Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2013, Article ID 761864, 21 pages
http://dx.doi.org/10.1155/2013/761864
Research Article
Composite Iterative Algorithms for Variational
Inequality and Fixed Point Problems in Real Smooth
and Uniformly Convex Banach Spaces
Lu-Chuan Ceng1 and Ching-Feng Wen2
1
Department of Mathematics, Shanghai Normal University and Scientific Computing Key Laboratory of Shanghai Universities,
Shanghai 200234, China
2
Center for Fundamental Science, Kaohsiung Medical University, Kaohsiung 807, Taiwan
Correspondence should be addressed to Ching-Feng Wen; cfwen@kmu.edu.tw
Received 30 April 2013; Accepted 6 June 2013
Academic Editor: Wei-Shih Du
Copyright © 2013 L.-C. Ceng and C.-F. Wen. 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 composite implicit and explicit iterative algorithms for solving a general system of variational inequalities and
a common fixed point problem of an infinite family of nonexpansive mappings in a real smooth and uniformly convex Banach
space. These composite iterative algorithms are based on Korpelevich’s extragradient method and viscosity approximation method.
We first consider and analyze a composite implicit iterative algorithm in the setting of uniformly convex and 2-uniformly smooth
Banach space and then another composite explicit iterative algorithm in a uniformly convex Banach space with a uniformly Gâteaux
differentiable norm. Under suitable assumptions, we derive some strong convergence theorems. The results presented in this paper
improve, extend, supplement, and develop the corresponding results announced in the earlier and very recent literatures.
1. Introduction
Let 𝑋 be a real Banach space whose dual space is denoted by
∗
𝑋∗ . The normalized duality mapping 𝐽 : 𝑋 → 2𝑋 is defined
by
󵄩 󵄩2
𝐽 (𝑥) = {𝑥∗ ∈ 𝑋∗ : ⟨𝑥, 𝑥∗ ⟩ = ‖𝑥‖2 = 󵄩󵄩󵄩𝑥∗ 󵄩󵄩󵄩 } ,
∀𝑥 ∈ 𝑋,
(1)
where ⟨⋅, ⋅⟩ denotes the generalized duality pairing. It is an
immediate consequence of the Hahn-Banach theorem that
𝐽(𝑥) is nonempty for each 𝑥 ∈ 𝑋. Let 𝐶 be a nonempty, closed,
and convex subset of 𝑋. A mapping 𝑇 : 𝐶 → C is called
nonexpansive if ‖𝑇𝑥 − 𝑇𝑦‖ ≤ ‖𝑥 − 𝑦‖ for every 𝑥, 𝑦 ∈ 𝐶.
The set of fixed points of 𝑇 is denoted by Fix(𝑇). We use the
notation ⇀ to indicate the weak convergence and the one →
to indicate the strong convergence. A mapping 𝐴 : 𝐶 → 𝑋 is
said to be accretive if for each 𝑥, 𝑦 ∈ 𝐶, there exists 𝑗(𝑥 − 𝑦) ∈
𝐽(𝑥 − 𝑦) such that
⟨𝐴𝑥 − 𝐴𝑦, 𝑗 (𝑥 − 𝑦)⟩ ≥ 0.
(2)
It is said to be 𝛼-strongly accretive if for each 𝑥, 𝑦 ∈ 𝐶, there
exists 𝑗(𝑥 − 𝑦) ∈ 𝐽(𝑥 − 𝑦) such that
󵄩2
󵄩
⟨𝐴𝑥 − 𝐴𝑦, 𝑗 (𝑥 − 𝑦)⟩ ≥ 𝛼󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,
(3)
for some 𝛼 ∈ (0, 1). The mapping is called 𝛽-inverse stronglyaccretive if for each 𝑥, 𝑦 ∈ 𝐶, there exists 𝑗(𝑥 − 𝑦) ∈ 𝐽(𝑥 − 𝑦)
such that
󵄩2
󵄩
⟨𝐴𝑥 − 𝐴𝑦, 𝑗 (𝑥 − 𝑦)⟩ ≥ 𝛽󵄩󵄩󵄩𝐴𝑥 − 𝐴𝑦󵄩󵄩󵄩 ,
(4)
for some 𝛽 > 0 and is said to be 𝜆-strictly pseudocontractive
if for each 𝑥, 𝑦 ∈ 𝐶, there exists 𝑗(𝑥 − 𝑦) ∈ 𝐽(𝑥 − 𝑦) such that
󵄩2
󵄩2
󵄩
󵄩
⟨𝐴𝑥 − 𝐴𝑦, 𝑗 (𝑥 − 𝑦)⟩ ≤ 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 − 𝜆󵄩󵄩󵄩𝑥 − 𝑦 − (𝐴𝑥 − 𝐴𝑦)󵄩󵄩󵄩
(5)
for some 𝜆 ∈ (0, 1).
2
Journal of Applied Mathematics
Let 𝑈 = {𝑥 ∈ 𝑋 :‖ 𝑥 ‖= 1} denote the unite sphere of 𝑋.
A Banach space 𝑋 is said to be uniformly convex if for each
𝜖 ∈ (0, 2], there exists 𝛿 > 0 such that for all 𝑥, 𝑦 ∈ 𝑈,
󵄩
󵄩󵄩
󵄩𝑥 + 𝑦󵄩󵄩󵄩
󵄩
󵄩󵄩
≤ 1 − 𝛿.
󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ≥ 𝜖 󳨐⇒ 󵄩
2
(6)
It is known that a uniformly convex Banach space is reflexive
and strict convex. A Banach space 𝑋 is said to be smooth if
the limit
‖ 𝑥 + 𝑡𝑦 ‖ − ‖ 𝑥 ‖
𝑡→0
𝑡
(7)
lim
exists for all 𝑥, 𝑦 ∈ 𝑈; in this case, 𝑋 is also said to have a
Gâteaux differentiable norm. 𝑋 is said to have a uniformly
Gâteaux differentiable norm if for each 𝑦 ∈ 𝑈, the limit
is attained uniformly for 𝑥 ∈ 𝑈. Moreover, it is said to be
uniformly smooth if this limit is attained uniformly for 𝑥, 𝑦 ∈
𝑈. The norm of 𝑋 is said to be the Fréchet differential if for
each 𝑥 ∈ 𝑈, this limit is attained uniformly for 𝑦 ∈ 𝑈. In
addition, we define a function 𝜌 : [0, ∞) → [0, ∞) called
the modulus of smoothness of 𝑋 as follows:
1 󵄩
󵄩
󵄩 󵄩
𝜌 (𝜏) = sup { (󵄩󵄩󵄩𝑥 + 𝑦󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩)
2
󵄩 󵄩
− 1 : 𝑥, 𝑦 ∈ 𝑋, ‖𝑥‖ = 1, 󵄩󵄩󵄩𝑦󵄩󵄩󵄩 = 𝜏} .
(8)
It is known that 𝑋 is uniformly smooth if and only if
lim𝜏 → 0 𝜌(𝜏)/𝜏 = 0. Let 𝑞 be a fixed real number with 1 < 𝑞 ≤
2. Then a Banach space 𝑋 is said to be 𝑞-uniformly smooth if
there exists a constant 𝑐 > 0 such that 𝜌(𝜏) ≤ 𝑐𝜏𝑞 for all 𝜏 > 0.
As pointed out in [1], no Banach space is 𝑞-uniformly smooth
for 𝑞 > 2. In addition, it is also known that 𝐽 is single-valued
if and only if 𝑋 is smooth, whereas if 𝑋 is uniformly smooth,
then the mapping 𝐽 is norm-to-norm uniformly continuous
on bounded subsets of 𝑋. If 𝑋 has a uniformly Gâteaux
differentiable norm, then the duality mapping 𝐽 is norm-toweak∗ uniformly continuous on bounded subsets of 𝑋.
Very recently, Cai and Bu [2] considered the following
general system of variational inequalities (GSVI) in a real
smooth Banach space 𝑋, which involves finding (𝑥∗ , 𝑦∗ ) ∈
𝐶 × 𝐶 such that
⟨𝜇1 𝐵1 𝑦∗ + 𝑥∗ − 𝑦∗ , 𝐽 (𝑥 − 𝑥∗ )⟩ ≥ 0,
∗
∗
∗
∗
⟨𝜇2 𝐵2 𝑥 + 𝑦 − 𝑥 , 𝐽 (𝑥 − 𝑦 )⟩ ≥ 0,
∀𝑥 ∈ 𝐶,
∀𝑥 ∈ 𝐶,
(9)
where 𝐶 is a nonempty, closed, and convex subset of 𝑋, 𝐵1 ,
and 𝐵2 : 𝐶 → 𝑋 are two nonlinear mappings, and 𝜇1 and 𝜇2
are two positive constants. Here the set of solutions of GSVI
(9) is denoted by GSVI(𝐶, 𝐵1 , 𝐵2 ). In particular, if 𝑋 = 𝐻,
a real Hilbert space, then GSVI (9) reduces to the following
GSVI of finding (𝑥∗ , 𝑦∗ ) ∈ 𝐶 × 𝐶 such that
⟨𝜇1 𝐵1 𝑦∗ + 𝑥∗ − 𝑦∗ , 𝑥 − 𝑥∗ ⟩ ≥ 0,
∀𝑥 ∈ 𝐶,
⟨𝜇2 𝐵2 𝑥∗ + 𝑦∗ − 𝑥∗ , 𝑥 − 𝑦∗ ⟩ ≥ 0,
∀𝑥 ∈ 𝐶,
(10)
which 𝜇1 and 𝜇2 are two positive constants. The set of
solutions of problem (10) is still denoted by GSVI(𝐶, 𝐵1 , 𝐵2 ).
It is clear that the problem (10) covers as special case the
classical variational inequality problem (VIP) of finding 𝑥∗ ∈
𝐶 such that
⟨𝐴𝑥∗ , 𝑥 − 𝑥∗ ⟩ ≥ 0,
∀𝑥 ∈ 𝐶.
(11)
The solution set of the VIP (11) is denoted by VI(𝐶, 𝐴).
Recently, Ceng et al. [3] transformed problem (10) into a
fixed point problem in the following way.
Lemma 1 (see [3]). For given 𝑥, 𝑦 ∈ 𝐶, (𝑥, 𝑦) is a solution of
problem (10) if and only if 𝑥 is a fixed point of the mapping
𝐺 : 𝐶 → 𝐶 defined by
𝐺 (𝑥) = 𝑃𝐶 [𝑃𝐶 (𝑥 − 𝜇2 𝐵2 𝑥) − 𝜇1 𝐵1 𝑃𝐶
× (𝑥 − 𝜇2 𝐵2 𝑥)] ,
∀𝑥 ∈ 𝐶,
(12)
where 𝑦 = 𝑃𝐶(𝑥−𝜇2 𝐵2 𝑥) and 𝑃𝐶 is the the projection of 𝐻 onto
𝐶.
In particular, if the mappings 𝐵𝑖 : 𝐶 → 𝐻 is 𝛽𝑖 -inverse
strongly monotone for 𝑖 = 1, 2, then the mapping 𝐺 is nonexpansive provided 𝜇𝑖 ∈ (0, 2𝛽𝑖 ) for 𝑖 = 1, 2.
Let 𝐶 be a nonempty, closed, and convex subset of a real
smooth Banach space 𝑋. Let Π𝐶 be a sunny nonexpansive
retraction from 𝑋 onto 𝐶, and let 𝑓 : 𝐶 → 𝐶 be a
contraction with coefficient 𝜌 ∈ (0, 1). In this paper we
introduce composite implicit and explicit iterative algorithms
for solving GSVI (9) and the common fixed point problem
of an infinite family {𝑆𝑛 } of nonexpansive mappings of 𝐶
into itself. These composite iterative algorithms are based
on Korpelevich’s extragradient method [4] and viscosity
approximation method [5]. Let the mapping 𝐺 be defined by
𝐺 (𝑥) := Π𝐶 (𝐼 − 𝜇1 𝐵1 ) Π𝐶 (𝐼 − 𝜇2 𝐵2 ) 𝑥,
∀𝑥 ∈ 𝐶.
(13)
We first propose a composite implicit iterative algorithm in
the setting of uniformly convex and 2-uniformly smooth
Banach space 𝑋:
𝑦𝑛 = 𝛼𝑛 𝑓 (𝑦𝑛 ) + (1 − 𝛼𝑛 ) 𝑆𝑛 𝐺 (𝑥𝑛 ) ,
𝑥𝑛+1 = 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝑦𝑛 + 𝛿𝑛 𝑆𝑛 𝐺 (𝑦𝑛 ) ,
∀𝑛 ≥ 0,
(14)
where 𝐵𝑖 : 𝐶 → 𝑋 is 𝛼𝑖 -inverse-strongly accretive with
0 < 𝜇𝑖 < 𝛼𝑖 /𝜅2 for 𝑖 = 1, 2 and {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, and {𝛿𝑛 }
are the sequences in (0, 1) such that 𝛽𝑛 + 𝛾𝑛 + 𝛿𝑛 = 1 for all
𝑛 ≥ 0. It is proven that under appropriate conditions, {𝑥𝑛 }
converges strongly to 𝑞 ∈ 𝐹 = ⋂∞
𝑖=0 Fix(𝑆𝑖 ) ∩ Ω, which solves
the following VIP:
⟨𝑞 − 𝑓 (𝑞) , 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,
∀𝑝 ∈ 𝐹.
(15)
On the other hand, we also propose another composite
explicit iterative algorithm in a uniformly convex Banach
space 𝑋 with a uniformly Gateaux differentiable norm:
𝑦𝑛 = 𝛼𝑛 𝐺 (𝑥𝑛 ) + (1 − 𝛼𝑛 ) 𝑆𝑛 𝐺 (𝑥𝑛 ) ,
𝑥𝑛+1 = 𝛽𝑛 𝑓 (𝑥𝑛 ) + 𝛾𝑛 𝑦𝑛 + 𝛿𝑛 𝑆𝑛 𝐺 (𝑦𝑛 ) ,
∀𝑛 ≥ 0,
(16)
Journal of Applied Mathematics
3
where 𝐵𝑖 : 𝐶 → 𝑋 is 𝜆 𝑖 -strictly pseudocontractive and 𝛼𝑖 strongly accretive with 𝛼𝑖 + 𝜆 𝑖 ≥ 1 for 𝑖 = 1, 2 and {𝛼𝑛 }, {𝛽𝑛 },
{𝛾𝑛 }, and {𝛿𝑛 } are the sequences in (0, 1) such that 𝛽𝑛 + 𝛾𝑛 +
𝛿𝑛 = 1 for all 𝑛 ≥ 0. It is proven that under mild conditions,
{𝑥𝑛 } also converges strongly to 𝑞 ∈ 𝐹 = ⋂∞
𝑖=0 Fix(𝑆𝑖 ) ∩ Ω,
which solves the VIP (15). The results presented in this paper
improve, extend, supplement, and develop the corresponding
results announced in the earlier and very recent literatures.
2. Preliminaries
We list some lemmas that will be used in the sequel. Lemma 2
can be found in [6]. Lemma 3 is an immediate consequence
of the subdifferential inequality of the function (1/2)‖ ⋅ ‖2 .
Lemma 2. Let {𝑠𝑛 } be a sequence of nonnegative real numbers
satisfying
𝑠𝑛+1 ≤ (1 − 𝛼𝑛 ) 𝑠𝑛 + 𝛼𝑛 𝛽𝑛 + 𝛾𝑛 ,
∀𝑛 ≥ 0,
(17)
(i) Π is sunny and nonexpansive;
(ii) ‖ Π(𝑥) − Π(𝑦)‖2 ≤ ⟨𝑥 − 𝑦, 𝐽(Π(𝑥) − Π(𝑦))⟩, ∀𝑥, 𝑦 ∈ 𝐶;
(iii) ⟨𝑥 − Π(𝑥), 𝐽(𝑦 − Π(𝑥))⟩ ≤ 0, ∀𝑥 ∈ 𝐶, 𝑦 ∈ 𝐷.
It is well known that if 𝑋 = 𝐻 a Hilbert space, then
a sunny nonexpansive retraction Π𝐶 is coincident with the
metric projection from 𝑋 onto 𝐶; that is, Π𝐶 = 𝑃𝐶. If 𝐶 is
a nonempty, closed, and convex subset of a strictly convex
and uniformly smooth Banach space 𝑋 and if 𝑇 : 𝐶 → 𝐶 is
a nonexpansive mapping with the fixed point set Fix(𝑇) ≠ 0,
then the set Fix(𝑇) is a sunny nonexpansive retract of 𝐶.
Lemma 6 (see [9]). Given a number 𝑟 > 0. A real Banach
space 𝑋 is uniformly convex if and only if there exists a
continuous strictly increasing function 𝑔 : [0, ∞) → [0, ∞),
𝑔(0) = 0, such that
󵄩2
󵄩󵄩
󵄩󵄩𝜆𝑥 + (1 − 𝜆) 𝑦󵄩󵄩󵄩
󵄩 󵄩2
󵄩
󵄩
≤ 𝜆‖𝑥‖2 + (1 − 𝜆) 󵄩󵄩󵄩𝑦󵄩󵄩󵄩 − 𝜆 (1 − 𝜆) 𝑔 (󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩)
where {𝛼𝑛 }, {𝛽𝑛 }, and {𝛾𝑛 } satisfy the conditions:
(21)
(i) {𝛼𝑛 } ⊂ [0, 1] and ∑∞
𝑛=0 𝛼𝑛 = ∞,
for all 𝜆 ∈ [0, 1] and 𝑥, 𝑦 ∈ 𝑋 such that ‖ 𝑥 ‖≤ 𝑟 and ‖ 𝑦 ‖≤ 𝑟.
(ii) lim sup𝑛 → ∞ 𝛽𝑛 ≤ 0,
Lemma 7 (see [10]). Let 𝐶 be a nonempty, closed, and convex
subset of a Banach space 𝑋. Let 𝑆0 , 𝑆1 , . . ., be a sequence of
mappings of 𝐶 into itself. Suppose that ∑∞
𝑛=1 sup{‖𝑆𝑛 𝑥−𝑆𝑛−1 𝑥‖ :
𝑥 ∈ 𝐶} < ∞. Then for each 𝑦 ∈ 𝐶, {𝑆𝑛 𝑦} converges strongly
to some point of 𝐶. Moreover, let 𝑆 be a mapping of 𝐶 into
itself defined by 𝑆𝑦 = lim𝑛 → ∞ 𝑆𝑛 𝑦 for all 𝑦 ∈ 𝐶. Then
lim𝑛 → ∞ sup{‖ 𝑆𝑥 − 𝑆𝑛 𝑥 ‖: 𝑥 ∈ 𝐶} = 0.
