Orlicz-Lorentz Spaces and Their Composition Operators

Proyecciones Journal of Mathematics Vol. 34, No 1, pp. 85-105, March 2015. Universidad Cat´olica del Norte Antofagasta - Chile

Orlicz-Lorentz Spaces and their Composition Operators

René Erlin Castillo Universidad Nacional de Colombial, Colombia Héctor Camilo Chaparro Universidad Nacional de Colombia, Colombia and Julio César Ramos Fernández Universidad de Oriente, Venezuela Received : November 2012. Accepted : March 2014

Abstract In a self-contained presentation, we discuss the Orlicz-Lorentz space. Also the boundedness of composition operators on Orlicz-Lorentz spaces are characterized in this paper.

2010 Mathematics Subject Clasification : Primary 47B33, 47B38, secondary 46E30. 86 RenéE.Castillo,Héctor C. Chaparro and Julio C. Ramos

1. Introduction

Let f a complex-valued measurable function defined on a σ-finite measure space (X, ,µ). For λ 0, define Df (λ) the distribution function of f as A ≥

(1.1) Df (λ)=µ ( x X : f(x) >λ ) . { ∈ | | }

Observe that Df depends only on the absolute value f of the function f | | and Df may assume the value + . ∞ The distribution function Df provides information about the size of f but not about the behavior of f itself near any given point. For instance, a function on Rn and each of its translates have the same distribution function. It follows from (1.1) that Df is a decreasing function of λ (not necessarily strictly) and continuous from the right. Let (X, µ) be a measurable space and f and g be a measurable functions on (X, µ)thenDf enjoy the following properties for all λ1,λ2 0: ≥

1. g f µ-a.e. implies that Dg Df ; | | ≤ | | ≤ λ 2. Dcf (λ)=Df c for all c C 0 ; | | ∈ { } ³ ´ 3. Df+g(λ1 + λ2) Df (λ1)+Dg(λ2); ≤

4. Dfg(λ1λ2) Df (λ1)+Dg(λ2). ≤ For more details on distribution function see [5]. By f ∗ we mean the non-increasing rearrangement of f given as

f ∗(t)=inf λ>0:Df (λ) t ,t 0 { ≤ } ≥

where we use the convention that inf = . f ∗ is decreasing and right- continuous. Notice ∅ ∞

f ∗(0) = inf λ>0:Df (λ) 0 = f , { ≤ } k k∞ since

f =inf α 0:µ( x X : f(x) >α )=0 . k k∞ { ≥ { ∈ | | } } Also observe that if Df is strictly decreasing, then

f ∗(Df (t)) = inf λ>0:Df (λ) D f)t = t. { ≤ ( } Orlicz-Lorentz Spaces and their Composition Operators 87

This fact demonstrates that f ∗ is the inverse function of the distribution function Df .Let (X, )denotethesetofall -measurable functions on F A A X.Let(X, 0,µ)and(Y, 1,ν)betwomeasurespaces. A A Two functions f F (X, 0)andg F (X, 1) are said to be equimea- surable if they have the∈ sameA distribution∈ function,A that is, if

µ ( x X : f(x) >λ )=ν ( y Y : g(y) >λ ) , for all λ 0. { ∈ | | } { ∈ | | } ≥ (1.2)

So then there exists only one right-continuous decreasing function f ∗ equimea- surable with f. Hence the decreasing rearrangement is unique. In what follows, we gather some useful properties of the decreasing rearrangement function:

a) f ∗ is decreasing.

b) f ∗(t) >λif and only if Df (λ) >t.

c) f and f are equimeasurables, that is Df (λ)=Df (λ) for all λ 0. ∗ ∗ ≥

d) If f lim infn fn then f ∗ lim infn fn∗. | | ≤ →∞ | | ≤ →∞

e) If E ,then(χE)∗ (t)=χ (t). ∈ A [0,µ(E))

f) If E ,then(fχE)∗ (t) f (t)χ (t). ∈ A ≤ ∗ [0,µ(E)) A weight is a nonnegative locally integrable function on Rn that takes values in (0, ) almost everywhere. Therefore, weights are allowed to be zero or inﬁnite∞ only on a set of Lebesgue measure zero. Let ϕ :[0, ) [0, ) be a convex function such that ∞ → ∞

1. ϕ(x) = 0 if and only if x =0;

2. limx ϕ(x)= . →∞ ∞ 88 Ren´eE.Castillo,H´ector C. Chaparro and Julio C. Ramos

Such as function is known as a Young function. A Young function is strictly x increasing, in fact, let 0

x x x = 1 0+ y. y y µ − ¶ Since ϕ is convex, we have

x x ϕ(x)=ϕ 1 0+ y y y µµ − ¶ ¶ x x 1 ϕ(0) + ϕ(y) y y ≤ µ − ¶ <ϕ(y).

