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 Β© 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 Korpelevichβs 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 π be a real Banach space whose dual space is denoted by β πβ . The normalized duality mapping π½ : π β 2π is defined by σ΅© σ΅©2 π½ (π₯) = {π₯β β πβ : β¨π₯, π₯β β© = βπ₯β2 = σ΅©σ΅©σ΅©π₯β σ΅©σ΅©σ΅© } , βπ₯ β π, (1) where β¨β
, β
β© denotes the generalized duality pairing. It is an immediate consequence of the Hahn-Banach theorem that π½(π₯) is nonempty for each π₯ β π. Let πΆ be a nonempty closed convex subset of π. A mapping π : πΆ β πΆ is called nonexpansive if βππ₯ β ππ¦β β€ βπ₯ β π¦β for every π₯, π¦ β πΆ. The set of fixed points of π is denoted by Fix(π). We use the notation β to indicate the weak convergence and the one β to indicate the strong convergence. A mapping π΄ : πΆ β π is said to be (i) accretive if for each π₯, π¦ β πΆ there exists π(π₯ β π¦) β π½(π₯ β π¦) such that β¨π΄π₯ β π΄π¦, π (π₯ β π¦)β© β₯ 0;
(2)
(ii) πΌ-strongly accretive if for each π₯, π¦ β πΆ there exists π(π₯ β π¦) β π½(π₯ β π¦) such that σ΅©2 σ΅© β¨π΄π₯ β π΄π¦, π (π₯ β π¦)β© β₯ πΌσ΅©σ΅©σ΅©π₯ β π¦σ΅©σ΅©σ΅© ,
(3)
for some πΌ β (0, 1); (iii) π½-inverse strongly accretive if, for each π₯, π¦ β πΆ, there exists π(π₯ β π¦) β π½(π₯ β π¦) such that σ΅©2 σ΅© β¨π΄π₯ β π΄π¦, π (π₯ β π¦)β© β₯ π½σ΅©σ΅©σ΅©π΄π₯ β π΄π¦σ΅©σ΅©σ΅© ,
(4)
for some π½ > 0; (iv) π-strictly pseudocontractive [1] (see also [2]) if for each π₯, π¦ β πΆ there exists π(π₯ β π¦) β π½(π₯ β π¦) such that σ΅©2 σ΅©2 σ΅© σ΅© β¨π΄π₯ β π΄π¦, π (π₯ β π¦)β© β€ σ΅©σ΅©σ΅©π₯ β π¦σ΅©σ΅©σ΅© β πσ΅©σ΅©σ΅©π₯ β π¦ β (π΄π₯ β π΄π¦)σ΅©σ΅©σ΅© (5) for some π β (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, [3β5]. Let π = {π₯ β π : βπ₯β = 1} denote the unite sphere of π. A Banach space π is said to be uniformly convex if, for each π β (0, 2], there exists πΏ > 0 such that, for all π₯, π¦ β π, σ΅© σ΅©σ΅© σ΅©π₯ + π¦σ΅©σ΅©σ΅© σ΅© σ΅©σ΅© β€ 1 β πΏ. σ΅©σ΅©π₯ β π¦σ΅©σ΅©σ΅© β₯ π σ³¨β σ΅© 2
(6)
It is known that a uniformly convex Banach space is reflexive and strict convex. A Banach space π is said to be smooth if the limit σ΅©σ΅©σ΅©π₯ + π‘π¦σ΅©σ΅©σ΅© β βπ₯β σ΅© (7) lim σ΅© π‘β0 π‘ exists for all π₯, π¦ β π; in this case, π is also said to have a Gateaux differentiable norm. π is said to have a uniformly, Gateaux differentiable norm if, for each π¦ β π, the limit is attained uniformly for π₯ β π. Moreover, it is said to be uniformly smooth if this limit is attained uniformly for π₯, π¦ β π. The norm of π is said to be the Frechet differential if for each π₯ β π, this limit is attained uniformly for π¦ β π. In the meantime, we define a function π : [0, β) β [0, β) called the modulus of smoothness of π as follows: 1 σ΅© σ΅© σ΅© σ΅© π (π) = sup { (σ΅©σ΅©σ΅©π₯ + π¦σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯ β π¦σ΅©σ΅©σ΅©) β 1 : π₯, π¦ β π, 2 σ΅© σ΅© βπ₯β = 1, σ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅© = π} .
(8)
(i) 1 β€ π½π β€ 1 β π, for all π β₯ π0 for some integer π0 β₯ 1, (ii) ββ π=0 π½π = β, (iii) 0 < lim inf π β β πΌπ β€ lim supπ β β πΌπ < 1, (iv) limπ β β (π½π+1 /(1 β (1 β π½π+1 )πΌπ+1 ) β π½π /(1 β (1 β π½π )πΌπ )) = 0. Ceng and Yao [8] proved that π₯π σ³¨β π ββ π½π (π (π₯π ) β π₯π ) σ³¨β 0,
(9)
where π β Fix(π) solves the variational inequality problem (VIP): β¨π β π (π) , π½ (π β π)β© β€ 0,
βπ β Fix (π) .
(10)
Such a result includes [7, Theorem 1] as a special case. Let πΆ be a nonempty closed convex subset of a real Banach space π and π β ΞπΆ with a contractive coefficient π β (0, 1), where ΞπΆ is the set of all contractive self-mappings on πΆ. Let {ππ }β π=0 be a sequence of nonexpansive self-mappings on πΆ and {π π }β π=0 a sequence of nonnegative numbers in [0, 1]. For any π β₯ 0, define a self-mapping ππ on πΆ as follows: ππ,π+1 = πΌ, ππ,π = π π ππ ππ,π+1 + (1 β π π ) πΌ,
It is known that π is uniformly smooth if and only if limπ β 0 π(π)/π = 0. Let π be a fixed real number with 1 < π β€ 2. Then, a Banach space π is said to be π-uniformly smooth if there exists a constant π > 0 such that π(π) β€ πππ for all π > 0. As pointed out in [6], no Banach space is π-uniformly smooth for π > 2. In addition, it is also known that π½ is single valued if and only if π is smooth, whereas if π is uniformly smooth, then the mapping π½ is norm-to-norm uniformly continuous on bounded subsets of π. If π has a uniformly Gateaux differentiable norm, then the duality mapping π½ is norm-toweakβ uniformly continuous on bounded subsets of π. Recently, Yao et al. [7] combined the viscosity approximation method and Mann iteration method and gave the following hybrid viscosity approximation method. Let πΆ be a nonempty closed convex subset of a real uniformly smooth Banach space π, π : πΆ β πΆ a nonexpansive mapping with Fix(π) =ΜΈ 0, and π : πΆ β πΆ a contraction with coefficient π β (0, 1). For an arbitrary π₯0 β πΆ, define {π₯π } in the following way: π¦π = πΌπ π₯π + (1 β πΌπ ) ππ₯π , π₯π+1 = π½π π (π₯π ) + (1 β π½π ) π¦π ,
a fixed point of π. Subsequently, under the following control conditions on {πΌπ } and {π½π }:
βπ β₯ 0,
(YCY)
where {πΌπ } and {π½π } are two sequences in (0, 1). They proved under certain control conditions on the sequences {πΌπ } and {π½π } that {π₯π } converges strongly to
ππ,πβ1 = π πβ1 ππβ1 ππ,π + (1 β π πβ1 ) πΌ, .. . ππ,π = π π ππ ππ,π+1 + (1 β π π ) πΌ,
(CY)
ππ,πβ1 = π πβ1 ππβ1 ππ,π + (1 β π πβ1 ) πΌ, .. . ππ,1 = π 1 π1 ππ,2 + (1 β π 1 ) πΌ, ππ = ππ,0 = π 0 π0 ππ,1 + (1 β π 0 ) πΌ. Such a mapping ππ is called the π-mapping generated by ππ , ππβ1 , . . . , π0 , and π π , π πβ1 , . . . , π 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 π be a strictly convex and reflexive Banach space with a uniformly Gateaux differentiable norm, πΆ a nonempty closed convex subset of π, {ππ }β π=0 a sequence of nonexpansive self-mappings on πΆ such that the common fixed point set πΉ := ββ π=0 Fix(ππ ) =ΜΈ 0, and π β ΞπΆ with a contractive
Abstract and Applied Analysis
3
coefficient π β (1/2, 1). For any given π₯0 β πΆ, let {π₯π }β π=0 be the iterative sequence defined by π¦π = (1 β πΎπ ) π₯π + πΎπ ππ π₯π , π₯π+1 = (1 β πΌπ β π½π ) π₯π + πΌπ π (π¦π ) + π½π ππ π¦π ,
βπ β₯ 0, (11)
β where {πΌπ }β π=0 and {π½π }π=0 are two sequences in (0, 1) with πΌπ + β π½π β€ 1 (π β₯ 0), {πΎπ }π=0 is a sequence in [0, 1], and ππ is the π-mapping generated by (CY). Assume that
(i) limπ β β πΌπ = 0, ββ = β and 0 π=0 πΌπ lim inf π β β π½π β€ lim supπ β β π½π < 1;
0 a given number, which is known as the extragradient method (see also [16]). The literature on the VIP is vast and Korpelevichβs extragradient method has received great attention given by many authors, who improved it in various ways; see, for example, [3, 11, 13, 17β33] and references therein, to name but a few. In particular, whenever π is still a real smooth Banach space, π΅1 = π΅2 = π΄ and π₯β = π¦β , then GSVI (13) reduces to the variational inequality problem (VIP) of finding π₯β β πΆ such that β¨π΄π₯β , π½ (π₯ β π₯β )β© β₯ 0,
βπ₯ β πΆ,
(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 π΄: π₯π+1 = πΌπ π₯π + (1 β πΌπ ) Ξ πΆ (π₯π β π π π΄π₯π ) ,
βπ β₯ 1, (19)
where Ξ πΆ is a sunny nonexpansive retraction from π onto πΆ. Then, they proved a weak convergence theorem. For the related work, see [37] and the references therein. Let πΆ be a nonempty convex subset of a real Banach space π. Let {ππ }π π=1 be a finite family of nonexpansive mappings of
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Abstract and Applied Analysis
πΆ into itself and let π 1 , . . . , π π be real numbers such that 0 β€ π π β€ 1 for every π = 1, . . . , π. Define a mapping πΎ : πΆ β πΆ as follows: π1 = π 1 π1 + (1 β π 1 ) πΌ, π2 = π 2 π2 π1 + (1 β π 2 ) π1 , π3 = π 3 π3 π2 + (1 β π 3 ) π2 , (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 π. Lemma 4 (see [39]). Let π be a 2-uniformly smooth Banach space. Then,
ππβ1 = π πβ1 ππβ1 ππβ2 + (1 β π πβ1 ) ππβ2 , πΎ = ππ = π πππππβ1 + (1 β π π) ππβ1 . Such a mapping πΎ is called the πΎ-mapping generated by π1 , . . . , ππ and π 1 , . . . , π π. Very recently, Kangtunyakarn [38] introduced and analyzed an iterative algorithm by the modification of Mannβs 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 π-strictly pseudocontractive mapping and a nonexpansive mapping in uniformly convex and 2uniformly smooth Banach spaces.
σ΅© σ΅©2 σ΅©2 σ΅©σ΅© 2 σ΅©σ΅©π₯ + π¦σ΅©σ΅©σ΅© β€ βπ₯β + 2 β¨π¦, π½ (π₯)β© + 2σ΅©σ΅©σ΅©π
π¦σ΅©σ΅©σ΅© ,
βπ₯, π¦ β π, (23)
where π
is the 2-uniformly smooth constant of π and π½ is the normalized duality mapping from π into πβ . Define the mapping πΊ : πΆ β πΆ as follows: πΊ (π₯) := Ξ πΆ (πΌ β π1 π΅1 ) Ξ πΆ (πΌ β π2 π΅2 ) π₯,
βπ₯ β πΆ.
(24)
The fixed point set of πΊ is denoted by Ξ©. Then, their strong convergence theorem on the proposed method is stated as follows.
Theorem 3 (see [38]). Let πΆ be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space π. Let Ξ πΆ be a sunny nonexpansive retraction from π onto πΆ. Let π΄ π : πΆ β π be an πΌπ -inverse-strongly accretive mapping for each π = 1, . . . , π. Define the mapping πΊπ : πΆ β πΆ by πΊπ = Ξ πΆ(πΌ β π π π΄ π ) for π = 1, . . . , π, where π π β (0, πΌπ /π
2 ) and π
is the 2-uniformly smooth constant of π. Let π΅ : πΆ β πΆ be the πΎ-mapping generated by πΊ1 , . . . , πΊπ and π1 , . . . , ππ, where ππ β (0, 1), for all π = 1, . . . , π β 1, and ππ β (0, 1]. Let π : πΆ β πΆ a contraction with coefficient π β (0, 1). Let π : πΆ β πΆ be an π-strictly pseudocontractive mapping and π : πΆ β πΆ be a nonexpansive mapping such that πΉ = Fix(π) β© Fix(π) β© (βπ π=1 VI(πΆ, π΄ π )) =ΜΈ 0. For arbitrarily given π₯0 β πΆ, let {π₯π } be the sequence generated by
Theorem 5 (see [11]). Let πΆ be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space π. Let Ξ πΆ be a sunny nonexpansive retraction from π onto πΆ. Let the mapping π΅π : πΆ β π be π½π -inverse-strongly accretive with 0 < ππ < π½π /π
2 for π = 1, 2. Let π be a contraction of πΆ into itself with coefficient πΏ β (0, 1). Let {ππ }β π=1 be a countable family of nonexpansive mappings of πΆ into itself such that πΉ = ββ π=1 Fix(ππ ) β© Ξ© =ΜΈ 0, where Ξ© is the fixed point set of the mapping πΊ defined by (24). For arbitrarily given π₯1 β πΆ, let {π₯π } be the sequence generated by
π₯π+1 = πΌπ π (π₯π ) + π½π π₯π + πΎπ π΅π₯π + πΏπ π ((1 β πΌ) πΌ + πΌπ) π₯π ,
π¦π = πΌπ π (π₯π ) + (1 β πΌπ ) π§π ,
βπ β₯ 0, (21) 2
where πΌ β (0, π/π
). Suppose that {πΌπ }, {π½π }, {πΎπ }, and {πΏπ } are the sequences in [0, 1], πΌπ + π½π + πΎπ + πΏπ = 1 and satisfy the following conditions:
π§π = Ξ πΆ (π’π β π1 π΅1 π’π ) , π’π = Ξ πΆ (π₯π β π2 π΅2 π₯π ) ,
(25)
βπ β₯ 1.
Suppose that {πΌπ } and {π½π } are two sequences in (0, 1) satisfying the following conditions: (i) limπ β β πΌπ = 0 and ββ π=1 πΌπ = β;
(i) limπ β β πΌπ = 0 and ββ π=0 πΌπ = β;
(ii) 0 < lim inf π β β π½π β€ lim supπ β β π½π < 1.
(ii) {πΎπ }, {πΏπ } β [π, π] for some π, π β (0, 1);
(iii) ββ π=1 (|π½π β π½πβ1 | + |πΎπ β πΎπβ1 | + |πΏπ β πΏπβ1 |) < β; (iv) 0 < lim inf π β β π½π β€ lim supπ β β π½π < 1. Then, {π₯π } converges strongly to π β πΉ, which solves the following VIP: β¨π β π (π) , π½ (π β π)β© β€ 0,
π₯π+1 = π½π π₯π + (1 β π½π ) ππ π¦π ,
βπ β πΉ.
(22)
Assume that ββ π=1 supπ₯βπ· βππ+1 π₯βππ π₯β < β for any bounded subset π· of πΆ and let π be a mapping of πΆ into π defined by ππ₯ = limπ β β ππ π₯ for all π₯ β πΆ and suppose that Fix(π) = ββ π=1 Fix(ππ ). Then, {π₯π } converges strongly to π β πΉ, which solves the following VIP: β¨π β π (π) , π½ (π β π)β© β€ 0,
βπ β πΉ.
(26)
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: π¦π = πΌπ π (π₯π ) + (1 β πΌπ ) πΊπ₯π , π₯π+1 = π½π π₯π + (1 β π½π ) ππ π¦π ,
βπ β₯ 1.
(27)
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 π. Proposition 6 (see [40]). Let π be a real smooth and uniform convex Banach space and let π > 0. Then, there exists a strictly increasing, continuous, and convex function π : [0, 2π] β R, π(0) = 0 such that σ΅© σ΅©2 σ΅© σ΅© π (σ΅©σ΅©σ΅©π₯ β π¦σ΅©σ΅©σ΅©) β€ βπ₯β2 β 2 β¨π₯, π½ (π¦)β© + σ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅© ,
βπ₯, π¦ β π΅π , (28)
where π΅π = {π₯ β π : βπ₯β β€ π}. Let πΆ be a nonempty closed convex subset of a real smooth Banach space π. Let Ξ πΆ be a sunny nonexpansive retraction from π onto πΆ and π : πΆ β πΆ a contraction with coefficient π β (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 π. Here, the hybrid and relaxed Mann iteration methods are based on Korpelevichβs 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 {π π } be a sequence of nonnegative real numbers satisfying π π+1 β€ (1 β πΌπ ) π π + πΌπ π½π + πΎπ ,
βπ β₯ 0,
where {πΌπ }, {π½π }, and {πΎπ } satisfy the following conditions: (i) {πΌπ } β [0, 1] and
ββ π=0
(ii) lim supπ β β π½π β€ 0;
πΌπ = β;
(iii) πΎπ β₯ 0, for all π β₯ 0, and ββ π=0 πΎπ < β. Then, lim supπ β β π π = 0.
(29)
The following lemma is an immediate consequence of the subdifferential inequality of the function (1/2)β β
β2 . Lemma 8 (see [42]). Let π be a real Banach space π. Then, for all π₯, π¦ β π (i) βπ₯ + π¦β2 β€ βπ₯β2 + 2β¨π¦, π(π₯ + π¦)β© for all π(π₯ + π¦) β π½(π₯ + π¦); (ii) βπ₯ + π¦β2 β₯ βπ₯β2 + 2β¨π¦, π(π₯)β© for all π(π₯) β π½(π₯). Let π· be a subset of πΆ and let Ξ be a mapping of πΆ into π·. Then, Ξ is said to be sunny if Ξ [Ξ (π₯) + π‘ (π₯ β Ξ (π₯))] = Ξ (π₯) ,
(30)
whenever Ξ (π₯) + π‘(π₯ β Ξ (π₯)) β πΆ for π₯ β πΆ and π‘ β₯ 0. A mapping Ξ of πΆ into itself is called a retraction if Ξ 2 = Ξ . If a mapping Ξ of πΆ into itself is a retraction, then Ξ (π§) = π§ for every π§ β π
(Ξ ) where π
(Ξ ) is the range of Ξ . A subset π· of πΆ is called a sunny nonexpansive retract of πΆ if there exists a sunny nonexpansive retraction from πΆ onto π·. The following lemma concerns the sunny nonexpansive retraction. Lemma 9 (see [43]). Let πΆ be a nonempty closed convex subset of a real smooth Banach space π. Let π· be a nonempty subset of πΆ. Let Ξ be a retraction of πΆ onto π·. Then, the following are equivalent: (i) Ξ is sunny and nonexpansive; (ii) βΞ (π₯) β Ξ (π¦)β2 β€ β¨π₯ β π¦, π½(Ξ (π₯) β Ξ (π¦))β©, for all π₯, π¦ β πΆ; (iii) β¨π₯ β Ξ (π₯), π½(π¦ β Ξ (π₯))β© β€ 0, for all π₯ β πΆ, π¦ β π·. It is well known that if π = π» a Hilbert space, then a sunny nonexpansive retraction Ξ πΆ is coincident with the metric projection from π onto πΆ; that is, Ξ πΆ = ππΆ. If πΆ is a nonempty closed convex subset of a strictly convex and uniformly smooth Banach space π and if π : πΆ β πΆ is a nonexpansive mapping with the fixed point set Fix(π) =ΜΈ 0, then the set Fix(π) is a sunny nonexpansive retract of πΆ. Lemma 10. Let πΆ be a nonempty closed convex subset of a smooth Banach space π. Let Ξ πΆ be a sunny nonexpansive retraction from π onto πΆ and let π΅1 , π΅2 : πΆ β π be nonlinear mappings. For given π₯β , π¦β β πΆ, (π₯β , π¦β ) is a solution of GSVI (13) if and only if π₯β = Ξ πΆ(π¦β β π1 π΅1 π¦β ), where π¦β = Ξ πΆ(π₯β β π2 π΅2 π₯β ). Proof. We can rewrite GSVI (13) as β¨π₯β β (π¦β β π1 π΅1 π¦β ) , π½ (π₯ β π₯β )β© β₯ 0,
βπ₯ β πΆ,
β¨π¦β β (π₯β β π2 π΅2 π₯β ) , π½ (π₯ β π¦β )β© β₯ 0,
βπ₯ β πΆ,
(31)
which is obviously equivalent to π₯β = Ξ πΆ (π¦β β π1 π΅1 π¦β ) , π¦β = Ξ πΆ (π₯β β π2 π΅2 π₯β ) , because of Lemma 9. This completes the proof.
(32)
6
Abstract and Applied Analysis In terms of Lemma 10, we observe that π₯β = Ξ πΆ [Ξ πΆ (π₯β β π2 π΅2 π₯β ) β π1 π΅1 Ξ πΆ (π₯β β π2 π΅2 π₯β )] , (33)
which implies that π₯β is a fixed point of the mapping πΊ. Throughout this paper, the set of fixed points of the mapping πΊ is denoted by Ξ©. Lemma 11 (see [44]). Let π be a uniformly convex Banach space and π΅π = {π₯ β π : βπ₯β β€ π}, π > 0. Then, there exists a continuous, strictly increasing, and convex function π : [0, β] β [0, β], π(0) = 0 such that σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅©σ΅© 2 2 σ΅©σ΅©πΌπ₯ + π½π¦ + πΎπ§σ΅©σ΅©σ΅© β€ πΌβπ₯β + π½σ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅© + πΎβπ§β β πΌπ½π (σ΅©σ΅©σ΅©π₯ β π¦σ΅©σ΅©σ΅©) (34)
a continuous strictly increasing function π : [0, β) β [0, β), π(0) = 0, such that σ΅© σ΅©2 σ΅©2 σ΅©σ΅© 2 σ΅©σ΅©ππ₯ + (1 β π) π¦σ΅©σ΅©σ΅© β€ πβπ₯β + (1 β π) σ΅©σ΅©σ΅©π¦σ΅©σ΅©σ΅© σ΅© σ΅© β π (1 β π) π (σ΅©σ΅©σ΅©π₯ β π¦σ΅©σ΅©σ΅©)
(37)
for all π β [0, 1] and π₯, π¦ β π such that βπ₯β β€ π and βπ¦β β€ π. Lemma 16 (see [48, Lemma 3.2]). Let πΆ be a nonempty closed convex subset of a strictly convex Banach space π. Let {ππ }β π=0 be a sequence of nonexpansive self-mappings on πΆ such that β βπ=0 Fix(ππ ) =ΜΈ 0 and let {π π }β π=0 be a sequence of positive numbers in (0, π] for some π β (0, 1). Then, for every π₯ β πΆ and π β₯ 0, the limit limπ β β ππ,π π₯ exists. Using Lemma 16, one can define a mapping π : πΆ β πΆ as follows:
for all π₯, π¦, π§ β π΅π , and all πΌ, π½, πΎ β [0, 1] with πΌ + π½ + πΎ = 1.
ππ₯ = lim ππ π₯ = lim ππ,0 π₯
Lemma 12 (see [45]). Let πΆ be a nonempty closed convex subset of a Banach space π. Let π0 , π1 , . . . be a sequence of mappings of πΆ into itself. Suppose that ββ π=1 sup{βππ π₯βππβ1 π₯β : π₯ β πΆ} < β. Then for each π¦ β πΆ, {ππ π¦} converges strongly to some point of πΆ. Moreover, let π be a mapping of πΆ into itself defined by ππ¦ = limπ β β ππ π¦ for all π¦ β πΆ. Then limπ β β sup{βππ₯ β ππ π₯β : π₯ β πΆ} = 0.
for every π₯ β πΆ. Such a π is called the π-mapping generated β by the sequences {ππ }β π=0 and {π π }π=0 . Throughout this paper, β we always assume that {π π }π=0 is a sequence of positive numbers in (0, π] for some π β (0, 1).
Let πΆ be a nonempty closed convex subset of a Banach space π and π : πΆ β πΆ a nonexpansive mapping with Fix(π) =ΜΈ 0. As previous, let ΞπΆ be the set of all contractions on πΆ. For π‘ β (0, 1) and π β ΞπΆ, let π₯π‘ β πΆ be the unique fixed point of the contraction π₯ σ³¨β π‘π(π₯) + (1 β π‘)ππ₯ on πΆ; that is, π₯π‘ = π‘π (π₯π‘ ) + (1 β π‘) ππ₯π‘ .
(35)
Lemma 13 (see [35, 46]). Let π be a uniformly smooth Banach space, or a reflexive and strictly convex Banach space with a uniformly Gateaux differentiable norm. Let πΆ be a nonempty closed convex subset of π, π : πΆ β πΆ a nonexpansive mapping with Fix(π) =ΜΈ 0, and π β ΞπΆ. Then, the net {π₯π‘ } defined by π₯π‘ = π‘π(π₯π‘ ) + (1 β π‘)ππ₯π‘ converges strongly to a point in Fix(π). If we define a mapping π : ΞπΆ β Fix(π) by π(π) := π β limπ‘ β 0 π₯π‘ , for all π β ΞπΆ, then π(π) solves the VIP: β¨(πΌ β π) π (π) , π½ (π (π) β π)β© β€ 0,
βπ β ΞπΆ, π β Fix (π) . (36)
Lemma 14 (see [47]). Let πΆ be a nonempty closed convex subset of a strictly convex Banach space π. Let {ππ }β π=0 be a sequence of nonexpansive mappings on πΆ. Suppose that ββ π=0 Fix(ππ ) is nonempty. Let {π π } be a sequence of positive numbers with ββ π=0 π π = 1. Then, a mapping π on πΆ defined by ππ₯ = ββ π=0 π π ππ π₯ for π₯ β πΆ is defined well; nonexpansive and Fix(π) = ββ π=0 Fix(ππ ) holds. Lemma 15 (see [39]). Given a number π > 0, A real Banach space π is uniformly convex if and only if there exists
πββ
πββ
(38)
Lemma 17 (see [48]). Let πΆ be a nonempty closed convex subset of a strictly convex Banach space π. Let {ππ }β π=0 be a sequence of nonexpansive self-mappings on πΆ such that β ββ π=0 Fix(ππ ) =ΜΈ 0 and let {π π }π=0 be a sequence of positive numbers in (0, π] for some π β (0, 1). Then, Fix(π) = ββ π=0 Fix(ππ ). Let π be a continuous linear functional on πβ and π = (π0 , π1 , . . .) β πβ . One writes ππ (ππ ) instead of π(π ). π is called a Banach limit if π satisfies βπβ = ππ (1) = 1 and ππ (ππ+1 ) = ππ (ππ ) for all (π0 , π1 , . . .) β πβ . If π is a Banach limit, then, there hold the following: (i) for all π β₯ 0, ππ β€ ππ implies ππ (ππ ) β€ ππ (ππ ); (ii) ππ (ππ+π ) = ππ (ππ ) for any fixed positive integer π; (iii) lim inf π β β ππ β€ ππ (ππ ) β€ lim supπ β β ππ for all (π0 , π1 , . . .) β πβ . Lemma 18 (see [49]). Let π β R be a real number and a sequence {ππ } β πβ satisfy the condition ππ (ππ ) β€ π for all Banach limit π. If lim supπ β β (ππ+π β ππ ) β€ 0, then lim supπ β β ππ β€ π. In particular, if π = 1 in Lemma 18, then we immediately obtain the following corollary. Corollary 19 (see [50]). Let π β R be a real number and a sequence {ππ } β πβ satisfy the condition ππ (ππ ) β€ π for all Banach limit π. If lim supπ β β (ππ+1 β ππ ) β€ 0, then, lim supπ β β ππ β€ π. Lemma 20 (see [51]). Let {π₯π } and {π§π } be bounded sequences in a Banach space π and let {π½π } be a sequence of nonnegative numbers in [0, 1] with 0 < lim inf π β β π½π β€ lim supπ β β π½π < 1. Suppose that π₯π+1 = π½π π₯π + (1 β π½π )π§π for all integers
Abstract and Applied Analysis
7
π β₯ 0 and lim supπ β β (βπ§π+1 β π§π β β βπ₯π+1 β π₯π β) β€ 0. Then, limπ β β βπ₯π β π§π β = 0.
0 < ππ < π½Μπ /π
2 for π = 1, 2. For arbitrarily given π₯0 β πΆ, let {π₯π } be the sequence generated by
Lemma 21 (see [34]). Let πΆ be a nonempty closed convex subset of a smooth Banach space π. Let Ξ πΆ be a sunny nonexpansive retraction from π onto πΆ and π΄ : πΆ β π an accretive mapping. Then for all π > 0,
π¦π = π½π π₯π + πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π ,
VI (πΆ, π΄) = Fix (Ξ πΆ (πΌ β ππ΄)) .
(39)
Lemma 22 (see [11]). Let πΆ be a nonempty closed convex subset of a real 2-uniformly smooth Banach space π. Let the mapping π΅π : πΆ β π be π½Μπ -inverse-strongly accretive. Then, one has σ΅©2 σ΅©σ΅© σ΅©σ΅©(πΌ β ππ π΅π ) π₯ β (πΌ β ππ π΅π ) π¦σ΅©σ΅©σ΅© σ΅©2 σ΅© σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©π₯ β π¦σ΅©σ΅©σ΅© + 2ππ (ππ π
2 β π½Μπ ) σ΅©σ΅©σ΅©π΅π π₯ β π΅π π¦σ΅©σ΅©σ΅© ,
for π = 1, 2 where ππ > 0. In particular, if 0 < ππ β€ π½Μπ /π
2 , then πΌ β ππ π΅π is nonexpansive for π = 1, 2. Lemma 23 (see [11]). Let πΆ be a nonempty closed convex subset of a real 2-uniformly smooth Banach space π. Let Ξ πΆ be a sunny nonexpansive retraction from π onto πΆ. Let the mapping π΅π : πΆ β π be π½Μπ -inverse-strongly accretive for π = 1, 2. Let πΊ : πΆ β πΆ be the mapping defined by
βπ₯ β πΆ.
βπ β₯ 0,
(42)
where {πΌπ }, {π½π }, {πΎπ }, and {πΏπ } are the sequences in [0, 1] such that π½π + πΎπ + πΏπ = 1 for all π β₯ 0. Suppose that the following conditions hold: (i) ββ π=0 πΌπ = β and 0 β€ πΌπ β€ 1 β π, for all π β₯ π0 for some integer π0 β₯ 0; (ii) lim inf π β β πΎπ > 0 and lim inf π β β πΏπ > 0; (iii) limπ β β (|πΌπ+1 /(1 β (1 β πΌπ+1 )π½π+1 ) β πΌπ /(1 β (1 β πΌπ )π½π )| + |πΏπ+1 /(1 β π½π+1 ) β πΏπ /(1 β π½π )|) = 0; (iv) 0 < lim inf π β β π½π β€ lim supπ β β π½π < 1.
