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Available online at www.tjnsa.com J. Nonlinear Sci. Appl. 9 (2016), 2789–2797 Research Article Viscosity splitting methods for variational inclusion and fixed point problems in Hilbert spaces Xiaolong Qina , B. A. Bin Dehaishb , Sun Young Choc,∗ a Institute of Fundamental and Frontier Sciences, University of Electronic Science and Technology of China, Sichuan, China. b Department of mathematics, Faculty of Science, AL Faisaliah Campus, King Abdulaziz University, Jeddah, Saudi Arabia. c Department of Mathematics, Gyeongsang National University, Jinju 660-701, Korea. Communicated by Y. Yao Abstract In this paper, a viscosity splitting method is investigated for treating variational inclusion and fixed point problems. Strong convergence theorems of common solutions are established in the framework of Hilbert c spaces. Applications are also provided to support the main results. 2016 All rights reserved. Keywords: Convex feasibility problem, fixed point, iterative process, monotone operator, splitting algorithm. 2010 MSC: 47H05, 47H09. 1. Introduction and Preliminaries p Let H be a real Hilbert space with inner product hx, yi and induced norm kxk = hx, xi for x, y ∈ H. Let C be a nonempty convex and closed subset of H. Let T : C → C be a mapping. In this paper, we use F ix(T ) to stand for the set of fixed points of T . Recall that T is said to be an α-contractive mapping iff there exists a constant α with 0 < α < 1 such that kT x − T yk ≤ αkx − yk, ∀x, y ∈ C. T is said to be nonexpansive iff kT x − T yk ≤ kx − yk, ∗ ∀x, y ∈ C. Corresponding author Email addresses: qxlxajh@163.com (Xiaolong Qin), bbendehaish@kau.edu.sa (B. A. Bin Dehaish), ooly61@hotmail.com (Sun Young Cho) Received 2015-12-27 X. Qin, B. A. Bin Dehaish, S. Y. Cho, J. Nonlinear Sci. Appl. 9 (2016), 2789–2797 2790 If C is also bounded, then the set of fixed points of S is not empty; see [5] and the references therein. In the real world, many important problems have reformulations which require finding fixed points of nonexpansive mapping. Mann iteration is powerful to study fixed points of nonexpansive mappings. However it is only weakly convergent. Recently, many authors studied the problem of modifying Mann iteration so that strong convergence is guaranteed without any compactness assumption; see [7, 8, 12, 13, 14, 16, 24, 25, 26, 27] and the references therein. T is said to be a λ-strict pseudocontraction iff there exists a constant λ with 0 ≤ λ < 1 such that kT x − T yk2 ≤ kx − yk + λkx − T x − y + T yk2 , ∀x, y ∈ C. The class of λ-strict pseudocontractions was introduced by Browder and Petryshyn [6] in 1967. It is clear that the class of λ-strict pseudocontractions strictly include the class of nonexpansive mappings as a special cases. It is also known that every λ-strict pseudocontraction is Lipschitz continuous; see [6] and the references therein. Let A : C → H be a mapping. Recall that A is said to be monotone iff hAx − Ay, x − yi ≥ 0, ∀x, y ∈ C. A is said to be λ-strongly monotone iff there exists a positive constant λ > 0 such that hAx − Ay, x − yi ≥ λkx − yk2 , ∀x, y ∈ C. A is said to be inverse λ-strongly monotone iff there exists a positive constant λ > 0 such that hAx − Ay, x − yi ≥ λkAx − Ayk2 , ∀x, y ∈ C. From the above, we see that A is inverse λ-strongly monotone iff A−1 is strongly monotone. A is said to be L-Lipschitz continuous iff there exists a positive constant L > 0 such that kAx − Ayk ≤ Lkx − yk, ∀x, y ∈ C. It is obvious that A is inverse λ-strongly monotone, then A is also monotone and Recall that the classical variational inequality is to find an x ∈ C such that hAx, y − xi ≥ 0, ∀y ∈ C. 1 λ -Lipschitz continuous. (1.1) The solution set of variational inequality (1.1) is denoted by V I(C, A). Projection methods have been recently investigated for solving variational inequality (1.1). Let P rojC be the metric projection from H onto C and I the identity on H. It is known that x is a solution to (1.1) iff x is a fixed point of mapping P rojC (I − rA). If A is inverse λ-strongly monotone, then P rojC (I − rA) is nonexpansive. If C is bounded, closed and convex, then the existence of solutions of variational inequality (1.1) is guaranteed by the nonexpansivity of mapping P rojC (I − rA). Recall that an operator B : H ⇒ H is said to be monotone iff, for all x, y ∈ H, f ∈ Bx and g ∈ By imply hx − y, f − gi ≥ 0. In this paper, we use B −1 (0) to stand for the zero point of B. A monotone mapping B : H ⇒ H is maximal iff the graph G(B) of B is not properly contained in the graph of any other monotone mapping. It is known that a monotone mapping B is maximal if and only if, for any (x, f ) ∈ H × H, hx − y, f − gi ≥ 0, for all (y, g) ∈ G(B) implies f ∈ Bx. For a maximal monotone operator B on H, and r > 0, we may define the single-valued resolvent JrB = (I + rB)−1 , where D(B) denote the domain of B. It is known that JrB : H → D(B) is firmly nonexpansive, and B −1 (0) = F ix(JrB ). The property of the resolvent ensures that the Picard iterative algorithm xn+1 = JrB xn converge weakly to a fixed point of JrB , which is necessarily a zero point of B. Rockafellar introduced this iteration method and call it the proximal point algorithm (PPA); for more detail, see [20, 23] and the references therein. The PPA and its dual version in the context of convex programming, the method of multipliers of Hesteness and X. Qin, B. A. Bin Dehaish, S. Y. Cho, J. Nonlinear Sci. Appl. 9 (2016), 2789–2797 2791 Powell, have been extensively studied and are known to yield as special cases decomposition methods such as the method of partial inverses [22], the Douglas-Rachford splitting method, and the alternating direction method of multipliers [10]. In the case of B = B1 + B2 , where B1 and B2 are maximal monotone on H, the forward-backward splitting method xn+1 = (I + rn B1 )−1 (I − rn B2 )xn , n = 0, 1, · · · , where rn > 0, was proposed by Lions and Mercier [15], and in a dual form for convex programming, by Han and Lou [11]. In the case where B1 = NC , this method reduces to a projection method proposed by Sibony [21] for monotone variational inequalities (1.1). Recently, many authors have studied the splitting algorithm; see [2, 3, 4, 9, 17, 18, 19] and the references therein. In this paper, a viscosity splitting method is investigated for treating a inclusion problem with two monotone operators and a fixed point problem of λ-strict pseudocontractions. Strong convergence theorems of common solutions are established in the framework of Hilbert spaces. Applications