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Onno van Gaans
sem ι norms on ordered vector spaces
een wetenschappelijke proeve op het gebied van de Natuurwetenschappen, Wiskunde en Informatica
Proefschrift
ter verkrijging van de graad van doctor aan de Katholieke Universiteit Nijmegen, volgens besluit van het College van Decanen in het openbaar te verdedigen op maandag 22 februari 1999 des namiddags om 3.30 uur precies door Onno Ward van Gaans
geboren op 15 oktober 1971 te Waddinxveen Promotor: Prof. dr. A.C.M, van Rooij
Manuscriptcommissie: Prof. dr. CD. Aliprantis, Purdue University, West Lafayette, Indiana Prof. dr. G.J.H.M. Buskes, University of Mississippi, Oxford, Mississippi Dr. B. de Pagter, Technische Universiteit Delft, Delft, The Netherlands
ISBN 90-9012410-1 Contents
Summary ν
Samenvatting vii
1 Seminorms and Topologies on Ordered Spaces 1 1.1 Partially Ordered Vector Spaces 1 1.2 Seminorms on Ordered Spaces 3 1.2.1 Monotone, Monotone*, and Fremlin Seminorms 3 1.2.2 Monotone Seminorms and Relatively Uniform Convergence . . 7 1.2.3 Additivity on the Positive Cone and Full Unit Balls 7 1.2.4 Monotone Norms and Closedness of the Positive Cone 8 1.2.5 Quotients Over Kernels of Monotone Seminorms 9 1.3 Locally Full Topologies 12 1.4 Topologies such that the Positive Cone is Closed 15 1.5 Locally Solid Topologies 16 1.6 Dual Spaces 21
2 Extending Monotone Seminorms 27 2.1 Introduction 27 2.2 Existence of Extensions 28 2.3 Uniqueness of Extensions 31 2.4 Extensions and Norm Completeness 34
3 Riesz Seminorms on Non-Riesz Spaces 37 3.1 Solid Sets in Partially Ordered Vector Spaces 37 3.2 Solvex Sets and Pre-Riesz Seminorms 42 3.3 Restricting Pre-Riesz Seminorms to Pre-Riesz Subspaces 44 3.3.1 Riesz* Homomorphisms and Pre-Riesz Spaces 44 3.3.2 Pre-Riesz Seminorms and Riesz* Homomorphisms 46 3.4 Extending Pre-Riesz Seminorms 46 3.5 Extensions and Non-Uniqueness 50
i Contents
3.6 Extensions and L- and M-Seminorms 51 3.7 Regular Seminorms: Introduction 53 3.8 Extending Regular Seminorms and Regularizations 56 3.9 Regularizations and Norm Completeness 57 3.10 Norm Completions , 59 3.10.1 Norm Completions of Normed Ordered Vector Spaces 59 3.10.2 On the Norm Completion of a Monotonely Normed Riesz Space 60
4 Subspaces of Normed Riesz Spaces 65 4.1 Introduction 65 4.2 Dual Characterization of Fremlin Seminorms 66 4.3 Embedding in a Normed Riesz Space 69 4.3.1 Construction 69 4.3.2 Preservation of Lattice Structure 71 4.3.3 No Minimality 76 4.4 On Embedding of Spaces with Unclosed Positive Cones 77 4.4.1 Embedding in a Regularly Normed Space 77 4.4.2 Embedding in a Riesz Space with a Riesz Seminorm 81 4.5 The Subspaces of an M-space 84 4.5.1 Dual Characterization of Seminorms with Full Unit Balls ... 84 4.5.2 Embedding in an M-normed Riesz Space 86
5 Spaces of Operators 89 5.1 Introduction 89 5.2 Norm Dual Spaces 91 5.2.1 Krein's Lemma and the Dual of a Regularly Normed Space . . 91 5.2.2 Counterexamples 93 5.3 Bonsall's Theorem and a Riesz Space Valued Version of Krein's Lemma 95 5.4 Operators Between Normed Partially Ordered Vector Spaces 98 5.5 An Alternative for the r-Norm 101
Commentary 103
List of Symbols 107
Bibliography 109
Index 111
Curriculum Vitae 115
ii Dankwoord
Velen hebben direct of indirect bijgedragen aan het totstandkomen van dit proef schrift en aan het plezier dat ik voortdurend in mijn werk heb gehad. Een aantal van hen wil ik met name noemen: Arnoud voor al zijn steun en hartelijkheid op elk gewenst moment; de collega's van de vakgroep voor de prettige werksfeer en hun persoonlijke belangstelling; mijn vrienden voor hun steun, belangstelling en afleid ing; de studenten voor het regelmatig aanwakkeren van mijn enthousiasme voor het vak; Peter en Dick, Ernie en mijnheer Bakker voor 'de muzikale omlijsting'; mijn kamergenoten Frank en Danny en Paul voor de gezellige huiskamersfeer op de werk plek en hun niet aflatende zorg om mijn caffeïnespiegel; and thanks to the members of the manuscript committee for carefully reading the manuscript in a very short time. Ten slotte mijn ouders en zus die mij door de dalen heen sleepten en tijdens de pieken op de grond hielden. Mijn hartelijke dank aan allen die mijn AiO-jaren tot zo'n geweldige tijd hebben gemaakt.
Onno van Gaans
111 iv Summary
Many of the real vector spaces occurring in analysis have a natural norm or topology as well as a natural ordering. The relation between the norm or topology and the or dering usually fits in the theory of normed Riesz spaces, locally full partially ordered vector spaces, or related notions. Especially for normed Riesz spaces —and Ba- nach lattices in particular— a rich theory has been developed (see e.g. Aliprantis & Burkinshaw[2], Meyer-Nieberg[17], Schaefer[22], Vulikh[23]). For normed partially ordered vector spaces, the theory includes duality (see e.g. Davies[7], Ng[19]) and spaces of operators (see e.g. Wickstead[24]). There is also theory on ordered topolog ical spaces (see e.g. Kelley L· Namioka[14], Namioka[18], Peressini[20], Schaefer[21] and also Aliprantis L· Burkinshaw[l] and Fremlin[8]). It is not without interest to extend theory of normed Riesz spaces to the more general setting of partially ordered vector spaces. Not merely to cover a wider range of spaces, but also because such spaces appear naturally in the theory of normed Riesz spaces itself. Indeed, linear subspaces of normed Riesz spaces are not always Riesz spaces —every partially ordered vector space is in fact a linear subspace of a Riesz space— and, also, spaces of operators between normed Riesz spaces need not be Riesz spaces. The study of normed partially ordered vector spaces is not new and neither is the study of embedding them in Riesz spaces. What seems to be new in this text is the study of seminorms on partially ordered vector spaces by viewing these spaces as subspaces of Riesz spaces. Thus, special attention has been paid to restriction, extension, and isometrical embedding problems and new results have been obtained. The properties of a seminorm ρ on a partially ordered vector space E that play important roles from many points of view are what we call the Fremlin property {—y < x < У =* p(x) < p(y))j regularity (p(x) = inf{p(r/): — y < χ < y}), and closedness of E+. Part of their significance can be found in the literature, mainly concerning spaces of operators. This text provides a more systematic treatment, dealing also with extension, isometrical embedding, norm completeness, and norm completions. Chapter 1 starts with a glossary of the basic theory of locally full and locally
ν Summary solid topologies. Also, basic theory of various types of seminorms is developed, dealing with, for instance, quotients with respect to kernels and closures of positive cones. Most results are accompanied by (counter)examples. Chapter 2 studies the question whether every monotone seminorm on a par tially ordered vector space is the restriction of a monotone seminorm on a Riesz space. The answer is negative. A condition on a seminorm ρ that is necessary for extendability to a monotone seminorm on a Riesz space is the monotone* property (—11 < χ < v, u,v > 0 ^ p(x) < p(u) +p(v)). It is sufficient for extendability to a monotone seminorm on every partially ordered vector space, in particular every Riesz space, in which the initial space is majorizing (2.1). Fremlin seminorms on majorizing subspaces can always be extended to Fremlin seminorms on the whole space (2.3). Explicit formulas for greatest extensions are given and preservation of norm completeness is investigated (2.12). Chapter 3 studies a reasonable notion of 'Riesz seminorm' on a partially ordered vector space. The first approach is to consider seminorms that are restrictions of Riesz seminorms on the Dedekind completion. A characterization of such seminorms is given (3.15, 3.29), based on a notion of 'solid set' in a partially ordered vector space (3.7). There is, as before, an explicit construction of extensions. The ensuing notion of 'Riesz seminorm' turns out to be rather weak. For example, there may exist two such seminorms that coincide on the positive cone (3.38). A stronger notion is regularity, generalizing Riesz seminorms in a very convenient way. Theorems about norm completeness are given, generalizing results concerning the r-norm of the theory of spaces of operators (3.46). The last section studies norm completions (3.51, 3.55). Chapter 4 investigates what norms are restrictions of Riesz norms. Since every partially ordered vector space is (isomorphic to) a subspace of a Riesz space, it is very natural to ask which norms on partially ordered vector spaces allow embedding in normed Riesz spaces. The norms that can be obtained this way turn out to be Fremlin norms such that the positive cone is closed (4.8). A similar result is obtained for restriction of M-norms (4.34). Under additional conditions, the results can be extended to unclosed cones (4.21) and to seminorms (4.28). Chapter 5 applies the concepts and results developed in Chapters 1-4 to the theory of spaces of operators. Many of the results known in the literature are obtained as straightforward generalizations of results for spaces of operators between normed Riesz spaces (5.16). The theory of Chapter 4 leads to a new proof of Krein's lemma (5.3). Every chapter starts with a short abstract, outlining its contents and main re sults. A commentary with conclusions is included after Chapter 5.
vi Samenvatting
Titel: Halfnormen op geordende vectorruimten Veel van de reële vectorruimten die in de analyse voorkomen hebben zowel een natuurlijke norm of topologie als een natuurlijke ordening. Het verband tussen de norm of topologie en de ordening valt gewoonlijk binnen het kader van genormeerde Rieszruimten, lokaal gevulde vectorruimten en dergelijke begrippen. Vooral voor genormeerde Rieszruimten —en Banachtralies (-roosters) in het bijzonder— is er een rijke theorie ontwikkeld (zie bijv. Aliprantis & Burkinshaw[2], Meyer-Nieberg[17], Schaefer[22], Vulikh[23]). Voor genormeerde partieel geordende vectorruimten is er theorie over dualiteit (zie bijv. Davies[7], Ng[19]) en ruimten van operatoren (zie bijv. Wickstead[24]) en er is theorie over geordende topologische ruimten (zie bijv. Kelley &¿ Namioka[14], Namioka[18], Peressini[20], Schaefer[21] en ook Alipran tis L· Burkinshaw[l] en Premlin[8]). Het lijkt interessant om theorie van genormeerde Rieszruimten uit te breiden naar de meer algemene context van partieel geordende vectorruimten. Niet alleen om een breder scala aan ruimten te beslaan, maar ook omdat zulke ruimten op een natuurlijke manier voorkomen binnen de theorie van genormeerde Rieszruimten. Zo zijn lineaire deelruimten van genormeerde Rieszruimten niet altijd Rieszruimten — iedere partieel geordende vectorruimte is in feite een lineaire deelruimte van een Rieszruimte— en ook ruimten van operatoren tussen genormeerde Rieszruimten hoeven geen Rieszruimten te zijn. Het bestuderen van genormeerde partieel geordende vectorruimten is niet nieuw en hetzelfde geldt voor inbedding in Rieszruimten. Wat nieuw lijkt in deze tekst is het bestuderen van halfnormen op partieel geordende vectorruimten waarbij de ruimten worden gezien als deelruimten van Rieszruimten. Op deze manier is er extra aandacht besteed aan restricties, uitbreidingen en isometrische inbeddingen en zijn er nieuwe resultaten verkregen. De eigenschappen van een halfnorm ρ op een partieel geordende vectorruimte E die in veel opzichten een belangrijke rol spelen zijn wat we de Fremlin-eigenschap noemen (—y < χ < y => p(x) < p{y)), regulariteit (p{x) = m{{p(y): —y < χ < y}) en geslotenheid van E+. Een gedeelte van het belang van deze eigenschappen kan in de literatuur gevonden worden, met name waar het
vu Samenvatting gaat om operatoren. Deze tekst biedt een meer systematische behandeling, inclusief uitbreiding, isometrische inbedding, normvolledigheid en normcompleteringen. Hoofdstuk 1 begint met een overzicht van de basistheorie van lokaal gevulde en lokaal solide topologieën. Daarnaast wordt basistheorie van diverse soorten halfnor men ontwikkeld, ondermeer over quotiënten over kernen en afsluiting van positieve kegels. Bij de meeste resultaten worden (tegen)voorbeelden gegeven. Hoofdstuk 2 bestudeert de vraag of iedere monotone halfnorm op een par tieel geordende vectorruimte de restrictie is van een monotone halfnorm op een Rieszruimte. Het antwoord is ontkennend. Een voorwaarde voor een halfnorm ρ die nodig is voor uitbreidbaarheid tot een monotone halfnorm op een Rieszruimte is de monotone* eigenschap (—и < χ < υ, и, ν >0 => р{х) < р(и) +p(v)). Deze eigen schap is voldoende voor uitbreidbaarheid tot een monotone halfnorm op elke partieel geordende vectorruimte, dus ook elke Rieszruimte, waar de gegeven ruimte major erend in ligt (2.1). Fremlinhalfnormen op majorerende deelruimten kunnen altijd worden uitgebreid tot Fremlinhalfnormen op de hele ruimte (2.3). Voor de groot ste uitbreidingen worden expliciete formules gegeven en er is onderzocht wanneer normvolledigheid behouden blijft (2.12). Hoofdstuk 3 gaat over een zinvolle uitbreiding van het begrip 'Rieszhalfnorm' naar partieel geordende vectorruimten. De eerste aanpak gaat uit van halfnormen die de restrictie zijn van Rieszhalfnormen op de Dedekindcompletering. Er wordt een karakterisering van zulke normen gegeven (3.15, 3.29) op basis van een begrip 'solide verzameling' in partieel geordende vectorruimten (3.7). Er is weer een ex pliciete constructie van uitbreidingen. De verkregen eigenschap blijkt nogal zwak te zijn. Zo kunnen er twee halfnormen van dit soort zijn die niet overal gelijk zijn maar wel op de positieve kegel (3.38). Regulariteit is een sterkere eigenschap die het begrip Rieszhalfnorm op een zeer bruikbare manier generaliseert. Er worden stellingen gegeven over normvolledigheid die resultaten voor de r-norm in de the orie van ruimten van operatoren uitbreiden (3.46). De laatste paragraaf gaat over normcompleteringen (3.51, 3.55). Hoofdstuk 4 onderzoekt welke normen restrictie zijn van Riesznormen. Om dat elke partieel geordende vectorruimte isomorf is met een deelruimte van een Rieszruimte, is het een heel natuurlijke vraag welke normen op een partieel geor- dene vectorruimte inbedding in een genormeerde Rieszruimte mogelijk maken. Deze normen blijken de Fremlinnormen te zijn waarvoor de positieve kegel gesloten is (4.8). Een vergelijkbaar resultaat is verkregen voor restricties van M-normen (4.34). Onder extra voorwaarden kunnen de resultaten worden uitgebreid naar normen zon der gesloten positieve kegel (4.21) en halfnormen (4.28). Hoofdstuk 5 past de begrippen en resultaten uit de Hoofdstukken 1-4 toe in de
vin theorie van ruimten van operatoren. Veel van de resultaten bekend uit de literatuur worden verkregen als rechtstreekse generalisaties van resultaten voor ruimten van operatoren tussen genormeerde Rieszruimten (5.16). De theorie van Hoofdstuk 4 leidt tot een nieuw bewijs van Krein's lemma (5.3). Ieder hoofdstuk begint met een korte beschrijving van de inhoud en de belang rijkste resultaten. Er is een nabeschouwing met conclusies (in het Engels) achter Hoofdstuk 5.
IX χ Chapter 1
Seminorms and Topologies on Ordered Spaces
The main objective of this chapter is to introduce the basic concepts and results that will be used throughout the text. Many of the definitions and statements are well- known and can be found in almost any textbook on Banach lattices or topological partially ordered vector spaces. They are included for reference and to leave no room for ambiguity in meaning or notation. It is not intended to give a survey of the literature or to point out the origin of the results. Detailed references are included for the readers' convenience, only. There also are new concepts. They are presented in this chapter, because they will be needed at several places in the text, or simply because they fit well in the context of the glossary. The survey will treat partially ordered vector spaces, monotone seminorms and derived notions, locally full and locally solid topologies, closedness of positive cones, and dual spaces, including an elementary example of an order bounded linear func tion that is not the difference of two positive linear functions.
1.1 Partially Ordered Vector Spaces
This section lists definitions of notions that will be used freely throughout the text. For the standard terminology on Riesz spaces we refer to Aliprantis & Burkinshaw[2], Luxemburg & Zaanen[16], Zaanen[25], De Jonge &: van Rooij[12], and Vulikh[23], on Banach lattices to Meyer-Nieberg[17] and Schaefer[22], and on topological partially ordered vector spaces to Fremlin[8], Namioka[18], Peressini[20], and Schaefer[21]. For more information on partially ordered vector spaces and lattices in general, see Jameson[ll] and Birkhoff[4], respectively, and for more information on topological
1 Chapter 1. Seminorms and Topologies on Ordered Spaces
Ricsz spaces, sec Aliprantis & Burkinsha\v[l] and Fremlin[8]. Notice that all vector spaces appearing in this text are assumed to be vector spaces over the real numbers.
Definition 1.1 Let £ be a vector space.
(i) A partial ordering < on E is called a vector space ordering if
(a) x, у, ζ e Ε, χ < у =*· χ + ζ < у + г, (b) χ e E, 0 < χ, λ 6 [0, oo) =• 0 < λχ.
(E, <), or, less precisely, E is then called a partially ordered vector space. An element χ 6 E is called positive (negative) if 0 < χ (χ < 0, respectively). The set E+ consisting of all positive elements is called the positive cone of E. χ < у will also be written as у > χ.
Let < be a vector space ordering on E
(ii) E is called directed if for every x,y € E there exists ζ 6 E such that ζ > χ, у.
(ili) E is called integrally closed if for every x,y 6 E such that nx < у for all η € N one has that χ < 0.
(iv) £ is called Archimedean if for every x,y £ E such that nx < у for all η € Ζ one has that χ = 0
(ν) E is called a Riesz space if < is a lattice ordering on E, i e for every x,y ε E the set {x, j/} has a least upper bound (supiemum) and a largest lower bound (mfimum), denoted by χ V у and χ Л у, respectively. For χ € E, x+ .= χ V 0 is the positive part of x, x~ := (—x)+ the negative part, and |x| := χ V (-χ) is the absolute value of x. V, Λ, | |, .+, and .~ are called the lattice operations.
(vi) E is called Dedektnd complete (σ-Dedekind complete) if every nonempty set that is bounded above (and countable, respectively) has a supremum.
(vii) E has the decomposition property if for every x, j/i, j/2 € E+ with χ < yi + У2 there exist Xi,X2 € E+ such that χ = Xi + X2 and ii < j/i, I2 < У2-
(vni) E is called pre-Riesz if for every χ e E and every finite non-empty subset S of £• such that every upper bound of χ + S is an upper bound of 5 one has that χ is positive (see Van Haandel[10]).
Let F be a partially ordered vector space.
2 1.2. Seminorms on Ordered Spaces
(ix) A linear mapping A : E —> F is called positive if for every χ 6 Ε: χ > О =*· Αχ > 0. .4 is called bipositive if for every χ € Ε: χ > 0 «=>· Ла; > 0.
(χ) If £ and F are Riesz spaces, a linear map A : E —¥ F is called a Ліелг homomorphism if Л|і| = |Лх| for all χ Ç E. A linear subspace of a partially ordered vector space is a partially ordered vector space with respect to the inherited ordering. To stress that the inherited ordering is considered, it will sometimes be called an ordered subspace. Thus, every ordered subspace of a Riesz space is a partially ordered vector space. In this way all partially ordered vector spaces are obtained up to isomorphism: Theorem 1.2 Let E be a partially ordered vector space. Then E can linearly, bipositively be embedded in a Riesz space (i.e. there exist a Riesz space F and a linear, bipositive mapping г : E —> F). Proof. See Luxemburg[15].
It follows that an ordered subspace of a Riesz space need not be Riesz. An explicit example is the subspace Cl[0,1] of C[0,1]: the continuously differentiable functions in the space of continuous functions on the interval [0,1]. A subspace of a Riesz space is called a Riesz subspace if it is closed under the lattice operations. A subspace of a Riesz space may be a Riesz space in its own right, without being a Riesz subspace: Example 1.3 Л subspace of a Riesz space that is a Riesz space and not a Riesz subspace. Take E = C{0,1] and take for EQ the subspace of all affine functions. Then E and Eo are Riesz spaces and £Ό is an ordered subspace of E. EQ is not a Riesz subspace of Ε, since the absolute value in £Ό of 11-» 2i — 1 is 1, whereas the lattice operations in E are the pointwise operations. A linear subspace EQ of a partially ordered vector space E is called majorizing if for every χ € E there is а у 6 EQ with у > x.
1.2 Seminorms on Ordered Spaces
1.2.1 Monotone, Monotone*, and Fremlin Seminorms
An interesting theory of seminorms on ordered spaces requires some relations be tween the seminorm and the ordering. Two important and natural types of semi- norms on partially ordered vector spaces and Riesz spaces, respectively, are mono tone seminorms (i.e. seminorms that are increasing on the positive cone) and Riesz
3 Chapter 1. S«minorms and Topologies on Ordered Spaces scminorms (i.e. seminorms ρ that are monotone and satisfy p(|x|) = p(x) for all x). Restriction of such seminorms on Riesz spaces leads to two other concepts.
Lemma 1.4 Let E be a Riesz space with a seminorm p.
(i) If ρ is Riesz, then for every x,y € E such that —y (it) If ρ is monotone, then for every χ € E, and u, ν € E+ such that —u < χ < ν one has that p(x) < p(u) +p(v). Proof, (i) If -y < χ < y, then \x\ < \y\, so p(x) = p(\x\) < p(\y\) = p(y). (ii) If и, ν > 0 and — и < χ < ν, then 0 < x~ < ν and 0 < x+ < и, so p(x) < p(x~) +p(x+) < p(u) +p(v). The properties in the conclusions of the above lemma are inherited by restrictions to subspaces. They will show to be useful enough to get names. Definition 1.5 Let E be a partially ordered vector space with a seminorm p. (i) ρ is called monotone if for every x,y € E such that 0 < χ < у one has that p(x) < p(y), (or, equivalently, p(x) < p(x + y) for all x, у e E+). (ii) ρ is called monotone * if for every χ ζ E and u,v £ E+ such that — и < χ < ν one has that p(x) < p(u) + p{v). (iii) ρ is called Fremlin1 if for every x,y € E such that — у < χ < у one has that p(x) < p(y), (or, equivalently, p(x - y) < p(x + y) for all x,y 6 E+). (iv) If £ is a Riesz space, ρ is called Riesz if for every x,y 6 E such that |x| < \y\ one has that p(x) < p(y) (or, equivalently, ρ is monotone and p{\x\) = p(x) for all xeE). The implications in Proposition 1.6 are immediate. Proposition 1.6 Let E be a partially ordered vector space with a seminorm p. Then: ρ is Riesz ==>· ρ is Fremlin ==• ρ is monotone* =ϊ ρ is monotone, where the first implication only makes sense if E is a Riesz space. 4 1.2. Sem i norms on Ordered Spaces ./ \|a| /: •\ y ÄM\ / Figure 1.1: unit balls of monotone, Fremlin, and Riesz norms on R2 For illustration, consider the following examples. Unless stated otherwise, the or dering is pointwise. Example 1.7 (i) Let E = R2 and let ρ be a seminorm on E. The properties of Definition 1.5 have geometrical interpretations for the unit ball В of ρ (see fig. 1.1). For a, 6 6 R2 with a < b denote R(a,b) = [ai,&i] x [аг,^]· Ρ ÌS monotone if and only if R(a, b) С В for all a, b € В with 0 < a < b. ρ is Fremlin if and only if R(—a,a) С В for all α e В with a > 0. ρ is Riesz if and only if R(—\a\, \a\) С В for all a € B. (ii) Let E = ί°°(Ν), i.e. the space of all bounded sequences. The norm ι >-> ||z+||oo + llI_l|oo is monotone*, χ »-> ||z||oo + 15Z„2_"x„| is Fremlin, and χ »-> lklloo + E„2-nkn| is Riesz. (iii) Let E = C[0,1]. χ n· (ƒ i+) V (ƒ x~) is monotone*, χ ^ \\x\\i + |i(0) + i(l)| is Fremlin, χ ι-> ||x||i + |i(0)| is Riesz. (iv) Let E be a partially ordered vector space and let ƒ be a positive linear function on E. Then χ i-> | ƒ (ar)[ is a Fremlin seminorm on E. If E is a Riesz space, then χ н-у /(|i|) is a Riesz seminorm. Note that sums and suprema of monotone (or monotone*, or Fremlin, or, on a Riesz space, Riesz) seminorms are monotone (or monotone*, or Fremlin, or Riesz, respectively). Let us present some examples to show that the notions of Definition 1.5 are all distinct. 2 Example 1.8 (i) A Fremlin norm that is not Riesz: Take E = R and p(x\, i2) := 2 |ζι| + |χ2| + |χι + χ2|, (xi,i2)eR . 4 use D.H. FVemlin's name, because the place where I have first encountered this property in the literature was in his article[9]. It is not clear to me to whom honour is due. The property is discussed more extensively in earlier papers, e.g. Davies[7], Ng[19], Wickstead[24]. 5 Chapter 1. Seminorms and Topologies on Ordered Spaces (ii) A monotone* norm that іч not Fremlin: χ ι-+ ||i+||oo + II1 lloo °n R2· 2 (Hi) A norm that is not monotone: m |xi| V |xj — x2| on R . An example of a monotone norm that is not monotone* cannot be found on a Riesz space, because of Lemma 1.4. Example 1.9 A monotone norm that із not monotone*. Take E = R3, ordered by χ > 0 <=• X\, x2, хз > 0 and x3 < xj +x2 and take p(x) = max{|xi + x3|, |xt + 2x2 - 3x3|, |xi + 2x2 - хз|і \xi - 2x2 + хз\], 3 χ = (хі,х2,хэ) € R . Then £ is a directed partially ordered vector space and ρ is a norm on E. To show monotonicity, observe that if xi,x2,x3 > 0 and x3 < X\ + x2, then |xi-2x2+x3| < тах{хі+х3,хі+2х2-х3} and |xi+2x2-3x3| < max{ii+x3,Xi + — 2x2—x3}. Hence, if (xi,X2,x3) > 0 thenp(x) = max{xi+x3,xi + 2x2 ^з}· Thus, if x,y 6 E, x,y > 0, thenp(x+y) = тах{хі+х3+уі+у3,хі+2х2-хз+Уі+2у2-Уз} > max{xi + x3, ii + 2x2 - x3} = p{x), because yi + y3 > 0 and Уі+2у2-уз> 0. That ρ is not monotone* can be seen by taking y = (0,0,1), u = (1,0,0) and ν = (0,1,1). Then u, ν > 0 and — и < у < ν, whereas p(u) = 1, p(v) = 1 and p{y) = 3. Non-Archimedean spaces do not allow monotone norms: Proposition 1.10 Let E be a partially ordered vector space. If there exists a mono tone norm on E, then E is Archimedean. Proof. Let ρ be a monotone norm on E and let x, у e E be such that nx < у for all η e Ζ. Then 0 < nx + у < 2y for all η 6 Ν, so that np(x) < Zp(y) for all η e N. Therefore p(x) = 0, hence χ = 0 and E is Archimedean. 2 2 R with the ordering determined by the cone {(xi,x2) £ R : Xi,x2 > 0} U {(0,0)} and with Euclidean norm is an example of a partially ordered vector space with a monotone norm that is not integrally closed. 6 1.2. Seminorms on Ordered Spaces 1.2.2 Monotone Seminorms and Relatively Uniform Con vergence The notion of relatively uniform convergence in Riesz spaces and related terminol ogy (see e.g. Aliprantis & Burkinshaw[l],[2], Birkhoff[4], Luxemburg & Zaanen[16], Peressini[20], Schaefer[22]) can be extended to partially ordered vector spaces. Definition 1.11 Let E be a partially ordered vector space. Let (x,),g/ be a net in E and let χ € J?, (x,), is called relatively uniformly convergent to χ if there exists a vi € E such that for every ε > 0 there is an ¿o € / with — ew < x, — χ < ew for all г > ÌQ. (Χ,), is called uniformly Cauchy if there exists a w € E such that for every ε > 0 there is an i0 € I with — ew < x, — Xj < ew for all i, j > ÌQ. E is called uniformly complete if every uniformly Cauchy sequence is relatively uniformly convergent. Relatively uniform convergence implies convergence in every monotone seminorm: Proposition 1.12 Let E be a partially ordered vector space with a monotone semi- norm p. Let (ii)ie/ be a net in E and let χ € E. If (χ,), г* uniformly Cauchy, then it is Cauchy relative to p. If x, —>· χ relatively uniformly, then p(x, — x) —• 0. Proof. Observe that p(x¿ — x3) < 3ep(w) if —ew < x, — Xj < ew, or, in other words, 0 < x, — Xj + ew < lew. The remainder of the proof is straightforward. 1.2.3 Additivity on the Positive Cone and Full Unit Balls In Riesz space theory, two special types of Riesz seminorms, namely abstract L- and M-seminorms, play an important role. (A seminorm ρ on a Riesz space E is an L-seminorm if p(|x| + \y\) = p(x) + p(y) for all x,y € E and it is called an M-seminorm if p(|x| V \y\) = p(x) V p(y) for all x,y € E). Restriction of these seminorms to ordered subspaces leads to the notions of seminorms that are additive on the positive cone and, respectively, seminorms that have full unit balls. Definition 1.13 Let E be & partially ordered vector space. A subset S of E is called full if for all χ 6 E and y, ζ ζ S such that y < χ < z one has that χ € 5. For a set S С E, full 5 = {χ e E: there are y,z € S such that y < χ < z} is the smallest full set containing 5, called the full hull of S. (See Peressini[20]. Schaefer[21, p.215],[22] uses the term saturated instead of 'full'). The following is worth observing. 7 Chapter 1. Seminorms and Topologies on Ordered Spaces Lemma 1.14 Let E be a partially ordered vector space with a seminorm p. The following statements are equivalent: (i) ρ has full unit ball. (ii) Ifx,y,z€E are such that y < χ < ζ, then p(x) < p(y) Vp{z). Clearly, any restriction of an L-seminorm is additive on the positive cone. If ρ is an M-seminorm on a Riesz space E, then for y < χ < ζ one has that \x\ < \y\ V \z\, so p{x) < p(y) Vp(z); hence it has a full unit ball. These properties are inherited by restrictions. There is a connection with the concepts introduced in Definition 1.5. Proposition 1.15 Let E be a partially ordered vector space with a seminorm p. (i) If ρ has a full unit ball, then ρ is Fremlin. (ii) If ρ is additive on E+, then ρ is Fremlin. Proof, (i) Directly. (ii) If i, y € £ are such that — y < χ < y, then 0 < χ + у < 2χ and 0 < у - χ < 2х, so р(2х) =р(х + у-{у- х)) < р(х + у)+ р(у - х) = р{2у). Conversely, the Euclidean norm on R2 is a Riesz norm (hence Fremlin) but it is not additive on the positive cone and its unit ball is not full. The properties introduced via restrictions may seem somewhat arbitrary. In Chap ter 2 we will show that every monotone* seminorm on a directed partially ordered vector space is the restriction of a monotone seminorm on a larger Riesz space. In Chapter 4 we will show that every Fremlin norm which renders the positive cone closed is the restriction of a Riesz norm on a larger Riesz space. A similar result will be shown for a norm with a full unit ball. 1.2.4 Monotone Norms and Closedness of the Positive Cone Like monotonicity, a natural property for a norm on a partially ordered vector space is closedness of the positive cone. Some authors choose this property as the basic link between the ordering and the norm (e.g. Ordered topological vector space' in Schaefer[21], 'compatible topology' in Fremlin[8]). A seminorm with closed positive cone has to be be norm, because otherwise the cone would contain a linear subspace (see also 1.30). Thus for closedness of the positive cone it does not make sense to 8 1.2. Seminorms on Ordered Spaces consider a single seminorm that is not a norm. However, for collections of seminorms, and vector space topologies in general, closedness of the positive cone is very useful. There are no implications between it and the properties of Definition 1.5 except that in a normed Riesz space (i.e. a Riesz space with a Riesz norm) the positive cone is closed (see also Proposition 4.1). The norm in Example 1.8(iii) is not monotone and the positive cone is closed. Conversely, in R2 with Euclidean norm and the 2 ordering induced by the cone К := {(ii,i2) S R : X\,i2 > 0} U {(0,0)}, the norm is Fremlin while К is not closed. If the positive cone is not closed, what can be said about the ordering generated by the closure? Proposition 1.16 Let E be a partially ordered vector space with a monotone norm p. The closure E+ ofE+ is the positive cone of a vector space ordering on E and ρ is monotone for this ordering. Furthermore, if ρ is Fremlin, then ρ is E+-Fremlin, and if ρ has an E+-full unit ball, then it has an E+-full unit ball. If E is E+-directed, then E is E+ -directed. Proof. (See also Peressini[20, p.63, Cor.1.6], or Schaefer[21, p.216, Cor.3]). (a) E+ is a cone and E+ Π (—E+) = {0}: If х, у 6 E+, then there are sequences + + (xn)„ and (yn)n in E such that x„ —ν χ and yn —• y. Then xn + y„ Ç E for all + + η and x„ + yn —• χ + y, so χ + у e E . Similarly, if χ S E and λ € [0, oo), then Ax € E+. So E+ is a cone. If χ € E+ Π (—£+), then there are sequences (x„)„ and (y„)n in E+ such that x„ -¥ χ and yn -* -x. Then 0 < x„ < x„ + yn, so p(x„) < p{xn + Уп) -* 0, so that x„ —¥ 0 and therefore χ = 0. Thus, E+ is the positive cone of a vector space ordering. + + (b) ρ is E -monotone: If x, у € f?