# Characterization of Banach Lattices in Terms of Quasi-Interior Points

Authors: Egor A. Alekhno
Journal: International Journal of Mathematics and Mathematical Sciences
 (2013)

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## Abstract

In terms of quasi-interior points, criteria that a Banach latticeE has order continuous norm or is an AM-space with a unit are given. For example, if E is Dedekind complete and has a weak order unit, then E has order continuous norm if and only if the set of quasi-interior points of E coincides with the set of weak order units of E; a Banach lattice E is an AM-space with a unit x if and only if the set of all quasi-interior points of E coincides with the set {z:x≤λzfor someλ>0}. Analogous questions are considered for the case of an ordered Banach space Z with a cone K. Moreover, it is shown that every nonzero point of a cone K≠{0} is quasi-interior if and only if dimZ=1. We also study various 𝒟-properties of a cone K; in particular, the conditions for which the relation a∧b=0 with a>0 implies that b is not a quasi-interior point are considered.


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## Body

## 1. Introduction

One of natural notions from the theory of Riesz spaces is a notion of a weak order unit: an elementx>0in a Riesz spaceEis said to be [1, page 36] a weak order unit whenever for eachz∈E, the relationz⊥ximpliesz=0(or, equivalently, wheneverz∧nx↑zfor eachz≥0). A topological analog of a weak order unit is a quasi-interior point. Namely, an elementx>0in a Banach latticeEis said to be [1, page 259] a quasi-interior point, whenever for eachz≥0, the sequencez∧nxnorm converges toz. This is equivalent to the assertion thatx*x>0for each nonzero positive functionalx*∈E*. In the case of an ordered Banach spaceZwith a coneK, the last assertion is taken up as a definition of a quasi-interior point ofK(see [2, page 24]).The main idea of using quasi-interior points is the following. If an elementxof a Banach latticeEis quasi-interior, then, on the one hand, the (order) idealIxgenerated byxis norm dense inE; on the other hand, the idealIxunder the norm∥z∥x=min{λ≥0:|z|≤λx}is anAM-space with a unitxand so that, by Kakutani-Bohnenblust-Kreins' theorem [1, page 194],Ixis lattice isometric onto a spaceC(S)of all continuous functions on some compact topological spaceS. Thus, it is often possible to reduce the investigation of number properties of a Banach latticeEto the study of these properties, or closed to them, in the space of continuous functions, and then, sinceIxis dense inE, we can return to the spaceE. In other words, from all Banach lattices, Banach lattices having quasi-interior points are the “closest” toAM-spaces with a unit.In spite of the considerable progress in the theory of quasi-interior points (see [3, pages 14–17] for precise and detailed reference), several aspects of this theory have received almost no attention. Up to now, conditions which give the characterization of an order structure of an arbitrary Banach latticeEwith help of quasi-interior points were not studied. For example, conditions in terms of quasi-interior points guaranteeing that a Banach latticeEis anAM-space with a unit were not known. The class of Banach lattices with order continuous norms which have been researched well and are important was not even considered with the point of view of quasi-interior points. Moreover, the following question raised to Schaefer [4, page 270, Remark (iii)] were remained without an answer: Does a coneKof some Banach spaceZ,dimZ>1, such that every nonzero elementz∈Kis a quasi-interior point ofK, exist?The principal purpose of this paper is to give in terms of quasi-interior points the characterizations of two most important classes of Banach lattices, namely, Banach lattices with order continuous norms andAM-spaces with a unit. The second section is devoted to these problems. In the third section, analogous questions are considered for ordered Banach spaces, and a negative answer to Schaefer's problem stated above is provided (Theorem 6). The objective of Section 4 is to introduce various𝒟-properties of a cone and to study them in conjunction with the notion of quasi-interior point. The last section presents some applications of the theory of ordered Banach algebras to the scope of questions discussed in the preceding sections.For any unexplained terminology, notations, and elementary properties of Banach lattices, we refer the reader to [1, 5]. For information on the theory of ordered Banach spaces, we suggest [2] (see also [6]). Throughout the paper, unless stated otherwise,Ewill stand for an arbitrary Banach lattice, whileZwill be an arbitrary ordered Banach space with a (closed) coneK. We assume thatE≠{0}orK≠{0}.