are also provided to support the main results. The following lemmas are essential to prove our main results. Lemma 1.1 ([3]). Let C be a nonempty convex and closed subset of a real Hilbert space H. Let A : C → H be a mapping, and B : H ⇒ H a maximal monotone operator. Then F (Jr (I − rA)) = (A + B)−1 (0). Lemma 1.2 ([25]). Let {an } be a sequence of nonnegative numbers satisfying the condition an+1 P∞≤ (1 − tn )an + tn bn + cn , ∀n ≥ 0, where {tn } is a number sequence in (0, 1) such that limn→∞ tn = 0 and n=0 tn = ∞, {b Pn } is a number sequence such that lim supn→∞ bn ≤ 0, and {cn } is a positive number sequence such that ∞ n=0 cn < ∞. Then limn→∞ an = 0. Lemma 1.3 ([1]). Let Hbe a Hilbert space, and A an maximal monotone operator. For λ > 0, µ > 0, and µ µ x ∈ E, we have Jλ x = Jµ λ x + 1 − λ Jλ x , where Jλ = (I + λA)−1 and Jµ = (I + µA)−1 . Lemma 1.4 ([6]). Let C be a nonempty convex and closed subset of a real Hilbert space H. Let T : C → C be a λ-strict pseudocontraction. Define a mapping S by S = βI + (1 − β)T . If beta ∈ [λ, 1), then S is nonexpansive and F ix(T ) = F ix(S). Lemma 1.5 ([6]). Let C be a nonempty convex and closed subset of a real Hilbert space H. Let T : C → C be a λ-strict pseudocontraction. Then T is Lipschitz continuous and I − T is demiclosed at zero. 2. Main results Theorem 2.1. Let C be a nonempty convex closed subset of a real Hilbert space H. Let A : C → H be an inverse κ-strongly monotone mapping and let B be a maximal monotone operator on H. Let f : C → C be a fixed α-contraction and let T : C → C be a λ-strict pseudocontraction. Assume that (A + B)−1 (0) ∩ F ix(T ) is not empty. Let {αn }, {βn } be real number sequences in (0, 1) and let {rn } be a real number sequence in (0, 2κ). Let {xn } be a sequence in C generated in the following process: x0 ∈ C, xn+1 = βn yn + (1 − βn )T yn , −1 ∀n ≥ 0, where {yn } is a sequence in C such that kyn − (I + rn B) αn f (xn ) + (1 − αn )xn − rn A αn f (xn ) + (1 − αn )xn k ≤Pen . Assume that the control sequences satisfy the following restrictions: P limn→∞ αn = 0, P∞ P∞ ∞ 0 αn = ∞, n=1 |αn − αn−1 | < ∞, 0 < a ≤ rn ≤ a < 2κ, n=1 |rn − rn−1 | < ∞, ∞ n=0 ken k < ∞, Pn=0 ∞ 00 < 1, where a, a0 and a00 are three real numbers. Then {x } |β − β | < ∞, and λ ≤ β ≤ a n−1 n n n=1 n converges strongly to a point x̄ ∈ F ix(T ) ∩ (A + B)−1 (0), where x̄ = P rojF ix(T )∩(A+B)−1 (0) f (x̄), that is, x̄ solves the following variational inequality hf (x̄) − x̄, x̄ − xi ≥ 0, ∀x ∈ F ix(T ) ∩ (A + B)−1 (0). Proof. Since A is inverse κ-strongly monotone, one has k(I − rn A)x − (I − rn A)yk2 = kx − yk2 − 2rn hx − y, Ax − Ayi + rn 2 kAx − Ayk2 ≤ kx − yk2 − rn (2κ − rn )kAx − Ayk2 . From the restriction imposed on {rn }, one has I − rn A is nonexpansive. Setting Tn = βn I + (1 − βn )T, X. Qin, B. A. Bin Dehaish, S. Y. Cho, J. Nonlinear Sci. Appl. 9 (2016), 2789–2797 2792 where I is the identity, one sees from Lemma 1.4 that Tn is nonexpansive with F ix(Tn ) = F ix(T ). Fixing p ∈ F ix(T ) ∩ (A + B)−1 (0), one has from Lemma 1.1 that p = Tn p = (I + rn B)−1 (p − rn Ap). Setting zn = αn f (xn ) + (1 − αn )xn , one has kzn − pk ≤ αn kf (xn ) − pk + (1 − αn )kxn − pk ≤ 1 − αn (1 − α) kxn − pk + αn kf (p) − pk. Hence, one has kxn+1 − pk ≤ kyn − pk ≤ kyn − (I + rn B)−1 zn − rn Azn k + k zn − rn Azn − p − rn Ap k ≤ kzn − pk + en ≤ 1 − αn (1 − α) kxn − pk + αn kf (p) − pk + en kf (p) − pk ≤ max{kxn − pk, } + en . 