A Young function is said to satisfy the ∆2-condition if there exists a nonnegative constant x0 and k such that

(1.3) ϕ(2x) kϕ(x)forx x0. ≤ ≥ If x0 =0,wesaythatϕ satisfy globally the ∆2-condition. The smaller constant k which satisfy (1.3) is denoted by k∆.

Claim 1.1. If ϕ is a Young function such that satisfy the ∆2-condition, then for each r 0 there exists a constant k∆(r) such that ≥

(1.4) ϕ(rx) k∆(r)ϕ(x) ≤ for x>0 large enough.

Proof. [Proof of the claim.] If r>0, we can choose n N such that r 2n. Then we can applied (1.3) n-times and use the∈ fact that ϕ is increasing≤ to obtain

ϕ(rx) ϕ(2nx) knϕ(x), ≤ ≤ and hence we have (1.4). 2

xp Example 1.2. The function ϕ1(x)= p with p>1 is a Young function 2p which satisfy globally the ∆2-condition with k∆ = p . Orlicz-Lorentz Spaces and their Composition Operators 89

p Example 1.3. The function ϕ2(t)=t log(1 + t) with p 1 and t 0 is ≥ ≥ a Young function which satisfy the ∆2-condition, indeed, since p p ϕ2(2t) 2 t log(1 + 2t) p 1 lim =lim =2 − . t ϕ (t) t tp log(1 + t) →∞ 2 →∞

Also, ϕ2 satisfy globally the ∆2-condition. In fact, since for each t 0 we have (1 + t)2 1+2t,then ≥ ≥ p p ϕ2(2t)=2 t log(1 + 2t) 2p+1tp log(1 + 2t) ≤ p+1 2 ϕ2(2t). ≤

Lemma 1.4. A Young function ϕ satisfy the ∆2-condition if and only if there exist constants λ>1 and t0 > 0 such that tp(t) <λ ϕ(t) for all t t0,wherep is the right derivate of ϕ. ≥

Proof. Suppose that ϕ satisfy the ∆2-condition, then there exists a constant k>0 such that 2t 2t kϕ(t) ϕ(2t)= p(s) ds > p(s) ds ≥ 0 t Z Z for t large enough, since p is increasing, then we have

2t p(s) ds > tp(t); Zt hence, for t large enough, we obtain tp(t) k. ϕ(t) ≤ Conversely, if tp(t) <λ ϕ(t) for all t t0,then ≥ 2t p(s) 2t ds ds < λ = λ log 2. ϕ(s) s Zt Zt 90 Ren´eE.Castillo,H´ector C. Chaparro and Julio C. Ramos

Since p(s)=ϕ0(s), we have ϕ(2t) log <λlog 2, ϕ(t) µ ¶ which implies that ϕ(2t) < 2λϕ(t). 2 The following result show us that the Young functions which satisfy the p ∆2-condition have a cross rate less than the function t for some p>1.

Theorem 1.5. If ϕ is a Young function which satisfy the ∆2-condition, then there exists constants λ>1 and C>0 such that

ϕ(t) Ctλ ≤ for t large enough.

Proof. By (1.4) we can write

t p(s) t ds ds < λ ϕ(s) s Zt0 Zt0

where t t0.Then ≥ ϕ(t) t log <λlog , ϕ(t ) t µ 0 ¶ µ 0 ¶ therefore ϕ(t0) λ ϕ(t) < λ t . t0 And the proof is complete. 2

ψ(t)=sup ts ϕ(s):s 0 . { − ≥ } Orlicz-Lorentz Spaces and their Composition Operators 91

1 p Example 1.7. If ϕ(t)= p t with p>1 and t 0,thenitscomplementary 1 q 1 1 ≥ function is ψ(t)= q t where p + q =1. Indeed, by deﬁnition we have 1 ψ(t)=sup ts sp : s 0 , p ½ − ≥ ¾ next, for t>0 ﬁxed, we can consider the function 1 g(s)=ts sp, with s 0. − p ≥

1 p 1 It is not hard to check that g achieves its maximum at s = t − which is given by 1 1 g t p 1 = tq. − q ³ ´ Hence 1 1 ψ(t)=sup ts sp : s 0 = tq. p q ½ − ≥ ¾ Proposition 1.8. If ϕ is a Young function, then its complementary function ψ is also a Young function.

