Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2013, Article ID 531859, 19 pages
http://dx.doi.org/10.1155/2013/531859
Research Article
Relaxed Viscosity Approximation Methods with
Regularization for Constrained Minimization Problems
Lu-Chuan Ceng,1 Hong-Kun Xu,2,3 and Ching-Feng Wen4
1
Department of Mathematics, Shanghai Normal University, Scientific Computing Key Laboratory of Shanghai Universities,
Shanghai 200234, China
2
Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung 80424, Taiwan
3
Department of Mathematics, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia
4
Center for Fundamental Science, Kaohsiung Medical University, Kaohsiung 807, Taiwan
Correspondence should be addressed to Ching-Feng Wen; cfwen@kmu.edu.tw
Received 13 January 2013; Accepted 26 March 2013
Academic Editor: Luigi Muglia
Copyright © 2013 Lu-Chuan 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 a new relaxed viscosity approximation method with regularization and prove the strong convergence of the method
to a common fixed point of finitely many nonexpansive mappings and a strict pseudocontraction that also solves a convex
minimization problem and a suitable equilibrium problem.
1. Introduction
Let 𝐻 be a real Hilbert space with inner product ⟨⋅, ⋅⟩ and
norm ‖ ⋅ ‖, 𝐶 a nonempty closed convex subset of 𝐻, and 𝑃𝐶
the metric projection of 𝐻, onto 𝐶. Let 𝑇 : 𝐶 → 𝐶 be selfmapping on 𝐶. We denote by Fix(𝑇) the set of fixed points of
𝑇 and by R the set of all real numbers. A mapping 𝑇 : 𝐶 → 𝐶
is called 𝜁-strictly pseudocontractive if there exists a constant
𝜁 ∈ [0, 1) such that
󵄩2
󵄩
󵄩2
󵄩2 󵄩
󵄩󵄩
󵄩󵄩𝑇𝑥 − 𝑇𝑦󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 + 𝜁󵄩󵄩󵄩(𝐼 − 𝑇) 𝑥 − (𝐼 − 𝑇) 𝑦󵄩󵄩󵄩 ,
∀𝑥, 𝑦 ∈ 𝐶.
(1)
In particular, if 𝜁 = 0, then 𝑇 is called a nonexpansive
mapping. A mapping 𝐴 : 𝐶 → 𝐻 is called 𝛼-inverse strongly
monotone, if there exists a constant 𝛼 > 0 such that
󵄩2
󵄩
⟨𝐴𝑥 − 𝐴𝑦, 𝑥 − 𝑦⟩ ≥ 𝛼󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,
∀𝑥, 𝑦 ∈ 𝐶.
(2)
Let 𝑓 : 𝐶 → R be a convex and a continuous Fréchet differentiable functional. Consider the minimization problem
(MP) of minimizing 𝑓 over the constraint set 𝐶
min𝑓 (𝑥) ,
𝑥∈𝐶
(3)
where we assume the existence of minimizers. We denote by
Γ the set of minimizers of (3). The gradient-projection algorithm (GPA) generates a sequence {𝑥𝑛 } determined by the
gradient ∇𝑓 and the metric projection 𝑃𝐶 as follows:
𝑥𝑛+1 := 𝑃𝐶 (𝑥𝑛 − 𝜆∇𝑓 (𝑥𝑛 )) ,
∀𝑛 ≥ 0,
(4)
∀𝑛 ≥ 0,
(5)
or more generally,
𝑥𝑛+1 := 𝑃𝐶 (𝑥𝑛 − 𝜆 𝑛 ∇𝑓 (𝑥𝑛 )) ,
where, in both (4) and (5), the initial guess 𝑥0 is taken from 𝐶
arbitrarily, the parameters 𝜆 or 𝜆 𝑛 are positive real numbers.
The convergence of algorithms (4) and (5) depends on the
behavior of the gradient ∇𝑓. As a matter of fact, it is known
that if ∇𝑓 is strongly monotone and Lipschitz continuous,
then, for 0 < 𝜆 < 2𝛼/𝐿2 , the operator
𝑆 := 𝑃𝐶 (𝐼 − 𝜆∇𝑓)
(6)
is a contraction. Hence, the sequence {𝑥𝑛 } defined by the GPA
(4) converges in norm to the unique solution of (3). More
generally, if the sequence {𝜆 𝑛 } is chosen to satisfy the property
0 < lim inf 𝜆 𝑛 ≤ lim sup𝜆 𝑛 <
𝑛→∞
𝑛→∞
2𝛼
,
𝐿2
(7)
2
Journal of Applied Mathematics
then the sequence {𝑥𝑛 } defined by the GPA (5) converges in
norm to the unique minimizer of (3). If the gradient ∇𝑓 is
only assumed to be a Lipschitz continuous, then {𝑥𝑛 } can
only be weakly convergent if 𝐻 is infinite dimensional. A
counterexample is given by Xu in [1].
Since the Lipschitz continuity of the gradient ∇𝑓 implies
that it is inverse strongly monotone (ism), it can be expressed
as a proper convex combination of the identity mapping
and a nonexpansive mapping. Consequently, the GPA can be
rewritten as the composite of a projectionand an averaged
mapping which is again an averaged mapping. This shows that
averaged mappings play an important role in the GPA. Very
recently, Xu [1] used averaged mappings to study the convergence analysis of the GPA which is an operator-oriented
approach.
We observe that the regularization, in particular, the
traditional Tikhonov regularization, is usually used to solve
ill-posed optimization problems. Consider the following regularized minimization problem:
𝛼
min𝑓𝛼 (𝑥) := 𝑓 (𝑥) + ‖𝑥‖2 ,
(8)
𝑥∈𝐶
2
where 𝛼 > 0 is the regularization parameter and again 𝑓 is
convex with an 𝐿-Lipschitz continuous gradient ∇𝑓.
The advantage of a regularization method is that it is possible to get strong convergence to the minimum-norm solution of the optimization problem under investigation. The
disadvantage is however its implicity, and hence explicit
iterative methods seem more attractive. See, for example, [1].
Given a mapping 𝐴 : 𝐶 → 𝐻, the classical variational
inequality problem (VIP) is to find 𝑥∗ ∈ 𝐶 such that
⟨𝐴𝑥∗ , 𝑥 − 𝑥∗ ⟩ ≥ 0,
∀𝑥 ∈ 𝐶.
(9)
The solution set of VIP (9) is denoted by VI(𝐶, 𝐴). It is well
known that 𝑥∗ ∈ VI(𝐶, 𝐴) if and only if 𝑥∗ = 𝑃𝐶(𝑥∗ − 𝜆𝐴𝑥∗ )
for some 𝜆 > 0. The variational inequality was first discussed
by Lions [2] and now is well known. The variational inequality
theory has been studied quite extensively and has emerged
as an important tool in the study of a wide class of obstacle,
unilateral, free, moving, and equilibrium problems arising in
several branches of pure and applied sciences in a unified
and general framework. See, for example, [3–10] and the
references therein.
In this paper, we study the following equilibrium problem
(EP) which is to find 𝑥∗ ∈ 𝐶 such that
𝐹 (𝑥∗ , 𝑦) + ℎ (𝑥∗ , 𝑦) ≥ 0,
∀𝑦 ∈ 𝐶.
(10)
The solution set of EP (10) is denoted by EP(𝐹, ℎ). We will
introduce and consider a relaxed viscosity iterative scheme
with regularization for finding a common element of the
solution set Γ of the minimization problem (3), the solution
set EP(𝐹, ℎ) of the equilibrium problem (10), and the common
fixed point set Fix(𝑇) ∩ (⋂𝑖 Fix(𝑆𝑖 )) of finitely many nonexpansive mappings 𝑆𝑖 : 𝐶 → 𝐶, 𝑖 = 1, . . . , 𝑁, and a strictly
pseudocontractive mapping 𝑇 in the setting of the infinitedimensional Hilbert space. We will prove that this iterative
scheme converges strongly to a common fixed point of the
mappings 𝑇, 𝑆𝑖 : 𝐶 → 𝐶, 𝑖 = 1, . . . , 𝑁, which is both a
minimizer of MP (3) and an equilibrium point of EP (10).
2. Preliminaries
Let 𝐻 be a real Hilbert space whose inner product and norm
are denoted by ⟨⋅, ⋅⟩ and ‖⋅‖, respectively. Let 𝐾 be a nonempty
closed convex subset of 𝐻. We write 𝑥𝑛 ⇀ 𝑥 to indicate
that the sequence {𝑥𝑛 } converges weakly to 𝑥 and 𝑥𝑛 → 𝑥
to indicate that the sequence {𝑥𝑛 } converges strongly to 𝑥.
Moreover, we use 𝜔𝑤 (𝑥𝑛 ) to denote the weak 𝜔-limit set of
the sequence {𝑥𝑛 } and 𝜔𝑠 (𝑥𝑛 ) to denote the strong 𝜔-limit set
of the sequence {𝑥𝑛 }; that is,
𝜔𝑤 (𝑥𝑛 ) := {𝑥 ∈ 𝐻 : 𝑥𝑛𝑖 ⇀ 𝑥
for some subsequence {𝑥𝑛𝑖 } of {𝑥𝑛 }} ,
𝜔𝑠 (𝑥𝑛 ) :={𝑥 ∈ 𝐻 : 𝑥𝑛𝑖 󳨀→ 𝑥
for some subsequence {𝑥𝑛𝑖 } of {𝑥𝑛 }} .
(11)
The metric (or nearest point) projection from 𝐻 onto 𝐾 is
the mapping 𝑃𝐾 : 𝐻 → 𝐾 which assigns to each point 𝑥 ∈ 𝐻
the unique point 𝑃𝐾 𝑥 ∈ 𝐾 satisfying the property
󵄩
󵄩󵄩
󵄩
󵄩
󵄩󵄩𝑥 − 𝑃𝐾 𝑥󵄩󵄩󵄩 = inf 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 =: 𝑑 (𝑥, 𝐾) .
𝑦∈𝐾
(12)
Some important properties of projections are gathered in
the following.
Proposition 1. For given 𝑥 ∈ 𝐻 and 𝑧 ∈ 𝐾
(i) 𝑧 = 𝑃𝐾 𝑥 ⇔ ⟨𝑥 − 𝑧, 𝑦 − 𝑧⟩ ≤ 0, for all 𝑦 ∈ 𝐾;
(ii) 𝑧 = 𝑃𝐾 𝑥 ⇔ ‖𝑥−𝑧‖2 ≤ ‖𝑥−𝑦‖2 −‖𝑦−𝑧‖2 , for all 𝑦 ∈ 𝐾;
(iii) ⟨𝑃𝐾 𝑥 − 𝑃𝐾 𝑦, 𝑥 − 𝑦⟩ ≥ ‖𝑃𝐾 𝑥 − 𝑃𝐾 𝑦‖2 , for all 𝑦 ∈
𝐻, which hence implies that 𝑃𝐾 is nonexpansive and
monotone.
Definition 2. A mapping 𝑇 : 𝐻 → 𝐻 is said to be
(a) nonexpansive if
󵄩
󵄩 󵄩
󵄩󵄩
󵄩󵄩𝑇𝑥 − 𝑇𝑦󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,
∀𝑥, 𝑦 ∈ 𝐻;
(13)
(b) firmly nonexpansive if 2𝑇 − 𝐼 is nonexpansive, or
equivalently,
󵄩2
󵄩
⟨𝑥 − 𝑦, 𝑇𝑥 − 𝑇𝑦⟩ ≥ 󵄩󵄩󵄩𝑇𝑥 − 𝑇𝑦󵄩󵄩󵄩 ,
∀𝑥, 𝑦 ∈ 𝐻;
(14)
alternatively, 𝑇 is firmly nonexpansive if and only if 𝑇
can be expressed as
𝑇=
1
(𝐼 + 𝑆) ,
2
(15)
where 𝑆 : 𝐻 → 𝐻 is nonexpansive; projections are
firmly nonexpansive.
Definition 3. Let 𝑇 be a nonlinear operator with domain
𝐷(𝑇) ⊆ 𝐻 and range 𝑅(𝑇) ⊆ 𝐻.
Journal of Applied Mathematics
3
(a) 𝑇 is said to be monotone if
⟨𝑥 − 𝑦, 𝑇𝑥 − 𝑇𝑦⟩ ≥ 0,
∀𝑥, 𝑦 ∈ 𝐷 (𝑇) .
(16)
(b) Given a number 𝛽 > 0, 𝑇 is said to be 𝛽 strongly
monotone if
󵄩2
󵄩
(17)
⟨𝑥 − 𝑦, 𝑇𝑥 − 𝑇𝑦⟩ ≥ 𝛽󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 , ∀𝑥, 𝑦 ∈ 𝐷 (𝑇) .
(c) Given a number ] > 0, 𝑇 is said to be ]-inverse
strongly monotone (]-ism) if
󵄩2
󵄩
⟨𝑥 − 𝑦, 𝑇𝑥 − 𝑇𝑦⟩ ≥ ]󵄩󵄩󵄩𝑇𝑥 − 𝑇𝑦󵄩󵄩󵄩 , ∀𝑥, 𝑦 ∈ 𝐷 (𝑇) . (18)
It can be easily seen that if 𝑇 is nonexpansive, then
𝐼 − 𝑇 is monotone. It is also easy to see that a projection
𝑃𝐾 is 1-ism. Inverse strongly monotone (also referred to as
cocoercive) operators have been applied widely in solving
practical problems in various fields.
Definition 4. A mapping 𝑇 : 𝐻 → 𝐻 is said to be an averaged mapping if it can be written as the average of the identity
𝐼 and a nonexpansive mapping; that is,
𝑇 ≡ (1 − 𝛼) 𝐼 + 𝛼𝑆,
(19)
where 𝛼 ∈ (0, 1) and 𝑆 : 𝐻 → 𝐻 is nonexpansive. More
precisely, when the last equality holds, we say that 𝑇 is 𝛼averaged. Thus, firmly nonexpansive mappings (in particular,
projections) are (1/2)-averaged maps.
Proposition 5 (see [11]). Let 𝑇 : 𝐻 → 𝐻 be a given mapping.
The notation Fix(𝑇) denotes the set of all fixed points of the
mapping 𝑇, that is, Fix(𝑇) = {𝑥 ∈ 𝐻 : 𝑇𝑥 = 𝑥}.
It is clear that, in a real Hilbert space 𝐻, 𝑇 : 𝐶 → 𝐶
is 𝜁-strictly pseudocontractive if and only if there holds the
following inequality:
󵄩2 1 − 𝜁
󵄩
⟨𝑇𝑥 − 𝑇𝑦, 𝑥 − 𝑦⟩ ≤ 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 −
2
󵄩2
󵄩󵄩
× 󵄩󵄩(𝐼 − 𝑇)𝑥 − (𝐼 − 𝑇)𝑦󵄩󵄩󵄩 ,
∀𝑥, 𝑦 ∈ 𝐶.
(21)
This immediately implies that if 𝑇 is a 𝜁-strictly pseudocontractive mapping, then 𝐼 − 𝑇 is ((1 − 𝜁)/2)-inverse
strongly monotone; for further detail, we refer to [12] and
the references therein. It is well known that the class of strict
pseudocontractions strictly includes the class of nonexpansive mappings.
Lemma 7 (see [12, Proposition 2.1]). Let 𝐶 be a nonempty
closed convex subset of a real Hilbert space 𝐻 and 𝑇 : 𝐶 → 𝐶
be a mapping.
(i) If 𝑇 is a 𝜁-strictly pseudocontractive mapping, then 𝑇
satisfies the Lipschitz condition where
󵄩
󵄩 1 + 𝜁 󵄩󵄩
󵄩󵄩
󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,
󵄩󵄩𝑇𝑥 − 𝑇𝑦󵄩󵄩󵄩 ≤
1−𝜁󵄩
∀𝑥, 𝑦 ∈ 𝐶.
(22)
(ii) If 𝑇 is a 𝜁-strictly pseudocontractive mapping, then the
mapping 𝐼 − 𝑇 is semiclosed at 0; that is, if {𝑥𝑛 } is a
sequence in 𝐶 such that 𝑥𝑛 → 𝑥̃ weakly and (𝐼 −
𝑇)𝑥𝑛 → 0 strongly, then (𝐼 − 𝑇)𝑥̃ = 0.
(i) 𝑇 is nonexpansive if and only if the complement 𝐼−𝑇 is
(1/2)-ism.
(ii) If 𝑇 is ]-ism, then for 𝛾 > 0, 𝛾𝑇 is (]/𝛾)-ism.
(iii) 𝑇 is averaged if and only if the complement 𝐼 − 𝑇 is ]ism for some ] > 1/2. Indeed, for 𝛼 ∈ (0, 1), 𝑇 is 𝛼averaged if and only if 𝐼 − 𝑇 is (1/2𝛼)-ism.
(iii) If 𝑇 is a 𝜁-(quasi-)strict pseudocontraction, then the
fixed point set Fix(𝑇) of 𝑇 is closed and convex so that
the projection 𝑃Fix(𝑇) is well defined.
Proposition 6 (see [11]). Let 𝑆, 𝑇, 𝑉 : 𝐻 → 𝐻 be given
operators.
The following lemma is an immediate consequence of an
inner product.
(i) If 𝑇 = (1 − 𝛼)𝑆 + 𝛼𝑉 for some 𝛼 ∈ (0, 1) and if 𝑆 is
averaged and 𝑉 is nonexpansive, then 𝑇 is averaged.
(ii) 𝑇 is firmly nonexpansive if and only if the complement 𝐼 − 𝑇 is firmly nonexpansive.
(iii) If 𝑇 = (1 − 𝛼)𝑆 + 𝛼𝑉 for some 𝛼 ∈ (0, 1) and if 𝑆 is
firmly nonexpansive and 𝑉 is nonexpansive, then 𝑇 is
averaged.
(iv) The composite of finitely many averaged mappings is
averaged. That is, if each of the mappings {𝑇𝑖 }𝑁
𝑖=1 is
averaged, then so is the composite 𝑇1 ⋅ ⋅ ⋅ 𝑇𝑁. In particular, if 𝑇1 is 𝛼1 -averaged and 𝑇2 is 𝛼2 -averaged,
where 𝛼1 , 𝛼2 ∈ (0, 1), then the composite 𝑇1 𝑇2 is 𝛼averaged, where 𝛼 = 𝛼1 + 𝛼2 − 𝛼1 𝛼2 .
Lemma 8. In a real Hilbert space 𝐻, there holds the following
inequality:
(v) If the mappings {𝑇𝑖 }𝑁
𝑖=1 are averaged and have a common fixed point, then
𝑁
⋂ Fix (𝑇𝑖 ) = Fix (𝑇1 ⋅ ⋅ ⋅ 𝑇𝑁) .
𝑖=1
(20)
󵄩2
󵄩󵄩
2
󵄩󵄩𝑥 + 𝑦󵄩󵄩󵄩 ≤ ‖𝑥‖ + 2 ⟨𝑦, 𝑥 + 𝑦⟩ ,
∀𝑥, 𝑦 ∈ 𝐻.
(23)
The following elementary result on real sequences is quite
well known.
