A NEW SYSTEM OF GENERALIZED NONLINEAR RELAXED COCOERCIVE VARIATIONAL INEQUALITIES
KE DING, WEN-YONG YAN, AND NAN-JING HUANG
Received 21 November 2004; Revised 13 April 2005; Accepted 28 June 2005
We introduce and study a new system of generalized nonlinear relaxed cocoercive in- equality problems and construct an iterative algorithm for approximating the solutions of the system of generalized relaxed cocoercive variational inequalities in Hilbert spaces. We prove the existence of the solutions for the system of generalized relaxed cocoercive variational inequality problems and the convergence of iterative sequences generated by the algorithm. We also study the convergence and stability of a new perturbed iterative algorithm for approximating the solution.
Copyright © 2006 Ke Ding et al. This is an open access article distributed under the Cre- ative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
Variational inequality problems have various applications in mechanics and physics, opti- mization and control, linear and nonlinear programming, economics and finance, trans- portation equilibrium and engineering science, and so forth. Consequently considerable attention has been devoted to the study of the theory and efficient numerical methods for variational inequality problems (see, e.g., [2–17] and the references therein). In [15], Verma introduced a new system of nonlinear strongly monotone variational inequalities and studied the approximate of this system based on the projection method, and in [16], Verma discussed the approximate solvability of a system of nonlinear relaxed cocoercive variational inequalities in Hilbert spaces. Recently, Kim and Kim [14] introduced and studied a system of nonlinear mixed variational inequalities in Hilbert spaces, and ob- tained some approximate solvability results. In the recent paper [6], Cho et al. introduced and studied a new system of nonlinear variational inequalities in Hilbert spaces. They proved some existence and uniqueness theorems of solutions for the system of nonlinear variational inequalities. They also constructed an iterative algorithm for approximating the solution of the system of nonlinear variational inequalities. Some related works, we refer to [2, 3, 5, 7–10, 12, 13]. Motivated and inspired by these works, in this paper, we introduce and study a new system of generalized nonlinear relaxed cocoercive variational
Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2006, Article ID 40591, Pages 1–14 DOI 10.1155/JIA/2006/40591
2 Nonlinear relaxed cocoercive variational inequalities
inequality problems and construct an iterative algorithm for approximating the solutions of the system of generalized relaxed cocoercive variational inequalities in Hilbert spaces. We prove the existence of the solutions for the system of generalized relaxed cocoercive variational inequality problems and the convergence of iterative sequences generated by the algorithm. We also study the convergence and stability of a new perturbed iterative algorithm for approximating the solution. The results presented in this paper improve and extend the previously known results in this area.
2. Preliminaries
(cid:3) ≥ 0, ∀z ∈ K1(x),
Let H be a Hilbert space endowed with a norm (cid:2) · (cid:2) and inner product (·, ·), respectively. Let CB(H) be the family of all nonempty subsets of H and K1, K2 be two convex and closed subsets of H. Let g1, g2, m1, m2 : H → H and F, G : H × H → H be mappings. We consider the following system of generalized nonlinear variational inequality problems: find x, y ∈ H such that gi(x) ∈ Ki(x) for i = 1, 2, and (cid:2) F(x, y),z − g1(x) (cid:3)
(2.1)
(cid:2) G(x, y),z − g2(y) ≥ 0, ∀z ∈ K2(y),
where Ki(x) = mi(x) + Ki for i = 1, 2.
When K1 and K2 are both convex cones of H, it is easy to see that problem (2.1) is equivalent to the following system of generalized nonlinear co-complementarity prob- lems: find x, y ∈ H such that gi(x) ∈ Ki(x) for i = 1, 2, and
(cid:3)∗
,
F(x, y) ∈
(cid:2) K1(x) − g1(x)
(2.2)
(cid:3)∗
,
G(x, y) ∈
(cid:2) K2(y) − g2(y)
where Ki(x) = mi(x) + Ki and (Ki(x) − gi(x))∗ is the dual of Ki(x) − gi(x) for i = 1, 2, that is,
(cid:3)∗ =
(2.3)
(cid:5) . (cid:2) Ki(x) − gi(x) (cid:4) u ∈ H | (u,v) ≥ 0, ∀v ∈ Ki(x) − gi(x)
Some examples of problems (2.1) and (2.2) are as follows. (I) If G = 0 and F(x, y) = Tx + Ax for all x, y ∈ X, where T, A : H → H are two map-
pings, then problem (2.2) reduces to finding x ∈ H such that
(cid:3)∗
,
(2.4)
Tx + Ax ∈
(cid:2) K1(x) − g1(x)
which is called the generalized complementarity problem. The problem (2.4) was ex- tended and studied by Jou and Yao [11] in Hilbert spaces, and by Chen et al. [5] in the setting of Banach spaces.
(II) Let T : H × H → H be a mapping. If F(x, y) = ρT(y,x) + x − y, G(x, y) = ηT(x, y) + y − x for all x, y ∈ H, m1 = m2 = 0, K1 = K2 = K, and g1 = g2 = I, where I is an identity
Ke Ding et al.
3
mapping and ρ > 0, η > 0, then problem (2.1) reduces to finding x, y ∈ K such that
(cid:3) (cid:2) ρT(y,x) + x − y,z − x ≥ 0, ∀z ∈ K, (cid:3)
(2.5)
(cid:2) ηT(x, y) + y − x,z − y ≥ 0, ∀z ∈ K,
which is called the system of nonlinear variational inequality problems considered by Verma [16]. The special case of problem (2.5) was studied by Verma [15]. The problem (2.5) was extended and studied by Agarwal et al. [1], Kim and Kim [14], and Cho et al. [6].
(III) If m1 = m2 = 0, and g1 = g2 = I, then problem (2.1) reduces to finding x ∈ K1 and
y ∈ K2 such that
(cid:3) (cid:2) F(x, y),z − x ≥ 0, ∀z ∈ K1, (cid:3)
(2.6)
(cid:2) G(x, y),z − y ≥ 0, ∀z ∈ K2,
which is just the problem considered in [12] with F, G being single-valued mappings.
Definition 2.1. A mapping N : H × H → H is said to be
(i) α-strongly monotone with respect to first argument if there exists some α > 0 such
that
(cid:3)
(2.7)
(cid:2) N(x, ·) − N(y, ·),x − y ≥ α(cid:2)x − y(cid:2)2, ∀(x, y) ∈ H × H;
(ii) ξ-Lipschitz continuous with respect to the first argument, if there exists a constant
ξ > 0 such that
(2.8)
(cid:6) (cid:6)N(x, ·) − N(y, ·) (cid:6) (cid:6) ≤ ξ (cid:6) (cid:6)x − y (cid:6) (cid:6), ∀(x, y) ∈ H × H.
