©2021 Ada Academica https://adac.eeEur. J. Math. Anal. 1 (2021) 19-33doi: 10.28924/ada/ma.1.19 The Generalized Viscosity Implicit Rules of Asymptotically Nonexpansive Mappings in Hilbert Spaces Sang B Mendy, John T Mendy∗ , Alieu Jobe University of the Gambia, Brikama Campus, Gambia sangbm1@gmail.com, jt.mendy@utg.edu.gm, alieueejobe@gmail.com ∗Correspondence: jt.mendy@utg.edu.gm Abstract. The generalized viscosity implicit rules of nonexpansive asymptotically mappings in Hilbertspaces are considered. The strong convergence theorems of the rules are proved under certain as-sumptions imposed on the sequences of parameters. An application of it in the convex minimizationproblem is considered. The results presented in this paper improve and extend some recent corre-sponding results in the literature. 1. Background Let H be a real Hilbert space and M be a nonempty closed convex subset of H,T : M→Mbe a nonexpansive mapping with a nonempty fixed point set F(T )The following iteration method is known as the viscosity approximation method: for arbitrarilychosen u0 ∈M un+1 = αnψ(un) + (1 −αn)Tun,n ≥ 0, (1.1)where ψ : M → M is a contraction and {αn} is a sequence in (0, 1). Under some certainconditions, the sequence {un} converges strongly to a point z ∈ F (T ) which solves the variationalinequality (V I) 〈(I −ψ)z,u −z〉≥ 0,u ∈ F (T ), (1.2) where I is the identity of H. Many authors studied iterative sequence for the implicit midpointrule because of it’s significant for solving ordinary differential equations; see [?]- [12], John T [9], [7]and the references therein. Recently, Xu et al [3] proposed the following viscosity implicit midpointrule (VIMR) for nonexpansive mappings: un+1 = αnψ(un) + (1 −αn)T (un + un+1 2 ) ,n ≥ 0, (1.3) Received: 23 Aug 2021. Key words and phrases. viscosity; Hilbert space; convex minimization; asymptotically nonexpansive mapping; varia-tional inequality; fixed point. 19 https://adac.ee https://doi.org/10.28924/ada/ma.1.19 https://orcid.org/0000-0002-3774-0761 Eur. J. Math. Anal. 1 (2021) 20 In 2015, Ke and Ma [4] proposed the generalized viscosity implicit rules of nonexpansive mappingsin Hilbert spaces as follows: un+1 = αnψ(un) + (1 −αn)T (snun + (1 − sn)un+1),n ≥ 0, (1.4) and un+1 = αnun + βnψ(un) + γnT (snun + (1 − sn)un+1),n ≥ 0, (1.5) They proved that the generalized viscosity implicit rules 1.4 and 1.5 converge strongly to a fixedpoint of T under certain assumptions, which also solved the V I(1.1). In 2016, motivated by the work of Xu [3], Zhao et al [5] proposed the following implicit midpointrule for asymptotically nonexpansive