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This article is cited in 2 scientific papers (total in 2 papers) Diffuse orthogonally additive operators N. M. Abasova, N. A. Dzhusoevab, M. A. Plievcd a Bauman Moscow State Technical University, Moscow, Russia b North Ossetian State University after Kosta Levanovich Khetagurov, Vladikavkaz, Russia c Southern Mathematical Institute of the Vladikavkaz Scientific Center of the Russian Academy of Sciences, Vladikavkaz, Russia d North Caucasus Center for Mathematical Research, Vladikavkaz Scientific Centre of the Russian Academy of Sciences, Vladikavkaz, Russia
Russian version:
Matematicheskii Sbornik, 2024, Volume 215, Number 1, DOI: https://doi.org/10.4213/sm9909 
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    § 1. Introduction and preliminariesOrthogonally additive operators (OAOs) appear naturally in the theory of nonlinear differential and integral equations, in the theory of dynamical systems and in convex geometry (see [21], [42], [45] and [46]). First results on OAOs date back to Drewnowski, Orlicz, Marcus, and Mizel (see [20], [29] and [30]). In the context of vector lattices, orthogonally additive operators were originally considered by Mazón and Segura de León in the 1990s (see [28] and [45]). In the recent years the general theory of OAOs in ordered spaces was actively developed both in Russia and other countries (see [3], [13], [23]–[25], [36] and [43]). It was shown in [35] that the space of regular orthogonally additive operators $\mathcal{OA}_{\mathrm r}(E,F)$ acting from a vector lattice $E$ to an order-complete vector lattice $F$ is an order-complete vector lattice. Hence the problem arises of studying the lattice order structure of the space of orthogonally additive operators. This problem is interesting in its own right in the framework of abstract ‘order’ analysis, but it also has a wide range of applications. As an illustration of this thesis in the case of linear regular operators, we mention the solution of the problem of the characterization of linear continuous operators in the space $L_2[0,1]$ — this problem was stated by von Neumann in 1932, in relation to the question of a rigorous mathematical foundation for quantum mechanics. The above problem of characterization was solved by Bukhvalov [1] in 1974. The solution of the von Neumann problem depends substantially on the presence of ‘rich’ order structure on the set of regular linear integral operators on the $L_2[0,1]$. Various order properties of the space $\mathcal{OA}_{\mathrm r}(E,F)$ were considered in [10]–[12], [26] and [32]. The purpose of the present paper is to study the new class of diffuse orthogonally additive operators in vector lattices. Diffuse operators are closely related to another important class of disjointness preserving OAOs. The diffuse OAOs form an order-closed ideal in the vector lattice $\mathcal{OA}_{\mathrm r}(E,F)$, which is the disjoint band generated by the disjointness preserving orthogonally additive operators. In the linear case the first studies of this class go back to [4], [27] and [47]. Diffuse-like narrow OAOs were investigated in [38]–[40]. The paper is organized as follows. First we present the necessary preliminary facts and notation used below. In § 2 we give basic examples of OAOs, and show that linear regular operators can naturally be embedded in the space of regular OAOs. Next we consider disjointness preserving orthogonally additive operators and establish a criterion for principal lateral projections onto an order-complete vector lattice $E$ to be disjoint. In § 3 we study diffuse OAOs, which constitute the main subject of the paper. We show that the zero operator is the only OAO on a finite-dimensional vector space. We also present a criterion for a regular orthogonally additive operator to be diffuse. This result extends Theorem 2.2 in [27] to the nonlinear case. Next we use this criterion to show that laterally continuous OAOs are diffuse. Integral OAOs are considered in § 4. We obtain a criterion for the regularity of an integral Urysohn operator and verify that the set of regular OAOs forms a sublattice of the vector lattice of all regular OAOs. We also show that an integral Urysohn operator defined on an atomless Banach function space is diffuse. In § 5 we find the general form of an order projection operator onto the band in $\mathcal{OA}_{\mathrm r}(E,F)$ generated by the disjointness preserving operators. The necessary facts on vector lattices and Banach ideal spaces can be found in [2], [14] and [15]. All vector lattices considered in the paper are assumed to be Archimedean. A net $(x_\alpha)_{\alpha\in A}$ of elements of a vector lattice $E$ is said to order converge to an element $x\in E$ if there exists a net $e_{\alpha}\downarrow 0$ in $E_+$ such that ${|x-x_{\alpha}|\leqslant e_{\alpha}}$ for any $\alpha\in A$ such that $\alpha\geqslant\alpha_{0}$ for some fixed index $\alpha_0\in A$. Two elements $x$ and $y$ of a vector lattice $E$ are said to be disjoint (written $x\perp y$) if $|x|\wedge|y|=0$. The notation $x=\bigsqcup_{i=1}^{n}x_{i}$ means that $x=\sum_{i=1}^{n}x_{i}$ and $x_{i}\perp x_{j}$ for all $i\neq j$. For $n=2$ we write $x=x_{1}\sqcup x_{2}$. We say that an element $y$ is a fragment $x \in E$ (written $y \sqsubseteq x$) if $y\perp(x-y)$. The set of all fragments of $x\in E$ is denoted by $\mathfrak{F}_{x}$. Proposition 1 (Proposition 3.4 in [31]). Let $E$ be a vector lattice, and let $x,y\in E$. Then $y\sqsubseteq x$ if and only if $y^{+}\sqsubseteq x^{+}$ and $y^{-}\sqsubseteq x^{-}$, where $\mathfrak{F}_{x}$ is the Boolean algebra with respect to the partial order $\sqsubseteq$ with minimal element $0$ and maximal element $x$. The Boolean operations are as follows: A Boolean algebra $\mathfrak{B}$ is said to be order complete if each subset $D$ of $\mathfrak{B}$ has a supremum and an infimum. The binary relation $\sqsubseteq$ on $E$ is a partial order relation known as the lateral order (see [31]). A linear subspace $E_0$ of a vector lattice $E$ is an order ideal of $E$ if for all $x\in E_0$ and $y\in E$ the inequality $|y|\leqslant|x|$ implies that $y\in E_0$. Let $D$ be an arbitrary subset of a vector lattice $E$. Then we set
$$
\begin{equation*}
D^{\perp}:=\{x\in E\colon x\perp y\text{ for any }y\in D\}\quad\text{and} \quad D^{\perp\perp}:=(D^{\perp})^{\perp}.
\end{equation*}
\notag
$$
The set $D^{\perp}$ is the disjoint complement of $D$. A subset $B$ of a vector lattice $E$ is called a band if $B=D^{\perp}$ for some $D\subset E$. Note that, for an arbitrary set $D\subset E$, its disjoint complement (the band) $D^{\perp}$ is an order closed order ideal in $E$. A band $B$ in $E$ is generated by a set $D$ if $B=D^{\perp\perp}$. A band $B$ in $E$ is a principal band if $B=\{x\}^{\perp\perp}$ for some $x\in E$. Definition 1. Let $E$ and $F$ be vector lattices. Then a linear operator $T\colon E\to F$ is said to be A linear operator $T\colon E\to F$ is a lattice homomorphism if, for all $x,y\in E$, at least one of the following equalities holds:
Note that each lattice homomorphism is a positive operator; in addition, each of equalities 1)–5) implies the other equalities. By a lattice monomorphism we mean an injective lattice homomorphism. A (not necessarily linear) operator $\pi\colon E\to E$ is called a projection onto $E$ if $\pi=\pi^2$. A positive linear operator $\pi\colon E\to E$ is said to be an order projection if
The set of all order projections onto $E$ is denoted by $\mathfrak{B}(E)$. The set $\mathfrak{B}(E)$ is a Boolean algebra with respect to the partial order
$$
\begin{equation*}
\pi\leqslant \rho\quad \Longleftrightarrow\quad \pi\circ\rho=\pi,
\end{equation*}
\notag
$$
the Boolean operations being as follows:
$$
\begin{equation*}
\pi\wedge\rho:=\pi\circ\rho, \qquad \pi\vee\rho:=\pi+\rho-\pi\circ\rho, \qquad \neg\,\pi=\mathrm{Id}-\pi.
\end{equation*}
\notag
$$
A band $B$ in a vector lattice $E$ admits an order projection if each element $x$ of $E$ has a unique decomposition
$$
\begin{equation*}
x=x_{1}+x_2, \qquad x_1\in B, \quad x_2\in B^{\perp}.
\end{equation*}
\notag
$$
In this case the linear operator $\pi\colon E\to E$ defined by $\pi x=x_1$ is an order projection onto the band $B$. We say that $E$ is a vector lattice with projections (onto principal bands) if there exists an order projection of $E$ onto any band (a principal band). Note that each band $B$ in an order complete vector lattice $E$ admits an order projection (see Theorem 1.42 in [15]). Let $E$ be a vector lattice with projections and $D$ be a subset of $E$. Then we denote by $\pi_{D}$ the order projection of $E$ onto the band $\{D\}^{\perp\perp}$. The characteristic function of a set $D$ is denoted by $1_{D}$. Let $(A,\Sigma,\mu)$ and $(B,\Xi,\nu)$ be spaces with $\sigma$-finite measures. Then we denote by $L_{0}(\mu)$ and $L_{0}(\nu)$ the vector spaces of equivalence classes of measurable, almost everywhere finite functions on $A$ and $B$, respectively. There exists a natural partial order in the space $L_{0}(\nu)$: $f\leqslant g$ if $f(t)\leqslant g(t)$ for almost all $t\in A$. It is known that $L_{0}(\nu)$ is an order complete vector lattice with respect to this partial order (see Theorem 1.80 in [14]). Let $\mathcal{L}_{\infty}(B,\Xi,\nu)$ (or, simply, $\mathcal{L}_{\infty}(\nu)$) denote the Banach space of bounded $\nu$-measurable real-valued functions on $B$. The space $L_{\infty}(\nu)$ of essentially bounded measurable functions (equivalence classes) is a vector sublattice of $L_{0}(\nu)$. Note that $L_{\infty}(\nu)$ is a quotient space of $\mathcal{L}_{\infty}(\nu)$, where the canonical quotient map $\pi\colon {\mathcal{L}_{\infty}(\nu)\to L_{\infty}(\nu)}$ takes a measurable function $f\in\mathcal{L}_{\infty}(\nu)$ to the class of functions equivalent to $f$. A positive linear operator $l\colon L_{\infty}(\nu)\to\mathcal{L}_{\infty}(\nu)$ is called a lifting if the following conditions are satisfied: The image $l(f)$ of an element $f\in L_{\infty}(\nu)$ under the map $l\colon L_{\infty}(\nu)\to\mathcal{L}_{\infty}(\nu)$ is denoted by $\widehat{f}$. For a detailed account of the theory of lifting, see [7]. § 2. Regular orthogonally additive operatorsIn this section we define regular OAOs and give some basic examples. We also continue the study of projections onto lateral bands, which was begun in [22], and establish a criterion for principal lateral projections to be disjoint. Definition 2. Let $E$ be a vector lattice and $X$ be a real vector space. A map $T\colon E\to X$ is an orthogonally additive operator if $T(x\sqcup y)=Tx+Ty$ for all disjoint $x,y\in E$. Note that $T(0)=0$ by definition. Definition 3. Let $E$ and $F$ be vector lattices. An orthogonally additive operator $T\colon E\to F$ is said to be: It is clear that each order-bounded OAO is $C$-bounded. The following example shows that, in contrast to the linear case, the positivity of an orthogonally additive operator does not imply its order boundedness. Let the map $T\colon \mathbb{R}\to\mathbb{R}$ be defined by
$$
\begin{equation*}
T(x)=\begin{cases} \dfrac{1}{x^{2}} &\text{for } x\neq 0, \\ 0 &\text{for }x=0. \end{cases}
\end{equation*}
\notag
$$
Note that two elements $x,y\in\mathbb{R}$ are disjoint if and only if at least one of them is zero; in this case $T$ is an orthogonally additive operator. An operator cannot be order bounded, since the image $T(-1,1)$ of the interval $(-1,1)$ is not bounded in $\mathbb{R}$. For any positive OAO $T\colon E\to F$ and all $x,y\in E$, where $y\in\mathfrak{F}_x$, we have the inequality and so positive regular OAOs are $C$-bounded. The sets of all positive, regular, order- and fragment-bounded OAOs from $E$ to $F$ are denoted by $\mathcal{OA}_{+}(E,F)$, $\mathcal{OA}_{\mathrm r}(E,F)$, $\mathcal{OA}_{\mathrm b}(E,F)$ and $\mathcal{P}(E,F)$, respectively. For brevity we will write $\mathcal{OA}_{\mathrm r}(E)$ in place of $\mathcal{OA}_{\mathrm r}(E,E)$. There exists a natural partial order on $\mathcal{OA}_{\mathrm r}(E,F)$ defined by the relation
$$
\begin{equation*}
S\leqslant T\quad\Longleftrightarrow\quad (T-S)\in\mathcal{OA}_{+}(E,F).