(40)
βπ₯, π¦ β πΆ,
πΊπ₯ = Ξ πΆ [Ξ πΆ (π₯ β π2 π΅2 π₯) β π1 π΅1 Ξ πΆ (π₯ β π2 π΅2 π₯)] ,
π₯π+1 = πΌπ π (π₯π ) + (1 β πΌπ ) π¦π ,
(41)
If 0 < ππ β€ π½Μπ /π
2 for π = 1, 2, then πΊ : πΆ β πΆ 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 πΆ be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space π. Let Ξ πΆ be a sunny nonexpansive retraction from π onto πΆ. Let {ππ }β π=0 be a sequence of positive numbers in (0, π] for some π β (0, 1) and π΄ π : πΆ β πΈ an πΌΜπ -inverse strongly accretive mapping for each π = 0, 1, . . .. Define a mapping πΊπ : πΆ β πΆ by Ξ πΆ(πΌ β π π π΄ π )π₯ = πΊπ π₯ for all π₯ β πΆ and π = 0, 1, . . ., where π π β (0, πΌΜπ /π
2 ], π
is the 2-uniformly smooth constant of π. Let π΅π : πΆ β πΆ be the π-mapping generated by πΊπ , πΊπβ1 , . . . , πΊ0 and ππ , ππβ1 , . . . , π0 . Let the mapping π΅π : πΆ β π be π½Μπ inverse strongly accretive for π = 1, 2. Let π : πΆ β πΆ be a contraction with coefficient π β (0, 1). Let {ππ }β π=0 be a countable family of nonexpansive mappings of πΆ into itself such that πΉ = β (ββ π=0 Fix(ππ )) β© Ξ© β© (βπ=0 VI(πΆ, π΄ π )) =ΜΈ 0, where Ξ© is the fixed point set of the mapping πΊ = Ξ πΆ(πΌ β π1 π΅1 )Ξ πΆ(πΌ β π2 π΅2 ) with
Assume that ββ π=0 supπ₯βπ· βππ+1 π₯ β ππ π₯β < β for any bounded subset π· of πΆ and let π be a mapping of πΆ into itself defined by ππ₯ = limπ β β ππ π₯ for all π₯ β πΆ and suppose that Fix(π) = ββ π=0 Fix(ππ ). Then, there hold the following: (I) limπ β β βπ₯π+1 β π₯π β = 0; (II) π₯π β π β πΌπ (π(π₯π ) β π₯π ) β 0 provided π½π β‘ π½ for some fixed π½ β (0, 1), where π β πΉ solves the following VIP: β¨π β π (π) , π½ (π β π)β© β€ 0,
βπ β πΉ.
(43)
Proof. First of all, since 0 < π π < πΌΜπ /π
2 for π = 0, 1, . . ., it is easy to see that πΊπ is a nonexpansive mapping for each π = 0, 1, . . .. Since π΅π : πΆ β πΆ is the π-mapping generated by πΊπ , πΊπβ1 , . . . , πΊ0 and ππ , ππβ1 , . . . , π0 , by Lemma 16 we know that, for each π₯ β πΆ and π β₯ 0, the limit limπ β β ππ,π π₯ exists. Moreover, one can define a mapping π΅ : πΆ β πΆ as follows: π΅π₯ = lim π΅π π₯ = lim ππ,0 π₯ πββ
πββ
(44)
for every π₯ β πΆ. That is, such a π΅ is the π-mapping generated β by the sequences {πΊπ }β π=0 and {ππ }π=0 . According to Lemma 17, β we know that Fix(π΅) = βπ=0 Fix(πΊπ ). From Lemma 15 and the definition of πΊπ , we have Fix(πΊπ ) = VI(πΆ, π΄ π ) for each π = 0, 1, . . .. Hence, we have β
β
π=0
π=0
Fix (π΅) = β Fix (πΊπ ) = β VI (πΆ, π΄ π ) .
(45)
Next, let us show that the sequence {π₯π } is bounded. Indeed, take a fixed π β πΉ arbitrarily. Then, we get π = πΊπ, π = π΅π π, and π = ππ π for all π β₯ 0. By Lemma 23 we know that πΊ is nonexpansive. Then, from (42), we have σ΅© σ΅© σ΅© σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©πΊπ₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© , (46)
8
Abstract and Applied Analysis
and hence
=(
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅©σ΅© σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + (1 β πΌπ ) σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅©
σ΅© σ΅© σ΅© σ΅© β€ πΌπ (σ΅©σ΅©σ΅©π (π₯π ) β π (π)σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅©)
β
σ΅© σ΅© + (1 β πΌπ ) σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅©
+ πΏπ+1 ππ+1 πΊπ₯π+1 ]
σ΅© σ΅© + (1 β πΌπ ) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
β ππ+1 π₯π+1 Γ (1 β ππ+1 )
(47)
σ΅© σ΅© = (1 β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
=(
σ΅© σ΅©σ΅© σ΅©π (π) β πσ΅©σ΅©σ΅© + πΌπ (1 β π) σ΅© 1βπ σ΅© σ΅© σ΅© σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅© β€ max {σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© , σ΅© }. 1βπ
β =(
βπ β₯ 0. (48)
+
Thus, {π₯π } is bounded, and so are the sequences {π¦π }, {πΊπ₯π } and {π(π₯π )}. Let us show that σ΅© σ΅© lim σ΅©σ΅©π₯ β π₯π σ΅©σ΅©σ΅© = 0. π β β σ΅© π+1
(1 β πΌπ+1 ) (1 β π½π+1 ) 1 β ππ+1
+[ Γ
βπ β₯ π0 , (50)
(51)
(52)
πΎπ+1 π΅π+1 π₯π+1 + πΏπ+1 ππ+1 πΊπ₯π+1 1 β π½π+1 πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π ] 1 β π½π
(1 β πΌπ+1 ) (1 β π½π+1 ) (1 β πΌπ ) (1 β π½π ) β ] 1 β ππ+1 1 β ππ
πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π 1 β π½π
πΌπ+1 (π (π₯π+1 ) β π (π₯π )) 1 β ππ+1 +( +
Define π₯π+1 = ππ π₯π + (1 β ππ ) π§π .
πΌπ+1 π (π₯π+1 ) πΌπ π (π₯π ) β ) 1 β ππ+1 1 β ππ
(49)
and hence πββ
1 β πΌπ (πΎ π΅ π₯ + πΏπ ππ πΊπ₯π ) 1 β ππ π π π
β
= 0 < lim inf ππ β€ lim sup ππ < 1.
πΌπ+1 π (π₯π+1 ) πΌπ π (π₯π ) 1 β πΌπ+1 β )+ 1 β ππ+1 1 β ππ 1 β ππ+1
Γ[
As a matter of fact, put ππ = (1βπΌπ )π½π , for all π β₯ 0. Then, it follows from (i) and (iv) that π½π β₯ ππ = (1 β πΌπ ) π½π β₯ (1 β (1 β π)) π½π = ππ½π ,
πΌπ+1 πΌπ β ) π (π₯π ) 1 β ππ+1 1 β ππ
(1 β πΌπ+1 ) (1 β π½π+1 ) 1 β ππ+1
Γ[
Observe that
πΎπ+1 (π΅ π₯ β π΅π π₯π ) πΎπ+1 + πΏπ+1 π+1 π+1 +(
π§π+1 β π§π π₯π+2 β ππ+1 π₯π+1 π₯π+1 β ππ π₯π β 1 β ππ+1 1 β ππ
+
=
πΌπ+1 π (π₯π+1 ) + (1 β πΌπ+1 ) π¦π+1 β ππ+1 π₯π+1 1 β ππ+1
+(
πΌπ π (π₯π ) + (1 β πΌπ ) π¦π β ππ π₯π 1 β ππ
β(
πΎπ πΎπ+1 β ) π΅π π₯π πΎπ+1 + πΏπ+1 πΎπ + πΏπ
πΏπ+1 (π πΊπ₯ β ππ πΊπ₯π ) πΎπ+1 + πΏπ+1 π+1 π+1
=
β
β1
Γ (πΎπ+1 π΅π+1 π₯π+1 + πΏπ+1 ππ+1 πΊπ₯π+1 )
By induction, we obtain
πββ
(1 β πΌπ ) [π½π π₯π + πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π ] β ππ π₯π 1 β ππ
+ (1 β πΌπ+1 ) [π½π+1 π₯π+1 + πΎπ+1 π΅π+1 π₯π+1
σ΅© σ΅© σ΅© σ΅© β€ πΌπ (π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅©)
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅©σ΅© }, σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β€ max {σ΅©σ΅©σ΅©π₯0 β πσ΅©σ΅©σ΅© , σ΅© 1βπ
πΌπ+1 π (π₯π+1 ) πΌπ π (π₯π ) β ) 1 β ππ+1 1 β ππ
πΏπ+1 πΏπ β ) π πΊπ₯ ] πΎπ+1 + πΏπ+1 πΎπ + πΏπ π π
πΎ π΅ π₯ + πΏπ ππ πΊπ₯π πΌπ+1 πΌπ β ) π π π 1 β ππ+1 1 β ππ πΎπ + πΏπ
Abstract and Applied Analysis =
9
πΌπ+1 (π (π₯π+1 ) β π (π₯π )) 1 β ππ+1 +(
πΌπ+1 πΌπ β ) 1 β ππ+1 1 β ππ
Γ (π (π₯π ) β +
πΏπ+1 σ΅©σ΅© σ΅© σ΅©π πΊπ₯ β ππ πΊπ₯π σ΅©σ΅©σ΅© πΎπ+1 + πΏπ+1 σ΅© π+1 π+1 σ΅¨σ΅¨ πΏ πΏπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅©σ΅© σ΅¨ σ΅© π+1 + σ΅¨σ΅¨σ΅¨ β σ΅¨ σ΅©π πΊπ₯ σ΅©σ΅©] . σ΅¨σ΅¨ πΎπ+1 + πΏπ+1 πΎπ + πΏπ σ΅¨σ΅¨σ΅¨ σ΅© π π σ΅©
+
πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π ) πΎπ + πΏπ
(54)
1 β ππ+1 β πΌπ+1 1 β ππ+1
Γ[
On the other hand, we note that, for all π β₯ 0,
πΎπ+1 (π΅ π₯ β π΅π π₯π ) πΎπ+1 + πΏπ+1 π+1 π+1
σ΅©σ΅© σ΅© σ΅©σ΅©ππ+1 πΊπ₯π+1 β ππ πΊπ₯π σ΅©σ΅©σ΅©
σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©ππ+1 πΊπ₯π+1 β ππ+1 πΊπ₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ+1 πΊπ₯π β ππ πΊπ₯π σ΅©σ΅©σ΅©
πΎπ πΎπ+1 β ) π΅π π₯π +( πΎπ+1 + πΏπ+1 πΎπ + πΏπ +
σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©πΊπ₯π+1 β πΊπ₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ+1 πΊπ₯π β ππ πΊπ₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ+1 πΊπ₯π β ππ πΊπ₯π σ΅©σ΅©σ΅© .
πΏπ+1 (π πΊπ₯ β ππ πΊπ₯π ) πΎπ+1 + πΏπ+1 π+1 π+1
+(
πΏπ+1 πΏπ β ) π πΊπ₯ ] , πΎπ+1 + πΏπ+1 πΎπ + πΏπ π π (53)
and hence σ΅© σ΅©σ΅© σ΅©σ΅©π§π+1 β π§π σ΅©σ΅©σ΅© β€
πΌπ+1 σ΅©σ΅© σ΅© σ΅©π (π₯π+1 ) β π (π₯π )σ΅©σ΅©σ΅© 1 β ππ+1 σ΅© σ΅¨σ΅¨ πΌ πΎ π΅ π₯ + πΏπ ππ πΊπ₯π σ΅©σ΅©σ΅©σ΅© πΌπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©σ΅© σ΅¨ + σ΅¨σ΅¨σ΅¨ π+1 β σ΅¨σ΅¨ σ΅©σ΅©π (π₯π ) β π π π σ΅©σ΅© σ΅¨σ΅¨ 1 β ππ+1 1 β ππ σ΅¨σ΅¨ σ΅©σ΅© σ΅©σ΅© πΎπ + πΏπ 1 β ππ+1 β πΌπ+1 1 β ππ+1 σ΅©σ΅© πΎ σ΅© π+1 Γ σ΅©σ΅©σ΅© (π΅ π₯ β π΅π π₯π ) σ΅©σ΅© πΎπ+1 + πΏπ+1 π+1 π+1 +
πΎπ πΎπ+1 β ) π΅π π₯π +( πΎπ+1 + πΏπ+1 πΎπ + πΏπ πΏπ+1 (π πΊπ₯ β ππ πΊπ₯π ) πΎπ+1 + πΏπ+1 π+1 π+1 σ΅©σ΅© πΏπ+1 πΏπ σ΅© β ) ππ πΊπ₯π σ΅©σ΅©σ΅© +( σ΅©σ΅© πΎπ+1 + πΏπ+1 πΎπ + πΏπ σ΅¨ ππΌπ+1 σ΅©σ΅© πΌπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅© σ΅¨σ΅¨ πΌ β€ σ΅¨σ΅¨ σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + σ΅¨σ΅¨σ΅¨ π+1 β 1 β ππ+1 σ΅¨σ΅¨ 1 β ππ+1 1 β ππ σ΅¨σ΅¨ σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© Γ (σ΅©σ΅©σ΅©π (π₯π )σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯π σ΅©σ΅©σ΅©) +
1 β ππ+1 β πΌπ+1 + 1 β ππ+1 Γ[
πΎπ+1 σ΅©σ΅© σ΅© σ΅©π΅ π₯ β π΅π π₯π σ΅©σ΅©σ΅© πΎπ+1 + πΏπ+1 σ΅© π+1 π+1 σ΅¨σ΅¨σ΅¨ πΎπ+1 πΎπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅©σ΅© σ΅© + σ΅¨σ΅¨σ΅¨ β σ΅¨σ΅¨ σ΅©σ΅©π΅π π₯π σ΅©σ΅©σ΅© σ΅¨σ΅¨ πΎπ+1 + πΏπ+1 πΎπ + πΏπ σ΅¨σ΅¨
(55)
Furthermore, by (CY), since πΊπ and ππ,π are nonexpansive, we deduce that for each π β₯ 0 σ΅©σ΅© σ΅© σ΅©σ΅©π΅π+1 π₯π+1 β π΅π π₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π΅π+1 π₯π+1 β π΅π+1 π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π+1 π₯π β π΅π π₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π+1 π₯π β π΅π π₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π 0 πΊ0 ππ+1,1 π₯π β π 0 πΊ0 ππ,1 π₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + π 0 σ΅©σ΅©σ΅©ππ+1,1 π₯π β ππ,1 π₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + π 0 σ΅©σ΅©σ΅©π 1 πΊ1 ππ+1,2 π₯π β π 1 πΊ1 ππ,2 π₯π σ΅©σ΅©σ΅© (56) σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + π 0 π 1 σ΅©σ΅©σ΅©ππ+1,2 π₯π β ππ,2 π₯π σ΅©σ΅©σ΅© .. . π
σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + (βπ π ) σ΅©σ΅©σ΅©ππ+1,π+1 π₯π β ππ,π+1 π₯π σ΅©σ΅©σ΅© π=0
π
σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + π0 βπ π , π=0
for some constant π0 > 0. Utilizing (54)β(56), we have σ΅©σ΅© σ΅© σ΅©σ΅©π§π+1 β π§π σ΅©σ΅©σ΅© β€
σ΅¨ ππΌπ+1 σ΅©σ΅© πΌπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅© σ΅¨σ΅¨ πΌ σ΅¨ σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + σ΅¨σ΅¨σ΅¨ π+1 β σ΅¨σ΅¨ 1 β ππ+1 1 β ππ σ΅¨σ΅¨σ΅¨ 1 β ππ+1 σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© Γ (σ΅©σ΅©σ΅©π (π₯π )σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯π σ΅©σ΅©σ΅©)
+
1 β ππ+1 β πΌπ+1 1 β ππ+1
10
Abstract and Applied Analysis Γ[
π πΎπ+1 σ΅© σ΅© (σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + π0 βπ π ) πΎπ+1 + πΏπ+1 π=0
where supπβ₯0 {βπ(π₯π )β+βπ΅π π₯π β+βππ πΊπ₯π β+π0 } β€ π for some π > 0. So, from (58), condition (iii), and the assumption on {ππ }, it follows that (noting that 0 < π π β€ π < 1, for all π β₯ 0)
σ΅¨σ΅¨ πΎ πΎπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅©σ΅© σ΅¨ σ΅© π+1 + σ΅¨σ΅¨σ΅¨ β σ΅¨ σ΅©π΅ π₯ σ΅©σ΅© σ΅¨σ΅¨ πΎπ+1 + πΏπ+1 πΎπ + πΏπ σ΅¨σ΅¨σ΅¨ σ΅© π π σ΅© +
σ΅© σ΅© σ΅© σ΅© lim sup (σ΅©σ΅©σ΅©π§π+1 β π§π σ΅©σ΅©σ΅© β σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅©) β€ 0.
πΏπ+1 σ΅© σ΅© σ΅© σ΅© (σ΅©σ΅©π₯ β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ+1 πΊπ₯π β ππ πΊπ₯π σ΅©σ΅©σ΅©) πΎπ+1 + πΏπ+1 σ΅© π+1
lim σ΅©π§ πββ σ΅© π
1 β ππ+1 β πΌπ+1 (1 β π) σ΅©σ΅© σ΅© = σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© 1 β ππ+1 σ΅¨σ΅¨ πΌ πΌπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅©σ΅© σ΅¨ σ΅© σ΅© σ΅© σ΅© σ΅© + σ΅¨σ΅¨σ΅¨ π+1 β σ΅¨ (σ΅©π (π₯π )σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯π σ΅©σ΅©σ΅©) σ΅¨σ΅¨ 1 β ππ+1 1 β ππ σ΅¨σ΅¨σ΅¨ σ΅©
σ΅© β π₯π σ΅©σ΅©σ΅© = 0.
(60)
It follows from (51) and (52) that σ΅©σ΅©
lim σ΅©π₯ π β β σ΅© π+1
σ΅© σ΅© σ΅© β π₯π σ΅©σ΅©σ΅© = lim (1 β ππ ) σ΅©σ΅©σ΅©π§π β π₯π σ΅©σ΅©σ΅© = 0. πββ
(61)
From (42), we have π₯π+1 β π₯π = πΌπ (π (π₯π ) β π₯π ) + (1 β πΌπ ) (π¦π β π₯π ) ,
1 β ππ+1 β πΌπ+1 1 β ππ+1
(59)
Consequently, by Lemma 20, we have σ΅©σ΅©
σ΅¨σ΅¨ πΏ πΏπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅©σ΅© σ΅¨ σ΅© π+1 + σ΅¨σ΅¨σ΅¨ β σ΅¨ σ΅©π πΊπ₯ σ΅©σ΅© ] σ΅¨σ΅¨ πΎπ+1 + πΏπ+1 πΎπ + πΏπ σ΅¨σ΅¨σ΅¨ σ΅© π π σ΅©
+
πββ
(62)
which hence implies that
π πΎπ+1 Γ[ π0 βπ π πΎπ+1 + πΏπ+1 π=0
σ΅© σ΅© σ΅© σ΅© π σ΅©σ΅©σ΅©π¦π β π₯π σ΅©σ΅©σ΅© = (1 β (1 β π)) σ΅©σ΅©σ΅©π¦π β π₯π σ΅©σ΅©σ΅© σ΅© σ΅© β€ (1 β πΌπ ) σ΅©σ΅©σ΅©π¦π β π₯π σ΅©σ΅©σ΅©
πΏπ+1 σ΅©σ΅© σ΅© + σ΅©π πΊπ₯ β ππ πΊπ₯π σ΅©σ΅©σ΅© πΎπ+1 + πΏπ+1 σ΅© π+1 π σ΅¨σ΅¨ πΏ πΏπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅©σ΅© σ΅¨ σ΅© σ΅© σ΅© π+1 + σ΅¨σ΅¨σ΅¨ β σ΅¨ (σ΅©π΅ π₯ σ΅©σ΅© + σ΅©σ΅©π πΊπ₯ σ΅©σ΅©) ] σ΅¨σ΅¨ πΎπ+1 + πΏπ+1 πΎπ + πΏπ σ΅¨σ΅¨σ΅¨ σ΅© π π σ΅© σ΅© π π σ΅© σ΅¨ πΌπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅© σ΅© σ΅¨σ΅¨ πΌ β€ σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + σ΅¨σ΅¨σ΅¨ π+1 β σ΅¨π σ΅¨σ΅¨ 1 β ππ+1 1 β ππ σ΅¨σ΅¨σ΅¨ π
σ΅© σ΅© + πβπ π + σ΅©σ΅©σ΅©ππ+1 πΊπ₯π β ππ πΊπ₯π σ΅©σ΅©σ΅©
σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π+1 β π₯π β πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅©
(63)
σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅© . Since π₯π+1 β π₯π β 0 and πΌπ (π(π₯π ) β π₯π ) β 0, we get σ΅©σ΅©
lim σ΅©π¦ πββ σ΅© π
σ΅© β π₯π σ΅©σ΅©σ΅© = 0.
(64)
Next, we show that βπ₯π β πΊπ₯π β β 0 as π β β. Indeed, for simplicity, put π = Ξ πΆ(π β π2 π΅2 π), π’π = Ξ πΆ(π₯π β π2 π΅2 π₯π ) and Vπ = Ξ πΆ(π’π β π1 π΅1 π’π ). Then, Vπ = πΊπ₯π for all π β₯ 0. From Lemma 22, we have
π=0
σ΅¨σ΅¨ πΏ πΏπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨ π+1 + σ΅¨σ΅¨σ΅¨ β σ΅¨π σ΅¨σ΅¨ πΎπ+1 + πΏπ+1 πΎπ + πΏπ σ΅¨σ΅¨σ΅¨ σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© σ΅¨σ΅¨ πΌ πΌπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨ + π ( σ΅¨σ΅¨σ΅¨ π+1 β σ΅¨σ΅¨ σ΅¨σ΅¨ 1 β ππ+1 1 β ππ σ΅¨σ΅¨
σ΅©2 σ΅© σ΅©2 σ΅©σ΅© σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© = σ΅©σ΅©σ΅©Ξ πΆ (π₯π β π2 π΅2 π₯π ) β Ξ πΆ (π β π2 π΅2 π)σ΅©σ΅©σ΅© σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©π₯π β π β π2 (π΅2 π₯π β π΅2 π)σ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β 2π2 (π½Μ2 β π
2 π2 ) σ΅©σ΅©σ΅©π΅2 π₯π β π΅2 πσ΅©σ΅©σ΅© , (65)
σ΅¨σ΅¨ πΏ πΏπ σ΅¨σ΅¨σ΅¨σ΅¨ π σ΅¨ π+1 + σ΅¨σ΅¨σ΅¨ β σ΅¨ + βπ ) σ΅¨σ΅¨ πΎπ+1 + πΏπ+1 πΎπ + πΏπ σ΅¨σ΅¨σ΅¨ π=0 π σ΅© σ΅© + σ΅©σ΅©σ΅©ππ+1 πΊπ₯π β ππ πΊπ₯π σ΅©σ΅©σ΅© , (57) which hence yields σ΅©σ΅©σ΅©π§π+1 β π§π σ΅©σ΅©σ΅© β σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅¨σ΅¨ πΌ πΌπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨σ΅¨ πΏπ+1 πΏπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨ β€ π ( σ΅¨σ΅¨σ΅¨ π+1 β β σ΅¨σ΅¨ + σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨ 1 β ππ+1 1 β ππ σ΅¨σ΅¨ σ΅¨σ΅¨ πΎπ+1 + πΏπ+1 πΎπ + πΏπ σ΅¨σ΅¨σ΅¨ π
σ΅© σ΅© +βπ π ) + σ΅©σ΅©σ΅©ππ+1 πΊπ₯π β ππ πΊπ₯π σ΅©σ΅©σ΅© , π=0
(58)
σ΅©σ΅© σ΅©2 σ΅© σ΅©2 σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© = σ΅©σ΅©σ΅©Ξ πΆ (π’π β π1 π΅1 π’π ) β Ξ πΆ (π β π1 π΅1 π)σ΅©σ΅©σ΅© σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©π’π β π β π1 (π΅1 π’π β π΅1 π)σ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© β 2π1 (π½Μ1 β π
2 π1 ) σ΅©σ΅©σ΅©π΅1 π’π β π΅1 πσ΅©σ΅©σ΅© . (66) Substituting (65) for (66), we obtain σ΅©σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅©2 2 σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β 2π2 (π½Μ2 β π
π2 ) σ΅©σ΅©σ΅©π΅2 π₯π β π΅2 πσ΅©σ΅©σ΅© σ΅© σ΅©2 β 2π1 (π½Μ1 β π
2 π1 ) σ΅©σ΅©σ΅©π΅1 π’π β π΅1 πσ΅©σ΅©σ΅© .
(67)
Abstract and Applied Analysis
11
From (42) and (67), we have
In the same way, we derive
σ΅©σ΅© σ΅©2 σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅©2 σ΅©2 σ΅©2 β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅©2 σ΅©2 σ΅©2 β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅©2 β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅©2 + πΏπ [σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β 2π2 (π½Μ2 β π
2 π2 ) σ΅©2 σ΅© Γ σ΅©σ΅©σ΅©π΅2 π₯π β π΅2 πσ΅©σ΅©σ΅©
σ΅©2 σ΅©σ΅© σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© σ΅© σ΅©2 = σ΅©σ΅©σ΅©Ξ πΆ (π’π β π1 π΅1 π’π ) β Ξ πΆ (π β π1 π΅1 π)σ΅©σ΅©σ΅© β€ β¨π’π β π1 π΅1 π’π β (π β π1 π΅1 π) , π½ (Vπ β π)β© = β¨π’π β π, π½ (Vπ β π)β© + π1 β¨π΅1 π β π΅1 π’π , π½ (Vπ β π)β© β€ (68)
σ΅© σ΅©2 β 2π1 (π½Μ1 β π
2 π1 ) σ΅©σ΅©σ΅©π΅1 π’π β π΅1 πσ΅©σ΅©σ΅© ]
1 σ΅©σ΅© σ΅©2 σ΅© σ΅©2 [σ΅©π’ β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© 2 σ΅© π σ΅© σ΅© β π2 (σ΅©σ΅©σ΅©π’π β Vπ + (π β π)σ΅©σ΅©σ΅©) ] σ΅© σ΅©σ΅© σ΅© + π1 σ΅©σ΅©σ΅©π΅1 π β π΅1 π’π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© ,
σ΅©2 σ΅© = σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© which implies that
σ΅© σ΅©2 β 2πΏπ [π2 (π½Μ2 β π
2 π2 ) σ΅©σ΅©σ΅©π΅2 π₯π β π΅2 πσ΅©σ΅©σ΅©
σ΅©σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅© σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© β π2 (σ΅©σ΅©σ΅©π’π β Vπ + (π β π)σ΅©σ΅©σ΅©) σ΅© σ΅©σ΅© σ΅© + 2π1 σ΅©σ΅©σ΅©π΅1 π β π΅1 π’π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© .
σ΅© σ΅©2 +2π1 (π½Μ1 β π
2 π1 ) σ΅©σ΅©σ΅©π΅1 π’π β π΅1 πσ΅©σ΅©σ΅© ] , which hence implies that
σ΅©2 σ΅© σ΅©2 σ΅© β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ (σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅©) σ΅©σ΅©σ΅©π₯π β π¦π σ΅©σ΅©σ΅© .
(69)
σ΅©σ΅©
σ΅© β π΅2 πσ΅©σ΅©σ΅© = 0,
σ΅©σ΅©
lim σ΅©π΅ π’ πββ σ΅© 1 π
σ΅© β π΅1 πσ΅©σ΅©σ΅© = 0. (70)
Utilizing Proposition 6 and Lemma 9, we have σ΅©σ΅© σ΅©2 σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© σ΅© σ΅©2 = σ΅©σ΅©σ΅©Ξ πΆ (π₯π β π2 π΅2 π₯π ) β Ξ πΆ (π β π2 π΅2 π)σ΅©σ΅©σ΅© β€ β¨π₯π β π2 π΅2 π₯π β (π β π2 π΅2 π) , π½ (π’π β π)β© = β¨π₯π β π, π½ (π’π β π)β© + π2 β¨π΅2 π β π΅2 π₯π , π½ (π’π β π)β© β€
1 σ΅©σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅© [σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© β π1 (σ΅©σ΅©σ΅©π₯π β π’π β (π β π)σ΅©σ΅©σ΅©)] 2 σ΅© σ΅©σ΅© σ΅© + π2 σ΅©σ΅©σ΅©π΅2 π β π΅2 π₯π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© , (71)
(75)
σ΅© σ΅© σ΅©σ΅© + 2π2 σ΅©σ΅©σ΅©π΅2 π β π΅2 π₯π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅©
σ΅© σ΅©σ΅© σ΅© + 2π1 σ΅©σ΅©σ΅©π΅1 π β π΅1 π’π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© .
By Lemma 8(i), we have from (68) and (75) σ΅©σ΅© σ΅©2 σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅©2 σ΅©2 σ΅©2 β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅©2 β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅© + πΏπ [σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β π1 (σ΅©σ΅©σ΅©π₯π β π’π β (π β π)σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅© σ΅© β π2 (σ΅©σ΅©σ΅©π’π β Vπ + (π β π)σ΅©σ΅©σ΅©) + 2π2 σ΅©σ΅©σ΅©π΅2 π β π΅2 π₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅©σ΅© σ΅© σ΅© Γ σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© + 2π1 σ΅©σ΅©σ΅©π΅1 π β π΅1 π’π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© ] σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
σ΅© σ΅© β πΏπ [π1 (σ΅©σ΅©σ΅©π₯π β π’π β (π β π)σ΅©σ΅©σ΅©)
σ΅© σ΅© +π2 (σ΅©σ΅©σ΅©π’π β Vπ + (π β π)σ΅©σ΅©σ΅©)]
which implies that σ΅©σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅© σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β π1 (σ΅©σ΅©σ΅©π₯π β π’π β (π β π)σ΅©σ΅©σ΅©) σ΅© σ΅©σ΅© σ΅© + 2π2 σ΅©σ΅©σ΅©π΅2 π β π΅2 π₯π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© .
σ΅©σ΅© σ΅©2 σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β π1 (σ΅©σ΅©σ΅©π₯π β π’π β (π β π)σ΅©σ΅©σ΅©) σ΅© σ΅© β π2 (σ΅©σ΅©σ΅©π’π β Vπ + (π β π)σ΅©σ΅©σ΅©)
Since 0 < ππ < π½Μπ /π
2 for π = 1, 2, and {π₯π }, {π¦π } are bounded, we obtain from (64), (69), and condition (ii) that lim σ΅©π΅ π₯ πββ σ΅© 2 π
(74)
Substituting (72) for (74), we get
σ΅© σ΅©2 2πΏπ [π2 (π½Μ2 β π
2 π2 ) σ΅©σ΅©σ΅©π΅2 π₯π β π΅2 πσ΅©σ΅©σ΅© σ΅© σ΅©2 +π1 (π½Μ1 β π
2 π1 ) σ΅©σ΅©σ΅©π΅1 π’π β π΅1 πσ΅©σ΅©σ΅© ]
(73)
σ΅© σ΅©σ΅© σ΅© + 2π2 σ΅©σ΅©σ΅©π΅2 π β π΅2 π₯π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© (72)
σ΅© σ΅©σ΅© σ΅© + 2π1 σ΅©σ΅©σ΅©π΅1 π β π΅1 π’π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© ,
(76)
12
Abstract and Applied Analysis σ΅© σ΅©2 = πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β πσ΅©σ΅©σ΅© + (π½π + πΎπ )
which hence leads to σ΅© σ΅© σ΅© σ΅© πΏπ [π1 (σ΅©σ΅©σ΅©π₯π β π’π β (π β π)σ΅©σ΅©σ΅©) + π2 (σ΅©σ΅©σ΅©π’π β Vπ + (π β π)σ΅©σ΅©σ΅©)] σ΅© σ΅© σ΅©σ΅© σ΅©2 σ΅© σ΅©2 σ΅© β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© + 2π2 σ΅©σ΅©σ΅©π΅2 π β π΅2 π₯π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© σ΅© σ΅©σ΅© σ΅© + 2π1 σ΅©σ΅©σ΅©π΅1 π β π΅1 π’π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅©
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ (σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅©) σ΅©σ΅©σ΅©π₯π β π¦π σ΅©σ΅©σ΅©
σ΅©σ΅© π½ σ΅©σ΅©2 πΎπ σ΅© σ΅© π Γ σ΅©σ΅©σ΅© (π₯π β π) + (π΅π π₯π β π)σ΅©σ΅©σ΅© σ΅©σ΅© π½π + πΎπ σ΅©σ΅© π½π + πΎπ σ΅© σ΅©2 β€ πΏπ σ΅©σ΅©σ΅©πΊπ₯π β πσ΅©σ΅©σ΅© + (π½π + πΎπ ) Γ[
σ΅© σ΅©σ΅© σ΅© + 2π2 σ΅©σ΅©σ΅©π΅2 π β π΅2 π₯π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅©
σ΅© σ΅©σ΅© σ΅© + 2π1 σ΅©σ΅©σ΅©π΅1 π β π΅1 π’π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© .
