Volume 2011, Article ID 915629, 17 pagesdoi:10.1155/2011/915629 Research Article Iterative Algorithms for Finding Common Solutions to Variational Inclusion Equilibrium and Fixed Point Pr
Trang 1Volume 2011, Article ID 915629, 17 pages
doi:10.1155/2011/915629
Research Article
Iterative Algorithms for Finding Common
Solutions to Variational Inclusion Equilibrium and Fixed Point Problems
J F Tan and S S Chang
Department of Mathematics, Yibin University, Yibin, Sichuan 644007, China
Correspondence should be addressed to S S Chang,changss@yahoo.cn
Received 30 October 2010; Accepted 9 November 2010
Academic Editor: Qamrul Hasan Ansari
Copyrightq 2011 J F Tan and S S Chang 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
The main purpose of this paper is to introduce an explicit iterative algorithm to study the existence problem and the approximation problem of solution to the quadratic minimization problem Under suitable conditions, some strong convergence theorems for a family of nonexpansive mappings are proved The results presented in the paper improve and extend the corresponding results announced by some authors
1 Introduction
Throughout this paper, we assume thatH is a real Hilbert space with inner product ·, · and
norm · , C is a nonempty closed convex subset of H, and FT {x ∈ H : Tx x} is the
set of fixed points of mappingT.
A mappingS : C → C is called nonexpansive if
Let A : H → H be a single-valued nonlenear mapping and M : H → 2 H be a multivalued mapping The so-called quasivariational inclusion problemsee 1 3 is to find
u ∈ H such that
θ ∈ Au Mu. 1.2 The set of solutions to quasivariational inclusion problem1.2 is denoted by VIH, A, M.
Trang 2Special Cases
I If M ∂φ : H → 2 H, whereφ : H → Ê∪{∞} is a proper convex lower semi-continuous function and∂φ is the subdifferential of φ, then the quasivariational inclusion problem 1.2
is equivalent to findingu ∈ H such that
which is called the mixed quasivariational inequalitysee 4
II If M ∂δ C, whereC is a nonempty closed convex subset of H and δ C : H →
0, ∞ is the indicator function of C, that is,
δ C x
⎧
⎨
⎩
0, x ∈ C,
then the quasivariational inclusion problem1.2 is equivalent to finding u ∈ C such that
This problem is called the Hartman-Stampacchia variational inequalitysee 5 The set of solutions to variational inequality1.5 is denoted by VIA, C.
LetB : C → H be a nonlinear mapping and F : C × C → Ê be a bifunction The so-called generalized equilibrium problem is to find a pointu ∈ C such that
Fu, yBu, y − u ≥ 0, ∀y ∈ C. 1.6
The set of solutions to1.6 is denoted by GEP see 5,6 If B 0, then 1.6 reduces to the following equilibrium problem: to findu ∈ C such that
The set of solutions to1.7 is denoted by EP
Iterative methods for nonexpansive mappings and equilibrium problems have been applied to solve convex minimization problemssee 7 9 A typical problem is to minimize
a quadratic function over the set of the fixed points of a nonexpansive mapping on a real Hilbert spaceH:
min
x∈F
1
whereF is the fixed point set of a nonexpansive mapping T on H.
In 2010, Zhang et al.see 10 proposed the following iteration method for variational inclusion problem1.5 and equilibrium problem 1.6 in a Hilbert space H:
x t SP C
1 − tJ M,λ I − λAT μ
I − μBx t , t ∈ 0, 1. 1.9
Trang 3Under suitable conditions, they proved the sequence {x n} generated by 1.9 converges strongly to the fixed pointx∗, which solves the quadratic minimization problem1.8 Motivated and inspired by the researches going on in this direction, especially inspired
by Zhang et al.10, the purpose of this paper is to introduce an explicit iterative algorithm
to studying the existence problem and the approximation problem of the solution to the quadratic minimization problem 1.8 and prove some strong convergence theorems for a family of nonexpansive mappings in the setting of Hilbert spaces
2 Preliminaries
LetH be a real Hilbert space, and C be a nonempty closed convex subset of H For any x ∈ H,
there exists a unique nearest point inC, denoted by P C x, such that
Such a mappingP C fromH onto C is called the metric projection It is well-known that the
metric projectionP C :H → C is nonexpansive.
In the sequel, we usex n x and x n → x to denote the weak convergence and the
strong convergence of the sequence{x n}, respectively
Definition 2.1 A mapping A : H → H is called α-inverse strongly monotone if there exists
anα > 0 such that
Ax − Ay, x − y ≥ αAx − Ay2, ∀x, y ∈ H. 2.2
A multivalued mappingM : H → 2 His called monotone if∀x, y ∈ H, u ∈ Mx, v ∈ My,
u − v, x − y ≥ 0. 2.3
A multivalued mappingM : H → 2 His called maximal monotone if it is monotone and for anyx, u ∈ H × H, when
u − v, x − y ≥ 0 for everyy, v∈ GraphM, 2.4
thenu ∈ Mx.
the following statements hold:
i A is an 1/α-Lipschitz continuous and monotone mapping;
ii if λ is any constant in 0, 2α, then the mapping I − λA is nonexpansive, where I is the
identity mapping on H.