mappings: un+1 = αnψ(un) + (1 −αn)T n (un + un+1 2 ) ,n ≥ 0, (1.6) where T is an asymptotically nonexpansive mapping. They proved that the sequence {un} con-verges strongly to a fixed point of T , which, in addition, also solves the V I(1.1). In 2017, He et la [14] studied the following iterative un+1 = αnψ(un) + (1 −αn)T n(βnun + (1 −βn)un+1),n ≥ 0 (1.7) in the setting of a Hilbert space and proved that the sequence {un} converges strongly to u∗ = PF(T )ψ(u ∗) which is also the unique solution of the following V I 〈(I −ψ)u,v −u〉≥ 0,∀v ∈ F (T ) (1.8) In this paper, we introduce and study the generalized viscosity implicit rules of asymptoticallynonexpansive mappings in Hilbert spaces. More precisely, we consider the following implicititerative algorithm: u1 ∈Mun+1 = αnun + βnψ(un) + γnT n(snun + (1 − sn)un+1) ∀n ∈N (1.9) Under suitable conditions, we proved that the sequence {un} converge strongly to a fixed point ofthe asymptotically nonexpansive mapping T , which also solves the variational inequality 〈(I −ψ)u,p−u〉≥ 0 p ∈ F (T ). As applications, we apply our results to solve convexly constrained minimization problem. This wayresults in 1.5 are complemented, extended and generalized. Eur. J. Math. Anal. 1 (2021) 21 2. Preliminaries In the sequel, we always assume that H is a real Hilbert space and M is a nonempty, closed,and convex subset of H. The nearest point projection from H onto M,PM, is defined by PM(u) := arg min z∈M ∥∥∥u −z∥∥∥2, u ∈H. (2.1) Namely, PM(u) is the only point in M that minimizes the objective ∥∥∥u − z∥∥∥ over z ∈ M. and PM(u) is characterized as follows: PM(u) ∈M and 〈 u −PM(u),z −PM(u) 〉 ≤ 0 f or all z ∈M. (2.2) Definition 2.1. . A mapping T : M→M is said to be: a): α-inverse strongly monotone if there exists α > 0 satisfying 〈u −v,Tu −T v〉≥ α‖Au −Av‖2 ∀u,v ∈M; (2.3) b): L-Lipschitz continuous if there exists L ≥ 0 satisfying ‖Tu −T v‖≤ L‖u −v‖ ∀u,v ∈M; (2.4) c): nonexpansive if ‖Tu −T v‖≤‖u −v‖ ∀u,v ∈M; (2.5) d): asymptotically nonexpansive if there exists a sequence {kn} ⊂ [1,∞) with lim n→∞ kn = 1 suchthat ‖T nu −T nv‖≤ kn‖u −v‖ ∀u,v ∈M and ∀n ∈N; (2.6) e): contraction if there exists the contractive constant α ∈ [0, 1) such that ‖Tu −T v‖≤ α‖u −v‖ ∀u,v ∈M; (2.7) Lemma 2.2. (The demiclosedness principle [10]) . Let H be a Hilbert space, M be a nonempty closed convex subset of H, and T : M → M be a asymptotically nonexpansive mapping with Fix(T ) 6= ∅. If {un} is a sequence in M such that {un} weakly converges to u and {(I −T )un} converges strongly to 0, then u = T (u) Lemma 2.3. Let H be a Hilbert space. Then for all