Characterization of Banach Lattices in Terms of Quasi-Interior Points

In terms of quasi-interior points, criteria that a Banach lattice E 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 ≤ λz for some λ > 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


Introduction
One of natural notions from the theory of Riesz spaces is a notion of a weak order unit: an element  > 0 in a Riesz space  is said to be [1, page 36] a weak order unit whenever for each  ∈ , the relation  ⊥  implies  = 0 (or, equivalently, whenever  ∧  ↑  for each  ≥ 0).A topological analog of a weak order unit is a quasi-interior point.Namely, an element  > 0 in a Banach lattice  is said to be [1, page 259] a quasiinterior point, whenever for each  ≥ 0, the sequence  ∧  norm converges to .This is equivalent to the assertion that  *  > 0 for each nonzero positive functional  * ∈  * .In the case of an ordered Banach space  with 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 element  of a Banach lattice  is quasiinterior, then, on the one hand, the (order) ideal   generated by  is norm dense in ; on the other hand, the ideal   under the norm ‖‖  = min{ ≥ 0 : || ≤ } is an -space with a unit  and so that, by Kakutani-Bohnenblust-Kreins' theorem [1, page 194],   is lattice isometric onto a space () of all continuous functions on some compact topological space .Thus, it is often possible to reduce the investigation of number properties of a Banach lattice  to the study of these properties, or closed to them, in the space of continuous functions, and then, since   is 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 lattice  with help of quasi-interior points were not studied.For example, conditions in terms of quasi-interior points guaranteeing that a Banach lattice  is 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 , dim  > 1, such that every nonzero element  ∈  is a quasiinterior 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.

International Journal of Mathematics and Mathematical Sciences
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 D-properties of a cone and to study them in conjunction with the notion of quasiinterior 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, while  will be an arbitrary ordered Banach space with a (closed) cone .We assume that  ̸ = {0} or  ̸ = {0}.

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.Let  be a Banach lattice with the countable sup property and let  be a closed ideal in , such that  d = {0}.If  has a weak order unit , then ideal  also has a weak order unit.
In order to proceed further, we recall the well-known fact that every Banach lattice  with order continuous norm has the countable sup property.