+, then there are (x„)n, (Уп)п in E such that x„ —> x, y„ —» y. Then p(x) = Iimp(xn) < limp(xn + j/„) = p(x + y). Similar arguments work for Fremlin norms and for norms with full unit balls, by 1.14. E+ Э E+, hence if E is £'+-directed, it surely is E+-directed. 1.2.5 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. 9 Chapter 1. Seminoms and Topologies on Ordered Spaces Proposition 1.17 Let E be a partially ordered vector space with a monotone semi- norm ρ and let N :— {i € E: p(x) = 0}. Let Es be the quotient space E/N and let q : E —> E\ be the quotient mapping. Then: (i) q(E+) is the positive cone of a vector space ordering on EN and q is positive with respect to this ordering. If E is directed, then so is EN- (it) The quotient norm ρ of Ey is monotone with respect to the q(E+)-ordering. If ρ is Fremlin, then so is p. If ρ has a full unit ball, then so has p. Proof. (See also Peressini[20, p.64, Prop.1.8] or Schaefer[21, p.250, Ex.3]). First, note that EN is a normed space and that q is a linear, surjective isometry. (i): Let К := q{E+). Clearly, A" is a cone with vertex at 0 and Кп(-К) D {0}. Let χ 6 EN be such that x, ~x € K. Then there are x, у 6 E+ with q(x) = -q(y) = x. q(x + y) = 0, so p(x + y) = 0, hence p(x) < p(x + y) = 0, so that χ € Ν, which means that χ = 0. Thus ΚΠ(-Κ) = {0}, yielding that К determines a vector space ordering on Ец. It is immediate that q is positive with respect to this ordering. Surjectivity and positivity of q yield that EN is directed if E is directed, (ii): Let x, у e EN be such that 0 < χ < у. Then there are x, и 6 E+ with q(x) = χ and q(u) = у - χ, hence q(x + u) =y, and therefore p{x) = p(x) < p(x + u) = p(y). Assume that ρ is Fremlin. Let x,y £ E^, then there are x,y € E+ with q(x) = S and q(y) = y. Then p(x — y)= p{x — y) < p(x + y) = p{S + y), so ρ is Fremlin. Assume that ρ has a full unit ball. Let x, y, ζ € EN be such that y < χ < ζ. Then there are y G E and и, υ & Ε+ such that q(y) = у, q(y + u) = χ, and q(y+u+v) = ζ, so that p(x) = p{y + u) < p(y) Vp(y + и + ν) = p(y) V p(z). Hence ρ has a full unit ball. The ç(£'+)-ordering on EN will be called the quotient ordering. The norm duals of E and EN are isomorphic, as will be shown in Proposition 1.55. It follows from Proposition 1.10 that EN is Archimedean, whether E is Archimedean or not. Many properties may be lost by the quotient construction. Indeed, even if £ is a Dedekind complete Riesz space, EN may not be integrally closed: Proposition 1.18 Let E be a directed partially ordered vector space with a mono tone norm p. Then there are a set S and a monotone seminorm po on CQO(S) such that the quotient ofcoo(S) with respect to the kernel ofpç equipped with the quotient ordering and quotient norm is isomorphic to {E,p). + Proof. Take S := {s e E : p(s) — 1}, and let E0 := coo(5), i.e. the space of x s s x all functions on S with finite support. Define j(x) := ^,,es ( ) ^ ^ Eo- Then j : E0 —> E is a positive, linear mapping, j is surjective, since E is directed. Let 10 1.2. Seminorms on Ordered Spaces Po(x) '•= p(j(x)), x ζ Ei). Then po is a seminorm on EQ and it is monotone, because j is positive and ρ is monotone. Let N := {x € E0: po(x) = 0} and let q : Eo -+ E0/N be the quotient mapping. If x,y € EQ are such that q(x) = q(y), then p0(x - y) = 0, so p(j(x - y)) = 0, hence j(x) = j(y). Therefore i(q(x)) := j(x), ζ 6 E0, defines a mapping from Eo/N to E. (a) г is linear, positive, and surjective: Immediate. + _1 (b) г is bipositive: Let χ 6 E0 be such that j(x) e £ \{0}. Then s := a j(x) e 5, where a := p(j(x)), so that j(x) = as = j(al[a)), which means that q(x) = + q(al(,)) 6 (E0/N) . Hence г is bipositive. x (c) г is isometrical: For χ e E0 one has that p(i(q(x))) = p(j( )) = Po(x), and po(x) is equal to the quotient norm of x. So г is isometrical. Thus, г is an isomorphism from E0/N to E. Proposition 1.17 states that the quotient construction preserves monotonicity, the Fremlin property, and the full unit ball property. It follows from the previous proposition that the monotone* property may be lost: Example 1.19 A Riesz space E with a monotone* seminorm ρ such that the quo tient norm of E/{x e E: p(x) = 0} is not monotone*. Take F = R3 with the norm and ordering of Example 1.9. Then F is a directed partially ordered vector space and the norm is monotone, but not monotone*. The previous proposition yields a set S and a monotone seminorm ρ on E := CQQ(S) such that E/{x 6 E: p(x) = 0} is isomorphic to F. ρ is monotone on the Riesz space E, hence monotone*. Throughout this text, terms such as norm completeness will be used for seminorms as well as for norms. In fact, a seminorm induces a uniformity and a topology (that may not be Hausdorff). Words like 'norm convergent', 'norm closed', and 'norm complete' refer to these structures. If a partially ordered vector space is norm complete with respect to a monotone seminorm, then the quotient with respect to the kernel is complete relative to the quotient norm. Norm completeness of the positive cone results in closedness of the cone in the quotient space. These assertions are immediate consequences of the following observation. Lemma 1.20 Let E be a vector space with a seminorm p. Let N := {x € E: p(x) = 0}, let q : E —• Ε /Ν be the quotient mapping, and ¡et ρ be the quotient norm. Let S be a subset of E. Then S is p-complete if and only if q(S) is p-complete. 11 Chapter 1. Seminorms and Topologies on Ordered Spaces Proof. Let (x„)n be a sequence in S and let χ € S. Then (x„)„ is p-Cauchy if and х only if (q(xn))„ is p-Cauchy, and p(xn - x) -* 0 if and only if p{q(xn) — ч( )) ~~• 0· Tims the assertion follows. Proposition 1.21 Let E be a partially ordered vector space mth a monotone semi- norm p. Let N := {x € E: p(x) = 0}. Then: (%) E is p-complete if and only if E/N is p-complete. (%i) E+ is p-complete if and only if (E/N)+ is p-complete. (in) If E+ is p-complete, then (E/N)+ is closed m E/N. If {E/N)+ is closed m E/N and E is p-complete, then E+ is p-complete. Proof, (i): Apply the lemma to 5 = E. (ii): Apply the lemma to E+ and note that (E/N)+ = q{E+) by definition of the ordering on E/N. (iii). If E+ is p-complete, then (E/N)+ is p-complete, hence closed in the normed space E/N. If (E/N)+ is closed in E/N and E/N is p-complete, then {E/N)+ is p-complete, so that E+ is p-complete. 1.3 Locally Full Topologies For the types of seminorm defined in §1.2, the topology generated by a collection of such seminorms can be studied. It turns out that the topologies generated by collections of monotone, monotone*, Fremlin seminorms, or seminorms with full unit balls are the same. A collection of Riesz seminorms generates a different topology. The former will be studied in this section, the latter in §1.5. Lemma 1.22 Let E be a partially ordered vector space with a monotone seminorm p. Define q(x) := inf {p(y) V p(z) : y,z e E such that y < χ < ζ}, χ e E. Then q is a seminorm with a full unit ball and q < ρ < 3q. Proof. Let Xi,i2 € E. If y\,yi,z\,zi ε E are such that y\ < x\ < Z\ and У2<Х2< 22, then j/i+î/2 < Z1+I2 < zl+z2, so q{xl+x2) < р(уі+У2) р(г1+г2) < р(Уі) р(г1)+р(у2) р(22). Hence, q(x1+x2) < q{x1)+q{x2). Clearly, q(Xx) = Xq{x) and q[-x) = q(x) for all a; e E and λ e [0, oo). So g is a seminorm. 12 1.3. Locally Full Topologies Let ι,ι/, ζ € E bc such that y < χ < ζ. If и, υ € E are such that u < у and ζ < ν, then и < χ < υ, so 9(1) < q(u) V ç(v). Thus, q(x) < q(y) V д(г). Hence q has a full unit ball. Obviously, q < p. Ux,y,z e E are such that у < χ < ζ, then 0 < x-y < z-y, so p(x) < p(x -y)+ p(y) < p{z -y)+ p(y) < 3 (p(y) V p(z)). Hence p(x) < 3q{x). Thus, every monotone seminorm is equivalent to a seminorm with a full unit ball. Then the same is true for every monotone* or Fremlin seminorm. Hence, a topology generated by a collection of Fremlin, monotone*, or monotone seminorms is also generated by a collection of seminorms with full unit balls. For a brief while, let us broaden our scope and consider vector space topologies that are not necessarily locally convex. Definition 1.23 Let E be a partially ordered vector space and r a vector space topology (not necessarily Hausdorff ) on Ε. τ is called locally full if it has a neigh bourhood base of 0 consisting of full sets. If U is a full neighbourhood of 0, then so is U Π (—U) and it is contained in U. Therefore, every locally full vector space topology has a neighbourhood base of 0 consisting of balanced, full sets. (A set S С E is called balanced if —5 = S). Locally full vector space topologies have been studied extensively, see e.g. Jameson[ll], Kelly & Namioka[14], Namioka[18], Peressini[20], Schaefer[21]. Some authors prefer to see locally fullness as a property of the ordering instead of the topology. They call the positive cone normal if it makes, in our terminology, the topology locally full (e.g. in [14], [20], [21]). The following lemma implies that a locally convex and locally full topology has a neighbourhood base of 0 consisting of sets that are both full and convex. Lemma 1.24 Let E be a partially ordered vector space. The full hull of a convex set in E is convex. Proof. Let 5 be a convex set in E. Let x, у e full S and λ € [0,1]. There are ^11^2.2/1,3/2 € S such that ii < χ < x-¿ and y\ < у < j/2· Then \x\ + (1 — A)j/i < λχ + (1 — X)y < A12 + (1 — λ)ι/2 and Ax¿ + (1 — A)y, € 5, i = 1,2, because S is convex, so Xx + (1 — A)y 6 full 5. Hence full 5 is convex. Example 1.25 A full set in a Riesz space with a convex hull that is not full. Take E = R3 and S := {(0,0,0), (-1,2,1), (2, -1,1)}. Then S is full and the convex hull of S contains y := (0,0,0) and ζ := (1/2,1/2,1). The element χ := (1/2,1/2,0) 13 Chapter 1 Seminorms and Topologies on Ordered Spaces satisfies y < χ < г and it is not contained in the convex hull со 5, because for every element ш со S the third coordinate is the sum of the other two Hence co S is not full Combination of Lemmas 1 22, 1 24, and Proposition 1 15 yields Proposition 1.26 Let E be a partially ordered vector space with a vector space topology τ The following statements are equivalent (ι) τ is locally convex and locally full (u) τ is generated by a collection of monotone seminorms (ui) τ is generated by a collection of seminorms with full unit balL· Results similar to the above can be found in almost every book dealing with topolo gies on ordered spaces (see e g Namioka[18, Thm 6 7], Peressim[20, ρ 63, Prop 1 7] or Schaefer[21, ρ 215, 3 1]) The same can be said about the following important property of locally full topologies (see e g Peressini[20, ρ 62, Prop 1 3]) Proposition 1.27 (Sandwich Theorem) Let E be a partially ordered vector space with a vector space topology The following statements are equivalent (ι) τ is locally full (n) If (Xi),ci, (yt)ie/, and (z,),e/ are nets m E such that yt < x, < z% for all ι and yt ->0, г,-> 0, then xt —> 0 Proof. (i)=>(n) τ has a neighbourhood base U of 0 consisting of full sets Let t/ € W For large г one has that y„ г, 6 U, so, x, 6 U, because U is full Hence i, -»• 0 (ιι)=>·(ι) Suppose that r is not locally full Then there is a neighbourhood V of 0 that contains no full neigbourhoods Let U be the collection of all neighbourhoods of 0 contained in V Let ω e U If for all y,z e ω and χ € E such that y < χ < ζ one would have χ € V, then the full hull of ω would be contained in V, contradicting the assumption of V, hence there are yu, zu 6 ω and χω € E\ V with yu < χω < ζω With U directed under inclusion {u)\ <ω2«ωι3 ыг), the nets {уш)шси and {гш)иеи converge to 0, whereas (хи)шеи does not, contradicting (и) Example 1.28 A Riesz space with a norm that is not equivalent to a monotone norm Take E = Coo(N*) and p(x) = ΕΓ=ιλ_ΊΙ(Α:) ~ Φ + 1)1. x £ E Then E is a Riesz space and ρ is a norm on E Let xn = l{i, іП}, η e Ν* Then ρ(χη) = l/n —• 0 and 0 < χι < хп for all η, but χι φ 0 14 1.4. Topologies such that the Positive Cone is Closed 1.4 Topologies such that the Positive Cone is Closed Let us summarize some elementary results for topologies with closed positive cones (see Fremlin[8]. Schaefer[21, Ch.V,§4], Namioka[18]). Lemma 1.29 Let E be a partially ordered vector space with a vector space topology т. The following statements are equivalent: (ι) E+ is closed. an are ne s m (n) If (i,)ie/ d (Уг)г€і ^ E and x,y € E are such that x, —¥ x, yt —• y, and x, < y, for all ι, then χ < y. Proposition 1.30 Let E be a partially ordered vector space and let τ be a vector space topology on E. If E+ is closed, then τ is Hausdorff. Proof. (See Fremlin[8, p.37, Prop.21B]). Suppose that r is not Hausdorff. Then there is a non-trivial subspace that is adherent to 0 and therefore contained in E+. This contradicts the axioms of a partial ordering. Proposition 1.31 Let E be a partially ordered vector space. If there is a vector space topology on E such that E+ is sequentially closed, then E is integrally closed. Proof. (See also Schaefer[21, p.222, 4.1] and Fremlin[8, p.37, Prop.21B]). Let x,y € E be such that nx < y for all η 6 N*. Then n~ly — χ e E+ for all η and n~ly — χ -¥ —i, so —x e E+, hence χ < 0. Thus, it is proved that E is integrally closed. Example 1.32 A directed, integrally closed partially ordered vector space with a monotone norm such that E+ is not closed. Let E be the subspace of /°°(N) consisting of linear combinations of «o := (-1,0,1,1,1,...), Un := (1,ì,i,...,1,0,0,...) = 1{0) + Ì1{1, ,n}, η>1, and let p(x) := 8υρλ>1 |x(A:)|, χ 6 E0. (This is not Ц.Ц»!). (a) E is directed: щ + 2ui = (1,2,1,2_1,3_1,...) is a strong unit in E. (b) ρ is a monotone norm on Ε: ρ is clearly a monotone seminorm. Let χ € E be such that p(x) = 0. Then χ = X0u0 + Х\Щ + 1- A„un for certain AQ, ..., λη € R. 15 Chapter 1. Seminoms and Topologies on Ordered Spaces x(n + 1) = 0, so Λ0 = 0; χ(ή) = 0, so that A„ = 0, and then by induction λ„_ι = • · • = Αι = 0. Therefore χ = 0, hence ρ is a norm. (e) E is integrally closed: E is an ordered subspace of the integrally closed space Iх (N). + + 1 (d) E is not norm closed: щ + un € E for all η and p(un) = n" —> 0, so + щ + ип —>щ, whereas щ &. E . An example of a Riesz space E with a norm ρ such that E+ is closed and ρ is not equivalent to a monotone norm is postponed to §1.5 (Examples 1.41 and 1.42). 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 1.16. Proposition 1.33 Let E be a partially ordered vector space with a locally full Haus- dorff vector space topology т. The closure E+ of E+ is the positive cone of a vector space ordering and τ is locally full for this ordering. Proof. (See also Peressini[20, p.63, Cor.1.6] or Schaefer[21, p.126, Cor.3]). Let U be a neighbourhood base of 0. To show that E+ is a cone, let x, у 6 E+. For every + + ω 6 U there are χω, уш e E with χω—χ, уы—у 6 ω. Then хы+уш € Е for all ω and Хи + Уш —> х + У, hence х + у € Е+. By a similar argument for scalar multiplication, it follows that E+ is a cone with its vertex at 0. If χ e E+ П (—£+), then for every + ω e U there are χω, уш e E with χω, -уш € χ + ω. Then 0 < χω < χω + уш -¥ 0, so that χω —> 0, because τ is locally full. Since τ is Hausdorff, it follows that χ = 0. Thus, E+ is the positive cone of a vector space ordering on E. To show that τ is locally full for the i?+-ordering, let U be a E+-full neighbour hood of 0. Choose a full, balanced neighbourhood V of 0 such that V+V CU. Let V' be the £+-full hull of V. We prove V' cU. Letx€E and и, е Ъе such that r+ + и < χ < υ in £ -sense. There are nets (s,),e/ and (ij);gj· in E with s,4i-ti + and t3 -> ν -x. Then χ — s, < χ < χ + i, in £' -sense and x-u — s„ t}— v + x €. V for large ι and j, so χ — s„ χ + tj 6 V + V С U for large г and j. Thus χ e U. It follows that V' С U. 1.5 Locally Solid Topologies On a partially ordered vector space, local fullness is the topological generaliza tion of monotonicity. On a Riesz space, topologies generated by collections of Riesz seminorms lead to the notion of locally solid topologies (see Aliprantis ii Burkinshaw[l],[2], Fremlin[8], Namioka[18], Peressini[20], Schaefer[21]). 16 1.5. Locally Solid Topologies Definition 1.34 Let E be a Riesz space. A subset 5 of £ is called solid if for all χ ζ E and y € S such that |x| < |y| one has that χ € S. For a set S С E, sol S = {χ g E : there exists y € S such that \x\ < \y\} is the smallest solid set containing S, called the solid hull of S. A vector space topology on E is called locally solid if there exists a neighbourhood base of 0 consisting of solid sets. Lemma 1.35 Let E be a Riesz space with a seminorm p. Then ρ is R%esz if and only if the unit ball of ρ is solid. Note that every solid set is balanced. As a counterpart of lemma 1.24 one has: Lemma 1.36 Let E be a Riesz space. The convex hull of a solid set in E is solid. Proof. (See Aliprantis & Burkinshaw[l, p.4, Thm.1.3], Fremlin[8, p.14, Lem.l4J(d)], or Peressini[20, p.101, Prop.4.1]). First, we derive a generalization of the decomposition property in Riesz spaces: Ifχ € E and уі,...,у„ € E+ are such that \x\ < yi+.. .+yn, thenx = xH \-xn for certainxb... ,xn € E with |x,| < y¿, + i = 1,..., п. Indeed, for such χ and y„ x < \x\ < yi Η 1- yn, so by the standard + + decomposition property there are щ,..., u„ € E such that i = щ + • • • + un + and u, < y„ г — 1,...,п. Then x~ = |x| — x < (y\ — щ) + 1- (yn — w„), so, + again by decomposition, there are v\,... ,vn e E such that x~ = v\ + • • • + vn and v, < yt — Ui, г = 1,..., п. Let χ, := u, — vlt then χ = Χι + • · • + xn and |x,| < u, + w, < y„ » = l,...,n. Let now 5 be a solid set in E. Let χ & E and у € со S be such that |x| < \y]. There are Уі,...,уп 6 S and λι,..., A„ 6 (0,1], λι + · • • + λ„ = 1, such that У = λιΐ/i + 1- Xny„, thus |x| < |Aij/i| + 1- |Aij/i|. Then by the first part of the proof, there are x\,...,xn 6 E with χ = x\ Η l· xn and |x,| < \X,y,\. Then ΙΑ,-1^·! < \Уг\, so A,-1x¿ e S, because S is solid, and therfore χ 6 coS. Figure 1.2 shows that the solid hull of a convex set need not be convex. The above lemma yields that every locally convex, locally solid topology has a neighbourhood base of 0 consisting of convex, solid sets. Hence, by lemma 1.35: Proposition 1.37 Let E be a Riesz space with a vector space topology т. The following statements are equivalent: (i) τ is locally convex and locally solid (ii) τ is generated by a collection of Riesz semmorms. 17 Chapter 1. Seminorms and Topologies on Ordered Spaces Figure 1.2: square ABCD is convex and balanced and its solid hull is not convex (See Aliprantis к Burkinshaw[l, p.38, Thm.6.1], Namioka[18, Thm.8.4], or Peressini[20, p.105, Prop.4.10], also Fremlin[8, p.43, Exerc.22G(c)]). Proposition 1.38 Let E be a Riesz space with a locally full vector space topology т. Then the lattice operations are continuous if and only if τ is locally solid. Proof. (See Aliprantis L· Burkinshaw[l, p.34, Thm.5.2], Namioka[18, Thm.8.1], Peressini[20, p.104, Prop.4.7], Schaefer[21, p.234, 7.1], also Fremlin[8, p.38, Prop.22B(b)]). =•) Let U be a balanced, full neighbourhood of 0. χ <-¥ |x| is continuous, so there is a full neighbourhood V of 0 such that {|x|: χ € V} с U. Let V' be the solid hull of V. Then V D V' and for у 6 V one has \y\ 6 U, so -\y\ e U and then χ 6 U for all χ e E with -\y\ < χ < \y\, hence V' С U. <=) See Peressini[20, p.104, Prop.4.7]. Not every solid set is full: the unit disk D in R2 is solid and not full; indeed, (0,-1) < (1,-1) < (1,0) and (0,-1), (1,0) e D, but (1,-1) g D. However: Proposition 1.39 Let E be a Riesz space. Every locally solid vector space topology on E is locally full. Proof. (See also Peressini[20, p.104, Prop.4.7] or Schaefer[21, p.234, 7.1]). Let τ be a locally solid vector space topology on E and let U be a neighbourhood base of 0 consisting of solid sets. Let V £ U. Then there is U e li such that U + U CV. Let U' be the full hull of U. Then U' D U, so U' is a neighbourhood of 0. If χ € £ is such that y < χ < ζ for some y, ζ e U, then |x| < |i/|v|z| < |ì/| + |z| e U + U С V, so χ 6 V. Hence U' С V. Thus, τ has a neighbourhood base of 0 consisting of full sets. Locally full topologies need not be locally solid, not even normable ones. 18 1.5. Locally Solid Topologies Example 1.40 A Riesz space E with a monotone norm ρ that is not equivalent to a Riesz norm. Take E = см (Ν*) and p(x) := ||x||œ + | Σ™=1 x(k)\, χ 6 E. For η 6 Ν* let xn(Jfc) := (-1)*1{1 2n}(fc),fc € Ν*. Then p{xn) = 1/n -> 0, whereas p(|x„|) = 2 + 1/n > 2 for all n. In the above example, the norm ρ is bracketed between two Riesz norms. The following, more complicated, example presents a different situation. Example 1.41 A Riesz space E with a monotone norm ρ that does not majorize a nonzero Riesz seminorm, such that E+ is closed. -n n Define the subintervals In¿ '·= [(к—l)2 , k2~ ) of [0,1] and their indicator functions u„,t := !ƒ„„, к € {1,...,2"}, η e Ν*. Take for the space E the linear hull in Ш°-У of {u„¿: к = 1,..., 2",η e Ν*}, with pointwise ordering. Then £ is a Riesz space. Define for χ € E: ƒ„,*(*) := / x{t)un,k{t)dt, Jo and take p(i) := sup{|/„,t(x)| : Ы,...,п,пе№}. {χ ι-» |/n,*(^)|}n,«: is a separating collection of monotone seminorms on E, so ρ is a monotone norm. Let ρ be a Riesz seminorm on E such that ρ 2" к 1п:=^(-1) -'Чк, n€N\ m m Then fmJ{xn) = 0 for all m < η, Í = 1,..., 2 , and \fm,iM\ < ƒ„ um,, < 2" < m n m 2-" for all m > η, I = 1,..., 2 . Thus, |/m^(i„)| < 2" for all/ € {1,..., 2 } and n m,n € №, hence p(xn) < 2~ , n G Ν*. |x„| = 1 for all n, so p(l) = p(|x„|) = PÍ^n) < p(in) —• 0, so ρ = 0, because 1 is a strong unit in E. Remains to show that E+ is closed. Let a sequence (x¿)¿ in E+ and χ & E be such that p{xi - x) ->· 0. Suppose χ JÍ 0. Then there are η e Ν*,fc e {1,...,2"}, and a € (0, оо) with x(t) < -a for all t € ƒ„,*. Then /„,*(ХІ -Χ) = f¿ fa -x)un¡k > n - J0 xun¡k > <*|Л»,к| = 2~"a, sop(ii—x) > |/„(x¡— x)| > 2~ a forali i, contradicting p(xj — x) —> 0. Henee χ 6 £+. It follows that E+ is closed. Example 1.42 A Riesz space E with a norm ρ such that χ ι-+ |x| is continuous but not uniformly continuous. Then ρ is not equivalent to a Riesz norm and therefore not equivalent to a monotone norm. 19 Chapter 1. Seminorms and Topologies on Ordered Spaces Take E = Coo(N') and OO p(x) = Σ (И*) - x(k + 1)| + 2-*|x(*)|) , χ E E. к-1 Ε is Ά Riesz space and ρ is a norm on E. (a) | | is continuous- Let x Ç. E. Then there is an η € N such that x{k) = 0 for all к > п. For у € E one has (for the sake of readability, put x* := x{k), etc.): P(IVI-W) = ¿(|l»*l-N-|y*+il + kt+il|+2-*||»t|-|ifc||) + *=i oo + E (lwb-y*+il + 2-*lv*l)< fc=n+l < ¿ (\Ук - Xk\ + \Vk+i - H+il + 2~к\Ук - Xk\) + /t=i oo + Σ (іУ*-У*+іІ+2"*Ы)< *=n+fc=n+ll n+1 oo k < (2^ + l)^2- \yk-xk\+ ¿ (|yfc-yt+1| + 2-*|»|)< < 2(2n+1 + l)p(y - JE). Hence |.| is continuous at the point x. a (b) |.| is not umformly continuous: Let xn := 1{і,з, ,2n+i} nd for δ € (0,1) let Уп,6 ••= (1 - ¿)1{1,3, .,2n+l} - ¿1(2,4, ,2η}, η e Ν*. Then Χη - y„,í = ¿1{ι,. ,2„+1}, so 1 P(xn - yn,s) < (1+E^t 2-*)i < 2J for all δ and п. \xn\ - \y„,s\ = ¿(1{ІД ,2n+1} - 1{2,4, ,2η}), SO p(|x„| — |Уп,*|) •> 4nJ for all J and n. Thus, for ε > 0 there is no δ > 0 — such that for every xn one has that p(xn — y) < S implies that p(|xn| |j/|) < ε; in other words, |.| is not uniformly continuous. Then | | is surely not Lipschitz continuous and therefore ρ is not equivalent to a Riesz norm By Proposition 1.38, it follows that ρ is not equivalent to a monotone norm. If E is a Riesz space with a monotone norm p, then р:іи p(|x|) is a Riesz norm on E and p(x) < p(x+) +p(x~) < 2p(|x|) = 2p(x) for all χ e E, so ρ is stronger than p. Similarly, for topologies: Lemma 1.43 Let E be a Riesz space with a locally full vector space topology r. Let ƒ (x) = |x|, χ € E, and let τ' be the vector space topology generated by {f~l{U) : U e τ). Then τ1 is the coarsest locally solid vector space topology finer than τ, and for 20 1.6. Dual Spaces every τ-neighbourhood base U of 0, the collection {ƒ ' (U) :UçU}is neighbourhood base of 0 for τ'. If τ is Hausdorff, so is τ'. Proof. Let U be the collection of all balanced, full r-neighbourhoods of 0 and let V := {f~l(U): U € U). Then V is a directed collection of subsets. Let U e U. Then /_1(i/) is a solid, balanced set containing 0. It is absorbing, because |x| is absorbed by U, for every χ € E. Choose a τ-neighbourhood V of 0 such that V + V с U. Let l V' := /~ (V). If x,y € V', then |x|, |V| € V, so 0 < \x + y\ < \x\ + \y\ € V + V С U, hence |i + y\ e U, yielding that χ + у € }~1{U). So V' + V' с f'l{U). Thus, V is a neighbourhood base of 0 for a locally solid vector space topology. Because li is a neighbourhood base of 0 for r, the topology generated by V is precisely τ'. Clearly', the inverse images under ƒ of the sets in any r-neighbourhood base of 0 constitute a τ'-neighbourhood base of 0. For any balanced, full set U one has f~l(U) С U, because if χ 6 /-1(^)> tnen \x\,— \x\ € U and —|x| < χ < \x\, so ι € U. Hence τ' is finer than r. If τ is Hausdorff, this implies that r' is Hausdorff, too. If σ is a locally solid vector space topology finer than r, then for any balanced, full r-neighbourhood U of 0, there is an S e σ balanced and solid with 0 € S С U. Then S = /_1(5) С f~l{U). Thus, a is finer than τ'. 1.6 Dual Spaces Similar to topological structures, ordering structures give rise to dual spaces. Let us begin with notations and terminology. Denote for a vector space E the algebraic dual (i.e. the vector space of all linear functions from E to R) by E*. For a directed partially ordered vector space E, the cone of all positive linear functions generates a vector space ordering on Ε*, coinciding with the pointwise ordering on E+. Unless otherwise stated, we will consider this ordering on E*. Denote the topological dual (i.e. the subspace of E* consisting of all continuous functions) of a topological vector space E by E'. If E is also a partially ordered vector space, E'+ is the set of all continuous linear functions that are positive. If E is directed, then E'+ is the positive cone of E' with the ordering inherited from E*. For a topology generated by a seminorm p, the topological dual is also called the norm dual, and it is equipped with the operator norm: ƒ .-> Il/ll = sup{|/(x)|: p(x) < 1}. Definition 1.44 Let E be a partially ordered vector space. The subsets of the form [a, b] := {x € E : a < χ < b}, a, b € E, a < b, are called the order intervals 21 Chapter 1. Seminoms and Topologies on Ordered Spaces of E. A subset of E is called order bounded if it is contained in an order interval. A function ƒ 6 E* is called order bounded if it is bounded on every order bounded set (or, equivalently, on every order interval). The order dual of E, denoted by E~, is the subspace of E* consisting of all order bounded functions. E~ is a partially ordered vector space if E is directed. Every positive linear function on a partially ordered vector space E is order bounded. It is a well-known fact that not every order bounded linear function is equal to the difference of two positive linear functions. However, examples do not seem to abound in the literature (see Namioka[18, p.33, Ex.6.10]). Next, an example is given that is fairly elementary. Example 1.45 A directed, integrally closed partially ordered vector space E and an order bounded linear function on E that is not equal to the difference of two positive linear functions. Consider the following functions on [0, oo) (see fig. 1.3): en := l[n,n+i)i η € Ν, «„,*(') := «%_!](*) +il{B+_Lj(í), «€[0,оо),п,*€№. "3.4 F-= 1 1 Η 1 Figure 1.3: the functions unik Take for E the subspace of /°°[0, cc) consisting of all linear combinations of {e„}„ and {un,*}n,*· Every element χ of E is right differentiable at 0. Let f(x) be its right derivative: f(x) := x'(0), ι € E. Then £ is a partially ordered vector space and ƒ is a linear function on E. (a) E is directed and integrally closed: Every element of E is bounded and has a bounded support, so for every x,y e E there is an η e N such that x, у < n(ei Η h e„), so that E is directed. E is integrally closed, since it is a subspace of an integrally closed space. (b) ƒ is order bounded: Let a £ E+. It has to be shown that ƒ is bounded on [0, a]. There are ΛΓ e Ν* and С € (Ο,οο) such that suppo с [0,JV] and α < Cl. Let χ e [0, α]. Then also suppi С [0, ΛΓ] and χ < Cl. By definition of E, there are {К,к)п,к € coo (Ν*2) and (μ„)„ € Coo(N) such that x = Ση,* K¿Un,k + En №„• 22 1.6. Dual Spaces г χ = Ü on (Л , oo), so μη = μηβη(η + 3/4) = 0, η > Ν, and λ„,* = kXn¡kUntk(n + (к + I)'1) = 0, η > Ν, к € Ν*. It follows that χ is linear on [0,l/N]. Because 0 < x < CI, this yields that |/(x)| = |x'(0)| < NC. Hence |Дх)| < NC for all χ € [0, о]. This establishes that ƒ is order bounded. (c) There is no positive linear function g on E with g > f, hence f is not the difference of two positive linear functions: Suppose that there exists such a function g. Let kn e N be such that k„ > g(en), η e Ν*. Then k^gfa) < 1 and u„,fc < 1 1 eo+fc-'en.foralln, sog(e0) = д{ей+к- еп)-к- д{е„) > g(unjkn)-l > /(u„,jtj-l = n— 1 for all n, which is a contradiction. Thus, such a g does not exist, and therefore ƒ is not the difference of two positive linear functions. Even for a Riesz space, the order dual may be trivial. Example 1.46 A Riesz space E such that E~ = {0}. (See Namioka[18, p.25, Ex.5.7(3)] or Zaanen[25, p.132, Ex.85.2]). Take E = Щ0, l], for any ρ € (0,1). From the theory of topological vector spaces it is known that 0 is the only continuous linear function on E. We show that every positive linear function on E is continuous, then it follows that E~ = {0}, because E~ is a Riesz space. The sets {x e E: ƒ |x|p < 1/n}, η e Ν*, constitute a neighbourhood base of 0, so the topology of E is locally solid. Let ƒ € E~+. Suppose that ƒ is not continuous. Then there are (i„)„ in E with /(x„) —• oo and ƒ |x„|p < 2~", or, in other words, n 1 1 || \xn\ ||i < 2" for all n. Let у := £n |i„|»' 6 L and χ := у /" e E. Then χ > |χ„| χ for all η, so f (χ) > /(|xn|) > ί{ η) —* °°¡ contradiction. Hence ƒ is continuous. There are many results about relations between positivity, continuity, and order boundedness of linear functions. Proposition 1.47 Let E be a partially ordered vector space with о vector space topology such that every order bounded set is bounded. Then E' С E~. Proof. Let ƒ € E'. Then ƒ is bounded on every bounded set. Every order bounded set is bounded, so ƒ is bounded on every order bounded set, hence ƒ e E~. In particular, E' С E~ for locally full spaces, because: Lemma 1.48 In a partially ordered vector space with a locally full vector space topology, every order bounded set is bounded. (See also Peressini[20, p.62, Prop. 1.4] or Schaefer[21, p.216, Cor.2]). 