Pre-Riesz Spaces 3110475391, 9783110475395

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Anke Kalauch, Onno van Gaans Pre-Riesz Spaces

De Gruyter Expositions in Mathematics

| Edited by Lev Birbrair, Fortaleza, Brasilia Victor P. Maslov, Moscow, Russia Walter D. Neumann, New York City, New York, USA Markus J. Pflaum, Boulder, Colorado, USA Dierk Schleicher, Bremen, Germany Katrin Wendland, Freiburg, Germany

Volume 66

Anke Kalauch, Onno van Gaans

Pre-Riesz Spaces

|

Mathematics Subject Classification 2010 Primary: 06F50, 46A40, 46B40; Secondary: 06F30, 47D06 Authors Dr. Anke Kalauch Technische Universität Dresden Fakultät Mathematik Institut für Analysis 01062 Dresden Germany [email protected]

Dr. Onno van Gaans Leiden University Mathematical Institute P.O. Box 9512 2300 RA Leiden The Netherlands [email protected]

The artwork at the beginning of each chapter is by Ricardo Pacheco. 1 – Untitled [detail] 01-2016, charcoal on paper 100 × 70 cm 2 – Wandering [detail] 02-2015, charcoal on paper 100 × 70 cm 3 – Untitled [detail] 09-2014, charcoal on paper 100 × 70 cm 4, 5 – Elephant graveyard [detail] 09-2015, charcoal on paper 100 × 70 cm Ricardo Pacheco is a Portuguese artist, with a degree in painting by the Faculty of Fine Arts of the University of Lisbon. [email protected]/www.ricardopacheco.info

ISBN 978-3-11-047539-5 e-ISBN (PDF) 978-3-11-047629-3 e-ISBN (EPUB) 978-3-11-047544-9 ISSN 0938-6572 Library of Congress Cataloging-in-Publication Data Names: Kalauch, Anke, 1972- author. | van Gaans, Onno, 1971- author. Title: Pre-Riesz spaces / Anke Kalauch, Onno van Gaans. Description: Berlin ; Boston : Walter de Gruyter GmbH, [2018] | Series: De Gruyter expositions in mathematics ; Band/Volume 66 | Includes bibliographical references and index. Identifiers: LCCN 2018031534 (print) | LCCN 2018039652 (ebook) | ISBN 9783110476293 (electronic Portable Document Format (pdf) | ISBN 9783110475395 (print : alk. paper) | ISBN 9783110475449 (e-book epub) | ISBN 9783110476293 (e-book pdf) Subjects: LCSH: Riesz spaces. | Partially ordered spaces. | Vector spaces. Classification: LCC QA322 (ebook) | LCC QA322 .K337 2018 (print) | DDC 515/.73–dc23 LC record available at https://lccn.loc.gov/2018031534 Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2019 Walter de Gruyter GmbH, Berlin/Boston Typesetting: le-tex publishing services GmbH, Leipzig Printing and binding: CPI books GmbH, Leck www.degruyter.com

Contents Preface | IX 1 1.1 1.1.1 1.1.2 1.1.3 1.1.4 1.2 1.2.1 1.2.2 1.3 1.3.1 1.3.2 1.4 1.5 1.5.1 1.5.2 1.5.3 1.5.4 1.5.5 1.6 1.7 1.7.1 1.7.2 1.8

A primer on ordered vector spaces | 1 Ordered vector spaces | 2 Vector space orders and cones | 2 Vector lattices | 6 The Riesz decomposition property | 8 Order convergence | 11 Operators on ordered vector spaces | 12 Order-bounded and regular operators | 12 The Riesz–Kantorovich formulas | 15 Ideals and bands in vector lattices | 17 Definitions and basic properties | 17 Ideals and bands in C(Ω) | 20 Riesz homomorphisms and disjointness preserving operators in vector lattices | 22 Norm and order | 26 Closed cones | 27 Semimonotone norms | 28 Order unit spaces | 29 Normed vector lattices | 35 Relatively uniform convergence | 36 Order denseness | 40 Examples | 42 The ice cream cone and friends | 42 Polyhedral cones | 46 Odds and ends from functional analysis and topology | 50

2 2.1 2.1.1 2.1.2 2.1.3 2.2 2.2.1 2.2.2 2.3 2.3.1 2.3.2 2.3.3

Embeddings, covers, and completions | 53 Dedekind completion | 54 Dedekind cuts | 54 Addition and scalar multiplication on the set of Dedekind cuts | 56 The Dedekind completion of an Archimedean space | 61 Pre-Riesz spaces | 68 Definition and basic properties | 68 Some lattice-like formulas | 71 Riesz* homomorphisms | 74 Definition and basic properties | 74 Extension and restriction | 76 Comparison with Riesz homomorphisms | 80

VI | Contents

2.4 2.5 2.6 2.6.1 2.6.2 2.6.3 2.6.4 2.7 2.7.1 2.7.2 2.7.3 2.8 3 3.1 3.2 3.2.1 3.2.2 3.2.3 3.2.4 3.3 3.3.1 3.3.2 3.3.3 3.3.4 3.3.5 3.3.6 3.4 3.4.1 3.4.2 3.4.3 3.5 3.6 3.6.1 3.6.2 3.6.3 3.7 3.7.1 3.7.2 3.7.3

Vector lattice cover and Riesz completion | 87 Functional representation | 94 Examples of vector lattice covers | 101 Polyhedral cones | 102 Lorentz cones | 103 Positive semidefinite matrices | 104 Finite dimensional examples by means of a dual base | 107 The Fremlin tensor product as Riesz completion | 108 The vector space tensor product | 108 The Fremlin tensor product | 110 The projective cone and its relatively uniform closure | 111 Extension and restriction method | 115 Seminorms on pre-Riesz spaces | 119 Basic properties of seminorms | 119 Pre-Riesz seminorms | 121 Generalization of Riesz seminorms: plan of attack | 121 Solid sets | 123 Solvex sets | 128 Pre-Riesz seminorms: seminorms with solvex unit balls | 132 Extension and restriction of monotone seminorms | 138 Monotone seminorms and related concepts | 139 Extension of seminorms | 144 Extension and norm completeness | 150 Uniqueness of extensions | 153 Quotients over kernels of monotone seminorms | 157 Extension to M-seminorms and L-seminorms | 160 Regular seminorms | 163 Definition and examples | 164 Extension of regular seminorms | 165 Regularization of seminorms | 168 Semimonotone norms revisited | 171 Topologies on pre-Riesz spaces | 173 Locally full topologies | 174 Topologies such that the positive cone is closed | 179 Locally solid topologies | 180 Order convergence and unbounded order convergence | 183 Order convergence revisited | 184 Extension and restriction of order convergence | 187 Unbounded order convergence | 190

Contents |

4 4.1 4.1.1 4.1.2 4.1.3 4.1.4 4.1.5 4.2 4.2.1 4.2.2 4.2.3 4.3 4.3.1 4.3.2 4.3.3 4.4 4.4.1 4.4.2 4.4.3 4.4.4 4.4.5 5 5.1 5.1.1 5.1.2 5.2 5.2.1 5.2.2 5.2.3 5.2.4 5.3 5.4 5.4.1 5.4.2 5.4.3 5.5 5.5.1

Disjointness, bands, and ideals in pre-Riesz spaces | 193 Disjointness | 193 Disjointness in partially ordered vector spaces | 193 Disjointness and embedding | 195 Disjointness in spaces with the Riesz decomposition property | 196 Disjointness in antilattices | 198 Fordable pre-Riesz spaces | 200 Bands | 203 Bands in pre-Riesz spaces | 204 Extension of bands | 204 Restriction of bands in fordable pre-Riesz spaces | 206 Ideals | 206 Ideals in partially ordered vector spaces | 207 Extension and restriction of ideals | 209 Directed ideals | 215 Bands in spaces with order units | 219 Carriers of bands | 219 Characterization of bands by bisaturated sets | 222 Bands in spaces with polyhedral cones | 225 Bands in finite dimensional spaces | 228 Carriers of extension bands | 234 Operators on pre-Riesz spaces | 237 Disjointness preserving operators | 237 Definitions and examples | 237 Riesz* homomorphisms on pre-Riesz spaces of continuous functions | 243 Disjointness preserving inverse | 247 Inverses of bijective Riesz* homomorphisms | 247 On d-isomorphisms in pre-Riesz spaces | 250 Disjointness preserving bijections in finite dimensions | 252 Disjointness preserving inverse in infinite dimensions | 254 Disjointness preserving C0 -semigroups | 256 Spaces of operators | 261 Operators between ordered normed spaces | 262 Operators with Dedekind complete range spaces | 264 The space of order continuous operators: an example | 269 Spectral theory for operators on ordered Banach spaces | 279 Definitions and basic properties | 279

VII

VIII | Contents

5.5.2 5.5.3

Spectral properties of positive operators | 281 Some spectral properties for operators on Banach lattices | 286

Bibliography | 289 Index | 297

Preface A major part of functional analysis revolves around the study of the abstract structures of function spaces. The linear structure and the norm structure have become standard topics in any mathematics program. Considerations concerning a more general topological structure or a multiplicative structure are also widespread. Many function spaces also have a natural order structure, where for two functions f and g one says that f ≤ g if f(s) ≤ g(s) for every s of the underlying set S. For compatibility of a partial order ≤ with the linear structure of a real vector space X, a mild condition is required, namely f ≤ g implies f + h ≤ g + h and αf ≤ αg for every positive scalar α. Then X is called a partially ordered vector space. Order structures have been studied right from the early years in functional analysis. In particular, their relation to norms and topologies has led to an interesting theory; see, e.g., [14, 102, 127, 162]. The theory of general partially ordered vector spaces is meagre, since almost any result beyond very basic observations requires additional conditions. The situation is much better for partially ordered vector spaces X which have the property that for every two elements x, y ∈ X the set {x, y} has a least upper bound (supremum), denoted by x ∨ y. Such spaces are called Riesz spaces or vector lattices. A prototype example is the space C(S) of real valued continuous functions on some topological space S. The supremum of two functions is then the pointwise maximum. A rich theory for Riesz spaces has been developed in the past century, most notably in the 1970s and 1980s. This theory includes compatible norm and topological structures on Riesz spaces and operators between Riesz spaces. There are several excellent monographs on the topic, for instance [2, 12, 111, 119, 165, 166]. Interestingly, in many results in the theory of Riesz spaces the condition of existence of suprema seems unnecessarily strong. For instance, two elements x and y in a Riesz space X are said to be disjoint if |x| ∧ |y| = 0, where |x| = x ∨ (−x) and x ∧ y = −((−x) ∨ (−y)). Then in the Riesz space C(S) two functions f and g are disjoint if and only if {s; f(s) ≠ 0} and {s; g(x) ≠ 0} are disjoint sets. The latter also makes sense for functions in C1 [0, 1], that is, differentiable functions, although C1 [0, 1] is not a Riesz space. More than that, partially ordered vector spaces that are not Riesz spaces cannot be avoided in the theory of Riesz spaces. Indeed, linear subspaces of Riesz spaces with the inherited order are only rarely Riesz spaces themselves. A similar situation appears for tensor products of Riesz spaces. Also spaces of operators between Riesz spaces are always partially ordered vector spaces but only rarely Riesz spaces. In conclusion, the conditions of a Riesz space are more restrictive than is desirable, whereas the general condition of being a partially ordered vector space is too weak to yield an interesting theory. Pre-Riesz spaces provide a suitable setting for a rich theory of partially ordered vector spaces beyond Riesz spaces. They are the spaces that allow an appropriate embedding into Riesz spaces. The idea is then to make use of the existing theory for the https://doi.org/10.1515/9783110476293-201

X | Preface

larger Riesz space and then to restrict again to the pre-Riesz space. For instance, two elements x and y of a pre-Riesz space X will be disjoint in X if they are disjoint in the Riesz space Y in which X is embedded. Of course, this approach would be impracticable if the ensuing notion of disjointness would depend on the choice of the larger Riesz space Y. It turns out that this is not the case if the embedding is order dense, meaning that every y ∈ Y is the infimum of the set {x ∈ X; x ≥ y}. Pre-Riesz spaces are precisely the partially ordered vector spaces that allow an order dense embedding into a Riesz space. Such ambient Riesz spaces are called vector lattice covers. For the special case of Archimedean directed partially ordered vector spaces, one could use the classical Dedekind completion as a vector lattice cover, but in many instances there are much smaller and more convenient vector lattice covers. The concept of pre-Riesz spaces has been developed by Maris van Haandel in [151]. An earlier version of his theory for Archimedean spaces, which emphasizes the role of order denseness, is due to Gerard Buskes and Arnoud van Rooij; see [36]. A study of seminorms on pre-Riesz spaces by means of embedding in Riesz spaces is given in [146]. In the past decade, the authors of the present book have generalized central concepts of Riesz space theory to pre-Riesz spaces: disjointness, bands, and ideals. Meanwhile, more researchers are involved in a systematic study of pre-Riesz spaces, and a theory of operators on pre-Riesz spaces is emerging. The present book is a comprehensive exposition of the theory of pre-Riesz spaces. As every directed Archimedean partially ordered vector space is a pre-Riesz space, the theory presented in the book covers almost every ordered space appearing in the fields of monotone dynamical systems, game theory, economic equilibrium theory, optimization, and Markov chains. Although these applications are beyond the scope of the book, an overview of the abstract theory may be beneficial to experts in these areas. Of special interest for applications are ordered Banach spaces, which are pre-Riesz spaces. They naturally appear in several parts of the book, where we look at spectral theory and strongly continuous operator semigroups. Typical examples of pre-Riesz spaces that are not Riesz spaces are spaces of smooth functions, higher-order Sobolev spaces or function spaces with linear constraints. The book is meant to serve several purposes. First, it is a convenient up-to-date account of the present state of the theory of pre-Riesz spaces for experienced researchers. Second, we tried to make the book a friendly introduction to pre-Riesz spaces for researchers starting in the field of positivity. Third, the material in Chapter 1 is a brief overview of the more classical theory of partially ordered vector spaces, and can be used as a source for an introductory graduate course. Throughout the book, we elaborate proofs in quite some detail, so that the exposition is accessible for students with some training in functional analysis. The book is organized as follows. In Chapter 1, a primer on partially ordered vector spaces is given, where all preliminaries for the book are collected. It serves as a first brief course on partially ordered vector spaces and vector lattices. We do not pretend to give a balanced overview of

Preface | XI

historic developments of the subject. The theory is enriched by some important examples. Chapter 2 treats the embedding of partially ordered vector spaces into vector lattices. We give the construction of the Dedekind completion in full detail. We derive van Haandel’s theory of pre-Riesz spaces and vector lattice covers. In particular, we consider the Riesz completion, which is the smallest vector lattice cover, and is unique up to Riesz isomorphisms. The theory involves several generalized concepts of Riesz homomorphisms, of which Riesz* homomorphisms turn out to be the most useful. For Archimedean partially ordered vector spaces with order units, Kadison’s functional representation (1951) yields a convenient explicit construction of ambient vector lattices. We show that this embedding is order dense and illustrate the method by several examples. We also obtain the Fremlin tensor product as a Riesz completion. The strategy to obtain properties of structures in pre-Riesz spaces by means of the embedding technique is discussed at the end of Chapter 2. Sometimes the pre-Riesz space needs to be dense in a vector lattice cover in a stronger way. Such pre-Riesz spaces are called pervasive and play a role in different parts of the book. Chapter 3 is dedicated to the question of how seminorms on pre-Riesz spaces are linked to seminorms on their vector lattice covers. Riesz seminorms in a vector lattice cover lead to pre-Riesz seminorms in a pre-Riesz space. It turns out that the unit balls of pre-Riesz seminorms are solvex sets. This notion combines in an appropriate way solidness and convexity. We further study monotone seminorms on vector lattice covers and the corresponding notion of monotone* seminorm on the pre-Riesz space. Also, we discuss regular seminorms on pre-Riesz spaces, whose properties resemble the properties of Riesz seminorms. Moreover, we give a brief survey of locally full and locally solid topologies. We conclude the chapter with a glance at unbounded order convergence in pre-Riesz spaces. In Chapter 4, notions that are at the core of vector lattice theory are introduced and investigated in partially ordered vector spaces. In the definitions of the notions, we replace lattice operations by sets of upper bounds. More precisely, instead of the modulus |x| = x ∨ (−x) of an element x we deal with the set of upper bounds of x and −x. In this way, we define disjointness in pre-Riesz spaces. This intrinsic definition corresponds to disjointness in a vector lattice cover. Bands are defined to be subspaces that equal their double disjoint complement, just as in Archimedean vector lattices. In contrast to bands in vector lattices, bands in pre-Riesz spaces need not be directed. Bands in a pre-Riesz space can be extended to bands in a vector lattice cover. Conversely, bands in a vector lattice cover restrict to order closed solvex subspaces in the pre-Riesz space. If the pre-Riesz space is, in addition, pervasive, bands restrict to bands. The notion of a directed ideal in a partially ordered vector space traces back to the work of Kadison, Bonsall, Kist, and Fuchs in the 1950s and 1960s. This notion is a special case of our notion of a solid subspace. Solid subspaces in a vector lattice cover are restricted to solid subspaces in the pre-Riesz space, whereas solid subspaces in the

XII | Preface

pre-Riesz space can be extended appropriately provided they are, in addition, solvex. Directed ideals in pre-Riesz spaces are characterized similarly to ideals in vector lattices, e.g., as full subspaces or kernels of positive operators. Using the functional representation of Archimedean pre-Riesz spaces with order unit, i.e., the order dense embedding into a space of continuous functions on a compact Hausdorff space Ω, we characterize bands in the pre-Riesz space by means of certain subsets of Ω. In finite dimensions, this leads to a geometrical description using the base of the dual cone. We show that, for a given dimension, the number of bands is bounded. Some first steps towards a theory of operators on pre-Riesz spaces are collected in Chapter 5. In particular, we list basic properties of disjointness preserving and band preserving operators and discuss Riesz* homomorphisms on pre-Riesz spaces of continuous functions. As a rule, techniques that are well-known from vector lattice theory can not be applied straightforwardly to pre-Riesz spaces. One reason is that orderbounded disjointness preserving operators on pre-Riesz spaces are not regular, in general. Another restriction appears due to the fact that only Riesz* homomorphisms on pre-Riesz spaces can be, so far, appropriately extended to the Riesz completion. In vector lattice theory, there is some treatment of conditions under which the inverse of a disjointness preserving bijection is disjointness preserving. We show that in finite-dimensional pre-Riesz spaces this is always the case, and also give sufficient conditions in the infinite-dimensional case. Again motivated by vector lattice theory, we study disjointness preserving strongly continuous operator semigroups on ordered Banach spaces. We show that the generator of such a semigroup is band preserving. We also discuss conditions on the generator such that the semigroup consists of band preserving operators. Spaces of operators on vector lattices are typical examples of partially ordered vector spaces. They are vector lattices, provided the range space is Dedekind complete. In this setting, many results are obtained involving subspaces of operators, which are, e.g., bands. It is a challenging task to study similar questions when the range space is only Archimedean. We give an illustrative example of this approach, considering a space of order continuous operators. Finally, we list well-known results for positive operators on ordered normed spaces. Under mild conditions, every positive operator is continuous. Moreover, positive operators on ordered Banach spaces with semimonotone norms have important spectral properties, which are used in a broad variety of applications. We present Krein’s theorems that the spectral radius is contained in the spectrum, and that in case of a strictly positive spectral radius for a positive compact operator there is an associated eigenvector in the cone. The theory of operators on pre-Riesz spaces, in particular disjointness preserving operators, band preserving operators and Riesz* homomorphisms, is a developing area, and Chapter 5 may serve as an inspiration for further research.

Preface |

XIII

Our common interest in partially ordered vector spaces started many years ago. We thank Arnoud van Rooij and Martin R. Weber who introduced us to the theory of partially ordered vector spaces. They support us with their continuing interest in our research and are always there for good advice. Also, many other members of the positivity community have influenced our work by asking interesting questions and making helpful remarks, and we gratefully acknowledge this exchange of ideas. In particular, the Positivity Conferences were scientific highlights as well as heartwarming meetings with friends. We hope that experts on Riesz spaces will browse the book to see how their work is applied. We would be thrilled if they find inspiration to join in the investigation of pre-Riesz spaces. Parts of the book rely on a close and enriching cooperation with Bas Lemmens. There are also results in the book that have been obtained by our PhD students Hent van Imhoff, Helena Malinowski, and Feng Zhang. We thank our colleagues for the inspiring collaboration. We are grateful to Helena Malinowski for preparing the figures and schemes for this book, and to Marcus Waurick (Moppi) and all other colleagues who improved the book by corrections and remarks. We thank the Portuguese artist Ricardo Pacheco for allowing us to use details of his drawings to ornament our book at the beginning of each chapter. Dresden and Leiden, Anke Kalauch and Onno van Gaans

July 2018

1 A primer on ordered vector spaces This chapter provides an introduction to ordered vector spaces and vector lattices and fixes notation and terminology. For the theory on vector lattices we mainly refer to [12, 111, 119]. For the theory of partially ordered vector spaces see also [155, 156], their edited translation into German [162], and [14]. We will discuss topics such as cones, order convergence, dual cones, and positive operators. We investigate the space of regular operators and give a proof of the classical Riesz–Kantorovich formulas. Ideals and bands in vector lattices are reviewed and bands are characterized in the space of continuous functions on a compact Hausdorff space. Moreover, we collect properties of Riesz homomorphisms and disjointness preserving operators. Relations between norms and the order structure are investigated, with emphasis on cones with nonempty interiors. This includes a treatment of order unit spaces. We introduce order denseness, which is a central concept throughout the book. Several important examples serve as illustrations. We conclude the chapter in Section 1.8 with a choice of classical theorems from real and functional analysis that are used throughout the book. For all vector spaces appearing in this book, the scalar field is the field ℝ of real numbers, unless otherwise stated. Let X be a real vector space. For sets A, B ⊆ X, we use standard notations such as the sum A + B := {a + b; a ∈ A, b ∈ B} and the scalar multiple λA := {λa; a ∈ A} for λ ∈ ℝ, where we denote −A := (−1)A and A − B := A + (−B). Furthermore, we will use the linear span n

span(A) := { ∑ λ i a i ; n ∈ ℕ, i ∈ {1, . . . , n}, λ i ∈ ℝ, a i ∈ A} , i=1

the affine hull n

n

aff(A) := { ∑ λ i a i ; n ∈ ℕ, i ∈ {1, . . . , n}, λ i ∈ ℝ, a i ∈ A, ∑ λ i = 1} , i=1

i=1

the convex hull n

n

co(A) := { ∑ λ i a i ; n ∈ ℕ, i ∈ {1, . . . , n}, λ i ∈ [0, 1], a i ∈ A, ∑ λ i = 1} , i=1 https://doi.org/10.1515/9783110476293-001

i=1

2 | 1 A primer on ordered vector spaces

and the positive-linear hull n

pos(A) := { ∑ λ i a i ; n ∈ ℕ, i ∈ {1, . . . , n}, λ i ∈ [0, ∞), a i ∈ A} . i=1

Here ℕ denotes the set of natural numbers {1, 2, 3, . . .}. Sometimes we omit the brackets and simply write span A, aff A, co A, pos A. The main object of interest in this book is a vector space endowed with a partial order that is suitably compatible with the linear structure. Recall that a partially ordered set S is a nonempty set with a reflexive, transitive, and antisymmetric binary relation ≤. We use the standard terminology of partially ordered sets, e.g., the notions of upper and lower bound, or the notation y ≥ x for x ≤ y, etc. For a ∈ S we denote S ≥a := {s ∈ S; s ≥ a}.

1.1 Ordered vector spaces 1.1.1 Vector space orders and cones On a vector space, we consider partial orders that are appropriately linked to the linear operations. Definition 1.1.1. Let X be a vector space. (i) A partial order ≤ on X is called a vector space order if (a) x, y, z ∈ X and x ≤ y imply x + z ≤ y + z, (b) x ∈ X, 0 ≤ x and λ ∈ [0, ∞) imply 0 ≤ λx. Instead of x ≤ y, x ≠ y we write x < y. An element x ∈ X is called positive if 0 ≤ x. For given a, b ∈ X, a ≤ b, we denote [a, b] := {x ∈ X; a ≤ x ≤ b} and call [a, b] an order interval. For M ⊆ X the sets of all upper and lower bounds, respectively, are denoted by M u = {x ∈ X; ∀m ∈ M : x ≥ m}

and

M l = {x ∈ X; ∀m ∈ M : x ≤ m} .

We use abbreviations such as M ul for (M u )l , etc. A set M ⊆ X is called orderbounded if there are a, b ∈ X such that M ⊆ [a, b]. Further, a set M ⊆ X is called full if for every a, b ∈ M we have [a, b] ⊆ M. (ii) A nonempty set K ⊆ X is called a wedge in X if x, y ∈ K, λ, μ ∈ [0, ∞) imply λx + μy ∈ K. (iii) If K is a wedge in X with the additional property K ∩ (−K) = {0}, then K is called a cone in X. (iv) A wedge K in X is called generating if X = K − K.

1.1 Ordered vector spaces | 3

For a set M ⊆ X the (by inclusion) smallest wedge that contains M is pos(M). The following relation between vector space orders and cones is straightforward. Proposition 1.1.2. Let X be a vector space. (i) Let K be a cone in X and ≤ the binary relation on X defined by means of x ≤ y :⇔ y − x ∈ K .

(1.1)

Then ≤ is a vector space order. (ii) Let ≤ be a vector space order on X. Then the set K0 := {x ∈ X; 0 ≤ x}

(1.2)

of all positive elements in X is a cone in X. (iii) Let K be a cone in X, ≤ the binary relation in (1.1), and K0 the corresponding cone in (1.2). Then K = K0 . Remark 1.1.3. If K is a wedge in the vector space X and ≤ the binary relation defined by (1.1), then the relation ≤ is reflexive and transitive and satisfies the properties (a) and (b) of Definition 1.1.1 (i). On the other hand, if ≤ is a reflexive and transitive relation on X that satisfies the properties (a) and (b) of Definition 1.1.1 (i), then the set K0 defined by (1.2) is a wedge in X. If in a vector space X a cone K is given, then we equip X with the vector space order ≤ introduced in (1.1) and call (X, K) a (partially) ordered vector space. Occasionally we loosely write X instead of (X, K), provided that K is fixed in advance. The next proposition provides simple constructions with partially ordered vector spaces. Proposition 1.1.4. (i) Let (X, K) be a partially ordered vector space and let D be a linear subspace of X. Then (D, D ∩ K) is a partially ordered vector space. (ii) Let I be a nonempty set and for every i ∈ I, let (X i , K i ) be a partially ordered vector space. Then the Cartesian product (Π i∈I X i , Π i∈I K i ) is a partially ordered vector space. In (i), the order on D given by the cone D ∩ K is called the induced order. Example 1.1.5. (i) For every vector space X we have that (X, {0}) is a partially ordered vector space. This order is sometimes called the trivial order on X. (ii) The space of real numbers (ℝ, [0, ∞)) is a partially ordered vector space. Its order is referred to as the standard order on ℝ. (iii) Let S be a nonempty set. Consider the vector space ℝS := Π i∈S ℝ = {f; f : S → ℝ} and the subset ℝ+S := {f ∈ ℝS ; ∀s ∈ S : f(s) ≥ 0}. Then (ℝS , ℝ+S ) is a partially ordered vector space. The partial order is called the pointwise order or the standard order and the cone ℝ+S is called the standard cone. One can view this example as a special case of Proposition 1.1.4, using (ii).

4 | 1 A primer on ordered vector spaces (iv) As a special case of (iii), we obtain ℝn := ℝ{1,...,n} with the cone ℝ+n = {(x1 , . . . , x n )T ; ∀i ∈ {1, . . . , n} : x i ≥ 0} , which yields the standard order on ℝn . (v) Let ℝS be as in (iii). We consider for p > 0 the subspaces ℓp (S) := {x ∈ ℝS ; ∑ |x(s)|p < ∞}

and

s∈S

cc (S) := {x ∈ ℝS ; {s ∈ S; x(s) ≠ 0} is finite} with the induced order. Usually we consider these spaces for S = ℕ. Definition 1.1.6. Let (X, K) be an ordered vector space. (i) A subset M ⊆ X is called directed if for every x, y ∈ M there is z ∈ M such that x ≤ z, y ≤ z. (ii) An element u > 0 is called an order unit if for every x ∈ X there is λ ∈ (0, ∞) such that x ∈ [−λu, λu]. Note that a directed subspace M of X is also downward directed, i.e., for every x, y ∈ M there is z ∈ M such that x ≥ z, y ≥ z. An element u is an order unit if and only if for every x ∈ X there is λ ∈ (0, ∞) such that x ≤ λu. The following observations on directed partially ordered vector spaces will be used frequently without reference. Proposition 1.1.7. Let (X, K) be an ordered vector space. (i) The cone K is generating if and only if X is directed. (ii) If there exists an order unit u > 0 in X, then K is generating. Proof. The statement in (i) is straightforward. For a proof of (ii), observe that for every x ∈ X there is a λ ∈ (0, ∞) such that λu − x ∈ K; moreover, λu ∈ K and x = λu − (λu − x). Note that in Example 1.1.5 (i) with X ≠ {0}, the space X is not directed, whereas ℝS with the standard order is directed. Example 1.1.8. (i) Let S be a nonempty set and let B(S) be the subspace of ℝS consisting of all bounded functions. For every subspace X of B(S) that contains the constant one function 𝟙, the element 𝟙 is an order unit in X. We sometimes write ℓ∞ (S) instead of B(S), particularly if S is countable. (ii) For a nonempty compact Hausdorff space Ω, denote the vector space of all real continuous functions on Ω with C(Ω). We endow C(Ω) with the pointwise order, i.e., we consider in C(Ω) the order induced from ℝΩ . Every x ∈ C(Ω) with x(ω) > 0 for every ω ∈ Ω is an order unit in C(Ω). (iii) In the spaces ℓp (ℕ) (for p finite) and cc (ℕ), respectively, there does not exist an order unit.

1.1 Ordered vector spaces |

5

Definition 1.1.9. A partially ordered vector space (X, K) is called Archimedean¹ if for every x, y ∈ X with nx ≤ y for every n ∈ ℕ one has that x ≤ 0. The following statement is immediate. Proposition 1.1.10. Let (X, K) be an ordered vector space and let D be a linear subspace of X. If X is Archimedean, then (D, D ∩ K) is Archimedean as well. Example 1.1.11. The space ℝS with standard order (see Example 1.1.5) is Archimedean. Therefore the space ℝn and also the spaces B(S) and C(Ω) in Example 1.1.8 are Archimedean. Example 1.1.12. In ℝn with n ≥ 2, we consider the so-called lexicographical order, i.e., for u = (u 1 , . . . , u n )T and v = (v1 , . . . , v n )T we define that u ≤ v if u 1 < v1 , or there is k ∈ {2, . . . , n} such that ∀j ∈ {1, . . . , k − 1} : u j = v j and u k < v k , or u = v. The space ℝn endowed with this order is not Archimedean. Indeed, if e(i) denotes the ith unit vector in ℝn , then me(2) ≤ e(1) for every m ∈ ℕ, but e(2) ≰ 0. Let us now consider quotients of partially ordered vector spaces. Let (X, K) be a partially ordered vector space and let I be a linear subspace of X. For x ∈ X denote [x] := {y ∈ X; y − x ∈ I}. From the theory of vector spaces it is known that the quotient space X̂ := X/I := {[x]; x ∈ X} (1.3) equipped with the canonical operations is a vector space and that the quotient map q : X → X,̂ x 󳨃→ [x], is linear and surjective. Let K̂ := {[x]; x ∈ K}. It is straightforward that K̂ is a wedge in X.̂ Clearly, q maps K onto K.̂ If, in addition, I is full, we show next that K̂ is a cone in X.̂ Proposition 1.1.13. Let (X, K) be a partially ordered vector space and let I be a full linear subspace of X. Then the quotient space (X,̂ K)̂ is a partially ordered vector space. If K is generating, then K̂ is generating as well. ̂ so then there is x1 ∈ [x] with x1 ∈ K and x2 ∈ [x] with Proof. Let [x] ∈ K̂ ∩ (−K), −x2 ∈ K, and thus 0 ≤ x1 ≤ x1 − x2 ∈ I. Since I is full, one gets x1 ∈ I and hence x ∈ I, which implies [x] = I = [0]. Let K be generating and let y ∈ X.̂ Take x ∈ X such that y = q(x). There are x1 , x2 ∈ K with x = x1 − x2 . Then y = q(x1 ) − q(x2 ), so that K̂ is generating. Given the assumptions in Proposition 1.1.13, we call K̂ the quotient cone and the induced order in X̂ the quotient order. 1 Our definition of Archimedean follows [14, 72]. Note that there are different notions of Archimedean in the literature. In [151] our notion is called integrally closed, and Archimedean there means that nx ≤ y for every n ∈ ℤ implies x = 0, which is a weaker notion.

6 | 1 A primer on ordered vector spaces

1.1.2 Vector lattices A partially ordered vector space (X, K) is called a vector lattice or Riesz space if for every x, y in X the set {x, y} has a least upper bound (supremum) and a greatest lower bound (infimum), denoted by x ∨ y and x ∧ y, respectively. We list basic properties of vector lattices, mostly without proofs. For the next result see, e.g., [14, Lemma 1.15]. Proposition 1.1.14. Let (X, K) be a partially ordered vector space. The following statements are equivalent. (i) For every x, y in X the supremum of the set {x, y} exists. (ii) (X, K) is a vector lattice. (iii) Every nonempty finite subset of X has a supremum and an infimum. Every vector lattice X is distributive, i.e., for every x, y, z ∈ X one has x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z) and x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z); see, e.g., [111, Corollary 12.3]. In vector lattices many formulas are valid that are familiar from the real numbers. To illustrate how such formulas are typically proven, we include the proof of one such formula in detail. For x, y ∈ X we have x+y=x∧y+x∨y.

(1.4)

Indeed, x + y − x ∧ y ≥ x + y − x = y and x + y − x ∧ y ≥ x, hence x + y − x ∧ y is an upper bound of the set {x, y}. If v is another upper bound of {x, y}, then x + y − v ≤ x and x + y − v ≤ y so x + y − v ≤ x ∧ y. Therefore, v ≥ x + y − x ∧ y. We conclude that x + y − x ∧ y is the least upper bound of {x, y}. The following formulas will be used throughout the book without reference. The statements in (i), (ii), (iii) are straightforward. Note that (iv) extends the distributive property; see also [12, Theorem 1.5]. Proposition 1.1.15. Let (X, K) be an ordered vector space and let M ⊆ X be such that sup M exists. Then the following statements hold: (i) The infimum of the set {−m; m ∈ M} exists and inf{−m; m ∈ M} = − sup M. (ii) For every x ∈ X the supremum of the set {x + m; m ∈ M} exists and sup{x + m; m ∈ M} = x + sup M . (iii) For every λ ∈ ℝ+ the supremum of the set {λm; m ∈ M} exists and sup{λm; m ∈ M} = λ sup M . (iv) If, in addition, (X, K) is a vector lattice, then for every x ∈ X the supremum of the set {x ∧ m; m ∈ M} exists and sup{x ∧ m; m ∈ M} = x ∧ sup M . Similarly, if M ⊆ X is such that inf M exists, then for every x ∈ X the infimum of the set {x ∨ m; m ∈ M} exists and inf{x ∨ m; m ∈ M} = x ∨ inf M .

1.1 Ordered vector spaces |

7

As usual, for an element x of a vector lattice X, the element x+ := x ∨ 0 is called the positive part of x, the element x− := (−x) ∨ 0 the negative part and |x| := x ∨ (−x) the modulus of x. We list properties of these elements; see [12, Theorems 1.3, 1.4]. Proposition 1.1.16. Let (X, K) be a vector lattice and x, y, z ∈ X. Then the following statements hold. (i) |x| = x+ + x− , (ii) x+ ∧ x− = 0, (iii) x = x+ − x− , (iv) if x = y − z with y, z ∈ K, then x+ ≤ y, x− ≤ z, (v) the decomposition in (iii) is unique in the sense that if x = y − z with y, z ∈ K and y ∧ z = 0, then y = x+ and z = x− , (vi) |x| ∧ |y| = 12 (|x + y| − |x − y|) and |x| ∨ |y| = 12 (|x + y| + |x − y|), (vii) | |x| − |y| | ≤ |x + y| ≤ |x| + |y|, (viii) |x ∨ z − y ∨ z| ≤ |x − y| and |x ∧ z − y ∧ z| ≤ |x − y|, (ix) if, in addition, x, y, z ∈ K, then x ∧ (y + z) ≤ x ∧ y + x ∧ z. The maps (x, y) 󳨃→ x ∧ y, (x, y) 󳨃→ x ∨ y, x 󳨃→ x+ , x 󳨃→ x− , x 󳨃→ |x| are called the lattice operations. Infimum and supremum can be expressed by means of the modulus with the formulas x ∧ y = 12 (x + y − |x − y|)z , x ∨ y = 12 (x + y + |x − y|) . (1.5) This implies |x − y| = x ∨ y − x ∧ y. Moreover, from Proposition 1.1.15 it follows that (x − y)+ = (−y + x) ∨ (−y + y) = −y + (x ∨ y)

and

−

(x − y) = (−x + y) ∨ (−x + x) = −x + (y ∨ x) .

(1.6)

The following notion will be used frequently throughout the book. Definition 1.1.17. A linear subspace D of a Riesz space X is called a Riesz subspace if for every x, y ∈ D one has x ∨ y ∈ D. Clearly, a Riesz subspace of a Riesz space with the induced order is itself a Riesz space. Example 1.1.18. Consider ℝS as in Example 1.1.5. This space is a Riesz space, and the lattice operations are pointwise, e.g., for s ∈ S and x, y ∈ ℝS one has (x ∨ y)(s) = max{x(s), y(s)}. As a special case, ℝn is a Riesz space. The space B(S) as in Example 1.1.8 is a Riesz subspace of ℝS and, therefore, a Riesz space. Similarly, the spaces C(Ω), l p (S), and cc (S) are Riesz spaces. Example 1.1.19. We consider the vector space C1 [0, 1] of all continuously differentiable functions on [0, 1]. With pointwise order, this space is a partially ordered vector space. We show that C1 [0, 1] is not a vector lattice. Indeed, let x(s) = s, y(s) = 1 − s, where s ∈ [0, 1], and suppose that u is an upper bound of {x, y}. We will show that there is a strictly smaller upper bound, i.e., the supremum of {x, y} does not exist. The

8 | 1 A primer on ordered vector spaces function s 󳨃→ max{x(s), y(s)} is continuous, but not differentiable. Therefore it is not equal to u, hence there exist ε > 0 and an open interval I ⊆ [0, 1] such that for every s ∈ I we have u(s)−max{x(s), y(s)} ≥ ε. According to elementary calculus, there exists a positive nonzero function v ∈ C1 [0, 1], which vanishes outside I. We may assume that v is bounded on I by ε. Now u − v ∈ C1 [0, 1] is an upper bound of {x, y}, which is strictly below u. In the same fashion it can be shown that for every k ∈ ℕ the vector space Ck [0, 1] of all k-times continuously differentiable functions is not a Riesz space. Example 1.1.20. Let X be the subspace of C[0, 1] consisting of all affine functions. Then X is a Riesz space, but the lattice operations in X are not pointwise. The subspace X is not a Riesz subspace of C[0, 1]. With the infimum, one can characterize the property of being Archimedean as follows. Proposition 1.1.21. An ordered vector space (X, K) is Archimedean if and only if for every x ∈ K one has that inf{ 1n x; n ∈ ℕ} exists and equals zero. A vector lattice X is Archimedean if and only if for every x, y ∈ X with 0 ≤ nx ≤ y for every n ∈ ℕ one has that x = 0; see [10, Definition 7.2]. Next we discuss an important completeness notion in vector lattices. Definition 1.1.22. A vector lattice (X, K) is called Dedekind complete, if every nonempty set that is bounded from above has a supremum. Since in a Dedekind complete vector lattice for every x ∈ K the element u = inf{ 1n x; n ∈ ℕ} exists and satisfies 12 u = u, we obtain u = 0, which yields the next result. Proposition 1.1.23. If a vector lattice (X, K) is Dedekind complete, then it is Archimedean. We conclude that every subspace of a Dedekind complete vector lattice is an Archimedean ordered vector space. In Theorem 2.1.13 below we will show that every directed Archimedean ordered vector space is a subspace of a Dedekind complete vector lattice. Example 1.1.24. The Riesz spaces ℝS and B(S) are Dedekind complete. The Riesz space C[0, 1] is not Dedekind complete. Indeed, the set {x ∈ C[0, 1]; ∀s ∈ [0, 12 ] : x(s) = 0, ∀s ∈ ( 12 , 1] : x(s) ≤ 1} is bounded above by 𝟙 and does not have a supremum.

1.1.3 The Riesz decomposition property In a partially ordered vector space (X, K), for every v1 , v2 ∈ K the implication [0, v1 ] + [0, v2 ] ⊆ [0, v1 + v2 ]

1.1 Ordered vector spaces | 9

holds. The converse inclusion is not true, in general, and leads to the following definition. Definition 1.1.25. A partially ordered vector space (X, K) has the Riesz decomposition property if for every x, y1 , y2 ∈ K with x ≤ y1 + y2 there exist x1 , x2 ∈ K such that x = x1 + x2 and x1 ≤ y1 , x2 ≤ y2 . The following characterizations of the Riesz decomposition property² are due to F. Riesz [131]. For a more recent exposition, see [14, Secion 1.8]. Proposition 1.1.26. Given a partially ordered vector space (X, K), the following statements are equivalent. (i) X has the Riesz decomposition property. (ii) For every v1 , v2 ∈ K one has [0, v1 ] + [0, v2 ] = [0, v1 + v2 ] . (iii) For every x, y1 , . . . , y n ∈ K with x ≤ ∑ni=1 y i there exist x1 , . . . , x n ∈ K such that x = ∑ni=1 x i and x i ≤ y i for every i. (iv) For every four elements a1 , a2 , b 1 , b 2 ∈ X that satisfy the inequalities a i ≤ b j (i, j ∈ {1, 2}) there is an element c ∈ X such that ai ≤ c ≤ bj

(i, j ∈ {1, 2}) .

(v) Let m, n ∈ ℕ and a1 , . . . , a m , b 1 , . . . , b n ∈ X be such that for every i ∈ {1, . . . , m} and j ∈ {1, . . . , n} the inequality a i ≤ b j holds. Then there exists c ∈ X with a i ≤ c ≤ b j for every i and j. n (vi) Let x1 , . . . , x m , y1 , . . . , y n ∈ K be such that ∑m i=1 x i = ∑ j=1 y j . Then there exists for every i ∈ {1, . . . , m} and j ∈ {1, . . . , n} an element z ij ∈ K such that x i = ∑nj=1 z ij for every i and y j = ∑m i=1 z ij for every j. An ordered vector space with the Riesz decomposition property can be viewed as a generalization of a vector lattice, due to the next result. Proposition 1.1.27. Every vector lattice has the Riesz decomposition property. Proof. Let (X, K) be a Riesz space and let x, y1 , y2 ∈ K be such that x ≤ y1 + y2 . Take x1 := x ∧ y1 and x2 := x − x1 , then x1 and x2 have the desired properties. Next we provide an example of an ordered vector space X with the Riesz decomposition property, where X is not a vector lattice. This is in contrast to the finite-dimensional case, which we will discuss in Theorem 1.7.8 below. Note that X is constructed as a subspace of an Archimedean vector lattice, where X is the kernel of a functional.

2 In [161], the property in Proposition 1.1.26 (iv) is called the Riesz separation property, whereas in [14] the property in (v) is called the Riesz interpolation property.

10 | 1 A primer on ordered vector spaces

Example 1.1.28. This example is due to I. Namioka [124]. We consider the subspace X of C[−1, 1] consisting of all functions x with x(−1) + x(1) = x(0), endowed with the induced order. The space X is not a Riesz space, since the functions t 󳨃→ t and t 󳨃→ −t do not have a supremum in X. Nevertheless, X possesses the Riesz decomposition property. We show condition (iv) in Proposition 1.1.26. Let a1 , a2 , b 1 , b 2 ∈ X be such that the inequalities a i ≤ b j hold for every i, j ∈ {1, 2}. Let v ∈ C[−1, 1] be such that v is linear on [−1, 0] and on [0, 1] with v(−1) := max{a1 (−1), a2 (−1)}, v(1) := max{a1 (1), a2 (1)}, v(0) := v(−1) + v(1), i.e., v ∈ X. Define c : [−1, 1] → ℝ , t 󳨃→ min {b 1 (t), b 2 (t), max {a1 (t), a2 (t), v(t)}} . Clearly, c ∈ C[−1, 1] and a i ≤ c ≤ b j hold for every i, j ∈ {1, 2}. It remains to show that c ∈ X, i.e., that c(0) = c(−1) + c(1). We have c(1) = min {b 1 (1), b 2 (1), max {a1 (1), a2 (1), v(1)}} = min {b 1 (1), b 2 (1), v(1)} = v(1) and, similarly, c(−1) = v(−1). In order to show that c(0) = v(0), we observe that max {a1 (0), a2 (0)} = max{a1 (−1) + a1 (1), a2 (−1) + a2 (1)} ≤ max{a1 (−1), a2 (−1)} + max{a1 (1), a2 (1)} = v(0) ≤ min{b 1 (−1), b 2 (−1)} + min{b 1 (1), b 2 (1)} ≤ min{b 1 (−1) + b 1 (1), b 2 (−1) + b 2 (1)} = min{b 1 (0), b 2 (0)} and get c(0) = max{a1 (0), a2 (0), v(0)} ≤ v(0) ≤ min{b 1 (0), b 2 (0)} ≤ c(0). Since v ∈ X, we obtain c ∈ X. Example 1.1.29. We refer to Example 1.1.19, where it was established that C1 [0, 1] is not a Riesz space. The space C1 [0, 1] has the Riesz decomposition property, for details see [14, Example 1.58]. Similarly, C1 (ℝ) and C2 (ℝ) have the Riesz decomposition property, see [120, Theorem 1]. Interestingly, C k (ℝ) does not have the Riesz decomposition property for k ≥ 3, which is shown in [120, Example 1]. Here, Ck (ℝ) denotes the space of k-times continuously differentiable functions on ℝ (for k ∈ ℕ), endowed with pointwise order. Example 1.1.30. Let Pn [0, 1] be the subspace of ℝ[0,1] that consists of all polynomials of degree n at most and consider the induced order. The space P n [0, 1] does not have the Riesz decomposition property. For n = 2, consider the parabolas a : t 󳨃→ 1 − kt2 and b : t 󳨃→ kt2 , where we take the constant k ≥ 36. Define a1 := a ,

a2 : t 󳨃→ a(t − 23 ) ,

b 1 : t 󳨃→ b(t − 13 ) and P2 [0, 1]

b 2 : t 󳨃→ b(t − 1) .

Then a i ≤ b j for every i and j. Suppose there is c ∈ such that a i ≤ c ≤ b j for every i and j, so then c is not constant and has at least two critical points, which is a contradiction. For n > 2, one can use appropriately many similar a i and b j and apply Proposition 1.1.26 (v). The same argument works for P n (ℝ) (as a subspace of ℝℝ ).

1.1 Ordered vector spaces |

11

1.1.4 Order convergence Order convergence in a partially ordered vector space X is defined for nets. Recall that a net x in X is a mapping x : A → X, where on the index set A a reflexive, transitive, and directed relation ⊲ is given. As usual, we write (x α )α∈A . Definition 1.1.31. Let (X, K) be a partially ordered vector space. (i) A net (x α )α∈A in X is called decreasing (denoted by x α ↓, or, if the index is stressed, x α ↓ α ), whenever for every α, β ∈ A with α ⊲ β one has x α ≥ x β , and increasing if the net (−x α )α∈A is decreasing. For x ∈ X the notation x α ↓ x means that x α ↓ and inf { x α ; α ∈ A} = x. The meanings of x α ↑ and x α ↑ x are analogous. o (ii) A net (x α )α∈A in X order converges or o-converges to x ∈ X (denoted by x α → 󳨀 x), if there is a net (y α )α∈A in X such that y α ↓ 0 and for every α ∈ A one has ±(x α − x) ≤ y α . (iii) A subset M ⊆ X is called order closed or o-closed, if for each net (x α )α∈A in M which o-converges to x ∈ X one has that x ∈ M. Note that the intersection of o-closed sets is o-closed. We collect elementary properties of o-convergence. Proposition 1.1.32. Let (X, K) be a partially ordered vector space. (i) If a net (x α ) o-converges in X to both u and v, then u = v. o (ii) Let (x α )α∈A and (y β )β∈B be nets in X such that there are x, y ∈ X with x α → 󳨀 x and o

󳨀 y. Then the net (x α + y β )(α,β)∈A×B o-converges to x + y, where on A × B the yβ → o

󳨀 λx for every λ ∈ ℝ. entrywise relation is considered. Moreover, λx α → Proof. (i) From ±(x α − u) ≤ y α ↓ 0 and ±(x α − v) ≤ z α ↓ 0 it follows that ±(u − v) = ±(u − x α + x α − v) ≤ y α + z α ↓ 0 , hence u = v. (ii) The statement is a direct consequence of Proposition 1.1.15 (ii). The notion of o-convergence leads to a corresponding notion of o-continuity. Definition 1.1.33. Let (X1 , K1 ) and (X2 , K2 ) be ordered vector spaces. A map f : X1 → X2 is called order continuous or o-continuous if for every net (x α )α∈A in X1 and x ∈ X1 o o with x α → 󳨀 x it follows that f(x α ) → 󳨀 f(x). The lattice operations are o-continuous; see also [10, Lemma 7.14]. Proposition 1.1.34. Let (X, K) be a vector lattice and let (x α )α∈A and (y β )β∈B be nets in o

o

󳨀 x and y β → 󳨀 y. Then one has (with entrywise X such that there are x, y ∈ X with x α → o

o

o

o

󳨀 x ∨ y, x α ∧ y β → 󳨀 x ∧ y, x+α → 󳨀 x+ , x−α → 󳨀 x− and order relation on A × B) that x α ∨ y β → o

󳨀 |x|. |x α | →

12 | 1 A primer on ordered vector spaces Proof. If |x α − x| ≤ v α ↓ 0 and |y β − y| ≤ w α ↓ 0, then it follows that |x α ∨ y β − x ∨ y| = |x α ∨ y β − x α ∨ y + x α ∨ y − x ∨ y| ≤ |x α ∨ y β − x α ∨ y| + |x α ∨ y − x ∨ y| ≤ |y β − y| + |x α − x| ≤ w β + v α ↓ 0 . The other statements follow with the aid of (1.4) and Proposition 1.1.32. For linear maps, it is sufficient to consider o-continuity in 0. Proposition 1.1.35. Let (X1 , K1 ) and (X2 , K2 ) be ordered vector spaces and let f : X1 → X2 be a linear map. Then f is o-continuous if and only if for every net (x α )α∈A in X1 with o o xα → 󳨀 0 it follows that f(x α ) → 󳨀 0. A further discussion on properties of o-convergence, o-closedness, and o-continuity can be found in Subsection 3.7.1 below.

1.2 Operators on ordered vector spaces 1.2.1 Order-bounded and regular operators For nonempty sets X1 and X2 , a map j : X1 → X2 and subsets U ⊆ X1 , V ⊆ X2 , we denote the image³ of U with respect to j by j[U] := {j(x); x ∈ U} , and the pre-image of V with respect to j by [V]j := {x ∈ X1 ; j(x) ∈ V} . Now let X1 and X2 be vector spaces. As usual, L(X1 , X2 ) denotes the vector space of all linear operators from X1 into X2 . For L(X1 , X1 ) we write L(X1 ) for short. Definition 1.2.1. Let (X1 , K1 ) and (X2 , K2 ) be partially ordered vector spaces and S ∈ L(X1 , X2 ). The operator S is called – positive if for every x ∈ K1 one has Sx ∈ K2 , – bipositive (or an embedding) if for every x ∈ X1 one has x ∈ K1 if and only if Sx ∈ K2 , – an order isomorphism if it is surjective and bipositive, – order-bounded if S maps order-bounded subsets of X1 into order-bounded subsets of X2 , – regular if S equals the difference of two positive operators.

3 This notation is used according to [128].

1.2 Operators on ordered vector spaces | 13

Observe that bipositivity implies injectivity. The spaces X1 and X2 are called order isomorphic if there is an order isomorphism S : X1 → X2 . The set of all positive linear operators in L(X1 , X2 ) is denoted by L+ (X1 , X2 ). This set is a wedge. In the sense of Remark 1.1.3, we write S ≤ T if T − S is positive. Note that if T ∈ L+ (X1 , X2 ) and x, y ∈ X1 are such that x ≤ y, then Tx ≤ Ty, in other words, T is increasing (or monotone). Similarly, if S ≤ T and x ∈ K1 , then Sx ≤ Tx. Proposition 1.2.2. Let (X1 , K1 ) and (X2 , K2 ) be partially ordered vector spaces. The cone K1 is generating if and only if L+ (X1 , X2 ) is a cone (i.e., (L(X1 , X2 ), L+ (X1 , X2 )) is a partially ordered vector space). Proof. Let K1 be generating. If T, −T ∈ L+ (X1 , X2 ), then for every x ∈ K1 we have Tx, −Tx ∈ K2 , hence Tx = 0. Then T = 0. Conversely, assume that K1 is not generating, so for the linear subspace D := K1 − K1 there exists a subspace E ≠ {0} such that X1 = D ⊕ E. Let T ∈ L(X1 , X2 ) be the projection on E, then T, −T ∈ L+ (X1 , X2 ), and T ≠ 0. Note that if X1 has the trivial order as in Example 1.1.5 (i), then L+ (X1 , X2 ) = L(X1 , X2 ), which is a wedge, but not a cone. Clearly, every positive operator is regular, and every regular operator is orderbounded. The sets of all order-bounded or regular operators, respectively, are denoted by Lb (X1 , X2 ), Lr (X1 , X2 ), respectively, and their intersections with L+ (X1 , X2 ) by Lb+ (X1 , X2 ), Lr+ (X1 , X2 ), respectively. For a partially ordered vector space (X, K) we introduce the algebraic dual X ∗ := L(X, ℝ) ordered by the wedge K ∗ := L+ (X, ℝ) of all positive linear functionals (here the space ℝ is assumed to be equipped with the standard cone ℝ+ = {t ∈ ℝ; t ≥ 0}). A nonempty subset M ⊆ K ∗ is called total if x ∈ X and f(x) ≥ 0 for every f ∈ M imply x ∈ K. If K is generating, then K ∗ is a cone, i.e., (X ∗ , K ∗ ) is a partially ordered vector space. The order dual of (X, K) is the space X ∼ := Lb (X, ℝ) ordered by the wedge K ∼ := b L+ (X, ℝ). If K is generating, then by Proposition 1.2.2 the space (X ∼ , K ∼ ) is a partially ordered vector space. In general, K ∼ is not generating⁴ Example 1.2.3. We present a directed Archimedean partially ordered vector space (X, K) and an order-bounded linear functional on X that is not equal to the difference of two positive linear functionals⁵. Consider the following functions on [0, ∞), where for a set M ⊂ [0, ∞) the indicator function of M is denoted by 𝟙M : e n := 𝟙[n,n+1) ,

n ∈ ℕ ∪ {0} ,

4 An involved example is also given in [124, Example 6.10]. 5 This example is due to our dear friend and colleague Otto van Gaans, who published this result in the paper An elementary example of an order bound dual space that is not directed, Positivity 9(2), 2005, 265–267.

14 | 1 A primer on ordered vector spaces u n,k (t) := nt𝟙

1 (t) [0, n ]

+ 1k 𝟙

1 (t) {n+ k+1 }

,

t ∈ [0, ∞), n, k ∈ ℕ .

Let X be the subspace of B[0, ∞) consisting of all linear combinations of {e n }n and {u n,k }n,k ordered by the induced cone K. Every element x ∈ X is right differentiable at 0. Define f : X → ℝ, x 󳨃→ x󸀠 (0), where x󸀠 (0) denotes the right derivative of x at 0. Clearly, f ∈ X ∗ . We list properties of X and f . (i) The space X is directed and Archimedean. Indeed, every element of X is a bounded function and has a bounded support, so for every x, y ∈ X there is an n ∈ ℕ such that x, y ≤ n(e1 + ⋅ ⋅ ⋅ + e n ), hence X is directed. The space X is Archimedean, since it is a subspace of the Archimedean space ℝ[0,∞) . (ii) The functional f is order-bounded. Indeed, let a ∈ K. It has to be shown that f is bounded on [−a, a]. There are N ∈ ℕ and C ∈ (0, ∞) such that the support of a satisfies supp a ⊆ [0, N] and a ≤ C𝟙. Let x ∈ [−a, a]. Then also supp x ⊆ [0, N] and −C𝟙 ≤ x ≤ C𝟙. By definition of X, there are (λ n,k )n,k ∈ cc (ℕ2 ) and (μ n )n ∈ cc (ℕ ∪ {0}) such that x = ∑ λ n,k u n,k + ∑ μ n e n . n,k

n

To establish that the right derivative of x at 0 is bounded, we show that for every n ∈ ℕ≥N and k ∈ ℕ we have λ n,k = μ n = 0. Indeed, since x = 0 on (N, ∞), for every n ∈ ℕ≥N we obtain 0 = μ n e n (n + 34 ) = μ n . Moreover, for every k ∈ ℕ we λ

1 ) = n,k have 0 = λ n,k u n,k (n + k+1 k , hence λ n,k = 0. It follows that x is linear on [0, N1 ]. Because of −C𝟙 ≤ x ≤ C𝟙, this yields that |f(x)| = |x󸀠 (0)| ≤ 2NC. Hence, |f(x)| ≤ 2NC for every x ∈ [−a, a]. This establishes that f is order-bounded. (iii) There is no positive linear functional g on X with g ≥ f , hence f is not the difference of two positive linear functionals. For a proof, suppose that there exists such a functional g. For every n ∈ ℕ, let the number k n ∈ ℕ ∪ {0} be such that k n > g(e n ). Then k1n g(e n ) < 1 and u n,k n ≤ e0 + k1n e n , thus

g(e0 ) = g (e0 +

1 1 k n e n ) − k n g(e n )

≥ g(u n,k n ) − 1 ≥ f(u n,k n ) − 1 = n − 1

for every n, which is a contradiction. Thus, such a g does not exist, and therefore f is not the difference of two positive linear functionals. Proposition 1.2.4. Let (X1 , K1 ) and (X2 , K2 ) be partially ordered vector spaces such that K1 is generating and X2 is Archimedean. Then (L(X1 , X2 ), L+ (X1 , X2 )) is an Archimedean partially ordered vector space. Proof. By Proposition 1.2.2, the space (L(X1 , X2 ), L+ (X1 , X2 )) is a partially ordered vector space. To show that it is Archimedean, let S, T ∈ L(X1 , X2 ) be such that for every n ∈ ℕ we have that nS ≤ T. Then for every x ∈ K1 we obtain nSx ≤ Tx. Since X2 is Archimedean, it follows that Sx ≤ 0. Thus, S ≤ 0. Since ℝ with standard order is Archimedean, for every directed ordered vector space X the algebraic dual X ∗ and the order dual X ∼ are Archimedean.

1.2 Operators on ordered vector spaces | 15

1.2.2 The Riesz–Kantorovich formulas In this section we study conditions on partially ordered vector spaces (X1 , K1 ) and (X2 , K2 ) such that the space Lb (X1 , X2 ) of all order-bounded operators turns out to be a Riesz space. The following theorems are due to L. V. Kantorovich [89]. Theorem 1.2.5. Let (X1 , K1 ) and (X2 , K2 ) be partially ordered vector spaces such that K1 is generating. Let S : K1 → X2 be additive and positively homogeneous⁶. Then there exists a unique linear operator T : X1 → X2 such that T = S on K1 . Moreover, if S[K1 ] ⊆ K2 , then T is positive. Proof. First observe that for u, v, x, y ∈ K1 with v − u = y − x we have that Sv − Su = Sy − Sx. Indeed, from v + x = u + y it follows that Sv + Sx = S(v + x) = S(u + y) = Su + Sy. For x ∈ X1 there are u, v ∈ K1 such that x = u − v. Define Tx := Su − Sv and note that the definition is independent of the choice of u and v. T is additive. Indeed, let x, y ∈ X1 be such that x = v − u and y = z − w with u, v, w, z ∈ K1 . Since S(v) + S(z) + S(u + w) = S(v + z) + S(u) + S(w), we have T(x + y) = T(v − u + z − w) = S(v + z) − S(u + w) = S(v) − S(u) + S(z) − S(w) = T(v − u) + T(z − w) = T(x) + T(y) . For x ∈ X 1 and λ ∈ [0, ∞) we have T(λx) = λT(x). For a proof, let u, v ∈ K1 be such that x = v − u. Then T(λx) = S(λv) − S(λu) = λ(S(v) − S(u)) = λT(x). With the same x, u, and v, T(−x) = S(u) − S(v) = −T(x). It follows that T is linear. Moreover, T is unique. The next theorem contains the crucial conditions under which the space Lb (X1 , X2 ) is a Riesz space. Theorem 1.2.6. Let (X1 , K1 ) be a partially ordered vector space such that K1 is generating, let X1 have the Riesz decomposition property, and let (X2 , K2 ) be a Dedekind complete Riesz space. For T ∈ Lb (X1 , X2 ) and x ∈ K1 define S(x) := sup{T(u); u ∈ [0, x]} . Then there exists a unique linear operator R ∈ L+ (X1 , X2 ) such that R = S on K1 . Moreover, the supremum of T and 0 exists in Lb (X1 , X2 ) and equals R. Proof. As T is order-bounded and X2 is Dedekind complete, S is well-defined. Moreover, S[K1 ] ⊆ K2 .

6 Let X and Y be vector spaces, M ⊆ X and A : M → Y a map. The map A is called additive if for every x, y ∈ M with x + y ∈ M we have A(x + y) = A(x) + A(y), and called positively homogeneous if for every x ∈ M and λ ∈ [0, ∞) with λx ∈ M it holds that A(λx) = λA(x).

16 | 1 A primer on ordered vector spaces To show that S is additive, let x, y ∈ K1 . For u ∈ [0, x] and v ∈ [0, y] we have u + v ∈ [0, x + y] and T(u + v) = T(u) + T(v), hence S(x + y) ≥ T(u) + T(v). Then, by taking the supremum over all u, we have S(x+y) ≥ S(x)+T(v). Similarly, the supremum over v yields S(x + y) ≥ S(x) + S(y). Next, for w ∈ [0, x + y] the Riesz decomposition property of X 1 provides us with u ∈ [0, x] and v ∈ [0, y] such that w = u + v. Then T(w) = T(u) + T(v) ≤ S(x) + S(y). The supremum over w results in S(x + y) ≤ S(x) + S(y). Since the multiplication by positive scalars and the supremum can be interchanged, it is straightforward that S is positively homogeneous. According to Theorem 1.2.5, there exists R ∈ L+ (X1 , X2 ) such that R = S on K1 . Now we show that R is the supremum of T and 0. Indeed, for x ∈ K1 we have R(x) = S(x) ≥ T(x), hence R is an upper bound of T and 0. Let Q ∈ L+ (X1 , X2 ) be an upper bound of T. Then for x ∈ K1 and u ∈ [0, x] we have Q(x) ≥ Q(u) ≥ T(u), so that Q(x) ≥ S(x) = R(x), thus Q ≥ R. Hence, R = T ∨ 0. In fact, Theorem 1.2.6 yields the positive part T + := R of T, hence Lb (X1 , X2 ) is a Riesz space. There are similar formulas for the other lattice operations. Corollary 1.2.7. Under the conditions of Theorem 1.2.6, the space Lb (X1 , X2 ) is a Riesz space, where for every x ∈ K1 and S, T ∈ Lb (X1 , X2 ) we have T − (x) = sup{−T(u); u ∈ [0, x]} , |T|(x) = sup{|T(u)|; u ∈ [−x, x]} , (S ∨ T)(x) = sup{S(x − u) + T(u); u ∈ [0, x]} , (S ∧ T)(x) = inf{S(x − u) + T(u); u ∈ [0, x]} . The formulas in Theorem 1.2.6 and Corollary 1.2.7 are called the Riesz–Kantorovich formulas. Theorem 1.2.8. Let (X 1 , K1 ) be a partially ordered vector space such that K1 is generating, X1 has the Riesz decomposition property and let (X2 , K2 ) be a Dedekind complete Riesz space. Then Lb (X1 , X2 ) is a Dedekind complete Riesz space. Proof. It remains to show that Lb (X1 , X2 ) is Dedekind complete. Let A be a nonempty subset of Lb (X1 , X2 ) that is bounded above. Let U be an upper bound of A. Denote by B the set of all suprema of finite nonempty subsets of A. Note that U is also an upper bound of B. For x ∈ K1 define S(x) := sup{T(x); T ∈ B} . To show that S is additive, let x, y ∈ K1 . For every T ∈ B we have T(x+y) = T(x)+T(y) ≤ S(x) + S(y), hence S(x + y) ≤ S(x) + S(y). Conversely, for every T, R ∈ B we have T ∨ R ∈ B, hence S(x + y) ≥ (T ∨ R)(x + y) = (T ∨ R)(x) + (T ∨ R)(y) ≥ T(x) + R(y). By taking supremum first over T and then over R we obtain S(x + y) ≥ S(x) + S(y). We conclude that S is additive. It is simple to check that S is positively homogeneous.

1.3 Ideals and bands in vector lattices | 17

According to Theorem 1.2.5 there exists a unique operator R ∈ L(X1 , X2 ) with R = S on K1 . From the definition of S it is clear that R is an upper bound of B, and hence of A. As A is nonempty, there is T ∈ A such that T ≤ R. Moreover, R ≤ U, hence R ∈ Lb (X1 , X2 ). As U is an arbitrary upper bound of A, it follows that R is the supremum of A. Since ℝ is a Dedekind complete Riesz space, we arrive at the following statement on the order dual, which is due to F. Riesz [130]. Corollary 1.2.9. If (X, K) is a partially ordered vector space such that K is generating and X has the Riesz decomposition property, then the order dual X ∼ is a Dedekind complete Riesz space. For the Archimedean directed partially ordered vector space X in Example 1.2.3 the order dual is not directed, hence X does not have the Riesz decomposition property.

1.3 Ideals and bands in vector lattices 1.3.1 Definitions and basic properties Next we define disjointness, ideals, and bands in vector lattices, as usual. In Chapter 4, these notions will be defined and studied in the more general setting of ordered vector spaces. Definition 1.3.1. Let (X, K) be a vector lattice. (i) Two elements x, y ∈ X are called disjoint, denoted by x ⊥ y, if |x| ∧ |y| = 0. The disjoint complement of a set M ⊆ X is the set M d = {x ∈ X; ∀m ∈ M : x ⊥ m} . (ii) A set M ⊆ X is called solid if for every x ∈ X and v ∈ M with |x| ≤ |v| it follows that x ∈ M. (iii) A set I ⊆ X is called an ideal if I is a solid linear subspace of X. Observe that the intersection of solid sets is solid. For a nonempty set M ⊆ X, define IM := ⋂ {I ⊆ X; I is an ideal, M ⊆ I} . The set IM is the smallest ideal in X that contains M. We call IM the ideal generated by M. Every ideal in X is a Riesz subspace of X. As an example of a Riesz subspace of X that is not an ideal, consider the space of all piecewise affine functions in C[0, 1] with the induced order. Further note that the intersection of a solid set in X with a Riesz subspace D of X is solid in D. As a motivation for Definitions 4.1.1 and 4.1.7 below, we state equivalent conditions for disjointness.

18 | 1 A primer on ordered vector spaces Proposition 1.3.2. Let (X, K) be a vector lattice and let x, y ∈ X. The following conditions are equivalent. (i) x ⊥ y. (ii) |x + y| = |x − y|. (iii) {x + y, −x − y}u = {x − y, −x + y}u . If, in addition, x, y ∈ K, then each of the conditions (i), (ii), (iii) is equivalent to (iv) [0, x] ∩ [0, y] = {0}. Proof. Proposition 1.1.16 (vi) yields the equivalence of (i) and (ii). The condition in (iii) is a reformulation of (ii). For x, y ∈ K, the equivalence of (i) and (iv) is straightforward. Let us investigate properties of disjoint complements. Lemma 1.3.3. Let X be a vector lattice and M ⊆ X a solid set. Then the following statements are equivalent. (i) M is o-closed. (ii) For every net (x α )α∈A in M and x ∈ X with 0 ≤ x α ↑ x it follows that x ∈ M. Proof. The implication (i) ⇒ (ii) is clear. To show (ii) ⇒ (i), let (z α )α∈A be a net in M o and z ∈ X with z α → 󳨀 z, i.e., there is a net (y α )α∈A in X such that ±(z − z α ) ≤ y α ↓ 0. Due to | |z| − |z α | | ≤ |z − z α | ≤ y α we have 0 ≤ (|z| − y α )+ ≤ |z α |, i.e., for every α ∈ A we obtain x α := (|z| − y α )+ ∈ M, since M is solid. Moreover, x α ↑ |z|, and by the assumption we conclude |z| ∈ M and, hence, z ∈ M. Proposition 1.3.4. Let X be a vector lattice and M ⊆ X. Then M d is an o-closed ideal. Proof. To show that M d is solid, let y ∈ M d and x ∈ X be such that |x| ≤ |y|. Then for every m ∈ M we have 0 ≤ |m| ∧ |x| ≤ |m| ∧ |y| = 0, hence m ⊥ x. Thus, x ∈ M d . To see that M d is a vector subspace, observe that for x, y ∈ M d and m ∈ M one has |x + y| ∧ |m| ≤ (|x| + |y|) ∧ |m| ≤ |x| ∧ |m| + |y| ∧ |m| = 0, hence x + y ∈ M d . As M d is solid, we can use Lemma 1.3.3. Consider a net (x α )α∈A in M d and x ∈ X such that 0 ≤ x α ↑ x. Let m ∈ M. For every α ∈ A we have |m| ∧ x α = 0, therefore sup{|m| ∧ x α ; α ∈ A} = 0. Hence, |m| ∧ x = |m| ∧ sup{x α ; α ∈ A} = sup{|m| ∧ x α ; α ∈ A} = 0. We conclude x ∈ M d . Definition 1.3.5. Let X be a vector lattice. A set B ⊆ X is called a band if B is an o-closed ideal in X. Since the intersection of bands is a band, we define for a nonempty set M ⊆ X the set BM := ⋂ {B ⊆ X; B is a band, M ⊆ B} , which is the smallest band in X that contains M. We call BM the band generated by M. By Proposition 1.3.4, for a set M ⊆ X the set M d is a band in X. The next discussion will show that in Archimedean vector lattices disjoint complements and bands coin-

1.3 Ideals and bands in vector lattices | 19

cide. We start with a preliminary proposition. The statement in (iii) can be found, e.g., in [12, Theorem 3.4], the other statements are straightforward. Proposition 1.3.6. Let X be a vector lattice and let M ⊆ X be nonempty. (i) The ideal I M generated by M satisfies n

IM = {x ∈ X; ∃x1 , . . . , x n ∈ M and λ1 , . . . , λ n ∈ ℝ+ with |x| ≤ ∑ λ i |x i |} . i=1

(ii) We have M d = (IM )d . (iii) If I is an ideal in X, then the band B I generated by I satisfies BI = {x ∈ X; ∃ a net (x α )α∈A in I with 0 ≤ x α ↑ |x|} . (iv) The band generated by M equals the band generated by the ideal generated by M, i.e., BM = BIM . Proposition 1.3.7. Let I be an ideal in X. (i) For every x ∈ I dd with x > 0 there is y ∈ I such that 0 < y ≤ x. (ii) If X is, in addition, Archimedean, then for every x ∈ I dd there is a net (x α )α∈A in I such that 0 ≤ x α ↑ |x|. Proof. (i) Let x ∈ I dd with x > 0. If for every z ∈ I one would have |z| ∧ x = 0, then x ∈ I d ∩ I dd = {0}, which is a contradiction. Therefore there is z ∈ I with x ≥ y := |z| ∧ x > 0. (ii) Let x ∈ I dd . It suffices to consider the case |x| > 0. Due to (i), the set A := {y ∈ I; 0 < y ≤ |x|} is nonempty, moreover A is directed. We consider the net (x α )α∈A with x α := α. Then 0 ≤ x α ↑, and |x| ∈ I dd is an upper bound of the set {x α ; α ∈ A} = A. It remains to show that sup A = |x|. Let z ∈ X be an upper bound of A, then z ∧ |x| ∈ I dd is an upper bound of A. Assume that z ∧ |x| < |x|. Due to (i), there is u ∈ I with 0 < u ≤ |x| − z ∧ |x| ≤ |x|, so u ∈ A and hence, u ≤ z ∧ |x|. Therefore 0 < 2u = u + u ≤ |x| − z ∧ |x| + z ∧ |x| = |x|. Inductively, for every n ∈ ℕ we get 0 < nu ≤ |x|, and as X is Archimedean, we obtain u ≤ 0, which is a contradiction. We conclude z ∧ |x| = |x|, i.e., |x| ≤ z. Consequently, sup A = |x|. For the following result, see also [10, Theorem 7.18]. Theorem 1.3.8. In an Archimedean vector lattice (X, K) the following statements hold. (i) The band BM generated by a nonempty set M ⊆ X equals the set M dd . (ii) A nonempty set B ⊆ X is a band in X if and only if B = Bdd . Proof. (i) Since M dd is a band, we obtain BM ⊆ M dd . Conversely, by Proposition 1.3.6 (ii) we have M dd = (I M )dd . Proposition 1.3.7 (ii) in combination with Proposition 1.3.6 (iii) yields that (IM )dd ⊆ BIM . Since the latter set equals BM due to Proposition 1.3.6 (iv), we obtain M dd ⊆ BM .

20 | 1 A primer on ordered vector spaces (ii) If B is a band in X, then B equals the band generated by B, and (i) implies B = Bdd . If B = Bdd , then by Proposition 1.3.4 the set B is a band.

1.3.2 Ideals and bands in C(Ω) An important example of a vector lattice is the space C(Ω), where Ω is a compact Hausdorff space. Typically, properties of Ω characterize properties of C(Ω). For example, C(Ω) is Dedekind complete if and only if Ω is extremely disconnected⁷, see [13, Theorem 1.50]. In this subsection, we present results concerning ideals and bands in C(Ω), which can also be found in [166, Example 9.4]. In particular, we characterize bands by means of certain subsets of Ω. Observe that disjointness in C(Ω) can easily be checked, since x, y ∈ C(Ω) are disjoint if and only if for every ω ∈ Ω one has x(ω) = 0 or y(ω) = 0. For a set S ⊆ C(Ω) the carrier of S is defined by car(S) = {ω ∈ Ω; ∃s ∈ S : s(ω) ≠ 0} .

(1.7)

The carrier of an ideal in C(Ω) is an open subset of Ω. Different ideals in C(Ω) may have the same carrier; see Example 1.3.9 below. For an open set O ⊆ Ω the set I O := {s ∈ C(Ω); ∀ω ∈ Ω \ O : s(ω) = 0}

(1.8)

is the largest ideal having O as its carrier. If P is an open subset of Ω, then I O = I P if and only if O = P. Example 1.3.9. Consider Ω = [0, 1], x : [0, 1] → ℝ, t 󳨃→ t, and let I be the ideal in C[0, 1] generated by {x}. For O := (0, 1], the ideal I O = {x ∈ C[0, 1]; x(0) = 0} has the same carrier as I, but I O ≠ I. Note that I d = I dO = {0}, hence I dd = I dd O = C[0, 1]. Theorem 1.3.8 (ii) yields that I and I O are not bands in C[0, 1]. In contrast to ideals, for every band B in C(Ω) there is an open subset O of Ω such that B = I O , as follows from the next statement. This means that a band is determined whenever its carrier is known. Proposition 1.3.10. (i) For every set S ⊆ C(Ω) we have S ⊆ Icar(S) ⊆ Sdd . (ii) For every band B in C(Ω) we have B = Icar(B) .

7 A topological space is called extremely disconnected if the closure of every open set in it is open.

1.3 Ideals and bands in vector lattices

| 21

Proof. (i) To show that S ⊆ Icar(S) , let s ∈ S. For every ω ∈ Ω \ car(S) we have s(ω) = 0, so s ∈ Icar(S) . Next we show that Icar(S) ⊆ Sdd . Let x ∈ Icar(S) and let y ∈ Sd . For every ω ∈ car(S) there exists s ∈ S with s(ω) ≠ 0, hence y(ω) = 0. For every ω ∈ Ω \ car(S) we have that x(ω) = 0, since x ∈ Icar(S) . Hence x ⊥ y and therefore x ∈ Sdd . (ii) By Theorem 1.3.8 (ii), we have B = Bdd , so the statement follows directly from (i).

We intend to specify those open subsets O of Ω for which I O turns out to be a band. Definition 1.3.11. An open set O ⊆ Ω is called regularly open if it equals the interior of its closure, i.e., O = int(O). Lemma 1.3.12. For every regularly open set O ⊆ Ω one has Ω \ O = Ω \ O. Proof. In order to show that Ω \ O ⊆ Ω \ O, note that O ⊆ O, so Ω \ O ⊇ Ω \ O, hence Ω \ O ⊇ Ω \ O. For a proof of Ω \ O ⊇ Ω\O, denote C := Ω \ O. Then C is a closed set and C ⊇ Ω\O, so Ω \ C ⊆ O. As Ω \ C is open and O is regularly open, we obtain Ω \ C ⊆ int(O) = O, which yields that C ⊇ Ω \ O. Thus, Ω \ O ⊇ Ω \ O. Bands in C(Ω) are characterized as follows. Proposition 1.3.13. For an open set O ⊆ Ω the ideal I O is a band if and only if O is regularly open. Proof. Let O ⊆ Ω be an open set and let I O be the ideal as in (1.8). dd Assume that O is regularly open. Since I dd O ⊇ I O , it suffices to show that I O ⊆ I O . dd Let x ∈ I O ; we intend to show that x ∈ I O , i.e., we have to show that x = 0 on Ω \ O. First, we establish x = 0 on Ω \ O. Indeed, let ω ∈ Ω \ O, then by Urysohn’s lemma 1.8.9 below there is a y ∈ C(Ω) such that y(ω) = 1 and y = 0 on O. Then y ∈ I dO and therefore x ⊥ y. It follows that x(ω) = 0 for every ω ∈ Ω \ O. We conclude that x = 0 on Ω \ O, so x = 0 on Ω \ O, due to Lemma 1.3.12. Hence, I O is a band. Now, conversely, assume that I O is a band. In order to show that O is regularly open, it suffices to show that O ⊇ int(O). We start with two preliminary observations. (a) For every y ∈ I dO we have y = 0 on O. Indeed, let ω ∈ O, then {ω} and Ω \ O are disjoint closed sets, so by Urysohn’s lemma there is x ∈ C(Ω) such that x(ω) = 1 and x = 0 on Ω \ O. Then x ∈ I O , so x ⊥ y, hence y(ω) = 0. Therefore y = 0 on O and thus y = 0 on O by continuity. d (b) For every x ∈ C(Ω) with x = 0 on Ω \ O we have that x ∈ I dd O . Indeed, let y ∈ I O , then y = 0 on O by (a), hence x ⊥ y. Therefore x ∈ I dd O .

22 | 1 A primer on ordered vector spaces Now let ω ∈ int(O). Then {ω} and Ω \ int(O) are disjoint closed sets, so by Urysohn’s lemma there is x ∈ C(Ω) such that x(ω) = 1 and x = 0 on Ω \ int(O). Since Ω \ O ⊆ Ω \ int(O), we obtain x = 0 on Ω \ O. Hence, by (b), we infer x ∈ I dd O . By assumption, I dd = I , so x ∈ I . Hence x = 0 on Ω \ O. As x(ω) = 1 we get ω ∈ O. We conclude that O O O int(O) ⊆ O. We conclude the discussion by characterizing the ideals and bands in ℝk with standard order. Example 1.3.14. We show that for every ideal J in ℝk there is a set N ⊆ {1, . . . , k} such that J = {(y1 , . . . , y k ); ∀i ∈ N : y i = 0} , (1.9) hence J is a band. Note that ℝk = C(Ω), where Ω = {1, . . . , k} endowed with the discrete topology. Let J be an ideal in ℝk . Define N := Ω \ car(J). Clearly, J ⊆ {(y1 , . . . , y k ); ∀i ∈ N : y i = 0}. Conversely, let y ∈ {(y1 , . . . , y k ); ∀i ∈ N : y i = 0} (i) with y ≠ 0. For every i ∈ Ω \ N there is x(i) ∈ J such that x i ≠ 0. Set x := ∑i∈Ω\N |x(i) |. i ; i∈Ω\N} Then |y| ≤ λx for λ := min{x max{|y i |; i∈Ω} . Therefore y ∈ J. In conclusion, ideals and bands in ℝk coincide and are of the form (1.9).

1.4 Riesz homomorphisms and disjointness preserving operators in vector lattices In this section, let (X1 , K1 ) and (X2 , K2 ) be vector lattices. We discuss operators that preserve the lattice structure or other structures as, e.g., disjointness. Definition 1.4.1. An operator S ∈ L(X1 , X2 ) is called 1. disjointness preserving if for every x, y ∈ X1 with x ⊥ y it follows that Sx ⊥ Sy, 2. a Riesz homomorphism or lattice homomorphism if for every x, y ∈ X1 one has S(x ∨ y) = (Sx) ∨ (Sy), 3. a Riesz isomorphism if S is a bijective Riesz homomorphism. The spaces X1 and X2 are called Riesz isomorphic if there is a Riesz isomorphism S : X1 → X2 . Note that the inverse of a Riesz isomorphism is a Riesz isomorphism too, which will be discussed in Theorem 5.2.1 below. Lemma 1.4.2. For T ∈ L(X1 , X2 ) the following statements are equivalent. (i) T is a Riesz homomorphism. (ii) For every x ∈ X1 we have T(x+ ) = (Tx)+ . (iii) For every x, y ∈ X1 we have T(x ∧ y) = T(x) ∧ T(y). (iv) If for x, y ∈ X1 we have that x ∧ y = 0, then T(x) ∧ T(y) = 0. (v) For every x ∈ X1 we have T(|x|) = |T(x)|.

1.4 Riesz homomorphisms and disjointness preserving operators in vector lattices |

23

Proof. To establish (ii) ⇒ (iii), first observe that for x, y ∈ X1 from the formula x + y = x ∧ y + x ∨ y it follows that x = x ∨ y − y + x ∧ y = (x − y) ∨ (y − y) + x ∧ y = (x − y) ∨ 0 + x ∧ y = (x − y)+ + x ∧ y . Therefore, T(x ∧ y) = T(x − (x − y)+ ) = T(x) − T(x − y)+ = T(x) − (Tx − Ty)+ = T(x) ∧ T(y) . To show (iv) ⇒ (v), observe that |T(x)| = |T(x+ ) − T(x− )| = T(x+ ) ∨ T(x− ) − T(x+ ) ∧ T(x− ) = T(x+ ) ∨ T(x− ) = T(x+ ) + T(x− ) = T(x+ + x− ) = T(|x|) . For the proof (v) ⇒ (i) use formula (1.5). The remaining implications are straightforward. Riesz homomorphisms and disjointness preserving operators play an important role in the sequel. In vector lattices, they are related as follows. Proposition 1.4.3. T ∈ L(X1 , X2 ) is a Riesz homomorphism if and only if T is a positive disjointness preserving operator. Proof. Let T ∈ L(X1 , X2 ) be a Riesz homomorphism. Then for x ∈ K1 we have T(x) = T(x ∨ 0) = T(x) ∨ T(0) = T(x) ∨ 0 ≥ 0, i.e T is positive. Let x, y ∈ X 1 be such that |x| ∧ |y| = 0, and then due to Lemma 1.4.2 we obtain |Tx| ∧ |Ty| = T(|x|) ∧ T(|y|) = 0, hence T is disjointness preserving. Conversely, if T ∈ L(X1 , X2 ) is a positive disjointness preserving operator, then condition (iv) in Lemma 1.4.2 is satisfied, i.e., T is a Riesz homomorphism. We collect some further properties of Riesz homomorphisms. Proposition 1.4.4. (i) For a Riesz homomorphism T ∈ L(X1 , X2 ), the kernel Ker(T) = {x ∈ X1 ; T(x) = 0} is an ideal in X1 . (ii) If for T ∈ L+ (X1 , X2 ) the kernel Ker(T) is a Riesz subspace of X1 , then Ker(T) is an ideal. Proof. (i) Using Lemma 1.4.2, we obtain from Tx = 0 and |y| ≤ |x| that |Ty| = T|y| ≤ T|x| = |Tx| = 0. Hence, Ty = 0. (ii) Let Tx = 0 and |y| ≤ |x|. First observe that T|x| = T(x ∨ (−x)) = 0, since Ker(T) is a Riesz subspace. As T is positive, −|x| ≤ y ≤ |x| implies −T|x| ≤ Ty ≤ T|x|, therefore Ty = 0.

24 | 1 A primer on ordered vector spaces

We consider o-continuity of linear maps as discussed in Proposition 1.1.35. Proposition 1.4.5. Let S ∈ L(X1 , X2 ). Then S is an o-continuous Riesz homomorphism if and only if for every nonempty set A ⊆ X1 with inf A = 0 one has inf S[A] = 0. Proof. Let S be an o-continuous Riesz homomorphism and let A ⊆ X 1 be such that inf A = 0. As S is positive, the element 0 is a lower bound of S[A]. Define B := {inf F; ⌀ ≠ F ⊆ A, F is finite} and endow B with the reversed order. Then B is directed. The net (x b )b∈B with x b := b is decreasing and o-converges to 0. Thus the net (Sx b )b∈B is decreasing, and as S is o-continuous, we obtain that (Sx b )b∈B o-converges to 0. Hence inf S[B] = 0. Let v be a lower bound of S[A]. As S is a Riesz homomorphism, we have v ≤ Sx b for every b ∈ B. Thus v ≤ 0. We conclude that inf S[A] = 0. Conversely, it is straightforward that S is positive and disjointness preserving. By Proposition 1.4.3 we obtain that S is a Riesz homomorphism. Let (x α )α∈I be a net in X1 that o-converges to 0. Then there exists a net (y α )α∈I in X1 , which decreases to 0 such that |x α | ≤ y α . Since S is a Riesz homomorphism, we obtain |Sx α | ≤ Sy α . Let A := {y α ; α ∈ I}. As inf A = 0, we have inf S[A] = 0. Thus the net (Sy α )α∈I decreases to 0, therefore the net (Sx α )α∈I converges to 0. Next we mention some results on Riesz homomorphisms on spaces of continuous functions. For a compact Hausdorff space Ω, we endow the space C(Ω) with the supremum norm. The subsequent statement can be found in [2, Theorem 4.25]. Proposition 1.4.6. Let Ω1 and Ω2 be compact Hausdorff spaces and let α : Ω2 → Ω1 be a homeomorphism. Then A : C(Ω1 ) → C(Ω2 ), f 󳨃→ f ∘ α, is an isometric Riesz isomorphism. The next result states that the real valued Riesz homomorphisms on a space of continuous functions are precisely the multiples of point evaluations; see, e.g., [12, Theorem 7.21]. Theorem 1.4.7. Let Ω be a compact Hausdorff space and let h : C(Ω) → ℝ be a linear map. The map h is a Riesz homomorphism if and only if there exist ω ∈ Ω and α ∈ ℝ+ such that for every x ∈ C(Ω) one has h(x) = αx(ω) . Theorem 1.4.7 can be used to prove the next result; see also [43, Corollary 12.4]. Theorem 1.4.8. Let Ω1 and Ω2 be nonempty compact Hausdorff spaces. If C(Ω1 ) and C(Ω2 ) are isomorphic as Riesz spaces, then Ω1 and Ω2 are homeomorphic. Let us turn our attention to the set of all disjointness preserving operators, which we denote by Ldp (X1 , X2 ). This set is not a vector space, in general.

1.4 Riesz homomorphisms and disjointness preserving operators in vector lattices |

25

Example 1.4.9. Let S, T : C[0, 1] → C[0, 1] be defined by S : x 󳨃→ x(0)𝟙

and

T : x 󳨃→ x(1)𝟙 .

Then S and T are Riesz homomorphisms, but the operator T + S is not disjointness preserving. To show that an operator is disjointness preserving, it suffices to check the condition on positive elements. Lemma 1.4.10. Let T ∈ L(X1 , X2 ). If for every x, y ∈ K1 with x ⊥ y we have Tx ⊥ Ty, then T is disjointness preserving. Proof. Let x, y ∈ X1 with x ⊥ y. Since {y}d is an ideal and, hence, a Riesz subspace, we have x+ ⊥ y and x− ⊥ y. Similarly, x+ ⊥ y+ and x− ⊥ y+ . By assumption we obtain Tx+ ⊥ Ty+ and Tx− ⊥ Ty+ , and therefore Tx = Tx+ − Tx− ⊥ Ty+ . Analogously, Tx ⊥ Ty− , hence Tx ⊥ Ty+ − Ty− = Ty. The set of disjointness preserving operators has the following structure. Theorem 1.4.11. Let X1 and X2 be vector lattices and let X2 be Dedekind complete. (i) If T ∈ Ldp (X1 , X2 ) ∩ Lr (X1 , X2 ), then |T| ∈ Ldp (X1 , X2 ). In particular, |T| is a Riesz homomorphism. (ii) The set Ldp (X1 , X2 ) ∩ Lr (X1 , X2 ) is solid in Lr (X1 , X2 ). Proof. As X2 is Dedekind complete, due to Theorem 1.2.8 the space Lb (X1 , X2 ) = Lr (X1 , X2 ) is a vector lattice, where the lattice operations are given by the Riesz– Kantorovich formulas. (i) Let T ∈ Ldp (X1 , X2 ) ∩ Lr (X1 , X2 ). Due to Lemma 1.4.10 it remains to show that |T| is disjointness preserving on K1 . Let x, y ∈ K1 be such that x ⊥ y. Recall that |T|(x) = sup{|Tu|; u ∈ [−x, x]}. Let v ∈ [−y, y]. For every u ∈ [−x, x] it follows that 0 ≤ |u| ∧ |v| ≤ x ∧ y = 0, hence |Tu| ∧ |Tv| = 0, as T is disjointness preserving. Thus, |T|(x) ∧ |Tv| = sup {|Tu| ∧ |Tv|; u ∈ [−x, x]} = 0 . We conclude that |T|(x) ∧ |T|(y) = |T|(x) ∧ sup{|Tv|; v ∈ [−y, y]} = 0. (ii) Let S ∈ Lr (X1 , X2 ) and T ∈ Ldp (X1 , X2 ) ∩ Lr (X1 , X2 ) be such that |S| ≤ |T|. To show that S is disjointness preserving, we use Lemma 1.4.10. Let x, y ∈ K1 with x ⊥ y. Using the Riesz–Kantorovich formula we have 0 ≤ |Sx| ∧ |Sy| ≤ |S|(x) ∧ |S|(y) ≤ |T|(x) ∧ |T|(y) = 0 , where the last equality is due to (i). In Archimedean vector lattices, order-bounded disjointness preserving operators have surprisingly strong properties. For example, Meyer showed in [118] the following result: If (X1 , K1 ) and (X2 , K2 ) are Archimedean vector lattices and S ∈ Lb (X1 , X2 ) ∩

26 | 1 A primer on ordered vector spaces Ldp (X1 , X2 ), then one has for every x, y ∈ K1 that (Sx)+ ∧ (Sy)− = 0 . This is used to prove the next statement, which can be found, e.g., in [12, Theorem 8.6]. Theorem 1.4.12. Let (X1 , K1 ) and (X2 , K2 ) be Archimedean vector lattices and let S ∈ Lb (X1 , X2 ) ∩ Ldp (X1 , X2 ). Then the supremum of S and −S exists in (L(X1 , X2 ), L+ (X1 , X2 )), i.e., S has a modulus |S|. In particular, S is regular. Moreover, for every x ∈ X1 we have |S|(|x|) = |S(x)|. Note that the codomain X2 in Theorem 1.4.12 is not assumed to be Dedekind complete. Nevertheless, for every x ∈ K1 we have |S|(x) = sup {|S(v)|; v ∈ [−x, x]} , which is the same as the Riesz–Kantorovich formula for the modulus in Corollary 1.2.7. This result relates to the following more general questions: For which operators with a modulus does the modulus satisfy the Riesz–Kantorovich formula? Under which conditions on the underlying spaces is this true for every operator with modulus? These questions have drawn considerable attention. For a further discussion see Notes and Remarks to Section 5.4. As a special case of disjointness preserving operators, band preserving operators are defined next. Definition 1.4.13. Let (X, K) be a vector lattice and S ∈ L(X). The operator S is called – band preserving if for every band B in X it follows that S[B] ⊆ B, – an orthomorphism if S is band preserving and order-bounded. Now let X be, in addition, Archimedean. By Theorem 1.3.8 (ii), a set B ⊆ X is a band if and only if B = Bdd . Therefore, an operator S ∈ L(X) is band preserving if and only if⁸ for every x, y ∈ X with x ⊥ y one has Sx ⊥ y. This implies that every band preserving operator is disjointness preserving. In contrast to disjointness preserving operators, the set of all band preserving operators on X is a vector space, and the set of orthomorphisms on X is a linear subspace. For more properties on band preserving operators and orthomorphisms in Archimedean vector lattices, see, e.g., [12, Section 8].

1.5 Norm and order In the present section we consider a partially ordered vector space (X, K) that is equipped with a norm ‖⋅‖. We call (X, K, ‖⋅‖) an ordered normed space. A priori, there is no relation between the partial order and the norm. Nevertheless, in examples they are linked in several ways. We collect the relevant notions and their consequences. 8 A generalization is given in Proposition 5.1.2 below.

1.5 Norm and order |

27

1.5.1 Closed cones Ordered normed spaces with closed cones are convenient, as one can take norm limits of inequalities. Proposition 1.5.1. The following statements are equivalent. (i) K is closed.

‖⋅‖

‖⋅‖

(ii) For each sequence (x i )i∈ℕ and (y i )i∈ℕ in X and x, y ∈ X with x i 󳨀󳨀→ x and y i 󳨀󳨀→ y and x i ≤ y i for every i ∈ ℕ, one has that x ≤ y. Note that the statement in Proposition 1.5.1 (ii) is equivalent to an analogous statement for nets. Proposition 1.5.2. If K is closed, then (X, K) is Archimedean. Proof. Let x, y ∈ X be such that nx ≤ y for every n ∈ ℕ. This means x ≤ sequence

( 1n y)n∈ℕ

we have

1 ny

‖⋅‖

1 n y.

For the

󳨀󳨀→ 0, hence x ≤ 0.

The converse of Proposition 1.5.2 is not true, in general; see Example 3.6.16 below. For ordered normed spaces (X1 , K1 , ‖⋅‖1 ) and (X2 , K2 , ‖⋅‖2 ), we denote L(X1 , X2 ) := {T ∈ L(X1 , X2 ); T is norm-bounded} and endow L(X1 , X2 ) with the operator norm ‖⋅‖. Let L+ (X1 , X2 ) := L(X1 , X2 ) ∩ L+ (X1 , X2 ). Clearly, the set L+ (X1 , X2 ) is a wedge. If X1 = X2 , we abbreviate L(X1 ) := L(X1 , X2 ) and L+ (X1 ) := L+ (X1 , X2 ). If X2 = ℝ we obtain the norm dual of X1 , which we denote by X1󸀠 := L(X1 , ℝ), and the dual wedge K1󸀠 := X1󸀠 ∩ K1∗ . Proposition 1.5.3. (i) If the norm closure of K1 − K1 equals X1 , then L+ (X1 , X2 ) is a cone. (ii) Let X2 ≠ {0}. If L+ (X1 , X2 ) is a cone, then the norm closure of K1 − K1 equals X1 . (iii) If K2 is closed, then L+ (X1 , X2 ) is closed. Proof. To show (i), assume that the norm closure of K1 − K1 equals X1 . Let ±T ∈ L+ (X1 , X2 ), then for every x ∈ K1 we have that Tx = 0, hence T = 0 on K1 − K1 . Due to the continuity of T we obtain T = 0 on X1 . Hence, L+ (X1 , X2 ) is a cone. For a proof of (ii), assume that the norm closure X0 of K1 − K1 does not equal X1 . By the Hahn–Banach theorem 1.8.2 below, there is a functional f ∈ X1󸀠 \ {0} such that f = 0 on X0 . Consequently, ±f ∈ K1󸀠 . Fix z ∈ X2 \ {0}, then for the operator T : X1 → X2 , x 󳨃→ f(x)z we obtain ±T ∈ L+ (X1 , X2 ) \ {0}, such that L+ (X1 , X2 ) is not a cone. The statement in (iii) is straightforward, as the convergence in the operator norm implies pointwise convergence. Under the condition in Proposition 1.5.3 (i), the space (L(X1 , X2 ), L+ (X1 , X2 ), ‖⋅‖) is an ordered normed space.

28 | 1 A primer on ordered vector spaces Corollary 1.5.4. Let (X, K, ‖⋅‖) be an ordered normed space. (i) The dual wedge K 󸀠 is a cone if and only if the norm closure of K − K equals X. (ii) The dual wedge K 󸀠 is closed in X 󸀠 . Under the condition in Corollary 1.5.4 (i), we call K 󸀠 the dual cone. In this case, (X 󸀠 , K 󸀠 , ‖⋅‖) is an ordered normed space with a closed cone. It turns out that K 󸀠 is total. Proposition 1.5.5. Let (X, K, ‖⋅‖) be an ordered normed space such that K is closed. (i) For every x ∈ X \ K there is a functional f ∈ K 󸀠 such that f(x) < 0. (ii) Let x ∈ X. Then x ∈ K if and only if f(x) ≥ 0 for every f ∈ K 󸀠 . (iii) For every x ∈ K \ {0} there is a functional f ∈ K 󸀠 such that f(x) > 0. Proof. To show (i), let x ∈ X \ K. Since K is a nonempty convex closed set and x ∈ ̸ K, by Theorem 1.8.2 below there is f ∈ X 󸀠 and c ∈ ℝ such that f ≥ c on K and f(x) < c. As 0 ∈ K we obtain 0 = f(0) ≥ c and, consequently, f(x) < 0. It remains to show that f ∈ K 󸀠 . Indeed, let v ∈ K and n ∈ ℕ, then f(v) = 1n f(nv) ≥ nc . We conclude f(v) ≥ 0 and, hence, f ∈ K 󸀠 . The statement in (ii) is a direct consequence of (i). To establish (iii), observe that for every x ∈ K \ {0} one has −x ∉ K, i.e., (i) implies that there is a functional f ∈ K 󸀠 such that f(−x) < 0, and hence, f(x) > 0.

1.5.2 Semimonotone norms An important compatibility condition between order and norm is defined next. Later on, we will need this concept also for seminorms. An extensive discussion on seminorms is given in Chapter 3 below. Definition 1.5.6. Let (X, K) be a partially ordered vector space. A seminorm (or norm, respectively) p is called semimonotone if there is a constant N > 0 such that for every x, y ∈ X with 0 ≤ x ≤ y one has p(x) ≤ Np(y), and monotone if for every x, y ∈ X with 0 ≤ x ≤ y one has p(x) ≤ p(y). In an ordered vector space X with a seminorm p we say that the set M ⊆ X is p-bounded if there exists a constant R such that for every x ∈ M we have p(x) ≤ R. Lemma 1.5.7. Let (X, K) be a partially ordered vector space and let p be a semimonotone seminorm on X. Then every order-bounded set in X is p-bounded. Proof. As p is semimonotone, there is a constant N ∈ (0, ∞) such that for every x, y ∈ X with 0 ≤ x ≤ y we have p(x) ≤ Np(y). Let M ⊆ X be order-bounded, i.e., there are a, b ∈ X such that M ⊆ [a, b]. Let x ∈ M. Then 0 ≤ x − a ≤ b − a, hence p(x − a) ≤ Np(b − a) and, therefore, p(x) = p(x − a + a) ≤ p(x − a) + p(a) ≤ Np(b − a) + p(a).

1.5 Norm and order

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If in an ordered vector space a seminorm is equivalent to a monotone seminorm, then it is semimonotone. Conversely, every semimonotone seminorm is equivalent to a monotone seminorm, as will be shown in Proposition 3.5.1 below. The cone of an ordered normed space with monotone norm need not be closed. For the closure of the cone we have the following property. Proposition 1.5.8. Let (X, K, ‖⋅‖) be an ordered vector space with a monotone norm. The closure C of K is a cone, and the norm is monotone on (X, C). Proof. It is straightforward that C is a wedge. Let x ∈ C ∩ (−C), so then there are sequences (x n ) and (y n ) in K such that x n → x and y n → −x. Then 0 ≤ x n ≤ x n + y n in (X, K), hence ‖x n ‖ ≤ ‖x n + y n ‖. Since ‖x n + y n ‖ → 0, we have x n → 0 and, therefore, x = 0. Thus C is a cone. To show that the norm is monotone on (X, C), let x, y ∈ C. Then there are sequences (x n ) and (y n ) in K such that x n → x and y n → y. Since ‖x n ‖ ≤ ‖x n + y n ‖ for every n, we obtain ‖x‖ ≤ ‖x + y‖. It follows that the norm is monotone.

1.5.3 Order unit spaces In a partially ordered vector space (X, K) that contains an order unit u there is a natural way to introduce a seminorm on X by means of ‖⋅‖ u : X → [0, ∞),

x 󳨃→ ‖x‖u := inf{λ ∈ (0, ∞); −λu ≤ x ≤ λu} .

(1.10)

If, in addition, X is Archimedean, then ‖⋅‖u is a norm, and the infimum in (1.10) is attained, i.e., for every x ∈ X there is a smallest λ ∈ (0, ∞) such that ±x ≤ λu. This norm is called an order unit norm or u-norm. The ordered normed space (X, K, ‖⋅‖u ) is then called an order unit space. Observe that ‖u‖u = 1 and that the u-norm is monotone. For order units u, v ∈ K \ {0}, the corresponding norms ‖⋅‖u and ‖⋅‖v are equivalent. Lemma 1.5.9. Let (X, K, ‖⋅‖u ) be an order unit space.

‖⋅‖u

(i) For a sequence (x i )i∈ℕ in X and x ∈ X we have x i 󳨀󳨀󳨀→ x if and only if there is a sequence (λ i )i∈ℕ in (0, ∞) such that λ i → 0, and ±(x i − x) ≤ λ i u for every i ∈ ℕ. (ii) K is closed with respect to ‖⋅‖u . Proof. The statement in (i) is straightforward. ‖⋅‖u

To show (ii), let (x i )i∈ℕ be a sequence in K and let x ∈ X be such that x i 󳨀󳨀󳨀→ x. Due to (i), there is a sequence (λ i )i∈ℕ in (0, ∞) such that λ i → 0, and ±(x i − x) ≤ λ i u for every i ∈ ℕ. In particular, −x ≤ λ i u. As X is Archimedean, it follows that −x ≤ 0, i.e., x ≥ 0. Example 1.5.10. For a nonempty compact Hausdorff space Ω, the space C(Ω) is Archimedean, and the constant function u := 𝟙 is an order unit; see also Example 1.1.8. The corresponding order unit norm ‖⋅‖u equals the supremum norm.

