#### Abstract

In this paper, a common solution problem is investigated based on a Bregman projection. Strong convergence of the monotone projection algorithm for monotone operators and bifunctions is obtained in a reflexive Banach space.

#### 1. Introduction

Fixed point theory as an important branch of nonlinear analysis theory has been applied in the study of nonlinear phenomena. Lots of problems arising in economics, engineering, and physics can be studied based on fixed point techniques; see [1–21] and the references therein. Many well-known problems can be studied by using algorithms which are iterative in their nature. As an example, in computer tomography with limited data, each piece of information implies the existence of a convex set in which the required solution lies. The problem of finding a point in the intersection , where is some positive integer, is then of crucial interest, and it cannot be usually solved directly. Therefore, an iterative algorithm must be used to approximate such point. The well-known convex feasibility problem which captures applications in various disciplines such as image restoration and radiation therapy treatment planning is to find a point in the intersection of common fixed point sets of a family of nonlinear mappings see, for example, [22–28].

Interest in monotone operators stems mainly from their firm connection with equations of evolution which is an important class of nonlinear operators. It is known that many physically significant problems can be modelled by initial value problems of the form: where is an monotone operator in an appropriate Banach space. Typical examples, where such evolution equations occur, can be found in the heat, wave, or Schrödinger’s equations. If is dependent on , then (1) is reduced to whose solutions correspond to the equilibrium points of the system (1). Consequently, considerable research efforts have been devoted, especially within the past 40 years or so, to methods for finding approximate solutions (when they exist) of (2). An early fundamental result in the theory of monotone operators, due to Browder [29], states that the initial value problem (1) is solvable if is locally Lipschitz and accretive on .

The Krasnoselskii-Mann iterative algorithm is efficient for treating zero points of monotone operators. However, it is weak convergence only. In many disciplines, including economics, image recovery, and control theory, problems arise in infinite dimension spaces. In such problems, strong convergence (norm convergence) is often much more desirable than weak convergence for it translates the physically tangible property that the energy of the error between the iterate and the solution eventually becomes arbitrarily small. The importance of strong convergence is also underlined, where a convex function is minimized via the proximal-point algorithm; it is shown that the rate of convergence of the value sequence is better when converges strongly than when it converges weakly. Such properties have a direct impact when the process is executed directly in the underlying infinite dimensional space. Projection methods which were first introduced by Haugazeau [30] have been considered for treating zero points of monotone operators. The advantage of projection methods is that strong convergence of iterative sequences can be guaranteed without any compact assumptions.

In this paper, we study common zero points of a family of maximal monotone operators and common solutions of a system of equilibrium problems based on the Bregman projection. Strong convergence of the monotone projection algorithm is obtained in a reflexive Banach space.

#### 2. Preliminaries

Let be a real Banach space, the dual space of , and a nonempty subset of . Let be a bifunction from to , where denotes the set of real numbers. Recall the following equilibrium problem. Find such that We use to denote the solution set of the equilibrium problem (3). That is, Given a mapping , let Then, if and only if is a solution of the following variational inequality. Find such that

In order to study the solution of the equilibrium problem (3), we assume that satisfies the following conditions: (A1), for all ; (A2) is monotone; that is, , for all ; (A3)(A4) for each , is convex and weakly lower semicontinuous.

For any convex function , we denote the domain of by . For any and , we denote by the right-hand derivative of at in the direction ; that is, The function is said to be Gâteaux differentiable at if exists for any . In this case, coincides with , the value of the gradient of at . The function is said to be Gâteaux differentiable if it is Gâteaux differentiable everywhere. The function is said to be Fréchet differentiable at if the limit is attained uniformly in . The function is said to be Fréchet differentiable if it is Fréchet differentiable everywhere. It is well known that if a continuous convex function is Gâteaux differentiable, then is norm-to-weak* continuous. If is Fréchet differentiable, then is norm-to-norm continuous; for more details, see [31] and the references therein; for more details, see [32] and the references therein. The function is said to be strongly coercive [32] if . It is said to be bounded on bounded subsets of if is bounded for each bounded subset of . Let be a reflexive Banach space. For any proper, lower semicontinuous, and convex function: , the conjugate function of is defined by It is well known that for all . It is also known that , where is the subdifferential of , is equivalent to If is a proper, lower semicontinuous, and convex function, then is a proper, weak* lower semicontinuous, and convex function.

