W.M. Schouten

The Riesz-Kantorovich formula forlexicographically ordered spaces

Master Thesis

Supervisor: Dr. O.W. van Gaans

01-07-2016

Mathematisch Instituut, Universiteit Leiden

1

Contents

Introduction 21. General theory 31.1. Partially ordered vector spaces and Riesz spaces 31.2. Linear operators between partially ordered vector spaces 71.3. Strong order units and ideals 81.4. Pre-Riesz spaces 81.5. Normed Riesz spaces 91.6. Ordinal numbers 102. The Riesz completion of the space Lr(E,F ) 133. The Riesz-Kantorovich formula 214. Main theorem 26Discussion 42References 43

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Introduction

If E and F are partially ordered vector spaces, then one can consider regularlinear maps from E to F , i.e. linear maps which can be written as the differenceof two positive linear maps. If the space E is directed, then the space Lr(E,F )of all regular linear operators becomes a partially ordered vector space itself. Wewill mainly concern ourselves with the questions when the space Lr(E,F ) is itself aRiesz space and how, even if it is not a Riesz space, its lattice operations work. Theso-called Riesz-Kantorovich theorem gives sufficient conditions for which Lr(E,F )is a Riesz space and it also specifies the lattice operations by means of the Riesz-Kantorovich formula: if S, T ∈ Lr(E,F ) and x ∈ E with x ≥ 0 then it holdsthat

(S ∨ T )(x) = sup{S(y) + T (x− y) : 0 ≤ y ≤ x}.It is still an open problem whether if the supremum of two regular operators ex-ists in Lr(E,F ), it automatically satisfies the Riesz-Kantorovich formula. We willdiscuss several new results on this subject, including our main theorem, which isdiscussed in Chapter 4.

In Chapter 1 we will discuss some general theory on partially ordered vectorspaces, (pre-)Riesz spaces and even some basic results on ordinal numbers.In Chapter 2 we will state two conjectures and give counter-examples to them. Theconjectures are concerned with the following situation: if E,F are Riesz spaces,but F is not Dedekind complete, then in general Lr(E,F ) is only a pre-Riesz spaceand not a Riesz space. As with all pre-Riesz spaces it is of interest to find a moreexplicit description of its Riesz completion. Naively it might seem logical that thisRiesz completion might have something to do with the space Lr(E,F δ), where F δ

is the Dedekind completion of F , or, if F is a normed Riesz space, with Lr(E,F ),where F is the norm completion of F . However, we will show that, in general, thereis no clear connection between the two.After that in Chapter 3, we will formulate and prove the Riesz-Kantorovich the-orem. Subsequently we prove a theorem which combines the result of Chapter 2with the Riesz-Kantorovich theorem.Finally in Chapter 4 we will state and prove our main theorem. This theorem statesthat if the image space is an arbitrary product of the real line, equipped with thelexicographic ordering, we only need very minor conditions on the domain space forthe Riesz-Kantorovich formula to hold. We will present two versions of this the-orem, one using ordinal numbers and one only considering the finite-dimensionalcase.

The new results in this thesis are the major part of Chapter 2 as well as Theorem3.4 and of course our main theorem in its two versions Theorem 4.5 and Theorem4.6.

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1. General theory

1.1. Partially ordered vector spaces and Riesz spaces.We will start by giving some of the most important definitions and results, whichwe will need in the rest of the thesis.

Definition 1.1. A vector space V over the real numbers with a partial order ≤ iscalled a (partially) ordered vector space if for all x, y, z ∈ V and all λ ∈ R≥0we have:

• If x ≤ y then also x+ z ≤ y + z.• If x ≤ y then also λx ≤ λy.

Vectors v ∈ V which satisfy v ≥ 0 are called positive. The set of all positivevectors is usually denoted by V+, so we have V+ = {v ∈ V : v ≥ 0}, and is calledthe positive cone of V . The positive cone is said to be generating if for all x ∈ Vthere exist y, z ∈ V+ such that x = y − z or, equivalently, if for any x ∈ V thereexists y ∈ V+ with x ≤ y. If the positive cone is generating then we call the partiallyordered vector space directed.

The positive cone of a partially ordered vector space V is always convex andsatisfies the following three properties.

(1) V+ + V+ ⊂ V+(2) λV+ ⊂ V+ for all λ ∈ R≥0(3) V+ ∩ (−V+) = {0}

Any convex subset of a vector space V which satisfies the properties (1), (2) and(3) above is called a convex cone in V . Instead of defining partially ordered vectorspaces as a vector space with a suitable partial order one can also define it in termsof a convex cone: if L is a convex cone in a real vector space V then we define theorder ≤ on V by saying that x ≤ y if and only if y − x ∈ L. Then (V,≤) is a par-tially ordered vector space in the sense of the previous definition. This means thata partially ordered vector space is in fact completely characterized by its positivecone.

From now on we always write V instead of (V,≤) when considering a partiallyordered vector space. Furthermore, if V is a partially ordered vector space and Uis a linear subspace of V then we will view U as a partially ordered vector spacewith the ordering inherited from V .

We will now give some other basic definitions:

Definition 1.2. Let V be a partially ordered vector space. If a, b ∈ V then wedefine the order interval [a, b] := {x ∈ V : x ≥ a, x ≤ b}. A subset B ⊂ Vis said to be bounded from above if there exists u ∈ V such that u ≥ b for allb ∈ B. Similarly B is bounded from below if there exists l ∈ V such that l ≤ bfor all b ∈ B. The elements u and l in the previous definitions are called an upperbound of B and a lower bound of B, respectively. B is called order-boundedif it is bounded from above and below or, equivalently, if it is contained in an order-interval. We will denote for a subset B which is bounded from above, the set of allupper bounds of B with Bu = {x ∈ V : x ≥ b for all b ∈ B}.

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Now we will define suprema and infima in a general partially ordered vectorspace, which will lead to the definition of a Riesz space.

Definition 1.3. A subset B of a partially ordered vector space V is said to havea least upper bound or supremum if there exists an element s ∈ V such that sis an upper bound for B and such that for all x ∈ V for which it holds that x is anupper bound of B we must have that s ≤ x. This element is necessarily unique ifit exists and we will denote it with supB. Similarly, B is said to have a greatestlower bound or infimum if there exists an element r ∈ V which is a lower boundof B and such that for all y ∈ V for which it holds that y is a lower bound wemust have that y ≤ r. This r is again unique if it exists and we will denote itwith inf B. A partially ordered vector space for which supB and inf B exist for allorder-bounded sets B is said to be Dedekind complete. We will use the followingnotation for the suprema and infima of sets with only two elements

x ∨ y = sup{x, y} and x ∧ y = inf{x, y}for x, y ∈ V .

Note that in general supB and inf B need not to be elements of B. If they are,they are sometimes referred to as the maximum and minimum of B respectively.Now we are ready to define a Riesz space.

Definition 1.4. A partially ordered vector space V is said to be a Riesz space ora vector lattice if for all x, y ∈ V we have that x ∨ y and x ∧ y exist in V .

If V is a Riesz space and x ∈ V then we can define the following vectors

x+ := x ∨ 0, x− := (−x) ∨ 0, |x| := x ∨ (−x),

called the positive part, negative part and absolute value of x respectively. We willalso use this terminology if V is not a Riesz space, whenever the suprema in thesedefinitions exist anyway. The positive part, negative part and absolute value satisfythe following important identities

x = x+ − x− and |x| = x+ + x−.

Note that this implies that every Riesz space is automatically directed. Using theabove notation we obtain the following list of basic results, which we will use lateron without reference.

Lemma 1.5. Let V be a partially ordered vector space. We have for u, v, w ∈ Vand λ ∈ R≥0, whenever, in (2)-(4), one of the two sides of the equality sign exists

(1) u = 0 if and only if u,−u ∈ V+;(2) (u+ w) ∨ (v + w) = (u ∨ v) + w; (u+ w) ∧ (v + w) = (u ∧ v) + w;(3) (λv) ∨ (λw) = λ(v ∨ w);(4) ((u ∨ v) ∨ w) = (u ∨ (v ∨ w));(5) u+, u− ∈ V+;u+ ∧ u− = 0; |u| ∈ V+;u+ ∨ u− = |u|;(6) |u| ∧ |v| = 0 if and only if |u− v| = |u+ v|.

Proof See [8, Theorem 1.4].

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Now for a slightly more interesting result, we obtain the following well-knownproposition.

Proposition 1.6. Let V be a partially ordered vector space. Suppose that x+ existsfor all x ∈ V . Then V is a Riesz space.

Proof Let x, y ∈ V . Then (x− y)+ and (x− y)− = (y−x)+ exist by assumptionand thus |x−y| = (x−y)++(x−y)− exists. We claim that x∨y = 1

2

(x+y+|x−y|

).

We have1

2

(x+ y + |x− y|

)≥ 1

2(x+ y + x− y) = x

and1

2

(x+ y + |x− y|

)≥ 1

2(x+ y + y − x) = y.

Now let z ∈ V with z ≥ x and z ≥ y. Then we have

z − 1

2(x+ y) =

(1

2z − 1

2x

)+

1

2z − 1

2y

≥ 0 +1

2z − 1

2y ≥ 1

2x− 1

2y =

1

2(x− y)

and

z − 1

2(x+ y) =

(1

2z − 1

2y

)+

1

2z − 1

2x

≥ 0 +1

2z − 1

2x ≥ 1

2y − 1

2x =

1

2(y − x).

So we must have that

z − 1

2(x+ y) ≥ 1

2|x− y|

and therefore

z ≥ 1

2

(x+ y + |x− y|

).

Hence x∨y exists and is equal to 12

(x+y+|x−y|

). Therefore V is a Riesz space.

An important definition is that of an Archimedean space. It reads as follows.

Definition 1.7. A partially ordered vector space V is said to be Archimedean iffor elements u, v ∈ V with nv ≤ u for all n ∈ Z>0 we must have that v ≤ 0. If V isa Riesz space then this is equivalent to the statement that if for elements u, v ∈ V+satisfying 0 ≤ nv ≤ u for all n ∈ Z>0 we must have that v = 0.

A Riesz space is not necessarily Dedekind complete, but it is always containedin a Riesz space that is, as can be seen from the following result.

Proposition 1.8. Let V be an Archimedean Riesz space. Then there exists aDedekind complete Riesz space W such that V ⊂W and such that for any Dedekindcomplete Riesz space U with V ⊂ U ⊂ W we must have that U = W . This spaceW is unique up to isomorphism and is usually denoted by V δ and is called theDedekind completion of V .

Proof See [9, Theorem 32.5].

A slightly weaker version of being a Riesz space is having the so-called Riesz-decomposition property. It is defined as follows.

Definition 1.9. A partially ordered vector space V is said to have the Riesz de-composition property if for any x1, x2, y ∈ V+ with y ≤ x1 + x2 there existy1, y2 ∈ V with 0 ≤ y1 ≤ x1, 0 ≤ y2 ≤ x2 and y1 + y2 = y.

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We obtain the following basic result, which can be found in [8].

Proposition 1.10. Every Riesz space has the Riesz decomposition property.

Proof Let V be a Riesz space and let x1, x2, y ∈ V+ with 0 ≤ y ≤ x1 +x2. Definey1 = y ∧ x1 and y2 = y − y ∧ x1. Then we immediately obtain that 0 ≤ y1 ≤ x1and y1 + y2 = y. Since y1 ≤ y we obtain that y2 ≥ 0. Finally we have that

y2 = y − y ∧ x1 = y + (−y ∨ −x1) = 0 ∨ (y − x1) ≤ 0 ∨ x2 = x2.

So this gives 0 ≤ y2 ≤ x2.

Now we are ready to give a few examples.

Example 1.11. On the vector space Rn there are two often-used partial orderswhich turn Rn into a partially ordered vector space. The first one is the com-ponentwise ordering and is defined by (x1, ..., xn) ≤ (y1, ..., yn) if and only ifx1 ≤ y1, ..., xn ≤ yn. The other one is called the lexicographic ordering and is de-fined by (x1, ..., xn) ≤ (y1, ..., yn) if and only if (x1, ..., xn) = (y1, ..., yn) or there ex-ists j ∈ {1, ..., n−1} such that xi = yi for all i ∈ {1, ..., j} and xj+1 < yj+1. One caneasily check that in both cases Rn becomes a Riesz space. For the componentwiseordering it is also Archimedean and even Dedekind complete. However, Rn withthe lexicographic ordering is neither, unless of course n = 1 in which case the lexico-graphic and the componentwise ordering coincide. It is not Archimedean for n > 1since (1, 0, ..., 0), (0, ..., 0, 1) ∈ Rn+ and 0 ≤ m(0, ..., 0, 1) = (0, ..., 0,m) ≤ (1, 0, ..., 0)for all m ∈ Z>0, while (0, ..., 0, 1) 6= 0. Also the set {(0, ..., 0, λ) : λ ∈ R} is boundedfrom above by (1, 0, ..., 0) and from below by (−1, 0, ..., 0), but it clearly does nothave a supremum.

Example 1.12. If (X, τ) is any topological space then the space C(X) of allcontinuous functions from X to R is a Riesz space if we say that f ≤ g if and onlyif f(x) ≤ g(x) for all x ∈ X. This follows from the fact that if f, g ∈ C(X) thenthe functions x 7→ max{f(x), g(x)} and x 7→ min{f(x), g(x)} are also continuous.The space C(X) is Archimedean, but in general it is not Dedekind complete. Forexample, if X = [0, 1] with the Euclidean topology and for n ∈ Z>1 we define fnto be 0 on [0, 12 ], linearly increasing on [ 12 ,

12 + 1

n ] and 1 on [12 + 1n , 1], then each fn

is continuous, but the set {fn : n ∈ Z>1} has no supremum, as the only functionfrom X to R which potentially could be a supremum is the function which is 0 on[0, 12 ] and 1 on ( 1

2 , 1], which is not a continuous function.

Example 1.13. Let (S,Σ, µ) be a measure space and let 1 ≤ p < ∞. Let ∼denote the equivalence relation on M(S), the measurable functions from (S,Σ) to(R,B(R)), where B(R) is the Borel-σ-algebra on R, given by f ∼ g if and only iff = g µ-almost everywhere. Now let Lp = {f ∈ M(S) :

∫|f |pdµ < ∞}/ ∼. We

define the ordering ≤ on Lp by f ≤ g if and only if f ≤ g almost everywhere. Now

if f ∈ Lp then we let f+ be given by f+(x) =

{f(x) if f(x) ≥ 0

0 if f(x) < 0. Then f+ is

clearly measurable and we have that∫|f+|pdµ ≤

∫|f |pdµ <∞,

since |f+|(x) ≤ |f |(x) for all x ∈ S. Hence f+ ∈ Lp. Clearly f+ ≥ f , f+ ≥ 0 andf+ is the smallest function which satisfies f+ ≥ f and f+ ≥ 0. Hence f+ is indeedthe positive part of f . So by Proposition 1.6 we get that Lp is a Riesz space.

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It is also Archimedean. In particular we get that the space

`p = {(an)n∈N : an ∈ R for all n ∈ N,∞∑n=1

|an|p <∞}

is an Archimedean Riesz space.Also for p =∞ we let

L∞ = {f ∈M(S) : there exists B ∈ R≥0 such that |f | ≤ B a.e.}.Again this space is an Archimedean Riesz space. In particular the space

`∞ = {(an)n∈N : an ∈ R for all n ∈ N, supn∈N|an| <∞}

is an Archimedean Riesz space.