β (77)
From (70), (77), condition (ii), and the boundedness of {π₯π }, {π¦π }, {π’π }, and {Vπ }, we deduce that lim π πββ 1
σ΅© σ΅© (σ΅©σ΅©σ΅©π₯π β π’π β (π β π)σ΅©σ΅©σ΅©) = 0,
σ΅© σ΅© lim π (σ΅©σ΅©π’ β Vπ + (π β π)σ΅©σ΅©σ΅©) = 0. πββ 2 σ΅© π
(78)
σ΅© σ΅© lim σ΅©σ΅©π’ β Vπ + (π β π)σ΅©σ΅©σ΅© = 0. πββ σ΅© π
π½π πΎπ
2
(π½π + πΎπ )
σ΅© σ΅© π3 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©)]
σ΅© σ΅© σ΅© σ΅©2 σ΅©2 σ΅©2 β€ πΏπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β
π½π πΎπ σ΅© σ΅© π (σ΅©σ΅©π₯ β π΅π π₯π σ΅©σ΅©σ΅©) π½π + πΎπ 3 σ΅© π
π½πΎ σ΅© σ΅© σ΅©2 σ΅© = σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β π π π3 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©) , π½π + πΎπ (83) which immediately implies that
Utilizing the properties of π1 and π2 , we deduce that σ΅© σ΅© lim σ΅©σ΅©π₯ β π’π β (π β π)σ΅©σ΅©σ΅© = 0, πββ σ΅© π
π½π σ΅©σ΅© πΎπ σ΅©σ΅© σ΅©2 σ΅©2 σ΅©π₯ β πσ΅©σ΅©σ΅© + σ΅©π΅ π₯ β πσ΅©σ΅©σ΅© π½π + πΎπ σ΅© π π½π + πΎπ σ΅© π π
σ΅© σ΅© π½π πΎπ π3 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©) (79)
β€
σ΅© σ΅©2 σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ (σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅©) σ΅©σ΅©σ΅©π₯π β π¦π σ΅©σ΅©σ΅© .
From (79), we get σ΅©σ΅© σ΅© σ΅© σ΅© σ΅©σ΅©π₯π β Vπ σ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©π₯π β π’π β (π β π)σ΅©σ΅©σ΅© σ΅© σ΅© + σ΅©σ΅©σ΅©π’π β Vπ + (π β π)σ΅©σ΅©σ΅© σ³¨β 0 as π σ³¨β β. (80) That is,
π½π πΎπ σ΅© σ΅© π (σ΅©σ΅©π₯ β π΅π π₯π σ΅©σ΅©σ΅©) π½π + πΎπ 3 σ΅© π
(84)
So, from (64), the boundedness of {π₯π }, {π¦π }, and conditions (ii), (iv), it follows that lim π πββ 3
σ΅© σ΅© (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©) = 0.
(85)
From the properties of π3 , we have σ΅©σ΅©
lim σ΅©π₯ πββ σ΅© π
σ΅© β πΊπ₯π σ΅©σ΅©σ΅© = 0.
σ΅©σ΅©
σ΅© β π₯π σ΅©σ΅©σ΅© = 0,
σ΅© β π΅π π₯π σ΅©σ΅©σ΅© = 0.
(86)
Taking into account that
Next, let us show that lim σ΅©π΅ π₯ πββ σ΅© π π
σ΅©σ΅©
lim σ΅©π₯ πββ σ΅© π
(81)
σ΅© σ΅© lim σ΅©σ΅©ππ₯π β π₯π σ΅©σ΅©σ΅© = 0.
πββ σ΅©
(82)
π¦π β π₯π = πΎπ (π΅π π₯π β π₯π ) + πΏπ (ππ πΊπ₯π β π₯π ) , we have σ΅© σ΅© πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π₯π σ΅©σ΅©σ΅© σ΅© σ΅© = σ΅©σ΅©σ΅©π¦π β π₯π β πΎπ (π΅π π₯π β π₯π )σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π¦π β π₯π σ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β π₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π¦π β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π π₯π β π₯π σ΅©σ΅©σ΅© .
Indeed, utilizing Lemma 15 and (42), we have σ΅©σ΅© σ΅©2 σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© σ΅©σ΅© σ΅©σ΅©2 π½ π₯ + πΎπ π΅π π₯π σ΅© σ΅© = σ΅©σ΅©σ΅©πΏπ (ππ πΊπ₯π β π) + (π½π + πΎπ ) ( π π β π)σ΅©σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© π½π + πΎπ σ΅© σ΅©2 β€ πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β πσ΅©σ΅©σ΅© + (π½π + πΎπ ) σ΅©σ΅©σ΅©2 σ΅©σ΅©σ΅© π½ π₯ + πΎπ π΅π π₯π β πσ΅©σ΅©σ΅© Γ σ΅©σ΅©σ΅© π π π½π + πΎπ σ΅©σ΅© σ΅©σ΅©
(87)
(88)
From (64), (86), and condition (ii), it follows that σ΅©σ΅©
lim σ΅©π πΊπ₯π πββ σ΅© π
σ΅© β π₯π σ΅©σ΅©σ΅© = 0,
σ΅©σ΅©
lim σ΅©π πΊπ₯π πββ σ΅© π
σ΅© β π΅π π₯π σ΅©σ΅©σ΅© = 0. (89)
Abstract and Applied Analysis
13 σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π₯π σ΅©σ΅©σ΅© + πΌπ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β πΌπ )
Note that
σ΅© σ΅© Γ [π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅©
σ΅©σ΅© σ΅© σ΅©σ΅©π₯π β ππ₯π σ΅©σ΅©σ΅©
σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β ππ πΊπ₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯π β ππ π₯π σ΅©σ΅©σ΅© σ΅© σ΅© + σ΅©σ΅©σ΅©ππ π₯π β ππ₯π σ΅©σ΅©σ΅©
(90)
σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β ππ πΊπ₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©πΊπ₯π β π₯π σ΅©σ΅©σ΅© σ΅© σ΅© + σ΅©σ΅©σ΅©ππ π₯π β ππ₯π σ΅©σ΅©σ΅© .
So, in terms of (81), (89), and Lemma 12, we have σ΅©σ΅©
lim σ΅©π₯ πββ σ΅© π
σ΅© β ππ₯π σ΅©σ΅©σ΅© = 0.
(91)
Suppose that π½π β‘ π½ for some fixed π½ β (0, 1) such that π½ + πΎπ + πΏπ = 1 for all π β₯ 0. Define a mapping ππ₯ = (1 β π1 β π2 )ππ₯ + π1 π΅π₯ + π2 πΊπ₯, where π1 , π2 β (0, 1) are two constants with π1 + π2 < 1. Then, by Lemmas 14 and 17, we have that Fix(π) = Fix(π) β© Fix(π΅) β© Fix(πΊ) = πΉ. For each π β₯ 1, let {ππ } be a unique element of πΆ such that ππ =
1 1 π (ππ ) + (1 β ) πππ . π π
(92)
From Lemma 13, we conclude that ππ β π β Fix(π) = πΉ as π β β. Observe that for every π, π σ΅© σ΅©σ΅© σ΅©σ΅©π¦π β π΅ππ σ΅©σ΅©σ΅©
σ΅© σ΅© = σ΅©σ΅©σ΅©π½ (π₯π β π΅ππ ) + πΎπ (π΅π π₯π β π΅ππ ) + πΏπ (ππ πΊπ₯π β π΅ππ )σ΅©σ΅©σ΅©
σ΅© σ΅© σ΅© σ΅© β€ π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β π΅ππ σ΅©σ΅©σ΅©
σ΅© σ΅© σ΅© σ΅© + πΏπ (σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π π₯π β π΅ππ σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅© σ΅© = π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β π½) σ΅©σ΅©σ΅©π΅π π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅© ,
σ΅© σ΅© σ΅© σ΅© + (1 β π½) (σ΅©σ΅©σ΅©π΅π π₯π β π΅π ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π ππ β π΅ππ σ΅©σ΅©σ΅©) σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©] σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π₯π σ΅©σ΅©σ΅© + πΌπ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© + (1 β πΌπ ) [π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© + (1 β π½) (σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π ππ β π΅ππ σ΅©σ΅©σ΅©) σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©] σ΅© σ΅© = πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π₯π σ΅©σ΅©σ΅© + (π½ + πΌπ (1 β π½)) σ΅© σ΅© σ΅© σ΅© Γ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β πΌπ ) (1 β π½) σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© σ΅© σ΅© + (1 β πΌπ ) [(1 β π½) σ΅©σ΅©σ΅©π΅π ππ β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©] σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π₯π σ΅©σ΅©σ΅© + (π½ + πΌπ (1 β π½)) σ΅© σ΅© σ΅© σ΅© Γ (σ΅©σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π+1 β π΅ππ σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅© σ΅© + (1 β πΌπ ) (1 β π½) (σ΅©σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π+1 β ππ σ΅©σ΅©σ΅©) σ΅© σ΅© + (1 β πΌπ ) [(1 β π½) σ΅©σ΅©σ΅©π΅π ππ β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅© ] σ΅© σ΅© = πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π₯π σ΅©σ΅©σ΅© + (π½ + πΌπ (1 β π½)) σ΅© σ΅© σ΅© σ΅© Γ σ΅©σ΅©σ΅©π₯π+1 β π΅ππ σ΅©σ΅©σ΅© + (1 β πΌπ ) (1 β π½) σ΅©σ΅©σ΅©π₯π+1 β ππ σ΅©σ΅©σ΅© σ΅© σ΅© + (1 β πΌπ ) [(1 β π½) σ΅©σ΅©σ΅©π΅π ππ β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©] + σ΅©σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅© . (94) So, it immediately follows from 0 β€ πΌπ β€ 1 β π, for all π β₯ π0 , that
(93)
and hence σ΅©σ΅© σ΅© σ΅©σ΅©π₯π+1 β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π΅ππ σ΅©σ΅©σ΅© + (1 β πΌπ ) σ΅©σ΅©σ΅©π¦π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ (σ΅©σ΅©σ΅©π (π₯π ) β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅©) σ΅© σ΅© + (1 β πΌπ ) σ΅©σ΅©σ΅©π¦π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π₯π σ΅©σ΅©σ΅© + πΌπ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β πΌπ ) σ΅© σ΅© Γ [π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© + (1 β π½) σ΅©σ΅©σ΅©π΅π π₯π β π΅ππ σ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©]
σ΅©σ΅© σ΅© σ΅©σ΅©π₯π+1 β π΅ππ σ΅©σ΅©σ΅© 1 (1 β πΌπ ) (1 β π½) σ΅© σ΅© σ΅© σ΅© Γ (σ΅©σ΅©σ΅©πΌπ (π₯π β π (π₯π ))σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅©)
σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π ππ β π΅ππ σ΅©σ΅©σ΅© +
πΏπ σ΅©σ΅© σ΅© σ΅©π πΊπ₯ β π΅π π₯π σ΅©σ΅©σ΅© 1βπ½σ΅© π π σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π ππ β π΅ππ σ΅©σ΅©σ΅© +
1 π (1 β π½) σ΅© σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© Γ (σ΅©σ΅©πΌπ (π₯π β π (π₯π ))σ΅©σ΅© + σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅©) σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π+1 β ππ σ΅©σ΅©σ΅© + ππ , βπ β₯ π0 , σ΅© σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅© +
(95)
14
Abstract and Applied Analysis
where ππ = βπ΅π ππ β π΅ππ β + βππ πΊπ₯π β π΅π π₯π β + (1/π(1 β π½))(βπΌπ (π₯π β π(π₯π ))β + βπ₯π β π₯π+1 β). Since limπ β β βπ΅π ππ β π΅ππ β = limπ β β βππ πΊπ₯π βπ΅π π₯π β = limπ β β βπΌπ (π₯π βπ(π₯π ))β = limπ β β βπ₯π β π₯π+1 β = 0, we know that ππ β 0 as π β β. From (95), we obtain
1 1 (1 β ) (π₯π β πππ ) = π₯π β ππ β (π₯π β π (ππ )) . π π
(102)
It follows from Lemma 8 (ii) and (102) that
σ΅©2 σ΅© σ΅©2 σ΅©σ΅© σ΅©σ΅©π₯π+1 β π΅ππ σ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β ππ σ΅©σ΅©σ΅©
σ΅© σ΅© + ππ (2 σ΅©σ΅©σ΅©π₯π+1 β ππ σ΅©σ΅©σ΅© + ππ ) ,
that is,
βπ β₯ π0 .
(96)
For any Banach limit π, from (96), we derive σ΅© σ΅©2 σ΅© σ΅©2 ππ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© = ππ σ΅©σ΅©σ΅©π₯π+1 β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅©2 β€ ππ σ΅©σ΅©σ΅©π₯π+1 β ππ σ΅©σ΅©σ΅© = ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© .
1 2σ΅© σ΅©2 (1 β ) σ΅©σ΅©σ΅©π₯π β πππ σ΅©σ΅©σ΅© π σ΅©2 2 σ΅© β₯ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© β β¨π₯π β ππ + ππ β π (ππ ) , π½ (π₯π β ππ )β© π
(97)
2 σ΅© σ΅©2 2 = (1 β ) σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + β¨π (ππ ) β ππ , π½ (π₯π β ππ )β© . π π (103)
In addition, note that σ΅©σ΅© σ΅©2 σ΅©σ΅©π₯π β πΊππ σ΅©σ΅©σ΅©
So by (100) and (103), we have
σ΅©2 σ΅© β€ σ΅©σ΅©σ΅©π₯π β πΊπ₯π + πΊπ₯π β πΊππ σ΅©σ΅©σ΅©
1 2 σ΅© σ΅©2 (1 β ) ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© π
σ΅© σ΅© σ΅©2 σ΅© β€ (σ΅©σ΅©σ΅©π₯π β πΊπ₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅©) σ΅©2 σ΅© σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πΊπ₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© Γ (2 σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πΊπ₯π σ΅©σ΅©σ΅©) , σ΅©σ΅© σ΅©2 σ΅©σ΅©π₯π β πππ σ΅©σ΅©σ΅© σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©π₯π β ππ₯π + ππ₯π β πππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅© β€ (σ΅©σ΅©σ΅©π₯π β ππ₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅©) σ΅©2 σ΅© σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β ππ₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© Γ (2 σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β ππ₯π σ΅©σ΅©σ΅©) .
2 σ΅© σ΅©2 β₯ (1 β ) ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© π
σ΅© σ΅©2 σ΅© σ΅©2 ππ σ΅©σ΅©σ΅©π₯π β πππ σ΅©σ΅©σ΅© β€ ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© .
and hence 1 σ΅©σ΅© σ΅©2 2 π σ΅©π₯ β ππ σ΅©σ΅©σ΅© β₯ ππ β¨π (ππ ) β ππ , π½ (π₯π β ππ )β© . π2 π σ΅© π π
1 σ΅©σ΅© σ΅©2 π σ΅©π₯ β ππ σ΅©σ΅©σ΅© β₯ ππ β¨π (ππ ) β ππ , π½ (π₯π β ππ )β© . 2π π σ΅© π
(99) ππ β¨π (π) β π, π½ (π₯π β π)β© β€ 0.
σ΅¨ σ΅¨ lim σ΅¨σ΅¨β¨π (π) β π, π½ (π₯π+1 β π)β© β β¨π (π) β π, π½ (π₯π β π)β©σ΅¨σ΅¨σ΅¨ = 0.
πββ σ΅¨
(108) (100)
So, utilizing Lemma 18 we deduce from (107) and (108) that lim sup β¨π (π) β π, π½ (π₯π β π)β© β€ 0, πββ
σ΅© σ΅©2 β€ ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© .
(109)
which together with (49) and the norm-to-norm uniform continuity of π½ on bounded subsets of π, implies that
Also, observe that 1 1 (π₯π β π (ππ )) + (1 β ) (π₯π β πππ ) ; π π
(107)
On the other hand, from (49) and the norm-to-norm uniform continuity of π½ on bounded subsets of π, it follows that
σ΅© σ΅©2 σ΅© σ΅©2 + π1 ππ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + π2 ππ σ΅©σ΅©σ΅©π₯π β πΊππ σ΅©σ΅©σ΅©
π₯π β ππ =
(106)
Since ππ β π β Fix(π) = πΉ as π β β, by the uniform Frechet differentiability of the norm of π we have
σ΅© σ΅©2 ππ σ΅©σ΅©σ΅©π₯π β πππ σ΅©σ΅©σ΅© σ΅© = ππ σ΅©σ΅©σ΅©(1 β π1 β π2 ) (π₯π β πππ ) σ΅© σ΅©2 β€ (1 β π1 β π2 ) ππ σ΅©σ΅©σ΅©π₯π β πππ σ΅©σ΅©σ΅©
(105)
This implies that
Utilizing (97) and (99), we deduce that
σ΅©2 +π1 (π₯π β π΅ππ ) + π2 (π₯π β πΊππ )σ΅©σ΅©σ΅©
2 + ππ β¨π (ππ ) β ππ , π½ (π₯π β ππ )β© , π
(98)
It is easy to see from (81) and (91) that σ΅© σ΅©2 σ΅© σ΅©2 ππ σ΅©σ΅©σ΅©π₯π β πΊππ σ΅©σ΅©σ΅© β€ ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© ,
(104)
(101)
lim sup β¨π (π) β π, π½ (π₯π+1 β π)β© β€ 0. πββ
(110)
Abstract and Applied Analysis
15
Finally, let us show that π₯π β π as π β β. Utilizing Lemma 8 (i), from (42) and the convexity of β β
β2 , we get σ΅©σ΅© σ΅©2 σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅©2 β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅©2 + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β πσ΅©σ΅©σ΅© (111) σ΅© σ΅© σ΅©2 σ΅©2 β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅©2 + πΏπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅©2 = σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© , σ΅©σ΅© σ΅©2 σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© = σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π (π)) + (1 β πΌπ ) (π¦π β π) σ΅©2 + πΌπ (π (π) β π)σ΅©σ΅©σ΅© σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π (π)) + (1 β πΌπ ) (π¦π β π)σ΅©σ΅©σ΅© + 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β© σ΅© σ΅©2 σ΅© σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π (π)σ΅©σ΅©σ΅© + (1 β πΌπ ) σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© + 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β© σ΅© σ΅©2 σ΅© σ΅©2 β€ πΌπ πσ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + (1 β πΌπ ) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
(112)
+ 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β© σ΅© σ΅©2 = (1 β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β© σ΅© σ΅©2 = (1 β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© 2 β¨π (π) β π, π½ (π₯π+1 β π)β© + πΌπ (1 β π) . 1βπ Applying Lemma 7 to (112), we obtain that π₯π β π as π β β. Conversely, if π₯π β π β πΉ as π β β, then from (42) it follows that σ΅©σ΅© σ΅©σ΅© σ΅©σ΅©π¦π β πσ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β πσ΅©σ΅©σ΅© (113) σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© β€ π½π σ΅©σ΅©π₯π β πσ΅©σ΅© + πΎπ σ΅©σ΅©π₯π β πσ΅©σ΅© + πΏπ σ΅©σ΅©π₯π β πσ΅©σ΅© σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ³¨β 0 as π σ³¨β β, that is, π¦π β π. Again from (42) we obtain that σ΅©σ΅© σ΅©σ΅© σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅© σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π+1 β π₯π β (1 β πΌπ ) (π¦π β π₯π )σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + (1 β πΌπ ) σ΅©σ΅©σ΅©π¦π β π₯π σ΅©σ΅©σ΅©
σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
σ΅© σ΅© σ΅© σ΅© + (1 β πΌπ ) (σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©)
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© + 2 σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© . (114) Since π₯π β π and π¦π β π, we get πΌπ (π(π₯π ) β π₯π ) β 0. This completes the proof. Corollary 25. Let πΆ be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space π. Let Ξ πΆ be a sunny nonexpansive retraction from π onto πΆ. Let {ππ }β π=0 be a sequence of positive numbers in (0, π] for some π β (0, 1) and π΄ π : πΆ β πΈ an πΌΜπ -inverse strongly accretive mapping for each π = 0, 1, . . .. Define a mapping πΊπ : πΆ β πΆ by Ξ πΆ(πΌ β π π π΄ π )π₯ = πΊπ π₯ for all π₯ β πΆ and π = 0, 1, . . ., where π π β (0, πΌΜπ /π
2 ] and π
is the 2-uniformly smooth constant of π. Let π΅π : πΆ β πΆ be the π-mapping generated by πΊπ , πΊπβ1 , . . . , πΊ0 and ππ , ππβ1 , . . . , π0 . Let π : πΆ β πΆ be an πΌ-strictly pseudocontractive mapping. Let π : πΆ β πΆ be a contraction with coefficient π β (0, 1). Let {ππ }β π=0 be a countable family of nonexpansive mappings of πΆ into itself such β that πΉ = (ββ π=0 Fix(ππ )) β© Fix(π) β© (βπ=0 VI(πΆ, π΄ π )) =ΜΈ 0. For arbitrarily given π₯0 β πΆ, let {π₯π } be the sequence generated by π¦π = π½π π₯π + πΎπ π΅π π₯π + πΏπ ππ ((1 β π) πΌ + ππ) π₯π , π₯π+1 = πΌπ π (π₯π ) + (1 β πΌπ ) π¦π ,
βπ β₯ 0,
(115)
where 0 < π < πΌ/π
2 , {πΌπ }, {π½π }, {πΎπ }, and {πΏπ } are the sequences in [0, 1] such that π½π + πΎπ + πΏπ = 1 for all π β₯ 0. Suppose that the following conditions hold: (i) ββ π=0 πΌπ = β and 0 β€ πΌπ β€ 1 β π, for all π β₯ π0 for some integer π0 β₯ 0; (ii) lim inf π β β πΎπ > 0 and lim inf π β β πΏπ > 0; (iii) limπ β β (|πΌπ+1 /(1 β (1 β πΌπ+1 )π½π+1 ) β πΌπ /(1 β (1 β πΌπ )π½π )| + |πΏπ+1 /(1 β π½π+1 ) β πΏπ /(1 β π½π )|) = 0; (iv) 0 < lim inf π β β π½π β€ lim supπ β β π½π < 1. Assume that ββ π=0 supπ₯βπ· βππ+1 π₯ β ππ π₯β < β for any bounded subset π· of πΆ and let π be a mapping of πΆ into itself defined by ππ₯ = limπ β β ππ π₯ for all π₯ β πΆ and suppose that Fix(π) = ββ π=0 Fix(ππ ). Then, there hold the following: (I) limπ β β βπ₯π+1 β π₯π β = 0; (II) π₯π β π β πΌπ (π(π₯π ) β π₯π ) β 0 provided π½π β‘ π½ for some fixed π½ β (0, 1), where π β πΉ solves the following VIP β¨π β π (π) , π½ (π β π)β© β€ 0,
βπ β πΉ.
(116)
Proof. In Theorem 24, we put π΅1 = πΌ β π, π΅2 = 0, and π1 = π, where 0 < π < πΌ/π
2 . Then, GSVI (13) is equivalent to the VIP of finding π₯β β πΆ such that β¨π΅1 π₯β , π½ (π₯ β π₯β )β© β₯ 0,
βπ₯ β πΆ.
(117)
16
Abstract and Applied Analysis
In this case, π΅1 : πΆ β π is πΌ-inverse strongly accretive. It is not hard to see that Fix(π) = VI(πΆ, π΅1 ). As a matter of fact, we have, for π > 0, π’ β VI (πΆ, π΅1 )
σ΅© σ΅©σ΅© σ΅© 1 σ΅©σ΅© σ΅©π₯ β π¦σ΅©σ΅©σ΅© . σ΅©σ΅©(πΌ β π΅π ) π₯ β (πΌ β π΅π ) π¦σ΅©σ΅©σ΅© β€ ππ σ΅©
βπ¦ β πΆ
β€ (1 +
ββ π’ = Ξ πΆ (π’ β ππ΅1 π’) ββ π’ = Ξ πΆ (π’ β ππ’ + πππ’)
(118)
ββ β¨π’ β ππ’ + πππ’ β π’, π½ (π’ β π¦)β© β₯ 0
βπ¦ β πΆ
(123)
1 σ΅©σ΅© σ΅© ) σ΅©π₯ β π¦σ΅©σ΅©σ΅© . ππ σ΅©
Utilizing the πΌπ -strong accretivity and π π -strict pseudocontractivity of π΅π , we get σ΅© σ΅©2 π π σ΅©σ΅©σ΅©(πΌ β π΅π ) π₯ β (πΌ β π΅π ) π¦σ΅©σ΅©σ΅©
ββ β¨π’ β ππ’, π½ (π’ β π¦)β© β€ 0 βπ¦ β πΆ
σ΅©2 σ΅© β€ σ΅©σ΅©σ΅©π₯ β π¦σ΅©σ΅©σ΅© β β¨π΅π π₯ β π΅π π¦, π½ (π₯ β π¦)β©
ββ π’ = ππ’
(124)
σ΅©2 σ΅© β€ (1 β πΌπ ) σ΅©σ΅©σ΅©π₯ β π¦σ΅©σ΅©σ΅© .
ββ π’ β Fix (π) . Accordingly, we know that πΉ = (ββ π=0 Fix(ππ )) β© Ξ© β© β VI(πΆ, π΄ )) = (β Fix(π )) β© Fix(π) β© (ββ (ββ π π π=0 π=0 π=0 VI(πΆ, π΄ π )), and Ξ πΆ (πΌ β π1 π΅1 ) Ξ πΆ (πΌ β π2 π΅2 ) π₯π = Ξ πΆ (πΌ β π1 π΅1 ) π₯π = Ξ πΆ ((1 β π) π₯π + πππ₯π )
So, we have 1 β πΌπ σ΅©σ΅© σ΅© σ΅©σ΅© σ΅© σ΅©σ΅©(πΌ β π΅π ) π₯ β (πΌ β π΅π ) π¦σ΅©σ΅©σ΅© β€ β σ΅©π₯ β π¦σ΅©σ΅©σ΅© . π σ΅©
(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 πΆ be a nonempty closed convex subset of a smooth Banach space π and let the mapping π΅π : πΆ β π be π π -strictly pseudocontractive and πΌπ -strongly accretive with πΌπ + π π β₯ 1 for π = 1, 2. Then, for ππ β (0, 1] one has σ΅© σ΅©σ΅© σ΅©σ΅©(πΌ β ππ π΅π ) π₯ β (πΌ β ππ π΅π ) π¦σ΅©σ΅©σ΅©
σ΅© σ΅©σ΅© σ΅©σ΅©(πΌ β ππ π΅π ) π₯ β (πΌ β ππ π΅π ) π¦σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©(πΌ β π΅π ) π₯ β (πΌ β π΅π ) π¦σ΅©σ΅©σ΅© + (1 β ππ ) σ΅©σ΅©σ΅©π΅π π₯ β π΅π π¦σ΅©σ΅©σ΅© β€β
1 β πΌπ σ΅©σ΅© 1 σ΅© σ΅© σ΅© σ΅©π₯ β π¦σ΅©σ΅©σ΅© + (1 β ππ ) (1 + ) σ΅©σ΅©σ΅©π₯ β π¦σ΅©σ΅©σ΅© ππ σ΅© ππ
= {β
1 β πΌπ 1 σ΅© σ΅© + (1 β ππ ) (1 + )} σ΅©σ΅©σ΅©π₯ β π¦σ΅©σ΅©σ΅© . ππ ππ (126)
Since 1 β (π π /(1 + π π ))(1 β β(1 β πΌπ )/π π ) β€ ππ β€ 1, it follows immediately that β
1 β πΌπ 1 σ΅© σ΅© + (1 β ππ ) (1 + )} σ΅©σ΅©σ΅©π₯ β π¦σ΅©σ΅©σ΅© , ππ ππ
(120)
βπ₯, π¦ β πΆ, for π = 1, 2. In particular, if 1β(π π /(1+π π ))(1ββ(1 β πΌπ )/π π ) β€ ππ β€ 1, then πΌ β ππ π΅π is nonexpansive for π = 1, 2. Proof. Taking into account the π π -strict pseudocontractivity of π΅π , we derive for every π₯, π¦ β πΆ
σ΅© σ΅©σ΅© σ΅© β€ σ΅©σ΅©σ΅©(πΌ β π΅π ) π₯ β (πΌ β π΅π ) π¦σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯ β π¦σ΅©σ΅©σ΅© ,
1 β πΌπ 1 + (1 β ππ ) (1 + ) β€ 1. ππ ππ
(127)
This implies that πΌ β ππ π΅π is nonexpansive for π = 1, 2. Lemma 27. Let πΆ be a nonempty closed convex subset of a smooth Banach space π. Let Ξ πΆ be a sunny nonexpansive retraction from π onto πΆ and let the mapping π΅π : πΆ β π be π π -strictly pseudocontractive and πΌπ -strongly accretive with πΌπ + π π β₯ 1 for π = 1, 2. Let πΊ : πΆ β πΆ be the mapping defined by πΊ (π₯) = Ξ πΆ [Ξ πΆ (π₯ β π2 π΅2 π₯) β π1 π΅1 Ξ πΆ (π₯ β π2 π΅2 π₯)] , βπ₯ β πΆ. (128)
σ΅© σ΅©2 π π σ΅©σ΅©σ΅©(πΌ β π΅π ) π₯ β (πΌ β π΅π ) π¦σ΅©σ΅©σ΅© β€ β¨(πΌ β π΅π ) π₯ β (πΌ β π΅π ) π¦, π½ (π₯ β π¦)β©
(125)
π
Therefore, for ππ β (0, 1] we have
= ((1 β π) πΌ + ππ) π₯π .
β€ {β
(122)
Hence, σ΅© σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅©σ΅©π΅π π₯ β π΅π π¦σ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©(πΌ β π΅π ) π₯ β (πΌ β π΅π ) π¦σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯ β π¦σ΅©σ΅©σ΅©
ββ β¨π΅1 π’, π½ (π¦ β π’)β© β₯ 0 βπ¦ β πΆ ββ β¨π’ β ππ΅1 π’ β π’, π½ (π’ β π¦)β© β₯ 0
which implies that
(121)
If 1β(π π /(1+π π ))(1ββ(1 β πΌπ )/π π ) β€ ππ β€ 1, then πΊ : πΆ β πΆ is nonexpansive.