30 | 1 A primer on ordered vector spaces In an order unit space (X, K, ‖⋅‖u ), we have u ∈ int(K), where int(K) denotes the interior of K. Indeed, the closed ball centered at u with radius 12 is the set [ 12 u, 32 u], which is contained in K. Conversely, every ordered normed space where the cone has a nonempty interior possesses an order unit, as the next proposition shows. For its proof we use the following basic observations. Given an ordered normed space (X, K, ‖⋅‖) with int(K) ≠ ⌀, for every x ∈ int(K) and λ ∈ (0, ∞) we have λx ∈ int(K). Moreover, if x ∈ int(K) and y ≥ x, then y ∈ int(K). Proposition 1.5.11. Let (X, K, ‖⋅‖) be an ordered normed space such that int(K) ≠ ⌀. (i) An element u ∈ K \ {0} is an order unit if and only if u ∈ int(K). (ii) For u ∈ int(K) the seminorm ‖⋅‖ u is weaker than the norm ‖⋅‖. (iii) If X is, in addition, Archimedean, then K is closed. Proof. (i) Let u > 0 be an order unit. Since int(K) ≠ ⌀, there is v ∈ int(K) and λ ∈ (0, ∞) such that −λu ≤ v ≤ λu, in particular 1λ v ≤ u. Hence, u ∈ int(K). Conversely, let u ∈ int(K), i.e., there is r ∈ (0, ∞) such that the ball Br (u) centered r x ∈ Br (u) ⊆ K, at u with radius r is contained in K. Let x ∈ X \ {0}. Then u ± 2‖x‖ i.e., r u ± 2‖x‖ x≥0. (1.11) For λ := 2‖x‖ r we obtain −λu ≤ x ≤ λu. Therefore, u is an order unit. (ii) We just showed in (i) that for every x ∈ X we have ‖x‖u ≤ 2r ‖x‖. ‖⋅‖

(iii) If (x i )i∈ℕ is a sequence in K and x ∈ X is such that x i 󳨀󳨀→ x, then ‖x i − x‖u → 0 by (ii). As X is Archimedean, due to Lemma 1.5.9 the cone K is closed with respect to ‖⋅‖u , i.e., x ∈ K. Note that even if X is Archimedean, the norms in Proposition 1.5.11 (ii) are not equivalent, in general, as the next example shows. This means that an Archimedean ordered normed space with a cone with nonempty interior need not be an order unit space. Example 1.5.12. We consider the space C[0, 1] with the standard cone K and the constant function u := 𝟙 as order unit; see Example 1.5.10. As the corresponding order unit norm ‖⋅‖u is the supremum norm, the interior of the cone with respect to ‖⋅‖u is the set U := {x ∈ C[0, 1]; ∀t ∈ [0, 1] : x(t) > 0}. There exists a functional f on C[0, 1] which is not continuous. Hence the norm x 󳨃→ ‖x‖ := ‖x‖u + |f(x)| is not equivalent to ‖⋅‖u . Nevertheless, the interiors of the cone with respect to ‖⋅‖ and ‖⋅‖u are equal. Indeed, let V be the interior of the cone with respect to ‖⋅‖. Since the topology for ‖⋅‖ is stronger than the one for ‖⋅‖u , we get U ⊆ V. Suppose that there is x ∈ V \ U. Then x ∈ K and there is t ∈ [0, 1] such that x(t) = 0. The map g : ℝ → C[0, 1], α 󳨃→ x + α𝟙 is continuous with respect to ‖⋅‖, hence the inverse image M of V under g is an open set in ℝ that contains 0. Therefore there is α 0 ∈ M with α 0 < 0. Thus (x + α 0 𝟙)(t) < 0, which contradicts x + α0 𝟙 ∈ V ⊆ K.

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We turn our attention to finite-dimensional spaces. Proposition 1.5.13. Let (X, K, ‖⋅‖) be a finite-dimensional ordered normed space. Then the following statements are equivalent. (i) K is generating. (ii) int(K) ≠ ⌀. Proof. The implication (ii) ⇒ (i) follows immediately from the Propositions 1.5.11 and 1.1.7. To show (i) ⇒ (ii), let n be the dimension of X. Let K be generating, i.e., for i ∈ {1, . . . , n} there is x(i) ∈ K such that x(1) , . . . , x(n) are linearly independent. Set u := ∑ni=1 x(i) , then clearly u ∈ K. We show u ∈ int(K). As X is finite-dimensional, the norm convergence coincides with the coordinate-wise convergence with respect to the basis (x(1) , . . . , x(n) ) of the vector space X. The set n

U := { ∑ λ i x(i) ; ∀i ∈ {1, . . . , n} : |λ i − 1| < 1} i=1

is a neighborhood of u. As for every i ∈ {1, . . . , n} we have λ i > 0, we obtain U ⊆ K. A combination of the Propositions 1.5.11 (iii) and 1.5.13 has the following consequence. Corollary 1.5.14. In a finite-dimensional directed Archimedean partially ordered vector space the cone is closed. We use Corollary 1.5.14 to characterize Archimedean ordered vector spaces. Proposition 1.5.15. A partially ordered vector space (X, K) is Archimedean if and only if for every 2-dimensional linear subspace D of X the set D ∩ K is closed in the standard topology of D. Proof. Let (X, K) be Archimedean and let D be a 2-dimensional linear subspace of X. If D0 := D ∩ K − D ∩ K is 0-dimensional or 1-dimensional, then D ∩ K is closed. If D0 is 2-dimensional, the space (D, D∩K) is directed. As D is Archimedean, by Corollary 1.5.14 we obtain that D ∩ K is closed. Let x, y ∈ X be such that for every n ∈ ℕ we have that nx ≤ y. Set D := span{x, y}. As D ∩ K is closed, by Proposition 1.5.2 the space D is Archimedean, therefore, x ≤ 0. Consequently, X is Archimedean. Proposition 1.5.16. Let (X, K, ‖⋅‖) be a finite-dimensional ordered normed space such that K is generating and closed. Then the norm is semimonotone. Proof. As K is generating, by Proposition 1.5.13 there is u ∈ int(K). Due to Proposition 1.5.11 the element u is an order unit. As K is closed, the space X is Archimedean by Proposition 1.5.2, hence the u-seminorm is a norm. As X is finite-dimensional, the norm ‖⋅‖ is equivalent to the u-norm, which is monotone. Hence ‖⋅‖ is semimonotone.

32 | 1 A primer on ordered vector spaces

Next we set up properties of ordered normed spaces where the cone has a nonempty interior. Recall that these results apply to order unit spaces. Proposition 1.5.17. Let (X, K, ‖⋅‖) be an ordered normed space such that int(K) ≠ ⌀. (i) Every positive linear functional on X is continuous. (ii) For every u ∈ int(K) and f ∈ K 󸀠 \ {0} one has f(u) > 0. Proof. Let u ∈ int(K). To show (i), let f : X → ℝ be a positive linear functional. For every x ∈ X with ‖x‖u ≤ 1 we have |f(x)| ≤ f(u). Hence, f is continuous with respect to the u-seminorm. By Proposition 1.5.11 (ii), it follows that f is continuous. For a proof of (ii), let f ∈ K 󸀠 \ {0}. Then there is x ∈ X with ‖x‖u ≤ 1 such that f(x) ≠ 0. Hence, 0 < |f(x)| ≤ f(u). Before we proceed to discuss properties of cones with nonempty interior, we introduce some further geometric notions for cones such as bases, faces, and extremal elements. We begin by recalling the notion of an extreme point. Let X be a vector space and M ⊆ X a nonempty convex set. An element x ∈ M is called an extreme point of M if for every x1 , x2 ∈ M and λ ∈ (0, 1) with x = λx1 + (1 − λ)x2 it follows that x = x1 = x2 . Definition 1.5.18. Let (X, K) be a partially ordered vector space. (i) A nonempty set S ⊆ K is called a base of K if S is a convex set such that every element x ∈ K \ {0} has a unique representation x = λy with y ∈ S and λ ∈ (0, ∞). (ii) Let H ⊆ K be a cone. We call H a face of K if for every y ∈ H and x ∈ K with x ≤ y one has x ∈ H. The faces {0} and K are called the trivial faces of K. (iii) An element y ∈ K \ {0} is called extremal if for every x ∈ K with x ≤ y there is λ ∈ [0, ∞) such that x = λy, i.e., the positive-linear hull pos{y} of y is a face of K. In this case, pos{y} is called an extreme ray. Lemma 1.5.19. Let (X, K) be a partially ordered vector space such that K has a base S. An element y ∈ S is extremal if and only if y is an extreme point of S. Proof. Let y ∈ S be extremal. Let y1 , y2 ∈ S and λ ∈ (0, 1) be such that y = λy1 + (1 − λ)y2 . Then λy1 ≤ y and (1 − λ)y2 ≤ y. As y is extremal, there are λ1 , λ2 ∈ (0, ∞) such that λ1 λy1 = y = λ2 (1 − λ)y2 . The uniqueness of the representation of y with respect to S implies y1 = y = y2 , hence y is an extreme point of S. Conversely, let y be an extreme point of S and let x ∈ K be such that 0 ≤ x ≤ y. We show that x = λy for some λ ∈ [0, ∞). Clearly, if x = 0 or x = y, we have x = 0y or x = 1y, respectively. We assume x ≠ 0 and x ≠ y, hence there are x0 , x1 ∈ S and λ, μ ∈ (0, ∞) such that x = λx0 and y − x = μx1 . Then 1 (λ+μ) y

=

λ (λ+μ) x 0

+

μ (λ+μ) x 1

=: v ,

1.5 Norm and order

| 33

where v ∈ S, since S is convex. For y we have two representations with respect to S, namely y = 1y and y = (λ + μ)v. The uniqueness of the representation implies y = v and 1 = λ + μ. Since y is an extreme point of S, we obtain y = x0 = x1 . Thus x = λy. Next we show that faces of codimension one entail extremals of the dual cone. Lemma 1.5.20. Let (X, K) be a directed partially ordered vector space, let H be a nontrivial face of K, and let f ∈ K ∗ be such that {x ∈ X; f(x) = 0} = H − H. Then f is extremal in K ∗ . Proof. As H is a nontrivial face, H − H ≠ X. Hence f ≠ 0. Let g ∈ K ∗ be such that g ≤ f . For x ∈ H we obtain 0 ≤ g(x) ≤ f(x) = 0. Hence g = 0 on the kernel H − H of f . Thus there is λ ∈ ℝ such that g = λf . As f ≠ 0 and X is directed, there is x ∈ K such that f(x) > 0. As g(x) ≥ 0 we obtain λ ∈ [0, ∞). The following important result relating interior points of the cone and bases of the dual cone will be used frequently later on. Theorem 1.5.21. Let (X, K, ‖⋅‖) be an ordered normed space such that K is closed. The following statements are equivalent. (i) int(K) ≠ ⌀, (ii) The dual cone K 󸀠 has a base Σ that is norm-bounded and weakly-∗ closed, i.e., Σ is weakly-∗ compact. Proof. To show that (i) implies (ii), let u ∈ int(K) and set Σ := {f ∈ K 󸀠 ; f(u) = 1} .

(1.12)

Using the functional F u : X 󸀠 → ℝ, f 󳨃→ f(u), we have Σ = K 󸀠 ∩ [{1}]F u . Clearly, Σ is 1 convex. By Proposition 1.5.17, for every g ∈ K 󸀠 \ {0} one has g(u) > 0 and g(u) g ∈ Σ. 1 󸀠 As for c ∈ ℝ we have cg ∈ Σ only if c = g(u) , Σ is a base of K . Observe that [{1}]F u is weakly-∗ closed. K 󸀠 is weakly-∗ closed as well. Indeed, a net (f α )α∈A in K 󸀠 converges to f ∈ X 󸀠 in the weak-∗ topology if and only if for every x ∈ X we have f α (x) → f(x). If x ∈ K, then f α (x) ≥ 0, hence f(x) ≥ 0, so f ∈ K 󸀠 . We conclude that Σ = K 󸀠 ∩ [{1}]F u is weakly-∗ closed. According to (1.11), there is r > 0 such that for every f ∈ K 󸀠 and x ∈ X \ {0} 2f(u) we have ±f(x) ≤ 2f(u) r ‖x‖. Hence, ‖f‖ ≤ r . We conclude that Σ is contained in the 2 󸀠 ball {f ∈ X ; ‖f‖ ≤ r }. By the Banach–Alaoglu theorem 1.8.6 the set Σ is weakly-∗ compact. Next we show that (ii) implies (i). As 0 ∈ ̸ Σ, by Theorem 1.8.2 the sets {0} and Σ can be separated by an appropriate hyperplane. By Theorem 1.8.5 this means that there is u ∈ X such that for every f ∈ Σ one has F u (f) = f(u) ≥ 1. Since Σ is normbounded, there is M > 0 such that for every f ∈ Σ it holds that ‖f‖ ≤ M. To prove that 1 u ∈ int(K), we show that the open ball centered at u with radius M is contained in K. 1 Let x ∈ X be such that ‖x − u‖ < M . For every f ∈ Σ we obtain |f(x − u)| ≤ 1; moreover,

34 | 1 A primer on ordered vector spaces f(x) = f(u) + f(x − u) ≥ 0. We conclude that for every f ∈ K 󸀠 it holds that f(x) ≥ 0, so Proposition 1.5.5 implies x ∈ K. The assertion that (i) implies (ii) in Theorem 1.5.21 is true without assuming that K is closed. Next we give conditions on an ordered normed space such that the spaces of continuous operators and order-bounded operators coincide. Lemma 1.5.22. Let (X, K, ‖⋅‖) be an ordered normed space such that the interior of K is nonempty and the norm is semimonotone. Then a set M ⊆ X is norm-bounded if and only if it is order-bounded. Proof. By Lemma 1.5.7, every order-bounded set is norm-bounded. To show the converse, fix u ∈ int(K) and let M ⊆ X be norm-bounded. By Proposition 1.5.11, the element u is an order unit and its order unit norm ‖⋅‖u is weaker than ‖⋅‖. Therefore, M is bounded with respect to ‖⋅‖u . Since closed balls for the ‖⋅‖u -norm are order intervals, we obtain that M is order-bounded. The next statement is a direct consequence of Lemma 1.5.22. Proposition 1.5.23. Let (X, K X , ‖⋅‖X ) and (Y, K Y , ‖⋅‖Y ) be ordered normed spaces such that the interiors of K X and K Y are nonempty and both norms are semimonotone. Then L(X, Y) = Lb (X, Y). Corollary 1.5.24. Let X and Y be finite-dimensional Archimedean directed ordered spaces. Then L(X, Y) = L(X, Y) = Lb (X, Y). Proof. As the underlying spaces are finite-dimensional, we have L(X, Y) = L(X, Y). It remains for us to verify the conditions of Proposition 1.5.23. By Proposition 1.5.13, the cones in X and Y have nonempty interiors. Due to Corollary 1.5.14 the cones are closed, hence Proposition 1.5.16 yields that the norms in X and Y are semimonotone. In the remainder of this subsection we discuss the uniform completion of order unit spaces. Recall that in an order unit space the norm is monotone. We even have the following slightly stronger property. Lemma 1.5.25. Let (X, K, ‖⋅‖u ) be an order unit space. For every x, y ∈ K we have ‖y − x‖u ≤ ‖x + y‖u . Proof. If λ > ‖x + y‖u , then −λu ≤ x + y ≤ λu, so y − x ≤ y ≤ x + y ≤ λu and y − x ≥ −x ≥ −(x + y) ≥ −λu. Hence, ‖y − x‖u ≤ λ. Therefore, ‖y − x‖u ≤ ‖x + y‖u . Definition 1.5.26. For an order unit space (X, K, ‖⋅‖u ), the completion Z of X with respect to ‖⋅‖u ordered by the closure C of K in Z is called the uniform completion of X. Proposition 1.5.27. If (X, K, ‖⋅‖u ) is an order unit space, then the uniform completion (Z, C) of X is an order unit space, and the inclusion map i : X → Z is bipositive.

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Proof. The norm ‖⋅‖ in Z is monotone in (Z, K), so by Proposition 1.5.8 the set C is a cone in Z. As C is closed, the partially ordered vector space (Z, C) is Archimedean. By Lemma 1.5.9 the inclusion map i : X → Z is bipositive. We show that u is an order unit in Z and that the norm in Z is the u-norm. Let z ∈ Z and let λ > ‖z‖. There exists a sequence (x n ) in X such that ‖x n − z‖ → 0. Then ‖x n ‖u = ‖x n ‖ → ‖z‖, so there is an N1 ∈ ℕ such that ‖x n ‖u < λ for every n ∈ ℕ≥N1 . Let ε > 0. As (x n ) is ‖⋅‖-Cauchy, we have that (x n ) is ‖⋅‖u -Cauchy. This means that there exists N2 ∈ ℕ such that for every m, n ∈ ℕ≥N2 we have −εu ≤ x n − x m ≤ εu. Define N := N1 ∨ N2 . Then x n ≤ x N + εu. Also, ‖x N ‖u < λ, so −λu ≤ x N ≤ λu and therefore, x n ≤ (λ + ε)u for every n ∈ ℕ≥N . Define v n := (λ + ε)u − x n . Observe that v n ∈ K and v n = (λ + ε)u − z − (x n − z), i.e., the sequence (v n )n∈ℕ≥N converges to (λ + ε)u − z. Therefore, (λ + ε)u − z ∈ C, which means that z ≤ (λ + ε)u in (Z, C). It follows that u is an order unit in (Z, C). Similarly, we obtain −z ≤ (λ + ε)u, so that ‖z‖u ≤ λ + ε. We infer that ‖z‖u ≤ ‖z‖. To show that ‖z‖u ≥ ‖z‖, let λ > ‖z‖u . Then −λu ≤ z ≤ λu, so λu − z, λu + z ∈ C, so there exist sequences (x n ) and (y n ) in K such that x n → λu − z and y n → λu + z. Due to Lemma 1.5.25, we have ‖y n − x n ‖ = ‖y n − x n ‖u ≤ ‖y n + x n ‖u = ‖y n + x n ‖ for every n ∈ ℕ. By taking limits, we infer that ‖2z‖ ≤ ‖2λu‖. As ‖u‖u = 1, it follows that ‖z‖ ≤ λ. Therefore, ‖z‖ ≤ ‖z‖u . Thus, the norm on (Z, C) is the u-norm.

1.5.4 Normed vector lattices The most useful property of a norm on a vector lattice is given in the next definition. Definition 1.5.28. Let (X, K) be a vector lattice. A seminorm ρ on X is called a lattice seminorm or Riesz seminorm if for every x, y ∈ X with |x| ≤ |y| one has that ρ(x) ≤ ρ(y). If ρ is, in addition, a norm, then ρ is called a lattice norm or Riesz norm. Observe that every Riesz seminorm ρ is monotone, and that the lattice operations are continuous with respect to ρ. If D is a Riesz subspace of a vector lattice X and ρ is a Riesz seminorm on X, then the restriction of ρ to D is a Riesz seminorm on D. As usual, we call a vector lattice (X, K) with a lattice norm a normed vector lattice or normed Riesz space. In this case, the cone K is closed. The following observations are straightforward. Proposition 1.5.29. Let X be a Riesz space and let ρ and η be Riesz seminorms on X. Then, for every λ ∈ [0, ∞), λρ, ρ + η and the pointwise maximum ρ ∨ η of ρ and η are Riesz seminorms as well. Proposition 1.5.30. Let X be a Riesz space and let ρ be a seminorm on X. Then ρ is a Riesz seminorm if and only if the unit ball of ρ is a solid set.

36 | 1 A primer on ordered vector spaces

As unit balls are convex, Proposition 1.5.30 leads to the investigation of convex solid sets and their generalizations in Section 3.2 below. The next result due to Namioka [124, Lemma 7.6] will be needed. Proposition 1.5.31. Let X be a Riesz space. The convex hull of a solid set in X is solid. Proof. Let S ⊆ X be solid and let T be the convex hull of S. Let x ∈ X and y ∈ T be such that |x| ≤ |y|. We show that x ∈ T. There are s1 , . . . , s n ∈ S and λ1 , . . . , λ n ∈ (0, ∞) such that ∑ni=1 λ i = 1 and y = ∑ni=1 λ i s i . Define z := ∑ni=1 λ i |s i |. Then x+ ≤ |x| ≤ |y| ≤ z and, similarly, x− ≤ z. Due to Proposition 1.1.27 the Riesz decomposition property holds, hence by Proposition 1.1.26 there exist u 1 , . . . , u n , v1 , . . . , v n ≥ 0 such that x+ = ∑ni=1 u i , x− = ∑ni=1 v i , and 0 ≤ u i , v i ≤ λ i |s i | for every i ∈ {1, . . . , n}. As S is solid, s i ∈ S, and 󵄨󵄨 1 󵄨 󵄨󵄨 λ (u i − v i )󵄨󵄨󵄨 = 󵄨 i 󵄨 for every i ∈ {1, . . . , n}, we obtain

1 λi

|u i − v i | ≤

1 λ i (u i

1 λ i (u i

∨ v i ) ≤ |s i |

− v i ) ∈ S and therefore,

n

n

i=1

i=1

x = x+ − x− = ∑ (u i − v i ) = ∑ λ i λ1i (u i − v i ) ∈ T. The next example shows that the solid hull of a convex set need not be convex. Example 1.5.32. Consider the space ℝ2 with standard order. Let S ⊆ ℝ2 be the convex hull of the set {(1, 2)T , (−2, 1)T , (−1, −2)T , (2, −1)T }. The solid hull of S is the set S ∪ {(x, y)T ; (y, x)T ∈ S} ∪ {(x, y)T ; |x| ≤ 1, |y| ≤ 2} ∪ {(x, y)T ; |x| ≤ 2, |y| ≤ 1} , which is not convex. If a normed Riesz space X is norm complete, then X is called a Banach lattice. There is an extensive literature on Banach lattices; see, e.g., [2, 119, 134]. Example 1.5.33. The space B(S) as in Example 1.1.8 with supremum norm is a Banach lattice. Moreover, the spaces C(Ω) (with supremum norm) and ℓp (S) (with p-norm, p ≥ 1) are Banach lattices. The space cc (ℕ) (with supremum norm or p-norm) is a normed vector lattice, but not a Banach lattice. For the following result, see [119, Proposition 1.3.7], or Proposition 5.4.13 below. Proposition 1.5.34. For a normed Riesz space X, the dual space X 󸀠 is a Banach lattice.

1.5.5 Relatively uniform convergence A convenient norm such as the order unit norm given in (1.10) does not exist in ordered vector spaces, in general. However, a similar concept of convergence can be defined.

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| 37

Definition 1.5.35. Let (X, K) be a partially ordered vector space, let (x i )i∈I be a net in X, and let x ∈ X. (i) We say that (x i )i converges relatively uniformly to x if there exists a u ∈ K, i0 ∈ I, and a decreasing net (λ i )i∈I≥i0 in [0, ∞) such that inf{λ i ; i ∈ I≥i 0 } = 0 and for every i ∈ I≥i 0 one has −λ i u ≤ x i − x ≤ λ i u. (ii) We say that (x i )i is relatively uniformly Cauchy if there exists a u ∈ K, i0 ∈ I, and a decreasing net (λ i )i∈I≥i0 in [0, ∞) such that inf{λ i ; i ∈ I≥i 0 } = 0 and for every i, j, k ∈ I≥i 0 with j, k ≥ i one has −λ i u ≤ x j − x k ≤ λ i u. (iii) We say that (X, K) is uniformly complete if every relatively uniformly Cauchy net in X is relatively uniformly convergent in X. (iv) A set M ⊆ X is called relatively uniformly closed if for every net in M that converges relatively uniformly to x ∈ X we have that x ∈ M. If the space is Archimedean, then relatively uniform limits are unique, according to the next observation. Proposition 1.5.36. Let (X, K) be an Archimedean partially ordered vector space, let (x i )i∈I be a net in X, and let x, y ∈ X. (i) If (x i )i∈I converges relatively uniformly to x, then (x i )i∈I o-converges to x. (ii) If (x i )i∈I converges relatively uniformly to x and to y, then x = y. Proof. (i) As (x i )i∈I converges relatively uniformly to x, there is u ∈ K, i0 ∈ I, and a decreasing net (λ i )i∈I≥i0 in [0, ∞) such that inf{λ i ; i ∈ I≥i 0 } = 0 and for every i ∈ I≥i 0 one has −λ i u ≤ x i − x ≤ λ i u. As X is Archimedean, by Proposition 1.1.21 we obtain inf{λ i u; i ∈ I≥i 0 } = 0. Hence, (x i )i∈I o-converges to x. (ii) According to Proposition 1.1.32 (i) the limit with respect to o-convergence is unique. Now the implication in (i) completes the proof. The next example shows that a relatively uniformly convergent net may have distinct limits if the space is not Archimedean. Example 1.5.37. Let X = ℝ2 ordered by the cone K := {(x1 , x2 )T ; x1 > 0 or x1 = 0 and x2 ≥ 0} . Define u := (1, 0), x := (0, 1), y := (0, 0), and for n ∈ ℕ, x(n) := (0, 0). Then for every n ∈ ℕ we have − 1n u ≤ x(n) − x ≤ 1n u, so (x(n) )n∈ℕ converges relatively uniformly to x. Also, (x(n) )n∈ℕ converges relatively uniformly to y and, clearly, x ≠ y. The definitions of relatively uniform convergence and relatively uniformly Cauchy net in (i) and (ii) of Definition 1.5.35 can be reformulated in the following way. Proposition 1.5.38. Let (X, K) be an Archimedean partially ordered vector space, let (x i )i∈I be a net in X, and let x ∈ X.

38 | 1 A primer on ordered vector spaces (i) The net (x i )i∈I converges relatively uniformly to x if and only if there exists a u ∈ K with the property that for every ε > 0 there exists i0 ∈ I such that for every i ∈ I with i ≥ i0 one has −εu ≤ x i − x ≤ εu. (ii) The net (x i )i∈I is relatively uniformly Cauchy if and only if there exists u ∈ K with the property that for every ε > 0 there exists i0 ∈ I such that for every i, j ∈ I with i, j ≥ i0 one has −εu ≤ x i − x j ≤ εu. Proof. (i) Suppose that (x i )i∈I converges relatively uniformly to x. Let u ∈ K and let i0 ∈ I be such that −λ i u ≤ x i − x ≤ λ i u for every i ≥ i0 . Let ε > 0. Take i1 ∈ I≥i 0 such that for every i ≥ i1 we have λ i < ε. For i ≥ i1 we then have x i − x ≤ λ i u ≤ εu and, similarly, x i − x ≥ −εu. For the converse implication, let u ∈ K be an element with the property that for every ε > 0 there exists i0 ∈ I such that for every i ∈ I with i ≥ i0 we have −εu ≤ x i − x ≤ εu. We construct a net (λ i )i as follows. Take i0 ∈ I such that for every i ≥ i0 we have −u ≤ x i − x ≤ u. Define for i ∈ I≥i 0 , λ i := inf {λ ∈ [0, 1]; ∀j ∈ I≥i : − λu ≤ x j − x ≤ λu} . By definition, the net (λ i )i∈I≥i0 is decreasing. For each i ∈ I≥i 0 we have for every n ∈ ℕ that −(λ i + 1n )u ≤ x i − x ≤ (λ i + 1n )u. As X is Archimedean, we obtain −λ i u ≤ x i − x ≤ λ i u. It remains to show that the infimum of the λ i is zero. Let ε > 0 and take i1 ∈ I such that for every i ∈ I with i ≥ i1 we have −εu ≤ x i − x ≤ εu. Then for every i ≥ i1 we have λ i ≤ λ i1 ≤ ε. Hence, inf{λ i ; i ∈ I≥i 0 } = 0. Thus, (x i )i∈I≥i0 converges relatively uniformly to x. (ii) Suppose that (x i )i∈I is relatively uniformly Cauchy, let u ∈ K, and let i0 ∈ I and (λ i )i∈I≥i0 be such that for every i ≥ i0 and j, k ≥ i we have −λ i u ≤ x j − x k ≤ λ i u. Let ε > 0. Take i1 ∈ I≥i 0 such that for every i ≥ i1 we have λ i < ε. For every j, k ≥ i1 we then have x j − x k ≤ λ i1 u ≤ εu and, similarly, x j − x k ≥ −εu. For the converse implication, suppose that u ∈ K has the property that for every ε > 0 there exists i0 ∈ I such that for every i, j ∈ I with i, j ≥ i0 one has −εu ≤ x i − x j ≤ εu. As in the proof of (i), take i0 ∈ I such that for every i, j ≥ i0 we have −u ≤ x i − x j ≤ u. Define for i ∈ I≥i 0 , λ i := inf {λ ∈ [0, 1]; ∀j, k ∈ I≥i : − λu ≤ x j − x k ≤ λu} . The net (λ i )i∈I≥i0 is decreasing. For every i ≥ i0 , every j, k ≥ i, and every n ∈ ℕ we have −(λ i + 1n )u ≤ x j − x k ≤ (λ i + 1n )u, so that the fact that X is Archimedean yields that −λ i u ≤ x j −x k ≤ λ i u. A similar argument as in (i) show that inf{λ i ; i ∈ I≥i 0 } = 0. Thus, (x i )i∈I is relatively uniformly Cauchy. Consider an Archimedean partially ordered vector space (X, K) with an order unit u and the corresponding order unit norm ‖⋅‖u as in (1.10). With the aid of Proposition 1.5.38 the statement in Lemma 1.5.9 (i) can be extended to nets.

1.5 Norm and order

|

39

Proposition 1.5.39. Let (X, K) be an Archimedean partially ordered vector space with an order unit u, let (x i )i∈I be a net in X, and let x ∈ X. (i) The net (x i )i∈I converges relatively uniformly to x if and only if (x i )i∈I converges to x in (X, ‖⋅‖u ). (ii) The net (x i )i∈I is relatively uniformly Cauchy if and only if (x i )i∈I is a Cauchy net in (X, ‖⋅‖u ). (iii) (X, K) is uniformly complete if and only if the normed space (X, ‖⋅‖u ) is complete. Proof. Observe first that for every ε > 0 and i, j ∈ I we have −εu ≤ x i − x ≤ εu ⇐⇒ ‖x i − x‖u ≤ ε and −εu ≤ x i − x j ≤ εu ⇐⇒ ‖x i − x j ‖u ≤ ε . With the aid of Proposition 1.5.38, (i) and (ii) and one implication of (iii) are clear. Assume that (X, K) is uniformly complete. Let (x i )i∈I be a Cauchy net in (X, ‖⋅‖u ). Then, with Proposition 1.5.38, there is x ∈ X such that (x i )i converges relatively uniformly to x. Let v ∈ X be such that for every n ∈ ℕ there is j n ∈ I with − 1n v ≤ x j − x ≤ 1n v for every j ∈ I≥j n . Let ε > 0. Take i ε ∈ I such that −εu ≤ x i − x j ≤ εu for every i, j ∈ I≥i ε . Then, for i, j ∈ I≥i ε with j ≥ j n we have x i − x = x i − x j + x j − x ≤ εu + 1n v , so x i − x ≤ εu, since X is Archimedean. Similarly, −εu ≤ x i − x. Hence x i → x in (X, ‖⋅‖u ). Example 1.5.40. We consider the space C(Ω) as in Example 1.5.10. As C(Ω) with the supremum norm is a Banach space, Proposition 1.5.39 (iii) yields that C(Ω) is uniformly complete. Dedekind complete Riesz spaces are uniformly complete, as we show next. Proposition 1.5.41. Let (X, K) be a Riesz space. If X is Dedekind complete, then X is uniformly complete. Proof. Let (x i )i∈I be a relatively uniformly Cauchy net in X and let u ∈ K, i0 ∈ I, and (λ i )i∈I≥i0 a decreasing net in [0, ∞) with inf{λ i ; i ∈ I≥i 0 } = 0 be such that for every i ∈ I≥i 0 and j, k ≥ i we have −λ i u ≤ x j − x k ≤ λ i u. We show that the net (x i )i∈I is relatively uniformly convergent and begin by constructing a candidate limit x as a ‘limsup’ of the net. For i ∈ I≥i 0 , the set {x j ; j ≥ i} is nonempty and bounded above by x i + λ i u. As X is Dedekind complete, we can define y i := sup {x j ; j ∈ I≥i } . Then y i ≤ x i + λ i u. Note that we also have that y i ≥ x i ≥ x i0 − λ i0 u. Define x := inf {y i ; i ∈ I≥i 0 } .

40 | 1 A primer on ordered vector spaces We show that (x i )i∈I converges relatively uniformly to x. For that purpose, let i ∈ I≥i 0 . Since y i ≤ x i + λ i u, we have x ≤ x i + λ i u, so x i − x ≥ −λ i u. In order to see that also x i − x ≤ λ i u, note that for j ∈ I≥i we have y j ≥ x j ≥ x i − λ i u. Since the net (y j )j∈I≥i0 is decreasing, it follows that x ≥ x i − λ i u, hence x i − x ≤ λ i u. Thus, (x i )i∈I converges relatively uniformly to x and we have shown that X is uniformly complete.

1.6 Order denseness We will often study partially ordered vector spaces by viewing them as subspaces of Riesz spaces. Here, a central notion will be order denseness of the subspace as defined next. Definition 1.6.1. Let (X, K) be an ordered vector space and let D be a linear subspace of X. (i) D is called majorizing if for every x ∈ X there is y ∈ D such that x ≤ y. (ii) D is called order dense⁹ if for every x ∈ X one has that x = inf{y ∈ D; y ≥ x} . It is straightforward that D is order dense in X if and only if for every x ∈ X one has that x = sup{y ∈ D; y ≤ x} . Clearly, every order dense subspace is majorizing. To see that the converse is not true, consider, e.g., the space of all constant functions in C[0, 1], which is a majorizing subspace, but clearly not order dense. The next statement shows that lattice operations carry over to order dense subspaces appropriately. Proposition 1.6.2. Let (X, K) be an ordered vector space, Z a linear subspace of X, and D a subset of Z. (i) If the infimum inf X D in X exists and is contained in Z, then the infimum inf Z D in Z exists and equals inf X D. (ii) Let Z be, in addition, order dense in X. If the infimum inf Z D in Z exists, then the infimum inf X D in X exists and equals inf Z D. Proof. (i) Let w := inf X D. As w ∈ Z, w is a lower bound of D in Z. If v ∈ Z is also a lower bound of D, then v is a lower bound of D in X, and hence v ≤ w.

9 The order denseness property in Definition 1.6.1 appears in the classical Dedekind completion of Archimedean directed ordered spaces, as discussed in Section 2.1 below. The name ‘order denseness’ in the general setting discussed here is due to [36].

1.6 Order denseness | 41

(ii) Define w := inf Z D, then w ∈ Z ⊆ X and w ≤ d for every d ∈ D. Let v ∈ X be such that v ≤ d for every d ∈ D. Since Z is order dense in X, one has v = sup{z ∈ Z; z ≤ v}. Let z ∈ Z be such that z ≤ v, then z ≤ d for every d ∈ D, hence z ≤ inf Z D = w. Therefore w is an upper bound of the set {z ∈ Z; z ≤ v}, which implies v ≤ w. Consequently, w = inf X D. Clearly, for the supremum, similar statements to the ones in Proposition 1.6.2 are valid. If Z is not order dense in Y, then the conclusion in Proposition 1.6.2 (ii) may fail; see Example 1.1.20. The next result is a fundamental property of order denseness, which is due to van Waaij; see [153, Theorem 1.44]. Proposition 1.6.3. Let (X, K) be a partially ordered vector space and let W and Z be subspaces of X with W ⊆ Z. If W is order dense in Z and Z is order dense in X, then W is order dense in X. Proof. Let x ∈ X and let A := {u ∈ W; u ≥ x}. We have to show that x = inf A. Clearly, x is a lower bound of A. Let v be another lower bound of A. For every z ∈ Z with z ≥ x we have z ≥ v. Indeed, as W is order dense in Z, we have that z = inf Z {u ∈ W; u ≥ z}. As Z is order dense in X, by Proposition 1.6.2 (ii) we obtain z = inf X {u ∈ W; u ≥ z}. For every u ∈ W with u ≥ z we have u ≥ x, so u ∈ A, hence u ≥ v. This implies z ≥ v. Since Z is order dense in X, we have x = inf X {z ∈ Z; z ≥ x} ≥ v. Hence x is the infimum of A. The subsequent result states that uniformly dense subspaces are order dense. Proposition 1.6.4. Let (X, K) be an Archimedean partially ordered vector space that contains an order unit u ∈ K. If a subspace Z of X is dense in X with respect to the u-norm, then Z is order dense in X. Proof. Let Z be a subspace of X that is dense in X with respect to the u-norm, let x ∈ X, and let A = {z ∈ Z; z ≥ x}. We have to show that x = inf A. Clearly, x is a lower bound of A. Let v ∈ X be another lower bound of A. For every ε > 0, there exists z ε ∈ Z such that ‖x + εu − z ε ‖u ≤ ε. Then z ε ≥ x, hence z ε ∈ A. Also, we have that z ε ≤ x + 2εu, so that v ≤ x + 2εu. As X is Archimedean, it follows that v ≤ x and therefore, x = inf A. We conclude that Z is order dense in X. For a nonempty compact Hausdorff space Ω, we consider in C(Ω) the constant function 𝟙 as an order unit; see Example 1.5.10, i.e., the order unit norm equals the supremum norm. In this case we obtain the following direct consequence of Proposition 1.6.4. Corollary 1.6.5. Let Ω be a nonempty compact Hausdorff space and let C(Ω) be endowed with the supremum norm. Every norm dense subspace of C(Ω) is order dense.

42 | 1 A primer on ordered vector spaces

1.7 Examples In the present section we discuss several types of examples of partially ordered vector spaces. In particular, we focus on finite-dimensional directed Archimedean ordered vector spaces. Recall that such spaces are precisely those ones that have a closed cone with a nonempty interior.

1.7.1 The ice cream cone and friends An important example of an ordered vector space without lattice structure is ℝ3 with the ice cream cone. It turns out that this space has different representations: polynomials of degree at most two, ℝ3 with the three-dimensional Lorentz cone, symmetric 2 × 2-matrices, and affine functions on a circle. We introduce similar ordered vector spaces of general dimension, if possible. We already dealt with examples of partially ordered function spaces, where the space was constructed as a subspace of a Banach lattice. The next example provides a finite-dimensional ordered vector space that is order dense in an infinite-dimensional Riesz space. Example 1.7.1. We consider C(ℝ) with standard order and its subspace X = P2 (ℝ) of all real polynomial functions on ℝ of degree at most 2, endowed with the induced order. As C(ℝ) is Archimedean, X is Archimedean as well. Moreover, X is directed. Indeed, if p1 : x 󳨃→ a2 x2 + a1 x + a0 and p2 : x 󳨃→ b 2 x2 + b 1 x + b 0 are given, then in the case a2 ≥ b 2 > 0 for the polynomial p3 : 2a2 x2 + a1 x + max {a0 , b 0 +

(a 1 −b 1 )2 4(2a 2 −b 2 ) }

one has p3 ≥ p1 and p3 ≥ p2 (and the other cases are straightforward as well). The space V = {v ∈ C(ℝ); ∃ p ∈ X : |v| ≤ p} (1.13) is a Riesz space. We show that X is order dense in V. Indeed, let v ∈ V and p ∈ X be such that |v(t)| < p(t) for every t ∈ ℝ. Fix an arbitrary point t0 ∈ ℝ and ε > 0 such that v(t0 ) + ε < p(t0 ). Since v is continuous, there is a δ > 0 such that for every t ∈ (t0 − δ, t0 + δ) one has v(t) ≤ v(t0 ) + ε. One obtains a polynomial p ε ∈ X such that p ε (t0 ) = v(t0 ) + ε, p ε (t0 − δ) = p ε (t0 + δ) = max{p(t0 − δ), p(t0 + δ)}. The polynomial p ε attains its minimum in t0 and satisfies p ε (t) ≥ p(t) for every t ∈ ℝ \ (t0 − δ, t0 + δ). Hence, p ε ≥ v. It follows that inf {p(t0 ); p ∈ X, p ≥ v} ≤ p ε (t0 ) = v(t0 ) + ε and, as ε tends to zero, v(t0 ) = inf{p(t0 ); p ∈ X, p ≥ v} . Hence, v = inf{p ∈ X; p ≥ v}.

1.7 Examples | 43

Example 1.7.2. The space P2 [0, 1] of polynomial functions on [0, 1] of degree at most two is an order dense subspace of C[0, 1]. The proof is similar to the one in Example 1.7.1. The space P[0, 1] of all polynomial functions on [0, 1] is order dense in C[0, 1], as it contains P2 [0, 1]. As the space P2 (ℝ) in Example 1.7.1 is a 3-dimensional vector space, the question arises as to which geometric properties the cone in P2 (ℝ) has. Moreover, as the dual space can be identified with ℝ3 as well, the same question arises for the dual cone. Before we answer these geometrical questions on P2 (ℝ), we discuss the more general setting of Lorentz cones. Let H be a (possibly infinite dimensional) Hilbert space with inner product ⟨⋅, ⋅⟩ and the corresponding norm ‖x‖ := √⟨x, x⟩. On X = ℝ × H we consider the inner product ⟨x | y⟩ = rs + ⟨z, v⟩ , (1.14) where x = (r, z) and y = (s, v), which turns X into a Hilbert space. Let X be ordered by means of the Lorentz cone LH = {(r, z) ∈ ℝ × H; r2 − ⟨z, z⟩ ≥ 0 and r ≥ 0} .

(1.15)

The cone LH is closed in X. We show that u = (1, 0) is an interior point of LH . Indeed, let (r, z) ∈ X be such that ‖u − (r, z)‖ < 12 , and we prove (r, z) ∈ LH . Note that ⟨u − (r, z) | u − (r, z)⟩
0 and hence, (r, z) ∈ LH . By Proposition 1.5.11, the point u is an order unit in X, which implies that LH is generating and L∗H is a cone. As LH has a nonempty interior, by Proposition 1.5.17 every positive linear functional on X is continuous, i.e., L∗H = L󸀠H . Using the identification of the Hilbert space X with X 󸀠 , we show that LH is selfdual, i.e., LH = L󸀠H . Indeed, consider y = (s, v) ∈ LH . For every x = (r, z) ∈ LH the Cauchy–Schwarz inequality yields that ⟨x | y⟩ = rs + ⟨z, v⟩ ≥ rs − ‖z‖‖v‖ ≥ 0 , due to s ≥ ‖v‖ and r ≥ ‖z‖. Thus y ∈ L󸀠H . Conversely, if y = (s, v) ∈ L󸀠H , then s ≥ 0. If v = 0, then clearly y ∈ LH . If v ≠ 0, then consider x = (r, z), where r = ‖v‖ and z = −v. As r2 − ⟨z, z⟩ = 0, one has x ∈ LH , and hence ⟨x | y⟩ ≥ 0. This implies that 0 ≤ ⟨x | y⟩ = rs − ⟨v, v⟩ = ‖v‖s − ‖v‖2 , so that s − ‖v‖ ≥ 0. Therefore, y ∈ LH . We conclude that by (1.12) the set Σ := {(s, v) ∈ L󸀠H ; ⟨(1, 0) | (s, v)⟩ = 1} = {(r, z) ∈ LH ; r = 1} L󸀠H

(1.16)

is a base of = LH . As a special case of a Lorentz cone, for H = ℝ2 one gets the ice cream cone in ℝ3 ; see the next example. We follow the convention in the literature to identify r with x3 and z with (x1 , x2 ), so that the ice cream does not drop out of the ice cream cone.

44 | 1 A primer on ordered vector spaces Example 1.7.3. Let the space X = ℝ3 be equipped with the cone Lℝ2 = {(x1 , x2 , x3 )T ; x21 + x22 ≤ x23 , x3 ≥ 0} . The set Σ := {(x1 , x2 , 1)T ; x21 + x22 ≤ 1} is a base of the cone Lℝ2 , so that the geometry of this cone and its dual cone are available. All nontrivial faces of Lℝ2 are rays. We claim that X is order isomorphic to the space P2 (ℝ) in Example 1.7.1. Let a, b, c ∈ ℝ and consider the element p(t) = at2 + bt + c in P2 (ℝ). In the case a ≠ 0 the zeros of p 1 (−b ± √b 2 − 4ac). Therefore p is positive if and only if either are given by t1/2 = 2a 2 (i) 0 < a and b − 4ac ≤ 0, or (ii) a = b = 0 and 0 ≤ c. Consider the linear bijection j : ℝ3 → P2 (ℝ),

(x1 , x2 , x3 )T 󳨃→ (p : ℝ → ℝ, t 󳨃→ (x1 + x3 )t2 + (2x2 )t + (x3 − x1 )) . (1.17) Denote a := x1 + x3 , b := 2x2 , c := x3 − x1 . We show that j is bipositive. Let x = b (x1 , x2 , x3 )T ∈ Lℝ2 , i.e., x21 + x22 ≤ x23 and 0 ≤ x3 . Due to x1 = a−c 2 , x 2 = 2 and 2 2 2 2 x3 = a+c 2 , the inequality x 1 + x 2 ≤ x 3 is satisfied if and only if b ≤ 4ac . In particular, a+c one has 0 ≤ ac. In combination with 0 ≤ x3 = 2 one gets 0 ≤ a and 0 ≤ c. If 0 < a, we have the condition in (i). On the other hand, if a = 0 we obtain b = 0, and, hence, the condition in (ii). Therefore, the polynomial j(x) is positive with respect to the pointwise order. Now let x = (x1 , x2 , x3 )T ∈ ℝ3 be such that j(x) is a positive element in P2 (ℝ). In the case (i) one obtains x21 + x22 ≤ x23 , in particular |x1 | ≤ |x3 |, and from 0 < a = x1 + x3 it follows that 0 < x3 , hence x ∈ Lℝ2 . In the case (ii), one has x2 = 0 and x3 = −x1 , consequently x21 = x23 . Now 0 ≤ c = 2x3 yields x ∈ Lℝ2 . Another important finite-dimensional example is the vector space V n of all symmetric n × n-matrices with the inner product ⟨A, B⟩ = tr(AB) .

(1.18)

Let Posn be the cone of all positive semidefinite matrices in Vn . It is well known that Posn is self-dual in Vn and that the identity matrix I is an order unit; see [53]. According to (1.12), the set Σ = {A ∈ Pos n ; tr(A) = 1} (1.19) is a base of Posn . In the next example we consider the case n = 2. Example 1.7.4. Let V2 be the space of all symmetric 2 × 2-matrices, ordered by the cone Pos2 of all matrices in V2 , which are positive semidefinite. We show that (V2 , Pos2 ) is order isomorphic to P2 (ℝ) in Example 1.7.1. Consider the linear bijection h : P2 (ℝ) → V2 ,

a (p : ℝ → ℝ, t 󳨃→ at2 + bt + c) 󳨃→ ( 1 2b

1 2 b)

c

.

(1.20)

1.7 Examples |

a

The matrix A = ( 1 b 2 by

1 2b

c

45

) is in Pos2 if and only if the eigenvalues of A, which are given

λ1/2 =

1 2

(a + c ± √(a + c)2 + b 2 − 4ac) ,

are both nonnegative. Furthermore, one has 0 ≤ λ1 , λ2 if and only if 0≤a+c

and

b 2 − 4ac ≤ 0 .

(1.21)

Indeed, if 0 ≤ λ1 , λ2 , then 0 ≤ λ1 + λ2 = a + c. Moreover, the inequality a + c ≥ √(a + c)2 + b 2 − 4ac implies 0 ≥ b 2 − 4ac. On the other hand, assuming (1.21), it is straightforward that 0 ≤ λ1 , λ2 . Next we show that (1.21) is equivalent to (i) or (ii) in Example 1.7.3. Indeed, the condition (i) yields 0 ≤ ac and hence 0 ≤ c, and therefore (1.21) is satisfied. Clearly, (ii) implies (1.21) as well. On the other hand, (1.21) implies (i) or (ii). Consequently, p ∈ P2 (ℝ) is positive if and only if A = h(p) ∈ Pos2 , i.e., j is bipositive. We conclude that V2 and P2 (ℝ) are order isomorphic. By Example 1.7.3, (V2 , Pos2 ) is order isomorphic to (ℝ3 , Lℝ2 ). Note that a similar statement in higher dimensions is not true, e.g., the spaces (V3 , Pos3 ) and (ℝ6 , Lℝ5 ) are not order isomorphic¹⁰. Next we consider the partially ordered vector space of affine functions on the unit circle and show that this space is order isomorphic to the space ℝ3 endowed with the ice cream cone. Example 1.7.5. Let S := {(ξ, η)T ; ξ 2 + η2 = 1} be the unit circle in ℝ2 and let X be the space of restrictions to S of all affine functions from ℝ2 into ℝ, that is, X := {x : S → ℝ, (ξ, η)T 󳨃→ μ1 ξ + μ 2 η + μ 3 ; μ 1 , μ2 , μ3 ∈ ℝ} . We endow X with the pointwise order, i.e., we consider the cone K := X ∩ ℝ+S in X. Consequently, X is Archimedean. As X contains the constant function 𝟙, the space X is directed. We will show that the space (X, K) is order isomorphic to the space (ℝ3 , Lℝ2 ) in Example 1.7.3. Indeed, consider the linear bijection q : ℝ3 → X,

(μ 1 , μ2 , μ3 )T 󳨃→ (x : S → ℝ, (ξ, η)T 󳨃→ μ1 ξ + μ 2 η + μ3 ) .

(1.22)

A straightforward calculation yields min{x(ξ, η); (ξ, η)T ∈ S} = μ 3 − √μ 21 + μ22 . Hence x is pointwise nonnegative on S if and only if μ 3 ≥ 0 and μ23 ≥ μ 21 + μ 22 . Therefore q is bipositive.

10 Note that the base of Lℝ5 is strictly convex, whereas the base of Pos3 is not; see, e.g., [106, Section 2.4].

46 | 1 A primer on ordered vector spaces By Example 1.7.3, the space (X, K) is also order isomorphic to the space P2 (ℝ) in Example 1.7.1. We conclude that X is order isomorphic to an order dense subspace of the infinite-dimensional Riesz space given in (1.13). We show that X is also order dense in C(S), which is due to [146, Example 3.6 (b)]. Denote for an angle φ ∈ (−π, π] the corresponding point in S by s φ , that is, s φ = (cos φ, sin φ)T . For every α ∈ (0, 2π ) and M ∈ [0, ∞) there is a z ∈ X such that z(s0 ) = −1, z ≥ −1 on S, and z(s φ ) ≥ M for every φ ∈ (−π, π] \ (−α, α). Indeed, define z : S → ℝ,

(ξ, η)T 󳨃→

M+1 1−cos α (1

− ξ) − 1 .

(1.23)

For every φ ∈ (−π, π] \ (−α, α) we have cos φ ≤ cos α and hence z(s φ ) ≥ M. The other properties of z are clear. We show that for every x ∈ C(S) one has x(s) = inf{u(s); u ∈ X, u ≥ x}

(1.24)

for every s ∈ S. Let x ∈ C(S). First, assume x ≥ 0. Because of rotation symmetry, it suffices to prove the equation for s = s0 , where s0 = (1, 0)T . Let ε ∈ (0, ∞). As x is continuous, there exists an α ∈ (0, 2π ) such that x(s φ ) ≤ x(s0 ) + ε for every φ ∈ (−α, α). Observe that x is bounded on S and take M ∈ [0, ∞) such that εM ≥ x on S. For these M and α, take a z ∈ X as in (1.23). Let y := (x(s0 ) + 2ε)𝟙 + εz. Then y ∈ X, y(s0 ) = x(s0 ) + ε, y(s φ ) ≥ x(s0 ) + ε ≥ x(s φ ) for every φ ∈ (−α, α), and y(s φ ) ≥ εz(s φ ) ≥ εM ≥ x(s φ ) for every φ ∈ (−π, π] \ (−α, α), so y ≥ x on S. Thus, inf{u(s0 ); u ∈ X, u ≥ x} ≤ y(s0 ) = x(s0 ) + ε . Hence we obtain (1.24) for s = s0 , and therefore for every s ∈ S. Now let x ∈ C(S) be arbitrary. As X is majorizing in C(S), there is v ∈ X such that v ≥ −x. Then x+v ≥ 0, hence formula (1.24) holds for x+v instead of x. By subtracting v, we have established (1.24) for x. It follows that X is order dense in C(S). So far, we have provided several different representations of the space ℝ3 endowed with the ice cream cone Lℝ2 , and we have used them to embed (ℝ3 , Lℝ2 ) order densely into a Riesz space. A higher-dimensional generalization will be considered in Theorem 2.6.4 below.

1.7.2 Polyhedral cones We consider in X := ℝn cones with finitely many faces. We endow ℝn with the Euclidean norm ‖⋅‖ and identify X 󸀠 with ℝn , as usual. Definition 1.7.6. A cone K in X is called polyhedral if there are finitely many elements x(1) , . . . , x(r) ∈ K \ {0} such that K = pos {x(1) , . . . , x(r) } ,

(1.25)

1.7 Examples | 47

where r and x(1) , . . . , x(r) are such that for every i ∈ {1, . . . , r} the ray pos{x(i)} is a face of K, and pos{x(i)} ≠ pos{x(j) } whenever i ≠ j. Note that the standard cone ℝ+n is a polyhedral cone, where r = n and x(1) , . . . , x(n) can be chosen to be the standard unit vectors in ℝn . In this subsection, let K be a generating polyhedral cone in X, which means that r ≥ n. We list properties of the ordered normed space (X, K, ‖⋅‖) without proofs; for details see [14]. Note that basic results in polyhedral geometry can also be found in [167]. As K is closed, by Proposition 1.5.2 the space X is Archimedean. Since K is generating, the dual wedge K 󸀠 is a cone. Moreover, due to Proposition 1.5.13 one has int(K) ≠ ⌀. We fix an element u ∈ int(K). Thus we have an order unit space, and by Theorem 1.5.21 the set Σ = {f ∈ K 󸀠 ; f(u) = 1} is a base of K 󸀠 . The convex set Σ has finitely many extreme points f (1) , . . . , f (k) , where k ≥ n. Therefore the cone K 󸀠 is polyhedral with K 󸀠 = pos{f (1) , . . . , f (k) }. We also have that K 󸀠 is generating and, hence, has a nonempty interior. The polyhedral cone K can be represented by means of the functionals f (1) , . . . , f (k) as the intersection of halfspaces, namely K = {x ∈ ℝn ; ∀i ∈ {1, . . . , k} : f (i) (x) ≥ 0} . (1.26) The representation (1.26) can also be used to construct polyhedral cones. Indeed, if one fixes finitely many linear functionals f (1) , . . . , f (k) that span X 󸀠 , then K defined by (1.26) is either {0} or a polyhedral cone. It is only true in special cases that X and X 󸀠 are order isomorphic. Even the number of extreme rays of K and K 󸀠 need not be the same, in general. More precisely, if n ∈ {1, 2, 3}, then one always has r = k. For n = 4, in [31] an example is given where r = 5 and k = 6, confer also [14, p. 147]. Next we present another example; see also [87]. Example 1.7.7. Let X = ℝn with n ≥ 2 and let e(1) , . . . , e(n) be the standard unit vectors. For i ∈ {1, . . . , n − 1} define x(i) := e(i) + e(n)

and

y(i) := −e(i) + e(n) ,

and consider the cone K in X consisting of their positive linear combinations, i.e., K := pos {x(1) , . . . , x(n−1) , y(1) , . . . , y(n−1) } . Then K is a closed cone in X and e(n) is an interior point of K, so e(n) is an order unit. We show that the dual cone K 󸀠 satisfies K 󸀠 = pos {φ ε ; ε ∈ {1, −1} n−1 } , where n−1

φ ε (x) := x n + ∑ ε j x j , j=1

x = (x1 , . . . , x n )T ∈ ℝn ,

ε = (ε1 , . . . , ε n−1 ) ∈ {1, −1} n−1 .

48 | 1 A primer on ordered vector spaces Indeed, it is clear that for every ε ∈ {1, −1} n−1 and i ∈ {1, . . . , n − 1} we have φ ε (x(i) ) ≥ 0 and φ ε (y(i) ) ≥ 0, so φ ε ∈ K 󸀠 . Hence, L := pos {φ ε ; ε ∈ {1, −1} n−1 } ⊆ K 󸀠 . For a proof that L ⊇ K 󸀠 , we first show that every x ∈ X with ψ(x) ≥ 0 for every ψ ∈ L satisfies x ∈ K. Consider such an x. Since for every ε ∈ {1, −1} n−1 we have n−1

0 ≤ φ ε (x) = x n + ∑ ε i x i , i=1

it follows that x n ≥ ∑n−1 i=1 −ε i x i . Therefore, n−1

x n ≥ ∑ |x i | . i=1

As x(i) − y(i) = 2e(i) and x(i) + y(i) = 2e(n) for every i ∈ {1, . . . , n − 1}, we obtain that n

n−1

2x = 2 ∑ x i e(i) = 2x n e(n) + ∑ x i (x(i) − y(i) ) i=1

i=1 n−1

n−1

= 2x n e(n) + ∑ ((x i + |x i |) x(i) + (|x i | − x i ) y(i) ) − ∑ |x i | (x(i) + y(i) ) i=1

i=1

n−1

n−1

n−1

i=1

i=1

i=1

= 2 (x n − ∑ |x i |) e(n) + ∑ (x i + |x i |) x(i) + ∑ (|x i | − x i ) y(i) ∈ K . Hence, x ∈ K. Now suppose that L ⊇ K 󸀠 does not hold and let φ ∈ K 󸀠 \ L. According to Theorem 1.8.2 below, there exists x ∈ X and c ∈ ℝ such that ψ(x) ≥ c for every ψ ∈ L and φ(x) < c. Then c ≤ 0. For every ψ ∈ L and m ∈ ℕ we have mψ ∈ L, so mψ(x) ≥ c, hence ψ(x) ≥ mc and therefore ψ(x) ≥ 0. Thus, x ∈ K, which contradicts φ(x) < c ≤ 0. We conclude that K 󸀠 = L. Observe that the base Σ := {f ∈ K 󸀠 ; f(e(n) ) = 1} of K 󸀠 has the set Λ = {φ ε ; ε ∈ {1, −1} n−1 } as set of extreme points. Therefore, the cone K 󸀠 has 2n−1 extreme rays, whereas K has 2n − 2 extreme rays. For n = 2 these numbers both equal 2 and for n = 3 they are 4. For n ≥ 4 the numbers of extreme rays of K and K 󸀠 are not the same. Next we consider a subspace D of X with the induced order. Then by (1.26) either D ∩ K equals {0}, or D ∩ K is a polyhedral cone in D. In particular, subspaces of (ℝk , ℝ+k ) have either trivial order or are spaces with polyhedral cones. The converse, namely an appropriate embedding of spaces with polyhedral cones into (ℝk , ℝ+k ), will be discussed in Section 2.6 below. The polyhedral cones in ℝn with precisely n one-dimensional faces, i.e., r = n, correspond to lattice cones in ℝn , as is stated next.

1.7 Examples |

49

Theorem 1.7.8. Let (X, K) be an n-dimensional Archimedean directed partially ordered vector space. Then the following statements are equivalent. (i) There are x(1) , . . . , x(n) ∈ X such that K = pos{x(1) , . . . , x(n) }. (ii) (X, K) has the Riesz decomposition property. (iii) (X, K) is a Riesz space. (iv) (X, K) is order isomorphic to (ℝn , ℝ+n ). Note that in Theorem 1.7.8 (i) the vectors x(1) , . . . , x(n) are linearly independent, as otherwise K would not be generating. The implication from (i) to (iv) is straightforward. The equivalence of (iii) and (iv) is due to Yudin [163], and the equivalence of (i) and (ii) is shown in [46]. The implication from (iii) to (i) can be found in [105, Theorem 2.4]. Example 1.7.9. For n ≥ 3, the Lorentz cone Lℝn−1 in ℝn is not polyhedral. Therefore, by Theorem 1.7.8, the space (ℝn , Lℝn−1 ) does not have the Riesz decomposition property. Spaces of linear operators on spaces with polyhedral cones reflect the properties of the underlying space. We list some related statements. Recall that by Corollary 1.5.24 every linear operator between finite-dimensional Archimedean directed spaces is orderbounded. Remark 1.7.10. (i) If (X, K X ) and (Y, K Y ) are finite-dimensional Archimedean vector lattices, then by Theorem 1.7.8 the space Y is Dedekind complete, thus (L(X, Y), L+ (X, Y)) is a Dedekind complete vector lattice by Theorem 1.2.8. (ii) For X := Y := ℝ2 , there are no generating Archimedean cones K X in X and K Y in Y such that the cone Pos2 of positive semidefinite matrices in V2 ⊆ L(X, Y) is induced by the cone of positive operators. Indeed, assume the contrary. By Theorem 1.7.8 every generating Archimedean cone in ℝ2 is a lattice cone, thus, by (i), the cone L+ (X, Y) is a lattice cone in L(X, Y). Hence, on the one hand, L+ (X, Y)∩V2 is polyhedral. On the other hand, by Example 1.7.4, the space (V2 , Pos2 ) is order isomorphic to (ℝ3 , Lℝ2 ), where Lℝ2 is not a polyhedral cone, by Example 1.7.9. We conclude that V2 is not order isomorphic to a subspace of L(X, Y). (iii) In the vector lattices (ℝn , ℝ+n ) and (ℝm , ℝ+m ) we consider the standard algebraic basis of unit vectors, respectively, and for A ∈ L(ℝn , ℝm ) the corresponding matrix representation (a ij )i,j . Then A ∈ L+ (ℝn , ℝm ) if and only if a ij ≥ 0 for every i, j. (iv) Let ℝn be endowed with a generating polyhedral cone K and ℝm with a polyhedral cone C. Schneider and Vidyasagar showed in [135, Lemma 9] that then L+ (ℝn , ℝm ) is a polyhedral cone in L(ℝn , ℝm ). The idea of the proof is as follows. Using for elements of Z := L(ℝn , ℝm ) matrix representations as in (iii), a scalar product in Z is defined by ⟨A, B⟩ = tr(AT B) for A, B ∈ Z. (Note that (1.18) is a special case.) We represent the cone K as in (1.25), i.e., K = pos{x(1) , . . . , x(r) }, and the cone C as in (1.26), i.e., C = {x ∈ ℝm ; ∀i ∈ {1, . . . , k} : f (i) (x) ≥ 0}, where we take x(j) and f (i) such that they have the appropriate properties as discussed above. Note that A ∈ Z is positive if and only if for every j ∈ {1, . . . , r} and i ∈ {1, . . . , k} one has

50 | 1 A primer on ordered vector spaces f (j) (Ax(i) ) ≥ 0, which means for A := (a qp )q,p that m

n

(i) (j)

∑ ∑ a qp f q x p ≥ 0 .

(1.27)

q=1 p=1 (i) (j)

Consider in Z the finitely many elements A(ij) := (f q x p )q,p for j ∈ {1, . . . , r} and i ∈ {1, . . . , k}. Then (1.27) is satisfied if and only if ⟨A(ij) , A⟩ ≥ 0. We conclude that the cone L+ (ℝn , ℝm ) has a representation as in (1.26), hence it is polyhedral.

1.8 Odds and ends from functional analysis and topology In this section we collect well-known results from real and functional analysis. These results will be used freely throughout the book. For the following two versions of the Hahn–Banach theorem, see [116, Theorems 1.9.5 and 2.2.28]. For a vector space V, a map p : V → ℝ is called subadditive if for every x, y ∈ V we have p(x+y) ≤ p(x)+p(y), and positively homogeneous if for every x ∈ V and λ ∈ ℝ+ we have p(λx) = λp(x). Theorem 1.8.1. Let V be a vector space and p : V → ℝ a subadditive positively homogeneous functional. Let D be a linear subspace of V and f : D → ℝ a linear functional such that f ≤ p on D. Then there exists a linear functional F : V → ℝ such that F = f on D, and F ≤ p on V. Theorem 1.8.2. Let (V, τ) be a locally convex topological vector space. Let A and B be disjoint nonempty convex subsets of V such that A is compact and B is closed. Then there exist a continuous linear functional f : V → ℝ and α, β ∈ ℝ with α < β such that f ≤ α on A and f ≥ β on B. The following related statement in [116, Corollary 2.2.20] is due to Mazur. Theorem 1.8.3. Let (V, τ) be a locally convex topological vector space, let D be a closed linear subspace of V, and let x ∈ V \ D. Then there exists a continuous linear functional f : V → ℝ such that f(x) = 1 and f = 0 on D. The next result is sometimes called Banach’s isomorphism theorem; see [116, Corollary 1.6.6]. Theorem 1.8.4. Let X be a Banach space and let T : X → X a bounded linear bijection. Then T −1 is bounded. For an overview of weak topologies we refer to [116, Chapter 2]. The next result is given in [116, Proposition 2.6.4]. Theorem 1.8.5. Let X be a normed space and let X 󸀠 be its dual space. For every functional F : X 󸀠 → ℝ, which is weakly-∗ continuous there is x ∈ X such that for every f ∈ X 󸀠 we have F(f) = f(x).

1.8 Odds and ends from functional analysis and topology

|

51

The following Banach–Alaoglu theorem can be found, e.g., in [116, Theorem 2.6.18]. Theorem 1.8.6. Let X be a normed space and let X 󸀠 be its dual space. Then the unit ball in X 󸀠 is weakly-∗ compact. In a topological vector space (V, τ) consider for a set C ⊆ V the convex hull co(C) and its closure co(C). Recall that an element v ∈ C is an extreme point of C if whenever there are v1 , v2 ∈ C and t ∈ (0, 1) such that v = tv1 + (1 − t)v2 , then v = v1 = v2 . The set of extreme points of C is denoted by ext(C). The following Krein–Milman theorem is given in [116, Theorem 2.10.6]. Theorem 1.8.7. Let (V, τ) be a locally convex topological vector space that is Hausdorff. Let C be a nonempty compact convex subset of V. Then the set ext(C) is nonempty, and co(ext(C)) = C. For the next result by Milman, see [116, Theorem 2.10.15]. Theorem 1.8.8. Let (V, τ) be a locally convex topological vector space that is Hausdorff. Let C be a nonempty compact subset of V such that co(C) is compact. Then ext(co(C)) ⊆ C. Next we state the version of Urysohn’s lemma needed in this book; see, e.g., [11, Theorem 7.7]. Theorem 1.8.9. Let Ω be a nonempty compact Hausdorff space and let A and B be nonempty disjoint closed subsets of Ω. Then there is a continuous function f : Ω → ℝ with 0 ≤ f ≤ 1, f = 0 on A and f = 1 on B. For the following Stone–Weierstrass theorem, see, e.g., [11, Theorem 8.3]. Theorem 1.8.10. Let Ω be a nonempty compact Hausdorff space and let C(Ω) be endowed with the supremum norm. Let X be a Riesz subspace of C(Ω) that contains the constant function 𝟙 and separates the points of Ω, i.e., for every ω1 , ω2 ∈ Ω with ω1 ≠ ω2 there is x ∈ X such that x(ω1 ) ≠ x(ω2 ). Then X is norm dense in C(Ω). The next fact from topology in [37, Corollary 13.23] is sometimes useful. Proposition 1.8.11. Let Ω and Γ be nonempty topological spaces such that Ω is compact and Γ is Hausdorff. Then every continuous bijection from Ω to Γ is a homeomorphism. Next we state Brouwer’s fixpoint theorem; see, e.g., [141, Theorem 17.14]. Theorem 1.8.12. Let F be a compact convex subset of a finite-dimensional Banach space, and let a map A : F → F be continuous. Then there is y ∈ F such that A(y) = y. We conclude the section by a version of the monotone convergence theorem. Theorem 1.8.13. Let (S, A, μ) be a measure space and let (f n )n∈ℕ be a sequence of Ameasurable functions f n : S → ℝ, where ℝ is endowed with the Borel σ-algebra. If for

52 | 1 A primer on ordered vector spaces every n ∈ ℕ and for every s ∈ S we have 0 ≤ f n (s) ≤ f n+1 (s), then the function f : S → ℝ, s 󳨃→ lim n→∞ f n (s) is A-measurable and satisfies lim ∫ f n dμ = ∫ f dμ .

n→∞

S

S

Notes and Remarks Some of the early works on partially ordered vector spaces are due to F. Riesz (1929 and 1936), L. V. Kantorovich (1935) and H. Freudenthal (1936). For an overview on the history of the subject see the commentary by Martin Weber in [162]. Let us briefly sketch some of the motivations leading to the development of the theory of partially ordered vector spaces. The early years of functional analysis have been devoted to the abstraction of joint structures in sequence and function spaces. The most attention has been addressed to the metric structure, but also an abstraction of the order structure has been investigated. Motivated by the study of integral operators with positive kernels and matrices with nonnegative entries, the notion of a positive operator has been introduced. The foundations of the theory on positive operators were, e.g., laid in the monographs [25, 90]. Spectral properties of positive operators have later on led to a long list of applications. In [105] a generalization of the classical Perron–Frobenius theorem to a setting of ordered Banach spaces is given, which we will discuss in Section 5.5 below. An area in which the spectral theory of positive operators is applied is the vast field of positive dynamical systems; see, e.g., [23, 68, 102, 106] and references therein. In an abstract functional analytic setting, this has led to the theory of positive operator semigroups; see [38, 123]. Positivity is also a natural ingredient in economic theory; for a modern treatment see [13, Chapter 8]. The topics in the present book are inspired by the rich theory of vector lattices, of which we like to mention [12, 13, 43, 111, 119, 134, 154]. Also the literature on ordered topological vector spaces has been an inspiration; see, e.g., [55, 124, 127, 133]. From the monographs on general partially ordered vector spaces we mainly used [14, 72, 162]. Our theory bases on [151], which is elaborated in the subsequent Chapter 2.

2 Embeddings, covers, and completions The great advantage of pre-Riesz spaces is that they can be studied with the aid of order dense embeddings into vector lattices. In this chapter, we discuss three types of order dense embedding. We begin by recalling the classical Dedekind completion, which amounts to the following statement: 1. If (X, K) is an Archimedean directed ordered vector space, then there is a Dedekind complete Riesz space X δ and an embedding J : X → X δ such that J[X] is order dense in X δ . In many examples the Dedekind completion is difficult to describe explicitly. For instance, the Dedekind completion of C[0, 1] is involved; see, e.g., [41]. For partially ordered vector spaces, embedding into a (not necessarily Dedekind complete) Riesz space is of interest. For this purpose, van Haandel introduced pre-Riesz spaces in [151] (see Definition 2.2.1 below) and studied their vector lattice covers. The main result is as follows: 2. If (X, K) is a pre-Riesz space, then there is a Riesz space V and an embedding i : X → V such that i[X] is order dense in V. There is a smallest vector lattice cover called the Riesz completion. An embedding into a Riesz space of continuous functions is even more convenient, since there the pointwise order is available. We show that the classical functional representation of Kadison [75] provides an order dense embedding and therefore a vector lattice cover. The result reads as follows: 3. If (X, K) is an Archimedean ordered vector space with order unit, then there is a compact Hausdorff space Ω and an embedding Φ : X → C(Ω) such that Φ[X] is order dense in C(Ω). We derive explicit functional representations of several familiar examples, such as spaces with polyhedral cones, Lorentz cones, and the cone of positive semidefinite matrices. The statements in 1., 2. and 3. are given in Theorems 2.1.13, 2.4.5, and 2.5.9 below.

https://doi.org/10.1515/9783110476293-002

54 | 2 Embeddings, covers, and completions

2.1 Dedekind completion The construction of the Dedekind completion can be traced back to Richard Dedekind’s construction of the real numbers in 1872; see [47]. He defines the real numbers as the set of Dedekind cuts of rational numbers endowed with suitable arithmetic operations. In 1937, MacNeille observed in [112] that Dedekind cuts can be considered in any partially ordered set X and that they constitute a Dedekind complete set X δ in which X can be embedded. The set X δ is called the Dedekind completion of X. For a partially ordered Abelian group X, the set X δ can be endowed with an extension of the group operation. In [39], Clifford discusses that X δ is a group itself if and only if X is Archimedean. Similarly, the Dedekind completion of an Archimedean directed partially ordered vector space is described in [164]; see also [162, Section X.3]. Although this construction is well-known, we give a detailed proof here, where we highlight the role of the Archimedean property. Later on in Section 2.4, the construction of vector lattice covers will use the same procedure, and the Archimedean condition will be replaced by the pre-Riesz condition.

2.1.1 Dedekind cuts Let (X, K) be a partially ordered vector space. Recall that for a set A ⊆ X we denote the set of upper bounds by Au and the set of lower bounds by Al . The next lemma presents elementary properties, which we use freely in the sequel. Lemma 2.1.1. For A ⊆ X and x, y ∈ X we have {x}ul = {x}l , {x + y}ul = {x}l + {y}l , (−A)u = −Al

and

(−A)l = −Au ,

A ⊆ Aul , Aulu = Au , Alul = Al . The observation (Aul )ul = Aul motivates the following definition. Definition 2.1.2. A subset A of X is called a (Dedekind) cut if Aul = A. We denote X δ = {A ⊆ X; Aul = A} \ {⌀, X} . Note that for every x ∈ X the set {x}l is a cut. Since the inclusion relation on the power set P(X) is a partial order, it is clear that the set X δ ⊂ P(X) endowed with the inclusion ⊆ is a partially ordered set.

2.1 Dedekind completion |

55

A partially ordered set is a Dedekind complete lattice whenever the supremum and infimum exist for any two elements, and for every subset that is bounded above the supremum exists, and for every subset that is bounded below the infimum¹. Proposition 2.1.3. Consider the partially ordered set X δ ordered by inclusion ⊆ and let A ⊆ X δ be nonempty. (i) If A is bounded above, we have sup A = (⋃ A)ul . (ii) If A is bounded below, we have inf A = ⋂ A. (iii) If (X, K) is directed, then the set X δ ordered by inclusion ⊆ is a Dedekind complete lattice. Proof. (i) Suppose that A is bounded above and that A0 ∈ X δ is such that A ⊆ A0 for every A ∈ A. To see that (⋃ A)ul ∈ X δ , note that ⋃ A ⊆ A0 , so that (⋃ A)ul ⊆ Aul 0 = A0 . As the elements of A are nonempty, (⋃ A)ul ⊇ ⋃ A is nonempty, so (⋃ A)ul ∈ X δ . For a proof that (⋃ A)ul is the supremum of A, observe that for every A ∈ A we have (⋃ A)ul ⊇ ⋃ A ⊇ A. Hence, (⋃ A)ul is an upper bound of A. If U ∈ X δ is another upper bound of A, then U ⊇ A for every A ∈ A, so U ⊇ ⋃ A, hence U = U ul ⊇ (⋃ A)ul . Thus, (⋃ A)ul is the supremum of A. (ii) Let A be bounded below. Take A0 ∈ X δ such that A ⊇ A0 for every A ∈ A. Then (⋂ A)ul ⊇ ⋂ A ⊇ A0 , so (⋂ A)ul is nonempty. For A ∈ A we have A ⊇ ⋂ A, so ul

A = Aul ⊇ (⋂ A)

for every A ∈ A .

(2.1)

This yields firstly that (⋂ A)ul ≠ X and therefore, (⋂ A)ul ∈ X δ and, secondly, that (⋂ A)ul is a lower bound of A. If V ∈ X δ is another lower bound of A, then V ⊆ A for every A ∈ A, so V ⊆ ⋂ A, hence V = V ul ⊆ (⋂ A)ul . Thus, (⋂ A)ul is the infimum of A. By (2.1), ⋂ A ⊇ (⋂ A)ul ⊇ ⋂ A, hence (⋂ A)ul = ⋂ A. (iii) Assume that (X, K) is directed. Let A, B ∈ X δ . We show that {A, B} is bounded below and bounded above in X δ . We have A = Aul , A ≠ ⌀ and Aul ≠ X, so there exist a ∈ A and u ∈ Au . Similarly, there exist b ∈ B and v ∈ Bu . Since X is directed, there are c, w ∈ X with c ≤ a, b and w ≥ u, v. Let C = {c}l and W = {w}l . Then C, W ∈ X δ . We have A ⊆ W and B ⊆ W, so {A, B} is bounded above in X δ . For x ∈ C we have x ≤ c ≤ a and x ≤ c ≤ b, so x ∈ Aul = A and x ∈ Bul = B. Hence, C ⊆ A and C ⊆ B, so {A, B} is bounded below in X δ . It follows from (i) and (ii) that the supremum and infimum of {A, B} exist in X δ . From now on, we assume X to be a directed partially ordered vector space. As usual, we denote the lattice operations in X δ by ∨ and ∧. Next we discuss the embedding of X

1 Note the difference with the notion of a complete lattice, which includes, in addition, the existence of a largest and a smallest element.

56 | 2 Embeddings, covers, and completions into X δ . In the subsequent proposition, a map j : V → Y between partially ordered sets V, Y is called bipositive if for every u, v ∈ V we have u ≤ v if and only if j(u) ≤ j(v). Proposition 2.1.4. The map J : X → Xδ ,

x 󳨃→ {x}l

(2.2)

is bipositive and J[X] is order dense in X δ . Proof. Let x, y ∈ X. If x ≤ y, then {x}l ⊆ {y}l by transitivity of the order relation. Conversely, if {x}l ⊆ {y}l , then x ∈ {x}l yields that x ≤ y. Hence J is bipositive. Next we show that J[X] is order dense in X δ . Let A ∈ X δ . We have to show that A = inf{J(x); x ∈ X, A ⊆ J(x)} , that is, A = ⋂ {x}l , x∈U

{x}l }.

First we establish that U = Au . Indeed, if x ∈ U, then where U = {x ∈ X; A ⊆ lu u u x ∈ {x} ⊆ A . If x ∈ A , then A ⊆ {x}l , so x ∈ U. Next, for x ∈ U we have {x}l ⊇ A, so A ⊆ ⋂x∈U {x}l . For y ∈ ⋂x∈U {x}l we have y ≤ x for all x ∈ U = Au , so y ∈ Aul = A, so ⋂ x∈U {x}l ⊆ A. Hence, A = ⋂ x∈U {x}l . Thus, J[X] is order dense in X δ .

2.1.2 Addition and scalar multiplication on the set of Dedekind cuts Let X again be a directed partially ordered vector space. We aim to endow X δ with a vector space structure. For A, B ∈ X δ and λ ∈ ℝ define A ⊕ B := (A + B)ul , ⊖A := −Au , λA, { { { λ ∗ A := {{0}l , { { u {λA ,

λ >0, λ =0, λ 0, and μ < 0, we have (A + b)u = Au + b , λAu = (λA)u , μAu = (μA)l

(A + b)l = Al + b , λAl = (λA)l ,

and

μAl = (μA)u .

2.1 Dedekind completion | 57

Lemma 2.1.6. For every nonempty A, B ⊆ X we have (A + B)ul = (Aul + B)ul = (Aul + Bul )ul ul

ul

ul

ul

and

ul ul

(A ∪ B) = (A ∪ B) = (A ∪ B ) . Proof. By Lemma 2.1.5, we have (A + B)u = ⋂ (A + b)u = ⋂ (Au + b) = ⋂ (Aulu + b) b∈B

b∈B ul

u

b∈B ul

u

= ⋂ (A + b) = (A + B) b∈B

and therefore (A + B)ul = (Aul + B)ul . By application of the previous argument to B and Aul instead of A and B, it follows that (Aul + B)ul = (Aul + Bul )ul . For the second identity, note that (A ∪ B)ul ⊆ (Aul ∪ B)ul ⊆ (Aul ∪ Bul )ul . It remains to show that (A ∪ B)u ⊆ (Aul ∪ Bul )u . If u ∈ (A ∪ B)u , then u ∈ Au = Aulu , so u ≥ x for every x ∈ Aul . Similarly, u ≥ x for every x ∈ Bul , so that u ∈ (Aul ∪ Bul )u . Lemma 2.1.7. For every A, B, C ∈ X δ and λ, μ ∈ ℝ, one has (i) (A ⊕ B) ⊕ C = A ⊕ (B ⊕ C), (ii) A ⊕ B = B ⊕ A, (iii) A ⊕ {0}l = A, (iv) ⊖(⊖A) = A, (v) A ⊕ (⊖A) ⊆ {0}l . (vi) λ ∗ (μ ∗ A) = (λμ) ∗ A, (vii) 1 ∗ A = A (viii) (−1) ∗ A = ⊖A. Proof. (i) Lemma 2.1.6 yields (A ⊕ B) ⊕ C = ((A + B)ul + C)ul = ((A + B) + C)ul = (A + (B + C))ul = (A + (B + C)ul )ul = A ⊕ (B ⊕ C) . (ii) By definition, A ⊕ B = (A + B)ul = B ⊕ A. (iii) For a ∈ A and x ∈ X with x ≤ 0, we have a + x ≤ a, so a + x ∈ Aul = A. Hence, A + {0}l ⊆ A and therefore, ul

A ⊕ {0}l = (A + {0}l ) ⊆ Aul = A . Since 0 ∈ {0}l , we also have A ⊕ {0}l = (A + {0}l )ul ⊇ Aul = A. Thus, A ⊕ {0}l = A .

58 | 2 Embeddings, covers, and completions

(iv) By definition, ⊖(⊖A) = −(−Au )u . For a ∈ A and u ∈ Au , we have a ≤ u, so −a ≥ −u, so that −a ∈ (−Au )u , hence a ∈ −(−Au )u . Thus, A ⊆ ⊖(⊖A) . If x ∈ −(−Au )u , then for every y ∈ −Au we have −x ≥ y, so x ≤ −y. Hence, x ∈ Aul = A. Thus, ⊖(⊖A) ⊆ A, and therefore, ⊖(⊖A) = A . (v)

Note that A ⊕ (⊖A) = (A + (−Au ))

ul

.

If a ∈ A and x ∈ (−Au ), then −x ≥ a, so a + x ≤ 0. Hence, 0 ∈ (A + (−Au ))u and therefore, ul (A + (−Au )) ⊆ {0}l . (vi) If λ, μ > 0, then λ ∗ (μ ∗ A) = λ(μA) = (λμ)A = (λμ) ∗ A. For the other cases we make use of Lemma 2.1.5. If λ > 0, we have λ{0}l = {λ0}l = {0}l , so λ ∗ (0 ∗ A) = λ{0}l = {0}l = (λ0) ∗ A. If λ < 0, then λ ∗ (0 ∗ A) = λ({0}l )u = λ{0}u = {λ0}l = {0}l = (λ0) ∗ A. Similarly, for any μ, 0 ∗ (μ ∗ A) = {0}l = (0μ) ∗ A. Further, if λ > 0 and μ < 0, then λ ∗ (μ ∗ A) = λ(μAu ) = λμAu = (λμ) ∗ A. If λ < 0 and μ > 0, then λ ∗ (μ ∗ A) = λ(μA)u = λ(μAu ) = (λμ)Au = (λμ) ∗ A. If λ < 0 and μ < 0, then λ ∗ (μ ∗ A) = λ(μAu )u = λ((μA)l )u = (λ(μA)l )l = (λμA)ul = (λμ)Aul = (λμ)A = (λμ) ∗ A. (vii) By definition, 1 ∗ A = 1A = A. (viii) We have (−1) ∗ A = (−1)Au = −Au = ⊖A. It is in general not true that A ⊕ (⊖A) = {0}l if X is not Archimedean. We postpone the discussion of this issue to Lemma 2.1.12 and first proceed with properties of the operations ⊕, ⊖ and ∗ that hold in general. For convenience, for A, B ∈ X δ we write A ⊕ ⊖B := A ⊕ (⊖B). Lemma 2.1.8. For A, B ∈ X δ and λ, μ ∈ ℝ, one has (i) if A ⊕ B = {0}l , then B = ⊖A, (ii) if λ ≥ 0, then λ ∗ (A ⊕ B) = (λ ∗ A) ⊕ (λ ∗ B), (iii) if A ⊕ ⊖A = {0}l and B ⊕ ⊖B = {0}l , then λ ∗ (A ⊕ B) = (λ ∗ A) ⊕ (λ ∗ B) and ⊖(A ⊕ B) = (⊖A) ⊕ (⊖B), (iv) if λ, μ ≥ 0, then (λ + μ) ∗ A = (λ ∗ A) ⊕ (μ ∗ A), (v) if A ⊕ ⊖A = {0}l , then (λ + μ) ∗ A = (λ ∗ A) ⊕ (μ ∗ A).

2.1 Dedekind completion |

59

Proof. (i) The assumption A ⊕ B = {0}l yields that (A + B)ul = {0}l . So for each a ∈ A and b ∈ B we have a + b ≤ 0, hence b ≤ −a. Therefore, B ⊆ (−A)l = −Au = ⊖A. For the converse inclusion, by Lemma 2.1.7 we get ⊖A = (⊖A) ⊕ {0}l = (⊖A) ⊕ (A ⊕ B) = ((⊖A) ⊕ A) ⊕ B = (A ⊕ (⊖A)) ⊕ B ⊆ {0}l ⊕ B = B ⊕ {0}l = B . (ii) If λ > 0, then λ ∗ (A ⊕ B) = λ(A + B)ul = (λ(A + B))ul = (λA + λB)ul = (λ ∗ A + λ ∗ B)ul = (λ ∗ A) ⊕ (λ ∗ B) and ul

0 ∗ (A ⊕ B) = {0}l = ({0}l + {0}l ) ul

= ((0A)l + (0B)l ) = (0 ∗ A + 0 ∗ B)ul = (0 ∗ A) ⊕ (0 ∗ B) . (iii) If λ ≥ 0, the assertion follows from (ii). Consider the case that λ < 0. According to Lemma 2.1.7 (vi) and (viii) we have λ ∗ A = (−λ) ∗ ((−1) ∗ A) = (−λ) ∗ (⊖A) and, similarly, λ∗ B = (−λ)∗(⊖B). With the aid of Lemma 2.1.7 (i) and (ii) we obtain that (A ⊕ B) ⊕ ((⊖A) ⊕ (⊖B)) = (A ⊕ (⊖A)) ⊕ (B ⊕ (⊖B)) = {0}l ⊕ {0}l ul

ul

= ({0}l + {0}l ) = ({0}l ) = {0}l , so, by (i), ⊖(A ⊕ B) = (⊖A) ⊕ (⊖B). Therefore, (ii) yields that λ ∗ (A ⊕ B) = (−λ) ∗ ⊖(A ⊕ B) = (−λ) ∗ ((⊖A) ⊕ (⊖B)) = ((−λ) ∗ (⊖A)) ⊕ ((−λ) ∗ (⊖B)) = (λ ∗ A) ⊕ (λ ∗ B) . (iv) On the one hand, (λ + μ) ∗ A = (λ + μ)A ⊆ (λA) + (μA) = λ ∗ A + μ ∗ A ⊆ (λ ∗ A + μ ∗ A)ul = (λ ∗ A) ⊕ (μ ∗ A) .

60 | 2 Embeddings, covers, and completions On the other hand, if u ∈ Au , then for every a1 , a2 ∈ A we have (λ+μ)u ≥ λa1 +μa2 , so (λ + μ)u ∈ (λA + μA)u . Hence, (λ + μ)Au ⊆ (λA + μA)u and thus (λ + μ)A = (λ + μ)Aul ⊇ (λA + μA)ul = (λA) ⊕ (μA) = (λ ∗ A) ⊕ (μ ∗ A) . (v) If λ, μ ≥ 0, then the assertion follows from (iv). Next, consider the case that λ + μ ≥ 0, λ ≥ 0 and μ < 0. Then by (iv) we have λ ∗ A = ((λ + μ) ∗ A) ⊕ ((−μ) ∗ A) , so, by Lemma 2.1.7, (λ ∗ A) ⊕ (μ ∗ A) = ((λ + μ) ∗ A) ⊕ ((−μ) ∗ A) ⊕ (μ ∗ A) = ((λ + μ) ∗ A) ⊕ ((−μ) ∗ A) ⊕ ((−μ) ∗ (⊖A)) = ((λ + μ) ∗ A) ⊕ (−μ) ∗ (A ⊕ (⊖A)) = ((λ + μ) ∗ A) ⊕ (−μ) ∗ {0}l = ((λ + μ) ∗ A) ⊕ {0}l = (λ + μ) ∗ A .

(2.3)

By commutativity of ⊕, we also obtain the desired identity if λ + μ ≥ 0, λ < 0 and μ ≥ 0. If λ + μ < 0, then −λ ≥ 0 or −μ ≥ 0 and we use (2.3) to obtain (λ + μ) ∗ A = (−(λ + μ)) ∗ (⊖A) = ((−λ) ∗ (⊖A)) ⊕ ((−μ) ∗ (⊖A)) = (λ ∗ A) ⊕ (μ ∗ A). Compatibility properties of the partial order ⊆ in X δ with the operations ⊕ and ∗ are contained in the following lemma. Lemma 2.1.9. Let A, B, C ∈ X δ and λ ∈ ℝ. (i) If A ⊆ B, then A ⊕ C ⊆ B ⊕ C. (ii) If A ⊆ B and λ ≥ 0, then λ ∗ A ⊆ λ ∗ B. Proof. (i) If A ⊆ B, then A ⊕ C = (A + C)ul ⊆ (B + C)ul = B ⊕ C. (ii) If A ⊆ B and λ ≥ 0, then λ ∗ A = λA ⊆ λB = λ ∗ B. The next lemma will be used in the proof of Theorem 2.4.2. Lemma 2.1.10. If X is directed, then for A, B, C ∈ X δ we have (A∨B)⊕C = (A⊕C)∨(B⊕C). Proof. As X is directed, Proposition 2.1.3 yields that any two elements A, B ∈ X δ have a supremum in X δ satisfying A ∨ B = (A ∪ B)ul . Due to Lemma 2.1.6 we have

2.1 Dedekind completion |

61

ul

(A ∨ B) ⊕ C = ((A ∪ B)ul + C) = ((A ∪ B) + C)ul = ((A + C) ∪ (B + C))ul ul

⊆ ((A + C)ul ∪ (B + C)ul ) = ((A ⊕ C) ∪ (B ⊕ C))ul = (A ⊕ C) ∨ (B ⊕ C) . Also, ul

(A ∨ B) ⊕ C = ((A ∪ B)ul + C) ul

⊇ (Aul + C) = (A + C)ul = A ⊕ C and, similarly, (A ∨ B) ⊕ C ⊇ B ⊕ C, so that (A ∨ B) ⊕ C = ((A ∨ B) ⊕ C)ul ⊇ ((A ⊕ C) ∪ (B ⊕ C))ul = (A ⊕ C) ∨ (B ⊕ C). In the event that X δ is a vector space, the embedding map from X to X δ will be linear. Lemma 2.1.11. The map J : X → X δ , x 󳨃→ {x}l , satisfies J(x + y) = J(x) ⊕ J(y)

J(λx) = λ ∗ J(x)

and

for every x, y ∈ X and every λ ∈ ℝ. Proof. For x, y ∈ X we have ul

ul

J(x) ⊕ J(y) = {x}l ⊕ {y}l = ({x}l + {y}l ) = ({x + y}l ) = {x + y}l = J(x + y) . For x ∈ X and λ ∈ ℝ, λ{x}l { { λ ∗ J(x) = λ ∗ {x}l = { {0}l { l u { λ ({x} )

= {λx}l , = {λx}l ,

λ >0, λ =0, l

l

= (λ{x}l ) = ({λx}u ) = {λx}l ,

λ 0} ∪ {(0, 0)T }. The partially ordered vector space (ℝ2 , K) is directed and not pre-Riesz. Indeed, consider x = (−1, 0)T , y = (0, 0)T and z = (1, 0)T . If u = (u 1 , u 2 )T ∈ {x + z, y + z}u , then u ≥ (0, 0)T and u ≥ (1, 0)T . The latter inequality yields u ≠ (0, 0)T and then the former inequality yields u 2 > 0. Then u ∈ {x, y}u . Hence, {x + z, y + z}u ⊆ {x, y}u . However, z ∈ ̸ K. (ii) If K = {(x1 , x2 )T ; x2 > 0} ∪ {(x1 , 0)T ; x1 ≥ 0}, then (ℝ2 , K) is the lexicographically ordered plane, which is a (non-Archimedean) Riesz space and hence pre-Riesz. (iii) If K = {(x1 , x2 )T ; x1 ≥ 0, x2 ≥ 0}, then K induces the standard order on ℝ2 , so that (ℝ2 , K) is a Riesz space and hence a pre-Riesz space. (iv) Let K = {(x1 , x2 )T ; x1 > 0, x2 > 0}∪{(0, 0)T }. Then (ℝ2 , K) is directed and not preRiesz. Consider, as in (i), the elements x = (−1, 0)T , y = (0, 0)T and z = (1, 0)T . If u = (u 1 , u 2 )T ∈ {x + z, y + z}u , then u ≥ (0, 0)T and u ≥ (1, 0)T . This yields first that u 1 ≥ 1 and u 2 ≥ 0. Then, as u ≠ (0, 0)T , it follows that u 2 > 0 and then also that u 1 > 1, as u ≠ (1, 0)T . We obtain that u − x = (u 1 + 1, u 2 )T ∈ K and u − y = u ∈ K, so u ∈ {x, y}u . However, z ∈ ̸ K. (v) Let K = {(x1 , x2 )T ; x1 > 0, x2 ≥ 0} ∪ {(0, 0)T }. As in (i) and (iv), (ℝ2 , K) is directed and not pre-Riesz. Consider x = (1, 0)T , y = (1, 1)T and z = (0, 1)T . If

2.2 Pre-Riesz spaces | 71

u = (u 1 , u 2 )T ∈ {x + z, y + z}u , then u ≥ (1, 1)T and u ≥ (1, 2)T , so u 1 ≥ 1 and u 2 ≥ 2. As u ≠ (1, 1)T , it follows that u 1 > 1. Then u − x = (u 1 − 1, u 2 )T ∈ K and u − y = (u 1 − 1, u 2 − 1)T ∈ K, so u ∈ {x, y}u . However, z ∈ ̸ K. For each generating cone K in ℝ2 there exists a linear bijection on ℝ2 that transforms K to one of the cones of the examples (i)–(v) above. Thus we have established which twodimensional partially ordered vector spaces are pre-Riesz. If the cone is generating and closed or if the cone is a half plane with one of the bounding half lines included, then the space is pre-Riesz and otherwise it is not. Since every subspace of an Archimedean partially ordered vector space is Archimedean, Proposition 2.2.3 has the following consequence. Corollary 2.2.7. Every directed subspace of an Archimedean partially ordered vector space is pre-Riesz. A directed subspace of a pre-Riesz space need not be pre-Riesz, as the following example shows. Example 2.2.8. We consider the space X = ℝ3 with the cone K = {(x1 , x2 , x3 )T ; x1 > 0} ∪ {(0, x2 , x3 )T ; x2 ≥ 0, x3 ≥ 0} . Observe that (X, K) is a non-Archimedean vector lattice. The subspace D := {(x1 , x2 , −x2 )T ; x1 , x2 ∈ ℝ} of (X, K) is order isomorphic to the space in Example 2.2.6 (i), i.e., (D, D ∩ K) is directed but not a pre-Riesz space.

2.2.2 Some lattice-like formulas Let us illustrate the structure present in pre-Riesz spaces by considering some formulas. In the theory of Riesz spaces, the formula a∨b+a∧b=a+b

(2.11)

is often used. Formulas such as (2.11) can be rephrased in terms of the operations ⊕, ⊖, ∗, ∨, and ∧ of Section 2.1 without explicit use of the lattice operations of the Riesz space. Formula (2.11) then reads ul

({a, b}ul + {a, b}l ) = {a + b}ul .

(2.12)

It turns out that (2.12) holds in a partially ordered vector space X for every a, b ∈ X if and only if X is a pre-Riesz space. The ‘if’ part of this statement can be proved easily by means of the Riesz completion that will be constructed in Section 2.4. We give an explicit proof here. Note that Proposition 2.2.9 will not be used in the construction of the Riesz completion.

72 | 2 Embeddings, covers, and completions

Proposition 2.2.9. Let X be a partially ordered vector space. The formula (2.12) holds for every a, b ∈ X if and only if X is a pre-Riesz space. Proof. Assume that X is a pre-Riesz space and let a, b ∈ X. We show first that ul

({a, b}ul + {a, b}l ) ⊇ {a + b}ul . Let u ∈ ({a, b}ul + {a, b}l )u . We will show that u − (a + b) ≥ 0 by showing that {a + u − (a + b), b + u − (a + b)}u ⊆ {a, b}u and applying that X is a pre-Riesz space. Indeed, let w ∈ {u−b, u−a}u. Then w ≥ u−b, u−a, so u−w ≤ a, b, hence u−w ∈ {a, b}l . Therefore, with the choice of u, for every z ∈ {a, b}ul , we have u ≥ z + u − w, so w ≥ z. Hence, w ∈ {a, b}ulu = {a, b}u . Thus we have shown that {a+u−(a+b), b+u−(a+b)}u ⊆ {a, b}u and the assumption that X is a pre-Riesz space then yields that u − (a + b) ≥ 0. Hence, u ∈ {a + b}u . Next we show ({a, b}ul + {a, b}l )ul ⊆ {a + b}ul . Let x ∈ {a, b}ul and y ∈ {a, b}l . We show that x+y ≤ a+b, by proving that {a+(a+b)−(x+y), b+(a+b)−(x+y)}u ⊆ {a, b}u and using that X is a pre-Riesz space. Indeed, if u ∈ {a + (a + b) − (x + y), b + (a + b) − (x + y)}u , then u − (a + b) + (x + y) ∈ {a, b}u , so u − (a + b) + (x + y) ≥ x, hence u ≥ a+b−y ≥ a, b as y ∈ {a, b}l , and therefore u ∈ {a, b}u . Since X is a pre-Riesz space, it follows that a+ b −(x + y) ≥ 0, so x + y ≤ a+ b. We infer that {a, b}ul +{a, b}l ⊆ {a+ b}ul and hence ({a, b}ul + {a, b}l )ul ⊆ {a + b}ul . Thus we have established (2.12). Assume that X is such that (2.12) holds for every a, b ∈ X. We show that X is a pre-Riesz space. Let x, y, z ∈ X be such that {x + z, y + z}u ⊆ {x, y}u . Then {x, y}ul ⊆ {x + z, y + z}ul . We have, by applying formula (2.12) twice, ul

{x + y + z}ul = {x + y}ul + z = ({x, y}ul + {x, y}l ) + z ul

ul

= ({x, y}ul + {x, y}l + z) = ({x, y}ul + {x + z, y + z}l ) ul

⊆ ({x + z, y + z}ul + {x + z, y + z}l ) = {x + y + 2z}ul , so x + y + z ≤ x + y + 2z and therefore z ≥ 0. Thus, X is a pre-Riesz space. The next formula that we consider is a formula that resembles the distributive property of Riesz spaces. For a Riesz space X and nonempty subsets A, B ⊆ X that are bounded from below, the identity inf A ∨ inf B = inf{a ∨ b; a ∈ A, b ∈ B}

(2.13)

is called the distributive property of the lattice X. In pre-Riesz spaces a similar formula holds. For its proof we need the next lemma. Lemma 2.2.10. Let X be a pre-Riesz space, let A ⊆ X be a finite nonempty subset, and let x ∈ X. (i) If x ≤ a − v for every a ∈ A and v ∈ Al , then x ≤ 0. (ii) If x ≤ u − a for every u ∈ Au and a ∈ A, then x ≤ 0.

2.2 Pre-Riesz spaces | 73

Proof. (i) By assumption, for every v ∈ Al we have x + v ∈ Al , hence x + Al ⊆ Al . Then −x − Al ⊆ −Al , so (−x − A)u ⊆ (−A)u and then −x ≥ 0 due to the fact that X is a pre-Riesz space and Lemma 2.2.5. (ii) For every a ∈ A we have a ≤ u − x for every u ∈ Au , so a ∈ (Au − x)l = (−x + A)ul , hence A ⊆ (−x + A)ul and therefore (−x + A)u = (−x + A)ulu ⊆ Au . Lemma 2.2.5 then yields that −x ≥ 0, so that x ≤ 0. Formula (2.14) in the next lemma is the distributive property in pre-Riesz spaces similar to (2.13). Lemma 2.2.11. Let X be a partially ordered vector space. (i) For all nonempty subsets A, B ⊆ X that are bounded below one has ul

(Al ∪ Bl ) ⊆

⋂

{a, b}ul .

a∈A, b∈B

(ii) If X is a pre-Riesz space and A, B ⊆ X are finite nonempty sets, then ul

(Al ∪ Bl ) =

⋂

{a, b}ul .

(2.14)

a∈A, b∈B

(iii) If X is a directed Archimedean partially ordered vector space, then (2.14) holds for all nonempty subsets A, B ⊆ X that are bounded below. Proof. (i) Let x ∈ (Al ∪ Bl )ul and let a ∈ A and b ∈ B. Let u ∈ {a, b}u . Then, for every v ∈ Al we have u ≥ a ≥ v and for every w ∈ Bl we have u ≥ b ≥ w. Hence u ∈ (Al ∪ Bl )u , so x ≤ u. Thus, x ∈ {a, b}ul . It follows that (Al ∪ Bl )ul ⊆ ⋂a∈A, b∈B {a, b}ul . (ii) Let x ∈ ⋂ a∈A, b∈B{a, b}ul . Let y ∈ Al and z ∈ Bl and let a ∈ A and b ∈ B. We show that x + y + z − (a + b) ∈ (Al ∪ Bl )ul and then conclude that x ∈ (Al ∪ Bl )ul . Proposition 2.2.9 yields that ul

x + {y, z}l ⊆ {a, b}ul + {y, z}l ⊆ {a, b}ul + {a, b}l ⊆ ({a, b}ul + {a, b}l ) = {a + b}ul , so x + {y, z}l + {y, z}ul ⊆ {a + b}ul + {y, z}ul = {y, z}ul + a + b , hence ul

ul

ul

x + ({y, z}l + {y, z}ul ) = (x + {y, z}l + {y, z}ul ) ⊆ ({y, z}ul + a + b) = {y, z}ul + a + b . Due to Proposition 2.2.9 we obtain ul

x + {y + z}ul ⊆ {y, z}ul + a + b = ({y} ∪ {z})ul + a + b ⊆ (Al ∪ Bl ) + a + b , which yields that x + y + z − (a + b) ∈ (Al ∪ Bl )ul .

74 | 2 Embeddings, covers, and completions Let now u ∈ (Al ∪ Bl )u . Then we obtain that x + y + z − (a + b) ≤ u, so x + z − b − u ≤ a − y. Since this holds for every a ∈ A and y ∈ A l , Lemma 2.2.10 yields that x + z − b − u ≤ 0, so that x − u ≤ b − z. Then it follows similarly that x − u ≤ 0. Hence x ≤ u and thus we have shown that x ∈ (Al ∪ Bl )ul . (iii) We use the Dedekind completion (X δ , J) of X as given in Theorem 2.1.13. We will denote the Riesz space operations of X δ with the standard symbols, instead of ⊕, ⊖, and ∗. Let x ∈ ⋂ a∈A, b∈B{a, b}ul . Let y = inf J[A] and z = inf J[B]. Then for every a ∈ A and b ∈ B we have J(x) + y ∧ z ≤ J(a) ∨ J(b) + y ∧ z ≤ J(a) ∨ J(b) + J(a) ∧ J(b) = J(a) + J(b) , hence J(x)+ y ∧ z ≤ y + z, and thus J(x) ≤ y + z − y ∧ z = y ∨ z. For every u ∈ (A l ∪ Bl )u we have y = sup{J(x); x ∈ X, J(x) ≤ y} ≤ sup{J(x); x ∈ Al } ≤ J(u) , so J(u) ≥ y and, similarly, J(u) ≥ z, so J(u) ≥ y ∨ z, hence we obtain J(x) ≤ J(u). It follows that x ∈ (A l ∪ Bl )ul .

2.3 Riesz* homomorphisms In the theory of partially ordered vector spaces various generalizations of the notion of a Riesz homomorphism from Riesz space theory have emerged. The most useful one in the context of pre-Riesz spaces is the notion of a Riesz* homomorphism, due to van Haandel [151]. This notion has been considered by van Haandel as a modification of the notion of a Riesz homomorphism in Archimedean directed partially ordered vector spaces introduced somewhat earlier by Buskes and van Rooij [36]. It turns out that Riesz* homomorphisms have convenient intrinsic properties. Moreover, they are precisely the linear maps between pre-Riesz spaces that extend to Riesz homomorphisms between their Riesz completions.

2.3.1 Definition and basic properties Although not intuitively appealing at first sight, the next definition is one of the core ingredients of the theory of pre-Riesz spaces; see [151, Definition 5.1]. Definition 2.3.1. Let X and Y be partially ordered vector spaces. A linear map h : X → Y is called a Riesz* homomorphism if for every nonempty finite subset F of X one has h[F ul ] ⊆ h[F]ul . Let us collect some easy properties in a lemma.

