Abstract
In this paper, criteria for non-squareness and uniform non-squareness of Orlicz-Lorentz function spaces are given. Since degenerated Orlicz functions φ and degenerated weight functions ω are also admitted, this investigation concerns the most possible wide class of Orlicz-Lorentz function spaces.
It is worth recalling that uniform non-squareness is an important property, because it implies super-reflexivity as well as the fixed point property (see James in Ann. Math. 80:542-550, 1964; Pacific J. Math. 41:409-419, 1972 and García-Falset et al. in J. Funct. Anal. 233:494-514, 2006).
MSC:46B20, 46B42, 46A80, 46E30.
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Uniform non-squareness of Banach spaces has been defined by James as the geometric property which implies super-reflexivity (see [1, 2]). So, after proving this property for a Banach space, we know, without any characterization of the dual space, that it is super-reflexive, so reflexive as well. Recently, García-Falset, Llorens-Fuster and Mazcuñan-Navarro have shown that uniformly non-square Banach spaces have the fixed point property (see [3]).
Therefore, it was natural and interesting to look for criteria of non-squareness properties in various well-known classes of Banach spaces. Among a great number of papers concerning this topic, we list here [4–12].
The problem of uniform non-squareness of Calderón-Lozanovskiĭ spaces was initiated by Cerdà, Hudzik and Mastyło in [13]. Since the class of Orlicz-Lorentz spaces is a subclass of Calderón-Lozanovskiĭ spaces, we can say that also the problem of uniform non-squareness of Orlicz-Lorentz spaces was initiated in [13]. However, the results of our paper show that those results were only some sufficient conditions for uniform non-squareness which were very far from being necessary and sufficient. Analogous results for Orlicz-Lorentz sequence spaces were presented in [14], but the techniques of the proofs in the function case are different (in some parts completely different) than in the sequence case.
1 Preliminaries
We say that a Banach space is non-square if for any x and y from (the unit sphere of X). A Banach space X is said to be uniformly non-square if there exists such that for any (the unit ball of X). In the last definition, the unit ball can be replaced, equivalently, by the unit sphere .
Let be the space of all (equivalence classes of) Lebesgue measurable real-valued functions defined on the interval , where . For any , we write if almost everywhere with respect to the Lebesgue measure m on the interval .
Given any , we define its distribution function by
(see [15, 16] and [17]) and the non-increasing rearrangement of x as
(under the convention ). We say that two functions are equimeasurable if for all . Then we obviously have .
Let and be totally σ-finite measure spaces. A map σ from into is called a measure preserving transformation if for each -measurable subset A from , the set is a -measurable subset of and (see [15]). It is well known that a measure preserving transformation induces equimeasurability, that is, if σ is a measure preserving transformation, then x and are equimeasurable functions. The converse is false (see [15]).
A Banach space , where , is said to be a Köthe space if the following conditions are satisfied:
-
(i)
if , and , then and ,
-
(ii)
there exists a function x in E that is strictly positive on the whole .
Recall that a Köthe space E is called a symmetric space if E is rearrangement invariant which means that if , and , then and (see [18]). For basic properties of symmetric spaces, we refer to [15, 16] and [17].
In the whole paper, φ denotes an Orlicz function (see [19–21]), that is, (our definition is extended from R into by assuming ) and φ is convex, even, vanishing and continuous at zero, left continuous on and not identically equal to zero on . Let us denote
and
Let us note that if , then , while left continuity of φ on is equivalent to the fact that .
Recall that an Orlicz function φ satisfies the condition for all ( for short) if there exists a constant such that the inequality
holds for any (then we have and ). Analogously, we say that an Orlicz function φ satisfies the condition at infinity ( for short) if there exist a constant and a constant such that and inequality (1) holds for any (then we obtain ).
For any Orlicz function φ, we define its complementary function in the sense of Young by the formula
for all . It is easy to show that ψ is also an Orlicz function.
Let be a non-increasing and locally integrable function called a weight function. Let us define
We say that a weight function ω is regular if there exists such that
for any (see [22–25]). Note that if the weight function ω is regular, then in the case when and whenever .
Now we recall the definition of Orlicz-Lorentz spaces. These spaces were introduced by Kamińska (see [26, 27] and [24]) at the beginning of 1990s. Her investigations gave an impulse to further investigations of the spaces, results of which have been published, among others, in the papers [14, 28–42].