Lemma 9 (see [13]). Let {𝑎𝑛 } be a sequence of nonnegative real
numbers satisfying the property
𝑎𝑛+1 ≤ (1 − 𝑠𝑛 ) 𝑎𝑛 + 𝑠𝑛 𝑡𝑛 + 𝜖𝑛 ,
∀𝑛 ≥ 0,
(24)
where {𝑠𝑛 } ⊂ (0, 1] and {𝑡𝑛 } are the real sequences such that
(i) ∑∞
𝑛=0 𝑠𝑛 = ∞;
(ii) either lim sup𝑛 → ∞ 𝑡𝑛 ≤ 0 or ∑∞
𝑛=0 𝑠𝑛 |𝑡𝑛 | < ∞;
(iii) ∑∞
𝑛=0 𝜖𝑛 < ∞ where 𝜖𝑛 ≥ 0, for all 𝑛 ≥ 0.
Then, lim𝑛 → ∞ 𝑎𝑛 = 0.
4
Journal of Applied Mathematics
Lemma 10 (see [14]). Let 𝐶 be a nonempty closed convex
subset of a real Hilbert space 𝐻. Let 𝑇 : 𝐶 → 𝐶 be a 𝜁-strictly
pseudocontractive mapping. Let 𝛾 and 𝛿 be two nonnegative
real numbers such that (𝛾 + 𝛿)𝜁 ≤ 𝛾. Then,
󵄩󵄩󵄩𝛾 (𝑥 − 𝑦) + 𝛿 (𝑇𝑥 − 𝑇𝑦)󵄩󵄩󵄩 ≤ (𝛾 + 𝛿) 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 , ∀𝑥, 𝑦 ∈ 𝐶.
󵄩
󵄩
󵄩
󵄩
(25)
The following lemma appears implicitly in the paper of
Reinermann [15].
Lemma 11 (see [15]). Let 𝐻 be a real Hilbert space. Then, for
all 𝑥, 𝑦 ∈ 𝐻 and 𝜆 ∈ [0, 1],
󵄩2
󵄩󵄩
󵄩󵄩𝜆𝑥 + (1 − 𝜆) 𝑦󵄩󵄩󵄩
󵄩2
󵄩 󵄩2
󵄩
= 𝜆‖𝑥‖2 + (1 − 𝜆) 󵄩󵄩󵄩𝑦󵄩󵄩󵄩 − 𝜆 (1 − 𝜆) 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 .
(26)
(a) if ‖𝑥𝑛 − 𝑇𝑟𝑛 𝑥𝑛 ‖ → 0 as 𝑛 → ∞, each weak cluster
point of {𝑥𝑛 } satisfies the problem:
𝐹 (𝑥, 𝑦) + ℎ (𝑥, 𝑦) ≥ 0,
∀𝑦 ∈ 𝐶,
(29)
that is, 𝜔𝑤 (𝑥𝑛 ) ⊆ EP(𝐹, ℎ).
(b) The demiclosedness principle holds in the sense that, if
𝑥𝑛 ⇀ 𝑥∗ and ‖𝑥𝑛 − 𝑇𝑟𝑛 𝑥𝑛 ‖ → 0 as 𝑛 → ∞, then
(𝐼 − 𝑇𝑟𝑘 )𝑥∗ = 0 for all 𝑘 ≥ 1.
3. Main Results
Lemma 12 (see [16]). Let 𝐶 be a nonempty closed convex
subset of a real Hilbert space 𝐻. Let 𝐹 : 𝐶 × 𝐶 → R be a
bifunction such that
(f1) 𝐹(𝑥, 𝑥) = 0 for all 𝑥 ∈ 𝐶;
(f2) 𝐹 is monotone and upper hemicontinuous in the first
variable;
(f3) 𝐹 is lower semicontinuous and convex in the second
variable.
We now propose the following relaxed viscosity iterative
scheme with regularization:
𝐹 (𝑢𝑛 , 𝑦) + ℎ (𝑢𝑛 , 𝑦) +
1
⟨𝑦 − 𝑢𝑛 , 𝑢𝑛 − 𝑥𝑛 ⟩ ≥ 0,
𝑟𝑛
∀𝑦 ∈ 𝐶,
𝑦𝑛,1 = 𝛽𝑛,1 𝑆1 𝑢𝑛 + (1 − 𝛽𝑛,1 ) 𝑢𝑛 ,
𝑦𝑛,𝑖 = 𝛽𝑛,𝑖 𝑆𝑖 𝑢𝑛 + (1 − 𝛽𝑛,𝑖 ) 𝑦𝑛,𝑖−1 ,
Let ℎ : 𝐶 × 𝐶 → R be a bifunction such that
(h1) ℎ(𝑥, 𝑥) = 0 for all 𝑥 ∈ 𝐶;
(h2) ℎ is monotone and weakly upper semicontinuous in the
first variable;
(h3) ℎ is convex in the second variable.
Moreover, let one suppose that
(H) for fixed 𝑟 > 0 and 𝑥 ∈ 𝐶, there exists a bounded 𝐾 ⊂ 𝐶
̂ 𝑧) +
and 𝑥̂ ∈ 𝐾 such that for all 𝑧 ∈ 𝐶 \ 𝐾, −𝐹(𝑥,
̂ + (1/𝑟)⟨𝑥̂ − 𝑧, 𝑧 − 𝑥⟩ < 0.
ℎ(𝑧, 𝑥)
For 𝑟 > 0 and 𝑥 ∈ 𝐻, let 𝑇𝑟 : 𝐻 → 2𝐶 be a mapping defined
by
𝑇𝑟 𝑥
= {𝑧 ∈ 𝐶 : 𝐹 (𝑧, 𝑦) + ℎ (𝑧, 𝑦)
Lemma 14 (see [17]). Suppose that the hypotheses of
Lemma 12 are satisfied. Let {𝑟𝑛 } be a sequence in (0, ∞) with
lim inf 𝑛 → ∞ 𝑟𝑛 > 0. Suppose that {𝑥𝑛 } is a bounded sequence.
Then, the following statements are equivalent and true.
(27)
1
+ ⟨𝑦 − 𝑧, 𝑧 − 𝑥⟩ ≥ 0, ∀𝑦 ∈ 𝐶}
𝑟
called the resolvent of 𝐹 and ℎ. Then,
(1) 𝑇𝑟 𝑥 ≠ 0;
(2) 𝑇𝑟 𝑥 is a singleton;
(3) 𝑇𝑟 is firmly nonexpansive;
(4) 𝐸𝑃(𝐹, ℎ) = Fix(𝑇𝑟 ) and it is closed and convex.
Lemma 13 (see [16]). Let one suppose that (f1)–(f3), (h1)–(h3)
and (H) hold. Let 𝑥, 𝑦 ∈ 𝐻, 𝑟1 , 𝑟2 > 0. Then,
󵄨
󵄨
󵄩󵄩󵄩𝑇 𝑦 − 𝑇 𝑥󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑦 − 𝑥󵄩󵄩󵄩 + 󵄨󵄨󵄨󵄨 𝑟2 − 𝑟1 󵄨󵄨󵄨󵄨 󵄩󵄩󵄩𝑇 𝑦 − 𝑦󵄩󵄩󵄩 .
(28)
󵄩 󵄩
𝑟1 󵄩
󵄩 󵄨󵄨 𝑟 󵄨󵄨 󵄩󵄩 𝑟2
󵄩󵄩 𝑟2
󵄩󵄩
󵄨 2 󵄨
𝑖 = 2, . . . , 𝑁,
(30)
𝑦𝑛 = 𝛽𝑛 𝑄𝑦𝑛,𝑁 + (1 − 𝛽𝑛 ) 𝑃𝐶 (𝑦𝑛,𝑁 − 𝜆∇𝑓𝛼𝑛 (𝑦𝑛,𝑁)) ,
𝑥𝑛+1 = 𝜎𝑛 𝑦𝑛 + 𝛾𝑛 𝑃𝐶 (𝑦𝑛 − 𝜆∇𝑓𝛼𝑛 (𝑦𝑛 ))
+ 𝛿𝑛 𝑇𝑃𝐶 (𝑦𝑛 − 𝜆∇𝑓𝛼𝑛 (𝑦𝑛 )) ,
for all 𝑛 ≥ 0, where the mapping 𝑄 : 𝐶 → 𝐶 is a 𝜌contraction; the mapping 𝑇 : 𝐶 → 𝐶 is a 𝜁-strict pseudocontraction; 𝑆𝑖 : 𝐶 → 𝐶 is a nonexpansive mapping for each 𝑖 =
1, . . . , 𝑁; ∇𝑓 : 𝐶 → 𝐻 satisfies the Lipschitz condition (10)
with 0 < 𝜆 < (2/𝐿); 𝐹, ℎ : 𝐶 × 𝐶 → R are two bifunctions
satisfying the hypotheses of Lemma 12; {𝛼𝑛 } is a sequence in
(0, ∞) with ∑∞
𝑛=0 𝛼𝑛 < ∞; {𝛽𝑛 }, {𝜎𝑛 } are sequences in (0, 1)
with 0 < lim inf 𝑛 → ∞ 𝜎𝑛 ≤ lim sup𝑛 → ∞ 𝜎𝑛 < 1; {𝛾𝑛 }, {𝛿𝑛 } are
sequences in [0, 1] with 𝜎𝑛 + 𝛾𝑛 + 𝛿𝑛 = 1, for all 𝑛 ≥ 0; {𝛽𝑛,𝑖 }𝑁
𝑖=1
are sequences in (0, 1) and (𝛾𝑛 + 𝛿𝑛 )𝜁 ≤ 𝛾𝑛 , for all 𝑛 ≥ 0;
{𝑟𝑛 } is a sequence in (0, ∞) with lim inf 𝑛 → ∞ 𝑟𝑛 > 0 and
lim inf 𝑛 → ∞ 𝛿𝑛 > 0.
Before stating and proving the main convergence results,
we first establish the following lemmas.
Lemma 15. Let one suppose that Ω = Fix(𝑇) ∩ (⋂𝑖 Fix(𝑆𝑖 )) ∩
EP(𝐹, ℎ) ∩ Γ ≠ 0. Then, the sequences {𝑥𝑛 }, {𝑦𝑛 }, {𝑦𝑛,𝑖 } for all 𝑖,
and {𝑢𝑛 } are bounded.
Proof. First of all, we can show as in [18] that 𝑃𝐶(𝐼 − 𝜆∇𝑓𝛼 )
is nonexpansive for 𝜆 ∈ (0, 2/(𝛼 + 𝐿)), and 𝑃𝐶(𝐼 − 𝜆∇𝑓𝛼𝑛 ) is
nonexpansive for all 𝑛 ≥ 0 and 𝜆 ∈ (0, 2/𝐿). We observe that
if 𝑝 ∈ Ω, then
󵄩 󵄩
󵄩 󵄩
󵄩
󵄩󵄩
(31)
󵄩󵄩𝑦𝑛,1 − 𝑝󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑢𝑛 − 𝑝󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 .
Journal of Applied Mathematics
5
For all, from 𝑖 = 2 to 𝑖 = 𝑁, by induction, one proves that
󵄩󵄩
󵄩
󵄩
󵄩
󵄩
󵄩
󵄩󵄩𝑦𝑛,𝑖 − 𝑝󵄩󵄩󵄩 ≤ 𝛽𝑛,𝑖 󵄩󵄩󵄩𝑢𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛,𝑖 ) 󵄩󵄩󵄩𝑦𝑛,𝑖−1 − 𝑝󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩𝑢𝑛 − 𝑝󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 .
(32)
Thus, we obtain that for every 𝑖 = 1, . . . , 𝑁,
󵄩󵄩
󵄩 󵄩
󵄩 󵄩
󵄩
󵄩󵄩𝑦𝑛,𝑖 − 𝑝󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑢𝑛 − 𝑝󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 .
(33)
For simplicity, put 𝑦̃𝑛,𝑁 = 𝑃𝐶(𝑦𝑛,𝑁 − 𝜆∇𝑓𝛼𝑛 (𝑦𝑛,𝑁)) and 𝑦̃𝑛 =
𝑃𝐶(𝑦𝑛 − 𝜆∇𝑓𝛼𝑛 (𝑦𝑛 )) for every 𝑛 ≥ 0. Then, 𝑦𝑛 = 𝛼𝑛 𝑄𝑦𝑛,𝑁 +
(1 − 𝛼𝑛 )𝑦̃𝑛,𝑁 and 𝑥𝑛+1 = 𝜎𝑛 𝑦𝑛 + 𝛾𝑛 𝑦̃𝑛 + 𝛿𝑛 𝑇𝑦̃𝑛 for every 𝑛 ≥ 0.
Taking into consideration that 𝑇𝑝 = 𝑝 and 𝑃𝐶(𝐼 − 𝜆∇𝑓)𝑝 = 𝑝
for 𝜆 ∈ (0, 2/𝐿), we have
󵄩
󵄩󵄩 ̃
󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
= 󵄩󵄩󵄩󵄩𝑃𝐶 (𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − 𝑃𝐶 (𝐼 − 𝜆∇𝑓) 𝑝󵄩󵄩󵄩󵄩
󵄩
󵄩
≤ 󵄩󵄩󵄩󵄩𝑃𝐶 (𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − 𝑃𝐶 (𝐼−𝜆∇𝑓𝛼𝑛 ) 𝑝󵄩󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩󵄩𝑃𝐶 (𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑝 − 𝑃𝐶 (𝐼 − 𝜆∇𝑓) 𝑝󵄩󵄩󵄩󵄩
󵄩
󵄩
󵄩 󵄩
≤ 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝑃𝐶 (𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑝 − 𝑃𝐶 (𝐼 − 𝜆∇𝑓) 𝑝󵄩󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 .
(34)
Similarly, we get ‖𝑦̃𝑛 − 𝑝‖ ≤ ‖𝑦𝑛 − 𝑝‖ + 𝜆𝛼𝑛 ‖𝑝‖. Thus, from
(34) we have
󵄩 󵄩
󵄩
󵄩󵄩
󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 = 󵄩󵄩󵄩𝛽𝑛 (𝑄𝑦𝑛,𝑁 − 𝑝) + (1 − 𝛽𝑛 ) (𝑦̃𝑛,𝑁 − 𝑝)󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑦̃𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑄𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑄𝑝 − 𝑝󵄩󵄩󵄩
󵄩 󵄩
󵄩
󵄩
+ (1 − 𝛽𝑛 ) (󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩
󵄩
󵄩
≤ 𝛽𝑛 𝜌 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑄𝑝 − 𝑝󵄩󵄩󵄩
󵄩 󵄩
󵄩
󵄩
+ (1 − 𝛽𝑛 ) (󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩
= (1 − 𝛽𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩 󵄩
+ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑝 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 ) 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩
󵄩
󵄩
= (1 − 𝛽𝑛 (1 − 𝜌)) 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩
󵄩󵄩
󵄩𝑄𝑝 − 𝑝󵄩󵄩󵄩
󵄩 󵄩
+ 𝛽𝑛 (1 − 𝜌) 󵄩
+ (1 − 𝛽𝑛 ) 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩
1−𝜌
󵄩
󵄩
󵄩
󵄩 󵄩
󵄩 󵄩󵄩𝑄𝑝 − 𝑝󵄩󵄩󵄩
≤ max {󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 , 󵄩
} + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩
1−𝜌
Since (𝛾𝑛 + 𝛿𝑛 )𝜁 ≤ 𝛾𝑛 for all 𝑛 ≥ 0, utilizing Lemma 10, we
derive from (35)
󵄩
󵄩󵄩
󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
= 󵄩󵄩󵄩𝜎 (𝑦𝑛 − 𝑝) + 𝛾𝑛 (𝑦̃𝑛 − 𝑝) + 𝛿𝑛 (𝑇𝑦̃𝑛 − 𝑝)󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩
≤ 𝜎𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝛾𝑛 (𝑦̃𝑛 − 𝑝) + 𝛿𝑛 (𝑇𝑦̃𝑛 − 𝑝)󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
≤ 𝜎𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + (𝛾𝑛 + 𝛿𝑛 ) 󵄩󵄩󵄩𝑦̃𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩
󵄩
󵄩
≤ 𝜎𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + (𝛾𝑛 + 𝛿𝑛 ) (󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩 󵄩
󵄩 󵄩
󵄩 󵄩󵄩𝑄𝑝 − 𝑝󵄩󵄩󵄩
≤ max {󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 , 󵄩
} + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩
1−𝜌
󵄩
󵄩
󵄩
󵄩 󵄩
󵄩 󵄩󵄩𝑄𝑝 − 𝑝󵄩󵄩󵄩
= max {󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 , 󵄩
} + 2𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 .
1−𝜌
(36)
By induction, we get
󵄩
󵄩
󵄩󵄩
󵄩
󵄩
󵄩 󵄩󵄩𝑄𝑝 − 𝑝󵄩󵄩󵄩
}
󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 ≤ max {󵄩󵄩󵄩𝑥0 − 𝑝󵄩󵄩󵄩 , 󵄩
1−𝜌
𝑛
󵄩 󵄩
+ 2𝜆 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 ⋅ ∑𝛼𝑖 ,
(37)
∀𝑛 ≥ 0.
𝑖=0
This implies that {𝑥𝑛 } is bounded and so are {𝑦𝑛 }, {𝑢𝑛 }, and
{𝑦𝑛,𝑖 } for each 𝑖 = 1, . . . , 𝑁. It is clear that both {𝑦̃𝑛,𝑁} and {𝑦̃𝑛 }
are also bounded. Since ‖𝑇𝑦̃𝑛 − 𝑝‖ ≤ ((1 + 𝜁)/(1 − 𝜁))‖𝑦̃𝑛 − 𝑝‖,
{𝑇𝑦̃𝑛 } is also bounded.
Lemma 16. Let one suppose that Ω ≠ 0. Moreover, let one
suppose that the following hold:
(H1) lim𝑛 → ∞ 𝛽𝑛 = 0 and ∑∞
𝑛=0 𝛽𝑛 = ∞;
(H2) ∑∞
𝑛=1 |𝛽𝑛 − 𝛽𝑛−1 | < ∞ or lim𝑛 → ∞ (|𝛽𝑛 − 𝛽𝑛−1 |/𝛽𝑛 ) = 0;
(H3) ∑∞
𝑛=1 |𝛽𝑛,𝑖 −𝛽𝑛−1,𝑖 | < ∞ or lim𝑛 → ∞ (|𝛽𝑛,𝑖 −𝛽𝑛−1,𝑖 |/𝛽𝑛 ) =
0 for each 𝑖 = 1, . . . , 𝑁;
(H4) ∑∞
𝑛=1 |𝑟𝑛 − 𝑟𝑛−1 | < ∞ or lim𝑛 → ∞ (|𝑟𝑛 − 𝑟𝑛−1 |/𝛽𝑛 ) = 0;
(H5) ∑∞
𝑛=1 |𝜎𝑛 − 𝜎𝑛−1 | < ∞ or lim𝑛 → ∞ (|𝜎𝑛 − 𝜎𝑛−1 |/𝛽𝑛 ) = 0;
(H6) ∑∞
< ∞ or
𝑛=1 |𝛾𝑛 /(1 − 𝜎𝑛 ) − 𝛾𝑛−1 /(1 − 𝜎𝑛−1 )|
lim𝑛 → ∞ (1/𝛽𝑛 )|𝛾𝑛 /(1 − 𝜎𝑛 ) − 𝛾𝑛−1 /(1 − 𝜎𝑛−1 )| = 0.
󵄩
󵄩
󵄩
󵄩 󵄩
󵄩 󵄩󵄩𝑄𝑝 − 𝑝󵄩󵄩󵄩
≤ max {󵄩󵄩󵄩𝑢𝑛 − 𝑝󵄩󵄩󵄩 , 󵄩
} + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩
1−𝜌
Then, lim𝑛 → ∞ ‖𝑥𝑛+1 − 𝑥𝑛 ‖ = 0, that is, {𝑥𝑛 } is asymptotically
regular.
