Research Article | Open Access
Kadec-Klee Properties of Calderón-Lozanovskiĭ Function Spaces
We study Kadec-Klee properties with respect to global (local) convergence in measure. First, we present some results concerning Köthe spaces and Orlicz functions. Next, we shall give full criteria for Kadec-Klee properties with respect to global (local) convergence in measure in Calderón-Lozanovskiĭ function spaces. In particular, we obtain the full characterizations of Kadec-Klee properties in Orlicz-Lorentz function spaces, which have not been presented until now.
The Kadec-Klee properties are fundamental notions in the theory of Banach function spaces (see [1, 2]). On the other hand, the Calderón-Lozanovskiĭ spaces are important class of Banach lattices, especially because of their role in the interpolation theory. Geometry of Calderón-Lozanovskiĭ spaces has been intensively developed during the last decades (see, e.g., [3–9]). The complete characterization of Kadec-Klee properties with respect to local (global ) convergence in measure for Orlicz function spaces has been presented in . The Orlicz spaces are special case of Calderón-Lozanovskiĭ spaces . Thus, it is natural to investigate properties and in the spaces generated by arbitrary Köthe spaces . Since and when , so finding full criteria of and in Orlicz spaces is simpler than in general Calderón-Lozanovskiĭ spaces. The sufficient conditions of in Calderón-Lozanovskiĭ function spaces have been proved in [4, 6]. Note that the implication follows from the fact that property implies order continuity in Köthe spaces and from results concerning copy of in (see [5, 10]). The implication has not been discussed at all. Moreover, property in has not been considered early. We present full criteria of properties and in Calderón-Lozanovskiĭ function spaces . Furthermore, in the case of property , we prove some general results concerning Köthe spaces which will be essential in studying this property in spaces . The crucial difficulty in the proofs concerning property in spaces is the fact that property needs not imply order continuity in general. Note that if , then (see ), so looking for complete characterizations in the spaces , the case of property is easier than . As an application, we get the criteria of properties and in Orlicz-Lorentz function spaces . A sufficient condition for property of over a finite, atomless measure space has been presented in [6, Theorem 1]. Note also that the case of Kadec-Klee properties with respect to pointwise (uniform) convergence in Calderón-Lozanovskiĭ sequence spaces has been solved in .
Let , , be the sets of real, nonnegative real, and positive integer numbers, respectively. As usual, (resp., ) stands for the unit sphere (resp., the closed unit ball) of a real Banach space .
Let be a -finite and complete measure space. By we mean the set of all -equivalence classes of real-valued measurable functions defined on .
Every Köthe space is a Banach lattice under the natural partial order ( if for -a.e. ). In particular, if we consider the space over the nonatomic measure , then we will say that is a Köthe function space. If we replace the measure space by the counting measure space , then we will say that is a Köthe sequence space.
The set is called the positive cone of . For any subset , define .
By the symmetric Köthe function space (symmetric Banach function space) on , where or with the Lebesgue measure , we mean a Köthe space with the additional property that for any two equimeasurable functions , (i.e., they have the same distribution functions , where , ) and we have and . In particular, , where For basic properties of symmetric spaces and rearrangements, we refer to [13, 14]. In the sequel, considering symmetric Köthe function space over measure space () we mean or with the Lebesgue measure .
A point is said to have order continuous norm if for any sequence in such that and -a.e. we have . A Köthe space is called order continuous () if every element of has an order continuous norm (see [2, 12, 15]). As usual, stands for the subspace of order continuous elements of . It is known that if and only if for any sequence satisfying (i.e., and . Clearly, if and only if -a.e. in .
Recall that is said to have Kadec-Klee property ( for short) whenever for any and in satisfying in the weak topology and (see ). This property is sometimes called the Radon-Riesz property or property . It has been considered in many classes of Banach spaces (see [1, 4, 16–18]). If we replace the weak convergence by the convergence in measure (), by the convergence in measure on every set of finite measure ( locally), or by the uniform convergence (), then we say that has the Kadec-Klee property with respect to convergence in measure, local convergence in measure, or uniform convergence, respectively (we will write , , and ). Clearly, . Moreover, the converse of any of these implications is not true in general (see Sections 3.1 and 3.2 below).
Given a property , we will say that a Köthe space satisfies property if satisfies property restricted only to nonnegative elements.
In the whole paper, denotes an Orlicz function, that is, , it is convex, even, vanishing, and continuous at zero, left continuous on and not identically equal to zero. Denote We write when and if . Let , where Define on a convex semimodular by where , . By the Calderón-Lozanovskiĭ space we mean equipped with so called Luxemburg norm defined by We generally assume that if , then , because when , then and .
If , then is the Orlicz function (sequence) space equipped with the Luxemburg norm. If —the Lorentz function space (—the Lorentz sequence space), then is the corresponding Orlicz-Lorentz function (sequence) space denoted by and equipped with the Luxemburg norm (see [5, 8]).
We will assume in the whole paper that has the Fatou property, that is, if with in and , then and . Since has the Fatou property, has also this property, whence is a Banach space.