Example 1.14. Finally we give an example of a space which is not a Riesz space.Let C1[0, 1] denote the set of functions from [0, 1] to R which are continuouslydifferentiable and we say that f ≤ g if and only f(x) ≤ g(x) for all x ∈ [0, 1]. Thenthe functions x 7→ x and x 7→ 1 − x do not have a supremum. Intuitively this isbecause the pointwise maximum of the two functions is not differentiable, but aslightly more rigorous argument is needed to fully prove it. It does, however, havethe Riesz decomposition property: if 0 ≤ f, g1, g2 and f ≤ g1+g2 then for i ∈ {1, 2}we define the function

fi : [0, 1]→ R, x 7→

{gi(x)f(x)g1(x)+g2(x)

if g1(x) + g2(x) 6= 0

0 else.

One can check that these f1 and f2 have the desired properties.

1.2. Linear operators between partially ordered vector spaces.Now we will discuss several spaces of operators between two partially ordered vectorspaces. We start with the following definition.

Definition 1.15. Let V and W be two partially ordered vector spaces. A linearmap f : V → W is called positive if f(x) ∈ W+ for all x ∈ V+. The map f iscalled order-bounded if f carries order-bounded subsets of V into order-boundedsubsets of W . Finally, f is called regular if it can be written as the difference oftwo positive linear maps.

If in the setting of Definition 1.15 the space V is directed then we can definethe ordering ≤ on the space of linear operators from V to W , which we denoteby L(V,W ), by saying that f ≤ g if and only if g − f is positive. Then L(V,W )with this ordering becomes a partially ordered vector space itself. Similarly, thisordering turns the space of all order-bounded operators and the space of all regularoperators into partially ordered vector spaces, which we will denote by L∼(V,W )and Lr(V,W ) respectively. We always have the following sequence of inclusions

Lr(V,W ) ⊂ L∼(V,W ) ⊂ L(V,W ).

These inclusions hold because any positive operator is clearly order-bounded andtherefore any regular operator is order-bounded as well. Furthermore if L∼(V,W )is directed, in particular if it is a Riesz space, then we have Lr(V,W ) = L∼(V,W ).Finally we also have the following definition.

Definition 1.16. A map f ∈ L(V,W ) is called bipositive if for all x ∈ V we havex ∈ V+ if and only if f(x) ∈W+.

Note that a bipositive map is always positive. Bipositive maps are also auto-matically injective: let f : V → W be bipositive and suppose f(x) = 0 for somex ∈ V . Then f(x) ≥ 0 so x ≥ 0. Then also f(−x) = −f(x) ≥ 0 and thus −x ≥ 0.

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Hence we obtain that x = 0, so f is injective.

1.3. Strong order units and ideals.Next we will define what a strong order unit is.

Definition 1.17. Let V be a partially ordered vector space. An element e ∈ V+ isa strong order unit if for all x ∈ V there exists a λ ∈ R>0 such that x ≤ λe.

If e is a strong order unit then for all x ∈ V we have x ≤ λe for some λ ∈ R>0

and since λe ≥ 0 we obtain that then V is automatically directed. Note that if eis a strong order unit, then so are λe for all λ ∈ R>0 and x+ e for all x ∈ V+. Wealso have the following definition:

Definition 1.18. If X is a vector space over R and A is a subset of X, then avector a ∈ A is called an internal point of A if for each x ∈ X there exists aλ0 ∈ R>0 such that a+ λx ∈ A for each −λ0 ≤ λ ≤ λ0.

The following lemma shows the connection between strong order units and in-ternal points, which we will need later on.

Lemma 1.19. A vector e ∈ V+ is a strong order unit if and only if it is an internalpoint of V+.

Proof See [1, Lemma 2.1].

Definition 1.20. Let V be a partially ordered vector space and let x ∈ V+. Then we

define the ideal generated by the element x to be the linear subspace∞⋃n=1

[−nx, nx]

and we denote this subspace by Vx.

Although more general ideals exist, they are not of use for us and hence wewill only consider ideals generated by a specific element. Note that if x ∈ V andy ∈ Vx with y ≥ 0 then for all z ∈ V with 0 ≤ z ≤ y we have z ∈ Vx. Further-more x is a strong order unit in Vx, which is an important fact we will need later on.

1.4. Pre-Riesz spaces.Now we will introduce the concept of a pre-Riesz space. For this we need someterminology.

Definition 1.21. We call a linear subspace W of a partially ordered vector spaceV order-dense in V if for all v ∈ V we have v = inf{w ∈W : w ≥ v}, where theinfimum is taken in V .

Definition 1.22. We say that a subspace W of a Riesz space V generates V iffor every v ∈ V there exist a1, ..., am, b1, ..., bn ∈W such that v = ∨mi=1ai − ∨mi=1bi.

Now we can define the concept of a pre-Riesz space.

Definition 1.23. Let V be a partially ordered vector space. Then V is called apre-Riesz space if one of the following equivalent conditions hold.

(1) For every x, y, z ∈ V the inclusion {x+y, x+z}u ⊂ {y, z}u implies x ∈ V+.(2) There exists a Riesz space W and a bipositive linear map ι : V → W such

that ι(V ) is order-dense in W .(3) There exists a Riesz space W and a bipositive linear map ι : V → W such

that ι(V ) is order-dense in W and generates W as a Riesz space.

All spaces W as in (3) are isomorphic as Riesz spaces. A pair (W, ι) as in (3) iscalled, since it is unique up to isomorphism, the Riesz completion of X.

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The equivalence of these definitions can be found in [6, Corollaries 4.9-11, The-orems 3.5, 3.7, 4.13]. We have the following observations.

Proposition 1.24. (1) Every pre-Riesz space is directed.(2) Every directed Archimedean partially ordered vector space is a pre-Riesz

space.(3) Every Riesz space is a pre-Riesz space.

Proofs of these results can be found in [6]. However, the construction of theRiesz completion of a pre-Riesz space is far from explicit. It is often an interestingquestion whether we can explicitly describe it. We have the following importanttheorem (see [6]).

Theorem 1.25. Let E,F be Riesz spaces, such that F is Archimedean. ThenLr(E,F ) is directed and Archimedean. In particular, Lr(E,F ) is a pre-Riesz space.

Proof For T ∈ Lr(E,F ) we have since T is regular that we can find positiveT1, T2 such that T = T1 − T2 and therefore Lr(E,F ) is directed.Let S, T ∈ Lr(E,F ) and suppose that nS ≤ T for all n ∈ Z>0. Suppose that S 6≤ 0.Since E is directed we can let x ∈ E+ with S(x) 6≤ 0. Then nS(x) ≤ T (x) for alln ∈ Z>0 since x ∈ E+. Since F is Archimedean this implies that S(x) ≤ 0 whichgives a contradiction. So S ≤ 0. Hence Lr(E,F ) is Archimedean. By Proposition1.24 this means that Lr(E,F ) is a pre-Riesz space.

In Chapter 2 we will consider the question if we can determine the Riesz com-pletion of the space of regular operators.

Now we will introduce the concept of disjointness.

Definition 1.26. Let V be a partially ordered vector space. The elements v, w ∈ Vare called disjoint, which we denote by v ⊥ w, if {v+w,−v−w}u = {v−w,−v+w}u. If V is a Riesz space then v ⊥ w is equivalent to the equality |v| ∧ |w| = 0.

We have the following useful result which shows under what circumstances dis-jointness is preserved in subspaces. It will be used later on.

Proposition 1.27. Let V and W be partially ordered vector spaces and x, y ∈W .If W is a subspace of V then x ⊥ y in V implies x ⊥ y in W . Furthermore, if Wis an order dense subspace of V , then x ⊥ y in V if and only if x ⊥ y in W .

Proof See [5, proposition 2.1].

1.5. Normed Riesz spaces.To conclude our discussion of partially ordered vector spaces we consider the conceptof normed Riesz spaces. We start with the following definition.

Definition 1.28. Let V be a Riesz space and ‖·‖ be a norm on V . Then ‖·‖ iscalled a Riesz norm if for x, y ∈ V with |x| ≤ |y| we have ‖x‖ ≤ ‖y‖. The pair(V, ‖·‖) is called a normed Riesz space. If V is complete with respect to the norm‖·‖ then the pair (V, ‖·‖) is called a Banach lattice.

We will first discuss a few examples.

Example 1.29. Let 1 ≤ p ≤ ∞ and consider the space `p from Example 1.13. For

p < ∞ we define the norm ‖(an)n∈N‖p :=

( ∞∑n=1|an|p

) 1p

and for p = ∞ we define

the norm ‖(an)n∈N‖∞ := supn∈N|an| for (an)n∈N ∈ `p. Then the space

(`p, ‖·‖p

)is a

Banach lattice.

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Example 1.30. If X is a compact and Hausdorff topological space then the space(C(X), ‖·‖∞), where ‖·‖∞ denotes the usual supremum-norm on C(X), is a Banachlattice.

Example 1.31. The space Rn with the lexicographic ordering and the usual Eu-clidean norm ‖·‖ is not a normed Riesz space. To see this note that |(0, 2)| =(0, 2) ≤ (1, 0) = |(1, 0)|, while ‖(0, 2)‖ = 2 > 1 = ‖(0, 1)‖.

We first have the following elementary result.

Proposition 1.32. Let (V, ‖·‖) be a normed Riesz space. Then we have:

(1) The maps (x, y) 7→ x ∨ y, (x, y) 7→ x ∧ y, x 7→ x+, x 7→ x− and x 7→ |x| areall continuous as maps either from V × V to V or from V to V .

(2) The positive cone V+ is closed with respect to the topology induced by ‖·‖.

Proof See [11, proposition 5.2].Now we obtain a more interesting result (see [2]).

Proposition 1.33. Let (V, ‖·‖) be a Banach lattice and let (W, ‖·‖′) be a normedRiesz space. Let T ∈ Lr(V,W ). Then T is norm-bounded, i.e. ‖T‖ <∞.

Proof Since the difference of two norm-bounded operators is norm-bounded werestrict ourselves to the case where T ≥ 0. We will argue by contradiction. Sup-pose that T is not norm-bounded. Then we can find a sequence (xn)n∈N in V suchthat ‖xn‖ = 1 and ‖Tx′n‖ ≥ n3 for all n ∈ N. Since V is a Banach lattice and

both |x| ≤∣∣∣(|x|)∣∣∣ and

∣∣∣(|x|)∣∣∣ ≤ |x| (since∣∣∣(|x|)∣∣∣ = |x|) for all x ∈ V we get that

‖|xn|‖ = ‖xn‖ = 1 for all n ∈ N. Since T is positive and |xn| ≥ xn ≥ −|xn|we obtain that T |xn| ≥ Txn ≥ −T |xn| and in particular T |xn| ≥ |Txn| for alln ∈ N. Hence ‖T |xn|‖′ ≥ ‖Txn‖′ ≥ n3 since W is a normed Riesz space. Now we

let y =∑∞n=1

|xn|n2 , which is a well-defined element of V since the series converges

absolutely and V is norm complete. For m ∈ N we see that y ≥ |xm|m2 ≥ 0, since by

Proposition 1.32 the positive cone of V is closed, so we obtain that Ty ≥ T |xm|m2 ≥ 0.

Hence we get ‖Ty‖′ ≥∣∣∣∣∣∣T |xm|m2

∣∣∣∣∣∣′ ≥ m for all m ∈ N. This gives a contradiction

since naturally ‖Ty‖′ <∞ and thus ‖Ty‖′ ≥ m cannot hold for all m ∈ N. Hencewe must have that T is norm-bounded.

1.6. Ordinal numbers.Although this subject has (almost) nothing to do with the concept of partially or-dered vector spaces, we will introduce the concept of ordinal numbers and show afew basic results, since one version of our main theorem is formulated in terms ofordinal numbers and uses transfinite induction.

The concept of an ordinal number originates in set theory and that is where itis used mostly. We do not assume that the reader has any knowledge of axiomaticset theory, but we also do not need that: all concepts can be defined and discussedusing ’intuitive’ set theory. For a more detailed discussion and proofs of the results,we refer to [7]. We start with the following definition.

Definition 1.34. A set S is called transitive if for all x ∈ S it holds that x ⊂ S.An ordinal number is a transitive set such that if we say that x ≤ y if and onlyif x = y or x ∈ y that then S is well-ordered by ≤.

The most basic example of an ordinal number is the empty set. We have thefollowing result.

11

Proposition 1.35. If α is a non-empty ordinal number and β ∈ α then β is alsoan ordinal number. The class of all ordinal numbers is itself well-ordered by ∈(although it is not a set). In particular, any two ordinal numbers α and β withα 6= β satisfy either α ∈ β or β ∈ α.

Instead of β ∈ α we also often write β < α. ∅ is the smallest ordinal number,followed by {∅}, {∅, {∅}}, .... ∅ is often identified with the natural number 0, {∅}with 1 etc. In this way we can view each natural number as an ordinal number.The ordinal numbers are an important class of well-ordered sets. In fact we get thatevery well-ordered set is isomorphic to a unique ordinal number in the followingsense.

Proposition 1.36. Let (V,�) be a well-ordered set. Then there exists a uniqueordinal number α and a bijective map f : V → α such that f(x) < f(y) if and onlyif x � y for every x, y ∈ V .

We also use the following terminology.

Definition 1.37. Let α be an ordinal number. The immediate successor ofα, notation α + 1, is the unique ordinal number min{β : α ∈ β} and it satisfiesα + 1 = α ∪ {α}. The ordinal α is called a successor ordinal if α = β + 1 forsome ordinal β and α is called a limit ordinal if it is not 0 and it is no successorordinal.

The set of all natural numbers is the first limit ordinal. If α is a limit ordinalthen one has the following identity:

α =⋃β∈α

β.

Using this terminology we can say what we mean by transfinite induction.

Theorem 1.38. (Transfinite induction) Let A be a class of ordinal numbers.Suppose A has the following three properties:

(1) 0 ∈ A.(2) If α ∈ A then also α+ 1 ∈ A.(3) If α is a limit ordinal and β ∈ A for all β ∈ α then also α ∈ A.

Then A is the class of all ordinal numbers.

This allows us to prove statements in the following way: first we prove that forsome minimal ordinal number the property holds; then we assume the propertyholds for some ordinal α and we prove that this implies it holds for α + 1 andfinally we assume the property holds for all β ∈ α where α is a limit ordinal thatthis implies that the property holds for α. Since natural numbers are also ordinalnumbers this allows us to include ’normal’ induction in transfinite induction. Toconclude we give an example of a partially ordered vector space which uses ordinalnumbers.

