Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2011, Article ID 921835, 15 pages
doi:10.1155/2011/921835
Research Article
Sensitivity Analysis for a System of Generalized
Nonlinear Mixed Quasi Variational Inclusions with
H-Monotone Operators
Han-Wen Cao
Department of Science, Nanchang Institute of Technology, Nanchang 330099, China
Correspondence should be addressed to Han-Wen Cao, chwhappy@163.com
Received 20 March 2011; Revised 2 June 2011; Accepted 13 June 2011
Academic Editor: Ya Ping Fang
Copyright q 2011 Han-Wen Cao. 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 existence of the solution for a new system of generalized nonlinear mixed quasi variational
inclusions with H-monotone operators is proved by using implicit resolvent technique, and the
sensitivity analysis of solution in Hilbert spaces is given. Our results improve and generalize some
results of the recent ones.
1. Introduction
Sensitivity analysis of solution for variational inequalities and variational inclusions has been
studied by many authors via quite different technique. See 1–7 and the reference therein.
In 2004, Agarwal et al. 1 introduced and studied the following problem which is
called the system of parametric generalized nonlinear mixed quasi variational inclusions.
Let H be a real Hilbert space endowed with the product ·, · and norm · ,
respectively. Let Ω and Λ be two nonempty open subsets of H in which the parametric ω
and λ take values. Let M : H × Ω → 2H and N : H × Λ → 2H be two maximal monotone
mappings with respect to the first argument. H1 , S : H × Ω → H and H2 , T : H × Λ → H
be nonlinear single-valued mappings. The system of parametric generalized nonlinear mixed
quasi variational inclusions problem 1 is to find x, y ∈ H × H such that
0 ∈ x − y ρ H1 y, ω , S y, ω ρMx, ω,
0 ∈ y − x γH2 x, λ T x, λ γN y, λ ,
where ρ > 0 and γ > 0 are two constants.
1.1
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In this paper, we introduce a new system of parametric generalized nonlinear mixed
quasi variational inclusions problem.
For each ω ∈ Ω, λ ∈ Λ, find x xω, λ, y yω, λ such that
0 ∈ f x, y, ω ρ1 F x, y, ω Mx, ω ,
0 ∈ g x, y, λ ρ2 G x, y, λ N y, λ ,
1.2
where f, F : H × H × Ω → H, g, G : H × H × Λ → H are nonlinear single-valued mappings,
M : H × Ω → 2H , N : H × Λ → 2H are multivalued mappings, ρ1 > 0, ρ2 > 0 are constants.
By using implicit resolvent equations technique of H-monotone operator, the existence of
solution is proved and the sensitivity analysis of solution for the problem 1.2 is given. Our
results improve and generalize the known results of 1, 2, 8, 9.
2. Preliminaries
Let H be a real Hilbert space, Ω, Λ be two nonempty open subsets of H.
Definition 2.1 see 1.
i A mapping T : H × Ω → H is said to be monotone with
respect to the first argument if
T x, ω − T y, ω , x − y ≥ 0,
∀x, ω, y, ω ∈ H × Ω.
2.1
ii T is said to be κ-strongly monotone with respect to the first argument if there exists
a constant κ > 0 such that
2
T x, ω − T y, ω , x − y ≥ κx − y ,
∀x, ω, y, ω ∈ H × Ω.
2.2
iii T is said to be ξ, η-Lipschitz continuous if there exist ξ > 0, η > 0 such that
T x, ω1 − T y, ω2 ≤ ξx − y ηω1 − ω2 ,
∀x, ω1 , y, ω2 ∈ H × Ω.
2.3
Definition 2.2 see 1. A mapping F : H × H × Ω → H is said to be ξ, η, ζ-Lipschitz
continuous if there exist ξ > 0, η > 0, ζ > 0 such that
F x1 , y1 , ω1 − F x2 , y2 , ω2 ≤ ξx1 − x2 ηy1 − y2 ζω1 − ω2 ,
∀ x1 , y1 , ω1 , x2 , y2 , ω2 ∈ H × H × Ω.
2.4
Definition 2.3 see 1. A mapping F : H × H × Ω → H is said to be α-strongly monotone
with respect to H in the first argument if there exists α > 0 such that
F x1 , y, ω − F x2 , y, ω , Hx1 − Hx2 ≥ αx1 − x2 2 , ∀ x1 , y, ω , x2 , y, ω ∈ H × H × Ω.
