y n y 2 γ n [2 Ax n By n 2 γ n ( B (Ax n By n ) 2 + A (Ax n By n ) 2 )] x n x 2 + y n y 2. (3.8) This implies that x n+1 x 2 + y n+1 y 2 (1 α n )( z n x 2 + w n y 2 ) + α n ( ϑ x 2 + ζ y 2 ) (1 α n )( x n x 2 + y n y 2 ) + α n ( ϑ x 2 + ζ y 2 ) max{ x n x 2 + y n y 2, ϑ x 2 + ζ y 2 }. max{ x 0 x 2 + y 0 y 2, ϑ x 2 + ζ y 2 }. Thus x n+1 x 2 + y n+1 y 2 is bounded. Therefore {x n } and {y n } are bounded. Consequently {z n }, {w n }, {v n }, and {t n } are all bounded. From (3.6), (3.7), and (3.8) we have that x n+1 x 2 + y n+1 y 2 (1 α n )( z n x 2 + w n y 2 ) + α n ( ϑ x 2 + ζ y 2 ) β n δ n T(r n )v n v n 2 β n δ n S(ι n )t n t n 2 (1 α n ) z n η n 2 (1 α n ) w n σ n 2 (1 α n )(1 λ n L) η n u n 2 (1 α n )(1 λ n L) u n v n 2, (1 α n )(1 ρ n K) σ n s n 2 (1 α n )(1 ρ n K) t n s n 2 (1 α n )( x n x 2 + y n y 2 ) + α n ( ϑ x 2 + ζ y 2 ) (1 α n )γ n [2 Ax n By n 2 γ n ( B (Ax n By n ) 2 + A (Ax n By n ) 2 )] β n δ n T(r n )v n v n 2 β n δ n S(ι n )t n t n 2 (1 α n ) z n η n 2 (1 α n ) w n σ n 2 (1 α n )(1 λ n L) η n u n 2 (1 α n )(1 λ n L) u n v n 2, (1 α n )(1 ρ n K) σ n s n 2 (1 α n )(1 ρ n K) t n s n 2. (3.9) From above inequality we have that (1 α n )(1 λ n L) η n u n 2 (1 α n )( x n x 2 + y n y 2 ) x n+1 x 2 y n+1 y 2 + α n ( ϑ x 2 + ζ y 2 ). (3.10) By our assumption that γ n (ɛ, 2 Ax n By n 2 B (Ax n By n ) 2 + A (Ax n By n ) 2 ɛ),

8 A. Latif, M. Eslamian, J. Nonlinear Sci. Appl., 10 (2017), we have that (γ n + ɛ) B (Ax n By n ) 2 + A (Ax n By n ) 2 2 Ax n By n 2. From above inequality and inequality (3.9) we have that (1 α n )γ 2 n( B (Ax n By n ) 2 + A (Ax n By n ) 2 ) (1 α n )γ n [2 Ax n By n 2 γ n ( B (Ax n By n ) 2 + A (Ax n By n ) 2 )] (1 α n )( x n x 2 + y n y 2 ) x n+1 x 2 y n+1 y 2 + α n ( ϑ x 2 + ζ y 2 ). (3.11) We finally analyze the inequalities (3.10) and (3.11) by considering the following two cases. Case A. Put Γ n = x n x 2 + y n y 2 for all n N. Suppose that Γ n+1 Γ n for all n n 0 (for n 0 large enough). In this case, since Γ n is bounded, the limit lim Γ n exists. Since lim α n = 0, from (3.11) and by our assumption on {γ n }, we have lim ( B (Ax n By n ) 2 + A (Ax n By n ) 2 ) = 0. So we obtain that lim B (Ax n By n ) = 0 and lim A (Ax n By n ) = 0. This implies that lim Ax n By n = 0. Also from (3.10) we deduce lim (1 α n)(1 λ n L) η n u n 2 = 0. By our assumption that λ n [a, b] (0, 1 L ), we obtain that By similar argument we get that and lim η n u n = 0. (3.12) lim u n v n = σ n s n = t n s n = 0, (3.13) lim S(ι n)t n t n = lim T(r n)v n v n = z n η n = w n σ n = 0. Further, for all h 0 and n 0, we see that t n S(h)t n t n S(ι n )t n + S(ι n )t n S(h)S(ι n )t n + S(h)S(ι n )t n S(h)t n 2 t n S(ι n )t n + sup S(ι n )t n S(h)S(ι n )t n x {t n } Since {S(h)} is u.a.r. nonexpansive semigroup and lim ι n =, so we have lim t n S(h)t n = 0. By similar argument for all h 0, we can obtain that lim T(h)v n v n = 0. Since z n x n = γ n A (Ax n By n ) and {γ n } is bounded, we have From (3.12), (3.13), and (3.14) we have lim z n x n = 0. (3.14) x n v n x n z n + z n η n + η n u n + u n v n 0, as n.