θ,u,v ∈H, the following inequality holds ‖u −θ‖2 ≤‖v −θ‖2 + 2〈u −v,u −θ〉 Lemma 2.4. [11]). Assume that {αn} is a sequence of nonnegative real numbers such that αn+1 ≤ (1 −λn)αn + δn for all n ∈ N, where {λn} ⊆ (0, 1) and {δn} ⊆ R are two sequences satisfying the following conditions: Eur. J. Math. Anal. 1 (2021) 22 (i): ∞∑ n=1 λn = ∞ (ii): lim sup n→∞ δn λn ≤ 0 or ∞∑ n=1 |δn| < ∞ Then lim n→∞ αn = 0 Then the sequence {αn} converges to 0. 3. Main Result We now prove the following new result. Theorem 3.1. Let M be a nonempty closed convex subset a real Hilbert space H,T : M → M be asymptotically nonexpansive mappings with the same sequence {kn} ⊆ [1,∞) such that limn→∞kn = 1,Fix(T ) 6= ∅ and ψ : M → M be a contraction mapping with the contractive constant α ∈ [0, 1). Define a sequence {un} in M as follows: u1 ∈Mun+1 = αnun + βnψ(un) + γnT n(snun + (1 − sn)un+1) ∀n ∈N (3.1) where αn,βn,γn,sn ∈ (0, 1) satisfying the following conditions, A1: αn + βn + γn = 1 A2: ∞∑ n=0 αn = ∞ A3: 0 < � ≤ sn ≤ sn+1 < 1 for all n ≥ 0 A4: lim n→∞ γn = 1 and lim n→∞ αn = lim n→∞ βn = lim n→∞ sn = 0 lim n→∞ ‖un −T nun‖ = 0 Then the sequence {un} strongly converges to a common fixed point q of T , which is also the unique solution of the following variational inequality 〈(I −ψ)u,p−u〉≥ 0 p ∈ F (T ). We now show that algorithm 3.1 is well posed. Letting Bn(u) = αnun + βnψ(un) + γnT n ( snun + (1 − sn)un ) ‖Bn(u) −Bn(v)‖ = ‖γnT n ( snun + (1 − sn)u ) −γnT n ( snun + (1 − sn)v ) ‖ = ‖γnT n(1 − sn)u −γnT n(1 − sn)v‖ ≤ γnkn(1 − sn)‖u −v‖ Since lim n→∞ sn = 0, lim n→∞ kn = 1, lim n→∞ γn = 1 and 0 < � ≤ sn ≤ sn+1 < 1 for all n > 0, we mayassume that γnkn(1 − sn) ≤ 1 − � for all n > 0. This implies that Bn is a contraction for each Eur. J. Math. Anal. 1 (2021) 23 n. Therefore there exists a unique fixed point for Bn by Banach contraction principle, which alsoimplies that (3.1) is well-defined. We now show that the sequence {un} is bounded. Rewriting 3.1, we have un+1 = βnψ(un) + αnun + (1 −βn)vn (3.2) where vn = γnT n(snun + (1 − sn)un+1) 1 −βn Remark 3.2. The real sequences that satisfies the above conditions are αn = 1 n , βn = 1 n and γn = 1 − 2 n Proof. Our prove are in six steps. First we prove that the sequence {un} defined by 3.1 is bounded. Step 1: Letting p ∈ Fix(T ), we have the following estimates ‖un+1 −p‖ = ‖βnψ(un) + αnun + (1 −βn)vn −p‖ ≤ βn‖ψ(un) −ψ(p)‖ + βn‖ψ(p) −p‖ + αn‖un −p‖ + (1 −βn)‖vn −p‖ ≤ (αβn + αn)‖un −p‖ + βn‖ψ(p) −p‖ + (1 −βn)‖vn −p‖ (3.3) ‖vn −p‖ = ‖ γnT n(snun + (1 − sn)un+1) 1 −βn −p‖ = γnT nsn(un −p) 1 −βn + γnT n(1 − sn)(un+1 −p) 1 −βn ‖ ≤ γnknsn 1 −βn ‖un −p‖ + γnkn(1 − sn) 1 −βn ‖un+1 −p‖ (3.4) Putting 3.4 in 3.3, gives the following ‖un+1 −p‖ ≤ (αβn + αn)‖un −p‖ + βn‖ψ(p) −p‖ + γnknsn‖un −P‖ + γnkn(1 − sn)‖un+1 −p‖ (1 −γnkn(1 − sn))‖un+1 −p‖ ≤ (αβn + αn + γnknsn)‖un −p‖ + βn‖ψ(p) −p‖ ‖un+1 −p‖ ≤ (αβn + αn + γnknsn) 1 −γnkn(1 − sn) ‖un −p‖ + βn 1 −γnkn(1 − sn) ‖ψ(p) −p‖ Eur. J. Math. Anal. 1 (2021) 24 Since γn,sn ∈ (0, 1), 1 −γnkn(1 − sn) > 0 and lim n→∞ kn = 1. From the condition (A1), wehave ‖un+1 −p‖ ≤ 1 − 1 −αβn −αn −γnkn 1 −γnkn(1 − sn) ‖un −p‖ + βn 1 −γnkn(1 − sn) ‖ψ(p) −p‖ ] ‖un+1 −p‖ ≤ 1 − βn(1 −α) 1 −γnkn(1 − sn) ‖un −p‖ + βn(1 −α) 1 −γnkn(1 − sn) 1 (1 −α) ‖ψ(p) −p‖ ] ‖un+1 −p‖ ≤ max { ‖un −p‖, 1 (1 −α) ‖ψ(p) −p‖ } Therefore by mathematical induction, we have ‖un+1 −p‖≤ max { ‖u0 −p‖, 1 (1 −α) ‖ψ(p) −p‖ } for all n ≥ N. Therefore {un} is bounded. Consequently, {ψ(un)} and {vn} are also bounded. Step 2: We now prove that the sequence {un+1} converges to {un} as n →∞. That is lim n→∞ ‖un+1− un‖ = 0 ‖un+1 −un‖ = ‖un+1 −T nun + T nun −un‖ = ‖βnψ(un) + αnun + (1 −βn)vn − (βn + αn + γn)T n + T nun −un‖ ≤ ‖βnψ(un) −βnT nun‖ + ‖αnun −αnT nun‖ +‖(1 −βn)vn −γnT n + T nun −un‖ ≤ βn‖ψ(un) −T nun‖ + αn‖un −T nun‖ +(1 −βn)‖vn −γnT n‖ + ‖T nun −un‖ (3.5) ‖vn −γnT nun‖ = ‖ γnsn 1 −βn T nun + γn(1 − sn) 1 −βn T nun+1 −γnT nun‖ ≤ ‖ γnsn 1 −βn ‖T nun −T nun‖ + γn(1 − sn) 1 −βn ‖T nun+1 −T nun‖ ≤ γn(1 − sn)kn 1 −βn ‖un+1 −un‖ (3.6) Eur. J. Math. Anal. 1 (2021) 25 Now putting 3.6 in 3.5, we have the following ‖un+1 −un‖ ≤ βn‖ψ(un) −T nun‖ + αn‖un −T nun‖ +(1 −βn) [γn(1 − sn)kn 1 −βn ‖un+1 −un‖ ] + ‖T nun −un‖ ≤ βn‖ψ(un) −T nun‖ + αn‖un −T nun‖ +γn(1 − sn)kn‖un+1 −un‖ ] + ‖T nun −un‖ ≤ βn‖ψ(un) −T nun‖ + (αn + 1)‖un −T nun‖ + γn(1 − sn)kn‖un+1 −un‖[ 1 −γn(1 − sn)kn ] ‖un+1 −un‖ ≤ βn‖ψ(un) −T nun‖ + (αn + 1)‖un −T nun‖ ‖un+1 −un‖ ≤ βn 1 −γn(1 − sn)kn ‖ψ(un) −T nun‖ + (αn + 1) 1 −γn(1 − sn)kn ‖un −T nun‖ Let M :> max {‖ψ(un) −T nun‖}, then we have ‖un+1 −un‖≤ βnM 1 −γn(1 − sn)kn + (αn + 1) 1 −γn(1 − sn)kn ‖un −T nxn‖ ‖un+1 −un‖≤ βnM 1 −γn(1 − sn)(1 + �αn) + (αn + 1) 1 −γn(1 − sn)(1 + �αn) ‖un −T nun‖ Since lim n→∞ αn = lim n→∞ βn = lim n→∞ ‖un −T nun‖ = 0, we then conclude that lim n→∞ ‖un+1 − un‖ = 0 Step 3: Again we then show that lim n→∞ ∥∥∥un −T (un)∥∥∥ = 0. Estimating as follows we have ‖un −T nun‖ = ‖un −un+1 + un+1 −T nun‖ ≤ ‖un −un+1‖ + ‖un+1 −T nun‖ ≤ ‖un −un+1‖ + ‖βnψ(un) + αnun + (1 −βn)vn −T nun ∥∥∥ ≤ ‖un −un+1‖ + βn‖ψ(un) −T nun‖ + αn‖un −T nun‖ + (1 −βn)‖vn −γnT nun‖ (3.7) ‖vn −γnT nun‖ = ‖ γnT n(snun + (1 − sn)un+1) 1 −βn −γnT nun‖ ≤ ‖ γnsn 1 −βn ‖T nun −T nun‖ + (1 − sn)γn 1 −βn ‖T nun+1 −T nun‖ ≤ (1 − sn)γnkn 1 −βn ‖un+1 −un‖ (3.8) Eur. J. Math. Anal. 1 (2021) 26 Now substituting 3.8 into 3.7, gives the following estimation ‖un −T nun‖ ≤ ‖un −un+1‖ + βn‖ψ(un) −T nun‖ + αn‖un −T nun‖ + (1 −βn) ((1 − sn)γnkn 1 −βn ‖un+1 −un‖ ) ≤ ( 1 + (1 − sn)γnkn ) ‖un −un+1‖ + βn‖ψ(un) −T nun‖ + αn‖un −T nun‖ ≤ ( 1 + (1 − sn)γnkn ) 1 −αn ‖un −un+1‖ + βn 1 −αn ‖ψ(un) −T nun‖ ‖un −T nun‖ ≤ ( 1 + (1 − sn)γnkn ) 1 −αn ‖un+1 −xn‖ + βnM 1 −αn Therefore from 3.1 condition A4, with lim n→∞ ‖un+1 −un‖ = 0, we can conclude that lim n→∞ ‖un −T nun‖ = 0 (3.9) But we know that from the following fact lim n→∞ ‖un −T (un)‖ ≤ lim n→∞ ‖un −T nun‖ + lim n→∞ ‖T nun −T xn‖ ≤ lim n→∞ ‖un −T nun‖ + lim n→∞ k1‖T n−1un −un‖ (3.10) Proving that lim n→∞ ‖Tn−1un −un‖ = 0, we have the following estimation ‖T n−1(un) −un‖ = ‖un −T n−1(un)‖ = ‖βn−1ψ(un−1) + αn−1un−1 + (1 −βn−1)vn−1 −(βn−1 + αn−1 + γn−1)T n−1un‖ = ‖βn−1ψ(un−1) −βn−1T n−1un + αn−1un−1 −αn−1T n−1un +(1 −βn−1)vn−1 −γn−1T n−1un‖ ≤ βn−1‖ψ(un−1) −T n−1un‖ + αn−1‖un−1 −T n−1un‖ +(1 −βn−1)‖vn−1 −γn−1Tn−1xn‖ (3.11) ‖vn−1 −γn−1T n−1un‖ = ‖ γn−1T n−1(sn−1un−1 + (1 − sn−1)un) 1 −βn−1 −γn−1T n−1un‖ ≤ γn−1kn−1sn−1 1 −βn−1 ‖|un −un−1‖ (3.12) Eur. J. Math. Anal. 1 (2021) 27 combining 3.12 and 3.11 we have the following ‖T n−1(un) −un‖ ≤ βn−1‖ψ(un−1) −T n−1un‖ + αn−1‖un−1 −T n−1un‖ +(1 −βn−1) [γn−1kn−1sn−1 1 −βn−1 ‖|un −un−1‖ ] ≤ βn−1‖ψ(un−1) −T n−1un‖ + αn−1‖un−1 −T n−1un‖ +γn−1kn−1sn−1‖un −un−1‖ ≤ βn−1‖ψ(un−1) −T n−1un‖ + αn−1‖un−1 −T n−1un‖ +γn−1kn−1sn−1‖|un −un−1‖ With the assumption of {αn},{βn} and lim n→∞ ‖un+1 −un‖ = 0, we can conclude that lim n→∞ ‖T n−1un −un‖ = 0 (3.13) Therefore from 3.9 and 3.13, we can see from inequality 3.10, that lim n→∞ ‖un −T (un)‖ = 0 (3.14) Step 4: In this step, we will show that wω(xn) ⊆ Fix(T ), where wω(un) := {u ∈H : there exist a subsequence of {un} converges weakly to u}.Suppose that u ∈ wω(un). Then there exists a subsequence {uni} of {un} such that uni ⇀ xas i →∞ . From 3.14, we have lim i→∞ ∥∥∥(I −T )xni∥∥∥ = lim n→∞ ∥∥∥uni −Tuni∥∥∥ = 0 . This implies that {(I −T )uni} converges strongly to 0. By using Lemma 2.2, we have Tu = u, and so u ∈ Fix(T ). Step 5: In this step, we will show that lim sup n→∞ 〈q −ψ(q),q −un〉≤ 0, (3.15) where q ∈ F (T ) is the unique fixed point of PF(T ) ◦ψ, that is, q = PF(T )(ψ(z)). Since {un} is bounded, there exists a subsequence {uni} of {un} such that uni ⇀ u as i →∞ forsome u ∈H and lim sup n→∞ 〈q −ψ(q),q −un〉 = lim i→∞ 〈q −ψ(q),q −uni〉 (3.16) From Step 4, we get x ∈ F (T ). By using inequality 2.2, we obtain lim sup n→∞ 〈q −ψ(q),q −un〉 = lim i→∞ 〈q −ψ(q),q −uni〉 = 