Next, we recall some facts about Bregman distance. Let be a convex and Gâteaux differentiable function. Then, Bregman distance corresponding to is the function defined by It is clear that for all . If is smooth and for all , we obtain that , where is the generalized duality mapping. If is a nonempty, closed and convex subset of a reflexive Banach space and is a strongly coercive Bregman function, then for each , there exists a unique such that . Bregman projection from onto is defined by for all . It is also well known [32] that has the following property: Let . A function is said to be uniformly convex on bounded subsets of if for all , where is defined by The function is called the gauge of the uniform convexity of . If is a convex function which is uniformly convex on bounded subsets, then for all and , where is the gauge of the uniform convexity of .

Let be a mapping. In this paper, we use to denote the fixed point set of . is said to be closed; if for any sequence such that and , then .

Let be a proper, lower semicontinuous and convex function. Recall that is said to be Bregman quasi-nonexpansive, if and only if

A point in is said to be an asymptotic fixed point of if and only if contains a sequence , which converges weakly to such that . The set of asymptotic fixed points of will be denoted by . is said to be Bregman relatively nonexpansive if and only if and for all and .

Let be a multivalued operator from to with domain and range . An operator is said to be monotone if and only if for each and , . A monotone operator is said to be maximal if its graph is not properly contained in the graph of any other monotone operator. We know that if is a maximal monotone operator, then is closed and convex.

Let be a reflexive, strictly convex, and smooth Banach space, and let be a maximal monotone operator from to . From Rockafellar [33], we find that and ; there exists a unique such that . If , then we can define a single-valued mapping by and such a is called the resolvent of . We know that is closed and for all . From [34], we know that is a Bregman quasi-nonexpansive.

In this paper, we study a common solution problem based on Bregman projections. Strong convergence of the monotone projection algorithm for monotone operators and bifunctions is obtained in a reflexive Banach space.

In order to introduce our main results, we also need the following lemmas.

Lemma 1 (see [34]). *Let be a Banach space and let be a Gâteaux differentiable function which is uniformly convex on bounded subsets of . Let and be bounded sequences in . Then,
*

Lemma 2 (see [35]). *Let be a reflexive Banach space and let be a convex, continuous, strongly coercive, and Gâteaux differentiable function which is bounded on bounded subsets and uniformly convex on bounded subsets of . Let be a nonempty, closed, and convex subset of . Let be a Bregman relatively nonexpansive mapping. Then, is closed and convex.*

Lemma 3 (see [36]). *Let be a reflexive Banach space and let be a convex, continuous, and strongly coercive function which is bounded on bounded subsets and uniformly convex on bounded subsets of . Let be a nonempty, closed, and convex subset of and let be a bifunction from to satisfying (A1)–(A4). Let and . Then, there exists such that , for all . Define a mapping by , for all . Then, the following conclusions hold: *(1)* is single-valued; *(2)* is a Bregman firmly nonexpansive-type mapping; that is, for all ,
*(3)*;
*(4)* is Bregman quasi-nonexpansive;*(5)*(6)** is closed and convex.*

#### 3. Main Results

Theorem 4. *Let be a reflexive Banach space and let be a strongly coercive Bregman function which is bounded on bounded subsets and uniformly convex and smooth on bounded subsets of . Let be a nonempty, closed, and convex subset of and let be an index set. Let be a positive real number sequence. Let be a bifunction from to satisfying (A1)–(A4) and let be a maximal monotone operator such that for every . Assume that the common solution set is nonempty. Let be a sequence generated in the following manner:
**
where is the right-hand derivative of , , is a real sequence in such that , and is a real sequence in , where is some positive real number, for every . Then, the sequence converges strongly to , where is the Bregman projection from onto .*

*Proof. *In view of Lemma 3, we find that the common solution set CSS is closed and convex. Next, we prove that is closed and convex. It suffices to show that, for each fixed but arbitrary , is closed and convex. This can be proved by induction into . It is obvious that is closed and convex. Assume that is closed and convex for some . For , we see that . It follows that , where . Notice that
The above inequalities are equivalent to
Multiplying and on both sides of (21), respectively, yields that
That is,
where . This implies that is closed and convex. We conclude that is closed and convex. This in turn implies that is closed and convex. This implies that is well defined.