Similarly, we can define the strong monotonicity and Lipschitzian continuity with re-
spect to the second argument of N.
Definition 2.2. A Mapping N : H × H → H is said to be relaxed (a,b)-cocoercive with respect to the first argument if there exists constants a > 0 and b > 0 such that
(cid:3)
(2.9)
(cid:2) N(x, ·) − N(y, ·),x − y ≥ (−a)(cid:2)x − y(cid:2)2 + b(cid:2)x − y(cid:2)2, ∀(x, y) ∈ H × H.
If a = 0, then N is b-strongly monotone. Similarly, we can define the relaxed (a,b)-co- coercivity with respect to the second argument of N.
Lemma 2.3 [4]. If K ⊂ H is a closed convex subset and z ∈ H is a given point, then there exists x ∈ K such that
(2.10)
(x − z, y − x) ≥ 0, ∀y ∈ K
if and only if x = PK z, where PK is the projection of H onto K.
Lemma 2.4 [4]. The projection PK is nonexpansive, that is,
(2.11)
(cid:6) (cid:6) ≤ (cid:2)u − v(cid:2), ∀u,v ∈ H. (cid:6) (cid:6)PK u − PK v
4 Nonlinear relaxed cocoercive variational inequalities
Lemma 2.5 [18]. Let {Kn} be a sequence of closed convex subsets of H such that H(Kn,K) → 0 as n → ∞, where H(·, ·) is the Hausdorff metric, that is, for any A,B ∈ CB(H),
(cid:7) (cid:8)
(2.12)
H(A,B) = max
(cid:2)a − b(cid:2)
.
sup a∈A
(cid:2)a − b(cid:2), sup b∈B
inf b∈B
inf a∈A
Then
(cid:6) (cid:6) −→ 0
(2.13)
(n −→ ∞), ∀v ∈ H.
(cid:6) (cid:6)PKnv − PK v
Lemma 2.6 [4]. If K(u) = m(u) + K for all u ∈ H, then
(2.14)
(cid:2) v − m(u) (cid:3) .
PK(u)v = m(u) + PK
From Lemmas 2.3 and 2.6, we have the following lemma.
Lemma 2.7. If K1,K2 ⊂ H are two closed convex cones, and Ki(·) = m(·) + Ki (i = 1,2), then x, y ∈ H solve problem (2.1) if and only if x, y ∈ H such that
x = x − g1(x) + m1(x) + PK1
(2.15)
(cid:3) , (cid:3) , (cid:2) g1(x) − ρF(x, y) − m1(x) (cid:2) g2(y) − ρG(x, y) − m2(y)
y = y − g2(y) + m2(y) + PK2
n=0
(cid:9)∞
(2.16)
where ρ > 0 is a constant. Lemma 2.8 [17]. Let {μn} be a real sequence of nonnegative numbers and {νn} be a real sequence of numbers in [0,1] with (cid:2) 1 − νn
νn = ∞. If there exists a constant n1 such that (cid:3) μn + νnδn, ∀n ≥ n1,
μn+1 ≤
where δn ≥ 0 for all n ≥ 0, and δn → 0 (n → ∞), then limn→∞ μn = 0.
3. Existence and convergence
In this section, we construct an iterative algorithm to approximate the solution of prob- lem (2.1) and study the convergence of the sequence generated by the algorithm.
(cid:3) (cid:3) (cid:3)
+ m1
xn+1 = xn − g1
+ PK1
Algorithm 3.1. For any given x0, y0 ∈ H, we compute (cid:3) (cid:2) xn (cid:2)
(cid:2) xn (cid:2) (cid:2) xn (cid:2) (cid:3) (cid:3) − ρF (cid:3) − m1 (cid:3) (cid:2) xn (cid:2)
(3.1)
− ρG (cid:3)(cid:3) , (cid:3)(cid:3) .
+ m2
(cid:2) g1 (cid:2) g2 − m2
yn+1 = yn − g2
yn
yn
yn
(cid:2) xn, yn (cid:2) xn, yn
yn
+ PK2
Theorem 3.2. Let gi : H → H be ηi-strongly monotone and ζi-Lipschitz continuous and mi : H → H be γi-Lipschitz continuous (i = 1,2). Let F : H × H → H be l1, l2-Lipschitz con- tinuous with respect to the first, second arguments, respectively, and relaxed (a,b)-cocoercive with respect to the first argument. Let G : H × H → H be n1, n2-Lipschitz continuous with respect to the first, second arguments, respectively, and relaxed (c,d)-cocoercive with respect
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5
to the second argument. If
(cid:10) (cid:10)
1 + ρ2l2
− 2η1 + 2γ1 + − 2ρb + ρn1 < 1,
1 + ζ 2 1
1 + 2ρal2 1
2 (cid:10)
(cid:10)
(3.2)
2
1 + ρ2n2
− 2η2 + 2γ2 + − 2ρd + ρl2 < 1.
1 + ζ 2 2
2 + 2ρcn2 2
then there exist x∗, y∗ ∈ H, which solve problem (2.1). Moreover, the iterative sequences {xn} and {yn} generated by Algorithm 3.1 converge to x∗ and y∗, respectively.
Proof. From (3.1) and Lemma 2.6, we have
(cid:3) (cid:2) (cid:3) (cid:3) (cid:3) (cid:3)(cid:3) (cid:6) (cid:6) = − ρF − m1 (cid:6) (cid:6)xn+1 − xn
xn
(cid:2) xn (cid:2) xn, yn (cid:2) xn (cid:2) xn (cid:3) (cid:2) g1 (cid:3) −
+ m1
(cid:6) (cid:6)xn − g1 (cid:11) xn−1 − g1
+ m1 (cid:2) xn−1
(cid:2) (cid:3) (cid:3) (cid:3)(cid:3)(cid:12)(cid:6) (cid:6)
− ρF (cid:3)
xn−1 (cid:3)
(3.3)
(cid:3)(cid:6) (cid:6) ≤ − g1 − m1
xn
(cid:2) xn−1 (cid:2) xn−1 (cid:2) xn (cid:2) xn−1 (cid:3) − m1 (cid:6) (cid:2) (cid:6)m1 (cid:3)(cid:3)
+
− m1 (cid:6) (cid:6)PK1
+ PK1 (cid:2) xn−1 (cid:2) xn−1, yn−1 (cid:3)(cid:3)(cid:6) (cid:6) + (cid:2) xn (cid:3)
− ρF (cid:3) − ρF (cid:3)(cid:6) (cid:6). (cid:2) xn, yn (cid:2) xn−1, yn−1 − m1(xn−1) (cid:2) + PK1 g1 (cid:6) (cid:2) (cid:6)xn − xn−1 − g1 (cid:3) (cid:2) (cid:2) xn g1 (cid:2) (cid:2) xn−1 g1 − PK1
Since g1 is ζ1-Lipschitz continuous and η1-strongly monotone,
(cid:3) (cid:3)(cid:3)(cid:6) (cid:6) (cid:6)2 (cid:6)2 ≤
(3.4)
.
