Báo cáo hóa học: " Strong convergence theorems for equilibrium problems and fixed point problems: A new iterative method, some comments and applications" pptx

RESEARC H Open Access Strong convergence theorems for equilibrium problems and fixed point problems: A new iterative method, some comments and applications Zhenhua He 1 and Wei-Shih Du 2* * Correspondence: wsdu@nknucc. nknu.edu.tw 2 Department of Mathematics, National Kaohsiung Normal University, Kaohsiung 824, Taiwan Full list of author information is available at the end of the article Abstract In this paper, we introduce a new approach method to find a common element in the intersection of the set of the solutions of a finite family of equilibrium problems and the set of fixed points of a nonexpansive mapping in a real Hilbert space. Under appropriate conditions, some strong convergence theorems are established. The results obtained in this paper are new, and a few examples illustrating these results are given. Finally, we point out that some ‘so-called’ mixed equilibrium problems and generalized equilibrium problems in the literature are still usual equilibrium problems. 2010 Mathematics Subject Classification: 47H09; 47H10, 47J25. Keywords: strong convergence, iterative method, equilibrium problem, fixed point problem 1 Introduction and preliminaries Throughout this paper, we assume that H is a real Hilbert space with zero vector θ, whose inner product and norm are denoted by 〈·, ·〉 and || · ||, respect ively. The sym- bols N and ℝ are used to denote the sets of positive integers and real numbers, respec- tively. Let K be a nonempty closed convex subset of H and T : K ® H be a mapping. In this paper, the set of fixed points of T is denoted by F(T). We use symbols ® and ⇀ to denote strong and weak convergence, respectively. For each point x Î H, there exists a unique nearest point in K, denoted by P K x, such that  x − P K x  ≤  x − y  , ∀ y ∈ K . The mapping P K is called the metric projection from H onto K. It is well known that P K satisfies x − y , P K x − P K y ≥P K x − P K y  2 for every x, y Î H. Moreover, P K x is characterized by the properties: for x Î H, and z Î K, He and Du Fixed Point Theory and Applications 2011, 2011:33 http://www.fixedpointtheoryandapplications.com/content/2011/1/33 © 2011 He and Du; licensee S pringer. This i s an Open Access article distribu ted under the terms of the Creative Common s Attribution License (http://creativecommons .org/license s/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. z = P K ( x ) ⇔x − z, z − y≥0, ∀ y ∈ K . Let f be a bi-function from K × K into ℝ. The classical equilibrium problem is to find x Î K such that f ( x, y ) ≥ 0, ∀ y ∈ K . (1:1) Let EP(f) denote the set of all solutions of the problem (1.1). Since several probl ems in phys ics, optimization, and economics r educe to find a solution of (1.1) (see, e.g., [1,2]), some authors had proposed some methods to find the solution of e quilibrium problem (1.1); for instance, see [1-4]. We know that a mapping S is said to be nonex- pansive mapping if for all x, y Î K,||Sx - Sy|| ≤ ||x - y||. Recently, some authors used iterative method including composite iterative , CQ iterative, viscosity iterative etc. to find a common element in the intersection of EP(f) and F(S); see, e.g., [5-11]. Let I be an index set. For each i Î I,letf i be a bi-function from K × K into ℝ.The system of equilibrium problem is to find x Î K such that f i ( x, y ) ≥ 0, ∀ y ∈ K and ∀i ∈ I . (1:2) We know that  i ∈ I EP( f i ) is the set of all solutions of the system of equilibrium pro- blem (1.2). For each i Î I,iff i (x, y)=〈A i x, y - x〉, where A i : K ® K is a nonlinear operator, then the problem (1.2) becomes the following system of variational inequality problem: Find an element x ∈ K such that A i x, y − x≥0, ∀ y ∈ K . (1:3) It is obvious that the problem (1.3) is a special case of the problem (1.2). The following Lemmas are crucial to our main results. Lemma 1.1 (Demicloseness principle [12]) LetHbearealHilbertspaceandKa closed convex subset of H. S : K ® H is a nonexpansive mapping. Then the mapping I - S is demiclosed on K, where I is the identity mapping, i.e., x n ⇀ x in K and (I - S)x n ® y implies that × Î K and (I - S)x = y. Lemma 1.2 [13] Let {x n }and {y n } be bounded sequences in a Banach space E and let {b n } be a sequence in [0,1] with 0 < lim inf n®∞ b n ≤ lim sup n®∞ b n <1.Suppose x n+1 = b n y n +(1-b n )x n for al l 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 1.3 [5] Let H be a real Hilbert space. Then the following hold. (a) ||x + y|| 2 ≤ ||y|| 2 +2〈x, x + y〉 for all x, y Î H; (b) ||ax +(1-a)y|| 2 = a||x|| 2 +(1-a)||y|| 2 - a(1 - a)||x - y|| 2 for all x, y Î H and a Î ℝ; (c) ||x - y|| 2 =||x|| 2 +||y|| 2 -2〈x, y 〉 for all x, y Î H. Lemma 1.4. [14] Let {a n } be a sequence of nonnegative real numbers satisfying the following relation: a n+1 ≤ ( 1 − λ n ) a n + γ n , n ≥ 0 . If He and Du Fixed Point Theory and Applications 2011, 2011:33 http://www.fixedpointtheoryandapplications.com/content/2011/1/33 Page 2 of 15 (i) l n Î [0,1], ∞  n=0 λ n = ∞ or, equivalently,  ∞ n = 0 (1 − λ n )=0 ; (ii) lim sup n→∞ γ n λ n ≤ 0 or ∞  n=0 |γ n | < ∞ , then lim n →∞ a n = 0 . Lemma 1.5 [1] Let K be a nonempty closed convex subset of H and F be a bi-function of K × K into ℝ satisfying the following conditions. (A1) F(x, x)=0for all × Î K; (A2) F is monotone, that is, F(x, y)+F(y, x) ≤ 0 for all x, y Î K; (A3) for each x, y, z Î K, lim t ↓ 0 F( tz +(1− t)x, y) ≤ F(x, y) ; (A4) for each × Î K, y ® F (x, y) is convex and lower semi-continuous.Let r >0and × Î H. Then, there exists z Î K such that F( z , y)+ 1 r y − z, z − x≥0, for all y ∈ K . Lemma 1.6 [3] Let K be a nonempty closed convex subset of H and let F be a bi- function of K × K into R satisfying (A1) - (A4).Forr>0 and × Î H, define a mapping T r : H ® K as follows: T r (x)=  z ∈ K : F(z, y)+ 1 r y − z, z − x≥0, ∀ y ∈ K  for all × Î H. Then the following hold: (i) T r is single-valued; (ii) T r is firmly nonexpansive, that is, for any x, y Î H,  T r x − T r y  2 ≤T r x − T r y , x − y  ; (iii) F(T r )=EP (F); (iv) EP(F) is closed and convex. 