Hybrid and Relaxed Mann Iterations for General Systems of ...

Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2013, Article ID 102820, 44 pages http://dx.doi.org/10.1155/2013/102820

Research Article Hybrid and Relaxed Mann Iterations for General Systems of Variational Inequalities and Nonexpansive Mappings L. C. Ceng,1 A. E. Al-Mazrooei,2 A. A. N. Abdou,2 and A. Latif2 1

Department of Mathematics, Shanghai Normal University and Scientific Computing Key Laboratory of Shanghai Universities, Shanghai 200234, China 2 Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia Correspondence should be addressed to A. Latif; [email protected] Received 4 August 2013; Accepted 10 August 2013 Academic Editor: Jen-Chih Yao Copyright © 2013 L. C. Ceng et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We introduce hybrid and relaxed Mann iteration methods for a general system of variational inequalities with solutions being also common solutions of a countable family of variational inequalities and common fixed points of a countable family of nonexpansive mappings in real smooth and uniformly convex Banach spaces. Here, the hybrid and relaxed Mann iteration methods are based on Korpelevich’s extragradient method, viscosity approximation method, and Mann iteration method. Under suitable assumptions, we derive some strong convergence theorems for hybrid and relaxed Mann iteration algorithms not only in the setting of uniformly convex and 2-uniformly smooth Banach space but also in a uniformly convex Banach space having a uniformly Gateaux differentiable norm. The results presented in this paper improve, extend, supplement, and develop the corresponding results announced in the earlier and very recent literature.

1. Introduction Let 𝑋 be a real Banach space whose dual space is denoted by ∗ 𝑋∗ . The normalized duality mapping 𝐽 : 𝑋 → 2𝑋 is defined by 󵄩 󵄩2 𝐽 (𝑥) = {𝑥∗ ∈ 𝑋∗ : ⟨𝑥, 𝑥∗ ⟩ = ‖𝑥‖2 = 󵄩󵄩󵄩𝑥∗ 󵄩󵄩󵄩 } , ∀𝑥 ∈ 𝑋, (1) where ⟨⋅, ⋅⟩ denotes the generalized duality pairing. It is an immediate consequence of the Hahn-Banach theorem that 𝐽(𝑥) is nonempty for each 𝑥 ∈ 𝑋. Let 𝐶 be a nonempty closed convex subset of 𝑋. A mapping 𝑇 : 𝐶 → 𝐶 is called nonexpansive if ‖𝑇𝑥 − 𝑇𝑦‖ ≤ ‖𝑥 − 𝑦‖ for every 𝑥, 𝑦 ∈ 𝐶. The set of fixed points of 𝑇 is denoted by Fix(𝑇). We use the notation ⇀ to indicate the weak convergence and the one → to indicate the strong convergence. A mapping 𝐴 : 𝐶 → 𝑋 is said to be (i) accretive if for each 𝑥, 𝑦 ∈ 𝐶 there exists 𝑗(𝑥 − 𝑦) ∈ 𝐽(𝑥 − 𝑦) such that ⟨𝐴𝑥 − 𝐴𝑦, 𝑗 (𝑥 − 𝑦)⟩ ≥ 0;

(2)

(ii) 𝛼-strongly accretive if for each 𝑥, 𝑦 ∈ 𝐶 there exists 𝑗(𝑥 − 𝑦) ∈ 𝐽(𝑥 − 𝑦) such that 󵄩2 󵄩 ⟨𝐴𝑥 − 𝐴𝑦, 𝑗 (𝑥 − 𝑦)⟩ ≥ 𝛼󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,

(3)

for some 𝛼 ∈ (0, 1); (iii) 𝛽-inverse strongly accretive if, for each 𝑥, 𝑦 ∈ 𝐶, there exists 𝑗(𝑥 − 𝑦) ∈ 𝐽(𝑥 − 𝑦) such that 󵄩2 󵄩 ⟨𝐴𝑥 − 𝐴𝑦, 𝑗 (𝑥 − 𝑦)⟩ ≥ 𝛽󵄩󵄩󵄩𝐴𝑥 − 𝐴𝑦󵄩󵄩󵄩 ,

(4)

for some 𝛽 > 0; (iv) 𝜆-strictly pseudocontractive [1] (see also [2]) if for each 𝑥, 𝑦 ∈ 𝐶 there exists 𝑗(𝑥 − 𝑦) ∈ 𝐽(𝑥 − 𝑦) such that 󵄩2 󵄩2 󵄩 󵄩 ⟨𝐴𝑥 − 𝐴𝑦, 𝑗 (𝑥 − 𝑦)⟩ ≤ 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 − 𝜆󵄩󵄩󵄩𝑥 − 𝑦 − (𝐴𝑥 − 𝐴𝑦)󵄩󵄩󵄩 (5) for some 𝜆 ∈ (0, 1).

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Abstract and Applied Analysis

It is worth emphasizing that the definition of the inverse strongly accretive mapping is based on that of the inverse strongly monotone mapping, which was studied by so many authors; see, for example, [3–5]. Let 𝑈 = {𝑥 ∈ 𝑋 : ‖𝑥‖ = 1} denote the unite sphere of 𝑋. A Banach space 𝑋 is said to be uniformly convex if, for each 𝜖 ∈ (0, 2], there exists 𝛿 > 0 such that, for all 𝑥, 𝑦 ∈ 𝑈, 󵄩 󵄩󵄩 󵄩𝑥 + 𝑦󵄩󵄩󵄩 󵄩 󵄩󵄩 ≤ 1 − 𝛿. 󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ≥ 𝜖 󳨐⇒ 󵄩 2

(6)

It is known that a uniformly convex Banach space is reflexive and strict convex. A Banach space 𝑋 is said to be smooth if the limit 󵄩󵄩󵄩𝑥 + 𝑡𝑦󵄩󵄩󵄩 − ‖𝑥‖ 󵄩 (7) lim 󵄩 𝑡→0 𝑡 exists for all 𝑥, 𝑦 ∈ 𝑈; in this case, 𝑋 is also said to have a Gateaux differentiable norm. 𝑋 is said to have a uniformly, Gateaux differentiable norm if, for each 𝑦 ∈ 𝑈, the limit is attained uniformly for 𝑥 ∈ 𝑈. Moreover, it is said to be uniformly smooth if this limit is attained uniformly for 𝑥, 𝑦 ∈ 𝑈. The norm of 𝑋 is said to be the Frechet differential if for each 𝑥 ∈ 𝑈, this limit is attained uniformly for 𝑦 ∈ 𝑈. In the meantime, we define a function 𝜌 : [0, ∞) → [0, ∞) called the modulus of smoothness of 𝑋 as follows: 1 󵄩 󵄩 󵄩 󵄩 𝜌 (𝜏) = sup { (󵄩󵄩󵄩𝑥 + 𝑦󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩) − 1 : 𝑥, 𝑦 ∈ 𝑋, 2 󵄩 󵄩 ‖𝑥‖ = 1, 󵄩󵄩󵄩𝑦󵄩󵄩󵄩 = 𝜏} .

(8)

(i) 1 ≤ 𝛽𝑛 ≤ 1 − 𝜌, for all 𝑛 ≥ 𝑛0 for some integer 𝑛0 ≥ 1, (ii) ∑∞ 𝑛=0 𝛽𝑛 = ∞, (iii) 0 < lim inf 𝑛 → ∞ 𝛼𝑛 ≤ lim sup𝑛 → ∞ 𝛼𝑛 < 1, (iv) lim𝑛 → ∞ (𝛽𝑛+1 /(1 − (1 − 𝛽𝑛+1 )𝛼𝑛+1 ) − 𝛽𝑛 /(1 − (1 − 𝛽𝑛 )𝛼𝑛 )) = 0. Ceng and Yao [8] proved that 𝑥𝑛 󳨀→ 𝑞 ⇐⇒ 𝛽𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 ) 󳨀→ 0,

(9)

where 𝑞 ∈ Fix(𝑇) solves the variational inequality problem (VIP): ⟨𝑞 − 𝑓 (𝑞) , 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,

∀𝑝 ∈ Fix (𝑇) .

(10)

Such a result includes [7, Theorem 1] as a special case. Let 𝐶 be a nonempty closed convex subset of a real Banach space 𝑋 and 𝑓 ∈ Ξ𝐶 with a contractive coefficient 𝜌 ∈ (0, 1), where Ξ𝐶 is the set of all contractive self-mappings on 𝐶. Let {𝑇𝑛 }∞ 𝑛=0 be a sequence of nonexpansive self-mappings on 𝐶 and {𝜆 𝑛 }∞ 𝑛=0 a sequence of nonnegative numbers in [0, 1]. For any 𝑛 ≥ 0, define a self-mapping 𝑊𝑛 on 𝐶 as follows: 𝑈𝑛,𝑛+1 = 𝐼, 𝑈𝑛,𝑛 = 𝜆 𝑛 𝑇𝑛 𝑈𝑛,𝑛+1 + (1 − 𝜆 𝑛 ) 𝐼,

It is known that 𝑋 is uniformly smooth if and only if lim𝜏 → 0 𝜌(𝜏)/𝜏 = 0. Let 𝑞 be a fixed real number with 1 < 𝑞 ≤ 2. Then, a Banach space 𝑋 is said to be 𝑞-uniformly smooth if there exists a constant 𝑐 > 0 such that 𝜌(𝜏) ≤ 𝑐𝜏𝑞 for all 𝜏 > 0. As pointed out in [6], no Banach space is 𝑞-uniformly smooth for 𝑞 > 2. In addition, it is also known that 𝐽 is single valued if and only if 𝑋 is smooth, whereas if 𝑋 is uniformly smooth, then the mapping 𝐽 is norm-to-norm uniformly continuous on bounded subsets of 𝑋. If 𝑋 has a uniformly Gateaux differentiable norm, then the duality mapping 𝐽 is norm-toweak∗ uniformly continuous on bounded subsets of 𝑋. Recently, Yao et al. [7] combined the viscosity approximation method and Mann iteration method and gave the following hybrid viscosity approximation method. Let 𝐶 be a nonempty closed convex subset of a real uniformly smooth Banach space 𝑋, 𝑇 : 𝐶 → 𝐶 a nonexpansive mapping with Fix(𝑇) ≠ 0, and 𝑓 : 𝐶 → 𝐶 a contraction with coefficient 𝜌 ∈ (0, 1). For an arbitrary 𝑥0 ∈ 𝐶, define {𝑥𝑛 } in the following way: 𝑦𝑛 = 𝛼𝑛 𝑥𝑛 + (1 − 𝛼𝑛 ) 𝑇𝑥𝑛 , 𝑥𝑛+1 = 𝛽𝑛 𝑓 (𝑥𝑛 ) + (1 − 𝛽𝑛 ) 𝑦𝑛 ,

a fixed point of 𝑇. Subsequently, under the following control conditions on {𝛼𝑛 } and {𝛽𝑛 }:

∀𝑛 ≥ 0,

(YCY)

where {𝛼𝑛 } and {𝛽𝑛 } are two sequences in (0, 1). They proved under certain control conditions on the sequences {𝛼𝑛 } and {𝛽𝑛 } that {𝑥𝑛 } converges strongly to

𝑈𝑛,𝑛−1 = 𝜆 𝑛−1 𝑇𝑛−1 𝑈𝑛,𝑛 + (1 − 𝜆 𝑛−1 ) 𝐼, .. . 𝑈𝑛,𝑘 = 𝜆 𝑘 𝑇𝑘 𝑈𝑛,𝑘+1 + (1 − 𝜆 𝑘 ) 𝐼,

(CY)

𝑈𝑛,𝑘−1 = 𝜆 𝑘−1 𝑇𝑘−1 𝑈𝑛,𝑘 + (1 − 𝜆 𝑘−1 ) 𝐼, .. . 𝑈𝑛,1 = 𝜆 1 𝑇1 𝑈𝑛,2 + (1 − 𝜆 1 ) 𝐼, 𝑊𝑛 = 𝑈𝑛,0 = 𝜆 0 𝑇0 𝑈𝑛,1 + (1 − 𝜆 0 ) 𝐼. Such a mapping 𝑊𝑛 is called the 𝑊-mapping generated by 𝑇𝑛 , 𝑇𝑛−1 , . . . , 𝑇0 , and 𝜆 𝑛 , 𝜆 𝑛−1 , . . . , 𝜆 0 ; see [9]. In 2008, Ceng and Yao [10] introduced and analyzed the following relaxed viscosity approximation method for finding a common fixed point of an infinite family of nonexpansive mappings in a strictly convex and reflexive Banach space with a uniformly Gateaux differentiable norm. Theorem 1 (see [10]). Let 𝑋 be a strictly convex and reflexive Banach space with a uniformly Gateaux differentiable norm, 𝐶 a nonempty closed convex subset of 𝑋, {𝑇𝑛 }∞ 𝑛=0 a sequence of nonexpansive self-mappings on 𝐶 such that the common fixed point set 𝐹 := ⋂∞ 𝑛=0 Fix(𝑇𝑛 ) ≠ 0, and 𝑓 ∈ Ξ𝐶 with a contractive

Abstract and Applied Analysis

3

coefficient 𝜌 ∈ (1/2, 1). For any given 𝑥0 ∈ 𝐶, let {𝑥𝑛 }∞ 𝑛=0 be the iterative sequence defined by 𝑦𝑛 = (1 − 𝛾𝑛 ) 𝑥𝑛 + 𝛾𝑛 𝑊𝑛 𝑥𝑛 , 𝑥𝑛+1 = (1 − 𝛼𝑛 − 𝛽𝑛 ) 𝑥𝑛 + 𝛼𝑛 𝑓 (𝑦𝑛 ) + 𝛽𝑛 𝑊𝑛 𝑦𝑛 ,

∀𝑛 ≥ 0, (11)

∞ where {𝛼𝑛 }∞ 𝑛=0 and {𝛽𝑛 }𝑛=0 are two sequences in (0, 1) with 𝛼𝑛 + ∞ 𝛽𝑛 ≤ 1 (𝑛 ≥ 0), {𝛾𝑛 }𝑛=0 is a sequence in [0, 1], and 𝑊𝑛 is the 𝑊-mapping generated by (CY). Assume that

(i) lim𝑛 → ∞ 𝛼𝑛 = 0, ∑∞ = ∞ and 0 𝑛=0 𝛼𝑛 lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ 𝛽𝑛 < 1;


0 a given number, which is known as the extragradient method (see also [16]). The literature on the VIP is vast and Korpelevich’s extragradient method has received great attention given by many authors, who improved it in various ways; see, for example, [3, 11, 13, 17–33] and references therein, to name but a few. In particular, whenever 𝑋 is still a real smooth Banach space, 𝐵1 = 𝐵2 = 𝐴 and 𝑥∗ = 𝑦∗ , then GSVI (13) reduces to the variational inequality problem (VIP) of finding 𝑥∗ ∈ 𝐶 such that ⟨𝐴𝑥∗ , 𝐽 (𝑥 − 𝑥∗ )⟩ ≥ 0,

∀𝑥 ∈ 𝐶,

(18)

which was considered by Aoyama et al. [34]. Note that VIP (18) is connected with the fixed point problem for nonlinear mapping (see, e.g., [35]), the problem of finding a zero point of a nonlinear operator (see, e.g., [36]), and so on. It is clear that VIP (18) extends VIP (15) from Hilbert spaces to Banach spaces. In order to find a solution of VIP (18), Aoyama et al. [34] introduced the following Mann-type iterative scheme for an accretive operator 𝐴: 𝑥𝑛+1 = 𝛼𝑛 𝑥𝑛 + (1 − 𝛼𝑛 ) Π𝐶 (𝑥𝑛 − 𝜆 𝑛 𝐴𝑥𝑛 ) ,

∀𝑛 ≥ 1, (19)

where Π𝐶 is a sunny nonexpansive retraction from 𝑋 onto 𝐶. Then, they proved a weak convergence theorem. For the related work, see [37] and the references therein. Let 𝐶 be a nonempty convex subset of a real Banach space 𝑋. Let {𝑇𝑖 }𝑁 𝑖=1 be a finite family of nonexpansive mappings of

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Abstract and Applied Analysis

𝐶 into itself and let 𝜆 1 , . . . , 𝜆 𝑁 be real numbers such that 0 ≤ 𝜆 𝑖 ≤ 1 for every 𝑖 = 1, . . . , 𝑁. Define a mapping 𝐾 : 𝐶 → 𝐶 as follows: 𝑈1 = 𝜆 1 𝑇1 + (1 − 𝜆 1 ) 𝐼, 𝑈2 = 𝜆 2 𝑇2 𝑈1 + (1 − 𝜆 2 ) 𝑈1 , 𝑈3 = 𝜆 3 𝑇3 𝑈2 + (1 − 𝜆 3 ) 𝑈2 , (20)

.. .

Beyond doubt, it is an interesting and valuable problem of constructing some algorithms with strong convergence for solving GSVI (13) which contains VIP (18) as a special case. Very recently, Cai and Bu [11] constructed an iterative algorithm for solving GSVI (13) and a common fixed point problem of a countable family of nonexpansive mappings in a uniformly convex and 2-uniformly smooth Banach space. They proved the strong convergence of the proposed algorithm by virtue of the following inequality in a 2uniformly smooth Banach space 𝑋. Lemma 4 (see [39]). Let 𝑋 be a 2-uniformly smooth Banach space. Then,

𝑈𝑁−1 = 𝜆 𝑁−1 𝑇𝑁−1 𝑈𝑁−2 + (1 − 𝜆 𝑁−1 ) 𝑈𝑁−2 , 𝐾 = 𝑈𝑁 = 𝜆 𝑁𝑇𝑁𝑈𝑁−1 + (1 − 𝜆 𝑁) 𝑈𝑁−1 . Such a mapping 𝐾 is called the 𝐾-mapping generated by 𝑇1 , . . . , 𝑇𝑁 and 𝜆 1 , . . . , 𝜆 𝑁. Very recently, Kangtunyakarn [38] introduced and analyzed an iterative algorithm by the modification of Mann’s iteration process for finding a common element of the set of solutions of a finite family of variational inequalities and the set of fixed points of an 𝜂-strictly pseudocontractive mapping and a nonexpansive mapping in uniformly convex and 2uniformly smooth Banach spaces.

󵄩 󵄩2 󵄩2 󵄩󵄩 2 󵄩󵄩𝑥 + 𝑦󵄩󵄩󵄩 ≤ ‖𝑥‖ + 2 ⟨𝑦, 𝐽 (𝑥)⟩ + 2󵄩󵄩󵄩𝜅𝑦󵄩󵄩󵄩 ,

∀𝑥, 𝑦 ∈ 𝑋, (23)

where 𝜅 is the 2-uniformly smooth constant of 𝑋 and 𝐽 is the normalized duality mapping from 𝑋 into 𝑋∗ . Define the mapping 𝐺 : 𝐶 → 𝐶 as follows: 𝐺 (𝑥) := Π𝐶 (𝐼 − 𝜇1 𝐵1 ) Π𝐶 (𝐼 − 𝜇2 𝐵2 ) 𝑥,

∀𝑥 ∈ 𝐶.

(24)

The fixed point set of 𝐺 is denoted by Ω. Then, their strong convergence theorem on the proposed method is stated as follows.

Theorem 3 (see [38]). Let 𝐶 be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space 𝑋. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶. Let 𝐴 𝑖 : 𝐶 → 𝑋 be an 𝛼𝑖 -inverse-strongly accretive mapping for each 𝑖 = 1, . . . , 𝑁. Define the mapping 𝐺𝑖 : 𝐶 → 𝐶 by 𝐺𝑖 = Π𝐶(𝐼 − 𝜆 𝑖 𝐴 𝑖 ) for 𝑖 = 1, . . . , 𝑁, where 𝜆 𝑖 ∈ (0, 𝛼𝑖 /𝜅2 ) and 𝜅 is the 2-uniformly smooth constant of 𝑋. Let 𝐵 : 𝐶 → 𝐶 be the 𝐾-mapping generated by 𝐺1 , . . . , 𝐺𝑁 and 𝜌1 , . . . , 𝜌𝑁, where 𝜌𝑖 ∈ (0, 1), for all 𝑖 = 1, . . . , 𝑁 − 1, and 𝜌𝑁 ∈ (0, 1]. Let 𝑓 : 𝐶 → 𝐶 a contraction with coefficient 𝜌 ∈ (0, 1). Let 𝑉 : 𝐶 → 𝐶 be an 𝜂-strictly pseudocontractive mapping and 𝑆 : 𝐶 → 𝐶 be a nonexpansive mapping such that 𝐹 = Fix(𝑆) ∩ Fix(𝑉) ∩ (⋂𝑁 𝑖=1 VI(𝐶, 𝐴 𝑖 )) ≠ 0. For arbitrarily given 𝑥0 ∈ 𝐶, let {𝑥𝑛 } be the sequence generated by

Theorem 5 (see [11]). Let 𝐶 be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space 𝑋. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶. Let the mapping 𝐵𝑖 : 𝐶 → 𝑋 be 𝛽𝑖 -inverse-strongly accretive with 0 < 𝜇𝑖 < 𝛽𝑖 /𝜅2 for 𝑖 = 1, 2. Let 𝑓 be a contraction of 𝐶 into itself with coefficient 𝛿 ∈ (0, 1). Let {𝑇𝑛 }∞ 𝑛=1 be a countable family of nonexpansive mappings of 𝐶 into itself such that 𝐹 = ⋂∞ 𝑖=1 Fix(𝑇𝑖 ) ∩ Ω ≠ 0, where Ω is the fixed point set of the mapping 𝐺 defined by (24). For arbitrarily given 𝑥1 ∈ 𝐶, let {𝑥𝑛 } be the sequence generated by

𝑥𝑛+1 = 𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑥𝑛 + 𝛿𝑛 𝑆 ((1 − 𝛼) 𝐼 + 𝛼𝑉) 𝑥𝑛 ,

𝑦𝑛 = 𝛼𝑛 𝑓 (𝑥𝑛 ) + (1 − 𝛼𝑛 ) 𝑧𝑛 ,

∀𝑛 ≥ 0, (21) 2

where 𝛼 ∈ (0, 𝜂/𝜅 ). Suppose that {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, and {𝛿𝑛 } are the sequences in [0, 1], 𝛼𝑛 + 𝛽𝑛 + 𝛾𝑛 + 𝛿𝑛 = 1 and satisfy the following conditions:

𝑧𝑛 = Π𝐶 (𝑢𝑛 − 𝜇1 𝐵1 𝑢𝑛 ) , 𝑢𝑛 = Π𝐶 (𝑥𝑛 − 𝜇2 𝐵2 𝑥𝑛 ) ,

(25)

∀𝑛 ≥ 1.

Suppose that {𝛼𝑛 } and {𝛽𝑛 } are two sequences in (0, 1) satisfying the following conditions: (i) lim𝑛 → ∞ 𝛼𝑛 = 0 and ∑∞ 𝑛=1 𝛼𝑛 = ∞;

(i) lim𝑛 → ∞ 𝛼𝑛 = 0 and ∑∞ 𝑛=0 𝛼𝑛 = ∞;

(ii) 0 < lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ 𝛽𝑛 < 1.

(ii) {𝛾𝑛 }, {𝛿𝑛 } ⊂ [𝑐, 𝑑] for some 𝑐, 𝑑 ∈ (0, 1);

(iii) ∑∞ 𝑛=1 (|𝛽𝑛 − 𝛽𝑛−1 | + |𝛾𝑛 − 𝛾𝑛−1 | + |𝛿𝑛 − 𝛿𝑛−1 |) < ∞; (iv) 0 < lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ 𝛽𝑛 < 1. Then, {𝑥𝑛 } converges strongly to 𝑞 ∈ 𝐹, which solves the following VIP: ⟨𝑞 − 𝑓 (𝑞) , 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,

𝑥𝑛+1 = 𝛽𝑛 𝑥𝑛 + (1 − 𝛽𝑛 ) 𝑆𝑛 𝑦𝑛 ,

∀𝑝 ∈ 𝐹.

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Assume that ∑∞ 𝑛=1 sup𝑥∈𝐷 ‖𝑇𝑛+1 𝑥−𝑇𝑛 𝑥‖ < ∞ for any bounded subset 𝐷 of 𝐶 and let 𝑇 be a mapping of 𝐶 into 𝑋 defined by 𝑇𝑥 = lim𝑛 → ∞ 𝑇𝑛 𝑥 for all 𝑥 ∈ 𝐶 and suppose that Fix(𝑇) = ⋂∞ 𝑛=1 Fix(𝑇𝑛 ). Then, {𝑥𝑛 } converges strongly to 𝑞 ∈ 𝐹, which solves the following VIP: ⟨𝑞 − 𝑓 (𝑞) , 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,

∀𝑝 ∈ 𝐹.

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Abstract and Applied Analysis

5

It is easy to see that the iterative scheme in Theorem 5 is essentially equivalent to the following two-step iterative scheme: 𝑦𝑛 = 𝛼𝑛 𝑓 (𝑥𝑛 ) + (1 − 𝛼𝑛 ) 𝐺𝑥𝑛 , 𝑥𝑛+1 = 𝛽𝑛 𝑥𝑛 + (1 − 𝛽𝑛 ) 𝑇𝑛 𝑦𝑛 ,

∀𝑛 ≥ 1.

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For the convenience of implementing the argument techniques in [14], the authors of [11] have used the following inequality in a real smooth and uniform convex Banach space 𝑋. Proposition 6 (see [40]). Let 𝑋 be a real smooth and uniform convex Banach space and let 𝑟 > 0. Then, there exists a strictly increasing, continuous, and convex function 𝑔 : [0, 2𝑟] → R, 𝑔(0) = 0 such that 󵄩 󵄩2 󵄩 󵄩 𝑔 (󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩) ≤ ‖𝑥‖2 − 2 ⟨𝑥, 𝐽 (𝑦)⟩ + 󵄩󵄩󵄩𝑦󵄩󵄩󵄩 ,

∀𝑥, 𝑦 ∈ 𝐵𝑟 , (28)

where 𝐵𝑟 = {𝑥 ∈ 𝑋 : ‖𝑥‖ ≤ 𝑟}. Let 𝐶 be a nonempty closed convex subset of a real smooth Banach space 𝑋. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶 and 𝑓 : 𝐶 → 𝐶 a contraction with coefficient 𝜌 ∈ (0, 1). Motivated and inspired by the research going on this area, we consider and introduce hybrid and relaxed Mann iteration methods for finding solutions of the GSVI (13) which are also common solutions of a countable family of variational inequalities and common fixed points of a countable family of nonexpansive mappings in 𝑋. Here, the hybrid and relaxed Mann iteration methods are based on Korpelevich’s extragradient method, viscosity approximation method, and Mann iteration method. Under suitable assumptions, we derive some strong convergence theorems for hybrid and relaxed Mann iteration algorithms not only in the setting of uniformly convex and 2-uniformly smooth Banach space but also in a uniformly convex Banach space having a uniformly Gateaux differentiable norm. The results presented in this paper improve, extend, supplement, and develop the corresponding results announced in the earlier and very recent literature; see, for example, [8, 10, 11, 14, 33, 38].

2. Preliminaries We list some lemmas that will be used in the sequel. Lemma 7 (see [41]). Let {𝑠𝑛 } be a sequence of nonnegative real numbers satisfying 𝑠𝑛+1 ≤ (1 − 𝛼𝑛 ) 𝑠𝑛 + 𝛼𝑛 𝛽𝑛 + 𝛾𝑛 ,

∀𝑛 ≥ 0,

where {𝛼𝑛 }, {𝛽𝑛 }, and {𝛾𝑛 } satisfy the following conditions: (i) {𝛼𝑛 } ⊂ [0, 1] and

∑∞ 𝑛=0

(ii) lim sup𝑛 → ∞ 𝛽𝑛 ≤ 0;

𝛼𝑛 = ∞;

(iii) 𝛾𝑛 ≥ 0, for all 𝑛 ≥ 0, and ∑∞ 𝑛=0 𝛾𝑛 < ∞. Then, lim sup𝑛 → ∞ 𝑠𝑛 = 0.

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The following lemma is an immediate consequence of the subdifferential inequality of the function (1/2)‖ ⋅ ‖2 . Lemma 8 (see [42]). Let 𝑋 be a real Banach space 𝑋. Then, for all 𝑥, 𝑦 ∈ 𝑋 (i) ‖𝑥 + 𝑦‖2 ≤ ‖𝑥‖2 + 2⟨𝑦, 𝑗(𝑥 + 𝑦)⟩ for all 𝑗(𝑥 + 𝑦) ∈ 𝐽(𝑥 + 𝑦); (ii) ‖𝑥 + 𝑦‖2 ≥ ‖𝑥‖2 + 2⟨𝑦, 𝑗(𝑥)⟩ for all 𝑗(𝑥) ∈ 𝐽(𝑥). Let 𝐷 be a subset of 𝐶 and let Π be a mapping of 𝐶 into 𝐷. Then, Π is said to be sunny if Π [Π (𝑥) + 𝑡 (𝑥 − Π (𝑥))] = Π (𝑥) ,

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whenever Π(𝑥) + 𝑡(𝑥 − Π(𝑥)) ∈ 𝐶 for 𝑥 ∈ 𝐶 and 𝑡 ≥ 0. A mapping Π of 𝐶 into itself is called a retraction if Π2 = Π. If a mapping Π of 𝐶 into itself is a retraction, then Π(𝑧) = 𝑧 for every 𝑧 ∈ 𝑅(Π) where 𝑅(Π) is the range of Π. A subset 𝐷 of 𝐶 is called a sunny nonexpansive retract of 𝐶 if there exists a sunny nonexpansive retraction from 𝐶 onto 𝐷. The following lemma concerns the sunny nonexpansive retraction. Lemma 9 (see [43]). Let 𝐶 be a nonempty closed convex subset of a real smooth Banach space 𝑋. Let 𝐷 be a nonempty subset of 𝐶. Let Π be a retraction of 𝐶 onto 𝐷. Then, the following are equivalent: (i) Π is sunny and nonexpansive; (ii) ‖Π(𝑥) − Π(𝑦)‖2 ≤ ⟨𝑥 − 𝑦, 𝐽(Π(𝑥) − Π(𝑦))⟩, for all 𝑥, 𝑦 ∈ 𝐶; (iii) ⟨𝑥 − Π(𝑥), 𝐽(𝑦 − Π(𝑥))⟩ ≤ 0, for all 𝑥 ∈ 𝐶, 𝑦 ∈ 𝐷. It is well known that if 𝑋 = 𝐻 a Hilbert space, then a sunny nonexpansive retraction Π𝐶 is coincident with the metric projection from 𝑋 onto 𝐶; that is, Π𝐶 = 𝑃𝐶. If 𝐶 is a nonempty closed convex subset of a strictly convex and uniformly smooth Banach space 𝑋 and if 𝑇 : 𝐶 → 𝐶 is a nonexpansive mapping with the fixed point set Fix(𝑇) ≠ 0, then the set Fix(𝑇) is a sunny nonexpansive retract of 𝐶. Lemma 10. Let 𝐶 be a nonempty closed convex subset of a smooth Banach space 𝑋. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶 and let 𝐵1 , 𝐵2 : 𝐶 → 𝑋 be nonlinear mappings. For given 𝑥∗ , 𝑦∗ ∈ 𝐶, (𝑥∗ , 𝑦∗ ) is a solution of GSVI (13) if and only if 𝑥∗ = Π𝐶(𝑦∗ − 𝜇1 𝐵1 𝑦∗ ), where 𝑦∗ = Π𝐶(𝑥∗ − 𝜇2 𝐵2 𝑥∗ ). Proof. We can rewrite GSVI (13) as ⟨𝑥∗ − (𝑦∗ − 𝜇1 𝐵1 𝑦∗ ) , 𝐽 (𝑥 − 𝑥∗ )⟩ ≥ 0,

∀𝑥 ∈ 𝐶,

⟨𝑦∗ − (𝑥∗ − 𝜇2 𝐵2 𝑥∗ ) , 𝐽 (𝑥 − 𝑦∗ )⟩ ≥ 0,

∀𝑥 ∈ 𝐶,

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which is obviously equivalent to 𝑥∗ = Π𝐶 (𝑦∗ − 𝜇1 𝐵1 𝑦∗ ) , 𝑦∗ = Π𝐶 (𝑥∗ − 𝜇2 𝐵2 𝑥∗ ) , because of Lemma 9. This completes the proof.

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6

Abstract and Applied Analysis In terms of Lemma 10, we observe that 𝑥∗ = Π𝐶 [Π𝐶 (𝑥∗ − 𝜇2 𝐵2 𝑥∗ ) − 𝜇1 𝐵1 Π𝐶 (𝑥∗ − 𝜇2 𝐵2 𝑥∗ )] , (33)

which implies that 𝑥∗ is a fixed point of the mapping 𝐺. Throughout this paper, the set of fixed points of the mapping 𝐺 is denoted by Ω. Lemma 11 (see [44]). Let 𝑋 be a uniformly convex Banach space and 𝐵𝑟 = {𝑥 ∈ 𝑋 : ‖𝑥‖ ≤ 𝑟}, 𝑟 > 0. Then, there exists a continuous, strictly increasing, and convex function 𝑔 : [0, ∞] → [0, ∞], 𝑔(0) = 0 such that 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩󵄩 2 2 󵄩󵄩𝛼𝑥 + 𝛽𝑦 + 𝛾𝑧󵄩󵄩󵄩 ≤ 𝛼‖𝑥‖ + 𝛽󵄩󵄩󵄩𝑦󵄩󵄩󵄩 + 𝛾‖𝑧‖ − 𝛼𝛽𝑔 (󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩) (34)

a continuous strictly increasing function 𝑔 : [0, ∞) → [0, ∞), 𝑔(0) = 0, such that 󵄩 󵄩2 󵄩2 󵄩󵄩 2 󵄩󵄩𝜆𝑥 + (1 − 𝜆) 𝑦󵄩󵄩󵄩 ≤ 𝜆‖𝑥‖ + (1 − 𝜆) 󵄩󵄩󵄩𝑦󵄩󵄩󵄩 󵄩 󵄩 − 𝜆 (1 − 𝜆) 𝑔 (󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩)

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for all 𝜆 ∈ [0, 1] and 𝑥, 𝑦 ∈ 𝑋 such that ‖𝑥‖ ≤ 𝑟 and ‖𝑦‖ ≤ 𝑟. Lemma 16 (see [48, Lemma 3.2]). Let 𝐶 be a nonempty closed convex subset of a strictly convex Banach space 𝑋. Let {𝑇𝑛 }∞ 𝑛=0 be a sequence of nonexpansive self-mappings on 𝐶 such that ∞ ⋂𝑛=0 Fix(𝑇𝑛 ) ≠ 0 and let {𝜆 𝑛 }∞ 𝑛=0 be a sequence of positive numbers in (0, 𝑏] for some 𝑏 ∈ (0, 1). Then, for every 𝑥 ∈ 𝐶 and 𝑘 ≥ 0, the limit lim𝑛 → ∞ 𝑈𝑛,𝑘 𝑥 exists. Using Lemma 16, one can define a mapping 𝑊 : 𝐶 → 𝐶 as follows:

for all 𝑥, 𝑦, 𝑧 ∈ 𝐵𝑟 , and all 𝛼, 𝛽, 𝛾 ∈ [0, 1] with 𝛼 + 𝛽 + 𝛾 = 1.

𝑊𝑥 = lim 𝑊𝑛 𝑥 = lim 𝑈𝑛,0 𝑥

Lemma 12 (see [45]). Let 𝐶 be a nonempty closed convex subset of a Banach space 𝑋. Let 𝑆0 , 𝑆1 , . . . be a sequence of mappings of 𝐶 into itself. Suppose that ∑∞ 𝑛=1 sup{‖𝑆𝑛 𝑥−𝑆𝑛−1 𝑥‖ : 𝑥 ∈ 𝐶} < ∞. Then for each 𝑦 ∈ 𝐶, {𝑆𝑛 𝑦} converges strongly to some point of 𝐶. Moreover, let 𝑆 be a mapping of 𝐶 into itself defined by 𝑆𝑦 = lim𝑛 → ∞ 𝑆𝑛 𝑦 for all 𝑦 ∈ 𝐶. Then lim𝑛 → ∞ sup{‖𝑆𝑥 − 𝑆𝑛 𝑥‖ : 𝑥 ∈ 𝐶} = 0.

for every 𝑥 ∈ 𝐶. Such a 𝑊 is called the 𝑊-mapping generated ∞ by the sequences {𝑇𝑛 }∞ 𝑛=0 and {𝜆 𝑛 }𝑛=0 . Throughout this paper, ∞ we always assume that {𝜆 𝑛 }𝑛=0 is a sequence of positive numbers in (0, 𝑏] for some 𝑏 ∈ (0, 1).

Let 𝐶 be a nonempty closed convex subset of a Banach space 𝑋 and 𝑇 : 𝐶 → 𝐶 a nonexpansive mapping with Fix(𝑇) ≠ 0. As previous, let Ξ𝐶 be the set of all contractions on 𝐶. For 𝑡 ∈ (0, 1) and 𝑓 ∈ Ξ𝐶, let 𝑥𝑡 ∈ 𝐶 be the unique fixed point of the contraction 𝑥 󳨃→ 𝑡𝑓(𝑥) + (1 − 𝑡)𝑇𝑥 on 𝐶; that is, 𝑥𝑡 = 𝑡𝑓 (𝑥𝑡 ) + (1 − 𝑡) 𝑇𝑥𝑡 .