Abstract and Applied Analysis
17
Proof. According to Lemma 26, we know that πΌ β ππ π΅π is nonexpansive for π = 1, 2. Hence, for all π₯, π¦ β πΆ, we have σ΅© σ΅©σ΅© σ΅©σ΅©πΊ (π₯) β πΊ (π¦)σ΅©σ΅©σ΅© σ΅© = σ΅©σ΅©σ΅©Ξ πΆ [Ξ πΆ (π₯ β π2 π΅2 π₯) β π1 π΅1 Ξ πΆ (π₯ β π2 π΅2 π₯)]
(iv) 0 < lim inf π β β π½π β€ lim supπ β β π½π < 1.
σ΅© β Ξ πΆ [Ξ πΆ (π¦ β π2 π΅2 π¦) β π1 π΅1 Ξ πΆ (π¦ β π2 π΅2 π¦)]σ΅©σ΅©σ΅©
σ΅© = σ΅©σ΅©σ΅©Ξ πΆ (πΌ β π1 π΅1 ) Ξ πΆ (πΌ β π2 π΅2 ) π₯
(II) π₯π β π β πΌπ (π(π₯π ) β π₯π ) β 0 provided π½π β‘ π½ for some fixed π½ β (0, 1), where π β πΉ solves the following VIP
σ΅© β€ σ΅©σ΅©σ΅©(πΌ β π1 π΅1 ) Ξ πΆ (πΌ β π2 π΅2 ) π₯
σ΅© β (πΌ β π1 π΅1 ) Ξ πΆ (πΌ β π2 π΅2 ) π¦σ΅©σ΅©σ΅©
β¨π β π (π) , π½ (π β π)β© β€ 0,
σ΅© σ΅© β€ σ΅©σ΅©σ΅©Ξ πΆ (πΌ β π2 π΅2 ) π₯ β Ξ πΆ (πΌ β π2 π΅2 ) π¦σ΅©σ΅©σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©(πΌ β π2 π΅2 ) π₯ β (πΌ β π2 π΅2 ) π¦σ΅©σ΅©σ΅©
(129)
This shows that πΊ : πΆ β πΆ is nonexpansive. This completes the proof. Theorem 28. Let πΆ be a nonempty closed convex subset of a uniformly convex Banach space π which has a uniformly Gateaux differentiable norm. Let Ξ πΆ be a sunny nonexpansive retraction from π onto πΆ. Let {ππ }β π=0 be a sequence of positive numbers in (0, π] for some π β (0, 1) and π΄ π : πΆ β π be ππ -strictly pseudocontractive and πΌΜπ -strongly accretive with ππ + πΌΜπ β₯ 1 for each π = 0, 1, . . .. Define a mapping πΊπ : πΆ β πΆ by Ξ πΆ(πΌ β π π π΄ π )π₯ = πΊπ π₯ for all π₯ β πΆ and π = 0, 1, . . ., where 1 β (ππ /(1 + ππ ))(1 β β(1 β πΌΜπ )/ππ ) β€ π π β€ 1 for all π = 0, 1, . . .. Let π΅π : πΆ β πΆ be the π-mapping generated by πΊπ , πΊπβ1 , . . . , πΊ0 and ππ , ππβ1 , . . . , π0 . Let the mapping π΅π : πΆ β π ππ -strictly pseudocontractive and π½Μπ -strongly accretive with ππ + π½Μπ β₯ 1 for π = 1, 2. Let π : πΆ β πΆ be a contraction with coefficient π β (0, 1). Let {ππ }β π=0 be a countable family of nonexpansive mappings of πΆ into itself such that πΉ = β (ββ π=0 Fix(ππ )) β© Ξ© β© (βπ=0 VI(πΆ, π΄ π )) =ΜΈ 0, where Ξ© is the fixed point set of the mapping πΊ = Ξ πΆ(πΌ β π1 π΅1 )Ξ πΆ(πΌ β π2 π΅2 ) with 1 β (π /(1 + π ))(1 β β(1 β π½Μ )/π ) β€ π β€ 1 for π = 1, 2. For π
π
π
arbitrarily given π₯0 β πΆ, let {π₯π } be the sequence generated by π¦π = π½π π₯π + πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π , π₯π+1 = πΌπ π (π₯π ) + ππ πΊπ₯π + (1 β πΌπ β ππ ) π¦π ,
βπ β₯ 0, (130)
where {ππ }, {πΌπ }, {π½π }, {πΎπ }, and {πΏπ } are the sequences in [0, 1] such that π½π + πΎπ + πΏπ = 1 and πΌπ + ππ β€ 1 for all π β₯ 0. Suppose that the following conditions hold: (i) ββ π=0 πΌπ = β and 0 β€ πΌπ + ππ β€ 1 β π, for all π β₯ π0 for some integer π0 β₯ 0; (ii) lim inf π β β ππ > lim inf π β β πΏπ > 0;
0, lim inf π β β πΎπ
βπ β πΉ.
(131)
Proof. First of all, take a fixed π β πΉ arbitrarily. Then we obtain π = πΊπ, π = π΅π π and ππ π = π for all π β₯ 0. By Lemma 27, we get from (130)
σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯ β π¦σ΅©σ΅©σ΅© .
π
Assume that ββ π=0 supπ₯βπ· βππ+1 π₯ β ππ π₯β < β for any bounded subset π· of πΆ and let π be a mapping of πΆ into itself defined by ππ₯ = limπ β β ππ π₯ for all π₯ β πΆ and suppose that Fix(π) = ββ π=0 Fix(ππ ). Then there hold the following: (I) limπ β β βπ₯π+1 β π₯π β = 0;
σ΅© β Ξ πΆ (πΌ β π1 π΅1 ) Ξ πΆ (πΌ β π2 π΅2 ) π¦σ΅©σ΅©σ΅©
π
(iii) limπ β β (|πΌπ+1 /(1β(1βπΌπ+1 βππ+1 )π½π+1 )β(πΌπ /(1β(1β πΌπ βππ )π½π ))|+|ππ+1 /(1β(1βπΌπ+1 βππ+1 )π½π+1 )β(ππ /(1β (1 β πΌπ β ππ )π½π ))| + |πΏπ+1 /(1 β π½π+1 ) β πΏπ /(1 β π½π )|) = 0;
>
0 and
σ΅© σ΅©σ΅© σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅©
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
(132)
σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© ,
and hence σ΅©σ΅© σ΅© σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + ππ σ΅©σ΅©σ΅©πΊπ₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© + (1 β πΌπ β ππ ) σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ (σ΅©σ΅©σ΅©π (π₯π ) β π (π)σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅©) σ΅© σ΅© σ΅© σ΅© + ππ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + (1 β πΌπ β ππ ) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ (π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅©) σ΅© σ΅© σ΅© σ΅© + ππ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + (1 β πΌπ β ππ ) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© = (1 β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅© σ΅©σ΅© σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅© σ΅© = (1 β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΌπ (1 β π) σ΅© 1βπ σ΅© σ΅© σ΅© σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅© β€ max {σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© , σ΅© }. 1βπ (133) By induction, we have σ΅© σ΅© σ΅© σ΅©σ΅© σ΅© σ΅© σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© }, σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β€ max {σ΅©σ΅©σ΅©π₯0 β πσ΅©σ΅©σ΅© , σ΅© 1βπ
βπ β₯ 0, (134)
which implies that {π₯π } is bounded and so are the sequences {π¦π }, {πΊπ₯π }, and {π(π₯π )}.
18
Abstract and Applied Analysis Let us show that σ΅©σ΅©
lim σ΅©π₯ π β β σ΅© π+1
Γ[
σ΅© β π₯π σ΅©σ΅©σ΅© = 0.
(135)
β
As a matter of fact, put ππ = (1 β πΌπ β ππ )π½π , for all π β₯ 0. Then, it follows from (i) and (iv) that π½π β₯ ππ = (1 β πΌπ β ππ ) π½π β₯ (1 β (1 β π)) π½π = ππ½π , βπ β₯ π0 ,
+[ (136)
πββ
πββ
Γ
(137)
Define
= π₯π+1 = ππ π₯π + (1 β ππ ) π§π .
(138)
π§π+1 β π§π =
π₯π+2 β ππ+1 π₯π+1 π₯π+1 β ππ π₯π β 1 β ππ+1 1 β ππ
Γ (1 β ππ+1 ) β
πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π 1 β π½π
ππ+1 (πΊπ₯π+1 β πΊπ₯π ) 1 β ππ+1
+(
πΌπ+1 πΌπ β ) π (π₯π ) 1 β ππ+1 1 β ππ
+(
ππ+1 ππ β ) πΊπ₯π 1 β ππ+1 1 β ππ
= (πΌπ+1 π (π₯π+1 ) + ππ+1 πΊπ₯π+1 +
+ (1 β πΌπ+1 β ππ+1 ) π¦π+1 β ππ+1 π₯π+1 ) β1
(1 β πΌπ+1 β ππ+1 ) (1 β π½π+1 ) 1 β ππ+1
Γ[
πΌπ π (π₯π ) + ππ πΊπ₯π + (1 β πΌπ β ππ ) π¦π β ππ π₯π 1 β ππ
(1 β πΌπ β ππ ) (1 β π½π ) ] 1 β ππ
πΌπ+1 (π (π₯π+1 ) β π (π₯π )) 1 β ππ+1 +
Observe that
πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π ] 1 β π½π
(1 β πΌπ+1 β ππ+1 ) (1 β π½π+1 ) 1 β ππ+1 β
and hence 0 < lim inf ππ β€ lim sup ππ < 1.
πΎπ+1 π΅π+1 π₯π+1 + πΏπ+1 ππ+1 πΊπ₯π+1 1 β π½π+1
πΎπ+1 (π΅ π₯ β π΅π π₯π ) πΎπ+1 + πΏπ+1 π+1 π+1 +(
πΌ π (π₯π+1 + ππ+1 πΊπ₯π+1 ) πΌπ π (π₯π ) + ππ πΊπ₯π = ( π+1 β ) 1 β ππ+1 1 β ππ
+
(1 β πΌπ β ππ ) [π½π π₯π + πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π ] β ππ π₯π β 1 β ππ
πΏπ+1 (π πΊπ₯ β ππ πΊπ₯π ) πΎπ+1 + πΏπ+1 π+1 π+1
+(
+ (1 β πΌπ+1 β ππ+1 ) Γ [π½π+1 π₯π+1 + πΎπ+1 π΅π+1 π₯π+1 + πΏπ+1 ππ+1 πΊπ₯π+1 ]
β(
β1
β ππ+1 π₯π+1 Γ (1 β ππ+1 )
πΎπ πΎπ+1 β ) π΅π π₯π πΎπ+1 + πΏπ+1 πΎπ + πΏπ
πΏπ+1 πΏπ β ) π πΊπ₯ ] πΎπ+1 + πΏπ+1 πΎπ + πΏπ π π
πΌπ+1 + ππ+1 πΌπ + ππ πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π β ) 1 β ππ+1 1 β ππ πΎπ + πΏπ
πΌπ+1 (π (π₯π+1 ) β π (π₯π )) 1 β ππ+1
=(
πΌπ+1 π (π₯π+1 ) + ππ+1 πΊπ₯π+1 πΌπ π (π₯π ) + ππ πΊπ₯π β ) 1 β ππ+1 1 β ππ
+
1 β πΌπ+1 β ππ+1 (πΎπ+1 π΅π+1 π₯π+1 + πΏπ+1 ππ+1 πΊπ₯π+1 ) 1 β ππ+1
β
1 β πΌπ β ππ (πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π ) 1 β ππ
+(
πΎ π΅ π₯ + πΏπ ππ πΊπ₯π πΌπ+1 πΌπ β ) (π (π₯π ) β π π π ) 1 β ππ+1 1 β ππ πΎπ + πΏπ
πΌπ+1 π (π₯π+1 ) + ππ+1 πΊπ₯π+1 πΌπ π (π₯π ) + ππ πΊπ₯π β ) 1 β ππ+1 1 β ππ
+(
πΎ π΅ π₯ + πΏπ ππ πΊπ₯π ππ+1 ππ β ) (πΊπ₯π β π π π ) 1 β ππ+1 1 β ππ πΎπ + πΏπ
(1 β πΌπ+1 β ππ+1 ) (1 β π½π+1 ) 1 β ππ+1
+
=( +
=
+
ππ+1 (πΊπ₯π+1 β πΊπ₯π ) 1 β ππ+1
1 β πΌπ+1 β ππ+1 β ππ+1 1 β ππ+1
Abstract and Applied Analysis Γ[
19 σ΅¨σ΅¨ πΌ πΌπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨ + σ΅¨σ΅¨σ΅¨ π+1 β σ΅¨ σ΅¨σ΅¨ 1 β ππ+1 1 β ππ σ΅¨σ΅¨σ΅¨ σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© Γ (σ΅©σ΅©σ΅©π (π₯π )σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯π σ΅©σ΅©σ΅©) σ΅¨σ΅¨ π ππ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨ + σ΅¨σ΅¨σ΅¨ π+1 β σ΅¨ σ΅¨σ΅¨ 1 β ππ+1 1 β ππ σ΅¨σ΅¨σ΅¨ σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© Γ (σ΅©σ΅©σ΅©πΊπ₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯π σ΅©σ΅©σ΅©)
πΎπ+1 (π΅ π₯ β π΅π π₯π ) πΎπ+1 + πΏπ+1 π+1 π+1
+( +
πΎπ πΎπ+1 β ) π΅π π₯π πΎπ+1 + πΏπ+1 πΎπ + πΏπ
πΏπ+1 (π πΊπ₯ β ππ πΊπ₯π ) πΎπ+1 + πΏπ+1 π+1 π+1
+(
πΏπ+1 πΏπ β ) π πΊπ₯ ] , πΎπ+1 + πΏπ+1 πΎπ + πΏπ π π
+ (139)
and hence σ΅©σ΅©σ΅©π§π+1 β π§π σ΅©σ΅©σ΅© σ΅© σ΅© πΌπ+1 σ΅©σ΅© σ΅© β€ σ΅©π (π₯π+1 ) β π (π₯π )σ΅©σ΅©σ΅© 1 β ππ+1 σ΅©
1 β πΌπ+1 β ππ+1 β ππ+1 1 β ππ+1
Γ[
σ΅¨σ΅¨ πΎ πΎπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅©σ΅© σ΅¨ σ΅© π+1 + σ΅¨σ΅¨σ΅¨ β σ΅¨ σ΅©π΅ π₯ σ΅©σ΅© σ΅¨σ΅¨ πΎπ+1 + πΏπ+1 πΎπ + πΏπ σ΅¨σ΅¨σ΅¨ σ΅© π π σ΅©
ππ+1 σ΅©σ΅© σ΅© σ΅©πΊπ₯ β πΊπ₯π σ΅©σ΅©σ΅© 1 β ππ+1 σ΅© π+1 σ΅¨σ΅¨ πΌ πΎ π΅ π₯ + πΏπ ππ πΊπ₯π σ΅©σ΅©σ΅©σ΅© πΌπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©σ΅© σ΅¨ + σ΅¨σ΅¨σ΅¨ π+1 β σ΅¨σ΅¨ σ΅©σ΅©π (π₯π ) β π π π σ΅©σ΅© σ΅¨σ΅¨ 1 β ππ+1 1 β ππ σ΅¨σ΅¨ σ΅©σ΅© σ΅©σ΅© πΎπ + πΏπ σ΅¨σ΅¨ π πΎ π΅ π₯ + πΏπ ππ πΊπ₯π σ΅©σ΅©σ΅©σ΅© ππ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©σ΅© σ΅¨ + σ΅¨σ΅¨σ΅¨ π+1 β σ΅¨σ΅¨ σ΅©σ΅©πΊπ₯π β π π π σ΅©σ΅© σ΅¨σ΅¨ 1 β ππ+1 1 β ππ σ΅¨σ΅¨ σ΅©σ΅© σ΅©σ΅© πΎπ + πΏπ
+
+
+
1 β πΌπ+1 β ππ+1 β ππ+1 1 β ππ+1 πΎπ+1 σ΅©σ΅© σ΅© σ΅©π΅ π₯ β π΅π π₯π σ΅©σ΅©σ΅© πΎπ+1 + πΏπ+1 σ΅© π+1 π+1 σ΅¨σ΅¨ πΎ πΎπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅©σ΅© σ΅¨ σ΅© π+1 + σ΅¨σ΅¨σ΅¨ β σ΅¨ σ΅©π΅ π₯ σ΅©σ΅© σ΅¨σ΅¨ πΎπ+1 + πΏπ+1 πΎπ + πΏπ σ΅¨σ΅¨σ΅¨ σ΅© π π σ΅©
=
1 β πΌπ+1 (1 β π) β ππ+1 σ΅©σ΅© σ΅© σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© 1 β ππ+1 1 β πΌπ+1 β ππ+1 β ππ+1 1 β ππ+1
Γ[
π σ΅¨σ΅¨ πΏ πΎπ+1 πΏπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨ π+1 π0 βπ π + σ΅¨σ΅¨σ΅¨ β σ΅¨ σ΅¨σ΅¨ πΎπ+1 + πΏπ+1 πΎπ + πΏπ σ΅¨σ΅¨σ΅¨ πΎπ+1 + πΏπ+1 π=0
σ΅© σ΅© σ΅© σ΅© Γ (σ΅©σ΅©σ΅©π΅π π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯π σ΅©σ΅©σ΅©)
πΏπ+1 σ΅©σ΅© σ΅© σ΅©π πΊπ₯ β ππ πΊπ₯π σ΅©σ΅©σ΅© πΎπ+1 + πΏπ+1 σ΅© π+1 π+1 σ΅¨σ΅¨ πΏ πΏπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅©σ΅© σ΅¨ σ΅© π+1 + σ΅¨σ΅¨σ΅¨ β σ΅¨σ΅¨ σ΅©σ΅©ππ πΊπ₯π σ΅©σ΅©σ΅©] . σ΅¨σ΅¨ πΎπ+1 + πΏπ+1 πΎπ + πΏπ σ΅¨σ΅¨
+
+
(140) On the other hand, repeating the same arguments as those of (55) and (56) in the proof of Theorem 24, we can get σ΅©σ΅©σ΅©ππ+1 πΊπ₯π+1 β ππ πΊπ₯π σ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ+1 πΊπ₯π β ππ πΊπ₯π σ΅©σ΅©σ΅© , σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© π
σ΅©σ΅© σ΅© σ΅© σ΅© σ΅©σ΅©π΅π+1 π₯π+1 β π΅π π₯π σ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + π0 βπ π ,
πΏπ+1 σ΅©σ΅© σ΅© σ΅©π πΊπ₯ β ππ πΊπ₯π σ΅©σ΅©σ΅© ] πΎπ+1 + πΏπ+1 σ΅© π+1 π
σ΅¨σ΅¨ πΌ πΌπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨ + σ΅¨σ΅¨σ΅¨ π+1 β σ΅¨ σ΅¨σ΅¨ 1 β ππ+1 1 β ππ σ΅¨σ΅¨σ΅¨ σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© Γ (σ΅©σ΅©σ΅©π (π₯π )σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯π σ΅©σ΅©σ΅©) σ΅¨σ΅¨ π ππ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨ + σ΅¨σ΅¨σ΅¨ π+1 β σ΅¨σ΅¨ σ΅¨σ΅¨ 1 β ππ+1 1 β ππ σ΅¨σ΅¨ σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© Γ (σ΅©σ΅©σ΅©πΊπ₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯π σ΅©σ΅©σ΅©)
σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅©
π=0
(141) for some constant π0 > 0. Utilizing (140)-(141), we have σ΅©σ΅©σ΅©π§π+1 β π§π σ΅©σ΅©σ΅© σ΅© σ΅© πΌπ+1 σ΅© σ΅© β€ π σ΅©σ΅©π₯ β π₯π σ΅©σ΅©σ΅© 1 β ππ+1 σ΅© π+1 ππ+1 σ΅©σ΅© σ΅© σ΅©π₯ β π₯π σ΅©σ΅©σ΅© 1 β ππ+1 σ΅© π+1
πΏπ+1 σ΅© σ΅© σ΅© σ΅© (σ΅©σ΅©π₯ β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ+1 πΊπ₯π β ππ πΊπ₯π σ΅©σ΅©σ΅©) πΎπ+1 + πΏπ+1 σ΅© π+1
σ΅¨σ΅¨ πΏ πΏπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅©σ΅© σ΅¨ σ΅© π+1 + σ΅¨σ΅¨σ΅¨ β σ΅¨ σ΅©π πΊπ₯ σ΅©σ΅© ] σ΅¨σ΅¨ πΎπ+1 + πΏπ+1 πΎπ + πΏπ σ΅¨σ΅¨σ΅¨ σ΅© π π σ΅©
+
Γ[
+
π πΎπ+1 σ΅© σ΅© (σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + π0 βπ π ) πΎπ+1 + πΏπ+1 π=0
π σ΅¨σ΅¨ πΏ πΏπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨ + π (βπ π + σ΅¨σ΅¨σ΅¨ π+1 β σ΅¨ σ΅¨σ΅¨ 1 β π½π+1 1 β π½π σ΅¨σ΅¨σ΅¨ π=0
σ΅¨σ΅¨ πΌ πΌπ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨σ΅¨σ΅¨σ΅¨ ππ+1 ππ σ΅¨σ΅¨σ΅¨σ΅¨ σ΅¨ + σ΅¨σ΅¨σ΅¨ π+1 β β σ΅¨σ΅¨ + σ΅¨σ΅¨ σ΅¨) σ΅¨σ΅¨ 1 β ππ+1 1 β ππ σ΅¨σ΅¨ σ΅¨σ΅¨ 1 β ππ+1 1 β ππ σ΅¨σ΅¨σ΅¨ σ΅© σ΅© + σ΅©σ΅©σ΅©ππ+1 πΊπ₯π β ππ πΊπ₯π σ΅©σ΅©σ΅© , (142)
20
Abstract and Applied Analysis
where supπβ₯0 {βπ(π₯π )β + βπΊπ₯π β + βπ΅π π₯π β + βππ πΊπ₯π β + π0 } β€ π for some π > 0. So, from (142), condition (iii), and the assumption on {ππ } it follows that (noting that 0 < π π β€ π < 1, for all π β₯ 0) σ΅© σ΅© σ΅© σ΅© lim sup (σ΅©σ΅©σ΅©π§π+1 β π§π σ΅©σ΅©σ΅© β σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅©) β€ 0. πββ
(143)
Consequently, by Lemma 20, we have σ΅©σ΅©
lim σ΅©π§ πββ σ΅© π
σ΅© β π₯π σ΅©σ΅©σ΅© = 0.
(144)
It follows from (137) and (138) that σ΅©σ΅©
lim σ΅©π₯ π β β σ΅© π+1
σ΅© σ΅© σ΅© β π₯π σ΅©σ΅©σ΅© = lim (1 β ππ ) σ΅©σ΅©σ΅©π§π β π₯π σ΅©σ΅©σ΅© = 0. πββ
σ΅© Γ [2 σ΅©σ΅©σ΅©πΌπ (π₯π β π) + ππ (πΊπ₯π β π) σ΅© + (1 β πΌπ β ππ ) (π¦π β π)σ΅©σ΅©σ΅© σ΅© σ΅© + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅©]
σ΅© σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅©σ΅© π σ΅©σ΅©2 1 β πΌπ β ππ σ΅© σ΅© + (1 β πΌπ ) σ΅©σ΅©σ΅© π (πΊπ₯π β π) + (π¦π β π)σ΅©σ΅©σ΅© σ΅©σ΅© 1 β πΌπ σ΅©σ΅© 1 β πΌπ σ΅© σ΅© + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© Γ [2 (πΌπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + ππ σ΅©σ΅©σ΅©πΊπ₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© + (1 β πΌπ β ππ ) σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅©) + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅©] σ΅© σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + (1 β πΌπ )
(145) Γ[
Next, we show that βπ₯π β πΊπ₯π β β 0 as π β β. Indeed, in terms of Lemma 11, from (130), we have σ΅©2 σ΅©σ΅© σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅©2 β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅©2 + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅©2 σ΅©2 σ΅©2 β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅©2 = σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© , σ΅©2 σ΅©σ΅© σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© = σ΅©σ΅©σ΅©πΌπ (π₯π β π) + ππ (πΊπ₯π β π) + (1 β πΌπ β ππ ) (π¦π β π) σ΅©2 + πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅©
σ΅© σ΅© β€ [σ΅©σ΅©σ΅©πΌπ (π₯π β π) + ππ (πΊπ₯π β π) + (1 β πΌπ β ππ ) (π¦π β π)σ΅©σ΅©σ΅© σ΅© σ΅©2 + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅©] σ΅© σ΅©2 = σ΅©σ΅©σ΅©πΌπ (π₯π β π) + ππ (πΊπ₯π β π) + (1 β πΌπ β ππ ) (π¦π β π)σ΅©σ΅©σ΅© σ΅© σ΅© + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅©
σ΅© Γ [2 σ΅©σ΅©σ΅©πΌπ (π₯π β π) + ππ (πΊπ₯π β π) σ΅© σ΅© σ΅© + (1 β πΌπ β ππ ) (π¦π β π)σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅©] σ΅©σ΅© σ΅© = σ΅©σ΅©σ΅©πΌπ (π₯π β π) + (1 β πΌπ ) σ΅©σ΅© σ΅©σ΅©2 1 β πΌπ β ππ ππ σ΅© (πΊπ₯π β π) + (π¦π β π)]σ΅©σ΅©σ΅© σ΅©σ΅© 1 β πΌπ 1 β πΌπ σ΅© σ΅© + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅© Γ[
ππ σ΅©σ΅© σ΅©2 1 β πΌπ β ππ σ΅©σ΅© σ΅©2 σ΅©πΊπ₯ β πσ΅©σ΅©σ΅© + σ΅©π¦ β πσ΅©σ΅©σ΅© 1 β πΌπ σ΅© π 1 β πΌπ σ΅© π β
ππ (1 β πΌπ β ππ ) 2
(1 β πΌπ )
σ΅© σ΅© π (σ΅©σ΅©σ΅©πΊπ₯π β π¦π σ΅©σ΅©σ΅©)]
σ΅© σ΅© + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅©
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© Γ [2 (πΌπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + ππ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + (1 β πΌπ β ππ ) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©) σ΅© σ΅© + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅©] σ΅© σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + (1 β πΌπ ) Γ[
ππ σ΅©σ΅© σ΅©2 1 β πΌπ β ππ σ΅©σ΅© σ΅©2 σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©π₯ β πσ΅©σ΅©σ΅© 1 β πΌπ 1 β πΌπ σ΅© π β
ππ (1 β πΌπ β ππ ) 2
(1 β πΌπ )
σ΅© σ΅© π (σ΅©σ΅©σ΅©πΊπ₯π β π¦π σ΅©σ΅©σ΅©)]
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅© (2 σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅©2 π (1 β πΌπ β ππ ) σ΅©σ΅© = σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β π π (σ΅©σ΅©πΊπ₯π β π¦π σ΅©σ΅©σ΅©) 1 β πΌπ σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© + σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅© (2 σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅©) . (146) Then, it immediately follows from 0 β€ πΌπ + ππ β€ 1 β π, for all π β₯ π0 that σ΅© σ΅© πππ π (σ΅©σ΅©σ΅©πΊπ₯π β π¦π σ΅©σ΅©σ΅©) β€
ππ (1 β πΌπ β ππ ) σ΅©σ΅© σ΅© π (σ΅©σ΅©πΊπ₯π β π¦π σ΅©σ΅©σ΅©) 1 β πΌπ
σ΅© σ΅©2 σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅©
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅© (2 σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅©)
Abstract and Applied Analysis
21
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ (σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅©) σ΅©σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅© σ΅© σ΅© + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅©
σ΅© σ΅© σ΅© σ΅© Γ (2 σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅©) , (147) for all π β₯ π0 . Since βπΌπ (π(π₯π ) β π₯π )β β 0 and {π₯π } is bounded, we deduce from (145) and condition (ii) that σ΅© σ΅© lim π (σ΅©σ΅©σ΅©πΊπ₯π β π¦π σ΅©σ΅©σ΅©) = 0. (148) πββ Utilizing the properties of π, we have σ΅© σ΅© lim σ΅©σ΅©πΊπ₯π β π¦π σ΅©σ΅©σ΅© = 0. πββ σ΅©
(149)
Also, from (130) we have π₯π+1 β π₯π = πΌπ (π (π₯π ) β π₯π ) + ππ (πΊπ₯π β π₯π ) + (1 β πΌπ β ππ ) (π¦π β π₯π ) = πΌπ (π (π₯π ) β π₯π ) + ππ (πΊπ₯π β π¦π + π¦π β π₯π )
(150)
+ (1 β πΌπ β ππ ) (π¦π β π₯π ) = πΌπ (π (π₯π ) β π₯π ) + ππ (πΊπ₯π β π¦π ) + (1 β πΌπ ) (π¦π β π₯π ) , which hence leads to σ΅© σ΅© π σ΅©σ΅©σ΅©π¦π β π₯π σ΅©σ΅©σ΅©
σ΅© σ΅© β€ (1 β πΌπ β ππ ) σ΅©σ΅©σ΅©π¦π β π₯π σ΅©σ΅©σ΅© σ΅© σ΅© β€ (1 β πΌπ ) σ΅©σ΅©σ΅©π¦π β π₯π σ΅©σ΅©σ΅©
σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π+1 β π₯π β πΌπ (π (π₯π ) β π₯π ) β ππ (πΊπ₯π β π¦π )σ΅©σ΅©σ΅©
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅© + ππ σ΅©σ΅©σ΅©πΊπ₯π β π¦π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©πΊπ₯π β π¦π σ΅©σ΅©σ΅© .
(151)
So, it is easy to see from (145), (149), and βπΌπ (π(π₯π ) β π₯π )β β 0 that σ΅© σ΅© lim σ΅©σ΅©π¦ β π₯π σ΅©σ΅©σ΅© = 0. (152) πββ σ΅© π We note that σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅©σ΅©πΊπ₯π β π₯π σ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©πΊπ₯π β π¦π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π¦π β π₯π σ΅©σ΅©σ΅© .
