Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2012, Article ID 572326, 18 pages
doi:10.1155/2012/572326
Research Article
Hybrid Iterative Scheme by a Relaxed
Extragradient Method for Equilibrium Problems,
a General System of Variational Inequalities and
Fixed-Point Problems of a Countable Family of
Nonexpansive Mappings
Qiao-Li Dong,1 Yan-Ni Guo,1 and Fang Su2
1
2
College of Science, Civil Aviation University of China, Tianjin 300300, China
College of Science, National University of Defense Technology, Changsha 410073, China
Correspondence should be addressed to Qiao-Li Dong, dongql@lsec.cc.ac.cn
Received 29 November 2011; Accepted 5 December 2011
Academic Editor: Yonghong Yao
Copyright q 2012 Qiao-Li Dong 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.
Based on the relaxed extragradient method and viscosity method, we introduce a new iterative
method for finding a common element of solution of equilibrium problems, the solution set of
a general system of variational inequalities, and the set of fixed points of a countable family of
nonexpansive mappings in a real Hilbert space. Furthermore, we prove the strong convergence
theorem of the studied iterative method. The results of this paper extend and improve the results
of Ceng et al., 2008 , W. Kumam and P. Kumam, 2009 , Yao et al., 2010 and many others.
1. Introduction
Let H be a real Hilbert space with the inner product ·, · and the norm · . Let C be a closed
convex subset of H. Let F be a bifunction of C × C into R, where R is the set of real numbers.
The equilibrium problem for F : C × C → R is to find x ∈ C such that
F x, y ≥ 0,
∀y ∈ C.
1.1
The set of solutions of 1.1 is denoted by EPF. The equilibrium problems covers, as
special cases, monotone inclusion problems, saddle point problems, minimization problems,
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optimization problems, variational inequality problems, Nash equilibria in noncooperative
games, and various forms of feasibility problems see 1–4 and the references therein.
A mapping A : C → H is called α-inverse-strongly monotone if there exists a positive
real number α such that
Au − Av, u − v ≥ αAu − Av2 ,
∀u, v ∈ C.
1.2
It is obvious that any α-inverse-strongly monotone mapping A is monotone and Lipschitz
continuous. A mapping T : C → C is said to be nonexpansive if
T x − T y ≤ x − y,
∀x, y ∈ C.
1.3
We denote by FT the set of fixed points of T . Recently, Wang and Guo 5 introduced an
iterative scheme for a countable family of nonexpansive mappings.
Let C be a nonempty closed convex subset of a real Hilbert space H. For a given
nonlinear operator A : C → H, consider the following variational inequality problem of
finding x∗ ∈ C such that
Ax∗ , x − x∗ ≥ 0,
x ∈ C.
1.4
The set of solutions of the variational inequality 1.4 is denoted by VIC, A see 6–9 and
the references therein.
Let A, B : C → H be two mappings. Consider the following problem of finding
x∗ , y∗ ∈ C × C such that
λAy∗ x∗ − y∗ , x − x∗ ≥ 0,
μBx∗ y∗ − x∗ , x − y∗ ≥ 0,
∀x ∈ C,
∀x ∈ C,
1.5
which is called a general system of variational inequalities, where λ > 0 and μ > 0 are two
constants. The set of solutions of 1.5 is denoted by GSVIA, B, and C. In particular, if
A B, then problem 1.5 reduces to finding x∗ , y∗ ∈ C × C such that
λAy∗ x∗ − y∗ , x − x∗ ≥ 0,
μAx∗ y∗ − x∗ , x − y∗ ≥ 0,
∀x ∈ C,
∀x ∈ C,
1.6
which is defined by Verma 7 see also 10 and is called the new system of variational
inequalities. Further, if we add up the requirement that x∗ y∗ , then problem 1.6 reduces
to the classical variational inequality problem 1.4. Recently, Yao et al. 11 presented system
of variational inequalities in Banach space. For solving problem 1.5, recently, Ceng et al. 12
introduced and studied a relaxed extragradient method. Based on the relaxed extragradient
method and the viscosity approximation method, W. Kumam and P. Kumam 13 constructed
a new viscosity-relaxed extragradient approximation method. Very recently, based on the
extragradient method, Yao et al. 14 proposed an iterative method for finding a common
element of the set of a general system of variational inequalities and the set of fixed points of
a strictly pseudocontractive mapping in a real Hilbert space.
