Available online at www.tjnsa.com J. Nonlinear Sci. Appl. 5 (2012), 475–494 Research Article Approximation algorithm for fixed points of nonlinear operators and solutions of mixed equilibrium problems and variational inclusion problems with applications Uamporn Witthayarata , Yeol Je Chob,∗, Poom Kumama,∗ a Department of Mathematics, Faculty of Science, King Mongkut’s University of Technology Thonburi, KMUTT, Bangkok 10140, Thailand. b Department of Mathematics Education and the RINS, Gyeongsang National University, Chinju 660-701, Korea. Dedicated to George A Anastassiou on the occasion of his sixtieth birthday Communicated by Professor R. Saadati Abstract The purpose of this paper is to introduce an iterative algorithm for finding a common element of the set of fixed point of nonexpansive mappings, set of a mixed equilibrium problem and the set of variational inclusions in a real Hilbert space. We prove that the sequence xn which is generated by the proposed iterative algorithm converges strongly to a common element of four sets above. Furthermore, we give an application to optimization and some numerical examples which support our main theorem in the last part. Our result extended and improve the existing result of Yao et al. [19] and references therein. Keywords: Common fixed point; Equilibrium problem; Iterative algorithm; Nonexpansive mapping; Variational inequality. ∗ Corresponding author Email addresses: [email protected] (Uamporn Witthayarat), [email protected] (Yeol Je Cho), [email protected] (Poom Kumam) Received 2011-9-11 U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 476 1. Introduction Throughout this paper, we assume that H be a real Hilbert space with the inner product h·, ·i and norm k · k and C is a nonempty closed convex subset of H. Let F : C → H be a nonlinear mapping, ϕ : C → R be a function and Θ : C × C → R be a bifunction. First, we consider the following mixed equilibrium problem: find x∗ ∈ C such that Θ(x∗ , y) + ϕ(y) − ϕ(x∗ ) + hF x∗ , y − x∗ i ≥ 0, ∀y ∈ C. (1.1) If F = 0, then the mixed equilibrium problem (1.1) becomes the following mixed equilibrium problem, which was studied by Ceng and Yao [2]: Find x∗ ∈ C such that Θ(x∗ , y) + ϕ(y) − ϕ(x∗ ) ≥ 0, ∀y ∈ C. (1.2) If ϕ = 0, then the mixed equilibrium problem (1.1) becomes the following equilibrium problem, which was considered by Takahashi and Takahashi [16]: Find x∗ ∈ C such that Θ(x∗ , y) + hF x∗ , y − x∗ i ≥ 0, ∀y ∈ C. (1.3) If ϕ = 0 and F = 0, then the mixed equilibrium problem (1.1) becomes the following equilibrium problem: Find x∗ ∈ C such that Θ(x∗ , y) ≥ 0, ∀y ∈ C. (1.4) If Θ(x, y) = 0 for all x, y ∈ C, then the mixed equilibrium problem (1.1) becomes the following variational inequality problem: Find x∗ ∈ C such that ϕ(y) − ϕ(x∗ ) + hF x∗ , y − x∗ i ≥ 0, ∀y ∈ C. (1.5) We denote the sets of solutions of the problems (1.1)-(1.5) by EP (1)-EP (5), respectively. Equilibrium problem theory is one of most interesting and useful branch for the existence of solutions of many problems arising in economics, physics, operation research and other fields. The mixed equilibrium problems include fixed point problems, optimization problems, variational inequality problems, Nash equlibrium problems as the special cases. Also, many authors have introduced some kind of methods to solve these problems. In 1997, Combettes and Hirstoaga [4] introduced an iterative method to find the best approximation to the initial data and also prove some strong convergence theorems by using the proposed method. Consequently, Takahashi and Takahashi [17] presented an another iterative scheme for finding a common element of the set of solutions of the equilibrium problem (1.2) and the set of fixed points of a nonexpansive mappings. Moreover, Yao, Liou and Yao [20], [21] also introduced new iterative schemes for finding a common element of the set of solutions of the equilibrium problem and the set of common fixed points of finitely (infinitely) nonexpansive mappings. Recently, Ceng and Yao [2] introduced a new iterative scheme for finding a common element of the set of solution of the mixed equilibrium problem and the set of common fixed points of infinitely nonexpansive mappings. Furthermore, Peng and Yao [12] applied the CQ method to solve the mixed equilibrium problem and variational inequality problem and they also obtained the strong convergence results. Their result extended and improved the corresponding results in [3], [9], [16] and [21]. Recall that the mapping f : C → C is called a ρ-contraction if there exists a constant ρ ∈ [0, 1) such that kf (x) − f (y)k ≤ ρkx − yk, ∀x, y ∈ C. A mapping