(cid:2) g1 − g1 − 2η1 (cid:6) (cid:6)xn − xn−1 − (cid:2) xn (cid:2) xn−1 (cid:3)(cid:6) (cid:6)xn − xn−1 (cid:2) 1 + ζ 2 1
From the γ1-Lipschitzian continuity of m1, we have
(cid:3)
(3.5)
(cid:6) (cid:6). (cid:6) (cid:6)m1 (cid:3)(cid:6) (cid:6) ≤ γ1 − m1 (cid:2) xn (cid:2) xn−1 (cid:6) (cid:6)xn − xn−1
Lemma 2.4 implies that PK1 is nonexpansive and it follows from the strong monotonicity of g1 that
(cid:3)(cid:3) (cid:3) (cid:3) (cid:3) (cid:3) (cid:3)(cid:3)(cid:6) (cid:6) (cid:2) g1 (cid:2) xn (cid:6) (cid:6)PK1 − m1 (cid:3) − PK1 (cid:3)(cid:3) − ρF (cid:3) − m1 (cid:3) (cid:3)(cid:3)(cid:6) (cid:6) ≤ − ρF (cid:2) xn (cid:6) (cid:2) (cid:6) g1 − m1 (cid:2) xn (cid:2) xn−1 (cid:2) xn−1 − m1 (cid:2) − (cid:3)(cid:3)(cid:6) (cid:3) ≤ (cid:6) + − g1 (cid:2) g1 (cid:2) g1 (cid:6) (cid:6)m1 (cid:2) xn−1, yn−1 (cid:2) xn−1, yn−1 (cid:3)(cid:6) (cid:2) (cid:6) xn−1
xn−1 (cid:3)
(cid:3)
+
F
− F (cid:2) xn−1 (cid:2) xn−1 (cid:2) xn (cid:3)(cid:3)(cid:6) (cid:6) + ρ − F − ρF (cid:3) (cid:2) − ρF xn (cid:6) (cid:2) (cid:6)xn − xn−1 − g1 (cid:6) (cid:6)xn − xn−1 − ρ (cid:2) xn, yn (cid:2) xn, yn (cid:3) (cid:2) xn (cid:2) (cid:2) xn, yn (cid:2) xn−1, yn − m1 (cid:6) (cid:2) (cid:6)F xn−1, yn (cid:2) xn−1, yn−1 (cid:3)(cid:6) (cid:6). (3.6)
6 Nonlinear relaxed cocoercive variational inequalities
Since F is relaxed (a,b)-cocoercive and l1-Lipschitz continuous with respect to the first argument,
(cid:3) (cid:2) (cid:3)(cid:3)(cid:6) (cid:6)2
F
(cid:2) xn−1, yn (cid:3) (cid:3)(cid:6) (cid:6)2 = (cid:2) xn−1, yn (cid:2) xn, yn (cid:6) (cid:6)2 + ρ2 (cid:2) (cid:3)(cid:3)(cid:3)
(3.7)
− F (cid:3) (cid:3)(cid:6) (cid:6)2 ≤
F (cid:6) (cid:6)F
(cid:6) (cid:6)xn − xn−1 − ρ (cid:6) (cid:6)xn − xn−1 (cid:2) xn − xn−1,ρ (cid:6) (cid:6)2 + ρ2 (cid:3) (cid:6) (cid:6)2 − F − 2 (cid:6) (cid:6)xn − xn−1 (cid:6) (cid:2) (cid:6)F xn, yn − F (cid:2) xn−1, yn (cid:2) xn−1, yn (cid:6) (cid:6)xn − xn−1
=
+ 2ρa (cid:2) 1 + l2
1ρ2 + 2ρal2 1
− F (cid:3)(cid:6) (cid:6)2 − 2ρb (cid:6) (cid:6)2. − 2ρb − F (cid:6) (cid:2) (cid:6)F xn, yn (cid:3) (cid:2) xn, yn (cid:2) xn, yn (cid:2) xn−1, yn (cid:3)(cid:6) (cid:6)xn − xn−1
Since F is l2-Lipschitz continuous with respect to the second argument,
(cid:3)
(3.8)
(cid:6) (cid:6)F (cid:6) (cid:6). − F (cid:3)(cid:6) (cid:6) ≤ l2 (cid:2) xn−1, yn (cid:2) xn−1, yn−1 (cid:6) (cid:6)yn − yn−1
(cid:6) (cid:6)
(cid:10) (cid:10)
It follows from (3.3)–(3.8) that (cid:6) (cid:6)xn+1 − xn (cid:13)
≤
2
1 + ρ2l2
− 2ρb − 2η1 + 2γ1 + (cid:6) (cid:6) + ρl2 (cid:14)(cid:6) (cid:6)xn − xn−1 (cid:6) (cid:6)yn − yn−1
1 + ζ 2 1
1 + 2ρal2 1
(cid:6) (cid:6). (3.9)
(cid:6) (cid:6)
(cid:10) ≤
Similarly, we have (cid:6) (cid:6)yn+1 − yn (cid:10) (cid:13) 2
1 + ρ2n2
− 2ρd − 2η2 + 2γ2 + (cid:6) (cid:6) + ρn1 (cid:14)(cid:6) (cid:6)yn − yn−1 (cid:6) (cid:6)xn − xn−1
1 + ζ 2 2
2 + 2ρcn2 2
(cid:6) (cid:6). (3.10)
(cid:6) (cid:6)
Now (3.9) and (3.10) imply (cid:6) (cid:6)yn+1 − yn
1 + 2ρal2 1
(cid:6) (cid:6) + (cid:10) (cid:10) (cid:6) (cid:6)xn+1 − xn (cid:13) (cid:6) (cid:6) ≤ − 2η1 + 2γ1 + − 2ρb + ρn1
1 + ζ 2 1 (cid:10)
1 + ρ2l2 (cid:10)
(3.11)
2 (cid:13)
(cid:14)(cid:6) (cid:6)xn − xn−1 (cid:14)
+
2
,
− 2ρd + ρl2 (cid:6) (cid:3) (cid:6) ,
1 + ζ 2 2 (cid:6) (cid:6)yn − yn−1
2 + 2ρcn2 1 + ρ2n2 2 (cid:6) (cid:6) (cid:6)yn − yn−1 (cid:6) +
− 2η2 + 2γ2 + (cid:6) (cid:2)(cid:6) (cid:6) ≤ ω (cid:6)xn − xn−1
where
(cid:15) (cid:10) (cid:10)
2
1 + ρ2l2
ω = max
− 2η1 + 2γ1 +
1 + ζ 2 1
1 + 2ρal2 1
(3.12)
− 2ρb + ρn1, (cid:16) (cid:10) (cid:10)
2
1 + ρ2n2
.