Example 1.39. Let α 6= 0 be an ordinal number. Let M =∏

1≤j≤α R. If x ∈ Mand β < α then we let x(β) denote the βth coordinate of x and x(1),...,(β) the vectorconsisting of the first β coordinates of x. Then we can give M the lexicographicordering by saying that x ≤ y if and only if either x = y or there exists someβ ≤ α such that for all γ < β we have x(γ) = y(γ) and x(β) < y(β). Just as for thelexicographic ordering on Rn we see that M becomes a Riesz space as it is totallyordered. If α > 1 then M not Archimedean and not Dedekind complete. It alsohas the following useful property. We give M the product topology. Let K ⊂ Mbe non-empty and compact. We claim that K has an order-maximum and showthis by transfinite induction. For α = 1 the result is well-known. Now let α ≥ 1

12

and assume each non-empty compact set in∏

1≤j≤α R has an order-maximum. LetK ⊂M be compact. Let πα be the projection from M onto its first α coordinates.Then we know that πα is continuous. Hence πα(K) is compact in

∏1≤j≤α R and

thus has an order-maximum, say x. Then {y(α+1) : y ∈ K,πα(y) = x} is a compactand non-empty subset of R and thus has an order-maximum y. Then (x, y) ∈ K andif z ∈ K then παz ≤ x and if παz = x then z(α+1) ≤ y. Hence z ≤ (x, y). So K hasan order-maximum. Finally let α be a limit ordinal and assume that for each β < αwe have that every non-empty compact subset of

∏1≤j≤β R has an order-maximum.

Let K ⊂M be compact. Let β < α and let πβ be the projection from M on the firstβ coordinates. As before we can let xβ ∈ πβ(K) be the order-maximum. If γ < βthen clearly the first γ coordinates of xβ must be xγ since we use the lexicographic

ordering. Hence we can let x ∈ K be defined by x(β) = x(β)β and by construction

we indeed have x ∈ K. Now if z ∈ K with z 6= x then we let β < α be minimalsuch that z(β) 6= x(β) and thus we must have that πβz < xβ = πβx. Hence alsoz < x. So indeed x is an order-maximum of K. So we get that for all α ≥ 1 andall compact K in

∏1≤j≤α R we must have that K has an order-maximum.

13

2. The Riesz completion of the space Lr(E,F )

Let E and F be Archimedean Riesz spaces. In Theorem 1.25 we saw thatLr(E,F ) is always a pre-Riesz space. As with all pre-Riesz spaces it is interesting towonder if we can find an explicit expression for its Riesz completion. The followingtheorem, which is part of the theorem we will discuss and prove in Chapter 3, mightgive us a tool in that.

Theorem 2.1. If F is Dedekind complete, then Lr(E,F ) is a Riesz space and itis even Dedekind complete.

Since any Riesz space F has a Dedekind completion, which we will denote byF δ, it is a natural question to wonder whether the Riesz completion of Lr(E,F )might have something to do with the Riesz space Lr(E,F δ), since there is clearly alinear order-preserving injection from Lr(E,F ) to Lr(E,F δ). It might be too muchto ask that the Riesz completion of Lr(E,F ) is isomorphic to Lr(E,F δ), but atleast it would help if Lr(E,F ) would be an order dense subset of Lr(E,F δ), sincethen the Riesz completion of Lr(E,F ) would be isomorphic to the Riesz subspaceof Lr(E,F δ) generated by Lr(E,F ).

Conjecture 2.2. The space Lr(E,F ) lies order dense in the space Lr(E,F δ),where F δ is the Dedekind completion of F .

In [3], Deckers studies this question for the simplest cases where E = Rn orE = c00, the space of sequences which are eventually constant and he showed thatin these cases, this conjecture is true. In the following part we will investigate somemore difficult examples and discuss if this result remains true.

We let 1 ≤ p ≤ ∞ and let E = `p be the space of all p-summable real sequenceswith the componentwise ordering. Let q ∈ [1,∞] satisfy the usual relation 1

p+ 1q = 1.

Let F be any Archimedean Riesz space, which has a Riesz norm, i.e. a norm ‖·‖which satisfies

‖f‖ ≤ ‖g‖ if |f | ≤ |g|.For now, we assume that this norm is complete, i.e. we assume that F is a Banachlattice. We define the following space, which will turn out to be very important inthe upcoming part,

X(F ) = {f : N→ F |∞∑n=1

αnf(n) converges for all (αn) ∈ `p

and there exist g ≥ 0, h ≥ 0 with f = g − h

such that

∞∑n=1

αng(n),

∞∑n=1

αnh(n) converge for all (αn) ∈ `p}.

Note that if f ≥ 0 satisfies that∞∑n=1

αnf(n) converges for all (αn) ∈ `p that then

f ∈ X(F ), since we can write f = f − 0 and both sequences have the desired prop-erty. At first sight it is not clear what F -valued sequences are elements of X(F ).Luckily, by Holder’s inequality we at least get the following inclusion.

14

Lemma 2.3. We have

{f : N→ F |∞∑n=1

‖f(n)‖q <∞} ⊂ X(F )

if p > 1 and{f : N→ F |f is norm-bounded} ⊂ X(F )

if p = 1.

Proof If f : N→ F satisfies∑∞n=1‖f(n)‖q <∞ for p > 1 or if f is norm-bounded

for p = 1, then that means that ‖(f(n))‖q <∞, so then by Holder’s inequality we

have for (αn) ∈ `p that∞∑n=1

‖αnf(n)‖ ≤ ‖(αn)‖p‖(f(n))‖q <∞.

So since F is Banach, we get that absolute convergence implies convergence. Since|f+(n)| ≤ |f(n)|, |f−(n)| ≤ |f(n)| for all n ∈ N we get since ‖·‖ is a Riesz normthat

∞∑n=1

‖f+(n)‖q ≤∞∑n=1

‖f(n)‖q <∞

if p > 1 andsupn∈N‖f+(n)‖ ≤ sup

n∈N‖f(n)‖ <∞

if p = 1 and a similar result holds for f−. Hence we get, using the same argumentas before, that

∑∞n=1 αnf

+(n) and∑∞n=1 αnf

−(n) converge for all (αn) ∈ `p. Sincef = f+ − f− we get that f ∈ X(F ).

Note that if p > 1 then X(F ) need not be equal to the set of F -valued sequencesof which the series of q-th power of norms converges. The simplest counterexampleis to look at F = `p, which is a Banach lattice, and look at f(n) = en for n ∈ N,where en is the sequence which has a 1 at the n-th coordinate and is 0 elsewhere.Then for all (αn) ∈ `p we see that

∞∑n=1

αnf(n) = (αn),

∞∑n=1

αnf+(n) = (αn),

∞∑n=1

αnf−(n) = (0).

Hence f ∈ X(F ), but we clearly have∞∑n=1

‖f(n)‖q =

∞∑n=1

1 =∞.

Now we get the following result, which shows the relation between X(F ) andLr(E,F ), where we still assume that E = `p and that F is a Banach lattice.

Theorem 2.4. The map

j : X(F )→ Lr(E,F ), f 7→ ((αn) 7→∞∑n=1

αnf(n))

is a bipositive isomorphism.

Proof First we show that j is well-defined. Let f ∈ X(F ) and g, h ≥ 0 withf = g− h such that for all (αn) ∈ `p we have that

∑∞n=1 αng(n) and

∑∞n=1 αnh(n)

converge. Clearly we see that (αn) 7→∑∞n=1 αng(n) is well-defined and linear.

Since g ≥ 0 we get if (αn) ≥ 0 that∑∞n=1 αng(n) ≥ 0, since in Banach lattices

the positive cone is closed. So (αn) 7→∑∞n=1 αng(n) is a positive operator from E

15

to F . Similarly (αn) 7→∑∞n=1 αnh(n) is a positive operator from E to F . Since

f = g − h we see for (αn) ∈ `p that

∞∑n=1

αnf(n) =

∞∑n=1

αng(n)−∞∑n=1

αnh(n)

So j(f) is the difference of two positive operators and is therefore a regular operatorand we see that j(f) = j(g)− j(h). So j is well-defined.

The linearity of j is clear. We have already seen that if f ≥ 0 that then j(f) ≥ 0.Now suppose that for some f ∈ X(F ) we have that j(f) ≥ 0. For m ∈ N we canlook at (αn) = em. Since em ≥ 0 and j(f) ≥ 0 we get that j(f)(em) = f(m) ≥ 0.So f ≥ 0. Hence we see that j is bipositive. As remarked below Definition 1.16, weobtain that j is injective.

Finally we let T : E → F be a positive operator. Let for i ∈ N xi = T (ei)and since T ≥ 0 we get xi ≥ 0. Since positive operators on Banach lattices arecontinuous (see Proposition 1.33), we get for (αn) ∈ `p that

T ((αn)) = T (

∞∑n=1

αnen) =

∞∑n=1

αnT (en) =

∞∑n=1

αnxn.

Since xn ≥ 0 and xn = xn − 0 for all n ∈ N, we see that (xn)n∈N ∈ X(F ). Fur-thermore we see that j((xn)n∈N) = T . Now if T = T1− T2 with T1, T2 ≥ 0, then asshown above we can find f1, f2 ∈ X(F ) with T1 = j(f1) and T2 = j(f2). So thenby linearity we must have that T = j(f1 − f2). So T is surjective.

Hence j is a bipositive isomorphism.

Clearly this representation of Lr(E,F ) as the sequence space X(F ) is much eas-ier to grasp than the operator space Lr(E,F ). One application is immediately animportant one: it turns out that for this choice of E and F , Lr(E,F ) is already aRiesz space!

Theorem 2.5. The space X(F ) as described before is a Riesz space.

Proof We use Proposition 1.6 to obtain that for X(F ) to be a Riesz space, it issufficient to check that for each f ∈ X(F ) we have that f+ := f ∨ 0 ∈ X(F ) exists.Let f ∈ X(F ). Now if we can show that the F valued sequence f+ := (f(n)+)n∈Nis again an element of X(F ) then we are done, since this element clearly is thepositive part of f since the ordering on X(F ) is pointwise.Since f ∈ X(F ) we can let g, h ∈ X(F ) with g ≥ 0, h ≥ 0 and f = g − h. We seethat 0 ≤ f+ ≤ g. Let (αn) ∈ (`p)+ then we see that for k ≤ m ∈ N we have that

0 ≤m∑n=k

αnf(n)+ ≤m∑n=k

αng(n).

Now since the norm on F is a Riesz norm, we get that for k ≤ m ∈ N we have that∣∣∣∣∣∣∣∣∣∣m∑n=k

αnf(n)+

∣∣∣∣∣∣∣∣∣∣ ≤

∣∣∣∣∣∣∣∣∣∣m∑n=k

αng(n)

∣∣∣∣∣∣∣∣∣∣ .

Since the series∞∑n=1

αng(n) converges, we get that the righthandside of the previous

equation converges to 0 as k,m→∞ and therefore the lefthandside must converge

16

to 0 as well. So (N∑n=1

αnf(n)+)N∈N is a Cauchy sequence in F . Since F is a Banach

lattice, it is complete, so we get that∞∑n=1

αnf(n)+ converges.

Now if (αn) ∈ `p, then we write (αn) = (α+n )− (α−n ). Since (α+

n ), (α−n ) ∈ (`p)+, we

get that∞∑n=1

α+n f(n)+ and

∞∑n=1

α−n f(n)+ converge. Hence we obtain that

∞∑n=1

αnf(n)+ =

∞∑n=1

α+n f(n)+ −

∞∑n=1

α−n f(n)+

converges. So since f+ ≥ 0 we see that f+ ∈ X(F ). So X(F ) is a Riesz space.

An immediate corollary is the following.

Corollary 2.6. If E = `p and F is a Banach lattice, then Lr(E,F ) is a Rieszspace.

Proof Since X(F ) is a Riesz space and j is a bipositive isomorphism from X(F )to Lr(E,F ) we get that X(F ) is Riesz if and only if Lr(E,F ) is Riesz.

Now we separate the cases where p = 1 and where p > 1. We start with p = 1and obtain the following lemma:

Lemma 2.7. Suppose that p = 1. Then we have that

{f : N→ F |f is norm-bounded} = X(F ).

Proof The inclusion {f : N → F |f is norm-bounded} ⊂ X(F ) follows fromLemma 2.3. Now let f ∈ X(F ). Let T = j(f). Since T is regular, it is continuous.For j ∈ N we see for (αn) = ej ∈ `p that

T ((αn)) =

∞∑n=1

αnf(n) = f(j).

Since T is continuous, it is bounded, so we obtain that

‖f(j)‖ = ‖T ((αn))‖ ≤ ‖T‖‖(αn)‖ = ‖T‖.So we see that f is norm-bounded. So X(F ) ⊂ {f : N→ F |f is norm-bounded}.Therefore we get {f : N→ F |f is norm-bounded} = X(F ).

Now we will return to our main question: is Lr(E,F ) order dense in Lr(E,F δ)?It is well-known that F δ is again a Banach lattice and we let ‖·‖ denote its norm,which is an extension of the norm on F . It will turn out that in the case wherep = 1, the answer to our question is indeed yes, which is the statement of thefollowing theorem.

Theorem 2.8. If p = 1, then Lr(E,F ) is order dense in Lr(E,F δ).

Proof Let f ∈ X(F δ). We define

ρ : F δ → [0,∞), y 7→ inf{‖x‖ : x ∈ F, x ≥ y}.By, for instance, [4, Theorem 3.43] we have that ρ defines a norm on F δ and thatρ(y) ≥ ‖y‖ for all y ∈ F δ. Hence the identity map from (F, ‖·‖) to (F, ρ) is boundedand therefore continuous. Since F δ is complete with respect to ‖·‖ and (F δ)+ is ‖·‖-closed, (F δ)+ is complete with respect to ‖·‖ as well. Now with [4, Theorem 3.46]

17

we get that F δ is complete with respect to ρ. So then by the Banach isomorphismtheorem the inverse of the identity is bounded as well and thus we can concludethat ‖·‖ and ρ are equivalent norms. Hence we can let C > 0 such that ρ(y) ≤ C‖y‖for all y ∈ F δ.Now for n ∈ N we get that ρ(f(n)) ≤ C‖f(n)‖ and by definition of ρ this meansthat we can find hn ∈ F with ‖hn‖ ≤ C‖f(n)‖ + 1 and hn ≥ f(n). Since f isnorm-bounded we let M ≥ 0 such that ‖f(n)‖ ≤M for all n ∈ N. Then for n ∈ Nwe have that

‖hn‖ ≤ C‖f(n)‖+ 1 ≤ CM + 1.

So (hn)n∈N is norm-bounded. So (hn)n∈N ∈ X(F ) by Lemma 2.7. Furthermore wesee that f ≤ (hn)n∈N. So {h ∈ X(F ) : h ≥ f} 6= ∅. Now by Theorem 2.1 we havethat X(F δ) ∼= Lr(E,F δ) is Dedekind complete. Since {h ∈ X(F ) : h ≥ f} is non-empty and bounded from below by f , its infimum exists. Suppose that g ∈ X(F δ)satisfies g ≤ h for all h ∈ X(F ) with h ≥ f . Suppose that g 6≤ f . Then we letn ∈ N with g(n) 6≤ f(n). Since F is order dense in F δ we can let h(n) ∈ F suchthat f(n) ≤ h(n) and g(n) 6≤ h(n), since f(n) = inf{x ∈ F : x ≥ f(n)}. Now foreach m ∈ N with m 6= n we saw that we can find h(m) ∈ F with h(m) ≥ f(m)such that h = (h(m))m∈N is norm-bounded and thus h ∈ X(F ). Then h ≥ f , buth 6≥ g, since h(n) 6≥ g(n). So that gives a contradiction. So X(F ) is order dense inX(F δ) and therefore Lr(E,F ) is order dense in Lr(E,F δ).

Note that it took quite some effort to prove that for each f ∈ X(F δ) there isan element in X(F ) majorizing f . We will see that precisely this is what might gowrong in other cases, see Counterexamples 2.9 and 2.10.