23 Chapter 1 Seminorms and Topologies on Ordered Spaces Example 1.49 A Riesz space with a norm that is not equivalent to a monotone norm (so its topology is not locally full) such that every order bounded set is bounded. Take, as in Example 1.28, E = Coo and p(x) := Σΐ=ι к~1\х{к) - x(k + 1)|. It suffices to show that [—a, a] is bounded for every a e E+. Let a € E+. Then there is an η € N such that a(k) = 0 for all к > n. Then for any χ 6 [—α, α] one has P(x) < Ylk=i[a(^) + a(k + 1)]· so P([~aia]) ls bounded. Example 1.50 A Riesz space with a norm and a continuous linear function that is not order bounded. Take E = R2 with lexicographical ordering and Euclidean norm. Then the vertical line V = {(0,t)-t e Щ is contained in [(-1,0), (1,0)], hence order bounded. The linear function (11,12) ·-• %г is continuous, but not bounded on V, hence not order bounded. If E is a Riesz space, then E~ is a Dedekind complete Riesz space. For locally solid vector space topologies, the positive part of a continuous linear function is continuous as well: Proposition 1.51 Let E be a Riesz space with a locally solid topology. Then E' is a Riesz ideal in E~. Proof. (See Aliprantis к Burkinshaw[l, p.36, Thm.5.7],[2, p.135, Thm.11.5], Fremlin[8, p.40, Prop.22D], Peressini[20, p.108, Prop.4.17]). Let ƒ e E'. There is a solid neighbourhood U of 0 such that ƒ is bounded on U. f+ is given by /+(i) = sup{/(u) : 0 < u < χ}, χ 6 E+, so ƒ+ is bounded on U+ = U Π E+. Because U is solid, |i| e U for every χ 6 U, so from |/+(x)| < /+(|x|) it follows that /+ is bounded on U, hence continuous. If g € E~ is such that |p| < ƒ+ in E~, then \g(x)\ < \g\(\x\) < f+(\x\), so g is bounded on U, hence continuous. Thus, E' is an Riesz ideal in E~. Example 1.52 A Riesz space E with a monotone norm ρ such that E' is not a Riesz space. We try to obtain that E' resembles a space of differentiable functions, then there exist functions with their pointwise positive part not in that space. Let J: L°°[0,1] —• L°°[0,1] be the primitivation operator: {Jw)(t) = f*w(s)ds, t e [0,1], w € L°°; J is linear and positive. Can we choose E and ρ such that {E,p)' = J2(Z.°°)? Take E = Ll[Q,l], define (Hx)(s) := fix(t)dt, s € [0,1], χ e E, and take p(x) := ||#2x||i, χ 6 E. Because H:E —• Ll is positive, linear, and injective, ρ is a monotone norm on E. 24 1.6 Dual Spaces (a) E' is isomorphic to J2(L°°) as partially ordered vector spaces: Define for w € L°°: i{J2w)(x) ·= f {J2w)(t)x(t)dt = f w{t){H2x)(t)dt, χ(ΞΕ. Jo Jo The last expression shows that i^w) e E', from the other it can be seen that ι is a linear, bipositive, injective mapping from J^L00) to E'. To prove surjectivity, let ƒ 6 E'. Because Я is injective, H2x i-+ f{x), x ε E, defines a continuous, linear function on H2(E) с Ll. H2(E) is dense in Lx (because it contains all twice differentiable functions ι with i(0) = i'(0) = 0), so this function uniquely extends to an ƒ € Ll[0,1]'. Then, by standard ¿''-space theory, there is a unique w e L°° such that f(y) = f wy for all у € Ll. In particular, f(x) = f(H2x) = t(J2w)(x), χ € E. Thus, it is proved that г is surjective. (b) If an element of J2(L°°) has a positive part, then that is the pomtwise positive part: Let x,у € J2(L°°) be such that y > x,Q. If there is a t € [0,1] where y(t) > x(t)+, then y(s) — x(s)+ > a > 0 on a small interval [o, 6] containing t, because χ and y are continuous. Let m := (За + ό)/4 and η := (a + 36)/4 and take 2 w := a(l[0im)u[n>fc] — l[m,n)), then у > у - J w > x,0, so then у is not the positive part of x. w Jw fiw Z\ Figure 1.4: a w 6 L°° such that (J2w)+ is not differentiable (c) J2(L°°) is not a Riesz space: The elements of ^(L00) are differentiable, so, according to (b), it suffices to find aw & L°° such that the pointwise positive part of J2w is not differentiable. Let w := 1[ο,ι/4Ι — l(i/4,i]· ^ can easily be calculated that (J2w)+ is not differentiable at ί = (2 + \/2)/4 (see fig. 1.4). In this example, t(J2w)+ is not continuous. However, i(J2l) is a positive linear function larger than i^w) - so larger than ί(Λ)+ - that is continuous! With aid of Hahn-Banach, it can be shown that E' is directed for every partially ordered vector space E with a locally convex, locally full topology. This result is usually refered to as 'Krein's lemma' (see Schaefer[21, p.218, Lem.l], KelleyL· Namioka[14 , p.227, Thm 23.5], Peressini[20, ρ 72, Prop.1.21]). In this text, the term 'Krein's lemma' is also used for the following, stronger, result. 25 Chapter 1 Seminorms and Topologies on Ordered Spaces Proposition 1.53 (Krein's Lemma) Let E be a partially ordered vector space with a monotone semtnorm p. Let ƒ € E" be such that f < ρ on E. Then there exists a g g E* such that g > f, 0 on Е+ and g < ρ on E. In Chapter 5 a closely related result and a generalization of the above proposition will be proved, see Theorem 5.3 and Corollaries 5.12 and 5.13. It will be said that E'+ determines the ordering of E if for any χ € E one has: if ƒ (x) > 0 for all ƒ e E'+ then χ 6 E+. Proposition 1.54 Let E be a partially ordered vector space with a locally convex topology. Then the following two statements are equivalent: (t) E+ is closed. (n) E'+ determines the ordering of E. Proof. (See Namioka[18, Cor.4.2]). (ii)=>(i): ƒ > 0 on E~+ for all ƒ e E'+, so if E'+ determines the ordering of E, then E+ is closed. (i)=>(ii): Let χ e E\E+. By Mazur's lemma, there is an ƒ € E' with f(x) < 0 and ƒ > 0 on E+. So E'+ determines the ordering of E. The kernel of a monotone seminorm can be factored out, yielding a monotonely normed quotient space (see Proposition 1.17). Let us complete this result by showing that the norm duals are essentially the same. Proposition 1.55 Let E be a partially ordered vector space with a monotone semi- norm ρ and let N := {x € E: p(x) = 0}. Let EN := E/N be equipped with the quotient ordering and the quotient norm, and let q : E —> E¡v be the quotient map 1 ping. Then the adjoint mapping q : E'N —> E' is a bipositive, surjecttve isometry. Proof. The adjoint mapping of q is given by q'(f) := ƒ09, ƒ g E'N. q' is a surjective isometry. For ƒ € E'fi, q'(f) is the composition of two positive mappings, hence + positive. To prove bipositivity, let ƒ € E'N be such that 0. Hence ƒ is positive and thus q' is bipositive. 26 Chapter 2 Extending Monotone Seminorms Partially ordered vector spaces can be embedded in Riesz spaces (see Theorem 1.2). For the study of seminorms on partially ordered vector spaces, it is interesting to know whether seminorms on such spaces can be extended to certain kinds of semi- norms on Riesz spaces in which the spaces can be embedded, so-called 'larger Riesz spaces'. Chapter 3 studies seminorms that can be extended to Riesz seminorms on every larger Riesz space. Chapter 4 deals with extension to Riesz seminorms on some larger Riesz space. This chapter investigates extension to monotone seminorms. 2.1 Introduction The theory of normed Riesz spaces includes a formula extending Riesz norms to Dedekind completions: if ρ is a Riesz norm on a Riesz space E, then χ ι-> inf{p(y): y e E, y > |i|} defines a Riesz norm on the Dedekind completion Εδ (see e.g. Schaefer[22, p.146, Ex.16]). The formula works equally well for extension to any Riesz space of which E is a majorizing subspace. In a setting of partially ordered vector spaces with monotone norms, this result raises the question whether a monotone norm on a partially ordered vector space can be extended to a mono tone norm on the Dedekind completion. For our approach it turns out to be more suitable to consider extension of monotone seminorms and drop the condition that the extension should be a norm. The main result in this chapter is that if a seminorm can be extended to a mono tone seminorm on some Riesz space, then it is monotone*, and if it is monotone*, then it can be extended to a monotone seminorm on any space in which the original space is majorizing. Uniqueness, extensions of Fremlin seminorms and of seminorms with full unit balls, and preservation of norm completeness are studied as well. 27 Chapter 2. Extending Monotone Seminorms 2.2 Existence of Extensions Which seminorms on a partially ordered vector space can be extended to monotone seminorms on a larger Riesz space? According to Lemma 1.4 a monotone seminorm on a Riesz space is monotone* and the monotone* property is inherited by restric tions. So, to be extendable to a monotone seminorm on a Riesz space, a seminorm should be monotone*. This turns out to be sufficient, too. The extension formula mentioned in §2.1 suggests that an extension of a monotone seminorm on a majoriz ing subspace EQ of a Riesz space E is given by p(x) = inf{p(j/): y € Eg, y > χ} for χ e E+. However, monotone seminorms are not determined by their behaviour on the positive cone, so an extension involving only positive elements will not work. The following theorem presents an explicit formula for the greatest extension. The larger space may be any partially ordered vector space as long as the embedding is majorizing. Theorem 2.1 Let E be a partially ordered vector space with a directed, majorizing linear subspace EQ and let ρ be a seminorm on EQ. Define for χ € E: p(x) := ini{p(y) + p(u)+p(v): y € E0, u,v 6 E¿, -и < χ - у < ν}. Then p(x) — mî{p(w): w € Eg, w > x} for χ € E+, ρ is the greatest monotone* seminorm on E that is less than or equal to ρ on EQ, and ρ extends ρ if and only if ρ is monotone* on EQ. Proof. First, note that the infimum in the definition of ρ exists, because EQ is directed and majorizing in E. (a) p(x) = 'mî{p(u>): w € Eo,w > x) for every χ € E+: If w € EQ is such that w > x, then 0 < ι — 0 < ui, so p{w) > p{x). If u, ν € Ej¡, y 6 EQ are such that —u (b) ρ is a seminorm: Let x,ii,i2 € E and A € (0, co). If 2/1,3/2 G -ΕΌ and Ui,tii,u2,ii2 6 EQ are such that — щ < χι — y\ < vi and — u2 < 12 - У2 < f2, then -(ui + u2) < (xi + I2) - (Уі + 2/г) < (щ + v2), so p(xi + i2) < p(j/i + Уг) + р{щ + u2) +p(vi + v2) < (p(yi) +p(«i) +p(vi)) + {р{у2) +P{U2)+PM). Hence p(l!+l2) 28 2.2. Existence of Extensions (d) p is the greatest monotone* semmorm on E that is less than or equal to ρ on EQ: It is clear from the definition that p(x) < p(x) for all χ 6 Eo (take и = υ — О, у = χ) Furthermore, if q is any monotone* seminorm on E and q < ρ on EQ, then for χ g E and y € Eo, и, ν ζ Ε$ with — и < χ — у < ν one has q(x) < q(y) + q(u) + q(v) < p(y) + p(u) + p{v); hence q(x) < p(x). (e) If ρ is monotone*, then p> ρ on EQ: If χ 6 EQ and у € Eo, u, ν G E$ are such that — и < χ — у < ν, then p(y) +p(u) +p{v) > p(x), so, by taking infimum over u, ν and y, p(x) > p{x). Then p extends p. This completes the proof of the theorem. Because the notions 'monotone' and 'monotone*' for seminorms coincide on Riesz spaces, the above theorem has the following consequence. Corollary 2.2 Let E be a directed partially ordered vector space and let ρ be a monotone semmorm on E. If ρ can be extended to a monotone semmorm on some Riesz space containing E, then ρ extends to a monotone seminorm on any partially ordered vector space containing E as a majorizing linear subspace. Example 1.9 presents a monotone norm that is not monotone* and therefore not extendable to a monotone seminorm on a Riesz space. According to Lemma 1.22, ev ery monotone seminorm is equivalent to a monotone* seminorm, so every monotone seminorm has an equivalent seminorm that can be extended. In fact, Lemma 1 22 is a special case of the following extension theorem for Fremlin seminorms and semi- norms with full unit balls. On its turn, it is very similar to Theorem 2.1. Theorem 2.3 Let E be a partially ordered vector space, with a directed, majorizing linear subspace Eo and let ρ be a semmorm on Eo- Define for χ € E: Ρι{χ) ·= mî{p(y)+p{u): y,ue E0,-u P2(x) := inf{p(u) Vp(i'): u,v 6 E0, и < χ < ν}, and let p be as m Theorem 2.1. Then (ι) p\ is the greatest Fremlin seminorm on E that is less than or equal to ρ on Eo and it extends ρ if and only if ρ is a Fremlin seminorm. ('îîj p2 '5 the greatest semmorm with a full unit ball on E that is less than or equal to ρ on E0 and it extends ρ if and only if ρ has a full unit ball. (m) Pi < p\ < ρ < 2pi and ρ < 3p2,' in particular, ρ, ρχ, and p2 are equivalent. If ρ is monotone, then ρ < \\p, ρ < 2p1( and ρ < Зрг on Eo- For the special case E = Eo, this means that ρ is equivalent to all three p, p\, and p2- 29 Chapter 2. Extending Monotone Seminorms Proof. The proofs of (i) and (ii) are very similar to the one of Theorem 2.1 and are omitted. For the inequalities in (iii), first observe that, because every semi- norm with a full unit ball is Fremlin and every Fremlin seminorm is monotone* (Proposition 1.6), P2 < Pi < p. To prove p < 2pi, let ι e E and и,у e E0 be such that — и < χ — у < и. Then р(х) < 2р(и) + р(у) < 2(р(и) + р{у)), hence р(х) < 2pi(x). For ρ < Зр2> let χ € Ε and и, υ e EQ be such that и < χ < v. Then 0<χ — и< — и, so р{х) < p(u)+p(v — u) < 3(p(u) Vp(ii)). Thus p(x) < 3p2{x). If ρ is monotone, let x, у e Eo, и, υ 6 EQ be such that —u < χ — у < ν, then 0 < χ — у + и < и + ν and — (и + ν) < χ — у — υ < 0, so p(2x — 2y) < p(x — у + и) + p(x — у — υ) + ρ(υ — и) < 2р(и + υ) + ρ(υ — и) < Зр(и) + 3ρ(υ), hence p(x) < p(y) +1\ (p(u) + p(v))\ thus p(x) < l^p(x). If i, и € E0 are such that —u < χ < u, then p(2x) < p(x + u) + p(x — u) < 4p(u), hence p(x) < 2pi(x) For the last inequality, let x,u,v 6 EQ be such that и < χ < v. Then 0 Example 1.9 presents a monotone norm ρ on a partially ordered vector space E + (= E0) and у e E, и, υ e E with — и < у < ν, р(и) = ρ(ν) = 1, and p(y) = 3, so that p(y) < p(u) + p(v) < 2 and p^(y) < p(u) V p(v) = 1. This means that the inequalities ρ < \\p and ρ < Зрг are sharp in this example. The inequality ρ < 2pi is sharp for E = C[0,1] and p(x) := ||ι+[|οο + ||ζ~||οοι χ & E, since χ : t n· 2ί — 1 satisfies — 1 < χ < 1 and p(x) = 2, and p(l) = 1. For the case E = E0, the above theorem gives formulas to 'monotone*-izc', 'Fremlinize', and to fill the unit ball of seminorms. According to the inequalities, in (iii) such procedures yield smaller but equivalent seminorms. The formula to fill a unit ball has already been applied in Chapter 1, see Lemma 1.22 and Proposi tion 1.15. The theorem can be used to extend locally convex, locally full topologies. Corollary 2.4 Let E be a partially ordered vector space with a directed, majorinng subspace EQ. If Τ is a locally convex locally full topology on EQ, then there is a locally convex locally full topology f on E that extends r. Proof. Take a collection of monotone seminorms generating the topology and apply Theorem 2.3(iii) to switch to equivalent Fremlin seminorms. Then use part (i) of the theorem to extend these seminorms and thereby with the topology. A monotone norm may have monotone seminorm extensions none of which are norms: 30 2.3. Uniqueness of Extensions Example 2.5 A majorizing Riesz subspace £Ό of a Riesz space E and a monotone norm ρ on EQ that does not have a monotone seminorm extension on E that is a norm. Let E be the space of bounded functions on [0,1] and let E0 = C[0,1]. Take ρ = ||.||i on EQ. EQ is a majorizing Riesz subspace of the Riesz space E and ρ is a monotone (even a Riesz) norm on EQ. For the greatest monotone extension ρ of ρ, using the formula of Theorem 2.1 it is easy to see that, for example, р{1щ) = 0, so that ρ is not a norm on Ε. Because ρ is the greatest monotone seminorm on E extending p, no extending monotone seminorm on E is a norm. At the end of this section, let us look at an example that shows that existence of a monotone extension may fail if the small space is not majorizing in the large one. Example 2.6 A Riesz space E with a Riesz subspace Eo and a Riesz norm on Eç, that cannot be extended to a monotone seminorm on E. Take E = ί°°(Ν), E0 = Coo(N) and p(x) := £n |ι„|, ι € E0. Then E is a Riesz space, E0 is a Riesz subspace and ρ is a Riesz norm on EQ. Suppose that q is a monotone seminorm on E extending ρ, then q(l) > q(l{o,...,n}) = p(l{o n}) = n+1 for all n, which is a contradiction. 2.3 Uniqueness of Extensions Let us now consider the problem of uniqueness of extensions, or rather non- uniqueness: extensions are seldom unique. Because every positive linear function generates a monotone seminorm (Example 1.7(iv)), existence of distinct monotone seminorms is closely related to existence of distinct positive linear functions. Ac cordingly, the main tools to demonstrate non-uniqueness will be Hahn-Banach the orems or Mazur's lemma to obtain linear functions and Krein's lemma (see Propo sition 1.53) to split them up into positive linear functions. Let us start with an explicit example. Example 2.7 A Riesz space E and a majorizing, order dense Riesz subspace Eo with a Riesz norm on EQ which can be extended to inequivalent Riesz norms on E. Take E = C[0,1] and E0 = {x € Ε: ι(0) = x(l)}. Let ||x|| := sup{|x(í)|í(l -í): t e [0,1]}, χ e E, and take p{x) := ||i|| V |x(0)| and q{x) := \\x\\ V |x(l)|, χ £ E. Then Eo is a majorizing, order dense Riesz subspace of the Riesz space E while ρ and q are Riesz norms on E. On Eo they are equal, on E they are not even equivalent. There is, of course, a trivial situation in which there is only one extension, that is if the subspace is norm dense: 31 Chapter 2 Extending Monotone Semmorms Proposition 2.8 Let E be a partially ordered vector space with a directed, majoriz ing subspace EQ and let ρ be a monotone* seminorm on EQ. Let p be the greatest monotone * semmorm on E extending p. If EQ IS norm dense m E with respect to p, then p is the only monotone* semmorm extending p. Proof. Every monotone* seminorm that extends ρ is continuous with respect to p, hence equal to p if Eo is norm dense. If the subspace is not norm dense, Mazur's lemma can be used to find a nonzero linear function on the large space that vanishes on the closure of the subspace. However, such a linear function cannot always be chosen to be positive. With the aid of Krein's lemma, it is possible to find two distinct positive linear functions which are equal on the subspace. The precise statements follow. Lemma 2.9 Let E be a directed partially ordered vector space with a seminorm ρ and let EQ be a linear subspace of E which is not norm dense m E. Then there exist a continuous linear function f : E —• R and an χ ζ E+ such that f(x) φ 0 and f -0 on E0. Proof. Take χ ζ E \ ÊQ. By Mazur's lemma, there exists a continuous linear function ƒ : E —> К such that f(x) φ 0 and ƒ = 0 on ÈQ. Since E is directed, ƒ cannot vanish identically on E+. Theorem 2.10 Let E be a directed partially ordered vector space, let Eg be a ma jorizing linear subspace and let ρ be a monotone* seminorm on Eo, such that EQ IS not norm dense with respect to the greatest monotone* semmorm on E extending p. Then there is a monotone seminorm on EQ which is equivalent to ρ and which has two distinct monotone* extensions. Proof. By the lemma, there exist a linear function ƒ : E —» R that is continuous with respect to the greatest monotone* extension p, and an io € E+ with ƒ (io) / 0 and ƒ = 0 on Eo- By Proposition 1.53 (Krein's lemma), there are fu f2 : E -> R, positive linear and p-continuous, such that f — f\ — fi- Then f χ = /2 on EQ and ЛЫ Φ h(xo)· Take qi{x) := p(x) + |/i(x)| and q2(x) := p{x) + \f2(x)\, x € E. Then gì and q2 are monotone* seminorms on E. Because f\ and fi are continuous, <7i and Ç2 are equivalent top on EQ. Furthermore, c¡ = q2 on EQ and The above theorem shows that a simple uniqueness theorem more general than Proposition 2.8 cannot be expected. That the theorem involves an equivalent semi- norm is not superfluous: the seminorm itself may have a unique extension. 32 2 3 Uniqueness of Extensions Example 2.11 A Riesz space E with a majorizing linear subspace EQ and a mono tone* seminorm ρ on EQ, such that there is only one monotone semmorm ρ on E extending ρ and such that EQ IS not p-dense m E. Такс E := {χ S C[0,1] : χ is piecewise affine} with pointwise ordering, Eg := {x 6 Ε · φ(χ) = 0}, where φ(χ) := Ι χ- Ι χ, χ e E, and p(x) := Ці+ІІ«, + ||x"1U, x e E, Ρ •= P\E0- (a) E is a Riesz space and p ís a monotone norm on E: Obvious. (b) £Ό is a p-closed, majorizing linear subspace of E and EQ φ E: Indeed, £Ό is a linear subspace of E, 1 € EQ and 1 is a strong unit in E, so EQ is majorizing, φ is 11.1 loo-continuous and therefore p-continuous, so EQ is closed, φ φ 0, so EQ φ E. (c) p is the only monotone seminorm on E that extends p: Let q also be a monotone seminorm on E extending p. E is a Riesz space, so q is monotone*. Let χ Ç. E. It has to be shown that q(x) = p(x). Because — ЦагЦооІ < χ < ||ι+||οο1, one has q(x) < (||x_||oo + ||з:+||оо)9(1) = P(X)PW = p{x)- It remains to show that q(x) > p(x). Take e(t) := 2 - 4i, t € [0,1]. Then e e E, ф{е) = 1 and р(е) = 4. For y € E, observe that y — ф(у)е e £o, so p(y - ф(у)е) = q{y — ф(у)е), whence Р(У) - q(y) < Р(ФІУ)е) + q^{y)e) < 2\ф{у)\р(е) = 8\ф(у)\. If χ > 0, choose τ € [0,1] such that χ(τ) = \\х\\ж- Let ε > 0. Because χ is piecewise affine, there exists a 'narrow peak' и e E+ such that 0 < II^Hoo^ ^ x> U(T) — 1 and 8p(x) ƒ„ и <ε. Then p(x) = p(||x||ooti) < 0 < ||х+||оо« < x+, 0<||ζ-||οοΐ7<ϊ-, (2.1) 33 Chapter 2. Extending Monotone Seminorms y - χ Figure 2.1: the functions x, y and y — χ и(т) = u{a) = 1, u,v are disjoint, and 16p(x) fQ u, 16p(x) ƒ„ υ < ε. Define у := p(x)(u — ν) (seefig. 2.1) . We check that у has the desired properties. Clearly, \ф{у)\ < p(x) {\4>{u)\ + \Φ{ν)\) < ρ(χ)2ε/16ρ(χ) < ε/8 and because и and + υ are disjoint we have p{y) = \\y \\œ + ЦіГЦоо = P(x) (u(r) + υ(σ)) = 2p{x). (2.1) and disjointness of и and x~ yield that y-x < p(x)u - χ < ||χ+||οοΙί- x+ + ^"IIOQU + X- < < (Цж-Цоо«) VX" < Ці-Цооі, and, similarly, y-x>- (ll*+ll«»v) ν X+ > -||x+||ooi. Because q is monotone*, it follows that q(y — x) < p(x)q(l) = p(x), so q(y) — q(x) < p(x). As noted above, then ç(x) > p(x) — ε, which completes the proof of assertion (c). 2.4 Extensions and Norm Completeness If a directed, integrally closed partially ordered vector space E with a monotone seminorm ρ is norm complete, is then its Dedekind completion complete with respect to the greatest monotone* extension of p? To investigate this problem, let (xn)„ be δ χ a sequence in Ε such that (p(xn))n is summable. Is the series Σ π convergent? + According to Theorem 2.1, yn € E and un,vn € E with — un < xn — yn < vn can be chosen such that Σ{ρ(νη) + p("n) +p{vn)) < oo. By norm completeness of E, there are y, u, υ e E with J^ì 3/* ~~* У> Σ" ик ""* и> Σι υ* ~~* ν· ^ ^+ 's closed —and, consequently, ρ a norm, see Proposition 1.30— then u, υ € E+ and u v r n so и > Σ" k, ν > Σ? k f° all > that — (u+v) < x„ — yn < u+v. In the same way, 34 2 4 Extensions and Norm Completeness n n n by assuming that Σ 2"p(a„) < oo, the inequalities — 2~ w < xn — 2~ yn < 2~ w, η e N can be obtained for a certain w 6 E+ To derive norm convergence of Σ x„ from this, the Dedekind completeness of Es can be used Actually, we only need uniform completeness Theorem 2.12 Let Ε be a uniformly complete partially ordered vector space with a directed, majorizing linear subspace Ες, and let ρ be a monotone* norm on Eg such that EQ IS closed Let p on E be the greatest monotone* extension of ρ If EQ IS norm complete with respect to p, then E is norm complete with respect to p Proof. It suffices to prove norm convergence of ^i„ for every sequence {x„)„ in n E with Σ 2 p(:cn) < oo So, let (x„)n be such a sequence From the formula for ρ (Theorem 2 1) it follows that there exist (yn)n m Eo and (u„)„, (vn)n in EQ such that —u„ < xn — yn < v„ for all η and Σ 2" (p(j/n) + p(un) + p(vn)) < oo Because E0 is norm complete, w = ΣΊ?=ι 2"(u„ + vn) (norm convergent) exists in E0 Eg is closed, so w € Eg and w > Σ£=12*(ω* + vk) > ±2"(i„ - yn) for all η Write _ _n Zn — χ*. — Уп, then —2 "w < zn < 2 № for all η The uniform completeness of E yields that the series Σ zn 1S relatively uniformly convergent in E, and then also norm convergent in the monotone semmorm ρ Also, Σ 3M ls norm convergent, χ iS norm because ^p(yn) < oo and E0 is norm complete Thus Σ η convergent, proving the theorem Example 2 5 shows that ρ in the above theorem may not be a norm If ρ is Fremlin, then its greatest Fremhn extension is equivalent to its greatest monotone* extension (Theorem 2 3(in)), so that norm completeness means the same for these extensions Therefore, the words 'monotone*' in the above theorem may be replaced by 'Fremhn' In the same way, the monotone* condition can be replaced by the 'full unit ball' condition Because Dedekind complete spaces are uniformly complete, the theorem has the following consequence Corollary 2.13 Let E be a directed, integrally closed partially ordered vector space with a monotone* norm (or Fremhn norm, or norm with a full unit ball) ρ such that E+ is closed If E is norm complete with respect to p, then so is Es with respect to the greatest monotone* semmorm (or Fremhn semmorm, or semmorm with a full unit ball) extending ρ Is Theorem 2 12 also true if ρ is a seminorm? Closedness of EQ only makes sense if ρ is a norm, but norm completeness of EQ IS a sensible property for seminorms as well To obtain an extension of the theorem for seminorms, one can quotient the 35 Chapter 2. Extending Monotone Seminorms kernel Proposition 1.21 states that norm completeness of the positive cone implies closedness of the positive cone of the quotient space. To apply Theorem 2.12, uniform completeness of the quotient is needed. Proposition 1.18 shows that this does not follow from uniform completeness of the space itself. Therefore, it will be an explicit hypothesis. Since the quotient norm may not inherit the monotone* property of ρ (see Example 1.19), the first step in the proof is a change to Fremlin seminorms. Corollary 2.14 Let E be a partially ordered vector space with a directed, majorizing linear subspace D and let ρ be a monotone* semmorm on D. Let p\ be the greatest monotone* extension on E and denote N := {x € E: Px(x) = 0}. Assume that E/N is uniformly complete. If D and D+ are p-complete, then E is ρχ-complete. Proof. Denote the quotient mapping from E to E/N by q. Let ρ be the 'Fremlin- ization' of p, i.e. p(x) = mf{p(y) + p(u): y,u 6 D, — и < χ — у < и], χ e D (see Theorem 2.3), and let p\ be the greatest Fremlin seminorm on E extending p. Ac cording to Theorem 2.3, ρ and ρ are equivalent and so are ρχ and ρχ. In particular, N = {x S E: pi (ι) = 0}. Thus, we may and do alter the words 'monotone*' in the conditions of the corollary to 'Fremlin'. (a) The natural embedding ofDN := Df(NnD) in EN '•= E/N is linear, isometncal, and bipositwe, so DN can be interpreted as a subspace of Ец: Observe that {x € D:p(x) =0} = NDD. (b) Dti is directed and majorizing m EN, and the quotient norm px on ΕΝ is Fremlin: From Proposition 1.17. + (c) If D and D are p-complete, then DN IS ρχ-complete and DN is closed m DN-' From Proposition 1.21. (d) px is the greatest Fremlin extension of P\\DN'• Denote , for abbreviation, г := PX\DN- By (a) and (b), p\ is a Fremlin extension of r. For χ € Εχ, there is an χ € E with q(x) = x, and then Pi(x) = Pi(x) = inf{p(y): у e D, — y < χ < y}. For y ξ. D with —y 36 Chapter 3 Riesz Seminorms on Non-Riesz Spaces For a generalization of the notion of Riesz seminorm to partially ordered vector spaces, we would like to have a notion that has similar properties and that on Riesz spaces is the same as Riesz seminorm. The first idea presented in this chapter is to consider restrictions of Riesz seminorms. One cannot just take all restrictions, because it can happen that the restriction of a Riesz norm to a subspace that is a Riesz space (but not a Riesz subspace) is not Riesz. To avoid this problem, only relatively large subspaces are considered. It will be more convenient to look at extensions than restrictions: consider the seminorms that can be extended to Riesz seminorms on a small larger Riesz space, for instance the enveloping Riesz space (see Buskes L· van Rooij[6]) or the Dedekind completion. If there exists a smallest Riesz space containing the partially ordered vector space, then seminorms that are extendable to Riesz seminorms on that Riesz space can be extended to every Riesz space in which it is majorizing, by means of the formula in §2.1. Thus, the idea comes down to considering seminorms that can be extended to Riesz seminorms on every larger Riesz space that is majorized. It turns out that this kind of seminorms need not have the important property of being determined by the values on the positive cone. This leads to the second idea, which is a generalization of the r-norm in operator theory. 