International Journal of Mathematics and Mathematical Sciences

International Journal of Mathematics and Mathematical Sciences / 2013 / Article

Research Article | Open Access

Volume 2013 |Article ID 507482 | https://doi.org/10.1155/2013/507482

Egor A. Alekhno, "Characterization of Banach Lattices in Terms of Quasi-Interior Points", International Journal of Mathematics and Mathematical Sciences, vol. 2013, Article ID 507482, 11 pages, 2013. https://doi.org/10.1155/2013/507482

Characterization of Banach Lattices in Terms of Quasi-Interior Points

Academic Editor: Gideon Schechtman
Received01 Apr 2013
Revised26 Aug 2013
Accepted11 Sep 2013
Published27 Nov 2013

Abstract

In terms of quasi-interior points, criteria that a Banach lattice has order continuous norm or is an -space with a unit are given. For example, if is Dedekind complete and has a weak order unit, then has order continuous norm if and only if the set of quasi-interior points of coincides with the set of weak order units of ; a Banach lattice is an -space with a unit if and only if the set of all quasi-interior points of coincides with the set . Analogous questions are considered for the case of an ordered Banach space with a cone . Moreover, it is shown that every nonzero point of a cone is quasi-interior if and only if . We also study various -properties of a cone ; in particular, the conditions for which the relation with implies that is not a quasi-interior point are considered.

1. Introduction

One of natural notions from the theory of Riesz spaces is a notion of a weak order unit: an elementin a Riesz spaceis said to be [1, page 36] a weak order unit whenever for each, the relationimplies(or, equivalently, wheneverfor each). A topological analog of a weak order unit is a quasi-interior point. Namely, an elementin a Banach latticeis said to be [1, page 259] a quasi-interior point, whenever for each, the sequencenorm converges to. This is equivalent to the assertion thatfor each nonzero positive functional. In the case of an ordered Banach spacewith a cone, the last assertion is taken up as a definition of a quasi-interior point of(see [2, page 24]).

The main idea of using quasi-interior points is the following. If an elementof a Banach latticeis quasi-interior, then, on the one hand, the (order) idealgenerated byis norm dense in; on the other hand, the idealunder the normis an-space with a unitand so that, by Kakutani-Bohnenblust-Kreins' theorem [1, page 194],is lattice isometric onto a spaceof all continuous functions on some compact topological space. Thus, it is often possible to reduce the investigation of number properties of a Banach latticeto the study of these properties, or closed to them, in the space of continuous functions, and then, sinceis dense in, we can return to the space. In other words, from all Banach lattices, Banach lattices having quasi-interior points are the “closest” to-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 latticewith help of quasi-interior points were not studied. For example, conditions in terms of quasi-interior points guaranteeing that a Banach latticeis an-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 cone of some Banach space , , such that every nonzero element is a quasi-interior point of , 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 and-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 various-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,will stand for an arbitrary Banach lattice, whilewill be an arbitrary ordered Banach space with a (closed) cone. We assume thator.

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. Letbe a Banach lattice with the countable sup property and letbe a closed ideal in, such that. Ifhas a weak order unit, then idealalso has a weak order unit.

Proof. From the relation, it follows that the idealis order dense in. Consequently, there exists a sequencesatisfying. Then, the elementis a weak order unit (in). Indeed, ifwith, thenfor all. Since, we obtainor.

In order to proceed further, we recall the well-known fact that every Banach latticewith order continuous norm has the countable sup property.

Theorem 2. A Banach latticewith 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 ofcoincides with the set of all weak order units of.

Proof. The necessity is valid without the assumption thathas the countable sup property and can be found, for example, in [1, page 260]. We will prove the sufficiency. Assume by way of contradiction thatdoes not have order continuous norm. This is equivalent to the fact that the bandof all order continuous functionals does not coincide with the norm dualof. Therefore, there exists a functionalsatisfying the relation, where. Next,is a closed ideal; hence, by Lemma 1,has a weak order unit. By our hypothesis, the elementis a quasi-interior point. However,, which is impossible.

The preceding theorem is also valid if we replace the property thathas the countable sup property by the Dedekind completeness of(see Theorem 14 later). Moreover, Theorem 2 can be obtained with help of the next Andô theorem [1, page 173]: a Banach lattice   has order continuous norm if and only if every closed ideal is a band.