Proof. It is clear that ψ(0) = 0 if and only if x = 0. Now, we just need to show that ψ is a convex function. To this end, let us choose t1,t2 [0, + ) and λ [0, 1]. Then, by deﬁnition of ψ we have ∈ ∞ ∈

ψ(λt1 +(1 λ)t2)=sup s(λt1 +(1 λ)t2) ϕ(s):s 0 . − { − − ≥ } On the other hand

λψ(t1)=λ sup st1 ϕ(s):s 0 λ(st1 ϕ(s)) s 0 { − ≥ } ≥ − ∀ ≥ and

(1 λ)ψ(t2)=(1 λ)sup st2 ϕ(s):s 0 (1 λ)(st2 ϕ(s)) s 0. − − { − ≥ } ≥ − − ∀ ≥ From the last two inequalities, we have

s(λt1 +(1 λ)t2) ϕ(s)=λ(st1 ϕ(s)) + (1 λ)(st2 ϕ(s)) − − − − −

λψ(t1)+(1 λ)ψ(t2) ≤ − 92 Ren´eE.Castillo,H´ector C. Chaparro and Julio C. Ramos

for all s 0. Which means that λψ(t1)+(1 λ)ψ(t2) is an upper bound of the set≥ − s(λt1 +(1 λ)t2) ϕ(s):s 0 , { − − ≥ } then

ψ(λt1 +(1 λ)t2)) ψ(t1)+(1 λ)ψ(t2), − ≤ − and so ψ is convex. 2

Theorem 1.9 (Young’s Inequality). Let ψ be the complementary function of ϕ.Then ts ϕ(s)+ψ(t) ≤ where t, s [0, + ). ∈ ∞ Proof. Let t, s [0, + ). Then ψ(t)=sup st ϕ(s):s 0 st ϕ(s) s ∈0, then∞ { − ≥ } ≥ − ∀ ≥ ψ(t)+ϕ(s) st, ≥ and the proof is complete. 2 For more details on Young functions see [13].

2. Weighted Lorentz-Orlicz Spaces

The aim of this section is to present basic results about Lorentz-Orlicz spaces. We have tried to make the proofs as self-contained and synthetic as possible. Deﬁnition 2.1 (Luxemburg norm). Let ϕ be a Young function. For any measurable function f on X,

∞ f ∗(t) f ϕ,w =inf ε>0: ϕ w(t) dt 1 [0, ), k k 0 ε ≤ ∈ ∞ ½ Z µ ¶ ¾ where it is understood that inf( )=+ . ∅ ∞ Remark 2.2. In this article, we will not always require that the Luxem- burg norm actually be a norm. ϕ,w is indeed a quasinorm. A quasinorm is a functional that is like a normk·k except that it does only satisfy the tri- angle inequality with a constant C 1,thatis, f + g C( f + g ) where C 1. ≥ k k ≤ k k k k ≥ Lemma 2.3. For any measurable function f on X, f ϕ,w =0if and only if f =0µ-almost everywhere. k k Orlicz-Lorentz Spaces and their Composition Operators 93

f ∗(t) Proof. Clearly f ϕ,w =0ifandonlyif ∞ ϕ w(t) dt 1 ε>0. k k 0 ε ≤ ∀ It follows that R ³ ´

∞ f ϕ,w =0ifandonlyif ϕ (αf ∗(t)) w(t) dt =0 α>0 k k 0 ∀ Z

if and only if ϕ (αf ∗(t)) w(t)=0µ a.e. α>0 − ∀

if and only if f ∗(t)=0µ a.e. −

if and only if Df (λ)=0µ a.e. − if and only if f =0µ a.e. − 2 Identification of almost everywhere equal functions. As with Lp spaces, one identifies the function which are µ-almost everywhere equal. This means that one works with the equivalence classes of the equivalence rela- tion defined by the µ-almost everywhere equality. From now on, this will be done without further mention. Consequently, one write:

(2.1) f ϕ,w =0ifandonlyiff =0. k k

f ∗(t) Lemma 2.4. If 0 < f < then ∞ ϕ w(t) dt 1.In ϕ,w 0 f ϕ,w k k ∞ k k ≤ particular, f ϕ,w 1 is equivalent to ∞ ϕ (f (³t)) w(t)´dt 1. k k ≤ 0 R ∗ ≤ R

Proof. For all b> f ϕ,w,wehave k k f (t) ∞ ϕ ∗ w(t) dt 1. 0 b ≤ Z µ ¶

Letting b decrease to f ϕ,w, one obtains the ﬁrst result by monotone con- vergence. The secondk statementk follows from this and lemma 2.8. 2

Proposition 2.5. The gauge ϕ,w is a quasinorm on the vector space k·k of all the measurable functions f such that f ϕ,w < . k k ∞ 94 Ren´eE.Castillo,H´ector C. Chaparro and Julio C. Ramos

Proof. It is already seen that (2.1) holds under identiﬁcation of a.e. equal functions. It is clear that for all real λ, λf ϕ,w = λ f ϕ,w. It remains to prove the trianglek inequality.k | |k Letk f and g be two measurable functions such that 0 < f ϕ,w + g ϕ,w < .Then k k k k ∞

(f + g) (t) ∞ ϕ ∗ w(t) dt 0 2( f ϕ,w + g ϕ,w) Z Ã k k k k ! f (t/2) + g (t/2) ∞ ϕ ∗ ∗ w(t) dt ≤ 0 2( f ϕ,w + g ϕ,w) Z Ã k k k k ! f ϕ,w f (t/2) g ϕ,w g (t/2) = ∞ ϕ k k ∗ + k k ∗ w(t) dt 0 2( f ϕ,w + g ϕ,w) f ϕ,w 2( f ϕ,w + g ϕ,w) g ϕ,w Z Ã k k k k k k k k k k k k ! f ϕ,w f (t/2) k k ∞ ϕ ∗ w(t) dt ≤ 2( f ϕ,w + g ϕ,w) 0 f ϕ,w k k k k Z Ã k k ! g ϕ,w g (t/2) + k k ∞ ϕ ∗ w(t) dt 2( f ϕ,w + g ϕ,w) 0 f ϕ,w k k k k Z Ã k k ! f ϕ,w f (t) = k k 2 ∞ ϕ ∗ w(2t) dt 2( f ϕ,w + g ϕ,w) 0 f ϕ,w k k k k Z Ãk k ! g ϕ,w g (t) + k k 2 ∞ ϕ ∗ w(2t) dt 2( f ϕ,w + g ϕ,w) 0 f ϕ,w k k k k Z Ãk k ! f ϕ,w f (t) k k ∞ ϕ ∗ w(t) dt ≤ f ϕ,w + g ϕ,w 0 f ϕ,w k k k k Z Ãk k ! g ϕ,w g (t) + k k ∞ ϕ ∗ w(t) dt f ϕ,w + g ϕ,w 0 f ϕ,w k k k k Z Ãk k ! 1, ≤ where the last but one inequality follows from the convexity of ϕ and the fact that w is nonincreasing and the last inequality from lemma 2.4. There- fore f + g ϕ,w 2( f ϕ,w + g ϕ,w) . k k ≤ k k k k As a consequence, the set of all measurable functions f such that f ϕ,w < is a vector space. 2 k k ∞ Orlicz-Lorentz Spaces and their Composition Operators 95

Deﬁnition 2.6. Let ϕ be a Young function. We deﬁne the weighted Lorenz-Orlicz spaces

∞ Lϕ,w = f : X C measurable : ϕ(αf ∗(t))w(t) dt < , for some α>0 . → 0 ∞ ½ Z ¾

It follows from proposition 1.8 that if Lϕ,w is a weighted Lorentz-Orlicz space, then Lψ,w is also a weighted Lorenz-Orlicz space.