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Lemma 13 (see [35, 46]). Let 𝑋 be a uniformly smooth Banach space, or a reflexive and strictly convex Banach space with a uniformly Gateaux differentiable norm. Let 𝐶 be a nonempty closed convex subset of 𝑋, 𝑇 : 𝐶 → 𝐶 a nonexpansive mapping with Fix(𝑇) ≠ 0, and 𝑓 ∈ Ξ𝐶. Then, the net {𝑥𝑡 } defined by 𝑥𝑡 = 𝑡𝑓(𝑥𝑡 ) + (1 − 𝑡)𝑇𝑥𝑡 converges strongly to a point in Fix(𝑇). If we define a mapping 𝑄 : Ξ𝐶 → Fix(𝑇) by 𝑄(𝑓) := 𝑠 − lim𝑡 → 0 𝑥𝑡 , for all 𝑓 ∈ Ξ𝐶, then 𝑄(𝑓) solves the VIP: ⟨(𝐼 − 𝑓) 𝑄 (𝑓) , 𝐽 (𝑄 (𝑓) − 𝑝)⟩ ≤ 0,

∀𝑓 ∈ Ξ𝐶, 𝑝 ∈ Fix (𝑇) . (36)

Lemma 14 (see [47]). Let 𝐶 be a nonempty closed convex subset of a strictly convex Banach space 𝑋. Let {𝑇𝑛 }∞ 𝑛=0 be a sequence of nonexpansive mappings on 𝐶. Suppose that ⋂∞ 𝑛=0 Fix(𝑇𝑛 ) is nonempty. Let {𝜆 𝑛 } be a sequence of positive numbers with ∑∞ 𝑛=0 𝜆 𝑛 = 1. Then, a mapping 𝑆 on 𝐶 defined by 𝑆𝑥 = ∑∞ 𝑛=0 𝜆 𝑛 𝑇𝑛 𝑥 for 𝑥 ∈ 𝐶 is defined well; nonexpansive and Fix(𝑆) = ⋂∞ 𝑛=0 Fix(𝑇𝑛 ) holds. Lemma 15 (see [39]). Given a number 𝑟 > 0, A real Banach space 𝑋 is uniformly convex if and only if there exists

𝑛→∞

𝑛→∞

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Lemma 17 (see [48]). Let 𝐶 be a nonempty closed convex subset of a strictly convex Banach space 𝑋. Let {𝑇𝑛 }∞ 𝑛=0 be a sequence of nonexpansive self-mappings on 𝐶 such that ∞ ⋂∞ 𝑛=0 Fix(𝑇𝑛 ) ≠ 0 and let {𝜆 𝑛 }𝑛=0 be a sequence of positive numbers in (0, 𝑏] for some 𝑏 ∈ (0, 1). Then, Fix(𝑊) = ⋂∞ 𝑛=0 Fix(𝑇𝑛 ). Let 𝜇 be a continuous linear functional on 𝑙∞ and 𝑠 = (𝑎0 , 𝑎1 , . . .) ∈ 𝑙∞ . One writes 𝜇𝑛 (𝑎𝑛 ) instead of 𝜇(𝑠). 𝜇 is called a Banach limit if 𝜇 satisfies ‖𝜇‖ = 𝜇𝑛 (1) = 1 and 𝜇𝑛 (𝑎𝑛+1 ) = 𝜇𝑛 (𝑎𝑛 ) for all (𝑎0 , 𝑎1 , . . .) ∈ 𝑙∞ . If 𝜇 is a Banach limit, then, there hold the following: (i) for all 𝑛 ≥ 0, 𝑎𝑛 ≤ 𝑐𝑛 implies 𝜇𝑛 (𝑎𝑛 ) ≤ 𝜇𝑛 (𝑐𝑛 ); (ii) 𝜇𝑛 (𝑎𝑛+𝑟 ) = 𝜇𝑛 (𝑎𝑛 ) for any fixed positive integer 𝑟; (iii) lim inf 𝑛 → ∞ 𝑎𝑛 ≤ 𝜇𝑛 (𝑎𝑛 ) ≤ lim sup𝑛 → ∞ 𝑎𝑛 for all (𝑎0 , 𝑎1 , . . .) ∈ 𝑙∞ . Lemma 18 (see [49]). Let 𝑎 ∈ R be a real number and a sequence {𝑎𝑛 } ∈ 𝑙∞ satisfy the condition 𝜇𝑛 (𝑎𝑛 ) ≤ 𝑎 for all Banach limit 𝜇. If lim sup𝑛 → ∞ (𝑎𝑛+𝑟 − 𝑎𝑛 ) ≤ 0, then lim sup𝑛 → ∞ 𝑎𝑛 ≤ 𝑎. In particular, if 𝑟 = 1 in Lemma 18, then we immediately obtain the following corollary. Corollary 19 (see [50]). Let 𝑎 ∈ R be a real number and a sequence {𝑎𝑛 } ∈ 𝑙∞ satisfy the condition 𝜇𝑛 (𝑎𝑛 ) ≤ 𝑎 for all Banach limit 𝜇. If lim sup𝑛 → ∞ (𝑎𝑛+1 − 𝑎𝑛 ) ≤ 0, then, lim sup𝑛 → ∞ 𝑎𝑛 ≤ 𝑎. Lemma 20 (see [51]). Let {𝑥𝑛 } and {𝑧𝑛 } be bounded sequences in a Banach space 𝑋 and let {𝛽𝑛 } be a sequence of nonnegative numbers in [0, 1] with 0 < lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ 𝛽𝑛 < 1. Suppose that 𝑥𝑛+1 = 𝛽𝑛 𝑥𝑛 + (1 − 𝛽𝑛 )𝑧𝑛 for all integers

Abstract and Applied Analysis

7

𝑛 ≥ 0 and lim sup𝑛 → ∞ (‖𝑧𝑛+1 − 𝑧𝑛 ‖ − ‖𝑥𝑛+1 − 𝑥𝑛 ‖) ≤ 0. Then, lim𝑛 → ∞ ‖𝑥𝑛 − 𝑧𝑛 ‖ = 0.

0 < 𝜇𝑖 < 𝛽̂𝑖 /𝜅2 for 𝑖 = 1, 2. For arbitrarily given 𝑥0 ∈ 𝐶, let {𝑥𝑛 } be the sequence generated by

Lemma 21 (see [34]). Let 𝐶 be a nonempty closed convex subset of a smooth Banach space 𝑋. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶 and 𝐴 : 𝐶 → 𝑋 an accretive mapping. Then for all 𝜆 > 0,

𝑦𝑛 = 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 ,

VI (𝐶, 𝐴) = Fix (Π𝐶 (𝐼 − 𝜆𝐴)) .

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Lemma 22 (see [11]). Let 𝐶 be a nonempty closed convex subset of a real 2-uniformly smooth Banach space 𝑋. Let the mapping 𝐵𝑖 : 𝐶 → 𝑋 be 𝛽̂𝑖 -inverse-strongly accretive. Then, one has 󵄩2 󵄩󵄩 󵄩󵄩(𝐼 − 𝜇𝑖 𝐵𝑖 ) 𝑥 − (𝐼 − 𝜇𝑖 𝐵𝑖 ) 𝑦󵄩󵄩󵄩 󵄩2 󵄩 󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 + 2𝜇𝑖 (𝜇𝑖 𝜅2 − 𝛽̂𝑖 ) 󵄩󵄩󵄩𝐵𝑖 𝑥 − 𝐵𝑖 𝑦󵄩󵄩󵄩 ,

for 𝑖 = 1, 2 where 𝜇𝑖 > 0. In particular, if 0 < 𝜇𝑖 ≤ 𝛽̂𝑖 /𝜅2 , then 𝐼 − 𝜇𝑖 𝐵𝑖 is nonexpansive for 𝑖 = 1, 2. Lemma 23 (see [11]). Let 𝐶 be a nonempty closed convex subset of a real 2-uniformly smooth Banach space 𝑋. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶. Let the mapping 𝐵𝑖 : 𝐶 → 𝑋 be 𝛽̂𝑖 -inverse-strongly accretive for 𝑖 = 1, 2. Let 𝐺 : 𝐶 → 𝐶 be the mapping defined by

∀𝑥 ∈ 𝐶.

∀𝑛 ≥ 0,

(42)

where {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, and {𝛿𝑛 } are the sequences in [0, 1] such that 𝛽𝑛 + 𝛾𝑛 + 𝛿𝑛 = 1 for all 𝑛 ≥ 0. Suppose that the following conditions hold: (i) ∑∞ 𝑛=0 𝛼𝑛 = ∞ and 0 ≤ 𝛼𝑛 ≤ 1 − 𝜌, for all 𝑛 ≥ 𝑛0 for some integer 𝑛0 ≥ 0; (ii) lim inf 𝑛 → ∞ 𝛾𝑛 > 0 and lim inf 𝑛 → ∞ 𝛿𝑛 > 0; (iii) lim𝑛 → ∞ (|𝛼𝑛+1 /(1 − (1 − 𝛼𝑛+1 )𝛽𝑛+1 ) − 𝛼𝑛 /(1 − (1 − 𝛼𝑛 )𝛽𝑛 )| + |𝛿𝑛+1 /(1 − 𝛽𝑛+1 ) − 𝛿𝑛 /(1 − 𝛽𝑛 )|) = 0; (iv) 0 < lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ 𝛽𝑛 < 1.

(40)

∀𝑥, 𝑦 ∈ 𝐶,

𝐺𝑥 = Π𝐶 [Π𝐶 (𝑥 − 𝜇2 𝐵2 𝑥) − 𝜇1 𝐵1 Π𝐶 (𝑥 − 𝜇2 𝐵2 𝑥)] ,

𝑥𝑛+1 = 𝛼𝑛 𝑓 (𝑥𝑛 ) + (1 − 𝛼𝑛 ) 𝑦𝑛 ,

(41)

If 0 < 𝜇𝑖 ≤ 𝛽̂𝑖 /𝜅2 for 𝑖 = 1, 2, then 𝐺 : 𝐶 → 𝐶 is nonexpansive.

3. Hybrid Mann Iterations and Their Convergence Criteria In this section, we introduce our hybrid Mann iteration algorithms in real smooth and uniformly convex Banach spaces and present their convergence criteria. Theorem 24. Let 𝐶 be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space 𝑋. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶. Let {𝜌𝑛 }∞ 𝑛=0 be a sequence of positive numbers in (0, 𝑏] for some 𝑏 ∈ (0, 1) and 𝐴 𝑖 : 𝐶 → 𝐸 an 𝛼̂𝑖 -inverse strongly accretive mapping for each 𝑖 = 0, 1, . . .. Define a mapping 𝐺𝑖 : 𝐶 → 𝐶 by Π𝐶(𝐼 − 𝜆 𝑖 𝐴 𝑖 )𝑥 = 𝐺𝑖 𝑥 for all 𝑥 ∈ 𝐶 and 𝑖 = 0, 1, . . ., where 𝜆 𝑖 ∈ (0, 𝛼̂𝑖 /𝜅2 ], 𝜅 is the 2-uniformly smooth constant of 𝑋. Let 𝐵𝑛 : 𝐶 → 𝐶 be the 𝑊-mapping generated by 𝐺𝑛 , 𝐺𝑛−1 , . . . , 𝐺0 and 𝜌𝑛 , 𝜌𝑛−1 , . . . , 𝜌0 . Let the mapping 𝐵𝑖 : 𝐶 → 𝑋 be 𝛽̂𝑖 inverse strongly accretive for 𝑖 = 1, 2. Let 𝑓 : 𝐶 → 𝐶 be a contraction with coefficient 𝜌 ∈ (0, 1). Let {𝑆𝑖 }∞ 𝑖=0 be a countable family of nonexpansive mappings of 𝐶 into itself such that 𝐹 = ∞ (⋂∞ 𝑖=0 Fix(𝑆𝑖 )) ∩ Ω ∩ (⋂𝑖=0 VI(𝐶, 𝐴 𝑖 )) ≠ 0, where Ω is the fixed point set of the mapping 𝐺 = Π𝐶(𝐼 − 𝜇1 𝐵1 )Π𝐶(𝐼 − 𝜇2 𝐵2 ) with

Assume that ∑∞ 𝑛=0 sup𝑥∈𝐷 ‖𝑆𝑛+1 𝑥 − 𝑆𝑛 𝑥‖ < ∞ for any bounded subset 𝐷 of 𝐶 and let 𝑆 be a mapping of 𝐶 into itself defined by 𝑆𝑥 = lim𝑛 → ∞ 𝑆𝑛 𝑥 for all 𝑥 ∈ 𝐶 and suppose that Fix(𝑆) = ⋂∞ 𝑖=0 Fix(𝑆𝑖 ). Then, there hold the following: (I) lim𝑛 → ∞ ‖𝑥𝑛+1 − 𝑥𝑛 ‖ = 0; (II) 𝑥𝑛 → 𝑞 ⇔ 𝛼𝑛 (𝑓(𝑥𝑛 ) − 𝑥𝑛 ) → 0 provided 𝛽𝑛 ≡ 𝛽 for some fixed 𝛽 ∈ (0, 1), where 𝑞 ∈ 𝐹 solves the following VIP: ⟨𝑞 − 𝑓 (𝑞) , 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,

∀𝑝 ∈ 𝐹.

(43)

Proof. First of all, since 0 < 𝜆 𝑖 < 𝛼̂𝑖 /𝜅2 for 𝑖 = 0, 1, . . ., it is easy to see that 𝐺𝑖 is a nonexpansive mapping for each 𝑖 = 0, 1, . . .. Since 𝐵𝑛 : 𝐶 → 𝐶 is the 𝑊-mapping generated by 𝐺𝑛 , 𝐺𝑛−1 , . . . , 𝐺0 and 𝜌𝑛 , 𝜌𝑛−1 , . . . , 𝜌0 , by Lemma 16 we know that, for each 𝑥 ∈ 𝐶 and 𝑘 ≥ 0, the limit lim𝑛 → ∞ 𝑈𝑛,𝑘 𝑥 exists. Moreover, one can define a mapping 𝐵 : 𝐶 → 𝐶 as follows: 𝐵𝑥 = lim 𝐵𝑛 𝑥 = lim 𝑈𝑛,0 𝑥 𝑛→∞

𝑛→∞

(44)

for every 𝑥 ∈ 𝐶. That is, such a 𝐵 is the 𝑊-mapping generated ∞ by the sequences {𝐺𝑛 }∞ 𝑛=0 and {𝜌𝑛 }𝑛=0 . According to Lemma 17, ∞ we know that Fix(𝐵) = ⋂𝑖=0 Fix(𝐺𝑖 ). From Lemma 15 and the definition of 𝐺𝑖 , we have Fix(𝐺𝑖 ) = VI(𝐶, 𝐴 𝑖 ) for each 𝑖 = 0, 1, . . .. Hence, we have ∞

∞

𝑖=0

𝑖=0

Fix (𝐵) = ⋂ Fix (𝐺𝑖 ) = ⋂ VI (𝐶, 𝐴 𝑖 ) .

(45)

Next, let us show that the sequence {𝑥𝑛 } is bounded. Indeed, take a fixed 𝑝 ∈ 𝐹 arbitrarily. Then, we get 𝑝 = 𝐺𝑝, 𝑝 = 𝐵𝑛 𝑝, and 𝑝 = 𝑆𝑛 𝑝 for all 𝑛 ≥ 0. By Lemma 23 we know that 𝐺 is nonexpansive. Then, from (42), we have 󵄩 󵄩 󵄩 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 , (46)

8

Abstract and Applied Analysis

and hence

=(

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩 󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩

󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 (󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑓 (𝑝)󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩)

−

󵄩 󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩

+ 𝛿𝑛+1 𝑆𝑛+1 𝐺𝑥𝑛+1 ]

󵄩 󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩

− 𝜎𝑛+1 𝑥𝑛+1 × (1 − 𝜎𝑛+1 )

(47)

󵄩 󵄩 = (1 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩

=(

󵄩 󵄩󵄩 󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 + 𝛼𝑛 (1 − 𝜌) 󵄩 1−𝜌 󵄩 󵄩 󵄩 󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩 ≤ max {󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 , 󵄩 }. 1−𝜌

− =(

∀𝑛 ≥ 0. (48)

+

Thus, {𝑥𝑛 } is bounded, and so are the sequences {𝑦𝑛 }, {𝐺𝑥𝑛 } and {𝑓(𝑥𝑛 )}. Let us show that 󵄩 󵄩 lim 󵄩󵄩𝑥 − 𝑥𝑛 󵄩󵄩󵄩 = 0. 𝑛 → ∞ 󵄩 𝑛+1

(1 − 𝛼𝑛+1 ) (1 − 𝛽𝑛+1 ) 1 − 𝜎𝑛+1

+[ ×

∀𝑛 ≥ 𝑛0 , (50)

(51)

(52)

𝛾𝑛+1 𝐵𝑛+1 𝑥𝑛+1 + 𝛿𝑛+1 𝑆𝑛+1 𝐺𝑥𝑛+1 1 − 𝛽𝑛+1 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 ] 1 − 𝛽𝑛

(1 − 𝛼𝑛+1 ) (1 − 𝛽𝑛+1 ) (1 − 𝛼𝑛 ) (1 − 𝛽𝑛 ) − ] 1 − 𝜎𝑛+1 1 − 𝜎𝑛

𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 1 − 𝛽𝑛

𝛼𝑛+1 (𝑓 (𝑥𝑛+1 ) − 𝑓 (𝑥𝑛 )) 1 − 𝜎𝑛+1 +( +

Define 𝑥𝑛+1 = 𝜎𝑛 𝑥𝑛 + (1 − 𝜎𝑛 ) 𝑧𝑛 .

𝛼𝑛+1 𝑓 (𝑥𝑛+1 ) 𝛼𝑛 𝑓 (𝑥𝑛 ) − ) 1 − 𝜎𝑛+1 1 − 𝜎𝑛

(49)

and hence 𝑛→∞

1 − 𝛼𝑛 (𝛾 𝐵 𝑥 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 ) 1 − 𝜎𝑛 𝑛 𝑛 𝑛

−

= 0 < lim inf 𝜎𝑛 ≤ lim sup 𝜎𝑛 < 1.

𝛼𝑛+1 𝑓 (𝑥𝑛+1 ) 𝛼𝑛 𝑓 (𝑥𝑛 ) 1 − 𝛼𝑛+1 − )+ 1 − 𝜎𝑛+1 1 − 𝜎𝑛 1 − 𝜎𝑛+1

×[

As a matter of fact, put 𝜎𝑛 = (1−𝛼𝑛 )𝛽𝑛 , for all 𝑛 ≥ 0. Then, it follows from (i) and (iv) that 𝛽𝑛 ≥ 𝜎𝑛 = (1 − 𝛼𝑛 ) 𝛽𝑛 ≥ (1 − (1 − 𝜌)) 𝛽𝑛 = 𝜌𝛽𝑛 ,

𝛼𝑛+1 𝛼𝑛 − ) 𝑓 (𝑥𝑛 ) 1 − 𝜎𝑛+1 1 − 𝜎𝑛

(1 − 𝛼𝑛+1 ) (1 − 𝛽𝑛+1 ) 1 − 𝜎𝑛+1

×[

Observe that

𝛾𝑛+1 (𝐵 𝑥 − 𝐵𝑛 𝑥𝑛 ) 𝛾𝑛+1 + 𝛿𝑛+1 𝑛+1 𝑛+1 +(

𝑧𝑛+1 − 𝑧𝑛 𝑥𝑛+2 − 𝜎𝑛+1 𝑥𝑛+1 𝑥𝑛+1 − 𝜎𝑛 𝑥𝑛 − 1 − 𝜎𝑛+1 1 − 𝜎𝑛

+

=

𝛼𝑛+1 𝑓 (𝑥𝑛+1 ) + (1 − 𝛼𝑛+1 ) 𝑦𝑛+1 − 𝜎𝑛+1 𝑥𝑛+1 1 − 𝜎𝑛+1

+(

𝛼𝑛 𝑓 (𝑥𝑛 ) + (1 − 𝛼𝑛 ) 𝑦𝑛 − 𝜎𝑛 𝑥𝑛 1 − 𝜎𝑛

−(

𝛾𝑛 𝛾𝑛+1 − ) 𝐵𝑛 𝑥𝑛 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛

𝛿𝑛+1 (𝑆 𝐺𝑥 − 𝑆𝑛 𝐺𝑥𝑛 ) 𝛾𝑛+1 + 𝛿𝑛+1 𝑛+1 𝑛+1

=

−

−1

× (𝛾𝑛+1 𝐵𝑛+1 𝑥𝑛+1 + 𝛿𝑛+1 𝑆𝑛+1 𝐺𝑥𝑛+1 )

By induction, we obtain

𝑛→∞

(1 − 𝛼𝑛 ) [𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 ] − 𝜎𝑛 𝑥𝑛 1 − 𝜎𝑛

+ (1 − 𝛼𝑛+1 ) [𝛽𝑛+1 𝑥𝑛+1 + 𝛾𝑛+1 𝐵𝑛+1 𝑥𝑛+1

󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 (𝜌 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩)

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩 }, 󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 ≤ max {󵄩󵄩󵄩𝑥0 − 𝑝󵄩󵄩󵄩 , 󵄩 1−𝜌

𝛼𝑛+1 𝑓 (𝑥𝑛+1 ) 𝛼𝑛 𝑓 (𝑥𝑛 ) − ) 1 − 𝜎𝑛+1 1 − 𝜎𝑛

𝛿𝑛+1 𝛿𝑛 − ) 𝑆 𝐺𝑥 ] 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 𝑛 𝑛

𝛾 𝐵 𝑥 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 𝛼𝑛+1 𝛼𝑛 − ) 𝑛 𝑛 𝑛 1 − 𝜎𝑛+1 1 − 𝜎𝑛 𝛾𝑛 + 𝛿𝑛

Abstract and Applied Analysis =

9

𝛼𝑛+1 (𝑓 (𝑥𝑛+1 ) − 𝑓 (𝑥𝑛 )) 1 − 𝜎𝑛+1 +(

𝛼𝑛+1 𝛼𝑛 − ) 1 − 𝜎𝑛+1 1 − 𝜎𝑛

× (𝑓 (𝑥𝑛 ) − +

𝛿𝑛+1 󵄩󵄩 󵄩 󵄩𝑆 𝐺𝑥 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 𝛾𝑛+1 + 𝛿𝑛+1 󵄩 𝑛+1 𝑛+1 󵄨󵄨 𝛿 𝛿𝑛 󵄨󵄨󵄨󵄨 󵄩󵄩 󵄨 󵄩 𝑛+1 + 󵄨󵄨󵄨 − 󵄨 󵄩𝑆 𝐺𝑥 󵄩󵄩] . 󵄨󵄨 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 󵄨󵄨󵄨 󵄩 𝑛 𝑛 󵄩

+

𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 ) 𝛾𝑛 + 𝛿𝑛

(54)

1 − 𝜎𝑛+1 − 𝛼𝑛+1 1 − 𝜎𝑛+1

×[

On the other hand, we note that, for all 𝑛 ≥ 0,

𝛾𝑛+1 (𝐵 𝑥 − 𝐵𝑛 𝑥𝑛 ) 𝛾𝑛+1 + 𝛿𝑛+1 𝑛+1 𝑛+1

󵄩󵄩 󵄩 󵄩󵄩𝑆𝑛+1 𝐺𝑥𝑛+1 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩

󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑆𝑛+1 𝐺𝑥𝑛+1 − 𝑆𝑛+1 𝐺𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛+1 𝐺𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩

𝛾𝑛 𝛾𝑛+1 − ) 𝐵𝑛 𝑥𝑛 +( 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 +

󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝐺𝑥𝑛+1 − 𝐺𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛+1 𝐺𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛+1 𝐺𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 .

𝛿𝑛+1 (𝑆 𝐺𝑥 − 𝑆𝑛 𝐺𝑥𝑛 ) 𝛾𝑛+1 + 𝛿𝑛+1 𝑛+1 𝑛+1

+(

𝛿𝑛+1 𝛿𝑛 − ) 𝑆 𝐺𝑥 ] , 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 𝑛 𝑛 (53)

and hence 󵄩 󵄩󵄩 󵄩󵄩𝑧𝑛+1 − 𝑧𝑛 󵄩󵄩󵄩 ≤

𝛼𝑛+1 󵄩󵄩 󵄩 󵄩𝑓 (𝑥𝑛+1 ) − 𝑓 (𝑥𝑛 )󵄩󵄩󵄩 1 − 𝜎𝑛+1 󵄩 󵄨󵄨 𝛼 𝛾 𝐵 𝑥 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩󵄩 𝛼𝑛 󵄨󵄨󵄨󵄨 󵄩󵄩󵄩󵄩 󵄨 + 󵄨󵄨󵄨 𝑛+1 − 󵄨󵄨 󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑛 𝑛 𝑛 󵄩󵄩 󵄨󵄨 1 − 𝜎𝑛+1 1 − 𝜎𝑛 󵄨󵄨 󵄩󵄩 󵄩󵄩 𝛾𝑛 + 𝛿𝑛 1 − 𝜎𝑛+1 − 𝛼𝑛+1 1 − 𝜎𝑛+1 󵄩󵄩 𝛾 󵄩 𝑛+1 × 󵄩󵄩󵄩 (𝐵 𝑥 − 𝐵𝑛 𝑥𝑛 ) 󵄩󵄩 𝛾𝑛+1 + 𝛿𝑛+1 𝑛+1 𝑛+1 +

𝛾𝑛 𝛾𝑛+1 − ) 𝐵𝑛 𝑥𝑛 +( 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 𝛿𝑛+1 (𝑆 𝐺𝑥 − 𝑆𝑛 𝐺𝑥𝑛 ) 𝛾𝑛+1 + 𝛿𝑛+1 𝑛+1 𝑛+1 󵄩󵄩 𝛿𝑛+1 𝛿𝑛 󵄩 − ) 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 +( 󵄩󵄩 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 󵄨 𝜌𝛼𝑛+1 󵄩󵄩 𝛼𝑛 󵄨󵄨󵄨󵄨 󵄩 󵄨󵄨 𝛼 ≤ 󵄨󵄨 󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄨󵄨󵄨 𝑛+1 − 1 − 𝜎𝑛+1 󵄨󵄨 1 − 𝜎𝑛+1 1 − 𝜎𝑛 󵄨󵄨 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 × (󵄩󵄩󵄩𝑓 (𝑥𝑛 )󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩) +

1 − 𝜎𝑛+1 − 𝛼𝑛+1 + 1 − 𝜎𝑛+1 ×[

𝛾𝑛+1 󵄩󵄩 󵄩 󵄩𝐵 𝑥 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 𝛾𝑛+1 + 𝛿𝑛+1 󵄩 𝑛+1 𝑛+1 󵄨󵄨󵄨 𝛾𝑛+1 𝛾𝑛 󵄨󵄨󵄨󵄨 󵄩󵄩 󵄩 + 󵄨󵄨󵄨 − 󵄨󵄨 󵄩󵄩𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 󵄨󵄨 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 󵄨󵄨

(55)

Furthermore, by (CY), since 𝐺𝑖 and 𝑈𝑛,𝑖 are nonexpansive, we deduce that for each 𝑛 ≥ 0 󵄩󵄩 󵄩 󵄩󵄩𝐵𝑛+1 𝑥𝑛+1 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝐵𝑛+1 𝑥𝑛+1 − 𝐵𝑛+1 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛+1 𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛+1 𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝜆 0 𝐺0 𝑈𝑛+1,1 𝑥𝑛 − 𝜆 0 𝐺0 𝑈𝑛,1 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 𝜆 0 󵄩󵄩󵄩𝑈𝑛+1,1 𝑥𝑛 − 𝑈𝑛,1 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 𝜆 0 󵄩󵄩󵄩𝜆 1 𝐺1 𝑈𝑛+1,2 𝑥𝑛 − 𝜆 1 𝐺1 𝑈𝑛,2 𝑥𝑛 󵄩󵄩󵄩 (56) 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 𝜆 0 𝜆 1 󵄩󵄩󵄩𝑈𝑛+1,2 𝑥𝑛 − 𝑈𝑛,2 𝑥𝑛 󵄩󵄩󵄩 .. . 𝑛

󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + (∏𝜆 𝑖 ) 󵄩󵄩󵄩𝑈𝑛+1,𝑛+1 𝑥𝑛 − 𝑈𝑛,𝑛+1 𝑥𝑛 󵄩󵄩󵄩 𝑖=0

𝑛

󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 𝑀0 ∏𝜆 𝑖 , 𝑖=0

for some constant 𝑀0 > 0. Utilizing (54)–(56), we have 󵄩󵄩 󵄩 󵄩󵄩𝑧𝑛+1 − 𝑧𝑛 󵄩󵄩󵄩 ≤

󵄨 𝜌𝛼𝑛+1 󵄩󵄩 𝛼𝑛 󵄨󵄨󵄨󵄨 󵄩 󵄨󵄨 𝛼 󵄨 󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄨󵄨󵄨 𝑛+1 − 󵄨󵄨 1 − 𝜎𝑛+1 1 − 𝜎𝑛 󵄨󵄨󵄨 1 − 𝜎𝑛+1 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 × (󵄩󵄩󵄩𝑓 (𝑥𝑛 )󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩)

+

1 − 𝜎𝑛+1 − 𝛼𝑛+1 1 − 𝜎𝑛+1

10

Abstract and Applied Analysis ×[

𝑛 𝛾𝑛+1 󵄩 󵄩 (󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 𝑀0 ∏𝜆 𝑖 ) 𝛾𝑛+1 + 𝛿𝑛+1 𝑖=0

where sup𝑛≥0 {‖𝑓(𝑥𝑛 )‖+‖𝐵𝑛 𝑥𝑛 ‖+‖𝑆𝑛 𝐺𝑥𝑛 ‖+𝑀0 } ≤ 𝑀 for some 𝑀 > 0. So, from (58), condition (iii), and the assumption on {𝑆𝑛 }, it follows that (noting that 0 < 𝜆 𝑖 ≤ 𝑏 < 1, for all 𝑖 ≥ 0)

󵄨󵄨 𝛾 𝛾𝑛 󵄨󵄨󵄨󵄨 󵄩󵄩 󵄨 󵄩 𝑛+1 + 󵄨󵄨󵄨 − 󵄨 󵄩𝐵 𝑥 󵄩󵄩 󵄨󵄨 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 󵄨󵄨󵄨 󵄩 𝑛 𝑛 󵄩 +

󵄩 󵄩 󵄩 󵄩 lim sup (󵄩󵄩󵄩𝑧𝑛+1 − 𝑧𝑛 󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩) ≤ 0.

𝛿𝑛+1 󵄩 󵄩 󵄩 󵄩 (󵄩󵄩𝑥 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛+1 𝐺𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩) 𝛾𝑛+1 + 𝛿𝑛+1 󵄩 𝑛+1

lim 󵄩𝑧 𝑛→∞ 󵄩 𝑛

1 − 𝜎𝑛+1 − 𝛼𝑛+1 (1 − 𝜌) 󵄩󵄩 󵄩 = 󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 1 − 𝜎𝑛+1 󵄨󵄨 𝛼 𝛼𝑛 󵄨󵄨󵄨󵄨 󵄩󵄩 󵄨 󵄩 󵄩 󵄩 󵄩 󵄩 + 󵄨󵄨󵄨 𝑛+1 − 󵄨 (󵄩𝑓 (𝑥𝑛 )󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩) 󵄨󵄨 1 − 𝜎𝑛+1 1 − 𝜎𝑛 󵄨󵄨󵄨 󵄩

󵄩 − 𝑥𝑛 󵄩󵄩󵄩 = 0.

(60)

It follows from (51) and (52) that 󵄩󵄩

lim 󵄩𝑥 𝑛 → ∞ 󵄩 𝑛+1

󵄩 󵄩 󵄩 − 𝑥𝑛 󵄩󵄩󵄩 = lim (1 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑧𝑛 − 𝑥𝑛 󵄩󵄩󵄩 = 0. 𝑛→∞

(61)

From (42), we have 𝑥𝑛+1 − 𝑥𝑛 = 𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 ) + (1 − 𝛼𝑛 ) (𝑦𝑛 − 𝑥𝑛 ) ,

1 − 𝜎𝑛+1 − 𝛼𝑛+1 1 − 𝜎𝑛+1

(59)

Consequently, by Lemma 20, we have 󵄩󵄩

󵄨󵄨 𝛿 𝛿𝑛 󵄨󵄨󵄨󵄨 󵄩󵄩 󵄨 󵄩 𝑛+1 + 󵄨󵄨󵄨 − 󵄨 󵄩𝑆 𝐺𝑥 󵄩󵄩 ] 󵄨󵄨 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 󵄨󵄨󵄨 󵄩 𝑛 𝑛 󵄩

+

𝑛→∞

(62)

which hence implies that

𝑛 𝛾𝑛+1 ×[ 𝑀0 ∏𝜆 𝑖 𝛾𝑛+1 + 𝛿𝑛+1 𝑖=0

󵄩 󵄩 󵄩 󵄩 𝜌 󵄩󵄩󵄩𝑦𝑛 − 𝑥𝑛 󵄩󵄩󵄩 = (1 − (1 − 𝜌)) 󵄩󵄩󵄩𝑦𝑛 − 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 ≤ (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑥𝑛 󵄩󵄩󵄩

𝛿𝑛+1 󵄩󵄩 󵄩 + 󵄩𝑆 𝐺𝑥 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 𝛾𝑛+1 + 𝛿𝑛+1 󵄩 𝑛+1 𝑛 󵄨󵄨 𝛿 𝛿𝑛 󵄨󵄨󵄨󵄨 󵄩󵄩 󵄨 󵄩 󵄩 󵄩 𝑛+1 + 󵄨󵄨󵄨 − 󵄨 (󵄩𝐵 𝑥 󵄩󵄩 + 󵄩󵄩𝑆 𝐺𝑥 󵄩󵄩) ] 󵄨󵄨 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 󵄨󵄨󵄨 󵄩 𝑛 𝑛 󵄩 󵄩 𝑛 𝑛 󵄩 󵄨 𝛼𝑛 󵄨󵄨󵄨󵄨 󵄩 󵄩 󵄨󵄨 𝛼 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄨󵄨󵄨 𝑛+1 − 󵄨𝑀 󵄨󵄨 1 − 𝜎𝑛+1 1 − 𝜎𝑛 󵄨󵄨󵄨 𝑛

󵄩 󵄩 + 𝑀∏𝜆 𝑖 + 󵄩󵄩󵄩𝑆𝑛+1 𝐺𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩

󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 − 𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩

(63)

󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩 . Since 𝑥𝑛+1 − 𝑥𝑛 → 0 and 𝛼𝑛 (𝑓(𝑥𝑛 ) − 𝑥𝑛 ) → 0, we get 󵄩󵄩

lim 󵄩𝑦 𝑛→∞ 󵄩 𝑛

󵄩 − 𝑥𝑛 󵄩󵄩󵄩 = 0.

(64)

Next, we show that ‖𝑥𝑛 − 𝐺𝑥𝑛 ‖ → 0 as 𝑛 → ∞. Indeed, for simplicity, put 𝑞 = Π𝐶(𝑝 − 𝜇2 𝐵2 𝑝), 𝑢𝑛 = Π𝐶(𝑥𝑛 − 𝜇2 𝐵2 𝑥𝑛 ) and V𝑛 = Π𝐶(𝑢𝑛 − 𝜇1 𝐵1 𝑢𝑛 ). Then, V𝑛 = 𝐺𝑥𝑛 for all 𝑛 ≥ 0. From Lemma 22, we have

𝑖=0

󵄨󵄨 𝛿 𝛿𝑛 󵄨󵄨󵄨󵄨 󵄨 𝑛+1 + 󵄨󵄨󵄨 − 󵄨𝑀 󵄨󵄨 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 󵄨󵄨󵄨 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 󵄨󵄨 𝛼 𝛼𝑛 󵄨󵄨󵄨󵄨 󵄨 + 𝑀 ( 󵄨󵄨󵄨 𝑛+1 − 󵄨󵄨 󵄨󵄨 1 − 𝜎𝑛+1 1 − 𝜎𝑛 󵄨󵄨

󵄩2 󵄩 󵄩2 󵄩󵄩 󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 = 󵄩󵄩󵄩Π𝐶 (𝑥𝑛 − 𝜇2 𝐵2 𝑥𝑛 ) − Π𝐶 (𝑝 − 𝜇2 𝐵2 𝑝)󵄩󵄩󵄩 󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝 − 𝜇2 (𝐵2 𝑥𝑛 − 𝐵2 𝑝)󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 2𝜇2 (𝛽̂2 − 𝜅2 𝜇2 ) 󵄩󵄩󵄩𝐵2 𝑥𝑛 − 𝐵2 𝑝󵄩󵄩󵄩 , (65)

󵄨󵄨 𝛿 𝛿𝑛 󵄨󵄨󵄨󵄨 𝑛 󵄨 𝑛+1 + 󵄨󵄨󵄨 − 󵄨 + ∏𝜆 ) 󵄨󵄨 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 󵄨󵄨󵄨 𝑖=0 𝑖 󵄩 󵄩 + 󵄩󵄩󵄩𝑆𝑛+1 𝐺𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 , (57) which hence yields 󵄩󵄩󵄩𝑧𝑛+1 − 𝑧𝑛 󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄨󵄨 𝛼 𝛼𝑛 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 𝛿𝑛+1 𝛿𝑛 󵄨󵄨󵄨󵄨 󵄨 ≤ 𝑀 ( 󵄨󵄨󵄨 𝑛+1 − − 󵄨󵄨 + 󵄨󵄨 󵄨 󵄨󵄨 1 − 𝜎𝑛+1 1 − 𝜎𝑛 󵄨󵄨 󵄨󵄨 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 󵄨󵄨󵄨 𝑛

󵄩 󵄩 +∏𝜆 𝑖 ) + 󵄩󵄩󵄩𝑆𝑛+1 𝐺𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 , 𝑖=0

(58)

󵄩󵄩 󵄩2 󵄩 󵄩2 󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 = 󵄩󵄩󵄩Π𝐶 (𝑢𝑛 − 𝜇1 𝐵1 𝑢𝑛 ) − Π𝐶 (𝑞 − 𝜇1 𝐵1 𝑞)󵄩󵄩󵄩 󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝑢𝑛 − 𝑞 − 𝜇1 (𝐵1 𝑢𝑛 − 𝐵1 𝑞)󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 − 2𝜇1 (𝛽̂1 − 𝜅2 𝜇1 ) 󵄩󵄩󵄩𝐵1 𝑢𝑛 − 𝐵1 𝑞󵄩󵄩󵄩 . (66) Substituting (65) for (66), we obtain 󵄩󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 2 󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 2𝜇2 (𝛽̂2 − 𝜅 𝜇2 ) 󵄩󵄩󵄩𝐵2 𝑥𝑛 − 𝐵2 𝑝󵄩󵄩󵄩 󵄩 󵄩2 − 2𝜇1 (𝛽̂1 − 𝜅2 𝜇1 ) 󵄩󵄩󵄩𝐵1 𝑢𝑛 − 𝐵1 𝑞󵄩󵄩󵄩 .