Next, we show that . is clear. Suppose that for some positive integer . For any , we see that
which shows that . This implies that . This yields that . This completes the proof that .

Next, we show that the sequence is bounded. It follows from (12) that
This implies that the sequence is bounded. It follows that the sequence is also bounded. In view of the construction of the set , we find that and for any positive integer . It follows from (12) that
It follows that
This shows that the sequence is nondecreasing. Hence, the limit exists. In view of , we find that
This concludes that exists. Let in (26); we find that . In view of Lemma 1, we obtain that as . This yields that is a Cauchy sequence. Since is closed and convex, we see that there exists an ; that is,
Next, we show that . It follows from (26) that
Therefore, we obtain that
Since , we find that
In view of (30), we find that
It follows that
Notice that
It follows from (29) and (34) that In view of Lemma 1, we find that
Since is uniformly norm-to-norm continuous on any bounded subset of , we find that
Notice that
It follows from (36) and (37) that
In view of (14), we find that
where is the gauge of the uniform convexity of . It follows that
In view of the restrictions on the sequence , we find from (39) that . Since is uniformly norm-to-norm continuous on bounded subsets of , we obtain that
Since is closed Bregman quasi-nonexpansive, we find that . This proves that .

Next, we show that . Notice that
It follows from (39) that
Therefore, we find that . Since is uniformly norm-to-norm continuous on any bounded sets, we have . From the assumption , we see that
In view of , we see that
It follows from that
In view of , we find from (45) that
For and , define . It follows that , which yields that . It follows from and that
That is,
Letting , we obtain from that , for all . This implies that for every . This shows that .

Finally, we prove that . In view of , we conclude that
Since , we arrive at
Letting in the above inequality, we see that
This yields that . This completes the proof.

If the index set is singleton, then we have the following result.

Corollary 5. *Let be a reflexive Banach space and let be a strongly coercive Bregman function which is bounded on bounded subsets and uniformly convex and smooth on bounded subsets of . Let be a nonempty, closed, and convex subset of and let be a positive real number. Let be a bifunction from to satisfying (A1)–(A4) and let be a maximal monotone operator such that . Assume that the common solution set is nonempty. Let be a sequence generated in the following manner:
**
where is the right-hand derivative of . , is a real sequence in such that , and is a real sequence in , where is some positive real number. Then, the sequence converges strongly to , where is the Bregman projection from onto .*

*Remark 6. *Corollary 5 improves Theorem 4 of Qin et al. [24] in the following aspects: (1) the framework of spaces is relaxed; (2) a more general notion of Bregman function is considered instead of the generalized duality mapping; (3) a more general notion of Bregman projection is considered instead of the generalized projection operator; (4) families of operators extend from a finite family to an infinite family.

Finally, we give some results in the framework of Hilbert spaces.

Theorem 7. *Let be a Hilbert space and let be a nonempty, closed, and convex subset of . Let be an index set. Let be a positive real number sequence. Let be a bifunction from to satisfying (A1)–(A4) and let be a maximal monotone operator such that for every . Assume that the common solution set is nonempty. Let be a sequence generated in the following manner:
**
where , is a real sequence in such that , and is a real sequence in , where is some positive real number, for every . Then, the sequence converges strongly to , where is the metric projection from onto .*

*Proof. *Putting , we find that for all . It follows that for all . From Theorem 4, we can immediately obtain the desired conclusion.

If the index set is singleton, then we have the following result.

Corollary 8. *Let be a Hilbert space and let be a nonempty, closed, and convex subset of . Let be a positive real number. Let be a bifunction from to satisfying (A1)–(A4) and let be a maximal monotone operator such that . Assume that the common solution set is nonempty. Let be a sequence generated in the following manner:
**
where , is a real sequence in such that and is a real sequence in , where is some positive real number. Then, the sequence converges strongly to , where is the metric projection from onto .*