2 Main results and their applications Let I = {1, 2, , k} be a finite index set, where k Î N. For each i Î I, let f i be a bi-func- tions from K × K into ℝ satisfying the conditions (A1)-(A4). Denote T i r n : H → K by T i r n (x)=  z ∈ K : f i (z, y)+ 1 r n y − z, z − x≥0, ∀ y ∈ K  . For each (i, n) Î I×N, applying Lemmas 1.5 and 1.6, T i r n is a firmly nonexpansive single-valued mapping such that F( T i r n )=EP(f i ) is closed and convex. For each i Î I, let u i n = T i r n x n , n Î N. First, let us consider the following example. He and Du Fixed Point Theory and Applications 2011, 2011:33 http://www.fixedpointtheoryandapplications.com/content/2011/1/33 Page 3 of 15 Example A Let f i :[-1,0]×[-1,0] ®ℝ be defined by f i (x, y) = (1+x 2i )(x-y), i =1,2,3. It is easy to see that for any i Î {1, 2, 3}, f i (x, y) satisfies the conditions (A1)-(A4) and  3 i =1 EP( f i )={0 } .LetSx = x 3 and g x = 1 2 x , ∀ x Î [-1, 0] Then g is a 1 2 -contraction from K into itself and S : K ® K is a nonexpansive mapping with   3 i=1 EP( f i ))   F( S)={0 } . Let l Î (0, 1), {r n } ⊂ [1, + ∞)and{a n } ⊂ (0,1) satisfy the conditions (i) lim n® ∞ a n =0,and(ii)  ∞ n =1 α n =+ ∞ , or equivalently,  ∞ n =1 ( 1 − α n ) = 0 ; e.g., let λ = 1 3 ,{a n } ⊂ (0, 1) and {r n } ⊂ [1, + ∞) be given by α n =  0, if n is even; 1 n ,ifn is odd. and r n =  2, if n is even ; 2 − 1 n ,ifn is odd. Define a sequence {x n }by ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ x 1 ∈ [−1, 0], u i n = T i r n x n , i = 1,2,3, x n+1 = α n g(x n )+(1− α n )y n , y n =(1− λ)x n + λSz n , z n = u 1 n + u 2 n + u 3 n 3 , ∀n ∈ N . (2:1) Then the sequences {x n } and {u i n } , i = 1, 2, 3, defined by (2.1) all strongly converge to 0. Proof (a) By Lemmas 1.5 and 1.6, (2.1) is well defined. (b) Let K = [-1, 0]. For each i Î {1, 2, 3}, define L i (y, z, v, r)=(z − y)  (1 + z 2i ) − 1 r (z − v)  ∀y, z, v ∈ K, ∀r ≥ 1 . We claim that for each v Î K and any i Î {1, 2, 3}, there exists a unique z =0Î K such that ( P ) L i ( y, z, v, r ) ≥ 0 ∀y ∈ K, ∀ r ≥ 1 or, equivalently, (1+z 2i )(z−y)+ 1 r y− z, z− v =(1+z 2i )(z−y)+ 1 r (y− z)(z− v) ≥ 0 ∀y ∈ K, ∀r ≥ 1 . Obviously, z = 0 is a solution of the problem ( P ) . On the other hand, there does not exist z Î [-1, 0) such that z-y≤ 0 and (1 + z 2i ) − 1 r (z − v) ≤ 0 .Soz = 0 is the uniq ue solution of the problem ( P ) . (c) We notice that (2.1) is equivalent with (2.2), where ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ x 1 ∈ [−1, 0], f i (u i n , y)+ i r n y − u i n , u i n − x n ≥0, ∀ y ∈ K, ∀i =1,2,3 , x n+1 = α n g(x n )+(1− α n )y n , y n =(1− λ)x n + λSz n , z n = u 1 n + u 2 n + u 3 n 3 , n ∈ N. (2:2) He and Du Fixed Point Theory and Applications 2011, 2011:33 http://www.fixedpointtheoryandapplications.com/content/2011/1/33 Page 4 of 15 It is easy to see that {x n } ⊂ [-1, 0], so, by (b), u 1 n = u 2 n = u 3 n = 0 for all n Î N. We need to prove x n ® 0asn ® ∞. Since z n = 0 for all n Î N, we have y n =(1-l)x n and x n+1 = α n g(x n )+(1−α n )y n = 1 2 α n x n +(1−α n )(1−λ)x n =  1 − 1 2 α n  − (1 − α n )λ  x n (2:3) for all n Î N. For any n Î N, from (2.3), we have | x n+1 | =  1 − 1 2 α n  − (1 − α n )λ  | x n | ≤  