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Volume 2011, Article ID 635030, 13 pagesdoi:10.1155/2011/635030 Research Article Variational-Like Inclusions and Resolvent Equations Involving Infinite Family of Set-Valued Mappings Rais

Volume 2011, Article ID 635030, 13 pages

doi:10.1155/2011/635030

Research Article

Variational-Like Inclusions and Resolvent

Equations Involving Infinite Family of Set-Valued Mappings

Rais Ahmad and Mohd Dilshad

Department of Mathematics, Aligarh Muslim University,

Aligarh 202002, India

Correspondence should be addressed to Rais Ahmad,raisain 123@rediffmail.com

Received 18 December 2010; Accepted 23 December 2010

Academic Editor: Qamrul Hasan Ansari

Copyrightq 2011 R Ahmad and M Dilshad 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

We study variational-like inclusions involving infinite family of set-valued mappings and their equivalence with resolvent equations It is established that variational-like inclusions in real Banach spaces are equivalent to fixed point problems This equivalence is used to suggest an iterative algorithm for solving resolvent equations Some examples are constructed

1 Introduction

The important generalization of variational inequalities, called variational inclusions, have been extensively studied and generalized in different directions to study a wide class of problems arising in mechanics, optimization, nonlinear programming, economics, finance and applied sciences, and so forth; see, for example 1 7 and references theirin The resolvent operator technique for solving variational inequalities and variational inclusions

is interesting and important The resolvent operator technique is used to establish an equivalence between variational inequalities and resolvent equations The resolvent equation technique is used to develop powerful and efficient numerical techniques for solving various classes of variational inequalitiesinclusions and related optimization problems

In this paper, we established a relationship between variational-like inclusions and resolvent equations We propose an iterative algorithm for computing the approximate solutions which converge to exact solution of considered resolvent equations Some examples are constructed

2 Formulation and Preliminaries

Throughout the paper, unless otherwise specified, we assume that E is a real Banach space

with its norm · , E∗ is the topological dual of E, ·, · is the pairing between E and E∗, d

is the metric induced by the norm · , 2E resp., CBE is the family of nonempty resp.,

nonempty closed and bounded subsets of E, and H·, · is the Housdorff metric on CBE defined by

HP, Q  max

 sup

x ∈P d x, Q, sup

y ∈Q d

P, y

where dx, Q  inf y ∈Q d x, y and dP, y  inf x ∈P d x, y The normalized duality

mappingJ : E → 2 E∗ is defined by

Jx f ∈ E∗:

x, f

Definition 2.1 Let E be a real Banach space Let η : E × E → E; g, A : E → E be the single-valued mapping, and let M : E → 2Ebe a set-valued mapping Then,

i the mapping g is said to be accretive if



g x − gy

, j

ii the mapping g is said to be strictly accretive if



g x − gy

, j

and the equality hold if and only if x  y,

iii the mapping g is said to be k-strongly accretive k ∈ 0, 1 if for any x, y ∈ E, there exists jx − y ∈ Jx − y such that



g x − gy

, j

x − y≥ k x − y 2

iv the mapping A is said to be r-strongly η-accretive, if there exists a constant r >

0 such that



A x − Ay

, j

η

x, y

≥ r x − y 2

v the mapping M is said to be m-relaxed η-accretive, if there exists a constant m >

0 such that



u − v, jη

x, y

≥ −m x − y 2

, ∀x, y ∈ E, u ∈ Mx, v ∈ My

Definition 2.2 Let A : E → E, η : E × E → E be the single-valued mappings Then, a set-valued mapping M : E → 2E is called A, η-accretive if M is m-relaxed

η-accretive andA ρME  E, for every ρ > 0.

Proposition 2.3 see 8,9 Let E be a real Banach space, and let J : E → 2 E∗be the normalized duality mapping Then, for any x, y ∈ E

x y 2≤ x2 2y, j

x y, ∀jx y∈ Jx y. 2.8

Definition 2.4 Let A : E → E, W : E × E → E, and let N : E∞  E × E × E · · · → E be the

mappings Then,

i the mapping A is said to be Lipschitz continuous with constant λ Aif

A x − A

y  ≤ λ A x − y , ∀x, y ∈ E, 2.9

ii the mapping W is said to be Lipschitz continuous in the first argument with constant λ W1 if

Wx1, · − Wx2, · ≤ λ W1x1− x2, ∀x1, x2∈ E. 2.10

Similarly, we can define Lipschitz continuity in the second argument

iii the mapping N is said to be Lipschitz continuous in the ith argument with constant β iif

N ·, , x i ,  − N

·, , y i ,  ≤ β i x i − y i , ∀x i , y i ∈ E. 2.11

Definition 2.5 Let A : E → E be a strictly η-accretive mapping, and let M : E → 2 E be an

A, η-accretive mapping Then, the resolvent operator J ρ,A

η,M : E → E is defined by

J η,M ρ,A u A ρM−1u, ∀u ∈ E. 2.12

Proposition 2.6 see 10 Let E be a real Banach space, and let η : E × E → E be τ-Lipschitz

continuous; let A : E → E be an r-strongly η-accretive mapping, and let M : E → 2 E be an A,

η-accretive mapping Then the resolvent operator J η,M ρ,A : E → E is τ/r − ρm-Lipschitz continuous,

that is,

J ρ,A η,M x − J ρ,A

η,M



y ≤ τ

r − ρm x − y , ∀x, y ∈ E, 2.13

where ρ ∈ 0, r/m is a constant.

