Bin Dehaish et al. Journal of Inequalities and Applications (2015) 2015:51 DOI 10.1186/s13660-014-0541-z R E S E A R C H Open Access A regularization projection algorithm for various problems with nonlinear mappings in Hilbert spaces Buthinah A Bin Dehaish 1,AbdulLatif 2,HudaOBakodah 1 and Xiaolong Qin 2* * Correspondence: qxlxajh@163.com 2 Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia Full list of author information is available at the end of the article Abstract In this paper, a regularization projection algorithm is investigated for solving common elements of an equilibrium problem, a variational inequality problem and a fixed point problem of a strictly pseudocontractive mapping. Strong convergence theorems are established in the framework of real Hilbert spaces. Keywords: Hilbert space; equilibrium problem; variational inequality; nonexpansive mapping; fixed point 1 Introduction and preliminaries Let H be a real Hilbert space with the inner product, and the norm.letc be a nonempty closed convex subset of H. P C stands for the metric projection from H onto C. Let F be a bifunction of C C into R,whereR denotes the set of real numbers. Consider the following equilibrium problem in the terminology of Blum and Oettli [1]: Find x C such that F(x, y) 0, y C. (1.1) In this paper, the solution set of problem (1.1)isdenotedbyFP(F). To study problem (1.1), we may assume that F satisfies the following conditions: (A1) F(x, x)=0for all x C; (A2) F is monotone, i.e., F(x, y)+f(y, x) 0 for all x, y C; (A3) for each x, y, z C, lim sup F ( tz +(1 t)x, y ) F(x, y); t 0 (A4) for each x C, y F(x, y) is convex and lower semi-continuous. Let S be a self-mapping of C. F(S) stands for the fixed point set of S.RecallthatS is said to be β-contractive if there exists a constant β [0, 1) such that Sx Sy β x y, x, y C. S is said to be nonexpansive if Sx Sy x y, x, y C. 2015 Bin Dehaish et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.
Bin Dehaish et al. Journal of Inequalities and Applications (2015) 2015:51 Page 2 of 14 S is said to be κ-strictly pseudocontractive if there exists a constant κ [0, 1) such that Sx Sy 2 x y 2 + κ (x Sx) (y Sy) 2, x, y C. The class of κ-strictly pseudocontractive mappings was introduced by Browder and Petryshyn [2]. Itisclearthateverynonexpansivemappingisa 0-strictlypseudocontractive mapping. Let A : C H be a mapping. Recall that A is said to be monotone if Ax Ay, x y 0, x, y C. A is said to be strongly monotone if there exists a constant α >0suchthat Ax Ay, x y α x y 2, x, y C. For such a case, we say that A is α-strongly monotone. A is said to be inverse-strongly monotone if there exists a constant α >0 such that Ax Ay, x y α Ax Ay 2, x, y C. For such a case, we say that A is α-inverse-strongly monotone. It is clear that A is α-inverse-strongly monotone if and only if A 1 is α-strongly monotone. Recall that the classical variational inequality is to find x C such that Ax, y x 0, y C. (1.2) It is known that x C is a solution of (1.2) ifandonlyifx is a fixed point of P C (I ra), where r > 0 is a constant and I is an identity mapping. From now on, the solution set of (1.2)isdenotedbyVI(C, A). A set-valued mapping T : H 2 H is said to be monotone if for all x, y H, f Tx and g Ty imply x y, f g 0. A monotone mapping T : H 2 H is maximal if the graph G(T) of T is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping T is maximal if and only if, for any (x, f ) H H, x y, f g 0 for all (y, g) G(T)impliesf Tx. Recently, Iiduka and Takahshi [3] investigated variational inequality (1.2) andfixed points of a nonexpansive mapping based on a Halpern-like algorithm. To be more clear, they proved the following result. Theorem IT Let C be a closed convex subset of a real Hilbert space H. Let A be an α-inverse-strongly monotone mapping of C into H, and let S be a nonexpansive mapping of C into itself such that F(S) VI(C, A). Suppose x 1 = x Cand{x n } is given by x n+1 = α n x +(1 α n )SP C (x n λ n Ax n ) for every n =1,2,..., where {α n } is a sequence in [0, 1) and {λ n } is a sequence in [0, 2α]. If {α n } and {λ n } are chosen so that λ n [a, b] for some a, bwith0<a < b <2α, lim α n =0, n=1 α n =, n=1 α n+1 α n < and n=1 λ n+1 λ n <, then {x n } converges strongly to P F(S) VI(C,A) x.