## 2. The Case of a Banach Lattice

We start with a simple result that will be needed later for the characterization of Banach lattices, having order continuous norms with help of quasi-interior points.Lemma 1.
LetEbe a Banach lattice with the countable sup property and letJbe a closed ideal inE, such thatJd={0}. IfEhas a weak order unitx, then idealJalso has a weak order unit.Proof.
From the relationJd={0}, it follows that the idealJis order dense inE. Consequently, there exists a sequencexn∈Jsatisfying0<xn↑x. Then, the elementx0=∑i=1∞(1/2i∥xi∥)xi∈Jis a weak order unit (inE). Indeed, ifz⊥x0withz∈E, then|z|∧xn=0for alln. Since|z|∧xn↑|z|∧x, we obtain|z|∧x=0orz=0.In order to proceed further, we recall the well-known fact that every Banach latticeEwith order continuous norm has the countable sup property.Theorem 2.
A Banach latticeEwith the countable sup property and with a weak order unit has order continuous norm if and only if the set of all quasi-interior points ofEcoincides with the set of all weak order units ofE.Proof.
The necessity is valid without the assumption thatEhas the countable sup property and can be found, for example, in [1, page 260]. We will prove the sufficiency. Assume by way of contradiction thatEdoes not have order continuous norm. This is equivalent to the fact that the bandEn~of all order continuous functionals does not coincide with the norm dualE*ofE. Therefore, there exists a functionalx*>0satisfying the relationNx*d={0}, whereNx*={z∈E:x*|z|=0}. Next,Nx*is a closed ideal; hence, by Lemma 1,Nx*has a weak order unitx. By our hypothesis, the elementxis a quasi-interior point. However,x*x=0, which is impossible.The preceding theorem is also valid if we replace the property thatEhas the countable sup property by the Dedekind completeness ofE(see Theorem 14 later). Moreover, Theorem 2 can be obtained with help of the next Andô theorem [1, page 173]: a Banach latticeE  has order continuous norm if and only if every closed ideal is a band.The following examples show that, in Lemma1 and Theorem 2, the condition thatEhas the countable sup property is essential.Example 3.
Consider the Banach latticeB(ℝ)of all bounded functionsz:ℝ→ℝwith the norm∥z∥=sups∈ℝ|z(s)|, and letJbe the closed ideal inB(ℝ)consisting from all functions which have countable supports. ThenJd={0}andJdoes not have a weak order unit.Example 4.
LetB(ℝ)be the space from the preceding example and lete∈B(ℝ)be the constant function one onℝ. Throughc0(ℝ), we denote a closed subspace ofB(ℝ)consisting from all functionsz, such that for everyϵ>0the set{s:|z(s)|>ϵ}is finite. Consider the space
(1)E={i+λe∈B(ℝ):i∈c0(ℝ),λ∈ℝ}.
It is closed inB(ℝ). Actually, if a sequencein+λne→zwithin∈c0(ℝ),λn∈ℝ, andz∈B(ℝ), thenλnis a bounded sequence. Consequently, we can assume thatλnconverges to someλ0. In this case,z-λ0e∈c0(ℝ). Next,Eis also closed under the lattice operations. Thus,Eis a Banach lattice. Nevertheless,Edoes not have order continuous norm, and, on the other hand, the set of all quasi-interior points ofEcoincides with the set of all weak order units ofE. We mention thatEis anAM-space with a unit which is not Dedekind complete (see Theorem 14 and Corollary 15).The most classical Banach lattice which does not have order continuous norm is an infinite dimensional
A
M
-space with a unit. In the following theorem, the characterization of such spaces in terms of quasi-interior points is given. Recall that a Riesz spaceFhas [1, page 35] the principal projection property if for every elementz∈Fthe bandBzgenerated byzis a projection band; that is,Bz⊕Bzd=For, in other words, there exists an order projectionPzfrom the spaceFontoBz.Theorem 5.
For a Banach latticeEand an elementx∈E+, the following two statements are equivalent: (a)
Eis anAM-space with a unitx;(b)
the set of all quasi-interior points ofEcoincides with the set of all elementsz, such thatx≤λzfor someλ>0.Moreover, in caseEhas the principal projection property, each from the conditions (a) and (b) is equivalent to the next: (c)
the relationinfz≠0∥Pzx∥>0holds.Proof.
First of all, we recall that an elementyis a unit of some Banach latticeFif and only ifyis an interior point ofF+. On the other hand, if in some ordered Banach spaceZwith a coneKthe interiorK°ofKis nonempty, then it is well known (see, e.g., [2, page 22]) thatK°coincides with the set of all quasi-interior points ofK. Now, the implication (a)⇒(b) is obvious. For a proof of (a)⇒(c), it suffices to observe that the norms∥·∥xand∥·∥are equivalent, and the identities∥Pzx∥x=∥x∥x=1hold. Indeed, if∥Pzx∥x<1, thenPzx≤λxfor some nonnegative scalarλ<1; hence,Pzx≤λPzxorPzx=0. Thus,z⊥x, which is impossible (in the next section, this result will be stated for the case of an ordered Banach space (Theorem5′), and in the same place another approach to a proof of the implication (a)⇒(c) can be found).
(c)⇒(a) Suppose by way of contradiction thatxis not an interior point ofE+. Then, there exists a sequencexnsatisfyingxn→xandxn∉E+. The order projectionsP(xn)-are nonzero. Clearly,P(xn)-(xn-x)→0. Hence, using the inequalityP(xn)-x≤P(xn)-x-P(xn)-xn, we inferP(xn)-x→0. The latter is contrary to (c).
(b)⇒(a) Again suppose by way of contradiction thatxis not an interior point of the setE+. This set is ideally convex; therefore [2, pages 10, 11],xis not an interior point ofE+if and only ifxis notc-interior point ofE+. Consequently, there exists an elementy∈Esatisfyingx-(1/n)y∉E+for alln. Sincex-(1/n)y+≤x-(1/n)y, we can assume thaty>0. Putxn=x-(1/n)y. Clearly,xn↑xandxn(norm) converge tox. Define the elementx0=∑i=1∞(1/2i)(xi)+. We will establish thatx0is a quasi-interior point. To this end, proceeding by contradiction, we find a functionalx*>0satisfyingx*x0=0. Therefore,x*(xn)+=0for alln. Whence, using the relation
(2)(xn)+⟶x,
we getx*x=0, which is impossible. By the condition (b), for some natural numberk, the inequalityx≤2kx0holds. Taking into account the relationP(xk+1)-(xj)+=0for allj=1,…,k+1, we have
(3)P(xk+1)-x≤∑i=1∞‍2k2iP(xk+1)-(xi)+=∑i=k+2∞‍2k2iP(xk+1)-(xi)+≤∑i=k+2∞‍2k2iP(xk+1)-x=12P(xk+1)-x.
Consequently,P(xk+1)-x=0; hence,x⊥(xk+1)-. Finally,(xk+1)-=0orxk+1≥0, and we have a contradiction.For the case of Banach function spaces, Theorems2 and 5 were obtained in [7], and then in [3] they were generalized for the case of Banach lattices in some other view. Analogues of these theorems for cones can be found in the next section (Theorems2′and5′, resp.).We will close this section with the following remarks. As a matter of fact, in the proof of Theorem5, the validity of the next assertion was shown: ifx  is a quasi-interior point of a Banach latticeE, a sequencexnconverges to  x, and a sequence  α=(α1,α2,…)∈ℓ1, such that  αn≠0  for all  n, then the element  x0=∑i=1∞|αi|(xi)+is quasi-interior. Clearly, if  |xn|≤x, then  x0∈Ix. The converse statement is also true: if an element  x∈E+  and an element  x0∈Ix  are quasi-interior points of  E, then there exist an increasing sequence  xn≥0  converging to  x  and a sequence  α=(α1,α2,…)∈ℓ1  satisfying  αn>0  for all  n, such that  x0=∑i=1∞αixi. In fact, we can assume that  0<x0≤x. For natural numbers  n≥2, we put  xn=x∧(nx0)  and  αn=1/2n-1n. Then,  ∑i=2∞αixi≤x0. Define  x1=x0-∑i=2∞αixi  and  α1=1. It remains to observe the validity of the relations  0≤x1≤x0≤x2  and  x0=∑i=1∞αixi.An analogous result holds for the case of weak order units in a Riesz spaceFsatisfying Axiom (OS) (recall that [5, page 54] an ordered linear spaceXsatisfies Axiom (OS) if the inequalities0≤zn≤λnz, wherezn,z∈Xandλ=(λ1,λ2,…)∈ℓ1, imply the existence ofsupn∑i=1nzi): if  x  is a weak order unit inF, thenx0∈Ix  is also a weak order unit in  F  if and only if there exist sequences0≤xn↑x  andα=(α1,α2,…)∈ℓ1  withαn>0  for alln, such that∑i=1nαixi↑nx0. Obviously, everyσ-Dedekind complete ordered linear spaceXand, in particular, every ordered Banach spaceZwith aσ-minihedral coneK(see the next section) satisfy Axiom (OS).