(iii) 𝛾𝑛 ≥ 0, ∀𝑛 ≥ 0, and ∑∞
𝑛=0 𝛾𝑛 < ∞.
Then lim sup𝑛 → ∞ 𝑠𝑛 = 0.
Lemma 3. In a smooth Banach space 𝑋, there holds the
inequality
󵄩2
󵄩󵄩
2
󵄩󵄩𝑥 + 𝑦󵄩󵄩󵄩 ≤ ‖𝑥‖ + 2 ⟨𝑦, 𝐽 (𝑥 + 𝑦)⟩ ,
∀𝑥, 𝑦 ∈ 𝑋.
(18)
Lemma 4 (see [7]). Let {𝑥𝑛 } and {𝑧𝑛 } be bounded sequences
in a Banach space 𝑋, and let {𝛼𝑛 } be a sequence in [0, 1] which
satisfies the following condition:
0 < lim inf 𝛼𝑛 ≤ lim sup 𝛼𝑛 < 1.
𝑛→∞
𝑛→∞
(19)
Suppose that 𝑥𝑛+1 = 𝛼𝑛 𝑥𝑛 + (1 − 𝛼𝑛 )𝑧𝑛 , ∀𝑛 ≥ 0, and
lim sup𝑛 → ∞ (‖ 𝑧𝑛+1 − 𝑧𝑛 ‖ − ‖ 𝑥𝑛+1 − 𝑥𝑛 ‖) ≤ 0. Then
lim𝑛 → ∞ ‖ 𝑧𝑛 − 𝑥𝑛 ‖= 0.
Let 𝐷 be a subset of 𝐶, and let Π be a mapping of 𝐶 into
𝐷. Then Π is said to be sunny if
Π [Π (𝑥) + 𝑡 (𝑥 − Π (𝑥))] = Π (𝑥) ,
(20)
whenever Π(𝑥) + 𝑡(𝑥 − Π(𝑥)) ∈ 𝐶 for 𝑥 ∈ 𝐶 and 𝑡 ≥ 0. A
mapping Π of 𝐶 into itself is called a retraction if Π2 = Π. If
a mapping Π of 𝐶 into itself is a retraction, then Π(𝑧) = 𝑧 for
every 𝑧 ∈ 𝑅(Π) where 𝑅(Π) is the range of Π. A subset 𝐷 of
𝐶 is called a sunny nonexpansive retract of 𝐶 if there exists a
sunny nonexpansive retraction from 𝐶 onto 𝐷. The following
lemma concerns the sunny nonexpansive retraction.
Lemma 5 (see [8]). Let 𝐶 be a nonempty, closed, and convex
subset of a real smooth Banach space 𝑋. Let 𝐷 be a nonempty
subset of 𝐶. Let Π be a retraction of 𝐶 onto 𝐷. Then the
following are equivalent:
Let 𝐶 be a nonempty, closed, and convex subset of a
Banach space 𝑋, and let 𝑇 : 𝐶 → 𝐶 be a nonexpansive mapping with Fix(𝑇) ≠ 0. As previous, let Ξ𝐶 be the set of all
contractions on 𝐶. For 𝑡 ∈ (0, 1) and 𝑓 ∈ Ξ𝐶, let 𝑥𝑡 ∈ 𝐶 be the
unique fixed point of the contraction 𝑥 󳨃→ 𝑡𝑓(𝑥) + (1 − 𝑡)𝑇𝑥
on 𝐶; that is,
𝑥𝑡 = 𝑡𝑓 (𝑥𝑡 ) + (1 − 𝑡) 𝑇𝑥𝑡 .
(22)
Lemma 8 (see [11, 12]). Let 𝑋 be a uniformly smooth Banach
space or a reflexive and strictly convex Banach space with a
uniformly Gateaux differentiable norm. Let 𝐶 be a nonempty,
closed, and convex subset of 𝑋, let 𝑇 : 𝐶 → 𝐶 be a nonexpansive mapping with Fix(𝑇) ≠ 0, and 𝑓 ∈ Ξ𝐶. Then the net
{𝑥𝑡 } defined by 𝑥𝑡 = 𝑡𝑓(𝑥𝑡 ) + (1 − 𝑡)𝑇𝑥𝑡 converges strongly to a
point in Fix(𝑇). If we define a mapping 𝑄 : Ξ𝐶 → Fix(𝑇) by
𝑄(𝑓) := 𝑠 − lim𝑡 → 0 𝑥𝑡 , ∀𝑓 ∈ Ξ𝐶, then 𝑄(𝑓) solves the VIP:
⟨(𝐼 − 𝑓) 𝑄 (𝑓) , 𝐽 (𝑄 (𝑓) − 𝑝)⟩ ≤ 0,
∀𝑓 ∈ Ξ𝐶, 𝑝 ∈ Fix (𝑇) .
(23)
Lemma 9 (see [13]). Let 𝐶 be a nonempty, closed, and convex
subset of a strictly convex Banach space 𝑋. Let {𝑇𝑛 }∞
𝑛=0 be
a sequence of nonexpansive mappings on 𝐶. Suppose that
⋂∞
𝑛=0 Fix(𝑇𝑛 ) is nonempty. Let {𝜆 𝑛 } be a sequence of positive
numbers with ∑∞
𝑛=0 𝜆 𝑛 = 1. Then a mapping 𝑆 on 𝐶 defined by
𝜆
𝑇
𝑥
for 𝑥 ∈ 𝐶 is defined well, nonexpansive, and
𝑆𝑥 = ∑∞
𝑛
𝑛
𝑛=0
Fix(𝑆) = ⋂∞
𝑛=0 Fix(𝑇𝑛 ) holds.
4
Journal of Applied Mathematics
3. Implicit Iterative Schemes
In this section, we introduce our implicit iterative schemes
and show the strong convergence theorems. We will use the
following useful lemmas in the sequel.
Lemma 10 (see [2, Lemma 2.8]). Let 𝐶 be a nonempty, closed,
and convex subset of a real 2-uniformly smooth Banach space
𝑋. Let the mapping 𝐵𝑖 : 𝐶 → 𝑋 be 𝛼𝑖 -inverse-strongly
accretive. Then, one has
󵄩󵄩󵄩(𝐼 − 𝜇𝑖 𝐵𝑖 ) 𝑥 − (𝐼 − 𝜇𝑖 𝐵𝑖 ) 𝑦󵄩󵄩󵄩2
󵄩
󵄩
󵄩󵄩2
󵄩󵄩
(24)
≤ 󵄩󵄩𝑥 − 𝑦󵄩󵄩 + 2𝜇𝑖 (𝜇𝑖 𝜅2 − 𝛼𝑖 )
󵄩2
󵄩
× 󵄩󵄩󵄩𝐵𝑖 𝑥 − 𝐵𝑖 𝑦󵄩󵄩󵄩 ,
∀𝑥, 𝑦 ∈ 𝐶,
for 𝑖 = 1, 2, where 𝜇𝑖 > 0. In particular, if 0 < 𝜇𝑖 ≤ 𝛼𝑖 /𝜅2 , then
𝐼 − 𝜇𝑖 𝐵𝑖 is nonexpansive for 𝑖 = 1, 2.
Lemma 11 (see [2, Lemma 2.9]). Let 𝐶 be a nonempty, closed,
and convex subset of a real 2-uniformly smooth Banach space
𝑋. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶.
Let the mapping 𝐵𝑖 : 𝐶 → 𝑋 be 𝛼𝑖 -inverse-strongly accretive
for 𝑖 = 1, 2. Let 𝐺 : 𝐶 → 𝐶 be the mapping defined by
𝐺 (𝑥) = Π𝐶 [Π𝐶 (𝑥 − 𝜇2 𝐵2 𝑥)
−𝜇1 𝐵1 Π𝐶 (𝑥 − 𝜇2 𝐵2 𝑥)] ,
∀𝑥 ∈ 𝐶.
(25)
If 0 < 𝜇𝑖 ≤ 𝛼𝑖 /𝜅2 for 𝑖 = 1, 2, then 𝐺 : 𝐶 → 𝐶 is nonexpensive.
Lemma 12 (see [2, Lemma 2.10]). Let 𝐶 be a nonempty, closed,
and convex subset of a real 2-uniformly smooth Banach space
𝑋. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶.
Let 𝐵1 , 𝐵2 : 𝐶 → 𝑋 be two nonlinear mappings. For given
𝑥∗ , 𝑦∗ ∈ 𝐶, (𝑥∗ , 𝑦∗ ) is a solution of GSVI (9) if and only if
𝑥∗ = Π𝐶(𝑦∗ − 𝜇1 𝐵1 𝑦∗ ) where 𝑦∗ = Π𝐶(𝑥∗ − 𝜇2 𝐵2 𝑥∗ ).
(i) lim𝑛 → ∞ 𝛼𝑛 = 0 and ∑∞
𝑛=0 𝛼𝑛 = ∞,
(ii) 0 < lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ (𝛽𝑛 + 𝛾𝑛 ) < 1 and
lim inf 𝑛 → ∞ 𝛾𝑛 > 0,
(iii) lim𝑛 → ∞ |𝛾𝑛 /(1 − 𝛽𝑛 ) − 𝛾𝑛−1 /(1 − 𝛽𝑛−1 )| = 0.
Assume that ∑∞
𝑛=1 sup𝑥∈𝐷 ‖ 𝑆𝑛 𝑥−𝑆𝑛−1 𝑥 ‖< ∞ for any bounded
subset 𝐷 of 𝐶, and let 𝑆 be a mapping of 𝐶 into itself defined
by 𝑆𝑥 = lim𝑛 → ∞ 𝑆𝑛 𝑥 for all 𝑥 ∈ 𝐶. Suppose that Fix(𝑆) =
⋂∞
𝑛=0 Fix(𝑆𝑛 ). Then {𝑥𝑛 } converges strongly to 𝑞 ∈ 𝐹, which
solves the following VIP:
⟨𝑞 − 𝑓 (𝑞) , 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,
∀𝑝 ∈ 𝐹.
(28)
Proof. Take a fixed 𝑝 ∈ 𝐹 arbitrarily. Then by Lemma 12, we
know that 𝑝 = 𝐺(𝑝) and 𝑝 = 𝑆𝑛 𝑝 for all 𝑛 ≥ 0. Moreover, by
Lemma 11, we have
󵄩
󵄩󵄩
󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
= 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑦𝑛 ) − 𝑝) + (1 − 𝛼𝑛 ) (𝑆𝑛 𝐺 (𝑥𝑛 ) − 𝑝)󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑦𝑛 ) − 𝑓 (𝑝)󵄩󵄩󵄩 + 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩
󵄩
󵄩
+ (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
≤ 𝛼𝑛 𝜌 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩
󵄩
󵄩
+ (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
≤ 𝛼𝑛 𝜌 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩
󵄩
󵄩
+ (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 ,
(29)
which hence implies that
Remark 13. By Lemma 12, we observe that
𝑥∗ = Π𝐶 [Π𝐶 (𝑥∗ − 𝜇2 𝐵2 𝑥∗ ) − 𝜇1 𝐵1 Π𝐶 (𝑥∗ − 𝜇2 𝐵2 𝑥∗ )] ,
(26)
We now state and prove our first result on the implicit
iterative scheme.
Theorem 14. Let 𝐶 be a nonempty, closed, and convex subset
of a uniformly convex and 2-uniformly smooth Banach space
𝑋. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶.
Let the mapping 𝐵𝑖 : 𝐶 → 𝑋 be 𝛼𝑖 -inverse-strongly accretive
for 𝑖 = 1, 2. Let 𝑓 : 𝐶 → 𝐶 be a contraction with coefficient
𝜌 ∈ (0, 1). Let {𝑆𝑛 }∞
𝑛=0 be an infinite family of nonexpansive
mappings of 𝐶 into itself such that 𝐹 = ⋂∞
𝑛=0 Fix(𝑆𝑛 ) ∩ Ω ≠ 0,
where Ω is the fixed point set of the mapping 𝐺 = Π𝐶(𝐼 −
𝜇1 𝐵1 )Π𝐶(𝐼 − 𝜇2 𝐵2 ). For arbitrarily given 𝑥0 ∈ 𝐶, let {𝑥𝑛 } be
the sequence generated by
𝑦𝑛 = 𝛼𝑛 𝑓 (𝑦𝑛 ) + (1 − 𝛼𝑛 ) 𝑆𝑛 𝐺 (𝑥𝑛 ) ,
∀𝑛 ≥ 0,
󵄩󵄩
󵄩
󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
≤ (1 −
which implies that 𝑥∗ is a fixed point of the mapping 𝐺.
𝑥𝑛+1 = 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝑦𝑛 + 𝛿𝑛 𝑆𝑛 𝐺 (𝑦𝑛 ) ,
where 0 < 𝜇𝑖 < 𝛼𝑖 /𝜅2 for 𝑖 = 1, 2 and {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, and {𝛿𝑛 }
are the sequences in (0, 1) such that 𝛽𝑛 + 𝛾𝑛 + 𝛿𝑛 = 1, ∀𝑛 ≥ 0.
Suppose that the following conditions hold:
(27)
+
1−𝜌
󵄩
󵄩
𝛼 ) 󵄩󵄩𝑥 − 𝑝󵄩󵄩󵄩
1 − 𝛼𝑛 𝜌 𝑛 󵄩 𝑛
1
󵄩
󵄩
𝛼 󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩
1 − 𝛼𝑛 𝜌 𝑛 󵄩
󵄩
󵄩
≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 +
(30)
𝛼𝑛 󵄩󵄩
󵄩
󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 .
1 − 𝛼𝑛 𝜌 󵄩
Thus, from (27), we have
󵄩󵄩
󵄩
󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
= 󵄩󵄩󵄩𝛽𝑛 (𝑥𝑛 − 𝑝) + 𝛾𝑛 (𝑦𝑛 − 𝑝) + 𝛿𝑛 (𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑝)󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
󵄩
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝐺 (𝑦𝑛 ) − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
󵄩
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
= 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
Journal of Applied Mathematics
5
󵄩
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 )
× {(1 −
+
= [1 −
which hence yields
1−𝜌
󵄩
󵄩
𝛼 ) 󵄩󵄩𝑥 − 𝑝󵄩󵄩󵄩
1 − 𝛼𝑛 𝜌 𝑛 󵄩 𝑛
󵄩
󵄩󵄩
󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩
1
󵄩
󵄩
𝛼 󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩}
1 − 𝛼𝑛 𝜌 𝑛 󵄩
≤
(1 − 𝛽𝑛 ) (1 − 𝜌)
󵄩
󵄩
𝛼𝑛 ] 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
1 − 𝛼𝑛 𝜌
󵄩
󵄩
(1 − 𝛽𝑛 ) (1 − 𝜌) 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩
+
𝛼𝑛
1 − 𝛼𝑛 𝜌
1−𝜌
󵄩
󵄩
󵄩 󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩
󵄩
≤ max {󵄩󵄩󵄩𝑥0 − 𝑝󵄩󵄩󵄩 , 󵄩
}.
1−𝜌
(31)
It immediately follows that {𝑥𝑛 } is bounded, and so are the
sequences {𝑦𝑛 }, {𝐺(𝑥𝑛 )}, and {𝐺(𝑦𝑛 )} due to (30) and the
nonexpansivity of 𝐺.
Let us show that ‖𝑥𝑛+1 −𝑥𝑛 ‖ → 0 as 𝑛 → ∞. As a matter
of fact, from (27), we have
𝑦𝑛 = 𝛼𝑛 𝑓 (𝑦𝑛 ) + (1 − 𝛼𝑛 ) 𝑆𝑛 𝐺 (𝑥𝑛 ) ,
𝑦𝑛−1 = 𝛼𝑛−1 𝑓 (𝑦𝑛−1 ) + (1 − 𝛼𝑛−1 ) 𝑆𝑛−1 𝐺 (𝑥𝑛−1 ) ,
1 − 𝛼𝑛 󵄩󵄩
󵄩
(󵄩𝑥 − 𝑥𝑛−1 󵄩󵄩󵄩
1 − 𝛼𝑛 𝜌 󵄩 𝑛
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩)
󵄨󵄨
󵄨
󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩
󵄩
+
󵄩𝑓 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
1 − 𝛼𝑛 𝜌 󵄩
󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄨󵄨
󵄨
󵄨𝛼 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩
󵄩
+󵄨 𝑛
󵄩𝑓 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩 .
1−𝛼 𝜌 󵄩
𝑛
Now, we write 𝑥𝑛 = 𝛽𝑛−1 𝑥𝑛−1 + (1 − 𝛽𝑛−1 )𝑧𝑛−1 , ∀𝑛 ≥ 1, where
𝑧𝑛−1 = (𝑥𝑛 − 𝛽𝑛−1 𝑥𝑛−1 )/(1 − 𝛽𝑛−1 ). It follows that for all 𝑛 ≥ 1,
𝑧𝑛 − 𝑧𝑛−1 =
=
∀𝑛 ≥ 1.
(32)
+ (𝛼𝑛 − 𝛼𝑛−1 ) (𝑓 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 ))
=
(33)
+ (1 − 𝛼𝑛 ) (𝑆𝑛 𝐺 (𝑥𝑛 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )) .
It follows that
󵄩󵄩
󵄩
󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩
󵄩
󵄨
󵄩 󵄨
≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑦𝑛 ) − 𝑓 (𝑦𝑛−1 )󵄩󵄩󵄩 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨
󵄩
󵄩
× 󵄩󵄩󵄩𝑓 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩 󵄨
󵄨
󵄩
≤ 𝛼𝑛 𝜌 󵄩󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨
󵄩
󵄩
× 󵄩󵄩󵄩𝑓 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩 + (1 − 𝛼𝑛 )
(34)
󵄩
󵄩
× (󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛 ) − 𝑆𝑛 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩)
󵄩 󵄨
󵄨
󵄩
≤ 𝛼𝑛 𝜌 󵄩󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨
󵄩
󵄩
× 󵄩󵄩󵄩𝑓 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩 + (1 − 𝛼𝑛 )
󵄩
󵄩
× (‖ 𝑥𝑛 − 𝑥𝑛−1 ‖ + 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩) ,
𝑥𝑛+1 − 𝛽𝑛 𝑥𝑛 𝑥𝑛 − 𝛽𝑛−1 𝑥𝑛−1
−
1 − 𝛽𝑛
1 − 𝛽𝑛−1
𝛾𝑛 𝑦𝑛 + 𝛿𝑛 𝑆𝑛 𝐺 (𝑦𝑛 )
1 − 𝛽𝑛
−
Simple calculations show that
𝑦𝑛 − 𝑦𝑛−1 = 𝛼𝑛 (𝑓 (𝑦𝑛 ) − 𝑓 (𝑦𝑛−1 ))
(35)
𝛾𝑛−1 𝑦𝑛−1 + 𝛿𝑛−1 𝑆𝑛−1 𝐺 (𝑦𝑛−1 )
1 − 𝛽𝑛−1
𝛾𝑛 (𝑦𝑛 − 𝑦𝑛−1 ) + 𝛿𝑛 (𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑆𝑛−1 𝐺 (𝑦𝑛−1 ))
1 − 𝛽𝑛
+(
𝛾𝑛
𝛾𝑛−1
−
)𝑦
1 − 𝛽𝑛 1 − 𝛽𝑛−1 𝑛−1
+(
𝛿𝑛−1
𝛿𝑛
−
) 𝑆 𝐺 (𝑦𝑛−1 ) .
