A Hybrid Extragradient Method for
Bilevel Pseudomonotone Variational
Inequalities with Multiple Solutions
Lu-Chuan Ceng1 , Yeong-Cheng Liou2 and Ching-Feng Wen3
1
Department of Mathematics, Shanghai Normal University, Shanghai 200234, China. This research was
partially supported by the Innovation Program of Shanghai Municipal Education Commission (15ZZ068),
Ph.D. Program Foundation of Ministry of Education of China (20123127110002), and Program for Outstanding
Academic Leaders in Shanghai City (15XD1503100). Email: zenglc@hotmail.com
2
Department of Information Management, Cheng Shiu University, Kaohsiung 833, Taiwan; and Research
Center of Nonlinear Analysis and Optimization and Center for Fundamental Science, Kaohsiung Medical
University, Kaohsiung 807, Taiwan (simplex liou@hotmail.com). This research was supported in part by
MOST 103-2923-E-037-001-MY3, MOST 104-2221-E-230-004 and Kaohsiung Medical University ‘Aim for the
Top Universities Grant, grant No. KMU-TP103F00’.Email: simplexliou@hotmail.com
3
Corresponding author. Center for Fundamental Science, and Center for Nonlinear Analysis and Optimization, Kaohsiung Medical University, Kaohsiung 80708, Taiwan. Email: cfwen@kmu.edu.tw
1
Abstract. In this paper, we introduce and analyze a hybrid extragradient algorithm for
solving bilevel pseudomonotone variational inequalities with multiple solutions in a real Hilbert
space. The proposed algorithm is based on Korpelevich’s extragradient method, Mann’s iteration method, hybrid steepest-descent method and viscosity approximation method (including
Halpern’s iteration method). Under mild conditions, the strong convergence of the iteration
sequences generated by the algorithm is derived.
Keywords: Bilevel variational inequality; Hybrid extragradient algorithm; Pseudomonotonicity; Lipschitz continuity; Global convergence.
AMS subject classifications. 49J30; 47H09; 47J20; 49M05
2
1. Introduction
Let H be a real Hilbert space with inner product h·, ·i and norm k · k. C be a nonempty
closed convex subset of H and PC be the metric projection of H onto C. If {xn } is a sequence
in H, then we denote by xn → x (respectively, xn * x) the strong (respectively, weak)
convergence of the sequence {xn } to x. Let S : C → H be a nonlinear mapping on C. We
denote by Fix(S) the set of fixed points of S and by R the set of all real numbers. A mapping
S : C → H is called L-Lipschitz continuous if there exists a constant L ≥ 0 such that
kSx − Syk ≤ Lkx − yk,
∀x, y ∈ C.
In particular, if L = 1 then S is called a nonexpansive mapping; if L ∈ [0, 1) then S is called
a contraction.
Let A : C → H be a nonlinear mapping on C. The classical variational inequality problem
(VIP) is to find x ∈ C such that
hAx, y − xi ≥ 0,
∀y ∈ C.
(1.1)
The solution set of VIP (1.1) is denoted by VI(C, A).
The VIP (1.1) was first discussed by Lions [13]. There are many applications of VIP (1.1)
in various fields; see e.g., [6,7,9,30]. In 1976, Korpelevich [2] proposed an iterative algorithm
for solving the VIP (1.1) in Euclidean space Rn :
(
yk = PC (xk − τ Axk ),
xk+1 = PC (xk − τ Ayk ),
∀k ≥ 0,
with τ > 0 a given number, which is known as the extragradient method. The literature on
the VIP is vast and Korpelevich’s extragradient method has received great attention given by
many authors, who improved it in various ways; see e.g., [10-12,14-16,18,24-25,33-35,37-38,40]
and references therein.
Let A : C → H and B : H → H be two mappings. Consider the following bilevel
variational inequality problem (BVIP):
Problem AKM. We find x∗ ∈ VI(C, B) such that
hAx∗ , x − x∗ i ≥ 0,
∀x ∈ VI(C, B),
(1.2)
where VI(C, B) denotes the set of solutions of the VIP: Find y ∗ ∈ C such that
hBy ∗ , y − y ∗ i ≥ 0,
∀y ∈ C,
(1.3)
In particular, whenever H = Rn , the BVIP was recently studied by Anh, Kim and Muu
[40].
Bilevel variational inequalities are special classes of quasivariational inequalities (see [1,3,4,39])
and of equilibrium with equilibrium constraints considered in [17,29]. However it covers some
3
classes of mathematical programs with equilibrium constraints (see [29]), bilevel minimization
problems (see [36]), variational inequalities (see [20,22,26,28]) and complementarity problems.
In what follows, suppose that A and B satisfy the following conditions:
(C1) B is pseudomonotone on H and A is β-strongly monotone on C;
(C2) A is L1 -Lipschitz continuous on C;
(C3) B is L2 -Lipschitz continuous on H;
(C4) VI(C, B) 6= ∅.
It is remarkable that under conditions (C1)-(C4), Problem AKM has only a solution because A is β-strongly monotone and L1 -Lipschitz continuous on C. In 2012, Anh, Kim and
Muu [40] introduced the following extragradient iterative algorithm for solving the above
bilevel variational inequality.
Algorithm AKM (see [40, Algorithm 2.1]). Initialization. Choose u ∈ Rn , x0 ∈ C, k =
0, 0 < λ ≤ L2β2 , positive sequences {δk }, {λk }, {αk }, {βk }, {γk } and {¯k } such that
1

















lim δk = 0,
k→∞
∞
X
¯k < ∞,
k=0
αk + βk + γk = 1 ∀k ≥ 0,
∞
X
αk = ∞,
k=0
lim αk = 0, lim βk = ξ ∈ (0, 12 ], lim λk = 0, λk ≤
k→∞
k→∞
k→∞
1
L2
∀k ≥ 0.
Step 1. Compute
(
yk := PC (xk − λk Bxk ),
zk := PC (xk − λk Byk ).
Step 2. Inner loop j = 0, 1, .... Compute















xk,0 := zk − λAzk ,
yk,j := PC (xk,j − δj Bxk,j ),
xk,j+1 := αj xk,0 + βj xk,j + γj PC (xk,j − δj Byk,j ).
If kxk,j+1 − PVI(C,B) xk,0 k ≤ ¯k then set hk := xk,j+1 and go to Step 3.
Otherwise, increase j by 1 and repeat the inner loop Step 2.
Step 3. Set xk+1 := αk u + βk xk + γk hk . Then increase k by 1 and go to Step 1.
Theorem AKM (see [40, Theorem 3.1]). Suppose that the assumptions (C1)-(C4) hold.
Then the two sequences {xk } and {zn } in Algorithm AKM converges to the same point x∗
which is a solution of the BVIP.
It is well known than an important approach to the BVIP is the Tikhonov regularization method. The main idea of this method for monotone variational inequalities is to add
a strongly monotone operator depending on a parameter to the cost operator to obtain a
parameterized strongly monotone variational inequality, which is uniquely solved. By letting
4
the parameter to a suitable limit, the sequence of the solutions of the regularized problems
will tend to the solution of the original problem. This result allows that the Tikhonov regularization method can be used to solve bilevel monotone variational inequalities. Recently,
in [5,21,27] the Tikhonov method with generalized regularization operators and bifunctions is
edtended to pseudomonotone variational inequalities and equilibrium problems, respectively.