## 3. The Case of an Ordered Banach Space

The main purpose of this section is to obtain with help of quasi-interior points the characterizations of some classes of cones in Banach spaces and, in particular, to discuss possible generalizations of Theorems2 and 5 for the case of ordered Banach spaces.We will show first that if a Banach spaceZwith a coneKis not one-dimensional, then there exists a nonzero elementwofKwhich is not a quasi-interior point ofK(see also [3, Theorem1] and [8, Proposition7]). We need to recall the following result obtained in [9] (see also [10]). LetM  be a closed subset of a Banach spaceY, and lety∈Y∖M. Let0<r<d(y,M)<R. Then, there exists a pointw∈M, such that||w-y||≤R  andM∩co(B(y,r)∪{w})={w}. Here, as usual,d(·,M)is the distance function,co(N)is the convex hull of the setN, andB(y,r)is the closed ball centered at the pointywith radiusr(below the ballB(0,1)will be simply denoted byB).Theorem 6.
LetZbe a Banach space, and letKbe a cone inZ. Then, the set of quasi-interior points ofKcoincides with the set of nonzero elements ofKif and only if the spaceZis one-dimensional.Proof.
The sufficiency is obvious. We will prove the necessity. Assume by way of contradiction that the dimensionaldimZ>1. We will show first thatKhas a nonzero boundary pointz. To this end, we find two linearly independent elementsz1∈Kandz2∉K. Let
(4)λ0=max{λ∈[0,1]:(1-λ)z1+λz2∈K}.
Clearly, the nonzero pointz=(1-λ0)z1+λ0z2is a boundary point ofK.
We can suppose that∥z∥=1. Choose an elementywhich does not belong to the coneK, such that∥z-y∥<1/2. Then,d(y,K)<1/2. Fix arbitrary numbersrandRare satisfying the inequalities0<r<d(y,K)<R≤1/2. There exists an elementw∈Kwith the properties∥w-y∥≤RandK∩W={w}, where the setW=co(B(y,r)∪{w}). The elementw≠0. By the classical separation theorem, there exists a nonzero functionalz*∈Z*, such thatz*x≥z*yfor allx∈Kandy∈W. Clearly,z*∈K*. Moreover,z*λw≥z*wfor allλ≥0; hence,z*w=0. Therefore, the elementwis not a quasi-interior point.Recall that a coneKin a Banach spaceZis called [2, page 8] solid if it contains some ball of positive radius and is called (see [6, page 4] or [2, page 9]) generating (or reproducing) ifK-K=Z. Obviously, every solid cone is generating. A coneKis called [2, page 57] minihedral if every finite order bounded from above subsetS⊂Zhas a supremum and strongly minihedral if every order bounded from above subsetS⊂Zhas a supremum. If an arbitrary countable order bounded from above subsetS⊂Zhas a supremum, thenKis calledσ-minihedral. A coneKis called [2, page 37] normal if for everyy∈Kthe order interval[0,y]={z∈Z:0≤z≤y}is norm bounded. It can be shown that [6, page 109, Exercise 12]Kis normal if and only if the spaceZsatisfies Axiom (OS); in particular, everyσ-minihedral cone is normal. If inZevery increasing and order bounded above sequenceznconverges in norm to some elementz0, then a coneKis called [2, page 47] regular; in this case, it is easy to see thatz0=supnzn. Every regular cone [2, page 48] is also normal.An arbitrary ordered Banach spaceZwith a coneKis a Riesz space (of course, under the partial ordering induced byK) if and only ifKis generating and minihedral. In this case, all notions from the theory of Riesz spaces can be used for the ordered Banach spaceZ, in particular, the notion of weak order unit. Moreover, as in the case of a Banach lattice, every quasi-interior pointxof the coneKis a weak order unit. Indeed, ifx⊥ywithy∈Z, then
(5)x+|y|=|y|∨x=0∨(x-|y|)+|y|,
whencex=(x-|y|)+. Therefore,(x-|y|)+is a quasi-interior point ofK. Consequently ([2, page 112]; see also Example 10 (c) below),x≥|y|, so that|y|=0ory=0.The following result is an analogue of Theorem2 for the case of a space with a cone.Theorem  2′.LetZ  be an ordered Banach space with a generating, minihedral, and normal coneK. Moreover, letZ  have a weak order unit and have the countable sup property. ThenK  is regular if and only if the set of all quasi-interior points ofK  coincides with the set of all weak order units ofZ.Proof.
First of all, we mention that [5, page 145, Exercise 13(a)] the coneKis generating, minihedral, and normal if and only if there exists an equivalent norm on the spaceZ, such thatZ, under this norm, is a Banach lattice. Now, it remains to remember that an arbitrary Banach latticeEhas order continuous norm if and only if the coneE+is regular and to use Theorem 2.Remark 7.
Most of the constructions using the notion of quasi-interior point, for example, the representation theory (see [5, Chapter III, sections 4–6]), approximation properties of operators on Banach lattices (see [1, section 15]), or the spectral theory of ideal irreducible operators (see [5, Chapter V, sections 5, 6]) actually require that a coneKenjoys the assumptions of the preceding theorem; that is, it is generating, minihedral, and normal. As it follows from the remarks done in the proof of this theorem, these assumptions actually mean that the ordered Banach spaceZwith the coneKis a Banach lattice. The given circumstance explains the reason why the notion of quasi-interior point turns out more natural and more useful for the case of a Banach lattice.In the preceding theorem, the assumption about the normality of the coneKis unnecessary ifKisσ-minihedral. Moreover, the necessity also holds without the explicit assumption about the normality ofK. The next example shows that for the validity of the sufficiency, this assumption is essential.Example 8.
Consider the spacebv0of all real sequencesz=(z1,z2,…)converging to zero, such that the norm∥z∥bv0=∑i=1∞|zi+1-zi|is finite. Under the natural ordering, that is, for elementsy,z∈bv0, we puty≥zifyn≥znfor alln,bv0is a Riesz space; in particular, the cone(bv0)+is generating and minihedral. Indeed, fory,z∈bv0, defining the sequencewbywn=yn∨zn, we have
(6)∥w∥bv0=∑i=1∞‍|yi+1∨zi+1-yi∨zi|≤∑i=1∞‍(|yi+1-yi|+|zi+1-zi|)<+∞.
Thus,w∈bv0. Now, the identityw=y∨zis obvious. Moreover, the latter means thatbv0is a Riesz subspace of the spacesof all real sequences; in particular, a sequencex∈bv0is a weak order unit if and only ifxn>0for alln. Next, it is easy to see that the cone(bv0)+is closed; hence,bv0is an ordered Banach space. We assert that every weak order unitx∈bv0is a quasi-interior point of(bv0)+. Indeed, letx*x=0for some nonzero positive functionalx*∈(bv0)*. The relations0=x*x≥x*(xnen)=xnx*en≥0hold; hence,x*en=0for alln, whereenis thenthunit coordinate vector. Consider an arbitrary elementz∈(bv0)+and definez(n)=(z1,…,zn,0,0,…). Clearly,x*z(n)=0. On the other hand,
(7)∥z-z(n)∥bv0=|zn+1|+∑i=n+1∞‍|zi+1-zi|⟶0
asn→∞, whencex*z=0. Sincezis arbitrary, we getx*=0, which is impossible. Finally, inbv0, the set of quasi-interior points coincides with the set of weak order units. Nevertheless, the cone(bv0)+is not normal and, in particular, is not regular. For this, we consider the sequenceyn=∑i=1n(1/(2i-1))e2i-1inbv0and the elementy=(1,1/2,1/3,…)∈bv0and remark that0≤yn≤yand∥yn∥bv0=1+2∑i=2n(1/(2i-1))→+∞asn→∞. This establishes the validity of our last claim.
We also do the following remark. If we consider the spacebvof all (convergent) real sequencesz, such that the norm∥z∥bv=|z1|+∑i=1∞|zi+1-zi|is finite, then under the natural ordering,bvis also an ordered Banach space and, simultaneously, a Riesz space. Moreover,bv0is a Riesz subspace ofbv. Evidently, ifz∈bv0, then the inequalities∥z∥bv0≤∥z∥bv≤2∥z∥bv0hold; in particular, the cone(bv)+is not also normal. Nevertheless, the cone(bv)+is solid. Indeed, it suffices to show that ife=(1,1,…), thenB(e,1)⊂(bv)+. For this, ifz∈B(e,1), then for arbitrary numberk∈ℕ, we have
(8)1≥|1-z1|+∑i=1∞‍|zi+1-zi|≥|1-z1|+∑i=1k‍|zi+1-zi|≥|1-zk+1|;
hence,zk+1≥0, so thatz≥0.In the case of an ordered Banach space, Theorem5 has the next form.Theorem5′. For an ordered Banach spaceZ  with a generating and minihedral coneK  and an elementx∈K, the following two statements are equivalent:(a)
K  is a solid cone with an interior point  x;(b)
the set of all quasi-interior points of  K  coincides with the set of all elements  z, such that  x≤λz  for some  λ>0.Moreover, in case  K  is normal and  Z  has the principal projection property, each from the conditions (a) and (b) is equivalent to the next:(c)
the relation  infz≠0∥Pzx∥>0  holds.Proof.
As it follows from the remarks done in the proof of Theorem2′once more, ifKis generating, minihedral, and normal, then the required assertion is a consequence of Theorem 5. Nevertheless, since the statement (a) implies thatK∘coincides with the set of all quasi-interior points, it is easy to see that the implication (a)⇒(b) is true for an arbitrary ordered Banach space. Next, the implications (a)⇒(c) and (b)⇒(a) remain valid without the assumption about the normality of the coneK. Indeed, let the statement (a) hold. Assume that for some sequencezn∈Zwithzn≠0for alln, we have∥Pznx∥→0asn→∞. For some sufficiently largen0, the element(I-Pzn0)xis an interior point of the coneK. Puty=(I-Pzn0)x-Pzn0x. Obviously,y∉K. Therefore ([2, page 112]; see also Example 10 (c)),y+=(I-Pzn0)xis not quasi-interior, which is impossible.
Now, we will show the validity of (b)⇒(a). In the proof of the analogous implication of Theorem 5, the normality of the coneKwas only used for the conclusion of the convergence(xn)+→x(see (2)), that is, in the proof of the fact thatx0is a quasi-interior point (we use the notations from the proof of Theorem 5). Again, ifx*x0=0for somex*>0, thenx*(xn)+=0for alln, whencex*xn=-x*(xn)-≤0. Consequently,x*x=0, which is impossible. Finally,x0is a quasi-interior point.The author is grateful to the referee for an improvement of this theorem and other useful remarks and suggestions.It is not known whether the implication (b)⇒(a) of the preceding theorem is valid for an arbitrary ordered Banach space and the implication (c)⇒(a) is valid without the assumption about the normality of the coneK. Moreover, the author does not know any example of an ordered Banach spaceZwith a coneK, such thatZis a Riesz space with the principal projection property, whileKis not normal. Nevertheless, if the coneKisσ-minihedral, then the implication (c)⇒(a) remains valid without the assumption about the normality ofK.