− 2η2 + 2γ2 + − 2ρd + ρl2
1 + ζ 2 2
2 + 2ρcn2 2
Ke Ding et al.
7
It follows from (3.2) that ω < 1. Thus (3.11) implies that {xn} and {yn} are both Cauchy sequences in H, and {xn} converges to x∗ ∈ H, {yn} converges to y∗ ∈ H. Since m1, m2, g1, g2, PK1, PK2, F, G are all continuous, we have
(cid:3) (cid:3) (cid:3) (cid:3)
x∗ = x∗ − g1
+ m1
(cid:3)(cid:3) , (cid:3)(cid:3) (cid:3) (cid:2) x∗ (cid:2) (cid:2) x∗ (cid:2) (cid:2) g1(x∗ (cid:2) (cid:2) − ρF (cid:3)
+ PK (cid:3)
− m1 (cid:3) (cid:2) x∗ (cid:2)
(3.13)
,
y∗
y∗
y∗
− ρG (cid:2) x∗, y∗ (cid:2) x∗, y∗
y ∗
y∗ = y∗ − g2
g2
− m2
+ m2
+ PK
(cid:2)
The result follows then from Lemma 2.7. This completes the proof.
Remark 3.3. Let ρ > 0 be a number satisfying the conditions.
(cid:2) 1 − e1 (cid:3) n1 (cid:17) (cid:17) (cid:17) (cid:17) (cid:17) (cid:17) (cid:17)
(cid:13)(cid:2) (cid:2) (cid:3) (cid:3)2 − (cid:17)ρ − b − al2 − 1 − n2 l2 1 1 (cid:3)2 − 1 + (cid:14) / (cid:2) 1 − e1 (cid:2) 1 − e1 (cid:3) n1
l2 1
− n2 1
,
<
ρn1 < 1 − e1, n1 < l1,
b − al2 1 − n2 l2 1 1
(cid:3) (cid:2) 1 − e2
l2
(cid:17) (cid:17) (cid:17) (cid:17) (cid:17) (cid:17) (cid:17)
(cid:14) (cid:13)(cid:2) (cid:3) (cid:3)2 − (cid:17)ρ − d − cn2 − 2 − l2 n2 2 2 (cid:3)2 − 1 +
/
(cid:2) 1 − e2 (cid:2) 1 − e2 (cid:3) l2 (cid:2) n2 2 − l2 2
,
<
ρl2 < 1 − e2, l2 < n2,
d − cn2 2 − l2 n2 2 2
(3.14)
(cid:10) (cid:10)
where e1 = 2
− 2η1 + 2γ1 and e2 = 2 − 2η2 + 2γ2. Then (3.2) holds.
1 + ζ 2 1
1 + ζ 2 2
4. Perturbed algorithm and stability
In this section, we construct a new perturbed iterative algorithm for solving problem (2.1) and prove the convergence and stability of the iterative sequence generated by the algorithm.
Definition 4.1. Let T be a self-map of H, x0 ∈ H and let xn+1 = f (T,xn) define an iteration procedure which yields a sequence of points {xn}∞ n=0 in H. Suppose that {x ∈ H : Tx = x} (cid:11)= ∅ and {xn}∞ n=0 converge to a fixed point x∗ of T. Let {un} ⊂ H and let (cid:2)n = (cid:2)un+1 − f (T,un)(cid:2). If lim (cid:2)n = 0 implies that limun = x∗, then the iteration procedure defined by xn+1 = f (T,xn) is said to be T-stable or stable with respect to T. Some results for the stability of various iterative processes, we refer to [1, 10] and the references therein.
Let {K 1 n
} and {K 2 } be two sequences of closed convex subsets of H such that n n ,K) → 0, when n → ∞. Now we consider the following perturbed n ,K) → 0, H(K 2
H(K 1 algorithm for solving problem (2.1).
8 Nonlinear relaxed cocoercive variational inequalities
(cid:3) (cid:3) (cid:3) (cid:3) (cid:3)(cid:3)(cid:3) − ρF − m1 (cid:2) xn, yn (cid:2) xn
+ tnen,
Algorithm 4.2. For any given x0, y0 ∈ H, we compute (cid:2) xn (cid:3)
(cid:2) xn − g1(xn (cid:3) (cid:3) xn + tn (cid:2) (cid:2) 1−tn (cid:3) (cid:2) xn (cid:3)
+ m1 (cid:2)
(cid:2) g1 (cid:2) (cid:2) (cid:3) (cid:2) (cid:3)(cid:3)(cid:3) − ρG
+ m2
+ PK 1 n (cid:2) g2
− m2
xn+1 = (cid:2) 1−tn
yn+1 =
yn + tn
yn − g2
yn
yn
+ PK 2
yn
(cid:2) xn, yn
yn
n
+ tn jn, (4.1)
for all n = 0,1,2,..., where {en} and { jn} are two sequences of the elements of H, and the sequence {tn} satisfies the following conditions
∞(cid:18)
(4.2)
0 ≤ tn ≤ 1, ∀n ≥ 0,
tn = ∞.
n=0
Let {un} and {vn} be any sequences in H and define (cid:2)n = (cid:2)1
n + (cid:2)2
n by
(cid:4)(cid:2) (cid:11) (cid:3) (cid:3) = (cid:6) (cid:6)un+1 −
1 − tn
(cid:3) un + tn (cid:2)1 n
+ m1 (cid:3)
(cid:3) (cid:3)(cid:3)(cid:12) (cid:5)(cid:6) (cid:6)
+ m1
(cid:2) un
+ tnen
+ PK1
(4.3)
(cid:4)(cid:2) (cid:11) (cid:2) un (cid:2) un (cid:3) (cid:2) un (cid:2) un,vn (cid:3) = (cid:6) (cid:6)vn+1 −
1 − tn
(cid:3) vn + tn (cid:2)2 n
+ m2 (cid:3)
(cid:3) (cid:3)(cid:3)(cid:12) − ρG (cid:5)(cid:6) (cid:6).
