# Some remarks on orthogonally additive operators on Riesz spaces

Olena Fotiy  Department of Mathematics and Informatics
Chernivtsi National University (Ukraine<EMAIL_ADDRESS>,  Vladimir Kadets
Department of Mathematics Holon Institute of Technology (Israel) and
V.N.Karazin Kharkiv National University (Ukraine<EMAIL_ADDRESS>and
Mikhail Popov  Institute of Exact and Technical Sciences, Pomeranian
University in Słupsk, Słupsk (Poland) and Vasyl Stefanyk Precarpathian
National University, Ivano-Frankivsk (Ukraine<EMAIL_ADDRESS>
###### Abstract.

We study orthogonally additive operators between Riesz spaces without the
Dedekind completeness assumption on the range space. Our first result gives
necessary and sufficient conditions on a pair of Riesz spaces $(E,F)$ for
which every orthogonally additive operator from $E$ to $F$ is laterally-to-
order bounded. Second result provides sufficient conditions on a pair of
orthogonally additive operators $S$ and $T$ to have $S\vee T$, as well as to
have $S\wedge T$, and consequently, for an orthogonally additive operator $T$
to have $T^{+}$, $T^{-}$ or $|T|$ without any assumption on the domain and
range spaces. Finally we prove an analogue of Meyer’s theorem on the existence
of modules of disjointness preserving operator for the setting of orthogonally
additive operators.

###### Key words and phrases:

Riesz space; Orthogonally additive operator; Disjointness preserving operator

###### 2010 Mathematics Subject Classification:

Primary 47H30; Secondary 47B38; 47B65

##  1\. Introduction

Orthogonally additive operators (OAOs in short) between Riesz spaces
generalize linear operators. Nonlinear OAOs naturally appeared in different
fields of analysis, including Uryson and Nemytsky integral operators.
Fortunately, lots of tools applied for linear operators, which use the
additivity only on disjoint vectors, maintain applicability to OAOs (see [9]
for a recent survey on OAOs). For familiarly used notions and facts on Riesz
spaces we refer the reader to [2].

Let $E,F$ be Riesz spaces. A function $T\colon E\to F$ is called an _OAO_
provided $T(x+y)=T(x)+T(y)$ for all disjoint vectors $x,y\in E$. An OAO
$T\colon E\to F$ is said to be

  * •

_positive_ (write $T\geq 0$) provided $T(x)\geq 0$ for all $x\in E$;

  * •

_order bounded_ or an _abstract Uryson operator_ provided $T$ sends order
bounded subsets of $E$ to order bounded subsets of $F$.

The symbols $\mathcal{O}(E,F)$, $\mathcal{O}^{+}(E,F)$ and $\mathcal{U}(E,F)$
stand to denote the sets of all OAOs, the set of all positive OAOs and the set
of all abstract Uryson operators respectively.

Obviously, $\mathcal{O}(E,F)$ is an ordered vector space with respect to the
order $S\leq T$ provided $T-S\geq 0$, that is, $S(x)\leq T(x)$ for all $x\in
E$. Remark that the standard order on the vector space $\mathcal{L}(E,F)$ of
all linear operators from $E$ to $F$, which is a linear subspace of
$\mathcal{O}(E,F)$, is different, and the only linear operator which is
positive as an OAO is zero.

Let $E$ be a Riesz space. The notation $x=y\sqcup z$ for $x,y,z\in E$ means
that $x=y+z$ and $y\perp z$. An element $x\in E$ is called a _fragment_
111component in the terminology of [2] of $y\in E$ provided $x\perp y-x$. In
this case we write $x\boldsymbol{\sqsubseteq}y$. The binary relation
$\boldsymbol{\sqsubseteq}$ is a (non-strict) partial order on $E$, called the
lateral order. The set of all fragments of a given element $e\in E$ will be
denoted by $\mathfrak{F}_{e}$. The set $\mathfrak{F}_{e}$ is a Boolean algebra
with zero $0$, unit $e$ with respect to the operations $\boldsymbol{\cup}$ and
$\boldsymbol{\cap}$ of taking the lateral supremum and infimum respectively.
Moreover, $x\boldsymbol{\cup}y=(x^{+}\vee y^{+})-(x^{-}\vee y^{-})$ and
$x\boldsymbol{\cap}y=(x^{+}\wedge y^{+})-(x^{-}\wedge y^{-})$ for all
$x,y\in\mathfrak{F}_{e}$ (see [6] for details).

