Orthogonally Biadditive Operators

Dzhusoeva, Nonna; Kulaev, Ruslan; Pliev, Marat

doi:https://doi.org/10.1155/2021/2593884

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Research Article | Open Access

Volume 2021 | Article ID 2593884 | https://doi.org/10.1155/2021/2593884

Orthogonally Biadditive Operators

Nonna Dzhusoeva,¹Ruslan Kulaev,^1,2and Marat Pliev²

Academic Editor: Ismat Beg

Received12 Sept 2021

Accepted03 Nov 2021

Published24 Nov 2021

Abstract

In this article, we introduce and study a new class of operators defined on a Cartesian product of ideal spaces of measurable functions. We use the general approach of the theory of vector lattices. We say that an operator defined on a Cartesian product of vector lattices and and taking values in a vector lattice is orthogonally biadditive if all partial operators and are orthogonally additive. In the first part of the article, we prove that, under some mild conditions, a vector space of all regular orthogonally biadditive operators is a Dedekind complete vector lattice. We show that the set of all horizontally-to-order continuous regular orthogonally biadditive operators is a projection band in . In the last section of the paper, we investigate orthogonally biadditive operators on a Cartesian product of ideal spaces of measurable functions. We show that an integral Uryson operator which depends on two functional variables is orthogonally biadditive and obtain a criterion of the regularity of an orthogonally biadditive Uryson operator.

1. Introduction and Preliminaries

Orthogonally additive operators in vector lattices first were introduced by Mazón and Segura de León in [1]. Today, the theory of these operators is an active field of the modern analysis (see [2–9]). We note that the study of orthogonally additive operators has useful applications in different areas of modern mathematics, e.g., convex geometry [10, 11], dynamical systems [12], and nonlinear integral equations [13, 14]. In applications, it is often necessary to study integral equations depending on several variables. Nonlinear operators in two variables satisfying the natural condition of the orthogonal additivity with respect to each variable are often appear in applications ([15]). Such operators in the literature are called orthogonally biadditive. We note that this notion is traced back to paper [16] by Mizel and Sundaresan. In present note, we investigate orthogonally biadditive operators in the general setting of the theory of vector lattices. We note that the tools of the theory of vector lattices turned out to be useful and effective in solving a number of problems of the theory of linear integral operators in ideal spaces [17]. The nonlinear integral operators of Uryson and Hammerstein were investigated by methods of the theory of ordered spaces in [14, 18].

Let us describe the content of the article. In the following section, we briefly present a necessary information on vector lattices and orthogonally additive operators. In the next section, we investigate the vector space of all orthogonally biadditive operators defined on a Cartesian product of vector lattices and and taking values in a vector lattice . It turned out that there is a natural partial order on . We get the lattice calculus of orthogonally biadditive operators and prove the first main result of the paper stated that for a Dedekind complete vector lattice the vector space of all regular orthogonally biadditive operators defined on a Cartesian product of vector lattices and and taking values in a Dedekind complete vector lattice is a Dedekind complete vector lattice. Then, we explore the special type of horizontally-to-order continuous regular orthogonally biadditive operators. We prove that the vector space of these operators is a projection band in .

In the last section, we investigate orthogonally biadditive operators defined on a Cartesian product of ideal spaces of measurable functions. We show that a nonlinear superposition operator and a Uryson integral operator depending on two variables are orthogonally biadditive in appropriate function spaces. We note that in the classical theory of integral operators, all information concerning an integral operator is encoded by the properties of its kernel. In the final section of the paper, we show that the same is true for integral orthogonally biadditive operators and obtain the second main result of the article which is a criteria for the regularity of an orthogonally biadditive Uryson operator. It is worth noting that the general theory of orthogonally biadditive operators developed below is aimed at getting an additional information on the abovementioned particular operators. This article is the beginning of a project devoted to the study of analytic, algebraic, and order properties of orthogonally biadditive operators.

Now we state our main results. All unexplained notions are defined in next sections.