Trang 4Lemma 2.3 see 12 Let X be a strictly convex Banach space, C be a closed convex subset of X,
and {T n :C → C} be a sequence of nonexpansive mappings Suppose ∞
n1 FT n / ∅ Let {λ n } be a
sequence of positive numbers withΣ∞
n1 λ n 1 Then the mapping S : C → C defined by
Sx Σ∞
n1 λ n T n x, x ∈ C 2.5
is well defined And it is nonexpansive and
FS ∞
Definition 2.4 Let H be a Hilbert space and M : H → 2 H be a multivalued maximal monotone mapping Then, the single-valued mappingJ M,λ:H → H defined by
is called the resolvent operator associated with M, where λ is any positive number and I is the
identity mapping
Proposition 2.5 see 11 i The resolvent operator J M,λ associated with M is single-valued and nonexpansive for all λ > 0, that is,
J M,λ x − J M,λy ≤ x − y, ∀x,y ∈ H, ∀λ > 0. 2.8
ii The resolvent operator J M,λ is 1-inverse strongly monotone, that is,
J M,λ x − J M,λ y2≤x − y, J M,λ x − J M,λ
y , ∀x, y ∈ H. 2.9
Definition 2.6 A single-valued mapping A : H → H is said to be hemicontinuous if for any
x, y, z ∈ H, function t → Ax ty, z is continuous at 0.
It is well-known that every continuous mapping must be hemicontinuous
Lemma 2.7 see 13 Let {x n } and {y n } be bounded sequences in a Banach space X Let {β n } be a
sequence in 0, 1 with
0< lim inf
n → ∞ β n ≤ lim sup
n → ∞ β n < 1. 2.10 Suppose that
x n11− β n
y n β n x n , ∀n ≥ 0,
lim sup
n → ∞ y n1 − y n − x n1 − x n≤ 0. 2.11
Trang 5lim
n → ∞ y n − x n 0. 2.12
Lemma 2.8 see 14 Let X be a real Banach space, X∗be the dual space of X, T : X → 2 X∗
be a maximal monotone mapping, and P : X → X∗be a hemicontinuous bound monotone mapping with DP X Then, the mapping S T P : X → 2 X∗
is a maximal monotone mapping.
Lemma 2.9 see 15 Let X be a uniformly convex Banach space, let C be a nonempty closed convex
subset of X, and T : C → C be a nonexpansive mapping with a fixed point Then, I − T is demiclosed
in the sense that if {x n } is a sequence in C satisfying
x n x, I − T n −→ 0, 2.13
then
Throughout this paper, we assume that the bifunction F : C × C → Ê satisfies the following conditions:
H1 Fx, x 0 for all x ∈ C;
H2 F is monotone, that is,
H3 for each x, y, z ∈ C,
lim
t↓0 Ftz 1 − tx, y≤ Fx, y, 2.16
H4 for each x ∈ C, y → Fx, y is convex and lower semi-continuous.
and F : C × C → Êbe a bifunction satisfying the conditions (H 1 )–(H 4 ) Let μ > 0 and x ∈ H Then, there exists a point z ∈ C such that
Fz, y 1
μ
y − z, z − x ≥ 0, ∀y ∈ C. 2.17
Moreover, if T μ:H → C is a mapping defined by
T μ x
z ∈ C : Fz, y1
μ
y − z, z − x ≥ 0, ∀y ∈ C , x ∈ H, 2.18
then the following results hold:
Trang 6i T μ is single-valued and firmly nonexpansive, that is, for any x, y ∈ H,
T μ x − T μ y2≤T μ x − T μ y, x − y , 2.19
ii EP is closed and convex, and EP FT μ .
Lemma 2.11 i see 11 u ∈ H is a solution of variational inclusion 1.2 if and only if
u J M,λ u − λAu, ∀λ > 0, 2.20
that is,
ii see 10 u ∈ C is a solution of generalized equilibrium problem 1.6 if and only if
u T μu − μBu, ∀μ > 0, 2.22
that is,
GEP FT μ
u − μBu, ∀μ > 0. 2.23
iii see 10 Let A : H → H be an α-inverse strongly monotone mapping and B : C → H
be a β-inverse strongly monotone mapping If λ ∈ 0, 2α and μ ∈ 0, 2β, then VIH, A, M is a closed convex subset in H and GEP is a closed convex subset in C.