Let $E,F$ be Riesz spaces. An OAO $T\colon E\to F$ is said to be _laterally-
to-order bounded_ provided for every $e\in E$ the image $T(\mathfrak{F}_{e})$
under $T$ of the set $\mathfrak{F}_{e}$ of all fragments of $e$ is order
bounded in $F$. The set of laterally-to-order bounded OAOs from $E$ to $F$ is
denoted by $\mathcal{P}(E,F)$. The notion of laterally-to-order bounded OAOs
was introduced and first studied by Pliev and Ramdane in [10].

Obviously,
$\mathcal{U}(E,F)\subseteq\mathcal{P}(E,F)\subseteq\mathcal{O}(E,F)$.

##  2\. Laterally-to-order bounded OAOs

The first study of the Riesz space structure of the ordered vector space
$\mathcal{O}(E,F)$ of all OAOs between Riesz spaces $E$ and $F$ was presented
by Mazón and Segura de León in [4] and [5](1990). One of the main structural
results in [4] asserts that, if the Riesz space $F$ is Dedekind complete then
$\mathcal{U}(E,F)$ is a Dedekind complete Riesz space as well, and some
natural formulas for lattice operations on $\mathcal{U}(E,F)$ were provided.
In 2018 Pliev and Ramdane [10] generalized the above mentioned result of Mazón
and Segura de León to the class $\mathcal{P}(E,F)$ of laterally-to-order
bounded OAOs; see the next theorem.

######  Theorem 2.1 ([10]).

Let $E,F$ be Riesz spaces with $F$ Dedekind complete. Then $\mathcal{P}(E,F)$
is a Dedekind complete Riesz space and the following assertions hold.

  1. (1)

For every $S,T\in\mathcal{P}(E,F)$ and every $x\in E$ one has

    1. (a)

$(S\vee T)(x)=\sup\\{S(u)+T(v):\,x=u\sqcup v\\}$;

    2. (b)

$(S\wedge T)(x)\,=\,\inf\\{S(u)+T(v):\,x=u\sqcup v\\}$;

    3. (c)

$T^{+}(x)=\sup\\{T(u):\,u\sqsubseteq x\\}$;

    4. (d)

$T^{-}(x)=-\inf\\{T(u):\,u\sqsubseteq x\\}$;

    5. (e)

$|T(x)|\leq|T|(x)$.

  2. (2)

The set $\mathcal{U}(E,F)$ is an order ideal of $\mathcal{P}(E,F)$ and hence
itself is a Dedekind complete Riesz space with properties (a)-(e).

It is interesting to note that formulas (a) and (b) coincide with Abramovich’s
formulas for join and meet of linear operators, which are true under the
additional assumption on $E$ to have the principal projection property and
false without this assumption [2, Theorem 1.50].

So the Riesz space $\mathcal{P}(E,F)$ is a much more natural object for the
study of OAOs than $\mathcal{U}(E,F)$. Remark that $\mathcal{U}(E,F)$ need not
be a band of $\mathcal{P}(E,F)$ [10].

The linear subspace $\mathcal{P}(E,F)$ of $\mathcal{O}(E,F)$ is so “large”
that these two ordered vector spaces have common positive cone, that is, every
positive OAO is laterally-to-order bounded, which is easy to see. However, the
inclusion $\mathcal{P}(E,F)\subset\mathcal{O}(E,F)$ can be strict due to the
following example.