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3
Motivated and inspired by the above works, in this paper, we introduce an iterative
method based on the extragradient method and viscosity method for finding a common
element of solution of equilibrium problems, the solution set of a general system of
variational inequalities, and the set of fixed points of a countable family of nonexpansive
mappings in a real Hilbert space. Furthermore, we prove the strong convergence theorem of
the proposed iterative method.
2. Preliminaries
Let C be a closed convex subset of H, and let T : C → C be nonexpansive such that FT /
∅.
For all x ∈ FT and all x ∈ C, we have
2 T x − x
2 T x − x x − x
2
2 ≥ T x − T x
x − x
T x − x2 x − x
2 2T x − x, x − x,
2.1
and hence
T x − x2 ≤ 2x − T x, x − x,
∀x ∈ FT , ∀x ∈ C.
2.2
Remark 2.1. Let A be α-inverse-strongly monotone. For all x, y ∈ C, λ > 0, we have
I − λAx − I − λAy2 x − y2 − 2λ Ax − Ay, x − y λ2 Ax − Ay2
2
2
≤ x − y λλ − 2αAx − Ay .
2.3
So, if λ ≤ 2α, then I − λA is a nonexpansive mapping from C to H.
Recall that the nearest point projection PC from H onto C assigns to each x ∈ H the
unique point PC x ∈ C satisfying the property
x − PC x minx − y.
y∈C
2.4
The following characterizes the projection PC .
Lemma 2.2. Given that x ∈ H and y ∈ C, then PC x y if and only if there holds the inequality
x − y, y − z ≥ 0
∀z ∈ C.
2.5
Lemma 2.3 see 15. Let H be a Hilbert space, C a closed convex subset of H, and T : C → C a
nonexpansive mapping with FT /
∅. If {xn } is a sequence in C weakly converging to x ∈ C and if
{1 − T xn } converges strongly to y, then1 − T x y.
Lemma 2.4 see 16. Assume that {an } is a sequence of nonnegative real numbers such that
αn1 ≤ 1 − γn αn δn ,
∀n ≥ 0,
2.6
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where two sequences {γn } ⊂ 0, 1 and {δn } satisfy
1 ∞
n1 γn ∞;
2 lim supn → ∞ δn /γn ≤ 0 or ∞
n1 |δn | < ∞.
Then limn → ∞ αn 0.
Lemma 2.5 see 12. For given x∗ , y∗ ∈ C, x∗ , y∗ is a solution of problem 1.5 if and only if x∗
is a fixed point of the mapping G : C → C defined by
Gx PC PC x − μBx − λAPC x − μBx ,
∀x ∈ C,
2.7
where y∗ PC x∗ − μBx∗ .
Note that the mapping G is nonexpansive provided that λ ∈ 0, 2α and μ ∈ 0, 2β.
Throughout this paper, the set of fixed points of the mapping G is denoted by Γ.
Lemma 2.6 see 1. Let C be a nonempty closed convex subset of H and F : C × C → R satisfy
following conditions:
A1 Fx, x 0, ∀x ∈ C;
A2 F is monotone, that is, Fx, y Fy, x ≤ 0, ∀x, y ∈ C;
A3 lim supt → 0 Ftz 1 − t x, y ≤ Fx, y, ∀x,y, z ∈ C;
A4 for each x ∈ C, Fx, · is convex and lower semicontinuous.
For x ∈ C and r > 0, set TrF : H → C to be
TrF x
1
z ∈ C : F z, y y − z, z − x ≥ 0, ∀y ∈ C .
r
2.8
Then TrF is well defined and the following holds:
1 TrF is single valued;
2 TrF is firmly nonexpansive [17], that is, for any x, y ∈ E,
2 F
Tr x − TrF y ≤ TrF x − TrF y, x − y ;
2.9
3 FTrF EP F;
4 EP F is closed and convex.
By the proof of Lemma 5 in 2 also see 3, we have following lemma.
Lemma 2.7. Let C be a nonempty closed convex subset of a Hilbert space H and F : C × C → R be
a bifunction. Let x ∈ C and r1 , r2 ∈ 0, ∞. Then
r2 Tr1 x − Tr2 x ≤ 1 − Tr1 x x.