However in this case, the regularized subproblems, may fail to be strongly monotone, even
pseudomonotone, since the sum of a strongly monotone operator and a pseudomonotone operator, in general, is not pseudomonotone. In our opinion, the existing methods that require some
monotonicity properties cannot be applied to solve the regularized subvariational inequalities.
Therefore the above Theorem AKM shows that the Algorithm AKM (i.e., an extragradienttyle algorithm) is an efficient approach for directly solving bilevel pseudomonotone variational
inequalities.
Motivated and inspired by the above facts, we introduce and analyze a hybrid extragradient
algorithm for solving bilevel pseudomonotone variational inequalities with multiple solutions
in a real Hilbert space. The proposed algorithm is based on Korpelevich’s extragradient
method (see [2]), Mann’s iteration method, hybrid steepest-descent method (see [19,31]) and
viscosity approximation method (see [15,41]) (including Halpern’s iteration method). Under
some mild conditions, the strong convergence of the iteration sequences generated by the
proposed algorithm is derived. Our results improve and extend the corresponding results
announced by some others, e.g., Anh, Kim amd Muu [40, Theorem 3.1].
2. Preliminaries
Throughout this paper, we assume that H is a real Hilbert space whose inner product and
norm are denoted by h·, ·i and k · k, respectively. Let C 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 }, i.e.,
ωw (xn ) := {x ∈ H : xni * x for some subsequence {xni } of {xn }}.
Recall that a mapping A : C → H is called
(i) monotone if
hAx − Ay, x − yi ≥ 0,
∀x, y ∈ C;
(ii) η-strongly monotone if there exists a constant η > 0 such that
hAx − Ay, x − yi ≥ ηkx − yk2 ,
∀x, y ∈ C;
(iii) α-inverse-strongly monotone if there exists a constant α > 0 such that
hAx − Ay, x − yi ≥ αkAx − Ayk2 ,
5
∀x, y ∈ C.
It is obvious that if A is α-inverse-strongly monotone, then A is monotone and α1 -Lipschitz
continuous.
The metric (or nearest point) projection from H onto C is the mapping PC : H → C
which assigns to each point x ∈ H the unique point PC x ∈ C satisfying the property
kx − PC xk = inf kx − yk =: d(x, C).
y∈C
Some important properties of projections are gathered in the following proposition.
Proposition 2.1. For given x ∈ H and z ∈ C:
(i) z = PC x ⇔ hx − z, y − zi ≤ 0, ∀y ∈ C;
(ii) z = PC x ⇔ kx − zk2 ≤ kx − yk2 − ky − zk2 , ∀y ∈ C;
(iii) hPC x − PC y, x − yi ≥ kPC x − PC yk2 , ∀y ∈ H.
Consequently, PC is nonexpansive and monotone.
If A is an α-inverse-strongly monotone mapping of C into H, then it is obvious that A is
continuous. We also have that, for all u, v ∈ C and λ > 0,
1
-Lipschitz
α
k(I − λA)u − (I − λA)vk2 ≤ ku − vk2 + λ(λ − 2α)kAu − Avk2 .
So, if λ ≤ 2α, then I − λA is a nonexpansive mapping from C to H.
Definition 2.1. A mapping T : H → H is said to be:
(a) nonexpansive if
kT x − T yk ≤ kx − yk, ∀x, y ∈ H;
(b) firmly nonexpansive if 2T − I is nonexpansive, or equivalently, if T is 1-inverse strongly
monotone (1-ism),
hx − y, T x − T yi ≥ kT x − T yk2 , ∀x, y ∈ H;
alternatively, T is firmly nonexpansive if and only if T can be expressed as
1
T = (I + S),
2
where S : H → H is nonexpansive; projections are firmly nonexpansive.
It can be easily seen that if T is nonexpansive, then I − T is monotone. It is also easy to
see that a projection PC is 1-ism. Inverse strongly monotone (also referred to as co-coercive)
operators have been applied widely in solving practical problems in various fields.
We need some facts and tools in a real Hilbert space H which are listed as lemmas below.
Lemma 2.1. Let X be a real inner product space. Then there holds the following
inequality
kx + yk2 ≤ kxk2 + 2hy, x + yi, ∀x, y ∈ X.
6
It is not hard to prove the following lemma which will be used in the sequel. Here we omit
its proof.
Lemma 2.2. Let F : H → H be a κ-Lipschitzian and η-strongly monotone operator with
positive constants κ, η > 0 and V : H → H be an l-Lipschitzian mapping. If µη − γl > 0 for
µ, γ ≥ 0, then µF − γV is (µη − γl)-strongly monotone, that is,
h(µF − γV )x − (µF − γV )y, x − yi ≥ (µη − γl)kx − yk2 ,
∀x, y ∈ H.
Lemma 2.3. Let H be a real Hilbert space. Then the following hold:
(a) kx − yk2 = kxk2 − kyk2 − 2hx − y, yi for all x, y ∈ H;
(b) kλx + µyk2 = λkxk2 + µkyk2 − λµkx − yk2 for all x, y ∈ H and λ, µ ∈ [0, 1] with
λ + µ = 1;
(c) If {xk } is a sequence in H such that xk * x, it follows that
lim sup kxk − yk2 = lim sup kxk − xk2 + kx − yk2 ,
k→∞
∀y ∈ H.
k→∞
Let C be a nonempty closed convex subset of a real Hilbert space H. We introduce some
notations. Let λ be a number in (0, 1] and let µ > 0. Associating with a nonexpansive
mapping S : C → H, we define the mapping S λ : C → H by
S λ x := Sx − λµF (Sx),
∀x ∈ C,
where F : H → H is an operator such that, for some positive constants κ, η > 0, F is
κ-Lipschitzian and η-strongly monotone on H; that is, F satisfies the conditions:
kF x − F yk ≤ κkx − yk and hF x − F y, x − yi ≥ ηkx − yk2
for all x, y ∈ H.
Lemma 2.4 (see [19, Lemma 3.1]). S λ is a contraction provided 0 < µ <
kS λ x − S λ yk ≤ (1 − λτ )kx − yk,
where τ = 1 −
q
2η
;
κ2
that is,
∀x, y ∈ C,
1 − µ(2η − µκ2 ) ∈ (0, 1].
Lemma 2.5 (see [7, Demiclosedness principle]). Let C be a nonempty closed convex subset
of a real Hilbert space H. Let S be a nonexpansive self-mapping on C with Fix(S) 6= ∅. Then
I − S is demiclosed. That is, whenever {xn } is a sequence in C weakly converging to some
x ∈ C and the sequence {(I −S)xn } strongly converges to some y, it follows that (I −S)x = y.
Here I is the identity operator of H.
Lemma 2.6 (see [19, Lemma 2.1]). Let {an } be a sequence of nonnegative numbers
satisfying the condition
an+1 ≤ (1 − αn )an + αn βn , ∀n ≥ 0,
7
where {αn } and {βn } are sequences of real numbers such that
P
(i) {αn } ⊂ [0, 1] and ∞
n=0 αn = ∞, or equivalently,
∞
Y
(1 − αn ) := lim
n→∞
n=0
(ii) lim supn→∞ βn ≤ 0, or
Then, limn→∞ an = 0.