We say an Orlicz function satisfies condition (resp., ) if there exist and such that (resp., ), and the inequality holds for all (resp., ). If there exists such that for all , then we say that satisfies condition . We write for short , , and , respectively.
For a Köthe space and an Orlicz function , we say that satisfies condition ( for short) if(1) whenever ,(2) whenever ,(3) whenever neither nor (see ),
where the symbol stands for the continuous embedding of the space into the space .
Relationships between the modular and the norm are collected in .
3.1. Köthe Spaces and Orlicz Functions
Given a Köthe function space , we set
Remark 3.1. It is easy to see that if is a symmetric Köthe function space, then
Recall that if , then . For a Köthe space , let be a subset. By the support of , we mean the set satisfying(i)for each , there is , with , such that ,(ii)there is with .
Lemma 3.2. Suppose that is a Köthe function space. Consider the following statements:(i),(ii),(iii)For any and each sequence in the implication (iv), (v), (vi). Then (i) ⇒ (iii) ⇒ (iv) ⇒ (v) ⇒ (vi) and (ii) ⇒ (iii). For a symmetric Köthe function space we have additionally (vi) ⇒ (iv). Furthermore, for a nonsymmetric Köthe function space, the implication (vi) ⇒ (iv) is not true in general. Finally, the implications (i) ⇒ (iii) ⇒ (iv) cannot be reversed in general.
Proof. The implications (i) ⇒ (iii) ⇒ (iv) and (v) ⇒ (vi) are obvious.
(iv) ⇒ (v). If , then for each with we have . Let be such that and . Then and consequently there is a sequence of measurable pairwise disjoint sets with and a number satisfying for each (see ). Since , we get .
(ii) ⇒ (iii). Assume that (iii) does not hold, then we find an element , a number , and a sequence with such that . Without loss of generality, we may assume that . Set Let . Clearly, . Put and . Since is decreasing, so is increasing. Next, in measure, because, for each , In particular, by , . Finally, . Thus, .
Assume that is symmetric. We prove that (vi) ⇒ (iv). If , then we find a set with and a sequence , such that . Let . We prove that . There is a number such that the set has positive measure. We find a sequence in of pairwise disjoint sets and a subsequence of satisfying and . Then, by the symmetry of , for any , whence .
Now we will prove that the implication (vi) ⇒ (iv) is not true in general in nonsymmetric case. Take Obviously, and .
Now we will see that implications (i) ⇒ (iii) ⇒ (iv) cannot be reversed in general. If —the Lorentz space over with Lebesgue measure and , then and (see [1, 8]), whence (iii) ⇏ (i) in general.
Finally, let —the Orlicz space with and . By the proof of Theorem 1 from , we find an element and a sequence with and . On the other hand, for each with , we claim that , whence . Indeed, for any sequence satisfying and for each , we have whence . Thus, the implication (iii) ⇒ (iv) cannot be reversed in general.
Recall that if , then (see ) and, as we have seen above, property needs not imply in general. However, from the above result, it follows that if , then and consequently . Note also that the implication (vi) ⇒ (v) for a symmetric Köthe function space has been proved in [8, Proposition 2.2].
Remark 3.3. It may happen that and only in a nonsymmetric Köthe space. However, if and , then there is an increasing sequence of subsets of satisfying and for each .
Proof. Let be a weak unit of , that is, for each . It is enough to take for each .
Remark 3.4. Let be a Köthe space. If , then for each , where .
Proof. Let . Clearly, and . By , we get . Moreover, uniformly. Then , because .
It is easy to see that the above is not true if . It is enough to take where and . Then for each . On the other hand, taking , we get , uniformly and , whence .
It is known that if and only if whenever (see ). Since (see ), so if and only if . The same proof as in  shows that if , then if and only if . However, the assumption is now too strong in this case because there are Köthe spaces satisfying and . Hence, the following result is more natural and precise.
Lemma 3.5. Suppose is a Köthe space. Then if and only if .
Proof. The necessity is obvious. We prove the sufficiency. Let . Assume that is a Köthe function space. Let and in measure. For each , we get
whence in measure. By the assumption , we conclude that
Then we find a subsequence of , an element , and a sequence such that for each (see ). Denote still by and by . Thus, we may assume that for all . Set . Let . Applying Remark 3.4, with respective sets
take big enough to satisfy . Moreover, for sufficiently large . Setting
Note that and consequently as . Then, by Lemma 3.2 (iii),
for sufficiently large . Finally, divide set into two subsets
Then, by (3.15),
Combining (3.14), (3.17), and (3.20), we get .
If is a Köthe sequence space, the proof is the same, because for a counting measure space for each and any sequence in with , we have .
The straightforward calculation shows that we may equivalently take and in in the definition of property (see also the proof below). However, the analogous problem for property is a little more complicated. The crucial point is described in the following lemma.
Lemma 3.6. Suppose that is a symmetric Köthe function space. If in measure and , then in measure. Moreover, the above implication is not true in general if is a nonsymmetric Köthe function space.
Proof. Having our assumptions, for each , we get
Since , so for , whence
for . Moreover, applying the symmetry of , we get
Considering the second assertion, take Set and , then , , and in measure. However, for each , whence in measure.