r1
2.10
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5
3. Main Results
Theorem 3.1. Let C be a nonempty closed convex subset of a real Hilbert space H. Let the mappings
A, B : C → H be α-inverse strongly monotone and β-inverse strongly monotone, respectively. Let F
be a bifunction from C × C to R satisfying (A1)–(A4) and {Tn }∞
n1 : C → C be a countable family of
∅. Let f : C → C be a contraction
nonexpansive mappings such that Ω : ∞
n1 FTn ∩ EP F∩ Γ /
with coefficient ρ ∈ 0, 1/2. Set β0 1. For given x1 ∈ C arbitrarily, let the sequences {xn }, {yn },
{zn }, and {un } be generated by
1
F un , y y − un , un − xn ≥ 0,
rn
zn PC un − μBun ,
xn1
∀y ∈ C,
yn αn fxn 1 − αn PC zn − λAzn ,
n
βn xn σn
βi−1 − βi Ti yn 1 − βn 1 − σn PC zn − λAzn ,
3.1
i1
where λ ∈ 0, 2α, μ ∈ 0, 2β, and sequences {αn } ⊂ 0, 1, {βn } ⊂ 0, 1, {σn } ⊂ 0, 1, and
{rn } ⊂ r, ∞, r > 0, are such that
i {βn } is strictly decreasing,
ii 0 < lim infn → ∞ βn ≤ lim supn → ∞ βn < 1,
iii limn → ∞ αn 0 and ∞
n1 αn ∞,
∞
iv σn > 1/21 − ρ, n1 |σn − σn−1 | < ∞,
v ∞
n1 |rn − rn−1 | < ∞.
Then the sequence {xn } generated by 3.1 converges strongly to x∗ PΩ · fx∗ , and x∗ , y∗ is a
solution of the general system of variational inequalities 1.5, where y∗ PC x∗ − μBx∗ .
Proof. The proof is divided into several steps.
Step 1. The sequence {xn } defined by 3.1 is bounded.
For each x∗ ∈ Ω, from Lemma 2.6, we have un Trn xn and hence
un − x∗ Trn xn − Trn x∗ ≤ xn − x∗ .
3.2
Since f is a ρ-contraction mapping, using 2.3 and 3.2, we have
zn − y∗ 2 PC un − μBun − PC x∗ − μBx∗ 2
2
≤ un − μBun − x∗ − μBx∗ ≤ un − x∗ 2 μ μ − 2β Bun − Bx∗ 2
≤ xn − x∗ 2 μ μ − 2β Bun − Bx∗ 2
≤ xn − x∗ 2 .
3.3
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Set vn PC zn − λAzn . Since PC is nonexpansive, from 2.3, we have
2
vn − x∗ 2 PC zn − λAzn − PC y∗ − λAy∗ 2
2
≤ zn − y∗ λλ − 2αAzn − Ay∗ 2
≤ zn − y∗ .
3.4
yn − x∗ αn fxn 1 − αn vn − x∗ ≤ αn fxn − x∗ 1 − αn zn − y∗ ≤ αn fxn − fx∗ fx∗ − x∗ 1 − αn xn − x∗ ≤ 1 − αn 1 − ρ xn − x∗ αn fx∗ − x∗ .
3.5
Hence we get
From 3.1 and 3.3–3.5, we get
n
∗
βi−1 − βi Ti yn 1 − βn 1 − σn vn − x xn1 − x βn xn σn
i1
∗
≤ βn xn − x∗ σn
n
βi−1 − βi Ti yn − x∗ 1 − βn 1 − σn vn − x∗ i1
≤ βn xn − x∗ σn
n
βi−1 − βi yn − x∗ 1 − βn 1 − σn zn − y∗ i1
≤ 1 − σn 1 − βn xn − x∗ σn 1 − βn yn − y∗ ≤ 1 − σn 1 − βn xn − x∗ σn 1 − βn 1 − αn 1 − ρ xn − x∗ αn fx∗ − x∗ ≤ 1 − σn 1 − βn αn 1 − ρ xn − x∗ 1 fx∗ − x∗ σn 1 − βn αn 1 − ρ
1−ρ
3.6
≤ M,
where M max{x1 − x∗ , 1/1 − ρfx∗ − x∗ }. Hence, {xn } is bounded and therefore
{un }, {zn }, {vn }, and {yn } are also bounded.