󵄩
󵄩
󵄩
󵄩 󵄩
󵄩 󵄩󵄩𝑄𝑝 − 𝑝󵄩󵄩󵄩
≤ max {󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 , 󵄩
} + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 .
1−𝜌
(35)
≤
Proof. Taking into account 0 < lim inf 𝑛 → ∞ 𝜎𝑛
lim sup𝑛 → ∞ 𝜎𝑛 < 1, we may assume, without loss of generality, that {𝜎𝑛 } ⊂ [𝑐, 𝑑] for some 𝑐, 𝑑 ∈ (0, 1). First, we write
6
Journal of Applied Mathematics
𝑥𝑛 = 𝜎𝑛−1 𝑦𝑛−1 + (1 − 𝜎𝑛−1 )V𝑛−1 , for all 𝑛 ≥ 1, where V𝑛−1 =
(𝑥𝑛 − 𝜎𝑛−1 𝑦𝑛−1 )/(1 − 𝜎𝑛−1 ). It follows that for all 𝑛 ≥ 1
V𝑛 − V𝑛−1 =
𝑥𝑛+1 − 𝜎𝑛 𝑦𝑛 𝑥𝑛 − 𝜎𝑛−1 𝑦𝑛−1
−
1 − 𝜎𝑛
1 − 𝜎𝑛−1
󵄨
󵄨󵄩
󵄩
󵄩
󵄩
+ 󵄨󵄨󵄨𝛽𝑛 −𝛽𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑄𝑦𝑛−1,𝑁 − 𝑦̃𝑛−1,𝑁󵄩󵄩󵄩 +𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑄𝑦𝑛−1,𝑁󵄩󵄩󵄩
󵄩 󵄨
󵄨󵄩
󵄩
󵄩
≤ (1 − 𝛽𝑛 ) (󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑦𝑛−1,𝑁󵄩󵄩󵄩 +𝜆 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑦𝑛−1,𝑁󵄩󵄩󵄩)
𝛾𝑛 (𝑦̃𝑛 − 𝑦̃𝑛−1 ) + 𝛿𝑛 (𝑇𝑦̃𝑛 − 𝑇𝑦̃𝑛−1 )
1 − 𝜎𝑛
󵄨
󵄨󵄩
󵄩
+ 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑄𝑦𝑛−1,𝑁 − 𝑦̃𝑛−1,𝑁󵄩󵄩󵄩
𝛾
𝛾𝑛−1
+( 𝑛 −
) 𝑦̃
1 − 𝜎𝑛 1 − 𝜎𝑛−1 𝑛−1
+(
𝛿𝑛−1
𝛿𝑛
−
) 𝑇𝑦̃𝑛−1 .
1 − 𝜎𝑛 1 − 𝜎𝑛−1
󵄩
󵄩
+ 𝛽𝑛 𝜌 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑦𝑛−1,𝑁󵄩󵄩󵄩
󵄩
󵄩
≤ (1 − (1 − 𝜌) 𝛽𝑛 ) 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑦𝑛−1,𝑁󵄩󵄩󵄩
(38)
Since (𝛾𝑛 + 𝛿𝑛 )𝜁 ≤ 𝛾𝑛 for all 𝑛 ≥ 0, utilizing Lemma 10, we
have
󵄩󵄩
󵄩󵄩
󵄩󵄩𝛾𝑛 (𝑦̃𝑛 − 𝑦̃𝑛−1 ) + 𝛿𝑛 (𝑇𝑦̃𝑛 − 𝑇𝑦̃𝑛−1 )󵄩󵄩
(39)
󵄩
󵄩
≤ (𝛾𝑛 + 𝛿𝑛 ) 󵄩󵄩󵄩𝑦̃𝑛 − 𝑦̃𝑛−1 󵄩󵄩󵄩 .
Next, we estimate ‖𝑦𝑛 −𝑦𝑛−1 ‖. Observe that for every 𝑛 ≥ 1
󵄩󵄩 ̃
󵄩
󵄩󵄩𝑦𝑛,𝑁 − 𝑦̃𝑛−1,𝑁󵄩󵄩󵄩
󵄩
󵄩
≤ 󵄩󵄩󵄩󵄩𝑃𝐶 (𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − 𝑃𝐶 (𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛−1,𝑁󵄩󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩󵄩𝑃𝐶 (𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛−1,𝑁 − 𝑃𝐶 (𝐼 − 𝜆∇𝑓𝛼𝑛−1 ) 𝑦𝑛−1,𝑁󵄩󵄩󵄩󵄩
󵄩
󵄩
≤ 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑦𝑛−1,𝑁󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩󵄩𝑃𝐶 (𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛−1,𝑁 − 𝑃𝐶 (𝐼 − 𝜆∇𝑓𝛼𝑛−1 ) 𝑦𝑛−1,𝑁󵄩󵄩󵄩󵄩
󵄩
󵄩
≤ 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑦𝑛−1,𝑁󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩󵄩(𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛−1,𝑁 − (𝐼 − 𝜆∇𝑓𝛼𝑛−1 ) 𝑦𝑛−1,𝑁󵄩󵄩󵄩󵄩
󵄩
󵄩
󵄩 󵄩
= 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑦𝑛−1,𝑁󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝜆∇𝑓𝛼𝑛 (𝑦𝑛−1,𝑁) − 𝜆∇𝑓𝛼𝑛−1 (𝑦𝑛−1,𝑁)󵄩󵄩󵄩󵄩
󵄨
󵄩
󵄩
󵄨󵄩
󵄩
= 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑦𝑛−1,𝑁󵄩󵄩󵄩 + 𝜆 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑦𝑛−1,𝑁󵄩󵄩󵄩 ,
(40)
and similarly,
󵄩󵄩 ̃
󵄨
󵄩 󵄩
󵄩
󵄨󵄩
󵄩
󵄩󵄩𝑦𝑛 − 𝑦̃𝑛−1 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩 + 𝜆 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑦𝑛−1 󵄩󵄩󵄩 .
Also, from (30), we have
󵄩
󵄩󵄩
󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩
󵄩
󵄩
≤ (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑦̃𝑛,𝑁 − 𝑦̃𝑛−1,𝑁󵄩󵄩󵄩
𝛾 𝑦̃ + 𝛿𝑛 𝑇𝑦̃𝑛 𝛾𝑛−1 𝑦̃𝑛−1 + 𝛿𝑛−1 𝑇𝑦̃𝑛−1
= 𝑛 𝑛
−
1 − 𝜎𝑛
1 − 𝜎𝑛−1
=
Then, passing to the norm we get from (40) that
(41)
󵄨
󵄨󵄩
󵄩
+ 𝜆 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑦𝑛−1,𝑁󵄩󵄩󵄩
󵄨
󵄨󵄩
󵄩
+ 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑄𝑦𝑛−1,𝑁 − 𝑦̃𝑛−1,𝑁󵄩󵄩󵄩
󵄩
󵄩
≤ (1 − (1 − 𝜌) 𝛽𝑛 ) 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑦𝑛−1,𝑁󵄩󵄩󵄩
󵄨 󵄨
󵄨
󵄨
+ 𝑀1 (󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨) ,
where 𝜆‖𝑦𝑛,𝑁‖ + ‖𝑄𝑦𝑛,𝑁 − 𝑦̃𝑛,𝑁‖ ≤ 𝑀1 , for all 𝑛 ≥ 0 for some
𝑀1 ≥ 0. Furthermore, by the definition of 𝑦𝑛,𝑖 one obtains
that, for all 𝑖 = 𝑁, . . . , 2
󵄩
󵄩
󵄩
󵄩󵄩
󵄩󵄩𝑦𝑛,𝑖 − 𝑦𝑛−1,𝑖 󵄩󵄩󵄩 ≤ 𝛽𝑛,𝑖 󵄩󵄩󵄩𝑢𝑛 − 𝑢𝑛−1 󵄩󵄩󵄩
󵄩
󵄩󵄨
󵄨
+ 󵄩󵄩󵄩𝑆𝑖 𝑢𝑛−1 − 𝑦𝑛−1,𝑖−1 󵄩󵄩󵄩 󵄨󵄨󵄨𝛽𝑛,𝑖 − 𝛽𝑛−1,𝑖 󵄨󵄨󵄨
󵄩
󵄩
+ (1 − 𝛽𝑛,𝑖 ) 󵄩󵄩󵄩𝑦𝑛,𝑖−1 − 𝑦𝑛−1,𝑖−1 󵄩󵄩󵄩 .
(42)
∀𝑛 ≥ 1.
󵄩󵄩
󵄩
󵄩󵄩𝑦𝑛,1 − 𝑦𝑛−1,1 󵄩󵄩󵄩
󵄩
󵄩
≤ 𝛽𝑛,1 󵄩󵄩󵄩𝑢𝑛 − 𝑢𝑛−1 󵄩󵄩󵄩
󵄩
󵄩󵄨
󵄨
+ 󵄩󵄩󵄩𝑆1 𝑢𝑛−1 − 𝑢𝑛−1 󵄩󵄩󵄩 󵄨󵄨󵄨𝛽𝑛,1 − 𝛽𝑛−1,1 󵄨󵄨󵄨
󵄩
󵄩
+ (1 − 𝛽𝑛,1 ) 󵄩󵄩󵄩𝑢𝑛 − 𝑢𝑛−1 󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩󵄨
󵄨
= 󵄩󵄩󵄩𝑢𝑛 − 𝑢𝑛−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆1 𝑢𝑛−1 − 𝑢𝑛−1 󵄩󵄩󵄩 󵄨󵄨󵄨𝛽𝑛,1 − 𝛽𝑛−1,1 󵄨󵄨󵄨 .
Substituting (46) in all (45) type one obtains for 𝑖 = 2, . . . , 𝑁
𝑖
󵄩
󵄩󵄨
󵄨
+ ∑ 󵄩󵄩󵄩𝑆𝑘 𝑢𝑛−1 − 𝑦𝑛−1,𝑘−1 󵄩󵄩󵄩 󵄨󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
𝑦𝑛 − 𝑦𝑛−1 = (1 − 𝛽𝑛 ) (𝑦̃𝑛,𝑁 − 𝑦̃𝑛−1,𝑁)
+ 𝛽𝑛 (𝑄𝑦𝑛,𝑁 − 𝑄𝑦𝑛−1,𝑁) .
(46)
󵄩󵄩
󵄩 󵄩
󵄩
󵄩󵄩𝑦𝑛,𝑖 − 𝑦𝑛−1,𝑖 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑢𝑛 − 𝑢𝑛−1 󵄩󵄩󵄩
Simple calculations show that
+ (𝛽𝑛 − 𝛽𝑛−1 ) (𝑄𝑦𝑛−1,𝑁 − 𝑦̃𝑛−1,𝑁)
(45)
In the case of 𝑖 = 1, we have
𝑦𝑛 = 𝛽𝑛 𝑄𝑦𝑛,𝑁 + (1 − 𝛽𝑛 ) 𝑦̃𝑛,𝑁,
𝑦𝑛−1 = 𝛽𝑛−1 𝑄𝑦𝑛−1,𝑁+ (1 − 𝛽𝑛−1 ) 𝑦̃𝑛−1,𝑁,
(44)
𝑘=2
(43)
󵄩
󵄩󵄨
󵄨
+ 󵄩󵄩󵄩𝑆1 𝑢𝑛−1 − 𝑢𝑛−1 󵄩󵄩󵄩 󵄨󵄨󵄨𝛽𝑛,1 − 𝛽𝑛−1,1 󵄨󵄨󵄨 .
(47)
Journal of Applied Mathematics
7
󵄨
󵄨󵄨
󵄨𝑟 − 𝑟 󵄨󵄨 𝑁 󵄨
󵄨 󵄨
󵄨
+ 𝑀2 [ 󵄨 𝑛 𝑛−1 󵄨 + ∑ 󵄨󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛,1 − 𝛽𝑛−1,1 󵄨󵄨󵄨
𝑟𝑛
𝑘=2
This together with (44) implies that
󵄩󵄩
󵄩
󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩
󵄨
󵄨 󵄨
󵄨
+ 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 ]
≤ (1 − (1 − 𝜌) 𝛽𝑛 )
󵄩
󵄩
≤ (1 − (1 − 𝜌) 𝛽𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩
󵄨
󵄨󵄨
󵄨𝑟 − 𝑟 󵄨󵄨 𝑁 󵄨
󵄨
+ 𝑀2 [ 󵄨 𝑛 𝑛−1 󵄨 + ∑ 󵄨󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
𝑏
𝑘=1
󵄩
󵄩
× [ 󵄩󵄩󵄩𝑢𝑛 − 𝑢𝑛−1 󵄩󵄩󵄩
𝑁
󵄩
󵄩󵄨
󵄨
+ ∑ 󵄩󵄩󵄩𝑆𝑘 𝑢𝑛−1 − 𝑦𝑛−1,𝑘−1 󵄩󵄩󵄩 󵄨󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
󵄨
󵄨 󵄨
󵄨
+ 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 ] ,
𝑘=2
󵄨󵄨
󵄨󵄨
󵄩󵄩
󵄨󵄨
󵄩󵄩 󵄨󵄨󵄨
+ 󵄩󵄩𝑆1 𝑢𝑛−1 − 𝑢𝑛−1 󵄩󵄩 󵄨󵄨𝛽𝑛,1 − 𝛽𝑛−1,1 󵄨󵄨󵄨]
󵄨󵄨
󵄨󵄨
󵄨
󵄨
󵄨 󵄨
󵄨
󵄨
+ 𝑀1 (󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨)
󵄩
󵄩
≤ (1 − (1 − 𝜌) 𝛽𝑛 ) 󵄩󵄩󵄩𝑢𝑛 − 𝑢𝑛−1 󵄩󵄩󵄩
𝑁
󵄩
󵄩󵄨
󵄨
+ ∑ 󵄩󵄩󵄩𝑆𝑘 𝑢𝑛−1 − 𝑦𝑛−1,𝑘−1 󵄩󵄩󵄩 󵄨󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
𝑘=2
󵄩
󵄩󵄨
󵄨
+ 󵄩󵄩󵄩𝑆1 𝑢𝑛−1 − 𝑢𝑛−1 󵄩󵄩󵄩 󵄨󵄨󵄨𝛽𝑛,1 − 𝛽𝑛−1,1 󵄨󵄨󵄨
󵄨 󵄨
󵄨
󵄨
+ 𝑀1 (󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨) .
By Lemma 13, we know that
󵄨󵄨
𝑟 󵄨󵄨󵄨
󵄨
󵄩󵄩
󵄩 󵄩
󵄩
󵄩󵄩𝑢𝑛 − 𝑢𝑛−1 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 𝜅 󵄨󵄨󵄨1 − 𝑛−1 󵄨󵄨󵄨 ,
𝑟𝑛 󵄨󵄨
󵄨󵄨
(49)
where 𝜅 = sup𝑛≥0 ‖𝑢𝑛 − 𝑥𝑛 ‖. So, substituting (49) in (48) we
obtain
󵄩
󵄩󵄩
󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩
󵄨󵄨
𝑟 󵄨󵄨󵄨
󵄨
󵄩
󵄩
≤ (1 − (1 − 𝜌) 𝛽𝑛 ) (󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 𝜅 󵄨󵄨󵄨1 − 𝑛−1 󵄨󵄨󵄨)
󵄨󵄨
𝑟𝑛 󵄨󵄨
𝑁
󵄩
󵄩󵄨
󵄨
+ ∑ 󵄩󵄩󵄩𝑆𝑘 𝑢𝑛−1 − 𝑦𝑛−1,𝑘−1 󵄩󵄩󵄩 󵄨󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
𝑘=2
󵄩
󵄩󵄨
󵄨
+ 󵄩󵄩󵄩𝑆1 𝑢𝑛−1 − 𝑢𝑛−1 󵄩󵄩󵄩 󵄨󵄨󵄨𝛽𝑛,1 − 𝛽𝑛−1,1 󵄨󵄨󵄨
󵄨 󵄨
󵄨
󵄨
+ 𝑀1 (󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨)
󵄨󵄨
󵄨
󵄨𝑟 − 𝑟 󵄨󵄨
󵄩
󵄩
≤ (1 − (1 − 𝜌) 𝛽𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 + 𝜅 󵄨 𝑛 𝑛−1 󵄨
𝑟
𝑛
𝑁
󵄩
󵄩󵄨
󵄨
+ ∑ 󵄩󵄩󵄩𝑆𝑘 𝑢𝑛−1 − 𝑦𝑛−1,𝑘−1 󵄩󵄩󵄩 󵄨󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
𝑘=2
󵄩
󵄩󵄨
󵄨
+ 󵄩󵄩󵄩𝑆1 𝑢𝑛−1 − 𝑢𝑛−1 󵄩󵄩󵄩 󵄨󵄨󵄨𝛽𝑛,1 − 𝛽𝑛−1,1 󵄨󵄨󵄨
󵄨 󵄨
󵄨
󵄨
+ 𝑀1 (󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨)
󵄩
󵄩
≤ (1 − (1 − 𝜌) 𝛽𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩
(50)
(48)
where 𝑏 > 0 is a minorant for {𝑟𝑛 } and 𝜅 + 𝑀1 + ∑𝑁
𝑘=2 ‖𝑆𝑘 𝑢𝑛 −
𝑦𝑛,𝑘−1 ‖ + ‖𝑆1 𝑢𝑛 − 𝑢𝑛 ‖ ≤ 𝑀2 , for all 𝑛 ≥ 0 for some 𝑀2 ≥ 0.