Bin Dehaish et al. Journal of Inequalities and Applications (2015) 2015:51 Page 3 of 14 Since equilibrium problem (1.1) provides a unified model of several problems such as variational inequalities, fixed point problems and inclusion problems. In [4], Takahashi and Takahashi further studied fixed points of a nonexpansive mapping and equilibrium problem (1.1) based on the viscosity approximation method, which was introduced by Moudafi [5]. To be moreclear, they provedthe followingresult. Theorem TT Let C be a nonempty closed convex subset of a real Hilbert space H. Let F be a bifunction from C C to R satisfying (A1)-(A4), and let S be a nonexpansive mapping of C into H such that F(S) EP(F). Let f be a contraction of H into itself, and let {x n } and {u n } be sequences generated by x 1 Hand F(u n, y)+ 1 y u n, u n x n 0, y C, x n+1 = α n f (x n )+(1 α n )Su n for every n =1,2,...,where {α n } [0, 1] and { } (0, ) satisfy lim α n =0, n=1 α n =, n=1 α n+1 α n <, lim inf >0and n=1 +1 <. Then {x n } and {u n } converge strongly to z F(S) EP(F), where z = P F(S) EP(F) f (z). Subsequently, many authors investigated common solution problems based on hybrid projection methods due to real world applications, for example, image restoration and digital signal processing; see [6 20] and the references therein. In view of the complexity of convex, hybrid projection methods, they are not easy to implement. In this paper, we study a regularization projection algorithm for solving equilibrium problem (1.1), variational inequality (1.2) and a fixed point problem of a κ-strictly pseudocontractive mapping. Possible computation errors are taken into account. Strong convergence theorems are established in the framework of real Hilbert spaces. In order to prove our main results, we also need the following lemmas. Lemma 1.1 [21] Let C be a nonempty closed convex subset of H. Let A be a monotone mapping of C into H, and let N C v be a normal cone to C at v C, i.e., N C v = { w H : v u, w 0, u C } and define a mapping T on C by Av + N C v, v C, Tv =, v / C. Then T is maximal monotone and 0 Tv if and only if Av, u v 0 for all u C. Lemma 1.2 [2] Let C be a nonempty closed convex subset of H, and let S : C Hbea κ-strictly pseudocontractive mapping. Then I S is demi-closed atzero. Lemma 1.3 [1] Let C be a nonempty closed convex subset of H, and let F : C C R be a bifunction satisfying (A1)-(A4). Then, for any r >0and x H, there exists z Csuchthat F(z, y)+ 1 y z, z x 0, y C. r