〈q −ψ(q),q −u〉≤ 0 Eur. J. Math. Anal. 1 (2021) 28 Step 6: Finally, setting ϕn = βnq + αnq + (1 −βn)vn we show that un → q as n → ∞. Again,take q ∈ F (T ) to be the unique fixed point of the contraction PF(T ) ◦ψ. For each n ∈ N,consider ‖un+1 −q‖2 ≤ ‖ϕn −q‖2 + 2〈un+1 −ϕn,un+1 −q〉 = (1 −βn)2‖vn −q‖2 + 2〈βn(ψ(un) −q) + αn(un −q),un+1 −q〉 ≤ (1 −βn)‖vn −q‖2 + 2〈βn(ψ(un) −ψ(q)) + βn(ψ(q) −q) + αn(un −q),un+1 −q〉 ≤ (1 −βn)2‖vn −q‖2 + 2βn‖ψ(xn) −ψ(q)‖‖un+1 −q‖ + 2αn‖un −q‖‖un+1 −q‖ +2βn〈ψ(q) −q,un+1 −q〉 ≤ (1 −βn)2‖vn −q‖2 + 2βnα‖un −q‖‖un+1 −q‖ + 2αn‖un −q‖‖un+1 −q‖ +2βn〈ψ(q) −q,un+1 −q〉 ≤ (1 −βn)2‖vn −q‖2 + (2βnα + 2αn)‖un −q‖‖un+1 −q‖ +2βn〈ψ(q) −q,un+1 −q〉 (3.17) For the fact that ‖vn −q‖2 = ∥∥∥γnTn(snun + (1 − sn)un+1) (1 −βn) −q ∥∥∥2 ≤ γ2ns 2 nk 2 n (1 −βn)2 ‖un −q‖2 + γ2n(1 − sn)2k2n (1 −βn)2 ‖un+1 −q‖2 + γ2nsn(1 − sn)k2n (1 −βn)2 〈un −q,un+1 −q〉 ≤ γ2ns 2 nk 2 n (1 −βn)2 ‖un −q‖2 + γ2n(1 − sn)2k2n (1 −βn)2 ‖un+1 −q‖2 + γ2nsn(1 − sn)k2n (1 −βn)2 ‖un −q‖‖un+1 −q‖ +2αnβn 〈 ψ(un) −ψ(q),Tn (un + un+1 2 ) −q 〉 (3.18) Now substituting 3.18 into 3.17, we have the following estimation ‖un+1 −q‖2 ≤ γ2ns 2 nk 2 n‖un −q‖ 2 + γ2n(1 − sn) 2k2n‖un+1 −q‖ 2 +γ2nsn(1 − sn)k 2 n‖un −q‖‖un+1 −q‖ +2(βnα + αn)‖un −q‖‖un+1 −q‖ + 2βn〈ψ(q) −q,un+1 −q〉 ≤ γ2ns 2 nk 2 n‖un −q‖ 2 + γ2n(1 − sn) 2k2n‖un+1 −q‖ 2 + [ γ2nsn(1 − sn)k 2 n + 2(βnα + αn) ] ‖un −q‖‖un+1 −q‖ +2βn〈ψ(q) −q,un+1 −q〉 (3.19) Eur. J. Math. Anal. 1 (2021) 29 Again using the fact that( ‖un −q‖−‖un+1 −q‖ )2 ≤ ‖un −q‖2 − 2‖un −q‖‖un+1 −q‖ +‖un+1 −q‖2 Setting the left hand to zero, we have the following estimate 2‖un −q‖‖un+1 −q‖ ≤ ‖un −q‖2 + ‖un+1 −q‖2 ‖un −q‖‖un+1 −q‖ ≤ 1 2 ‖un −q‖2 + 1 2 ‖un+1 −q‖2 (3.20) Putting inequality 3.20 in inequality 3.19, gives the following ‖un+1 −q‖2 ≤ γ2ns 2 nk 2 n‖un −q‖ 2 + γ2n(1 − sn) 2k2n‖un+1 −q‖ 2 + γ2nsn(1 − sn)k2n 2 ‖un −q‖2 + (βnα + αn)‖un −q‖2 + γ2nsn(1 − sn)k2n 2 ‖un+1 −q‖2 + (βnα + αn)‖un+1 −q‖2 +2βn〈ψ(q) −q,un+1 −q〉 ‖un+1 −q‖2 ≤ [γ2nsnk2n(sn + 1) + 2(βnα + αn) 2 ] ‖un −q‖2 + [γ2n(1 − sn)2k2n(2 − sn) + 2(βnα + αn) 2 ] ‖un+1 −q‖2 +2βn〈ψ(q) −q,un+1 −q〉 Thus we have( 1 − [γ2n(1 − sn)2k2n(2 − sn) + 2(βnα + αn) 2 ]) ‖un+1 −q‖2 ≤ [γ2nsnk2n(sn + 1) + 2(βnα + αn) 2 ] ‖un −q‖2 + 2βn〈ψ(q) −q,un+1 −q〉 ‖un+1 −q‖2 ≤ γ2nsnk 2 n(sn + 1) + 2(βnα + αn) 2 − [ γ2n(1 − sn)2k2n(2 − sn) + 2(βnα + αn) ]‖un −q‖2 + 4βn 2 − [ γ2n(1 − sn)2k2n(2 − sn) + 2(βnα + αn) ]〈ψ(q) −q,un+1 −q〉 ‖un+1 −q‖2 ≤ ( 1 − 2 −γ2n(1 − sn)2k2n(2 − sn) −γ2nsnk2n(sn + 1) 2 − [ γ2n(1 − sn)2k2n(2 − sn) + 2(βnα + αn) ])‖un −q‖2 + 4βn 2 − [ γ2n(1 − sn)2k2n(2 − sn) + 2(βnα + αn) ]〈ψ(q) −q,un+1 −q〉 Eur. J. Math. Anal. 1 (2021) 30 