(67)

Abstract and Applied Analysis

11

From (42) and (67), we have

In the same way, we derive

󵄩󵄩 󵄩2 󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩2 󵄩2 󵄩2 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩2 󵄩2 󵄩2 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩2 + 𝛿𝑛 [󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 2𝜇2 (𝛽̂2 − 𝜅2 𝜇2 ) 󵄩2 󵄩 × 󵄩󵄩󵄩𝐵2 𝑥𝑛 − 𝐵2 𝑝󵄩󵄩󵄩

󵄩2 󵄩󵄩 󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩2 = 󵄩󵄩󵄩Π𝐶 (𝑢𝑛 − 𝜇1 𝐵1 𝑢𝑛 ) − Π𝐶 (𝑞 − 𝜇1 𝐵1 𝑞)󵄩󵄩󵄩 ≤ ⟨𝑢𝑛 − 𝜇1 𝐵1 𝑢𝑛 − (𝑞 − 𝜇1 𝐵1 𝑞) , 𝐽 (V𝑛 − 𝑝)⟩ = ⟨𝑢𝑛 − 𝑞, 𝐽 (V𝑛 − 𝑝)⟩ + 𝜇1 ⟨𝐵1 𝑞 − 𝐵1 𝑢𝑛 , 𝐽 (V𝑛 − 𝑝)⟩ ≤ (68)

󵄩 󵄩2 − 2𝜇1 (𝛽̂1 − 𝜅2 𝜇1 ) 󵄩󵄩󵄩𝐵1 𝑢𝑛 − 𝐵1 𝑞󵄩󵄩󵄩 ]

1 󵄩󵄩 󵄩2 󵄩 󵄩2 [󵄩𝑢 − 𝑞󵄩󵄩󵄩 + 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 2 󵄩 𝑛 󵄩 󵄩 − 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩) ] 󵄩 󵄩󵄩 󵄩 + 𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 ,

󵄩2 󵄩 = 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 which implies that

󵄩 󵄩2 − 2𝛿𝑛 [𝜇2 (𝛽̂2 − 𝜅2 𝜇2 ) 󵄩󵄩󵄩𝐵2 𝑥𝑛 − 𝐵2 𝑝󵄩󵄩󵄩

󵄩󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩 󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 − 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩) 󵄩 󵄩󵄩 󵄩 + 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 .

󵄩 󵄩2 +2𝜇1 (𝛽̂1 − 𝜅2 𝜇1 ) 󵄩󵄩󵄩𝐵1 𝑢𝑛 − 𝐵1 𝑞󵄩󵄩󵄩 ] , which hence implies that

󵄩2 󵄩 󵄩2 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩 .

(69)

󵄩󵄩

󵄩 − 𝐵2 𝑝󵄩󵄩󵄩 = 0,

󵄩󵄩

lim 󵄩𝐵 𝑢 𝑛→∞ 󵄩 1 𝑛

󵄩 − 𝐵1 𝑞󵄩󵄩󵄩 = 0. (70)

Utilizing Proposition 6 and Lemma 9, we have 󵄩󵄩 󵄩2 󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 󵄩 󵄩2 = 󵄩󵄩󵄩Π𝐶 (𝑥𝑛 − 𝜇2 𝐵2 𝑥𝑛 ) − Π𝐶 (𝑝 − 𝜇2 𝐵2 𝑝)󵄩󵄩󵄩 ≤ ⟨𝑥𝑛 − 𝜇2 𝐵2 𝑥𝑛 − (𝑝 − 𝜇2 𝐵2 𝑝) , 𝐽 (𝑢𝑛 − 𝑞)⟩ = ⟨𝑥𝑛 − 𝑝, 𝐽 (𝑢𝑛 − 𝑞)⟩ + 𝜇2 ⟨𝐵2 𝑝 − 𝐵2 𝑥𝑛 , 𝐽 (𝑢𝑛 − 𝑞)⟩ ≤

1 󵄩󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩 [󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 − 𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩)] 2 󵄩 󵄩󵄩 󵄩 + 𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 , (71)

(75)

󵄩 󵄩 󵄩󵄩 + 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩

󵄩 󵄩󵄩 󵄩 + 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 .

By Lemma 8(i), we have from (68) and (75) 󵄩󵄩 󵄩2 󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩2 󵄩2 󵄩2 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩 + 𝛿𝑛 [󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩) 󵄩 󵄩 󵄩 󵄩 − 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩) + 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩󵄩 󵄩 󵄩 × 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 + 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 ] 󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩

󵄩 󵄩 − 𝛿𝑛 [𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩)

󵄩 󵄩 +𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩)]

which implies that 󵄩󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩 󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩) 󵄩 󵄩󵄩 󵄩 + 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 .

󵄩󵄩 󵄩2 󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩) 󵄩 󵄩 − 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩)

Since 0 < 𝜇𝑖 < 𝛽̂𝑖 /𝜅2 for 𝑖 = 1, 2, and {𝑥𝑛 }, {𝑦𝑛 } are bounded, we obtain from (64), (69), and condition (ii) that lim 󵄩𝐵 𝑥 𝑛→∞ 󵄩 2 𝑛

(74)

Substituting (72) for (74), we get

󵄩 󵄩2 2𝛿𝑛 [𝜇2 (𝛽̂2 − 𝜅2 𝜇2 ) 󵄩󵄩󵄩𝐵2 𝑥𝑛 − 𝐵2 𝑝󵄩󵄩󵄩 󵄩 󵄩2 +𝜇1 (𝛽̂1 − 𝜅2 𝜇1 ) 󵄩󵄩󵄩𝐵1 𝑢𝑛 − 𝐵1 𝑞󵄩󵄩󵄩 ]

(73)

󵄩 󵄩󵄩 󵄩 + 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 (72)

󵄩 󵄩󵄩 󵄩 + 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 ,

(76)

12

Abstract and Applied Analysis 󵄩 󵄩2 = 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (𝛽𝑛 + 𝛾𝑛 )

which hence leads to 󵄩 󵄩 󵄩 󵄩 𝛿𝑛 [𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩) + 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩)] 󵄩 󵄩 󵄩󵄩 󵄩2 󵄩 󵄩2 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 󵄩 󵄩󵄩 󵄩 + 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩

󵄩󵄩 𝛽 󵄩󵄩2 𝛾𝑛 󵄩 󵄩 𝑛 × 󵄩󵄩󵄩 (𝑥𝑛 − 𝑝) + (𝐵𝑛 𝑥𝑛 − 𝑝)󵄩󵄩󵄩 󵄩󵄩 𝛽𝑛 + 𝛾𝑛 󵄩󵄩 𝛽𝑛 + 𝛾𝑛 󵄩 󵄩2 ≤ 𝛿𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (𝛽𝑛 + 𝛾𝑛 ) ×[

󵄩 󵄩󵄩 󵄩 + 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩

󵄩 󵄩󵄩 󵄩 + 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 .

− (77)

From (70), (77), condition (ii), and the boundedness of {𝑥𝑛 }, {𝑦𝑛 }, {𝑢𝑛 }, and {V𝑛 }, we deduce that lim 𝑔 𝑛→∞ 1

󵄩 󵄩 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩) = 0,

󵄩 󵄩 lim 𝑔 (󵄩󵄩𝑢 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩) = 0. 𝑛→∞ 2 󵄩 𝑛

(78)

󵄩 󵄩 lim 󵄩󵄩𝑢 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩 = 0. 𝑛→∞ 󵄩 𝑛

𝛽𝑛 𝛾𝑛

2

(𝛽𝑛 + 𝛾𝑛 )

󵄩 󵄩 𝑔3 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩)]

󵄩 󵄩 󵄩 󵄩2 󵄩2 󵄩2 ≤ 𝛿𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 −

𝛽𝑛 𝛾𝑛 󵄩 󵄩 𝑔 (󵄩󵄩𝑥 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) 𝛽𝑛 + 𝛾𝑛 3 󵄩 𝑛

𝛽𝛾 󵄩 󵄩 󵄩2 󵄩 = 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝑛 𝑛 𝑔3 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) , 𝛽𝑛 + 𝛾𝑛 (83) which immediately implies that

Utilizing the properties of 𝑔1 and 𝑔2 , we deduce that 󵄩 󵄩 lim 󵄩󵄩𝑥 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩 = 0, 𝑛→∞ 󵄩 𝑛

𝛽𝑛 󵄩󵄩 𝛾𝑛 󵄩󵄩 󵄩2 󵄩2 󵄩𝑥 − 𝑝󵄩󵄩󵄩 + 󵄩𝐵 𝑥 − 𝑝󵄩󵄩󵄩 𝛽𝑛 + 𝛾𝑛 󵄩 𝑛 𝛽𝑛 + 𝛾𝑛 󵄩 𝑛 𝑛

󵄩 󵄩 𝛽𝑛 𝛾𝑛 𝑔3 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) (79)

≤

󵄩 󵄩2 󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩 .

From (79), we get 󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝑥𝑛 − V𝑛 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩 󵄩 󵄩 + 󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩 󳨀→ 0 as 𝑛 󳨀→ ∞. (80) That is,

𝛽𝑛 𝛾𝑛 󵄩 󵄩 𝑔 (󵄩󵄩𝑥 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) 𝛽𝑛 + 𝛾𝑛 3 󵄩 𝑛

(84)

So, from (64), the boundedness of {𝑥𝑛 }, {𝑦𝑛 }, and conditions (ii), (iv), it follows that lim 𝑔 𝑛→∞ 3

󵄩 󵄩 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) = 0.

(85)

From the properties of 𝑔3 , we have 󵄩󵄩

lim 󵄩𝑥 𝑛→∞ 󵄩 𝑛

󵄩 − 𝐺𝑥𝑛 󵄩󵄩󵄩 = 0.

󵄩󵄩

󵄩 − 𝑥𝑛 󵄩󵄩󵄩 = 0,

󵄩 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 = 0.

(86)

Taking into account that

Next, let us show that lim 󵄩𝐵 𝑥 𝑛→∞ 󵄩 𝑛 𝑛

󵄩󵄩

lim 󵄩𝑥 𝑛→∞ 󵄩 𝑛

(81)

󵄩 󵄩 lim 󵄩󵄩𝑆𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 = 0.

𝑛→∞ 󵄩

(82)

𝑦𝑛 − 𝑥𝑛 = 𝛾𝑛 (𝐵𝑛 𝑥𝑛 − 𝑥𝑛 ) + 𝛿𝑛 (𝑆𝑛 𝐺𝑥𝑛 − 𝑥𝑛 ) , we have 󵄩 󵄩 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝑦𝑛 − 𝑥𝑛 − 𝛾𝑛 (𝐵𝑛 𝑥𝑛 − 𝑥𝑛 )󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑦𝑛 − 𝑥𝑛 󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑦𝑛 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 .

Indeed, utilizing Lemma 15 and (42), we have 󵄩󵄩 󵄩2 󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩2 𝛽 𝑥 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 󵄩 󵄩 = 󵄩󵄩󵄩𝛿𝑛 (𝑆𝑛 𝐺𝑥𝑛 − 𝑝) + (𝛽𝑛 + 𝛾𝑛 ) ( 𝑛 𝑛 − 𝑝)󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 𝛽𝑛 + 𝛾𝑛 󵄩 󵄩2 ≤ 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (𝛽𝑛 + 𝛾𝑛 ) 󵄩󵄩󵄩2 󵄩󵄩󵄩 𝛽 𝑥 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 − 𝑝󵄩󵄩󵄩 × 󵄩󵄩󵄩 𝑛 𝑛 𝛽𝑛 + 𝛾𝑛 󵄩󵄩 󵄩󵄩

(87)

(88)

From (64), (86), and condition (ii), it follows that 󵄩󵄩

lim 󵄩𝑆 𝐺𝑥𝑛 𝑛→∞ 󵄩 𝑛

󵄩 − 𝑥𝑛 󵄩󵄩󵄩 = 0,

󵄩󵄩

lim 󵄩𝑆 𝐺𝑥𝑛 𝑛→∞ 󵄩 𝑛

󵄩 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 = 0. (89)

Abstract and Applied Analysis

13 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑥𝑛 󵄩󵄩󵄩 + 𝛼𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛼𝑛 )

Note that

󵄩 󵄩 × [𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩

󵄩󵄩 󵄩 󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩

󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑆𝑛 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩

(90)

󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩 .

So, in terms of (81), (89), and Lemma 12, we have 󵄩󵄩

lim 󵄩𝑥 𝑛→∞ 󵄩 𝑛

󵄩 − 𝑆𝑥𝑛 󵄩󵄩󵄩 = 0.

(91)

Suppose that 𝛽𝑛 ≡ 𝛽 for some fixed 𝛽 ∈ (0, 1) such that 𝛽 + 𝛾𝑛 + 𝛿𝑛 = 1 for all 𝑛 ≥ 0. Define a mapping 𝑉𝑥 = (1 − 𝜃1 − 𝜃2 )𝑆𝑥 + 𝜃1 𝐵𝑥 + 𝜃2 𝐺𝑥, where 𝜃1 , 𝜃2 ∈ (0, 1) are two constants with 𝜃1 + 𝜃2 < 1. Then, by Lemmas 14 and 17, we have that Fix(𝑉) = Fix(𝑆) ∩ Fix(𝐵) ∩ Fix(𝐺) = 𝐹. For each 𝑘 ≥ 1, let {𝑝𝑘 } be a unique element of 𝐶 such that 𝑝𝑘 =

1 1 𝑓 (𝑝𝑘 ) + (1 − ) 𝑉𝑝𝑘 . 𝑘 𝑘

(92)

From Lemma 13, we conclude that 𝑝𝑘 → 𝑞 ∈ Fix(𝑉) = 𝐹 as 𝑘 → ∞. Observe that for every 𝑛, 𝑘 󵄩 󵄩󵄩 󵄩󵄩𝑦𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩

󵄩 󵄩 = 󵄩󵄩󵄩𝛽 (𝑥𝑛 − 𝐵𝑝𝑘 ) + 𝛾𝑛 (𝐵𝑛 𝑥𝑛 − 𝐵𝑝𝑘 ) + 𝛿𝑛 (𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑝𝑘 )󵄩󵄩󵄩

󵄩 󵄩 󵄩 󵄩 ≤ 𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩

󵄩 󵄩 󵄩 󵄩 + 𝛿𝑛 (󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩) 󵄩 󵄩 󵄩 󵄩 = 𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛽) 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 ,

󵄩 󵄩 󵄩 󵄩 + (1 − 𝛽) (󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑛 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 󵄩󵄩󵄩) 󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩] 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑥𝑛 󵄩󵄩󵄩 + 𝛼𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 + (1 − 𝛼𝑛 ) [𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + (1 − 𝛽) (󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 󵄩󵄩󵄩) 󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩] 󵄩 󵄩 = 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑥𝑛 󵄩󵄩󵄩 + (𝛽 + 𝛼𝑛 (1 − 𝛽)) 󵄩 󵄩 󵄩 󵄩 × 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 + (1 − 𝛼𝑛 ) [(1 − 𝛽) 󵄩󵄩󵄩𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩] 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑥𝑛 󵄩󵄩󵄩 + (𝛽 + 𝛼𝑛 (1 − 𝛽)) 󵄩 󵄩 󵄩 󵄩 × (󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝐵𝑝𝑘 󵄩󵄩󵄩) 󵄩 󵄩 󵄩 󵄩 + (1 − 𝛼𝑛 ) (1 − 𝛽) (󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝𝑘 󵄩󵄩󵄩) 󵄩 󵄩 + (1 − 𝛼𝑛 ) [(1 − 𝛽) 󵄩󵄩󵄩𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 ] 󵄩 󵄩 = 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑥𝑛 󵄩󵄩󵄩 + (𝛽 + 𝛼𝑛 (1 − 𝛽)) 󵄩 󵄩 󵄩 󵄩 × 󵄩󵄩󵄩𝑥𝑛+1 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 + (1 − 𝛼𝑛 ) [(1 − 𝛽) 󵄩󵄩󵄩𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩] + 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 . (94) So, it immediately follows from 0 ≤ 𝛼𝑛 ≤ 1 − 𝜌, for all 𝑛 ≥ 𝑛0 , that

(93)

and hence 󵄩󵄩 󵄩 󵄩󵄩𝑥𝑛+1 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 (󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩) 󵄩 󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑥𝑛 󵄩󵄩󵄩 + 𝛼𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩 󵄩 × [𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + (1 − 𝛽) 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩]

󵄩󵄩 󵄩 󵄩󵄩𝑥𝑛+1 − 𝐵𝑝𝑘 󵄩󵄩󵄩 1 (1 − 𝛼𝑛 ) (1 − 𝛽) 󵄩 󵄩 󵄩 󵄩 × (󵄩󵄩󵄩𝛼𝑛 (𝑥𝑛 − 𝑓 (𝑥𝑛 ))󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩)

󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 󵄩󵄩󵄩 +

𝛿𝑛 󵄩󵄩 󵄩 󵄩𝑆 𝐺𝑥 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 1−𝛽󵄩 𝑛 𝑛 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 󵄩󵄩󵄩 +

1 𝜌 (1 − 𝛽) 󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 × (󵄩󵄩𝛼𝑛 (𝑥𝑛 − 𝑓 (𝑥𝑛 ))󵄩󵄩 + 󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩) 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝𝑘 󵄩󵄩󵄩 + 𝜃𝑛 , ∀𝑛 ≥ 𝑛0 , 󵄩 󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 +

(95)

14

Abstract and Applied Analysis

where 𝜃𝑛 = ‖𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 ‖ + ‖𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 ‖ + (1/𝜌(1 − 𝛽))(‖𝛼𝑛 (𝑥𝑛 − 𝑓(𝑥𝑛 ))‖ + ‖𝑥𝑛 − 𝑥𝑛+1 ‖). Since lim𝑛 → ∞ ‖𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 ‖ = lim𝑛 → ∞ ‖𝑆𝑛 𝐺𝑥𝑛 −𝐵𝑛 𝑥𝑛 ‖ = lim𝑛 → ∞ ‖𝛼𝑛 (𝑥𝑛 −𝑓(𝑥𝑛 ))‖ = lim𝑛 → ∞ ‖𝑥𝑛 − 𝑥𝑛+1 ‖ = 0, we know that 𝜃𝑛 → 0 as 𝑛 → ∞. From (95), we obtain

1 1 (1 − ) (𝑥𝑛 − 𝑉𝑝𝑘 ) = 𝑥𝑛 − 𝑝𝑘 − (𝑥𝑛 − 𝑓 (𝑝𝑘 )) . 𝑘 𝑘

(102)

It follows from Lemma 8 (ii) and (102) that

󵄩2 󵄩 󵄩2 󵄩󵄩 󵄩󵄩𝑥𝑛+1 − 𝐵𝑝𝑘 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝𝑘 󵄩󵄩󵄩

󵄩 󵄩 + 𝜃𝑛 (2 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝𝑘 󵄩󵄩󵄩 + 𝜃𝑛 ) ,

that is,

∀𝑛 ≥ 𝑛0 .

(96)

For any Banach limit 𝜇, from (96), we derive 󵄩 󵄩2 󵄩 󵄩2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 = 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛+1 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 ≤ 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝𝑘 󵄩󵄩󵄩 = 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 .

1 2󵄩 󵄩2 (1 − ) 󵄩󵄩󵄩𝑥𝑛 − 𝑉𝑝𝑘 󵄩󵄩󵄩 𝑘 󵄩2 2 󵄩 ≥ 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 − ⟨𝑥𝑛 − 𝑝𝑘 + 𝑝𝑘 − 𝑓 (𝑝𝑘 ) , 𝐽 (𝑥𝑛 − 𝑝𝑘 )⟩ 𝑘

(97)

2 󵄩 󵄩2 2 = (1 − ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + ⟨𝑓 (𝑝𝑘 ) − 𝑝𝑘 , 𝐽 (𝑥𝑛 − 𝑝𝑘 )⟩ . 𝑘 𝑘 (103)

In addition, note that 󵄩󵄩 󵄩2 󵄩󵄩𝑥𝑛 − 𝐺𝑝𝑘 󵄩󵄩󵄩

So by (100) and (103), we have

󵄩2 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑥𝑛 + 𝐺𝑥𝑛 − 𝐺𝑝𝑘 󵄩󵄩󵄩

1 2 󵄩 󵄩2 (1 − ) 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 𝑘

󵄩 󵄩 󵄩2 󵄩 ≤ (󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩) 󵄩2 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 × (2 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑥𝑛 󵄩󵄩󵄩) , 󵄩󵄩 󵄩2 󵄩󵄩𝑥𝑛 − 𝑆𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 + 𝑆𝑥𝑛 − 𝑆𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩 ≤ (󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩) 󵄩2 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 × (2 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩) .

2 󵄩 󵄩2 ≥ (1 − ) 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 𝑘

󵄩 󵄩2 󵄩 󵄩2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑝𝑘 󵄩󵄩󵄩 ≤ 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 .

and hence 1 󵄩󵄩 󵄩2 2 𝜇 󵄩𝑥 − 𝑝𝑘 󵄩󵄩󵄩 ≥ 𝜇𝑛 ⟨𝑓 (𝑝𝑘 ) − 𝑝𝑘 , 𝐽 (𝑥𝑛 − 𝑝𝑘 )⟩ . 𝑘2 𝑛 󵄩 𝑛 𝑘

1 󵄩󵄩 󵄩2 𝜇 󵄩𝑥 − 𝑝𝑘 󵄩󵄩󵄩 ≥ 𝜇𝑛 ⟨𝑓 (𝑝𝑘 ) − 𝑝𝑘 , 𝐽 (𝑥𝑛 − 𝑝𝑘 )⟩ . 2𝑘 𝑛 󵄩 𝑛

(99) 𝜇𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩ ≤ 0.

󵄨 󵄨 lim 󵄨󵄨⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ − ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩󵄨󵄨󵄨 = 0.

𝑛→∞ 󵄨

(108) (100)

So, utilizing Lemma 18 we deduce from (107) and (108) that lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩ ≤ 0, 𝑛→∞

󵄩 󵄩2 ≤ 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 .

(109)

which together with (49) and the norm-to-norm uniform continuity of 𝐽 on bounded subsets of 𝑋, implies that

Also, observe that 1 1 (𝑥𝑛 − 𝑓 (𝑝𝑘 )) + (1 − ) (𝑥𝑛 − 𝑉𝑝𝑘 ) ; 𝑘 𝑘

(107)

On the other hand, from (49) and the norm-to-norm uniform continuity of 𝐽 on bounded subsets of 𝑋, it follows that

󵄩 󵄩2 󵄩 󵄩2 + 𝜃1 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝜃2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑝𝑘 󵄩󵄩󵄩

𝑥𝑛 − 𝑝𝑘 =

(106)

Since 𝑝𝑘 → 𝑞 ∈ Fix(𝑉) = 𝐹 as 𝑘 → ∞, by the uniform Frechet differentiability of the norm of 𝑋 we have

󵄩 󵄩2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑉𝑝𝑘 󵄩󵄩󵄩 󵄩 = 𝜇𝑛 󵄩󵄩󵄩(1 − 𝜃1 − 𝜃2 ) (𝑥𝑛 − 𝑆𝑝𝑘 ) 󵄩 󵄩2 ≤ (1 − 𝜃1 − 𝜃2 ) 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑝𝑘 󵄩󵄩󵄩

(105)

This implies that

Utilizing (97) and (99), we deduce that

󵄩2 +𝜃1 (𝑥𝑛 − 𝐵𝑝𝑘 ) + 𝜃2 (𝑥𝑛 − 𝐺𝑝𝑘 )󵄩󵄩󵄩

2 + 𝜇𝑛 ⟨𝑓 (𝑝𝑘 ) − 𝑝𝑘 , 𝐽 (𝑥𝑛 − 𝑝𝑘 )⟩ , 𝑘

(98)

It is easy to see from (81) and (91) that 󵄩 󵄩2 󵄩 󵄩2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑝𝑘 󵄩󵄩󵄩 ≤ 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 ,

(104)

(101)

lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ ≤ 0. 𝑛→∞

(110)

Abstract and Applied Analysis

15

Finally, let us show that 𝑥𝑛 → 𝑞 as 𝑛 → ∞. Utilizing Lemma 8 (i), from (42) and the convexity of ‖ ⋅ ‖2 , we get 󵄩󵄩 󵄩2 󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑞󵄩󵄩󵄩 󵄩 󵄩2 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑞󵄩󵄩󵄩 (111) 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 󵄩 󵄩2 + 𝛿𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 󵄩 󵄩2 = 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 , 󵄩󵄩 󵄩2 󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩 󵄩 = 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑓 (𝑞)) + (1 − 𝛼𝑛 ) (𝑦𝑛 − 𝑞) 󵄩2 + 𝛼𝑛 (𝑓 (𝑞) − 𝑞)󵄩󵄩󵄩 󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑓 (𝑞)) + (1 − 𝛼𝑛 ) (𝑦𝑛 − 𝑞)󵄩󵄩󵄩 + 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ 󵄩 󵄩2 󵄩 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑓 (𝑞)󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 + 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ 󵄩 󵄩2 󵄩 󵄩2 ≤ 𝛼𝑛 𝜌󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩

(112)

+ 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ 󵄩 󵄩2 = (1 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ 󵄩 󵄩2 = (1 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 2 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ + 𝛼𝑛 (1 − 𝜌) . 1−𝜌 Applying Lemma 7 to (112), we obtain that 𝑥𝑛 → 𝑞 as 𝑛 → ∞. Conversely, if 𝑥𝑛 → 𝑞 ∈ 𝐹 as 𝑛 → ∞, then from (42) it follows that 󵄩󵄩 󵄩󵄩 󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑞󵄩󵄩󵄩 󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑞󵄩󵄩󵄩 (113) 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 ≤ 𝛽𝑛 󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩 + 𝛾𝑛 󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩 + 𝛿𝑛 󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 󳨀→ 0 as 𝑛 󳨀→ ∞, that is, 𝑦𝑛 → 𝑞. Again from (42) we obtain that 󵄩󵄩 󵄩󵄩 󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 − (1 − 𝛼𝑛 ) (𝑦𝑛 − 𝑥𝑛 )󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑥𝑛 󵄩󵄩󵄩

󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩

󵄩 󵄩 󵄩 󵄩 + (1 − 𝛼𝑛 ) (󵄩󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩)

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩 + 2 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 . (114) Since 𝑥𝑛 → 𝑞 and 𝑦𝑛 → 𝑞, we get 𝛼𝑛 (𝑓(𝑥𝑛 ) − 𝑥𝑛 ) → 0. This completes the proof. Corollary 25. Let 𝐶 be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space 𝑋. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶. Let {𝜌𝑛 }∞ 𝑛=0 be a sequence of positive numbers in (0, 𝑏] for some 𝑏 ∈ (0, 1) and 𝐴 𝑖 : 𝐶 → 𝐸 an 𝛼̂𝑖 -inverse strongly accretive mapping for each 𝑖 = 0, 1, . . .. Define a mapping 𝐺𝑖 : 𝐶 → 𝐶 by Π𝐶(𝐼 − 𝜆 𝑖 𝐴 𝑖 )𝑥 = 𝐺𝑖 𝑥 for all 𝑥 ∈ 𝐶 and 𝑖 = 0, 1, . . ., where 𝜆 𝑖 ∈ (0, 𝛼̂𝑖 /𝜅2 ] and 𝜅 is the 2-uniformly smooth constant of 𝑋. Let 𝐵𝑛 : 𝐶 → 𝐶 be the 𝑊-mapping generated by 𝐺𝑛 , 𝐺𝑛−1 , . . . , 𝐺0 and 𝜌𝑛 , 𝜌𝑛−1 , . . . , 𝜌0 . Let 𝑉 : 𝐶 → 𝐶 be an 𝛼-strictly pseudocontractive mapping. Let 𝑓 : 𝐶 → 𝐶 be a contraction with coefficient 𝜌 ∈ (0, 1). Let {𝑆𝑖 }∞ 𝑖=0 be a countable family of nonexpansive mappings of 𝐶 into itself such ∞ that 𝐹 = (⋂∞ 𝑖=0 Fix(𝑆𝑖 )) ∩ Fix(𝑉) ∩ (⋂𝑖=0 VI(𝐶, 𝐴 𝑖 )) ≠ 0. For arbitrarily given 𝑥0 ∈ 𝐶, let {𝑥𝑛 } be the sequence generated by 𝑦𝑛 = 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 ((1 − 𝑙) 𝐼 + 𝑙𝑉) 𝑥𝑛 , 𝑥𝑛+1 = 𝛼𝑛 𝑓 (𝑥𝑛 ) + (1 − 𝛼𝑛 ) 𝑦𝑛 ,

∀𝑛 ≥ 0,

(115)

where 0 < 𝑙 < 𝛼/𝜅2 , {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, and {𝛿𝑛 } are the sequences in [0, 1] such that 𝛽𝑛 + 𝛾𝑛 + 𝛿𝑛 = 1 for all 𝑛 ≥ 0. Suppose that the following conditions hold: (i) ∑∞ 𝑛=0 𝛼𝑛 = ∞ and 0 ≤ 𝛼𝑛 ≤ 1 − 𝜌, for all 𝑛 ≥ 𝑛0 for some integer 𝑛0 ≥ 0; (ii) lim inf 𝑛 → ∞ 𝛾𝑛 > 0 and lim inf 𝑛 → ∞ 𝛿𝑛 > 0; (iii) lim𝑛 → ∞ (|𝛼𝑛+1 /(1 − (1 − 𝛼𝑛+1 )𝛽𝑛+1 ) − 𝛼𝑛 /(1 − (1 − 𝛼𝑛 )𝛽𝑛 )| + |𝛿𝑛+1 /(1 − 𝛽𝑛+1 ) − 𝛿𝑛 /(1 − 𝛽𝑛 )|) = 0; (iv) 0 < lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ 𝛽𝑛 < 1. Assume that ∑∞ 𝑛=0 sup𝑥∈𝐷 ‖𝑆𝑛+1 𝑥 − 𝑆𝑛 𝑥‖ < ∞ for any bounded subset 𝐷 of 𝐶 and let 𝑆 be a mapping of 𝐶 into itself defined by 𝑆𝑥 = lim𝑛 → ∞ 𝑆𝑛 𝑥 for all 𝑥 ∈ 𝐶 and suppose that Fix(𝑆) = ⋂∞ 𝑖=0 Fix(𝑆𝑖 ). Then, there hold the following: (I) lim𝑛 → ∞ ‖𝑥𝑛+1 − 𝑥𝑛 ‖ = 0; (II) 𝑥𝑛 → 𝑞 ⇔ 𝛼𝑛 (𝑓(𝑥𝑛 ) − 𝑥𝑛 ) → 0 provided 𝛽𝑛 ≡ 𝛽 for some fixed 𝛽 ∈ (0, 1), where 𝑞 ∈ 𝐹 solves the following VIP ⟨𝑞 − 𝑓 (𝑞) , 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,

∀𝑝 ∈ 𝐹.

(116)

Proof. In Theorem 24, we put 𝐵1 = 𝐼 − 𝑉, 𝐵2 = 0, and 𝜇1 = 𝑙, where 0 < 𝑙 < 𝛼/𝜅2 . Then, GSVI (13) is equivalent to the VIP of finding 𝑥∗ ∈ 𝐶 such that ⟨𝐵1 𝑥∗ , 𝐽 (𝑥 − 𝑥∗ )⟩ ≥ 0,

∀𝑥 ∈ 𝐶.

(117)

16

Abstract and Applied Analysis

In this case, 𝐵1 : 𝐶 → 𝑋 is 𝛼-inverse strongly accretive. It is not hard to see that Fix(𝑉) = VI(𝐶, 𝐵1 ). As a matter of fact, we have, for 𝑙 > 0, 𝑢 ∈ VI (𝐶, 𝐵1 )

󵄩 󵄩󵄩 󵄩 1 󵄩󵄩 󵄩𝑥 − 𝑦󵄩󵄩󵄩 . 󵄩󵄩(𝐼 − 𝐵𝑖 ) 𝑥 − (𝐼 − 𝐵𝑖 ) 𝑦󵄩󵄩󵄩 ≤ 𝜆𝑖 󵄩

∀𝑦 ∈ 𝐶

≤ (1 +

⇐⇒ 𝑢 = Π𝐶 (𝑢 − 𝑙𝐵1 𝑢) ⇐⇒ 𝑢 = Π𝐶 (𝑢 − 𝑙𝑢 + 𝑙𝑉𝑢)

(118)

⇐⇒ ⟨𝑢 − 𝑙𝑢 + 𝑙𝑉𝑢 − 𝑢, 𝐽 (𝑢 − 𝑦)⟩ ≥ 0

∀𝑦 ∈ 𝐶

(123)

1 󵄩󵄩 󵄩 ) 󵄩𝑥 − 𝑦󵄩󵄩󵄩 . 𝜆𝑖 󵄩

Utilizing the 𝛼𝑖 -strong accretivity and 𝜆 𝑖 -strict pseudocontractivity of 𝐵𝑖 , we get 󵄩 󵄩2 𝜆 𝑖 󵄩󵄩󵄩(𝐼 − 𝐵𝑖 ) 𝑥 − (𝐼 − 𝐵𝑖 ) 𝑦󵄩󵄩󵄩

⇐⇒ ⟨𝑢 − 𝑉𝑢, 𝐽 (𝑢 − 𝑦)⟩ ≤ 0 ∀𝑦 ∈ 𝐶

󵄩2 󵄩 ≤ 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 − ⟨𝐵𝑖 𝑥 − 𝐵𝑖 𝑦, 𝐽 (𝑥 − 𝑦)⟩

⇐⇒ 𝑢 = 𝑉𝑢

(124)

󵄩2 󵄩 ≤ (1 − 𝛼𝑖 ) 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 .

