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Non-squareness properties of Orlicz-Lorentz function spaces

Journal of Inequalities and Applications20132013:32

https://doi.org/10.1186/1029-242X-2013-32

  • Received: 2 October 2012
  • Accepted: 8 January 2013
  • Published:

Abstract

In this paper, criteria for non-squareness and uniform non-squareness of Orlicz-Lorentz function spaces Λ φ , ω 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.

Keywords

  • uniform non-squareness
  • non-squareness
  • Orlicz-Lorentz space
  • Lorentz space
  • Orlicz function
  • Luxemburg norm
  • strict monotonicity
  • uniform monotonicity
  • reflexivity
  • super-reflexivity
  • fixed point property

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 [412].

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 ( X , ) is non-square if min ( x y 2 , x + y 2 ) < 1 for any x and y from S ( X ) (the unit sphere of X). A Banach space X is said to be uniformly non-square if there exists δ ( 0 , 1 ) such that min ( x y 2 , x + y 2 ) 1 δ for any x , y B ( X ) (the unit ball of X). In the last definition, the unit ball B ( X ) can be replaced, equivalently, by the unit sphere S ( X ) .

Let L 0 = L 0 ( [ 0 , γ ) ) be the space of all (equivalence classes of) Lebesgue measurable real-valued functions defined on the interval [ 0 , γ ) , where γ . For any x , y L 0 , we write x y if x ( t ) y ( t ) almost everywhere with respect to the Lebesgue measure m on the interval [ 0 , γ ) .

Given any x L 0 , we define its distribution function μ x : [ 0 , + ) [ 0 , γ ] by
μ x ( λ ) = m ( { t [ 0 , γ ) : | x ( t ) | > λ } )
(see [15, 16] and [17]) and the non-increasing rearrangement x : [ 0 , γ ) [ 0 , ] of x as
x ( t ) = inf { λ 0 : μ x ( λ ) t }

(under the convention inf = ). We say that two functions x , y L 0 are equimeasurable if μ x ( λ ) = μ y ( λ ) for all λ 0 . Then we obviously have x = y .

Let ( R 1 , Σ 1 , μ 1 ) and ( R 2 , Σ 2 , μ 2 ) be totally σ-finite measure spaces. A map σ from R 1 into R 2 is called a measure preserving transformation if for each Σ 2 -measurable subset A from R 2 , the set σ 1 ( A ) = { t R 1 : σ ( t ) A } is a Σ 1 -measurable subset of R 1 and μ 1 ( σ 1 ( A ) ) = μ 2 ( A ) (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 x σ are equimeasurable functions. The converse is false (see [15]).

A Banach space E = ( E , , ) , where E L 0 , is said to be a Köthe space if the following conditions are satisfied:
  1. (i)

    if x E , y L 0 and | y | | x | , then y E and y x ,

     
  2. (ii)

    there exists a function x in E that is strictly positive on the whole [ 0 , γ ) .

     

Recall that a Köthe space E is called a symmetric space if E is rearrangement invariant which means that if x E , y L 0 and x = y , then y E and x = y (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 [1921]), that is, φ : [ , ] [ 0 , ] (our definition is extended from R into R e by assuming φ ( ) = φ ( ) = ) and φ is convex, even, vanishing and continuous at zero, left continuous on ( 0 , ) and not identically equal to zero on ( , ) . Let us denote
a φ = sup { u 0 : φ ( u ) = 0 } , b φ = sup { u 0 : φ ( u ) < }
and
δ = sup { u 0 : φ ( u 2 ) = 1 2 φ ( u ) } .

Let us note that if a φ > 0 , then δ = a φ , while left continuity of φ on ( 0 , ) is equivalent to the fact that lim u ( b φ ) φ ( u ) = φ ( b φ ) .

Recall that an Orlicz function φ satisfies the condition Δ 2 for all u R ( φ Δ 2 ( R ) for short) if there exists a constant K > 0 such that the inequality
φ ( 2 u ) K φ ( u )
(1)

holds for any u R (then we have a φ = 0 and b φ = ). Analogously, we say that an Orlicz function φ satisfies the condition Δ 2 at infinity ( φ Δ 2 ( ) for short) if there exist a constant K > 0 and a constant u 0 0 such that φ ( u 0 ) < and inequality (1) holds for any u u 0 (then we obtain b φ = ).

For any Orlicz function φ, we define its complementary function in the sense of Young by the formula
ψ ( u ) = sup v > 0 { | u | v φ ( v ) }

for all u R . It is easy to show that ψ is also an Orlicz function.

Let ω : [ 0 , γ ) R + be a non-increasing and locally integrable function called a weight function. Let us define
γ 0 = sup { t 0 : ω  is constant on  ( 0 , t ) } , α = sup { t 0 : ω ( t ) > 0 } .
We say that a weight function ω is regular if there exists η > 0 such that
0 2 t ω ( t ) d t ( 1 + η ) 0 t ω ( t ) d t

for any t [ 0 , γ / 2 ) (see [2225]). Note that if the weight function ω is regular, then 0 ω ( t ) d t = in the case when γ = and α > γ / 2 whenever γ < .

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, 2842].

Given any Orlicz function φ and a weight function ω, we define on L 0 the convex modular
I φ , ω ( x ) = 0 γ φ ( x ( t ) ) ω ( t ) d t
(see [26] and [28]) and the Orlicz-Lorentz function space
Λ φ , ω = Λ φ , ω ( [ 0 , γ ) ) = { x L 0 : I φ , ω ( λ x ) <  for some  λ > 0 }
(see [26] and [28]), which becomes a Banach symmetric space under the Luxemburg norm
x = inf { λ > 0 : I φ , ω ( x / λ ) 1 } .
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 Λ ω (see [10, 18, 22, 23, 4550]) are defined by the formula
Λ ω = Λ ω ( [ 0 , γ ) ) = { x L 0 : x ω = 0 γ x ( t ) ω ( t ) d t < } .

A Banach lattice E = ( E , , ) is said to be strictly monotone if x , y E , 0 y x and y x imply that y < x . We say that E is uniformly monotone if for any ε ( 0 , 1 ) , there is δ ( ε ) ( 0 , 1 ) such that x y 1 δ ( ε ) whenever x , y E , 0 y x , x = 1 and y ε (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 E + only.

Theorem 1.1 ([25], Theorem 2 and [43], Lemma 3.1)

The Lorentz function space Λ ω is strictly monotone if and only if ω is positive on [ 0 , γ ) and 0 γ ω ( t ) d t = whenever γ = .

The following theorem has been proved in [[25], Theorem 1] for γ = . 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 γ < .

Theorem 1.2 The Lorentz function space Λ ω is uniformly monotone if and only if the weight function ω is regular and ω is positive on [ 0 , γ ) whenever γ < .

In our further investigations, we will also apply Lemma 1.1 and Remark 1.1. By convexity of the modular I φ , ω , 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 Δ 2 , that is, φ Δ ( R ) if γ = and 0 ω ( t ) d t = , and φ Δ ( ) otherwise. Then, for any ε ( 0 , 1 ) , there exists δ = δ ( ε ) ( 0 , 1 ) such that x 1 δ for any x Λ φ , ω whenever I φ , ω ( x ) 1 ε .

In particular, for any x Λ φ , ω , we then have that x = 1 if and only if I φ , ω ( x ) = 1 .