(153)
Therefore, from (149) and (152) it follows that σ΅© σ΅© lim σ΅©σ΅©πΊπ₯π β π₯π σ΅©σ΅©σ΅© = 0. πββ σ΅©
(154)
Repeating the same arguments as those of (86), (89), and (91) in the proof of Theorem 24, we can obtain σ΅© σ΅© σ΅© σ΅© lim σ΅©σ΅©π₯ β π΅π π₯π σ΅©σ΅©σ΅© = lim σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅© πββ σ΅© π πββ (155) σ΅© σ΅© = lim σ΅©σ΅©σ΅©π₯π β ππ₯π σ΅©σ΅©σ΅© = 0. πββ
Suppose that π½π β‘ π½ for some fixed π½ β (0, 1) such that π½ + πΎπ + πΏπ = 1 for all π β₯ 0. Define a mapping ππ₯ = (1 β π1 β π2 )ππ₯ + π1 π΅π₯ + π2 πΊπ₯, where π1 , π2 β (0, 1) are two constants with π1 + π2 < 1. Then, by Lemmas 14 and 17, we have that Fix(π) = Fix(π) β© Fix(π΅) β© Fix(πΊ) = πΉ. For each π β₯ 1, let {ππ } be a unique element of πΆ such that ππ =
1 1 π (ππ ) + (1 β ) πππ . π π
(156)
From Lemma 13, we conclude that ππ β π β Fix(π) = πΉ as π β β. Observe that for every π, π σ΅© σ΅©σ΅© σ΅©σ΅©π¦π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© + πΏπ (σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π π₯π β π΅ππ σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅© σ΅© = π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β π½) σ΅©σ΅©σ΅©π΅π π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅© ,
(157)
and hence σ΅©σ΅© σ΅© σ΅©σ΅©π₯π+1 β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π΅ππ σ΅©σ΅©σ΅© + ππ σ΅©σ΅©σ΅©πΊπ₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© + (1 β πΌπ β ππ ) σ΅©σ΅©σ΅©π¦π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ (σ΅©σ΅©σ΅©π (π₯π ) β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅© σ΅© + ππ (σ΅©σ΅©σ΅©πΊπ₯π β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅©) + (1 β πΌπ β ππ ) σ΅© σ΅© σ΅© σ΅© Γ [π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β π½) σ΅©σ΅©σ΅©π΅π π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©] σ΅© σ΅© σ΅© σ΅© β€ πΌπ (σ΅©σ΅©σ΅©π (π₯π ) β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅© σ΅© + ππ (σ΅©σ΅©σ΅©πΊπ₯π β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅©) + (1 β πΌπ β ππ ) σ΅© σ΅© Γ [π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© + (1 β π½) (σ΅©σ΅©σ΅©π΅π π₯π β π΅π ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π ππ β π΅ππ σ΅©σ΅©σ΅©) σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©] σ΅© σ΅© σ΅© σ΅© β€ πΌπ (σ΅©σ΅©σ΅©π (π₯π ) β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅© σ΅© + ππ (σ΅©σ΅©σ΅©πΊπ₯π β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅©) + (1 β πΌπ β ππ ) σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© Γ [π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β π½) (σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π ππ β π΅ππ σ΅©σ΅©σ΅©) σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©] σ΅© σ΅© σ΅© σ΅© = πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π₯π σ΅©σ΅©σ΅© + ππ σ΅©σ΅©σ΅©πΊπ₯π β π₯π σ΅©σ΅©σ΅© σ΅© σ΅© + [π½ + (πΌπ + ππ ) (1 β π½)] σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© + (1 β πΌπ β ππ ) (1 β π½) σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + (1 β πΌπ β ππ ) σ΅© σ΅© σ΅© σ΅© Γ [(1 β π½) σ΅©σ΅©σ΅©π΅π ππ β π΅ππ σ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©]
22
Abstract and Applied Analysis σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π₯π σ΅©σ΅©σ΅© + ππ σ΅©σ΅©σ΅©πΊπ₯π β π₯π σ΅©σ΅©σ΅©
Repeating the same arguments as those of (99), in the proof of Theorem 24, we can get
+ [π½ + (πΌπ + ππ ) (1 β π½)] σ΅© σ΅© σ΅© σ΅© Γ (σ΅©σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π+1 β π΅ππ σ΅©σ΅©σ΅©) + (1 β πΌπ β ππ ) (1 β π½) σ΅© σ΅© σ΅© σ΅© Γ (σ΅©σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π+1 β ππ σ΅©σ΅©σ΅©) + (1 β πΌπ β ππ ) σ΅© σ΅© σ΅© σ΅© Γ [(1 β π½) σ΅©σ΅©σ΅©π΅π ππ β π΅ππ σ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©] σ΅© σ΅© σ΅© σ΅© = πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π₯π σ΅©σ΅©σ΅© + ππ σ΅©σ΅©σ΅©πΊπ₯π β π₯π σ΅©σ΅©σ΅© σ΅© σ΅© + [π½ + (πΌπ + ππ ) (1 β π½)] σ΅©σ΅©σ΅©π₯π+1 β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© + (1 β πΌπ β ππ ) (1 β π½) σ΅©σ΅©σ΅©π₯π+1 β ππ σ΅©σ΅©σ΅© + (1 β πΌπ β ππ ) σ΅© σ΅© σ΅© σ΅© Γ [(1 β π½) σ΅©σ΅©σ΅©π΅π ππ β π΅ππ σ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©] σ΅© σ΅© + σ΅©σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅© . (158) So, it immediately follows from 0 β€ πΌπ β€ 1 β π, for all π β₯ π0 that σ΅©σ΅© σ΅© σ΅©σ΅©π₯π+1 β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π ππ β π΅ππ σ΅©σ΅©σ΅© 1 (1 β πΌπ β ππ ) (1 β π½) σ΅© σ΅© σ΅© σ΅© Γ (σ΅©σ΅©σ΅©πΌπ (π₯π β π (π₯π ))σ΅©σ΅©σ΅© + ππ σ΅©σ΅©σ΅©πΊπ₯π β π₯π σ΅©σ΅©σ΅©
+
σ΅© σ΅©2 σ΅© σ΅©2 ππ σ΅©σ΅©σ΅©π₯π β πΊππ σ΅©σ΅©σ΅© β€ ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© , σ΅© σ΅©2 σ΅© σ΅©2 ππ σ΅©σ΅©σ΅©π₯π β πππ σ΅©σ΅©σ΅© β€ ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© .
(162)
Utilizing (161) and (162), we deduce that σ΅© σ΅©2 ππ σ΅©σ΅©σ΅©π₯π β πππ σ΅©σ΅©σ΅© σ΅© σ΅©2 β€ (1 β π1 β π2 ) ππ σ΅©σ΅©σ΅©π₯π β πππ σ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅©2 + π1 ππ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + π2 ππ σ΅©σ΅©σ΅©π₯π β πΊππ σ΅©σ΅©σ΅©
(163)
σ΅© σ΅©2 β€ ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© . Also, observe that 1 1 (1 β ) (π₯π β πππ ) = π₯π β ππ β (π₯π β π (ππ )) . π π
(164)
Repeating the same arguments as those of (106) in the proof of Theorem 24, we can get
πΏ σ΅© σ΅© σ΅© σ΅© + σ΅©σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅©) + π σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅© 1βπ½ σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π ππ β π΅ππ σ΅©σ΅©σ΅©
(159)
1 σ΅©σ΅© σ΅©2 π σ΅©π₯ β ππ σ΅©σ΅©σ΅© β₯ ππ β¨π (ππ ) β ππ , π½ (π₯π β ππ )β© . 2π π σ΅© π
(165)
Since ππ β π β Fix(π) = πΉ as π β β, by the uniform Gateaux differentiability of the norm of π, we have
1 π (1 β π½) σ΅© σ΅© Γ (σ΅©σ΅©σ΅©πΌπ (π₯π β π (π₯π ))σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© + σ΅©σ΅©σ΅©πΊπ₯π β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅© ) σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π+1 β ππ σ΅©σ΅©σ΅© + ππ , βπ β₯ π0 , σ΅© σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅© +
ππ β¨π (π) β π, π½ (π₯π β π)β© β€ 0.
(166)
On the other hand, from (135) and the norm-to-weakβ uniform continuity of π½ on bounded subsets of π, it follows that
where ππ = βπ΅π ππ β π΅ππ β + βππ πΊπ₯π β π΅π π₯π β + 1/(π(1 β π½))(βπΌπ (π₯π β π(π₯π ))β + βπΊπ₯π β π₯π β + βπ₯π β π₯π+1 β). Since limπ β β βπ΅π ππ β π΅ππ β = limπ β β βππ πΊπ₯π β π΅π π₯π β = limπ β β βπΌπ (π₯π β π(π₯π ))β = limπ β β βπΊπ₯π β π₯π β = limπ β β βπ₯π β π₯π+1 β = 0, we know that ππ β 0 as π β β. From (159), we obtain σ΅©σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅© σ΅©σ΅©π₯π+1 β π΅ππ σ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β ππ σ΅©σ΅©σ΅© + ππ (2 σ΅©σ΅©σ΅©π₯π+1 β ππ σ΅©σ΅©σ΅© + ππ ) , βπ β₯ π0 . (160)
σ΅© σ΅©2 σ΅© σ΅©2 β€ ππ σ΅©σ΅©σ΅©π₯π+1 β ππ σ΅©σ΅©σ΅© = ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© .
So, utilizing Lemma 18, we deduce from (166) and (167) that lim sup β¨π (π) β π, π½ (π₯π β π)β© β€ 0, πββ
(168)
which, together with (135) and the norm-to-norm uniform continuity of π½ on bounded subsets of π, implies that
For any Banach limit π, from (160) we derive σ΅© σ΅©2 σ΅© σ΅©2 ππ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© = ππ σ΅©σ΅©σ΅©π₯π+1 β π΅ππ σ΅©σ΅©σ΅©
σ΅¨ σ΅¨ lim σ΅¨σ΅¨β¨π (π) β π, π½ (π₯π+1 β π)β© β β¨π (π) β π, π½ (π₯π β π)β©σ΅¨σ΅¨σ΅¨ = 0. (167)
πββ σ΅¨
(161)
lim sup β¨π (π) β π, π½ (π₯π+1 β π)β© β€ 0. πββ
(169)
Abstract and Applied Analysis
23
Finally, let us show that π₯π β π as π β β. Utilizing Lemma 8 (i), from (130) and the convexity of β β
β2 , we get σ΅©2 σ΅©2 σ΅©2 σ΅© σ΅© σ΅©σ΅© σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅©2 + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β πσ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© , σ΅©2 σ΅©σ΅© σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© = σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π (π)) + ππ (πΊπ₯π β π)
(170)
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + ππ (σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©) σ΅© σ΅© σ΅© σ΅© + (1 β πΌπ β ππ ) (σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©)
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© + 3 σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© .
(173) Since π₯π β π and π¦π β π, we get πΌπ (π(π₯π ) β π₯π ) β 0. This completes the proof.
σ΅©2 + (1 β πΌπ β ππ ) (π¦π β π) + πΌπ (π (π) β π) σ΅©σ΅©σ΅©
σ΅© β€ σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π (π)) + ππ (πΊπ₯π β π) σ΅©2 + (1 β πΌπ β ππ ) (π¦π β π) σ΅©σ΅©σ΅© + 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β©
σ΅© σ΅© σ΅©2 σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π (π)σ΅©σ΅©σ΅© + ππ σ΅©σ΅©σ΅©πΊπ₯π β πσ΅©σ΅©σ΅© σ΅© σ΅©2 + (1 β πΌπ β ππ ) σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© + 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β© σ΅© σ΅© σ΅©2 σ΅©2 β€ πΌπ πσ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + ππ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅©2 + (1 β πΌπ β ππ ) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
Corollary 29. Let πΆ be a nonempty closed convex subset of a uniformly convex Banach space π which has a uniformly Gateaux differentiable norm. Let Ξ πΆ be a sunny nonexpansive retraction from π onto πΆ. Let {ππ }β π=0 be a sequence of positive numbers in (0, π] for some π β (0, 1) and π΄ π : πΆ β πππ strictly pseudocontractive and πΌΜπ -strongly accretive with ππ + πΌΜπ β₯ 1 for each π = 0, 1, . . .. Define a mapping πΊπ : πΆ β πΆ by Ξ πΆ(πΌ β π π π΄ π )π₯ = πΊπ π₯ for all π₯ β πΆ and π = 0, 1, . . ., where 1β(ππ /(1+ππ ))(1ββ(1 β πΌΜπ )/ππ ) β€ π π β€ 1 for all π = 0, 1, . . .. Let π΅π : πΆ β πΆ be the π-mapping generated by πΊπ , πΊπβ1 , . . . , πΊ0 and ππ , ππβ1 , . . . , π0 . Let π : πΆ β πΆ be a self-mapping such that πΌ β π : πΆ β π is π-strictly pseudocontractive and πΌstrongly accretive with πΌ + π β₯ 1. Let π : πΆ β πΆ be a contraction with coefficient π β (0, 1). Let {ππ }β π=0 be a countable family of nonexpansive mappings of πΆ into itself such that πΉ = β (ββ π=0 Fix(ππ )) β© Fix(π) β© (βπ=0 VI(πΆ, π΄ π )) =ΜΈ 0. For arbitrarily given π₯0 β πΆ, let {π₯π } be the sequence generated by π¦π = π½π π₯π + πΎπ π΅π π₯π + πΏπ ππ ((1 β π) πΌ + ππ) π₯π ,
+ 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β©
π₯π+1 = πΌπ π (π₯π ) + ππ ((1 β π) πΌ + ππ) π₯π + (1 β πΌπ β ππ ) π¦π ,
σ΅© σ΅©2 = (1 β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
βπ β₯ 0, (174)
+ 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β© σ΅© σ΅©2 = (1 β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΌπ (1 β π)
2 β¨π (π) β π, π½ (π₯π+1 β π)β© . 1βπ (171)
Applying Lemma 7 to (171), we obtain that π₯π β π as π β β. Conversely, if π₯π β π β πΉ as π β β, then from (130) it follows that σ΅© σ΅© σ΅© σ΅© σ΅© σ΅©σ΅© σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β πσ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ³¨β 0
(172)
as π β β; that is, π¦π β π. Again from (130) we obtain that σ΅©σ΅© σ΅© σ΅©σ΅©πΌπ (π (π₯π ) β π₯π )σ΅©σ΅©σ΅© σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π+1 β π₯π β ππ (πΊπ₯π β π₯π ) β (1 β πΌπ β ππ ) (π¦π β π₯π )σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© + ππ σ΅©σ΅©σ΅©πΊπ₯π β π₯π σ΅©σ΅©σ΅© + (1 β πΌπ β ππ ) σ΅©σ΅©σ΅©π¦π β π₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + ππ (σ΅©σ΅©σ΅©πΊπ₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©) σ΅© σ΅© σ΅© σ΅© + (1 β πΌπ β ππ ) (σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©)
where 1 β (π/(1 + π))(1 β β(1 β πΌ)/π) β€ π β€ 1 and {ππ }, {πΌπ }, {π½π }, {πΎπ }, and {πΏπ } are the sequences in [0, 1] such that π½π + πΎπ + πΏπ = 1 and πΌπ + ππ β€ 1 for all π β₯ 0. Suppose that the following conditions hold: (i) ββ π=0 πΌπ = β and 0 β€ πΌπ + ππ β€ 1 β π, for all π β₯ π0 for some integer π0 β₯ 0; (ii) lim inf π β β ππ > lim inf π β β πΏπ > 0;
0, lim inf π β β πΎπ
>
0 and
(iii) limπ β β (|πΌπ+1 /(1β(1βπΌπ+1 βππ+1 )π½π+1 )βπΌπ /(1β(1β πΌπ βππ )π½π )|+|ππ+1 /(1β(1βπΌπ+1 βππ+1 )π½π+1 )βππ /(1β (1 β πΌπ β ππ )π½π )| + |πΏπ+1 /(1 β π½π+1 ) β πΏπ /(1 β π½π )|) = 0; (iv) 0 < lim inf π β β π½π β€ lim supπ β β π½π < 1. Assume that ββ π=0 supπ₯βπ· βππ+1 π₯ β ππ π₯β < β for any bounded subset π· of πΆ and let π be a mapping of πΆ into itself defined by ππ₯ = limπ β β ππ π₯ for all π₯ β πΆ and suppose that Fix(π) = ββ π=0 Fix(ππ ). Then there hold the following: (I) limπ β β βπ₯π+1 β π₯π β = 0; (II) π₯π β π β πΌπ (π(π₯π ) β π₯π ) β 0 provided π½π β‘ π½ for some fixed π½ β (0, 1), where π β πΉ solves the following VIP β¨π β π (π) , π½ (π β π)β© β€ 0,
βπ β πΉ.
(175)
24
Abstract and Applied Analysis
Proof. In Theorem 28, we put π΅1 = πΌ β π, π΅2 = 0, and π1 = π, where 1 β (π/(1 + π))(1 β β(1 β πΌ)/π) β€ π β€ 1. Then, GSVI (13) is equivalent to the VIP of finding π₯β β πΆ such that β¨π΅1 π₯β , π½ (π₯ β π₯β )β© β₯ 0,
βπ₯ β πΆ.
(176)
In this case, π΅1 : πΆ β π is π-strictly pseudocontractive and πΌ-strongly accretive. Repeating the same arguments as those in the proof of Corollary 25, we can infer that Fix(π) = VI(πΆ, π΅1 ). Accordingly, πΉ = ββ π=0 Fix(ππ ) β© Ξ© β© β β (ββ π=0 VI(πΆ, π΄ π )) = βπ=0 Fix(ππ ) β© Fix(π) β© (βπ=0 VI(πΆ, π΄ π )), and πΊπ₯π = ((1 β π) πΌ + ππ) π₯π ,
βπ β₯ 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 Yaoβs [10, Theorem 3.2], Cai and Buβs [11, Theorem 3.1], Kangtunyakarnβs [38, Theorem 3.1], and Ceng and Yaoβs [8, Theorem 3.1], in the following aspects. (i) The problem of finding a point π β (ββ π=0 Fix(ππ )) β© Ξ© β© (ββ π=0 VI(πΆ, π΄ π )) in our Theorems 24 and 28 is more general and more subtle than every one of the problem of finding a point π β ββ π=0 Fix(ππ ) in [10, Theorem 3.2], the problem of finding a point π β ββ π=1 Fix(ππ ) β© Ξ© in [11, Theorem 3.1], the problem of finding a point π β Fix(π) β© Fix(π) β© (βπ π=1 VI(πΆ, π΄ π )) in [38, Theorem 3.1], and the problem of finding a point π β Fix(π) 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 {πΌπ }, {π½π }, {πΎπ }, {πΏπ }, (and {ππ }). (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 Yaoβs [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 Buβs [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 πΊ and ππ in [11, Theorem 3.1] and the mapping π in [8, Theorem 3.1] are replaced with the same composite mapping ππ πΊ in the iterative schemes (42) and (130) and the mapping ππ in [10, Theorem 3.2] is replaced with π΅π . (v) Cai and Buβs 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 ππ πΊ 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 π-mapping and the Banach limit (see Lemmas 16β18 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 π-mapping and the Banach limit (see Lemmas 16β18 in Section 2 of this paper). (vi) The assumption of the uniformly convex and 2uniformly smooth Banach space π in [11, Theorem 3.1] is weakened to the one of the uniformly convex Banach space π having a uniformly Gateaux differentiable norm in our Theorem 28. Moreover, the assumption of the uniformly smooth Banach space π in [8, Theorem 3.1] is replaced with the one of the uniformly convex Banach space π 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 {πΌπ }, {π½π }, {πΎπ }, and {πΏπ } (and {ππ }) 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 πΆ be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space π. Let Ξ πΆ be a sunny nonexpansive retraction from π onto πΆ. Let {ππ }β π=0 be a sequence of positive numbers in (0, π] for some π β (0, 1) and π΄ π : πΆ β π an πΌΜπ -inverse strongly accretive mapping for each π = 0, 1, . . .. Define a mapping πΊπ : πΆ β πΆ by Ξ πΆ(πΌ β π π π΄ π )π₯ = πΊπ π₯ for all π₯ β πΆ and π = 0, 1, . . .,
Abstract and Applied Analysis
25
where π π β (0, πΌΜπ /π
2 ] and π
is the 2-uniformly smooth constant of π. Let π΅π : πΆ β πΆ be the π-mapping generated by πΊπ , πΊπβ1 , . . . , πΊ0 and ππ , ππβ1 , . . . , π0 . Let the mapping π΅π : πΆ β π be π½Μπ -inverse strongly accretive for π = 1, 2. Let π : πΆ β πΆ be a contraction with coefficient π β (0, 1). Let {ππ }β π=0 be a countable family of nonexpansive mappings of πΆ into itself β such that πΉ = (ββ π=0 Fix(ππ )) β© Ξ© β© (βπ=0 VI(πΆ, π΄ π )) =ΜΈ 0, where Ξ© is the fixed point set of the mapping πΊ = Ξ πΆ(πΌβπ1 π΅1 )Ξ πΆ(πΌβ π2 π΅2 ) with 0 < ππ < π½Μπ /π
2 for π = 1, 2. For arbitrarily given π₯0 β πΆ, let {π₯π } be the sequence generated by π₯π+1 = πΌπ π (π₯π ) + π½π π₯π + πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π ,
βπ β₯ 0, (178)
where {πΌπ }, {π½π }, {πΎπ }, and {πΏπ } are the sequences in (0, 1) such that πΌπ +π½π +πΎπ +πΏπ = 1 for all π β₯ 0. Suppose that the following conditions hold: (i) limπ β β πΌπ = 0 and ββ π=0 πΌπ = β; (ii) {πΎπ }, {πΏπ } β [π, π] for some π, π β (0, 1); (iii) limπ β β (|π½π β π½πβ1 | + |πΎπ β πΎπβ1 | + |πΏπ β πΏπβ1 |) = 0; (iv) 0 < lim inf π β β π½π β€ lim supπ β β π½π < 1. Assume that ββ π=1 supπ₯βπ· βππ π₯ β ππβ1 π₯β < β for any bounded subset π· of πΆ and let π be a mapping of πΆ into itself defined by ππ₯ = limπ β β ππ π₯ for all π₯ β πΆ and suppose that Fix(π) = ββ π=0 Fix(ππ ). Then, there hold the following: (I) limπ β β βπ₯π+1 β π₯π β = 0; (II) the sequence {π₯π }β π=0 converges strongly to some π β πΉ which is the unique solution of the variational inequality problem (VIP) β¨(πΌ β π) π, π½ (π β π)β© β€ 0,
βπ β πΉ,
πββ
(180)
for every π₯ β πΆ. That is, such a π΅ is the π-mapping generated β by the sequences {πΊπ }β π=0 and {ππ }π=0 . According to Lemma 17, β we know that Fix(π΅) = βπ=0 Fix(πΊπ ). From Lemma 21 and the definition of πΊπ , we have Fix(πΊπ ) = VI(πΆ, π΄ π ) for each π = 0, 1, . . .. Hence, we have β
β
π=0
π=0
Fix (π΅) = β Fix (πΊπ ) = β VI (πΆ, π΄ π ) .
σ΅© σ΅© σ΅© σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅© β€ max {σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© , σ΅© }. 1βπ (182) By induction, we obtain σ΅© σ΅© σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅©σ΅© { }, π₯ β€ max , π₯ β π β π σ΅©σ΅© 0 σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© π 1βπ
βπ β₯ 0. (183)
Hence, {π₯π } is bounded, and so are the sequences {πΊπ₯π } and {π(π₯π )}. Let us show that σ΅©σ΅©
lim σ΅©π₯ π β β σ΅© π+1
σ΅© β π₯π σ΅©σ΅©σ΅© = 0.
(184)
As a matter of fact, observe that π₯π+1 can be rewritten as follows:
πΌπ /π
2 ] for π = 0, 1, . . ., it Proof. First of all, since 0 < π π < [Μ is easy to see that πΊπ is a nonexpansive mapping for each π = 0, 1, . . .. Since π΅π : πΆ β πΆ is the π-mapping generated by πΊπ , πΊπβ1 , . . . , πΊ0 , and ππ , ππβ1 , . . . , π0 , by Lemma 16 we know that, for each π₯ β πΆ and π β₯ 0, the limit limπ β β ππ,π π₯ exists. Moreover, one can define a mapping π΅ : πΆ β πΆ as follows: πββ
σ΅© σ΅©σ΅© σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ (σ΅©σ΅©σ΅©π (π₯π ) β π (π)σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅©) σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©πΊπ₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ (π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅©) σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅©σ΅© σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅© σ΅© = (1 β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΌπ (1 β π) σ΅© 1βπ
(179)
provided π½π β‘ π½ for some fixed π½ β (0, 1).
π΅π₯ = lim π΅π π₯ = lim ππ,0 π₯
π = π΅π π, and π = ππ π for all π β₯ 0. By Lemma 23, we know that πΊ is nonexpansive. Then, from (178), we have
(181)
Next, let us show that the sequence {π₯π } is bounded. Indeed, take a fixed π β πΉ arbitrarily. Then, we get π = πΊπ,
π₯π+1 = π½π π₯π + (1 β π½π ) π§π ,
(185)
where π§π = (πΌπ π(π₯π ) + πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π )/(1 β π½π ). Observe that σ΅© σ΅©σ΅© σ΅©σ΅©π§π β π§πβ1 σ΅©σ΅©σ΅© σ΅©σ΅© πΌ π (π₯ ) + πΎ π΅ π₯ + πΏ π πΊπ₯ σ΅© π π π π π π π = σ΅©σ΅©σ΅©σ΅© π 1 β π½π σ΅©σ΅© β
πΌπβ1 π (π₯πβ1 ) + πΎπβ1 π΅πβ1 π₯πβ1 + πΏπβ1 ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© 1 β π½πβ1 σ΅©
σ΅©σ΅© π₯ β π½ π₯ π₯ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅©σ΅© σ΅© π π = σ΅©σ΅©σ΅© π+1 β π σ΅© σ΅©σ΅© 1 β π½π 1 β π½πβ1 σ΅©σ΅©σ΅© σ΅©σ΅© π₯ β π½ π₯ π₯ β π½πβ1 π₯πβ1 σ΅© π π = σ΅©σ΅©σ΅© π+1 β π 1 β π½π σ΅©σ΅© 1 β π½π +
π₯π β π½πβ1 π₯πβ1 π₯π β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅©σ΅© β σ΅© 1 β π½π 1 β π½πβ1 σ΅©σ΅©σ΅©
26
Abstract and Applied Analysis σ΅©σ΅© π₯ β π½ π₯ π₯ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅©σ΅© σ΅© π π β π β€ σ΅©σ΅©σ΅© π+1 σ΅©σ΅© σ΅©σ΅© 1 β π½π σ΅©σ΅© 1 β π½π σ΅©σ΅© π₯ β π½ π₯ π₯ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅©σ΅© σ΅© πβ1 πβ1 + σ΅©σ΅©σ΅© π β π σ΅© σ΅©σ΅© 1 β π½π 1 β π½πβ1 σ΅©σ΅©σ΅© =
=
=
σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + π 0 σ΅©σ΅©σ΅©ππ,1 π₯πβ1 β ππβ1,1 π₯πβ1 σ΅©σ΅©σ΅©
σ΅© σ΅© σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + π 0 σ΅©σ΅©σ΅©π 1 πΊ1 ππ,2 π₯πβ1 β π 1 πΊ1 ππβ1,2 π₯πβ1 σ΅©σ΅©σ΅©
σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + π 0 π 1 σ΅©σ΅©σ΅©ππ,2 π₯πβ1 β ππβ1,2 π₯πβ1 σ΅©σ΅©σ΅©
1 σ΅©σ΅© σ΅© σ΅©π₯ β π½π π₯π β (π₯π β π½πβ1 π₯πβ1 )σ΅©σ΅©σ΅© 1 β π½π σ΅© π+1 σ΅¨σ΅¨ 1 σ΅¨σ΅¨ 1 σ΅¨ σ΅¨σ΅¨ σ΅©σ΅© σ΅© + σ΅¨σ΅¨σ΅¨ β σ΅¨ σ΅©π₯ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅© σ΅¨σ΅¨ 1 β π½π 1 β π½πβ1 σ΅¨σ΅¨σ΅¨ σ΅© π
.. . πβ1
σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + (βπ π ) σ΅©σ΅©σ΅©ππ,π π₯πβ1 β ππβ1,π π₯πβ1 σ΅©σ΅©σ΅© π=0
1 σ΅©σ΅© σ΅© σ΅©π₯ β π½π π₯π β (π₯π β π½πβ1 π₯πβ1 )σ΅©σ΅©σ΅© 1 β π½π σ΅© π+1 σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ σ΅© σ΅©σ΅© + σ΅©π₯ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅© (1 β π½πβ1 ) (1 β π½π ) σ΅© π
πβ1
σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + πβπ π , π=0
(188)
1 σ΅©σ΅© σ΅©πΌ π (π₯π ) + πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π β πΌπβ1 π (π₯πβ1 ) 1 β π½π σ΅© π σ΅© β πΎπβ1 π΅πβ1 π₯πβ1 β πΏπβ1 ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ σ΅© σ΅©σ΅© + σ΅©π₯ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅© (1 β π½ ) (1 β π½ ) σ΅© π πβ1
π
1 σ΅© σ΅© β€ [πΌ σ΅©σ΅©π (π₯π ) β π (π₯πβ1 )σ΅©σ΅©σ΅© 1 β π½π π σ΅© σ΅© σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅© σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΌπ β πΌπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π (π₯πβ1 )σ΅©σ΅©σ΅© σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΎπ β πΎπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅© σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΏπ β πΏπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅©] σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ σ΅© σ΅©σ΅© + σ΅©π₯ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅© . (1 β π½πβ1 ) (1 β π½π ) σ΅© π (186)
for some constant π > 0. Taking into account 0 < lim inf π β β π½π β€ lim supπ β β π½π < 1, we may assume, Μ Utilizing (186)β without loss of generality, that {π½π } β [Μπ, π]. (188), we have σ΅© σ΅©σ΅© σ΅©σ΅©π§π β π§πβ1 σ΅©σ΅©σ΅© β€
β€
On the other hand, we note that, for all π β₯ 1,
1 σ΅© σ΅© [πΌ σ΅©σ΅©π (π₯π ) β π (π₯πβ1 )σ΅©σ΅©σ΅© 1 β π½π π σ΅© σ΅© σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅© σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΌπ β πΌπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π (π₯πβ1 )σ΅©σ΅©σ΅© σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΎπ β πΎπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅© σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΏπ β πΏπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅©] σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ σ΅© σ΅©σ΅© + σ΅©π₯ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅© (1 β π½πβ1 ) (1 β π½π ) σ΅© π πβ1 1 σ΅© σ΅© σ΅© σ΅© {πΌπ π σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + πΎπ [σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + πβπ π ] 1 β π½π π=0
σ΅© σ΅© σ΅© σ΅© + πΏπ [σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯πβ1 β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅©]
σ΅©σ΅© σ΅© σ΅©σ΅©ππ πΊπ₯π β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅©
σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©ππ πΊπ₯π β ππ πΊπ₯πβ1 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯πβ1 β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©πΊπ₯π β πΊπ₯πβ1 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯πβ1 β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯πβ1 β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© .