Example 2.7 Let EÊ, Ax √x, M y  √y, and ηx, y  √x −√y for all x, y ≥ 0 ∈ E Then, M is η-accretive.

Example 2.8 Let M ·, · : E × E → 2 E be r-strongly η-accretive in the first argument Then, M is m-relaxed η-accretive for m ∈ 1, r r2, for r > 0.6180.

Let T i : E → CBE, i  1, 2, , ∞ be an infinite family of set-valued mappings, and let N : E∞ E×E×E · · · → E be a nonlinear mapping Let η, W : E×E → E; A, g, m : E → E

be single-valued mappings, let and B, C, D:E → CBE be set-valued mappings Suppose that M·, · : E × E → 2 EisA, η-accretive mapping in the first argument We consider the

following problem

Find u ∈ E, w i ∈ T i u, i  1, 2, , ∞, a ∈ Bu, x ∈ Cu, and y ∈ Du such that

0∈ Nw1, w2,  − Wx, y

ma Mg u − ma, u. 2.14 The problem2.14 is called variational-like inclusions problem

Special Cases

i If W  0, m  0, then problem 2.14 reduces to the problem of finding u ∈ E, w i∈

T i u, i  1, 2, , ∞ such that

0∈ Nw1, w2,  Mg u, u. 2.15

Problem2.15 is introduced and studied by Wang 11

ii If W  0, m  0, N·,   N·, ·, then problem 2.14 reduces to a problem considered by Chang, et al.12,13 that is, find u ∈ H, w1 ∈ T1u, w2 ∈ T2u

such that

0∈ Nw1, w2 Mg u, u. 2.16

It is now clear that for a suitable choice of maps involved in the formulation of problem 2.14, we can drive many known variational inclusions considered and studied

in the literature

In connection with problem 2.14, we consider the following resolvent equation problem

Find z, u ∈ E, w i ∈ T i u, i  1, 2, , ∞; a ∈ Bu, x ∈ Cu, y ∈ Du such that

N w1, w2,  − Wx, y

ma ρ−1R ρ,A η,M ·,u z  0, 2.17

where ρ is a constant and R ρ,A η,M ·,u  I − AJ ρ,A

η,M ·,u , where AJ ρ,A

η,M ·,u z  AJ ρ,A

η,M ·,u z and I is the identity mapping Equation2.17 is called the resolvent equation problem

In support of problem2.17, we have the following example.

Example 2.9 Let us suppose that EÊ, T i u  −i, i, i  1, 2, , ∞, Cu  {π/2}, Bu 

0, 1, and Du  {1}.

We define for w i ∈ T i u, i  1, 2, , ∞, a ∈ Bu, x ∈ Cu and y ∈ Du.

i Nw1, w2,   min{−1, sin w1, sin w2, },

ii ma  sin−1a cos−1a,

iii Wx, y  xy,

iv Ax  x − 1, for all x ∈Ê,

v M·, x  1, for all x ∈Ê,

Then, for ρ 1, it is easy to check that the resolvent equation problem 2.17 is satisfied

3 An Iterative Algorithm and Convergence Result

We mention the following equivalence between the problem2.14 and a fixed point problem which can be easily proved by using the definition of resolvent operator

Lemma 3.1 Let u, a, x, y, w1, w2,  where u ∈ E, w i ∈ T i u, i  1, 2, , ∞, a ∈ Bu, x ∈

C u, and y ∈ Du, is a solution of 2.14 if and only if it is a solution of the following equation:

g u  ma J ρ,A

η,M ·,u

A

g u − ma− ρN w1, w2,  − Wx, y

ma 3.1

Now, we show that the problem2.14 is equivalent to a resolvent equation problem

Lemma 3.2 Let u ∈ E, w i ∈ T i u, i  1, 2, , ∞, a ∈ Bu, x ∈ Cu, y ∈ Du, then the

following are equivalent:

i u, a, x, y, w1, w2,  is a solution of variational inclusion problem 2.14,

ii z, u, a, x, y, w1, w2,  is a solution of the problem 2.17,

where

z  Ag u − ma− ρN w1, w2,  − Wx, y

ma ,

g u  ma J ρ,A

η,M ·,u

A

g u − ma− ρN w1, w2,  − Wx, y

ma 3.2

Proof Let u, a, x, y, w1, w2,  be a solution of the problem 2.14, then byLemma 3.1, it is

a solution of the problem

g u  ma J ρ,A

η,M ·,u

A

g u − ma− ρN w1, w2,  − Wx, y

ma , 3.3 using the fact that

R ρ,A η,M ·,u  I − A J η,M ρ,A ·,u

,

R ρ,A η,M ·,u z  R ρ,A η,M ·,uA

g u − ma− ρN w1, w2,  − Wx, y

ma

 I − A J η,M ρ,A ·,u

A

g u − ma−ρN w1, w2,  − Wx, y

ma

 Ag u − ma− ρN w1, w2,  − Wx, y

ma

− A J η,M ρ,A ·,u

A

g u − ma− ρN w1, w2,  − Wx, y

ma

 Ag u − ma− ρN w1, w2,  − Wx, y

ma

− Ag u − ma,

3.4 which implies that

N w1, w2,  − Wx, y

ma ρ−1R ρ,A η,M ·,u z  0, 3.5 with

z  Ag u − ma− ρN w1, w2,  − Wx, y

that is,z, u, a, x, y, w1, w2,  is a solution of problem 2.17

Conversly, letz, u, a, x, y, w1, w2,  be a solution of problem 2.17, then

ρ

N w1, w2,  − Wx, y

ma  −R ρ,A

η,M ·,u z,

ρ

N w1, w2,  − Wx, y

ma  AJ η,M ρ,A ·,u z− z, 3.7

from3.2 and 3.7, we have

ρ

N w1, w2,  − Wx, y

ma

 AJ η,M ρ,A ·,u

A

g u − ma− ρN w1, w2,  − Wx, y

ma

−A

g u − ma− ρN w1, w2,  − Wx, y

ma ,

3.8

which implies that

g u  ma J ρ,A

η,M ·,u

A

g u − ma− ρN w1, w2,  − Wx, y

ma , 3.9

that is,u, a, x, y, w1, w2,  is a solution of 2.14

We now invoke Lemmas3.1and3.2to suggest the following iterative algorithm for solving resolvent equation problem2.17

Algorithm 3.3 For a given z0, u0∈ E, w0

i ∈ T i u0, i  1, 2, , ∞, a0∈ Bu0, x0∈ Cu0, and

y0∈ Du0 Let

z1 Ag u0 − ma0− ρN w01, w02, 

− Wx0, y0



ma0. 3.10

Take z1, u1∈ E such that

g u1  ma1 J ρ,A

Since for each i, w0

i ∈ T i u0, a0 ∈ Bu0, x0 ∈ Cu0, and y0 ∈ Du0 by Nadler’s theorem14 there exist w1

i ∈ T i u1, a1∈ Bu1, x1∈ Cu1, and y1∈ Du1 such that

w0

i − w1

i ≤ HT

i u0, T i u1,

a0− a1 ≤ HBu0, Bu1,

x0− x1 ≤ HCu0, Cu1,

y0− y1 ≤ HDu0, Du1,

3.12

whereH is the Housdorff metric on CBE.

Let

z2 Ag u1 − ma1− ρN w11, w12, 

− Wx1, y1



ma1, 3.13

and take any u2∈ E such that

g u2  ma2 J η,M ρ,A ·,u2 z2. 3.14

Continuing the above process inductively, we obtain the following

For any z0, u0 ∈ E, w0

i ∈ T i u0, i  1, 2, , ∞, a0 ∈ Bu0, x0 ∈ Cu0, and

y0 ∈ Du0 Compute the sequences {z n }, {u n }, {w n

i }, i  1, 2, , ∞, {a0}, {x0}, {y0} by the following iterative schemes:

i gu n   ma n  J ρ,A

ii a n ∈ Bu n , a n − a n 1 ≤ HBu n , Bu n 1, 3.16

iii x n ∈ Cu n , x n − x n 1 ≤ HCu n , Cu n 1, 3.17

iv y n ∈ Du n , y n − y n 1 ≤ HDu n , Du n 1, 3.18

v for each i, w n

i ∈ T i u n , w n

i − w n 1

i ≤ HT

i u n , T i u n 1, 3.19

vi z n 1 Ag un − ma n− ρN

w n1, w2n , 

− Wx n , y n



ma n , 3.20

where ρ > 0 is a constant and n  0, 1, 2,

Theorem 3.4 Let E be a real Banach space Let T i , B, C, D : E → CBE be H-Lipschitz

continuous mapping with constants δ i , α, t, γ , respectively Let N  E∞  E × E × E · · · → E be

Lipschitz continuous with constant β i , let A, g, m : E → E be Lipschitz continuous with constants