######  Example 2.2 ([10], Example 3.4).

Let $\ell_{0}^{\infty}$ be the Riesz space of all real eventually constant
sequences $x=(x_{n})_{n=1}^{\infty}$, that is, $(\exists
k\in\mathbb{N})(\forall n\geq k)\,x_{n}=x_{k}$ with the coordinate-wise order.
Then the map $T\colon E\to\mathbb{R}$ defined by setting
$T(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n}|x_{n}|}{n}$ for an arbitrary $x\in E$,
belongs to $\mathcal{O}(E,\mathbb{R})\setminus\mathcal{P}(E,\mathbb{R})$.

Observe, that the Riesz space $\ell_{0}^{\infty}$ is not Dedekind complete,
and the idea used in Example 2.2 cannot help to construct an example of the
kind with Dedekind complete domain space.

The following result characterizes pairs of Riesz spaces $(E,F)$ for which the
inclusion $\mathcal{P}(E,F)\subset\mathcal{O}(E,F)$ is strict.

######  Theorem 2.3.

Let $E,F$ be Riesz spaces with $F$ Archimedean. Then the following assertions
are equivalent.

  1. (1)

$\mathcal{P}(E,F)=\mathcal{O}(E,F)$.

  2. (2)

Either $F=\\{0\\}$ or for every $e\in E$ the set $\mathfrak{F}_{e}$ is finite.

###### Proof.

(1) $\Rightarrow$ (2). Assuming the contrary, let $f\in F\setminus\\{0\\}$ and
let $e$ be an element of $E$ with infinite $\mathfrak{F}_{e}$. Our goal is to
construct an OAO $T\colon E\to F$ which is not laterally-to-order bounded (it
is going to be even linear). Since $\mathfrak{F}_{e}$ is an infinite Boolean
algebra, we can choose a disjoint sequence $(e_{n})_{n=1}^{\infty}$ of nonzero
elements of $\mathfrak{F}_{e}$, which is linearly independent in $E$. Extend
$(e_{n})_{n=1}^{\infty}$ to a Hamel basis $(e_{i})_{i\in I}$,
$\mathbb{N}\subseteq I$ in the linear space $E$. Then define a linear operator
$T\colon E\to F$ using the Hamel basis as follows. First define $Te_{n}=nf$
for all $n\in\mathbb{N}$ and somehow define $T$ on the rest of the basis, say,
$Te_{i}=0$ for $i\in I\setminus\mathbb{N}$. Finally we extend $T$ from the
basis to the entire $E$ by linearity. It remains to observe that
$T(\mathfrak{F}_{e})$ is not order bounded in $F$, because $nf\in
T(\mathfrak{F}_{e})$ for all $n\in\mathbb{N}$ and $F$ is Archimedean.

(2) $\Rightarrow$ (1) is obvious. ∎

So the inclusion $\mathcal{P}(E,F)\subset\mathcal{O}(E,F)$ is strict, except
for somewhat trivial cases.

######  Remark 1.

The infinite dimensional Dedekind complete Riesz space $c_{00}$ of all
eventually zero sequences possesses the second part of (2) in Theorem 2.3, and
so $\mathcal{P}(c_{00},F)=\mathcal{O}(c_{00},F)$ for every Archimedean Riesz
space $F$.

##  3\. Existence of a supremum $S\vee T$ for OAOs $S$ and $T$

In this section, we generalize Pliev and Ramdane’s Theorem 2.1 and find
sufficient condition on OAOs $S$ and $T$ to have $S\vee T$ as well as to have
$S\wedge T$, and consequently, for an OAO $T$ to have $T^{+}$, $T^{-}$ or
$|T|$ without any assumption on the domain and range spaces.

We need the following Pliev Lemma [7, Proposition 3.11], which is a lateral
analogue of the Riesz decomposition property [2, Theorem 1.20].