\end{equation*}
\notag
$$
Note that the ordered spaces $L_{\mathrm r}(E,F)$ and $\mathcal{OA}_{\mathrm r}(E,F)$ are substantially different. In particular, $T=0$ is the only linear operator $T\colon E\to F$ which is positive both in the sense of Definition 1 and in the sense of Definition 3. If the vector lattice $F$ of images is order complete, then the ordered vector space $\mathcal{OA}_{\mathrm r}(E,F)$ is an order-complete vector lattice. Lemma 1 (see Theorem 3.6 in [35]). Let $E$ and $F$ be vector lattices, where $F$ is order complete. Then $\mathcal{OA}_{\mathrm r}(E,F) = \mathcal{P}(E,F)$ and $\mathcal{OA}_{\mathrm r}(E,F)$ is an order-complete vector lattice. In addition, for all $S$, $T\in \mathcal{OA}_{\mathrm r}(E,F)$ and $x\in E$: The set of all disjointness preserving orthogonally additive operators from $E$ to $F$ is denoted by $\mathcal{OA}_{\mathrm{dpo}}(E,F)$. Note that, in contrast to the linear case, any disjointness preserving orthogonally additive operator is necessarily regular. Proposition 2. Let $E$ and $F$ be vector lattices. Then $\mathcal{OA}_{\mathrm{dpo}}(E,F)\subset\mathcal{OA}_{\mathrm r}(E,F)$. Proof. Let $T\in\mathcal{OA}_{\mathrm{dpo}}(E,F)$, and let the map $R\colon E\to F$ be defined by Consider disjoint $x,y\in E$. Then we have and $R\in \mathcal{OA}_{+}(E,F)$. It is clear that $T\leqslant R$. That the operator $T$ is regular follows from the representation $T=R-(R-T)$, where $R$ and $R-T$ are positive OAOs.
This proves Proposition 2. We require the following auxiliary results. Proposition 3 (Theorem 1.50 in [15]). Let $E$ be a lattice with projections onto principal bands, let $F$ be an order complete vector lattice and $T\colon E\to F$ be a regular linear operator. Then Proposition 4. Let $E$ be a vector lattice, and let $x\in E$. Then the sets $A:=\{z\colon z\sqsubseteq |x|\}$ and $B:=\{|z|\colon z\sqsubseteq x\}$ are equal. Proof. We prove the inclusion $A\subset B$. Let $z\sqsubseteq |x|=x_{+}\sqcup x_-$. By Lemma 3.5 in [35], there exists a representation $z=z_1\sqcup z_2$, where $z_1\sqsubseteq x_{+}$ and $z_2\sqsubseteq x_-$. Let $y=z_1-z_2$. Then $y\sqsubseteq x$ and $|y|=z$. Let us prove the reverse inclusion $B\subset A$. Let $y\sqsubseteq x$. Then by Proposition 3.1 in [31] we have $y_+\sqsubseteq x_{+}$, $y_-\sqsubseteq x_-$ and $|y|=(y_{+}\sqcup y_-)\sqsubseteq|x|$, which proves Proposition 4. Below we consider some examples of regular orthogonally additive operators. The following result shows that, for broad classes of vector lattices $E$ and $F$, there exists an isomorphic embedding of $L_{\mathrm r}(E,F)$ to $\mathcal{OA}_{\mathrm r}(E,F)$. Lemma 2. Let $E$ be a vector lattice with projections onto principal bands, and let $E$ be an order complete vector lattice. Then there exists a lattice monomorphism $J\colon L_{\mathrm r}(E,F)\to \mathcal{OA}_{\mathrm r}(E,F)$. Proof. Let $T\in L_{+}(E,F)$, and let the map $j(T)\colon E\to F$ be defined by Then for all disjoint $x_1,x_2\in E$,
$$
\begin{equation*}
j(T)(x_1\sqcup x_2)=T|x_1\sqcup x_2|=T(|x_1|\sqcup |x_2|)=j(T)x_1+j(T)x_2,
\end{equation*}
\notag
$$
and, since the linear operator $T$ is positive, it follows that $j(T)$ is a positive OAO. We have $j(T_1+T_2)=j(T_1)+j(T_2)$ for all $T_1,T_2\in L_{+}(E,F)$, and therefore, by Kantorovich’s theorem (see Theorem 1.10 in [15]) there exists a unique linear positive operator $J\colon L_{\mathrm r}(E,F)\to \mathcal{OA}_{\mathrm r}(E,F)$ such that and
Let us show that the operator $J$ is injective. First we note that for all $T,S\in L_{+}(E,F)$ the equality implies the identity $T=S$. Now consider $T,S\in L_{+}(E,F)$, and let $J(T)=J(S)$. Then for any $x\in E_+$ we have Hence $T=S$, and the operator $J$ is injective on the cone $L_{+}(E,F)$. Now let $T,S\in L_{\mathrm r}(E,F)$ and $J(T)=J(S)$. Then
$$
\begin{equation*}
\begin{aligned} \, &J(T)=J(T_+)-J(T_-), \qquad J(S)=J(S_+)-J(S_-) \\ &\qquad\Longrightarrow \quad J(T_{+}+S_-)=J(S_{+}+T_-) \quad \Longrightarrow \quad T_{+}+S_-=S_{+}+T_-. \end{aligned}
\end{equation*}
\notag
$$
Hence $T=S$, and the operator $J$ is injective on the whole space $L_{\mathrm r}(E,F)$.
Now let us show that $J\colon L_{\mathrm r}(E,F)\to \mathcal{OA}_{\mathrm r}(E,F)$ is a lattice homomorphism. By the definition of a lattice homomorphism it suffices to show that Consider an arbitrary element $x\in E$ and a regular linear operator $T\colon E\to F$. An application of Propositions 3 and 4 and Lemma 1 shows that
$$
\begin{equation*}
J(T_+)x=T_+|x|\stackrel{\rm \text{Prop. }3}{=} \sup\{Tz\colon z\sqsubseteq |x|\}\stackrel{\rm \text{Prop. }4}{=} \sup\{T|z|\colon z\sqsubseteq x\} \stackrel{\rm \text{Lemma }1}{=}J(T)_{+}x.
\end{equation*}
\notag
$$
This proves Lemma 2. The theory of linear regular operators in vector lattices is a highly developed part of contemporary analysis; there is an extensive literature on this theory (see [14] and [15]). Regular OAOs were originally considered in [35]. The following example is of great value for applications. Definition 4. Let $(A,\Sigma,\mu)$ and $(B,\Xi,\nu)$ be finite measure spaces. Then the product of these spaces is denoted by $(A\times B,\mu\otimes\nu)$. A map $K: A\times B\times\mathbb{R}\to\mathbb{R}$ is a Carathéodory function if the following conditions are satisfied: $(\mathrm C_{1})$ $K(\,{\cdot}\,{,}\,\cdot\,,r)$ is $\mu\otimes\nu$-measurable for each $r\in\mathbb{R}$; $(\mathrm C_{2})$ $K(s,t,\cdot\,)$ is continuous on $\mathbb{R}$ for $\mu\otimes\nu$-almost all $(s,t)\in A\times B$. We say that $K$ is a normalized Carathéodory function if $K(s,t,0)=0$ for ${\mu\otimes\nu}$-almost all $(s,t)\in A\times B$. Proposition 5 (Proposition 3.2 in [22]). Let $E$ be an order ideal in the space $L_{0}(\nu)$, let $K\colon A\times B\times\mathbb{R}\to\mathbb{R}$ be a normalized Carathéodory function, and for each $f\in E$ let the inequality holds for almost all $s\in A$. Then the map $T\colon E\to L_{0}(\mu)$ defined by is a regular orthogonally additive operator from $E$ to $L_{0}(\mu)$. The map $T\colon E\to L_{0}(\mu)$ defined by (2.1) is known as the integral Urysohn operator (see [5]). The third class of examples is related to superposition operators. Definition 5. Let $(A,\Sigma,\mu)$ be a space with finite measure. Then a function $N\colon {A\times \mathbb{R}}\to \mathbb{R}$ is called: Proposition 6 (see Example 1 in [22]). Let $E$ be an order ideal of the space $L_{0}(\nu)$, and let $N\colon A\times \mathbb{R}\to \mathbb{R}$ be a sup-measurable function. Then the formula defines a regular orthogonally additive operator $\mathcal{N}\colon E\to L_{0}(\mu)$. The operator $\mathcal{N}$ is known as the nonlinear superposition operator, or the Nemytskii operator (see [16]). The following class of examples is related to the lateral order in vector lattices. Definition 6. Let $E$ be a vector lattice. A subset $\mathcal{I}$ of $E$ is a lateral ideal if the following conditions are satisfied: Note that each order ideal of a vector lattice is also a lateral ideal. A typical example of a ‘nonlinear’ lateral ideal is given by the set of fragments $\mathfrak{F}_x$ of an arbitrary $x\in E$. The following fact demonstrates the importance of lateral ideals for the theory of orthogonally additive operators. Proposition 7 (Proposition 6.4 in [31]). Let $E$ and $F$ be vector lattices, and let $T\colon E \to F$ be a orthogonally additive operator. Then $\ker T: = \{x \in E\colon Tx = 0\}$ is a lateral ideal in $E$. A net $(x_\alpha)_{\alpha\in A}$ (a sequence $(x_n)_{n\in \mathbb{N}}$) of elements of a vector lattice $E$ is said to converge laterally to $x\in E$ if $(x_\alpha)_{\alpha\in A}$ (respectively, $(x_n)_{n\in \mathbb{N}}$) order converges to $x$ and if the condition $\alpha\leqslant\beta$ ($n\leqslant m$) implies that $x_{\alpha}\sqsubseteq x_{\beta}$ (respectively, $x_{n}\sqsubseteq x_{m}$) for all $\alpha,\beta\in A$ (for $n,m\in\mathbb{N}$). A subset $D$ of a vector lattice $E$ is laterally closed (laterally $\sigma$-closed) if it contains the limits of all laterally convergent nets (respectively, sequences) of elements of $D$. Definition 7. Let $E$ be a vector lattice. A laterally closed lateral ideal $\mathcal{I}$ in $E$ is called a lateral band. Each band in a vector lattice $E$ is a lateral band. The set of fragments $\mathfrak{F}_x$ of an arbitrary $x\in E$ is a lateral band. A vector lattice $E$ is $C$-complete if, for any $x\in E_+$, the Boolean algebra $\mathfrak{F}_{x}$ is order complete. Each order complete vector lattice is $C$-complete (see Theorem 1.49 in [15]). Note that the class of $C$-complete lattices includes the class of order complete vector lattices (see Proposition 4.2 in [34]). Consider an example of an order incomplete vector lattice which is $C$-complete. Let $A$ be an open bounded subset of $\mathbb{R}^n$, let $\Sigma$ be the $\sigma$-algebra of Lebesgue measurable subsets of $A$, and let $\mu$ be the Lebesgue measure on $A$. Then $S(A,\Sigma,\mu)$ (or, briefly, $S(\mu)$) denotes the vector space of all measurable step functions on $A$. Thus,
$$
\begin{equation*}
f\in S(\mu) \quad\Longleftrightarrow \quad f=\sum_{i=1}^{n}\lambda_i1_{A_i}, \qquad \lambda_i\in\mathbb{R}, \quad A_i\in\Sigma, \quad 1\leqslant i\leqslant n, \quad n\in\mathbb{N}.