## 4. Various𝒟-Properties of a Cone

LetZbe an arbitrary ordered Banach space with a coneK, and let a set𝒟⊆K×K. A coneKis said to have the weak𝒟-property if for every pair(a,b)∈𝒟witha>0there exists a functionalx*>0satisfyingx*b=0. We mention at once the following simple result:K  has the weak𝒟-property if and only if the relations(a,b)∈𝒟  and  a>0  imply that  b  is not a quasi-interior point of  K. A cone  K  is said to have the𝒟-property if for every pair  (a,b)∈𝒟  with  a>0  there exists a functional  x*≥0  satisfying  x*a>0  and  x*b=0. Observe that if  K  has the𝒟-property, then for every pair  (a,b)∈𝒟  with  a,b>0, the elements  a  and  b  are linearly independent. Next, a cone  K  is said to have the strong  𝒟-property if for every pair  (a,b)∈𝒟  there exists a functional  x*≥0  satisfying  ∥x*∥=1,  x*a=∥a∥, and  x*b=0. Obviously, the strong  𝒟-property implies the  𝒟-property and the  𝒟-property implies the weak  𝒟-property.The following result characterizes cones having the weak𝒟-property.Theorem 9.
For an ordered Banach spaceZwith a coneKand a set𝒟⊆K×K, the following statements are equivalent: (a)
Khas the weak𝒟-property;(b)
if a pair(a,b)∈𝒟, and for every nonzero elementc∈Z, there exist two sequencescn∈Zconverging to zero andλn∈ℝwhich satisfy the inequalityc-λnb≤cnfor alln, thena=0;(c)
for every pair(a,b)∈𝒟witha>0there exist a numberϵ>0and an elementc∈Zsatisfying the relation
(9)(⋃λ∈ℝ(ϵB+λb))∩(c+K)=∅.Proof.
(a)⇒(c) Let(a,b)∈𝒟witha>0. There exists a functionalx*>0satisfyingx*b=0. Fix an arbitrary elementcenjoyingx*c>0. We claim thatcis required; that is, for some numberϵ>0the relation (9) holds. To see this, proceeding by contradiction, we find sequencesbn∈B,λn∈ℝ, andkn∈K, such that(1/n)bn+λnb=c+kn≥c; hence,x*c≤(1/n)x*bn+λnx*b=(1/n)x*bn→0asn→∞, which is impossible.
(c)⇒(a) Again let(a,b)∈𝒟witha>0. Pick a numberϵ>0and an elementc≠0satisfying (9). The sets⋃λ∈ℝ(ϵB+λb)andc+Kare convex, and the interior of the first one is nonempty. By the classical separation theorem, there exists a nonzero functionalx*∈Z*, such thatx*(ϵb′+λb)≤x*c+x*kfor allb′∈B,λ∈ℝ, andk∈K. It is easy to see thatx*>0andx*b=0, as desired. We also remark thatϵx*b′≤x*c+x*kfor allb′∈Bandk∈K; hence,0<ϵ∥x*∥≤x*c+x*kfor allk∈K. Thus,x*c≠0.
The check of the equivalence (b)⇔(c) is trivial.The validity of the following theorem can be established by analogy with the preceding one. Nevertheless, in the proof given below another approach is suggested.Theorem9′. For an ordered Banach space  Z  with a cone  K  and a set  𝒟⊆K×K, the following statements are equivalent:(a)
K  has the  𝒟-property;(b)
if a pair  (a,b)∈𝒟  and there exist two sequences  cn∈Z  converging to zero and  λn∈ℝ  which satisfy the inequality  a-λnb≤cn  for all  n, then  a=0;(b′)
for every pair  (a,b)∈𝒟, there exists a number  m>0, such that the inequality  a-λb≤c  with  λ∈ℝ  and  c∈Z  implies  ∥a∥≤m∥c∥;(c)
for every pair  (a,b)∈𝒟  with  a>0, there exists a number  ϵ>0  satisfying the relation(10)(⋃λ∈ℝ(ϵB+λb))∩(a+K)=∅.Proof.
(a)⇒(b) Leta-λnb≤cnwithcn→0asn→∞andλn∈ℝ, and leta>0. Then, for a functionalx*≥0satisfyingx*a>0andx*b=0, we havex*a=x*(a-λnb)≤x*cn→0asn→∞, which is impossible.
(b)⇒(a) Let(a,b)∈𝒟witha>0. We can also assume thatb>0. Then, the elementsaandbare linearly independent. Indeed, if not then, for some scalarλ>0, we havea=λb, and so thata-λb=0. Therefore, according to our hypothesis, we obtaina=0, a contradiction. Consider the two-dimensional subspaceZ0={αa+βb:α,β∈ℝ}ofZand define a functionalz0*onZ0via the formulaz0*(αa+βb)=α. The desired assertion will be proved if we can verify thatz0*can be extended to a positive functionalz*on the whole spaceZ. Suppose that such extension does not exist. By the Schaefer theorem [2, page 21, Exercise 2.3], the latter is equivalent to an unboundedness below ofz0*on the set(B+K)∩Z0. Consequently, we can find three sequencesbn∈Bandαn,βn∈ℝfor whichαna+βnb≥bn, andαn→-∞asn→∞. We can assume thatαn<0for alln. The inequalitya+(βn/αn)b≤(1/αn)bnis valid. Obviously,(1/αn)bn→0asn→∞; hence,a=0, which is impossible.
(b)⇒(b′) Proceeding by contradiction, for some pair(a,b)∈𝒟, we find two sequencesλn∈ℝandcn∈Zsatisfying the inequalitiesa-λnb≤cnand∥cn∥<(1/n)∥a∥. Then, the second inequality impliesa>0andcn→0asn→∞, so that from the first one, we concludea=0, which is impossible.
The check of the implications (b′)⇒(b)⇔(c) is trivial.Observe that the condition (b) of the preceding theorem implies the condition (b) of Theorem9 explicitly. Actually, assume that(a,b)∈𝒟, and for everyc≠0, there exist two sequencescnandλnfor whichc-λnb≤cn, andcn→0asn→∞. Therefore, ifa>0, then, in particular, we can find two sequencescn′andλn′for whicha-λn′b≤cn′, andcn′→0asn→∞. A glance at the condition (b) of Theorem9′guarantees thata=0, a contradiction.Next, the results of Theorems9 and9′can be applied to the case of the set  𝒟={(a,b)}  with  a,b∈K. For example, we have the following result: for elements  a,b∈K  with  a>0, there exists a functional  x*≥0  satisfying  x*a>0  and  x*b=0  if and only if for every sequence  cn  converging to zero and  λn∈ℝ, the relation  cn+λnb-a∉K  holds for sufficiently large  n.In the following example, we consider another important case of a set𝒟.Example 10.
Let the set(11)𝒟={(a,b)∈Z×Z:a∧b=0},
the set of all pairs(a,b)∈Z×Z, such that the infimuma∧bexists and is equal to zero (in other words, we consider all pairs of disjoint elements inZ). (a)
For a cone  K, the following statements are equivalent:
(i)
K  has the weak  𝒟-property;(ii)
if  z∈Z∖K, then the element  z+=z∨0,  if it exists, is not a quasi-interior point of K;(iii)