9 A. Latif, M. Eslamian, J. Nonlinear Sci. Appl., 10 (2017), Therefore x n+1 x n α n ϑ x n + β n v n x n + δ n T(r n )v n x n 0, as n. Similarly we have that lim y n+1 y n = 0. Now we claim that (ω w (x n ), ω w (y n )) Ω, where ω w (x n ) = {x H 1 : x ni x for some subsequences {x ni } of {x n }}. Since the sequences {x n } and {y n } are bounded we have ω w (x n ) and ω w (y n ) are nonempty. Now, take x ω w (x n ) and ŷ ω w (y n ). Thus, there exists a subsequence {x ni } of {x n } which converges weakly to x. Without loss of generality, we can assume that x n x. Since lim η n x n = 0, we have η n x. From u n = P C (η n λ n F(η n )), for each x C we have that Since, F is monotone, for each x C we have Utilizing the inequalities (3.15) and (3.16) we have Hence λ n F(x), η n x λ n F(η n ), η n x x u n, η n λ n F(η n ) u n 0. (3.15) λ n F(x), η n x λ n F(η n ), η n x. (3.16) = λ n F(η n ), η n u n + λ n F(η n ), u n x = λ n F(η n ), η n u n + λ n F(η n ) η n + u n, u n x + η n u n, u n x λ n F(η n ), η n u n + η n u n, u n x λ n F(η n ) η n u n + η n u n u n x. Fx, η n x F(η n ) η n u n + 1 λ n η n u n u n x. Since {F(η n )} is bounded, η n u n 0 and η n x, we have F(x), x x = lim F(x), η n x 0, x C. This implies that x VI(C, F). By similar argument we can obtain that ŷ VI(Q, G). Next we show that x Fix(T) and ŷ Fix(S). Since lim v n x n = 0, we have v n x. Now, for all r 0 we have This implies that v n T(r) x v n T(r)v n + T(r)v n T(r) x v n T(r)v n + v n x. lim inf v n T(r) x lim inf v n x. By the Opial property of the Hilbert space H 1 we obtain that T(r)( x) = x for all r 0, which implies that x Fix(T). By similar argument we obtain that ŷ Fix(S). Since lim η n z n = lim U Φ κ n,1 z n z n = 0, and z n x, we have x EP(Φ), (see [18] for details). Similarly we have ŷ EP(Ψ). On the other hand, A x Bŷ ω w (Ax n By n ) and weakly lower semi continuity of the norm imply that A x Bŷ lim inf Ax n By n = 0. Thus ( x, ŷ) Ω. We also have that the uniqueness of the weak cluster point of {x n } are {y n }, (see [42] for details) which implies that the whole sequence {(x n, y n )} weakly converges to a point ( x, ŷ) Ω. Next

10 A. Latif, M. Eslamian, J. Nonlinear Sci. Appl., 10 (2017), we prove that the sequence {(x n, y n )} converges strongly to (ϑ, ζ ) where ϑ = P Ω ϑ and ζ = P Ω ζ. First we show that lim sup ϑ ϑ, x n ϑ 0. To show this inequality, we choose a subsequence {x nk } of {x n } such that Since {x nk } converges weakly to x, it follows that lim ϑ k ϑ, x nk ϑ = lim sup ϑ ϑ, x n ϑ. lim sup ϑ ϑ, x n ϑ = lim ϑ ϑ, x nk ϑ = ϑ ϑ, x ϑ 0. k By similar argument we obtain that lim sup ζ ζ, y n ζ 0. From the inequality x + y 2 x y, x + y, ( x, y H 1 ), we find that x n+1 ϑ 2 β n v n + δ n T(r n )v n (1 α n )ϑ 2 + 2α n ϑ ϑ, x n+1 ϑ = (1 α n ) 2 β n (1 α n ) v δ n n + (1 α n ) T(r n)v n ϑ 2 + 2α n ϑ ϑ, x n+1 ϑ Similarly we obtain that β n (1 α n ) v n ϑ ) 2 + δ n (1 α n ) T(r n )v n ϑ 2 + 2α n ϑ ϑ, x n+1 ϑ (1 α n ) 2 v n ϑ 2 + 2α n ϑ ϑ, x n+1 ϑ. y n+1 ζ 2 (1 α n ) 2 w n ζ 2 + 2α n ζ ζ, y n+1 ζ. By adding the two last inequalities we have that x n+1 ϑ 2 + y n+1 ζ 2 (1 α n ) 2 ( x n ϑ 2 + y n ζ 2 ) + 2α n ( ϑ ϑ, x n+1 ϑ + ζ ζ, y n+1 ζ ). It immediately follows that Γ n+1 (1 α n ) 2 Γ n + 2α