######  Lemma 3.1 ([7]).

Let $E$ be a Riesz space, $u_{1},\ldots,u_{m},v_{1},\ldots v_{n}\in E$ and
$e:=\bigsqcup_{i=1}^{m}u_{i}=\bigsqcup_{k=1}^{n}v_{k}$. Then there exists a
(disjoint) double sequence ${{(w_{i,k})}_{i=1}^{m}}_{k=1}^{n}$ in $E$ such
that

  1. (1)

$u_{i}=\bigsqcup_{k=1}^{n}w_{i,k}$ for any $i\in\\{1,\ldots,m\\}$;

  2. (2)

$v_{k}=\bigsqcup_{i=1}^{m}w_{i,k}$ for any $k\in\\{1,\ldots,n\\}$.

###### Sketch of proof of Lemma 3.1.

Since $u_{i},v_{k}$ are elements of the Boolean algebra $\mathfrak{F}_{e}$,
the elements $w_{i,k}:=u_{i}\boldsymbol{\cap}v_{k}$ are well defined and
possess the desired properties. ∎

######  Theorem 3.2.

Let $E,F$ be Riesz spaces and $S,T\in\mathcal{O}(E,F)$. If for every $x\in E$
the supremum $R(x):=\sup\\{S(u)+T(v):\,x=u\sqcup v\\}$ exists in $F$ then
$S\vee T$ exists in $\mathcal{O}(E,F)$ and $S\vee T=R$. Moreover, in case of
existence the following hold.

  1. (1)

If $S,T\in\mathcal{P}(E,F)$ then $S\vee T\in\mathcal{P}(E,F)$.

  2. (2)

If $S,T\in\mathcal{U}(E,F)$ then $S\vee T\in\mathcal{U}(E,F)$.

###### Proof.

Show that $R$ is an OAO. Fix any $x,y\in E$ with $x\perp y$. Let $x+y=u\sqcup
v$ be any decomposition into disjoint fragments. Choose by Lemma 3.1 disjoint
vectors $w_{i}$, $i=1,\ldots,4$ so that $x=w_{1}\sqcup w_{3}$, $y=w_{2}\sqcup
w_{4}$, $u=w_{1}\sqcup w_{2}$ and $v=w_{3}\sqcup w_{4}$. Then

$S(u)+T(v)=S(w_{1})+S(w_{2})+T(w_{3})+T(w_{4})\leq R(x)+R(y).$

By the arbitrariness of a decomposition of $x+y$, we obtain $R(x+y)\leq
R(x)+R(y)$. On the other hand,

$\displaystyle R(x)+R(y)$ $\displaystyle=\sup\\{S(u)+T(v):\,x=u\sqcup
v\\}+\sup\\{S(w)+T(z):\,y=w\sqcup z\\}$
$\displaystyle=\sup\\{S(u+w)+T(v+z):\,x=u\sqcup v,\,y=w\sqcup z\\}$
$\displaystyle\leq R(x+y).$

Thus, $R$ is an OAO. Show that $R=S\vee T$ in $\mathcal{O}(E,F)$. Indeed,
obviously $S\leq R$ and $T\leq R$. Let $P\in\mathcal{O}(E,F)$ satisfy $S\leq
P$ and $T\leq P$. Fix any $x\in E$ and consider any decomposition $x=u\sqcup
v$. Then $S(u)+T(v)\leq P(u)+T(v)=P(x)$. By the arbitrariness of the
decomposition of $x$, $R(x)\leq P(x)$ and so $R\leq P$. Thus, $R=S\vee T$.

Items (1) and (2) easily follow from the obtained formula for $S\vee T$. ∎

Theorem 3.2 gives the following consequences for the existence of meet of two
OAOs, positive and negative parts and modulus of an OAO.

######  Corollary 3.3.

Let $E,F$ be Riesz spaces and $S,T\in\mathcal{O}(E,F)$. If for every $x\in E$
the infimum $R(x):=\inf\\{S(u)+T(v):\,x=u\sqcup v\\}$ exists in $F$ then
$S\wedge T$ exists in $\mathcal{O}(E,F)$ and $S\wedge T=R$. Moreover, in case
of existence the following hold.