whenever  x  is a quasi-interior point of  K  and  z∈Z, the relation  z∧x=0  implies  z=0.
We will check first that (i)⇒(ii). Evidently,(z-,z+)∈𝒟, wherez-=(-z)∨0>0. Therefore, for some functionalx*>0, we havex*z+=0. For the converse, let (ii) hold and let(a,b)∈𝒟witha>0. Putz=b-a. Clearly,
(12)b=b-a∧b=b+((-a)∨(-b)),=(b-a)∨0=z∨0=z+;
in particular, the elementz+exists. On the other hand, ifz∈K, thenb≥a; hence,a=b∧a=0, which is impossible. Thus,z∉K. Consequently,z+=bis not a quasi-interior point; that is,Khas the weak𝒟-property. The equivalence (i)⇔(iii) is obvious.
Next, the following assertion holds.(b)
If the linear space  K-K  is closed, then the cone  K  has the  𝒟-property; in particular, every cone in a finite-dimensional space and every generating cone have the  𝒟-property.
Without loss of generality, we can assume thatKis generating. Let a pair(a,b)∈𝒟. First of all, we mention that if for an elementu∈Kthe inequalitya-λb≤uholds withλ∈ℝ, thena≤u. Indeed, sinceu≥a-|λ|b, we can assume thatλ=1. Using the inequalitiesa≥a-uandb≥a-u, we have0≥a-u. Consequently,u≥a. Next, suppose that for two sequencescn∈Zconverging to zero andλn∈ℝ, the inequalitya-λnb≤cnholds. Since the coneKis generating, we find two sequencesun,vn∈Kfor whichcn=un-vnandun,vn→0asn→∞. Obviously,un≥a-λnb; hence, from the above, we concludeun≥a. Therefore,a=0. By the part (b) of the preceding theorem,Khas the𝒟-property.
From the parts (a) and (b), we at once obtain the following assertion (for the case of a generating and minihedral coneK, the validity of this assertion was mentioned in [2, page 112]).(c)
If a cone  K  is generating and an element  z∈Z∖K, then the element  z+=z∨0, if it exists, is not a quasi-interior point of  K.
Nevertheless, it is important to observe that not every cone has the weak𝒟-property, and, as a consequence, not every cone has the𝒟-property (see Examples 12 and12′).We now turn our attention to the strong𝒟-property. A proof of the next theorem is analogous to proofs of Theorems 9 and9′and will be omitted.Theorem9''. For an ordered Banach space  Z  with a cone  K  and a set  𝒟⊆K×K, the following statements are equivalent:(a)
K  has the strong  𝒟-property;(b)
if a pair  (a,b)∈𝒟, and an element  c∈Z  and a number  λ∈ℝ  satisfy the inequality  a-λb≤c, then  ∥a∥≤∥c∥;(c)
for every pair  (a,b)∈𝒟, with  a>0, the relation(13)(⋃λ∈ℝ‍(B<∥a∥+λb))∩(a+K)=∅holds, where  B<∥a∥is the open ball centered at zero with radius  ∥a∥.Example 11.
LetZbe an arbitrary ordered Banach space with a coneK, and let𝒟be a set, such thatK×{0}⊆𝒟⊆K×K.(a)
IfKhas the strong𝒟-property, then the norm ofZis monotone; in particular,Kis normal.
To see this, let0≤a≤c. Since(a,0)∈𝒟, from the part (b) of the preceding theorem, we infer∥a∥≤∥c∥, which means the monotonicity of the norm (see also the part (c) below).
Now, let𝒟defined by (11) be the set. First of all, it should be noticed that even in this case the monotonicity of the norm of a spaceZdoes not imply the strong𝒟-property. For instance, letZ=C[0,1]with the sup norm, and letKbe a cone of all nonnegative and increasing functions ofZ. Clearly, for a functionz∈K, the identity∥z∥=z(1)holds. Therefore, if0≤Ky<Kz, then∥y∥<∥z∥; that is, the norm ofZis strictly monotone. However, we claim that the cone  K  does not have the strong  𝒟-property (a more careful analysis shows thatKdoes not even have the weak𝒟-property (see Examples 12 and12′), but we present here a direct proof thatKdoes not have the strong𝒟-property). To see this, we notice that for every functionz∈Kwithz(0)=0, the pair(z,e)∈𝒟, whereeis the constant function one on[0,1]. Now, fixz∈Kwithz(0)=0andz(1)>0, and suppose thatKhas the strong𝒟-property. Then, there exists a functionalx*∈K*satisfying
(14)∥x*∥=1,x*z=∥z∥,x*e=0.
It is well known thatx*can be written as a difference of two functionalsx1*andx2*, such thatx1*andx2*are positive, andx1*⊥x2*under the natural ordering of the spaceC[0,1]and the dual of this space. Then, using the first identity in (14), we have
(15)1=∥x*∥=∥x1*-x2*∥=∥|x1*-x2*|∥,=∥x1*+x2*∥=∥x1*∥+∥x2*∥,
and using the third one, we have
(16)0=x*e=x1*e-x2*e=∥x1*∥-∥x2*∥,
whence∥x1*∥=∥x2*∥=1/2. Therefore,x*z=x1*z-x2*z≤x1*z≤∥z∥/2, which contradicts the second identity in (14) (the conditionx*∈K*was not used).
In some more generality, the next assertion holds:if  E  is an  AM-space with a unit  e  and  K  is a cone in  E, such that  K⊆E+,  e∈K, and for some nonzero element  z∈K, the infimumz∧eexists under the ordering induced byKand is equal to zero, thenKdoes not have the strong𝒟-property.
Nevertheless, it is well known that an arbitrary coneK  in some Banach space  Z  is normal if and only if the space  Z  admits an equivalent monotone norm. It is not known whenever, for a normal cone  K, we can assert the existence of an equivalent norm under which  K  has the strong  𝒟-property.
As the following assertion shows, under some additional assumptions, the monotonicity of the norm implies the strong𝒟-property.(b)
Suppose that the norms of Banach spaces  Z  and  Z*  are monotone, and a cone  K  is minihedral. Then,  K  has the strong  𝒟-property. In particular, if  E  is a Banach lattice, then the cone  E+  has the strong  𝒟-property.
Before the check of the assertion (b), we will establish the next fact.(c)
For everyz∈K, there exists a functionalx*∈K*satisfying∥x*∥=1  andx*z=∥z∥  if and only if the norm of  Z  is monotone.