This together with (38)-(39), implies that
󵄩
󵄩
󵄩󵄩
󵄩 󵄩󵄩𝛾𝑛 (𝑦̃𝑛 − 𝑦̃𝑛−1 ) + 𝛿𝑛 (𝑇𝑦̃𝑛 − 𝑇𝑦̃𝑛−1 )󵄩󵄩󵄩
󵄩󵄩V𝑛 − V𝑛−1 󵄩󵄩󵄩 ≤ 󵄩
1 − 𝜎𝑛
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨 󵄩𝑦̃ 󵄩󵄩
󵄨󵄨 1 − 𝜎𝑛 1 − 𝜎𝑛−1 󵄨󵄨󵄨 󵄩 𝑛−1 󵄩
󵄨󵄨 𝛿
𝛿𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨󵄨 󵄩󵄩𝑇𝑦̃𝑛−1 󵄩󵄩󵄩
󵄨󵄨 1 − 𝜎𝑛 1 − 𝜎𝑛−1 󵄨󵄨
󵄩
󵄩
(𝛾𝑛 + 𝛿𝑛 ) 󵄩󵄩󵄩𝑦̃𝑛 − 𝑦̃𝑛−1 󵄩󵄩󵄩
≤
1 − 𝜎𝑛
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨 󵄩𝑦̃ 󵄩󵄩
󵄨󵄨 1 − 𝜎𝑛 1 − 𝜎𝑛−1 󵄨󵄨󵄨 󵄩 𝑛−1 󵄩
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨 󵄩𝑇𝑦̃ 󵄩󵄩
󵄨󵄨 1 − 𝜎𝑛 1 − 𝜎𝑛−1 󵄨󵄨󵄨 󵄩 𝑛−1 󵄩
󵄩
󵄩
= 󵄩󵄩󵄩𝑦̃𝑛 − 𝑦̃𝑛−1 󵄩󵄩󵄩
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩 󵄩
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨 (󵄩𝑦̃ 󵄩󵄩 + 󵄩󵄩𝑇𝑦̃ 󵄩󵄩)
󵄨󵄨 1 − 𝜎𝑛 1 − 𝜎𝑛−1 󵄨󵄨󵄨 󵄩 𝑛−1 󵄩 󵄩 𝑛−1 󵄩
󵄨
󵄩
󵄩
󵄨󵄩
󵄩
≤ 󵄩󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩 + 𝜆 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑦𝑛−1 󵄩󵄩󵄩
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩 󵄩
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨 (󵄩𝑦̃ 󵄩󵄩 + 󵄩󵄩𝑇𝑦̃ 󵄩󵄩)
󵄨󵄨 1 − 𝜎𝑛 1 − 𝜎𝑛−1 󵄨󵄨󵄨 󵄩 𝑛−1 󵄩 󵄩 𝑛−1 󵄩
󵄩
󵄩
≤ (1 − (1 − 𝜌) 𝛽𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩
󵄨󵄨󵄨𝑟 − 𝑟 󵄨󵄨󵄨 𝑁 󵄨
󵄨
+ 𝑀2 [ 󵄨 𝑛 𝑛−1 󵄨 + ∑ 󵄨󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
𝑏
𝑘=1
󵄨
󵄨 󵄨
󵄨
+ 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 ]
󵄨
󵄨󵄩
󵄩
+ 𝜆 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑦𝑛−1 󵄩󵄩󵄩
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨 󵄩󵄩
󵄨
󵄩 󵄩
󵄩
+ 󵄨󵄨󵄨 𝑛 −
󵄨 (󵄩𝑦̃ 󵄩󵄩 + 󵄩󵄩𝑇𝑦̃ 󵄩󵄩)
󵄨󵄨 1 − 𝜎𝑛 1 − 𝜎𝑛−1 󵄨󵄨󵄨 󵄩 𝑛−1 󵄩 󵄩 𝑛−1 󵄩
󵄩
󵄩
≤ (1 − (1 − 𝜌) 𝛽𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩
8
Journal of Applied Mathematics
󵄨
󵄨󵄨
󵄨𝑟 − 𝑟 󵄨󵄨 𝑁 󵄨
󵄨
+ 𝑀3 [ 󵄨 𝑛 𝑛−1 󵄨 + ∑ 󵄨󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
𝑏
𝑘=1
󵄨󵄩
󵄩
󵄨
+ 󵄨󵄨󵄨𝜎𝑛 − 𝜎𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑦𝑛−1 − V𝑛−1 󵄩󵄩󵄩
󵄩
󵄩
+ 𝜎𝑛 { (1 − (1 − 𝜌) 𝛽𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩
󵄨
󵄨 󵄨
󵄨
+ 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 ]
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨
󵄨
󵄨
󵄨
+ 𝑀3 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 𝑀3 󵄨󵄨󵄨 𝑛 −
󵄨
󵄨󵄨 1 − 𝜎𝑛 1 − 𝜎𝑛−1 󵄨󵄨󵄨
󵄩
󵄩
= (1 − (1 − 𝜌) 𝛽𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩
󵄨
󵄨󵄨
󵄨󵄨𝑟𝑛 − 𝑟𝑛−1 󵄨󵄨󵄨 𝑁 󵄨󵄨
󵄨
+ 𝑀3 [
+ ∑ 󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
𝑏
𝑘=1
󵄨
󵄨 󵄨
󵄨
+ 2 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨
󵄨
+ 󵄨󵄨󵄨 𝑛 −
󵄨],
󵄨󵄨 1 − 𝜎𝑛 1 − 𝜎𝑛−1 󵄨󵄨󵄨
where 𝑀2 + 𝜆‖𝑦𝑛 ‖ + ‖𝑦̃𝑛 ‖ + ‖𝑇𝑦̃𝑛 ‖ ≤ 𝑀3 , for all 𝑛 ≥ 0 for some
𝑀3 ≥ 0.
Further, we observe that
𝑥𝑛+1 = 𝜎𝑛 𝑦𝑛 + (1 − 𝜎𝑛 ) V𝑛 ,
(52)
𝑥𝑛+1 − 𝑥𝑛 = (1 − 𝜎𝑛 ) (V𝑛 − V𝑛−1 )
Then, passing to the norm, we get from (51)
󵄩󵄩
󵄩
󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩
󵄩
󵄩
≤ (1 − 𝜎𝑛 ) 󵄩󵄩󵄩V𝑛 − V𝑛−1 󵄩󵄩󵄩
󵄨
󵄨󵄩
󵄩
󵄩
󵄩
+ 󵄨󵄨󵄨𝜎𝑛 − 𝜎𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑦𝑛−1 − V𝑛−1 󵄩󵄩󵄩 + 𝜎𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑦𝑛−1 󵄩󵄩󵄩
≤ (1 − 𝜎𝑛 )
󵄨
󵄨󵄨
󵄨𝑟 − 𝑟 󵄨󵄨 𝑁 󵄨
󵄨
+ 𝑀3 [ 󵄨 𝑛 𝑛−1 󵄨 + ∑ 󵄨󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
𝑏
𝑘=1
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨
󵄨
+ 󵄨󵄨󵄨 𝑛 −
󵄨]
󵄨󵄨 1 − 𝜎𝑛 1 − 𝜎𝑛−1 󵄨󵄨󵄨
󵄨
󵄨󵄩
󵄩
+ 󵄨󵄨󵄨𝜎𝑛 − 𝜎𝑛−1 󵄨󵄨󵄨 󵄩󵄩󵄩𝑦𝑛−1 − V𝑛−1 󵄩󵄩󵄩
󵄩
󵄩
≤ (1 − (1 − 𝜌) 𝛽𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩
󵄨󵄨
󵄨
󵄨𝑟 − 𝑟 󵄨󵄨 𝑁 󵄨
󵄨
+ 𝑀 [ 󵄨 𝑛 𝑛−1 󵄨 + ∑ 󵄨󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
𝑏
𝑘=1
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨
󵄨
󵄨
󵄨
+ 󵄨󵄨󵄨 𝑛 −
󵄨 ] + 2𝑀 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 ,
󵄨󵄨 1 − 𝜎𝑛 1 − 𝜎𝑛−1 󵄨󵄨󵄨
Simple calculations show that
+ 𝜎𝑛 (𝑦𝑛 − 𝑦𝑛−1 ) .
󵄩
󵄩
≤ (1 − (1 − 𝜌) 𝛽𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩
󵄨
󵄨 󵄨
󵄨
+ 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝜎𝑛 − 𝜎𝑛−1 󵄨󵄨󵄨
∀𝑛 ≥ 1.
+ (𝜎𝑛 − 𝜎𝑛−1 ) (𝑦𝑛−1 − V𝑛−1 )
󵄨
󵄨 󵄨
󵄨
+ 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 ]}
󵄨
󵄨 󵄨
󵄨
+ 2 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨
(51)
𝑥𝑛 = 𝜎𝑛−1 𝑦𝑛−1 + (1 − 𝛽𝑛−1 ) V𝑛−1 ,
󵄨
󵄨󵄨
󵄨󵄨𝑟𝑛 − 𝑟𝑛−1 󵄨󵄨󵄨 𝑁 󵄨󵄨
󵄨
+ 𝑀2 [
+ ∑ 󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
𝑏
𝑘=1
(54)
(53)
where 𝑀3 + ‖𝑦𝑛 − V𝑛 ‖ ≤ 𝑀, for all 𝑛 ≥ 0 for some 𝑀 ≥ 0. By
hypotheses (H1)–(H6) and Lemma 9, from ∑∞
𝑛=0 𝛼𝑛 < ∞, we
obtain the claim.
Lemma 17. Let one suppose that Ω ≠ 0. Let one suppose that
{𝑥𝑛 } is asymptotically regular. Then, ‖𝑥𝑛 − 𝑦𝑛 ‖ → 0 and ‖𝑥𝑛 −
𝑢𝑛 ‖ = ‖𝑥𝑛 − 𝑇𝑟𝑛 𝑥𝑛 ‖ → 0 as 𝑛 → ∞.
Proof. We recall that, by the firm nonexpansivity of 𝑇𝑟𝑛 , a
standard calculation (see [17]) shows that if V ∈ EP(𝐹, ℎ), then
󵄩󵄩
󵄩2
󵄩2 󵄩
󵄩2 󵄩
󵄩󵄩𝑢𝑛 − V󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − V󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 󵄩󵄩󵄩 .
󵄩
󵄩
× { (1 − (1 − 𝜌) 𝛽𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩
󵄨
󵄨󵄨
󵄨𝑟 − 𝑟 󵄨󵄨 𝑁 󵄨
󵄨
+ 𝑀3 [ 󵄨 𝑛 𝑛−1 󵄨 + ∑ 󵄨󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
𝑏
𝑘=1
󵄨
󵄨 󵄨
󵄨
+ 2 󵄨󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨 + 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 +
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨
󵄨
+ 󵄨󵄨󵄨 𝑛 −
󵄨 ]}
󵄨󵄨 1 − 𝜎𝑛 1 − 𝜎𝑛−1 󵄨󵄨󵄨
(55)
Let 𝑝 ∈ Ω. Then by Lemma 11, we have from (33)–(34) the
following
󵄩2 󵄩
󵄩2
󵄩󵄩 ̃
󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 = 󵄩󵄩󵄩𝛽𝑛 (𝑄𝑦𝑛,𝑁 − 𝑝) + (1 − 𝛽𝑛 ) (𝑦̃𝑛,𝑁 − 𝑝)󵄩󵄩󵄩
󵄩
󵄩2
󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑦̃𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2 󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦̃𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩
󵄩 󵄩2
󵄩2 󵄩
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + [󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩]
Journal of Applied Mathematics
9
󵄩
󵄩2 󵄩
󵄩2
= 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩2 󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩) ,
󵄩
󵄩2
󵄩2
󵄩
= 𝜎𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑦̃𝑛 − 𝑝󵄩󵄩󵄩
−
󵄩
󵄩2
≤ 𝜎𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝜎𝑛 )
(56)
󵄩2
󵄩󵄩 ̃
󵄩2 󵄩
󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 = 󵄩󵄩󵄩󵄩𝑃𝐶 (𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛 − 𝑃𝐶 (𝐼 − 𝜆∇𝑓) 𝑝󵄩󵄩󵄩󵄩
󵄩 󵄩2
󵄩
󵄩
≤ (󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
󵄩2
󵄩
= 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩) .
(57)
Since (𝛾𝑛 + 𝛿𝑛 )𝜁 ≤ 𝛾𝑛 for all 𝑛 ≥ 0, utilizing Lemma 10, we
have
󵄩󵄩
󵄩2
󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
= 󵄩󵄩󵄩𝜎𝑛 (𝑦𝑛 − 𝑝) + 𝛾𝑛 (𝑦̃𝑛 − 𝑝) + 𝛿𝑛 (𝑇𝑦̃𝑛 − 𝑝)󵄩󵄩󵄩
󵄩󵄩
1
󵄩
= 󵄩󵄩󵄩𝜎𝑛 (𝑦𝑛 − 𝑝) + (𝛾𝑛 + 𝛿𝑛 )
󵄩󵄩
𝛾𝑛 + 𝛿𝑛
󵄩󵄩2
󵄩
× [𝛾𝑛 (𝑦̃𝑛 − 𝑝) + 𝛿𝑛 (𝑇𝑦̃𝑛 − 𝑝)] 󵄩󵄩󵄩
󵄩󵄩
󵄩
󵄩2
= 𝜎𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + (𝛾𝑛 + 𝛿𝑛 )
󵄩󵄩2
󵄩󵄩󵄩 1
󵄩
[𝛾𝑛 (𝑦̃𝑛 − 𝑝) + 𝛿𝑛 (𝑇𝑦̃𝑛 − 𝑝)]󵄩󵄩󵄩
× 󵄩󵄩󵄩
󵄩󵄩
󵄩󵄩 𝛾𝑛 + 𝛿𝑛
󵄩󵄩
1
󵄩
− 𝜎𝑛 (𝛾𝑛 + 𝛿𝑛 ) 󵄩󵄩󵄩(𝑦𝑛 − 𝑝) −
󵄩󵄩
𝛾𝑛 + 𝛿𝑛
󵄩󵄩2
󵄩
× [𝛾𝑛 (𝑦̃𝑛 − 𝑝) + 𝛿𝑛 (𝑇𝑦̃𝑛 − 𝑝)] 󵄩󵄩󵄩
󵄩󵄩
󵄩
󵄩2
= 𝜎𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + (𝛾𝑛 + 𝛿𝑛 )
󵄩󵄩2
󵄩󵄩 1
󵄩
󵄩
[𝛾𝑛 (𝑦̃𝑛 − 𝑝) + 𝛿𝑛 (𝑇𝑦̃𝑛 − 𝑝)]󵄩󵄩󵄩
× 󵄩󵄩󵄩
󵄩󵄩
󵄩󵄩 𝛾𝑛 + 𝛿𝑛
󵄩󵄩 1
󵄩󵄩2
󵄩
󵄩
− 𝜎𝑛 (𝛾𝑛 + 𝛿𝑛 ) 󵄩󵄩󵄩
[𝛾𝑛 (𝑦̃𝑛 − 𝑦𝑛 ) + 𝛿𝑛 (𝑇𝑦̃𝑛 − 𝑦𝑛 )]󵄩󵄩󵄩
󵄩󵄩 𝛾𝑛 + 𝛿𝑛
󵄩󵄩
󵄩
󵄩2
= 𝜎𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + (𝛾𝑛 + 𝛿𝑛 )
󵄩󵄩 1
󵄩󵄩2
󵄩
󵄩
× 󵄩󵄩󵄩
[𝛾𝑛 (𝑦̃𝑛 − 𝑝) + 𝛿𝑛 (𝑇𝑦̃𝑛 − 𝑝)]󵄩󵄩󵄩
󵄩󵄩 𝛾𝑛 + 𝛿𝑛
󵄩󵄩
𝜎𝑛 󵄩󵄩
󵄩2
−
󵄩𝑥 − 𝑦𝑛 󵄩󵄩󵄩
𝛾𝑛 + 𝛿𝑛 󵄩 𝑛+1
󵄩
󵄩2
󵄩
󵄩2
≤ 𝜎𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + (𝛾𝑛 + 𝛿𝑛 ) 󵄩󵄩󵄩𝑦̃𝑛 − 𝑝󵄩󵄩󵄩
−
𝜎𝑛 󵄩󵄩
󵄩2
󵄩𝑥 − 𝑦𝑛 󵄩󵄩󵄩
𝛾𝑛 + 𝛿𝑛 󵄩 𝑛+1
𝜎𝑛 󵄩󵄩
󵄩2
󵄩𝑥 − 𝑦𝑛 󵄩󵄩󵄩
1 − 𝜎𝑛 󵄩 𝑛+1
󵄩2
󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
× [󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)]
−
𝜎𝑛 󵄩󵄩
󵄩2
󵄩𝑥 − 𝑦𝑛 󵄩󵄩󵄩
1 − 𝜎𝑛 󵄩 𝑛+1
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
󵄩2
󵄩
≤ 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
−
𝜎𝑛 󵄩󵄩
󵄩2
󵄩𝑥 − 𝑦𝑛 󵄩󵄩󵄩
1 − 𝜎𝑛 󵄩 𝑛+1
󵄩2 󵄩
󵄩2
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
−
𝜎𝑛 󵄩󵄩
󵄩2
󵄩𝑥 − 𝑦𝑛 󵄩󵄩󵄩
1 − 𝜎𝑛 󵄩 𝑛+1
󵄩
󵄩
󵄩2
󵄩2
= 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩 󵄩
󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
+ 2𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
𝜎𝑛 󵄩󵄩
󵄩2
−
󵄩𝑥 − 𝑦𝑛 󵄩󵄩󵄩 .
1 − 𝜎𝑛 󵄩 𝑛+1
(58)
Taking into account 0 < lim inf 𝑛 → ∞ 𝜎𝑛 ≤ lim sup𝑛 → ∞ 𝜎𝑛 <
1, we may assume that {𝜎𝑛 } ⊂ [𝑐, 𝑑] for some 𝑐, 𝑑 ∈ (0, 1). So,
we deduce that
𝑐 󵄩󵄩
󵄩2
󵄩𝑥 − 𝑦𝑛 󵄩󵄩󵄩
1 − 𝑐 󵄩 𝑛+1
𝜎𝑛 󵄩󵄩
󵄩2
≤
󵄩𝑥 − 𝑦𝑛 󵄩󵄩󵄩
1 − 𝜎𝑛 󵄩 𝑛+1
󵄩
󵄩
󵄩2 󵄩
󵄩2
󵄩2
≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩 󵄩
󵄩
+ 2𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩
󵄩
󵄩 󵄩
󵄩 󵄩
󵄩2
≤ (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 + 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩 󵄩
󵄩
+ 2𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩) .
(59)
Since 𝛼𝑛 → 0, 𝛽𝑛 → 0 and ‖𝑥𝑛 − 𝑥𝑛+1 ‖ → 0 as 𝑛 → ∞, we
conclude from the boundedness of {𝑥𝑛 }, {𝑦𝑛 }, and {𝑦𝑛,𝑁} that
‖𝑥𝑛+1 −𝑦𝑛 ‖ → 0 as 𝑛 → ∞. This together with ‖𝑥𝑛 −𝑥𝑛+1 ‖ →
0, implies that
󵄩󵄩
lim 󵄩𝑥
𝑛→∞ 󵄩 𝑛
󵄩
− 𝑦𝑛 󵄩󵄩󵄩 = 0.
(60)
10
Journal of Applied Mathematics
Furthermore, from (33), (55), and (56), we have
󵄩󵄩
󵄩
󵄩2
󵄩2 󵄩
󵄩2
󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩2 󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢𝑛 − 𝑝󵄩󵄩󵄩
(61)
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩2
󵄩2 󵄩
󵄩2 󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩) ,
which hence implies that
󵄩󵄩
󵄩2
󵄩
󵄩2 󵄩
󵄩2 󵄩
󵄩2
󵄩󵄩𝑥𝑛 − 𝑢𝑛 󵄩󵄩󵄩 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩2
(62)
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩 󵄩
󵄩 󵄩
+ (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩) .
(H8) there exists a constant 𝜏 > 0 such that (1/𝛽𝑛 )|1/𝛽𝑛,𝑘0 −
1/𝛽𝑛−1,𝑘0 | < 𝜏 for all 𝑛 ≥ 1. Then,
󵄩󵄩
󵄩
󵄩𝑥 − 𝑥𝑛 󵄩󵄩󵄩
lim 󵄩 𝑛+1
= 0.