2.5
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In a similar way, we can define the strong monotonicity of F with respect to H in the second
argument.
Definition 2.4 see 9. Let H : H × Ω → H be a single-valued mapping and M : H × Ω →
2H be a multi-valued mapping. M is said to be H-monotone if M is monotone with respect
to the first argument and H·, ω ρM·, ωH H holds for all ρ > 0 and ω ∈ Ω.
Definition 2.5 see 9. Let H : H×Ω → H be a strictly monotone mapping and M : H×Ω →
H
: H → H is defined by
2H be an H-monotone mapping. The resolvent operator JM·,ω,ρ
−1
H
JM·,ω,ρ
u H·, ω ρM·, ω u,
∀u ∈ H.
2.6
Lemma 2.6 see 9. Let H1 : H × Ω → H be γ1 -strongly monotone with respect to the first
argument, H2 : H × Λ → H be γ2 -strongly monotone with respect to the first argument, and
M : H × Ω → 2H be H1 -monotone, andN : H × Λ → 2H be H2 -monotone. Then for any fixed
H1
H2
and JN·,λ,ρ
are Lipschitz continuous:
ω ∈ Ω, λ ∈ Λ, the resolvent operator JM·,ω,ρ
1
2
1
H1
H1
v
≤ u − v, ∀u, v ∈ H,
JM·,ω,ρ1 u − JM·,ω,ρ
1
γ1
1
H2
H2
v ≤ u − v, ∀u, v ∈ H,
JN·,λ,ρ2 u − JN·,λ,ρ
2
γ2
2.7
where γ1 > 0, γ2 > 0 are constants.
Lemma 2.7. Let M : H × Ω → 2H be H1 -monotone and N : H × Λ → 2H be H2 -monotone. For
any fixed ω, λ ∈ Ω × Λ. xω, λ, yω, λ is a solution of 1.2 if and only if
H1
H1 xω, λ, ω − f xω, λ, yω, λ, ω − ρ1 F xω, λ, yω, λ, ω
xω, λ JM·,ω,ρ
1
T x, y, ω, λ ,
yω, λ H2
JN·,λ,ρ
H2 yω, λ, λ
2
− g xω, λ, yω, λ, λ − ρ2 G xω, λ, yω, λ, λ
S x, y, ω, λ ,
2.81 2.82 here x xω, λ, y yω, λ.
H1
Proof. Assume that xω, λ, yω, λ satisfies relations 2.81 and 2.82 , since JM·,ω,ρ
1
H2
H2 ·, ω ρ2 N·, ω−1 , then 2.81 and 2.82 holds if
H1 ·, ω ρ1 M·, ω−1 , JM·,ω,ρ
2
and only if
0 ∈ f xω, λ, yω, λ, ω ρ1 F xω, λ, yω, λ, ω ρ1 Mxω, λ, ω,
0 ∈ g xω, λ, yω, λ, λ ρ2 G xω, λ, yω, λ, λ ρ2 N yω, λ, λ ,
and hence xω, λ, yω, λ is a solution of 2.81 and 2.82 .
2.9
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3. Main Results
Lemma 3.1. Let T : H × H → H, S : H × H → H be two continuous mappings. If there exist Θ1 ,
Θ2 , 0 < Θ1 , Θ2 < 1, such that
T x1 , y1 − T x2 , y2 S x1 , y1 − S x2 , y2 ≤ Θ1 x1 − x2 Θ2 y1 − y2 , ∀x1 , x2 , y1 , y2 ∈ H.
3.1
then there exist x∗ , y∗ ∈ H such that
x∗ T x∗ , y ∗ ,
y ∗ S x∗ , y ∗ .