  1. (1)

If $S,T\in\mathcal{P}(E,F)$ then $S\wedge T\in\mathcal{P}(E,F)$.

  2. (2)

If $S,T\in\mathcal{U}(E,F)$ then $S\wedge T\in\mathcal{U}(E,F)$.

###### Proof.

For every $x\in E$ one has

$\displaystyle(S\wedge T)(x)=-\bigl{(}(-S)\vee(-T)\bigr{)}(x)$
$\displaystyle=-\sup\\{-S(u)-T(v):\,x=u\sqcup v\\}\,\,\,\,\,\,\,\,\,\,\,\,$
$\displaystyle=\inf\\{S(u)+T(v):\,x=u\sqcup v\\}.$

Items (1) and (2) follow from the obtained formula for $S\wedge T$. ∎

Next three corollaries directly follow from Theorem 3.2.

######  Corollary 3.4.

Let $E,F$ be Riesz spaces and $T\in\mathcal{O}(E,F)$. If for every $x\in E$
the supremum $R(x):=\sup\\{T(u):\,u\sqsubseteq x\\}$ exists in $F$ then
$T^{+}$ exists in $\mathcal{O}(E,F)$ and $T^{+}=R$. Moreover, in case of
existence the following hold.

  1. (1)

If $T\in\mathcal{P}(E,F)$ then $T^{+}\in\mathcal{P}(E,F)$.

  2. (2)

If $T\in\mathcal{U}(E,F)$ then $T^{+}\in\mathcal{U}(E,F)$.

######  Corollary 3.5.

Let $E,F$ be Riesz spaces and $T\in\mathcal{O}(E,F)$. If for every $x\in E$
the infimum $-R(x):=\inf\\{T(u):\,u\sqsubseteq x\\}$ exists in $F$ then
$T^{-}$ exists in $\mathcal{O}(E,F)$ and $T^{-}=R$. Moreover, in case of
existence the following hold.

  1. (1)

If $T\in\mathcal{P}(E,F)$ then $T^{-}\in\mathcal{P}(E,F)$.

  2. (2)

If $T\in\mathcal{U}(E,F)$ then $T^{-}\in\mathcal{U}(E,F)$.

######  Corollary 3.6.

Let $E,F$ be Riesz spaces and $T\in\mathcal{O}(E,F)$. If for every $x\in E$
the supremum $R(x):=\sup\\{T(u)-T(v):\,x=u\sqcup v\\}$ exists in $F$ then
$|T|$ exists in $\mathcal{O}(E,F)$ and $|T|=R$. Moreover, in case of existence
the following hold.

  1. (1)

If $T\in\mathcal{P}(E,F)$ then $|T|\in\mathcal{P}(E,F)$.

  2. (2)

If $T\in\mathcal{U}(E,F)$ then $|T|\in\mathcal{U}(E,F)$.

Remark that Theorem 3.2 together with all its corollaries entirely yield
Pliev-Ramdane Theorem 2.1.

##  4\. The existence of modules of a disjointness preserving OAO

As usual, $\mathcal{L}_{b}(E,F)$ denotes the ordered vector space of all order
bounded linear operators between Riesz spaces $E$ and $F$, which is a Riesz
space once $F$ is Dedekind complete [2, Theorem 1.18]. Recall that an OAO
$T\colon E\to F$ is said to preserve disjointness provided $T(x)\perp T(y)$
for all $x,y\in E$ with $x\perp y$.

In this section, we show that the following Meyer theorem on the existence of
a modulus $|T|$ of a disjointness preserving linear operator $T$ holds true
for OAOs.

######  Theorem 4.1 (Theorem 2.40, [2]).

Let $E,F$ be Riesz spaces with $F$ Archimedean. Then every disjointness
preserving linear operator $T\in\mathcal{L}_{b}(E,F)$ has the modulus
$|T|:=T\vee(-T)$. Moreover, for every $x\in E$

$\bigl{|}T\bigr{|}|x|=\bigl{|}T|x|\bigr{|}=\bigl{|}Tx\bigr{|}.$

We also prove an analogue of Meyer’s lemma on disjointness preserving
operators [2, Lemma 2.39] for the setting of OAOs.