Bin Dehaish et al. Journal of Inequalities and Applications (2015) 2015:51 Page 4 of 14 Further, define T r x = {z C : F(z, y)+ 1r } y z, z x 0, y C for all r >0and x H. Then the following hold: (a) T r is single-valued; (b) 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 ; (c) F(T r )=EP(F); (d) EP(F) is closed and convex. Lemma 1.4 [2] Let C be a nonempty closed convex subset of H, and let S : C Hbea κ-strictly pseudocontractive mapping. Define a mapping T by T = δi +(1 δ)s, where δ is aconstantin(0, 1). If δ [κ,1)then T is nonexpansive and F(T)=F(S). Lemma 1.5 [22] Let {x n } and {y n } be bounded sequences in H, and let {β n } be a sequence in [0, 1] with 0<lim inf β n lim sup β n <1.Suppose x n+1 =(1 β n )y n + β n x n for all integers n 0 and ( lim sup yn+1 y n x n+1 x n ) 0. Then lim y n x n =0. Lemma 1.6 [23] Assume that {α n } is a sequence of nonnegative real numbers such that α n+1 (1 γ n )α n + δ n + μ n, where {γ n } is a sequence in (0, 1) and {μ n }, {δ n } are real sequences such that (i) n=1 γ n = and n=1 μ n < ; (ii) lim sup δ n /γ n 0 or n=1 δ n <. Then lim α n =0. 2 Main results Theorem 2.1 Let C be a closed convex subset of a real Hilbert space H. Let A : C Hbe an α-inverse-strongly monotone mapping, and let F be a bifunction from C CtoR which satisfies (A1)-(A4). Let S : C H be a κ-strictly pseudocontractive mapping, and let f be a β-contraction on H. Assume that = F(S) VI(C, A) EP(F). Let { } and {s n } be positive real number sequences. Let {α n }, {β n }, {γ n } and {δ n } be real number sequences in (0, 1) such that α n + β n + γ n =1.Let {x n } be a sequence generated in the following process: x 1 H, F(z n, z)+ 1 z z n, z n x n 0, z C, y n = P C (z n s n Az n + e n ), x n+1 = α n f (x n )+β n x n + γ n (δ n y n +(1 δ n )Sy n ),
Bin Dehaish et al. Journal of Inequalities and Applications (2015) 2015:51 Page 5 of 14 where {e n } is a sequence in H. Assume that the control sequences satisfy the following conditions: (a) lim α n =0and n=1 α n = ; (b) 0<lim inf β n lim sup β n <1; (c) n=1 e n <, lim δ n+1 δ n =0and κ δ n δ <1; (d) lim +1 =0and lim inf >0; (e) lim s n+1 s n =0, 0<s s n s <2α, where δ, s, s are real constants. Then {x n } converges strongly to q, which is also a unique solution to the variational inequality f (x) x, x y 0, y C. Proof For any x, y C,weseethat (I sn A)x (I s n A)y 2 = x y 2 2s n x y, Ax Ay + s 2 n Ax Ay 2 x y 2 s n (2α s n ) Ax Ay 2. By using condition (e), we see that (I s n A)x (I s n A)y x y. This proves that I s n A is nonexpansive. Put S n = δ n I +(1 δ n )S. It follows from Lemma 1.4 that S n is nonexpansive and F(S n )=F(S). Let p be fixed arbitrarily. Hence, we have x n+1 p α n f (xn ) p + βn x n p + γ n S n y n p α n β x n p + α n f (p) p + β n x n p + γ n y n p ( 1 α n (1 β) ) x n p + α n f (p) p + en { } f (p) p max x n p, + e n. 1 β f (p) p It follows that x n p max{ x 1 p, } + 1 β n=1 e n. This shows that {x n } is bounded, so are {y n } and {z n }.Letλ n = x n+1 β n x n 1 β n. It follows that λ n+1 λ n = α n+1f (x n+1 )+γ n+1 S n+1 y n+1 1 β n+1 α nf (x n )+γ n S n y n 1 β n Hence, we have = α n+1(f (x n+1 ) S n+1 y n+1 )+(1 β n+1 )S n+1 y n 1 β n+1 α n(f (x n ) S n y n )+(1 β n )S n y n 1 β n = α n+1(f (x n+1 ) S n+1 y n+1 ) 1 β n+1 α n(f (x n ) S n y n ) 1 β n + S n+1 y n+1 S n y n. λ n+1 λ n α n+1 f (x n+1 ) S n+1 y n+1 1 β n+1 + α n f (x n ) S n y n 1 β n + S n+1 y n+1 S n y n. (2.1)