3.2
Proof. For any x0 , y0 ∈ H, let xn1 T xx , yn , yn1 Sxn , yn , n 0, 1, 2, . . ., then by
3.1, for all x1 , x2 , . . . , xn , xn1 , y1 , y2 , . . . , yn , yn1 ∈ H,we have
xn1 − xn yn1 − yn ≤ Θ1 xn − xn−1 Θ2 yn − yn−1 ≤ Θ xn − xn−1 yn − yn−1 ,
3.3
where Θ max{Θ1 , Θ2 }. Let a x1 − x0 y1 − y0 , then
xn1 − xn yn1 − yn ≤ Θ xn − xn−1 yn − yn−1 ≤ Θ2 xn−1 − xn−2 yn−1 − yn−2 ≤ · · · ≤ Θn x1 − x0 y1 − y0 Θn a,
3.4
0 ≤ xn1 − xn ≤ Θn a,
0 ≤ yn1 − yn ≤ Θn a.
3.5
and hence,
Since 0 < Θ max{Θ1 , Θ2 } < 1, 3.5 implies that {xn } and {yn } are both cauchy sequences.
Therefore, there exist x∗ , y∗ ∈ H such that xn → x∗ , yn → y∗ n → ∞. By continuity of T
and S, x∗ T x∗ , y∗ , y∗ Sx∗ , y∗ .
Theorem 3.2. Let H1 : H × Ω → H be ξ1 , η1 -Lipschitz continuous and H2 : H × Λ → H be
ξ2 , η2 -Lipschitz continuous. Let f : H × H × Ω → H be ξf , ηf , ζf -Lipschitz continuous and
γf -strongly monotone with respect to H in the first argument, F : H × H × Ω → H be ξF , ηF , ζF Lipschitz continuous, g : H × H × Λ → H be ξg , ηg , ζg -Lipschitz continuous and γg -strongly
monotone with respect to H2 in the second argument, G : H × H × Λ → H be ξG , ηG , ζG -Lipschitz
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continuous. Suppose that M : H × Ω → 2H is H1 -monotone and N : H × Λ → 2H is H2 -monotone
If
ξg ρ2 ξG
1 2
ξ1 − 2γf ξf2 ρ1 ξF < 1,
γ1
γ2
ηf ρ1 ηF
1 2
2
ξ2 − 2γg ηg ρ2 ηG < 1,
Θ2 γ2
γ1
Θ1 3.6
then, for each ω, λ ∈ Ω × Λ, the problem 1.2 has an unique solution x∗ ω, λ, y∗ ω, λ.
Proof. Let x1 x1 ω, λ, x2 x2 ω, λ, y1 y1 ω, λ, y2 y2 ω, λ. By defining 2.81 of T
and Lemma 2.6, we have
T x1 , y1 , ω, λ − T x2 , y2 , ω, λ ≤ 1 H1 x1 , ω − f x1 , y1 , ω − ρ1 F x1 , y1 , ω − H1 x2 , ω
γ1
f x2 , y2 , ω ρ1 F x2 , y2 , ω ≤
1
H1 x1 , ω − H1 x2 , ω − f x1 , y1 , ω f x2 , y2 , ω γ1
1 ρ1 F x1 , y1 , ω − F x2 , y2 , ω .
γ1
3.7
Since f is ξf , ηf , ζf -Lipschitz continuous and γf -strongly monotone with respect to H1 in
the first argument, F is ξF , ηF , ζF -Lipschitz continuous, we have
H1 x1 , ω − H1 x2 , ω − f x1 , y1 , ω f x2 , y2 , ω ≤ H1 x1 , ω − H1 x2 , ω − f x1 , y1 , ω f x2 , y1 , ω f x2 , y1 , ω − f x2 , y2 , ω ≤ ξ12 − 2γf ξf2 x1 − x2 ηf y1 − y2 ,
F x1 , y1 , ω − F x2 , y2 , ω ≤ ξF x1 − x2 ηF y1 − y2 .
3.8
3.9
Combining 3.7–3.9, we have
T x1 , y1 , ω, λ − T x2 , y2 , ω, λ ≤ 1
ξ12 − 2γf ξf2 ρ1 ξF x1 − x2 γ1
ηf ρ1 ηF y1 − y2 .
γ1
3.10
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By defining 2.82 of S and Lemma 2.6, we have
S x1 , y1 , ω, λ − S x2 , y2 , ω, λ ≤ 1 H2 y1 , λ − g x1 , y1 , λ − ρ2 G x1 , y1 , λ
γ2
−H2 y2 , λ g x2 , y2 , λ ρ2 G x2 , y2 , λ ≤
1
H2 y1 , λ − H2 y2 , λ − g x1 , y1 , λ g x2 , g2 , λ γ2
ρ2 G x1 , y1 , λ − G x2 , y2 , λ .
