J. Appl. Math. & Informatics Vol. 29(2011), No. 1-2, pp. 227-245 Website: http://www.kcam.biz A VISCOSITY APPROXIMATIVE METHOD TO CESÀRO MEANS FOR SOLVING A COMMON ELEMENT OF MIXED EQUILIBRIUM, VARIATIONAL INEQUALITIES AND FIXED POINT PROBLEMS THANYARAT JITPEERA, PHAYAP KATCHANG AND POOM KUMAM Abstract. In this paper, we introduce a new iterative method for finding a common element of the set of solutions for mixed equilibrium problem, the set of solutions of the variational inequality for a β-inverse-strongly monotone mapping and the set of fixed points of a family of finitely nonexpansive mappings in a real Hilbert space by using the viscosity and Cesàro mean approximation method. We prove that the sequence converges strongly to a common element of the above three sets under some mind conditions. Our results improve and extend the corresponding results of Kumam and Katchang [A viscosity of extragradient approximation method for finding equilibrium problems, variational inequalities and fixed point problems for nonexpansive mapping, Nonlinear Analysis: Hybrid Systems, 3(2009), 475 486], Peng and Yao [Strong convergence theorems of iterative scheme based on the extragradient method for mixed equilibrium problems and fixed point problems, Mathematical and Computer Modelling, 49(2009), 1816 1828], Shimizu and Takahashi [Strong convergence to common fixed points of families of nonexpansive mappings, Journal of Mathematical Analysis and Applications, 211(1) (1997), 71 83] and some authors. AMS Mathematics Subject Classification : 46C05, 47D03, 47H09, 47H10. Key words and phrases : Nonexpansive mapping, Fixed point, Variational inequality, Mixed equilibrium problem, Viscosity approximation method, Cesàro mean approximation method. 1. Introduction Throughout this paper, we assume that H is a real Hilbert space with inner product and norm are denoted by.,. and., respectively and let C Received January 10, 2010. Revised May 17, 2010. Accepted May 20, 2010. Corresponding author. This Research was Financially Supported by Faculty of Science, King Mongkut s University of Technology Thonburi. c 2011 Korean SIGCAM and KSCAM. 227
228 Thanyarat Jitpeera, Phayap Katchang and Poom Kumam be a nonempty closed convex subset of H. A mapping T : C C is called nonexpansive if T x T y x y, x, y C. We use F (T ) to denote the set of fixed points of T, that is, F (T ) = {x C : T x = x}. It is assumed throughout the paper that T is a nonexpansive mapping such that F (T ). Recall that a self-mapping f : C C is a contraction on C if there exists a constant α [0, 1) and x, y C such that f(x) f(y) α x y. Let ϕ : C R {+ } be a proper extended real-valued function and φ be a bifunction of C C into R, where R is the set of real numbers. Ceng and Yao [4] considered the following mixed equilibrium problem for finding x C such that φ(x, y) + ϕ(y) ϕ(x), for all y C. (1) The set of solutions of (1) is denoted by MEP (φ, ϕ). We see that x is a solution of problem (1) implies that x domϕ = {x C ϕ(x) < + }. If ϕ = 0, then the mixed equilibrium problem (1) becomes the following equilibrium problem is to find x C such that φ(x, y) 0, for all y C. (2) The set of solutions of (2) is denoted by EP (φ). The mixed equilibrium problems include fixed point problems, variational inequality problems, optimization problems, Nash equilibrium problems and the equilibrium problem as special cases. Numerous problems in physics, optimization and economics reduce to find a solution of (2). Some methods have been proposed to solve the equilibrium problem (see [2, 3, 5, 6, 7, 8, 9, 10, 12, 21, 22, 24, 25]). Let B : C H be a mapping. The variational inequality problem, denoted by V I(C, B), is to find x C such that Bx, y x 0, (3) for all y C. The variational inequality problem has been extensively studied in the literature. See, e.g. [28, 29] and the references therein. A mapping B of C into H is called monotone if Bx By, x y 0, (4) for all x, y C. B is called β-inverse-strongly monotone if there exists a positive real number β > 0 such that for all x, y C Bx By, x y β Bx By 2. (5) Let A be a strongly positive bounded linear operator on H: that is, there is a constant γ > 0 with property Ax, x γ x 2, for all x H. (6) A typical problem is to minimize a quadratic function over the set of the fixed points of a nonexpansive mapping on a real Hilbert space H: 1 min Ax, x x, b, (7) x C 2
A viscosity approximative method to cesàro means for solving a common element 229 where A is a linear bounded operator, C is the fixed point set of a nonexpansive mapping S on H, and b is a given point in H. Moreover, it is shown in [11] that the sequence {x n } defined by the scheme x n+1 = ɛ n γf(x n ) + (1 ɛ n A)Sx n, (8) converges strongly to the unique solution of the variational inequality (A γf)x, x x 0, x F (S), which is the optimality condition for the minimization problem min x F (S) 1 Ax, x h(x), 2 where h is a potential function for γf (i.e., h (x) = γf(x) for x H). Recently, Kumam and Katchang [10] introduced a viscosity extragradient approximation method which solved the problem for finding the set of solutions for equilibrium problems, the set of solutions of the variational inequalites for k-lipschitz continuous mapping and the set of fixed point problems for nonexpansive mapping in a