Theorem 1. Let be vector lattices and be a Dedekind complete vector lattice. Then, is a Dedekind complete vector lattice, and for all and , the following relations hold: (1)(2)(3)(4)(5)(6)

Theorem 2. Let , , and be finite measure spaces; , , and be ideal subspaces of , , and , respectively; be a normalized Carathéodory function; and be an integral Uryson operator with the kernel . Then, the following statements are equivalent: (1) is a regular operator(2) is a positive integral Uryson operator with the kernel

Here, we provide some necessary facts and notations that we need in the further presentation. The standard reference book on the theory of vector lattices is [19]. All vector lattices we consider below are supposed to be Archimedean. The term “operator” between vector spaces and means in general a nonlinear map . We say that two elements of a vector lattice are disjoint and write , if . We write if and for all . In particular, for , we use the notation . We say that is a fragment (a component) of and use the symbol , if . The set of all fragments of an element is denoted by . We say that are mutually complemented, if . For vector lattices and by , we denote the Cartesian product of and . We observe that is a vector lattice with the pointwise algebraic and lattice operations. Namely, for all and , we have that

Let be a finite measure space. By (or for shortness), we denote the vector space of all real valued measurable functions on . More precisely, consists of equivalence classes of such functions, where two functions and are said to be equivalent if for -almost all . We note that is equipped with the natural partial order, that is

It is worth noting that is a Dedekind complete vector lattice (see [20], page 52). We say that a vector subspace of is an ideal space if for every , the relation implies that . In particular, the classical -spaces are typical examples of ideal spaces. For a given by , we denote the measurable set . The characteristic function of a set is denoted by . The union of two disjoint sets and we denote by . The set of all maps from to we denote by .

Definition 3. Let be a vector lattice and let be a real vector space. An operator is said to be orthogonally additive if for all disjoint elements . It follows from the definition that .

We observe that classical operators of nonlinear analysis such as Uryson, Hammerstein, and Nemytskii operators are orthogonally additive in suitable function spaces (see [1]).

2. The Vector Lattice of Regular Orthogonally Biadditive Operators

In this section, we introduce a notion of an orthogonally biadditive operator and prove that the vector space of all regular orthogonally biadditive operators defined on the Cartesian product of vector lattice and and taking values in a Dedekind complete vector lattice is a Dedekind complete vector lattice.

Definition 4. Let be vector lattices and be a vector space. With an operator is associated two families of partial operators , , and , defined by setting:

We say that is an orthogonally biadditive operator (OBAO) if all , , and , are orthogonally additive operators from to and to , respectively. The vector space of all orthogonally biadditive operators from to we denote by .

It is clear from the definition that for all and . We note that an OBAO need not be orthogonally additive as an operator defined on the vector lattice . Indeed, if , then the operator defined by setting is an OBAO; however, for disjoint elements and of , one has

Now we present some examples of OBAOs.

Example 5. Every bilinear operator is orthogonally biadditive.

Example 6. Suppose that . Then, coincides with the vector space of all function such that for all .

Definition 7. Let be vector lattices. An orthogonally biadditive operator is said to be: (i)Positive if for all (ii)-bounded, if it maps to order bounded sets in for every (iii)Regular, if , where are positive orthogonally biadditive operators from to

The sets of all positive, -bounded, and regular orthogonally biadditive operators from to we denote by , , and , respectively. There is a natural partial order on , namely, .

Proposition 8. Let be vector lattices. Then, every is -bounded.

Proof. Suppose that with . Fix and take . Then, , and for every, , we have that It follows that , and therefore for all .☐

Now we need the following auxiliary statement.

Proposition 9 (see [21], Prop. 3.11). Let be a vector lattice and for some and Then, there exist a family of pairwise disjoint elements , where and such that (i) for any (ii) for any (iii)

Now we ready to prove the first main result of the article.

Proof of Theorem 1. First we prove (1). Put, by definition Since , then for all decompositions , , and all maps , we have that where are positive orthogonally biadditive operators such that and . Thus, is an order bounded subset of and by the Dedekind completeness of there exists . We show that is an orthogonally biadditive operator. Fix , disjoint elements and partitions and . By Proposition 9, for every , there exists a decomposition such that and . Take . Then Since and , we have that , and therefore, . Let us prove that converse inequality. Pick and , where Suppose that . Adding, if necessary zero fragments to the sum , we may assume that and By Proposition 9, there is a family of pairwise disjoint elements , where and such that (i) for every (ii) for every (iii)Then, we may write Thus Then, , and we have that . Hence, . Since , we have that and . Suppose is an orthogonally biadditive operator with and for all . Then for all disjoint decompositions , , , and all functions . Hence, and . Now we are in the position to derive the other formulas of the lattice calculus. Assuming that and we get formulas for the positive and the negative parts of . The formula for the modulus is obvious. Now we prove inequality (6). Take trivial decomposition , , and with and . Then It remains to show the Dedekind completeness of the vector lattice . Take a family of positive OBAOs with . Without a loss of generality, we may assume that is an upward directed set. Define an operator as . Since the vector lattice is Dedekind complete, the operator is well defined. Let us show the orthogonal biadditivity of . Fix and with . Then, we have that and we deduce that is an orthogonally additive operator. Similar arguments are valid for for all . Clearly, .

