Abstract

In this paper, using a new shrinking projection method and new generalized -demimetric mappings, we consider the strong convergence for finding a common point of the sets of zero points of maximal monotone mappings, common fixed points of a finite family of Bregman -demimetric mappings, and common zero points of a finite family of Bregman inverse strongly monotone mappings in a reflexive Banach space. To the best of our knowledge, such a theorem for Bregman -demimetric mapping is the first of its kind in a Banach space. This manuscript is online on arXiv by the link http://arxiv.org/abs/2107.13254.

1. Introduction

Let be a Hilbert space and let be a nonempty, closed, and convex subset of . Let be a mapping. Then, we denote by the set of fixed points of . For a real number with , a mapping is said to be a -strict pseudocontraction [1] iffor all . In particular, if , then is nonexpansive, i.e.,

If is a -strict pseudocontraction and , then we get that, for and ,

From this inequality, we get that

Then, we get that

A mapping is said to be a generalized hybrid [2] if there exist real numbers such thatfor all . Such a mapping is said to be a -generalized hybrid. The class of generalized hybrid mappings covers several well-known mappings. A -generalized hybrid mapping is nonexpansive. For and , it is nonspreading [3, 4], i.e.,

For and , it is also a hybrid [5], i.e.,

In general, nonspreading mappings and hybrid mappings are not continuous (see [6]). If is a generalized hybrid and , then we get that, for and ,and hence, . From this, we have that

Let be a smooth Banach space and let be a maximal monotone mapping with . Then, for the metric resolvent of for a positive number , we obtain from [7, 8] that, for and ,

Then, we getand hence,where is the duality mapping on . Motivated by (5), (10), and (13), Takahashi [9] introduced a nonlinear mapping in a Banach space as follows: let be a nonempty, closed, and convex subset of a smooth Banach space and let be a real number with . A mapping with is said to be -demimetric if, for and ,

According to this definition, we have that a -strict pseudocontraction with is -demimetric, an -generalized hybrid mapping with is -demimetric, and the metric resolvent with is -demimetric.

On the other hand, in 1967, Bregman [10] discovered an effective technique using the so-called Bregman distance function in the process of designing and analyzing feasibility and optimization algorithms. This led to a growing area of research in which Bregman’s technique is applied in various ways in order to design and analyze iterative algorithms for solving feasibility problems, equilibrium problems, fixed point problems for nonlinear mappings, and so on (see, e.g., [11, 12] and the references therein).

In 2010, Reich and Sabach [11] using the Bregman distance function introduced the concept of Bregman strongly nonexpansive mappings and studied the convergence of two iterative algorithms for finding common fixed points of finitely many Bregman strongly nonexpansive operators in reflexive Banach spaces.

In this paper, motivated by Takahashi [13], we generalize -demimetric mappings by the Bregman distance, and using a new shrinking projection method, we deal with the strong convergence for finding a common point of the sets of zero points of maximal monotone mappings, common fixed points of a finite family of Bregman -demimetric mappings, and common zero points of a finite family of Bregman inverse strongly monotone mappings in a reflexive Banach space (see [14]).

2. Preliminaries

Let be a reflexive real Banach space and be a nonempty, closed, and convex subset of . Throughout this paper, the dual space of is denoted by . The norm and duality pairing between and are, respectively, denoted by and . Let be a sequence in , and we denote the strong convergence of to as by and the weak convergence by .

Throughout this paper, is a proper, lower semicontinuous, and convex function. We denote by , the domain of . The function is said to be strongly coercive if . Let , and the subdifferential of at is the convex mapping set defined byand is the Fenchel conjugate of defined by

It is well known that is equivalent to

For any and , we denote by the right-hand derivative of at in the direction , that is,

The function is called differentiable at , if the limit in (18) exists for any . In this case, the gradient of at is the linear function which is defined by for any . The function is said to be differentiable if it is differentiable at each . The function is said to be Fréchet differentiable at , if the limit in (18) is attained uniformly in , for any . Finally, is said to be uniformly Fréchet differentiable on a subset of , if the limit in (18) is attained uniformly for and .

Lemma 1 (see [11]). If is uniformly Fréchet differentiable and bounded on bounded subsets of , then is uniformly continuous on bounded subsets of and is uniformly continuous on bounded subsets of from the strong topology of to the strong topology of .

