# Orthogonally Biadditive Operators

Authors: Nonna Dzhusoeva; Ruslan Kulaev; Marat Pliev
Journal: Journal of Function Spaces
 (2021)

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## 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 operatorT:E×F⟶W defined on a Cartesian product of vector lattices E and F and taking values in a vector lattice W is orthogonally biadditive if all partial operators Ty:E⟶W and Tx:F⟶W 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 OBArE,F;W 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 OBArE,F;W. 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.


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## Body

## 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 spaceOBAE,F;W of all orthogonally biadditive operators defined on a Cartesian product of vector lattices E and F and taking values in a vector lattice W. It turned out that there is a natural partial order on OBAE,F;W. 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 W the vector space OBArE,F;W of all regular orthogonally biadditive operators defined on a Cartesian product of vector lattices E and F and taking values in a Dedekind complete vector lattice W 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 OBArE,F;W.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.
LetE,F be vector lattices and W be a Dedekind complete vector lattice. Then, OBArE,F;W is a Dedekind complete vector lattice, and for all T,T1,T2∈OBArE,F;W and x,y∈E×F, the following relations hold:
(1)
T1∨T2x,y≔sup∑i=1n∑j=1mTki,jxi,yj:x=∐i=1nxi;y=∐j=1myj;n,m∈N;k∈1,21,⋯,n×1,⋯,m(2)
T1∧T2x,y≔inf∑i=1n∑j=1mTki,jxi,yj:x=∐i=1nxi;y=∐j=1myj;n,m∈N;k∈1,21,⋯,n×1,⋯,m(3)
T+x,y≔sup∑i=1n∑j=1mTki,jxi,yj:∐i=1nxi⊑x;∐j=1myj⊑y;n,m∈N;k∈1,21,⋯,n×1,⋯,m(4)
T−x,y≔−inf∑i=1n∑j=1mTki,jxi,yj:∐i=1nxi⊑x;∐j=1myj⊑y;n,m∈N;k∈1,21,⋯,n×1,⋯,m(5)
∣T∣x,y≔sup∑i=1n∑j=1m−1ki,jTxi,yj:x=∐i=1nxi;y=∐j=1myj;n,m∈N;k∈1,21,⋯,n×1,⋯,m(6)
∣Tx,y∣≤∣T∣x,yTheorem 2.
LetC,Θ,λ, A,Σ,μ, and B,Ξ,ν be finite measure spaces; E, F, and J be ideal subspaces of L0μ, L0ν, and L0λ, respectively; K:C×A×B×R2 be a normalized Carathéodory function; and T:E×F⟶J be an integral Uryson operator with the kernel K. Then, the following statements are equivalent:
(1)
T is a regular operator(2)
∣T∣:E×F⟶J is a positive integral Uryson operator with the kernel ∣K∣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 E and W means in general a nonlinear map T:E⟶W. We say that two elements x,y of a vector lattice E are disjoint and write x⊥y, if x∧y=0. We write x=∐i=1nxi if x=∑i=1nxi and xi⊥xj for all i≠j. In particular, for n=2, we use the notation x=x1⊔x2. We say that y is a fragment (a component) of x∈E and use the symbol y⊑x, if y⊥x−y. The set of all fragments of an element x∈E is denoted by Fx. We say that x1,x2∈Fx are mutually complemented, if x=x1⊔x2. For vector lattices E and F by E×F, we denote the Cartesian product E×F≔x,y:x∈E,y∈F of E and F. We observe that E×F is a vector lattice with the pointwise algebraic and lattice operations. Namely, for all x,u∈E and y,v∈F, we have that
(1)x,y≤u,v⇔x≤uandy≤v,x,y∨u,v=x∨u,y∨v;x,y∧u,v=x∧u,y∧v,∣x,y∣=∣x∣,∣y∣.LetA,Σ,μ be a finite measure space. By L0A,Σ,μ (or L0μ for shortness), we denote the vector space of all real valued measurable functions on A. More precisely, L0μ consists of equivalence classes of such functions, where two functions f1 and f2 are said to be equivalent if f1s=f2s for μ-almost all s∈A. We note that L0μ is equipped with the natural partial order, that is
(2)f≤h⇔fs≤hsμ−a.e.s∈A;f,h∈L0μ.It is worth noting thatL0μ is a Dedekind complete vector lattice (see [20], page 52). We say that a vector subspace E of L0μ is an ideal space if for every f∈L0μ, h∈E the relation ∣f∣≤∣h∣ implies that f∈E. In particular, the classical Lpμ-spaces are typical examples of ideal spaces. For a given f∈L0μ by suppf, we denote the measurable set t∈A:ft≠0. The characteristic function of a set D is denoted by 1D. The union H∪D of two disjoint sets H and D we denote by H⊔D. The set of all maps from H to D we denote by DH.Definition 3.
LetE be a vector lattice and let X be a real vector space. An operator T:E⟶X is said to be orthogonally additive if Tx+y=Tx+Ty for all disjoint elements x,y∈E. It follows from the definition that T0=0.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 productE×F of vector lattice E and F and taking values in a Dedekind complete vector lattice W is a Dedekind complete vector lattice.Definition 4.