The following examples show that, in Lemma 1 and Theorem 2, the condition thathas the countable sup property is essential.

Example 3. Consider the Banach latticeof all bounded functionswith the norm, and letbe the closed ideal inconsisting from all functions which have countable supports. Thenanddoes not have a weak order unit.

Example 4. Letbe the space from the preceding example and letbe the constant function one on. Through, we denote a closed subspace ofconsisting from all functions, such that for everythe setis finite. Consider the space It is closed in. Actually, if a sequencewith,, and, thenis a bounded sequence. Consequently, we can assume thatconverges to some. In this case,. Next,is also closed under the lattice operations. Thus,is a Banach lattice. Nevertheless,does not have order continuous norm, and, on the other hand, the set of all quasi-interior points ofcoincides with the set of all weak order units of. We mention thatis an-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 -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 spacehas [1, page 35] the principal projection property if for every elementthe bandgenerated byis a projection band; that is,or, in other words, there exists an order projectionfrom the spaceonto.

Theorem 5. For a Banach latticeand an element, the following two statements are equivalent: (a)is an-space with a unit;(b)the set of all quasi-interior points ofcoincides with the set of all elements, such thatfor some.Moreover, in casehas the principal projection property, each from the conditions (a) and (b) is equivalent to the next: (c)the relationholds.

Proof. First of all, we recall that an elementis a unit of some Banach latticeif and only ifis an interior point of. On the other hand, if in some ordered Banach spacewith a conethe interiorofis nonempty, then it is well known (see, e.g., [2, page 22]) thatcoincides with the set of all quasi-interior points of. Now, the implication (a)(b) is obvious. For a proof of (a)(c), it suffices to observe that the normsandare equivalent, and the identitieshold. Indeed, if, thenfor some nonnegative scalar; hence,or. Thus,, which is impossible (in the next section, this result will be stated for the case of an ordered Banach space (Theorem), 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 thatis not an interior point of. Then, there exists a sequencesatisfyingand. The order projectionsare nonzero. Clearly,. Hence, using the inequality, we infer. The latter is contrary to (c).
(b)(a) Again suppose by way of contradiction thatis not an interior point of the set. This set is ideally convex; therefore [2, pages 10, 11],is not an interior point ofif and only ifis not-interior point of. Consequently, there exists an elementsatisfyingfor all. Since, we can assume that. Put. Clearly,and(norm) converge to. Define the element. We will establish thatis a quasi-interior point. To this end, proceeding by contradiction, we find a functionalsatisfying. Therefore,for all. Whence, using the relation we get, which is impossible. By the condition (b), for some natural number, the inequalityholds. Taking into account the relationfor all, we have Consequently,; hence,. Finally,or, and we have a contradiction.

For the case of Banach function spaces, Theorems 2 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 (Theoremsand, resp.).

We will close this section with the following remarks. As a matter of fact, in the proof of Theorem 5, the validity of the next assertion was shown: if   is a quasi-interior point of a Banach lattice , a sequence converges to  , and a sequence  , such that    for all  , then the element  is quasi-interior. Clearly, if  , then  . The converse statement is also true: if an element    and an element    are quasi-interior points of  , then there exist an increasing sequence    converging to    and a sequence    satisfying    for all  , such that  . In fact, we can assume that  . For natural numbers  , we put    and  . Then,  . Define    and  . It remains to observe the validity of the relations    and  .

An analogous result holds for the case of weak order units in a Riesz spacesatisfying Axiom (OS) (recall that [5, page 54] an ordered linear spacesatisfies Axiom (OS) if the inequalities, whereand, imply the existence of): if    is a weak order unit in , then   is also a weak order unit in    if and only if there exist sequences   and   with   for all , such that . Obviously, every-Dedekind complete ordered linear spaceand, in particular, every ordered Banach spacewith a-minihedral cone(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 Theorems 2 and 5 for the case of ordered Banach spaces.

We will show first that if a Banach spacewith a coneis not one-dimensional, then there exists a nonzero elementofwhich is not a quasi-interior point of(see also [3, Theorem] and [8, Proposition]). We need to recall the following result obtained in [9] (see also [10]). Let   be a closed subset of a Banach space , and let . Let . Then, there exists a point , such that   and . Here, as usual,is the distance function,is the convex hull of the set, andis the closed ball centered at the pointwith radius(below the ballwill be simply denoted by).