Proposition 2 (see [15]). Let be a convex function which is bounded on bounded subsets of . Then, the following assertions are equivalent:(i) is strongly coercive and uniformly convex on bounded subsets of (ii) is Fréchet differentiable, and is uniformly norm-to-norm continuous on bounded subsets of

Definition 3. The function is said to be “Legendre” if it satisfies the following two conditions:
(L1): , and is single-valued on its domain.
(L2): , and is single-valued on its domain.

Because here the space is assumed to be reflexive, we always have ([16], p. 83). This fact, when combined with the conditions (L1) and (L2), implies the following equalities:

In addition, the conditions (L1) and (L2), in conjunction with Theorem 5.4 of [17], imply that the functions and are strictly convex on the interior of their respective domains and is Legendre if and only if is Legendre.

One important and interesting Legendre function is . When is a uniformly convex and -uniformly smooth Banach space with , the generalized duality mapping is defined by

In this case, the gradient of coincides with the generalized duality mapping of , . Several interesting examples of Legendre functions are presented in [1719].

From now on, we always assume that the convex function is Legendre.

Definition 4 (see [20]). Let be a convex and differentiable function. The bifunction defined byis called the Bregman distance with respect to .

It should be noted that is not a distance in the usual sense of the term. Clearly, , but may not imply . In our case, when is Legendre, this indeed holds ([17], Theorem 7.3 (vi), p. 642). In general, satisfies the three-point identityand the four-point identityfor any and . Over the last 30 years, Bregman distances have been studied by many researchers (see [17, 2123]).

Let be a convex function on which is differentiable on . The function is said to be totally convex at a point if its modulus of total convexity at , defined byis positive whenever . The function is said to be totally convex when it is totally convex at every point of . The function is said to be totally convex on bounded sets, if for any nonempty bounded set , the modulus of total convexity of on is positive for any , where is defined by

We remark in passing that is totally convex on bounded sets if and only if is uniformly convex on bounded sets (see [24, 25]).

Proposition 5 (see [24]). Let be a convex function that its domain contains at least two points. If is lower semicontinuous, then is totally convex on bounded sets if and only if is uniformly convex on bounded sets.

Lemma 6 (see [11]). If , then the following statements are equivalent:(i)The function is totally convex at (ii)For any sequence ,

Recall that the function is called sequentially consistent [25], if for any two sequences and in such that is bounded,

Lemma 7 (see [14]). If contains at least two points, then the function is totally convex on bounded sets if and only if the function is sequentially consistent.

Lemma 8 (see [26]). Let be a differentiable and totally convex function. If and the sequence is bounded, then the sequence is also bounded.

Lemma 9 (see [12]). Let be a Legendre function such that is bounded on bounded subsets of . Let , and if is bounded, then the sequence is bounded too.

Recall that the Bregman projection [10] with respect to of onto a nonempty, closed, and convex set is the unique vector satisfying

Similar to the metric projection in Hilbert spaces, the Bregman projection with respect to totally convex and differentiable functions has a variational characterization ([25], corollary 4.4, p. 23).

Lemma 10 (see [25]). Suppose that is differentiable and totally convex on . Let and be a nonempty, closed, and convex set. Then, the following Bregman projection conditions are equivalent:(i)(ii) is the unique solution of the following variational inequality:(iii) is the unique solution of the following variational inequality:

Let be a real Banach space and be a nonempty subset of . An element is called a fixed point of a single-valued mapping , if . The set of fixed points of is denoted by .

A point is called an asymptotic fixed point of if contains a sequence which converges weakly to and . We denote the asymptotic fixed points of by .

Let be a nonempty, closed, and convex subset of and be a mapping. Now, is said to be Bregman quasi-nonexpansive, if and

Let be a nonempty, closed, and convex subset of . An operator is said to be Bregman strongly nonexpansive with respect to a nonempty , ifand for any bounded sequence withit follows that

A mapping is called Bregman inverse strongly monotone on the set , if , and for any , , and , we have that

Let be a mapping. Then, the mapping defined byis called an antiresolvent associated with and for any .

Suppose that is a mapping of into for the real reflexive Banach space . The effective domain of is denoted by , that is, . A multivalued mapping on is said to be monotone if for all , , and . A monotone operator on is said to be maximal if graph , the graph of , is not a proper subset of the graph of any monotone operator on .

Let be a real reflexive Banach space, uniformly Fréchet differentiable and bounded on bounded subsets of . Then for any , the resolvent of defined byis a single-valued Bregman quasi-nonexpansive mapping from onto and . We denote by the Yosida approximation of for any . We get from [26] (prop. 2.7, p. 10) that

See [11], too.