Indeed, if0≤z≤yand a functionalx*∈K*satisfies∥x*∥=1andx*z=∥z∥, then∥z∥=x*z≤x*y≤∥y∥. For the converse, for a nonzero elementz∈K, we consider the one-dimensional subspaceZ0={αz:α∈ℝ}ofZand define a functionalz0*onZ0via the formulaz0*(αz)=α∥z∥. Let us show thatz0*can be extended to a positive functionalz*on the whole spaceZwithout increasing the norm. By the Schaefer theorem [2, page 21, Exercise 2.3], we must show that for every elementy∈(B+K)∩Z0, the inequalityz0*y≥-1holds. To this end, proceeding by contradiction, we find a scalarα<-1/∥z∥and an elementb∈Bsatisfyingαz≥b, so that-b≥(-α)z≥0. Consequently,1≥∥b∥≥|α|∥z∥>1must be the case, which is absurd.
Now, we are in a position to prove the assertion (b). Since the norm ofZ*is monotone,K*is a normal cone. Therefore, by Andô theorem [2, page 40, Theorem4.6], the coneKis generating; hence,Zis a Riesz space. Let a pair(a,b)∈𝒟; that is,a,b∈Kanda∧b=0. Using the assertion (c), we findx*∈K*satisfying∥x*∥=1andx*a=∥a∥. Next, there exists [1, page 17] a functionaly*, such that0≤y*≤x*,y*a=x*a, andy*b=0. Using the monotonicity of the norm ofZ*once more, we infer∥y*∥=1, as required.LetZbe an arbitrary ordered Banach space with a cone  K, and let a set𝒟⊆K×K. A cone  K  is said to have the  𝒟𝒟-property if for every pair  (a,b)∈𝒟, there exists a functional  x*>0  satisfying  x*a·x*b=0. It is clear that a cone  K  has the  𝒟𝒟-property if and only if the relation  (a,b)∈𝒟  implies that either  a  or  b  is not a quasi-interior point of  K. Moreover, if  K  has the weak  𝒟-property, then  K  has the  𝒟𝒟-property.Nevertheless, the next example shows that a coneKcannot have the𝒟𝒟-property, even when the set𝒟is defined by (11).Example 12.
Consider the setY=C0[0,1]of all continuous functionsyon the segment[0,1]withy(0)=0which is an ordered Banach space equipped with a coneK0of all nonnegative and increasing functions ofYand under the sup norm. Obviously,Yis a closed subspace ofC[0,1]having codimension one. It is well known that the dual spaceC*[0,1]can be identified with the spaceca([0,1])of all (signed,σ-additive) measures (of bounded variation) defined on theσ-algebra of all Borel sets of[0,1]. From this, it can be easily obtained that the dual spaceC0*[0,1]can be identified with the space{μ∈ca([0,1]):μ({0})=0}; that is, in other words, for every functionalx*∈C0*[0,1], there exists a unique measureμx*∈ca([0,1])satisfyingμx*({0})=0andx*y=∫01‍ydμx*for ally∈C0[0,1]; moreover, this representation preserves the natural order.
In the following assertion the description of the dual coneK0*is given.(a)
The equality(17)K0*={x*∈C0*[0,1]:μx*([a,1])≥0∀a∈[0,1]}holds.
To see this, letx*∈K0*anda∈(0,1]. Fixϵ∈(0,a]and define the function
(18)yϵ(t)={00≤t≤a-ϵ,ϵ-1(t-(a-ϵ))a-ϵ≤t≤a,1a≤t≤1.
Obviously,yϵ∈K0. Hence,∫01‍yϵdμx*=x*yϵ≥0. Lettingϵ↓0yieldsμx*([a,1])≥0. From this (remember thatμx*({0})=0), we also haveμx*([0,1])≥0. For the converse, assume that for a measureμx*defining a functionalx*, the inequalityμx*([a,1])≥0is valid for alla∈[0,1]. Lety∈K0. For an arbitrary partitionτ={τ0,τ1,…,τn},n>1, of the segment[0,1], where, of course,0=τ0<τ1<⋯<τn=1, we define a functionyτon[0,1]via the formulaeyτ(t)=y(τi)witht∈[τi,τi+1)andi=0,…,n-1andyτ(1)=y(τn-1). Since the functionycan be approximated uniformly by functions of the formyτ, it suffices to show that∫01‍yτdμx*≥0. To this end, letAi=[τi,τi+1)fori=0,…,n-2andAn-1=[τn-1,τn]. Using an Abel transformation, we have
(19)∫01‍yτdμx*=∑i=0n-1‍y(τi)μx*(Ai)=y(0)∑j=0n-1‍μx*(Aj)+∑i=1n-1‍((y(τi)-y(τi-1))∑j=in-1‍μx*(Aj))=∑i=1n-1‍(y(τi)-y(τi-1))μx*([τi,1])≥0,
as desired.
Now, we can give a characterization of quasi-interior points ofK0.(b)
A functionx∈K0  is a quasi-interior point of  K0  if and only if  x  is strictly increasing on[0,1].
Indeed, letxbe a quasi-interior point ofK0. Fix two pointst1,t2∈[0,1]witht1<t2and define a functionalx*onC0[0,1]byx*=δt2-δt1; that is,x*y=y(t2)-y(t1)for ally∈C0[0,1]. Evidently,x*∈K0*, so thatx(t2)-x(t1)=x*x>0. For the converse, letx∈C0[0,1]be a strictly increasing function, and letx*∈K0*be a nonzero functional. There exist numbersa,b∈(0,1]and a numberϵ>0, such thata<b, and the inequalityμx*([c,1])≥ϵholds for everyc∈[a,b]. Then, for an arbitrary partitionτ={τ0,τ1,…,τn}of the segment[0,1]for whichτia=aandτib=bfor some indicesia,ib=1,…,n-1, using (19), we have
(20)∫01‍xτdμx*≥∑i=ia+1ib‍(x(τi)-x(τi-1))μx*([τi,1])≥ϵ∑i=ia+1ib‍(x(τi)-x(τi-1))=ϵ(x(b)-x(a)).
Therefore,x*x=∫01‍xdμx*≥ϵ(x(b)-x(a))>0, and the assertion (b) is proven.
Next, it is well known that an arbitrary functionz∈K0-K0defines the atomless measureμz∈ca([0,1]); that is,μz({t})=0for each pointt∈[0,1], such that for alla,b∈[0,1],a<b, the relationμz((a,b])=z(b)-z(a)is valid. Conversely, every atomless measureμz∈ca([0,1])defines the functionz∈K0-K0via the formulaz(a)=μ((0,a])fora>0and, obviously,μ=μz. Forz∈K0-K0, we putRz=μz. It is easy to see thatz≥K00if and only ifμz≥0. Therefore, the vector spaceK0-K0with an order generated by the coneK0is a Riesz space, and the operatorRis a lattice isomorphism betweenK0-K0, and the band ofca([0,1])consisting from all atomless measures. From this, we conclude easily that ify,z∈K0-K0andwis an infimum of{y,z}inK0-K0, then an infimumy∧zalso exists in the ordered Banach spaceC0[0,1]with a coneK0and is equal tow; in particular, ify,z∈K0-K0andy∧z=0inK0-K0then(y,z)∈𝒟, where𝒟is defined by (11); that is,𝒟={(a,b)∈Y×Y:a∧K0b=0}. In fact, evidently,w≤K0yandw≤K0z. If for some elementv∈C0[0,1], the inequalitiesv≤K0yandv≤K0zare true, then, sincev=y-(y-v)∈K0-K0, we obtainv≤K0w, as required. An analogous statement is also valid for a supremumy∨z.
We are now ready to prove the following assertion.(c)
There exists a family  {xp}p∈[0,1]  of elements of the cone  K0, such that  xp  is a quasi-interior point of  K0  for all  p, and for any two indices  p′  and  p′′,  p′≠p′′, the relation  xp′∧xp′′=0  holds; that is, the pair  (xp′,xp′′)∈𝒟.