LetE,F be vector lattices and W be a vector space. With an operator T:E×F⟶W is associated two families of partial operators Tx:F⟶W, x∈E, and Ty:E⟶W, y∈F defined by setting:
(3)Txv≔Tx,v,v∈F;Tyu≔Tu,y,u∈E.We say thatT:E×F⟶W is an orthogonally biadditive operator (OBAO) if all Tx:F⟶W, x∈E, and Ty:E⟶W, y∈F are orthogonally additive operators from E to W and F to W, respectively. The vector space of all orthogonally biadditive operators from E×F to W we denote by OBAE,F;W.It is clear from the definition thatT0,y=Tx,0=0 for all x∈E and y∈F. We note that an OBAO T:E×F⟶W need not be orthogonally additive as an operator defined on the vector lattice E×F. Indeed, if E=F=W=R, then the operator T:E×F⟶W defined by setting
(4)Tx,y≔xy,x,y∈ℝ2,is an OBAO; however, for disjoint elements s=0,1 and t=1,0 of E×F, one has
(5)Ts+t=1≠0=Ts+Tt.Now we present some examples of OBAOs.Example 5.
Every bilinear operatorT:E×F⟶W is orthogonally biadditive.Example 6.
Suppose thatE=F=W=R. Then, OBAE,F;W coincides with the vector space of all function f:R2⟶R such that f0,y=fx,0=0 for all x,y∈R.Definition 7.
LetE,F,W be vector lattices. An orthogonally biadditive operator T:E×F⟶W is said to be:
(i)
Positive if Tx,y≥0 for all x,y∈E×F(ii)
C-bounded, if it maps Fx,y to order bounded sets in W for every x,y∈E×F(iii)
Regular, if T=S1−S2, where S1,S2 are positive orthogonally biadditive operators from E×F to WThe sets of all positive,C-bounded, and regular orthogonally biadditive operators from E×F to W we denote by OBA+E,F;W, OBAcbE,F;W, and OBArE,F;W, respectively. There is a natural partial order on OBArE,F;W, namely, S≤T⇔T−S∈OBA+E,F;W.Proposition 8.
LetE,F,W be vector lattices. Then, every T∈OBArE,F;W is C-bounded.Proof.
Suppose thatT=S1−S2 with S1,S2∈OBA+E,F;W. Fix x,y∈E×F and take v,u∈Fx,y. Then, x−v,y−u⊥v,u, and for every, i∈1,2, we have that
(6)Six,y=Sivòx−v,uòy−u=Siv,u+Siv,y−u+Six−v,u+Six−v,y−u.
It follows thatSiv,u≤Six,y, and therefore
(7)Tv,u=S1v,u−S2v,u≤S1x,y+S2x,y,for all v,u∈Fx,y.☐Now we need the following auxiliary statement.Proposition 9 (see [21], Prop. 3.11).
LetE be a vector lattice and ∐i=1nxi=∐k=1myk for some xii=1n and ykk=1m⊂E. Then, there exist a family of pairwise disjoint elements zik⊂E, where i∈1,⋯,n and k∈1,⋯,m such that
(i)
xi=∐k=1mzik for any i∈1,⋯,n(ii)
yk=∐i=1nzik for any k∈1,⋯,m(iii)
∐i=1n∐k=1mzik=∐i=1nxi=∐k=1mykNow we ready to prove the first main result of the article.Proof of Theorem 1.
First we prove (1). Put, by definition(8)Rx,y≔∑i=1n∑j=1mTki,jxi,yj:x=∐i=1nxi;;y=∐j=1myj;n,m∈N;k∈1,21,⋯,n×1,⋯,m.
SinceT1,T2∈OBArE,F;W, then for all decompositions x=∐i=1nxi, y=∐j=1myj, and all maps k:1,⋯,n×1,⋯,m⟶1,2, we have that
(9)∑i=1n∑j=1mTki,jxi,yj≤∑i=1n∑j=1mS11+S21+S12+S22xi,yj=S11+S21+S12+S22x,y,where S11,S21,S12,S22 are positive orthogonally biadditive operators such that T1=S11−S21 and T2=S12−S22. Thus, Rx,y is an order bounded subset of W and by the Dedekind completeness of W there exists Rx,y≔supRx,y. We show that R:E×F⟶W is an orthogonally biadditive operator. Fix y∈F, disjoint elements u,v∈E and partitions uòv=∐i=1nxi and y=∐j=1myj. By Proposition 9, for every i∈1,⋯,n, there exists a decomposition xi=xi1òxi2 such that u=∐i=1nxi1 and v=∐i=1nxi2. Take ∑i=1n∑j=1mTki,jxi,yj∈Ruòv,y. Then
(10)∑i=1n∑j=1mTki,jxi,yj=∑i=1n∑j=1mTki,jxi1òxi2,yj=∑i=1n∑j=1mTki,jxi1,yj+∑i=1n∑i=1nTki,jxi2,yj.