Bin Dehaish et al. Journal of Inequalities and Applications (2015) 2015:51 Page 6 of 14 Since z n = T rn x n,wefindthatf(z n, z) + 1 z z n, z n x n 0, z C and F(z n+1, z) + 1 +1 z z n+1, z n+1 x n+1 0, z C. It follows that F(z n, z n+1 )+ 1 z n+1 z n, z n x n 0 and F(z n+1, z n )+ 1 +1 z n z n+1, z n+1 x n+1 0, z C. By using condition (A2), we find that z n+1 z n, z n x n 0. Hence, we have z n+1 x n+1 +1 z n+1 z n, z n+1 x n r n (z n+1 x n+1 ) z n+1 z n 2. +1 This implies that z n+1 z n, x n+1 x n +(1 +1 )(z n+1 x n+1 ) z n+1 z n 2.Hence,wehave z n+1 z n x n+1 x n + +1 T rn+1 x n x n+1. +1 It follows that y n+1 y n PC (z n+1 s n+1 Az n+1 + e n+1 ) P C (z n s n+1 Az n + e n+1 ) This implies that S n+1 y n+1 S n y n + PC (z n s n+1 Az n + e n+1 ) P C (z n s n Az n + e n ) (I sn+1 A)z n+1 (I s n+1 A)z n + (zn s n+1 Az n + e n+1 ) (z n s n Az n + e n ) z n+1 z n + s n+1 s n Az n + e n+1 + e n x n+1 x n + +1 T rn+1 x n x n +1 + s n+1 s n Az n + e n+1 + e n. S n+1 y n+1 S n+1 y n + S n+1 y n S n y n y n+1 y n + S n+1 y n S n y n x n+1 x n + +1 T rn+1 x n x n +1 + s n+1 s n Az n + e n+1 + e n + δ n+1 δ n Sy n y n. (2.2) Substituting (2.2)into(2.1), wefindthat λ n+1 λ n x n+1 x n α n+1 f (x n+1 ) S n+1 y n+1 1 β n+1 + α n f (x n ) S n y n 1 β n + +1 T rn+1 x n x n +1 + s n+1 s n Az n + e n+1 + e n + δ n+1 δ n Sy n y n. It follows from conditions (a)-(e) that ( lim sup λn+1 λ n x n+1 x n ) 0.
Bin Dehaish et al. Journal of Inequalities and Applications (2015) 2015:51 Page 7 of 14 By using Lemma 1.5,weseethatlim λ n x n = 0, which in turn implies that lim x n+1 x n =0. (2.3) Since T rn is firmly nonexpansive, we find that z n p 2 = T rn x n T rn p 2 x n p, z n p = 1 2( xn p 2 + z n p 2 x n z n 2). That is, z n p 2 x n p 2 x n z n 2. It follows that x n+1 p 2 α n f (xn ) p 2 + β n x n p 2 + γ n S n y n p 2 α n f (xn ) p 2 + β n x n p 2 + γ n y n p 2 α n f (x n ) p 2 + β n x n p 2 + γ n z n p 2 + f n α n f (x n ) p 2 + x n p 2 γ n x n z n 2 + f n, where f n = e n ( e n +2 z n p ). This further implies that γ n x n z n 2 α n f (xn ) p 2 + x n p 2 x n+1 p 2 + f n α n f (xn ) p 2 + ( x n p + x n+1 p ) x n x n+1 + f n. By using conditions (a) and (b), we find from (2.3)that lim z n x n =0. (2.4) Since A is α-inverse-strongly monotone, we find that y n p 2 (z n s n Az n ) (p s n Ap)+e n 2 (z n p) s n (Az n Ap) 2 + f n x n p 2 s n (2α n s n ) Az n Ap 2 + f n. It follows that x n+1 p 2 α n f (x n ) p 2 + β n x n p 2 + γ n S n y n p 2 α n f (x n ) p 2 + β n x n p 2 + γ n y n p 2 α n f (x n ) p 2 + x n p 2 s n (2α s n )γ n Az n Ap 2 + f n.