Lemma 2.12 see 17 Assume that {a n } is a sequence of nonnegative real numbers such that
where {γ n } is a sequence in 0, 1 and {δ n } is a sequence such that:
i∞n1 γ n ∞;
ii lim supn → ∞ δ n /γ n ≤ 0 or∞
n1 |δ n | < ∞.
Then, lim n → ∞ a n 0.
3 Main Results
Theorem 3.1 Let H be a real Hilbert space, C be a nonempty closed convex subset of H, A : H → H
be an α-inverse strongly monotone mapping and B : C → H be a β-inverse strongly monotone
Trang 7mapping Let M : H → 2 H be a maximal monotone mapping, {T n : C → C} be a sequence of nonexpansive mappings with ∞n1 FT n / ∅, S : C → C be the nonexpansive mapping defined by
2.5, and F : C × C → Êbe a bifunction satisfying conditions (H 1 )–(H 4 ) Let {x n } be the sequence
defined by
x n1 α n x n 1 − α nSP C
1 − t n J M,λ I − λAT μ
I − μBx n
where the mapping T μ:H → C is defined by 2.18 , and λ, μ are two constants with λ ∈ 0, 2α, μ ∈
0, 2β, and
t n ∈ 0, 1, t n −→ 0n −→ ∞, ∞
n1
t n ∞, 0 < a < α n < b < 1. 3.2
If
where VI H, A, M and GEP is the set of solutions of variational inclusion 1.2 and generalized
equilibrium problem1.6, respectively, then the sequence {x n } defined by 3.1 converges strongly to
x∗∈ Ω, which is the unique solution of the following quadratic minimization problem:
x∗2 min
Proof We divide the proof ofTheorem 3.1into four steps
Step 1 The sequence {x n} is bounded Set
u n T μI − μBx n , y n J M,λ I − λAu n , z n SP C1 − t n y n. 3.5
Takingz ∈ Ω, then it follows fromLemma 2.11that
z T μ
z − μBz J M,λ z − λAz SP C z. 3.6
Since bothT μ andJ M,λ are nonexpansive,A and B are α-inverse strongly monotone and
β-inverse strongly monotone, respectively, fromProposition 2.2, we have
Trang 8u n − z2 T μ I − μBx n − T μ z − μBz2
≤I − μBx n − z − μBz2
≤ x n − z2 μμ − 2βBx n − Bz2,
3.7
y n − z2
J M,λ I − λAu n − J M,λ z − λAz2
≤ I − λAu n − z − λAz2
≤ u n − z2 λλ − 2αAu n − Az2
≤ x n − z2 λλ − 2αAu n − Az2 μμ − 2βBx n − Bz2.
3.8
This implies that
It follows from3.1 and 3.9 that
x n1 − z α n x n 1 − α n SP C
1 − t n y n
− z
α n x n − z 1 − α nSP C1 − t n y n− SP C z
≤ α n x n − z 1 − α nSPC
1 − t n y n
− SP C z
≤ α n x n − z 1 − α n1− tn y n − z
≤ α n x n − z 1 − α n1 − t nyn − z t n z
≤ α n x n − z 1 − α n 1 − t n x n − z t n z
≤ 1 − t n 1 − α n x n − z t n 1 − α n z
≤ max{x n − z, z}
≤ max{x n−1 − z, z}
≤ · · · ≤ max{x1− z, z} M,
3.10
whereM max{x1− z, z} This shows that {x n} is bounded Hence, it follows from 3.9 that the sequence{u n } and {y n} are also bounded
It follows from3.5, 3.6, and 3.9 that
z n − z SP C 1 − t n y n − SP C z ≤ 1 − t n y n − z
≤ 1 − t nyn − z t n z ≤ 1 − t n x n − z t n z ≤ M. 3.11
This shows that{z n} is bounded
Trang 9Step 2 Now, we prove that
lim
n → ∞ x n − u n lim
n → ∞ u n − y n limn → ∞ x n − y n 0,
lim
n → ∞ x n − Sx n 0. 3.12
SinceSP Cis nonexpansive, from3.5 and 3.9, we have that
z n1 − z n SP C
1 − t n1 y n1− SP C1 − t n y n
≤1− t n1 y n1 − 1 − t n y n
1− t n1y n1 − y n 1 − t n1 − 1 − t n y n
≤ 1 − t n1yn1 − y n |t n1 − t n|yn
≤y n1 − y n |t n1 − t n|yn
≤ u n1 − u n |t n1 − t n|yn
≤ x n1 − x n |t n1 − t n|yn .
3.14
Letn → ∞ in 3.14 , in view of condition t n → 0n → ∞, we have
lim
n → ∞ z n1 − z n − x n1 − x n 0. 3.15
By virtue ofLemma 2.7, we have
lim
n → ∞ x n − z n 0. 3.16
This implies that
lim
n → ∞ x n1 − x n lim
n → ∞ 1 − α n z n − x n 0. 3.17
Trang 10We derive from3.17 that
lim
n → ∞
x n − z2− x n1 − z2
lim
n → ∞
x n − x n12 2x n − x n1 , x n1 − z
≤ lim
n → ∞
x n − x n12 2x n − x n1 · x n1 − z 0.