𝑛→∞
𝛽𝑛,𝑘0
Proof. We start by (54). Dividing both the terms by 𝛽𝑛,𝑘0 we
have
󵄩
󵄩󵄩
󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩
𝛽𝑛,𝑘0
󵄩
󵄩󵄩
󵄩𝑥 − 𝑥𝑛−1 󵄩󵄩󵄩
≤ [1 − (1 − 𝜌) 𝛽𝑛 ] 󵄩 𝑛
𝛽𝑛,𝑘0
󵄨
𝑁 󵄨
󵄨
󵄨󵄨
󵄨󵄨𝑟𝑛 − 𝑟𝑛−1 󵄨󵄨󵄨 ∑𝑘=1 󵄨󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
+𝑀 [
+
𝑏𝛽𝑛,𝑘0
𝛽𝑛,𝑘0
󵄨 󵄨
󵄨 󵄨
󵄨
󵄨
2 󵄨󵄨𝛼 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄨󵄨󵄨𝜎𝑛 − 𝜎𝑛−1 󵄨󵄨󵄨
+ 󵄨 𝑛
+
+
𝛽𝑛,𝑘0
𝛽𝑛,𝑘0
𝛽𝑛,𝑘0
Since 𝛼𝑛 → 0, 𝛽𝑛 → 0 and ‖𝑥𝑛 − 𝑦𝑛 ‖ → 0 as 𝑛 → ∞, we
deduce from the boundedness of {𝑥𝑛 }, {𝑦𝑛 }, and {𝑦𝑛,𝑁} that
󵄩󵄩
lim 󵄩𝑥
𝑛→∞ 󵄩 𝑛
󵄩
− 𝑢𝑛 󵄩󵄩󵄩 = 0.
(66)
󵄨
󵄨󵄨
󵄨𝛾 / (1 − 𝜎𝑛 ) − 𝛾𝑛−1 / (1 − 𝜎𝑛−1 )󵄨󵄨󵄨
+󵄨 𝑛
].
𝛽𝑛,𝑘0
(63)
(67)
So, by (H8) we have
Remark 18. By the last lemma we have 𝜔𝑤 (𝑥𝑛 ) = 𝜔𝑤 (𝑢𝑛 ) and
𝜔𝑠 (𝑥𝑛 ) = 𝜔𝑠 (𝑢𝑛 ); that is, the sets of strong/weak cluster points
of {𝑥𝑛 } and {𝑢𝑛 } coincide.
Of course, if 𝛽𝑛,𝑖 → 𝛽𝑖 ≠ 0, as 𝑛 → ∞, for all index 𝑖, the
assumptions of Lemma 16 are enough to assure that
󵄩󵄩
󵄩
󵄩𝑥 − 𝑥𝑛 󵄩󵄩󵄩
= 0,
lim 󵄩 𝑛+1
𝑛→∞
𝛽𝑛,𝑖
∀𝑖 ∈ {1, . . . , 𝑁} .
(64)
In the next lemma, we examine the case in which at least one
sequence {𝛽𝑛,𝑘0 } is a null sequence.
Lemma 19. Let one suppose that Ω ≠ 0. Let one suppose
that (H1) holds. Moreover, for an index 𝑘0 ∈ {1, . . . , 𝑁},
lim𝑛 → ∞ 𝛽𝑛,𝑘0 = 0, and the following hold:
(H7) for all 𝑖,
󵄨󵄨
󵄨
󵄨󵄨
󵄨
󵄨󵄨
󵄨
󵄨󵄨𝛽𝑛,𝑖 − 𝛽𝑛−1,𝑖 󵄨󵄨󵄨
󵄨󵄨𝛼𝑛 − 𝛼𝑛−1 󵄨󵄨󵄨
󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨
= lim
= lim
lim
𝑛→∞
𝑛→∞ 𝛽 𝛽
𝑛→∞ 𝛽 𝛽
𝛽𝑛 𝛽𝑛,𝑘0
𝑛 𝑛,𝑘0
𝑛 𝑛,𝑘0
󵄨󵄨
󵄨
󵄨󵄨
󵄨
󵄨𝜎 − 𝜎𝑛−1 󵄨󵄨󵄨
󵄨𝑟 − 𝑟 󵄨󵄨
= lim 󵄨 𝑛
= lim 󵄨 𝑛 𝑛−1 󵄨
𝑛→∞ 𝛽 𝛽
𝑛→∞ 𝛽 𝛽
𝑛 𝑛,𝑘0
𝑛 𝑛,𝑘0
1
𝑛→∞𝛽 𝛽
𝑛 𝑛,𝑘0
= lim
󵄨󵄨 𝛾
𝛾𝑛−1 󵄨󵄨󵄨󵄨
󵄨󵄨 𝑛
−
󵄨󵄨
󵄨 = 0;
󵄨󵄨 1 − 𝜎𝑛 1 − 𝜎𝑛−1 󵄨󵄨󵄨
(65)
󵄩󵄩
󵄩
󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩
𝛽𝑛,𝑘0
󵄩
󵄩󵄩
󵄩𝑥 − 𝑥𝑛−1 󵄩󵄩󵄩
≤ [1 − (1 − 𝜌) 𝛽𝑛 ] 󵄩 𝑛
𝛽𝑛−1,𝑘0
󵄨
󵄨
1 󵄨󵄨󵄨
󵄩 󵄨󵄨 1
󵄩
󵄨󵄨
+ [1 − (1 − 𝜌) 𝛽𝑛 ] 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 󵄨󵄨󵄨󵄨
−
󵄨󵄨 𝛽𝑛,𝑘0 𝛽𝑛−1,𝑘0 󵄨󵄨󵄨
󵄨
𝑁 󵄨
󵄨󵄨
󵄨
󵄨𝑟 − 𝑟 󵄨󵄨 ∑ 󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
+ 𝑀 [ 󵄨 𝑛 𝑛−1 󵄨 + 𝑘=1 󵄨
𝑏𝛽𝑛,𝑘0
𝛽𝑛,𝑘0
󵄨
󵄨󵄨
󵄨𝛾 / (1 − 𝜎𝑛 ) − 𝛾𝑛−1 / (1 − 𝜎𝑛−1 )󵄨󵄨󵄨
+󵄨 𝑛
𝛽𝑛,𝑘0
󵄨 󵄨
󵄨 󵄨
󵄨
󵄨
2 󵄨󵄨𝛼 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄨󵄨󵄨𝜎𝑛 − 𝜎𝑛−1 󵄨󵄨󵄨
+ 󵄨 𝑛
+
+
]
𝛽𝑛,𝑘0
𝛽𝑛,𝑘0
𝛽𝑛,𝑘0
󵄩
󵄩󵄩
󵄩𝑥 − 𝑥𝑛−1 󵄩󵄩󵄩
≤ [1 − (1 − 𝜌) 𝛽𝑛 ] 󵄩 𝑛
𝛽𝑛−1,𝑘0
󵄨
󵄨
1 󵄨󵄨󵄨
󵄩 󵄨󵄨 1
󵄩
󵄨󵄨
+ 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩 󵄨󵄨󵄨󵄨
−
󵄨󵄨 𝛽𝑛,𝑘0 𝛽𝑛−1,𝑘0 󵄨󵄨󵄨
󵄨󵄨󵄨𝑟 − 𝑟 󵄨󵄨󵄨 ∑𝑁 󵄨󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
󵄨
+ 𝑀 [ 󵄨 𝑛 𝑛−1 󵄨 + 𝑘=1 󵄨
𝑏𝛽𝑛,𝑘0
𝛽𝑛,𝑘0
Journal of Applied Mathematics
11
󵄨
󵄨󵄨
󵄨𝛾 / (1 − 𝜎𝑛 ) − 𝛾𝑛−1 / (1 − 𝜎𝑛−1 )󵄨󵄨󵄨
+󵄨 𝑛
𝛽𝑛,𝑘0
Proof. Let 𝑝 ∈ Ω. Then, by Lemma 11 we have
󵄨
󵄨
2 󵄨󵄨𝛼 − 𝛼𝑛−1 󵄨󵄨󵄨
+ 󵄨 𝑛
𝛽𝑛,𝑘0
󵄩2
󵄩󵄩
󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
= 󵄩󵄩󵄩𝛽𝑛 (𝑄𝑦𝑛,𝑁 − 𝑝) + (1 − 𝛽𝑛 ) (𝑦̃𝑛,𝑁 − 𝑝)󵄩󵄩󵄩
󵄩
󵄩2
󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑦̃𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
= 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 )
󵄩
󵄩2
× 󵄩󵄩󵄩󵄩𝑃𝐶 (𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − 𝑃𝐶(𝐼 − 𝜆∇𝑓)𝑝󵄩󵄩󵄩󵄩
󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 )
󵄨 󵄨
󵄨
󵄨󵄨
󵄨𝛽 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄨󵄨󵄨𝜎𝑛 − 𝜎𝑛−1 󵄨󵄨󵄨
+󵄨 𝑛
+
]
𝛽𝑛,𝑘0
𝛽𝑛,𝑘0
󵄩
󵄩󵄩
󵄩𝑥 − 𝑥𝑛−1 󵄩󵄩󵄩
󵄩
󵄩
+ 𝛽𝑛 𝜏 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩
≤ [1 − (1 − 𝜌) 𝛽𝑛 ] 󵄩 𝑛
𝛽𝑛−1,𝑘0
󵄨
𝑁 󵄨
󵄨󵄨
󵄨
󵄨𝑟 − 𝑟 󵄨󵄨 ∑ 󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
+ 𝑀 [ 󵄨 𝑛 𝑛−1 󵄨 + 𝑘=1 󵄨
𝑏𝛽𝑛,𝑘0
𝛽𝑛,𝑘0
󵄩
󵄩2
× 󵄩󵄩󵄩(𝐼 − 𝜆∇𝑓) 𝑦𝑛,𝑁 − (𝐼 − 𝜆∇𝑓)𝑝 − 𝜆𝛼𝑛 𝑦𝑛,𝑁󵄩󵄩󵄩
󵄨
󵄨󵄨
󵄨𝛾 / (1 − 𝜎𝑛 ) − 𝛾𝑛−1 / (1 − 𝜎𝑛−1 )󵄨󵄨󵄨
+󵄨 𝑛
𝛽𝑛,𝑘0
󵄨 󵄨
󵄨 󵄨
󵄨
󵄨
2 󵄨󵄨𝛼 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨 󵄨󵄨󵄨𝜎𝑛 − 𝜎𝑛−1 󵄨󵄨󵄨
+ 󵄨 𝑛
+
+
]
𝛽𝑛,𝑘0
𝛽𝑛,𝑘0
𝛽𝑛,𝑘0
󵄩
󵄩󵄩
󵄩𝑥 − 𝑥𝑛−1 󵄩󵄩󵄩
1
+ (1 − 𝜌) 𝛽𝑛 ⋅
= [1 − (1 − 𝜌) 𝛽𝑛 ] 󵄩 𝑛
𝛽𝑛−1,𝑘0
1−𝜌
󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 )
󵄩2
󵄩
× [󵄩󵄩󵄩(𝐼 − 𝜆∇𝑓) 𝑦𝑛,𝑁 − (𝐼 − 𝜆∇𝑓) 𝑝󵄩󵄩󵄩
−2𝜆𝛼𝑛 ⟨𝑦𝑛,𝑁, (𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − (𝐼 − 𝜆∇𝑓) 𝑝⟩ ]
󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 )
2 󵄩
󵄩
󵄩2
󵄩2
× [󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 2𝜆 (𝜆 − ) 󵄩󵄩󵄩∇𝑓(𝑦𝑛,𝑁) − ∇𝑓(𝑝)󵄩󵄩󵄩
𝐿
󵄩
󵄩
× {𝜏 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛−1 󵄩󵄩󵄩
󵄩
󵄩
󵄩󵄩
+2𝜆𝛼𝑛 󵄩󵄩󵄩𝑦𝑛,𝑁󵄩󵄩󵄩 󵄩󵄩󵄩󵄩(𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − (𝐼 − 𝜆∇𝑓) 𝑝󵄩󵄩󵄩󵄩 ]
󵄨
𝑁 󵄨
󵄨󵄨
󵄨
󵄨𝑟 − 𝑟 󵄨󵄨 ∑ 󵄨󵄨𝛽𝑛,𝑘 − 𝛽𝑛−1,𝑘 󵄨󵄨󵄨
+ 𝑀 [ 󵄨 𝑛 𝑛−1 󵄨 + 𝑘=1 󵄨
𝑏𝛽𝑛 𝛽𝑛,𝑘0
𝛽𝑛 𝛽𝑛,𝑘0
󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 )
󵄨
󵄨󵄨
󵄨𝛾 / (1 − 𝜎𝑛 ) − 𝛾𝑛−1 / (1 − 𝜎𝑛−1 )󵄨󵄨󵄨
+󵄨 𝑛
𝛽𝑛 𝛽𝑛,𝑘0
2 󵄩
󵄩
󵄩2
󵄩2
× [󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 2𝜆 (𝜆 − ) 󵄩󵄩󵄩∇𝑓(𝑦𝑛,𝑁) − ∇𝑓(𝑝)󵄩󵄩󵄩
𝐿
󵄨 󵄨
󵄨
󵄨
2 󵄨󵄨𝛼 − 𝛼𝑛−1 󵄨󵄨󵄨 󵄨󵄨󵄨𝛽𝑛 − 𝛽𝑛−1 󵄨󵄨󵄨
+ 󵄨 𝑛
+
𝛽𝑛 𝛽𝑛,𝑘0
𝛽𝑛 𝛽𝑛,𝑘0
󵄩
󵄩
󵄩󵄩
+2𝜆𝛼𝑛 󵄩󵄩󵄩𝑦𝑛,𝑁󵄩󵄩󵄩 󵄩󵄩󵄩󵄩(𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − (𝐼 − 𝜆∇𝑓) 𝑝󵄩󵄩󵄩󵄩 ] .
(71)
󵄨
󵄨󵄨
󵄨𝜎 − 𝜎𝑛−1 󵄨󵄨󵄨
+󵄨 𝑛
]} .
𝛽𝑛 𝛽𝑛,𝑘0
So, we obtain
(68)
Therefore, utilizing Lemma 9, from (H1), (H7), and the
asymptotical regularity of {𝑥𝑛 } (due to Lemma 16), we deduce
that
󵄩󵄩
󵄩
󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩
= 0.
lim
𝑛→∞
𝛽𝑛,𝑘0
(69)
Lemma 20. Let one suppose that Ω ≠ 0. Let one suppose that
(H1)–(H6) hold. Then,
󵄩
󵄩
󵄩
󵄩
lim 󵄩󵄩𝑦̃ − 𝑦𝑛,𝑁󵄩󵄩󵄩 = lim 󵄩󵄩󵄩𝑦̃𝑛 − 𝑦𝑛 󵄩󵄩󵄩 = 0.
𝑛 → ∞ 󵄩 𝑛,𝑁
𝑛→∞
(70)
2
󵄩2
󵄩
− 𝜆) 󵄩󵄩󵄩∇𝑓(𝑦𝑛,𝑁) − ∇𝑓(𝑝)󵄩󵄩󵄩
𝐿
󵄩
󵄩2
󵄩
󵄩2 󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
+ (1 − 𝛽𝑛 ) 2𝜆𝛼𝑛 󵄩󵄩󵄩𝑦𝑛,𝑁󵄩󵄩󵄩
󵄩
󵄩
× 󵄩󵄩󵄩󵄩(𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − (𝐼 − 𝜆∇𝑓) 𝑝󵄩󵄩󵄩󵄩
󵄩
󵄩2 󵄩
󵄩 󵄩
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩)
󵄩
󵄩 󵄩
󵄩
× (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩)
󵄩
󵄩
󵄩󵄩
+ 2𝜆𝛼𝑛 󵄩󵄩󵄩𝑦𝑛,𝑁󵄩󵄩󵄩 󵄩󵄩󵄩󵄩(𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − (𝐼 − 𝜆∇𝑓) 𝑝󵄩󵄩󵄩󵄩
󵄩
󵄩
󵄩2 󵄩
󵄩 󵄩
󵄩 󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩) (󵄩󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩)
(1 − 𝛽𝑛 ) 2𝜆 (
󵄩
󵄩
󵄩󵄩
+ 2𝜆𝛼𝑛 󵄩󵄩󵄩𝑦𝑛,𝑁󵄩󵄩󵄩 󵄩󵄩󵄩󵄩(𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − (𝐼 − 𝜆∇𝑓) 𝑝󵄩󵄩󵄩󵄩 .
(72)
12
Journal of Applied Mathematics
Since 𝛼𝑛 → 0, 𝛽𝑛 → 0, ‖𝑥𝑛 − 𝑦𝑛 ‖ → 0, and 0 < 𝜆 < 2/𝐿,
from the boundedness of {𝑥𝑛 }, {𝑦𝑛 }, and {𝑦𝑛,𝑁} it follows that
lim𝑛 → ∞ ‖∇𝑓(𝑦𝑛,𝑁) − ∇𝑓(𝑝)‖ = 0, and hence
󵄩
󵄩
lim 󵄩󵄩󵄩∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓 (𝑝)󵄩󵄩󵄩󵄩 = 0.
(73)
𝑛→∞ 󵄩
Moreover, from the firm nonexpansiveness of 𝑃𝐶 we obtain
󵄩󵄩 ̃
󵄩2
󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
= 󵄩󵄩󵄩󵄩𝑃𝐶(𝐼 − 𝜆∇𝑓𝛼𝑛 )𝑦𝑛,𝑁 − 𝑃𝐶(𝐼 − 𝜆∇𝑓)𝑝󵄩󵄩󵄩󵄩
1 󵄩
󵄩
󵄩
󵄩2
≤ {󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 2𝜆 󵄩󵄩󵄩󵄩∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓 (𝑝)󵄩󵄩󵄩󵄩
2
󵄩
󵄩
× 󵄩󵄩󵄩󵄩(𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − (𝐼 − 𝜆∇𝑓) 𝑝󵄩󵄩󵄩󵄩
󵄩
󵄩2
󵄩2 󵄩
+ 󵄩󵄩󵄩𝑦̃𝑛,𝑁 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑦̃𝑛,𝑁󵄩󵄩󵄩
󵄩
󵄩2
− (1 − 𝛽𝑛 ) 𝜆2 󵄩󵄩󵄩󵄩∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓 (𝑝)󵄩󵄩󵄩󵄩
󵄩
󵄩2 󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩2
− 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 2𝜆 󵄩󵄩󵄩󵄩∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓 (𝑝)󵄩󵄩󵄩󵄩
󵄩
󵄩
× 󵄩󵄩󵄩󵄩(𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − (𝐼 − 𝜆∇𝑓) 𝑝󵄩󵄩󵄩󵄩
󵄩
󵄩2
− (1 − 𝛽𝑛 ) 𝜆2 󵄩󵄩󵄩󵄩∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓 (𝑝)󵄩󵄩󵄩󵄩
󵄩
󵄩
󵄩2 󵄩
󵄩 󵄩
󵄩 󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩
󵄩
󵄩
+ 2𝜆 󵄩󵄩󵄩󵄩∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓 (𝑝)󵄩󵄩󵄩󵄩
2󵄩
󵄩
(74)
󵄩
󵄩
× 󵄩󵄩󵄩󵄩(𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − (𝐼 − 𝜆∇𝑓) 𝑝󵄩󵄩󵄩󵄩
󵄩
󵄩
󵄩󵄩
+ 2𝜆 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑦̃𝑛,𝑁󵄩󵄩󵄩 󵄩󵄩󵄩󵄩∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓 (𝑝)󵄩󵄩󵄩󵄩 .
and so
󵄩󵄩 ̃
󵄩2
󵄩2 󵄩
󵄩2 󵄩
󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑦̃𝑛,𝑁󵄩󵄩󵄩
󵄩
󵄩
+ 2𝜆 󵄩󵄩󵄩󵄩∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓 (𝑝)󵄩󵄩󵄩󵄩
󵄩
󵄩
× 󵄩󵄩󵄩󵄩(𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − (𝐼 − 𝜆∇𝑓) 𝑝󵄩󵄩󵄩󵄩
(77)
Since 𝛽𝑛 → 0, ‖𝑥𝑛 − 𝑦𝑛 ‖ → 0, and ‖∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓(𝑝)‖ →
0, from the boundedness of {𝑥𝑛 }, {𝑦𝑛 }, {𝑦𝑛,𝑁}, and {𝑦̃𝑛,𝑁}, it
follows that lim𝑛 → ∞ ‖𝑦𝑛,𝑁 − 𝑦̃𝑛,𝑁‖ = 0. Observe that
+ 2𝜆 ⟨𝑦𝑛,𝑁 − 𝑦̃𝑛,𝑁, ∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓 (𝑝)⟩
(75)
Thus, we have
󵄩󵄩
󵄩
󵄩2
󵄩2
󵄩
󵄩2
󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑦̃𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2 󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩2
󵄩
− (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑦̃𝑛,𝑁󵄩󵄩󵄩
󵄩
󵄩
+ 2𝜆 󵄩󵄩󵄩󵄩∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓 (𝑝)󵄩󵄩󵄩󵄩
󵄩
󵄩
× 󵄩󵄩󵄩󵄩(𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − (𝐼 − 𝜆∇𝑓) 𝑝󵄩󵄩󵄩󵄩
󵄩󵄩
󵄩
󵄩
󵄩
󵄩󵄩𝑦𝑛 − 𝑦̃𝑛,𝑁󵄩󵄩󵄩 = 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑦̃𝑛,𝑁󵄩󵄩󵄩 󳨀→ 0 as 𝑛 󳨀→ ∞, (78)
and hence
󵄩󵄩 ̃
󵄩
󵄩󵄩𝑦𝑛 − 𝑦𝑛 󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩𝑦̃𝑛 − 𝑦̃𝑛,𝑁󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦̃𝑛,𝑁 − 𝑦𝑛 󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩
= 󵄩󵄩󵄩󵄩𝑃𝐶 (𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛 −𝑃𝐶 (𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦̃𝑛,𝑁 − 𝑦𝑛 󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩𝑦𝑛 − 𝑦𝑛,𝑁󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦̃𝑛,𝑁 − 𝑦𝑛 󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩𝑦𝑛 − 𝑦̃𝑛,𝑁󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦̃𝑛,𝑁 − 𝑦𝑛,𝑁󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦̃𝑛,𝑁 − 𝑦𝑛 󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩
= 2 󵄩󵄩󵄩𝑦𝑛 − 𝑦̃𝑛,𝑁󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦̃𝑛,𝑁 − 𝑦𝑛,𝑁󵄩󵄩󵄩 󳨀→ 0 as 𝑛 󳨀→ ∞.