Theorem 6. Letbe a Banach space, and letbe a cone in. Then, the set of quasi-interior points ofcoincides with the set of nonzero elements ofif and only if the spaceis one-dimensional.

Proof. The sufficiency is obvious. We will prove the necessity. Assume by way of contradiction that the dimensional. We will show first thathas a nonzero boundary point. To this end, we find two linearly independent elementsand. Let Clearly, the nonzero pointis a boundary point of.
We can suppose that. Choose an elementwhich does not belong to the cone, such that. Then,. Fix arbitrary numbersandare satisfying the inequalities. There exists an elementwith the propertiesand, where the set. The element. By the classical separation theorem, there exists a nonzero functional, such thatfor alland. Clearly,. Moreover,for all; hence,. Therefore, the elementis not a quasi-interior point.

Recall that a conein a Banach spaceis 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) if. Obviously, every solid cone is generating. A coneis called [2, page 57] minihedral if every finite order bounded from above subsethas a supremum and strongly minihedral if every order bounded from above subsethas a supremum. If an arbitrary countable order bounded from above subsethas a supremum, thenis called-minihedral. A coneis called [2, page 37] normal if for everythe order intervalis norm bounded. It can be shown that [6, page 109, Exercise 12]is normal if and only if the spacesatisfies Axiom (OS); in particular, every-minihedral cone is normal. If inevery increasing and order bounded above sequenceconverges in norm to some element, then a coneis called [2, page 47] regular; in this case, it is easy to see that. Every regular cone [2, page 48] is also normal.

An arbitrary ordered Banach spacewith a coneis a Riesz space (of course, under the partial ordering induced by) if and only ifis generating and minihedral. In this case, all notions from the theory of Riesz spaces can be used for the ordered Banach space, in particular, the notion of weak order unit. Moreover, as in the case of a Banach lattice, every quasi-interior pointof the coneis a weak order unit. Indeed, ifwith, then whence. Therefore,is a quasi-interior point of. Consequently ([2, page 112]; see also Example 10 (c) below),, so thator.

The following result is an analogue of Theorem 2 for the case of a space with a cone.

Theorem  2′. Let   be an ordered Banach space with a generating, minihedral, and normal cone . Moreover, let   have a weak order unit and have the countable sup property. Then   is regular if and only if the set of all quasi-interior points of   coincides with the set of all weak order units of .

Proof. First of all, we mention that [5, page 145, Exercise 13(a)] the coneis generating, minihedral, and normal if and only if there exists an equivalent norm on the space, such that, under this norm, is a Banach lattice. Now, it remains to remember that an arbitrary Banach latticehas order continuous norm if and only if the coneis 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 coneenjoys 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 spacewith the coneis 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 coneis unnecessary ifis-minihedral. Moreover, the necessity also holds without the explicit assumption about the normality of. The next example shows that for the validity of the sufficiency, this assumption is essential.

Example 8. Consider the spaceof all real sequencesconverging to zero, such that the normis finite. Under the natural ordering, that is, for elements, we putiffor all,is a Riesz space; in particular, the coneis generating and minihedral. Indeed, for, defining the sequenceby, we have Thus,. Now, the identityis obvious. Moreover, the latter means thatis a Riesz subspace of the spaceof all real sequences; in particular, a sequenceis a weak order unit if and only iffor all. Next, it is easy to see that the coneis closed; hence,is an ordered Banach space. We assert that every weak order unitis a quasi-interior point of. Indeed, letfor some nonzero positive functional. The relationshold; hence,for all, whereis theunit coordinate vector. Consider an arbitrary elementand define. Clearly,. On the other hand, as, whence. Sinceis arbitrary, we get, which is impossible. Finally, in, the set of quasi-interior points coincides with the set of weak order units. Nevertheless, the coneis not normal and, in particular, is not regular. For this, we consider the sequenceinand the elementand remark thatandas. This establishes the validity of our last claim.
We also do the following remark. If we consider the spaceof all (convergent) real sequences, such that the normis finite, then under the natural ordering,is also an ordered Banach space and, simultaneously, a Riesz space. Moreover,is a Riesz subspace of. Evidently, if, then the inequalitieshold; in particular, the coneis not also normal. Nevertheless, the coneis solid. Indeed, it suffices to show that if, then. For this, if, then for arbitrary number, we have hence,, so that.