3.1 Solid Sets in Partially Ordered Vector Spaces Which seminorms on a partially ordered vector space can be extended to Riesz seminorms on every Riesz space in which it is majorizing? To obtain a description without explicit use of larger Riesz spaces, we will reformulate the Riesz seminorm 37 Chapter 3 Riesz Seminorms on Non-Riesz Spaces property in a way that makes sense in partially ordered vector spaces, too, and that entalles the desired extendability Our approach is via unit balls Riesz seminorms are precisely the seminorms with solid unit balls, so let us start with developing a definition of solidness in partially ordered vector spaces For χ and y in a Riesz space E, the assertion '|x| < \y\' means the same as 'every upper bound of {y, —y} is an upper bound of {x, —χ}' The latter formulation also makes sense if E is a partially ordered vector space that is not a lattice We will denote it shortly as χ <¡ y Further on, such a relation for sets will appear frequently, too It would be a burden to write out the full sentences every time Therefore we will use the symbol < for sets as well Notation. Let E be a partially ordered vector space and let A and В be subsets of E The assertion 'Every upper bound of of A U — A is an upper bound of В U —B' will be denoted by A > В or В < A 1 If A = {a}, then A > В will also be written as о > В and likewise notations as В < a, b < a, etc will be used Remark that the relation < strongly depends on the space in which the upper bounds are taken If there is risk of confusion, the specification 'in E' will be added Using upper bounds, solidness can be defined in partially ordered vector spaces Definition 3.1 Let E be a partially ordered vector space A set S С £ is called solid if for every χ e E and у £ S such that χ < y one has that x g S For S С E the set sol S = {χ 6 E there exists y G S such that χ < y} ¡s the smallest solid set in E containing S and it is called the solid hull of S This definition corresponds to the definition for Riesz spaces (Definition 1 34) In a Riesz space, the intersection of a solid set and a Riesz subspace is solid in the subspace In partially ordered vector spaces more prudence is needed Example 3.2 A Riesz space E with an ordered subspace E0 and a Riesz norm ρ on E such that EQ is a Riesz space and such that the restriction of ρ to EQ IS not Riesz Consequently, the unit ball В of ρ is solid m E and EQ Л В is not solid m Eo Take, as in Example 1 3, E = C[0,1] and Eo = Aff[0,1], ι e the subspace of all affine functions Take ρ = || ||i, then ρ is a Riesz norm on E To show that the restriction of ρ to E0 is not Riesz, let x(t) = 2t — 1, t € [0,1] Then p(x) = 1/2, whereas the absolute value of χ in £Ό is the function 1 and p(l) = 1 So ρ is not Riesz on E0 'I pronounce A > В as 'A is wider than B' 38 3 1. Solid Sets ¡η Partially Ordered Vector Spaces Solidness is preserved under restriction to subspaces that are order dense in a Riesz subspace. Lemma 3.3 Let E be a Riesz space with an ordered subspace EQ such that Eg is order dense m the Riesz subspace that it generates. If S is a solid subset of E, then EQD S is solid m £Ό- Proof. Let χ € E0 and y G £Ό Π S be such that χ < y in EQ. Then ζ > \x\ for all ζ 6 £Ό with ζ > \y\. Because of the order denseness, it follows that |i| = inf{z € EQ: Z > \x\} < inf{z e EQ·. Ζ > \y\} = \y\, so that χ € S, by solidness of S. Thus χ € Eg Π S, proving that EQ Π S is solid in EQ. Let us now consider seminorms with solid unit balls on partially ordered vector spaces. Proposition 3.4 Let E be a partially ordered vector space with a seminorm p. (t) ρ has a solid unit ball if and only if p(x) < p(y) for all x,y € E such that χ < y. f«^ If ρ has a solid unit ball, then ρ is Fremiva. Proof, (i): Directly from the definition. (ii): If x,y e E are such that —y In a Riesz space, a Riesz seminorm on a majorizing Riesz subspace can be extended to the whole space (see §2.1). In general, if the ordered subspace is not a Riesz subspace, extension of a seminorm with a solid unit ball to a Riesz seminorm is not always possible, as the next example shows. First, we show that if there is an extending Riesz seminorm, then there is also a greatest one. Lemma 3.5 Let E be a Riesz space with a majorizing ordered subspace EQ and let ρ be a seminorm on Eo with unit ball B. Then co sol В is the unit ball of the greatest seminorm q with a solid unit ball on E such that q < ρ on EQ. If there exists a seminorm with a solid unit ball on E extending p, then q extends p, too. Proof. В is absorbing in E0 and E0 is majorizing in E, so sol В is absorbing in Е. Hence, со sol ß is the unit ball of a seminorm q on E and, according to Lemma 1.36, co sol В is solid. Moreover, со sol В is the smallest convex, solid set in E containing B, so q is the greatest seminorm on E with a solid unit ball that is less than or equal to ρ on EQ. If B' is the solid unit ball of a seminorm on E extending p, then 39 Chapter 3. Riesz Seminorms on Non-Riesz Spaces E0 Π В' = В, and В' Э со sol В, so В С Е0 П со sol В С Е0ПВ' = В. This yields that q extends p. If, in the situation of the above lemma, Eo Π со sol Β φ В, then there is no Riesz seminorm on E extending p. The following example presents such a ball B. It also shows that the solid hull of a convex set that is solid in a subspace need not be convex. Example 3.6 A Riesz space E with on order dense subspace E0 that generates E as a Riesz space and a seminorm with a solid unit ball on EQ that cannot be extended to a Riesz seminorm on Е. 2 Let D be the unit disk in R and let S be the unit circle. Take for E0 the space of all restrictions to D of affine functions from K2 to R, with pointwise ordering on D. Then EQ is a directed, integrally closed partially ordered vector space. Note that the ordering is the same as the pointwise ordering on S, so £Ό can be interpreted as an ordered subspace of C(S). Let Ε be the Riesz subspace of C(S) generated by Eo. + Take p(x) := (JDx(t) dt) V (fDx(t)~dt), χ € E0. Then ρ is a Premlin norm on Eo- Denote for an angle φ ζ (—π, π] the corresponding point in S by s¿: s¿ — (cos φ,sin.φ). Let Яф\ Eo —ϊ Eo be the operator rotating over the angle φ: 2 (ηψζ)(ίι,ί2) := χ(t\ cos φ + Í2 sin φ, <2cos0 - ¿isin^), (ti, ¿2) € Κ , ζ € Εΰ, φ € R. (a) For every a 6 (0, π/2) and M € [Ο,οο) there is a z € E0 such that z(s0) = —1, z > — 1 on S, and ζ(εφ) > M for all φ e (—η,π] \ (—a, a): Take z(t t ):= Μ + 1 (1-tO-l, (tbí )eR2. u 2 Λ1 — cos a 2 and verify that ζ € E0, 2(1,0) = —1, ζ > — 1 on S, and го(эф) > M for all φ such that cos φ < cos a, i.e., for all φ € (—π, π] \ (—α, a). (b) For efen/ ι € C(S) one has x(t) = inf{u(t):w £ E0, и > χ on S} for all t g S: Let χ € C(S)+. Because of rotation symmetry, it suffices to prove the equation for t = so = (1,0). Let ε e (Ο,οο). Because t 1-+ ι(ί) is continuous on S, there exists an α e (0, π/2) such that x(s¿) < a;(so) + ε for all 0 Ç. (—α, a). Observe that χ is bounded on S and take M g [Ο,οο) such that εΜ > χ on 5. For these M and a, take а г G E0 as in (a). Let y := (x(s0) + 2e)l + εζ. Then y e So, 2/(«o) = x(sa) + e, î/(s0) > i(s0) + ε > х(зф) for all 6 (-α,a), and j/(s¿) > εΜ > х(эф) for all <£ G (—π,π] \ (—α,α), so y > χ on S. Thus, inf{u(so):ii e £θι^ > ι on 5} < y(so) = i(so)+e. From this, the assertion follows 40 3 1 Solid Sets in Partially Ordered Vector Spaces + for χ € C(S) To complete the proof, note that E0 is majorizing in C(S), so that to every χ 6 C(S) an element of EQ can be added such that the sum is positive (c) EQ IS order dense in E By (b), Eo is order dense in C(S), and Eg is contained in E, so order dense in E (d) The unit ball В of ρ is solid m E0 Let x,y £ E0 be such that χ < y If y > О, then -у < χ < у on D, hence p(x) < p(y) Similarly, if y < 0, then y < χ < —y, so p(x) < p(y) If y is neither nonpositive nor nonnegative, then there are two points of 5 where y is zero Then χ is zero in those points, too, because, by (b), χ < y implies that \x(t)\ < \y(t)\ for all t € S Thus, χ is a scalar multiple of y Because χ < y this yields that p{x) < p(y) Hence, В is solid + + Figure 3 1 the functions хг = y , X2 — (Rn-ay) , and ΐχ + x2 (e) sol В = {χ e E there is у 6 E0 such that \x\ < \y\ on D} is not convex Let 2 y{t\, h) = (3/2)Í!, (ti, t2) € R With the rotations Яф defined above, we find that 2 V(R*V) = Р(У) = fD(y2)ttd(ti,t2) = j^ /2(3/2)j¿rcos {βφ φ € [0, π — a]}, so ζ has the same sign m the end points so and s„-a, and there N > 3/2, because xi(s0) = x2(s*-.a) = 3/2 ζ is affine, so \z\ cannot be lower than 3/2 on both sides of the line through s0 and s„_Q, hence p(z) > 3(π/2 —sina)/2 > 2 _1 It follows that 2 (ii + x2) & sol В This shows that sol В is not convex (f) Eo Π cosolB Э В Let у be as m (e) and let ii = \y\ and x2 = IR^ßyl _1 Then Xi,x2 € sol ß, soi = 2 (i! + x2) e co sol В On S, χ is of the form (зф H4 3(|sin0| + |cos0|)/4, so χ > 3/4 on S Thus, 0 < (3/4)1 < i, so (3/4)1 € со sol B, because in a Riesz space the convex hull of a solid set is solid (Lemma 1 36) Furthermore, (3/4)1 € E0, whereas p((3/4)l) = 3π/4 > 1, so (3/4)1 £ В 41 Chapter 3. Riesz Seminorms on Non-Riesz Spaces 3.2 Solvex Sets and Pre-Riesz Seminorms As seen in Example 3 6, a solid unit ball is not sufficient for extendability to a Riesz seminorm. The idea presented in this subsection is to consider sets that are both solid and convex, instead of generalizing the notion of solid set on its own. It leads to a property that we will call solvexity. Definition 3.7 Let £ be a partially ordered vector space. A set S С E is called solvei if for every χ € Ε, Χι,...,χη € S, and λι,..., An e (0,1] with J2k ^k = 1 such that χ < {Σ}Λε*λ*2:*:£* = ±1} one has that x Ç. S. Lemma 3.8 Let E be a partially ordered vector space. Every solvex set in E is solid and convex. If E is a Riesz space, then a set is solvex in E if and only if it is solid and convex. Proof. Let S С £ be solvex. It is clear that S is solid, by taking η = 1 in the definition of solvex sets. To prove that S is convex, let Χχ,..., xn € S and λ L λ χ £ x let Ab...,An e (0,1] with Σ* * = Σ* * * < {T,k k kXk- £k = ±1}, so J2k XkXk € S, by solvexity of S. So S is convex. Assume that E is a Riesz space and let S be a solid, convex set in E. Let χ € Ε, χγ,... ,χη e 5, and Аь..., λ„ 6 (0,1] with Σ* λ* = 1 be such that χ < {Σ*ε*λ*ι*:ε* = ±1}· & 's solid, so \xk\ € S for all к and then Σ*^!1*! ε ε λ λ S by convexity of S. Because |Σ* * *:τ*Ι < Σ* *|ι*| for any ek 6 {-1,1}, A; = 1,... ,n, every upper bound of {£* At|xt|, - ΣΗ λ*|ζ*|} is an upper bound of {Σ*ε*-^χ*:ε^ = Ü} an(l therefore one of {x, — x]. S is solid, so it follows that χ e S. Thus, S is solvex. Example 3.9 A solid, convex set that is not solvex 2 Consider Example 3.6: Let D be the unit disk in R , S the unit circle, E0 the space of restrictions to D of affine functions from R2 to R with pointwise ordering on D, p{x) = ¡Dx(t)+dt V ¡Dx(t)-dt, χ £ EQ, and Β = {χ ζ E0:p{x) < 1}. It has been proved that В is solid and convex in E0. To show that В is not solvex, let 2 Уі(*і,*г) := (3/2)*! and y2(*i,*2) := (3/2)*2, (*i,<2) € R . Then yuy2 6 В and 1 (3/4)1 < {2~ (y1 + y2), 2'Цуі - y2)}. Indeed, yi + y2> 3/2 on the first quadrant, andj/2—у\,У\— y2, — 2/1 — y2 > 3/2 on the second, third, fouth quadrant, respectively. However, (3/4)1 0 B, so В is not solvex. There is a result for restrictions similar to Lemma 3.3 42 3.2. Solvex Sets and Pre-Riesz Seminorms Lemma 3.10 Let E be a Riesz space with an ordered subspace £Ό such that for every xl,... ,i„ € E one has \x\\ + · · · + |x„| = inf{j/ € EQ: y > |xi| + · · · + |xn|}· If S is a solvei subset of Ε, then EQHS is solvei in E<¡. Proof. Let χ € E0, Xi,..., xn € E0 Π S, and λι,..., λ„ € (0,1] with ^jt λ* = 1 be such that χ < (ΣΑε*λ*χ*: ε* = ±1}· Then: |x| < iní{y£E0: 2/>sup{^etA/tifc:£i = ±1}} = = inf{y e £0: у > |λιΐι| + ··· + |λ„ι„|} = = |λιχι| + l-|A„z„| for every и ξ. E that is an upper bound of \^kek\xk' ε* = ±1}· S is solvex, so it follows that χ 6 S and therefore χ G EQ Π S. Thus Eg П S is solvex in EQ. We will see that the seminorms with a solvex unit ball are precisely the seminorms that can be extended to Riesz seminorms on every Riesz space that is majorized. After a definition, this section continues with some elementary properties. Then restrictions are studied, and finally the desired extension properties will be proved. Definition 3.11 Let E be a partially ordered vector space. A seminorm on E is called pre-Riesz if its unit ball is solvex. Proposition 3.12 Let E be a partially ordered vector space with a seminorm p. (i) If ρ is pre-Riesz, then ρ is Fremlin. (ii) If E is a Riesz space, then ρ is pre-Riesz if and only if it is Riesz. Proof, (i) The unit ball of ρ is solvex, hence solid, so ρ is Fremlin by Proposition 3.4. (ii) By Lemma 3.8. Proposition 3.13 Let E be a partially ordered vector space with a seminorm p. Then ρ is pre-Riesz if and only ifp{x) = inf{p(xj) V • · · Vp(x„): X\,..., x„ 6 E such that there are λι,...,λ„ e (0,1] with Σ*^* = 1 and χ < {Σ^ε^λ^χ*: ε* = ±1}} for all χ e E. Proof. Let В be the unit ball of p. ^•) Let χ € E, X\,...xn € E, and λι,..., λ„ 6 (0,1] with ^2k λ* = 1 be such that χ < {Σ^ε*λ*χ*: ε* = ±1}. χι,.. .,x„ € (p(xi)V.. .Vp(xn))B and this set is solvex, because В is solvex. It follows that χ € (p(xi) V ... V p(xn))B, or, in other words, 43 Chapter 3. Riesz Seminoms on Non-Riesz Spaces p{x) < p{x\) V ... Vj)(i„), proving that p(x) is less than or equal to the infimum at the right hand side. By taking η = 1 and X\ = x, it is clear that the infimum is not strictly greater than p(x). Thus, the equality is established. = an( <= ) Let χ e Ε, Αι,..., λη € (0,1] with Σ/c ^* 1> ^ Χι,- • • ,xn £ В Ъе such that χ < {Y2k ek\kxk: Ek — ±1}. Then, by assumption, p(x) < p(\\x{) V... Vp(Anin) < 1, so χ € В. Hence В is Solvex, thus ρ is pre-Riesz. 3.3 Restricting Pre-Riesz Seminorms to Pre- Riesz Subspaces The following proposition is a direct consequence of Lemma 3.10. Proposition 3.14 Let E be a Riesz space with an ordered subspace EQ such that ¡or every xi,...,xn e E0 one has \xi\ 4 1- \xn\ = inf{y 6 E0: у > \x\\ 4 1- |i„|}. If ρ is a Riesz seminorm on E, then its restriction to EQ is pre-Riesz. Corollary 3.15 Let E be a Riesz space with an order dense subspace £Ό and let ρ be a Riesz seminorm on E. Then the restriction of ρ to EQ is pre-Riesz. Thus, restrictions of Riesz seminorms to Riesz subspaces and to order dense sub- spaces are pre-Riesz. What can be said about restrictions of pre-Riesz seminorms on spaces that are not Riesz? Restrictions of pre-Riesz seminorms will be pre-Riesz if the subspace is decent enough. An interesting class of such subspaces can be described using the notion of Riesz* homomorphisms, see Van Haandel[10]. 3.3.1 Riesz* Homomorphisms and Pre-Riesz Spaces Definition 3.16 Let E and F be directed partially ordered vector spaces and let h: E —> F be a linear mapping, h is called a Riesz* homomorphisrn if for every finite subset S of E (or, equivalently, every S consisting of two elements of E) and every χ € E one has: if χ < и for every upper bound и of 5, then h(x) < ν for every upper bound ν of h(S). (Van Haandel[10, p.26-27, Def.5.1, Thm.5.3]). Riesz* homomorphisms are closely related to pre-Riesz spaces (Definition 1.1). Let us collect some properties. Lemma 3.17 Let E be a Riesz space, let EQ be a partially ordered vector space, and let i: EQ -+ E be a bipositive linear mapping. Then i is a Riesz* homomorphisrn in both of the following cases: 44 3 3 Restricting Pre-Riesz Seminorms to Pre-Riesz Subspaces (г) г(Е0) is order dense m E. (η) I(EQ) IS a Riesz subspace of E Proof. (1) Let x,y € E<¡ If ζ 6 E0 is such that ζ < и for every и € EQ with и > χ,у, then i(z) < i(u) for all such u, so, because г(Е0) is order dense and ι is bipositive, i(x) V г(у) = lnf^u)- и € Ea such that i(u) > i(x),i(y)} = m{{i(u). uÇ. E0 such that u > x,y) > i(z), so that ι(ζ) < υ for every upper bound υ of {i{x),i{y)}. (ii) i is a Riesz homomorphism, hence a Riesz* homomorphism (Van Haandel[10, p.26, Rem.5.2(i)]). Proposition 3.18 (ι) A directed partially ordered vector space E0 is pre-Riesz if and only if there exist a Riesz space E and a bipositwe linear mapping г:Ео —> E such that i(Eo) is order dense m E. Such a mapping г is a Riesz* homomorphism. ^м^ Let E and F be directed partially ordered vector spaces and let h:E—*Fbea positive linear mapping. If F is pre-Riesz and h is a Riesz* homomorphism, then E is pre-Riesz. Proof, (i) This equivalence is precisely Cor. 4.10 in Van Haandel(10, p.22]. The last assertion follows from the previous lemma. (ii) Let x,y,z 6 E be such that every upper bound of {i + y, x + z] is an upper bound of {y, z}, or, in other words, y, ζ < и for every upper bound и of {x+y, x+z}. Because Л is a Riesz* homomorphism, it follows that h(y), h(z) < υ for every upper bound ν of {h(x) + h(y),h(x) + h(z)}. The assumption that F is pre-Riesz yields that h(x) < 0, so χ < 0, by bipositivity of h. Thus, E is pre-Riesz. For examples of pre-Riesz spaces, remark that every directed integrally closed par tially ordered vector space is pre-Riesz (Van Haandel[10, p.4, Thm.l.7(ii)]). An ordered subspace of a Riesz space is a Riesz subspace if and only if the inclusion mapping is a Riesz homomorphism. Proposition 3.18 justifies the following definition. Definition 3.19 Let E be a directed partially ordered vector space. An ordered subspace Eo of E is called a pre-Riesz subspace if the inclusion mapping from Ea to E is a Riesz* homomorphism. 45 Chapter 3 Riesz Seminorms on Non-Riesz Spaces 3.3.2 Pre-Riesz Seminorms and Riesz* Homomorphisms Lemma 3.20 Let E and F be directed partially ordered vector spaces and let h: E —> F be a Riesz* homomorphism Let S be a finite subset of E and let χ & E If χ < S m E, then h(x) < h(S) m F Proof. Assume that χ < S. Then χ, — χ < и for every upper bound и of S U —S, hence h(x), h(—x) < υ for every upper bound υ of h(SU —S), because h is a Riesz* homomorphism. From this it follows that h(x) < h(S). Theorem 3.21 Let E and F be directed partially ordered vector spaces and let h:E—*Fbea Riesz* homomorphism. If ρ is a pre-Riesz semmorm on F, then poh is a pre-Riesz semmorm on E. Proof. Observe that ρ о h is a seminorm on E. It remains to be shown that its unit ball В is solvex. Let B' be the unit ball of p, then В — {χ € E:h(x) € В'}. Let χ e Ε, χι,..., χη € Β, and Аь..., λ„ € (0,1] with Ylk λ* = 1 be such that χ < {Y2k£k^kXk' ε к = ±1}· According to the previous lemma, it follows that h(x) < {Ylk£k^kh(xk)'· £k = ±1} h(xk) € В' for к = 1,... , π and В' is solvex, so /i(x) e В'. Hence χ £ В, proving that В is solvex. Corollary 3.22 Let E be a directed partially ordered vector space with a pre-Riesz subspace EQ- If ρ IS a pre-Riesz seminorm on E, then its restriction to Eo is pre- Riesz, too. 3.4 Extending Pre-Riesz Seminorms Let us now consider extensions to Riesz spaces. Lemma 3.23 Let E be a Riesz space and let x\,...,xn e E. Then Ylk\xk\ < £ {T,k k^k- £k = ±1}. Proof. By induction with respect to n. First, observe that |xi| < {χι,— χι} holds trivially. Then, assume that Σ£=ι |xit| < {^k6-^^· ε* = ±1}· Let ζ be an upper £ bound of {53£=j kXk- ε к = ±1}· Then ζ — xn+i and ζ + xn+ì are upper bounds of 1 £ {ELi^ *: k = ±1}, so they are greater than £*=i \xk\. Thus, ζ > ££+} \xk\. In other words: Ztl Ы < {Σ£Ν*ι*: ** = ±1}· 46 3 4 Extending Pre-Riesz Seminorms Lemma 3.24 Let E be a Riesz space with an ordered subspace EQ. If S С E<¡ is solvei m Eg, then for the convex hull of the solid hull in E one has EQHCO sol S = S. Proof. Let χ e EQÍICO sol S. Then there are x\,..., x„ S sol S and Ai..., A„ € (0,1] with J^Ajt = 1 such that χ = ]Γ]λλ*χ*. There are Ui,.. ,u„ € S such that l^itl < \щ\ for all k. If 2 6 EQ is an upper bound of {Σ*ε*^*η*: ε* = ^Ь *пеп the previous lemma yields that ζ > Σ*^*Ιω*Ι — J2k ^*Iх*I — χι~χ· Because S is solvex in EQ, it follows that χ e S, proving that EQ Π co sol S С S. Conversely, со sol S D S, completing the proof. Theorem 3.25 Let E be a Riesz space with a majorizing ordered subspace Eg. Let ρ be a pre-Riesz semmorm on EQ with unit ball B. There exists a Riesz semmorm on E extending p. Among such extensions there is a greatest one and its unit ball is co sol 5. Proof. Let B' be the convex hull of the solid hull in E of B. Then B' is convex, it contains 0, and it is absorbing in E, because В is absorbing in EQ and EQ is majorizing in E. Hence B' is the unit ball of a seminorm ρ on E. By Lemma 1.36, B' is solid, so ρ is Riesz. В is solvex in EQ, SO, according to the previous lemma, EQ Π В' = ß, which means that the restriction of ρ to E0 is equal to p. Finally, B' is the smallest convex, solid set in E containing B, so ρ is the greatest Riesz seminorm on E that is less than or equal to ρ on EQ. This theorem can be used to prove properties of pre-Riesz seminorms on pre-Riesz spaces via embedding in Riesz spaces and Riesz seminorms. Corollary 3.26 Let E be a pre-Ritsz space. If ρ and q are pre-Riesz seminorms on E, then ρ V q and p + q are pre-Riesz, too. Proof. Because of Proposition 3.18, E is an order dense subspace of a Riesz space, ρ and q can be extended to Riesz seminorms ρ and q on this Riesz space. Then pV q and ρ + q are also Riesz seminorms, so their restrictions to E are pre-Riesz, by Corollary 3.15. Therefore ρ V q and ρ + q are pre-Riesz. Because every directed partially ordered vector space can be embedded as a majoriz ing subspace of a Riesz space, Theorem 3.25 leads to embedding in a Riesz space with a Riesz seminorm. Theorem 3.27 Let E be a directed partially ordered vector space with a pre-Riesz seminorm p. Then E can linearly, isometrically, bipositively be embedded m a Riesz space with a Riesz seminorm. 47 Chapter 3 Riesz Seminorms on Non-Riesz Spaces Proof. According to Theorem 1.2, there is a Riesz space F with a linear bipositivc mapping ι : E —> F. Because E is directed, ι{Ε) is majorizing in the ideal FQ that it generates in F. By Theorem 3.25 the pre-Riesz seminorm рог can be extended to a Riesz seminorm on FQ. In this way, F0 and г : E —> F0 have the desired properties. An explicit formula for the greatest Riesz extension may be more convenient than the construction involving unit balls. Being careful with notations, such a formula can be formulated for arbitrary directed partially ordered vector spaces, yielding the largest pre-Riesz extension. Thus, an extension theorem is derived that is entirely similar to those of Chapter 2. First, a lemma. Lemma 3.28 Let E be a partially ordered vector space. If x, x\,... ,xn 6 E and yk¡l e E, г = 1,..., тк, к = 1,..., η, are such that xk < {^™* ¿¡УкУ- ε, = ±1} for к = 1,..., η, and χ < {Σ*=ι £^k- ε* = ±1}, then χ < {Σ1=ι Σ.=Ί ε*,.ΐ/*,.: ε*,, = ±1}. Proof. Let и be an upper bound of {Σ*=ι Σ™\ ε*,ι2/*,«: ε*,. = ±1}- Then и — е is an £ Σ*=ί Σ)«ι *,іУ*,і upper bound of {£)™"i i3/n,.: ε, = ±1} and therefore of {xn, — xn}, for any ε к,ι 6 {—1,1}, г = 1,... ,m,k, к = 1,. ..,η — 1. This yields that : и - xn, и + xn are upper bounds of {£t=i Σ™\ ε*,«2/*,ι ε*,, = ±1}. By induction, it follows that и is an upper bound of {£)*=i £k^k'- ε* = ±1} and therefore of {x, —x}, completing the proof. Theorem 3.29 Let E be a directed partially ordered vector space with a majorizing ordered subspace E¡¡. Let ρ be a seminorm on EQ. Define for χ 6 E: η p{x) :=inf{p(¡/i) + --- + p(y„): yu .. • ,yn € E0 such that χ < {^СкУк- ε* = ±1}}. /fc=l Then p is the largest pre-Riesz seminorm on E that is on E0 less than or equal to p. p extends ρ if and only if ρ is pre-Riesz. Proof. It follows directly from the definition that ρ is a seminorm on E. For χ € £Ό it is clear that p(x) < p(x). To prove that ρ is pre-Riesz, it suffices to establish for every χ ζ E that p(x) = ïnÎ{p(Xi) V ... V p(xn): xl}... ,xn € E such that there are А ь..., A„ e (0,1] with ^2kXk = 1 and ι < {^,kekXkxk:ek = ±1}} and to use Proposi tion 3.13. Let χ 6 E. It is clear that p(x) is greater than the infimum at the right hand side. To prove that it is not strictly greater, let x\,..., xn e E and λι,...,Α„ G (0,1] with Σ/t^* = 1 be such that χ < {^,kek\kxk: ek = ±1}, and 48 3.4. Extending Pre-Riesz Seminorms let ε > 0. By the definition of p, for each к there are Ук,\,- •• ,Ук,тк € E0 such that xk < {Σ?=\ ekyk¡t: ε, = ±1} and p(xk) > p(yk,i) + l· РІУк,тк) - ε. Then, by the lemma, χ < {Σί=ι Σ™\ ε*,,λ*2/*,.: ε*,, = ±1}, so ρ(χ) < Σ* Σ. Ρ(λ*2/*,.) = λ Λ χ ε Σ* *Σ,ί>(2/*,.) ^ Σ* *(ρ( *)+ ) ^ ρ(ΐι) V... νρ(ιη)+ε. It follows that ρ is рге-Riesz. Let g be a pre-Riesz seminorm on E with q < ρ on Eo- 1ΐ χ £ E and ¡/ι,...,y„ € E0 are such that χ < {J^ε*2/*: ε* = ±1} and p(yi) Η h p{yn) < 1> then there are Ai,...,An € К with λ* > p{yk) > ЧІУк) for all к and Σλλ* = 1· Then £ 1 £ χ < {^2к к^к(^к Ук)'- к = ±1}, so that q(x) < 1, because the unit ball of q is solvex. It follows that q(x) < p(x), so ρ is the largest pre-Riesz seminorm on E that is less than or equal to ρ on Eo. If ρ is pre-Riesz, the same argument shows for an χ € £Ό that p(x) < p{x), so that then ρ = ρ on E0. Conversely, if ρ extends p, then for every 2 e ΕΌ and = Xi,...,x„ € E0 with p(xfc) < 1 for all λ and Ab...,A„ 6 (0,1] with 2*^* 1 £ such that ι < {^2k k^k^k- ε* = ±1} the definition of ρ implies that p(x) = p(x) < ρ(λιΐι) Η l-p(Anin) < 1, so that the unit ball of ρ is solvex, or, in other words, ρ pre-Riesz. The following example illustrates the extension theorem. Example 3.30 χ >-> |x(0)| + |x(l)| is a Riesz seminorm on Aff[0,1] and its largest Riesz extension to C[0,1] is χ >->· 2infneRsup{|x(í)| - a(t - 1/2): t 6 [0,1]}. Let E = C[0,1] and let E0 be the subspace of all affine functions: E0 = Aff[0,1]. Let p(x) := |χ(0)| + |χ(1)|, χ e Ea. E and E0 are Riesz spaces and E0 is majorizing in E, but Eo is not a Riesz subspace of E. To avoid confusion, denote the absolute value in E by |.| and the one in £Ό by |.|0. For every χ € E0 one has |x|o(0) = |^(0)| and |x|o(l) = |x(l)|. Therefore ρ is a Riesz seminorm on E0. Clearly, χ ь+ |x(0)| + |x(l)| is a Riesz seminorm on E extending p, but it is not the largest extension of p. Indeed, if χ 6 E and y\,...,yn ε Eo are such that |x| < |yi| + · · · + |yn|, then W < ІУіІо + · · • + ІУпІо, and р{уг) + •••+ p(yn) = p(|yi|0 + · · • + |yn|0), so that the formula of Theorem 3.29 reduces to p(x) = inf{p(y): у e E0,y > |x|}, χ S E. \у{0)\ + |y(l)| = 2у(1/2) for every у e EQ, SO every у € EQ can be written as y{t) = 2_1p(î/) + ce(t - 1/2), ί € [0,1], for some a € R. Hence y > \x\ if and only if 2_1p(y) > |x(i)| - a(t - 1/2) for all ί € [0,1], with a € R such that y(t) = p{y) + a(t - 1/2), t € [0,1]. Thus p{x) = 2 infaeRsup{|x(<)| - a(t - 1/2): t € [0,1]}, xe E. In the example, ρ is not equal toi4 |x(0)| + |x(l)|, because p(x) > 2|x(l/2)| for χ € E. This shows once more that there can be distinct Riesz extensions (see also 49 Chapter 3 Riesz Seminorms on Non-Riesz Spaces Example 2.7) The special circumstances in this example which provide the reduction of the formula of Theorem 3.29 will be studied in §3.7 and 3.8. 3.5 Extensions and Non-Uniqueness Every pre-Riesz seminorm on a majorizing subspace of a Riesz space can be ex tended to a Riesz seminorm on the whole space. Theorem 3.29 gives a formula for the greatest extension. Example 3.30 shows that there can be distinct extensions. Example 2.7 shows that even a Riesz norm on an order dense Riesz subspace may be extendable to distinct, even inequivalent, Riesz seminorms. Just as in Proposition 2.8, norm denseness with respect to the greatest extension guarantees uniqueness: Proposition 3.31 Let E be a Riesz space with a majorizing ordered subspace E0 and let ρ be a pre-Riesz seminorm on Eo- If EQ IS norm dense with respect to the greatest Riesz norm ρ on E extending p, then ρ is the only Riesz seminorm on E extending p. Proof. Every Riesz extension of ρ is p-continuous, so if EQ is p-dense in E, then the extensions coincide. If the subspace is not norm dense and it is a Riesz subspace, then there is, as in Theorem 2.10, an equivalent Riesz seminorm with distinct extensions: Proposition 3.32 Let E be a Riesz space with a majorizing Riesz subspace Eo and let ρ be a Riesz seminorm on Eo. If Eo is not norm dense in E with respect to the greatest Riesz seminorm on E extending p, then there is a Riesz seminorm which is equivalent to ρ and which has distinct extensions. Proof. Let ρ be the greatest Riesz extension of p. According to Lemma 2.9, there + exist a continuous linear function f:E—>R and an x0 e E such that f{xo) Φ 0 + _ and ƒ = 0 on EQ. Take p^x) = p{x) + / (|x|), p(x) + / (|ι|), χ e E. Then ρλ and P2 are Riesz seminorms on E and they are equivalent to p, because /+ and f~ are continuous, by Proposition 1.51. /+ = f~ on Eo, so pi = p-¿ on .Eo, and + / (zo) / /~(x0), so pi(xo) Φ Pì{xo). 