\documentstyle{SibMatJ} %\proofmodetrue \topmatter \Author Gorokhova \Initial S. \Initial G. \Email lanagor71\@gmail.com \AffilRef 1 \endAuthor \Author Emelyanov \Initial E. \Initial Yu. \Email emelanov\@math.nsc.ru \AffilRef 2 \endAuthor \Affil 1 \Organization Southern Mathematical Institute \City Vladikavkaz \Country Russia \endAffil \Affil 2 \Organization Sobolev Institute of Mathematics \City Novosibirsk \Country Russia \endAffil \translator S.~G.~Gorokhova\endtranslator %\? \iauthor Gorokhova~S.G. and Emelyanov~E.Yu.\endiauthor \UDclass 517.98\endUDclass \Origin \Journal \VMZh \Year 2021 \CopyrightYear 2021 \Volume 24 \Issue 3 \Pages 55--61 \DOI 10.46698/f5525-0005-3031-h \endOrigin \pforename S.G.\endpforename \psurname Gorokhova\endpsurname \pemail lanagor71\@gmail.com\endpemail \pforename E.Yu.\endpforename \psurname Emelyanov\endpsurname \pemail emelanov\@math.nsc.ru\endpemail \author S.~G.~Gorokhova and E.~Yu.~Emelyanov\endauthor \title On Operators Dominated by the Kantorovich--Banach and the L\'evi Operators in Locally Solid Lattices \endtitle \thanks The work was carried out in the framework of the State Task to the Sobolev Institute of Mathematics (Project FWNF--2022--0004). \endthanks \datesubmitted October 10, 2021\enddatesubmitted \dateaccepted March 4, 2023\enddateaccepted %\? \address Vladikavkaz; Novosibirsk\endaddress \affil S.~G.~Gorokhova\\ Southern Mathematical Institute, Vladikavkaz, Russia \endaffil \email lanagor71\@gmail.com\endemail %\orcid\endorcid \affil E.~Yu.~Emelyanov\\ Sobolev Institute of Mathematics, Novosibirsk, Russia \endaffil \email emelanov\@math.nsc.ru\endemail %\orcid\endorcid \keywords locally solid lattice, Lebesgue operator, L\'evi operator, $KB$-operator, lattice homomorphism \endkeywords \abstract A~linear operator $T$ acting in a locally solid vector lattice $(E,\tau)$ is said to be a~Lebesgue operator, if $Tx_\alpha\overset{\tau}\to{\rightarrow}0$ for every net in $E$ satisfying $x_\alpha\downarrow 0$; a $KB$-operator, if, for every $\tau$-bounded increasing net $x_\alpha$ in $E_+$, there exists an $x\in E$ with $Tx_\alpha\overset{\tau}\to{\rightarrow}Tx$; a quasi $KB$-operator, if $T$ takes $\tau$-bounded increasing nets in $E_+$ to $\tau$-Cauchy ones; a Levi operator, if, for every $\tau$-bounded increasing net $x_\alpha$ in $E_+$, there exists an $x\in E$ such that $Tx_\alpha\overset{o}\to{\rightarrow}Tx$; a quasi Levi operator, if $T$ takes $\tau$-bounded increasing nets in $E_+$ to $o$-Cauchy ones. The present article is devoted to the domination problem for the quasi $KB$-operators and the quasi Levi operators in locally solid vector lattices. Moreover, some properties of Lebesgue operators, Levi operators, and $KB$-operators are investigated. In particularly, it is proved that the vector space of regularly Lebesgue operators is a subalgebra of the algebra of all regular operators. \endabstract \endtopmatter \noindent In the present paper all vector lattices are real and Archimedean, and linear topologies are Hausdorff. The main subject of the paper is the domination problem for operators in locally solid vector lattices dominated by Kantorovich--Banach lattice homomorphisms, quasi $KB$-operators, and quasi Levi operators. Recently, this topic has attracted attention (see~[1] and the references therein). Recall that a locally solid vector lattice $(E,\tau)$ is said to be (i) a {\it Lebesgue lattice\/} ($\sigma$-{\it Lebesgue lattice}) if, for every net (sequence) $x_\alpha\downarrow 0$ in $E$, we have $x_\alpha\overset{\tau}\to{\rightarrow} 0$; (ii) a {\it Levi lattice\/} ($\sigma$-{\it Levi lattice}) if any increasing $\tau$-bounded net (sequence) in $E_+$ has a supremum in $E$. In (i), the condition $x_\alpha\downarrow 0$ can be replaced by $x_\alpha\overset{o}\to{\rightarrow} 0$. A normed lattice $(E,\|\cdot\|)$ is called a~{\it Kantorovich--Banach\/} or {\it $KB$-space\/} if every norm bounded upward directed set in $E_+$ converges in norm. Each $KB$-space is a Levi lattice with order continuous complete norm; and every normed Levi ($\sigma$-Levi) lattice is Dedekind ($\sigma$-Dedekind) complete. (iii) A locally solid lattice $(E,\tau)$ is called a $KB$ ($\sigma$-$KB$) {\it lattice}, if every increasing $\tau$-bounded net (sequence) in $E_+$ is $\tau$-convergent. It is well known that any $KB$ ($\sigma$-$KB$) lattice is Levi ($\sigma$-Levi) lattice, and every Levi ($\sigma$-Levi) lattice is Dedekind ($\sigma$-Dedekind) complete (see Theorem~2.21(c) in~[2]). A linear operator $T$ acting in a locally solid lattice $(E,\tau)$ is called {\it $o\tau$-continuous\/} ({\it $\sigma$-$o\tau$-continuous}), if $Tx_\alpha\overset{\tau}\to{\rightarrow} 0$ for all net (sequence) $x_\alpha$ such that $x_\alpha\overset{o}\to{\rightarrow} 0$ (see~[3]). Various locally solid versions of properties of Banach lattices, like the property to be a $KB$-space, has been studied recently by many authors (see, a.e.,~[3--6]). The main idea of operator versions of topological vector lattices is the redistribution of topological and order properties between domain and range of the operator. Since the order convergence is not topological in general, the most interesting versions arise when the $o$- and the $\tau$-convergence are involved simultaneously. For the more general form of the following definition we refer to~[1]. \goodbreak \demo{Definition~1} Let $T$ be a linear operator in a locally solid lattice $(E,\tau)$. Then (a) $T$ is a {\it Lebesgue\/} ({\it $\sigma$-Lebesgue\/} operator), if $Tx_\alpha\overset{\tau}\to{\rightarrow} 0$ for every net (sequence) $x_\alpha\downarrow 0$ in $E$; (b) $T$ is {\it $o\tau$-bounded\/} ({\it $o\tau$-compact}), if the set $T[0,x]$ is a $\tau$-bounded ($\tau$-totally bounded) in $E$ for each $x\in E_+$; (c) $T$ is a ($\sigma$-) {\it Kantorovich--Banach\/} (or ($\sigma$-) $KB$) operator, if, for every $\tau$-bounded increasing net (sequence) $x_\alpha$ in $E_+$, there exists $x\in E$ such that $Tx_\alpha\overset{\tau}\to{\rightarrow} Tx$; (d) $T$ is a {\it quasi $KB$\/} ({\it quasi $\sigma$-$KB$}) operator, if it takes $\tau$-bounded increasing nets (sequences) in $E_+$ onto $\tau$-Cauchy ones; (e) $T$ is a {\it Levi\/} ({\it $\sigma$-Levi}) operator, if, for every $\tau$-bounded increasing net (sequence) $x_\alpha$ in $E_+$, there exists $x\in E$ such that $Tx_\alpha\overset{o}\to{\rightarrow} Tx$; (f) $T$ is a {\it quasi Levi\/} ({\it quasi $\sigma$-Levi}) operator, if it takes $\tau$-bounded increasing nets (sequences) in $E_+$ onto $o$-Cauchy ones. \enddemo \proclaim{Proposition~1 \rm (cf. Proposition~2.7 in [1])} Every continuous linear operator $T$ in a locally solid lattice $(E,\tau)$ is $o\tau$-bounded. \endproclaim The