In the case of an ordered Banach space, Theorem 5 has the next form.

Theorem  . For an ordered Banach space   with a generating and minihedral cone   and an element , the following two statements are equivalent: (a)  is a solid cone with an interior point  ; (b)the set of all quasi-interior points of    coincides with the set of all elements  , such that    for some  . Moreover, in case    is normal and    has the principal projection property, each from the conditions (a) and (b) is equivalent to the next: (c)the relation    holds.

Proof. As it follows from the remarks done in the proof of Theoremonce more, ifis generating, minihedral, and normal, then the required assertion is a consequence of Theorem 5. Nevertheless, since the statement (a) implies thatcoincides 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 cone. Indeed, let the statement (a) hold. Assume that for some sequencewithfor all, we haveas. For some sufficiently large, the elementis an interior point of the cone. Put. Obviously,. Therefore ([2, page 112]; see also Example 10 (c)),is 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 conewas only used for the conclusion of the convergence(see (2)), that is, in the proof of the fact thatis a quasi-interior point (we use the notations from the proof of Theorem 5). Again, iffor some, thenfor all, whence. Consequently,, which is impossible. Finally,is 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 cone. Moreover, the author does not know any example of an ordered Banach spacewith a cone, such thatis a Riesz space with the principal projection property, whileis not normal. Nevertheless, if the coneis-minihedral, then the implication (c)(a) remains valid without the assumption about the normality of.

4. Various  -Properties of a Cone

Letbe an arbitrary ordered Banach space with a cone, and let a set. A coneis said to have the weak -property if for every pairwiththere exists a functionalsatisfying. We mention at once the following simple result:  has the weak -property if and only if the relations   and    imply that    is not a quasi-interior point of  . A cone    is said to have the-property if for every pair    with    there exists a functional    satisfying    and  . Observe that if    has the-property, then for every pair    with  , the elements    and    are linearly independent. Next, a cone    is said to have the strong  -property if for every pair    there exists a functional    satisfying  ,  , and  . Obviously, the strong  -property implies the  -property and the  -property implies the weak  -property.

The following result characterizes cones having the weak-property.

Theorem 9. For an ordered Banach spacewith a coneand a set, the following statements are equivalent: (a)has the weak-property;(b)if a pair, and for every nonzero element, there exist two sequencesconverging to zero andwhich satisfy the inequalityfor all, then;(c)for every pairwiththere exist a numberand an elementsatisfying the relation

Proof. (a)(c) Letwith. There exists a functionalsatisfying. Fix an arbitrary elementenjoying. We claim thatis required; that is, for some numberthe relation (9) holds. To see this, proceeding by contradiction, we find sequences,, and, such that; hence,as, which is impossible.
(c)(a) Again letwith. Pick a numberand an elementsatisfying (9). The setsandare convex, and the interior of the first one is nonempty. By the classical separation theorem, there exists a nonzero functional, such thatfor all,, and. It is easy to see thatand, as desired. We also remark thatfor alland; hence,for all. Thus,.
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.

Theorem  . For an ordered Banach space    with a cone    and a set  , the following statements are equivalent: (a)  has the  -property; (b)if a pair    and there exist two sequences    converging to zero and    which satisfy the inequality    for all  , then  ; (b′)for every pair  , there exists a number  , such that the inequality    with    and    implies  ; (c)for every pair    with  , there exists a number    satisfying the relation

Proof. (a)(b) Letwithasand, and let. Then, for a functionalsatisfyingand, we haveas, which is impossible.
(b)(a) Letwith. We can also assume that. Then, the elementsandare linearly independent. Indeed, if not then, for some scalar, we have, and so that. Therefore, according to our hypothesis, we obtain, a contradiction. Consider the two-dimensional subspaceofand define a functionalonvia the formula. The desired assertion will be proved if we can verify thatcan be extended to a positive functionalon the whole space. 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 ofon the set. Consequently, we can find three sequencesandfor which, andas. We can assume thatfor all. The inequalityis valid. Obviously,as; hence,, which is impossible.
(b)() Proceeding by contradiction, for some pair, we find two sequencesandsatisfying the inequalitiesand. Then, the second inequality impliesandas, so that from the first one, we conclude, which is impossible.
The check of the implications ()(b)(c) is trivial.