1−α By mathematical induction, one finds that sequence {xn } is bounded, so are {yn } and {zn }. Notice that kzn − zn−1 k ≤ |αn − αn−1 |kxn−1 − f (xn−1 )k + 1 − αn (1 − α) kxn−1 − xn k. (2.1) Putting wn = zn − rn Azn , we find from (2.1) that kwn − wn−1 k ≤ kzn − zn−1 k + krn − rn−1 kkAzn−1 k ≤ |αn − αn−1 |kxn−1 − f (xn−1 )k + 1 − αn (1 − α) kxn−1 − xn k (2.2) + |rn − rn−1 |kAzn−1 k. Set JrBn = (I + rn B)−1 . Using Lemma 1.3, one has kxn − xn+1 k = kTn−1 yn−1 − Tn yn k ≤ kyn−1 − yn k + |βn − βn−1 |kyn − T yn k ≤ kJrBn wn − JrBn−1 wn−1 k + |βn − βn−1 |kyn − T yn k + en−1 + en rn−1 rn−1 ≤ k(1 − )(JrBn wn − wn−1 ) + (wn − wn−1 )k rn rn + |βn − βn−1 |kyn − T yn k + en−1 + en rn−1 )(JrBn wn − wn ) + (wn − wn−1 )k ≤ k(1 − rn + |βn − βn−1 |kyn − T yn k + en−1 + en |rn − rn−1 | ≤ kwn − JrBn wn k + kwn−1 − wn k rn + |βn − βn−1 |kyn − T yn k + en−1 + en . (2.3) Combining (2.2) with (2.3), one has kxn − xn+1 k ≤ |rn − rn−1 | kwn − JrBn wn k + |αn − αn−1 |kxn−1 − f (xn−1 )k rn + 1 − αn (1 − α) kxn−1 − xn k + |rn − rn−1 |kAzn−1 k + |βn − βn−1 |kyn − T yn k + en−1 + en . Using the restrictions imposed on {rn }, {en }, {αn } and {βn } and Lemma 1.1, we find limn→∞ kxn −xn+1 k = 0. Since αn → 0 as n → ∞, we find lim kxn − zn k = 0. (2.4) n→∞ X. Qin, B. A. Bin Dehaish, S. Y. Cho, J. Nonlinear Sci. Appl. 9 (2016), 2789–2797 2793 Since k · k2 is convex, we have kzn − pk2 ≤ αn kf (xn ) − pk2 + (1 − αn )kxn − pk2 ≤ kxn − pk2 + αn kf (xn ) − pk2 . This in turn implies kxn+1 − pk2 ≤ kyn − JrBn (zn − rn Azn )k2 + kJrBn (zn − rn Azn ) − pk2 + 2kJrBn (zn − rn Azn ) − pkkyn − JrBn (zn − rn Azn )k ≤ k(zn − rn Azn ) − (p − rn Ap)k2 + 2en kJrBn (zn − rn Azn ) − pk + e2n ≤ kzn − pk2 − rn (2κ − rn )kAzn − Apk2 + 2en kzn − pk + e2n ≤ kxn − pk2 + αn kf (xn ) − pk2 − rn (2κ − rn )kAzn − Apk2 + 2en kzn − pk + e2n . It follows that rn (2κ − rn )kAzn − Apk2 ≤ kxn − pk2 + αn kf (xn ) − pk2 − kxn+1 − pk2 + (2kzn − pk + en )en . Therefore, one finds lim kAp − Azn k = 0. n→∞ (2.5) Since JrBn is firmly nonexpansive, one has kJrBn (zn − rn Azn ) − pk2 ≤ h(zn − rn Azn ) − (p − rn Ap), JrBn (zn − rn Azn ) − pi 1 k(zn − rn Azn ) − (p − rn Ap)k2 + kJrBn (zn − rn Azn ) − pk2 ≤ 2 − kzn − JrBn (zn − rn Azn ) − rn (Azn − Ap)k2 . It follows that kJrBn (zn − rn Azn ) − pk2 ≤ kzn − pk2 − kzn − JrBn (zn − rn Azn )k2 − rn kAzn − Apk2 + 2rn kzn − JrBn (zn − rn Azn )kkAzn − Apk. Hence, one has kxn+1 − pk2 ≤ kyn − JrBn (zn − rn Azn )k2 + kJrBn (zn − rn Azn ) − pk2 + 2kJrBn (zn − rn Azn ) − pkkyn − JrBn (zn − rn Azn )k ≤ kJrBn (zn − rn Azn ) − pk2 + en (2kJrBn (zn − rn Azn ) − pk + en ) ≤ kzn − pk2 − kzn − JrBn (zn − rn Azn )k2 + 2rn kzn − JrBn (zn − rn Azn )kkAzn − Apk + en (2kJrBn (zn − rn Azn ) − pk + en ) ≤ kxn − pk2 + αn kf (xn ) − pk2 − kzn − JrBn (zn − rn Azn )k2 + 2rn kzn − JrBn (zn − rn Azn )kkAzn − Apk + en (2kJrBn (zn − rn Azn ) − pk + en ), which further implies from (2.5) lim kzn − JrBn (zn − rn Azn )k = 0. n→∞ (2.6) X. Qin, B. A. Bin Dehaish, S. Y. Cho, J. Nonlinear Sci. Appl. 9 (2016), 2789–2797 2794 Since Tn is nonexpansive, one has kβn xn + (1 − βn )T xn − xn k ≤ kβn xn + (1 − βn )T xn − βn yn − (1 − βn )T yn k + kxn − βn yn − (1 − βn )T yn k ≤ kxn − yn k + kxn − xn+1 k ≤ kxn − zn k + kzn − JrBn (zn − rn Azn )k + kxn − xn+1 k + en . In view of (2.4) and (2.6), one has limn→∞ kβn xn + (1 − βn )T xn − xn k = 0. Note that kT xn − xn k ≤ kT xn − βn xn − (1 − βn )T xn k + kβn xn + (1 − βn )T xn − xn k ≤ βn kxn − T xn k + kβn xn + (1 − βn )T xn − xn k. From the restriction imposed on sequence {βn }, one finds that limn→∞ kT xn − xn k = 0. Next, we show that lim suphf (x̄) − x̄, zn − x̄i ≤ 0, (2.7) n→∞ where x̄ is the unique fixed point of the mapping P roj(A+B)−1 (0)∩F ix(T ) f. To show this inequality, we choose a subsequence {zni } of {zn } such that lim suphf (x̄) − x̄, zn − x̄i = lim hf (x̄) − x̄, zni − x̄i ≤ 0. n→∞ i→∞ Since {zni } is bounded, we find that there exists a subsequence {znij } of {zni } which converges weakly to x̂. Without loss of generality, we assume that zni * x̂. Putting µn = JrBn (zn − rn Azn ), we find that µni * x̂. Next, we show x̂ ∈ (A + B)−1 (0). Notice that zn − rn Azn ∈ µn + rn Bµn ; that is, zn − rn Azn − µn ∈ Bµn . rn Let µ ∈ Bν. Since B is maximal monotone, we find zn − µ n − Azn − µ, µn − ν ≥ 0. rn It follows that h−Ax̂ − µ, x̂ − νi ≥ 0. This in turn implies that −Ax̂ ∈ B x̂, that is, x̂ ∈ (A + B)−1 (0). Now, we are in a position to show that x̂ is also in F ix(T ). Since xni * x̂, we find from Lemma 1.5 that x̂ ∈ F ix(T ) immediately. This proves that (2.7) holds. Finally, we show that {xn } converges strongly to x̄, where x̄ is the unique fixed point of mapping P roj(A+B)−1 (0)∩F ix(T ) f. Note that kzn − x̄k2 ≤ αn hf (x̄) − x̄, zn − x̄i + 1 − αn (1 − α) kzn − x̄kkxn − x̄k. It follows that kzn − x̄k2 ≤ 2αn hf (x̄) − x̄, zn − x̄i + 1 − αn (1 − α) kxn − x̄k2 . Hence, one has kxn+1 − x̄k2 ≤ kyn − x̄k2 ≤ kJrBn (zn − rn Azn ) − x̄k2 + en (2kJrBn (zn − rn Azn ) − x̄k + en ) ≤ kzn − x̄k2 + en (2kJrBn (zn − rn Azn ) − x̄k + en ) ≤ 2αn hf (x̄) − x̄, zn − x̄i + 1 − αn (1 − α) kxn − x̄k2 + en (2kJrBn (zn − rn Azn ) − x̄k + en ). An application of Lemma 1.2 to the above inequality yields that limn→∞ kxn − x̄k = 0. This completes the proof. X. Qin, B. A. Bin Dehaish, S. Y. Cho, J. Nonlinear Sci. Appl. 9 (2016), 2789–2797 2795 From Theorem 2.1, the following results are not hard to derive. Corollary 2.2. Let C be a nonempty convex closed subset of a real Hilbert space H. Let A : C → H be an inverse κ-strongly monotone mapping and let B be a maximal monotone operator on H. Let f : C → C be a fixed α-contraction and let T : C → C be a nonexpansive mapping. Assume that (A + B)−1 (0) ∩ F ix(T ) is not empty. Let {αn }, {βn } be real number sequences in (0, 1) and let {rn } be a real number sequence in (0, 2κ). Let {xn } be a sequence in C generated in the following process: x0 ∈ C, xn+1 = βn yn + (1 − βn )T yn , −1 ∀n ≥ 0, where {yn } is a sequence in C such that kyn − (I + rn B) αn f (xn ) + (1 − αn )xn − rn A αn f (xn ) + (1 − αn )xn k ≤Pen . Assume that the control sequences satisfy the following restrictions: P limn→∞ αn = 0, P∞ P∞ ∞ 0 αn = ∞, n=1 |αn − αn−1 | < ∞, 0 < a ≤ rn ≤ a < 2κ, n=1 |rn − rn−1 | < ∞, ∞ n=0 ken k < ∞, Pn=0 ∞ 00 < 1, where a, a0 and a00 are three real numbers. Then {x } |β − β | < ∞, and 0 ≤ β ≤ a n−1 n n n=1 n converges strongly to a point x̄ ∈ F ix(T ) ∩ (A + B)−1 (0), where x̄ = P rojF ix(T )∩(A+B)−1 (0) f (x̄), that is, x̄ solves the