⇐⇒ 𝑢 ∈ Fix (𝑉) . Accordingly, we know that 𝐹 = (⋂∞ 𝑖=0 Fix(𝑆𝑖 )) ∩ Ω ∩ ∞ VI(𝐶, 𝐴 )) = (⋂ Fix(𝑆 )) ∩ Fix(𝑉) ∩ (⋂∞ (⋂∞ 𝑖 𝑖 𝑖=0 𝑖=0 𝑖=0 VI(𝐶, 𝐴 𝑖 )), and Π𝐶 (𝐼 − 𝜇1 𝐵1 ) Π𝐶 (𝐼 − 𝜇2 𝐵2 ) 𝑥𝑛 = Π𝐶 (𝐼 − 𝜇1 𝐵1 ) 𝑥𝑛 = Π𝐶 ((1 − 𝑙) 𝑥𝑛 + 𝑙𝑉𝑥𝑛 )

So, we have 1 − 𝛼𝑖 󵄩󵄩 󵄩 󵄩󵄩 󵄩 󵄩󵄩(𝐼 − 𝐵𝑖 ) 𝑥 − (𝐼 − 𝐵𝑖 ) 𝑦󵄩󵄩󵄩 ≤ √ 󵄩𝑥 − 𝑦󵄩󵄩󵄩 . 𝜆 󵄩

(119)

So, the scheme (42) reduces to (115). Therefore, the desired result follows from Theorem 24. Here, we prove the following important lemmas which will be used in the sequel. Lemma 26. Let 𝐶 be a nonempty closed convex subset of a smooth Banach space 𝑋 and let the mapping 𝐵𝑖 : 𝐶 → 𝑋 be 𝜆 𝑖 -strictly pseudocontractive and 𝛼𝑖 -strongly accretive with 𝛼𝑖 + 𝜆 𝑖 ≥ 1 for 𝑖 = 1, 2. Then, for 𝜇𝑖 ∈ (0, 1] one has 󵄩 󵄩󵄩 󵄩󵄩(𝐼 − 𝜇𝑖 𝐵𝑖 ) 𝑥 − (𝐼 − 𝜇𝑖 𝐵𝑖 ) 𝑦󵄩󵄩󵄩

󵄩 󵄩󵄩 󵄩󵄩(𝐼 − 𝜇𝑖 𝐵𝑖 ) 𝑥 − (𝐼 − 𝜇𝑖 𝐵𝑖 ) 𝑦󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩(𝐼 − 𝐵𝑖 ) 𝑥 − (𝐼 − 𝐵𝑖 ) 𝑦󵄩󵄩󵄩 + (1 − 𝜇𝑖 ) 󵄩󵄩󵄩𝐵𝑖 𝑥 − 𝐵𝑖 𝑦󵄩󵄩󵄩 ≤√

1 − 𝛼𝑖 󵄩󵄩 1 󵄩 󵄩 󵄩 󵄩𝑥 − 𝑦󵄩󵄩󵄩 + (1 − 𝜇𝑖 ) (1 + ) 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 𝜆𝑖 󵄩 𝜆𝑖

= {√

1 − 𝛼𝑖 1 󵄩 󵄩 + (1 − 𝜇𝑖 ) (1 + )} 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 . 𝜆𝑖 𝜆𝑖 (126)

Since 1 − (𝜆 𝑖 /(1 + 𝜆 𝑖 ))(1 − √(1 − 𝛼𝑖 )/𝜆 𝑖 ) ≤ 𝜇𝑖 ≤ 1, it follows immediately that √

1 − 𝛼𝑖 1 󵄩 󵄩 + (1 − 𝜇𝑖 ) (1 + )} 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 , 𝜆𝑖 𝜆𝑖

(120)

∀𝑥, 𝑦 ∈ 𝐶, for 𝑖 = 1, 2. In particular, if 1−(𝜆 𝑖 /(1+𝜆 𝑖 ))(1−√(1 − 𝛼𝑖 )/𝜆 𝑖 ) ≤ 𝜇𝑖 ≤ 1, then 𝐼 − 𝜇𝑖 𝐵𝑖 is nonexpansive for 𝑖 = 1, 2. Proof. Taking into account the 𝜆 𝑖 -strict pseudocontractivity of 𝐵𝑖 , we derive for every 𝑥, 𝑦 ∈ 𝐶

󵄩 󵄩󵄩 󵄩 ≤ 󵄩󵄩󵄩(𝐼 − 𝐵𝑖 ) 𝑥 − (𝐼 − 𝐵𝑖 ) 𝑦󵄩󵄩󵄩 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,

1 − 𝛼𝑖 1 + (1 − 𝜇𝑖 ) (1 + ) ≤ 1. 𝜆𝑖 𝜆𝑖

(127)

This implies that 𝐼 − 𝜇𝑖 𝐵𝑖 is nonexpansive for 𝑖 = 1, 2. Lemma 27. Let 𝐶 be a nonempty closed convex subset of a smooth Banach space 𝑋. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶 and let the mapping 𝐵𝑖 : 𝐶 → 𝑋 be 𝜆 𝑖 -strictly pseudocontractive and 𝛼𝑖 -strongly accretive with 𝛼𝑖 + 𝜆 𝑖 ≥ 1 for 𝑖 = 1, 2. Let 𝐺 : 𝐶 → 𝐶 be the mapping defined by 𝐺 (𝑥) = Π𝐶 [Π𝐶 (𝑥 − 𝜇2 𝐵2 𝑥) − 𝜇1 𝐵1 Π𝐶 (𝑥 − 𝜇2 𝐵2 𝑥)] , ∀𝑥 ∈ 𝐶. (128)

󵄩 󵄩2 𝜆 𝑖 󵄩󵄩󵄩(𝐼 − 𝐵𝑖 ) 𝑥 − (𝐼 − 𝐵𝑖 ) 𝑦󵄩󵄩󵄩 ≤ ⟨(𝐼 − 𝐵𝑖 ) 𝑥 − (𝐼 − 𝐵𝑖 ) 𝑦, 𝐽 (𝑥 − 𝑦)⟩

(125)

𝑖

Therefore, for 𝜇𝑖 ∈ (0, 1] we have

= ((1 − 𝑙) 𝐼 + 𝑙𝑉) 𝑥𝑛 .

≤ {√

(122)

Hence, 󵄩 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝐵𝑖 𝑥 − 𝐵𝑖 𝑦󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩(𝐼 − 𝐵𝑖 ) 𝑥 − (𝐼 − 𝐵𝑖 ) 𝑦󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩

⇐⇒ ⟨𝐵1 𝑢, 𝐽 (𝑦 − 𝑢)⟩ ≥ 0 ∀𝑦 ∈ 𝐶 ⇐⇒ ⟨𝑢 − 𝑙𝐵1 𝑢 − 𝑢, 𝐽 (𝑢 − 𝑦)⟩ ≥ 0

which implies that

(121)

If 1−(𝜆 𝑖 /(1+𝜆 𝑖 ))(1−√(1 − 𝛼𝑖 )/𝜆 𝑖 ) ≤ 𝜇𝑖 ≤ 1, then 𝐺 : 𝐶 → 𝐶 is nonexpansive.

Abstract and Applied Analysis

17

Proof. According to Lemma 26, we know that 𝐼 − 𝜇𝑖 𝐵𝑖 is nonexpansive for 𝑖 = 1, 2. Hence, for all 𝑥, 𝑦 ∈ 𝐶, we have 󵄩 󵄩󵄩 󵄩󵄩𝐺 (𝑥) − 𝐺 (𝑦)󵄩󵄩󵄩 󵄩 = 󵄩󵄩󵄩Π𝐶 [Π𝐶 (𝑥 − 𝜇2 𝐵2 𝑥) − 𝜇1 𝐵1 Π𝐶 (𝑥 − 𝜇2 𝐵2 𝑥)]

(iv) 0 < lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ 𝛽𝑛 < 1.

󵄩 − Π𝐶 [Π𝐶 (𝑦 − 𝜇2 𝐵2 𝑦) − 𝜇1 𝐵1 Π𝐶 (𝑦 − 𝜇2 𝐵2 𝑦)]󵄩󵄩󵄩

󵄩 = 󵄩󵄩󵄩Π𝐶 (𝐼 − 𝜇1 𝐵1 ) Π𝐶 (𝐼 − 𝜇2 𝐵2 ) 𝑥

(II) 𝑥𝑛 → 𝑞 ⇔ 𝛼𝑛 (𝑓(𝑥𝑛 ) − 𝑥𝑛 ) → 0 provided 𝛽𝑛 ≡ 𝛽 for some fixed 𝛽 ∈ (0, 1), where 𝑞 ∈ 𝐹 solves the following VIP

󵄩 ≤ 󵄩󵄩󵄩(𝐼 − 𝜇1 𝐵1 ) Π𝐶 (𝐼 − 𝜇2 𝐵2 ) 𝑥

󵄩 − (𝐼 − 𝜇1 𝐵1 ) Π𝐶 (𝐼 − 𝜇2 𝐵2 ) 𝑦󵄩󵄩󵄩

⟨𝑞 − 𝑓 (𝑞) , 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,

󵄩 󵄩 ≤ 󵄩󵄩󵄩Π𝐶 (𝐼 − 𝜇2 𝐵2 ) 𝑥 − Π𝐶 (𝐼 − 𝜇2 𝐵2 ) 𝑦󵄩󵄩󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩(𝐼 − 𝜇2 𝐵2 ) 𝑥 − (𝐼 − 𝜇2 𝐵2 ) 𝑦󵄩󵄩󵄩

(129)

This shows that 𝐺 : 𝐶 → 𝐶 is nonexpansive. This completes the proof. Theorem 28. Let 𝐶 be a nonempty closed convex subset of a uniformly convex Banach space 𝑋 which has a uniformly Gateaux differentiable norm. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶. Let {𝜌𝑛 }∞ 𝑛=0 be a sequence of positive numbers in (0, 𝑏] for some 𝑏 ∈ (0, 1) and 𝐴 𝑖 : 𝐶 → 𝑋 be 𝜉𝑖 -strictly pseudocontractive and 𝛼̂𝑖 -strongly accretive with 𝜉𝑖 + 𝛼̂𝑖 ≥ 1 for each 𝑖 = 0, 1, . . .. Define a mapping 𝐺𝑖 : 𝐶 → 𝐶 by Π𝐶(𝐼 − 𝜆 𝑖 𝐴 𝑖 )𝑥 = 𝐺𝑖 𝑥 for all 𝑥 ∈ 𝐶 and 𝑖 = 0, 1, . . ., where 1 − (𝜉𝑖 /(1 + 𝜉𝑖 ))(1 − √(1 − 𝛼̂𝑖 )/𝜉𝑖 ) ≤ 𝜆 𝑖 ≤ 1 for all 𝑖 = 0, 1, . . .. Let 𝐵𝑛 : 𝐶 → 𝐶 be the 𝑊-mapping generated by 𝐺𝑛 , 𝐺𝑛−1 , . . . , 𝐺0 and 𝜌𝑛 , 𝜌𝑛−1 , . . . , 𝜌0 . Let the mapping 𝐵𝑖 : 𝐶 → 𝑋 𝜁𝑖 -strictly pseudocontractive and 𝛽̂𝑖 -strongly accretive with 𝜁𝑖 + 𝛽̂𝑖 ≥ 1 for 𝑖 = 1, 2. Let 𝑓 : 𝐶 → 𝐶 be a contraction with coefficient 𝜌 ∈ (0, 1). Let {𝑆𝑖 }∞ 𝑖=0 be a countable family of nonexpansive mappings of 𝐶 into itself such that 𝐹 = ∞ (⋂∞ 𝑖=0 Fix(𝑆𝑖 )) ∩ Ω ∩ (⋂𝑖=0 VI(𝐶, 𝐴 𝑖 )) ≠ 0, where Ω is the fixed point set of the mapping 𝐺 = Π𝐶(𝐼 − 𝜇1 𝐵1 )Π𝐶(𝐼 − 𝜇2 𝐵2 ) with 1 − (𝜁 /(1 + 𝜁 ))(1 − √(1 − 𝛽̂ )/𝜁 ) ≤ 𝜇 ≤ 1 for 𝑖 = 1, 2. For 𝑖

𝑖

𝑖

arbitrarily given 𝑥0 ∈ 𝐶, let {𝑥𝑛 } be the sequence generated by 𝑦𝑛 = 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 , 𝑥𝑛+1 = 𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝜎𝑛 𝐺𝑥𝑛 + (1 − 𝛼𝑛 − 𝜎𝑛 ) 𝑦𝑛 ,

∀𝑛 ≥ 0, (130)

where {𝜎𝑛 }, {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, and {𝛿𝑛 } are the sequences in [0, 1] such that 𝛽𝑛 + 𝛾𝑛 + 𝛿𝑛 = 1 and 𝛼𝑛 + 𝜎𝑛 ≤ 1 for all 𝑛 ≥ 0. Suppose that the following conditions hold: (i) ∑∞ 𝑛=0 𝛼𝑛 = ∞ and 0 ≤ 𝛼𝑛 + 𝜎𝑛 ≤ 1 − 𝜌, for all 𝑛 ≥ 𝑛0 for some integer 𝑛0 ≥ 0; (ii) lim inf 𝑛 → ∞ 𝜎𝑛 > lim inf 𝑛 → ∞ 𝛿𝑛 > 0;

0, lim inf 𝑛 → ∞ 𝛾𝑛

∀𝑝 ∈ 𝐹.

(131)

Proof. First of all, take a fixed 𝑝 ∈ 𝐹 arbitrarily. Then we obtain 𝑝 = 𝐺𝑝, 𝑝 = 𝐵𝑛 𝑝 and 𝑆𝑛 𝑝 = 𝑝 for all 𝑛 ≥ 0. By Lemma 27, we get from (130)

󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 .

𝑖

Assume that ∑∞ 𝑛=0 sup𝑥∈𝐷 ‖𝑆𝑛+1 𝑥 − 𝑆𝑛 𝑥‖ < ∞ for any bounded subset 𝐷 of 𝐶 and let 𝑆 be a mapping of 𝐶 into itself defined by 𝑆𝑥 = lim𝑛 → ∞ 𝑆𝑛 𝑥 for all 𝑥 ∈ 𝐶 and suppose that Fix(𝑆) = ⋂∞ 𝑖=0 Fix(𝑆𝑖 ). Then there hold the following: (I) lim𝑛 → ∞ ‖𝑥𝑛+1 − 𝑥𝑛 ‖ = 0;

󵄩 − Π𝐶 (𝐼 − 𝜇1 𝐵1 ) Π𝐶 (𝐼 − 𝜇2 𝐵2 ) 𝑦󵄩󵄩󵄩

𝑖

(iii) lim𝑛 → ∞ (|𝛼𝑛+1 /(1−(1−𝛼𝑛+1 −𝜎𝑛+1 )𝛽𝑛+1 )−(𝛼𝑛 /(1−(1− 𝛼𝑛 −𝜎𝑛 )𝛽𝑛 ))|+|𝜎𝑛+1 /(1−(1−𝛼𝑛+1 −𝜎𝑛+1 )𝛽𝑛+1 )−(𝜎𝑛 /(1− (1 − 𝛼𝑛 − 𝜎𝑛 )𝛽𝑛 ))| + |𝛿𝑛+1 /(1 − 𝛽𝑛+1 ) − 𝛿𝑛 /(1 − 𝛽𝑛 )|) = 0;

>

0 and

󵄩 󵄩󵄩 󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩

(132)

󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 ,

and hence 󵄩󵄩 󵄩 󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 𝜎𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 + (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 (󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑓 (𝑝)󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩) 󵄩 󵄩 󵄩 󵄩 + 𝜎𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 (𝜌 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩) 󵄩 󵄩 󵄩 󵄩 + 𝜎𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 = (1 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩 󵄩󵄩 󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩 󵄩 = (1 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛼𝑛 (1 − 𝜌) 󵄩 1−𝜌 󵄩 󵄩 󵄩 󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩 ≤ max {󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 , 󵄩 }. 1−𝜌 (133) By induction, we have 󵄩 󵄩 󵄩 󵄩󵄩 󵄩 󵄩 󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 }, 󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 ≤ max {󵄩󵄩󵄩𝑥0 − 𝑝󵄩󵄩󵄩 , 󵄩 1−𝜌

∀𝑛 ≥ 0, (134)

which implies that {𝑥𝑛 } is bounded and so are the sequences {𝑦𝑛 }, {𝐺𝑥𝑛 }, and {𝑓(𝑥𝑛 )}.

18

Abstract and Applied Analysis Let us show that 󵄩󵄩

lim 󵄩𝑥 𝑛 → ∞ 󵄩 𝑛+1

×[

󵄩 − 𝑥𝑛 󵄩󵄩󵄩 = 0.

(135)

−

As a matter of fact, put 𝜃𝑛 = (1 − 𝛼𝑛 − 𝜎𝑛 )𝛽𝑛 , for all 𝑛 ≥ 0. Then, it follows from (i) and (iv) that 𝛽𝑛 ≥ 𝜃𝑛 = (1 − 𝛼𝑛 − 𝜎𝑛 ) 𝛽𝑛 ≥ (1 − (1 − 𝜌)) 𝛽𝑛 = 𝜌𝛽𝑛 , ∀𝑛 ≥ 𝑛0 ,

+[ (136)

𝑛→∞

𝑛→∞

×

(137)

Define

= 𝑥𝑛+1 = 𝜃𝑛 𝑥𝑛 + (1 − 𝜃𝑛 ) 𝑧𝑛 .

(138)

𝑧𝑛+1 − 𝑧𝑛 =

𝑥𝑛+2 − 𝜃𝑛+1 𝑥𝑛+1 𝑥𝑛+1 − 𝜃𝑛 𝑥𝑛 − 1 − 𝜃𝑛+1 1 − 𝜃𝑛

× (1 − 𝜃𝑛+1 ) −

𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 1 − 𝛽𝑛

𝜎𝑛+1 (𝐺𝑥𝑛+1 − 𝐺𝑥𝑛 ) 1 − 𝜃𝑛+1

+(

𝛼𝑛+1 𝛼𝑛 − ) 𝑓 (𝑥𝑛 ) 1 − 𝜃𝑛+1 1 − 𝜃𝑛

+(

𝜎𝑛+1 𝜎𝑛 − ) 𝐺𝑥𝑛 1 − 𝜃𝑛+1 1 − 𝜃𝑛

= (𝛼𝑛+1 𝑓 (𝑥𝑛+1 ) + 𝜎𝑛+1 𝐺𝑥𝑛+1 +

+ (1 − 𝛼𝑛+1 − 𝜎𝑛+1 ) 𝑦𝑛+1 − 𝜃𝑛+1 𝑥𝑛+1 ) −1

(1 − 𝛼𝑛+1 − 𝜎𝑛+1 ) (1 − 𝛽𝑛+1 ) 1 − 𝜃𝑛+1

×[

𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝜎𝑛 𝐺𝑥𝑛 + (1 − 𝛼𝑛 − 𝜎𝑛 ) 𝑦𝑛 − 𝜃𝑛 𝑥𝑛 1 − 𝜃𝑛

(1 − 𝛼𝑛 − 𝜎𝑛 ) (1 − 𝛽𝑛 ) ] 1 − 𝜃𝑛

𝛼𝑛+1 (𝑓 (𝑥𝑛+1 ) − 𝑓 (𝑥𝑛 )) 1 − 𝜃𝑛+1 +

Observe that

𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 ] 1 − 𝛽𝑛

(1 − 𝛼𝑛+1 − 𝜎𝑛+1 ) (1 − 𝛽𝑛+1 ) 1 − 𝜃𝑛+1 −

and hence 0 < lim inf 𝜃𝑛 ≤ lim sup 𝜃𝑛 < 1.

𝛾𝑛+1 𝐵𝑛+1 𝑥𝑛+1 + 𝛿𝑛+1 𝑆𝑛+1 𝐺𝑥𝑛+1 1 − 𝛽𝑛+1

𝛾𝑛+1 (𝐵 𝑥 − 𝐵𝑛 𝑥𝑛 ) 𝛾𝑛+1 + 𝛿𝑛+1 𝑛+1 𝑛+1 +(

𝛼 𝑓 (𝑥𝑛+1 + 𝜎𝑛+1 𝐺𝑥𝑛+1 ) 𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝜎𝑛 𝐺𝑥𝑛 = ( 𝑛+1 − ) 1 − 𝜃𝑛+1 1 − 𝜃𝑛

+

(1 − 𝛼𝑛 − 𝜎𝑛 ) [𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 ] − 𝜃𝑛 𝑥𝑛 − 1 − 𝜃𝑛

𝛿𝑛+1 (𝑆 𝐺𝑥 − 𝑆𝑛 𝐺𝑥𝑛 ) 𝛾𝑛+1 + 𝛿𝑛+1 𝑛+1 𝑛+1

+(

+ (1 − 𝛼𝑛+1 − 𝜎𝑛+1 ) × [𝛽𝑛+1 𝑥𝑛+1 + 𝛾𝑛+1 𝐵𝑛+1 𝑥𝑛+1 + 𝛿𝑛+1 𝑆𝑛+1 𝐺𝑥𝑛+1 ]

−(

−1

− 𝜃𝑛+1 𝑥𝑛+1 × (1 − 𝜃𝑛+1 )

𝛾𝑛 𝛾𝑛+1 − ) 𝐵𝑛 𝑥𝑛 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛

𝛿𝑛+1 𝛿𝑛 − ) 𝑆 𝐺𝑥 ] 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 𝑛 𝑛

𝛼𝑛+1 + 𝜎𝑛+1 𝛼𝑛 + 𝜎𝑛 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 − ) 1 − 𝜃𝑛+1 1 − 𝜃𝑛 𝛾𝑛 + 𝛿𝑛

𝛼𝑛+1 (𝑓 (𝑥𝑛+1 ) − 𝑓 (𝑥𝑛 )) 1 − 𝜃𝑛+1

=(

𝛼𝑛+1 𝑓 (𝑥𝑛+1 ) + 𝜎𝑛+1 𝐺𝑥𝑛+1 𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝜎𝑛 𝐺𝑥𝑛 − ) 1 − 𝜃𝑛+1 1 − 𝜃𝑛

+

1 − 𝛼𝑛+1 − 𝜎𝑛+1 (𝛾𝑛+1 𝐵𝑛+1 𝑥𝑛+1 + 𝛿𝑛+1 𝑆𝑛+1 𝐺𝑥𝑛+1 ) 1 − 𝜃𝑛+1

−

1 − 𝛼𝑛 − 𝜎𝑛 (𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 ) 1 − 𝜃𝑛

+(

𝛾 𝐵 𝑥 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 𝛼𝑛+1 𝛼𝑛 − ) (𝑓 (𝑥𝑛 ) − 𝑛 𝑛 𝑛 ) 1 − 𝜃𝑛+1 1 − 𝜃𝑛 𝛾𝑛 + 𝛿𝑛

𝛼𝑛+1 𝑓 (𝑥𝑛+1 ) + 𝜎𝑛+1 𝐺𝑥𝑛+1 𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝜎𝑛 𝐺𝑥𝑛 − ) 1 − 𝜃𝑛+1 1 − 𝜃𝑛

+(

𝛾 𝐵 𝑥 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 𝜎𝑛+1 𝜎𝑛 − ) (𝐺𝑥𝑛 − 𝑛 𝑛 𝑛 ) 1 − 𝜃𝑛+1 1 − 𝜃𝑛 𝛾𝑛 + 𝛿𝑛

(1 − 𝛼𝑛+1 − 𝜎𝑛+1 ) (1 − 𝛽𝑛+1 ) 1 − 𝜃𝑛+1

+

=( +

=

+

𝜎𝑛+1 (𝐺𝑥𝑛+1 − 𝐺𝑥𝑛 ) 1 − 𝜃𝑛+1

1 − 𝛼𝑛+1 − 𝜎𝑛+1 − 𝜃𝑛+1 1 − 𝜃𝑛+1

Abstract and Applied Analysis ×[

19 󵄨󵄨 𝛼 𝛼𝑛 󵄨󵄨󵄨󵄨 󵄨 + 󵄨󵄨󵄨 𝑛+1 − 󵄨 󵄨󵄨 1 − 𝜃𝑛+1 1 − 𝜃𝑛 󵄨󵄨󵄨 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 × (󵄩󵄩󵄩𝑓 (𝑥𝑛 )󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩) 󵄨󵄨 𝜎 𝜎𝑛 󵄨󵄨󵄨󵄨 󵄨 + 󵄨󵄨󵄨 𝑛+1 − 󵄨 󵄨󵄨 1 − 𝜃𝑛+1 1 − 𝜃𝑛 󵄨󵄨󵄨 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 × (󵄩󵄩󵄩𝐺𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩)

𝛾𝑛+1 (𝐵 𝑥 − 𝐵𝑛 𝑥𝑛 ) 𝛾𝑛+1 + 𝛿𝑛+1 𝑛+1 𝑛+1

+( +

𝛾𝑛 𝛾𝑛+1 − ) 𝐵𝑛 𝑥𝑛 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛

𝛿𝑛+1 (𝑆 𝐺𝑥 − 𝑆𝑛 𝐺𝑥𝑛 ) 𝛾𝑛+1 + 𝛿𝑛+1 𝑛+1 𝑛+1

+(

𝛿𝑛+1 𝛿𝑛 − ) 𝑆 𝐺𝑥 ] , 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 𝑛 𝑛

+ (139)

and hence 󵄩󵄩󵄩𝑧𝑛+1 − 𝑧𝑛 󵄩󵄩󵄩 󵄩 󵄩 𝛼𝑛+1 󵄩󵄩 󵄩 ≤ 󵄩𝑓 (𝑥𝑛+1 ) − 𝑓 (𝑥𝑛 )󵄩󵄩󵄩 1 − 𝜃𝑛+1 󵄩

1 − 𝛼𝑛+1 − 𝜎𝑛+1 − 𝜃𝑛+1 1 − 𝜃𝑛+1

×[

󵄨󵄨 𝛾 𝛾𝑛 󵄨󵄨󵄨󵄨 󵄩󵄩 󵄨 󵄩 𝑛+1 + 󵄨󵄨󵄨 − 󵄨 󵄩𝐵 𝑥 󵄩󵄩 󵄨󵄨 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 󵄨󵄨󵄨 󵄩 𝑛 𝑛 󵄩

𝜎𝑛+1 󵄩󵄩 󵄩 󵄩𝐺𝑥 − 𝐺𝑥𝑛 󵄩󵄩󵄩 1 − 𝜃𝑛+1 󵄩 𝑛+1 󵄨󵄨 𝛼 𝛾 𝐵 𝑥 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩󵄩 𝛼𝑛 󵄨󵄨󵄨󵄨 󵄩󵄩󵄩󵄩 󵄨 + 󵄨󵄨󵄨 𝑛+1 − 󵄨󵄨 󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑛 𝑛 𝑛 󵄩󵄩 󵄨󵄨 1 − 𝜃𝑛+1 1 − 𝜃𝑛 󵄨󵄨 󵄩󵄩 󵄩󵄩 𝛾𝑛 + 𝛿𝑛 󵄨󵄨 𝜎 𝛾 𝐵 𝑥 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩󵄩 𝜎𝑛 󵄨󵄨󵄨󵄨 󵄩󵄩󵄩󵄩 󵄨 + 󵄨󵄨󵄨 𝑛+1 − 󵄨󵄨 󵄩󵄩𝐺𝑥𝑛 − 𝑛 𝑛 𝑛 󵄩󵄩 󵄨󵄨 1 − 𝜃𝑛+1 1 − 𝜃𝑛 󵄨󵄨 󵄩󵄩 󵄩󵄩 𝛾𝑛 + 𝛿𝑛

+

+

+

1 − 𝛼𝑛+1 − 𝜎𝑛+1 − 𝜃𝑛+1 1 − 𝜃𝑛+1 𝛾𝑛+1 󵄩󵄩 󵄩 󵄩𝐵 𝑥 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 𝛾𝑛+1 + 𝛿𝑛+1 󵄩 𝑛+1 𝑛+1 󵄨󵄨 𝛾 𝛾𝑛 󵄨󵄨󵄨󵄨 󵄩󵄩 󵄨 󵄩 𝑛+1 + 󵄨󵄨󵄨 − 󵄨 󵄩𝐵 𝑥 󵄩󵄩 󵄨󵄨 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 󵄨󵄨󵄨 󵄩 𝑛 𝑛 󵄩

=

1 − 𝛼𝑛+1 (1 − 𝜌) − 𝜃𝑛+1 󵄩󵄩 󵄩 󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 1 − 𝜃𝑛+1 1 − 𝛼𝑛+1 − 𝜎𝑛+1 − 𝜃𝑛+1 1 − 𝜃𝑛+1

×[

𝑛 󵄨󵄨 𝛿 𝛾𝑛+1 𝛿𝑛 󵄨󵄨󵄨󵄨 󵄨 𝑛+1 𝑀0 ∏𝜆 𝑖 + 󵄨󵄨󵄨 − 󵄨 󵄨󵄨 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 󵄨󵄨󵄨 𝛾𝑛+1 + 𝛿𝑛+1 𝑖=0

󵄩 󵄩 󵄩 󵄩 × (󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩)

𝛿𝑛+1 󵄩󵄩 󵄩 󵄩𝑆 𝐺𝑥 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 𝛾𝑛+1 + 𝛿𝑛+1 󵄩 𝑛+1 𝑛+1 󵄨󵄨 𝛿 𝛿𝑛 󵄨󵄨󵄨󵄨 󵄩󵄩 󵄨 󵄩 𝑛+1 + 󵄨󵄨󵄨 − 󵄨󵄨 󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩] . 󵄨󵄨 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 󵄨󵄨

+

+

(140) On the other hand, repeating the same arguments as those of (55) and (56) in the proof of Theorem 24, we can get 󵄩󵄩󵄩𝑆𝑛+1 𝐺𝑥𝑛+1 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛+1 𝐺𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 , 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 𝑛

󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝐵𝑛+1 𝑥𝑛+1 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 𝑀0 ∏𝜆 𝑖 ,

𝛿𝑛+1 󵄩󵄩 󵄩 󵄩𝑆 𝐺𝑥 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 ] 𝛾𝑛+1 + 𝛿𝑛+1 󵄩 𝑛+1 𝑛

󵄨󵄨 𝛼 𝛼𝑛 󵄨󵄨󵄨󵄨 󵄨 + 󵄨󵄨󵄨 𝑛+1 − 󵄨 󵄨󵄨 1 − 𝜃𝑛+1 1 − 𝜃𝑛 󵄨󵄨󵄨 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 × (󵄩󵄩󵄩𝑓 (𝑥𝑛 )󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩) 󵄨󵄨 𝜎 𝜎𝑛 󵄨󵄨󵄨󵄨 󵄨 + 󵄨󵄨󵄨 𝑛+1 − 󵄨󵄨 󵄨󵄨 1 − 𝜃𝑛+1 1 − 𝜃𝑛 󵄨󵄨 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 × (󵄩󵄩󵄩𝐺𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩)

󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩

𝑖=0

(141) for some constant 𝑀0 > 0. Utilizing (140)-(141), we have 󵄩󵄩󵄩𝑧𝑛+1 − 𝑧𝑛 󵄩󵄩󵄩 󵄩 󵄩 𝛼𝑛+1 󵄩 󵄩 ≤ 𝜌 󵄩󵄩𝑥 − 𝑥𝑛 󵄩󵄩󵄩 1 − 𝜃𝑛+1 󵄩 𝑛+1 𝜎𝑛+1 󵄩󵄩 󵄩 󵄩𝑥 − 𝑥𝑛 󵄩󵄩󵄩 1 − 𝜃𝑛+1 󵄩 𝑛+1

𝛿𝑛+1 󵄩 󵄩 󵄩 󵄩 (󵄩󵄩𝑥 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛+1 𝐺𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩) 𝛾𝑛+1 + 𝛿𝑛+1 󵄩 𝑛+1

󵄨󵄨 𝛿 𝛿𝑛 󵄨󵄨󵄨󵄨 󵄩󵄩 󵄨 󵄩 𝑛+1 + 󵄨󵄨󵄨 − 󵄨 󵄩𝑆 𝐺𝑥 󵄩󵄩 ] 󵄨󵄨 𝛾𝑛+1 + 𝛿𝑛+1 𝛾𝑛 + 𝛿𝑛 󵄨󵄨󵄨 󵄩 𝑛 𝑛 󵄩

+

×[

+

𝑛 𝛾𝑛+1 󵄩 󵄩 (󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 𝑀0 ∏𝜆 𝑖 ) 𝛾𝑛+1 + 𝛿𝑛+1 𝑖=0

𝑛 󵄨󵄨 𝛿 𝛿𝑛 󵄨󵄨󵄨󵄨 󵄨 + 𝑀 (∏𝜆 𝑖 + 󵄨󵄨󵄨 𝑛+1 − 󵄨 󵄨󵄨 1 − 𝛽𝑛+1 1 − 𝛽𝑛 󵄨󵄨󵄨 𝑖=0

󵄨󵄨 𝛼 𝛼𝑛 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 𝜎𝑛+1 𝜎𝑛 󵄨󵄨󵄨󵄨 󵄨 + 󵄨󵄨󵄨 𝑛+1 − − 󵄨󵄨 + 󵄨󵄨 󵄨) 󵄨󵄨 1 − 𝜃𝑛+1 1 − 𝜃𝑛 󵄨󵄨 󵄨󵄨 1 − 𝜃𝑛+1 1 − 𝜃𝑛 󵄨󵄨󵄨 󵄩 󵄩 + 󵄩󵄩󵄩𝑆𝑛+1 𝐺𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 , (142)

20

Abstract and Applied Analysis

where sup𝑛≥0 {‖𝑓(𝑥𝑛 )‖ + ‖𝐺𝑥𝑛 ‖ + ‖𝐵𝑛 𝑥𝑛 ‖ + ‖𝑆𝑛 𝐺𝑥𝑛 ‖ + 𝑀0 } ≤ 𝑀 for some 𝑀 > 0. So, from (142), condition (iii), and the assumption on {𝑆𝑛 } it follows that (noting that 0 < 𝜆 𝑖 ≤ 𝑏 < 1, for all 𝑖 ≥ 0) 󵄩 󵄩 󵄩 󵄩 lim sup (󵄩󵄩󵄩𝑧𝑛+1 − 𝑧𝑛 󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩) ≤ 0. 𝑛→∞

(143)

Consequently, by Lemma 20, we have 󵄩󵄩

lim 󵄩𝑧 𝑛→∞ 󵄩 𝑛

󵄩 − 𝑥𝑛 󵄩󵄩󵄩 = 0.

(144)

It follows from (137) and (138) that 󵄩󵄩

lim 󵄩𝑥 𝑛 → ∞ 󵄩 𝑛+1

󵄩 󵄩 󵄩 − 𝑥𝑛 󵄩󵄩󵄩 = lim (1 − 𝜃𝑛 ) 󵄩󵄩󵄩𝑧𝑛 − 𝑥𝑛 󵄩󵄩󵄩 = 0. 𝑛→∞

󵄩 × [2 󵄩󵄩󵄩𝛼𝑛 (𝑥𝑛 − 𝑝) + 𝜎𝑛 (𝐺𝑥𝑛 − 𝑝) 󵄩 + (1 − 𝛼𝑛 − 𝜎𝑛 ) (𝑦𝑛 − 𝑝)󵄩󵄩󵄩 󵄩 󵄩 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩]

󵄩 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩󵄩 𝜎 󵄩󵄩2 1 − 𝛼𝑛 − 𝜎𝑛 󵄩 󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩 𝑛 (𝐺𝑥𝑛 − 𝑝) + (𝑦𝑛 − 𝑝)󵄩󵄩󵄩 󵄩󵄩 1 − 𝛼𝑛 󵄩󵄩 1 − 𝛼𝑛 󵄩 󵄩 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 × [2 (𝛼𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝜎𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩) + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩] 󵄩 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝛼𝑛 )

(145) ×[

Next, we show that ‖𝑥𝑛 − 𝐺𝑥𝑛 ‖ → 0 as 𝑛 → ∞. Indeed, in terms of Lemma 11, from (130), we have 󵄩2 󵄩󵄩 󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩2 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩2 󵄩2 󵄩2 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩2 = 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 , 󵄩2 󵄩󵄩 󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 󵄩 = 󵄩󵄩󵄩𝛼𝑛 (𝑥𝑛 − 𝑝) + 𝜎𝑛 (𝐺𝑥𝑛 − 𝑝) + (1 − 𝛼𝑛 − 𝜎𝑛 ) (𝑦𝑛 − 𝑝) 󵄩2 + 𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩

󵄩 󵄩 ≤ [󵄩󵄩󵄩𝛼𝑛 (𝑥𝑛 − 𝑝) + 𝜎𝑛 (𝐺𝑥𝑛 − 𝑝) + (1 − 𝛼𝑛 − 𝜎𝑛 ) (𝑦𝑛 − 𝑝)󵄩󵄩󵄩 󵄩 󵄩2 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩] 󵄩 󵄩2 = 󵄩󵄩󵄩𝛼𝑛 (𝑥𝑛 − 𝑝) + 𝜎𝑛 (𝐺𝑥𝑛 − 𝑝) + (1 − 𝛼𝑛 − 𝜎𝑛 ) (𝑦𝑛 − 𝑝)󵄩󵄩󵄩 󵄩 󵄩 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩

󵄩 × [2 󵄩󵄩󵄩𝛼𝑛 (𝑥𝑛 − 𝑝) + 𝜎𝑛 (𝐺𝑥𝑛 − 𝑝) 󵄩 󵄩 󵄩 + (1 − 𝛼𝑛 − 𝜎𝑛 ) (𝑦𝑛 − 𝑝)󵄩󵄩󵄩 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩] 󵄩󵄩 󵄩 = 󵄩󵄩󵄩𝛼𝑛 (𝑥𝑛 − 𝑝) + (1 − 𝛼𝑛 ) 󵄩󵄩 󵄩󵄩2 1 − 𝛼𝑛 − 𝜎𝑛 𝜎𝑛 󵄩 (𝐺𝑥𝑛 − 𝑝) + (𝑦𝑛 − 𝑝)]󵄩󵄩󵄩 󵄩󵄩 1 − 𝛼𝑛 1 − 𝛼𝑛 󵄩 󵄩 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩 ×[

𝜎𝑛 󵄩󵄩 󵄩2 1 − 𝛼𝑛 − 𝜎𝑛 󵄩󵄩 󵄩2 󵄩𝐺𝑥 − 𝑝󵄩󵄩󵄩 + 󵄩𝑦 − 𝑝󵄩󵄩󵄩 1 − 𝛼𝑛 󵄩 𝑛 1 − 𝛼𝑛 󵄩 𝑛 −

𝜎𝑛 (1 − 𝛼𝑛 − 𝜎𝑛 ) 2

(1 − 𝛼𝑛 )

󵄩 󵄩 𝑔 (󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩)]

󵄩 󵄩 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 × [2 (𝛼𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝜎𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩) 󵄩 󵄩 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩] 󵄩 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) ×[

𝜎𝑛 󵄩󵄩 󵄩2 1 − 𝛼𝑛 − 𝜎𝑛 󵄩󵄩 󵄩2 󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩𝑥 − 𝑝󵄩󵄩󵄩 1 − 𝛼𝑛 1 − 𝛼𝑛 󵄩 𝑛 −

𝜎𝑛 (1 − 𝛼𝑛 − 𝜎𝑛 ) 2

(1 − 𝛼𝑛 )

󵄩 󵄩 𝑔 (󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩)]

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩) 󵄩 󵄩 󵄩2 𝜎 (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩󵄩 = 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝑛 𝑔 (󵄩󵄩𝐺𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩) 1 − 𝛼𝑛 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + 󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩) . (146) Then, it immediately follows from 0 ≤ 𝛼𝑛 + 𝜎𝑛 ≤ 1 − 𝜌, for all 𝑛 ≥ 𝑛0 that 󵄩 󵄩 𝜌𝜎𝑛 𝑔 (󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩) ≤

𝜎𝑛 (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩󵄩 󵄩 𝑔 (󵄩󵄩𝐺𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩) 1 − 𝛼𝑛

󵄩 󵄩2 󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩)

Abstract and Applied Analysis

21

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 󵄩 󵄩 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩

󵄩 󵄩 󵄩 󵄩 × (2 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩) , (147) for all 𝑛 ≥ 𝑛0 . Since ‖𝛼𝑛 (𝑓(𝑥𝑛 ) − 𝑥𝑛 )‖ → 0 and {𝑥𝑛 } is bounded, we deduce from (145) and condition (ii) that 󵄩 󵄩 lim 𝑔 (󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩) = 0. (148) 𝑛→∞ Utilizing the properties of 𝑔, we have 󵄩 󵄩 lim 󵄩󵄩𝐺𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩 = 0. 𝑛→∞ 󵄩

(149)

Also, from (130) we have 𝑥𝑛+1 − 𝑥𝑛 = 𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 ) + 𝜎𝑛 (𝐺𝑥𝑛 − 𝑥𝑛 ) + (1 − 𝛼𝑛 − 𝜎𝑛 ) (𝑦𝑛 − 𝑥𝑛 ) = 𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 ) + 𝜎𝑛 (𝐺𝑥𝑛 − 𝑦𝑛 + 𝑦𝑛 − 𝑥𝑛 )

(150)

+ (1 − 𝛼𝑛 − 𝜎𝑛 ) (𝑦𝑛 − 𝑥𝑛 ) = 𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 ) + 𝜎𝑛 (𝐺𝑥𝑛 − 𝑦𝑛 ) + (1 − 𝛼𝑛 ) (𝑦𝑛 − 𝑥𝑛 ) , which hence leads to 󵄩 󵄩 𝜌 󵄩󵄩󵄩𝑦𝑛 − 𝑥𝑛 󵄩󵄩󵄩

󵄩 󵄩 ≤ (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 ≤ (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑥𝑛 󵄩󵄩󵄩

󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 − 𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 ) − 𝜎𝑛 (𝐺𝑥𝑛 − 𝑦𝑛 )󵄩󵄩󵄩

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩 + 𝜎𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩 .

(151)

So, it is easy to see from (145), (149), and ‖𝛼𝑛 (𝑓(𝑥𝑛 ) − 𝑥𝑛 )‖ → 0 that 󵄩 󵄩 lim 󵄩󵄩𝑦 − 𝑥𝑛 󵄩󵄩󵄩 = 0. (152) 𝑛→∞ 󵄩 𝑛 We note that 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝐺𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑥𝑛 󵄩󵄩󵄩 .

(153)

Therefore, from (149) and (152) it follows that 󵄩 󵄩 lim 󵄩󵄩𝐺𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 = 0. 𝑛→∞ 󵄩

(154)

Repeating the same arguments as those of (86), (89), and (91) in the proof of Theorem 24, we can obtain 󵄩 󵄩 󵄩 󵄩 lim 󵄩󵄩𝑥 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 = lim 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 𝑛→∞ 󵄩 𝑛 𝑛→∞ (155) 󵄩 󵄩 = lim 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩 = 0. 𝑛→∞

Suppose that 𝛽𝑛 ≡ 𝛽 for some fixed 𝛽 ∈ (0, 1) such that 𝛽 + 𝛾𝑛 + 𝛿𝑛 = 1 for all 𝑛 ≥ 0. Define a mapping 𝑉𝑥 = (1 − 𝜃1 − 𝜃2 )𝑆𝑥 + 𝜃1 𝐵𝑥 + 𝜃2 𝐺𝑥, where 𝜃1 , 𝜃2 ∈ (0, 1) are two constants with 𝜃1 + 𝜃2 < 1. Then, by Lemmas 14 and 17, we have that Fix(𝑉) = Fix(𝑆) ∩ Fix(𝐵) ∩ Fix(𝐺) = 𝐹. For each 𝑘 ≥ 1, let {𝑝𝑘 } be a unique element of 𝐶 such that 𝑝𝑘 =

1 1 𝑓 (𝑝𝑘 ) + (1 − ) 𝑉𝑝𝑘 . 𝑘 𝑘

(156)

From Lemma 13, we conclude that 𝑝𝑘 → 𝑞 ∈ Fix(𝑉) = 𝐹 as 𝑘 → ∞. Observe that for every 𝑛, 𝑘 󵄩 󵄩󵄩 󵄩󵄩𝑦𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + 𝛿𝑛 (󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩) 󵄩 󵄩 󵄩 󵄩 = 𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛽) 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 ,

(157)

and hence 󵄩󵄩 󵄩 󵄩󵄩𝑥𝑛+1 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝜎𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 + (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 (󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩) 󵄩 󵄩 󵄩 󵄩 + 𝜎𝑛 (󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩) + (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩 󵄩 󵄩 󵄩 × [𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛽) 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩] 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 (󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩) 󵄩 󵄩 󵄩 󵄩 + 𝜎𝑛 (󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩) + (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩 󵄩 × [𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + (1 − 𝛽) (󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑛 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 󵄩󵄩󵄩) 󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩] 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 (󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩) 󵄩 󵄩 󵄩 󵄩 + 𝜎𝑛 (󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩) + (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 × [𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛽) (󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 󵄩󵄩󵄩) 󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩] 󵄩 󵄩 󵄩 󵄩 = 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑥𝑛 󵄩󵄩󵄩 + 𝜎𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 + [𝛽 + (𝛼𝑛 + 𝜎𝑛 ) (1 − 𝛽)] 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 + (1 − 𝛼𝑛 − 𝜎𝑛 ) (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩 󵄩 󵄩 󵄩 × [(1 − 𝛽) 󵄩󵄩󵄩𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩]

22

Abstract and Applied Analysis 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑥𝑛 󵄩󵄩󵄩 + 𝜎𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩

Repeating the same arguments as those of (99), in the proof of Theorem 24, we can get

+ [𝛽 + (𝛼𝑛 + 𝜎𝑛 ) (1 − 𝛽)] 󵄩 󵄩 󵄩 󵄩 × (󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝐵𝑝𝑘 󵄩󵄩󵄩) + (1 − 𝛼𝑛 − 𝜎𝑛 ) (1 − 𝛽) 󵄩 󵄩 󵄩 󵄩 × (󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝𝑘 󵄩󵄩󵄩) + (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩 󵄩 󵄩 󵄩 × [(1 − 𝛽) 󵄩󵄩󵄩𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩] 󵄩 󵄩 󵄩 󵄩 = 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑥𝑛 󵄩󵄩󵄩 + 𝜎𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 + [𝛽 + (𝛼𝑛 + 𝜎𝑛 ) (1 − 𝛽)] 󵄩󵄩󵄩𝑥𝑛+1 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 + (1 − 𝛼𝑛 − 𝜎𝑛 ) (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩 󵄩 󵄩 󵄩 × [(1 − 𝛽) 󵄩󵄩󵄩𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩] 󵄩 󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 . (158) So, it immediately follows from 0 ≤ 𝛼𝑛 ≤ 1 − 𝜌, for all 𝑛 ≥ 𝑛0 that 󵄩󵄩 󵄩 󵄩󵄩𝑥𝑛+1 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 󵄩󵄩󵄩 1 (1 − 𝛼𝑛 − 𝜎𝑛 ) (1 − 𝛽) 󵄩 󵄩 󵄩 󵄩 × (󵄩󵄩󵄩𝛼𝑛 (𝑥𝑛 − 𝑓 (𝑥𝑛 ))󵄩󵄩󵄩 + 𝜎𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩

+

󵄩 󵄩2 󵄩 󵄩2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑝𝑘 󵄩󵄩󵄩 ≤ 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 , 󵄩 󵄩2 󵄩 󵄩2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑝𝑘 󵄩󵄩󵄩 ≤ 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 .

(162)

Utilizing (161) and (162), we deduce that 󵄩 󵄩2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑉𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩2 ≤ (1 − 𝜃1 − 𝜃2 ) 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 + 𝜃1 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝜃2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑝𝑘 󵄩󵄩󵄩

(163)

󵄩 󵄩2 ≤ 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 . Also, observe that 1 1 (1 − ) (𝑥𝑛 − 𝑉𝑝𝑘 ) = 𝑥𝑛 − 𝑝𝑘 − (𝑥𝑛 − 𝑓 (𝑝𝑘 )) . 𝑘 𝑘

(164)

Repeating the same arguments as those of (106) in the proof of Theorem 24, we can get

𝛿 󵄩 󵄩 󵄩 󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩) + 𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 1−𝛽 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 󵄩󵄩󵄩

(159)

1 󵄩󵄩 󵄩2 𝜇 󵄩𝑥 − 𝑝𝑘 󵄩󵄩󵄩 ≥ 𝜇𝑛 ⟨𝑓 (𝑝𝑘 ) − 𝑝𝑘 , 𝐽 (𝑥𝑛 − 𝑝𝑘 )⟩ . 2𝑘 𝑛 󵄩 𝑛

(165)

Since 𝑝𝑘 → 𝑞 ∈ Fix(𝑉) = 𝐹 as 𝑘 → ∞, by the uniform Gateaux differentiability of the norm of 𝑋, we have

1 𝜌 (1 − 𝛽) 󵄩 󵄩 × (󵄩󵄩󵄩𝛼𝑛 (𝑥𝑛 − 𝑓 (𝑥𝑛 ))󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 ) 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝𝑘 󵄩󵄩󵄩 + 𝜏𝑛 , ∀𝑛 ≥ 𝑛0 , 󵄩 󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 +

𝜇𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩ ≤ 0.

(166)

On the other hand, from (135) and the norm-to-weak∗ uniform continuity of 𝐽 on bounded subsets of 𝑋, it follows that

where 𝜃𝑛 = ‖𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 ‖ + ‖𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 ‖ + 1/(𝜌(1 − 𝛽))(‖𝛼𝑛 (𝑥𝑛 − 𝑓(𝑥𝑛 ))‖ + ‖𝐺𝑥𝑛 − 𝑥𝑛 ‖ + ‖𝑥𝑛 − 𝑥𝑛+1 ‖). Since lim𝑛 → ∞ ‖𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 ‖ = lim𝑛 → ∞ ‖𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 ‖ = lim𝑛 → ∞ ‖𝛼𝑛 (𝑥𝑛 − 𝑓(𝑥𝑛 ))‖ = lim𝑛 → ∞ ‖𝐺𝑥𝑛 − 𝑥𝑛 ‖ = lim𝑛 → ∞ ‖𝑥𝑛 − 𝑥𝑛+1 ‖ = 0, we know that 𝜏𝑛 → 0 as 𝑛 → ∞. From (159), we obtain 󵄩󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩 󵄩󵄩𝑥𝑛+1 − 𝐵𝑝𝑘 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝𝑘 󵄩󵄩󵄩 + 𝜏𝑛 (2 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝𝑘 󵄩󵄩󵄩 + 𝜏𝑛 ) , ∀𝑛 ≥ 𝑛0 . (160)

󵄩 󵄩2 󵄩 󵄩2 ≤ 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝𝑘 󵄩󵄩󵄩 = 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 .

So, utilizing Lemma 18, we deduce from (166) and (167) that lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩ ≤ 0, 𝑛→∞

(168)

which, together with (135) and the norm-to-norm uniform continuity of 𝐽 on bounded subsets of 𝑋, implies that

For any Banach limit 𝜇, from (160) we derive 󵄩 󵄩2 󵄩 󵄩2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 = 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛+1 − 𝐵𝑝𝑘 󵄩󵄩󵄩

󵄨 󵄨 lim 󵄨󵄨⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ − ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩󵄨󵄨󵄨 = 0. (167)

𝑛→∞ 󵄨

(161)

lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ ≤ 0. 𝑛→∞

(169)

Abstract and Applied Analysis

23

Finally, let us show that 𝑥𝑛 → 𝑞 as 𝑛 → ∞. Utilizing Lemma 8 (i), from (130) and the convexity of ‖ ⋅ ‖2 , we get 󵄩2 󵄩2 󵄩2 󵄩 󵄩 󵄩󵄩 󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑞󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑞󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 , 󵄩2 󵄩󵄩 󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩 󵄩 = 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑓 (𝑞)) + 𝜎𝑛 (𝐺𝑥𝑛 − 𝑞)

(170)

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝜎𝑛 (󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩) 󵄩 󵄩 󵄩 󵄩 + (1 − 𝛼𝑛 − 𝜎𝑛 ) (󵄩󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩)

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩 + 3 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 .

(173) Since 𝑥𝑛 → 𝑞 and 𝑦𝑛 → 𝑞, we get 𝛼𝑛 (𝑓(𝑥𝑛 ) − 𝑥𝑛 ) → 0. This completes the proof.

󵄩2 + (1 − 𝛼𝑛 − 𝜎𝑛 ) (𝑦𝑛 − 𝑞) + 𝛼𝑛 (𝑓 (𝑞) − 𝑞) 󵄩󵄩󵄩

󵄩 ≤ 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑓 (𝑞)) + 𝜎𝑛 (𝐺𝑥𝑛 − 𝑞) 󵄩2 + (1 − 𝛼𝑛 − 𝜎𝑛 ) (𝑦𝑛 − 𝑞) 󵄩󵄩󵄩 + 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩

󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑓 (𝑞)󵄩󵄩󵄩 + 𝜎𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑞󵄩󵄩󵄩 󵄩 󵄩2 + (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 + 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛼𝑛 𝜌󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝜎𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 󵄩 󵄩2 + (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩

Corollary 29. Let 𝐶 be a nonempty closed convex subset of a uniformly convex Banach space 𝑋 which has a uniformly Gateaux differentiable norm. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶. Let {𝜌𝑛 }∞ 𝑛=0 be a sequence of positive numbers in (0, 𝑏] for some 𝑏 ∈ (0, 1) and 𝐴 𝑖 : 𝐶 → 𝑋𝜉𝑖 strictly pseudocontractive and 𝛼̂𝑖 -strongly accretive with 𝜉𝑖 + 𝛼̂𝑖 ≥ 1 for each 𝑖 = 0, 1, . . .. Define a mapping 𝐺𝑖 : 𝐶 → 𝐶 by Π𝐶(𝐼 − 𝜆 𝑖 𝐴 𝑖 )𝑥 = 𝐺𝑖 𝑥 for all 𝑥 ∈ 𝐶 and 𝑖 = 0, 1, . . ., where 1−(𝜉𝑖 /(1+𝜉𝑖 ))(1−√(1 − 𝛼̂𝑖 )/𝜉𝑖 ) ≤ 𝜆 𝑖 ≤ 1 for all 𝑖 = 0, 1, . . .. Let 𝐵𝑛 : 𝐶 → 𝐶 be the 𝑊-mapping generated by 𝐺𝑛 , 𝐺𝑛−1 , . . . , 𝐺0 and 𝜌𝑛 , 𝜌𝑛−1 , . . . , 𝜌0 . Let 𝑉 : 𝐶 → 𝐶 be a self-mapping such that 𝐼 − 𝑉 : 𝐶 → 𝑋 is 𝜆-strictly pseudocontractive and 𝛼strongly accretive with 𝛼 + 𝜆 ≥ 1. Let 𝑓 : 𝐶 → 𝐶 be a contraction with coefficient 𝜌 ∈ (0, 1). Let {𝑆𝑖 }∞ 𝑖=0 be a countable family of nonexpansive mappings of 𝐶 into itself such that 𝐹 = ∞ (⋂∞ 𝑖=0 Fix(𝑆𝑖 )) ∩ Fix(𝑉) ∩ (⋂𝑖=0 VI(𝐶, 𝐴 𝑖 )) ≠ 0. For arbitrarily given 𝑥0 ∈ 𝐶, let {𝑥𝑛 } be the sequence generated by 𝑦𝑛 = 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 ((1 − 𝑙) 𝐼 + 𝑙𝑉) 𝑥𝑛 ,

+ 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩

𝑥𝑛+1 = 𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝜎𝑛 ((1 − 𝑙) 𝐼 + 𝑙𝑉) 𝑥𝑛 + (1 − 𝛼𝑛 − 𝜎𝑛 ) 𝑦𝑛 ,

󵄩 󵄩2 = (1 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩

∀𝑛 ≥ 0, (174)

+ 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ 󵄩 󵄩2 = (1 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝛼𝑛 (1 − 𝜌)

2 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ . 1−𝜌 (171)

Applying Lemma 7 to (171), we obtain that 𝑥𝑛 → 𝑞 as 𝑛 → ∞. Conversely, if 𝑥𝑛 → 𝑞 ∈ 𝐹 as 𝑛 → ∞, then from (130) it follows that 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩 󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑞󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑞󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 󳨀→ 0

(172)

as 𝑛 → ∞; that is, 𝑦𝑛 → 𝑞. Again from (130) we obtain that 󵄩󵄩 󵄩 󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑥𝑛 )󵄩󵄩󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 − 𝜎𝑛 (𝐺𝑥𝑛 − 𝑥𝑛 ) − (1 − 𝛼𝑛 − 𝜎𝑛 ) (𝑦𝑛 − 𝑥𝑛 )󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 𝜎𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 + (1 − 𝛼𝑛 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝜎𝑛 (󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩) 󵄩 󵄩 󵄩 󵄩 + (1 − 𝛼𝑛 − 𝜎𝑛 ) (󵄩󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩)

where 1 − (𝜆/(1 + 𝜆))(1 − √(1 − 𝛼)/𝜆) ≤ 𝑙 ≤ 1 and {𝜎𝑛 }, {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, and {𝛿𝑛 } are the sequences in [0, 1] such that 𝛽𝑛 + 𝛾𝑛 + 𝛿𝑛 = 1 and 𝛼𝑛 + 𝜎𝑛 ≤ 1 for all 𝑛 ≥ 0. Suppose that the following conditions hold: (i) ∑∞ 𝑛=0 𝛼𝑛 = ∞ and 0 ≤ 𝛼𝑛 + 𝜎𝑛 ≤ 1 − 𝜌, for all 𝑛 ≥ 𝑛0 for some integer 𝑛0 ≥ 0; (ii) lim inf 𝑛 → ∞ 𝜎𝑛 > lim inf 𝑛 → ∞ 𝛿𝑛 > 0;

0, lim inf 𝑛 → ∞ 𝛾𝑛

>

0 and

(iii) lim𝑛 → ∞ (|𝛼𝑛+1 /(1−(1−𝛼𝑛+1 −𝜎𝑛+1 )𝛽𝑛+1 )−𝛼𝑛 /(1−(1− 𝛼𝑛 −𝜎𝑛 )𝛽𝑛 )|+|𝜎𝑛+1 /(1−(1−𝛼𝑛+1 −𝜎𝑛+1 )𝛽𝑛+1 )−𝜎𝑛 /(1− (1 − 𝛼𝑛 − 𝜎𝑛 )𝛽𝑛 )| + |𝛿𝑛+1 /(1 − 𝛽𝑛+1 ) − 𝛿𝑛 /(1 − 𝛽𝑛 )|) = 0; (iv) 0 < lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ 𝛽𝑛 < 1. Assume that ∑∞ 𝑛=0 sup𝑥∈𝐷 ‖𝑆𝑛+1 𝑥 − 𝑆𝑛 𝑥‖ < ∞ for any bounded subset 𝐷 of 𝐶 and let 𝑆 be a mapping of 𝐶 into itself defined by 𝑆𝑥 = lim𝑛 → ∞ 𝑆𝑛 𝑥 for all 𝑥 ∈ 𝐶 and suppose that Fix(𝑆) = ⋂∞ 𝑖=0 Fix(𝑆𝑖 ). Then there hold the following: (I) lim𝑛 → ∞ ‖𝑥𝑛+1 − 𝑥𝑛 ‖ = 0; (II) 𝑥𝑛 → 𝑞 ⇔ 𝛼𝑛 (𝑓(𝑥𝑛 ) − 𝑥𝑛 ) → 0 provided 𝛽𝑛 ≡ 𝛽 for some fixed 𝛽 ∈ (0, 1), where 𝑞 ∈ 𝐹 solves the following VIP ⟨𝑞 − 𝑓 (𝑞) , 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,

∀𝑝 ∈ 𝐹.

(175)

24

Abstract and Applied Analysis

Proof. In Theorem 28, we put 𝐵1 = 𝐼 − 𝑉, 𝐵2 = 0, and 𝜇1 = 𝑙, where 1 − (𝜆/(1 + 𝜆))(1 − √(1 − 𝛼)/𝜆) ≤ 𝑙 ≤ 1. Then, GSVI (13) is equivalent to the VIP of finding 𝑥∗ ∈ 𝐶 such that ⟨𝐵1 𝑥∗ , 𝐽 (𝑥 − 𝑥∗ )⟩ ≥ 0,

∀𝑥 ∈ 𝐶.

(176)

In this case, 𝐵1 : 𝐶 → 𝑋 is 𝜆-strictly pseudocontractive and 𝛼-strongly accretive. Repeating the same arguments as those in the proof of Corollary 25, we can infer that Fix(𝑉) = VI(𝐶, 𝐵1 ). Accordingly, 𝐹 = ⋂∞ 𝑖=0 Fix(𝑆𝑖 ) ∩ Ω ∩ ∞ ∞ (⋂∞ 𝑖=0 VI(𝐶, 𝐴 𝑖 )) = ⋂𝑖=0 Fix(𝑆𝑖 ) ∩ Fix(𝑉) ∩ (⋂𝑖=0 VI(𝐶, 𝐴 𝑖 )), and 𝐺𝑥𝑛 = ((1 − 𝑙) 𝐼 + 𝑙𝑉) 𝑥𝑛 ,

∀𝑛 ≥ 0.

(177)

So, scheme (130) reduces to (174). Therefore, the desired result follows from Theorem 31. Remark 30. Our Theorems 24 and 28 improve, extend, supplement and develop Ceng and Yao’s [10, Theorem 3.2], Cai and Bu’s [11, Theorem 3.1], Kangtunyakarn’s [38, Theorem 3.1], and Ceng and Yao’s [8, Theorem 3.1], in the following aspects. (i) The problem of finding a point 𝑞 ∈ (⋂∞ 𝑖=0 Fix(𝑆𝑖 )) ∩ Ω ∩ (⋂∞ 𝑖=0 VI(𝐶, 𝐴 𝑖 )) in our Theorems 24 and 28 is more general and more subtle than every one of the problem of finding a point 𝑞 ∈ ⋂∞ 𝑖=0 Fix(𝑇𝑖 ) in [10, Theorem 3.2], the problem of finding a point 𝑞 ∈ ⋂∞ 𝑖=1 Fix(𝑇𝑖 ) ∩ Ω in [11, Theorem 3.1], the problem of finding a point 𝑞 ∈ Fix(𝑆) ∩ Fix(𝑉) ∩ (⋂𝑁 𝑖=1 VI(𝐶, 𝐴 𝑖 )) in [38, Theorem 3.1], and the problem of finding a point 𝑞 ∈ Fix(𝑇) in [8, Theorem 3.1]. (ii) The iterative scheme in [8, Theorem 3.1] is extended to develop the iterative schemes (42) and (130) in our Theorems 24 and 28 by virtue of the iterative schemes of [11, Theorem 3.1] and [10, Theorems 3.2]. The iterative schemes (42) and (130) in our Theorems 24 and 28 are more advantageous and more flexible than the iterative scheme of [8, Theorem 3.1] because they can be applied to solving three problems (i.e., GSVI (13), fixed point problem and infinitely many VIPs), and involve several parameter sequences {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, {𝛿𝑛 }, (and {𝜎𝑛 }). (iii) Our Theorems 24 and 28 extend and generalize Ceng and Yao [8, Theorem 3.1] from a nonexpansive mapping to a countable family of nonexpansive mappings, and Ceng and Yao’s [10, Theorems 3.2], to the setting of the GSVI (13) and infinitely many VIPs, Kangtunyakarn [38, Theorem 3.1], from finitely many VIPs to infinitely many VIPs, from a nonexpansive mapping to a countable family of nonexpansive mappings and from a strict pseudocontraction to the GSVI (13). In the meantime, our Theorems 24 and 28 extend and generalize Cai and Bu’s [11, Theorem 3.1], to the setting of infinitely many VIPs. (iv) The iterative schemes (42) and (130) in our Theorems 24 and 28 are very different from every one in [10, Theorem 3.2], [11, Theorem 3.1], [38, Theorem 3.1],

and [8, Theorem 3.1] because the mappings 𝐺 and 𝑇𝑛 in [11, Theorem 3.1] and the mapping 𝑇 in [8, Theorem 3.1] are replaced with the same composite mapping 𝑆𝑛 𝐺 in the iterative schemes (42) and (130) and the mapping 𝑊𝑛 in [10, Theorem 3.2] is replaced with 𝐵𝑛 . (v) Cai and Bu’s proof in [11, Theorem 3.1] depends on the argument techniques in [14], the inequality in 2uniformly smooth Banach spaces (see Lemma 4), and the inequality in smooth and uniform convex Banach spaces (see Proposition 6). Because the composite mapping 𝑆𝑛 𝐺 appears in the iterative scheme (42) of our Theorem 24, the proof of our Theorem 24 depends on the argument techniques in [14], the inequality in 2-uniformly smooth Banach spaces (see Lemma 4), the inequality in smooth and uniform convex Banach spaces (see Proposition 6), the inequality in uniform convex Banach spaces (see Lemma 15 in Section 2 of this paper), and the properties of the 𝑊-mapping and the Banach limit (see Lemmas 16–18 in Section 2 of this paper). However, the proof of our Theorem 28 does not depend on the argument techniques in [14], the inequality in 2-uniformly smooth Banach spaces (see Lemma 4), and the inequality in smooth and uniform convex Banach spaces (see Proposition 6). It depends on only the inequality in uniform convex Banach spaces (see Lemma 15 in Section 2 of this paper) and the properties of the 𝑊-mapping and the Banach limit (see Lemmas 16–18 in Section 2 of this paper). (vi) The assumption of the uniformly convex and 2uniformly smooth Banach space 𝑋 in [11, Theorem 3.1] is weakened to the one of the uniformly convex Banach space 𝑋 having a uniformly Gateaux differentiable norm in our Theorem 28. Moreover, the assumption of the uniformly smooth Banach space 𝑋 in [8, Theorem 3.1] is replaced with the one of the uniformly convex Banach space 𝑋 having a uniformly Gateaux differentiable norm in our Theorem 28. It is worth emphasizing that there is no assumption on the convergence of parameter sequences {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, and {𝛿𝑛 } (and {𝜎𝑛 }) to zero in our Theorems 24 and 28.

4. Relaxed Mann Iterations and Their Convergence Criteria In this section, we introduce our relaxed Mann iteration algorithms in real smooth and uniformly convex Banach spaces and present their convergence criteria. Theorem 31. Let 𝐶 be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space 𝑋. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶. Let {𝜌𝑛 }∞ 𝑛=0 be a sequence of positive numbers in (0, 𝑏] for some 𝑏 ∈ (0, 1) and 𝐴 𝑖 : 𝐶 → 𝑋 an 𝛼̂𝑖 -inverse strongly accretive mapping for each 𝑖 = 0, 1, . . .. Define a mapping 𝐺𝑖 : 𝐶 → 𝐶 by Π𝐶(𝐼 − 𝜆 𝑖 𝐴 𝑖 )𝑥 = 𝐺𝑖 𝑥 for all 𝑥 ∈ 𝐶 and 𝑖 = 0, 1, . . .,

Abstract and Applied Analysis

25

where 𝜆 𝑖 ∈ (0, 𝛼̂𝑖 /𝜅2 ] and 𝜅 is the 2-uniformly smooth constant of 𝑋. Let 𝐵𝑛 : 𝐶 → 𝐶 be the 𝑊-mapping generated by 𝐺𝑛 , 𝐺𝑛−1 , . . . , 𝐺0 and 𝜌𝑛 , 𝜌𝑛−1 , . . . , 𝜌0 . Let the mapping 𝐵𝑖 : 𝐶 → 𝑋 be 𝛽̂𝑖 -inverse strongly accretive for 𝑖 = 1, 2. Let 𝑓 : 𝐶 → 𝐶 be a contraction with coefficient 𝜌 ∈ (0, 1). Let {𝑆𝑖 }∞ 𝑖=0 be a countable family of nonexpansive mappings of 𝐶 into itself ∞ such that 𝐹 = (⋂∞ 𝑖=0 Fix(𝑆𝑖 )) ∩ Ω ∩ (⋂𝑖=0 VI(𝐶, 𝐴 𝑖 )) ≠ 0, where Ω is the fixed point set of the mapping 𝐺 = Π𝐶(𝐼−𝜇1 𝐵1 )Π𝐶(𝐼− 𝜇2 𝐵2 ) with 0 < 𝜇𝑖 < 𝛽̂𝑖 /𝜅2 for 𝑖 = 1, 2. For arbitrarily given 𝑥0 ∈ 𝐶, let {𝑥𝑛 } be the sequence generated by 𝑥𝑛+1 = 𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 ,

∀𝑛 ≥ 0, (178)

where {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, and {𝛿𝑛 } are the sequences in (0, 1) such that 𝛼𝑛 +𝛽𝑛 +𝛾𝑛 +𝛿𝑛 = 1 for all 𝑛 ≥ 0. Suppose that the following conditions hold: (i) lim𝑛 → ∞ 𝛼𝑛 = 0 and ∑∞ 𝑛=0 𝛼𝑛 = ∞; (ii) {𝛾𝑛 }, {𝛿𝑛 } ⊂ [𝑐, 𝑑] for some 𝑐, 𝑑 ∈ (0, 1); (iii) lim𝑛 → ∞ (|𝛽𝑛 − 𝛽𝑛−1 | + |𝛾𝑛 − 𝛾𝑛−1 | + |𝛿𝑛 − 𝛿𝑛−1 |) = 0; (iv) 0 < lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ 𝛽𝑛 < 1. Assume that ∑∞ 𝑛=1 sup𝑥∈𝐷 ‖𝑆𝑛 𝑥 − 𝑆𝑛−1 𝑥‖ < ∞ for any bounded subset 𝐷 of 𝐶 and let 𝑆 be a mapping of 𝐶 into itself defined by 𝑆𝑥 = lim𝑛 → ∞ 𝑆𝑛 𝑥 for all 𝑥 ∈ 𝐶 and suppose that Fix(𝑆) = ⋂∞ 𝑖=0 Fix(𝑆𝑖 ). Then, there hold the following: (I) lim𝑛 → ∞ ‖𝑥𝑛+1 − 𝑥𝑛 ‖ = 0; (II) the sequence {𝑥𝑛 }∞ 𝑛=0 converges strongly to some 𝑞 ∈ 𝐹 which is the unique solution of the variational inequality problem (VIP) ⟨(𝐼 − 𝑓) 𝑞, 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,

∀𝑝 ∈ 𝐹,

𝑛→∞

(180)

for every 𝑥 ∈ 𝐶. That is, such a 𝐵 is the 𝑊-mapping generated ∞ by the sequences {𝐺𝑛 }∞ 𝑛=0 and {𝜌𝑛 }𝑛=0 . According to Lemma 17, ∞ we know that Fix(𝐵) = ⋂𝑖=0 Fix(𝐺𝑖 ). From Lemma 21 and the definition of 𝐺𝑖 , we have Fix(𝐺𝑖 ) = VI(𝐶, 𝐴 𝑖 ) for each 𝑖 = 0, 1, . . .. Hence, we have ∞

∞

𝑖=0

𝑖=0

Fix (𝐵) = ⋂ Fix (𝐺𝑖 ) = ⋂ VI (𝐶, 𝐴 𝑖 ) .

󵄩 󵄩 󵄩 󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩 ≤ max {󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 , 󵄩 }. 1−𝜌 (182) By induction, we obtain 󵄩 󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩 { }, 𝑥 ≤ max , 𝑥 − 𝑝 − 𝑝 󵄩󵄩 0 󵄩󵄩 󵄩󵄩 󵄩󵄩 𝑛 1−𝜌

∀𝑛 ≥ 0. (183)

Hence, {𝑥𝑛 } is bounded, and so are the sequences {𝐺𝑥𝑛 } and {𝑓(𝑥𝑛 )}. Let us show that 󵄩󵄩

lim 󵄩𝑥 𝑛 → ∞ 󵄩 𝑛+1

󵄩 − 𝑥𝑛 󵄩󵄩󵄩 = 0.

(184)

As a matter of fact, observe that 𝑥𝑛+1 can be rewritten as follows:

𝛼𝑖 /𝜅2 ] for 𝑖 = 0, 1, . . ., it Proof. First of all, since 0 < 𝜆 𝑖 < [̂ is easy to see that 𝐺𝑖 is a nonexpansive mapping for each 𝑖 = 0, 1, . . .. Since 𝐵𝑛 : 𝐶 → 𝐶 is the 𝑊-mapping generated by 𝐺𝑛 , 𝐺𝑛−1 , . . . , 𝐺0 , and 𝜌𝑛 , 𝜌𝑛−1 , . . . , 𝜌0 , by Lemma 16 we know that, for each 𝑥 ∈ 𝐶 and 𝑘 ≥ 0, the limit lim𝑛 → ∞ 𝑈𝑛,𝑘 𝑥 exists. Moreover, one can define a mapping 𝐵 : 𝐶 → 𝐶 as follows: 𝑛→∞

󵄩 󵄩󵄩 󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 (󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑓 (𝑝)󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩) 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 (𝜌 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩) 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩󵄩 󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩 󵄩 = (1 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛼𝑛 (1 − 𝜌) 󵄩 1−𝜌

(179)

provided 𝛽𝑛 ≡ 𝛽 for some fixed 𝛽 ∈ (0, 1).

𝐵𝑥 = lim 𝐵𝑛 𝑥 = lim 𝑈𝑛,0 𝑥

𝑝 = 𝐵𝑛 𝑝, and 𝑝 = 𝑆𝑛 𝑝 for all 𝑛 ≥ 0. By Lemma 23, we know that 𝐺 is nonexpansive. Then, from (178), we have

(181)

Next, let us show that the sequence {𝑥𝑛 } is bounded. Indeed, take a fixed 𝑝 ∈ 𝐹 arbitrarily. Then, we get 𝑝 = 𝐺𝑝,

𝑥𝑛+1 = 𝛽𝑛 𝑥𝑛 + (1 − 𝛽𝑛 ) 𝑧𝑛 ,

(185)

where 𝑧𝑛 = (𝛼𝑛 𝑓(𝑥𝑛 ) + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 )/(1 − 𝛽𝑛 ). Observe that 󵄩 󵄩󵄩 󵄩󵄩𝑧𝑛 − 𝑧𝑛−1 󵄩󵄩󵄩 󵄩󵄩 𝛼 𝑓 (𝑥 ) + 𝛾 𝐵 𝑥 + 𝛿 𝑆 𝐺𝑥 󵄩 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 = 󵄩󵄩󵄩󵄩 𝑛 1 − 𝛽𝑛 󵄩󵄩 −

𝛼𝑛−1 𝑓 (𝑥𝑛−1 ) + 𝛾𝑛−1 𝐵𝑛−1 𝑥𝑛−1 + 𝛿𝑛−1 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 1 − 𝛽𝑛−1 󵄩

󵄩󵄩 𝑥 − 𝛽 𝑥 𝑥 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩󵄩 󵄩 𝑛 𝑛 = 󵄩󵄩󵄩 𝑛+1 − 𝑛 󵄩 󵄩󵄩 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄩󵄩󵄩 󵄩󵄩 𝑥 − 𝛽 𝑥 𝑥 − 𝛽𝑛−1 𝑥𝑛−1 󵄩 𝑛 𝑛 = 󵄩󵄩󵄩 𝑛+1 − 𝑛 1 − 𝛽𝑛 󵄩󵄩 1 − 𝛽𝑛 +

𝑥𝑛 − 𝛽𝑛−1 𝑥𝑛−1 𝑥𝑛 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩󵄩 − 󵄩 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄩󵄩󵄩

26

Abstract and Applied Analysis 󵄩󵄩 𝑥 − 𝛽 𝑥 𝑥 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩󵄩 󵄩 𝑛 𝑛 − 𝑛 ≤ 󵄩󵄩󵄩 𝑛+1 󵄩󵄩 󵄩󵄩 1 − 𝛽𝑛 󵄩󵄩 1 − 𝛽𝑛 󵄩󵄩 𝑥 − 𝛽 𝑥 𝑥 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩󵄩 󵄩 𝑛−1 𝑛−1 + 󵄩󵄩󵄩 𝑛 − 𝑛 󵄩 󵄩󵄩 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄩󵄩󵄩 =

=

=

󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 𝜆 0 󵄩󵄩󵄩𝑈𝑛,1 𝑥𝑛−1 − 𝑈𝑛−1,1 𝑥𝑛−1 󵄩󵄩󵄩

󵄩 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 𝜆 0 󵄩󵄩󵄩𝜆 1 𝐺1 𝑈𝑛,2 𝑥𝑛−1 − 𝜆 1 𝐺1 𝑈𝑛−1,2 𝑥𝑛−1 󵄩󵄩󵄩

󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 𝜆 0 𝜆 1 󵄩󵄩󵄩𝑈𝑛,2 𝑥𝑛−1 − 𝑈𝑛−1,2 𝑥𝑛−1 󵄩󵄩󵄩

1 󵄩󵄩 󵄩 󵄩𝑥 − 𝛽𝑛 𝑥𝑛 − (𝑥𝑛 − 𝛽𝑛−1 𝑥𝑛−1 )󵄩󵄩󵄩 1 − 𝛽𝑛 󵄩 𝑛+1 󵄨󵄨 1 󵄨󵄨 1 󵄨 󵄨󵄨 󵄩󵄩 󵄩 + 󵄨󵄨󵄨 − 󵄨 󵄩𝑥 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 𝑛

.. . 𝑛−1

󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + (∏𝜆 𝑖 ) 󵄩󵄩󵄩𝑈𝑛,𝑛 𝑥𝑛−1 − 𝑈𝑛−1,𝑛 𝑥𝑛−1 󵄩󵄩󵄩 𝑖=0

1 󵄩󵄩 󵄩 󵄩𝑥 − 𝛽𝑛 𝑥𝑛 − (𝑥𝑛 − 𝛽𝑛−1 𝑥𝑛−1 )󵄩󵄩󵄩 1 − 𝛽𝑛 󵄩 𝑛+1 󵄨󵄨 󵄨 󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 󵄩󵄩 + 󵄩𝑥 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 (1 − 𝛽𝑛−1 ) (1 − 𝛽𝑛 ) 󵄩 𝑛

𝑛−1

󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 𝑀∏𝜆 𝑖 , 𝑖=0

(188)

1 󵄩󵄩 󵄩𝛼 𝑓 (𝑥𝑛 ) + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 − 𝛼𝑛−1 𝑓 (𝑥𝑛−1 ) 1 − 𝛽𝑛 󵄩 𝑛 󵄩 − 𝛾𝑛−1 𝐵𝑛−1 𝑥𝑛−1 − 𝛿𝑛−1 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 󵄨󵄨 󵄨 󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 󵄩󵄩 + 󵄩𝑥 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 (1 − 𝛽 ) (1 − 𝛽 ) 󵄩 𝑛 𝑛−1

𝑛

1 󵄩 󵄩 ≤ [𝛼 󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑓 (𝑥𝑛−1 )󵄩󵄩󵄩 1 − 𝛽𝑛 𝑛 󵄩 󵄩 󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑓 (𝑥𝑛−1 )󵄩󵄩󵄩 󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛾𝑛 − 𝛾𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛿𝑛 − 𝛿𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩] 󵄨󵄨 󵄨 󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 󵄩󵄩 + 󵄩𝑥 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 . (1 − 𝛽𝑛−1 ) (1 − 𝛽𝑛 ) 󵄩 𝑛 (186)

for some constant 𝑀 > 0. Taking into account 0 < lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ 𝛽𝑛 < 1, we may assume, ̂ Utilizing (186)– without loss of generality, that {𝛽𝑛 } ⊂ [̂𝑐, 𝑑]. (188), we have 󵄩 󵄩󵄩 󵄩󵄩𝑧𝑛 − 𝑧𝑛−1 󵄩󵄩󵄩 ≤

≤

On the other hand, we note that, for all 𝑛 ≥ 1,

1 󵄩 󵄩 [𝛼 󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑓 (𝑥𝑛−1 )󵄩󵄩󵄩 1 − 𝛽𝑛 𝑛 󵄩 󵄩 󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑓 (𝑥𝑛−1 )󵄩󵄩󵄩 󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛾𝑛 − 𝛾𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛿𝑛 − 𝛿𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩] 󵄨󵄨 󵄨 󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 󵄩󵄩 + 󵄩𝑥 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 (1 − 𝛽𝑛−1 ) (1 − 𝛽𝑛 ) 󵄩 𝑛 𝑛−1 1 󵄩 󵄩 󵄩 󵄩 {𝛼𝑛 𝜌 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 𝛾𝑛 [󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 𝑀∏𝜆 𝑖 ] 1 − 𝛽𝑛 𝑖=0

󵄩 󵄩 󵄩 󵄩 + 𝛿𝑛 [󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛−1 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩]

󵄩󵄩 󵄩 󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩

󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛−1 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝐺𝑥𝑛−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛−1 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛−1 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 .