The proof of the next theorem, which is an analogue of Meyer’s theorem for
OAOs, based on the results of the previous section, is much more simpler than
the proof of the original Meyer Theorem 4.1 for linear operators. Although
OAOs generalize linear operators, we cannot consider our theorem as a
generalization of Meyer’s theorem for linear operators, because spaces of
linear operators and OAOs have different orders.

######  Theorem 4.2.

Let $E,F$ be Riesz spaces and let $T\in\mathcal{O}(E,F)$ preserve
disjointness. Then the following assertions hold.

  1. (1)

$T\in\mathcal{P}(E,F)$.

  2. (2)

$|T|$ exists in $\mathcal{O}(E,F)$ and $|T|\in\mathcal{P}(E,F)$. Moreover, for
every $x\in E$ one has $\bigl{|}T\bigr{|}(x)=\bigl{|}T(x)\bigr{|}.$

  3. (3)

$T$ is regular; $T^{+}$ and $T^{-}$ exist and both preserve disjointness.
Moreover, for every $x\in E$ one has $T^{+}(x)=\bigl{(}T(x)\bigr{)}^{+}$ and
$T^{+}(x)=\bigl{(}T(x)\bigr{)}^{+}.$

  4. (4)

Let $\\{x,y\\}\subset E$ be a laterally bounded set. Then
$\bigl{(}T(x)\bigr{)}^{+}\wedge\bigl{(}T(y)\bigr{)}^{-}=0$.

Remark that Theorem 4.2 was partially proved in [8, Lemma 2.2] under the
additional assumptions of Dedekind completeness of $F$, and formulas in (2)
and (3) for the modulus, positive and negative parts were discovered earlier
for an order bounded $T$ in [1, Lemma 3.1]. Item (4) is an analogue of Meyer’s
lemma (see Lemma 4.1 below).

For the proof of Theorem 4.2, we need the following two known simple facts.

######  Lemma 4.3 ([6]).

Let $E,F$ be Riesz spaces and $T\colon E\to F$ a disjointness preserving OAO.
Then for every $e\in E$ and every $x\sqsubseteq e$ one has $T(x)\sqsubseteq
T(e)$.

The proof is straightforward and easy. Note that the converse implication also
holds: every lateral order preserving OAO preserves disjointness [6, Theorem
4.9].

######  Lemma 4.4.

[6] Let $E$ be a Riesz spaces and $x,y\in E$.

  1. (1)

If $x\sqsubseteq y$ then $x^{+}\sqsubseteq y^{+}$, $x^{-}\sqsubseteq y^{-}$,
$|x|\sqsubseteq|y|$.

  2. (2)

If $|x|\sqsubseteq|y|$ then $|x|\leq|y|$.

See [6, Proposition 3.1] for details about Lemma 4.4.

###### Proof of Theorem 4.2.

(1) follows from Lemma 4.3.

(2). By Corollary 3.6, for the existence of $|T|$ it is enough to prove that,
for every $x\in E$ the supremum $\sup D_{x}$ exists in $F$, where
$D_{x}:=\\{T(u)-T(v):\,x=u\sqcup v\\}$. Fix any $x\in E$. Then for every $t\in
D_{x}$, say, $t=T(u)-T(v)$, where $x=u\sqcup v$, since $T$ preserves
disjointness, one has by Lemma 4.3

$|t|=|T(u)|\sqcup|T(v)|\stackrel{{\scriptstyle\tiny{\rm
Lemma}\,\ref{th:jgnbfgy}}}{{\sqsubseteq}}|T(x)|.$

By Lemma 4.4 (2), $|t|\leq|T(x)|$. On the other hand, $T(x)\in D_{x}$ and
$-T(x)\in D_{x}$. By the above, $\sup D_{x}=|T(x)|$. By Corollary 3.6, $|T|$
exists and $\bigl{|}T\bigr{|}(x)=\bigl{|}T(x)\bigr{|}$ for all $x\in E$. The
fact that $|T|\in\mathcal{P}(E,F)$ easily follows from the formula for $|T|$,
see Corollary 3.6.