n η n = (1 2α n )Γ n + α 2 nγ n + 2α n η n (1 2α n )Γ n + 2α n { α nn + η n ) 2 (1 ζ n ) x n x 2 + ζ n δ n, where η n = ϑ ϑ, x n+1 ϑ + ζ ζ, y n+1 ζ, N = sup{ x n x 2 + y n y 2 : n 0}, ζ n = 2α n and δ n = α nn 2 + η n. It is easy to see that ζ n 0, n=1 ζ n =, and lim sup δ n 0. Hence, all conditions of Lemma 2.1 are satisfied. Therefore, we immediately deduce that lim Γ n = 0. Consequently, lim x n ϑ = lim y n ζ = 0, that is (x n, y n ) (ϑ, ζ ). Case B. Assume that {Γ n } is not a monotone sequence. Then, we can define an integer sequence {τ(n)} for all n n 0 (for some n 0 large enough) by τ(n) = max{k n : Γ k < Γ k+1 }. Clearly, τ is a nondecreasing sequence such that τ(n) as n and for all n n 0, Γ τ(n) < Γ τ(n)+1. Following an argument similar to that in Case A we have Γ τ(n)+1 (1 ζ τ(n) )Γ τ(n) + ζ τ(n) η τ(n),

11 A. Latif, M. Eslamian, J. Nonlinear Sci. Appl., 10 (2017), where ζ τ(n) 0, n=1 ζ τ(n) =, and lim sup ζ τ(n) 0. Hence, by Lemma 2.1, we obtain lim Γ τ(n) = 0. Since lim x n+1 x n =lim y n+1 y n = 0, we also have lim Γ τ(n)+1 = 0. Now Lemma 2.2 implies 0 Γ n max{γ τ(n), Γ n } Γ τ(n)+1. Therefore (x n, y n ) (ϑ, ζ ). This completes the proof. From Theorem 3.1, we obtain the following result for split equality equilibrium problem and fixed point problem. Theorem 3.2. Let H 1, H 2, and H 3 be real Hilbert spaces, A : H 1 H 3 and B : H 2 H 3 be bounded linear operators, and let C and Q be two nonempty closed convex subsets of H 1 and H 2, respectively. Let T := {T(t) : t 0} and S := {S(t) : t 0} be two u.a.r. nonexpansive semigroups on H 1 and H 2, respectively. Let Φ : C C R and Ψ : Q Q R be functions satisfying conditions (A1)-(A4). Suppose Ω = {x Fix(T) EP(Φ), y Fix(S) EP(Ψ) : Ax = By}. Let {x n } and {y n } be sequences generated by x 0, ϑ H 1, y 0, ζ H 2, and by z n = x n γ n A (Ax n By n ), v n = U Φ κ n,1 z n, x n+1 = α n ϑ + β n v n + δ n T(r n )v n, w n = y n + γ n B (Ax n By n ), where the step-size γ n is chosen in such a way that γ n (ɛ, t n = U Ψ κ n,2 w n, y n+1 = α n ζ + β n t n + δ n S(ι n )t n, n 0, 2 Ax n By n 2 B (Ax n By n ) 2 + A ɛ), n Π, (Ax n By n ) 2 otherwise γ n = γ (γ being any nonnegative value), where the index set Π = {n : Ax n By n 0}. Let the sequences {r n }, {ι n }, {κ n,1 }, {κ n,2 }, {α n }, {β n }, and {δ n } satisfy the following conditions: (i) α n + β n + δ n = 1 and lim inf n β n δ n > 0; (ii) lim r n = and lim ι n = ; (iii) lim α n = 0, n=0 α n = ; (iv) lim inf n κ n,1 > 0 and lim inf n κ n,2 > 0. Then, the sequence {(x n, y n )} converges strongly to (x, y ) Ω. As a corollary we obtain the following result for split equality fixed point problem of nonexpansive semigroups. Corollary 3.3. Let H 1, H 2, and H 3, be real Hilbert spaces, A : H 1 H 3 and B : H 2 H 3 be bounded linear operators. Let T := {T(t) : t 0} and S := {S(t) : t 0} be two u.a.r. nonexpansive semigroups on H 1 and H 2, respectively. Suppose Ω = {x Fix(T), y Fix(S) : Ax = By}. Let {x n } and {y n } be sequences generated by x 0, ϑ H 1, y 0, ζ H 2, and by z n = x n γ n A (Ax n By n ), x n+1 = α n ϑ + β n z n + δ n T(r n )z n, where the step-size γ n is chosen in such a way that γ n (ɛ, w n = y n + γ n B (Ax n By n ), y n+1 = α n ζ + β n w n + δ n S(ι n )w n, n 0, 2 Ax n By n 2 B (Ax n By n ) 2 + A ɛ), n Π, (Ax n By n ) 2