real Hilbert space. The sequences were generated by x 1 H and F (u n, y) + 1 r n y u n, u n x n 0, y C, y n = P C (u n λ n Bu n ), x n+1 = α nγf(x n) + β nx n + ((1 β n)i α na)sp C(x n λ nby n), for all n 1, where B is a monotone k-lipschitz continuous mapping, S is a nonexpansive mapping and A is a strongly bounded linear operator. They proved the strong convergence theorems under suitable conditions. In 2008, Peng and Yao [15] introduced an iterative algorithm based on extragradient method which solves the problem of finding a common element of the set of solutions of a mixed equilibrium problem, the set of fixed points of a family of finitely nonexpansive mappings and the set of the variational inequality for a monotone, Lipschitz continuous mapping in a real Hilbert space. The sequences generated by v C, x 1 = x C, F (u n, y) + ϕ(y) ϕ(u n) + 1 r n y u n, u n x n 0, y C, y n = P C (u n γ n Bu n ), x n+1 = α n v + β n x n + (1 α n β n )W n P C (u n λ n By n ), for all n 0, where W n is W -mapping. They proved the strong convergence theorems under some mind conditions. On the other hand, Shimizu and Takahashi [19] originally studied the convergence of an iteration process {x n } for a family of nonexpansive mappings in the framework of a real Hilbert space. They restate the sequence {x n } as follows: 1 n x n+1 = α n x + (1 α n ) T j x n, for n = 0, 1, 2,... (11) n + 1 j=0 (9) (10)
230 Thanyarat Jitpeera, Phayap Katchang and Poom Kumam where x 0 and x are all elements of C and α n is an appropriate in [0, 1]. They proved that {x n } converges strongly to an element of fixed point of T which is the nearest to x. In this paper, motivated by the above results and the iterative schemes considered in [10], [15] and [19], we introduce a new iterative process (17) below base on viscosity and Cesàro means approximation method for finding a common element of the set of fixed points of a family of finitely nonexpansive mappings, the set of solutions of the variational inequality problem for a β-inverse-strongly monotone mapping and the set of solutions of a mixed equilibrium problem in a real Hilbert space. Then, we prove strong convergence theorems which are connected with [6, 18, 20, 26, 27] and extend and improve the corresponding results of Kumam and Katchang [10], Peng and Yao [15] and Shimizu and Takahashi [19]. 2. Preliminaries Let H be a real Hilbert space with inner product, and norm and let C be a nonempty closed convex subset of H. Then and x y 2 = x 2 y 2 2 x y, y, (12) λx + (1 λ)y 2 = λ x 2 + (1 λ) y 2 λ(1 λ) x y 2, (13) for all x, y H and λ [0, 1]. For every point x H, there exists a unique nearest point in C, denoted by P C x, such that x P C x x y, for all y C. P C is called the metric projection of H onto C. It is well known that P C is a nonexpansive mapping of H onto C and satisfies x y, P C x P C y P C x P C y 2, (14) for every x, y H. Moreover, P C x is characterized by the following properties: P C x C and x P C x, y P C x 0, (15) x y 2 x P C x 2 + y P C x 2, (16) for all x H, y C. Let B be a monotone mapping of C into H. In the context of the variational inequality problem the characterization of projection (15) implies the following: u V I(C, B) u = P C (u λbu), λ > 0. It is also known that H satisfies the Opial condition [13], i.e., for any sequence {x n } H with x n x, the inequality lim inf n x n x < lim inf n x n y,
A viscosity approximative method to cesàro means for solving a common element 231 holds for every y H with x y. A set-valued mapping U : H 2 H is called monotone if for all x, y H, f Ux and g Uy imply x y, f g 0. A monotone mapping U : H 2 H is maximal if the graph of G(U) of U is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping U is maximal if and only if for (x, f) H H, x y, f g 0 for every (y, g) G(U) implies f Ux. Let B be a monotone mapping of C into H and let N C ȳ be the normal cone to C at ȳ C, i.e., N C ȳ = {w H : u ȳ, w 0, u C} and define { Bȳ + NC ȳ, ȳ C; Uȳ =, ȳ / C. Then U is the maximal monotone and 0 Uȳ if and only if ȳ V I(C, B); see [16]. For solving the mixed equilibrium problem, let us give the following assumptions for a bifunction φ : C C R and a proper extended real-valued function ϕ : C R {+ } 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 all x, y C; (A3) for each x, y, z C, lim t 0 φ(tz + (1 t)x, y) φ(x, y); (A4) for each x C, y φ(x, y) is convex and lower semicontinuous; (A5) for each y C, x φ(x, y) is weakly upper semicontinuous; (B1) for each x H and r > 0, there exist abounded subset D x C and y x C such that for any z C \ D x, (B2) C is a bounded set. φ(z, y x ) + ϕ(y x ) + 1 r y x z, z x < ϕ(z); We need the following lemmas for proving our main results. Lemma 1. (Peng and Yao [15]) Let C be a nonempty closed convex subset of H. Let φ : C C R be a bifunction satisfies (A1)-(A5) and let ϕ : C R {+ } be a proper lower semicontinuous and convex function. Assume that either (B1) or (B2) holds. For r > 0 and x H, define a mapping T r : H C as follows: T r (x) = {z C : φ(z, y) + ϕ(y) + 1 } r y z, z x ϕ(z), y C, for all z H. Then, the following hold: (1) For each x H, T r (x) ; (2) T r is single-valued; (3) T r is firmly nonexpansive, i.e., for any x, y H, T r x T r y 2 T r x T r y, x y ; (4) F (T r ) = MEP (φ, ϕ); (5) MEP (φ, ϕ) is closed and convex.