Observe that the condition (b) of the preceding theorem implies the condition (b) of Theorem 9 explicitly. Actually, assume that, and for every, there exist two sequencesandfor which, andas. Therefore, if, then, in particular, we can find two sequencesandfor which, andas. A glance at the condition (b) of Theoremguarantees that, a contradiction.

Next, the results of Theorems 9 andcan be applied to the case of the set    with  . For example, we have the following result: for elements    with  , there exists a functional    satisfying    and    if and only if for every sequence    converging to zero and  , the relation    holds for sufficiently large  .

In the following example, we consider another important case of a set.

Example 10. Let the set the set of all pairs, such that the infimumexists and is equal to zero (in other words, we consider all pairs of disjoint elements in). (a)For a cone  , the following statements are equivalent: (i)  has the weak  -property; (ii)if  , then the element  ,  if it exists, is not a quasi-interior point of; (iii)whenever    is a quasi-interior point of    and  , the relation    implies  .
We will check first that (i)(ii). Evidently,, where. Therefore, for some functional, we have. For the converse, let (ii) hold and letwith. Put. Clearly, in particular, the elementexists. On the other hand, if, then; hence,, which is impossible. Thus,. Consequently,is not a quasi-interior point; that is,has the weak-property. The equivalence (i)(iii) is obvious.
Next, the following assertion holds.(b)If the linear space    is closed, then the cone    has the  -property; in particular, every cone in a finite-dimensional space and every generating cone have the  -property.
Without loss of generality, we can assume thatis generating. Let a pair. First of all, we mention that if for an elementthe inequalityholds with, then. Indeed, since, we can assume that. Using the inequalitiesand, we have. Consequently,. Next, suppose that for two sequencesconverging to zero and, the inequalityholds. Since the coneis generating, we find two sequencesfor whichandas. Obviously,; hence, from the above, we conclude. Therefore,. By the part (b) of the preceding theorem,has the-property.
From the parts (a) and (b), we at once obtain the following assertion (for the case of a generating and minihedral cone, the validity of this assertion was mentioned in [2, page 112]).(c)If a cone    is generating and an element  , then the element  , if it exists, is not a quasi-interior point of  .
Nevertheless, it is important to observe that not every cone has the weak-property, and, as a consequence, not every cone has the-property (see Examples 12 and).

We now turn our attention to the strong-property. A proof of the next theorem is analogous to proofs of Theorems 9 andand will be omitted.

Theorem  . For an ordered Banach space    with a cone    and a set  , the following statements are equivalent: (a)  has the strong  -property; (b)if a pair  , and an element    and a number    satisfy the inequality  , then  ; (c)for every pair  , with  , the relationholds, where  is the open ball centered at zero with radius  .

Example 11. Letbe an arbitrary ordered Banach space with a cone, and letbe a set, such that.(a)Ifhas the strong-property, then the norm ofis monotone; in particular,is normal.
To see this, let. Since, from the part (b) of the preceding theorem, we infer, which means the monotonicity of the norm (see also the part (c) below).
Now, letdefined by (11) be the set. First of all, it should be noticed that even in this case the monotonicity of the norm of a spacedoes not imply the strong-property. For instance, letwith the sup norm, and letbe a cone of all nonnegative and increasing functions of. Clearly, for a function, the identityholds. Therefore, if, then; that is, the norm ofis strictly monotone. However, we claim that the cone    does not have the strong  -property (a more careful analysis shows thatdoes not even have the weak-property (see Examples 12 and), but we present here a direct proof thatdoes not have the strong-property). To see this, we notice that for every functionwith, the pair, whereis the constant function one on. Now, fixwithand, and suppose thathas the strong-property. Then, there exists a functionalsatisfying It is well known thatcan be written as a difference of two functionalsand, such thatandare positive, andunder the natural ordering of the spaceand the dual of this space. Then, using the first identity in (14), we have and using the third one, we have whence. Therefore,, which contradicts the second identity in (14) (the conditionwas not used).
In some more generality, the next assertion holds: if    is an  -space with a unit    and    is a cone in  , such that  ,  , and for some nonzero element  , the infimum exists under the ordering induced by and is equal to zero, then does not have the strong -property.
Nevertheless, it is well known that an arbitrary cone    in some Banach space    is normal if and only if the space    admits an equivalent monotone norm. It is not known whenever, for a normal cone  , we can assert the existence of an equivalent norm under which    has the strong  -property.
As the following assertion shows, under some additional assumptions, the monotonicity of the norm implies the strong-property.(b)Suppose that the norms of Banach spaces    and    are monotone, and a cone    is minihedral. Then,    has the strong  -property. In particular, if    is a Banach lattice, then the cone    has the strong  -property.
Before the check of the assertion (b), we will establish the next fact.(c)For every , there exists a functional satisfying   and   if and only if the norm of    is monotone.
Indeed, ifand a functionalsatisfiesand, then. For the converse, for a nonzero element, we consider the one-dimensional subspaceofand define a functionalonvia the formula. Let us show thatcan be extended to a positive functionalon the whole spacewithout increasing the norm. By the Schaefer theorem [2, page 21, Exercise 2.3], we must show that for every element, the inequalityholds. To this end, proceeding by contradiction, we find a scalarand an elementsatisfying, so that. Consequently,must be the case, which is absurd.
Now, we are in a position to prove the assertion (b). Since the norm ofis monotone,is a normal cone. Therefore, by Andô theorem [2, page 40, Theorem], the coneis generating; hence,is a Riesz space. Let a pair; that is,and. Using the assertion (c), we findsatisfyingand. Next, there exists [1, page 17] a functional, such that,, and. Using the monotonicity of the norm ofonce more, we infer, as required.