\end{equation*}
\notag
$$
Proof. We show that $S(\mu)$ is a vector lattice. Consider $f,g\in S(\mu)$, where $f=\bigsqcup_{i=1}^{n}\lambda_{i}1_{A_i}$ and $g=\bigsqcup_{j=1}^{m}\kappa_{j}1_{B_j}$. Then we have where
Now we show that the vector lattice $S(\mu)$ is $C$-complete. Let $f \in S_{+}$ and $f=\sum_{i=1}^{n}\lambda_{i}1_{A_i}$. Consider an arbitrary set $D\subset \mathfrak{F}_{f}$. It suffices to show that $z=\sup{D}$ exists. Let $g\in D$. By Proposition 3.1 in [34], the relation $g\sqsubseteq \bigl(\bigsqcup_{i=1}^{n}\lambda_{i}1_{A_i}\bigr)$ implies the existence of a disjoint decomposition $g=\bigsqcup_{i=1}^{n}g_i$, where $g_i\sqsubseteq \lambda_{i}1_{A_i}$ for any $1\leqslant i\leqslant n$. Note that $g_i=\lambda_{i}1_{A_i^{g}}$, where $A_i^{g}\in\Sigma$ and $A_i^{g}\subset A_i$ for any $1\leqslant i\leqslant n$. The set $D_i=\{1_{A_{i}^{g}}\colon g\in D\}$, $1\leqslant i\leqslant n$, is order bounded in $L_{0}(\mu)$, and therefore, since $L_{0}(\mu)$ is order complete (see Theorem 1.80 in [14]), there exists a finite set of measurable functions $G_i=\sup D_i$. In addition, $G_i=\sup_{k}\{1_{A_{i}^{g_k}}\}$ for $1\leqslant i\leqslant n$, where $(1_{A_{i}^{g_k}})_{k\in\mathbb{N}}$ is an increasing sequence in $D_i$. We set $A_{D_i}=\bigcup_{k=1}^{\infty}A_{i}^{g_k}$. Note that $A_{D_i}\in\Sigma$ and $A_{D_i}\subset A_i$. The element $z:=\sum_{i=1}^{n}\lambda_{i}1_{A_{D_i}}$ is the required supremum. This completes the proof of Proposition 8. Proposition 9 (Theorem 5.1 in [22]). Let $E$ be a $C$-complete vector lattice, and $\mathcal{B}$ be a lateral band in $E$. Then there exists a map $\mathfrak{p}_{\mathcal{B}}\colon E\to E$ such that: Moreover, the operator $\mathfrak{p}_{\mathcal{B}}$ is given by where the supremum is taken with respect to the lateral order $\sqsubseteq$.If the lateral band $\mathcal{B}$ is equal to $\mathfrak{F}_{x}$ for some $x\in E$, then we write $\mathfrak{p}_{x}$ for the operator $\mathfrak{p}_{\mathcal{B}}$. Operators $\mathfrak{p}_{\mathcal{B}}$ are called lateral projections. Lateral projections of the form $\mathfrak{p}_{x}$ are known as principal projections. If a lateral band $\mathcal{B}$ in the vector lattice $E$ is a ‘classical’ band, then the operator $\mathfrak{p}_{\mathcal{B}}$ coincides with the order projection onto the band. Below we give an example of a nonlinear projection onto a lateral band. Proposition 10 (Proposition 5.4 in [22]). Let $(A,\Sigma,\mu)$ be a finite measure, and let $f\in L_{0}(\mu)$. Then there exists a normalized sup-measurable function $N_{f}\colon A\times\mathbb{R}\to\mathbb{R}$ such that Definition 8. Let $E$ be a vector lattice, and let $x,y\in E$. Then we say that $x$ is laterally disjoint from $y$ (written $x\mathbin{\unicode{8224}} y$) if $\{z\in\mathcal{F}_{x}\cap\mathcal{F}_y\}=0$. Subsets $H$ and $D$ of a vector lattice $E$ are called laterally disjoint (written $H\mathbin{\unicode{8224}} D$) if $x\mathbin{\unicode{8224}} y$ for all $x\in H$ and $y\in D$. The following theorem is the main result of this section. Proof. That the operator $\mathfrak{p}_x$ is regular is a direct consequence of Proposition 2.
First we prove equality 1). By Lemma 1 we have
$$
\begin{equation*}
(\mathfrak{p}_x)_{+}u=\sup\{\mathfrak{p}_{x}z\colon z\sqsubseteq u\}.
\end{equation*}
\notag
$$
Since $\mathfrak{p}_{x}z\sqsubseteq x=x_{+}\sqcup (-x_{-})$, we have the disjoint partition
$$
\begin{equation*}
\mathfrak{p}_{x}z=w_1^z\sqcup w_2^z, \quad \text{where } w_1^z\sqsubseteq x_{+}\quad\text{and}\quad w_2^z\sqsubseteq -x_{-}.
\end{equation*}
\notag
$$
As a result,
$$
\begin{equation*}
\begin{aligned} \, (\mathfrak{p}_x)_{+}u &=\sup\{\mathfrak{p}_{x}z\colon z\sqsubseteq u\}=\sup\{w_1^z\sqcup w_2^z\colon z\sqsubseteq u\}=\sup\{w_1^z\colon z\sqsubseteq u\} \\ &=\sup\{z\in\mathfrak{F}_{x_+}\cap \mathfrak{F}_u\}=\mathfrak{p}_{x_+}u. \end{aligned}
\end{equation*}
\notag
$$
Let us prove equality 2). Arguing as above, we find that
$$
\begin{equation*}
\begin{aligned} \, (\mathfrak{p}_x)_{-}u &=-\inf\{\mathfrak{p}_{x}z\colon z\sqsubseteq u\}=-\inf\{w_1^z\sqcup w_2^z\colon z\sqsubseteq u\} \\ &=-\inf\{w_2^z\colon z\sqsubseteq u\} =\sup\{-z\in\mathfrak{F}_{-x_-}\cap\mathfrak{F}_u\}=\sup\{z\in\mathfrak{F}_{x_-} \,{\cap}\,\mathfrak{F}_u\}=\mathfrak{p}_{x_-}u. \end{aligned}
\end{equation*}
\notag
$$
Let us prove 3). Let $u$ be an arbitrary element of $E$, and let $v\in\mathfrak{F}_{x_+}\cap\mathfrak{F}_u$ and $w\in\mathfrak{F}_{x_-}\cap\mathfrak{F}_u$. Then $u\sqcup w\in \mathfrak{F}_{|x|}\cap\mathfrak{F}_u$. Hence
$$
\begin{equation*}
\{z\colon z\in\mathfrak{F}_{|x|}\cap \mathfrak{F}_u\}\supset\bigl\{\{v\colon v\in\mathfrak{F}_{x_{+}}\cap \mathfrak{F}_u\}\sqcup\{w\colon w\in\mathfrak{F}_{x_{-}}\cap \mathfrak{F}_u\}\bigr\}.
\end{equation*}
\notag
$$
Consequently, $v\sqcup w\sqsubseteq \mathfrak{p}_{|x|}u$ for all $v\in\mathfrak{F}_{x_{+}}\cap \mathfrak{F}_u$ and $w\in\mathfrak{F}_{x_{-}}\cap \mathfrak{F}_u$. Taking the lateral supremum, we get that $\mathfrak{p}_{x_{+}}\sqcup\mathfrak{p}_{x_{-}}\sqsubseteq \mathfrak{p}_{|x|}$. We prove the reverse inclusion. Let $z\in\mathfrak{F}_{|x|}\cap\mathfrak{F}_u$. Then $z\sqsubseteq (x_{+}\sqcup x_{-})$, $z\sqsubseteq u$, and there exists a representation $z=z_1\,{\sqcup}\, z_2$, where $z_1\in\mathfrak{F}_{x_{+}}\cap\mathfrak{F}_u$ and $z_2\in\mathfrak{F}_{x_{-}}\cap\mathfrak{F}_u$. Hence $z\sqsubseteq \mathfrak{p}_{x_{+}}u\sqcup \mathfrak{p}_{x_{-}}u$ for all $z\in\mathfrak{F}_{|x|}\cap \mathfrak{F}_u$, and so, taking the lateral supremum, we get that $\mathfrak{p}_{|x|}u\sqsubseteq \mathfrak{p}_{x_{+}}u\sqcup\mathfrak{p}_{x_{-}}u$. As a result,
$$
\begin{equation*}
\mathfrak{p}_{|x|}u=\mathfrak{p}_{x_{+}}u\sqcup\mathfrak{p}_{x_{-}}u =(\mathfrak{p}_x)_{+}u\sqcup(\mathfrak{p}_x)_{-}u=|\mathfrak{p}_x|u
\end{equation*}
\notag
$$
for any $u\in E$, which verifies equality 3).
Now we prove equivalence 4). Let $|x|\mathbin{\unicode{8224}} |y|$, and let $z$ be an arbitrary element of $E$. Note that $|\mathfrak{p}_x|z\sqsubseteq |x|$ and $|\mathfrak{p}_y|z\sqsubseteq |y|$, and, since $|x|$ and $|y|$ are laterally disjoint, we have $|\mathfrak{p}_x|z\,{\wedge}\, |\mathfrak{p}_y|z=0$. Now an application of Lemma 1 shows that
$$
\begin{equation*}
|\mathfrak{p}_x|\wedge|\mathfrak{p}_{y}|(z)=\inf\{|\mathfrak{p}_x|u+|\mathfrak{p}_y|v\colon z=u\sqcup v\}\leqslant |\mathfrak{p}_x|z\wedge|\mathfrak{p}_y|z.
\end{equation*}
\notag
$$
This proves the implication $|x|\mathbin{\unicode{8224}} |y|$ $\Longrightarrow$ $\mathfrak{p}_x\perp\mathfrak{p}_{y}$. To prove the reverse implication we suppose for a contradiction that there exists $0\neq z\in \mathfrak{F}_{|x|}\cap\mathfrak{F}_{|y|}$. In this case we have
$$
\begin{equation*}
|\mathfrak{p}_x|\wedge|\mathfrak{p}_{y}|(z)=\inf\{|\mathfrak{p}_x|u+|\mathfrak{p}_y|v\colon z=u\sqcup v\}= u\sqcup v=z>0.