Remark 1.1 Let x , y Λ φ , ω and t ( 0 , γ ) be such that ( x + y 2 ) ( t ) > lim s ( x + y 2 ) ( s ) = ( x + y 2 ) ( ) . By [[16], Property 7, p.64], there exists a set e t = e t ( x + y 2 ) such that m ( e t ) = t and
0 t ( x + y 2 ) ( s ) d s = e t | x + y 2 | ( s ) d s .
Defining t ( x ) = m ( supp x e t ) and t ( y ) = m ( supp y e t ) , by convexity of the modular I φ , ω , we have
0 t φ ( ( x + y 2 ) ( s ) ) ω ( s ) d s = I φ , ω ( ( x + y 2 ) χ e t ) 1 2 I φ , ω ( x χ e t ) + 1 2 I φ , ω ( y χ e t ) = 1 2 0 t ( x ) φ ( ( x χ e t ) ( s ) ) ω ( s ) d s + 1 2 0 t ( y ) φ ( ( y χ e t ) ( s ) ) ω ( s ) d s .
Denoting A t = [ 0 , γ ) e t , a ( x ) = m ( supp x A t ) , a ( y ) = m ( supp y A t ) and applying convexity of the modular I φ , t , defined by the formula
I φ , t ( x ) = 0 γ φ ( x ( s ) ) ω ( t + s ) d s
(if γ < , we assume that ω ( t + s ) = 0 for s γ t ), we get
t γ φ ( ( x + y 2 ) ( s ) ) ω ( s ) d s = 0 γ φ ( ( ( x + y 2 ) χ A t ) ( s ) ) ω ( t + s ) d s = I φ , t ( ( x + y 2 ) χ A t ) 1 2 I φ , t ( x χ A t ) + 1 2 I φ , t ( y χ A t ) = 1 2 0 a ( x ) φ ( ( x χ A t ) ( s ) ) ω ( t + s ) d s + 1 2 0 a ( y ) φ ( ( y χ A t ) ( s ) ) ω ( t + s ) d s = 1 2 t t + a ( x ) φ ( ( x χ A t ) ( s t ) ) ω ( s ) d s + 1 2 t t + a ( y ) φ ( ( y χ A t ) ( s t ) ) ω ( s ) d s .
(2)

2 Results

We start with the following

Theorem 2.1 Let γ = . Then the Orlicz-Lorentz function space Λ φ , ω is non-square if and only if 0 ω ( t ) d t = , φ Δ 2 ( R ) and 0 γ 0 / 2 φ ( δ ) ω ( t ) d t < 1 .

Proof Necessity. If 0 ω ( t ) d t < or φ Δ 2 ( R ) , then Λ φ , ω contains an order isometric copy of l (see [[26], Theorem 2.4]). Finally, suppose that 0 γ 0 / 2 φ ( δ ) ω ( t ) d t 1 . Taking
x = a χ [ 0 , γ 0 / 2 ) and y = a χ [ γ 0 / 2 , γ 0 ) ,

where a δ is such that 0 γ 0 / 2 φ ( a ) ω ( t ) d t = 1 , we get I φ , ω ( x ) = I φ , ω ( y ) = I φ , ω ( x + y 2 ) = I φ , ω ( x y 2 ) = 1 and, consequently, x = y = x + y 2 = x y 2 = 1 . Thus, Λ φ , ω is not non-square.

Sufficiency. Let x , y S ( Λ φ , ω ) . Since φ satisfies the condition Δ 2 ( R ) , by Lemma 1.1, it is enough to show that min ( I φ , ω ( x y 2 ) , I φ , ω ( x + y 2 ) ) < 1 . Let us denote
A 1 = { t ( 0 , ) : x ( t ) y ( t ) > 0 } , A 2 = { t ( 0 , ) : x ( t ) y ( t ) < 0 } , A 3 = { t ( 0 , ) : x ( t ) y ( t ) = 0  and  max ( | x ( t ) | , | y ( t ) | ) > δ } , A 4 = { t ( 0 , ) : x ( t ) y ( t ) = 0  and  0 < max ( | x ( t ) | , | y ( t ) | ) δ } .
(3)
By φ Δ 2 ( R ) , we have a φ = 0 and b φ = . Therefore,
φ ( u v 2 ) < φ ( max ( | u | , | v | ) 2 ) < 1 2 { φ ( u ) + φ ( v ) }
if u v > 0 and
φ ( u + v 2 ) < φ ( max ( | u | , | v | ) 2 ) < 1 2 { φ ( u ) + φ ( v ) }
whenever u v < 0 . Moreover, if u > δ , then φ ( u 2 ) < 1 2 φ ( u ) . Consequently,
φ ( x y 2 ) 1 2 { φ ( x ) + φ ( y ) } if  m ( A 1 ) > 0 , φ ( x + y 2 ) 1 2 { φ ( x ) + φ ( y ) } if  m ( A 2 A 3 ) > 0 .
Hence, by strict monotonicity of the Lorentz space Λ ω (see Theorem 1.1), we get
I φ , ω ( x y 2 ) = φ ( x y 2 ) ω < 1 2 φ x + 1 2 φ y ω 1 if  m ( A 1 ) > 0 , I φ , ω ( x + y 2 ) = φ ( x + y 2 ) ω < 1 2 φ x + 1 2 φ y ω 1 if  m ( A 2 A 3 ) > 0 .

Therefore, if m ( A 1 A 2 A 3 ) > 0 , we have min ( I φ , ω ( x y 2 ) , I φ , ω ( x + y 2 ) ) < 1 .

Finally, suppose that m ( A 1 A 2 A 3 ) = 0 . Then δ > 0 and I φ , ω ( x y 2 ) = I φ , ω ( x + y 2 ) . We will prove that
I φ , ω ( x ± y 2 ) = 0 φ ( ( x ± y 2 ) ( t ) ) ω ( t ) d t < 1 2 0 φ ( x ( t ) ) ω ( t ) d t + 1 2 0 φ ( y ( t ) ) ω ( t ) d t = 1 .
(4)

In order to do this, we will consider two cases.

Case 1. Suppose that γ 0 > 0 . Since I φ , ω ( x ) = I φ , ω ( y ) = 1 , by the condition 0 γ 0 / 2 φ ( δ ) ω ( t ) d t < 1 , we have m ( supp x ) > γ 0 / 2 and m ( supp y ) > γ 0 / 2 . Hence, by m ( supp x supp y ) = 0 , we obtain m ( supp x supp y ) > γ 0 . By the condition 0 ω ( t ) d t = , we have lim t ( x + y 2 ) ( t ) = ( x + y 2 ) ( ) = 0 , whence we get ( x + y 2 ) ( γ 0 ) > ( x + y 2 ) ( ) . Then there exists a set e γ 0 = e γ 0 ( x + y 2 ) with m ( e γ 0 ) = γ 0 and
0 γ 0 ( x + y 2 ) ( t ) d t = e γ 0 | x + y 2 | ( t ) d t
(see [[16], Property 7, p.64]). Defining
γ 0 ( x ) = m ( e γ 0 supp x ) and γ 0 ( y ) = m ( e γ 0 supp y ) ,
we have γ 0 ( x ) + γ 0 ( y ) = γ 0 and, by convexity of the modular I φ , ω ,
0 γ 0 φ ( ( x + y 2 ) ( t ) ) ω ( t ) d t = I φ , ω ( ( x + y 2 ) χ e γ 0 ) 1 2 I φ , ω ( x χ e γ 0 ) + 1 2 I φ , ω ( y χ e γ 0 ) = 1 2 0 γ 0 ( x ) φ ( x ( t ) ) ω ( t ) d t + 1 2 0 γ 0 ( y ) φ ( y ( t ) ) ω ( t ) d t .
(5)
Setting A γ 0 = [ 0 , γ ) e γ 0 , by inequality (2) from Remark 1.1, we get
(6)
Since φ ( ( x + y 2 ) ( γ 0 ) ) > 0 , we may assume without loss of generality that
γ 0 φ ( ( x χ A γ 0 ) ( t γ 0 ) ) ω ( t ) d t > 0 .
Denote ω ( t ) = ω for t ( 0 , γ 0 ) . If γ 0 ( x ) < γ 0 , applying the inequality ω ( t ) < ω for t > γ 0 , we get
(7)
Suppose now that γ 0 ( x ) = γ 0 . Then γ 0 ( y ) = 0 , whence supp y A γ 0 and consequently,
0 < 1 2 γ 0 φ ( ( y χ A γ 0 ) ( t γ 0 ) ) ω ( t ) d t < 1 2 0 φ ( y ( t ) ) ω ( t ) d t .
(8)

Applying inequalities (5), (6), (7) and (8), we obtain (4).