Now we look at the case where p > 1. Interestingly enough, it will turn out thatin this case Lr(E,F ) is not necessarily order dense in Lr(E,F δ) and we will showthis using a few counterexamples.

Counterexample 2.9. In this counterexample we assume that 1 < p <∞. Con-sider F = P2[0, 1], the set of polynomials of degree at most 2 on the interval [0, 1]with the pointwise ordering. For n ∈ N we let fn ∈ C[0, 1] be a piecewise linearfunction, with values between 0 and 1, which is 1 at three points, one of which is inthe interval (0, 13 ), one in ( 1

3 ,23 ) and one in ( 2

3 , 1) such that fn( 13 ) = fn( 2

3 ) = 0. Wealso need that if for n ∈ N and x ∈ [0, 1] we have fn(x) > 0 that then fm(x) = 0 forall m ∈ N with m 6= n. Then for each x ∈ [0, 1] we have that fn(x) 6= 0 for at most

one n ∈ N so we see that for each (αn) ∈ `p we have that∞∑n=1

αnfn converges to a

continuous function: the only points x where it possibly would not be continuousis where for each δ > 0 we have that there are infinitely many n such that fn isnon-zero on [x, x + δ) or (x − δ, x]. Without loss of generality we assume it is on[x, x + δ). In such a case we must clearly have that fn(x) = 0 for all n ∈ N, sinceotherwise we can find δ > 0 such that fn > 0 on [x, x+δ) and by the disjointness ofthe supports of the functions (fm)m we get that only fn 6= 0 on [x, x+δ). However,for each ε > 0 we can find N ∈ N with |αn| < ε for all n ≥ N , since lim

n→∞αn = 0

and if we chose δ > 0 small enough such that there are only fn with n ≥ N non-zero

on [x, x+ δ), then clearly |∞∑n=1

αnfn(y)| < ε for all y ∈ [x, x+ δ). So then∞∑n=1

αnfn

is continuous in x and therefore it is continuous on [0, 1]. So since (fn) ≥ 0 wesee that (fn) ∈ X(C[0, 1]). Let (pn) be a sequence in F such that pn ≥ fn for alln ∈ N. Since non-constant polynomials of degree at most two either have preciselyone local minimum on [0, 1] or they have two local minima at 0 and at 1 and since

18

for each n ∈ N we have that fn equals 1 on three different points, we get thatpn( 1

3 ) ≥ 1 or pn( 23 ) ≥ 1. Since p > 1 we have that (αn) = ( 1

n ) ∈ `p and we have

∞∑n=1

1

npn(

1

3) +

∞∑n=1

1

npn(

2

3) =

∞∑n=1

1

n(pn(

1

3) + pn(

2

3)) ≥

∞∑n=1

1

n=∞.

So we see that∞∑n=1

1npn does not converge, since it is infinite at x = 1

3 or at x = 23 .

Hence there is no sequence (pn) ∈ X(F ) with (pn) ≥ (fn). So in particular we havethat

(fn) 6= inf{(pn) ∈ X(F ) : (pn) ≥ (fn)},since the righthandside does not exist, since the infimum of the empty set does notexist. So X(F ) is not order dense in X(C[0, 1]). Since F is order dense in C[0, 1],we see that C[0, 1] ⊂ F δ and thus X(F ) is in particular not order dense in X(F δ).Therefore Lr(E,F ) is not order dense in Lr(E,F δ) by Theorem 2.4.

Now since P2[0, 1] is not a Riesz space, it is not too surprising that our conjectureis not true in this case. However, we can find another counterexample in which Fis indeed a Banach lattice:

Counterexample 2.10. Let 1 < p ≤ ∞. Consider F = C[0, 1], which is a Banachlattice. We have already seen that F is not Dedekind complete, and it is well-known that F δ contains all piecewise continuous bounded functions on [0, 1]. Foreach n ∈ N let fn ∈ F δ be continuous and strictly increasing on [0, 12 ) such that

fn(0) = 0 and limx↑ 1

2

fn(x) = 1, such that fn(x) = 0 for x ≥ 12 and such that for

each x ∈ [0, 12 ) we have that (fn(x))n∈N is q-summable with ‖fn(x)‖q ≤ 2. Then

by Holder’s inequality we have for all x ∈ [0, 12 ) and all (αn) ∈ `p that

∞∑n=1

|αn||fn(x)| ≤ ‖(αn)‖p‖(fn(x))n∈N‖q ≤ 2‖(αn)‖p.

So∞∑n=1

αnfn converges pointwise (and even uniformly) and the limit is a bounded

piecewise continuous function and therefore an element of F δ. Since f ≥ 0 we getthat f ∈ X(F δ). Now for each n ∈ N and each g ∈ F with g ≥ fn we must havethat g( 1

2 ) ≥ 1, since g is continuous and limx↑ 1

2

fn(x) = 1. Hence if g = (gn)n∈N is a

majorizing sequence in F of f , then by taking (αn) = ( 1n ) ∈ `p we get that

∞∑n=1

αngn(1

2) ≥

∞∑n=1

1

n=∞.

So there is no sequence g ∈ X(F ) with g ≥ f . So just as in the previous counterex-ample we get that X(F ) is not order-dense in X(F δ). Hence by Theorem 2.4 weobtain that Lr(E,F ) is not order-dense in Lr(E,F δ).

Now we have shown that in general Lr(E,F ) need not be an order dense subspaceof Lr(E,F δ) and thus that we cannot say anything about the Riesz completion ofLr(E,F ) in terms of Lr(E,F δ). So Conjecture 2.2 is not true in general.

19

Now one might wonder what happens if F is a normed space with Riesz norm,which is not norm-complete. It might be possible that Lr(E,F ) is related toLr(E,F ), where F is the norm completion of F , which is known to be a Banachlattice. Therefore we can formulate our second conjecture.

Conjecture 2.11. If F is a normed Riesz space and F is its norm completion,then Lr(E,F ) is order dense in Lr(E,F ).

In the remainder of the chapter we again let 1 ≤ p ≤ ∞ and let E = `p. Let Fbe a normed Riesz space, which is not necessarily norm-complete. We again wantto establish a bipositive isomorphism from X(F ), which we can also define for thisF , and Lr(E,F ). If we look at the proof of Theorem 2.4, then we see that the onlytime we used that F is a Banach lattice is to show that F+ is norm-closed. Nowlet (xn) be a sequence in F+ which converges to some x ∈ F . Then (xn) is also a

sequence in F+

which converges to x. Since F is a Banach lattice, we get x ∈ F+.

Since x ∈ F we get x ∈ F+, so also in this case we have that F+ is closed. Thuswe again have that there is a bipositive isomorphism from X(F ) to Lr(E,F ). Notethat we cannot lift Lemma 2.3 to this case, since if F is not norm-complete, thenthere are always sequences which converge absolutely, but do not converge, so wecannot conclude that the q-summable F -valued sequences are contained in X(F )any more. However, we do get the following result.

Theorem 2.12. If F is a normed Riesz space, then Lr(E,F ) is a Riesz space.

Proof Let f ∈ X(F ). Then f ∈ X(F ), since X(F ) ⊂ X(F ). Hence we get

since X(F ) is Riesz that f+ ∈ X(F ), so for all (αn) ∈ `p we get that∞∑n=1

αnf(n)+

converges in F . Since lattice operations are continuous we see that for (αn) ∈ (`p)+

we have that∞∑n=1

αnf(n)+ =

( ∞∑n=1

αnf(n)

)+

and that for (αn) ∈ −(`p)+ we have that

∞∑n=1

αnf(n)+ = −

( ∞∑n=1

|αn|f(n)

)+

.

Hence for (αn) ∈ `p we have that

∞∑n=1

αnf(n)+ =

( ∞∑n=1

α+n f(n)

)+

−

( ∞∑n=1

α−n f(n)

)+

.

Since (α+n ) ∈ `p and (α−n ) ∈ `p and since f ∈ X(F ) we get that

∞∑n=1

α+n f(n) and

∞∑n=1

α−n f(n) converge in F . Since F is Riesz, their positive parts are also in F and

thus we get that∑∞n=1 αnf(n)+ ∈ F . Hence f+ ∈ X(F ). So X(F ) is a Riesz space

and therefore Lr(E,F ) is a Riesz space as well.

Unfortunately we can still find a counterexample for our conjecture, which isgiven by the following example.

20

Counterexample 2.13. Let F = C[0, 1], but now we give it the 1-norm instead ofthe supremum norm, i.e. for f ∈ F we let ‖f‖1 =

∫|f |dλ where λ is the Lebesgue

measure on [0, 1]. Then it is well-known that F is norm dense in the Banach latticeF = L1[0, 1], which has the ordering that f ≤ g if and only if f(x) ≤ g(x) for almostevery x ∈ [0, 1]. Let f ∈ F be any unbounded function. Since [0, 1] is compact, weget that every g ∈ F is bounded, so there is no g ∈ F with g ≥ f . Now if we let(fn) be a sequence in F given by f1 = f, fn = 0 for n ≥ 2, then for each (αn) ∈ `pwe get that

∞∑n=1

αnfn = α1f,

∞∑n=1

αnf+n = α1f

+,

∞∑n=1

αnf−n = α1f

−.

So (fn) ∈ X(F ). Since there are no g ∈ F with g ≥ f there are no h ∈ X(F )with h ≥ (fn). So using the same argument as before, X(F ) is not order dense inX(F ).

So we conclude that Conjecture 2.11 is also not true in general.

21

3. The Riesz-Kantorovich formula

In this chapter we will look at the main subject of this thesis: the Riesz-Kantorovich formula. It is stated as follows (see [10, Theorem 1.3.2]).

Theorem 3.1. (The Riesz-Kantorovich formula) Let E,F be Riesz spaces.Assume that F is Dedekind complete. Then Lr(E,F ) is a Riesz space and forS, T ∈ Lr(E,F ) and x ∈ E+ we have the so-called Riesz-Kantorovich formula

(S ∨ T )(x) = sup{S(y) + T (z) : 0 ≤ y ≤ x, 0 ≤ z ≤ x, y + z = x}.

Our main object of interest is the following long-standing open problem.

If for some partially ordered vector spaces E,F and regular operators S, T : E →F the supremum S ∨ T exists in the space Lr(E,F ), does it necessarily satisfy theRiesz-Kantorovich formula?

As before, it is easier to only consider T ∈ Lr(E,F ) for which T+ exists. Forx ∈ E+ the Riesz-Kantorovich formula then states that

T+(x) = sup{T (y) : 0 ≤ y ≤ x}.Similar to the above, if we have Riesz-Kantorovich for all positive parts of operatorsfor which the positive part exists, we have it for all pairs of operators for which thesupremum exists.

First we will give a proof of Theorem 3.1, but we will formulate it slightly differ-ently, so that it might give an insight how to prove the general Riesz-Kantorovichformula for regular operators between two Riesz spaces.

Theorem 3.2. Let E,F be Riesz spaces. Let T ∈ Lr(E,F ). Then T has a positivepart in Lr(E,F ) if for every x ∈ E+ it holds that sup{T (u) : 0 ≤ u ≤ x} exists in F ,which, in particular, is always the case if F is Dedekind complete, and furthermorethe Riesz-Kantorovich formula holds.

Of course the existence of sup{T (u) : 0 ≤ u ≤ x} for x ∈ E+ is necessary for theRiesz-Kantorovich formula to hold. If one can prove the reverse statement of thistheorem, namely that if the positive part of some operator T ∈ Lr(E,F ) exists,that then sup{T (u) : 0 ≤ u ≤ x} exists for x ∈ E+, then the Riesz-Kantorovichformula always holds in Lr(E,F ). We will now prove the theorem.

Proof Let T ∈ Lr(E,F ) such that sup{T (u) : 0 ≤ u ≤ x} exists for all x ∈ E+.Define for x ∈ E+ S(x) = sup{T (u) : 0 ≤ u ≤ x}. We will first show linearity of S.Let x, y ∈ E+ and λ > 0. Then we have that

S(λx) = sup{T (u) : 0 ≤ u ≤ λx} = sup{T (u) : 0 ≤ u

λ≤ x}

= sup{T (λv) : 0 ≤ v ≤ x} = λ sup{T (v) : 0 ≤ v ≤ x} = λS(x),

where we used the linearity of T .Let 0 ≤ u ≤ x + y. Since E is a Riesz space, it has the Riesz decompositionproperty, so we can let u1 ≥ 0, u2 ≥ 0 with u1 + u2 = u and u1 ≤ x, u2 ≤ y. Thenwe get that

T (u) = T (u1) + T (u2) ≤ S(x) + S(y).

Hence we also obtain that

S(x+ y) ≤ S(x) + S(y).

We also have if 0 ≤ u1 ≤ x and 0 ≤ u2 ≤ y then 0 ≤ u1 + u2 ≤ x+ y and thus that

S(x+ y) ≥ T (u1 + u2) = T (u1) + T (u2).

22

Therefore we must also have that

S(x+ y) ≥ S(x) + T (u2).

and thus we obtain thatS(x+ y) ≥ S(x) + S(y).

Hence it holds that S(x+ y) = S(x) + S(y). Finally we also have that

S(0) = sup{T (u) : 0 ≤ u ≤ 0} = T (0) = 0.

So S is linear on E+. Furthermore we see if x ≥ 0 that then

S(x) = sup{T (u) : 0 ≤ u ≤ x} ≥ T (0) = 0,

since 0 ≤ 0 ≤ x.

Now we extend S to E by defining S(e) = S(e+) − S(e−) for e ∈ E. Then ife ≥ 0 we have e− = 0 and e+ = e so indeed this definition gives an extension of S.Now we will show linearity of S. Let e, x, y ∈ E and λ ∈ R. Then we have if λ ≥ 0that

S(λe) = S((λe)+)− S((λe)−) = S(λe+)− S(λe−)

= λS(e+)− λS(e−) = λS(e)

and if λ < 0 that

S(λe) = S((λe)+)− S((λe)−) = S(|λ|e−)− S(|λ|e+)

= |λ|S(e−)− |λ|S(e+) = |λ|(−S(e)) = λS(e).

Now if z ≥ 0 then we have that

S(e+ + z)− S(e− + z) = S(e+) + S(z)− S(e−)− S(z) = S(e).

So if we can write e = f − g with f ≥ 0 and g ≥ 0 then we can let z ≥ 0 withf = e+ + z and g = e− + z and then it holds that

S(f)− S(g) = S(e+ + z)− S(e− + z) = S(e).

Hence we obtain since x+ y = x+ + y+− (x−+ y−) and x+ + y+ ≥ 0, x−+ y− ≥ 0that

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

Hence we see that S is linear. Furthermore we saw that S(e) ≥ 0 for e ∈ E+ sowe must have that S ≥ 0. In particular S ∈ Lr(E,F ). Furthermore for e ∈ E+ wehave that

S(e) = sup{T (u) : 0 ≤ u ≤ x} ≥ T (x).

Hence we also obtain S ≥ T .

Now if R ∈ Lr(E,F ) also satisfies R ≥ 0 and R ≥ T , then for e ∈ E+ and0 ≤ u ≤ x we must have that

R(x) ≥ R(u) ≥ T (u),

where the first inequality follows since R ≥ 0 and the second since R ≥ T . Hencewe must have that R(x) ≥ S(x) and thus R ≥ S.Thus we obtain that indeed S = sup{0, T} and thus S = T+. So T+ exists inLr(E,F ) and satisfies the Riesz-Kantorovich formula.