σ΅¨ σ΅¨σ΅© σ΅¨σ΅© σ΅© σ΅© σ΅¨ + σ΅¨σ΅¨σ΅¨πΌπ β πΌπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π (π₯πβ1 )σ΅©σ΅©σ΅© + σ΅¨σ΅¨σ΅¨πΎπ β πΎπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅© (187)
σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΏπ β πΏπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© } +
Furthermore, by (CY), since πΊπ and ππ,π are nonexpansive, we deduce that for each π β₯ 1 σ΅© σ΅©σ΅© σ΅©σ΅©π΅π π₯π β π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π΅π π₯π β π΅π π₯πβ1 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π π₯πβ1 β π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π π₯πβ1 β π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π 0 πΊ0 ππ,1 π₯πβ1 β π 0 πΊ0 ππβ1,1 π₯πβ1 σ΅©σ΅©σ΅©
=
σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ σ΅© σ΅©σ΅© σ΅©π₯ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅© (1 β π½πβ1 ) (1 β π½π ) σ΅© π
1 σ΅© σ΅© { (1 β π½π β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© 1 β π½π πβ1
σ΅© σ΅© + πΎπ πβπ π + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯πβ1 β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© π=0
σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΌπ β πΌπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π (π₯πβ1 )σ΅©σ΅©σ΅©
Abstract and Applied Analysis
27
σ΅¨σ΅© σ΅© σ΅¨ + σ΅¨σ΅¨σ΅¨πΎπ β πΎπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅©
Next we show that βπ₯π β πΊπ₯π β β 0 as π β β. Indeed, for simplicity, put π = Ξ πΆ(π β π2 π΅2 π), π’π = Ξ πΆ(π₯π β π2 π΅2 π₯π ) and Vπ = Ξ πΆ(π’π β π1 π΅1 π’π ). Then, Vπ = πΊπ₯π for all π β₯ 0. From Lemma 26 we have
σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΏπ β πΏπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© } +
σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ σ΅© σ΅©σ΅© σ΅©π₯ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅© (1 β π½πβ1 ) (1 β π½π ) σ΅© π
= (1 β
σ΅©2 σ΅©σ΅© σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© σ΅© σ΅©2 = σ΅©σ΅©σ΅©Ξ πΆ (π₯π β π2 π΅2 π₯π ) β Ξ πΆ (π β π2 π΅2 π)σ΅©σ΅©σ΅© (193) σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©π₯π β π β π2 (π΅2 π₯π β π΅2 π)σ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β 2π2 (π½Μ2 β π
2 π2 ) σ΅©σ΅©σ΅©π΅2 π₯π β π΅2 πσ΅©σ΅©σ΅© ,
πβ1 πΌπ (1 β π) σ΅©σ΅© σ΅© πΎπ ) σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + π βπ π 1 β π½π 1 β π½π π=0
πΏπ σ΅©σ΅© 1 σ΅© σ΅©π πΊπ₯ β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© + 1 β π½π σ΅© π πβ1 1 β π½π σ΅¨σ΅© σ΅¨σ΅© σ΅© σ΅¨ σ΅© σ΅¨ Γ [σ΅¨σ΅¨σ΅¨πΌπ β πΌπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π (π₯πβ1 )σ΅©σ΅©σ΅© + σ΅¨σ΅¨σ΅¨πΎπ β πΎπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅© σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΏπ β πΏπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅©] σ΅¨σ΅¨ σ΅¨σ΅¨ σ΅© σ΅©σ΅© σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨ + σ΅©π₯ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅© (1 β π½ ) (1 β π½ ) σ΅© π +
πβ1
σ΅©σ΅© σ΅©2 σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© σ΅© σ΅©2 = σ΅©σ΅©σ΅©Ξ πΆ (π’π β π1 π΅1 π’π ) β Ξ πΆ (π β π1 π΅1 π)σ΅©σ΅©σ΅© σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©π’π β π β π1 (π΅1 π’π β π΅1 π)σ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© β 2π1 (π½Μ1 β π
2 π1 ) σ΅©σ΅©σ΅©π΅1 π’π β π΅1 πσ΅©σ΅©σ΅© .
π
πβ1
σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + πβπ π + σ΅©σ΅©σ΅©ππ πΊπ₯πβ1 β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© π=0
Substituting (193) for (194), we obtain
1 σ΅¨ σ΅© σ΅¨ σ΅¨σ΅© σ΅¨ + [σ΅¨σ΅¨πΌ β πΌπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π (π₯πβ1 )σ΅©σ΅©σ΅© + σ΅¨σ΅¨σ΅¨πΎπ β πΎπβ1 σ΅¨σ΅¨σ΅¨ 1 β π½π σ΅¨ π σ΅© σ΅© σ΅¨ σ΅¨σ΅© σ΅© Γ σ΅©σ΅©σ΅©π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅© + σ΅¨σ΅¨σ΅¨πΏπ β πΏπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅©] σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ + (1 β π½πβ1 ) (1 β π½π ) σ΅© σ΅© Γ σ΅©σ΅©σ΅©πΌπβ1 π (π₯πβ1 ) + πΎπβ1 π΅πβ1 π₯πβ1 + πΏπβ1 ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅©
σ΅©σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅©2 2 σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β 2π2 (π½Μ2 β π
π2 ) σ΅©σ΅©σ΅©π΅2 π₯π β π΅2 πσ΅©σ΅©σ΅© σ΅© σ΅©2 β 2π1 (π½Μ1 β π
2 π1 ) σ΅©σ΅©σ΅©π΅1 π’π β π΅1 πσ΅©σ΅©σ΅© .
By Lemma 8, we have from (178) and (195) σ΅©σ΅© σ΅©2 σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© = σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π (π)) + π½π (π₯π β π)
πβ1
σ΅¨ σ΅¨ + π1 [βπ π + σ΅¨σ΅¨σ΅¨πΌπ β πΌπβ1 σ΅¨σ΅¨σ΅¨ π=0
σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ + σ΅¨σ΅¨σ΅¨πΎπ β πΎπβ1 σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨πΏπ β πΏπβ1 σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ ]
+ πΎπ (π΅π π₯π β π) + πΏπ (ππ πΊπ₯π β π) σ΅©2 + πΌπ (π (π) β π)σ΅©σ΅©σ΅©
σ΅© σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯πβ1 β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© , (189) Μ 2 )(βπ(π₯ )β+βπ΅ π₯ β+βπ πΊπ₯ β+π)} β€ where supπβ₯0 {(1/(1β π) π π π π π π1 for some π1 > 0. Thus, from (189), conditions (i), (iii) and the assumption on {ππ }, it follows that (noting that 0 < π π β€ π < 1, for all π β₯ 0) σ΅© σ΅© σ΅© σ΅© lim (σ΅©σ΅©π§ β π§πβ1 σ΅©σ΅©σ΅© β σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅©) β€ 0. (190) πββ σ΅© π Since 0 < lim inf π β β π½π β€ lim supπ β β π½π < 1, by Lemma 20 we get σ΅© σ΅© lim σ΅©σ΅©π₯ β π§π σ΅©σ΅©σ΅© = 0. (191) πββ σ΅© π Consequently, σ΅© σ΅© σ΅© σ΅© lim σ΅©σ΅©π₯ β π₯π σ΅©σ΅©σ΅© = lim (1 β π½π ) σ΅©σ΅©σ΅©π§π β π₯π σ΅©σ΅©σ΅© = 0. π β β σ΅© π+1 πββ
(194)
σ΅© β€ σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π (π)) + π½π (π₯π β π) σ΅©2 + πΎπ (π΅π π₯π β π) + πΏπ (ππ πΊπ₯π β π)σ΅©σ΅©σ΅© + 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β© σ΅© σ΅© σ΅©2 σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π (π)σ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅©2 + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β πσ΅©σ΅©σ΅© + 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β© σ΅© σ΅© σ΅©2 σ΅©2 β€ πΌπ π2 σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅©2 + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅©
(192)
+ 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β©
(195)
28
Abstract and Applied Analysis σ΅© σ΅© σ΅© σ΅©2 σ΅©2 σ΅©2 β€ πΌπ πσ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
which implies that
σ΅© σ΅©2 σ΅© σ΅©2 + πΏπ [σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β 2π2 (π½Μ2 β π
2 π2 ) σ΅©σ΅©σ΅©π΅2 π₯π β π΅2 πσ΅©σ΅©σ΅© σ΅© σ΅©2 β2π1 (π½Μ1 β π
2 π1 ) σ΅©σ΅©σ΅©π΅1 π’π β π΅1 πσ΅©σ΅©σ΅© ] + 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β© σ΅© σ΅©2 = (1 β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
σ΅©σ΅© σ΅©2 σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅©
σ΅© σ΅©2 +π1 (π½Μ1 β π
2 π1 ) σ΅©σ΅©σ΅©π΅1 π’π β π΅1 πσ΅©σ΅©σ΅© ]
σ΅© σ΅©2 = σ΅©σ΅©σ΅©Ξ πΆ (π’π β π1 π΅1 π’π ) β Ξ πΆ (π β π1 π΅1 π)σ΅©σ΅©σ΅©
+ 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β© σ΅©2 σ΅© σ΅©2 σ΅© β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β 2πΏπ [π2 (π½Μ2 β π
2 π2 ) σ΅©σ΅©σ΅©π΅2 π₯π β π΅2 πσ΅©σ΅©σ΅© σ΅© σ΅©2 + π1 (π½Μ1 β π
2 π1 ) σ΅©σ΅©σ΅©π΅1 π’π β π΅1 πσ΅©σ΅©σ΅© ] σ΅©σ΅© σ΅© σ΅© + 2πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© , which hence implies that σ΅© σ΅©2 2πΏπ [π2 (π½Μ2 β π
2 π2 ) σ΅©σ΅©σ΅©π΅2 π₯π β π΅2 πσ΅©σ΅©σ΅© σ΅© σ΅©2 + π1 (π½Μ1 β π
2 π1 ) σ΅©σ΅©σ΅©π΅1 π’π β π΅1 πσ΅©σ΅©σ΅© ] (197)
σ΅© σ΅© lim σ΅©σ΅©π΅ π’ β π΅1 πσ΅©σ΅©σ΅© = 0. (198) πββ σ΅© 1 π
Utilizing Proposition 6 and Lemma 9, we have
σ΅©σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅© σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© β π2 (σ΅©σ΅©σ΅©π’π β Vπ + (π β π)σ΅©σ΅©σ΅©) σ΅© σ΅©σ΅© σ΅© + 2π1 σ΅©σ΅©σ΅©π΅1 π β π΅1 π’π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© .
(201)
= β¨π₯π β π, π½ (π’π β π)β© + π2 β¨π΅2 π β π΅2 π₯π , π½ (π’π β π)β©
(199)
(202)
Substituting (200) for (202), we get σ΅©σ΅© σ΅©2 σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β π1 (σ΅©σ΅©σ΅©π₯π β π’π β (π β π)σ΅©σ΅©σ΅©) σ΅© σ΅©σ΅© σ΅© + 2π2 σ΅©σ΅©σ΅©π΅2 π β π΅2 π₯π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© σ΅© σ΅©σ΅© σ΅© + 2π1 σ΅©σ΅©σ΅©π΅1 π β π΅1 π’π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© .
β€ β¨π₯π β π2 π΅2 π₯π β (π β π2 π΅2 π) , π½ (π’π β π)β© 1 σ΅©σ΅© σ΅©2 σ΅© σ΅©2 [σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© 2 σ΅© σ΅© βπ1 (σ΅©σ΅©σ΅©π₯π β π’π β (π β π)σ΅©σ΅©σ΅©) ] σ΅© σ΅©σ΅© σ΅© + π2 σ΅©σ΅©σ΅©π΅2 π β π΅2 π₯π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© ,
1 σ΅©σ΅© σ΅©2 σ΅© σ΅©2 [σ΅©π’ β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© 2 σ΅© π σ΅© σ΅© βπ2 (σ΅©σ΅©σ΅©π’π β Vπ + (π β π)σ΅©σ΅©σ΅©) ] σ΅© σ΅©σ΅© σ΅© + π1 σ΅©σ΅©σ΅©π΅1 π β π΅1 π’π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© ,
σ΅© σ΅© β π2 (σ΅©σ΅©σ΅©π’π β Vπ + (π β π)σ΅©σ΅©σ΅©)
σ΅©σ΅© σ΅©2 σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© σ΅© σ΅©2 = σ΅©σ΅©σ΅©Ξ πΆ (π₯π β π2 π΅2 π₯π ) β Ξ πΆ (π β π2 π΅2 π)σ΅©σ΅©σ΅©
β€
= β¨π’π β π, π½ (Vπ β π)β© + π1 β¨π΅1 π β π΅1 π’π , π½ (Vπ β π)β©
which implies that
Since βπ₯π β π₯π+1 β β 0, 0 < ππ < π½Μπ /π
2 for π = 1, 2, and {π₯π } is bounded, we obtain from conditions (i), (ii) that σ΅© σ΅© lim σ΅©σ΅©π΅ π₯ β π΅2 πσ΅©σ΅©σ΅© = 0, πββ σ΅© 2 π
β€ β¨π’π β π1 π΅1 π’π β (π β π1 π΅1 π) , π½ (Vπ β π)β©
β€ (196)
σ΅©σ΅© σ΅© σ΅© + 2πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ (σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅©) σ΅©σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© + 2πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© .
(200)
In the same way, we derive
σ΅© σ΅©2 β 2πΏπ [π2 (π½Μ2 β π
2 π2 ) σ΅©σ΅©σ΅©π΅2 π₯π β π΅2 πσ΅©σ΅©σ΅©
σ΅©2 σ΅© σ΅©2 σ΅© β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅©
σ΅©2 σ΅© σ΅©2 σ΅© σ΅© σ΅©σ΅© σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β π1 (σ΅©σ΅©σ΅©π₯π β π’π β (π β π)σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅©σ΅© + 2π2 σ΅©σ΅©σ΅©π΅2 π β π΅2 π₯π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© .
By Lemma 8, we have from (196) and (203) σ΅©σ΅© σ΅©2 σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅©2 σ΅©2 σ΅©2 β€ πΌπ π2 σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅©2 + πΏπ σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© + 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β©
(203)
Abstract and Applied Analysis
29
σ΅© σ΅© σ΅© σ΅©2 σ΅©2 σ΅©2 β€ πΌπ πσ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
Utilizing the properties of π1 and π2 , we deduce that
σ΅© σ΅©2 σ΅© σ΅© + πΏπ [σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β π1 (σ΅©σ΅©σ΅©π₯π β π’π β (π β π)σ΅©σ΅©σ΅©)
+ 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β©
σ΅©σ΅©
σ΅© β π’π β (π β π)σ΅©σ΅©σ΅© = 0,
σ΅©σ΅©
σ΅© β Vπ + (π β π)σ΅©σ΅©σ΅© = 0.
lim σ΅©π₯ πββ σ΅© π
σ΅© σ΅© σ΅© σ΅© β π2 (σ΅©σ΅©σ΅©π’π β Vπ + (π β π)σ΅©σ΅©σ΅©) + 2π2 σ΅©σ΅©σ΅©π΅2 π β π΅2 π₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅©σ΅© σ΅© σ΅© Γ σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© + 2π1 σ΅©σ΅©σ΅©π΅1 π β π΅1 π’π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© ]
lim σ΅©π’ πββ σ΅© π
(207)
From (207), we get
σ΅© σ΅©2 β€ (1 β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
σ΅©σ΅© σ΅© σ΅© σ΅© σ΅©σ΅©π₯π β Vπ σ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©π₯π β π’π β (π β π)σ΅©σ΅©σ΅© σ΅© σ΅© + σ΅©σ΅©σ΅©π’π β Vπ + (π β π)σ΅©σ΅©σ΅© σ³¨β 0
σ΅© σ΅© β πΏπ [π1 (σ΅©σ΅©σ΅©π₯π β π’π β (π β π)σ΅©σ΅©σ΅©) σ΅© σ΅© + π2 (σ΅©σ΅©σ΅©π’π β Vπ + (π β π)σ΅©σ΅©σ΅©)] σ΅© σ΅©σ΅© σ΅© σ΅© σ΅© + 2π2 σ΅©σ΅©σ΅©π΅2 π β π΅2 π₯π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© + 2π1 σ΅©σ΅©σ΅©π΅1 π β π΅1 π’π σ΅©σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© Γ σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© + 2πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β πΏπ [π1 (σ΅©σ΅©σ΅©π₯π β π’π β (π β π)σ΅©σ΅©σ΅©) σ΅© σ΅© + π2 (σ΅©σ΅©σ΅©π’π β Vπ + (π β π)σ΅©σ΅©σ΅©)] σ΅© σ΅©σ΅© σ΅© σ΅© σ΅© + 2π2 σ΅©σ΅©σ΅©π΅2 π β π΅2 π₯π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© + 2π1 σ΅©σ΅©σ΅©π΅1 π β π΅1 π’π σ΅©σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© Γ σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© + 2πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© ,
as π β β. (208)
That is, σ΅©σ΅©
lim σ΅©π₯ πββ σ΅© π
σ΅© β πΊπ₯π σ΅©σ΅©σ΅© = 0.
(209)
Next, let us show that σ΅©σ΅©
lim σ΅©π πΊπ₯π πββ σ΅© π
σ΅© β π₯π σ΅©σ΅©σ΅© = 0,
σ΅©σ΅©
lim σ΅©π΅ π₯ πββ σ΅© π π
σ΅© β π₯π σ΅©σ΅©σ΅© = 0. (210)
(204) Indeed, observe that π₯π+1 can be rewritten as follows: which hence leads to
π₯π+1 = πΌπ π (π₯π ) + π½π π₯π + πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π
σ΅© σ΅© σ΅© σ΅© πΏπ [π1 (σ΅©σ΅©σ΅©π₯π β π’π β (π β π)σ΅©σ΅©σ΅©) + π2 (σ΅©σ΅©σ΅©π’π β Vπ + (π β π)σ΅©σ΅©σ΅©)] σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅© β σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅© σ΅© σ΅©σ΅© σ΅© + 2π2 σ΅©σ΅©σ΅©π΅2 π β π΅2 π₯π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© σ΅© σ΅©σ΅© σ΅© + 2π1 σ΅©σ΅©σ΅©π΅1 π β π΅1 π’π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© + 2πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ (σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅©) σ΅©σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅© σ΅© σ΅©σ΅© σ΅© + 2π2 σ΅©σ΅©σ΅©π΅2 π β π΅2 π₯π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π’π β πσ΅©σ΅©σ΅© σ΅© σ΅©σ΅© σ΅© + 2π1 σ΅©σ΅©σ΅©π΅1 π β π΅1 π’π σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©Vπ β πσ΅©σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© + 2πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© . σ΅©σ΅©2
= πΌπ π (π₯π ) + π½π π₯π + (πΎπ + πΏπ ) Γ
σ΅©σ΅©2
πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π πΎπ + πΏπ
(211)
= πΌπ π (π₯π ) + π½π π₯π + ππ π§Μπ , where ππ = πΎπ + πΏπ and π§Μπ = (πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π )/(πΎπ + πΏπ ). Utilizing Lemma 11 and (211), we have σ΅©2 σ΅©σ΅© σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© σ΅©2 = σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π) + π½π (π₯π β π) + ππ (Μπ§π β π)σ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© (205)
σ΅© σ΅© σ΅©2 σ΅© + ππ σ΅©σ΅©σ΅©π§Μπ β πσ΅©σ΅©σ΅© β π½π ππ π3 (σ΅©σ΅©σ΅©π§Μπ β π₯π σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅©2 σ΅©2 = πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
From (198), (205), conditions (i), (ii) and the boundedness of {π₯π }, {π’π }, and {Vπ }, we deduce that
lim π πββ 1
σ΅© σ΅© (σ΅©σ΅©σ΅©π₯π β π’π β (π β π)σ΅©σ΅©σ΅©) = 0,
lim π πββ 2
σ΅© σ΅© (σ΅©σ΅©σ΅©π’π β Vπ + (π β π)σ΅©σ΅©σ΅©) = 0.
(206)
σ΅©σ΅©2 σ΅©σ΅© πΎ π΅ π₯ + πΏ π πΊπ₯ σ΅© σ΅© σ΅© σ΅© π π π β π½π ππ π3 (σ΅©σ΅©σ΅©π§Μπ β π₯π σ΅©σ΅©σ΅©) + ππ σ΅©σ΅©σ΅© π π π β πσ΅©σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© πΎπ + πΏπ σ΅© σ΅© σ΅© σ΅©2 σ΅©2 σ΅© = πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β π½π ππ π3 (σ΅©σ΅©σ΅©π§Μπ β π₯π σ΅©σ΅©σ΅©) σ΅©σ΅©2 σ΅©σ΅©σ΅© πΎπ πΏπ σ΅© + ππ σ΅©σ΅©σ΅© (π΅π π₯π β π) + (ππ πΊπ₯π β π)σ΅©σ΅©σ΅© πΎπ + πΏπ σ΅©σ΅© σ΅©σ΅© πΎπ + πΏπ
30
Abstract and Applied Analysis σ΅© σ΅© σ΅© σ΅©2 σ΅© σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β π½π ππ π3 (σ΅©σ΅©σ΅©π§Μπ β π₯π σ΅©σ΅©σ΅©) + ππ [
β€
πΎπ σ΅©σ΅© πΏπ σ΅©σ΅© σ΅©2 σ΅©2 σ΅©π΅ π₯ β πσ΅©σ΅©σ΅© + σ΅©π πΊπ₯ β πσ΅©σ΅©σ΅© ] πΎπ + πΏπ σ΅© π π πΎπ + πΏπ σ΅© π π
β
σ΅© σ΅© σ΅© σ΅©2 σ΅©2 σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β π½π ππ π3 (σ΅©σ΅©σ΅©π§Μπ β π₯π σ΅©σ΅©σ΅©) β€
πΎπ σ΅©σ΅© πΏπ σ΅©σ΅© σ΅© σ΅©2 + ππ [ σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©πΊπ₯ β πσ΅©σ΅©σ΅© ] πΎπ + πΏπ πΎπ + πΏπ σ΅© π σ΅© σ΅© σ΅© σ΅©2 σ΅©2 σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β π½π ππ π3 (σ΅©σ΅©σ΅©π§Μπ β π₯π σ΅©σ΅©σ΅©) + ππ [
πΎπ σ΅©σ΅© πΏπ σ΅©σ΅© σ΅© σ΅©2 σ΅©π₯ β πσ΅©σ΅©σ΅© + σ΅©π₯ β πσ΅©σ΅©σ΅© ] πΎπ + πΏπ σ΅© π πΎπ + πΏπ σ΅© π
σ΅© σ΅© π4 (σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©)
πΎπ σ΅©σ΅© πΏπ σ΅©σ΅© σ΅©2 σ΅©2 σ΅©π₯ β πσ΅©σ΅©σ΅© + σ΅©π₯ β πσ΅©σ΅©σ΅© πΎπ + πΏπ σ΅© π πΎπ + πΏπ σ΅© π β
πΎπ πΏπ
2
(πΎπ + πΏπ )
σ΅© σ΅© π4 (σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©) πΎπ πΏπ
2
(πΎπ + πΏπ )
σ΅© σ΅© π4 (σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©) , (216)
which hence yields πΎπ πΏπ (πΎπ + πΏπ )
2
σ΅© σ΅© π4 (σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©) (217)
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ (σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π§Μπ β πσ΅©σ΅©σ΅©) σ΅©σ΅©σ΅©π₯π β π§Μπ σ΅©σ΅©σ΅© .
σ΅© σ΅© π½π ππ π3 (σ΅©σ΅©σ΅©π§Μπ β π₯π σ΅©σ΅©σ΅©) σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + (σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅©) σ΅© σ΅© Γ σ΅©σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅© .
(213)
σ΅© σ΅© lim π (σ΅©σ΅©π§Μ β π₯π σ΅©σ΅©σ΅©) = 0. πββ 3 σ΅© π
(214)
σ΅©σ΅©
σ΅© β π₯π σ΅©σ΅©σ΅© = 0.
(215)
Utilizing Lemma 15 and the definition of π§Μπ , we have σ΅©2 σ΅©σ΅©Μ σ΅©σ΅©π§π β πσ΅©σ΅©σ΅© σ΅©σ΅© πΎ π΅ π₯ + πΏ π πΊπ₯ σ΅©σ΅©2 σ΅© σ΅© π π π = σ΅©σ΅©σ΅© π π π β πσ΅©σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© πΎπ + πΏπ σ΅©σ΅© πΎ σ΅©σ΅©2 πΏπ σ΅© σ΅© π = σ΅©σ΅©σ΅© (π΅π π₯π β π) + (ππ πΊπ₯π β π)σ΅©σ΅©σ΅© σ΅©σ΅© πΎπ + πΏπ σ΅©σ΅© πΎπ + πΏπ πΎπ σ΅©σ΅© πΏπ σ΅©σ΅© σ΅©2 σ΅©2 σ΅©π΅ π₯ β πσ΅©σ΅©σ΅© + σ΅©π πΊπ₯ β πσ΅©σ΅©σ΅© πΎπ + πΏπ σ΅© π π πΎπ + πΏπ σ΅© π π 2
(πΎπ + πΏπ )
σ΅© σ΅© π4 (σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©)
(β ππ πΊπ₯π β π΅π π₯π β) = 0.
(218)
From the properties of π4 , we have σ΅©σ΅©
lim σ΅©π πΊπ₯π πββ σ΅© π
σ΅© β π΅π π₯π σ΅©σ΅©σ΅© = 0.
(219)
On the other hand, π₯π+1 can also be rewritten as follows:
From the properties of π3 , we have lim σ΅©π§Μ πββ σ΅© π
Since {π₯π } and {Μπ§π } are bounded and βΜπ§π β π₯π β β 0 as π β β, we deduce from condition (ii) that lim π πββ 4
Utilizing (184), conditions (i), (ii), (iv), and the boundedness of {π₯π } and {π(π₯π )}, we get
πΎπ πΏπ
2
(πΎπ + πΏπ )
σ΅© σ΅©2 σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β σ΅©σ΅©σ΅©π§Μπ β πσ΅©σ΅©σ΅©
which hence implies that
β
πΎπ πΏπ
σ΅© σ΅©2 = σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β
σ΅© σ΅©2 σ΅© σ΅©2 = πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + (1 β πΌπ ) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© β π½π ππ π3 (σ΅©σ΅©σ΅©π§Μπ β π₯π σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅©2 σ΅© σ΅©2 σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β π½π ππ π3 (σ΅©σ΅©σ΅©π§Μπ β π₯π σ΅©σ΅©σ΅©) , (212)
β€
πΎπ σ΅©σ΅© πΏπ σ΅©σ΅© σ΅©2 σ΅©2 σ΅©π₯ β πσ΅©σ΅©σ΅© + σ΅©πΊπ₯ β πσ΅©σ΅©σ΅© πΎπ + πΏπ σ΅© π πΎπ + πΏπ σ΅© π
π₯π+1 = πΌπ π (π₯π ) + π½π π₯π + πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π = π½π π₯π + πΎπ π΅π π₯π + (πΌπ + πΏπ )
πΌπ π (π₯π ) + πΏπ ππ πΊπ₯π πΌπ + πΏπ
= π½π π₯π + πΎπ π΅π π₯π + ππ π§Μπ , (220) where ππ = πΌπ + πΏπ and π§Μπ = (πΌπ π(π₯π ) + πΏπ ππ πΊπ₯π )/(πΌπ + πΏπ ). Utilizing Lemma 11 and the convexity of β β
β2 , we have σ΅©2 σ΅©σ΅© σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© σ΅©2 = σ΅©σ΅©σ΅©π½π (π₯π β π) + πΎπ (π΅π π₯π β π) + ππ (Μπ§π β π)σ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅©2 β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅© + ππ σ΅©σ΅©σ΅©π§Μπ β πσ΅©σ΅©σ΅© β π½π πΎπ π5 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©)
Abstract and Applied Analysis
31
σ΅© σ΅© σ΅©2 σ΅©2 = π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅©
which, together with (219), implies that
σ΅©σ΅©2 σ΅©σ΅© πΌ π (π₯ ) + πΏ π πΊπ₯ σ΅© σ΅© π π π π + ππ σ΅©σ΅©σ΅©σ΅© π β πσ΅©σ΅©σ΅©σ΅© πΌπ + πΏπ σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© β π½π πΎπ π5 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅©2 σ΅©2 = π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© σ΅©σ΅©2 σ΅©σ΅© πΌ πΏπ σ΅© σ΅© π + ππ σ΅©σ΅©σ΅© (π (π₯π ) β π) + (ππ πΊπ₯π β π)σ΅©σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© πΌπ + πΏπ πΌπ + πΏπ σ΅© σ΅© β π½π πΎπ π5 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅©2 σ΅©2 β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© πΌπ σ΅©σ΅© πΏπ σ΅©σ΅© σ΅©2 σ΅©2 σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + σ΅©π πΊπ₯ β πσ΅©σ΅©σ΅© ] πΌπ + πΏπ σ΅© πΌπ + πΏπ σ΅© π π σ΅© σ΅© β π½π πΎπ π5 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅©2 σ΅©2 β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + ππ [
πΌπ σ΅©σ΅© πΏπ σ΅©σ΅© σ΅©2 σ΅©2 σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + σ΅©πΊπ₯ β πσ΅©σ΅©σ΅© ] πΌπ + πΏπ πΌπ + πΏπ σ΅© π σ΅© σ΅© β π½π πΎπ π5 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©)
+ ππ [
σ΅© σ΅© σ΅©2 σ΅©2 β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© πΌπ σ΅©σ΅© πΏπ σ΅©σ΅© σ΅©2 σ΅©2 σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© ] πΌπ + πΏπ πΌπ + πΏπ σ΅©σ΅© σ΅©σ΅© β π½π πΎπ π5 (σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©) σ΅© σ΅©2 σ΅© σ΅©2 = πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + (1 β πΌπ ) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© β π½π πΎπ π5 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅©2 σ΅© σ΅©2 σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β π½π πΎπ π5 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©) , (221) + ππ [
as π β β. (225) That is, σ΅©σ΅©
lim σ΅©π πΊπ₯π πββ σ΅© π
σ΅© β π₯π σ΅©σ΅©σ΅© = 0.
(226)
We note that σ΅©σ΅© σ΅© σ΅©σ΅©π₯π β ππ₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β ππ πΊπ₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯π β ππ π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ π₯π β ππ₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β ππ πΊπ₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©πΊπ₯π β π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ π₯π β ππ₯π σ΅©σ΅©σ΅© . (227) So, in terms of (209), (226), and Lemma 12, we have σ΅©σ΅©
lim σ΅©π₯ πββ σ΅© π
σ΅© β ππ₯π σ΅©σ΅©σ΅© = 0.
(228)
Suppose that π½π β‘ π½ for some fixed π½ β (0, 1) such that πΌπ + π½ + πΎπ + πΏπ = 1 for all π β₯ 0. Define a mapping ππ₯ = (1 β π1 β π2 )ππ₯ + π1 π΅π₯ + π2 πΊπ₯, where π1 , π2 β (0, 1) are two constants with π1 + π2 < 1. Then by Lemmas 14 and 17, we have that Fix(π) = Fix(π) β© Fix(π΅) β© Fix(πΊ) = πΉ. For each π β₯ 1, let {ππ } be a unique element of πΆ such that ππ =
1 1 π (ππ ) + (1 β ) πππ . π π
(229)
From Lemma 13, we conclude that ππ β π β Fix(π) = πΉ as π β β. Observe that for every π, π
which hence implies that σ΅© σ΅© π½π πΎπ π5 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©) σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + (σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅©) σ΅© σ΅© Γ σ΅©σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅© .
(222)
σ΅© σ΅© (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©) = 0.
(223)
Utilizing the properties of π5 , we have σ΅©σ΅©
lim σ΅©π₯ πββ σ΅© π
σ΅© β π΅π π₯π σ΅©σ΅©σ΅© = 0,
σ΅©σ΅© σ΅© σ΅©σ΅©π₯π+1 β π΅ππ σ΅©σ΅©σ΅© σ΅© = σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π΅ππ ) + π½ (π₯π β π΅ππ )
σ΅© + πΎπ (π΅π π₯π β π΅ππ ) + πΏπ (ππ πΊπ₯π β π΅ππ )σ΅©σ΅©σ΅©
From (184), conditions (i), (ii), (iv), and the boundedness of {π₯π } and {π(π₯π )}, we have lim π πββ 5
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅©σ΅© σ΅©σ΅©ππ πΊπ₯π β π₯π σ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π π₯π β π₯π σ΅©σ΅©σ΅© σ³¨β 0
(224)
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π΅ππ σ΅©σ΅©σ΅© + π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© + πΏπ (σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π π₯π β π΅ππ σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅© σ΅© = πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π΅ππ σ΅©σ΅©σ΅© + π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© + (πΎπ + πΏπ ) σ΅©σ΅©σ΅©π΅π π₯π β π΅ππ σ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© = πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π΅ππ σ΅©σ΅©σ΅© + π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© + (1 β πΌπ β π½) σ΅©σ΅©σ΅©π΅π π₯π β π΅ππ σ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©
32
Abstract and Applied Analysis σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π΅ππ σ΅©σ΅©σ΅© + π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β πΌπ β π½)
σ΅©2 σ΅©σ΅© σ΅©σ΅©π₯π β πππ σ΅©σ΅©σ΅© σ΅©2 σ΅© β€ σ΅©σ΅©σ΅©π₯π β ππ₯π + ππ₯π β πππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅© β€ (σ΅©σ΅©σ΅©π₯π β ππ₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅©) σ΅©2 σ΅© σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β ππ₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© Γ (2 σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β ππ₯π σ΅©σ΅©σ΅©) .