T : C → C is said to be nonexpansive if kT x − T yk ≤ kx − yk, ∀x, y ∈ C. U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 477 A mapping B : C → C is said to be β -inverse strongly monotone if there exists a constant β > 0 such that hBx − By, x − yi ≥ βkBx − Byk2 , ∀x, y ∈ C. A mapping G is said to be strongly positive on H if there exists a constant µ > 0 such that hAx, xi ≥ µkxk2 , ∀x ∈ H. (1.6) Let A : H → H be a single-valued nonlinear mapping and R : H → 2H be a set-valued mapping. The variational inclusion problem is as follows: Find x ∈ H such that θ ∈ B(x) + R(x), (1.7) where θ is the zero vector in H. We denote the set of solution of this problem by I(A, R). Remark 1.1. (1) If R = ∂φ : H → 2H in (1.7), where φ : H → R is proper convex lower semi-continuous and ∂φ is the sub-differential of φ, then the variational inclusion (1.7) is equivalent to the following problem : Find x ∈ H such that hBx, v − xi + φ(y) − φ(x) ≥ 0, ∀v, y ∈ H, which is called the mixed quasi-variational inequality in Noor [10]. (2) Let M = ∂δC in (1.7), where C is a nonempty closed convex subset of H and δC : H → [0, ∞) is the indicator function of C, i.e., 0, x ∈ C; δC (x) = +∞, otherwise. Then the variational inclusion (1.7) is equivalent to the following problem: Find x ∈ H such that hBx, v − xi ≥ 0, ∀v ∈ H, which is called Hartman-Stampacchia’s variational inequality. Remark 1.2. (1) If H = Rm , then the problem (1.7) becomes the generalized equation introduced by Robinson [13]. (2) If B = 0, then the problem (1.7) becomes the inclusion problem introduced by Rockafellar [14]. The problem (1.7) is the most widely use for the study of optimal solutions in many related areas including mathematical programming, complementarity, variational inequalities, optimal control and many other fields. Many kinds of variational inclusions problems have been improved, extended and generalized in recent years by many authors. In 2008, Zhang et al. [22] introduced an algorithm for finding a common solutions for quasi variational in clusion and fixed point problems, they also prove a strong convergence theorems for approximating this common elements under the suitable condition. Moreover, Kocourek et al. [7] presented a new iterative scheme for finding a common element of the set of solution to the problem (1.7) and the set of fixed points of nonexpansive, nonspreading and hybrid mappings in Hilbert spaces. Peng et al. [11] introduced another iterative algorithm by the viscosity approximate method for finding a common element of the set of solutions of a variational inclusion with a set-valued maximal monotone mapping and inverse strongly monotone mappings, the set of solutions of an equilibrium problem and the set of fixed points of a nonexpansive mapping. Very recently, Yao et al. [19], proposed the following iterative algorithm: Θ(u , y) + ϕ(y ) − ϕ(u ) + 1 hy − u , u − (x − rF x )i ≥ 0, n n n n n n n r (1.8) xn+1 = αn (u + γf (xn )) + βn xn + [(1 − βn )I − αn (I + νA)]Wn JR,λ (zn − rAzn ), U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 478 where {αn }, {βn } are two real sequences in [0, 1]. Furthermore, they also proved that the proposed above algorithm converges strongly to a common element of the set of solution of mixed equilibrium problem and the set of fixed point of nonexpansive mapping and the set of a variational inclusion in a real Hilbert space. Motivated and inspired by the above works, in this paper, we propose an iterative algorithm which extend from Yao et al. [19] as follows: 1 Θ(un , y) + ϕ(yn ) − ϕ(un ) + hy − un , un − (xn − rF xn )i ≥ 0, r zn = JR,s2 (un − s2 Bun ), (1.9) (z − s Az ), y = J n 1 n n R,s 1 xn+1 = αn (u + γf (xn )) + βn xn + [(1 − βn )I − αn (I + νG)]Wn yn , where {αn } and {βn } are two real sequences in [0, 1]. Furthermore, we prove the strong convergence theorem and give an illustrative example to support our main theorem. 2. Preliminaries Let H be a real Hilbert space and C be a nonempty closed convex subset of H. It follows that kx − yk2 = kxk2 + kyk2 − 2hx, yi, ∀x, y ∈ H. (2.1) Recall that the nearest projection PC from H to C assigns to each x ∈ H, the unique point PC x ∈ C satisfying the property kx − PC xk = min kx − yk, y∈C which is equivalent to the following inequality hx − PC x, PC x − yi ≥ 0, ∀y ∈ C. A set-valued mapping T : H → 2H is called monotone if, for all x, y ∈ H, f ∈ T x and g ∈ T y imply hx − y, f − gi ≥ 0. A monotone mapping T : H → 2H is said to be maximal if its graph G(T ) is not properly contained in the graph of any other monotone mapping. It is well known that a monotone mapping T is maximal if and only if, for