Since∑i=1n∑j=1mTki,jxi1,yj∈Ru,y and ∑i=1n∑j=1mTki,jxi2,yj∈Rv,y, we have that Ru⊔v,y⊂Ru,y+Rv,y, and therefore, Ru⊔v,y≤Ru,y+Rv,y. Let us prove that converse inequality. Pick ∑i=1n∑j=1mTki,jxi1,yj∈Ru,y and ∑t=1l∑s=1rTkt,sxt2,ys∈Rv,y, where
(11)u=∐i=1nxi1;v=∐t=1lxi2;y=∐j=1myj;y=∐s=1rys.
Suppose thatl≤n. Adding, if necessary zero fragments to the sum ∐t=1lxi2, we may assume that l=n and
(12)u=∐i=1nxi1;v=∐i=1nxi2.
By Proposition9, there is a family of pairwise disjoint elements wjs⊂E, where j∈1,⋯,m and s∈1,⋯,r such that
(i)
yj=∐s=1rwjs for every j∈1,⋯,m(ii)
ys=∐i=1nwjs for every s∈1,⋯,r(iii)
y=∐j=1m∐s=1rwjs
Then, we may write(13)∑i=1n∑j=1mTki,jxi1,yj=∑i=1n∑j=1mTki,jxi1,∐s=1rwjs=∑i=1n∑j=1m∑s=1rTki,jxi1,wjs,∑i=1n∑s=1rTkt,sxt2,ys=∑t=1l∑s=1rTki,sxi2,∐j=1mwjs=∑i=1n∑s=1r∑j=1mTki,sxi2,wjs.
Thus(14)∑i=1n∑j=1m∑s=1rTki,jxi1,wjs+∑i=1n∑s=1r∑j=1mTki,sxt2,wjs=∑i=1n∑s=1r∑j=1mTki,sxi1⊔xi2,wjs∈Ru⊔v,y.
Then,Ru,y+Rv,y⊂Ru⊔v,y, and we have that Ru,y+Rv,y≤Ru⊔v,y. Hence, Ru⊔v,y=Ru,y+Rv,y. Since T1x,y,T2x,y∈Rx,y, we have that T1x,y≤Rx,y and T2x,y≤Rx,y. Suppose H:E×F⟶W is an orthogonally biadditive operator with T1x,y≤Hx,y and T2x,y≤Hx,y for all x,y∈E×F. Then
(15)Hx,y=H∐i=1nxi,∐j=1myj=∑i=1n∑j=1mHxi,yj≥∑i=1n∑j=1mTki,jxi,yj,for all disjoint decompositions x=∐i=1nxi, y=∐j=1myj, n,m∈N, and all functions k:1,⋯,n×1,⋯,m⟶1,2. Hence, H≥R and R=T1∨T2. Now we are in the position to derive the other formulas of the lattice calculus.
(16)T1∧T2x,y=−−T1∨−T2x,y=−sup∑i=1n∑i=1n−Tki,jxi,yj:x=∐i=1nxi;y=∐j=1myj;n,m∈N;k∈1,21,⋯,n×1,⋯,m=inf∑i=1n∑i=1nTki,jxi,yj:x=∐i=1nxi;y=∐j=1myj;n,m∈N;k∈1,21,⋯,n×1,⋯,m.
Assuming thatT1=T and T2=0 we get formulas for the positive and the negative parts of T. The formula for the modulus T is obvious. Now we prove inequality (6). Take trivial decomposition x=x, y=y, and k,k′∈1,21×1 with k1,1=1 and k′1,1=2. Then
(17)∣Tx,y∣=Tx,y∨−Tx,y=−1k′1,1Tx,y∨−1k1,1Tx,y≤sup∑i=1n∑i=1n−1ki,jTxi,yj:x=∐i=1nxi;y=∐j=1myj;n,m∈N;k∈1,21,⋯,n×1,⋯,m=∣T∣x,y.
It remains to show the Dedekind completeness of the vector latticeOBArE,F;W. Take a family Tαα∈A of positive OBAOs with 0≤Tα≤T∈OBA+E,F;W. Without a loss of generality, we may assume that Tαα∈A is an upward directed set. Define an operator G:E×F⟶W as Gx,y=supα∈ATαx,y. Since the vector lattice W is Dedekind complete, the operator G is well defined. Let us show the orthogonal biadditivity of G. Fix y∈F and x,v∈E with x⊥v. Then, we have that
(18)Gx⊔v,y=supαTαx⊔v,y=supαTαx,y+supαTαv,y=Gx,y+Gv,y,and we deduce that Gy is an orthogonally additive operator. Similar arguments are valid for Gx for all x∈E. Clearly, G=supα∈ATα.Definition 10.
LetE be a vector lattice. A net eαα∈A in E horizontally converges (or laterally converges in another terminology) to an element e∈E (notation eα⟶he) if the net eαα∈A order converges to e and eα⊑eβ⊑e for all α,β∈A with α≤β.Definition 11.