Case 2. Let now γ 0 = 0 . Then there exists v such that ( x + y 2 ) ( v ) > 0 and ω ( t ) > ω ( s ) for any t and s satisfying t < v < s . Proceeding similarly as in the above Case 1, but with v instead of γ 0 , we get again inequality (4). □

Theorem 2.2 If γ < , then the Orlicz-Lorentz function space Λ φ , ω is non-square if and only if γ 2 < α γ , φ Δ 2 ( ) and 0 γ 0 / 2 φ ( δ ) ω ( t ) d t < 1 .

Proof Necessity. The necessity of conditions φ Δ 2 ( ) and 0 γ 0 / 2 φ ( δ ) ω ( t ) d t < 1 can be shown similarly as in Theorem 2.1. Suppose that α γ 2 . Since φ Δ 2 ( ) , so b φ = , whence we can find a > 0 such that 0 α φ ( a ) ω ( t ) d t = 1 . Putting
x = a χ [ 0 , 2 α ) , y = a χ [ 0 , α ) a χ [ α , 2 α ) ,
we have
I φ , ω ( x ) = I φ , ω ( y ) = I φ , ω ( x + y 2 ) = I φ , ω ( x y 2 ) = 1 ,

which means that Λ φ , ω is not non-square.

Sufficiency. Let x , y S ( Λ φ , ω ) . Analogously as in the proof of Theorem 2.1, it is enough to show that min ( I φ , ω ( x y 2 ) , I φ , ω ( x + y 2 ) ) < 1 . We divide the proof into several parts.

Case 1. Assume that α = γ . Let us define the sets A i , i = 1 , , 4 as in (3) and
A 1 = { t A 1 : max ( | x ( t ) | , | y ( t ) | ) > a φ } , A 2 = { t A 2 : max ( | x ( t ) | , | y ( t ) | ) > a φ } .
If m ( A 1 ) > 0 , then
0 = φ ( x ( t ) y ( t ) 2 ) = φ ( max ( | x ( t ) | , | y ( t ) | ) 2 ) < 1 2 φ ( max ( | x ( t ) | , | y ( t ) | ) ) 1 2 { φ ( x ( t ) ) + φ ( y ( t ) ) }
for t A 1 whenever max ( | x ( t ) | , | y ( t ) | ) / 2 a φ and
φ ( x ( t ) y ( t ) 2 ) < φ ( max ( | x ( t ) | , | y ( t ) | ) 2 ) 1 2 { φ ( x ( t ) ) + φ ( y ( t ) ) }
for t A 1 whenever max ( | x ( t ) | , | y ( t ) | ) / 2 > a φ . Analogously as in Theorem 2.1, by strict monotonicity of the Lorentz space Λ ω (see Theorem 1.1), we have I φ , ω ( x y 2 ) < 1 . Similarly, I φ , ω ( x + y 2 ) < 1 provided m ( A 2 ) > 0 . Notice that if 0 = m ( A 1 A 2 ) < m ( A 1 A 2 ) , then δ = a φ > 0 , whence m ( A 3 ) > 0 (because I φ , ω ( x ) = I φ , ω ( y ) = 1 ). Now we will consider the case m ( A 3 ) > 0 . Then
φ ( x ( t ) ± y ( t ) 2 ) = φ ( max ( | x ( t ) | , | y ( t ) | ) 2 ) < 1 2 φ ( max ( | x ( t ) | , | y ( t ) | ) ) = 1 2 { φ ( x ( t ) ) + φ ( y ( t ) ) }
for t A 3 , whence by strict monotonicity of the Lorentz space Λ ω , we have again I φ , ω ( x ± y 2 ) < 1 . Finally, suppose that m ( A 1 A 2 A 3 ) = 0 . Then 0 = a φ < δ and I φ , ω ( x χ A 4 ) = I φ , ω ( y χ A 4 ) = 1 . Analogously as in the proof of Theorem 2.1, we can show
I φ , ω ( x ± y 2 ) < 1 2 0 γ φ ( x ( t ) ) ω ( t ) d t + 1 2 0 γ φ ( y ( t ) ) ω ( t ) d t = 1 .
(9)
Case 2. Now suppose that γ 2 < α < γ and denote
A x , y = { t [ 0 , γ ) : max ( | x ( t ) | , | y ( t ) | ) > a φ } .
Case 2.1. If m ( A x , y ) α , then we define
x ˜ = x χ A x , y σ and y ˜ = y χ A x , y σ ,
where σ : [ 0 , m ( A x , y ) ) A x , y is a measure preserving transformation (see [[54], Theorem 17, p.410]). Obviously, φ x ˜ , φ y ˜ , φ x ˜ + y ˜ 2 and φ x ˜ y ˜ 2 are equimeasurable with φ x χ A x , y , φ y χ A x , y , φ x + y 2 χ A x , y and φ x y 2 χ A x , y , respectively. Since Λ ω ( [ 0 , α ) ) is strictly monotone, repeating the proof from Case 1, we get
min ( I φ , ω ( x y 2 ) , I φ , ω ( x + y 2 ) ) = min ( I φ , ω ( ( x y 2 ) χ A x , y ) , I φ , ω ( ( x + y 2 ) χ A x , y ) ) = min ( I φ , ω ( x ˜ y ˜ 2 ) , I φ , ω ( x ˜ + y ˜ 2 ) ) < 1 .
Case 2.2. Assume now that m ( A x , y ) > α , that is,
( 1 2 φ x + 1 2 φ y ) ( α ) > 0 .
(10)
By convexity of φ and appropriate properties of the rearrangement (see [[15], Proposition 1.7, p.41]), we obtain
φ ( ( x ± y 2 ) ( t ) ) = ( φ ( x ± y 2 ) ) ( t ) ( 1 2 φ x + 1 2 φ y ) ( t )
(11)
for any t [ 0 , γ ) . If there exists t [ 0 , α ) such that inequality (11) is sharp for the sum or for the difference, then by the right continuity of the rearrangement, we get
min ( I φ , ω ( x y 2 ) , I φ , ω ( x + y 2 ) ) < 1 .

Consequently, in the remaining part of the proof, we will assume that for any t [ 0 , α ) in formula (11), we have equality for both the sum and the difference.