all (x, f ) ∈ H × H, hx − y, f − gi ≥ 0 for all (y, g) ∈ G(T ) implies f ∈ T x. Let a set valued mapping R : H → 2H be a maximal monotone. We define a resolvent operator JR,λ generated by R and λ as follows: JR,λ = (I + λR)−1 (x), ∀x ∈ H, where λ is a positive number. It is easily to see that the resolvent operator JR,λ is single-valued, nonexpansive and 1-inverse strongly monotone and moreover, a solution of the problem (1.7) is a fixed point of the operator JR,λ (I − λB) for all λ > 0 (see, for example, [8]). In this paper, we assume that a bifunction Θ : H × H → R and a convex function ϕ : H → R satisfy the following conditions: (H1) Θ(x, x) = 0 for all x ∈ H; (H2) Θ is monotone, i.e., Θ(x, y) + Θ(y, x) ≤ 0 for all x, y ∈ H; (H3) for any y ∈ H, x → Θ(x, y) is weakly upper semi-continuous; (H4) for any x ∈ H, y → Θ(x, y) is convex and lower semi-continuous; U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 479 (H5) for any x ∈ H and r > 0, there exists a bounded subset Dx ⊂ H and yx ∈ H such that, for any z ∈ H \ Dx , 1 Θ(z, yx ) + ϕ(yx ) − ϕ(z) + hyx − z, z − xi < 0. r Next, we recall some lemmas which will be needed in the rest of this paper. Lemma 2.1. [12] Let H be a real Hilber space and Θ : H × H → R be a bifunction. Let ϕ : H → R be a proper lower semicontinuous and convex function. For any r > 0 and x ∈ H, define a mapping Sr : H → H as follow: for all x ∈ H, 1 Sr (x) = {z ∈ H : Θ(z, y) + ϕ(y) − ϕ(z) + hy − z, z − xi ≥ 0, ∀y ∈ H}. r Assume that the condition (H1)-(H5) hold. Then we have the following: (1) For each x ∈ H, Sr (x) 6= 0 and Sr is single valued; (2) Sr is firmly nonexpansive, i.e., for any x, y ∈ H, kSr x − Sr yk2 ≤ hSr x − Sr y, x − yi; (3) F ix(Sr ) = EP (1); (4) EP (1) is closed and convex. Lemma 2.2. [15] Let {xn } and {zn } be bounded sequences in a Banach space E and {βn } be a sequence in [0, 1] satisfying the following condition: 0 < lim inf βn ≤ lim sup βn < 1. n→∞ n→∞ Suppose that xn+1 = βn xn + (1 − βn )zn for all n ≥ 0 and lim supn→∞ (kzn+1 − zn k − kxn+1 − xn k) ≤ 0. Then limn→∞ kzn − xn k = 0. Now, we define the mapping Wn by Un,n+1 = I, Un,n = λn Tn Un,n+1 + (1 − λn )I, Un,n−1 = λn−1 Tn−1 Un,n + (1 − λn−1 )I, ··· Un,k = λk Tk Un,k+1 + (1 − λk )I, Un,k−1 = λk−1 Tk−1 Un,k + (1 − λk−1 )I, ··· Un,2 = λ2 T2 Un,3 + (1 − λ2 )I, Wn = Un,1 = λ1 T1 Un,2 + (1 − λ1 )I, (2.2) where λ1 , λ2 , · · · are real numbers such that 0 ≤ λn ≤ 1 for all n ≥ 1 and {Ti }∞ n=1 is an infinite family of nonexpansive mappings Tn : H → H. Note that Wn is usually called the W -mapping generated by Tn , Tn−1 , · · · , T1 and λn , λn−1 , · · · , λ1 . It can be easily seen that Wn is also nonexpansive mapping. U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 480 Lemma 2.3. [15] Let C be a nonempty closed convex subset of a real Hilbert space H. Let {Tn }∞ n=1 be an ∞ infinite family of nonexpanxive mappings Tn : H → H such that ∩n=1 F ix(Tn ) 6= 0. Let λ1 , λ2 , · · · be a real number such that 0 < λn ≤ b < 1 for all n ∈ N . Then the following statements hold: (1) For all x ∈ H and k ∈ N , the limit limn→∞ Un,k x exists; (2) F ix(W ) = ∩∞ n=1 F ix(Tn ) where W x = limn→∞ Wn x = limn→∞ Un,1 x for all x ∈ C; (3) For any bounded sequence {xn } in H, lim kW xn − Wn xn k = 0. n→∞ Lemma 2.4. [1] Let R : H → 2H be a maximal monotone mapping and B : H → H be a Lipschitz and continuous monotone mapping. Then the mapping R + B : H → 2H is maximal monotone. Lemma 2.5. [5],[6] Let {an } be a sequence of non-negative real numbers satisfying an+1 ≤ (1 − γn )an + δn , n ≥ 1, where {γn } is a sequence in (0, 1) and {δn } satisfy the following conditions: P∞ (i) n=1 = ∞; P (ii) lim supn→∞ γδnn ≤ 0 or ∞ n=1 |δn | < ∞; Then limn→∞ an = 0. Lemma 2.6. [5],[6] Let C be a nonempty closed convex subset of a real Hilbert space H and g : C → R∪{∞} be a proper lower-semicontinuous differentiable convex function. If x∗ is a solution to the minimization problem: g(x∗ ) = inf g(x), x∈C then hg 0 (x), x − x∗ i ≥ 0, ∀x ∈ C. In particular, if x∗ solves the optimization problem: ν 1 min hAx, xi + kx − uk2 − h(x), x∈C 2 2 then hu + (γf − (I + νA))x∗ , x − x∗ i ≤ 0. 