Corollary 3.7. Assume that is a symmetric Köthe function space. Then if and only if for any and in satisfying in measure.
Proof. The necessity is obvious. We prove the sufficiency. Let in measure and . Taking and , we get in measure, by Lemma 3.6. By the assumption, we conclude that . Notice that Since , so as desired.
We will need in the sequel the following.
Lemma 3.8. If , then for any , there are and such that
Proof. Suppose conversely that there is such that for any and each , there holds Taking and , we get . We may assume that for each . Hence, for each , one may choose and with . Consequently, we get for each . Taking , one has , whence .
(i) Suppose that is a symmetric Köthe space. If in measure and , are finitely valued a.e., then in measure. Furthermore, for a nonsymmetric Köthe space, the above implication is not true in general.
(ii) If in measure, and , are well-defined functions, then in measure.
Proof. (i) Let .
(A) Suppose that . Set Then, by symmetry of , . Let and . Take such that . We have Since is uniformly continuous on the interval , so for each there is such that the implication is true for all . Set and Consequently for sufficiently large , because . Thus .
(B) Assume . The proof is simpler than above because and for -a.e. and is uniformly continuous on the interval .
(C) Suppose that , . Then and for -a.e. . Set Then , because is symmetric. Let and . Take such that . We have Since is uniformly continuous on the interval for each , so for each and there is such that the implication is true for all . Put and Consequently for sufficiently large , because . Thus .
Now we will show that in nonsymmetric case the above implication is not true in general. Consider the Lebesgue measure space (). Take Then in measure. On the other hand, , whence in measure. Clearly, such element can be well defined only in some nonsymmetric Köthe space .
(ii) First, we claim that for each , there is such that the implication is true for all . First, notice that for each , there is such that the implication is true for all , because is continuous at zero. Next, defining , we see that is subadditive on , because , where is an Orlicz function with . Consequently, Hence, if , then , whence , by (3.41). Thus, the claim is proved. Assume that , and , are well-defined functions and in measure, then, applying (3.40), for each , we get
3.2. Calderón-Lozanovskiĭ Function Spaces
We still assume below that is a Köthe function space.
It is known that if and only if and (see Corollary 12 from ). Considering some weaker property than , we get the following.
Lemma 3.10. Suppose that , then if and only if .
Let . We find a number such that . Taking , we get . Moreover, since , satisfies the local condition with respect to the element (see ). Consequently, , by Theorem 11 from .
We apply Theorem 11 from . Let , then . If , then for some , where . Thus, , and we are done, if then , by Theorem 11 from .
Theorem 3.11. if and only if(a), (b) whenever and provided .
Proof. Since the sufficiency has been proved in [4, Theorem 11] (we need also to apply [7, Lemma 2]), we need only to prove the necessity. Since , so (see [10, Proposition 2.1]), and consequently, , , and (see [9, Corollary 12]). Notice that implies that . Indeed, if , then there is a decreasing sequence in with and , whence , so . Thus, (b) is proved. Suppose that . Recall that property can be equivalently considered only on the positive cone [6, Proposition 1]. Consequently, we find and in with locally in measure and . Set and , then , whence . Moreover, locally in measure, because, by finiteness of the measure space , the local convergence in measure can be equivalently replaced by convergence -a.e. (also Lemma 3.9(ii) can be applied). It is enough to prove that there is with
for infinitely many . The similar condition has been proved in [8, Theorem 2.11], but under additional assumption that . Here, the lack of this assumption requires new techniques in comparison with the respective proof in [8, Theorem 2.11]. We have
By (3.45), we get
for each . We assume that for infinitely many , because otherwise the proof is analogous. We divide the proof into two cases.(I)Suppose that and . Let . Denoting
The remaining proof of the case (I) we divide into two parts.(1)Suppose that . By Lemma 3.8, for , there are and such that
(a)Assume that . Taking , we get
Thus, .(b)Let . Thus, by Lemma 3.8,
and consequently (cf. [8, Lemma 1.1]).(2) Assume that , then, by (3.50),
Since , for and , there is such that
for every (see [20, Theorem 1.13(4)]). Moreover, we can choose satisfying
then, by (3.57) and (3.58), using convexity of , we get
Note that is a decreasing function of . Moreover, by convexity of and inequality (3.57). Hence, by (3.58), . Then, by (3.56),
Since , for and , there is such that for each (see ). Taking and applying (3.57), we get
Then , and consequently (see [8, Lemma 1.1]).
Combining cases (1) and (2), we get (3.44) with .
(II) Suppose that . Since , so for every , there is such that for every (see [20, Theorem 1.13(4)]). Moreover, for every , there is such that for each . Then the proof is analogous as in case (I) (it is simpler and shorter).
Now we will consider the Kadec-Klee property with respect to global convergence in measure. Theorem 3.19 below presents the full characterization of property . We divide the whole proof into several lemmas for readers convenience.
Lemma 3.12. If , then .
Proof. If , then there are , a set , and a sequence with and . Without loss of generality, we may assume that . Set Let . Clearly, . Put and . Choose such that