1 − 𝛽𝑛 1 − 𝛽𝑛−1 𝑛−1
(36)
This together with (35) implies that
󵄩
󵄩󵄩
󵄩󵄩𝑧𝑛 − 𝑧𝑛−1 󵄩󵄩󵄩
󵄩󵄩
󵄩
󵄩𝛾 (𝑦 − 𝑦𝑛−1 ) + 𝛿𝑛 (𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑆𝑛−1 𝐺 (𝑦𝑛−1 ))󵄩󵄩󵄩
≤󵄩 𝑛 𝑛
1 − 𝛽𝑛
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨 󵄩𝑦 󵄩󵄩
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 𝑛−1 󵄩
󵄨󵄨 𝛿
𝛿𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨 󵄩𝑆 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 𝑛−1
󵄩
󵄩
≤ (𝛾𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩
󵄩
󵄩
+ 𝛿𝑛 (󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑆𝑛 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩))
6
Journal of Applied Mathematics
−1
× (1 − 𝛽𝑛 )
Hence we obtain
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨 󵄩𝑦 󵄩󵄩
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 𝑛−1 󵄩
󵄨󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨󵄨 󵄩󵄩𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨
󵄩
󵄩
󵄩
󵄩
𝛾 󵄩󵄩𝑦 − 𝑦𝑛−1 󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩
≤ 𝑛󵄩 𝑛
𝛾𝑛 + 𝛿𝑛
󵄩󵄩
lim 󵄩𝑥
𝑛 → ∞ 󵄩 𝑛+1
𝛿𝑛 󵄩󵄩
󵄩
󵄩𝑆 𝐺 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩
𝛾𝑛 + 𝛿𝑛 󵄩 𝑛
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩 󵄩
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨 (󵄩𝑦 󵄩󵄩 + 󵄩󵄩𝑆 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩)
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 𝑛−1 󵄩 󵄩 𝑛−1
󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩 󵄩
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨 (󵄩𝑦 󵄩󵄩 + 󵄩󵄩𝑆 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩)
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 𝑛−1 󵄩 󵄩 𝑛−1
󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄨󵄨
󵄨
󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩
󵄩
+
󵄩𝑓 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
1 − 𝛼𝑛 𝜌 󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩 󵄩
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨 (󵄩𝑦 󵄩󵄩 + 󵄩󵄩𝑆 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩)
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 𝑛−1 󵄩 󵄩 𝑛−1
󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄨󵄨
󵄨
󵄨𝛼 − 𝛼𝑛−1 󵄨󵄨󵄨
󵄩
󵄩
+󵄨 𝑛
𝑀 + 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩
1−𝛼 𝜌
𝑛
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨
󵄨
+ 󵄨󵄨󵄨 𝑛 −
󵄨𝑀
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽n−1 󵄨󵄨󵄨
󵄩
󵄩
= 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩
󵄨
𝛾𝑛−1 󵄨󵄨󵄨󵄨
1
󵄨󵄨
󵄨 󵄨󵄨 𝛾
󵄨)
󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨 𝑛 −
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨󵄨
1 − 𝛼𝑛 𝜌
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩 ,
(37)
+ 𝑀(
where sup𝑛≥0 {‖𝑓(𝑦𝑛 )‖ + ‖𝑆𝑛 𝐺(𝑥𝑛 )‖ + ‖𝑦𝑛 ‖ + ‖𝑆𝑛 𝐺(𝑦𝑛 )‖} ≤ 𝑀
for some 𝑀 > 0. So, from 𝛼𝑛 → 0, condition (iii), and the
assumption on {𝑆𝑛 }, it immediately follows that
𝑛→∞
(38)
󵄩󵄩
󵄩
− 𝑥𝑛 󵄩󵄩󵄩 = 0.
󵄩2 󵄩
󵄩2
󵄩󵄩
󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 = 󵄩󵄩󵄩Π𝐶 (𝑥𝑛 − 𝜇2 𝐵2 𝑥𝑛 ) − Π𝐶 (𝑝 − 𝜇2 𝐵2 𝑝)󵄩󵄩󵄩
󵄩
󵄩2
≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝 − 𝜇2 (𝐵2 𝑥𝑛 − 𝐵2 𝑝)󵄩󵄩󵄩
󵄩
󵄩2
󵄩
󵄩2
≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 2𝜇2 (𝛼2 − 𝜅2 𝜇2 ) 󵄩󵄩󵄩𝐵2 𝑥𝑛 − 𝐵2 𝑝󵄩󵄩󵄩 ,
(41)
󵄩󵄩
󵄩2 󵄩
󵄩2
󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 = 󵄩󵄩󵄩Π𝐶 (𝑢𝑛 − 𝜇1 𝐵1 𝑢𝑛 ) − Π𝐶 (𝑞 − 𝜇1 𝐵1 𝑞)󵄩󵄩󵄩
󵄩
󵄩2
≤ 󵄩󵄩󵄩𝑢𝑛 − 𝑞 − 𝜇1 (𝐵1 𝑢𝑛 − 𝐵1 𝑞)󵄩󵄩󵄩
󵄩
󵄩2
󵄩
󵄩2
≤ 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 − 2𝜇1 (𝛼1 − 𝜅2 𝜇1 ) 󵄩󵄩󵄩𝐵1 𝑢𝑛 − 𝐵1 𝑞󵄩󵄩󵄩 .
(42)
Substituting (41) into (42), we obtain
󵄩󵄩
󵄩2 󵄩
󵄩2
󵄩
󵄩2
2
󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 2𝜇2 (𝛼2 − 𝜅 𝜇2 ) 󵄩󵄩󵄩𝐵2 𝑥𝑛 − 𝐵2 𝑝󵄩󵄩󵄩
󵄩
󵄩2
− 2𝜇1 (𝛼1 − 𝜅2 𝜇1 ) 󵄩󵄩󵄩𝐵1 𝑢𝑛 − 𝐵1 𝑞󵄩󵄩󵄩 .
(39)
(43)
According to Lemma 3, we have from (27)
󵄩󵄩
󵄩2
󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
󵄩
= 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑦𝑛 ) − 𝑓 (𝑝)) + (1 − 𝛼𝑛 ) (𝑆𝑛 V𝑛 − 𝑝)
󵄩2
+ 𝛼𝑛 (𝑓 (𝑝) − 𝑝) 󵄩󵄩󵄩
󵄩
󵄩2
≤ 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑦𝑛 ) − 𝑓 (𝑝)) + (1 − 𝛼𝑛 ) (𝑆𝑛 V𝑛 − 𝑝)󵄩󵄩󵄩
+ 2𝛼𝑛 ⟨𝑓 (𝑝) − 𝑝, 𝐽 (𝑦𝑛 − 𝑝)⟩
(44)
󵄩
󵄩2
󵄩
󵄩2
≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑦𝑛 ) − 𝑓 (𝑝)󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑆𝑛 V𝑛 − 𝑝󵄩󵄩󵄩
+ 2𝛼𝑛 ⟨𝑓 (𝑝) − 𝑝, 𝐽 (𝑦𝑛 − 𝑝)⟩
󵄩
󵄩2
󵄩
󵄩2
≤ 𝛼𝑛 𝜌󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩
󵄩󵄩
󵄩
󵄩
+ 2𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 ,
which hence yields
󵄩󵄩
󵄩2
󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
≤ (1 −
In terms of condition (ii) and Lemma 4, we get
lim 󵄩𝑧
𝑛→∞ 󵄩 𝑛
(40)
Next we show that ‖ 𝑥𝑛 − 𝐺(𝑥𝑛 ) ‖ → 0 as 𝑛 → ∞.
For simplicity, put 𝑞 = Π𝐶(𝑝 − 𝜇2 𝐵2 𝑝), 𝑢𝑛 = Π𝐶(𝑥𝑛 −
𝜇2 𝐵2 𝑥𝑛 ), and V𝑛 = Π𝐶(𝑢𝑛 − 𝜇1 𝐵1 𝑢𝑛 ). Then V𝑛 = 𝐺(𝑥𝑛 ). From
Lemma 10, we have
+
󵄩 󵄩
󵄩
󵄩
lim sup (󵄩󵄩󵄩𝑧𝑛 − 𝑧𝑛−1 󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩) ≤ 0.
󵄩
󵄩
󵄩
− 𝑥𝑛 󵄩󵄩󵄩 = lim (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑧𝑛 − 𝑥𝑛 󵄩󵄩󵄩 = 0.
𝑛→∞
+
1−𝜌
󵄩
󵄩2
𝛼 ) 󵄩󵄩V − 𝑝󵄩󵄩󵄩
1 − 𝛼𝑛 𝜌 𝑛 󵄩 𝑛
2𝛼𝑛 󵄩󵄩
󵄩󵄩
󵄩
󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 .
1 − 𝛼𝑛 𝜌 󵄩
(45)
Journal of Applied Mathematics
7
󵄩2 󵄩
󵄩2
󵄩
≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 + 𝛼𝑛 𝑀1
This together with (43) and the convexity of ‖ ⋅ ‖2 , we have
󵄩
󵄩
󵄩 󵄩
󵄩 󵄩
≤ (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 + 𝛼𝑛 𝑀1 .
󵄩󵄩
󵄩2
󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
= 󵄩󵄩󵄩𝛽𝑛 (𝑥𝑛 − 𝑝) + 𝛾𝑛 (𝑦𝑛 − 𝑝) + 𝛿𝑛 (𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑝)󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩2
󵄩2
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩2
󵄩2
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
󵄩
󵄩2
= 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 )
× {(1 −
+
(47)
Since 0 < 𝜇𝑖 < 𝛼𝑖 /𝜅2 for 𝑖 = 1, 2, from conditions (i), (ii), and
(40), we obtain
󵄩
󵄩
󵄩
󵄩
lim 󵄩󵄩𝐵 𝑢 − 𝐵1 𝑞󵄩󵄩󵄩 = 0.
lim 󵄩󵄩𝐵 𝑥 − 𝐵2 𝑝󵄩󵄩󵄩 = 0,
𝑛→∞ 󵄩 2 𝑛
𝑛→∞ 󵄩 1 𝑛
(48)
Utilizing [14, Proposition 1] and Lemma 5, we have
󵄩󵄩
󵄩2
󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩
󵄩
󵄩2
= 󵄩󵄩󵄩Π𝐶 (𝑥𝑛 − 𝜇2 𝐵2 𝑥𝑛 ) − Π𝐶 (𝑝 − 𝜇2 𝐵2 𝑝)󵄩󵄩󵄩
1−𝜌
󵄩
󵄩2
𝛼𝑛 ) 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩
1 − 𝛼𝑛 𝜌
≤ ⟨𝑥𝑛 − 𝜇2 𝐵2 𝑥𝑛 − (𝑝 − 𝜇2 𝐵2 𝑝) , 𝐽 (𝑢𝑛 − 𝑞)⟩
2𝛼𝑛 󵄩󵄩
󵄩󵄩
󵄩
󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩}
1 − 𝛼𝑛 𝜌 󵄩
= ⟨𝑥𝑛 − 𝑝, 𝐽 (𝑢𝑛 − 𝑞)⟩
+ 𝜇2 ⟨𝐵2 𝑝 − 𝐵2 𝑥𝑛 , 𝐽 (𝑢𝑛 − 𝑞)⟩
󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 )
× (1 −
≤
1−𝜌
󵄩
󵄩2
𝛼 ) 󵄩󵄩V − 𝑝󵄩󵄩󵄩 + 𝛼𝑛 𝑀1
1 − 𝛼𝑛 𝜌 𝑛 󵄩 𝑛
1−𝜌
󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 ) (1 −
𝛼)
1 − 𝛼𝑛 𝜌 𝑛
(46)
󵄩
󵄩2
󵄩
󵄩2
× [󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 2𝜇2 (𝛼2 − 𝜅2 𝜇2 ) 󵄩󵄩󵄩𝐵2 𝑥𝑛 − 𝐵2 𝑝󵄩󵄩󵄩
󵄩
󵄩2
− 2𝜇1 (𝛼1 − 𝜅2 𝜇1 ) 󵄩󵄩󵄩𝐵1 𝑢𝑛 − 𝐵1 𝑞󵄩󵄩󵄩 ] + 𝛼𝑛 𝑀1
= (1 −
(1 − 𝛽𝑛 ) (1 − 𝜌)
󵄩
󵄩2
𝛼𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 2 (1 − 𝛽𝑛 )
1 − 𝛼𝑛 𝜌
× (1 −
1−𝜌
𝛼)
1 − 𝛼𝑛 𝜌 𝑛
󵄩
󵄩2
× [𝜇2 (𝛼2 − 𝜅 𝜇2 ) 󵄩󵄩󵄩𝐵2 𝑥𝑛 − 𝐵2 𝑝󵄩󵄩󵄩
2
󵄩
󵄩2
+ 𝜇1 (𝛼1 − 𝜅2 𝜇1 ) 󵄩󵄩󵄩𝐵1 𝑢𝑛 − 𝐵1 𝑞󵄩󵄩󵄩 ] + 𝛼𝑛 𝑀1
1−𝜌
󵄩
󵄩2
≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 2 (1 − 𝛽𝑛 ) (1 −
𝛼)
1 − 𝛼𝑛 𝜌 𝑛
󵄩
󵄩2
× [𝜇2 (𝛼2 − 𝜅2 𝜇2 ) 󵄩󵄩󵄩𝐵2 𝑥𝑛 − 𝐵2 𝑝󵄩󵄩󵄩
󵄩
󵄩2
+ 𝜇1 (𝛼1 − 𝜅2 𝜇1 ) 󵄩󵄩󵄩𝐵1 𝑢𝑛 − 𝐵1 𝑞󵄩󵄩󵄩 ] + 𝛼𝑛 𝑀1 ,
where sup𝑛≥0 {2(1−𝛽𝑛 )/(1−𝛼𝑛 𝜌) ‖ 𝑓(𝑝)−𝑝 ‖‖ 𝑦𝑛 −𝑝 ‖} ≤ 𝑀1
for some 𝑀1 > 0. So, it follows that
2 (1 − 𝛽𝑛 ) (1 −
1−𝜌
𝛼)
1 − 𝛼𝑛 𝜌 𝑛
󵄩
󵄩2
× [𝜇2 (𝛼2 − 𝜅2 𝜇2 ) 󵄩󵄩󵄩𝐵2 𝑥𝑛 − 𝐵2 𝑝󵄩󵄩󵄩
󵄩
󵄩2
+ 𝜇1 (𝛼1 − 𝜅2 𝜇1 ) 󵄩󵄩󵄩𝐵1 u𝑛 − 𝐵1 𝑞󵄩󵄩󵄩 ]
1 󵄩󵄩
󵄩2 󵄩
󵄩2
󵄩
󵄩
[󵄩𝑥 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 −𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩) ]
2 󵄩 𝑛
󵄩
󵄩󵄩
󵄩
+ 𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 ,
(49)
which implies that
󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩2 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩2 − 𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩)
󵄩
󵄩
󵄩
󵄩
󵄩
󵄩
󵄩󵄩
󵄩󵄩 󵄩󵄩
󵄩󵄩
+ 2𝜇2 󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩 󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩 .
In the same way, we derive
󵄩󵄩
󵄩2
󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
= 󵄩󵄩󵄩Π𝐶 (𝑢𝑛 − 𝜇1 𝐵1 𝑢𝑛 ) − Π𝐶 (𝑞 − 𝜇1 𝐵1 𝑞)󵄩󵄩󵄩
(50)
≤ ⟨𝑢𝑛 − 𝜇1 𝐵1 𝑢𝑛 − (𝑞 − 𝜇1 𝐵1 𝑞) , 𝐽 (V𝑛 − 𝑝)⟩
= ⟨𝑢𝑛 − 𝑞, 𝐽 (V𝑛 − 𝑝)⟩ + 𝜇1 ⟨𝐵1 𝑞 − 𝐵1 𝑢𝑛 , 𝐽 (V𝑛 − 𝑝)⟩ (51)
1 󵄩
󵄩2 󵄩
󵄩2
≤ [󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 + 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩
2
󵄩
󵄩
− 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩) ]
󵄩
󵄩󵄩
󵄩
+ 𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 ,
which implies that
󵄩󵄩
󵄩2 󵄩
󵄩2
󵄩
󵄩
󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩 − 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩)
(52)
󵄩
󵄩󵄩
󵄩
+ 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 .
Substituting (50) into (52), we get
󵄩󵄩
󵄩2 󵄩
󵄩2
󵄩
󵄩
󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩)
󵄩
󵄩
− 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩)
(53)
󵄩
󵄩󵄩
󵄩
+ 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩
󵄩
󵄩󵄩
󵄩
+ 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 .
8
Journal of Applied Mathematics
From (46) and (53), we have
Utilizing conditions (i), (ii), from (40) and (48), we have
󵄩
󵄩2
󵄩2
󵄩󵄩
󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 ≤ 𝛼𝑛 𝑀1 + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
lim 𝑔
𝑛→∞ 1
󵄩
󵄩
lim 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩) = 0.
1−𝜌
+ (1 − 𝛽𝑛 ) (1 −
𝛼)
1 − 𝛼𝑛 𝜌 𝑛
Utilizing the properties of 𝑔1 and 𝑔2 , we deduce that
󵄩
− 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩 = 0,
󵄩󵄩
󵄩
− V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩 = 0.
lim 󵄩𝑢
𝑛→∞ 󵄩 𝑛
(1 − 𝛽𝑛 ) (1 − 𝜌)
󵄩
󵄩2
𝛼𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
1 − 𝛼𝑛 𝜌
1−𝜌
𝛼)
1 − 𝛼𝑛 𝜌 𝑛
󵄩
󵄩
× [𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩)
󵄩
󵄩
+ 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩)]
󵄩
󵄩󵄩
󵄩
+ 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩
󵄩
󵄩󵄩
󵄩
+ 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
≤ 𝛼𝑛 𝑀1 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − (1 − 𝛽𝑛 )
− (1 − 𝛽𝑛 ) (1 −
(57)
From (57), we obtain
󵄩󵄩
󵄩 󵄩
󵄩
󵄩󵄩𝑥𝑛 − V𝑛 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩 󳨀→ 0 as 𝑛 → ∞.
(58)
That is,
󵄩󵄩
lim 󵄩𝑥
𝑛→∞ 󵄩 𝑛
󵄩
− 𝐺 (𝑥𝑛 )󵄩󵄩󵄩 = 0.
(59)
On the other hand, since {𝑦𝑛 } and {𝑆𝑛 𝐺(𝑦𝑛 )} are bounded,
by Lemma 6, there exists a continuous strictly increasing
function 𝑔3 : [0, ∞) → [0, ∞), 𝑔3 (0) = 0 such that for 𝑝 ∈ 𝐹
󵄩󵄩
󵄩2
󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
= 󵄩󵄩󵄩𝛽𝑛 (𝑥𝑛 − 𝑝) + 𝛾𝑛 (𝑦𝑛 − 𝑝) + 𝛿𝑛 (𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑝)󵄩󵄩󵄩
󵄩󵄩󵄩
𝛾𝑛
= 󵄩󵄩󵄩(𝛾𝑛 + 𝛿𝑛 ) [
(𝑦 − 𝑝)
󵄩󵄩
𝛾𝑛 + 𝛿𝑛 𝑛
× (1 −
+
(54)
which implies that
1−𝜌
𝛼)
1 − 𝛼𝑛 𝜌 𝑛
󵄩
󵄩
× [𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩)
󵄩󵄩
lim 󵄩𝑥
𝑛→∞ 󵄩 𝑛
󵄩
󵄩
− 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩)
󵄩
󵄩󵄩
󵄩
+ 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩
󵄩
󵄩󵄩
󵄩
+ 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 ]
1−𝜌
𝛼)
1 − 𝛼𝑛 𝜌 𝑛
󵄩
󵄩
× [𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩)
󵄩
󵄩
+ 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩)]
󵄩
󵄩󵄩
󵄩
+ 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩
󵄩
󵄩󵄩
󵄩
+ 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 ,
(56)
𝑛→∞
󵄩
󵄩2
󵄩
󵄩
× [󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝑔1 (󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩)
≤ 𝛼𝑛 𝑀1 + (1 −
󵄩
󵄩
(󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 − (𝑝 − 𝑞)󵄩󵄩󵄩) = 0,
󵄩󵄩 𝛾
󵄩󵄩2
𝛿𝑛
󵄩
󵄩
𝑛
≤ (𝛾𝑛 + 𝛿𝑛 ) 󵄩󵄩󵄩
(𝑦𝑛 − 𝑝) +
(𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑝)󵄩󵄩󵄩
󵄩󵄩 𝛾𝑛 + 𝛿𝑛
󵄩󵄩
𝛾𝑛 + 𝛿𝑛
󵄩
󵄩2
+ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
≤ (𝛾𝑛 + 𝛿𝑛 ) [
(1 − 𝛽𝑛 ) (1 −
󵄩
󵄩
+ 𝑔2 (󵄩󵄩󵄩𝑢𝑛 − V𝑛 + (𝑝 − 𝑞)󵄩󵄩󵄩)]
󵄩
󵄩2 󵄩
󵄩2
≤ 𝛼𝑛 𝑀1 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩
(55)
󵄩
󵄩󵄩
󵄩
+ 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩
󵄩
󵄩󵄩
󵄩
+ 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩 󵄩
󵄩 󵄩
≤ 𝛼𝑛 𝑀1 + (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩
󵄩
󵄩󵄩
󵄩
+ 2𝜇2 󵄩󵄩󵄩𝐵2 𝑝 − 𝐵2 𝑥𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩𝑢𝑛 − 𝑞󵄩󵄩󵄩
󵄩
󵄩󵄩
󵄩
+ 2𝜇1 󵄩󵄩󵄩𝐵1 𝑞 − 𝐵1 𝑢𝑛 󵄩󵄩󵄩 󵄩󵄩󵄩V𝑛 − 𝑝󵄩󵄩󵄩 .
󵄩󵄩2
𝛿𝑛
󵄩
(𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑝)] + 𝛽𝑛 (𝑥𝑛 − 𝑝) 󵄩󵄩󵄩
󵄩󵄩
𝛾𝑛 + 𝛿𝑛
𝛾𝑛 󵄩󵄩
𝛿𝑛 󵄩󵄩
󵄩2
󵄩2
󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 +
󵄩𝑆 𝐺(𝑦𝑛 ) − 𝑝󵄩󵄩󵄩
𝛾𝑛 + 𝛿𝑛
𝛾𝑛 + 𝛿𝑛 󵄩 𝑛
−
𝛾𝑛 𝛿𝑛
2
(𝛾𝑛 + 𝛿𝑛 )
󵄩
󵄩
𝑔3 (󵄩󵄩󵄩𝑦𝑛 − 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩)]
󵄩
󵄩2
+ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩2
󵄩2
≤ 𝛾𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
−
𝛾𝑛 𝛿𝑛
󵄩
󵄩
󵄩
󵄩2
𝑔 (󵄩󵄩𝑦 − 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩) + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
𝛾𝑛 + 𝛿𝑛 3 󵄩 𝑛
𝛾𝛿
󵄩
󵄩2
= (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 − 𝑛 𝑛
𝛾𝑛 + 𝛿𝑛
󵄩
󵄩
󵄩
󵄩2
× 𝑔3 (󵄩󵄩󵄩𝑦𝑛 − 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩) + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 ,
(60)
Journal of Applied Mathematics
9
which together with (30) implies that
So,
󵄩
󵄩
(1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩
󵄩󵄩
󵄩2
󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
≤ (1 − 𝛽𝑛 ) (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 +
−
󵄩
󵄩
= 󵄩󵄩󵄩𝛿𝑛 (𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑦𝑛 ) − (𝑥𝑛+1 − 𝑥𝑛 )󵄩󵄩󵄩
2
𝛼𝑛 󵄩󵄩
󵄩
󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩)
1 − 𝛼𝑛 𝜌 󵄩
󵄩
󵄩 󵄩
󵄩
≤ 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑦𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩
𝛼𝑛 󵄩󵄩
󵄩
󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩)
1 − 𝛼𝑛 𝜌 󵄩
That is,
󵄩󵄩
lim 󵄩𝑥
𝑛→∞ 󵄩 𝑛
󵄩
󵄩
𝛾𝑛 𝛿𝑛 𝑔3 (󵄩󵄩󵄩𝑦𝑛 − 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩)
𝛾𝑛 𝛿𝑛
󵄩
󵄩
𝑔 (󵄩󵄩𝑦 − 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩)
𝛾𝑛 + 𝛿𝑛 3 󵄩 𝑛
󵄩
󵄩
≤ (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 +
2
𝛼𝑛 󵄩󵄩
󵄩
󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩)
1 − 𝛼𝑛 𝜌
󵄩
󵄩2
− 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩
(62)
󵄩
󵄩 󵄩
󵄩
≤ ( 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩
𝛼𝑛 󵄩󵄩
󵄩
󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩) .
1 − 𝛼𝑛 𝜌
lim inf 𝛿𝑛 = lim inf (1 − 𝛽𝑛 − 𝛾𝑛 )
𝑛→∞
= 1 − lim sup (𝛽𝑛 + 𝛾𝑛 ) > 0.
(63)
𝑛→∞
Since 𝛼𝑛 → 0, ‖ 𝑥𝑛+1 − 𝑥𝑛 ‖ → 0, and lim inf 𝑛 → ∞ 𝛾𝑛 > 0, we
conclude that
󵄩
󵄩
lim 𝑔 (󵄩󵄩𝑦 − 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩) = 0.
𝑛→∞ 3 󵄩 𝑛
(64)
Utilizing the property of 𝑔3 , we have
󵄩󵄩
lim 󵄩𝑦
𝑛→∞ 󵄩 𝑛
󵄩
− 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩 = 0.
Thus, from (59)–(68), we obtain that
󵄩
󵄩
lim 󵄩󵄩𝐺 (𝑥𝑛 ) − 𝑆𝑛 𝐺 (𝑥𝑛 )󵄩󵄩󵄩 = 0.
𝑛→∞ 󵄩
󵄩󵄩
󵄩
󵄩󵄩𝑆𝐺 (𝑥𝑛 ) − 𝐺 (𝑥𝑛 )󵄩󵄩󵄩
󵄩
󵄩
≤ 󵄩󵄩󵄩𝑆𝐺 (𝑥𝑛 ) − 𝑆𝑛 𝐺 (𝑥𝑛 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 G (𝑥𝑛 ) − 𝐺 (𝑥𝑛 )󵄩󵄩󵄩 󳨀→ 0 as 𝑛 󳨀→ ∞.
According to condition (ii), we get
𝑛→∞
(68)
(70)
By (70) and Lemma 7, we have
𝛼𝑛 󵄩󵄩
󵄩
󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩)
1 − 𝛼𝑛 𝜌 󵄩
󵄩
󵄩
× (󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 +
󵄩
− 𝑦𝑛 󵄩󵄩󵄩 = 0.
We observe that
󵄩󵄩
󵄩
󵄩󵄩𝐺 (𝑥𝑛 ) − 𝑆𝑛 𝐺 (𝑥𝑛 )󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑆𝑛 𝐺 (𝑥𝑛 )󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝑥𝑛 󵄩󵄩󵄩 + 2 󵄩󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩 .
(69)
It immediately follows that
+
󵄩
󵄩
+ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 󳨀→ 0 as 𝑛 󳨀→ ∞.
2
𝛾𝛿
󵄩
󵄩
− 𝑛 𝑛 𝑔3 (󵄩󵄩󵄩𝑦𝑛 − 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩) .
𝛾𝑛 + 𝛿𝑛
≤
󵄩
󵄩
≤ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑦𝑛 󵄩󵄩󵄩
𝛾𝑛 𝛿𝑛
󵄩
󵄩
󵄩
󵄩2
𝑔 (󵄩󵄩𝑦 − 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩) + 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
(61)
𝛾𝑛 + 𝛿𝑛 3 󵄩 𝑛
󵄩
󵄩
≤ (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 +
(67)
(65)
We note that
𝑥𝑛+1 − 𝑥𝑛 + 𝑥𝑛 − 𝑦𝑛 = 𝑥𝑛+1 − 𝑦𝑛
= 𝛽𝑛 (𝑥𝑛 − 𝑦𝑛 ) + 𝛿𝑛 (𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑦𝑛 ) .
(66)
(71)
In terms of (59) and (71), we have
󵄩󵄩
󵄩 󵄩
󵄩 󵄩
󵄩
󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝐺 (𝑥𝑛 )󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝑆𝐺 (𝑥𝑛 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝐺 (𝑥𝑛 ) − 𝑆𝑥𝑛 󵄩󵄩󵄩
󵄩
󵄩
≤ 2 󵄩󵄩󵄩𝑥𝑛 − 𝐺 (𝑥𝑛 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝑆𝐺 (𝑥𝑛 )󵄩󵄩󵄩 󳨀→ 0 as 𝑛 󳨀→ ∞.
(72)
Define a mapping 𝑊𝑥 = (1 − 𝜃)𝑆𝑥 + 𝜃𝐺(𝑥), where 𝜃 ∈ (0, 1) is
a constant. Then by Lemma 9, we have that Fix(𝑊) = Fix(𝑆)∩
Fix(𝐺) = 𝐹. We observe that
󵄩󵄩
󵄩 󵄩
󵄩
󵄩󵄩𝑥𝑛 − 𝑊𝑥𝑛 󵄩󵄩󵄩 = 󵄩󵄩󵄩(1 − 𝜃) (𝑥𝑛 − 𝑆𝑥𝑛 ) + 𝜃 (𝑥𝑛 − 𝐺 (𝑥𝑛 ))󵄩󵄩󵄩
(73)
󵄩
󵄩
󵄩
󵄩
≤ (1 − 𝜃) 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩 + 𝜃 󵄩󵄩󵄩𝑥𝑛 − 𝐺 (𝑥𝑛 )󵄩󵄩󵄩 .
From (59) and (72), we obtain
󵄩
󵄩
lim 󵄩󵄩𝑥 − 𝑊𝑥𝑛 󵄩󵄩󵄩 = 0.
𝑛→∞ 󵄩 𝑛
(74)
Now, we claim that
lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩ ≤ 0,
𝑛→∞
(75)
10
Journal of Applied Mathematics
where 𝑞 = 𝑠 − lim𝑡 → 0 𝑥𝑡 with 𝑥𝑡 being the fixed point of the
contraction
𝑥 󳨃󳨀→ 𝑡𝑓 (𝑥) + (1 − 𝑡) 𝑊𝑥.
(76)
Then 𝑥𝑡 solves the fixed point equation 𝑥𝑡 = 𝑡𝑓(𝑥𝑡 ) + (1 −
𝑡)𝑊𝑥𝑡 . Thus we have
󵄩󵄩
󵄩 󵄩
󵄩
󵄩󵄩𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩 = 󵄩󵄩󵄩(1 − 𝑡) (𝑊𝑥𝑡 − 𝑥𝑛 ) + 𝑡 (𝑓 (𝑥𝑡 ) − 𝑥𝑛 )󵄩󵄩󵄩 .
(77)
On the other hand, we have
⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩
= ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩ − ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩
+ ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩ − ⟨𝑓 (𝑞) − 𝑥𝑡 , 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩
+ ⟨𝑓 (𝑞) − 𝑥𝑡 , 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩ − ⟨𝑓 (𝑥𝑡 ) − 𝑥𝑡 , 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩
+ ⟨𝑓 (𝑥𝑡 ) − 𝑥𝑡 , 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩
By Lemma 3, we conclude that
= ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞) − 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩
󵄩󵄩
󵄩2
󵄩󵄩𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩
󵄩
󵄩2
= 󵄩󵄩󵄩(1 − 𝑡)(𝑊𝑥𝑡 − 𝑥𝑛 ) + 𝑡(𝑓(𝑥𝑡 ) − 𝑥𝑛 )󵄩󵄩󵄩
󵄩
󵄩2
≤ (1 − 𝑡)2 󵄩󵄩󵄩𝑊𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩
+ ⟨𝑥𝑡 − 𝑞, 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩ + ⟨𝑓 (𝑞) − 𝑓 (𝑥𝑡 ) , 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩
+ ⟨𝑓 (𝑥𝑡 ) − 𝑥𝑡 , 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩ .
(83)
It follows that
+ 2𝑡 ⟨𝑓 (𝑥𝑡 ) − 𝑥𝑛 , 𝐽 (𝑥𝑡 − 𝑥𝑛 )⟩
󵄩 󵄩
󵄩2
󵄩
≤ (1 − 𝑡)2 (󵄩󵄩󵄩𝑊𝑥𝑡 − 𝑊𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑊𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩)
lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩
𝑛→∞
+ 2𝑡 ⟨𝑓 (𝑥𝑡 ) − 𝑥𝑛 , 𝐽 (𝑥𝑡 − 𝑥𝑛 )⟩
≤ lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞) − 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩
𝑛→∞
󵄩 󵄩
󵄩
󵄩
≤ (1 − 𝑡) (󵄩󵄩󵄩𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑊𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩)
2
2
󵄩
󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑥𝑡 − 𝑞󵄩󵄩󵄩 lim sup 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑡 󵄩󵄩󵄩
(78)
+ 2𝑡 ⟨𝑓 (𝑥𝑡 ) − 𝑥𝑛 , 𝐽 (𝑥𝑡 − 𝑥𝑛 )⟩
𝑛→∞
(84)
󵄩
󵄩
󵄩
󵄩
+ 𝜌 󵄩󵄩󵄩𝑞 − 𝑥𝑡 󵄩󵄩󵄩 lim sup 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑡 󵄩󵄩󵄩
󵄩2
󵄩
󵄩
󵄩
= (1 − 𝑡) [󵄩󵄩󵄩𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩 + 2 󵄩󵄩󵄩𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩
󵄩
󵄩
× 󵄩󵄩󵄩𝑊𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 + ‖ 𝑊𝑥𝑛 − 𝑥𝑛 ‖2 ]
𝑛→∞
2
+ lim sup ⟨𝑓 (𝑥𝑡 ) − 𝑥𝑡 , 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩ .
𝑛→∞
Taking into account that 𝑥𝑡 → 𝑞 as 𝑡 → 0, we have from
(82)
+ 2𝑡 ⟨𝑓 (𝑥𝑡 ) − 𝑥𝑡 , 𝐽 (𝑥𝑡 − 𝑥𝑛 )⟩
+ 2𝑡 ⟨𝑥𝑡 − 𝑥𝑛 , 𝐽 (𝑥𝑡 − 𝑥𝑛 )⟩
lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩
󵄩2
󵄩
= (1 − 2𝑡 + 𝑡2 ) 󵄩󵄩󵄩𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩 + 𝑓𝑛 (𝑡)
𝑛→∞
= lim sup lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩
󵄩2
󵄩
+ 2𝑡 ⟨𝑓 (𝑥𝑡 ) − 𝑥𝑡 , 𝐽 (𝑥𝑡 − 𝑥𝑛 )⟩ + 2𝑡󵄩󵄩󵄩𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩 ,
𝑡→0
𝑛→∞
≤ lim sup lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞) − 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩ .
where
𝑡→0
󵄩 󵄩
󵄩
󵄩
𝑓𝑛 (𝑡) = (1 − 𝑡)2 (2 󵄩󵄩󵄩𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑊𝑥𝑛 󵄩󵄩󵄩)
󵄩
󵄩
× 󵄩󵄩󵄩𝑥𝑛 − 𝑊𝑥𝑛 󵄩󵄩󵄩 󳨀→ 0, as 𝑛 󳨀→ ∞.