Step 2. limn → ∞ xn1 − xn 0. Since un Trn xn and un−1 Trn−1 xn−1 , using Lemma 2.7, we
have
un − un−1 Trn xn − Trn−1 xn−1 ≤ Trn xn − Trn xn−1 Trn xn−1 − Trn−1 xn−1 rn ≤ xn − xn−1 1 −
un−1 xn−1 rn−1 1
≤ xn − xn−1 |rn−1 − rn |L,
r
3.7
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where L sup{un xn : n 1, 2, . . .}. From 3.1 and 3.7, it follows that
zn − zn−1 PC I − μB un − PC I − μB un−1 ≤ un − un−1 1
≤ xn − xn−1 |rn−1 − rn |L.
r
3.8
From 3.1 and 3.8, we have
yn − yn−1 ≤ αn fxn − fxn−1 1 − αn PC I − λAzn − PC I − λAzn−1 ≤ αn ρxn − xn−1 1 − αn zn − zn−1 1
≤ αn ρxn − xn−1 1 − αn xn − xn−1 |rn−1 − rn |L
r
1
≤ 1 − αn 1 − ρ xn − xn−1 1 − αn |rn−1 − rn |L.
r
3.9
By definition of scheme 3.1, we have
n
xn1 − xn βn xn − xn−1 βn − βn−1 xn−1 σn
βi−1 − βi Ti yn − Ti yn−1
i1
n−1
σn − σn−1 βi−1 − βi Ti yn−1 σn βn−1 − βn Tn yn−1
i1
1 − βn 1 − σn PC zn − λAzn − PC zn−1 − λAzn−1 − 1 − βn−1 1 − σn−1 PC zn−1 − λAzn−1 1 − βn 1 − σn PC zn−1 − λAzn−1 .
3.10
Thus, from 3.8–3.10,
n
βi−1 − βi Ti yn − Ti yn−1 xn1 − xn ≤ βn xn − xn−1 βn−1 − βn xn−1 σn
i1
|σn − σn−1 |
n−1
βi−1 − βi Ti yn−1 σn βn−1 − βn Tn yn−1 i1
1 − βn 1 − σn PC zn − λAzn − PC zn−1 − λAzn−1 1 − βn 1 − σn − 1 − βn−1 1 − σn−1 PC zn−1 − λAzn−1 n
≤ βn xn − xn−1 βn−1 − βn xn−1 σn
βi−1 − βi yn − yn−1 i1
|σn − σn−1 |
n−1
βi−1 − βi Ti yn−1 βn−1 − βn Tn yn−1 i1
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1 − βn 1 − σn zn − zn−1 βn−1 − βn 1 − σn PC I − λAzn−1 1 − βn−1 |σn − σn−1 |PC I − λAzn−1 ≤ βn xn − xn−1 βn−1 − βn xn−1 σn 1 − βn
1
× 1 − αn 1 − ρ xn − xn−1 1 − αn |rn−1 − rn |L
r
|σn − σn−1 |
n−1
βi−1 − βi Ti yn−1 i1
1
βn−1 − βn Tn yn−1 1 − βn 1 − σn xn − xn−1 |rn−1 − rn |L
r
βn−1 − βn PC I − λAzn−1 |σn − σn−1 |PC I − λAzn−1 ≤ 1 − σn 1 − βn αn 1 − ρ xn − xn−1 βn−1 − βn M
n−1
1
βi−1 − βi M βn−1 − βn M
|rn−1 − rn |L |σn − σn−1 |
r
i1
βn−1 − βn |σn − σn−1 | M
1
≤ 1 − σn 1 − βn αn 1 − ρ xn − xn−1 |rn−1 − rn |L
r
3 βn−1 − βn M 2|σn − σn−1 |M,
3.11
where M max{supn≥1 xn , supi≥1,n≥1 Ti yn , supn≥1 PC I −λAzn }. Since {βn } is strictly
decreasing, we have ∞
n2 βn−1 − βn β1 < ∞. Further, by assumption conditions iv-v,
we have
∞ 1
n2
r
|rn1 − rn |L 3 βn−1 − βn M 2|σn − σn−1 |M
< ∞.
3.12
Thus, using Lemma 2.4, we have limn → ∞ xn1 − xn 0.
Step 3. limn → ∞ xn − un 0.
For any x∗ ∈ Ω, it follows from Lemma 2.6 that
un − x∗ 2 Trn xn − Trn x∗ 2 ≤ Trn xn − Trn x∗ , xn − x∗ un − x∗ , xn − x∗ 1
un − x∗ 2 xn − x∗ 2 − un − xn 2 .
2
3.13
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From 2.3, 3.1–3.4, we can get
yn − x∗ 2 ≤ αn fxn − x∗ 2 1 − αn vn − x∗ 2
2
2
2 ≤ αn fxn − x∗ 1 − αn zn − y∗ λλ − 2αAzn − Ay∗ 2
2
≤ αn fxn − x∗ 1 − αn zn − y∗ 2
≤ αn fxn − x∗ 1 − αn un − x∗ 2 .