(3). Since $T(x)=\bigl{(}T(x)\bigr{)}^{+}-\bigl{(}T(x)\bigr{)}^{-}$, to prove
the regularity of $T$, it is enough to show that the functions
$\phi,\psi\colon E\to F$ defined by $\phi(x)=\bigl{(}T(x)\bigr{)}^{+}$ and
$\psi(x)=\bigl{(}T(x)\bigr{)}^{-}$ for all $x\in E$ are OAOs. Indeed, fix any
$x,y\in E$ with $x\perp y$. Then

$\phi(x+y)=\bigl{(}T(x\sqcup y)\bigr{)}^{+}=\bigl{(}T(x)\sqcup
T(y)\bigr{)}^{+}=\bigl{(}T(x)\bigr{)}^{+}\sqcup\bigl{(}T(y)\bigr{)}^{+}=\phi(x)\sqcup\phi(y).$

So, $\phi$ is a disjointness preserving OAO. Similarly, $\psi$ is. It is left
to show that $\phi=T\vee 0$ and $\psi=(-T)\vee 0$. Obviously, $T\leq\phi$ and
$0\leq\phi$. Let $S\in\mathcal{O}(E,F)$ satisfy $T\leq S$ and $0\leq S$. Then
for every $x\in E$ one has $\phi(x)=\bigl{(}T(x)\bigr{)}^{+}\vee 0\leq S(x)$,
that is, $\phi\leq S$ and so $T^{+}=\phi$. Analogously, $T^{-}=\psi$.

(4) Let $\\{x,y\\}$ be laterally bounded by $e\in E$, that is, $x\sqsubseteq
e$ and $y\sqsubseteq e$. By Lemma 4.3, $T(x)\sqsubseteq T(e)$ and
$T(y)\sqsubseteq T(e)$. By Lemma 4.4,
$\bigl{(}T(x)\bigr{)}^{+}\sqsubseteq\bigl{(}T(e)\bigr{)}^{+}$ and
$\bigl{(}T(y)\bigr{)}^{-}\sqsubseteq\bigl{(}T(e)\bigr{)}^{-}$ and therefore
$\bigl{(}T(x)\bigr{)}^{+}\leq\bigl{(}T(e)\bigr{)}^{+}$ and
$\bigl{(}T(y)\bigr{)}^{-}\leq\bigl{(}T(e)\bigr{)}^{-}$. Taking into account
that $\bigl{(}T(e)\bigr{)}^{+}\wedge\bigl{(}T(e)\bigr{)}^{-}=0$, we obtain
that $\bigl{(}T(x)\bigr{)}^{+}\wedge\bigl{(}T(y)\bigr{)}^{-}=0$. ∎

Now we come back to the following Meyer lemma, which precedes Meyer’s theorem
for linear operators.

######  Lemma 4.5 (Lemma 2.39, [2]).

Let $E,F$ be Riesz spaces with $F$ Archimedean. Then for every disjointness
preserving linear operator $T\in\mathcal{L}_{b}(E,F)$ and every $x,y\in E^{+}$
one has $(Tx)^{+}\wedge(Ty)^{-}=0$.

The above Meyer’s lemma is not longer true for OAOs due to the following
simple example.