50 3 6 Extensions and L- and M-Seminorms 3.6 Extensions and L- and M-Seminorms In what cases is the greatest Riesz extension of a pre-Riesz semmorm an L- or M-semmorm' It will be proved that the greatest Riesz extension of a pre-Riesz ьсшшоші is hardly ever an L-seminorm, but is an M-seminorm if its unit ball is in a certain sense dnected Proposition 3.33 Let E0 be a pre-Riesz space with a pre-Riesz seminorm ρ The following statements are equivalent (ι) There is a Riesz space E that contains EQ as a majorizing ordered subspace such that the greatest Riesz seminorm on E extending ρ is an M-semmorm (u) For every Riesz space E that contains Ες, as a majorizing ordered subspace, the greatest Riesz seminorm on E extending ρ is an M-semmorm (m) For every X\, ,xm G EQ with p{xi), ,p(xm) < 1 there are yi, ,yn€ EQ s such that {xu ,xm} < {J2k kVk ek = ±1} andp{yi) + +p(yn) < 1 Proof. (n)=»(i) £o is pre-Riesz, so there is a Riesz space Ε that contains EQ as a majorizing ordered subspace The greatest Riesz extension of ρ on this space is, an M-seminorm, by assumption (n) (і)=>(ш) Let E be a Riesz space that contains E0 as a majorizing ordered subspace and let ρ be the greatest Riesz extension of ρ Let xlt ,xm (Ξ E0 be such that P(x\), iPfrm) < 1 Then p{\Xl\ V V |im|) = р(ц) V V p{xm) = р(ц) V V p(xm) < 1 By the formula for ρ in Theorem 3 29, it follows that there exist 2/1, ,Vn € besuch that p(yx)+ +p{yn) < 1 and |xi|V V|xm| < \yx\+ +\yn\, so {τ,, ,im} < {Y,kekyk £k = ±1} (ni)=>(ii) Let E be a Riesz space that contains E0 as a majorizing ordered subspace and let ρ be the greatest Riesz extension of ρ For any x, y € E, p(\x\ V |y|) > p(\x\) V р(Ы), because ρ is Riesz To prove that p(\x\ V \y\) < p{\x\) V p(\y\), let x,y € E be such that p(x),p(y) < 1 It suffices to show that p(\x\ V \y\) < 1 There exist ib ,XJ,IÌ+], ,xm e E0 such that \x\ < \xi\ + + \xi\, \y\ < \xt+i\ + + \xm\, p{xi) + +p{xi) < 1, and p{yt+i) + + p{xm) < 1 Then there exist Αχ, ,\¡,X¡+1, Am e К such that λχ + +λ|<1, λ|+ι+ + Xm < 1, and A, > р{хг) for all ι By assumption, there are yb ,yn g E0 such that 1 р(Уі) + +р(Уп) < 1 -ìnd {Af'ari, ,λ- ^,,,} < {Σ}ε,ν} ε3 = ±1} Then \х,\<М\Уі\+ + І2/„|) for every г, so |ι| < (λ! + + А,)(ІУі| + + |у„|) < ІЗ/1І+ +\Уп\, and \у\ < \уі\ + + \уп\, hence |і| V \у\ < \уг\ + + \уп\ This yields that р(\х\ V \у\) < р(у\) + +р(уп) < 1, which completes the proof 51 Chapter 3. Riesz Seminorms on Non-Riesz Spaces Because the property in (iii) is not very illuminating, some special cases are men tioned. Corollary 3.34 Let E be a Riesz space with a majorizing ordered subspace E(¡ and let ρ be a pre-Riesz seminorm on E. Then the greatest Riesz seminorm on E ex tending ρ is an M-seminorm if ρ has at least one of the following properties: (i) The unit ball of ρ is directed. (ii) The unit ball В of ρ is directed in the sense that f or every x\,X2 € В there is ay £ В such that у > {ii,^}. Proof. Observe that (i) implies (ii). If (ii) is satisfied, it follows by induction that for every i\,... ,xm S ΕΌ with p{x\) + • • • + p(xm) < 1 there is a y € В such that у t> {x\,... ,xm} and p(y) < 1. Now apply Proposition 3.33. Note that the unit ball of an M-seminorm on a Riesz space is directed, so that the greatest Riesz extension of an M-seminorm is an M-seminorm. There may be distinct M-extensions. This can be seen by noting that ρ and q in Example 2.7 are M-norms on EQ and E. The greatest Riesz extension of an L-seminorm is only an L-seminorm in case of norm denseness. This will be proved via dual spaces, so let us have a look at them, first. Lemma 3.35 Let E be a Riesz space with a Riesz seminorm p. The following statements are equivalent: (i) ρ is an L-seminorm (ii) The operator norm on E' is an M-norm. Proof. This result is well known in case ρ is a norm (see e.g. Meyer-Nieberg[17, p.35, Prop.1.4.7], ). Let N := {x € E: p(x) = 0}. N is a Riesz ideal in E, so the quotient space EQ := E/N is a Riesz space. Let q: E —> EQ be the quotient mapping. Define a seminorm p0 on EQ by po{q(x)) := p{x), x € E0. po is a Riesz norm. The mapping q': E' —¥ E'0 defined by q'(f)(x) := f(q(x)), f € E'0, χ € E, is linear, bijective, bipositive, and isometrical. If ρ is an L-seminorm, then po(q(x) +q(y)) = p(x + y) = p(x)+p(y) = p(q(x)) + p(q{y)), for every x,y G E+, so po is an L-norm on EQ. Conversely, if po is an х L-seminorm on E0, then p(x + y) = Po{q{x + y)) = Мя( ) + ЯІУ)) - Ра{я(х)) + Ро{я(у)) = р{х) + р(у) f°r а'1 χι У € Е+ > so Ρ is an L-seminorm on E. Hence: ρ is 52 3 7 Regular Seminorms Introduction 1S an L-scmmoim if and only if p0 an L-norm, and this is the case if and only if the operator norm on Etí is an M-norm, which is equivalent to the operator norm on E' being an M-noim Proposition 3.36 Let E be a Riesz space with a majorizing Riesz subspace EQ and let ρ be a Riesz semmorm on EQ Let ρ be the greatest Riesz semmorm on E extending ρ For every ƒ € (E,p)'+ one has that f]^ 6 (EQ,P)H and ll/lfollfí, = Proof. Because EQ is a Riesz subspace of E, the formula for ρ reduces to p(x) = mf{p(y) y€E0,y> \x\}, xeE Let ƒ e (Ε,ρ)* Clearly, f\bo € (E0,p)'+ and x 6 Wltn x < ||/|¿olbó — II^IU' For E P( ) 1' there is у e E0 with у > \x\ and p(y) < 1 Then \f(x)\ < f(\x\) < f(y) = f\Eo(y) Hence ||/||E< = sup{[/(x)| χ <Ξ E,p(x) < 1} < sup{/U0 у e Eo, p(y) < 1} = \\/\Е0\\Е'0 Theorem 3.37 Let E be a Riesz space with a majorizing Riesz subspace EQ and let ρ be an L semmorm on EQ Let ρ be the greatest Riesz semmorm on Ε extending ρ Then ρ is an L-semmorm if and only if Eo is p-dense m Ε + Proof. p{xn-x) -»• 0, p{yn-y) -+ 0 Then p(x+y) = hmnp(a;n+y„) = Umn(p(xn)+p(yn)) = p{x) + p{y) =>) Suppose that EQ IS not p-dense in E Because of Lemma 3 35 it suffices to prove that the operator norm on E' is not an M-norm There is an ƒ e E', f φ 0, with / = 0οη£ο Then ƒ+,ƒ-€£'and/+V/-=ƒ++ƒ- Denote ƒ„ = /+|E, =/-|£¿ Because of the previous proposition and Lemma 3 35 again, it follows that + ІІГ Гіи- = ||/ + r||E- = ||/o + /o||Ei = 2(||/o||£-v|l/0|U¿) = 2(||Л|Я, ν ||ли > l№'ν iiriu,, because /+ or f~ φ 0 on E So || \\E' is not an M-norm, hence ρ is not an L- seminorm 3.7 Regular Seminorms: Introduction In the first part of this chapter the notion of Riesz semmorm has been generalized to pre-Riesz seminorm on directed partially ordered vector spaces, by considering 53 Chapter 3. Riesz Seminorms on Non-Riesz Spaces extendability to a Riesz seminorm on any larger Riesz space. The notion of pre- Riesz seminorm has a disadvantage that turns out to be a rather crucial obstruction to generalizing results about Riesz seminorms: a pre-Riesz seminorm need not be deteimined by its restriction to the positive cone. Example 3.38 A pre-Riesz space E with pre-Riesz seminorms ρ and q such that ρ = q on E+ and such that ρ and q are not equivalent. Let S be the unit circle in K2, D the unit disk, and take for E the space of restrictions to D of affine functions from R2 to R. In Example 3.6 it is proved that E can be seen as an order dense subspace of a Riesz subspace of C(S), so E is pre-Riesz. Take p(x) := \x(l,0)\ + \x(-l,0)\ and q{x) := |z(0,1)| + |z(0, -1)|, χ € E. Clearly, ρ and q are restrictions of Riesz seminorms on C(S) and therefore they are pre-Riesz seminorms, by Corollary 3.15. Any χ £ E+ is affine, so p(x) = i(l,0) + x(—1,0) = x(0,0) and, similarly, q(x) = x(0,0), hence ρ = q on E+. To show that ρ and q are not equivalent on E, take x(s,t) := s, (s,t) € D. Then χ € E, p(x) = 2, and q(x) = 0, and therefore ρ and q are not equivalent. In the rest of this chapter we will study a generalization of Riesz seminorms based on the property of being determined by the restrictions to the positive cone. Such a generalization is obtained by extending the idea of the regular norm of operator theory to more general situations. If E and F are normed Riesz spaces, then L := C(E, F) is a partially ordered vector space with respect to the ordering associated with the cone of positive oper ators (see Definition 1.1), but it is in general not a Riesz space. The operator norm ||.|| is monotone (it is Premlin, see Theorem 5.16(iii)), but it need not be Riesz, not even if L is a Riesz space. In that case, more general if U := L+ - L+ is a Riesz space, the regular norm or r-norm is defined as А м· || \A\ ||, Α ε U. Meyer- Nieberg[17, p.27, Prop.1.3.6] defines the r-norm also if U is not Riesz, namely as + ||Л||Г := inf{||B||: \Ax\ < Bx for all χ e E ), A e U. It is this definition that can be rephrased in a much more general context. The notion thus introduced also appears in the literature on duality of ordered normed spaces, see e.g. Davies[7], Ng[19], or Wickstead[24]. It is very useful in the theory of spaces of operators between normed partially ordered vector spaces. One of the main properties of the r-norm is that, if (L, ||.||) is complete, then so is T (L , \\.\\T). In §3.9 it will be shown that such a result is not restricted to operator norms. Definition 3.39 Let E be a directed partially ordered vector space with a semi- norm p. ρ is called regular if p(x) = mï{p(y): y € E such that — y < χ < y} for all χ e E. 54 3 7. Regular Seminorms: Introduction Proposition 3.40 Let E be a directed partially ordered vector space. Every regular semmorm on E is pre-Riesz. Consequently, ij E is a Riesz space, then a seminorm on E is regular if and only if it is Riesz. Proof. Let ρ be a regular seminorm on E. Let χ € E. If Χχ,... ,xn 6 E and Ai,..., A„ € (0,1] with J2k ^k = 1 are such that ζ < {J2k£k^k^k'- £* = ±1}>tnen *°г every iji,..., yn € E such that yk > Xk, -Xk, A; = 1,..., τι, it follows that £fc Afcj/fc > x, -x, hence p(x) < p(J2k ^кУк) < РІУі) V ... V p(yn). Thus, p(x) < inf{p(zi) V ... Vp(i„): ii,...,in € £ such that there are Ai,... ,An 6 (0,1] with ^2k Xk = 1 and χ <¡ {Y^kEkXk'^k'· £k = ±1}}, and the last expression is clearly not greater than p{x). According to Proposition 3.13, it follows that ρ is pre-Riesz. The proof is completed by reference to Proposition 3.12(ii). Example 3.38 shows that there are pre-Riesz seminorms that are not regular. This chapter continues with some properties of regular seminorms and their topologies. Those that have to do with extensions, norm completeness, and norm completions are very convenient, as is shown in the last sections of this chapter. In Chapter 5 the notion of regular seminorms will be applied in operator theory. The next lemma illustrates their use in that respect. It is a special case of Lemma 5.14(ii). Lemma 3.41 Let E be a directed partially ordered vector space with a regular semi- norm p. For every f 6 E~ and g e E' such that 0 < ƒ < g one has that f is continuous and \\f\\ < \\g\\. Proof. Let χ 6 E and ε > 0. Because ρ is regular, there is a y € E+ with —y < χ < y and p(y) < p(x) + ε. Then | ƒ (x)| < f(y) < g{y) < \\g\\p(y) < \\g\\(p(x) + ε). The following fact is an inconvenient one: the set of regular seminorms need not be closed under addition. Example 3.42 A directed partially ordered vector space with two regular seminorms Pi and p-i such that p\ + Рг is not regular. Let S be the unit circle in R2, D the unit disk, and E the space of restrictions to D of affine functions from R2 to R (see also Examples 3.6 and 3.38). Take Pi(x) := |x(l,0)| and Рг(а;) := \x(—1,0)|, χ e E. Then £ is a directed partially ordered vector space and p\ and p2 are seminorms on E. Example 3.38 shows that Pi + P2 is not regular. To see that pi and pi are regular, let χ £ E. Then |i| € C(S) and, from Example 3.6(b), |i(i)| = inf{u(i): и 6 E, и > \x\} for any t € 5, in particular for t = (1,0) and t = (—1,0). This yields that pi and p-¿ are regular. 55 Chapter 3 Riesz Seminorms on Non-Riesz Spaces 3.8 Extending Regular Seminorms and Regular- izations Fiom Definition 3 39 it is dear that regular seminoims are determined by their ipstnctions to the positive cone Can the restriction of an arbitrary seminorm to the positive cone be extended to a iegular scramorm' Obviously, monotonicity is needed We will see that it is also sufficient The construction can as well be foimuldted more generally, yielding extensions to larger spaces Theorem 3.43 Let E be a directed partially ordered vector space with a majorizing ordered tubspace EQ and let ρ be a seminorm on E0 Define pT{x) = uii{p{y) y e E0 such that — y Then pT is the greatest regular seminorm on E that is less than or equal to ρ on EQ pT = ρ on EQ if and only if ρ is monotone, pT > ρ on EQ if and only if ρ is Fremhn, pT = ρ on EQ if and only if ρ is regular Proof, (a) pr is a seminorm Let Xi,x2 6 E and 3/1,2/2 6 EQ be such that — j/i < x\ < 3/1, -2/2 < хг < 2/2 Then -(уг + y2) < Τι + x2 < 2/i + 2/2 and рт{хл + x2) < РІУі+Уг) < р{уі)+рЫ), s°Pr(xi+X2) (c) pr is regular Let χ g E Let y € E Ъе such that —y < χ < y If ζ S EQ IS such that — z < y < z, then — z < χ < z, so pT{v) < pT{y) For every ε € (Ο,οο) there is y G EQ such that —y Pr{y) < Pr{x) + ε Hence pr(x) = inf{pr(y) y e Ε, -y < χ < y} (d) If ρ is a regular seminorm on E and ρ < ρ on Eg , then ρ < ρτ For χ € Ε one hai, ρ(χ) = mf{p(y) y e E,-y (e) pT = ρ on EQ if and only if ρ is monotone If pr = ρ on EQ, then for x, y 6 EQ with χ < y we see that p(x) = pr{x) < p{y) If ρ is monotone, then pT > ρ on EQ, so pT — ρ on EQ, from (b) (f) Pr > Ρ on EQ if and only if ρ is Fremhn Straightforward (e) Pr = ρ on EQ if and only if ρ is regular Directly from the definition of regularity for ρ This theorem justifies the following definition Definition 3.44 Let E be a partially ordered vector space with a seminorm ρ The regular seminorm pT defined by pT{x) = inf{p(y) y 6 E, — y < χ < y}, χ ζ Ε, іь called the regularization of ρ 56 3.9. Regularizations and Norm Completeness For EQ = E, Theorem 3.43 yields some properties of the regularizaron. Let us state them explicitly. (Compare with Davies[7, Lem.2.1]). Corollary 3.45 Let E be a directed partially ordered vector space with a seminorm p. Let pT be the regularization of p. Then: (i) ρ = Pr on E+ if and only if ρ is monotone. _1 (ii) If ρ is Kremlin, then pT > p. If ρ is monotone, then pT > 2 p· Consequently, if ρ is a monotone norm, then pr is a norm. Proof, (i) and the first assertion in (ii) follow directly from the theorem. To prove the rest of (ii), let p\ be the 'Fremlinization' of ρ (Theorem 2.3), i.e. the greatest Fremlin seminorm on E that is less than p. Then ρ < 2pi on E and ρ = p\ on E+, _1 so their regularizations are equal. Thus, pr = (pi)r > Pi > 2 p, which proves the assertion. Restrictions of regular seminorms need not be regular, because restrictions of Riesz seminorms to subspaces that are Riesz in their own right but not Riesz subspaces, need not be Riesz (see Example 3.2). 3.9 Regularizations and Norm Completeness A surprising fact about the r-norm is that for a normed Riesz space E and a Dedekind complete Banach lattice F the subspace of L := C{E, F) generated by the positive operators is norm complete with respect to the r-norm (see, e.g., Aliprantis & Burkinshaw[2, p.248, Thm.15.2], Meyer-Nieberg[17, p.27, Prop.1.3.6], Schaefer[22, p.230, Prop.1.4]). The standard proof is based on norm completeness of L and closedness of L+ with respect to the unrcgularized operator norm. By noting that the operator norm is monotone (see Theorem 5.16(iii)), the result follows from a more general theorem. The proof consists of the same arguments. Recall that a subset of a normed space is called norm complete (or, more explicitly, p-complete, if ρ is the norm) if it is complete with respect to the metric induced by the norm. Theorem 3.46 Let E be a directed partially ordered vector space with a monotone norm p. Let ρ be the regularization of p. If E+ is p-complete, then E is p-complete and E+ is closed in {E,p). Proof. Remark that, by Corollary 3.45, monotonicity of ρ implies that ρ < 2p, so ρ is a norm and every p-closed set is p-closed. In particular, E+ is p-closed. 57 Chapter 3. Riesz Seminorms on Non-Riesz Spaces Remark also that, because ρ < ρ on E+ and E+ is p-closed, every increasing x p-convergent sequence is p-convergent to the same limit. Indeed, if xn t (p- convergent) in E, then χ > xn for all n, so p(x - xn) < p(x — xn) —> 0. Let (xn)n in E be such that Σρ(χη) < сю. Then there are yn e E such that —2/n < xn < yn and p(yn) < p(^n) for all n. Define un := yn - xn, then + p(u„) < p{yn) + 2p(i„) < 3p(.-r„). E is p-complete, so j/ := Y^yn and u := + Y^un (p-convergent) exist in E . Because yn and un are positive for all n, the sequences of partial sums are increasing and therefore also p-convergent. Then their difference is p-convergent; in other words: £x„ (p-convergent) exists in E, proving p-completeness of E. The above theorem shows that it is the p-completeness of E+ that leads to the p- completeness of E. For the special case that ρ is regular, this yields an interesting property of regular norms. Corollary 3.47 Let E be a directed partially ordered vector space with a regular norm. Then E+ is norm complete if and only if E is norm complete and E+ is closed. Proof. <=) is clear. =*·) follows from the theorem by taking ρ = p. Corollary 3.48 Let E be a directed partially ordered vector space with a monotone norm ρ such that E+ is closed. If E is norm complete, then ρ is equivalent to a regular norm. Proof. According to the theorem, the regularization ρ of ρ makes E norm complete, too. Furthermore, ρ < 2p, by Corollary 3.45. Then it follows from the Closed Graph Theorem that ρ and ρ are equivalent. The equivalence constant in the above corollary may be arbitrary large: Example 3.49 A Riesz space E such that for any number С > 1 there is a mono tone norm ρ on E such that E is norm complete with respect to both ρ and its regularization ρ and such that p(x) = Cp(x) for some nonzero χ g E. 2 1 2 Take E = К and р(х) := {(С - l)' )]^ + \χγ + х2\, х = (xi,x2) e К · Then ρ is a Fremlin norm on E, E+ is closed and E is p-complete. The regularization of ρ is given by p(x) = p(\x\) = ((C - l)"1 + 1)||і||г = C(C - іГЧІхЦ, < Cp(x), χ G E. Take χ = (1, -1). Then p{x) = C{C - l)"1 = Cp(x). 58 3 10 Norm Completions 3.10 Norm Completions In an investigation of norms on partially ordered vector spaces, it is a natural ques tion whether the norm completion of a normed partially ordered vector space can be equipped with a sensible ordering and, if so, what properties the new space has There are satisfactory answers for the following cases monotonely or Fremlin normed partially ordered vector spaces with closed positive cones, regularly normed directed partially ordered vector spaces with closed positive cones, and, of course, normed Ries¿ spaces For the question whether the norm completion of a Riesz space with a monotone norm and a closed positive cone is a Riesz space, there are no general results Detailed consideration of a special case illustrates the kind of complications that may arise in the general case 3.10.1 Norm Completions of Normed Ordered Vector Spaces Let E be a partially ordered vector space with a norm ρ It seems natural to make an ordering on the norm completion E of E by taking the limits of sequences in E+ as positive elements This yields a decent ordering if ρ is equivalent to a monotone norm and E+ is closed Proposition 3.50 Let E be a partially ordered vector space with a norm ρ and let E be the norm completion of E Let К be the closure of E+ in E If ρ is equivalent to a monotone norm, then К is the positive cone of a vector space ordering on E If, m addition, E+ is closed m E, then the К-ordering extends the ordering of E Proof. It is straightforward that A" is a cone with its vertex at 0 Assume that ρ is equivalent to a monotone norm To prove that Κπ(-Κ) = {0}, let χ € Κ Π (—К) + Then there are sequences (un)n and (vn)n in E with un -* χ and — vn —» χ Then 0 < un < un + vn —> 0, so that, by Proposition 1 27, χ = hm„ un = 0 Thus, К is the positive cone of a vector space ordering on E If E+ is closed, then clearly КПЕ = E+ In this text, norm completions will be equipped with the ordering presented m the above proposition Note that the natural isomorphism between the norm dual of a partially ordered vector space with a monotone norm and closed positive cone and the norm dual of its norm completion is norm preserving as well as order preserving What are the properties of the ordering and of the norm on the norm completion7 For certain types of norms and ordenngs of E, the norm completion of E is of the same type as E 59 Chapter 3 Riesz Seminorms on Non-Riesz Spaces Theorem 3.51 Let E be a partially ordered vector space with a monotone norm ρ such that E+ is closed Let E be the norm completion of E and denote its norm by ρ Then (ι) ρ is monotone on E If ρ is Fremhn on E, then so is ρ on E (ιι) If E is directed and ρ is regular, then so are E and ρ (m) If (E, p) is a normed Riesz space, then so is (E, p) Proof. (1) Let x,y ζ E Ъе such that 0 < χ < у Then there are (x„)„ and (un)n in + E withxn —> χ and un —> y — x, so thatp(x) — limnp(xn) < Ιιτηηρ(χη+υ,η) =p(y) In the same fashion it can be proved that ρ is Fremhn if ρ is (n) Let χ € E Then there is a sequence (xn)„ in E with x„ —> χ It may be + assumed that p(xn+\ —xn) < 2~" for all η ρ is regular, so there are {yn)n m E with n ~Уп < Гп+і - Xn < Уп and p{yn) < 2 Then Σηρ{νη) < oo, so uN = Ση=Ν У* or a + (norm convergent) exists in E for all N u^ > Ση=/ν У* ^ ^ M > N, because E is closed, so that -u^ < x« - Хлг < «лг for all M > N Applying the closedness of + E once more, it follows that —uN <х — хц<и^ for all Ν E is directed and ρ is regular, so there exist w^ e E with — w^ < ijv < Wjv and p(wN) < p(xN) + l/N In the first place, this yields that и^ + г ^ > x, —x, so that E is directed In the second place, it follows that P(UN +%) < ρ(ω^) +p(wpf) < р{и^) +р(х^) + 1/N —> p(x) Hence inf{p(u) —u < χ < u} < p(x) The opposite inequality follows from the fact that ρ is Fremhn, by (ι), and Propositions 3 40, 3 12(i) Thus, ρ is regular (in) Well-known, see e g Aliprantis к Burkinshaw[2, ρ 175, Thm 12 2], Schaefer[22, ρ 84, Cor 2], also Aliprantis & Burkinshaw[l, ρ 43, Thm 7 1], Fremhn[8, ρ 41, Prop22F] 3.10.2 On the Norm Completion of a Monotonely Normed Riesz Space Is the norm completion of a Riesz space with a monotone norm and a closed positive cone a Riesz space7 The answer is negative For norms of the form χ t-> p(x) + |/(x)|, where ρ is a Riesz norm and ƒ a positive linear function, it is shown what cases may appear, including one m which the norm completion is not directed It turns out that the properties of the norm completion depend on the set of continuous, positive lineai functions that are majorized by ƒ Let us consider such sets, first Let E be a partially ordered space with a seminorm and let ƒ e E~+ For abbreviation, ƒ will be called completely discontinuous if it is nonzero and does not 60 3.10. Norm Completions majorize any nonzero continuous, positive linear function, i.e. {g 6 E': 0 < g < /} = {0}· Lemma 3.52 Let {E,p) be a normed Riesz space and let ƒ € E~+ be completely + discontinuous. Then for every x0 g E and every a € [0, oo) there is a sequence + (xn)n in E with xn —> XQ and f(xn) —> a. + Proof, (a) There is a sequence (yn)n in E with yn —¥ XQ, and f{yn) —> 0: Define p(x) := inf{p(u) + f{v): u, υ 6 E+, \x\ < u + v}, χ e E. Then ρ is a Riesz sominorm on Ε, ρ < ρ on E, and ρ < f on E+. By Hahn-Banach, there is a ρ € E~+ with g < ρ and 5(10) = p(^o)· Then g is continuous and g < ƒ, so g = 0 and therefore + p(io) = 0. Hence there exist (un)n and (vn)n in U with io < u„ + wn for all η + and p(un),f(vn) —» 0. Take ;/„ := (x 0 < f{yn) < IM -» 0. (b) If f(xo) φ 0 on¿a € [0, f{xo)), then the assertion of the lemma holds true: Take (y„)n as in (a). Let λη := (ƒ(ar0) - a)/f(x0 - yn), for η so large that f{yn) < ƒ (i0), r — and let in := X„yn + (1 — Λη)χο f° such n. Then in —» x0 and /(zn) —> /(^o) (ƒ(x0) - α) = a. (c) If a e [/(xo)iOo), i/ien i/ie assertion /ioWs £тае: ƒ is discontinuous and ρ is a + Riesz norm, so there is a sequence (un)n in E with p{un) < 1 for all η and ƒ (un) —» 00. Multiplying with appropriate positive scalars (λ„)„ yields that Anu„ —» 0 and /(Anun) —> α — f(xo)· Take i„ := xo + A„un, η G Ν, then (xn)n has the desired properties. Lemma 3.53 Let (E, p) be a normed Riesz space and let ƒ € E~+ be discontinuous. Then the following two statements are equivalent: fi) {llfllh 9 € E'+,9 < ƒ} ls bounded. + + (ή) There are a continuous fc € E~ and a completely discontinuous fs € E~ such that f = fc + fs- + Proof. (i)=>(ii): Let fc be the supremum of the set {g 6 E' : g < ƒ} in E~. Then + + fc is positive and fc(x) := sup{<7(a:): g e E' ,g < ƒ} for χ € E . Hence, for any + x e E, \fc{x)\ < fc(\x\) < sup{\\g\\: g e £'+,g for all g e E' + with g < f. Take ƒ,:=ƒ- /c. Then ƒ, € E~ and if ρ € £'+ is such that g < ƒ„ then g + fs < ƒ, so that 5 + fc < fc and therefore g = 0. Thus, /s is completely discontinuous. 61 Chapter 3 Riesz Semmorms on Non-Riesz Spaces (n)=>(i) By Lemma 3 41, it suffices to prove the following if g € E'+ is such that g < f, then g < fc Let g G E'+ be such that g < ƒ and let χ € E+ According to the previous lemma, there is a sequence (x„)„ in E+ with x„ -» τ and f¡{x„) —> 0 By continuity of 9, g(x) = hmn £(xn) < 1ітп(/с(хп) + fs[xn)) = /c(x) Hence ρ < ƒ An example of a completely discontinuous, positive, linear function is a point eval uation on (C[0,1], || ||i) The lemma yields that examples of discontinuous, linear functions ƒ on a space E such that {||p|| g e E'+,g < ƒ} is bounded are sums of continuous and completely discontinuous, linear functions An example of a linear + function ƒ on a space E with {\\g\\ g € E' ,g < ƒ} unbounded isin^ifcon ls (coo(N), || ||oo), since gn χ ь* Y^=0Xk continuous for every π 6 N Lemma 3.54 Let (E,p) be a normed Riesz space and let ƒ e E~+ If {\\g\\ g € Ε'+, g < f} is unbounded, then there exists an χ ζ E+ such that for every sequence + (xn)n ín E with p(xn — x) —> 0 one has that f(xn) —t oo + Proof. There are gn € E' with gn < f and ||5n|| > 4", η e Ν, and then there + _n are xn e E with p(xn) < 1 and дп(з-п) > 4™ for all η Then (Ση=0 2 a;n)jv is + n + increasing and Cauchy in E , so χ = J^n 2~ xn exists in E For η 6 N one has n _ + χ > 2 xn, so ffn(^) > Sn(2 "xn) > 2" If (j/„)n m E is such that yn —> x, then 5n(2/Jt) > 2" for large fc, η e Ν, so that f(yn) -> oo Proposition 3.55 Let E be a Riesz space with a Riesz norm ρ and let ƒ € E~+ Letp(x) = p(x) + \f(x)\, χ € E, and let E be the norm completion of E with respect to ρ Then (ι) If f = 0, then E is a normed Riesz space (n) If f is continuous, then Ë is a Riesz space with a monotone norm (m) If f is completely discontinuous, then E is a normed Riesz space (w) If f is discontinuous and {\\g\\ g £ E'+,g < ƒ} is bounded, then E is a Riesz space with a monotone norm (v) If f is discontinuous and {\\g\\ g 6 E'+,g < ƒ} is unbounded, then E is not directed Proof. Note that ρ > ρ, so that E+ is closed and therefore E is a partially ordered vector space with a monotone norm Denote the norm completion with respect to ρ bj Ep and its norm by ρ This space is a normed Riesz space 62 3 10 Norm Completions (ι) Theorem 3 51(ш) (u) ρ extends uniquely to a monotone norm ρ on Ep that is equivalent to ρ The natural mapping t {E,p) —> {Ep,p) is linear and isometrical, i(E) is dense in Ep + and i[E ) is dense in E+ Thus, {Ep,p) is the norm completion of E (in) Let г be the mapping from E to Ep χ R defined by г(х) = (χ, f (χ)) Then г іь a positive, linear ìsometry from E to (Ê+ χ R+,p+ | |) г(£) is dense in Ep χ R and г(£+) is dense in l+xl+, by Lemma 3 52 So £,, χ R is the norm completion off; + + (ìv) By Lemma 3 53, ƒ = fc + fs, where fc e E' and fs € E~ is completely discontinuous The mapping г χ ι-4 (χ,/s(x)) is a positive, linear ìsometry from £ to Êp χ R with ρ (χ, α) ι-> p(x) + \fc{x) + a\ For every (x,a) € Êp χ R there is а a sequence (i„)„ in £ with p(xn — x) —> 0 and fs{xn) —• Q, so р(г(хп) - (я> )) = p(xn - χ) + \fc{xn - x) + (/s(in) — a)\ —> 0 Hence, i(E) is dense in (Ep x R,p) Using Lemma 3 52, it can be shown in a similar fashion that i(E+) is dense in + (Èp χ R , p) Thus, (Êp χ Μ, ρ) is the norm completion of E (ν) Denote the norm of E by ρ According to Lemma 3 54, there is an x g Ep such + that for every sequence (i„)„ in E with p(x„ — x) —> 0 one has that ƒ (xn) —> co Since ƒ is discontinuous, there are (yn)n ш .E with p(yn — x) —> 0 and ƒ (j/n) —> 0 and then (i/n)n іь a p-Cauchy sequence, so that there is an i e £ with p(yn — x) —> 0 + 1 Suppose that there is a û € E with û > χ Then there is a sequence (itn)„ in E" " 1 with p(un — û) —» 0 and, because χ < û, there is a sequence (xn)n in E with xn < un for all η and p(x„ — x) —• 0 Then (x„)n and {yn)n are sequences in £; with the same p-hmit, which yields that they have the same p-limit Hence p(xn — x) —> 0 and therefore p(x„ — x) -¥ 0 Furthermore, 0 < /(x*) < }{un) < p(un) is bounded This contradicts the properties of χ Thus such a û does not exist, which means that E is not directed Case (v) of the proposition shows that E need not be directed If it is, then it must be a Riesz space, though the proof involves different cases If E is a partially ordered vector space with a monotone norm such that E+ is closed and E is directed, then the norm of E is equivalent to a regular norm, by Corollary 3 48 In Proposition 3 57, it will be shown that therefore, if E is a Riesz space, it can be found as a Riesz subspace of E" Consider the following lemma, first Lemma 3.56 Let E be a directed partially ordered vector space with a regular norm ρ Then for every ƒ S E'+ and every x g E one has that mf{/(y) yeE, y>T,0}=sup{5(x) geE', 0 < g < f} 63 Chapter 3 Riesz Seminorms on Non-Riesz Spaces Proof. If y e E is such that y > χ, 0 and g ζ E' is such that 0 < g < ƒ, then ƒ('/) > 9(y) > Φ) Hence mf{/(y) y e E,y > χ,Ο} > sup{?(i) 5 e Ê',0 < ff < ƒ} Let p(u) = inf{/(y) y Ç. E,y > u,Q], и e E Then ρ is a subadditive, positively homogeneous function on E and ρ = ƒ on i?+ By Halin-Banach, there exists a linear function g E -> R with 5 < ρ on £ and 5(1) = p(i) Then g(u) < p(u) = f(u) for all и e E+ and g(u) < p(u) = 0 for all и 6 — £+, so 0 < g < ƒ By Lemma 3 41, it follows that g is continuous The definition of ρ yields the existence of a sequence (yn)n m E with yn > x, 0 and f{yn) —• р(я) = s(i) Then inf{/(j/) ¡/É£,}>I,(I}< ƒ(?/„) -> g{x), and sup{/i(z) h e E',0 < h < ƒ} > ρ(ι) Hence, inf{/(7/) y Ê £,j/ > 1,0} < sup{5(a;) g € E',0 < g < ƒ}, establishing the assertion Proposition 3.57 Let E be a partially ordered vector space with a monotone norm such that E+ is closed Then the norm completion E is a Riesz space if and only if E" 10 a Riesz space and E is a Riesz subspace of E" Proof. <=) Directly =>) By Corollary 3 48, the norm of E' is equivalent to a Riesz norm, so E" is a Riesz space To show that E is a Riesz subspace, define 4>x{f) = f{x), f £ Ε', χ € E χ >-* φχ is a bipositive, linear mapping from E to E" Let χ € E It has to be + shown that φχ+ = φ£ Since x > x,0, one has that φχ+ > φχ,0, hence φχ+ > φ+ For ƒ € E'+ and и > χ,Ο one has φχ+if) — f{x+) < /(w), so, according to the lemma, it follows that φχ+{ί) < mî{f{y) y > χ,0} = sup{ 8ир{<Ш 0 < S < ƒ} = <#(ƒ) 64 Chapter 4 Subspaces of Normed Riesz Spaces The goal of this chapter is to find an intrinsic description of the class of all normed partially ordered vector spaces that are isomorphic to ordered subspaces of normed Riesz spaces. The description turns out to be fairly elegant: all partially ordered vector spaces with Fremlin norms and closed positive cones. The second part of this chapter presents a very similar result for embeddability in a Riesz space with an M-norm. There is an interesting application in operator theory, see Theorem 5.17. 