Given any Orlicz function φ and a weight function ω, we define on the convex modular
(see [26] and [28]) and the Orlicz-Lorentz function space
(see [26] and [28]), which becomes a Banach symmetric space under the Luxemburg norm
In our investigations, we apply the results concerning the monotonicity properties of Lorentz function spaces that were presented in [25, 43, 44]. Let us recall that the Lorentz function spaces (see [10, 18, 22, 23, 45–50]) are defined by the formula
A Banach lattice is said to be strictly monotone if , and imply that . We say that E is uniformly monotone if for any , there is such that whenever , , and (see [51]). Recall (see [52]) that in Banach lattices E, strict monotonicity and uniform monotonicity are restrictions of rotundity and uniform rotundity (respectively) to couples of comparable elements in the positive cone only.
Theorem 1.1 ([25], Theorem 2 and [43], Lemma 3.1)
The Lorentz function space is strictly monotone if and only if ω is positive on and whenever .
The following theorem has been proved in [[25], Theorem 1] for . Moreover, applying some ideas from the proof of Theorem 3.7 (see Case 2 on p.2722) in [53], this theorem can be also shown for .
Theorem 1.2 The Lorentz function space is uniformly monotone if and only if the weight function ω is regular and ω is positive on whenever .
In our further investigations, we will also apply Lemma 1.1 and Remark 1.1. By convexity of the modular , Lemma 1.1 can be proved analogously as in the case of Orlicz spaces (cf. also [43] for considering a more general case).
Lemma 1.1 Suppose that the Orlicz function φ satisfies a suitable condition , that is, if and , and otherwise. Then, for any , there exists such that for any whenever .
In particular, for any , we then have that if and only if .
Remark 1.1 Let and be such that . By [[16], Property 7∘, p.64], there exists a set such that and
Defining and , by convexity of the modular , we have
Denoting , , and applying convexity of the modular , defined by the formula
(if , we assume that for ), we get
2 Results
We start with the following
Theorem 2.1 Let . Then the Orlicz-Lorentz function space is non-square if and only if , and .
Proof Necessity. If or , then contains an order isometric copy of (see [[26], Theorem 2.4]). Finally, suppose that . Taking
where is such that , we get and, consequently, . Thus, is not non-square.
Sufficiency. Let . Since φ satisfies the condition , by Lemma 1.1, it is enough to show that . Let us denote
By , we have and . Therefore,
if and
whenever . Moreover, if , then . Consequently,
Hence, by strict monotonicity of the Lorentz space (see Theorem 1.1), we get
Therefore, if , we have .
Finally, suppose that . Then and . We will prove that
In order to do this, we will consider two cases.
Case 1. Suppose that . Since , by the condition , we have and . Hence, by , we obtain . By the condition , we have , whence we get . Then there exists a set with and
(see [[16], Property 7∘, p.64]). Defining
we have and, by convexity of the modular ,
Setting , by inequality (2) from Remark 1.1, we get
Since , we may assume without loss of generality that
Denote for . If , applying the inequality for , we get
Suppose now that . Then , whence and consequently,
Applying inequalities (5), (6), (7) and (8), we obtain (4).
Case 2. Let now . Then there exists v such that and for any t and s satisfying . Proceeding similarly as in the above Case 1, but with v instead of , we get again inequality (4). □
Theorem 2.2 If , then the Orlicz-Lorentz function space is non-square if and only if , and .
Proof Necessity. The necessity of conditions and can be shown similarly as in Theorem 2.1. Suppose that . Since , so , whence we can find such that . Putting
we have
which means that is not non-square.
Sufficiency. Let . Analogously as in the proof of Theorem 2.1, it is enough to show that . We divide the proof into several parts.
Case 1. Assume that . Let us define the sets , as in (3) and
If , then
for whenever and
for whenever . Analogously as in Theorem 2.1, by strict monotonicity of the Lorentz space (see Theorem 1.1), we have . Similarly, provided . Notice that if , then , whence (because ). Now we will consider the case . Then
for , whence by strict monotonicity of the Lorentz space , we have again . Finally, suppose that . Then and . Analogously as in the proof of Theorem 2.1, we can show
Case 2. Now suppose that and denote
Case 2.1. If , then we define
where is a measure preserving transformation (see [[54], Theorem 17, p.410]). Obviously, , , and are equimeasurable with , , and , respectively. Since is strictly monotone, repeating the proof from Case 1, we get
Case 2.2. Assume now that , that is,
By convexity of φ and appropriate properties of the rearrangement (see [[15], Proposition 1.7, p.41]), we obtain
for any . If there exists such that inequality (11) is sharp for the sum or for the difference, then by the right continuity of the rearrangement, we get
Consequently, in the remaining part of the proof, we will assume that for any in formula (11), we have equality for both the sum and the difference.