Proposition 2.7 (H¨older’s type inequality). For f Lϕ,1 and g ∈ ∈ Lψ,1

fg dµ 2 f ϕ,1 g ψ,1. X | | ≤ k k k k Z In particular, fg L1. ∈

Proof. If f ϕ,1 =0or g ψ,1 = 0, one concludes with lemma 2.3. k k k k Assume now that 0 < f ϕ,1, g ψ,1. Because of Young’s inequality: st ϕ(s)+ϕ(t)wehave k k k k ≤ fg f (t)g (t) | | dµ ∞ ∗ ∗ dt X f ϕ,1 g ψ,1 ≤ 0 f ϕ,1 g ψ,1 Z k k k k Z k k k k f (t) g (t) ∞ ϕ ∗ dt + ∞ ψ ∗ dt ≤ 0 f ϕ,1 0 g ψ,1 Z Ãk k ! Z Ãk k ! 2. ≤ Therefore

fg dµ 2 f ϕ,1 g ψ,1. X | | ≤ k k k k Z 2

Lemma 2.8. Let fn n N be a sequence in Lϕ,w. Then, the following assertions are equivalent:{ } ∈

a) limn fn ϕ,w =0; →∞ k k

b) For all α>0, lim supn 0∞ ϕ(αfn∗(t))w(t) dt 1; →∞ ≤ R c) For all α>0, limn 0∞ ϕ(αfn∗(t))w(t) dt =0. →∞ R 96 Ren´eE.Castillo,H´ector C. Chaparro and Julio C. Ramos

Proof. The equivalence (a) (b) is a direct consequence of the deﬁni- ⇔ tion of ϕ,w.Oﬀ course (c) (b)isobvious.Asϕ is convex and ϕ(0) = 0 for all tk·k0and0<ε 1, we⇒ have ≥ ≤ t t ϕ(t)=ϕ (1 ε)0 + ε (1 ε)ϕ(0) + εϕ , ε ε µ − ¶ ≤ − µ ¶ that is t ϕ(t) εϕ t 0, 0 <ε 1. ε ≤ µ ¶ ≥ ≤ From which (b) (c) follows easily. 2 ⇒

Theorem 2.9. The space Lϕ,w is a quasi-Banach space.

Proof. Let fn n N be a Cauchy sequence in Lϕ,w.Letuschoose˜ε>0 1{ ε} ∈ 1 such thatεϕ ˜ − < for n, m N and ε>0,k0 > 0. For suchε ˜ k0 n+m ∈ there exists n0 ³ N´such that ∈

fn fm ϕ,w < ε.˜ k − k

If n, m n0. By the deﬁnition of the Luxemburg quasi-norm we can use ≥ k0 > 0insuchawaythatk0 < ε˜ and

(fn fm) (t) ∞ ϕ − ∗ w(t) dt 1. 0 k ≤ Z µ 0 ¶

Let E = x X : fn(x) fm(x) >ε ,then { ∈ | − | }

εχE(x) fn(x) fm(x) . ≤ | − | Hence εχ∗ (t) (fn fm)∗(t), E ≤ − εχ (t) (fn fm)∗(t). (0,µ(E)) ≤ − Therefore

∞ ε ∞ (fn fm)∗(t) ϕ χ(0,µ(E))(t) w(t) dt ϕ − w(t) dt. 0 k ≤ 0 k Z µ 0 ¶ Z µ 0 ¶ Then

µ(E) ε (fn fm) (t) ϕ w(t) dt ∞ ϕ − ∗ w(t) dt 0 k ≤ 0 k Z µ 0 ¶ Z µ 0 ¶ Orlicz-Lorentz Spaces and their Composition Operators 97

Dfn fm(ε) − 1 ε ∞ (fn fm)∗(t) ε˜ w(t) dt εϕ˜ − ϕ − w(t) dt ⇒ 0 ≤ k 0 k Z µ 0 ¶ Z µ 0 ¶ D fn fm(ε) 1 ε˜ − w(t) dt ⇒ 0 ≤ n + m Z Dfn fm(ε) ε˜ lim − w(t)=0. ⇒ n,m 0 →∞ Z

Since w>0, we must have limn,m Dfn fm (ε)=0whichmeans →∞ − that fn n N is a Cauchy sequence in measure. Then some subsequence { } ∈ fnk k N converges almost everywhere to a measurable function f,thatis, { } ∈ fnk fµ-a.e. Let→ α>0. By lemma 2.8 there exists a large enough integer n(α)such that ∞ ϕ (α(fn fm)∗(t)) w(t) dt 1, m, n n(α). 0 − ≤ ∀ ≥ Z With Fatou’s lemma this gives