Case 2.2.1. Let ( 1 2 φ x + 1 2 φ y ) ( 0 ) > ( 1 2 φ x + 1 2 φ y ) ( t ) for all t > α and let us set in this case
t 0 = sup { s : ( 1 2 φ x + 1 2 φ y ) ( s ) > ( 1 2 φ x + 1 2 φ y ) ( t )  for each  t > α } .
By the right continuity of the rearrangement, we have 0 < t 0 α and
( 1 2 φ x + 1 2 φ y ) ( t 0 ) = ( 1 2 φ x + 1 2 φ y ) ( α ) > 0 .
(12)
Moreover, if t 0 = α , then ( 1 2 φ x + 1 2 φ y ) ( s ) > ( 1 2 φ x + 1 2 φ y ) ( α ) for any s < α or ( 1 2 φ x + 1 2 φ y ) ( α ) > ( 1 2 φ x + 1 2 φ y ) ( t ) for all t > α . In the case when t 0 < α , there exists t > α such that ( 1 2 φ x + 1 2 φ y ) ( s ) > ( 1 2 φ x + 1 2 φ y ) ( t 0 ) = ( 1 2 φ x + 1 2 φ y ) ( t ) for any s < t 0 . Let e t 0 = e t 0 ( 1 2 φ x + 1 2 φ y ) be the set such that m ( e t 0 ) = t 0 and
0 t 0 ( 1 2 φ x + 1 2 φ y ) ( t ) d t = e t 0 ( 1 2 φ x + 1 2 φ y ) ( t ) d t
(13)
(see [[16], Property 7, p.64]). By the proof of Property 7 from [16], we conclude that
( 1 2 φ x + 1 2 φ y ) ( s ) lim t t 0 ( 1 2 φ x + 1 2 φ y ) ( t )
for m-a.e. s e t 0 . Hence, by the definition of t 0 , we obtain
( 1 2 φ x + 1 2 φ y ) ( s ) > ( 1 2 φ x + 1 2 φ y ) ( t )
(14)
for m-a.e. s e t 0 and each t > t 0 . Moreover, using again the definition of t 0 , we get that for m-a.e. s [ 0 , γ ) e t 0 , there exists t ( s ) > t 0 such that
( 1 2 φ x + 1 2 φ y ) ( s ) ( 1 2 φ x + 1 2 φ y ) ( t ( s ) ) .
(15)
Since for any t [ 0 , α ) we have equality in formula (11) for both the sum and the difference, we can find sets e t 0 ( + ) = e t 0 ( φ ( x + y 2 ) ) and e t 0 ( ) = e t 0 ( φ ( x y 2 ) ) such that m ( e t 0 ( + ) ) = m ( e t 0 ( ) ) = t 0 and
0 t 0 ( 1 2 φ x + 1 2 φ y ) ( t ) d t = e t 0 ( + ) φ ( x + y 2 ) ( t ) d t = e t 0 ( ) φ ( x y 2 ) ( t ) d t .
(16)
Similarly as in the case of the set e t 0 , for m-a.e. s e t 0 ( + ) and for each t > t 0 , we get
φ ( x + y 2 ) ( s ) > ( 1 2 φ x + 1 2 φ y ) ( t ) .
Hence, by convexity of the function φ and inequalities (14) and (15), we obtain e t 0 ( + ) e t 0 . Since m ( e t 0 ) = t 0 = m ( e t 0 ( + ) ) , so e t 0 ( + ) = e t 0 . Analogously, we derive the equality e t 0 ( ) = e t 0 . Note also that convexity of the function φ and equations (13) and (16) imply the equalities
φ ( x + y 2 ) χ e t 0 = φ ( x y 2 ) χ e t 0 = ( 1 2 φ x + 1 2 φ y ) χ e t 0 ,
whence, by inequality (10), we get m ( supp ( x χ e t 0 ) supp ( y χ e t 0 ) ) = 0 and
0 = a φ < ( 1 2 φ x + 1 2 φ y ) ( 0 ) δ .
(17)
Denoting t 0 ( x ) = m ( e t 0 supp x ) and t 0 ( y ) = m ( e t 0 supp y ) , we have
t 0 ( x ) + t 0 ( y ) = t 0 .
(18)
Case 2.2.1.1. Suppose t 0 = α . By convexity of the modular I φ , ω , we get
0 t 0 φ ( ( x + y 2 ) ( t ) ) ω ( t ) d t = I φ , ω ( ( x + y 2 ) χ e t 0 ) 1 2 I φ , ω ( x χ e t 0 ) + 1 2 I φ , ω ( y χ e t 0 ) = 1 2 0 t 0 ( x ) φ ( x ( t ) ) ω ( t ) d t + 1 2 0 t 0 ( y ) φ ( y ( t ) ) ω ( t ) d t .
If t 0 ( y ) = 0 ( t 0 ( x ) = 0 ), then I φ , ω ( x + y 2 ) 1 2 I φ , ω ( x ) = 1 2 ( I φ , ω ( x + y 2 ) 1 2 I φ , ω ( y ) = 1 2 ). So, 0 < t 0 ( x ) < t 0 and 0 < t 0 ( y ) < t 0 . Furthermore, by equation (10), we may assume without loss of generality that β ( x ) : = m ( ( A x , y e t 0 ) supp x ) > 0 . Thus
0 t 0 ( x ) φ ( x ( t ) ) ω ( t ) d t < 0 t 0 ( x ) φ ( x ( t ) ) ω ( t ) d t + t 0 ( x ) t 0 ( x ) + β ( x ) φ ( ( x χ A x , y e t 0 ) ( t t 0 ( x ) ) ) ω ( t ) d t = 0 α φ ( x ( t ) ) ω ( t ) d t = 1 ,

whence we get I φ , ω ( x + y 2 ) < 1 .

Case 2.2.1.2. Let now t 0 < α . Then, by the definition of t 0 , there exists t > α satisfying
( 1 2 φ x + 1 2 φ y ) ( t ) = ( 1 2 φ x + 1 2 φ y ) ( t 0 ) .
Define
t 1 = sup { t > α : ( 1 2 φ x + 1 2 φ y ) ( t ) = ( 1 2 φ x + 1 2 φ y ) ( t 0 ) } , A t 0 = { t [ 0 , γ ) : ( 1 2 φ x + 1 2 φ y ) ( t ) = ( 1 2 φ x + 1 2 φ y ) ( t 0 ) }
and
A t 0 , x , y = { t A t 0 : min ( | x ( t ) | , | y ( t ) | ) = 0 } .
Since for any t [ 0 , α ) we have equality in formula (11) for both the sum and the difference, we can find a set e α = e α ( 1 2 φ x + 1 2 φ y ) such that m ( e α ) = α and
0 α ( 1 2 φ x + 1 2 φ y ) ( t ) d t = e α ( 1 2 φ x + 1 2 φ y ) ( t ) d t = e α φ ( x + y 2 ) ( t ) d t .
(19)

If m ( A t 0 , x , y ) α t 0 , then we can assume without loss of generality that e t 0 e α e t 0 A t 0 , x , y , whence we get the equality m ( supp x χ e α supp y χ e α ) = 0 . Proceeding analogously as in Case 2.2.1.1, we obtain I φ , ω ( x + y 2 ) < 1 .

Let now m ( A t 0 , x , y ) < α t 0 . Then we will suppose that e t 0 A t 0 , x , y e α e t 0 A t 0 and consequently
m ( ( A t 0 e α ) supp x ) = m ( ( A t 0 e α ) supp y ) = m ( A t 0 e α ) = t 1 α = : d > 0 .
Putting α ( x ) = m ( e α supp x ) , α ( y ) = m ( e α supp y ) and applying again convexity of the modular I φ , ω , we obtain
0 α φ ( ( x + y 2 ) ( t ) ) ω ( t ) d t = I φ , ω ( ( x + y 2 ) χ e α ) 1 2 I φ , ω ( x χ e α ) + 1 2 I φ , ω ( y χ e α ) = 1 2 0 α ( x ) φ ( ( x χ e α ) ( t ) ) ω ( t ) d t + 1 2 0 α ( y ) φ ( ( y χ e α ) ( t ) ) ω ( t ) d t .
Simultaneously, by equality (18), we may assume without loss of generality that α ( x ) = t 0 ( x ) + m ( ( e α e t 0 ) supp x ) < α , whence
0 α ( x ) φ ( ( x χ e α ) ( t ) ) ω ( t ) d t < 0 α ( x ) φ ( ( x χ e α ) ( t ) ) ω ( t ) d t + α ( x ) α ( x ) + d φ ( ( x χ A t 0 e α ) ( t α ( x ) ) ) ω ( t ) d t 0 α φ ( x ( t ) ) ω ( t ) d t = 1 .