2.3 Riesz* homomorphisms

| 75

Lemma 2.3.2. Let X and Y be partially ordered vector spaces and let h : X → Y be a linear map. (i) If h is a Riesz* homomorphism, then h is positive. (ii) The map h is a Riesz* homomorphism if and only if for every nonempty finite subset F of X the equality h[F ul ]ul = h[F]ul holds. (iii) If X and Y are Riesz spaces, then h is a Riesz* homomorphism if and only if h is a Riesz homomorphism. Proof. (i) Take F = {0}. For every x ∈ X with x ≥ 0 we have −x ∈ F ul , hence h(−x) ∈ h[F ul ] ⊆ h[F]ul = {0}l , so that h(x) ≥ 0. (ii) If for every nonempty finite subset F of X the equality h[F ul ]ul = h[F]ul holds, then h[F ul ] ⊆ h[F ul ]ul = h[F]ul for every nonempty finite subset F of X, so that h is a Riesz* homomorphism. For the converse implication, note that for every set F ⊆ X we have F ul ⊇ F, so h[F ul ] ⊇ h[F], hence h[F ul ]ul ⊇ h[F]ul , which shows one inclusion. If h is a Riesz* homomorphism, then for every nonempty finite subset F of X we have h[F ul ] ⊆ h[F]ul , hence h[F ul ]ul ⊆ h[F]ul , so that the desired equality follows. (iii) If X and Y are Riesz spaces, then for every a1 , . . . , a n ∈ X and F = {a1 , . . . , a n }, the inclusion h[F ul ] ⊆ h[F]ul comes down to h[{a1 ∨⋅ ⋅ ⋅∨a n }l ] ⊆ {h(a1 )∨⋅ ⋅ ⋅∨h(a n )}l . If h is a Riesz homomorphism, it is immediate that the latter inclusion holds and it follows that h is a Riesz* homomorphism. If h is a Riesz* homomorphism, from the inclusion we obtain h(a1 ∨ ⋅ ⋅ ⋅ ∨ a n ) ≤ h(a1 ) ∨ ⋅ ⋅ ⋅ ∨ h(a n ). As h is positive, also h(a1 ∨ ⋅ ⋅ ⋅ ∨ a n ) ≥ h(a1 ) ∨ ⋅ ⋅ ⋅ ∨ h(a n ) holds. Thus, h is a Riesz homomorphism. It is important to note that the spaces in which the lower and upper bounds are taken matter in the definition of a Riesz* homomorphism. On the one hand, if a map h is a Riesz* homomorphism from X to Y and Y ⊆ Z, then h need not be a Riesz* homomorphism from X to Z, as the next example shows. On the other hand, for a map h from X to Y with Z ⊆ X, it may be that h is a Riesz* homomorphism, but its restriction to Z is not, as is shown in the second part of the next example. Example 2.3.3. By Example 1.1.20, the subspace Aff[0, 1] of C[0, 1] consisting of all affine functions is a Riesz space but not a Riesz subspace of C[0, 1], as the lattice operations are different. 1. The map h : Aff[0, 1] → Aff[0, 1], f 󳨃→ f , is a Riesz homomorphism and hence a Riesz* homomorphism. However, viewed as a map from Aff[0, 1] to C[0, 1], h is not a Riesz homomorphism, hence not a Riesz* homomorphism. 2. The map h : C[0, 1] → C[0, 1], f 󳨃→ f , is a Riesz homomorphism and hence a Riesz* homomorphism, but viewed as a map from Aff[0, 1] to C[0, 1], h is not a Riesz homomorphism, hence not a Riesz* homomorphism.

76 | 2 Embeddings, covers, and completions

The following two lemmas are especially useful for inclusion maps. An improvement of Lemma 2.3.4 will be given in Proposition 2.3.27 below. Lemma 2.3.4. Let X and Y be partially ordered vector spaces, and let h : X → Y be a bipositive linear map. Then h is a Riesz* homomorphism in both of the following cases: (i) h[X] is order dense in Y. (ii) Y is a Riesz space and h[X] is a Riesz subspace of Y. Proof. (i) Let F ⊆ X be a nonempty finite set. If F ul = ⌀ or h[F]u = ⌀, then the statement is clear. Let v ∈ F ul and y ∈ h[F]u . We wish to show that h(v) ≤ y. As h[X] is order dense in Y, we have y = inf{h(u); u ∈ X, h(u) ≥ y}. For u ∈ X with h(u) ≥ y we have h(u) ≥ h(x) for every x ∈ F, so that the bipositivity of h yields that u ≥ x for every x ∈ F. Hence, u ∈ F u and therefore v ≤ u. Then h(u) ≥ h(v). It follows that y = inf{h(u); u ∈ X, h(u) ≥ y} ≥ h(v). Thus, h(v) ∈ h[F]ul . We conclude that h[F ul ] ⊆ h[F]ul , so that h is a Riesz* homomorphism. (ii) Let F ⊆ X be a nonempty finite set, let v ∈ F ul , and let y ∈ h[F]u . As h[X] is a Riesz subspace of Y, we have that sup h[F] ∈ h[X], so there exists u ∈ X such that h(u) = sup h[F]. As h is bipositive, we obtain u ≥ x for every x ∈ F, so that u ∈ F u and therefore v ≤ u. Then h(v) ≤ h(u) = sup h[F] ≤ y, so that h(v) ∈ h[F]ul . It follows that h[F ul ] ⊆ h[F]ul , which yields that h is a Riesz* homomorphism. Note that in Lemma 2.3.4 (ii) the space X is a Riesz space, since h is a bipositive linear bijection between X and the Riesz space h[X]. Moreover, h is a Riesz homomorphism. Lemma 2.3.5. Let X and Y be directed partially ordered vector spaces and let h : X → Y be a bipositive linear map. If Y is a pre-Riesz space and h is a Riesz* homomorphism, then X is a pre-Riesz space. Proof. Let x, y, z ∈ X be such that {x + z, y + z}u ⊆ {x, y}u . Then {x, y}ul ⊆ {x + z, y + z}ul . Because h is a Riesz* homomorphism, it follows that h[{x, y}] ⊆ h[{x, y}ul ] ⊆ h[{x + z, y + z}ul ] ⊆ {h(x + z), h(y + z)}ul . Hence, {h(x) + h(z), h(y) + h(z)}u ⊆ {h(x), h(y)}u . Since Y is a pre-Riesz space, we obtain that h(z) ≥ 0. Bipositivity of h then yields that z ≥ 0, which shows that X is pre-Riesz.

2.3.2 Extension and restriction The most important property of Riesz* homorphisms between pre-Riesz spaces is that they are precisely the linear maps that extend to Riesz homomorphisms on their Riesz completions. Equivalently, a linear map between order dense subspaces of two Riesz spaces that generate them is a Riesz* homomorphism if and only if it extends to a Riesz homomorphism on the Riesz spaces. We will discuss such an extension and restriction of Riesz* homomorphisms below, with quite some care for the generality of the con-

2.3 Riesz* homomorphisms

| 77

ditions. We begin with restriction, which needs the order denseness of the domain to which we want to restrict. Proposition 2.3.6. Let X1 and X2 be partially ordered vector spaces and let h : X1 → X2 be a linear map. Assume that there exist Riesz spaces Y1 and Y 2 and bipositive linear maps i1 : X1 → Y 1 and i2 : X2 → Y 2 such that i1 [X1 ] is order dense in Y 1 . If there exists a Riesz homomorphism ĥ : Y 1 → Y 2 such that i2 ∘ h = ĥ ∘ i1 , then h is a Riesz* homomorphism. Proof. Let a1 , . . . , a n ∈ X1 and let F := {a1 , . . . , a n }. We show that h[F ul ] ⊆ h[F]ul . Let x ∈ F ul . If the set of upper bounds of h[F] is empty there is nothing to prove, so suppose that it is nonempty. Let y ∈ h[F]u . Since i1 [X1 ] is order dense in Y 1 , we have i1 (a1 ) ∨ ⋅ ⋅ ⋅ ∨ i1 (a n ) = inf {i1 (u); u ∈ {a1 , . . . , a n }u } ≥ i1 (x) . Then ̂ 1 (x)) ≤ h(i ̂ 1 (a1 ) ∨ ⋅ ⋅ ⋅ ∨ i1 (a n )) i2 (h(x)) = h(i ̂ 1 (a n )) ̂ 1 (a1 )) ∨ ⋅ ⋅ ⋅ ∨ h(i = h(i = i2 (h(a1 )) ∨ ⋅ ⋅ ⋅ ∨ i2 (h(a n )) ≤ i2 (y) , hence h(x) ≤ y. Thus, h(x) ∈ h[F]ul , which yields that h is a Riesz* homomorphism. The message of Proposition 2.3.6 is easier to read if we consider subspaces instead of embeddings. Corollary 2.3.7. Let Y1 and Y 2 be Riesz spaces, let X1 be an order dense subspace of Y 1 and let X2 be a subspace of Y2 . If h : Y 1 → Y2 is a Riesz homomorphism and h[X1 ] ⊆ X2 , then h|X1 : X1 → X2 is a Riesz* homomorphism. Corollary 2.3.8. Let Y be a Riesz space and let X be a subspace of Y. If X is order dense in Y, then the inclusion map h : X → Y, x 󳨃→ x, is a Riesz* homomorphism. For the extension of Riesz* homomorphisms we need Theorem 2.1.15. The condition in this theorem holds for a Riesz* homomorphism due to the next lemma. Lemma 2.3.9. Let X and Y be partially ordered vector spaces. If h : X → Y is a Riesz* homomorphism, then for all nonempty finite subsets F, G ⊆ X one has F u = Gu 󳨐⇒ h[F]u = h[G]u . Proof. With the aid of Lemma 2.3.2 (ii) we have ul

ul

h[F]ul = h[F ul ] = h [Gul ] = h[G]ul . Thus, h[F]u = h[F]ulu = h[G]ulu = h[G]u .

78 | 2 Embeddings, covers, and completions

A combination of Lemma 2.3.9 and Theorem 2.1.15 yields the following result. Proposition 2.3.10. Let X1 and X2 be partially ordered vector spaces and let h : X1 → X2 be a Riesz* homomorphism. Assume that there exist Riesz spaces Y1 and Y 2 and bipositive linear maps i1 : X1 → Y 1 and i2 : X2 → Y 2 . If i1 [X1 ] generates Y 1 as a Riesz space and i2 [X2 ] is order dense in Y 2 , then there exists a Riesz homomorphism ĥ : Y 1 → Y 2 such that ĥ ∘ i1 = i2 ∘ h . Moreover, the range of ĥ then equals the Riesz subspace of Y2 generated by i2 [h[X1 ]]. If we consider subspaces rather than embeddings, then Proposition 2.3.10 reads as follows. Corollary 2.3.11. Let Y 1 and Y 2 be Riesz spaces, let X1 be a subspace of Y1 that generates Y1 as a Riesz space, and let X2 be an order dense subspace of Y 2 . If h : X1 → X2 is a Riesz* homomorphism, then there exists a Riesz homomorphism ĥ : Y 1 → Y 2 that extends h. In the extension and restriction of Riesz* homomorphisms discussed above, we need the order denseness of the domain for restriction and order denseness of the range for extension. For the latter we also need the domain to generate the Riesz space. If we have both, then we can characterize Riesz* homomorphisms as restrictions of Riesz homomorphisms on the Riesz spaces; see also [151, Theorem 5.6]. Theorem 2.3.12. Let X1 and X2 be partially ordered vector spaces and let h : X1 → X2 be a linear map. Assume that there exist Riesz spaces Y1 and Y 2 and bipositive linear maps i1 : X1 → Y1 and i2 : X2 → Y2 such that i1 [X1 ] is order dense in Y 1 and generates Y 1 as a Riesz space, such that i2 [X2 ] is order dense in Y2 . The following three statements are equivalent. (a) h is a Riesz* homomorphism. (b) For all nonempty finite subsets F, G ⊆ X1 one has F u = Gu 󳨐⇒ h[F]u = h[G]u .

(2.15)

(c) There exists a Riesz homomorphism ĥ : Y 1 → Y2 with i2 ∘ h = ĥ ∘ i1 . Moreover, the map ĥ as in (c) is unique. Proof. The implication (a) ⇒ (b) is contained in Lemma 2.3.9. Theorem 2.1.15 yields that (b) implies (c), and that ĥ is unique. Proposition 2.3.6 says that (c) implies (a). Corollary 2.3.13. Let Y1 and Y 2 be Riesz spaces, let X1 be an order dense subspace of Y 1 that generates Y 1 as a Riesz space and let X2 be an order dense subspace of Y 2 . A linear map h : X1 → X2 is a Riesz* homomorphism if and only if there exists a Riesz homomorphism ĥ : Y 1 → Y2 that extends h.

2.3 Riesz* homomorphisms

| 79

In the definition of a Riesz homomorphism between two Riesz spaces it is sufficient to consider two elements rather than arbitrary nonempty finite subsets². The next example shows that in condition (b) of Theorem 2.3.12 it is not enough to restrict to sets consisting of two elements. Example 2.3.14. Let X = P2 [0, 1] be the subspace of C[0, 1] consisting of polynomials of degree at most two and let Y be the Riesz subspace of C[0, 1] generated by X. According to Example 1.7.2, the space X is order dense in C[0, 1] and hence in Y. Let h : X → X be defined by h(x 󳨃→ c2 x2 + c1 x + c0 ) = (x 󳨃→ c2 x2 + c1 x + 2c0 ) , c0 , c1 , c2 ∈ ℝ. Then h : X → X is linear and positive, since if p : x 󳨃→ c2 x2 + c1 x + c0 is such that p(x) ≥ 0 for every x ∈ [0, 1], then c0 ≥ 0, so h(p) = p + c0 ≥ 0. In the arguments below we use the following observation: if p1 , p2 , and p3 are polynomials on [0, 1] such that p1 = p2 ∨ p3 on some nonempty open subinterval I of [0, 1], then p1 = p2 or p1 = p3 . For a proof, note that if p1 (x) = p2 (x) for every x ∈ I then p1 = p2 on [0, 1], as p1 and p2 are polynomials. Suppose that p1 (x) = p2 (x) does not hold for every x ∈ I. Then there exists x ∈ I with p2 (x) < p3 (x). Then p2 < p3 on a neighborhood U of x, so p1 = p3 on U, so that p1 = p3 on [0, 1]. We show that (2.15) holds for subsets F, G ⊆ X consisting of two elements. Let a1 , a2 , b 1 , b 2 ∈ X be such that {a1 , a2 }u = {b 1 , b 2 }u . Then a1 ∨ a2 = b 1 ∨ b 2 holds in C[0, 1], as X is order dense in C[0, 1]. We observe that either 1. a1 ∨ a2 equals a1 or a2 , or 2. a1 ∨a2 equals a1 on some nonempty open subinterval I of [0, 1] and a2 on another nonempty open subinterval J of [0, 1]. In case 1. we have a1 ≥ a2 or a2 ≥ a1 . Without loss of generality we assume a1 ≥ a2 . Then a1 = b 1 ∨ b 2 , so a1 = b 1 and b 1 ≥ b 2 or a1 = b 2 and b 2 ≥ a1 . As h is positive, it follows that {h(a1 ), h(a2 )}u = {h(a1 )}u = {h(b 1 ), h(b 2 )}u . In case 2. we have a1 = b 1 ∨ b 2 on I and a2 = b 1 ∨ b 2 on J. Hence, a1 = b 1 or a1 = b 2 and a2 = b 1 or a2 = b 2 . It follows that {h(a1 ), h(a2 )} = {h(b 1 ), h(b 2 )}, so that {h(a1 ), h(a2 )}u = {h(b 1 ), h(b 2 )}u . Next, we show that (2.15) does not hold for all subsets F, G ⊆ X consisting of three elements. Let a1 (x) = x ,

a2 (x) = 1 − x ,

2

a3 (x) = 4 (x − 12 ) = 4x2 − 4x + 1 ,

x ∈ [0, 1] ,

and let b 1 = a1 , b 2 = a2 and b 3 = 0. Then a1 , a2 , a3 , b 1 , b 2 , b 3 ∈ X and a1 ∨ a2 ∨ a3 = a1 ∨ a2 = b 1 ∨ b 2 = b 1 ∨ b 2 ∨ b 3 , so {a1 , a2 , a3 }u = {b 1 , b 2 , b 3 }u .

2 The definition of Riesz* homomorphisms in [151, Definition 5.1] is given by means of two elements. In [151, Theorem 5.3 (ii)] it is stated that the similar formula is true for arbitrary nonempty finite subsets, but in the proof there is a step that is incorrect.

80 | 2 Embeddings, covers, and completions We have h(a1 ) = a1 , h(a2 )(x) = 2 − x and h(a3 )(x) = 4x2 − 4x + 2 for x ∈ [0, 1]. Hence, h(a3 )(1) = 2 > 1 = h(a1 )(1) ∨ h(a2 )(1). Thus, h(a1 ) ∨ h(a2 ) ∨ h(a3 ) > h(a1 ) ∨ h(a2 ) = h(b 1 ) ∨ h(b 2 ) ∨ h(b 3 ) , so that {h(a1 ), h(a2 ), h(a3 )}u ≠ {h(b 1 ), h(b 2 ), h(b 3 )}u . So far we have shown that h does not satisfy (b) of Theorem 2.3.12, i.e., h is not a Riesz* homomorphism. One could wonder whether h satisfies the formula in the definition of Riesz* homomorphisms for sets consisting of two elements. This is not the case, as a3 ∈ {a1 , a2 }ul , but h(a3 ) ≰ h(a1 ), h(a2 ), so h(a3 ) ∈ ̸ {h(a1 ), h(a2 )}ul .

2.3.3 Comparison with Riesz homomorphisms We will briefly compare Riesz* homomorphisms with Riesz homomorphisms as defined by Buskes and van Rooij in [36]. We also give the definition of a complete Riesz homomorphism, which generalizes the notion of an order continuous Riesz homomorphism in vector lattices as discussed in Proposition 1.4.5. Definition 2.3.15. Let X and Y be partially ordered vector spaces. (i) A linear map h : X → Y is called a Riesz homomorphism if for every x, y ∈ X one has l h [{x, y}u ] = h[{x, y}]ul . (2.16) (ii) A linear map h : X → Y is called a complete Riesz homomorphism if for every nonempty set A ⊆ X we have inf A = 0 󳨐⇒ inf h[A] = 0 . If X and Y are Riesz spaces, then the notion of Riesz homomorphism of Definition 2.3.15 coincides with that of Riesz space theory. This can be seen from (2.16), of which the left-hand side then equals {h(x ∨ y)}l and the right-hand side {h(x) ∨ h(y)}l . Proposition 2.3.16. Let X be a pre-Riesz space and let Y be a partially ordered vector space. If h : X → Y is a complete Riesz homomorphism, then h is a Riesz homomorphism. Proof. Let x, y ∈ X. We wish to show that h[{x, y}u ]l = {h(x), h(y)}ul . To show ⊇, we first observe that h is positive. Indeed, take A = {x ∈ X; x ≥ 0}. Then inf A = 0, so inf h[A] = 0, hence h(x) ≥ 0 for every x ≥ 0, which means that h is positive. Next, as h is positive, we have h[{x, y}u ] ⊆ {h(x), h(y)}u and then h[{x, y}u ]l ⊇ {h(x), h(y)}ul . For a proof of ⊆, take A = {x, y}u − {x, y}. Clearly, 0 is a lower bound of A. If v is a lower bound of A, then Lemma 2.2.10 (ii) applied to {x, y} yields that v ≤ 0, as X is a pre-Riesz space. Hence, inf A = 0. Since h is a complete Riesz homomorphism, it

81

2.3 Riesz* homomorphisms |

follows that inf h[A] = 0, so that (h[{x, y}u ] − {h(x), h(y)})l = {0}l . If v ∈ h[{x, y}u ]l and u ∈ {h(x), h(y)}u , then v − u ∈ (h[{x, y}u ] − {h(x), h(y)})l , so v − u ≤ 0, hence v ≤ u. Therefore, h[{x, y}u ]l ⊆ {h(x), h(y)}ul . In order to show that every Riesz homomorphism is a Riesz* homomorphism, we need to involve arbitrary nonempty finite sets rather than sets consisting of two points, which is the content of Lemma 2.3.17; see also [151, Theorem 5.3 (i)]. The proof of that lemma uses the observation that ⋃ {x}u = Au

⋃ {x}l = Bl ,

and

x∈A u

(2.17)

x∈Bl

which holds for all nonempty subsets A and B of a partially ordered vector space X. For a proof of the first identity in (2.17), observe that the inclusion ⊆ is immediate from the transitivity of the partial order. To show ⊇, note that if u ∈ Au , then u ∈ {u}u , so u ∈ ⋃x∈A u {x}u . The second identity has a similar proof. Lemma 2.3.17. Let X be a partially ordered vector space and let Y be a pre-Riesz space. If h : X → Y is a Riesz homomorphism, then for every nonempty finite set F ⊆ X one has h[F u ]l = h[F]ul . Proof. We prove the desired equality by induction on the number of elements of the set F, which we denote with |F|. By definition, the assertion is true if |F| ≤ 2. Suppose n ∈ ℕ, n ≥ 2 is such that ∀G ⊆ X, G ≠ ⌀ ,

with

|G| ≤ n : h[Gu ]l = h[G]ul

(2.18)

and suppose that F = {a1 , . . . , a n+1 }. Let G = {a1 , . . . , a n } and V = {a n+1 }. With the aid of Lemma 2.2.11 (ii), which can be applied since Y is a pre-Riesz space, we obtain ul

ul

h[F]ul = h[G ∪ V]ul = (h[G] ∪ h[V])ul = (h[G]ul ∪ h[V]ul ) = (h[Gu ]l ∪ h[V u ]l ) =

{a, b}ul =

⋂ a∈h[G u ], b∈h[V u]

{h(x), h(y)}ul =

⋂ x∈G u , y∈V u

l

=(

⋃

⋂

h[{x, y}u ]l

x∈G u , y∈V u l

h[{x, y}u ]) = h[

x∈G u , y∈V u

⋃

l

{x, y}u ] = h[

x∈G u , y∈V u

⋃

{x}u ∩ {y}u ]

x∈G u , y∈V u

l l

l

l

= h[ ⋃ {x}u ∩ ⋃ {y}u ] = h [Gu ∩ V u ] = h [(G ∪ V)u ] = h [F u ] , x∈G u

y∈V u

where we have used (2.17). Thus we have established that (2.18) also holds if n is replaced by n + 1, and the induction proof is complete. Proposition 2.3.18. Let X be a partially ordered vector space, let Y be a pre-Riesz space, and let h : X → Y be a linear map. If h is a Riesz homomorphism, then h is a Riesz* homomorphism.

82 | 2 Embeddings, covers, and completions Proof. We begin by noting that h is positive by taking x = y = 0, so that K := {u ∈ X; u ≥ 0} = {x, y}u and therefore, h[K]l = h[{x, y}u ]l = h[{x, y}]ul = {0}ul , which yields that 0 ∈ h[K]l and hence, h[K] ⊆ {v ∈ Y; v ≥ 0}. Next, if A ⊆ X is nonempty, then the positivity of h implies h[Al ] ⊆ h[A]l , so l h[A ]ul ⊆ h[A]lul = h[A]l . Now let F ⊆ X be a nonempty finite set. We wish to show that h[F ul ]ul = h[F]ul . The previous step yields h[F ul ]ul ⊆ h[F u ]l . According to Lemma 2.3.17, we have h[F u ]l = h[F]ul , and thus we obtain h[F ul ]ul ⊆ h[F]ul . As F ul ⊇ F, we have h[F ul ]ul ⊇ h[F]ul . Therefore, h[F ul ]ul = h[F]ul . Due to Lemma 2.3.2 (ii), it follows that h is a Riesz* homomorphism. Combining Lemma 2.3.2 (i), Proposition 2.3.16, and Proposition 2.3.18 amounts to the following list of implications between the different types of homomorphisms. Theorem 2.3.19. Let X and Y be pre-Riesz spaces and let h : X → Y be a linear map. Consider the following four properties of h: (i) h is a complete Riesz homomorphism. (ii) h is a Riesz homomorphism. (iii) h is a Riesz* homomorphism. (iv) h is positive. Then (i) ⇒ (ii) ⇒ (iii) ⇒ (iv). Proposition 2.3.20. Let X and Y be partially ordered vector spaces and let h : X → Y be a complete Riesz homomorphism. Then h is o-continuous. o

Proof. Let (x α )α∈A be a net in X with x α → 󳨀 x, i.e., there is a net (y α )α∈A in X such that y α ↓ 0 and for every α ∈ A one has ±(x α − x) ≤ y α . As h is a complete Riesz homomorphism, we obtain inf{h(y α ); α ∈ A} = 0. Since h is positive and linear, we o 󳨀 h(x). get h(y α ) ↓ 0 and ±(h(x α ) − h(x)) ≤ h(y α ), hence h(x α ) → Later on we will often consider bipositive linear maps. Note that a complete Riesz homomorphism need not be bipositive, as the zero map shows. A discussion on Riesz* isomorphisms can be found in Subsection 5.2.1 below. The composition of two Riesz* homomorphisms is a Riesz* homomorphism, whereas the composition of two Riesz homorphisms need not be a Riesz homomorphism. Proposition 2.3.21 and Example 2.3.22 give the details. Proposition 2.3.21. Let X, Y, and Z be partially ordered vector spaces. If h : X → Y and k : Y → Z are Riesz* homomorphisms, then k ∘ h : X → Z is a Riesz* homomorphism. Proof. Let F be a nonempty finite subset of X. Then h[F] is a nonempty finite subset of Y, and since h and k are Riesz* homomorphisms, we have: (k ∘ h)[F ul ] = k [h [F ul ]] ⊆ k [h[F]ul ] ⊆ k[h[F]]ul = ((k ∘ h)[F])ul . The next example is due to [36].

2.3 Riesz* homomorphisms

| 83

Example 2.3.22. The composition of two Riesz homomorphisms need not be a Riesz homomorphism. Let X = {x ∈ C[−1, 1]; x(−1) + x(1) = 2x(0)} and Y = C[−1, 1]. It is straightforward that X is order dense in Y. Let h : X → Y be the inclusion map and let k : Y → ℝ be given by k(x) = x(0), x ∈ Y. By Theorem 1.4.7, the functional k is a Riesz homomorphism. For a proof that h is a Riesz homomorphism, let x, y ∈ X. If z ∈ h[{x, y}u ]l , then z ≤ h(u) = u for every u ∈ X with u ≥ x, y. Since X is order dense in Y, it follows that h(x) ∨ h(y) = inf{u ∈ X; u ≥ h(x), h(y)} ≥ z , so z ∈ {h(x), h(y)}ul . Conversely, since h is positive, h[{x, y}u ] ⊆ {h(x), h(y)}u , so h[{x, y}u ]l ⊇ {h(x), h(y)}ul . Hence, h is a Riesz homomorphism. The composition k∘h is not a Riesz homomorphism. Indeed, let x(s) = s, y(s) = −s, s ∈ [−1, 1]. Then x, y ∈ X and k(h(x)) = k(h(y)) = 0. If u ∈ X is such that u ≥ x and u ≥ y, then u(1) ≥ 1 and u(−1) ≥ 1, so u(0) = 12 (u(−1) + u(1)) ≥ 1. Therefore, for every z ∈ h[{x, y}u ] we have z(0) ≥ 1 and hence, (k ∘ h)[{x, y}u ] ⊆ [1, ∞). Thus, 1 ∈ ((k ∘ h)[{x, y}u ])l , but 1 ∈ ̸ {0}ul = {(k ∘ h)(x), (k ∘ h)(y)}ul . More results on the compositions of (complete) Riesz homomorphism are known. For the sake of completeness we include them here. The next two results are due to [151, Theorem 2.5 and Corollary 2.6]. Lemma 2.3.23. Let X be a directed partially ordered vector space, let Y be a pre-Riesz space, and let h : X → Y be a linear map. Then h is a Riesz homomorphism if and only if for every a, b ∈ X one has inf h [{a, b}u − {a, b}] = 0 .

(2.19)

Proof. Assume that h is a Riesz homomorphism. Let a, b ∈ X and denote A := {a, b}u − {a, b}. The set A is nonempty, since X is directed. Clearly, h(x) ≥ 0 for every x ∈ A, since h is positive. Let v ∈ Y be a lower bound of h[A]. For every u ∈ {a, b}u , we have h(u − a) ≥ v and h(u − b) ≥ v, so h(u) ≥ h(a) + v and h(u) ≥ h(b) + v. Hence, h[{a, b}u ] ⊆ {h(a) + v, h(b) + v}u . Because h is a Riesz homomorphism, we obtain l

{h(a), h(b)}ul = h [{a, b}u ] ⊇ {h(a) + v, h(b) + v}ul . Subtracting v yields {h(a) − v, h(b) − v}ul ⊇ {h(a), h(b)}ul , so that {h(a) − v, h(b) − v}u ⊆ {h(a), h(b)}u . As Y is pre-Riesz, it follows that −v ≥ 0. Thus, inf h[A] = 0. Next we show the converse implication. Assume that (2.19) holds for every a, b ∈ X. First we note that h is positive. Indeed, with the choice a := b := 0, we have for every x ∈ X with x ≥ 0 that x ∈ {a, b}u − {a, b}, so that h(x) ≥ 0. Hence, h is positive.

84 | 2 Embeddings, covers, and completions Let a, b ∈ X. Since h is positive, we have that h[{a, b}u ] ⊆ {h(a), h(b)}u and hence, h[{a, b}u ]l ⊇ {h(a), h(b)}ul . It remains to show that h[{a, b}u ]l ⊆ {h(a), h(b)}ul . Let v ∈ h[{a, b}u ]l and let u ∈ {h(a), h(b)}u . We wish to show that v ≤ u. For every x ∈ {a, b}u , we have that v ≤ h(x), so v − u ≤ h(x) − h(a) and v − u ≤ h(x) − h(b). Therefore, v − u is a lower bound of h[{a, b}u − {a, b}]. Hence, by assumption, v − u ≤ 0, so that v ≤ u. It follows that h[{a, b}u ]l ⊆ {h(a), h(b)}ul . Thus, h is a Riesz homomorphism. Proposition 2.3.24. Let X, Y and Z be partially ordered vector spaces and let h : X → Y and k : Y → Z be linear maps. (i) If h and k are complete Riesz homomorphisms, then k ∘ h is a complete Riesz homomorphism. (ii) If X is directed, Y is a pre-Riesz space, h is a Riesz homomorphism, and k is a complete Riesz homomorphism, then k ∘ h is a Riesz homomorphism. Proof. (i) Let A ⊆ X be such that inf A = 0. Then inf h[A] = 0 and therefore, inf k[h[A]] = 0. Hence, k ∘ h is a complete Riesz homomorphism. (ii) We use Lemma 2.3.23. Let a, b ∈ X and denote A := {a, b}u − {a, b}. Since h is a Riesz homomorphism, Lemma 2.3.23 yields that inf h[A] = 0. Then inf k[h[A]] = 0, as k is a complete Riesz homomorphism. Therefore, by Lemma 2.3.23 once more, k ∘ h is a Riesz homomorphism. Next we present some counterexamples concerning the different types of homomorphisms. First we consider the implications (i) ⇒ (ii) and (ii) ⇒ (iii) of Theorem 2.3.19 and show that the converse implications do not hold in general. Example 2.3.25. (i) We give a Riesz homomorphism that is not a complete Riesz homomorphism. Such an example is known from the theory of Riesz spaces. Take X = C[0, 1] and Y = ℝ, and take h(x) = x(1), x ∈ X. Observe that X and Y are Riesz spaces. It is clear that h(x ∨ y) = h(x) ∨ h(y) for every x, y ∈ X, so h is a Riesz homomorphism. Take x n = (t 󳨃→ t n ) ∈ X, n ∈ ℕ, and take A = {x n ; n ∈ ℕ}. Then inf A = 0, but h(x n ) = 1 for every n ∈ ℕ, so inf h[A] = 1, hence h is not a complete Riesz homomorphism. (ii) A Riesz* homomorphism that is not a Riesz homomorphism is presented next. Take h and k as in Example 2.3.22. Then h and k are Riesz* homomorphisms due to Proposition 2.3.18. Proposition 2.3.21 then yields that k ∘ h is a Riesz* homomorphism. In Example 2.3.22 it is shown that k ∘ h is not a Riesz homomorphism. Note that in Theorem 2.3.19 we have used that the spaces X and Y are pre-Riesz spaces. In a setting of directed spaces that are not pre-Riesz, an example of a complete Riesz homomorphism that is not a Riesz homomorphism is given after [151, Corollary 2.7]. The following proposition is sometimes useful, particularly if the image space is ℝ.

2.3 Riesz* homomorphisms

| 85

Proposition 2.3.26. Let X be a partially ordered vector space, let Y be a Riesz space, and let h : X → Y be a linear map. (i) h is a Riesz* homomorphism if and only if n

sup {h(x); x ∈ {a1 , . . . , a n }ul } = ⋁ h(a k ) k=1

for every a1 , . . . , a n ∈ X and n ∈ ℕ. (ii) h is a Riesz homomorphism if and only if n

inf {h(x); x ∈ {a1 , . . . , a n }u } = ⋁ h(a k ) k=1

for every a1 , . . . , a n ∈ X and n ∈ ℕ. Proof. (i) The stated identity can be reformulated as sup h[F ul ] = sup h[F], where F = {a1 , . . . , a n }, and the latter is equivalent to h[F ul ]u = h[F]u , which is equivalent to h[F ul ]ul = h[F]ul . According to Lemma 2.3.2 (ii), h[F ul ]ul = h[F]ul holds for every nonempty finite set F ⊆ X if and only if h is a Riesz* homomorphism. (ii) The stated identity can be reformulated as inf h[F u ] = sup h[F], where F = {a1 , . . . , a n }. The latter equality holds if and only if inf h[F u ] = inf h[F]u , which is equivalent to h[F u ]l = h[F]ul . This holds for every nonempty finite set F ⊆ X if and only if h is a Riesz homomorphism. In Lemma 2.3.4 it is discussed under which conditions embedding maps are Riesz* homomorphisms. Lemma 2.3.4 (i) can be improved to the subsequent statement, which is due to [151, Theorem 2.11]. Proposition 2.3.27. Let X and Y be partially ordered vector spaces and let h : X → Y be a bipositive linear map such that h[X] is order dense in Y. Then h is a complete Riesz homomorphism. Proof. Let A ⊆ X be such that inf A = 0. As h is positive, 0 is a lower bound of h[A]. Let v ∈ Y be a lower bound of h[A]. Since h[X] is order dense in Y, we have that v = sup{h(x); x ∈ X, h(x) ≤ v}. For x ∈ X with h(x) ≤ v, we have h(x) ≤ v ≤ h(a) for every a ∈ A, hence x ≤ a due to the bipositivity of h. As inf A = 0, we obtain x ≤ 0. Hence, v ≤ 0. Therefore, inf h[A] = 0. Riesz homomorphisms with range space ℝ have been studied earlier than in [36]. An important result, which we state next, is due to Hayes [67, Corollary 1]; see also [72, Theorem 1.8.1]. Proposition 2.3.28. Let (X, K) be a directed partially ordered vector space and let h : X → ℝ be a positive linear map. The following statements are equivalent. (a) The map h is a Riesz homomorphism.

86 | 2 Embeddings, covers, and completions (b) For every x, y ∈ X one has inf{h(u); u ∈ {x, y}u } = h(x) ∨ h(y). (c) The element h ∈ K ∗ is extremal, i.e., for every linear map g : X → ℝ with 0 ≤ g ≤ h, there exists λ ∈ ℝ such that g = λh. (d) For every a ∈ X such that h(a) = 0 and every ε > 0 there exists u ∈ K with u ≥ a such that h(u) < ε. Proof. The implication (a) ⇒ (b) follows from Proposition 2.3.26 (ii). To see that (b) ⇒ (a), let x, y ∈ X and observe that (b) yields that l

{h(u); u ∈ {x, y}u } = {h(x) ∨ h(y)}l = {h(x), h(y)}ul , which means that h is a Riesz homomorphism. To show (b) ⇒ (c), let g : X → ℝ be linear with 0 ≤ g ≤ h. Let x ∈ X be such that h(x) = 0. We show that g(x) = 0. Let ε > 0. By (b), there exists u ∈ {x, 0}u such that h(u) < h(x) ∨ h(0) + ε = ε, so g(x) = g(u) − g(u − x) ≤ g(u) ≤ h(u) < ε . Hence, g(x) ≤ 0. By the same argument for −x, we obtain {x ∈ X; h(x) = 0} ⊆ {x ∈ X; g(x) = 0} . It follows that there exists λ ∈ ℝ such that g = λh. For a proof of (c) ⇒ (d), define for x ∈ X, p(x) := inf {h(u); u ∈ {x, 0}u } . Note that for every x ∈ X the set {h(u); u ∈ {x, 0}u } is nonempty and bounded below as X is directed and h is positive. Hence, p is well-defined. It is routine to check that p : X → ℝ is positively homogeneous and subadditive and that h(x) ≤ p(x) for every x ∈ X. Moreover, for every x ∈ K we have h(x) = p(x)

and

p(−x) = 0 .

Let a ∈ X be such that h(a) = 0. According to the Hahn–Banach theorem, there exists a linear map g : X → ℝ with g(x) ≤ p(x) for every x ∈ X, such that g(a) = p(a). For x ∈ K we have g(x) ≤ p(x) = h(x) and g(−x) ≤ p(−x) = 0 , so 0 ≤ g ≤ h. By (c), there exists λ ∈ ℝ such that g = λh and therefore g(a) = λh(a) = 0. It follows that p(a) = g(a) = 0, which implies that for every ε > 0 there exists u ∈ {a, 0}u such that h(u) < ε. Thus, we have established (d). Finally, we prove (d) ⇒ (b). Let x, y ∈ X. For u ∈ {x, y}u , positivity of h yields h(u) ≥ h(x) ∨ h(y), so that inf {h(u); u ∈ {x, y}u } ≥ h(x) ∨ h(y) .

2.4 Vector lattice cover and Riesz completion |

87

It remains to show the converse inequality. If h = 0, then the inequality is trivially true, so we may assume that h ≠ 0. Without loss of generality we may also assume that h(x) ≥ h(y) and we take λ = h(x) − h(y). As h ≠ 0, there exists v ∈ X with h(v) ≠ 0 and since X is directed there exist v1 , v2 ∈ K with v = v1 − v2 . Then h(v1 ) ≠ 0 or h(v2 ) ≠ 0, so after suitable scaling we find a z ∈ K with h(z) = 1. Define a = x − y − λz. Then h(a) = h(x) − h(y) − λ = 0 . Let ε > 0. By (d), there exists v ∈ K with v ≥ a and h(v) < ε. Take u = v + y + λz. Then u ≥ y and u ≥ a + y + λz = x, so u ∈ {x, y}u . Further, h(u) = h(v + y + λz) = h(v + x − a) = h(x) + h(v) < h(x) + ε = h(x) ∨ h(y) + ε . It follows that inf {h(u); u ∈ {x, y}u } ≤ h(x) ∨ h(y) and the proof is complete. Related results for Riesz* homomorphisms on order unit spaces are given in Proposition 2.5.5 below.

Notes and Remarks Before van Haandel introduced Riesz* homomorphisms, there were several generalizations of Riesz homomorphisms to certain classes of partially ordered vector spaces. Riesz homomorphisms as in Definition 2.3.15 were considered by Buskes and van Rooij in [36]. An earlier version for operators with values in Riesz spaces appeared in [34], where the definition was in the spirit of Proposition 2.3.26 (ii). A notion of Riesz homomorphism in ordered Banach spaces with the Riesz decomposition property was introduced by Wickstead in [158]. In a setting of Choquet theory, Jellett gave a notion of Riesz homomorphism in [73].

2.4 Vector lattice cover and Riesz completion We proceed to the discussion of Section 2.1. Let us consider a partially ordered vector space (X, K) that is directed, the partially ordered set X δ of its Dedekind cuts, and the operations ⊕, ⊖, and ∗ on X δ . The main aim is to establish a result similar to Theorem 2.1.13, where the embedding of an Archimedean space X in its Dedekind completion is given. We want to embed X order densely in a Riesz space, which need not be Dedekind complete, and we wish to relax the condition that X be Archimedean to a condition that is as weak as possible. It turns out that X can be embedded order densely in a Riesz space if and only if X is pre-Riesz. We will construct the Riesz space

88 | 2 Embeddings, covers, and completions as a suitable subset X ρ of X δ ; see (2.21) below. On the set X ρ the operations ⊕ and ∗ are vector space operations if and only if X is a pre-Riesz space. The space X ρ will then be a Riesz space. For any finite subset F ⊆ X δ the supremum sup F exists in X δ according to Proposition 2.1.3. Recall the embedding map J : X → X δ , x 󳨃→ {x}l . For a1 , . . . , a m ∈ X we have m

m

m

ul

A := ⋁ J(a i ) = ⋁{a i }l = (⋃{a i }l ) i=1

and note that

i=1 m

i=1

u

Au = (⋃{a i }l ) = {a1 , . . . , a m }u .

(2.20)

i=1

Proposition 2.4.1. Let X be a directed partially ordered vector space. For every a1 , . . . , l a m ∈ X and A = ⋁ m i=1 {a i } the identity A ⊕ ⊖A = {0}l holds true if and only if X is a pre-Riesz space. l Proof. First assume that X is pre-Riesz. Consider a1 , . . . , a m ∈ X and A = ⋁m i=1 {a i } = ul u u ulu u u {a1 , . . . , a m } . Let x ∈ (A ⊕ ⊖A) = (A + (−A )) = (A − A ) . For a ∈ A and u ∈ Au we have x ≥ a − u, so a − x ≤ u, so a − x ∈ Aul = A, hence A − x ⊆ A. Then Au ⊆ (A − x)u , so (A+x)u ⊆ Au . Hence, ({a1 , . . . , a m }+x)u = (A+x)u ⊆ Au = {a1 , . . . , a m }u . Since X is pre-Riesz, it follows with the aid of Lemma 2.2.5 (i) that x ≥ 0. Hence (A ⊕ ⊖A)u ⊆ {0}u and thus, A ⊕ ⊖A = (A ⊕ ⊖A)ul ⊇ {0}ul = {0}l .

Lemma 2.1.7 (v) now yields that A ⊕ ⊖A = {0}l . For the converse statement, let a1 , . . . , a m ∈ X, let F = {a1 , . . . , a m } and let x ∈ X l be such that (F + x)u ⊆ F u . We will show that x ≥ 0. Let A = ⋁m i=1 {a i } . By assumption, {0}l = A ⊕ ⊖A = (A + (−Au ))ul = (Aul − Au )ul , so by (2.20), u

{0}u = ({a1 , . . . , a m }ul − {a1 , . . . , a m }u ) . For every u ∈ {a1 , . . . , a m }u and v ∈ {a1 , . . . , a m }ul , we have u + x ≥ a i + x for every i, so u + x ∈ (F + x)u ⊆ F u , so u + x ≥ a i for every i. Hence, u + x ∈ {a1 , . . . , a m }u and therefore, u + x ≥ v, so x ≥ v − u. Thus, x ∈ ({a1 , . . . , a m }ul − {a1 , . . . , a m }u )u , which yields that x ≥ 0. For a directed partially ordered vector space (X, K) we consider the subset X ρ of X δ given by m

n

X ρ = {⋁{a i }l ⊕ ⊖ ⋁{b j }l ; a1 , . . . , a m , b 1 , . . . , b n ∈ X, m, n ∈ ℕ} . i=1

j=1

On X ρ we consider the operations ⊕ and ∗ and the partial order ⊆.

(2.21)

2.4 Vector lattice cover and Riesz completion | 89

Theorem 2.4.2. Let X be a pre-Riesz space and let X ρ be given by (2.21). Then X ρ is a Riesz space, J : X → X ρ given by x 󳨃→ {x}l , is a bipositive linear map, J[X] is order dense in X ρ , and J[X] generates X ρ as a Riesz space. Moreover, J is a complete Riesz homomorphism. Proof. Since X is directed, the suprema in (2.21) exist in X δ according to Proposition 2.1.3. Let us denote m

V = {⋁{a i }l ; a1 , . . . , a m ∈ X, m ∈ ℕ} . i=1

Then X ρ = {V ⊕ ⊖W; V, W ∈ V}. We first show that for every C ∈ X ρ we have C ⊕ ⊖C = {0}l . Indeed, for A ∈ V, Proposition 2.4.1 yields that A ⊕ ⊖A = {0}l . If A, B ∈ V, then by Lemma 2.1.7 and 2.1.8, (A ⊕ ⊖B) ⊕ ⊖(A ⊕ ⊖B) = (A ⊕ ⊖A) ⊕ (B ⊕ ⊖B) = {0}l ⊕ {0}l = {0}l . It follows that for every C ∈ X ρ we have C ⊕ ⊖C = {0}l . Next, we will show that X ρ is a vector space. Our major concern is showing that λ ∗ C ∈ X ρ for every λ ∈ ℝ and C ∈ X ρ and that X ρ is closed under addition ⊕. Let n l l a1 , . . . , a m , b 1 , . . . , b n ∈ X, A = ⋁m i=1 {a i } , B = ⋁j=1 {b j } , and C = A ⊕ ⊖B. Then for λ > 0, by Lemma 2.1.7 and Lemma 2.1.8 (iii), λ ∗ C = (λ ∗ A) ⊕ (λ ∗ (⊖B)) = (λ ∗ A) ⊕ (λ ∗ ((−1) ∗ B)) = (λ ∗ A) ⊕ ((−λ) ∗ B) = (λ ∗ A) ⊕ ((−1) ∗ (λ ∗ B)) = (λ ∗ A) ⊕ ⊖(λ ∗ B) = (λA) ⊕ ⊖(λB) m

n

= ⋁{λa i }l ⊕ ⊖ ⋁{λb j }l ∈ X ρ . i=1

j=1

Moreover, 0 ∗ C = {0}l ∈ X ρ and (−1) ∗ C = ((−1) ∗ A) ⊕ ((−1) ∗ (⊖B)) = ⊖A ⊕ (⊖ ⊖ B) = ⊖A ⊕ B = B ⊕ ⊖A ∈ X ρ . Hence, λ ∗ C ∈ X ρ for every λ ∈ ℝ and every C ∈ X ρ . n l l Let a1 , . . . , a m , b 1 , . . . , b n ∈ X and let A = ⋁m i=1 {a i } , B = ⋁j=1 {b j } . Then, by Lemma 2.1.6 and (2.20), A ⊕ B = (A + B)ul = (Aul + Bul )ul ul

= ({a1 , . . . , a m }ul + {b 1 , . . . , b n }ul )

90 | 2 Embeddings, covers, and completions = {a i + b j ; i = 1, . . . , m, j = 1, . . . , n}ul m

ul

n

m

n

= (⋃ ⋃{a i + b j }l ) = ⋁ ⋁{a i + b j }l ∈ V . i=1 j=1

i=1 j=1

Let A, B, C, D ∈ V. Then A ⊕ C, B ⊕ D ∈ V. Hence, by Lemma 2.1.8 (iii), (A ⊕ ⊖B) ⊕ (C ⊕ ⊖D) = (A ⊕ C) ⊕ (⊖B ⊕ ⊖D) = (A ⊕ C) ⊕ ⊖(B ⊕ D) ∈ X ρ . Together with Lemma 2.1.7 (i)–(iii) and (vi)–(vii) and Lemma 2.1.8 (iii) and (v) it follows that the set X ρ with addition ⊕, scalar multiplication ∗, and zero element {0}l is a vector space. Lemma 2.1.11 yields that the map J is linear. According to Lemma 2.1.9, X ρ is a partially ordered vector space. The space X ρ is closed under supremum, as we show next. For a1 , . . . , a m , a m+1 , n l l . . . , a n ∈ X and A = ⋁m i=1 {a i } , B = ⋁i=m+1 {a i } , we have m

n

n

A ∨ B = ⋁{a i }l ∨ ⋁ {a i }l = ⋁{a i }l ∈ V . i=1

i=m+1

i=1

Hence, for A, B, C, D ∈ V we obtain with the aid of Lemma 2.1.10 that (A ⊕ ⊖B) ∨ (C ⊕ ⊖D) = [(A ⊕ ⊖B) ∨ (C ⊕ ⊖D)] ⊕ (B ⊕ D) ⊕ ⊖(B ⊕ D) = [(A ⊕ ⊖B ⊕ (B ⊕ D)) ∨ (C ⊕ ⊖D ⊕ (B ⊕ D))] ⊕ ⊖(B ⊕ D) = [(A ⊕ D) ∨ (C ⊕ B)] ⊕ ⊖(B ⊕ D) ∈ X ρ , since A ⊕ D, C ⊕ B, B ⊕ D ∈ V. It follows that X ρ is closed under the lattice operations of X δ and therefore, X ρ is a Riesz space. According to Proposition 2.1.4, the map J is bipositive and J[X] is order dense in X ρ . It is immediately clear from (2.21) that J[X] generates X ρ as a Riesz space. Due to Proposition 2.3.27 the map J is a complete Riesz homomorphism. Next we claim that in the special case of a Riesz space X the construction of X ρ in fact produces X itself. Proposition 2.4.3. If X is a Riesz space and X ρ and J are as in Theorem 2.4.2, then J : X → X ρ is a Riesz isomorphism. Proof. Theorem 2.4.2 in combination with Theorem 2.3.19 yields that J is a Riesz homomorphism. If X is a Riesz space, then from the definition of X ρ in (2.21) it is easily observed that J is surjective. Thus, J is a Riesz isomorphism. The space X ρ is unique in the following sense. Proposition 2.4.4. Let X be a pre-Riesz space and let X ρ and J be as in Theorem 2.4.2. If Y is a Riesz space and i : X → Y is a bipositive linear map such that i[X] is order dense in Y and generates Y as a Riesz space, then Y and X ρ are isomorphic Riesz spaces.

2.4 Vector lattice cover and Riesz completion | 91

Proof. Due to Theorem 2.4.2 and the assumptions, we have that X1 := X2 := X, Y1 := X ρ , Y 2 := Y, i1 := J, and i2 := i satisfy the conditions of Theorem 2.3.12. The identity map h : X → X, x 󳨃→ x, is a Riesz* homomorphism. According to Theorem 2.3.12 there exists a Riesz homomorphism ĥ : X ρ → Y such that ĥ ∘ J = i ∘ h = i. It remains to show that ĥ is surjective. Let y ∈ Y. Since i[X] generates Y as a Riesz space, there are a1 , . . . , a m , b 1 , . . . , n b n ∈ X such that y = ⋁m j=1 i(a j ) − ⋁k=1 i(b k ). Then m

n

m

n

j=1

k=1

j=1

k=1

̂ ̂ ̂ y = ⋁ h(J(a j )) − ⋁ h(J(b k )) = h(⋁ J(a j ) − ⋁ J(b k )) , so that y is in the range of h.̂ Thus, ĥ : X ρ → Y is surjective and hence a Riesz isomorphism. We summarize the results of this section in the following theorem, which is originally due to van Haandel; see [151, Corollaries 4.9–11 and Theorems 3.5, 3.7, 4.13]. Theorem 2.4.5. Let X be a partially ordered vector space. The following statements are equivalent. (i) X is a pre-Riesz space. (ii) There exist a Riesz space Y and a bipositive linear map i : X → Y such that i[X] is order dense in Y. (iii) There exist a Riesz space Y and a bipositive linear map i : X → Y such that i[X] is order dense in Y and generates Y as a Riesz space. Moreover, all Riesz spaces Y as in (iii) are isomorphic as Riesz spaces. Proof. The implication (i) ⇒ (iii) is the content of Theorem 2.4.2. It is trivial that (iii) implies (ii). According to Proposition 2.2.4, from (ii) follows that i[X] is a pre-Riesz space, which implies (i). Proposition 2.4.4 yields that all Riesz spaces Y as in (iii) are isomorphic as Riesz spaces. A central method for studying pre-Riesz spaces is to embed them into Riesz spaces with the aid of Theorem 2.4.5. Therefore, the following important notions are introduced. Definition 2.4.6. A pair (Y, i) as in Theorem 2.4.5 (ii) is called a vector lattice cover of X, and a pair (Y, i) as in (iii) is called a Riesz completion of X. As all spaces Y as in (iii) are Riesz isomorphic, we will speak of the Riesz completion of X when we mean a realization of it. Occasionally the Riesz completion is denoted by (X ρ , i), as the space X ρ defined in (2.21) is one of the realizations. Beside the Riesz completion, other vector lattice covers will also be useful. The results in Theorem 2.1.13 and Proposition 1.5.41 have the following consequence. Corollary 2.4.7. Every Archimedean directed partially ordered vector space has a Dedekind complete and, hence, a uniformly complete vector lattice cover.

92 | 2 Embeddings, covers, and completions

We compute explicit representations of the Riesz completions of the examples discussed in Subsection 1.7.1. Example 2.4.8. Recall that Example 1.7.1 considers the space X = P2 (ℝ) with the order induced from C(ℝ). We have shown that X is Archimedean and directed, hence X is a pre-Riesz space by Proposition 2.2.3. Moreover, X is order dense in the vector lattice V given by (1.13). Hence, (V, i) is a vector lattice cover of X, where i : X → V, x 󳨃→ x. The Riesz completion of X can be obtained as a Riesz subspace of V. Let us describe its elements. A function y on ℝ is called a piecewise polynomial function of degree at most 2 if there are n ∈ ℕ and t1 , . . . , t n ∈ ℝ such that t1 < t2 < . . . < t n and y is a polynomial function of degree at most 2 on (−∞, t1 ], [t n , ∞) and [t i , t i+1 ] for every i ∈ {1, . . . , n − 1}. For such a function y we define L(y) = lim

t→−∞

y(t) t2

and

R(y) = lim

t→∞

y(t) . t2

Denote Y = {y ∈ C(ℝ); y is a piecewise polynomial function of degree at most 2 and L(y) = R(y)} . Clearly, X ⊂ Y ⊆ V, hence X is also order dense in Y. In addition, Y is a Riesz subspace of V. We show that X generates Y as a vector lattice. Indeed, consider a function y ∈ Y with the corresponding t1 < t2 < . . . < t n and p i ∈ X, i ∈ {0, . . . , n}, so that y = p0 on (−∞, t1 ], y = p i on [t i , t i+1 ], i ∈ {1, . . . , n − 1}, and y = p n on [t n , ∞). For a fixed i ∈ {1, . . . , n − 1}, choose a parabola z ∈ X through (t i , y(t i )) and (t i+1 , y(t i+1 )) with the coefficient of t2 sufficiently large, so that y i := p i ∨ z satisfies the conditions y i ≥ y on ℝ and y i = y on [t i , t i+1 ]. Similarly, one can construct y0 as a pointwise maximum of at most three elements of X with y0 = p0 on (−∞, t1 ] and y0 ≥ y. Indeed, if a = L(y) = R(y) ≤ 0, one can take for y0 the pointwise maximum of p0 and an appropriate affine function. If a > 0 one can take y0 = p0 ∨ v ∨ w, where v(t) = at2 + pt + q is such that v(t n ) = y(t n ) and v󸀠 (t n ) is so large that v ≤ p0 on (−∞, t1 ] and v ≥ p n on [t n , ∞). Further, w is a parabola through (t1 , y(t1 )) and (t n , y(t n )) such that w ≥ y on [t1 , t n ] and w ≤ p0 on (−∞, t1 ]. One can construct y n analogously. Now y = ⋀ni=0 y i = − ⋁ni=0 (−y i ). We conclude that (Y, i) is the Riesz completion of X. Example 2.4.9. By Example 1.7.2, the space P2 [0, 1] of polynomial functions on [0, 1] of degree at most 2 is an order dense subspace of C[0, 1]. Similar to the arguments in Example 2.4.8, the smallest Riesz subspace of C[0, 1] that contains P2 [0, 1] is the space Y of all continuous piecewise polynomial functions on [0, 1] of degree at most two. Again using the identity map i, we obtain (Y, i) as the Riesz completion of P 2 [0, 1]. Example 2.4.10. By means of Y and i as in Example 2.4.8, we obtain the Riesz completion of spaces that are order isomorphic to P2 (ℝ). Example 1.7.3 considers the space

2.4 Vector lattice cover and Riesz completion |

93

with ice cream cone (ℝ3 , Lℝ2 ). Recall that there is an order isomorphism j : ℝ3 → P2 (ℝ) given by (1.17). Hence, (Y, i ∘ j) is the Riesz completion of (ℝ3 , Lℝ2 ). Moreover, Example 1.7.4 deals with the space of symmetric matrices (V2 , Pos2 ), and an order isomorphism h : P2 (ℝ) → V2 is given by (1.20). So, the Riesz completion of (V2 , Pos2 ) is (Y, i ∘ h−1 ). Finally, Example 1.7.5 presents the space of affine functions X on the unit circle S and an order isomorphism q given in (1.22) from (ℝ3 , Lℝ2 ) to X. Therefore, (Y, i ∘ j ∘ q−1 ) is the Riesz completion of X. Moreover, we have shown in Example 1.7.5 that X is an order dense subspace of C(S). Therefore the Riesz subspace of C(S) generated by X is another representation of the Riesz completion of X. Next we consider Riesz* homomorphisms between pre-Riesz spaces and Riesz homomorphisms between their vector lattice covers. They correspond in the following way, which is a result due to van Haandel; see [151, Theorem 5.6]. Theorem 2.4.11. Let X1 and X2 be pre-Riesz spaces and let (Y1 , i1 ) and (Y 2 , i2 ) be vector lattice covers of X1 and X2 , respectively. Let h : X1 → X2 be a linear map. (i) If there exists a Riesz homomorphism ĥ : Y 1 → Y2 such that ĥ ∘ i1 = i2 ∘ h, then h is a Riesz* homomorphism. (ii) If (Y 1 , i1 ) is the Riesz completion of X1 and h is a Riesz* homomorphism, then there exists a unique Riesz homomorphism ĥ : Y 1 → Y2 with ĥ ∘ i1 = i2 ∘ h. Moreover, for every a1 , . . . , a m , b 1 , . . . , b n ∈ X1 , the map ĥ satisfies m

n

m

n

j=1

k=1

j=1

k=1

̂ ⋁ i1 (a j ) − ⋁ i1 (b k )) = ⋁ i2 (h(a j )) − ⋁ i2 (h(b k )) . h( Proof. (i) By the definition of a vector lattice cover, the result is immediately obtained due to Proposition 2.3.6. (ii) The existence of ĥ with the desired properties follows from Theorem 2.3.12. This is a good place to hide the following treasure. Proposition 2.4.12. If a pre-Riesz space (X, K) is Archimedean and (Y, i) is a vector lattice cover of X, then Y is Archimedean as well. Proof. Let z, y ∈ Y be such that for every n ∈ ℕ one has that nz ≤ y. Since i[X] is majorizing in Y, there is v ∈ X such that y ≤ i(v). Let x ∈ X be such that i(x) ≤ z, so for every n ∈ ℕ one has ni(x) ≤ nz ≤ y ≤ i(v) , hence, nx ≤ v. Since X is Archimedean, it follows that x ≤ 0. Due to the order denseness of i[X] in Y we obtain z = sup{i(x); x ∈ X, i(x) ≤ z} ≤ 0.

94 | 2 Embeddings, covers, and completions

Notes and remarks If X is a pre-Riesz space and (X ρ , i) its Riesz completion, then by Theorem 2.4.2 the map i is an injective Riesz* homomorphism. The Riesz completion has the so-called universal property of a completion. Indeed, due to Theorem 2.4.11, for every Riesz space Y and Riesz* homomorphism h : X → Y there exists a unique Riesz* homomorphism h0 : X ρ → Y such that h = h0 ∘ i. For Archimedean directed ordered vector spaces the ingredients of the Riesz completion are also given in [36]. If one does not require order denseness of the embedding, every partially ordered vector space can be embedded into a Riesz space. Indeed, Luxemburg showed in [109] that for every partially ordered vector space X there exist a Riesz space Y and a bipositive linear map j : X → Y such that for every x1 , x2 ∈ X for which x1 ∨ x2 exists we have j(x1 ∨ x2 ) = j(x1 ) ∨ j(x2 ).

2.5 Functional representation The Riesz completion is a powerful tool in the study of pre-Riesz spaces. However, its construction by means of Dedekind cuts is cumbersome in most examples. A vector lattice cover consisting of continuous functions would be more convenient, as then the cover is endowed with pointwise order. A classical method to associate a space of continuous functions to an Archimedean ordered vector space (X, K) with order unit u is the functional representation as defined next. In this case, X is directed and, by Proposition 2.2.3, a pre-Riesz space. We endow X with the u-norm ‖⋅‖u as defined in (1.10), i.e., (X, K, ‖⋅‖u ) is an order unit space. The subsequent two definitions are due to [8, p. 387]. Definition 2.5.1. Let (X, K) be an Archimedean ordered vector space with order unit u. A pair (σ, Ω), where Ω is a compact Hausdorff space and σ : X → C(Ω) a bipositive linear map, is called a functional representation of X if σ maps u to the constant function 𝟙 and σ(X) separates the points of Ω, i.e., for every ω1 ≠ ω2 in Ω there is an x ∈ X such that σ(x)(ω1 ) ≠ σ(x)(ω2 ). Note that the map σ is in fact an isometry, since the supremum norm in C(Ω) equals the 𝟙-norm. Definition 2.5.2. Let (X, K) be an Archimedean ordered vector space with order unit and let (σ 1 , Ω1 ) and (σ 2 , Ω2 ) be two functional representations of X. (i) The functional representation (σ 1 , Ω1 ) is called smaller than (σ 2 , Ω2 ) if there exists Ω ⊆ Ω2 and a homeomorphism α : Ω1 → Ω such that for all x ∈ X and all ω ∈ Ω1 one has σ 1 (x)(ω) = σ 2 (x)(α(ω)) .

2.5 Functional representation |

95

(ii) If (σ 1 , Ω1 ) is smaller than (σ 2 , Ω2 ) with the additional property Ω = Ω2 , then (σ 1 , Ω1 ) and (σ 2 , Ω2 ) are called isomorphic. A functional representation can be constructed explicitly in the following way; see [75]. For an order unit space (X, K, ‖⋅‖u ), consider the (norm) dual space X 󸀠 of X and the dual cone K 󸀠 . Recall that by Proposition 1.5.17 we have K ∗ = K 󸀠 . The set Σ := Σ X := {φ ∈ K 󸀠 ; φ(u) = 1}

(2.22)

is a base of K 󸀠 , and specifically, it is convex. By the Banach–Alaoglu theorem, the closed unit ball B󸀠 of X 󸀠 is weakly-∗ compact. As Σ is a weakly-∗ closed subset of B󸀠 , the set Σ is weakly-∗ compact in X 󸀠 . By the Krein–Milman theorem, the set Σ equals the weak-∗ closure of the convex hull of the extreme points of Σ. Denote the set of all extreme points³ of Σ by Λ := Λ X := ext(Σ) . (2.23) In general, Λ need not be weakly-∗ closed, not even if X is finite dimensional; see Example 2.6.10 below. Let Λ denote the weak-∗ closure of Λ in Σ. The set Λ with weak∗ topology is a compact Hausdorff space. We consider the map Φ := Φ X : X → C (Λ) ,

x 󳨃→ (φ 󳨃→ φ(x)) .

(2.24)

Lemma 2.5.3. Let x ∈ X. Then x ∈ K if and only if for every φ ∈ Λ one has φ(x) ≥ 0. Proof. Recall that K is ‖⋅‖u -closed; see Lemma 1.5.9 (ii). Due to Proposition 1.5.5 (ii), we have that x ∈ K is equivalent to φ(x) ≥ 0 for all φ ∈ K 󸀠 . Σ equals the weak-∗ closure of the convex hull of Λ. Now suppose that φ(x) ≥ 0 for every φ ∈ Λ. As x acts as a weakly-∗ continuous functional on X 󸀠 , we get φ(x) ≥ 0 for all φ ∈ Σ, hence for all φ ∈ K 󸀠 . We are now in the position to state Kadison’s well-known result on the functional representation of X; see [75, Theorem 2.1]. Theorem 2.5.4. Let (X, K, ‖⋅‖u ) be an order unit space and let Λ and Φ be defined as in (2.23) and (2.24). Then (Φ, Λ) is a functional representation of X. Proof. First observe that Φ is linear and maps u to the constant function 𝟙. The space Φ[X] separates the points of Λ, as for φ1 , φ2 ∈ Λ with φ1 ≠ φ2 there is x ∈ X such that φ1 (x) ≠ φ2 (x). Lemma 2.5.3 ensures that Φ is bipositive. In the next proposition we observe that the elements of Λ and Λ are precisely the Riesz homomorphisms and the Riesz* homomorphisms, respectively. The statement in (i) is based on Hayes’s theorem (Proposition 2.3.28), and the result in (iii) is stated in [151, Theorem 5.10 (iii)]. 3 The elements of Σ are also called states of X, and the elements of Λ pure states of X; see, e.g., [9, Definition 1.17].

96 | 2 Embeddings, covers, and completions Proposition 2.5.5. Let (X, K, ‖⋅‖u ) be an order unit space, let Σ and Λ be defined as in (2.22) and (2.23), and let φ ∈ Σ. (i) One has φ ∈ Λ if and only if φ is a Riesz homomorphism. (ii) If X is, in addition, a Riesz space, then Λ is weakly-∗ compact. (iii) One has φ ∈ Λ if and only if φ is a Riesz* homomorphism. Proof. (i) Let φ ∈ Σ. By Lemma 1.5.19 we have φ ∈ Λ if and only if φ is extremal in K 󸀠 = K ∗ . Then Proposition 2.3.28 yields (i). (ii) It remains to show that Λ is a weakly-∗ closed subset of Σ. Let (φ i )i∈I be a net in Λ and let φ ∈ Σ be such that φ i → φ in the weak-∗ topology. Let x1 , x2 ∈ X. Then on the one hand φ i (x1 ∨ x2 ) → φ(x1 ∨ x2 ), and on the other hand φ i (x1 ∨ x2 ) = φ i (x1 ) ∨ φ i (x2 ) → φ(x1 ) ∨ φ(x2 ). Hence, φ is a Riesz homomorphism. Due to (i), we obtain φ ∈ Λ. (iii) Let φ ∈ Λ. Then due to (i) there exists a net (φ i ) of Riesz homomorphisms that converges to φ in the weak-∗ topology. To show that φ is a Riesz* homomorphism, let A = {a1 , . . . , a m } be a nonempty finite subset of X and let x ∈ Aul . Then x ≤ v for every upper bound v of A, so, according to Proposition 2.3.26 (ii), for every i we get φ i (a1 ) ∨ ⋅ ⋅ ⋅ ∨ φ i (a m ) = inf {φ i (v); v ∈ Au } ≥ φ i (x) . Then φ(x) ≤ φ(a1 ) ∨ ⋅ ⋅ ⋅ ∨ φ(a m ) and φ is a Riesz* homomorphism, by Proposition 2.3.26 (i). To prove the converse, let (Y, i) be the Riesz completion of (X, K) and let A denote the set of Riesz homomorphisms ψ : Y → ℝ with ψ(i(u)) = 1. Due to (i), the set A consists of all extremal points of the weakly-∗ compact set Σ Y := {ψ : Y → ℝ; ψ is positive and ψ(i(u)) = 1} . By (ii), the set A is weakly-∗ compact. Let C be the set of Riesz* homomorphisms φ : X → ℝ with φ(u) = 1. According to Theorem 2.3.12, the Riesz* homomorphisms on X and the Riesz homomorphisms on Y correspond, that is, the map η : A → C given by η(ψ) = ψ ∘ i is a bijection. Since η is continuous with respect to the weak-∗ topologies, the set C = η[A] is weakly-∗ compact. As η is a continuous bijection between compact spaces, the map η is a homeomorphism with respect to the weak-∗ topologies; see Proposition 1.8.11. Due to (i) and Theorem 2.3.19, the set Λ is contained in C, hence Λ ⊆ C as C is weakly-∗ closed. Consequently, Λ is a weakly-∗ compact subset of the weakly-∗ compact set C. It remains to show that C ⊆ Λ. First we show that η−1 [Λ] is a total subset of Y +󸀠 . Let z ∈ Y be such that ψ(z) ≥ 0 for every ψ ∈ η−1 [Λ]. Let φ ∈ Λ and ψ := η−1 (φ). Then for every x ∈ X with i(x) ≥ z we have φ(x) = η(ψ(x)) = ψ(i(x)) ≥ ψ(z) ≥ 0 .

2.5 Functional representation | 97

As Λ is a total subset of K 󸀠 , we obtain x ≥ 0. By order denseness of i[X] in Y, it follows that z ≥ 0. We conclude that η−1 [Λ] is a total set. Further we consider the weak-∗ closure of the convex hull of the set η −1 [Λ], denoted by co η−1 [Λ]. This set is total as well, as it contains η−1 [Λ]. Next we show that co η−1 [Λ] = Σ Y . Since Λ ⊆ C, we have η−1 [Λ] ⊆ A ⊆ Σ Y and hence co η−1 [Λ] ⊆ Σ Y . To show the converse inclusion, suppose that there exists ψ0 ∈ Σ Y \ co η−1 [Λ]. By the Hahn–Banach theorem, there are y ∈ Y and α ∈ ℝ such that ψ0 (y) < α and ψ(y) ≥ α for every ψ ∈ co η−1 [Λ]. Let z := y − αi(u). Then ψ0 (z) < 0 and ψ(z) ≥ 0 for every ψ ∈ co η−1 [Λ]. As co η−1 [Λ] is a total set, we obtain z ≥ 0, which contradicts ψ0 (z) < 0. Hence, co η−1 [Λ] = Σ Y . As η−1 is continuous, the set η−1 [Λ] is weakly-∗ compact. Due to Theorem 1.8.8, the extremal points of Σ Y = co η−1 [Λ] are contained in η−1 [Λ], so that A ⊆ η−1 [Λ]. Hence, C = η[A] ⊆ Λ. Let us now consider the case that (X, K, ‖⋅‖u ) is, in addition, a Riesz space. If we apply Theorem 2.5.4, we reproduce Kakutani’s famous theorem on representation of abstract M-spaces⁴; see [76, Theorem 2]. Theorem 2.5.6. Let (X, K, ‖⋅‖u ) be an order unit space that is a Riesz space, and let Λ be defined as in (2.23). Then Λ is weakly-∗ compact, Φ[X] is a norm dense Riesz subspace of C(Λ), and Φ : X → Φ[X] is an isometric Riesz isomorphism. Proof. Proposition 2.5.5 yields that Λ is weakly-∗ compact, hence Λ = Λ. From Theorem 2.5.4 it follows that (Φ, Λ) is a functional representation, that is Φ : X → C(Λ) is linear and bipositive, the set Φ[X] separates the points of Λ, and Φ maps u to the constant function 𝟙, hence Φ is an isometry. Next we show that Φ is a Riesz homomorphism. Let x1 , x2 ∈ X and φ ∈ Λ. Then φ is a Riesz homomorphism according to Proposition 2.5.5, hence Φ(x1 ∨ x2 )(φ) = φ(x1 ∨ x2 ) = φ(x1 ) ∨ φ(x2 ) = (Φ(x1 ) ∨ Φ(x2 ))(φ) . We conclude that Φ is a Riesz homomorphism and that Φ[X] is a Riesz subspace of C(Λ). We apply the Stone–Weierstrass theorem 1.8.10 and obtain that Φ[X] is norm dense in C(Λ). In the remaining part of this section we consider again an order unit space (X, K, ‖⋅‖u ) and Λ and Φ defined as in (2.23) and (2.24). We investigate the question whether Φ[X] is order dense in C(Λ), i.e., whether (C(Λ), Φ) is a vector lattice cover of the pre-Riesz space X. The subsequent results in Lemma 2.5.7 and Theorems 2.5.8 and 2.5.9 are a joint work with Bas Lemmens; see [77]. Denote by Lat(X, K) the Riesz subspace generated

4 An abstract M-space is a Riesz space with an M-norm such that the space is norm complete. Then in Theorem 2.5.6 we have Φ[X] = C(Λ). M-norms are studied in Subsection 3.3.6 below.

98 | 2 Embeddings, covers, and completions

by Φ[X] inside C(Λ), i.e., m

n

Lat(X, K) = {⋁ f j − ⋁ g j ; f1 , . . . , f m , g1 , . . . , g n ∈ Φ[X]} . j=1

(2.25)

j=1

The following lemma shows that (Φ, Λ) is smaller than any order dense functional representation (σ, Ω) of X. Lemma 2.5.7. Let (σ, Ω) be a functional representation of X such that σ[X] is order dense in C(Ω). (i) For every φ ∈ Λ there exists a unique ω ∈ Ω such that for each x ∈ X one has φ(x) = σ(x)(ω). (ii) Let α: Λ → Ω (2.26) be the unique map such that for every φ ∈ Λ and for every x ∈ X one has that φ(x) = σ(x)(α(φ)), and let Γ := α [Λ] . (2.27) Then α : Λ → Γ is a homeomorphism. Proof. In this proof, let R be the Riesz subspace of C(Ω) generated by σ[X]. The set σ[X] is order dense in C(Ω) and hence in R, therefore R is the Riesz completion of X. We have that R separates the points of X, since R ⊇ σ[X], and σ[X] separates the points of X. Moreover, R contains the constant function 𝟙. By the Stone–Weierstrass theorem 1.8.10 the subspace R is norm dense in C(Ω). (i) Let φ ∈ Λ. Due to Proposition 2.5.5, the map φ is a Riesz* homomorphism on X. Since σ is bipositive, the map σ : X → σ[X] is a bijection, hence φ ∘ σ −1 : σ[X] → ℝ is a Riesz* homomorphism. Due to Theorem 2.3.12 there exists a unique Riesz homomorphism ψ : R → ℝ extending φ ∘ σ −1 . As the constant function 𝟙 is an interior point of the cone in R and ψ is positive, by Proposition 1.5.17 the map ψ is continuous. Hence ψ extends uniquely to a continuous linear functional η : C(Ω) → ℝ, as R is norm dense in C(Ω). Since the lattice operations are continuous, η is a Riesz homomorphism⁵. By Theorem 1.4.7 there exist ω ∈ Ω and α ∈ ℝ+ such that for every f ∈ C(Ω) one has that η(f) = αf(ω). As η(𝟙) = ψ(𝟙) = φ(u) = 1, it follows that α = 1. We obtain φ(x) = ψ(σ(x)) = η(σ(x)) = σ(x)(ω) for every x ∈ X. Finally we verify the uniqueness of ω. Let ρ ∈ Ω be such that ρ ≠ ω. Since σ[X] separates the points of Ω, there exists x ∈ X such that σ(x)(ρ) ≠ σ(x)(ω) = φ(x). Hence ω is the only element in Ω with the property that for every x ∈ X one has that σ(x)(ω) = φ(x).

5 To extend ψ to a Riesz homomorphism η, one could alternatively use the Lipecki–Luxemburg–Schep extension in Theorem 2.1.17.

2.5 Functional representation | 99

(ii) Note that the existence and uniqueness of α are given by (i). We show that α is injective. Indeed, if φ1 , φ2 ∈ Λ are such that α(φ1 ) = α(φ2 ), then for every x ∈ X one has that φ1 (x) = φ2 (x), and therefore φ1 = φ2 . Hence, α is injective, and so α : Λ → Γ is a bijection. Next we show that α is continuous. Let (φ i )i∈I be a net in Λ and let φ ∈ Λ be such that φ i → φ. Then for every x ∈ X one has φ i (x) → φ(x), hence σ(x)(α(φ i )) → σ(x)(α(φ)). As the lattice operations in C(Ω) are pointwise, one has for every g ∈ R that g(α(φ i )) → g(α(φ)). Suppose that the net (α(φ i ))i∈I does not converge to α(φ). Then there is an open set U ⊆ Ω with α(φ) ∈ U such that for every i ∈ I there is j ∈ I with j ≥ i and α(φ j ) ∈ ̸ U. According to Urysohn’s lemma there exists a function f ∈ C(Ω) with 0 ≤ f ≤ 1, f(α(φ)) = 1 and f(ω) = 0 for every ω ∈ Ω \ U. Since R is norm dense in C(Ω), there exists g ∈ R such that g(α(φ)) > 12 and g(ω) < 12 for every ω ∈ Ω \ U. But then for every i ∈ I there exists j ∈ I with j ≥ i such that g(α(φ j )) < 12 , which contradicts g(α(φ i )) → g(α(φ)). Thus α is continuous. Since α is a continuous bijection between a compact space and a Hausdorff space, by Proposition 1.8.11 we conclude that α is a homeomorphism. The next theorem shows for an Archimedean partially ordered vector space with an order unit, that all functional representations in which the space is order densely embedded are isomorphic. Theorem 2.5.8. Let (X, K, ‖⋅‖u ) be an order unit space and let Λ = Λ X and Φ = Φ X be defined as in (2.23) and (2.24), respectively. (i) Let (σ, Ω) be a functional representation of (X, K). If σ[X] is order dense in C(Ω), then (σ, Ω) is isomorphic to (Φ, Λ). (ii) Let (Y, i) be the Riesz completion of (X, K) and consider the functional representation (Φ Y , Λ Y ) of Y. Then Φ Y [i[X]] is order dense in C(Λ Y ). Moreover, the compact Hausdorff spaces Λ and Λ Y are homeomorphic. Proof. (i) Due to Lemma 2.5.7 there exist Γ ⊆ Ω and a homeomorphism α : Λ → Γ such that for every x ∈ X and every φ ∈ Λ one has Φ(x)(φ) = σ(x)(α(φ)). We show that Γ = Ω. Let ω ∈ Ω. Define ψ : C(Ω) → ℝ, f 󳨃→ f(ω). Then ψ is a Riesz homomorphism by Theorem 1.4.7, and ψ(1) = 1. Since σ[X] is order dense in C(Ω), Proposition 2.3.6 yields that the map γ : X → ℝ given by γ := ψ ∘ σ is a Riesz* homomorphism. Moreover, γ(u) = 1. According to Proposition 2.5.5 one obtains γ ∈ Λ. For every x ∈ X one has σ(x)(α(γ)) = Φ(x)(γ) = γ(x) = ψ(σ(x)) = σ(x)(ω) . Since σ[X] separates the points of Ω, it follows that α(γ) = ω. (ii) First note that by Proposition 2.4.12 the space Y is an Archimedean Riesz space. Moreover, as i[X] is majorizing in Y, the element i(u) is an order unit in Y. According to Proposition 2.5.5 (ii), the topological space Λ Y is weakly-∗ compact. Next

100 | 2 Embeddings, covers, and completions we show that (Φ Y ∘ i, Λ Y ) is a functional representation of (X, K). Clearly, Φ Y ∘ i is bipositive and maps u to the constant function 𝟙. Furthermore, (Φ Y ∘ i)[X] separates the points of Λ Y . Indeed, let φ, ψ ∈ Λ Y with φ ≠ ψ. By Proposition 2.5.5, the maps φ and ψ are Riesz homomorphisms on Y. Due to Theorem 2.4.11 (i), the maps φ0 := φ ∘ i and ψ0 := ψ ∘ i are Riesz* homomorphisms on X. Since (Y, i) is the Riesz completion of X, by Theorem 2.4.11 (ii) the maps φ and ψ are the unique Riesz homomorphisms on Y extending φ0 and ψ0 . Hence, φ0 ≠ ψ0 , i.e., there is x ∈ X such that φ0 (x) ≠ ψ0 (x). Thus, Φ Y (i(x))(φ) = φ(i(x)) = φ0 (x) ≠ ψ0 (x) = ψ(i(x)) = Φ Y (i(x))(ψ) . We conclude that (Φ Y ∘ i, Λ Y ) is a functional representation of (X, K). To establish that Φ Y [i[X]] is order dense in C(Λ Y ), we apply the Kakutani theorem 2.5.6 and obtain that Φ Y [Y] is a norm dense Riesz subspace of C(Λ Y ). By Corollary 1.6.5, the space Φ Y [Y] is order dense in C(Λ Y ). As i[X] is order dense in Y and Φ Y is bipositive, we obtain that Φ Y [i[X]] is order dense in Φ Y [Y], and due to Proposition 1.6.3 also in C(Λ Y ). By (i), the topological spaces Λ Y and Λ are homeomorphic. With the notations of the previous theorem, we formulate the main result of this section and obtain the functional representation as a vector lattice cover. Theorem 2.5.9. Let (X, K, ‖⋅‖u ) be an order unit space. Then (C(Λ), Φ) is a vector lattice cover of X. Moreover, Lat(X, K) is the Riesz completion of (X, K). Proof. According to Theorem 2.5.4, the pair (Φ, Λ) is a functional representation of X. In particular, the map Φ : X → C(Λ) is a bipositive linear map. Next we show that Φ[X] is order dense in C(Λ). We use the Riesz completion (Y, i) of X and Theorem 2.5.8 (ii), which says that Φ Y [i[X]] is order dense in C(Λ Y ). Further, there exists a homeomorphism α : Λ → Λ Y . Define A : C(Λ Y ) → C(Λ), f 󳨃→ f ∘ α. By Proposition 1.4.6, the map A is an isometric Riesz isomorphism. Hence, A[Φ Y [i[X]]] is order dense in C(Λ). For every x ∈ X and every φ ∈ Λ one has A(Φ Y (i(x)))(φ) = Φ Y (i(x))(α(φ)) = Φ(x)(φ) , therefore A(Φ Y (i(x))) = Φ(x). This shows that A[Φ Y [i[X]]] = Φ[X], hence Φ[X] is order dense in C(Λ). Thus, (C(Λ), Φ) is a vector lattice cover of X. Then, clearly, Lat(X, K) is the Riesz completion of (X, K). We summarize this section in the subsequent scheme.

2.6 Examples of vector lattice covers

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i X

i[X] order isomorphism

⊆

Y

order dense

Φ

ΦY

order isomorphism

Riesz isomorphism

Φ[X]

⊆

ΦY [Y ]

⊆

A C(Λ)

⊆

Lat(X, K)

norm dense

C(ΛY ) isometric Riesz isomorphism

Finally we note that the space C(Λ) is in fact the uniform completion of the Riesz completion of X. Corollary 2.5.10. Let (X, K, ‖⋅‖u ) be an order unit space. Then the uniform completion of the Riesz completion of X is isometric and Riesz isomorphic to C(Λ). Proof. Due to Theorem 2.5.9 we have that (Lat(X, K), Φ) is the Riesz completion of X. By Theorem 2.5.4 the pair (Φ, Λ) is a functional representation of X, hence Lat(X, K) is a Riesz subspace of C(Λ) that contains the constant function 𝟙 and separates the points of Λ. According to the Stone–Weierstrass theorem 1.8.10, we obtain that Lat(X, K) is norm dense in C(Λ) with respect to the supremum norm. Note that the function Φ(u) = 𝟙 is an order unit in the Riesz completion Lat(X, K), and that the 𝟙-norm on Lat(X, K) equals the supremum norm. Recall that the uniform completion is the completion with respect to the 𝟙-norm; see Proposition 1.5.27. Hence the uniform completion of Lat(X, K) is C(Λ).

2.6 Examples of vector lattice covers We list three examples that show that one obtains the Riesz completion by Theorem 2.5.9 in a very convenient way; see also [77].

102 | 2 Embeddings, covers, and completions

2.6.1 Polyhedral cones Let us consider functional representations of finite dimensional spaces with generating polyhedral cones. We use the notations of Subsection 1.7.2, i.e., X = ℝn is endowed with the Euclidean norm and K is a generating polyhedral cone in X. We fix an order unit u in K and obtain an order unit space (X, K, ‖⋅‖u ). Let Σ, Λ and Φ be defined as in (2.22), (2.23) and (2.24), respectively. The set Σ has finitely many distinct extreme points f (1) , . . . , f (k) , where k ≥ n, so that Λ = Λ = {f (1) , . . . , f (k) } .

(2.28)

As usual, we identify C(Λ) with ℝk endowed with the standard cone ℝ+k , and obtain Φ : ℝn → ℝk ,

T

x 󳨃→ (f (1) (x), . . . , f (k) (x)) .

(2.29)

We show that the elements of Λ characterize the n − 1-dimensional faces of K. Here, by the dimension of a face H of K we mean the dimension of the linear hull of H. Lemma 2.6.1. A nonempty subcone H of K is a face of K with dim(span(H)) = n − 1 if and only if there is i ∈ {1, . . . , k} such that H = {x ∈ K; f (i) (x) = 0}. Proof. First observe that if H is a face of K, then (H − H) ∩ K = H. Indeed, let x ∈ K and x1 , x2 ∈ H be such that x = x1 − x2 . Then x ≤ x1 , hence x ∈ H. Now let H be a face of K with dim(span(H)) = n − 1. As H is a subcone, we have span(H) = H − H. Note that u ∈ ̸ H, as otherwise K ⊆ H and then dim(span(H)) = n. Hence, u ∈ ̸ H − H. It follows that there is f ∈ X 󸀠 such that H − H = {x ∈ X; f(x) = 0} and f(u) = 1. Next we show that f is positive. Indeed, assume that there is x ∈ K such that f(x) = −1. Then f(u + x) = 0. Hence, u + x ∈ (H − H) ∩ K = H. Since 0 ≤ x ≤ u + x and H is a face, we obtain x ∈ H, a contradiction. If n = 1, then k = 1 and H = {0}. If n ≥ 2, then H is a nontrivial face, so by Lemma 1.5.20 we have that f is extremal in K 󸀠 . Due to Lemma 1.5.19 we get that f is an extreme point of Σ, i.e., f ∈ Λ. Conversely, let i ∈ {1, . . . , k} and H := {x ∈ K; f (i) (x) = 0}. Clearly, dim(span(H)) = n − 1. For y ∈ H and x ∈ K with x ≤ y we obtain 0 ≤ f (i) (x) ≤ f (i) (y) = 0, hence x ∈ H. As a consequence of Theorem 2.5.9, the pair (ℝk , Φ) is a vector lattice cover of X. It even turns out that (ℝk , Φ) is the Riesz completion of X, which is obtained in [77, Proposition 13]. Theorem 2.6.2. Let K be a generating polyhedral cone in ℝn with k faces of dimension n − 1. Then (ℝk , Φ) is the Riesz completion of (ℝn , K). Proof. By Theorem 2.5.9, the pair (ℝk , Φ) is a vector lattice cover of X. To establish that (ℝk , Φ) is the Riesz completion of X, we show that Φ[ℝn ] generates ℝk as a Riesz space. Fix m ∈ {1, . . . , k}. There exists x ∈ K such that f (m) (x) = 0 and f (i) (x) > 0 whenever i ≠ m (i.e., f (m) is exposed in K 󸀠 ). Indeed, as f (m) is an extreme point of Σ,

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we have that f (m) is not contained in the convex hull M of Λ \ {f (m) }. By the Hahn– Banach theorem, there is v ∈ X such that f (m) (v) ≤ 0 and g(v) > 0 for every g ∈ M. Let x := v − f (m) (v)u. Then x ≥ v, hence g(x) > 0 for every g ∈ Λ \ {f (m) }. Moreover, f (m) (x) = 0. As Λ is a total set, we get x ∈ K. Next we consider the unit vector e(m) in ℝk and show that e(m) ∈ Φ[ℝn ]. Set α := min {f (i) (x); i ∈ {1, . . . , k} \ {m}} . Then α > 0. Define a := u − 1α x and note that f (m) (a) = 1 and g(a) ≤ 0 for g ∈ Λ \ {f (m)}. Therefore, Φ(a) ∨ 0 = e(m) . We conclude that the smallest Riesz subspace of ℝk containing Φ[ℝn ] equals ℝk .

2.6.2 Lorentz cones We use the functional representation to provide a vector lattice cover for spaces with Lorentz cones as discussed in Subsection 1.7.1. We consider a Hilbert space H with inner product ⟨⋅, ⋅⟩ and the associated norm ‖⋅‖. Let X := ℝ × H be endowed with the inner product as in (1.14) and ordered by means of the Lorentz cone LH given in (1.15). Recall that the point u := (1, 0) is an order unit in (X, LH ) and that the corresponding base of L󸀠H = LH is Σ = {(1, z) ∈ X; (1, z) ∈ LH } = {(1, z) ∈ X; ‖z‖ ≤ 1} ; see (1.16). The next lemma describes the extreme points of Σ. Lemma 2.6.3. The set Λ of extreme points of Σ is Λ = {(1, z) ∈ LH ; ‖z‖ = 1} .

(2.30)

Proof. It is straightforward that the extreme points of Σ are contained in {(1, z) ∈ LH ; ‖z‖ = 1}. Conversely, let (1, z) ∈ LH with ‖z‖ = 1. In order to prove that (1, z) is an extreme point of Σ, let x = (1, w) and y = (1, v) be such that ‖w‖ ≤ 1 and ‖v‖ ≤ 1. Suppose that (1, z) = λx + (1 − λ)y with λ ∈ (0, 1). As 1 = ‖z‖ = ‖λw + (1 − λ)v‖ ≤ λ‖w‖ + (1 − λ)‖v‖ ≤ λ + (1 − λ) = 1 , we obtain ‖w‖ = ‖v‖ = 1. Therefore, ‖λw + (1 − λ)v‖2 = 1 − 2λ + 2λ2 + 2⟨λw, (1 − λ)v⟩ , hence, ⟨λw, (1 − λ)v⟩ = λ(1 − λ) = ‖λw‖‖(1 − λ)v‖. Thus, in the Cauchy–Schwarz inequality we have equality, therefore v = cw for some constant c ∈ ℝ. Moreover, |c| = 1. As z = (λ + c(1 − λ))w, one has |λ + c(1 − λ)| = 1, hence, c = 1, i.e., x = y. The functional representation (Φ, Λ) with Φ as in (2.24) yields a vector lattice cover for X, where we distinguish between the infinite-dimensional and the finite-dimensional case.

104 | 2 Embeddings, covers, and completions

Theorem 2.6.4. (i) Let H be an infinite-dimensional Hilbert space and let B := {z ∈ H; ‖z‖ ≤ 1} be the unit ball in H endowed with the weak topology. Then (C(B), Φ) is a vector lattice cover of (ℝ × H, LH ). (ii) Let S n−1 := {z ∈ ℝn ; ‖z‖ = 1} be the (n − 1)-sphere in ℝn with Euclidean norm. Then (C(S n−1 ), Φ) is a vector lattice cover of (ℝ × ℝn , Lℝn ). Proof. Let S := {z ∈ H; ‖z‖ = 1} be the unit sphere in H. Due to Lemma 2.6.3, the map S → Λ, z 󳨃→ (1, z) is a homeomorphism. In the following we identify S and Λ. (i) As H is an infinite-dimensional Hilbert space, it is a standard exercise⁶ to see that S = B. Therefore, Λ = B, and Theorem 2.5.9 yields that (C(B), Φ) is a vector lattice cover of (ℝ × H, LH ). (ii) According to (2.30), for H = ℝn we have Λ = Λ. Moreover, Λ can be identified with the (n − 1)-sphere S n−1 . Then Theorem 2.5.9 implies that (C(Sn−1 ), Φ) is a vector lattice cover of (ℝ × ℝn , Lℝn ). Recall that the Lorentz cone for H = ℝ2 is the ice cream cone Lℝ2 in ℝ3 as discussed in Example 1.7.3. In this case, Theorem 2.6.4 says that (C(S1 ), Φ) is a vector lattice cover of (ℝ × ℝ2 , Lℝ2 ). Observe that for every x ∈ ℝ3 the map Φ(x) : S1 → ℝ ,

φ 󳨃→ ⟨x, φ⟩

is an affine function on S1 . Therefore, Φ : ℝ3 → Φ[ℝ3 ] is the order isomorphism between (ℝ × ℝ2 , Lℝ2 ) and the space X of affine functions on S1 as described in Example 1.7.5. The proof in Example 1.7.5 that X is order dense in C(S1 ) is a special case of Theorem 2.6.4 (ii).

2.6.3 Positive semidefinite matrices By means of the functional representation, we compute a vector lattice cover of the space (V n , Posn ) of symmetric n × n-matrices, which was already discussed above Example 1.7.4. Recall that Posn is the cone of all positive semidefinite matrices in V n , and that V n is endowed with the inner product defined in (1.18). Moreover, Posn is selfdual and the identity matrix I is an order unit. Note that the order unit norm on V n with respect to I generates the natural topology on V n , which coincides with the entrywise topology. Moreover, the weak-∗ topology on V n󸀠 = V n also equals the entrywise topology. According to (1.19), we have Σ = {A ∈ Posn ; tr(A) = 1} . The following lemma characterizes the set Λ of extreme points of Σ. 6 See, e.g., [Conway, A course in functional analysis, 2nd edition, Exercise V.1.10].

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105

Lemma 2.6.5. Let ∆ denote the set of those A ∈ Σ of which the rank is 1 and the only nonzero eigenvalue equals 1. Then Λ = Λ = ∆. Proof. As usual, let e(i) be the i-th unit vector in ℝn . For x = (x1 , x2 , . . . , x n )T ∈ ℝn the diagonal matrix with x1 , x2 , . . . , x n on the diagonal is denoted by Diag(x1 , x2 , . . . , x n ) = Diag(xT ). For A ∈ V n , we have A ∈ Σ if and only if there exists an orthogonal matrix S and λ n ≥ λ n−1 ≥ . . . ≥ λ1 ≥ 0 with ∑ni=1 λ i = 1 such that n

A = ∑ λ i ST Diag((e(i) )T )S .

(2.31)

i=1

Indeed, for A ∈ Σ the spectral theorem yields that there exists an orthogonal matrix S such that A = ST Diag(λ1 , . . . , λ n )S, where λ1 , . . . , λ n are the eigenvalues of A in the appropriate order. Hence, (2.31) is satisfied, and ∑ni=1 λ i = tr A = 1. Conversely, for every i ∈ {1, . . . , n} the matrix ST Diag((e(i) )T )S is in Σ, and therefore also convex combinations of such matrices. Now we show Λ ⊆ ∆. Indeed, if A ∈ Σ \ ∆ has the representation (2.31), then there is i ∈ {1, . . . , n} such that 0 < λ i < 1. Hence, A = λ i ST Diag((e(i) )T )S + (1 − λ i )

∑ j∈{1,...,n}\{i}

and ∑j∈{1,...,n}\{i}

λj 1−λ i

= ∑j∈{1,...,n}\{i}

λj ∑k∈{1,... ,n}\{i} λ k

λj T 1−λ i S

Diag((e(j) )T )S

= 1, so that A ∈ Σ \ Λ.

Next we establish ∆ ⊆ Λ. First, observe that for A ∈ Σ with representation (2.31) and x ∈ ℝn we have n

⟨Ax, x⟩ = ∑ λ i ((Sx)i )2 . i=1

Let A ∈ ∆ and B, C ∈ Σ be such that A = λB + (1 − λ)C for some λ ∈ (0, 1). Let A have the representation (2.31) and let B and C have similar representations n

B = ∑ μ i QT Diag((e(i) )T )Q

n

and

i=1

C = ∑ ν i RT Diag((e(i) )T )R . i=1

For x ∈ ℝn with ‖x‖ = 1, on the one hand we obtain n

n

⟨Ax, x⟩ = ⟨(λB + (1 − λ)C)x, x⟩ = λ ∑ μ i ((Qx)i )2 + (1 − λ) ∑ ν i ((Rx)i )2 i=1 2

(2.32)

i=1

≤ λμ n ‖Qx‖2 + (1 − λ)ν n ‖Rx‖ = λμ n + (1 − λ)ν n . On the other hand, as A ∈ ∆, we get ⟨Ax, x⟩ = ((Sx)n )2 . Let x := S−1 e(n) . Then by (2.32) we have 1 = ((Sx)n )2 ≤ λμ n + (1 − λ)ν n . Since μ n , ν n ≤ 1 we conclude 1 = μ n = v n , hence 0 = μ j = ν j for j ∈ {1, . . . , n − 1}. Moreover, from (2.32) it follows that 1 = λ((Qx)n )2 + (1 − λ)((Rx)n )2 .

106 | 2 Embeddings, covers, and completions As ((Qx)n )2 , ((Rx)n )2 ≤ 1, we get ((Qx)n )2 = ((Rx)n )2 = 1. Since Q and R are orthogonal, we have (Qx)j = (Rx)j = 0 for j ∈ {1, . . . , n − 1}. We obtain ⟨Bx, x⟩ = ⟨Cx, x⟩ = 1. Moreover, as B is symmetric, we have ‖B‖ = μ n = 1. By the Cauchy–Schwarz inequality, we get 1 = ⟨Bx, x⟩ ≤ ‖Bx‖ ‖x‖ ≤ ‖B‖ ‖x‖2 = 1 . (2.33) From the equality in (2.33) it follows that there is α ∈ ℝ such that Bx = αx and, moreover, α ≠ 0. As x ≠ 0, we conclude that x is an eigenvector for the eigenvalue α = μ n = 1 of B. The symmetry of B implies that the orthogonal complement of x is the eigenspace of the eigenvector 0 of B. The same properties are true for the matrix C. Consequently, B = C, hence A ∈ Λ. It is straightforward that ∆ is closed. Recall that S n−1 = {(q1 , . . . , q n )T ∈ ℝn ; q21 + . . . + q2n = 1} denotes the (n − 1)-sphere in ℝn . Lemma 2.6.6. Define the equivalence relation ∼ in Sn−1 by q∼p

⇐⇒

(q = p or q = −p) .

(2.34)

Consider on S n−1/∼ the quotient topology of the Euclidean topology on S n−1 , and on ∆ the entrywise topology. Then the map h : S n−1/∼ → ∆, [q] 󳨃→ qqT , is a homeomorphism. Proof. Note that for every q ∈ S n−1 we have qT q = 1. First we establish that h is welldefined. Indeed, for every q ∈ S n−1 the matrix qqT has rank one, and 1 is an eigenvalue of qqT with eigenvector q. Hence, qqT ∈ ∆. If q, p ∈ S n−1 are such that q ∼ p, then either q = p or q = −p, and in both cases we have qqT = ppT . Next we show that h is injective. Indeed, if p, q ∈ S n−1 and ppT = qqT , then p = p(pT p) = (ppT )p = (qqT )p = q(qT p) . Therefore, p is a multiple of q, and as both are in S n−1 , we obtain p = q or p = −q, so that p ∼ q. To prove that h is surjective, let A ∈ ∆. By the spectral theorem there exists an orthogonal matrix Q such that A = Q Diag(e (1) )QT . Let q denote the first column of Q. Then q ∈ Sn−1 and A = qqT = h([q]). In order to show that h is continuous, let (q i ) be a sequence in Sn−1 and let q ∈ S n−1 be such that q i → q. Then q i converges to q entrywise, hence, q i qTi → qqT in ∆. As S n−1 is compact and the quotient map is continuous, we have that S n−1/ ∼ is compact. By Proposition 1.8.11, the map h is a homeomorphism. We arrive at a convenient representation of a vector lattice cover of (V n , Pos n ) by means of the functional representation.

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Theorem 2.6.7. Define the map Ψ : V n → C(S n−1/∼) ,

A 󳨃→ ([q] 󳨃→ tr (A (qqT ))) .

(2.35)

Then (C(S n−1/∼), Ψ) is a vector lattice cover of (V n , Posn ). Proof. By the functional representation given in Theorem 2.5.9, the pair (C(Λ), Φ) is a vector lattice cover of (V n , Posn ), where Φ : V n → C (Λ) ,

A 󳨃→ (B 󳨃→ tr(AB)) .

By Lemma 2.6.5 we have Λ = ∆, and the weak-∗ topology on Λ and the entrywise topology on ∆ coincide. With the homeomorphism h of Lemma 2.6.6, the map T h : C(∆) → C(S n−1/∼), x 󳨃→ x ∘ h is an isometric Riesz isomorphism. As Ψ = T h ∘ Φ, we have that Ψ is bipositive and that Ψ[V n ] is order dense in C(S n−1/∼).

2.6.4 Finite dimensional examples by means of a dual base In this subsection, let ℝn be endowed with the Euclidean norm. As usual, we identify (ℝn )󸀠 with ℝn . We provide a convenient method to construct finite dimensional order unit spaces by means of a base of the dual cone. Lemma 2.6.8. Let K be a closed cone in ℝn . Then K 󸀠󸀠 = K. Proof. For every x ∈ K and y ∈ K 󸀠 we have ⟨x, y⟩ ≥ 0, hence x ∈ K 󸀠󸀠 . If x ∈ K 󸀠󸀠 , then ⟨x, y⟩ ≥ 0 for every y ∈ K 󸀠 . Hence, by Proposition 1.5.5 (ii), we have x ∈ K. Subsequently, we write elements of ℝn+1 in the form (s, v), where s ∈ ℝ and v ∈ ℝn . Proposition 2.6.9. Let S be a nonempty convex compact subset of ℝn . Define C := {(r, z); r ∈ ℝ+ , z ∈ rS} , K := {x ∈ ℝn+1 ; ∀y ∈ C : ⟨x, y⟩ ≥ 0} and Σ := {(1, z); z ∈ S}. Then K is a closed cone in ℝn+1 and the element u := (1, 0) ∈ ℝ × ℝn is an order unit in K. Moreover, C = K 󸀠 and Σ = {φ ∈ C; ⟨u, φ⟩ = 1}. Proof. Clearly, K is a closed cone in ℝn+1 . To show that u is an order unit, let x = (s, v) ∈ K and choose λ := s + max{⟨v, w⟩; w ∈ Σ}. Then for every y := (r, z) ∈ C \ {0} we have ⟨λu − x, y⟩ = (λ − s)r − ⟨v, z⟩ = r (max{⟨v, w⟩; w ∈ Σ} − ⟨v, 1r z⟩) ≥ 0 , hence λu ≥ x. Since K = C󸀠 , by Lemma 2.6.8 we obtain K 󸀠 = C󸀠󸀠 = C. Finally, for every (r, z) ∈ n+1 ℝ we have (r, z) ∈ Σ if and only if (r, z) ∈ C and ⟨(1, 0), (r, z)⟩ = 1.

108 | 2 Embeddings, covers, and completions

We use Proposition 2.6.9 to construct an example of an order unit space where the set Λ of extreme points of Σ is not closed. Example 2.6.10. In the space ℝ3 with Euclidean norm let S be the convex combination of the closed unit ball with the line segment between the points (1, 0, 1)T and (−1, 0, 1)T . Let the sets C, K and Σ be as in Proposition 2.6.9, i.e., they are subsets of ℝ4 . For the set Λ of extreme points of Σ, we have (1, 0, 0, 1)T ∈ ̸ Λ, but {(1, 0, x2 , x3 )T ; x22 + x23 = 1} \ {(1, 0, 0, 1)T } ⊆ Λ. Hence, Λ is not closed.

Notes and Remarks In view of Theorem 2.6.2 it is natural to ask for which Archimedean partially ordered vector spaces (X, K) with an order unit u the vector lattice generated by Φ[X] in C(Λ) equals C(Λ). Besides polyhedral cones there exist infinite-dimensional examples for which this is satisfied; see, e.g., [149, Example 2.2]. So far, however, no general characterization of such partially ordered vector spaces exists.

2.7 The Fremlin tensor product as Riesz completion The aim in the present section is to describe the classical Fremlin tensor product as a Riesz completion of the vector space tensor product with a suitable cone. Fremlin’s tensor product is motivated by tensor products of functions spaces. These tensor products often have a natural lattice order, just as their components. For instance, if X = Y = C[0, 1], then the vector space tensor product T = X ⊗ Y can be viewed as a subspace of C([0, 1] × [0, 1]), and the pointwise order induces a natural order on T. However, the canonical order given by the cone K T generated by the elementary tensors of positive elements (the so-called projective cone; see (2.36) below) does not coincide with the pointwise order. In a seminal paper [54], Fremlin gives a construction of a tensor product of Archimedean vector lattices. This construction reproduces the pointwise order in the tensor product of function spaces. It is quite involved and relies on representation theory for Archimedean vector lattices with order units. One obtains an ambient vector lattice of the vector space tensor product, which yields the appropriate order. The projective cone is in general non-Archimedean and smaller than the cone of Fremlin’s order.

2.7.1 The vector space tensor product For (real) vector spaces X and Y, we recall basic definitions and well-known facts concerning the vector space tensor product X ⊗ Y; see also [33, Section 3.1 in Chapter 2].

2.7 The Fremlin tensor product as Riesz completion | 109

Definition 2.7.1. Let X and Y be (real) vector spaces. A pair (T, τ) is called a (vector space) tensor product of X and Y provided (i) T is a vector space and τ : X × Y → T is a bilinear map; and (ii) if S is a vector space and σ : X × Y → S is a bilinear map, then there is a unique linear map σ ∗ : T → S such that σ(x, y) = σ ∗ (τ(x, y)) for all x ∈ X and y ∈ Y. Remark 2.7.2. If (T, τ) is a tensor product of X and Y, then {τ(x, y); x ∈ X, y ∈ Y} spans T. Indeed, for S = span{τ(x, y); x ∈ X, y ∈ Y} and σ = τ there is a linear map σ ∗ : T → S such that τ(x, y) = σ ∗ (τ(x, y)) for every x ∈ X and y ∈ Y. Suppose that T \ S ≠ ⌀. By choosing first a Hamel basis of S and then extending it to a basis of T, one can construct a linear map ϕ : T → ℝ with ϕ ≠ 0 and ϕ = 0 on S. Choose s ∈ S with s ≠ 0 and define σ ∗2 : T → S by σ ∗2 (t) = σ ∗ (t) + ϕ(t)s. Then σ ∗2 = σ ∗ on S, in particular τ(x, y) = σ ∗2 (τ(x, y)) for every x ∈ X and y ∈ Y, and for u ∈ T with ϕ(u) ≠ 0 one has σ ∗2 (u) = σ ∗ (u) + ϕ(u)s ≠ σ ∗ (u), which contradicts the uniqueness of σ ∗ . Theorem 2.7.3. Let X and Y be vector spaces. (i) There exists a tensor product (T, τ) of X and Y. (ii) If (T,̂ τ)̂ is another tensor product of X and Y, then there is a linear bijection τ ∗ : T̂ → T ̂ y)) for every x ∈ X, y ∈ Y. such that τ(x, y) = τ∗ (τ(x, Due to Theorem 2.7.3 there exists an essentially unique tensor product (T, τ) of X and Y. We denote it as usual by X ⊗ Y, and for x ∈ X and y ∈ Y we use the notation x ⊗ y := τ(x, y). The next lemma will be needed in the sequel to relate the tensor product of preRiesz spaces and the tensor product of their Riesz completions. Lemma 2.7.4. Let X, Y, U, and V be vector spaces and let ρ X : X → U and ρ Y : Y → V be linear injections. Let ρ(x, y) := ρ X (x) ⊗ ρ Y (y) ,

x ∈ X, y ∈ Y .