232 Thanyarat Jitpeera, Phayap Katchang and Poom Kumam Lemma 2. (Xu [23]) Assume {a n } is a sequence of nonnegative real numbers such that a n+1 (1 α n )a n + δ n, n 0 where {α n } is a sequence in (0, 1) and {δ n } is a sequence in R such that (1) n=1 α n =, δ (2) lim sup n n α n 0 or n=1 δ n <. Then lim n a n = 0. Lemma 3. (Osilike and Igbokwe [14]) Let (C,.,. ) be an inner product space. Then for all x, y, z C and α, β, γ [0, 1] with α + β + γ = 1, we have αx+βy+γz 2 = α x 2 +β y 2 +γ z 2 αβ x y 2 αγ x z 2 βγ y z 2. Lemma 4. (Suzuki [17]) Let {x n } and {y n } be bounded sequences in a Banach space X and let {β n } be a sequence in [0, 1] with 0 < lim inf n β n lim sup n β n < 1. Suppose x n+1 = (1 β n )y n + β n x n for all integers n 0 and lim sup n ( y n+1 y n x n+1 x n ) 0. Then, lim n y n x n = 0. Lemma 5. (Marino and Xu [11]) Assume A is a strongly positive linear bounded operator on a Hilbert space H with coefficient γ > 0 and 0 < ρ A 1. Then I ρa 1 ρ γ. Lemma 6. (Bruck [1]) Let C be a nonempty bounded closed convex subset of a uniformly convex Banach space E and T : C C a nonexpansive mapping. For each x C and the Cesàro means T n x = 1 n n+1 T i x, then lim sup n T n x T (T n x) = 0. 3. Main results In this section, we show a strong convergence theorem for finding a common element of the set of fixed points of a family of finitely nonexpansive mappings, the set of solutions of mixed equilibrium problem and the set of solutions of a variational inequality problem for a β-inverse strongly monotone mapping in a real Hilbert space by using the viscosity of Cesàro mean approximation method. Theorem 1. Let C be a nonempty closed convex subset of a real Hilbert Space H. Let φ be a bifunction of C C into real numbers R satisfying (A1) (A5) and let ϕ : C R {+ } be a proper lower semicontinuos and convex function. Let T i : H H be a nonexpansive mappings for all i = 1, 2, 3,..., n, such that Θ := F (T ) V I(C, B) MEP (φ, ϕ). Let f be a contraction of H into itself with coefficient α (0, 1) and let B be a β-inverse-strongly monotone mapping of C into H. Let A be a strongly positive bounded linear self-adjoint on H with coefficient γ > 0 and 0 < γ < γ α. Assume that either B 1 or B 2 holds. Let {x n }, {y n } and {u n } be sequences generated by x 0 H, u n C and φ(u n, y) + ϕ(y) ϕ(u n) + 1 r n y u n, u n x n 0, y C, y n = δ n u n + (1 δ n )P C (u n λ n Bu n ), x n+1 = α nγf(x n) + β nx n + ((1 β n)i α na) 1 n+1 n T i y n, n 0, (17)
A viscosity approximative method to cesàro means for solving a common element 233 where {α n }, {β n } and δ n (0, 1), {λ n } (0, 2β) and {r n } (0, ) satisfy the following conditions: (i). n=0 α n = and lim n α n = 0, (ii). lim n δ n = 0, (iii). lim inf n r n > 0 and lim n r n+1 r n = 0, (iv). 0 < lim inf n β n lim sup n β n < 1, (v). {λ n } [a, b], a, b (0, 2β) and lim n λ n+1 λ n = 0, (vi). lim n β n+1 β n = 0. Then, {x n } converges strongly to z Θ where z = P Θ (I A + γf)(z), which is the unique solution of the variational inequality (A γf)z, x z 0, x Θ, which is the optimality condition for the minimization problem min x Θ where h is a potential function for γf. 1 Ax, x h(x), 2 Proof. Now, we have (1 β n )I α n A 1 β n α n γ (see [10] p.479). Since λ n (0, 2β) and B is a β-inverse-strongly monotone mapping. For any