P∞
n=0
n
Y
(1 − αk ) = 0;
k=0
|αn βn | < ∞.
3. Iterative Algorithm and Convergence Criteria
Let C be a nonempty closed convex subset of a real Hilbert space H. Throughout this
section, we always assume the following
F : H → H is a κ-Lipschitzian and η-strongly monotone operator with positive constants
κ, η > 0, and V : H → H is an l-Lipschitzianqmapping;
0 < µ < κ2η2 and 0 ≤ γl < τ with τ = 1 − 1 − µ(2η − µκ2 );
A : C → H and B : H → H are two mappings such that the hypotheses (H1)-(H4) hold:
(H1) B is pseudomonotone on H;
(H2) A is β-inverse-strongly monotone on C;
(H3) B is L-Lipschitz continuous on H;
(H4) VI(C, B) 6= ∅.
Next, we introduce and consider the following BVIP, which may have multiple solutions.
Problem 3.1. We find x∗ ∈ VI(C, B) such that
hAx∗ , x − x∗ i ≥ 0,
∀x ∈ VI(C, B),
(3.1)
where VI(C, B) denotes the set of solutions of the VIP: Find y ∗ ∈ C such that
hBy ∗ , y − y ∗ i ≥ 0,
∀y ∈ C.
(3.2)
Algorithm 3.1. Initialization. Choose u ∈ H, x0 ∈ H, k = 0, 0 < λ ≤ 2β, positive
sequences {δk }, {λk }, {αk }, {βk }, {γk } and {¯k } such that

















lim δk = 0,
k→∞
∞
X
¯k < ∞,
k=0
αk + βk + γk = 1 ∀k ≥ 0,
∞
X
αk = ∞,
k=0
lim αk = 0, lim βk = ξ ∈ (0, 12 ], lim λk = 0, λk ≤
k→∞
k→∞
k→∞
8
1
L
∀k ≥ 0.
Step 1. Compute





vk = αk γV xk + γk xk + ((1 − γk )I − αk µF )xk ,
yk := PC (vk − λk Bvk ),
zk := PC (vk − λk Byk ).
Step 2. Inner loop j = 0, 1, .... Compute















xk,0 := zk − λAzk ,
yk,j := PC (xk,j − δj Bxk,j ),
xk,j+1 := αj xk,0 + βj xk,j + γj PC (xk,j − δj Byk,j ).
If kxk,j+1 − PVI(C,B) xk,0 k ≤ ¯k then set hk := xk,j+1 and go to Step 3.
Otherwise, increase j by 1 and repeat the inner loop Step 2.
Step 3. Set xk+1 := αk u + βk xk + γk hk . Then increase k by 1 and go to Step 1.
Let C be a nonempty closed convex subset of H, B : C → H be monotone and L-Lipschitz
continuous on C, and S : C → C be a nonexpansive mapping such that VI(C, B)∩Fix(S) 6= ∅.
Let the sequences {xn } and {yn } be generated by





x0 ∈ C chosen arbitrarily,
yk = PC (xk − δk Bxk ),
xk+1 = αk x0 + βk xk + γk SPC (xk − δk Byk ) ∀k ≥ 0,
where {αk }, {βk }, {γk } and {δk } satisfy the following conditions:



















δk > 0 ∀k ≥ 0, lim δk = 0,
k→∞
αk + βk + γk = 1 ∀k ≥ 0,
∞
X
αk = ∞, lim αk = 0,
k→∞
k=0
0 < lim inf βk ≤ lim supβk < 1.
k→∞
k→∞
Under these conditions, Yao, Liou and Yao [23] proved that the sequences {xk } and {yk }
converge to the same point PVI(C,B)∩Fix(S) x0 .
Applying these iteration sequences with S being the identity mapping, we have the following lemma.
Lemma 3.1. Suppose that the hypotheses (H1)-(H4) hold. Then the sequence {xk,j }
generated by Algorithm 3.1 converges strongly to the point PVI(C,B) (zk − λAzk ) as j → ∞.
Consequently, we have
khk − PVI(C,B) (zk − λAzk )k ≤ ¯k
∀k ≥ 0.
In the sequel we always suppose that the inner loop in the Algorithm 3.1 terminates after
a finite number of steps. This assumption, by Lemma 3.1, is satisfied when B is monotone on
C.
9
Lemma 3.2. Let sequences {vk }, {yk } and {zk } be generated by Algorithm 3.1, B be
L-Lipschitzian and pseudomonotone on H, and p ∈ VI(C, B). Then, we have
kzk − pk2 ≤ kvk − pk2 − (1 − λk L)kvk − yk k2 − (1 − λk L)kyk − zk k2 .
(3.3)
Proof. Let p ∈ VI(C, B). That means
hBp, x − pi ≥ 0,
∀x ∈ C.
Then, for each λk > 0, p satisfies the fixed point equation
p = PC (p − λk Bp).
Since B is pseudomonotone on H and p ∈ VI(C, B), we have
hBp, yk − pi ≥ 0 ⇒ hByk , yk − pi ≥ 0.
Then, applying Proposition 2.1 (ii) with vk − λk Byk and p, we obtain
kzk − pk2 ≤ kvk − λk Byk − pk2 − kvk − λk Byk − zk k2
= kvk − pk2 − 2λk hByk , vk − pi + λ2k kByk k2 − kvk − zk k2
− λ2k kByk k2 + 2λk hByk , vk − zk i
= kvk − pk2 − kvk − zk k2 + 2λk hByk , p − zk i
= kvk − pk2 − kvk − zk k2 + 2λk hByk , p − yk i + 2λk hByk , yk − zk i
≤ kvk − pk2 − kvk − zk k2 + 2λk hByk , yk − zk i.
(3.4)
Applying Proposition 2.1 (i) with vk − λk Bvk and zk , we also have
hvk − λk Bvk − yk , zk − yk i ≤ 0.
Combining this inequality with (3.4) and observing that B is L-Lipschitz continuous on H,
we obtain
kzk − pk2 ≤ kvk − pk2 − k(vk − yk ) + (yk − zk )k2 + 2λk hByk , yk − zk i
= kvk − pk2 − kvk − yk k2 − kyk − zk k2 − 2hvk − yk , yk − zk i
+ 2λk hByk , yk − zk i
= kvk − pk2 − kvk − yk k2 − kyk − zk k2 − 2hvk − λk Byk − yk , yk − zk i
= kvk − pk2 − kvk − yk k2 − kyk − zk k2 − 2hvk − λk Bvk − yk , yk − zk i
+ 2λk hBvk − Byk , zk − yk i
≤ kvk − pk2 − kvk − yk k2 − kyk − zk k2 + 2λk hBvk − Byk , zk − yk i
≤ kvk − pk2 − kvk − yk k2 − kyk − zk k2 + 2λk kBvk − Byk kkzk − yk k
≤ kvk − pk2 − kvk − yk k2 − kyk − zk k2 + 2λk Lkvk − yk kkzk − yk k
≤ kvk − pk2 − kvk − yk k2 − kyk − zk k2 + λk L(kvk − yk k2 + kzk − yk k2 )
≤ kvk − pk2 − (1 − λk L)kvk − yk k2 − (1 − λk L)kyk − zk k2 .