(79)
Thus, lim𝑛 → ∞ ‖𝑦̃𝑛 − 𝑦𝑛 ‖ = 0.
+ 2 (1−𝛽𝑛 ) 𝜆 ⟨𝑦𝑛,𝑁 − 𝑦̃𝑛,𝑁, ∇𝑓𝛼𝑛 (𝑦𝑛,𝑁)−∇𝑓 (𝑝)⟩
󵄩
󵄩2
− (1 − 𝛽𝑛 ) 𝜆2 󵄩󵄩󵄩󵄩∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓 (𝑝)󵄩󵄩󵄩󵄩 ,
󵄩
󵄩
󵄩
󵄩2
− 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 2𝜆 󵄩󵄩󵄩󵄩∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓 (𝑝)󵄩󵄩󵄩󵄩
+ 2 (1 − 𝛽𝑛 ) 𝜆 ⟨𝑦𝑛,𝑁 − 𝑦̃𝑛,𝑁, ∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓 (𝑝)⟩
+ 2𝜆 ⟨𝑦𝑛,𝑁 − 𝑦̃𝑛,𝑁, ∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓 (𝑝)⟩
󵄩
󵄩2
− 𝜆2 󵄩󵄩󵄩󵄩∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓 (𝑝)󵄩󵄩󵄩󵄩 .
󵄩
󵄩2 󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩
+ 2 (1 − 𝛽𝑛 ) 𝜆⟨𝑦𝑛,𝑁 − 𝑦̃𝑛,𝑁, ∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓 (𝑝)⟩
1 󵄩󵄩
󵄩2 󵄩
󵄩2
{󵄩󵄩(𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − (𝐼 − 𝜆∇𝑓) 𝑝󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦̃𝑛,𝑁 − 𝑝󵄩󵄩󵄩
2 󵄩
󵄩
󵄩2
− 󵄩󵄩󵄩󵄩(𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − (𝐼 − 𝜆∇𝑓) 𝑝 − (𝑦̃𝑛,𝑁 − 𝑝)󵄩󵄩󵄩󵄩 }
󵄩2
− 𝜆 󵄩󵄩󵄩∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓 (𝑝)󵄩󵄩󵄩󵄩 } ,
󵄩2
󵄩
(1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑦̃𝑛,𝑁󵄩󵄩󵄩
󵄩
󵄩
× 󵄩󵄩󵄩󵄩(𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − (𝐼 − 𝜆∇𝑓) 𝑝󵄩󵄩󵄩󵄩
≤ ⟨(𝐼 − 𝜆∇𝑓𝛼𝑛 ) 𝑦𝑛,𝑁 − (𝐼 − 𝜆∇𝑓) 𝑝, 𝑦̃𝑛,𝑁 − 𝑝⟩
=
which implies that
(76)
Lemma 21. Let one suppose that Ω ≠ 0. Let one suppose that
0 < lim inf 𝑛 → ∞ 𝛽𝑛,𝑖 ≤ lim sup𝑛 → ∞ 𝛽𝑛,𝑖 < 1 for each 𝑖 =
1, . . . , 𝑁. Moreover, suppose that (H1)–(H6) are satisfied. Then,
lim𝑛 → ∞ ‖𝑆𝑖 𝑢𝑛 − 𝑢𝑛 ‖ = 0 for each 𝑖 = 1, . . . , 𝑁.
Journal of Applied Mathematics
13
󵄩
󵄩2 󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
󵄩2
󵄩
− 𝛽𝑛,𝑁 (1 − 𝛽𝑛,𝑁) 󵄩󵄩󵄩𝑆𝑁𝑢𝑛 − 𝑦𝑛,𝑁−1 󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩) .
Proof. First of all, observe that
𝑥𝑛+1 − 𝑦𝑛 = 𝛾𝑛 (𝑦̃𝑛 − 𝑦𝑛 ) + 𝛿𝑛 (𝑇𝑦̃𝑛 − 𝑦𝑛 )
= 𝛾𝑛 (𝑦̃𝑛 − 𝑦𝑛 ) + 𝛿𝑛 (𝑇𝑦̃𝑛 − 𝑇𝑦𝑛 ) + 𝛿𝑛 (𝑇𝑦𝑛 − 𝑦𝑛 ) .
(80)
By Lemmas 16 and 20, we know that ‖𝑥𝑛+1 − 𝑥𝑛 ‖ → 0 and
‖𝑦̃𝑛 − 𝑦𝑛 ‖ → 0 as 𝑛 → ∞. Hence, utilizing Lemma 7(i), we
have
󵄩
󵄩2 󵄩
󵄩2 󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩
󵄩2 󵄩
󵄩 󵄩
󵄩 󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩 󵄩
󵄩 1 + 𝜁 󵄩󵄩 ̃
󵄩
≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦̃𝑛 − 𝑦𝑛 󵄩󵄩󵄩 +
󵄩𝑦 − 𝑦𝑛 󵄩󵄩󵄩
1−𝜁󵄩 𝑛
2 󵄩󵄩
󵄩
󵄩𝑦̃ − 𝑦𝑛 󵄩󵄩󵄩 ,
1−𝜁󵄩 𝑛
(81)
which together with ‖𝑥𝑛 − 𝑦𝑛 ‖ → 0 implies that
lim𝑛 → ∞ ‖𝛿𝑛 (𝑇𝑦𝑛 − 𝑦𝑛 )‖ = 0. Taking into account
lim inf 𝑛 → ∞ 𝛿𝑛 > 0, we have
󵄩
󵄩
lim 󵄩󵄩𝑇𝑦𝑛 − 𝑦𝑛 󵄩󵄩󵄩 = 0.
𝑛→∞ 󵄩
So, we have
󵄩2
󵄩
𝛽𝑛,𝑁 (1 − 𝛽𝑛,𝑁) 󵄩󵄩󵄩𝑆𝑁𝑢𝑛 − 𝑦𝑛,𝑁−1 󵄩󵄩󵄩
󵄩󵄩
󵄩
󵄩󵄩𝛿𝑛 (𝑇𝑦𝑛 − 𝑦𝑛 )󵄩󵄩󵄩
󵄩
󵄩
= 󵄩󵄩󵄩𝑥𝑛+1 − 𝑦𝑛 − 𝛾𝑛 (𝑦̃𝑛 − 𝑦𝑛 ) − 𝛿𝑛 (𝑇𝑦̃𝑛 − 𝑇𝑦𝑛 )󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
󵄩
󵄩
≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑦𝑛 󵄩󵄩󵄩 + 𝛾𝑛 󵄩󵄩󵄩𝑦̃𝑛 − 𝑦𝑛 󵄩󵄩󵄩 + 𝛿𝑛 󵄩󵄩󵄩𝑇𝑦̃𝑛 − 𝑇𝑦𝑛 󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩𝑥𝑛+1 − 𝑦𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦̃𝑛 − 𝑦𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑇𝑦̃𝑛 − 𝑇𝑦𝑛 󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩
= 󵄩󵄩󵄩𝑥𝑛+1 − 𝑥𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩 +
(83)
(82)
Let us show that for each 𝑖 ∈ {1, . . . , 𝑁}, one has ‖𝑆𝑖 𝑢𝑛 −
𝑦𝑛,𝑖−1 ‖ → 0 as 𝑛 → ∞. Let 𝑝 ∈ Ω. When 𝑖 = 𝑁, by
Lemma 11, we have from (33)-(34) the following:
󵄩
󵄩2
󵄩2
󵄩
󵄩2
󵄩󵄩
󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 ≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑦̃𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2 󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦̃𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩
󵄩 󵄩2
󵄩2 󵄩
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + (󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩2 󵄩
󵄩2
= 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩
󵄩2
󵄩2
= 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝛽𝑛,𝑁󵄩󵄩󵄩𝑆𝑁𝑢𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
+ (1 − 𝛽𝑛,𝑁) 󵄩󵄩󵄩𝑦𝑛,𝑁−1 − 𝑝󵄩󵄩󵄩
󵄩2
󵄩
− 𝛽𝑛,𝑁 (1 − 𝛽𝑛,𝑁) 󵄩󵄩󵄩𝑆𝑁𝑢𝑛 − 𝑦𝑛,𝑁−1 󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩2 󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑢𝑛 − 𝑝󵄩󵄩󵄩
󵄩2
󵄩
− 𝛽𝑛,𝑁 (1 − 𝛽𝑛,𝑁) 󵄩󵄩󵄩𝑆𝑁𝑢𝑛 − 𝑦𝑛,𝑁−1 󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩) .
(84)
Since 𝛼𝑛 → 0, 𝛽𝑛 → 0, 0 < lim inf 𝑛 → ∞ 𝛽𝑛,𝑁 ≤
lim sup𝑛 → ∞ 𝛽𝑛,𝑁 < 1, and lim𝑛 → ∞ ‖𝑥𝑛 − 𝑦𝑛 ‖ = 0, it is known
that {‖𝑆𝑁𝑢𝑛 − 𝑦𝑛,𝑁−1 ‖} is a null sequence.
Let 𝑖 ∈ {1, . . . , 𝑁 − 1}. Then, one has
󵄩2
󵄩󵄩
󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2 󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦̃𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩
󵄩 󵄩2
󵄩2 󵄩
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + (󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩2 󵄩
󵄩2
= 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩
󵄩2
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝛽𝑛,𝑁󵄩󵄩󵄩𝑆𝑁𝑢𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
+ (1 − 𝛽𝑛,𝑁) 󵄩󵄩󵄩𝑦𝑛,𝑁−1 − 𝑝󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩
󵄩2
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝛽𝑛,𝑁󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
+ (1 − 𝛽𝑛,𝑁) 󵄩󵄩󵄩𝑦𝑛,𝑁−1 − 𝑝󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩2
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩2
+ 𝛽𝑛,𝑁󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
+ (1 − 𝛽𝑛,𝑁) [𝛽𝑛,𝑁−1 󵄩󵄩󵄩𝑆𝑁−1 𝑢𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
+ (1 − 𝛽𝑛,𝑁−1 ) 󵄩󵄩󵄩𝑦𝑛,𝑁−2 − 𝑝󵄩󵄩󵄩 ]
󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩2
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
14
Journal of Applied Mathematics
󵄩2
󵄩
+ (𝛽𝑛,𝑁 + (1 − 𝛽𝑛,𝑁) 𝛽𝑛,𝑁−1 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩
To conclude, we have that
𝑁
󵄩
󵄩2
+ ∏ (1 − 𝛽𝑛,𝑘 ) 󵄩󵄩󵄩𝑦𝑛,𝑁−2 − 𝑝󵄩󵄩󵄩 ,
𝑘=𝑁−1
(85)
and so, after (𝑁 − 𝑖 + 1) iterations,
󵄩2
󵄩󵄩
󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
󵄩2
󵄩
󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
𝑁
𝑁
𝑗=𝑖+2
𝑙=𝑗
+ (𝛽𝑛,𝑁 + ∑ (∏ (1 − 𝛽𝑛,𝑙 )) 𝛽𝑛,𝑗−1 )
𝑁
󵄩
󵄩2
󵄩
󵄩2
× 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + ∏ (1 − 𝛽𝑛,𝑘 ) 󵄩󵄩󵄩𝑦𝑛,𝑖 − 𝑝󵄩󵄩󵄩
𝑘=𝑖+1
󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩2
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
𝑁
𝑁
𝑗=𝑖+2
𝑙=𝑗
+ (𝛽𝑛,𝑁 + ∑ (∏ (1 − 𝛽𝑛,𝑙 )) 𝛽𝑛,𝑗−1 )
󵄩 󵄩
󵄩 󵄩
󵄩
󵄩󵄩
󵄩󵄩𝑆2 𝑢𝑛 − 𝑢𝑛 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑆2 𝑢𝑛 − 𝑦𝑛,1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛,1 − 𝑢𝑛 󵄩󵄩󵄩
󵄩
󵄩
󵄩
󵄩
= 󵄩󵄩󵄩𝑆2 𝑢𝑛 − 𝑦𝑛,1 󵄩󵄩󵄩 + 𝛽𝑛,1 󵄩󵄩󵄩𝑆1 𝑢𝑛 − 𝑢𝑛 󵄩󵄩󵄩 ,
(89)
from which ‖𝑆2 𝑢𝑛 − 𝑢𝑛 ‖ → 0. Thus, by induction ‖𝑆𝑖 𝑢𝑛 −
𝑢𝑛 ‖ → 0 for all 𝑖 = 2, . . . , 𝑁 since it is enough to observe that
󵄩
󵄩󵄩
󵄩󵄩𝑆𝑖 𝑢𝑛 − 𝑢𝑛 󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩𝑆𝑖 𝑢𝑛 − 𝑦𝑛,𝑖−1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛,𝑖−1 − 𝑆𝑖−1 𝑢𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑆𝑖−1 𝑢𝑛 − 𝑢𝑛 󵄩󵄩󵄩
(90)
󵄩
󵄩
󵄩
󵄩
≤ 󵄩󵄩󵄩𝑆𝑖 𝑢𝑛 − 𝑦𝑛,𝑖−1 󵄩󵄩󵄩 + (1 − 𝛽𝑛,𝑖−1 ) 󵄩󵄩󵄩𝑆𝑖−1 𝑢𝑛 − 𝑦𝑛,𝑖−2 󵄩󵄩󵄩
󵄩
󵄩
+ 󵄩󵄩󵄩𝑆𝑖−1 𝑢𝑛 − 𝑢𝑛 󵄩󵄩󵄩 .
Remark 22. As an example, we consider 𝑁 = 2 and the following sequences:
(a) 𝜎𝑛 = 1/2 + 2/𝑛, 𝛾𝑛 = 𝛿𝑛 = 1/4 − 1/𝑛 for all 𝑛 > 4;
(b) 𝛽𝑛 = 1/√𝑛, 𝑟𝑛 = 2 − 1/𝑛, for all 𝑛 > 1;
(c) 𝛽𝑛,1 = 1/2 − 1/𝑛, 𝛽𝑛,2 = 1/2 − 1/𝑛2 , for all 𝑛 > 2.
𝑁
󵄩
󵄩2
× 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + ∏ (1 − 𝛽𝑛,𝑘 )
Then, they satisfy the hypotheses on the parameter sequences
in Lemma 21.
𝑘=𝑖+1
󵄩
󵄩2
󵄩
󵄩2
× [𝛽𝑛,𝑖 󵄩󵄩󵄩𝑆𝑖 𝑢𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝛽𝑛,𝑖 ) 󵄩󵄩󵄩𝑦𝑛,𝑖−1 − 𝑝󵄩󵄩󵄩
󵄩2
󵄩
−𝛽𝑛,𝑖 (1 − 𝛽𝑛,𝑖 ) 󵄩󵄩󵄩𝑆𝑖 𝑢𝑛 − 𝑦𝑛,𝑖−1 󵄩󵄩󵄩 ]
󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩2
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
𝑁
Lemma 23. Let one suppose that Ω ≠ 0 and 𝛽𝑛,𝑖 → 𝛽𝑖 for all
𝑖 as 𝑛 → ∞. Suppose there exists 𝑘 ∈ {1, . . . , 𝑁} such that
𝛽𝑛,𝑘 → 0 as 𝑛 → ∞. Let 𝑘0 ∈ {1, . . . , 𝑁} be the largest index
such that 𝛽𝑛,𝑘0 → 0 as 𝑛 → ∞. Suppose that
(i) (𝛼𝑛 + 𝛽𝑛 )/𝛽𝑛,𝑘0 → 0 as 𝑛 → ∞;
󵄩
󵄩2
󵄩2
󵄩
+ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝛽𝑛,𝑖 ∏ (1 − 𝛽𝑛,𝑘 ) 󵄩󵄩󵄩𝑆𝑖 𝑢𝑛 − 𝑦𝑛,𝑖−1 󵄩󵄩󵄩 .
𝑘=𝑖
(86)
Again, we obtain that
(ii) if 𝑖 ≤ 𝑘0 and 𝛽𝑛,𝑖 → 0, then 𝛽𝑛,𝑘0 /𝛽𝑛,𝑖 → 0 as 𝑛 →
∞;
(iii) if 𝛽𝑛,𝑖 → 𝛽𝑖 ≠ 0, then 𝛽𝑖 lies in (0, 1).
𝑁
󵄩2
󵄩
𝛽𝑛,𝑖 ∏ (1 − 𝛽𝑛,𝑘 ) 󵄩󵄩󵄩𝑆𝑖 𝑢𝑛 − 𝑦𝑛,𝑖−1 󵄩󵄩󵄩
Moreover, suppose that (H1), (H7), and (H8) hold. Then,
lim𝑛 → ∞ ‖𝑆𝑖 𝑢𝑛 − 𝑢𝑛 ‖ = 0 for each 𝑖 = 1, . . . , 𝑁.
𝑘=𝑖
󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩2
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩2 󵄩
󵄩2
+ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩2
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩
󵄩 󵄩
󵄩 󵄩
+ (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩) 󵄩󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩 .
(87)
Since 𝛼𝑛 → 0, 𝛽𝑛 → 0, 0 < lim inf 𝑛 → ∞ 𝛽𝑛,𝑖 ≤
lim sup𝑛 → ∞ 𝛽𝑛,𝑖 < 1 for each 𝑖 = 1, . . . , 𝑁 − 1, and
lim𝑛 → ∞ ‖𝑥𝑛 − 𝑦𝑛 ‖ = 0, it is known that
󵄩󵄩
lim 󵄩𝑆 𝑢
𝑛→∞ 󵄩 𝑖 𝑛
󵄩
− 𝑦𝑛,𝑖−1 󵄩󵄩󵄩 = 0.
Obviously, for 𝑖 = 1, we have ‖𝑆1 𝑢𝑛 − 𝑢𝑛 ‖ → 0.
(88)
Proof. First of all, we note that if (H7) holds, then also (H2)–
(H6) are satisfied. So {𝑥𝑛 } is asymptotically regular. Let 𝑘0
be as in the hypotheses. As in Lemma 21, for every index
𝑖 ∈ {1, . . . , 𝑁} such that 𝛽𝑛,𝑖 → 𝛽𝑖 ≠ 0 (which leads to
0 < lim inf 𝑛 → ∞ 𝛽𝑛,𝑖 ≤ lim sup𝑛 → ∞ 𝛽𝑛,𝑖 < 1), one has ‖𝑆𝑖 𝑢𝑛 −
𝑦𝑛,𝑖−1 ‖ → 0 as 𝑛 → ∞.
For all the other indexes 𝑖 ≤ 𝑘0 , we can prove that ‖𝑆𝑖 𝑢𝑛 −
𝑦𝑛,𝑖−1 ‖ → 0 as 𝑛 → ∞ in a similar manner. By the following
relation (due to (86)):
󵄩2
󵄩󵄩
󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
󵄩
󵄩2
≤ 𝜎𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + (𝛾𝑛 + 𝛿𝑛 ) 󵄩󵄩󵄩𝑦̃𝑛 − 𝑝󵄩󵄩󵄩
󵄩
󵄩2
󵄩
󵄩2
= 𝜎𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑦̃𝑛 − 𝑝󵄩󵄩󵄩
Journal of Applied Mathematics
15
󵄩
󵄩 󵄩2
󵄩
󵄩2
󵄩
≤ 𝜎𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + (1 − 𝜎𝑛 ) (󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
Proof. By Remark 18, we have 𝜔𝑤 (𝑥𝑛 ) = 𝜔𝑤 (𝑢𝑛 ) and 𝜔𝑠 (𝑥𝑛 ) =
𝜔𝑠 (𝑢𝑛 ). Observe that
󵄩
󵄩
󵄩 󵄩2
≤ (󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩 󵄩
󵄩 󵄩
󵄩
󵄩󵄩
󵄩󵄩𝑥𝑛 − 𝑦𝑛,1 󵄩󵄩󵄩 ≤ 󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛,1 − 𝑢𝑛 󵄩󵄩󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
󵄩2
󵄩
= 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩
󵄩
󵄩
= 󵄩󵄩󵄩𝑥𝑛 − 𝑢𝑛 󵄩󵄩󵄩 + 𝛽𝑛,1 󵄩󵄩󵄩𝑆1 𝑢𝑛 − 𝑢𝑛 󵄩󵄩󵄩 .
󵄩
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩2
󵄩
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
𝑁
󵄩
󵄩2
󵄩2
󵄩
+ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝛽𝑛,𝑖 ∏ (1 − 𝛽𝑛,𝑘 ) 󵄩󵄩󵄩𝑆𝑖 𝑢𝑛 − 𝑦𝑛,𝑖−1 󵄩󵄩󵄩
By Lemmas 17 and 21, ‖𝑥𝑛 − 𝑢𝑛 ‖ → 0 and ‖𝑆1 𝑢𝑛 − 𝑢𝑛 ‖ → 0
as 𝑛 → ∞, and hence
𝑘=𝑖
󵄩 󵄩 󵄩
󵄩 󵄩
󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩󵄩
lim 󵄩𝑥
𝑛→∞ 󵄩 𝑛
󵄩
󵄩 󵄩
󵄩2
≤ 𝛽𝑛 󵄩󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 2𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩
󵄩
󵄩 󵄩
󵄩 󵄩
󵄩
× (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
𝑁
󵄩
󵄩2
󵄩2
󵄩
+ 󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 − 𝛽𝑛,𝑖 ∏ (1 − 𝛽𝑛,𝑘 ) 󵄩󵄩󵄩𝑆𝑖 𝑢𝑛 − 𝑦𝑛,𝑖−1 󵄩󵄩󵄩 ,
𝑘=𝑖
(91)
we immediately obtain that
󵄩2
󵄩
∏ (1 − 𝛽𝑛,𝑘 ) 󵄩󵄩󵄩𝑆𝑖 𝑢𝑛 − 𝑦𝑛,𝑖−1 󵄩󵄩󵄩
(96)
So, we get 𝜔𝑤 (𝑥𝑛 ) = 𝜔𝑤 (𝑦𝑛,1 ) and 𝜔𝑠 (𝑥𝑛 ) = 𝜔𝑠 (𝑦𝑛,1 ).
Let 𝑝 ∈ 𝜔𝑤 (𝑥𝑛 ). Since 𝑝 ∈ 𝜔𝑤 (𝑢𝑛 ), by Lemma 21
and Lemma 7(ii) (demiclosedness principle), we have 𝑝 ∈
Fix(𝑆𝑖 ) for all index 𝑖, that is, 𝑝 ∈ ⋂𝑖 Fix(𝑆𝑖 ). Taking
into consideration that 𝑇 is 𝜁-strictly pseudocontractive, by
Lemma 7(i), we get
󵄩
󵄩 󵄩
󵄩 󵄩
󵄩
≤ 󵄩󵄩󵄩𝑇𝑥𝑛 − 𝑇𝑦𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑇𝑦𝑛 − 𝑦𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑥𝑛 󵄩󵄩󵄩
𝑘=𝑖
𝛽𝑛 󵄩󵄩
󵄩2 𝛼
󵄩 󵄩
󵄩󵄩𝑄𝑦𝑛,𝑁 − 𝑝󵄩󵄩󵄩 + 𝑛 2𝜆 󵄩󵄩󵄩𝑝󵄩󵄩󵄩
𝛽𝑛,𝑖
𝛽𝑛,𝑖
󵄩
− 𝑦𝑛,1 󵄩󵄩󵄩 = 0.
󵄩
󵄩󵄩
󵄩󵄩𝑇𝑥𝑛 − 𝑥𝑛 󵄩󵄩󵄩
𝑁
≤
(95)
(92)
󵄩
󵄩 󵄩
󵄩 󵄩
󵄩
× (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑝󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩
󵄩
󵄩 󵄩
󵄩 󵄩󵄩𝑥 − 𝑥𝑛+1 󵄩󵄩󵄩
+ (󵄩󵄩󵄩𝑥𝑛 − 𝑝󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛+1 − 𝑝󵄩󵄩󵄩) 󵄩 𝑛
.
𝛽𝑛,𝑖
By Lemma 19 or by hypothesis (ii) of the sequences, we have
󵄩󵄩
󵄩 󵄩
󵄩
󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 󵄩󵄩󵄩𝑥𝑛 − 𝑥𝑛+1 󵄩󵄩󵄩 𝛽𝑛,𝑘0
=
⋅
󳨀→ 0.
(93)
𝛽𝑛,𝑖
𝛽𝑛,𝑘0
𝛽𝑛,𝑖
So, the thesis follows.
Remark 24. Let us consider 𝑁 = 3 and the following
sequences:
(a) 𝛼𝑛 = 1/𝑛5/4 , 𝛽𝑛 = 1/𝑛1/2 , 𝑟𝑛 = 2 − 1/𝑛2 , for all 𝑛 > 1;
(b) 𝜎𝑛 = 1/2 + 2/𝑛2 , 𝛾𝑛 = 𝛿𝑛 = 1/4 − 1/𝑛2 , for all 𝑛 > 2;
(c) 𝛽𝑛,1 = 1/𝑛1/4 , 𝛽𝑛,2 = 1/2 − 1/𝑛2 , 𝛽𝑛,3 = 1/𝑛1/3 , for all
𝑛 > 1.
It is easy to see that all hypotheses (i)–(iii), (H1), (H7), and
(H8) of Lemma 23 are satisfied.
Remark 25. Under the hypotheses of Lemma 23, analogously
to Lemma 21, one can see that
󵄩
󵄩
lim 󵄩󵄩𝑆 𝑢 − 𝑦𝑛,𝑖−1 󵄩󵄩󵄩 = 0, ∀𝑖 ∈ {2, . . . , 𝑁} .
(94)
𝑛→∞ 󵄩 𝑖 𝑛
Corollary 26. Let one suppose that the hypotheses of either
Lemma 21 or Lemma 23 are satisfied. Then, 𝜔𝑤 (𝑥𝑛 ) =
𝜔𝑤 (𝑢𝑛 ) = 𝜔𝑤 (𝑦𝑛,1 ), 𝜔𝑠 (𝑥𝑛 ) = 𝜔𝑠 (𝑢𝑛 ) = 𝜔𝑠 (𝑦𝑛,1 ), and
𝜔𝑤 (𝑥𝑛 ) ⊂ Ω.
≤
1 + 𝜁 󵄩󵄩
󵄩 󵄩
󵄩 󵄩
󵄩
󵄩𝑥 − 𝑦𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑇𝑦𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥𝑛 − 𝑦𝑛 󵄩󵄩󵄩
1−𝜁󵄩 𝑛
=
2 󵄩󵄩
󵄩 󵄩
󵄩
󵄩𝑥 − 𝑦𝑛 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑦𝑛 − 𝑇𝑦𝑛 󵄩󵄩󵄩 ,
1−𝜁󵄩 𝑛
(97)
which together with ‖𝑥𝑛 − 𝑦𝑛 ‖ → 0 (by Lemma 17) and ‖𝑦𝑛 −
𝑇𝑦𝑛 ‖ → 0 (by (82)) implies that
󵄩󵄩
lim 󵄩𝑥
𝑛→∞ 󵄩 𝑛
󵄩
− 𝑇𝑥𝑛 󵄩󵄩󵄩 = 0.
(98)
Utilizing Lemma 7(ii) (demiclosedness principle), we
have 𝑝 ∈ Fix(𝑇). Furthermore, by Lemmas 14 and 17, we
know that 𝑝 ∈ EP(𝐹, ℎ). Finally, by similar argument as in
[18], we can show that 𝑝 ∈ Γ, and as a result 𝑝 ∈ Ω.
Theorem 27. Let one suppose that Ω ≠ 0. Let {𝛽𝑛 }, {𝛽𝑛,𝑖 }, 𝑖 =
1, . . . , 𝑁, be sequences in (0, 1) such that 0 < lim inf 𝑛 → ∞ 𝛽𝑛,𝑖 ≤
lim sup𝑛 → ∞ 𝛽𝑛,𝑖 < 1 for all index 𝑖. Moreover, Let one suppose
that (H1)–(H6) hold. Then, the sequences {𝑥𝑛 }, {𝑦𝑛 }, and {𝑢𝑛 },
explicitly defined by scheme (30), all converge strongly to the
unique solution 𝑥∗ ∈ Ω of the following variational inequality:
⟨𝑄𝑥∗ − 𝑥∗ , 𝑧 − 𝑥∗ ⟩ ≤ 0,
∀𝑧 ∈ Ω.
(99)
Proof. Since the mapping 𝑃Ω 𝑄 is a 𝜌-contraction, it has a
unique fixed point 𝑥∗ ; it is the unique solution of (99). Since
(H1)–(H6) hold, the sequence {𝑥𝑛 } is asymptotically regular
(by Lemma 16). In terms of Lemma 17, ‖𝑥𝑛 − 𝑦𝑛 ‖ → 0 and
16
Journal of Applied Mathematics
‖𝑥𝑛 − 𝑢𝑛 ‖ → 0 as 𝑛 → ∞. Moreover, utilizing Lemmas 8
and 10, we have from (33)-(34) the following:
we know that 𝜔𝑤 (𝑥𝑛 ) ⊂ Ω, and hence 𝑧 ∈ Ω. Taking into
consideration that 𝑥∗ = 𝑃Ω 𝑄𝑥∗ we obtain from (101) that
lim sup ⟨𝑄𝑥∗ − 𝑥∗ , 𝑦𝑛 − 𝑥∗ ⟩
󵄩󵄩
∗ 󵄩2
󵄩󵄩𝑥𝑛+1 − 𝑥 󵄩󵄩󵄩
󵄩
󵄩2
󵄩2
󵄩
≤ 𝜎𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑥∗ 󵄩󵄩󵄩 + (1 − 𝜎𝑛 ) 󵄩󵄩󵄩𝑦̃𝑛 − 𝑥∗ 󵄩󵄩󵄩
󵄩
󵄩2
󵄩
󵄩 󵄩2
󵄩
≤ 𝜎𝑛 󵄩󵄩󵄩𝑦𝑛 − 𝑥∗ 󵄩󵄩󵄩 + (1 − 𝜎𝑛 ) (󵄩󵄩󵄩𝑦𝑛 − 𝑥∗ 󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
𝑛→∞
= lim sup [⟨𝑄𝑥∗ − 𝑥∗ , 𝑥𝑛 − 𝑥∗ ⟩ + ⟨𝑄𝑥∗ − 𝑥∗ , 𝑦𝑛 − 𝑥𝑛 ⟩]
𝑛→∞
= lim sup ⟨𝑄𝑥∗ − 𝑥∗ , 𝑥𝑛 − 𝑥∗ ⟩
𝑛→∞
󵄩
󵄩 󵄩2
󵄩
≤ (󵄩󵄩󵄩𝑦𝑛 − 𝑥∗ 󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
= lim ⟨𝑄𝑥∗ − 𝑥∗ , 𝑥𝑛𝑘 − 𝑥∗ ⟩
󵄩 󵄩 󵄩
󵄩
󵄩 󵄩
󵄩
= 󵄩󵄩󵄩𝑦𝑛 − 𝑥 󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛 − 𝑥∗ 󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩
󵄩2
≤ 󵄩󵄩󵄩𝛽𝑛 (𝑄𝑦𝑛,𝑁 − 𝑄𝑥∗ ) + (1 − 𝛽𝑛 ) (𝑦̃𝑛,𝑁 − 𝑥∗ )󵄩󵄩󵄩
= ⟨𝑄𝑥∗ − 𝑥∗ , 𝑧 − 𝑥∗ ⟩ ≤ 0.
𝑘→∞
∗󵄩
󵄩2
∗
∗
(102)
∞
Since ∑∞
𝑛=0 𝛼𝑛 < ∞ and ∑𝑛=0 𝛽𝑛 = ∞, we deduce that
∞
∗
∑𝑛=0 𝜆𝛼𝑛 ‖𝑝‖(2‖𝑦𝑛 − 𝑥 ‖ + 𝜆𝛼𝑛 ‖𝑝‖) < ∞ and ∑∞
𝑛=0 (1 − 𝜌)𝛽𝑛 =
∞. In terms of Lemma 9 we derive 𝑥𝑛 → 𝑥∗ as 𝑛 → ∞.
∗
+ 2𝛽𝑛 ⟨𝑄𝑥 − 𝑥 , 𝑦𝑛 − 𝑥 ⟩
󵄩 󵄩 󵄩
󵄩
󵄩 󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛 − 𝑥∗ 󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
In a similar way, we can derive the following result.
󵄩
󵄩2
󵄩
󵄩
≤ (𝛽𝑛 𝜌 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑥∗ 󵄩󵄩󵄩 + (1 − 𝛽𝑛 ) 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑥∗ 󵄩󵄩󵄩)
+ 2𝛽𝑛 ⟨𝑄𝑥∗ − 𝑥∗ , 𝑦𝑛 − 𝑥∗ ⟩
󵄩 󵄩 󵄩
󵄩
󵄩 󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛 − 𝑥∗ 󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
2󵄩
󵄩2
= (1 − (1 − 𝜌) 𝛽𝑛 ) 󵄩󵄩󵄩𝑦𝑛,𝑁 − 𝑥∗ 󵄩󵄩󵄩
(100)
+ 2𝛽𝑛 ⟨𝑄𝑥∗ − 𝑥∗ , 𝑦𝑛 − 𝑥∗ ⟩
(i) (𝛼𝑛 + 𝛽𝑛 )/𝛽𝑛,𝑘0 → 0 as 𝑛 → ∞;
󵄩 󵄩 󵄩
󵄩
󵄩 󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛 − 𝑥∗ 󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩2
󵄩
≤ (1 − (1 − 𝜌) 𝛽𝑛 ) 󵄩󵄩󵄩𝑢𝑛 − 𝑥∗ 󵄩󵄩󵄩
(ii) if 𝑖 ≤ 𝑘0 and 𝛽𝑛,𝑖 → 0, then 𝛽𝑛,𝑘0 /𝛽𝑛,𝑖 → 0 as 𝑛 →
∞;
(iii) if 𝛽𝑛,𝑖 → 𝛽𝑖 ≠ 0, then 𝛽𝑖 lies in (0, 1).
+ 2𝛽𝑛 ⟨𝑄𝑥∗ − 𝑥∗ , 𝑦𝑛 − 𝑥∗ ⟩
Then, the sequences {𝑥𝑛 }, {𝑦𝑛 }, and {𝑢𝑛 } explicitly defined by
scheme (30) all converge strongly to the unique solution 𝑥∗ ∈ Ω
of the following variational inequality:
󵄩 󵄩 󵄩
󵄩
󵄩 󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛 − 𝑥∗ 󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩2
󵄩
≤ (1 − (1 − 𝜌) 𝛽𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑥∗ 󵄩󵄩󵄩
⟨𝑄𝑥∗ − 𝑥∗ , 𝑧 − 𝑥∗ ⟩ ≤ 0,
+ 2𝛽𝑛 ⟨𝑄𝑥∗ − 𝑥∗ , 𝑦𝑛 − 𝑥∗ ⟩
∀𝑧 ∈ Ω.
(103)
Remark 29. According to the above argument processes for
Theorems 27 and 28, we can readily see that if in scheme (30),
the iterative step 𝑦𝑛 = 𝛽𝑛 𝑄𝑦𝑛,𝑁 +(1−𝛽𝑛 )𝑃𝐶(𝑦𝑛,𝑁 −𝜆∇𝑓𝛼𝑛 (𝑦𝑛,𝑁))
is replaced by the iterative one 𝑦𝑛 = 𝛽𝑛 𝑄𝑥𝑛 +(1−𝛽𝑛 )𝑃𝐶(𝑦𝑛,𝑁 −
𝜆∇𝑓𝛼𝑛 (𝑦𝑛,𝑁)), then Theorems 27 and 28 remain valid.
󵄩 󵄩 󵄩
󵄩
󵄩 󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛 − 𝑥∗ 󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩)
󵄩2
󵄩
= (1 − (1 − 𝜌) 𝛽𝑛 ) 󵄩󵄩󵄩𝑥𝑛 − 𝑥∗ 󵄩󵄩󵄩
2
⟨𝑄𝑥∗ − 𝑥∗ , 𝑦𝑛 − 𝑥∗ ⟩
1−𝜌
󵄩 󵄩 󵄩
󵄩
󵄩 󵄩
+ 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩 (2 󵄩󵄩󵄩𝑦𝑛 − 𝑥∗ 󵄩󵄩󵄩 + 𝜆𝛼𝑛 󵄩󵄩󵄩𝑝󵄩󵄩󵄩) .