Theorem 2. A Banach lattice 𝐸 with 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 of 𝐸 coincides with the set of all weak order units of 𝐸.
Proof.The necessity is valid without the assumption that  has the countable sup property and can be found, for example, in [1, page 260].We will prove the sufficiency.Assume by way of contradiction that  does not have order continuous norm.This is equivalent to the fact that the band  ∼  of all order continuous functionals does not coincide with the norm dual  * of .Therefore, there exists a functional  * > 0 satisfying the relation    * = {0}, where   * = { ∈  :  * || = 0}.Next,   * is a closed ideal; hence, by Lemma 1,   * has a weak order unit .By our hypothesis, the element  is a quasiinterior point.However,  *  = 0, which is impossible.
The preceding theorem is also valid if we replace the property that  has 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 that  has the countable sup property is essential.
Example 3. Consider the Banach lattice (R) of all bounded functions  : R → R with the norm ‖‖ = sup ∈R |()|, and let  be the closed ideal in (R) consisting from all functions which have countable supports.Then  d = {0} and  does not have a weak order unit.
Example 4. Let (R) be the space from the preceding example and let  ∈ (R) be the constant function one on R. Through  0 (R), we denote a closed subspace of (R) consisting from all functions , such that for every  > 0 the set { : |()| > } is finite.Consider the space It is closed in (R).Actually, if a sequence   +    →  with   ∈  0 (R),   ∈ R, and  ∈ (R), then   is a bounded sequence.Consequently, we can assume that   converges to some  0 .In this case,  −  0  ∈  0 (R).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 of  coincides with the set of all weak order units of .We mention that  is 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 space  has [1, page 35] the principal projection property if for every element  ∈  the band   generated by  is a projection band; that is,   ⊕    =  or, in other words, there exists an order projection   from the space  onto   .
Theorem 5.For a Banach lattice  and an element  ∈  + , the following two statements are equivalent: (a)  is an -space with a unit ; (b) the set of all quasi-interior points of  coincides with the set of all elements , such that  ≤  for some  > 0.
Moreover, in case  has the principal projection property, each from the conditions (a) and (b) is equivalent to the next: Proof.First of all, we recall that an element  is a unit of some Banach lattice  if and only if  is an interior point of  + .On the other hand, if in some ordered Banach space  with a cone  the interior ∘  of  is nonempty, then it is well known (see, e.g., [2, page 22]) that ∘  coincides 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 norms ‖ ⋅ ‖  and ‖ ⋅ ‖ are equivalent, and the identities ‖  ‖  = ‖‖  = 1 hold.Indeed, if ‖  ‖  < 1, then    ≤  for some nonnegative scalar  < 1; hence,    ≤    or    = 0. Thus,  ⊥ , which is impossible (in the next section, this result will be stated for the case of an ordered Banach space (Theorem 5  ), and in the same place another approach to a proof of the implication (a) ⇒ (c) can be found).
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 (Theorems 2  and 5  , 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  = ( 1 ,  2 , . ..) ∈ ℓ 1 , such that   ̸ = 0 for all , then the element The converse statement is also true: if an element  ∈  + and an element  0 ∈   are quasi-interior points of , then there exist an increasing sequence   ≥ 0 converging to  and a sequence  = ( 1 ,  2 , . ..) ∈ ℓ 1 satisfying   > 0 for all , such that  0 = ∑ ∞ =1     .In fact, we can assume that 0 <  0 ≤ .For natural numbers  ≥ 2, we put   =  ∧ ( 0 ) and and  1 = 1.It remains to observe the validity of the relations 0 ≤  1 ≤  0 ≤  2 and  0 = ∑ ∞ =1     .An analogous result holds for the case of weak order units in a Riesz space  satisfying Axiom (OS) (recall that [5, page 54] an ordered linear space  satisfies Axiom (OS) if the inequalities 0 ≤   ≤   , where   ,  ∈  and  = ( 1 ,  2 , . ..) ∈ ℓ 1 , imply the existence of sup  ∑  =1   ): if  is a weak order unit in , then  0 ∈   is also a weak order unit in  if and only if there exist sequences 0 ≤   ↑  and  = ( 1 ,  2 , . ..) ∈ ℓ 1 with   > 0 for all , such that ∑  =1     ↑   0 .Obviously, every -Dedekind complete ordered linear space  and, in particular, every ordered Banach space  with a -minihedral cone  (see the next section) satisfy Axiom (OS).

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 space  with a cone  is not one-dimensional, then there exists a nonzero element  of  which is not a quasi-interior point of  (see also [3,Theorem 1] and [8,Proposition 7]).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 0 <  < (, ) < .Then, there exists a point  ∈ , such that || − || ≤  and  ∩ co((, ) ∪ {}) = {}.Here, as usual, (⋅, ) is the distance function, co() is the convex hull of the set , and (, ) is the closed ball centered at the point  with radius  (below the ball (0, 1) will be simply denoted by ).Theorem 6.Let  be a Banach space, and let  be a cone in .Then, the set of quasi-interior points of  coincides with the set of nonzero elements of  if and only if the space  is onedimensional.
Recall that a cone  in a Banach space  is called [2, page 8] solid if it contains some ball of positive radius and is called (see [6, page 4]  An arbitrary ordered Banach space  with a cone  is a Riesz space (of course, under the partial ordering induced by ) if and only if  is 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 point  of the cone  is a weak order unit.Indeed, if  ⊥  with  ∈ , then 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 cone  is 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 lattice  has order continuous norm if and only if the cone  + 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 cone  enjoys 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 space  with the cone  is 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 cone  is unnecessary if  is -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.
We also do the following remark.If we consider the space V of all (convergent) real sequences , such that the norm is finite, then under the natural ordering, V is also an ordered Banach space and, simultaneously, a Riesz space.Moreover, V 0 is a Riesz subspace of V.Evidently, if  ∈ V 0 , then the inequalities ‖‖ V 0 ≤ ‖‖ V ≤ 2‖‖ V 0 hold; in particular, the cone (V) + is not also normal.Nevertheless, the cone (V) + is solid.Indeed, it suffices to show that if  = (1, 1, . ..), then (, 1) ⊂ (V) + .For this, if  ∈ (, 1), then for arbitrary number  ∈ N, we have hence,  +1 ≥ 0, so that  ≥ 0.
In the case of an ordered Banach space, Theorem 5 has the next form.Theorem 5  .For an ordered Banach space  with a generating and minihedral cone  and an element  ∈ , the following two statements are equivalent: ) 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 cone  was only used for the conclusion of the convergence (  ) + →  (see ( 2)), that is, in the proof of the fact that  0 is a quasi-interior point (we use the notations from the proof of Theorem 5).Again, if  *  0 = 0 for some  * > 0, then  * (  ) + = 0 for all , whence  *   = − * (  ) − ≤ 0. Consequently,  *  = 0, which is impossible.Finally,  0 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 space  with a cone , such that  is a Riesz space with the principal projection property, while  is not normal.Nevertheless, if the cone  is -minihedral, then the implication (c) ⇒ (a) remains valid without the assumption about the normality of .