\end{equation*}
\notag
$$
This contradiction proves Theorem 1. Remark 1. Note that a lateral order appears naturally in the study of orthogonally additive maps of vector lattices, and also in the theory of operator algebras (see [6], [17], [19], [31], [22] and [34]). § 3. Criterion for a regular OAO to be diffuseIn this section, we introduce diffuse orthogonally additive operators and give a criterion for a regular OAO acting from a vector lattice $E$ to an order-complete vector lattice $F$ to be diffuse. Definition 9. Let $E$ and $F$ be vector lattices such that $F$ is order complete. A regular OAO $T\colon E\to F$ is diffuse if $T\perp S$ for each $S\in \mathcal{OA}_{\mathrm{dpo}}(E,F)$. The set of all diffuse orthogonally additive operators from $E$ into $F$ is denoted by $\mathcal{OA}_{\mathrm{dif}}(E,F)$. It is clear from the definition that the set $\mathcal{OA}_{\mathrm{dif}}(E,F)$ is a band in $\mathcal{OA}_{\mathrm r}(E,F)$. The complemented band $\{\mathcal{OA}_{\mathrm{dpo}}(E,F)\}^{\perp\perp}$ of $\mathcal{OA}_{\mathrm{dif}}(E,F)$ is denoted by $\mathcal{OA}_{\mathrm a}(E,F)$. Elements of the band $\mathcal{OA}_{\mathrm a}(E,F)$ are commonly called atomic operators. That $\mathcal{OA}_{\mathrm r}(E,F)$ is order complete follows from the disjoint decomposition
$$
\begin{equation*}
\mathcal{OA}_{\mathrm r}(E,F)=\mathcal{OA}_{\mathrm a}(E,F)\oplus\mathcal{OA}_{\mathrm{dif}}(E,F)
\end{equation*}
\notag
$$
of the vector lattice $\mathcal{OA}_{\mathrm r}(E,F)$ into the direct sum of the disjoint bands $\mathcal{OA}_{\mathrm a}(E,F)$ and $\mathcal{OA}_{\mathrm{dif}}(E,F)$. Consider some examples of diffuse operators. Recall that a positive element $x$ of a vector lattice $E$ is an atom if the conditions $0\leqslant y\leqslant x$, $0\leqslant z\leqslant x$ and $y\perp z$ imply that at least one of $y$ or $z$ is zero. A vector lattice $E$ is atomless if any atom of $E$ is zero. A typical example of an atomless vector lattice is the space $L_{p}([0,1],\Sigma,\mu)$, $0\leqslant p\leqslant \infty$, where $\mu$ is the Lebesgue measure. Definition 10. Let $E$ and $F$ be vector lattices. An orthogonally additive operator $T\colon E\to F$ is said to be laterally continuous (laterally $\sigma$-continuous) if, for each net $(x_{\alpha})_{\alpha\in A}$ (respectively, each sequence $(x_n)_{n\in \mathbb{N}A}$) laterally converging to an element $x\in E$ in $E$, the net $(Tx_{\alpha})_{\alpha\in A}$ (the sequence $(Tx_n)_{n\in \mathbb{N}A}$, respectively) order converges in $F$ to $Tx\in F$. The sets of regular laterally continuous and laterally $\sigma$-continuous OAOs from $E$ to $F$ are denoted by $\mathcal{OA}_{\mathrm c}(E,F)$ and $\mathcal{OA}_{\sigma \mathrm c}(E,F)$, respectively. Proposition 11 (see Theorem 3.13 in [35]). Let $E$ and $F$ be vector lattices such that $F$ is order complete. Then $\mathcal{OA}_{\mathrm c}(E,F)$ and $\mathcal{OA}_{\sigma \mathrm c}(E,F)$ are bands of the order-complete vector lattice $\mathcal{OA}_{\mathrm r}(E,F)$. Proposition 12. Let $E$ be an atomless vector lattice. Then each regular laterally $\sigma$-continuous orthogonally additive functional $T\colon E\to \mathbb{R}$ is diffuse. Proof. By Proposition 11 the set $\mathcal{OA}_{\sigma \mathrm c}(E,\mathbb{R})$ is a band, and therefore it is an order-complete sublattice of $\mathcal{OA}_{\mathrm r}(E,\mathbb{R})$. Hence it suffices to verify that if $0\leqslant T\in\mathcal{OA}_{\mathrm{dpo}}(E,F)$ is a laterally $\sigma$-continuous operator, then $T=0$. Assume that there exists $x\in E$ such that $Tx=r>0$. Since $E$ is atomless, there exists a disjoint partition $x=x_1\sqcup x_2$, where $x_1$ and $x_2$ are distinct from zero. Since $x_1\perp x_2$, we have $Tx_1\perp Tx_2$, and so either $Tx_1=0$ or $Tx_2=0$. Let $Tx_1=0$ and $Tx_2=Tx$. We consider the disjoint partition $x_2=x_3\sqcup x_4$, where $Tx_3=0$ and $Tx_4=Tx$, and construct a sequence $(x_n)_{n\in \mathbb{N}}$ such that $x_{2k}=x_{2k+1}\sqcup x_{2k+2}$, $Tx_{2k+1}=0$ and $Tx_{2k+2}=Tx$ for each integer natural $k\geqslant 1$. Let the sequence $(y_n)_{n\in\mathbb{N}}$ be defined by
$$
\begin{equation*}
y_1=x_1, \quad y_2=x_1\sqcup x_3, \quad \dots, \quad y_n=x_1\sqcup x_3\sqcup\dots\sqcup x_{2n-1}, \quad \dots\,.
\end{equation*}
\notag
$$
The sequence $(y_{n})$ converges laterally to $x$, and $Ty_{n}=0$ for any $n\in\mathbb{N}$. This contradiction proves Proposition 12. The following proposition shows that a diffuse OAO between finite-dimensional spaces can only be the zero operator. Proposition 13. Let $n,m\in\mathbb{N}$. Then $\mathcal{OA}_{\mathrm{dif}}(\mathbb{R}^n,\mathbb{R}^m)=\{0\}$. Proof. Suppose for a contradiction that there exist a positive diffuse OAO $T\colon {\mathbb{R}^n\to \mathbb{R}^m}$ and a coefficient $\xi=(\xi_1,\dots,\xi_n)$ such that
$$
\begin{equation*}
T\xi=T(\xi_1,\dots,\xi_n)=\sum_{i=1}^{n}T(\xi_i)= \biggl(\sum_{i=1}^{n}T_{1i}(\xi_i),\dots,\sum_{i=1}^{n}T_{mi}(\xi_i)\biggr)>0,
\end{equation*}
\notag
$$
where $T_{ki}\colon\mathbb{R}\to \mathbb{R}_{+}$, $k\in\{1,\dots,m\}$, $i\in\{1,\dots,n\}$ and $T_{ki}(0)=0$. Since
$$
\begin{equation*}
\biggl(\sum_{i=1}^{n}T_{1i}(\xi_i),\dots,\sum_{i=1}^{n}T_{mi}(\xi_i)\biggr)>0,
\end{equation*}
\notag
$$
there exists $k_{0}\in \{1,\dots,m\}$ such that $\sum_{i=1}^{n}T_{k_{0}i}(\xi_i)>0$. Assume that $k_{0} = 1$. We can assume without loss of generality that $\xi_1 \neq 0$ and $T_{11}(\xi_1)\,{>}\,0$. Consider the operator $S\in\mathcal{OA}_{\mathrm r}(\mathbb{R}^n,\mathbb{R}^m)$ defined by
$$
\begin{equation*}
S(x_1,\dots,x_{n})=\biggl(\sum_{i=1}^{n}S_{1i}(\xi_i),\dots,\sum_{i=1}^{n}S_{mi}(\xi_i)\biggr),
\end{equation*}
\notag
$$
where for $x\in \mathbb{R}$ the functions $S_{ki}$, $k\in\{1,\dots,m\}$, $i\in\{1,\dots,n\}$, are defined by
$$
\begin{equation*}
S_{ki}(x)=\begin{cases} x^{2} &\text{if }(k,i)=(1,1), \\ 0 &\text{otherwise}. \end{cases}
\end{equation*}
\notag
$$
It is clear that $S$ is a positive disjointness preserving orthogonally additive operator from $\mathbb{R}^n$ to $\mathbb{R}^m$, and since $T$ is diffuse, we have $T\wedge S=0$. Consider the vector $\xi'=(\xi_1,0,\dots,0)$. By Lemma 1,
$$
\begin{equation*}
\begin{aligned} \, (T\wedge S) (\xi') &=(T\wedge S)(\xi_1,0,\dots,0)=(\xi_1^2,0,\dots,0)\wedge (T_{11}(\xi_1),0,\dots,0) \\ &=T_{11}(\xi_1)\wedge (\xi_1^2,0,\dots,0)>0. \end{aligned}
\end{equation*}
\notag
$$
This contradiction proves Proposition 13. Proposition 14 (Proposition 3.11 in [34]). Let $E$ be a vector lattice, and let $\bigsqcup_{i=1}^{n}x_{i}=\bigsqcup_{k=1}^{m}y_{k}$ for some families $(x_{i})_{i=1}^{n}$ and $(y_{k})_{k=1}^{m}$ of elements of $E$. Then there exists a family $(z_{ik})$ of pairwise disjoint elements of $E$, where $i\in\{1,\dots,n\}$ and $k\in\{1,\dots,m\}$, such that: Let $E$ be a vector lattice, and let $x\in E$. A family $(x_i)_{i=1}^{n}$ of pairwise disjoint elements of $E$ such that $x=\bigsqcup_{i=1}^{n}x_i$ is known as a finite disjoint partition of $x$. The set of all finite disjoint partitions of $x$ is denoted by $\Theta(x)$. Note that there exists a natural partial order $\preceq$ on $\Theta(x)$; namely,
$$
\begin{equation*}
\xi=(x_i)_{i=1}^{n}\preceq (y_j)_{j=1}^{m}=\zeta \quad\Longleftrightarrow\quad x_{i}=\bigsqcup_{j(i)}^{k(i)}y_{j_{i}}
\end{equation*}
\notag
$$
for any $i\in\{1,\dots, n\}$ and some set $\{j(i),\dots, k(i)\}\subset\{1,\dots, m\}$. Recall that a partially ordered set $(X,\leqslant)$ is upward directed if, for all $x,y\in X$, there exists $z\in X$ such that $x\leqslant z$ and $y\leqslant z$. From Proposition 14 it readily follows that the set $(\Theta(x),\preceq)$ is upward directed for any element $x$ of an arbitrary vector lattice $E$. Now let $F$ be an order-complete vector lattice. With each operator $T\in\mathcal{OA}_{\mathrm r}(E,F)$ we associate the map $\mathfrak{d}_{T}\colon E\to F$ defined by
$$
\begin{equation*}
\mathfrak{d}_{T}(x):=\bigwedge_{(x_i)_{i=1}^{n}\in\Theta(x)}\,\bigvee_{i=1}^{n}|T|x_{i}, \qquad x\in E.