Definition 10. Let be a vector lattice. A net in horizontally converges (or laterally converges in another terminology) to an element (notation ) if the net order converges to and for all with .

Definition 11. Let and be vector lattices. An operator is said to be: (i)Horizontally-to-order continuous (or laterally-to-order continuous) if every horizontally convergent net in with maps to an order convergent net in with (ii)Horizontally-to-order-continuous (or laterally-to-order-continuous) if every horizontally convergent sequence in with maps to an order convergent sequence in with

The vector space of all horizontally-to-order continuous (-continuous) orthogonally additive operators from to is denoted by ().

We observe that this class of operators has been studied in [22–25]. It is worth noting that the Dedekind completeness of a vector lattice implies the relation ([24], Theorem 3.6., Lemma 3.12).

Definition 12. Let , , and be vector lattices. An orthogonally biadditive operator is called separately horizontally-to-order continuous (-continuous), if partial operators and are horizontally-to-order continuous (-continuous), for all , . The sets of all horizontally-to-order continuous (-continuous) and separately horizontally-to-order continuous regular OBAOs we denote by () and (), respectively.

Example 13. Suppose that . Then, . Indeed, since , we have that every horizontally convergent net in with is the constant one, that is for all where is some index.

The next theorem has its own interest.

Theorem 14. Let , , and be vector lattices with Dedekind complete. Then, and are projection bands in .

Proof. We prove the assertion for ; the proof for is similar. It is clear that is a vector space. We show that is an order ideal of . Suppose that . We show that too. Indeed, take and a horizontally convergent net in with . We need to prove that . Since , we have that , and therefore On the other hand, by Theorem 1 we have that Put, by definition Pick . By Proposition 9, for every , there exists a decomposition such that for all . Since by assumptions above all partial operators are horizontally-to-order continuous, we have that Passing to the supremum in the left-hand side of the above inequality over all elements of , we deduce that and therefore, . The horizontal-to-order continuity of a partial operator for can be proved by the same way. Now we prove that is an order ideal of . Suppose that , , and . Then, () for every () and by ([24], Theorem 3.13) we have that . It remains to show that is a band in . Pick a net in with for some . Then, we have that () for every (). Now, applying ([24], Theorem 3.13), we obtain that . Finally, taking into account the Dedekind completeness of , we deduce that is the projection band in .
Now we are ready to prove that the Dedekind completeness of a vector lattice implies the horizontal-to-order continuity (-continuity) of a regular separately horizontal-to-order continuous (-continuous) operator .☐

Proposition 15. Let and be vector lattices, be a Dedekind complete vector lattice, and . Then, the following statements hold: (1)(2)

Proof. We prove statement (1). The implication is obvious. Suppose that . We need to show horizontal-to-order continuity of . Pick a horizontally convergent net with . Then, and . Now we may write Taking into account the separate horizontal-to-order continuity of , we have that . Now, suppose that is an arbitrary element of . Then, by Theorem 14, every has the representation , where . Hence, by above, we have that .☐

3. Orthogonally Biadditive Operators on a Cartesian Product of Ideal Spaces of Measurable Functions

In this section, we consider orthogonally biadditive operators in lattices of measurable functions and obtain a criteria of the regularity of an integral Uryson operator.

Definition 16. Suppose that and are finite measure space and is the product measure on . We say that is a superpositionally measurable (or sup-measurable for shortness) function, if is -measurable for each . A sup-measurable function is said to be normalized if for -almost all .

The following proposition provides an important example of an orthogonally biadditive operator.

Proposition 17. Let be a normalized sup-measurable function and and be order ideals of and , respectively. Then, the map defined by is an orthogonally biadditive operator from to .

Proof. Take and . Put and . We note that the relations and imply that , . Then, , and therefore, the operator is well defined. Fix . We show that the partial operator is orthogonally additive. Indeed, take disjoint . Then, and are disjoint elements of , and we may write Noting that similar arguments are valid for a partial operator , for all , we finish the proof.☐

We observe that is known as the nonlinear superposition operator or Nemytskii operator. The basic constructions of the theory of Nemytskii operators are presented in [26]. Recently, nonlinear superposition operators were investigated in [2, 27, 28].