Case 2.2.1. Let for all and let us set in this case
By the right continuity of the rearrangement, we have and
Moreover, if , then for any or for all . In the case when , there exists such that for any . Let be the set such that and
(see [[16], Property 7∘, p.64]). By the proof of Property 7∘ from [16], we conclude that
for m-a.e. . Hence, by the definition of , we obtain
for m-a.e. and each . Moreover, using again the definition of , we get that for m-a.e. , there exists such that
Since for any we have equality in formula (11) for both the sum and the difference, we can find sets and such that and
Similarly as in the case of the set , for m-a.e. and for each , we get
Hence, by convexity of the function φ and inequalities (14) and (15), we obtain . Since , so . Analogously, we derive the equality . Note also that convexity of the function φ and equations (13) and (16) imply the equalities
whence, by inequality (10), we get and
Denoting and , we have
Case 2.2.1.1. Suppose . By convexity of the modular , we get
If (), then (). So, and . Furthermore, by equation (10), we may assume without loss of generality that . Thus
whence we get .
Case 2.2.1.2. Let now . Then, by the definition of , there exists satisfying
Define
and
Since for any we have equality in formula (11) for both the sum and the difference, we can find a set such that and
If , then we can assume without loss of generality that , whence we get the equality . Proceeding analogously as in Case 2.2.1.1, we obtain .
Let now . Then we will suppose that and consequently
Putting , and applying again convexity of the modular , we obtain
Simultaneously, by equality (18), we may assume without loss of generality that , whence
So, we get .
Case 2.2.2. Finally, assume that for some and define
Applying convexity of the Orlicz function and the equality in formula (11), we get the conditions , , and . Since , the set has positive measure. If , we can assume that (where is defined analogously as in (19)); in the opposite case, we can assume that . Proceeding analogously as in Case 2.2.1, we obtain . □
Theorem 2.3 In the case when , the Orlicz-Lorentz function space is uniformly non-square if and only if , and ω is regular.
Proof Necessity. The necessity of the condition follows from Theorem 2.1. If , then contains an order isomorphic copy of (see [[38], Theorem 7.18] or [[29], Theorem 2]), whence it is not reflexive. Finally, suppose that ω is not regular. Then we can find a sequence of positive numbers such that
for any . Since , for every , there exists satisfying
Define
Then and
whence we have .
Sufficiency. Let . By we conclude that there is such that for all (see [55]). Let us set
Since , we have . Suppose that
. Since the inequality
holds for m-a.e. , we get
Hence, by uniform monotonicity of the Lorentz space (see Theorem 1.2), we obtain
where is the constant from the definition of uniform monotonicity of the Lorentz space corresponding to . Analogously, we get in the case when . Finally, by Lemma 1.1, we obtain
where depends only on . □
Theorem 2.4 If , then the Orlicz-Lorentz function space is uniformly non-square if and only if , , ω is regular and .
Proof Necessity. The necessity of the conditions and has been shown in Theorem 2.2, whereas the necessity of the conditions and regularity of ω can be shown analogously as in Theorem 2.3.
Sufficiency. Let . If we show the inequality
for some independent of x and y, then Lemma 1.1 will give the inequality
with some depending only on q, and the proof will be finished. In order to show (20), we consider three cases.
Case 1. First assume that (in particular, this holds if or ). Then we can find such that . Since for any there holds
by , there exists such that
for all (see [55]). Define
We have , whence and consequently
If , analogously as in the proof of Theorem 2.3, we get
Hence, by uniform monotonicity of the Lorentz space (see Theorem 1.2), we obtain
where is the constant from the definition of uniform monotonicity of the Lorentz space corresponding to . If , then we get similarly that . Therefore, if , we obtain inequality (20) with .
Case 2. Now assume that and . Then for
by the condition , we have . Moreover, we can find a constant such that
Applying again the condition , we get that there exists such that inequality (21) holds for any . Denote
Now we divide the proof of this case into several parts.
Case 2.1. If , then proceeding analogously as in the Case 1, we get
where is the constant from the definition of uniform monotonicity of the Lorentz space corresponding to .
Case 2.2. Now assume that and define and by the formulas
By the definition of and the inequality , we have and , respectively.
Now we will show that
where
Indeed, by the equalities and the definition of , we have and , whence by and the definition of , we get (25).