Various D-Properties of a Cone
Let  be an arbitrary ordered Banach space with a cone , and let a set D ⊆  × .A cone  is said to have the weak Dproperty if for every pair (, ) ∈ D with  > 0 there exists a functional  * > 0 satisfying  *  = 0. We mention at once the following simple result:  has the weak D-property if and only if the relations (, ) ∈ D and  > 0 imply that  is not a quasiinterior point of .A cone  is said to have the D-property if for every pair (, ) ∈ D with  > 0 there exists a functional  * ≥ 0 satisfying  *  > 0 and  *  = 0. Observe that if  has the D-property, then for every pair (, ) ∈ D with ,  > 0, the elements  and  are linearly independent.Next, a cone  is said to have the strong D-property if for every pair (, ) ∈ D there exists a functional  * ≥ 0 satisfying ‖ * ‖ = 1,  *  = ‖‖, and  *  = 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 weak D-property.
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 9  .For an ordered Banach space  with a cone  and a set D ⊆  × , the following statements are equivalent: We can also assume that  > 0.Then, the elements  and  are linearly independent.Indeed, if not then, for some scalar  > 0, we have  = , and so that  −  = 0. Therefore, according to our hypothesis, we obtain  = 0, a contradiction.Consider the two-dimensional subspace  0 = { +  : ,  ∈ R} of  and define a functional  * 0 on  0 via the formula  * 0 ( + ) = .The desired assertion will be proved if we can verify that  * 0 can be extended to a positive functional  * on 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 of  * 0 on the set ( + ) ∩  0 .Consequently, we can find three sequences   ∈  and   ,   ∈ R for which    +    ≥   , and   → −∞ as  → ∞.We can assume that   < 0 for all .The inequality  + (  /  ) ≤ (1/  )  is valid.Obviously, (1/  )  → 0 as  → ∞; hence,  = 0, which is impossible.
Then, the second inequality implies  > 0 and   → 0 as  → ∞, so that from the first one, we conclude  = 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 Theorem 9 explicitly.Actually, assume that (, ) ∈ D, and for every  ̸ = 0, there exist two sequences   and   for which  −    ≤   , and   → 0 as  → ∞.Therefore, if  > 0, then, in particular, we can find two sequences    and    for which  −     ≤    , and    → 0 as  → ∞.A glance at the condition (b) of Theorem 9  guarantees that  = 0, a contradiction.Next, the results of Theorems 9 and 9  can be applied to the case of the set D = {(, )} with ,  ∈ .For example, we have the following result: for elements ,  ∈  with  > 0, there exists a functional  * ≥ 0 satisfying  *  > 0 and  *  = 0 if and only if for every sequence   converging to zero and   ∈ R, the relation   +    −  ∉  holds for sufficiently large .
In the following example, we consider another important case of a set D.
Example 10.Let the set the set of all pairs (, ) ∈  × , such that the infimum  ∧  exists 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 D-property; (ii) if  ∈  \ , then the element  + =  ∨ 0, if it exists, is not a quasi-interior point of ; (iii) whenever  is a quasi-interior point of  and  ∈ , the relation  ∧  = 0 implies  = 0.
Next, the following assertion holds.
(b) If the linear space  −  is closed, then the cone  has the D-property; in particular, every cone in a finitedimensional space and every generating cone have the D-property.
Without loss of generality, we can assume that  is generating.Let a pair (, ) ∈ D. First of all, we mention that if for an element  ∈  the inequality  −  ≤  holds with  ∈ R, then  ≤ .Indeed, since  ≥  − ||, we can assume that  = 1.Using the inequalities  ≥  −  and  ≥  − , we have 0 ≥  − .Consequently,  ≥ .Next, suppose that for two sequences   ∈  converging to zero and   ∈ R, the inequality  −    ≤   holds.Since the cone  is generating, we find two sequences   , V  ∈  for which   =   − V  and   , V  → 0 as  → ∞.Obviously,   ≥  −   ; hence, from the above, we conclude   ≥ .Therefore,  = 0.By the part (b) of the preceding theorem,  has the D-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 Nevertheless, it is important to observe that not every cone has the weak D-property, and, as a consequence, not every cone has the D-property (see Examples 12 and 12  ).
We now turn our attention to the strong D-property.A proof of the next theorem is analogous to proofs of Theorems 9 and 9  and will be omitted.