3.14
From 3.1, 3.3-3.4, and 3.13-3.14, it follows that
2
n
∗
βi−1 − βi Ti yn 1 − βn 1 − σn vn − x xn1 − x βn xn σn
i1
2
n
1 ∗ 2
∗
≤ βn xn − x 1 − βn σn
βi−1 − βi Ti yn 1 − σn vn − x 1 − βn i1
2
n
1 βi−1 − βi Ti yn − x∗ ≤ βn xn − x∗ 2 1 − βn σn 1 − βn i1
1 − βn 1 − σn vn − x∗ 2
n
2
1 3.15
≤ βn xn − x∗ 2 1 − βn σn
βi−1 − βi Ti yn − x∗ 1 − βn i1
1 − βn 1 − σn vn − x∗ 2
2 2
≤ βn xn − x∗ 2 1 − βn σn yn − x∗ 1 − βn 1 − σn zn − y∗ 2 ≤ βn xn − x∗ 2 1 − βn σn 1 − αn un − x∗ 2 αn fxn − x∗ 1 − βn 1 − σn un − x∗ 2
2
≤ xn − x∗ 2 1 − βn σn αn fxn − x∗ − xn − x∗ 2
− 1 − βn 1 − σn αn un − xn 2
∗ 2
which implies that
1 − βn 1 − σn αn un − xn 2 ≤ xn − x∗ xn1 − x∗ xn − xn1 2
αn fxn − x∗ .
3.16
Since {xn } is both bounded, using Step 2 and conditions ii–iv, we conclude the result.
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Step 4. limn → ∞ Azn − Ay∗ 0 and limn → ∞ Bxn − Bx∗ 0.
Using 3.3, 3.14, and 3.15, we have
2 2
xn1 − x∗ 2 ≤ βn xn − x∗ 2 1 − βn σn yn − x∗ 1 − βn 1 − σn zn − y∗ ≤ βn xn − x∗ 2 1 − βn σn
2
2
2 × αn fxn − x∗ 1 − αn zn − y∗ λλ − 2αAzn − Ay∗ 2
1 − βn 1 − σn zn − y∗ 2
≤ βn xn − x∗ 2 1 − βn σn 1 − αn λλ − 2αAzn − Ay∗ 3.17
2 2
αn fxn − x∗ 1 − βn zn − y∗ 2
≤ βn xn − x∗ 2 λλ − 2αAzn − Ay∗ 2 αn fxn − x∗ 1 − βn xn − x∗ 2 μ μ − 2β Bun − Bx∗ 2
2
≤ xn − x∗ 2 λλ − 2αAzn − Ay∗ 2
αn fxn − x∗ μ μ − 2β Bun − Bx∗ 2 .
Therefore,
2
λ2α − λAzn − Ay∗ μ 2β − μ Bun − Bx∗ 2
2
≤ xn − x∗ xn1 − x∗ xn − xn1 αn fxn − y∗ .
3.18
From Step 2, using condition iii, we get limn → ∞ Azn −Ay∗ 0 and limn → ∞ Bun −Bx∗ 0.
From the fact that the B is β-inverse strongly monotone operator, it follows that
Bxn − Bx∗ ≤ Bxn − Bun Bun − Bx∗ ≤
1
xn − un Bun − Bx∗ .
β
Applying Step 3, we have limn → ∞ Bxn − Bx∗ 0.
Step 5. limn → ∞ xn − Ti xn 0, for all i 1, 2, . . ..
Noting that PC is firmly nonexpansive, we have
zn − y∗ 2 PC un − μBun − PC x∗ − μBx∗ 2
≤ un − μBun − x∗ − μBx∗ , zn − y∗
2 2
1 un − x∗ − μBun − Bx∗ zn − y∗ 2
2 −un − zn − μBun − Bx∗ − x∗ − y∗ 3.19
Journal of Applied Mathematics
11
2 2
1
un − x∗ 2 zn − y∗ − un − zn − x∗ − y∗ 2
2μ un − zn − x∗ − y∗ , Bun − Bx∗ − μ2 Bun − Bx∗ 2
2 2
1
≤
xn − x∗ 2 zn − y∗ − un − zn − x∗ − y∗ 2
2μ un − zn − x∗ − y∗ , Bun − Bx∗ ,
≤
2
vn − x∗ 2 PC zn − λAzn − PC y∗ − λAy∗ ≤ zn − λAzn − y∗ − λAy∗ , vn − x∗
2
1 zn − y∗ − λ Azn − Ay∗ vn − x∗ 2
2
2 −zn − vn − λ Azn − Ay∗ − y∗ − x∗ 2
2
1 ≤ zn − y∗ vn − x∗ 2 − zn − vn − y∗ − x∗ 2
2 2λ zn − vn − y∗ − x∗ , Azn − Ay∗ − λ2 Azn − Ay∗ 2
1
≤
xn − x∗ 2 vn − x∗ 2 − zn − vn − y∗ − x∗ 2
∗
∗
∗
2λ zn − vn − y − x Azn − Ay .