\end{equation*}
\notag
$$
The map $\mathfrak{d}_{T}$ is well defined, since $F$ is an order-complete vector lattice. We need the following auxiliary constructions. Proposition 15. Let $E$ be a vector lattice, and let $x \in E$ and $y,z\in\mathfrak{F}_x$. Then $y+z=y\cup z+y\cap z$. Proof. We have
$$
\begin{equation*}
\begin{aligned} \, y+z&=y^{+}-y^{-}+z^{+}-z^{-}=(y^{+}+z^{+})-(y^{-}+z^{-}) \\ &=(y^{+}\vee z^{+}+y^{+}\wedge z^{+})-(y^{-}\vee z^{-}+y^{-}\wedge z^{-}) \\ &=(y^{+}\vee z^{+})-(y^{-}\vee z^{-})+(y^{+}\wedge z^{+})-(y^{-}\wedge z^{-}) =y\cup z+y\cap z, \end{aligned}
\end{equation*}
\notag
$$
which proves Proposition 14. Definition 11. Let $E$ and $F$ be vector lattices and $\mathcal{I}$ be a lateral ideal of $E$. A set $D\subset \mathcal{I}$ is order bounded if there exists an element $y\in E_{+}$ such that $|x|\leqslant y$ for each $x\in D$. A map $R\colon \mathcal{I}\to F$ is said to be:
Proposition 16 (Theorem 1 in [8]). Let $E$ and $F$ be vector lattices, where the lattice $F$ is order complete, let $\mathcal{I}$ be a lateral ideal in $E$, and let $R\colon \mathcal{I}\to F$ be a positive order-bounded orthogonally additive map. Then there exists a positive orthogonally additive operator $\widetilde{R}_{\mathcal{I}}\colon E\to F$ such that $\widetilde{R}_{\mathcal{I}}x=Rx$ for each $x\in \mathcal{I}$. The operator $\widetilde{R}_{\mathcal{I}}\in\mathcal{OA}_{+}(E,F)$ (or simply $\widetilde{R}$ for brevity) is called the minimal extension (with respect to $\mathcal{I}$) of an orthogonally additive map $R\colon \mathcal{I}\to F$. Note that this operator $\widetilde{R}$ is defined by
$$
\begin{equation*}
\widetilde{R}x=\sup\{Ry\colon y\in \mathcal{I}\cap\mathfrak{F}_{x}\}, \qquad x\in E.
\end{equation*}
\notag
$$
Proposition 17. Let $E$, $F$ and $\mathcal{I}$ be as in Proposition 16, and let $R\colon\mathcal{I}\to F$ be a positive order-bounded disjointness preserving orthogonally additive map. Then $\widetilde{R} \in \mathcal{OA}_{+}(E,F)$ is a disjointness preserving operator. Definition 12. Let $\mathfrak{A}$ and $\mathfrak{B}$ be lattices (which are not necessarily vector spaces). A function $\varphi\colon\mathfrak{A}\to \mathfrak{B}$ is called: If $\mathfrak{A}$ and $\mathfrak{B}$ are Boolean algebras and, in addition, $\varphi(0_{\mathfrak{A}})=0_{\mathfrak{B}}$ ($\varphi(1_{\mathfrak{A}})=1_{\mathfrak{A}}$), then we say that $\varphi$ is a Boolean $\cup$-map (a Boolean $\cap$-map). Proposition 18. Let $E$ be a vector lattice, and let $x\in E$, $y\in \mathfrak{F}_x$ and $\xi=(x_{i})_{i=1}^{n}\in \Theta(x)$. Then there exists a finite disjoint partition $(y_{i}^{\xi})_{i=1}^{n}$ of $y$ such that $y_i\sqsubseteq x_i$ for any $1\leqslant i\leqslant n$. Proof. Consider the double partition of the element $x$. The existence of the required finite disjoint partition is secured by Proposition 14.
This proves Proposition 18. Let $E$ and $F$ be vector lattices such that $F$ is order complete, let $x\in E$ and $T\in\mathcal{OA}_{\mathrm r}(E,F)$. With each element $\xi=(x_i)_{i=1}^{n}\in\Theta(x)$ we can associate the map $\mathfrak{d}_{T,\xi}\colon \mathfrak{F}_x\to F$ defined by
$$
\begin{equation*}
\mathfrak{d}_{T,\xi}(y):=\bigvee_{i=1}^{n}|T|y_{i}^{\xi}, \qquad y\in \mathfrak{F}_x.
\end{equation*}
\notag
$$
It follows from the definition of the maps $\mathfrak{d}_{T,\xi}$ and $\mathfrak{d}_{T}$ that for $x\in E$,
$$
\begin{equation*}
\mathfrak{d}_{T}(x)=\inf_{\xi\in\Theta(x)}\mathfrak{d}_{T,\xi}(x).
\end{equation*}
\notag
$$
Note that for all $y,z\in\mathfrak{F}_x$ and $\xi\in\Theta(x)$ the relation $y\sqsubseteq z$ implies that ${\mathfrak{d}_{T,\xi}(y)\leqslant \mathfrak{d}_{T,\xi}(z)}$. Hence $\mathfrak{d}_{T}(y)\leqslant \mathfrak{d}_{T}(z)$. Proposition 19. Let $E$ and $F$ be vector lattices such that $F$ is order complete, let $x\in E$ and $T\in\mathcal{OA}_{\mathrm r}(E,F)$. Then $\mathfrak{d}_{T}\colon \mathfrak{F}_x\to F$ is a $\cup$-map. Proof. We must show that for all $y,z\in\mathfrak{F}_x$
$$
\begin{equation*}
\mathfrak{d}_{T}(y\cup z)=\mathfrak{d}_{T}(y)\vee\mathfrak{d}_{T}(z).
\end{equation*}
\notag
$$
First we consider the particular case where $y\perp z$. Let $\xi\in \Theta(x)$. Then
$$
\begin{equation*}
y=\bigvee_{i=1}^{n}y_{i}^{\xi}, \qquad z=\bigvee_{i=1}^{n}z_{i}^{\xi}\quad\text{and} \quad y\sqcup z=\bigvee_{i=1}^{n}(y_{i}^{\xi}\sqcup z_{i}^{\xi}).
\end{equation*}
\notag
$$
The lateral infimum $y\cap z$ of the disjoint elements $y$ and $z$ is zero, and therefore
$$
\begin{equation*}
\begin{aligned} \, \mathfrak{d}_{T,\xi}(y\cup z) &\stackrel{\rm \text{Prop. }15}{=} \mathfrak{d}_{T,\xi}(y\sqcup z)= \bigvee_{i=1}^{n}|T|(y_{i}^{\xi}\sqcup z_{i}^{\xi}) \\ &\quad\ \geqslant\bigvee_{i=1}^{n}|T|y_{i}^{\xi}\vee\bigvee_{i=1}^{n}|T|z_{i}^{\xi} \geqslant\mathfrak{d}_{T}(y)\vee\mathfrak{d}_{T}(z) \\ &\!\!\!\!\!\!\!\!\!\Longrightarrow\quad \mathfrak{d}_{T}(y\cup z)\geqslant \mathfrak{d}_{T}(y)\vee\mathfrak{d}_{T}(z). \end{aligned}
\end{equation*}
\notag
$$
On the other hand, for any $\xi\in \Theta(x)$,
$$
\begin{equation*}
\mathfrak{d}_{T}(y\cup z)\leqslant \bigvee_{i=1}^{n}|T|y_{i}^{\xi}\vee\bigvee_{i=1}^{n}|T|z_{i}^{\xi}.
\end{equation*}
\notag
$$
So, taking the infimum over all $\xi\in \Theta(x)$ on the right, we have
$$
\begin{equation*}
\mathfrak{d}_{T}(y\cup z)\leqslant \mathfrak{d}_{T}(y)\vee\mathfrak{d}_{T}(z) \quad\Longrightarrow\quad \mathfrak{d}_{T}(y\cup z)=\mathfrak{d}_{T}(y)\vee\mathfrak{d}_{T}(z).
\end{equation*}
\notag
$$
Now let $y$ and $z$ be arbitrary. In this case $y\cup z=y\sqcup (z-y\cap z)$. Since $\mathfrak{d}_{T}(y)\vee \mathfrak{d}_{T}(y\cap z)=\mathfrak{d}_{T}(y)$, we have
$$
\begin{equation*}
\begin{aligned} \, \mathfrak{d}_{T}(y\cup z) &=\mathfrak{d}_{T}(y\sqcup (z-y\cap z)) \\ &= \mathfrak{d}_{T}(y)\vee\mathfrak{d}_{T}(z-y\cap z) \\ &=\mathfrak{d}_{T}(y)\vee\mathfrak{d}_{T}(y\cap z)\vee\mathfrak{d}_{T}(z-y\cap z) \\ &= \mathfrak{d}_{T}(y)\vee\mathfrak{d}_{T}((y\cap z)\cup(z-y\cap z)) \\ &=\mathfrak{d}_{T}(y)\vee\mathfrak{d}_{T}((y\cap z)\sqcup (z-y\cap z))= \mathfrak{d}_{T}(y)\vee\mathfrak{d}_{T}(z). \end{aligned}
\end{equation*}
\notag
$$
Recall that a compact set $Q$ is extremally disconnected or simply extremal if the closure of each open subset $D$ of $Q$ is clopen in $Q$. For an extremal compact set $Q$, we denote the space of continuous functions that take infinite values on nowhere dense subsets of $Q$ by $C_{\infty}(Q)$. It is known that $C_{\infty}(Q)$ is an order-complete vector lattice (see Theorem 5.2.2 in [2]). An order-complete ideal $F$ of $C_{\infty}(Q)$ is called an order-dense ideal in $C_{\infty}(Q)$. Each order-complete vector lattice $F$ is an order-dense ideal of the vector lattice $C_{\infty}(Q)$ for some extremally disconnected compact set $Q$ (see Theorem 5.4.2 in [2]). Proposition 20 (see Theorem 10.32 in [44]). Let $\mathfrak{A}$ and $\mathfrak{B}$ be Boolean algebras such that $\mathfrak{B}$ is order complete, and let $\varphi\colon\mathfrak{A}\to \mathfrak{B}$ be a Boolean $\cup$-map. Then each Boolean homomorphism $\psi_{0}\colon \mathfrak{A}_{0}\to \mathfrak{B}$ defined on a Boolean subalgebra $\mathfrak{A}_{0}$ of $\mathfrak{A}$ and satisfying $\psi_{0}(x)\leqslant\varphi(x)$ for all $x\in \mathfrak{A}_0$ can be extended to a Boolean homomorphism $\psi\colon\mathfrak{A}\to \mathfrak{B}$ such that $\psi(x)\leqslant\varphi(x)$ for each $x\in \mathfrak{A}$. Below we assume that $F$ is an order-dense ideal of the order-complete vector lattice $C_{\infty}(Q)$ for some extremally disconnected compact set $Q$. In this case the order projection operator onto the band $C_{\infty}(Q)$ is the operator of multiplication by the characteristic function $1_{G}$ of a clopen set $G\subset Q$. Let $x\in E$. We let $\{\mathfrak{d}_{T}(y)=\mathfrak{d}_{T}(x)\}$ denote the closed clopen set $\{q\in Q\colon \mathfrak{d}_{T}(y)(q)=\mathfrak{d}_{T}(x)(q)\}$. Next, let the map $\phi\colon\mathfrak{F}_x\to\mathfrak{F}_{\mathfrak{d}_{T}(x)}$ be defined by
$$
\begin{equation*}
\phi(y):=1_{\{\mathfrak{d}_{T}(y)=\mathfrak{d}_{T}(x)\}}\mathfrak{d}_{T}(x).