Remark 3.3. The condition in the previous theorem that E should be a Rieszspace can be slightly relaxed to only requiring that E is directed and has the Rieszdecomposition property.

23

Our first new result is immediately linked to the contents of Chapter 2.

Theorem 3.4. Let E,F be Riesz spaces. We do not assume F to be Dedekindcomplete and let F δ be its Dedekind completion. If Lr(E,F ) is order-dense inLr(E,F δ) and for S, T ∈ Lr(E,F ) the supremum S ∨ T exists in Lr(E,F ), then itsatisfies the Riesz-Kantorovich formula.

Proof We only consider T ∈ Lr(E,F ) for which T+ exists in Lr(E,F ). First wewill show that T+ is also the positive part of T in the Riesz space Lr(E,F δ). Let Sbe the positive part of T in Lr(E,F δ). Then S ≤ T+ since T+ still satisfies T+ ≥ Tand T+ ≥ 0. Let R be the negative part of T in Lr(E,F δ) and T− the negativepart of T in Lr(E,F ) (which exists since it equals T+ − T ). We let U = T+ − S.Then U ≥ 0. Then we have that

T = T+ − T− = S −R = T+ − U −R.Hence we obtain that

T− = U +R.

Let V ∈ Lr(E,F ) such that V ≥ T+ + T−, V ≥ −T+ − T−. Then it also holdsthat

V ≥ T+ + T− ≥ T+ ≥ T+ − T−

andV ≥ T+ + T− ≥ −T+ + T−.

So {T+ + T−,−T+ − T−}u ⊂ {T+ − T−,−T+ − T−}u.Now let W ∈ Lr(E,F ) such that W ≥ T+−T− = T,W ≥ −T+ +T− = −T . Thenit also holds that

W ≥ T ∨ (−T ) = |T | = T+ + T− ≥ −T+ − T−.So we obtain {T+ + T−,−T+ − T−}u ⊃ {T+ − T−,−T+ − T−}u and thus that

{T+ + T−,−T+ − T−}u = {T+ − T−,−T+ − T−}u.So T+ ⊥ T− in Lr(E,F ) by definition. By Proposition 1.27 we get that T+ ⊥ T−in Lr(E,F δ) since Lr(E,F ) is order-dense in Lr(E,F δ) by assumption. Now inLr(E,F δ) we have that

|T+ − T−| = |S + U − (U +R)| = |S −R| = |T |and

|T+ + T−| = |S + U +R+ U | = ||T |+ 2U | = |T |+ 2U,

since |T | + 2U ≥ 0. However, since T+ and T− are disjoint in the Riesz spaceLr(E,F δ), we saw that this means that |T+| ∧ |T−| = 0 and thus |T+ − T−| =|T+ + T−|. Hence we must have that

|T | = |T |+ 2U.

which implies U = 0. Hence S = T+ and thus T+ is the positive part of T inLr(E,F δ).

Now let x ∈ E+. Since the Riesz-Kantorovich formula holds in Lr(E,F δ) we seethat

T+(x) = sup{T (u) : 0 ≤ u ≤ x},where the supremum is taken in F δ. Hence T+(x) ≥ T (u) for all u ∈ E with0 ≤ u ≤ x and if y ∈ F satisfies y ≥ T (u) for all 0 ≤ u ≤ x then in F δ we shouldhave that y ≥ T+(x), but since y, T+(x) ∈ F it follows that y ≥ T+(x) in F . HenceT+(x) = sup{T (u) : 0 ≤ u ≤ x} in F .

24

Corollary 3.5. If E = `1 and F any Banach lattice, then for all S, T ∈ Lr(E,F )for which S ∨ T exists, the Riesz-Kantorovich formula holds.

Proof This follows immediately from Theorem 3.4 and Theorem 2.8.

To show the non-triviality of taking suprema in the space of regular operators,we consider the following example.

Example 3.6. Recall that a function f : R → R is called Lipschitz con-tinuous if there exists a constant L ≥ 0 such that for all x, y ∈ R we have|f(x) − f(y)| ≤ L|x − y|. The smallest such constant is called the Lipschitzconstant of f and is denoted by Lip(f). Then one can show that the spaceof all Lipschitz continuous functions is a Riesz space under the order it inher-its from C(R). However, for this example we consider a new ordering on thespace E = {f ∈ C([0, 4]) : f is Lipschitz continuous}, namely we say that g � 0if and only if g ≥ 0 and Lip(g) ≤ g( 3

2 ). Then E becomes a partially orderedvector space and we will show it is directed. Let g ∈ E. Let c ∈ R suchthat c − g( 3

2 ) ≥ Lip(g) and c ≥ 0. Let f = c. Then f is Lipschitz continu-ous and Lip(f) = 0 ≤ c. So f � 0. Furthermore since f is constant we getLip(f − g) = Lip(−g) = Lip(g) ≤ c − g( 3

2 ) = (f − g)( 32 ). So f − g � 0 and thus

f � g. Hence E is directed.We let F = C([0, 4]). For h ∈ C([0, 4]) we consider the following map

Th : E → F, g 7→ (x 7→∫ x

0

h(t)g(t)dt).

It is clear that Th = Th+ − Th− and that Th+ ≥ 0, Th− ≥ 0. In particular Th ∈Lr(E,F ). We define the following function

f : x 7→

1 if x ∈ [0, 1]

−2x+ 3 if 1 ≤ x ≤ 2

−1 if 2 ≤ x ≤ 4.

Then indeed f ∈ C[0, 4]. It would make sense that Tf+ = (Tf )+ but we will showthat that cannot be the case assuming that the Riesz-Kantorovich formula holds.We let g = 1. Then g is constant so Lip(g) = 0 ≤ 1 = g( 3

2 ), so g � 0. Let 0 � u � g.Since g is constant we have Lip(g − u) = Lip(−u) = Lip(u). Since also 0 ≤ u ≤ gwe get 0 ≤ u( 3

2 ) ≤ 1. Suppose 1 ≥ u( 32 ) ≥ 1

2 . Then since u � g we get that

Lip(u) = Lip(g − u) ≤ (g − u)(3

2) = 1− u(

3

2) ≤ 1

2.

Suppose that u( 32 ) ≤ 1

2 . Then since 0 � u we get Lip(u) ≤ 12 . So in each case we

have Lip(u) ≤ 12 . Now some estimations give that we must have that

Tfu(2) ≤ 3

2

for every u with 0 � u � g and that the maximum of Tfu(2) over such functions uis attained for a piecewise linear function u satisfying u( 3

2 ) = 34 , which is as large

as allowed on [0, 32 ] and as small as allowed on [ 32 , 4]. Since Tfu is decreasing on

[ 32 , 4] since u ≥ 0 we get that Tfu(2 + ε) ≤ Tfu(2) ≤ 1312 for all u with 0 � u � g

and all 0 ≤ ε ≤ 2. So if (Tf )+ would exist and would satisfy the Riesz-Kantorovichformula then we should have, since (Tf )+g is continuous, that

(Tf )+g(2 + ε) ≤ 13

12

25

for all 0 ≤ ε ≤ 2. However, it holds that

(Tf+)g(2 + ε) =

∫ 1

0

1dx+

∫ 32

1

(−2x+ 3)dx =5

4>

13

12

for all 0 ≤ ε ≤ 2 and that we have equality for ε = 0. So Tf+ cannot be equal to(Tf )+ if the latter exists and satisfies the Riesz-Kantorovich formula. However, itis unknown if this is a counter-example for the Riesz-Kantorovich formula or that(Tf )+ does not exist or that it exists, but is simply not equal to Tf+ . In any case,it may be interesting to investigate this example in any further research.

26

4. Main theorem

To start we will formulate the Riesz-Kantorovich formula for the supremum offinitely many regular operators:If S1, ..., Sn ∈ Lr(E,F ) and

∨ni=1 Si exists in Lr(E,F ) and x ∈ E+ then the Riesz-

Kantorovich formula is given byn∨i=1

Si(x) = sup{n∑i=1

S(xi) : 0 ≤ xi for all i ∈ {1, ..., n} and

n∑i=1

xi = x}

In this chapter we will discuss our main theorem. In their article [1], Aliprantiset al. show the following theorem.

Theorem 4.1. Let L be an ordered vector space with an order unit and a Haus-dorff linear topology such that all order intervals are compact. If for some linearcontinuous functionals f1, ..., fm ∈ L∼(L,R) the supremum g = f1 ∨ ... ∨ fm existsin L∼(L,R) then g satisfies the Riesz-Kantorovich formula.

In other words, they show the Riesz-Kantorovich formula for linear functionals,where the domain space has a strong order unit and where we assume some conti-nuity properties. Of course R is a Dedekind complete Riesz space, so in that senseit is a ’good space’. However, the domain space need not be a Riesz space; it doesnot even need to have the Riesz decomposition property as can be seen from thefollowing example, which is based on [1, Example 2.3].

Example 4.2. Consider the vector space R3 with the following ’ice cream’ cone:

C = {(x, y, z) ∈ R3 : z ≥ 0, z2 ≥ 4(x2 + y2)}= {λ(x, y, 2) : λ ≥ 0, x2 + y2 ≤ 1}.

Since C is a convex cone, it induces a partial order on R3, which we will denote by≤C , by x ≤C y if and only if y − x ∈ C. We equip the space with the Euclideantopology. Then we obtain the following properties:

(1) C is a closed cone.(2) The space (R3,≤C) has a strong order unit; for example e = (0, 0, 2) is

a strong order unit. This can be seen as follows: let u = (x, y, z) ∈ R3.

Choose α > 0 large enough such that λ = −z+2α2 satisfies λ >

√x2 + y2.

Then we see that

−u+ αe = (−x,−y,−z) + α(0, 0, 2) = λ

(−xλ,−yλ, 2

)∈ C

and thus u ≤C αe.(3) The order intervals of (R3,≤C) are closed bounded subsets of R3 and thus

by Heine-Borel theorem they are compact.(4) (R3,≤C) does not have the Riesz decomposition property and is thus in

particular is not a Riesz space. To show this we first show the following: ifu = λ(x, y, 2) ∈ C and x2 +y2 = 1 then v ∈ R3 satisfies 0 ≤C u ≤C v if andonly if there exists 0 ≤ µ ≤ 1 such that v = µu. The ’if’ part is clear, so weonly show the ’only if’ part. So assume we have 0 ≤C λ(a, b, 2) ≤C (x, y, 2)with x2 + y2 = 1. Then 0 ≤ λ ≤ 1 and if either λ = 0 or λ = 1 thenthe result follows immediately. Hence we assume 0 < λ < 1. Since(x, y, 2) − λ(a, b, 2) ∈ C we get that there exists (α, β) with α2 + β2 ≤ 1and some µ ≥ 0 such that (x, y, 2) − λ(a, b, 2) = µ(α, β, 2). Hence we ob-tain that (x, y) = λ(a, b) + µ(α, β). Since x2 + y2 = 1, the point (x, y)is an extreme point of the unit disk in R2 and thus we must have that(x, y) = (a, b)(= (α, β)). Hence λ(x, y, 2) = λ(a, b, 2).

27

Now we show that (R3,≤C) does not have the Riesz decomposition prop-erty by considering the vectors (1, 0, 2), (−1, 0, 2), (0, 1, 2) ∈ C. We beginby observing that

0 ≤C (0, 1, 2) ≤C (0, 0, 4) = (1, 0, 2) + (−1, 0, 2).

If (R3,≤C) would have the Riesz decomposition property then we wouldbe able to find u, v ∈ R3 with 0 ≤C u ≤C (1, 0, 2), 0 ≤C v ≤C (−1, 0, 2)and u + v = (0, 1, 2). Now by what we showed we have 0 ≤ λ, µ ≤ 1 withu = λ(1, 0, 2) and v = µ(−1, 0, 2). However, then we would get that

(0, 1, 2) = u+ v = λ(1, 0, 2) + µ(−1, 0, 2) = (λ+ µ, 0, 2(λ+ µ)),

which is impossible. Hence (R3, C) indeed does not have the Riesz decom-position property and is thus also not a Riesz space.

In particular this means that (R3, C) is a space we can apply Theorem 4.1 and ourupcoming main theorem to, but not the standard results on the Riesz-Kantorovichformula.

Before we go on we introduce some notation.

Notation 4.3. Let L be an ordered vector space. We use the short-hand notationL∼ := L∼(L,R). For any integer m > 0 and x ∈ L+ we define the followingnon-empty convex sets

Amx = {(y1, ..., ym) ∈ Lm+ :

m∑i=1

yi ≤ x}

and

Fmx = {(y1, ..., ym) ∈ Lm+ :

m∑i=1

yi = x}.

If x ∈∏j∈J R for some index set J then we write x(j) for the j-th component of

x and similarly if f : L →∏j∈J R is any map then we write f (j) for the map

f (j) : L → R, x 7→ (f(x))(j). Similarly, if α is an ordinal, then for x ∈∏

1≤β≤α Rand γ < α we mean by x(1),...,(γ) the vector in

∏1≤β≤γ R which corresponds to the

first γ coordinates of x and if f : L →∏

1≤β≤α R is any map and γ < α then we

mean by f (1),...,(γ) the map f (1),...,(γ) : L→∏

1≤β≤γ R, x 7→ (f(x))(1),...,(γ).

We have to prove one more lemma before we can state our main theorem.

Lemma 4.4. Let L be a directed partially ordered vector space. Let α ≥ 1 bean ordinal number and let M =

∏1≤ j≤α R with the lexicographic ordering. Let

f1, ..., fm ∈ L∼(L,M) such that f(j)i is regular for all i ∈ {1, ...,m} and all 1 ≤

j ≤ α and such that g = f1 ∨ ... ∨ fm exists in L∼(L,M). Let β < α. Then

f(1),...,(β)1 ∨ ... ∨ f (1),...,(β)m exists and equals g(1),...,(β).

Proof Suppose this is not the case. Then we can let h : L →∏

1≤j≤β R be

order-bounded and x ∈ L+ such that h ≥ f(1),...,(β)i for all 1 ≤ i ≤ m and h(x) <

g(1),...,(β)(x). By regularity we can write

f(j)i = g

(j)i − h

(j)i

where g(j)i ≥ 0, h

(j)i ≥ 0 for 1 ≤ i ≤ m and 1 ≤ j ≤ α. We define

h′ : L→M,y 7→ (h(1)(y), ..., h(β)(y),

m∑i=1

g(β+1)i (y), ...,

m∑i=1

g(α)i (y)).

28

Then clearly h′ ∈ L∼(L,M) and we see that h′ ≥ fi for all i ∈ {1, ...,m} since

we use the lexicographic ordering on M and h ≥ f(1),...,(β)i for all 1 ≤ i ≤ m.

Furthermore we see that

h′(x) = (h(1)(x), ..., h(β)(x),

m∑i=1

g(β+1)i (x), ...,

m∑i=1

g(α)i (x)) < g(x)

since h(x) < g(1),...,(β)(x) and we use the lexicographic ordering on M . Henceh′ 6≥ g and that gives a contradiction with the definition of g.

Now we are ready to state our main theorem, which is an improved versionof Theorem 4.1. Since we can imagine that not every reader appreciates ordinalnumbers, we present this theorem in two versions: the first version is for the spacesRn and uses only ’ordinary’ induction, while the second one is for arbitrary productsof copies of R and uses transfinite induction. Other than that, the differences inthe proofs are very minor.