LetE and F be vector lattices. An operator T:E⟶F is said to be:
(i)
Horizontally-to-order continuous (or laterally-to-order continuous) if every horizontally convergent net eαα∈A in E with eα⟶heT maps to an order convergent net Teαα∈A in F with Teα⟶oTe(ii)
Horizontally-to-orderσ-continuous (or laterally-to-orderσ-continuous) if every horizontally convergent sequence enn∈N in E with en⟶heT maps to an order convergent sequence Tenn∈N in F with Ten⟶oTeThe vector space of all horizontally-to-order continuous (σ-continuous) orthogonally additive operators from E to F is denoted by OAcE,F (OAσcE,F).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 F implies the relation OAcE,F⊂OArE,F ([24], Theorem 3.6., Lemma 3.12).Definition 12.
LetE, F, and W be vector lattices. An orthogonally biadditive operator T:E×F⟶W is called separately horizontally-to-order continuous (σ-continuous), if partial operators Tx and Ty are horizontally-to-order continuous (σ-continuous), for all x∈E, y∈F. The sets of all horizontally-to-order continuous (σ-continuous) and separately horizontally-to-order continuous regular OBAOs we denote by OBAcE,F;W (OBAσcE,F;W) and OBAscE,F;W (OBAsσcE,F;W), respectively.Example 13.
Suppose thatE=F=W=R. Then, OBArE,F;W=OBAscE,F;W=OBAscE,F;W. Indeed, since Fe=0,e, we have that every horizontally convergent net eαα∈A in E with eα⟶he is the constant one, that is eα=e for all α≥α0 where α0∈A is some index.The next theorem has its own interest.Theorem 14.
LetE, F, and W be vector lattices with W Dedekind complete. Then, OBAscE,F;W and OBAsσcE,F;W are projection bands in OBArE,F;W.Proof.
We prove the assertion forOBAscE,F;W; the proof for OBAsσcE,F;W is similar. It is clear that OBAscE,F;W is a vector space. We show that OBAscE,F;W is an order ideal of OBArE,F;W. Suppose that T∈OBAscE,F;W. We show that ∣T∣∈OBAscE,F;W too. Indeed, take y∈F and a horizontally convergent net xαα∈A in E with xα⟶hx. We need to prove that Tyxα⟶oTyx. Since ∣T∣∈OBA+E,F;W, we have that Ty∈OA+E,W, and therefore
(19)o−limαTyxα=supαTyxα≤Tyx.
On the other hand, by Theorem1 we have that
(20)Tyx=∣T∣x,y=sup∑i=1n∑i=1n−1ki,jTxi,yj:x=∐i=1nxi;y=∐j=1myj,n,m∈N;k∈1,21,⋯,n×1,⋯,m.
Put, by definition(21)R≔∑i=1n∑j=1m−1ki,jTxi,yj:x=∐i=1nxi;y=∐j=1myj,n,m∈N;k∈1,21,⋯,n×1,⋯,m.
Pick∑i=1n∑j=1m−1ki,jTxi,yj∈R. By Proposition 9, for every α∈A, there exists a decomposition xα=∐i=1nxαi such that xαi⟶hxi for all 1,⋯,n. Since by assumptions above all partial operators Tyj:F⟶W are horizontally-to-order continuous, we have that
(22)∑i=1n∑i=1n−1ki,jTxi,yj=∑i=1n∑j=1m−1ki,jTyjxi=o−limα∈A∑i=1n∑j=1m−1ki,jTyjxαi=∑i=1n∑j=1m−1ki,jTyjh−limα∈Axαi≤∑i=1n∑j=1msupα∈ATyjxαi=supα∈ATyxα.
Passing to the supremum in the left-hand side of the above inequality over all elements ofR, we deduce that
(23)T∣yx≤supα∈ATyxα=o−limα∈ATyxα,and therefore, Tyx=o−limα∈ATyxα. The horizontal-to-order continuity of a partial operator Tx for x∈E can be proved by the same way. Now we prove that OBAscE,F;W is an order ideal of OBArE,F;W. Suppose that 0≤T∈OBAscE,F;W, 0≤S∈OBArE,F;W, and 0≤S≤T. Then, 0≤Sy≤Ty (0≤Sx≤Tx) for every y∈F (x∈E) and by ([24], Theorem 3.13) we have that S∈OBAscE,F;W. It remains to show that OBAscE,F;W is a band in OBArE,F;W. Pick a net Tαα∈A in OBAscE,F;W with 0≤Tα↑T for some T∈OBA+E,F;W. Then, we have that 0≤Tαy↑Ty (0≤Tαx↑Tx) for every y∈F (x∈E). Now, applying ([24], Theorem 3.13), we obtain that T∈OBAscE,F;W. Finally, taking into account the Dedekind completeness of OBArE,F;W, we deduce that OBAscE,F;W is the projection band in T∈OBArE,F;W.
Now we are ready to prove that the Dedekind completeness of a vector latticeW implies the horizontal-to-order continuity (σ-continuity) of a regular separately horizontal-to-order continuous (σ-continuous) operator T:E×W.☐Proposition 15.