Lemma 11 (see [27]). Let be a real reflexive Banach space and be a Legendre function which is totally convex on bounded subsets of . Also, let be a nonempty, closed, and convex subset of and be a multivalued Bregman quasi-nonexpansive mapping. Then, the fixed point set of is a closed and convex subset of .

Lemma 12 (see[28]). Assume that is a Legendre function which is uniformly Fréchet differentiable and bounded on bounded subsets of . Let be a nonempty, closed, and convex subset of . Also, let be Bregman strongly nonexpansive mappings which satisfy for each and let . If and are nonempty, then is also Bregman strongly nonexpansive with .

Lemma 13 (see [29]). Let be a maximal monotone operator and be a Bregman inverse strongly monotone mapping such that . Also, let be a Legendre function which is uniformly Fréchet differentiable and bounded on bounded subsets of . Then,(i)(ii) is a Bregman strongly nonexpansive mapping such that(iii), , , and

Lemma 14 (see [30]). Let be a proper convex and lower semicontinuous Legendre function. Then, for any , for any and with , the following holds

Proposition 15 (see [26], prop. 2.8, p. 10). Let be differentiable and be a maximal monotone operator such that . Thenfor all , , and .

Next, we generalize the -demimetric notation introduced in [15].

Definition 16. Let be a reflexive Banach space, be a Legendre function which is differentiable, be a nonempty, closed and convex subset of and let . A mapping with is said to be Bregman -demimetric, if for and ,

Example 1. Every Bregman quasi-nonexpansive mapping with the required conditions in Definition 16 is a Bregman -demimetric mapping. Let and , and we haveTherefore,

Example 2. From [31] (Lemma 2.1), every Bregman quasi-strictly pseudocontractive mapping with the required conditions is a Bregman -demimetric mapping for .

Example 3. Let be a reflexive Banach space, be a Legendre function which is differentiable, and be a maximal monotone operator with and . Then, the -resolvent is Bregman -demimetric. In fact, from (22) and Proposition 15, we have thatfor any and . Then, we obtainTherefore,and hence, from (22), we have thatThus,and then, we get that is Bregman -demimetric.

3. Main Results

The following lemma is important and crucial in the proof of Theorem 18.

Lemma 17. Let be a reflexive Banach space and be a Legendre function which is differentiable, and let be a nonempty, closed, and convex subset of and let be a real number with and let be a Bregman -demimetric mapping of into . Then, is closed and convex.

Proof. First, we show that is closed. Consider a sequence such that and . We conclude from the definition of thatSince , we have . Then, from , we have that , and hence, , and therefore, . This implies that is closed.
Next, we show that is convex. Suppose and set , where . Then, we have thatThus, from and , we have thatFrom these inequalities, we get thatThus, we get that ; hence, ; therefore, . We conclude that is convex.

Theorem 18. Let be a real reflexive Banach space. Suppose that is a proper, convex, lower semicontinuous, strongly coercive, Legendre function which is bounded on bounded subsets of , uniformly Fréchet differentiable, and totally convex on bounded subsets of . Let be a nonempty, closed, and convex subset of . Let and be a finite family of Bregman -demimetric and Bregman quasi-nonexpansive and demiclosed mappings of into itself. Suppose that is a finite family of Bregman inverse strongly monotone mappings of into and is the family of antiresolvent mappings of . Let and be maximal monotone mappings on and let and be the resolvents of and for and , respectively. Assume thatFor and , let be a sequence defined bywhere , , , , , and satisfy the following:(1), (2), (3) and Then, converges strongly to a point where .