Letbe an arbitrary ordered Banach space with a cone  , and let a set. A cone    is said to have the  -property if for every pair  , there exists a functional    satisfying  . It is clear that a cone    has the  -property if and only if the relation    implies that either    or    is not a quasi-interior point of  . Moreover, if    has the weak  -property, then    has the  -property.

Nevertheless, the next example shows that a conecannot have the-property, even when the setis defined by (11).

Example 12. Consider the setof all continuous functionson the segmentwithwhich is an ordered Banach space equipped with a coneof all nonnegative and increasing functions ofand under the sup norm. Obviously,is a closed subspace ofhaving codimension one. It is well known that the dual spacecan be identified with the spaceof all (signed,-additive) measures (of bounded variation) defined on the-algebra of all Borel sets of. From this, it can be easily obtained that the dual spacecan be identified with the space; that is, in other words, for every functional, there exists a unique measuresatisfyingandfor all; moreover, this representation preserves the natural order.
In the following assertion the description of the dual coneis given.(a)The equalityholds.
To see this, letand. Fixand define the function Obviously,. Hence,. Lettingyields. From this (remember that), we also have. For the converse, assume that for a measuredefining a functional, the inequalityis valid for all. Let. For an arbitrary partition,, of the segment, where, of course,, we define a functiononvia the formulaewithandand. Since the functioncan be approximated uniformly by functions of the form, it suffices to show that. To this end, letforand. Using an Abel transformation, we have as desired.
Now, we can give a characterization of quasi-interior points of.(b)A function   is a quasi-interior point of    if and only if    is strictly increasing on .
Indeed, letbe a quasi-interior point of. Fix two pointswithand define a functionalonby; that is,for all. Evidently,, so that. For the converse, letbe a strictly increasing function, and letbe a nonzero functional. There exist numbersand a number, such that, and the inequalityholds for every. Then, for an arbitrary partitionof the segmentfor whichandfor some indices, using (19), we have Therefore,, and the assertion (b) is proven.
Next, it is well known that an arbitrary functiondefines the atomless measure; that is,for each point, such that for all,, the relationis valid. Conversely, every atomless measuredefines the functionvia the formulaforand, obviously,. For, we put. It is easy to see thatif and only if. Therefore, the vector spacewith an order generated by the coneis a Riesz space, and the operatoris a lattice isomorphism between, and the band ofconsisting from all atomless measures. From this, we conclude easily that ifandis an infimum ofin, then an infimumalso exists in the ordered Banach spacewith a coneand is equal to; in particular, ifandinthen, whereis defined by (11); that is,. In fact, evidently,and. If for some element, the inequalitiesandare true, then, since, we obtain, as required. An analogous statement is also valid for a supremum.
We are now ready to prove the following assertion.(c)There exists a family    of elements of the cone  , such that    is a quasi-interior point of    for all  , and for any two indices    and  ,  , the relation    holds; that is, the pair  .
To see this, note first that [11, section] there exists a familyinof probability and atomless measures, such that for any two indicesand,, we havein the Riesz space; moreover,<