Bin Dehaish et al. Journal of Inequalities and Applications (2015) 2015:51 Page 8 of 14 This yields that s n (2α s n )γ n Az n Ap 2 α n f (xn ) p 2 + x n p 2 x n+1 p 2 + f n. By using (2.3), we find from conditions (a), (c) and (d) that lim Az n Ap =0. (2.5) Since P C is firmly nonexpansive, we find that y n p 2 (I s n A)z n + e n (I s n A)p, y n p = 1 { (I sn A)z n + e n (I s n A)p 2 + y n p 2 2 (I s n A)z n + e n (I s n A)p (y n p) 2 } 1 { zn p 2 + f n + y n p 2 z n y n ( ) s n (Az n Ap) e n 2} 2 = 1 { zn p 2 + f n + y n p 2 z n y n 2 2 +2 z n y n, s n (Az n Ap) e n s n (Az n Ap) e n 2} 1 { xn p 2 + f n + y n p 2 z n y n 2 2 +2 z n y n s n (Az n Ap) e n s n (Az n Ap) e n 2}, which yields that y n p 2 x n p 2 + f n z n y n 2 +2s n z n y n Az n Ap +2 z n y n e n. It follows that x n+1 p 2 α n f (x n ) p 2 + β n x n p 2 + γ n S n y n p 2 α n f (xn ) p 2 + β n x n p 2 + γ n y n p 2 α n f (xn ) p 2 + x n p 2 + f n γ n z n y n 2 +2s n γ n z n y n Az n Ap +2 z n y n e n. This in turn leads to γ n z n y n 2 α n f (xn ) p 2 + x n p 2 x n+1 p 2 + f n +2s n γ n z n y n Az n Ap +2 z n y n e n. By use of (2.3)and(2.5), we find from restrictions (a), (b), (c) and (e) that lim z n y n =0. (2.6)
Bin Dehaish et al. Journal of Inequalities and Applications (2015) 2015:51 Page 9 of 14 On the other hand, we have γ n S n y n x n x n+1 x n + α n xn f (x n ). By using conditions (a) and (b), we find from (2.3)that lim x n S n y n =0. (2.7) Since S n is nonexpansive, we find that S n x n x n S n x n S n y n + S n y n x n x n z n + z n y n + S n y n x n. It follows from (2.4), (2.6)and(2.7)that lim x n S n x n =0. (2.8) Notice that Sx n x n Sxn ( δ n x n +(1 δ n )Sx n ) + Sn x n x n δ n Sx n x n + S n x n x n. By using condition (c), we find that lim x n Sx n =0. (2.9) Now, we are in a position to show lim sup f (q) q, x n q 0, where q = P f (q). To show it, we can choose a subsequence {x ni } of {x n } such that lim sup f (q) q, xn q = lim f (q) q, xni q. i Since {x ni } is bounded, we can choose a subsequence {x nij } of {x ni } which converges weakly to some point x. Wemayassume,withoutlossofgenerality,that{x ni } converges weakly to x. Next, we prove x.first,weshowx EP(F). Notice that F(y n, y)+ 1 y y n, y n x n 0, y C. By using condition (A2), we see that 1 y y n, y n x n F(y, y n ), y C. Replacingn by n i, we arrive at y y ni, y n i x ni F(y, y ni ), y C. i By using (2.4)and(2.6), we find that {y ni } converges weakly to x. It follows that 0 F(y, x). For each t with 0 < t 1, let z t = tz +(1 t) x, wherez C. It follows that z t C. Hence,
Bin Dehaish et al. Journal of Inequalities and Applications (2015) 2015:51 Page 10 of 14 we have F(z t, x) 0. It follows that 0=F(z t, z t ) tf(z t, z)+(1 t)f(z t, x) tf(z t, z), which yields that F(z t, z) 0, z C. Letting t 0, we obtain from condition (A3) that F( x, z) 0, z C.Thisimpliesthat x EP(F). Next, we show that x VI(C, A). Let T be a maximal monotone mapping defined by Ax + N C x, x C, Tx =, x / C. For any given (x, y) Graph(T), we have y Ax N C x.sincey n C, wehavex y n, y Ax 0. Since y n = P C (z n s n Az n + e n ), we see that x y n, y n (I s n A)z n e n 0and hence x y n, y n z n e n + Az n 0. s n It follows that x y ni, y x y ni, Ax x y ni, Ax x y ni, y n i z ni e ni + Az ni s ni = x y ni, Ax Ay ni + x y ni, Ay ni Az ni x y ni, y n i z ni e ni s ni x y ni, Ay ni Az ni x y ni, y n i z ni e ni. s ni Since A is Lipschitz continuous, we see that x x, y 0. Notice that T is maximal monotone and hence 0 T x. This shows that x VI(C, A). By using Lemma 1.2, wefindthat x F(S). It follows that lim sup f (q) q, x n q 0, x n+1 q 2 α n f (xn ) q, x n+1 q + β n x n q x n+1 q + γ n S n y n q x n+1 q α n f (xn ) f (q), x n+1 q + α n f (q) q, xn+1 q + β n x n x x n+1 q ( ) + γ n xn q + e n xn+1 q α nβ + β n + γ n ( xn q 2 + x n+1 q 2) + α n f (q) q, xn+1 q + e n x n+1 q. 2 It follows that x n+1 q 2 ( 1 α n (1 β) ) x n q 2 +2α n f (q) q, xn+1 q +2K n e n, where K = sup n 1 { x n q }. By using Lemma 1.6, wefindthatlim x n q =0.This completes the proof.