3. Main Results In this section, we prove some strong convergence theorems for finding a common element of the set of solution of a mixed equilibrium problem, the set of solution of variational inclusions and the set of fixed points of nonexpansive mappings. U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 481 Theorem 3.1. Let H a real Hilbert space and {Tn }∞ n=1 be an infinite family of nonexpansive mappings Tn : H → H. Let ϕ : H → R be a lower semicontinuous and convex function and Θ : H × H → R be a bifunction satisfying the condition (H1)-(H5). Let G be a strongly positive bounded linear operator with coefficient µ > 0 and R : H → 2H be a maximal monotone mapping. Let A, B, F : H → H be α, β, η-inverse strongly monotone mappings, respectively. Let f : H → H be a ρ-contraction and r > 0, s1 , s2 > 0, γ > 0 be four constants such that s1 < 2α, s2 < 2β, r < 2ρ and γ < (1+ν)µ . Assume that ρ ∞ Ω := ∩n=1 F ix(Tn ) ∩ EP (1) ∩ I(A, R) ∩ I(B, R) 6= ∅. Let Wn be the mapping defined by (2.2). If {xn } is the sequence generated by x1 ∈ H and 1 Θ(un , y) + ϕ(yn ) − ϕ(un ) + hy − un , un − (xn − rF xn )i ≥ 0, r zn = JR,s2 (un − s2 Bun ), (3.1) (z − s Az ), y = J n 1 n n R,s 1 xn+1 = αn (u + γf (xn )) + βn xn + [(1 − βn )I − αn (I + νG)]Wn yn , where {αn } and {βn } are two real sequences in [0, 1] satisfying the following conditions: (C1) limn→∞ αn = 0 and Σ∞ n=1 αn = ∞; (C2) 0 < lim inf n→∞ βn < lim supn→∞ βn < 1. Then the sequence {xn } converges strongly to a point x∗ ∈ Ω, which solve the following optimization problem: ν 1 min hGx, xi + kx − uk2 − h(x), x∈Ω 2 2 (3.2) where h is a potential function for γf . Proof. We divide the proof of Theorem 3.1 into several steps. Step 1. We show that {xn } is bounded. First, we note that I − s1 A and I − s2 B are nonexpansive for all x, y ∈ C and k(I − s1 A)x − (I − s1 A)yk2 = kx − yk2 − 2s1 hx − y, Ax − Ayi + s21 kAx − Ayk2 ≤ kx − yk2 − 2s1 αkAx − Ayk2 + s21 kAx − Ayk2 = kx − yk2 − s1 (2α − s1 )kAx − Ayk2 ≤ kx − yk2 . Thus I − s1 A is nonexpansive and so are I − s2 B and I − rF . Let p ∈ Ω. From (C1) and (C2), we assume that αn ≤ (1 − βn )(1 + νkGk)−1 for all n ≥ 1. Since G is a linear bounded self-adjoint operator on H, we have kGk = sup{|hGx, xi| : x ∈ H, kxk = 1} and h((1 − βn )I − αn (I + νG))x, xi = h(1 − βn )x − αn x − αn νGx, xi = 1 − βn − αn − αn νhGx, xi ≥ 1 − βn − αn − αn νkGk ≥ 0. Thus we can see that (1 − βn )I − αn (I + νG) is positive. Further, it follows that k(1 − βn )I − αn (I + νG)k = sup{h((1 − βn )I − αn (I + νG))x, xi x ∈ H, kxk = 1} = sup{1 − βn − αn − αn νhGx, xi, x ∈ H, kxk = 1} ≤ 1 − βn − αn (1 + νµ). U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 482 From zn = JR,s (un − s2 Bun ) for all n ≥ 0, we can compute kzn − pk = kJR,s2 (un − s2 Bun ) − JR,s2 (p − s2 Bp)k ≤ k(un − p) − s2 (Bun − Bp)k ≤ kun − pk and kyn − pk = kJR,s1 (zn − s1 Azn ) − JR,s1 (p − s1 Ap)k ≤ k(zn − p) − s1 (Azn − Ap)k ≤ kzn − pk ≤ kun − pk. Letting un = Sr (xn − rF xn ) for all n ≥ 0, by Lemma 2.1, we have kun − pk2 = kSr (xn − rF xn ) − Sr (p − rF p)k2 ≤ kxn − pk2 − 2rhxn − p, F xn − F pi + r2 kF xn − F pk2 ≤ kxn − pk2 − r(2ρ − r)kF xn − F pk2 ≤ kxn − pk2 (3.3) and so kzn − pk ≤ kxn − pk and kyn − pk ≤ kxn − pk. Thus we have kxn+1 − pk = kαn u + αn (γf (xn ) − (I + νG)p) + βn (xn − p) +[(1 − βn )I − αn (I + νG)](Wn yn − p)k ≤ (1 − βn − αn (1 + νµ))kyn − pk + βn kxn − pk +αn kuk + αn kγf (x − n) − (I + νG)pk ≤ (1 − αn (1 + νµ))kxn − pk + αn kuk + αn γηkxn − pk +αn kγf (p) − (I + νG)pk = [1 − αn (1 + νµ + γη)]kxn − pk + αn [kγf (p) − (I + νG)pk + kuk] n kγf (p) − (I + νG)pk + kuk o ≤ max kx0 − pk, . 1 + νµ − γp Hence {xn } is bounded and so are {un }, {zn }, {yn }, {Wn yn }, {f (xn )} and {GWn yn }. Step 2. We show that limn→∞ kxn+1 − xn k = 0. Define xn+1 = βn + (1 − βn )vn for all n ≥ 0, that is, xn+1 − βn xn vn = for all n ≥ 0. It follows that 1 − βn xn+2 − βn+1 xn+1 xn+1 − βn xn − 1 − βn+1 1 − βn αn+1 αn = [u + γf (xn+1 )] − [u + γf (xn )] + Wn+1 yn+1 − Wn yn 1 − βn+1 1 − βn αn+1 αn − (I + νG)Wn+1 yn+1 + (I + νG)Wn yn 1 − βn+1 1 − βn αn+1 = [u + γf (xn+1 ) − (I + νG)Wn+1 yn+1 ] 1 − βn+1 αn + [u + γf (xn ) + (I + νG)Wn yn ] 1 − βn +Wn+1 yn+1 − Wn+1 yn + Wn+1 yn − Wn yn vn+1 − vn = U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 483 and so kvn+1 − vn k − kxn+1 − xn k ≤ αn+1 [kuk + kγf (xn+1 )k + k(I + νG)Wn+1 yn+1 k] 1 − βn+1 αn + [kuk + kγf (xn )k + k(I + νG)Wn yn k] 1 − βn +kyn+1 − yn k + kWn+1 yn − Wn yn k − kxn+1 − xn k. Since Ti and Un,i are nonexpansive, it follows that kWn+1 yn − Wn yn k = kλ1 Tq Un+1,2 yn − λ1 T1 Un,2 yn k ≤ λ1 kUn+1,2 yn − Un,2 yn k = λ1 kλ2 T2 Un+1,3 yn − λ2 T2 Un,3 yn k ≤ λ1 λ2 kUn+1,3 yn − Un,3 yn k .. . ≤ λ1 λ2 . . . λn kUn+1,n+1 yn − Un,n+1 yn k n Y ≤ M λi , i=1 where M > 0 is a constant such that sup{kUn+1,n+1 yn − Un,n+1 yn k : n ≥ 0} ≤ M . Observe that kyn+1 − yn k = kJR,s1 (zn+1 − s1 Azn+1 ) − JR,s1 (zn − s1 Azn )k ≤ k(I − s1 A)zn+1 − (I − s1 A)zn k ≤ kzn+1 − zn k ≤ kJR,s2 (un+1 − s2 Bun+1 ) − JR,s2 (un − s2 Bun )k ≤ kun+1 − un k ≤ kSr (xn+1 − rF xn+1 ) − Sr (xn − rF xn )k ≤ kxn+1 − xn k. Therefore, we have kvn+1 − vn k − kxn+1 − xn k ≤ αn+1 [kuk + kγf (xn+1 )k + k(I + νG)Wn+1 yn+1 