Let
and
Case 2.2.1. First assume that and define
Denoting by the right derivative of φ at a point u, we have for . Note that for m-a.e. , we have
Hence, by and , we get
where for any . Analogously, if , for
we obtain
Therefore, if , by uniform monotonicity of the Lorentz space , we have
where is the constant from the definition of uniform monotonicity of the Lorentz space corresponding to .
Case 2.2.2. Finally, suppose that . Then for
we have
and by (25) and definition of ,
whence we get
Define
Let and be such that , ,
and
(see [[16], Property 7∘, p.64]). Moreover, by the proof of Property 7∘, we can assume that . Denoting and , by Remark 1.1, we have
By formulas (33) and (35), we have
and, in consequence, we can assume without loss of generality that . If , then
where p denotes as above the right derivative of φ on the interval and for any ; note that by the definition of , we have . Hence,
Now assume that . Then
whence we get
Therefore, by the inequality , we obtain
and, in consequence, . Simultaneously, by formulas (34) and (38) and the equality , we have
Thus, , which gives a possibility to repeat the investigations from (36) and (37) for y. In consequence, we have
Recapitulating Case 2, by inequalities (24), (30) and (39), we get inequality (20) for
Case 3. Finally, assume that and . For arbitrary fixed , we define the sets and by formulas (22) and (23). If , then proceeding analogously as in Case 2, we get inequality (24) with the constant .
If , then we define and by the equalities
We have , and , where the sets and are defined by formulas (26) and (27). By the assumption , we can find two positive constants and such that and . Let
If or , where the sets and are defined by formulas (28) and (29), then analogously as in Case 2, we obtain inequality (30) with the constant .
In the case when , we define the sets and by formulas (31) and (32). We have
Defining and repeating the procedure from Case 2, putting in place of , we get inequality (39) with the constant .
Summarizing Case 3, we get inequality (20) with
□
Theorem 2.5 Let and . Then the Orlicz-Lorentz function space is uniformly non-square if and only if , , ω is regular and .
Proof Necessity. Condition follows from Theorem 2.2, while the necessity of remaining conditions can be proved as in Theorem 2.4.
Sufficiency. Analogously as in Theorem 2.4, it is enough to show that there exists such that inequality (20) holds for any .
First note that the space , in opposite to the space , is uniformly monotone (see Theorem 1.2). Hence, by [[52], Theorem 6], for all there exists such that for any and any with and , we have
Now, for any fixed , we denote
In order to show (20), we will consider two cases.
Case 1. If , then we define
where is a measure preserving transformation (see [[54], Theorem 17, p.410]). Obviously , , and are equimeasurable with , , and , respectively. Therefore, by Theorem 2.4, there exists independent of x and y such that
Case 2. Let now . Denote by the smallest possible number satisfying and let be the constant from inequality (40) for . Fix satisfying
Since , for satisfying the equality
analogously as in Case 1 of Theorem 2.4, we can find such that
for all . We may assume without loss of generality that
Applying [[16], Property 7∘, p.64], we can find sets and of measure α such that
Let us define the sets
and
From [[15], Theorem 2.6, p.49] it follows that there are functions and both equimeasurable with and satisfying the equalities
By the Hardy-Littlewood inequality, we have
whence by (43), we conclude that
Similarly, we get
The remaining part of the proof of Case 2 will be divided into three subcases.
Case 2.1. Suppose . Then
for m-a.e. . Hence, by equality (44), we get
Case 2.2. If , then analogously as above, we can show that
Case 2.3. Finally, we will prove that the remaining case
is not possible. In the opposite case, by (46) and (47), we get
Since , we can assume without loss of generality that . Moreover, by the Hardy-Littlewood inequality and convexity of the modular , we obtain
Since , so
Similarly,
Let be such that and
Then, by the definition of , the first inequality in (50) and the second inequality in (41), we get
Consequently, by (40) (note that ) and first inequalities of formulas (49) and (41), we obtain
which is a contradiction.
Summarizing both cases, we get inequality (20) with . □
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The authors thank the referees for valuable comments and suggestions. All authors are supported partially by the State Committee for Scientific Research, Poland, Grant No. N N201 362236.
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Foralewski, P., Hudzik, H. & Kolwicz, P. Non-squareness properties of Orlicz-Lorentz function spaces. J Inequal Appl 2013, 32 (2013). https://doi.org/10.1186/1029-242X-2013-32
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DOI: https://doi.org/10.1186/1029-242X-2013-32