following variational inequality hf (x̄) − x̄, x̄ − xi ≥ 0, ∀x ∈ F ix(T ) ∩ (A + B)−1 (0). Corollary 2.3. Let C be a nonempty convex closed subset of a real Hilbert space H. Let A : C → H be an inverse κ-strongly monotone mapping and let B be a maximal monotone operator on H. Let f : C → C be a fixed α-contraction. Assume that (A + B)−1 (0) is not empty. Let {αn }, {βn } be real number sequences in (0, 1) and let {rn } be a real number sequence in (0, 2κ). Let x0 ∈ C and {xn } be a sequence in C such that kxn+1 − (I + rn B)−1 αn f (xn ) + (1 − αn )xn − rn A αn f (xn ) +P (1 − αn )xn k P ≤ en . Assume that the ∞ control sequences satisfy P the following restrictions:Plimn→∞ αn = 0, n=0 αn = ∞, ∞ n=1 |αn − αn−1 | < ∞, ∞ 0 are three real numbers. ke k < ∞, where a and a |r − r | < ∞, 0 < a ≤ rn ≤ a0 < 2κ, ∞ n n−1 n=0 n=1 n Then {xn } converges strongly to a point x̄ ∈ (A + B)−1 (0), where x̄ = P roj(A+B)−1 (0) f (x̄), that is, x̄ solves the following variational inequality hf (x̄) − x̄, x̄ − xi ≥ 0, ∀x ∈ (A + B)−1 (0). Let C be a nonempty closed and convex subset of a Hilbert space H. Let iC be the indicator function of C, that is, ( 0, x ∈ C, iC (x) = ∞, x ∈ / C. Since iC is a proper lower and semicontinuous convex function on H, the subdifferential ∂iC of iC is maximal monotone. So, we can define the resolvent Jr∂iC of ∂iC for r > 0, i.e., Jr∂iC := (I+r∂iC )−1 . Letting x = Jr∂iC y, we find that y ∈ x + r∂iC x ⇐⇒ y ∈ x + rNC x ⇐⇒ hy − x, v − xi ≤ 0, ∀v ∈ C ⇐⇒ x = P rojC y, where P rojC is the metric projection from H onto C and NC x := {e ∈ H : he, v − xi, ∀v ∈ C}. From Theorem 2.1, we have the following results on variational inequality (1.1). Corollary 2.4. Let C be a nonempty convex closed subset of a real Hilbert space H. Let A : C → H be an inverse κ-strongly monotone mapping. Let f : C → C be a fixed α-contraction and let T : C → C be a λ-strict pseudocontraction. Assume that V I(C, A) ∩ F ix(T ) is not empty. Let {αn }, {βn } be real number sequences in (0, 1) and let {rn } be a real number sequence in (0, 2κ). Let {xn } be a sequence in C generated in the following process: x0 ∈ C, xn+1 = βn yn + (1 − βn )T yn , ∀n ≥ 0,where {yn } is a sequence in C such that kyn − P rojC αn f (xn ) + (1 − αn )xn − rn A αn f (xn ) + − αn )xn kP≤ en . Assume that the P(1 ∞ ∞ control sequences satisfy the following restrictions: lim α = 0, n→∞ n n=0 αn = ∞, n=1 |αn − αn−1 | < ∞, P∞ P∞ P∞ 0 0 < a ≤ rn ≤ a < 2κ, n=1 |rn −rn−1 | < ∞, n=0 ken k < ∞, n=1 |βn −βn−1 | < ∞, and λ ≤ βn ≤ a00 < 1, where a, a0 and a00 are three real numbers. Then {xn } converges strongly to a point x̄ ∈ F ix(T ) ∩ V I(C, A), where x̄ = P rojF ix(T )∩V I(C,A) f (x̄), that is, x̄ solves the following variational inequality hf (x̄) − x̄, x̄ − xi ≥ 0, ∀x ∈ F ix(T ) ∩ V I(C, A). X. Qin, B. A. Bin Dehaish, S. Y. Cho, J. Nonlinear Sci. Appl. 9 (2016), 2789–2797 2796 Acknowledgements This article was supported by the National Natural Science Foundation of China under grant No.11401152. References [1] V. Barbu, Nonlinear semigroups and differential equations in Banach spaces, Translated from the Romanian, Editura Academiei Republicii Socialiste România, Bucharest; Noordhoff International Publishing, Leiden (1976). 1.3 [2] H. H. 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