LetE and F be vector lattices, W be a Dedekind complete vector lattice, and T∈OBArE,F;W. Then, the following statements hold:
(1)
T∈OBAscE,F;W⇔T∈OBAcE,F;W(2)
T∈OBAsσcE,F;W⇔T∈OBAσcE,F;WProof.
We prove statement (1). The implicationT∈OBAcE,F;W⇒T∈OBAscE,F;W is obvious. Suppose that 0≤T∈OBAscE,F;W. We need to show horizontal-to-order continuity of T. Pick a horizontally convergent net x,yαα∈A with x,yα⟶hx,y. Then, xα⟶hx and yα⟶hy. Now we may write
(24)∣Tx,y−Txα,yα∣=∣Tx−xα+xα,y−yα+yα−Txα,yα∣≤∣Tx−xα,yα∣+∣Txα,y−yα∣+∣Tx−xα,y−yα∣=Tx−xα,yα+Txα,y−yα+Tx−xα,y−yα≤Tx−xα,y+Tx,y−yα+Tx,y−yα=Tx−xα,y+2Tx,y−yα.
Taking into account the separate horizontal-to-order continuity ofT, we have that Tx,yα⟶oTx,y. Now, suppose that T is an arbitrary element of OBAscE,F;W. Then, by Theorem 14, every T∈OBAscE,F;W has the representation T=T+−T−, where 0≤T+,−T−∈OBAscE,F;W. Hence, by above, we have that T∈OBAcE,F;W.☐


## 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 thatA,Σ,μ and B,Ξ,ν are finite measure space and μ⊗ν is the product measure on Σ⊗Ξ. We say that N:A×B×ℝ⟶ℝ is a superpositionally measurable (or sup-measurable for shortness) function, if N·,·,f·,· is μ⊗ν-measurable for each f∈L0μ⊗ν. A sup-measurable function N is said to be normalized if Ns,t,0=0 for μ⊗ν-almost all s,t∈A×B.The following proposition provides an important example of an orthogonally biadditive operator.Proposition 17.
LetN:A×B×ℝ⟶ℝ be a normalized sup-measurable function and E and F be order ideals of L0μ and L0μ, respectively. Then, the map N defined by
(25)Nf,gs,t≔Ns,t,fsgt,f∈E,g∈F,is an orthogonally biadditive operator from E×F to L0λ.Proof.
Takef∈E and g∈F. Put f~≔f1B and g~≔g1A. We note that the relations f∈L0μ and g∈L0ν imply that f~, g~∈L0μ⊗ν. Then, Ns,t,fsgt=Ns,t,f~s,tg~s,t, and therefore, the operator N is well defined. Fix g∈F. We show that the partial operator Ng:E⟶W is orthogonally additive. Indeed, take disjoint f,e∈E. Then, f~ and e~ are disjoint elements of L0μ⊗ν, and we may write
(26)N·,·,f⊔eg=N·,·,f~⊔e~g~=N·,·,f~⊔e~1suppf~⊔e~g~=N·,·,f~⊔e~g~1suppf~⊔e~=N·,·,f~⊔e~g~1suppf~+N·,·,f~⊔e~g~1suppe~=N·,·,f~g~+N·,·,e~g~=N·,·,fg+N·,·,eg.
Noting that similar arguments are valid for a partial operatorNf:F⟶W, for all f∈E, we finish the proof.☐We observe thatN 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.
LetC,Θ,λ, A,Σ,μ, and B,Ξ,ν be finite measure spaces. By C×A×B,λ⊗μ⊗ν, we denote the completion of their product measure space. A map K:C×A×B×R2⟶R is said to be a Carathéodory function if it is satisfies the following conditions:
(C1) K·,·,·,r,q is μ⊗ν⊗λ-measurable for all r,q∈R2
(C2) Kp,s,t,·,· is continuous on R2 for λ⊗μ⊗ν-almost all p,s,t∈C×A×B
We say that a Carathéodory functionK is normalized if Kp,s,t,0,q=Kp,s,t,r,0=0 for λ⊗μ⊗ν-almost all p,s,t∈C×A×B and all q,r∈R.Proposition 19.
LetK:C×A×B×ℝ2 be a normalized Carathéodory function, f∈L0μ and g∈L0ν. Then, K·,·,·,f·,g·∈L0λ⊗μ⊗ν.Proof.
First we note thatλ⊗μ⊗ν-measurable null sets
(27)Dq≔p,s,t∈C×A×B:Kp,s,t,0,q≠0,q∈ℝ,Dr≔p,s,t∈C×A×B:Kp,s,t,r,0≠0,r∈ℝ,depend of q and r, respectively. We claim that there exists λ⊗μ⊗ν-measurable set H⊂C×A×B such that
(28)λ⊗μ⊗νC×A×B=λ⊗μ⊗νH,Kp,s,t,0,q=Kp,s,t,r,0=0,for all p,s,t∈H and all q,r∈ℝ. Indeed, consider two sequences of λ⊗μ⊗ν-measurable sets Dqq∈Q and Drr∈Q and put
(29)G≔p,s,t∈C×A×B:Kp,s,t,·,·isnotcontinuous,H=C×A×B\G∪⋃q∈QDq∪⋃r∈QDr.