vn − g2 (cid:2) g2
+ m2
un − g1 (cid:2) g1 (cid:2) vn (cid:2) vn
− ρF (cid:2) vn (cid:2) un,vn (cid:2) vn
+ tn jn
+ PK2
(cid:2) 1 − e1 (cid:3) n1 (cid:17) (cid:17) (cid:17) (cid:17)
(cid:14) (cid:13)(cid:2) (cid:3) (cid:3)2 −
Theorem 4.3. Let gi : X → X be ηi-strongly monotone and ζi-Lipschitz continuous, and mi : X → X be τi-Lipschitz continuous for i = 1,2. Let F : X × X → X be l1, l2-Lipschitz continuous with respect to the first and second arguments, respectively, and relaxed (a,b)- cocoercive with respect to the first argument. Let G : X × X → X be n1, n2-Lipschitz continu- ous with respect to the first and second arguments, respectively, and relaxed (c,d)-cocoercive with respect to the second argument. Suppose H(Kn,K) → 0 (n → ∞) and (cid:17) (cid:17) (cid:17)ρ − b − al2 − (cid:17) 1 − n2 l2 1 1 (cid:3)2 − 1 +
/
(cid:2) 1 − e1 (cid:2) 1 − e1 (cid:3) n1 − n2 1 (cid:2) l2 1
,
<
ρn1 < 1 − e1, n1 < l1,
b − al2 1 − n2 l2 1 1
(4.4)
(cid:2) 1 − e2 (cid:3) l2 (cid:17) (cid:17) (cid:17) (cid:17) (cid:17) (cid:17) (cid:17)
(cid:13)(cid:2) (cid:3) (cid:3)2 − (cid:17)ρ − d − cn2 − 2 − l2 n2 2 2 (cid:3)2 − 1 + (cid:14) / (cid:2) 1 − e2 (cid:2) 1 − e2 (cid:3) l2 (cid:2) n2 2 − l2 2
,
<
ρl2 < 1 − e2, l2 < n2,
d − cn2 2 − l2 n2 2 2
(cid:10) − 2ηi + 2γi for i = 1,2. If limn→∞ (cid:2)en(cid:2) = 0 and limn→∞ (cid:2) jn(cid:2) = 0, then
1 + ζ 2 where ei = 2 i we have the following conclusions.
(I) The iterative sequences generated by Algorithm 4.2 converge to the unique solution of
(2.1).
(II) Moreover, if 0 < t ≤ tn, then limun = x∗, limvn = y∗ if and only if lim((cid:2)1
n + (cid:2)2
n) = 0,
where (cid:2)1
n and (cid:2)2
n are defined by (4.3).
Ke Ding et al.
9
Proof. By Theorem 3.2, problem (2.1) admits a solution (x∗, y∗). It is easy to prove that (x∗, y∗) is the unique solution of (4.1). From Lemma 2.7, we have (cid:3)
(cid:3) (cid:3) (cid:3)
x∗ =
(cid:3)(cid:3)(cid:3) , (cid:3)(cid:3)(cid:3) (cid:3) x∗ + tn (cid:3) (cid:2) x∗ − g1 (cid:2) (cid:2) x∗ (cid:2) (cid:2) x∗ (cid:2)
+ m1 (cid:3)
(cid:2) x∗ (cid:2) − ρF (cid:3) − m1 (cid:3) (cid:2) x∗ (cid:2)
+ PK1 (cid:3)
y∗
y∗
y∗
− ρG (cid:2) x∗, y∗ (cid:2) x∗, y∗
y ∗
y∗ =
y∗ − g2
+ m2
(cid:2) g1 (cid:2) g2 − m2 (cid:2) 1 − tn (cid:2) 1 − tn
y∗ + tn
+ PK2
, (4.5)
(cid:6) (cid:6)
n
(cid:3) (cid:3) (cid:11) (cid:3) (cid:3)(cid:3)(cid:12) (cid:3) = (cid:2) xn
+PK 1 (cid:3)
+m1 (cid:3)
−m1 (cid:3) −ρF (cid:3)
+tnen (cid:3)(cid:3)(cid:12)(cid:6) (cid:6)
− (cid:2) xn, yn (cid:2) x∗, y∗
+PK1 (cid:3)
+m1 (cid:3)
(cid:6) (cid:6)
Since PK is nonexpansive and it follows from (4.1) and (4.5) that (cid:6) (cid:6)xn+1 − x∗ (cid:6) (cid:2) (cid:6) 1−tn (cid:2) 1−tn (cid:3)(cid:6) (cid:6)
≤ (cid:6) (cid:6)en −ρF (cid:6) (cid:6)m1(x)−m1 (cid:3) (cid:2) xn (cid:2) x∗ (cid:3)(cid:6) (cid:6)+tn (cid:3) (cid:2) xn (cid:2) x∗ (cid:2) xn −x∗ (cid:3) (cid:2) xn (cid:2) x∗ (cid:2) xn (cid:3)(cid:3) (cid:2) x∗ (cid:3)(cid:6) (cid:6)+tn (cid:3)(cid:3)(cid:6) (cid:6) (cid:2) x∗, y∗ (cid:3) xn+tn (cid:3) x∗ −tn (cid:2) xn −x∗ (cid:3) (cid:2) (cid:2) g1
xn
−PK1 (cid:3) −m1 (cid:3) (cid:6) (cid:6) (cid:2) 1−tn (cid:6) (cid:6)PK 1 n (cid:3)(cid:6) (cid:6) ≤ (cid:6) (cid:6)en −ρF (cid:6) (cid:6)m1(x)−m1 (cid:3) (cid:2) g1 (cid:2) g1 (cid:2) x∗ (cid:2) x∗ (cid:3)(cid:6) (cid:6)+tn (cid:3)
xn −g1 (cid:11) x∗ −g1 (cid:6) (cid:3)(cid:6) (cid:6) (cid:6)+tn (cid:2) xn, yn (cid:6) (cid:2) (cid:6) xn −x∗ (cid:3)
+g1 (cid:2) xn (cid:2) xn (cid:3)(cid:3)
−m1 (cid:2) x∗ (cid:2) x∗ (cid:3)(cid:6) (cid:6)+tn (cid:3)(cid:3)(cid:6) (cid:6)
xn
+tn
n
n
(cid:2) −m1 (cid:3) −ρF (cid:3) −m1 (cid:3) −PK 1 (cid:3)(cid:3) −ρF (cid:3) (cid:3)(cid:3)(cid:6) (cid:6)
x∗
(cid:2) x∗, y∗ (cid:2) x∗, y∗ −m1 (cid:2) x∗ (cid:2) x∗ (cid:2) x∗ (cid:2) xn −x∗ (cid:3) (cid:2) (cid:2) g1 (cid:2) g1 −g1 (cid:2) g1 (cid:2) g1 −m1 (cid:3) (cid:6) (cid:6)