Then the unique linear map ρ ∗ : X ⊗ Y → U ⊗ V satisfying ρ ∗ (x ⊗ y) = ρ(x, y) for every x ∈ X and y ∈ Y is injective. Proof. Let w, z ∈ X ⊗ Y be such that ρ ∗ (w) = ρ ∗ (z). Due to Remark 2.7.2, there are x i ∈ X, y i ∈ Y, and α i ∈ ℝ, i ∈ {1, . . . , n} such that n

z = ∑ αi xi ⊗ yi . i=1

Choose a Hamel basis (e i )i∈I of ρ X (X) and extend it to a Hamel basis (e i )i∈I 󸀠 of U and choose a Hamel basis (f j )j∈J of ρ Y (Y) and extend it to a Hamel basis (f j )j∈J 󸀠 of V. Define

110 | 2 Embeddings, covers, and completions η : U × V → X ⊗ Y by −1 η( ∑ λ i e i , ∑ μ j f j ) := ∑ ∑ λ i μ j ρ −1 X (e i ) ⊗ ρ Y (f j ) . i∈I 󸀠

j∈J 󸀠

i∈I j∈J

Then η is well-defined and bilinear and η(ρ X (x), ρ Y (y)) = x ⊗ y for every x ∈ X and y ∈ Y. Hence, there is a unique linear map η ∗ : U ⊗ V → X ⊗ Y such that η ∗ (u ⊗ v) = η(u, v) for every u ∈ U, v ∈ V. In particular, η∗ (ρ X (x) ⊗ ρ Y (y)) = x ⊗ y for every x ∈ X and y ∈ Y. We obtain that n

n

i=1 n

i=1

η ∗ (ρ ∗ (z)) = ∑ α i η∗ (ρ ∗ (x i ⊗ y i )) = ∑ α i η∗ (ρ(x i , y i )) n

= ∑ α i η∗ (ρ X (x i ) ⊗ ρ Y (y i )) = ∑ α i x i ⊗ y i = z . i=1

i=1

Similarly, η∗ (ρ ∗ (w)) = w. From ρ ∗ (z) = ρ ∗ (w) it follows that z = w.

2.7.2 The Fremlin tensor product We will first define the Fremlin tensor product and then establish it as the Riesz completion of the vector space tensor product. Let E, F, and H be vector lattices. A map ψ : E × F → H is called a Riesz bimorphism if for every e ∈ E the map from F to H given by f 󳨃→ ψ(e, f) is a Riesz homomorphism, and for every f ∈ F the map from E to H given by e 󳨃→ ψ(e, f) is a Riesz homomorphism as well. We recall Fremlin’s main result on the tensor product of Archimedean vector lattices in [54, 4.2]. Theorem 2.7.5. Let (E, E + ) and (F, F+ ) be Archimedean Riesz spaces. Then there is an Archimedean Riesz space G and a Riesz bimorphism ϕ : E × F → G such that (i) whenever H is an Archimedean Riesz space and ψ : E × F → H is a Riesz bimorphism, there is a unique Riesz homomorphism T : G → H such that Tϕ = ψ; (ii) ϕ induces an embedding ϕ̂ : E ⊗ F → G; ̂ ⊗ F] is dense in G in the sense that for every w ∈ G there exist x0 ∈ E and y0 ∈ F (iii) ϕ[E ̂ ⊗ F] such that |w − v| ≤ δ ϕ(x ̂ 0 ⊗ y0 ); such that for every δ > 0 there is a v ∈ ϕ[E + + ̂ (iv) if w > 0 in G, then there exist x ∈ E and y ∈ F such that 0 < ϕ(x ⊗ y) ≤ w. All such Archimedean Riesz spaces G are isomorphic as Riesz spaces. The essentially unique Archimedean Riesz space G in Theorem 2.7.5 is called the Frem̄ lin tensor product of E and F and is denoted by E⊗F. The next result states that the Fremlin tensor product is the Riesz completion of the vector space tensor product equipped with the order induced by the Fremlin tensor product. This fact is contained in [35, Theorem 4]. A proof based on the density property in Theorem 2.7.5 (iii) is given next.

2.7 The Fremlin tensor product as Riesz completion | 111

̄ the emTheorem 2.7.6. Let E and F be Archimedean Riesz spaces, ϕ̂ : E ⊗ F → E⊗F ̂ bedding map into the Fremlin tensor product, and K := {z ∈ E ⊗ F; ϕ(z) ≥ 0}. Then ̄ ϕ)̂ is its Riesz completion. Moreover, ϕ̂ is a (E ⊗ F, K) is a pre-Riesz space and (E⊗F, complete Riesz homomorphism. Proof. By definition of the order in E ⊗ F, the map ϕ̂ is bipositive. In order to show ̂ ⊗ F] is order dense in G = E⊗F. ̄ that E ⊗ F is a pre-Riesz space, we show that D := ϕ[E Let w ∈ G. According to Theorem 2.7.5 (iii) there exist x ∈ E and y ∈ F such that for ̂ ⊗ y). Since every δ > 0 there exists a v ∈ D such that |w − v| ≤ δz, where z = ϕ(x z ∈ D, one has v + δz ∈ D and w ≤ v + δz. Also v − w ≤ δz, consequently v ≤ w + δz. Therefore, if a is a lower bound of {u ∈ D; u ≥ w}, then a ≤ v + δz ≤ w + δz + δz. As G is Archimedean, it follows that a ≤ w. Hence, w = inf{u ∈ D; u ≥ w} in G. Thus, D is order dense in G. Consequently, D is a pre-Riesz space. Let G0 be the Riesz subspace of G generated by D. The embedding map ϕ̂ : E ⊗ F → G0 is bipositive and ̂ is the Riesz its range is order dense and generates G0 as a Riesz space, hence (G0 , ϕ) ̂ completion of E ⊗ F. Due to Proposition 2.3.27 the embedding ϕ is a complete Riesz homomorphism. Finally we show that G0 and G are isomorphic Riesz spaces. We verify that G0 satisfies the conditions of Theorem 2.7.5 just as G does, so that the essential uniqueness yields that G0 and G are isomorphic Riesz spaces. Since ϕ : E × F → G is a Riesz bimorphism and ϕ[E × F] is contained in G0 , the map ϕ is a Riesz bimorphism from E × F into G0 . If H is an Archimedean Riesz space and ψ : E × F → H is a Riesz bimorphism, then there exists a Riesz homomorphism T : G → H such that T ∘ ϕ = ψ. Then the restriction T0 of T to G0 is a Riesz homomorphism from G0 to H and T0 ∘ ϕ = ψ. If T1 is another Riesz homomorphism from G0 into H with T1 ∘ ϕ = ψ, then T1 = T0 , as G0 is the Riesz subspace generated by ϕ[E × F]. Hence, (ii) of Theorem 2.7.5 is satisfied by G0 . Clearly also (iii) and (iv) are satisfied by G0 . Thus G and G0 are isomorphic Riesz spaces and therefore G is the Riesz completion of D.

2.7.3 The projective cone and its relatively uniform closure An intrinsic construction of the Fremlin tensor product is given by Grobler and Labuschagne in [63]. They consider two directed Archimedean partially ordered vector spaces X and Y with the Riesz decomposition property and show that the relatively uniform closure of the projective cone already yields an order on the tensor product X ⊗ Y with a suitable universal property. If X and Y are Archimedean Riesz spaces, their order on X ⊗ Y coincides with the order induced from the Fremlin tensor product of X and Y. We present a construction without assuming the Riesz decomposition property. The results in this section are from [85]. In the proofs we use the Fremlin tensor product of the Riesz completions of X and Y.

112 | 2 Embeddings, covers, and completions Let (X, K X ) and (Y, K Y ) be directed Archimedean partially ordered vector spaces, let T = X ⊗ Y be their vector space tensor product, and define n

K T := { ∑ x i ⊗ y i ; n ∈ ℕ, i ∈ {1, . . . , n}, x i ∈ K X , y i ∈ K Y } .

(2.36)

i=1

It is straightforward to show that K T is a cone in T, called the projective cone, and that (T, K T ) is a directed ordered vector space; see [63] and [150, Proposition 3.4]. Next we define the Archimedean tensor cone as in [63, Section 2], where we use the notation of Definition 2.7.1. Definition 2.7.7. A cone K in X ⊗ Y is called an Archimedean tensor cone if K T ⊆ K and the following universal mapping property is satisfied. For every directed Archimedean partially ordered vector space (S, K S ) and every positive bilinear map σ: X × Y → S the induced linear map σ ∗ : (X ⊗ Y, K) → (S, K S ) is positive. If an Archimedean tensor cone in X ⊗ Y exists, then it is unique [63, Section 2]. We will show that the Archimedean tensor cone can be obtained as the relatively uniform closure of K T . As a technical tool, we consider the Riesz completions (X ρ , ρ X ) and (Y ρ , ρ Y ) of X and Y, respectively. Due to Proposition 2.4.12, the spaces X ρ and Y ρ are Archimē ρ be the Fremlin tensor product. Then there exists a Riesz dean Riesz spaces. Let X ρ ⊗Y ̄ ρ with the properties as in Theorem 2.7.5. In particbimorphism ϕ F : X ρ × Y ρ → X ρ ⊗Y ̄ ρ such that ular, there is an embedding h F : X ρ ⊗ Y ρ → X ρ ⊗Y h F (u ⊗ v) = ϕ F (u, v) for every u ∈ X ρ , v ∈ Y ρ . Define ρ : X × Y → X ρ ⊗ Y ρ , (x, y) 󳨃→ ρ(x, y) := ρ X (x) ⊗ ρ Y (y) . Then ρ is bilinear, hence a unique linear map ρ ∗ : X ⊗ Y → X ρ ⊗ Y ρ is induced such that ρ(x, y) = ρ ∗ (x ⊗ y) for every x ∈ X, y ∈ Y . ̄ ρ by the By Lemma 2.7.4, the map ρ ∗ is injective. Hence, X ⊗ Y is embedded into X ρ ⊗Y ∗ ρ ̄ injective linear map h F ∘ ρ . The order in the Fremlin tensor product X ⊗Y ρ induces an order in X ⊗ Y by means of the cone ̄ ρ )+ } . K F := {w ∈ X ⊗ Y; h F (ρ ∗ (w)) ∈ (X ρ ⊗Y Lemma 2.7.8. K F is a cone in X ⊗ Y, K T ⊆ K F , and (X ⊗ Y, K F ) is Archimedean. Proof. Since h F and ρ ∗ are linear and injective, K F is a cone. For x ∈ K X and y ∈ K Y , one has ̄ ρ )+ , h F (ρ ∗ (x ⊗ y)) = h F (ρ(x, y)) = h F (ρ X (x) ⊗ ρ Y (y)) = ϕ F (ρ X (x), ρ Y (y)) ∈ (X ρ ⊗Y

2.7 The Fremlin tensor product as Riesz completion | 113

as ϕ F is a Riesz bimorphism and therefore positive. Hence, x ⊗ y ∈ K F , which implies that K T ⊆ K F . If w1 , w2 ∈ X ⊗ Y are such that w1 − nw2 ∈ K F for every n ∈ ℕ, then ̄ ρ )+ h F (ρ ∗ (w1 )) − nh F (ρ ∗ (w2 )) ∈ (X ρ ⊗Y ̄ ρ )+ , as X ρ ⊗Y ̄ ρ is Archimedean. One obfor every n. Therefore, −h F (ρ ∗ (w2 )) ∈ (X ρ ⊗Y tains −w2 ∈ K F . Hence, (X ⊗ Y, K F ) is Archimedean. Due to Lemma 2.7.8, the cone K T is contained in an Archimedean cone. Let K be the intersection of all Archimedean cones in X ⊗ Y that contain K T . We will show that K is the Archimedean tensor cone in X ⊗ Y. For the alternative construction of the Archimedean tensor cone as the relatively uniform closure of K T we need the following preliminary result. Lemma 2.7.9. K F is relatively uniformly closed in (X ⊗ Y, K T ). Proof. Let (w α )α∈A be a net in K F and let w ∈ X ⊗ Y be such that w α converges relatively uniformly to w. Then there are an α 0 ∈ A, an a ∈ K T and a decreasing net (λ α )A ≥α0 in ℝ+ such that inf{λ α ; α ∈ A≥α0 } = 0, and with respect to K T -order one has for every α ∈ A≥α0 that −λ α a ≤ w α − w ≤ λ α a. Let w̃ := h F (ρ ∗ (w)), w̃ α := h F (ρ ∗ (w α )), α ∈ A≥α0 , and ã := h F (ρ ∗ (a)). Since h F ∘ ρ ∗ is positive, one obtains −λ α ã ≤ w̃ α − w̃ ≤ λ α ã with ̄ ρ )+ -order for every α ∈ A≥α0 . Then w̃ ≥ w̃ α − λ α ã ≥ −λ α a.̃ As X ρ ⊗Y ̄ ρ respect to (X ρ ⊗Y ρ ρ + ̄ ) , therefore w ∈ K F . is Archimedean, one gets w̃ ∈ (X ⊗Y Denote by K T the intersection of all relatively uniformly closed wedges in (X ⊗ Y, K T ) that contain K T . It is clear that K T is relatively uniformly closed. The Archimedean tensor cone in X ⊗ Y can now be characterized as follows; see also [85, Theorem 4.4]. Theorem 2.7.10. For a cone K in X ⊗ Y the following four statements are equivalent. (i) K is the Archimedean tensor cone. (ii) Let (S, K S ) be a directed Archimedean ordered vector space and let ϕ : X ⊗ Y → S be a linear map such that ϕ(w) ∈ K S for all w ∈ K T . Then ϕ(w) ∈ K S for all w ∈ K. (iii) K is the smallest Archimedean cone in X ⊗ Y with K T ⊆ K. (iv) K = K T . Proof. (i) ⇒ (iii): Let L be an Archimedean cone in X⊗Y with K T ⊆ L. With the notation as in Definition 2.7.7, let S := X ⊗ Y, K S := L, and σ : X × Y → S, (x, y) 󳨃→ σ(x, y) := x ⊗ y. Since σ is positive bilinear and K is the Archimedean tensor cone, it follows that σ ∗ : (X ⊗ Y, K) → (X ⊗ Y, L) is positive. As σ ∗ is the identity map, one obtains that K ⊆ L. (iii) ⇒ (iv): We first show that K is relatively uniformly closed. Let (w α )α∈A be a net in K and w ∈ X ⊗ Y such that (w α ) converges relatively uniformly to w. Then there are an α0 ∈ A, an a ∈ K T and a decreasing net (λ α )α∈A ≥α0 in ℝ+ such that inf{λ α ; α ∈ A≥α0 } = 0, and in K T -order one has for every α ∈ A≥α0 that −λ α a ≤ w α − w ≤ λ α a .

114 | 2 Embeddings, covers, and completions Then w + λ α a − w α ∈ K T ⊆ K, hence w + λ α a ∈ K for every α ∈ A≥α0 . As K is Archimedean, one has w ∈ K. Hence, K is relatively uniformly closed, and therefore K T ⊆ K. Since K T ⊆ K F and K F is relatively uniformly closed according to Lemma 2.7.9, we have K T ⊆ K F . As K F is a cone by Lemma 2.7.8 and K T is a wedge, it follows that K T is a cone. Next we show that K T is Archimedean. Let w1 , w2 ∈ X ⊗ Y be such that w2 − nw1 ∈ K T for every n ∈ ℕ. Since K T is directed, there is a w3 ∈ K T with w3 − w2 ∈ K T and w3 + w2 ∈ K T . Then in K T -order one has for every n that − 1n w3 ≤ 1n w2 = 1n w2 − w1 − (−w1 ) = 1n w2 ≤ 1n w3 , consequently the sequence ( 1n w2 − w1 )n∈ℕ converges relatively uniformly to −w1 . As K T is relatively uniformly closed, we get −w1 ∈ K T , i.e., K T is Archimedean. Thus, K T ⊇ K. (iv) ⇒ (ii): Let (S, K S ) be a directed Archimedean partially ordered vector space and let ϕ : X ⊗ Y → S be linear and such that ϕ(w) ∈ K S for all w ∈ K T . Let C := {w ∈ X ⊗ Y; ϕ(w) ∈ K S } . Then C ⊇ K T and C is a wedge. We show that C is relatively uniformly closed. Let (w α )α∈A be a net in C and w ∈ X ⊗ Y such that (w α ) converges relatively uniformly to w. Then there are an α 0 ∈ A, an a ∈ K T and a decreasing net (λ α )α∈A ≥α0 in ℝ+ such that inf{λ α ; α ∈ A≥α0 } = 0, and in K T -order one has for every α ∈ A≥α0 that −λ α a ≤ w α − w ≤ λ α a. Then for every α ∈ A≥α0 we obtain in K S -order −λ α ϕ(a) ≤ ϕ(w α − w) ≤ λ α ϕ(a) , therefore ϕ(w) ≥ ϕ(w α ) − λ α ϕ(a) ≥ −λ α ϕ(a). As S is Archimedean, we get ϕ(w) ∈ K S , consequently w ∈ C. Hence, C is relatively uniformly closed, which implies K = K T ⊆ C. (ii) ⇒ (i): Let (S, K S ) be a directed Archimedean partially ordered vector space and let σ : X×Y → S be a positive bilinear map. Then the induced linear map σ ∗ : X⊗Y → S is such that σ ∗ (w) ∈ K S for every w ∈ K T . Hence, σ ∗ (w) ∈ K S for every w ∈ K, which means that σ ∗ : (X ⊗ Y, K) → (S, K S ) is positive. Consequently, K is the Archimedean tensor cone. Remark 2.7.11. It follows from Theorem 2.7.10 that the Archimedean tensor cone K in X ⊗ Y is contained in the cone K F induced by the order of the Fremlin tensor product of the Riesz completions of X and Y. It remains open whether K and K F coincide. For further investigations of tensor products of partially ordered vector spaces see the references in [63] and [85]. Tensor products of order unit spaces are considered in [65].

2.8 Extension and restriction method |

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2.8 Extension and restriction method The main method used throughout this book to investigate partially ordered vector spaces is to embed them in suitable Riesz spaces and then use Riesz space theory. It turns out to be essential that the embedding is order dense. In this case, the lattice structure of the ambient Riesz space is uniquely determined by the partially ordered vector space. Due to Theorem 2.4.5, the pre-Riesz spaces are our scope, as those are precisely the order dense subspaces of Riesz spaces. In many cases we can use the embedding into any vector lattice cover. Sometimes it is appropriate to choose a specific one, for instance the Riesz completion or the Dedekind completion. In examples where the space contains an order unit, the functional representation is often a convenient vector lattice cover, due to its pointwise order. Let us describe abstractly how notions from Riesz space theory can be generalized to pre-Riesz spaces. One approach is to reformulate the notion in terms of relations of sets of upper and lower bounds instead of using the lattice operations. For instance, the modulus |x| of an element x in the formulation of a definition can be replaced by the set {−x, x}u . Thus a relation like |x| = |y| would become {−x, x}u = {−y, y}u . Another approach is to embed the pre-Riesz space into a vector lattice and use the vector lattice operations of that ambient vector lattice to formulate the definitions. The two approaches raise the question of compatibility. Under which conditions do they lead to the same definition? Also, if the pre-Riesz space is itself a vector lattice, the definitions should coincide with the usual ones. The compatibility question concerning the two approaches can be described explicitly as follows, if the notion concerns a specific class of subsets of the space. Let (X, K) be a pre-Riesz space and (Y, i) a vector lattice cover according to Theorem 2.4.5. We consider a class M of subsets of Y and define the corresponding class L of subsets of X by means of sets of upper bounds instead of lattice operations. The classes should match under the embedding i in two directions. The pair (L, M) ⊆ P(X) × P(Y) is said to satisfy – the restriction property (R), if whenever J ∈ M, then [J]i ∈ L, and – the extension property (E), if whenever I ∈ L, then there is J ∈ M such that I = [J]i. In Chapters 3 and 4 we will investigate those properties for classes of subsets such as solid sets, ideals, and bands. In the same spirit, we study generalizations of other notions from Riesz space theory, for instance disjointness and Riesz norms. Order denseness is not always sufficient for the extension and restriction property. A useful additional property introduced in [149, Definition 2.3] is contained in the next definition. Definition 2.8.1. A linear subspace D of a partially ordered vector space (Z, K) is called pervasive in Z if for every z ∈ K \ {0} there exists an d ∈ (K ∩ D) \ {0} such that d ≤ z.

116 | 2 Embeddings, covers, and completions A pre-Riesz space X is called pervasive if i[X] is pervasive in X ρ , where (X ρ , i) is the Riesz completion of X. The notion of pervasiveness is a modification of the familiar denseness notion known for Riesz subspaces; see, e.g., [12, p.29]. Typical examples of pervasive pre-Riesz spaces are given next. Example 2.8.2. Consider the space X = Ck [0, 1] of k times continuously differentiable functions on [0, 1], equipped with the pointwise order. Similarly to Example 2.4.8, X is order dense in C[0, 1], and the Riesz completion of X is given by X ρ = {y ∈ C[0, 1]; y is a piecewise k times continuously differentiable function} . It is straightforward that X is pervasive. Moreover, the space C∞ [0, 1] of smooth functions is pervasive as well. Observe that a pervasive subspace need not be majorizing, in general. For an example, take the Riesz subspace of finite sequences in the Riesz space of bounded sequences. Next we discuss the relation between order denseness and pervasiveness. Proposition 2.8.3. If Z is a Riesz space and D is an order dense Riesz subspace of Z, then D is pervasive in Z. Proof. Let y ∈ Z, y > 0. Then −y = inf{x ∈ D; x ≥ −y}, hence there exists x ∈ D with x ≥ −y and x ≱ 0, which implies x− ≠ 0. Then 0 ∧ x ≥ 0 ∧ (−y) = −y, and since x− = −(x ∧ 0) one obtains −x− ≥ −y and therefore 0 < x− ≤ y. Consequently D is pervasive in Z. For a converse statement, we need Archimedean spaces. Lemma 2.8.4. Let (Z, K) be an Archimedean Riesz space and D a pervasive subspace of Z. For every z ∈ K one has z = sup{d ∈ D; 0 ≤ d ≤ z}. Proof. Suppose there is z ∈ K such that z ≠ sup{d ∈ D; 0 ≤ d ≤ z}, i.e., there is w ∈ {d ∈ D; 0 ≤ d ≤ z}u such that w ≱ z. Put v := w ∧ z, then v ∈ {d ∈ D; 0 ≤ d ≤ z}u and v < z. As D is pervasive in Z, there is x ∈ D such that 0 < x ≤ z − v. Since x ≤ z, we obtain x ≤ w, hence x ≤ v. Hence, 2x ≤ (z − v) + v = z. Now 2x ∈ {d ∈ D; 0 ≤ d ≤ z}, so 2x ≤ v. Inductively, we find nx ≤ z for every n ∈ ℕ, which contradicts that Z is Archimedean. Proposition 2.8.5. If Z is an Archimedean Riesz space and D is a majorizing pervasive subspace of Z, then D is order dense in Z. Proof. Let z ∈ Z and set A := {d ∈ D; d ≥ z}. As D is majorizing, A is nonempty. Let x ∈ A. Clearly, z is a lower bound of A. By Lemma 2.8.4 we have x − z = sup{d ∈ D; 0 ≤ d ≤ x − z}. Then z = inf{x − d ∈ D; 0 ≤ d ≤ x − z} . (2.37)

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Let v be a lower bound of A. For every d ∈ D with 0 ≤ d ≤ x − z we have x − d ∈ A and hence v ≤ x − d. By (2.37) we obtain v ≤ z. Hence, inf A exists and equals z. Thus D is order dense in Z. Propositions 2.8.3 and 2.8.5 yield the following link between order denseness and pervasiveness⁷. Corollary 2.8.6. Let Z be an Archimedean Riesz space and let D be a Riesz subspace. Then D is order dense in Z if and only if D is majorizing and pervasive in Z. Observe that Proposition 2.8.3 is not true, in general, if D is not a Riesz subspace. Indeed, an example of a majorizing order dense subspace D of an Archimedean Riesz space Z such that D is not pervasive in Z is given in Example 4.1.19 below. The following example shows that Lemma 2.8.4 and Proposition 2.8.5 are not valid, in general, if Z is not Archimedean. Example 2.8.7. Let Z = ℝ3 be ordered by the lexicographic cone K = {(x1 , x2 , x3 )T ; x1 > 0, or x1 = 0 and x2 > 0, or x1 = x2 = 0 and x3 ≥ 0} . Recall that (Z, K) is a Riesz space that is not Archimedean. The Riesz subspace D = {(x1 , 0, x3 )T ; x1 , x3 ∈ ℝ} has the following properties: (i) D is majorizing in Z. Indeed, for (x1 , x2 , x3 )T ∈ Z we have (x1 , x2 , x3 )T ≤ (x1 + 1, 0, 0)T ∈ D. (ii) D is pervasive in Z. Indeed, let x := (x1 , x2 , x3 )T > 0. If x1 > 0 or x2 > 0, then 0 < (0, 0, 1)T < x. Otherwise x ∈ D. (iii) There exists z ∈ Z such that the supremum of the set A := {d ∈ D; 0 ≤ d ≤ z} does not exist. Indeed, take z = (0, 1, 0)T , then A = {(0, 0, x3 )T ; x3 ∈ ℝ+ }. For every ε > 0 the element u ε := (0, ε, 0)T is an upper bound of A, but there is no upper bound u of A such that u ≤ u ε for every ε > 0. (iv) D is not order dense in Z. Indeed take z = (0, 1, 0)T . With a similar argumentation as in (iii), the set {d ∈ D; d ≤ z} = {(0, 0, x3 )T ; x3 ∈ ℝ} does not have a supremum. We conclude the section by observing that a pervasive pre-Riesz space is pervasive in each of its vector lattice covers. Proposition 2.8.8. Let X be a pre-Riesz space and (Y, i) a vector lattice cover of X. If X is pervasive, then i[X] is pervasive in Y.

7 In vector lattice theory, the property that we call pervasiveness is usually referred to as order denseness, which is in general not the same as order denseness as defined in this book. Corollary 2.8.6 shows that for majorizing Riesz subspaces both terminologies are compatible.

118 | 2 Embeddings, covers, and completions Proof. First observe that the Riesz completion X ρ of X can be considered as an order dense Riesz subspace of Y. Due to Proposition 2.8.3, the space X ρ is pervasive in Y, i.e., for every y ∈ Y with 0 < y there is v ∈ X ρ such that 0 < v ≤ y. Since X is pervasive, there is x ∈ X such that 0 < i(x) ≤ v ≤ y. A further discussion on pervasive pre-Riesz spaces and a method to show intrinsically that a pre-Riesz space is not pervasive is given in Lemma 5.1.7 and beyond.

3 Seminorms on pre-Riesz spaces Many partially ordered vector spaces are endowed with a natural norm, and the norm and order are often related; see the results in Section 1.5. The present chapter presents a detailed investigation of norms and seminorms on pre-Riesz spaces. In particular, we study seminorms that can be extended to Riesz seminorms on vector lattice covers. A special case are the regular seminorms. Also we consider monotone seminorms and their extensions to vector lattice covers. Topologies generated by collections of the mentioned seminorms are discussed as well.

3.1 Basic properties of seminorms In this chapter, X is a real vector space. Definition 3.1.1. A map p : X → ℝ is called a seminorm on X if for every x, y ∈ X and λ ∈ ℝ we have p(x + y) ≤ p(x) + p(y) and p(λx) = |λ|p(x). For a seminorm p, for every x ∈ X it holds that p(x) = p(−x) and p(x) = 12 p(x) + 12 p(−x) ≥ 12 p(x − x) = 0 . Moreover, a map p : X → ℝ is a seminorm if and only if p is subadditive, positively homogeneous and for every x ∈ X it holds that p(x) = p(−x). Recall that a norm p is a seminorm with the additional property that p(x) = 0 implies x = 0. Every linear map f : X → ℝ is subadditive and positively homogeneous, and the map x 󳨃→ |f(x)| is a seminorm. By the unit ball of a seminorm p on X we mean the set {x ∈ X; p(x) ≤ 1}. We call {x ∈ X; p(x) < 1} the open unit ball. A subset M ⊆ X is called circled if for every m ∈ M and λ ∈ [−1, 1] we have λm ∈ M. Note that a convex set M is circled if and only if −M ⊆ M. Indeed, if −M ⊆ M, then for every m ∈ M we have −m, m ∈ M, and then by convexity also 0 ∈ M. For t ∈ [0, 1] it then follows that tm = tm + (1 − t)0 ∈ M and, similarly, for t ∈ [−1, 0] we have tm = (−t)(−m)+ (1+ t)0 ∈ M. A subset M ⊆ X is called absorbing if for every x ∈ X there is m ∈ M and λ ∈ ℝ+ such that x = λm. If M ⊆ X is absorbing, then p : X → ℝ , x 󳨃→ inf{λ ∈ ℝ+ ; x ∈ λM} https://doi.org/10.1515/9783110476293-003

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is called the Minkowski functional of M. Proposition 3.1.2. Let M ⊆ X be circled and absorbing and let p be the Minkowski functional of M. Then {x ∈ X; p(x) < 1} ⊆ M ⊆ {x ∈ X; p(x) ≤ 1} . Proof. If x ∈ M, then clearly p(x) ≤ 1. Now let x ∈ X be such that p(x) < 1. Then there is λ ∈ ℝ+ with λ < 1 such that x ∈ λM. Then x ∈ M as M is circled. Proposition 3.1.3. (i) For every seminorm p on X the unit ball of p is a convex circled absorbing set. (ii) If M ⊆ X is a convex circled absorbing set, then the Minkowski functional p of B := ⋂ α>1 αM is a seminorm on X, and B is the unit ball of p. Moreover, p is the Minkowski functional of M. Proof. We only prove (ii). Since M is absorbing, M is nonempty, and as it is circled, we have 0 ∈ M. Then it follows that B ⊇ M is absorbing. Let p be the Minkowski functional of B. First we show that p is a seminorm. Indeed, let x, y ∈ X. For λ > p(x) and μ > p(y) we have x ∈ λB and y ∈ λB, thus there exist a, b ∈ B with x = λa and y = μb. Then λ x + y = (λ + μ) ( λ+μ a+

μ λ+μ b)

∈ (λ + μ)B ,

since B is convex. Hence, p(x + y) ≤ λ + μ. Thus, p(x + y) ≤ p(x) + p(y). For λ, μ ∈ ℝ+ from x ∈ μB it follows that λx ∈ λμB, so p(λx) ≤ λp(x). If we replace λ by 1λ and then x by λx, we obtain p(λx) ≥ λp(x). As B is circled, we have p(−x) = p(x). It follows that p is a seminorm. Next we show that B is the unit ball of p. Indeed, if x ∈ X is such that p(x) ≤ 1, then for every α > 1 we have p(x) < α, so x ∈ αB. Hence, x ∈ ⋂ α>1 αB = B. Conversely, if x ∈ B, then for every α > 1 we have x ∈ αM ⊆ αB, so 1α x ∈ B. Hence, p( 1α x) ≤ 1 and therefore, p(x) ≤ α. Thus, p(x) ≤ 1. It follows that B is the unit ball of p. Finally we show that p is the Minkowski functional of M. Let q be the Minkowski functional of M. We show that p = q. Indeed, since M is circled, we have B ⊇ M, and therefore, p ≤ q. Let x ∈ X and let t > p(x). Then there is λ ∈ ℝ+ with λ < t and x ∈ λB. If λ = 0, then q(x) = 0 = p(x). If λ ≠ 0, then λt > 1, thus B ⊆ λt M. Then λB ⊆ tM, therefore x ∈ tM, and hence q(x) ≤ t. Consequently, q(x) ≤ p(x). On the set of all seminorms on X we will consider the pointwise order, where p ≤ q is defined by p(x) ≤ q(x) for every x ∈ X. Let p be a seminorm on X, let (x α )α∈I be a net in X and let x ∈ X. We say that the net (x α )α∈I p-converges to x if ∀ε > 0 ∃α0 ∈ I ∀α ≥ α 0 : p(x α − x) < ε , and that it is p-Cauchy if ∀ε > 0 ∃α 0 ∈ I ∀α, β ≥ α 0 : p(x α − x β ) < ε .

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We call a set M ⊆ X p-complete if every p-Cauchy sequence in M p-converges to a limit in M. We also use notions as p-convergent, p-open, p-closed, p-dense in the obvious meaning. Alternatively, we sometimes write convergent in (X, p), etc. We will skip the prefix p whenever the seminorm under consideration is clear from the context. For two seminorms p1 and p2 on X, we say that p1 and p2 are equivalent if there exist c, C ∈ (0, ∞) such that cp1 (x) ≤ p2 (x) ≤ Cp1 (x) for every x ∈ X. For equivalent seminorms p1 and p2 we have that in (X, p1 ) and (X, p2 ) convergence, Cauchy nets and completeness coincide. A map f : X → ℝ is p-continuous if for every net that converges to x ∈ X the image of the net under f converges to f(x). Equivalently, f is p-continuous if for every open set U ⊆ ℝ the set [U]f is p-open in X. If p and q are seminorms on X such that q ≤ p, then q is p-continuous. The dual of X with respect to the seminorm p (or, briefly, the norm dual of (X, p)) is defined by X 󸀠 := {f ∈ X ∗ ; f is p-continuous}. On X 󸀠 we define the dual of p by p󸀠 (f) := sup{|f(x)|; x ∈ X, p(x) ≤ 1} for f ∈ X 󸀠 . Note that p󸀠 is a norm on X 󸀠 (even though p is only a seminorm). Moreover, if X is a directed partially ordered vector space, then the order in X 󸀠 is given by the cone K 󸀠 := K ∗ ∩ X 󸀠 . One of the main concerns in the next sections will be extension and restriction of seminorms. If (X, K X ) and (Y, K Y ) are partially ordered vector spaces, i : X → Y is a bipositive linear map, p is a seminorm on X, and q is a seminorm on Y, then we say that q extends p if q ∘ i = p. Similarly, q ∘ i is called the restriction of q to X.

3.2 Pre-Riesz seminorms 3.2.1 Generalization of Riesz seminorms: plan of attack A major part of the theory of norms on Riesz spaces is devoted to Riesz norms. Recall that a (semi)norm ρ on a Riesz space X is called a Riesz (semi)norm if for every x, y ∈ X we have 0 ≤ x ≤ y 󳨐⇒ ρ(x) ≤ ρ(y) ρ(|x|) = ρ(x) .

(monotonicity) , and

(3.1) (3.2)

As the notion of Riesz norm involves absolute values, it has no direct generalization to partially ordered vector spaces. Monotonicity alone does make sense on partially ordered vector spaces, but is in general much weaker. In particular, results concerning dual spaces and operators rely heavily on condition (3.2). We will consider three ways to extend the notion of Riesz norm or Riesz seminorm to pre-Riesz spaces. It will turn out that they all lead to the same class of seminorms.

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The first approach is based on a formulation of the notion of ‘solid unit ball’ by means of upper and lower bounds without using lattice operations. A seminorm on a Riesz space is a Riesz seminorm if and only if its unit ball is solid; see Proposition 1.5.30. We will introduce the notion of a solvex set in an arbitrary partially ordered vector space, which in a Riesz space coincides with being solid and convex. The seminorms with solvex unit balls can serve as a suitable generalization of Riesz seminorms. The second approach considers extensions to Riesz seminorms on larger Riesz spaces. A Riesz seminorm ρ on a Riesz space X can be extended to a Riesz seminorm η on each Riesz space Y that contains X as a majorizing subspace, by Lemma 3.2.1 below. Hence, a seminorm on X is a Riesz seminorm if and only if it can be extended to a Riesz seminorm on every Riesz space that contains X as a majorizing subspace. The latter formulation also makes sense if X is only a partially ordered vector space and yields a second way to generalize the notion of Riesz seminorms. Lemma 3.2.1. Let Y be a Riesz space, let X be a majorizing Riesz subspace and let ρ be a Riesz seminorm on X. For the map η: Y → ℝ ,

y 󳨃→ inf{ρ(x); x ∈ X, x ≥ |y|}

the following statements hold. (i) η is a Riesz seminorm on Y that extends ρ. (ii) If X is order dense in Y and ρ is a norm, then η is a Riesz norm in Y. (iii) Let X be order dense in Y and let Y be Dedekind complete. If X is ρ-complete then Y is η-complete. Proof. (i) We begin by showing that η is a seminorm. Let y1 , y2 ∈ Y. If x1 , x2 ∈ X are such that x1 ≥ |y1 | and x2 ≥ |y2 |, then x1 + x2 ≥ |y1 + y2 |, so η(y1 + y2 ) ≤ ρ(x1 + x2 ) ≤ ρ(x1 ) + ρ(x2 ). By taking infimum over x1 and x2 we obtain that η(y1 + y2 ) ≤ η(y1 ) + η(y2 ). By definition, η(−y) = η(y) for every y ∈ Y. For y ∈ Y and λ ∈ ℝ+ we have for every x ∈ X with x ≥ |y| that λx ≥ |λy|, so η(λy) ≤ ρ(λx) = λρ(x) and therefore, η(λy) ≤ λη(y). If we first replace λ by 1λ and then y by λy, we also obtain η(λy) ≥ λη(y). Thus, η is a seminorm. Next we show that η is a Riesz seminorm. Let y1 , y2 ∈ Y be such that |y1 | ≤ |y2 |. For every x ∈ X with x ≥ |y2 | we have x ≥ |y1 |, so η(y1 ) ≤ ρ(x) and hence, η(y1 ) ≤ η(y2 ). In order to show that η extends ρ, let x ∈ X. We have η(x) ≤ ρ(|x|) = ρ(x), as ρ is a Riesz seminorm. For every v ∈ X with v ≥ |x| we have that ρ(x) = ρ(|x|) ≤ ρ(v). Hence, ρ(x) ≤ η(x). Thus, η extends ρ. (ii) Assume that X is order dense in Y and that ρ is a norm. We show that η is a norm. Let y ∈ Y \ {0}. Then, as X is an order dense Riesz subspace of Y, X is pervasive in Y; see Proposition 2.8.3. So there exists u ∈ X with 0 < u ≤ |y|. Then for every x ∈ X with x ≥ |y| we have x ≥ u, so ρ(x) ≥ ρ(u), hence, η(y) ≥ ρ(u) > 0. Thus, η is a norm on Y.

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(iii) The proof of this statement is as involved as the proof of the more general result in Theorem 3.4.8 below and therefore omitted here. The third approach concerns restriction of Riesz seminorms on Riesz completions. Trivially, a seminorm on a Riesz space X is a Riesz seminorm if it is the restriction of a Riesz seminorm on the Riesz completion of X. If we consider a partially ordered vector space X that is a pre-Riesz space, then we can consider those seminorms on X that are restrictions of Riesz seminorms on the Riesz completion of X as a suitable generalization of Riesz seminorms. Our three approaches lead to a class of seminorms on pre-Riesz spaces that we will call pre-Riesz seminorms. The material in this section can be found in [147].

3.2.2 Solid sets A seminorm on a Riesz space is a Riesz seminorm if and only if its unit ball is a solid set; see Proposition 1.5.30. Therefore, extending the notion of a Riesz seminorm to preRiesz spaces comes down to extending the notion of solid unit ball. Note that a unit ball is also convex. It turns out that a straightforward generalization of the notion of a solid set does not have the desired properties, and therefore we present the notion of a solvex set below, which jointly generalizes the notions of solid and convex. As solidness will nevertheless play an important role in the analysis, we investigate it first. Recall that a subset S of a Riesz space X is solid if |x| ≤ |y| with x ∈ X and y ∈ S implies that x ∈ S. The assertion |x| ≤ |y| means the same as {x, −x}u ⊇ {y, −y}u . The latter formulation also makes sense if X is a partially ordered vector space that is not a lattice. Definition 3.2.2. Let X be a partially ordered vector space. A set S ⊆ X is called solid if for every x ∈ X and y ∈ S such that {x, −x}u ⊇ {y, −y}u one has that x ∈ S. Remark that the meaning of {x, −x}u ⊇ {y, −y}u strongly depends on the space in which the upper bounds are taken. For instance, for x, y : [−1, 1] → ℝ given by x(t) = 1 and y(t) = t, t ∈ [−1, 1], the assertion {x, −x}u ⊇ {y, −y}u is true in the space of affine functions on [−1, 1] but fails in C[−1, 1]. If there is risk of confusion, the space will be explicitly specified. Lemma 3.2.3. Let X be a partially ordered vector space and S ⊆ X a solid set. Then S is circled. Proof. Let y ∈ S and λ ∈ [−1, 1]. First assume that λ ∈ [0, 1]. For u ∈ {y, −y}u we have u ≥ λu ≥ λy, and, similarly, u ≥ −λy. Hence, u ∈ {−λy, λy}u . As S is solid, it follows that λy ∈ S. It is clear that then also −λy ∈ S.

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The intersection of a collection of solid sets in a partially ordered vector space X is solid and the set X itself is solid. Therefore, every subset A ⊆ X is contained in a smallest solid set S in X, which is the intersection of all solid sets containing A. The set S is called the solid hull of A. Alternatively, we can construct the solid hull of A explicitly from A. Lemma 3.2.4. Let X be a partially ordered vector space and let A ⊆ X. The solid hull of A equals {x ∈ X; ∃a ∈ A : {x, −x}u ⊇ {a, −a}u } . (3.3) Proof. Denote by S the set in formula (3.3). Clearly, if a ∈ A, then {a, −a}u ⊇ {a, −a}u , so a ∈ S, hence A ⊆ S. Next, we show that S is solid. Let x ∈ X and y ∈ S be such that {x, −x}u ⊇ {y, −y}u . We wish to show that x ∈ S. There exists a ∈ A such that {y, −y}u ⊇ {a, −a}u . It follows that {x, −x}u ⊇ {a, −a}u and therefore, x ∈ S. Finally, we show that S is the smallest solid set containing A. Let T be a solid set containing A and let x ∈ S. Then there exists a ∈ A such that {x, −x}u ⊇ {a, −a}u . As a ∈ A ⊆ T, the solidness of T yields that x ∈ T. Hence, S ⊆ T. In a Riesz space, the intersection of a solid set and a Riesz subspace is solid in the subspace. It turns out that a similar result holds in a setting of partially ordered vector spaces for order dense subspaces. Both results can be combined by assuming that the inclusion map is a Riesz* homomorphism. Indeed, let Y be a partially ordered vector space and X a subspace of Y. Lemma 2.3.4 says that the inclusion map i : X → Y is a Riesz* homomorphism if Y is a Riesz space and X is a Riesz subspace of Y or if Y is a pre-Riesz space and X is an order dense subspace of Y. We will show below that the intersection of a solid set in Y and the subspace X is solid in X whenever the inclusion map i is a Riesz* homomorphism. We first give a direct argument for the case of order dense subspaces; see also [84, Proposition 5.3]. It has the advantage that it avoids the condition of Y being a pre-Riesz space. Proposition 3.2.5. Let Y be a directed partially ordered vector space, X an order dense subspace of Y, and S a solid subset of Y. Then the set S ∩ X is solid in X. Proof. Let x ∈ X and y ∈ S ∩ X be such that {x, −x}u ∩ X ⊇ {y, −y}u ∩ X. Since Y is directed, the set {y, −y}u is nonempty. If v ∈ Y is such that v ≥ y, v ≥ −y, then {u ∈ X; u ≥ v} ⊆ {y, −y}u ∩ X ⊆ {x, −x}u ∩ X , which implies, by order denseness of X in Y, that v = inf{u ∈ X; u ≥ v} ≥ x and, analogously, v ≥ −x. Hence, {x, −x}u ⊇ {y, −y}u . As S is solid in Y, it follows that x ∈ S, and therefore x ∈ S ∩ X. Hence, S ∩ X is solid in X. Next we consider the case where the inclusion map of X in Y is a Riesz* homomorphism. The key observation in the proof is the following lemma.

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Lemma 3.2.6. Let X and Y be directed partially ordered vector spaces and let h : X → Y be a Riesz* homomorphism. Let S and T be nonempty finite subsets of X. If Su ⊇ T u in X, then h[S]u ⊇ h[T]u in Y. Proof. According to Lemma 2.3.2 (ii), we have ul

ul

h[S]ul = h [Sul ] ⊆ h [T ul ] = h[T]ul , which yields the desired results. Proposition 3.2.7. Let X and Y be partially ordered vector spaces and let h : X → Y be a bipositive Riesz* homomorphism. If S is a solid set in Y, then [S]h is a solid set in X. Proof. Let a ∈ [S]h and let x ∈ X be such that {x, −x}u ⊇ {a, −a}u . Then {h(x), −h(x)}u ⊇ {h(a), −h(a)}u by Lemma 3.2.6. Since h(a) ∈ S and S is solid, it follows that h(x) ∈ S, so that x ∈ [S]h. Thus, [S]h is solid. In general, the intersection of a solid set and a linear subspace need not be solid in the subspace. Example 3.2.8. Let S be the closed unit ball of the 1-norm ‖⋅‖1 on C[0, 1] given by 1 ‖f‖1 = ∫0 |f(t)| dt, f ∈ C[0, 1]. Then S is solid, as it is the unit ball of a Riesz norm. However, the intersection of S with the subspace A of affine functions is not solid in A, since ‖⋅‖1 is not a Riesz norm on A. As a consequence of Proposition 3.2.5 we get the restriction property (R) as defined in Section 2.8 for solid sets. Proposition 3.2.9. Let X be a pre-Riesz space and let (Y, i) be a vector lattice cover of X. If J is a solid subset of Y, then [J]i is a solid subset of X. By means of the solid hull, a solid set in a subspace can be extended to a solid set in the space. Proposition 3.2.10. Let Y be a partially ordered vector space, X a subspace of Y, and A a solid subset of X. The solid hull S of A in Y satisfies S ∩ X = A. Proof. We prove S ∩ X ⊆ A. Let x ∈ S ∩ X. By Lemma 3.2.4, there exists a ∈ A such that {x, −x}u ⊇ {a, −a}u in Y. Then also {x, −x}u ⊇ {a, −a}u in X. As A is solid in X, it follows that x ∈ A. As a consequence of Proposition 3.2.10 we obtain the extension property (E) for solid sets as defined in Section 2.8. Proposition 3.2.11. Let X be a pre-Riesz space and let (Y, i) a vector lattice cover of X. If I is a solid subset of X, then the solid hull J of i[I] in Y is a solid set such that I = [J]i. Let us now consider seminorms with solid unit balls on partially ordered vector spaces.

126 | 3 Seminorms on pre-Riesz spaces

Proposition 3.2.12. Let X be a partially ordered vector space with a seminorm p. (i) p has a solid unit ball if and only if p(x) ≤ p(y) for every x, y ∈ X such that {x, −x}u ⊇ {y, −y}u . (ii) If p has a solid unit ball, then for every x, y ∈ X with −y ≤ x ≤ y one has p(x) ≤ p(y). Proof. Assertion (i) follows directly from the definition of solidness. For a proof of (ii), let x, y ∈ X be such that −y ≤ x ≤ y. Then {x, −x}u ⊇ {y, −y}u , so p(x) ≤ p(y) by (i). In a Riesz space, a Riesz seminorm on a majorizing Riesz subspace can be extended to the whole space; see Lemma 3.2.1. One could hope that also a seminorm with a solid unit ball on a majorizing linear subspace of a Riesz space can be extended to a Riesz seminorm on the Riesz space. In general, however, if the subspace is not a Riesz subspace, such an extension may fail, even if the subspace is order dense. Indeed, let us reformulate the question on extension of seminorms in terms of unit balls. Let X be an order dense subspace of a Riesz space Y and let p be a seminorm on X with a solid unit ball B. Suppose there exists a Riesz seminorm ρ on Y extending p. Then the unit ball C of ρ is convex and solid and C ∩ X = B. The convex hull W of the solid hull V of B in Y then satisfies W ⊇ B and W ⊆ C. Hence, W ∩ X = B. However, the subsequent Example 3.2.13 presents a situation where W ∩ X ⊋ B, so that a Riesz seminorm on Y extending p cannot exist. Note that the example also shows that the solid hull of a convex set that is solid in a subspace need not be convex. Example 3.2.13 is first given in [147, Example 2.8]. Example 3.2.13. We give an example of a Riesz space Y with an order dense linear subspace X that generates Y as a Riesz space and a seminorm with a solid unit ball on X that cannot be extended to a Riesz seminorm on Y. We consider the space X of affine functions from the unit circle S to ℝ as in Example 1.7.5. Let Y be the Riesz subspace of C(S) generated by X. As shown in Example 1.7.5, X is order dense in C(S), and hence in Y. Let D be the closed unit disk in ℝ2 and observe that every affine function on S extends uniquely to an affine function on D. The pointwise order on D corresponds to the order in X. Define p(x) := (∫ x(t)+ dt) ∨ (∫ x(t)− dt) D

D

for x ∈ X. Then p is a norm on X such that −y ≤ x ≤ y implies that p(x) ≤ p(y) for every x, y ∈ X. Denote for an angle φ ∈ (−π, π] the corresponding point in S by s φ , that is, s φ = (cos φ, sin φ)T . Let R φ : X → X be the operator of rotation over the angle φ ∈ ℝ: (R φ x) ((t1 , t2 )T ) := x ((t1 cos φ + t2 sin φ, t2 cos φ − t1 sin φ)T ) , (t1 , t2 )T ∈ S, x ∈ X. The example is completed by the three claims below. (a) The unit ball B of p is solid in X. For a proof, let x, y ∈ X be such that {x, −x}u ⊇ {y, −y}u . If y ≥ 0, then −y ≤ x ≤ y on D, hence p(x) ≤ p(y). Similarly, if y ≤ 0, then

3.2 Pre-Riesz seminorms | 127

y ≤ x ≤ −y, so p(x) ≤ p(y). If y is neither nonpositive nor nonnegative, then there are two points of S where y is zero. Then x is zero in those points, too. Indeed, by the pointwise approximation given in (1.24) the inclusion {x, −x}u ⊇ {y, −y}u implies that |x(t)| ≤ |y(t)| for every t ∈ S. Thus, x must be a scalar multiple of y with a scalar in [−1, 1]. This yields that p(x) ≤ p(y). Hence, B is solid. (b) The solid hull V of B given by V = {y ∈ Y; ∃x ∈ X : |y| ≤ |x|} is not convex. Indeed, let y((t1 , t2 )T ) := 32 t1 for (t1 , t2 )T ∈ S. With the rotations R φ defined above, we find that π 2

p(R φ y) = p(y) = ∫

T 3 + 2 t1 d(t 1 , t2 )

=

3 2

1

∫ ∫ r cos φ rdrdφ = 1 , − π2 0

D

so R φ y ∈ B for every φ ∈ ℝ, hence (R φ y)+ ∈ V for every φ. Choose an α ∈ (0, 15 ) and take x1 := y+ and x2 := (R π−α y)+ ; see Figure 3.1. To show that 12 (x1 + x2 ) ∈ ̸ V, let z ∈ X be such that |z| ≥ x1 + x2 in Y. Then |z| vanishes nowhere on the set {s φ ; φ ∈ [0, π − α]}, so z has the same sign in the end points s0 and s π−α , and there |z| ≥ 32 , because x1 (s0 ) = x2 (s π−α ) = 32 . The function z is affine, so |z| cannot be lower than 32 on both sides of the line through s0 and s π−α , hence p(z) ≥ 32 ( 2π − sin α) > 2. It follows that 12 (x1 + x2 ) ∈ ̸ V. This shows that V is not convex. x2 t2

x1 + x 2

x1

t1

Fig. 3.1: The functions x1 = y + , x2 = (R π−α y)+ , and x1 + x2 .

(c) The convex hull W of the solid hull V of B satisfies W ∩ X ⊋ B. Let y be as in (b) and let x1 := |y| and x2 := |R π2 y|. Then x1 , x2 ∈ V, so x := 12 (x1 + x2 ) ∈ W. On S, x is of the form s φ 󳨃→ 34 (| sin φ| + | cos φ|), so x ≥ 34 on S. Thus, 0 ≤ 34 𝟙 ≤ x, so 34 𝟙 ∈ W, because in a Riesz space the convex hull of a solid set is solid; see 3 Proposition 1.5.31. Furthermore, 34 𝟙 ∈ X, whereas p( 34 𝟙) = 3π 4 > 1, so 4 𝟙 ∈ ̸ B. In view of the discussion preceding this example, we have established that there is no Riesz seminorm on Y extending p. We proceed by showing that if there is an extending Riesz seminorm, then there is also a greatest one. Here we use pointwise order on the set of seminorms.

128 | 3 Seminorms on pre-Riesz spaces

Proposition 3.2.14. Let Y be a Riesz space with a majorizing linear subspace X and let p be a seminorm on X with unit ball B. Then the convex hull of the solid hull of B is the unit ball of the greatest seminorm q with a solid unit ball on Y such that q ≤ p on X. If there exists a seminorm with a solid unit ball on Y extending p, then q extends p as well. Proof. The set B is absorbing in X and X is majorizing in Y, so B is absorbing in Y. As the convex hull S of the solid hull of B contains B, the set S is also absorbing in Y. By Lemma 3.2.3, solid sets are circled, hence the solid hull of B is circled. It follows that S is circled. Clearly, S is convex. Since B = ⋂ α>1 αB, it follows S = ⋂α>1 αS. Hence, by Proposition 3.1.3, the set S is the unit ball of a seminorm q on Y. The set S is solid, due to Proposition 1.5.31. Moreover, S is the smallest convex solid set in Y containing B, so q is the greatest seminorm on Y with a solid unit ball that is less than or equal to p on X. If C is the solid unit ball of a seminorm on Y extending p, then C ∩ X = B, and C ⊇ S, so B ⊆ X ∩ S ⊆ X ∩ C = B. This yields that q extends p.

3.2.3 Solvex sets Rather than generalizing the notion of solid set on its own, we will consider seminorms with unit balls with a property that is a combined generalization of solidity and convexity. As solvex sets will also play an important role in Section 4.3 on ideals, we investigate more properties than strictly needed for the desired properties of pre-Riesz seminorms. Definition 3.2.15. Let X be a partially ordered vector space. A subset S of X is called solvex if for every x ∈ X, x1 , . . . , x n ∈ S, and λ1 , . . . , λ n ∈ (0, 1] with ∑nk=1 λ k = 1 such that u n

{x, −x}u ⊇ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}} k=1

one has that x ∈ S.

Observe that every solvex set is circled. Lemma 3.2.16. Let X be a partially ordered vector space. (i) Every solvex set in X is solid and convex. (ii) If X is a Riesz space, then a subset of X is solvex if and only if it is solid and convex. Proof. (i) Let S ⊆ X be solvex. It is clear that S is solid, by taking n = 1 in the definition of solvex sets. To prove that S is convex, let x1 , . . . , x n ∈ S and let λ1 , . . . , λ n ∈ (0, 1] with ∑nk=1 λ k = 1. Then n

n

u

n

u

{ ∑ λ k x k , − ∑ λ k x k } ⊇ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}} , k=1

so

∑nk=1

k=1

k=1

λ k x k ∈ S, by solvexity of S. Hence, S is convex.

3.2 Pre-Riesz seminorms

| 129

(ii) Assume that X is a Riesz space and let S be a solid and convex set in X. Let x ∈ X, x1 , . . . , x n ∈ S, and λ1 , . . . , λ n ∈ (0, 1] with ∑nk=1 λ k = 1 be such that u

n

{x, −x}u ⊇ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}} . k=1

As S is solid, |x k | ∈ S for every k and then ∑nk=1 λ k |x k | ∈ S by convexity of S. Because | ∑nk=1 ε k λ k x k | ≤ ∑nk=1 λ k |x k | for every ε1 , . . . , ε n ∈ {1, −1}, every upper bound of the set {∑nk=1 λ k |x k |, − ∑nk=1 λ k |x k |} is an upper bound of {∑nk=1 ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}} and therefore one of {x, −x}. As S is solid, it follows that x ∈ S. Thus, S is solvex. Example 3.2.17. We provide a solid, convex set that is not solvex. Consider the situation of Example 3.2.13. It has been proven that B = {x ∈ X; p(x) ≤ 1} is solid and convex in X. To show that B is not solvex, let y1 ((t1 , t2 )T ) := 32 t1 and y2 ((t1 , t2 )T ) := 32 t2 for (t1 , t2 )T ∈ ℝ2 . Then y1 , y2 ∈ B and u

u

{ 34 𝟙, − 34 𝟙} ⊇ { 12 (y1 + y2 ), 12 (y1 − y2 ), − 12 (y1 + y2 ), − 12 (y1 − y2 )} . Indeed, y1 + y2 ≥ 32 on the first quadrant, and y2 − y1 , y1 − y2 , −y1 − y2 ≥ 32 on the second, third, and fourth quadrant, respectively. However, 34 𝟙 ∈ ̸ B, so B is not solvex. It can be observed from the definition of a solvex set that the intersection of a collection of solvex sets in a partially ordered vector space X is solvex and that the space X itself is solvex. Therefore, every A ⊆ X is contained in a smallest solvex set, which is called the solvex hull of A. It is convenient to have an explicit representation of the solvex hull. For this, we consider the following preliminary statement; see also [146, Lemma 3.28]. Lemma 3.2.18. Let X be a partially ordered vector space. If x, x k ∈ X and y ki ∈ X, i ∈ {1, . . . , m k }, k ∈ {1, . . . , n}, are such that u

mk

{x k , −x k }u ⊇ { ∑ ε i y ki ; ε1 , . . . , ε m k ∈ {1, −1}} i=1

for k ∈ {1, . . . , n}, and n

u

{x, −x}u ⊇ { ∑ ε k x k ; ε1 , . . . , ε n ∈ {1, −1}} , k=1

then n mk

u

{x, −x}u ⊇ { ∑ ∑ ε ki y ki ; ε ki ∈ {1, −1} for i ∈ {1, . . . , m k }, k ∈ {1, . . . , n}} . k=1 i=1

Proof. Let u be an upper bound of n mk

{ ∑ ∑ ε ki y ki ; ε ki ∈ {1, −1}, i ∈ {1, . . . , m k }, k ∈ {1, . . . , n}} . k=1 i=1

130 | 3 Seminorms on pre-Riesz spaces m

m

k n Then u − ∑n−1 k=1 ∑ i=1 ε ki y ki is an upper bound of {∑i=1 ε i y ni ; ε1 , . . . , ε m n ∈ {1, −1}} and therefore of {x n , −x n }, for every ε ki ∈ {−1, 1}, i ∈ {1, . . . , m k }, k ∈ {1, . . . , n − 1}. This yields that u − x n and u + x n are upper bounds of

n−1 m k

{ ∑ ∑ ε ki y ki ; ε ki ∈ {1, −1} for i ∈ {1, . . . , m k }, k ∈ {1, . . . , n − 1}} . k=1 i=1

By induction, it follows that u is an upper bound of {∑nk=1 ε k x k ; ε1 , . . . , ε n ∈ {1, −1}} and therefore of {x, −x}. As a consequence, we obtain the following representation of the solvex hull; see [84, Section 3]. Proposition 3.2.19. Let X be a partially ordered vector space and A ⊆ X. The solvex hull of A is given by n

S = {x ∈ X; ∃ x1 , . . . , x n ∈ A, λ1 , . . . , λ n ∈ (0, 1] with ∑ λ k = 1 k=1 u

n

such that {x, −x}u ⊇ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}} } . k=1

Proof. Clearly, A ⊆ S, and if T is a solvex set containing A, then S ⊆ T. It remains to show that S is solvex. Let x ∈ X, x1 , . . . , x n ∈ S, λ1 , . . . , λ n ∈ (0, 1] with ∑nk=1 λ k = 1 be such that u n

{x, −x}u ⊇ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}} . k=1

For each k ∈ {1, . . . , n} there are y k1 , . . . , y km k ∈ A, λ k1 , . . . , λ km k ∈ (0, 1] with m ∑i=1k λ ki = 1 such that u

mk

{x k , −x k }u ⊇ { ∑ ε i λ ki y ki ; ε1 , . . . , ε m k ∈ {1, −1}} . i=1

Then

u

mk

{λ k x k , −λ k x k }u ⊇ { ∑ ε i λ k λ ki y ki ; ε1 , . . . , ε m k ∈ {1, −1}} . i=1

Due to Lemma 3.2.18, one has u

n mk

{x, −x}u ⊇ { ∑ ∑ ε ki λ k λ ki y ki ; ε ki ∈ {1, −1} for i ∈ {1, . . . , m k }, k ∈ {1, . . . , n}} . k=1 i=1

Since y ki ∈ A and λ k λ ki ∈ (0, 1] for k ∈ {1, . . . , n}, i ∈ {1, . . . , m k }, and n mk

n

k=1 i=1

k=1

mk

n

i=1

k=1

∑ ∑ λ k λ ki = ∑ λ k ∑ λ ki = ∑ λ k = 1 , one gets x ∈ S. Hence, S is solvex.

3.2 Pre-Riesz seminorms

| 131

Corollary 3.2.20. Let X be a Riesz space and A ⊆ X. The solvex hull S of A equals the convex hull W of the solid hull V of A. Proof. First we show that S ⊆ W. Due to Proposition 1.5.31, the set W is solid and convex and contains A. With the aid of Lemma 3.2.16 (ii), we obtain that W is a solvex set containing A, hence S ⊆ W. We proceed by showing that S ⊇ W. According to Lemma 3.2.16, the set S is solid and convex, so V ⊆ S and then W ⊆ S. Next we investigate restriction of solvex sets to subspaces. Just as for solid sets, the intersection of a solvex set and an order dense subspace is solvex in the subspace. We also consider extension. Let Y be a directed ordered vector space and X be an order dense subspace of Y. It turns out that the solvex hull in Y of a solvex set A in X extends A. Recall that the convex hull of the solid hull in Y of a solid and convex set A in X does not always extend A; see Example 3.2.13. For the next result see also [84, Proposition 5.5]. Proposition 3.2.21. Let Y be a directed partially ordered vector space and let X be an order dense subspace of Y. (i) If S is a solvex subset of Y, then S ∩ X is solvex in X. (ii) If A is a solvex subset of X and S is its solvex hull in Y, then S ∩ X = A. Proof. (i) Let x ∈ Y, x1 , . . . , x n ∈ S ∩ X and λ1 , . . . , λ n ∈ (0, 1] with ∑nk=1 λ k = 1 be such that u n

{x, −x}u ∩ X ⊇ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}} ∩ X . k=1

If v ∈ Y is an upper bound of the set {∑nk=1 ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}}, then u

n

{u ∈ X; u ≥ v} ⊆ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}} ∩ X ⊆ {x, −x}u ∩ X , k=1

which implies, by order denseness of X in Y, v = inf{u ∈ X; u ≥ v} ≥ x and, analogously, v ≥ −x. Therefore n

u

{x, −x}u ⊇ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}} . k=1

As S is solvex in Y, it follows that x ∈ S, consequently x ∈ S ∩ X. Hence, S ∩ X is solvex in X. (ii) We assume that A is solvex in X. According to Proposition 3.2.19, the set S is given by n

S = {y ∈ Y; ∃ x1 , . . . , x n ∈ A, λ1 , . . . , λ n ∈ (0, 1] with ∑ λ k = 1 k=1

132 | 3 Seminorms on pre-Riesz spaces u

n u

such that {y, −y} ⊇ { ∑ ε k λ k x k : ε1 , . . . , ε n ∈ {1, −1}} } . k=1

Clearly, A ⊆ S, hence A ⊆ S ∩ X. It remains to establish S ∩ X ⊆ A. Let x ∈ S ∩ X. Then there are x1 , . . . , x n ∈ A and λ1 , . . . , λ n ∈ (0, 1] with ∑nk=1 λ k = 1 such that u

n

{x, −x}u ⊇ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}} . k=1

Therefore, u

n

{x, −x}u ∩ X ⊇ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}} ∩ X , k=1

and, as A is solvex in X, it follows that x ∈ A. There is also a restriction result for solvex sets similar to Proposition 3.2.7. Proposition 3.2.22. Let X and Y be partially ordered vector spaces and let h : X → Y be a bipositive Riesz* homomorphism. If S is a solvex set in Y, then [S]h is a solvex set in X. Proof. Let x ∈ X and let x1 , . . . , x n ∈ [S]h, and λ1 , . . . , λ n ∈ (0, 1] with ∑nk=1 λ k = 1 such that u n

{x, −x}u ⊇ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}} . k=1

By Lemma 3.2.6, we have n

u

{h(x), −h(x)}u ⊇ { ∑ ε k λ k h(x k ); ε1 , . . . , ε n ∈ {1, −1}} . k=1

Since S is solvex and h(x k ) ∈ S for every k ∈ {1, . . . , n}, it follows that h(x) ∈ S. Then x ∈ [S]h, which yields that [S]h is solvex. Proposition 3.2.21 yields the extension property (E) and the restriction property (R) as discussed in Section 2.8 for solvex sets. Proposition 3.2.23. Let X be a pre-Riesz space and (Y, i) a vector lattice cover of X. (i) If I is a solvex subset of X, then the solvex hull J of i[I] in Y is a solvex set such that [J]i = I. (ii) If J is a solvex subset of Y, then [J]i is solvex in X.

3.2.4 Pre-Riesz seminorms: seminorms with solvex unit balls We define pre-Riesz seminorms by means of solvex sets and show that they have the desired extension and restriction properties as discussed in Subsection 3.2.1.

3.2 Pre-Riesz seminorms |

133

Definition 3.2.24. Let X be a partially ordered vector space. A seminorm on X is called a pre-Riesz seminorm if its unit ball is solvex. Since a subset of a Riesz space is solvex if and only if it is solid and convex, a seminorm on a Riesz space is a pre-Riesz seminorm if and only if it is a Riesz seminorm. We have the following explicit reformulation of Definition 3.2.24. Proposition 3.2.25. Let X be a partially ordered vector space with a seminorm p. Then p is a pre-Riesz seminorm if and only if p(x) = inf {p(x1 ) ∨ ⋅ ⋅ ⋅ ∨ p(x n ); x1 , . . . , x n ∈ X such that n

there are λ1 , . . . , λ n ∈ (0, 1] with ∑ λ k = 1 and k=1 u

n

{x, −x} ⊇ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}} } u

(3.4)

k=1

for every x ∈ X. Proof. Let B be the unit ball of p. Assume that p is a pre-Riesz seminorm. Let x ∈ X, x1 , . . . x n ∈ X, and λ1 , . . . , λ n ∈ (0, 1] with ∑nk=1 λ k = 1 be such that n

u

u

{x, −x} ⊇ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}} . k=1

Clearly, x1 , . . . , x n ∈ (p(x1 ) ∨ . . . ∨ p(x n ))B and this set is solvex, because B is solvex. It follows that x ∈ (p(x1 ) ∨ . . . ∨ p(x n ))B, or, in other words, p(x) ≤ p(x1 ) ∨ . . . ∨ p(x n ), which proves that p(x) is less than or equal to the infimum at the right-hand side. By taking n = 1 and x1 = x, it is clear that the infimum is not strictly greater than p(x). Thus, the equality is established. For the converse implication, let x ∈ X, λ1 , . . . , λ n ∈ (0, 1] with ∑nk=1 λ k = 1, and x1 , . . . , x n ∈ B be such that n

u

{x, −x}u ⊇ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}} . k=1

Then, by assumption, p(x) ≤ p(λ1 x1 ) ∨ . . . ∨ p(λ n x n ) ≤ 1, so x ∈ B. Hence, B is solvex and p is a pre-Riesz seminorm. Let us now consider the extension of pre-Riesz seminorms. We have two ways of extending a pre-Riesz seminorm p on a majorizing subspace X of a partially ordered vector space Y. The first one extends the unit ball of p and the second one extends p directly by an extension formula. The methods are actually reformulations of each other and the ensuing extension results are the same. Nevertheless, we will consider them both.

134 | 3 Seminorms on pre-Riesz spaces

Theorem 3.2.26. Let Y be a directed partially ordered vector space with a majorizing linear subspace X. Let p be a seminorm on X with unit ball B. Then the solvex hull S of B in Y is the unit ball of a pre-Riesz seminorm p̄ and B ⊆ S. Moreover, p̄ extends p if and only if p is a pre-Riesz seminorm. Proof. Recall that the solvex hull S of B is the smallest solvex set in Y containing B. By Lemma 3.2.16, S is solid and convex. Hence, S is circled. Since the unit ball B is absorbing in X and X is majorizing in Y, it follows that B is absorbing in Y and therefore S is absorbing in Y. Since B = ⋂ α>1 αB, it follows that S = ⋂α>1 αS. With the aid of Proposition 3.1.3, the set S is the unit ball of a pre-Riesz seminorm p̄ on Y. If p is a pre-Riesz seminorm, then B is solvex, so that Proposition 3.2.21 (ii) yields that S ∩ X = B, which means that the unit ball of p̄ restricted to X equals B. Thus, p̄ extends p. Theorem 3.2.27. Let Y be a directed partially ordered vector space with a majorizing linear subspace X. Let p be a seminorm on X. Define for y ∈ Y, ̄ p(y) := inf{p(x1 ) + ⋅ ⋅ ⋅ + p(x n ); x1 , . . . , x n ∈ X such that u

n

{y, −y}u ⊇ { ∑ ε k x k ; ε1 , . . . , ε n ∈ {1, −1}} } .

(3.5)

k=1

Then p̄ is the greatest pre-Riesz seminorm on Y that is less than or equal to p on X. Moreover, p̄ extends p if and only if p is a pre-Riesz seminorm. Proof. It follows directly from the definition that p̄ is a seminorm on Y. For x ∈ X it is ̄ clear that p(x) ≤ p(x). To prove that p̄ is a pre-Riesz seminorm, it suffices to establish for every y ∈ Y that ̄ ̄ 1 ) ∨ . . . ∨ p(y ̄ n ); y1 , . . . , y n ∈ Y such that there are p(y) = inf{p(y n

λ1 , . . . , λ n ∈ (0, 1] with ∑ λ k = 1 and k=1 u

n

{y, −y}u ⊇ { ∑ ε k λ k y k ; ε1 , . . . , ε n ∈ {1, −1}} }

(3.6)

k=1

and to use Proposition 3.2.25. ̄ Let y ∈ Y. It is clear that p(y) is greater than the infimum at the right-hand side of (3.6). To prove that it is not strictly greater, let y1 , . . . , y n ∈ Y and λ1 , . . . , λ n ∈ (0, 1] with ∑nk=1 λ k = 1 be such that n

u

{y, −y}u ⊇ { ∑ ε k λ k y k ; ε1 , . . . , ε n ∈ {1, −1}} , k=1

3.2 Pre-Riesz seminorms |

135

and let ε > 0. By the definition of p,̄ for each k there are x k,1 , . . . , x k,m k ∈ X such that u

mk

{y k , −y k }u ⊇ { ∑ ε k x k,i ; ε1 , . . . , ε m k ∈ {1, −1}} i=1

̄ k ) ≥ p(x k,1 ) + ⋅ ⋅ ⋅ + p(x k,m k ) − ε. Then, by Lemma 3.2.18, and p(y u

n mk

{y, −y}u ⊇ { ∑ ∑ ε k,i λ k x k,i ; ε k,i ∈ {1, −1}, i ∈ {1, . . . , m k }, k ∈ {1, . . . , n}} , k=1 i=1

so n mk

n

k=1 i=1

k=1

n

̄ k ) + ε) ≤ p(y ̄ 1 ) ∨ . . . ∨ p(y ̄ n) + ε . ̄ p(y) ≤ ∑ ∑ p(λ k x k,i ) = ∑ λ k ∑ p(x k,i ) ≤ ∑ λ k (p(y i

k=1

It follows that p̄ is a pre-Riesz seminorm. ̄ Let q be a pre-Riesz seminorm on Y with q ≤ p on X. Let y ∈ Y with p(y) < 1. Then there are x1 , . . . , x n ∈ X such that n

u

{y, −y}u ⊇ { ∑ ε k x k ; ε1 , . . . , ε n ∈ {1, −1}} k=1

and p(x1 ) + ⋅ ⋅ ⋅ + p(x n ) < 1. Choose λ1 , . . . , λ n ∈ ℝ with λ k > p(x k ) ≥ q(x k ) for every k ∈ {1, . . . , n} and ∑nk=1 λ k = 1. Then u

n

{y, −y}u ⊇ { ∑ ε k λ k (λ−1 k x k ); ε1 , . . . , ε n ∈ {1, −1}} , k=1

̄ so that q(y) < 1, because the unit ball of q is solvex. It follows that q(y) ≤ p(y) for all y ∈ Y, so p̄ is the greatest pre-Riesz seminorm on Y that is less than or equal to p on X. If p is a pre-Riesz seminorm, the same argument shows for every x ∈ X that p(x) ≤ ̄ p(x). Hence, p̄ = p on X. Conversely, if p̄ extends p, then for every x ∈ X and x 1 , . . . , x n ∈ X with p(x k ) ≤ 1 for every k and λ1 , . . . , λ n ∈ (0, 1] with ∑nk=1 λ k = 1 such that n

u

{x, −x}u ⊇ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}} , k=1

̄ the definition of p̄ implies that p(x) = p(x) ≤ p(λ1 x1 ) + ⋅ ⋅ ⋅ + p(λ n x n ) ≤ 1. Hence, the unit ball of p is solvex, or, in other words, p is a pre-Riesz seminorm. Observe that if Y is a Riesz space, then formula (3.5) comes down to ̄ p(y) = inf{p(x1 ) + ⋅ ⋅ ⋅ + p(x n ); x1 , . . . , x n ∈ X such that |y| ≤ |x1 | + ⋅ ⋅ ⋅ + |x n |} . Both of the above extension theorems have the following consequence.

136 | 3 Seminorms on pre-Riesz spaces Corollary 3.2.28. Let X be a pre-Riesz space and let p be a seminorm on X. Let (Y, i) be a vector lattice cover of X. If p is a pre-Riesz seminorm, then there exists a Riesz seminorm p̄ on Y such that p̄ ∘ i = p. The restriction of a pre-Riesz seminorm on a partially ordered vector space Y to a subspace X is a pre-Riesz seminorm if X is order dense in Y. Proposition 3.2.29. Let Y be a partially ordered vector space and let X be an order dense subspace of Y. If p is a pre-Riesz seminorm on Y, then the restriction of p to X is a pre-Riesz seminorm on X. Proof. Let S be the unit ball of p. By the definition of a pre-Riesz seminorm, S is solvex in Y. According to Proposition 3.2.21 (i), the intersection S ∩ X is solvex in X. Since S ∩ X is the unit ball of the restriction of p to X, this restriction is a pre-Riesz seminorm. In a similar fashion, Proposition 3.2.22 yields the following result. Proposition 3.2.30. Let X and Y be directed partially ordered vector spaces and let h : X → Y be a Riesz* homomorphism. If p is a pre-Riesz seminorm on Y, then p ∘ h is a pre-Riesz seminorm on X. Proof. As h is linear and p is a seminorm, p ∘ h is a seminorm on X. Let S denote the unit ball of p. Then S is a solvex set in Y. According to Proposition 3.2.22, the set [S]h is a solvex set in X. Since [S]h is the unit ball of p ∘ h, we obtain that p ∘ h is a pre-Riesz seminorm. The results in this subsection amount to the following characterizations of pre-Riesz seminorms. Theorem 3.2.31. Let X be a pre-Riesz space and let p be a seminorm on X. The following statements are equivalent. (a) p is a pre-Riesz seminorm, that is, the unit ball of p is solvex. (b) p satisfies the explicit characterization formula (3.4). (c) For every Riesz space Y such that X is a majorizing subspace of Y the seminorm p extends to a Riesz seminorm on Y. (d) For every vector lattice cover (Y, i) of X, there exists a Riesz seminorm p̄ on Y such that p = p̄ ∘ i. (e) There exists a Riesz seminorm p̄ on the Riesz completion (X ρ , i) of X such that p = p̄ ∘ i. Proof. The equivalence of (a) and (b) is the content of Proposition 3.2.25. The implication (a) ⇒ (c) is contained in Corollary 3.2.28. For a proof of (c) ⇒ (d) it suffices to note that p∘ i−1 is a Riesz seminorm on i[X] and that i[X] is majorizing in Y. The implication (d) ⇒ (e) is immediate. Finally, for the implication (e) ⇒ (a), we use that i[X] is order dense in the Riesz completion X ρ of X and apply Proposition 3.2.29 to obtain that p∘i−1 is a pre-Riesz seminorm on i[X], whence p is a pre-Riesz seminorm on X.

3.2 Pre-Riesz seminorms

| 137

If p and q are Riesz seminorms on a Riesz space, then the pointwise maximum p ∨ q and the sum p + q are Riesz seminorms as well; see Proposition 1.5.29. Therefore, Theorem 3.2.31 has the following consequence. Corollary 3.2.32. Let X be a pre-Riesz space. If p and q are pre-Riesz seminorms on X, then p ∨ q and p + q are pre-Riesz seminorms as well. Proof. Let (X ρ , i) be the Riesz completion of X. Due to Theorem 3.2.31, there exist Riesz seminorms p̄ and q̄ on X ρ such that p = p̄ ∘ i and q = q̄ ∘ i. Then p̄ ∨ q̄ and p̄ + q̄ are Riesz ̄ i seminorms due to Proposition 1.5.29, and therefore Theorem 3.2.31 yields that (p̄ ∨ q)∘ and (p̄ + q)̄ ∘ i are pre-Riesz seminorms on X, which means that p ∨ q and p + q are pre-Riesz seminorms. It is a natural question whether a pre-Riesz norm on a pre-Riesz space can be extended to a Riesz norm on the Riesz completion and not just to a seminorm. In general, extension to a Riesz norm need not be possible. If the pre-Riesz space is pervasive, then the situation improves; see Proposition 3.2.34 below. Example 3.2.33. We give an example of a pre-Riesz norm on a pre-Riesz space X which cannot be extended to a Riesz norm on the Riesz completion. Let X = P[0, 1] be the space of all polynomial functions on [0, 1]. Then X is an order dense subspace of C[0, 1], see Example 1.7.2. The Riesz completion X ρ of X is the subspace of C[0, 1] consisting of all piecewise polynomial functions. For x ∈ X, define p(x) := sup {|x(t)| ; t ∈ [0, 12 ]} . Then p is a pre-Riesz seminorm on X as it is the restriction of a Riesz seminorm on C[0, 1]. Moreover, if x ∈ X is such that p(x) = 0, then x = 0 on [0, 12 ], hence x = 0 since x is a polynomial. Let p̄ be the greatest Riesz seminorm on X ρ extending p. Let {0, y(t) = { t − 12 , {

t ∈ [0, 12 ] , t ∈ ( 12 , 1] .

̄ Then y ∈ X ρ , y ≥ 0, and y ≠ 0. We show that p(y) = 0. Let ε > 0. Due to the Stone– Weierstrass theorem 1.8.10, there exists x ∈ X such that |(y(t) + ε) − x(t)| < ε for every t ∈ [0, 1]. Then x ≥ (y + ε) − ε = y, so {y, −y}u ⊇ {x, −x}u . Therefore, ̄ p(y) ≤ p(x) = sup {|x(t)|; t ∈ [0, 12 ]} ≤ sup {(y(t) + ε) + ε; t ∈ [0, 12 ]} = 2ε . ̄ It follows that p(y) = 0. Hence, p̄ is not a norm on Y. Let us now consider pervasive subspaces as defined in Definition 2.8.1. Recall that a majorizing pervasive linear subspace of an Archimedean Riesz space is order dense; see Proposition 2.8.5. The idea of the next result is due to Feng Zhang.

138 | 3 Seminorms on pre-Riesz spaces

Proposition 3.2.34. Let Y be an Archimedean Riesz space, let X be a majorizing linear subspace, and let p be a pre-Riesz norm on X. Let p̄ be the greatest Riesz seminorm on Y extending p. If X is pervasive in Y, then p̄ is a norm on Y. Proof. Let y ∈ Y be such that y ≠ 0. Then |y| ≠ 0, so there exists x ∈ X \ {0} with ̄ ̄ ̄ 0 ≤ x ≤ |y|, since X is pervasive in Y. Then p(x) > 0, hence p(y) = p(|y|) ≥ p(x) = p(x) > 0. With the aid of Proposition 2.4.12 we have the following consequence. Corollary 3.2.35. Let X be a pervasive Archimedean pre-Riesz space with Riesz completion (X ρ , i) and let p be a norm on X. Then p is a pre-Riesz norm if and only if there exists a Riesz norm ρ on X ρ such that p = ρ ∘ i. We conclude this subsection with an example that shows that pre-Riesz seminorms are not determined by their restrictions to the positive cone of the space. This fact is a major difference with Riesz seminorms. A notion of seminorms on partially ordered vector spaces based on this property of Riesz seminorms is that of a regular seminorm, which will be discussed in Section 3.4 below. Example 3.2.36. We give an example of an Archimedean pre-Riesz space (X, K) with pre-Riesz seminorms p and q such that p = q on K and such that p and q are not equivalent. Let X be the space of affine functions on the unit circle S as discussed in Example 1.7.5, where it is proven that X is an order dense subspace of C(S). The space X also appears in Example 3.2.13. For our purpose, define 󵄨 󵄨 󵄨 󵄨 p(x) := 󵄨󵄨󵄨x((1, 0)T )󵄨󵄨󵄨 + 󵄨󵄨󵄨x((−1, 0)T )󵄨󵄨󵄨 and 󵄨 󵄨 󵄨 󵄨 q(x) := 󵄨󵄨󵄨x((0, 1)T )󵄨󵄨󵄨 + 󵄨󵄨󵄨x((0, −1)T )󵄨󵄨󵄨 for x ∈ X. Clearly, p and q are restrictions of Riesz seminorms on C(S) and therefore they are pre-Riesz seminorms, by Proposition 3.2.29. Let x ∈ K and let c = x((0, 0)T ). Since x is affine, the function v = x − c𝟙 is linear, so p(x) = x((1, 0)T ) + x((−1, 0)T ) = v((1, 0)T ) + v((−1, 0)T ) + 2c = v((0, 1)T ) + v((0, −1)T ) + 2c = x((0, 1)T ) + x((0, −1)T ) = q(x) . Thus, p = q on K. To show that p and q are not equivalent on X, take x((s, t)T ) := s for (s, t)T ∈ S. Then x ∈ X, p(x) = 2, and q(x) = 0, and therefore p and q are not equivalent. Note that if ρ is the supremum norm on X, then ρ + p and ρ + q are two pre-Riesz norms on X, which are equal on K but not equal on X.

3.3 Extension and restriction of monotone seminorms One of the most natural classes of seminorms on a partially ordered vector space is the class of monotone seminorms; see Definition 1.5.6. If we consider a monotone semi-

3.3 Extension and restriction of monotone seminorms

| 139

norm on a pre-Riesz space X, it is a natural question whether it extends to a monotone seminorm on the Riesz completion of X. It turns out that a slightly stronger property than mere monotonicity is needed for being extendable to a monotone seminorm on the Riesz completion. We call seminorms with that property monotone* seminorms. We show that a seminorm on a pre-Riesz space X can be extended to a monotone seminorm on any Riesz space Y that contains X as a majorizing subspace if and only if it is a monotone* seminorm. We also show that every monotone seminorm is equivalent to a monotone* seminorm, where the equivalence constants are between 1 and 32 .

3.3.1 Monotone seminorms and related concepts Let us recall the definition of a monotone seminorm and introduce two related concepts. Definition 3.3.1. Let (X, K) be a partially ordered vector space and let p be a seminorm on X. (i) p is called monotone if for every x, y ∈ X such that 0 ≤ x ≤ y one has that p(x) ≤ p(y). (ii) p is called monotone* if for every x ∈ X and u, v ∈ K such that −u ≤ x ≤ v one has that p(x) ≤ p(u) + p(v). (iii) p is called symmetrically monotone if for every x, y ∈ X such that −y ≤ x ≤ y one has that p(x) ≤ p(y). Definition 3.3.1 is illustrated by the following simple example.

b

a

a

|a|

a −a

−|a|

Fig. 3.2: Unit balls of monotone, symmetrically monotone, and Riesz norms on ℝ2 .

Example 3.3.2. Let X be (ℝ2 , ℝ2+ ) and let p be a seminorm on X. The properties of Definition 3.3.1 have geometrical interpretations for the unit ball B of p (see Figure 3.2). For a = (a1 , a2 )T , b = (b 1 , b 2 )T ∈ ℝ2 with a ≤ b denote R(a, b) = [a1 , b 1 ] × [a2 , b 2 ], which is the rectangle with lower left corner at a and upper right corner at b. A straightforward verification yields that p is monotone if and only if R(a, b) ⊆ B for every a, b ∈ B with 0 ≤ a ≤ b. Further, p is symmetrically monotone if and only if R(−a, a) ⊆ B for

140 | 3 Seminorms on pre-Riesz spaces every a ∈ B with a ≥ 0. Moreover, p is a Riesz seminorm if and only if R(−|a|, |a|) ⊆ B for every a ∈ B. Note that X is a Riesz space, so monotone and monotone* seminorms coincide due to Proposition 3.3.6 below. Let us collect some simple properties of monotone, monotone*, and symmetrically monotone seminorms and seminorms with full unit balls. Recall that a subset M of a partially ordered vector space is full if for every a, b ∈ M we have [a, b] ⊆ M. Proposition 3.3.3. Let (X, K) be a partially ordered vector space and let p be a seminorm on X. (i) p is monotone if and only if for every x, y ∈ K one has p(x) ≤ p(x + y). (ii) p is symmetrically monotone if and only if for every x, y ∈ K one has p(x − y) ≤ p(x + y). (iii) p has a full unit ball if and only if for every x, u, v ∈ X with u ≤ x ≤ v one has p(x) ≤ p(u) ∨ p(v). Proof. (i) If p is monotone, from x ≤ x + y it follows that p(x) ≤ p(x + y). For a proof of the converse implication, let x, y ∈ X be such that 0 ≤ x ≤ y. Then x ∈ K and y − x ∈ K, so p(x) ≤ p(x + y − x) = p(y). Hence, p is monotone. (ii) Assume that p is symmetrically monotone. For x, y ∈ K, we have −(x + y) ≤ x − y ≤ x + y and hence p(x − y) ≤ p(x + y). For the converse implication, let x, y ∈ X with −y ≤ x ≤ y. We have 12 (x + y), 1 2 (y − x) ∈ K, so that p(x) = p ( 12 (x + y) − 12 (y − x)) ≤ p ( 12 (x + y) + 12 (y − x)) = p(y) . (iii) Assume that p has a full unit ball. Let x, u, v ∈ X be such that u ≤ x ≤ v. Let α > p(u) ∨ p(v). Then 1α u and 1α v are in the unit ball of p. As 1α u ≤ 1α x ≤ 1α v and the unit ball of p is full, we obtain 1α x is in the unit ball of p and hence p(x) ≤ α. It follows that p(x) ≤ p(u) ∨ p(v). For the converse implication, let u and v be in the unit ball B of p and let u ≤ x ≤ v. Then p(x) ≤ p(u) ∨ p(v) ≤ 1, so x ∈ B. Hence, B is full. In the next proposition we relate the following properties of a seminorm p on a partially ordered vector space: (A) p has a full unit ball. (B) p is symmetrically monotone. (C) p is monotone*. (D) p is monotone. Proposition 3.3.4. Let (X, K) be a partially ordered vector space and let p be a seminorm on X. We have (A) ⇒ (B), (B) ⇒ (C), and (C) ⇒ (D).

3.3 Extension and restriction of monotone seminorms |

141

Proof. If p has a full unit ball B and x, u ∈ X are such that −u ≤ x ≤ u, then for every α > p(u) = p(−u) we have 1α u ∈ B and − 1α u ∈ B. As − 1α u ≤ 1α x ≤ 1α u we obtain 1α x ∈ B, so p(x) ≤ α. Hence, p(x) ≤ p(u). Thus, p is symmetrically monotone. If p is symmetrically monotone and x ∈ X and u, v ∈ K are such that −u ≤ x ≤ v, then −(u + v) ≤ x ≤ u + v, so p(x) ≤ p(u + v) ≤ p(u) + p(v). Hence, p is monotone*. If p is monotone*, then it is immediate that p is monotone, by taking u = 0. Remark 3.3.5. In Lemma 1.5.25 it is shown that an order unit norm ‖⋅‖u has the property of (ii) in Proposition 3.3.3, i.e., it is symmetrically monotone. Since its unit ball is the order interval [−u, u], it is clear that ‖⋅‖u has a full unit ball. Proposition 3.3.6. Let X be a Riesz space and let p be a seminorm on X. (i) If p is a Riesz seminorm then p is symmetrically monotone. (ii) p is monotone if and only if p is monotone*. Proof. (i) If p is a Riesz seminorm and x, y ∈ X are such that −y ≤ x ≤ y, then |x| ≤ y, hence p(x) = p(|x|) ≤ p(y), so that p is symmetrically monotone. (ii) Assume that p is monotone. If x ∈ X and u, v ∈ K are such that −u ≤ x ≤ v, then x+ ≤ v and x− ≤ u, so p(x) = p(x+ − x− ) ≤ p(x+ ) + p(x− ) ≤ p(u) + p(v) . Thus, p is monotone*. The converse implication is contained in Proposition 3.3.4. The next result is straightforward. Proposition 3.3.7. Let (X, K) be a partially ordered vector space and let Q be a set of seminorms on X. (i) Suppose that for every x ∈ X the sum s(x) := ∑q∈Q q(x) exists. If for every q ∈ Q property (A) with p = q holds, then property (A) with p = s holds. Similar statements are satisfied for (B) and (C). (ii) Suppose that for every x ∈ X the supremum s(x) := sup{q(x); q ∈ Q} exists. If for every q ∈ Q property (A) with p = q holds, then property (A) with p = s holds. Similar statements are satisfied for (B), (C), and (D). The sum of two norms with full unit balls need not be a norm with a full unit ball, as the next example shows. Example 3.3.8. Let X = ℝ2 with standard order and consider the order units u = (1, 12 )T and v = ( 12 , 1)T in X. The u-norm ‖⋅‖u and the v-norm ‖⋅‖v both have a full unit ball. Take x = (0, −1)T , y = (1, −1)T , and z = (1, 0)T . Then x ≤ y ≤ z. We have ‖x‖u = 2 ,

‖x‖v = 1 ,

‖y‖u = 2 ,

‖y‖v = 2 ,

‖z‖u = 1 ,

and ‖z‖v = 2 .

142 | 3 Seminorms on pre-Riesz spaces For the sum p of the u-norm and the v-norm we therefore have p(x) = 3, p(y) = 4 and p(z) = 3, so p(y) ≰ p(x) ∨ p(z), which yields by Proposition 3.3.3 (iii) that p does not have a full unit ball. The next proposition is convenient for constructing examples. Proposition 3.3.9. Let (X, K) be a partially ordered vector space, let p be a monotone seminorm on X, and let f : X → ℝ be a positive linear functional. (i) If X is a Riesz space, then x 󳨃→ p(x + ) + p(x− ) is a monotone* seminorm and x 󳨃→ p(x+ ) ∨ p(x− ) is a symmetrically monotone seminorm on X. (ii) x 󳨃→ |f(x)| is a symmetrically monotone seminorm on X. (iii) If X is a Riesz space, then x 󳨃→ f(x+ ) ∨ f(x− ) is a monotone* seminorm on X and x 󳨃→ f(|x|) is a Riesz seminorm on X. Proof. (i) Denote for x ∈ X, q(x) := p(x+ ) + p(x− ) and r(x) := p(x+ ) ∨ p(x− ). It is easy to check that q is a seminorm. We verify that r satisfies the triangle inequality. Let x1 , x2 ∈ X. Then r(x1 + x2 ) = p ((x1 + x2 )+ ) ∨ p ((x1 + x2 )− ) ≤ p (x+1 + x+2 ) ∨ p (x−1 + x−2 ) ≤ (p (x+1 ) + p (x+2 )) ∨ (p (x−1 ) + p (x−2 )) ≤ (p (x+1 ) ∨ p (x−1 ) + p (x+2 ) ∨ p (x−2 )) ∨ (p (x−1 ) ∨ p (x+1 ) + p (x−2 ) ∨ p (x+2 )) = r(x1 ) + r(x2 ) . By straightforward arguments, it follows that r is a seminorm. Since q is monotone, Proposition 3.3.6 (ii) yields that q is monotone*. To see that r is symmetrically monotone, let x, u ∈ X be such that −u ≤ x ≤ u. Then x+ ≤ u and x− ≤ u, so r(x) = p(x+ ) ∨ p(x− ) ≤ p(u) ∨ p(u) = p(u) = r(u). Hence, r is symmetrically monotone. The proofs of (ii) and (iii) are straightforward and omitted. The next example illustrates the constructions of Proposition 3.3.9. Example 3.3.10. (i) Let X = C[0, 1]. The map x 󳨃→ (∫ x+ ) ∨ (∫ x− ) is a monotone* norm, the norm x 󳨃→ ‖x‖1 + |x(0) + x(1)| is symmetrically monotone, and x 󳨃→ ‖x‖1 + |x(0)| is a Riesz seminorm. (ii) Let X = ℓ∞ (ℕ) with the standard order. The norm x 󳨃→ ‖x+ ‖∞ + ‖x− ‖∞ is mono−n tone*, x 󳨃→ ‖x‖∞ + | ∑∞ n=1 2 x n | is symmetrically monotone, and x 󳨃→ ‖x‖∞ + ∞ ∑n=1 2−n |x n | is a Riesz norm. Let us present some examples to show that not every symmetrically monotone seminorm is a Riesz seminorm, and that the notions of Definition 3.3.1 are all distinct. Example 3.3.11. Let X = ℝ2 with the standard order. (i) The norm (x1 , x2 )T 󳨃→ |x1 | + |x2 | + |x1 + x2 | on X is symmetrically monotone but not a Riesz norm.

3.3 Extension and restriction of monotone seminorms

| 143

(ii) The norm x 󳨃→ ‖x+ ‖∞ +‖x− ‖∞ on X is monotone* but not symmetrically monotone. (iii) The norm (x1 , x2 )T 󳨃→ |x1 | ∨ |x1 − x2 | on X is not monotone. An example of a monotone norm that is not monotone* cannot be found on a Riesz space, because of Proposition 3.3.6 (ii).

Fig. 3.3: The unit ball of p.

Example 3.3.12. We give an example of a monotone norm that is not monotone*. Take X = ℝ3 , ordered by the cone K := {(x1 , x2 , x3 )T ∈ X; x1 , x2 , x3 ≥ 0 and x3 ≤ x1 + x2 } , and for x = (x1 , x2 , x3 )T ∈ X define p(x) := max{|x1 + x3 |, |x1 + 2x2 − 3x3 |, |x1 + 2x2 − x3 |, |x1 − 2x2 + x3 |} . Note that X is a directed partially ordered vector space and that p is a norm on X. The unit ball¹ of p is depicted in Figure 3.3. We first show that for every x = (x1 , x2 , x3 )T ∈ K we have p(x) = max{x1 + x3 , x1 + 2x2 − x3 } . Indeed, x1 , x2 , x3 ≥ 0 and x3 ≤ x1 + x2 . Then x1 +2x2 − x3 = (x1 + x2 )+ x2 − x3 ≥ x2 ≥ 0. Further, x1 −2x2 + x3 ≤ x1 + x3 , −(x1 −2x2 + x3 ) ≤ x1 +2x2 − x3 , and x1 +2x2 −3x3 ≤ x1 + 2x2 −x3 . Since −x2 ≤ x1 −x3 , we have −(x1 +2x2 −3x3 ) ≤ −x1 +2(x1 −x3 )+3x3 = x1 +x3 . Therefore, p(x) = max{x1 + x3 , x1 + 2x2 − x3 }. For a proof that p is monotone, let x = (x1 , x2 , x3 )T , y = (y1 , y2 , y3 )T ∈ K. Then, since y1 + y3 ≥ 0 and y1 + 2y2 − y3 ≥ 0, we have p(x + y) = max{x1 + x3 + y1 + y3 , x1 + 2x2 − x3 + y1 + 2y2 − y3 } ≥ max{x1 + x3 , x1 + 2x2 − x3 } = p(x) . 1 The authors are grateful to Prof. Dr. Arnoud van Rooij who made the beautiful model on the photos in Figure 3.3 and kindly gave it as a present to his PhD student Onno van Gaans.

144 | 3 Seminorms on pre-Riesz spaces To see that p is not monotone*, take y = (0, 0, 1)T , u = (1, 0, 0)T and v = (0, 1, 1)T . Then u, v ≥ 0 and −u ≤ y ≤ v, whereas p(u) = 1, p(v) = 1 and p(y) = 3. Remark 3.3.13. Example 3.3.12 provides a pre-Riesz space with a monotone norm that cannot be extended to a monotone seminorm on any vector lattice cover. Indeed, the cone K is closed and generating, hence X is pre-Riesz. If q is a monotone seminorm on a vector lattice cover Y of X, then q is monotone* by Proposition 3.3.6. Hence, the restriction of q to X is monotone*, and therefore not equal to p. It follows that there is no monotone seminorm on Y that extends p. To conclude the subsection, we consider the following straightforward result on restriction of seminorms. Proposition 3.3.14. Let Y be a partially ordered vector space and let X be a linear subspace of Y. Let q be a seminorm on Y and let r be the restriction of q to X. If one of the properties (A), (B), (C) or (D) holds with p = q, then the corresponding property holds with p = r.

3.3.2 Extension of seminorms If X is a pre-Riesz space and p is the restriction to X of a monotone seminorm on the Riesz completion of X, then it follows from Proposition 3.3.6 that p is a monotone* seminorm. Next we consider extension of monotone* seminorms. Recall that on the set of seminorms on a vector space X we consider the pointwise order. Theorem 3.3.15. Let (X, K X ) and (Y, K Y ) be directed partially ordered vector spaces such that there exists a bipositive linear map i : X → Y with i[X] majorizing in Y. Let p be a seminorm on X. Define for y ∈ Y, p1 (y) := inf{p(x) + p(u) + p(v); x ∈ X, u, v ∈ K X , −i(u) ≤ y − i(x) ≤ i(v)} . (i) For every y ∈ K Y one has p1 (y) = inf{p(w); w ∈ K X , i(w) ≥ y}. (ii) p1 is the greatest monotone* seminorm on X with p1 ∘ i ≤ p. (iii) p1 ∘ i = p if and only if p is monotone*. Proof. First, note that for every y ∈ Y the set in the definition of p1 (y) is nonempty, because X is directed and i[X] is majorizing in Y. Hence, p1 (y) ∈ [0, ∞) is well defined. (i) Let y ∈ K Y . If w ∈ K X is such that i(w) ≥ y, then 0 ≤ y − 0 ≤ i(w), so p1 (y) ≤ p(w). Hence, p1 (y) ≤ inf{p(w); w ∈ K X , i(w) ≥ y}. For the converse inequality, let u, v ∈ K X and x ∈ X be such that −i(u) ≤ y − i(x) ≤ i(v). Then 0 ≤ y ≤ i(x + v), so p(x) + p(u) + p(v) ≥ p(x + v) ≥ inf{p(w); w ∈ K X , i(w) ≥ y}. Hence, p1 (y) ≥ inf{p(w); w ∈ K X , i(w) ≥ y}. (ii) We start by showing that p1 is a seminorm on Y. We have already observed that p1 ≥ 0 and it is immediate that p1 (0) = 0. To prove the triangle inequality, let

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y1 , y2 ∈ Y. If x1 , x2 ∈ X and u 1 , v1 , u 2 , v2 ∈ K X are such that −i(u 1 ) ≤ y1 − i(x1 ) ≤ i(v1 ) and −i(u 2 ) ≤ y2 − i(x2 ) ≤ i(v2 ), then −i(u 1 + u 2 ) ≤ (y1 + y2 ) − i(x1 + x2 ) ≤ i(v1 + v2 ), so p1 (y1 + y2 ) ≤ p(x1 + x2 ) + p(u 1 + u 2 ) + p(v1 + v2 ) ≤ (p(x1 ) + p(u 1 ) + p(v1 )) + (p(x2 ) + p(u 2 ) + p(v2 )) . Hence, by taking infimum twice, p1 (y1 + y2 ) ≤ p1 (y1 ) + p1 (y2 ). Next, let y ∈ Y and let λ ∈ (0, ∞). If x ∈ X and u, v ∈ K X are such that −i(u) ≤ y − i(x) ≤ i(v), then λu, λv ∈ K X and −i(λu) ≤ λy − i(λx) ≤ i(λv), so p1 (λy) ≤ p(λx) + p(λu) + p(λv) = λ (p(x) + p(u) + p(v)) . Hence, p1 (λy) ≤ λp1 (y). If we replace λ by 1λ and y by λy, we also obtain that p1 (y) ≤ 1λ p1 (λy). Therefore, p1 (λy) = λp1 (y). From −i(u) ≤ y − i(x) ≤ i(v) it also follows that i(u) ≥ −y + i(x) ≥ −i(v) which means that −i(v) ≤ −y − i(−x) ≤ i(u). Hence, p 1 (−y) ≤ p(−x) + p(v) + p(u) = p(x) + p(u) + p(v). We infer that p1 (−y) ≤ p1 (y). Replacing y by −y yields p1 (y) ≤ p1 (−y) and therefore p1 (−y) = p1 (y). Thus, we have established that p1 is a seminorm. We proceed by showing that p1 is monotone*. Let y ∈ Y and a, b ∈ K Y be such that −a ≤ y ≤ b. If u, v ∈ K X are such that a ≤ i(u) and b ≤ i(v), then −i(u) ≤ y ≤ i(v), so that p1 (y) ≤ p(u) + p(v). Taking infimum over u and v and applying (i) yields that p1 (y) ≤ p1 (a) + p1 (b). Thus, p1 is monotone*. To complete the proof of (ii), it remains to show that p1 is the greatest monotone* seminorm on Y with p1 ∘ i ≤ p. For x ∈ X, by taking u = v = 0, it is clear from the definition of p1 that p1 (i(x)) ≤ p(x). Further, if q is any monotone* seminorm on Y and q ∘ i ≤ p, then for y ∈ Y, x ∈ X, and u, v ∈ K X with −i(u) ≤ y − i(x) ≤ i(v) we have that q(y) ≤ q(i(x)) + q(y − i(x)) ≤ q(i(x)) + q(i(u)) + q(i(v)) ≤ p(x) + p(u) + p(v) . It follows that q(y) ≤ p 1 (y). (iii) According to Proposition 3.3.14, it follows from (ii) that p1 is monotone* on i[X]. Hence, if p1 ∘ i = p, then p is monotone*. For the converse implication, assume that p is monotone*. If x, y ∈ X and u, v ∈ K X are such that −u ≤ i(y)− i(x) ≤ v, then p(y) ≤ p(x)+ p(y − x) ≤ p(x)+ p(u)+ p(v), so, by taking infimum over u, v and x, p(y) ≤ p1 (i(y)). Combination with (ii) yields that p = p1 ∘ i. Propositions 3.3.6, 3.3.14 and Theorem 3.3.15 together yield restriction and extension for monotone* seminorms. Theorem 3.3.16. Let X be a pre-Riesz space and let p be a seminorm on X. The following statements are equivalent.