1 − 1 2 α n  | x n | . (2:4) Hence {|x n |} is a strictly deceasing sequence and |x n | ≥ 0foralln Î N.So lim n →∞ | x n | exists. On the other hand, for any n, m Î N with n>m, using (2.4), we obtain | x n+1 |≤  1 − 1 2 α n  | x n | ≤  1 − 1 2 α n  1 − 1 2 α n−1  | x n−1 | ≤···≤ n  j =m  1 − 1 2 α j  | x m | , which implies lim sup n →∞ | x n | ≤ 0 ≤ lim inf n→∞ | x n | . Therefore {x n } strongly converges to 0. □ In this paper, motivated by the preceding Example A, we introduce a new iterative algorithm for the problem of finding a common element in the set of sol utions to the system of equilibrium problem and the set of fixed points of a nonexpansive mapping. The following new strong convergence theorem is established in the framework of a real Hilbert space H. Theorem 2.1 Let K be a nonempty closed c onvex subset o f a real Hilbert space H and I = {1, 2, , k} be a finit e inde x set. For e ach i Î I, let f i be a bi-function from K × Kintoℝ satisfying (A1)-(A4). Let S : K ® K be a nonexpansive mapping with  =   k i=1 EP( f i )   F( S) = ∅ .Letl, r Î (0, 1) and g : K ® Kisar-contraction. Let {x n } be a sequence generated in the following manner: ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ x 1 ∈ K, u i n = T i r n x n , ∀i ∈ I. x n+1 = α n g(x n )+(1− α n )y n , y n =(1− λ)x n + λSz n , z n = u 1 n + ···+ u k n k , ∀n ∈ N. (D H ) If the above control coefficient sequences {a n } ⊂ (0, 1) and {r n } ⊂ (0, +∞) satisfy the following restrictions: (D1) lim n →∞ α n = 0 , ∞  n =1 α n =+ ∞ and lim n →∞ |α n+1 − α n | = 0 ; (D2) lim inf n → ∞ r n > 0 and lim n →∞ |r n+1 − r n | = 0 . then the sequences {x n } and {u i n } , for all i Î I, converge strongly to an element c = P Ω g (c) Î Ω. The following conclusion is immediately drawn from Theorem 2.1. He and Du Fixed Point Theory and Applications 2011, 2011:33 http://www.fixedpointtheoryandapplications.com/content/2011/1/33 Page 5 of 15 Corollary 2.1 LetKbeanonemptyclosedconvexsubsetofarealHilbertspaceH. Let f be a bi-function from K × K in to ℝ satisfying (A1)-(A4) and S : K ® Kbeanon- expansive mapping with Ω = EP(f) ∩F(S) ≠ ∅. Let l, r Î (0,1) and g : K ® Kisar- contraction. Let {x n } be a sequence generated in the following manner: ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ x 1 ∈ K, u n = T r n x n , x n+1 = α n g(x n )+(1− α n )y n , y n =(1− λ)x n + λSu n , ∀n ∈ N . If the above control coefficient sequences {a n } ⊂ (0, 1) and {r n } ⊂ (0, +∞) satisfy all the restrictions in Theorem 2.1, then the sequences {x n } and {u n } converge strongly to an element c = P Ω g(c) Î Ω, respectively. If f i (x, y) ≡ 0 for all (x, y) Î K × K in Theorem 2.1 and all i Î I, then, from the algo- rithm (D H ), we obtain u i n ≡ P K (x n ) , ∀ i Î I. So we have the following result . Corollary 2.2 LetKbeanonemptyclosedconvexsubsetofarealHilbertspaceH. Let S : K ® K be a nonexpansive mapping with F(S) ≠ ∅. Let l, r Î (0, 1) and g : K ® Kisar-contraction. Let {x n } be a sequence generated in the following manner: ⎧ ⎨ ⎩ x 1 ∈ K, x n+1 = α n g(x n )+(1− α n )y n , y n =(1− λ)x n + λSP K (x n ), ∀n ∈ N . If the