(85)
(79)
It follows from (78) that
𝑡󵄩
󵄩2 1
⟨𝑥𝑡 − 𝑓 (𝑥𝑡 ) , 𝐽 (𝑥𝑡 − 𝑥𝑛 )⟩ ≤ 󵄩󵄩󵄩𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩 + 𝑓𝑛 (𝑡) . (80)
2
2𝑡
Letting 𝑛 → ∞ in (80) and noticing (79), we derive
𝑡
lim sup ⟨𝑥𝑡 − 𝑓 (𝑥𝑡 ) , 𝐽 (𝑥𝑡 − 𝑥𝑛 )⟩ ≤ 𝑀2 ,
2
𝑛→∞
Since 𝑋 has a uniformly Fréchet differentiable norm, the
duality mapping 𝐽 is norm-to-norm uniformly continuous
on bounded subsets of 𝑋. Consequently, the two limits are
interchangeable, and hence (75) holds. From (68), we get
(𝑦𝑛 − 𝑞) − (𝑥𝑛 − 𝑞) → 0. Noticing that 𝐽 is norm-to-norm
uniformly continuous on bounded subsets of 𝑋, we deduce
from (75) that
lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑦𝑛 − 𝑞)⟩
𝑛→∞
(81)
= lim sup (⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩
𝑛→∞
2
where 𝑀2 > 0 is a constant such that ‖ 𝑥𝑡 − 𝑥𝑛 ‖ ≤ 𝑀2 for all
𝑡 ∈ (0, 1) and 𝑛 ≥ 0. Taking 𝑡 → 0 in (81), we have
lim sup lim sup ⟨𝑥𝑡 − 𝑓 (𝑥𝑡 ) , 𝐽 (𝑥𝑡 − 𝑥𝑛 )⟩ ≤ 0.
𝑡→0
𝑛→∞
𝑛→∞
(82)
+ ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑦𝑛 − 𝑞) − 𝐽 (𝑥𝑛 − 𝑞)⟩)
= lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩ ≤ 0.
𝑛→∞
(86)
Journal of Applied Mathematics
11
Finally, let us show that 𝑥𝑛 → 𝑞 as 𝑛 → ∞. We observe
that
󵄩2
󵄩󵄩
󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩
󵄩
= 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑦𝑛 ) − 𝑓 (𝑞)) + (1 − 𝛼𝑛 )
󵄩2
× (𝑆𝑛 𝐺 (𝑥𝑛 ) − 𝑞) + 𝛼𝑛 (𝑓 (𝑞) − 𝑞)󵄩󵄩󵄩
󵄩
≤ 󵄩󵄩󵄩𝛼𝑛 (𝑓 (𝑦𝑛 ) − 𝑓 (𝑞))
󵄩2
+ (1 − 𝛼𝑛 ) (𝑆𝑛 𝐺 (𝑥𝑛 ) − 𝑞) 󵄩󵄩󵄩
+ 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑦𝑛 − 𝑞)⟩
(87)
󵄩2
󵄩
≤ 𝛼𝑛 󵄩󵄩󵄩𝑓 (𝑦𝑛 ) − 𝑓 (𝑞)󵄩󵄩󵄩
󵄩
󵄩2
󵄩
󵄩2
≤ 𝛼𝑛 𝜌󵄩󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩
(iii) lim𝑛 → ∞ |(𝛾𝑛 /(1 − 𝛽𝑛 )) − (𝛾𝑛−1 /(1 − 𝛽𝑛−1 ))| = 0.
(88)
By (27) and the convexity of ‖ ⋅ ‖2 , we get
󵄩󵄩
󵄩2
󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩2
󵄩2
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑞󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩2
󵄩2
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩
󵄩
󵄩2
󵄩
󵄩2
= 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 ,
(89)
which together with (88) leads to
󵄩󵄩
󵄩2
󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩
󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + (1 − 𝛽𝑛 )
+
= [1 −
(91)
(ii) 0 < lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ (𝛽𝑛 + 𝛾𝑛 ) < 1 and
lim inf 𝑛 → ∞ 𝛾𝑛 > 0,
which implies that
× { (1 −
∀𝑛 ≥ 0,
(i) lim𝑛 → ∞ 𝛼𝑛 = 0 and ∑∞
𝑛=0 𝛼𝑛 = ∞,
+ 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑦𝑛 − 𝑞)⟩ ,
𝛼𝑛 (1 − 𝜌) 2 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑦𝑛 − 𝑞)⟩
⋅
.
1 − 𝛼𝑛 𝜌
1−𝜌
𝑦𝑛 = 𝛼𝑛 𝑓 (𝑦𝑛 ) + (1 − 𝛼𝑛 ) 𝑆𝐺 (𝑥𝑛 ) ,
where 0 < 𝜇𝑖 < 𝛼𝑖 /𝜅2 for 𝑖 = 1, 2 and {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, and {𝛿𝑛 }
are the sequences in (0, 1) such that 𝛽𝑛 + 𝛾𝑛 + 𝛿𝑛 = 1, ∀𝑛 ≥ 0.
Suppose that the following conditions hold:
+ 2𝛼𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑦𝑛 − 𝑞)⟩
+
Corollary 15. Let 𝐶 be a nonempty, closed, and convex subset
of a uniformly convex and 2-uniformly smooth Banach space
𝑋. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶.
Let the mapping 𝐵𝑖 : 𝐶 → 𝑋 be 𝛼𝑖 -inverse-strongly accretive
for 𝑖 = 1, 2. Let 𝑓 : 𝐶 → 𝐶 be a contraction with coefficient
𝜌 ∈ (0, 1). Let 𝑆 be a nonexpansive mapping of 𝐶 into itself such
that 𝐹 = Fix(𝑆) ∩ Ω ≠ 0, where Ω is the fixed point set of the
mapping 𝐺 = Π𝐶(𝐼 − 𝜇1 𝐵1 )Π𝐶(𝐼 − 𝜇2 𝐵2 ). For arbitrarily given
𝑥0 ∈ 𝐶, let {𝑥𝑛 } be the sequence generated by
𝑥𝑛+1 = 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝑦𝑛 + 𝛿𝑛 𝑆𝐺 (𝑦𝑛 ) ,
󵄩
󵄩2
+ (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛 ) − 𝑞󵄩󵄩󵄩
1−𝜌
󵄩2
󵄩
󵄩2
󵄩󵄩
𝛼 ) 󵄩󵄩𝑥 − 𝑞󵄩󵄩󵄩
󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 ≤ (1 −
1 − 𝛼𝑛 𝜌 𝑛 󵄩 𝑛
Applying Lemma 2 to (88), we obtain that 𝑥𝑛 → 𝑞 as 𝑛 →
∞. This completes the proof.
1−𝜌
󵄩
󵄩2
𝛼 ) 󵄩󵄩𝑥 − 𝑞󵄩󵄩󵄩
1 − 𝛼𝑛 𝜌 𝑛 󵄩 𝑛
𝛼𝑛 (1 − 𝜌) 2 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑦𝑛 − 𝑞)⟩
⋅
}
1 − 𝛼𝑛 𝜌
1−𝜌
(1 − 𝛽𝑛 ) (1 − 𝜌)
󵄩
󵄩2
𝛼𝑛 ] 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩
1 − 𝛼𝑛 𝜌
(1 − 𝛽𝑛 ) (1 − 𝜌)
2 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑦𝑛 − 𝑞)⟩
+
𝛼𝑛 ⋅
.
1 − 𝛼𝑛 𝜌
1−𝜌
(90)
Then {𝑥𝑛 } converges strongly to 𝑞 ∈ 𝐹, which solves the following VIP:
⟨𝑞 − 𝑓 (𝑞) , 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,
∀𝑝 ∈ 𝐹.
(92)
Further, we illustrate Theorem 14 by virtue of an example,
that is, the following corollary.
Corollary 16. Let 𝐶 be a nonempty, closed, and convex subset
of a uniformly convex and 2-uniformly smooth Banach space
𝑋. Let Π𝐶 be a sunny nonexpansive retraction from 𝑋 onto 𝐶.
Let 𝑓 : 𝐶 → 𝐶 be a contraction with coefficient 𝜌 ∈ (0, 1). Let
𝑇 be an 𝜂-strictly pseudocontractive mapping of 𝐶 into itself,
and let 𝑆 be a nonexpansive mapping of 𝐶 into itself such that
𝐹 = Fix(𝑆) ∩ Fix(𝑇) ≠ 0. For arbitrarily given 𝑥0 ∈ 𝐶, let {𝑥𝑛 }
be the sequence generated by
𝑦𝑛 = 𝛼𝑛 𝑓 (𝑦𝑛 ) + (1 − 𝛼𝑛 ) 𝑆 (𝐼 − 𝜆 (I − 𝑇)) 𝑥𝑛 ,
𝑥𝑛+1 = 𝛽𝑛 𝑥𝑛 + 𝛾𝑛 𝑦𝑛 + 𝛿𝑛 𝑆 (𝐼 − 𝜆 (𝐼 − 𝑇)) 𝑦𝑛 ,
∀𝑛 ≥ 0,
(93)
where 0 < 𝜆 < max{1, 𝜂/𝜅2 } and {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, and {𝛿𝑛 } are
the sequences in (0, 1) such that 𝛽𝑛 + 𝛾𝑛 + 𝛿𝑛 = 1, ∀𝑛 ≥ 0.
Suppose that the following conditions hold:
(i) lim𝑛 → ∞ 𝛼𝑛 = 0 and ∑∞
𝑛=0 𝛼𝑛 = ∞,
(ii) 0 < lim inf 𝑛 → ∞ 𝛽𝑛 ≤ lim sup𝑛 → ∞ (𝛽𝑛 + 𝛾𝑛 ) < 1 and
lim inf 𝑛 → ∞ 𝛾𝑛 > 0,
(iii) lim𝑛 → ∞ |(𝛾𝑛 /(1 − 𝛽𝑛 )) − (𝛾𝑛−1 /(1 − 𝛽𝑛−1 ))| = 0.
Then {𝑥𝑛 } converges strongly to 𝑞 ∈ 𝐹, which solves the following VIP:
⟨𝑞 − 𝑓 (𝑞) , 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,
∀𝑝 ∈ 𝐹.
(94)
12
Journal of Applied Mathematics
Proof. In Corollary 15, put 𝐵1 = 𝐼 − 𝑇, 𝐵2 = 0, 𝜇1 = 𝜆, and
𝛼1 = 𝜂. Since 𝑇 is an 𝜂-strictly pseudocontractive mapping,
it is clear that 𝐵1 = 𝐼 − 𝑇 is an 𝜂-inverse strongly accretive
mapping. Hence, the GSVI (9) is equivalent to the following
VIP of finding 𝑥∗ ∈ 𝐶 such that
(𝐵1 𝑥∗ , 𝐽 (𝑥 − 𝑥∗ )) ≥ 0,
∀𝑥 ∈ 𝐶,
(95)
which leads to Ω = VI(𝐶, 𝐵1 ). In the meantime, we have
𝐺𝑥𝑛 = Π𝐶 (𝐼 − 𝜇1 𝐵1 ) Π𝐶 (𝐼 − 𝜇2 𝐵2 ) 𝑥𝑛
= Π𝐶 (𝐼 − 𝜇1 𝐵1 ) 𝑥𝑛
= Π𝐶 [(1 − 𝜆) 𝑥𝑛 + 𝜆𝑇𝑥𝑛 ]
(96)
= 𝑥𝑛 − 𝜆 (𝐼 − 𝑇) 𝑥𝑛 .
In the same way, we get 𝐺𝑦𝑛 = 𝑦𝑛 − 𝜆(𝐼 − 𝑇)𝑦𝑛 . In this case, it
is easy to see that (91) reduces to (93). We claim that Fix(𝑇) =
VI(𝐶, 𝐵1 ). As a matter of fact, we have, for 𝜆 > 0,
𝑢 ∈ VI (𝐶, 𝐵1 )
⇐⇒ ⟨𝐵1 𝑢, 𝐽 (𝑦 − 𝑢)⟩ ≥ 0 ∀𝑦 ∈ 𝐶
iterative scheme of [2, Theorem 3.1] and the mapping
𝑆𝑃𝐶(𝐼 − 𝜆 𝑛 𝐴) in the iterative scheme of [5, Theorem 3.1] are replaced by the same composite mapping
𝑆𝑛 𝐺 in the iterative scheme (27) of Theorem 14.
(iv) The proof in [2, Theorem 3.1] depends on the argument techniques in [3], the inequality in 2-uniformly
smooth Banach spaces ([9]), and the inequality in
smooth and uniform convex Banach spaces ([14,
Proposition 1]). Because the composite mapping 𝑆𝑛 𝐺
appears in the iterative scheme (27) of Theorem 14,
the proof of Theorem 14 depends on the argument
techniques in [3], the inequality in 2-uniformly
smooth Banach spaces, the inequality in smooth and
uniform convex Banach spaces, and the inequality in
uniform convex Banach spaces (Lemma 6).
(v) The iterative scheme in [2, Corollary 3.2] is extended
to develop the new iterative scheme in Corollary 15
because the mappings 𝑆 and 𝐺 are replaced by the
same composite mapping 𝑆𝐺 in Corollary 15.
4. Explicit Iterative Schemes
⇐⇒ ⟨𝑢 − 𝜆𝐵1 𝑢 − 𝑢, 𝐽 (𝑢 − 𝑦)⟩ ≥ 0 ∀𝑦 ∈ 𝐶
In this section, we introduce our explicit iterative schemes
and show the strong convergence theorems. First, we give
several useful lemmas.
⇐⇒ 𝑢 = Π𝐶 (𝑢 − 𝜆𝐵1 𝑢)
⇐⇒ 𝑢 = Π𝐶 (𝑢 − 𝜆𝑢 + 𝜆𝑇𝑢)
⇐⇒ ⟨𝑢 − 𝜆𝑢 + 𝜆𝑇𝑢 − 𝑢, 𝐽 (𝑢 − 𝑦)⟩ ≥ 0 ∀𝑦 ∈ 𝐶
⇐⇒ ⟨𝑢 − 𝑇𝑢, 𝐽 (𝑢 − 𝑦)⟩ ≤ 0 ∀𝑦 ∈ 𝐶
⇐⇒ 𝑢 = 𝑇𝑢
Lemma 18. Let 𝐶 be a nonempty, closed, and convex subset of
a smooth Banach space 𝑋, and let the mapping 𝐵𝑖 : 𝐶 → 𝑋
be 𝜆 𝑖 -strictly pseudocontractive and 𝛼𝑖 -strongly accretive with
𝛼𝑖 + 𝜆 𝑖 ≥ 1 for 𝑖 = 1, 2. Then, for 𝜇𝑖 ∈ (0, 1], we have
󵄩󵄩
󵄩
󵄩󵄩(𝐼 − 𝜇𝑖 𝐵𝑖 ) 𝑥 − (𝐼 − 𝜇𝑖 𝐵𝑖 ) 𝑦󵄩󵄩󵄩
⇐⇒ 𝑢 ∈ Fix (𝑇) .
(97)
So, we conclude that 𝐹 = Fix(𝑆) ∩ Ω = Fix(𝑆) ∩ Fix(𝑇).
Therefore, the desired result follows from Corollary 15.
Remark 17. Theorem 14 improves, extends, supplements, and
develops Cai and Bu [2, Theorem 3.1 and Corollary 3.2] and
Jung [5, Theorem 3.1] in the following aspects.
(i) The problem of finding a point 𝑞 ∈ ⋂𝑛 Fix(𝑆𝑛 ) ∩ Ω in
Theorem 14 is more general and more subtle than the
problem of finding a point 𝑞 ∈ Fix(𝑆) ∩ VI(𝐶, 𝐴) in
Jung [5, Theorem 3.1].
(ii) The iterative scheme in [2, Theorem 3.1] is extended
to develop the iterative scheme (27) of Theorem 14
by virtue of the iterative scheme of [5, Theorem 3.1].
The iterative scheme (27) of Theorem 14 is more
advantageous and more flexible than the iterative
scheme of [2, Theorem 3.1] because it involves several
parameter sequences {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, and {𝛿𝑛 }.
(iii) The iterative scheme (27) in Theorem 14 is very different from everyone in both [2, Theorem 3.1] and [5,
Theorem 3.1] because the mappings 𝑆𝑛 and 𝐺 in the
≤ {√
1 − 𝛼𝑖
1
󵄩
󵄩
+ (1 − 𝜇𝑖 ) (1 + )} 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,
𝜆𝑖
𝜆𝑖
(98)
∀𝑥, 𝑦 ∈ 𝐶,
for 𝑖 = 1, 2. In particular, if 1−(𝜆 𝑖 /(1+𝜆 𝑖 ))(1−√(1 − 𝛼𝑖 )/𝜆 𝑖 ) ≤
𝜇𝑖 ≤ 1, then 𝐼 − 𝜇𝑖 𝐵𝑖 is nonexpansive for 𝑖 = 1, 2.
Proof. Taking into account the 𝜆 𝑖 -strict pseudocontractivity
of 𝐵𝑖 , we derive for every 𝑥, 𝑦 ∈ 𝐶
󵄩
󵄩2
𝜆 𝑖 󵄩󵄩󵄩(𝐼 − 𝐵𝑖 ) 𝑥 − (𝐼 − 𝐵𝑖 ) 𝑦󵄩󵄩󵄩
≤ ⟨(𝐼 − 𝐵𝑖 ) 𝑥 − (𝐼 − 𝐵𝑖 ) 𝑦, 𝐽 (𝑥 − 𝑦)⟩
(99)
󵄩
󵄩󵄩
󵄩
≤ 󵄩󵄩󵄩(𝐼 − 𝐵𝑖 ) 𝑥 − (𝐼 − 𝐵𝑖 ) 𝑦󵄩󵄩󵄩 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,
which implies that
󵄩󵄩
󵄩
󵄩 1 󵄩󵄩
󵄩󵄩(𝐼 − 𝐵𝑖 ) 𝑥 − (𝐼 − 𝐵𝑖 ) 𝑦󵄩󵄩󵄩 ≤
󵄩𝑥 − 𝑦󵄩󵄩󵄩 .
𝜆𝑖 󵄩
(100)
Journal of Applied Mathematics
13
Proof. According to Lemma 10, we know that 𝐼 − 𝜇𝑖 𝐵𝑖 is
nonexpansive for 𝑖 = 1, 2. Hence, for all 𝑥, 𝑦 ∈ 𝐶, we have
Hence,
󵄩󵄩
󵄩
󵄩󵄩𝐵𝑖 𝑥 − 𝐵𝑖 𝑦󵄩󵄩󵄩
󵄩
󵄩
󵄩 󵄩
≤ 󵄩󵄩󵄩(𝐼 − 𝐵𝑖 ) 𝑥 − (𝐼 − 𝐵𝑖 ) 𝑦󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩
(101)
1 󵄩
󵄩
≤ (1 + ) 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 .