To see this, note first that [11, section25] there exists a family{μp}p∈[0,1]inca([0,1])of probability and atomless measures, such that for any two indicesp′andp′′,p′≠p′′, we haveμp′⊥μp′′in the Riesz spaceca([0,1]); moreover,μp((a,b])>0for alla,b∈[0,1],a<b, and any indexp(we can also assume thatμ1/2is a Lebesgue measure). Now, a required family{xp}is obtained if we putxp=R-1μp.
The next assertion at once follows from the part (c).(d)
The cone  K0  does not have the  𝒟𝒟-property; in particular,  K0  does not have the weak  𝒟-property.Example  12′. Consider the Banach spaceZ=C[0,1]equipped with the coneKof all nonnegative and increasing functions on[0,1](see Example 11) and the set𝒟defined by the relation (11). The proofs of the following assertions are analogous to the proofs of the assertions (a) and (b) of the preceding example (again we identify functionalsx*∈C*[0,1]with respective measuresμx*∈ca([0,1])).(a)
The equality(21)K*={x*∈C*[0,1]:μx*([a,1])≥0∀a∈[0,1]}holds.(b)
A function  x∈K  is a quasi-interior point of  K  if and only if  x  is strictly increasing on[0,1]and  x(0)>0.Nevertheless, from (b), we at once obtain the next result.(c)
The cone  K  has the  𝒟𝒟-property.Indeed, ifxandyare quasi-interior points ofK, thenx,y≥Kmin{x(0),y(0)}e>K0. Hence,(x,y)∉𝒟.On the other hand, as it was mentioned in Example11, the coneKdoes not have the strong𝒟-property. In reality, we can assert more.(d)
The cone  K  does not have the weak  𝒟-property.The validity of this result can be easily checked by the use of the assertion (c) of the preceding example. To see this, letxp′andxp′′be two (strictly increasing) functions from this assertion, and assume thatp′≠p′′. Clearly,xp′,xp′′∈K. We claim thatxp′∧xp′′=0, where the infimum is taken in the spaceC[0,1]with the coneK. To this end, note first that ifz≤Kxandx(0)=0, thenz-z(0)e≤Kx. In fact, since the nonnegative functionx-zis increasing, we havex(t)-z(t)≥-z(0)for allt∈[0,1], so that the increasing functionx-z+z(0)eis nonnegative. Now, assume that for some functionz∈Z, the inequalitiesz≤Kxp′andz≤Kxp′′hold. By the above, we havez-z(0)e≤Kxp′andz-z(0)e≤Kxp′′. Sincez-z(0)e∈C0[0,1], we obtainz-z(0)e≤K0. Finally,z≤K0, as required. Next, fix an arbitrary numberc>0, and putx=xp′′+ce. Then, it follows from the part (b) of this example and the part (b) of the preceding one thatxis a quasi-interior point ofK, and it is not difficult to see that(xp′,x)∈𝒟.Nevertheless, two functionsx,y∈K, such thatxis a quasi-interior point ofK,y>K0, andy∧Kx=0can be chosen more “explicitly.” Indeed, the functionx(t)=t+cwith a numberc>0and the Cantor functiony(t)satisfy the desired properties. To this end, first of all, we mention that for every functionsz1,z2∈Zand numberst1,t2∈[0,1], the inequalitiesz1≤Kz2andt1≤t2implyz1(t2)-z1(t1)≤z2(t2)-z2(t1). Next, letJn,1,…,Jn,2n-1be the intervals deleted from the segment[0,1]on thenth step of the construction of the Cantor set. PutJn,k=(an,k,bn,k), wherek=1,…,2n-1. Now, we suppose that the functionz∈Zsatisfies
(22)z≤Ky,z≤Kx.
Obviously,z(0)≤y(0)=0. Sinceyis constant on the segments[an,k,bn,k], from the remarks done above, the functionzis decreasing on these segments. Consider two pointss1ands2belonging to the closure of two different intervals of the formJn,k, and show that ifs1<s2, thenz(s1)≥z(s2). We can assume without loss of generality thats1=bn′,k′<an′′,k′′=s2. Fix a numberϵ>0. LetJi=(ai,bi)withi=1,…,mbe an arbitrary collection of intervals from the set{Jn,k}, such that the inclusionJi⊆(s1,s2)holds and∑i=1m(bi-ai)>s2-s1-ϵ. We can also assume thatbi<ai+1for alli=1,…,m-1. In view of (22), we havez(bi)-z(ai)≤0for alli=1,…,mandz(ai+1)-z(bi)≤ai+1-bifor alli=1,…,m-1; moreover,z(a1)-z(s1)≤a1-s1andz(s2)-z(bm)≤s2-bm. Hence,
(23)z(s2)-z(s1)=z(s2)-z(bm)+∑i=1m‍(z(bi)-z(ai))+∑i=1m-1‍(z(ai+1)-z(bi))+z(a1)-z(s1)≤s2-bm+∑i=1m-1‍(ai+1-bi)+a1-s1=s2-s1+∑i=1m‍(ai-bi)<ϵ.Lettingϵ↓0, we obtainz(s2)≤z(s1). Since the set⋃n=1∞⋃k=12n-1Jn,kis dense in[0,1], the last inequality implies the decrease of the functionzon this segment. Finally,z≤K0, so thaty∧Kx=0, as desired.We close this section by the following result, showing a dependence between the𝒟𝒟-property and the weak𝒟-property. Let us denote for a set𝒟⊆K×Kby𝒟Tthe collection{(a,b):(b,a)∈𝒟}.Proposition 13.
For an ordered Banach spaceZwith a coneKand a set𝒟⊆K×K, the following statements are equivalent: (a)
Khas the𝒟𝒟-property;(b)
there exists a set𝒟0⊆K×K, such that the inclusions
(24)𝒟0⊆𝒟∪𝒟T,𝒟⊆𝒟0∪𝒟0T
hold, andKhas the weak𝒟0-property;(c)
there exists a set𝒟0⊆K×K, such that the inclusion𝒟⊆𝒟0∪𝒟0Tholds, andKhas the weak𝒟0-property.Proof.
Let us verify first that (a)⇒(b). Define the set𝒟0by the collection of all pairs(a,b)such that either(a,b)∈𝒟andbis not a quasi-interior point ofKor(b,a)∈𝒟and againbis not a quasi-interior point ofK. Clearly, the first inclusion in (24) holds. Next, the second one is also valid asKhas the𝒟𝒟-property. Moreover, from the construction of𝒟0, it follows immediately that if(a,b)∈𝒟0, thenbis not a quasi-interior point ofK. Thus,Khas the weak𝒟0-property. The implication (b)⇒(c) is obvious. For the check of (c)⇒(a), let(a,b)∈𝒟. We can assume thata,b≠0. Then, either(a,b)∈𝒟0so thatbis not a quasi-interior point ofKor(b,a)∈𝒟0so thatais not a quasi-interior point ofK.We mention also that, as it is easy to see from the proof, in the preceding proposition under a more punctual choice of a set𝒟0, the inclusion𝒟⊆𝒟0∪𝒟0Tcan be replaced by the inclusion𝒟∖(({0}×K)∪(K×{0}))⊆𝒟0∪𝒟0T.