󸀠󸀠
. For an ordered Banach space  with a cone  and a set D ⊆  × , the following statements are equivalent: Example 11.Let  be an arbitrary ordered Banach space with a cone , and let D be a set, such that  × {0} ⊆ D ⊆  × .
(a) If  has the strong D-property, then the norm of  is monotone; in particular,  is normal.
To see this, let 0 ≤  ≤ .Since (, 0) ∈ D, from the part (b) of the preceding theorem, we infer ‖‖ ≤ ‖‖, which means the monotonicity of the norm (see also the part (c) below).Now, let D defined by (11) be the set.First of all, it should be noticed that even in this case the monotonicity of the norm of a space  does not imply the strong D-property.For instance, let  = [0, 1] with the sup norm, and let  be a cone of all nonnegative and increasing functions of .Clearly, for a function  ∈ , the identity ‖‖ = (1) holds.Therefore, if 0 ≤  <  , then ‖‖ < ‖‖; that is, the norm of  is strictly monotone.However, we claim that the cone  does not have the strong D-property (a more careful analysis shows that  does not even have the weak D-property (see Examples 12 and 12  ), but we present here a direct proof that  does not have the strong D-property).To see this, we notice that for every function  ∈  with (0) = 0, the pair (, ) ∈ D, where  is the constant function one on [0, 1].Now, fix  ∈  with (0) = 0 and (1) > 0, and suppose that  has the strong Dproperty.Then, there exists a functional  * ∈  * satisfying      *     = 1,  *  = ‖‖ ,  *  = 0.
It is well known that  * can be written as a difference of two functionals  * 1 and  * 2 , such that  * 1 and  * 2 are positive, and  * 1 ⊥  * 2 under the natural ordering of the space [0, 1] and the dual of this space.Then, using the first identity in (14), we have and using the third one, we have whence 1  ≤ ‖‖/2, which contradicts the second identity in (14) (the condition  * ∈  * was 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 D-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 D-property.
As the following assertion shows, under some additional assumptions, the monotonicity of the norm implies the strong D-property.
(b) Suppose that the norms of Banach spaces  and  * are monotone, and a cone  is minihedral.Then,  has the strong D-property.In particular, if  is a Banach lattice, then the cone  + has the strong D-property.
Before the check of the assertion (b), we will establish the next fact.
(c) For every  ∈ , there exists a functional  * ∈  * satisfying ‖ * ‖ = 1 and  *  = ‖‖ if and only if the norm of  is monotone.
Let  be an arbitrary ordered Banach space with a cone , and let a set D ⊆  × .A cone  is said to have the DDproperty if for every pair (, ) ∈ D, there exists a functional  * > 0 satisfying  *  ⋅  *  = 0.It is clear that a cone  has the DD-property if and only if the relation (, ) ∈ D implies that either  or  is not a quasi-interior point of .Moreover, if  has the weak D-property, then  has the DD-property.Nevertheless, the next example shows that a cone  cannot have the DD-property, even when the set D is defined by (11).
In the following assertion the description of the dual cone  * 0 is given.(a) The equality holds.
We mention also that, as it is easy to see from the proof, in the preceding proposition under a more punctual choice of a set D 0 , the inclusion D ⊆ D 0 ∪ D  0 can be replaced by the inclusion D \ (({0} × ) ∪ ( × {0})) ⊆ D 0 ∪ D  0 .