+tn (cid:2) 1−tn (cid:6) (cid:6)PK 1 (cid:6) (cid:6)PK 1 n (cid:3)(cid:6) (cid:6)
≤ −ρF (cid:6) (cid:6)m1(x)−m1
+g1
−m1 (cid:3)(cid:6) (cid:2) (cid:6)+tn x∗ (cid:6) (cid:6)en −PK1 (cid:2) (cid:3) x∗ (cid:3) (cid:3)(cid:3)(cid:6) (cid:3)(cid:6) (cid:6)
+ tn
−g1 (cid:2) xn (cid:6) + tn −g1 (cid:2) g1 (cid:2) x∗ (cid:2) x∗ (cid:2) x∗ (cid:3)(cid:6) (cid:6)+tn (cid:2) x∗ (cid:3)
+g1 (cid:2) xn (cid:2) x∗ (cid:2) xn (cid:6) (cid:6)m1 (cid:3)(cid:3)(cid:6)
− g1 (cid:3) (cid:3)(cid:6) (cid:6) (cid:2) xn −x∗ (cid:2) g1 (cid:2) F − F (cid:3) (cid:2) x∗, y∗ (cid:3)
x∗
−ρF (cid:3)(cid:6) (cid:6)+tn (cid:2) xn, yn (cid:2) x∗, y∗ (cid:6) (cid:2) (cid:6) xn −x∗ (cid:2) x∗ (cid:2) x∗, yn (cid:2) x∗ −ρF (cid:3)(cid:6) (cid:6)+tn (cid:3) (cid:2) xn (cid:2) xn, yn (cid:2) x∗, y∗ −ρF − F (cid:2) x∗, y∗ (cid:2) x∗ −ρF − m1 (cid:6) (cid:2) (cid:6)F (cid:6) + ρtn x∗, yn (cid:3) (cid:2) (cid:2) (cid:3)(cid:3) x∗ g1 −m1 −m1
+tn (cid:2) 1−tn (cid:6) (cid:6)xn − x∗ − (cid:6) (cid:6)xn − x∗ − ρ + tn (cid:6) (cid:3) (cid:2) (cid:2) (cid:6)PK 1 g1
+tn
n
−PK1 (cid:3)(cid:3)(cid:6) (cid:6). (4.6)
Since F is l2-Lipschitz continuous with respect to the second argument, (cid:3)
(4.7)
(cid:6) (cid:6)F (cid:2) x∗, y∗ (cid:6) (cid:6). − F (cid:3)(cid:6) (cid:6) ≤ l2 (cid:2) x∗, yn (cid:6) (cid:6)yn − y∗
From the strong monotonicity and Lipschitzian continuity of g1, we obtain
(cid:3) (cid:3)(cid:3)(cid:6) (cid:6) (cid:6)2 (cid:6)2 ≤
(4.8)
(cid:2) x∗
.
(cid:2) g1 − g1 − 2η1 (cid:6) (cid:6)xn − x∗ − (cid:2) xn (cid:3)(cid:6) (cid:6)xn − x∗ (cid:2) 1 + ζ 2 1
(cid:3)
(4.9)
The Lipschitzian continuity of m1 implies (cid:2) x∗
(cid:6) (cid:6). (cid:6) (cid:6)m1 − m1 (cid:3)(cid:6) (cid:6) ≤ γ1 (cid:2) xn (cid:6) (cid:6)xn − x∗
10 Nonlinear relaxed cocoercive variational inequalities
Since F is relaxed (a,b)-cocoercive and l1-Lipschitz continuous with respect to the first argument,
(cid:10) (cid:3) (cid:3)(cid:3)(cid:6) (cid:6) ≤
(4.10)
(cid:2) F − F
1 + ρ2l2
1 + 2ρal2 1
− 2ρb (cid:6) (cid:6). (cid:6) (cid:6)xn − x∗ − ρ (cid:2) xn, yn (cid:2) x∗, yn (cid:6) (cid:6)xn − x∗
It follows from (4.6)–(4.10) that
1 + 2ρal2 1
(cid:10) (cid:10) (cid:6) (cid:6) (cid:6) (cid:6) ≤ (cid:6) (cid:6)xn+1 − x∗ (cid:13) 2tn − 2ρb + 1 − tn (cid:14)(cid:6) (cid:6)xn − x∗
1 + ρ2l2 (cid:6) (cid:6),
1 + ζ 2 1 (cid:6) (cid:6)yn − y∗
− 2η1 + 2tnγ1 + tn (cid:6) (cid:6)en (cid:6) (cid:6) + tnbn + tn
+ tnρl2
(4.11)
where
(cid:3) (cid:3) (cid:3)(cid:3) (cid:3) (cid:3) (cid:2) x∗ (cid:2) x∗, y∗ − ρF (cid:2) x∗ (cid:2) x∗ (cid:2) x∗, y∗ − ρF (cid:2) x∗ (cid:2) g1 − m1 (cid:2) g1 − m1
bn =
n
n ,K1) → 0 and Lemma 2.5, we know that bn → 0.
(cid:6) (cid:6)PK 1 − PK1 (cid:3)(cid:3)(cid:6) (cid:6). (4.12)
From the fact of H(K 1 Similarly, we have
2 + 2ρcn2 2
(cid:10) (cid:10) (cid:6) (cid:6) (cid:6) (cid:6) ≤ (cid:6) (cid:6)yn+1 − y∗ (cid:13) 2tn − 2ρd + 1 − tn (cid:14)(cid:6) (cid:6)yn − y∗
1 + ρ2n2 (cid:6) (cid:6),
1 + ζ 2 2 (cid:6) (cid:6)xn − x∗
− 2η2 + 2tnγ2 + tn (cid:6) (cid:6) jn (cid:6) (cid:6) + tncn + tn
+ tnρn1
(4.13)
where
(cid:2) (cid:3) (cid:3) (cid:2) (cid:3)(cid:3) (cid:2) (cid:2) (cid:3) (cid:3) (cid:2)
y∗
(cid:2) x∗, y∗ − ρG
y∗
y∗
− ρF (cid:2) x∗, y∗
y∗
(cid:2) g2 − m2
g2
− m2 (cid:6) (cid:6)PK 2
cn =
n
− PK2 (cid:3)(cid:3)(cid:6) (cid:6), (4.14)
and cn → 0. Now (4.11) and (4.13) imply
(cid:6) (cid:6) (cid:6) (cid:6)yn+1 − y∗ (cid:6) (cid:6) + (cid:10) (cid:10) (cid:6) (cid:6)xn+1 − x∗ (cid:13) (cid:6) (cid:6) ≤
2tn
1 + 2ρal2 1
− 2η1 + 2tnγ1 + 1 − tn + tn − 2ρb + tnρn1
1 + ζ 2 1 (cid:10)
1 + ρ2l2 (cid:10)
(cid:6) (cid:6)
+
1+ ρ2n2
(cid:13) 2tn − 2ρd + tnρl2 (cid:14)(cid:6) (cid:6)xn − x∗ (cid:14)(cid:6) (cid:6)yn − y∗
1+ ζ 2 2
2 + 2ρcn2 2
(cid:6) (cid:6). − 2η2+ 2tnγ2+ 1 − tn+ tn (cid:6) (cid:6) jn (cid:6) (cid:6) + tn (cid:6) (cid:6)en
+ tncn + tnbn + tn
(4.15)
Ke Ding et al.