\end{equation*}
\notag
$$
Proposition 21. Let $E$, $F$ and $T$ be as in Proposition 19. Then $\phi\colon\mathfrak{F}_x\to\mathfrak{F}_{\mathfrak{d}_{T}(x)}$ is a $\cup$-map. Proof. The required result is secured by Proposition 19 and the following chain of equalities:
$$
\begin{equation*}
\begin{aligned} \, \phi(y\cup z)&=1_{\{\mathfrak{d}_{T}(y\cup z)=\mathfrak{d}_{T}(x)\}}\mathfrak{d}_{T}(x) =1_{\{\mathfrak{d}_{T}(y)=\mathfrak{d}_{T}(x)\cup\mathfrak{d}_{T}(z) =\mathfrak{d}_{T}(x)\}}\mathfrak{d}_{T}(x) \\ &=1_{\{\mathfrak{d}_{T}(y)=\mathfrak{d}_{T}(x)\}}\mathfrak{d}_{T}(x)\vee 1_{\{\mathfrak{d}_{T}(z)=\mathfrak{d}_{T}(x)\}}\mathfrak{d}_{T}(x)= \phi(y)\vee\phi(z). \end{aligned}
\end{equation*}
\notag
$$
This proves Proposition 21. Proposition 22. Let $E$ and $F$ be vector lattices such that $F$ is order complete, and let $x\in E$ and $T\in \mathcal{OA}_+(E,F)$. Then Proof. Consider a disjointness preserving OAO $S\colon E\to F$ such that $0\leqslant S\leqslant T$ and fix an arbitrary $\xi=(x_{i})_{i=1}^{n}\in \Theta(x)$. Then
$$
\begin{equation*}
\begin{aligned} \, & \bigvee_{i=1}^{n}Tx_{i}\geqslant\bigvee_{i=1}^{n}Sx_{i}=\sum_{i=1}^{n}Sx_{i} =S\biggl(\sum_{i=1}^{n}x_{i}\biggr)=Sx \\ & \quad\Longrightarrow\quad \mathfrak{d}_{T}(x)\geqslant Sx \quad\Longrightarrow\quad \mathfrak{d}_{T}(x)\geqslant\sup\{Sx\colon S\in [0,T]\cap\mathcal{OA}_{\mathrm{dpo}}(E,F)\}. \end{aligned}
\end{equation*}
\notag
$$
Assume that $\mathfrak{d}_{T}(x)-\sup\{Sx\colon S\in [0,T]\cap\mathcal{OA}_{\mathrm{dpo}}(E,F)\}>0$. By Proposition 21, $\phi\colon\mathfrak{F}_x\to\mathfrak{F}_{\mathfrak{d}_{T}(x)}$ is a $\cup$-map. Let the Boolean homomorphism $\psi_{0}\colon \{0,x\}\to \mathfrak{F}_{\mathfrak{d}_{T}(x)}$ be defined by
$$
\begin{equation*}
\psi_{0}(0)=0\qquad\text{and}\qquad \psi_{0}(x)=\mathfrak{d}_{T}(x).
\end{equation*}
\notag
$$
Note that $\psi_{0}(y)\leqslant \phi(y)$ for all elements of the Boolean subalgebra $\{0,x\}$. By Proposition 20 the Boolean homomorphism $\psi_0$ can be extended to a Boolean homomorphism $\psi\colon\mathfrak{F}_x\to \mathfrak{F}_{\mathfrak{d}_{T}(x)}$ such that $\psi(y)\leqslant\varphi(y)$ for any $y\in \mathfrak{F}_x$. Note that $\psi\colon \mathfrak{F}_x\to F$ is a positive order-bounded disjointness preserving orthogonally additive map. By Proposition 17, $\widetilde{\varphi}\colon E\to F$ is a positive disjointness preserving OAO such that $0\leqslant \varphi\leqslant T$, and $\varphi(x)=\mathfrak{d}_{T}(x)$. This contradiction proves Proposition 22. The following theorem is the main result of this section. Theorem 2. Let $E$ and $F$ be vector lattices such that $F$ is order complete, and let $T\colon E\to F$ be a regular, orthogonally additive operator. Then the following conditions are equivalent: Proof. 1) $\Longrightarrow$ 2). Since the operator $T$ is diffuse, the set $[0,|T|]\cap \mathcal{OA}_{\mathrm{dpo}}(E,F)$ consists of the zero operator. Now $\mathfrak{d}_{T}(x)=0$ by Proposition 22.
2) $\Longrightarrow$ 1). The equality
$$
\begin{equation*}
\mathfrak{d}_{T}(x)=\sup\{Sx\colon S\in [0,|T|]\cap\mathcal{OA}_{\mathrm{dpo}}(E,F)\}=0
\end{equation*}
\notag
$$
holds for all $x\in E$, and therefore $[0,|T|]\cap \mathcal{OA}_{\mathrm{dpo}}(E,F)=0$. Consequently, $[0,|T|]\cap \mathcal{OA}_{\mathrm{dpo}}(E,F)^{\perp\perp}=0$, and $T$ is a diffuse operator.
This proves Theorem 2. Remark 2. Note that a particular case of Theorem 2, when $F$ is an order ideal of a Banach lattice with order continuous norm, was established earlier by Pliev and Popov (see Theorem 2.6 in [37]). Having a diffusion criterion for orthogonally additive operators at our disposal, we can substantially strengthen Proposition 13. We let $s$ denote the vector lattice of all coordinatewise ordered real number sequences. Proposition 23. Let $E$ be an order ideal of the vector lattice $s$, $F$ be an order-complete vector lattice, and let $T\colon E\to F$ be a laterally $\sigma$-continuous diffuse orthogonally additive operator. Then $T=0$. Proof. Assume for a contradiction that there exists a nonzero laterally $\sigma$-continuous diffuse OAO $T\colon E\to F$. By Theorem 3.13 in [35] and the definition of the diffuse operator, the modulus $|T|$ of the operator $T$ is also a laterally $\sigma$-continuous diffuse OAO from $E$ to $F$. Hence we can assume that $T$ is positive. By assumption the operator $T$ is nonzero, and therefore there exists a sequence $x=(x_i)_{i=1}^{\infty}\in E$ such that $f=Tx>0$. Consider the family of fragments $x^1,\dots, x^n,\dots$ of $x$ defined by
$$
\begin{equation*}
x^{1}=(x_1,0,\dots,0,\dots), \quad \dots,\quad x^{n}=(x_1,x_2,\dots,x_n,0,\dots), \quad\dots\,.
\end{equation*}
\notag
$$
The sequence $(x^{n})_{n=1}^{\infty}$ converges laterally to $x$, and therefore the sequence $(Tx^{n})_{n=1}^{\infty}$ order converges to $Tx$. Hence there exists a sequence $(e_n)_{n=1}^{\infty}\subset F_+$ such that $e_n\downarrow 0$ and $|Tx\,{-}\,Tx^{n}|\leqslant e_n$ for all $n\in\mathbb{N}$ such that $n\geqslant n_0$ for some $n_0\in\mathbb{N}$. Then there exist an index $k_0\in\mathbb{N}$, $k_0>n_{0}$, and an order projection $\pi\in\mathfrak{B}(F)$ such that $\pi e_{k_0}< \pi f$. We have
$$
\begin{equation*}
Tx-Tx^{k_0}=|Tx-Tx^{k_0}|\leqslant e_{k_0} \quad\Longrightarrow\quad 0\leqslant Tx\leqslant Tx^{k_0}+e_{k_0}.
\end{equation*}
\notag
$$
Let $e^i=(e_{n}^{i})_{n=1}^{\infty}$ be the sequence with 1 at the $i$th position and zeros elsewhere. Then
$$
\begin{equation*}
0\leqslant Tx^{k_0}=T\biggl(\sum_{n=1}^{k_0}x_ne^{n}\biggr)=\sum_{n=1}^{k_0}T(x_ne^{n})\leqslant k_0 \bigvee_{n=1}^{k_0}T(x_ne^{n}).
\end{equation*}
\notag
$$
Note that
$$
\begin{equation*}
\bigvee_{n=1}^{k_0}T(x_ne^{n})=\mathfrak{d}_{T}(x^{k_0})=0.
\end{equation*}
\notag
$$
As a result, $Tx^{k_0}=0$, and therefore $f=Tx\leqslant e_{k_0}$. This contradiction completes the proof of Proposition 23. § 4. Orthogonally additive integral operatorsIn this section, we present a regularity criterion for an integral Urysohn operator acting on order ideals of spaces of measurable functions, and then, using this criterion, we show that an orthogonally additive integral operator is diffuse. Consider an integral Urysohn operator $T\colon E\to F$ with kernel $K$ acting on order ideals $E$ and $F$ of spaces of measurable functions $L_{0}(\nu)$ and $L_{0}(\mu)$, respectively. Note that the operator $R\colon E\to F$ defined by
$$
\begin{equation}
Rf(s):=\int_{B}|K(s,t,f(t))|\,d\nu(t), \qquad f\in E,
\end{equation}
\tag{4.1}
$$
acts from $E$ to $L_{0}(\mu)$. Below we present necessary and sufficient conditions for the regularity of an integral operator. Theorem 3. Let $(A,\Sigma,\mu)$ and $(B,\Theta,\nu)$ be finite measure spaces, $E$ and $F$ be order ideals in $L_{0}(\nu)$ and $L_{0}(\mu)$, respectively, and $T\colon E\to F$ be an integral Urysohn operator with kernel $K$. Then the following conditions are equivalent: In addition, for each $f\in E$, and Proof. 2) $\Longrightarrow$ 1). The kernel $K$ of the integral operator $T$ is normalized, and therefore, for each $f\in E$,
$$
\begin{equation*}
\int_{B}K(\,\cdot\,,t,f1_{\operatorname{supp}f}(t))\,d\nu(t) =\int_{B}K(\,\cdot\,,t,f(t))1_{\operatorname{supp}f}\,d\nu(t)
\end{equation*}
\notag
$$
(see Proposition 3.2 in [22]). Let $f=g\sqcup h$ be a partition of $f$ into two disjoint fragments. For disjoint $g,h\in E$ we have $\nu\{t\in\operatorname{supp}g\cap \operatorname{supp}h\}=0$. Next,
$$
\begin{equation*}
\int_{B}K(\,\cdot\,,t,g(t))\,d\nu(t) \leqslant\int_{B}|K(\,\cdot\,,t,g(t))|\,d\nu(t),
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\begin{aligned} \, &\int_{B} (-K(\,\cdot\,,t,h(t)))\,d\nu(t) \leqslant\int_{B}|K(\,\cdot\,,t,h(t))|\,d\nu(t) \\ &\quad\Longrightarrow\quad \int_{B}K(\,\cdot\,,t,g(t)1_{\operatorname{supp}g})\,d\nu(t) -\int_{B}K(\,\cdot\,,t,h(t)1_{\operatorname{supp}h})\,d\nu(t) \\ &\qquad= \int_{B}K(\,\cdot\,,t,g(t))1_{\operatorname{supp}g}\,d\nu(t) -\int_{B}K(\,\cdot\,,t,h(t))1_{\operatorname{supp}h}\,d\nu(t) \\ &\qquad= \int_{\operatorname{supp}g}K(\,\cdot\,,t,g(t))\,d\nu(t) -\int_{\operatorname{supp}h}K(\,\cdot\,,t,h(t))\,d\nu(t) \\ &\qquad= \int_{\operatorname{supp}g}|K(\,\cdot\,,t,g(t))|\,d\nu(t) +\int_{\operatorname{supp}h}|K(\,\cdot\,,t,h(t))|\,d\nu(t) \\ &\qquad= \int_{\operatorname{supp}f}|K(\,\cdot\,,t,f(t))|\,d\nu(t) =\int_{B}|K(\,\cdot\,,t,f(t))|\,d\nu(t). \end{aligned}
\end{equation*}
\notag
$$
Taking the supremum over all disjoint partitions $f=g\sqcup h$ of $f$ on the right, we obtain This shows that the operator $|T|$ is well defined, and
2) $\Longrightarrow$ 1). Since $T\in \mathcal{OA}_{\mathrm r}(E,F)$, $T^{-},T^+, |T|\in \mathcal{OA}_{\mathrm r}(E,F)$ and $|T|=T^{+}+T^-$, it suffices to verify that for almost all $s\in A$ andFirst we prove (i). Since $T^{+}=T\vee 0$, we have
$$
\begin{equation*}
T^{+}f(\,\cdot\,)\leqslant \int_{B}K(\,\cdot\,,t,f(t))^{+}\,d\nu(t), \qquad f\in E.