𝜆𝑖
󵄩
− Π𝐶 [Π𝐶 (𝑦 − 𝜇2 𝐵2 𝑦) − 𝜇1 𝐵1 Π𝐶 (𝑦 − 𝜇2 𝐵2 𝑦)]󵄩󵄩󵄩
Utilizing the 𝛼𝑖 -strong accretivity and 𝜆 𝑖 -strict pseudocontractivity of 𝐵𝑖 , we get
󵄩2
󵄩
≤ 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 − ⟨𝐵𝑖 𝑥 − 𝐵𝑖 𝑦, 𝐽 (𝑥 − 𝑦)⟩
(102)
󵄩
− Π𝐶 (𝐼 − 𝜇1 𝐵1 ) Π𝐶 (𝐼 − 𝜇2 𝐵2 ) 𝑦󵄩󵄩󵄩
󵄩
− (𝐼 − 𝜇1 𝐵1 ) Π𝐶 (𝐼 − 𝜇2 𝐵2 ) 𝑦󵄩󵄩󵄩
󵄩
󵄩
≤ 󵄩󵄩󵄩Π𝐶 (𝐼 − 𝜇2 𝐵2 ) 𝑥 − Π𝐶 (𝐼 − 𝜇2 𝐵2 ) 𝑦󵄩󵄩󵄩
󵄩2
󵄩
≤ (1 − 𝛼𝑖 ) 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 .
󵄩
󵄩
≤ 󵄩󵄩󵄩(𝐼 − 𝜇2 𝐵2 ) 𝑥 − (𝐼 − 𝜇2 𝐵2 ) 𝑦󵄩󵄩󵄩
So, we have
1 − 𝛼𝑖 󵄩󵄩
󵄩
󵄩󵄩
󵄩
󵄩󵄩(𝐼 − 𝐵𝑖 ) 𝑥 − (𝐼 − 𝐵𝑖 ) 𝑦󵄩󵄩󵄩 ≤ √
󵄩𝑥 − 𝑦󵄩󵄩󵄩 .
𝜆 󵄩
󵄩
󵄩
≤ 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 .
This shows that 𝐺 : 𝐶 → 𝐶 is nonexpansive. This completes
the proof.
Therefore, for 𝜇𝑖 ∈ (0, 1], we have
󵄩
󵄩󵄩
󵄩󵄩(𝐼 − 𝜇𝑖 𝐵𝑖 ) 𝑥 − (𝐼 − 𝜇𝑖 𝐵𝑖 ) 𝑦󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
≤ 󵄩󵄩󵄩(𝐼 − 𝐵𝑖 ) 𝑥 − (𝐼 − 𝐵𝑖 ) 𝑦󵄩󵄩󵄩 + (1 − 𝜇𝑖 ) 󵄩󵄩󵄩𝐵𝑖 𝑥 − 𝐵𝑖 𝑦󵄩󵄩󵄩
1 − 𝛼𝑖 󵄩󵄩
1 󵄩
󵄩
󵄩
󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 + (1 − 𝜇𝑖 ) (1 + ) 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩
𝜆𝑖
𝜆𝑖
1 − 𝛼𝑖
1
󵄩
󵄩
+ (1 − 𝜇𝑖 ) (1 + )} 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 .
𝜆𝑖
𝜆𝑖
Lemma 20. Let 𝐶 be a nonempty, closed, and convex subset
of a smooth Banach space 𝑋. Let Π𝐶 be a sunny nonexpansive
retraction from 𝑋 onto 𝐶, and let the mapping 𝐵𝑖 : 𝐶 → 𝑋 be
𝜆 𝑖 -strictly pseudocontractive and 𝛼𝑖 -strongly accretive for 𝑖 =
1, 2. For given 𝑥∗ , 𝑦∗ ∈ 𝐶, (𝑥∗ , 𝑦∗ ) is a solution of GSVI (9)
if and only if 𝑥∗ = Π𝐶(𝑦∗ − 𝜇1 𝐵1 𝑦∗ ) where 𝑦∗ = Π𝐶(𝑥∗ −
𝜇2 𝐵2 𝑥∗ ).
Proof. We can rewrite GSVI (9) as
(104)
Since 1 − (𝜆 𝑖 /(1 + 𝜆 𝑖 ))(1 − √(1 − 𝛼i )/𝜆 𝑖 ) ≤ 𝜇𝑖 ≤ 1, it follows
immediately that
1 − 𝛼𝑖
1
√
+ (1 − 𝜇𝑖 ) (1 + ) ≤ 1.
𝜆𝑖
𝜆𝑖
(107)
(103)
𝑖
= {√
󵄩
= 󵄩󵄩󵄩Π𝐶 (𝐼 − 𝜇1 𝐵1 ) Π𝐶 (𝐼 − 𝜇2 𝐵2 ) 𝑥
󵄩
≤ 󵄩󵄩󵄩(𝐼 − 𝜇1 𝐵1 ) Π𝐶 (𝐼 − 𝜇2 𝐵2 ) 𝑥
󵄩
󵄩2
𝜆 𝑖 󵄩󵄩󵄩(𝐼 − 𝐵𝑖 ) 𝑥 − (𝐼 − 𝐵𝑖 ) 𝑦󵄩󵄩󵄩
≤√
󵄩
󵄩󵄩
󵄩󵄩𝐺 (𝑥) − 𝐺 (𝑦)󵄩󵄩󵄩
󵄩
= 󵄩󵄩󵄩Π𝐶 [Π𝐶 (𝑥 − 𝜇2 𝐵2 𝑥) − 𝜇1 𝐵1 Π𝐶 (𝑥 − 𝜇2 𝐵2 𝑥)]
⟨𝑥∗ − (𝑦∗ − 𝜇1 𝐵1 𝑦∗ ) , 𝐽 (𝑥 − 𝑥∗ )⟩ ≥ 0,
∀𝑥 ∈ 𝐶,
⟨𝑦∗ − (𝑥∗ − 𝜇2 𝐵2 𝑥∗ ) , 𝐽 (𝑥 − 𝑦∗ )⟩ ≥ 0,
∀𝑥 ∈ 𝐶,
which is obviously equivalent to
𝑥∗ = Π𝐶 (𝑦∗ − 𝜇1 𝐵1 𝑦∗ ) ,
(105)
𝑦∗ = Π𝐶 (𝑥∗ − 𝜇2 𝐵2 𝑥∗ ) ,
This implies that 𝐼 − 𝜇𝑖 𝐵𝑖 is nonexpansive for 𝑖 = 1, 2.
because of Lemma 5. This completes the proof.
Lemma 19. Let 𝐶 be a nonempty, closed, and convex subset
of a smooth Banach space 𝑋. Let Π𝐶 be a sunny nonexpansive
retraction from 𝑋 onto 𝐶, and let the mapping 𝐵𝑖 : 𝐶 → 𝑋
be 𝜆 𝑖 -strictly pseudocontractive and 𝛼𝑖 -strongly accretive with
𝛼𝑖 + 𝜆 𝑖 ≥ 1 for 𝑖 = 1, 2. Let 𝐺 : 𝐶 → 𝐶 be the mapping defined
by
Remark 21. By Lemma 20, we observe that
𝐺 (𝑥) = Π𝐶 [Π𝐶 (𝑥 − 𝜇2 𝐵2 𝑥)
(108)
(109)
𝑥∗ = Π𝐶 [Π𝐶 (𝑥∗ − 𝜇2 𝐵2 𝑥∗ ) − 𝜇1 𝐵1 Π𝐶 (𝑥∗ − 𝜇2 𝐵2 𝑥∗ )] ,
(110)
which implies that 𝑥∗ is a fixed point of the mapping 𝐺.
Throughout this paper, the set of fixed points of the mapping
𝐺 is denoted by Ω.
(106)
We are now in a position to state and prove our result on
the explicit iterative scheme.
If 1−(𝜆 𝑖 /(1+𝜆 𝑖 ))(1−√(1 − 𝛼𝑖 )/𝜆 𝑖 ) ≤ 𝜇𝑖 ≤ 1, then 𝐺 : 𝐶 → 𝐶
is nonexpansive.
Theorem 22. Let 𝐶 be a nonempty, closed, and convex subset
of a uniformly convex Banach space 𝑋 which has a uniformly
− 𝜇1 𝐵1 Π𝐶 (𝑥 − 𝜇2 𝐵2 𝑥)] ,
∀𝑥 ∈ 𝐶.
14
Journal of Applied Mathematics
Gâteaux differentiable norm. Let Π𝐶 be a sunny nonexpansive
retraction from 𝑋 onto 𝐶. Let the mapping 𝐵𝑖 : 𝐶 → 𝑋 be 𝜆 𝑖 strictly pseudocontractive and 𝛼𝑖 -strongly accretive with 𝛼𝑖 +
𝜆 𝑖 ≥ 1 for 𝑖 = 1, 2. Let 𝑓 : 𝐶 → 𝐶 be a contraction with
coefficient 𝜌 ∈ (0, 1). Let {𝑆𝑛 }∞
𝑛=0 be an infinite family of
nonexpansive mappings of 𝐶 into itself such that 𝐹 =
⋂∞
𝑛=0 Fix(𝑆𝑛 ) ∩ Ω ≠ 0, where Ω is the fixed point set of the
mapping 𝐺 = Π𝐶(𝐼 − 𝜇1 𝐵1 )Π𝐶(𝐼 − 𝜇2 𝐵2 ). For arbitrarily given
𝑥0 ∈ 𝐶, let {𝑥𝑛 } be the sequence generated by
𝑦𝑛 = 𝛼𝑛 𝐺 (𝑥𝑛 ) + (1 − 𝛼𝑛 ) 𝑆𝑛 𝐺 (𝑥𝑛 ) ,
𝑥𝑛+1 = 𝛽𝑛 𝑓 (𝑥𝑛 ) + 𝛾𝑛 𝑦𝑛 + 𝛿𝑛 𝑆𝑛 𝐺 (𝑦𝑛 ) ,
∀𝑛 ≥ 0,
(111)
where 1 − (𝜆 𝑖 /(1 + 𝜆 𝑖 ))(1 − √(1 − 𝛼𝑖 )/𝜆 𝑖 ) ≤ 𝜇𝑖 ≤ 1 for 𝑖 = 1, 2
and {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, and {𝛿𝑛 } are the sequences in (0, 1) such
that 𝛽𝑛 + 𝛾𝑛 + 𝛿𝑛 = 1, ∀𝑛 ≥ 0. Suppose that the following
conditions hold:
(i) 0 < lim inf 𝑛 → ∞ 𝛼𝑛 ≤ lim sup𝑛 → ∞ 𝛼𝑛 < 1,
(ii) lim𝑛 → ∞ 𝛽𝑛 = 0 and ∑∞
𝑛=0 𝛽𝑛 = ∞,
(iii) ∑∞
𝑛=1 |𝛼𝑛 − 𝛼𝑛−1 | < ∞ or lim𝑛 → ∞ |𝛼𝑛 − 𝛼𝑛−1 |/𝛽𝑛 = 0,
(iv) ∑∞
𝑛=1 |𝛽𝑛 − 𝛽𝑛−1 | < ∞ or lim𝑛 → ∞ 𝛽𝑛−1 /𝛽𝑛 = 1,
(v) ∑∞
𝑛=1 |(𝛾𝑛 /(1 − 𝛽𝑛 )) − (𝛾𝑛−1 /(1 − 𝛽𝑛−1 ))| < ∞ or
lim𝑛 → ∞ (1/𝛽𝑛 )|(𝛾𝑛 /(1 − 𝛽𝑛 )) − (𝛾𝑛−1 /(1 − 𝛽𝑛−1 ))| = 0,
(vi) 0 < lim inf 𝑛 → ∞ 𝛾𝑛 ≤ lim sup𝑛 → ∞ 𝛾𝑛 < 1.
∀𝑝 ∈ 𝐹.
(112)
Proof. Take a fixed 𝑝 ∈ 𝐹 arbitrarily. Then by Lemma 20, we
know that 𝑝 = 𝐺(𝑝) and 𝑝 = 𝑆𝑛 𝑝 for all 𝑛 ≥ 0. Moreover, by
Lemma 19, we have
󵄩
󵄩󵄩
󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
≤ 𝛼𝑛 󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
≤ 𝛼𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
= 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 .
󵄩
󵄩󵄩
󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
+ 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩 󵄩
≤ 𝛽𝑛 (󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑓 (𝑝)󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩)
󵄩
󵄩
󵄩
󵄩
+ 𝛾𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
≤ 𝛽𝑛 𝜌 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩
󵄩
󵄩
+ (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
≤ 𝛽𝑛 𝜌 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩
󵄩
󵄩
+ (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
= (1 − 𝛽𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
+ 𝛽𝑛 󵄩󵄩󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩
󵄩
󵄩
= (1 − 𝛽𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩󵄩
󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩
+ 𝛽𝑛 (1 − 𝜌) ⋅ 󵄩
1−𝜌
󵄩
󵄩󵄩
󵄩 󵄩𝑓 (𝑝) − 𝑝󵄩󵄩󵄩
󵄩
≤ max {󵄩󵄩󵄩𝑥0 − 𝑝󵄩󵄩󵄩 , 󵄩
},
1−𝜌
(114)
which implies that {𝑥𝑛 } is bounded. By Lemma 19 we know
from (113) that {𝑦𝑛 }, {𝐺(𝑥𝑛 )}, and {𝐺(𝑦𝑛 )} are bounded.
Let us show that ‖ 𝑥𝑛+1 − 𝑥𝑛 ‖ → 0 and ‖ 𝑥𝑛 − 𝑦𝑛 ‖ → 0
as 𝑛 → ∞. As a matter of fact, from (113), we have
𝑦𝑛 = 𝛼𝑛 𝐺 (𝑥𝑛 ) + (1 − 𝛼𝑛 ) 𝑆𝑛 𝐺 (𝑥𝑛 ) ,
Assume that ∑∞
𝑛=1 sup𝑥∈𝐷 ‖ 𝑆𝑛 𝑥−𝑆𝑛−1 𝑥 ‖< ∞ for any bounded
subset 𝐷 of 𝐶, and let 𝑆 be a mapping of 𝐶 into itself defined
by 𝑆𝑥 = lim𝑛 → ∞ 𝑆𝑛 𝑥 for all 𝑥 ∈ 𝐶. Suppose that Fix(𝑆) =
⋂∞
𝑛=0 Fix(𝑆𝑛 ). Then {𝑥𝑛 } converges strongly to 𝑞 ∈ 𝐹, which
solves the following VIP:
⟨𝑞 − 𝑓 (𝑞) , 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,
From (113) we obtain
(113)
𝑦𝑛−1 = 𝛼𝑛−1 𝐺 (𝑥𝑛−1 ) + (1 − 𝛼𝑛−1 ) 𝑆𝑛−1 𝐺 (𝑥𝑛−1 ) ,
∀𝑛 ≥ 1.
(115)
Simple calculations show that
𝑦𝑛 − 𝑦𝑛−1 = 𝛼𝑛 (𝐺 (𝑥𝑛 ) − 𝐺 (𝑥𝑛−1 ))
+ (𝛼𝑛 − 𝛼𝑛−1 ) (𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )) (116)
+ (1 − 𝛼𝑛 ) (𝑆𝑛 𝐺 (𝑥n ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )) .
It follows that
󵄩󵄩
󵄩
󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩
󵄩
󵄩
≤ 𝛼𝑛 󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄨
󵄨󵄩
󵄩
+ 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
Journal of Applied Mathematics
15
󵄩
󵄩
≤ 𝛼𝑛 󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
≤ (𝛾𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩
󵄨
󵄨󵄩
󵄩
+ 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ 𝛿𝑛 ( 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑆𝑛 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩))
󵄩
󵄩
+ (1 − 𝛼𝑛 ) (󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛 ) − 𝑆𝑛 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩)
󵄩
󵄩
≤ 𝛼𝑛 󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄨
󵄨󵄩
󵄩
+ 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ (1 − 𝛼𝑛 ) (󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩)
󵄩
󵄩
≤ 󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄨
󵄨󵄩
󵄩
+ 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩 󵄨
󵄨
≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨
󵄩
󵄩
× 󵄩󵄩󵄩𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩 .
−1
× (1 − 𝛽𝑛 )
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨 󵄩𝑦 󵄩󵄩
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 𝑛−1 󵄩
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨󵄨 󵄩󵄩𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨
󵄩
󵄩
󵄩
󵄩
𝛾𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩
≤
𝛾𝑛 + 𝛿𝑛
(117)
Now, we write 𝑥𝑛 = 𝛽𝑛−1 𝑓(𝑥𝑛−1 ) + (1 − 𝛽𝑛−1 )V𝑛−1 , ∀𝑛 ≥ 1,
where V𝑛−1 = (𝑥𝑛 − 𝛽𝑛−1 𝑓(𝑥𝑛−1 ))/(1 − 𝛽𝑛−1 ). It follows that,
for all 𝑛 ≥ 1,
V𝑛 − V𝑛−1
=
𝑥𝑛+1 − 𝛽𝑛 𝑓 (𝑥𝑛 ) 𝑥𝑛 − 𝛽𝑛−1 𝑓 (𝑥𝑛−1 )
−
1 − 𝛽𝑛
1 − 𝛽𝑛−1
=
𝛾𝑛−1 𝑦𝑛−1 + 𝛿𝑛−1 𝑆𝑛−1 𝐺 (𝑦𝑛−1 )
1 − 𝛽𝑛−1
𝛾𝑛 (𝑦𝑛 − 𝑦𝑛−1 ) + 𝛿𝑛 (𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑆𝑛−1 𝐺 (𝑦𝑛−1 ))
1 − 𝛽𝑛
+(
𝛾𝑛
𝛾𝑛−1
−
)𝑦
1 − 𝛽𝑛 1 − 𝛽𝑛−1 𝑛−1
𝛿𝑛−1
𝛿
+( 𝑛 −
) 𝑆 𝐺 (𝑦𝑛−1 ) .
1 − 𝛽𝑛 1 − 𝛽𝑛−1 𝑛−1
This together with (117) implies that
󵄩󵄩
󵄩
󵄩󵄩V𝑛 − V𝑛−1 󵄩󵄩󵄩
󵄩󵄩
󵄩
󵄩𝛾 (𝑦 − 𝑦𝑛−1 ) + 𝛿𝑛 (𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑆𝑛−1 𝐺 (𝑦𝑛−1 ))󵄩󵄩󵄩
≤󵄩 𝑛 𝑛
1 − 𝛽𝑛
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨󵄨 󵄩󵄩𝑦𝑛−1 󵄩󵄩󵄩
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨
󵄨󵄨󵄨 𝛿
𝛿𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨󵄨 󵄩󵄩𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩 󵄩
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨 (󵄩𝑦 󵄩󵄩 + 󵄩󵄩𝑆 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩) .
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 𝑛−1 󵄩 󵄩 𝑛−1
(119)
𝛾 𝑦 + 𝛿𝑛 𝑆𝑛 𝐺 (𝑦𝑛 )
= 𝑛 𝑛
1 − 𝛽𝑛
−
𝛿𝑛 󵄩󵄩
󵄩
󵄩𝑆 𝐺 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩
𝛾𝑛 + 𝛿𝑛 󵄩 𝑛
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨
󵄨
+ 󵄨󵄨󵄨 𝑛 −
󵄨
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨󵄨
󵄩
󵄩 󵄩
󵄩
× (󵄩󵄩󵄩𝑦𝑛−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩)
󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩 󵄩
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨 (󵄩𝑦 󵄩󵄩 + 󵄩󵄩𝑆 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩)
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 𝑛−1 󵄩 󵄩 𝑛−1
󵄩
󵄩 󵄨
󵄨󵄩
󵄩
≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
+
(118)
Furthermore, we note that
𝑥𝑛+1 = 𝛽𝑛 𝑓 (𝑥𝑛 ) + (1 − 𝛽𝑛 ) V𝑛 ,
𝑥𝑛 = 𝛽𝑛−1 𝑓 (𝑥𝑛−1 ) + (1 − 𝛽𝑛−1 ) V𝑛−1 ,
∀𝑛 ≥ 1.
(120)
Also, simple calculations show that
𝑥𝑛+1 − 𝑥𝑛 = 𝛽𝑛 (𝑓 (𝑥𝑛 ) − 𝑓 (𝑥𝑛−1 )) + (𝛽𝑛 − 𝛽𝑛−1 )
× (𝑓 (𝑥𝑛−1 ) − V𝑛−1 ) + (1 − 𝛽𝑛 ) (V𝑛 − V𝑛−1 ) .
(121)
This together with (119) implies that
󵄩󵄩
󵄩
󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩
󵄩
󵄨
󵄩 󵄨
≤ 𝛽𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑓 (𝑥𝑛−1 )󵄩󵄩󵄩 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨
󵄩
󵄩
󵄩
󵄩
× 󵄩󵄩󵄩𝑓 (𝑥𝑛−1 ) − V𝑛−1 󵄩󵄩󵄩 + (1 − 𝛽𝑛 ) 󵄩󵄩󵄩V𝑛 − V𝑛−1 󵄩󵄩󵄩
16
Journal of Applied Mathematics
󵄩
󵄩
≤ 𝛽𝑛 𝜌 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩
Taking into account the boundedness of {𝐺(𝑥𝑛 )} and
{𝑆𝑛 𝐺(𝑥𝑛 )}, by Lemma 6, we know that there exists a continuous strictly increasing function 𝑔1 : [0, ∞) → [0, ∞),
𝑔1 (0) = 0 such that for 𝑝 ∈ 𝐹
󵄨
󵄨󵄩
󵄩
+ 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑓 (𝑥𝑛−1 ) − V𝑛−1 󵄩󵄩󵄩 + (1 − 𝛽𝑛 )
󵄩
󵄩
× [ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩
󵄩󵄩
󵄩2
󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
󵄨
󵄨󵄩
󵄩
+ 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨
󵄨
+ 󵄨󵄨󵄨 𝑛 −
󵄨
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨󵄨
󵄩
󵄩2
󵄩
󵄩2
≤ 𝛼𝑛 󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩
󵄩
󵄩
− 𝛼𝑛 (1 − 𝛼𝑛 ) 𝑔1 (󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝑆𝑛 𝐺 (𝑥𝑛 )󵄩󵄩󵄩)
󵄩
󵄩2
󵄩
󵄩2
≤ 𝛼𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
− 𝛼𝑛 (1 − 𝛼𝑛 ) 𝑔1 (󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝑆𝑛 𝐺 (𝑥𝑛 )󵄩󵄩󵄩)
󵄩
󵄩 󵄩
󵄩
× (󵄩󵄩󵄩𝑦𝑛−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩) ]
󵄩
󵄩2
󵄩
󵄩
= 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝛼𝑛 (1 − 𝛼𝑛 ) 𝑔1 (󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝑆𝑛 𝐺 (𝑥𝑛 )󵄩󵄩󵄩) .
(124)
󵄩
󵄩
≤ (1 − 𝛽𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩
󵄨
󵄨󵄩
󵄩
+ 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑓 (𝑥𝑛−1 ) − V𝑛−1 󵄩󵄩󵄩
Since {𝑦𝑛 } and {𝑆𝑛 𝐺(𝑦𝑛 )} are bounded, by Lemma 6, there
exists a continuous strictly increasing function 𝑔2 : [0, ∞) →
[0, ∞), 𝑔2 (0) = 0 such that for 𝑝 ∈ 𝐹
󵄨
󵄨󵄩
󵄩
+ 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩 󵄩
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨 (󵄩𝑦 󵄩󵄩 + 󵄩󵄩𝑆 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩)
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩 𝑛−1 󵄩 󵄩 𝑛−1
󵄩
󵄩
≤ (1 − 𝛽𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩
󵄨
󵄨
󵄨
󵄨
+ 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 𝑀 + 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 𝑀
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑦𝑛−1 )󵄩󵄩󵄩
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨
󵄨
+ 󵄨󵄨󵄨 𝑛 −
󵄨𝑀
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨󵄨
󵄩
󵄩
= (1 − 𝛽𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩
󵄨
󵄨
+ 𝑀 ( 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨
󵄨
𝛾𝑛−1 󵄨󵄨󵄨󵄨
󵄨
󵄨 󵄨󵄨 𝛾
+ 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨 𝑛 −
󵄨)
󵄨󵄨 1 − 𝛽𝑛 1 − 𝛽𝑛−1 󵄨󵄨󵄨
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛−1 ) − 𝑆𝑛−1 𝐺 (𝑥𝑛−1 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛−1 ) − 𝑆𝑛−1 𝐺 (y𝑛−1 )󵄩󵄩󵄩 ,
×[
𝛾𝑛
(𝑦 − 𝑝)
𝛾𝑛 + 𝛿𝑛 𝑛
+
𝛿𝑛
(𝑆 𝐺 (𝑦𝑛 ) − 𝑝)]
𝛾𝑛 + 𝛿𝑛 𝑛
󵄩󵄩2
󵄩
+ 𝛽𝑛 (𝑓 (𝑥𝑛 ) − 𝑝) 󵄩󵄩󵄩
󵄩󵄩
󵄩 𝛾
2󵄩
󵄩
𝑛
≤ (𝛾𝑛 + 𝛿𝑛 ) 󵄩󵄩󵄩
(𝑦𝑛 − 𝑝)
󵄩󵄩 𝛾𝑛 + 𝛿𝑛
+
󵄩󵄩2
𝛿𝑛
󵄩
(𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑝)󵄩󵄩󵄩
󵄩󵄩
𝛾𝑛 + 𝛿𝑛
+ 2𝛽𝑛 ⟨𝑓 (𝑥𝑛 ) − 𝑝, 𝐽 (𝑥𝑛+1 − 𝑝)⟩
2
(122)
where sup𝑛≥0 {‖ 𝑓(𝑥𝑛 ) ‖ + ‖ V𝑛 ‖ + ‖ 𝐺(𝑥𝑛 ) ‖ + ‖ 𝑆𝑛 𝐺(𝑥𝑛 ) ‖ +
‖ 𝑦𝑛 ‖ + ‖ 𝑆𝑛 𝐺(𝑦𝑛 ) ‖} ≤ 𝑀 for some 𝑀 > 0. Utilizing
Lemma 2, from conditions (ii)–(v) and the assumption on
{𝑆𝑛 }, we deduce that
󵄩
󵄩
lim 󵄩󵄩𝑥 − 𝑥𝑛 󵄩󵄩󵄩 = 0.
𝑛 → ∞ 󵄩 𝑛+1
󵄩󵄩
󵄩2
󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩
󵄩
= 󵄩󵄩󵄩𝛽𝑛 (𝑓 (𝑥𝑛 ) − 𝑝) + 𝛾𝑛 (𝑦𝑛 − 𝑝)
󵄩2
+ 𝛿𝑛 (𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑝)󵄩󵄩󵄩
󵄩󵄩
󵄩
= 󵄩󵄩󵄩 (𝛾𝑛 + 𝛿𝑛 )
󵄩󵄩
(123)
≤ (𝛾𝑛 + 𝛿𝑛 ) [
𝛾𝑛 󵄩󵄩
󵄩2
󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
𝛾𝑛 + 𝛿𝑛
+
−
𝛿𝑛 󵄩󵄩
󵄩2
󵄩𝑆 𝐺 (𝑦𝑛 ) − 𝑝󵄩󵄩󵄩
𝛾𝑛 + 𝛿𝑛 󵄩 𝑛
𝛾𝑛 𝛿𝑛
2
󵄩
󵄩
𝑔2 (󵄩󵄩󵄩𝑦𝑛 − 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩)]
(𝛾𝑛 + 𝛿𝑛 )
󵄩󵄩
󵄩󵄩
󵄩
+ 2𝛽𝑛 󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩
Journal of Applied Mathematics
17
󵄩
󵄩
󵄩2
󵄩2
≤ 𝛾𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
Thus, from (123), (130), and 𝛽𝑛 → 0, it follows that
󵄩
󵄩
− 𝛾𝑛 𝛿𝑛 𝑔2 (󵄩󵄩󵄩𝑦𝑛 − 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩)
󵄩󵄩
lim 󵄩𝑦
𝑛→∞ 󵄩 𝑛
󵄩
󵄩󵄩
󵄩
+ 2𝛽𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
󵄩
󵄩
≤ 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 − 𝛾𝑛 𝛿𝑛 𝑔2 (󵄩󵄩󵄩𝑦𝑛 − 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩)
󵄩󵄩
lim 󵄩𝑦
𝑛→∞ 󵄩 𝑛
󵄩
− 𝐺 (𝑥𝑛 )󵄩󵄩󵄩 = lim (1 − 𝛼𝑛 )
𝑛→∞
󵄩
󵄩
× 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛 ) − 𝐺 (𝑥𝑛 )󵄩󵄩󵄩 = 0.
(125)
󵄩󵄩
󵄩2
󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
󵄩
󵄩
≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝛼𝑛 (1 − 𝛼𝑛 ) 𝑔1 (󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝑆𝑛 𝐺 (𝑥𝑛 )󵄩󵄩󵄩)
󵄩
󵄩
− 𝛾𝑛 𝛿𝑛 𝑔2 (󵄩󵄩󵄩𝑦𝑛 − 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩)
󵄩
󵄩󵄩
󵄩
+ 2𝛽𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 .
It immediately follows that
󵄩
󵄩
𝛼𝑛 (1 − 𝛼𝑛 ) 𝑔1 (󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝑆𝑛 𝐺 (𝑥𝑛 )󵄩󵄩󵄩)
󵄩
󵄩
+ 𝛾𝑛 𝛿𝑛 𝑔2 (󵄩󵄩󵄩𝑦𝑛 − 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩)
󵄩
󵄩2 󵄩
󵄩2
≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩
󵄩
󵄩󵄩
󵄩
+ 2𝛽𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩 󵄩
󵄩 󵄩
≤ (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩
󵄩
󵄩󵄩
󵄩
+ 2𝛽𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 .
(126)
(127)
𝑛→∞
(128)
Since 𝛽𝑛 → 0 and ‖ 𝑥𝑛+1 − 𝑥𝑛 ‖ → 0, we conclude from conditions (i) and (vi) that
(𝐺 (𝑥𝑛 ) − 𝑆𝑛 𝐺 (𝑥𝑛 )) = 0,
Utilizing the properties of 𝑔1 and 𝑔2 , we have
󵄩
󵄩
lim 󵄩󵄩𝐺 (𝑥𝑛 ) − 𝑆𝑛 𝐺 (𝑥𝑛 )󵄩󵄩󵄩 = 0,
𝑛→∞ 󵄩
󵄩
󵄩
lim 󵄩󵄩𝑦 − 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩 = 0.
𝑛→∞ 󵄩 𝑛
Note that
󵄩󵄩
󵄩
󵄩󵄩𝑦𝑛 − 𝑥𝑛 󵄩󵄩󵄩
󵄩
= 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 − 𝛽𝑛 (𝑓 (𝑥𝑛 ) − 𝑦𝑛 )
󵄩
− 𝛿𝑛 (𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑦𝑛 )󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑦𝑛 󵄩󵄩󵄩
󵄩
󵄩
+ 𝛿𝑛 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑦𝑛 󵄩󵄩󵄩 .
󵄩
− 𝐺 (𝑥𝑛 )󵄩󵄩󵄩 = 0.
(134)
By (130) and Lemma 7, we have
󵄩
󵄩
≤ 󵄩󵄩󵄩𝑆𝐺 (𝑥𝑛 ) − 𝑆𝑛 𝐺 (𝑥𝑛 )󵄩󵄩󵄩
(135)
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑛 𝐺 (𝑥𝑛 ) − 𝐺 (𝑥𝑛 )󵄩󵄩󵄩 󳨀→ 0 as 𝑛 󳨀→ ∞.
In terms of (134) and (135), we have
lim inf 𝛿𝑛 = lim inf (1 − 𝛽𝑛 − 𝛾𝑛 ) = 1 − lim sup (𝛽𝑛 + 𝛾𝑛 ) > 0.
󵄩
󵄩
lim 𝑔 (󵄩󵄩𝑦 − 𝑆𝑛 𝐺 (𝑦𝑛 )󵄩󵄩󵄩) = 0.
𝑛→∞ 2 󵄩 𝑛
󵄩󵄩
lim 󵄩𝑥
𝑛→∞ 󵄩 𝑛
󵄩󵄩
󵄩
󵄩󵄩𝑆𝐺 (𝑥𝑛 ) − 𝐺 (𝑥𝑛 )󵄩󵄩󵄩
According to condition (vi), we get
lim 𝑔
𝑛→∞ 1
(133)
This together with (132) implies that
which together with (124) implies that
𝑛→∞
(132)
On the other hand, from (130), we get
󵄩
󵄩󵄩
󵄩
+ 2𝛽𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑝󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 ,
𝑛→∞
󵄩
− 𝑥𝑛 󵄩󵄩󵄩 = 0.
(129)
󵄩󵄩
󵄩
󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩𝑥𝑛 − 𝐺 (𝑥𝑛 )󵄩󵄩󵄩 + 󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝑆𝐺 (𝑥𝑛 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝐺 (𝑥𝑛 ) − 𝑆𝑥𝑛 󵄩󵄩󵄩
󵄩
󵄩
≤ 2 󵄩󵄩󵄩𝑥𝑛 − 𝐺 (𝑥𝑛 )󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝐺 (𝑥𝑛 ) − 𝑆𝐺 (𝑥𝑛 )󵄩󵄩󵄩 󳨀→ 0 as 𝑛 󳨀→ ∞.
(136)
Define a mapping 𝑊𝑥 = (1 − 𝜃)𝑆𝑥 + 𝜃𝐺(𝑥), where 𝜃 ∈ (0, 1) is
a constant. Then by Lemma 9, we have that Fix(𝑊) = Fix(𝑆)∩
Fix(𝐺) = 𝐹. We observe that
󵄩
󵄩󵄩
󵄩󵄩𝑥𝑛 − 𝑊𝑥𝑛 󵄩󵄩󵄩
󵄩
󵄩
= 󵄩󵄩󵄩(1 − 𝜃) (𝑥𝑛 − 𝑆𝑥𝑛 ) + 𝜃 (𝑥𝑛 − 𝐺 (𝑥𝑛 ))󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
≤ (1 − 𝜃) 󵄩󵄩󵄩𝑥𝑛 − 𝑆𝑥𝑛 󵄩󵄩󵄩 + 𝜃 󵄩󵄩󵄩𝑥𝑛 − 𝐺 (𝑥𝑛 )󵄩󵄩󵄩 .
(137)
From (134) and (136), we obtain
lim ‖ 𝑥𝑛 − 𝑊𝑥𝑛 ‖= 0.
𝑛→∞
(130)
(138)
Now, we claim that
lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩ ≤ 0,
𝑛→∞
(139)
where 𝑞 = 𝑠 − lim𝑡 → 0 𝑥𝑡 with 𝑥𝑡 being the fixed point of the
contraction
𝑥 󳨃󳨀→ 𝑡𝑓 (𝑥) + (1 − 𝑡) 𝑊𝑥.
(131)
(140)
Then 𝑥𝑡 solves the fixed point equation 𝑥𝑡 = 𝑡𝑓(𝑥𝑡 ) + (1 −
𝑡)𝑊𝑥𝑡 . Thus we have
󵄩󵄩
󵄩 󵄩
󵄩
󵄩󵄩𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩 = 󵄩󵄩󵄩(1 − 𝑡) (𝑊𝑥𝑡 − 𝑥𝑛 ) + 𝑡 (𝑓 (𝑥𝑡 ) − 𝑥𝑛 )󵄩󵄩󵄩 .
(141)
18
Journal of Applied Mathematics
By Lemma 3, we conclude that
Hence it follows that
󵄩󵄩
󵄩2
󵄩󵄩𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩
lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩
𝑛→∞
2󵄩
󵄩
󵄩2
≤ (1 − 𝑡) 󵄩󵄩𝑊𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩
≤ lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞) − 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩
𝑛→∞
+ 2𝑡 ⟨𝑓 (𝑥𝑡 ) − 𝑥𝑛 , 𝐽 (𝑥𝑡 − 𝑥𝑛 )⟩
󵄩
󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑥𝑡 − 𝑞󵄩󵄩󵄩 lim sup 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑡 󵄩󵄩󵄩
󵄩 󵄩
󵄩2
󵄩
≤ (1 − 𝑡) (󵄩󵄩󵄩𝑊𝑥𝑡 − 𝑊𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑊𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩)
2
𝑛→∞
(148)
󵄩
󵄩
󵄩
󵄩
+ 𝜌 󵄩󵄩󵄩𝑞 − 𝑥𝑡 󵄩󵄩󵄩 lim sup 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑡 󵄩󵄩󵄩
+ 2𝑡 ⟨𝑓 (𝑥𝑡 ) − 𝑥𝑛 , 𝐽 (𝑥𝑡 − 𝑥𝑛 )⟩
𝑛→∞
󵄩
󵄩
󵄩
≤ (1 − 𝑡)2 [󵄩󵄩󵄩𝑥𝑡 − 𝑥𝑛 󵄩󵄩 + 2 󵄩󵄩󵄩𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩
󵄩󵄩2
+ lim sup ⟨𝑓 (𝑥𝑡 ) − 𝑥𝑡 , 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩ .
(142)
󵄩
󵄩 󵄩
󵄩2
× 󵄩󵄩󵄩𝑊𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑊𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩 ]
𝑛→∞
Taking into account that 𝑥𝑡 → 𝑞 as 𝑡 → 0, we have from
(146)
+ 2𝑡 ⟨𝑓 (𝑥𝑡 ) − 𝑥𝑡 , 𝐽 (𝑥𝑡 − 𝑥𝑛 )⟩
+ 2𝑡 ⟨𝑥𝑡 − 𝑥𝑛 , 𝐽 (𝑥𝑡 − 𝑥𝑛 )⟩
󵄩2
󵄩
= (1 − 2𝑡 + 𝑡2 ) 󵄩󵄩󵄩𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩 + 𝑓𝑛 (𝑡)
lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩
𝑛→∞
󵄩2
󵄩
+ 2𝑡 ⟨𝑓 (𝑥𝑡 ) − 𝑥𝑡 , 𝐽 (𝑥𝑡 − 𝑥𝑛 )⟩ + 2𝑡󵄩󵄩󵄩𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩 ,
= lim sup lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩
𝑡→0
𝑛→∞
≤ lim sup lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞) − 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩ .
where
󵄩 󵄩
󵄩
󵄩
𝑓𝑛 (𝑡) = (1 − 𝑡) (2 󵄩󵄩󵄩𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑊𝑥𝑛 󵄩󵄩󵄩)
󵄩
󵄩
× 󵄩󵄩󵄩𝑥𝑛 − 𝑊𝑥𝑛 󵄩󵄩󵄩 󳨀→ 0 as 𝑛 󳨀→ ∞.
𝑡→0
𝑛→∞
(149)
2
(143)
It follows from (142) that
𝑡󵄩
󵄩2 1
⟨𝑥𝑡 − 𝑓 (𝑥𝑡 ) , 𝐽 (𝑥𝑡 − 𝑥𝑛 )⟩ ≤ 󵄩󵄩󵄩𝑥𝑡 − 𝑥𝑛 󵄩󵄩󵄩 + 𝑓𝑛 (𝑡) .
2
2𝑡
(144)
Letting 𝑛 → ∞ in (144) and noticing (143), we derive
𝑡
lim sup ⟨𝑥𝑡 − 𝑓 (𝑥𝑡 ) , 𝐽 (𝑥𝑡 − 𝑥𝑛 )⟩ ≤ 𝑀2 ,
2
𝑛→∞
Since 𝑋 has a uniformly Gâteaux differentiable norm, the
duality mapping 𝐽 is norm-to-weak∗ uniformly continuous
on bounded subsets of 𝑋. Consequently, the two limits are
interchangeable, and hence (139) holds. From (123), we get
(𝑥𝑛+1 − 𝑞) − (𝑥𝑛 − 𝑞) → 0. Noticing the norm-to-weak∗
uniform continuity of 𝐽 on bounded subsets of 𝑋, we deduce
from (139) that
lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩
𝑛→∞
(145)
= lim sup (⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞) − 𝐽 (𝑥𝑛 − 𝑞)⟩
𝑛→∞
2
where 𝑀2 > 0 is a constant such that ‖ 𝑥𝑡 − 𝑥𝑛 ‖ ≤ 𝑀2 for all
𝑡 ∈ (0, 1) and 𝑛 ≥ 0. Taking 𝑡 → 0 in (145), we have
lim sup lim sup ⟨𝑥𝑡 − 𝑓 (𝑥𝑡 ) , 𝐽 (𝑥𝑡 − 𝑥𝑛 )⟩ ≤ 0.
𝑡→0
𝑛→∞
+ ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩)
= lim sup ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩ ≤ 0.
𝑛→∞
(150)
(146)
Finally, let us show that 𝑥𝑛 → 𝑞 as 𝑛 → ∞. We observe
that
On the other hand, we have
⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞)⟩
󵄩󵄩
󵄩 󵄩
󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 = 󵄩󵄩󵄩𝛼𝑛 (𝐺 (𝑥𝑛 ) − 𝑞)
= ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛 − 𝑞) − 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩
+ ⟨𝑥𝑡 − 𝑞, 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩
+ ⟨𝑓 (𝑞) − 𝑓 (𝑥𝑡 ) , 𝐽 (𝑥𝑛 − 𝑥𝑡 )⟩
+ ⟨𝑓 (𝑥𝑡 ) − 𝑥𝑡 , 𝐽 (x𝑛 − 𝑥𝑡 )⟩ .
(147)
󵄩
+ (1 − 𝛼𝑛 ) (𝑆𝑛 𝐺 (𝑥𝑛 ) − 𝑞)󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
≤ 𝛼𝑛 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + (1 − 𝛼𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩
󵄩
󵄩
= 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 ,
(151)
Journal of Applied Mathematics
19
󵄩󵄩
󵄩2
󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩 = 𝛽𝑛 ⟨𝑓 (𝑥𝑛 ) − 𝑓 (𝑞) + 𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩
+ ⟨𝛾𝑛 (𝑦𝑛 − 𝑞)
+ 𝛿𝑛 (𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑞) , 𝐽 (𝑥𝑛+1 − 𝑞)⟩
󵄩
󵄩󵄩
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑓 (𝑥𝑛 ) − 𝑓 (𝑞)󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩
+ 𝛽𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩
󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝛾𝑛 (𝑦𝑛 − 𝑞) + 𝛿𝑛 (𝑆𝑛 𝐺 (𝑦𝑛 ) − 𝑞)󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩
󵄩
󵄩󵄩
󵄩
≤ 𝛽𝑛 𝜌 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩
(i) 0 < lim inf 𝑛 → ∞ 𝛼𝑛 ≤ lim sup𝑛 → ∞ 𝛼𝑛 < 1,
+ 𝛽𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩
(ii) lim𝑛 → ∞ 𝛽𝑛 = 0 and ∑∞
𝑛=0 𝛽𝑛 = ∞,
󵄩
󵄩󵄩
󵄩
+ (𝛾𝑛 + 𝛿𝑛 ) 󵄩󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩
(iii) ∑∞
𝑛=1 |𝛼𝑛 − 𝛼𝑛−1 | < ∞ or lim𝑛 → ∞ |𝛼𝑛 − 𝛼𝑛−1 |/𝛽𝑛 = 0,
(iv) ∑∞
𝑛=1 |𝛽𝑛 − 𝛽𝑛−1 | < ∞ or lim𝑛 → ∞ 𝛽𝑛−1 /𝛽𝑛 = 1,
󵄩
󵄩󵄩
󵄩
≤ 𝛽𝑛 𝜌 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩
(v) ∑∞
𝑛=1 |(𝛾𝑛 /(1 − 𝛽𝑛 )) − (𝛾𝑛−1 /(1 − 𝛽𝑛−1 ))| < ∞ or
lim𝑛 → ∞ (1/𝛽𝑛 )|(𝛾𝑛 /(1 − 𝛽𝑛 )) − (𝛾𝑛−1 /(1 − 𝛽𝑛−1 ))| = 0,
+ 𝛽𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩
󵄩
󵄩󵄩
󵄩
+ (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩
(vi) 0 < lim inf 𝑛 → ∞ 𝛾𝑛 ≤ lim sup𝑛 → ∞ 𝛾𝑛 < 1.
󵄩
󵄩󵄩
󵄩
= (1 − 𝛽𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩
Then {𝑥𝑛 } converges strongly to 𝑞 ∈ 𝐹, which solves the
following VIP:
+ 𝛽𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩
⟨𝑞 − 𝑓 (𝑞) , 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,
1 − 𝛽𝑛 (1 − 𝜌) 󵄩󵄩
󵄩2 󵄩
󵄩2
(󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩 )
2
1 󵄩󵄩
󵄩2
󵄩𝑥 − 𝑞󵄩󵄩󵄩
2 󵄩 𝑛+1
+ 𝛽𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩ .
(152)
So, we have
󵄩󵄩
󵄩2
󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩
󵄩
󵄩2
≤ (1 − 𝛽𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩
+ 2𝛽𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩
󵄩
󵄩2
= (1 − 𝛽𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩
+ 𝛽𝑛 (1 − 𝜌)
∀𝑝 ∈ 𝐹.
(155)
Further, we illustrate Theorem 22 by virtue of an example,
that is, the following corollary.
+ 𝛽𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩
1 − 𝛽𝑛 (1 − 𝜌) 󵄩󵄩
󵄩2
󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 +
2
(154)
∀𝑛 ≥ 0,
where 1 − (𝜆 𝑖 /(1 + 𝜆 𝑖 ))(1 − √(1 − 𝛼𝑖 )/𝜆 𝑖 ) ≤ 𝜇𝑖 ≤ 1 for 𝑖 = 1, 2.
Suppose that {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, and {𝛿𝑛 } are the sequences in (0, 1)
satisfying the following conditions:
󵄩
󵄩
󵄩
󵄩 󵄩
󵄩
+ (𝛾𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑞󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩
󵄩
󵄩󵄩
󵄩
= 𝛽𝑛 𝜌 󵄩󵄩󵄩𝑥𝑛 − 𝑞󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛+1 − 𝑞󵄩󵄩󵄩
≤
𝑦𝑛 = 𝛼𝑛 𝐺 (𝑥𝑛 ) + (1 − 𝛼𝑛 ) 𝑆𝐺 (𝑥𝑛 ) ,
𝑥𝑛+1 = 𝛽𝑛 𝑓 (𝑥𝑛 ) + 𝛾𝑛 𝑦𝑛 + 𝛿𝑛 𝑆𝐺 (𝑦𝑛 ) ,
+ 𝛽𝑛 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩
≤
Gâteaux differentiable norm. Let Π𝐶 be a sunny nonexpansive
retraction from 𝑋 onto 𝐶. Let the mapping 𝐵𝑖 : 𝐶 → 𝑋 be 𝜆 𝑖 strictly pseudocontractive and 𝛼𝑖 -strongly accretive with 𝛼𝑖 +
𝜆 𝑖 ≥ 1 for 𝑖 = 1, 2. Let 𝑓 : 𝐶 → 𝐶 be a contraction with
coefficient 𝜌 ∈ (0, 1). Let 𝑆 be a nonexpansive mapping of 𝐶
into itself such that 𝐹 = Fix(𝑆) ∩ Ω ≠ 0, where Ω is the fixed
point set of the mapping 𝐺 = Π𝐶(𝐼 − 𝜇1 𝐵1 )Π𝐶(𝐼 − 𝜇2 𝐵2 ). For
arbitrarily given 𝑥0 ∈ 𝐶, let {𝑥𝑛 } be the sequence generated by
Corollary 24. Let 𝐶 be a nonempty, closed, and convex subset
of a uniformly convex Banach space 𝑋 which has a uniformly
Gâteaux differentiable norm. Let Π𝐶 be a sunny nonexpansive
retraction from 𝑋 onto 𝐶. Let 𝑓 : 𝐶 → C be a contraction with
coefficient 𝜌 ∈ (0, 1). Let 𝑇 : 𝐶 → 𝐶 be a self-mapping on 𝐶
such that 𝐼 − 𝑇 is 𝜁-strictly pseudocontractive and 𝜃-strongly
accretive with 𝜁 + 𝜃 ≥ 1, and let 𝑆 be a nonexpansive mapping
of 𝐶 into itself such that 𝐹 = Fix(𝑆) ∩ Fix(T) ≠ 0. For arbitrarily
given 𝑥0 ∈ 𝐶, let {𝑥𝑛 } be the sequence generated by
𝑦𝑛 = 𝛼𝑛 𝐺 (𝑥𝑛 ) + (1 − 𝛼𝑛 ) 𝑆 (𝐼 − 𝜆 (𝐼 − 𝑇)) 𝑥𝑛 ,
(153)
2 ⟨𝑓 (𝑞) − 𝑞, 𝐽 (𝑥𝑛+1 − 𝑞)⟩
.
1−𝜌
Since ∑∞
𝑛=0 𝛽𝑛 = ∞ and lim sup𝑛 → ∞ ⟨𝑓(𝑞) − 𝑞, 𝐽(𝑥𝑛+1 − 𝑞)⟩ ≤
0, by Lemma 2, we conclude from (153) that 𝑥𝑛 → 𝑞 as 𝑛 →
∞. This completes the proof.
Corollary 23. Let 𝐶 be a nonempty, closed, and convex subset
of a uniformly convex Banach space 𝑋 which has a uniformly
𝑥𝑛+1 = 𝛽𝑛 𝑓 (𝑥𝑛 ) + 𝛾𝑛 𝑦𝑛 + 𝛿𝑛 𝑆 (𝐼 − 𝜆 (𝐼 − 𝑇)) 𝑦𝑛 ,
∀𝑛 ≥ 0,
(156)
where 1 − (𝜁/(1 + 𝜁))(1 − √(1 − 𝜃)/𝜁) ≤ 𝜆 ≤ 1. Suppose that
{𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, and {𝛿𝑛 } are the sequences in (0, 1) satisfying
the following conditions:
(i) 0 < lim inf 𝑛 → ∞ 𝛼𝑛 ≤ lim sup𝑛 → ∞ 𝛼𝑛 < 1,
(ii) lim𝑛 → ∞ 𝛽𝑛 = 0 and ∑∞
𝑛=0 𝛽𝑛 = ∞,
(iii) ∑∞
𝑛=1 |𝛼𝑛 − 𝛼𝑛−1 | < ∞ or lim𝑛 → ∞ |𝛼𝑛 − 𝛼𝑛−1 |/𝛽𝑛 = 0,
(iv) ∑∞
𝑛=1 |𝛽𝑛 − 𝛽𝑛−1 | < ∞ or lim𝑛 → ∞ 𝛽𝑛−1 /𝛽𝑛 = 1,
20
Journal of Applied Mathematics
(v) ∑∞
𝑛=1 |(𝛾𝑛 /(1 − 𝛽𝑛 )) − (𝛾𝑛−1 /(1 − 𝛽𝑛−1 ))| < ∞ or
lim𝑛 → ∞ (1/𝛽𝑛 )|(𝛾𝑛 /(1 − 𝛽𝑛 )) − (𝛾𝑛−1 /(1 − 𝛽𝑛−1 ))| = 0,
(vi) 0 < lim inf 𝑛 → ∞ 𝛾𝑛 ≤ lim sup𝑛 → ∞ 𝛾𝑛 < 1.
Then {𝑥𝑛 } converges strongly to 𝑞 ∈ 𝐹, which solves the following VIP:
⟨𝑞 − 𝑓 (𝑞) , 𝐽 (𝑞 − 𝑝)⟩ ≤ 0,
∀𝑝 ∈ 𝐹.
(157)
Proof. Utilizing the arguments similar to those in the proof
of Corollary 16, we can obtain the desired result.
Remark 25. As previous, we emphasize that our composite
iterative algorithms (i.e., the iterative schemes (27) and
(111)) are based on Korpelevich’s extragradient method and
viscosity approximation method. It is well known that the
so-called viscosity approximation method must contain a
contraction 𝑓 on 𝐶. In the meantime, it is worth pointing
out that our proof of Theorems 14 and 22 must make use of
Lemma 8 for implicit viscosity approximation method; that
is, Lemma 8 plays a key role in our proof of Theorems 14
and 22. Therefore, there is no doubt that the contraction 𝑓
in Theorems 14 and 22 cannot be replaced by a general 𝑘Lipschitzian mapping with constant 𝑘 ≥ 0.
Remark 26. Theorem 22 improves, extends, supplements,
and develops [2, Theorem 3.1 and Corollary 3.2] and [5,
Theorems 3.1] in the following aspects.
(i) The problem of finding a point 𝑞 ∈ ⋂𝑛 Fix(𝑆𝑛 ) ∩ Ω in
Theorem 22 is more general and more subtle than the
problem of finding a point 𝑞 ∈ Fix(𝑆) ∩ VI(𝐶, 𝐴) in
Jung [5, Theorem 3.1].
(ii) The iterative scheme in [2, Theorem 3.1] is extended
to develop the iterative scheme (111) of Theorem 22
by virtue of the iterative scheme of [5, Theorem 3.1].
The iterative scheme (111) in Theorem 22 is more
advantageous and more flexible than the iterative
scheme in [2, Theorem 3.1] because it involves several
parameter sequences {𝛼𝑛 }, {𝛽𝑛 }, {𝛾𝑛 }, and {𝛿𝑛 }.
(iii) The iterative scheme (111) in Theorem 22 is very
different from everyone in both [2, Theorem 3.1]
and [5, Theorem 3.1] because the mappings 𝑆𝑛 and
𝐺 in the iterative scheme of [2, Theorem 3.1] and
the mapping 𝑆𝑃𝐶(𝐼 − 𝜆 𝑛 𝐴) in the iterative scheme
of [5, Theorem 3.1] are replaced by the same composite mapping 𝑆𝑛 𝐺 in the iterative scheme (111) of
Theorem 22.
(iv) The proof in [2, Theorem 3.1] depends on the argument techniques in [3], the inequality in 2-uniformly
smooth Banach spaces, and the inequality in smooth
and uniform convex Banach spaces. However, the
proof of Theorem 22 does not depend on the argument techniques in [3], the inequality in 2-uniformly
smooth Banach spaces, and the inequality in smooth
and uniform convex Banach spaces. It depends on
only the inequality in uniform convex Banach spaces.
(v) The assumption of the uniformly convex and 2uniformly smooth Banach space 𝑋 in [2, Theorem 3.1] is weakened to the one of the uniformly
convex Banach space 𝑋 having a uniformly Gateaux
differentiable norm in Theorem 22.
(vi) The iterative scheme in [2, Corollary 3.2] is extended
to develop the new iterative scheme in Corollary 15
because the mappings 𝑆 and 𝐺 are replaced by the
same composite mapping 𝑆𝐺 in Corollary 23.
Finally, we observe that related results can be found in
recent papers, for example, [15–24] and the references therein.
Acknowledgments
This research was partially supported by the National Science
Foundation of China (11071169), Innovation Program of
Shanghai Municipal Education Commission (09ZZ133), and
Ph.D. Program Foundation of Ministry of Education of China
(20123127110002). This research was partially supported by a
Grant from NSC 101-2115-M-037-001.
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21
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