γ2
3.11
Since g is ξg , ηg , ζg -Lipschitz continuous and γg -strongly monotone with respect to H2 in
the second argument, G is ξG , ηG , ζG -Lipschitz continuous. We have
H2 y1 , λ − H2 y2 , λ − g x1 , y1 , λ g x2 , y2 , λ ≤ H2 y1 , λ − H2 y2 , λ − g x1 , y1 , λ g x1 , y2 , λ g x1 , y2 , λ − g x2 , y2 , λ ≤ ξ22 − 2γg ηg2 y1 − y2 ξg x1 − x2 ,
3.12
G x1 , y1 , λ − G x2 , y2 , λ ≤ ξG x1 − x2 ηG y1 − y2 .
3.13
Combining 3.11–3.13, we have
S x1 , y1 , ω, λ − S x2 , y2 , ω, λ ≤ 1
ξ22 − 2γg ηg2 ρ2 ηG y1 − y2 γ2
ξg ρ2 ξG
x1 − x2 .
γ2
3.14
By 3.10 and 3.14, we have
T x1 , y1 , ω, λ − T x2 , y2 , ω, λ S x1 , y1 , ω, λ − S x2 , y2 , ω, λ ξg ρ2 ξG
1 2
2
≤
ξ1 − 2γf ξf ρ1 ξF x1 − x2 γ1
γ2
ηf ρ1 ηF 1
y1 − y2 ξ22 − 2γg ηg2 ρ2 ηG γ2
γ1
Θ1 x1 − x2 Θ2 y1 − y2 .
3.15
By 3.6 and Lemma 3.1, there exist x∗ x∗ ω, λ, y∗ y∗ ω, λ. x∗ , y∗ such that x∗ T x∗ , y∗ , ω, λ, y∗ Sx∗ , y∗ , ω, λ. By Lemma 2.7, x∗ , y∗ is a solution of 1.2. From 3.15,
we easily see that the solution of 1.2 is unique.
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H1
H2
, JN·,λ,ρ
, there are two constants ξ > 0
Assumption 1. For implicit resolvent operator JM·,ω,ρ
1
2
and η > 0 such that
H1
H1
u
≤ ξω − ω,
JM·,ω,ρ1 u − JM·,ω,ρ
1
H2
H2
J
≤ η
−
λ
−
J
v
v
λ
,
N·,λ,ρ2
N·,λ,ρ2
3.16
∀u, v ∈ H.
Theorem 3.3. Suppose that the mappings H1 , f, F, M, H2 , g, G, and N are the same as in
Theorem 3.2, and for any fixed x, y ∈ H, the mappings ω → H1 x, ω, ω → fx, y, ω, ω →
Fx, y, ω,λ → H2 y, λ, λ → gx, y, λ, λ → Gx, y, λ are continuous (or Lipschitz
continuous.) If Assumption 1 and condition 3.6 hold, then the solution xω, λ, yω, λ for the
problem 1.2 is continuous (or Lipschitz continuous).
Proof. Suppose that ω, ω ∈ Ω, λ, λ ∈ Λ such that ω → ω, λ → λ. From Theorem 3.2, we know
that the problem 1.2 has solution xω, λ, yω, λ, xω, λ, yω, λ, xω, λ, yω, λ and
xω, λ, yω, λ.
A Estimate xω, λ − xω, λ and yω, λ − yω, λ. By Lemma 2.7,we have
H1
xω, λ JM·,ω,ρ
H1 xω, λ, ω − f xω, λ, yω, λ, ω − ρ1 F xω, λ, yω, λ, ω
1
H1
JM·,ω,ρ
p ,
1
H1
H1 xω, λ, ω − f xω, λ, yω, λ, ω − ρ1 F xω, λ, yω, λ, ω
xω, λ JM·,ω,ρ
1
H1
JM·,ω
q .
,ρ1
3.17
It follows from Assumption 1, Lemmas 2.6 and 2.7 that
H1
H1
p
−
J
q xω, λ − xω, λ JM·,ω,ρ
M·,ω,ρ1
1
H1
H1
1
H1
H1
≤ JM·,ω,ρ
p
−
J
q
q
−
J
q ≤ p − q ξω − ω.