(3.5)
2
This completes the proof.
10
Lemma 3.3. Suppose that the hypotheses (H1)-(H4) hold and that VI(VI(C, B), A) 6= ∅.
Then the sequence {xk } generated by Algorithm 3.1 is bounded.
Proof. Since limk→∞ αk = 0, limk→∞ βk = ξ ∈ (0, 12 ] and αk + βk + γk = 1, we get
lim (1 − γk ) = lim (αk + βk ) = ξ.
k→∞
k→∞
Hence, we may assume, without loss of generality, that 0 <
Take an arbitrary p ∈ VI(VI(C, B), A). Then we have
hAp, x − pi ≥ 0,
αk
1−γk
≤ 1 for all k ≥ 0.
∀x ∈ VI(C, B),
which implies
p = PVI(C,B) (p − λAp).
Then, it follows from (2.1), Proposition 2.1 (iii), β-inverse strong monotonicity of A, and
0 < λ ≤ 2β that
kPVI(C,B) (zk − λAzk ) − pk2 = kPVI(C,B) (zk − λAzk ) − PVI(C,B) (p − λAp)k2
≤ kzk − λAzk − p + λApk2
≤ kzk − pk2 + λ(λ − 2β)kAzk − Apk2
≤ kzk − pk2 .
(3.6)
Furthermore, from Algorithm 3.1 and Lemma 2.4, we have
kvk − pk = kαk γV xk + γk xk + ((1 − γk )I − αk µF )xk − pk
= kαk (γV xk − µF p) + γk (xk − p) + ((1 − γk )I − αk µF )xk
− ((1 − γk )I − αk µF )pk
≤ αk γkV xk − V pk + αk k(γV − µF )pk + γk kxk − pk
+ k((1 − γk )I − αk µF )xk − ((1 − γk )I − αk µF )pk
≤ αk γlkxk − pk + αk k(γV − µF )pk + γk kxk − pk
αk µ
αk µ
+ (1 − γk )k(I − 1−γ
F )xk − (I − 1−γ
F )pk
k
k
≤ αk γlkxk − pk + αk k(γV − µF )pk + γk kxk − pk
αk
+ (1 − γk )(1 − 1−γ
τ )kxk − pk
k
= αk γlkxk − pk + αk k(γV − µF )pk + γk kxk − pk
+ (1 − γk − αk τ )kxk − pk
= (1 − αk (τ − γl))kxk − pk + αk k(γV − µF )pk
)pk
= (1 − αk (τ − γl))kxk − pk + αk (τ − γl) k(γVτ−µF
−γl
)pk
≤ max{kxk − pk, k(γVτ−µF
}.
−γl
11
(3.7)
Utilizing (3.5)-(3.7) and the assumptions 0 < λ ≤ 2β,
P∞
¯k
k=0 < ∞ we obtain that
kxk+1 − pk = kαk u + βk xk + γk hk − pk
≤ αk ku − pk + βk kxk − pk + γk khk − pk
≤ αk ku − pk + βk kxk − pk + γk khk − PVI(C,B) (zk − λAzk )k
+ γk kPVI(C,B) (zk − λAzk ) − pk
≤ αk ku − pk + βk kxk − pk + γk ¯k + γk kzk − pk
≤ αk ku − pk + βk kxk − pk + γk ¯k + γk kvk − pk
)pk
≤ αk ku − pk + βk kxk − pk + γk ¯k + γk max{kxk − pk, k(γVτ−µF
}
−γl
k(γV −µF )pk
≤ αk ku − pk + (βk + γk ) max{kxk − pk,
} + ¯k
τ −γl
k(γV −µF )pk
} + ¯k
= αk ku − pk + (1 − αk ) max{kxk − pk,
τ −γl
k(γV −µF )pk
≤ max{kxk − pk, ku − pk,
} + ¯k
τ −γl
)pk
}+
≤ max{kx0 − pk, ku − pk, k(γVτ−µF
−γl
)pk
}+
≤ max{kx0 − pk, ku − pk, k(γVτ−µF
−γl
k
X
¯j
j=0
∞
X
¯k
k=0
< ∞,
which shows that the sequence {xk } is bounded, and so are the sequences {vk }, {yk } and {zk }.
2
Lemma 3.4 (see [32]). Let {xk } and {yk } be two bounded sequences in a real Banach
space X. Let {βk } be a sequence in [0, 1]. Suppose that









0 < lim inf βk ≤ lim supβk < 1,
k→∞
k→∞
xk+1 = (1 − βk )yk + βk xk ,
lim sup(kyk+1 − yk k − kxk+1 − xk k) ≤ 0.
k→∞
Then, limk→∞ kyk − xk k = 0.
Lemma 3.5. Suppose that the hypotheses (H1)-(H4) and that the sequences {vk }, {yk }
and {zk } are generated by Algorithm 3.1. Then, we have
kzk+1 − zk k ≤ (1 + λk+1 L)kvk+1 − vk k + λk kByk k
+ λk+1 (kBvk+1 k + kByk+1 k + kBvk k).
Moreover
lim kzk+1 − zk k = lim kvk+1 − vk k = 0.
k→∞
k→∞
Proof. Since B is L-Lipschitzian on H, for each x, y ∈ H, we have
k(I − λk B)x − (I − λk B)yk = kx − y − λk (Bx − By)k
≤ kx − yk + λk kBx − Byk
≤ (1 + λk L)kx − yk.
12
(3.8)
Combining this inequality with Proposition 2.1 (iii), we have
kzk+1 − zk k = kPC (vk+1 − λk+1 Byk+1 ) − PC (vk − λk Byk )k
≤ k(vk+1 − λk+1 Byk+1 ) − vk + λk Byk k
= k(vk+1 − λk+1 Bvk+1 ) − (vk − λk+1 Bvk )
+ λk+1 (Bvk+1 − Byk+1 − Bvk ) + λk Byk k
≤ (1 + λk+1 L)kvk+1 − vk k + λk kByk k
+ λk+1 (kBvk+1 k + kByk+1 k + kBvk k).
(3.9)
This is the desired result (3.8).
Now we denote xk+1 = (1 − βk )wk + βk xk . Then, we have
u+γk+1 hk+1
u+γk hk
− αk 1−β
wk+1 − wk = αk+11−β
k+1
k
αk+1
γk+1
αk
= ( 1−β
−
)u
+
(
−
1−βk
1−βk+1
k+1
γk
)hk
1−βk
+
γk+1
(hk+1
1−βk+1
− hk ).
(3.10)
Note that, for 0 < λ ≤ 2β, we have from (2.1) that
kPVI(C,B) (zk+1 − λAzk+1 ) − PVI(C,B) (zk − λAzk )k2
≤ k(zk+1 − λAzk+1 ) − (zk − λAzk )k2
≤ kzk+1 − zk k2 + λ(λ − 2β)kAzk+1 − Azk k2
≤ kzk+1 − zk k2 .