λ A , λ g , λ m , respectively, and let A be r-strongly η-accretive mapping Suppose that η,W : E ×E → E

are mappings such that η is Lipschitz continuous with constant τ and W is Lipschitz continuous in both the argument with constant λ W1and λ W2, respectively Let M : E × E → 2 E be A, η-accretive

mapping in the first argument such that the following holds for μ > 0:

J ρ,η

M ·,u nz n  − J ρ,η

M ·,u n−1z n ≤ μu

Suppose there exists a ρ > 0 such that

λ A λ g λ m α

λ A ρ ρ∞

i1

β i δ i ρλ W1t λ W2γ

<



r − ρm

τ



1−λ2

m α2 μ2− 2k, m < r

ρ , λ

2

m α2< 1 2k − μ2.

3.22

Then, there exist z, u, ∈ E, a ∈ BE, and x ∈ CE, y ∈ DE, and w i ∈ T i u that

satisfy resolvent equation problem2.17 The iterative sequences {z n }, {u n }, {a n } {x n }, {y n}, and {w n

i }, i  1, 2, , ∞, n  0, 1, generated by Algorithm 3.3 converge strongly to

z, u, a, x, y, w i, respectively

Proof FromAlgorithm 3.3, we have

z n 1− z n  A

g u n  − ma n− ρN

w1n , w2n , 

− Wx n , y n

ma n

−A

g u n−1 − ma n−1

− ρN w n1−1, w n2−1, 

− Wx n−1, y n−1

ma n−1

≤ A

g u n  − ma n−A

g u n−1 − ma n−1

ρ N

w n1, w n2, 

− N w1n−1, w n2−1, 

ρ W

x n , y n

− Wx n−1, y n−1 ρma n  − ma n−1.

3.23

By using the Lipschitz continuty of A, g, and m with constants λ A , λ g , and λ m, respectively, and byAlgorithm 3.3, we have

A

g u n  − ma n−A

g u n−1 − ma n−1

≤ λ A g u n  − gu n−1 λA ma n  − ma n−1

≤ λ A λ g u n − u n−1 λ A λ m a n − a n−1

≤ λ A λ g u n − u n−1 λ A λ m HBu n , Bu n−1

≤ λ A λ g u n − u n−1 λ A λ m α u n − u n−1

λ A λ g λ A λ m α

u n − u n−1.

3.24

Since N is Lipschitz continuous in all the arguments with constant β i , i  1, 2, ,

respectively, and usingH-Lipschitz continuity of T i ’s with constant δ i, we have

Nw1n , w2n , 

− N w n1−1, w2n−1, 

 N

w n

1, w n

2, 

− N w n−1

2, 

N w n−1

2,  · · ·

≤ N

w1n , w2n , 

− N w n1−1, w n2, 

N w n−1

1 , w n2,



− N w n1−1, w2n−1,  · · ·

≤ β1 w n

1 − w n−1

1 β

2 w n

2− w n−1

2 · · ·

i1

β i w n

i − w n−1

i

≤∞

i1

β i HT i u n , T i u n−1

≤∞

i1

β i δ i u n − u n−1, n  0, 1, 2,

3.25

Since W is a Lipschitz continuous in both the arguments with constant λ W1, λ W2

respectively, and C and D are H-Lipschitz continuous with constant t and γ, respectively,

we have

W

x n , y n

− Wx n−1, y n−1 ≤ λ W2 y n − y n−1 λ W1x n − x n−1

≤ λ W2γ u n − u n−1 λ W1t u n − u n−1

λ W1t λ W2γ

u n − u n−1.

3.26

Combining3.24, 3.25, and 3.26 with 3.23, we have

z n 1− z n ≤λ A λ g λ A λ m α

u n − u n−1 ρ∞

i1

β i δ i u n − u n−1

ρλ W1t λ W2γ

u n − u n−1 ρλ m α u n − u n−1







λ A λ g λ m α

λ A ρ ρ∞

i1

β i δ i ρλ W1t λ W2γ

u n − u n−1.

3.27

By usingProposition 2.3and k-strong accretiveness of g, we have

u n − u n−12 ma

n  J M ρ,η ·,u nz n  − ma n−1

−J ρ,η

M ·,u n−1z n−1 −g u n  − u n−g u n−1 − u n−1 2

≤ ma n  − ma n−12 J ρ,η

M ·,u nz n  − J ρ,η

M ·,u n−1z n−1 2

− 2g u n  − u n−g u n−1 − u n−1

, j u n − u n−1

≤ ma n  − ma n−12

J ρ,η

M ·,u nz n  − J ρ,η

M ·,u n−1z n  J ρ,η

M ·,u n−1z n  − J ρ,η

M ·,u n−1z n−1 2

− 2g u n  − u n−g u n−1 − u n−1

, j u n − u n−1,