Sinceλ⊗μ⊗νG∪⋃q∈ℚDq∪⋃r∈ℚDr=0, we have that λ⊗μ⊗νC×A×B=λ⊗μ⊗νH. Fix p,s,t∈H and q∈ℝ (r∈ℝ). Then, there exists a sequence qnn∈N (rkk∈N) in ℚ that converges to q (r) with Kp,s,t,0,qn=0 (Kp,s,t,rk,0=0) for all n∈ℕ (n∈ℕ). Then, by C2, we have that Kp,s,t,0,q=0 (Kp,s,t,r,0=0).
Now, we show thatK·,·,·,r1A1·,q1B1·∈L0λ⊗μ⊗ν for arbitrary r,q∈ℝ, A1∈Σ, and B1∈Ξ. We claim that
(30)Kp,s,t,r1A1s,q1B1t=Kp,s,t,r,q1A1s1B1t,for λ⊗μ⊗ν almost all p,s,t∈C×A×B. Indeed, pick p,s,t∈H. If s∈A1 and t∈B1, we have
(31)Kp,s,t,r1A1s,q1B1t=Kp,s,t,r,q=Kp,s,t,r,q1A1s1B1t.
If eithers∉A1 or t∉B1, then
(32)Kp,s,t,r1A1s,q1B1t=0=Kp,s,t,r,q1A1s1B1t.
Suppose thatA1,⋯,An and B1,⋯,Bm are pairwise disjoint measurable subsets of A and B, respectively, f=∑i=1nri1Ai and g=∑j=1mqj1Bj are simple functions in L0μ and L0ν, respectively. Then
(33)Kp,s,t,fs,qt=Kp,s,t,∑i=1nri1Ais,∑j=1mqj1Bjt=∑i=1n∑j=1mKp,s,t,ri1Ais,qj1Bjt,and we deduce that K·,·,·,f·,q·∈L0λ⊗μ⊗ν. Finally, assume that f and g are arbitrary elements of L0μ and L0ν, fnn∈N and gkk∈N are sequences of simple functions in L0μ and L0ν, respectively, such that fnn∈N converges to fμ-a.e. and gnk∈N converges to gν-a.e. Put by definition
(34)A0≔s∈A:fnsdoesnotconvergestof,B0≔t∈B:gktdoesnotconvergestog.
Clearly,(35)λ⊗μ⊗νG=λ⊗μ⊗νC×A0×B=λ⊗μ⊗νC×A×B0=0.
Then,Kp,s,t,fns,gkt converges to Kp,s,t,fs,gt for all p,s,t∈C×A×B\G∪C×A0×B∪C×A×B0, and therefore, K·,·,·,f·,g·∈L0λ⊗μ⊗ν.☐Remark 20.
Using similar arguments as above, we get the following useful equalities:(36)Kp,s,t,fs,q=Kp,s,t,fs,q1suppfs,Kp,s,t,r,gt=Kp,s,t,r,gt1suppgt,for λ⊗μ⊗ν almost all p,s,t∈C×A×B and all q∈R (r∈R).The next proposition provides an important example of an orthogonally biadditive operator.Proposition 21.
LetK:C×A×B×ℝ2 be a normalized Carathéodory function, E and F be order ideals of L0μ and L0ν, respectively, and Kp,·,·,f·,g·∈L1μ⊗ν for all f∈E, g∈F, and λ-almost all p∈C. Then, the map T defined by setting
(37)Tf,gp≔∫A×BKp,s,t,fs,gtdμ⊗ν;f∈E,g∈F,is an orthogonally biadditive operator from E×F to L0λ.Proof.
By Proposition19, T is a well-defined operator from E×F to L0λ. We show the orthogonal additivity of a partial operator Tg, where g∈F. Fix disjoint f1,f2∈E. Then, taking into account considerations above, we may write
(38)Tf1⊔f2,gp=∫A×BKp,s,t,f1s⊔f2s,gtdμ⊗ν=∫A×BKp,s,t,f1s+f2s,gt1suppf1⊔f2dμ⊗ν=∫A×BKp,s,t,f1s⊔f2s,gt1suppf1dμ⊗ν+∫A×BKp,s,t,f1s⊔f2s,gt1suppf2dμ⊗ν=∫A×BKp,s,t,f1s,gtdμ⊗ν+∫A×BKp,s,t,f2s,gtdμ⊗ν=Tf1,gp+Tf2,gp.☐The orthogonal additivity of a partial operatorTf, f∈E can be proved analogously.We observe that an operatorT above can be considered as the Uryson integral operator that depends on two variables. We say that a function K is a kernel of an operator T. 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.