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(a) p is monotone*. (b) For every Riesz space Y such that X is a majorizing subspace of Y the seminorm p extends to a monotone seminorm on Y. (c) For every vector lattice cover (Y, i) of X there exists a monotone seminorm p1 on Y such that p1 ∘ i = p. (d) There exists a monotone seminorm p1 on the Riesz completion (X ρ , i) of X such that p1 ∘ i = p. For symmetrically monotone seminorms and seminorms with a full unit ball, there is an extension result similar to Theorem 3.3.15. Theorem 3.3.17. Let (X, K X ) and (Y, K Y ) be directed partially ordered vector spaces such that there exists a bipositive linear map i : X → Y with i[X] majorizing in Y. Let p be a seminorm on X. Define for y ∈ Y, p2 (y) := inf{p(x) + p(u); x, u ∈ X, −i(u) ≤ y − i(x) ≤ i(u)} , p3 (y) := inf{p(u) ∨ p(v); u, v ∈ X, i(u) ≤ y ≤ i(v)}, and let p1 be as in Theorem 3.3.15. (i) For every y ∈ K Y we have p2 (y) = inf{p(w); w ∈ X, i(w) ≥ y} and p3 (y) = inf{p(w); w ∈ X, i(w) ≥ y}. (ii) p2 is the greatest symmetrically monotone seminorm on X with p2 ∘ i ≤ p. (iii) p2 ∘ i = p if and only if p is symmetrically monotone. (iv) p3 is the greatest seminorm with a full unit ball on X with p3 ∘ i ≤ p. (v) p3 ∘ i = p if and only if p has a full unit ball. (vi) p3 ≤ p2 ≤ p1 ≤ 2p2 and p1 ≤ 3p3 . In particular, p1 , p2 , and p3 are equivalent. (vii) If p is monotone, then p ≤ 32 p1 ∘ i, p ≤ 2p2 ∘ i, and p ≤ 3p3 ∘ i. Proof. Many parts of the proof are similar to the arguments given in the proof of Theorem 3.3.15. We will only point out the differences. (i) Let y ∈ K Y . If w ∈ K X is such that i(w) ≥ y, then −i(w) ≤ y − 0 ≤ i(w) and 0 ≤ y ≤ i(w), so p2 (y) ≤ p(w) and p3 (y) ≤ p(w). Hence, p2 (y) ≤ inf{p(w); w ∈ K X , i(w) ≥ y} and p3 (y) ≤ inf{p(w); w ∈ K X , i(w) ≥ y}. For the converse inequality for p2 , let u, x ∈ X be such that −i(u) ≤ y − i(x) ≤ i(u). Then 0 ≤ y ≤ i(x + u), so p(x) + p(u) = p(x + u) ≥ inf{p(w); w ∈ K X , i(w) ≥ y}. Hence, p2 (y) ≥ inf{p(w); w ∈ K X , i(w) ≥ y}. For the converse inequality for p3 , let u, v ∈ X be such that i(u) ≤ y ≤ i(v). Then p(u) ∨ p(v) ≥ p(v) ≥ inf{p(w); w ∈ K X , i(w) ≥ y}. Hence, p3 (y) ≥ inf{p(w); w ∈ K X , i(w) ≥ y}. (ii) The proof that p2 and p3 are seminorms is analogous to the proof that p1 is a seminorm in Theorem 3.3.15 and therefore omitted. We show that p2 is symmetrically monotone. Let y ∈ Y and a ∈ Y be such that −a ≤ y ≤ a. If u ∈ X is such that a ≤ i(u), then −i(u) ≤ y ≤ i(u), so that

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(iii)

(iv)

(v)

(vi)

(vii)

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p1 (y) ≤ p(u). Taking infimum over u and applying (i) yields that p2 (y) ≤ p2 (a). Thus, p2 is symmetrically monotone. To see that p2 ∘ i ≤ p, let y ∈ X and take x = y and u = 0, then −i(u) ≤ i(y) − i(x) ≤ i(u), so p2 (i(y)) ≤ p(x) + p(u) = p(y). To complete the proof of (ii), let q be a symmetrically monotone seminorm on Y with q ∘ i ≤ p. We show that q ≤ p2 . Indeed, let y ∈ Y and x, u ∈ X be such that −i(u) ≤ y − i(x) ≤ i(u). Then q(y) ≤ q(i(x)) + q(y − i(x)) ≤ q(i(x)) + q(i(u)) ≤ p(x) + p(u). Hence, q(y) ≤ p 2 (y). If p2 ∘ i = p, then Proposition 3.3.14 yields that p is symmetrically monotone. Now assume that p is symmetrically monotone. Let y ∈ X and let x, u ∈ X be such that −i(u) ≤ i(y) − i(x) ≤ i(u). Then p(y) ≤ p(x) + p(u), so p(y) ≤ p2 (i(y)). Hence, p2 ∘ i = p. For a proof that the unit ball of p3 is full, let a, b ∈ Y be such that p3 (a) ≤ 1 and p3 (b) ≤ 1 and let y ∈ Y be such that a ≤ y ≤ b. We have to show that p3 (y) ≤ 1. For every ε > 0 there are u 1 , u 2 , v1 , v2 ∈ X such that i(u 1 ) ≤ a ≤ i(v1 ), i(u 2 ) ≤ b ≤ i(v2 ), and p(u 1 ), p(u 2 ), p(v1 ), p(v2 ) < 1 + ε. Then i(u 1 ) ≤ a ≤ y ≤ b ≤ i(v2 ), so p3 (y) ≤ p(u 1 ) ∨ p(v2 ) < 1 + ε. Hence, p3 (y) ≤ 1. For x ∈ X the choice u = v = x readily yields that p3 (i(x)) ≤ p(u) ∨ p(v) = p(x), so that p3 ∘ i ≤ p. If q is a seminorm with a full unit ball and q ∘ i ≤ p, then for every y ∈ Y and u, v ∈ X with i(u) ≤ y ≤ i(v) we have with the aid of Proposition 3.3.3 that q(y) ≤ q(i(u)) ∨ q(i(v)), so q(y) ≤ p3 (y). Hence, q ≤ p3 . Assume that p3 ∘ i = p. According to Proposition 3.3.14, it follows that p has a full unit ball. For the converse implication, assume that p has a full unit ball. If x ∈ X and u, v ∈ X are such that i(u) ≤ i(x) ≤ i(v), then p(x) ≤ p(u)∨p(v), so p(x) ≤ p3 (i(x)). Combination with (iv) yields that p = p3 ∘ i. Proposition 3.3.4 says that every seminorm with a full unit ball is symmetrically monotone and every symmetrically monotone seminorm is monotone*. Therefore Theorem 3.3.15 (ii) and (ii) and (iv) above yield that p3 ≤ p2 ≤ p1 . To prove p1 ≤ 2p2 , let y ∈ Y and x, u ∈ X be such that −i(u) ≤ y − i(x) ≤ i(u). Then p1 (y) ≤ p(x) + 2p(u) ≤ 2(p(x) + p(u)), hence p1 (y) ≤ 2p2 (y). For a proof that p1 ≤ 3p3 , let y ∈ Y and u, v ∈ X be such that i(u) ≤ y ≤ i(v). Then 0 ≤ y − i(u) ≤ i(v − u), so p1 (y) ≤ p(u) + p(v − u) ≤ 3(p(u) ∨ p(v)). Thus, p1 (y) ≤ 3p3 (y). Assume that p is monotone. Let x, y ∈ X, u, v ∈ K X be such that −u ≤ y − x ≤ v. Then 0 ≤ y − x + u ≤ u + v and −(u + v) ≤ y − x − v ≤ 0, so p(2y − 2x) ≤ p(y − x + u) + p(y − x − v) + p(v − u) ≤ 2p(u + v) + p(v − u) ≤ 3p(u) + 3p(v), hence p(y) ≤ p(x) + 32 (p(u) + p(v)) ≤ 32 (p(x) + p(u) + p(v)). Thus, p(y) ≤ 32 p1 (i(y)). For a proof of the second inequality, note that if x, u ∈ X are such that −u ≤ x ≤ u, then p(2x) ≤ p(x + u) + p(x − u) ≤ 4p(u), hence p(x) ≤ 2p2 (i(x)).

148 | 3 Seminorms on pre-Riesz spaces For a proof of the third inequality, let x, u, v ∈ X be such that u ≤ x ≤ v. Then 0 ≤ x − u ≤ v − u, so p(x) ≤ p(u)+p(v−u) ≤ 3(p(u)∨p(v)), hence p(x) ≤ 3p 2 (i(x)).

Corollary 3.3.18. Consider the setting of Theorem 3.3.17. (i) If p is a monotone* seminorm on X, then p1 is the greatest monotone* seminorm on Y extending p. (ii) If p is a symmetrically monotone seminorm on X, then p2 is the greatest symmetrically monotone seminorm on Y extending p. (iii) If the seminorm p on X has a full unit ball, then p3 is the greatest seminorm with a full unit ball on Y extending p. A combination of Proposition 3.3.14 and Corollary 3.3.18 gives the restriction and extension property for symmetrically monotone seminorms and seminorms with full unit balls. Theorem 3.3.19. Let X be a pre-Riesz space and let p be a seminorm on X. The following statements are equivalent. (a) p is symmetrically monotone (or has a full unit ball, respectively). (b) For every Riesz space Y such that X is a majorizing subspace of Y the seminorm p extends to a symmetrically monotone seminorm on Y (a seminorm with full unit ball on Y, respectively). (c) For every vector lattice cover (Y, i) of X there exists a symmetrically monotone seminorm p̄ on Y (a seminorm p̄ with full unit ball on Y, respectively) such that p̄ ∘ i = p. (d) There exists a symmetrically monotone seminorm p̄ (a seminorm p̄ with full unit ball, respectively) on the Riesz completion (X ρ , i) of X such that p̄ ∘ i = p. Given a monotone seminorm p on a partially ordered vector space (X, K), with Y = X and i : X → Y the identity map, Theorem 3.3.17 describes how to construct out of p a monotone* seminorm p1 , a symmetrically monotone seminorm p2 , and a seminorm p3 with a full unit ball on X. Corollary 3.3.20. Let (X, K) be a partially ordered vector space and let p be a monotone seminorm on X. Define for y ∈ X, p1 (y) := inf {p(x) + p(u) + p(v); x ∈ X, u, v ∈ K, −u ≤ y − x ≤ v} , p2 (y) := inf {p(x) + p(u); x, u ∈ X, −u ≤ y − x ≤ u} , p3 (y) := inf {p(u) ∨ p(v); u, v ∈ X, u ≤ y ≤ v} . Then p1 is a monotone* seminorm, p2 is a symmetrically monotone seminorm, and p3 is a seminorm with a full unit ball. The seminorms p1 , p2 , and p3 are equivalent to p and p1 , p2 , p3 ≤ p. The inequalities in Theorem 3.3.17 (vii) are sharp, as the next example shows.

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Example 3.3.21. We continue Example 3.3.12, which presents a monotone norm p on a partially ordered vector space (X, K) and elements y ∈ X, u, v ∈ K with −u ≤ y ≤ v, p(u) = p(v) = 1, and p(y) = 3. Let Y = X, let i : X → Y be the identity map, and let p1 , p2 , and p3 be as in Theorem 3.3.17. Then p1 (y) ≤ p(u) + p(v) ≤ 2 and p3 (y) ≤ p(u) ∨ p(v) = 1. This means that the inequalities p ≤ 32 p1 ∘ i and p ≤ 3p3 ∘ i are sharp. The inequality p ≤ 2p2 ∘ i is sharp if we take X = Y = C[0, 1], i : X → Y the identity map, and p(x) := ‖x+ ‖∞ + ‖x− ‖∞ , for x ∈ X. Indeed, if we define x(t) := 2t − 1 and u(t) := 1 for t ∈ [0, 1], then −u ≤ x ≤ u, p(x) = 2, and p(u) = 1, so that p2 (x) ≤ p(u) = 1. A monotone norm may have monotone seminorm extensions none of which are norms, as the next example shows. Example 3.3.22. We provide a majorizing Riesz subspace X of a Riesz space Y and a monotone norm p on X such that every monotone seminorm on Y that extends p is not a norm. Let Y = B[0, 1] and let X = C[0, 1]. Let p = ‖⋅‖1 on X. Then X is a majorizing Riesz subspace of the Riesz space Y and p is a monotone (even a Riesz) norm on X. Since every monotone seminorm on Y is monotone* due to Proposition 3.3.6, the greatest monotone seminorm on Y that extends p is the seminorm p1 given by Theorem 3.3.15. If we let y ∈ Y be given by y(t) = 0 for t ∈ (0, 1] and y(0) = 1, then p1 (y) = 0, so that p1 is not a norm on Y. Because p1 is the greatest monotone seminorm on Y extending p, no monotone seminorm on Y extending p is a norm. If a subspace X of a partially ordered vector space Y is not majorizing in Y then the existence of a monotone seminorm on Y extending a monotone seminorm on X may fail. Example 3.3.23. We provide a Riesz space Y with a Riesz subspace X and a Riesz norm on X that cannot be extended to a monotone seminorm on Y. Let Y = ℓ∞ (ℕ), X = cc (ℕ), and p(x) := ∑∞ n=1 |x n |, for x ∈ X. Then Y is a Riesz space, X is a Riesz subspace, and p is a Riesz norm on X. Suppose that q is a monotone seminorm on Y extending p. Consider x n , y ∈ Y given by x n (k) = 1 for k ∈ {1, . . . , n}, x n (k) = 0 for k ∈ ℕ with k ≥ n + 1, and y = 𝟙. For every n ∈ ℕ we have 0 ≤ x n ≤ y, so q(y) ≥ q(x n ) = p(x n ) = n, which is a contradiction. Hence, there is no monotone seminorm on Y extending p. The next example is merely an illustration of the extension formulas of Theorem 3.3.17. Example 3.3.24. Let X be a linear subspace of C[0, 1] containing the function 𝟙 and let Y be a linear subspace of C[0, 1] containing X. Let p be the restriction of the supremum norm on C[0, 1] to X and let ρ be the restriction of the supremum norm to Y. (i) The greatest symmetrically monotone seminorm p2 on Y extending p equals ρ. Indeed, ρ is a symmetrically monotone seminorm on Y, so p2 ≥ ρ. To see that p2 ≤ ρ, let y ∈ Y. Take u := ρ(y)𝟙. Then −u ≤ y ≤ u, so p2 (y) ≤ p(u) = ρ(y). Hence, p2 = ρ.

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(ii) The greatest seminorm p3 on Y extending p with a full unit ball equals ρ. Indeed, ρ has a full unit ball, so p3 ≥ ρ. For y ∈ Y, just as in (ii), u := ρ(y)𝟙 satisfies −u ≤ y ≤ y and therefore p3 (y) ≤ p(−u) ∨ p(u) = ρ(y). Hence, p3 = ρ. (iii) If X consists of the constant functions, the greatest monotone* seminom p1 on Y extending p is such that for every y ∈ Y we have p1 (y) = ‖y+ ‖∞ + ‖y− ‖∞ .

(3.7)

Indeed, y 󳨃→ ‖y+ ‖∞ +‖y− ‖∞ is a monotone* seminorm on Y extending p, so p1 (y) ≥ ‖y+ ‖∞ + ‖y− ‖∞ . For y ∈ Y, take u = ‖y+ ‖∞ 𝟙 and v := ‖y− ‖∞ 𝟙. Then −v ≤ y ≤ u, so p1 (y) ≤ p(u) + p(v), which yields formula (3.7).

3.3.3 Extension and norm completeness If Y is a Riesz space with a Riesz norm ρ such that Y is ρ-complete, then it is known that the Dedekind completion Y δ is complete for the greatest Riesz norm on Y δ extending ρ; see Lemma 3.2.1. We will explore similar results for extensions of monotone seminorms on partially ordered vector spaces. We begin by an example which shows that norm completeness may get lost if we extend the norm to the Riesz completion. Example 3.3.25. Let X be the subspace P2 [0, 1] of C[0, 1] consisting of all polynomial functions of degree at most two as in Example 1.7.2, endowed with the supremum norm p. As the supremum norm is an order unit norm, its unit ball is full. Moreover, X is p-complete as X is finite-dimensional. The Riesz completion X ρ of X is the subspace of C[0, 1] consisting of all piecewise polynomial functions of degree at most two. Note that X ρ is not complete with respect to the supremum norm. We show that the greatest monotone* seminorm p1 on X ρ extending p is equivalent to the supremum norm, which means that X ρ is not p1 -complete. Indeed, let q be the restriction of the supremum norm to the space of constant functions and let q1 be the greatest monotone* seminorm on X ρ extending q. Then q1 (y) ≥ p1 (y) ≥ ‖y‖∞ for every y ∈ X ρ . According to Example 3.3.24 (iii), for y ∈ X ρ we have q1 (y) = ‖y+ ‖∞ + ‖y− ‖∞ ≤ 2‖y‖∞ . Hence, p1 is equivalent to the supremum norm on X ρ . With the aid of Example 3.3.24 we see that the situation does not improve if we consider the greatest symmetrically monotone seminorm p2 on X ρ extending p or the greatest seminorm p3 with a full unit ball, as p2 and p3 both equal the supremum norm. The situation of Example 3.3.25 is typical for many examples of pre-Riesz spaces. If we consider a pre-Riesz space X consisting of a certain type of polynomial or differentiable functions, then the Riesz completion X ρ consists of functions that are piecewise of that type, and due to the Stone–Weierstrass theorem such a space X ρ will typically be norm dense for the supremum norm and therefore not complete. A much better choice with regard to norm completeness would be the space C[0, 1] as vector lattice cover of X rather than its subspace X ρ . Note that according to Corollary 2.5.10 the space

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C[0, 1] is the uniform completion of X ρ . In the more general setting of Archimedean pre-Riesz spaces, we will consider a uniformly complete vector lattice cover, which exists due to Corollary 2.4.7. Before stating a result on norm completeness for extensions of seminorms, we consider two preliminary statements. Lemma 3.3.26. Let (X, K) be a partially ordered vector space and let p be a monotone seminorm on X. Let (x i )i∈I be a net in X and let x ∈ X. If (x i )i∈I converges relatively uniformly to x, then p(x i − x) → 0. Proof. Let u ∈ K, i0 ∈ I, and (λ i )i∈I≥i0 in [0, ∞) with inf{λ i ; i ∈ I≥i 0 } = 0 be such that −λ i u ≤ x i − x ≤ λ i u for every i ≥ i0 . Then 0 ≤ (x i − x) + λ i u ≤ 2λ i u, so p(x i − x) ≤ p(x i − x + λ i u) + p(λ i u) ≤ p(2λ i u) + p(λ i u) = 3λ i p(u) , which converges to 0. Let X be a vector space and let p be a seminorm on X. For a sequence (x n )n in X we say N that the series ∑∞ n=1 x n is p-convergent if there exists s ∈ X such that p(∑ n=1 x n −s) → 0 as N → ∞. Lemma 3.3.27. Let X be a vector space and let p be a seminorm on X. If for every se∞ n quence (x n )n in X with ∑∞ n=1 2 p(x n ) < ∞ the series ∑ n=1 x n is p-convergent, then X is p-complete. Proof. Let (x n )n be a p-Cauchy sequence. The main step of the proof is to construct a p-convergent subsequence of (x n )n . For every ε > 0 there exists N ∈ ℕ such that for every n, m ≥ N we have p(x n − x m ) < ε. Choose n1 ∈ ℕ such that for every n, m ≥ n1 we have p(x n − x m ) < 4−1 . Choose n2 > n1 such that for every n, m ≥ n2 we have p(x n − x m ) < 4−2 . Inductively, choose a strictly increasing sequence (n k )k in ℕ such that for every k ∈ ℕ and every n, m ≥ n k we have p(x n − x m ) < 4−k . Then for every k k ∈ ℕ it follows that p(x n k+1 − x n k ) < 4−k . Hence, ∑∞ k=1 2 p(x n k+1 − x n k ) < ∞ and therefore, by assumption, there exists s ∈ X such that p(∑ Nk=1 (x n k+1 − x n k ) − s) → 0 as N → ∞. The latter means that p(x n N+1 − (s + x1 )) → 0 as N → ∞, so that the subsequence (x n k )k p-converges to x := s + x1 . To complete the proof, let ε > 0. Take N1 , N2 ∈ ℕ such that for every n, m ≥ N1 one has p(x n − x m ) < 12 ε and for every k ≥ N2 one has p(x n k − x) < 12 ε. Then for every k ≥ max{N1 , N2 }, we obtain p(x k − x) ≤ p(x k − x n k ) + p(x n k − x) < ε . Thus, (x n )n is p-convergent. It follows that X is p-complete. Theorem 3.3.28. Let (X, K X ) and (Y, K Y ) be directed partially ordered vector spaces such that there exists a bipositive linear map i : X → Y with i[X] majorizing in Y. Assume that Y is uniformly complete. Let p be a monotone* norm on X such that K X is closed.

152 | 3 Seminorms on pre-Riesz spaces Let p1 be the greatest monotone* seminorm on Y with p1 ∘ i = p. If X is p-complete, then Y is p1 -complete. Proof. Due to Lemma 3.3.27, it suffices to prove p1 -convergence of ∑∞ n=1 y n for every n p (y ) < ∞. So, let (y ) be such a sequence. From sequence (y n )n in Y with ∑∞ 2 1 n n n n=1 the formula for p1 in Theorem 3.3.15 it follows that there exist (x n )n in X and (u n )n , (v n )n in K X such that −i(u n ) ≤ y n − i(x n ) ≤ i(v n ) and p(x n ) + p(u n ) + p(v n ) < p1 (y n ) + n 4−n for every n. Then ∑∞ n=1 2 (p(x n ) + p(u n ) + p(v n )) < ∞. Since X is p-complete, the ∞ n n series ∑n=1 2 (u n + v n ) p-converges in X. Define w := ∑∞ n=1 2 (u n + v n ). As K X is closed, we have that w ∈ K X and, for every n ∈ ℕ, n

i(w) ≥ ∑ 2k (i(u k ) + i(v k )) ≥ 2n (i(u n ) + i(v n )) ≥ 2n (y n − i(x n )) . k=1

Similarly, i(w) ≥ −2n (y n − i(x n )). Then −2−n i(w) ≤ y n − i(x n ) ≤ 2−n i(w). Hence, for every N, M ∈ ℕ with M < N we have N

−2−M i(w) ≤ ∑ (y n − i(x n )) ≤ 2−M i(w) , n=M+1

i.e., the sequence of partial sums is relatively uniformly Cauchy. The uniform completeness of Y yields that the series ∑∞ n=1 (y n − i(x n )) is relatively uniformly convergent in Y. By Lemma 3.3.26, the series ∑∞ n=1 (y n − i(x n )) is then p1 -convergent, as p1 is a ∞ monotone seminorm. We also have that ∑∞ n=1 x n is p-convergent, since ∑ n=1 p(x n ) < ∞ ∞ and X is p-complete. As p1 ∘ i = p, it follows that ∑n=1 i(x n ) is p1 -convergent. We conclude that ∑∞ n=1 y n is p1 -convergent. Recall that the seminorm p1 in Theorem 3.3.28 need not be a norm; see Example 3.3.22. Similar statements as in Theorem 3.3.28 for a symmetrically monotone norm or a norm with a full unit ball are contained in the next corollary. Corollary 3.3.29. Let (X, K X ) and (Y, K Y ) be directed partially ordered vector spaces such that there exists a bipositive linear map i : X → Y with i[X] majorizing in Y. Assume that Y is uniformly complete. (i) If p is a symmetrically monotone norm on X such that K X is closed and X is p-complete and if q is the greatest symmetrically monotone seminorm on Y with q ∘ i = p, then Y is q-complete. (ii) If p is a norm on X with a full unit ball such that K X is closed and X is p-complete and if q is the greatest seminorm on Y with a full unit ball such that q ∘ i = p, then Y is q-complete. Proof. (i) The greatest symmetrically monotone extension q of p is equivalent to the greatest monotone* extension p1 of p, due to Theorem 3.3.17. Hence, Y is q-complete if and only if it is p1 -complete. Therefore, Theorem 3.3.28 yields that Y is q-complete.

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(ii) In this case Theorem 3.3.17 also says that q and p1 are equivalent and similar arguments as in (i) complete the proof. A particular case of Theorem 3.3.28 is the case where (Y, i) is a uniformly complete vector lattice cover of X. Corollary 3.3.30. Let (X, K) be an Archimedean pre-Riesz space and let (Y, i) be a uniformly complete vector lattice cover of X. Let p be a monotone* norm (or symmetrically monotone norm, or a norm with a full unit ball) on X such that K is closed. Let q be the greatest monotone* seminorm (or symmetrically monotone seminorm, or seminorm with a full unit ball) on Y such that q ∘ i = p. If X is p-complete, then Y is q-complete. According to Proposition 1.5.41, Dedekind complete spaces are uniformly complete. Therefore, Corollary 3.3.30 holds in particular for the Dedekind completion (X δ , i) of X.

3.3.4 Uniqueness of extensions Let us now consider uniqueness of extensions of monotone seminorms, or rather nonuniqueness, as extensions are almost never unique. Since every positive linear functional generates a monotone seminorm by Proposition 3.3.9 (ii), existence of distinct monotone seminorms is closely related to existence of distinct positive linear functionals. Accordingly, the main tools to demonstrate nonuniqueness are Hahn– Banach theorems 1.8.1 and 1.8.2 or Mazur’s Theorem 1.8.3 to obtain linear functionals, and a result by Grosberg and Krein (Corollary 5.4.11 below) to split them into positive linear functionals. Let us start with an explicit example. Example 3.3.31. We provide a Riesz space X and an order dense Riesz subspace D of X with a Riesz norm on D, which can be extended to inequivalent Riesz norms on X. Take X = C[0, 1] and D = {x ∈ X; x(0) = x(1)}. Then X is a Riesz space and D is an order dense Riesz subspace of X. Define for x ∈ X, ‖x‖ := sup {|x(t)|t(1 − t); t ∈ [0, 1]} and define p(x) := ‖x‖ ∨ |x(0)| and q(x) := ‖x‖ ∨ |x(1)|. The maps p and q are Riesz norms on X. On the subspace D the norms p and q coincide, whereas on X they are not equivalent. A situation in which there is only one extension is the case where the subspace is norm dense with respect to the greatest extension. Proposition 3.3.32. Let X be a partially ordered vector space with a directed, majorizing subspace D and let p be a monotone* seminorm on D. Let p̄ be the greatest monō tone* seminorm on X extending p. If D is p-dense in X, then p̄ is the unique monotone* seminorm on X extending p.

154 | 3 Seminorms on pre-Riesz spaces Proof. Every monotone* seminorm q that extends p is such that q ≤ p̄ and, hence, ̄ continuous with respect to p.̄ As D is p-dense, we obtain q = p.̄ If a subspace D of a partially ordered vector space X with a norm p is not norm dense, then Mazur’s theorem can be used to find a nonzero continuous linear functional that vanishes on the closure of D. However, such a linear functional cannot always be chosen to be positive. If X and p are such that the norm dual of (X, p) is directed, then it is possible to find two distinct positive linear functionals on X that are equal on D. We thus obtain the following result on nonuniqueness of extensions. Theorem 3.3.33. Let (X, K) be a directed partially ordered vector space, and let D be a majorizing linear subspace of X. Let p be a monotone* seminorm on D and let p̄ be ̄ the greatest monotone* seminorm on X extending p. If D is not p-dense in X, then there exists a monotone* seminorm on D that is equivalent to p and which has two distinct monotone* extensions. ̄ Proof. Since D is not p-dense in X, by Mazur’s Theorem 1.8.3 there exists a nonzero ̄ p-continuous linear functional φ : X → ℝ with φ = 0 on D. Since X is directed, there is an x 0 ∈ K with φ(x0 ) ≠ 0. According to 3.3.20, the seminorm p̄ is equivalent to a symmetrically monotone seminorm. Then by the result of Grosberg–Krein ̄ as stated in Corollary 5.4.11 below, there are positive p-continuous linear functionals φ1 , φ2 : X → ℝ such that φ = φ1 − φ2 . It follows that φ1 = φ2 on D and φ1 (x0 ) ≠ ̄ ̄ φ2 (x0 ). For x ∈ X define q1 (x) := p(x) + |φ1 (x)| and q2 (x) := p(x) + |φ2 (x)|. Then ̄ q1 and q2 are monotone* seminorms on X. Since φ1 and φ2 are p-continuous, we have that q1 and q2 are both equivalent to p on D. Furthermore, q1 = q2 on D and q1 (x0 ) ≠ q2 (x0 ). Theorem 3.3.33 shows that a simple uniqueness theorem more general than Proposition 3.3.32 cannot be expected. It is not superfluous that the theorem involves an equivalent seminorm. The seminorm itself may have a unique extension, as is shown by the next example. Example 3.3.34. We provide a Riesz space X with a majorizing linear subspace D and a monotone* seminorm p on D such that there is only one monotone seminorm ̄ p̄ on X extending p and such that D is not p-dense in X. Take X := {x ∈ C[0, 1]; x is piecewise affine} with the standard pointwise order. Let D := {x ∈ X; φ(x) = 0}, where 1 2

1

φ(x) := ∫ x(t) dt − ∫ x(t) dt , 0

and for x ∈ X define

1 2

̄ p(x) := ‖x+ ‖∞ + ‖x− ‖∞ .

It is straightforward that X is a Riesz space and that p̄ is a monotone norm on X. Let p ̄ be the restriction of p̄ to D. We show that D is a p-closed majorizing linear subspace

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̄ of X and that D is not p-dense in X. Indeed, D is a linear subspace of X, 𝟙 ∈ D and ̄ 𝟙 is an order unit in X, hence D is majorizing. Note that p(x) ≥ ‖x‖∞ for every x ∈ X. ̄ The functional φ is ‖⋅‖∞ -continuous and therefore p-continuous, consequently D is ̄ ̄ p-closed. As φ ≠ 0, we obtain that D is not p-dense in X. Next we show that p̄ is the only monotone seminorm on X that extends p. Indeed, let q also be a monotone seminorm on X extending p. Since X is a Riesz space, ̄ the seminorm q is monotone*. Let x ∈ X. It has to be shown that q(x) = p(x). Since − + −‖x ‖∞ 𝟙 ≤ x ≤ ‖x ‖∞ 𝟙, we have ̄ ̄ q(x) ≤ (‖x− ‖∞ + ‖x+ ‖∞ ) q(𝟙) = p(x)p(𝟙) = p(x) . ̄ It remains to show that q(x) ≥ p(x). Before we proceed, we do some preparations. ̄ Define w(t) := 2 − 4t for t ∈ [0, 1]. Then w ∈ X, φ(w) = 1, and p(w) = 4. For any y ∈ X, ̄ − φ(y)w) = q(y − φ(y)w), hence observe that y − φ(y)w ∈ D, thus p(y ̄ ̄ − φ(y)w) + p(φ(y)w)) ̄ p(y) − q(y) ≤ (p(y − (q(y − φ(y)w) − q(φ(y)w)) ̄ ̄ = p(φ(y)w) + q(φ(y)w) ≤ 2|φ(y)|p(w) = 8|φ(y)| , and therefore ̄ p(y) ≤ q(y) + 8|φ(y)| .

(3.8)

1 ∫0

Also, for every u ∈ X with u ≥ 0 we have |φ(u)| ≤ u(t) dt. Now we proceed and distinguish three cases. If x ≥ 0, choose τ ∈ [0, 1] such that ̄ x(τ) = ‖x‖∞ = p(x). Let ε > 0. As x is piecewise affine, there exists a ‘narrow peak’ u ∈ X, u ≥ 0, such that 1

u(τ) = 1, 0 ≤ ‖x‖∞ u ≤ x,

and

̄ ∫ u(t) dt < ε . 8p(x) 0

̄ Then ‖ ‖x‖∞ u ‖∞ = ‖x‖∞ u(τ) = ‖x‖∞ = p(x). With the aid of (3.8) we get 1

̄ ∫ u(t) dt < q(x) + ε , ̄ p(x) = p̄ (‖x‖∞ u) ≤ q(‖x‖∞ u) + 8‖x‖∞ |φ(u)| ≤ q(x) + 8p(x) 0

̄ and thus p(x) ≤ q(x). ̄ If x < 0, apply the above to −x, then it follows that p(x) ≤ q(x). + − If x ≠ 0 and x ≠ 0, a somewhat more complicated strategy is needed to show ̄ that p(x) ≤ q(x). Let ε > 0. An element y ∈ X will be constructed such that |φ(y)| < 8ε , ̄ ̄ ̄ ̄ p(y) = 2p(x), and q(y) − q(x) ≤ p(x). Then it follows by (3.8) that q(x) ≥ q(y) − p(x) ≥ ̄ ̄ ̄ p(y) − 8|φ(y)| − p(x) ≥ p(x) − ε. To provide such a y, choose σ, τ ∈ [0, 1] such that x + (τ) = ‖x+ ‖∞ and x− (σ) = ‖x− ‖∞ . Note that τ ≠ σ. Then, because x is piecewise affine, there exist two positive narrow peaks u, v ∈ X such that 0 ≤ ‖x+ ‖∞ u ≤ x+ ,

0 ≤ ‖x− ‖∞ v ≤ x− ,

(3.9)

156 | 3 Seminorms on pre-Riesz spaces 1

1

̄ ∫0 u(t) dt < ε, and 16p(x) ̄ ∫0 v(t) dt < ε. Note that u and v are u(τ) = u(σ) = 1, 16p(x) disjoint and, more than that, x+ and v are disjoint and, similarly, u and x− are disjoint. Define ̄ y := p(x)(u − v) . We check that y has the desired properties. Clearly, ̄ (|φ(u)| + |φ(v)|) < p(x) ̄ 162ε ≤ |φ(y)| ≤ p(x) ̄ p(x)

ε . 8

̄ ̄ ̄ Since u and v are disjoint, we have p(y) = ‖y+ ‖∞ + ‖y− ‖∞ = p(x)(u(τ) + v(σ)) = 2p(x). − Formula (3.9) and disjointness of u and x yield that ̄ y − x ≤ p(x)u − x = ‖x+ ‖∞ u − x+ + ‖x− ‖∞ u + x− ≤ ‖x− ‖∞ u + x− = (‖x− ‖∞ u) ∨ x− ≤ ‖x− ‖∞ 𝟙 . Similarly, y − x ≥ − ((‖x+ ‖∞ v) ∨ x+ ) ≥ −‖x+ ‖∞ 𝟙 . ̄ ̄ Because q is monotone*, it follows that q(y − x) ≤ p(x)q(𝟙) = p(x), hence q(y) − q(x) ≤ ̄ ̄ p(x). As noted above we obtain q(x) ≥ p(x) − ε, which completes the example. The uniqueness of extensions of pre-Riesz seminorms is similar to the case of monotone seminorms as given in Proposition 3.3.32 and Theorem 3.3.33. Recall that for a given pre-Riesz seminorm p on a majorizing subspace D of a directed partially ordered vector space X there exists a greatest pre-Riesz seminorm ρ on X extending p; see Theorem 3.2.27. Proposition 3.3.35. Let (X, K) be a Riesz space with a majorizing linear subspace D. Let p be a pre-Riesz seminorm on D and let ρ be the greatest Riesz seminorm on X extending p. (i) If D is ρ-dense in X, then ρ is the unique Riesz seminorm on X extending p. (ii) If D is a Riesz subspace of X that is not ρ-dense in X, then there is a Riesz seminorm on D, which is equivalent to p and which can be extended to distinct Riesz seminorms on X. Proof. (i) Every Riesz seminorm on X extending p is ρ-continuous. If D is ρ-dense in X, then all Riesz seminorms on X extending p coincide. (ii) As a consequence of Mazur’s Theorem 1.8.3, there exist a ρ-continuous linear functional φ : X → ℝ and x0 ∈ K such that φ(x0 ) ≠ 0 and φ = 0 on D. The functional φ is order-bounded, since for every a, b ∈ X with a ≤ b and a ≤ x ≤ b we have 0 ≤ x+ , x− ≤ |a| ∨ |b|, hence |φ(x)| = |φ(x+ ) − φ(x− )| ≤ ‖φ‖(ρ(x+ ) + ρ(x− )) ≤ 2‖φ‖ρ(|a| ∨ |b|) . Due to the Riesz–Kantorovich theorem 1.2.6, the positive and negative parts φ+ and φ− of φ exist. Moreover, the Riesz–Kantorovich formulas yield that φ+ and

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φ− are ρ-continuous. For x ∈ X define ρ 1 (x) := ρ(x) + φ+ (|x|) and ρ 2 (x) := ρ(x) + φ− (|x|). Then ρ 1 and ρ 2 are Riesz seminorms on X and they are equivalent to ρ, as φ+ and φ− are ρ-continuous. Since φ+ = φ− on D, we have that ρ 1 = ρ 2 on D. Further, φ+ (x0 ) ≠ φ− (x0 ), therefore ρ 1 (x0 ) ≠ ρ 2 (x0 ).

3.3.5 Quotients over kernels of monotone seminorms The kernel of a monotone seminorm on a partially ordered vector space can be factored out. The quotient space is a partially ordered vector space and the quotient norm is monotone. Let us summarize some properties of this construction. For a vector space X and a seminorm p on X, the set N := {x ∈ X; p(x) = 0} is a linear subspace. Let X̂ := X/N be the quotient space and let q : X → X̂ be the quotient map; see also (1.3). The map p̂ : X̂ → [0, ∞), [x] 󳨃→ p(x) is well-defined. It is straightforward that p̂ is a norm. We call p̂ the quotient norm. If K is a cone in X, we denote K̂ = q[K]. Proposition 3.3.36. Let (X, K) be a partially ordered vector space with a monotone seminorm p and let N := {x ∈ X; p(x) = 0}. Let X,̂ K̂ and q be as above and let p̂ be the quotient norm. (i) (X,̂ K)̂ is a partially ordered vector space. (ii) p̂ is monotone. (iii) If p is symmetrically monotone, then p̂ is symmetrically monotone. (iv) If p has a full unit ball, then p̂ has a full unit ball. (v) If X is a Riesz space and p is a Riesz seminorm, then (X,̂ K)̂ is a Riesz space and p̂ is a Riesz norm. Proof. (i) The subspace N is full. Indeed, if a, b ∈ N and x ∈ X are such that a ≤ x ≤ b, then 0 ≤ x − a ≤ b − a, so p(x − a) ≤ p(b − a) ≤ p(a) + p(b) = 0, hence p(x) ≤ p(x − a) + p(a) ≤ 0, so that x ∈ N. According to Proposition 1.1.13, the set K̂ is then a cone and the quotient map q is positive. (ii) To see that p̂ is monotone, let a, b ∈ X̂ be such that 0 ≤ a ≤ b. Then there are ̂ x, u ∈ K with q(x) = a and q(u) = b − a, hence q(x + u) = b. Therefore, p(a) = ̂ p(x) ≤ p(x + u) = p(b). Thus, p̂ is monotone. (iii) Let p be symmetrically monotone. Let a, b ∈ K.̂ Then there are x, y ∈ K with ̂ − b) = p(x − y) ≤ p(x + y) = p(a ̂ + b), so q(x) = a and q(y) = b. We have that p(a ̂p is symmetrically monotone. (iv) Let p have a full unit ball. Let a, b, c ∈ X̂ be such that b ≤ a ≤ c. Then there are y ∈ X and u, v ∈ K such that q(y) = b, q(y + u) = a, and q(y + u + v) = c, so that ̂ ̂ ̂ p(a) = p(y + u) ≤ p(y) ∨ p(y + u + v) = p(b) ∨ p(c). Hence, p̂ has a full unit ball. (v) Let X be a Riesz space and p a Riesz seminorm. Then N is an ideal in X. Indeed, if a ∈ N and x ∈ X are such that |x| ≤ |a|, then p(x) = p(|x|) ≤ p(|a|) = p(a) = 0, so x ∈ N. It is a standard result in the theory of Riesz spaces that the quotient space

158 | 3 Seminorms on pre-Riesz spaces (X,̂ K)̂ is then a Riesz space; see, e.g., [111, Theorem 18.9]. Moreover, the quotient ̂ ̂ map q is a Riesz homomorphism. For x ∈ X we have p(|q(x)|) = p(q(|x|)) = p(|x|) = ̂ p(x) = p(q(x)). By (ii) the norm p̂ is a Riesz norm. If a partially ordered vector space is norm complete with respect to a monotone seminorm with kernel N, then the quotient with respect to N is complete relative to the quotient norm. Norm completeness of the cone results in closedness of the quotient cone. These assertions in Proposition 3.3.38 below are consequences of the following observation. Lemma 3.3.37. Let X be a vector space with a seminorm p. Let N := {x ∈ X; p(x) = 0}, let q : X → X/N be the quotient map, and let p̂ be the quotient norm. ̂ (i) q is an isometry in the sense that for every x ∈ X we have p(q(x)) = p(x). ̂ (ii) Let S be a subset of X. Then S is p-complete if and only if q[S] is p-complete. Proof. The definition of p̂ implies (i). Hence, a sequence (x n )n in X is p-Cauchy if and ̂ only if (q(x n ))n is p-Cauchy. Moreover, for a sequence (x n )n in X and x ∈ X we have ̂ p(x n − x) → 0 if and only if p(q(x n ) − q(x)) → 0. Thus, the assertion in (ii) follows. Proposition 3.3.38. Let (X, K) be a partially ordered vector space with a monotone seminorm p. Let N := {x ∈ X; p(x) = 0}. ̂ (i) X is p-complete if and only if X/N is p-complete. ̂ (ii) K is p-complete if and only if q[K] is p-complete. ̂ ̂ (iii) If K is p-complete, then q[K] is p-closed in X/N. If q[K] is p-closed in X/N and X is p-complete, then K is p-complete. Proof. We obtain (i) and (ii) by applying Lemma 3.3.37 to S = X and S = K, respectively. ̂ It remains for us to show (iii). If K is p-complete, then q[K] is p-complete, hence closed ̂ If q[K] is closed in (X/N, p)̂ and X/N is p-complete, ̂ in the normed space (X/N, p). then ̂ q[K] is p-complete, so that K is p-complete. If p is a monotone seminorm on a directed partially ordered vector space X with kernel N and p̂ is the quotient norm, then the norm duals of (X, p) and (X/N, p)̂ are norm and order isomorphic, as we show next. Proposition 3.3.39. Let X be a directed partially ordered vector space with a monotone seminorm p and let N := {x ∈ X; p(x) = 0}. Let X/N be endowed with the quotient order and the quotient norm p,̂ and let q : X → X/N be the quotient map. Then the adjoint map q󸀠 : (X/N, p)̂ 󸀠 → (X, p)󸀠 is a bipositive linear surjective isometry. Proof. By Proposition 1.1.13 the quotient space is a directed partially ordered vector space. Due to Proposition 1.2.2 the norm dual spaces (X, p)󸀠 and (X/N, p)̂ 󸀠 are partially ordered vector spaces. The adjoint map of q is for every f ∈ (X/N, p)̂ 󸀠 defined by q󸀠 (f) := f ∘ q. It is straightforward that q󸀠 is linear and surjective. To see that q󸀠 is an isometry, ̂ = p(x). let f ∈ (X/N, p)̂ 󸀠 . For every x ∈ X we have |(q󸀠 (f))(x)| = |f(q(x))| and p(q(x))

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Hence 󵄨 󵄨 󵄩󵄩 󸀠 󵄩󵄩 󵄩󵄩 q (f)󵄩󵄩 = sup {󵄨󵄨󵄨(q󸀠 (f))(x)󵄨󵄨󵄨 ; p(x) ≤ 1} = sup {|(f(q(x))| ; p(q(x)) ̂ ≤ 1} = ‖f‖ . 󵄨 󵄩 󵄨 󵄩 For every positive f ∈ (X/N, p)̂ 󸀠 , the functional q󸀠 (f) is the composition of the two positive maps q and f , so therefore q󸀠 (f) is positive. Thus, q󸀠 is positive. To prove that q󸀠 is bipositive, let f ∈ (X/N, p)̂ 󸀠 be such that q󸀠 (f) is a positive linear functional on X. Then for every a ∈ q[K] there is an x ∈ K with q(x) = a, so that f(a) = f(q(x)) = (q󸀠 (f))(x) ≥ 0. Hence, f is positive and thus q󸀠 is bipositive. We proceed by showing that every directed partially ordered vector space with a monotone norm is a quotient of cc (S) with the standard order for some suitably chosen monotone seminorm and some set S. It entails that some properties of an ordered space may get lost by taking a quotient. For instance, the space cc (S) is a Dedekind complete Riesz space, whereas X could be any directed partially ordered vector space that can be endowed with a monotone norm. Proposition 3.3.40. Let (X, K, p) be a directed ordered normed space with a monotone norm. Then there exist a nonempty set S and a monotone seminorm p0 on cc (S) such that the quotient of cc (S) with respect to the kernel N := {x ∈ X; p0 (x) = 0} of p0 endowed with the quotient order and quotient norm is isomorphic to (X, K, p) as an ordered normed space. Proof. Take S := {s ∈ K; p(s) = 1} and let X0 := cc (S) with standard cone K0 . For x ∈ X0 , define j(x) := ∑ x(s)s s∈S

and p0 (x) := p(j(x)). Then j : X 0 → X is a positive linear map. Clearly, for every s ∈ S and x := 𝟙{s} ∈ X0 we have j(x) = s. As K is generating, the map j is surjective. The map p0 is a seminorm on X0 and it is monotone, since j is positive and p is monotone. Let N := {x ∈ X0 ; p0 (x) = 0} and let q : X0 → X0 /N be the quotient map. By Proposition 3.3.36 the space (X0 /N, q[K0 ], p0̂ ) is an ordered normed space with monotone norm. We will give a map i : X0 /N → X that is linear, bijective, bipositive, and such that ̂ p(i(a)) = p(a) for every a ∈ X0 /N. Note that if x, y ∈ X0 are such that q(x) = q(y), then p0 (x − y) = 0, so p(j(x − y)) = 0, hence j(x) = j(y). Therefore, for x ∈ X0 , we can define i(q(x)) := j(x). Thus, we obtain a map i : X0 /N → X. It is immediate that i is linear, positive, and surjective. To see that i is bipositive, let x ∈ X0 be such that j(x) ∈ K \ {0}. Put α := p(j(x)). Then s := 1α j(x) ∈ S and j(x) = αs = j(α𝟙{s} ), which means that q(x) = q(α𝟙{s} ) ∈ q[K0 ]. Hence, i is bipositive. For a proof that i is an isometry, let x ∈ X0 and observe that p(i(q(x))) = p(j(x)) = p0 (x) = p0̂ (q(x)).

160 | 3 Seminorms on pre-Riesz spaces

Proposition 3.3.36 states that the quotient norm of a monotone norm is monotone and that also symmetric monotonicity and a full unit ball are inherited by the quotient norm. It follows from Proposition 3.3.40 that the monotone* property may be lost, as is observed in the following example. Example 3.3.41. We give a Riesz space X with a monotone* seminorm p such that the quotient norm of X/{x ∈ X; p(x) = 0} is not monotone*. Take Y = ℝ3 with the norm and order of Example 3.3.12. Then Y is a directed partially ordered vector space and its norm is monotone, but not monotone*. Proposition 3.3.40 yields a nonempty set S and a monotone seminorm p on X := cc (S) such that X/{x ∈ X; p(x) = 0} is isomorphic to Y. The seminorm p is monotone on the Riesz space X, hence monotone*.

3.3.6 Extension to M-seminorms and L-seminorms In the theory of Banach lattices, an important role is played by M-norms and L-norms. A Riesz seminorm (or, norm) ρ on a Riesz space (X, X + ) is called an M-seminorm (M-norm) if ρ(x ∨ y) = ρ(x) ∨ ρ(y) for every x, y ∈ X+ , and an L-seminorm (L-norm) if ρ(x + y) = ρ(x) + ρ(y) for every x, y ∈ X+ . It is a natural question whether there are extension results for M-norms and L-norms similar to Lemma 3.2.1 for Riesz norms. We discuss when the greatest Riesz seminorm extending a pre-Riesz seminorm is an M-seminorm. It follows that the greatest extension of an M-norm is an M-norm. Proposition 3.3.42. Let (X, K) be a pre-Riesz space space with Riesz completion (X ρ , i) and let p be a pre-Riesz seminorm on X. The following statements are equivalent. (a) For every Riesz space Y that contains X as a majorizing linear subspace, the greatest Riesz seminorm on Y extending p is an M-seminorm. (b) The greatest Riesz seminorm on X ρ extending p is an M-seminorm. (c) For every x1 , . . . , x m ∈ X with p(x1 ), . . . , p(x m ) < 1 there exist y1 , . . . , y n ∈ X with p(y1 ) + ⋅ ⋅ ⋅ + p(y n ) < 1 such that u

n

{x1 , . . . , x m }u ⊇ { ∑ ε k y k ; ε1 , . . . , ε n ∈ {1, −1}} . k=1

Proof. The implication (a) ⇒ (b) is immediate, since i[X] is majorizing in X ρ . For a proof of (b) ⇒ (c), let x1 , . . . , x m ∈ X be such that p(x1 ), . . . , p(x m ) < 1. Let p̄ be the greatest Riesz seminorm on X ρ with p̄ ∘ i = p. Then ̄ ̄ ̄ p(|i(x 1 )| ∨ . . . ∨ |i(x m )|) = p(i(x 1 )) ∨ . . . ∨ p(i(x m )) = p(x 1 ) ∨ . . . ∨ p(x m ) < 1 . By the formula for p̄ given by Theorem 3.2.27, it follows that there exist y1 , . . . , y n ∈ X such that p(y1 ) + ⋅ ⋅ ⋅ + p(y n ) < 1 and i(|x1 | ∨ . . . ∨ |x m |) ≤ |i(y1 )| + ⋅ ⋅ ⋅ + |i(y n )|. The latter inequality implies n

u

{x1 , . . . , x m }u ⊇ { ∑ ε k y k ; ε1 , . . . , ε n ∈ {1, −1}} . k=1

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| 161

In order to show (c) ⇒ (a), let Y be a Riesz space that contains X as a majorizing linear subspace and let p̄ be the greatest Riesz seminorm on Y extending p. Let u, v ∈ Y + . ̄ ∨ v) ≥ p(u) ̄ Then u ∨ v ≥ u, v, so p(u ∨ p̄ (v), because p̄ is a Riesz seminorm. To ̄ ∨ v) ≤ p(u) ̄ ̄ prove that p(u ∨ p̄ (v), we assume that p(u), p̄ (v) < 1 and show that ̄ ∨ v) < 1. The general case then follows by consideration of 1α u and 1α v for every p(u ̄ α > p(u) ∨ p̄ (v). According to the formula for p̄ given by Theorem 3.2.27, there exist x1 , . . . , x l , x l+1 , . . . , x m ∈ X such that |u| ≤ |x1 | + ⋅ ⋅ ⋅ + |x l |,

|v| ≤ |x l+1 | + ⋅ ⋅ ⋅ + |x m | ,

p(x1 ) + ⋅ ⋅ ⋅ + p(x l ) < 1, and p(x l+1 ) + ⋅ ⋅ ⋅ + p(x m ) < 1. Then there exist λ1 , . . . , λ l , λ l+1 , . . . λ m ∈ ℝ such that λ1 + ⋅ ⋅ ⋅ + λ l < 1, λ l+1 + ⋅ ⋅ ⋅ + λ m < 1, and λ i > p(x i ) for every i ∈ {1, . . . , m}. By (c), there exist y1 , . . . , y n ∈ X such that p(y1 ) + ⋅ ⋅ ⋅ + p(y n ) < 1 and { λ11 x1 , . . . ,

u 1 λm xm }

n

u

⊇ { ∑ ε k y k ; ε1 , . . . , ε n ∈ {1, −1}} . k=1

Then |x i | ≤ λ i (|y1 | + ⋅ ⋅ ⋅ + |y n |) for every i ∈ {1, . . . , m}, so |u| ≤ (λ1 + ⋅ ⋅ ⋅ + λ l )(|y1 | + ⋅ ⋅ ⋅ + |y n |) ≤ |y1 | + ⋅ ⋅ ⋅ + |y n | , and, similarly, |v| ≤ |y1 | + ⋅ ⋅ ⋅ + |y n |. Hence, |u| ∨ |v| ≤ |y1 | + ⋅ ⋅ ⋅ + |y n |. This yields that ̄ n |) = p(y ̄ 1 ) + ⋅ ⋅ ⋅ + p(y ̄ n ) = p(y1 ) + ⋅ ⋅ ⋅ + p(y n ) < 1 , ̄ ̄ 1 |) + ⋅ ⋅ ⋅ + p(|y p(|u| ∨ |v|) ≤ p(|y which completes the proof. Corollary 3.3.43. Let X be a pre-Riesz space and let Y be a Riesz space that contains X as a majorizing linear subspace. Let p be a pre-Riesz seminorm on X and let p̄ be the greatest Riesz seminorm on Y extending p. (i) If the open unit ball of p is directed, then p̄ is an M-seminorm. (ii) If the open unit ball B of p is such that for every x1 , x2 ∈ B there exists a y ∈ B such that {x1 , x2 , −x1 , −x2 }u ⊇ {y, −y}u , then p̄ is an M-seminorm. Proof. Observe that (ii) implies (i). To prove (ii), assume that the open unit ball B of p has the property mentioned in (ii). We verify condition (c) of Proposition 3.3.42. Let x1 , . . . , x m ∈ X be such that p(x1 ), . . . , p(x m ) < 1. There exists a u ∈ B with {x1 , x2 , −x1 , −x2 }u ⊇ {u, −u}u and then there exists a v ∈ B with {x1 , x2 , x3 , −x1 , −x2 , −x3 }u ⊇ {u, −u, x3 , −x3 }u ⊇ {v, −v}u . By induction, it follows that there exists a y ∈ B such that {x1 , . . . , x m , −x1 , . . . , −x m }u ⊇ {y, −y}u . In particular, {x1 , . . . , x m }u ⊇ {y, −y}u . Now apply Proposition 3.3.42. Since the open unit ball of an M-norm on a Riesz space is directed, we have the next consequence of Corollary 3.3.43 (ii).

162 | 3 Seminorms on pre-Riesz spaces

Corollary 3.3.44. Let Y be a Riesz space and let X be a majorizing Riesz subspace of Y. Let p be a norm on X. The greatest Riesz seminorm p̄ on Y extending p is an M-norm if and only if p is an M-norm. Let us also investigate extension to an L-seminorm. The greatest Riesz seminorm ρ on a Riesz space Y extending an L-seminorm p on a subspace X is only an L-seminorm in case the subspace is ρ-dense. This will be proven below via dual spaces. The following result is well-known for norms; see e.g., [119, Proposition 1.4.7]. For a proof for the seminorm case, one can factor out the kernel of the seminorm and apply the result for the norm case to the quotient space. Lemma 3.3.45. Let X be a Riesz space with a Riesz seminorm ρ. Then ρ is an L-seminorm if and only if the dual norm ρ 󸀠 on (X, ρ)󸀠 is an M-norm. Proof. Let N := {x ∈ X; ρ(x) = 0}, let q : X → X/N be the quotient map, and consider the quotient cone q[K] in X/N. By Proposition 3.3.36 (v), the quotient space X/N is a Riesz space and the quotient norm ρ̂ is a Riesz norm on X/N. According to Proposition 3.3.39, the dual spaces (X, ρ)󸀠 and (X/N, ρ)̂ 󸀠 are isomorphic as ordered normed spaces. Assume that ρ is an L-seminorm. For x, y ∈ K we have q(x), q(y) ∈ q[K], so ̂ ̂ ̂ ρ(q(x) + q(y)) = ρ(x + y) = ρ(x) + ρ(y) = ρ(q(x)) + ρ(q(y)) , which means that ρ̂ is an L-norm on X/N. Due to [119, Proposition 1.4.7], the norm of the dual space (X/N, ρ)̂ 󸀠 is then an M-norm. It follows that the norm of (X, ρ)󸀠 is an M-norm. Conversely, assume that the norm of (X, ρ)󸀠 is an M-norm. Then the norm of (X/N, ρ)̂ 󸀠 is an M-norm and therefore ρ̂ is an L-norm, due to [119, Proposition 1.4.7]. ̂ ̂ ̂ For x, y ∈ K we then have ρ(x + y) = ρ(q(x + y)) = ρ(q(x)) + ρ(q(y)) = ρ(x) + ρ(y), so that ρ is an L-seminorm. Lemma 3.3.46. Let Y be a Riesz space and let X be a majorizing Riesz subspace of Y. Let p be a Riesz seminorm on X and let ρ be the greatest Riesz seminorm on Y extending p. For every positive ρ-continuous linear functional φ on Y, the restriction of φ to X is p-continuous, and ρ 󸀠 (φ) = p󸀠 (φ|X ). Proof. By Lemma 3.2.1, for y ∈ Y we have ρ(y) = inf{p(x); x ∈ X, x ≥ |y|} . Let φ be a positive ρ-continuous linear functional on Y. Clearly, φ|X is a positive p-continuous linear functional on X and ρ 󸀠 (φ) ≥ p󸀠 (φ|X ). For y ∈ Y with ρ(y) < 1, there is x ∈ X with x ≥ |y| and p(x) < 1. Then |φ(y)| ≤ φ(|y|) ≤ φ(x) = φ|X (x). Hence, ρ 󸀠 (φ) = sup{|φ(y)|; y ∈ Y, ρ(y) < 1} ≤ sup{φ|X (x); x ∈ X, p(x) < 1} = p󸀠 (φ|X ).

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Proposition 3.3.47. Let Y be a Riesz space with a majorizing Riesz subspace X and let p be an L-seminorm on X. Let ρ be the greatest Riesz seminorm on Y extending p. Then ρ is an L-seminorm if and only if X is ρ-dense in Y. Proof. Assume that X is ρ-dense in Y. Let x, y ∈ Y + . Then there are sequences (x n )n and (y n )n in X ∩ Y+ with ρ(x n − x) → 0, ρ(y n − y) → 0. Then ρ(x + y) = limn p(x n + y n ) = limn (p(x n ) + p(y n )) = ρ(x) + ρ(y). Hence, ρ is an L-seminorm. Suppose that X is not ρ-dense in Y. Because of Lemma 3.3.45 it suffices to prove that ρ 󸀠 is not an M-norm. There is a φ ∈ Y 󸀠 , φ ≠ 0, with φ = 0 on X. Then φ+ , φ− ∈ Y 󸀠 and φ+ ∨ φ− = φ+ + φ− . Denote φ0 := φ+ |X = φ− |X . Because of Lemma 3.3.45 again and Lemma 3.3.46, it follows that ρ 󸀠 (φ+ ∨ φ− ) = ρ 󸀠 (φ+ + φ− ) = p󸀠 (φ0 + φ0 ) = 2(p󸀠 (φ0 ) ∨ p󸀠 (φ0 )) = 2(ρ 󸀠 (φ+ ) ∨ ρ 󸀠 (φ− )) > ρ 󸀠 (φ+ ) ∨ ρ 󸀠 (φ− ) , because φ+ ≠ 0 or φ− ≠ 0 on Y. So ρ 󸀠 is not an M-norm, hence ρ is not an L-seminorm. As a generalization of the notion of an M-norm, Tzschichholtz and Weber consider in [144] a norm p on a directed partially ordered vector space with the property that {x ∈ K; p(x) < 1} is directed. They call such a norm p then an m≤ -norm; see also the proof of [144, Proposition 7]. They show, among other results, the following two duality theorems; see [144, Theorems 8 and 10]. Theorem 3.3.48. Let (X, K, p) be an ordered normed space such that K is generating and closed. Let p󸀠 be the dual norm of p. (i) If {x ∈ K; p(x) < 1} is directed, then p󸀠 is additive on K 󸀠 . (ii) Assume, in addition, that X is p-complete and K is p-closed. If p󸀠 is additive on K 󸀠 , then {x ∈ K; p(x) < 1} is directed. Theorem 3.3.49. Let (X, K, p) be an ordered normed space such that K is generating. Let p󸀠 be the dual norm of p. Then p is additive on K if and only if {f ∈ K 󸀠 ; p󸀠 (f) < 1} is directed. The ‘if’ part of Theorem 3.3.49 is due to [125, Proposition 4].

3.4 Regular seminorms The operator norm on the space Lr (X, Y) of all regular operators between normed Riesz spaces X and Y is in general not a Riesz norm, even if Lr (X, Y) is a Riesz space. Therefore, another norm on Lr (X, Y) is often considered, called the regular norm or the r-norm; see, e.g., [119, Proposition 1.3.6]. One of the main properties is that Lr (X, Y) is norm complete relative to the regular norm whenever Y is norm complete. Both the

164 | 3 Seminorms on pre-Riesz spaces

definition of the regular norm and the norm completeness result are not restricted to spaces of operators and can be extended to a general setting of directed partially ordered vector spaces. The ensuing notion of norm is a generalization of the notion of Riesz norm to partially ordered vector spaces. It is a stronger property than the notion of pre-Riesz norm discussed in Section 3.2. It turns out to be a useful concept, especially in investigations of spaces of operators; see Section 5.4.

3.4.1 Definition and examples Definition 3.4.1. Let (X, K) be a directed partially ordered vector space. A seminorm p on X is called regular if for every x ∈ X, p(x) = inf{p(y); y ∈ X, −y ≤ x ≤ y} . We collect basic properties of regular seminorms, where in (ii) we use the order unit seminorm given in (1.10). Proposition 3.4.2. (i) On a Riesz space, a seminorm is regular if and only if it is a Riesz seminorm. (ii) On a partially ordered vector space X with order unit u, the seminorm induced by the order unit is regular. (iii) On a directed partially ordered vector space X, every regular seminorm p is a preRiesz seminorm. Proof. The statement in (i) is straightforward. (ii) For every x ∈ X one has −‖x‖u u ≤ x ≤ ‖x‖u u. Therefore, s := inf{‖v‖u ; v ∈ X, −v ≤ x ≤ v} ≤ ‖‖x‖u u‖u = ‖x‖u . On the other hand, if v ∈ X is such that −v ≤ x ≤ v, then ‖x‖u ≤ ‖v‖u , which implies ‖x‖u ≤ s. (iii) We show that B := {x ∈ X; p(x) ≤ 1} is solvex. Let x ∈ X, x1 , . . . , x n ∈ B, and λ1 , . . . , λ n ∈ (0, 1] with ∑nk=1 λ k = 1 be such that n

u

{x, −x}u ⊇ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}} . k=1

Let α > 1. As p is regular, there are u 1 , . . . , u n ∈ X with −u k ≤ x k ≤ u k and p(u k ) < α for every k ∈ {1, . . . , n}. Then u := ∑nk=1 λ k u k is an upper bound of {∑nk=1 ε k λ k x k ; ε1 , . . . , ε n ∈ {1, −1}}, so that u ≥ x, −x. Then p(x) ≤ p(u) < α. Hence, p(x) ≤ 1, which means that x ∈ B. Thus, B is solvex and p is a pre-Riesz seminorm. It is an inconvenient truth that, in contrast to Riesz seminorms, the sum of two regular seminorms need not be regular.

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Example 3.4.3. We give a pre-Riesz space with two regular seminorms ρ 1 and ρ 2 such that ρ 1 + ρ 2 is not regular. Moreover, ρ 1 + ρ 2 is the restriction of a regular seminorm on a vector lattice cover. Let S be the unit circle in ℝ2 and let (X, K) be the pre-Riesz space of affine functions on S with pointwise order as in Example 3.2.13. Define for x ∈ X, ρ 1 (x) := |x(1, 0)|

and

ρ 2 (x) := |x(−1, 0)| .

To see that ρ 1 and ρ 2 are regular, let x ∈ X. Then |x| ∈ C(S) and, from the pointwise denseness shown in (1.24), for every t ∈ S we know that |x(t)| = inf{u(t); u ∈ X, u ≥ x, −x}. Since for t = (1, 0)T we have |v(t)| = ρ 1 (v) for every v ∈ X, it follows that ρ 1 is regular. Similarly, with t = (−1, 0)T we obtain that ρ 2 is regular. We complete the example by showing that p := ρ 1 + ρ 2 is not regular. Consider x ∈ X given by x(t1 , t2 ) = t2 , for every (t1 , t2 )T ∈ S. Then p(x) = 0. If u ∈ X is such that −u ≤ x ≤ u, then u ≥ 0 and, if we use that u is affine, we have p(u) = ρ 1 (u) + ρ 2 (u) = u(1, 0) + u(−1, 0) = u(0, 1) + u(0, −1) . Since u(0, 1) ≥ x(0, 1) = 1 and u(0, −1) ≥ −x(0, −1) = 1, we obtain that p(u) ≥ 2. Thus, p is not regular. This example also shows that the restriction property does not hold for regular seminorms. Indeed, p = ρ 1 + ρ 2 is clearly the restriction of the Riesz seminorm x 󳨃→ |x(1, 0)| + |x(−1, 0)| on the vector lattice cover C(S) of X. Hence, p is the restriction of a regular seminorm on a vector lattice cover, but p itself is not a regular seminorm.

3.4.2 Extension of regular seminorms For regular seminorms, we have an extension theorem similar to Theorem 3.3.15. Theorem 3.4.4. Let (X, K X ) and (Y, K Y ) be directed partially ordered vector spaces and let i : X → Y be a bipositive linear map such that i[X] is majorizing in Y. Let p be a seminorm on X. Define for y ∈ Y, ρ(y) := inf{p(x); x ∈ X, −i(x) ≤ y ≤ i(x)} . Then (i) ρ is the greatest regular seminorm on Y with ρ ∘ i ≤ p on K X . (ii) ρ ∘ i = p on K X if and only if p is monotone. (iii) ρ ∘ i ≥ p on X if and only if p is symmetrically monotone. (iv) ρ ∘ i = p on X if and only if p is regular. Proof. Since Y is directed and i[X] is majorizing in Y, for every y ∈ Y the set of which the infimum is taken in the definition of ρ(y) is not empty, so that ρ(y) ∈ [0, ∞) is well-defined.

166 | 3 Seminorms on pre-Riesz spaces (i) We first show that ρ is a seminorm. Let y1 , y2 ∈ Y and x1 , x2 ∈ X be such that −i(x1 ) ≤ y1 ≤ i(x1 ) and −i(x2 ) ≤ y2 ≤ i(x2 ). Then −i(x1 + x2 ) ≤ y1 + y2 ≤ i(x1 + x2 ) and ρ(y1 + y2 ) ≤ p(x1 + x2 ) ≤ p(x1 ) + p(x2 ). Hence, ρ(y1 + y2 ) ≤ ρ(y1 ) + ρ(y2 ). For y ∈ Y and λ ∈ [0, ∞) it is clear that ρ(−y) = ρ(y) and that ρ(λy) = λρ(y). Secondly, we note that ρ ∘ i ≤ p on K X , since −x ≤ x ≤ x for every x ∈ K X . Thirdly, we show that ρ is regular. Let y ∈ Y. Let u ∈ Y be such that −u ≤ y ≤ u. If x ∈ X is such that −i(x) ≤ u ≤ i(x), then −i(x) ≤ y ≤ i(x), so ρ(y) ≤ p(x). Hence, ρ(y) ≤ ρ(u). For every ε ∈ (0, ∞) there is x ∈ X such that −i(x) ≤ y ≤ i(x) and p(x) < ρ(y) + ε. Therefore, as ρ ∘ i ≤ p on K X , we have ρ(i(x)) < ρ(y) + ε. Hence, ρ(y) ≥ inf{ρ(i(x)); x ∈ X, −i(x) ≤ y ≤ i(x)} ≥ inf{ρ(u); u ∈ Y, −u ≤ y ≤ u} . It follows that ρ(y) = inf{ρ(u); u ∈ Y, −u ≤ y ≤ u}, which establishes that ρ is regular. Fourthly, we show that if η is a regular seminorm on Y and η ∘ i ≤ p on K X , then η ≤ ρ. Indeed, for y ∈ Y one has η(y) = inf{η(u); u ∈ Y, −u ≤ y ≤ u} ≤ inf{η(i(x)); x ∈ X, −i(x) ≤ y ≤ i(x)} = ρ(y) . Thus, ρ is a regular seminorm on Y with ρ ∘ i ≤ p on K X and it is the greatest seminorm on Y with these properties. (ii) If ρ ∘ i = p on K X , then for every x1 , x2 ∈ K X with x1 ≤ x2 we see that p(x1 ) = ρ(i(x1 )) ≤ p(x2 ). For the converse implication, if p is monotone, then for every x, y ∈ K X with −i(x) ≤ i(y) ≤ i(x) we have p(x) ≤ p(y), so p(x) ≤ ρ(i(x)). From (i) it then follows that ρ ∘ i = p on K X . (iii) Assume that p is symmetrically monotone and let y ∈ X. For every x ∈ X with −i(x) ≤ i(y) ≤ i(x) we have p(y) ≤ p(x). Hence, p(y) ≤ ρ(i(y)). Conversely, assume that ρ ∘ i ≥ p on X. For every x, y ∈ X with −x ≤ y ≤ x we then have p(y) ≤ ρ(i(y)) ≤ p(x), so that p is symmetrically monotone. (iv) If p is regular, then for every y ∈ X we have ρ(i(y)) = inf{p(x); x ∈ X, −x ≤ y ≤ x} = p(y) . Conversely, if ρ ∘ i = p on X, then for every y ∈ X we have p(y) = inf{p(x); x ∈ X, −x ≤ y ≤ x} = ρ(i(y)) , which means that p is regular. Theorem 3.4.4 implies the extension property for regular seminorms. Recall that on a Riesz space Y regular seminorms and Riesz seminorms coincide. Theorem 3.4.5. Let X be a pre-Riesz space, (Y, i) a vector lattice cover of X, and p a regular seminorm on X. Then there exists a regular seminorm ρ on Y such that ρ ∘ i = p.

3.4 Regular seminorms

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The extension formula in Theorem 3.4.4 resembles the extension formula for Riesz seminorms given in Lemma 3.2.1. Such a formula has also been used to construct seminorms on larger spaces starting from other seminorms than regular seminorms; see, e.g., [162, X.3.1]. Vulikh’s proof there yields the following result. Proposition 3.4.6. Let (X, K) be a pre-Riesz space and let (Y, i) be a vector lattice cover of X. Let p be a regular norm on X and let ρ be the greatest Riesz seminorm on Y with ρ ∘ i = p. Then ρ is a norm if and only if K is p-closed. Proof. If ρ is a norm, then ρ is a Riesz norm on Y, hence Y + is ρ-closed. Then K = [Y + ]i is p-closed. Conversely, assume that K is p-closed. Let y ∈ Y be such that ρ(y) = 0. From the formula for ρ in Theorem 3.4.4, we see that there is a sequence (u n )n∈ℕ in K with |y| ≤ i(u n ) for every n ∈ ℕ and p(u n ) → 0. Let x ∈ K be such that i(x) ≤ |y|. Then u n − x ∈ K for every n ∈ ℕ and, since K is p-closed, we obtain that −x ∈ K, so i(x) ≤ 0. Then, by order denseness of i[X] in Y, we have |y| = sup{i(x); x ∈ X, x ≤ |y|} ≤ 0, hence y = 0. Thus, ρ is a norm. For a regular seminorm, the greatest symmetrically monotone extension and the greatest regular extension coincide. Proposition 3.4.7. Let (X, K X ) and (Y, K Y ) be directed partially ordered vector spaces such that there exists a bipositive linear map i : X → Y with i[X] majorizing in Y. Let p be a regular seminorm on X, let p2 be the greatest symmetrically monotone seminorm on Y with p2 ∘ i = p and let ρ be the greatest regular seminorm on Y with ρ ∘ i = p. Then ρ = p2 . Proof. As ρ is symmetrically monotone, we have ρ ≤ p2 . By the extension formulas of Theorems 3.3.17 and 3.4.4 we obtain for every y ∈ Y that p2 (y) = inf {p(x) + p(u); x, u ∈ X, −i(u) ≤ y − i(x) ≤ i(u)} ≤ inf {p(u); u ∈ X, −i(u) ≤ y ≤ i(u)} = ρ(y) , so that ρ = p 2 . Concerning norm completeness we have the next result, similar to Theorem 3.3.28 and Corollary 3.3.29. Theorem 3.4.8. Let (X, K X ) and (Y, K Y ) be directed partially ordered vector spaces such that there exists a bipositive linear map i : X → Y with i[X] majorizing in Y. Assume that Y is uniformly complete. Let p be a regular norm on X such that K X is closed and let ρ be the greatest regular seminorm on Y with ρ ∘ i = p. If X is p-complete, then Y is ρ-complete. Proof. Due to Proposition 3.4.7, the seminorm ρ equals the greatest symmetrically monotone seminorm p2 on Y with p2 ∘ i = p. According to Corollary 3.3.29, the space Y is p2 -complete and therefore Y is ρ-complete.

168 | 3 Seminorms on pre-Riesz spaces

The next proposition proceeds the discussion on extension to M-seminorms given in Proposition 3.3.42 and Corollary 3.3.43. Proposition 3.4.9. Let (X, K) be a pre-Riesz space and let p be a regular seminorm on X. The following statements are equivalent. (a) The open unit ball of p is directed. (b) The set {x ∈ K; p(x) < 1} is directed. (c) For every vector lattice cover (Y, i) of X the greatest Riesz seminorm ρ on Y with ρ ∘ i = p is an M-seminorm. Proof. We first observe that (a) implies (c). Indeed, assume that the open unit ball of p is directed. As p is a pre-Riesz seminorm by Proposition 3.4.2 (iii), Corollary 3.3.43 yields (c). Next, assume (c). We show (b). Let x 1 , x2 ∈ {x ∈ K; p(x) < 1}. The greatest Riesz seminorm ρ on Y with ρ ∘ i = p satisfies for every y ∈ Y, ρ(y) = inf{p(x); x ∈ X, −i(x) ≤ y ≤ i(x)} . Since ρ is an M-seminorm, we have ρ(|i(x 1 )| ∨ |i(x2 )|) = ρ(|i(x1 )|) ∨ ρ(|i(x2 )|) = ρ(i(x1 )) ∨ ρ(i(x2 )) = p(x1 ) ∨ p(x2 ) < 1 . Hence, there is x ∈ X with −i(x) ≤ |i(x1 )| ∨ |i(x2 )| ≤ i(x) and p(x) < 1. Then x ∈ K and x ≥ x1 , x2 , so we arrive at (b). Finally, we assume (b) and show (a). Let x1 , x2 ∈ X with p(x1 ) < 1 and p(x2 ) < 1. As p is regular, there are u 1 , u 2 ∈ X with −u 1 ≤ x1 ≤ u 1 , −u 2 ≤ x2 ≤ u 2 , p(u 1 ) < 1, and p(u 2 ) < 1. Since u 1 , u 2 ∈ {x ∈ K; p(x) < 1}, there is u ∈ {x ∈ K; p(x) < 1} such that u ≥ u 1 , u 2 . Then u ≥ x1 , x2 , which yields (a). In [144, Theorem 5] it is shown that for every semimonotone norm p on a directed ordered vector space (X, K), where the positive part of the open unit ball is directed and K is closed, there exists an M-norm on the Dedekind completion of X, which is equivalent to p on X. This result is closely related to the implication (b) ⇒ (c) in Proposition 3.4.9.

3.4.3 Regularization of seminorms From every seminorm on a directed partially ordered vector space a regular seminorm can be constructed. The construction is actually a special case of Theorem 3.4.4 with X = Y. Definition 3.4.10. Let (X, K) be a directed partially ordered vector space and let p be a seminorm on X. The regular seminorm ρ on X defined by ρ(y) = inf{p(x); x ∈ X, −x ≤ y ≤ x}, y ∈ X ,

3.4 Regular seminorms |

169

is called the regularization of p. Proposition 3.4.11. Let (X, K) be a directed partially ordered vector space, let p be a monotone seminorm on X, and let ρ be the regularization of p. (i) ρ is the greatest regular seminorm on X with ρ ≤ p on K. (ii) If p is symmetrically monotone, then p ≤ ρ. (iii) p ≤ 2ρ. (iv) If p is a norm, then ρ is a norm. (v) ρ = p if and only if p is regular. (vi) If X is a Riesz space, then ρ(x) = p(|x|) for every x ∈ X. Proof. The statement in (i) follows directly from Theorem 3.4.4 (i). (ii) For x ∈ X and every u ∈ X with −u ≤ x ≤ u we have p(x) ≤ p(u), so that p(x) ≤ ρ(x). (iii) As p is monotone, Theorem 3.3.17 yields existence of a symmetrically monotone seminorm p2 on X with p2 ≤ p ≤ 2p2 . With the aid of (ii) we then obtain for every x ∈ X that ρ(x) = inf{p(u); u ∈ X, −u ≤ x ≤ u} ≥ inf{p2 (u); u ∈ X, −u ≤ x ≤ u} ≥ p2 (x) ≥ 12 p(x) . The statement in (iv) follows from (iii), and (v) is the content of Theorem 3.4.4 (iv). (vi) Let x ∈ X. For every u ∈ X with −u ≤ x ≤ u we have |x| ≤ u, so p(|x|) ≤ p(u) and hence, p(|x|) ≤ ρ(x). Since −|x| ≤ x ≤ |x|, we also have ρ(x) ≤ p(|x|). The regularization of a monotone norm inherits norm completeness in the following way. Theorem 3.4.12. Let (X, K) be a directed partially ordered vector space with a monotone norm p. Let ρ be the regularization of p. If K is p-complete, then X is ρ-complete and K is ρ-closed. Proof. Due to Proposition 3.4.11, we have that ρ is a norm and p ≤ 2ρ, so every p-closed set is ρ-closed. In particular, K is ρ-closed. Further, we note that every increasing p-convergent sequence in X is ρ-convergent to the same limit. Indeed, if x ∈ X and (x n )n is an increasing sequence in (X, K) which p-converges to x, then x ≥ x n for all n since K is p-closed, so with the aid of Theorem 3.4.11 we obtain ρ(x − x n ) ≤ p(x − x n ) → 0. Next, assume that K is p-complete and let (y n )n be a sequence in X such that ∞ ∑∞ n=1 ρ(y n ) < ∞. We show that ∑ n=1 y n is ρ-convergent in X. By definition of the regularization, for every n ∈ ℕ there is x n ∈ X such that −x n ≤ y n ≤ x n and p(x n ) ≤ ρ(y n ) + 2−n . Define for n ∈ ℕ, u n := x n − y n . Then x n , u n ∈ K and p(u n ) ≤ p(x n ) + ∞ 2ρ(y n ) ≤ 3ρ(y n ) + 2−n . Hence, ∑∞ n=1 p(x n ) < ∞ and ∑ n=1 p(u n ) < ∞. The cone K ∞ ∞ is p-complete, so ∑n=1 x n and ∑n=1 u n are both p-convergent in K. The sequences of ∞ partial sums of ∑∞ n=1 x n and ∑ n=1 u n are increasing in X, so it follows that they are

170 | 3 Seminorms on pre-Riesz spaces ∞ ρ-convergent, as noted above. Therefore, ∑∞ n=1 y n = ∑ n=1 (x n − u n ) is ρ-convergent. Thus, X is ρ-complete.

Corollary 3.4.13. (i) Let (X, K) be a directed partially ordered vector space with a regular norm ρ. Then K is ρ-complete if and only if X is ρ-complete and K is ρ-closed. (ii) Let (X, K) be a directed partially ordered vector space with a monotone norm p such that K is p-closed. If X is p-complete, then p is equivalent to a regular norm. Proof. (i) If K is ρ-complete, then the statement follows from Theorem 3.4.12 by taking p = ρ. The converse implication is straightforward. (ii) The conditions imply that K is p-complete. According to Theorem 3.4.12, the regularization ρ of p is then such that X is ρ-complete, as well. Furthermore, p ≤ 2ρ. Then it follows from Banach’s isomorphism theorem 1.8.4 that p and ρ are equivalent. The equivalence constant in Corollary 3.4.13 (ii) may be arbitrary large, as the next example shows. Example 3.4.14. We provide a Riesz space X such that for any real number C > 1 there is a monotone norm p on X such that X is norm complete with respect to both p and its regularization ρ and such that ρ(x) = Cp(x) for some nonzero x ∈ X. Take X = ℝ2 with the standard order. Let C > 1 and define for x = (x1 , x2 )T ∈ ℝ2 , p(x) :=

1 C−1 ‖x‖1

+ |x1 + x2 | .

Then p is a symmetrically monotone norm on X. As X is finite-dimensional, the positive cone K of X is p-closed and X is p-complete. The regularization of p is given by ρ(x) = p(|x|) =

1 C−1 ‖x‖1

+ ‖x‖1 =

for every x ∈ X. For x = (1, −1)T it is clear that ρ(x) =

C ‖x‖1 , C−1 C C−1 ‖x‖1

= Cp(x).

Let us summarize the restriction and extension results for seminorms of this chapter. For this, let X be a pre-Riesz space and (Y, i) be a vector lattice cover of X. Let (P) be a property of a seminorm, for instance monotone or regular. We say that restriction for (P) holds if for every seminorm p on Y with property (P) the seminorm p ∘ i on X also has property (P). We say that extension holds for (P), if for every seminorm p on X with property (P) there exists a seminorm p̄ on Y with property (P) such that p = p̄ ∘ i. For this setting, the restriction and extension results can be listed as follows.

3.5 Semimonotone norms revisited

monotone seminorm monotone* seminorm symmetrically monotone seminorm seminorm with a full unit ball pre-Riesz seminorm regular seminorm

restriction

extension

yes yes yes yes yes no

no yes yes yes yes yes

(clear) (Theorem 3.3.16) (Theorem 3.3.19) (Theorem 3.3.19) (Theorem 3.2.31) (Example 3.4.3)

| 171

(Remark 3.3.13) (Theorem 3.3.16) (Theorem 3.3.19) (Theorem 3.3.19) (Theorem 3.2.31) (Theorem 3.4.5)

Notes and Remarks Davies [42] studies regular norms on directed partially ordered vector spaces under the extra condition that the cone is closed. He presents results similar to our regularization and investigates duality properties. A formula similar to the regularization formula in Definition 3.4.10 has been used for operators by Lipecki in [108]. The notion of nonflatness of a cone, see, e.g., [102, Section 1.8], in combination with semimonotonicity of the norm is closely related to regular norms, see Proposition 3.5.2 below. Theorem 3.4.12 is a generalization of a well-known fact concerning the regular norm on spaces of regular operators, see [12, Theorem 15.2] or [119, Proposition 1.3.6]. For a slightly more general setting, see also Theorem 5.4.4 below. Much attention has been given to dual characterizations of monotonicity properties, where also conditions appear that are related to regular norms. It is straightforward that a norm ‖⋅‖ on a directed partially ordered vector space (X, K) is regular if and only if it is symmetrically monotone and ∀x ∈ X ∀ε ∈ (0, ∞) ∃y ∈ K :

−y≤x≤y

and ‖y‖ ≤ ‖x‖ + ε .

(3.10)

Ng has shown in [125] that if X is a Banach space and K is closed, then the norm of X is symmetrically monotone if and only if its dual norm satisfies (3.10). This result belongs to an extensive theory of duality relations between properties in the spirit of monotonicity and properties similar to (3.10); see, e.g., [22, 50]. A modern perspective on this topic with a thorough discussion of earlier literature is given by Messerschmidt in [117]. In Proposition 3.2.29, the condition that X is order dense in Y is important. It is shown in [148] that an ordered normed space can be embedded into a normed Riesz space by an isometric bipositive linear map if and only if the norm is symmetrically monotone and the cone is closed.

3.5 Semimonotone norms revisited Many of the results for monotone norms have an analogue for semimonotone norms. We collect them here. We start by showing that every semimonotone norm is equiv-

172 | 3 Seminorms on pre-Riesz spaces

alent to a monotone norm. We use Theorem 3.3.15 with the ordered vector space (X, K X ) = (Y, K Y ), i.e., for a semimonotone seminorm p on X we define p1 : X → ℝ,

y 󳨃→ inf{p(x) + p(u) + p(v); x ∈ X, u, v ∈ K X , −u ≤ y − x ≤ v} . (3.11)

Note that p1 is well-defined, as the infimum is taken over a nonempty set. Therefore, we do not have to assume that X is directed, as in Theorem 3.3.15. Proposition 3.5.1. Let (X, K X ) be a partially ordered vector space and let p be a semimonotone seminorm on X. Then p1 as in (3.11) is a monotone seminorm on X which is equivalent to p. Proof. Let p be semimonotone with constant N ∈ ℝ+ . By Theorem 3.3.15 we have that p1 is a monotone* seminorm on X and, hence, monotone by Proposition 3.3.4. Moreover, p1 ≤ p on X. Let y, x ∈ X and u, v ∈ K X be such that −u ≤ y − x ≤ v. Then 0 ≤ y − x + u ≤ u + v, thus p(y − x + u) ≤ Np(u + v). Hence, p(y) = p(y − x + u + x − u) ≤ p(y − x + u) + p(x) + p(u) ≤ Np(u + v) + p(x) + p(u) ≤ (N + 1)(p(u) + p(v) + p(x)) , so that p(x) + p(u) + p(v) ≥

1 N+1 p(y).

It follows that p1 (y) ≥

1 N+1 p(y).

In the literature, the cone of an ordered vector space with semimonotone norm is called normal; see, e.g., [162, Chapter IV], where also basic properties are collected. The result that every semimonotone norm is equivalent to a monotone norm is originally due to Krein [103]. Let us consider norm complete spaces. An ordered Banach space (X, K, ‖⋅‖) is defined to be a Banach space (X, ‖⋅‖) with a closed generating cone K. Proposition 3.5.2. Let (X, K, ‖⋅‖) be an ordered Banach space where the norm is semimonotone. (i) The norm is equivalent to a regular norm. (ii) There is a constant M ∈ ℝ+ such that for every x ∈ X there are x1 , x2 ∈ K such that x = x1 − x2 with ‖x1 ‖ ≤ M‖x‖ and ‖x2 ‖ ≤ M‖x‖. Proof. (i) By Proposition 3.5.1 the norm is equivalent to a monotone norm. Therefore, the statement follows from Corollary 3.4.13 (ii). (ii) By (i), it suffices to prove the statement for a regular norm ρ that is equivalent to ‖⋅‖. Fix C ∈ ℝ>1 . Let x ∈ X. The statement is clear for x = 0. Assume x ≠ 0. As ρ is regular, by definition there exists y ∈ X such that −y ≤ x ≤ y and ρ(y) < Cρ(x). Let x1 := 12 (x + y) and x2 := 12 (y − x). Then x1 , x2 ∈ K with x = x1 − x2 and ρ(x1 ) ≤ 12 (ρ(x) + ρ(y)) ≤ Cρ(x) and, similarly, ρ(x2 ) ≤ Cρ(x).

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In the literature, a cone in an ordered normed space that has the property in Proposition 3.5.2 (ii) is called non-flat; see, e.g., [162, Chapter III]. The statement in Proposition 3.5.2 (ii) is due to [104].

3.6 Topologies on pre-Riesz spaces For the types of seminorms defined in Subsection 3.3.1, the locally convex topologies generated by a collection of such seminorms can be studied. It turns out that we obtain two types of topologies. A locally convex topology is generated by a collection of monotone, monotone*, symmetrically monotone seminorms, or seminorms with full unit balls if and only if the topology has a neighborhood base consisting of full sets. A locally convex topology on a Riesz space is generated by a collection of Riesz seminorms if and only if it has a neighborhood base consisting of solid sets. In this section we present a brief overview of properties of locally full and locally solid topologies. Let X be a vector space over ℝ. Recall that a topology on X is called a vector space topology if addition and scalar multiplication are continuous maps from X × X to X and ℝ × X to X, respectively, where X × X and ℝ × X are endowed with their product topologies. Note that we do not require vector space topologies to be Hausdorff. We will frequently describe vector space topologies by means of neighborhoods of 0. A neighborhood base of an x ∈ X for a topology τ on X is a collection U of subsets of X such that every element of U is a neighborhood of x and for every neighborhood V of x there exists U ∈ U with U ⊆ V. If τ is a vector space topology on X, then every neighborhood base U of 0 has the following three properties: (i) For every U 1 , U2 ∈ U there exists U ∈ U such that U ⊆ U1 ∩ U2 . (ii) Every U ∈ U is absorbing. (iii) For every U ∈ U there exists V ∈ U with V + V ⊆ U. Moreover, there exists a neighborhood base U of 0 that also satisfies the following condition. (iv) Every U ∈ U is circled. Conversely, if U is a collection of subsets of X that contain 0 and U satisfies (i), then τ := {V ⊆ X; ∀x ∈ V ∃U ∈ U : x + U ⊆ V} is a topology on X. If U also satisfies (ii), (iii) and (iv), then τ is a vector space topology on X and U is a neighborhood base of 0 for τ. For more details, see [157, Section VIII.1].

174 | 3 Seminorms on pre-Riesz spaces

3.6.1 Locally full topologies Locally full topologies on partially ordered vector spaces are studied, e.g., in [72, 94, 124, 127, 133]. We discuss basic properties of such topologies and link the topologies to the appropriate seminorms. We consider a partially ordered vector space X and a subset M ⊆ X. Recall that the set M is full if for every a, b ∈ M we have [a, b] ⊆ M. The full hull full(M) of M is defined to be the intersection of all full subsets of X containing M. As the intersection of full sets is clearly full, full(M) is the smallest full set in X containing M. We have the following representation. Lemma 3.6.1. Let X be a partially ordered vector space and let M ⊆ X. The full hull of M satisfies full(M) = {x ∈ X; ∃a, b ∈ M : a ≤ x ≤ b} . Proof. Denote N := {x ∈ X; ∃a, b ∈ M : a ≤ x ≤ b}. If x ∈ N and a, b ∈ M are such that a ≤ x ≤ b, then a, b ∈ full(M) and as this set is full we obtain x ∈ full(M). Hence, N ⊆ full(M). To prove that full(M) ⊆ N, we show that N is a full set containing M. For every x ∈ M, with a := b := x we have a, b ∈ M and a ≤ x ≤ b, so x ∈ N. Hence, M ⊆ N. To see that N is full, let c, d ∈ N and let x ∈ X be such that c ≤ x ≤ d. There are a, b ∈ M with a ≤ c and d ≤ b and then a ≤ x ≤ b, so x ∈ N. Hence, N is full. It follows that full(M) ⊆ N. Definition 3.6.2. Let X be a partially ordered vector space. A vector space topology τ on X is called locally full if it has a neighborhood base of 0 consisting of full sets. The next theorem characterizes locally full topologies as those topologies for which the ‘sandwich theorem’ holds. Theorem 3.6.3. Let X be a partially ordered vector space with a vector space topology τ. The following statements are equivalent. (a) τ is locally full. (b) If (x i )i∈I , (y i )i∈I , and (z i )i∈I are nets in X such that y i ≤ x i ≤ z i for every i ∈ I and y i → 0, z i → 0, then x i → 0. Proof. Suppose that τ is locally full. We prove (b). Let (x i )i∈I , (y i )i∈I , and (z i )i∈I be nets in X such that y i ≤ x i ≤ z i for every i ∈ I and y i → 0, z i → 0. The topology τ has a neighborhood base U of 0 consisting of full sets. Let U ∈ U. There exists i0 ∈ I such that for every i ∈ I i≥i0 one has that y i , z i ∈ U, so, x i ∈ U, because U is full. Hence, x i → 0. Next, suppose that τ is not locally full. We prove that (b) is not true. There is a neighborhood V of 0 that contains no full neighborhoods of 0. Let U be the collection of all neighborhoods of 0 contained in V. Let ω ∈ U. If for every y, z ∈ ω and x ∈ X such that y ≤ x ≤ z one would have x ∈ V, then the full hull of ω would be contained in V, contradicting the assumption of V. Hence, there are y ω , z ω ∈ ω and x ω ∈ X \ V

3.6 Topologies on pre-Riesz spaces | 175

with y ω ≤ x ω ≤ z ω . If we take U as index set ordered by reverse inclusion (i.e., ω1 ≤ ω2 ⇔ ω1 ⊇ ω2 ), then the nets (y ω )ω∈U and (z ω )ω∈U converge to 0, whereas (x ω )ω∈U does not. Thus, (b) is not true. Part (ii) of the following lemma is needed to show that a locally convex and locally full topology has a neighborhood base of 0 consisting of sets that are both full and convex. Lemma 3.6.4. Let X be a partially ordered vector space. (i) The full hull of a circled set in X is circled. (ii) The full hull of a convex set in X is convex. Proof. For a proof of (i), let S be a circled set in X, let x ∈ full(S) and t ∈ [−1, 1]. By Lemma 3.6.1, there are a, b ∈ S such that a ≤ x ≤ b. As S is circled, we have ta, tb ∈ S. If t ∈ [0, 1], then ta ≤ tx ≤ tb, so tx ∈ full(S). If t ∈ [−1, 0), then tb ≤ tx ≤ ta, so tx ∈ full(S). Hence, full(S) is circled. To prove (ii), let S be a convex set in X. Let x, y ∈ full(S) and λ ∈ [0, 1]. There are x1 , x2 , y1 , y2 ∈ S such that x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2 . Then λx1 + (1 − λ)y1 ≤ λx + (1 − λ)y ≤ λx2 + (1 − λ)y2 and λx i + (1 − λ)y i ∈ S for every i ∈ {1, 2}, because S is convex. Hence, λx + (1 − λ)y ∈ full(S). Hence, full(S) is convex. Conversely to (ii) of Lemma 3.6.4, the convex hull of a full set need not be full, as the next example shows. Example 3.6.5. We provide a full set in a Riesz space with a convex hull that is not full. Take X = ℝ3 with the standard order and S := {(0, 0, 0)T , (−1, 2, 1)T , (2, −1, 1)T }. Then S is full and the convex hull of S contains y := (0, 0, 0)T and z := ( 12 , 12 , 1)T . The element x := ( 12 , 12 , 0)T satisfies y ≤ x ≤ z and it is not contained in the convex hull co S, because for every element in co S the third coordinate is the sum of the other two. Hence, co S is not full. Proposition 3.6.6. Let X be a partially ordered vector space. (i) A locally full topology τ on X has a neighborhood base of 0 consisting of circled full sets. (ii) A locally convex locally full topology τ on X has a neighborhood base of 0 consisting of circled convex full sets. Proof. To see (i), let U be a neighborhood base of 0 for τ consisting of full sets. Let V be the collection of all neighborhoods of 0 that are full and circled. Let U ∈ U. Since τ is a vector space topology, there is a circled neighborhood W of 0 with W ⊆ U. Then V := full(W) ⊇ W is a neighborhood of 0 and W is circled due to Lemma 3.6.4 (i), hence V ∈ V. Moreover, as U is full, we have V ⊆ W ⊆ U. Therefore, V is a neighborhood base of 0 for τ. Next we show (ii). As τ is locally full, τ has a neighborhood base U of 0 consisting of full sets. Let V be the collection of all neighborhoods of 0 that are circled, convex, and full. Let U ∈ U. Since τ is locally convex, there is a convex neighborhood W of 0

176 | 3 Seminorms on pre-Riesz spaces

contained in U. The full hull full(W) of W is full and convex, due to Lemma 3.6.4 (ii), and full(W) ⊆ U. Then V := full(W) ∩ (− full(W)) is a circled convex full neighborhood of 0, hence V ∈ V and V ⊆ U. Thus, V is a neighborhood base of 0 for τ consisting of circled convex full sets. Let us link locally convex locally full topologies to the corresponding classes of seminorms. According to Corollary 3.3.20, every monotone seminorm on a partially ordered vector space X is equivalent to a seminorm with a full unit ball. Then the same is true for every monotone* or symmetrically monotone seminorm. Hence, a topology generated by a collection of symmetrically monotone, monotone*, or monotone seminorms is also generated by a collection of seminorms with full unit balls. Theorem 3.6.8 below states that such topologies are precisely the locally convex locally full ones. We first need a lemma. Lemma 3.6.7. Let X be a partially ordered vector space and let U be a circled convex absorbing full subset of X. Then the Minkowski functional p of U is a seminorm on X with a full unit ball. Proof. Let V = ⋂α>1 αU and let p be the Minkowski functional of V. According to Proposition 3.1.3, the map p is a seminorm with unit ball V, and p is the Minkowski functional of U. It remains to show that V is full. If a, b ∈ V, a ≤ b and x ∈ X are such that a ≤ x ≤ b, then for every t ∈ (1, ∞) we have a ∈ tU and b ∈ tU. Since tU is full, it follows that x ∈ tU. Hence, x ∈ V. Theorem 3.6.8. Let X be a partially ordered vector space with a vector space topology τ. The following statements are equivalent. (i) τ is locally convex and locally full. (ii) τ is generated by a collection of monotone seminorms. (iii) τ is generated by a collection of seminorms with full unit balls. Proof. By Proposition 3.3.4, every seminorm with a full unit ball is monotone, so (iii) implies (ii). Due to Corollary 3.3.20, every monotone seminorm is equivalent to a seminorm with a full unit ball, so (ii) implies (iii). To see that (iii) implies (i), observe that for a seminorm p on X with a full unit ball, for every r ∈ [0, ∞) the set {x ∈ X; p(x) ≤ r} is a convex and full set. Moreover, intersections of convex and full sets are convex and full. Hence, if (iii) holds, then τ has a neighborhood base of 0 consisting of convex full sets, which means that (i) is true. Finally, assume (i). We show (iii). By Proposition 3.6.6, the topology τ has a neighborhood base U of 0 consisting of circled convex full sets. For every U ∈ U, by Lemma 3.6.7 there is a seminorm p U on X with a full unit ball and {x ∈ X; p U (x) < 1} ⊆ U ⊆ {x ∈ X; p U (x) ≤ 1}, by Proposition 3.1.2. Then the set of seminorms {p U ; U ∈ U} generates τ and (iii) is true. To illustrate that there are vector space topologies that are not locally full, we give an example where the norm is not equivalent to a monotone norm.

3.6 Topologies on pre-Riesz spaces | 177

Example 3.6.9. Consider the space X = cc (ℕ) of sequences that are eventually zero and define for every x ∈ X, ∞

p(x) := ∑ 1k |x(k) − x(k + 1)| . k=1

Then X is a Riesz space and p is a norm on X. Define x n := 𝟙{1,...,n} for every n ∈ ℕ. Then p(x n ) = 1n → 0 and 0 ≤ x1 ≤ x n for every n, but x1 ≠ 0. Hence p is not equivalent to a monotone norm. In view of the extension and restriction method explained in Section 2.8, we investigate extension and restriction of locally full topologies and locally convex locally full topologies. If D is a subset of a set X and τ is a topology on D, then we will say that a topology τ on X extends τ if τ = {U ∩ D; U ∈ τ}. For a topology σ on X the topology {U ∩ D; U ∈ σ} in D is called the restriction of σ to D. Lemma 3.6.10. Let X be a partially ordered vector space and let D be a linear subspace of X. (i) If M ⊆ D is full in D, then N := full(M) is a full set in X such that M = N ∩ D. (ii) If N ⊆ X is full, then N ∩ D is full in D. Proof. For a proof of (i), let M be a full subset of D and let N := full(M) in X. Then N is a full set in X and M ⊆ N ∩ D. To show that M ⊇ N ∩ D, let x ∈ N ∩ D. By Lemma 3.6.1 there exist a, b ∈ M such that a ≤ x ≤ b. As x ∈ D, M ⊆ D and M is full in D, it follows that x ∈ M. Hence, N ∩ D ⊆ M. To prove (ii), let x ∈ D and a, b ∈ N ∩ D be such that a ≤ x ≤ b. As N is full, we get x ∈ N and therefore x ∈ N ∩ D. Hence, N ∩ D is full in D. Proposition 3.6.11. Let X be a partially ordered vector space and let D be a directed majorizing linear subspace of X. (i) If τ is a locally full vector space topology on D, then there is a locally full vector space topology τ on X that extends τ. (ii) If σ is a locally full vector space topology on X, then the restriction of σ to D is a locally full vector space topology on D. Proof. For a proof of (i), we first use Proposition 3.6.6 (i) to obtain a neighborhood base U of 0 for τ consisting of full and circled sets. Let V := {full(U); U ∈ U}, where the full hull is taken in X. We show that V is a neighborhood base of a vector space topology on X. Firstly, for every V1 , V2 ∈ V, there are U1 , U2 ∈ U with V1 = full(U1 ) and V2 = full(U2 ). Then there is a U ∈ U such that U ⊆ U1 ∩ U2 . Let V := full(U) ∈ V. As U ⊆ V1 ∩ V2 and V1 ∩ V2 is full in X, we have V ⊆ V1 ∩ V2 . Secondly, we show that every V ∈ V is absorbing. Indeed, for V ∈ V and U ∈ U such that V = full(U), we have that U is absorbing in D. Let x ∈ X. As D is majorizing and directed, there is a y ∈ D with y ≥ x and y ≥ −x. Then there exists t ∈ ℝ>0 such that y ∈ tU. As U is circled, the set tU

178 | 3 Seminorms on pre-Riesz spaces is circled and therefore also −y ∈ tU. Since U ⊆ V, we obtain y, −y ∈ tV. Since V is full in X, we have that tV is full in X and from −y ≤ x ≤ y it then follows that x ∈ tV. Hence, V is absorbing in X. Thirdly, we show that for every V ∈ V there is W ∈ V with W + W ⊆ V. Indeed, let V ∈ V and let U ∈ U be such that V = full(U). There is a U0 ∈ U such that U0 + U0 ⊆ U. Let W := full(U0 ). For x1 , x2 ∈ W, there are a1 , b 1 , a2 , b 2 ∈ U0 such that a1 ≤ x1 ≤ b 1 and a2 ≤ x2 ≤ b 2 . Then a1 + a2 ≤ x1 + x2 ≤ b 1 + b 2 and a1 + a2 , b 1 + b 2 ∈ U0 + U0 ⊆ U ⊆ V, so x1 + x2 ∈ V, as V is full in X. Thus, W + W ⊆ V. Fourthly, every V ∈ V is circled, since every U ∈ U is circled in D and, hence, in X, and therefore full(U) is circled in X by Lemma 3.6.4. It follows that V is a neighborhood base of 0 for a vector space topology τ on X given by τ = {S ⊆ X; ∀x ∈ S ∃V ∈ V : x + V ⊆ S} . It is clear that τ is locally full. It remains to show that τ extends τ. Let S ∈ τ and let S0 := S ∩ D. Let x ∈ S0 . Then there exists V ∈ V such that x + V ⊆ S. Take U ∈ U such that V = full(U). Then V ∩ D = U by Lemma 3.6.10, since U is full in D. Hence, x + U = x + (V ∩ D) = (x + V) ∩ D ⊆ S ∩ D = S0 . Therefore, S0 ∈ τ. Thus, τ extends τ. To show (ii), let σ be a locally full vector space topology on X and let τ := {V ∩ D; V ∈ σ}. Then τ is a vector space topology in D and it remains to show that τ has a neighborhood base of 0 consisting of full sets. Let V be a neighborhood base of 0 for σ consisting of full sets. As for every V ∈ V the set V ∩ D is full in D due to Lemma 3.6.10, we obtain that {V ∩ D; V ∈ V} is a neighborhood base of 0 for τ consisting of full sets. A similar result holds for locally convex locally full topologies. Proposition 3.6.12. Let X be a partially ordered vector space and let D be a directed majorizing linear subspace of X. (i) If τ is a locally convex locally full topology on D, then there is a locally convex locally full topology τ on X that extends τ. (ii) If σ is a locally convex locally full topology on X, then the restriction of σ to D is a locally convex locally full topology on D. Proof. (i) According to Theorem 3.6.8, there is a collection P of seminorms with full unit balls on D that generates τ. Each element of P can be extended to a seminorm with a full unit ball on X, due to Corollary 3.3.17. The collection of those seminorms generates a locally convex locally full topology on X and that topology extends τ. (ii) If U is a convex and full subset of X, then U ∩ D is a convex and full subset of D, and the statement follows.