Then, combining (3.11) with λ ≤ 2β and (3.10) we get
kwk+1 − wk k − kxk+1 − xk k
αk+1
γk+1
γk+1
γk
αk
− 1−β
|kuk + | 1−β
− 1−β
|khk k + 1−β
khk+1 − hk k
≤ | 1−β
k+1
k
k+1
k
k+1
− kxk+1 − xk k
γk+1
αk+1
γk
αk
− 1−β
|kuk + | 1−β
− 1−β
|(kPVI(C,B) (zk − λAzk )k + ¯k )
≤ | 1−β
k+1
k
k+1
k
γk+1
+ 1−βk+1 kPVI(C,B) (zk+1 − λAzk+1 ) − PVI(C,B) (zk − λAzk )k
γk+1
+ 1−β
(kPVI(C,B) (zk+1 − λAzk+1 ) − hk+1 k
k+1
+ kPVI(C,B) (zk − λAzk ) − hk k) − kxk+1 − xk k
γk+1
αk+1
γk
αk
≤ | 1−β
− 1−β
|kuk + | 1−β
− 1−β
|(kPVI(C,B) (zk − λAzk )k + ¯k )
k+1
k
k+1
k
γk+1
γk+1
+ 1−βk+1 kzk+1 − zk k + 1−βk+1 (¯k+1 + ¯k ) − kxk+1 − xk k.
Hence,
kwk+1 − wk k − kxk+1 − xk k
γk+1
αk+1
αk
− 1−β
|kuk + | 1−β
−
≤ | 1−β
k+1
k
k+1
γk
|(kPVI(C,B) (zk
1−βk
γk+1
(¯
+ ¯k )
1−βk+1 k+1
γk+1 (1+λk+1 L)
kvk+1 − vk k +
1−βk+1
γk+1
(λk+1 (kBvk+1 k + kByk+1 k
1−βk+1
+
+
− kxk+1 − xk k
αk+1
γk+1
αk
≤ | 1−β
− 1−β
|kuk + | 1−β
−
k+1
k
k+1
+
+
− λAzk )k + ¯k )
+ kBvk k) + λk kByk k)
γk
|(kPVI(C,B) (zk − λAzk )k + ¯k )
1−βk
γk+1
kxk+1 − xk k + 1−β
(¯k+1 + ¯k )
k+1
γk+1 (1+λk+1 L)
kvk+1 − vk k −
1−βk+1
γk+1
(λk+1 (kBvk+1 k + kByk+1 k
1−βk+1
13
+ kBvk k) + λk kByk k).
(3.11)
On the other hand, we define w̃k =
calculations show that
w̃k+1 − w̃k =
=
=
=
vk −γk xk
,
1−γk
which implies that vk = (1 − γk )w̃k + γk xk . Simple
vk+1 −γk+1 xk+1
−γk xk
− vk1−γ
1−γk+1
k
αk+1 γV xk+1 +((1−γk+1 )I−αk+1 µF )xk+1
k )I−αk µF )xk
− αk γV xk +((1−γ
1−γk+1
1−γk
αk+1
αk+1
αk
αk
γV xk+1 − 1−γ
γV xk + xk+1 − xk + 1−γ
µF xk − 1−γ
µF xk+1
1−γk+1
k
k
k+1
αk+1
αk
(γV xk+1 − µF xk+1 ) + 1−γk (µF xk − γV xk ) + xk+1 − xk .
1−γk+1
So, it follows that
kw̃k+1 − w̃k k
αk+1
αk
≤ 1−γ
(kγV xk+1 k + kµF xk+1 k) + 1−γ
(kµF xk k + kγV xk k) + kxk+1 − xk k
k+1
k
αk+1
αk
≤ kxk+1 − xk k + ( 1−γk+1 + 1−γk )M0 ,
(3.12)
where supk≥0 {kµF xk k + kγV xk k} ≤ M0 for some M0 > 0. In the meantime, from vk =
(1 − γk )w̃k + γk xk , together with (3.12), we get
kvk+1 − vk k = kγk+1 xk+1 + (1 − γk+1 )w̃k+1 − (γk xk + (1 − γk )w̃k )k
= k(1 − γk+1 )(w̃k+1 − w̃k ) − (γk+1 − γk )w̃k
+ γk+1 (xk+1 − xk ) + (γk+1 − γk )xk k
≤ (1 − γk+1 )kw̃k+1 − w̃k k + γk+1 kxk+1 − xk k + |γk+1 − γk |kxk − w̃k k
αk+1
αk
+ 1−γ
)M0 ]
≤ (1 − γk+1 )[kxk+1 − xk k + ( 1−γ
k+1
k
+ γk+1 kxk+1 − xk k + |γk+1 − γk |kxk − w̃k k
αk+1
αk
+ 1−γ
)M0 + |γk+1 − γk |kxk − w̃k k.
≤ kxk+1 − xk k + ( 1−γ
k+1
k
(3.13)
Combining (3.11) and (3.13) we have
kwk+1 − wk k − kxk+1 − xk k
αk+1
γk+1
αk
≤ | 1−β
− 1−β
|kuk + | 1−β
−
k+1
k
k+1
+ ¯k )
=
+ ¯k )
γk
|(kPVI(C,B) (zk − λAzk )k
1−βk
γk+1 (1+λk+1 L)
αk+1
αk
+
[kxk+1 − xk k + ( 1−γk+1 + 1−γ
)M0
1−βk+1
k
γk+1
+ |γk+1 − γk |kxk − w̃k k] − kxk+1 − xk k + 1−βk+1 (¯k+1 + ¯k )
γk+1
+ 1−β
(λk+1 (kBvk+1 k + kByk+1 k + kBvk k) + λk kByk k)
k+1
αk+1
γk+1
γk
αk
|kuk + | 1−β
− 1−β
|(kPVI(C,B) (zk − λAzk )k
| 1−βk+1 − 1−β
k
k+1
k
(1+λk+1 L)
αk+1
αk
+ γk+11−β
[( 1−γ
+ 1−γ
)M0 + |γk+1 − γk |kxk − w̃k k]
k+1
k+1
k
γk+1 (1+λk+1 L)
γk+1
+ ( 1−βk+1
− 1)kxk+1 − xk k + 1−β
(¯k+1 + ¯k )
k+1
γk+1
+ 1−βk+1 (λk+1 (kBvk+1 k + kByk+1 k + kBvk k) + λk kByk k).
(3.14)
From the assumptions αk + βk + γk = 1, limk→∞ βk = ξ ∈ (0, 12 ], limk→∞ αk = 0 and
limk→∞ λk = 0, it follows that limk→∞ |γk+1 − γk | = 0,
lim |
k→∞
αk+1
αk
γk+1
γk
−
| = lim |
−
| = 0 and
k→∞
1 − βk+1 1 − βk
1 − βk+1 1 − βk
14
lim
k→∞
γk+1 (1 + λk+1 L)
= 1.
1 − βk+1
Combining these equalities with (3.14), we obtain from Lemma 3.3 and limk→∞ ¯k = 0 that
lim sup(kwk+1 − wk k − kxk+1 − xk k) ≤ 0.
k→∞
Now applying Lemma 3.4, we have
lim kwk − xk k = 0.
k→∞
Hence by xk+1 = (1 − βk )wk + βk xk , we deduce that
lim kxk+1 − xk k = lim (1 − βk )kwk − xk k = 0,
k→∞
k→∞
(3.15)
which together with limk→∞ λk = 0, (3.8) and (3.13), implies that
lim kvk+1 − vk k = 0 and
k→∞
lim kzk+1 − zk k = 0.
k→∞
(3.16)
Lemma 3.6. Suppose that the hypotheses (H1)-(H4) hold and that VI(VI(C, B), A) 6= ∅.