12 A. Latif, M. Eslamian, J. Nonlinear Sci. Appl., 10 (2017), otherwise γ n = γ (γ being any nonnegative value), where the index set Π = {n : Ax n By n 0}. Let the sequences {r n }, {ι n }, {α n }, {β n }, and {δ n } satisfy the following conditions: (i) α n + β n + γ n = 1 and lim inf n β n δ n > 0; (ii) lim r n = and lim ι n = ; (iii) lim α n = 0, n=0 α n =. Then, the sequence {(x n, y n )} converges strongly to (x, y ) Ω. As another corollary of our main result we obtain the following result for split equality variational inequality problem for monotone and Lipschitz continuous operators. Theorem 3.4. Let H 1, H 2, and H 3, be real Hilbert spaces, A : H 1 H 3 and B : H 2 H 3 be bounded linear operators, and let C and Q be two nonempty closed convex subsets of H 1 and H 2, respectively. Let F : H 1 H 1 be a monotone and L-Lipschitz continuous operator on C and G : H 2 H 2 be a monotone and K-Lipschitz continuous operator on Q. Suppose Ω = {x VI(C, F), y VI(Q, G) : Ax = By}. Let {x n } and {y n } be sequences generated by x 0, ϑ H 1, y 0, ζ H 2, and by z n = x n γ n A (Ax n By n ), u n = P C (z n λ n F(z n )), v n = P C (z n λ n F(u n )), x n+1 = α n ϑ + (1 α n )v n, where the step-size γ n is chosen in such a way that γ n (ɛ, w n = y n + γ n B (Ax n By n ), s n = P Q (w n η n G(w n )), t n = P Q (w n η n G(s n )), y n+1 = α n ζ + (1 α n )t n, n 0, 2 Ax n By n 2 B (Ax n By n ) 2 + A ɛ), n Π, (Ax n By n ) 2 otherwise γ n = γ (γ being any nonnegative value), where the index set Π = {n : Ax n By n 0}. Let the sequences {α n }, {β n }, {λ n }, and {η n } satisfy the following conditions: (i) λ n [a, b] (0, 1 L ) and η n [c, d] (0, 1 K ); (ii) lim α n = 0, n=0 α n =. Then, the sequence {(x n, y n )} converges strongly to (x, y ) Ω. We also have the following algorithm for finding a common element of the set of solution of an equilibrium problem, solution of a variational inequality problem, and common fixed point of a nonexpansive semigroup. Theorem 3.5. Let C be a nonempty closed convex subset of a Hilbert space H. Let T := {T(t) : t 0} be u.a.r. nonexpansive semigroup on H. Let F : H H be a monotone and L-Lipschitz continuous operator on C and Φ : C C R be a function satisfying conditions (A1)-(A4). Suppose Ω = Fix(T) VI(C, F) EP(Φ). Let {x n } be sequence generated by x 0, ϑ H, and by z n = U Φ κ n x n, u n = P C (z n λ n F(z n )), v n = P C (z n λ n F(u n )), x n+1 = α n ϑ + β n v n + δ n T(r n )v n, n 0. Let the sequences {r n }, {κ n }, {α n }, {β n }, {δ n }, and {λ n } satisfy the following conditions:

13 A. Latif, M. Eslamian, J. Nonlinear Sci. Appl., 10 (2017), (i) α n + β n + δ n = 1 and lim inf n β n δ n > 0; (ii) lim r n =, lim inf n κ n > 0, and λ n [a, b] (0, 1 L ); (iii) lim α n = 0, n=0 α n =. Then, the sequence {x n } converges strongly to x Ω. Remark 3.6. In [12], the authors presented some algorithm for solving split variational inequality problem for inverse strongly monotone operators, but in this paper (Theorem 3.4) we consider some algorithm for solving split equality variational inequality problem for monotone and Lipschitz continuous operators which does not require any knowledge of the operator