3.20
It follows that
zn − y∗ 2 ≤ xn − x∗ 2 − un − zn − x∗ − y∗ 2
2μ un − zn − x∗ − y∗ , Bun − Bx∗ ,
2
vn − x∗ 2 ≤ xn − x∗ 2 − zn − vn − y∗ − x∗ 2λ zn − vn − y∗ − x∗ , Azn − Ay∗ .
3.21
3.22
By 3.3, 3.14-3.15, and 3.22, we have
2
xn1 − x∗ 2 ≤ 1 − 1 − βn σn xn − x∗ 2 1 − βn σn yn − x∗ 2
≤ 1 − 1 − βn σn xn − x∗ 2 1 − βn σn αn fxn − x∗ vn − x∗ 2
≤ 1 − 1 − βn σn xn − x∗ 2
2
2
1 − βn σn αn fxn − x∗ xn − x∗ 2 − zn − vn − y∗ − x∗ 3.23
2λ zn − vn − y∗ − x∗ Azn − Ay∗
2 xn − x∗ 2 αn fxn − x∗ 1 − βn 1 − σn 2
× −zn − vn − y∗ − x∗ 2λ zn − vn − y∗ − x∗ , Azn − Ay∗ .
12
Journal of Applied Mathematics
It follows that
2
1 − βn σn vn − z − x∗ − y∗ 2
≤ xn − x∗ xn1 − x∗ xn − xn1 αn fxn − x∗ 2λ 1 − βn σn zn − vn − y∗ − x∗ , Azn − Ay∗ .
3.24
From conditions ii–iv, Steps 2 and 4, we get the following:
lim vn − zn − x∗ − y∗ 0.
n−→∞
3.25
On the other hand, from 3.14–3.21, we have
2 2 xn1 − x∗ 2 ≤ βn xn − x∗ 2 1 − βn σn αn fxn − x∗ zn − y∗ 2
1 − βn 1 − σn zn − y∗ 2 2
≤ βn xn − x∗ 2 αn fxn − x∗ 1 − βn zn − y∗ 2 ≤ βn xn − x∗ 2 αn fxn − x∗ 1 − βn
2
× xn − x∗ 2 − un − zn − x∗ − y∗ 2μ un − zn − x∗ − y∗ , Bun − Bx∗
2
xn − x∗ 2 αn fxn − x∗ 2
1 − βn −un − zn − x∗ − y∗ 2μ un − zn − x∗ − y∗ , Bun − Bx∗ ,
3.26
which implies that
1 − βn un − zn − x∗ − y∗ 2 ≤ xn − x∗ xn1 − x∗ xn − xn1 2
αn fxn − x∗ 2μ 1 − βn
× un − zn − x∗ − y∗ , Bun − Bx∗ .
3.27
From ii-iii, Steps 2 and 4, we get the following:
lim un − zn − x∗ − y∗ 0.
n−→∞
3.28
Combining 3.25 and 3.28, we get the following:
lim un − vn 0.
n−→∞
3.29
Using Step 3, we obtain the following:
lim xn − vn 0.
n−→∞
3.30
Journal of Applied Mathematics
13
This together with yn − vn αn fxn − vn → 0 implies that
lim xn − yn 0.
3.31
n−→∞
For any x∗ ∈ Ω, we have from 3.1 that
σn
n
βi−1 − βi
i1
σn
xn − Ti xn , xn − x∗ Ti xn − Ti yn , xn − x∗
n
βi−1 − βi
xn − Ti yn , xn − x∗
3.32
i1
xn − xn1 , xn − x∗ 1 − βn 1 − σn vn − xn , xn − x∗ .
Since each Ti is nonexpansive, from 2.2, we have
Ti xn − xn 2 ≤ 2xn − Ti xn , xn − x∗ .