3.6 Topologies on pre-Riesz spaces |

179

3.6.2 Topologies such that the positive cone is closed Let us summarize some elementary results for topologies with closed positive cones. For topologies induced by norms, most of these results have already been discussed in Subsection 1.5.1. The first property is an immediate observation. Proposition 3.6.13. Let (X, K) be a partially ordered vector space with a vector space topology τ. The following statements are equivalent. (a) K is closed. (b) If (x i )i∈I and (y i )i∈I are nets in X and x, y ∈ X are such that x i → x, y i → y, and x i ≤ y i for every i, then x ≤ y. Next, we consider the Hausdorff property. Proposition 3.6.14. Let (X, K) be a partially ordered vector space and let τ be a vector space topology on X. If K is closed, then τ is Hausdorff. Proof. Suppose that τ is not Hausdorff. Then there is a nontrivial subspace D that is contained in the closure of {0}. As K is closed and K ⊇ {0}, it follows that D ⊆ K, which contradicts that K is a cone. Closed cones induce Archimedean partial orders, as we show next. Proposition 3.6.15. Let (X, K) be a partially ordered vector space. If there is a vector space topology on X such that K is sequentially closed, then X is Archimedean. Proof. Let x, y ∈ X be such that nx ≤ y for every n ∈ ℕ. Then 1n y − x ∈ K for every n and 1n y − x → −x, so −x ∈ K, hence x ≤ 0. Local fullness of the topology does not imply that the cone is closed nor the other way around, not even for norm topologies. Example 3.6.16. We provide a directed Archimedean partially ordered vector space (X, K) with a monotone norm such that K is not closed. Let X be the subspace of ℓ∞ (ℤ≥0 ) consisting of linear combinations of u 0 := (−1, 0, 1, 12 , 13 , . . .) , u n := (1, 1n , 1n , . . . , 1n , 0, 0, . . .) = 𝟙{0} + 1n 𝟙{1,...,n} , n ≥ 1 , and let p(x) := supk≥1 |x(k)| for x ∈ X. Note that the supremum is taken over indices k ≥ 1, so that p is not the same as ‖⋅‖∞ . The element u 0 + 2u 1 = (1, 2, 1, 12 , 13 , . . .) is an order unit in X and therefore X is directed. As a subspace of an Archimedean space, X is Archimedean. Clearly, p is a monotone seminorm on X. We show that p is a norm on X. Indeed, let x ∈ X be such that p(x) = 0. Then there are λ0 , . . . , λ n ∈ ℝ such that x = λ0 u 0 + λ1 u 1 + ⋅ ⋅ ⋅ + λ n u n . Since n + 1 ≥ 1, we have x(n + 1) = 0 and therefore λ0 = 0. Then, if n ≥ 1, we get

180 | 3 Seminorms on pre-Riesz spaces x(n) = 0, so that λ n = 0. By induction, we obtain that λ n−1 = ⋅ ⋅ ⋅ = λ1 = 0. Therefore, x = 0, and hence p is a norm. Next we show that K is not p-closed in X. Indeed, u 0 + u n ∈ K for every n ∈ ℕ and p(u n ) = 1n → 0, so u 0 + u n → u 0 , whereas u 0 ∈ ̸ K. An example of a Riesz space X with a norm p such that X+ is closed and p is not equivalent to a monotone norm is postponed to Example 3.6.22. Concerning the closure of a positive cone in a partially ordered vector space with a locally full vector space topology, there is a result similar to Proposition 1.5.8. Proposition 3.6.17. Let (X, K) be a partially ordered vector space with a locally full Hausdorff vector space topology τ. The closure K of K is the positive cone of a vector space order and τ is locally full for this order. Proof. Let U be a neighborhood base of 0 for τ consisting of full sets. With the aid of Lemma 3.6.4, we may assume that every element of U is circled. To show that K is a wedge in X, let x, y ∈ K. For every ω ∈ U there are x ω , y ω ∈ K with x ω − x, y ω − y ∈ ω. Then x ω + y ω ∈ K for every ω ∈ U and x ω + y ω → x + y, hence x + y ∈ K. By a similar argument for the scalar multiplication, it follows that K is a wedge in X. In order to prove that K is a cone in X, let x ∈ K ∩ (−K). Then for every ω ∈ U there are x ω , y ω ∈ K with x ω , −y ω ∈ x + ω. Since 0 ≤ x ω ≤ x ω + y ω → 0 and τ is locally full, Theorem 3.6.3 yields that x ω → 0. As τ is Hausdorff, we obtain that x = 0. Thus, K is a cone. Next we show that τ is locally full for the partial order on X induced by K. In order to distinguish carefully between the two orders, we will add the specification ‘in (X, K)’ or ‘in (X, K)’ where appropriate. Let U ∈ U. Since τ is a vector space topology, there is a neighborhood U0 of 0 such that U0 + U0 ⊆ U. We choose a V ∈ U such that V ⊆ U0 . Let W be the full hull of V in (X, K). Then W ⊇ V, so W is a neighborhood of 0 and W is full in (X, K). We prove W ⊆ U. Indeed, let x ∈ W. Then there are u, v ∈ V such that u ≤ x ≤ v in (X, K). That means that there are nets (a i )i∈I and (b j )j∈J in K with a i → x − u and b j → v − x. Then there exist i0 ∈ I and j0 ∈ J such that for every i ∈ I≥i 0 and j ∈ J ≥j0 we have x−u−a i , b j −v+x ∈ V, and then x−a i , x+b j ∈ (−V)+V = V+V ⊆ U. Since x − a i ≤ x ≤ x + b j in (X, K) for every i ∈ I and j ∈ J and U is full in (X, K), it follows that x ∈ U. Thus, W ⊆ U. We conclude that τ is locally full in (X, K).

3.6.3 Locally solid topologies On a partially ordered vector space, local fullness is the topological counterpart of a monotone norm. On a Riesz space, topologies generated by collections of Riesz seminorms lead to the notion of locally solid topologies; see [13, 55, 124, 127, 133]. Let X be a Riesz space. Recall that a subset S of X is called solid if for every x ∈ X and y ∈ S such that |x| ≤ |y| we have that x ∈ S. A vector space topology on X is called locally solid if there exists a neighborhood base of 0 consisting of solid sets.

3.6 Topologies on pre-Riesz spaces | 181

Recall that every solid set is circled. We already showed in Proposition 1.5.31 that the convex hull of a solid set in X is solid. We obtain that every locally convex, locally solid topology has a neighborhood base of 0 consisting of convex solid sets. With the aid of Proposition 1.5.30 we obtain the following characterization. Proposition 3.6.18. Let X be a Riesz space with a vector space topology τ. The following statements are equivalent. (a) τ is locally convex and locally solid. (b) τ is generated by a collection of Riesz seminorms. Next we show that among the locally full topologies, the locally solid ones are those for which the lattice operations are continuous. Proposition 3.6.19. Let X be a Riesz space with a locally full vector space topology τ. Then the lattice operations are continuous if and only if τ is locally solid. Proof. Assume that the lattice operations are continuous. Let U be a circled full neighborhood of 0. Since x 󳨃→ |x| is continuous, there is a full neighborhood V of 0 such that {|x|; x ∈ V} ⊆ U. Let W be the solid hull of V. Then W ⊇ V, so W is a solid neighborhood of 0. For y ∈ V we have |y| ∈ U, hence −|y| ∈ U and then x ∈ U for every x ∈ X with −|y| ≤ x ≤ |y|, as U is full. Therefore, W ⊆ U. It follows that τ is locally solid. For the converse implication, assume that τ is locally solid. We show that x 󳨃→ x+ is continuous. The continuity of the other lattice operations then follows. Let x ∈ X and let V ∈ τ be such that x+ ∈ V. As τ is locally solid, there exists a solid neighborhood W of 0 such that x+ + W ⊆ V. Let U := x + W. Then U is a neighborhood of x. Let y ∈ U. We show that y+ ∈ V. Since y − x ∈ W, W is solid, and |y+ − x+ | ≤ |y − x|, we have that y+ − x+ ∈ W. Hence, y+ ∈ x+ + W ⊆ V. Thus, x 󳨃→ x+ is continuous. It is easy to observe that not every solid set is full, as the unit disk D in ℝ2 is solid and not full. Indeed, (0, −1)T ≤ (1, −1)T ≤ (1, 0)T and (0, −1)T , (1, 0)T ∈ D, but (1, −1)T ∈ ̸ D. Nevertheless, the next proposition says that every locally solid topology is locally full. Proposition 3.6.20. Let X be a Riesz space. Every locally solid vector space topology on X is locally full. Proof. Let τ be a locally solid vector space topology on X and let U be a neighborhood base of 0 consisting of solid sets. Let V ∈ U. Then there is U ∈ U such that U + U ⊆ V. Let W be the full hull of U. Then W ⊇ U, so W is a neighborhood of 0. Next, we show that W ⊆ V. Indeed, if x ∈ W, then y ≤ x ≤ z for some y, z ∈ U, so |x| ≤ |y| ∨ |z| ≤ |y| + |z| ∈ U + U ⊆ V, hence x ∈ V. Therefore, W ⊆ V. Thus, τ has a neighborhood base of 0 consisting of full sets. The next example shows that locally full topologies need not be locally solid, not even normable ones.

182 | 3 Seminorms on pre-Riesz spaces

Example 3.6.21. We consider a Riesz space X with a monotone norm p that is not equivalent to a Riesz norm. Take X = cc (ℕ), and for x ∈ X define p(x) := ‖x‖∞ + (n) 1 k | ∑∞ k=1 x k |. Clearly, p is a monotone norm. For n ∈ ℕ, let x k be defined as n (−1) for 1 ≤ k ≤ 2n and 0 for k > 2n. Then p(x(n) ) = 1n → 0, whereas p(|x(n) |) = 2 + 1n ≥ 2 for every n. Therefore, p is not equivalent to a Riesz norm. On a normed Riesz space (X, ‖⋅‖), the absolute value is Lipschitz and hence uniformly continuous, since for every x, y ∈ X we have ||x|−|y|| ≤ |x−y|, so that ‖|x|−|y|‖ ≤ ‖x−y‖. The next example presents a norm on a Riesz space for which the absolute value is continuous but not uniformly continuous. It follows that the norm is not equivalent to a monotone one, due to Proposition 3.6.19. Example 3.6.22. Let X = cc (ℕ) with the standard order, and for x ∈ X define ∞

p(x) := ∑ (|x k − x k+1 | + 2−k |x k |) . k=1

Then X is a Riesz space and p is a norm on X. As p-convergence implies entrywise convergence, the cone is p-closed. We show that | ⋅ | is continuous on X. Indeed, let x ∈ X. Then there is n ∈ ℕ such that x k = 0 for every k > n. For y ∈ X one has n 󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 p(|y| − |x|) = ∑ (󵄨󵄨󵄨|y k | − |x k | − |y k+1 | + |x k+1 |󵄨󵄨󵄨 + 2−k 󵄨󵄨󵄨|y k | − |x k |󵄨󵄨󵄨) 󵄨 󵄨 󵄨 󵄨 k=1

∞

+ ∑ (|y k − y k+1 | + 2−k |y k |) k=n+1 n

≤ ∑ (|y k − x k | + |y k+1 − x k+1 | + 2−k |y k − x k |) k=1 ∞

+ ∑ (|y k − y k+1 | + 2−k |y k |) k=n+1 n+1

∞

k=1

k=n+1

≤ (2n+1 + 1) ∑ 2−k |y k − x k | + ∑ (|y k − y k+1 | + 2−k |y k |) ≤ 2(2n+1 + 1)p(y − x) . Hence, | ⋅ | is continuous at x. Next we show that | ⋅ | is not uniformly continuous. Indeed, for n ∈ ℕ and δ ∈ (0, 1), let x(n) := 𝟙{1,3,...,2n+1} and y(n,δ) := (1 − δ)𝟙{1,3,...,2n+1} − δ𝟙{2,4,...,2n} . Then x(n) − y(n,δ) = δ𝟙{1,...,2n+1} , thus 2n+1

p (x(n) − y(n,δ)) ≤ (1 + ∑ 2−k )δ ≤ 2δ . k=1

Since |x(n) | − |y(n,δ) | = δ(𝟙{1,3,...,2n+1} − 𝟙{2,4,...,2n} ), we have p(|x(n) | − |y(n,δ) |) ≥ 4nδ. Thus, | ⋅ | is not uniformly continuous. By Proposition 3.6.19, the norm p is not equivalent to a monotone norm.

3.7 Order convergence and unbounded order convergence | 183

For a Riesz space X with a monotone norm p, the map ρ : X → ℝ, x 󳨃→ p(|x|) is a Riesz norm on X. Moreover, p(x) ≤ p(x+ ) + p(x− ) ≤ 2p(|x|) = 2ρ(x) for every x ∈ X, so ρ is stronger than p. A similar statement for topologies is contained in the next proposition. Proposition 3.6.23. Let X be a Riesz space with a locally full vector space topology τ. For x ∈ X define f(x) = |x|. Let σ be the vector space topology generated by {[U]f ; U ∈ τ}. Then σ is the coarsest locally solid vector space topology finer than τ. Moreover, for every τ-neighborhood base U of 0, the collection {[U]f; U ∈ U} is a neighborhood base of 0 for σ. If τ is Hausdorff, so is σ. Proof. Let U be the collection of all circled full τ-neighborhoods of 0 and let V := {[U]f; U ∈ U}. Let U ∈ U. Then [U]f is a solid circled set containing 0. It is absorbing, because for every x ∈ X there is a t ∈ ℝ+ such that |x| ∈ tU, as U is absorbing, and then x ∈ t[U]f . Choose a τ-neighborhood V of 0 such that V + V ⊆ U. Let W := [V]f . If x, y ∈ W, then |x|, |y| ∈ V, so 0 ≤ |x + y| ≤ |x| + |y| ∈ V + V ⊆ U, hence |x + y| ∈ U. Thus, x + y ∈ [U]f . It follows that W + W ⊆ [U]f . Therefore, V is a neighborhood base of 0 for a locally solid vector space topology. Because U is a neighborhood base of 0 for τ, the topology generated by V is precisely σ. Clearly, the inverse images under f of the sets in any τ-neighborhood base of 0 constitute a σ-neighborhood base of 0. For any circled, full set U one has [U]f ⊆ U, because if x ∈ [U]f , then |x|, −|x| ∈ U and −|x| ≤ x ≤ |x|, so x ∈ U. Hence σ is finer than τ. If τ is Hausdorff, this implies that σ is Hausdorff as well. If ν is a locally solid vector space topology finer than τ, then for any circled, full τ-neighborhood U of 0, there is S ∈ ν circled and solid with 0 ∈ S ⊆ U. Then S = [S]f ⊆ [U]f . Thus, ν is finer than σ.

Notes and remarks Some authors prefer to see local fullness of a topology as a property of the order instead of the topology. They call the positive cone normal if it makes, in our terminology, the topology locally full; see, e.g., [94, 127, 133]. Results similar to Theorem 3.6.8 can be found in almost every book dealing with topologies on ordered spaces; see, e.g., [124, Theorem 6.7] or [127, Proposition 1.7]. The same can be said about Theorem 3.6.3; see, e.g., [127, Proposition 1.3].

3.7 Order convergence and unbounded order convergence In this section we reconsider o-convergence as introduced in Definition 1.1.31. It turns out that o-convergence does not have good topological properties. We briefly address the topological shortcomings of o-convergence and discuss some alternatives.

184 | 3 Seminorms on pre-Riesz spaces

3.7.1 Order convergence revisited First we present [6, Example 1.3], where it is observed that o-convergence of a net does not only depend on the tails of the net. Example 3.7.1. Let X be an Archimedean Riesz space with X ≠ {0} and fix a ∈ X \ {0}. Consider the index set I := ℤ≥0 ∪ { 1n ; n ∈ ℕ} endowed with the standard order. Define x0 := a and for α ∈ I \ {0} define x α := 1α a . Then {x α }α∈I is a net in X for which there does not exist x ∈ X and a net (y α )α∈I with y α ↓ 0 such that −y α ≤ x α − x ≤ y α for every α ∈ I. Indeed, such a net (y α )α∈I would satisfy y0 ≥ y 1n ≥ x 1n − x = na − x for every n ∈ ℕ, which is impossible by the Archimedean property of X. The net (x α )α∈I≥1 clearly o-converges to 0. Hence, (x α )α∈I is a net with an o-convergent tail, whereas the net itself is not o-convergent. It should be noted that a situation as in Example 3.7.1 cannot occur for sequences. Every tail of a sequence is preceded by only finitely many elements and therefore a sequence is o-convergent if and only if any of its tails is o-convergent. Historically, the definition of o-convergence has been mostly used for sequences, where the problem with the tail behavior does not occur. There are several ways to resolve the problem described in Example 3.7.1. Abramovich and Sirotkin [6] propose the following modification of the definition of o-convergence: o2

A net (x α )α∈I o2-converges to x, denoted by x α 󳨀󳨀→ x, if there exists α 0 ∈ I and a net (y α )α∈I≥α0 with y α ↓ 0 such that −y α ≤ x α − x ≤ y α for every α ∈ I≥α0 . Vulikh [162, Section I.4] allows even more freedom in the choice of nets: o3

A net (x α )α∈I o3-converges to x, denoted by x α 󳨀󳨀→ x, if there exist nets (y β )β∈J and (z γ )γ∈H with y β ↓ 0 and z γ ↓ 0 such that for every β ∈ J and γ ∈ H there exists α 0 ∈ I such that for every α ∈ I≥α0 we have −y β ≤ x α − x ≤ z γ . It is clear that o-convergence of a net implies o2-convergence and that o2-convergence implies o3-convergence. For o3-convergence, one dominating net is sufficient, as we observe next. Proposition 3.7.2. Let X be a partially ordered vector space, (x α )α∈I a net in X, and x ∈ X. The following statements are equivalent. (a) The net (x α )α∈I o3-converges to x. (b) There exists a net (y β )β∈J with y β ↓ 0 such that for every β ∈ J there exists an α 0 ∈ I such that for every α ∈ I≥α0 we have −y β ≤ x α − x ≤ y β .

3.7 Order convergence and unbounded order convergence | 185

Proof. Clearly, (b) implies (a). For the converse implication, assume that (x α )α∈I o3-converges to x. Take (y β )β∈J and (z γ )γ∈H with y β ↓ 0 and z γ ↓ 0 such that for every β ∈ J and γ ∈ H there exists an α 0 ∈ I such that for every α ∈ I≥α0 we have −y β ≤ x α − x ≤ z γ . Consider the index set J × H and the net (y β + z γ )(β,γ)∈J×H . We have that (y β + z γ )(β,γ)∈J×H is decreasing and that 0 is a lower bound. For every lower bound v of (y β + z γ )(β,γ)∈J×H we have v ≤ y β + z γ for every β ∈ J and γ ∈ H, hence by taking infimum over γ we obtain v ≤ y β and then v ≤ 0. Thus, y β + z γ ↓ 0. Let (β, γ) ∈ J × H. Then there exists α 0 ∈ I such that for every α ∈ I≥α0 we have −y β ≤ x α − x ≤ z γ and then −(y β + z γ ) ≤ x α − x ≤ (y β + z γ ). For positive bounded nets in Dedekind complete Riesz spaces the following characterization for o-convergence to 0 will be useful. Proposition 3.7.3. Let X be a Dedekind complete Riesz space and let (x α )α∈I be a bounded net in X+ . The following statements are equivalent. o 󳨀 0. (a) x α → o3

(b) x α 󳨀󳨀→ 0. (c) inf α∈I supβ∈I≥α x β = 0. Proof. Note that for every α ∈ I we have that supβ∈I≥α x β exists and that, likewise, inf α∈I supβ∈I≥α x β exists. It is immediately apparent that (a) implies (b). To show that (b) implies (c), aso3 sume that x α 󳨀󳨀→ 0 and suppose that inf α∈I supβ∈I≥α x β = 0 does not hold. Then x := o3 inf α∈I supβ∈I≥α x β > 0. As x α 󳨀󳨀→ 0, there is a net (y γ )γ∈J with y γ ↓ 0 such that for every γ ∈ J there is α 0 ∈ I such that for every β ∈ I≥α0 we have x β ≤ y γ . Then y γ ≥ supβ∈I≥α0 x β ≥ x > 0, which contradicts y γ ↓ 0. Thus, inf α∈I supβ∈I≥α x β = 0. It remains to show that (c) implies (a). Assume that inf α∈I supβ∈I≥α x β = 0. For α ∈ I, define y α := supβ∈I≥α x β . Then y α ↓ 0 and for every α ∈ I we have 0 ≤ x α ≤ y α . o

󳨀 0. Hence, x α → Example 3.7.1 shows that o2-convergence of a net does not imply o-convergence. An example due to Fremlin discussed in [6, Example 1.4] shows that o3-convergence does not imply o2-convergence. Nevertheless, we have the next remarkable result due to [6, Theorem 1.7]. Theorem 3.7.4. Let X and Y be Archimedean Riesz spaces and let T : X → Y be a positive operator. Then T preserves o2-convergence of nets if and only if T preserves o3-convergence of nets. Another problem of the definition of o-convergence is the absence of a corresponding topology. Consider a nonempty set X. A notion of convergence in X is a boolean function c on {((x α ) α∈I , x) ; (x α ) α∈I is a net in X, x ∈ X} .

186 | 3 Seminorms on pre-Riesz spaces We say that the net (x α )α∈I in X c-converges to x ∈ X if c((x α )α∈I , x) is TRUE. Clearly, o-convergence, o2-convergence, and o3-convergence are examples of notions of convergence. Definition 3.7.5. A notion of convergence c in X is called topological if there exists a topology τ on X such that for every net (x α )α∈I and every x ∈ X we have that (x α )α∈I c-converges to x if and only if (x α )α∈I τ-converges to x. For a notion of convergence c in X, let us call a set A ⊆ X c-closed if for every net (x α )α∈I in A and for every x ∈ X such that (x α )α∈I converges to x we have that x ∈ A. It is straightforward to verify that τ := {X \ A; A is c-closed}

(3.12)

is a topology on X. Moreover, τ-convergence and c-convergence of nets in X are equivalent if and only if c is topological. If a notion of convergence c is not topological, familiar properties of convergence from topology may fail. For instance, if A ⊆ X and C ⊆ X is the set of all x ∈ X for which there exists a net (x α )α∈I in A that c-converges to x, then C need not be c-closed. The notions of o-convergence, o2-convergence, and o3-convergence introduced above are not topological, in general; see, e.g., [66, Example 8.4]. However, the following surprising result holds; see [66, Theorem 3.14]. The topology τo is the topology defined in (3.12), where c is the notion of o-convergence. Theorem 3.7.6. Let X be a partially ordered vector space and let A ⊆ X. The following statements are equivalent. (a) A is τ o -closed. (b) A is o-closed. (c) A is o2-closed. (d) A is o3-closed. Theorem 3.7.4 is contained in the subsequent result², which is given in [66, Theorem 4.4]. Theorem 3.7.7. Let X and Y be partially ordered vector spaces and let T : X → Y be a positive operator. The following statements are equivalent. (a) T is τ o -continuous. (b) T is o-continuous. (c) T preserves o2-convergence. (d) T preserves o3-convergence. In most parts of the book, we simply use the definition of o-convergence. Theorems 3.7.6 and 3.7.7 yield that results involving o-closed sets or o-continuity of positive 2 The statement in [66, Theorem 4.4] is in fact formulated for monotone maps on partially ordered sets.

3.7 Order convergence and unbounded order convergence | 187

operators are equally true with o-convergence replaced by o2-convergence or o3-convergence.

3.7.2 Extension and restriction of order convergence In this subsection we address the question how order convergence in pre-Riesz spaces relates to order convergence in their vector lattice covers. We present a few partial results. For the next statement see also [84, Proposition 5.1]. Proposition 3.7.8. Let Z be an order dense subspace of a partially ordered vector space Y. Let oi denote o2 or o3. o o 󳨀 x in Z, then x α → 󳨀 x in Y. (i) If (x α ) is a net in Z and x ∈ Z is such that x α → oi

oi

(ii) If (x α ) is a net in Z and x ∈ Z is such that x α 󳨀→ x in Z, then x α 󳨀→ x in Y. (iii) If J ⊆ Y is o-closed (oi-closed) in Y, then J ∩ Z is o-closed (oi-closed) in Z. o

Proof. (i) Assume that x α → 󳨀 x in Z, i.e., there is a net (y α ) in Z such that y α ↓ 0 and for all α one has ±(x α −x) ≤ y α . Then y α ↓ in Y and y α ≥ 0 for all α. Due to Proposition 1.6.2, o one gets inf{y α } = 0 in Y, i.e., y α ↓ 0 in Y. Hence, x α → 󳨀 x in Y. The proof of (ii) is similar. The statements in (iii) are direct consequences of (i) and (ii). As an immediate consequence of Proposition 3.7.8 we obtain the following result on the restriction of o-closed sets. Theorem 3.7.9. Let X be a pre-Riesz space and (Y, i) a vector lattice cover of X. If J is an o-closed subset of Y, then the set [J]i is o-closed in X. Theorem 3.7.9 provides the restriction property (R) discussed in Section 2.8 for o-closed sets, i.e., (R) holds for the pair (L, M) where L = {I ⊆ X; I is o-closed}

and

M = {J ⊆ Y; J is o-closed} .

The converse statement of Proposition 3.7.8 (i) is not true, in general. We give an example where the o-convergence in Y does not imply the o-convergence in Z, even if the limit of the sequence is in Z; see also [84, Example 5.2]. Example 3.7.10. Consider the vector lattice Y = {y = (y i )i∈ℤ ∈ ℓ∞ (ℤ); lim y i exists} i→∞

and its subspace

∞

Z = {x = (x i )i∈ℤ ∈ Y; ∑ k=1

x−k = lim x i } . i→∞ 2k

188 | 3 Seminorms on pre-Riesz spaces (i)

(i)

(i)

As usual, for i ∈ ℤ denote e(i) = (e j )j∈ℤ with e j = 1 if i = j and e j = 0 otherwise. We first show that Z is order dense in Y. Indeed, let y = (y i )i∈ℤ ∈ Y and ∞

α := lim y i − ∑ i→∞

If α ≥ 0, define

x(1) := y + 2αe(−1)

and

k=1

y−k . 2k

x(2) := y + 4αe(−2) .

Then x(1) , x(2) ∈ Z, and y = x(1) ∧ x(2) . If α < 0, then for every m ∈ ℕ take ∞

x(m) := y − α ∑ e(i) . i=m

We obtain x(m) ∈ Z, and, moreover, y = inf{x(m) ; m ∈ ℕ}. Next we show that the sequence (e(m) )m∈ℕ in Z o-converges to 0 in Y, but it does (i) (m) ∈ Y for not o-converge in Z. Indeed, for m ∈ ℕ let y(m) := ∑∞ i=m e . Clearly, y (m) (m) (m) every m ∈ ℕ, and y ↓ m 0 in Y. Since ±e ≤ y for every m ∈ ℕ, one has o 󳨀 0 in Y. We show by way of contradiction that the sequence (e(m) )m∈ℕ does not e(m) → o o o-converge in Z. Suppose that there is v ∈ Z such that e(m) → 󳨀 v in Z. Then e(m) → 󳨀 v in Y, hence v = 0. Consequently, there is a sequence (v(m) )m∈ℕ in Z such that v(m) ↓ m 0 (m) and e(m) ≤ v(m) for every m ∈ ℕ. If i ≥ m, then v(m) ≥ v(i) ≥ e(i) . Hence, v i ≥ 1 for (m) ∞ v (m) −k all i ≥ m. This implies limi→∞ v i ≥ 1, and, consequently, ∑k=1 2k ≥ 1 for all (m)

m ∈ ℕ. We do not have v−k ↓ m 0 for every k ∈ ℕ \ {0}, since otherwise we would get v (m) −k ∑∞ ↓ m 0 by Theorem 1.8.13. Hence, there is k ∈ ℕ \ {0} and δ > 0 such that k=1 2k (m)

v−k ≥ δ for every m ∈ ℕ. Put w = δe(−k) − 2δe(−k−1) , then w ∈ Z and w ≤ δe(−k) ≤ v(m) for every m ∈ ℕ. Moreover, w ≰ 0, which contradicts inf{v(m) ; m ∈ ℕ} = 0.

The subsequent Proposition 3.7.12 yields a converse to Proposition 3.7.8 (ii) for o3-convergence, provided the pre-Riesz space is Archimedean and pervasive. The key ingredient of its proof is the following statement, which emerged in a discussion with Yang Deng. Lemma 3.7.11. Let X be a pervasive Archimedean pre-Riesz space and let (Y, i) be a vector lattice cover of X. Let (y α )α∈I be a net in Y such that y α ↓ 0. Then there exists a net (x β )β∈J in X with x β ↓ 0 and such that for every β ∈ J there exists α ∈ I with i(x β ) ≥ y α . Proof. If y α = 0 for some α ∈ I, then we can take J := I and x β := 0 for every β ∈ J. It remains for us to consider the case where y α > 0 for every α ∈ I. We will obtain the net (x β )β∈J by means of Zorn’s lemma. Denote X>0 := {x ∈ X; x > 0} and define A := {y α ; α ∈ I}. Then A is a nonempty subset of {y ∈ Y; y > 0}. Further, A is downward directed, since for every a1 , a2 ∈ A there are α1 , α 2 ∈ I with a1 = y α1 and a2 = y α2 . Then there is α 3 ∈ I with α 3 ≥ α 1 , α 2 and then a3 := y α3 ≤ y α1 , y α2 , so a3 ∈ A and a3 ≤ a1 , a2 . Moreover, inf A = inf{y α ; α ∈ I} = 0.

3.7 Order convergence and unbounded order convergence | 189

Define V := {B ⊆ X>0 ; B is nonempty, downward directed, and ∀b ∈ B ∃a ∈ A with i(b) ≥ a} . We order the set V by inclusion. First we show that V is nonempty. Indeed, fix a ∈ A. As X is majorizing in Y, there is u ∈ X with i(u) ≥ a. Let B := {u}. Then B ⊆ X>0 , B is downward directed, and a is such that i(u) ≥ a. Hence, B ∈ V. Let B be a nonempty chain in V. Define B 0 := ⋃ B. We show that B0 ∈ V. Indeed, B0 ≠ ⌀ and B0 ⊆ X>0 . For every b 1 , b 2 ∈ B0 , there are B1 , B2 ∈ B with b 1 ∈ B1 and b 2 ∈ B2 . As B is a chain, we have B1 ⊆ B2 or B1 ⊇ B2 . Without loss of generality we assume B1 ⊇ B2 . Then b 1 , b 2 ∈ B1 and as B1 is downward directed, there exists b 3 ∈ B1 such that b 3 ≤ b 1 , b 2 . Then b 3 ∈ B0 , so that B0 is downward directed. Let b ∈ B0 and take B ∈ B such that b ∈ B. Then there is a ∈ A with i(b) ≥ a. Hence, B0 ∈ V. Thus, B0 is an upper bound of B in V. Due to Zorn’s lemma, there exists a maximal element M in V. We proceed by showing that inf M = 0. Suppose not. As 0 is clearly a lower bound of M, there exists a lower bound v ∈ X of M with v ≰ 0. Then i(v)+ > 0 and i(v)+ is a lower bound of i[M]. Since inf A = 0, there exists a0 ∈ A with a0 ≱ i(v)+ . Then 0 < (i(v) − a0 )+ < i(v)+ . Since X is pervasive, there exists w ∈ X with 0 < i(w) ≤ (i(v) − a0 )+ . Consider N := M ∪ {b − w; b ∈ M}. We show that N ∈ V. Indeed, for every b ∈ M we have i(b − w) ≥ i(v)+ − i(w) ≥ i(v)+ − (i(v) − a0 )+ > 0, so N is a subset of X>0 . Clearly, N is nonempty. Since M is downward directed and w ≥ 0, it follows that N is downward directed. Let b ∈ M. Take a ∈ A such that i(b) ≥ a and take a1 ∈ A such that a1 ≤ a0 , a. Then a1 + i(w) ≤ a0 ∧ a + i(w) ≤ a0 ∧ i(b) + i(w) ≤ a0 ∧ i(b) + (i(v) − a0 )+ ≤ a0 ∧ i(b) + (i(b) − a0 )+ = i(b) , so i(b − w) ≥ a1 . It follows that for every c ∈ N there exists a1 ∈ A with i(c) ≥ a1 . Thus, N ∈ V. Since N ⊇ M and M is maximal in V, it follows that N = M. Hence, for every b ∈ M we have b − w ∈ M. Then we obtain for every b ∈ M and n ∈ ℕ that b − nw ∈ M, hence b − nw ≥ 0, which contradicts that X is Archimedean and w > 0. We conclude that inf M = 0. Finally, take J := M with reverse order and for β ∈ J define x β := β. Then (x β )β∈J is a net in X>0 , (x β )β∈J is decreasing, and x β ↓ 0. For every β ∈ J = M ∈ V, there is a ∈ A such that i(β) ≥ a. Then there is α ∈ I with y α = a and then i(x β ) ≥ y α . Proposition 3.7.12. Let X be a pervasive Archimedean pre-Riesz space and let (Y, i) be o3

a vector lattice cover of X. For a net (x α )α∈I in X and x ∈ X we have x α 󳨀󳨀→ x if and only o3

if i(x α ) 󳨀󳨀→ i(x). Proof. One implication follows immediately from Proposition 3.7.8. Let a net (x α )α∈I o3

in X and x ∈ X be such that i(x α ) 󳨀󳨀→ i(x). Then there are nets (y β )β∈J and (z γ )γ∈H

190 | 3 Seminorms on pre-Riesz spaces with y β ↓ 0 and z γ ↓ 0 such that for every β ∈ J and γ ∈ H there exists α 0 ∈ I such that for every α ∈ I≥α0 we have −y β ≤ i(x α ) − i(x) ≤ z γ . By Lemma 3.7.11 there are nets (u μ )μ∈M and (v ν )ν∈N in X with u μ ↓ 0, v ν ↓ 0 such that for every μ ∈ M there exists β ∈ J with i(u μ ) ≥ y β , and for every ν ∈ N there exists γ ∈ H with i(v ν ) ≥ z γ . Hence, for every μ ∈ M and ν ∈ N there exists α 0 ∈ I such that for every α ∈ I≥α0 we have o3

−i(u μ ) ≤ i(x α )− i(x) ≤ i(v μ ), which implies u μ ≤ x α − x ≤ v μ . We conclude x α 󳨀󳨀→ x.

3.7.3 Unbounded order convergence The shortcoming of o-convergence shown by Example 3.7.1 does obviously not occur if only bounded nets are considered. A notion of convergence that is based on order convergence of bounded nets is the so-called unbounded order convergence or uoconvergence for short. It was introduced by DeMarr in [48] in a setting of partially ordered vector spaces and has become an active topic of research in the theory of vector lattices more recently; see [40, 49, 57–59, 88, 140, 145, 159]. It turns out that uo-convergence coincides with pointwise or almost everywhere convergence in most common function spaces. It has generally more convenient properties than o-convergence. DeMarr’s definition comes down to the following. Definition 3.7.13. Let (X, K) be a partially ordered vector space, let (x α )α∈I be a net in X and x ∈ X. The net (x α )α∈I uoDM -converges to x if there exists a net (z α )α∈I in K such that – for every α ∈ I one has −z α ≤ x α − x ≤ z α , and – for every bounded net (y α )α∈I with 0 ≤ y α ≤ z α for every α ∈ I one has that (y α )α∈I o3-converges to 0. Actually, DeMarr uses the characterization of o3-convergence in Proposition 3.7.2 (b). In vector lattices, the following equivalent formulation of uoDM -convergence is wellknown. Proposition 3.7.14. Let X be a Riesz space, let (x α )α∈I be a net in X and x ∈ X. Then (x α )α∈I uoDM -converges to x if and only if for every u ∈ X+ one has that (|x α − x| ∧ u)α∈I o3-converges to 0. Proof. Assume that (x α )α∈I uoDM -converges to x. Then there exists a net (z α )α∈I in X+ such that for every α ∈ I we have −z α ≤ x α − x ≤ z α and such that for every bounded net (y α )α∈I with 0 ≤ y α ≤ z α for every α ∈ I we have that (y α )α∈I o3-converges to 0. That means that there is a net (v β )β∈J with v β ↓ 0 such that for every β ∈ J there is α 0 ∈ I such that for every α ∈ I≥α0 we have y α ≤ v β . Let u ∈ X+ . For α ∈ I, define y α := |x α − x| ∧ u. Then (y α )α∈I is a bounded net with 0 ≤ y α ≤ z α for every α ∈ I and therefore there is a net (v β )β∈J with v β ↓ 0 such that for every β ∈ J there is α 0 ∈ I such that for every α ∈ I≥α0 we have y α ≤ v β . Then for every α ∈ I≥α0 we have 0 ≤ |x a − x| ∧ u ≤ v β , which means that (|x α − x| ∧ u)α o3-converges to 0.

3.7 Order convergence and unbounded order convergence |

191

Conversely, assume that for every u ∈ X+ we have that (|x α − x| ∧ u)α o3-converges to 0. For α ∈ I, define z α := |x α − x|. Then clearly z α ≥ 0 and −z α ≤ x α − x ≤ z α . Let (y α )α∈I be a bounded net with 0 ≤ y α ≤ z α for every α ∈ I. Take u ∈ X+ such that y α ≤ u for every α ∈ I. Since (|x α − x| ∧ u)α o3-converges to 0, there is a net (v β )β∈J with v β ↓ 0 such that for every β ∈ J there is α0 ∈ I such that for every α ∈ I≥α0 we have |x α − x| ∧ u ≤ v β . Then 0 ≤ y α ≤ |x α − x| ∧ u ≤ v β for every α ∈ I≥α0 , so that (y α )α∈I o3-converges to 0. Thus, (x α )α∈I uoDM -converges to x. In view of the embedding of pre-Riesz spaces into their vector lattice covers, it is desirable to have a definition of uo-convergence in pre-Riesz spaces that is equivalent to uo-convergence in a vector lattice cover. Based on the characterization in Proposition 3.7.14, we propose the following definition and show its compatibility with respect to the embedding, provided the pre-Riesz space is pervasive³ and Archimedean. Definition 3.7.15. Let (X, K) be a partially ordered vector space, let (x α )α∈I be a net in X and x ∈ X. The net (x α )α∈I uo-converges to x if for every u ∈ K and for every net (y α )α∈I in K with y α ∈ (⋃ δ∈I≥α {x δ − x, x − x δ }ul ∩ {u}l )ul for every α ∈ I one has that (y α )α∈I o3-converges to 0. Lemma 3.7.16. Let X be a Riesz space, let (x α )α∈I be a net in X and x ∈ X. If (x α )α∈I uo-converges to x in X then (x α )α∈I uoDM -converges to x in X. Proof. Let (x α )α∈I be a net in X and x ∈ X such that (x α )α∈I uo-converges to x. For α ∈ I, define z α := |x α − x|. Let (y α )α∈I be a bounded net with 0 ≤ y α ≤ z α for every α ∈ I. We wish to show that (y α )α∈I o3-converges to 0. Choose u ∈ X+ such that y α ≤ u for every α ∈ I. For every α ∈ I and b ∈ (⋃ δ∈I≥α {x δ − x, x − x δ }ul ∩ {u}l )u , we have b ∈ ({x α − x, x − x α }ul ∩ {u}l )u . As |x α − x| ∧ u ∈ {x α − x, x − x α }ul ∩ {u}l , we obtain b ≥ |x α − x| ∧ u ≥ y α , so that y α ∈ (⋃ δ∈I≥α {x δ − x, x − x δ }ul ∩ {u}l )ul . Thus, (y α )α∈I o3-converges to 0. Theorem 3.7.17. Let (X, K) be a pervasive Archimedean pre-Riesz space and let (Y, i) be a vector lattice cover. Let (x α )α∈I be a net in X and x ∈ X. Then (x α )α∈I uo-converges to x in X if and only if (i(x α ))α∈I uoDM -converges to i(x) in Y. Proof. Assume that (i(x α ))α∈I uoDM -converges to i(x) in Y. Let u ∈ K and let (y α )α∈I be a net in K with y α ∈ (⋃ δ∈I≥α {x δ −x, x−x δ }ul ∩{u}l )ul for every α ∈ I. By Proposition 3.7.14, the net (|i(x α ) − i(x)| ∧ i(u))α∈I o3-converges to 0. Hence, there exists a net (v β )β∈J in Y with v β ↓ 0 such that for every β ∈ J there is α β ∈ I such that for every α ∈ I≥α β we have |i(x α ) − i(x)| ∧ i(u) ≤ v β . As X is Archimedean and pervasive, Lemma 3.7.11 yields a net (u γ )γ∈H in X with u γ ↓ 0 such that for every γ ∈ H there is β γ ∈ J such that for every β ∈ J≥β γ we have v β ≤ i(u γ ). Let γ ∈ H. For every α ∈ I≥α βγ and δ ∈ I≥α we have

3 The value of the definition beyond the setting of pervasive pre-Riesz spaces is unclear, as then the set of lower bounds in the definition need not contain nontrivial positive elements.

192 | 3 Seminorms on pre-Riesz spaces |i(x δ ) − i(x)| ∧ i(u) ≤ i(u γ ), hence u γ ∈ (⋃ δ∈I≥α {x δ − x, x − x δ }ul ∩ {u}l )u . Therefore, y α ≤ u γ . It follows that (y α )α∈I o3-converges to 0. Thus, (x α )α∈I uo-converges to x in X. For the converse implication, we first consider the case that Y is Dedekind complete. Assume that (x α )α∈I uo-converges to x in X. We use Proposition 3.7.14. Let u ∈ Y+ . We show that (|i(x α ) − i(x)| ∧ u)α∈I o3-converges to 0. Suppose not. Define z α := |i(x α ) − i(x)| ∧ u. By Proposition 3.7.3, we have z := inf α∈I supδ∈I≥α z δ > 0. As X is pervasive, there exists a ∈ X with 0 < i(a) ≤ z. For every α ∈ I, define y α := a. Then (y α )α∈I is a net in K. Choose w ∈ K such that i(w) ≥ u. For every α ∈ I and b ∈ (⋃ δ∈I≥α {x δ − x, x − x δ }ul ∩ {w}l )u we have i(b) ≥ |i(x δ ) − i(x)| ∧ i(w) ≥ z δ for every δ ∈ I≥α . Consequently, i(b) ≥ supδ∈I≥α z δ ≥ z ≥ i(a). Therefore, b ≥ a = y α . Hence, y α ∈ (⋃ δ∈I≥α {x δ − x, x − x δ }ul ∩ {w}l )ul . By assumption, we obtain that (y α )α∈I o3-converges to 0. Then there is a net (v β )β∈J in X with v β ↓ 0 and for every β ∈ J there is α β ∈ I such that for every α ∈ I≥α β we have y α ≤ v β . Then v β ≥ y α β = a for every β ∈ J, which yields a contradiction to v β ↓ 0. Thus, (i(x α ))α∈I uoDM -converges to i(x) in Y. Finally, we consider an arbitrary vector lattice cover (Y, i) of X. Assume that (x α )α∈I uo-converges to x in X. Since X is Archimedean, the Riesz space Y is Archimedean by Proposition 2.4.12. Therefore, the Dedekind completion (Y δ , j) of Y is a vector lattice cover of Y. By Lemma 1.6.3 we obtain that (Y δ , j ∘ i) is a vector lattice cover of X. It follows from the previous part of the proof that (j(i(x α )))α∈I uoDM -converges to j(i(x)) in Y δ . The first part of the proof then yields that (i(x α ))α∈I uo-converges to i(x) in Y. Hence, by Lemma 3.7.16 we obtain that (i(x α ))α∈I uoDM -converges to i(x) in Y. Corollary 3.7.18. Let X be an Archimedean Riesz space, let (x α )α∈I be a net in X and x ∈ X. The following statements are equivalent. (a) The net (x α )α∈I uo-converges to x. (b) The net (x α )α∈I uoDM -converges to x. (c) For every u ∈ X+ one has that (|x α − x| ∧ u)α∈I o3-converges to 0.

4 Disjointness, bands, and ideals in pre-Riesz spaces The use of vector lattice structure in functional analysis often appears in the form of disjointness or special subspaces, such as ideals and bands. Basic properties of these notions in vector lattices have been discussed in Section 1.3. Several notions of ideals in partially ordered vector spaces were used in the literature from the 1950s on, independently of vector lattice theory. We develop a systematic theory of disjointness, bands and ideals in pre-Riesz spaces and transfer well-known properties from the vector lattice theory to pre-Riesz spaces as far as possible. The embedding methods presented in Chapter 2 serve as fundamental tools.

4.1 Disjointness In Section 2.8 we discussed two approaches to generalize notions from the theory of Riesz spaces to pre-Riesz spaces. One approach replaces lattice operations by sets of upper bounds, the second one applies the embedding into a vector lattice cover. We use the first approach to define disjointness in pre-Riesz spaces and show compatibility with the second approach. An early version of this work appeared in [83]. We explore properties of disjointness and relate it to known concepts from the literature, for example to antilattices.

4.1.1 Disjointness in partially ordered vector spaces Recall that in a vector lattice (X, K) two elements x, y ∈ X are disjoint if and only if |x + y| = |x − y| ; see Proposition 1.3.2. With the method discussed in Section 2.8, we replace this identity of moduli by an identity for corresponding sets of upper bounds. https://doi.org/10.1515/9783110476293-004

194 | 4 Disjointness, bands, and ideals in pre-Riesz spaces Definition 4.1.1. In a partially ordered vector space (X, K), two elements x, y ∈ X are called disjoint, denoted by x ⊥ y, if {x + y, −x − y}u = {x − y, −x + y}u . The disjoint complement of a nonempty set M ⊆ X is the set M d = {x ∈ X; x ⊥ v for all v ∈ M} . We collect some simple properties of disjointness. For x, y ∈ X one has y ⊥ x if and only if x ⊥ y, and, obviously, x ⊥ 0. Clearly, x ⊥ y implies −x ⊥ y. If x ⊥ x, then {x + x, −x − x}u = {x − x}u = K, hence x, −x ≤ 0, which yields x = 0. Similarly, x ⊥ (−x) implies x = 0. If x, y ∈ K are such that x ⊥ y, and z ≥ x, y, then z ∈ {x − y, −x + y}u = {x + y}u , therefore z ≥ x + y. For M ⊆ X the set M d is nonempty since one has 0 ∈ M d . Further, if M and N are subsets of X such that M ⊆ N, then M d ⊇ N d . Clearly, M ⊆ M dd and hence M d = M ddd . We continue with a geometric characterization of disjointness; see also [84, Section 3]. Proposition 4.1.2. Let (X, K) be a partially ordered vector space. For x, y ∈ X one has x ⊥ y if and only if the four inclusions (K ± x) ∩ (K ± y) ⊆ K

(4.1)

are satisfied. Proof. Let x, y ∈ X with x ⊥ y, i.e., (K + x + y) ∩ (K − x − y) = (K + x − y) ∩ (K − x + y) .

(4.2)

Adding x + y, using that K = 2K, and dividing by 2, respectively, (4.2) is equivalent to (K + 2x + 2y) ∩ K = (K + 2x) ∩ (K + 2y) ⇔

(2K + 2x + 2y) ∩ (2K) = (2K + 2x) ∩ (2K + 2y) ⇔

(K + x + y) ∩ K = (K + x) ∩ (K + y) .

(4.3)

Moreover, one has x ⊥ −y, −x ⊥ −y and −x ⊥ y. Using analogous calculations as in (4.3), one gets the four inclusions in (4.1). Conversely, assume (4.1). From (K + x) ∩ (K − y) ⊆ K one gets (K + x + y) ∩ K ⊆ (K + y) , furthermore (K − x) ∩ (K + y) ⊆ K implies K ∩ (K + x + y) ⊆ (K + x), hence (K + x + y) ∩ K ⊆ (K + x) ∩ (K + y) . On the other hand (K + x)∩(K + y) ⊆ K, and (K − x)∩(K − y) ⊆ K implies (K + y)∩(K + x) ⊆ (K + x + y). Thus, (K + x) ∩ (K + y) ⊆ (K + x + y) ∩ K . Hence, (4.3) is satisfied, which is equivalent to x ⊥ y.

4.1 Disjointness

| 195

4.1.2 Disjointness and embedding If a partially ordered vector space X is a linear subspace of a vector lattice Y, then we could consider the notion of disjointness in X induced by Y and call two elements disjoint in X if they are disjoint in Y. We will show that the notion of disjointness in X induced by Y coincides with the intrinsic definition given in Subsection 4.1.1, provided X is an order dense subspace of Y. We start by considering order dense subspaces of partially ordered vector spaces; see also [83, Proposition 2.1]. Here one has to pay attention in which space the upper bounds according to Definition 4.1.1 are taken. Proposition 4.1.3. Let Y be a partially ordered vector space, X a linear subspace of Y with the induced order, and let x, y ∈ X. (i) If x ⊥ y in Y, then x ⊥ y in X. (ii) If X is, in addition, order dense in Y, then x ⊥ y in X if and only if x ⊥ y in Y. Proof. (i) Recall that x ⊥ y in Y means that {v ∈ Y; v ≥ x + y, v ≥ −x − y} = {v ∈ Y; v ≥ x − y, v ≥ −x + y} . If we intersect these sets with X, we immediately obtain that x ⊥ y in X. (ii) We assume {z ∈ X; z ≥ x + y, z ≥ −x − y} = {z ∈ X; z ≥ x − y, z ≥ −x + y} . Let v ∈ Y be such that v ≥ x + y, v ≥ −x − y. If z ∈ X is such that z ≥ v, then z ≥ x + y, z ≥ −x − y, and due to the assumption we have z ≥ x − y, z ≥ −x + y. By order denseness, v = inf{z ∈ X; z ≥ v} ≥ x − y

and, similarly,

v ≥ −x + y ,

hence {v ∈ Y; v ≥ x + y, v ≥ −x − y} ⊆ {v ∈ Y; v ≥ x − y, v ≥ −x + y}. The converse inclusion is proven analogously. If (X, K) is a pre-Riesz space and (Y, i) an associated vector lattice cover, then Proposition 4.1.3 and the order denseness of i[X] in Y yield the following statement. Proposition 4.1.4. Let X be a pre-Riesz space and let (Y, i) be a vector lattice cover of X. Then for x, y ∈ X one has x ⊥ y in X if and only if i(x) ⊥ i(y) in Y. Embedding a pre-Riesz space X in its vector lattice cover (Y, i) does not only yield an equivalent way of describing disjointness, it also yields properties of disjoint complements. Here we use the notation introduced at the beginning of Subsection 1.2.1. Proposition 4.1.5. Let X be a pre-Riesz space, (Y, i) a vector lattice cover of X, and M ⊆ X. Then

196 | 4 Disjointness, bands, and ideals in pre-Riesz spaces (i) M d = [i[M]d ]i, and (ii) M d is a linear subspace of X. Proof. (i) The formula is obtained by Proposition 4.1.4, since for an element x ∈ X one has x ∈ M d if and only if i(x) ∈ i[M]d . (ii) Recall that i[M]d is a band in the vector lattice Y, hence a linear subspace. Therefore, by (i), M d is a linear subspace of X. If the partially ordered vector space X is not pre-Riesz, then the disjoint complement of a subset of X need not be a subspace, as the next example shows. Example 4.1.6. Let X = ℝ2 and let K = {0} ∪ {(x1 , x2 )T ∈ ℝ2 ; x2 > 0} . In Example 2.2.6 it is observed that (X, K) is a directed partially ordered vector space, which is not a pre-Riesz space. Let x = (1, 0)T . We show that {x}d = {(α, 0)T ; α ∈ ℝ \ {1, −1}} .

(4.4)

Let y = (α, 0)T with α ∈ ̸ {−1, 1}. Then {x + y, −x − y}u = [(1 + α, 0)T + K] ∩ [(−1 − α, 0)T + K] = {(u 1 , u 2 )T ; u 2 > 0} , because α ≠ −1. Further, as α ≠ 1, we obtain {x − y, −x + y}u = [(1 − α, 0)T + K] ∩ [(−1 + α, 0)T + K] = {(u 1 , u 2 )T ; u 2 > 0} . Hence, x ⊥ y, which shows ⊇ in (4.4). It remains to show that the elements of the complement of the right-hand side of (4.4) are not disjoint to x. Clearly, (1, 0)T = x ⊥̸ x, and (−1, 0)T = −x ⊥̸ x. Moreover, if y = (y1 , y2 )T ∈ ℝ2 is such that y2 > 0, then x + y ∈ {x + y, −x − y}u , but x + y ≱ −x + y, so x ⊥̸ y. Similarly, if y2 < 0, then we also obtain x ⊥̸ y. Now from (4.4) it follows that {x}d is not a linear subspace of X.

4.1.3 Disjointness in spaces with the Riesz decomposition property The literature contains approaches to disjointness in partially ordered vector spaces other than the one in Definition 4.1.1. In [93, Definition 8] a notion of disjointness is suggested, which employs Proposition 1.3.2 (iv). We use the notion D-disjointness to avoid confusion. Definition 4.1.7. In a partially ordered vector space (X, K), two elements x, y ∈ K are called D-disjoint if [0, x] ∩ [0, y] = {0} .

4.1 Disjointness

|

197

Note that D-disjointness is only defined for positive elements. An advantage of D-disjointness is that it can be easily checked. In [93] the notion of D-disjointness is only used in partially ordered vector spaces that have the Riesz decomposition property. In partially ordered vector spaces without the Riesz decomposition property, D-disjoint complements lack appropriate properties, as the next example illustrates. Example 4.1.8. We continue Example 1.7.3, i.e., X = ℝ3 is equipped with the cone Lℝ2 = {(x1 , x2 , x3 )T ; x21 + x22 ≤ x23 , x3 ≥ 0} . For x = (1, 0, 1)T the set of all (positive) elements which are D-disjoint to x is given by {λ(y1 , y2 , 1)T ; y21 + y22 = 1, y1 ≠ 1, λ ≥ 0} . This set is not convex. In general, D-disjointness does not imply disjointness. Indeed, in Example 4.1.8 there exist nontrivial D-disjoint elements, whereas there are no nontrivial disjoint elements¹. The following result establishes that the two notions coincide in a preRiesz space provided the space has the Riesz decomposition property; see also [84, Proposition 6.1]. Proposition 4.1.9. Let (X, K) be a pre-Riesz space and let x, y ∈ K. (i) If x ⊥ y, then x and y are D-disjoint. (ii) If (X, K) has, in addition, the Riesz decomposition property, then x ⊥ y if and only if x and y are D-disjoint. Proof. (i) Let (Y, i) be a vector lattice cover of (X, K). If x ⊥ y in X, then i(x) ⊥ i(y) in Y due to Proposition 4.1.4. Proposition 1.3.2 yields that [0, i(x)] ∩ [0, i(y)] = {0}, by intersecting with i[X] we obtain that i(x) and i(y) are D-disjoint in i[X], hence x and y are D-disjoint in X. (ii) Let X have the Riesz decomposition property and let x, y ∈ K be such that x and y are D-disjoint. Since x and y are positive, one has {x + y, −x − y}u = {x + y}u ⊆ {x − y, −x + y}u . We show the converse inclusion. Let z ∈ {x − y, −x + y}u , then 0 ≤ x ≤ z + y. The Riesz decomposition property implies that there are elements x1 , x2 ∈ K with x = x1 + x2 , 0 ≤ x1 ≤ z and 0 ≤ x2 ≤ y. Now 0 ≤ x2 ≤ x in combination with the D-disjointness of x and y implies x2 = 0, i.e., z ≥ x. Analogously, one gets z ≥ y.

1 This can e.g., be deduced from Proposition 4.1.11, as the spaces in Example 4.1.8 and Example 1.7.4 are order isomorphic.

198 | 4 Disjointness, bands, and ideals in pre-Riesz spaces Now, 0 ≤ x ≤ z = (z − y) + y, hence x = x3 + x4 with 0 ≤ x3 ≤ z − y and 0 ≤ x4 ≤ y. From 0 ≤ x4 ≤ x one obtains x4 = 0 and therefore x ≤ z − y, which shows that z ∈ {x + y}u . Hence, x ⊥ y.

4.1.4 Disjointness in antilattices Anti-lattices attracted attention in the theory of operator algebras. A partially ordered vector space (X, K) is called an antilattice if x and y are comparable in X whenever x and y have a greatest lower bound (or, equivalently, a least upper bound). This notion was introduced by Kadison [74], who showed that the space of all bounded self-adjoint operators on a Hilbert space is an antilattice, and wrote: A moment’s thought shows that this is as strongly nonlattice as a partially ordered vector space can be. It is natural to ask which conditions ensure that a pre-Riesz space is an antilattice. Moreover, is there any lattice structure in an antilattice in the sense that there exist nontrivial disjoint elements? The following theorem characterizes antilattices by means of the existence of nontrivial disjoint elements in the cone; see also [77, Theorem 14]. Theorem 4.1.10. Let (X, K) be a pre-Riesz space. (i) If x, y ∈ K and x ⊥ y, then x ∨ y exists and equals x + y. (ii) X is an antilattice if and only if there are no nontrivial disjoint elements in K. Proof. Let (X ρ , i) be the Riesz completion of X. (i) Let x, y ∈ K be such that x ⊥ y. By Proposition 4.1.4 one has i(x) ⊥ i(y) in X ρ , hence i(x) ∨ i(y) = i(x) + i(y) = i(x + y). Consequently, i(x) ∨ i(y) ∈ i(X), which implies that x ∨ y exists in X. Moreover, x ∨ y = x + y. (ii) Assume that there are elements x, y ∈ K \ {0} with x ⊥ y. Due to (i), x ∨ y exists, and x and y are not comparable (otherwise x ≤ y would imply i(x) ≤ i(y), hence i(x) ⊥̸ i(y), which yields x ⊥̸ y). Assume that X is not an antilattice, i.e., there are x, y ∈ X that are not comparable, but x ∨ y exists. The elements x ∨ y − y and x ∨ y − x are positive and nonzero, as otherwise x and y would be comparable. We now show that they are disjoint. First observe that i(x ∨ y) = i(x) ∨ i(y). Indeed, i is bipositive, and therefore i(x ∨ y) ≥ i(x) ∨ i(y). Moreover, one has {i(v); v∈X, i(v) ≥ i(x) ∨ i(y)} ⊆ {i(v); v∈X, v ≥ x, v ≥ y} = {i(v); v∈X, v ≥ x ∨ y} and, since i[X] is order dense in X ρ , i(x) ∨ i(y) = inf {i(v); v ∈ X, i(v) ≥ i(x) ∨ i(y)} ≥ inf {i(v); v ∈ X, i(v) ≥ i(x ∨ y)} = i(x ∨ y) .

| 199

4.1 Disjointness

Due to (1.6) one obtains i(x ∨ y − y) ∧ i(x ∨ y − x) = (i(x) ∨ i(y) − i(y)) ∧ (i(x) ∨ i(y) − i(x)) = (i(x) − i(y))+ ∧ (i(x) − i(y))− = 0 . Hence, i(x ∨ y − y) ⊥ i(x ∨ y − x) in Y, which implies x ∨ y − y ⊥ x ∨ y − x in X. If (X, K) is an Archimedean partially ordered vector space with order unit, then the existence of nontrivial disjoint elements in K can be checked by means of the vector lattice cover of X given in Theorem 2.5.9, since there the disjointness can be investigated easily due to the pointwise order. To illustrate this method, we consider the space (V, Pos n ) of symmetric n × n-matrices ordered by the cone of positive semidefinite matrices and equipped with the inner product (1.18). We show that (V, Pos n ) has no nontrivial disjoint elements. Then Theorem 4.1.10 implies that (V, Pos n ) is an antilattice (which is also shown in [75]). Proposition 4.1.11. There are no nontrivial disjoint elements in (V, Pos n ). Proof. Recall that Σ, Λ and Φ are given by (1.19), Lemma 2.6.5 and (2.35). We deal with the vector lattice cover in Proposition 2.6.7. Let A, B ∈ V \ {0}; we show that A and B are not disjoint. Recall that A ⊥ B is equivalent to Φ(A) ⊥ Φ(B) and note that Φ(A) ⊥ Φ(B) ⇐⇒ min{|tr(AC)|, |tr(BC)|} = 0 T

for every C ∈ Λ T

⇐⇒ min{|tr((Aq)q )|, |tr((Bq)q )|} = 0 for every q ∈ S n−1 ⇐⇒ min{|⟨Aq, q⟩|, |⟨Bq, q⟩|} = 0

for every q ∈ S n−1 .

As A ≠ 0 and A is symmetric, it follows from the spectral theorem that A has an eigenvalue λ ≠ 0 with a normalized eigenvector v. Likewise, B has an eigenvalue μ ≠ 0 with normalized eigenvector w. Put q = αv + βw with α, β ∈ ℝ. Then ⟨Aq, q⟩ = ⟨λαv + βAw, αv + βw⟩ = λα 2 + αβ⟨Aw, v⟩ + λαβ⟨v, w⟩ + β 2 ⟨Aw, w⟩ and ⟨Bq, q⟩ = μβ 2 + αβ⟨Bv, w⟩ + μαβ⟨v, w⟩ + α 2 ⟨Bv, v⟩ . If ⟨Bv, v⟩ ≠ 0, then |⟨Av, v⟩| ∧ |⟨Bv, v⟩| ≠ 0. If ⟨Bv, v⟩ = 0, we can take β = |⟨Aq, q⟩| = |λα 2 + ⟨Aw, v⟩ + λ⟨v, w⟩ +

1 ⟨Aw, w⟩| α2

→∞

1 α

> 0 to get

as α → ∞ ,

and ⟨Bq, q⟩ =

μ α2

+ ⟨Bv, w⟩ + μ⟨v, w⟩,

which is nonzero for α large. Therefore, there is a q ∈ ℝn such that |⟨Aq, q⟩| ∧ |⟨Bq, q⟩| ≠ 0. Thus, A and B are not disjoint. The next example shows that there exist antilattices with nontrivial disjoint elements; see also [77, Section 6]. We make use of Proposition 2.6.9, where a cone is constructed starting from a convex set, which becomes the base of the dual cone.

200 | 4 Disjointness, bands, and ideals in pre-Riesz spaces Example 4.1.12. Let T = {(cos θ, sin θ, 0)T ∈ ℝ3 ; 0 ≤ θ < 2π} and let S be the convex hull of the set T ∪ {(0, 0, −1)T , (0, 0, 1)T }, see Figure 4.1.

Fig. 4.1: The set S.

Let C, K and Σ be defined as in Proposition 2.6.9, then u := (1, 0, 0, 0)T is an order unit in K. The set of extreme points of Σ is Λ = {(1, 0, 0, −1)T , (1, 0, 0, 1)T } ∪ {(1, cos θ, sin θ, 0)T ; 0 ≤ θ < 2π} , which is a closed set. Hence, a vector lattice cover of (ℝ4 , K) is given by the functional representation (C(Λ), Φ) according to Theorem 2.5.9. There are disjoint elements in (ℝ4 , K). Take, e.g., x = (0, 0, 0, 1)T and y = (0, 0, 1, 0)T . Then it is straightforward that Φ(x) ⊥ Φ(y), hence x ⊥ y. Moreover, both x and y are not in K. We now show that there are no nontrivial disjoint elements in K, i.e., (ℝ4 , K) is an antilattice by Theorem 4.1.10. Let x, y ∈ K be such that x ≠ 0 and x ⊥ y. We show that y = 0. Let s x = {φ ∈ Λ; φ(x) = 0} and s y = {φ ∈ Λ; φ(y) = 0}. Then s x ∪ s y = Λ. According to the possible supporting hyperplanes for S, there are three essential cases to consider (up to multiplication of the fourth coordinate by −1): (1) s x = {(1, 0, 0, 1)T }, (2) there is θ ∈ [0, 2π) such that s x = {(1, 0, 0, 1)T , (1, cos θ, sin θ, 0)T }, and (3) there is θ ∈ [0, 2π) such that s x = {(1, cos θ, sin θ, 0)T }. In each of these cases we get span({s y }) ⊇ span({φ; φ ∈ Λ \ s x }) = ℝ4 . We conclude that y = 0. Apparently, among the antilattices as defined by Kadison some have nontrivial disjoint elements, and others do not. Thus, even antilattices may have some traces of lattice structure.

4.1.5 Fordable pre-Riesz spaces Subsequent results² concerning bands or disjointness preserving operators can be established if the underlying pre-Riesz space contains ‘sufficiently many’ disjoint ele2 See for example the Theorems 4.2.6 and 5.2.21 below.

4.1 Disjointness

| 201

ments. More precisely, if X is a pre-Riesz space and (Y, i) a vector lattice cover of X, one wants to replace the disjoint complement of an element of Y with the disjoint complement of a subset of X. We define fordable pre-Riesz spaces with respect to the Riesz completion and show in Proposition 4.1.18 below that the analogous statement is satisfied for an arbitrary vector lattice cover. Definition 4.1.13. Let X be a pre-Riesz space and (X ρ , i) its Riesz completion. X is called fordable if for every y ∈ X ρ , y ≥ 0, there is M ⊆ X such that {y}d = i[M]d . The notion of a fordable pre-Riesz space first appeared as a technical condition in [149]. We show that every pervasive pre-Riesz space is fordable; refer to [149, Lemma 2.4]. For a discussion of pervasive subspaces, see Section 2.8. Lemma 4.1.14. Let (Z, K) be a vector lattice and D a pervasive order dense subspace of Z. Then for every y ∈ Y, y ≥ 0, there is S ⊆ D ∩ K with S d = {y}d . Proof. Let y ∈ Z, y ≥ 0. We assume y ≠ 0. Let S := {x ∈ D; 0 ≤ x ≤ y}. Note that S is nonempty, since D is pervasive. On the one hand, if z ∈ Z is such that |z| ∧ y = 0, then 0 ≤ |z| ∧ x ≤ |z| ∧ y = 0 for every x ∈ S, hence {y}d ⊆ Sd . On the other hand, assume that there exists z ∈ Sd \ {y}d . Put w = |z| ∧ y, then w ∈ Z and w > 0. As D is pervasive, there is x ∈ X such that 0 < x ≤ w. As w ≤ y we obtain x ∈ S and therefore, x ∧ |z| = 0. Hence, x ≤ x ∧ w ≤ x ∧ |z| = 0 , which is a contradiction. We conclude that {y}d = Sd . For a pre-Riesz space X we obtain the subsequent statement as an immediate consequence of Lemma 4.1.14, where Z = X ρ and D = i[X]. Proposition 4.1.15. Every pervasive pre-Riesz space is fordable. Proposition 4.1.15 is useful for examples, as it seems to be slightly more handy to check whether a pre-Riesz space is pervasive than fordable. Before we show that for a pre-Riesz space X the property in the Definition 4.1.13 carries over to an arbitrary vector lattice cover instead of X ρ , we provide two technical results. ρ

Lemma 4.1.16. Let X be a fordable pre-Riesz space and let S ⊆ X+ . Then (Sdd ∩ i[X])d = Sd . Proof. Because of Sdd ∩ i[X] ⊆ Sdd we have (Sdd ∩ i[X])d ⊇ Sddd = Sd . Vice versa, let w ∈ (Sdd ∩ i[X])d and s ∈ S. Since X is fordable, there is M ⊆ X such that {s}d = i[M]d . This implies Sd ⊆ i[M]d and, hence, i[M]dd ⊆ Sdd . Therefore, i[M] ⊆ i[M]dd ∩ i[X] ⊆ Sdd ∩ i[X], which yields w ∈ i[M]d = {s}d . We conclude that w ∈ Sd .

202 | 4 Disjointness, bands, and ideals in pre-Riesz spaces

Lemma 4.1.17. Let Z be a Riesz space and let D be a pervasive Riesz subspace of Z. If A, B ⊆ D are such that their disjoint complements in D coincide, then Ad = Bd in Z. Proof. Let A, B ⊆ D be such that Ad ∩ D = Bd ∩ D. Let y ∈ Ad . Assume that there is b ∈ B such that y ⊥̸ b, i.e., |b| ∧ |y| ≠ 0. As D is pervasive in Z, there is x ∈ D with 0 < x ≤ |b| ∧ |y|. This implies x ∈ Ad ∩ D = Bd ∩ D. Because of x ≤ |b| one has x = x ∧ |b| = 0, which yields a contradiction. Therefore, Ad ⊆ Bd , and, by symmetry, Ad = Bd . Proposition 4.1.18. Let X be a pre-Riesz space and (Y, i) a vector lattice cover of X. If X is fordable, then for every y ∈ Y, y ≥ 0, there is M ⊆ X such that {y}d = i[M]d . Proof. Observe that X ρ can be considered as an order dense vector sublattice of Y. By Proposition 2.8.3, the space X ρ is pervasive in Y. ρ Let y ∈ Y, y ≥ 0. Due to Lemma 4.1.14, there is S ⊆ X + with {y}d = Sd in Y. Set M := [Sdd ]i, i.e., i[M] = Sdd ∩ i[X]. Lemma 4.1.16 implies that i[M]d = Sd in X ρ . From Lemma 4.1.17 it follows that i[M]d = Sd in Y, hence {y}d = i[M]d . The subsequent heartbreaking example shows that a fordable pre-Riesz space is not pervasive, in general. Example 4.1.19. Let Y = C[0, 1] and consider the linear subspace 1

X := {α𝟙 + v; α ∈ ℝ, v ∈ C[0, 1], v(0) = 0, ∫ v(t)dt = 0} , 0

where 𝟙 denotes the constant-one function on [0, 1]. To establish that X is order dense in Y, we show that for every y ∈ Y, t ∈ (0, 1), ε > 0 there is x ∈ X with x ≥ y and x(t) < y(t) + ε. Indeed, let y ∈ Y, t ∈ (0, 1), ε > 0, α > max{y(s); s ∈ [0, 1]}, δ := 14 min{t, 1 − t}. Choose z ∈ C[0, 1] such that 0 ≤ z ≤ α𝟙 − y, z(t) > α − y(t) − ε and z(s) = 0 for all s ∈ [0, t − δ] ∪ [t + δ, 1]. Set y(s) := z(s + 2t ) for s ∈ [ 2t − δ, 2t + δ] and y(s) := 0 for s ∈ [0, 2t − δ] ∪ [ 2t + δ, 1]. Let x := α𝟙 − z + y, then x ∈ X, x ≥ α𝟙 − z ≥ α𝟙 − (α𝟙 − y) = y, and x(t) = α − z(t) < y(t) + ε. Hence, X is order dense in Y, and X ρ is the vector lattice generated by X in Y. X is not pervasive. Indeed, there is y ∈ X ρ such that y > 0 and y(0) = 0. If x ∈ X is such that 0 ≤ x ≤ y, then x(0) = 0. Since x = α𝟙 + v with α ∈ ℝ, v ∈ C[0, 1], 1 1 ∫0 v(t)dt = 0, v(0) = 0, one obtains α = 0, hence ∫0 x(t)dt = 0, and therefore x = 0. Finally we show that X is fordable with respect to (Y, i). Let y ∈ Y be such that y > 0. There are finitely many or countably many mutually disjoint intervals (a n , b n ) such that Ω := {t ∈ (0, 1); y(t) ≠ 0} = ⋃(a n , b n ) . n

4.2 Bands |

203

Let z n ∈ X be the piecewise linear function with z n (s) = 0 for s ∈ [0, a n ] ∪ [b n , 1], z n ( 34 a n + 14 b n ) = 1, z n ( 14 a n + 34 b n ) = −1. Then x := ∑n 2z nn is well-defined, and x ∈ X. One obtains {t ∈ (0, 1); x(t) ≠ 0} = ⋃ (a n , 12 (a n + b n )) ∪ ( 12 (a n + b n ), b n ) ⊆ Ω , n

{x}d

{y}d .

hence ⊇ If u ∈ Y is such that u ⊥ x, then u(s) = 0 for every s ∈ Ω, therefore u ⊥ y. We conclude {y}d = {x}d , which implies that X is fordable with respect to (Y, i). Finally we give an example of a pre-Riesz space that is not fordable (and, hence, not pervasive). Example 4.1.20. The pre-Riesz space X = P2 (ℝ) of all real polynomial functions on ℝ of at most degree 2 in Example 1.7.1 is not fordable. Indeed, consider the vector lattice cover (V, i) of X, where V is given in 1.7.1, and i is the identity. For S ⊆ X one has either i[S]d = {0} or i[S]d = V, whereas there are y ∈ V with nontrivial {y}d . Another example of a nonfordable pre-Riesz space is given in Example 4.4.18 below.

Notes and Remarks In 2003 Arkady Kitover asked us whether a definition of disjointness in partially ordered vector spaces can be given with appropriate properties. The present chapter is our answer. Antilattices are considered in the context of (operator) algebras, so far. Proposition 4.1.11 is a special case of Kadison’s antilattice theorem, which reads as follows. For a real or complex Hilbert space H, the vector space of all bounded self-adjoint operators on H endowed with the cone of all positive semidefinite operators is an antilattice; see [74, Theorem 6]. Kadison also pointed out that a von Neumann algebra is a factor if and only if it is an antilattice; see [74, Corollary 11 and Theorem 12]. Archbold showed in [16, Theorem 3.2] that a C∗ -algebra is prime if and only if its selfadjoint part is an antilattice. In [138], Sherman proved that the self-adjoint part of a C∗ -algebra is a lattice if and only if the C∗ -algebra is commutative. In normed pre-Riesz spaces, properties of disjoint complements are also studied in [142] and [143].

4.2 Bands In Archimedean vector lattices, bands are the linear subspaces that equal their doubledisjoint complement. This property will be used to define bands in pre-Riesz spaces. We investigate the extension property (E) and the restriction property (R) as discussed in Section 2.8 for bands.

204 | 4 Disjointness, bands, and ideals in pre-Riesz spaces

4.2.1 Bands in pre-Riesz spaces Recall that in an Archimedean vector lattice Y for a set M ⊆ Y the following statements are equivalent; see Theorem 1.3.8: (i) M is a band, (ii) M equals its double disjoint complement, i.e., M = M dd . We use the property (ii) to define bands in a pre-Riesz space (X, K); see also [83, Definition 5.4]. Observe that for M ⊆ X the inclusion M ⊆ M dd always holds. Definition 4.2.1. A subset M of a pre-Riesz space (X, K) is called a band if M = M dd . Due to Proposition 4.1.5, every band is a linear subspace. Clearly, for M ⊆ X one has (M dd )d ⊆ M d ⊆ (M d )dd , which gives the subsequent statement. Proposition 4.2.2. Let X be a pre-Riesz space and M ⊆ X. Then M d is a band. The question arises under which conditions bands in a pre-Riesz space are compatible with bands in an associated vector lattice cover. Let (X, K) be a pre-Riesz space and (Y, i) its vector lattice cover. In view of the extension and restriction property for a pair (L, M) ⊆ P(X) × P(Y) in Section 2.8, now let L := {I ⊆ X; I is a band}

and

M := {J ⊆ Y; J is a band} .

(4.5)

We will show that (E) is always satisfied, whereas (R) holds under the condition that X is fordable. Further, we will deal with the set L of o-closed subsets of X and the set M of o-closed subsets of Y. In this case we show that (R) is true. We will conclude that every band in a pre-Riesz space is o-closed.

4.2.2 Extension of bands Let (X, K) be a pre-Riesz space and (Y, i) an associated vector lattice cover. For a band ̂ in Y. B in X, we establish two natural procedures to construct an extension band B For the first one, see also [84, Proposition 5.12]; for the second one see [78]. The next theorem yields the extension property (E) for bands. Theorem 4.2.3. Let X be a pre-Riesz space, (Y, i) a vector lattice cover of X, and B a band in X. Then the sets ̂ := i[B d ]d , and (I) B ̂ := i[B]dd (II) B are bands in Y, respectively, with the property ̂ . B = [B]i ̂ in (II) is the smallest band in Y extending B in the sense of (4.6). Moreover, B

(4.6)

4.2 Bands | 205

̂ is a band in Y due to Proposition 4.2.2. Recall that Proof. In both cases (I) and (II), B u ⊥ v in X is equivalent to i(u) ⊥ i(v) in Y, and that for M ⊆ X one has M d = [i[M]d ]i, due to Proposition 4.1.4. To show (4.6) for (I), let M := Bd , which implies ̂ i = [i[Bd ]d ] i = Bdd = B . [B] Next we show (4.6) for (II). First observe that for M ⊆ X one has M ⊆ [i[M]dd ] i ⊆ M dd . The first inclusion is obvious since i[M] ⊆ i[M]dd ∩ i[X]. To show the second inclusion, let x ∈ [i[M]dd ]i, that is, i(x) ∈ i[M]dd . For y ∈ M d = [i[M]d ]i one has i(y) ∈ i[M]d , hence y ⊥ x. Therefore, x ∈ M dd . ̂ = B for Now, if B is a band in X, then B ⊆ [i[B]dd ]i ⊆ Bdd = B, which implies [B]i (II). Finally let D be a band in Y with [D]i = B, then i[B] ⊆ D, hence Dd ⊆ i[B]d , which implies D = Ddd ⊇ i[B]dd . ̂ in Y that satisfies (4.6). In general, for a band B in X there is no largest band B Example 4.2.4. We continue Example 1.7.5, i.e., S is the unit circle in ℝ2 and X is the space of restrictions to S of affine functions from ℝ2 to ℝ with pointwise order on S. The space X is an order dense subspace of C(S), i.e., C(S) is a vector lattice cover of X. Let O1 := {(a, b)T ∈ S; a > 0 or b > 0} and O2 := {(a, b)T ∈ S; a < 0 or b < 0}. Then O1 and O2 are regularly open subsets of S. Recall that for a regularly open set O ⊆ S we denote I O := {x ∈ C(S); ∀s ∈ S \ O : x(s) = 0}. By Proposition 1.3.13, the ideals I O1 and I O2 are bands in C(S). Affine functions in I O1 or I O2 are identically zero, hence both I O1 and I O2 extend the band {0} in X. The only band in C(S) larger than I O1 and I O2 is C(S), which does not extend {0} ⊆ X. We conclude from Theorem 4.2.3 that the extension property (E) is satisfied for the pair (L, M) as in (4.5). As a consequence, we obtain the following; see also [84, Theorem 5.14]. Theorem 4.2.5. In a pre-Riesz space, every band is o-closed. Proof. Let X be a pre-Riesz space, (Y, i) a vector lattice cover of X, and let B be a band ̂ in Y such that B = [B]i. ̂ In particular, B ̂ is in X. Due to Theorem 4.2.3 there is a band B o-closed in Y. Because of Theorem 3.7.9, B is o-closed in X. Since in a pre-Riesz space X every disjoint complement is a band, it follows that every disjoint complement is o-closed. For the pair (L, M) as in (4.5), the restriction property (R) is not satisfied, in general; see Example 4.4.18 below.

206 | 4 Disjointness, bands, and ideals in pre-Riesz spaces

4.2.3 Restriction of bands in fordable pre-Riesz spaces We address the question of whether there are pre-Riesz spaces where the restriction property (R) for bands is valid. We show that it suffices to require that the pre-Riesz space is fordable, i.e., that there are ‘many’ disjoint elements; see also [149, Proposition 2.5]. The next theorem provides the restriction property (R) for bands in fordable pre-Riesz spaces. Theorem 4.2.6. Let X be a fordable pre-Riesz space, (Y, i) a vector lattice cover of X, and let B be a band in Y. Then [B]i is a band in X. Proof. Let B be a band in Y. Clearly, [B]i ⊆ [B]idd , hence we have to show that [B]idd ⊆ [B]i. Let x ∈ [B]idd . We prove that i(x) ∈ Bdd = B. Indeed, let y ∈ Bd . Then |y| ∈ Y, |y| ≥ 0, and due to Proposition 4.1.18 there is a set S ⊆ X such that {|y|}d = i[S]d . Let v ∈ [B]i. Then y ⊥ i(v), hence i(v) ∈ {|y|}d = i[S]d . Proposition 4.1.4 implies v ∈ Sd . Consequently, [B]i ⊆ Sd , hence [B]idd ⊆ Sddd = Sd , and, in particular, x ∈ Sd . We have i(x) ∈ i[Sd ] ⊆ i[S]d = {|y|}d , which implies i(x) ⊥ y. As y was an arbitrary element of Bd , we obtain i(x) ∈ Bdd = B, that is, x ∈ [B]i. We conclude [B]i = [B]idd . The restriction property (R) for bands in pervasive pre-Riesz spaces is obtained due to Proposition 4.1.15. Corollary 4.2.7. Let X be a pervasive pre-Riesz space, (Y, i) a vector lattice cover of X, and let B be a band in Y. Then [B]i is a band in X. In Theorem 4.2.6, the condition on X to be fordable is sufficient, but not necessary, as the next example shows. Example 4.2.8. We continue Example 4.1.20. Recall that the space X = P2 (ℝ) of all real polynomial functions on ℝ of at most degree 2 is not fordable. Nevertheless, (R) for bands is satisfied. For a nontrivial band B in V one has Bd ≠ {0}, i.e., for x ∈ [B]i there is y ∈ V, y ≠ 0, with i(x) ⊥ y, which implies i(x) = 0. We conclude [B]i = {0}, which is a band in X.

4.3 Ideals In the theory of vector lattices, there are many equivalent ways of stating that a subspace is an ideal. Most of these formulations have a counterpart in partially ordered vector spaces without the lattice property. We will consider such generalizations for partially ordered vector spaces and investigate their basic properties and relations. For pre-Riesz spaces we will study the subspaces that correspond to ideals in a vector lattice cover. For directed subspaces it turns out that all these generalizations are equivalent, just as in the case of vector lattices.

4.3 Ideals | 207

4.3.1 Ideals in partially ordered vector spaces We recall well-known characterizations of ideals in vector lattices. Proposition 4.3.1. Let X be a vector lattice and let I be a linear subspace of X. The following properties of I are equivalent. (a) I is a full Riesz subspace of X. (b) I is a solid subspace of X, i.e., I is an ideal. (c) For every y ∈ I, the convex solid hull of {y} in X is contained in I. (d) I is a Riesz subspace of X and there exist a Riesz space Y and a positive linear map T : X → Y such that I = [{0}]T. (e) There exist a Riesz space Y and a Riesz homomorphism T : X → Y such that I = [{0}]T. There are variations in formulation of the above stated properties of I. Instead of (a) the combination of I being a Riesz subspace and ‘for every x ∈ X, y ∈ I with −y ≤ x ≤ y one has x ∈ I,’ or ‘for every x ∈ X, y ∈ I with 0 ≤ x ≤ y one has x ∈ I’ are sometimes used. In more specific situations ideals are described by means of intersections of kernels of positive functionals or of Riesz homomorphisms into ℝ. These descriptions can be viewed as special cases of (d) and (e) with Y = ℝA for some set A. We will discuss generalizations of each of the formulations (a)–(e) and study their relations. We choose the formulation in Proposition 4.3.1(b) to define ideals in partially ordered vector spaces. Definition 4.3.2. Let (X, K) be a partially ordered vector space. A subset I ⊆ X is called an ideal of X if I is a solid linear subspace of X. As in the theory of vector lattices, every band will turn out to be an ideal; see Theorem 4.3.13 below. The remainder of this subsection is devoted to a discussion of the property in Proposition 4.3.1(a). The condition that I be a Riesz subspace cannot be stated in a partially ordered vector space. One generalization is obtained by omitting that condition completely. Note that there are full subspaces that are not directed. Even worse, every subspace without nontrivial positive elements is full. Lemma 4.3.3. Let (X, K) be a partially ordered vector space and let I be a linear subspace of X. The following three properties of I are equivalent. (a1) I is full. (a2) For every y ∈ I and x ∈ X with −y ≤ x ≤ y one has x ∈ I. (a3) For every y ∈ I and x ∈ X with 0 ≤ x ≤ y one has x ∈ I. Proof. Clearly, (a1) implies (a2), and (a2) implies (a3). If (a3) holds and y, z ∈ I and x ∈ X are such that y ≤ x ≤ z, then 0 ≤ x − y ≤ z − y and z − y ∈ I, hence x − y ∈ I, so x = (x − y) + y ∈ I, which yields (a1).

208 | 4 Disjointness, bands, and ideals in pre-Riesz spaces

Subspaces satisfying (a2) appear in Kadison’s paper [75, Definition 2.2] and are subsequently investigated in [26–28, 95]. In [56], Fuchs defines an ideal to be a directed subspace with property (a1). Since there are bands that are not directed (see Example 4.4.18), not every band is an ideal in the sense of Fuchs. For a subspace V of X define the directed part of V to be the subspace (V ∩ K) − (V ∩ K). Clearly, the directed part of V is a directed subspace of X. Lemma 4.3.4. Let (X, K) be a partially ordered vector space and let I be a linear subspace of X. I is a full subspace if and only if the directed part of I is a full subspace. Proof. Since I and the directed part of I have the same positive elements, it is clear that (a3) of Lemma 4.3.3 is satisfied for I if and only if it is satisfied for the directed part of I. We relate solid subspaces and full subspaces of a partially ordered vector space. Lemma 4.3.5. Let (X, K) be a partially ordered vector space and let I be a linear subspace of X. (i) If I is solid, then I is full. (ii) If I is full and directed, then I is solid. Proof. (i) If I is solid and y ∈ I and x ∈ X are such that 0 ≤ x ≤ y, then {−x, x}u ⊇ {−y, y}u , hence x ∈ I. Due to Lemma 4.3.3, it follows that I is full. (ii) Assume that I is full and directed. Let y ∈ I and x ∈ X be such that {−x, x}u ⊇ {−y, y}u . Since I is directed, there is v ∈ {−y, y}u ∩ I. Then v ≥ −x and v ≥ x, hence −v ≤ x ≤ v and therefore x ∈ I, as I is full. Thus I solid. Lemma 4.3.6. Let (X, K) be a partially ordered vector space and let I be a linear subspace of X. If I is solid then the directed part of I is solid. Proof. Assume that I is solid. Let y be in the directed part I0 of I and let x ∈ X be such that {−x, x}u ⊇ {−y, y}u . Then x ∈ I, and as I0 is directed there exists v ∈ I0 with v ≥ −y, y. Hence, −v ≤ x ≤ v, thus x = (x + v) − v ∈ (I ∩ K) − (I ∩ K) = I0 . Consequently, I0 is solid. The next example shows that there exist full subspaces that are not solid and that a subspace with a solid directed part need not be solid. Example 4.3.7. Consider in ℝ3 the generating polyhedral cone 1

0

1

1

K := pos {( 0 ) , ( 1 ) , (

−1 0 0 ) , ( −1 )} 1 1

.

In the space (ℝ3 , K) one easily finds a subspace that is full, but not solid. In view of Lemma 4.3.5, this subspace is not directed. Indeed, let I0 be the subspace of ℝ3 spanned by x0 := (1, 0, 0)T . Then I0 contains no positive elements other than the

4.3 Ideals |

209

origin, hence for every y, z ∈ I0 the order interval [y, z] is contained in I0 . Thus, I0 is full. Observe that {−x0 , x0 }u = K + {(0, 0, 1)T } = {−x1 , x1 }u = {−x2 , x2 }u , where x1 = (0, 1, 0)T and x2 = (0, 0, 1)T . Therefore, if I is a solid subspace containing x0 , then I = ℝ3 . Hence, I0 is not solid. The directed part of I0 equals {0}, which is trivially solid. More properties of the space (ℝ3 , K) will be shown in Example 4.4.18 below.

4.3.2 Extension and restriction of ideals A generalization of the subspaces with the property in Proposition 4.3.1(c) are solvex subspaces in partially ordered vector spaces. Indeed, in a vector lattice, a subset is solvex if and only if it is solid and convex; see Lemma 3.2.16. In partially ordered vector spaces, every solvex subspace is an ideal. We will show that for ideals the restriction property (R) as discussed in Section 2.8 holds, whereas the extension property (E) fails, in general. We show that solvex ideals satisfy both (R) and (E), i.e., solvex ideals are precisely those ideals in a pre-Riesz space that are compatible with ideals in a vector lattice cover. Since we will deal with the solvex hull of ideals, we consider the solvex hull of linear subspaces first; see also [84, Lemma 3.5]. Lemma 4.3.8. Let X be a partially ordered vector space and M a subspace of X. Then the solvex hull S of M is a subspace of X, and, hence, an ideal. Proof. Since M is a linear subspace, its solvex hull is given by S = {x ∈ X; ∃ x1 , . . . , x n ∈ M such that u

n

{x, −x}u ⊇ { ∑ ε k x k ; ε1 , . . . , ε n ∈ {1, −1}} } . k=1

Let x, y ∈ S; we show x + y ∈ S. There are x1 , . . . , x n , y1 , . . . , y m ∈ M such that n

u

{x, −x}u ⊇ { ∑ ε k x k ; ε1 , . . . , ε n ∈ {1, −1}} k=1

and

m

u

{y, −y}u ⊇ { ∑ δ j y j ; δ1 , . . . , δ m ∈ {1, −1}} . j=1

Suppose u is an upper bound of the set n

m

{ ∑ ε k x k + ∑ δ j y j ; ε1 , . . . , ε n , δ1 , . . . , δ m ∈ {1, −1}} , k=1

j=1

210 | 4 Disjointness, bands, and ideals in pre-Riesz spaces i.e., for any ε1 , . . . , ε n , δ1 , . . . , δ m ∈ {1, −1} we have u ≥ ∑nk=1 ε k x k +∑m j=1 δ j y j . Hence, m

n

u − ∑ δj yj ≥ ∑ εk xk , j=1

k=1

which implies that u − ∑m j=1 δ j y j is an upper bound of the set n

{ ∑ ε k x k ; ε1 , . . . , ε n ∈ {1, −1}} . k=1

Therefore, u

− ∑m j=1

δ j y j ≥ ±x. From this it follows that m

m

u − x ≥ ∑ δj yj

and

u + x ≥ ∑ δj yj ,

j=1

j=1

which means that u − x and u + x are upper bounds of the set m

{ ∑ δ j y j ; δ1 , . . . , δ m ∈ {1, −1}} j=1

and hence of {y, −y}. We conclude u − x ≥ y and u + x ≥ −y, which implies u ∈ {x + y, −x − y}u . Consequently, n

u

m

{x + y, −x − y}u ⊇ { ∑ ε k x k + ∑ δ j y j ; ε1 , . . . , ε n , δ1 , . . . , δ m ∈ {1, −1}} , j=1

k=1

and we infer that x + y ∈ S. Further, if x ∈ S, then −x ∈ S. Also, 0 ∈ S. Let x ∈ S and λ > 0, we show λx ∈ S. There are x 1 , . . . , x n ∈ M such that u

n

{x, −x}u ⊇ { ∑ ε k x k ; ε1 , . . . , ε n ∈ {1, −1}} . k=1

If u is an upper bound of the set {∑nk=1 ε k λx k ; ε1 , . . . , ε n ∈ {1, −1}}, then 1λ u is an upper bound of {∑nk=1 ε k x k ; ε1 , . . . , ε n ∈ {1, −1}}, therefore 1λ u ∈ {x, −x}u , which implies u ∈ {λx, −λx}u . Hence, n

u

{λx, −λx}u ⊇ { ∑ ε k λx k ; ε1 , . . . , ε n ∈ {1, −1}} k=1

and, therefore, λx ∈ S. For the next result, see also [84, Proposition 5.6]. Proposition 4.3.9. Let Y be a vector lattice and let Z be an order dense subspace of Y. For an ideal I in Z the following two statements are equivalent. (i) There is an ideal J in Y such that I = J ∩ Z. (ii) I is solvex in Z.

4.3 Ideals |

211

Proof. To show that (i) implies (ii), let J be an ideal in Y such that I = J ∩ Z. Then J is solid and convex in Y and, hence, solvex by Lemma 3.2.16. Due to Proposition 3.2.21 (i), the subspace J ∩ Z is solvex. For a proof that (ii) implies (i), let J be the solvex hull of I in Y. Due to Lemma 4.3.8, the set J is an ideal in Y. Further, according to Proposition 3.2.21 (ii), we obtain J ∩ Z = I. If X is a pre-Riesz space and (Y, i) a vector lattice cover of X, Proposition 3.2.5 gives the restriction property (R) for ideals, i.e., (R) is satisfied for the pair (L, M) where L = {I ⊆ X; I is an ideal}

and

M = {J ⊆ Y; J is an ideal} .

Proposition 3.2.21 yields (R) for solvex ideals, and Proposition 4.3.9 establishes the extension property (E) for solvex ideals, since in Y every ideal is solvex. We summarize these results in the subsequent theorem. Theorem 4.3.10. Let X be a pre-Riesz space and (Y, i) a vector lattice cover of X. (i) If J is an ideal in Y, then [J]i is a solvex ideal in X. (ii) If I is a solvex ideal in X, then there is an ideal J in Y such that I = [J]i. Moreover, one can take for J the solvex hull of i[I]. Next we present an example of a pre-Riesz space X and an ideal I in X such that I is not solvex in X; see also [84, Example 5.7]. According to Theorem 4.3.10, there is no ideal J in a vector lattice cover (Y, i) of X such that I = [J]i, i.e., the extension property (E) for ideals is not satisfied. The ideal I in the following example is even o-closed. Example 4.3.11. We continue Example 1.7.5, i.e., S is the unit circle in ℝ2 and X is the space of restrictions to S of all affine functions from ℝ2 into ℝ, equipped with the pointwise order. Recall that X is pre-Riesz and order dense in C(S). According to (1.24), for every s ∈ S and every x ∈ C(S) one has x(s) = inf{u(s); u ∈ X, u ≥ x}. Consider in X the subset I = {x : S → ℝ, (ξ, η)T 󳨃→ μ1 ξ + μ 2 η; μ 1 , μ2 ∈ ℝ} . (a) I is an ideal in X. Clearly, I is a linear subspace of X, hence it remains to show that I is solid. Let y = μ 1 ξ + μ 2 η ∈ I, y ≠ 0. The zeros of y are s1 = (

μ2

, 2

√ μ 21 +μ 2

−μ 1

T

) 2

√ μ 21 +μ 2

and

s2 = (

−μ 2

T

, 2

√ μ 21 +μ 2

μ1

) . 2

√ μ 21 +μ 2

212 | 4 Disjointness, bands, and ideals in pre-Riesz spaces Let x ∈ X be such that {x, −x}u ⊇ {y, −y}u . In the following calculation, for z ∈ X let |z|C(S) denote the absolute value of z taken in C(S). One obtains |x(s1 )| = inf{u(s1 ); u ∈ X, u ≥ |x|C(S) } = inf{u(s1 ); u ∈ X, u ≥ ±x} ≤ inf{u(s1 ); u ∈ X, u ≥ ±y} = inf{u(s1 ); u ∈ X, u ≥ |y|C(S) } = |y(s1 )| = 0 , consequently x(s1 ) = 0, and, similarly, x(s2 ) = 0. Hence, there is λ ∈ ℝ such that x = λy, i.e., x ∈ I. (b) I is not solvex. We consider X ρ as a subspace of C(S) as in Example 3.2.13. Let J be the smallest ideal in X ρ that contains I. In particular, for the functions x1 : S → ℝ, (ξ, η)T 󳨃→ ξ , and x2 : S → ℝ, (ξ, η)T 󳨃→ η, one obtains |x1 | ∨ |x2 | ∈ J. There is δ > 0 such that for all s ∈ S one has (|x1 | ∨ |x2 |)(s) ≥ δ, hence all constant functions belong to J. Since every element of X ρ is a bounded function, we obtain J = X ρ . Now, e.g., the function x3 : S → ℝ, s 󳨃→ 1, is an element of J ∩ X = X, but does not belong to I. Due to Proposition 4.3.9, I is not solvex. (c) I is o-closed. o Let (x α ) be a net in I and x ∈ X such that x α → 󳨀 x in X, i.e., there is a net (y α ) in X with y α ↓ 0 and ±(x α − x) ≤ y α for all α. There are μ 1 , μ2 , μ3 ∈ ℝ such that x : (ξ, η)T 󳨃→ μ 1 ξ + μ 2 η + μ 3 . We have to show that x ∈ I, which means μ 3 = 0. We will show that (y α ) converges pointwise to zero and then conclude that (x α ) converges pointwise to x. Then μ 3 = 0 follows. For each s ∈ S the net (y α (s)) is decreasing in [0, ∞), and hence convergent. Define y(s) := lim y α (s) . α

We claim that y is the restriction on S of an affine function on ℝ2 . The affine functions are determined by their values at three distinct points on the circle S. We choose the points (1, 0)T , (0, 1)T , and (−1, 0)T . Every point of S is a linear combination of these points with the sum of the coefficients equal to 1. That is, for (ξ, η)T ∈ S there are λ1 , λ2 , λ3 ∈ ℝ such that λ1 + λ2 + λ3 = 1 and (ξ, η)T = λ1 (1, 0)T + λ2 (0, 1)T + λ3 (−1, 0)T . Denote ν1 := 12 [y(1, 0) − y(−1, 0)] , ν2 := y(0, 1) − 12 [y(1, 0) + y(−1, 0)] , ν3 := 12 [y(1, 0) + y(−1, 0)] .

4.3 Ideals | 213

Then y(ξ, η) = lim y α (ξ, η) α

= lim y α (λ1 (1, 0)T + λ2 (0, 1)T + λ3 (−1, 0)T ) α

= lim(λ1 y α (1, 0) + λ2 y α (0, 1) + λ3 y α (−1, 0)) α

= λ1 y(1, 0) + λ2 y(0, 1) + λ3 y(−1, 0) = λ1 (ν1 + ν3 ) + λ2 (ν2 + ν3 ) + λ3 (ν3 − ν1 ) = ν1 (λ1 − λ3 ) + ν2 λ2 + ν3 (λ1 + λ2 + λ3 ) = ν1 ξ + ν2 η + ν3 . Consequently, y is an affine function, i.e., y ∈ X. One has 0 ≤ y and y ≤ y α for all α. Since inf α y α = 0, it follows that y = 0. Hence, limα y α (ξ, η) = 0 for all (ξ, η)T ∈ S. Therefore, as ±(x(ξ, η) − x α (ξ, η)) ≤ y α (ξ, η), one has x(ξ, η) = limα x α (ξ, η). From x(1, 0) = μ 1 + μ 3 and x(−1, 0) = −μ1 + μ 3 it follows that 2μ 3 = x(1, 0) + x(−1, 0) = lim(x α (1, 0) + x α (−1, 0)) . α

Since x α ∈ I one has x α (1, 0) = −x α (−1, 0), hence the last limit is equal to 0. We conclude x ∈ I. In C(Ω), where Ω is a compact Hausdorff space, an ideal is not characterized by means of a subset of Ω, in general, recall Example 1.3.9. In special cases, for instance if Ω is finite, such a characterization is possible; see Example 1.3.14. It carries over to a characterization of solvex subspaces in finite-dimensional spaces with polyhedral cones³, as the next example shows. Example 4.3.12. Let K be a generating polyhedral cone in ℝn as in Subsection 1.7.2. Then by (1.26) there exist k ≥ n linear functionals f (1) , . . . , f (k) on ℝn such that K = {x ∈ ℝn ; ∀i ∈ {1, . . . , k} : f (i) (x) ≥ 0} . By Lemma 2.6.1, for every i the functional f (i) defines a face {x ∈ K; f (i) (x) = 0} of K of dimension n − 1. Let Φ : ℝn → ℝk be given by Φ(x) = (f (1) (x), . . . , f (k) (x))T . By Theorem 2.6.2, the functional representation (ℝk , Φ) is the Riesz completion of (ℝn , K). By Example 1.3.14, every ideal J in ℝk is a band and there is a set N ⊆ {1, . . . , k} such that J = {(y1 , . . . , y k ); ∀i ∈ N : y i = 0} .

(4.7)

We show that a subspace I of (ℝn , K) is solvex if and only if there exists a set N ⊆ {1, . . . , k} such that I = {x ∈ ℝn ; ∀i ∈ N : f (i) (x) = 0} . (4.8) 3 This was first observed in the diploma thesis Solvexe Ideale in Räumen mit polyedralen Kegeln und finite Elemente of Manuela Berg, TU Dresden, 2010.

214 | 4 Disjointness, bands, and ideals in pre-Riesz spaces Indeed, for N ⊆ {1, . . . , k}, the subspace J given by (4.7) is a band and hence a solvex ideal. Now Theorem 4.3.10 (i) yields that I = [J]Φ is a solvex ideal in (ℝn , K). Conversely, let I be a solvex subspace of (ℝn , K). Let J be the solvex hull of Φ[I] in ℝk . Then J is an ideal in ℝk . Observe that {i ∈ {1, . . . , k}; ∀y ∈ J : y i = 0} = {i ∈ {1, . . . , k}; ∀y ∈ Φ[I] : y i = 0} = {i ∈ {1, . . . , k}; ∀x ∈ I : f (i) (x) = 0} . Hence, for N ⊆ {1, . . . , k} as in (4.7), we obtain [J]Φ = {x ∈ ℝn ; ∀i ∈ N : f (i) (x) = 0} . By Theorem 4.3.10 (ii) we get I = [J]Φ, which yields (4.8). In Definition 4.2.1, bands have been defined in pre-Riesz spaces by means of doubledisjoint complements. With the aid of the extension and restriction method, we show that a band is always a solvex ideal; see also [84, Theorem 5.14]. Theorem 4.3.13. In a pre-Riesz space, every band is an o-closed solvex ideal. Proof. Let X be a pre-Riesz space, (Y, i) a vector lattice cover of X, and let B be a band in X. Theorem 4.2.5 yields that B is o-closed in X. Due to Theorem 4.2.3 there is a band ̂ in Y such that B = [B]i. ̂ In particular, B ̂ is an ideal in Y. Because of Theorem 4.3.10(i), B the set B is a solvex ideal in X. Since in a pre-Riesz space X every disjoint complement is a band, it follows that a disjoint complement is also an o-closed solvex ideal. Example 4.4.18 below shows that an o-closed solvex ideal in a pre-Riesz space need not be a band, in general. The statements and counterexamples concerning the restriction property (R) and the extension property (E) for ideals and bands are summarized as follows: (R) ideal o-closed ideal solvex ideal band

yes yes yes no

(E) (Theorem 4.3.10) (Theorem 3.7.9) (Theorem 4.3.10) (Example 4.4.18)

no no yes yes

(Example 4.3.11) (Example 4.3.11) (Theorem 4.3.10) (Theorem 4.2.3)

One crucial difference between vector lattices and pre-Riesz spaces is the fact that in a vector lattice every ideal is directed, whereas there exist nondirected ideals and bands in pre-Riesz spaces; see Example 4.4.18 below. Recall that an o-closed ideal that is not solvex was given in Example 4.3.11, whereas a solvex ideal that is not o-closed was given in Example 1.3.9. The following picture illustrates the relations between subspaces in pre-Riesz spaces, in contrast to Archimedean vector lattices.

4.3 Ideals | 215

Archimedean vector lattice

pre-Riesz space

band

band

= o-closed ideal

o-closed ideal

solvex ideal

ideal = solvex ideal

ideal

4.3.3 Directed ideals Let us discuss generalizations of the properties in Proposition 4.3.1 (d) and (e). The generalization of (d) to partially ordered vector spaces is straightforward. The statement in (e) says that ideals in a vector lattice are kernels of Riesz homomorphisms. In a partially ordered vector space there are different ways of replacing the Riesz homomorphisms by other operators. One could use Riesz* homomorphisms or Riesz homomorphisms as defined in Section 2.3. Proposition 4.3.14. Let (X, K) and (Y, L) be partially ordered vector spaces and let T : X → Y be a positive linear operator such that for each w ∈ L there is x ∈ K with T(x) = w. If the kernel [{0}]T of T is directed, then T is a Riesz homomorphism. Proof. Let x, y ∈ X. Since T is positive, we have T({x, y}u )l ⊇ {T(x), T(y)}ul . To show T({x, y}u )l ⊆ {T(x), T(y)}ul , we prove T({x, y}u ) ⊇ {T(x), T(y)}u . Let w ∈ Y be such that w ≥ T(x) and w ≥ T(y). Then w − T(x) ∈ L, hence there is x1 ∈ K with T(x1 ) = w − T(x). Put x2 := x + x1 , then x2 ≥ x and T(x2 ) = w. Analogously, there is y2 ≥ y with T(y2 ) = w. Clearly, y2 − x2 ∈ [{0}]T, and, since [{0}]T is directed, there is z ∈ [{0}]T with z ≥ 0 and z ≥ y2 − x2 . Put v := x2 + z and observe T(v) = w, v ≥ x2 ≥ x and v ≥ x2 + (y2 − x2 ) = y2 ≥ y. Hence, v ∈ {x, y}u with T(v) = w, therefore w ∈ T({x, y}u ). Corollary 4.3.15. Let (X, K) be a partially ordered vector space and let f : X → ℝ be a positive functional such that there exists x ∈ K with f(x) > 0. If [{0}]f is directed, then f is a Riesz homomorphism. If (X, K) is a directed partially ordered vector space and f : X → ℝ is a positive functional, then either f = 0 or there is x ∈ X such that f(x) ≠ 0, which by directedness of X implies that there is y ∈ K with f(y) > 0. We thus arrive at the following corollary.