󵄨 󵄨󵄩 󵄨󵄩 󵄩 󵄩 󵄨 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑓 (𝑥𝑛−1 )󵄩󵄩󵄩 + 󵄨󵄨󵄨𝛾𝑛 − 𝛾𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 (187)

󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛿𝑛 − 𝛿𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 } +

Furthermore, by (CY), since 𝐺𝑖 and 𝑈𝑛,𝑖 are nonexpansive, we deduce that for each 𝑛 ≥ 1 󵄩 󵄩󵄩 󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑛 𝑥𝑛−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛−1 − 𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛−1 − 𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝜆 0 𝐺0 𝑈𝑛,1 𝑥𝑛−1 − 𝜆 0 𝐺0 𝑈𝑛−1,1 𝑥𝑛−1 󵄩󵄩󵄩

=

󵄨󵄨 󵄨 󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 󵄩󵄩 󵄩𝑥 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 (1 − 𝛽𝑛−1 ) (1 − 𝛽𝑛 ) 󵄩 𝑛

1 󵄩 󵄩 { (1 − 𝛽𝑛 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 1 − 𝛽𝑛 𝑛−1

󵄩 󵄩 + 𝛾𝑛 𝑀∏𝜆 𝑖 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛−1 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 𝑖=0

󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑓 (𝑥𝑛−1 )󵄩󵄩󵄩

Abstract and Applied Analysis

27

󵄨󵄩 󵄩 󵄨 + 󵄨󵄨󵄨𝛾𝑛 − 𝛾𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩

Next we show that ‖𝑥𝑛 − 𝐺𝑥𝑛 ‖ → 0 as 𝑛 → ∞. Indeed, for simplicity, put 𝑞 = Π𝐶(𝑝 − 𝜇2 𝐵2 𝑝), 𝑢𝑛 = Π𝐶(𝑥𝑛 − 𝜇2 𝐵2 𝑥𝑛 ) and V𝑛 = Π𝐶(𝑢𝑛 − 𝜇1 𝐵1 𝑢𝑛 ). Then, V𝑛 = 𝐺𝑥𝑛 for all 𝑛 ≥ 0. From Lemma 26 we have

󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛿𝑛 − 𝛿𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 } +

󵄨󵄨 󵄨 󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 󵄩󵄩 󵄩𝑥 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 (1 − 𝛽𝑛−1 ) (1 − 𝛽𝑛 ) 󵄩 𝑛

= (1 −

󵄩2 󵄩󵄩 󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 󵄩 󵄩2 = 󵄩󵄩󵄩Π𝐶 (𝑥𝑛 − 𝜇2 𝐵2 𝑥𝑛 ) − Π𝐶 (𝑝 − 𝜇2 𝐵2 𝑝)󵄩󵄩󵄩 (193) 󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝 − 𝜇2 (𝐵2 𝑥𝑛 − 𝐵2 𝑝)󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 2𝜇2 (𝛽̂2 − 𝜅2 𝜇2 ) 󵄩󵄩󵄩𝐵2 𝑥𝑛 − 𝐵2 𝑝󵄩󵄩󵄩 ,

𝑛−1 𝛼𝑛 (1 − 𝜌) 󵄩󵄩 󵄩 𝛾𝑀 ) 󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 𝑛 ∏𝜆 𝑖 1 − 𝛽𝑛 1 − 𝛽𝑛 𝑖=0

𝛿𝑛 󵄩󵄩 1 󵄩 󵄩𝑆 𝐺𝑥 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 + 1 − 𝛽𝑛 󵄩 𝑛 𝑛−1 1 − 𝛽𝑛 󵄨󵄩 󵄨󵄩 󵄩 󵄨 󵄩 󵄨 × [󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑓 (𝑥𝑛−1 )󵄩󵄩󵄩 + 󵄨󵄨󵄨𝛾𝑛 − 𝛾𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛿𝑛 − 𝛿𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩] 󵄨󵄨 󵄨󵄨 󵄩 󵄩󵄩 󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨 + 󵄩𝑥 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 (1 − 𝛽 ) (1 − 𝛽 ) 󵄩 𝑛 +

𝑛−1

󵄩󵄩 󵄩2 󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩2 = 󵄩󵄩󵄩Π𝐶 (𝑢𝑛 − 𝜇1 𝐵1 𝑢𝑛 ) − Π𝐶 (𝑞 − 𝜇1 𝐵1 𝑞)󵄩󵄩󵄩 󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝑢𝑛 − 𝑞 − 𝜇1 (𝐵1 𝑢𝑛 − 𝐵1 𝑞)󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 − 2𝜇1 (𝛽̂1 − 𝜅2 𝜇1 ) 󵄩󵄩󵄩𝐵1 𝑢𝑛 − 𝐵1 𝑞󵄩󵄩󵄩 .

𝑛

𝑛−1

󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 𝑀∏𝜆 𝑖 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛−1 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 𝑖=0

Substituting (193) for (194), we obtain

1 󵄨 󵄩 󵄨 󵄨󵄩 󵄨 + [󵄨󵄨𝛼 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑓 (𝑥𝑛−1 )󵄩󵄩󵄩 + 󵄨󵄨󵄨𝛾𝑛 − 𝛾𝑛−1 󵄨󵄨󵄨 1 − 𝛽𝑛 󵄨 𝑛 󵄩 󵄩 󵄨 󵄨󵄩 󵄩 × 󵄩󵄩󵄩𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 + 󵄨󵄨󵄨𝛿𝑛 − 𝛿𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩] 󵄨󵄨 󵄨 󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 + (1 − 𝛽𝑛−1 ) (1 − 𝛽𝑛 ) 󵄩 󵄩 × 󵄩󵄩󵄩𝛼𝑛−1 𝑓 (𝑥𝑛−1 ) + 𝛾𝑛−1 𝐵𝑛−1 𝑥𝑛−1 + 𝛿𝑛−1 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩

󵄩󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 2 󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 2𝜇2 (𝛽̂2 − 𝜅 𝜇2 ) 󵄩󵄩󵄩𝐵2 𝑥𝑛 − 𝐵2 𝑝󵄩󵄩󵄩 󵄩 󵄩2 − 2𝜇1 (𝛽̂1 − 𝜅2 𝜇1 ) 󵄩󵄩󵄩𝐵1 𝑢𝑛 − 𝐵1 𝑞󵄩󵄩󵄩 .

By Lemma 8, we have from (178) and (195) 󵄩󵄩 󵄩2 󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 󵄩 = 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑓 (𝑝)) + 𝛽𝑛 (𝑥𝑛 − 𝑝)

𝑛−1

󵄨 󵄨 + 𝑀1 [∏𝜆 𝑖 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 𝑖=0

󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 + 󵄨󵄨󵄨𝛾𝑛 − 𝛾𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛿𝑛 − 𝛿𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 ]

+ 𝛾𝑛 (𝐵𝑛 𝑥𝑛 − 𝑝) + 𝛿𝑛 (𝑆𝑛 𝐺𝑥𝑛 − 𝑝) 󵄩2 + 𝛼𝑛 (𝑓 (𝑝) − 𝑝)󵄩󵄩󵄩

󵄩 󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛−1 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 , (189) ̂ 2 )(‖𝑓(𝑥 )‖+‖𝐵 𝑥 ‖+‖𝑆 𝐺𝑥 ‖+𝑀)} ≤ where sup𝑛≥0 {(1/(1− 𝑑) 𝑛 𝑛 𝑛 𝑛 𝑛 𝑀1 for some 𝑀1 > 0. Thus, from (189), conditions (i), (iii) and the assumption on {𝑆𝑛 }, it follows that (noting that 0 < 𝜆 𝑖 ≤ 𝑏 < 1, for all 𝑖 ≥ 0) 󵄩 󵄩 󵄩 󵄩 lim (󵄩󵄩𝑧 − 𝑧𝑛−1 󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩) ≤ 0. (190) 𝑛→∞ 󵄩 𝑛 Since 0 < lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ 𝛽𝑛 < 1, by Lemma 20 we get 󵄩 󵄩 lim 󵄩󵄩𝑥 − 𝑧𝑛 󵄩󵄩󵄩 = 0. (191) 𝑛→∞ 󵄩 𝑛 Consequently, 󵄩 󵄩 󵄩 󵄩 lim 󵄩󵄩𝑥 − 𝑥𝑛 󵄩󵄩󵄩 = lim (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑧𝑛 − 𝑥𝑛 󵄩󵄩󵄩 = 0. 𝑛 → ∞ 󵄩 𝑛+1 𝑛→∞

(194)

󵄩 ≤ 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑓 (𝑝)) + 𝛽𝑛 (𝑥𝑛 − 𝑝) 󵄩2 + 𝛾𝑛 (𝐵𝑛 𝑥𝑛 − 𝑝) + 𝛿𝑛 (𝑆𝑛 𝐺𝑥𝑛 − 𝑝)󵄩󵄩󵄩 + 2𝛼𝑛 ⟨𝑓 (𝑝) − 𝑝, 𝐽 (𝑥𝑛+1 − 𝑝)⟩ 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑓 (𝑝)󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩2 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 2𝛼𝑛 ⟨𝑓 (𝑝) − 𝑝, 𝐽 (𝑥𝑛+1 − 𝑝)⟩ 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛼𝑛 𝜌2 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩2 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩

(192)

+ 2𝛼𝑛 ⟨𝑓 (𝑝) − 𝑝, 𝐽 (𝑥𝑛+1 − 𝑝)⟩

(195)

28

Abstract and Applied Analysis 󵄩 󵄩 󵄩 󵄩2 󵄩2 󵄩2 ≤ 𝛼𝑛 𝜌󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩

which implies that

󵄩 󵄩2 󵄩 󵄩2 + 𝛿𝑛 [󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 2𝜇2 (𝛽̂2 − 𝜅2 𝜇2 ) 󵄩󵄩󵄩𝐵2 𝑥𝑛 − 𝐵2 𝑝󵄩󵄩󵄩 󵄩 󵄩2 −2𝜇1 (𝛽̂1 − 𝜅2 𝜇1 ) 󵄩󵄩󵄩𝐵1 𝑢𝑛 − 𝐵1 𝑞󵄩󵄩󵄩 ] + 2𝛼𝑛 ⟨𝑓 (𝑝) − 𝑝, 𝐽 (𝑥𝑛+1 − 𝑝)⟩ 󵄩 󵄩2 = (1 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩

󵄩󵄩 󵄩2 󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩

󵄩 󵄩2 +𝜇1 (𝛽̂1 − 𝜅2 𝜇1 ) 󵄩󵄩󵄩𝐵1 𝑢𝑛 − 𝐵1 𝑞󵄩󵄩󵄩 ]

󵄩 󵄩2 = 󵄩󵄩󵄩Π𝐶 (𝑢𝑛 − 𝜇1 𝐵1 𝑢𝑛 ) − Π𝐶 (𝑞 − 𝜇1 𝐵1 𝑞)󵄩󵄩󵄩

+ 2𝛼𝑛 ⟨𝑓 (𝑝) − 𝑝, 𝐽 (𝑥𝑛+1 − 𝑝)⟩ 󵄩2 󵄩 󵄩2 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 2𝛿𝑛 [𝜇2 (𝛽̂2 − 𝜅2 𝜇2 ) 󵄩󵄩󵄩𝐵2 𝑥𝑛 − 𝐵2 𝑝󵄩󵄩󵄩 󵄩 󵄩2 + 𝜇1 (𝛽̂1 − 𝜅2 𝜇1 ) 󵄩󵄩󵄩𝐵1 𝑢𝑛 − 𝐵1 𝑞󵄩󵄩󵄩 ] 󵄩󵄩 󵄩 󵄩 + 2𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 , which hence implies that 󵄩 󵄩2 2𝛿𝑛 [𝜇2 (𝛽̂2 − 𝜅2 𝜇2 ) 󵄩󵄩󵄩𝐵2 𝑥𝑛 − 𝐵2 𝑝󵄩󵄩󵄩 󵄩 󵄩2 + 𝜇1 (𝛽̂1 − 𝜅2 𝜇1 ) 󵄩󵄩󵄩𝐵1 𝑢𝑛 − 𝐵1 𝑞󵄩󵄩󵄩 ] (197)

󵄩 󵄩 lim 󵄩󵄩𝐵 𝑢 − 𝐵1 𝑞󵄩󵄩󵄩 = 0. (198) 𝑛→∞ 󵄩 1 𝑛

Utilizing Proposition 6 and Lemma 9, we have

󵄩󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩 󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 − 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩) 󵄩 󵄩󵄩 󵄩 + 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 .

(201)

= ⟨𝑥𝑛 − 𝑝, 𝐽 (𝑢𝑛 − 𝑞)⟩ + 𝜇2 ⟨𝐵2 𝑝 − 𝐵2 𝑥𝑛 , 𝐽 (𝑢𝑛 − 𝑞)⟩

(199)

(202)

Substituting (200) for (202), we get 󵄩󵄩 󵄩2 󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩) 󵄩 󵄩󵄩 󵄩 + 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 󵄩 󵄩󵄩 󵄩 + 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 .

≤ ⟨𝑥𝑛 − 𝜇2 𝐵2 𝑥𝑛 − (𝑝 − 𝜇2 𝐵2 𝑝) , 𝐽 (𝑢𝑛 − 𝑞)⟩ 1 󵄩󵄩 󵄩2 󵄩 󵄩2 [󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 2 󵄩 󵄩 −𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩) ] 󵄩 󵄩󵄩 󵄩 + 𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 ,

1 󵄩󵄩 󵄩2 󵄩 󵄩2 [󵄩𝑢 − 𝑞󵄩󵄩󵄩 + 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 2 󵄩 𝑛 󵄩 󵄩 −𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩) ] 󵄩 󵄩󵄩 󵄩 + 𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 ,

󵄩 󵄩 − 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩)

󵄩󵄩 󵄩2 󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 󵄩 󵄩2 = 󵄩󵄩󵄩Π𝐶 (𝑥𝑛 − 𝜇2 𝐵2 𝑥𝑛 ) − Π𝐶 (𝑝 − 𝜇2 𝐵2 𝑝)󵄩󵄩󵄩

≤

= ⟨𝑢𝑛 − 𝑞, 𝐽 (V𝑛 − 𝑝)⟩ + 𝜇1 ⟨𝐵1 𝑞 − 𝐵1 𝑢𝑛 , 𝐽 (V𝑛 − 𝑝)⟩

which implies that

Since ‖𝑥𝑛 − 𝑥𝑛+1 ‖ → 0, 0 < 𝜇𝑖 < 𝛽̂𝑖 /𝜅2 for 𝑖 = 1, 2, and {𝑥𝑛 } is bounded, we obtain from conditions (i), (ii) that 󵄩 󵄩 lim 󵄩󵄩𝐵 𝑥 − 𝐵2 𝑝󵄩󵄩󵄩 = 0, 𝑛→∞ 󵄩 2 𝑛

≤ ⟨𝑢𝑛 − 𝜇1 𝐵1 𝑢𝑛 − (𝑞 − 𝜇1 𝐵1 𝑞) , 𝐽 (V𝑛 − 𝑝)⟩

≤ (196)

󵄩󵄩 󵄩 󵄩 + 2𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 󵄩󵄩 󵄩 󵄩 + 2𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 .

(200)

In the same way, we derive

󵄩 󵄩2 − 2𝛿𝑛 [𝜇2 (𝛽̂2 − 𝜅2 𝜇2 ) 󵄩󵄩󵄩𝐵2 𝑥𝑛 − 𝐵2 𝑝󵄩󵄩󵄩

󵄩2 󵄩 󵄩2 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩

󵄩2 󵄩 󵄩2 󵄩 󵄩 󵄩󵄩 󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩) 󵄩 󵄩 󵄩󵄩 + 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 .

By Lemma 8, we have from (196) and (203) 󵄩󵄩 󵄩2 󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩2 󵄩2 󵄩2 ≤ 𝛼𝑛 𝜌2 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩2 + 𝛿𝑛 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 + 2𝛼𝑛 ⟨𝑓 (𝑝) − 𝑝, 𝐽 (𝑥𝑛+1 − 𝑝)⟩

(203)

Abstract and Applied Analysis

29

󵄩 󵄩 󵄩 󵄩2 󵄩2 󵄩2 ≤ 𝛼𝑛 𝜌󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩

Utilizing the properties of 𝑔1 and 𝑔2 , we deduce that

󵄩 󵄩2 󵄩 󵄩 + 𝛿𝑛 [󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩)

+ 2𝛼𝑛 ⟨𝑓 (𝑝) − 𝑝, 𝐽 (𝑥𝑛+1 − 𝑝)⟩

󵄩󵄩

󵄩 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩 = 0,

󵄩󵄩

󵄩 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩 = 0.

lim 󵄩𝑥 𝑛→∞ 󵄩 𝑛

󵄩 󵄩 󵄩 󵄩 − 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩) + 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩󵄩 󵄩 󵄩 × 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 + 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 ]

lim 󵄩𝑢 𝑛→∞ 󵄩 𝑛

(207)

From (207), we get

󵄩 󵄩2 ≤ (1 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩

󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝑥𝑛 − V𝑛 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩 󵄩 󵄩 + 󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩 󳨀→ 0

󵄩 󵄩 − 𝛿𝑛 [𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩) 󵄩 󵄩 + 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩)] 󵄩 󵄩󵄩 󵄩 󵄩 󵄩 + 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 + 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 × 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 + 2𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝛿𝑛 [𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩) 󵄩 󵄩 + 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩)] 󵄩 󵄩󵄩 󵄩 󵄩 󵄩 + 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 + 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 × 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 + 2𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 ,

as 𝑛 → ∞. (208)

That is, 󵄩󵄩

lim 󵄩𝑥 𝑛→∞ 󵄩 𝑛

󵄩 − 𝐺𝑥𝑛 󵄩󵄩󵄩 = 0.

(209)

Next, let us show that 󵄩󵄩

lim 󵄩𝑆 𝐺𝑥𝑛 𝑛→∞ 󵄩 𝑛

󵄩 − 𝑥𝑛 󵄩󵄩󵄩 = 0,

󵄩󵄩

lim 󵄩𝐵 𝑥 𝑛→∞ 󵄩 𝑛 𝑛

󵄩 − 𝑥𝑛 󵄩󵄩󵄩 = 0. (210)

(204) Indeed, observe that 𝑥𝑛+1 can be rewritten as follows: which hence leads to

𝑥𝑛+1 = 𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛

󵄩 󵄩 󵄩 󵄩 𝛿𝑛 [𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩) + 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩)] 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩 − 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩 󵄩 󵄩󵄩 󵄩 + 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 󵄩 󵄩󵄩 󵄩 + 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 󵄩󵄩 󵄩 󵄩 + 2𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 󵄩 󵄩󵄩 󵄩 + 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 󵄩 󵄩󵄩 󵄩 + 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 󵄩󵄩 󵄩 󵄩 + 2𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 . 󵄩󵄩2

= 𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝛽𝑛 𝑥𝑛 + (𝛾𝑛 + 𝛿𝑛 ) ×

󵄩󵄩2

𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 𝛾𝑛 + 𝛿𝑛

(211)

= 𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝛽𝑛 𝑥𝑛 + 𝑒𝑛 𝑧̂𝑛 , where 𝑒𝑛 = 𝛾𝑛 + 𝛿𝑛 and 𝑧̂𝑛 = (𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 )/(𝛾𝑛 + 𝛿𝑛 ). Utilizing Lemma 11 and (211), we have 󵄩2 󵄩󵄩 󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 󵄩 󵄩2 = 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑝) + 𝛽𝑛 (𝑥𝑛 − 𝑝) + 𝑒𝑛 (̂𝑧𝑛 − 𝑝)󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 (205)

󵄩 󵄩 󵄩2 󵄩 + 𝑒𝑛 󵄩󵄩󵄩𝑧̂𝑛 − 𝑝󵄩󵄩󵄩 − 𝛽𝑛 𝑒𝑛 𝑔3 (󵄩󵄩󵄩𝑧̂𝑛 − 𝑥𝑛 󵄩󵄩󵄩) 󵄩 󵄩 󵄩2 󵄩2 = 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩

From (198), (205), conditions (i), (ii) and the boundedness of {𝑥𝑛 }, {𝑢𝑛 }, and {V𝑛 }, we deduce that

lim 𝑔 𝑛→∞ 1

󵄩 󵄩 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩) = 0,

lim 𝑔 𝑛→∞ 2

󵄩 󵄩 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩) = 0.

(206)

󵄩󵄩2 󵄩󵄩 𝛾 𝐵 𝑥 + 𝛿 𝑆 𝐺𝑥 󵄩 󵄩 󵄩 󵄩 𝑛 𝑛 𝑛 − 𝛽𝑛 𝑒𝑛 𝑔3 (󵄩󵄩󵄩𝑧̂𝑛 − 𝑥𝑛 󵄩󵄩󵄩) + 𝑒𝑛 󵄩󵄩󵄩 𝑛 𝑛 𝑛 − 𝑝󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 𝛾𝑛 + 𝛿𝑛 󵄩 󵄩 󵄩 󵄩2 󵄩2 󵄩 = 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝛽𝑛 𝑒𝑛 𝑔3 (󵄩󵄩󵄩𝑧̂𝑛 − 𝑥𝑛 󵄩󵄩󵄩) 󵄩󵄩2 󵄩󵄩󵄩 𝛾𝑛 𝛿𝑛 󵄩 + 𝑒𝑛 󵄩󵄩󵄩 (𝐵𝑛 𝑥𝑛 − 𝑝) + (𝑆𝑛 𝐺𝑥𝑛 − 𝑝)󵄩󵄩󵄩 𝛾𝑛 + 𝛿𝑛 󵄩󵄩 󵄩󵄩 𝛾𝑛 + 𝛿𝑛

30

Abstract and Applied Analysis 󵄩 󵄩 󵄩 󵄩2 󵄩 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝛽𝑛 𝑒𝑛 𝑔3 (󵄩󵄩󵄩𝑧̂𝑛 − 𝑥𝑛 󵄩󵄩󵄩) + 𝑒𝑛 [

≤

𝛾𝑛 󵄩󵄩 𝛿𝑛 󵄩󵄩 󵄩2 󵄩2 󵄩𝐵 𝑥 − 𝑝󵄩󵄩󵄩 + 󵄩𝑆 𝐺𝑥 − 𝑝󵄩󵄩󵄩 ] 𝛾𝑛 + 𝛿𝑛 󵄩 𝑛 𝑛 𝛾𝑛 + 𝛿𝑛 󵄩 𝑛 𝑛

−

󵄩 󵄩 󵄩 󵄩2 󵄩2 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝛽𝑛 𝑒𝑛 𝑔3 (󵄩󵄩󵄩𝑧̂𝑛 − 𝑥𝑛 󵄩󵄩󵄩) ≤

𝛾𝑛 󵄩󵄩 𝛿𝑛 󵄩󵄩 󵄩 󵄩2 + 𝑒𝑛 [ 󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩𝐺𝑥 − 𝑝󵄩󵄩󵄩 ] 𝛾𝑛 + 𝛿𝑛 𝛾𝑛 + 𝛿𝑛 󵄩 𝑛 󵄩 󵄩 󵄩 󵄩2 󵄩2 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝛽𝑛 𝑒𝑛 𝑔3 (󵄩󵄩󵄩𝑧̂𝑛 − 𝑥𝑛 󵄩󵄩󵄩) + 𝑒𝑛 [

𝛾𝑛 󵄩󵄩 𝛿𝑛 󵄩󵄩 󵄩 󵄩2 󵄩𝑥 − 𝑝󵄩󵄩󵄩 + 󵄩𝑥 − 𝑝󵄩󵄩󵄩 ] 𝛾𝑛 + 𝛿𝑛 󵄩 𝑛 𝛾𝑛 + 𝛿𝑛 󵄩 𝑛

󵄩 󵄩 𝑔4 (󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩)

𝛾𝑛 󵄩󵄩 𝛿𝑛 󵄩󵄩 󵄩2 󵄩2 󵄩𝑥 − 𝑝󵄩󵄩󵄩 + 󵄩𝑥 − 𝑝󵄩󵄩󵄩 𝛾𝑛 + 𝛿𝑛 󵄩 𝑛 𝛾𝑛 + 𝛿𝑛 󵄩 𝑛 −

𝛾𝑛 𝛿𝑛

2

(𝛾𝑛 + 𝛿𝑛 )

󵄩 󵄩 𝑔4 (󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) 𝛾𝑛 𝛿𝑛

2

(𝛾𝑛 + 𝛿𝑛 )

󵄩 󵄩 𝑔4 (󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) , (216)

which hence yields 𝛾𝑛 𝛿𝑛 (𝛾𝑛 + 𝛿𝑛 )

2

󵄩 󵄩 𝑔4 (󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) (217)

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑧̂𝑛 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑧̂𝑛 󵄩󵄩󵄩 .

󵄩 󵄩 𝛽𝑛 𝑒𝑛 𝑔3 (󵄩󵄩󵄩𝑧̂𝑛 − 𝑥𝑛 󵄩󵄩󵄩) 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩) 󵄩 󵄩 × 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 .

(213)

󵄩 󵄩 lim 𝑔 (󵄩󵄩𝑧̂ − 𝑥𝑛 󵄩󵄩󵄩) = 0. 𝑛→∞ 3 󵄩 𝑛

(214)

󵄩󵄩

󵄩 − 𝑥𝑛 󵄩󵄩󵄩 = 0.

(215)

Utilizing Lemma 15 and the definition of 𝑧̂𝑛 , we have 󵄩2 󵄩󵄩̂ 󵄩󵄩𝑧𝑛 − 𝑝󵄩󵄩󵄩 󵄩󵄩 𝛾 𝐵 𝑥 + 𝛿 𝑆 𝐺𝑥 󵄩󵄩2 󵄩 󵄩 𝑛 𝑛 𝑛 = 󵄩󵄩󵄩 𝑛 𝑛 𝑛 − 𝑝󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 𝛾𝑛 + 𝛿𝑛 󵄩󵄩 𝛾 󵄩󵄩2 𝛿𝑛 󵄩 󵄩 𝑛 = 󵄩󵄩󵄩 (𝐵𝑛 𝑥𝑛 − 𝑝) + (𝑆𝑛 𝐺𝑥𝑛 − 𝑝)󵄩󵄩󵄩 󵄩󵄩 𝛾𝑛 + 𝛿𝑛 󵄩󵄩 𝛾𝑛 + 𝛿𝑛 𝛾𝑛 󵄩󵄩 𝛿𝑛 󵄩󵄩 󵄩2 󵄩2 󵄩𝐵 𝑥 − 𝑝󵄩󵄩󵄩 + 󵄩𝑆 𝐺𝑥 − 𝑝󵄩󵄩󵄩 𝛾𝑛 + 𝛿𝑛 󵄩 𝑛 𝑛 𝛾𝑛 + 𝛿𝑛 󵄩 𝑛 𝑛 2

(𝛾𝑛 + 𝛿𝑛 )

󵄩 󵄩 𝑔4 (󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩)

(‖ 𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 ‖) = 0.

(218)

From the properties of 𝑔4 , we have 󵄩󵄩

lim 󵄩𝑆 𝐺𝑥𝑛 𝑛→∞ 󵄩 𝑛

󵄩 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 = 0.

(219)

On the other hand, 𝑥𝑛+1 can also be rewritten as follows:

From the properties of 𝑔3 , we have lim 󵄩𝑧̂ 𝑛→∞ 󵄩 𝑛

Since {𝑥𝑛 } and {̂𝑧𝑛 } are bounded and ‖̂𝑧𝑛 − 𝑥𝑛 ‖ → 0 as 𝑛 → ∞, we deduce from condition (ii) that lim 𝑔 𝑛→∞ 4

Utilizing (184), conditions (i), (ii), (iv), and the boundedness of {𝑥𝑛 } and {𝑓(𝑥𝑛 )}, we get

𝛾𝑛 𝛿𝑛

2

(𝛾𝑛 + 𝛿𝑛 )

󵄩 󵄩2 󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑧̂𝑛 − 𝑝󵄩󵄩󵄩

which hence implies that

−

𝛾𝑛 𝛿𝑛

󵄩 󵄩2 = 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 −

󵄩 󵄩2 󵄩 󵄩2 = 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 − 𝛽𝑛 𝑒𝑛 𝑔3 (󵄩󵄩󵄩𝑧̂𝑛 − 𝑥𝑛 󵄩󵄩󵄩) 󵄩 󵄩 󵄩2 󵄩 󵄩2 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝛽𝑛 𝑒𝑛 𝑔3 (󵄩󵄩󵄩𝑧̂𝑛 − 𝑥𝑛 󵄩󵄩󵄩) , (212)

≤

𝛾𝑛 󵄩󵄩 𝛿𝑛 󵄩󵄩 󵄩2 󵄩2 󵄩𝑥 − 𝑝󵄩󵄩󵄩 + 󵄩𝐺𝑥 − 𝑝󵄩󵄩󵄩 𝛾𝑛 + 𝛿𝑛 󵄩 𝑛 𝛾𝑛 + 𝛿𝑛 󵄩 𝑛

𝑥𝑛+1 = 𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 = 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + (𝛼𝑛 + 𝛿𝑛 )

𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 𝛼𝑛 + 𝛿𝑛

= 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝑑𝑛 𝑧̃𝑛 , (220) where 𝑑𝑛 = 𝛼𝑛 + 𝛿𝑛 and 𝑧̃𝑛 = (𝛼𝑛 𝑓(𝑥𝑛 ) + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 )/(𝛼𝑛 + 𝛿𝑛 ). Utilizing Lemma 11 and the convexity of ‖ ⋅ ‖2 , we have 󵄩2 󵄩󵄩 󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 󵄩 󵄩2 = 󵄩󵄩󵄩𝛽𝑛 (𝑥𝑛 − 𝑝) + 𝛾𝑛 (𝐵𝑛 𝑥𝑛 − 𝑝) + 𝑑𝑛 (̃𝑧𝑛 − 𝑝)󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩 + 𝑑𝑛 󵄩󵄩󵄩𝑧̃𝑛 − 𝑝󵄩󵄩󵄩 − 𝛽𝑛 𝛾𝑛 𝑔5 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩)

Abstract and Applied Analysis

31

󵄩 󵄩 󵄩2 󵄩2 = 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑝󵄩󵄩󵄩

which, together with (219), implies that

󵄩󵄩2 󵄩󵄩 𝛼 𝑓 (𝑥 ) + 𝛿 𝑆 𝐺𝑥 󵄩 󵄩 𝑛 𝑛 𝑛 𝑛 + 𝑑𝑛 󵄩󵄩󵄩󵄩 𝑛 − 𝑝󵄩󵄩󵄩󵄩 𝛼𝑛 + 𝛿𝑛 󵄩󵄩 󵄩󵄩 󵄩 󵄩 − 𝛽𝑛 𝛾𝑛 𝑔5 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) 󵄩 󵄩 󵄩2 󵄩2 = 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩󵄩2 󵄩󵄩 𝛼 𝛿𝑛 󵄩 󵄩 𝑛 + 𝑑𝑛 󵄩󵄩󵄩 (𝑓 (𝑥𝑛 ) − 𝑝) + (𝑆𝑛 𝐺𝑥𝑛 − 𝑝)󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 𝛼𝑛 + 𝛿𝑛 𝛼𝑛 + 𝛿𝑛 󵄩 󵄩 − 𝛽𝑛 𝛾𝑛 𝑔5 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 𝛼𝑛 󵄩󵄩 𝛿𝑛 󵄩󵄩 󵄩2 󵄩2 󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 󵄩𝑆 𝐺𝑥 − 𝑝󵄩󵄩󵄩 ] 𝛼𝑛 + 𝛿𝑛 󵄩 𝛼𝑛 + 𝛿𝑛 󵄩 𝑛 𝑛 󵄩 󵄩 − 𝛽𝑛 𝛾𝑛 𝑔5 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝑑𝑛 [

𝛼𝑛 󵄩󵄩 𝛿𝑛 󵄩󵄩 󵄩2 󵄩2 󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 󵄩𝐺𝑥 − 𝑝󵄩󵄩󵄩 ] 𝛼𝑛 + 𝛿𝑛 𝛼𝑛 + 𝛿𝑛 󵄩 𝑛 󵄩 󵄩 − 𝛽𝑛 𝛾𝑛 𝑔5 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩)

+ 𝑑𝑛 [

󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 𝛼𝑛 󵄩󵄩 𝛿𝑛 󵄩󵄩 󵄩2 󵄩2 󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 ] 𝛼𝑛 + 𝛿𝑛 𝛼𝑛 + 𝛿𝑛 󵄩󵄩 󵄩󵄩 − 𝛽𝑛 𝛾𝑛 𝑔5 (󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩) 󵄩 󵄩2 󵄩 󵄩2 = 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 − 𝛽𝑛 𝛾𝑛 𝑔5 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) 󵄩 󵄩 󵄩2 󵄩 󵄩2 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝛽𝑛 𝛾𝑛 𝑔5 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) , (221) + 𝑑𝑛 [

as 𝑛 → ∞. (225) That is, 󵄩󵄩

lim 󵄩𝑆 𝐺𝑥𝑛 𝑛→∞ 󵄩 𝑛

󵄩 − 𝑥𝑛 󵄩󵄩󵄩 = 0.

(226)

We note that 󵄩󵄩 󵄩 󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑆𝑛 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩 . (227) So, in terms of (209), (226), and Lemma 12, we have 󵄩󵄩

lim 󵄩𝑥 𝑛→∞ 󵄩 𝑛

󵄩 − 𝑆𝑥𝑛 󵄩󵄩󵄩 = 0.

(228)

Suppose that 𝛽𝑛 ≡ 𝛽 for some fixed 𝛽 ∈ (0, 1) such that 𝛼𝑛 + 𝛽 + 𝛾𝑛 + 𝛿𝑛 = 1 for all 𝑛 ≥ 0. Define a mapping 𝑉𝑥 = (1 − 𝜃1 − 𝜃2 )𝑆𝑥 + 𝜃1 𝐵𝑥 + 𝜃2 𝐺𝑥, where 𝜃1 , 𝜃2 ∈ (0, 1) are two constants with 𝜃1 + 𝜃2 < 1. Then by Lemmas 14 and 17, we have that Fix(𝑉) = Fix(𝑆) ∩ Fix(𝐵) ∩ Fix(𝐺) = 𝐹. For each 𝑘 ≥ 1, let {𝑝𝑘 } be a unique element of 𝐶 such that 𝑝𝑘 =

1 1 𝑓 (𝑝𝑘 ) + (1 − ) 𝑉𝑝𝑘 . 𝑘 𝑘

(229)

From Lemma 13, we conclude that 𝑝𝑘 → 𝑞 ∈ Fix(𝑉) = 𝐹 as 𝑘 → ∞. Observe that for every 𝑛, 𝑘

which hence implies that 󵄩 󵄩 𝛽𝑛 𝛾𝑛 𝑔5 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩) 󵄩 󵄩 × 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 .

(222)

󵄩 󵄩 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) = 0.