norms. We also present a strong convergence theorem which is more desirable than weak convergence. Remark 3.7. Theorem 3.1 generalized the main result of Moudafi [34, 35]. Indeed, we present an algorithm for solving split equality problem for nonexpansive semigroups and monotone and Lipschitz continuous operator which does not require any knowledge of the operator norms. We also present a strong convergence theorem which is more desirable than weak convergence. Remark 3.8. In [42], Zhao presented a convergence theorem for solving split equality fixed-point problem of quasi-nonexpansive mapping. In this paper we extend the result for solving split equality problem with equilibrium problem, variational inequality problem, and fixed point problem of nonexpansive semigroups. Acknowledgment This article was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. Therefore, the authors acknowledge with thanks DSR for the technical and financial support. The authors thank the honorable reviewers for their valuable comments and suggestions. References [1] A. Aleyner, Y. Censor, Best approximation to common fixed points of a semigroup of nonexpansive operators, J. Nonlinear Convex Anal., 6 (2005), [2] P. N. Anh, A hybrid extragradient method extended to fixed point problems and equilibrium problems, Optimization, 62 (2013), , 2.4 [3] H. Attouch, J. Bolte, P. Redont, A. Soubeyran, Alternating proximal algorithms for weakly coupled convex minimization problems. Applications to dynamical games and PDE s, J. Convex Anal., 15 (2008), [4] H. Attouch, A. Cabot, P. Frankel, J. Peypouquet, Alternating proximal algorithms for linearly constrained variational inequalities: application to domain decomposition for PDE s, Nonlinear Anal., 74 (2011), [5] D. P. Bertsekas, E. M. Gafni, Projection methods for variational inequalities with application to the traffic assignment problem, Nondifferential and variational techniques in optimization, Lexington, Ky., (1980), Math. Programming Stud., 17 (1982), [6] E. Blum, W. Oettli, From optimization and variational inequalities to equilibrium problems, Math. Student, 63 (1994), , 2, 2.5 [7] F. E. Browder, Nonexpansive nonlinear operators in a Banach space, Proc. Nat. Acad. Sci. U.S.A., 54 (1965), [8] C. Byrne, Iterative oblique projection onto convex sets and the split feasibility problem, Inverse Problems, 18 (2002), [9] C. Byrne, A unified treatment of some iterative algorithms in signal processing and image reconstruction, Inverse Problems, 20 (2004), [10] Y. Censor, T. Bortfeld, B. Martin, A. Trofimov, A unified approach for inversion problems in intensity-modulated radiation therapy, Phys. Med. Biol., 51 (2006), [11] Y. Censor, T. Elfving, A multiprojection algorithm using Bregman projections in a product space, Numer. Algorithms, 8 (1994), [12] Y. Censor, A. Gibali, S. Reich, Algorithms for the split variational inequality problem, Numer. Algorithms, 59 (2012), , 3.6 [13] Y. Censor, A. Segal, The split common fixed point problem for directed operators, J. Convex Anal., 16 (2009), [14] R.-D. Chen, Y.-Y. Song, Convergence to common fixed point of nonexpansive semigroups, J. Comput. Appl. Math., 200 (2007),

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