######  Example 4.6.

Define a function $T\colon C[0,1]\to C[0,1]$ by setting
$\mathbf{1}=\mathbf{1}_{[0,1]}$ and

$T(x)=\left\\{\begin{array}[]{ll}\mathbf{1},&\hbox{if}\,\,x=\mathbf{1},\\\
-\mathbf{1},&\hbox{if}\,\,x=2\cdot\mathbf{1},\\\ 0,&\hbox{if}\,\,x\in
C[0,1]\setminus\\{\mathbf{1},2\cdot\mathbf{1}\\}.\end{array}\right.$

Then $T$ is a disjointness preserving OAO and
$\bigl{(}T(\mathbf{1})\bigr{)}^{+}\wedge\bigl{(}T(2\cdot\mathbf{1})\bigr{)}^{-}=\mathbf{1}\neq
0$.

There are also examples of the kind of OAOs acting between Dedekind complete
Riesz spaces, however somewhat involved. For instance, define a map $S\colon
L_{p}\to L_{p}$, $0\leq p\leq\infty$ by setting
$S(x)=x\boldsymbol{\cap}\mathbf{1}-x\boldsymbol{\cap}(2\cdot\mathbf{1})$. By
[3, Theorem 4], $S$ is an OAO. One can easily show that $|Sx|\leq|x|$ for all
$x\in L_{p}$, and hence $S$ preserves disjointness. Moreover,
$\bigl{(}S(\mathbf{1})\bigr{)}^{+}\wedge\bigl{(}S(2\cdot\mathbf{1})\bigr{)}^{-}=\mathbf{1}\wedge(2\cdot\mathbf{1})=\mathbf{1}\neq
0$.

Such examples exist, because an OAO may have independent behavior on collinear
vectors (see Proposition 4.14 and Theorem 6.1 of [11] for examples of OAOs
with independent behavior on fragments of different elements of a Riesz
space). So Meyer’s lemma for OAOs is true for laterally bounded elements only,
see Theorem 4.2 (4).

##  5\. Remarks and open problems

We do not know whether the sufficient condition in Theorem 3.2 for the
existence of $S\vee T$ is necessary for the case where the range space is not
Dedekind complete.

######  Problem 5.1.

Let $E,F$ be Riesz spaces and $S,T\in\mathcal{O}(E,F)$. Assume that $S\vee T$
exists in $\mathcal{O}(E,F)$. Does for every $x\in E$ there exists
$\sup\\{S(u)+T(v):\,x=u\sqcup v\\}$ in $F$?

Let $S\vee T$ exist in $\mathcal{O}(E,F)$, and let $\widehat{F}$ be a Dedekind
completion of $F$ so that $F$ is an order dense ideal of $\widehat{F}$. Let
$J\colon F\to\widehat{F}$ be the inclusion embedding. For every
$P\in\mathcal{O}(E,F)$ we define an operator
$\widehat{P}\in\mathcal{O}(E,\widehat{F})$ by setting $\widehat{P}=J\circ P$.
Then the set $\\{\widehat{S}(u)+\widehat{T}(v):\,x=u\sqcup v\\}$ is order
bounded in $F$ by $(S\vee T)(x)$ and by the Dedekind completeness of
$\widehat{F}$, $R_{1}(x):=\sup\\{\widehat{S}(u)+\widehat{T}(v):\,x=u\sqcup
v\\}$ exists in $\widehat{F}$. Then by Theorem 3.2,
$\widehat{S}\vee\widehat{T}$ exists in $\mathcal{O}(E,\widehat{F})$ and equals
$R_{1}$. Moreover, it is easily seen that $R_{1}\leq\widehat{S\vee T}$. One
can show that, if $R_{1}=\widehat{S\vee T}$ then $\sup\\{S(u)+T(v):\,x=u\sqcup
v\\}$ exists in $F$ for all $x\in E$, and the answer to Problem 5.1 is
affirmative.

######  Problem 5.2.

Is $R_{1}=\widehat{S\vee T}$ true in all cases?

##  6\. Declarations

###  6.1. Funding

The second-named author was supported by Weizmann Institute of Science
(Israel) Emergency program for scientists affected by the war in Ukraine.

The third-named author was supported by the Ministry of Education of Ukraine
Grant no 0122U000857.

###  6.2. Availability of data and materials

All items of the bibliography are available in accordance with publishing
agreements.

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