(223)

Utilizing the properties of 𝑔5 , we have 󵄩󵄩

lim 󵄩𝑥 𝑛→∞ 󵄩 𝑛

󵄩 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 = 0,

󵄩󵄩 󵄩 󵄩󵄩𝑥𝑛+1 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 = 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝐵𝑝𝑘 ) + 𝛽 (𝑥𝑛 − 𝐵𝑝𝑘 )

󵄩 + 𝛾𝑛 (𝐵𝑛 𝑥𝑛 − 𝐵𝑝𝑘 ) + 𝛿𝑛 (𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑝𝑘 )󵄩󵄩󵄩

From (184), conditions (i), (ii), (iv), and the boundedness of {𝑥𝑛 } and {𝑓(𝑥𝑛 )}, we have lim 𝑔 𝑛→∞ 5

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩 󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 󳨀→ 0

(224)

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + 𝛿𝑛 (󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩) 󵄩 󵄩 󵄩 󵄩 = 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + (𝛾𝑛 + 𝛿𝑛 ) 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 = 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + (1 − 𝛼𝑛 − 𝛽) 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩

32

Abstract and Applied Analysis 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛼𝑛 − 𝛽)

󵄩2 󵄩󵄩 󵄩󵄩𝑥𝑛 − 𝑆𝑝𝑘 󵄩󵄩󵄩 󵄩2 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 + 𝑆𝑥𝑛 − 𝑆𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩 ≤ (󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩) 󵄩2 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 × (2 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩) .

󵄩 󵄩 󵄩 󵄩 × [󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑛 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 󵄩󵄩󵄩]

󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + (1 − 𝛼𝑛 − 𝛽) [󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 󵄩󵄩󵄩] 󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + (1 − 𝛽) [󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 󵄩󵄩󵄩] 󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 = 𝜃𝑛 + 𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 ,

(233) It is easy to see from (209) and (228) that 󵄩 󵄩2 󵄩 󵄩2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑝𝑘 󵄩󵄩󵄩 ≤ 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 , 󵄩 󵄩2 󵄩 󵄩2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑝𝑘 󵄩󵄩󵄩 ≤ 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 . (230)

where 𝜃𝑛 = 𝛼𝑛 ‖𝑓(𝑥𝑛 ) − 𝐵𝑝𝑘 ‖ + (1 − 𝛽)‖𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 ‖ + 𝛿𝑛 ‖𝑆𝑛 𝐺𝑥𝑛 −𝐵𝑛 𝑥𝑛 ‖. Since lim𝑛 → ∞ 𝛼𝑛 = lim𝑛 → ∞ ‖𝐵𝑛 𝑝𝑘 −𝐵𝑝𝑘 ‖ = lim𝑛 → ∞ ‖𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 ‖ = 0, we know that 𝜃𝑛 → 0 as 𝑛 → ∞. From (230), we obtain 󵄩2 󵄩󵄩 󵄩󵄩𝑥𝑛+1 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩 ≤ (𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩) 󵄩 󵄩 󵄩 󵄩 + 𝜃𝑛 [2 (𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩) + 𝜃𝑛 ] 2󵄩 󵄩 󵄩2 󵄩2 = 𝛽2 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 󵄩󵄩 󵄩 󵄩 + 2𝛽 (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 𝜏𝑛 2󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛽2 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 󵄩2 󵄩 󵄩2 󵄩 + 𝛽 (1 − 𝛽) (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 ) + 𝜏𝑛 󵄩2 󵄩2 󵄩 󵄩 = 𝛽󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 𝜏𝑛 ,

In addition, note that 󵄩󵄩 󵄩2 󵄩󵄩𝑥𝑛 − 𝐺𝑝𝑘 󵄩󵄩󵄩 󵄩2 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑥𝑛 + 𝐺𝑥𝑛 − 𝐺𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩 ≤ (󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩) 󵄩2 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 × (2 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑥𝑛 󵄩󵄩󵄩) ,

Utilizing (232) and (234), we deduce that 󵄩 󵄩2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑉𝑝𝑘 󵄩󵄩󵄩 󵄩 = 𝜇𝑛 󵄩󵄩󵄩(1 − 𝜃1 − 𝜃2 ) (𝑥𝑛 − 𝑆𝑝𝑘 ) 󵄩2 + 𝜃1 (𝑥𝑛 − 𝐵𝑝𝑘 ) + 𝜃2 (𝑥𝑛 − 𝐺𝑝𝑘 )󵄩󵄩󵄩 󵄩 󵄩2 ≤ (1 − 𝜃1 − 𝜃2 ) 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑝𝑘 󵄩󵄩󵄩

(235)

󵄩 󵄩2 󵄩 󵄩2 + 𝜃1 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝜃2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩2 ≤ 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 . Also, observe that 1 1 (𝑥 − 𝑓 (𝑝𝑘 )) + (1 − ) (𝑥𝑛 − 𝑉𝑝𝑘 ) ; 𝑘 𝑛 𝑘

(236)

1 1 (1 − ) (𝑥𝑛 − 𝑉𝑝𝑘 ) = 𝑥𝑛 − 𝑝𝑘 − (𝑥𝑛 − 𝑓 (𝑝𝑘 )) . 𝑘 𝑘

(237)

𝑥𝑛 − 𝑝𝑘 = that is,

It follows from Lemma 8(ii) and (237) that (231)

where 𝜏𝑛 = 𝜃𝑛 [2(𝛽‖𝑥𝑛 − 𝐵𝑝𝑘 ‖ + (1 − 𝛽)‖𝑥𝑛 − 𝑝𝑘 ‖) + 𝜃𝑛 ] → 0 as 𝑛 → ∞. For any Banach limit 𝜇, from (231) we derive 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 = 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛+1 − 𝐵𝑝𝑘 󵄩󵄩󵄩 ≤ 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 .

(234)

(232)

2

1 󵄩 󵄩2 (1 − ) 󵄩󵄩󵄩𝑥𝑛 − 𝑉𝑝𝑘 󵄩󵄩󵄩 𝑘 󵄩2 2 󵄩 ≥ 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 − ⟨𝑥𝑛 − 𝑝𝑘 + 𝑝𝑘 − 𝑓 (𝑝𝑘 ) , 𝐽 (𝑥𝑛 − 𝑝𝑘 )⟩ 𝑘 2 󵄩 󵄩2 2 = (1 − ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + ⟨𝑓 (𝑝𝑘 ) − 𝑝𝑘 , 𝐽 (𝑥𝑛 − 𝑝𝑘 )⟩ . 𝑘 𝑘 (238) So by (235) and (238), we have 1 2 󵄩 󵄩2 (1 − ) 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 𝑘 2 󵄩 󵄩2 ≥ (1 − ) 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 𝑘 2 + 𝜇𝑛 ⟨𝑓 (𝑝𝑘 ) − 𝑝𝑘 , 𝐽 (𝑥𝑛 − 𝑝𝑘 )⟩ , 𝑘

(239)

Abstract and Applied Analysis

33 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑓 (𝑞)󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩

and hence 1 󵄩󵄩 󵄩2 2 𝜇𝑛 󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 ≥ 𝜇𝑛 ⟨𝑓 (𝑝𝑘 ) − 𝑝𝑘 , 𝐽 (𝑥𝑛 − 𝑝𝑘 )⟩ . (240) 2 𝑘 𝑘

(241)

Since 𝑝𝑘 → 𝑞 ∈ Fix(𝑉) = 𝐹 as 𝑘 → ∞, by the uniform Frechet differentiability of the norm of 𝑋, we have 𝜇𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩ ≤ 0.

(242)

On the other hand, from (184) and the norm-to-norm uniform continuity of 𝐽 on bounded subsets of 𝑋, it follows that 󵄨 󵄨 lim 󵄨󵄨⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ − ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩󵄨󵄨󵄨 = 0. (243)

𝑛→∞ 󵄨

So, utilizing Lemma 18 we deduce from (242) and (243) that lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩ ≤ 0, 𝑛→∞

(244)

which, together with (184) and the norm-to-norm uniform continuity of 𝐽 on bounded subsets of 𝑋, implies that lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ ≤ 0. 𝑛→∞

(245)

Finally, let us show that 𝑥𝑛 → 𝑞 as 𝑛 → ∞. Utilizing Lemma 8 (i), from (178) and the convexity of ‖ ⋅ ‖2 , we get 󵄩󵄩 󵄩2 󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩 󵄩 = 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑓 (𝑞)) + 𝛽𝑛 (𝑥𝑛 − 𝑞) + 𝛾𝑛 (𝐵𝑛 𝑥𝑛 − 𝑞) 󵄩2 + 𝛿𝑛 (𝑆𝑛 𝐺𝑥𝑛 − 𝑞) + 𝛼𝑛 (𝑓 (𝑞) − 𝑞)󵄩󵄩󵄩 󵄩 ≤ 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑓 (𝑞)) + 𝛽𝑛 (𝑥𝑛 − 𝑞) 󵄩2 + 𝛾𝑛 (𝐵𝑛 𝑥𝑛 − 𝑞) + 𝛿𝑛 (𝑆𝑛 𝐺𝑥𝑛 − 𝑞)󵄩󵄩󵄩 + 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑓 (𝑞)󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩2 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩

+ 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ 󵄩 󵄩 󵄩 󵄩2 󵄩2 󵄩2 ≤ 𝛼𝑛 𝜌󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩

This implies that 1 󵄩󵄩 󵄩2 𝜇 󵄩𝑥 − 𝑝𝑘 󵄩󵄩󵄩 ≥ 𝜇𝑛 ⟨𝑓 (𝑝𝑘 ) − 𝑝𝑘 , 𝐽 (𝑥𝑛 − 𝑝𝑘 )⟩ . 2𝑘 𝑛 󵄩 𝑛

󵄩 󵄩 󵄩2 󵄩2 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑞󵄩󵄩󵄩

󵄩 󵄩2 + 𝛿𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ 󵄩 󵄩2 = (1 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ 󵄩 󵄩2 = (1 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝛼𝑛 (1 − 𝜌)

2 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ . 1−𝜌 (246)

Applying Lemma 7 to (246), we obtain that 𝑥𝑛 → 𝑞 as 𝑛 → ∞. This completes the proof. Corollary 32. Let 𝐶 be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space 𝑋. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶. Let {𝜌𝑛 }∞ 𝑛=0 be a sequence of positive numbers in (0, 𝑏] for some 𝑏 ∈ (0, 1) and 𝐴 𝑖 : 𝐶 → 𝑋 an 𝛼̂𝑖 -inverse strongly accretive mapping for each 𝑖 = 0, 1, . . .. Define a mapping 𝐺𝑖 : 𝐶 → 𝐶 by Π𝐶(𝐼 − 𝜆 𝑖 𝐴 𝑖 )𝑥 = 𝐺𝑖 𝑥 for all 𝑥 ∈ 𝐶 and 𝑖 = 0, 1, . . ., where 𝜆 𝑖 ∈ (0, 𝛼̂𝑖 /𝜅2 ] and 𝜅 is the 2-uniformly smooth constant of 𝑋. Let 𝐵𝑛 : 𝐶 → 𝐶 be the 𝑊-mapping generated by 𝐺𝑛 , 𝐺𝑛−1 , . . . , 𝐺0 , and 𝜌𝑛 , 𝜌𝑛−1 , . . . , 𝜌0 . Let 𝑉 : 𝐶 → 𝐶 be an 𝛼-strictly pseudocontractive mapping. Let 𝑓 : 𝐶 → 𝐶 be a contraction with coefficient 𝜌 ∈ (0, 1). Let {𝑆𝑖 }∞ 𝑖=0 be a countable family of nonexpansive mappings of 𝐶 into itself such ∞ that 𝐹 = (⋂∞ 𝑖=0 Fix(𝑆𝑖 )) ∩ Fix(𝑉) ∩ (⋂𝑖=0 VI(𝐶, 𝐴 𝑖 )) ≠ 0. For arbitrarily given 𝑥0 ∈ 𝐶, let {𝑥𝑛 } be the sequence generated by 𝑥𝑛+1 = 𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 ((1 − 𝑙) 𝐼 + 𝑙𝑉) 𝑥𝑛 ,

∀𝑛 ≥ 0,

(247)

where 0 < 𝑙 < 𝛼/𝜅2 and {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, and {𝛿𝑛 } are the sequences in (0, 1) such that 𝛼𝑛 + 𝛽𝑛 + 𝛾𝑛 + 𝛿𝑛 = 1 for all 𝑛 ≥ 0. Suppose that the following conditions hold: (i) lim𝑛 → ∞ 𝛼𝑛 = 0 and ∑∞ 𝑛=0 𝛼𝑛 = ∞; (ii) {𝛾𝑛 }, {𝛿𝑛 } ⊂ [𝑐, 𝑑] for some 𝑐, 𝑑 ∈ (0, 1); (iii) lim𝑛 → ∞ (|𝛽𝑛 − 𝛽𝑛−1 | + |𝛾𝑛 − 𝛾𝑛−1 | + |𝛿𝑛 − 𝛿𝑛−1 |) = 0; (iv) 0 < lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ 𝛽𝑛 < 1. Assume that ∑∞ 𝑛=1 sup𝑥∈𝐷 ‖𝑆𝑛 𝑥 − 𝑆𝑛−1 𝑥‖ < ∞ for any bounded subset 𝐷 of 𝐶 and let 𝑆 be a mapping of 𝐶 into itself defined

34

Abstract and Applied Analysis

by 𝑆𝑥 = lim𝑛 → ∞ 𝑆𝑛 𝑥 for all 𝑥 ∈ 𝐶 and suppose that Fix(𝑆) = ⋂∞ 𝑖=0 Fix(𝑆𝑖 ). Then, there hold the following: (I) lim𝑛 → ∞ ‖𝑥𝑛+1 − 𝑥𝑛 ‖ = 0;

(II) the sequence {𝑥𝑛 }∞ 𝑛=0 converges strongly to some 𝑞 ∈ 𝐹 which is the unique solution of the variational inequality problem (VIP) ⟨(𝐼 − 𝑓) 𝑞, 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,

∀𝑝 ∈ 𝐹,

(248)

Proof. In Theorem 31, we put 𝐵1 = 𝐼 − 𝑉, 𝐵2 = 0 and 𝜇1 = 𝑙 where 0 < 𝑙 < 𝛼/𝜅2 . Then GSVI (13) is equivalent to the VIP of finding 𝑥∗ ∈ 𝐶 such that ∀𝑥 ∈ 𝐶.

(249)

In this case, 𝐵1 : 𝐶 → 𝑋 is 𝛼-inverse strongly accretive. Repeating the same arguments as those in the proof of Corollary 25, we can infer that Fix(𝑉) = VI(𝐶, 𝐵1 ). −1 Accordingly, we know that 𝐹 = ⋂∞ 𝑖=0 Fix(𝑇𝑖 ) ∩ Ω ∩ 𝐴 0 = ∞ −1 ⋂𝑖=0 Fix(𝑇𝑖 ) ∩ Fix(𝑉) ∩ 𝐴 0, and Π𝐶 (𝐼 − 𝜇1 𝐵1 ) Π𝐶 (𝐼 − 𝜇2 𝐵2 ) 𝑥𝑛 = Π𝐶 (𝐼 − 𝜇1 𝐵1 ) 𝑥𝑛 = Π𝐶 ((1 − 𝑙) 𝑥𝑛 + 𝑙𝑉𝑥𝑛 )

(i) lim𝑛 → ∞ 𝛼𝑛 = 0 and ∑∞ 𝑛=0 𝛼𝑛 = ∞; (ii) {𝛾𝑛 }, {𝛿𝑛 } ⊂ [𝑐, 𝑑] for some 𝑐, 𝑑 ∈ (0, 1); (iii) ∑∞ 𝑛=1 (|𝜎𝑛 − 𝜎𝑛−1 | + |𝛼𝑛 − 𝛼𝑛−1 | + |𝛽𝑛 − 𝛽𝑛−1 | + |𝛾𝑛 − 𝛾𝑛−1 | + |𝛿𝑛 − 𝛿𝑛−1 |) < ∞;

provided 𝛽𝑛 ≡ 𝛽 for some fixed 𝛽 ∈ (0, 1).

⟨𝐵1 𝑥∗ , 𝐽 (𝑥 − 𝑥∗ )⟩ ≥ 0,

where {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, {𝛿𝑛 }, and {𝜎𝑛 } are the sequences in (0, 1) such that 𝛼𝑛 + 𝛽𝑛 + 𝛾𝑛 + 𝛿𝑛 = 1 for all 𝑛 ≥ 0. Suppose that the following conditions hold:

(250)

(iv) 0 < lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ 𝛽𝑛 < 1 and 0 < lim inf 𝑛 → ∞ 𝜎𝑛 ≤ lim sup𝑛 → ∞ 𝜎𝑛 < 1. Assume that ∑∞ 𝑛=1 sup𝑥∈𝐷 ‖𝑆𝑛 𝑥 − 𝑆𝑛−1 𝑥‖ < ∞ for any bounded subset 𝐷 of 𝐶 and let 𝑆 be a mapping of 𝐶 into itself defined by 𝑆𝑥 = lim𝑛 → ∞ 𝑆𝑛 𝑥 for all 𝑥 ∈ 𝐶 and suppose that Fix(𝑆) = ⋂∞ 𝑖=0 Fix(𝑆𝑖 ). Then, there hold the following: (I) lim𝑛 → ∞ ‖𝑥𝑛+1 − 𝑥𝑛 ‖ = 0; (II) the sequence {𝑥𝑛 }∞ 𝑛=0 converges strongly to some 𝑞 ∈ 𝐹 which is the unique solution of the variational inequality problem (VIP) ⟨(𝐼 − 𝑓) 𝑞, 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,

= ((1 − 𝑙) 𝐼 + 𝑙𝑉) 𝑥𝑛 .

∀𝑝 ∈ 𝐹,

(252)

So, scheme (178) reduces to (247). Therefore, the desired result follows from Theorem 31. Theorem 33. Let 𝐶 be a nonempty closed convex subset of a uniformly convex Banach space 𝑋 which has a uniformly Gateaux differentiable norm. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶. Let {𝜌𝑛 }∞ 𝑛=0 be a sequence of positive numbers in (0, 𝑏] for some 𝑏 ∈ (0, 1) and 𝐴 𝑖 : 𝐶 → 𝑋𝜉𝑖 strictly pseudocontractive and 𝛼̂𝑖 -strongly accretive with 𝜉𝑖 + 𝛼̂𝑖 ≥ 1 for each 𝑖 = 0, 1, . . .. Define a mapping 𝐺𝑖 : 𝐶 → 𝐶 by Π𝐶(𝐼 − 𝜆 𝑖 𝐴 𝑖 )𝑥 = 𝐺𝑖 𝑥 for all 𝑥 ∈ 𝐶 and 𝑖 = 0, 1, . . ., where 1−(𝜉𝑖 /(1+𝜉𝑖 ))(1−√(1 − 𝛼̂𝑖 )/𝜉𝑖 ) ≤ 𝜆 𝑖 ≤ 1 for all 𝑖 = 0, 1, . . .. Let 𝐵𝑛 : 𝐶 → 𝐶 be the 𝑊-mapping generated by 𝐺𝑛 , 𝐺𝑛−1 , . . . , 𝐺0 and 𝜌𝑛 , 𝜌𝑛−1 , . . . , 𝜌0 . Let the mapping 𝐵𝑖 : 𝐶 → 𝑋 be 𝜁𝑖 -strictly pseudocontractive and 𝛽̂𝑖 -strongly accretive with 𝜁𝑖 + 𝛽̂𝑖 ≥ 1 for 𝑖 = 1, 2. Let 𝑓 : 𝐶 → 𝐶 be a contraction with coefficient 𝜌 ∈ (0, 1). Let {𝑆𝑖 }∞ 𝑖=0 be a countable family of nonexpansive mappings of 𝐶 into itself such that 𝐹 = (⋂∞ 𝑖=0 Fix(𝑆𝑖 )) ∩ VI(𝐶, 𝐴 )) = ̸ 0, where Ω is the fixed point set of the Ω ∩ (⋂∞ 𝑖 𝑖=0 mapping 𝐺 = Π𝐶(𝐼 − 𝜇1 𝐵1 )Π𝐶(𝐼 − 𝜇2 𝐵2 ) with 1 − (𝜁𝑖 /(1 + 𝜁 ))(1 − √(1 − 𝛽̂ )/𝜁 ) ≤ 𝜇 ≤ 1 for 𝑖 = 1, 2. For arbitrarily 𝑖

𝑖

𝑖

𝑖

given 𝑥0 ∈ 𝐶, let {𝑥𝑛 } be the sequence generated by

Proof. First of all, it is easy to see that (251) can be rewritten as follows: 𝑦𝑛 = 𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 , 𝑥𝑥+1 = 𝜎𝑛 𝐺𝑥𝑛 + (1 − 𝜎𝑛 ) 𝑦𝑛 ,

∀𝑛 ≥ 0.

∀𝑛 ≥ 0, (251)

(253)

Take a fixed 𝑝 ∈ 𝐹 arbitrarily. Then, we obtain 𝑝 = 𝐺𝑝, 𝑝 = 𝐵𝑛 𝑝 and 𝑆𝑛 𝑝 = 𝑝 for all 𝑛 ≥ 0. Thus, we get from (253) 󵄩 󵄩󵄩 󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 (𝜌 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩)

𝑥𝑛+1 = 𝜎𝑛 𝐺𝑥𝑛 + (1 − 𝜎𝑛 ) × [𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 ] ,

provided 𝛽𝑛 ≡ 𝛽 for some fixed 𝛽 ∈ (0, 1).

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 = (1 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 ,

(254)

Abstract and Applied Analysis

35

and hence 󵄩󵄩 󵄩 󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝜎𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 ≤ 𝜎𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝜎𝑛 ) 󵄩 󵄩 󵄩 󵄩 × [(1 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩] 󵄩 󵄩 = (1 − (1 − 𝜎𝑛 ) 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 + (1 − 𝜎𝑛 ) 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩 󵄩 = (1 − (1 − 𝜎𝑛 ) 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩󵄩 󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 + (1 − 𝜎𝑛 ) 𝛼𝑛 (1 − 𝜌) 󵄩 1−𝜌 󵄩 󵄩 󵄩 󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩 ≤ max {󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 , 󵄩 }. 1−𝜌 (255)

󵄩󵄩 𝑦 − 𝛽 𝑥 𝑦 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩󵄩 󵄩 𝑛 𝑛 − 𝑛−1 ≤ 󵄩󵄩󵄩 𝑛 󵄩󵄩 󵄩󵄩 1 − 𝛽𝑛 󵄩󵄩 1 − 𝛽𝑛 󵄩󵄩 𝑦 − 𝛽 𝑥 𝑦 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩󵄩 󵄩 𝑛−1 𝑛−1 + 󵄩󵄩󵄩 𝑛−1 − 𝑛−1 󵄩󵄩 󵄩󵄩 󵄩󵄩 1 − 𝛽𝑛 1 − 𝛽𝑛−1 =

=

=

1 󵄩󵄩 󵄩 󵄩𝑦 − 𝛽𝑛 𝑥𝑛 − (𝑦𝑛−1 − 𝛽𝑛−1 𝑥𝑛−1 )󵄩󵄩󵄩 1 − 𝛽𝑛 󵄩 𝑛 󵄨󵄨 1 󵄨󵄨 1 󵄨 󵄨󵄨 󵄩󵄩 󵄩 + 󵄨󵄨󵄨 − 󵄨 󵄩𝑦 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 𝑛−1 1 󵄩󵄩 󵄩 󵄩𝑦 − 𝛽𝑛 𝑥𝑛 − (𝑦𝑛−1 − 𝛽𝑛−1 𝑥𝑛−1 )󵄩󵄩󵄩 1 − 𝛽𝑛 󵄩 𝑛 󵄨󵄨 󵄨 󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 󵄩󵄩 + 󵄩𝑦 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 (1 − 𝛽𝑛−1 ) (1 − 𝛽𝑛 ) 󵄩 𝑛−1 1 󵄩󵄩 󵄩𝛼 𝑓 (𝑥𝑛 ) + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 1 − 𝛽𝑛 󵄩 𝑛 − 𝛼𝑛−1 𝑓 (𝑥𝑛−1 ) − 𝛾𝑛−1 𝐵𝑛−1 𝑥𝑛−1 󵄩 −𝛿𝑛−1 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩

By induction, we have 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩 }, 󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 ≤ max {󵄩󵄩󵄩𝑥0 − 𝑝󵄩󵄩󵄩 , 󵄩 1−𝜌

+ ∀𝑛 ≥ 0. (256)

which implies that {𝑥𝑛 } is bounded and so are the sequences {𝑦𝑛 }, {𝐺𝑥𝑛 } and {𝑓(𝑥𝑛 )}. Let us show that 󵄩󵄩

lim 󵄩𝑥 𝑛 → ∞ 󵄩 𝑛+1

󵄩 − 𝑥𝑛 󵄩󵄩󵄩 = 0.

(257)

As a matter of fact, observe that 𝑦𝑛 can be rewritten as follows: 𝑦𝑛 = 𝛽𝑛 𝑥𝑛 + (1 − 𝛽𝑛 ) 𝑧𝑛 ,

(258)

where 𝑧𝑛 = (𝛼𝑛 𝑓(𝑥𝑛 ) + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 )/(1 − 𝛽𝑛 ). Observe that 󵄩󵄩 󵄩 󵄩󵄩𝑧𝑛 − 𝑧𝑛−1 󵄩󵄩󵄩 󵄩󵄩 𝛼 𝑓 (𝑥 ) + 𝛾 𝐵 𝑥 + 𝛿 𝑆 𝐺𝑥 󵄩 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 = 󵄩󵄩󵄩󵄩 𝑛 1 − 𝛽𝑛 󵄩󵄩 −

𝛼𝑛−1 𝑓 (𝑥𝑛−1 ) + 𝛾𝑛−1 𝐵𝑛−1 𝑥𝑛−1 + 𝛿𝑛−1 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 1 − 𝛽𝑛−1 󵄩

󵄩󵄩 𝑦 − 𝛽 𝑥 𝑦 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩󵄩 󵄩 𝑛 𝑛 = 󵄩󵄩󵄩 𝑛 − 𝑛−1 󵄩󵄩 󵄩󵄩 1 − 𝛽𝑛 󵄩󵄩 1 − 𝛽𝑛−1 󵄩󵄩 𝑦 − 𝛽 𝑥 𝑦 − 𝛽𝑛−1 𝑥𝑛−1 󵄩 𝑛 𝑛 = 󵄩󵄩󵄩 𝑛 − 𝑛−1 1 − 𝛽𝑛 󵄩󵄩 1 − 𝛽𝑛 +

𝑦𝑛−1 − 𝛽𝑛−1 𝑥𝑛−1 𝑦𝑛−1 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩󵄩 − 󵄩󵄩 󵄩󵄩 1 − 𝛽𝑛 1 − 𝛽𝑛−1

≤

󵄨󵄨 󵄨 󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 󵄩󵄩 󵄩󵄩𝑦𝑛−1 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 (1 − 𝛽𝑛−1 ) (1 − 𝛽𝑛 )

1 󵄩 󵄩 [𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑓 (𝑥𝑛−1 )󵄩󵄩󵄩 1 − 𝛽𝑛 󵄩 󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 󵄩 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑓 (𝑥𝑛−1 )󵄩󵄩󵄩 󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛾𝑛 − 𝛾𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛿𝑛 − 𝛿𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩] 󵄨󵄨 󵄨 󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 󵄩󵄩 + 󵄩󵄩𝑦𝑛−1 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 . (1 − 𝛽 ) (1 − 𝛽 ) 𝑛−1

𝑛

(259)

On the other hand, repeating the same arguments as those of (52) and (54) in the proof of Theorem 24, we can deduce that for all 𝑛 ≥ 1 󵄩󵄩 󵄩 󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛−1 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 , 𝑛−1

(260)

󵄩󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 𝑀∏𝜆 𝑖 , 𝑖=0

for some constant 𝑀 > 0. Taking into account 0 < lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ 𝛽𝑛 < 1, we may assume,

36

Abstract and Applied Analysis

̂ Utilizing (259)without loss of generality, that {𝛽𝑛 } ⊂ [̂𝑐, 𝑑]. (260) we have 󵄩 󵄩󵄩 󵄩󵄩𝑧𝑛 − 𝑧𝑛−1 󵄩󵄩󵄩 ≤

󵄨󵄩 󵄨 󵄩 × [󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑓 (𝑥𝑛−1 )󵄩󵄩󵄩

󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛾𝑛 − 𝛾𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩

󵄨󵄨 󵄨 󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 (1 − 𝛽𝑛−1 ) (1 − 𝛽𝑛 ) 󵄩 󵄩󵄩 × 󵄩󵄩𝛼𝑛−1 𝑓 (𝑥𝑛−1 ) + 𝛾𝑛−1 𝐵𝑛−1 𝑥𝑛−1 + 𝛿𝑛−1 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛿𝑛 − 𝛿𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩] +

1 1 − 𝛽𝑛 𝑛−1

󵄩 󵄩 󵄩 󵄩 × {𝛼𝑛 𝜌 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 𝛾𝑛 [󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 𝑀∏𝜆 𝑖 ]

≤ (1 −

𝑖=0

󵄩 󵄩 󵄩 󵄩 + 𝛿𝑛 [󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛−1 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩]

𝑛−1

󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 + 𝑀1 [∏𝜆 𝑖 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛾𝑛 − 𝛾𝑛−1 󵄨󵄨󵄨 𝑖=0

󵄨 󵄨󵄩 󵄨󵄩 󵄩 󵄩 󵄨 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑓 (𝑥𝑛−1 )󵄩󵄩󵄩 + 󵄨󵄨󵄨𝛾𝑛 − 𝛾𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩

󵄨 󵄩 󵄨 󵄩 + 󵄨󵄨󵄨𝛿𝑛 − 𝛿𝑛−1 󵄨󵄨󵄨 ] + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛−1 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 ,

󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛿𝑛 − 𝛿𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 } +

(261)

󵄨󵄨 󵄨 󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 󵄩󵄩 󵄩𝑦 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 (1 − 𝛽𝑛−1 ) (1 − 𝛽𝑛 ) 󵄩 𝑛−1

1 󵄩 󵄩 = { (1 − 𝛽𝑛 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 1 − 𝛽𝑛 𝑛−1

󵄩 󵄩 + 𝛾𝑛 𝑀∏𝜆 𝑖 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛−1 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 𝑖=0

󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑓 (𝑥𝑛−1 )󵄩󵄩󵄩 󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛾𝑛 − 𝛾𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝛿𝑛 − 𝛿𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 } 󵄨󵄨 󵄨 󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 󵄩󵄩 + 󵄩𝑦 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 (1 − 𝛽𝑛−1 ) (1 − 𝛽𝑛 ) 󵄩 𝑛−1 = (1 −

𝛼𝑛 (1 − 𝜌) 󵄩󵄩 󵄩 ) 󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 1 − 𝛽𝑛

̂ 2 )(‖𝑓(𝑥 )‖ + ‖𝐵 𝑥 ‖ + ‖𝑆 𝐺𝑥 ‖ + where sup𝑛≥0 {(1/(1 − 𝑑) 𝑛 𝑛 𝑛 𝑛 𝑛 𝑀)} ≤ 𝑀1 for some 𝑀1 > 0. In the meantime, observe that 𝑥𝑛+1 − 𝑥𝑛 = 𝜎𝑛 (𝐺𝑥𝑛 − 𝐺𝑥𝑛−1 ) + (𝜎𝑛 − 𝜎𝑛−1 ) (𝐺𝑥𝑛−1 − 𝑧𝑛−1 ) + (1 − 𝜎𝑛 ) (𝑧𝑛 − 𝑧𝑛−1 ) . (262) This together with (261), implies that 󵄩󵄩 󵄩 󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄨 󵄨󵄩 󵄩 ≤ 𝜎𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝐺𝑥𝑛−1 󵄩󵄩󵄩 + 󵄨󵄨󵄨𝜎𝑛 − 𝜎𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐺𝑥𝑛−1 − 𝑧𝑛−1 󵄩󵄩󵄩 󵄩 󵄩 + (1 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑧𝑛 − 𝑧𝑛−1 󵄩󵄩󵄩 󵄩 󵄩 󵄨 󵄨󵄩 󵄩 ≤ 𝜎𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 󵄨󵄨󵄨𝜎𝑛 − 𝜎𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐺𝑥𝑛−1 − 𝑧𝑛−1 󵄩󵄩󵄩 + (1 − 𝜎𝑛 ) {(1 −

𝑛−1

󵄨 󵄨 󵄨 󵄨 + 𝑀1 [∏𝜆 𝑖 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨

𝑛−1

𝛼𝑛 (1 − 𝜌) 󵄩󵄩 󵄩 𝛾𝑀 ) 󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 𝑛 ∏𝜆 𝑖 1 − 𝛽𝑛 1 − 𝛽𝑛 𝑖=0

𝑖=0

󵄨 󵄨 󵄨 󵄨 + 󵄨󵄨󵄨𝛾𝑛 − 𝛾𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛿𝑛 − 𝛿𝑛−1 󵄨󵄨󵄨 ]

𝛿𝑛 󵄩󵄩 󵄩 + 󵄩𝑆 𝐺𝑥 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 1 − 𝛽𝑛 󵄩 𝑛 𝑛−1 1 󵄨󵄩 󵄨 󵄨 󵄩 󵄨 [󵄨󵄨𝛼 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑓 (𝑥𝑛−1 )󵄩󵄩󵄩 + 󵄨󵄨󵄨𝛾𝑛 − 𝛾𝑛−1 󵄨󵄨󵄨 1 − 𝛽𝑛 󵄨 𝑛 󵄩 󵄩 󵄨 󵄨󵄩 󵄩 × 󵄩󵄩󵄩𝐵𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 + 󵄨󵄨󵄨𝛿𝑛 − 𝛿𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩] 󵄨󵄨 󵄨 󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑦 − 𝛽𝑛−1 𝑥𝑛−1 󵄩󵄩󵄩 + 󵄩 (1 − 𝛽 ) (1 − 𝛽 ) 󵄩 𝑛−1

󵄩 󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛−1 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 }

+

𝑛−1

𝑛

𝑛−1 𝛼 (1 − 𝜌) 󵄩󵄩 󵄩 ≤ (1 − 𝑛 ) 󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 𝑀∏𝜆 𝑖 1 − 𝛽𝑛 𝑖=0

󵄩 󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛−1 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 +

1 1 − 𝛽𝑛

𝛼𝑛 (1 − 𝜌) 󵄩󵄩 󵄩 ) 󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 1 − 𝛽𝑛

≤ (1 −

(1 − 𝜎𝑛 ) 𝛼𝑛 (1 − 𝜌) 󵄩󵄩 󵄩 ) 󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 1 − 𝛽𝑛

󵄨 󵄨󵄩 󵄩 + 󵄨󵄨󵄨𝜎𝑛 − 𝜎𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐺𝑥𝑛−1 − 𝑧𝑛−1 󵄩󵄩󵄩 𝑛−1

󵄨 󵄨 󵄨 󵄨 + 𝑀1 [∏𝜆 𝑖 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 𝑖=0

󵄨 󵄨 󵄨 󵄨 + 󵄨󵄨󵄨𝛾𝑛 − 𝛾𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛿𝑛 − 𝛿𝑛−1 󵄨󵄨󵄨 ]

Abstract and Applied Analysis

37

󵄩 󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛−1 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 ≤ (1 −

Utilizing Lemma 15 we get from (253) and (265)

(1 − 𝜎𝑛 ) 𝛼𝑛 (1 − 𝜌) 󵄩󵄩 󵄩 ) 󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 1 − 𝛽𝑛 𝑛−1

󵄨 󵄨 󵄨 󵄨 + 𝑀2 [∏𝜆 𝑖 + 󵄨󵄨󵄨𝜎𝑛 − 𝜎𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 𝑖=0

󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛾𝑛 − 𝛾𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛿𝑛 − 𝛿𝑛−1 󵄨󵄨󵄨 ] 󵄩 󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛−1 − 𝑆𝑛−1 𝐺𝑥𝑛−1 󵄩󵄩󵄩 , (263)

where sup𝑛≥0 {𝑀1 + ‖𝐺𝑥𝑛 − 𝑧𝑛 ‖} ≤ 𝑀2 for some 𝑀2 > 0. Since ∑∞ 𝑛=0 𝛼𝑛 = ∞ and (1−𝜎𝑛 )𝛼𝑛 (1−𝜌)/(1−𝛽𝑛 ) ≥ (1−𝜎𝑛 )𝛼𝑛 (1−𝜌), we obtain from conditions (i) and (iv) that ∑∞ 𝑛=0 ((1−𝜎𝑛 )𝛼𝑛 (1− 𝜌)/(1 − 𝛽𝑛 )) = ∞. Thus, applying Lemma 7 to (263), we deduce from condition (iii) and the assumption on {𝑆𝑛 } that (noting that 0 < 𝜆 𝑖 ≤ 𝑏 < 1, for all 𝑖 ≥ 0) 󵄩󵄩

lim 󵄩𝑥 𝑛 → ∞ 󵄩 𝑛+1

󵄩 − 𝑥𝑛 󵄩󵄩󵄩 = 0.

󵄩2 󵄩󵄩 󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 󵄩 󵄩2 = 󵄩󵄩󵄩𝜎𝑛 (𝐺𝑥𝑛 − 𝑝) + (1 − 𝜎𝑛 ) (𝑦𝑛 − 𝑝)󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 ≤ 𝜎𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 − 𝜎𝑛 (1 − 𝜎𝑛 ) 𝑔 (󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩) 󵄩2 󵄩 ≤ 𝜎𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝜎𝑛 ) 󵄩󵄩 󵄩 󵄩 󵄩2 󵄩 × [󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 2𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩] 󵄩 󵄩 − 𝜎𝑛 (1 − 𝜎𝑛 ) 𝑔 (󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩) 󵄩󵄩 󵄩 󵄩 󵄩2 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 2𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 − 𝜎𝑛 (1 − 𝜎𝑛 ) 𝑔 (󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩) , which hence yields 󵄩 󵄩 𝜎𝑛 (1 − 𝜎𝑛 ) 𝑔 (󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩) 󵄩 󵄩2 󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩

󵄩󵄩 󵄩 󵄩 + 2𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩

(264)

(267)

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩

Next, we show that ‖𝑥𝑛 − 𝐺𝑥𝑛 ‖ → 0 as 𝑛 → ∞. Indeed, according to Lemma 8 we have from (253)

󵄩󵄩 󵄩 󵄩 + 2𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 .