J
M·,ω,ρ1
M·,ω,ρ1
M·,ω,ρ1
1
γ1
3.18
Since f is ξf , ηf , ζf -Lipschitz continuous, γf -strongly monotone with respect to H1 in the
first argument and F is ξF , ηF , ζF -Lipschitz continuous, we conclude
p − q H1 xω, λ, ω − f xω, λ, yω, λ, ω − ρ1 F xω, λ, yω, λ, ω
−H1 xω, λ, ω f xω, λ, yω, λ, ω ρ1 F xω, λ, yω, λ, ω ≤ H1 xω, λ, ω − H1 xω, λ, ω − f xω, λ, yω, λ, ω f xω, λ, yω, λ, ω H1 xω, λ, ω − H1 xω, λ, ω
f xω, λ, yω, λ, ω − f xω, λ, yω, λ, ω 8
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f xω, λ, yω, λ, ω − f xω, λ, yω, λ, ω ρ1 F xω, λ, yω, λ, ω − F xω, λ, yω, λ, ω ρ1 F xω, λ, yω, λ, ω − F xω, λ, yω, λ, ω ≤
ξ12 − 2γf ξf2 xω, λ − xω, λ H1 xω, λ, ω − H1 xω, λ, ω
ηf yω, λ − yω, λ f xω, λ, yω, λ, ω − f xω, λ, yω, λ, ω ρ1 ξF xω, λ − xω, λ ρ1 ηF yω, λ − yω, λ
ρ1 F xω, λ, yω, λ, ω − F xω, λ, yω, λ, ω .
3.19
It follows from 3.18 and 3.19 that
xω, λ − xω, λ ≤ θ1 xω, λ − xω, λ 1
ηf ρ1 ηF yω, λ − yω, λ
γ1
1
H1 xω, λ, ω − H1 xω, λ, ω
γ1
1
f xω, λ, yω, λ, ω − f xω, λ, yω, λ, ω γ1
ρ1 F xω, λ, yω, λ, ω − F xω, λ, yω, λ, ω ξω − ω,
γ1
3.20
that is,
xω, λ − xω, λ ≤
ηf ρ1 ηF yω, λ − yω, λ
1 − θ1 γ1
1
1
H1 xω, λ, ω − H1 xω, λ, ω
1 − θ1 γ1
1
f xω, λ, yω, λ, ω − f xω, λ, yω, λ, ω γ1
ρ1 F xω, λ, yω, λ, ω − F xω, λ, yω, λ, ω γ1
ξω − ω ,
3.21
where
θ1 1 2
ξ1 − 2γf ξf2 ρ1 ξF ,
γ1
∗
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9
By Lemma 2.7, we have
H2
yω, λ JN·,λρ
H2 yω, λ, λ − g xω, λ, yω, λ, λ − ρ2 G xω, λ, yω, λ, λ
2
H2
JN·,λ,ρ
s,
2
H2
H2 yω, λ, λ − g xω, λ, yω, λ, λ − ρ2 G xω, λ, yω, λ, λ
yω, λ JN·,λ,ρ
2
H2
JN·,λ,ρ
t.
2
3.22
It follows from Assumption 1, Lemmas 2.6 and 2.7 that
H2
yω, λ − yω, λ J
N·,λ,ρ2 s
1
H2
− JN·,λ,ρ
t ≤ s − t.
2
γ2
3.23
Since g is ξg , ηg , ζg -Lipschitz continuous and γg -strongly monotone with respect to H2 in
the second argument and G is ξG , ηG , ζG -Lipschitz continuous,we conclude
s − t H2 yω, λ, λ − g xω, λ, yω, λ, λ − ρ2 G xω, λ, yω, λ, λ
−H2 yω, λ, λ g xω, λ, yω, λ, λ ρ2 G xω, λ, yω, λ, λ ≤ H2 yω, λ, λ − H2 yω, λ, λ − g xω, λ, yω, λ, λ g xω, λ, yω, λ, λ g xω, λ, yω, λ, λ − g xω, λ, yω, λ, λ ρ2 G xω, λ, yω, λ, λ − G xω, λ, yω, λ, λ ≤ ξ22 − 2γg ηg2 yω, λ − yω, λ ξg xω, λ − xω, λ
ρ2 ξG xω, λ − xω, λ ρ2 ηG yω, λ − yω, λ.