3.33
Hence, combining this inequality with 3.32, we get the following:
n
n
1 σn
βi−1 − βi Ti xn − xn 2 ≤ −σn
βi−1 − βi Ti xn − Ti yn , xn − x∗
2 i1
i1
xn − xn1 , xn − x∗ 1 − βn 1 − σn 3.34
× vn − xn , xn − x∗ ,
that is noting that βn is strictly decreasing,
Ti xn − xn 2 ≤
2 1 − βn xn − yn xn − x∗ 2
xn − xn1 xn − x∗ βi−1 − βi
σn βi−1 − βi
2 1 − βn 1 − σn vn − xn xn − x∗ .
σn βi−1 − βi
3.35
Now from 3.30-3.31 and Step 2, we conclude that
lim Ti xn − xn 0,
n−→∞
∀i 1, 2, . . . ,
which completes the proof.
Step 6. lim supn → ∞ fx∗ − x∗ , xn − x∗ ≤ 0, where x∗ PΩ · fx∗ .
3.36
14
Journal of Applied Mathematics
As {xn } is bounded, there exists a subsequence {xni } of {xn } such that {xni } → x
weakly. First, it is clear from Step 5 and Lemma 2.3 that x ∈ ∞
n1 FTn . Next, we prove that
x ∈ Γ. From iii, Step 3 and 3.31, we note that
yn − G yn ≤ αn fxn − G yn 1 − αn PC PC un − μBun − λAPC un − μBun − G yn αn fxn − G yn 1 − αn Gun − G yn ≤ αn fxn − G yn 1 − αn un − yn 3.37
−→ 0.
According to 3.31 and Lemma 2.3, we obtain that x ∈ Γ. Next, we show that x ∈ EPF.
Indeed, by un Trn xn , we have
1
y − un , un − xn ≥ 0,
F un , y rn
∀y ∈ C.
3.38
From A2, we have also
1
y − un , un − xn ≥ −F un , y ≥ F y, un ,
rn
3.39
un − xnk
≥ F y, unk .
y − unk , k
rnk
3.40
and hence,
According to rn > r > 0 and un − xn → 0, we conclude that uni − xni /rni → 0 and unk x.
From A4, we obtain the following:
F y, x ≤ 0,
∀y ∈ C.
3.41
Since y ∈ C and x ∈ C due to uni → x as
For t with 0 < t ≤ 1 and y ∈ C, let yt ty 1 − tx.
≤ 0. So, from A1 and A4, we have
i → ∞, we have yt ∈ C and hence Fyt , x
tF yt , y ≥ tF yt , y 1 − tF yt , x ≥ tF yt , yt 0
3.42
and Fyt , y ≥ 0, for all t ∈ 0, 1 and y ∈ C. From A3, we obtain the following:
F x,
y ≥0
∀y ∈ C,
3.43
Journal of Applied Mathematics
15
and hence x ∈ EPF. Therefore, we obtain that x ∈ Ω. Hence, it follows from Lemma 2.2 that
lim sup fx∗ − x∗ , xn − x∗ lim fx∗ − x∗ , xni − x∗
n→∞
i→∞
fx∗ − x∗ , x − x∗
3.44
≤ 0.
Step 7. limn → ∞ xn x∗ .
From 3.1, 3.3-3.4, and the convexity of · , we have
yn − x∗ 2 αn fxn 1 − αn vn − x∗ 2
≤ 1 − αn vn − x∗ 2 2αn fxn − x∗ , yn − x∗
≤ 1 − αn xn − x∗ 2 2αn fxn − x∗ , yn − x∗ .
3.45
We can also get
n
βi−1 − βi Ti yn − x∗ 1 − βn 1 − σn yn − x∗
xn1 − x βn xn − x∗ σn
i1
2
1 − βn 1 − σn αn vn − fxn 2
n
∗
∗
∗ ≤ βn xn − x σn
βi−1 − βi Ti yn − x 1 − βn 1 − σn yn − x i1
2 1 − βn 1 − σn αn vn − fxn , xn1 − x∗
2
≤ βn xn − x∗ 2 1 − βn yn − x∗ 2 1 − βn 1 − σn αn vn − x∗ , xn1 − x∗ ∗
2 1 − βn 1 − σn αn x − fxn , xn1 − x∗ .