LetC,Θ,λ, A,Σ,μ, and B,Ξ,ν be as above, N:A×B×ℝ⟶ℝ be a normalized sup-measurable function, E and F be order ideals in L0μandL0ν, respectively, L:C×A×B⟶R be a λ⊗μ⊗ν-measurable function, and Lp,·,·N·,·,f·g·∈L1μ⊗ν for all f∈E, g∈F, and λ-almost all p∈C. Then, the following formula defines a OBAOT:E×F⟶L0τ(39)Tf,gp≔∫A×BLp,s,tNs,t,fsgtdμ⊗ν,f∈E,g∈F.We note that biorthogonal additivity of a superposition operatorNf,g≔N·,·,f·g· implies that T∈OBAE,F;L0λ. The operator T 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.
2⇒1. Since S∈OBA+E,F;W and T≤S, we have that T=S−S−T, and therefore, T∈OBArE,F;W.
1⇒2. By Theorem 1, the modulus ∣T∣ exists and can be calculated by the formula
(40)Tf,g≔sup∑i=1n∑j=1m−1ki,jTfi,gj:f=∐i=1nfi;g=∐j=1mgj;n,m∈N,k∈1,21,⋯,n×1,⋯,m;f∈E;g∈F.
Fixf∈E, g∈F. By Remark 20, for almost all p∈C, we have that
(41)∫A×BKp,s,t,fs,gtdμ⊗ν=∫suppf×suppgKp,s,t,fs,gtdμ⊗ν.
We also note that(42)∫suppf×suppgKp,s,t,fs,gtdμ⊗ν=∫suppf×suppgKp,s,t,fs1suppgt1Cp,gt1suppfs1Cpdμ⊗ν,for λ-almost all p∈C. Put by definition
(43)A≔p,s,t∈C×suppf×suppg:Kp,s,t,fs1suppgt1Cp,gt1suppfs1Cp>0,(44)B≔p,s,t∈C×A×B:Kp,s,t,fs1suppgt1Cp,gt1suppfs1Cp<0=C×suppf×suppg\A.
Clearly,A and B are λ⊗μ⊗ν-measurable sets and
(45)λ⊗μ⊗νp,s,t∈A∩B=0.
We may assume thatλ⊗μ⊗νA>0 and λ⊗μ⊗νB>0. Otherwise, if λ⊗μ⊗νA=0 (λ⊗μ⊗νB=0), then the following equalities
(46)Tf,gp=∫suppf×suppg∣Kp,s,t,fs1suppgt1Cp,gt1suppfs1Cp∣dμ⊗ν=∫suppf×suppgKp,s,t,fs,gtdμ⊗ν,−Tf,gp=∫suppf×suppgKp,s,t,fs1suppgt1Cp,gt1suppfs1Cpdμ⊗ν=∫suppf×suppg∣Kp,s,t,fs,gt∣dμ⊗ν,hold for λ-almost all p∈C, and it is nothing to prove. We observe that
(47)Kp,s,t,fs1suppgt1Cp,gt1suppfs1Cp1Ap,s,t=Kp,s,t,fs1suppgt1Cp1Ap,s,t,gt1suppfs1Cp1Ap,s,t=Kp,s,t,fs1As1Bt1Cp,gt1As1Bt1Cp,where A∈Σ, A⊂suppf, B∈Ξ, B⊂suppg, and C∈Θ. First, we assume that μA=0. Then, we see that
(48)Kp,s,t,fs,gt<0for,λ⊗μ⊗νalmostallp,s,t∈C×A×B.
Consider the trivial decompositionsf=f and g=g. Then, for λ-almost all p∈C, we have that
(49)−Tf,gp=∫suppf×suppg−Kp,s,t,fs,gtdμ⊗ν=∫suppf×suppg∣Kp,s,t,fs,gt∣dμ⊗ν.
The same arguments are valid for the caseνB=0. Now, suppose that μA>0 and νB>0. Then, there is a decomposition C=Cℴ̀C1òC2, where C×A×B⊂A, C1×A×B⊂B, and C2×A×B⊂A∩B. Clearly λC2=0. Put
(50)f1≔f1A;f2≔f−f1A,g1≔g1B;g2≔g−g1B.
We observe thatC×suppf\A×B⊂B, C×A×suppg\B⊂B, and C×suppf\A×suppg\B⊂B. Now we may write
(51)Tf1,g1p−Tf1,g2p−Tf2,g1p−Tf2,g2p=∫suppf×suppgKp,s,t,f1s,g1tdμ⊗ν−∫suppf×suppgKp,s,t,f1s,g2tdμ⊗ν−∫suppf×suppgKp,s,t,f2s,g1tdμ⊗ν−∫suppf×suppgKp,s,t,f2s,g2tdμ⊗ν=∫A×BKp,s,t,fs,gtdμ⊗ν+∫A×suppg\B−Kp,s,t,fs,gtdμ⊗ν+∫suppf\A×B−Kp,s,t,fs,gtdμ⊗ν+∫suppf\A×suppg\B−Kp,s,t,fs,gtdμ⊗ν.