+ (1 − 𝜌) 𝛽𝑛 ⋅
Remark 30. Theorems 27 and 28 improve, extend, supplement, and develop [17, Theorems 3.12 and 3.13] and [1,
Theorems 5.2 and 6.1] in the following aspects:
Now, let {𝑥𝑛𝑘 } be a subsequence of {𝑥𝑛 } such that
lim sup ⟨𝑄𝑥∗ − 𝑥∗ , 𝑥𝑛 − 𝑥∗ ⟩ = lim ⟨𝑄𝑥∗ − 𝑥∗ , 𝑥𝑛𝑘 − 𝑥∗ ⟩ .
𝑛→∞
Theorem 28. Let one suppose that Ω ≠ 0. Let {𝛽𝑛 }, {𝛽𝑛,𝑖 }, 𝑖 =
1, . . . , 𝑁, be sequences in (0, 1) such that 𝛽𝑛,𝑖 → 𝛽𝑖 for all 𝑖 as
𝑛 → ∞. Suppose that there exists 𝑘 ∈ {1, . . . , 𝑁} for which
𝛽𝑛,𝑘 → 0 as 𝑛 → ∞. Let 𝑘0 ∈ {1, . . . , 𝑁} the largest index for
which 𝛽𝑛,𝑘0 → 0. Moreover, let one suppose that (H1), (H7),
and (H8) hold, and
𝑘→∞
(101)
By the boundedness of {𝑥𝑛 }, we may assume, without loss of
generality, that 𝑥𝑛𝑘 ⇀ 𝑧 ∈ 𝜔𝑤 (𝑥𝑛 ). According to Corollary 26,
(a) the multistep iterative scheme (30) of [17] is extended
to develop our relaxed viscosity iterative scheme (30)
with regularization for MP (3), EP (10), and strict
pseudocontraction 𝑇 by virtue of Xu iterative schemes
in [1];
(b) the argument techniques in Theorems 27 and 28 are
very different from the ones in [17, Theorems 3.12 and
3.13] and the ones in [1, Theorems 5.2 and 6.1] because
we use the properties of strict pseudocontractive
mappings and maximal monotone mappings (see,
e.g., Lemmas 7 and 10);
Journal of Applied Mathematics
17
(c) compared with the proof of Theorems 5.2 and 6.1
in [1], the proof of Theorems 27 and 28 shows that
lim𝑛 → ∞ ‖𝑦𝑛,𝑁 − 𝑃𝐶(𝐼 − 𝜆∇𝑓𝛼𝑛 )𝑦𝑛,𝑁‖ = lim𝑛 → ∞ ‖𝑦𝑛 −
𝑃𝐶(𝐼 − 𝜆∇𝑓𝛼𝑛 )𝑦𝑛 ‖ = 0 via the argument of
lim𝑛 → ∞ ‖∇𝑓𝛼𝑛 (𝑦𝑛,𝑁) − ∇𝑓(𝑝)‖ = 0, for all 𝑝 ∈ Ω (see
Lemma 20 and its proof);
(d) the problem of finding an element of Fix(𝑇) ∩
(⋂𝑖 Fix(𝑆𝑖 )) ∩ EP(𝐹, ℎ) ∩ Γ in Theorems 27 and 28 is
more general than the one of finding an element of
Fix(𝑇) ∩ (⋂𝑖 Fix(𝑆𝑖 )) ∩ EP(𝐹, ℎ) in [17, Theorems 3.12
and 3.13] and the one of finding an element of Γ in [1,
Theorems 5.2 and 6.1].
us consider the following mixed problem of finding 𝑥∗ ∈
Fix(𝑇) ∩ EP(𝐹, ℎ) ∩ Γ such that
⟨(𝐼 − 𝑆̃1 ) 𝑥∗ , 𝑦 − 𝑥∗ ⟩ ≥ 0,
∀𝑦 ∈ Fix (𝑇) ∩ EP (𝐹, ℎ) ∩ Γ,
⟨(𝐼 − 𝑆̃2 ) 𝑥∗ , 𝑦 − 𝑥∗ ⟩ ≥ 0,
∀𝑦 ∈ Fix (𝑇) ∩ EP (𝐹, ℎ) ∩ Γ,
..
.
⟨(𝐼 − 𝑆̃𝑀) 𝑥∗ , 𝑦 − 𝑥∗ ⟩ ≥ 0,
⟨𝐴 1 𝑥∗ , 𝑦 − 𝑥∗ ⟩ ≥ 0,
∀𝑦 ∈ 𝐶,
⟨𝐴 2 𝑥∗ , 𝑦 − 𝑥∗ ⟩ ≥ 0,
∀𝑦 ∈ 𝐶,
..
.
4. Applications
For a given nonlinear mapping 𝐴 : 𝐶 → 𝐻, we consider
again the variational inequality problem (VIP) of finding 𝑥 ∈
𝐶 such that
⟨𝐴𝑥, 𝑦 − 𝑥⟩ ≥ 0,
∀𝑦 ∈ Fix (𝑇) ∩ EP (𝐹, ℎ) ∩ Γ,
∀𝑦 ∈ 𝐶.
(104)
Recall that if 𝑢 is a point in 𝐶, then the following relation
holds:
𝑢 ∈ VI (𝐶, 𝐴) ⇐⇒ 𝑢 = 𝑃𝐶 (𝐼 − 𝜆𝐴) 𝑢,
for some 𝜆 > 0,
(105)
from which we have the following relation:
⟨𝐴 𝑁𝑥∗ , 𝑦 − 𝑥∗ ⟩ ≥ 0,
(108)
We denote by (SVI) the set of solutions of the above (𝑁 +
𝑀) system. This problem is equivalent to finding a common
𝑀
fixed point of 𝑇, {𝑃Fix(𝑇)∩EP(𝐹,ℎ)∩Γ 𝑆̃𝑖 }𝑁
𝑖=1 , {𝑃𝐶 (𝐼 − 𝜇𝐴 𝑖 )}𝑖=1 . The
following results are then consequences of Theorems 27 and
28.
Theorem 31. Let one suppose that Ω = Fix(𝑇) ∩ (SVI) ∩
EP(F, h) ∩ Γ ≠ 0. Fix 𝜇 ∈ (0, 2𝛼], and 𝜆 ∈ (0, 2/𝐿). Let {𝛼𝑛 },
{𝛽𝑛,𝑖 }, 𝑖 = 1, . . . , (𝑀 + 𝑁), be sequences in (0, 1) such that
0 < lim inf 𝑛 → ∞ 𝛽𝑛,𝑖 ≤ lim sup𝑛 → ∞ 𝛽𝑛,𝑖 < 1 for all index
𝑖. Moreover, Let one suppose that (H1)–(H6) hold. Then the
sequences {𝑥𝑛 }, {𝑦𝑛 }, and {𝑢𝑛 } explicitly defined by the following
scheme:
𝐹 (𝑢𝑛 , 𝑦) + ℎ (𝑢𝑛 , 𝑦) +
𝑢 ∈ Γ ⇐⇒ 𝑢 ∈ VI (𝐶, ∇𝑓) ⇐⇒ 𝑢 = 𝑃𝐶 (𝐼 − 𝜆∇𝑓) 𝑢,
for some 𝜆 > 0.
(106)
∀𝑥, 𝑦 ∈ 𝐶.
1
⟨𝑦 − 𝑢𝑛 , 𝑢𝑛 − 𝑥𝑛 ⟩ ≥ 0,
𝑟𝑛
∀𝑦 ∈ 𝐶,
𝑦𝑛,1 = 𝛽𝑛,1 𝑃Fix(𝑇)∩EP(𝐹,ℎ)∩Γ 𝑆̃1 𝑢𝑛 + (1 − 𝛽𝑛,1 ) 𝑢𝑛 ,
𝑦𝑛,𝑖 = 𝛽𝑛,𝑖 𝑃Fix(𝑇)∩EP(𝐹,ℎ)∩Γ 𝑆̃𝑖 𝑢𝑛
An operator 𝐴 : 𝐶 → 𝐻 is said to be an 𝛼-inverse
strongly monotone operator if there exists a constant 𝛼 > 0
such that
󵄩2
󵄩
⟨𝐴𝑥 − 𝐴𝑦, 𝑥 − 𝑦⟩ ≥ 𝛼󵄩󵄩󵄩𝐴𝑥 − 𝐴𝑦󵄩󵄩󵄩 ,
∀𝑦 ∈ 𝐶.
(107)
As an example, we recall that the 𝛼-inverse strongly
monotone operators are firmly nonexpansive mappings if 𝛼 ≥
1 and that every 𝛼-inverse strongly monotone operator is also
a (1/𝛼) Lipschitz continuous (see [19]). We observe that, if 𝐴
is 𝛼-inverse strongly monotone, the mapping 𝑃𝐶(𝐼 − 𝜇𝐴) is
nonexpansive for all 𝜇 ∈ (0, 2𝛼] since they are compositions
of nonexpansive mappings (see [19, page 419]).
Let us consider 𝑆̃1 , . . . , 𝑆̃𝑀 a finite number of nonexpansive self-mappings on 𝐶 and 𝐴 1 , . . . , 𝐴 𝑁 be a finite number
of 𝛼-inverse strongly monotone operators. Let 𝑇 : 𝐶 → 𝐶
be a 𝜁-strict pseudocontraction on 𝐶 with fixed points. Let
+ (1 − 𝛽𝑛,𝑖 ) 𝑦𝑛,𝑖−1 ,
𝑖 = 2, . . . , 𝑀,
𝑦𝑛,𝑀+𝑗 = 𝛽𝑛,𝑀+𝑗 𝑃𝐶 (𝐼 − 𝜇𝐴 𝑗 ) 𝑢𝑛
+ (1 − 𝛽𝑛,𝑀+𝑗 ) 𝑦𝑛,𝑀+𝑗−1 ,
𝑗 = 1, . . . , 𝑁,
𝑦𝑛 = 𝛽𝑛 𝑄𝑦𝑛,𝑀+𝑁 + (1 − 𝛽𝑛 )
× 𝑃𝐶 (𝑦𝑛,𝑀+𝑁 − 𝜆∇𝑓𝛼𝑛 (𝑦𝑛,𝑀+𝑁)) ,
𝑥𝑛+1 = 𝜎𝑛 𝑦𝑛 + 𝛾𝑛 𝑃𝐶 (𝑦𝑛 − 𝜆∇𝑓𝛼𝑛 (𝑦𝑛 ))
+ 𝛿𝑛 𝑇𝑃𝐶 (𝑦𝑛 − 𝜆∇𝑓𝛼𝑛 (𝑦𝑛 )) ,
∀𝑛 ≥ 0,
(109)
all converge strongly to the unique solution 𝑥∗ ∈ Ω of the
following variational inequality:
⟨𝑄𝑥∗ − 𝑥∗ , 𝑧 − 𝑥∗ ⟩ ≤ 0,
∀𝑧 ∈ Ω.
(110)
18
Journal of Applied Mathematics
Theorem 32. Let one suppose that Ω ≠ 0. Fix 𝜇 ∈ (0, 2𝛼] and
𝜆 ∈ (0, 2/𝐿). Let {𝛽𝑛 }, {𝛽𝑛,𝑖 }, 𝑖 = 1, . . . , (𝑀 + 𝑁), be sequences
in (0, 1) and 𝛽𝑛,𝑖 → 𝛽𝑖 for all 𝑖 as 𝑛 → ∞. Suppose that there
exists 𝑘 ∈ {1, . . . , 𝑀 + 𝑁} such that 𝛽𝑛,𝑘 → 0 as 𝑛 → ∞. Let
𝑘0 ∈ {1, . . . , 𝑀 + 𝑁} be the largest index for which 𝛽𝑛,𝑘0 → 0.
Moreover, let one suppose that (H1), (H7), and (H8) hold, and
(⋂𝑖 Fix(𝑆𝑖 )) ∩ EP(𝐹, ℎ) ∩ Γ = (⋂𝑖 Fix(𝑆𝑖 )) ∩ EP(𝐹, ℎ) ∩ Γ ≠ 0,
(𝛾𝑛 + 𝛿𝑛 )𝜁 ≤ 𝛾𝑛 , for all 𝑛 ≥ 0, and
𝑥𝑛+1 = 𝜎𝑛 𝑦𝑛 + 𝛾𝑛 𝑃𝐶 (𝑦𝑛 − 𝜆∇𝑓𝛼𝑛 (𝑦𝑛 ))
+ 𝛿𝑛 𝑇𝑃𝐶 (𝑦𝑛 − 𝜆∇𝑓𝛼𝑛 (𝑦𝑛 ))
(i) (𝛼𝑛 + 𝛽𝑛 )/𝛽𝑛,𝑘0 → 0 as 𝑛 → ∞;
= 𝜎𝑛 𝑦𝑛 + 𝛿𝑛 𝑃𝐶 (𝑦𝑛 − 𝜆∇𝑓𝛼𝑛 (𝑦𝑛 ))
(ii) if 𝑖 ≤ 𝑘0 and 𝛽𝑛,𝑖 → 0, then 𝛽𝑛,𝑘0 /𝛽𝑛,𝑖 → 0 as 𝑛 →
∞;
= 𝜎𝑛 𝑦𝑛 + (1 − 𝜎𝑛 ) 𝑃𝐶 (𝑦𝑛 − 𝜆∇𝑓𝛼𝑛 (𝑦𝑛 )) .
(iii) if 𝛽𝑛,𝑖 → 𝛽𝑖 ≠ 0, then 𝛽𝑖 lies in (0, 1).
Then, the sequences {𝑥𝑛 }, {𝑦𝑛 }, and {𝑢𝑛 } explicitly defined by
scheme (109) all converge strongly to the unique solution 𝑥∗ ∈
Ω of the following variational inequality:
⟨𝑄𝑥∗ − 𝑥∗ , 𝑧 − 𝑥∗ ⟩ ≤ 0,
∀𝑧 ∈ Ω.
(111)
Remark 33. If we choose ∇𝑓 = 𝐴 1 = ⋅ ⋅ ⋅ = 𝐴 𝑁 = 0 in system
(108), we obtain a system of hierarchical fixed point problems
introduced by Moudafi and Maingé [20, 21].
Further, utilizing Theorems 27 and 28, we again give
the following strong convergence theorems for finding a
common element of the solution set Γ of MP (3), the solution
set EP(𝐹, ℎ) of EP (10), and the common fixed point set
(⋂𝑖 Fix(𝑆𝑖 )) of a finite family of nonexpansive mappings 𝑆𝑖 :
𝐶 → 𝐶, 𝑖 = 1, . . . , 𝑁.
Theorem 34. Let one suppose that Ω = (⋂𝑖 Fix(𝑆𝑖 )) ∩
EP(𝐹, ℎ) ∩ Γ ≠ 0. Let {𝛽𝑛 }, {𝛽𝑛,𝑖 }, 𝑖 = 1, . . . , 𝑁, be sequences
in (0, 1) such that 0 < lim inf 𝑛 → ∞ 𝛽𝑛,𝑖 ≤ lim sup𝑛 → ∞ 𝛽𝑛,𝑖 < 1
for all index 𝑖. Moreover, Let one suppose that there hold (H1)–
(H6) with 𝛾𝑛 = 0, for all 𝑛 ≥ 0. Then, the sequences {𝑥𝑛 }, {𝑦𝑛 },
and {𝑢𝑛 } generated explicitly by
𝐹 (𝑢𝑛 , 𝑦) + ℎ (𝑢𝑛 , 𝑦) +
1
⟨𝑦 − 𝑢𝑛 , 𝑢𝑛 − 𝑥𝑛 ⟩ ≥ 0,
𝑟𝑛
∀𝑦 ∈ 𝐶,
𝑦𝑛,1 = 𝛽𝑛,1 𝑆1 𝑢𝑛 + (1 − 𝛽𝑛,1 ) 𝑢𝑛 ,
𝑦𝑛,𝑖 = 𝛽𝑛,𝑖 𝑆𝑖 𝑢𝑛 + (1 − 𝛽𝑛,𝑖 ) 𝑦𝑛,𝑖−1 ,
𝑦𝑛 = 𝛽𝑛 𝑄𝑦𝑛,𝑁 + (1 − 𝛽𝑛 ) 𝑃𝐶 (𝑦𝑛,𝑁 − 𝜆∇𝑓𝛼𝑛 (𝑦𝑛,𝑁)) ,
∀𝑛 ≥ 0,
(112)
all converge strongly to the unique solution 𝑥∗ ∈ Ω of the
following variational inequality:
⟨𝑄𝑥∗ − 𝑥∗ , 𝑧 − 𝑥∗ ⟩ ≤ 0,
∀𝑧 ∈ Ω.
Thus, the conditions in Theorem 27 are all satisfied. and from
which we obtain the desired result.
Theorem 35. Let one suppose that Ω = (⋂𝑖 Fix(𝑆𝑖 )) ∩
EP(𝐹, ℎ) ∩ Γ ≠ 0. Let {𝛽𝑛 }, {𝛽𝑛,𝑖 }, 𝑖 = 1, . . . , 𝑁, be sequences
in (0, 1) such that 𝛽𝑛,𝑖 → 𝛽𝑖 for all 𝑖 as 𝑛 → ∞. Suppose that
there exists 𝑘 ∈ {1, . . . , 𝑁} for which 𝛽𝑛,𝑘 → 0 as 𝑛 → ∞.
Let 𝑘0 ∈ {1, . . . , 𝑁} be the largest index for which 𝛽𝑛,𝑘0 → 0.
Moreover, let one suppose that there hold (H1), (H7), and (H8)
with 𝛾𝑛 = 0, for all 𝑛 ≥ 0, and
(i) (𝛼𝑛 + 𝛽𝑛 )/𝛽𝑛,𝑘0 → 0 as 𝑛 → ∞;
(ii) if 𝑖 ≤ 𝑘0 and 𝛽𝑛,𝑖 → 0, then 𝛽𝑛,𝑘0 /𝛽𝑛,𝑖 → 0 as 𝑛 →
∞;
(iii) if 𝛽𝑛,𝑖 → 𝛽𝑖 ≠ 0, then 𝛽𝑖 lies in (0, 1).
Then the sequences {𝑥𝑛 }, {𝑦𝑛 }, and {𝑢𝑛 } generated explicitly by
(112) all converge strongly to the unique solution 𝑥∗ ∈ Ω of the
following variational inequality:
⟨𝑄𝑥∗ − 𝑥∗ , 𝑧 − 𝑥∗ ⟩ ≤ 0,
∀𝑧 ∈ Ω.
(115)
Acknowledgments
This research was partially supported by the National Science
Foundation of China (11071169), Innovation Program of
Shanghai Municipal Education Commission (09ZZ133), and
Leading Academic Discipline Project of Shanghai Normal
University (DZL707). This research was partially supported
by the Grant from the NSC 101-2115-M-037-001.
References
𝑖 = 2, . . . , 𝑁,
𝑥𝑛+1 = 𝜎𝑛 𝑦𝑛 + (1 − 𝜎𝑛 ) 𝑃𝐶 (𝑦𝑛 − 𝜆∇𝑓𝛼𝑛 (𝑦𝑛 )) ,
(114)
(113)
Proof. In Theorems 27, put 𝑇 = 𝐼 the identity mapping and
𝛾𝑛 = 0, for all 𝑛 ≥ 0. Then, 𝑇 is a 𝜁-strictly pseudocontractive
mapping with 𝜁 = 0. Hence, we deduce that Ω = Fix(𝑇) ∩
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19
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