11
Let
(cid:10) (cid:10)
1 + ρ2l2
− 2η1 + 2γ1 + − 2ρb + ρn1,
1 + ζ 2 1
1 + 2ρal2 1
(4.16)
h1 = 2 (cid:10)
(cid:10)
1 + ρ2n2
h2 = 2
− 2η2 + 2γ2 + − 2ρd + ρl3.
1 + ζ 2 2
2 + 2ρcn2 2
From (4.4), it is easy to see that 0 ≤ h1 < 1 and 0 ≤ h2 < 1. Let h = max{h1,h2}. Then h < 1 and so (4.15) reduces to
(cid:6) (cid:6) (cid:6) (cid:6) + (cid:3) (cid:3) (cid:6) (cid:6)yn+1 − y∗ (cid:3)(cid:2)(cid:6) (cid:6) (cid:6) (cid:6) (cid:6) ≤ (cid:6) (cid:6) + (cid:6)xn − x∗ (cid:2) bn + cn + (cid:6) (cid:6)en (cid:6) (cid:6) jn
+ tn
(4.17)
(cid:3) (cid:3)(cid:2)(cid:6) (cid:6) (cid:6) = (cid:6) (cid:6) + (cid:6) (cid:6) + (cid:6) (cid:6)xn+1 − x∗ (cid:2) 1 − (1 − h)tn (cid:2) 1 − (1 − h)tn (cid:6)xn − x∗ (cid:6) (cid:6)yn − y∗ (cid:6) (cid:6)yn − y∗
+ (1 − h)tnδn,
where
(cid:6) (cid:6) (cid:6) (cid:6) + (cid:6) (cid:6) jn
(4.18)
.
δn = bn + cn +
(cid:6) (cid:6)en 1 − h
From (4.12), (4.14) and Lemma 2.5, we have
(cid:6) (cid:6) (cid:6) (cid:6) + (cid:6) (cid:6) jn −→ 0
(4.19)
(n −→ ∞).
bn −→ 0,
cn −→ 0,
δn = bn + cn +
(cid:6) (cid:6)en 1 − h
It follows from (4.2), (4.17), (4.19) and Lemma 2.8 that
(4.20)
(n −→ ∞).
xn −→ x∗,
yn −→ y∗
This completes the proof of Conclusion I.
Next we prove Conclusion II. By using (4.1), we obtain
(cid:6) (cid:6)
(cid:6) (cid:6)un+1 − x∗ (cid:4)(cid:2) (cid:11) (cid:3) (cid:3) (cid:3) (cid:3) (cid:3)(cid:3)(cid:12) (cid:5)(cid:6) (cid:6) ≤ (cid:2) un (cid:2) un (cid:2) un (cid:2) un (cid:2) g1 (cid:3) (cid:2) un,vn (cid:3) (cid:3)(cid:3)(cid:12)
+tnen (cid:5)
(cid:6) (cid:6)un+1 − (cid:6) (cid:4)(cid:2) (cid:6) (cid:6) (cid:6)
+tnen
un −g1 (cid:3) (cid:2) un (cid:3)
+m1 (cid:3) (cid:2) un (cid:3)
+PK1 (cid:2) (cid:2) un g1 (cid:3)
−ρF (cid:2) un,vn (cid:3)
+m1 (cid:2) un (cid:3)(cid:3)(cid:12)
≤ −ρF
+m1
+PK1 (cid:2) g1
+m1
+ (cid:6) (cid:2) (cid:6) 1−tn
1−tn (cid:3) un+tn
(cid:3) un+tn (cid:11) un −g1 (cid:2) un
1−tn (cid:3) un+tn (cid:11) un −g1
+m1 (cid:2) un
(cid:2) un
+m1 (cid:2) un
−ρF (cid:2) un,vn
+PK1
−x∗ (cid:6) (cid:6)+(cid:2)1 +tnen −x∗ n. (4.21)
12 Nonlinear relaxed cocoercive variational inequalities
(cid:4)(cid:2) (cid:3) (cid:3) (cid:11) (cid:3) (cid:3) (cid:5) (cid:3)(cid:3)(cid:12)
As the proof of inequality (4.11), we have (cid:6) (cid:6)
(cid:6) (cid:6) −ρF −x∗
+m1
(cid:2) g1
+m1
(cid:2) un (cid:2) un
un −g1
(cid:2) un (cid:2) un,vn (cid:3) un+tn (cid:10)
+PK1 (cid:10)
1−tn (cid:13)
+tnen (cid:6) (cid:6)
≤
2tn
1 + 2ρal2 1
− 2ρb + 1 − tn (cid:2) un (cid:14)(cid:6) (cid:6)un − x∗
1 + ρ2l2 (cid:6) (cid:6),
1 + ζ 2 1 (cid:6) (cid:6)vn − y∗
− 2η1 + 2tnγ1 + tn (cid:6) (cid:6) (cid:6)en (cid:6) + tnbn + tn
+ tnρl2
(4.22)
where bn is defined by (4.12). From (4.21) and (4.22), we have
1 + 2ρal2 1
(cid:10) (cid:10) (cid:6) (cid:6) (cid:6) (cid:6) ≤ (cid:6) (cid:6)un+1 − x∗ (cid:13) 2tn − 2ρb + 1 − tn (cid:14)(cid:6) (cid:6)un − x∗
1 + ζ 2 1 (cid:6) (cid:6)vn − y∗
− 2η1 + 2tnγ1 + tn (cid:6) (cid:6)en (cid:6) (cid:6) + tnbn + tn
+ tnρl2
1 + ρ2l2 (cid:6) (cid:6) + (cid:2)1 n.