Therefore from condition lim n→∞ αn = lim n→∞ βn = lim n→∞ sn = 0 in 3.1, we concludes that ‖un+1 −q‖2 ≤ ( 1 − 2 − 2γ2nk2n 2 − 2γ2nk2n ) ‖un −q‖2 lim n→∞ ‖un+1 −q‖2 = 0 This complete the proof. � Theorem 3.3. Let M be a nonempty closed convex subset a real Hilbert space H,T : M → M be asymptotically nonexpansive mappings with the same sequence {kn} ⊆ [1,∞) such that limn→∞kn = 1,Fix(T ) 6= ∅ and ω be a constant. Define a sequence {un} in M as follows: u1 ∈Mun+1 = αnun + βnω + γnT n(snun + (1 − sn)un+1) ∀n ∈N (3.21) where αn,βn,γn,sn ∈ (0, 1) satisfying conditions A1 −A4 and ψ(un) = ω lim n→∞ ‖T nun −un‖ = 0 Then the sequence {un} strongly converges to a common fixed point q of T , which is also the unique solution of the following variational inequality 〈(I −ψ)u,p−u〉≥ 0 p ∈ F (T ). Taking sn = 0The following corollaries holds: Corollary 3.4. Let M be a nonempty closed convex subset a real Hilbert space H,T : M → M be asymptotically nonexpansive mappings with the same sequence {kn} ⊆ [1,∞) such that limn→∞kn = 1,Fix(T ) 6= ∅ and ψ : M → M be a contraction mapping with the contractive constant α ∈ [0, 1). Define a sequence {un} in M as follows:{ u1 ∈M un+1 = αnun + βnψ(un) + γnT n(un+1) ∀n ∈N (3.22) where αn,βn,γn ∈ (0, 1) satisfying conditions A1 −A4 without lim n→∞ sn = 0 lim n→∞ ‖T nun −un‖ = 0 Then the sequence {un} strongly converges to a common fixed point q of T , which is also the unique solution of the following variational inequality 〈(I −ψ)u,p−u〉≥ 0 p ∈ F (T ). Eur. J. Math. Anal. 1 (2021) 31 Corollary 3.5. Let M be a nonempty closed convex subset a real Hilbert space H,T : M → M be asymptotically nonexpansive mappings with the same sequence {kn} ⊆ [1,∞) such that limn→∞kn = 1,Fix(T ) 6= ∅ and u ∈M be a constant. Define a sequence {un} in M as follows:{ u1 ∈M un+1 = αnun + βnω + γnT n(un+1) ∀n ∈N (3.23) where αn,βn,γn ∈ (0, 1) satisfying conditions A1 −A4 without lim n→∞ sn = 0 lim n→∞ ‖T nun −un‖ = 0 Then the sequence {un} strongly converges to a common fixed point q of T , which is also the unique solution of the following variational inequality 〈(I −ψ)u,p−u〉≥ 0 p ∈ F (T ). 4. Application to convex minimization problems In this section, we study the problem of finding a minimizer of a convex function Φ defined from areal Hilbert space M to R.Consider the optimization problem min x∈C Φ(x) (4.1) where Φ : M → R is a convex and differentiable function. Assume 4.1 is consistent, and let Ω 6= ∅ be its set of solutions. The gradient projection algorithm generates a sequence {un} via theiterative procedure: un+1 = PM(un −δ∇Φ(u)) (4.2)if ∇Φ is θ−inverse