x, y C we have (I λ n B)x (I λ n B)y 2 = (x y) λ n (Bx By) 2 = x y 2 2λ n x y, Bx By + λ 2 n Bx By 2 x y 2 2λ nβ Bx By 2 + λ 2 n Bx By 2 = x y 2 + λ n (λ n 2β) Bx By 2 x y 2. It follow that (I λ n B)x (I λ n B)y x y, hence I λ n B is nonexpansive. Let x Θ, T rn be a sequence of mapping defined as in Lemma 1 and u n = T rn x n, for all n 0, we have u n x = T rn x n T rn x x n x. By the fact that P C and I λ n B are nonexpansive and x = P C (x λ n Bx ), we get y n x = δ n u n + (1 δ n )P C (u n λ n Bu n ) x δ n u n x + (1 δ n) P C(u n λ nbu n) P C(x λ nbx ) δ n u n x + (1 δ n ) (I λ n B)u n (I λ n B)x δ n u n x + (1 δ n ) u n x = u n x x n x. Let T n = 1 n+1 n T i, it follows that (18) (19)
234 Thanyarat Jitpeera, Phayap Katchang and Poom Kumam T n x T n y = 1 n T i x 1 n T i y n + 1 n + 1 1 n T i x T i y n + 1 1 n x y n + 1 = n + 1 x y = x y, n + 1 which implies that T n is nonexpansive. Since x Θ, we have T n x = 1 n T i x = 1 n x = x n + 1 n + 1 for all x, y C. By (19) and (20), we have x n+1 x = α n (γf(x n ) Ax ) + β n (x n x ) + ((1 β n )I α n A)(T n y n x ) α n γf(x n) Ax + β n x n x + (1 β n α n γ) y n x α n γf(x n ) Ax + β n x n x + (1 β n α n γ) x n x α n γ f(x n ) f(x ) + α n γf(x ) Ax + (1 α n γ) x n x α nγα x n x + α n γf(x ) Ax + (1 α n γ) x n x = (1 α n ( γ γα)) x n x + α n ( γ γα) γf(x ) Ax ( γ γα) { max x 0 x, γf(x ) Ax }. ( γ γα) (20) Hence {x n } is bounded and also {u n }, {y n } and {T n y n } are bounded. Next, we show that lim n x n+1 x n = 0. Observing that u n = T rn x n dom ϕ and u n+1 = T rn+1 x n+1 dom ϕ, we get and φ(u n, y) + ϕ(y) ϕ(u n ) + 1 r n y u n, u n x n 0, y C (21) φ(u n+1, y) + ϕ(y) ϕ(u n+1 ) + 1 r n+1 y u n+1, u n+1 x n+1 0, y C. (22) Take y = u n+1 in (21) and y = u n in (22), by using condition (A2), it follows that u n+1 u n, u n x n u n+1 x n+1 r n r n+1 0. Thus u n+1 u n, u n u n+1 +x n+1 x n +(1 rn r n+1 )(u n+1 x n+1 ) 0. Without loss of generality, let us assume that there exists a nonnegative real number c such that r n > c, n 1. Then, we have
A viscosity approximative method to cesàro means for solving a common element 235 { u n+1 u n 2 u n+1 u n x n+1 x n + and hence 1 r } n un+1 x n+1 r n+1 u n+1 u n x n+1 x n + 1 r n+1 r n+1 r n u n+1 x n+1 x n+1 x n + 1 c r n+1 r n M 1, (23) where M 1 = sup{ u n x n : n N}. On the other hand, setting v n = P C (u n λ n Bu n ), it follows from the definition of {y n } that y n+1 y n = {δ n+1 u n+1 + (1 δ n+1 )P C (u n+1 λ n+1 Bu n+1 )} {δ n u n + (1 δ n )P C (u n λ n Bu n )} = δ n+1 (u n+1 u n ) + (δ n+1 δ n )u n + (1 δ n+1 )P C (u n+1 λ n+1 Bu n+1 ) (1 δ n+1 )P C (u n λ n Bu n ) + (1 δ n+1 )P C (u n λ n Bu n ) (1 δ n )P C (u n λ n Bu n ) = δ n+1 (u n+1 u n ) + (δ n+1 δ n )u n + (1 δ n+1 ){P C (u n+1 λ n+1 Bu n+1 ) P C (u n λ n Bu n )} + (δ n δ n+1 )P C (u n λ n Bu n ) δ n+1 u n+1 u n + δ n+1 δ n ( u n + v n ) + (1 δ n+1 ) (u n+1 λ n+1 Bu n+1 ) (u n λ n Bu n ) = δ n+1 u n+1 u n + δ n+1 δ n ( u n + v n ) (24) + (1 δ n+1 ) (u n+1 λ n+1 Bu n+1 ) (u n λ n+1 Bu n ) + (u n λ n+1 Bu n ) (u n λ n Bu n ) δ n+1 u n+1 u n + δ n+1 δ n ( u n + v n ) + (1 δ n+1 ){ u n+1 u n + λ n+1 λ n Bu n } = δ n+1 δ n ( u n + v n ) + u n+1 u n + (1 δ n+1 ) λ n+1 λ n Bu n δ n+1 δ n ( u n + v n ) + u n+1 u n + λ n+1 λ n Bu n δ n+1 δ n ( u n + v n ) + x n+1 x n + 1 c r n+1 r n M 1 + λ n+1 λ n Bu n.