Theorem 4.5. (Main theorem) Let L be an ordered vector space with an orderunit e and a Hausdorff linear topology such that all order intervals are compact. LetM = Rn with the lexicographic ordering. Assume we have f1, f2, ..., fm continuous,

linear maps from L to M such that f(j)i is regular for all i ∈ {1, ...,m} and all

j ∈ {1, ..., n} and such that the supremum g = ∨mi=1fi exists in L∼(L,M). Then gsatisfies the Riesz-Kantorovich formula.

Proof The proof will be done by induction on n. The case n = 1 follows fromTheorem 4.1. So we let n ≥ 1 and assume the theorem holds for image spacesR1, ...,Rn−1. The induction step will consist of proving the following three state-ments:

(i) Assume we have f1, f2, ..., fm continuous, linear maps from L to M such

that f(j)i is regular for all i ∈ {1, ...,m} and all j ∈ {1, ..., n} and such

that the supremum (∨mi=1fi)+ exists in L∼(L,M). Then there exists some

(x∗1, x∗2, ..., x

∗m) ∈ Ame satisfying

m∑i=1

fi(x∗i ) ≥

m∑i=1

fi(xi)

for each (x1, x2, ..., xm) ∈ Ame , and then

(∨mi=1fi)+

(e) =

m∑i=1

fi(x∗i ).

(ii) Assume we have f1, f2, ..., fm continuous, linear maps from L to M such that

f(j)i is regular for all i ∈ {1, ...,m} and all j ∈ {1, ..., n} and such that the

supremum ∨mi=1fi exists in L∼(L,M). Then there exists some (x∗1, x∗2, ..., x

∗m) ∈

Fme satisfyingm∑i=1

fi(x∗i ) ≥

m∑i=1

fi(xi)

for each (x1, x2, ..., xm) ∈ Fme , and then

(∨mi=1fi) (e) =

m∑i=1

fi(x∗i ).

(iii) Assume we have f1, f2, ..., fm continuous, linear maps from L to M such that

f(j)i is regular for all i ∈ {1, ...,m} and all j ∈ {1, ..., n} and such that the

supremum ∨mi=1fi exists in L∼(L,M). Then the Riesz-Kantorovich formulaholds.

29

Proof of (i) The first statement in (i) follows from the fact that if K ⊂ Rn is

compact (and since all fi are continuous we get that K = {m∑i=1

fi(xi) : (x1, ..., xm) ∈

Ame } is compact) then K contains a maximum with respect to the lexicographicordering on Rn.

Write g = (f1 ∨ ... ∨ fm)+. We let (x∗1, ..., x∗m) ∈ Ame such that

m∑i=1

fi(x∗i ) ≥

m∑i=1

fi(xi)

for all (x1, ..., xm) ∈ Ame .Now we have

g(e) = g(

m∑i=1

x∗i ) =

m∑i=1

g(x∗i ) ≥m∑i=1

fi(x∗i ).

So we need to show that the other inequality holds.

First we assume that f(j)i ≥ 0, f

(n)i ≥ 0 and f

(n)i ≥ f (j)i for all i ∈ {1, ...,m} and

all j ∈ {1, ..., n − 1}. Then without loss of generality we assume that∑mi=1 x

∗i =

e, since all maps are positive. Suppose that∑mi=1 fi(x

∗i )

(n) = 0. Then also∑mi=1 fi(x

∗i )

(j) = 0 for all j ∈ {1, ..., n − 1}, since f(n)i ≥ f

(j)i for all i ∈ {1, ...,m}.

So since f(j)i ≥ 0, f

(n)i ≥ 0 this is only possible if fi = 0 for all i ∈ {1, ...,m}

by definition of (x∗1, ..., x∗m). In that case the result holds. So we assume that∑m

i=1 fi(x∗i )

(n) > 0.

We know from Lemma 4.4 that f(1),...,(n−1)1 ∨...∨f (1),...,(n−1)m = (f1∨...∨fm)(1),...,(n−1),

since we use the lexicographic ordering on Rn. Hence by the induction hypothesis

we get if we let h = f(1),...,(n−1)1 ∨ .... ∨ f (1),...,(n−1)m then h is positive, linear and

satisfies for x ∈ L+

h(x) = max{m∑i=1

fi(xi)(1),...,(n−1) : (x1, ..., xm) ∈ Fmx },

where the usual supremum is now a maximum by the continuity of the functionals

f(1),...,(n−1)i and the compactness of order intervals. Then we have

h(e) = h(

m∑i=1

x∗i ) =

m∑i=1

fi(x∗i )

(1),...,(n−1),

by our choice of x∗1, ..., x∗m. We define

Y =

{(y1, ..., ym) ∈ Lm+ : h(

m∑i=1

yi) =

m∑i=1

fi(yi)(1),...,(n−1),

m∑i=1

fi(yi)(n) >

m∑i=1

fi(x∗i )

(n)

}.

This set is clearly convex and non-empty (it contains t(x∗1, ..., x∗m) for all t > 1).

We also let

Z = {(z1, ..., zm) ∈ Lm :

m∑i=1

zi ≤ e}.

Also this set Z is non-empty and convex. We will show that Y and Z are disjoint.Suppose that Y ∩Z 6= ∅ and let (y1, ..., ym) ∈ Y ∩Z. Since in the definition of h weonly have maxima, we can let z1, ..., zm ∈ L+ such that

∑mi=1 zi = e−

∑mi=1 yi ≥ 0

and h(∑mi=1 zi) =

∑mi=1 fi(zi)

(1),...,(n−1). Then∑mi=1(yi + zi) = e, yi + zi ≥ 0 for

30

all i ∈ {1, ...,m} and we havem∑i=1

fi(yi + zi)(1),...,(n−1) =

m∑i=1

fi(yi)(1),...,(n−1) +

m∑i=1

fi(zi)(1),...,(n−1)

= h(

m∑i=1

yi) + h(e−m∑i=1

yi) = h(e) =

m∑i=1

fi(x∗i )

(1),...,(n−1)

and by positivity of f(n)i for i ∈ {1, ...,m} we also have

m∑i=1

fi(yi + zi)(n) ≥

m∑i=1

fi(yi)(n) >

m∑i=1

fi(x∗i )

(n).

So∑mi=1 fi(yi + zi) >

∑mi=1 fi(x

∗i ) which gives a contradiction with the definition

of (x∗1, ..., x∗m). So we must have that Y ∩ Z = ∅.

This set Z has an internal point in Lm by [1, Lemma 2.1]. By the Separation

Theorem [2, Theorem 5.61] there exists a non-zero linear functional (h(n)1 , ..., h

(n)m ) ∈

(L∗)m that separates Z and Y . That is

(1)

m∑i=1

h(n)i (yi) ≥

m∑i=1

h(n)i (zi) for all (y1, ..., ym) ∈ Y and (z1, ..., zm) ∈ Z.

Since (x∗1, ..., x∗m) ∈ Z it follows from (1) that

(2)

m∑i=1

h(n)i (yi) ≥

m∑i=1

h(n)i (x∗i ) for all (y1, ..., ym) ∈ Y.

Furthermore, it follows from (1) that

(3)

m∑i=1

h(n)i (x∗i ) = lim

α↓1

m∑i=1

h(n)i (αx∗i ) ≥

m∑i=1

h(n)i (zi) for all (z1, ..., zm) ∈ Z.

Next, we show that h(n)1 = ... = h

(n)m := h(n). Suppose, by way of contradiction,

that there exists some z ∈ L such that h(n)1 (z) > h

(n)2 (z). Then for some α > 1 we

have

h(n)1 (x∗1 + z) + h

(n)2 (x∗2 − z) +

m∑i=3

h(n)i (x∗i ) >

m∑i=1

h(n)i (αx∗i ).

However, (αx∗1, ..., αx∗m) ∈ Y since α > 1 and (x∗1 + z, x∗2− z, x∗3, ..., x∗m) ∈ Z, which

contradicts (1). By the symmetry of the situation we see that h(n)1 = h

(n)2 = · · · =

h(n)m := h(n).

Now we show that h(n) ≥ 0. Let x ∈ L+. Note that (e − αx, 0, 0, ....) ∈ Z for allα > 0. Therefore, from (3) we get

∑mi=1 h

(n)(x∗i ) ≥ h(n)(e)− αh(n)(x) or

h(n)(x) ≥h(n)(e)−

∑mi=1 h

(n)(x∗i )

α

for each α > 0. Letting α → ∞ yields h(n)(x) ≥ 0. Hence h(n) ≥ 0 and thush(n) ∈ L∼.Furthermore, since h(n) 6= 0 it must be the case that h(n)(e) > 0 since e is an orderunit. Since (e, 0, ..., 0) ∈ Z it follows from (3) that

∑mi=1 h

(n)(x∗i ) > 0. So we candefine

δ(n) =

∑mi=1 fi(x

∗i )

(n)∑mi=1 h

(n)(x∗i )∈ R>0.

We claim that δ(n)h(n)(x) ≥ f(n)i (x) for i ∈ {1, ...,m} and x ∈ L+ such that

h(x) = fi(x)(1),...,(n−1). To see this, fix i ∈ {1, ...,m} and x ∈ L+ such that

31

h(x) = fi(x)(1),...,(n−1). If fi(x)(n) ≤ 0 then δ(n)h(n)(x) ≥ 0 ≥ fi(x)(n) is triviallytrue. Assume, therefore, that fi(x)(n) > 0 and let

γ(n) =

∑mi=1 fi(x

∗i )

(n)

fi(x)(n)∈ R>0.

Now we have for α > 1

fi(αγ(n)x)(n) = α

∑mi=1 fi(x

∗i )

(n)

fi(x)(n)fi(x)(n) = α

m∑i=1

fi(x∗i )

(n) >

m∑i=1

fi(x∗i )

(n),

since we assumed that∑mi=1 fi(x

∗)(n) > 0. Hence by linearity of h and f(1),...,(n−1)i

we get that (0, ..., 0, αγ(n)x, 0, ..., 0) ∈ Y , where αγ(n)x takes the i-th position ofthis vector. Hence with equation (2) we get

δ(n)γ(n)h(n)(x) = δ(n) limα↓1

h(n)(αγ(n)x)

≥ δ(n)m∑i=1

h(n)(x∗i ) =

m∑i=1

fi(x∗i )

(n) = γ(n)fi(x)(n).

Hence δ(n)h(n)(x) ≥ fi(x)(n) for all x ∈ L+ with h(x) = fi(x)(1),...,(n−1). Thus, dueto the lexicographic ordering, for all x ≥ 0 we have

fi(x) ≤ (h(x), δ(n)h(n)(x)).

Since this holds for all i ∈ {1, ...,m}, since h ≥ 0 and δ(n)h(n) ≥ 0, we get that

g(x) ≤ (h(x), δ(n)h(n)(x))

for all x ≥ 0. In particular we get

g(e) ≤ (h(e), δ(n)h(n)(e)) =

m∑i=1

fi(x∗i )

since e =∑mi=1 x

∗i . Hence we obtain that

g(e) =

m∑i=1

fi(x∗i ).

Now we do not assume that f(j)i ≥ 0, f

(n)i ≥ 0 or f

(n)i ≥ f (j)i for i ∈ {1, ...,m} and

j ∈ {1, ..., n− 1}. Since the f(j)i are all regular for i ∈ {1, ...,m} and j ∈ {1, ..., n}

we can let h1, ..., hn, g(1)1 , ..., g

(1)m , ..., g

(n)1 , ..., g

(n)m be positive linear functionals such

that for i ∈ {1, ...m} we have:

fi = (g(1)i − h1, ..., g

(n)i − hn)

Due to compactness of Am+1e we can let (y∗1 , ..., y

∗m+1) ∈ Ame such that

m∑i=1

(g(1)i (y∗i ), ..., g

(n−1)i (y∗i ), g

(n)i (y∗i ) +

n−1∑j=1

m∑k=1

g(j)k (y∗i ) +

n−1∑j=1

hj(y∗i ))

+(h1, ..., hn−1,

n−1∑j=1

m∑k=1

g(j)k +

n∑j=1

hj)(ym+1)

≥m∑i=1

(g(1)i (xi), ..., g

(n−1)i (xi), g

(n)i (xi) +

n−1∑j=1

m∑k=1

g(j)k (xi) +

n−1∑j=1

hj(xi))

+(h1, ..., hn−1,

n−1∑j=1

m∑k=1

g(j)k +

n∑j=1

hj)(xm+1)

32

for each (x1, x2, ..., xm+1) ∈ Am+1e . By positivity we can assume without loss of

generality thatm+1∑j=1

y∗j = e so that y∗m+1 = e −m∑j=1

y∗j . Then we obtain that the

set of functions {(g(1)i , ..., g(n−1)i , g

(n)i +

n−1∑j=1

m∑k=1

g(j)k +

n−1∑j=1

hj) : i ∈ {1, ...,m}} ∪

{(h1, ..., hn−1,n−1∑j=1

m∑k=1

g(j)k +

n∑j=1

hj)} satisfies the assumptions we made in the first

half of this proof. Hence we get by our previous results

g(e) =

(m∨i=1

(g(1)i − h1, ..., g

(n)i − hn)

)+

=

((m∨i=1

((g

(1)i , ..., g

(n−1)i , g

(n)i +

n−1∑j=1

m∑k=1

g(j)k +

n−1∑j=1

hj)

− (h1, ..., hn−1,

n−1∑j=1

m∑k=1

g(j)k +

n∑j=1

hj)

))

∨

(h1, ..., hn−1,

n−1∑j=1

m∑k=1

g(j)k +

n∑j=1

hj)− (h1, ..., hn−1,

n−1∑j=1

m∑k=1

g(j)k +

n∑j=1

hj)

)(e)

=

m∨i=1

(g(1)i , ..., g

(n−1)i , g

(n)i +

n−1∑j=1

m∑k=1

g(j)k +

n−1∑j=1

hj)

(e)

−

h1, ..., hn−1, n−1∑j=1

m∑k=1

g(j)k +

n∑j=1

hj

(e)

=

m∑i=1

(g(1)i (y∗i ), ..., g

(n−1)i (y∗i ), g

(n)i (y∗i ) +

n−1∑j=1

m∑k=1

g(j)k (y∗i ) +

n−1∑j=1

hj(y∗i ))

+

h1(e−m∑i=1

y∗i ), ..., hn−1(e−m∑i=1

y∗i ),

n−1∑j=1

m∑k=1

g(j)k (e−

m∑i=1

y∗i ) +

n∑j=1

hj(e−m∑i=1

y∗i )

− (h1(e), ..., hn−1(e),

n−1∑j=1

m∑k=1

g(j)k (e) +

n∑j=1

hj(e))

=

m∑i=1

(g(1)i (y∗i ), ..., g

(n−1)i (y∗i ), g

(n)i (y∗i ) +

n−1∑j=1

m∑k=1

g(j)k (y∗i ) +

n−1∑j=1

hj(y∗i ))

−m∑i=1

(h1(y∗i ), ..., hn−1(y∗i ),

n−1∑j=1

m∑k=1

g(j)k (y∗i ) +

n∑j=1

hj(y∗i ))

=

m∑i=1

(g(1)i (y∗i )− h1(y∗i ), ..., g

(n)i (y∗i )− h2(y∗i ))

=

m∑i=1

(f(1)i (y∗i ), ..., f

(n)i (y∗i )) ≤

m∑i=1

fi(x∗i )

by our choice of x∗1, ..., x∗m. Since in the very first part of the proof we saw that

g(e) ≥m∑i=1

fi(x∗i ) we have g(e) =

m∑i=1

fi(x∗i ). That finishes the proof of part (i).