σ΅© σ΅© σ΅© σ΅© Γ [σ΅©σ΅©σ΅©π΅π π₯π β π΅π ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π ππ β π΅ππ σ΅©σ΅©σ΅©]
σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π΅ππ σ΅©σ΅©σ΅© + π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© + (1 β πΌπ β π½) [σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π ππ β π΅ππ σ΅©σ΅©σ΅©] σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π΅ππ σ΅©σ΅©σ΅© + π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© + (1 β π½) [σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π΅π ππ β π΅ππ σ΅©σ΅©σ΅©] σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© = ππ + π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β π½) σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© ,
(233) It is easy to see from (209) and (228) that σ΅© σ΅©2 σ΅© σ΅©2 ππ σ΅©σ΅©σ΅©π₯π β πΊππ σ΅©σ΅©σ΅© β€ ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© , σ΅© σ΅©2 σ΅© σ΅©2 ππ σ΅©σ΅©σ΅©π₯π β πππ σ΅©σ΅©σ΅© β€ ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© . (230)
where ππ = πΌπ βπ(π₯π ) β π΅ππ β + (1 β π½)βπ΅π ππ β π΅ππ β + πΏπ βππ πΊπ₯π βπ΅π π₯π β. Since limπ β β πΌπ = limπ β β βπ΅π ππ βπ΅ππ β = limπ β β βππ πΊπ₯π β π΅π π₯π β = 0, we know that ππ β 0 as π β β. From (230), we obtain σ΅©2 σ΅©σ΅© σ΅©σ΅©π₯π+1 β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅© β€ (π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β π½) σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅© σ΅© + ππ [2 (π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β π½) σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅©) + ππ ] 2σ΅© σ΅© σ΅©2 σ΅©2 = π½2 σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β π½) σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© + 2π½ (1 β π½) σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + ππ 2σ΅© σ΅© σ΅©2 σ΅©2 β€ π½2 σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β π½) σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© σ΅©2 σ΅© σ΅©2 σ΅© + π½ (1 β π½) (σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© ) + ππ σ΅©2 σ΅©2 σ΅© σ΅© = π½σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β π½) σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + ππ ,
In addition, note that σ΅©σ΅© σ΅©2 σ΅©σ΅©π₯π β πΊππ σ΅©σ΅©σ΅© σ΅©2 σ΅© β€ σ΅©σ΅©σ΅©π₯π β πΊπ₯π + πΊπ₯π β πΊππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅© β€ (σ΅©σ΅©σ΅©π₯π β πΊπ₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅©) σ΅©2 σ΅© σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πΊπ₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© Γ (2 σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πΊπ₯π σ΅©σ΅©σ΅©) ,
Utilizing (232) and (234), we deduce that σ΅© σ΅©2 ππ σ΅©σ΅©σ΅©π₯π β πππ σ΅©σ΅©σ΅© σ΅© = ππ σ΅©σ΅©σ΅©(1 β π1 β π2 ) (π₯π β πππ ) σ΅©2 + π1 (π₯π β π΅ππ ) + π2 (π₯π β πΊππ )σ΅©σ΅©σ΅© σ΅© σ΅©2 β€ (1 β π1 β π2 ) ππ σ΅©σ΅©σ΅©π₯π β πππ σ΅©σ΅©σ΅©
(235)
σ΅© σ΅©2 σ΅© σ΅©2 + π1 ππ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + π2 ππ σ΅©σ΅©σ΅©π₯π β πΊππ σ΅©σ΅©σ΅© σ΅© σ΅©2 β€ ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© . Also, observe that 1 1 (π₯ β π (ππ )) + (1 β ) (π₯π β πππ ) ; π π π
(236)
1 1 (1 β ) (π₯π β πππ ) = π₯π β ππ β (π₯π β π (ππ )) . π π
(237)
π₯π β ππ = that is,
It follows from Lemma 8(ii) and (237) that (231)
where ππ = ππ [2(π½βπ₯π β π΅ππ β + (1 β π½)βπ₯π β ππ β) + ππ ] β 0 as π β β. For any Banach limit π, from (231) we derive σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅©2 ππ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© = ππ σ΅©σ΅©σ΅©π₯π+1 β π΅ππ σ΅©σ΅©σ΅© β€ ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© .
(234)
(232)
2
1 σ΅© σ΅©2 (1 β ) σ΅©σ΅©σ΅©π₯π β πππ σ΅©σ΅©σ΅© π σ΅©2 2 σ΅© β₯ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© β β¨π₯π β ππ + ππ β π (ππ ) , π½ (π₯π β ππ )β© π 2 σ΅© σ΅©2 2 = (1 β ) σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + β¨π (ππ ) β ππ , π½ (π₯π β ππ )β© . π π (238) So by (235) and (238), we have 1 2 σ΅© σ΅©2 (1 β ) ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© π 2 σ΅© σ΅©2 β₯ (1 β ) ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© π 2 + ππ β¨π (ππ ) β ππ , π½ (π₯π β ππ )β© , π
(239)
Abstract and Applied Analysis
33 σ΅© σ΅© σ΅©2 σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π (π)σ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
and hence 1 σ΅©σ΅© σ΅©2 2 ππ σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© β₯ ππ β¨π (ππ ) β ππ , π½ (π₯π β ππ )β© . (240) 2 π π
(241)
Since ππ β π β Fix(π) = πΉ as π β β, by the uniform Frechet differentiability of the norm of π, we have ππ β¨π (π) β π, π½ (π₯π β π)β© β€ 0.
(242)
On the other hand, from (184) and the norm-to-norm uniform continuity of π½ on bounded subsets of π, it follows that σ΅¨ σ΅¨ lim σ΅¨σ΅¨β¨π (π) β π, π½ (π₯π+1 β π)β© β β¨π (π) β π, π½ (π₯π β π)β©σ΅¨σ΅¨σ΅¨ = 0. (243)
πββ σ΅¨
So, utilizing Lemma 18 we deduce from (242) and (243) that lim sup β¨π (π) β π, π½ (π₯π β π)β© β€ 0, πββ
(244)
which, together with (184) and the norm-to-norm uniform continuity of π½ on bounded subsets of π, implies that lim sup β¨π (π) β π, π½ (π₯π+1 β π)β© β€ 0. πββ
(245)
Finally, let us show that π₯π β π as π β β. Utilizing Lemma 8 (i), from (178) and the convexity of β β
β2 , we get σ΅©σ΅© σ΅©2 σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© = σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π (π)) + π½π (π₯π β π) + πΎπ (π΅π π₯π β π) σ΅©2 + πΏπ (ππ πΊπ₯π β π) + πΌπ (π (π) β π)σ΅©σ΅©σ΅© σ΅© β€ σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π (π)) + π½π (π₯π β π) σ΅©2 + πΎπ (π΅π π₯π β π) + πΏπ (ππ πΊπ₯π β π)σ΅©σ΅©σ΅© + 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β© σ΅© σ΅© σ΅©2 σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π (π)σ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅©2 + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β πσ΅©σ΅©σ΅© + 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β©
+ 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β© σ΅© σ΅© σ΅© σ΅©2 σ΅©2 σ΅©2 β€ πΌπ πσ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
This implies that 1 σ΅©σ΅© σ΅©2 π σ΅©π₯ β ππ σ΅©σ΅©σ΅© β₯ ππ β¨π (ππ ) β ππ , π½ (π₯π β ππ )β© . 2π π σ΅© π
σ΅© σ΅© σ΅©2 σ΅©2 + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©πΊπ₯π β πσ΅©σ΅©σ΅©
σ΅© σ΅©2 + πΏπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β© σ΅© σ΅©2 = (1 β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β© σ΅© σ΅©2 = (1 β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΌπ (1 β π)
2 β¨π (π) β π, π½ (π₯π+1 β π)β© . 1βπ (246)
Applying Lemma 7 to (246), we obtain that π₯π β π as π β β. This completes the proof. Corollary 32. Let πΆ be a nonempty closed convex subset of a uniformly convex and 2-uniformly smooth Banach space π. Let Ξ πΆ be a sunny nonexpansive retraction from π onto πΆ. Let {ππ }β π=0 be a sequence of positive numbers in (0, π] for some π β (0, 1) and π΄ π : πΆ β π an πΌΜπ -inverse strongly accretive mapping for each π = 0, 1, . . .. Define a mapping πΊπ : πΆ β πΆ by Ξ πΆ(πΌ β π π π΄ π )π₯ = πΊπ π₯ for all π₯ β πΆ and π = 0, 1, . . ., where π π β (0, πΌΜπ /π
2 ] and π
is the 2-uniformly smooth constant of π. Let π΅π : πΆ β πΆ be the π-mapping generated by πΊπ , πΊπβ1 , . . . , πΊ0 , and ππ , ππβ1 , . . . , π0 . Let π : πΆ β πΆ be an πΌ-strictly pseudocontractive mapping. Let π : πΆ β πΆ be a contraction with coefficient π β (0, 1). Let {ππ }β π=0 be a countable family of nonexpansive mappings of πΆ into itself such β that πΉ = (ββ π=0 Fix(ππ )) β© Fix(π) β© (βπ=0 VI(πΆ, π΄ π )) =ΜΈ 0. For arbitrarily given π₯0 β πΆ, let {π₯π } be the sequence generated by π₯π+1 = πΌπ π (π₯π ) + π½π π₯π + πΎπ π΅π π₯π + πΏπ ππ ((1 β π) πΌ + ππ) π₯π ,
βπ β₯ 0,
(247)
where 0 < π < πΌ/π
2 and {πΌπ }, {π½π }, {πΎπ }, and {πΏπ } are the sequences in (0, 1) such that πΌπ + π½π + πΎπ + πΏπ = 1 for all π β₯ 0. Suppose that the following conditions hold: (i) limπ β β πΌπ = 0 and ββ π=0 πΌπ = β; (ii) {πΎπ }, {πΏπ } β [π, π] for some π, π β (0, 1); (iii) limπ β β (|π½π β π½πβ1 | + |πΎπ β πΎπβ1 | + |πΏπ β πΏπβ1 |) = 0; (iv) 0 < lim inf π β β π½π β€ lim supπ β β π½π < 1. Assume that ββ π=1 supπ₯βπ· βππ π₯ β ππβ1 π₯β < β for any bounded subset π· of πΆ and let π be a mapping of πΆ into itself defined
34
Abstract and Applied Analysis
by ππ₯ = limπ β β ππ π₯ for all π₯ β πΆ and suppose that Fix(π) = ββ π=0 Fix(ππ ). Then, there hold the following: (I) limπ β β βπ₯π+1 β π₯π β = 0;
(II) the sequence {π₯π }β π=0 converges strongly to some π β πΉ which is the unique solution of the variational inequality problem (VIP) β¨(πΌ β π) π, π½ (π β π)β© β€ 0,
βπ β πΉ,
(248)
Proof. In Theorem 31, we put π΅1 = πΌ β π, π΅2 = 0 and π1 = π where 0 < π < πΌ/π
2 . Then GSVI (13) is equivalent to the VIP of finding π₯β β πΆ such that βπ₯ β πΆ.
(249)
In this case, π΅1 : πΆ β π is πΌ-inverse strongly accretive. Repeating the same arguments as those in the proof of Corollary 25, we can infer that Fix(π) = VI(πΆ, π΅1 ). β1 Accordingly, we know that πΉ = ββ π=0 Fix(ππ ) β© Ξ© β© π΄ 0 = β β1 βπ=0 Fix(ππ ) β© Fix(π) β© π΄ 0, and Ξ πΆ (πΌ β π1 π΅1 ) Ξ πΆ (πΌ β π2 π΅2 ) π₯π = Ξ πΆ (πΌ β π1 π΅1 ) π₯π = Ξ πΆ ((1 β π) π₯π + πππ₯π )
(i) limπ β β πΌπ = 0 and ββ π=0 πΌπ = β; (ii) {πΎπ }, {πΏπ } β [π, π] for some π, π β (0, 1); (iii) ββ π=1 (|ππ β ππβ1 | + |πΌπ β πΌπβ1 | + |π½π β π½πβ1 | + |πΎπ β πΎπβ1 | + |πΏπ β πΏπβ1 |) < β;
provided π½π β‘ π½ for some fixed π½ β (0, 1).
β¨π΅1 π₯β , π½ (π₯ β π₯β )β© β₯ 0,
where {πΌπ }, {π½π }, {πΎπ }, {πΏπ }, and {ππ } are the sequences in (0, 1) such that πΌπ + π½π + πΎπ + πΏπ = 1 for all π β₯ 0. Suppose that the following conditions hold:
(250)
(iv) 0 < lim inf π β β π½π β€ lim supπ β β π½π < 1 and 0 < lim inf π β β ππ β€ lim supπ β β ππ < 1. Assume that ββ π=1 supπ₯βπ· βππ π₯ β ππβ1 π₯β < β for any bounded subset π· of πΆ and let π be a mapping of πΆ into itself defined by ππ₯ = limπ β β ππ π₯ for all π₯ β πΆ and suppose that Fix(π) = ββ π=0 Fix(ππ ). Then, there hold the following: (I) limπ β β βπ₯π+1 β π₯π β = 0; (II) the sequence {π₯π }β π=0 converges strongly to some π β πΉ which is the unique solution of the variational inequality problem (VIP) β¨(πΌ β π) π, π½ (π β π)β© β€ 0,
= ((1 β π) πΌ + ππ) π₯π .
βπ β πΉ,
(252)
So, scheme (178) reduces to (247). Therefore, the desired result follows from Theorem 31. Theorem 33. Let πΆ be a nonempty closed convex subset of a uniformly convex Banach space π which has a uniformly Gateaux differentiable norm. Let Ξ πΆ be a sunny nonexpansive retraction from π onto πΆ. Let {ππ }β π=0 be a sequence of positive numbers in (0, π] for some π β (0, 1) and π΄ π : πΆ β πππ strictly pseudocontractive and πΌΜπ -strongly accretive with ππ + πΌΜπ β₯ 1 for each π = 0, 1, . . .. Define a mapping πΊπ : πΆ β πΆ by Ξ πΆ(πΌ β π π π΄ π )π₯ = πΊπ π₯ for all π₯ β πΆ and π = 0, 1, . . ., where 1β(ππ /(1+ππ ))(1ββ(1 β πΌΜπ )/ππ ) β€ π π β€ 1 for all π = 0, 1, . . .. Let π΅π : πΆ β πΆ be the π-mapping generated by πΊπ , πΊπβ1 , . . . , πΊ0 and ππ , ππβ1 , . . . , π0 . Let the mapping π΅π : πΆ β π be ππ -strictly pseudocontractive and π½Μπ -strongly accretive with ππ + π½Μπ β₯ 1 for π = 1, 2. Let π : πΆ β πΆ be a contraction with coefficient π β (0, 1). Let {ππ }β π=0 be a countable family of nonexpansive mappings of πΆ into itself such that πΉ = (ββ π=0 Fix(ππ )) β© VI(πΆ, π΄ )) = ΜΈ 0, where Ξ© is the fixed point set of the Ξ© β© (ββ π π=0 mapping πΊ = Ξ πΆ(πΌ β π1 π΅1 )Ξ πΆ(πΌ β π2 π΅2 ) with 1 β (ππ /(1 + π ))(1 β β(1 β π½Μ )/π ) β€ π β€ 1 for π = 1, 2. For arbitrarily π
π
π
π
given π₯0 β πΆ, let {π₯π } be the sequence generated by
Proof. First of all, it is easy to see that (251) can be rewritten as follows: π¦π = πΌπ π (π₯π ) + π½π π₯π + πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π , π₯π₯+1 = ππ πΊπ₯π + (1 β ππ ) π¦π ,
βπ β₯ 0.
βπ β₯ 0, (251)
(253)
Take a fixed π β πΉ arbitrarily. Then, we obtain π = πΊπ, π = π΅π π and ππ π = π for all π β₯ 0. Thus, we get from (253) σ΅© σ΅©σ΅© σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ (π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅©)
π₯π+1 = ππ πΊπ₯π + (1 β ππ ) Γ [πΌπ π (π₯π ) + π½π π₯π + πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π ] ,
provided π½π β‘ π½ for some fixed π½ β (0, 1).
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© = (1 β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© ,
(254)
Abstract and Applied Analysis
35
and hence σ΅©σ΅© σ΅© σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ ππ σ΅©σ΅©σ΅©πΊπ₯π β πσ΅©σ΅©σ΅© + (1 β ππ ) σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© σ΅© σ΅© β€ ππ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + (1 β ππ ) σ΅© σ΅© σ΅© σ΅© Γ [(1 β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅©] σ΅© σ΅© = (1 β (1 β ππ ) πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© + (1 β ππ ) πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅© σ΅© = (1 β (1 β ππ ) πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅©σ΅© σ΅©π (π) β πσ΅©σ΅©σ΅© + (1 β ππ ) πΌπ (1 β π) σ΅© 1βπ σ΅© σ΅© σ΅© σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅© β€ max {σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© , σ΅© }. 1βπ (255)
σ΅©σ΅© π¦ β π½ π₯ π¦ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅©σ΅© σ΅© π π β πβ1 β€ σ΅©σ΅©σ΅© π σ΅©σ΅© σ΅©σ΅© 1 β π½π σ΅©σ΅© 1 β π½π σ΅©σ΅© π¦ β π½ π₯ π¦ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅©σ΅© σ΅© πβ1 πβ1 + σ΅©σ΅©σ΅© πβ1 β πβ1 σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© 1 β π½π 1 β π½πβ1 =
=
=
1 σ΅©σ΅© σ΅© σ΅©π¦ β π½π π₯π β (π¦πβ1 β π½πβ1 π₯πβ1 )σ΅©σ΅©σ΅© 1 β π½π σ΅© π σ΅¨σ΅¨ 1 σ΅¨σ΅¨ 1 σ΅¨ σ΅¨σ΅¨ σ΅©σ΅© σ΅© + σ΅¨σ΅¨σ΅¨ β σ΅¨ σ΅©π¦ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅© σ΅¨σ΅¨ 1 β π½π 1 β π½πβ1 σ΅¨σ΅¨σ΅¨ σ΅© πβ1 1 σ΅©σ΅© σ΅© σ΅©π¦ β π½π π₯π β (π¦πβ1 β π½πβ1 π₯πβ1 )σ΅©σ΅©σ΅© 1 β π½π σ΅© π σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ σ΅© σ΅©σ΅© + σ΅©π¦ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅© (1 β π½πβ1 ) (1 β π½π ) σ΅© πβ1 1 σ΅©σ΅© σ΅©πΌ π (π₯π ) + πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π 1 β π½π σ΅© π β πΌπβ1 π (π₯πβ1 ) β πΎπβ1 π΅πβ1 π₯πβ1 σ΅© βπΏπβ1 ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅©
By induction, we have σ΅© σ΅© σ΅© σ΅© σ΅© σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅©σ΅© }, σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β€ max {σ΅©σ΅©σ΅©π₯0 β πσ΅©σ΅©σ΅© , σ΅© 1βπ
+ βπ β₯ 0. (256)
which implies that {π₯π } is bounded and so are the sequences {π¦π }, {πΊπ₯π } and {π(π₯π )}. Let us show that σ΅©σ΅©
lim σ΅©π₯ π β β σ΅© π+1
σ΅© β π₯π σ΅©σ΅©σ΅© = 0.
(257)
As a matter of fact, observe that π¦π can be rewritten as follows: π¦π = π½π π₯π + (1 β π½π ) π§π ,
(258)
where π§π = (πΌπ π(π₯π ) + πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π )/(1 β π½π ). Observe that σ΅©σ΅© σ΅© σ΅©σ΅©π§π β π§πβ1 σ΅©σ΅©σ΅© σ΅©σ΅© πΌ π (π₯ ) + πΎ π΅ π₯ + πΏ π πΊπ₯ σ΅© π π π π π π π = σ΅©σ΅©σ΅©σ΅© π 1 β π½π σ΅©σ΅© β
πΌπβ1 π (π₯πβ1 ) + πΎπβ1 π΅πβ1 π₯πβ1 + πΏπβ1 ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© 1 β π½πβ1 σ΅©
σ΅©σ΅© π¦ β π½ π₯ π¦ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅©σ΅© σ΅© π π = σ΅©σ΅©σ΅© π β πβ1 σ΅©σ΅© σ΅©σ΅© 1 β π½π σ΅©σ΅© 1 β π½πβ1 σ΅©σ΅© π¦ β π½ π₯ π¦ β π½πβ1 π₯πβ1 σ΅© π π = σ΅©σ΅©σ΅© π β πβ1 1 β π½π σ΅©σ΅© 1 β π½π +
π¦πβ1 β π½πβ1 π₯πβ1 π¦πβ1 β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅©σ΅© β σ΅©σ΅© σ΅©σ΅© 1 β π½π 1 β π½πβ1
β€
σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ σ΅© σ΅©σ΅© σ΅©σ΅©π¦πβ1 β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅© (1 β π½πβ1 ) (1 β π½π )
1 σ΅© σ΅© [πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π (π₯πβ1 )σ΅©σ΅©σ΅© 1 β π½π σ΅© σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅© σ΅© σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΌπ β πΌπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π (π₯πβ1 )σ΅©σ΅©σ΅© σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΎπ β πΎπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅© σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΏπ β πΏπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅©] σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ σ΅© σ΅©σ΅© + σ΅©σ΅©π¦πβ1 β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅© . (1 β π½ ) (1 β π½ ) πβ1
π
(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 π β₯ 1 σ΅©σ΅© σ΅© σ΅©σ΅©ππ πΊπ₯π β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯πβ1 β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© , πβ1
(260)
σ΅©σ΅© σ΅© σ΅© σ΅© σ΅©σ΅©π΅π π₯π β π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + πβπ π , π=0
for some constant π > 0. Taking into account 0 < lim inf π β β π½π β€ lim supπ β β π½π < 1, we may assume,
36
Abstract and Applied Analysis
Μ Utilizing (259)without loss of generality, that {π½π } β [Μπ, π]. (260) we have σ΅© σ΅©σ΅© σ΅©σ΅©π§π β π§πβ1 σ΅©σ΅©σ΅© β€
σ΅¨σ΅© σ΅¨ σ΅© Γ [σ΅¨σ΅¨σ΅¨πΌπ β πΌπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π (π₯πβ1 )σ΅©σ΅©σ΅©
σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΎπ β πΎπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅©
σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ (1 β π½πβ1 ) (1 β π½π ) σ΅© σ΅©σ΅© Γ σ΅©σ΅©πΌπβ1 π (π₯πβ1 ) + πΎπβ1 π΅πβ1 π₯πβ1 + πΏπβ1 ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΏπ β πΏπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅©] +
1 1 β π½π πβ1
σ΅© σ΅© σ΅© σ΅© Γ {πΌπ π σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + πΎπ [σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + πβπ π ]
β€ (1 β
π=0
σ΅© σ΅© σ΅© σ΅© + πΏπ [σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯πβ1 β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅©]
πβ1
σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ + π1 [βπ π + σ΅¨σ΅¨σ΅¨πΌπ β πΌπβ1 σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨πΎπ β πΎπβ1 σ΅¨σ΅¨σ΅¨ π=0
σ΅¨ σ΅¨σ΅© σ΅¨σ΅© σ΅© σ΅© σ΅¨ + σ΅¨σ΅¨σ΅¨πΌπ β πΌπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π (π₯πβ1 )σ΅©σ΅©σ΅© + σ΅¨σ΅¨σ΅¨πΎπ β πΎπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅©
σ΅¨ σ΅© σ΅¨ σ΅© + σ΅¨σ΅¨σ΅¨πΏπ β πΏπβ1 σ΅¨σ΅¨σ΅¨ ] + σ΅©σ΅©σ΅©ππ πΊπ₯πβ1 β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© ,
σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΏπ β πΏπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© } +
(261)
σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ σ΅© σ΅©σ΅© σ΅©π¦ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅© (1 β π½πβ1 ) (1 β π½π ) σ΅© πβ1
1 σ΅© σ΅© = { (1 β π½π β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© 1 β π½π πβ1
σ΅© σ΅© + πΎπ πβπ π + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯πβ1 β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© π=0
σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΌπ β πΌπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π (π₯πβ1 )σ΅©σ΅©σ΅© σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΎπ β πΎπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅© σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨πΏπ β πΏπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© } σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ σ΅© σ΅©σ΅© + σ΅©π¦ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅© (1 β π½πβ1 ) (1 β π½π ) σ΅© πβ1 = (1 β
πΌπ (1 β π) σ΅©σ΅© σ΅© ) σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© 1 β π½π
Μ 2 )(βπ(π₯ )β + βπ΅ π₯ β + βπ πΊπ₯ β + where supπβ₯0 {(1/(1 β π) π π π π π π)} β€ π1 for some π1 > 0. In the meantime, observe that π₯π+1 β π₯π = ππ (πΊπ₯π β πΊπ₯πβ1 ) + (ππ β ππβ1 ) (πΊπ₯πβ1 β π§πβ1 ) + (1 β ππ ) (π§π β π§πβ1 ) . (262) This together with (261), implies that σ΅©σ΅© σ΅© σ΅©σ΅©π₯π+1 β π₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅¨ σ΅¨σ΅© σ΅© β€ ππ σ΅©σ΅©σ΅©πΊπ₯π β πΊπ₯πβ1 σ΅©σ΅©σ΅© + σ΅¨σ΅¨σ΅¨ππ β ππβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©πΊπ₯πβ1 β π§πβ1 σ΅©σ΅©σ΅© σ΅© σ΅© + (1 β ππ ) σ΅©σ΅©σ΅©π§π β π§πβ1 σ΅©σ΅©σ΅© σ΅© σ΅© σ΅¨ σ΅¨σ΅© σ΅© β€ ππ σ΅©σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + σ΅¨σ΅¨σ΅¨ππ β ππβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©πΊπ₯πβ1 β π§πβ1 σ΅©σ΅©σ΅© + (1 β ππ ) {(1 β
πβ1
σ΅¨ σ΅¨ σ΅¨ σ΅¨ + π1 [βπ π + σ΅¨σ΅¨σ΅¨πΌπ β πΌπβ1 σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨
πβ1
πΌπ (1 β π) σ΅©σ΅© σ΅© πΎπ ) σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + π βπ π 1 β π½π 1 β π½π π=0
π=0
σ΅¨ σ΅¨ σ΅¨ σ΅¨ + σ΅¨σ΅¨σ΅¨πΎπ β πΎπβ1 σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨πΏπ β πΏπβ1 σ΅¨σ΅¨σ΅¨ ]
πΏπ σ΅©σ΅© σ΅© + σ΅©π πΊπ₯ β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© 1 β π½π σ΅© π πβ1 1 σ΅¨σ΅© σ΅¨ σ΅¨ σ΅© σ΅¨ [σ΅¨σ΅¨πΌ β πΌπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π (π₯πβ1 )σ΅©σ΅©σ΅© + σ΅¨σ΅¨σ΅¨πΎπ β πΎπβ1 σ΅¨σ΅¨σ΅¨ 1 β π½π σ΅¨ π σ΅© σ΅© σ΅¨ σ΅¨σ΅© σ΅© Γ σ΅©σ΅©σ΅©π΅πβ1 π₯πβ1 σ΅©σ΅©σ΅© + σ΅¨σ΅¨σ΅¨πΏπ β πΏπβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅©] σ΅¨σ΅¨ σ΅¨ σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©π¦ β π½πβ1 π₯πβ1 σ΅©σ΅©σ΅© + σ΅© (1 β π½ ) (1 β π½ ) σ΅© πβ1
σ΅© σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯πβ1 β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© }
+
πβ1
π
πβ1 πΌ (1 β π) σ΅©σ΅© σ΅© β€ (1 β π ) σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© + πβπ π 1 β π½π π=0
σ΅© σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯πβ1 β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© +
1 1 β π½π
πΌπ (1 β π) σ΅©σ΅© σ΅© ) σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© 1 β π½π
β€ (1 β
(1 β ππ ) πΌπ (1 β π) σ΅©σ΅© σ΅© ) σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© 1 β π½π
σ΅¨ σ΅¨σ΅© σ΅© + σ΅¨σ΅¨σ΅¨ππ β ππβ1 σ΅¨σ΅¨σ΅¨ σ΅©σ΅©σ΅©πΊπ₯πβ1 β π§πβ1 σ΅©σ΅©σ΅© πβ1
σ΅¨ σ΅¨ σ΅¨ σ΅¨ + π1 [βπ π + σ΅¨σ΅¨σ΅¨πΌπ β πΌπβ1 σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ π=0
σ΅¨ σ΅¨ σ΅¨ σ΅¨ + σ΅¨σ΅¨σ΅¨πΎπ β πΎπβ1 σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨πΏπ β πΏπβ1 σ΅¨σ΅¨σ΅¨ ]
Abstract and Applied Analysis
37
σ΅© σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯πβ1 β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© β€ (1 β
Utilizing Lemma 15 we get from (253) and (265)
(1 β ππ ) πΌπ (1 β π) σ΅©σ΅© σ΅© ) σ΅©σ΅©π₯π β π₯πβ1 σ΅©σ΅©σ΅© 1 β π½π πβ1
σ΅¨ σ΅¨ σ΅¨ σ΅¨ + π2 [βπ π + σ΅¨σ΅¨σ΅¨ππ β ππβ1 σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨πΌπ β πΌπβ1 σ΅¨σ΅¨σ΅¨ π=0
σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ σ΅¨ + σ΅¨σ΅¨σ΅¨π½π β π½πβ1 σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨πΎπ β πΎπβ1 σ΅¨σ΅¨σ΅¨ + σ΅¨σ΅¨σ΅¨πΏπ β πΏπβ1 σ΅¨σ΅¨σ΅¨ ] σ΅© σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯πβ1 β ππβ1 πΊπ₯πβ1 σ΅©σ΅©σ΅© , (263)
where supπβ₯0 {π1 + βπΊπ₯π β π§π β} β€ π2 for some π2 > 0. Since ββ π=0 πΌπ = β and (1βππ )πΌπ (1βπ)/(1βπ½π ) β₯ (1βππ )πΌπ (1βπ), we obtain from conditions (i) and (iv) that ββ π=0 ((1βππ )πΌπ (1β π)/(1 β π½π )) = β. Thus, applying Lemma 7 to (263), we deduce from condition (iii) and the assumption on {ππ } that (noting that 0 < π π β€ π < 1, for all π β₯ 0) σ΅©σ΅©
lim σ΅©π₯ π β β σ΅© π+1
σ΅© β π₯π σ΅©σ΅©σ΅© = 0.
σ΅©2 σ΅©σ΅© σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© σ΅©2 = σ΅©σ΅©σ΅©ππ (πΊπ₯π β π) + (1 β ππ ) (π¦π β π)σ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅©2 β€ ππ σ΅©σ΅©σ΅©πΊπ₯π β πσ΅©σ΅©σ΅© + (1 β ππ ) σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© σ΅© σ΅© β ππ (1 β ππ ) π (σ΅©σ΅©σ΅©πΊπ₯π β π¦π σ΅©σ΅©σ΅©) σ΅©2 σ΅© β€ ππ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + (1 β ππ ) σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅© Γ [σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + 2πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅©] σ΅© σ΅© β ππ (1 β ππ ) π (σ΅©σ΅©σ΅©πΊπ₯π β π¦π σ΅©σ΅©σ΅©) σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅© β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + 2πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© σ΅© β ππ (1 β ππ ) π (σ΅©σ΅©σ΅©πΊπ₯π β π¦π σ΅©σ΅©σ΅©) , which hence yields σ΅© σ΅© ππ (1 β ππ ) π (σ΅©σ΅©σ΅©πΊπ₯π β π¦π σ΅©σ΅©σ΅©) σ΅© σ΅©2 σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅©
σ΅©σ΅© σ΅© σ΅© + 2πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅©
(264)
(267)
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ (σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅©) σ΅©σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅©
Next, we show that βπ₯π β πΊπ₯π β β 0 as π β β. Indeed, according to Lemma 8 we have from (253)
σ΅©σ΅© σ΅© σ΅© + 2πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© .