3.18
From3.1 and 3.8, we have
x n1 − z2≤α n x n − z 1 − α n1 − tn y n − z2
≤ α n x n − z2 1 − α n1 − tn y n − z − t n z2
α n x n − z2 1 − α n1 − t n2y n − z2− 2t n 1 − t nz, y n − z t2
n z2
≤ α n x n − z2 1 − α ny
n − z2 t n M1
≤ α n x n − z2 1 − α n
×x n − z2 λλ − 2αAu n − Az2 μμ − 2βBx n − Bz2 t n M1
x n − z2 1 − α nλλ − 2αAu n − Az2 μμ − 2βBx n − Bz2 t n M1
,
3.19 where
M1 sup
n
z2 2λ u n − y n μx n − y n < ∞, 3.20
that is,
1 − α nλ2α − λAu n − Az2 μ2β − μBx n − Bz2
≤ x n − z2− x n1 − z2 1 − α n t n M1.
3.21
Letn → ∞, noting the assumptions that λ ∈ 0, 2α, μ ∈ 0, 2β, from 3.2 and 3.18, we have
lim
n → ∞ Au n − Az lim n → ∞ Bx n − Bz 0. 3.22
Trang 11By virtue ofLemma 2.10i and 3.1, we have
u n − z2T μ x n − μBx n − T μz − μBz2
≤x n − μBx n
−z − μBz, u n − z
1 2
x
n − μBx n
−z − μBz2
u n − z2
−xn − z − μBx n − Bz − u n − z2
≤ 1 2
x n − z2 u n − z2−x n − u n − μBx n − Bz2
1 2
x n − z2 u n − z2− x n − u n2
2μx n − u n , Bx n − Bz − μ2Bx n − Bz2
.
3.23
Simplifying it, we have
u n − z2≤ x n − z2− x n − u n2 2μx n − u n , Bx n − Bz − μ2Bx n − Bz2
≤ x n − z2− x n − u n2 2μx n − u n · Bx n − Bz
≤ x n − z2− x n − u n2 M1Bx n − Bz.
3.24
Similarly, in view ofProposition 2.5ii and 3.1, we have
y n − z2
J M,λ u n − λAu n − J M,λ z − λAz2
≤u n − λAu n − z − λAz, y n − z
1 2
u n − λAu n − z − λAz2y n − z2
−u n − λAu n − z − λAz − y n − z2
≤ 1 2
u n − z2y n − z2−u n − y n − λAu n − Az2
1 2
u n − z2y n − z2−u n − y n2
2λu n − y n , Au n − Az − λ2Au n − Az2
.
3.25
Trang 12Simplifying it, from3.24, we have
y n − z2≤ u n − z2−u n − y n2 2λu n − y n , Au n − Az − λ2Au n − Az2
≤ u n − z2−u n − y n2 2λu n − y n · Au n − Az
≤ x n − z2− x n − u n2 M1Bx n − Bz − u n − y n2
M1Au n − Az.
3.26
From3.19 and 3.26, we have
x n1 − z2≤ α n x n − z2 1 − α ny
n − z2 t n M1
≤ α n x n − z2 1 − α n
×x n − z2− x n − u n2−u n − y n2 M1Bx n − Bz Au n − Az t n
x n − z2 1 − α n
×M1Bx n − Bz Au n − Az t n − x n − u n2−u n − y n2
.
3.27 Letn → ∞ nd in view of 3.18 and 3.22, we have
lim
n → ∞
x n − u n2u n − y n2
This shows that
lim
n → ∞ x n − u n lim
n → ∞ u n − y n 0,
lim
n → ∞ x n − y n 0. 3.29
Then, we have
x n1 − Sx n1 x n1 − Sx n Sx n − Sx n1
≤ x n1 − Sx n Sx n − Sx n1
SP C
1 − t n y n
− SP C x n Sx n − Sx n1
≤ 1 − t nyn − x n t n x n x n − x n1
−→ 0n −→ ∞.
3.30
...If
where VI H, A, M and GEP is the set of solutions of variational inclusion 1.2 and generalized
equilibrium problem1.6, respectively,... α-inverse strongly monotone mapping and B : C → H
be a β-inverse strongly monotone mapping If λ ∈ 0, 2α and μ ∈ 0, 2β, then VIH, A, M is a closed convex subset in H and GEP is a closed...
ii EP is closed and convex, and EP FT μ .
Lemma 2.11 i see 11 u ∈ H is a solution of variational inclusion 1.2 if and only if
u