Definition 18. Let , , and be finite measure spaces. By , we denote the completion of their product measure space. A map is said to be a Carathéodory function if it is satisfies the following conditions:
(C₁) is -measurable for all
(C₂) is continuous on for -almost all
We say that a Carathéodory function is normalized if for -almost all and all .

Proposition 19. Let be a normalized Carathéodory function, and . Then, .

Proof. First we note that -measurable null sets depend of and , respectively. We claim that there exists -measurable set such that for all and all . Indeed, consider two sequences of -measurable sets and and put Since , we have that . Fix and (). Then, there exists a sequence () in that converges to () with () for all (). Then, by , we have that ().
Now, we show that for arbitrary , , and . We claim that for almost all . Indeed, pick . If and , we have If either or , then Suppose that and are pairwise disjoint measurable subsets of and , respectively, and are simple functions in and , respectively. Then and we deduce that . Finally, assume that and are arbitrary elements of and , and are sequences of simple functions in and , respectively, such that converges to -a.e. and converges to -a.e. Put by definition Clearly, Then, converges to for all , and therefore, .☐

Remark 20. Using similar arguments as above, we get the following useful equalities: for almost all and all ().

The next proposition provides an important example of an orthogonally biadditive operator.

Proposition 21. Let be a normalized Carathéodory function, and be order ideals of and , respectively, and for all , , and -almost all . Then, the map defined by setting is an orthogonally biadditive operator from to .

Proof. By Proposition 19, is a well-defined operator from to . We show the orthogonal additivity of a partial operator , where . Fix disjoint . Then, taking into account considerations above, we may write ☐

The orthogonal additivity of a partial operator , can be proved analogously.

We observe that an operator above can be considered as the Uryson integral operator that depends on two variables. We say that a function is a kernel of an operator . Classical integral Uryson operators were investigated by many mathematicians (see for instance monograph [29]).

The following example of an OBAO is a Hammerstein operator which depends on two variables.

Example 22. Let , , and be as above, be a normalized sup-measurable function, and be order ideals in and, respectively, be a -measurable function, and for all , , and -almost all . Then, the following formula defines a OBAO

We note that biorthogonal additivity of a superposition operator implies that . The operator can be treated as an integral Hammerstein operator that depends on two variables.

Now we are ready to prove the second main result of the article.

Proof of Theorem 2. . Since and , we have that , and therefore, .
. By Theorem 1, the modulus exists and can be calculated by the formula Fix , . By Remark 20, for almost all , we have that We also note that for -almost all . Put by definition Clearly, and are -measurable sets and We may assume that and . Otherwise, if (), then the following equalities hold for -almost all , and it is nothing to prove. We observe that where , , , , and . First, we assume that . Then, we see that Consider the trivial decompositions and . Then, for -almost all , we have that The same arguments are valid for the case . Now, suppose that and . Then, there is a decomposition , where , , and . Clearly . Put We observe that , , and . Now we may write Thus, for all , we have that On the other hand, for all , we have Consider the second sum . Then, for all , we have and for all , we have that Put by definition Then, for all , we have that Thus, and map to . On the other hand, since , , and , we have that .

Corollary 23. Let , , and be finite measure spaces and and be ideal subspaces of and , respectively. Then, every integral Uryson operator is regular.

Proof. Suppose is an integral Uryson operator with kernel . Taking into account that the function is -measurable for each and by Theorem 2 we deduce that is a regular operator.☐

Proposition 24. Let , , and be finite measure spaces, , , and be ideal subspaces of , , and , respectively, and be a regular integral Uryson operator with a kernel . Then, is a horizontally-to-order continuous operator.

Proof. By Proposition 15, it is enough to prove the separate horizontal-to-order continuity of . Fix . We show the horizontal-to-order continuity of the partial operator . It is worth noting that the countable sup property of (see [20], page 52) implies that the concept of a sequentially order continuity for functionals and operators coincides with the concept of order continuity. Pick a sequence which horizontally converges to . We need to show that the sequence order converges to . Taking into the account the regularity of , we may write ☐

Since converges to by ([30], Theorem 2.5.7), we have that order converges to , and therefore, is a horizontally-to-order continuous operator. Similar arguments are valid for a partial operator , .

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Acknowledgments

Nonna Dzhusoeva and Ruslan Kulaev were supported by the Ministry of the Sciences and Education of the Russian Federation (number of the agreement is 075-02-2021-1552).

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Copyright © 2021 Nonna Dzhusoeva et al. 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.

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