3.46
∗ 2
By 3.45, we have
xn1 − x∗ 2 ≤ βn xn − x∗ 2 1 − βn 1 − αn xn − x∗ 2 2αn fxn − x∗ , yn − x∗
2 1 − βn 1 − σn αn vn − x∗ xn1 − x∗ 2 1 − βn αn 1 − σn x∗ − fxn , xn1 − x∗
≤ 1 − 1 − βn αn xn − x∗ 2 2 1 − βn 1 − σn αn fxn − x∗ , yn − xn1
2 1 − βn 1 − σn αn zn − y∗ xn1 − x∗ 2 1 − βn αn σn fxn − x∗ , yn − x∗
≤ 1 − 1 − βn αn xn − x∗ 2 2 1 − βn 1 − σn αn fxn − x∗ xn1 − yn 2 1 − βn 1 − σn αn xn − x∗ xn1 − x∗ 2 1 − βn αn σn fxn − x∗ , yn − xn 2 1 − βn αn σn fxn − fx∗ , xn − x∗
2 1 − βn αn σn fx∗ − x∗ , xn − x∗
16
Journal of Applied Mathematics
≤ 1 − 1 − βn αn xn − x∗ 2 2 1 − βn 1 − σn αn fxn − x∗ xn1 − yn 1 − βn 1 − σn αn xn − x∗ 2 xn1 − x∗ 2
2 1 − βn αn σn fxn − x∗ yn − xn 2 1 − βn αn σn ρxn − x∗ 2
2 1 − βn αn σn fx∗ − x∗ , xn − x∗ ,
3.47
which implies that
2σn 1 − ρ − 1 1 − βn
2σn 1 − ρ − 1 1 − βn αn
∗ 2
αn xn − x xn1 − x ≤ 1 −
1 − 1 − βn 1 − σn αn
1 − 1 − βn 1 − σn αn
2σn
21 − σn fxn − x∗ xn1 − yn fxn − x∗ ×
2σn 1 − ρ − 1
2σn 1 − ρ − 1
2σ
n
∗
∗
×yn − xn fxn − x , xn − x .
2σn 1 − ρ − 1
3.48
∗ 2
From lim infn → ∞ 2σn 1 − ρ − 11 − βn /1 − 1 − βn 1 − σn > 0, it follows that
∞
n0 2σn 1 − ρ − 11 − βn /1 − 1 − βn 1 − σn αn αn ∞. It is clear that
lim sup
n→∞
2σn
21 − σn fxn − x∗ xn1 − yn fxn − x∗ 2σn 1 − ρ − 1
2σn 1 − ρ − 1
2σn
∗
∗
≤ 0.
× yn − xn fxn − x , xn − x
2σn 1 − ρ − 1
3.49
Therefore, all conditions of Lemma 2.4 are satisfied. Therefore, we immediately deduce that
xn → x∗ . This completes the proof.
Corollary 3.2. Let C be a nonempty closed convex subset of a real Hilbert space H. Let the mappings
A, B : C → H be α-inverse strongly monotone and β-inverse strongly monotone, respectively. Let F
be a bifunction from C × C to R satisfying (A1)–(A4) and T : C → C be a nonexpansive mapping
such that Ω : F T ∩ EP F ∩ Γ /
∅.Let f : C → C be a contraction with coefficient ρ ∈ 0, 1/2.
For given x0 ∈ C arbitrarily, let the sequences {xn }, {yn }, {zn } and, {un } be generated By
1
F un , y y − un , un − xn ≥ 0,
rn
zn PC un − μBun ,
xn1
∀y ∈ C,
yn αn fxn 1 − αn PC zn − λAzn ,
βn xn σn 1 − βn T yn 1 − βn 1 − σn PC zn − λAzn ,
3.50
Journal of Applied Mathematics
17
where λ ∈ 0, 2α, μ ∈ 0, 2β, and sequences {αn } ⊂ 0, 1, {βn } ⊂ 0, 1, {σn } ⊂ 0, 1, and
{rn } ⊂ r, ∞, r > 0, are such that
i 0 < lim infn → ∞ βn ≤ lim supn → ∞ βn < 1,
ii limn → ∞ αn 0and ∞
n1 αn ∞,
∞
iii σn > 1/21 − ρ , n1 |σn − σn−1 | < ∞,
iv ∞
n1 |rn − rn−1 | < ∞.
Then the sequence {xn } generated by 3.50 converges strongly to x∗ PΩ · fx∗ , and x∗ , y∗ is a
solution of the general system of variational inequalities 1.5, where y∗ PC x∗ − μBx∗ .
Acknowledgments
The authors would like to express their thanks to the referee for the valuable comments and
suggestions for improving this article. This paper was supported by the NSFC Tianyuan
Youth Foundation of Mathematics of China no. 11126136 and Fundamental Research Funds
for the Central Universities no. ZXH2011C002.
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