3.24
It follows from 3.23 and 3.24 that
2
2
yω, λ − yω, λ ≤ 1
ξ2 − 2γg ηg ρ2 ηG yω, λ − yω, λ
γ2
1
ξg ρ2 ξG xω, λ − xω, λ.
γ2
3.25
That is,
yω, λ − yω, λ ≤
1 1
ξg ρ2 ξG xω, λ − xω, λ,
1 − θ2 γ2
3.26
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where
θ2 1 2
ξ2 − 2γg ηg2 ρ2 ηG .
γ2
∗∗
By combining 3.21 and 3.26, we derive
ηf ρ1 ηF ξg ρ2 ξG
xω, λ − xω, λ ≤
xω, λ − xω, λ
1 − θ1 1 − θ2 γ1 γ2
1
1
H1 xω, λ, ω − H1 xω, λ, ω
1 − θ1 γ1
1 f xω, λ, yω, λ, ω − f xω, λ, yω, λ, ω γ1
ρ1 F xω, λ, yω, λ, ω − F xω, λ, yω, λ, ω γ1
ξω − ωs .
3.27
That is,
1
1
xω, λ − xω, λ ≤
1 − κ 1 − θ1
1
H1 xω, λ, ω − H1 xω, λ, ω
γ1
1 f xω, λ, yω, λ, ω − f xω, λ, yω, λ, ω γ1
ρ1 F xω, λ, yω, λ, ω − F xω, λ, yω, λ, ω γ1
ξω − ω ,
ξ ρ2 ξG 1
1
yω, λ − yω, λ ≤ g
1 − θ2 γ2 1 − κ 1 − θ1
3.28
1
H1 xω, λ, ω − H1 xω, λ, ω
γ1
1 f xω, λ, yω, λ, ω
γ1
−f xω, λ, yω, λ, ω ρ1 F xω, λ, yω, λ, ω
γ1
−F xω, λ, yω, λ, ω ξω − ω ,
3.29
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where
ηf ρ1 ηF ξg ρ2 ξG
κ
.
1 − θ1 1 − θ2 γ1 γ2
∗ ∗ ∗
By 3.6,
1 − θ1 >
1
1
ξg ρ2 ξG , 1 − θ2 >
ηf ρ1 ηF ,
γ2
γ1
3.30
and hence
ξg ρ2 ξG ηf ρ1 ηF
< 1, that is, 0 < κ < 1.
0<
1 − θ1 1 − θ2 γ1 γ2
3.31
B Estimate yω, λ − yω, λ and xω, λ − xω, λ.
By Lemma 2.7, we have
y ω, λ J H2
N·,λ,ρ2
J H2
N·,λ,ρ2
H2 y ω, λ , λ − g x ω, λ , y ω, λ , λ − ρ2 G x ω, λ , y ω, λ , λ
m,
H2
y ω, λ JN·,λ,ρ
H2 yω, λ, λ − g xω, λ, yω, λ, λ − ρ2 G xω, λ, yω, λ, λ
2
H2
JN·,λ,
ρ2 n.
3.32
It follows from Assumption 1, Lemmas 2.6 and 2.7 that
H2
y ω, λ − yω, λ J
N·,λ,ρ2
H2
≤
J
H2
m − JN·,λ,ρ
n
2
N·,λ,ρ2
≤
m − J H2
N·,λ,ρ2
H2
n
J
N·,λ,ρ2
H2
n − JN·,λ,ρ
n
2
3.33
1
m − n ηλ − λ.