236 Thanyarat Jitpeera, Phayap Katchang and Poom Kumam We compute that T n+1 y n+1 T n y n T n+1y n+1 T n+1y n + T n+1y n T ny n y n+1 y n + 1 n+1 T i y n 1 n T i y n n + 2 n + 1 = y n+1 y n + 1 n T i y n + 1 n + 2 n + 2 T n+1 y n 1 n T i y n n + 1 1 n = y n+1 y n + T i y n + 1 (n + 1)(n + 2) n + 2 T n+1 y n 1 n y n+1 y n + T i y n + 1 (n + 1)(n + 2) n + 2 T n+1 y n y n+1 y n + 1 (n + 1)(n + 2) n ( T i y n T i x + x ) + 1 n + 2 ( T n+1 y n T n+1 x + x ) 1 n y n+1 y n + ( y n x + x ) + 1 (n + 1)(n + 2) n + 2 ( y n x + x ) y n+1 y n + n + 1 (n + 1)(n + 2) ( y n x + x ) + 1 n + 2 y n x + 1 n + 2 x = y n+1 y n + 2 n + 2 y n x + 2 n + 2 x δ n+1 δ n ( u n + v n ) + x n+1 x n + 1 rn+1 rn M1 c + λ n+1 λ n Bu n + 2 n + 2 yn x + 2 n + 2 x. Let x n+1 = (1 β n )z n + β n x n, then z n = x n+1 β n x n 1 β n = α nγf(x n ) + ((1 β n )I α n A)T n y n 1 β n, and hence z n+1 z n = α n+1γf(x n+1 ) + ((1 β n+1 )I α n+1 A)T n+1 y n+1 1 β n+1 αnγf(xn) + ((1 βn)i αna)tnyn 1 β n = α n+1γf(x n+1 ) + (1 β n+1)t n+1 y n+1 α n+1at n+1 y n+1 1 β n+1 1 β n+1 1 β n+1 α nγf(x n ) (1 β n)t n y n + α nat n y n 1 β n 1 β n 1 β n
A viscosity approximative method to cesàro means for solving a common element 237 = α n+1 (γf(x n+1 ) AT n+1 y n+1 ) + α n (AT n y n γf(x n )) + T n+1 y n+1 T n y n 1 β n+1 1 β n α n+1 1 β n+1 γf(x n+1 ) AT n+1 y n+1 + α n 1 β n AT n y n γf(x n ) + T n+1 y n+1 T n y n α n+1 1 β n+1 γf(x n+1 ) AT n+1 y n+1 + α n 1 β n AT n y n γf(x n ) + δ n+1 δ n ( u n + v n ) + x n+1 x n + 1 c r n+1 r n M 1 + λ n+1 λ n Bu n + 2 n + 2 y n x + 2 n + 2 x. Therefore z n+1 z n x n+1 x n α n+1 1 β n+1 γf(x n+1 ) AT n+1 y n+1 + α n 1 β n AT n y n γf(x n ) + δ n+1 δ n ( u n + v n ) + 1 c r n+1 r n M 1 + λ n+1 λ n Bu n + 2 n + 2 y n x + 2 n + 2 x. It follows from n and the conditions (i)-(v), that lim sup( z n+1 z n x n+1 x n ) 0. (25) n From Lemma 4 and (25), we obtain lim n z n x n = 0 and also lim x n+1 x n = lim (1 β n) z n x n = 0. n n Next, we show that x n u n 0 as n. For x Θ, we obtain u n x 2 = T rn x n T rn x 2 T rn x n T rn x, x n x = u n x, x n x = 1 2 ( u n x 2 + x n x 2 u n x x n + x 2 ) = 1 2 ( u n x 2 + x n x 2 u n x n 2 ), and hence u n x 2 x n x 2 u n x n 2. (26) Since y n x u n x and from Lemma 3 and (26), we obtain
238 Thanyarat Jitpeera, Phayap Katchang and Poom Kumam x n+1 x 2 = α n γf(x n ) + β n x n + ((1 β n )I α n A)T n y n x 2 = α n(γf(x n) Ax ) + β n(x n x ) + ((1 β n)i α na)(t ny n x ) 2 α n γf(x n ) Ax 2 + β n x n x 2 + (1 β n α n γ) y n x 2 α n γf(x n ) Ax 2 + β n x n x 2 + (1 β n α n γ) u n x 2 α n γf(x n) Ax 2 + β n x n x 2 + (1 β n α n γ)( x n x 2 x n u n 2 ) α n γf(x n ) Ax 2 + x n x 2 (1 β n α n γ) x n u n 2. Then, we have (1 β n α n γ) x n u n 2 α n γf(x n ) Ax 2 + x n x 2 x n+1 x 2 = α n γf(x n ) Ax 2 + x n x n+1 ( x n x + x n+1 x ). By lim n x n+1 x n = 0, (i) and (iv), imply that Since lim inf n r n > 0, we obtain lim x n u n n r n lim u n x n = 0. n = lim n 1 r n x n u n = 0. Next, we show that lim n T n