Then for any p ∈ VI(VI(C, B), A) we have
kxk+1 − pk2 ≤ αk ku − pk2 + βk kxk − pk2 + γk kvk − pk2 + 2γk ¯k kzk − pk
+ γk ¯2k − γk (1 − λk L)(kvk − yk k2 + kyk − zk k2 ).
Moreover


k→∞

k→∞
(3.17)
lim kPVI(C,B) (zk − λk Azk ) − zk k = 0,
lim kPVI(C,B) (yk − λk Ayk ) − yk k = 0.
Proof. By Lemma 3.1, we know that
lim xk,j = PVI(C,B) (zk − λAzk ),
j→∞
which together with 0 < λ ≤ 2β, inequality (3.3), limk→∞ βk = ξ ∈ (0, 12 ] and p ∈ VI(VI(C, B), A),
implies that
kxk+1 − pk2 = kαk u + βk xk + γk hk − pk2
≤ αk ku − pk2 + βk kxk − pk2 + γk khk − pk2
≤ αk ku − pk2 + βk kxk − pk2 + γk (kPVI(C,B) (zk − λAzk ) − pk + ¯k )2
= αk ku − pk2 + βk kxk − pk2
+ γk (kPVI(C,B) (zk − λAzk ) − PVI(C,B) (p − λAp)k + ¯k )2
≤ αk ku − pk2 + βk kxk − pk2 + γk (k(zk − λAzk ) − (p − λAp)k + ¯k )2
≤ αk ku − pk2 + βk kxk − pk2 + γk (kzk − pk + ¯k )2
= αk ku − pk2 + βk kxk − pk2 + γk kzk − pk2 + 2γk ¯k kzk − pk + γk ¯2k
≤ αk ku − pk2 + βk kxk − pk2 + 2γk ¯k kzk − pk + γk ¯2k
+ γk (kvk − pk2 − (1 − λk L)kvk − yk k2 − (1 − λk L)kyk − zk k2 )
= αk ku − pk2 + βk kxk − pk2 + γk kvk − pk2 + 2γk ¯k kzk − pk + γk ¯2k
− γk (1 − λk L)(kvk − yk k2 + kyk − zk k2 ).
15
(3.18)
On the other hand, from Algorithm 3.1 we have
kvk − pk2 = kαk γV xk + γk xk + ((1 − γk )I − αk µF )xk − pk2
= kαk (γV xk − µF p) + γk (xk − p) + ((1 − γk )I − αk µF )xk
− ((1 − γk )I − αk µF )pk2
= kαk γ(V xk − V p) + αk (γV p − µF p) + γk (xk − p)
+ ((1 − γk )I − αk µF )xk − ((1 − γk )I − αk µF )pk2
≤ kαk γ(V xk − V p) + γk (xk − p) + ((1 − γk )I − αk µF )xk
− ((1 − γk )I − αk µF )pk2 + 2αk h(γV − µF )p, vk − pi
≤ [αk γkV xk − V pk + γk kxk − pk + k((1 − γk )I − αk µF )xk
− ((1 − γk )I − αk µF )pk]2 + 2αk h(γV − µF )p, vk − pi
≤ [αk γlkxk − pk + γk kxk − pk + (1 − γk − αk τ )kxk − pk]2
+ 2αk h(γV − µF )p, vk − pi
= [αk τ γlτ kxk − pk + γk kxk − pk + (1 − γk − αk τ )kxk − pk]2
+ 2αk h(γV − µF )p, vk − pi
2
≤ αk τ (γl)
kxk − pk2 + γk kxk − pk2 + (1 − γk − αk τ )kxk − pk2
τ2
+ 2αk h(γV − µF )p, vk − pi
2
2
= (1 − αk τ −(γl)
)kxk − pk2 + 2αk h(γV − µF )p, vk − pi
τ
2
≤ kxk − pk + 2αk h(γV − µF )p, vk − pi.
(3.19)
Combining (3.18) and (3.19), we get
kxk+1 − pk2
≤ αk ku − pk2 + βk kxk − pk2 + γk kvk − pk2 + 2γk ¯k kzk − pk
+ γk ¯2k − γk (1 − λk L)(kvk − yk k2 + kyk − zk k2 )
≤ αk ku − pk2 + βk kxk − pk2 + γk [kxk − pk2 + 2αk h(γV − µF )p, vk − pi]
+ 2γk ¯k kzk − pk + γk ¯2k − γk (1 − λk L)(kvk − yk k2 + kyk − zk k2 )
≤ αk ku − pk2 + kxk − pk2 + 2αk k(γV − µF )pkkvk − pk + 2γk ¯k kzk − pk
+ γk ¯2k − γk (1 − λk L)(kvk − yk k2 + kyk − zk k2 ),
which immediately yields
γk (1 − λk L)(kvk − yk k2 + kyk − zk k2 )
≤ αk ku − pk2 + kxk − pk2 − kxk+1 − pk2 + 2αk k(γV − µF )pkkvk − pk
+ 2γk ¯k kzk − pk + γk ¯2k
≤ αk ku − pk2 + kxk − xk+1 k(kxk − pk + kxk+1 − pk) + 2αk k(γV − µF )pkkvk − pk
+ 2γk ¯k kzk − pk + γk ¯2k .
Since αk + βk + γk = 1, αk → 0, βk → ξ ∈ (0, 21 ], ¯k → 0, λk → 0 and kxk − xk+1 k → 0, from
the boundedness of {xk }, {vk } and {zk } we obtain
lim kvk − yk k = 0 and
k→∞
lim kyk − zk k = 0.
k→∞
In addition, it is clear from αk → 0 that as k → ∞,
kvk − xk k = kαk γV xk + γk xk + ((1 − γk )I − αk µF )xk − xk k
= kαk (γV xk − µF xk )k → 0.
16
(3.20)
That is,
lim kvk − xk k = 0.
(3.21)
k→∞
Taking into consideration that kvk − zk k ≤ kvk − yk k + kyk − zk k and kzk − xk k ≤ kzk − vk k +
kvk − xk k, we deduce from (3.20) and (3.21) that
lim kvk − zk k = 0 and
k→∞
lim kzk − xk k = 0.
(3.22)
k→∞
It is clear from (3.20) and (3.22) that
lim kxk − yk k = 0.
(3.23)
k→∞
Since A is β-inverse-strongly monotone, it is known from (2.1) that I − λA is a nonexpansive
mapping for 0 < λ ≤ 2β. Again by Proposition 2.1 (iii) and Lemma 3.1 we have
kPVI(C,B) (yk − λAyk ) − xk+1 k
≤ kPVI(C,B) (yk − λAyk ) − PVI(C,B) (zk − λAzk )k
+ kPVI(C,B) (zk − λAzk ) − xk+1 k
≤ k(I − λA)yk − (I − λA)zk k + kPVI(C,B) (zk − λAzk ) − xk+1 k
≤ kyk − zk k + αk kPVI(C,B) (zk − λAzk ) − uk
+ βk kPVI(C,B) (zk − λAzk ) − xk k + γk ¯k
≤ kyk − zk k + αk kPVI(C,B) (zk − λAzk ) − uk + ¯k
+ βk kPVI(C,B) (zk − λAzk ) − PVI(C,B) (yk − λAyk )k
+ βk kPVI(C,B) (yk − λAyk ) − yk k + βk kyk − xk k
≤ kyk − zk k + αk kPVI(C,B) (zk − λAzk ) − uk
+ ¯k + βk k(I − λA)zk − (I − λA)yk k
+ βk kPVI(C,B) (yk − λAyk ) − yk k + βk kyk − xk k
≤ kyk − zk k + αk kPVI(C,B) (zk − λAzk ) − uk + ¯k
+ βk kzk − yk k + βk kPVI(C,B) (yk − λAyk ) − yk k + βk kyk − xk k.