proposition follows from the two classical facts: (1) In a locally solid lattice $(E,\tau)$, all order bounded subsets are $\tau$-bounded. (2) Each $\tau$-continuous linear operator takes $\tau$-bounded sets to $\tau$-bounded ones. \demo{Definition~2 \rm see Definition~2 in~[7]} Let {\rm P} be a set of linear operators between ordered vector spaces~$X$ and~$Y$. An operator $T:X\to Y$ is called a {\it regularly {\rm P}-operator\/} $(${\it {\rm r-P}-operator} for short$)$, if there exist two positive {\rm P}-operators $T_1,T_2:X\to Y$ such that $T=T_1-T_2$. The set of all \text{\rm r-P}-operators from $X$ to $Y$ is denoted by $\text{\rm r-P}(X,Y)$. \enddemo \proclaim{Proposition~2}\Label{P2} %\label{reg lebesgue is o-cont} Each regularly Lebesgue $($regularly $\sigma$-Lebesgue$)$ operator $T$ in a locally solid lattice $(E,\tau)$ is order continuous $($order $\sigma$-continuous$)$. \endproclaim \demo{Proof} As the regularly $\sigma$-Lebesgue case is similar, we restrict ourselves to the case of a regularly Lebesgue operator $T$. Without loss of generality suppose that $T\ge 0$. In order to show that $T$ is order continuous, it suffices to prove $Tx_\alpha\downarrow 0$ for each net $x_\alpha$ in $E$ such that $x_\alpha\downarrow 0$. Let $x_\alpha\downarrow 0$ in~$E$. Since~$T$ is positive, $Tx_\alpha\downarrow$. Since~$T$ is a~Lebesgue operator, $Tx_\alpha\overset{\tau}\to{\rightarrow} 0$. It follows that $Tx_\alpha\downarrow 0$ (e.g., by Theorem~2.21(c) of~[2]), and hence $T$ is order continuous. \qed \proclaim{Corollary~1}\Label{C1} %\label{Cr} The family $\text{\rm r-L}_{Leb}(E)$ $($$\text{\rm r-L}_{Leb}^\sigma (E)$$)$ of all regularly Lebesgue {\rm(}regularly $\sigma$-Lebesgue{\rm)} operators in a locally solid lattice $(E,\tau)$ is a subalgebra of the algebra $L_r(E)$ of all regular operators in~$E$. Moreover, $I \in \text{\rm r-L}_{Leb}(E)$ {\rm(}$I \in \text{\rm r-L}_{Leb}^\sigma (E)${\rm)} iff $(E,\tau)$ is a Lebesgue {\rm(}$\sigma$-Lebesgue{\rm)} lattice, where $I$ is the identity of $E$. \endproclaim \demo{Proof} It suffices to prove that the set $\text{\rm r-L}_{Leb}(E)$ is closed with respect to the composition of operators. Let $T,S \in\text{\rm r-L}_{Leb}(E)$. Without loss of generality, assume that $T,S\ge 0$. Take a net $x_\alpha\downarrow 0$. Then, by \Par{P2}{Proposition~2}, $Sx_\alpha \downarrow 0$ and hence $TSx_\alpha \overset{\tau}\to{\rightarrow} 0$. In the case of regularly $\sigma$-Lebesgue operators, the proof is similar. The rest of the proof of \Par{C1}{Corollary~1} is trivial. \qed \proclaim{Lemma~1 \rm (cf. Lemma~2.1 in~[1])}\Label{L1} %\label{PC1} A linear operator $T$ in a locally solid lattice $(E,\tau)$ is regularly Lebesgue $($regularly $\sigma$-Lebesgue$)$ iff $T$ is regularly $o\tau$-continuous $($regularly $\sigma$-$o\tau$-continuous$)$. \endproclaim \demo{Proof} Without loss of generality assume that $T\ge 0$. As the $\sigma$-Lebesgue case is similar, we consider only Lebesgue operators. The sufficiency is routine. For the necessity, assume that~$T$ is a~Lebesgue operator and $x_\alpha\overset{o}\to{\rightarrow} 0$ in $E$. Take a net $y_\beta\downarrow 0$ in $E$ such that, for each $\beta$, there exists $\alpha_\beta$ with $|x_\alpha|\le y_\beta$ for $\alpha\ge\alpha_\beta$. It follows from $T\ge 0$ that $|Tx_\alpha|\le T|x_\alpha|\le Ty_\beta$ for $\alpha\geq\alpha_\beta$. Since $T$ is a~Lebesgue operator, $Ty_\beta\overset{\tau}\to{\rightarrow} 0$. As the topology $\tau$ is locally solid, it follows that $Tx_\alpha\overset{\tau}\to{\rightarrow} 0$, and hence $T$ is $o\tau$-continuous. \qed \proclaim{Proposition~3 \rm (cf. Proposition~2.1 in [1])}%\label{regular are otau-bounded} Every regular operator $T$ in a locally solid lattice $(E,\tau)$ is $o\tau$-bounded. \endproclaim Here we include the list of useful properties of operators under the consideration: \medskip (a) The identity operator in $E$ is a Lebesgue/$KB$/Levi operator iff $E$ is a Lebesgue/$KB$/Levi lattice respectively. Every $o\tau$-continuous ($\sigma$-$o\tau$-continuous) operator is a Lebesgue ($\sigma$-Lebesgue) operator and, by \Par{L1}{Lemma~1}, every regularly Lebesgue (regularly $\sigma$-Lebesgue) operator in $E$ is regularly $o\tau$-continuous (regularly $\sigma$-$o\tau$-continuous). The authors do not know whether any regular Lebesgue ($\sigma$-Lebesgue) operator is $o\tau$-continuous ($\sigma$-$o\tau$-continuous). \medskip (b) The discontinuous operator $Tx:=(\sum_{k=1}^{\infty}x_k)e_1$ in the normed lattice $(c_{00},\|\cdot\|_\infty)$ is $o\tau$-compact and $o\tau$-continuous, yet $T$ is not compact. Every continuous linear operator $T$ in a discrete Dedekind complete locally solid locally convex vector lattice is $o\tau$-compact in view of Corollary~6.57 of~[2]. \medskip (c) Every $KB$-operator is quasi $KB$, and every continuous linear operator in a $KB$-space is a $KB$-operator. It is well known that the identity $I$ in a Banach lattice is a $KB$-operator iff $I$ is a $\sigma$-$KB$-operator iff $I$ is a quasi $KB$-operator. The notions ``quasi $KB$ operator" and ``quasi $\sigma$-$KB$ operator" coincide by \Par{P4}{Proposition~4} below. Every order bounded operator in a $KB$-space is a quasi $KB$-operator. \medskip (d) Every compact operator $T$ in a Banach lattice is $o\tau$-compact. By \Par{E1}{Example~1} below, a compact operator need not to be a Lebesgue operator. In particular, an $o\tau$-compact operator is not necessarily $o\tau$-continuous. \demo{Example~1 \rm (see Example~3.1 of [1])}\Label{E1} \Label{E1}%\label{c_w(R)} Let $E=(c_\omega(\Bbb{R}),\|\cdot\|_\infty)$ be a Banach lattice of all bounded real functions on $\Bbb{R}$ such that each $f\in E$ differs from a constant $a_f$ on at most countable subset of~$\Bbb{R}$. Define a positive operator $T$ in $E$ as follows: Let $Tf$ be a constant function $a_f\cdot\Bbb{1}_{\Bbb{R}}\in E$ such that $\{d\in\Bbb{R}: f(d)\ne a_f\}$ is at most countable. (1) $T$ is a continuous rank one operator $($and consequently $T$ is $KB$ operator, compact, and $o\tau$-compact$)$ in $E$. Let $f_n\overset{o}\to{\rightarrow} 0$. Since, for any $\varepsilon>0$, there exists $n_\varepsilon$ such that $\cup_{n\ge n_\varepsilon}\{d\in\Bbb{R}: |f_n(d)|\ge\varepsilon\}$ is at most countable; therefore, $\|Tf_n\|_\infty<\varepsilon$ for all $n\ge n_\varepsilon$. Thus, the operator $T$ is $\sigma$-$o\tau$-continuous and hence a $\sigma$-Lebesgue operator with respect to the norm topology on $E$. (2) $T$ is not a Lebesgue operator. Indeed, for the net $f_\alpha:=\Bbb{1}_{\Bbb{R}\setminus\alpha}\in E$, indexed by the directed set~$\Delta$ of all finite subsets of $\Bbb{R}$ ordered by inclusion, we have $f_\alpha\downarrow 0$. However $\|Tf_\alpha\|_\infty=\|\Bbb{1}_{\Bbb{R}}\|_\infty=1$ for all $\alpha\in\Delta$. \enddemo \proclaim{Proposition~4 \rm (cf. Proposition~1.2 of [1])}\Label{P4} %\label{prop4} An operator $T$ in a locally solid lattice is a quasi $KB$-operator iff $T$ is quasi $\sigma$-$KB$. \endproclaim It follows from \Par{P4}{Proposition~4} that every topologically complete $\sigma$-$KB$-lattice $(E,\tau)$ is a $KB$-lattice (see Corollary~1.1 in~[1]). Note that the sets $L_{o\tau}(E)$, $L_{o\tau{}b}(E)$, and $L_{o\tau{}c}(E)$ of $o\tau$-continuous, $o\tau$-bounded, and $o\tau$-compact operators in a locally solid lattice $E$ are vector spaces satisfying $L_{o\tau c}(E)\subseteq L_{o\tau b}(E)$. \medskip A vector lattice $E$ is said to be {\it laterally $($$\sigma$-$)$complete}, if every (countable) subset of mutually disjoint vectors in $E_+$ has a supremum. A laterally complete vector lattice is discrete iff it is lattice isomorphic to $\Bbb{R}^S$ for some set $S$. \demo{Definition~3 \rm (cf. Definition~2.1 of [1])} A locally solid lattice $(E,\tau)$ is {\it $\tau$-laterally}\ ({\it $\sigma$-}) {\it complete}, if every $\tau$-bounded (countable) subset of mutually disjoint vectors in $E_+$ possesses a supremum. \enddemo Every laterally ($\sigma$-) complete locally solid lattice $(E,\tau)$ is $\tau$-laterally ($\sigma$-) complete, and every order complete $AM$-space~$X$ with the order unit is $\tau$-laterally complete with respect to the norm. \demo{Example~2 \rm (cf. Example~2.1 of~[1])} Consider a vector lattice $E$ of real functions on $\Bbb{R}$ such that every $f\in E$ differs from a constant $a_f$ on at most countable subset of $\Bbb{R}$, and $f-a_f\Bbb{1}_\Bbb{R}\in\ell_1(\Bbb{R})$ for all $f\in E$. The vector lattice $E$ is complete with respect to the norm $$\|f\|:=|a_f|+\|f-a_f\Bbb{1}_\Bbb{R}\|_1.$$ Clearly, $E$ is not a $K_\sigma$-space since $$f_n:=\Bbb{1}_{\Bbb{R}\setminus\{1,2,\dots,n\}}{\downarrow} \ge 0, $$ yet $\inf\nolimits_{n \in \Bbb{N}}f_n$ does not exist in $E$. Note that $E$ is not $\tau$-laterally $\sigma$-complete with respect to the norm topology on $E$. Indeed, the norm bounded countable set of mutually disjoint orts $e_n=\Bbb{1}_{\{n\}} \in E_+$ has no supremum in $E$. \enddemo By Proposition~2.4 of [1], each Levi $($$\sigma$-Levi$)$ lattice is a $\tau$-laterally ($\sigma$-) complete ($\sigma$-) Dedekind complete vector lattice. \medskip Now, we turn to the domination problem. Namely, let $T$ and $S$ be operators on $E$ with $0\le S\le T$. Under what conditions does the assumption that $T$ is a Lebesgue operator, $o\tau$-bounded, $o\tau$-compact, $KB$-operator, or Levi operator imply that $S$ has the same property? It is clear that the problem has the positive solution for the Lebesgue, the $\sigma$-Lebesgue, and the $o\tau$-bounded operators. Recall that every order bounded disjointness preserving operator $T$ in a vector lattice $E$ has a modulus $|T|$ satisfying $|T||x|=|T|x||=|Tx|$ for all $x\in E$ (cf. Theorem~2.40 in~[8]); moreover, there exist two lattice homomorphisms $R_1,R_2:E\to E$ such that $T = R_1-R_2$ (cf. [8, Exercise~1, p.