Proof. We divide the proof into several steps:(Step 1)First, we prove that is a closed and convex subset of .Since is a finite family of -Bregman demimetric mappings, by Lemma 17 and the condition , is nonempty, closed, and convex for . Also, it follows from Lemma 13 (i)–(ii) that and is a Bregman strongly nonexpansive mapping, and therefore, from Lemma 12, we have thatThus, is a family of Bregman quasi-nonexpansive mappings. Using and Lemma 11, we see that is a nonempty, closed, and convex set. We also know that is closed and convex. Then, is nonempty, closed, and convex. Therefore, is well defined.(Step 2)We prove that and are closed and convex subsets of and .In fact, it is clear that is closed and convex. Suppose that is closed and convex for some . Note thatSimilarly, we have thatThus, from the fact that is continuous for each fixed and using the above inequalities and which are closed and convex. We have also which is closed and convex. Therefore, is closed and convex. From mathematical induction, we have that is a closed and convex subset in with for all .
Also, it is clear that is closed and convex. Suppose that is closed and convex for some . Hence, is closed and convex; i.e., is a closed and convex subset of . Therefore, by mathematical induction, is a closed and convex subset of with for all . Next, we show that for all . Clearly, . Assume that for some . Note from Lemma 14 thatfor all . Furthermore, since is a Bregman inverse strongly monotone mapping for all and hence from Lemmas 13 and 14, we have thatfor all . Also, since is the resolvent of and , we have from (22) and Proposition 15 thatfor all . From (59)–(61), we have that . Therefore, we have by mathematical induction that for all .
Now, we shall show that for all . Note that . Assume that for some . Thus, for some . From and Lemma 10, we have thatSince , we have thatThen, we get . By mathematical induction, we have that for all . This implies that is well defined.(Step 3)We show that exists.Since is nonempty, closed, and convex, there exists a such that . From , we get thatfor all . From , we obtain thatThis shows that is a bounded sequence. By Lemma 10 (iii), we have thatfor all . Therefore,This implies that is bounded and nondecreasing. Then, exists. In view of Lemma 8, we deduce that the sequence is bounded. Also, from (66), we have thatSince the function is totally convex on bounded sets, by Lemma 7, we have that(Step 4)We prove that is a Cauchy sequence in .We have and for any with . From and Lemma 10, we have thatLetting in (70), we deduce that . In view of Lemma 7, since the function is totally convex on bounded sets, we get that as . Thus, is a Cauchy sequence. Since is a Banach space and is closed and convex, we conclude that there exists such that(Step 5)We prove that .Using (68) and from , we get thatThen, . Since the function is totally convex on bounded sets and is bounded, by Lemma 7, we have thatUsing (69) and (73), we have thatThen,Since from Lemma 1, is uniformly continuous, we have that(Step 6)We prove that .Using (72) and from , we get thatThen, . Since function is totally convex on bounded sets and is bounded, by Lemma 7, we have thatUsing (69) and (78), we have thatApplying (75) and (79), we get thatSince is uniformly continuous, we have that(Step 7)We prove that .Using and from (22), we get thatThen,Therefore, from (77), we have thatand then, . Since the function is totally convex on bounded sets and from Proposition 15 and Lemma 9, is bounded; hence, by Lemma 7, we have thatUsing (69) and (85), we conclude thatNow, by (79) and (86), we get thatSince from Lemma 1, is uniformly continuous, we have that(Step 8)We prove that .Since is Bregman -demimetric for all , we get thatfor all . We have from (76) thatfor all . Since the function is totally convex on bounded sets, by Lemma 7, we have thatSince is demiclosed for all and from (71) where as , we have that . We now show that . From (22) and Lemma 13 (iii), we get thatfor all and . Using (81), from the above, we have thatfor all . Since the function is totally convex on bounded sets, by Lemma 7, we conclude thatOn the other hand, it follows from (71) and (75) thatNow, from (94) and (95), we have that for all . From Lemma 13, we conclude thatand therefore, .
We now show that . Using , we have from (88) thatSo applying , the Yosida approximation of , we haveSince , for , we have from the monotonicity of that for all . By (71) and (79), we have thatFrom (87) and the above, we have . Thus, we get . From the maximality of , we have that . Therefore, .
Since , , and , we have from (65) thatwhich implies thatTherefore,From (55), (71), and for all and also since is uniformly continuous on bounded subsets and is bounded on bounded subsets, we have thatNow, from (22) and (102), we get thatFrom (103) and the above, we conclude that , and therefore, . Thus, , and hence, . This completes the proof.

Next, we prove a proposition to extend Theorem 18.

Proposition 19. Let be a real reflexive Banach space. Suppose that is a proper, convex, lower semicontinuous, Legendre function and is differentiable of . Let be a nonempty, closed, and convex subset of . Suppose and a mapping with is Bregman -demimetric. Then, is a Bregman quasi-nonexpansive mapping.

Proof. Let and . Using (22) and Definition 16, we haveThen, is a Bregman quasi-nonexpansive mapping.

Remark 20. Using Proposition 19, for each , we may remove the condition that is Bregman quasi-nonexpansive in Theorem 18 when .

Open Problem 1. Can one generalize Proposition 19 to Bregman -demimetric mappings with ?

Data Availability

No data were used to support the study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.