Consequently, from (3.24), we have
kPVI(C,B) (yk − λAyk ) − yk k
≤ kPVI(C,B) (yk − λAyk ) − xk+1 k + kxk+1 − xk k + kxk − yk k
≤ kyk − zk k + αk kPVI(C,B) (zk − λAzk ) − uk + ¯k
+ βk kzk − yk k + βk kPVI(C,B) (yk − λAyk ) − yk k + βk kyk − xk k
+ kxk+1 − xk k + kxk − yk k
= (1 + βk )kyk − zk k + αk kPVI(C,B) (zk − λAzk ) − uk + ¯k
+ βk kPVI(C,B) (yk − λAyk ) − yk k + (1 + βk )kyk − xk k + kxk+1 − xk k,
which immediately yields
kPVI(C,B) (yk − λAyk ) − yk k
1+βk
αk
kyk − zk k + 1−β
kPVI(C,B) (zk − λAzk ) − uk +
≤ 1−β
k
k
1+βk
1
+ 1−βk kyk − xk k + 1−β
kxk+1 − xk k.
k
17
¯k
1−βk
(3.24)
Since αk + βk + γk = 1, αk → 0, βk → ξ ∈ (0, 21 ], ¯k → 0, kyk − zk k → 0, kyk − xk k → 0 and
kxk+1 − xk k → 0 (due to (3.15), (3.20) and (3.23)), we conclude that
lim kPVI(C,B) (yk − λAyk ) − yk k = 0.
k→∞
(3.25)
From (2.1) and Proposition 2.1 (iii), it follows that
kPVI(C,B) (zk − λAzk ) − zk k
≤ kPVI(C,B) (zk − λAzk ) − PVI(C,B) (yk − λAyk )k
+ kPVI(C,B) (yk − λAyk ) − yk k + kyk − zk k
≤ k(I − λA)zk − (I − λA)yk k + kPVI(C,B) (yk − λAyk ) − yk k + kyk − zk k
≤ kzk − yk k + kPVI(C,B) (yk − λAyk ) − yk k + kyk − zk k
≤ kPVI(C,B) (yk − λAyk ) − yk k + 2kyk − zk k.
Utilizing the last inequality we obtain from (3.20) and (3.25) that
lim kPVI(C,B) (zk − λAzk ) − zk k = 0.
k→∞
(3.26)
2
This completes the proof.
Theorem 3.1. Suppose that the hypotheses (H1)-(H4) hold and that VI(VI(C, B), A) 6=
∅. Then the two sequences {xk } and {zk } in Algorithm 3.1 converge strongly to the same
point x∗ ∈ VI(VI(C, B), A) provided kxk+1 − xk k + ¯k = o(αk ), which is a unique solution to
the VIP
¯ − ξγV
¯ )x∗ − u, p − x∗ i ≥ 0, ∀p ∈ VI(VI(C, B), A),
h(I + ξµF
(3.27)
where ξ¯ = 1 − ξ ∈ [ 1 , 1).
2
Proof. Note that Lemma 3.3 shows the boundedness of {xk }. Since H is reflexive,
there is at least a weak convergence subsequence of {xk }. First, let us assert that ωw (xk ) ⊂
VI(VI(C, B), A). As a matter of fact, take an arbitrary w ∈ VI(VI(C, B), A). Then there
exists a subsequence {xki } of {xk } such that xki * w. From (3.23), we know that yki * w.
It is easy to see that the mapping PVI(C,B) (I − λA) : C → VI(C, B) ⊂ C is nonexpansive
because PVI(C,B) is nonexpansive and I − λA is nonexpansive for β-inverse-strongly monotone
mapping A with 0 < λ ≤ 2β. So, utilizing Lemma 2.5 and (3.25), we obtain
w = PVI(C,B) (w − λAw),
which leads to w ∈ VI(VI(C, B), A). Thus, the assertion is valid.
Also, note that 0 ≤ γl < τ and
µη ≥ τ ⇔ µη ≥ 1 −
q
q
1 − µ(2η − µκ2 )
⇔ 1 − µ(2η − µκ2 ) ≥ 1 − µη
⇔ 1 − 2µη + µ2 κ2 ≥ 1 − 2µη + µ2 η 2
⇔ κ2 ≥ η 2
⇔ κ ≥ η.
18
It is clear that
h(µF − γV )x − (µF − γV )y, x − yi ≥ (µη − γl)kx − yk2 ,
∀x, y ∈ H.
Hence, it follows from 0 ≤ γl < τ ≤ µη that µF − γV is (µη − γl)-strongly monotone. In the
meantime, it is clear that µF − γV is Lipschitzian with constant µκ + γl > 0. We define the
mapping Γ : H → H as below
1
Γ x = (µF − γV )x + ¯(x − u),
ξ
∀x ∈ H,
where u ∈ H and ξ¯ = 1 − ξ ∈ [ 21 , 1). Then it is easy to see that Γ is (µη − γl + 1ξ̄ )-strongly
monotone and Lipschitzian with constant µκ+γl + 1ξ̄ > 0. Thus, there exists a unique solution
x∗ ∈ VI(VI(C, B), A) to the VIP
1
h(µF − γV )x∗ + ¯(x∗ − u), p − x∗ i ≥ 0,
ξ
∀p ∈ VI(VI(C, B), A).
(3.28)
Next, let us show that xk * x∗ . Indeed, take an arbitrary p ∈ VI(VI(C, B), A). In terms
of Algorithm 3.1 and Lemma 2.1, we conclude from (3.3), (3.5) and the β-inverse-strong
monotonicity of A with λ ≤ 2β, that
kxk+1 − pk2 = kαk u + βk xk + γk hk − pk2
≤ kβk (xk − p) + γk (hk − p)k2 + 2αk hu − p, xk+1 − pi
≤ βk kxk − pk2 + γk khk − pk2 + 2αk hu − p, xk+1 − pi
≤ βk kxk − pk2 + γk (kPVI(C,B) (zk − λAzk ) − pk + ¯k )2 + 2αk hu − p, xk+1 − pi
= βk kxk − pk2 + γk (kPVI(C,B) (zk − λAzk ) − PVI(C,B) (p − λAp)k + ¯k )2
+ 2αk hu − p, xk+1 − pi
≤ βk kxk − pk2 + γk (k(I − λA)zk − (I − λA)pk + ¯k )2 + 2αk hu − p, xk+1 − pi
≤ βk kxk − pk2 + γk (kzk − pk + ¯k )2 + 2αk hu − p, xk+1 − pi
= βk kxk − pk2 + γk kzk − pk2 + γk ¯k (2kzk − pk + ¯k ) + 2αk hu − p, xk+1 − pi
≤ βk kxk − pk2 + γk kvk − pk2 + γk ¯k (2kzk − pk + ¯k ) + 2αk hu − p, xk+1 − pi.