(4.23)
Similarly, we have
2 + 2ρcn2 2
(cid:10) (cid:10) (cid:6) (cid:6) (cid:6) (cid:6) ≤ (cid:6) (cid:6)vn+1 − y∗ (cid:13) 2tn − 2ρd + 1 − tn (cid:14)(cid:6) (cid:6)vn − y∗
1 + ζ 2 2 (cid:6) (cid:6)un − x∗
− 2η2 + 2tnγ2 + tn (cid:6) (cid:6) jn (cid:6) (cid:6) + tncn + tn
+ tnρn1
1 + ρ2n2 (cid:6) (cid:6) + (cid:2)2 n,
(4.24)
n + (cid:2)2 n
(cid:6) (cid:6) (cid:6) (cid:6) + (cid:3) (cid:6) (cid:6) ≤ (cid:6)un − x∗ (cid:3) (cid:3)(cid:2)(cid:6) (cid:12) (cid:6) (cid:6) ≤ (cid:6) (cid:6) + (cid:6) (cid:6) + (cid:3) /t
+ tn(bn + cn + (cid:2)en(cid:2) + (cid:2) jn(cid:2)) + (cid:2)1 (cid:6) (cid:6)en
(cid:11) bn + cn + (cid:6)un − x∗ (cid:6) (cid:6) jn
+ tn
(cid:2) n + (cid:2)2 (cid:2)1 n (cid:3) (cid:3)(cid:2)(cid:6) (cid:6) (cid:6) = (cid:6) (cid:6) + (cid:6) (cid:6) + (cid:6) (cid:6) +
where cn is defined by (4.14). As the proof of inequality (4.17), and since 0 < t ≤ tn, (4.23) and (4.24) yield (cid:6) (cid:6) (cid:6)vn+1 − y∗ (cid:6)un+1 − x∗ (cid:3)(cid:2)(cid:6) (cid:2) 1 − (1 − h)tn (cid:2) 1 − (1 − h)tn (cid:2) 1 − (1 − h)tn
(cid:6) (cid:6)vn − y∗ (cid:6) (cid:6)vn − y∗ (cid:6) (cid:6)vn − y∗
+ (1 − h)tnδn,
(cid:6)un − x∗
(4.25)
where
(cid:6) (cid:6) + (cid:6) (cid:6) + (cid:3) /t (cid:6) (cid:6)en (cid:2) n + (cid:2)2 (cid:2)1 n
(4.26)
.
δn = bn + cn +
(cid:6) (cid:6) jn 1 − h
(cid:9) = 0, then from bn → 0, cn → 0, (cid:2)en(cid:2) → 0 and (cid:2) jn(cid:2) → 0, we have tn = ∞, (4.25) and Lemma 2.8, we
Suppose that lim (cid:2)1 n + (cid:2)2 n δn → 0 (as n → ∞). Then from the fact of tn → 0, have limun = x∗ and limvn = y∗.
Ke Ding et al.
13
Conversely, suppose that lim un = x∗ and limvn = y∗. Then we have
(cid:4)(cid:2) (cid:3) (cid:3) = (cid:6) (cid:6)un+1 −
1 − tn
n + (cid:2)2 (cid:2)1 n
(cid:3) un + tn (cid:2) un (cid:3) (cid:3)(cid:3)(cid:12)
+ m1 (cid:3)
(cid:5)(cid:6) (cid:6) − ρF
+ m1
(cid:2) un
+ tnen
(cid:4)(cid:2)
+ PK1 (cid:11)
(cid:2) un (cid:2) un,vn (cid:3) (cid:3)
+
(cid:11) un − g1 (cid:2) g1(un (cid:2) vn (cid:6) (cid:6)vn+1 −
1 − tn
(cid:3) vn + tn (cid:3)
+ m2 (cid:3)
(cid:3)(cid:3)(cid:12) (cid:5)(cid:6) (cid:6) (cid:2) vn
vn − g2 (cid:2) g2(vn
+ tn jn
− ρG (cid:3)
+ m2 (cid:3)
(cid:6) (cid:6) ≤ (cid:6) (cid:6) + (cid:6) (cid:6)un+1 − x∗ (cid:2) 1 − tn (cid:3) un + tn
+ m1 (cid:3)
(cid:3) (cid:3)(cid:3)(cid:12) − ρF
+ m1
(cid:2) vn (cid:2) un,vn (cid:2) un (cid:2) un,vn (cid:2) un
(cid:6) (cid:6) + (cid:3) (cid:3) (cid:6) (cid:6)
+
(cid:2) 1 − tn (cid:3) vn + tn
+ m2 (cid:3)
(cid:3)(cid:3)(cid:12) (cid:3)
vn − g2 (cid:2)
(cid:6) (cid:6)
+ tn jn − y∗ (cid:6) (cid:6)
≤ (cid:6) (cid:6) +
+ PK2 (cid:11) (cid:2) un − g1 un (cid:2) (cid:2) + PK1 un g1 (cid:6) (cid:6) (cid:6) (cid:6)vn+1 − y∗ (cid:2) vn (cid:2) (cid:2) − ρG + m2 un,vn vn (cid:6) (cid:6) (cid:6) (cid:6)un − x∗ (cid:6) + tnρn1 (cid:6)vn − y∗
(cid:6) (cid:6)un+1 − x∗ (cid:19) (cid:20)
+ tnen − x∗ (cid:2) (cid:11) vn (cid:2) + PK2 g2 vn (cid:6) (cid:6) (cid:6) + tnρl2 (cid:6)vn+1 − y∗ (cid:10)
(cid:10) (cid:6) (cid:6)
+
1 + ρ2l2
2tn
− 2η1 + 2tnγ1 + tn − 2ρb + 1 − tn
1 + ζ 2 1
1 + 2ρal2 1
(cid:19) (cid:6) (cid:6)un − x∗ (cid:20) (cid:10) (cid:10) (cid:6) (cid:6)
+
1 + ρ2n2
2tn
2 + 2ρcn2 2
− 2ρd + 1 − tn (cid:6) (cid:6)vn − y∗
1 + ζ 2 − 2η2 + 2tnγ2 + tn 2 (cid:6) (cid:6) (cid:6) (cid:6) (cid:6) + tncn + tn (cid:6)en
(cid:6) (cid:6) jn
+ tnbn + tn
(4.27)
(cid:2)
and so (cid:2)1
n + (cid:2)2 n
→ 0 as n → ∞. This completes the proof.
Acknowledgments
The authors thank the referees for their valuable suggestions. This work was supported by the National Natural Science Foundation of China and the Educational Science Foun- dation of Chongqing (KJ051307).
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Ke Ding: Department of Mathematics, Sichuan University, Chengdu, Sichuan 610064, China E-mail address: keding@yahoo.com
Wen-Yong Yan: Department of Mathematics, Sichuan University, Chengdu, Sichuan 610064, China E-mail address: wenyongy@yahoo.com
Nan-Jing Huang: Department of Mathematics, Sichuan University, Chengdu, Sichuan 610064, China E-mail addresses: nanjinghuang@126.com; nanjinghuang@hotmail.com