strongly monotone mapping and δ(0, 2θ). The following basic results are wellknown. Remark 4.1. It is well known that if Φ : M → R be a real-valued differentiable convex functionand u∗ ∈M, then the point u∗ is a minimizer of Φ on M if and only if dΦ(u∗) = 0. Definition 4.2. A function Φ : M→R is said to be strongly convex if there exists α > 0 such thatfor every u,v ∈M and λ ∈ (0, 1), the following inequality holds: Φ(λu + (1 −λ)v) ≤ λΦ(u) + (1 −λ)Φ(v) −α‖u −v‖2. (4.3) Lemma 4.3. Let E be normed linear space and Φ : M → R a real-valued differentiable convex function. Assume that Φ is strongly convex. Then the differential map dΨ : M → M is strongly monotone, i.e., there exists a positive constant k such that 〈dΦ(u) −dΦ(v),u −v〉≥ k‖u −v‖2 ∀u,v ∈M. (4.4) The prove of the following theorem follows from 3.1 Eur. J. Math. Anal. 1 (2021) 32 Theorem 4.4. Let M be a nonempty closed convex subset a real Hilbert space H. For the minimization problem 4.1, assume that Φ is (Gateaux) differentiable and the gradient ∇Φ is a θ−inverse-strongly monotone mapping for some positive real number θ. Let ψ : M → M be a contraction with coefficient α ∈ [0, 1). For a given u1 ∈M, let {un} be a sequence generated by:{ u1 ∈M un+1 = αnun + βnψ(un) + γnPM(1 −δ∇Φ)(snun + (1 − sn)(un+1)) ∀n ∈N (4.5) where αn,βn,γn,sn ∈ (0, 1) satisfying the following conditions A1: αn + βn + γn = 1 A2: lim n→∞ k2n − 1 αn = 0 A3: ∞∑ n=0 αn = ∞ A4: lim n→∞ γn = 1 and lim n→∞ αn = lim n→∞ βn = lim n→∞ sn = 0 Then {un} converges strongly to a solution (u∗) of the minimization problem 4.1, which is also the unique solution of the variational inequality 〈(I −ψ)u,p−u〉≥ 0 p ∈ F (T ). Conflict of Interest: The authors declare that they have no competing interests. Availability of data and materials: No data were used to support this study. Funding: No funding was given towards this manuscript. Authors Contributions: All authors have contributed equally and significantly in writing this paper and also readand approved the final manuscript. 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Sci. 11 (2017), 549-560. https://doi.org/10.12988/ams.2017.718. https://doi.org/10.1186/s13663-015-0439-6 http://doi.org/10.22436/jnsa.009.06.86 http://doi.org/10.22436/jnsa.009.06.86 https://doi.org/10.12988/ams.2017.718 http://doi.org/10.30538/oms2017.0011 https://doi.org/10.1112/S0024610702003332 https://doi.org/10.1112/S0024610702003332 https://dx.doi.org/10.1073/pnas.53.5.1100 https://doi.org/10.12988/ams.2017.718 1. Background 2. Preliminaries 3. Main Result 4. Application to convex minimization problems References