(3.29)
Combining (3.19) and (3.29), we get
kxk+1 − pk2
≤ βk kxk − pk2 + γk kvk − pk2 + γk ¯k (2kzk − pk + ¯k ) + 2αk hu − p, xk+1 − pi
2
2
)kxk − pk2 + 2αk h(γV − µF )p, vk − pi]
≤ βk kxk − pk2 + γk [(1 − αk τ −(γl)
τ
+ γk ¯k (2kzk − pk + ¯k ) + 2αk hu − p, xk+1 − pi
2
2
= (βk + γk − αk γk τ −(γl)
)kxk − pk2 + 2αk γk h(γV − µF )p, vk − xk+1 i
τ
+ γk ¯k (2kzk − pk + ¯k ) + 2αk γk h(γV − µF )p + γ1k (u − p), xk+1 − pi
2
2
≤ (1 − αk γk τ −(γl)
)kxk − pk2 + 2αk k(γV − µF )pk(kvk − xk k + kxk − xk+1 k)
τ
+ γk ¯k (2kzk − pk + ¯k ) + 2αk γk h(γV − µF )p + γ1k (u − p), xk+1 − pi
≤ kxk − pk2 + 2αk k(γV − µF )pk(kvk − xk k + kxk − xk+1 k)
+ γk ¯k (2kzk − pk + ¯k ) + 2αk γk h(γV − µF )p + γ1k (u − p), xk+1 − pi,
19
(3.30)
which immediately yields
h(µF − γV )p + γ1k (p − u), xk+1 − pi
≤ 2α1k γk (kxk − pk2 − kxk+1 − pk2 ) + γ1k k(γV − µF )pk(kvk − xk k + kxk − xk+1 k)
¯k
+ 2α
(2kzk − pk + ¯k )
k
kxk −xk+1 k
≤ 2αk γk (kxk − pk + kxk+1 − pk) + γ1k k(γV − µF )pk(kvk − xk k + kxk − xk+1 k)
¯k
+ 2α
(2kzk − pk + ¯k ).
k
(3.31)
Since for any w ∈ ωw (xk ) there exists a subsequence {xki } of {xk } such that xki * w, we
deduce from (3.21), (3.31), γ1k → 1ξ̄ and kxk+1 − xk k + ¯k = o(αk ) that
h(µF − γV )p + 1ξ̄ (p − u), w − pi
= lim h(µF − γV )p + γ1k (p − u), xki − pi
i→∞
i
≤ lim suph(µF − γV )p +
k→∞
= lim suph(µF − γV )p +
k→∞
1
(p
γk
1
(p
γk
− u), xk − pi
− u), xk+1 − pi
−xk+1 k
≤ lim sup kxk2α
(kxk − pk + kxk+1 − pk)
k γk
k→∞
¯k
(2kzk − pk + ¯k )
+ lim sup γ1k k(γV − µF )pk(kvk − xk k + kxk − xk+1 k) + lim sup 2α
k
k→∞
k→∞
= 0.
So, it follows that
1
h(µF − γV )p + ¯(p − u), p − wi ≥ 0,
ξ
∀p ∈ VI(VI(C, B), A).
Since w ∈ ωw (xk ) ⊂ VI(VI(C, B), A), by Minty’s lemma [7] we have
1
h(µF − γV )w + ¯(w − u), p − wi ≥ 0,
ξ
∀p ∈ VI(VI(Ω , B), A);
that is, w is a solution of VIP (3.28). Utilizing the uniqueness of solutions of VIP (3.28),
we get w = x∗ , which hence implies that ωw (xk ) = {x∗ }. Therefore, it is known that {xk }
converges weakly to the unique solution x∗ ∈ VI(VI(C, B), A) of VIP (3.28).
Finally, let us show that kxk − x∗ k → 0 as k → ∞. Indeed, utilizing (3.30) with p = x∗ ,
we have
kxk+1 − x∗ k2
2
2
≤ (1 − αk γk τ −(γl)
)kxk − x∗ k2 + 2αk k(γV − µF )x∗ k(kvk − xk k + kxk − xk+1 k)
τ
+ γk ¯k (2kzk − x∗ k + ¯k ) + 2αk γk h(γV − µF )x∗ + γ1k (u − x∗ ), xk+1 − x∗ i
2
2
τ 2 −(γl)2
∗ 2
= (1 − αk γk τ −(γl)
)kx
−
x
k
+
α
γ
×
k
k
k
τ
τ
τ
2
∗
× τ 2 −(γl)
[
k(γV
−
µF
)x
k(kv
−
x
k
+
kx
k
k
k − xk+1 k)
2 γ
k
¯k
1
∗
∗
+ αk (2kzk − x k + ¯k ) + 2h(γV − µF )x + γk (u − x∗ ), xk+1 − x∗ i].
20
(3.32)
∞
1
Since αk → 0, αk + βk + γk = 1,
¯k = o(αk ) and xk * x∗ ,
k=0 αk = ∞, βk → ξ ∈ (0, 2 ], P
τ 2 −(γl)2
= ∞ and
from (3.15) and (3.21) we conclude that ∞
k=0 αk γk
τ
P
τ
2
∗
lim sup{ τ 2 −(γl)
2 [ γ k(γV − µF )x k(kvk − xk k + kxk − xk+1 k)
k
k→∞
+
¯k
(2kzk
αk
− x∗ k + ¯k ) + 2h(γV − µF )x∗ +
1
(u
γk
− x∗ ), xk+1 − x∗ i]} ≤ 0.
Therefore, applying Lemma 2.6 to (3.32), we obtain that kxk − x∗ k → 0 as k → ∞. Utilizing
(3.23) we also obtain that kzk − x∗ k → 0 as k → ∞. This completes the proof.
2
Remark 3.1. Theorem 3.1 extends, improves, supplements and develops Anh, Kim and
Muu [40, Theorem 3.1] in the following aspects.
(i) The problem of finding a solution x∗ of Problem 3.1 (BVIP) in Theorem 3.1 is very
different from the problem of finding a solution x∗ of Problem AKM because our BVIP generalizes Anh, Kim and Muu’s BVIP in [40, Theorem 3.1] from the space Rn to the general
Hilbert space, and extends Anh, Kim and Muu’s BVIP with only a solution to the setting of
the BVIP with multiple solutions.
(ii) The Algorithm 2.1 in [40] is extended to develop Algorithm 3.1 by virtue of Mann’s
iteration method, hybrid steepest descent method and viscosity approximation method. The
Algorithm 3.1 is more advantageous and more flexible than Algorithm 2.1 in [40] because it
involves solving the BVIP with multiple solutions, i.e., Problem 3.1.
(iii) The proof of our Theorem 3.1 is very different from the proof of Anh, Kim and Muu’s
Theorem 3.1 [40] because the proof of our Theorem 3.1 makes use of the nonexpansivity of
the combination mapping I − λA for inverse-strongly monotone mapping A (see inequality
(2.1)), the contraction coefficient estimate for the composite mapping S λ (see Lemma 2.4),
the demiclosedness principle for nonexpansive mappings (see Lemma 2.5), the convergence
criteria for nonnegative real sequences (see Lemma 2.6) and Suzuki’s lemma (see Lemma 3.4).
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