γ2
Since g is ξg , ηg , ζg -Lipschitz continuous and γg -strongly monotone with respect to H2 in
the second argument and G is ξG , ηG , ζG -Lipschitz continuous,we conclude
m − n H2 y ω, λ , λ − g x ω, λ , y ω, λ , λ − ρ2 G x ω, λ , y ω, λ , λ
−H2 yω, λ, λ g xω, λ, yω, λ, λ ρ2 G xω, λ, yω, λ, λ ≤ H2 y ω, λ , λ − H2 yω, λ, λ
−g x ω, λ , y ω, λ , λ − g x ω, λ , yω, λ, λ 12
Journal of Applied Mathematics
H2 yω, λ, λ − H2 yω, λ, λ g x ω, λ , yω, λ, λ − g xω, λ, yω, λ, λ g xω, λ, yω, λ, λ − g xω, λ, yω, λ, λ ρ2 G x ω, λ , y ω, λ , λ − G xω, λ, yω, λ, λ ρ2 G xω, λ, yω, λ, λ − G xω, λ, yω, λ, λ ≤
ξ22 − 2γg ηg2 y ω, λ − yω, λ ξg x ω, λ − xω, λ
ρ2 ξG x ω, λ − xω, λ ηG y ω, λ − yω, λ
g xω, λ, yω, λ, λ − g xω, λ, yω, λ, λ H2 yω, λ, λ − H2 yω, λ, λ ρ2 G xω, λ, yω, λ, λ − G xω, λ, yω, λ, λ .
3.34
It follows from 3.33 and 3.34 that
ξg ρ2 ξG y ω, λ − yω, λ ≤
x ω, λ − xω, λ
1 − θ2 γ2
1
1
H2 yω, λ, λ − H2 yω, λ, λ 1 − θ2 γ2
1
g xω, λ, yω, λ, λ
γ2
−g xω, λ, yω, λ, λ ρ2 G xω, λ, yω, λ, λ
γ2
−G xω, λ, yω, λ, λ ηλ − λ ,
where θ2 defined by ∗∗.
3.35
Journal of Applied Mathematics
13
By Lemma 2.7, we have
H1
H
x
x ω, λ JM·,ω,ρ
ω,
λ
,
ω
−f
x
ω,
λ
,
y
ω,
λ
,
ω
−ρ
F
x
ω,
λ
,
y
ω,
λ
,ω ,
1
1
1
H1
xω, λ JM·,ω,ρ
H1 xω, λ, ω − f xω, λ, yω, λ, ω − ρ1 F xω, λ, yω, λ, ω .
1
3.36
As the proof of 3.25,we have
1
2
2
ξ1 − 2γf ξf ρ1 ξF x ω, λ − xω, λ
x ω, λ − xω, λ ≤
γ1
1
ηf ρ1 ηF y ω, λ − yω, λ,
γ1
3.37
that is,
ηf ρ1 ηF x ω, λ − xω, λ ≤
y ω, λ − yω, λ,
1 − θ1 γ1
3.38
where θ1 is defined by ∗.
Therefore
y ω, λ − yω, λ ≤
1
1
1 − κ 1 − θ2
1
H2 yω, λ, λ − H2 yω, λ, λ γ2
1
g xω, λ, yω, λ, λ
γ2
−g xω, λ, yω, λ, λ ρ2 G xω, λ, yω, λ, λ
γ2
−G xω, λ, yω, λ, λ ηλ − λ ,
3.39
14
Journal of Applied Mathematics
ηf ρ1 ηF 1
1
1
x ω, λ − xω, λ ≤
H2 yω, λ, λ − H2 yω, λ, λ 1 − θ1 γ1 1 − κ 1 − θ2 γ2
1
g xω, λ, yω, λ, λ
γ2
−g xω, λ, yω, λ, λ ρ2 G xω, λ, yω, λ, λ
γ2
−G xω, λ, yω, λ, λ ηλ − λ ,
3.40
where κ is defined by ∗ ∗ ∗.
C Prove that the conclusion of Theorem 3.3.
From 3.28 and 3.40, by the assumptions for H1 , f, F, H2 , g, and G in Theorem 3.3
and relation
xω, λ − x ω, λ ≤ xω, λ − xω, λ xω, λ − x ω, λ ,
3.41
we know that xω, λ is continuous or Lipschitz continuous.
From 3.29 and 3.39, by the assumptions for H1 , f, F, H2 , g and G in Theorem 3.3
and relation
yω, λ − y ω, λ ≤ yω, λ − yω, λ yω, λ − y ω, λ ,
3.42
we know that yω, λ is continuous or Lipschitz continuous.
Acknowledgment
The author would like to thank the anonymous referee for reading this paper carefully,
providing valuable suggestions and comments. The work was supported by the National
Science Foundation of China no. 10561007 and Science and Technology Research Project of
Education Department in Jiangxi province GJJ10269.
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