σ΅©2 σ΅©σ΅© σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© σ΅© = σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π (π)) + π½π (π₯π β π) + πΎπ (π΅π π₯π β π) σ΅©2 + πΏπ (ππ πΊπ₯π β π) + πΌπ (π (π) β π)σ΅©σ΅©σ΅© σ΅© β€ σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π (π)) + π½π (π₯π β π) σ΅©2 + πΎπ (π΅π π₯π β π) + πΏπ (ππ πΊπ₯π β π)σ΅©σ΅©σ΅©
Since πΌπ β 0 and βπ₯π+1 β π₯π β β 0, from condition (iv) and the boundedness of {π₯π }, it follows that σ΅© σ΅© lim π (σ΅©σ΅©σ΅©πΊπ₯π β π¦π σ΅©σ΅©σ΅©) = 0.
πββ
(268)
Utilizing the properties of π, we have σ΅© σ΅© lim σ΅©σ΅©πΊπ₯π β π¦π σ΅©σ΅©σ΅© = 0,
πββ σ΅©
(269)
which, together with (253) and (257), implies that σ΅©σ΅© σ΅© σ΅©σ΅©π₯π β π¦π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π+1 β π¦π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π β π₯π+1 σ΅©σ΅©σ΅© + ππ σ΅©σ΅©σ΅©πΊπ₯π β π¦π σ΅©σ΅©σ΅© σ³¨β 0
+ 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β© σ΅© σ΅© σ΅©2 σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π (π)σ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅©2 + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β πσ΅©σ΅©σ΅©
(270)
as π σ³¨β β.
+ 2πΌπ β¨π (π) β π, π½ (π₯π+1 β π)β© That is,
σ΅© σ΅© σ΅© σ΅©2 σ΅©2 σ΅©2 β€ πΌπ πσ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅© + πΏπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + 2πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© σ΅©2 = (1 β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© + 2πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅© β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + 2πΌπ σ΅©σ΅©σ΅©π (π) β πσ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© .
(266)
σ΅©σ΅©
lim σ΅©π₯ πββ σ΅© π
σ΅© β π¦π σ΅©σ΅©σ΅© = 0.
(271)
Since σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅©σ΅©π₯π β πΊπ₯π σ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©π₯π β π¦π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π¦π β πΊπ₯π σ΅©σ΅©σ΅© ,
(272)
it immediately follows from (269) and (271) that (265)
σ΅©σ΅©
lim σ΅©π₯ πββ σ΅© π
σ΅© β πΊπ₯π σ΅©σ΅©σ΅© = 0.
(273)
38
Abstract and Applied Analysis
On the other hand, observe that π¦π can be rewritten as follows: π¦π = πΌπ π (π₯π ) + π½π π₯π + πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π πΎ π΅ π₯ + πΏπ ππ πΊπ₯π = πΌπ π (π₯π ) + π½π π₯π + (πΎπ + πΏπ ) π π π (274) πΎπ + πΏπ = πΌπ π (π₯π ) + π½π π₯π + ππ π§Μπ , where ππ = πΎπ + πΏπ and π§Μπ = (πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π )/(πΎπ + πΏπ ). Utilizing Lemma 11, we have σ΅©2 σ΅©σ΅© σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© σ΅© σ΅©2 = σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π) + π½π (π₯π β π) + ππ (Μπ§π β π)σ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅© + ππ σ΅©σ΅©σ΅©π§Μπ β πσ΅©σ΅©σ΅© β π½π ππ π1 (σ΅©σ΅©σ΅©π§Μπ β π₯π σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅©2 σ΅©2 = πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© β π½π ππ π1 (σ΅©σ΅©σ΅©π§Μπ β π₯π σ΅©σ΅©σ΅©) σ΅©σ΅©2 σ΅©σ΅© πΎ π΅ π₯ + πΏ π πΊπ₯ σ΅© σ΅© π π π + ππ σ΅©σ΅©σ΅© π π π β πσ΅©σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© πΎπ + πΏπ σ΅© σ΅© σ΅©2 σ΅©2 = πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© β π½π ππ π1 (σ΅©σ΅©σ΅©π§Μπ β π₯π σ΅©σ΅©σ΅©) σ΅©σ΅©2 σ΅©σ΅© πΎ πΏπ σ΅© σ΅© π + ππ σ΅©σ΅©σ΅© (π΅π π₯π β π) + (ππ πΊπ₯π β π)σ΅©σ΅©σ΅© πΎπ + πΏπ σ΅©σ΅© σ΅©σ΅© πΎπ + πΏπ σ΅© σ΅© σ΅©2 σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© β π½π ππ π1 (σ΅©σ΅©σ΅©π§Μπ β π₯π σ΅©σ΅©σ΅©) + ππ [
πΎπ σ΅©σ΅© πΏπ σ΅©σ΅© σ΅©2 σ΅©2 σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© + σ΅©π πΊπ₯ β πσ΅©σ΅©σ΅© ] πΎπ + πΏπ πΎπ + πΏπ σ΅© π π
σ΅© σ΅© σ΅©2 σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© β π½π ππ π1 (σ΅©σ΅©σ΅©π§Μπ β π₯π σ΅©σ΅©σ΅©)
σ΅© σ΅© σ΅©2 σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + (σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅©) σ΅©σ΅©σ΅©π₯π β π¦π σ΅©σ΅©σ΅© . (276) Utilizing (271), conditions (i), (ii), (iv), and the boundedness of {π₯π }, {π¦π } and {π(π₯π )}, we get σ΅© σ΅© lim π (σ΅©σ΅©π§Μ β π₯π σ΅©σ΅©σ΅©) = 0. (277) πββ 1 σ΅© π From the properties of π1 , we have σ΅© σ΅© lim σ΅©σ΅©π§Μ β π₯π σ΅©σ΅©σ΅© = 0. πββ σ΅© π
(278)
Utilizing Lemma 15 and the definition of π§Μπ , we have σ΅©σ΅©Μ σ΅©2 σ΅©σ΅©π§π β πσ΅©σ΅©σ΅© σ΅©σ΅© πΎ π΅ π₯ + πΏ π πΊπ₯ σ΅©σ΅©2 σ΅© σ΅© π π π = σ΅©σ΅©σ΅© π π π β πσ΅©σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© πΎπ + πΏπ σ΅©σ΅© πΎ σ΅©σ΅©σ΅©2 πΏπ σ΅© π = σ΅©σ΅©σ΅© (π΅π π₯π β π) + (ππ πΊπ₯π β π)σ΅©σ΅©σ΅© σ΅©σ΅© πΎπ + πΏπ σ΅©σ΅© πΎπ + πΏπ β€
πΎπ σ΅©σ΅© πΏπ σ΅©σ΅© σ΅©2 σ΅©2 σ΅©π΅ π₯ β πσ΅©σ΅©σ΅© + σ΅©π πΊπ₯ β πσ΅©σ΅©σ΅© πΎπ + πΏπ σ΅© π π πΎπ + πΏπ σ΅© π π β
πΎπ πΏπ
2
(πΎπ + πΏπ )
σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β which leads to πΎπ πΏπ
2
(πΎπ + πΏπ )
σ΅© σ΅© π2 (σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©) πΎπ πΏπ
2
(πΎπ + πΏπ )
σ΅© σ΅© π2 (σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©) , (279)
σ΅© σ΅© π2 (σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅π π₯π σ΅©σ΅©σ΅©)
σ΅© σ΅©2 σ΅© σ΅©2 β€ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β σ΅©σ΅©σ΅©π§Μπ β πσ΅©σ΅©σ΅©
(280)
σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ (σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π§Μπ β πσ΅©σ΅©σ΅©) σ΅©σ΅©σ΅©π₯π β π§Μπ σ΅©σ΅©σ΅© .
Since {π₯π } and {Μπ§π } are bounded, we deduce from (278) and condition (ii) that σ΅© σ΅© lim π (σ΅©σ΅©π΅ π₯ β ππ πΊπ₯π σ΅©σ΅©σ΅©) = 0. (281) πββ 2 σ΅© π π From the properties of π2 , we have σ΅© σ΅© lim σ΅©σ΅©σ΅©π΅π π₯π β ππ πΊπ₯π σ΅©σ΅©σ΅© = 0.
πΎπ σ΅©σ΅© πΏπ σ΅©σ΅© σ΅©2 σ΅©2 + ππ [ σ΅©π₯ β πσ΅©σ΅©σ΅© + σ΅©π₯ β πσ΅©σ΅©σ΅© ] πΎπ + πΏπ σ΅© π πΎπ + πΏπ σ΅© π
πββ
(282)
Furthermore, π¦π can also be rewritten as follows:
σ΅©2 σ΅© σ΅©2 σ΅© = πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + (1 β πΌπ ) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
π¦π = πΌπ π (π₯π ) + π½π π₯π + πΎπ π΅π π₯π + πΏπ ππ πΊπ₯π
σ΅© σ΅© β π½π ππ π1 (σ΅©σ΅©σ΅©π§Μπ β π₯π σ΅©σ΅©σ΅©)
σ΅© σ΅©2 σ΅© σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© β π½π ππ π1 (σ΅©σ΅©σ΅©π§Μπ β π₯π σ΅©σ΅©σ΅©) ,
which hence implies that σ΅© σ΅© π½π ππ π1 (σ΅©σ΅©σ΅©π§Μπ β π₯π σ΅©σ΅©σ΅©) σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅©
= π½π π₯π + πΎπ π΅π π₯π + (πΌπ + πΏπ )
πΌπ π (π₯π ) + πΏπ ππ πΊπ₯π πΌπ + πΏπ
= π½π π₯π + πΎπ π΅π π₯π + ππ π§Μπ , (275)
(283)
Abstract and Applied Analysis
39
where ππ = πΌπ + πΏπ and π§Μπ = (πΌπ π(π₯π ) + πΏπ ππ πΊπ₯π )/(πΌπ + πΏπ ). Utilizing Lemma 11 and the convexity of β β
β2 , we have σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅©2 σ΅© σ΅© σ΅© σ΅©2 = σ΅©σ΅©σ΅©π½π (π₯π β π) + πΎπ (π΅π π₯π β π) + ππ (Μπ§π β π)σ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅©2 β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅© + ππ σ΅©σ΅©σ΅©π§Μπ β πσ΅©σ΅©σ΅© β π½π πΎπ π3 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©)
Thus, from (282) and (287), we get σ΅© σ΅© σ΅© σ΅©σ΅© σ΅©σ΅©π₯π β ππ πΊπ₯π σ΅©σ΅©σ΅© β€ σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©
σ΅© σ΅© + σ΅©σ΅©σ΅©π΅π π₯π β ππ πΊπ₯π σ΅©σ΅©σ΅© σ³¨β 0
That is,
σ΅© σ΅© σ΅©2 σ΅©2 = π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅©
σ΅©σ΅©
σ΅©σ΅© πΌ π (π₯ ) + πΏ π πΊπ₯ σ΅©σ΅©σ΅©2 σ΅©σ΅© π π π π π σ΅© + ππ σ΅©σ΅© β πσ΅©σ΅©σ΅©σ΅© πΌπ + πΏπ σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© β π½π πΎπ π3 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅©2 σ΅©2 = π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© σ΅©σ΅©2 σ΅©σ΅© πΌ πΏπ σ΅© σ΅© π + ππ σ΅©σ΅©σ΅© (π (π₯π ) β π) + (ππ πΊπ₯π β π)σ΅©σ΅©σ΅© σ΅©σ΅© σ΅©σ΅© πΌπ + πΏπ πΌπ + πΏπ σ΅© σ΅© β π½π πΎπ π3 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅©2 σ΅©2 β€ π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© πΌπ σ΅©σ΅© πΏπ σ΅©σ΅© σ΅©2 σ΅©2 σ΅©π (π₯π )βπσ΅©σ΅©σ΅© + σ΅©π πΊπ₯ βπσ΅©σ΅© ] πΌπ + πΏπ σ΅© πΌπ + πΏπ σ΅© π π σ΅© σ΅© σ΅© β π½π πΎπ π3 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©) σ΅© σ΅©2 σ΅© σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + (π½π + πΎπ ) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅© + πΏπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β π½π πΎπ π3 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©) σ΅© σ΅©2 σ΅© σ΅©2 = πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + (1 β πΌπ ) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© β π½π πΎπ π3 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©) σ΅© σ΅©2 σ΅© σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© σ΅© σ΅© β π½π πΎπ π3 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©) , (284) + ππ [
as π σ³¨β β. (288)
lim σ΅©π₯ πββ σ΅© π
σ΅© β ππ πΊπ₯π σ΅©σ΅©σ΅© = 0.
(289)
Therefore, from Lemma 12, (273), and (289), it follows that σ΅©σ΅© σ΅© σ΅©σ΅©π₯π β ππ₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β ππ πΊπ₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ πΊπ₯π β ππ π₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©ππ π₯π β ππ₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ σ΅©σ΅©σ΅©π₯π β ππ πΊπ₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©πΊπ₯π β π₯π σ΅©σ΅©σ΅© σ΅© σ΅© + σ΅©σ΅©σ΅©ππ π₯π β ππ₯π σ΅©σ΅©σ΅© σ³¨β 0 as π σ³¨β β. (290) That is, σ΅©σ΅©
lim σ΅©π₯ πββ σ΅© π
σ΅© β ππ₯π σ΅©σ΅©σ΅© = 0.
(291)
Suppose that π½π β‘ π½ for some fixed π½ β (0, 1) such that πΌπ + π½ + πΎπ + πΏπ = 1 for all π β₯ 0. Define a mapping ππ₯ = (1 β π1 β π2 )ππ₯ + π1 π΅π₯ + π2 πΊπ₯, where π1 , π2 β (0, 1) are two constants with π1 + π2 < 1. Then, by Lemmas 14 and 17, we have that Fix(π) = Fix(π) β© Fix(π΅) β© Fix(πΊ) = πΉ. For each π β₯ 1, let {ππ } be a unique element of πΆ such that ππ =
1 1 π (ππ ) + (1 β ) πππ . π π
(292)
which hence implies that σ΅© σ΅© π½π πΎπ π3 (σ΅©σ΅©σ΅©π₯π β π΅π π₯π σ΅©σ΅©σ΅©) σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© β σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© (285) σ΅© σ΅©2 σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β πσ΅©σ΅©σ΅© + (σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅©) σ΅© σ΅© Γ σ΅©σ΅©σ΅©π₯π β π¦π σ΅©σ΅©σ΅© . Utilizing (271), conditions (i), (ii), (iv), and the boundedness of {π₯π }, {π¦π } and {π(π₯π )}, we get σ΅© σ΅© lim π (σ΅©σ΅©π₯ β π΅π π₯π σ΅©σ΅©σ΅©) = 0. (286) πββ 3 σ΅© π
From Lemma 13, we conclude that ππ β π β Fix(π) = πΉ as π β β. Repeating the same arguments as those of (81) in the proof of Theorem 24, we can conclude that for every π, π
From the properties of π3 , we have σ΅© σ΅© lim σ΅©σ΅©π₯ β π΅π π₯π σ΅©σ΅©σ΅© = 0. πββ σ΅© π
where ππ = πΌπ βπ(π₯π ) β π΅ππ β + (1 β π½)βπ΅π ππ β π΅ππ β + πΏπ βππ πΊπ₯π βπ΅π π₯π β. Since limπ β β πΌπ = limπ β β βπ΅π ππ βπ΅ππ β = limπ β β βππ πΊπ₯π β π΅π π₯π β = 0, we know that ππ β 0 as
(287)
σ΅© σ΅©σ΅© σ΅©σ΅©π¦π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π΅ππ σ΅©σ΅©σ΅© + π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β π΅ππ σ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© β€ ππ + π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β π½) σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© ,
(293)
40
Abstract and Applied Analysis
π β β. So, it immediately follows that
Utilizing (295) and (297), we deduce that
σ΅©σ΅© σ΅©2 σ΅©σ΅©π₯π+1 β π΅ππ σ΅©σ΅©σ΅© σ΅©2 σ΅© σ΅© σ΅© = σ΅©σ΅©σ΅©π¦π β π΅ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π+1 β π¦π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© Γ [2 σ΅©σ΅©σ΅©π¦π β π΅ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π+1 β π¦π σ΅©σ΅©σ΅©] σ΅© σ΅© σ΅© σ΅©2 β€ (π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β π½) σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅©) σ΅© σ΅© σ΅© σ΅© + ππ [2 (π½ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β π½) σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅©) + ππ ] σ΅© σ΅© σ΅© σ΅© σ΅© σ΅© + σ΅©σ΅©σ΅©π₯π+1 β π¦π σ΅©σ΅©σ΅© [2 σ΅©σ΅©σ΅©π¦π β π΅ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π+1 β π¦π σ΅©σ΅©σ΅©] 2σ΅© σ΅© σ΅©2 σ΅©2 = π½2 σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β π½) σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© σ΅©σ΅© σ΅© σ΅© + 2π½ (1 β π½) σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + ππ 2σ΅© σ΅© σ΅©2 σ΅©2 β€ π½2 σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β π½) σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© σ΅©2 σ΅© σ΅©2 σ΅© + π½ (1 β π½) (σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© ) + ππ σ΅©2 σ΅©2 σ΅© σ΅© = π½σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© + (1 β π½) σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + ππ ,
(294)
σ΅© σ΅©2 ππ σ΅©σ΅©σ΅©π₯π β πππ σ΅©σ΅©σ΅© σ΅© = ππ σ΅©σ΅©σ΅©(1 β π1 β π2 ) (π₯π β πππ ) σ΅©2 + π1 (π₯π β π΅ππ ) + π2 (π₯π β πΊππ )σ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅©2 β€ (1 β π1 β π2 ) ππ σ΅©σ΅©σ΅©π₯π β πππ σ΅©σ΅©σ΅© + π1 ππ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© σ΅© σ΅©2 + π2 ππ σ΅©σ΅©σ΅©π₯π β πΊππ σ΅©σ΅©σ΅© σ΅© σ΅©2 β€ ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© .
Repeating the same arguments as those of (99) in the proof of Theorem 24, we can obtain that 1 σ΅©σ΅© σ΅©2 π σ΅©π₯ β ππ σ΅©σ΅©σ΅© β₯ ππ β¨π (ππ ) β ππ , π½ (π₯π β ππ )β© . 2π π σ΅© π
(295)
In addition, note that σ΅©σ΅© σ΅©2 σ΅©σ΅©π₯π β πΊππ σ΅©σ΅©σ΅© σ΅©2 σ΅© β€ σ΅©σ΅©σ΅©π₯π β πΊπ₯π + πΊπ₯π β πΊππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅© β€ (σ΅©σ΅©σ΅©π₯π β πΊπ₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅©) σ΅©2 σ΅© σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πΊπ₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© Γ (2 σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β πΊπ₯π σ΅©σ΅©σ΅©) , σ΅©σ΅© σ΅©2 σ΅©σ΅©π₯π β πππ σ΅©σ΅©σ΅© σ΅©2 σ΅© β€ σ΅©σ΅©σ΅©π₯π β ππ₯π + ππ₯π β πππ σ΅©σ΅©σ΅© σ΅© σ΅© σ΅©2 σ΅© β€ (σ΅©σ΅©σ΅©π₯π β ππ₯π σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅©) σ΅©2 σ΅© σ΅© σ΅© = σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β ππ₯π σ΅©σ΅©σ΅© σ΅© σ΅© σ΅© σ΅© Γ (2 σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© + σ΅©σ΅©σ΅©π₯π β ππ₯π σ΅©σ΅©σ΅©) .
σ΅© σ΅©2 σ΅© σ΅©2 ππ σ΅©σ΅©σ΅©π₯π β πππ σ΅©σ΅©σ΅© β€ ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© .
ππ β¨π (π) β π, π½ (π₯π β π)β© β€ 0.
(300)
On the other hand, from (257) and the norm-to-weakβ uniform continuity of π½ on bounded subsets of π, it follows that σ΅¨ σ΅¨ lim σ΅¨σ΅¨β¨π (π) β π, π½ (π₯π+1 β π)β© β β¨π (π) β π, π½ (π₯π β π)β©σ΅¨σ΅¨σ΅¨ = 0. (301)
πββ σ΅¨
So, utilizing Lemma 18, we deduce from (300) and (301) that lim sup β¨π (π) β π, π½ (π₯π β π)β© β€ 0, πββ
(296)
(302)
which together with (271) and the norm-to-weakβ uniform continuity of π½ on bounded subsets of π, implies that lim sup β¨π (π) β π, π½ (π¦π β π)β© β€ 0. πββ
(303)
Finally, let us show that π₯π β π as π β β. Utilizing Lemma 8 (i), from (253) and the convexity of β β
β, we get σ΅©σ΅© σ΅©2 σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅© σ΅© β€ σ΅©σ΅©σ΅©πΌπ (π (π₯π ) β π (π)) + π½π (π₯π β π)
It is easy to see from (273) and (291) that σ΅© σ΅©2 σ΅© σ΅©2 ππ σ΅©σ΅©σ΅©π₯π β πΊππ σ΅©σ΅©σ΅© β€ ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© ,
(299)
Since ππ β π β Fix(π) = πΉ as π β β, by the uniform Gateaux differentiability of the norm of π we have
where ππ = ππ [2(π½βπ₯π βπ΅ππ β+(1βπ½)βπ₯π βππ β)+ππ ]+βπ₯π+1 β π¦π β[2βπ¦π β π΅ππ β + βπ₯π+1 β π¦π β] β 0 as π β β. For any Banach limit π, from (294), we derive σ΅© σ΅©2 σ΅© σ΅©2 σ΅© σ΅©2 ππ σ΅©σ΅©σ΅©π₯π β π΅ππ σ΅©σ΅©σ΅© = ππ σ΅©σ΅©σ΅©π₯π+1 β π΅ππ σ΅©σ΅©σ΅© β€ ππ σ΅©σ΅©σ΅©π₯π β ππ σ΅©σ΅©σ΅© .
(298)
(297)
σ΅©2 + πΎπ (π΅π π₯π β π) + πΏπ (ππ πΊπ₯π β π)σ΅©σ΅©σ΅© + 2πΌπ β¨π (π) β π, π½ (π¦π β π)β©
Abstract and Applied Analysis
41
σ΅© σ΅© σ΅©2 σ΅©2 β€ πΌπ σ΅©σ΅©σ΅©π (π₯π ) β π (π)σ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
arbitrarily given π₯0 β πΆ, let {π₯π } be the sequence generated by
σ΅© σ΅© σ΅©2 σ΅©2 + πΎπ σ΅©σ΅©σ΅©π΅π π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©ππ πΊπ₯π β πσ΅©σ΅©σ΅©
π₯π+1
+ 2πΌπ β¨π (π) β π, π½ (π¦π β π)β©
= ππ ((1 β π) πΌ + ππ) π₯π + (1 β ππ )
σ΅© σ΅© σ΅©2 σ΅©2 β€ πΌπ πσ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + π½π σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
Γ [πΌπ π (π₯π ) + π½π π₯π + πΎπ π΅π π₯π + πΏπ ππ ((1βπ) πΌ+ππ) π₯π ] ,
σ΅© σ΅© σ΅©2 σ΅©2 + πΎπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + πΏπ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅©
βπ β₯ 0, (306)
+ 2πΌπ β¨π (π) β π, π½ (π¦π β π)β© σ΅© σ΅©2 = (1 β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + 2πΌπ β¨π (π) β π, π½ (π¦π β π)β© , (304)
where 1 β (π/(1 + π))(1 β β(1 β π)/π) β€ π β€ 1 and {πΌπ }, {π½π }, {πΎπ }, {πΏπ }, and {ππ } are the sequences in (0, 1) such that πΌπ +π½π + πΎπ + πΏπ = 1 for all π β₯ 0. Suppose that the following conditions hold: (i) limπ β β πΌπ = 0 and ββ π=0 πΌπ = β;
and hence
(ii) {πΎπ }, {πΏπ } β [π, π] for some π, π β (0, 1);
(iii) ββ π=1 (|ππ β ππβ1 | + |πΌπ β πΌπβ1 | + |π½π β π½πβ1 | + |πΎπ β πΎπβ1 | + |πΏπ β πΏπβ1 |) < β;
σ΅©σ΅© σ΅©2 σ΅©σ΅©π₯π+1 β πσ΅©σ΅©σ΅© σ΅© σ΅©2 σ΅© σ΅©2 β€ ππ σ΅©σ΅©σ΅©πΊπ₯π β πσ΅©σ΅©σ΅© + (1 β ππ ) σ΅©σ΅©σ΅©π¦π β πσ΅©σ΅©σ΅©
(iv) 0 < lim inf π β β π½π β€ lim supπ β β π½π < 1 and 0 < lim inf π β β ππ β€ lim supπ β β ππ < 1.
σ΅© σ΅©2 β€ ππ σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + (1 β ππ ) σ΅© σ΅©2 Γ [(1 β πΌπ (1 β π)) σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + 2πΌπ β¨π (π) β π, π½ (π¦π β π)β© ]
(305)
σ΅© σ΅©2 = [1 β (1 β ππ ) πΌπ (1 β π)] σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© + 2 (1 β ππ ) πΌπ β¨π (π) β π, π½ (π¦π β π)β© σ΅© σ΅©2 = [1 β (1 β ππ ) πΌπ (1 β π)] σ΅©σ΅©σ΅©π₯π β πσ΅©σ΅©σ΅© 2 β¨π (π) β π, π½ (π¦π β π)β© + (1 β ππ ) πΌπ (1 β π) . 1βπ From conditions (i) and (iv), it is easy to see that ββ π=0 (1 β ππ )πΌπ (1 β π) = β. Applying Lemma 7 to (305), we infer that π₯π β π as π β β. This completes the proof. Corollary 34. Let πΆ be a nonempty closed convex subset of a uniformly convex Banach space π which has a uniformly Gateaux differentiable norm. Let Ξ πΆ be a sunny nonexpansive retraction from π onto πΆ. Let {ππ }β π=0 be a sequence of positive numbers in (0, π] for some π β (0, 1) and π΄ π : πΆ β πππ strictly pseudocontractive and πΌΜπ -strongly accretive with ππ + πΌΜπ β₯ 1 for each π = 0, 1, . . .. Define a mapping πΊπ : πΆ β πΆ by Ξ πΆ(πΌ β π π π΄ π )π₯ = πΊπ π₯ for all π₯ β πΆ and π = 0, 1, . . ., where 1β(ππ /(1+ππ ))(1ββ(1 β πΌΜπ )/ππ ) β€ π π β€ 1 for all π = 0, 1, . . .. Let π΅π : πΆ β πΆ be the π-mapping generated by πΊπ , πΊπβ1 , . . . , πΊ0 and ππ , ππβ1 , . . . , π0 . Let π : πΆ β πΆ be a self-mapping such that πΌ β π : πΆ β π is π-strictly pseudocontractive and πstrongly accretive with π + π β₯ 1. Let π : πΆ β πΆ be a contraction with coefficient π β (0, 1). Let {ππ }β π=0 be a countable family of nonexpansive mappings of πΆ into itself such that β πΉ = (ββ π=0 Fix(ππ )) β© Fix(π) β© (βπ=0 VI(πΆ, π΄ π )) =ΜΈ 0. For
Assume that ββ π=1 supπ₯βπ· βππ π₯ β ππβ1 π₯β < β for any bounded subset π· of πΆ and let π be a mapping of πΆ into itself defined by ππ₯ = limπ β β ππ π₯ for all π₯ β πΆ and suppose that Fix(π) = ββ π=0 Fix(ππ ). Then, there hold the following: (I) limπ β β βπ₯π+1 β π₯π β = 0;
(II) the sequence {π₯π }β π=0 converges strongly to some π β πΉ which is the unique solution of the variational inequality problem (VIP) β¨(πΌ β π) π, π½ (π β π)β© β€ 0,
βπ β πΉ,
(307)
provided π½π β‘ π½ for some fixed π½ β (0, 1). Proof. In Theorem 33, we put π΅1 = πΌ β π, π΅2 = 0 and π1 = π where 1 β (π/(1 + π))(1 β β(1 β π)/π) β€ π β€ 1. Then, GSVI (13) is equivalent to the VIP of finding π₯β β πΆ such that β¨π΅1 π₯β , π½ (π₯ β π₯β )β© β₯ 0,
βπ₯ β πΆ.
(308)
In this case, π΅1 : πΆ β π is π-strictly pseudocontractive and π-strongly accretive. Repeating the same arguments as those in the proof of Corollary 25, we can infer that Fix(π) = VI(πΆ, π΅1 ). Accordingly, πΉ = (ββ π=0 Fix(ππ )) β© Ξ© β© β VI(πΆ, π΄ )) = (β Fix(π )) β© Fix(π) β© (ββ (ββ π π π=0 π=0 π=0 VI(πΆ, π΄ π )), πΊπ₯π = ((1 β π) πΌ + ππ) π₯π ,
(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 Yaoβs [10, Theorem 3.2], Cai and Buβs [11, Theorem 3.1], Kangtunyakarnβs [38, Theorem 3.1], and Ceng and Yaoβs [8, Theorem 3.1], in the following aspects.
42
Abstract and Applied Analysis (i) The problem of finding a point π β (ββ π=0 Fix(ππ )) β© Ξ© β© (ββ π=0 VI(πΆ, π΄ π )) in our Theorems 31 and 33 is more general and more subtle than every one of the problem of finding a point π β ββ π=0 Fix(ππ ) in [10, Theorem 3.2], the problem of finding a point π β ββ π=1 Fix(ππ ) β© Ξ© in [11, Theorem 3.1], the problem of finding a point π β Fix(π) β© Fix(π) β© (βπ π=1 VI(πΆ, π΄ π )) in [38, Theorem 3.1], and the problem of finding a point π β Fix(π) 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 {πΌπ }, {π½π }, {πΎπ }, {πΏπ }, (and {ππ }).
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 π-mapping and the Banach limit (see Lemmas 16β18 in Section 2 of this paper). (vi) The assumption of the uniformly convex and 2uniformly smooth Banach space π in [11, Theorem 3.1] is weakened to the one of the uniformly convex Banach space π having a uniformly Gateaux differentiable norm in our Theorem 33. Moreover, the assumption of the uniformly smooth Banach space π in [8, Theorem 3.1] is replaced with the one of the uniformly convex Banach space π having a uniformly Gateaux differentiable norm in our Theorem 33.
Conflict of Interests
(iii) Our Theorems 31 and 33 extend and generalize Ceng and Yaoβs [8, Theorem 3.1] from a nonexpansive mapping to a countable family of nonexpansive mappings, and Ceng and Yaoβs [10, Theorems 3.2] to the setting of the GSVI (13) and infinitely many VIPs, Kangtunyakarnβs [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 Buβs [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 πΊ and ππ in [11, Theorem 3.1] and the mapping π in [8, Theorem 3.1] are replaced with the same composite mapping ππ πΊ in the iterative schemes (42) and (130) and the mapping ππ in [10, Theorem 3.2] is replaced by π΅π .
References
(v) Cai and Buβs 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 ππ πΊ 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 π-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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