Bin Dehaish et al. Journal of Inequalities and Applications (2015) 2015:51 Page 11 of 14 Foonexpansive mappings, we have the following result. Corollary 2.2 Let C be a closed convex subset of a real Hilbert space H. Let A : C Hbe an α-inverse-strongly monotone mapping, and let F be a bifunction from C CtoR which satisfies (A1)-(A4). Let S : C Hbenonexpansive, and let f be a β-contraction on H. Assume that = F(S) VI(C, A) EP(F). Let { } and {s n } be positive real number sequences. Let {α n }, {β n } and {γ n } be real number sequences in (0, 1) such that α n + β n + γ n =1. Let {x n } be a sequence generated in the following process: x 1 H, F(z n, z)+ 1 z z n, z n x n 0, z C, y n = P C (z n s n Az n + e n ), x n+1 = α n f (x n )+β n x n + γ n Sy n, where {e n } is a sequence in H. Assume that the control sequences satisfy the following conditions: (a) lim α n =0and n=1 α n = ; (b) 0<lim inf β n lim sup β n <1; (c) n=1 e n < ; (d) lim +1 =0and lim inf >0; (e) lim s n+1 s n =0, 0<s s n s <2α, where s, s are real constants. Then {x n } converges strongly to q, which is also a unique solution to the variational inequality f (x) x, x y 0, y C. Further, if S is an identity, we have the following result. Corollary 2.3 Let C be a closed convex subset of a real Hilbert space H. Let A : C Hbe an α-inverse-strongly monotone mapping, and let F be a bifunction from C CtoR which satisfies (A1)-(A4). Let f be a β-contraction on H. Assume that = VI(C, A) EP(F). Let { } and {s n } be positive real number sequences. Let {α n }, {β n } and {γ n } be real number sequences in (0, 1) such that α n +β n +γ n =1.Let {x n } be a sequence generated in the following process: x 1 H, F(z n, z)+ 1 z z n, z n x n 0, z C, x n+1 = α n f (x n )+β n x n + γ n P C (z n s n Az n + e n ), where {e n } is a sequence in H. Assume that the control sequences satisfy the following conditions: (a) lim α n =0and n=1 α n = ; (b) 0<lim inf β n lim sup β n <1; (c) n=1 e n < ; (d) lim +1 =0and lim inf >0; (e) lim s n+1 s n =0, 0<s s n s <2α,
Bin Dehaish et al. Journal of Inequalities and Applications (2015) 2015:51 Page 12 of 14 where s, s are real constants. Then {x n } converges strongly to q, which is also a unique solution to the variational inequality f (x) x, x y 0, y C. Putting F(x, y)=0and = 1, we find the following result. Corollary 2.4 Let C be a closed convex subset of a real Hilbert space H. Let A : C Hbe an α-inverse-strongly monotone mapping. Let S : C Hbeaκ-strictly pseudocontractive mapping, and let f be a β-contraction on H. Assume that = F(S) VI(C, A). Let {s n } be a positive real number sequence. Let {α n }, {β n }, {γ n } and {δ n } be real number sequences in (0, 1) such that α n + β n + γ n =1.Let {x n } be a sequence generated in the following process: x 1 H, z n = P C x n, y n = P C (z n s n Az n + e n ), x n+1 = α n f (x n )+β n x n + γ n (δ n y n +(1 δ n )Sy n ), where {e n } is a sequence in H. Assume that the control sequences satisfy the following conditions: (a) lim α n =0and n=1 α n = ; (b) 0<lim inf β n lim sup β n <1; (c) n=1 e n <, lim δ n+1 δ n =0and κ δ n δ <1; (d) lim s n+1 s n =0, 0<s s n s <2α, where δ, s, s are real constants. Then {x n } converges strongly to q, which is also a unique solution to the variational inequality f (x) x, x y 0, y C. Corollary 2.5 Let C be a closed convex subset of a real Hilbert space H. Let F be a bifunction from C CtoRwhich satisfies (A1)-(A4). Let S : C Hbeaκ-strict pseudocontraction, and let f be a β-contraction on H. Assume that = F(S) EP(F). Let { } be a positive real number sequence. Let {α n }, {β n }, {γ n } and {δ n } be real number sequences in (0, 1) such that α n + β n + γ n =1.Let {x n } be a sequence generated in the following process: x 1 H, F(z n, z)+ 1 z z n, z n x n 0, z C, x n+1 = α n f (x n )+β n x n + γ n (δ n z n +(1 δ n )Sz n ), where {e n } is a sequence in H. Assume that the control sequences satisfy the following conditions: (a) lim α n =0and n=1 α n = ; (b) 0<lim inf β n lim sup β n <1; (c) n=1 e n <, lim δ n+1 δ n =0and κ δ n δ <1; (d) lim +1 =0and lim inf >0,