k] 1 − βn+1 n Y αn [kuk + kγf (xn )k + k(I + νG)Wn yn k] + M λi , + 1 − βn i=1 which implies that lim sup(kvn+1 − vn k − kxn+1 − xn k) ≤ 0. n→∞ Hence, by Lemma 2.2, it follows that limn→∞ kvn − xn k = 0 and xn+1 − βn xn − xn k 1 − βn kxn+1 − xn k = lim , n→∞ 1 − βn lim kvn − xn k = n→∞ lim k n→∞ and so lim kxn+1 − xn k = 0. (3.4) n→∞ Step 3. We show that lim kF xn − F pk = 0, lim kBun − Bpk = 0, lim kAzn − Apk = 0. n→∞ n→∞ n→∞ U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 484 From (3.1), it follows that xn+1 = αn (u + γf (xn )) + βn xn + [(1 − βn )I − αn (I + νG)]Wn yn , which can be rewritten as xn+1 = αn (u + γf (xn ) − (I + νG)Wn yn ) + βn (xn − Wn yn ) + Wn yn . Observe that kxn − Wn yn k ≤ kxn − xn+1 k + kxn+1 − Wn yn k ≤ kxn − xn+1 k + αn ku + γg(xn ) − (I + νG)Wn yn k + βn kxn − Wn yn k and so kxn − Wn yn k ≤ αn 1 kxn − xn+1 k + ku + γf (xn ) − (I + νG)Wn yn k. 1 − βn 1 − βn Thus, from αn → 0 and limn→∞ kxn+1 − xn k = 0, it follows that lim kxn − Wn yn k = 0. n→∞ (3.5) Since I − rA, I − sB and I − rF are nonexpansive, it follows from (3.3) that kyn − pk2 = kJR,s1 (zn − s1 Azn ) − JR,s1 (p − s1 Ap)k2 ≤ kzn − pk2 + s1 (s1 − 2α)kAzn − Apk2 ≤ kJR,s2 (un − s2 Bun ) − JR,s2 (p − s2 Bp)k2 + s1 (s1 − 2α)kAzn − Apk2 ≤ kun − pk2 + s2 (s2 − 2β)kBun − Bpk2 + s1 (s1 − 2α)kAzn − Apk2 ≤ kxn − pk2 + r(r − 2ρ)kF xn − F pk2 + s2 (s2 − 2β)kBun − Bpk2 +s1 (s1 − 2α)kAzn − Apk2 . Thus we obtain kxn+1 − pk2 = kαn (u + γf (xn ) − (I + νG)p) + βn (xn − Wn yn ) + (I − αn (I + νG))(Wn yn − p)k2 ≤ k(I − αn (I + νG))(Wn yn − p) + βn (xn − Wn yn )k2 +2αn hu + γf (xn ) − (I + νG)p, xn+1 − pi ≤ [kI − αn (I + νG)kkyn − pk + βn kxn − Wn yn k]2 +2αn ku + γf (xn ) − (I + νG)p)kkxn+1 − pk = (1 − α − n(1 + νµ))2 kyn − pk2 + βn2 kxn − Wn yn k2 +2(1 − αn (1 + νµ))βn kyn − pkkxn − Wn yn k +2αn ku + γf (xn ) − (I + νG)pkkxn+1 − pk. Since (1 − αn (1 + νµ))2 < 1, it follows that kxn+1 − pk2 ≤ kyn − pk2 + βn2 kxn − Wn yn k2 + 2(1 − αn (1 + νµ))βn kyn − pkkxn − Wn yn k +2αn ku + γf (xn ) − (I + νG)pkkxn+1 − pk ≤ kxn − pk2 + r(r − 2ρ)kF xn − F pk2 + s2 (s2 − 2β)kBun − Bpk2 +s1 (s1 − 2α)kAzn − Apk2 βn2 kxn − Wn yn k2 +2(1 − αn (1 + νµ))βn kyn − pkkxn − Wn yn k +2αn ku + γf (xn ) − (I + νG)pkkxn+1 − pk, (3.6) U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 485 and so r(2ρ − r)kF xn − F pk2 + s2 (2β − s2 )kBun − Bpk2 + s1 (2α − s1 )kAzn − Apk2 ≤ (kxn − pk + kxn+1 − pk)(kxn+1 − xn k) + βn kxn − Wn yn k2 +2(1 − αn (1 + νµ))βn kyn − pkkxn − Wn yn k +2αn ku + γf (xn ) − (I + νG)pkkxn+1 − pk. (3.7) Therefore, from (3.4), (3.5) and (3.7) it follows that lim kF xn − F pk = 0, lim kBun − Bpk = 0, lim kAzn − Apk = 0. n→∞ n→∞ n→∞ Step 4. We show that kxn − W xn k → 0. Since Sr is firmly nonexpansive, we have kun − pk2 = kSr (xn − rF xn ) − Sr (p − rF p)k2 ≤ hxn − rF xn − (p − rF p), un − pi 1 = (kxn − pk2 + kun − pk2 − kxn − un − r(F xn − F p)k2 ) 2 1 (kxn − pk2 + kun − pk2 − kxn − un k2 + 2rhF xn − F p, xn − un i = 2 −r2 kF xn − F pk2 ), which implies that kun − pk2 ≤ kxn − pk2 − kxn − un k2 + 2rhF xn − F p, xn − un i − r2 kF xn − F pk2 = kxn − pk2 − kxn − un k2 + 2rkF xn − F pkkxn − un k − r2 kF xn − F pk2 ≤ kxn − pk2 − kxn − un k2 + 2rkF xn − F pkkxn − un k. Since JR,s is 1-inverse strongly monotone, we have kzn − pk2 = kJR,s2 (un − s2 Bun ) − JR,s2 (p − s2 Bp)k2 ≤ h(u − s2 Bun ) − (p − s2 Bp), zn − p)i 1 = (k(un − s2 Bun ) − (p − s2 Bp)k2 + kzn − pk2 2 −k(un − s2 Bun ) − (p − s2 Bp) − (zn − p)k2 ) 1 ≤ (kun − pk2 + kzn − pk2 − kun − zn − s2 (Bun − Bp)k2 ) 2 1 (kun − pk2 + kzn − pk2 − kun − zn k2 + 2s2 hBun − Bp, un − zn i = 2 −s22 kBun − Bpk2 ), which implies that kzn − pk2 ≤ kun − pk2 − ku − n − zn k2 + 2s2 kBun − Bpkkun − zn k. Now, we observe that kyn − pk2 = kJR,s1 (zn − s1 Azn ) − JR,s1 (p − s1 Ap)k2 ≤ hzn − s1 Azn − (p − s1 Ap), yn − pi 1 (kzn − pk2 + kyn − pk2 − kzn − yn − s1 (Azn − Ap)k2 ) = 2 1 = (kzn − pk2 + kyn − pk2 − kzn − yn k2 + 2s1 hAzn − Ap, zn − yn i 2 −s21 kAzn − Apk2 ), U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 486 that is, kyn − pk2 ≤ kzn − pk2 − kzn − yn k2 + 2s1 kAzn − Apkkzn − yn k ≤ kun − pk2 − kun − zn k2 + 2s2 kBun − Bpkkun − zn k − kzn − yn k2 +2s1 kAzn − Apkkzn − yn k ≤ kxn − pk2 − kxn − un k2 + 2rkF xn − F pkkxn − un k − kun − zn k2 +2s2 kBun − Bpkkun − zn k − kzn − yn k2 + 2s1 kAzn − Apkkzn − yn k. (3.8) Substituting (3.8) in (3.6), we have kxn+1 − pk2 ≤ (1 − αn (1 + νµ))2 {kxn − pk2 − kxn − un k2 − kun − zn k2 − kzn − yn k2 +2rkF xn − F pkkxn − un k + 2s2 kBun − Bpkkun − zn k +2s1 kAzn − Apkkzn − yn k} βn2 kxn − Wn yn k2 + 2(1 − αn (1 + νµ))βn kyn − pkkxn − Wn yn k +2αn ku + γf (xn ) − (I + νG)pkkxn+1 − pk. It follows that (1 − αn (1 + νµ))2 { kxn − un k2 + kun − zn k2 + kzn − yn k2 } ≤ (kxn − pk + kxn+1 − pk)kxn+1 − xn k + 2rkF xn − F pkkxn − un k +2s2 kBun − Bpkkun − zn k + 2s1 kAzn − Apkkzn − yn k +2(1 − αn (1 + νµ))βn kyn − pkkxn − Wn yn k +2αn ku + γf (xn ) − (I + νG)pkkxn+1 − pk. Thus we have lim kxn − un k = 0, lim kun − zn k = 0, and lim kzn − yn k = 0. n→∞ n→∞ n→∞ Note that kW yn − yn k ≤ kW yn − Wn yn k + kWn yn − yn k and so, from this and Lemma 2.3, it follows that limn→∞ kW yn − yn k = 0. Therefore, we have lim kxn − W xn k = 0. n→∞ Step 5. We show that lim suphu + γf (x∗ ) − (I + νG)x∗ , xn − x∗ i ≤ 0, n→∞ where x∗ is a solution of the optimization problem. First, we note that there exists a subsequence {xni } of {xn } such that lim suphu + (γf − (I + νG))x∗ , xn − x∗ i = n→∞ lim hu + (γf − (I + νG))x∗ , xnj − x∗ i. j→∞ Since {xnj } is bounded, there exists a subsequence {xnji } of {xnj } which converges weakly to ω. Without loss of generality, we can assume that {xnj } converges weakly to the point ω. Since kW xn − xn k → 0, it follows the demiclosed principle of nonexpansive mappings that ω ∈ F ix(W ). Now, we show that ω ∈ EP (1). By un = Sr (xn − rF xn ), it follows that 1 Θ(un , y) + ϕ(y) − ϕ(un ) + hy − un , un − (xn − rF xn )i ≥ 0, r ∀y ∈ H. U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 487 From (H2), we have 1 ϕ(y) − ϕ(un ) + hy − un , un − (xn − rF xn )i ≥ Θ(y, un ), ∀y ∈ H, r and hence 1 ϕ(y) − ϕ(uni ) + hy − uni , uni − (xni − rF xni )i ≥ Θ(y, uni ), ∀y ∈ H. r For any t ∈ (0, 1] and y ∈ H, let yt = ty + (1 − t)ω. From (3.9), we have (3.9) 1 ϕ(yt ) − ϕ(uni ) + hyt − uni , uni − (xni − rF xni )i ≥ Θ(yt , uni ), r uni − xni + rF xni i ≥ Θ(yt , uni ) − ϕ(yt ) + ϕ(uni ), hyt − uni , r un − xni + rF xni hyt − uni , i + F (yt ) − F( yt )i ≥ Θ(yt , uni ) − ϕ(yt ) + ϕ(uni ). r It follows that hyt − uni , F yt i ≥ ϕ(uni ) − ϕ(yt ) + hyt − uni , F yt i + Θ(yt − uni ) un − x − ni + rF xni −hyt − uni , i i r = hyt − uni , F yt − F uni i + hyt − uni , F uni − F xni i − hyt − uni , uni − xni i r +ϕ(uni ) − ϕ(yt ) + Θ(yt , uni ). Since kuni − xni k → 0, we have kF uni − F xni k → 0. Further, from the inverse strongly monotonicity of F , it follows that hyt − uni , F yt − F uni i ≥ 0 and so, from (H4), the weakly lower semi-continuity of ϕ, uni −xni → 0 and uni → ω weakly, it follows that r hyt − ω, F yt i ≥ −ϕ(yt ) + ϕ(ω) + Θ(yt , ω). (3.10) Thus, from (H1), (H4) and (3.10), we have 0 = Θ(yt , yt ) − ϕ(yt ) + ϕ(yt ) = Θ(yt , ty + (1 − t)ω) − ϕ(yt ) + ϕ(ty + (1 − t)ω) ≤ t[Θ(yt , y) + ϕ(y) − ϕ(yt )] + (1 − t)[Θ(yt , ω) + ϕ(ω) − ϕ(yt )] ≤ t[Θ(yt , y) + ϕ(y) − ϕ(yt )] + (1 − t)hyt − ω, F yt i = t[Θ(yt , y) + ϕ(y) − ϕ(yt )] + (1 − t)thy − ω, F yt i. Letting t → 0, we have, for any y ∈ H, 0 ≤ Θ(ω, y) + ϕ(y) − ϕ(ω) + hy − ω, F ωi, which implies that ω ∈ EP (1). Next, we show that ω ∈ I(B, R). Since A be an α-inverse strongly monotone, A is Lipschitz continuous monotone mapping. It follows from Lemma (2.4) that R+A is maximal monotone. Let (v, g) ∈ G(R+A), i.e., g−Av ∈ R(v). Since yni = JR,s1 (zni −s1 Azni ), we have zni −s1 zni ∈ (I +s1 R)yni , i.e., s11 (zni −yni −s1 Azni ) ∈ R(yni ). By maximal monotonicity of R + A, we have hv − yni , g − Bv − 1 (zn − yni − s1 Azni )i ≥ 0 s1 i and so 1 (zn − yni − s1 Azni )i s1 i 1 − Azni i + hv − yni , (zni − yni )i. s1 hv − yni , gi ≥ hv − yni , Av + ≥ hv − yni , Ayni U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 488 It follows from kzn − yn k → 0, kAzn − Ayn k → 0 and yni → ω weakly that lim hv − yni , gi = hv − ω, gi ≥ 0. ni →∞ Therefore, ω ∈ I(A, R) and, by the same method as in above, we can also find that ω ∈ I(B, R). Further, we have ω ∈ Ω and lim suphu + (γf − (I + νG))x∗ , xn − x∗ i = n→∞ lim hu + (γf − (I + νG))x∗ , xnj − x∗ i j→∞ = hu + (γf − (I + νG))x∗ , ω − x∗ i ≤ 0. Step 6. We show that the sequence {xn } converges to the point x∗ . Observe that kxn+1 − x∗ k2 = kαn (u + γf (xn )) + βn xn + [(1 − βn )I − αn (I + νG)]Wn yn − x∗ k2 = kαn (u + γf (xn ) − (I + νG)x∗ ) + βn (xn − x∗ ) +((1 − βn )I − αn (I + νG))(Wn yn − x∗ )k2 = kβn (xn − x∗ ) + ((1 − βn )I − αn (I + νG))(Wn yn − x∗ )k2 +2αn hu + γf (xn ) − (I + νG)x∗ , xn+1 − x∗ i ≤ [k((1 − βn )I − αn (I + νG))(Wn yn − x∗ )k + kβn (xn − x∗ )k]2 +2αn γhf (xn ) − f (x∗ ), xn+1 − x∗ i +2αn hu + γf (x∗ ) − (I + νG)x∗ , xn+1 − x∗ i ≤ [k((1 − βn )I − αn (I + νµ))kyn − x∗ k + βn kxn − x∗ k]2 +2αn γρkxn − x∗ kkxn+1 − x∗ k +2αn hu + γf (x∗ ) − (I + νG)x∗ , xn+1 − x∗ i ≤ (1 − αn (1 + ν)µ)2 kxn − x∗ k2 + αn γρ{kxn − x∗ k2 + kxn+1 − x∗ k2 } +2αn hu + γf (x∗ ) − (I + νG)x∗ , xn+1 − x∗ i and so αn γρ (1 − αn (1 + ν)µ)2 kxn − x∗ k2 + kxn − x∗ k2 1 − αn γρ 1 − αn γρ 2αn + hu + γf (x∗ ) − (I + νG)x∗ , xn+1 − x∗ i 1 − αn γρ 2αn ((1 + ν)µ − γρ) (αn (1ν )µ)2 kxn − x∗ k2 + kxn − x∗ k2 = [1 − 1 − αn γρ 1 − αn γρ 2αn + hu + γf (x∗ ) − (I + νG)x∗ , xn+1 − x∗ i 1 − αn γρ 2αn ((1 + ν)µ − γρ) 2αn ((1 + ν)µ − γρ) = [1 − kxn − x∗ k2 + 1 − αn γρ 1 − αn γρ 2 αn ((1ν )µ) 1 ×{ M1 + hu + γf (x∗ ) − (I + νG)x∗ , xn+1 − x∗ i} 1 − αn γρ (1ν )µ − γρ = (1 − δn )kxn − x∗ k2 + δn αn , kxn+1 − x∗ k ≤ where M1 = sup{kxn − x∗ k2 : n ≥ 1}, σn = δn = 2αn ((1 + ν)µ − γρ) , 1 − αn γρ αn ((1ν )µ)2 1 M1 + hu + γf (x∗ ) − (I + νG)x∗ , xn+1 − x∗ i. 1 − αn γρ (1ν )µ − γρ U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 489 P It can be easily seen that ∞ n=1 δn = ∞ and lim supn→∞ σn ≤ 0. Hence, by Lemma 2.5, we conclude that the sequence {xn } converges strongly to the point x∗ ∈ Ω. This completes the proof. Using Theorem 3.1, we can obtain the following corollaries: Corollary 3.2. For any x0 ∈ H, let {xn } be the sequence in H generated by the following iterative algorithm: for any y ∈ H and n ≥ 1 1 Θ(un , y) + ϕ(yn ) − ϕ(un ) + hy − un , un − (xn − rF xn )i ≥ 0, r zn = JR,s2 (un − s2 Bun ), (3.11) yn = JR,s1 (zn − s1 Azn ), xn+1 = αn (u + γf (xn )) + βn xn + (1 − βn − αn )Wn yn , where {αn } and {βn } are two real sequences in [0, 1] satisfying the following conditions: (C1) limn→∞ αn = 0 and Σ∞ n=1 αn = ∞; (C2) 0 < lim inf n→∞ βn < lim supn→∞ βn < 1. Suppose that Ω := ∩∞ n=1 F ix(Tn ) ∩ EP (1) ∩ I(A, R) ∩ I(B, R) 6= ∅ and the mapping Wn is defined by (2.2). Then the sequence {xn } converges strongly to a point x∗ ∈ Ω, which solves the following variational inequality: hu + γf (x∗ ) − x∗ , y − xi ≤ 0, ∀y ∈ Ω. Corollary 3.3. For any x0 ∈ H, let {xn } be the sequence in H generated by the following iterative algorithm: for any y ∈ H and n ≥ 1 1 Θ(un , y) + ϕ(yn ) − ϕ(un ) + hy − un , un − xn i ≥ 0, r zn = JR,s2 (un − s2 Bun ), (3.12) yn = JR,s1 (zn − s1 Azn ), xn+1 = αn γf (xn ) + βn xn + [(1 − βn )I − αn G]Wn yn , where {αn } and {βn } are two real sequences in [0, 1] satisfying the following conditions: (C1) limn→∞ αn = 0 and Σ∞ n=1 αn = ∞; (C2) 0 < lim inf n→∞ βn < lim supn→∞ βn < 1. Suppose that Ω := ∩∞ n=1 F ix(Tn ) ∩ EP (2) ∩ I(A, R) ∩ I(B, R) 6= ∅ and the mapping Wn is defined by (2.2). Then the sequence {xn } converges strongly to x∗ ∈ Ω, where x∗ = PΩ (γf (x∗ ) + (I − G)x∗ ), which solves the following variational inequality: hγf (x) − Gx, y − xi ≤ 0, ∀y ∈ Ω. Corollary 3.4. For any x0 ∈ H, let {xn } be the sequence in H generated by the following iterative algorithm: for any y ∈ H and n ≥ 1 1 Θ(un , y) + ϕ(yn ) − ϕ(un ) + hy − un , un − (xn − rF xn )i ≥ 0, r zn = JR,s2 (un − s2 Bun ), (3.13) y = J (z − s Az ), n 1 n R,s1 n xn+1 = αn γf (xn ) + βn xn + [(1 − βn )I − αn G]yn , where {αn } and {βn } are two real sequences in [0, 1], which satisfy the following conditions: U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 490 (C1) limn→∞ αn = 0 and Σ∞ n=1 αn = ∞; (C2) 0 < lim inf n→∞ βn < lim supn→∞ βn < 1. Suppose that Ω := EP (1) ∩ I(A, R) ∩ I(B, R) 6= ∅ and the mapping Wn is defined by (2.2). Then the sequence {xn } converges strongly to x∗ ∈ Ω, where x∗ = PΩ (γf (x∗ ) + (I − G)x∗ ), which solves the following variational inequality: hγf (x) − Gx, y − xi ≤ 0, ∀y ∈ Ω. Corollary 3.5. For any x0 ∈ H, let {xn } be the sequence in H generated by the following iterative algorithm: for any y ∈ H and n ≥ 1 1 Θ(un , y) + hy − un , un − (xn − rF xn )i ≥ 0, r zn = JR,s2 (un − s2 Bun ), (3.14) yn = JR,s1 (zn − s1 Azn ), xn+1 = αn γf (xn ) + βn xn + [(1 − βn )I − αn G]Wn yn , where {αn } and {βn } are two real sequences in [0, 1] satisfying the following conditions: (C1) limn→∞ αn = 0 and Σ∞ n=1 αn = ∞; (C2) 0 < lim inf n→∞ βn < lim supn→∞ βn < 1. Suppose that Ω := ∩∞ n=1 F ix(Tn ) ∩ EP (3) ∩ I(A, R) ∩ I(B, R) 6= ∅ and the mapping Wn is defined by (2.2). Then the sequence {xn } converges strongly to x∗ ∈ Ω, where x∗ = PΩ (γf (x∗ ) + (I − G)x∗ ), which solves the following variational inequality: hγf (x) − Gx, y − xi ≤ 0, ∀y ∈ Ω. Corollary 3.6. For any x0 ∈ H, let {xn } be the sequence in H generated by the following iterative algorithm: for any y ∈ H and n ≥ 1 1 Θ(un , y) + hy − un , un − (xn − rF xn )i ≥ 0, r zn = JR,s2 (un − s2 Bun ), (3.15) y = J (z − s Az ), n n 1 n R,s 1 xn+1 = αn γf (xn ) + βn xn + [(1 − βn )I − αn G]yn , where {αn } and {βn } are two real sequences in [0, 1] satisfying the following conditions: (C1) limn→∞ αn = 0 and Σ∞ n=1 αn = ∞; (C2) 0 < lim inf n→∞ βn < lim supn→∞ βn < 1. Suppose that Ω := EP (3) ∩ I(A, R) ∩ I(B, R) 6= ∅ and the mapping Wn is defined by (2.2). Then the sequence {xn } converges strongly to x∗ ∈ Ω, where x∗ = PΩ (γf (x∗ ) + (I − A)x∗ ), which solves the following variational inequality: hγf (x) − Gx, y − xi ≤ 0, ∀y ∈ Ω. U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 491 Corollary 3.7. For any x0 ∈ H, let {xn } be the sequence in H generated by the following iterative algorithm: for any y ∈ H and n ≥ 1 1 Θ(un , y) + hy − un , un − (xn − rF xn )i ≥ 0, r zn = JR,s2 (un − s2 Bun ), (3.16) (z − s Az ), y = J 1 n n R,s1 n xn+1 = αn f (xn ) + βn xn + (1 − βn − αn )yn , where {αn } and {βn } are two real sequences in [0, 1] satisfying the following conditions: (C1) limn→∞ αn = 0 and Σ∞ n=1 αn = ∞; (C2) 0 < lim inf n→∞ βn < lim supn→∞ βn < 1. Suppose that Ω := EP (3)∩I(A, R)∩I(B, R) 6= ∅ and the mapping Wn is defined by (2.2). Then the sequence {xn } converges strongly to x∗ ∈ Ω, where x∗ = PΩ f (x∗ ), which solves the following variational inequality: hf (x) − x, y − xi ≤ 0, ∀y ∈ Ω. 4. Numerical Examples In this section, we give a real numerical example of the Main Theorem as follows: Example 4.1. For simplicity, we assume H = R and C = [−1, 1]. Let Θ(z, y) = −7z 2 + zy + 6y 2 , F = I and φ(x) = x2 . Find z ∈ [−1, 1] such that 1 Θ(z, y) + φ(y) − φ(z) + hy − z, z − (x − rx)i ≥ 0, r ∀y ∈ [−1, 1]. Solution. It can easily be seen that Θ, φ and F are satisfied the conditions in Theorem 3.1. For any r > 0 and x ∈ [−1, 1], by Lemma 2.1, we can see that there exists z ∈ [−1, 1] such that, for any y ∈ [−1, 1], 1 Θ(z, y) + φ(y) − φ(z) + hy − z, z − (x − rx)i ≥ 0, r 1 2 2 2 −7z + 6y + zy + y − z 2 + hy − z, z − (x − rx)i ≥ 0, r 2 7ry + (rz + z − x + rx)y + (−8rz 2 − z 2 + xz − rxz) ≥ 0. Let H(y) = 7ry 2 + (rz + z − x + rx)y + (−8rz 2 − z 2 + xz − rxz). Then H is a quadratic function of y with coefficient a = 7r, b = rz + z − x + rx and c = −8rz 2 − z 2 + xz − rxz. Therefore, we can compute the discriminant ∆ of H as follows: ∆ = b2 − 4ac = [rz + z − x + rx]2 − 4(7r)(−8rz 2 − z 2 + xz − rxz) = 225r2 z 2 + 30rz 2 − 28rxz + 30r2 xz + z 2 − 2zx − 2rx2 + r2 x2 + x2 = (225r2 + 30r + 1)z 2 + (30r2 − 28r − 2)xz + (r2 − 2r + 1)x2 = (15r + 1)2 z 2 + 2(15r + 1)(r − 1)xz + (r − 1)2 x2 = [(15r + 1)z + (r − 1)x]2 . We know that H(y) ≥ 0 for all y ∈ [−1, 1] if it has at most one solution in [−1, 1]. Thus ∆ ≤ 0 and 1−r 1−r x. Then we have z = Sr xn = xn . Let {xn } be the sequence generated by hence z = 15r + 1 15r + 1 U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 x1 = x ∈ [−1, 1] and 1 Θ(un , y) + ϕ(yn ) − ϕ(un ) + hy − un , un − (xn − rF xn )i ≥ 0, r zn = JR,s2 (un − s2 Bun ), yn = JR,s1 (zn − s1 Azn ), xn+1 = αn (u + γf (xn )) + βn xn + [(1 − βn )I − αn (I + νG)]Wn yn . 492 (4.1) Assume that JR,s1 , JR,s2 , G = I, Ax = x3 , Bx = x2 and Wn = x2n . 1 n 1 Algorithm 1. Choose r = 0.5, αn = 10n , βn = 2n+1 , f (xn ) = 10 xn and s1 = s2 = 31 , γ = 1, ν = 12 , u = 0. Then the algorithm (4.1) becomes 3 un = xn , 17 1 zn = un 3 (4.2) 2 y = z n n 9 nxn n + 1 yn xn yn xn+1 = + +( ) − , ∀n ≥ 1. 100n 2n + 1 2n + 1 2 20n We can see that xn → 0 as n → ∞, where 0 is the unique solution of the optimization problem: min 7 2 x + C. 10 (4.3) Algorithm 2. Choose r = 0.5, αn = 1 50n , βn = n 20n+5 . Then the algorithm (4.1) becomes 3 un = xn , 17 1 zn = un 3 2 yn = zn 9 xn nxn 19n + 5 yn yn xn+1 = + +( ) − , ∀n ≥ 1. 500n 20n + 5 20n + 5 2 20n (4.4) Numerical result. In this part we give the numerical results that support our main theorem as shown by calculating and plotting graphs using Matlab 7.11.0. Algorithm 1. n 1 2 3 4 5 6 .. . xn 0.500000000000000 0.173518518518519 0.070898759380295 0.030870860056218 0.013904609765372 0.006395839590864 .. . 47 48 49 0.000000000000001 0.000000000000001 0.000000000000000 U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 Figure 1. This table shows the value of sequence {xn } on each iteration steps. Figure 2. This figure shows the graph of the above table, we can see that xn converges to zero. Algorithm 2. n 1 2 3 4 5 6 .. . xn 0.500000000000000 0.023810457516340 0.001222979105852 0.000064618469337 0.000003465087782 0.000000187506436 .. . 11 12 13 0.000000000000093 0.000000000000005 0.000000000000000 Figure 3. This table shows the value of sequence {xn } on each iteration steps. Figure 4. This figure shows the graph of the above table, we can see that xn converges to zero. 493 U. Witthayarat, Y.J. Cho, P. Kumam, J. Nonlinear Sci. Appl. 5 (2012), 475–494 494 5. Acknowledgements We would like to thank the Faculty of Science, King Mongkut’s University of Technology Thonburi (KMUTT) and the National Research Council of Thailand (NRCT-2555). This research is finished at Department of Mathematics Education, Gyeongsang National University. 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