So, we get I φ , ω ( x + y 2 ) < 1 .

Case 2.2.2. Finally, assume that ( 1 2 φ x + 1 2 φ y ) ( 0 ) = ( 1 2 φ x + 1 2 φ y ) ( α ) = ( 1 2 φ x + 1 2 φ y ) ( t ) > 0 for some t > α and define
A = { t [ 0 , γ ) : 1 2 φ ( x ( t ) ) + 1 2 φ ( y ( t ) ) = ( 1 2 φ x + 1 2 φ y ) ( 0 ) } , A + = { t [ 0 , γ ) : φ ( x + y 2 ) ( t ) = ( 1 2 φ x + 1 2 φ y ) ( 0 ) } , A = { t [ 0 , γ ) : φ ( x y 2 ) ( t ) = ( 1 2 φ x + 1 2 φ y ) ( 0 ) } .

Applying convexity of the Orlicz function and the equality in formula (11), we get the conditions m ( A ) > α , A + A , A A and min ( m ( A + ) , m ( A ) ) α . Since α > γ 2 , the set A x , y = A + A = { t A : min ( | x ( t ) | , | y ( t ) | ) = 0 } has positive measure. If m ( A x , y ) α , we can assume that e α A x , y (where e α is defined analogously as in (19)); in the opposite case, we can assume that A x , y e α A . Proceeding analogously as in Case 2.2.1, we obtain I φ , ω ( x + y 2 ) < 1 . □

Theorem 2.3 In the case when γ = , the Orlicz-Lorentz function space Λ φ , ω is uniformly non-square if and only if φ Δ 2 ( R ) , ψ Δ 2 ( R ) and ω is regular.

Proof Necessity. The necessity of the condition φ Δ 2 ( R ) follows from Theorem 2.1. If ψ Δ 2 ( R ) , then Λ φ , ω contains an order isomorphic copy of l 1 (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 ( t n ) of positive numbers such that
0 2 t n ω ( t ) d t ( 1 + 1 n ) 0 t n ω ( t ) d t
for any n N . Since b φ = , for every n N , there exists a n satisfying
φ ( a n ) 0 2 t n ω ( t ) d t = 1 .
Define
x n = a n χ [ 0 , 2 t n ) , y n = a n χ [ 0 , t n ) a n χ [ t n , 2 t n ) .
Then I φ , ω ( x n ) = I φ , ω ( y n ) = 1 and
I φ , ω ( x n + y n 2 ) = I φ , ω ( x n y n 2 ) = 0 t n φ ( a n ) ω ( t ) d t n n + 1 0 2 t n φ ( a n ) ω ( t ) d t 1 ,

whence we have min ( x n y n 2 , x n + y n 2 ) 1 .

Sufficiency. Let x , y S ( Λ φ , ω ) . By ψ Δ 2 ( R ) we conclude that there is η ( 0 , 1 ) such that φ ( u 2 ) 1 η 2 φ ( u ) for all u > 0 (see [55]). Let us set
A 1 = { t ( 0 , ) : x ( t ) y ( t ) > 0 } , A 2 = { t ( 0 , ) : x ( t ) y ( t ) < 0 } , A 3 = { t ( 0 , ) : | x ( t ) | > 0  and  y ( t ) = 0 } .

Since I φ , ω ( x ) = 1 , we have max ( I φ , ω ( x χ A 1 A 3 ) , I φ , ω ( x χ A 2 ) ) 1 / 2 . Suppose that

I φ , ω ( x χ A 1 A 3 ) 1 / 2 . Since the inequality
φ ( x ( t ) y ( t ) 2 ) φ ( max ( | x ( t ) | , | y ( t ) | ) 2 ) 1 η 2 φ ( max ( | x ( t ) | , | y ( t ) | ) ) 1 2 φ ( x ( t ) ) + 1 2 φ ( y ( t ) ) η 2 φ ( x ( t ) )
holds for m-a.e. t A 1 A 3 , we get
φ ( x y 2 ) 1 2 φ x + 1 2 φ y η 2 φ x χ A 1 A 3 .
Hence, by uniform monotonicity of the Lorentz space Λ ω (see Theorem 1.2), we obtain
I φ , ω ( x y 2 ) = φ ( x y 2 ) ω 1 2 φ x + 1 2 φ y η 2 φ x χ A 1 A 3 φ 1 δ ( η 4 ) ,
where δ ( η 4 ) is the constant from the definition of uniform monotonicity of the Lorentz space Λ ω corresponding to η 4 . Analogously, we get I φ , ω ( x + y 2 ) 1 δ ( η 4 ) in the case when I φ , ω ( x χ A 2 ) 1 / 2 . Finally, by Lemma 1.1, we obtain
min ( x y 2 , x + y 2 ) 1 r ,

where r = r ( δ ( η 4 ) ) depends only on δ ( η 4 ) . □

Theorem 2.4 If α = γ < , then the Orlicz-Lorentz function space Λ φ , ω is uniformly non-square if and only if φ Δ 2 ( ) , ψ Δ 2 ( ) , ω is regular and 0 γ 0 / 2 φ ( δ ) ω ( t ) d t < 1 .

Proof Necessity. The necessity of the conditions φ Δ 2 ( ) and 0 γ 0 / 2 φ ( δ ) ω ( t ) d t < 1 has been shown in Theorem 2.2, whereas the necessity of the conditions ψ Δ 2 ( ) and regularity of ω can be shown analogously as in Theorem 2.3.

Sufficiency. Let x , y S ( Λ φ , ω ) . If we show the inequality
min ( I φ , ω ( x y 2 ) , I φ , ω ( x + y 2 ) ) 1 q
(20)
for some q > 0 independent of x and y, then Lemma 1.1 will give the inequality
min ( x y 2 , x + y 2 ) 1 r ,

with some r > 0 depending only on q, and the proof will be finished. In order to show (20), we consider three cases.

Case 1. First assume that 0 γ φ ( δ ) ω ( t ) d t < 1 (in particular, this holds if δ = 0 or 0 < a φ = δ ). Then we can find u δ > δ such that 0 γ φ ( u δ ) ω ( t ) d t = : a δ < 1 . Since for any u > δ there holds
φ ( u 2 ) < 1 2 φ ( u ) ,
by ψ Δ 2 ( ) , there exists η = η ( u δ ) ( 0 , 1 ) such that
φ ( u 2 ) 1 η 2 φ ( u )
(21)
for all u u δ (see [55]). Define
A = { t [ 0 , γ ) : | x ( t ) | u δ } , A 1 = { t A : x ( t ) y ( t ) 0 } , A 2 = { t A : x ( t ) y ( t ) < 0 } .
We have I φ , ω ( x χ [ 0 , γ ) A ) < a δ , whence I φ , ω ( x χ A ) > 1 a δ and consequently
max ( I φ , ω ( x χ A 1 ) , I φ , ω ( x χ A 2 ) ) > 1 a δ 2 .
If I φ , ω ( x χ A 1 ) > ( 1 a δ ) / 2 , analogously as in the proof of Theorem 2.3, we get
φ x y 2 1 2 φ x + 1 2 φ y η 2 φ x χ A 1 .
Hence, by uniform monotonicity of the Lorentz space Λ ω (see Theorem 1.2), we obtain
I φ , ω ( x y 2 ) = φ ( x y 2 ) ω 1 2 φ x + 1 2 φ y η 2 φ x χ A 1 ω 1 δ ( η ( 1 a δ ) 4 ) ,

where δ ( η ( 1 a δ ) / 4 ) is the constant from the definition of uniform monotonicity of the Lorentz space Λ ω corresponding to η ( 1 a δ ) / 4 . If I φ , ω ( x χ A 2 ) > ( 1 a δ ) / 2 , then we get similarly that I φ , ω ( x + y 2 ) 1 δ ( η ( 1 a δ ) / 4 ) . Therefore, if 0 γ φ ( δ ) ω ( t ) d t < 1 , we obtain inequality (20) with q = δ ( η ( 1 a δ ) / 4 ) .