216 | 4 Disjointness, bands, and ideals in pre-Riesz spaces Corollary 4.3.16. Let (X, K) be a partially ordered vector space. Suppose that X is directed. Let f : X → ℝ be a positive functional. If [{0}]f is directed, then f is a Riesz homomorphism. In the theory of Riesz spaces, the equivalence of ideals and kernels of Riesz homomorphisms is established by means of quotient maps. A similar approach succeeds for partially ordered vector spaces. Recall that for a partially ordered vector space (X, K) and a subspace I of X the quotient space X̂ = X/I is a vector space and the quotient map q : X → X̂ is linear and surjective. According to Proposition 1.1.13 the set K̂ = q[K] is a cone in X,̂ provided I is full. The following is an immediate consequence of Proposition 4.3.14, as I is the kernel of q. Proposition 4.3.17. Let (X, K) be a partially ordered vector space and let I be a directed full linear subspace. Then the quotient map q : X → X/I is a Riesz homomorphism. The following simple example shows that the quotient map is not a complete Riesz homomorphism, in general. Example 4.3.18. We consider the vector lattice X := C[0, 1] and the ideal I := {x ∈ X; x(1) = 0} and show that the quotient map q : X → X/I is not order continuous. Indeed, for n ∈ ℕ define x(n) (t) = t n , for t ∈ [0, 1]. Then x(n) ↓ 0. For every n ∈ ℕ we have q(x(n) ) = q(𝟙) = [𝟙] ≠ [0], hence q is not order continuous. The next result concerns kernels of positive operators. Proposition 4.3.19. Let (X, K) be a partially ordered vector space and let I be a linear subspace of X. The subspace I is full if and only if there exist a partially ordered vector space (Y, L) and a positive operator T : X → Y such that I = [{0}]T. Proof. If I is a full subspace, then the quotient space Y := X/I is a partially ordered vector space, the quotient map T := q is positive, and its kernel equals I. If T : X → Y is positive, x ∈ X and y, z ∈ I := [{0}]T with y ≤ x ≤ z, then 0 = T(y) ≤ T(x) ≤ T(z) = 0, thus x ∈ I, which implies that I is a full subspace. If X is a Riesz space and f is a positive linear functional on X, then [{0}]f is directed if and only if f is a Riesz homomorphism. Indeed, Corollary 4.3.16 shows one implication. Conversely, let f be a Riesz homomorphism and x ∈ [{0}]f , then f(x+ ) = f(x ∨ 0) = f(x) ∨ f(0) = 0, hence x+ ∈ [{0}]f ∩ K. Analogously, x− ∈ [{0})]f ∩ K, hence [{0}]f is directed. In a partially ordered vector space the kernel of a Riesz homomorphism is not directed, in general, as the next example shows. Example 4.3.20. Let the space X = ℝ3 be equipped with the ice cream cone Lℝ2 = {(x1 , x2 , x3 )T ; x21 + x22 ≤ x23 , x3 ≥ 0}

4.3 Ideals | 217

as in Example 1.7.3. Recall that Lℝ2 is self-dual, that Σ := {(x1 , x2 , 1)T ; x21 + x22 ≤ 1} is a base of the cone Lℝ2 , and that the set of extreme points of Σ equals Λ = {(x1 , x2 , 1)T ; x21 + x22 = 1}, according to (2.30). Consider the functional f := (0, 1, 1) ∈ Λ. By Proposition 2.5.5, f is a Riesz homomorphism. The kernel of f is the subspace I = {(x1 , x2 , x3 )T ; x2 + x3 = 0}. The directed part of I equals J = (I ∩ Lℝ2 ) − (I ∩ Lℝ2 ) = {(0, −t, t)T ; t ∈ ℝ} , which is strictly smaller than I, thus I is not directed. It will be shown in Proposition 4.3.21 below that I is a solvex subspace and hence a solid subspace. According to Lemma 4.3.6 the subspace J is then directed and solid. This example also shows that the image of an Archimedean directed ordered vector space under a Riesz* homomorphism need not be pre-Riesz. Indeed, note that J is full by Lemma 4.3.5, and consider the quotient map q : X → X/J. By Proposition 4.3.17, q is a Riesz* homomorphism. The quotient space X/J with quotient cone can be identified with the space in Example 2.2.6(i), which is not a pre-Riesz space. Next we will show that the kernel of a Riesz* homomorphism is a solvex subspace. Proposition 4.3.21. Let (X, K) and (Y, L) be partially ordered vector spaces. If T : X → Y is a Riesz* homomorphism, then the kernel [{0}]T of T is a solvex subspace of X. Proof. We show that I := [{0}]T is solvex. Let x ∈ X, x1 , . . . , x n ∈ I and λ1 , . . . , λ n ∈ (0, 1] with ∑nk=1 λ k = 1 be such that u

n

{−x, x}u ⊇ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {−1, 1}} . k=1

Then

ul

n

{−x, x}ul ⊆ { ∑ ε k λ k x k ; ε1 , . . . , ε n ∈ {−1, 1}} . k=1

By applying T to this inclusion and using that T is a Riesz* homomorphism, we obtain n

ul

T({−x, x}ul ) ⊆ { ∑ ε k λ k T(x k ); ε1 , . . . , ε n ∈ {−1, 1}} = {0}ul = −L . k=1

Clearly, x ∈ {−x, x}ul and −x ∈ {−x, x}ul , hence T(x) ≤ 0 and T(−x) ≤ 0, which implies T(x) = 0, consequently x ∈ I. Thus, I is a solvex subspace. For a directed subspace I of a partially ordered vector space (X, K), the results presented above amount to five characterizations of I being an ideal, analogously to the Riesz space case. Theorem 4.3.22. Let (X, K) be a partially ordered vector space and let I be a linear subspace of X. Suppose that I is a directed subspace. The following six properties of I are equivalent.

218 | 4 Disjointness, bands, and ideals in pre-Riesz spaces

(i) I is full. (ii) I is solid, i.e., I is an ideal. (iii) I is solvex. (iv) There exist a partially ordered vector space (Y, L) and a positive operator T : X → Y such that I = [{0}]T. (v) There exist a partially ordered vector space (Y, L) and a Riesz* homomorphism T : X → Y such that I = [{0}]T. (vi) There exist a partially ordered vector space (Y, L) and a Riesz homomorphism T : X → Y such that I = [{0}]T. Proof. Clearly, (iii) implies (ii), (vi) implies (v), and (v) implies (iv). The equivalence of (i) and (ii) is contained in Lemma 4.3.5 and the equivalence of (i) and (iv) in Proposition 4.3.19. If (i) holds, then (vi) is satisfied with Y = X/I and T = q, according to Proposition 4.3.17. Proposition 4.3.21 yields that (v) implies (iii).

Notes and Remarks For directed ideals, in [114] conditions on a pre-Riesz space are given such that a converse of Theorem 4.3.13 is valid, that is, every order closed directed ideal is a band. In [82] the question is addressed how a directed ideal I in a pre-Riesz space X is related to the smallest extension ideal of I in a vector lattice cover of X. More precisely, it is investigated under which conditions the extension ideal provides a vector lattice cover of I. In a similar manner, bands and their extension bands are studied. In [27] one can find special cases of the equivalences in Theorem 4.3.22. For instance, it is used that the kernels of positive linear functionals are full subspaces. In addition to only being full, a subspace of a partially ordered vector space with an order unit u is called a maximal extreme ideal in [27] if it is the kernel of a positive functional which is an extreme point of the set of positive functionals φ with φ(u) = 1. Note that such extreme positive functionals are Riesz homomorphisms; see Proposition 2.5.5. Theorem 4.3.22 shows that directed ideals in partially ordered vector spaces have similar convenient properties as ideals in vector lattices. The situation is not as neat for nondirected ideals. Concerning order projections in pre-Riesz spaces, some first results are given by Jochen Glück in [61]. In particular, the following is shown: For a linear projection P, one has that P and I − P are positive if and only if P is a band projection. Band projections on ordered Banach spaces are considered in [62]. Projection bands in a pervasive pre-Riesz space and their extensions to projection bands in a vector lattice cover are discussed in [80].

4.4 Bands in spaces with order units

| 219

4.4 Bands in spaces with order units Many relevant examples of pre-Riesz spaces are Archimedean and have order units. By Theorem 2.5.9, such a space X has C(Ω) as a vector lattice cover, where Ω is an appropriate compact Hausdorff space. The question arises whether bands in X can be characterized by means of subsets of Ω. We answer this question in the affirmative, single out a class of subsets of Ω, and use it to characterize the bands in X. Considering the extension property for bands in this setting, carriers of bands are related to carriers of extension bands. The material in this section is due to [78].

4.4.1 Carriers of bands In Proposition 1.3.13 bands in C(Ω) (where Ω is a nonempty compact Hausdorff space) are characterized by means of their carriers, i.e., by regularly open subsets of Ω. In preRiesz spaces that have C(Ω) as a vector lattice cover, it is expected that bands can be characterized by means of subsets of Ω as well. In the present section we consider an Archimedean partially ordered vector space X with order unit and apply Theorem 2.5.9 to obtain (C(Λ), Φ) as a vector lattice cover, with Λ defined as in (2.23) and the embedding map Φ given in (2.24). For B ⊆ X we define the carrier of B as the carrier of Φ[B] in C(Λ) according to (1.7), i.e., car(B) := car(Φ[B]) .

(4.9)

The questions arise whether bands in X are determined by means of their carriers, and which subsets of Λ turn out to be carriers of bands in X. In the present section this problem will be completely solved by introducing bisaturated sets. For spaces with polyhedral cones, we obtain a convenient method to determine all bands. Remark 4.4.1. If X = C(Ω), where Ω is a nonempty compact Hausdorff space, then Λ = Λ, and Λ is homeomorphic to Ω. Hence, the carrier of B ⊆ C(Ω) defined as a subset of Λ corresponds to the carrier of B seen as a subset of Ω. In this sense the above definition of the carrier is compatible with the definition of the carrier in (1.7). For B ⊆ X denote N(B) := Λ \ car(B) = {φ ∈ Λ; ∀b ∈ B : φ(b) = 0} . For a set M ⊆ Λ we also use the notation M c := Λ \ M. In particular, for B ⊆ X one has car(B) = (N(B))c . For M ⊆ Λ, as usual span(M) denotes the linear subspace of X 󸀠 spanned by M, and the affine hull of M is given by n

n

aff(M) = { ∑ α i m i ; n ∈ ℕ, α i ∈ ℝ, ∑ α i = 1, m i ∈ M} . i=1

i=1

220 | 4 Disjointness, bands, and ideals in pre-Riesz spaces For M ⊆ Λ denote the zero set of M by Z(M) = {x ∈ X; ∀φ ∈ M : φ(x) = 0} . We will characterize those sets M ⊆ Λ that belong to a band B in X in the sense that B = Z(M) and M = N(B). For M ⊆ X 󸀠 let M denote the closure in the weak-∗ topology. We introduce the following notion. Definition 4.4.2. For M ⊆ Λ we define the saturation of M by sat(M) := Λ ∩ span(M) . A set M ⊆ Λ is called saturated if M = sat(M). Instead of the linear span one could also use the affine hull in the definition of saturation; see Corollary 4.4.6 below. We mention also that, obviously, any saturated set is weakly-∗ closed. It is straightforward that for any B ⊆ X the set N(B) is saturated. In the subsequent proposition we characterize the saturation by means of zero sets. First we state two basic properties. Lemma 4.4.3. For M ⊆ Λ the following holds: (i) sat(sat(M)) = sat(M). (ii) Z(sat(M)) = Z(M). Proof. (i) Clearly, M ⊆ sat(M), and therefore sat(M) ⊆ sat(sat(M)). Since sat(M) ⊆ span(M) it follows that span(sat(M)) ⊆ span(M) , so that sat(sat(M)) ⊆ sat(M). Hence, sat(sat(M)) = sat(M). (ii) Due to sat(M) ⊇ M one has Z(sat(M)) ⊆ Z(M). If x ∈ Z(M), then φ(x) = 0 for every φ ∈ M, therefore φ(x) = 0 for every φ ∈ span(M). Hence, x ∈ Z(sat(M)). Remark 4.4.4. The definitions of N(B) and Z(M), where B ⊆ X and M ⊆ Λ, are closely related to the notion of annihilator. Indeed, N(B) = Λ ∩ B⊥ and Z(M) = ⊥ M, where B⊥ = {φ ∈ X 󸀠 ; ∀b ∈ B : φ(b) = 0} is the annihilator of B in X 󸀠 and ⊥ M = {x ∈ X; ∀φ ∈ M : φ(x) = 0} the annihilator of M in X, cf. [116, Definition 1.10.14]. Note that the subsequent proposition is a variant of [116, Proposition 2.6.6 (b)]. Proposition 4.4.5. Let M ⊆ Λ. Then sat(M) = N(Z(M)) . Proof. If φ ∈ Λ ∩ span(M) and x ∈ Z(M), then φ(x) = 0, hence Λ ∩ span(M) ⊆ {φ ∈ Λ; ∀x ∈ Z(M) : φ(x) = 0} = N(Z(M)) .

4.4 Bands in spaces with order units

| 221

Since the right-hand side of the equality is a weakly-∗ closed set, it follows that sat(M) = Λ ∩ span(M) ⊆ N(Z(M)) . Now let ψ ∈ Λ \ span(M), i.e., ψ ∈ ̸ sat(M). By the Hahn–Banach theorem, there is an x ∈ X such that ψ(x) ≠ 0, and for every φ ∈ span(M) one has φ(x) = 0. In particular, x ∈ Z(M). Therefore, ψ ∈ ̸ N(Z(M)). Observe that the saturation of M ⊆ Λ is the largest subset N of Λ such that Z(N) = Z(M). Later on in examples it will turn out to be more natural to use the affine hull instead of the linear span in describing saturations. Corollary 4.4.6. Let M ⊆ Λ. For every subset L of X 󸀠 with aff(M) ⊆ L ⊆ span(M) one has sat(M) = Λ ∩ L. Proof. Since L ⊆ span(M) one has Λ ∩ L ⊆ Λ ∩ span(M) = sat(M). It remains to show that Λ ∩ aff(M) ⊇ sat(M). This is clear if M = ⌀, therefore we may assume that M is nonempty. Let ψ ∈ Λ \aff(M). By the Hahn–Banach theorem, there exist x ∈ X and γ ∈ ℝ such that ψ(x) > γ and φ(x) ≤ γ for all φ ∈ aff(M). If φ1 , φ2 ∈ aff(M), then φ1 + α(φ2 − φ1 ) ∈ aff(M) for all α ∈ ℝ, hence φ1 (x) = φ2 (x), as φ(x) ≤ γ for all φ ∈ aff(M). Take φ0 ∈ M and z := x−φ0 (x)u (where u is the order unit as in (2.22)). Then for every φ ∈ aff(M) we have φ(x) = φ0 (x), hence φ(z) = φ(x) − φ0 (x)φ(u) = 0, since aff(M) ⊆ Σ. In particular, z ∈ Z(M). Also, ψ(z) = ψ(x)−φ0 (x)ψ(u) = ψ(x)−φ0 (x) > γ−γ = 0, so that ψ ∈ ̸ N(Z(M)). Hence, Λ ∩ aff(M) ⊇ N(Z(M)) = sat(M), according to Proposition 4.4.5. As a preparation for the main result concerning the characterization of bands by means of subsets of Λ, we provide some lemmas. Lemma 4.4.7. Let B ⊆ X. Then Z(car(B)) = Bd . Proof. Let x ∈ Z(car(B)), then for every φ ∈ car(B) one has φ(x) = 0, which implies Φ(x)(φ) = 0. Let b ∈ B. Then for every ψ ∈ Λ \ car(B) it holds that ψ(b) = 0, therefore Φ(b)(ψ) = 0. Hence, Φ(b) ⊥ Φ(x), which implies b ⊥ x. Consequently, x ∈ Bd . Now, let x ∈ Bd , i.e., for every b ∈ B one has Φ(b) ⊥ Φ(x). Let φ ∈ car(B), then there is b ∈ B such that φ(b) ≠ 0. This implies Φ(b)(φ) ≠ 0, hence Φ(x)(φ) = φ(x) = 0. Consequently, x ∈ Z(car(B)). Lemma 4.4.8. Let B ⊆ X. Then B ⊆ Z(N(B)) ⊆ Bdd . Proof. If x ∈ B, then for every φ ∈ N(B) it holds that φ(x) = 0, hence x ∈ Z(N(B)). Let x ∈ Z(N(B)) and y ∈ Bd . If φ ∈ Λ\N(B), then there is a b ∈ B such that φ(b) ≠ 0, consequently Φ(b)(φ) ≠ 0, hence φ(y) = Φ(y)(φ) = 0. Thus for every φ ∈ Λ one has Φ(x)(φ) = φ(x) = 0 or Φ(y)(φ) = φ(y) = 0, therefore x ⊥ y. Hence, x ∈ Bdd .

222 | 4 Disjointness, bands, and ideals in pre-Riesz spaces

As a direct consequence of Lemma 4.4.8, we obtain the following result, which states that (in analogy to bands in C(Ω)) a band in X is determined whenever its carrier is known. Recall that N(B) = (car(B))c . Proposition 4.4.9. Let B be a band in X. Then B = Z(N(B)) . For a set that is determined by its carrier, we characterize the disjoint complement. Lemma 4.4.10. Let B ⊆ X be such that B = Z(N(B)). Then Bd = {x ∈ X; ∀φ ∈ sat((N(B))c ) : φ(x) = 0} = Z(sat((N(B))c )) . Proof. Due to Lemma 4.4.7, one has Bd = Z(car(B)) = {x ∈ X; ∀φ ∈ (N(B))c : φ(x) = 0} . Let x ∈ X be such that for every φ ∈ (N(B))c one has φ(x) = 0, and let ψ ∈ sat((N(B))c ) = Λ ∩ span((N(B))c ). Then it is straightforward that ψ(x) = 0, hence the assertion is proved.

4.4.2 Characterization of bands by bisaturated sets The next definition introduces those saturated subsets of Λ the zero sets of which will turn out to be bands. Definition 4.4.11. A set M ⊆ Λ is called bisaturated if M = sat((sat(M c ))c ) . Clearly, every bisaturated set is saturated. For an example of a saturated set that is not bisaturated, see Example 4.4.19 below. Next we give an example of a bisaturated set. Example 4.4.12. Consider X = C[0, 1] (then Λ consists of the point evaluations and can be identified with [0, 1]), and B = {x ∈ C[0, 1]; ∀t ∈ [ 12 , 1] : x(t) = 0} . We claim that the set N(B) = [ 12 , 1] is bisaturated. Indeed, sat((N(B))c ) = [0, 12 ] , then sat[(sat[(N(B))c ])c ] = [ 12 , 1] = N(B) . Observe that (N(B))c = [0, 12 ) is not saturated, since every saturated set is closed.

4.4 Bands in spaces with order units

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223

Just as a band and its disjoint complement, it will turn out that bisaturated sets appear in pairs, which leads to the following definition. Definition 4.4.13. (M1 , M2 ) ⊆ Λ × Λ is called a bisaturated pair if M1 = sat(M2 c ) and M2 = sat(M1 c ). Clearly, if (M1 , M2 ) is a bisaturated pair, then (M2 , M1 ) is a bisaturated pair as well. Observe that for a bisaturated pair (M1 , M2 ) one has M1 ∪ M2 = Λ ,

(4.10)

since M1 ∪ M2 = M1 ∪ sat(M1c ) ⊇ M1 ∪ M1 c = Λ. Remark 4.4.14. Sometimes it is convenient to extend Definition 4.4.13 to subsets F ⊆ X 󸀠 other than Λ. The pair (M1 , M2 ) ⊆ F × F is then said to be a bisaturated pair in F if M1 = F ∩ aff(F \ M2 ) and M2 = F ∩ aff(F \ M1 ). The next proposition relates bisaturated sets and bisaturated pairs, and provides a geometric description⁴ of bisaturated sets. Proposition 4.4.15. For two saturated sets M1 , M2 ⊆ Λ the following four statements are equivalent. (a) M1 is bisaturated and M2 = sat(M1 c ). (b) (M1 , M2 ) is a bisaturated pair. (c) aff(M1 ) = aff(Λ \ M2 ) and aff(M2 ) = aff(Λ \ M1 ). (d) There exist affine subspaces S 1 and S2 of X 󸀠 such that Λ ⊆ S1 ∪ S2 , S1 = aff(Λ \ S2 ), S2 = aff(Λ \ S1 ), M1 = Λ ∩ S1 , M2 = Λ ∩ S2 . Proof. (a) ⇒ (b): M1 is bisaturated, hence M1 = sat((sat(M1 c ))c ) = sat(M2 c ). (b) ⇒ (c): Using Corollary 4.4.6, one has M2 = sat (M1c ) = Λ ∩ aff(M1 c ) ⊆ aff(M1 c ) , therefore aff(M2 ) ⊆ aff(M1 c ). On the other hand, M2 = Λ ∩ aff(M1 c ) ⊇ Λ ∩ M1 c = M1 c , consequently aff(M2 ) ⊇ aff(M1 c ). Hence, aff(M2 ) = aff(M1 c ). Similarly, aff(M1 ) = aff(M2 c ). (c) ⇒ (d): Choose S i = aff(M i ), i ∈ {1, 2}. Since M i is saturated, M i = Λ ∩ aff(M i ) = Λ ∩ S i . Hence, S1 = aff(M1 ) = aff(Λ \ M2 ) = aff(Λ \ (Λ ∩ S2 )) = aff(Λ \ S2 ) 4 We thank Ulrich Brehm (TU Dresden) for an inspiring discussion on the geometrical meaning of bisaturated sets, which lead to the description in (d).

224 | 4 Disjointness, bands, and ideals in pre-Riesz spaces

and, similarly, S2 = aff(Λ \ S1 ). (d) ⇒ (a): Since Λ \ M1 = Λ \ (Λ ∩ S1 ) = Λ \ S1 , one has M1 = Λ ∩ S1 = Λ ∩ aff(Λ \ S2 ) = Λ ∩ aff (Λ \ aff(Λ \ S1 )) = Λ ∩ aff (Λ \ aff(Λ \ M1 )) = Λ ∩ aff (Λ \ (Λ ∩ aff(Λ \ M1 ))) = sat((sat(M1 c ))c ) , thus M1 is bisaturated. Next we relate bisaturated pairs to disjoint complements. Proposition 4.4.16. Let M1 , M2 ⊆ Λ be saturated sets and B1 = Z(M1 ) and B2 = Z(M2 ). Then (M1 , M2 ) is a bisaturated pair if and only if B1 = B2 d and B2 = B1 d . Proof. Let (M1 , M2 ) be a bisaturated pair. To show that B2 ⊆ B1 d , let x1 ∈ B1 , x2 ∈ B2 . For every φ ∈ M1 one has Φ(x1 )(φ) = 0 and for every φ ∈ M2 one has Φ(x2 )(φ) = 0. Since M1 ∪ M2 ⊇ Λ, it follows that Φ(x1 ) ⊥ Φ(x2 ), which implies that x1 ⊥ x2 . Hence, B2 ⊆ B1 d . Now, let x ∈ B1 d . Then for every y ∈ B1 one has x ⊥ y and hence Φ(x) ⊥ Φ(y). Therefore, Φ(x)(φ) = 0 for every φ ∈ car(B1 ) = Λ \ N(B1 ) = Λ \ M1 , due to Proposition 4.4.5. Thus Φ(x)(φ) = φ(x) = 0 for every φ ∈ sat(car(B1 )) = sat(Λ \ M1 ) = M2 . Hence, x ∈ Z(M2 ) = B2 . Therefore, B1 d = B2 and, by symmetry, B2 d = B1 . If B2 = B1 d , then due to Proposition 4.4.5, Λ \ M1 = Λ \ N(Z(M1 )) = Λ \ N(B1 ) = car(B1 ) , therefore, by Lemma 4.4.7, Z(Λ \ M1 ) = Z(car(B1 )) = B1 d . Again by Proposition 4.4.5, M2 = N(Z(M2 )) = N(B2 ) = N(B1 d ) = N(Z(Λ \ M1 )) = sat(Λ \ M1 ) . By symmetry, (M1 , M2 ) is a bisaturated pair. We arrive at the main result in this section. The assumption in the subsequent theorem is natural due to Proposition 4.4.9. Theorem 4.4.17. Let B ⊆ X be such that B = Z(N(B)). Then B is a band if and only if N(B) is bisaturated.

4.4 Bands in spaces with order units

|

225

Proof. Suppose that B1 := B = Z(N(B)) is a band. Define M1 := N(B) and M2 := sat(M1 c ). Then by Lemma 4.4.10 we have B2 := B1 d = Z(M2 ) and B2 d = B1 since B is a band. Hence, by Proposition 4.4.16, the pair (M1 , M2 ) is bisaturated. Therefore, N(B) is a bisaturated set, according to Proposition 4.4.15. Now, let M1 := N(B) be bisaturated and let B1 := B = Z(N(B)). Let M2 := sat(M1 c ). Then, by Proposition 4.4.15, (M1 , M2 ) is a bisaturated pair. Let B2 := Z(M2 ). Proposition 4.4.16 yields that B1 = B2 d = B1 dd , hence B = B1 is a band.

4.4.3 Bands in spaces with polyhedral cones A major motivation for Theorem 4.4.17 was the problem in determining the bands in finite dimensional spaces with polyhedral cones. We provide some examples to illustrate the convenience of using bisaturated sets. For a generating polyhedral cone K in ℝn , we use the notations for u, Σ, and Λ = Λ = {f (1) , . . . , f (k) } as in Subsection 2.6.1. By (2.29) and Theorem 2.6.2, an order dense embedding Φ of (ℝn , K) into (ℝk , ℝ+k ) is given by Φ : x 󳨃→ (f (1) (x), . . . , f (k) (x))T , and the Riesz completion is (ℝk , Φ). Recall ̂ is a band in the vector lattice (ℝk , ℝ+k ) if and only if there is N ⊆ {1, . . . , k} such that B that ̂ = {(x1 , . . . , x k )T ; ∀i ∈ N : x i = 0} ; B (4.11) see Example 1.3.14. Since every subset of Λ is weakly-∗ closed, we obtain for a set M ⊆ Λ that sat(M) = Λ ∩ aff(M) . Theorem 4.4.17 can be used to list all bands in (ℝn , K). Namely, after determining the linear dependence of the elements f (1) , . . . , f (k) in Λ, the saturation of every set M ⊆ Λ is known, and one can check whether M is bisaturated; see the subsequent example. In contrast to the vector lattice case, nondirected bands will also appear. Moreover, the example shows that the restriction property for bands is not satisfied, in general. Example 4.4.18. We continue Example 4.3.7, i.e., we consider in ℝ3 the generating polyhedral cone −1 0 1 0 K = pos {( 0 ) , ( 1 ) , ( 0 ) , ( −1 )} . 1

(0, 0, 1)T

1

1

1

is an order unit in (ℝ3 , K). In the functional representation

The element u := of (ℝ3 , K) one has Λ = {f (1) , f (2) , f (3) , f (4) } by (2.28), where f (1) = (−1, −1, 1) ,

f (2) = (1, −1, 1) ,

f (3) = (1, 1, 1) ,

f (4) = (−1, 1, 1) .

With Φ given by (2.29) the Riesz completion of (ℝ3 , K) is (ℝ4 , Φ); see Theorem 2.6.2. We want to determine all bands in (ℝ3 , K). Due to Proposition 4.4.9 in combination with Lemma 4.4.3, every band equals Z(N) for some saturated set N ⊆ {f (1) , . . . , f (4) }. Theorem 4.4.17 implies that for a saturated set N ⊆ {f (1) , . . . , f (4) } one has that Z(N)

226 | 4 Disjointness, bands, and ideals in pre-Riesz spaces is a band if and only if N is bisaturated. We list all sets M ⊆ {1, 2, 3, 4} such that {f (i); i ∈ M} is bisaturated: ⌀, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}, {1, 2, 3, 4}. We obtain 8 bands in (ℝ3 , K), where the nontrivial directed bands are 1

0

span {( 0 )} ,

span {( 1 )} ,

span {(

1

1

−1 0 )} 1

,

0

span {( −1 )} , 1

and the nontrivial nondirected bands are 1

span {( 1 )} 0

and

1

span {( −1 )} . 0

Now we are in a position to give a counterexample for the restriction property for bands, which was already announced at the end of Subsection 4.2.2. By (4.11), the set 0 0 0 ̂ = span {( 1 ) , ( 0 ) , ( 0 )} B 0 0

1 0

0 1

is a band in the Riesz completion ℝ4 . Because of the Theorems 3.7.9 and 4.3.10, the set ̂ [B]Φ = span {( 0 ) , ( 1 )} 1

0

1

1

is an o-closed solvex ideal in (ℝ3 , K), but it is not a band. Hence, the restriction property (R) for bands is not satisfied. As a consequence of Theorem 4.2.6, (ℝ3 , K) is not fordable and, hence, not pervasive. A typical method to construct illustrative examples is to consider Λ as the set of vertices of a polytope in ℝ3 and embed ℝ3 as an affine subspace into ℝ4 ; see the subsequent example. This method was already discussed in more generality in Proposition 2.6.9. Note that only nondirected nontrivial bands will appear in this example. Moreover, it turns out that the carriers of the extension bands in the procedures (I) and (II) in Theorem 4.2.3 differ. Example 4.4.19. Let X = ℝ4 , k = 9, f (1) = (1, 0, 0, 1) ,

f (2) = (0, 1, 0, 1) ,

f (3) = (−1, 0, 0, 1) ,

f (4) = (0, −1, 0, 1) ,

f (5) = (1, 0, 1, 1) ,

f (6) = (0, 1, 1, 1) ,

f (7) = (−1, 0, 1, 1) ,

f (8) = (0, −1, 1, 1) ,

f (9) = (0, 0, 2, 1) ,

and K according to (1.26), where the order unit u = (0, 0, 0, 1)T is used. The sets N1 := {f (1) , f (3) , f (5) , f (7) , f (9) }

and

N2 := {f (2) , f (4) , f (6) , f (8) , f (9) }

are bisaturated. By Theorem 4.4.17, the only nontrivial bands are B1 := Z(N1 ) and B2 := Z(N2 ). Observe that car(B1 ) = {f (2) , f (4) , f (6) , f (8) } = car (Φ[B1 ] ) dd

due to Theorem 4.4.31 below, whereas car(Φ[Bd1 ]d ) = {f (2) , f (4) , f (6) , f (8) , f (9) } due to Theorem 4.4.29 below.

4.4 Bands in spaces with order units |

227

Considering in a space of a fixed dimension all generating polyhedral cones, the question arises how large the number of bands can be. The next example presents a cone K in ℝn+1 , which yields a number of bands that is larger than the number of bands in a vector lattice of the same dimension. It is convenient to take the cone in ℝn+1 , since then the base of its dual cone can be considered as a subset of ℝn . Recall that in an (n + 1)-dimensional vector lattice the number of bands is 2n+1 . In the subsequent n+1 for n ≥ 3. example we obtain (2n n ) + 2 bands, which is greater than 2 Example 4.4.20. For every n ∈ ℕ we show that in ℝn+1 there is a polyhedral cone K n+1 , K). The with nonempty interior such that there are (2n n ) + 2 distinct bands in (ℝ proof relies on the following well-known statement in (i). (i) For each N ≥ n there are N distinct points x1 , . . . , x N in general position on the unit sphere in ℝn , i.e., (a) ‖x i ‖ = 1 for every i ∈ {1, . . . , N}, (b) for every V ⊆ {x1 , . . . , x N } with |V| ≤ n one has dim aff(V) = |V| − 1 , (c) if N ≥ n + 1 then for every V ⊆ {x1 , . . . , x N } with |V| = n + 1 one has aff(V) = ℝn . (ii) Suppose N := 2n, x1 , . . . , x2n ∈ ℝn as in (i), and W ⊆ {x1 , . . . , x2n } with |W| ≤ n. If x ∈ {x1 , . . . , x2n } is such that aff(W ∪ {x}) = aff(W), then x ∈ W. Indeed, (b) implies dim aff(W ∪ {x}) = dim aff(W) = |W| − 1 .

(4.12)

Assume x ∈ ̸ W. If |W| ≤ n − 1, then due to (b) one has dim aff(W ∪ {x}) = |W ∪ {x}| − 1 , a contradiction to (4.12). If |W| = n, then due to (c) it follows that dim aff(W ∪ {x}) = n , which contradicts (4.12). Thus, x ∈ W. (iii) We construct an appropriate cone K in ℝn+1 . Let N := 2n, x1 , . . . , x2n ∈ ℝn as in (i), g : ℝn → ℝn+1 , x 󳨃→ (x, 1), and y i := g(x i ) for i ∈ {1, . . . , 2n}. Define K := {z ∈ ℝn+1 ; ∀i ∈ {1, . . . , 2n} : ⟨z, y i ⟩ ≥ 0} . In (ℝn+1 , K), the element (0, . . . , 0, 1)T is an order unit. Since the x i are the extreme points of the convex hull of {x1 , . . . , x2n } in ℝn , one obtains Λ = {y1 , . . . , y2n } in ℝn+1 .

228 | 4 Disjointness, bands, and ideals in pre-Riesz spaces (iv) Every M ⊆ Λ with |M| = n is bisaturated. Indeed, take S1 := aff(M) and S2 := aff(Λ \ M). Observe that Λ ∩ S1 = M, since x ∈ Λ ∩ S1 implies aff(M ∪ {x}) = aff(M), and (ii) yields x ∈ M. Now, to show that aff(Λ \ S 2 ) = S1 , we prove S2 ∩ Λ = Λ \ M. Let x ∈ S2 ∩ Λ, then aff (Λ \ M) ⊆ aff ((Λ \ M) ∪ {x}) ⊆ aff (S2 ∪ {x}) = S2 = aff (Λ \ M) , hence aff(Λ \ M) = aff((Λ \ M) ∪ {x}), and (ii) implies x ∈ Λ \ M. Now we conclude aff (Λ \ S2 ) = aff (Λ \ (Λ \ M)) = aff(M) = S1 . Similarly, aff(Λ \ S1 ) = aff(Λ \ M) = S2 . In conclusion, M is bisaturated. (v) Every M ⊆ Λ with |M| > n satisfies aff(M) ∩ Λ = Λ. Indeed, for L ⊆ M with |L| = n + 1 from (c) it follows that Λ = aff(L) ∩ Λ ⊆ aff(M) ∩ Λ . Due to (iv) and (v), for every M ⊆ Λ with |M| = n the set {z ∈ ℝn+1 ; ∀y ∈ M : ⟨z, y⟩ = 0} n+1 . This yields the is a band. We obtain (2n n ) such bands, and the trivial ones {0} and ℝ 2n n+1 total number of ( n ) + 2 bands in (ℝ , K).

4.4.4 Bands in finite dimensional spaces We consider the set of all Archimedean pre-Riesz spaces of a fixed dimension n. The number of bands in such a space depends on the precise geometry of the cone, and the question arises whether one can find cones with an arbitrary large number of bands. We will show that this is not the case. It turns out that the number of bands is bounded by a number only depending on n. For this, the characterization of bands in terms of bisaturated subsets of Λ given in Theorem 4.4.17 is crucial. Notations as Λ, Φ, etc. are as in Subsection 4.4.1. The proof of the main result in Theorem 4.4.26 below consists of two parts. First it will be shown that the upper bound holds for partially ordered vector spaces with polyhedral cones. Subsequently we use the Plastering Lemma 4.4.25 to show that the same upper bound holds for general cones. Let X be a vector space of dimension n. For a subset M ⊆ X we mean by dim M the dimension of aff M. Let K be a generating polyhedral cone in X, and let Σ and Λ be as in Subsection 2.6.1. The number of bands in (X, K) equals the number of bisaturated sets in Λ by Theorem 4.3.22. Here, Λ is the set of vertices of the convex polytope Σ of dimension n − 1. Hence, we arrive at a purely combinatorial problem for polytopes. Recall that a convex polytope P in ℝn−1 is a compact convex subset of ℝn−1 with finitely many extreme points. Equivalently, P is the convex hull of a finite set V, where

4.4 Bands in spaces with order units

| 229

one can take as V the set of extreme points of P. Each extreme point of P is called a vertex of P. Let P be a convex polytope in ℝn−1 and let V be the set of its vertices. We carry over the Definitions 4.4.11 and 4.4.13 to sets S, T ⊆ V. The set S is called bisaturated in V if S = V ∩ aff(V \ (V ∩ aff(V \ S))) , and (S, T) is called a bisaturated pair in V if S = V ∩ aff(V \ T)

and

T = V ∩ aff(V \ S) .

Thus, to prove Theorem 4.4.22 it suffices to show the following combinatorial statement. The result and its ingenious proof are due to Bas Lemmens. Theorem 4.4.21. Let P be a polytope in ℝn−1 , where n ≥ 2. In the vertex set V of P there n are at most 14 22 bisaturated sets. Proof. Let P0 be a polytope in ℝn−1 and let V0 be its vertex set. Let q0 denote the n number of bisaturated sets in V0 . In order to show that q0 ≤ 14 22 , we consider the set Pq0 of all polytopes in ℝn−1 for which the number of bisaturated sets is at least q0 . We choose a polytope P ∈ Pq0 for which the number of vertices is minimal among all polytopes in Pq0 . Let V be the set of vertices of P and let q be the number of bisaturated n subsets of V. Since q ≥ q0 , it suffices to show q ≤ 14 22 . The proof comes down to showing the following statement for k ∈ {1, . . . , n − 1}: (H k ) For every k-dimensional affine subspace U in ℝn−1 we have that |U ∩ V| ≤ 2k+1 − 2 . Obviously, (H1 ) is true, as every line U contains at most two elements of V. By induction we will show that (H n−1 ) is true, which implies that V has at most 2n − 2 elements. n n So the number of subsets of V is at most 22 −2 , and therefore q ≤ 22 −2 . Let k ∈ {2, . . . , n − 1} be such that (H l ) holds for l < k. We show that (H k ) holds. To argue by contradiction, suppose that (H k ) is not true. Then there exists a k-dimensional affine subspace U such that |U ∩ V| > 2k+1 − 2 .

(4.13)

The following claim will make the reader happy many times. 

For each W ⊆ V with |U ∩ W| > 2k − 2, we have that U ⊆ aff W.

For a proof of , observe that U ⊈ aff W implies that U 󸀠 := U ∩ aff W has dimension l < k. Then by (H l ) we have |U 󸀠 ∩ V| ≤ 2l+1 − 2 ≤ 2k − 2, which proves . Fix u ∈ V ∩ U and let P󸀠 be the polytope with vertex set V 󸀠 := V \ {u}. We show that if (S, T) is a bisaturated pair in V, then (S󸀠 , T 󸀠 ) with S󸀠 := S \ {u} and T 󸀠 := T \ {u}

230 | 4 Disjointness, bands, and ideals in pre-Riesz spaces is a bisaturated pair in V 󸀠 . Indeed, since S ∪ T = V, we have |S ∩ U| > 2k − 1 or |T ∩ U| > 2k − 1. Without loss of generality we may assume that |S ∩ U| > 2k − 1. By  we obtain U ⊆ aff S and hence U ∩ V ⊆ V ∩ aff S = sat S = S. Therefore, u ∈ S. This implies that V \ S = V 󸀠 \ S󸀠 . So, T 󸀠 = T \ {u} = (V ∩ aff(V \ S)) \ {u} = V 󸀠 ∩ aff(V \ S) = V 󸀠 ∩ aff(V 󸀠 \ S󸀠 ) . To show S󸀠 = V 󸀠 ∩ aff(V 󸀠 \ T 󸀠 ), we distinguish two cases, namely u ∈ T and u ∈ ̸ T. If u ∈ T, then V \ T = V 󸀠 \ T 󸀠 , hence S󸀠 = S \ {u} = V 󸀠 ∩ aff(V \ T) = V 󸀠 ∩ aff(V 󸀠 \ T 󸀠 ) . If u ∈ ̸ T, then one obtains |T ∩ U| ≤ 2k − 2 .

(4.14)

Indeed, |T ∩ U| > 2k −2 implies by means of  that U ⊆ aff T, so that U ∩ V ⊆ V ∩aff T = T, which shows that u ∈ T. By (4.13) and (4.14) we get |(V\T)∩U| ≥ 2k , which implies that |(V 󸀠 \T 󸀠 )∩U| > 2k −2. Again by , it follows that U ⊆ aff(V 󸀠 \ T 󸀠 ) and, in particular, u ∈ aff(V 󸀠 \ T 󸀠 ). Hence, aff(V 󸀠 \ T 󸀠 ) = aff((V 󸀠 \ T 󸀠 ) ∪ {u}). As u ∈ ̸ T we have T 󸀠 = T, so V \ T = (V 󸀠 \ T 󸀠 ) ∪ {u}. Therefore, S󸀠 = S \ {u} = V 󸀠 ∩ aff(V \ T) = V 󸀠 ∩ aff(V 󸀠 \ T 󸀠 ) . We conclude that (S󸀠 , T 󸀠 ) is a bisaturated pair in V 󸀠 . It remains to show that P󸀠 has at least as many bisaturated sets as P. We have shown that for every bisaturated set S in V the set S󸀠 = S \ {u} is bisaturated in V 󸀠 . It suffices to show that different bisaturated sets in V induce different bisaturated sets in V 󸀠 . For that purpose, we have to show that if S is a bisaturated set in V and u ∈ S, then S󸀠 is not a bisaturated set in V. Assume that both S and S󸀠 are bisaturated sets in V. If |S ∩ U| > 2k − 1, then |S󸀠 ∩ U| > 2k − 2. By  we get U ⊆ aff(S󸀠 ), so V ∩ U ⊆ V ∩ aff S󸀠 = S󸀠 , which is a contradiction, as u ∈ ̸ S󸀠 . We conclude that |S ∩ U| ≤ 2k − 1. Then |(V \ S) ∩ U| ≥ 2k > 2k − 1. By  (for the last time) we obtain U ⊆ aff(V \ S). Let T := V ∩ aff(V \ S) and R := V ∩ aff(V \ S 󸀠 ). As u ∈ aff(V \ S) we get aff(V \ S) = aff(V \ S 󸀠 ) and hence T = R. Therefore, S = S󸀠 . We conclude that V 󸀠 contains at least q ≥ q0 bisaturated sets, hence P󸀠 ∈ Pq0 . But 󸀠 V has one element less than V, which contradicts the minimality of V. It follows that (H k ) is true. Next we state our main result on the number of bands in spaces with polyhedral cones. Theorem 4.4.22. Let (X, K) be an n-dimensional partially ordered vector space with a n generating polyhedral cone K and n ≥ 2. The number of bands in X is at most 14 22 . Proof. With the notations as in Subsection 2.6.1, the set Σ is a convex polytope in the (n − 1)-dimensional affine hyperplane H = {f ∈ X 󸀠 ; f(u) = 1} with vertex set Λ. From

4.4 Bands in spaces with order units

| 231

Theorem 4.3.22 we know that B is a band in X if and only if N(B) is a bisaturated set in Λ. Therefore, the number of bands in X equals the number of bisaturated sets in Λ, n which is at most 14 22 by Theorem 4.4.21. In the remainder of this subsection, (X, K) will be a finite-dimensional partially ordered vector space with a closed generating cone K. We will show that the upper bound in Theorem 4.4.22 for the number of bands in a partially ordered vector space with a polyhedral cone also holds in (X, K). We fix an order unit u in K and consider the corresponding vector lattice cover (C(Λ), Φ) given by the functional representation. We will prove the main result in Theorem 4.4.26 below with a Plastering Lemma that approximates K by polyhedral cones. More precisely, given bands B1 , B2 , . . . , B q in (X, K), the idea is to construct a polyhedral cone L in X containing K such that each B i is a band in (X, L). We start with two technical lemmas. In the first one, we consider a bisaturated pair (M1 , M2 ) in Λ. If we consider closed sets W1 ⊆ Λ \ M2 and W2 ⊆ Λ \ M1 that have the same affine hulls as M1 and M2 , respectively, then the pair of their saturations in W := W1 ∪ W2 turns out to be a bisaturated pair in W. Although we need the statement here only in finite dimensions, it holds in general Archimedean partially ordered vector spaces with an order unit. Lemma 4.4.23. Let (M1 , M2 ) be a bisaturated pair in Λ. If W1 ⊆ Λ\M2 and W2 ⊆ Λ\M1 are such that aff(W i ) = aff(M i ) , for i ∈ {1, 2} , and if F is a closed subset of Λ with W1 ∪ W2 ⊆ F, then for the sets N i := F ∩ aff(W i ) ,

with i ∈ {1, 2} ,

the pair (N1 , N2 ) is bisaturated⁵ in F. Proof. We have F \ N1 = F \ (F ∩ aff(W1 )) = F ∩ (Λ \ aff(W1 )) = F ∩ (Λ \ (Λ ∩ aff(M1 ))) = F ∩ (Λ \ M1 ) = F \ M1 , as M1 is saturated in Λ, so aff(F \ N1 ) = aff(F \ M1 ) . On the one hand, F \ M1 ⊆ Λ \ M1 , so, aff(F \ M1 ) ⊆ aff(Λ \ M1 ) = aff(M2 ) = aff(W2 ) , by Proposition 4.4.15. On the other hand, F \ M1 = F ∩ (Λ \ M1 ) ⊇ F ∩ W2 = W2 , 5 Recall the extended definition of bisaturated pair in Remark 4.4.14.

232 | 4 Disjointness, bands, and ideals in pre-Riesz spaces so that aff(F \ M1 ) ⊇ aff(W2 ). Thus, aff(F \ N1 ) = aff(F \ M1 ) = aff(W2 ) and therefore, F ∩ aff(F \ N1 ) = F ∩ aff(W2 ) = N2 . Similarly, F ∩ aff(F \ N2 ) = N1 . Recall that for a subset B of X the carrier of B is given by car(B) = {f ∈ Λ; ∃b ∈ B : f(b) ≠ 0}. Lemma 4.4.24. Let B be a subspace of X. Then aff(car(B)) = aff{f ∈ Λ; ∃b ∈ B : f(b) ≠ 0} . Proof. Denote C := aff{f ∈ Λ; ∃b ∈ B : f(b) ≠ 0}. Since Λ ⊆ Λ, we have that aff(C) ⊆ aff(car(B)). The affine subspace aff(C) is finite dimensional, hence closed, so it contains the closure of C. It remains to show that car(B) ⊆ C. If f ∈ Λ and x ∈ B are such that f(x) ≠ 0, then there is a net (f α )α∈I in Λ with f α → f (in the weak-∗ topology). Then there is α 0 ∈ I such that for every α ∈ I≥α0 we have f α (x) ≠ 0. Therefore, f α ∈ C, so f ∈ C. Thus aff(car(B)) ⊆ aff(C) = aff(C). Now we are in a position to present the Plastering Lemma. Lemma 4.4.25. Let (X, K) be a finite dimensional partially ordered vector space with a closed generating cone K. For any choice of bands B1 , . . . , B q in (X, K) there exists a closed generating polyhedral cone L in X with K ⊆ L such that B i is a band in (X, L) for every i ∈ {1, . . . , q}. Proof. Due to Proposition 4.4.9, we have B i = Z(N(B i )) and B i d = Z(N(B i d )) and then according to Proposition 4.4.5 the sets N(B i ) and N(B i d ) are saturated. By Proposition 4.4.16, it follows that (N(B i ), N(B i d )) is a bisaturated pair in Λ. By definition, car(B i d ) = Λ \ N (B i d ) .

car(B i ) = Λ \ N(B i ) and

Choose, with the aid of Lemma 4.4.24, finite sets U i ⊆ car(B i ) ∩ Λ

and

V i ⊆ car (B i d ) ∩ Λ

such that aff(U i ) = aff(car(B i )) and aff(V i ) = aff(car(B i d )). Define q

F := ⋃(U i ∪ V i ) i=1

and L := {x ∈ X; ∀f ∈ F : f(x) ≥ 0}. Since for every i ∈ {1, . . . , q} we have that (N(B i ), N(B i d )) is a bisaturated pair, we obtain Λ ∩ aff(U i ∪ V i ) ⊇ Λ ∩ aff(U i ) = Λ ∩ aff(car(B i )) = Λ ∩ aff(Λ \ N(B i )) = N(B i d )

4.4 Bands in spaces with order units

| 233

and, similarly, Λ ∩ aff(U i ∪ V i ) ⊇ N(B i ), which implies Λ = N(B i ) ∪ N (B i d ) ⊆ Λ ∩ aff(U i ∪ V i ) , due to (4.10). Hence, Λ ⊆ aff(F), so that L is a cone. Also, since F ⊆ Λ, we have L ⊇ K, so that L is generating. It is clear that L is a polyhedral cone. Moreover, the order unit u of (X, K) is an interior point of K, hence an interior point of L and therefore an order unit in (X, L). Every element of F is an extreme point of Σ L = {f ∈ X 󸀠 ; f(x) ≥ 0 for all x ∈ L, f(u) = 1} . Indeed, if φ ∈ F, 0 < λ < 1, and φ1 , φ2 ∈ Σ L are such that φ = λφ1 + (1 − λ)φ2 , then φ ∈ Λ and φ1 , φ2 ∈ Σ, so φ1 = φ2 = φ, as Λ consists of extreme points of Σ. Hence, φ is an extreme point of Σ L . It follows that co(F) ⊆ Σ L . In order to show that every extreme point of Σ L is contained in F, we show that Σ L is a subset of the convex hull co(F) of F. Suppose that is not the case. Then there is a ψ ∈ Σ L that is not an element of co(F). Since co(F) is a compact convex subset of X 󸀠 , we can separate co(F) and ψ by a hyperplane. This means that there exists an x ∈ X such that φ(x) ≥ 0 for every φ ∈ co(F) and ψ(x) < 0. Then x ∈ L by definition of L, so ψ(x) ≥ 0, which is a contradiction. Hence, co(F) = Σ L . Since F ⊇ ext(co(F)), it follows that F equals the set of extreme points of Σ L . It remains to show that for each i the subspace B i is a band in (X, L). For i ∈ {1, . . . , q} we have U i ⊆ car(B i ) = Λ \ N(B i ) , V i ⊆ car(B i d ) = Λ \ N(B i d ) , and, since (N(B i d ), N(B i )) is a bisaturated pair in Λ, aff(U i ) = aff(car(B i )) = aff(Λ \ N(B i )) = aff N(B i d ) , aff(V i ) = aff(N(B i )) . According to Lemma 4.4.23, the saturations in F of the sets U i and V i constitute a bisaturated pair in F, so that, according to Proposition 4.4.16, Z(U i ) and Z(V i ) are bands in (X, L) and Z(U i )d = Z(V i ). Finally, by Lemma 4.4.3 (ii), Z(U i ) = Z(sat(U i )) = Z(sat(car(B i ))) = Z(sat(Λ \ N(B i ))) , hence Z(U i ) = B i d by Lemma 4.4.10. Similarly, Z(V i ) = B i . It follows that every B i is a band in (X, L). Clearly the number of bands in an n-dimensional partially ordered vector space is 2 if n = 1. For n ≥ 2 we have the following result, which is a combination of Theorem 4.4.22 and Lemma 4.4.25.

234 | 4 Disjointness, bands, and ideals in pre-Riesz spaces

Theorem 4.4.26. The number of bands in an n-dimensional partially ordered vector space (X, K), where K is a closed generating cone and n ≥ 2, does not exceed 1 2n 42

.

The statement in Theorem 4.4.26 will be used in Subsection 5.2.3 below to investigate inverses of disjointness preserving bijections. The upper bound for the number of bands is expected to be far from sharp. It is an interesting problem to determine optimal upper bounds for the number of bands as well as for the number of directed bands.

4.4.5 Carriers of extension bands The extension property (E) for bands in pre-Riesz spaces is shown in Theorem 4.2.3, where two procedures (I) and (II) are given to obtain an extension band in a vector lattice cover. We investigate these procedures in a more specific setting. In this subsection let X be an Archimedean partially ordered vector space with order unit. The functional representation (C(Λ), Φ) of X is a vector lattice cover by Theorem 2.5.9, and the bands in X are characterized by means of bisaturated subsets of Λ in Theorem 4.3.22. ̂ in C(Λ) are determined by A band B in X and a corresponding extension band B ̂ ̂ their carriers, so it is natural to ask how car(B) and car(B) are related and how car(B) can be determined if car(B) is known. Note that for every band F in C(Λ) extending a band B in X one has car(B) = car(Φ[B]) ⊆ car(Φ[B]dd ) ⊆ car(F) , since Φ[B]dd according to (II) is the smallest band extending B. We already observed in Example 4.4.19 that for the extension band F := Φ[Bd ]d by procedure (I) the sets car(Φ[B]dd ) and car(F) may differ. In the next example we also show that car(B) and car(Φ[B]dd ) differ, in general. Example 4.4.27. The space X = {x ∈ C[−1, 1]; x(0) = 12 (x(−1) + x(1))} is not a vector lattice. It is Archimedean, and the function u = 1 is an order unit in X. The set Λ according to (2.23) can be identified with [−1, 0) ∪ (0, 1], and Λ with [−1, 1]. Let B = {x ∈ X; x(1) = 0 and ∀t ∈ [−1, 0] : x(t) = 0} . Then Bd = {x ∈ X; x(−1) = 0 and ∀t ∈ [0, 1] : x(t) = 0} . Furthermore, B = Bdd , therefore B is a band in X with car(B) = (0, 1). In C(Λ) (with Φ according to (2.24)) we obtain Φ[B]d = {x ∈ C(Λ); ∀t ∈ [0, 1] : x(t) = 0}

4.4 Bands in spaces with order units

| 235

and Φ[B]dd = {x ∈ C(Λ); ∀t ∈ [−1, 0] : x(t) = 0} , hence car(Φ[B]dd ) = (0, 1] ≠ car(B). Let us investigate procedure (I) in more detail. To determine car(Φ[Bd ]d ) by means of car(B), we need the following identity. Lemma 4.4.28. For a subset M ⊆ X one has car(Φ[M]d ) = (car(M))c . Proof. For s ∈ C(Λ) one has s ∈ Φ[M]d if and only if s(φ) = 0 for every φ ∈ car(M), which is equivalent to s(φ) = 0 for every φ ∈ car(M). Hence, car(Φ[M]d ) ⊆ (car(M))c . If ψ ∈ (car(M))c , then by Urysohn’s lemma there exists s ∈ C(Λ) such that s(ψ) ≠ 0 and s(φ) = 0 for every φ ∈ car(M), hence s ∈ Φ[M]d . Therefore, car(Φ[M]d ) ⊇ (car(M))c . Theorem 4.4.29. For a band B in X, the carrier of Φ[Bd ]d is given by d

c

car (Φ[Bd ] ) = ((sat(car(B)))c ) , and Φ[Bd ]d is the largest ideal in C(Λ) with carrier ((sat(car(B)))c )c . Proof. Due to Lemmas 4.4.10 and 4.4.3 (ii), we have Bd = Z(sat(N(B)c )) = Z(N(B)c ) . Then Proposition 4.4.5 implies N(Bd ) = sat(N(B)c ) = sat(car(B)) , hence car(Bd ) = (sat(car(B)))c . Applying Lemma 4.4.28 with M = Bd completes the proof. We proceed by discussing procedure (II). First we show that the carrier of B and the ̂ = Φ[B]dd have the same closure (in other words, the supports of B and B ̂ carrier of B are the same). Proposition 4.4.30. For a band B in X one has car(B) = car(Φ[B]dd ). Proof. We know that car(B) ⊆ car(Φ[B]dd ), hence car(B) ⊆ car(Φ[B]dd ). Now suppose that there is a φ0 ∈ car(Φ[B]dd )\car(B). By Urysohn’s lemma, there is an x ∈ C(Λ) such that x(φ0 ) ≠ 0 and x(φ) = 0 for every φ ∈ car(B). If b ∈ B, then Φ(b)(φ) = φ(b) = 0 for every φ ∈ Λ \ car(B), therefore x ⊥ Φ(b) for every b ∈ B. Hence, x ∈ Φ[B]d . Since φ0 ∈ car(Φ[B]dd ) and x(φ0 ) ≠ 0, there exists φ1 ∈ car(Φ[B]dd ) such that x(φ1 ) ≠ 0. Moreover, there is y ∈ Φ[B]dd such that y(φ1 ) ≠ 0. This implies x ⊥̸ y, which contradicts x ∈ Φ[B]d , y ∈ Φ[B]dd . Thus car(B) = car(Φ[B]dd ). Since Φ[B]dd is a band in C(Λ), the set car(Φ[B]dd ) is regularly open, i.e., car(Φ[B]dd ) = int(car(Φ[B]dd )). Now Proposition 4.4.30 yields the following result.

236 | 4 Disjointness, bands, and ideals in pre-Riesz spaces Theorem 4.4.31. For a band B in X, the carrier of Φ[B]dd is given by car (Φ[B]dd ) = int (car(B)) , and Φ[B]dd is the largest ideal in C(Λ) with carrier int(car(B)). Remark 4.4.32. Note that car(Φ[B]dd ) is the smallest regularly open set containing car(B). Indeed, int(car(B)) is a regularly open set, car(B) ⊆ car(B) and car(B) is open, therefore car(B) ⊆ int(car(B)). Hence, int(car(B)) is a regularly open set containing car(B). Let O be any regularly open set with O ⊇ car(B). Then O ⊇ car(B), hence O = int(O) ⊇ int(car(B)). Consequently car(Φ[B]dd ) = int(car(B)) is the smallest regularly open set containing car(B). In Example 4.4.27, car(Φ[B]dd ) = (0, 1] is indeed the smallest regularly open set in Λ = [−1, 1] containing car(B) = (0, 1). For a band B in X, we address the question whether the carrier of B can be recovered if the carrier of Φ[B]dd is known. Observe that every saturated subset of Λ is weakly*closed. On the contrary, a weakly*-closed subset of Λ is not saturated, in general (note that in Example 4.4.18 the set {f (1) , f (2) , f (3) } is weakly*-closed but not saturated). Nevertheless, a statement similar to Proposition 4.4.30 is true, concerning the saturations of car(B) and car(Φ[B]dd ). Proposition 4.4.33. For a band B in X, one has sat(car(B)) = sat ( car (Φ[B]dd )) . Proof. First observe that for sets M, N ⊆ Λ with M = N it follows that sat(M) = sat(N). Indeed, the continuity of the vector space operations yields span(M) ⊆ span(M), hence span(M) = span(M), which implies span(M) = span(N). Now the statement follows immediately from Proposition 4.4.30. Next we reconstruct car(B) provided car(Φ[B]dd ) is known. Proposition 4.4.34. For a band B in X, one has c

c

car(B) = ( sat (( sat ( car (Φ[B]dd ))) )) . Proof. Due to Theorem 4.4.17, (car(B))c is bisaturated. By Proposition 4.4.33 we obtain c

(car(B))c = sat((sat(car(B)))c ) = sat (( sat ( car (Φ[B]dd ))) ) , which implies the assertion. Summing up the above results, the extension method (II) gives that car(B) and car(Φ[B]dd ) have the same closure and the same saturation, and if one of the two sets is known then the other one is determined.

5 Operators on pre-Riesz spaces The success of the theory of vector lattices is largely due to the rich theory of operators on vector lattices; see for instance [2, 12, 118, 123, 134, 166]. It is a natural question to ask whether similar results are true for operators on pre-Riesz spaces. We choose a few problems to illustrate the developments in this expanding area. The notions of disjointness and band introduced in Chapter 4 lead to generalized notions of disjointness preserving and band preserving operator. We will discuss examples of such operators in Section 5.1. In Section 5.2, we investigate under which conditions the inverse of a disjointness preserving bijection is disjointness preserving. We give some properties of disjointness preserving C0 -semigroups in Section 5.3. Section 5.4 investigates properties of spaces of operators involving order and norm. Finally, classical spectral theory of positive operators is collected in Section 5.5.

5.1 Disjointness preserving operators In this section we study basic properties of disjointness preserving operators. In particular, we discuss band preserving operators and Riesz* homomorphisms. In vector lattices, disjointness preserving operators, band preserving operators, orthomorphisms, and Riesz homomorphisms are thoroughly studied in the literature; see, e.g., [1, 3–5, 12, 32, 45, 52, 69, 115, 160]. In pre-Riesz spaces, some aspects are very different to the vector lattice setting. For instance, we will show that an order-bounded disjointness preserving operator need not even be regular. On the other hand, for Riesz* homomorphisms there are promising results similar to the lattice case.

5.1.1 Definitions and examples In Section 1.4, the definitions of disjointness preserving and band preserving operator are given in vector lattices. We extend these definitions to pre-Riesz spaces. We also define local operators. They play a role in the theory of differential equations, where the https://doi.org/10.1515/9783110476293-005

238 | 5 Operators on pre-Riesz spaces

notion ‘local’ is used ambiguously (see Remark 5.1.5 below). In [18, (5.4)], local operators are defined on Banach lattices by means of disjointness, so that we can naturally extend this definition to pre-Riesz spaces. We consider operators T that are defined on a domain D(T), since we encounter unbounded generators in the discussion of C0 -semigroups below. Definition 5.1.1. Let X and Y be pre-Riesz spaces and let T : X ⊇ D(T) → Y be a linear operator. (i) T is called disjointness preserving if for every x, y ∈ D(T) from x ⊥ y it follows that Tx ⊥ Ty. Let, in addition, X = Y. (ii) T is called band preserving if for every band B in X one has T(B ∩ D(T)) ⊆ B. (iii) T is called local if for every x ∈ D(T), y ∈ X with x ⊥ y it follows that Tx ⊥ y. The notions in (i) and (ii) cover the ones given in Definitions 1.4.1 and 1.4.13. Observe that if T is local and x, y ∈ D(T) are such that x ⊥ y, then Tx ⊥ y and, moreover, Tx ⊥ Ty, i.e., T is disjointness preserving. It turns out that local operators and band preserving operators coincide. Proposition 5.1.2. Let X be a pre-Riesz space and T : X ⊇ D(T) → X a linear operator. Then T is local if and only if T is band preserving. Proof. Let T be local. For every x ∈ D(T) one obtains Tx ∈ {x}dd . Indeed, for every y ∈ {x}d it follows that Tx ⊥ y, therefore Tx ∈ {x}dd . Let B be a band in X and x ∈ B ∩ D(T). Then {x}dd ⊆ Bdd = B, hence Tx ∈ B. We conclude that T is band preserving. Now let T be band preserving, x ∈ D(T) and y ∈ X such that x ⊥ y. Then {y}d is a band in X, and x ∈ {y}d ∩ D(T), hence Tx ∈ {y}d , which yields Tx ⊥ y. Consequently T is local. Typical examples of local operators are differential operators and multiplication operators. We discuss some of these settings in the subsequent remarks. Recall that for an ordered normed space X we denote by L(X) the vector space of all bounded linear operators on X, endowed with the operator norm and the wedge of all positive operators. Remark 5.1.3. If X is a Banach lattice, every bounded local operator on X is contained in the center Z(X) := {T ∈ L(X); ∃α > 0 : − αI ≤ T ≤ αI} . More precisely, the following assertions are equivalent for a bounded linear operator T : X → X; see, e.g., [122, Section 9]: (i) T is local, (ii) −‖T‖I ≤ T ≤ ‖T‖I, (iii) for every ideal J in X one has T[J] ⊆ J.

5.1 Disjointness preserving operators

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239

In [122, Section 9] the following two examples are given to illustrate that local operators are closely related to multiplication operators. (a) Let (Ω, μ) be a σ-finite measure space and X = Lp (Ω, μ) (1 ≤ p ≤ ∞), then Z(X) is isomorphic to L∞ (Ω, μ) via the identification of y ∈ L∞ (Ω, μ) with the operator from X to X given by x 󳨃→ yx. (b) Let Ω be a nonempty locally compact Hausdorff space and X the space C0 (Ω) of all continuous functions x on Ω vanishing at infinity (i.e., for every ε > 0 there is a compact set C ⊆ Ω such that for every t ∈ Ω \ C one has |x(t)| < ε), endowed with the supremum norm. Z(X) is isomorphic to the space Cb (Ω) of bounded continuous functions via the identification of y ∈ Cb (Ω) with the operator from X to X given by x 󳨃→ yx. Remark 5.1.4. We continue the discussion of case (a) in Remark 5.1.3 for unbounded T. Let X = Lp (Ω, μ) (1 ≤ p ≤ ∞) and T : Lp (Ω, μ) ⊇ D(T) → Lp (Ω, μ). A notion of ‘locality’ in this setting is defined in [51, I.4.13(8)]. To avoid confusion, we denote this property by (L): (L) For every measurable set S ⊆ Ω and for every x ∈ D(T) with x = 0 almost everywhere on S it follows that Tx = 0 almost everywhere on S. The operator T is local if and only if (L) is satisfied. Indeed, suppose that (L) is satisfied and let x ∈ D(T), y ∈ X and x ⊥ y. Then S := {t ∈ Ω; y(t) ≠ 0} is measurable and x = 0 almost everywhere on S, hence Tx = 0 almost everywhere on S, which implies Tx ⊥ y. Therefore T is local. Now suppose that T is local and let S ⊆ Ω be measurable and x ∈ D(T) be such that x = 0 almost everywhere on S. Denote by 𝟙S the indicator function of S. Then x ⊥ 𝟙S , hence Tx ⊥ 𝟙S , which implies Tx = 0 on S. Consequently (L) is satisfied. In the present setting, a local operator need not be a multiplication operator, consider, e.g., the operator T : Lp [0, 1] ⊇ C1 [0, 1] → Lp [0, 1], x 󳨃→ x󸀠 . Remark 5.1.5. We continue the discussion of (b) in Remark 5.1.3, where we now consider the special case where Ω is a compact Hausdorff space and X = C(Ω). By Remark 5.1.3, for every bounded local operator T : C(Ω) → C(Ω) there is y ∈ C(Ω) such that T : x 󳨃→ yx. For unbounded operators, we discuss a further notion of ‘locality’ and its equivalent formulations from the literature. These notions turn out to be stronger than locality as defined in Definition 5.1.1. (1) In [19, Theorem 3.7] a linear operator T : C(Ω) ⊇ D(T) → C(Ω) is called ‘local’ if the following property is satisfied: ∀ x ∈ D(T) , x ≥ 0 , ω ∈ Ω with x(ω) = 0

󳨐⇒

(Tx)(ω) = 0 .

(5.1)

If an operator T satisfies (5.1), then T is local in the sense of Definition 5.1.1. The converse implication is not true, in general. Indeed, consider D(T) := {x ∈ C2 [0, 1]; x󸀠 (0) = x󸀠 (1) = 0} and T : C[0, 1] ⊇ D(T) → C[0, 1] , x 󳨃→ x󸀠󸀠 ,

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which generates the one-dimensional diffusion semigroup in [121, 2.7]. On the one hand, T does not satisfy (5.1), take, e.g., x(t) = 13 t3 − 12 t2 . On the other hand, T is local. Indeed, let x ∈ D(T), y ∈ C[0, 1] with x ⊥ y, and N := {t ∈ [0, 1]; x(t) = 0}. For t ∈ int(N) one has (Tx)(t) = 0, whereas for t ∈ ∂N and every n ∈ ℕ there is t n ∈ [t − 1n , t + 1n ] ∩ [0, 1] such that x(t n ) ≠ 0, i.e., y(t n ) = 0, which implies y(t) = 0. Hence, Tx ⊥ y, consequently T is local. (2) An operator T : C(Ω) ⊇ D(T) → C(Ω) satisfies (5.1) if and only if for every open set O ⊆ Ω one has T(I O ∩ D(T)) ⊆ I O , where I O := {x ∈ C(Ω); ∀ω ∈ Ω \ O : x(ω) = 0} , i.e., T preserves for every open set O ⊆ Ω the largest ideal having O as its carrier. (3) We relate (5.1) to a notion of locality given in [19, p.147, second Remark]. Let (X, K) be a partially ordered vector space with a norm ‖⋅‖. For a linear operator T : X ⊇ D(T) → X the following property¹ is considered: ∀ x ∈ D(T) ∩ K , f ∈ K 󸀠 with f(x) = 0

󳨐⇒

f(Tx) = 0 .

(5.2)

Before we link (5.1) and (5.2), we reformulate (5.2) for the case that X is an Archimedean partially ordered vector space with order unit u. In this case we can use the functional representation of X with the base Σ of the dual cone and its set of extreme points Λ given by (2.22) and (2.23), respectively. For x ∈ D(T) ∩ K the set N = {φ ∈ Σ; φ(x) = 0} is a weakly-∗ closed subset of the weakly-∗ compact set Σ and, therefore, N is weakly-∗ compact. Moreover, N is convex, thus, by the Krein– Milman theorem 1.8.7, the set N is the weak-∗ closure of the convex hull of the set ext(N). We show that ext(N) = N ∩ Λ. Indeed, let φ ∈ ext(N) and let φ1 , φ2 ∈ Σ and t ∈ (0, 1) be such that φ = tφ1 + (1 − t)φ2 . As 0 = φ(x) = tφ1 (x) + (1 − t)φ2 (x) ≥ 0, we have φ1 (x) = φ2 (x) = 0 and, hence, φ1 , φ2 ∈ N. Therefore, φ = φ1 = φ2 , so φ ∈ Λ, which yields the inclusion ext(N) ⊆ N ∩ Λ. The converse inclusion is immediate. We now conclude that (5.2) holds if and only if ∀ x ∈ D(T) ∩ K , f ∈ Λ with f(x) = 0

󳨐⇒

f(Tx) = 0 .

(5.3)

For X = C(Ω) as above and u the constant function 𝟙, by Corollary 2.5.10 the spaces C(Ω) and C(Λ) are isomorphic as Riesz spaces. By Theorem 1.4.8 we obtain that Ω and Λ are homeomorphic. Therefore, the condition (5.1) implies (5.3), and (5.2) implies (5.1), hence all three conditions are equivalent. An important result on disjointness preserving operators is already given in Theorem 1.4.12, which states that every order-bounded disjointness preserving operator between Archimedean vector lattices is regular. We show that for operators between

1 Note that (5.2) is satisfied if and only if T and −T are positive-off-diagonal. For a definition of this notion see [38, Definition 7.18] and also the positive minimum principle in [18, Definition 1.6].

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pre-Riesz spaces a similar statement is not true, in general. In Example 1.2.3 a linear functional on an Archimedean directed ordered vector space is given that is orderbounded, but not regular. The same space is used to construct an appropriate operator in the subsequent example. Example 5.1.6. For A ⊆ [0, ∞) let 𝟙A denote the corresponding indicator function. Define for n, k ∈ ℕ e n : [0, ∞) → ℝ,

t 󳨃→ 𝟙[n−1,n)(t) ,

u n,k : [0, ∞) → ℝ,

t 󳨃→ nt𝟙[0, 1 ] (t) + 1k 𝟙 1 (t) , n {n+ } k

and consider the subspace X := span{e n , u n,k ; n, k ∈ ℕ} of ℝ[0,∞) with pointwise order. By Example 1.2.3, the space X is directed and Archimedean, and, hence, a preRiesz space by Proposition 2.2.3. For every x ∈ X there exists t0 > 0 such that x is affine and, hence, differentiable on (0, t0 ). Define T: X → X ,

x 󳨃→ (lim x󸀠 (t)) e1 . t↓0

The operator T has the following properties: (i) T is disjointness preserving. Indeed, let x(1) , x(2) ∈ X with x(1) ⊥ x(2) . There is M ∈ ℕ such that for i ∈ {1, 2} M

(i)

M

(i)

x(i) = ∑ α n e n + ∑ λ n,k u n,k , n=1

x(1)

n,k=1

x(2)

and of generality, let x(1) = 0 on i.e., 1 [0, M ]. Then (ii) T is order-bounded. Indeed, for v, w ∈ X with v ≤ w there are N ∈ ℕ and C ∈ (0, ∞) such that ±v, ±w ≤ C ∑Ni=1 e i . An element x ∈ X with v ≤ x ≤ w is affine on [0, N1 ], hence −2NCe1 ≤ Tx ≤ 2NCe1 . (iii) T is not regular. Indeed, assume that there is a positive linear operator S : X → X such that S ≥ T. For n, k ∈ ℕ one has that 0 ≤ u n,k ≤ e1 + 1k e n+1 , hence 1 are affine on [0, M ]. Without loss (1) (1) Tx = 0 and hence Tx ⊥ Tx(2) .

ne1 = T(u n,k ) ≤ S(u n,k ) ≤ S(e1 ) + 1k S(e n+1 ) , and therefore k(ne1 − S(e1 )) ≤ S(e n+1 ) for every k ∈ ℕ. Since X is Archimedean, it follows that ne1 − S(e1 ) ≤ 0. From ne1 ≤ S(e1 ) for every n ∈ ℕ one obtains e1 ≤ 0, a contradiction. The question appears whether the pre-Riesz space X in Example 5.1.6 is pervasive. Since there is no obvious candidate for a vector lattice cover, we present a method to show intrinsically that a pre-Riesz space is not pervasive. We start with a simple observation on pervasive pre-Riesz spaces. Lemma 5.1.7. Let X be a pre-Riesz space and (Y, i) a vector lattice cover of X. Then the following are equivalent.

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(i) X is pervasive. (ii) For every b 1 , b 2 ∈ X with i(b 1 ) ∨ i(b 2 ) > 0 there exists x ∈ X such that 0 < i(x) ≤ i(b 1 ) ∨ i(b 2 ). Proof. The implication from (i) to (ii) is clear. To show the implication from (ii) to (i), let y ∈ Y with y > 0. Since X is order dense in Y, we have y = sup{z ∈ i[X]; z ≤ y}. Thus, there exists z ∈ i[X] with z ≤ y and z ≰ 0. For b := i−1 (z) it follows 0 < i(b) ∨ 0. By assumption there exists x ∈ X such that 0 < i(x) ≤ i(b) ∨ 0. We get 0 < i(x) ≤ i(b) ∨ 0 = z ∨ 0 ≤ y, i.e., X is pervasive. The question arises whether the supremum in Lemma 5.1.7 (ii) can be replaced by the infimum. The following definition is due to Helena Malinowski; see [81, Definition 9]. Definition 5.1.8. Let X be a pre-Riesz space and (Y, i) a vector lattice cover of X. Then X is called weakly pervasive in Y if for every b 1 , b 2 ∈ X with i(b 1 ) ∧ i(b 2 ) > 0 there exists x ∈ X such that 0 < i(x) ≤ i(b 1 ) ∧ i(b 2 ). It is clear that every pervasive pre-Riesz space is weakly pervasive². We give an intrinsic characterization of weakly pervasive pre-Riesz spaces. Lemma 5.1.9. Let X be a pre-Riesz space and let (Y, i) be a vector lattice cover of X. Then the following statements are equivalent. (i) X is weakly pervasive in Y. (ii) For every b 1 , b 2 ∈ X with b 1 , b 2 > 0 and b 1 ⊥̸ b 2 there exists an element x ∈ X such that 0 < x ≤ b 1 , b 2 . Proof. The implication from (i) to (ii) is immediate. To show the converse implication, let b 1 , b 2 ∈ X with i(b 1 ) ∧ i(b 2 ) > 0. Then it follows b 1 , b 2 > 0 and b 1 ⊥̸ b 2 . By assumption there exists x ∈ X with 0 < x ≤ b 1 , b 2 , i.e., we get 0 < i(x) ≤ i(b 1 ) ∧ i(b 2 ). Note that the statement in Lemma 5.1.9 (ii) does not depend on the vector lattice cover. Similarly to pervasiveness, X is weakly pervasive in X ρ if and only if X is weakly pervasive in any vector lattice cover of X. A weakly pervasive pre-Riesz space does not have the Riesz decomposition property, in general; see [81, Example 13]. The converse statement is valid. Proposition 5.1.10. If a pre-Riesz space X has the Riesz decomposition property, then X is weakly pervasive. Proof. Let X be a pre-Riesz space with Riesz decomposition property. We use Lemma 5.1.9 (ii). Let b1 , b 2 ∈ X be such that b 1 , b 2 > 0 and b 1 ⊥̸ b 2 , i.e., i(b 1 ) ∧ i(b 2 ) > 0 in a

2 The converse is not true, in general. In [81, Example 11] it is shown that the space Z in Example 3.7.10 is weakly pervasive, but not fordable. Hence, Z is not pervasive.

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vector lattice cover (Y, i) of X. We set a1 := 0. Since i[X] is order dense in Y, for the set S := {z ∈ i[X]; z ≤ i(b 1 ) ∧ i(b 2 )} we have i(b 1 ) ∧ i(b 2 ) = sup S. The element a1 = 0 is a lower bound of i(b 1 ) ∧ i(b 2 ). There exists a2 ∈ X with a2 ≰ 0 such that i(a2 ) ∈ S. We have now four elements with the relationships a1 , a2 ≤ b 1 , b 2 . Since X has the Riesz decomposition property, there exists x ∈ X such that a1 , a2 ≤ x ≤ b 1 , b 2 . Since a1 = 0 and a2 ≰ 0, it follows x > 0. We conclude that X is weakly pervasive. The observations in Lemma 5.1.9 and Proposition 5.1.10 can be found in [81], as well as the subsequent example. Example 5.1.11. We continue Example 5.1.6 and show that X is not weakly pervasive by using the characterization of Lemma 5.1.9. For b 1 := 2u 1,2 and b 2 := e2 we have b 1 , b 2 > 0. We show that b 1 ⊥̸ b 2 by establishing that {b 1 − b 2 , −b 1 + b 2 }u ≠ {b 1 + b 2 , −b 1 − b 2 }u . Consider the element v := 2u 1,3 + e2 . We show that v ∈ {b 1 − b 2 , −b 1 + b 2 }u . Indeed, for t ∈ [0, 1] we have v(t) = 2u 1,3 (t) = 2t ≥ ±2t = ±(b 1 − b 2 )(t). For t ∈ [1, 2] \ { 32 , 43 } we obtain v(t) = e2 (t) = 1 ≥ ±1 = ±(b 1 − b 2 )(t). Moreover, we get v( 43 ) = 53 ≥ ±1 = ±(b 1 − b 2 )( 43 ) and v( 32 ) = 1 ≥ ∓ 12 = ±(b 1 − b 2 )( 32 ). We conclude v ∈ {b 1 − b 2 , −b 1 + b 2 }u . On the other hand, v ∉ {b 1 + b 2 , −b 1 − b 2 }u , as v( 32 ) = 1 < 2 = (b 1 + b 2 )( 32 ). This establishes b 1 ⊥̸ b 2 . Assume there is x ∈ X with 0 < x ≤ b 1 , b 2 . Then for t ∈ [0, ∞) \ { 23 } we have x(t) = 0 and 0 < x( 32 ) ≤ 1. It follows that x is a nonzero multiple of u 1,2 , which is a contradiction. Consequently, X is not weakly pervasive, and, hence, not pervasive. By Proposition 5.1.10, the space X does not have the Riesz decomposition property.

5.1.2 Riesz* homomorphisms on pre-Riesz spaces of continuous functions In vector lattices, positive disjointness preserving operators are precisely the Riesz homomorphisms; see Proposition 1.4.3. In pre-Riesz spaces, Riesz* homomorphisms preserve disjointness. Theorem 5.1.12. Let X and Y be pre-Riesz spaces and T : X → Y a Riesz* homomorphism. Then T is a positive disjointness preserving operator. Proof. If (X ρ , i X ) and (Y ρ , i Y ) are the Riesz completions of X and Y, respectively, then due to Theorem 2.4.11 there is a Riesz homomorphism T ρ : X ρ → Y ρ that extends T in the sense that Tρ ∘ iX = iY ∘ T . In particular, T ρ is positive and disjointness preserving. Since i X and i Y are bipositive, T is positive. Moreover, due to Proposition 4.1.4 one obtains that T is disjointness preserving.

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On spaces of continuous functions, Riesz homomorphisms can be represented as follows; see also [2, Theorem 4.25]. Theorem 5.1.13. Let P and Q be nonempty compact Hausdorff spaces and let T : C(P) → C(Q). The map T is a Riesz homomorphism if and only if there exist a map α : Q → P and a weight function w ∈ C(Q), w ≥ 0, such that for every x ∈ C(P) and q ∈ Q we have (Tx)(q) = w(q)x(α(q)) . (5.4) Moreover, in this case, w = T𝟙 P and the map α is uniquely determined and continuous on the set {q ∈ Q; w(q) > 0}. An operator T : C(P) → C(Q) satisfying (5.4) is called a weighted composition operator. The result in Theorem 5.1.13 has been extended to pre-Riesz subspaces of continuous functions by Hent van Imhoff in [152]; see the subsequent theorem. We say that a subspace X of C(P) strongly separates the points of P if for every p1 , p2 ∈ P with p1 ≠ p2 there is x ∈ X such that x(p1 ) = 0 and x(p2 ) = 1. Note that X := {x ∈ C[0, 1]; x(1) = 2x(0)} separates the points of [0, 1], but does not strongly separate the points of [0, 1]. Theorem 5.1.14. Let P and Q be nonempty compact Hausdorff spaces and let X and Y be order dense subspaces of C(P) and C(Q), respectively. Let T : X → Y be linear. (i) If T is a Riesz* homomorphism, then there exist w : Q → ℝ+ and α : Q → P such that for every x ∈ X and q ∈ Q we have (Tx)(q) = w(q)x(α(q)) .

(5.5)

If, in addition, X strongly separates the points of P, then w can be taken such that w is continuous on Q, the map α is uniquely determined on {q ∈ Q; w(q) > 0}, and on this set α is continuous. (ii) If for T there exist w ∈ C(Q), w ≥ 0, and α : Q → P continuous on {q ∈ Q; w(q) > 0} such that (5.5) holds for every x ∈ X and q ∈ Q, then T is a Riesz* homomorphism. Proof. (i) Let T be a Riesz* homomorphism. As X is order dense in C(P), the Riesz subspace X ρ of C(P) generated by X is the Riesz completion of X. Similarly, we obtain the Riesz completion Y ρ of Y as the Riesz subspace of C(Q) generated by Y. By Theorem 2.4.11 there is a Riesz homomorphism T ρ : X ρ → Y ρ that extends T. Let us fix some q ∈ Q and define Tq : Xρ → ℝ ,

x 󳨃→ (T ρ x)(q) .

Then T q is a Riesz homomorphism as it is the composition of T ρ and the point evaluation, which are both Riesz homomorphisms. Since X ρ is a majorizing Riesz subspace in C(P) and ℝ is Dedekind complete, we can extend T q by means of Theorem 2.1.17. Hence, we obtain a Riesz homomorphism T̂ q : C(P) → ℝ that extends

5.1 Disjointness preserving operators

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245

T q . By Theorem 1.4.7 there exist v(q) ∈ ℝ+ and α(q) ∈ P such that for every x ∈ C(P) we have T̂ q x = v(q)x(α(q)). Thus, we obtain maps v : Q → ℝ+ and α : Q → P such that for every x ∈ X and q ∈ Q we have (Tx)(q) = (T ρ x)(q) = T q x = T̂ q x = v(q)x(α(q)) . Define w : Q → ℝ+ by w = v on {q ∈ Q; ∃x ∈ X : (Tx)(q) ≠ 0}, and w = 0 otherwise. Then (5.5) is satisfied. In order to show the second part of (i), suppose that X, in addition, strongly separates the points of P. First, we show that w is bounded. Indeed, there is x ∈ X with x ≥ 𝟙P . Then Tx is continuous and, hence, bounded. From (5.5) it follows for every q ∈ Q that w(q) ≤ ‖Tx‖, as x(α(q)) ≥ 1. Fix q ∈ Q. We aim to show that w is continuous at q and that α is continuous at q whenever w(q) > 0. To this end, we will show that every net (q γ )γ∈Γ in Q, which converges to q has a subnet (r β ) such that limβ w(r β ) = w(q), and that in the case w(q) > 0 we have limβ α(r β ) = α(q). Let (q γ )γ∈Γ be a net in Q, which converges to q. For every x ∈ X the function Tx is continuous and we get w(q)x(α(q)) = (Tx)(q) = lim(Tx)(q γ ) = lim w(q γ )x(α(q γ )) . γ

γ

(5.6)

Since P is compact and w[Q] is bounded in ℝ, there exists a subnet of (q γ ), which we will denote by (r β ), such that (w(r β ))β converges to say t ∈ ℝ and such that (α(r β ))β converges to say p ∈ P. Hence, by (5.6), we obtain w(q)x(α(q)) = tx(p) . First consider the case w(q) = 0. Since X is majorizing, there exists x ∈ X such that x(p) > 0. Therefore, t = 0. Thus, lim β w(r β ) = 0 = w(q). Next we deal with the case w(q) > 0. Since X is majorizing in C(P), there exists x ∈ X, x ≥ 0, so that x(α(q)) > 0. Hence, t > 0. We show that p = α(q). Suppose p ≠ α(q). Since X strongly separates the points of P, there exists x ∈ X such that x(p) = 1 and x(α(q)) = 0, which contradicts w(q) > 0. Therefore, p = α(q) and, hence, t = tx(p) = w(q)x(α(q)) = w(q)x(p) = w(q). It follows that limβ α(r β ) = p = α(q) and limβ w(r β ) = t = w(q). We conclude that w is continuous at q. If w(q) > 0, then α is continuous at q. Finally we show that α is uniquely determined on {q ∈ Q; w(q) > 0}. Indeed, let q ∈ Q be such that w(q) > 0. Suppose that there are α 1 , α 2 : Q → P such that (5.5) holds for α = α 1 and α = α 2 , respectively, and α 1 (q) ≠ α 2 (q). Since X strongly separates the points of P, there exists x ∈ X such that x(α1 (q)) = 0 and x(α 2 (q)) = 1. Then w(q) = w(q)x(α 2 (q)) = (Tx)(q) = w(q)x(α 1 (q)) = 0 , which contradicts w(q) > 0.

246 | 5 Operators on pre-Riesz spaces To show (ii), let T : X → Y satisfy (5.5) for some w ∈ C(Q), w ≥ 0, and α : Q → P, which is continuous on {q ∈ Q; w(q) > 0}. Define S : C(P) → C(Q) by (5.5) with T replaced by S. Then S is a Riesz homomorphism, and its restriction T to X is a Riesz* homomorphism due to Theorem 2.4.11. In [152], Theorem 5.1.14 is applied to Riesz* homomorphisms on Sobolev spaces and on spaces of differentiable functions on manifolds. The converse statement in Theorem 5.1.12 is not true, in general. The following exquisite example of a positive disjointness preserving operator that is not a Riesz* homomorphism is given by Hent van Imhoff. Example 5.1.15. We consider the pre-Riesz space P[0, 1] of all polynomial functions s on [0, 1] and the operator T : P[0, 1] → P[0, 1], x 󳨃→ (s 󳨃→ ∫0 x(t)dt). Clearly, T is positive. As there are no nontrivial disjoint elements in P[0, 1], the operator T is disjointness preserving. Suppose that T is a Riesz* homomorphism, then by Theorem 5.1.14 (i) there are w : [0, 1] → ℝ+ and α : [0, 1] → [0, 1] such that for every x ∈ P[0, 1] and q ∈ [0, 1] we have (Tx)(q) = w(q)x(α(q)). Choosing the polynomials x := 𝟙 and x : t 󳨃→ t, we find w(q) = q and α(q) = 12 q. The choice x : t 󳨃→ t2 then yields a contradiction. We conclude that T is not a Riesz* homomorphism.

Notes and remarks In view of Examples 5.1.6 and 5.1.11, it remains an open question under which additional conditions on a pre-Riesz space every order-bounded disjointness preserving operator is regular. Similar to Theorem 5.1.13, there is a representation result for disjointness preserving operators between spaces of continuous functions; see, e.g., [119, Theorem 3.2.10]. It is an interesting question whether this result can be extended to disjointness preserving operators between order dense subspaces in the spirit of Theorem 5.1.14. The characterization of Riesz* homomorphisms in Theorem 5.1.14 relies on the fact that Riesz* homomorphisms on pre-Riesz spaces can be extended to Riesz homomorphisms on the corresponding Riesz completions, due to Theorem 2.4.11. Similarly, one can ask which operators on a pre-Riesz space can be extended to disjointness preserving operators on the Riesz completion. Clearly, a necessary condition is that the operator itself is disjointness preserving. If the pre-Riesz space has only trivial disjoint elements, then clearly not every disjointness preserving operator can be extended to a disjointness preserving operator on the Riesz completion. The situation might improve if one considers pervasive pre-Riesz spaces. Up to now, there are only results that are far from a general answer. Differences of Riesz* homomorphisms are investigated in [83, 6.7], where a rather technical condition is given to characterize those operators that extend to order-bounded disjointness preserving operators. Results that involve

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strong norm conditions are provided in [87]. If an appropriate extension exists, then one easily transfers properties from the vector lattice case to pre-Riesz spaces. The scope of this method is presently limited.

5.2 Disjointness preserving inverse If X and Y are Banach lattices and T : X → Y is a linear bijective disjointness preserving operator, then in [70] and [100] it is shown that the inverse T −1 is also disjointness preserving. In [71] it is proven that an order-bounded disjointness preserving bijection between Archimedean vector lattices has an order-bounded disjointness preserving inverse, generalizing the corresponding result for Banach lattices in [17]. Various other conditions under which T −1 is disjointness preserving are given in [4], where it is also observed that for a disjointness preserving operator between arbitrary vector lattices the inverse is not disjointness preserving, in general. We now address the question under which conditions the inverse of a disjointness preserving bijection on a pre-Riesz space is disjointness preserving. For vector lattices, this problem is formulated in [5, Problem A]. As Riesz* homomorphisms are disjointness preserving, we first consider inverses of bijective Riesz* homomorphisms. On pervasive pre-Riesz spaces, they will turn out to be Riesz* homomorphisms and, hence, disjointness preserving operators. Further, we study d-isomorphisms on pre-Riesz spaces, i.e., disjointness preserving bijections with disjointness preserving inverse. This will lead to the remarkable fact that in finite-dimensional Archimedean pre-Riesz spaces the inverse of a disjointness preserving bijection is disjointness preserving. In infinite dimensions, a necessary condition is investigated, which is also sufficient provided the underlying preRiesz space is fordable.

5.2.1 Inverses of bijective Riesz* homomorphisms As a motivation for the discussion in this subsection, we recall the following classical observation on bijective Riesz homomorphisms. Theorem 5.2.1. Let X and Y be vector lattices and let T : X → Y be a bijective Riesz homomorphism. Then T −1 is a Riesz homomorphism. Proof. Let y1 , y2 ∈ Y, then T −1 (y1 ∨ y2 ) = T −1 (T (T −1 (y1 )) ∨ T (T −1 (y2 ))) = T −1 (T (T −1 (y1 ) ∨ T −1 (y2 ))) = T −1 (y1 ) ∨ T −1 (y2 ) , hence T −1 is a Riesz homomorphism.

248 | 5 Operators on pre-Riesz spaces

We present a generalization of Theorem 5.2.1 to Riesz* homomorphisms on pervasive pre-Riesz spaces. The results in this subsection are due to Hent van Imhoff and can be found in [152, Section 5]. In the next lemma we use Theorem 2.4.11. Lemma 5.2.2. Let X and Y be pre-Riesz spaces and let (X ρ , i X ) and (Y ρ , i Y ) be their Riesz completions, respectively. Let T : X → Y be a Riesz* homomorphism and T ρ : X ρ → Y ρ the Riesz homomorphism such that T ρ ∘ i X = i Y ∘ T. (i) If T is surjective, then T ρ is surjective. (ii) If X is pervasive and T is injective, then T ρ is injective. Proof. (i) Let w ∈ Y ρ be given and let c1 , . . . , c n , d1 , . . . , d m ∈ Y be such that n

m

w = ⋁ i Y (c i ) − ⋁ i Y (d j ) i=1

j=1

holds in Y ρ . As T is surjective, there are a1 , . . . , a n , b 1 , . . . , b m ∈ X with c i = T(a i ) and d j = T(b j ), i ∈ {1, . . . , n} and j ∈ {1, . . . , m}. We define v := ⋁ ni=1 i X (a i )− ρ ⋁m j=1 i X (b i ) ∈ X . Then n

m

T ρ v = T ρ (⋁ i X (a i ) − ⋁ i X (b j )) i=1

j=1

n

m

= ⋁ T ρ (i X (a i )) − ⋁ T ρ (i X (b j )) i=1 n

j=1 m

= ⋁ i Y (c i ) − ⋁ i Y (d j ) = w , i=1

j=1

Tρ

is surjective. hence (ii) Suppose that X is pervasive. We first show that for every v ∈ X ρ with v > 0 we have T ρ v > 0. Indeed, let v ∈ X ρ , v > 0, then there exists x ∈ X with 0 < i X (x) ≤ v. Since T and T ρ are positive, this yields 0 ≤ i Y (Tx) = T ρ (i X (x)) ≤ T ρ v. As T is injective, we have Tx ≠ 0. Hence, T ρ v > 0. Next, let v ∈ X ρ be such that T ρ v = 0. Since T ρ is a Riesz homomorphism, we obtain T ρ v+ = (T ρ v)+ = 0. By the previous step, we get v+ = 0. Similarly, we have T ρ v− = (T ρ v)− = 0 and, therefore, v− = 0. Hence, v = 0. We conclude that T ρ is injective. In particular, Lemma 5.2.2 shows that a bijective Riesz* homomorphism T on a pervasive pre-Riesz space extends to a bijective Riesz homomorphism on the Riesz completion. Next we study properties of the inverse of T.

5.2 Disjointness preserving inverse |

249

Theorem 5.2.3. Let X and Y be pre-Riesz spaces with X pervasive. Let T : X → Y be a bijective Riesz* homomorphism. Then T −1 is a Riesz* homomorphism and, hence, T is an order isomorphism. Proof. Let T be a bijective Riesz* homomorphism. With the notations as in Lemma 5.2.2, there is a bijective Riesz homomorphism T ρ : X ρ → Y ρ with T ρ ∘ i X = i Y ∘ T. By Theorem 5.2.1, the inverse of T ρ is a Riesz homomorphism. We show that T −1 is the restriction of (T ρ )−1 to Y in the sense that (T ρ )−1 ∘ i Y = i X ∘ T −1 . Indeed, let y ∈ Y, then T −1 (y) ∈ X, thus (T ρ ∘ i X ) (T −1 (y)) = (i Y ∘ T) (T −1 (y)) = i Y (y) , hence, by applying (T ρ )−1 , we get i X (T −1 (y)) = (T ρ )−1 (i Y (y)). By Theorem 2.4.11 we conclude that T −1 is a Riesz* homomorphism. As T and T −1 are positive, we get that T is an order isomorphism. If X is not pervasive, then the statement in Theorem 5.2.3 is not true, in general. Indeed, in [152, Example 5.4] an example is given where X is a nonpervasive pre-Riesz space and T : X → X is a bijective complete Riesz homomorphism such that T −1 is not a Riesz* homomorphism. As a converse to Theorem 5.2.3 we have the following result. Proposition 5.2.4. Let X and Y be partially ordered vector spaces and let T : X → Y be a linear operator. Then T is an order isomorphism if and only if T is bijective and both T and T −1 are complete Riesz homomorphisms. Proof. Let T be an order isomorphism. We show that T is a complete Riesz homomorphism. Indeed, let A ⊆ X be nonempty with inf A = 0. As T is positive, clearly 0 is a lower bound of T[A]. Let y ∈ Y be a lower bound of T[A], then T −1 y is a lower bound of A. Therefore, T −1 y ≤ 0. As T is positive, we get y = T(T −1 y) ≤ 0. Hence, inf T[A] = 0. We conclude that T is a complete Riesz homomorphism. By considering the order isomorphism T −1 instead of T, we obtain that T −1 is a complete Riesz homomorphism. The converse implication is clear, since every complete Riesz homomorphism is positive according to Theorem 2.3.19. A combination of the Theorems 2.3.19 and 5.2.3 and Proposition 5.2.4 yields the following list of equivalent statements. Corollary 5.2.5. Let X and Y be pervasive pre-Riesz spaces. For a linear bijection T : X → Y the following statements are equivalent: (i) T is a Riesz* homomorpism. (ii) T is a Riesz homomorpism. (iii) T is a complete Riesz homomorpism. (iv) T −1 is a Riesz* homomorpism. (v) T −1 is a Riesz homomorpism. (vi) T −1 is a complete Riesz homomorpism. (vii) T is an order isomorphism.

250 | 5 Operators on pre-Riesz spaces

The main implication of Corollary 5.2.5 is that a bijective Riesz* homomorphism between pervasive pre-Riesz spaces is an order isomorphism, just as in the vector lattice case. 5.2.2 On d-isomorphisms in pre-Riesz spaces In the present subsection, let X be a pre-Riesz space and T : X → X a linear operator. We call T a d-isomorphism if T is bijective, disjointness preserving, and has a disjointness preserving inverse. We intend to show that every d-isomorphism yields a bijection on the set of bands. For the results in the present subsection, see also [79]. Denote B := {B ⊆ X; B is a band in X} and define T: B → B ,

B 󳨃→ T[B]dd .

(5.7)

If B1 , B2 ∈ B are such that B1 ⊆ B2 , then T(B1 ) ⊆ T(B2 ). Proposition 5.2.6. If T is disjointness preserving, then for B1 , B2 ∈ B with B1 ⊥ B2 one has T(B1 ) ⊥ T(B2 ). Proof. For every B ∈ B one has T(B) ⊆ T(Bd )d . Indeed, we show d

T[B]dd ⊆ T[Bd ] = T[Bd ]

ddd

.

(5.8)

Let x ∈ T[B]dd . Let y ∈ T[Bd ] and v ∈ Bd be such that Tv = y. For u ∈ B one has u ⊥ v and, since T is disjointness preserving, Tu ⊥ Tv, hence y ∈ T[B]d . Therefore, x ⊥ y, which yields (5.8). As a consequence, one has T(Bd ) = T(Bd )dd ⊆ T(B)d . For B1 , B2 ∈ B with B1 ⊥ B2 , from B2 ⊆ Bd1 one obtains T(B2 ) ⊆ T(Bd1 ) ⊆ T(B1 )d , and we conclude T(B2 ) ⊥ T(B1 ). Remark 5.2.7. If T is, in addition, injective, and B is a nontrivial band, then T(B) is a nontrivial band. Indeed, B contains an element b ≠ 0, and since T is injective, one has T(b) ≠ T(0) = 0. Then T(b) ∈ T[B] ⊆ T[B]dd = T(B). Similarly, since Bd contains a nonzero element c, one has 0 ≠ T(c) ∈ T[Bd ] ⊆ T[Bd ]dd = T(Bd ) ⊆ T(B)d , by Proposition 5.2.6. Proposition 5.2.8. If T is a d-isomorphism and B ∈ B, then T(Bd ) = T(B)d . Proof. Let B ∈ B. Since B ⊥ Bd , by Proposition 5.2.6 one has T(B) ⊥ T(Bd ), hence T(Bd ) ⊆ T(B)d . Next we show T[B]d ⊆ T[Bd ]dd . Let x ∈ T[B]d and v ∈ X be such that Tv = x. Let y ∈ T[Bd ]d . For z ∈ B one has x ⊥ Tz and, since T is a d-isomorphism, v ⊥ z. Therefore, v ∈ Bd , which implies x = Tv ∈ T[Bd ]. Consequently y ⊥ x, which yields x ∈ T[Bd ]dd . It follows that T(B)d = T[B]ddd = T[B]d ⊆ T[Bd ]dd = T(Bd ).

5.2 Disjointness preserving inverse |

251

Proposition 5.2.9. Let T be a d-isomorphism. Then T is band preserving if and only if T equals the identity. Proof. Let T be band preserving and B ∈ B. Then T[B] ⊆ B, moreover T(B) = T[B]dd ⊆ B. Similarly, T(Bd ) ⊆ Bd . By Proposition 5.2.8, T(B)d ⊆ Bd and therefore B = Bdd ⊆ T(B)dd = T(B). We conclude T(B) = B. Conversely, let T be the identity and B ∈ B. Then T[B] ⊆ T[B]dd = T(B) = B, hence T is band preserving. Proposition 5.2.10. If T is a d-isomorphism, then for every k ∈ ℕ and every B ∈ B one has T k (B) = (T k [B])dd . Proof. We show T 2 (B) = (T 2 [B])dd , then the assertion follows by similar arguments. First observe that (T 2 [B])

dd

dd

= T[T[B]]dd ⊆ T[T[B]dd ]

= T 2 (B) .

Similarly, (T 2 [Bd ])dd ⊆ T 2 (Bd ). By Proposition 5.2.6, from B ⊥ Bd it follows that T 2 (B) ⊥ T 2 (Bd ). Therefore d

T 2 (B) ⊆ T 2 (Bd ) ⊆ (T 2 [Bd ])

ddd

d

= (T 2 [Bd ]) .

It remains to show (T 2 [Bd ])d ⊆ (T 2 [B])dd . Indeed, let x ∈ (T 2 [Bd ])d . Let y ∈ (T 2 [B])d and v ∈ X be such that T 2 v = y. For z ∈ B one has T 2 z ⊥ y, and, since T is a d-isomorphism, z ⊥ v. Therefore, v ∈ Bd , and hence y = T 2 v ∈ T 2 [Bd ]. Thus, x ⊥ y. If T is a d-isomorphism, then for k ∈ ℕ the operator T k is a d-isomorphism as well. Combining the Propositions 5.2.9 and 5.2.10, we obtain the following result. Proposition 5.2.11. Let T be a d-isomorphism and k ∈ ℕ. Then T k is band preserving if and only if T k equals the identity. Now define S: B → B ,

B 󳨃→ [B]T dd .

Proposition 5.2.12. If T is a d-isomorphism, then for every B ∈ B one has S(T(B)) = B and T(S(B)) = B. Proof. We show the first equality. Given B ∈ B, then B ⊆ S(T(B)) = [T[B]dd ]T

dd

.

(5.9)

d

Indeed, let x ∈ B and y ∈ [T[B]dd ]T . For z ∈ T[B]dd and v ∈ X with Tv = z, one has v ⊥ y. Since T is disjointness preserving, we obtain Tv ⊥ Ty, hence z ⊥ Ty. Thus, Ty ∈ T[B]ddd = T[B]d . In particular, Tx ⊥ Ty. Since T is a d-isomorphism, we get dd x ⊥ y, and hence x ∈ [T[B]dd ]T . One obtains (5.9), and, moreover, Bd ⊆ S(T(Bd )) .

(5.10)

252 | 5 Operators on pre-Riesz spaces By Proposition 5.2.6 we have that B ⊥ Bd implies T(B) ⊥ T(Bd ). Applying Proposition 5.2.6 to the inverse of T, we get S(T(B)) ⊥ S(T(Bd )). Together with (5.10), this boils down to d S(T(B)) ⊆ S(T(Bd )) ⊆ Bdd = B. Proposition 5.2.12 yields that T is a bijection on B, and that S is its inverse.

5.2.3 Disjointness preserving bijections in finite dimensions In this subsection, let K be a generating closed cone in ℝn . Note that (ℝn , K) is then an Archimedean directed partially ordered vector space, and, hence, a pre-Riesz space, so that we can apply the results of Subsection 5.2.2. We intend to show that the inverse of a disjointness preserving bijection on (ℝn , K) is disjointness preserving. The proof will depend on the fact that the number b of bands in (ℝn , K) is finite. Recall that n b ≤ 14 22 by Theorem 4.4.26. The material in this subsection can also be found in [79]. We define p : B → ℕ , B 󳨃→ dim(B) + dim(Bd ) , and collect properties of p in the subsequent lemmas. Let T : ℝn → ℝn be a disjointness preserving linear bijection, and let T be defined as in (5.7). Lemma 5.2.13. (i) For every B ∈ B we have p(T(B)) ≥ p(B). (ii) If B ∈ B is such that p(T(B)) = p(B), then T(B) = T[B] and T(B)d = T[Bd ]. Proof. (i) Clearly, T[B] ⊆ T[B]dd = T(B), and by Proposition 5.2.6 T[Bd ] ⊆ T[Bd ]dd = T(Bd ) ⊆ T(B)d . Hence, dim(T(B)) ≥ dim(T[B]) = dim(B) and dim(T(B)d ) ≥ dim(T[Bd ]) = dim(Bd ), therefore p(T(B)) ≥ p(B). (ii) Let B be a band such that p(T(B)) = p(B). Since T[B] is a subspace of T(B), the assumption T(B) ≠ T[B] implies dim(T(B)) > dim(T[B]) = dim(B). Analogously, T[Bd ] is a subspace of T(B)d , and the assumption T(B)d ≠ T[Bd ] implies dim(T(B))d > dim(T[Bd ]) = dim(Bd ). In both cases a contradiction to p(T(B)) = p(B) is obtained. Lemma 5.2.14. (i) If A, B ∈ B are such that T(A) = T[B] and T(A)d = T[Bd ], then A = B. (ii) If A, B ∈ B are such that T(A) = T(B) and p(T(B)) = p(B), then A = B.

5.2 Disjointness preserving inverse

| 253

Proof. (i) From T[B] = T(A) = T[A]dd ⊇ T[A] we obtain B ⊇ A. Similarly, as T is disjointness preserving, we get T[Bd ] = T(A)d = T[A]d ⊇ T[Ad ], hence Bd ⊇ Ad . As A and B are bands, it follows that B = Bdd ⊆ Add = A. (ii) By Lemma 5.2.13 (ii), we obtain T(A) = T(B) = T[B] and T(A)d = T(B)d = T[Bd ]. Now the statement follows from (i). Next we consider certain sets of nontrivial bands. For k ∈ ℕ we denote Bk := {B ∈ B \ {ℝn , {0}}; p(B) = k} . Furthermore, let m := max{p(B); B ∈ B \ {ℝn , {0}}}. Lemma 5.2.15. (i) For every k ∈ {2, . . . , m} one has T[Bk ] ⊆ Bk , and T : Bk → Bk is a bijection. (ii) For every B ∈ B we have T[B] ∈ B. For every B ∈ B there is A ∈ B such that T[A] = B and T[Ad ] = Bd . Proof. (i) We first consider the case k = m. Let B ∈ Bm . By Remark 5.2.7, the band T(B) is nontrivial. From Lemma 5.2.13 (i) we obtain m = p(B) ≤ p(T(B)) ≤ m ,

(5.11)

hence T(B) ∈ Bm . To show that T : Bm → Bm is injective, let A, B ∈ Bm be such that T(A) = T(B). By (5.11) and Lemma 5.2.14 (ii), we obtain A = B. As Bm is a finite set, it follows that T : Bm → Bm is a bijection. Next, assume that ℓ < m is such that for every k ∈ {ℓ + 1, . . . , m} we have T[Bk ] ⊆ Bk , and T : Bk → Bk is a bijection. Let A ∈ Bℓ . Suppose that T(A) ∉ Bℓ . Then, by Lemma 5.2.13 (i), there is k ∈ {ℓ+1, . . . , m} such that T(A) ∈ Bk . Since T : Bk → Bk is a bijection, there exists B ∈ Bk such that T(B) = T(A). Moreover, p(T(B)) = p(B) = k, therefore Lemma 5.2.14 (ii) implies that A = B. This contradicts p(A) = ℓ < k = p(B). Hence, T(A) ∈ Bℓ . We obtain T[Bℓ ] ⊆ Bℓ . If A, B ∈ Bℓ are such that T(A) = T(B), then by the previous step we get p(T(B)) = ℓ = p(B), so that Lemma 5.2.14 (ii) implies A = B. Thus, T : Bℓ → Bℓ is injective, and, moreover, it is a bijection. By consecutively considering ℓ = m − 1, ℓ = m − 2, . . . , ℓ = 1, the proof is complete. (ii) For trivial bands the assertion is clear. Let B ∈ B be nontrivial, i.e., there exists k ∈ ℕ such that B ∈ Bk . By (i), we obtain T(B) ∈ Bk , so p(T(B)) = p(B). Lemma 5.2.13 (ii) implies T(B) = T[B], hence T[B] ∈ B. Moreover, by (i) there exists A ∈ Bk such that T(A) = B. Then p(T(A)) = p(A), and Lemma 5.2.13 (ii) implies T(A) = T[A] and T(A)d = T[Ad ]. Hence, B = T[A] and Bd = T[Ad ]. The next result states that the inverse of a disjointness preserving bijection in a finitedimensional Archimedean pre-Riesz space is disjointness preserving.