4.1 Introduction To tie in this chapter with the previous chapters, recall that Chapter 2 is arranged around the question which norms can be extended to a monotone norm on any larger Riesz space. Chapter 3 is about a description of all norms that can be extended to a Riesz seminorm on every larger Riesz space. This chapter describes the norms for which there exists a larger Riesz space that has a Riesz norm extending the given norm. So in this chapter both the space and the norm must be extended, whereas in the second and third chapters the larger space was given. In the second and third chapters it was more convenient to consider seminorms instead of norms. The meth ods presented in this chapter will not work for seminorms. Some generalizations for seminorms will be given, but they involve rather unappealing, technical conditions. Intuitively speaking, if a monotone norm or rather a Fremlin norm on a Riesz space is not Riesz, then that is because there are elements with absolute values that are too large in the sense of the norm. By extending the space with positive elements, there will arise more upper bounds and therefore absolute values will decrease, making the gap between the norms of an element and its absolute value smaller. So, roughly speaking, embedding into a normed Riesz space comes down to finding a sufficiently 65 Chapter 4. Subspaces of Normed Riesz Spaces largo extension of the space. The construction piesented below uses Hahn-Banach theorems. Two properties of noi med Riesz spaces that are inherited by subspaces are: closed- ness of the positive cone and the norm being Fremlin. More precisely: Proposition 4.1 Let E be a Riesz space with a Riesz norm ρ and let Eo be an ordered subspace. Then (ι) EQ is closed in Eo- (n) The resctnction of ρ to Eo is a Fremlin norm. Proof, (i) Because E+ is closed in E. (ii) From Lemma 1.4(i). This proposition provides necessary conditions for a norm to be extendable to a Riesz norm: it must be Fremlin and the positive cone must be closed. It will be shown that these conditions are sufficient, too. That is to say that for any partially ordered vector space E with a Fremlin norm ρ such that E+ is closed there exist a Riesz space F and a Riesz norm ρ on F such that there is a bipositive, isometrical, linear mapping from E to F. The space F will be constructed as a space of functions on the unit sphere of £". Therefore the properties of E and ρ should be described in terms of E'. According to Proposition 1.54, E+ is closed if and only if E'+ determines the ordering of E. A dual characterization of Fremlin norms will be developed in the next section. 4.2 Dual Characterization of Fremlin Seminorms The Fremlin property for a seminorm ρ on a partially ordered vector space E can be described in terms of E'+ and p. The description will be derived in two steps. First, closedness of E+ will be assumed, inherently limiting the setting to norms. Then, via quotient with respect to the kernel and closure of the positive cone, the general result is obtained. The main idea is in the following lemma. Lemma 4.2 Let E be a partially ordered vector space with a norm ρ such that E+ is closed. If x, yu ..., yn e E are such that \f(x)\ < |/(j/i)| + ... + |ƒ(y„)| for all + f £ E' , then there are Ab ..., λη,μι,... , μη € [—1,1] such that Aij/i + ... + \nyn < x < μνΊΙι + ... + μη])η- Proof. Let x,yi,... ,yn be as in the assumptions. The numbers λ* and μ* will be obtained via a compactness argument and Hahn-Banach. 66 4.2. Dual Characterization of Fremlin Seminorms Let Bp := {z G E: p(z) < 1} and Y := {ХіУі + ... + ХпУп: Аь ..., λ„ € [-1,1]}. + Define for ε > 0 the set W€ := (—E ) + Υ + εΒρ. We is convex and absorbing; let qE be its Minkowski function. Then qe is subadditive and positively homogeneous, _1 4 4ε < ε Ρ on E, and q€ < 0 on — J5 ". Bp is open, hence We is open, so W£ = {z G £: і?£(л) < 1}, yielding that qe < 1 on У. (a) For ewen/ ε > 0 one has χ G VV£ and —ι € W£.· Let ε > 0. Suppose that ι (or -a.) 0 W£. Then, according to Hahn-Banach, there· exists a linear function ƒ:£->! with ƒ < qE and Дх) = qc{x) > 1 (or f{-x) = qc{-x) > 1), so _1 + \f(x)\ > 1· ƒ < 9e < г р on £ and ƒ < g£ < 0 on -E+, so ƒ e E' and therefore, by assumption, 1 < |/(i)| < |/(yi)| + - • - + |/Ы| = /(±ϊι + · • · + ±!Λ.) < ï.(±Vi + h±yn) for the right choice of+- and —signs. However, ±yiH \-±yn €Y CWe so ç£ (±1/1 + · · · + ±i/n) < 1, contradicting the previous inequality. Thus χ 6 Wc (and -x e W£). (b) There exist \\,..., λη, μι,..., μη G [—1,1] such that Хіуі + • • • + Xnyn < χ < μιj/i + h ßnVn-' By (a), ι 6 Wi/n for all η e Ν*, so for every η e Ν* there are + i α„ € (—E ) + Y and zn € n~ Bp with χ = α„ + zn. p{zn) < 1/n —У 0, so ζ = lim α„. Since i?"1" is closed and Y is compact, (—E+) + Y is closed, so that χ e (-£+) + V. The same arguments prove that also — χ € (—E+) + Y. This means that there are + u, ν € E and λι,..., An, μι,..., μ„ G [—1,1] with χ = — и + ХіУі Η 1- Xnyn and χ = υ- (μι2/ι + hß„yn). Thus: (-μι)ΐΛ Η Η (-μη)ΐ/η = χ-υ<χ<χ + ω = λι2/ι Η h Anj/n, proving the assertion. The following lemma is an elementary observation about Fremlin seminorms. Lemma 4.3 Let E be a partially ordered vector space with a Fremlin seminorm p. Then for any a,b e E one has: If a < χ < b, then p(x) < 2~l(p(a + 6) + p(b — a)). Proof, a - 2_1(a + 6) < χ - 2_1(a + ό) < ò - 2_1(o + b), or, in other words, -2-40 - a) < χ - 2-г(а + 6) < 2'l(b - a), so p(x - 2_1(a + 6)) < 2~lp(b - a) and therefore p(x) < 2_1(p(a + b) +p(b - a)). Combination of both lemmas leads to the most interesting part of the dual charac terization of Fremlin norms with closed positive cones. The condition of closedness of E+ can be eliminated by considering the closure of the cone. Then, with aid of quotient spaces, the result can be extended to Fremlin seminorms. These steps are carried out in the following lemma. Lemma 4.4 Let E be a partially ordered vector space with a Fremlin seminorm p. Ifx,yu...,ya€E are such that \f(x)\ < \f(yi)\ + ---+\f{yn)\forallf G E'+, then p(x) < p(yi) -\ + p(yn). 67 Chapter 4. Subspaces of Norrned Riesz Spaces Proof, (a) The assertion holds true if ρ ts a Fremhn norm and E+ is closed: Let x,u\ Уп e E be such that |/(i)| < \f(yx)\ + ·•· + \f(yn)\ for all ƒ € + E' . Then, according to Lemma 4.2, there are λι,..., λη,μι,... , μ„ € [—1,1] with \\iji + • • • + Xnyn < χ < ß\y\ + · · · + μη2/η- Ρ is Fremlin, so, using the previous lemma, it follows that 1 p(x) < 2- [p((A1 + μι)νι + • · · + (A„ + /i„)i/n) + Р((Мі - Ai)i/i + · · · + (μ„ - An)y„)] < Ί-'ΚΙΧ, + /л| + |Ai - ßi\)p[yi) + ••• + (|λη + μη\ + |An - μη\)ρ(νη)} since |Α + μ| + |λ — μ\ < 2 —simply because |λ + μ| + |λ — μ| = 2λ, —2λ,2μ, or -2μ— for any μ, λ € R with |λ|, |μ| < 1. (b) The assertion holds true if ρ is a Fremhn norm: By Proposition 1.16, E+ is the positive cone of a vector space ordering on E and ρ is Fremlin with respect to this ordering. For ƒ £ E' one has that ƒ > 0 on E+ if and only if ƒ > 0 on E+. So for ,+ any і,й,...,й,е ^one has. if \f{x)\ < \f{yi)\ + • • •+ \f(yn)\ for all ƒ € E , then also for all ƒ € (E,E+)'+, so, by (a), p(x) Proposotions 1.17 and 1.55, EN is a partially ordered vector space, the quotient norm is a Fremlin nom, and E'N is isomorphic to E'. So, if x, yit..., yn g E are such that |/(x)| < |/(yi)| + · · · +1/(3/„)| for all ƒ 6 ^'+, then |/(x)| < |/(»ι)| + ···|/(»»)| for all ƒ e E'N (where ö denotes the equivalence class in Ε μ of и ξ. E), so that, by (b), p(x) = p(x) < р(Уі) + · · • + p(yn) = p(yi) H + p(yn). Thus, the proof is complete. The lemma states a necessary condition for a norm to be Fremlin. By remarking that the condition is also sufficient, the announced dual characterization of the Fremlin property is established. The implication from (i) to (a) will be of great importance in the next section. Proposition 4.5 Let E be a partially ordered vector space with a seminorm p. The following statements are equivalent: (ι) ρ is Fremlin. (n) Whenever x,yu ..,yneE and \f{x)\ < \f(y,)\ + - •• + \f{yn)\ for all f e E'+, one has that p(x) < p(yi) Η РІУп)- 68 4.3. Embedding in a Normed Riesz Space (ivi) Whenever x,y € E and \f(x)\ < \f(y)\ for all ƒ € E'+, one has that p(x) < p(y)- Proof. (i)=>(ii) is in Lemma 4.4. (ii)=i-(iii) is a direct consequence (take n=l). (iii)=>(i): Let x, y € E be such that -y < χ < y. Then -f (y) < f{x) < f (y) for every ƒ € E'+, so p{x) < p(y), proving that ρ is Fremlin. 4.3 Embedding in a Normed Riesz Space This section presents a construction that, for every partially ordered vector space E with a Fremlin norm ρ such that E+ is closed, yields a normed Riesz space (F, p) in which E can isometrically, bipositively be embedded. F will in general be much larger than E, even if {E,p) is itself already a normed Riesz space. E need not even be a Riesz subspace of F. The construction of F can be modified to produce an enveloping space that is equal to E in case (E, p) is a normed Riesz space. This will be discussed in §4.3.2. The modified construction is, however, rather artificial. There does not always exist a smallest normed Riesz space F, as is shown in §4.3.3. 4.3.1 Construction F will be constructed as a space of functions on the positive part of the unit sphere of E', based on the idea of embedding E in E". Let us show with an example that E" itself need not be a Riesz space and that the norm of E" need not be pre-Riesz, so that the enveloping Riesz space of E" can not be equipped with a Riesz norm extending p. Example 4.6 A directed pariially ordered vector space E with a Fremlin norm ρ such that E+ is closed and such that E" is not a Riesz space and the norm of E" is not pre-Riesz. Take the reflexive partially ordered vector space with Fremlin norm that is presented in Example 5.8. The theorem presenting the embedding of E will be stated with an explicit descrip tion of F and p. The more concise but less informative plain existence statement is given as a corollary. The choice of the space F is rather obvious (a similar con struction is given in Davies[7]). The interesting part is the definition of the norm and the proof that it extends p. 69 Chapter 4. Subspaces of IMormed Riesz Spaces Theorem 4.7 Let E be a partially ordered vector space with a Fremhn norm ρ such that E+ ÍS closed. Let X := {y e £'+: IMI = 1}, where || || denotes the operator norm on E' and equip X with the topology induced by ||.||. Let F •— {f € C(X): there exist S\,.. .,sn 6 E with \f(x)\ < \x(Sl)\ +••· + \x(sn)\ for allxeX} and let p(f) := inf{p(si) + ---+p(sn): su...,sn e E with \f(x)\ < \x(Sl)\ + ••• + \x(sn)\ for all χ e X}, f e F. Then (F, ρ) is a normed Riesz space and i(s) := χ ι-> x(s), χ € X, s £ E, defines an isometncal, bipositwe, linear mapping from (E,p) to (F,p). Furthermore, if E is directed, then i(E) is majorizing m F. Proof. Denote the point evaluation at a point s Ç E by f¡: fs(x) •= x(s), χ € E'. (a) F is a Riesz subspace of C(X): It can straightforwardly be verified that F is a linear subspace of C(X). From the definition of F it is immediately clear that f & F implies |/| 6 F, so that F is a Riesz ideal in C(X), and, consequently, F is a Riesz space. (b) ||ƒ[|oo < p{f) for every f € F· Let f & F and S\,..., sn € £ be such that |/(i)|<|x(ai)| + --- + |^n)|forallieX. Then|/(i)|<||i||(p(el) + -..+p(e„)) = p(s\) 4 \-p(sn), from which it follows that ||/||oo < p{f). (c) ρ is a norm on F: Let f,g e F. Let si,...,sm,ti,...,tn e E be such that |/(i)| < |/„(Χ)| + ··· + |ΛΜ(Ι)|, \g(x)\ < |/tl(x) + --- + |/Én(ar)|foralla:eX. Then |/(x) +g(x)\ < |/(ι)| + \g(x)\ < \fSl(x)\ + • • •+ \fSm(x)\ + \ftl(x)\ + • · · |/(„(x)|, for all χ e X, hence p(f + g) < p(sx) 4 + p{sm) + p(ti) + •·· + p(t„). Taking the infimum over all such m,n,Si,..., sm, t\,...,tn yields that p(f + g) < p(f) + p{g). From the definition of ρ it is clear that p{\f) = |λ|ρ(/) for all λ € M, establishing that ρ is a seminorm. By (b) it follows that ρ is a norm. (d) ρ is a Riesz norm on F: If ƒ, g € F are such that | ƒ | < | (e) ι is linear and bipositwe: By definition, i(s) = fs\x, s £ E. That ι is linear needs no explanation. If s € E+, then f„(x) = x(s) > 0 for all χ 6 X, so i(s) > 0. Hence ι is positive. If, conversely, s £ E is such that i(s) > 0, then /s(x) > 0 for all χ 6 X, which means that x(s) > 0, χ 6 X. From this it follows that y(s) > 0 for all y € E'+. E+ is closed in E, so E'+ determines the ordering of E (see Proposition 1.54), so that s € E+. Thus, г is bipositive. 70 4 3. Embedding in a Normed Riesz Space (f) г is isometncal: Let s Ç. E. Then |/s(x)| = |i(s)|, so, by definition of p, p{i(s)) = р(Л) < p{s). To prove that p{i{s)) > p(s), let si,...,sn € E be such that |Л(х)| < \fst{x)\ + • • • + |/s„(z)| for all χ e λ', or, in other words, \x(s)\ < \x(sx)\ + ••• + \x(sn)\, χ € X. Then |y(Sl)| + · · · + |у(в„)| for all у 6 E'+, and therefore, by using that ρ is Fremlin and Proposition 4.5, p(s) < p(si) + · · -+p(sn). Taking the infimum over all such η and si,..., sn yields that p(s) < p(f3) = p(t(s)), proving that p{t(s)) = p(s). Thus г is isometrical. (g) If E is directed, then г(Е) ÍS majorizing in F: Let ƒ 6 F. Then there are for a11 su ..., sn € E with \f{x)\ < |i(si)| + l· |i(sn)l x € X- H E is directed, then there is an upper bound s € E of the set {e^Si + l·- en*n: ε* £ {—1,1}, fc ^ l,...,n}, so that i{s){x) = x(s) > \x(si)\ + ••• + \x(sn)\ > \f(x)\ for all χ € X, proving that i(E) is majorizing in F. (a), (d), (e), (f), and (g) together establish the assertions of the theorem. The fact that ρ is Fremlin does not play a role in the proof until the very last step. For any norm ρ on E, the proof shows that (F, p) is a normed Riesz space and that г:(Е,р) -> {F, ρ) is a positive, linear contraction. The fact that E+ is closed is necessary and sufficient for making г bipositive. The Fremlin property is precisely the property that provides that p{i(s)) is not less than p(s) for any s € E, establishing that г is isometncal. The choice of the topology on X is rather arbitrary. Any topology that renders the point evaluations continuous is equally suitable. Corollary 4.8 Let E be a partially ordered vector space with a norm p. Then the following two statements are equivalent: (ι) ρ is Fremlin, and E+ %s closed. (ιι) E can isometncally, bipositively, linearly be embedded m a normed Riesz space. Remark that, because the norm completion of a normed Riesz space is a Banach lattice, both (i) and (ii) in the corollary imply that E can isometrically, bipositively be embedded in a Banach lattice. 4.3.2 Preservation of Lattice Structure For a partially ordered vector space E with a Fremlin norm ρ and a closed positive cone, Theorem 4.7 describes a construction of a normed Riesz space (F, p) in which (E,p) can isometncally, bipositively. linearly be embedded. If E is already a Riesz space and ρ is a Riesz norm, then, of course, F and ρ could also be taken equal to 71 Chapter 4. Subspaces of Normed Riesz Spaces E and p, respectively. However, the construction of the theorem may yield a much larger space F. One could try to reduce the space by taking the Riesz subspace of F generated by E, but i(E) need not be a Riesz subspace of F — that is to say, the embedding need not be a Riesz homomorphism — so even the Riesz subspace of F generated by E may be larger than E itself. Example 4.9 A Riesz space E with a Riesz norm ρ such that the embedding i of Theorem 4-7 is not a Riesz homomorphism. 2 2 Take E = R with ρ = ||.||i. Then E' = R with \\.\\ж. Let X, F, and i be as in Theorem 4.7. Take f(x,y) := x, and g{x,y) := y, (x,y) 6 X. Then ¿(1,0) = (x,y) ь+ (1,0) · (x,y) = ƒ and ¿(0,1) = g. In E one has (1,0) Л (0,1) = 0, but ƒ Л g φ 0 in F, because (ƒ Л д)(х,у) — х Л у ψ 0 on X. So г is not a Riesz homomorphism. This example shows also that if instead of the whole set X only the extreme points are taken, the corresponding embedding still need not be a Riesz homomorphism. Of course it is more elegant to have a construction that leaves a space unchanged if it already is a normed Riesz space. In the following lemmas and theorem the con struction of Theorem 4.7 will be modified in such a fashion that the constructed space F is Riesz isomorphic to E in case (Ε, p) is a normed Riesz space. Exam ple 4.15 shows that the modified construction does not lead to a minimal Riesz space F. The idea of the modification is to factor out a suitable closed Riesz ideal N. N should be such that qoi:E-¥ F/N is isometrical, bipositive, and such that q o i is a Riesz isomorphism if (E, p) is a normed Riesz space. The presented choice of N is rather artificial. To obtain that q o i is a Riesz homomorphism in case (E, p) is a normed Riesz space, (q о i)(\x\) should be equal to \(q о i)(x)\ for all χ e E, or, in other words, i(\x\) — \i(x)\ should be in N for all x. The first attempt is to take the Riesz ideal generated by such elements. Without using absolute values in E this comes down to: : ^o = jyEf: there are αϊ,..., αη, щ,..., un e Ε with uk < ak for each к, η such that \y\ < ^(¿(u*) - |ί(ο*)|)|. /t=i Lemma 4.10 Let E, p, F, p, and г be as in Theorem 4-7. Let V0 be as above. Let q be the quotient mapping from E to E/VQ, where V0 is the p-closure of VQ in F. Then: 72 4 3. Embedding in a Normed Riesz Space (ι) V0 is a Riesz ideal in F. (n) If E is a Riesz space, then V0 = {y G F: there are 0.1,...,an € E with \y\ < г — г Σ*=ι ( (КІ) І ЮІ)}> E/V0 is a Riesz space, дог' is a Riesz homomorphism and then the norm on E/V0 is a Riesz norm. Proof, (i): The only point that may be not immediately clear is that V0 is closed under addition. If y\,y2 £ ^o. then there are αχ,... ,am,am+i,... ,an, г и - ub.. .,um,um+i,... ,un G E with uk < ak for each k, such that |з/і I < ΣΓ=ι ( ( *) |г(а*)|) and \y2\ < £Lm+iW«t) - l»K)l)> so that К + y2\ < Ы + \y2\ < Σϊ=ι (*(«*)-l*(o*)l)· (ii): If у G Vó, then there are α,ι,. ..,ап,щ,... ,un G E with u* < а* —hence Kl < Ы in E for each к— such that |y| < £Li (»(«*)- Іг(а*)І) < Σ*=ι (г(КІ) ~ |?Κ)|), \vhere the last inequality follows from the fact that г is a positive mapping. г _ г Conversely, if y G F is such that \y\ < Σ*=ι ( (І°*І) І ЮІ) for certain αχ,... ,an e E, then the choice u* := K|, к = 1,..., η, shows that y G Vó. Because ρ is a Riesz norm on F, it follows from (i) that Vó is a Riesz ideal in F and therefore E/Vo is a Riesz space and the norm of E/VQ is a Riesz norm. To prove that дог is a Riesz homomorphism, let ζ € E. Then t(\z\) — |г(г)| G Vó С V0, so that g(i(|z|)) = д(|г(г)|) = |д(г(.г))|, or, in other words, (д ο ι)(\ζ\) = \(q ο ι)(ζ)|. According to the lemma, if £ is a Riesz space, then the norm on E/Vo is a Riesz norm and the embedding ρ ο ι is a Riesz homomorphism. Therefore, if E is a Riesz space, but p is not a Riesz norm, then дог cannot be isometrical. In any case дог is a contraction, so to obtain that д о г is an isometry, the ideal to be factored out should be smaller than Vó but large enough to make дога Riesz homomorphism if E is a normed Riesz space. To take care of the former, Vó can be shrunk to: Vi := {y € V0: p(z) < p(si)-\ +p(s„) for all z, su ..., sn G E with |»(z)-y|< |»(βι)| + ···|*(*»)|}· Vi is probably not a Riesz ideal. Therefore define V2 := {уСЦ:[-|у|,Ы]с^}, V3 := {y £ V2: \y\ + V2C V2}. Lemma 4.11 Let E, p, F, p, and ι be as m Theorem 4-7- Let V0, V\, V2, and V3 be as above. Then: (1) V3 is a Riesz ideal m F. 73 Chapter 4. Subspaces of Normed Riesz Spaces (%%) Let N := V3, the p-closure of V3 in F. Let q : F —> F/N be the quotient mapping and let \\ \\q be the quotient norm of F/N. Then ||.||, is a Riesz norm and q ο ι : E -> F/N is isometncal. Proof, (i): (a) If у € V2 and и e F are such that \u\ < \y\, then и 6 V2: By définition of V2, y e V"2 and \u\ < \y\ yield that и G Vi. For any χ e F with |x| < |u| one has that |i| < \y\, hence χ 6 VI. This means that и € V2. (b) If уъу2 € V3, then yi + y2 € V3: Let χ € V2. Then, by (a), |x| 6 V2, so x 12/11 + \ \ € V"2, since j/2 € V3. Therefore |i/i| + |j/2| + \x\ e Vi, since yi e V3. Using (a) and II2/1I + Іг/гІ + a;| < lî/i| + ІУ2І + \x\, it follows that |з/ж +y2\+x e V2, establishing that 3/1+2/26 V3. (c) If у 6 V3 ond A € R, then Xy e V3: If χ e V0 and μ e R, then μχ e V0, by Lemma 4.10(i). Therefore, if χ € Vi and μ e R, then μι € Vi, so that if χ € Vi and μ € R, then μι 6 V"2. From this, the assertion follows. (d) Ify&V3 and и e F are such that \u\ < \y\, then и € V3.· Let 1 € V2. Then, by (a), \x\ e V2, hence \y\ + \x\ € V2, since у € V3. \\u\ + x\ < \u\ + \x\ < \y\ + \x\, so |it| + χ € V2. Therefore и € V3. (b),(c), and (d) together imply that V3 is a Riesz ideal, (ii): The first assertion follows from (i), because ρ is a Riesz norm on F. To prove that q о г is isometrical, note that for 2 e E one has by definition of the quotient norm that ||д(г(г))||, = inf{p(i(z) - у): у € Ν} = inf{p(î(z) - у): у 6 V3} < ρ{ι{ζ)) — p(z). Let у 6 V3. To show that p{i{z) - y) > p(z), let a > р(г(г) - у). It suffices to prove that p(z) < a. By definition of p, there are s\,...,sn € E such that |г(г) - 2/1 < |Î(SI)| + h |i(s„)| and p(sj) + · · · +p{s„) < a. Because y 6 V3 С Vi, it follows that p(z) < p(si) H l· p(sn) and therefore p(z) < a. Lemma 4.12 Let E, p, F, p, and ι be as m Theorem 4-7. Let V0, V1; V2, and V3 be as above. Then: (1) If Vi is a Riesz ideal m F, then V3 = V2 = Vi. (11) If (E,p) is a normed Riesz space, then Vi = Vj). Consequently, Vi is then a Riesz ideal and therefore V3 = V2 = V\ = Vó = {y € F: there are αϊ,..., on e E such that Ivi < £2=1 (ι(|ο*|) - |г(о*)|)}. (m) If {Ε, ρ) is a normed Riesz space, then the mapping q о г : E —¥ F/V3 is a Riesz homomorphism. Proof, (i): By definition, V3 с V2 с Vb To prove that Vi с V2, let у e Vi and let χ S F be such that \x\ < \y\ \\ is a Riesz ideal, so χ € Vi and therefore y e V"2. Thus V"2 = V*i, which means in particular that Vi is a Riesz ideal. 74 4.3. Embedding in a Normed Riesz Space To prove that V¿ С з, let y G V2. Since V2 is a Riesz ideal, \y\ +x € V2 for every 1 € V2, so y G V3. Hence V2 С V3, establishing that Vi = V2 = V3. (ii): By definition, Vi С VO. To prove that Vo С Vi, let у 6 V0 and let 2, S\,..., sn G £ be such that \i{z)-y\ < |t(si)| + .. . + |i(s„)| < г(|«і|) + · • - + ¿(|s„|) = г(|5і| + · · · + |s„|) = Î(S), where s := |si| + • • · + |sn|. y e Vo means that there are α,ι,... ,an £ E with |V| < ELi (»(|e*l) - l*K)l)· Then η г(г) < У + *(s) < 53 (ïflûfcl) + ejfci(ajb)) + t(s), and η г(-г) > У — «(s) > — 5Ζ (*(Ια*Ι) + e**(°*)) — *(s)> for every combination of signs e* 6 {—1,1}, к = Ι,.,.,η. Since г is linear and α ε α — s bipositive, it follows that 2 < ^¡J=1(|ajt|+etOt) + s and 2 > — Σ£=ι(Ι *Ι+ * Ό > for all et G {—1,1}, A; = 1,..., n. Thefirst inequalit y yields that 2 — s < 0, and the second one that 2 + s > 0, so that —s that p{z) < p(s) = p{\si\ + ••• + |β„|) < ρ(|βι|) + · · · +ρ(|β„|) = ρ(«ι) + · • · +p(sn), proving that y G Vi. Hence Vó С Vi and thus Vi = Vo- According to the previous lemma, Vo is a Riesz ideal in F, yielding that Vi is a Riesz ideal so that, by (i), V3 = V2 = Vi = V0. The description of Vo follows from the assumption that E is a Riesz space and Lemma 4.10. (iii): Since, by (ii), V3 = Vo, Lemma 4.10(ii) yields that g о г is a Riesz homomor- phism. Lemmas 4.11 and 4.12 combine to the announced modified embedding: Theorem 4.13 Let E, p, F, p, and г be as m Theorem 4-7 and let V3 be as above. Let G be the Riesz subspace of F/V3 generated by (qoi)(E), where q is the quotient mapping, and let ||.|| be the restriction to G of the quotient norm. Then (G, ||.||) is a normed Riesz space, the mapping qoi from E to G is linear, bipositive, and isometrical, and qoi is a Riesz isomorphism if {E,p) is a normed Riesz space. Proof. According to Lemma 4.11, (G, ||.||) is a normed Riesz space and g о г is an isometry. From Theorem 4.7 it follows that дог is linear, isometrical, and bipositive. Lemma 4.12 states that g о г is a Riesz homomorphism if (E,p) is a normed Riesz space, (g о г) (E) is then equal to G so that g о г is a Riesz isomorphism. Let us conclude this subsection by remarking that this theorem only claims preser vation of suprema in case the entire space (E,p) is a normed Riesz space. In view of some embedding theorems without norms (e.g. Theorem 4.3 in Luxemburg[15]), 75 Chapter 4 Subspaces of Normed Riesz Spaces the question arises whether a stronger, local, preservation result holds true. For example, with the notations of Theorem 4 13, is (q o i)(|x|) = \(q о г)(і)| for any element χ € E such that \x\ exists in E and p{\x\) = p(x)? The answer is negative. More than that, there may be no such an embedding at all. Example 4.14 A Rtesz space E with a Fremhn norm ρ such that E+ is closed and an element χ € E with p(\x\) = p(x) such that there is no normed Rtesz space F with a linear, tsometncal, biposihve embedding ι: E —> F such that i(\x\) = |t(i)|. Take E = R2 and p(xi,X2) = 2|xi| V |i! +І2І. (^ь^г) € E. ρ is Fremlin on E, E+ is closed, and ρ is not Riesz: p(l, —2) = 2 and p(l, 2) = 3. Take χ = (1, —1), then p(x) = 2 = p(|i|). Let F be a normed Riesz space with a linear, isometrical, bipositive embedding ν E —• F. Suppose that г(|і|) = |t(i)|, i.e. г(еі) + i(e2) = |г(ег) — г(е2)|, where d = (1,0) and e2 = (0,1). Then г(еі)Лг(е2) = 0 which means that г preserves disjointness and thus is a Riesz homomorphism, contradicting the fact that ρ is not Riesz. 4.3.3 No Minimality The Riesz space (F, p), as constructed in Theorem 4.7, in general is larger than necessary. The modified construction yields a normed Riesz space that is, trivially, the smallest one in case (E, p) is a normed Riesz space. Is there, for every Fremlin normed partially ordered vector space (E,p) with closed positive cone, a smallest normed Riesz space that contains E as a subspace? If 'smallest' is interpreted as being isometrically and Riesz isomorphically embeddable in every normed Riesz space containing (E,p), then the answer is negative: Example 4.15 A partially ordered vector space E with a Fremhn norm ρ such that E+ is closed and such that there does not exist a smallest normed Riesz space containing (E,p) as a subspace. Take E = R with the trivial ordering, i.e. ordered by the equality relation. Take for ρ the absolute value function. (R2,|| ||oo) and (R2,2-1||.||i) are two normed Riesz spaces in which E can linearly, isometrically, bipositively be embedded by Α Η-» λ(—1,1), A € E. Because (E,p) is not a normed Riesz space, a smallest normed Riesz space containing E should have at least dimension 2. Thus, it should be isometrically, Riesz isomorphic to R2 under || ||oo as well as 2-1||.||i, which is impossible. 76 4.4. On Embedding of Spaces with Unclosed Positive Cones 4.4 On Embedding of Spaces with Unclosed Pos itive Cones The embedding theory presented in §4.3 cannot be generalized straightforwardly to a seminorm setting. The presented method relies heavily on the closedness of the positive cone, which makes sense in normed spaces only. There, closedness of the positive cone comes in naturally, because it is necessary for embedding in a normed Riesz space. In order to make the method suitable for seminorms, we first handle positive cones that are not closed, including the case of a Fremlin normed space. This leads to two directions in which the embedding problem can be generalized: - Can every Fremlin normed partially ordered vector space be embedded in a directed partially ordered vector space with a regular norm? - Can every partially ordered vector space with a Fremlin seminorm be embed ded in a Riesz space with a Riesz seminorm? We do not obtain complete answers. By adapting the presented methods, some special cases can be dealt with. The following results will be proved: - Let E be a directed partially ordered vector space with a Fremlin norm. If E+ is i?'+-feebly closed, then E can linearly, isometrically, bipositively be embedded in a directed partially ordered vector space with a regular norm. (See Definition 4.16 and Theorem 4.21). - Let £ be a directed partially orderd vector space with a Fremlin seminorm. If E+ is £~+-feebly closed, then E can linearly, isometrically, bipositively be embedded in a Riesz space with a Riesz seminorm. (See Definition 4.24 and Theorem 4.28). 4.4.1 Embedding in a Regularly Normed Space Our first approach to extend the method of the previous section is to consider a Fremlin normed partially orderd vector space E and to take the closure E+ of the positive cone. The space E with the ¿^-ordering can be embedded into a normed Riesz space (F,p). Then one can try to shrink F+ to a cone that induces the E+- ordering on E. The resulting space will probably not be a Riesz space and ρ will not be a Riesz norm anymore, but maybe its cone can be kept large enough to leave the space directed and ρ regular. In that case, it can eventually be embedded into a Riesz space with a Riesz seminorm. A more general result for embedding in a Riesz space with a Riesz seminorm can be obtained via ideas presented in §4.4.2. 77 Chapter 4 Subspaces of Normed Riesz Spaces The main problem in the approach sketched above is to find a suitable cone in F. A more 01 less natural choice works if E+ has a property that is somewhat weaker than closedness. Lemma 4.23 shows that this property is not void. Definition 4.16 Let E be a partially ordered vector space with a seminorm. E+ is called E'+-feebly closed if for every s 6 E+ \ {0} and a G E such that f (a — s) > 0 for all ƒ € E'+ one has that α € E+. By noting that E'+ determines the ordering if E+ is closed (Proposition 1.54), the following assertion is clear. Proposition 4.17 Let E be a partially ordered vector space with a norm. If E+ is closed, then E+ is E'+-feebly closed. For monotone norms the definition can be reformulated. Lemma 4.18 Let E be a partially ordered vector space with a monotone norm p. The following two statements are equivalent: (ι) E+ is E'+-feebly closed. (n) If s € E+\ {0} and a e E are such that a- s € Έ~+, then a £ E+. Proof. (i)=>(ii): Let s € E+ \ {0} and α <= E be such that α-seF. Then there + + are (u„)„ in E with un—¥ a — s. Then f(a — s) — Iim,,-»«, f{u„) > 0 for all ƒ € E' and therefore a 6 E+. (ii)=>(i): According to Proposition 1.16, E+ is the positive cone of a vector space ordering on E. E'+ = (E, E+)'+ and this sei determines the £+-ordering on E, by Proposition 1.54. So, if a, s e £ are such that f(a — s) > 0 for all ƒ G E'+, then a — s e E+. From this, the assertion follows. A typical example is the following. Example 4.19 A partially ordered vector space E with a Fremhn norm, such thai E+ is E'+ -feebly closed and not closed. Take E = R2 equipped with the ordering defined by the 'open' cone: {(11,12) 6 E: ii, X2 > 0} U {(0,0)}, and, for example, the Euclidean norm. E+ corresponds to the ordinary ordering, so if s € E+ \ {0} and α e E are such that a — s S E+, then + Οι > Si > 0 and a2 > S2 > 0 and therefore о € £' . Any directed Fremlin normed partially ordered vector space E with an i?'