Case 2. Now assume that 0 γ φ ( δ ) ω ( t ) d t 1 and γ 0 > 0 . Then for
c : = 1 0 γ 0 / 2 φ ( δ ) ω ( t ) d t 8 ,
by the condition 0 γ 0 / 2 φ ( δ ) ω ( t ) d t < 1 , we have 0 < c < 1 8 . Moreover, we can find a constant v δ > δ such that
0 γ 0 / 2 φ ( v δ ) ω ( t ) d t = 1 4 c .
Applying again the condition ψ Δ 2 ( ) , we get that there exists η = η ( v δ ) ( 0 , 1 ) such that inequality (21) holds for any u v δ . Denote
(22)
(23)

Now we divide the proof of this case into several parts.

Case 2.1. If max ( I φ , ω ( x χ A x , v δ ) , I φ , ω ( y χ A y , v δ ) ) c , then proceeding analogously as in the Case 1, we get
min ( I φ , ω ( x y 2 ) , I φ , ω ( x + y 2 ) ) 1 δ ( η c 4 ) ,
(24)

where δ ( η c 4 ) is the constant from the definition of uniform monotonicity of the Lorentz space Λ ω corresponding to η c 4 .

Case 2.2. Now assume that max ( I φ , ω ( x χ A x , v δ ) , I φ , ω ( y χ A y , v δ ) ) < c and define t 0 > 0 and u 0 > 0 by the formulas
0 t 0 φ ( v δ ) ω ( t ) d t = 1 2 c and 0 γ φ ( u 0 ) ω ( t ) d t = c .

By the definition of v δ and the inequality 0 γ φ ( δ ) ω ( t ) d t 1 , we have t 0 > γ 0 2 and u 0 < δ , respectively.

Now we will show that
m ( A x , u 0 ) t 0 and m ( A y , u 0 ) t 0 ,
(25)
where
(26)
(27)

Indeed, by the equalities I φ , ω ( x ) = I φ , ω ( y ) = 1 and the definition of u 0 , we have I φ , ω ( x χ A x , u 0 ) 1 c and I φ , ω ( y χ A y , u 0 ) 1 c , whence by max ( I φ , ω ( x χ A x , v δ ) , I φ , ω ( y χ A y , v δ ) ) < c and the definition of t 0 , we get (25).

Let
t 1 = min ( t 0 γ 0 2 , γ 0 2 ) 4
and
(28)
(29)
Case 2.2.1. First assume that m ( A x , y , u 0 + ) t 1 and define
z 1 = ( 1 2 φ x + 1 2 φ y ) φ ( x y 2 ) .
Denoting by p ( u ) the right derivative of φ at a point u, we have p ( u ) = : p > 0 for u [ 0 , δ ) . Note that for m-a.e. t A x , y , u 0 + , we have
( 1 2 φ ( x ( t ) ) + 1 2 φ ( y ( t ) ) ) φ ( x ( t ) y ( t ) 2 ) φ ( x ( t ) + y ( t ) 2 ) φ ( x ( t ) y ( t ) 2 ) φ ( x ( t ) y ( t ) 2 ) φ ( x ( t ) + y ( t ) 2 ) p ( u ) d u 0 u 0 / 2 p d u = p u 0 2 .
Hence, by m ( A x , y , u 0 + ) t 1 and t 1 < γ 0 , we get
z 1 ω 0 t 1 p u 0 2 ω ( t ) d t = p u 0 ω 0 t 1 2 ,
where ω 0 = ω ( t ) for any t ( 0 , γ 0 ) . Analogously, if m ( A x , y , u 0 ) t 1 , for
z 2 = ( 1 2 φ x + 1 2 φ y ) φ ( x + y 2 )
we obtain
z 2 ω 0 t 1 p u 0 2 ω ( t ) d t = p u 0 ω 0 t 1 2 .
Therefore, if max ( m ( A x , y , u 0 + ) , m ( A x , y , u 0 ) ) t 1 , by uniform monotonicity of the Lorentz space Λ ω , we have
min ( I φ , ω ( x y 2 ) , I φ , ω ( x + y 2 ) ) 1 δ ( p u 0 ω 0 t 1 2 ) ,
(30)

where δ ( p u 0 ω 0 t 1 2 ) is the constant from the definition of uniform monotonicity of the Lorentz space Λ ω corresponding to p u 0 ω 0 t 1 2 .