## 5. A Sight from the Point of View of Ordered Banach Algebras

The purpose of this section is both to supplement and to make more precise results of the second section of the paper and to present another approach to this scope of questions. This approach uses some properties of the spaceℒ(E)of all (bounded, linear) operators onEas an ordered Banach algebra.We start by recalling that a Banach algebra𝒜with a uniteand with a coneKis called (see, e.g., [12]) an ordered Banach algebra ife≥0, and the inequalitiesy,z≥0implyyz≥0. An elementb∈𝒜is called [12] an order idempotent if0≤b≤eandb2=b. An ordered Banach algebra𝒜is said to have [12] a strongly disjunctive product if for anyy,z≥0withyz=0, there exists an order idempotentbsatisfyingyb=(e-b)z=0.The algebraℒ(E)under the natural algebraic operations and ordering is an ordered Banach algebra. An elementT∈ℒ(E)is an order idempotent if and only ifTis an order projection onE. It was shown in [12] that in the case of a Dedekind complete Banach latticeE, the algebra  ℒ(E)  has a strongly disjunctive product if and only if  E  has order continuous norm. This assertion implies the next result showing that Theorem 2 also remains valid if we will consider the Dedekind complete spaces instead of spaces with the countable sup property. We also recall a well-known fact that the order continuity of a norm implies the Dedekind completeness.Theorem 14.
A Dedekind complete Banach latticeEwith a weak order unit has order continuous norm if and only if the set of all quasi-interior points ofEcoincides with the set of all weak order units ofE.Proof.
The necessity is clear (see the beginning of the proof of Theorem2). From the remarks above, it follows that for the check of the converse assertion, it suffices to verify thatℒ(E)has the strongly disjunctive product. To this end, let positive operatorsSandTonEsatisfy the identityST=0. For a weak order unitx∈E, the equality(I-PTx)Tx=0holds. Therefore,
(25)(I-PTx)T=0.
Next, the elementTx+(I-PTx)xis a weak order unit. Indeed, ifz⊥Txandz⊥(I-PTx)xthen from the first relation, we obtain
(26)PTxz=0,
and from the second one, we get|z|∧(I-PTx)x=0. Hence,
(27)0=(I-PTx)(|z|∧(I-PTx)x)=(I-PTx)(((I-PTx)|z|)∧x).
Using the equalityPTx(((I-PTx)|z|)∧x)=0, we infer((I-PTx)|z|)∧x=0. Consequently,(I-PTx)|z|=0. This and (26) givez=0. Finally,Tx+(I-PTx)xis a quasi-interior point ofE. Therefore,Txis a quasi-interior point ofBTx. On the other hand, fromS(Tx)=0, we at once getSPTx=0. The latter and (25) mean that the algebraℒ(E)has a strongly disjunctive product, as required.As Example4 shows, the assumption about the Dedekind completeness in the preceding theorem is essential. Moreover, the next result holds.Corollary 15.
LetEbe a Dedekind completeAM-space with a unit. Then, the set of all quasi-interior points ofEcoincides with the set of all weak order units ofEif and only if the spaceEis finite-dimensional.In the case of an ordered Banach space, Theorem14 has the next form.Theorem14'. Let  Z  be an ordered Banach space with a generating and strongly minihedral cone  K. Moreover, let  Z  have a weak order unit. Then,  K  is regular if and only if the set of all quasi-interior points of  K  coincides with the set of all weak order units of  Z.



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