\end{equation*}
\notag
$$
We set $\displaystyle \psi_f(\,\cdot\,)\,{=}\int_{B}K(\,\cdot\,,t,f(t))^{+}\,d\nu(t)$ and $A_n:=\{s\in A\colon n-1\,{\leqslant}\,\psi_f(s) < n\}$, ${n\in\mathbb{N}}$. Note that $A=\bigcup_{n=1}^{\infty}A_n$. It suffices to show that $T^{+}f(\,{\cdot}\,)1_{A_n}\mkern-2mu=\psi_f(\,{\cdot}\,)1_{A_n}$ for an arbitrary $n\in\mathbb{N}$, and so we may assume that the function $\psi_f(\,{\cdot}\,)$ is integrable with respect to the measure $\mu$. Let $D\in \Sigma$ be arbitrary, and let $D_{\Sigma}:=\{H\in D\cap \Sigma \}$. We claim that for any $\mu\otimes\nu$-measurable set $V\subset D\times B$
$$
\begin{equation*}
\int_{D}T^{+}f(s)\,d\mu(s)=\int_{V}K(s,t,f(t))^{+}\,d\mu\otimes\nu(s,t), \qquad f\in E.
\end{equation*}
\notag
$$
We set $V=\bigcup_{i=1}^{n}D_i\times B_i$, where $D_i\in D_{\Sigma}$ and $B_i\in \Theta$. Refining the partitions if necessary, we can represent $V$ in the form
$$
\begin{equation*}
V=\bigcup_{i=1}^{n}\bigcup_{j=1}^{m(i)}D_i\times B_{ij}, \qquad D_i\subset D_{\Sigma}, \quad B_{ij}\in \Theta,
\end{equation*}
\notag
$$
where $(D_{i})_{i=1}^{n}$ are pairwise disjoint, and for each $i\in\{1,\dots,n\} $ the sets $(B_{ij})_{j=1}^{m(i)}$ are pairwise disjoint. Next, we have
$$
\begin{equation*}
\begin{aligned} \, &\begin{aligned} \, \int_{D_i}T^{+}f(s)\,d\mu(s) &\geqslant\int_{D_i}T^{+}\bigl(f1_{\bigcup_{j=1}^{m(i)}B_{ij}}(s)\bigr)\,d\mu(s) \\ &\geqslant \int_{D_i}K\bigl(s,t,f1_{\bigcup_{j=1}^{m(i)}B_{ij}}(s)\bigr)\,d\mu\otimes\nu(s,t) \end{aligned} \\ &\begin{aligned} \, \quad\Longrightarrow\quad\sum_{i=1}^{n}\int_{D_i}T^{+}f(s)\,d\mu(s) &\geqslant\sum_{i=1}^{n}\int_{D_i}T^{+}\bigl(f1_{\bigcup_{j=1}^{m(i)}B_{ij}}(s)\bigr)\,d\mu(s) \\ &=\int_{\bigcup_{i=1}^{n}\bigcup_{j=1}^{m(i)}D_i\times B_{ij}}K(s,t,f(s))\,d\mu\otimes\nu(s,t). \end{aligned} \end{aligned}
\end{equation*}
\notag
$$
As a result,
$$
\begin{equation*}
\int_{D}T^{+}f(s)\,d\mu(s)\geqslant\int_{V}K(s,t,f(s))\,d\mu\otimes\nu(s,t).
\end{equation*}
\notag
$$
Now let $V$ be an arbitrary $\mu\otimes\nu$-measurable subset of $D\times B$, and let $\varepsilon>0$. By the absolute continuity of the integral, there exists $\delta>0$ such that, for each $\mu\otimes\nu$-measurable set $H\subset D\times B$, the condition $\mu\otimes\nu(H)<\delta$ implies the inequality
$$
\begin{equation*}
\int_{H}|K(s,t,f(s))|\,d\mu\otimes\nu(s,t)<\varepsilon.
\end{equation*}
\notag
$$
Consider a family $\bigcup_{i=1}^{n}D_i\times B_i$ such that $\mu\otimes\nu\bigl(V\triangle(\bigcup_{i=1}^{n}D_i\times B_i) \bigr)<\delta$, where $\triangle$ denotes the symmetric difference operation. Now we have
$$
\begin{equation*}
\begin{aligned} \, &\begin{aligned} \, \int_{V}K(s,t,f(s))\,d\mu\otimes\nu(s,t) &\leqslant \int_{V\triangle(\bigcup_{i=1}^{n}D_i\times B_i)}|K(s,t,f(s))|\,d\mu\otimes\nu(s,t) \\ &\qquad+\int_{\bigcup_{i=1}^{n}D_i\times B_i}|K(s,t,f(s))|\,d\mu\otimes\nu(s,t) \end{aligned} \\ &\quad\Longrightarrow\quad \int_{V}K(s,t,f(s))\,d\mu\otimes\nu(s,t)\leqslant \int_{D}T^{+}f(s)\,d\mu(s). \end{aligned}
\end{equation*}
\notag
$$
We set $V:=\{(s,t)\in D\times B\colon K(s,t,f(t))\geqslant 0\}$. Note that $V$ is a $\mu\otimes\nu$-measurable set, and
$$
\begin{equation*}
\int_{V}K(s,t,f(s))\,d\mu\otimes\nu(s,t)=\int_{D\times B}K(s,t,f(s))^{+}\,d\mu\otimes\nu(s,t).
\end{equation*}
\notag
$$
Hence, by what we proved above,
$$
\begin{equation*}
\begin{aligned} \, &\int_{D}T^{+}f(s)\,d\mu(s)\geqslant\int_{D\times B}K(s,t,f(s))^{+}\,d\mu\otimes\nu(s,t) \\ &\quad\Longrightarrow\quad \int_{D}T^{+}f(s)\,d\mu(s)=\int_{D\times B}K(s,t,f(s))^{+}\,d\mu\otimes\nu(s,t). \end{aligned}
\end{equation*}
\notag
$$
This equality holds for each $D\in\Sigma$, and therefore
$$
\begin{equation*}
T^{+}f(\,{\cdot}\,)=\int_{B}K(\,\cdot\,,t,f(s))^{+}\,d\nu(t).
\end{equation*}
\notag
$$
An integral expression for the negative part $T^{-}$ of the operator $T$ can be found by the same arguments. As a result, we have
$$
\begin{equation*}
\begin{aligned} \, |T|f(\,{\cdot}\,) &=(T^{+}+T^{-})f(\,{\cdot}\,) \\ &=\int_{B}\bigl(K(\,\cdot\,,t,f(s))^{+}+K(\,\cdot\,,t,f(s))^{-}\bigr)\,d\nu(t) =\int_{B}|K(\,\cdot\,,t,f(s))|\,d\nu(t), \end{aligned}
\end{equation*}
\notag
$$
which proves Theorem 3. Remark 3. Theorem 3 shows that the regular integral Urysohn operators form a vector sublattice of the order-complete vector lattice $\mathcal{OA}_{\mathrm r}(E,F)$. We show below that each regular integral Urysohn operator on an ideal subspace of the space of measurable functions is diffuse. Theorem 4. Let $(A,\Sigma,\mu)$ be a finite measure space, $(B,\Theta,\nu)$ be a space with finite atomless measure, $E$ and $F$ be order ideals in $L_{0}(\nu)$ and $L_{0}(\mu)$, respectively, and $T\colon E\to F$ be a regular integral Urysohn operator with kernel $K$. Then $T\in\mathcal{OA}_{\mathrm{dif}}(E,F)$. Proof. Consider an arbitrary $f\in E$. We claim that $\mathfrak{d}_{T}(f)=0$. By Theorem 2,
$$
\begin{equation*}
\begin{aligned} \, \mathfrak{d}_{T}(f) &=\inf \biggl\{ \bigvee_{i=1}^m |T| f_i\colon f = \bigsqcup_{i=1}^m f_i, \ m \in \mathbb N\biggr\} \\ &=\inf \biggl\{ \bigvee_{i=1}^m \int_{B} |K(\,\cdot\,,t,f1_{B_i}(t))|\,d\nu(t)\colon f = \bigsqcup_{i=1}^m f1_{B_i}, \ B=\bigsqcup_{i=1}^m B_{i},\ m \in \mathbb N \biggr\}. \end{aligned}
\end{equation*}
\notag
$$
Assume that $\psi=:\mathfrak{d}_{T}(f)$ is nonzero. Then there exist a measurable set $D\in \Sigma$ and a positive constant $c$ such that $c1_{D}\leqslant \psi$. The space $L_{0}(\mu)$ is sequentially order complete (see [2]), and therefore there exists a decreasing sequence $(\varphi_{n})_{n\in\mathbb{N}}$ in $F$ such that $\psi=\inf_{n}(\varphi_n)$, where
$$
\begin{equation*}
\begin{gathered} \, \varphi_n(\,{\cdot}\,)=\bigvee_{i=1}^{m(n)} \int_{B} |K(\,\cdot\,,t,f1_{B_i^n}(t))|\,d\nu(t) \\ \text{for }f = \bigsqcup_{i=1}^{m(n)} f1_{B_i^n}\quad\text{and} \quad B=\bigsqcup_{i=1}^{m(n)} B_{i}^n, \qquad n, m(n)\in \mathbb{N}. \end{gathered}
\end{equation*}
\notag
$$
Consider $K>0$ such that $K>c$. Given $g\in F$, we define
$$
\begin{equation*}
g_{K}(s)=\begin{cases} g(s)&\text{if } g(s)\leqslant K, \\ K&\text{if } g(s)>K. \end{cases}
\end{equation*}
\notag
$$
Note that the inequality $c1_{D}\leqslant \psi$ implies that $c1_{D}\leqslant \psi_{K}$. It is clear that ${g_{K}\in L_{\infty}(\mu)}$ for each $g\in F$. Now consider some lifting $\mathrm{l}\colon L_{\infty}(\mu)\to\mathcal{L}(\mu)$. Let
$$
\begin{equation*}
\widehat{g}:=\mathrm{l}(g)\quad\text{for } g\in L_{\infty}(\mu).
\end{equation*}
\notag
$$
By the monotonicity of $\mathrm{l}$, we have $\widehat{c1_{D}}\leqslant \widehat{\psi_{K}}$. Hence there exists $s_0\in D$ such that $\widehat{\psi_{K}}(s_0)\geqslant c>0$. Consequently, there exists a sequence of measurable sets $(B^{n}_{k(n)})_{n\in\mathbb{N}}$ such that $\nu(B^{n}_{k(n)})\downarrow 0$ and
$$
\begin{equation}
c\leqslant\widehat{\psi_{K}}(s_0)\leqslant\inf_{n}\left\{\widehat{\xi^{n}_{K}}(s_0)\right\},
\end{equation}
\tag{4.5}
$$
where
$$
\begin{equation*}
\xi^{n}(\,{\cdot}\,):=\int_{B} |K(\,\cdot\,,t,f1_{B^n_{k(n)}}(t))|\,d\nu(t).