y n x n = 0. Indeed, observe that x n T n y n x n x n+1 + x n+1 T n y n and then = x n x n+1 + α n γf(x n ) + β n x n + ((1 β n )I α n A)T n y n T n y n = x n x n+1 + α nγf(x n) α nat ny n + α nat ny n + β nx n β n T n y n + β n T n y n + ((1 β n )I α n A)T n y n T n y n x n x n+1 + α n γf(x n ) AT n y n + β n x n T n y n x n T n y n 1 1 β n x n x n+1 + α n 1 β n γf(x n ) AT n y n. (27) Since lim n x n+1 x n = 0, (i) and (iv), we get lim n x n T n y n = 0. Next, we show that lim n y n v n = 0, where v n = P C (u n λ n Bu n ). From Lemma 3, (13) and (18), we obtain
A viscosity approximative method to cesàro means for solving a common element 239 x n+1 x 2 α n γf(x n) Ax 2 + β n x n x 2 + ((1 β n)i α na) T ny n x 2 α n γf(x n ) Ax 2 + β n x n x 2 + (1 β n α n γ) y n x 2 = α n γf(x n) Ax 2 + β n x n x 2 + (1 β n α n γ) δ n(u n x ) + (1 δ n){p C(u n λ nbu n) P C(x λ nbx )} 2 α n γf(x n ) Ax 2 + β n x n x 2 + (1 β n α n γ)δ n u n x 2 + (1 β n α n γ)(1 δ n ) (u n λ n Bu n ) (x λ n Bx ) 2 = α n γf(x n) Ax 2 + β n x n x 2 + (1 β n α n γ)δ n u n x 2 + (1 β n α n γ)(1 δ n ) (u n x ) λ n (Bu n Bx ) 2 α n γf(x n ) Ax 2 + β n x n x 2 + (1 β n α n γ)δ n u n x 2 + (1 β n α n γ)(1 δ n){ u n x 2 + λ n(λ n 2β) Bu n Bx 2 } α n γf(x n ) Ax 2 + (1 α n γ) x n x 2 + (1 β n α n γ)(1 δ n )λ n (λ n 2β) Bu n Bx 2 α n γf(x n) Ax 2 + x n x 2 it follows that + (1 β n α n γ)(1 δ n )a(b 2β) Bu n Bx 2, 0 (1 β n α n γ)(1 δ n)a(2β b) Bu n Bx 2 α n γf(x n ) Ax 2 + x n x 2 x n+1 x 2 α n γf(x n ) Ax 2 + x n+1 x n ( x n x + x n+1 x ). Since α n 0 and x n+1 x n 0, as n, we obtain Bu n Bx 0 as n. Using (12), we have v n x 2 = P C (u n λ n Bu n ) P C (x λ n Bx ) 2 (u n λ nbu n) (x λ nbx ), v n x = 1 } { (u n λ n Bu n ) (x λ n Bx ) 2 + v n x 2 2 1 } { (u n λ n Bu n ) (x λ n Bx ) (v n x ) 2 2 = 1 } { u n x 2 + v n x 2 (u n v n ) λ n (Bu n Bx ) 2 2 = 1 } { u n x 2 + v n x 2 1 { u n v n 2 λ 2 n Bu n Bx 2 2 2 } 2 u n v n λ n (Bu n Bx ), λ n (Bu n Bx ) 1 } { u n x 2 + v n x 2 1 { u n v n 2 λ 2 n Bu n Bx 2 2 2 2λ n u n v n, Bu n Bx + 2λ 2 n Bu n Bx 2 },
240 Thanyarat Jitpeera, Phayap Katchang and Poom Kumam so, we obtain v n x 2 u n x 2 u n v n 2 λ 2 n Bu n Bx 2 +2λ n u n v n, Bu n Bx, and hence x n+1 x 2 α n γf(x n ) Ax 2 + β n x n x 2 + (1 β n α n γ) y n x 2 = α n γf(x n ) Ax 2 + β n x n x 2 + (1 β n α n γ) δ n u n + (1 δ n )v n x 2 α n γf(x n) Ax 2 + β n x n x 2 + (1 β n α n γ)δ n u n x 2 + (1 β n α n γ)(1 δ n ) v n x 2 α n γf(x n ) Ax 2 + β n x n x 2 + (1 β n α n γ)δ n u n x 2 + (1 β n α n γ) v n x 2 α n γf(x n ) Ax 2 + β n x n x 2 + (1 β n α n γ)δ n u n x 2 { + (1 β n α n γ) u n x 2 u n v n 2 λ 2 n Bu n Bx 2 } + 2λ n u n v n, Bu n Bx α n γf(x n) Ax 2 + β n x n x 2 + (1 β n α n γ)δ n u n x 2 + (1 β n α n γ) x n x 2 (1 β n α n γ) u n v n 2 (1 β n α n γ)λ 2 n Bu n Bx 2 + (1 β n α n γ)2λ n u n v n, Bu n Bx α n γf(x n) Ax 2 + x n x 2 + (1 β n α n γ)δ n u n x 2 (1 β n α n γ) u n v n 2 (1 β n α n γ)λ 2 n Bu n Bx 2 + (1 β n α n γ)2λ n u n v n Bu n Bx, which