+-feebly closed positive cone can be embedded linearly, isometrically, and bipositively in a directed partially ordered vector space with a regular norm. 78 4 4 On Embedding of Spaces with Unclosed Positive Cones Lemma 4.20 Let E be a directed partially ordered vector space with a Fremhn norm ρ such that E+ is E'+-feebly closed Then there exist a directed partially ordered vector space F with a regular norm ρ such that F+ is F'+-feebly closed and a linear, isometrical, bipositwe mapping г E —> F Proof. It is convenient to assume that Ε φ {0} (For the case E = {0} the assertion is clear) Then E+ \ {0} is not empty, because E is directed By Proposition 1 16, E with the F+-ordenng is a partially ordered vector space and ρ is again Fremhn with respect to this ordering Denote, for convenience, E with the F+-ordermg as E\ E* is closed, so by Theorem 4 7 there exist a normed Ries7 space (F, p) and a linear, isometrical, bipositive mapping г E —> F, given by F = {feC(X) there exist si, ,s„eFwith \f{x)\ < |i(si)| + + |i(e„)| for all χ € X}, where X = {y € E[+ \\y\\ = 1}, p{f) = inf{p(si)+ + p(s„) si, ,sn£Ev/ith \f(x)\ < |x(si)| + + |x(sn)| for all χ e X}, f e F, i(s) = χ ь-> x(s), χ e X, s ζ E The ordering of F has to be adapted in order to make г bipositive with repect to the F+-ordenng on E Define К = {ƒ e F there exists ansE£+\ {0} with ƒ (x) > x{s) for all χ e X} U {0} (a) К is a cone in F, Κ Π {—К) = {0} and F is directed with respect to the K- ordenng The proof that A" is a cone is straightforward Because К С F ', Κ Π {-К) С F+Л (-F+) = {0} E is directed, so г(Е) is majorizing in F (Theorem 4 7), hence for ƒ e F there is an s e E+ \ {0} with i(s) > ƒ,0, so i(2s) - ƒ > i(s) and therefore t(2s) — f ζ К and i(2s) 6 К, which means that i(2s) > ƒ,0 in the A"-ordenng Thus, F is directed with repect to the A%ordering (b) Κ Π г(Е) - г{Е+) Let a e E be such that ι{α) G К Then there is an s e E+\ {0} with x(a) > x{s) for all χ € X Because E+ is £"+-feebly closed, it follows that a e E+ Hence Κ Π ι(Ε) С г{Е+), so that К П ι(Ε) = ι{Ε+) (c) г E -* F is bipoiitive with respect to the К-ordering on F From (b) (d) ρ is К-regular Let f ζ F It has to be shown that p(f) — inf {p(g) g + f, g — f € К} If g € F is such that g + f, g - f e К, then ρ > \f\ in F, so that p( р(Л For the opposite inequality, let e > 0 Choose an s 6 F+ \ {0} and 79 Chapter 4 Subspaces of Normed Riesz Spaces scale it such that 0 < p{s) < ε. Take g := |/| + i(s). Then g + f, g - f € К and Р(9) < P(\f\) + P(i(s)) = P(f) + P(s) < p(f) + ε. (e) {FJ Theorem 4.21 Let E be a directed partially ordered vector space with a Fremhn norm p. Then E can linearly, isometrxcally, bipositively be embedded in a directed partially ordered vector space F with a regular norm such that F+ is F'+-feebly closed if and only if E+ is E'+-feebly closed. Proof. -4=) Prom the previous lemma. =>) Let s e E+\ {0} and a e E be such that ƒ (a - β) > 0 for all ƒ € E'+. Let ƒ € F'+. Then fe € E'+, so f(a-s) > 0. F+ is F'+-feebly closed, so then o e F+, hence о e E+. The embedding part of the above theorem may be followed by embedding into a Riesz space with a Riesz seminorm. Corollary 4.22 Let E be a directed partially ordered vector space with a Fremhn norm such that E+ is E'+-feebly closed. Then E can be linearly, isometncally, bipositively be embedded m a Riesz space with a Riesz seminorm. Proof. By the previous theorem, E can be embedded into a directed partially ordered vector space FQ with a regular norm. By Proposition 3.40, this norm is pre-Riesz, so that Theorem 3.27 can be used to embed FQ into a Riesz space with a Riesz seminorm. The following lemma shows that the above results are of interest for not integrally closed spaces only Lemma 4.23 Let E be a directed partially ordered vector space with a monotone norm. Then the following two statements are equivalent: (ι) E+ is closed. 80 4 4 On Embedding of Spaces with Unclosed Positive Cones (n) E is integrally closed and E+ is E'+-feebly closed + + Proof. (n)=>(i) Suppose that E is not closed Then there are (u„)n in E and an α € E \ E+ such that u„ -> a E is directed, so there is a & € £ with b > 0, α and b φ 0 Because E is integrally closed and —o ¿ 0, there is an m 6 N with l l m~ b 2 -a Take 6' = m~ b and с = α + ò' = lim«^,» un + 6' Then V £E+\ {0}, с — b' e £+, and с £ E+, contradicting the £'+-feeble closedness of E+ Thus E+ is closed (i)=>(n) Propositions 1 31 and 4 17 Example 1 32 is an example of a directed, integrally closed partially ordered vector space E with a monotone norm and E+ not closed According to the above lemma, it also is an example of a positive cone that is not i?'+-feebly closed 4.4.2 Embedding in a Riesz Space with a Riesz Seminorm In §4 3 any partially ordered vector space E with a Fremhn norm ρ and E+ closed was embedded in asubspace F of C(X), where X = {x € E'+ \\x\\ = 1} Bipositiv- lty of the embedding followed from the fact that E'+ was large enough to determine the ordering of E, which fails if E+ is not closed In §4 4 1 bipositivity was obtained by making the ordering on F more restrictive In this subsection we deviate farther from the previous constructions We will be looking for only a Riesz seminorm on F Therefore, there is more freedom in the choice of the set X, for example X = E~+ The embedding will then be bipositive if E~+ determines the ordering of Ε, ι с if f(x) > 0 forali ƒ G E~+, then χ > 0, for every χ € E As before, this condition can be weakened to 'i?~+-feeble closedness of£+' Definition 4.24 Let E be a partially ordered vector space E+ is called E~+-feebly closed if for every s e E+\ {0} and α e £ such that ƒ (a - s) > 0 for all ƒ 6 E~+ one has that a s E+ Note that in a space E with a monotone seminorm £"+-feeble closedness of E+ im plies £~+-feeble closedness If E~+ determines the ordering of E, then E+ is E~+- feebly closed However, if E+ is £~+-feebly closed, then E~+ need not determine the ordering (e g , M2 with the 'open' cone) If, in addition, E is integrally closed, then the next lemma shows that E~+ separates the points It can be seen directly from the definition that any Archimedean Riesz space E(^ {0}) with E~+ — {0} has a positive cone that is not £'~+-feebly closed 81 Chapter 4. Subspaces of Normed Riesz Spaces Lemma 4.25 Let E be an integrally closed partially ordered vector space. If E+ is E~+-feebly closed, then E~+ separates the points of E. Proof. If E+ = {0}, then the assertion is clear. If E+ Φ {0}, suppose that E~+ does not separate the points of E. Then there are x,y € E such that χ φ y and f (χ) = f (y) for all ƒ e E~+. Either χ ~¿ y or y ~¿ χ; assume the latter. Take а и G E+ \ {0} Because E is integrally closed, there is an η € N such that n-1u ¿ χ — y Take s :— n_1u and о := у — χ + s. Then s e E+ \ {0} and ƒ (o — s) = f (y — x)=Q for ail ƒ 6 E~+. However, a = y — г + sJfO, contradicting the £~+-feeble closedness of E+. Therefore, E~+ separates the points of E. The following lemma lists results similar to Theorems 4.7, 4.13, and Lemma 4.20. Lemma 4.26 Let E be a partially ordered vector space such that E+ is E~+- feebly closed with a Fremhn norm p. Let X := E~+, F := {ƒ : X —» R: there are si,.. , sn 6 E such that \f(x)\ < \x(si)\-\ h|i(sn)| for all χ & X}, p(f) := inf{p(si) + · · · + p[sn): sb ..., s„ € E such that \f{x)\ < \x{si) + · · · + |i(s„)| for all χ 6 X}, ƒ g F, and i(s) :=IH x(s), χ € X, s e E. (ι) Then F is a Riesz space, F~+ determines the ordering of F, ρ is a Riesz semmorm on F and the embedding ι : E —* F is linear, isometrical and positive. (n) If E~+ determines the ordering of E, then ι is bipositive. (m) Let К .= {ƒ e F: there is an s € E+ \ {0} with f(x) > x(s) for all χ e X} U {0}. If E is directed, then К is the positive cone of a vector space ordering on F, F equipped with this ordering is directed, К is (F, K)~+-feebly closed, ρ is regular with respect to the К-ordering and ι . E —• (F, К) is bipositive. (F, К) is a pre-Riesz space if and only if E~+ determines the ordering of E. Proof. The proof is very similar to the proofs of Theorem 4.7 and Lemma 4.20 and therefore it is stated rather briefly. (i): (a.) F is a Riesz ideal ofRx : Clear from the definition of F. (b) F~+ determines the ordering of F: Note that F~+ contains the point evalua tions. (c) ρ is a semmorm on F: Straightforwardly, just as in Theorem 4.7(c). (d) ι is an isometry: Let s e E. Then p(i(s)) < p(s). Let si,...,sn G E be such + + that |a;(s)| < |i(si)| + h \x{sn)\ for all χ e X. Since E~ D E' , it follows from 82 4 4 On Embedding of Spaces with Unclosed Positive Cones Proposition 4.5 that p(s) < p(s\) + h p(sn). Take infimum over all such η and A'I, ..., sn then p(s) < p(i(s)) Thus г is an isometry. (ii): Let s 6 E with i(s) e F+. Then x(s) > 0 for all χ e E~+, so χ € E+. (iii): (а) К is a cone in F and г(Е+) С К С F+, so К determines a vector space ordering and F is directed with respect to this ordering: As in Lemma 4.20(a). (b) КПг{Е) = г[Е+): Analogous to Lemma 4.20(b) from the F~+-feeble closedness ofE+. (c) г : E —• {F, К) is isometncal and bipositive: From (i)(d) and (iii)(b). (d) ρ is К-regular: as in Lemma 4.20(d). (e) (F, K) is pre-Riesz if and only if К = F+: •$=) Is clear: F is a Riesz space. =>) (F, K) is pre-Riesz, so for every f,g,h € F one has that ƒ > 0 if every upper bound of {ƒ + g, ƒ + h} is also one of {g,h}, where all inequalities are in (F,K). Suppose that Κ φ F+, then there is an ƒ G F+\K. Take g := 0, h := -ƒ. If j € F is such that j > ƒ + /ι, ƒ + /ι (in A"-sense), then J > ƒ, 0 (in Ä"-sense). Since ƒ e F+, it follows that j > 0, — ƒ in A"-sense, so that j > g,h (in if-sense). This means that every /f-upper bound of {ƒ + g,f + h} is also one of {g, h}. Hence ƒ 6 K, which is a contradiction. Thus, К = F+. This lemma yields that every directed partially ordered vector space E in which E+ is F~+-feebly closed and that is equipped with a Fremlin seminorm can linearly, isometrically, bipositively be embedded in a directed partially ordered vector space with a regular seminorm. Consequently, it can be embedded in a Riesz space with a Riesz seminorm, though the last assertion of the lemma shows that this embedding will not be a Riesz* homomorphism. The precise statements are in the next two theorems. Theorem 4.27 Let E be a directed partially ordered vector space with a Fremlin seminorm. Then E can linearly, isometrically, bipositively be embedded m a directed partially ordered vector space F such that F+ is F~+-feebly closed with a regular seminorm, if and only if E+ is E~+-feebly closed. Proof. <=) From part (iii) of the lemma. =>) Observe that .~+-feeble closedness of the positive cone is inherited by subspaces. Theorem 4.28 Let E be a directed partially ordered vector space of which E+ is E~+-feebly closed with a Fremlin seminorm. Then E can linearly, isometrically, bipositively be embedded m a Riesz space with a Riesz seminorm. 83 Chapter 4. Subspaces of Normed Riesz Spaces Proof. From the previous theorem, Proposition 3.40, and Theorem 3.27. Another consequence of Lemma 4.26 is the following. Corollary 4.29 Let E be a directed partially ordered vector space with a Fremlin seminorm. Then E can linearly, isometrically, bipositively be embedded in a Riesz space F such that F~+ determines the ordering of F and equipped with a Riesz seminorm if and only if E~+ determines the ordering of E. Proof. Use part (ii) of Lemma 4.26 and remark that the property that the positive cone of the order dual determines the ordering is inherited by subspaces. 4.5 The Subspaces of an M-space The first part of this chapter presented a description of the subspaces of normed Riesz spaces. This last section develops a similar description of the subspaces of Riesz spaces with M-norms: they are precisely the partially ordered vector spaces with norms with full unit balls and closed positive cones. It follows from §1.2.3 and §1.2.4 that a norm on a partially ordered vector space must have full unit ball and that the positive cone must be closed in order for the space to be embeddable in a Riesz space with an M-norm. It will be proved that these properties are also sufficient, by presenting an embedding theory very similar to the one for embedding in a normed Riesz space. We start with a dual characterization of seminorms with full unit balls. 4.5.1 Dual Characterization of Seminorms with Full Unit Balls The statements of §4.2 can be adapted to results for seminorms with full unit balls. Lemma 4.30 Let E be a partially ordered vector space with a norm ρ such that E+ is closed. If x, ylt..., yn g E are such that \f{x)\ < |ƒ{yi)\ V ... V |ƒ[yn)\ for all + ƒ 6 E' , then there are Aj,..., A„, μι,,.., μη 6 R with Σ \Xk\ = 1 and Σ ІМ*І — 1 such that Ajj/j + · • · Xnyn Proof. (Similar to Lemma 4.2). Let В := {ζ € E: p(z) < 1} and Y := {A^i + h A„2/n: Αχ,..., A„ e R, |λι | + · · • + |λη| = 1}. Υ is compact. For ε > 0 define + We := {-E ) + Υ + εΒ and let q£ be the Minkowski function of W£. Let ε > 0. + Suppose that χ φ We. Then there is an ƒ € E' with f(x) = qc{x) > 1, so 84 4.5. The Subspaces of an M-space 1 < \f{x)\ < I ƒ ЫІV... V \f(Un)\ = f(±yk) < However, ±yk € Y, so q£(±yk) < 1, yielding a contradiction. Hence χ € We. In the same fashion it can be shown that —x 6 WE. Via compactness of У', closedness of E+, and e 4- 0, it follows that there are + u, ν £ E and λι,..., λ„, μι,..., μη € R with Σ |λ*| = 1 and Σ \ßk\ = 1 such that χ = -и + Аіуі + · · · + Хпуп and -χ = -ν + {μ^ι + · · · + ßnyn)- Thus, (-Мі)Уі + · • · + (-Уп)Уп < х < λι2/ι + · · · + Хпуп. Lemma 4.31 Let E be a partially ordered vector space with a seminorm ρ with a full unit ball. Ifx,yl,;..,ynçE are such that \f{x)} < \f{yi)\ V... V\f(yn)\ for all + f e E' , then p{x) < p{Vl) V... V p(yn). Proof. (Similar to Lemma 4.4). + (a) The assertion holds true if ρ is a norm and E is closed: Let x,yi,...,yn € E + be such that \f(x)\ < |/(yi)|V.. -V\f{yn)\ for all ƒ € E' . According to the previous lemma, there are Ab ..., λη, μι,..., μη £ R with J^ |λ*| = 1 and £ \μ^\ = 1 such that Аіуг + · · · + Xnyn < χ < μ^ι + •••+ μ„ι/η. Then ρ(χ) < р{\іуі + h Xnyn) V ρ(μινι + · · · + μ„2/„) < < ( Σ |λ*|) (p(yi) V ... V р(у„)) V ( Σ Ы) (Р(УІ) ν... V р(уп)) = = РІУі) ... р(з/„). (b) The assertion holds true if ρ is a norm: ρ is a norm with a full unit ball with respect to the ordering determined by E+ (see Proposition 1.16). The norm duals of E and (E, E+) are equal, so the assertion follows from (a). (c) The assertion holds true if ρ is a seminorm: Let N be the kernel of p. The quotient norm on the quotient space E/N has a full unit ball (see Proposition 1.17) and E' and E'N are isomorphic (see Proposition 1.55) so the assertion follows from (b). Proposition 4.32 Let E be a partially ordered vector space with a seminorm p. The following statements are equivalent: (i) ρ has a full unit ball. + (ii) Whenever x,yu ... ,yne E ond|/(i)| < |/Ы| .. .V|/(y„)| for all ƒ € E' , one has that p(x) < p(yi) V ... V p{yn). (Hi) Whenever x,y,z Ç. E and f(y) < f(x) < f(z) for all f e E'+, one has that p{x) 85 Chapter 4 Subspaces of Normed Riesz Spaces Proof. (i)=>(ii): The previous lemma. (ii)=>(iii)· Directly (iii)=>(i). Let x, y, ζ e E be such that y < χ < ζ. Then f (y) < f (χ) < f (ζ) for all ƒ 6 E'+, so p(x) < p(y) Vp(z), proving that ρ has a full unit ball. 4.5.2 Embedding in an M-normed Riesz Space By changing sums to suprema, Theorem 4.7 yields an embedding in a Riesz space with an M-norm. Theorem 4.33 Let E be a partially ordered vector space with a norm ρ with a full unit ball such that E+ is closed. Let X:={y€E'+:\\y\\ = l}, where ||.|| denotes the operator norm on E' and equip X with the topology induced by ||.||. Let F := {ƒ 6 C{X): there exist S\, ..., s„ 6 E with \f(x)\ < \x(Sl)\ V.. V |s(en)| for all χ € X} and let p{f) := inî{p(si)V ...Vp(sn): Si,...,s„eE with \f(x)\ < \x(Sl)\ V ... V |φη)| for all χ e X}, f € F. Then F is a Riesz space, ρ is an M-norm, and i(s) ·= χ ι-> x{s), χ £ X, s £ E, defines an isometrical, bipositwe, linear mapping from {E,p) to (F,p). Furthermore, if E is directed, then i(E) is majorizing in F. Proof. Some comments are needed to adapt the proof of Theorem 4.7. Uf, gZF, then f+g e F and p(f+g) < p{f)+p{g): Let su..., sm, tu...,tn€ E be such that |Дх)| < |x(st)| V ... V |i(sm)| and \g(x)\ < \x(ti)\ V ... V \x(tn)\ for all χ e X. For every χ e X there are к and / such that |/(a;)| < \x{sk)\ and |/(i)| < |(i(t,)|. Then |/(x)| + \g(x)\ < \x{sk + i,)| V \x(sk - t,)\, showing that \f(x)+g{x)\ all χ £ X, which means that f + g 6 F. It also follows that p{f+g) < supJti((p(sjt) + P(ti)) 86 4.5 The Subspaces of an M-space The mapping г is lineai and bipositive. That г is isometrical follows from the assumption that ρ has a full unit ball by applying Proposition 4.32, imitating 4.7(f). Obvious adaptations of 4 7 complete the proof. A result for embedding in a special class of M-normed Riesz spaces is obtained by considering the second dual of F. Corollary 4.34 Let E be a partially ordered vector space with a norm p. Then the following three statements are equivalent: (ι) ρ has a full unit ball and E+ is closed. (ιι) E can linearly, isometncally, bipositwely be embedded in a Riesz space with an M-norm. (m) E can linearly, isometncally, bipositively be embedded m a norm complete, Dedekmd complete Riesz space with a strong unit, equipped with the norm induced by the unit. Proof. (i)=>(ii): The previous theorem. (ii)=>(iii): Let F be a Riesz space with an M-norm in which E can linearly, isomet ncally, bipositively be embedded. The second dual F" of F is a norm complete, Dedekind complete Riesz space with a strong unit and its norm is the norm in duced by the unit. The embedding of F in F" is linear, isometrical, and bipositive. Composition of the two embeddings establishes the assertion. (iii)=>(i): The norm is an M-norm, hence its restrictions have full unit balls (§1.2.3) and the positive cone is closed by Proposition 4.1(i). 87 88 Chapter 5 Spaces of Operators This chapter aims to generalize some elementary results in the theory of spaces of operators between normed Riesz spaces to spaces of operators between normed partially ordered vector spaces. The concepts developed in the previous chapters turn out to be useful for extending results to a wider range of spaces as well as to describe properties of spaces of operators and of operator norms. Chapter 4 leads to an alternative proof of Krein's lemma. In fact, a slightly stronger result is proved. It yields that the norm dual of a regularly normed, directed partially ordered vector space is a regularly normed, directed partially ordered vector space, which is a theorem due to Ng[19]. Furthermore, an alternative for the r-norm will be given, where a regular norm on C(E, F) is obtained as an operator norm. 5.1 Introduction Let us first introduce some notations. Let E and F be partially ordered vector spaces. The vector space of linear mappings from E to F is denoted by L(E, F). If E is directed, then the cone of positive linear mappings (see Definition 1.1) generates a vector space ordering on L(E,F): the pointwise ordering on E+. Impose this ordering on L(E, F). The differences of positive linear mappings are called regular, and they constitute a subspace Lr{E, F) := L(E, F)+ - L(E, F)+. Assume that E and F are equipped with norms. The subspace of L(E, F) con sisting of continuous mappings is denoted by C(E,F). It inherits the ordering of L(E, F). The subspace of continuous, linear mappings that are differences of contin uous, positive linear mappings is denoted by Cr(E, F) = C(E, F)+—C(E, F)+. Note that Cr(E,F) is not the same as C{E, F) П LT(E, F) (see Example 5.9). CT{E,F) is usually equipped with the r-norm (or regular norm), which is the regularization of the operator norm on C(E,F) (see Definition 3.44). 89 Chapter 5 Spaces of Operators The theory of spaces of operators on normed partially ordered vector spaces studies the relation between properties of the space C(E, F) and the spaces E and F. Foi example, what kind of order properties (e.g. directed, Riesz) has the space C(E, F) and what kind of order properties (e g. monotone, Riesz) has the operator norm for certain classes of spaces E and F? Here is a list of some well-known results for operators between normed Riesz spaces. Proposition 5.1 Let E and F be normed Riesz spaces. Then: (ι) For every A € C(E, F)+ the operator norm of A is equal to \\A\\ = sup{||Ar||: χ 6 E+, \\x\\ < 1}. (u) If F is Dedekmd complete, then C(E,F) is a Riesz ideal m the Riesz space Lr(E,F). (m) The operator norm on C(E, F) is Fremhn and C(E, F)+ is closed m C(E, F). T (w) If F is Dedekmd complete, then (C (E, F), ||.||r) is a Dedekmd complete, normed Riesz space, where \\ \\T is the regulanzation of the operator norm. r (v) If F is Dedekmd complete and norm complete, then (C (E,F), ||.||r) м a Ba- nach lattice. (vi) If F is a Dedekmd complete Riesz space with a strong unit and if its norm is the norm generated by the unit, then C(E,F) is a Dedekmd complete Riesz space and the operator norm is a Riesz norm. (vi%) If E is norm complete, then every regular linear mapping from E to F is continuous: LT{E,F) С C{E,F). Proof. (i): see Schacfer[22, p.230], (ii)· see Schaefer[22, ρ 230, Prop.1.4], (iii): see Schaefer[22, p.230], (iv),(v): see Aliprantis & Burkinshaw[2, p.248, Thm 15.2], Schaefer[22, ρ 230, Prop.1.4], (vi): see Aliprantis & Burkmshaw[2, p.249], Schaefer[22, p.232, Thm.1.5], (vii): see Aliprantis к Burkinshaw[2, p.175, Thm 12 3], Schaefer[22, p.84, Thm.5.3]. The proofs of these statements seem to allow less restrictive conditions. Just replac ing the normed Riesz spaces by monotonely normed partially ordered vector spaces is much too crude. The previous chapters provide terminology for an appropriate setting. It turns out that, mainly by adapting notations, generalizatons can be ob tained for regularly normed directed partially ordered vector spaces E and Fremlin normed partially ordered vector spaces F. 90 5.2. Norm Dual Spaces The discussion begins with dual spaces. Then, Bonsall's theorem (in [5]) will be proved. Next, a generalization of Proposition 5.1 will be given, and, finally, an alternative for the r-norm will be presented. 5.2 Norm Dual Spaces The norm dual of a normed Riesz space is a (norm complete) normed Riesz space (Proposition 1.51). Is there a similar result for normed partially ordered vector spaces? According to Krein's lemma (Proposition 1.53), the norm dual of a mono- tonely normed partially ordered vector space is directed. The embedding theory of Chapter 4 can be used for an alternative proof of Krein's lemma. It leads to a result by Ng[19]: the norm dual of a regularly normed directed partially ordered vector space is a regularly normed directed partially ordered vector space. The second part of this section presents counterexamples. 5.2.1 Krein's Lemma and the Dual of a Regularly Normed Space A continuous, linear function on a subspace of a normed Riesz space can be extended to a continuous, linear function on the whole space. The extension splits up in a positive and a negative part and their restrictions yield a splitting for the original function. In this way, Krein's lemma is proved for all spaces that are subspaces of normed Riesz spaces. According to the theory in Chapter 4, these spaces are pre cisely the partially ordered vector spaces with Fremlin norms such that the positive cones are closed. Via quotient spaces and closures of positive cones, the result can be extended to every partially ordered vector space with a Fremlin seminorm. Let us make these statements precise. Lemma 5.2 Let (F, p) be a normed Riesz space and let E be an ordered subspace of F. Let ƒ : E -> R be linear and such that \f(x)\ < p(x) for all χ € E. Then there are positive, linear functions f\,fi • E —> R such that f = f\ — h and \{h+f2)(x)\ Proof. Using Hahn-Banach, ƒ can be extended to a linear function ƒ : F -» M with l/(z)l < p{x) for all χ 6 F, i.e. ƒ € E' and ||/|| < 1. E' is a normed Riesz space + (Proposition 1.51), so ||/ + ГЦ = |||/||| = Ц/Ц < 1. Take h := ƒ"+ and f2 := ƒ'", then the assertions are clear. 91 Chapter 5 Spaces of Operators Theorem 5.3 Let E be a partially ordered vector space with a Fremhn seminorm ρ + For every ƒ € E', there are Д, f2 G E' such that ƒ = ƒι - ƒ2 and Ц/Ί + /2|| < || /|| Proof. (See also Bonsall[5] and Ng[19]) It may be assumed that ||/|| = 1 Let N be the kernel of ρ and let EN = E/N According to Propositon 1 55, the norm duals E'N and E' are isomorphic, so ƒ corresponds to an element ƒ e E'N with ||/|| = 1 The quotient norm on Ец is Fremhn (Proposition 1 17) and therefore it is also Fremhn with respect to the ordering generated by E% Application of + an the lemma yields that there are f\,]2 € (E\,E^)' such that ƒ = Л — /г d ||/i + h\\ < 1 Then /i,/г € E'j$ and thus, via the isomorphism between E'N and E', there are fltf2 e E' such that ƒ = ƒι - /г and ||/i + f2\\ < 1 This completes the proof The usual version of Krein's lemma follows from the one above In §5 3 a version for Riesz space space valued functions will be proved For monotone norms, there is the following consequence Corollary 5.4 Let E be a partially ordered vector space with a monotone seminorm + ρ For every ƒ 6 E' there are flt f2 e E' such that ƒ = Д—/2 and || Λ+/2|| < 2|| ƒ || Proof. According to Theorem 2 3, there is a Fremhn seminorm p\ on E with 2~lp < pi < ρ Application of the previous theorem to p\ establishes the assertion According to the theorem, the norm dual of a partially ordered vector space E with a Fremhn seminorm ρ is directed and the norm estimate yields that the regulanzation of the operator norm is not greater than the operator norm itself If the operator norm would be Fremhn, then it could not be greater than its regulanzation, so then it would be regular It turns out that this is the case if ρ is regular This will be proved in a more general context in Theorem 5 16(in), therefore the proof will be omitted here Theorem 5.5 Let E be a directed partially ordered vector space with a regular semi- norm Then E' is directed and its norm is regular Proof. (See also Ng[19]) According to Theorem 5 3, E' is directed and for every + ƒ 6 E' there are ƒ,, f2 e E' such that ЦЛ + /2|| < ||/|| and ƒ = ƒ, - /2, so —(/i + j2) < ƒ < fi + f2 It follows that the regulanzation || ||r of the operator norm (see Definition 3 44) satisfies || ||r < || || By Theorem 5 16(in), || || is Fremhn, so that, by Corollary 3 45, one has || ||r > || || Thus, the operator norm is equal to its regulanzation and therefore regular 92 5 2 Norm Dual Spaces 5.2.2 Counterexamples The following statements about the norm dual of a partially ordered vector space E with a seminorm ρ hold true - If {E,p) is a normed Riesz space, then E' is a normed Riesz space (Proposi tion 151) and it is a Riesz ideal of E~ If E is norm complete, then E' = E~ - If E is a directed partially ordered vector space and ρ is a regular seminorm, then E' is directed and its norm is regular (Theorem 5 5) HE is norm complete, then E' = E~ (Theorem 5 16(vu)) - If E is a partially ordered vector space with a monotone seminorm, then E' is directed The following examples show that many similar assertions do not hold in general Example 5.6 A Riesz space E with a monotone norm such that the norm of E' ÍS not equivalent to a monotone norm x k x e E Then E ls a Riesz Take E = c00(N) and p(x) = ЦхЦ«, + \^k ( )\> space and ρ is a monotone norm on E Define f(x) = ^kx{k) and fn{x) = Σίΐη+ι2-^)' x € Ε, η S N f and ƒ„ are positive, linear functions on E and they are continuous, because |/(x)| < p{x) and |/„(i)| = | J2kx(k) ~ Σ^ο3-^)! — ΙΣ**(*)Ι + Σ*=ο l*(*)l < ("+I)PW, xeE,neN (ƒ - ƒ„)(*) = E!L>*(*). so ƒ — ƒ„ is positive, hence 0 < ƒ„ < ƒ for all η e Ν Let xn = -1{ο η-i} + l(n ,2n+i). " e N Then p(xn) = 1 and fn(xn) = η for all n, so ||/n|| > η, η e N Thus, 0 < ƒ„ < ƒ, ||/|| < 1, ||/n|| > η, for all η e Ν, which means that the norm of E' is not equivalent to a monotone norm Example 5.7 A Riesz space E with a monotone norm ρ that is not equivalent to a Riesz norm, such that E' is a Banach lattice Take E = C[0,1] and p{x) = ||x||, + |/(ι)|, χ e E, where f {χ) = χ(0) + χ(1), χ Ε Ε Then Ε is a Riesz space and ρ is a norm which is not equivalent to a Riesz norm E' is isomorphic to the norm dual of the norm completion of (E, p), which is, according to Proposition 3 55(ш), a Banach lattice Hence E' is a Banach lattice The norm dual of a norm complete, directed partially ordered vector space with a regular norm and closed positive cone is a Riesz space if and only if it has the decom position property (see Wickstead[24, Thm 2 8] or Andô[3, Thm 2]) The following is an example of a reflexive, regularly normed, directed partially ordered vector space that is not a Riesz space and an explicit example of a norm dual that is not a Riesz space 93 Chapter 5. Spaces of Operators Example 5.8 A directed partially ordered vector space E with a regular norm such that E is norm complete and E+ is closed and such that E' = E~ is not a Riesz space. Take E = R3 with the ordering generated by the circular cone К with vertex at 0, axis in the direction of e := (\/3)_1(l, 1,1) and the apical angle equal to 90°, and equip E with the Euclidean norm ||.||. Then £ is a directed partially ordered vector space and not a Riesz space. E+ is closed and E is norm complete. Denote the natural inner product in R3 by <, >. (a) ||.|| is Fremhn: Let x,y e К. Then < χ,у >> 0, so that ||x - у\\2 < \\x + y\\2. Hence ||.[| is Fremlin. (b) ||.|| is regular: Let χ € E. It suffices to be shown that there is а у e E with у + χ, у — χ € Κ and \\y\\ = \\x\\. It may be assumed that the angle α between χ and e is not obtuse (otherwise, consider —x) and that ||x|| = 1. If α € [0,45o], then χ 6 К, so take in this case у := χ. If α e (45°, 90o], take cos α cos 2α у :— χ : е. sin α sin α Then, since < χ, e >= cosa: \\y\\2 — (cos2 a- 2 cos2 a cos 2a + cos2 2a)/sin2 a = (cos2 a—cos 2a)/sin2a = l and < у, χ > = cos α/ sin α—cos 2α cos a/sina = sin2a, so that < у + χ, e >= cos a + sin α > 0, < у — χ, e >= sin a — cos α > 0, Цу + хЦ2 =2 + 2 This yields that the angles between у + χ and e and у — χ and e are precisely 45°, which means that у + χ and у — χ are on the surface of K. Thus, ||.|| is regular. (c) E' = E~ is naturally isomorphic to E: Every linear function is continuous, so E' = E~. It remains to show that the natural isomorphism between R3 and (R3)' is bipositive with respect to the .ff-ordering, i.e. for every χ € R3 one has that < x, и >> 0 for all и € К if and only if χ € A". If χ € A", then the angle between χ and any и e К is at most 90°, so < x,u >> 0 for all и € К. If χ e R3 is such that < χ, и >> 0 for all и 6 К, then the angle between χ and any element of К is not obtuse, which is only possible if χ e A'. Hence, E~ and E are isomorphic, and therefore E' = E~ is not a Riesz space. Example 5.9 A Riesz space E with a norm, such that E' is not directed. 