~130]). \proclaim{Proposition~5 \rm (cf. Theorem~2.5 in~[1])} %\label{KB disj pres dominated property} Let $T$ be an order bounded disjointness preserving $KB$ $($$\sigma$-$KB$$)$ operator in a locally solid lattice $(E,\tau)$. If $|S|\le|T|$, then $S$ is also a $KB$ $($$\sigma$-$KB$$)$ operator. \endproclaim \demo{Proof} Take a $\tau$-bounded increasing net (sequence) $x_\alpha$ in $E_+$. Then $T(x_\alpha-x)\overset{\tau}\to{\rightarrow} 0$ for some $x\in E$. Hence $$ |S(x_\alpha-x)|\le|S||x_\alpha-x|\le|T||x_\alpha-x|=|Tx_\alpha-Tx|\overset{\tau}\to{\rightarrow} 0. $$ Therefore $Sx_\alpha\overset{\tau}\to{\rightarrow} Sx$. \qed \noindent Since, for a lattice homomorphism $T$, every operator $S$ with $0\le S\le T$ is a lattice homomorphism (cf. e.g., [8, Theorem~2.14]), the next theorem follows from Proposition~5. \proclaim{Theorem~1 \rm (cf. Corollary~2.3 of~[1])} Let $T$ be a $KB$ {\rm(}$\sigma$-$KB${\rm)} lattice homomorphism in a~locally solid lattice. Then every operator~$S$ satisfying $0\le S\le T$ is also a~$KB$ {\rm(}$\sigma$-$KB${\rm)} lattice homomorphism. \endproclaim The following result generalizes Proposition~2.9 of [4] for locally solid lattices. \proclaim{Theorem~2 \rm (cf. Theorem~2.6 of~[1])} Let $T$ be a positive quasi $KB$-operator in a locally solid lattice $(E,\tau)$. Then every operator $S$ satisfying $0\le S\le T$ is also quasi $KB$. \endproclaim \demo{Proof} Let $x_\alpha$ be an increasing $\tau$-bounded net in $E_+$. Then $Tx_\alpha\uparrow$, and since~$T$ is a quasi $KB$-operator, the net $Tx_\alpha$ is $\tau$-Cauchy. Take $U\in\tau(0)$ and a solid neighborhood $V\in\tau(0)$ such that $V-V\subseteq U$. There exists $\alpha_0$ satisfying $T(x_\alpha-x_\beta)\in V$ for all $\alpha,\beta\ge\alpha_0$. In particular, $T(x_\alpha-x_{\alpha_0})\in V$ for all $\alpha\ge\alpha_0$. Since $0\le S\le T$ and $V$ is solid, $S(x_\alpha-x_{\alpha_0})\in V$ for all $\alpha\ge\alpha_0$. Thus $$ Sx_\alpha-Sx_\beta=S(x_\alpha-x_{\alpha_0})-S(x_\beta-x_{\alpha_0})\in V-V\subseteq U $$ for all $\alpha,\beta\ge\alpha_0$. Since $U\in\tau(0)$ was taken arbitrarily, $Sx_\alpha$ is a $\tau$-Cauchy net. \qed \proclaim{Theorem~3 \rm (see Theorem~2.7 of~[1])} Let $T$ be a positive quasi Levi operator in a locally solid lattice $(E,\tau)$. Then each operator $S$ with $0\le S\le T$ is also quasi Levi. \endproclaim \demo{Proof} Take an increasing $\tau$-bounded net $x_\alpha$ in $E_+$. Then the net $Tx_\alpha$ is $o$-Cauchy. Hence there exists a net $y_\beta\downarrow 0$ in $E$ such that given $\beta$ there is an $\alpha_\beta$ with $|Tx_{\alpha_1}-Tx_{\alpha_2}|\le y_\beta$ for all $\alpha_1, \alpha_2\ge \alpha_\beta$. Then, for $\alpha_1, \alpha_2\ge \alpha_\beta$, we have $$ Sx_{\alpha_1}-Sx_{\alpha_2}\le S(x_{\alpha_1}-x_{\alpha_\beta})\le T(x_{\alpha_1}-x_{\alpha_\beta}) \le y_\beta, $$ $$ Sx_{\alpha_2}-Sx_{\alpha_1}\le S(x_{\alpha_2}-x_{\alpha_\beta})\le T(x_{\alpha_2}-x_{\alpha_\beta}) \le y_\beta . $$ Thus $|Sx_{\alpha_1}-Sx_{\alpha_2}|\le y_\beta$ for all $\alpha_1, \alpha_2\ge \alpha_\beta$, and so $S$ is a quasi Levi operator. \qed \enddemo \Refs \ref\no 1 \by Alpay~S., Emelyanov~E., and Gorokhova~S. \paper $o\tau$-Continuous, Lebesgue, $KB$, and Levi operators between vector lattices and topological vector spaces \jour Results in Mathematics \yr 2022 \vol 77 \issue 3 \pages Article no.~117, 25~pp. \endref %DOI: 10.1007/s00025-022-01650-3. \ref\no 2 \by Aliprantis~C.D. and Burkinshaw~O. \book Locally Solid Riesz Spaces with Applications to Economics \bookinfo 2nd Edition \publaddr Providence \publ Amer. Math. 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