above control coefficient sequences {a n } ⊂ (0, 1) satisfy lim n →∞ α n = 0 , lim n →∞ |α n+1 − α n | = 0 and lim n →∞ |α n+1 − α n | = 0 , then the sequences {x n } converge strongly to an element c = P Ω g(c) Î F (S). As some interesting and important applications of Theorem 2.1 for optimization pro- blems and fixed point problems, we have the following. Application (I) of Theorem 2.1 We will give an iterative algorithm for the following optimization problem with a nonempty common solution set: min x ∈ K h i (x), i ∈{1, 2, , k},(OP ) where h i (x), i Î {1, 2, , k}, are convex and l ower semi-continuous funct ions defined on a closed convex subset K of a Hilbert space H (for example, h i (x)=x i , x Î K := [0, 1], i Î {1, 2, , k}). If we put f i (x, y)=h i (y)-h i (x), i Î {1, 2, , k}, then  k i =1 EP( f i ) is the common solu- tion set of the problem (OP), where  k i =1 EP( f i ) denote the common solution set of the following equilibrium: F ind x ∈ K such that f i ( x, y ) ≥ 0, ∀ y ∈ K and ∀ i ∈{1, 2, , k} . For i Î {1, 2, , k}, it is obvious that the f i (x, y) satisfies the conditions (A1)-(A4). Let S = I (identity mapping), then from (D H ), we have the following algorithm ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ h i (y) − h i (u i n )+ 1 r n y − u i n , u i n − x n ≥0, ∀ y ∈ K and ∀ i ∈{1, 2, , k} , x n+1 = α n g(x n )+(1− α n )y n , y n =(1− λ)x n + λz n , z n = u 1 n + ···+ u k n k , n ≥ 1. (2:5) He and Du Fixed Point Theory and Applications 2011, 2011:33 http://www.fixedpointtheoryandapplications.com/content/2011/1/33 Page 6 of 15 where x 1 Î K, l Î (0, 1), g : K ® K is a r-contraction. From Theorem 2.1, we kno w that {x n }and {u i n } , i Î{1,2, , k}, generated by (2.5), strongly converge to an element of  k i =1 EP( f i ) if the coefficients {a n } and {r n } satisfy the conditions of Theorem 2.1. Application ( II) of Theorem 2.1 Let H, K, I, l, r, g bethesameasTheorem2.1.Let A 1 , A 2 , , A k : K ® K be k nonlinear mappings with k  i =1 F( A i ) = ∅ .Foranyi Î I,putf i (x, y)=〈x-A i x, y-x〉, ∀ x, y Î K.Since  k i =1 EP( f i )=  k i =1 F( A i ) ,wehave k  i =1 EP( f i ) = ∅ .LetS = I (identity mapping) in the algorithm (D H ). Then the sequences { x n }and {u i n } , defined by the algorithm (D H ), converge str ongly to a common fixed point of {A 1 , A 2 , , A k }, respectively. The following result is important in this paper. Lemma 2.1 Let H be a real Hilbert space. Then for any x 1 , x 2 , x k Î H and a 1 , a 2 , , a k Î [0,1] with  k i =1 a i = 1 , k Î N, we have      k  i=1 a i x i      2 = k  i=1 a i  x i  2 − k−1  i=1 k  j =i+1 a i a j  x i − x j  2 . (2:6) Proof It is obvious that (2.6) is true if a j =1forsomej, so it suffices to show that (2.6) i s true for a j ≠ 1forallj. The proof is by mathematic induction on k.Clearly, (2.6) is true for k =1.Letx 1 , x 2 Î H and a 1 , a 2 Î [0,1] with a 1 + a 2 = 1. By Lemma 1.3, we obtain  a 1 x 1 + a 2 x 2  2 = a 1  x 1  2 + a 2  x 2  2 − a 1 a 2  x 1 − x 2  2 , which means that (2.6) hold for k = 2. Suppose that (2.6) is true for k = l Î N.Let x 1 , x 2 , , x l , x l+1 Î H and a 1 , a 2 , , a l , a l+1 Î [0, 1) with  l+1 i =1 a i = 1 .Let y =  l+1 i=2 a i 1−a 1 x i . Then applying the induction hypothesis we have      l+1  i=1 a i x i      2 = a 1 x 1 +(1− a 1 )y 2 = a 1  x 1  2 +(1− a 1 )  y 2 − a 1 (1 − a 1 )  x 1 − y 2 = l+1  i=1 a i  x i  2 − 1 1 − a 1 l  i=2 l+1  j=i+1 a i a j  x i − x j  2 − a 1 (1 − a 1 )      l+1  i=2 a i 1 − a 1 ( x i − x 1 )      2 = l+1  i=1 a i  x i  2 − 1 1 − a 1 l  i=2 l+1  j=i+1 a i a j  x i − x j  2 − a 1 (1 − a 1 ) l+1  i=2 a i 1 − a 1  x 1 − x i  2 + a 1 (1 − a 1 ) l  i=2 l+1  j=i+1 a i 1 − a 1 a j 1 − a 1  x i − x j  2 = l+1  i=1 a i  x i  2 − 1 1 − a 1 l  i=2 l+1  j=i+1 a i a j  x i − x j  2 − l+1  i=2 a 1 a i  x 1 − x i  2 + a 1 1 − a 1 l  i=2 l+1  j=i+1 a i a j  x i − x j  2 = l+1  i=1 a i  x i  2 − l+1  i=2 a 1 a i  x 1 − x i  2 − l  i=2 l+1  j=i+1 a i a j  x i − x j  2 = l+1  i=1 a i  x i  2 − l  i=1 l+1  j =i+1 a i a j  x i − x j  2 . He and Du Fixed Point Theory and Applications 2011, 2011:33 http://www.fixedpointtheoryandapplications.com/content/2011/1/33 Page 7 of 15 Hence, the equality (2.6) is also true for k = l + 1. This completes the induction. □ 3 Proof of Theorem 2.1 We will proceed with the following steps. Step 1: There exists a unique c Î Ω ⊂ H such that P Ω g(c)=c. Since P Ω g is a r-contraction on H, Banach contraction principle ensures that there exists a unique c Î H such that c = P Ω g(c) Î Ω. Step 2: We prove that the sequences {x n }, {y n }, {z n } and {u i n } , ∀i Î I, are all bounded. First, we notice that (D H ) is equivalent with (Z H ), where ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ x 1 ∈ K f 1 (u 1 n , y)+ 1 r n y − u 1 n , u 1 n − x n ≥0, ∀ y ∈ K, f 2 (u 2 n , y)+ 1 r n y − u 2 n , u 2 n − x n ≥0, ∀ y ∈ K, . . . f k (u k n , y)+ 1 r n y − u k n , u k n − x n ≥0, ∀ y ∈ K, x n+1 = α n g(x n )+(1− α n )y n , y n =(1− λ)x n + λSz n , z n = u 1 n + ···+ u k n k , n ∈ N. (Z H ) For each i Î I, we have ||u i n − c|| = ||T i r n x n − T i r n c|| ≤ ||x n − c||, ∀ n ∈ N . (3:1) For any n Î N, from (Z H ) we have  z n − c  ≤  x n − c  and  y n − c  ≤  x n − c  . (3:2) Since g is a r-contraction, it follows from (3.2) that  x n+1 − c ≤α n   g(x n ) − c   +(1− α n )   y n − c   ≤ α n   g(x n ) − g(c)   + α n   g(c) − c   +(1− α n )   y n − c   ≤ α n ρ  x n − c  + α n   g(c) − c   +(1− α n )  x n − c  =  1 − α n (1 − ρ)   x n − c  + α n (1 − ρ)   g(c) − c   1 − ρ ≤ max   x n − c ,  g(c) − c  1 − ρ  ,forn ∈ N . By induction, we obtain  x n − c ≤max   x 1 − c ,  g(c) − c  1 − ρ  for all n ∈ N , which shows that {x n } is bounded. Also, we know that {y n }, {z n }and { u i n } , ∀i Î I,are all bounded. Step 3: We prove lim n®∞ ||x n+1 - x n || = 0. He and Du Fixed Point Theory and Applications 2011, 2011:33 http://www.fixedpointtheoryandapplications.com/content/2011/1/33 Page 8 of 15 For each i Î I, since u i n − 1 , u i n ∈ K , from (Z H ), we have f i (u i n , u i n−1 )+ 1 r n u i n−1 − u i n , u i n − x n ≥0 , (3:3) and f i (u i n−1 , u i n )+ 1 r n −1 u i n − u i n−1 , u i n−1 − x n−1 ≥0 . (3:4) By (3.3) and (3.4) and (A2), 0 ≤ r n  f i (u i n , u i n−1 )+f i (u i n−1 , u i n )  + u i n−1 − u i n , u i n − x n − r n r n−1 (u i n−1 − x n−1 )  ≤u i n−1 − u i n , u i n − x n − r n r n −1 (u i n−1 − x n−1 ), which implies u i n−1 − u i n , u i n−1 − u i n + x n − x n−1 + x n−1 − u i n−1 + r n r n −1 (u i n−1 − x n−1 )≤0 . (3:5) It follows from (3.5) that  u i n − u i n−1 ≤x n − x n−1  +     r n − r n−1 r n−1      x n−1 − u i n−1  for all n ∈ N . (3:6) Let M := 1 k  k i=1  x n−1 − u i n−1 < ∞ .Foranyn Î N,since z n = 1 k (u 1 n + ···+ u k n ) ,by (3.6), we have  z n − z n−1 ≤ 1 k k  i =1  u i n − u i n−1 ≤x n − x n−1  +M     r n − r n−1 r n−1     . (3:7) Set v n = x n+1 − (1 − β n )x n β n , (3:8) where b n =1-(1-l)(1 - a n ), n Î N. Then for each n Î N, x n+1 − x n = β n ( v n − x n ) (3:9) and v n = α n g(x n )+λ(1 − α n )Sz n β n . (3:10) For any n Î N, since v n+1 − v n = α n+1 g(x n+1 ) β n+1 − α n g(x n ) β n − λ(1 − α n )Sz n β n + λ(1 − α n+1 )Sz n+1 β n+1 = α n+1 g(x n+1 ) β n+1 − α n g(x n ) β n − λ(1 − α n )(Sz n − Sz n+1 ) β n − λ( 1 − α n β n − 1 − α n+1 β n+1 )Sz n+1 , He and Du Fixed Point Theory and Applications 2011, 2011:33 http://www.fixedpointtheoryandapplications.com/content/2011/1/33 Page 9 of 15 by (3.7), it follows that  v n+1 − v n −x n+1 − x n ≤ α n+1  g ( x n+1 )  β n+1 + α n  g ( x n )  β n + λ ( 1 − α n )  z n − z n+1  β n +     1 − α n β n − 1 − α n+1 β n+1      Sz n+1 −x n+1 − x n  ≤ α n+1  g(x n+1 )  β n+1 + α n  g(x n )  β n +  λ(1 − α n ) β n − 1   x n+1 − x n  + M β n     r n+1 − r n r n     +     1 − α n β n − 1 − α n+1 β n+1      Sz n+1  . From this and (D1), (D2), we get lim sup n →∞ { v n+1 − v n −x n+1 − x n } ≤ 0 . (3:11) By Lemma 1.2 and (3.11), lim n → ∞  v n − x n  =0 . (3:12) Owing to (3.9) and (3.12), we obtain lim n → ∞  x n+1 − x n  =0 . (3:13) Step 4: We show lim n→∞  Su i n − u i n  = 0 . By (3.6), (3.13) and (D2), we have lim n →∞  u i n+1 − u i n  =0, ∀i ∈ I . From (Z H ), we get lim n → ∞  x n+1 − y n  = lim n → ∞ α n  g(x n ) − y n  =0 . (3:14) Since ||x n - y n || ≤ ||x n - x n+1 || + ||x n+1 - y n ||, by (3.13) and (3.14), lim n → ∞  y n − x n  =0 , which implies that lim n→∞  Sz n − x n  = lim n→∞ 1 λ  y n − x n  =0 . By Lemma 1.6,  u i n −c 2 =  T i r n x n −T i r n c 2 ≤T i r n x n −T i r n c, x n −c = 1 2   u i n − c 2 +  x n − c 2 −u i n − x n  2  , which yields that  u i n − c 2 ≤x n − c 2 −u i n − x n  2 . (3:15) From (3.15) and Lemma 2.1,  z n − c 2 =      k  i=1 1 k  u i n − c       2 ≤ 1 k k  i=1  u i n − c 2 ≤x n − c 2 − 1 k k  i=1  u i n − x n  2 . 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RESEARC H Open Access Strong convergence theorems for equilibrium problems and fixed point problems: A new iterative method, some comments and applications Zhenhua He 1 and Wei-Shih. Nonlinear Anal. 72, 1180–1202 (2010) 17. Imnang, S, Suantai, S: Strong convergence theorems for a general system of variational inequality problems, mixed equilibrium problems and fixed points problems. iterative method of solution for equilibrium and optimization problems. Nonlinear Anal. 69, 2709–2719 (2008). doi:10.1016/j.na.2007.08.045 9. Tada, A, Takahashi, W: Weak and strong convergence theorems