33

Proof of (ii) The first statement follows for the same reason as in part (i). So let

(x∗1, ..., x∗m) ∈ Fme such that

m∑i=1

fi(x∗i ) ≥

m∑i=1

fi(xi) for each (x1, x2, ..., xm) ∈ Fme .

First note since (x∗1, ..., x∗m) ∈ Fme that x∗1 = e −

m∑i=2

x∗i . Let g = ∨mi=1fi and start

by observing that g − f1 = [∨mi=2(fi − f1)]+ andm∑i=2

(fi − f1)(x∗i ) =

m∑i=1

fi(x∗i )− f1(e)

≥m∑i=2

fi(xi) + f1

(e−

m∑i=2

xi

)− f1(e) =

m∑i=2

(fi − f1)(xi)

for any (x2, ..., xm) ∈ Am−1e since if (x2, ..., xm) ∈ Am−1e then (e−m∑i=2

xi, x2, ..., xm) ∈

Fme . Therefore by part (i) we have

(g − f1)(e) =

m∑i=2

(fi − f1)(x∗i ).

In particular it holds that

g(e) = f1

(e−

m∑i=2

x∗i

)+

m∑i=2

fi(x∗i ) =

m∑i=1

fi(x∗i ),

which is the desired formula.

Proof of (iii) Now fix an arbitrary x ∈ L+ and then select an order unit e ∈ Lsuch that x ≤ e.For each 0 < α < 1 let xα = αx + (1 − α)e. Clearly, each xα is an order unit and0 ≤ xα ≤ e holds for each 0 < α < 1. We consider the index set (0, 1) directed by theincreasing order relation �, i.e. α � β in (0, 1) if and only if α ≥ β. Clearly xα → x.Since Fme is compact, the continuous function Fme →M, (x1, ..., xm) 7→

∑mi=1 fi(xi)

attains its maximum. By part (ii), we know that g satisfies the Riesz-Kantorovichformula for each order unit. Therefore, by our assumption on the space M and thecompactness of Fmeα , for each 0 < α < 1 there exists some (zα1 , ..., z

αm) ∈ Lm+ such

that∑mi=1 z

αi = xα and

(4) αg(x) + (1− α)g(e) = g(xα) =

m∑i=1

fi(zαi ).

Since zαi ∈ [0, e] for each α ∈ (0, 1) and each i ∈ {1, ...,m} and the order interval[0, e] is compact, there exists a subnet of {(zα1 , ..., zαm)} (which without loss of gener-ality we also denote by {(zα1 , ..., zαm)}) such that zαi → zi for some zi ∈ [0, e]. Fromxα − zαi ∈ [0, e] for each α and the closedness of [0, e], we see that 0 ≤ x− zi ≤ e.So 0 ≤ zi ≤ x and

∑mi=1 zi = x. Finally, letting α → 1 in equation (4) it follows

from the continuity of each fi that

g(x) =

m∑i=1

fi(zi).

Since we already know that g(x) ≥∑mi=1 fi(yi) for all (y1, ..., ym) ∈ Fmx we get that

sup{∑mi=1 fi(yi) : (y1, ..., ym) ∈ Fmx } = g(x). So the proof is finished.

34

Theorem 4.6. (Main theorem, ordinal version) Let L be an ordered vectorspace with an order unit e and a Hausdorff linear topology such that all orderintervals are compact. Let α be an ordinal number with α ≥ 1 and let M =∏

1≤j≤α R with the lexicographic ordering and the product topology. Assume we

have f1, f2, ..., fm continuous, linear maps from L to M such that f(j)i is regular

for all i ∈ {1, ...,m} and all ordinals j with 1 ≤ j ≤ α and such that the supremumg = ∨mi=1fi exists in L∼(L,M). Then g satisfies the Riesz-Kantorovich formula.

Proof The proof will be done by transfinite induction on α. The case α = 1follows from Theorem 4.1. So we let α ≥ 1 and assume the theorem holds for imagespaces R1, ...,

∏1≤j≤α R. Let M =

∏1≤j≤α R. The induction step for successor

ordinals will consist of proving the following three statements:

(i) Assume we have f1, f2, ..., fm continuous, linear maps from L to M such that

f(j)i is regular for all i ∈ {1, ...,m} and all ordinals j with 1 ≤ j ≤ α and such

that the supremum (∨mi=1fi)+ exists in L∼(L,M). Then there exists some

(x∗1, x∗2, ..., x

∗m) ∈ Ame satisfying

m∑i=1

fi(x∗i ) ≥

m∑i=1

fi(xi)

for each (x1, x2, ..., xm) ∈ Ame , and then

(∨mi=1fi)+

(e) =

m∑i=1

fi(x∗i ).

(ii) Assume we have f1, f2, ..., fm continuous, linear maps from L to M such that

f(j)i is regular for all i ∈ {1, ...,m} and all ordinals j with 1 ≤ j ≤ α and

such that the supremum ∨mi=1fi exists in L∼(L,M). Then there exists some(x∗1, x

∗2, ..., x

∗m) ∈ Fme satisfying

m∑i=1

fi(x∗i ) ≥

m∑i=1

fi(xi)

for each (x1, x2, ..., xm) ∈ Fme , and then

(∨mi=1fi) (e) =

m∑i=1

fi(x∗i ).

(iii) Assume we have f1, f2, ..., fm continuous, linear maps from L to M such that

f(j)i is regular for all i ∈ {1, ...,m} and all ordinals j with 1 ≤ j ≤ α and such

that the supremum ∨mi=1fi exists in L∼(L,M). Then the Riesz-Kantorovichformula holds.

Proof of (i) The first statement in (i) follows from the fact that if K ⊂∏

1≤j≤α R

is compact (and since all fi are continuous we get thatK = {m∑i=1

fi(xi) : (x1, ..., xm) ∈

Ame } is compact) then K contains a maximum with respect to the lexicographicordering on

∏1≤j≤α R.

Write g = (f1 ∨ ... ∨ fm)+. We let (x∗1, ..., x∗m) ∈ Ame such that

m∑i=1

fi(x∗i ) ≥

m∑i=1

fi(xi)

35

for all (x1, ..., xm) ∈ Ame .Now we have

g(e) = g(

m∑i=1

x∗i ) =

m∑i=1

g(x∗i ) ≥m∑i=1

fi(x∗i ).

So we need to show that the other inequality holds.

First we assume that f(α+1)i ≥ 0 for all i ∈ {1, ...,m}. We know from Lemma 4.4

that f(1),...,(α)1 ∨...∨f (1),...,(α)m = (f1∨...∨fm)(1),...,(α), since we use the lexicographic

ordering on∏

1≤j≤α R. Hence by the induction hypothesis we get if we let h =(f(1),...,(α)1 ∨ .... ∨ f (1),...,(α)m

)+then h is positive, linear and satisfies for x ∈ L+

h(x) = max{m∑i=1

fi(xi)(1),...,(α) : (x1, ..., xm) ∈ Amx },

where the usual supremum is now a maximum by the continuity of the functionals

f(1),...,(α)i and the compactness of order intervals. Then we have that

h(e) =

m∑i=1

fi(x∗i )

(1),...,(α) = h(

m∑i=1

x∗i )

by our choice of x∗1, ..., x∗m.

Suppose that∑mi=1 fi(x

∗i )

(α+1) = 0. Let j ∈ {1, ...,m} and y ≥ 0 and suppose

we have h(y) = fj(y)(1),...,(α) and fj(y)(α+1) > 0. Since in the definition of hwe only have maxima, we can let z1, ..., zm ∈ L+ such that

∑mi=1 zi ≤ e − y and

h(e − y) =∑mi=1 fi(zi)

(1),...,(α). Then y +∑mi=1 zi ≤ e, y + zj ≥ 0, zi ≥ 0 for all

i ∈ {1, ...,m} and we have

m∑i=1,i6=j

fi(zi)(1),...,(α) + fj(y + zj)

(1),...,(α) = fj(y)(1),...,(α) +

m∑i=1

fi(zi)(1),...,(α)

= h(y) + h(e− y) = h(e) =

m∑i=1

fi(x∗i )

(1),...,(α)

and by positivity of f(α+1)i for i ∈ {1, ...,m} we also have

m∑i=1,i6=j

fi(zi)(α+1) + fj(y + zj)

(α+1) ≥ fj(y)(α+1) > 0 =

m∑i=1

fi(x∗i )

(α+1).

So∑mi=1,i6=j fi(zi) + fj(y + zj) >

∑mi=1 fi(x

∗i ), which gives a contradiction with

the definition of (x∗1, ..., x∗m). So for all j ∈ {1, ...,m} and y ∈ L+ with h(y) =

fj(y)(1),...,(α) we have fj(y)(α+1) = 0. Clearly we also have for all j ∈ {1, ...,m}and y ∈ L+ that h(y) ≥ fj(y)(1),...,(α) by our definition of h. Hence we see that forall j ∈ {1, ...,m} and all y ∈ L+ we have

fj(y) ≤ (h(y), 0)

and thus

fj ≤ (h, 0).

This implies that g ≤ (h, 0). In particular we see that

g(e) ≤ (h(e), 0) = (

m∑i=1

fi(x∗i )

(1),...,(α),

m∑i=1

fi(x∗i )

(α+1)) =

m∑i=1

fi(x∗i ).

36

Therefore we obtain

g(e) =

m∑i=1

fi(x∗i ).

So it remains to consider the case where∑mi=1 fi(x

∗i )

(α+1) > 0. We define

Y = {(y1, ..., ym) ∈ Lm+ : h(

m∑i=1

yi) =

m∑i=1

fi(yi)(1),...,(α),

m∑i=1

fi(yi)(α+1) >

m∑i=1

fi(x∗i )

(α+1)}.

This set is clearly convex and non-empty (it contains t(x∗1, ..., x∗m) for all t > 1).

We also let

Z = {(z1, ..., zm) ∈ Lm :

m∑i=1

zi ≤ e}.

Also this set Z is non-empty and convex. We will show that Y and Z are disjoint.Suppose that Y ∩ Z 6= ∅ and let (y1, ..., ym) ∈ Y ∩ Z. Since in the definition of hwe only have maxima and e −

∑mi=1 yi ≥ 0, we can let z1, ..., zm ∈ L+ such that∑m

i=1 zi ≤ e−∑mi=1 yi and h(e−

∑mi=1 yi) =

∑mi=1 fi(zi)

(1),...,(α). Then∑mi=1(yi +

zi) ≤ e, yi + zi ≥ 0 for all i ∈ {1, ...,m} and we have

m∑i=1

fi(yi + zi)(1),...,(α) =

m∑i=1

fi(yi)(1),...,(α) +

m∑i=1

fi(zi)(1),...,(α)

= h(

m∑i=1

yi) + h(e−m∑i=1

yi) = h(e) =

m∑i=1

fi(x∗i )

(1),...,(α)

and by positivity of f(α+1)i for i ∈ {1, ...,m} we also have

m∑i=1

fi(yi + zi)(α+1) ≥

m∑i=1

fi(yi)(α+1) >

m∑i=1

fi(x∗i )

(α+1).

So∑mi=1 fi(yi + zi) >

∑mi=1 fi(x

∗i ), which gives a contradiction with the definition

of (x∗1, ..., x∗m). So we must have that Y ∩ Z = ∅.

This set Z has an internal point in Lm by [1, Lemma 2.1]. By the Separation The-

orem [2, Theorem 5.61] there exists a non-zero linear functional (h(α+1)1 , ..., h

(α+1)m ) ∈

(L∗)m that separates Z and Y . That is

(5)

m∑i=1

h(α+1)i (yi) ≥

m∑i=1

h(α+1)i (zi) for all (y1, ..., ym) ∈ Y and (z1, ..., zm) ∈ Z.

Since (x∗1, ..., x∗m) ∈ Z it follows from (5) that

(6)

m∑i=1

h(α+1)i (yi) ≥

m∑i=1

h(α+1)i (x∗i ) for all (y1, ..., ym) ∈ Y.

Furthermore, it follows from (5) that

(7)

m∑i=1

h(α+1)i (x∗i ) = lim

β↓1

m∑i=1

h(α+1)i (βx∗i ) ≥

m∑i=1

h(α+1)i (zi) for all (z1, ..., zm) ∈ Z.

Next, we show that h(α+1)1 = ... = h

(α+1)m := h(α+1). Suppose, by way of contradic-

tion, that there exists some z ∈ L such that h(α+1)1 (z) > h

(α+1)2 (z). Then for some

37

β > 1 we have

h(α+1)1 (x∗1 + z) + h

(α+1)2 (x∗2 − z) +

m∑i=3

h(α+1)i (x∗i ) >

m∑i=1

h(α+1)i (βx∗i ).

However, (βx∗1, ..., βx∗m) ∈ Y since β > 1 and (x∗1 + z, x∗2 − z, x∗3, ..., x∗m) ∈ Z, which

contradicts (5). By the symmetry of the situation we see that h(α+1)1 = h

(α+1)2 =

· · · = h(α+1)m := h(α+1).

Now we show that h(α+1) ≥ 0. Let x ∈ L+. Note that (e− βx, 0, 0, ....) ∈ Z for allβ > 0. Therefore, from (7) we get

∑mi=1 h

(α+1)(x∗i ) ≥ h(α+1)(e)− βh(α+1)(x) or

h(α+1)(x) ≥h(α+1)(e)−

∑mi=1 h

(α+1)(x∗i )

β

for each β > 0. Letting β → ∞ yields h(α+1)(x) ≥ 0. Hence h(α+1) ≥ 0 and thush(α+1) ∈ L∼.

Now since (e, 0, ..., 0) ∈ Z andm∑i=1

x∗i ≤ e we get from equation (7) that

h(α+1)(

m∑i=1

x∗i ) ≤ h(α+1)(e) ≤ h(α+1)(

m∑i=1

x∗i )

and thus we obtain

h(α+1)(

m∑i=1

x∗i ) = h(α+1)(e).

Furthermore, since h(α+1) 6= 0 it must be the case that h(α+1)(e) > 0 since e is anorder unit. Since (e, 0, ..., 0) ∈ Z it follows from (7) that

∑mi=1 h

(α+1)(x∗i ) > 0. Sowe can define

δ(α+1) =

∑mi=1 fi(x

∗i )

(α+1)∑mi=1 h

(α+1)(x∗i )∈ R>0.

We claim that δ(α+1)h(α+1)(x) ≥ f(α+1)i (x) for i ∈ {1, ...,m} and x ∈ L+ such

that h(x) = fi(x)(1),...,(α). To see this, fix i ∈ {1, ...,m} and x ∈ L+ such thath(x) = fi(x)(1),...,(α). If fi(x)(α+1) ≤ 0 then δ(α+1)h(α+1)(x) ≥ 0 ≥ fi(x)(α+1) istrivially true. Assume, therefore, that fi(x)(α+1) > 0 and let

γ(α+1) =

∑mi=1 fi(x

∗i )

(α+1)

fi(x)(α+1)∈ R>0.