94 5.3. Bonsall's Theorem and a Riesz Space Valued Version of Krein's Lemma x n _ x n x Take E = с00(Ы) and p(x) := £]n l ( + 1) ( )\, € E. E is a Riesz space and ρ is a norm on E (see also Example 1.28). The algebraic dual E' is isomorphic N to M via the isomorphism ƒ ь-> (/(en))n, ƒ € E', where e„ = !{„}, η 6 N. The positive functions correspond to positive sequences. Let /o G £* be the function corresponding to the sequence (—1,1, —1,1,...). (a) /o is continuous: For χ e E one has |/o(z)l = | Ση(ι(2η + 1) — i(2n))| < p(x). + (b) For ei/erj/ ƒ € £' i/ie sequence (ƒ (e„))n ¿s summable: For every N ÇN one has which means that Σ"=ο /Ы = /(Etoen) < ||/||ρ(Σΐοβη) = II/II. Ση/(en) < 11/11· (c) /o £ £'+ — £"+: The sequence (—1,1, —1,1,...) is not equal to the difference of any two positive, summable sequences, so, by (b), /o ^ E'+ — E'+. This shows that E' is not directed. 5.3 Bonsall's Theorem and a Riesz Space Valued Version of Krein's Lemma Krein's lemma is for real valued linear functions. A version for linear mappings that take values in a Dedekind complete Riesz space can be proved via Bonsall's theorem (see Bonsall[5]). It will be shown that, as a consequence, C(E,F) is directed if J? is a regularly normed directed partially ordered vector space and F a Dedekind complete Riesz space with a strong unit and equipped with the norm generated by the unit. First, an elementary lemma. Lemma 5.10 Let E and F be partially ordered vector spaces and let A : E —> F be a subadditive, positively homogeneous mapping. If A(—x) = — A(x) for all χ S E, then A is linear. Proof. Let x,y € E. 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. Theorem 5.11 (Bonsall) Let E be a directed partially ordered vector space, let F be a Dedekind complete Riesz space, and let Ρ : E —> F be a subadditive, positively homogeneous mapping. If Q : E+ -> F is a superadditive, positively homogeneous mapping such that Q < Ρ on E+, then there is a linear mapping В : E —¥ F such that B>QonE+ and В < Ρ on E. 95 Chapter 5. Spaces of Operators Proof. В will be found with aid of Zorn's lemma as a minimal element of the following set of subbadditive, positively homogeneous mappings: V := {A : E —¥ F: A is subadditive, positively homogeneous, A < Ρ on E and A > Q on E+}. (a) There is a minimal element В m V: Ρ e V, so V is not empty. Let С be a chain in V. For Л e V one has that 0 < A(x) + A(-x), so A(x) > -A(-x) > -P(-x) for all χ 6 E. F is Dedekind complete, so for every χ ζ E the infimum A<¡(x) •— inf{A(z): A e С} exists. Then: • A0 is subadditive: Let x,y € E. If A\, A2 € C, then one of the two, say A\ is the smallest, so A\(x) + A2(y) > Ai(x) + Ai(y) > Ax(x + y) > A0{x + y), hence A0(x) + A0(y) > A0(x + y). • До is positively homogeneous. • Ao < Ρ on E, since A(x) < P{x) for all A e V and χ & E. + • A0(x) > Q(x) for all χ e E . • AQ is a lower bound of C. According to Zorn's lemma, it follows that there is a minimal element В in V. (b) В is linear: Let С : E+ -»· F be an arbitrary mapping with С < В on E+. For every χ Ç. E and o e E+ one has that the function λ (->· B(x+Xa)—XC(a) from [0, co) to F is bounded below, because B(x + λα) — XC(a) > B(Xa) — B(—x) — XC{a) > —B(—x). So for every a e E+ we can define Ca(x) := inf (B(x + Xa) - AC(o)), χ e E. Ae[0,oo) Then for every a 6 E+: • Ca is subadditive: Let x,y Ç. E and let λι, λ2 6 [0, co). ß(i + X\a) — X\C{a) + ß(i/ + A2o) - X2C{a) > B(x + y+ (Χχ + X2)a) - (Ax + A2)C(o) > Ca(x + y), so Ca{x) + Ca(y) > Ca{x + y). • Ca is positively homogeneous: Let μ e (Ο,οο). Οα(μχ) = ίηίΑμ(Β(χ+ ^o) — ¿C(a)) = μΟα(χ). Co(0) = 0, because В > С on Е+ and 5(0) = 0. • Ca Xa) - XQ{a)) > infA(Q(i + Aa) - XQa) > Q(x), by superadditivity of Q, so Ca e V. + В is minimal in V and Ca < B, so Ca = B. Hence, for every a e E and χ 6 E one hasß(i+o) = infA(ß(i+Aa)-Ag(a)) < B{x+a)-Q{a), so B(x+a) > B{x)+Q(a). Thus, ß(Ao + χ) > B{Xa) + Q(x) for all χ G £+, α e E, A e R. + Take now С = S. Then for every ζ € E one has Ca(x) > infA(S(Aa) + Q{x) - XB(a)) — Q(x), so C, e V. Ca < В and В is minimal in V, hence Ca = В for all 96 5 3. Bonsall's Theorem and a Riesz Space Valued Version of Krein's Lemma + a e E , which yields that B{x) = mix{B(x + λα) - XB(a)) < B(x + a) - B(a), so B(x + a) > B{x) + B(a). By subadditivity of В it follows that B(x + a) = B(x) + B{a) for all χ € E, a e E+ and, in particular (ι = —a): В (-a) = -B(a) for all a £ E+. + Let χ € E. E \s directed, so there are X\,X2 ё E such that χ — x1 — x^. Then, by the previous equations, B(—x) = B(x2 — Xi) = B(x2) + B(x\) = — B{—12) — Β(χχ) — —B(—X2 + xi) = — B(x). Together with the subadditivity and the positive homogeneity this yields that В is linear, according to the previous lemma. Thus, 5 is a linear mapping and an element of V and therefore it has the desired properties. Corollary 5.12 Let E be a directed partially ordered vector space, let F be a Dedekmd complete Riesz space, and let Ρ : E —»· F be a subadditive, positively homogeneous mapping such that P(x) < P(y) for every x,y G E with —y Proof. Define Q(x) := sup^u: — χ < и < χ}, χ 6 Ε+. Q is superadditive and positively homogeneous, and Q < Ρ on E+. By the theorem, there is a linear mapping В : E -» F with В < Ρ on E and В > Q on E+, hence В > A, -A on E+. The result that the norm dual of a monotonely normed partially ordered vector space E is directed can, with aid of this corollary, be extended to directedness of the space C(E, F) for any Dedekind complete Riesz space F with a strong unit and the norm generated by the unit, under the additional condition that E be directed. Corollary 5.13 Let E be a directed partially ordered vector space with a Fremlin norm and let F be a Dedekind complete Riesz space with a strong unit, equipped with the norm generated by the unit. Let A : E —» F be a continuous, linear mapping with \\A\\ < 1. Then there is a positive, continuous, linear mapping В : E —> F such that -В Proof. Let ρ be the norm of E and let и be the unit of F. Define P(x) := p{x)u, χ € E. Then Ρ : E —» F is subadditive, positively homogeneous, ||P(i)|| < p(x)u, 97 Chapter 5 Spaces of Operators P(-x) = -P(x) for all ι 6 E, and Ρ is increasing on E+ \\A\\ < I, so A < Ρ on E+ It follows that the previous corollary can be applied, thus establishing the assertion 5.4 Operators Between Normed Partially Or dered Vector Spaces This section presents a generalization for partially ordered vector spaces of the results listed in Proposition 5 1 For most results it suffices that £ is a directed partially ordered vector space with a regular norm and that F is a partially ordered vetor space with a Fremlin norm For the assertion that CT(E, F) be a Riesz space, a certain kind of lattice structure is needed on E and F For that purpose, it will be assumed that F is Dedekind complete and that E has the decomposition property We start with a lemma and a proposition Lemma 5.14 Let E be a directed partially ordered vector space with a regular norm ρ and let F be a partially ordered vector space with a Fremlin norm ρ (ι) Let AQ E+ —> F+ be an additive, positively homogeneous mapping, such that + M = sup{p(i4ox) χ S E ,p(x) < 1} < oo ThenA0 extends to a continuous, positive, linear mapping A E —> F and \\A\\ = M (η) Let A e L(E, F) and В € C{E, F) be such that -B < A < В Then A is continuous and \\A\\ < ||ß|| Proof, (ι) Since E is directed, Л0 extends to a linear mapping A E -» F Clearly, A is positive To prove that A is continuous, let χ e E be such that p(x) < 1 ρ is regular, so there is a y € E+ with —y A mapping between partially ordered vector spaces is called order bounded if it maps order bounded sets (see Definition 1 44) to order bounded sets 98 5 4. Operators Between Normed Partially Ordered Vector Spaces Proposition 5.15 Let E be a directed partially ordered vector space with a regular norm ρ such that E is norm complete and E+ closed and let F be a partially ordered vector space with a norm p. Then: (ι) If A & L(E, F) maps order bounded sets to norm bounded sets, then A is continuous. (n) If order bounded sets in F are norm bounded, then every order bounded, linear mapping from E to F is continuous. Proof, (i): Suppose that A is not continuous. Then there is a sequence (xn)n in n E such that p{xn) < 2" for all η and p(Axn) —¥ oo. ρ is regular, so there are + _n [Уп)п in E with — yn < xn < yn and p(yn) < 2 for all n. Since E is norm + complete, у := J2nyn exists in E, and у > yn for all n, because E is closed. Then -y < xn < у for all n, which contradicts the assumption that A maps order bounded sets to norm bounded sets. Thus, A is continuous. (ii): Directly from (ι). Remark that the condition that E be norm complete and E+ closed is the same as that E+ be norm complete, by Corollary 3.47, and that the condition that order bounded sets be norm bounded is satisfied if the norm is monotone, by Lemma 1 48. The following is the main theorem in this section. Most of the results can be found in the literature, often considered from another point of view. Theorem 5.16 Let E be a directed partially ordered vector space with a regular norm ρ and let F be a partially ordered vector space with a Fremhn norm p. Then- (г) For any A £ C(E,F)+ the operator norm of A is equal to \\A\\ = sup{p(Ar): χ e E+,p(x) < 1}. (n) C(E,F) is a full subspace of L(E,F) and C(E,F) и a full subspace of L'(E,F). (m) The operator norm on C(E, F) is Fremhn. If F* is closed m F, then C(E, F)+ is closed m C(E,F). (iv) If F+ is norm complete, then C(E,F) is norm complete with respect to the regulanzation ||.||r of the operator norm. (v) If F is a Dedekmd complete Riesz space and E has the decomposition property, then Cr(E, F) is a Riesz space. 99 Chapter 5 Spaces of Operators (m) If F is a Dedekind complete Riesz space with a strong unit and if ρ is the norm generated by the unit, then C(E, F) is directed and the operator norm is regular If, m addition, E has the decomposition property, then C(E, F) is a normed Riesz space (vn) If E+ is norm complete, then every order bounded, linear mapping from E to F is continuous, m particular U(E, F) С C(E, F) Proof. (1) Follows from Lemma 5 14(i) (n) Let B,C € C(E,F) and A e L(E,F) be such that В < A < С Then 0 < A — В < С — В, so it follows from Lemma 5 14(H) that A — В is continuous and therefore A is continuous Hence, C(E, F) is a full subspace of L(E, F) If В, С € С (E, F) and A e L(E, F) are such that В < A < C, then A is continuous and —D C(E, F) is norm complete with respect to || ||r + + Let (An)n be a Cauchy sequence in C(E,F) Then for any χ e E , (Anx)n + + is Cauchy in F , so A0x = hmnAnx exists in F A0 is additive, positively + homogeneous, and p(Ax) = \imnp(Anx) < sup„ ||ЛП|| for all χ € E with p(x) < 1, so, according to Lemma 5 14(i), AQ extends to a continuous, positive, linear mapping A E -• F To show that An -> A in C{E, F), let ε > 0 Take N € N such that IIA» - Anil < ε for all m,n > N, then p({An - A)x) = hmmp((A - Am)x) < hmm ||ЛП - Лт||р(і) < ερ(χ) for any χ e E, so ЦД, — A\\ < ε, n> N (v) In this situation, Lr(E, F) is a Riesz space (see e g Wickstead[24, Prop 3 16]) С [E, F) is directed and it is a full subspace of Lr(E,F), according to (n) This yields that C(E,F) is a Riesz subspace of L(E, F) (vi) According to Corollary 5 13, C(E, F) is directed and || ||r < || || Since || || is Fremlin, by (in), it follows that || ||r = || || (Corollary 3 45), which means that || || is regular If E has the decomposition property, then, by (v), CT(E,F) is a Riesz space, so that it follows that C(E, F) — Cr(E, F) is a normed Riesz space, by Proposition 3 40 (vn) If E+ is norm complete, then, by Corollary 3 47, E is norm complete and E+ is closed Every order bounded set in F is norm bounded (Lemma 1 48), so, according to Proposition 5 15, every order bounded, linear mapping from E to F is continuous 100 5 5 An Alternative for the r-Norm 5.5 An Alternative for the r-Norm For a normed Ricsz space E and a Dedekind complete Riesz space F, the space C(E,F) is a Riesz space, but the operator norm on this space need not be Riesz Therefore, one usually considers the r-norm (or regular norm) || ||r on this space, which is the regularizaron of the operator norm, hence a Riesz norm Surprisingly, T {C (E, F), || ||r) is norm complete if F is However, || ||r is in general not an operator norm Is there an alternative construction that yields a Riesz norm that is an operator norm7 It turns out that such a construction is possible if E is a directed partially ordered vector space with the decomposition property, equipped with a regular norm and if F is a partially ordered vector space with a norm with a full unit ball such that F+ is closed The idea is to enlarge the space F If an operator maps to F, then it maps to any larger space as well So, if F is isomorphic to a subspace of a normed partially orderd vector space Fb then C(E, F) is isomorphic to subspace of C(E, F\) Since there are more positive, continuous mappings from E to Fi than to F, the regulanzation of the norm of C(E, Fi) will on C(E, F) be less than or equal to the regulanzation of the norm of C(E, F) It may be strictly less Can Fi be taken so large that, on C(E, F), the regulanzation of the norm of C(E,Fx) becomes equal to the operator norm of C(E, F)? Fi would be large enough if it is a Dedekind complete Riesz space with a strong unit, equipped with the norm generated by the unit (see Theorem 5 16(vi)) According to the embedding theory m Chapter 4, such an Fi exists precisely if F is a partially ordered vector space with a norm with a full unit ball such that F+ is closed Let us state this idea as a theorem Theorem 5.17 Let E be a directed partially ordered vector space with a regular norm ρ and let F be a partially ordered vector space with a norm ρ with a full unit ball such that F+ is closed Then there exists a Dedekind complete Riesz space Fi with a strong unit, equipped with the norm generated by the unit, such that F is isomorphic to an ordered subspace of F\ Then C(E, F) is isomorphic to a subspace ofC(E, F\) and the norm of'C(E, Fi) is regular If E has the decomposition property, then C(E, Fi) is a normed Riesz space Proof. According to Corollary 4 34, F can be embedded into such a space F\ By Theorem 5 16(vi), the norm of C(E, Fi) is a regular norm, and C(E, Fi) is a normed Riesz space if E has the decomposition property 101 Chapter 5 Spaces of Operators If the norm on F does not have a full unit ball, but is only a Fremlin norm such that F+ is closed, then a weaker consequence remains true Corollary 5.18 Let E be a directed partially ordered vector space with a regular norm and let F be a partially ordered vector space with a Fremlin norm such that F+ is closed Then there is a normed Riesz space F\ such that F is ismorphic to a subspace of Fi and therefore C(E,F) isomorphic to a subspace of L{E,F{), and such that the norm of C(E, Fi) is equivalent to a regular norm. Proof. According to Theorem 2.3, the Fremlin norm on F is equivalent to a norm with a full unit ball. Then application of the theorem establishes the assertion. 102 Commentary Inspired by the fact that every partially ordered vector space is a subspace of a Riesz space, this text has investigated seminorms on partially ordered vector spaces, focussing on extension and embedding problems and generalizations of the theory of normed Riesz spaces. Various new notions have been introduced and some existing concepts have been formulated in a more general context, serving as solutions to characterization problems or as suitable generalizations. Some notions turned out to be very useful, others are rather artificial and technical. This section aims to present a summary with comments on the main definitions and results. Of course, such a selection is highly subjective. The seminorms discussed in this text can roughly be divided into two classes: the seminorms that generate locally full topologies and those that generalize the notion of Riesz seminorm. Monotone, monotone*, Premlin seminorms, and seminorms with full unit balls belong to the first class, pre-Riesz and regular seminorms to the second one. The most useful notion in the first class is undoubtedbly that of a Fremlin seminorm. It turns out to be convenient in almost every respect. The definition of the Fremlin property is a symmetrized version of monotonicity. Therefore one may expect that results will look nicer for Fremlin seminorms. This is indeed the case, but the advantage is more than merely cosmetic. Unlike Fremlin seminorms, monotone seminorms cause troubles with extensions, monotone* semi- norms with quotients, and seminorms with full unit balls exclude many important examples such as restrictions of Riesz seminorms, χ м- \f (i)| for any positive, linear function ƒ, and operator norms. Regular seminorms on directed spaces are the most convenient choice in the sec ond class. Apparently, for many purposes the main property of a Riesz seminorm, in addition to its monotonicity, is that it is determined by its behaviour on the positive cone. The definition of a regular seminorm closely resembles the definition of a Riesz seminorm and also its properties turn out to be rather similar. A disadvantage is that the class of regular seminorms is not closed under addition. Pre-Riesz semi- norms —ensuing from generalizing the notion of solid set to partially ordered vector spaces— fit nicely in the context of pre-Riesz spaces, but are often inconvenient to 103 Commentary work with. Considering norms instead of seminorms, it is natural to require closedness of the positive cone. Thus, for normed partially ordered vector spaces, the two most useful objects are partially ordered vector spaces with Fremlin norms and closed positive cones, and directed partially ordered vector spaces with regular norms and closed positive cones. Let us now go over the main results. Chapter 1 presents basic properties. Fremlin seminorms turn out to be at least as convenient as monotone seminorms. Chapter 2 develops extension theorems. In Chapter 3, similar theorems are proved for regular and pre-Riesz seminorms. Greatest extensions exist as long as the large space is majorized, except for monotone seminorms. Chapter 3 presents generalizations of Riesz seminorms. The first approach by solid sets leads to a characterization of the seminorms that can be extended to Riesz seminorms on any majorized larger Riesz space: pre-Riesz seminorms. Their description is rather technical and easily superseded by the convenience of regular seminorms. Fremlin seminorms can in a natural way be 'regularized'. Regulariza- tions and regular seminorms in general behave nicely with respect to norm com pleteness. With respect to norm completions, Fremlin norms with closed positive cones fit to partially ordered vector spaces just as well as Riesz norms fit to Riesz spaces. The same is true for regular norms with closed positive cones on directed partially ordered vector spaces. Chapter 4 deals with embedding in normed Riesz spaces. It yields one of the strongest arguments in favour of Fremlin norms. It has already been argued that the most natural and convenient norm on a partially ordered vector space is a Fremlin norm with closed positive cone. Chapter 4 proves that the spaces with such norms are precisely the subspaces of normed Riesz spaces. This is another reason to define —in analogy with normed Riesz spaces— a normed partially ordered vector space as a partially ordered vector space with a Fremlin norm and a closed positive cone. The theorem that partially ordered vector spaces are precisely the subspaces of Riesz spaces then has a normed counterpart: the normed partially ordered vector spaces are precisely the subspaces of normed Riesz spaces. Partially ordered vector spaces with norms with full unit balls and closed positive cones are precisely the subspaces of M-spaces. Chapter 4 also presents partial results for embedding spaces of which the positive cone is not closed. They are considerably more general, but rather unappealing. Chapter 5 applies Fremlin and regular norms in the theory of spaces of operators. The results of Chapter 4 lead to an alternative proof of Krein's lemma. It yields that the norm dual of a partially ordered vector space with a Fremlin seminorm is 104 a legulaily normctl duceteci pai tidily oidercd \ector space, a result due to Ng It (.onfiims, once again, that Fremlin and regular seminorms are the appropnate kind of seminonns on pai tidily ordered vector spaces For operators between two paitially oideied vectoi spaces, the suitable setting is to consider operators from regularly noi med directed partially ordered vector spaces to Fremlin normed partially ordered vectoi spaces with closed positive cones One may conclude that a considerable part of the theory of Riesz seminorms on Riesz spaces can naturally be generalized to partially ordered vector spaces Embedding in (normed) Riesz spaces may be a convenient tool The appropriate seminorms are Fiemhn and regular seminorms Suitable generalizations of normed Riesz spaces are partially ordered vector spaces with Fremlin norms and closed positive cones and directed partially ordered vector spaces with regular norms with closed positive cones 105 106 List of Symbols N = {0,1,2,3,...}: natural numbers, 2 N* = {1,2,3,...}: natural numbers except 0, 14 <, >, V, .+,.", |.|, 2 [a, 6]: order interval, 21 <, t>, 38 E+: positive cone of E, 2 E": algebraic dual of E, 21 E': topological dual of E, 21 E~: order dual of E, 22 £"*: Dedekind completion of E, 27 E: norm completion of E, 59 L(E,F): linear mappings from E to F, 89 Lr(E,F): regular linear mappings from E to F, 89 C(E,F): continuous linear mappings from E to F, 89 £r(E,F): differences of continuous positive linear mappings from E to F, 89 C[0,1], C(X): continuous functions on the interval [0,1] or the topological space X, respectively, 3 Cl[Q, 1], C"*[0,1]: (n times) continuously differentiable functions on [0,1], 3 /°°(N), l°°(S): bounded sequences or function on 5, respectively, 5 Coo(N), Coo (S)'- sequences or functions, respectively, with support consisting of finitely many points, 10 R[0'4, Rs: functions from [0,1] or 5, respectively, to R, 19 L°°[0,1]: equivalence classes of bounded Lebesgue measurable functions on [0,1], 24 L^O, 1]: equivalence classes of functions on [0,1] with Lebesgue integrable p-th powers, 24 Aff[0,1]: affine functions on [0,1], 38 Ц.Цоо: supremum norm, ||x||oo = sup( |x(i)|, 5 107 List of Symbols p 1/i p I/p ||.||p: p-norm, ||ι||ρ = (^n |x„| ) ' on sequence spaces and \\x\\p = (ƒ |i| ) on function spaces, 5 1: constant 1 function, 3 1A: indicator function of A, i.e. 1 on A and 0 elsewhere, 11 f\A- restriction of ƒ to A, 36 A: closure of the set A, 9 full 5: full hull of 5, 7 co S: convex hull of 5, 14 sol S: solid hull of S, 17, 38 108 Bibliography [1] Charalambob D Ahprantis and Owen Burkmshaw, Locally Solid Riesz Spaces, Academic Press, New York - San Francisco - London, 1978 [2] Charalambos D Ahprantis and Owen Burkmshaw, Positive Operators, Aca demic Press, Ine , Orlando - San Diego - New York - London - Toronto - Mon treal - Sydney - Tokyo, 1985 [3] Τ Andò, On fundamental properties of a Banach space with a cone, Pacific J Math 12 (1962), 1163-1169 [4] Garrett Birkhoff, Lattice Theory (Amer Math Soc Colloq Pubi 25), 3rd ed , American Mathematical Society, Providence, Rhode Island, 1967 [5] F F Bonsall, The decomposition of continuous linear functionals into non- negative components, Ртос Umv Durham Philos Soc Ser A 13 (1957), 6-11 [6] G Buskes and ACM van Rooij, The vector lattice cover of certain partially ordered groups, J Austral Math Soc (Series A) 54 (1993), 352-367 [7] Ε В Davies, The structure and ideal theory of the pre-dual of a Banach lattice, Trans Amer Math Soc 131 (1968), 544-555 [8] D H Fremhn, Topological Riesz Spaces and Measure Theory, Cambridge Uni versity Press, 1974 [9] D H Fremhn, Tensor Products of Banach Lattices, Math Ann 211 (1974), 87-106 [10] Mans van Haandel, Completions in Riesz Space Theory (Ph D thesis University of Nijmegen), 1993 [11] Graham Jameson, Ordered Linear Spaces, Lecture Notes in Math 141, Springer-Verlag, Berlin - Heidelberg - New York, 1970 109 Bibliography [12] E de Jonge and А С M van Rooij, Introduction to Riesz spaces (Mathematical Center Tracts 78), Mathematisch Centrum, Amsterdam, 1977 [13] L V Kantorovich and G Ρ 4kilov, Functional Analysis, 2nd ed , Pergamon Press, Oxford - New York - Toronto - Sydney - Pans - Frankfurt, 1982 (Trans lation) [14] J L Kelley and Isaac Namioka (and co-authors), Linear Topological Spaces, D Van Nostrand Company, Ine , Princeton - New Jersey -Toronto - New York - London,1963 [15] W A J Luxemburg, Concurrent Binary Relations and Embedding Theorems for Partially Ordered Linear Spaces, Algebra and Order, Proc First Int Symp Or dered Algebraic Structures Luminy-Marseilles 1984 (ed by S Wolfenstein), Heldermann Verlag, Berlin, 1986, 223-229 [16] W A J Luxemburg and А С Zaanen, Riesz Spaces I, North-Holland Publishing Company, Amsterdam - London, 1971 [17] Peter Meyer-Nieberg, Banach Lattices, Springer-Verlag, Berlin - Heidelberg, 1991 [18] Isaac Namioka, Partially Ordered Linear Topological Spaces (Amer Math Soc Memoir 24), Providence, 1957 [19] Kung-Fu Ng, A duality theorem on partially ordered normed spaces, J London Math Soc (2) 3 (1971), 403-404 [20] Anthony L Peressini, Ordered Topological Vector Spaces, Harper к Row, New York, 1967 [21] H H Schaefer, Topological Vector Spaces, Springer-Verlag, Berlin - New York, 1971 [22] Helmut II Schaefer, Banach Lattices and Positive Operators, Springer-Verlag, Berlin - Heidelberg - New York, 1974 [23] Β Ζ Vulikh, Introduction to the Theory of Partially Ordered Spaces, Wolters- NoordhofT, Groningen, 1967 (English translation from the Russian) [24] A W Wickstead, Spaces of linear operators between partially ordered Banach spaces, Proc London Math Soc (3) 28 (1974), 141 158 [25] А С Zaanen, Riesz Spaces II, North-Holland Publishing Company, Amsterdam - New York - Oxford, 1983 110 Index σ-Dedekind complete, 2 in regularly normed space, 79 in Riesz space, 3, 45 algebraic dual, 21 in Riesz space with M-norm, 86 Archimedean, 2, б in Riesz space with M-seminorm, balanced, 13 51 bipositive linear mapping, 3 in Riesz space with Riesz semi- Bonsall's theorem, 95 norm, 47, 82 extension closed positive cone, 8, 15 Fremlin seminorm, 29 closure of positive cone, 9, 16 L-seminorm, 52 complete locally convex, full topology, 30 σ-Dedekind, 2 M-seminorm, 51 Dedekind, 2 monotone* seminorm, 28 uniformly, 7 norm completeness, 35 w.r.t. seminorm, 11 pre-Riesz seminorm, 47, 48 completely discontinuous, 60 regular seminorm, 56 continuous linear mapping, 89 Riesz seminorm, 27 seminorm with full unit ball, 29 decomposition property, 2 seminorm with solid unit ball, 40 Dedekind complete, 2 solvex set, 47 determine the ordering, 26 directed, 2 feebly closed discontinuous positive linear function, E'+-, 78 60 E~+-, 81 dual filling a unit ball, 30 algebraic, 21 Fremlin seminorm, 4, 68 norm, 21, 92 Fremlinize, 30 of quotient space, 26 order, 22 full topological, 21 hull, 7 locally full topology, 13 embedding set, 7 in normed Riesz space, 70, 75 unit ball, 8 111 Index homomorphism Riesz space with monotone norm, Riesz, 3 62 Riesz*, 44, 46 normal cone, 13 infimum, 2 operator integrally closed, 2, 80 between normed partially ordered vector spaces, 99 kernel of monotone seminorm, 9 between normed Riesz spaces, 90 Krein's lemma, 92, 97 continuous, 89 L-seminorm, 7 order bounded, 98 lattice positive, 3 operations, 2, 18 regular, 89 ordering, 2 operator norm, 21, 99 locally order convex, full, 14 dual, 22 convex, solid, 17 interval, 21 full, 13 order bounded solid, 17 linear function, 22 linear mapping, 98 M-seminorm, 7 set, 22 majorizing, 3 ordered monotone seminorm, 4 subspace, 3 monotone* seminorm, 4, 28 vector space, 2 negative, 2 partially ordered vector space, 2 negative part, 2 positive norm (see also seminorm) cone, 2 dual, 21, 92 element, 2 operator, 21 linear mapping, 3 norm complete part, 2 positive cone, 58, 100 pre-Riesz set, 11 seminorm, 43, 47 w.r.t. monotone norm, 57 space, 2, 45 w.r.t. regular norm, 57 subspace, 45 w.r.t. seminorm, 11 quotient norm completion norm, 10 Fremlin norm, 60 ordering, 10 monotone norm, 60 space, 10, 26 ordering, 59 regular norm, 60 r-norm, 54, 89, 101 112 regular set, 38 linear mapping, 89 unit ball, 39 seminorm, 54 solid (in Riesz space) regularization, 56 hull, 17 relatively uniform convergence, 7 locally solid topology, 17 restriction set, 17 L-seminorm, 8 unit ball, 17 M-seminorm, 8 solvex monotone seminorm, 4 set, 42 pre-Riesz seminorm, 44, 46 unit ball, 43 Riesz norm, 66 subspace Riesz seminorm, 4, 38 ordered, 3 solid set, 39 Riesz, 3 solvex set, 43 supremum, 2 Riesz topological dual, 21 homomorphism, 3 topology seminorm, 4 locally convex, full, 14 space, 2 locally convex, solid, 17 subspace, 3 locally full, 13 Riesz* homomorphism, 44, 46 locally solid, 17 sandwich theorem, 14 vector space, 13 saturated set, 7 uniformly seminorm Cauchy, 7 additive on positive cone, 8 complete, 7 Fremlin, 4, 68 uniqueness of extensions, 31, 32, 50 L-, 7 M-, 7 vector space monotone, 4 ordering, 2 monotone*, 4, 28 topology, 13 pre-Riesz, 43, 47 regular, 54 Riesz, 4 with full unit ball, 8, 85 with solid unit ball, 17 with solvex unit ball, 43 solid (in partially ordered vector space) hull, 38 113 114 Curriculum Vitae Op 15 oktober 1971 werd ik in Waddinxveen geboren. In 1984 begon ik op de Open bare Dalton Scholengemeenschap Voorburg-Leidschendam in Voorburg en behaalde er in 1990 mijn VWO-diploma. In de periode 1990-1994 studeerde ik Technische Wiskunde aan de Technische Universiteit Delft, de laatste twee jaar in de richting functionaalanalyse bij Philippe Clément. Ik behaalde er in 1991 mijn Propedeuse (met lof) en in 1994 mijn Inge nieursdiploma (met lof). Mijn afstudeerwerk ging over 'Multiresoluties en wavelet- bases voor Н"{Щ, sel'. Op 1 september 1994 kwam ik in dienst bij de Katholieke Universiteit Nijmegen als AiO bij de Vakgroep Wiskunde. Ik verrichtte er promotieonderzoek onder leiding van Arnoud van Rooij, uitmondend in dit proefschrift. Daarnaast gaf ik onderwijs in de vorm van practica en enkele hoorcolleges. Van januari tot en met mei 1998 werd mij verlof verleend om als Teaching Assistant aan de Texas Tech University in Lubbock, Texas te gaan werken. Tijdens mijn verblijf aldaar heb ik een kort bezoek gebracht aan de University of Mississippi in Oxford, Mississippi. In juni 1998 bezocht ik het congres 'Positivity' in Ankara, Turkije. Op 31 december 1998 is mijn AiO-aanstelling afgelopen. 115