󵄩2 󵄩󵄩 󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 󵄩 = 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑓 (𝑝)) + 𝛽𝑛 (𝑥𝑛 − 𝑝) + 𝛾𝑛 (𝐵𝑛 𝑥𝑛 − 𝑝) 󵄩2 + 𝛿𝑛 (𝑆𝑛 𝐺𝑥𝑛 − 𝑝) + 𝛼𝑛 (𝑓 (𝑝) − 𝑝)󵄩󵄩󵄩 󵄩 ≤ 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑓 (𝑝)) + 𝛽𝑛 (𝑥𝑛 − 𝑝) 󵄩2 + 𝛾𝑛 (𝐵𝑛 𝑥𝑛 − 𝑝) + 𝛿𝑛 (𝑆𝑛 𝐺𝑥𝑛 − 𝑝)󵄩󵄩󵄩

Since 𝛼𝑛 → 0 and ‖𝑥𝑛+1 − 𝑥𝑛 ‖ → 0, from condition (iv) and the boundedness of {𝑥𝑛 }, it follows that 󵄩 󵄩 lim 𝑔 (󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩) = 0.

𝑛→∞

(268)

Utilizing the properties of 𝑔, we have 󵄩 󵄩 lim 󵄩󵄩𝐺𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩 = 0,

𝑛→∞ 󵄩

(269)

which, together with (253) and (257), implies that 󵄩󵄩 󵄩 󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑦𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 + 𝜎𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩 󳨀→ 0

+ 2𝛼𝑛 ⟨𝑓 (𝑝) − 𝑝, 𝐽 (𝑥𝑛+1 − 𝑝)⟩ 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑓 (𝑝)󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩2 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑝󵄩󵄩󵄩

(270)

as 𝑛 󳨀→ ∞.

+ 2𝛼𝑛 ⟨𝑓 (𝑝) − 𝑝, 𝐽 (𝑥𝑛+1 − 𝑝)⟩ That is,

󵄩 󵄩 󵄩 󵄩2 󵄩2 󵄩2 ≤ 𝛼𝑛 𝜌󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩󵄩 󵄩 󵄩 󵄩2 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 2𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 󵄩 󵄩2 = (1 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩󵄩 󵄩 󵄩 + 2𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 󵄩󵄩 󵄩 󵄩 󵄩2 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 2𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 .

(266)

󵄩󵄩

lim 󵄩𝑥 𝑛→∞ 󵄩 𝑛

󵄩 − 𝑦𝑛 󵄩󵄩󵄩 = 0.

(271)

Since 󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩󵄩𝑥𝑛 − 𝐺𝑥𝑛 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝐺𝑥𝑛 󵄩󵄩󵄩 ,

(272)

it immediately follows from (269) and (271) that (265)

󵄩󵄩

lim 󵄩𝑥 𝑛→∞ 󵄩 𝑛

󵄩 − 𝐺𝑥𝑛 󵄩󵄩󵄩 = 0.

(273)

38

Abstract and Applied Analysis

On the other hand, observe that 𝑦𝑛 can be rewritten as follows: 𝑦𝑛 = 𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 𝛾 𝐵 𝑥 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 = 𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝛽𝑛 𝑥𝑛 + (𝛾𝑛 + 𝛿𝑛 ) 𝑛 𝑛 𝑛 (274) 𝛾𝑛 + 𝛿𝑛 = 𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝛽𝑛 𝑥𝑛 + 𝑒𝑛 𝑧̂𝑛 , where 𝑒𝑛 = 𝛾𝑛 + 𝛿𝑛 and 𝑧̂𝑛 = (𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 )/(𝛾𝑛 + 𝛿𝑛 ). Utilizing Lemma 11, we have 󵄩2 󵄩󵄩 󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩2 = 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑝) + 𝛽𝑛 (𝑥𝑛 − 𝑝) + 𝑒𝑛 (̂𝑧𝑛 − 𝑝)󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩 + 𝑒𝑛 󵄩󵄩󵄩𝑧̂𝑛 − 𝑝󵄩󵄩󵄩 − 𝛽𝑛 𝑒𝑛 𝑔1 (󵄩󵄩󵄩𝑧̂𝑛 − 𝑥𝑛 󵄩󵄩󵄩) 󵄩 󵄩 󵄩2 󵄩2 = 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 − 𝛽𝑛 𝑒𝑛 𝑔1 (󵄩󵄩󵄩𝑧̂𝑛 − 𝑥𝑛 󵄩󵄩󵄩) 󵄩󵄩2 󵄩󵄩 𝛾 𝐵 𝑥 + 𝛿 𝑆 𝐺𝑥 󵄩 󵄩 𝑛 𝑛 𝑛 + 𝑒𝑛 󵄩󵄩󵄩 𝑛 𝑛 𝑛 − 𝑝󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 𝛾𝑛 + 𝛿𝑛 󵄩 󵄩 󵄩2 󵄩2 = 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 − 𝛽𝑛 𝑒𝑛 𝑔1 (󵄩󵄩󵄩𝑧̂𝑛 − 𝑥𝑛 󵄩󵄩󵄩) 󵄩󵄩2 󵄩󵄩 𝛾 𝛿𝑛 󵄩 󵄩 𝑛 + 𝑒𝑛 󵄩󵄩󵄩 (𝐵𝑛 𝑥𝑛 − 𝑝) + (𝑆𝑛 𝐺𝑥𝑛 − 𝑝)󵄩󵄩󵄩 𝛾𝑛 + 𝛿𝑛 󵄩󵄩 󵄩󵄩 𝛾𝑛 + 𝛿𝑛 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 − 𝛽𝑛 𝑒𝑛 𝑔1 (󵄩󵄩󵄩𝑧̂𝑛 − 𝑥𝑛 󵄩󵄩󵄩) + 𝑒𝑛 [

𝛾𝑛 󵄩󵄩 𝛿𝑛 󵄩󵄩 󵄩2 󵄩2 󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩𝑆 𝐺𝑥 − 𝑝󵄩󵄩󵄩 ] 𝛾𝑛 + 𝛿𝑛 𝛾𝑛 + 𝛿𝑛 󵄩 𝑛 𝑛

󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 − 𝛽𝑛 𝑒𝑛 𝑔1 (󵄩󵄩󵄩𝑧̂𝑛 − 𝑥𝑛 󵄩󵄩󵄩)

󵄩 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩 . (276) Utilizing (271), conditions (i), (ii), (iv), and the boundedness of {𝑥𝑛 }, {𝑦𝑛 } and {𝑓(𝑥𝑛 )}, we get 󵄩 󵄩 lim 𝑔 (󵄩󵄩𝑧̂ − 𝑥𝑛 󵄩󵄩󵄩) = 0. (277) 𝑛→∞ 1 󵄩 𝑛 From the properties of 𝑔1 , we have 󵄩 󵄩 lim 󵄩󵄩𝑧̂ − 𝑥𝑛 󵄩󵄩󵄩 = 0. 𝑛→∞ 󵄩 𝑛

(278)

Utilizing Lemma 15 and the definition of 𝑧̂𝑛 , we have 󵄩󵄩̂ 󵄩2 󵄩󵄩𝑧𝑛 − 𝑝󵄩󵄩󵄩 󵄩󵄩 𝛾 𝐵 𝑥 + 𝛿 𝑆 𝐺𝑥 󵄩󵄩2 󵄩 󵄩 𝑛 𝑛 𝑛 = 󵄩󵄩󵄩 𝑛 𝑛 𝑛 − 𝑝󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 𝛾𝑛 + 𝛿𝑛 󵄩󵄩 𝛾 󵄩󵄩󵄩2 𝛿𝑛 󵄩 𝑛 = 󵄩󵄩󵄩 (𝐵𝑛 𝑥𝑛 − 𝑝) + (𝑆𝑛 𝐺𝑥𝑛 − 𝑝)󵄩󵄩󵄩 󵄩󵄩 𝛾𝑛 + 𝛿𝑛 󵄩󵄩 𝛾𝑛 + 𝛿𝑛 ≤

𝛾𝑛 󵄩󵄩 𝛿𝑛 󵄩󵄩 󵄩2 󵄩2 󵄩𝐵 𝑥 − 𝑝󵄩󵄩󵄩 + 󵄩𝑆 𝐺𝑥 − 𝑝󵄩󵄩󵄩 𝛾𝑛 + 𝛿𝑛 󵄩 𝑛 𝑛 𝛾𝑛 + 𝛿𝑛 󵄩 𝑛 𝑛 −

𝛾𝑛 𝛿𝑛

2

(𝛾𝑛 + 𝛿𝑛 )

󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − which leads to 𝛾𝑛 𝛿𝑛

2

(𝛾𝑛 + 𝛿𝑛 )

󵄩 󵄩 𝑔2 (󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) 𝛾𝑛 𝛿𝑛

2

(𝛾𝑛 + 𝛿𝑛 )

󵄩 󵄩 𝑔2 (󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) , (279)

󵄩 󵄩 𝑔2 (󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩)

󵄩 󵄩2 󵄩 󵄩2 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑧̂𝑛 − 𝑝󵄩󵄩󵄩

(280)

󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑧̂𝑛 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑧̂𝑛 󵄩󵄩󵄩 .

Since {𝑥𝑛 } and {̂𝑧𝑛 } are bounded, we deduce from (278) and condition (ii) that 󵄩 󵄩 lim 𝑔 (󵄩󵄩𝐵 𝑥 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩) = 0. (281) 𝑛→∞ 2 󵄩 𝑛 𝑛 From the properties of 𝑔2 , we have 󵄩 󵄩 lim 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 = 0.

𝛾𝑛 󵄩󵄩 𝛿𝑛 󵄩󵄩 󵄩2 󵄩2 + 𝑒𝑛 [ 󵄩𝑥 − 𝑝󵄩󵄩󵄩 + 󵄩𝑥 − 𝑝󵄩󵄩󵄩 ] 𝛾𝑛 + 𝛿𝑛 󵄩 𝑛 𝛾𝑛 + 𝛿𝑛 󵄩 𝑛

𝑛→∞

(282)

Furthermore, 𝑦𝑛 can also be rewritten as follows:

󵄩2 󵄩 󵄩2 󵄩 = 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩

𝑦𝑛 = 𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛

󵄩 󵄩 − 𝛽𝑛 𝑒𝑛 𝑔1 (󵄩󵄩󵄩𝑧̂𝑛 − 𝑥𝑛 󵄩󵄩󵄩)

󵄩 󵄩2 󵄩 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 − 𝛽𝑛 𝑒𝑛 𝑔1 (󵄩󵄩󵄩𝑧̂𝑛 − 𝑥𝑛 󵄩󵄩󵄩) ,

which hence implies that 󵄩 󵄩 𝛽𝑛 𝑒𝑛 𝑔1 (󵄩󵄩󵄩𝑧̂𝑛 − 𝑥𝑛 󵄩󵄩󵄩) 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩

= 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + (𝛼𝑛 + 𝛿𝑛 )

𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 𝛼𝑛 + 𝛿𝑛

= 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝑑𝑛 𝑧̃𝑛 , (275)

(283)

Abstract and Applied Analysis

39

where 𝑑𝑛 = 𝛼𝑛 + 𝛿𝑛 and 𝑧̃𝑛 = (𝛼𝑛 𝑓(𝑥𝑛 ) + 𝛿𝑛 𝑆𝑛 𝐺𝑥𝑛 )/(𝛼𝑛 + 𝛿𝑛 ). Utilizing Lemma 11 and the convexity of ‖ ⋅ ‖2 , we have 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩2 󵄩 󵄩 󵄩 󵄩2 = 󵄩󵄩󵄩𝛽𝑛 (𝑥𝑛 − 𝑝) + 𝛾𝑛 (𝐵𝑛 𝑥𝑛 − 𝑝) + 𝑑𝑛 (̃𝑧𝑛 − 𝑝)󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩 + 𝑑𝑛 󵄩󵄩󵄩𝑧̃𝑛 − 𝑝󵄩󵄩󵄩 − 𝛽𝑛 𝛾𝑛 𝑔3 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩)

Thus, from (282) and (287), we get 󵄩 󵄩 󵄩 󵄩󵄩 󵄩󵄩𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩

󵄩 󵄩 + 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 󳨀→ 0

That is,

󵄩 󵄩 󵄩2 󵄩2 = 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑝󵄩󵄩󵄩

󵄩󵄩

󵄩󵄩 𝛼 𝑓 (𝑥 ) + 𝛿 𝑆 𝐺𝑥 󵄩󵄩󵄩2 󵄩󵄩 𝑛 𝑛 𝑛 𝑛 𝑛 󵄩 + 𝑑𝑛 󵄩󵄩 − 𝑝󵄩󵄩󵄩󵄩 𝛼𝑛 + 𝛿𝑛 󵄩󵄩 󵄩󵄩 󵄩 󵄩 − 𝛽𝑛 𝛾𝑛 𝑔3 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) 󵄩 󵄩 󵄩2 󵄩2 = 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩󵄩2 󵄩󵄩 𝛼 𝛿𝑛 󵄩 󵄩 𝑛 + 𝑑𝑛 󵄩󵄩󵄩 (𝑓 (𝑥𝑛 ) − 𝑝) + (𝑆𝑛 𝐺𝑥𝑛 − 𝑝)󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 𝛼𝑛 + 𝛿𝑛 𝛼𝑛 + 𝛿𝑛 󵄩 󵄩 − 𝛽𝑛 𝛾𝑛 𝑔3 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) 󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 𝛼𝑛 󵄩󵄩 𝛿𝑛 󵄩󵄩 󵄩2 󵄩2 󵄩𝑓 (𝑥𝑛 )−𝑝󵄩󵄩󵄩 + 󵄩𝑆 𝐺𝑥 −𝑝󵄩󵄩 ] 𝛼𝑛 + 𝛿𝑛 󵄩 𝛼𝑛 + 𝛿𝑛 󵄩 𝑛 𝑛 󵄩 󵄩 󵄩 − 𝛽𝑛 𝛾𝑛 𝑔3 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) 󵄩 󵄩2 󵄩 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + (𝛽𝑛 + 𝛾𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝛽𝑛 𝛾𝑛 𝑔3 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) 󵄩 󵄩2 󵄩 󵄩2 = 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 − 𝛽𝑛 𝛾𝑛 𝑔3 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) 󵄩 󵄩2 󵄩 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 󵄩 󵄩 − 𝛽𝑛 𝛾𝑛 𝑔3 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) , (284) + 𝑑𝑛 [

as 𝑛 󳨀→ ∞. (288)

lim 󵄩𝑥 𝑛→∞ 󵄩 𝑛

󵄩 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 = 0.

(289)

Therefore, from Lemma 12, (273), and (289), it follows that 󵄩󵄩 󵄩 󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑆𝑛 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑛 𝐺𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩 󳨀→ 0 as 𝑛 󳨀→ ∞. (290) That is, 󵄩󵄩

lim 󵄩𝑥 𝑛→∞ 󵄩 𝑛

󵄩 − 𝑆𝑥𝑛 󵄩󵄩󵄩 = 0.

(291)

Suppose that 𝛽𝑛 ≡ 𝛽 for some fixed 𝛽 ∈ (0, 1) such that 𝛼𝑛 + 𝛽 + 𝛾𝑛 + 𝛿𝑛 = 1 for all 𝑛 ≥ 0. Define a mapping 𝑉𝑥 = (1 − 𝜃1 − 𝜃2 )𝑆𝑥 + 𝜃1 𝐵𝑥 + 𝜃2 𝐺𝑥, where 𝜃1 , 𝜃2 ∈ (0, 1) are two constants with 𝜃1 + 𝜃2 < 1. Then, by Lemmas 14 and 17, we have that Fix(𝑉) = Fix(𝑆) ∩ Fix(𝐵) ∩ Fix(𝐺) = 𝐹. For each 𝑘 ≥ 1, let {𝑝𝑘 } be a unique element of 𝐶 such that 𝑝𝑘 =

1 1 𝑓 (𝑝𝑘 ) + (1 − ) 𝑉𝑝𝑘 . 𝑘 𝑘

(292)

which hence implies that 󵄩 󵄩 𝛽𝑛 𝛾𝑛 𝑔3 (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 (285) 󵄩 󵄩2 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩) 󵄩 󵄩 × 󵄩󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩 . Utilizing (271), conditions (i), (ii), (iv), and the boundedness of {𝑥𝑛 }, {𝑦𝑛 } and {𝑓(𝑥𝑛 )}, we get 󵄩 󵄩 lim 𝑔 (󵄩󵄩𝑥 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩) = 0. (286) 𝑛→∞ 3 󵄩 𝑛

From Lemma 13, we conclude that 𝑝𝑘 → 𝑞 ∈ Fix(𝑉) = 𝐹 as 𝑘 → ∞. Repeating the same arguments as those of (81) in the proof of Theorem 24, we can conclude that for every 𝑛, 𝑘

From the properties of 𝑔3 , we have 󵄩 󵄩 lim 󵄩󵄩𝑥 − 𝐵𝑛 𝑥𝑛 󵄩󵄩󵄩 = 0. 𝑛→∞ 󵄩 𝑛

where 𝜃𝑛 = 𝛼𝑛 ‖𝑓(𝑥𝑛 ) − 𝐵𝑝𝑘 ‖ + (1 − 𝛽)‖𝐵𝑛 𝑝𝑘 − 𝐵𝑝𝑘 ‖ + 𝛿𝑛 ‖𝑆𝑛 𝐺𝑥𝑛 −𝐵𝑛 𝑥𝑛 ‖. Since lim𝑛 → ∞ 𝛼𝑛 = lim𝑛 → ∞ ‖𝐵𝑛 𝑝𝑘 −𝐵𝑝𝑘 ‖ = lim𝑛 → ∞ ‖𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑛 𝑥𝑛 ‖ = 0, we know that 𝜃𝑛 → 0 as

(287)

󵄩 󵄩󵄩 󵄩󵄩𝑦𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 ≤ 𝜃𝑛 + 𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 ,

(293)

40

Abstract and Applied Analysis

𝑛 → ∞. So, it immediately follows that

Utilizing (295) and (297), we deduce that

󵄩󵄩 󵄩2 󵄩󵄩𝑥𝑛+1 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩2 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝑦𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑦𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 × [2 󵄩󵄩󵄩𝑦𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑦𝑛 󵄩󵄩󵄩] 󵄩 󵄩 󵄩 󵄩2 ≤ (𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩) 󵄩 󵄩 󵄩 󵄩 + 𝜃𝑛 [2 (𝛽 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩) + 𝜃𝑛 ] 󵄩 󵄩 󵄩 󵄩 󵄩 󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑦𝑛 󵄩󵄩󵄩 [2 󵄩󵄩󵄩𝑦𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑦𝑛 󵄩󵄩󵄩] 2󵄩 󵄩 󵄩2 󵄩2 = 𝛽2 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 󵄩󵄩 󵄩 󵄩 + 2𝛽 (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 𝜏𝑛 2󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛽2 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 󵄩2 󵄩 󵄩2 󵄩 + 𝛽 (1 − 𝛽) (󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 ) + 𝜏𝑛 󵄩2 󵄩2 󵄩 󵄩 = 𝛽󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 + (1 − 𝛽) 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 𝜏𝑛 ,

(294)

󵄩 󵄩2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑉𝑝𝑘 󵄩󵄩󵄩 󵄩 = 𝜇𝑛 󵄩󵄩󵄩(1 − 𝜃1 − 𝜃2 ) (𝑥𝑛 − 𝑆𝑝𝑘 ) 󵄩2 + 𝜃1 (𝑥𝑛 − 𝐵𝑝𝑘 ) + 𝜃2 (𝑥𝑛 − 𝐺𝑝𝑘 )󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 ≤ (1 − 𝜃1 − 𝜃2 ) 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑝𝑘 󵄩󵄩󵄩 + 𝜃1 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩2 + 𝜃2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩2 ≤ 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 .

Repeating the same arguments as those of (99) in the proof of Theorem 24, we can obtain that 1 󵄩󵄩 󵄩2 𝜇 󵄩𝑥 − 𝑝𝑘 󵄩󵄩󵄩 ≥ 𝜇𝑛 ⟨𝑓 (𝑝𝑘 ) − 𝑝𝑘 , 𝐽 (𝑥𝑛 − 𝑝𝑘 )⟩ . 2𝑘 𝑛 󵄩 𝑛

(295)

In addition, note that 󵄩󵄩 󵄩2 󵄩󵄩𝑥𝑛 − 𝐺𝑝𝑘 󵄩󵄩󵄩 󵄩2 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑥𝑛 + 𝐺𝑥𝑛 − 𝐺𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩 ≤ (󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩) 󵄩2 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 × (2 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑥𝑛 󵄩󵄩󵄩) , 󵄩󵄩 󵄩2 󵄩󵄩𝑥𝑛 − 𝑆𝑝𝑘 󵄩󵄩󵄩 󵄩2 󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 + 𝑆𝑥𝑛 − 𝑆𝑝𝑘 󵄩󵄩󵄩 󵄩 󵄩 󵄩2 󵄩 ≤ (󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩) 󵄩2 󵄩 󵄩 󵄩 = 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩 󵄩 󵄩 󵄩 󵄩 × (2 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩) .

󵄩 󵄩2 󵄩 󵄩2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑝𝑘 󵄩󵄩󵄩 ≤ 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 .

𝜇𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩ ≤ 0.

(300)

On the other hand, from (257) and the norm-to-weak∗ uniform continuity of 𝐽 on bounded subsets of 𝑋, it follows that 󵄨 󵄨 lim 󵄨󵄨⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ − ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩󵄨󵄨󵄨 = 0. (301)

𝑛→∞ 󵄨

So, utilizing Lemma 18, we deduce from (300) and (301) that lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩ ≤ 0, 𝑛→∞

(296)

(302)

which together with (271) and the norm-to-weak∗ uniform continuity of 𝐽 on bounded subsets of 𝑋, implies that lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑦𝑛 − 𝑞)⟩ ≤ 0. 𝑛→∞

(303)

Finally, let us show that 𝑥𝑛 → 𝑞 as 𝑛 → ∞. Utilizing Lemma 8 (i), from (253) and the convexity of ‖ ⋅ ‖, we get 󵄩󵄩 󵄩2 󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 󵄩 ≤ 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑥𝑛 ) − 𝑓 (𝑞)) + 𝛽𝑛 (𝑥𝑛 − 𝑞)

It is easy to see from (273) and (291) that 󵄩 󵄩2 󵄩 󵄩2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐺𝑝𝑘 󵄩󵄩󵄩 ≤ 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 ,

(299)

Since 𝑝𝑘 → 𝑞 ∈ Fix(𝑉) = 𝐹 as 𝑘 → ∞, by the uniform Gateaux differentiability of the norm of 𝑋 we have

where 𝜏𝑛 = 𝜃𝑛 [2(𝛽‖𝑥𝑛 −𝐵𝑝𝑘 ‖+(1−𝛽)‖𝑥𝑛 −𝑝𝑘 ‖)+𝜃𝑛 ]+‖𝑥𝑛+1 − 𝑦𝑛 ‖[2‖𝑦𝑛 − 𝐵𝑝𝑘 ‖ + ‖𝑥𝑛+1 − 𝑦𝑛 ‖] → 0 as 𝑛 → ∞. For any Banach limit 𝜇, from (294), we derive 󵄩 󵄩2 󵄩 󵄩2 󵄩 󵄩2 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝐵𝑝𝑘 󵄩󵄩󵄩 = 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛+1 − 𝐵𝑝𝑘 󵄩󵄩󵄩 ≤ 𝜇𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝𝑘 󵄩󵄩󵄩 .

(298)

(297)

󵄩2 + 𝛾𝑛 (𝐵𝑛 𝑥𝑛 − 𝑞) + 𝛿𝑛 (𝑆𝑛 𝐺𝑥𝑛 − 𝑞)󵄩󵄩󵄩 + 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑦𝑛 − 𝑞)⟩

Abstract and Applied Analysis

41

󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑓 (𝑞)󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩

arbitrarily given 𝑥0 ∈ 𝐶, let {𝑥𝑛 } be the sequence generated by

󵄩 󵄩 󵄩2 󵄩2 + 𝛾𝑛 󵄩󵄩󵄩𝐵𝑛 𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺𝑥𝑛 − 𝑞󵄩󵄩󵄩

𝑥𝑛+1

+ 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑦𝑛 − 𝑞)⟩

= 𝜎𝑛 ((1 − 𝑙) 𝐼 + 𝑙𝑉) 𝑥𝑛 + (1 − 𝜎𝑛 )

󵄩 󵄩 󵄩2 󵄩2 ≤ 𝛼𝑛 𝜌󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩

× [𝛼𝑛 𝑓 (𝑥𝑛 ) + 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝐵𝑛 𝑥𝑛 + 𝛿𝑛 𝑆𝑛 ((1−𝑙) 𝐼+𝑙𝑉) 𝑥𝑛 ] ,

󵄩 󵄩 󵄩2 󵄩2 + 𝛾𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩

∀𝑛 ≥ 0, (306)

+ 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑦𝑛 − 𝑞)⟩ 󵄩 󵄩2 = (1 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑦𝑛 − 𝑞)⟩ , (304)

where 1 − (𝜁/(1 + 𝜁))(1 − √(1 − 𝜃)/𝜁) ≤ 𝑙 ≤ 1 and {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, {𝛿𝑛 }, and {𝜎𝑛 } are the sequences in (0, 1) such that 𝛼𝑛 +𝛽𝑛 + 𝛾𝑛 + 𝛿𝑛 = 1 for all 𝑛 ≥ 0. Suppose that the following conditions hold: (i) lim𝑛 → ∞ 𝛼𝑛 = 0 and ∑∞ 𝑛=0 𝛼𝑛 = ∞;

and hence

(ii) {𝛾𝑛 }, {𝛿𝑛 } ⊂ [𝑐, 𝑑] for some 𝑐, 𝑑 ∈ (0, 1);

(iii) ∑∞ 𝑛=1 (|𝜎𝑛 − 𝜎𝑛−1 | + |𝛼𝑛 − 𝛼𝑛−1 | + |𝛽𝑛 − 𝛽𝑛−1 | + |𝛾𝑛 − 𝛾𝑛−1 | + |𝛿𝑛 − 𝛿𝑛−1 |) < ∞;

󵄩󵄩 󵄩2 󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩 󵄩 󵄩2 󵄩 󵄩2 ≤ 𝜎𝑛 󵄩󵄩󵄩𝐺𝑥𝑛 − 𝑞󵄩󵄩󵄩 + (1 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩

(iv) 0 < lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ 𝛽𝑛 < 1 and 0 < lim inf 𝑛 → ∞ 𝜎𝑛 ≤ lim sup𝑛 → ∞ 𝜎𝑛 < 1.

󵄩 󵄩2 ≤ 𝜎𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + (1 − 𝜎𝑛 ) 󵄩 󵄩2 × [(1 − 𝛼𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑦𝑛 − 𝑞)⟩ ]

(305)

󵄩 󵄩2 = [1 − (1 − 𝜎𝑛 ) 𝛼𝑛 (1 − 𝜌)] 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 2 (1 − 𝜎𝑛 ) 𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑦𝑛 − 𝑞)⟩ 󵄩 󵄩2 = [1 − (1 − 𝜎𝑛 ) 𝛼𝑛 (1 − 𝜌)] 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 2 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑦𝑛 − 𝑞)⟩ + (1 − 𝜎𝑛 ) 𝛼𝑛 (1 − 𝜌) . 1−𝜌 From conditions (i) and (iv), it is easy to see that ∑∞ 𝑛=0 (1 − 𝜎𝑛 )𝛼𝑛 (1 − 𝜌) = ∞. Applying Lemma 7 to (305), we infer that 𝑥𝑛 → 𝑞 as 𝑛 → ∞. This completes the proof. Corollary 34. Let 𝐶 be a nonempty closed convex subset of a uniformly convex Banach space 𝑋 which has a uniformly Gateaux differentiable norm. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶. Let {𝜌𝑛 }∞ 𝑛=0 be a sequence of positive numbers in (0, 𝑏] for some 𝑏 ∈ (0, 1) and 𝐴 𝑖 : 𝐶 → 𝑋𝜉𝑖 strictly pseudocontractive and 𝛼̂𝑖 -strongly accretive with 𝜉𝑖 + 𝛼̂𝑖 ≥ 1 for each 𝑖 = 0, 1, . . .. Define a mapping 𝐺𝑖 : 𝐶 → 𝐶 by Π𝐶(𝐼 − 𝜆 𝑖 𝐴 𝑖 )𝑥 = 𝐺𝑖 𝑥 for all 𝑥 ∈ 𝐶 and 𝑖 = 0, 1, . . ., where 1−(𝜉𝑖 /(1+𝜉𝑖 ))(1−√(1 − 𝛼̂𝑖 )/𝜉𝑖 ) ≤ 𝜆 𝑖 ≤ 1 for all 𝑖 = 0, 1, . . .. Let 𝐵𝑛 : 𝐶 → 𝐶 be the 𝑊-mapping generated by 𝐺𝑛 , 𝐺𝑛−1 , . . . , 𝐺0 and 𝜌𝑛 , 𝜌𝑛−1 , . . . , 𝜌0 . Let 𝑉 : 𝐶 → 𝐶 be a self-mapping such that 𝐼 − 𝑉 : 𝐶 → 𝑋 is 𝜁-strictly pseudocontractive and 𝜃strongly accretive with 𝜃 + 𝜁 ≥ 1. Let 𝑓 : 𝐶 → 𝐶 be a contraction with coefficient 𝜌 ∈ (0, 1). Let {𝑆𝑖 }∞ 𝑖=0 be a countable family of nonexpansive mappings of 𝐶 into itself such that ∞ 𝐹 = (⋂∞ 𝑖=0 Fix(𝑆𝑖 )) ∩ Fix(𝑉) ∩ (⋂𝑖=0 VI(𝐶, 𝐴 𝑖 )) ≠ 0. For

Assume that ∑∞ 𝑛=1 sup𝑥∈𝐷 ‖𝑆𝑛 𝑥 − 𝑆𝑛−1 𝑥‖ < ∞ for any bounded subset 𝐷 of 𝐶 and let 𝑆 be a mapping of 𝐶 into itself defined by 𝑆𝑥 = lim𝑛 → ∞ 𝑆𝑛 𝑥 for all 𝑥 ∈ 𝐶 and suppose that Fix(𝑆) = ⋂∞ 𝑖=0 Fix(𝑆𝑖 ). Then, there hold the following: (I) lim𝑛 → ∞ ‖𝑥𝑛+1 − 𝑥𝑛 ‖ = 0;

(II) the sequence {𝑥𝑛 }∞ 𝑛=0 converges strongly to some 𝑞 ∈ 𝐹 which is the unique solution of the variational inequality problem (VIP) ⟨(𝐼 − 𝑓) 𝑞, 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,

∀𝑝 ∈ 𝐹,

(307)

provided 𝛽𝑛 ≡ 𝛽 for some fixed 𝛽 ∈ (0, 1). Proof. In Theorem 33, we put 𝐵1 = 𝐼 − 𝑉, 𝐵2 = 0 and 𝜇1 = 𝑙 where 1 − (𝜁/(1 + 𝜁))(1 − √(1 − 𝜃)/𝜁) ≤ 𝑙 ≤ 1. Then, GSVI (13) is equivalent to the VIP of finding 𝑥∗ ∈ 𝐶 such that ⟨𝐵1 𝑥∗ , 𝐽 (𝑥 − 𝑥∗ )⟩ ≥ 0,

∀𝑥 ∈ 𝐶.

(308)

In this case, 𝐵1 : 𝐶 → 𝑋 is 𝜁-strictly pseudocontractive and 𝜃-strongly accretive. Repeating the same arguments as those in the proof of Corollary 25, we can infer that Fix(𝑉) = VI(𝐶, 𝐵1 ). Accordingly, 𝐹 = (⋂∞ 𝑖=0 Fix(𝑆𝑖 )) ∩ Ω ∩ ∞ VI(𝐶, 𝐴 )) = (⋂ Fix(𝑆 )) ∩ Fix(𝑉) ∩ (⋂∞ (⋂∞ 𝑖 𝑖 𝑖=0 𝑖=0 𝑖=0 VI(𝐶, 𝐴 𝑖 )), 𝐺𝑥𝑛 = ((1 − 𝑙) 𝐼 + 𝑙𝑉) 𝑥𝑛 ,

(309)

So, the scheme (251) reduces to (306). Therefore, the desired result follows from Theorem 33. Remark 35. Our Theorems 31 and 33 improve, extend, supplement and develop Ceng and Yao’s [10, Theorem 3.2], Cai and Bu’s [11, Theorem 3.1], Kangtunyakarn’s [38, Theorem 3.1], and Ceng and Yao’s [8, Theorem 3.1], in the following aspects.

42

Abstract and Applied Analysis (i) The problem of finding a point 𝑞 ∈ (⋂∞ 𝑖=0 Fix(𝑆𝑖 )) ∩ Ω ∩ (⋂∞ 𝑖=0 VI(𝐶, 𝐴 𝑖 )) in our Theorems 31 and 33 is more general and more subtle than every one of the problem of finding a point 𝑞 ∈ ⋂∞ 𝑖=0 Fix(𝑇𝑖 ) in [10, Theorem 3.2], the problem of finding a point 𝑞 ∈ ⋂∞ 𝑖=1 Fix(𝑇𝑖 ) ∩ Ω in [11, Theorem 3.1], the problem of finding a point 𝑞 ∈ Fix(𝑆) ∩ Fix(𝑉) ∩ (⋂𝑁 𝑖=1 VI(𝐶, 𝐴 𝑖 )) in [38, Theorem 3.1], and the problem of finding a point 𝑞 ∈ Fix(𝑇) in [8, Theorem 3.1]. (ii) The iterative scheme in [38, Theorem 3.1] is extended to develop the iterative scheme (178) of our Theorem 31, and the iterative scheme in [11, Theorem 3.1] is extended to develop the iterative scheme (251) of our Theorem 33. Iterative schemes (178) and (181) in our Theorems 31 and 33 are more advantageous and more flexible than the iterative scheme of [11, Theorem 3.1] because they both are one-step iteration schemes and involve several parameter sequences {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, {𝛿𝑛 }, (and {𝜎𝑛 }).

argument techniques in [14], the inequality in 2uniformly smooth Banach spaces (see Lemma 4), and the inequality in smooth and uniform convex Banach spaces (see Proposition 6). It depends on only the inequalities in uniform convex Banach spaces (see Lemmas 11 and 15 in Section 2 of this paper) and the properties of the 𝑊-mapping and the Banach limit (see Lemmas 16–18 in Section 2 of this paper). (vi) The assumption of the uniformly convex and 2uniformly smooth Banach space 𝑋 in [11, Theorem 3.1] is weakened to the one of the uniformly convex Banach space 𝑋 having a uniformly Gateaux differentiable norm in our Theorem 33. Moreover, the assumption of the uniformly smooth Banach space 𝑋 in [8, Theorem 3.1] is replaced with the one of the uniformly convex Banach space 𝑋 having a uniformly Gateaux differentiable norm in our Theorem 33.

Conflict of Interests

(iii) Our Theorems 31 and 33 extend and generalize Ceng and Yao’s [8, Theorem 3.1] from a nonexpansive mapping to a countable family of nonexpansive mappings, and Ceng and Yao’s [10, Theorems 3.2] to the setting of the GSVI (13) and infinitely many VIPs, Kangtunyakarn’s [38, Theorem 3.1] from finitely many VIPs to infinitely many VIPs, from a nonexpansive mapping to a countable family of nonexpansive mappings and from a strict pseudocontraction to the GSVI (13). In the meantime, our Theorems 31 and 33 extend and generalize Cai and Bu’s [11, Theorem 3.1] to the setting of infinitely many VIPs.

The authors declare that there is no conflict of interests regarding the publication of this paper.

(iv) The iterative schemes (178) and (251) in our Theorems 31 and 33 are very different from every one in [10, Theorem 3.2], [11, Theorem 3.1], [38, Theorem 3.1], and [8, Theorem 3.1] because the mappings 𝐺 and 𝑇𝑛 in [11, Theorem 3.1] and the mapping 𝑇 in [8, Theorem 3.1] are replaced with the same composite mapping 𝑆𝑛 𝐺 in the iterative schemes (42) and (130) and the mapping 𝑊𝑛 in [10, Theorem 3.2] is replaced by 𝐵𝑛 .

References

(v) Cai and Bu’s proof in [11, Theorem 3.1] depends on the argument techniques in [14], the inequality in 2uniformly smooth Banach spaces (see Lemma 4), and the inequality in smooth and uniform convex Banach spaces (see Proposition 6). Because the composite mapping 𝑆𝑛 𝐺 appears in the iterative scheme (178) of our Theorem 31, the proof of our Theorem 31 depends on the argument techniques in [14], the inequality in 2-uniformly smooth Banach spaces (see Lemma 4), the inequality in smooth and uniform convex Banach spaces (see Proposition 6), the inequalities in uniform convex Banach spaces (see Lemmas 11 and 15 in Section 2 of this paper), and the properties of the 𝑊-mapping and the Banach limit (see Lemmas 16, 17, and 18 in Section 2 of this paper). However, the proof of our Theorem 33 does not depend on the

Acknowledgments This work was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University under Grant no. HiCi/15130-1433. The authors, therefore, acknowledge technical and financial support of KAU. The authors would like to thank Professor J. C. Yao for motivation and many fruitful discussions regarding this work.

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