implies that (1 β n α n γ) u n v n 2 α n γf(x n) Ax 2 + x n x 2 x n+1 x 2 + (1 β n α n γ)δ n u n x 2 (1 β n α n γ)λ 2 n Bu n Bx 2 + (1 β n α n γ)2λ n u n v n Bu n Bx α n γf(x n ) Ax 2 + x n x n+1 ( x n x + x n+1 x ) + (1 β n α n γ)δ n u n x 2 (1 β n α n γ)λ 2 n Bu n Bx 2 + (1 β n α n γ)2λ n u n v n Bu n Bx. Since Bu n Bx 0, x n+1 x n 0 as n 0 and the condition (i), (ii), (iv), we have u n v n 0 as n 0. At the same time, we note that y n v n = δ n (u n v n ) + (1 δ n )(v n v n ) = δ n u n v n, since δ n 0, we have y n v n 0 as n. Consequently, we observe that T n y n y n T n y n x n + x n u n + u n v n + v n y n 0, as n. It is easy to see that P Θ (I A + γf)(z) is a contradiction of H into itself. Hence H is complete, there exists a unique fixed point z H, such that z = P Θ (I A + γf)(z).
A viscosity approximative method to cesàro means for solving a common element 241 Next, we show that lim sup (A γf)z, z x n 0. (28) n Indeed, we can choose a subsequence {y ni } of {y n }, such that lim (A γf)z, z y n i = lim sup (A γf)z, z y n. i n Since {y ni } is bounded, there exists a subsequence {y nij } of {y ni } which converge weakly to ȳ C. Without loss of generality, we can assume that y ni ȳ. From T n y n y n 0, we obtain T n y ni ȳ. Let us show ȳ MEP (φ, ϕ). Since u n = T rn x n domϕ, we obtain φ(u n, y) + ϕ(y) ϕ(u n ) + 1 r n y u n, u n x n 0, y C. From (A2), we also have and hence ϕ(y) ϕ(u n ) + 1 r n y u n, u n x n F (y, u n ), y C ϕ(y) ϕ(u n ) + y u ni, u n i x ni r ni F (y, u ni ), y C. From u n x n 0, x n T n y n 0 and T n y n y n 0, we get u ni ȳ. Since un i xn i r ni 0, thus that from (A4) and the weakly lower semicontinuity of ϕ that φ(y, ȳ) + ϕ(ȳ) ϕ(y) 0, y C. For t with 0 < t 1 and x C, let x t = tx + (1 t)ȳ. Since x C and ȳ C, we have x t C and hence φ(x t, ȳ) + ϕ(ȳ) ϕ(x t ) 0. So, from (A1), (A4) and the convexity of ϕ, we have 0 = φ(x t, x t ) + ϕ(x t ) ϕ(x t ) tφ(x t, x) + (1 t)φ(x t, ȳ) + tϕ(x) + (1 t)ϕ(ȳ) ϕ(x t ) t(φ(x t, x) + ϕ(x) ϕ(x t )). (29) Dividing by t, we get φ(x t, x) + ϕ(x) ϕ(x t ) 0. From (A3) and the weakly lower semicontinuity of ϕ, we have φ(ȳ, y) + ϕ(y) ϕ(ȳ) 0, for all y C and hence ȳ MEP (φ, ϕ). Next, we show that ȳ F (T n ) = 1 n n+1 F (T i ). Assume that ȳ / 1 n n+1 F (T i ), since y ni ȳ and T n ȳ ȳ. From Opial s condition, we have lim inf i y n i ȳ < lim inf y n i T n ȳ i lim inf ( y n i T n y ni + T n y ni T n ȳ ) i lim inf i y n i ȳ,
242 Thanyarat Jitpeera, Phayap Katchang and Poom Kumam which is a contradiction. Thus, we obtain ȳ F (T n ) = 1 n+1 n F (T i ). Now, let us show that ȳ V I(C, B). Let U : H 2 H be a set-valued mapping is defined by { Bȳ + NC ȳ, ȳ C, Uȳ =, ȳ / C, where N C ȳ is the normal cone to C at ȳ C. We have U is maximal monotone and 0 Uȳ if and only if ȳ V I(C, B). Let (ȳ, w) G(U), hence w Bȳ N C ȳ and v n C, we have ȳ v n, w Bȳ 0. On the other hand, from v n = P C (u n λ n Bu n ), we have ȳ v n, v n (u n λ n Bu n ) 0, that is Therefor, we have ȳ v n, v n u n