254 | 5 Operators on pre-Riesz spaces Theorem 5.2.16. Let T : (ℝn , K) → (ℝn , K) be a disjointness preserving bijection. Then T is a d-isomorphism. Proof. We show that the inverse of T is disjointness preserving. Let u, v ∈ ℝn be such that u ⊥ v, and let x, y ∈ ℝn be such that Tx = u and Ty = v. Let B := {u}dd , then B ∈ B, u ∈ B and v ∈ Bd . By Lemma 5.2.15 (ii), there exists A ∈ B such that B = T[A] and Bd = T[Ad ]. Then x ∈ A and y ∈ Ad , hence x ⊥ y. For the number b of bands in (ℝn , K), let P(b) denote the set of orders of permutations on b symbols. Theorem 5.2.17. Let T : (ℝn , K) → (ℝn , K) be a disjointness preserving bijection. Then there is k ∈ P(b) such that T k is band preserving. Proof. By Theorem 5.2.16, the operator T is a d-isomorphism. Due to Proposition 5.2.12, the map T is a bijection, so T is a permutation on the finite set B. Hence, there is k ∈ P(b) such that T k is the identity. By Proposition 5.2.11, the operator T k is band preserving. Theorem 5.2.16 is an instance of a theory on inverses of disjointness preserving operators on suitably normed pre-Riesz spaces. The finite-dimensional spaces under consideration can be such that there are no nontrivial disjoint elements at all or such that there are even more disjoint elements than in a vector lattice of the same dimension. Apparently, that does not matter for Theorem 5.2.16.

5.2.4 Disjointness preserving inverse in infinite dimensions In [3] a condition (β) for a linear disjointness preserving bijection T between vector lattices is introduced which is equivalent to T −1 being disjointness preserving. The condition (β) is formulated by means of disjointness, hence it generalizes naturally to pre-Riesz spaces. Note that for elements x, y of a pre-Riesz space X one has {x}dd ⊆ {y}dd if and only if x ∈ {y}dd . Definition 5.2.18. Let X and Y be pre-Riesz spaces and let T : X → Y be a linear operator. T is said to satisfy condition (β) if for every x, y ∈ X with {x}dd ⊆ {y}dd it follows that {Tx}dd ⊆ {Ty}dd . The idea³ of this condition traces back to [129]. In function spaces with pointwise order, (β) means the following: If the support of a function is contained in the support of another function, then the same is true for the supports of their images.

3 The symbol (β) honors Beata Randrianantoanina.

255

5.2 Disjointness preserving inverse |

For a linear bijection T between pre-Riesz spaces X and Y, we will show that (β) for T implies T −1 being disjointness preserving, provided X is fordable; see also [87]. For this, we need the following technical observation. Lemma 5.2.19. (i) Let Z be a vector lattice and let D be a majorizing subspace of Z. If z ∈ Z is such that z ⊥ d for every d ∈ D, then z = 0. (ii) Let X be a pre-Riesz space and (Z, i) a vector lattice cover of X. If z ∈ Z is such that z ⊥ d for every d ∈ i[X], then z = 0. Proof. To show (i), let z ∈ Z be such that z ⊥ d for every d ∈ D. As D is majorizing, there is d ∈ D such that |z| ≤ d. Hence, |z| = |z| ∧ d = 0, consequently z = 0. The statement in (ii) follows directly from (i), as i[X] is order dense and, therefore, majorizing in Z. Next we relate disjointness and bands that are generated by a singleton. Lemma 5.2.20. Let X be a pre-Riesz space and let x, y ∈ X. (i) If x ⊥ y then {x}dd ∩ {y}dd = {0}. (ii) If X is, in addition, fordable, then from {x}dd ∩ {y}dd = {0} it follows that x ⊥ y. Proof. (i) Assume that x ⊥ y, which means y ∈ {x}d . Hence, {y}dd ⊆ {x}ddd = {x}d , which implies {x}dd ∩ {y}dd ⊆ {x}dd ∩ {x}d = {0} . (ii) Assume that {x}dd ∩ {y}dd = {0}. Let (X ρ , i) be the Riesz completion of X and define u := |i(x)| ∧ |i(y)|. Since X is fordable, there is S ⊆ X such that {u}d = i[S]d in X ρ . From Proposition 4.1.5 it follows that [{u}d ] i = [i[S]d ] i = Sd .

(5.12)

We show that Sdd ⊆ {x}dd . Indeed, let z ∈ {x}d , then i(z) ⊥ i(x), hence i(z) ⊥ u. Due to (5.12), z ∈ [{u}d ]i = Sd . It follows that {x}d ⊆ Sd , and therefore Sdd ⊆ {x}dd . Analogously, one obtains Sdd ⊆ {y}dd . The assumption yields S ⊆ Sdd ⊆ {0}. Now (5.12) implies that i[X] ⊆ {u}d , hence from Lemma 5.2.19 it follows that u = 0. Consequently, i(x) ⊥ i(y), which implies x ⊥ y, by Proposition 4.1.4. If X is not fordable, then Lemma 5.2.20 (ii) is not true, in general. Indeed, consider in Example 4.4.18 the elements x = (1, 0, 1)T and y = (0, 1, 1)T in ℝ3 , then {x}dd ∩ {y}dd = {0}, but x ⊥̸ y. Theorem 5.2.21. Let X and Y be pre-Riesz spaces and let T : X → Y be a linear injective operator. (i) If X is, in addition, fordable and T satisfies condition (β), then T −1 : T[X] → X is disjointness preserving.

256 | 5 Operators on pre-Riesz spaces (ii) Let T be surjective and disjointness preserving. If T −1 is disjointness preserving then T satisfies (β). Proof. (i) Let y1 , y2 ∈ T[X] be such that y1 ⊥ y2 in Y and let x1 , x2 ∈ X be such that Tx1 = y1 and Tx2 = y2 . Due to Lemma 5.2.20 (i) one obtains {y1 }dd ∩ {y2 }dd = {0} . Let u ∈ {x1 }dd ∩ {x2 }dd . From u ∈ {x1 }dd it follows that {u}dd ⊆ {x1 }dd , hence property (β) yields that {Tu}dd ⊆ {Tx1 }dd . Analogously, {Tu}dd ⊆ {Tx2 }dd , therefore {Tu}dd ⊆ {Tx1 }dd ∩ {Tx2 }dd = {y1 }dd ∩ {y2 }dd = {0} , which yields Tu = 0. As T is injective it follows that u = 0. Thus, we obtain that {x1 }dd ∩ {x2 }dd = {0}. Since X is fordable, Lemma 5.2.20 (ii) yields that x1 ⊥ x2 . Consequently, T −1 is disjointness preserving. (ii) The line of reasoning here is similar to the proof of [3, Proposition 3.3]. Let x1 , x2 ∈ X be such that {x1 }dd ⊆ {x2 }dd , and assume that {Tx1 }dd ⊈ {Tx2 }dd . This means Tx1 ∈ ̸ {Tx2 }dd , i.e., there is y ∈ {Tx2 }d , y ≠ 0, so that Tx1 ⊥̸ y. In particular, one has y ⊥ Tx2 . Since T is bijective, there is an x ∈ X, x ≠ 0, so that Tx = y. Since T −1 is disjointness preserving, one obtains x ⊥ x2 . On the other hand, since T is disjointness preserving, one gets x ⊥̸ x1 . This contradicts the assumption, since x ∈ {x2 }d ⊆ {x1 }d . Remark 5.2.22. A combination of the Theorems 5.2.16 and 5.2.21 yields that in finite dimensions every disjointness preserving bijection satisfies condition (β).

5.3 Disjointness preserving C0 -semigroups Disjointness preserving operators have been studied in the setting of C0 -semigroups on Banach lattices; see [18]. A major part of this theory depends on the existence of a modulus of order-bounded disjointness preserving operators. In view of Example 5.1.6, a similar theory for ordered Banach spaces is expected to be much more involved. Nevertheless, we will choose a specific result that can be extended to ordered Banach spaces with the help of methods developed in the previous chapters. We define strongly continuous operator semigroups on Banach spaces and list some basic properties; for an overview see, e.g., [38, 51]. Let (X, ‖⋅‖) be a real Banach space. A C0 -semigroup on X is a map T : [0, ∞) → L(X) that satisfies the following conditions: (i) T(0) = I, (ii) for every s, t ∈ [0, ∞) one has T(s + t) = T(s)T(t), (iii) for every x ∈ X one has lim ‖T(t)x − x‖ = 0. t↓0

5.3 Disjointness preserving C0 -semigroups

| 257

The conditions (i) and (ii) are referred to as the semigroup property, whereas (iii) is the strong continuity of T at 0. For a given C0 -semigroup T, let D(A) := {x ∈ X; lim t↓0

T(t)x−x t

exists} ,

and define A : D(A) → X ,

x 󳨃→ lim t↓0

T(t)x−x t

.

Then D(A) is a norm dense linear subspace of X and A is a linear closed⁴ operator, which is called the generator of T. For every t ∈ [0, ∞), the operator T(t) leaves D(A) invariant, and for every x ∈ D(A) one has AT(t)x = T(t)Ax. If A ∈ L(X), then ∞ k k t A T : t 󳨃→ etA := ∑ k! k=0 is a C0 -semigroup with generator A, and T is even uniformly continuous, i.e., lim ‖T(t) − I‖ = 0 . t↓0

Conversely, every uniformly continuous C0 -semigroup has a bounded generator. We will discuss C0 -semigroups on Banach spaces that are, in addition, pre-Riesz spaces. Recall that by an ordered Banach space (X, K, ‖⋅‖) we mean a Banach space (X, ‖⋅‖) with a closed generating cone K. Note that an ordered Banach space is a preRiesz space. Indeed, as K is closed, X is Archimedean due to Proposition 1.5.2, and as X is directed, we get that X is a pre-Riesz space by Proposition 2.2.3. On Banach lattices, there is a rich theory on C0 -semigroups, for instance on positive or disjointness preserving C0 -semigroups; see, e.g., [123]. A part of the theory of positive C0 -semigroups has been developed on ordered Banach spaces; see [22, 38]. Here we extend some results on disjointness preserving C0 -semigroups; see also [86]. For a given ordered Banach space (X, K, ‖⋅‖), a C0 -semigroup T : [0, ∞) → L(X) is called disjointness preserving if for every t ∈ [0, ∞) the operator T(t) is disjointness preserving. Disjointness preserving C0 -semigroups on Banach lattices are discussed, e.g., in [18, Section 5], where, in particular, it is shown that then the generator is local. We prove the analogous result for disjointness preserving C0 -semigroups on ordered Banach spaces with a semimonotone norm. Note that in this case the norm is equivalent to a regular norm; see Proposition 3.5.2 (i). In the next lemma we consider the extension of a norm in X similarly to Theorem 3.4.4 and use it to detect disjointness.

4 Let X be a Banach space, D a linear subspace of X and A : D → X a linear operator. A is closed if for every sequence (x n ) in D which converges to x ∈ X such that (Ax n ) converges to y ∈ X it follows that x ∈ D and Ax = y.

258 | 5 Operators on pre-Riesz spaces Lemma 5.3.1. Let (X, K, ‖⋅‖) be a directed ordered normed space such that K is closed. Let Y be a vector lattice and i : X → Y a bipositive linear map, such that i[X] is majorizing in Y. For y ∈ Y let ρ(y) := inf {‖x‖; x ∈ X, |y| ≤ i(x)} . If u, v ∈ X are such that ρ(|i(u)| ∧ |i(v)|) = 0, then u ⊥ v. Proof. Let u, v ∈ X be such that ρ(|i(u)| ∧ |i(v)|) = 0. As |i(u)| ∧ |i(v)| =

1 2

| |i(u) + i(v)| − |i(u) − i(v)| |

one obtains that ρ(|i(u) + i(v)| − |i(u) − i(v)|) = 0. Hence, for every n ∈ ℕ there exists an x n ∈ X such that − i(x n ) ≤ |i(u) + i(v)| − |i(u) − i(v)| ≤ i(x n )

(5.13)

and limn→∞ ‖x n ‖ = 0. The first inequality in (5.13) yields ±(i(u) − i(v)) − i(x n ) ≤ |i(u) + i(v)| , hence for x ≥ ±(u + v) it follows that x ≥ ±(u − v) − x n for every n ∈ ℕ. Since K is closed and x n → 0, one obtains x ≥ ±(u − v). We conclude {u + v, −u − v}u ⊆ {u − v, −u + v}u .

(5.14)

From the second inequality in (5.13) it follows that ±(i(u) + i(v)) − i(x n ) ≤ |i(u) − i(v)| , and an analogous argumentation yields equality in (5.14), which implies that u and v are disjoint. Theorem 5.3.2. Let (X, K, ‖⋅‖) be an ordered Banach space with a semimonotone norm, and let T : [0, ∞) → L(X) be a disjointness preserving C0 -semigroup with generator A. Then A is local. Proof. Due to Proposition 3.5.2 (i) there is a regular norm ‖⋅‖0 on X, which is equivalent to ‖⋅‖. Let (X ρ , i) be the Riesz completion of X and define ρ : Xρ → ℝ ,

y 󳨃→ inf {‖x‖0 ; x ∈ X, |y| ≤ i(x)} .

(5.15)

According to Theorem 3.4.4, the map ρ is a Riesz seminorm on X ρ such that ρ(i(x)) = ‖x‖0 for every x ∈ X. Let x ∈ D(A) and y ∈ X be such that x ⊥ y. Then i(x) ⊥ i(y) in X ρ due to Proposition 4.1.4. Now the line of reasoning is as in the proof of [18, Proposition 5.4]. For every

5.3 Disjointness preserving C0 -semigroups |

t > 0 one has 󵄨󵄨 1 󵄨 󵄨󵄨 t i(T(t)x − x)󵄨󵄨󵄨 ∧ |i(y)| ≤ 󵄨 󵄨 = ≤

1 t 1 t 1 t

|i(T(t)x)| ∧ |i(y)| +

1 t

259

|i(x)| ∧ |i(y)|

|i(T(t)x)| ∧ |i(T(t)y − y) − i(T(t)y)| |i(T(t)x)| ∧ |i(T(t)y − y)|

≤ |i(T(t)y − y)| . Here we used that X ρ is distributive and T is disjointness preserving. We conclude 󵄨 󵄨 ρ (󵄨󵄨󵄨󵄨 1t i(T(t)x − x)󵄨󵄨󵄨󵄨 ∧ |i(y)|) ≤ ρ(|i(T(t)y − y)|) . For t ↓ 0 one has T(t)y − y → 0 in X, hence ρ(|i(T(t)y − y)|) → 0, which implies 󵄨 󵄨 ρ (󵄨󵄨󵄨󵄨 1t i(T(t)x − x)󵄨󵄨󵄨󵄨 ∧ |i(y)|) → 0 . Further, 󵄨 󵄨󵄨 󵄨 󵄨 󵄨󵄨ρ (|i(Ax)| ∧ |i(y)|) − ρ (󵄨󵄨󵄨 1t i(T(t)x − x)󵄨󵄨󵄨 ∧ |i(y)|)󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 ≤ ρ (󵄨󵄨󵄨󵄨|i(Ax)| ∧ |i(y)| − 󵄨󵄨󵄨󵄨 1t i(T(t)x − x)󵄨󵄨󵄨󵄨 ∧ |i(y)|󵄨󵄨󵄨󵄨) 󵄨󵄨 󵄨 󵄨 ≤ ρ (󵄨󵄨󵄨󵄨|i(Ax)| − 󵄨󵄨󵄨󵄨i ( 1t (T(t)x − x))󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨) 󵄨 󵄨 ≤ ρ (󵄨󵄨󵄨󵄨i (Ax − 1t (T(t)x − x))󵄨󵄨󵄨󵄨) = ‖Ax − 1t (T(t)x − x)‖0 → 0 for t ↓ 0. Therefore, ρ(|i(Ax)| ∧ |i(y)|) = 0. Now Lemma 5.3.1 implies that Ax ⊥ y, hence A is local. Notice that the converse of the statement in Theorem 5.3.2 is not true, in general, not even in Banach lattices. We illustrate this by an example from [51, 121]. Example 5.3.3. Let A be the second derivative operator given in Remark 5.1.5(1). We have already shown that A is local. The one-dimensional diffusion semigroup generated by A is given by 1

T(t)f(x) = ∫ K t (x, y)f(x)dy , 0

with kernel ∞

K t (x, y) = 1 + 2 ∑ exp(−π2 n2 t) cos(πnx) ⋅ cos(πny) ; n=1

see [121, 2.7] or [51, II.2.12]. There it is also shown that K t (⋅, ⋅) is a positive, continuous function on [0, 1]2 . Obviously, for t ∈ (0, ∞), the operator T(t) is not disjointness preserving on C[0, 1]. It is nevertheless interesting to investigate a converse of Theorem 5.3.2. We will show that a bounded local generator yields a disjointness preserving C0 -semigroup. For this, we need some preliminary statements.

260 | 5 Operators on pre-Riesz spaces

Lemma 5.3.4. Let X be a pre-Riesz space. (i) If S : X ⊇ D(S) → X and T : X ⊇ D(T) → X are local operators and α, β ∈ ℝ, then αS + βT : X ⊇ D(S) ∩ D(T) → X is a local operator. (ii) If S : X ⊇ D(S) → X and T : X ⊇ D(T) → D(S) ⊆ X are local operators, then ST : X ⊇ D(T) → X is a local operator. Proof. (i) Let x ∈ D(S) ∩ D(T) and y ∈ X be such that x ⊥ y. Then Sx ⊥ y and Tx ⊥ y, so that Sx, Tx ∈ {y}d . Since {y}d is a linear subspace by Proposition 4.1.5, it follows that αSx + βTx ⊥ y. Hence, αS + βT is local. (ii) Let x ∈ D(T), y ∈ X be such that x ⊥ y. Then Tx ⊥ y as T is local, so that STx ⊥ y as S is local and Tx ∈ D(S). Hence, ST is local. Proposition 5.3.5. Let (X, K, ‖⋅‖) be an ordered normed space such that K is closed and generating and the norm ‖⋅‖ is regular. Then every band in X is closed. Proof. Let (X ρ , i) be the Riesz completion of X. We define ρ as in (5.15). By Theorem 3.4.4, the map ρ is a Riesz seminorm on X ρ such that for every x ∈ X we have ρ(i(x)) = ‖x‖. Let B be a band in X, let (x n )n∈ℕ be a sequence in B and let x ∈ X be such that ‖x n − x‖ → 0. We show that x ∈ Bdd . Let y ∈ Bd . Then for every n ∈ ℕ we have x n ⊥ y, hence i(x n ) ⊥ i(y), which means |i(x n )| ∧ |i(y)| = 0. Since ρ(i(x n )− i(x)) = ‖x n − x‖ → 0, the continuity of the lattice operations with respect to ρ yields that ρ(|i(x)|∧|i(y)|) = 0. Hence, by Lemma 5.3.1, we obtain x ⊥ y. It follows that x ∈ Bdd = B. Hence B is closed. Corollary 5.3.6. Let (X, K, ‖⋅‖) be an ordered Banach space with a semimonotone norm. Then every band in X is closed. Proof. According to Proposition 3.5.2 (i) the norm ‖⋅‖ is equivalent to a regular norm. Then Proposition 5.3.5 yields the statement. For bounded generators, we arrive at a converse to Theorem 5.3.2. Theorem 5.3.7. Let (X, K, ‖⋅‖) be an ordered Banach space with a semimonotone norm. If A ∈ L(X) is local, then e tA is local for every t ∈ ℝ. Proof. Let x, y ∈ X be such that x ⊥ y. Let t ∈ ℝ. For every N ∈ ℕ, Lemma 5.3.4 yields tn n tn n that the operator ∑Nn=0 n! A is local, thus ∑Nn=0 n! A x ⊥ y. According to Lemma 5.3.6, n t n x ∈ {y}d . Hence, e tA is local. the band {y}d is closed, therefore e tA x = ∑∞ A n=0 n! Next we consider an unbounded generator A of a C0 -semigroup T. It turns out that T(t) is local for every t ∈ [0, ∞) whenever both A and the resolvent operators are local. Here we denote the resolvent set ρ(A) := {λ ∈ ℂ; (λI − A)−1 ∈ L(X)}, as usual.

5.4 Spaces of operators

|

261

Theorem 5.3.8. Let (X, K, ‖⋅‖) be an ordered Banach space with a semimonotone norm and let T : [0, ∞) → L(X) be a C0 -semigroup with generator A. If A : X ⊇ D(A) → X is local and there exists λ0 ∈ ρ(A) ∩ ℝ such that for every λ ∈ ρ(A) with λ ≥ λ0 we have that the resolvent operator (λI − A)−1 : X → D(A) ⊆ X is local, then T(t) is local for every t ∈ [0, ∞). Proof. By Lemma 5.3.4, the Yosida approximation A λ := λA(λI − A)−1 is local for every λ ∈ ρ(A) with λ ≥ λ0 . Let t ∈ [0, ∞). Since A λ is bounded, due to Theorem 5.3.7 we obtain that e tA λ is local. For x ∈ X, we have T(t)x = limλ→∞ e tA λ x; see [51, II.3.5]. We infer that T(t) is local. Indeed, if x, y ∈ X are such that x ⊥ y, then e tA λ x ⊥ y. Since the band {y}d is closed by Lemma 5.3.6, it follows that T(t)x ⊥ y. Thus, T(t) is local. Note that a typical example for Theorem 5.3.8 is a multiplication operator A. For a differential operator A, the resolvent operators are not local, in general, and the generated C0 -semigroup need not be disjointness preserving; see Example 5.3.3. A further discussion can be found in [86, Section 5].

5.4 Spaces of operators In this section we present several results on spaces of operators between pre-Riesz spaces. There is a rich theory on spaces of operators if the underlying spaces are vector lattices with the range space Dedekind complete. Dedekind completeness is a strong condition that excludes many interesting examples, as, e.g., spaces of continuous functions. Without Dedekind completeness, spaces of operators between vector lattices are not vector lattices, in general. Here the theory of pre-Riesz spaces appears naturally. It turns out that basic results on spaces of operators even hold when the underlying spaces are pre-Riesz spaces instead of vector lattices. For considerations on the space of continuous operators, we make use of the norms studied in Chapter 3. We also discuss a few results on spaces of operators from a pre-Riesz space to a Dedekind complete vector lattice. In particular, we study dual spaces. In the theory of vector lattices, there are interesting results on relations of spaces of operators. For instance, the space of order continuous operators is a band in the space of regular operators, provided the range space is Dedekind complete. The theory developed in Chapter 4 provides the terminology to extend such statements to Archimedean range spaces. Whether or not such statements are true in this general setting is an open field of research. We illustrate the potential use of pre-Riesz theory and embedding methods in Example 5.4.15 below.

262 | 5 Operators on pre-Riesz spaces

5.4.1 Operators between ordered normed spaces In this subsection we discuss some properties of spaces of operators between ordered normed spaces (X, K X ) and (Y, K Y ). The typical setting is that X is a directed partially ordered vector space with a regular norm and that Y is a partially ordered vector space with a symmetrically monotone norm. As in the Sections 1.2 and 1.5, we denote by L(X, Y) the vector space of all linear operators, by L+ (X, Y) the wedge of all positive operators, by Lr (X, Y) the vector space of all regular operators, by L(X, Y) the vector space of all continuous linear operators and by L+ (X, Y) the wedge of all continuous positive linear operators. We define⁵ Lr (X, Y) := L+ (X, Y) − L+ (X, Y). Recall that L+ (X, Y) is a cone provided the norm closure of K X −K X equals X; see Proposition 1.5.3. Moreover, if K Y is closed then L+ (X, Y) is closed. The next lemma concerns extension of a map defined on a cone to a continuous operator on the generated vector space. Lemma 5.4.1. Let (X, K X , ρ) and (Y, K Y , p) be ordered normed spaces such that X is directed, ρ is regular and p is symmetrically monotone. (i) Let A0 : K X → K Y be an additive positively homogeneous map such that α := sup{p(A0 x); x ∈ K X , ρ(x) ≤ 1} < ∞ . Then A0 extends uniquely to a continuous positive linear operator A : X → Y with ‖A‖ = α. (ii) Let A ∈ L(X, Y) and B ∈ L(X, Y) be such that −B ≤ A ≤ B. Then A is continuous and ‖A‖ ≤ ‖B‖. Proof. (i) Since X is directed, for every x ∈ X there exist x1 , x2 ∈ K X such that x = x1 − x2 . We define A : X → Y, x 󳨃→ A0 x1 − A0 x2 . It is routine to show that the definition of A is independent of the choice of x1 and x2 , that A is linear and positive, and that such an A is unique. To prove that A is continuous, let x ∈ X be such that ρ(x) < 1. As ρ is regular, there is a y ∈ K X with −y ≤ x ≤ y and ρ(y) < 1. By positivity of A we get −Ay ≤ Ax ≤ Ay. Since p is symmetrically monotone, this yields that p(Ax) ≤ p(Ay) = p(A 0 y) ≤ α. It follows that A is continuous and that ‖A‖ ≤ α. Since also ‖A‖ ≥ α, the proof is complete. (ii) Let x ∈ X be such that ρ(x) < 1. Since ρ is regular, there is a y ∈ K X with −y ≤ x ≤ y and ρ(y) < 1. Then 2Ax = A(y + x) + (−A)(y − x) ≤ B(y + x) + B(y − x) = 2By, so Ax ≤ By. Similarly, Ax ≥ −By. As p is symmetrically monotone, we obtain p(Ax) ≤ p(By) ≤ ‖B‖. Thus, A is continuous and ‖A‖ ≤ ‖B‖.

5 Note that the vector space L(X, Y) ∩ Lr (X, Y) might differ from Lr (X, Y); see Example 5.4.12.

5.4 Spaces of operators

| 263

It is a well-known and important result in the theory of Banach lattices that every positive operator between two Banach lattices is continuous. There is a more general result for operators between ordered normed spaces, as we see next. Proposition 5.4.2. Let (X, K X , ρ) and (Y, K Y , p) be ordered normed spaces such that X is directed and ρ-complete, K X is closed and ρ is regular. (i) If A ∈ L(X, Y) maps order-bounded sets to norm-bounded sets, then A is continuous. (ii) If order-bounded sets in Y are norm-bounded, then every order-bounded linear operator from X to Y is continuous. Proof. For a proof of (i), suppose that A is not continuous. Then there is a sequence (x n )n∈ℕ in X such that ρ(x n ) < 2−n for every n ∈ ℕ and p(Ax n ) → ∞. The norm ρ is regular, so for every n ∈ ℕ there is a y n ∈ K X with −y n ≤ x n ≤ y n and ρ(y n ) < 2−n . Since X is norm complete, the sum y := ∑∞ n=1 y n exists in X, and since K X is closed we have y ≥ y n for every n ∈ ℕ. Then −y ≤ x n ≤ y for every n ∈ ℕ, which contradicts the assumption that A maps order-bounded sets to norm-bounded sets. Thus, A is continuous. The assertion in (ii) follows directly from (i). Recall that an ordered Banach space is a Banach space with a closed and generating cone. Corollary 5.4.3. Let (X, K X , p X ) be an ordered Banach space with a semimonotone norm and let (Y, K Y , p Y ) be an ordered normed space with a semimonotone norm. Then every positive operator from X to Y is continuous, and Lr (X, Y) ⊆ L(X, Y). Proof. Due to Proposition 3.5.2 (i) the norm p X is equivalent to a regular norm ρ. Then X is ρ-complete and K X is ρ-closed. As p Y is semimonotone, order-bounded sets in Y are norm-bounded by Lemma 1.5.7. Furthermore, every positive operator from X to Y is order-bounded. Proposition 5.4.2 now yields the desired result. For more results related to Corollary 5.4.3, see also [162, Section XI.2]. The following is the main theorem in this subsection. Most of the results can be found in the literature, sometimes considered from another point of view; see, e.g., [12, 134, 162]. The standard notation of the regularization of the operator norm according to Definition 3.4.10 is ‖⋅‖r . Theorem 5.4.4. Let (X, K X , ρ) and (Y, K Y , p) be ordered normed spaces such that X is directed, ρ is regular and p is symmetrically monotone. (i) For every A ∈ L+ (X, Y) the operator norm of A satisfies ‖A‖ = sup{p(Ax); x ∈ K X , ρ(x) ≤ 1} . (ii) The set L(X, Y) is a full subspace of L(X, Y). (iii) The set Lr (X, Y) is a directed ideal in Lr (X, Y).

264 | 5 Operators on pre-Riesz spaces (iv) The operator norm on L(X, Y) is symmetrically monotone, and ‖⋅‖r is a regular norm on Lr (X, Y). (v) If K Y is p-complete, then (Lr (X, Y), ‖⋅‖r ) is norm complete. Proof. (i) Let A0 be the restriction of A to K X , then apply Lemma 5.4.1 (i). (ii) Let B, C ∈ L(X, Y) and A ∈ L(X, Y) be such that B ≤ A ≤ C. Then 0 ≤ A − B ≤ C − B, so it follows from Lemma 5.4.1 (ii) that A−B is continuous. Hence, A is continuous. Therefore, L(X, Y) is a full subspace of L(X, Y). (iii) First we show that Lr (X, Y) is a full subspace of Lr (X, Y). Indeed, let B, C ∈ Lr (X, Y) and A ∈ L(X, Y) be such that B ≤ A ≤ C. Then A is continuous and there is D ∈ L(X, Y)+ such that −D ≤ B ≤ A ≤ C ≤ D. Hence, A = D − (D − A) ∈ Lr (X, Y). As the space Lr (X, Y) is directed, from Theorem 4.3.22 it follows that Lr (X, Y) is an ideal. (iv) The statements follow from Lemma 5.4.1 (ii) and Proposition 3.4.11. (v) We first show that L+ (X, Y) is norm complete with respect to the operator norm. Let (A n )n be a Cauchy sequence in L+ (X, Y). Let x ∈ K X . Then (A n x)n is a Cauchy sequence in K Y , hence A0 x := limn A n x exists in K Y . It is straightforward that A0 is additive and positively homogeneous. If ρ(x) < 1, then p(Ax) = limn p(A n x) ≤ supn ‖A n ‖. According to Lemma 5.4.1 (i), it follows that the map A0 extends to a continuous positive linear map A : X → Y. To show that A n → A in L(X, Y), let ε > 0. Take N ∈ ℕ such that ‖A n − A m ‖ < ε for every m, n ∈ ℕ≥N . Then, for every x ∈ X, we have p((A n − A)x) = limm p((A n − A m )x) ≤ limm ‖A n − A m ‖ρ(x) ≤ ερ(x), therefore ‖A n − A‖ < ε. Since the operator norm is monotone, Theorem 3.4.12 yields that Lr (X, Y) is norm complete with respect to ‖⋅‖r . Note that normed Riesz spaces satisfy the conditions of Theorem 5.4.4. Also order unit spaces are contained in this setting. Corollary 5.4.5. Let (X, K, ‖⋅‖u ) be an order unit space. For every T ∈ L+ (X) one has ‖T‖ = ‖Tu‖u . Proof. Since ‖u‖u = 1, we get ‖Tu‖u ≤ ‖T‖. On the other hand, as ‖⋅‖u is regular by Proposition 3.4.2 (ii), Theorem 5.4.4 implies that ‖T‖ = sup{‖Tx‖u ; x ∈ K, ‖x‖u ≤ 1}. Choose x ∈ K with ‖x‖u ≤ 1, then 0 ≤ x ≤ u. Since T ∈ L+ (X) and ‖⋅‖u is monotone, we get ‖Tx‖u ≤ ‖Tu‖u and hence ‖T‖ ≤ ‖Tu‖u .

5.4.2 Operators with Dedekind complete range spaces We intend to provide conditions on ordered normed spaces X and Y such that L(X, Y) is directed. For this, we collect some classical results for operators with Dedekind com-

5.4 Spaces of operators

| 265

plete range space. We need the following notions. For a vector space X, an ordered vector space Y and M ⊆ X, a map A : M → Y is called subadditive if for every x, y ∈ M with x + y ∈ M we have A(x + y) ≤ A(x) + A(y), and superadditive if for every x, y ∈ M with x + y ∈ M we have A(x + y) ≥ A(x) + A(y). Lemma 5.4.6. Let X be a vector space, Y an ordered vector space and let A : X → Y be a subadditive, positively homogeneous map. If A(−x) = −A(x) for every x ∈ X, then A is linear. Proof. Let x, y ∈ X. Then A(x + y) ≤ A(x) + A(y) and A(x + y) = −A(−x − y) ≥ −(A(−x) + A(−y)) = Ax + Ay, so A(x + y) = A(x) + A(y). It is clear that the assumptions imply homogeneity. Hence A is linear. The subsequent statement is an extended version of the Hahn–Banach theorem 1.8.1 and can be found, e.g., in [12, Theorem 2.1]. Theorem 5.4.7. Let V be a vector space, Y a Dedekind complete Riesz space, and P : V → Y a subadditive positively homogeneous map. Let D be a linear subspace of V and S : D → Y a linear operator such that S ≤ P on D. Then there exists a linear operator T : V → Y such that T = S on D, and T ≤ P on V. The next result is due to Bonsall [29]. Theorem 5.4.8. Let (X, K) be a directed partially ordered vector space, let Y be a Dedekind complete Riesz space, and let P : X → Y be a subadditive, positively homogeneous map. If Q : K → Y is a superadditive, positively homogeneous map such that Q ≤ P on K, then there is a linear map B : X → Y such that B ≥ Q on K and B ≤ P on X. Proof. The map B will be found with aid of Zorn’s lemma as a minimal element of the set V := {A : X → Y; A is subadditive, positively homogeneous, A ≤ P on X and A ≥ Q on K} . The first step is to show that there is a minimal element B in V. As P ∈ V, we have that V is not empty. Let C be a chain in V and let x ∈ X. For A ∈ V one has that 0 ≤ A(x)+A(−x), thus A(x) ≥ −A(−x) ≥ −P(−x). The space F is Dedekind complete, hence the infimum A 0 (x) := inf{A(x); A ∈ C} exists. Then A0 has the following properties. (i) A0 is subadditive. Indeed, let x, y ∈ X. If A1 , A2 ∈ C, then one of the two, say A1 , is the smallest. Therefore, A1 (x) + A2 (y) ≥ A1 (x) + A1 (y) ≥ A1 (x + y) ≥ A0 (x + y), hence A0 (x) + A0 (y) ≥ A0 (x + y). (ii) A0 is positively homogeneous. (iii) A0 ≤ P on X, since A(x) ≤ P(x) for every A ∈ V and x ∈ X. (iv) A 0 (x) ≥ Q(x) for every x ∈ K. (v) A0 is a lower bound of C. According to Zorn’s lemma, it follows that there is a minimal element B in V.

266 | 5 Operators on pre-Riesz spaces

The second step is to show that B is linear. We intend to apply Lemma 5.4.6 and show that for every x ∈ X we have that B(−x) = −B(x). Our argument involves two auxiliary positively homogeneous subadditive maps, one based on Q and one based on B. We will construct them at the same time by considering an arbitrary map C : K → Y with C ≤ B on K, and then apply the result below to C := Q|K and C := B|K , respectively. Let a ∈ K and x ∈ X. Then the function λ 󳨃→ B(x + λa) − λC(a) from [0, ∞) to Y is bounded below. Indeed, we have B(λa) = B(x + λa − x) ≤ B(x + λa) + B(−x). Therefore, B(x + λa) − λC(a) ≥ B(λa) − B(−x) − λC(a) ≥ −B(−x). As Y is Dedekind complete, we can define C a (x) := inf {B(x + λa) − λC(a); λ ∈ [0, ∞)} . Then for every a ∈ K the map C a : X → Y has the following properties. (i) C a is subadditive. Indeed, let x, y ∈ X and let λ1 , λ2 ∈ [0, ∞). Then B(x + λ1 a) − λ1 C(a) + B(y + λ2 a) − λ2 C(a) ≥ B(x + y + (λ1 + λ2 )a) − (λ1 + λ2 )C(a) ≥ C a (x + y), thus C a (x) + C a (y) ≥ C a (x + y). (ii) C a is positively homogeneous. Indeed, let μ ∈ (0, ∞). We have C a (μx) = inf {μ (B (x + μλ a) − μλ C(a)) ; λ ∈ [0, ∞)} = μC a (x) . Moreover, C a (0) = 0, since B ≥ C on K and B(0) = 0. (iii) C a ≤ B ≤ P, because (with λ = 0 in the definition of C a ) C a (x) ≤ B(x+0)−0 = B(x). Now we proceed with the proof of linearity of B. Let a ∈ K and apply the construction of C a with C := Q|K . Then for every x ∈ K one has that C a (x) = inf {B(x + λa) − λQ(a); λ ∈ [0, ∞)} ≥ inf {Q(x + λa) − λQ(a); λ ∈ [0, ∞)} ≥ Q(x) , by superadditivity of Q, so C a ∈ V. Since B is minimal in V and C a ≤ B, we obtain C a = B. Hence, for every x ∈ X we have B(x) = C a (x) = inf {B(x + λa) − λQ(a); λ ∈ [0, ∞)} ≤ B(x + a) − Q(a) , thus B(x + a) ≥ B(x) + Q(a). Next, let a ∈ K and apply the construction of C a with C := B|K . Then for every x ∈ K and λ ∈ [0, ∞) it follows from the previous step that B(λa + x) ≥ B(λa) + Q(x). Consequently, C a (x) ≥ inf {B(λa) + Q(x) − λB(a); λ ∈ [0, ∞)} = Q(x) , therefore C a ∈ V. Due to (iii) above, we have C a ≤ B. As B is minimal in V, we obtain C a = B. Hence, for every x ∈ X we get B(x) = C a (x) = inf{B(x + λa) − λB(a); λ ∈ [0, ∞)} ≤ B(x + a) − B(a) .

5.4 Spaces of operators

| 267

Therefore, B(x + a) ≥ B(x) + B(a). By the subadditivity of B it follows that B(x + a) = B(x) + B(a)

for every x ∈ X, a ∈ K

and, in particular, with x = −a we obtain B(−a) = −B(a)

for every a ∈ K .

Let x ∈ X. As X is directed, there are x 1 , x2 ∈ K such that x = x1 − x2 . Then, by the previous identities, we get B(−x + x1 ) = B(−x) + B(x1 ), hence B(−x) = B(−x + x1 ) − B(x1 ) = B(x2 ) − B(x1 ) = −B(−x2 ) − B(x1 ) = −B(−x2 + x1 ) = −B(x) . Together with the subadditivity and the positive homogeneity of B, this yields that B is linear, according to Lemma 5.4.6. Since B is an element of V, it has the desired properties. Corollary 5.4.9. Let (X, K X ) be a directed partially ordered vector space, let Y be a Dedekind complete Riesz space, and let P : X → Y be a subadditive, positively homogeneous map such that P(x) ≤ P(y) for every x, y ∈ X with −y ≤ x ≤ y. If A : X → Y is a linear map with A ≤ P on X, then there exists a linear map B : X → Y such that −B ≤ A ≤ B on K X and B ≤ P on X. Proof. Define the map Q : K X → Y, x 󳨃→ sup{Au; −x ≤ u ≤ x}. Then Q is superadditive and positively homogeneous, and Q ≤ P on K X . By Theorem 5.4.8, there is a linear map B : X → Y with B ≤ P on X and B ≥ Q on K X , hence B ≥ A, −A on K X . With the aid of Corollary 5.4.9 we next show that the wedge L+ (X, Y) is generating in L(X, Y), where we additionally assume that Y has an order unit and is endowed with the order unit norm. Corollary 5.4.10. Let (X, K X ) be a partially ordered vector space with a symmetrically monotone seminorm p and let Y be a Dedekind complete Riesz space with an order unit u, equipped with the order unit norm. (i) Let A : X → Y be a continuous linear operator with ‖A‖ ≤ 1. Then there is a positive continuous linear operator B : X → Y such that −Bx ≤ Ax ≤ Bx for every x ∈ K X and ‖B‖ ≤ 1. (ii) The wedge L+ (X, Y) is generating in L(X, Y). Proof. (i) Define P : X → Y, x 󳨃→ p(x)u. Then P : X → Y is subadditive, positively homogeneous, and P is increasing on K X . For every x ∈ X we have ‖P(x)‖u ≤ p(x)u and P(−x) = −P(x). Since ‖A‖ ≤ 1, we have A ≤ P on K X . Let X0 := K X − K X be the directed part of X. Now it follows from Corollary 5.4.9 that there is a positive continuous linear operator B0 : X0 → Y such that −B0 ≤ A ≤ B0 on K X and B0 ≤ P on X0 . Then ‖B0 ‖ ≤ 1. Due to the generalized Hahn–Banach theorem 5.4.7, the operator B0 extends to a linear operator B : X → Y with B ≤ P on X. Then ‖B‖ ≤ 1.

268 | 5 Operators on pre-Riesz spaces (ii) For A ∈ L(X, Y) as in (i), the corresponding operator B is positive, B − A is positive, and A = B − (B − A). If, in addition to the assumptions in Corollary 5.4.10, the space X is the norm closure of K X − K X , then by Proposition 1.5.3 we obtain that L(X, Y) is an ordered vector space. Now Corollary 5.4.10 (ii) says that L(X, Y) is directed. We recover a classical result of Grosberg and Krein; see [64]. Corollary 5.4.11. Let (X, K X ) be a directed partially ordered vector space with a symmetrically monotone seminorm. Then for every φ ∈ X 󸀠 there exist φ1 , φ2 ∈ K X 󸀠 such that φ = φ1 − φ2 , ‖φ1 ‖ ≤ ‖φ‖ and ‖φ2 ‖ ≤ ‖φ‖. In particular, X 󸀠 is directed. We give an example where the norm dual is not directed, which implies that the norm of the underlying space is not equivalent to a monotone one. The example provides a continuous functional that is regular, but not the difference of two positive continuous functionals. Example 5.4.12. We consider the Riesz space X = cc (ℤ≥0 ) with the norm p: X → ℝ ,

∞

x 󳨃→ ∑ |x n − x n+1 | . n=0

The algebraic dual is isomorphic to the space ℝℤ≥0 of all sequences with the natural ℤ≥0 . It is clear that positive isomorphism a 󳨃→ (x 󳨃→ ∑∞ n=0 a n x n ), where x ∈ X and a ∈ ℝ sequences correspond to positive functionals. Let f be the functional on X corresponding to the sequence (−1, 1, −1, 1, . . .). The functional f is continuous, since for x ∈ X one has 󵄨󵄨 󵄨󵄨 ∞ 󵄨 󵄨 |f(x)| = 󵄨󵄨󵄨󵄨 ∑ (x2n+1 − x2n )󵄨󵄨󵄨󵄨 ≤ p(x) . 󵄨󵄨 󵄨󵄨n=0 Note that every positive continuous functional g corresponds to a summable sequence b. Indeed, for every N ∈ ℕ we have N

N

N

N

n=0

n=0

n=0

n=0

∑ b n = ∑ g(e(n) ) = g( ∑ e(n) ) ≤ ‖g‖ p( ∑ e(n) ) = ‖g‖ , e(n)

denotes the nth unit sequence. The functional f is not the difference of where two positive continuous functionals, since the sequence (−1, 1, −1, 1, . . .) is not the difference of two positive summable sequences. Clearly, f is regular. We list further properties of spaces of continuous operators. Recall that by Proposition 3.4.2 order unit norms are regular and, hence, symmetrically monotone. Proposition 5.4.13. Let (X, K X , ρ) and (Y, K Y , p) be ordered normed spaces such that X is directed, ρ is regular, and p is symmetrically monotone. (i) If Y is a Dedekind complete Riesz space and X has the Riesz decomposition property, then (Lr (X, Y), ‖⋅‖r ) is a Dedekind complete normed Riesz space. If, in addition, K Y is norm complete, then (Lr (X, Y), ‖⋅‖r ) is a Banach lattice.

5.4 Spaces of operators

| 269

(ii) If Y is a Dedekind complete Riesz space with an order unit and the order unit norm, then L(X, Y) is directed and the operator norm is regular. If, in addition, X has the Riesz decomposition property, then L(X, Y) is a Dedekind complete normed Riesz space. Proof. (i) By Theorem 1.2.8, the space Lr (X, Y) is a Dedekind complete Riesz space. According to Theorem 5.4.4 (iii), Lr (X, Y) is a directed ideal in Lr (X, Y). In a Dedekind complete Riesz space, every ideal is Dedekind complete; see [154, Theorem IV.1.2]. Hence, Lr (X, Y) is a Dedekind complete Riesz subspace of Lr (X, Y). By Proposition 3.4.2, the norm ‖⋅‖r is a Riesz norm. Now Theorem 5.4.4 (v) implies that (Lr (X, Y), ‖⋅‖r ) is a Banach lattice. (ii) By Corollary 5.4.10, the space L(X, Y) is directed. Moreover, for A ∈ L(X, Y) with ‖A‖ = 1 it follows that ‖A‖r ≤ 1. Hence, ‖⋅‖r ≤ ‖⋅‖. By Theorem 5.4.4 (iv), the operator norm ‖⋅‖ is symmetrically monotone, hence Proposition 3.4.11 (ii) yields that ‖⋅‖r ≥ ‖⋅‖. Therefore, ‖⋅‖r = ‖⋅‖, which means that ‖⋅‖ is regular. If X has the Riesz decomposition property, then, by (i), the space Lr (X, Y) is a Dedekind complete Riesz space. According to Proposition 3.4.2 (i), it follows that the operator norm is a Riesz norm, thus L(X, Y) = Lr (X, Y) is a Dedekind complete normed Riesz space.

5.4.3 The space of order continuous operators: an example For vector lattices X and Y, in the literature subspaces of Lr (X, Y) are studied that consist of operators with certain additional properties. So far, mostly the setting is considered where Y is Dedekind complete, since then Lr (X, Y) is a vector lattice by Theorem 1.2.8. For ordered vector spaces X and Y, where X is directed and Y is Archimedean, the space Lr (X, Y) is an Archimedean directed ordered vector space and, hence, a pre-Riesz space. In particular, Lr (X, Y) is a pre-Riesz space if X and Y are vector lattices with Y Archimedean. It is a challenging question which results for Dedekind complete Y are also true in this general setting. We choose a result by Ogasawara [12, Theorem 4.4] to illustrate the approach and show how the notions and results developed in previous chapters are of use. For ordered vector spaces X and Y let Loc (X, Y) be the vector space of all linear order continuous operators. Ogasawara states that for vector lattices X and Y with Y Dedekind complete Loc (X, Y) is a band in Lr (X, Y). The following slightly more general version is established in [66, Theorem 7.8]. Theorem 5.4.14. Let X be a pre-Riesz space with the Riesz decomposition property and Y a Dedekind complete Riesz space. Then Loc (X, Y) is a band in Lr (X, Y).

270 | 5 Operators on pre-Riesz spaces We will present one particular vector lattice X such that Lr (X) is not a vector lattice and it is nevertheless true that Loc (X) is a band in Lr (X), where Definition 4.2.1 is used. Consider the space X := ℓ∞ 0 of all real sequences that are constant except for a finite number of terms. The space ℓ∞ 0 is an Archimedean vector lattice, but not Dedekind b ∞ r ∞ complete. In [7, Theorem 4.1] it is established that Lr (ℓ∞ 0 ) = L (ℓ0 ). The space L (ℓ0 ) ∞ r does not satisfy the Riesz decomposition property [7, Theorem 5.1], hence L (ℓ0 ) is not a vector lattice. Since every order continuous operator is automatically order-bounded r ∞ oc ∞ (see [6, Theorem 2.1]), the space Loc (ℓ∞ 0 ) is a subspace of L (ℓ0 ). We show that L (ℓ0 ) ∞ ∞ r r is a band in L (ℓ0 ). The proof is based on the embedding technique. As L (ℓ0 ) is a preRiesz space, we calculate its Riesz completion, show that Lr (ℓ∞ 0 ) is pervasive and apply the restriction property (R) for bands given in Corollary 4.2.7. Let us fix some notations. Denote ℕ0 := ℤ≥0 and cc := cc (ℕ0 ), where the latter is the space of all real sequences which are zero except for a finite number of terms. Let X be a vector space with a countable algebraic basis B = (b (i) )i∈ℕ0 , i.e., for every x ∈ X there is a unique sequence ξ = (ξ i )i∈ℕ0 ∈ cc such that ∞

x = ∑ ξ i b (i) . i=0

For i ∈ ℕ0 define the coordinate functional f (i) : X → ℝ by f (i) (x) = ξ i , thus one can write ∞

x = ∑ f (i) (x)b (i) . i=0

Let A : X → X be a linear operator. We denote the matrix representation of A with respect to B by A,̂ i.e., Â = (a ij )i,j∈ℕ0

with

a ij = f (i) (Ab (j) ) .

(5.16)

For every j ∈ ℕ0 one has (a ij )i∈ℕ0 ∈ cc . Conversely, every matrix (a ij )i,j∈ℕ0 with (a ij )i∈ℕ0 ∈ cc for every j ∈ ℕ0 corresponds to a linear operator A : X → X. We view  as a linear operator on cc , where  : cc → cc . Example 5.4.15. We consider the space of all eventually constant sequences X := ℓ∞ 0 := {(x i )i∈ℕ ; ∃β ∈ ℝ, k ∈ ℕ ∀i > k : x i = β} equipped with the standard cone X+ of the coordinatewise order. Then (X, X+ ) is an Archimedean vector lattice. For every j ∈ ℕ denote e(j) = (x i )i∈ℕ

with

x j = 1 and x i = 0 for every i ≠ j .

Moreover, denote 𝟙 = (1)i∈ℕ . The set B = {𝟙, e(1) , e(2) , . . .}

5.4 Spaces of operators

| 271

is an algebraic basis of X, where for x = (x i )i∈ℕ ∈ X one has the coordinate functionals f (0) (x) = lim x i i→∞

and

(j)

f (x) = x j − lim x i for every j ≥ 1 . i→∞

(5.17)

For a linear operator A : X → X we have the matrix representation  : cc → cc according to (5.16). Next we address the issue of the appropriate order in cc . Define K = {ξ = (ξ i )i∈ℕ0 ∈ cc ; ∀i ∈ ℕ : ξ0 ≥ 0, ξ i + ξ0 ≥ 0} . Then one has for every x ∈ X that x ∈ X+ if and only if (f (i) (x))i∈ℕ0 ∈ K . So, A is positive if and only if  is positive in the space (cc , K). We characterize the positivity of A.̂ (a)  = (a ij )i,j∈ℕ0 is positive in (cc , K) if and only if for every i ∈ ℕ the both conditions (i) a0j + a ij ≥ 0 for every j ∈ ℕ, and (ii) a00 + a i0 ≥ ∑∞ j=1 (a0j + a ij ) are satisfied. For a proof, assume first that  is positive in (cc , K). Denote ϵ(j) := (ϵ i )i∈ℕ0 with ϵ j = 1 ̂ (j) ∈ K for every j ∈ ℕ0 , which and ϵ i = 0 for every i ≠ j. Since ϵ(j) ∈ K, we have Aϵ means that (i) holds and that (iii) a0j ≥ 0 for every j ∈ ℕ0 . Further, let N ∈ ℕ and consider ξ = (ξ i )i∈ℕ0 ∈ K defined by ξ0 := 1, ξ i := −1 for ̂ ∈ K we obtain a00 ≥ ∑N a0j and 1 ≤ i ≤ N, and ξ i := 0 for i > N. As Aξ j=1 N

a00 + a i0 ≥ ∑ (a0j + a ij )

for every i ∈ ℕ .

j=1

We infer (ii) and (iv) a00 ≥ ∑∞ j=1 a0j . Conversely, assume that (i) and (ii) hold and let ξ ∈ K. For every i ∈ ℕ we have ∞

̂ )0 = (a i0 + a00 )ξ0 + ∑ (a ij + a0j )ξ j ̂ )i + (Aξ (Aξ j=1 ∞

≥ (a i0 + a00 )ξ0 − ∑ (a ij + a0j )ξ0 ≥ 0 . j=1

Since the columns of  are eventually zero, we see that (iii) and (iv) follow from (i) and (ii). Hence, ∞

∞

j=1

j=1

̂ )0 = a00 ξ0 + ∑ a0j ξ j ≥ a00 ξ0 − ∑ a0j ξ0 ≥ 0 . (Aξ Thus, Â is positive.

272 | 5 Operators on pre-Riesz spaces (b)  = (a ij )i,j∈ℕ0 is a regular operator in (cc , K) if and only if the sequence of absolute row sums (∑∞ j=1 |a ij |)i∈ℕ0 is bounded. Indeed, assume that  is regular. Then there exist positive operators G and H on (cc , K) with G = (g ij )i,j∈ℕ0 and H = (h ij )i,j∈ℕ0 such that  = G − H. Define B = (b ij )i,j∈ℕ0 by B = G + H. Then B is positive and B −  = 2H is positive, so that for every i ∈ ℕ, b 0j ≥ 0 ,

b 0j + b ij ≥ 0 for every j ∈ ℕ ,

∞

b 00 ≥ ∑ b 0j ,

∞

b 00 + b i0 ≥ ∑ (b 0j + b ij ) ,

j=1

b 0j − a0j ≥ 0 ,

j=1

b 0j − a0j + b ij − a ij ≥ 0 for every j ∈ ℕ ,

and

∞

b 00 − a00 ≥ ∑ (b 0j − a0j ) . j=1

Hence, a ij ≤ b 0j + b ij − a0j ≤ (b 0j + b ij ) + (b 0j − a0j )

for every i, j ∈ ℕ .

Both terms at the right-hand side are positive, so taking positive parts and summing over j yields ∞

∞

j=1

j=1

∑ a+ij ≤ ∑ ((b ij + b 0j ) + (b 0j − a0j )) ≤ b 00 + b i0 + b 00 − a00

for every i ∈ ℕ .

Similarly, if we consider − instead of  we obtain b 0j + a0j ≥ 0 for every j ∈ ℕ0 and ∞

∑ (−a ij )+ ≤ b 00 + b i0 + b 00 + a00

for every i ∈ ℕ .

j=1

Since (b i0 )i∈ℕ0 ∈ cc , the maximum b = max{b i0 ; i ∈ ℕ0 } exists, and ∑∞ j=1 |a ij | ≤ 4b 00 + 2b for every i ∈ ℕ. It also follows that |a0j | ≤ b 0j for every j ∈ ℕ0 and hence ∞ ∑∞ j=1 |a0j | ≤ b 00 . Thus, (∑ j=1 |a ij |)i∈ℕ0 is bounded by 4b 00 + 2b. ∞ Next assume that (∑∞ j=1 |a ij |)i∈ℕ0 is bounded. Let b ≥ 4 ∑ j=1 |a ij | for every i ∈ ℕ0 . Define B = (b ij )i,j∈ℕ0 by b 00 = b + |a00 | and

b ij = |a ij | for i, j ∈ ℕ0 with (i, j) ≠ (0, 0) .

With the aid of (a) it follows that B and B − Â are positive. Hence, Â = B − (B − A)̂ is regular. (c) The space Lr (cc , K) of regular operators in (cc , K) is not a lattice. The statement follows from the fact that (cc , K) and ℓ∞ 0 are order isomorphic and that Lr (ℓ∞ ) does not have the Riesz decomposition property. To be more explicit, we give 0

5.4 Spaces of operators

| 273

an operator A for which A+ does not exist. Consider A = (a ij )i,j∈ℕ0 with a ij = −1/i if i = j ≥ 1 and a ij = 0 otherwise. Due to (b), A is a regular operator on (cc , K). To see (n) that A ∨ 0 does not exist, define for n ∈ ℕ0 the operator B(n) = (b ij )i,j∈ℕ0 by 1

(n) b ij

{ { { n+1 = { 1i { { {0

if i = j = 0 , if j = 0 and 1 ≤ i ≤ n , otherwise .

It is easily checked that B(n) and B(n) − A are positive for every n ∈ ℕ0 . Further, let G = (g ij )i,j∈ℕ0 be an operator on (cc , K) such that 0 ≤ G ≤ B(n) for every n ∈ ℕ0 . If we show that G ≥ A cannot hold, then we know that there is no least upper bound of {0, A}. For i ∈ ℕ and n ∈ ℕ0 , the positivity of G and B(n) − G yields g00 + g i0 ≥ 0 (n) (n) and b 00 − g00 + b i0 − g i0 ≥ 0. Since the columns of B(n) and G are eventually zero, it (n) follows that 0 ≤ g00 ≤ b 00 for every n ∈ ℕ0 and therefore g00 = 0. If we now suppose that G ≥ A, then ∞

g00 − a00 + g i0 − a i0 ≥ ∑ (g0j − a0j + g ij − a ij )

for every i ∈ ℕ ,

j=1

so g i0 ≥ 1/i, which contradicts the fact that the columns of G are eventually zero. Thus, A ∨ 0 does not exist in Lr (cc , K). Recall that the space Lr (cc , K) consists of the matrix representations of the regular r operators on ℓ∞ 0 . Due to (b), we can describe the elements of L (cc , K) completely in terms of matrix entries, and get Lr (cc , K) = R := {(a ij )i,j∈ℕ0 ; ∀j ∈ ℕ0 : (a ij )i ∈ cc and ( ∑∞ j=1 |a ij |)i∈ℕ0 is bounded} . Note that the positive elements in R are the matrices satisfying (i) and (ii) of (a). It is our aim to provide an explicit vector lattice cover of R. Define the space Y as the space of all matrices (b ij )i∈ℕ,j∈ℕ0 that satisfy the following four conditions: (b ij )i∈ℕ is eventually constant for every j ≥ 1 ,

(5.18)

∞

∑ |β j | < ∞, where β j = lim b ij , i→∞

j=1

(b i0 )i∈ℕ is bounded ,

(5.19) (5.20)

∞

( ∑ |b ij |) j=1

i∈ℕ

is bounded .

(5.21)

We endow Y with the entrywise order. We define a map F on R by F(A) := (f ij (A)) i∈ℕ,j∈ℕ0 for A = (a ij )i,j∈ℕ0 ∈ R, where { a0j + a ij f ij (A) := { a + a i0 − ∑∞ p=1 (a0p + a ip ) { 00

for i ∈ ℕ, j ≥ 1 , for i ∈ ℕ, j = 0 .

274 | 5 Operators on pre-Riesz spaces (d) The space Y is a vector lattice and F : R → Y is a bipositive linear map. It is straightforward that Y is a vector lattice. We next show that F maps into Y. Let A ∈ R and put b ij := f ij (A) for i ∈ ℕ and j ∈ ℕ0 . For j ≥ 1 we have b ij = a0j for i ∞ large, so (5.18) holds. Further, β j := limi→∞ b ij = a0j and ∑∞ j=1 |β j | = ∑ j=1 |a0j | < ∞, ∞ so we have (5.19). We obtain (5.20) and (5.21) by using that (∑j=1 |a ij |)i is bounded and (a i0 )i ∈ cc . It is clear that the map F is linear and that F(A) is positive if and only if A is a positive element of R. (e) The subspace F[R] is order dense in Y. Indeed, let B = (b ij )i∈ℕ,j∈ℕ0 ∈ Y. We construct a sequence (A N )N∈ℕ in R such that for each N ∈ ℕ we have F(A N )ij = b ij F(A N )ij ≥ b ij

for i ∈ ℕ, j ∈ ℕ, and for i = 1, . . . , N, j = 0 , for j = 0, i > N .

Then it follows that B = inf{F(A); A ∈ R, F(A) ≥ B}. Fix N ∈ ℕ. The construction of A N is as follows. Denote β j := limi→∞ b ij for j ≥ 1 and choose β 0 ≥ 0 such that ∞

β 0 ≥ sup |b i0 | + sup ∑ |b ij | , i∈ℕ

i∈ℕ j=1

which is possible due to (5.20) and (5.21). Define βj { { { { { {b ij − β j a Nij := { { b i0 − β 0 + ∑∞ { p=1 b ip { { { {0

for i = 0, j ∈ ℕ0 , for i ∈ ℕ, j ∈ ℕ , for i ∈ ℕ with i ≤ N, j = 0 , for i ∈ ℕ with i > N, j = 0 ,

and put A N := (a Nij )i,j∈ℕ0 . It is straightforward that A N ∈ R and that F(A N ) is as desired. Thus, we have shown that R = Lr (cc , K) is a pre-Riesz space and that (Y, F) is a vector lattice cover of R. Note that Y is not the Dedekind completion of R, as we see next. (f) The space Y is not Dedekind complete⁶. (n)

For a proof, define B(n) := (b ij )i∈ℕ,j∈ℕ0 for n ∈ ℕ by {1 (n) b ij := { i 0 {

for i ∈ ℕ with 1 ≤ i ≤ n, j = 1 , otherwise.

Then B(n) ∈ Y for every n ∈ ℕ and {B(n) ; n ∈ ℕ} does not have a supremum in Y.

6 The proof actually shows that Y is not σ-Dedekind complete.

5.4 Spaces of operators

| 275

(g) The space Lr (cc , K) is pervasive. We show that for any B ∈ Y, B ≥ 0, B ≠ 0, there is A ∈ R such that F(A) ≥ 0, F(A) ≠ 0, F(A) ≤ B. Let B = (b ij )i∈ℕ,j∈ℕ0 , b ij ≥ 0, B ≠ 0. We examine two cases. (i) There is i ∈ ℕ such that b i0 > 0. Put a i0 := b i0 and a kl := 0 otherwise. Then A := (a kl )k,l∈ℕ0 ∈ R and { b i0 F(A)kl = { 0 {

for k = i, l = 0 , otherwise .

(ii) There are i, j ∈ ℕ such that b ij > 0. Put a i0 := a ij := b ij and a kl := 0 otherwise. Then A := (a kl )k,l∈ℕ0 ∈ R and { b ij F(A)kl = { 0 {

for k = i, l = j , otherwise .

In both cases A satisfies F(A) ≥ 0, F(A) ≠ 0, and F(A) ≤ B. We conclude from Corollary 4.2.7 that the space Lr (cc , K) (and, hence, Lr (l∞ 0 )) has the restriction property (R) for bands. Let us now consider the space Loc (cc , K) of order continuous operators in (cc , K), which is a subspace of Lr (cc , K). We will show that Loc (cc , K) is a directed band in Lr (cc , K). First, we characterize the positive order continuous operators and then we consider the arbitrary case. In order to prove the characterizations, we need two statements concerning o-convergence of sequences in (cc , K). (h) (i) A net (x α )α in (cc , K) satisfies x α ↓ 0 if and only if x0α ↓ and x0α + x αi ↓ 0 for every i ∈ ℕ. (ii) The sequence (x n )n∈ℕ in (cc , K) defined by x0n := 1, x ni := −1 for i = 1, . . . , n, and x ni := 0 for i > n, n ∈ ℕ, satisfies x n ↓ 0. To show (i), we first assume that x α ↓ 0. Then x0α ≥ 0 and x0α + x αi ≥ 0, and both x0α and x0α + x αi are decreasing in α for every i ∈ ℕ. Suppose that there exist k ∈ ℕ and δ > 0 such that x0α + x αk ≥ δ for every α. If we define y ∈ cc by y k := δ and y i := 0 for every i ≠ k, then y ≤ x α for every α. However, y ≰ 0, since y0 + y k = δ ≰ 0, which contradicts x α ↓ 0. Conversely, we assume that x0α ↓ and x0α + x αi ↓ 0 for every i ∈ ℕ. Then (x α )α is a decreasing net of positive elements. If y ∈ cc satisfies y ≤ x α for every α, then y0 + y i ≤ x0α + x αi for every i ∈ ℕ and every α. Hence, y0 + y i ≤ 0 for every i ∈ ℕ and then also y0 ≤ 0 as y ∈ cc . So y ≤ 0 and therefore x α ↓ 0. For a proof of assertion (ii), observe that for n > m, {1 if m < i ≤ n , (x m − x n )0 = 0 and (x m − x n )i + (x m − x n )0 = { 0 otherwise , {

276 | 5 Operators on pre-Riesz spaces thus x m ≥ x n . Further, x n ≥ 0 for every n. Finally, if y ≤ x n for every n, then for every i ∈ ℕ we have y i + y0 ≤ x ni + x0n for every n, therefore y i + y0 ≤ 0. Since y ∈ cc , we also obtain y0 ≤ 0 and y ≤ 0. Hence, x n ↓ 0. (j) A regular operator A = (a ij )i,j∈ℕ0 in (cc , K) is order continuous if and only if ∞

∑ (a ij + a0j ) = a i0 + a00

for every i ∈ ℕ .

j=1

Moreover, the space Loc (cc , K) is directed. We first assume that A is an order continuous operator on (cc , K) and show that the o algebraic condition holds. For a net x α ↓ 0 we have Ax α → 󳨀 0, so that there exists a net (v α )α in (cc , K) with ±Ax α ≤ v α for every α and v0α ↓ and v αi + v0α ↓ 0 for every i ∈ ℕ. α α α We then have ±((Ax α )i + (Ax α )0 ) ≤ v αi + v0α , thus | ∑∞ j=0 (a ij + a0j )x j | ≤ v i + v 0 ↓ 0 for every i ∈ ℕ. Further, for each i ∈ ℕ, 󵄨󵄨 ∞ 󵄨󵄨 ∞ 󵄨 󵄨󵄨 󵄨󵄨 ∑ (a ij + a0j ) (x αj + x0α )󵄨󵄨󵄨 ≤ ∑ (|a ij | + |a0j |) (x αj + x0α ) , 󵄨󵄨 󵄨󵄨 󵄨 j=1 󵄨j=1 which converges to 0 by Theorem 1.8.13, since ∑∞ j=1 |a ij | < ∞ for every i ∈ ℕ0 and α α x j + x0 ↓ 0 for every j ∈ ℕ. Hence, for each i ∈ ℕ, the right-hand side of the identity ∞

∞

∞

j=1

j=1

j=0

∑ (a ij + a0j )x0α − (a i0 + a00 )x0α = ∑ (a ij + a0j )(x αj + x0α ) − ∑ (a ij + a0j )x αj converges to 0. If we consider the sequence (x n )n∈ℕ of (ii) of (h), we have x0n = 1 for every n, and we infer that ∑∞ j=1 (a ij + a0j ) = a i0 + a00 for every i ∈ ℕ. Next, we show that any regular operator A = (a ij )i,j∈ℕ0 in (cc , K) that satisfies the algebraic condition is order continuous. We assume as a first step that A is positive. Let (x α )α be a net in (cc , K) with x α ↓ 0. Then Ax α ↓ and in particular (Ax α )0 ↓. Due to the property of A we have for every i ∈ ℕ, ∞

(Ax α )0 + (Ax α )i = ∑ (a0j + a ij )(x0α + x αj ) . j=1 α α α α Since ∑∞ j=1 (a0j + a ij ) < ∞ and x 0 + x j ↓ 0 for every i, it follows that (Ax )0 + (Ax )i ↓ 0. α By (i) of (h) we infer that Ax ↓ 0. Hence A is order continuous. Now, let A = (a ij )i,j∈ℕ0 be an arbitrary regular operator in (cc , K) that satisfies the algebraic condition. We show that there exists a positive order continuous operator B on (cc , K) such that B − A is positive and order continuous. Let ∞

+

d := sup( ∑ (|a0j | + |a ij |) − (|a00 | + |a i0 |)) , i∈ℕ

j=1

5.4 Spaces of operators

| 277

which is a finite number, as A is regular. Define d ij := |a ij | for i, j ∈ ℕ0 with (i, j) ≠ (0, 0) and d00 := |a00 | + d. Then ∞

β i = d00 + d i0 − ∑ (d0j + d ij ) ≥ 0

for every i ∈ ℕ .

j=1

Now define B := (b ij )i,j∈ℕ0 by d00 { { { b ij := { d ij { { {d ii + β i

for i = j = 0 , for i, j ∈ ℕ with i ≠ j , for i, j ∈ ℕ with i = j .

Then the columns of B are eventually zero, thus B is an operator on (cc , K). Further, B satisfies the algebraic condition, and so does B − A. Also, B and B − A are positive. Therefore, both B and B − A are order continuous. Thus, A = B − (B − A) is order continuous and the proof is complete. (k) The space Loc (cc , K) is a band in Lr (cc , K). Due to (j), we can describe the elements of Loc (cc , K) completely in terms of matrix entries, and get Loc (cc , K) = N := {(a ij )i,j∈ℕ0 ∈ R; ∀i ∈ ℕ : ∑∞ j=1 (a ij + a0j ) = a i0 + a00 } . We show that N is a band in R. We use the embedding in the vector lattice cover (Y, F) and the fact that disjointness in Y is the entrywise disjointness. Note that an element A ∈ R is in N if and only if F(A)i0 = 0 for every i ∈ ℕ. The set B := {(b ij )i∈ℕ,j∈ℕ0 ∈ Y; ∀i ∈ ℕ : b i0 = 0} is a band in Y. Since the restriction property (R) for bands holds, N = [B]F is a band in R. r ∞ Summing up, we conclude that Loc (ℓ∞ 0 ) is a band in the pre-Riesz space L (ℓ0 ). Example 5.4.15 may be seen as a first instance of the Ogasawara theorem 5.4.14 for X = Y not being Dedekind complete. For vector lattices X and Y with Y Archimedean, it is interesting to determine under which conditions the subspace Loc (X, Y) is a band in Lr (X, Y). There are many other statements in the theory of vector lattices where disjointness, ideals, or bands are involved. In each of these instances one could formulate a similar statement in pre-Riesz spaces and investigate under which conditions it holds. In particular, many results concerning the structure of spaces of operators between vector lattices require the range space to be Dedekind complete, since then the space of operators is a vector lattice, and the familiar notions from vector lattice theory are available. Now, after developing the similar notions in pre-Riesz spaces, it becomes a natural question which results remain true for an Archimedean vector lattice as range space.

278 | 5 Operators on pre-Riesz spaces

Notes and remarks In Example 5.4.15, the concept of a vector lattice cover shows itself to be an appropriate tool to deal with spaces of operators on a vector lattice that is not Dedekind complete. The construction of the vector lattice cover is very specific there. It is of interest to know under which conditions a vector lattice cover of a space of operators is obtained as a space of operators. For vector lattices X and Y with Y Archimedean, one approach to obtain a vector lattice cover of Lr (X, Y) would be to consider the ideal J generated by Lr (X, Y) in Lr (X, Y δ ). Then J is a Dedekind complete vector lattice and Lr (X, Y) is majorizing in J. It is of interest to identify cases where Lr (X, Y) is order dense in J. Partial results are given in [113]. If we have order denseness, then by Proposition 1.6.2 the Riesz–Kantorovich formula for the supremum holds for operators in Lr (X, Y), provided their supremum exists. This approach provides affirmative instances to the following question on the Riesz–Kantorovich formula, which has drawn attention for many years: Let X and Y be partially ordered vector spaces with X directed and Y Archimedean. Under which conditions on X and Y is, for every two operators in Lr (X, Y) for which the supremum exists, the supremum given by the Riesz–Kantorovich formula? Even for X and Y vector lattices, there are instances where the answer to the question is negative⁷. For functionals on a partially ordered vector space X, this question is dealt with in [15], where the problem is posed in the setting of economic theory, and the Riesz–Kantorovich formula is understood as a revenue function of an auction and linked to the Walrasian equilibrium. In [15, Theorem 3.3] it is shown for a large class of partially ordered vector spaces that the existence of the supremum of two functionals implies that it is given by the Riesz–Kantorovich formula. For an Archimedean vector lattice X, an important subset of Lb (X) is the set Orth(X) of orthomorphisms, i.e., the set of order-bounded band preserving operators. The set Orth(X) is a vector lattice and, moreover, under composition an Archimedean f-algebra with the identity operator I as a unit; see, e.g., [12, Theorem 8.24]. Furthermore, Orth(X) is contained in Loc (X); see, e.g., [12, Theorem 8.10]. If X is a Dedekind complete vector lattice, Orth(X) coincides with the band generated by I in Lr (X); see, e.g., [12, Theorem 8.11]. It is open whether similar properties hold in the setting of pre-Riesz spaces.

7 An example due to Michael Elliott was presented by Anthony Wickstead at the Positivity Conference IX in Edmonton 2017.

5.5 Spectral theory for operators on ordered Banach spaces | 279

5.5 Spectral theory for operators on ordered Banach spaces There is a large variety of applications where positive operators play a central role; see, e.g., [21, 23, 24, 38, 101, 102, 123, 126, 136, 137, 139]. In particular, there is an extensive treatment of positive dynamical systems or positive C0 -semigroups on Banach lattices and, more generally, on ordered Banach spaces. In these applications, spectral properties of positive operators are a fundamental tool for obtaining results, e.g., on the long-term behavior or on the stability of the system. The spectral theory traces back to the classical Perron–Frobenius theorem for positive matrices. There are several well-known generalizations, e.g., for positive operators on Banach lattices. Here we elaborate Krein’s spectral theory for positive operators on ordered Banach spaces. Provided the norm is semimonotone, we obtain that the spectral radius is contained in the spectrum, and that for compact operators there is an associated eigenvector in the cone. Furthermore we mention a few spectral properties of operators that are related to disjointness, ideals, or bands, where we only consider operators on Banach lattices.

5.5.1 Definitions and basic properties We give a short overview of the basic notions and results of spectral theory that we need in the sequel. First we recall the complexification of a Banach space, where we mainly follow [38, p. 268]. Let (X ℝ , ‖⋅‖X ℝ ) be a real Banach space. The product space X := X ℝ × X ℝ is a complex vector space with the linear structure defined by (x, y) + (v, w) = (x + v, y + w)

and (α + iβ)(x, y) = (αx − βy, βx + αy)

for x, y, v, w ∈ X and α, β ∈ ℝ. Via the real-linear isomorphism x 󳨃→ (x, 0) the space X ℝ is identified by a real-linear subspace of X, and one obtains X = X ℝ ⊕ iX ℝ . The space X equipped with the norm ‖x + iy‖X = sup {‖(cos θ)x + (sin θ)y‖ X ℝ ; 0 ≤ θ ≤ 2π} ,

x, y ∈ X ,

is called the complexification of X ℝ . The space X is a Banach space, and the embedding of X ℝ into X is an isometry⁸. Let Y and V be complexifications of ordered Banach spaces (Y ℝ , Y+ℝ , ‖⋅‖Y ℝ ) and ℝ (V , V+ℝ , ‖⋅‖V ℝ ), respectively. Denote by L(Y, V) the Banach space of all (complex) linear bounded operators from Y into V and by L(Y ℝ , V ℝ ) the Banach space of all (real) linear bounded operators from Y ℝ into V ℝ . An operator Tℝ ∈ L(Y ℝ , V ℝ ) can uniquely be extended to its complexification T ∈ L(Y, V) by means of T : x + iw 󳨃→ Tℝ x + iTℝ w .

(5.22)

8 If there exists already a norm ‖⋅‖0 on X, which makes X into a complex Banach space and the restriction of ‖⋅‖0 to X ℝ coincides with ‖⋅‖Xℝ , then ‖⋅‖0 and ‖⋅‖ are equivalent.

280 | 5 Operators on pre-Riesz spaces On the other hand, an operator T ∈ L(Y, V) is called real if T[Y ℝ ] ⊆ V ℝ , and positive if T is real and T[Y+ℝ ] ⊆ V+ℝ . We use the notation Lℝ (Y, V) := {T ∈ L(Y, V); T is real} , L+ (Y, V) := {T ∈ L(Y, V); T is positive} .

(5.23)

As usual, for S, T ∈ Lℝ (Y, V) we write S ≤ T whenever T − S ∈ L+ (Y, V). For T ∈ Lℝ (Y, V) one has ‖T‖L(Y,V) = ‖T|Y ℝ ‖L(Y ℝ ,V ℝ ) . We proceed by defining the spectrum of a bounded linear operator on a complex Banach space X, where we mainly refer to [157]. Let T ∈ L(X). The resolvent set of T is defined by ρ(T) := {λ ∈ ℂ; λI − T is bijective}, the spectrum of T by σ(T) := ℂ \ ρ(T), and the point spectrum by σ p (T) := {λ ∈ ℂ; λI − T is not injective}. An element λ ∈ σ p (T) is an eigenvalue of T, and a corresponding x ∈ X \ {0} with Tx = λx an eigenvector of T. The spectral radius of T is r(T) := sup {|λ|; λ ∈ σ(T)} . If λ ∈ ℂ is such that λI − T is bijective, then (λI − T)−1 ∈ L(X); see Theorem 1.8.4, since X is a Banach space and λI − T is a bounded operator. Therefore, the resolvent map λ 󳨃→ (λI − T)−1

R(⋅, T) : ρ(T) → L(X),

is well-defined. For the subsequent basic statements see, e.g., [157, Satz VII.2.15]. Proposition 5.5.1. Let X be a complex Banach space and T ∈ L(X). (i) The resolvent set ρ(T) is an open set, the spectrum σ(T) is nonempty, compact, and contained in {λ ∈ ℂ; |λ| ≤ ‖T‖}. (ii) The resolvent map is analytic, where for λ0 ∈ ρ(T) and λ ∈ ℂ with |λ − λ0 | < ‖R(λ10 ,T)‖ one has λ ∈ ρ(T) and ∞

R(λ, T) = ∑ (λ0 − λ)n (R(λ0 , T)) n+1 , n=0

where the series converges in operator norm. In particular, the resolvent map is continuous. (iii) For λ ∈ ℂ with |λ| > r(T) we have ∞

R(λ, T) = ∑ n=0

1 λ n+1

Tn ,

where the series converges in the operator norm. (iv) For the spectral radius the Gelfand formula 1

r(T) = lim ‖T n ‖ n n→∞

holds. Moreover, r(T k ) = (r(T))k and r(αT) = |α|r(T) for α ∈ ℂ.

(5.24)

5.5 Spectral theory for operators on ordered Banach spaces |

281

Remark 5.5.2. If X is the complexification of a real ordered Banach space and T ∈ L+ (X), then (5.24) implies for every λ ∈ (r(T), ∞) that R(λ, T) is a positive operator. Let us next consider spectral properties of compact operators. Recall that T ∈ L(X) is compact if T maps the unit ball B in X into a relatively compact set, i.e., the closure of T[B] is compact. The following theorem summarizes the main result of the Riesz– Schauder theory for compact operators; see also [132, Theorem 4.25]. Theorem 5.5.3. Let X be a complex Banach space and let T ∈ L(X) be compact. For every ε > 0 the set {λ ∈ σ(T); |λ| ≥ ε} is finite and consists of eigenvalues of T.

5.5.2 Spectral properties of positive operators For matrices with positive entries, spectral properties are given by the classical Perron–Frobenius theory; see e.g., [2, Chapter 8]. As a generalization for positive operators on ordered Banach spaces, we present a proof of the celebrated Krein– Bonsall–Karlin theorem. For the subsequent version, see [30, 91] or [102, Theorem 8.1]. Recall that an ordered Banach space is a Banach space with a closed generating cone. Theorem 5.5.4. Let (X, K, ‖⋅‖) be the complexification of an ordered Banach space with a semimonotone norm and let T ∈ L+ (X). Then r(T) ∈ σ(T). Proof. Since σ(T) is nonempty, the case r(T) = 0 is clear. We consider the case r := r(T) > 0. We argue by contradiction. Suppose r ∉ σ(T). Then R(r, T) is well-defined and in L(X). Due to Remark 5.5.2, for every n ∈ ℕ the operator R (r + 1n , T) is positive, and since the resolvent map is continuous, we have lim R (r + 1n , T) = R(r, T) .

n→∞

Since K is closed and, hence, L+ (X) is closed, we can pass to the limit in inequalities in L(X), thus R(r, T) is positive. Choose β ∈ (0, r) such that (r − β)‖R(r, T)‖ < 1 . We aim to obtain a contradiction by estimating the operator norm of 1β T k and will arrive at r ≤ β using the Gelfand formula. Since ‖(r − β)R(r, T)‖ = (r − β)‖R(r, T)‖ < 1, the operator (I − (r − β)R(r, T))−1 exists in L(X) and is given by the Neumann series⁹ ∞

(I − (r − β)R(r, T))−1 = ∑ (r − β)n R(r, T)n .

(5.25)

n=0

9 If X is a Banach space and T ∈ L(X) with ‖T‖ < 1, then (I − T)−1 exists in L(X) and is given by the n Neumann series (I − T)−1 = ∑∞ n=0 T ; see e.g., [157, Satz II.1.11].

282 | 5 Operators on pre-Riesz spaces We show that (βI − T)−1 exists in L(X) and is positive. Indeed, βI − T = rI − T − (r − β)I = (rI − T)(I − (r − β)R(r, T)), hence −1

(βI − T)

= (I − (r − β)R(r, T))−1 R(r, T) .

From (5.25) and the positivity of R(r, T) it follows that (βI − T)−1 is positive. For every k ∈ ℕ we have k

(βI − T)−1 = ∑

1 i−1 T βi

+

1 k T (βI βk

− T)−1 .

(5.26)

i=1

Indeed, I = (I − 1β T) + ( 1β T −

1 2 T ) + ( β12 T 2 β2

= (I − 1β T) + 1β T(I − 1β T) + = 1β (βI − T) +

1 T(βI β2

1 2 T (I β2

− T) + ⋅ ⋅ ⋅ +

−

1 3 T )+ β3

1 ⋅ ⋅ ⋅ + ( β k−1 T k−1 −

− 1β T) + ⋅ ⋅ ⋅ +

1 k−1 T (βI βk

1 T k−1 (I β k−1

− T) +

1 k T βk

1 k T ) βk

− 1β T) +

+

1 k T βk

1 k T βk

.

Multiplying this equation by (βI − T)−1 from the right gives (5.26). From (5.26) it follows, in particular, that (βI − T)−1 ≥

1 k−1 T βk

,

since T and (βI − T)−1 are positive. This means ( 1β T)

k−1

≤ β(βI − T)−1 .

(5.27)

Since X is an ordered Banach space with a semimonotone norm, Proposition 3.5.2 (ii) yields that there is a constant M ∈ (0, ∞) such that for every x ∈ X there are u, v ∈ K with x = u − v and ‖u‖ ≤ M‖x‖, ‖v‖ ≤ M‖x‖ . Fix x ∈ X. Due to (5.27), for the associated u and v we obtain k

( 1β T) u ≤ β(βI − T)−1 u,

k

( 1β T) v ≤ β(βI − T)−1 v .

The norm is semimonotone with a certain constant N ∈ (0, ∞). This implies 󵄩󵄩󵄩 1 k 󵄩󵄩󵄩 󵄩󵄩( T) u 󵄩󵄩 ≤ Nβ‖(βI − T)−1 ‖ ‖u‖ ≤ MNβ‖(βI − T)−1 ‖ ‖x‖ , 󵄩󵄩 β 󵄩󵄩 󵄩 󵄩 󵄩󵄩 󵄩 󵄩󵄩 1 k 󵄩󵄩󵄩 󵄩󵄩( β T) v󵄩󵄩 ≤ MNβ‖(βI − T)−1 ‖ ‖x‖ . 󵄩󵄩 󵄩󵄩 We obtain for every k ∈ ℕ that 󵄩󵄩 󵄩 󵄩 󵄩󵄩 󵄩󵄩 1 k 󵄩󵄩󵄩 󵄩󵄩󵄩 1 k 󵄩 󵄩󵄩( β T) x󵄩󵄩 = 󵄩󵄩( β T) (u − v)󵄩󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 ≤ 2MNβ‖(βI − T)−1 ‖ ‖x‖ .

5.5 Spectral theory for operators on ordered Banach spaces | 283 k

Hence, for the operator norm of ( 1β T) holds that 󵄩󵄩 󵄩 󵄩󵄩 k 󵄩 󵄩󵄩 󵄩󵄩 1 k 󵄩󵄩󵄩 󵄩 󵄩󵄩( β T) 󵄩󵄩 = sup {󵄩󵄩󵄩( 1β T) x󵄩󵄩󵄩 ; ‖x‖ ≤ 1} ≤ 2MNβ‖(βI − T)−1 ‖ . 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 Now, by the Gelfand formula (and since 2MNβ‖(βI − T)−1 ‖ does not depend on k), we obtain 1 β r(T)

= r ( 1β T) 1 󵄩󵄩 k󵄩 󵄩󵄩 k 󵄩 = lim 󵄩󵄩󵄩( 1β T) 󵄩󵄩󵄩 󵄩󵄩 k→∞ 󵄩 󵄩 1

≤ lim (2MNβ‖(βI − T)−1 ‖) k = 1 , k→∞

hence r = r(T) ≤ β, which is a contradiction. We reformulate Theorem 5.5.4 for a positive operator on a real ordered Banach space, where we use Corollary 5.4.3. Corollary 5.5.5. Let X ℝ be an ordered Banach space with a semimonotone norm and let Tℝ ∈ L+ (X ℝ ). Then for the complexification X of X ℝ and the complexification T : X → X of Tℝ given by (5.22) we have T ∈ L+ (X) and r(T) ∈ σ(T). Next we address the existence of a positive eigenvector. For the subsequent theorem, see [133, Appendix 2.4] or [102, Theorem 9.2]. Theorem 5.5.6. Let (X, K, ‖⋅‖) be the complexification of an ordered Banach space and let T ∈ L+ (X) be compact. If r(T) > 0, then r(T) is an eigenvalue of T with a corresponding eigenvector in K. Proof. There are finitely many eigenvalues λ of T with |λ| = r(T), denoted by λ i with i ∈ {1, . . . , k}. For λ i consider the generalized eigenspace E i := {x ∈ X; ∃n ∈ ℕ : (λ i I − T)n (x) = 0} . Define X0 := span{

⋃

Ei } ,

i∈{1,...,k}

then X0 is a finite-dimensional linear subspace of X, and T[X0 ] ⊆ X0 . Since r(T) > 0, we have X0 ≠ {0}. Define T0 := T|X0 , then σ(T0 ) = {λ i ; i ∈ {1, . . . , k}} and, hence, r(T0 ) = r(T). Let K0 := K ∩ X0 , then T[K0 ] ⊆ K0 , hence T0 [K0 ] ⊆ K0 .

284 | 5 Operators on pre-Riesz spaces We first consider the case that K0 ≠ {0}. Since K is a closed cone, K0 is a closed cone in X0 . Consider Y := K0 − K0 . Since Y is finite-dimensional, by Proposition 1.5.16 the space (Y, K0 , ‖⋅‖) is an ordered Banach space with a semimonotone norm. Moreover, T0 |Y is positive. By the Krein– Bonsall–Karlin theorem 5.5.4 we obtain r(T0 |Y ) ∈ σ(T0 |Y ) ⊆ σ(T0 ) = {λ i ; i ∈ {1, . . . , k}} , i.e., there is j ∈ {1, . . . , k} with λ j ∈ ℝ and λ j > 0, hence λ j = r(T). This means that r(T) ∈ σ(T). Next we show that we are always in the case that there is z0 ∈ K0 \ {0}. For this, we use the decomposition according to [92, Theorem III.6.17], i.e., there is a closed subspace X1 of X with X0 ∩ X1 = {0} and X0 + X1 = X such that T[X1 ] ⊆ X1 , and for T1 = T|X1 one has σ(T1 ) = σ(T) \ {λ i ; i ∈ {1, . . . , k}}. Hence, there is q ∈ (0, r(T)) such that σ(T1 ) ⊆ {z ∈ ℂ; |z| ≤ q} , i.e., r(T1 ) ≤ q. Every x ∈ X has a unique representation x = u + v, where u ∈ X0 and v ∈ X1 . We define P : X → X0 , x 󳨃→ u, which is a linear continuous projection onto X0 . As X0 ≠ {0}, there is x ∈ X such that Px ≠ 0. Since K is generating, there is x0 ∈ K such that Px0 =: u 0 ≠ 0. Since X0 is finite-dimensional, the closed unit ball in X0 is compact, i.e., the sequence T n (u 0 ) ( n ) ‖T (u 0 )‖ n∈ℕ has a subsequence that converges to some element z0 ∈ X0 , where ‖z0 ‖ = 1. We show that z0 ∈ K, i.e., z0 ∈ K0 \ {0}. We write T n (x0 ) T n (x0 − u 0 ) T n (u 0 ) = − . ‖T n (u 0 )‖ ‖T n (u 0 )‖ ‖T n (u 0 )‖

(5.28)

Observe that the first term at the right-hand side is in K, so if it converges, then the limit is in K, as K is closed. It remains to show that the second term tends to zero. We use that T n (u 0 ) = T0n (u 0 ), and x0 − u 0 = x0 − P(x0 ) ∈ X1 , hence T n (x0 − u 0 ) = T1n (x0 − u 0 ). Choose numbers q0 , q1 ∈ (q, r(T)) such that q1 < q0 . On the one hand, 1

r(T1 ) = lim ‖T1n ‖ n ≤ q < q1 , n→∞

i.e., there is n1 ∈ ℕ such that for every n > n1 one has that ‖T1n ‖ < q1n . On the other hand, 0 ∈ ̸ σ(T0 ), hence T0 is invertible, and¹⁰ σ (T0−1 ) = { λ1i ; i ∈ {1, . . . , k}} . 10 This is a special case of the spectral mapping theorem; see, e.g., [132, Theorem 10.33].

(5.29)

5.5 Spectral theory for operators on ordered Banach spaces | 285

We obtain 󵄨 󵄨 r(T0−1 ) = max {󵄨󵄨󵄨󵄨 λ1i 󵄨󵄨󵄨󵄨 ; i ∈ {1, . . . , k}} = r(T10 ) = 󵄩 󵄩1 lim 󵄩󵄩 T −n 󵄩󵄩 n < q10 , n→∞ 󵄩 0 󵄩

1 r(T)


n0 one has that ‖T0−n ‖
max{n1 , n2 } and combine (5.29), (5.30) and (5.31), and get ‖T1n (x0 − u 0 )‖ ‖T1n ‖ ‖T0−n ‖ ‖x0 − u 0 ‖ q1 n ‖x0 − u 0 ‖ n→∞ ≤( ) 󳨀󳨀󳨀󳨀→ 0 . ≤ n ‖u 0 ‖ q0 ‖u 0 ‖ ‖T0 u 0 ‖ From (5.28) it follows that z0 ∈ K. The existence of a positive eigenvector associated to r(T) is shown by means of Brouwer’s fixpoint theorem; see Theorem 1.8.12. We consider F := {x ∈ K0 ; ‖x‖ ≤ 1} , which is a closed bounded convex subset of the finite-dimensional space X0 . We use z0 ∈ K0 \ {0} as above and define A : F → F,

x 󳨃→

‖x‖T0 x + (1 − ‖x‖)z0 , ‖‖x‖T0 x + (1 − ‖x‖)z0 ‖

which is a continuous map. By Theorem 1.8.12 there is y ∈ F such that Ay = y. We obtain 1 = ‖Ay‖ = ‖y‖. Hence, y = Ay =

T0 y ‖T 0 y‖ ,

so T0 y = ‖T0 y‖y ,

i.e., y is an eigenvector of T0 and, therefore, of T. For the associated eigenvalue we obtain ‖T0 y‖ = r(T0 ) = r(T) , since the modulus of every eigenvalue of T0 is r(T). We conclude that r(T) is an eigenvalue of T with an eigenvector y that belongs to K. Corollary 5.5.7. Let K be a closed generating cone in ℝd and let T be the complexification of a linear positive operator in (ℝd , K) such that r(T) > 0. Then r(T) is an eigenvalue of T with a corresponding eigenvector in K. In the literature, there are further Perron–Frobenius type theorems. A recent account with emphasis on nonlinear Perron–Frobenius theory can be found in the outstanding monograph [106]; see also the references therein. On Banach lattices, a Perron– Frobenius theory is also developed for eventually positive operators; see [60].

286 | 5 Operators on pre-Riesz spaces

5.5.3 Some spectral properties for operators on Banach lattices This subsection contains a small selection of results on spectral properties of certain types of operators that are related to disjointness, ideals, or bands. So far, those results are known in the setting of Banach lattices. With the notions of disjointness, ideals, and bands developed in Chapter 4, similar statements can be formulated for operators in ordered Banach spaces, yielding a new area of research to be explored. The selection of results presented here is meant as a source of inspiration. First we consider operators that are related to ideals. In view of Theorem 5.5.6, it is of interest for which operators the spectral radius is strictly positive. We consider a Banach lattice X and T ∈ L(X). The operator T is called ideal irreducible if for every closed ideal J with T[J] ⊆ J we have J = {0} or J = X. The following result is due to Ben de Pagter; see [44]. Theorem 5.5.8. Let X be a Banach lattice and let T ∈ L(X) be ideal irreducible. Then r(T) > 0. More results in a similar spirit are given in [2, Sections 9.2 and 9.3], where band irreducible operators are also considered. We proceed by spectral properties of disjointness preserving operators. First we recall a classical result on Riesz homomorphisms, which is given in [134, Theorem V.4.4]. Hereby a set A ⊆ ℂ is called cyclic if for every α ∈ A with α = |α|γ (for some γ ∈ ℂ) we have for every k ∈ ℤ that |α|γ k ∈ A. Theorem 5.5.9. Let X be a Banach lattice and let T ∈ L(X) be a Riesz homomorphism. Then σ(T) is cyclic. As T in Theorem 5.5.9 is, in particular, positive, we have T ∈ L(X) by Corollary 5.4.3. Note that spectral properties of Riesz* homomorphisms, Riesz homomorphisms or complete Riesz homomorphisms on (appropriately normed) pre-Riesz spaces are not known, so far. For disjointness preserving operators on Banach lattices, we state Arendt’s result from [17, Theorem 3.5]. Hereby we call for r0 > 0 and θ0 , θ1 ∈ [0, 2π) the set {re iθ ∈ ℂ; 0 < r ≤ r0 , θ0 ≤ θ < θ0 + θ1 } a sector, and θ1 the angle of the sector. Theorem 5.5.10. Let X be a Banach lattice and let T ∈ L(X) be an order-bounded disjointness preserving operator such that σ(T) is contained in a sector of angle 2π 3 . Then T is an element of the center Z(X). As by Theorem 1.4.12 every operator T as in Theorem 5.5.10 has a modulus and, hence, is regular, we obtain T ∈ L(X) by Corollary 5.4.3. Further results on the spectrum of disjointness preserving operators can be found, e.g., in [20].

5.5 Spectral theory for operators on ordered Banach spaces |

287

For a compact Hausdorff space Ω and the space X := C(Ω) of all continuous complex-valued functions on Ω, every disjointness preserving operator T ∈ L(X) is a weighted composition operator, and a complete description of its spectrum is given in [96]. For different types of subspaces Z of X, spectral properties of weighted composition operators T ∈ L(Z) are investigated in [97–99], where for Z also spaces of differentiable functions and Sobolev spaces are considered, leading to first insights into nonlattice cases.

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Index absorbing set 119 additive map 15 affine hull 1, 219 algebraic dual 13 angle of a sector 286 annihilator 220 antilattice 198 Archimedean space 5 Archimedean tensor cone 112 (β), condition 254 Banach lattice 36 Banach–Alaoglu theorem 51 Banach’s isomorphism theorem 50 band – in pre-Riesz space 204 – in Riesz space 18 band preserving operator 254 – on pre-Riesz space 238 – on Riesz space 26 base of a cone 32 bipositive operator 12 bisaturated – pair 223 – set 222 Bonsall theorem 265 Brouwer’s fixpoint theorem 51 C0 -semigroup 257 carrier – of a band 219 – of a set of functions 20 center 238 circled set 119 closed cone 27, 179 closed operator 257 closure of positive cone 180 compact operator 281 complete Riesz homomorphism 80 complexification – of a Banach space 279 – of an operator 279 complexification of a Banach space 279 condition (β) 254 cone 2 convex hull 1 https://doi.org/10.1515/9783110476293-007

convex polytope 228 cut 54 D-disjoint 196 decreasing net 11 Dedekind complete – lattice 55 – vector lattice 8 Dedekind completion 63 Dedekind cut 54 directed ideal 215 directed part 208 directed set 4 disjoint – in ordered vector space 194 – in Riesz space 17 disjointness preserving C0 -semigroup 257 disjointness preserving inverse 253, 255 disjointness preserving operator – between pre-Riesz spaces 238 – between Riesz spaces 22 d-isomorphism 250 distributive lattice 6 distributive property 72 dual – cone 28 – norm 121 – of quotient space 158 – wedge 27 eigenvalue 280 eigenvector 280 equivalent seminorms 121 example – affine functions on unit circle 45, 126, 129, 138, 165, 205, 211 – C1 [0, 1] 7, 10, 116 – four ray cone 48, 208, 225 – ice cream cone 43, 92, 197, 216 – lexicographical order 5, 37, 117 – Lorentz cone 43, 103 – Namioka space 10, 83, 234 – Otto space 13, 241, 243 – polyhedral cone 46, 102, 213, 230 – polynomials of degree at most two 10, 42, 79, 92, 150, 203, 206 – positive semidefinite matrices 44, 92, 104

298 | Index

extension – of locally convex locally full topology 178 – of monotone* norm, completeness 151 – of monotone* seminorm 144 – of pre-Riesz seminorm 134 – of regular seminorm 165 – of Riesz* homomorphism 78 – of seminorm 121 – of seminorm with full unit ball 146 – of seminorm, uniqueness, counterexample 153 – of symmetrically monotone seminorm 146 extension band 234 extension property 115 – for bands 204 – for full sets 177 – for solid sets 125 – for solvex ideals 211 – for solvex sets 131 extremal element 32 extreme – point 32, 51 – ray 32 extremely disconnected 20 face of a cone 32 fordable pre-Riesz space 201, 206, 226 Fremlin tensor product 110 full – locally full topology 174 – set 2 – unit ball 140 functional representation 94 – isomorphic 95 – smaller 95 Gelfand formula 280 generated as a Riesz space 63 generating – cone 2 – wedge 2 generator of a C0 -semigroup 257 Grosberg–Krein theorem 268 Hahn–Banach theorem 50, 265 Hayes theorem 85, 95 ice cream cone 43 ideal – directed 215

– in ordered vector space 207 – in Riesz space 17 – order closed 214 – solvex 209, 214 ideal irreducible operator 286 image of a map 12 increasing net 11 infimum 6 integrally closed 5 interior of a cone 30 isomorphic functional representation 95 Kakutani theorem 97 kernel of monotone seminorm 157 kernel of Riesz* homomorphism 217 Krein–Bonsall–Karlin theorem 281 Krein–Milman theorem 51 lattice – homomorphism 22 – operations 7 – operations, continuity 181 – seminorm 35 lexicographical order 5 Lipecki–Luxemburg–Schep theorem 67 local operator 238, 258 locally – convex, full 176 – convex, solid 181 – full 174 – solid 180 Lorentz cone 43, 103 L-seminorm 160 m≤ -norm 163 majorizing subspace 40 Mazur theorem 50 Minkowski functional 120 modulus 7 monotone seminorm 28, 139 monotone* seminorm 139 M-seminorm 160 negative part 7 non-flat cone 173 norm see seminorm norm dual 27, 121 normal cone 172, 183 normed Riesz space 35 normed vector lattice 35

Index |

o-closed set 11 o-continuous map 11 o-convergent net 11 Ogasawara theorem 269 open unit ball 119 operator – band preserving 26, 238, 254 – between ordered normed spaces 263 – bipositive 12 – compact 281 – complete Riesz homomorphism 80 – disjointness preserving 22 – d-isomorphism 250 – embedding 12 – ideal irreducible 286 – local 238, 258 – order continuous 11, 186, 269 – order isomorphism 12 – order-bounded 12 – orthomorphism 26 – positive 12 – regular 12 – Riesz homomorphism 22, 80, 244 – Riesz isomorphism 22 – Riesz* homomorphism 74, 244 – weighted composition 244 operator norm 263, 264 order – lexicographical 5 – on set of seminorms 120 – partial order 2 – trivial order 3 – vector space order 2 order closed – ideal 214 – set 11 order continuous – map 11 – operator 186, 269 order convergent net 11 order dense subspace 40 order dual 13 order interval 2 order isomorphic spaces 13 order isomorphism 12 order unit 4 order unit norm 29 order unit space 29, 264 order-bounded operator 12

299

order-bounded set 2 ordered Banach space 172, 257 ordered normed space 26 ordered vector space 3 orthomorphism 26, 278 p-bounded 28 p-Cauchy 120 p-complete 121, 287 p-continuous 121 p-convergent 120 partial order 2 partially ordered vector space 3 pervasive pre-Riesz space 115, 137, 188, 201, 206, 226, 241, 249, 274 piecewise polynomial function 92 point spectrum 280 polyhedral cone 46 positive eigenvector 283 positive element 2 positive operator 12, 280 positive part 7 positive semidefinite matrices 44 positive-linear hull 2 positively homogeneous map 15, 50 pre-image of a map 12 pre-Riesz seminorm 133 pre-Riesz space 68 – fordable 201, 206, 226 – pervasive 115, 137, 188, 201, 206, 226, 241, 249, 274 – weakly pervasive 242 projective cone 112 pure state 95 quotient cone 5 quotient norm 157 quotient order 5 quotient space 5, 158 real operator 280 regular operator 12 regular seminorm 164, 258 regularization of a seminorm 169 regularly open set 21, 205, 236 relatively uniformly – Cauchy 37 – convergent 37 resolvent map 280

300 | Index

resolvent set 280 restriction – of locally convex locally full topology 178 – of monotone seminorm 144 – of monotone* seminorm 144 – of pre-Riesz seminorm 136 – of regular seminorm, counterexample 164 – of Riesz* homomorphism 77 – of seminorm 121 – of seminorm with full unit ball 144 – of symmetrically monotone seminorm 144 restriction property 115 – for bands in fordable spaces 206, 275 – for bands, counterexample 225 – for full sets 177 – for ideals 211 – for o-closed sets 187 – for solid sets 125 – for solvex sets 131 Riesz bimorphism 110 Riesz completion 91 Riesz decomposition property 9 Riesz homomorphism – between ordered vector spaces 80 – between Riesz spaces 22, 244 Riesz interpolation property 9 Riesz isomorphic 22 Riesz isomorphism 22 Riesz seminorm 35, 121 Riesz separation property 9 Riesz space 6 Riesz subspace 7 – generated by a subspace 63 Riesz* homomorphism 74, 244 Riesz–Kantorovich formulas 16, 278 Riesz–Schauder theorem 281 sandwich theorem 174 saturated set 220 saturation 220 sector 286 semimonotone seminorm 28 seminorm 119, 287 – lattice 35 – monotone 28, 139 – monotone* 139 – pre-Riesz 133 – regular 164, 258 – Riesz 35

– semimonotone 28 – symmetrically monotone 139 – with directed open unit ball 161 – with full unit ball 140 – with solid unit ball 125 – with solvex unit ball 133 smaller functional representation 95 solid – hull 124 – set, in ordered vector space 123 – set, in Riesz space 17 – unit ball 126 solvex – hull 129 – ideal 209, 214 – set 128 – unit ball 133 spectral radius 280 spectrum 280 standard cone in ℝn 3 standard order in ℝn 3 state 95 Stone–Weierstrass theorem 51 strongly separates points 244 subadditive map 50, 265 superadditive map 265 supremum 6 symmetrically monotone seminorm 139

tensor product 109 topology – locally convex, full 176 – locally convex, solid 181 – locally full 174 – locally solid 180 total set 13 trivial face 32 trivial order 3

u-norm 29, 94 uoDM -convergence 190 uniform completion of Riesz completion 101 uniformly complete 37, 91, 151, 167 uniformly continuous C0 -semigroup 257 unit ball 119 uo-convergence 191 Urysohn’s lemma 51

Index |

van Haandel theorem – Riesz completion 91 – Riesz* homomorphisms 93 vector lattice 6 vector lattice cover 91 vector space of linear operators 12 vector space order 2 vertex of a convex polytope 228 weakly pervasive pre-Riesz space 242 wedge 2 weighted composition operator 244

301

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