Case 2.2.2. Finally, suppose that max ( m ( A x , y , u 0 + ) , m ( A x , y , u 0 ) ) < t 1 . Then for
(31)
(32)
we have
B x , u 0 B y , u 0 =
(33)
and by (25) and definition of t 1 ,
min ( m ( B x , u 0 ) , m ( B y , u 0 ) ) t 0 2 t 1 t 0 1 2 ( t 0 γ 0 2 ) = t 0 2 + γ 0 4 > γ 0 2 ,
(34)
whence we get
m ( B x , u 0 B y , u 0 ) t 0 + γ 0 2 > γ 0 .
(35)
Define
a = min ( ( t 0 + γ 0 2 ) γ 0 8 , γ 0 4 ) and t 2 = γ 0 + a .
Let e γ 0 = e γ 0 ( x + y 2 ) and e t 2 = e t 2 ( x + y 2 ) be such that m ( e γ 0 ) = γ 0 , m ( e t 2 ) = t 2 ,
0 γ 0 ( x + y 2 ) ( t ) d t = e γ 0 | x + y 2 | ( t ) d t
and
0 t 2 ( x + y 2 ) ( t ) d t = e t 2 | x + y 2 | ( t ) d t
(see [[16], Property 7, p.64]). Moreover, by the proof of Property 7, we can assume that e γ 0 e t 2 . Denoting A γ 0 = e t 2 e γ 0 and A t 2 = [ 0 , γ ) e t 2 , by Remark 1.1, we have
I φ , ω ( x + y 2 ) = 0 γ 0 φ ( ( x + y 2 ) ( t ) ) ω ( t ) d t + γ 0 t 2 φ ( ( x + y 2 ) ( t ) ) ω ( t ) d t + t 2 γ φ ( ( x + y 2 ) ( t ) ) ω ( t ) d t = 0 γ 0 φ ( ( ( x + y 2 ) χ e γ 0 ) ( t ) ) ω ( t ) d t + γ 0 t 2 φ ( ( ( x + y 2 ) χ A γ 0 ) ( t γ 0 ) ) ω ( t ) d t + t 2 γ φ ( ( ( x + y 2 ) χ A t 2 ) ( t t 2 ) ) ω ( t ) d t 1 2 0 γ 0 φ ( ( x χ e γ 0 ) ( t ) ) ω ( t ) d t + 1 2 0 γ 0 φ ( ( y χ e γ 0 ) ( t ) ) ω ( t ) d t + 1 2 γ 0 t 2 φ ( ( x χ A γ 0 ) ( t γ 0 ) ) ω ( t ) d t + 1 2 γ 0 t 2 φ ( ( y χ A γ 0 ) ( t γ 0 ) ) ω ( t ) d t + 1 2 t 2 γ φ ( ( x χ A t 2 ) ( t t 2 ) ) ω ( t ) d t + 1 2 t 2 γ φ ( ( y χ A t 2 ) ( t t 2 ) ) ω ( t ) d t .
By formulas (33) and (35), we have
m ( ( B x , u 0 B y , u 0 ) A t 2 ) = m ( ( B x , u 0 B y , u 0 ) e t 2 ) t 0 + γ 0 2 t 2 = t 0 γ 0 2 a 7 a
and, in consequence, we can assume without loss of generality that ( x χ A t 2 ) ( a ) > u 0 . If ( x χ e γ 0 ) ( γ 0 a ) u 0 4 , then
(36)
where p denotes as above the right derivative of φ on the interval [ 0 , δ ) and ω 0 = ω ( t ) for any t ( 0 , γ 0 ) ; note that by the definition of γ 0 , we have ω 0 ω ( t 2 ) > 0 . Hence,
(37)
Now assume that ( x χ e γ 0 ) ( γ 0 a ) > u 0 4 . Then
m ( e γ 0 ( A x , y , u 0 + A x , y , u 0 B x , u 0 ) ) > γ 0 a 3 4 γ 0 ,
whence we get
m ( e γ 0 B y , u 0 ) < 1 4 γ 0 .
(38)
Therefore, by the inequality max ( m ( A x , y , u 0 + ) , m ( A x , y , u 0 ) ) < t 1 1 8 γ 0 , we obtain
m ( e γ 0 ( A x , y , u 0 + A x , y , u 0 B y , u 0 ) ) < 1 2 γ 0 < γ 0 a ,
and, in consequence, ( y χ e γ 0 ) ( γ 0 a ) < u 0 4 . Simultaneously, by formulas (34) and (38) and the equality t 2 = γ 0 + a , we have
m ( B y , u 0 A t 2 ) > t 0 2 + γ 0 4 γ 0 4 a > t 0 2 γ 0 4 a 3 a .
Thus, ( y χ A t 2 ) ( a ) > u 0 , which gives a possibility to repeat the investigations from (36) and (37) for y. In consequence, we have
I φ , ω ( x + y 2 ) 1 3 a p u 0 ( ω 0 ω ( t 2 ) ) 8 .
(39)
Recapitulating Case 2, by inequalities (24), (30) and (39), we get inequality (20) for
q = min ( δ ( η c 4 ) , δ ( p u 0 ω 0 t 1 2 ) , 3 a p u 0 ( ω 0 ω ( t 2 ) ) 8 ) .

Case 3. Finally, assume that 0 γ φ ( δ ) ω ( t ) d t 1 and γ 0 = 0 . For arbitrary fixed v δ > δ , we define the sets A x , v δ and A y , v δ by formulas (22) and (23). If max ( I φ , ω ( x χ A x , v δ ) , I φ , ω ( y χ A y , v δ ) ) 1 8 , then proceeding analogously as in Case 2, we get inequality (24) with the constant δ ( η 32 ) .

If max ( I φ , ω ( x χ A x , v δ ) , I φ , ω ( y χ A y , v δ ) ) < 1 8 , then we define t 0 > 0 and u 0 > 0 by the equalities
0 t 0 φ ( v δ ) ω ( t ) d t = 3 4 and 0 γ φ ( u 0 ) ω ( t ) d t = 1 8 .
We have t 0 < γ , u 0 < δ and min ( m ( A x , u 0 ) , m ( A y , u 0 ) ) t 0 , where the sets A x , u 0 and A x , u 0 are defined by formulas (26) and (27). By the assumption γ 0 = 0 , we can find two positive constants t 2 and t 3 such that 0 < t 3 < t 2 < t 0 2 and ω ( t 3 ) > ω ( t 2 ) . Let
t 1 = t 3 8 and ω 1 = 0 t 1 ω ( t ) d t .

If m ( A x , y , u 0 + ) t 1 or m ( A x , y , u 0 ) t 1 , where the sets A x , y , u 0 + and A x , y , u 0 are defined by formulas (28) and (29), then analogously as in Case 2, we obtain inequality (30) with the constant δ ( p u 0 ω 1 2 ) .

In the case when max ( m ( A x , y , u 0 + ) , m ( A x , y , u 0 ) ) < t 1 , we define the sets B x , u 0 and B y , u 0 by formulas (31) and (32). We have
min ( m ( B x , u 0 ) , m ( B y , u 0 ) ) t 0 2 t 1 7 8 t 0 .

Defining a = min ( t 3 , t 0 2 t 2 ) 4 and repeating the procedure from Case 2, putting t 3 in place of γ 0 , we get inequality (39) with the constant 3 a p u 0 ( ω ( t 3 ) ω ( t 2 ) ) 8 .

Summarizing Case 3, we get inequality (20) with
q = min ( δ ( η 32 ) , δ ( p u 0 ω 1 2 ) , 3 a p u 0 ( ω ( t 3 ) ω ( t 2 ) ) 8 ) .

 □

Theorem 2.5 Let 0 < α < γ < and 0 a φ = δ . Then the Orlicz-Lorentz function space Λ φ , ω is uniformly non-square if and only if φ Δ 2 ( ) , ψ Δ 2 ( ) , ω is regular and α ( γ 2 , γ ) .

Proof Necessity. Condition α ( γ 2 , γ ) 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 q > 0 such that inequality (20) holds for any x , y S ( Λ φ , ω ) .

First note that the space Λ φ , ω ( [ 0 , α ) ) , in opposite to the space Λ φ , ω = Λ φ , ω ( [ 0 , γ ) ) , is uniformly monotone (see Theorem 1.2). Hence, by [[52], Theorem 6], for all δ > 0 there exists p ( δ ) > 0 such that for any u B ( Λ φ , ω ( [ 0 , α ) ) ) and any v Λ φ , ω ( [ 0 , α ) ) with m { supp u supp v } = 0 and v δ , we have
u + v ( 1 + p ( δ ) ) u .
(40)
Now, for any fixed x , y ( Λ φ , ω ) , we denote
A x , y = { t [ 0 , γ ) : max { | x ( t ) | , | y ( t ) | } > a φ } .

In order to show (20), we will consider two cases.