λ n + Bu n 0. ȳ v ni, w ȳ v ni, Bȳ ȳ v ni, Bȳ ȳ v ni, v n i u ni + Bu ni λ ni = ȳ v ni, Bȳ v n i u ni Bu ni λ ni = ȳ v ni, Bȳ Bv ni + ȳ v ni, Bv ni Bu ni ȳ v ni, v n i u ni λ ni ȳ v ni, Bv ni Bu ni ȳ v ni, v n i u ni λ ni ȳ v ni Bv ni Bu ni ȳ v ni v n i u ni. λ ni (30) Noting that v ni u ni 0 as i and B is β-inverse-strongly monotone, hence from (30), we obtain ȳ z, w 0 as i. Since U is maximal monotone, we have ȳ U 1 0, and hence ȳ V I(C, B). Therefore ȳ Θ. Since z = P Θ (I A + γf)(z), we have lim sup n ((γf A)z, x n z = lim sup (γf A)z, T n y n z n = lim sup (γf A)z, T n y ni z i = (γf A)z, ȳ z 0. (31)
A viscosity approximative method to cesàro means for solving a common element 243 Finally we show that {x n } converge strongly to z, we obtain that x n+1 z 2 = α n γf(x n ) + β n x n + ((1 β n )I α n A)T n y n z 2 = α n(γf(x n) Az) + β n(x n z) + ((1 β n)i α na)(t ny n z) 2 = α 2 n γf(x n) Az 2 + β n(x n z) + ((1 β n)i α na)(t ny n z) 2 + 2 β n(x n z) + ((1 β n)i α na)(t ny n z), α n(γf(x n) Az) α 2 n γf(x n ) Az 2 + {β n x n z + (1 β n α n γ) y n z } 2 + 2α nβ n x n z, γf(x n) Az + 2α n(1 β n α n γ) T ny n z, γf(x n) Az α 2 n γf(x n ) Az 2 + {β n x n z + (1 β n α n γ) x n z } 2 + 2α nβ n x n z, γf(x n) γf(z) + 2α nβ n x n z, γf(z) Az + 2α n(1 β n α n γ) T ny n z, γf(x n) γf(z) + 2α n (1 β n α n γ) T n y n z, γf(z) Az α 2 n γf(x n) Az 2 + (1 α n γ) 2 x n z 2 + 2α nβ nγ x n z f(x n) f(z) + 2α n β n x n z, γf(z) Az + 2α n (1 β n α n γ)γ T n y n z f(x n ) f(z) + 2α n (1 β n α n γ) T n y n z, γf(z) Az α 2 n γf(x n) Az 2 + (1 α n γ) 2 x n z 2 + 2α nβ nγα x n z 2 + 2α nβ n x n z, γf(z) Az + 2α n(1 β n α n γ)γα x n z 2 + 2α n(1 β n α n γ) T ny n z, γf(z) Az = α 2 n γf(x n ) Az 2 + (1 2α n γ + α 2 n γ 2 + 2α n γα 2α 2 n γγα) x n z 2 + 2α nβ n x n z, γf(z) Az + 2α n(1 β n α n γ) T ny n z, γf(z) Az {1 α n (2 γ α n γ 2 2γα + 2α n γγα)} x n z 2 + α 2 n γf(x n ) Az 2 + 2α nβ n x n z, γf(z) Az + 2α n(1 β n α n γ) T ny n z, γf(z) Az {1 α n (2 γ α n γ 2 2γα + 2α n γγα)} x n z 2 + α n σ n, (32) where σ n = α n γf(x n ) Az 2 +2β n x n z, γf(z) Az +2(1 β n α n γ) T n y n z, γf(z) Az. By (31) and (i), we get lim sup n σ n 0. Applying Lemma 2 to (32) we conclude that x n z. This complete the proof. Using Theorem 1, we obtain the following corollary. Corollary 1. Let C be a nonempty closed convex subset of a real Hilbert Space H. Let φ be a bifunction of C C into real numbers R satisfying (A1) (A5) and let T : H H be a nonexpansive mapping such that Ω := F (T ) V I(C, B) EP (φ). Let B be a β-inverse-strongly monotone mapping of C into H. Let {x n }, {y n } and {u n } be sequences generated by x 0 H, u n C and φ(u n, y) + 1 r n y u n, u n x n 0, y C, y n = P C (u n λ n Bu n ), x n+1 = α n v + β n x n + (1 β n α n )T y n, n 0, (33) where {α n }, {β n } (0, 1), {λ n } (0, 2β), for all n 0 satisfy the condition (i), (iii) (v). Then, {x n } converges strongly to z Ω where z = P Ω z.
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