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 space L() of all (bounded, linear) operators on  as an ordered Banach algebra.
We start by recalling that a Banach algebra A with a unit  and with a cone  is called (see, e.g., [12]) an ordered Banach algebra if  ≥ 0, and the inequalities ,  ≥ 0 imply  ≥ 0. An element  ∈ A is called [12] an order idempotent if 0 ≤  ≤  and  2 = .An ordered Banach algebra A is said to have [12] a strongly disjunctive product if for any ,  ≥ 0 with  = 0, there exists an order idempotent  satisfying  = (−) = 0.
The algebra L() under the natural algebraic operations and ordering is an ordered Banach algebra.An element  ∈ L() is an order idempotent if and only if  is an order projection on .It was shown in [12] that in the case of a Dedekind complete Banach lattice , the algebra L() has a strongly disjunctive product if and only if  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 lattice  with a weak order unit has order continuous norm if and only if the set of all quasi-interior points of  coincides with the set of all weak order units of .
Proof.The necessity is clear (see the beginning of the proof of Theorem 2).From the remarks above, it follows that for the check of the converse assertion, it suffices to verify that L() has the strongly disjunctive product.To this end, let positive operators  and  on  satisfy the identity  = 0.For a weak order unit  ∈ , the equality ( −   ) = 0 holds.Therefore, ( −   )  = 0. (25) Next, the element +(−  ) is a weak order unit.Indeed, if  ⊥  and  ⊥ ( −   ) then from the first relation, we obtain give  = 0. Finally,  + ( −   ) is a quasi-interior point of .Therefore,  is a quasi-interior point of   .On the other hand, from () = 0, we at once get   = 0.The latter and (25) mean that the algebra L() has a strongly disjunctive product, as required.
As Example 4 shows, the assumption about the Dedekind completeness in the preceding theorem is essential.Moreover, the next result holds.
Corollary 15.Let  be a Dedekind complete -space with a unit.Then, the set of all quasi-interior points of  coincides with the set of all weak order units of  if and only if the space  is finite-dimensional.
In the case of an ordered Banach space, Theorem 14 has the next form.

󸀠
. Let  be an ordered Banach space with a generating and strongly minihedral cone .Moreover, let  have a weak order unit.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 .

( 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  > 0.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 inf z ̸ = 0 ‖  ‖ > 0 holds.Proof.As it follows from the remarks done in the proof of Theorem 2  once more, if  is generating, minihedral, and normal, then the required assertion is a consequence of Theorem 5. Nevertheless, since the statement (a) implies that ∘  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 cone .Indeed, let the statement (a) hold.Assume that for some sequence   ∈  with   ̸ = 0 for all , we have ‖   ‖ → 0 as  → ∞.For some sufficiently large  0 , the element ( −    0) is an interior point of the cone .Put  = ( −    0 ) −    0 .Obviously,  ∉ .Therefore ([2, page 112]; see also Example 10 (c)),  + = ( −    0

Theorem 9 .
For an ordered Banach space  with a cone  and a set D ⊆  × , the following statements are equivalent: (a)  has the weak D-property; (b) if a pair (, ) ∈ D, and for every nonzero element  ∈ , there exist two sequences   ∈  converging to zero and   ∈ R which satisfy the inequality  −    ≤   for all , then  = 0; (c) for every pair (, ) ∈ D with  > 0 there exist a number  > 0 and an element  ∈  satisfying the relation ( ⋃ ∈R ( + )) ∩ ( + ) = 0.
We close this section by the following result, showing a dependence between the DD-property and the weak Dproperty.Let us denote for a set D ⊆  ×  by D  the collection {(, ) : (, ) ∈ D}.Proposition 13.For an ordered Banach space  with a cone  and a set D ⊆  × , the following statements are equivalent: (a)  has the DD-property; (b) there exists a set D 0 ⊆  × , such that the inclusions D 0 ⊆ D ∪ D T , D ⊆ D 0 ∪ D  0 (24) hold, and  has the weak D 0 -property; (c) there exists a set D 0 ⊆  × , such that the inclusion D ⊆ D 0 ∪ D  0 holds, and  has the weak D 0 -property.Proof.Let us verify first that (a) ⇒ (b).Define the set D 0 by the collection of all pairs (, ) such that either (, ) ∈ D and  is not a quasi-interior point of  or (, ) ∈ D and again  is not a quasi-interior point of .