Case 1. If m ( A x , y ) α , then we define
x ˜ = x σ and y ˜ = y σ ,
where σ : [ 0 , m ( A x , y ) ) A x , y is a measure preserving transformation (see [[54], Theorem 17, p.410]). Obviously φ x ˜ , φ y ˜ , φ x ˜ + y ˜ 2 and φ x ˜ y ˜ 2 are equimeasurable with φ x χ A x , y , φ y χ A x , y , φ x + y 2 χ A x , y and φ x + y 2 χ A x , y , respectively. Therefore, by Theorem 2.4, there exists q ( α ) > 0 independent of x and y such that
min ( I φ , ω ( x y 2 ) , I φ , ω ( x + y 2 ) ) = min ( I φ , ω ( ( x y 2 ) χ A x , y ) , I φ , ω ( ( x + y 2 ) χ A x , y ) ) = min ( I φ , ω ( x ˜ y ˜ 2 ) , I φ , ω ( x ˜ + y ˜ 2 ) ) 1 q ( α ) .
Case 2. Let now m ( A x , y ) > α . Denote by m 0 N the smallest possible number satisfying m 0 ( α γ / 2 ) α and let p ( 1 / 2 m 0 ) be the constant from inequality (40) for δ = 1 / 2 m 0 . Fix ε > 0 satisfying
( 1 + p ( 1 2 m 0 ) ) ( 1 6 ε ) > 1 and ε < 1 12 .
(41)
Since ψ Δ 2 ( ) , for a ε satisfying the equality
0 α φ ( a ε ) ω ( t ) d t = ε ,
analogously as in Case 1 of Theorem 2.4, we can find η = η ( a ε ) ( 0 , 1 ) such that
φ ( u 2 ) < 1 η 2 φ ( u )
(42)
for all u a ε . We may assume without loss of generality that
min ( I φ , ω ( x y 2 ) , I φ , ω ( x + y 2 ) ) 1 ε .
(43)
Applying [[16], Property 7, p.64], we can find sets e α ( + ) = e α ( φ ( x + y 2 ) ) and e α ( ) = e α ( φ ( x y 2 ) ) of measure α such that
0 α ( φ ( x + y 2 ) ) ( t ) d t = e α ( + ) φ ( x + y 2 ) ( t ) d t , 0 α ( φ ( x y 2 ) ) ( t ) d t = e α ( ) φ ( x y 2 ) ( t ) d t .
Let us define the sets
A + = { t e α ( + ) : max ( | x ( t ) | , | y ( t ) | ) a ε } , A 1 + = { t A + : x ( t ) y ( t ) > 0 } , A 2 + = { t A + : x ( t ) y ( t ) 0 }
and
A = { t e α ( ) : max ( | x ( t ) | , | y ( t ) | ) a ε } , A 1 = { t A : x ( t ) y ( t ) 0 } , A 2 = { t A : x ( t ) y ( t ) < 0 } .
From [[15], Theorem 2.6, p.49] it follows that there are functions u + and u both equimeasurable with ω χ [ 0 , α ] and satisfying the equalities
(44)
(45)
By the Hardy-Littlewood inequality, we have
e α ( + ) A + φ ( x + y 2 ) ( t ) u + ( t ) d t 0 α ( φ ( ( x + y 2 ) χ e α ( + ) A + ) ) ( t ) ( u + χ e α ( + ) A + ) ( t ) d t < 0 α φ ( a ε ) ω ( t ) d t = ε ,
whence by (43), we conclude that
A + φ ( x + y 2 ) ( t ) u + ( t ) d t 1 2 ε .
(46)
Similarly, we get
A φ ( x y 2 ) ( t ) u ( t ) d t 1 2 ε .
(47)

The remaining part of the proof of Case 2 will be divided into three subcases.

Case 2.1. Suppose A 2 + φ ( x + y 2 ) ( t ) u + ( t ) d t ε . Then
φ ( ( x + y 2 ) ( t ) ) φ ( max ( | x ( t ) | , | y ( t ) | ) 2 ) 1 η 2 { φ ( x ( t ) ) + φ ( y ( t ) ) }
for m-a.e. t A 2 + . Hence, by equality (44), we get
I φ , ω ( x + y 2 ) = 0 α φ ( ( x + y 2 ) ( t ) ) ω ( t ) d t = e α ( + ) φ ( x + y 2 ) ( t ) u + ( t ) d t 1 2 e α ( + ) A 2 + { φ ( x ( t ) ) + φ ( y ( t ) ) } u + ( t ) d t + 1 η 2 A 2 + { φ ( x ( t ) ) + φ ( y ( t ) ) } u + ( t ) d t 1 2 { e α ( + ) φ ( x ( t ) ) u + ( t ) d t + e α ( + ) φ ( y ( t ) ) u + ( t ) d t } η ε 1 η ε .
(48)
Case 2.2. If A 1 φ ( x y 2 ) ( t ) u ( t ) d t ε , then analogously as above, we can show that
I φ , ω ( x y 2 ) 1 η ε .
Case 2.3. Finally, we will prove that the remaining case
A 2 + φ ( x + y 2 ) ( t ) u + ( t ) d t < ε and A 1 φ ( x y 2 ) ( t ) u ( t ) d t < ε
is not possible. In the opposite case, by (46) and (47), we get
A 1 + φ ( x + y 2 ) ( t ) u + ( t ) d t 1 3 ε and A 2 φ ( x + y 2 ) ( t ) u ( t ) d t 1 3 ε .
Since A 1 + A 2 = , we can assume without loss of generality that m ( A 1 + ) γ / 2 . Moreover, by the Hardy-Littlewood inequality and convexity of the modular I φ , ω , we obtain
1 3 ε A 1 + φ ( x + y 2 ) ( t ) u + ( t ) d t 0 m ( A 1 + ) ( φ ( ( x + y 2 ) χ A 1 + ) ) ( t ) ( u + χ A 1 + ) ( t ) d t 0 m ( A 1 + ) ( φ ( ( x + y 2 ) χ A 1 + ) ) ( t ) ω ( t ) d t = I φ , ω ( ( x + y 2 ) χ A 1 + ) 1 2 I φ , ω ( x χ A 1 + ) + 1 2 I φ , ω ( y χ A 1 + ) = 1 2 0 m ( A 1 + ) ( φ x χ A 1 + ) ( t ) ω ( t ) d t + 1 2 0 m ( A 1 + ) ( φ y χ A 1 + ) ( t ) ω ( t ) d t .
Since I φ , ω ( x ) = I φ , ω ( y ) = 1 , so
0 m ( A 1 + ) ( φ x χ A 1 + ) ( t ) ω ( t ) d t 1 6 ε and 0 m ( A 1 + ) ( φ y χ A 1 + ) ( t ) ω ( t ) d t 1 6 ε .
(49)
Similarly,
0 m ( A 2 ) ( φ x χ A 2 ) ( t ) ω ( t ) d t 1 6 ε and 0 m ( A 2 ) ( φ y χ A 2 ) ( t ) ω ( t ) d t 1 6 ε .
(50)
Let e ( α γ / 2 ) = e ( α γ / 2 ) ( φ x χ A 2 ) A 2 be such that m ( e ( α γ / 2 ) ) = α γ / 2 and
0 α γ / 2 ( φ x χ A 2 ) ( t ) d t = e ( α γ / 2 ) φ x χ A 2 ( t ) d t = e ( α γ / 2 ) φ x χ e ( α γ / 2 ) ( t ) d t .
Then, by the definition of m 0 , the first inequality in (50) and the second inequality in (41), we get
φ x χ e ( α γ / 2 ) ω = 0 α γ / 2 ( φ x χ A 2 ) ( t ) ω ( t ) d t 1 6 ε m 0 1 2 m 0 .
Consequently, by (40) (note that m ( supp ( φ x χ A 1 + + φ x χ e ( α γ / 2 ) ) ) γ / 2 + α γ / 2 = α ) and first inequalities of formulas (49) and (41), we obtain
1 = 0 α φ ( x ( t ) ) ω ( t ) d t 0 α ( φ x χ A 1 + + φ x χ e ( α γ / 2 ) ) ( t ) ω ( t ) d t = φ x χ A 1 + + φ x χ e ( α γ / 2 ) ω ( 1 + p ( 1 2 m 0 ) ) φ x χ A 1 + ω ( 1 + p ( 1 2 m 0 ) ) ( 1 6 ε ) > 1 ,

which is a contradiction.

Summarizing both cases, we get inequality (20) with q = min ( q ( α ) , η ε ) . □

Declarations

Acknowledgements

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.

Authors’ Affiliations

(1)
Faculty of Mathematics and Computer Science, Adam Mickiewicz University, Umultowska 87, Poznań, 61-614, Poland
(2)
Institute of Mathematics of Electric Faculty, Poznań University of Technology, Piotrowo 3a, Poznań, 60-965, Poland

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