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  • Research Article
  • Open Access

Variational-Like Inclusions and Resolvent Equations Involving Infinite Family of Set-Valued Mappings

Fixed Point Theory and Applications20102011:635030

  • Received: 18 December 2010
  • Accepted: 23 December 2010
  • Published:


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.


  • Variational Inequality
  • Iterative Algorithm
  • Real Banach Space
  • Lipschitz Continuity
  • Variational Inclusion

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 [17] 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 inequalities (inclusions) 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 is a real Banach space with its norm ,   is the topological dual of ,   is the pairing between and ,   is the metric induced by the norm ,   (resp., ) is the family of nonempty (resp., nonempty closed and bounded) subsets of   ,  and is the Housdorff metric on defined by
where   and . The normalized duality mapping     is defined by

Definition 2.1.

Let be a real Banach space. Let ;   be the single-valued mapping, and let be a set-valued mapping. Then,

(i)the mapping   is said to be accretive if
(ii)the mapping     is said to be strictly accretive if

and the equality hold if and only if ,

(iii)the mapping     is said to be -strongly accretive if for any   ,  there exists   such that
(iv)the mapping     is said to be   -strongly   -accretive, if there exists a constant   such that
(v)  the mapping     is said to be   -relaxed   -accretive, if there exists a constant   such that

Definition 2.2.

Let  ,     be the single-valued mappings. Then, a set-valued mapping is called -accretive if     is   -relaxed   -accretive  and   ,  for every   .

Proposition 2.3 (see [8, 9]).

Let be a real Banach space, and let be the normalized duality mapping. Then, for any

Definition 2.4.

Let ,   ,  and let   be the mappings. Then,

(i)the mapping is said to be Lipschitz continuous with constant if
(ii)the mapping     is said to be Lipschitz continuous in the first argument with constant     if

Similarly, we can define Lipschitz continuity in the second argument.

(iii)the mapping     is said to be Lipschitz continuous in the th argument with constant   if

Definition 2.5.

Let   be a strictly -accretive mapping, and let   be an -accretive mapping.  Then, the resolvent operator     is defined by

Proposition 2.6 (see [10]).

Let     be a real Banach space, and let     be   -Lipschitz continuous; let   be an -strongly -accretive mapping, and let   be an -accretive mapping. Then the resolvent operator   is -Lipschitz continuous, that is,

where    is a constant.

Example 2.7.

Let , ,  and   for all . Then, is -accretive.

Example 2.8.

Let   be -strongly   -accretive in the first argument. Then,     is -relaxed   -accretive for   , for .

Let ,   be an infinite family of set-valued mappings, and let   be a nonlinear mapping. Let   ; be single-valued mappings, let and be set-valued mappings. Suppose that is -accretive mapping in the first argument. We consider the following problem.

Find , , , , , and such that

The problem (2.14) is called variational-like inclusions problem.

Special Cases

(i)If   , ,   then problem (2.14) reduces to the problem of finding , ,  such that


Problem (2.15) is introduced and studied by Wang [11].

(ii)If , , , then problem (2.14) reduces to a problem considered by Chang, et al. [12, 13] that is, find , , such that


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 , , ; , , such that

where is a constant and , where and     is the identity mapping. Equation (2.17) is called the resolvent equation problem.

In support of problem (2.17), we have the following example.

Example 2.9.

Let us suppose that , ,   ,   ,   , and .

We define for , , , and .

(i) ,

(ii) ,

(iii) ,

(iv) , for all ,

(v) , for all ,

Then, for , 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 problem (2.14) and a fixed point problem which can be easily proved by using the definition of resolvent operator.

Lemma 3.1.

Let where , , ,   , , and , is a solution of (2.14) if and only if it is a solution of the following equation:

Now, we show that the problem (2.14) is equivalent to a resolvent equation problem.

Lemma 3.2.

Let , , ,   , , , then the following are equivalent:

(i) is a solution of variational inclusion problem (2.14),

(ii) is a solution of the problem (2.17),



Let be a solution of the problem (2.14), then by Lemma 3.1, it is a solution of the problem
using the fact that
which implies that

that is, is a solution of problem (2.17).

Conversly, let be a solution of problem (2.17), then
from (3.2) and (3.7), we have
which implies that

that is, is a solution of (2.14).

We now invoke Lemmas 3.1 and 3.2 to suggest the following iterative algorithm for solving resolvent equation problem (2.17).

Algorithm 3.3.

For a given , , ,   , , and . Let
Take such that
Since for each ,   , , and by Nadler's theorem [14] there exist , , , and such that

where     is the Housdorff metric on .

and take any such that

Continuing the above process inductively, we obtain the following.

For any , , ,   , , and Compute the sequences , , ,   , , , by the following iterative schemes:

where is a constant and .

Theorem 3.4.

Let be a real Banach space.  Let be   -Lipschitz continuous mapping with constants , , , , respectively. Let   be Lipschitz continuous with constant , let be Lipschitz continuous with constants , , , respectively, and let     be -strongly   -accretive mapping. Suppose that are mappings such that is Lipschitz continuous with constant and is Lipschitz continuous in both the argument with constant and , respectively. Let be -accretive mapping in the first argument such that the following holds for :
Suppose there exists a such that

Then, there exist ,   ,  and   , ,  and  that satisfy resolvent equation problem (2.17). The iterative sequences , , , , and , , generated by Algorithm 3.3 converge strongly to , , , , , , respectively.


From Algorithm 3.3, we have
By using the Lipschitz continuty of , , and with constants , , and , respectively, and by Algorithm 3.3, we have
Since is Lipschitz continuous in all the arguments with constant   , , respectively,  and using   -Lipschitz continuity of   's with constant   , we have
Since is a Lipschitz continuous in both the arguments with constant , respectively, and and are -Lipschitz continuous with constant and , respectively, we have
Combining (3.24), (3.25), and (3.26) with (3.23), we have
By using Proposition 2.3 and   -strong accretiveness of   , we have
Using (3.28), (3.27) becomes

From (3.22), we have , and consequently is a Cauchy sequence in . Since is a Banach space, there exists     such that   . From (3.28), we know that is also a Cauchy sequence in   . Therefore, there exists     such that   . Since the mappings 's, ,     and   are   -Lipschitz continuous, it follows from (3.16)–(3.19) of Algorithm 3.3 that , , , and are also Cauchy sequences. We can assume that , ,   , and .

Now, we prove that . In fact, since and

which implies that   .  As , we have , .

Finally, by the continuity of   , , , ,  and and by Algorithm 3.3, it follows that
From (3.32), and Lemma 3.2, it follows that

that is, is a solution of resolvent equation poblem (2.17).



This work is supported by Department of Science and Technology, Government of India, under Grant no. SR/S4/MS: 577/09.

Authors’ Affiliations

Department of Mathematics, Aligarh Muslim University, Aligarh, 202002, India


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© Rais Ahmad and Mohd Dilshad. 2011

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