Bin Dehaish et al. Journal of Inequalities and Applications (2015) 2015:51 Page 13 of 14 where δ is a real constant. Then {x n } converges strongly to q, which is also a unique solution to the variational inequality f (x) x, x y 0, y C. Remark 2.6 Comparing Theorem 2.1, Corollaries 2.2-2.5 with Theorems IT, TT and the corresponding results in [6 20], we have the following. (a) We extend the nonlinear mapping from the class of nonexpansive mappings to the class of κ-strictly pseudocontractive mappings. (b) Possible computation errors are taken into account. (c) The conditions imposed on control sequences {α n } and {s n } are relaxed. (d) The common element is also a unique solution of variational inequality (1.2). Theorem 2.7 Let C be a closed convex subset of a real Hilbert space H. Let A : C Hbe an α-inverse-strongly monotone mapping, and let f be a β-contraction on H. Assumethat VI(C, A). Let {s n } be a positive real number sequence. Let {α n }, {β n } and {γ n } be real number sequences in (0, 1) such that α n + β n + γ n =1.Let {x n } be a sequence generated in x 1 C, x n+1 = α n f (x n )+β n x n + γ n y n, where y n = P C (x n s n Ax n + e n ), where {e n } is a sequence in H. Assume that the control sequences satisfy the following conditions: (a) lim α n =0and n=1 α n = ; (b) 0<lim inf β n lim sup β n <1; (c) n=1 e n <, lim s n+1 s n =0, 0<s s n s <2α, where s and s are real constants. Then {x n } converges strongly to q = P VI(C,A) f (q). Remark 2.8 To construct a mathematical model which is as close as possible to a real complex problem, we often have to use more than one constraint. Solving such problems, we have to obtain some solution which is simultaneously the solution of two or more subproblems. This is a common problem in diverse areas of mathematics and physical sciences. It consists of trying to find a solution satisfying certain constraints. In this paper, we investigate the problem of solving common solutions of an equilibrium problem, a variational inequality problem and a fixed point problem of a strictly pseudocontractive mapping based on a regularization projection algorithm. Possible computation errors are also taken into account. Strong convergence theorems are established without compact assumptions and additional metric projections in the framework of real Hilbert spaces. Competing interests The authors declare that they have no competing interests. Authors contributions All authors have contributed equally to the work. All authors read and approved the final version of the manuscript. Author details 1 Department of Mathematics, Faculty of Science for Girls, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia. 2 Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah, 21589, Saudi Arabia. Acknowledgements This project was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University under grant No. (58-130-35-HiCi). The authors, therefore, acknowledge technical and financial support of KAU. The authors are grateful to the editor and the anonymous reviewers for useful suggestions which improved the contents of the article. Received: 16 August 2014 Accepted: 20 December 2014
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