Now we have for β > 1

fi(βγ(α+1)x)(α+1) = β

∑mi=1 fi(x

∗i )

(α+1)

fi(x)(α+1)fi(x)(α+1)

= β

m∑i=1

fi(x∗i )

(α+1) >

m∑i=1

fi(x∗i )

(α+1),

since we assumed that∑mi=1 fi(x

∗)(α+1) > 0. Hence by linearity of h and f(1),...,(α)i

we get that (0, ..., 0, βγ(α+1)x, 0, ..., 0) ∈ Y , where βγ(α+1)x takes the i-th positionof this vector. Hence with equation (6) we get

δ(α+1)γ(α+1)h(α+1)(x) = δ(α+1) limβ↓1

h(α+1)(βγ(α+1)x)

≥ δ(α+1)m∑i=1

h(α+1)(x∗i ) =

m∑i=1

fi(x∗i )

(α+1) = γ(α+1)fi(x)(α+1).

38

Hence δ(α+1)h(α+1)(x) ≥ fi(x)(α+1) for all x ∈ L+ with h(x) = fi(x)(1),...,(α). Thus,due to the lexicographic order, for all x ≥ 0 we have

fi(x) ≤ (h(x), δ(α+1)h(α+1)(x)).

Since this holds for all i ∈ {1, ...,m} we get that

g(x) ≤ (h(x), δ(α+1)h(α+1)(x))

for all x ≥ 0. In particular we get

g(e) ≤ (h(e), δ(α+1)h(α+1)(e)) =

m∑i=1

fi(x∗i )

since h(e) = h(∑mi=1 x

∗i ) and h(α+1)(e) = h(α+1)(

∑mi=1 x

∗i ). Hence we obtain

g(e) =

m∑i=1

fi(x∗i ).

Now we do not assume that f(α+1)i ≥ 0 for i ∈ {1, ...,m}. Since the f

(α+1)i are

all regular for i ∈ {1, ...,m} we can let gi, h be positive linear functionals such thatfor i ∈ {1, ...m} we have:

f(α+1)i = gi − h.

Due to compactness of Am+1e we can let (y∗1 , ..., y

∗m+1) ∈ Ame such that

m∑i=1

(f(1)i (y∗i ), ..., f

(α)i (y∗i ), gi(y

∗i )) + (0, ..., 0, h)(y∗m+1)

≥m∑i=1

(f(1)i (xi), ..., f

(α)i (xi), gi(xi)) + (0, ..., 0, h)(xm+1)

for each (x1, x2, ..., xm+1) ∈ Am+1e . By positivity of (0, ..., 0, h) we can assume

without loss of generality thatm+1∑j=1

y∗j = e so that y∗m+1 = e −m∑j=1

y∗j . Then we

obtain that the set of functions {(f (1)i , ..., f(α)i , gi) : i ∈ {1, ...,m}} ∪ {(0, ..., 0, h)}

satisfies the assumptions we made in the first half of this proof. Hence we get byour previous results

g(e) =

(m∨i=1

(f(1)i , ..., f

(α)i , g

(α+1)i − h)

)+

(e)

=

((m∨i=1

((f

(1)i , ..., f

(α)i , gi)− (0, ..., 0, h)

))∨ ((0, ..., 0, h)− (0, ..., 0, h))

)(e)

=

(m∨i=1

(f(1)i , ..., f

(α)i , gi) ∨ (0, ..., h)

)(e)− (0, ..., 0, h) (e)

=

m∑i=1

(f(1)i (y∗i ), ..., f

(α)i (y∗i ), gi(y

∗i )) + (0, ..., 0, h(e−

m∑i=1

y∗i ))− (0, ..., 0, h(e))

=

m∑i=1

(f(1)i (y∗i ), ..., f

(α)i (y∗i ), gi(y

∗i )− h(y∗i )) + (0, ..., 0, h(e)− h(e))

=

m∑i=1

(f(1)i (y∗i ), ..., f

(α)i (y∗i ), f

(α+1)i (y∗i ))

=

m∑i=1

fi(y∗i ) ≤

m∑i=1

fi(x∗i )

39

by our choice of x∗1, ..., x∗m. Since in the very first part of the proof we saw that

g(e) ≥∑mi=1 fi(x

∗i ) we must have g(e) =

∑mi=1 fi(x

∗i ) and that finishes the proof of

part (i).

Proof of (ii) The first statement follows for the same reason as in part (i). So let

(x∗1, ..., x∗m) ∈ Fme such that

m∑i=1

fi(x∗i ) ≥

m∑i=1

fi(xi) for each (x1, x2, ..., xm) ∈ Fme .

First note since (x∗1, ..., x∗m) ∈ Fme that x∗1 = e −

m∑i=2

x∗i . Let g = ∨mi=1fi and start

by observing that g − f1 = [∨mi=2(fi − f1)]+ and

m∑i=2

(fi − f1)(x∗i ) =

m∑i=1

fi(x∗i )− f1(e)

≥m∑i=2

fi(xi) + f1

(e−

m∑i=2

xi

)− f1(e) =

m∑i=2

(fi − f1)(xi)

for any (x2, ..., xm) ∈ Am−1e since if (x2, ..., xm) ∈ Am−1e then (e−m∑i=2

xi, x2, ..., xm) ∈

Fme . Therefore by part (i) we have that

(g − f1)(e) =

m∑i=2

(fi − f1)(x∗i ).

In particular it holds that

g(e) = f1

(e−

m∑i=2

x∗i

)+

m∑i=2

fi(x∗i ) =

m∑i=1

fi(x∗i ),

which is the desired formula.

Proof of (iii) Now fix an arbitrary x ∈ L+ and then select an order unit e ∈ Lsuch that x ≤ e.For each 0 < β < 1 let xβ = βx + (1 − β)e. Clearly, each xβ is an order unit and0 ≤ xβ ≤ e holds for each 0 < β < 1. We consider the index set (0, 1) directed by theincreasing order relation �, i.e. β � γ in (0, 1) if and only if β ≥ γ. Clearly xβ → x.Since Fme is compact, the continuous function Fme →M, (x1, ..., xm) 7→

∑mi=1 fi(xi)

attains its maximum. By part (ii), we know that g satisfies the Riesz-Kantorovichformula for each order unit. Therefore, by our assumption on the space M and the

compactness of Fmeβ , for each 0 < β < 1 there exists some (zβ1 , ..., zβm) ∈ Lm+ such

that∑mi=1 z

βi = xβ and

(8) βg(x) + (1− β)g(e) = g(xβ) =

m∑i=1

fi(zβi ).

Since zβi ∈ [0, e] for each β ∈ (0, 1) and each i ∈ {1, ...,m} and the order interval

[0, e] is compact, there exists a subnet of {(zβ1 , ..., zβm)} (which without loss of gener-

ality we also denote by {(zβ1 , ..., zβm)}) such that zβi → zi for some zi ∈ [0, e]. From

xβ − zβi ∈ [0, e] for each β and the closedness of [0, e], we see that 0 ≤ x − zi ≤ e.So 0 ≤ zi ≤ x and

∑mi=1 zi = x. Finally, letting β → 1 in equation (8) it follows

from the continuity of each fi that

g(x) =

m∑i=1

fi(zi).

40

Since we already know that g(x) ≥∑mi=1 fi(yi) for all (y1, ..., ym) ∈ Fmx we get that

sup{∑mi=1 fi(yi) : (y1, ..., ym) ∈ Fmx } = g(x). So the proof of this induction step is

finished.

Finally let α be a limit ordinal and assume that the theorem holds for all imagespaces

∏1≤j≤β for β < α. Let M =

∏1≤j≤α R and let f1, f2, ..., fm be continu-

ous, linear maps from L to M such that f(j)i is regular for all i ∈ {1, ...,m} and

all ordinals j with 1 ≤ j ≤ α and such that the supremum g :=∨mi=1 fi exists

in L∼(L,M). Let x ∈ L+. Let β < α. Since we know from Lemma 4.4 that∨mi=1(f

(1),...,(β)i ) = f

(1),...,(β)1 ∨ ... ∨ f (1),...,(β)m we can use the induction hypothesis

to obtain that

g(x)(1),...,(β) = sup{m∑i=1

fi(xi)(1),...,(β) : (x1, ..., xm) ∈ Fmx }.

Since we use the lexicographic ordering on M we know that sup{∑mi=1 fi(xi) :

(x1, ..., xm) ∈ Fmx } exists in M and that we have

sup{m∑i=1

fi(xi) : (x1, ..., xm) ∈ Fmx }(1),...,(β)

= sup{m∑i=1

fi(xi)(1),...,(β) : (x1, ..., xm) ∈ Fmx } = g(x)(1),...,(β).

Since this holds for all β < α we immediately obtain

g(x) = sup{m∑i=1

fi(xi) : (x1, ..., xm) ∈ Fmx }

and the proof is finished.

Remark 4.7. One might wonder why we used the lexicographic ordering and notthe componentwise ordering. The reason is as follows: throughout the proof weuse that in the lexicographic ordering each compact set has an order-maximum.This is not true for the componentwise ordering as can be seen from the followingexample. Let f : R → R2, λ 7→ λ(−1, 1) and g : R → R2, λ 7→ 0. Then it holdsthat {f(1 − y) + g(y) : 0 ≤ y ≤ 1} = {λ(−1, 1) : 0 ≤ λ ≤ 1}, which does not havean order maximum, since its order supremum equals (0, 1) which is not an elementof this set. Since sets of the form {f(1 − y) + g(y) : 0 ≤ y ≤ 1} are precisely thesets for which we need an order maximum to exist, there is no hope to prove asimilar statement for the componentwise ordering in the same way we did for thelexicographic ordering. Note, however, that [1, Lemma 3.1] and [1, Lemma 3.2],which form the base for steps (i) and (ii) in the proof above, can easily be provenfor the componentwise ordering.

The following corollary might give a more general solution to the Riesz-Kantorovichproblem than before. It is concerned with the situation of Theorem 4.6 but thenfor spaces without a strong order unit.

41

Corollary 4.8. Let L be a partially ordered vector spaces with a Hausdorff topologyfor which the order-intervals are compact. Let α ≥ 1 be an ordinal number andlet M =

∏1≤j≤α R equipped with the lexicographic ordering. Then the following

statements are equivalent.

(1) There is a pair f, g ∈ Lr(L,M) of continuous maps such that f (j) and g(j)

are regular for all 1 ≤ j ≤ α, such that f ∨ g exists in L∼(L,M), but doesnot satisfy the Riesz-Kantorovich formula for all x ∈ L+.

(2) There is a pair f, g ∈ Lr(L,M) of continuous maps and a point x ∈ L+

such that f (j) and g(j) are regular for all 1 ≤ j ≤ α, such that f ∨ g existsin L∼(L,M), but the restriction (f ∨ g)|Lx is not the supremum of f |Lxand g|Lx in L∼(Lx, V ).

Proof Assume that (1) holds and let f, g ∈ Lr(L,M) be a pair of continuousmaps such that f (j) and g(j) are regular for all 1 ≤ j ≤ α, such that f ∨ g existsin L∼(L,M), but does not satisfy the Riesz-Kantorovich formula. Then we can letx ∈ L+ such that

f ∨ g(x) > sup{f(y) + g(x− y) : 0 ≤ y ≤ x}.Consider the space Lx. As stated before we get that x is a strong order unit forLx. We will argue by contradiction. Assume that f |Lx ∨ g|Lx exists in L∼(Lx,M)and that

(f ∨ g)|Lx = f |Lx ∨ g|Lx .Then by Theorem 4.6, which we can apply since f |Lx and g|Lx satisfy the conditionsof Theorem 4.6 and the space Lx has a strong order unit, we obtain that

(f |Lx ∨ g|Lx)(x) = sup{f |Lx(y) + g|Lx(x− y) : y ∈ Lx, 0 ≤ y ≤ x}= sup{f |Lx(y) + g|Lx(x− y) : y ∈ L, 0 ≤ y ≤ x}

< (f ∨ g)(x) = (f |Lx ∨ g|Lx)(x),

where we used the definition of Lx and the assumptions. So we get a contradiction.Therefore (2) holds and we obtain that (1)⇒ (2).

Conversely, suppose that (2) is true. Let f, g ∈ Lr(L,M) be a pair of continuousmaps and let x ∈ L+ such that f (j) and g(j) are regular for all 1 ≤ j ≤ α, suchthat f ∨ g exists in L∼(L,M), but the restriction (f ∨ g)|Lx is not the supremum off |Lx and g|Lx in L∼(Lx, V ). Again we argue by contradiction. Assume that f ∨ gsatisfies the Riesz-Kantorovich formula for all z ∈ L+. Let h ∈ L∼(Lx,M) suchthat h ≥ f |Lx and h ≥ g|Lx . Let z ∈ L+

x and 0 ≤ y ≤ z. Then we obtain sincey, z − y ∈ Lx that

h(z) = h(y) + h(z − y) ≥ f |Lx(y) + g|Lx(z − y).

Therefore we must have that

h(z) ≥ sup{f |Lx(y) + g|Lx(z − y) : 0 ≤ y ≤ z}= sup{f(y) + g(z − y) : 0 ≤ y ≤ z} = (f ∨ g)(z) = (f ∨ g)|Lx(z)

since f ∨ g satisfies the Riesz-Kantorovich formula. Therefore we must have thath ≥ (f ∨ g)|Lx . Furthermore we obtain for z ∈ L+ that

(f ∨ g)|Lx(z) = (f ∨ g)(z) ≥ f(z) = f |Lx(z)

and similarly (f ∨ g)|Lx(z) ≥ g|Lx(z). Hence (f ∨ g)|Lx ≥ f |Lx ∨ g|Lx . Combiningthese two facts gives that (f ∨ g)|Lx = f |Lx ∨ g|Lx which gives a contradiction toour assumption. Therefore (1) must hold and we obtain (2)⇒ (1).

42

Discussion

In this thesis we discussed some new results on the space Lr(E,F ) of regularoperators from E to F , where E and F are partially ordered vector spaces. Weshowed that, if E and F are Riesz spaces, in general the space Lr(E,F ) need notbe an order-dense subspace of Lr(E,F δ), but if it is, that then its lattice operationsmust be given by the Riesz-Kantorovich formula. Furthermore we showed that ifthe space E has a strong order unit and F is an arbitrary product of R, equippedwith the lexicographic ordering, that, under a few continuity conditions, the Riesz-Kantorovich formula holds. We also showed, in this situation, how to deal with theabsence of a strong order unit in E.

There are several possibilities for future research, most of them dealing with theRiesz-Kantorovich formula, which in general is still an open problem. One could forinstance investigate the case when E and F are Riesz spaces, but Lr(E,F ) is notan order-dense subset of Lr(E,F δ). For example, one can try to prove the reversestatement of Theorem 3.2. It may also be interesting to look more into Example3.6, since this might give a counterexample to the Riesz-Kantorovich theorem andeven if it turns out to be no counterexample, it is still interesting to investigate itfurther.Since the componentwise ordering on products of R is used more often than thelexicographic ordering, it is of course of much interest to prove a statement similarto Theorem 4.6 or even Theorem 4.5 for the componentwise ordering. As statedbefore it is not difficult to show an equivalent for [1, Lemma 3.1] and [1, Lemma 3.2]for this case, but the problem lies in the fact that a compact set no longer needs tohave an order-maximum, so the proof of [1, Theorem 3.3] cannot be adapted easily.Finally it might be interesting to find an explicit condition for when the spaceLr(E,F ) is an order-dense subset of Lr(E,F δ), since then we have a more explicitclass of spaces we know we are allowed to apply the Riesz-Kantorovich formula to.

43

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