Thus, for allp∈C, we have that
(52)Tf1,g1p−Tf1,g2p−Tf2,g1p−Tf2,g2p=∫A×B∣Kp,s,t,fs,gt∣dμ⊗ν+∫A×suppg\B∣Kp,s,t,fs,gt∣dμ⊗ν+∫suppf\A×B∣Kp,s,t,fs,gt∣dμ⊗ν+∫suppf\A×suppg\B∣Kp,s,t,fs,gt∣dμ⊗ν.
On the other hand, for allp∈C1, we have
(53)Tf1,g1p−Tf1,g2p−Tf2,g1p−Tf2,g2p=∫A×B−∣Kp,s,t,fs,gt∣dμ⊗ν+∫A×suppg\B∣Kp,s,t,fs,gt∣dμ⊗ν+∫suppf\A×B∣Kp,s,t,fs,gt∣dμ⊗ν+∫suppf\A×suppg\B∣Kp,s,t,fs,gt∣dμ⊗ν.
Consider the second sum−Tf1,g1−Tf1,g2−Tf2,g1−Tf2,g2. Then, for all p∈C, we have
(54)−Tf1,g1p−Tf1,g2p−Tf2,g1p−Tf2,g2p=∫A×B−∣Kp,s,t,fs,gt∣dμ⊗ν+∫A×suppg\B∣Kp,s,t,fs,gt∣dμ⊗ν+∫suppf\A×B∣Kp,s,t,fs,gt∣dμ⊗ν+∫suppf\A×suppg\B∣Kp,s,t,fs,gt∣dμ⊗ν,and for all p∈C1, we have that
(55)−Tf1,g1p−Tf1,g2p−Tf2,g1p−Tf2,g2p=∫A×B∣Kp,s,t,fs,gt∣dμ⊗ν+∫A×suppg\B∣Kp,s,t,fs,gt∣dμ⊗ν+∫suppf\A×B∣Kp,s,t,fs,gt∣dμ⊗ν+∫suppf\A×suppg\B∣Kp,s,t,fs,gt∣dμ⊗ν.
Put by definition(56)R1=Tf1,g1−Tf1,g2−Tf2,g1−Tf2,g2,R2=−Tf1,g1−Tf1,g2−Tf2,g1−Tf2,g2.
Then, for allp∈C, we have that
(57)∣T∣f,gp≥R1∨R2p=∫A×B∣Kp,s,t,fs,gt∣dμ⊗ν+∫A×suppg\B∣Kp,s,t,fs,gt∣dμ⊗ν+∫suppf\A×B∣Kp,s,t,fs,gt∣dμ⊗ν+∫suppf\A×suppg\B∣Kp,s,t,fs,gt∣dμ⊗ν=∫suppf×suppg∣Kp,s,t,fs,gt∣dμ⊗ν.
Thus,S≤∣T∣ and S map E×F to J. On the other hand, since S∈OBA+E,F;W, S≥T, and S≥−T, we have that S≥T∨−T=∣T∣.Corollary 23.
LetC,Θ,λ, A,Σ,μ, and B,Ξ,ν be finite measure spaces and E and F be ideal subspaces of L0μ and L0ν, respectively. Then, every integral Uryson operator T:E×F⟶L0λ is regular.Proof.
SupposeT is an integral Uryson operator with kernel K. Taking into account that the function
(58)p↦∫A×B∣Kp,s,t,fs,gt∣dμ⊗ν,is λ-measurable for each f∈E and g∈F by Theorem 2 we deduce that T is a regular operator.☐Proposition 24.
LetC,Θ,λ, A,Σ,μ, and B,Ξ,ν be finite measure spaces, E, F, and J be ideal subspaces of L0μ, L0ν, and L0λ, respectively, and T:E×F⟶J be a regular integral Uryson operator T:E×F⟶L0λ with a kernel K. Then, T is a horizontally-to-order continuous operator.Proof.
By Proposition15, it is enough to prove the separate horizontal-to-order continuity of T. Fix g∈F. We show the horizontal-to-order continuity of the partial operator Tg:E⟶J. It is worth noting that the countable sup property of L0μ (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 fnn∈N which horizontally converges to f. We need to show that the sequence Tgfnn∈N order converges to Tgf. Taking into the account the regularity of T, we may write
(59)∣∫A×BKp,s,t,fs,gtdμ⊗ν−∫A×BKp,s,t,fns,gtdμ⊗ν∣=∣∫A×BKp,s,t,f−fn⊔fns,gtdμ⊗ν−∫A×BKp,s,t,fns,gtdμ⊗ν∣=∣∫A×BKp,s,t,f−fns,gtdμ⊗ν∣≤∫A×BKp,s,t,f−fns,gtdμ⊗ν=∫suppf−fn×suppgKp,s,t,fs,gtdμ⊗ν.☐Sinceμsuppf−fn converges to 0 by ([30], Theorem 2.5.7), we have that Tgf−fnn∈N order converges to 0, and therefore, Tg is a horizontally-to-order continuous operator. Similar arguments are valid for a partial operator Tf, f∈E.



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