\end{equation*}
\notag
$$
We have
$$
\begin{equation*}
\int_{B} |K(\,\cdot\,,t,f1_{B^n_{k(n)}}(t))|\,d\nu(t)\leqslant \int_{B} |K(\,\cdot\,,t,f(t))|\,d\nu(t)
\end{equation*}
\notag
$$
for each $n\in\mathbb{N}$, so that by the monotone convergence theorem, the infimum on the right-hand side of (4.5) is zero. This contradiction proves Theorem 4. § 5. Projection onto the band generated by disjointness preserving operatorsIn this section we establish a formula for the projection of a positive orthogonally additive operator onto the band generated by the disjointness preserving orthogonally additive operators. It is worth recalling again that elements of the band $\mathcal{OA}_{\mathrm{dpo}}(E,F)^{\perp\perp}$ generated by the disjointness preserving orthogonally additive operators are traditionally called atomic operators. The band of atomic OAOs will be denoted by $\mathcal{OA}_{\mathrm a}(E,F)$. Assume that the vector lattice $F$ is order complete. Since the vector lattice $\mathcal{OA}_{\mathrm r}(E,F)$ is order complete, it can be represented as the direct sum $\mathcal{OA}_{\mathrm a}(E,F)\oplus\mathcal{OA}_{\mathrm{dif}}(E,F)$ of the disjoint bands $\mathcal{OA}_{\mathrm a}(E,F)=\mathcal{OA}_{\mathrm{dpo}}(E,F)^{\perp\perp}$ and $\mathcal{OA}_{\mathrm{dif}}(E,F)$. Hence each regular orthogonally additive operator $T\colon E\to F$ is uniquely represented as a sum
$$
\begin{equation*}
T=T_{\mathrm a}+ T_{\mathrm{dif}}, \quad \text{where } T_{\mathrm a}\in \mathcal{OA}_{\mathrm a}(E,F), \quad T_{\mathrm{dif}}\in \mathcal{OA}_{\mathrm{dif}}(E,F).
\end{equation*}
\notag
$$
This defines the order projection $\pi_{\mathrm a}\colon \mathcal{OA}_{\mathrm r}(E,F)\to \mathcal{OA}_{\mathrm a}(E,F)$ onto the band $\mathcal{OA}_{\mathrm a}(E,F)$, where
$$
\begin{equation*}
\pi_{\mathrm a}T= T_{\mathrm a}\quad\text{for } T\in \mathcal{OA}_{\mathrm r}(E,F).
\end{equation*}
\notag
$$
We say that $\pi_{\mathrm a}T$ is the atomic component of a regular orthogonally additive operator $T$. Note that finding an analytic representation of an order projection operator onto a given band in the space of operators is a conventional problem of functional analysis in ordered spaces (see [4], [9], [18], [33] and [41]). The following lemma, which supplements Lemma 1, provides formulae for the supremum and infimum of an arbitrary bounded family of regular orthogonally additive operators. Lemma 3. Let $E$ and $ F$ be vector lattices such that $F$ is order complete, let $x\in E$, and let $\mathfrak{A}$ be an order-bounded subset of $\mathcal{OA}_{\mathrm r}(E,F)$. Then: Proof. For order-bounded families of operators the existence of a supremum and an infimum follows from the order completeness of the vector lattice $\mathcal{OA}_{\mathrm r}(E,F)$. Let $Rx$ be the right-hand side of the formula in assertion 1). We claim that $R$ is an orthogonally additive operator. Let $x,y\in E$ be disjoint, and let $x\sqcup y=\bigsqcup_{i=1}^{n}u_i$. By Proposition 14 there exist disjoint partitions $u_i=x_i\sqcup y_i$, $1\leqslant i\leqslant n$, such that $x\sqcup y=\bigsqcup_{i=1}^{n}x_i$ and $x\sqcup y=\bigsqcup_{i=1}^{n}y_i$. Next, for any finite set $T_1,\dots, T_n$ of elements of $\mathfrak{A}$ we have Taking the supremum over all finite sums of the form $\sum_{i=1}^{n}T_iu_i$, $T_i\in\mathfrak{A}$, $1\leqslant i\leqslant n$, $n\in\mathbb{N}$, on the left-hand side of this inequality we obtain the estimate
Let us prove the reverse inequality. Let $x=\bigsqcup_{i=1}^{n}x_i$, $y=\bigsqcup_{j=1}^{m}y_j$, and let $T_1,\dots,T_n$, $S_1,\dots,S_m$ be finite sets of elements of $\mathfrak{A}$. Consider the family of pairwise disjoint elements $u_1,\dots, u_{n+m}$, where $u_i=x_i$, $1\leqslant i\leqslant n$, and $u_{i=j}=y_j$, $1\leqslant j\leqslant m$. We also set $T_{n+j}:=S_j$, $1\leqslant j\leqslant m$. We have $x\sqcup y=\bigsqcup_{i=1}^{n+m}u_i$ and
$$
\begin{equation*}
\sum_{i=1}^{n}T_ix_i+\sum_{j=1}^{m}S_jy_j= \sum_{i=1}^{n+m}T_iu_i\leqslant R(x\sqcup y).
\end{equation*}
\notag
$$
Taking the supremum over all finite sums of the form $\sum_{i=1}^{n}T_ix_i$ and $\sum_{j=1}^{m}S_jy_j$ on the left-hand side of this inequality, we have
$$
\begin{equation*}
Rx+Ry\leqslant R(x\sqcup y) \quad\Longrightarrow\quad Rx+Ry=R(x\sqcup y).
\end{equation*}
\notag
$$
Thus, we have established that $R$ is a regular orthogonally additive operator and $T\leqslant R$ for all $T\in \mathfrak{A}$. We claim that $R=\sup\{\mathfrak{A}\}$. Indeed, let $x\in E$, $H\in \mathcal{OA}_{\mathrm r}(E,F)$, and assume that $T\leqslant H$ for all $T\in \mathfrak{A}$. Let $x=\bigsqcup_{i=1}^{n}x_i$ be an arbitrary disjoint partition of $x$, and let $T_1,\dots,T_n$ be a finite set of elements of $\mathfrak{A}$. Then
$$
\begin{equation*}
\sum_{i=1}^{n}T_ix_i\leqslant\sum_{i=1}^{n}Hx_i=H\biggl(\sum_{i=1}^{n}x_i\biggr)=Hx.
\end{equation*}
\notag
$$
Taking the supremum over all finite sums $\sum_{i=1}^{n}T_ix_i$ such that $x=\bigsqcup_{i=1}^{n}x_i$ and $\{T_1,\dots, T_n\}\subset \mathfrak{A}$ on the left-hand side of this inequality, we find that $R\leqslant H$. Now we use the well-known equality $\inf\{\mathfrak{A}\}=-\sup\{-\mathfrak{A}\}$, which holds in any vector lattice, provided that a supremum (infimum) of $\mathfrak{A}$ exists (see [15], p. 3). Recall that $-\mathfrak{A}:=\{-T\colon T\in\mathfrak{A}\}$. Then for each $x\in E$ we have
$$
\begin{equation*}
\begin{aligned} \, \inf\{\mathfrak{A}\}x &=-\sup\{-\mathfrak{A}\}x \\ &=-\sup\biggl\{\sum_{i=1}^{n}-T_ix_i\colon x=\bigsqcup_{i=1}^{n}x_i;\ T_i\in\mathfrak{A},\ 1\leqslant i\leqslant n,\ n\in\mathbb{N}\biggr\} \\ &=\inf\biggl\{\sum_{i=1}^{n}T_ix_i\colon x=\bigsqcup_{i=1}^{n}x_i;\ T_i\in\mathfrak{A},\ 1\leqslant i\leqslant n,\ n\in\mathbb{N}\biggr\}. \end{aligned}
\end{equation*}
\notag
$$
This proves Lemma 3. A family $H\subset \mathcal{OA}_{\mathrm{dpo}}(E,F)$ consisting of pairwise disjoint operators is called a disjoint system in $\mathcal{OA}_{\mathrm{dpo}}(E,F)$. Let $\mathfrak{D}$ be the set of all disjoint families in $\mathcal{OA}_{\mathrm{dpo}}(E,F)$. Then there is a natural partial order $\preceq$ on $\mathfrak{D}$: It can be shown with the help of Zorn’s lemma that $\mathfrak{D}$ has a maximal element $H^*$. It is clear that $\mathcal{OA}_{\mathrm a}(E,F)=\mathcal{OA}_{\mathrm{dpo}}(E,F)^{\perp\perp}=\{H^*\}^{\perp\perp}$.The main result of this section is as follows. Theorem 5. Let $E$ and $F$ be vector lattices such that $F$ is order complete, and let $x\in E$, and $T\in\mathcal{OA}_+(E,F)$. Then the atomic component $\pi_{\mathrm a}T$ of the operator $T$ is given by Proof. Let $g(T)x$ be the right-hand side of (5.1). Also set
$$
\begin{equation*}
\mathfrak{A}:=\{S\in \mathcal{OA}_{\mathrm{dpo}}(E,F)\colon 0\leqslant S\leqslant T\}.
\end{equation*}
\notag
$$
Let $H^*$ be a maximal disjoint system in $\mathcal{OA}_{\mathrm{dpo}}(E,F)$. By the above, we have $\mathcal{OA}_{\mathrm{dpo}}(E,F)^{\perp\perp}\cap[0,T]=\{H^*\}^{\perp\perp}\cap[0,T]$. We claim that $\pi_{\mathrm a}T=\sup\{\mathfrak{A}\}$. It is clear that $\pi_{\mathrm a}T\geqslant\sup\{\mathfrak{A}\}$. Let us prove the reverse inequality. Indeed, $\pi_{\mathrm a}T=\sup_{\alpha\in A}T_{\alpha}$, where $(T_{\alpha})_{\alpha\in A}$ is an increasing net of positive OAOs such that for each $\alpha\in A$ the operator $T_{\alpha}$ can be written as Hence $T_{\alpha}\leqslant \sup\{\mathfrak{A}\}$ for any $\alpha\in A$. As a result,
Now we show that $g(T)x=\sup\{\mathfrak{A}\}x$ for each $x\in E$. Consider an arbitrary finite disjoint partition $x=\bigsqcup_{i=1}^{n}x_i$ of $x$ and an arbitrary finite set $S_{1},\dots, S_{n}$ of elements of $\mathfrak{A}$. We have
$$
\begin{equation*}
\sum_{i=1}^{n}S_ix_i\leqslant \sum_{i=1}^{n}\mathfrak{d}_{T}(x_i)\leqslant g(T)x.
\end{equation*}
\notag
$$
Taking the supremum over all sums of the form $\sum_{i=1}^{n}S_ix_i$ on the left-hand side of this inequality and using Lemma 3, this gives On the other hand $\mathfrak{d}_{T}(x)=\sup_{S\in \mathfrak{A}}Sx\leqslant\sup\{\mathfrak{A}\}x$ by Proposition 22. Hence, for an arbitrary finite disjoint partition $x=\bigsqcup_{i=1}^{n}x_i$ of $x$, Taking the supremum over all sums of the form $\sum_{i=1}^{n}\mathfrak{d}_{T}(x_i)$ on the left-hand side of the last inequality, we find that $g(T)x\leqslant\sup\{\mathfrak{A}\}x$. As a result, which essentially completes the proof of the theorem. AcknowledgementThe authors are grateful to the referee, who read carefully the original version of the manuscript. 
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