∞ ∞ ϕ (α(fn f)∗(t)) w(t) dt lim inf ϕ (α(fn fm)∗(t)) w(t) dt 1 0 − ≤ 0 − ≤ Z Z m n(α). Therefore fn f belongs to Lϕ,w, but fn Lϕ,w,sothat ∀ ≥ − ∈ f Lϕ,w. ∈ Moreover, as lim supm 0∞ ϕ (α(fm f)∗(t)) w(t) dt 1 for all α> 0, we have lim f →∞f =0.Thisprovesthat− L ≤ is complete. m m ϕ,wR ϕ,w 2 →∞ k − k

Theorem 2.10. Simple functions are dense in Lϕ,w.

Proof. Suppose f Lϕ,w. We may assume that f 0. Note that if ∈ ≥ Df (λ)= ,thenlimt f ∗(t) = 0. It follows that Df (λ) < . ∞ →∞ ∞ Hence, given ε, δ > 0, we can ﬁnd a simple function sn 0 such that ≥ sn(x)=0whenf(x) ε and f(x) ε sn(x) f(x)whenf(x) >ε except on a set of measure≤ less than δ−. It≤ follows that≤

µ ( x X : f(x) sn(x) >ε ) <δ. { ∈ | − | } Next, choose n N such that n 1 ,then ∈ ≥ ε

(f sn)∗(t)=inf ε>0:Df sn (ε) <δ t . − { − ≤ } Thus 1 (f sn)∗(t) for t δ, − ≤ n ≥ 98 Ren´eE.Castillo,H´ector C. Chaparro and Julio C. Ramos

1 since sn f,thens (t) f (t), for each t>0. Since n> ,wehave ≤ n∗ ≤ ∗ ε 1 (f sn)∗(t) <ε, − ≤ n next, (f sn) (t) 1 ∞ ϕ − ∗ w(t) dt ∞ ϕ w(t) dt. 0 k ≤ 0 nk Z µ ¶ Z µ ¶ Let a = 0∞ w(t) dt,then R

∞ (f sn)∗(t) f sn ϕ,w =inf k>0: ϕ − w(t) dt 1 k − k 0 k ≤ ½ Z µ ¶ ¾ 1 = 0asn . 1 1 → →∞ nϕ− a 2 ³ ´

3. Composition Operator

Let (X, ,µ)beaσ-finite complete measure space and let T : X X be A 1 → a measurable transformation, that is, T − (A) for any A . 1 ∈ A ∈ A If µ T − (A) =0forallA with µ(A)=0,thenT is said to be ∈ A 1 nonsingular. This condition means that the measure µ T − ,defined by 1 ¡ ¢ 1 ◦ µ T − (A)=µ T − (A) for A is absolutely continuous with respect to µ◦(it is usually denoted µ T 1∈ A µ). Then the Radon-Nikodym theorem ¡ ¢ ◦ − ¿ ensure the existence of a non-negative locally integrable function fT on X such that 1 µ T − (A)= fT dµ for A . ◦ A ∈ A Z Any measurable nonsingular transformation T induces a linear operator (composition operator) CT from F (X, ,µ) into itself defined by A CT (f)(x)=f (T (x)) ,x X, f F (X, ,µ), ∈ ∈ A where F (X, ,µ) denotes the linear space of all equivalence classes of - measurable functionsA on X, where we identify any two functions that areA equal µ-almost everywhere on X. Here the nonsingularty of T guarantees that the operator CT is well defined as a mapping of equivalence classes of functions into itself since f = gµ-a.e. implies CT (f)=CT (g) µ-a.e. Orlicz-Lorentz Spaces and their Composition Operators 99

Example 3.1. Let ([0, 1], ,m) be a Lebesgue measure space, stand for the Borel’s σ-algebra and TB :[0, 1] [0, 1] a transformation deﬁBned by → 2x, if 0 x 1 T (x)= 2 1, if 1

CT (f)=CT χ[0,1) = χ ³T ´ [0,1) ◦ = χ 1 [0, 2 ) and

CT (g)=CT χ[0,1] = χ ³T ´ [0,1] ◦ = χ[0,1].

Then CT (f) = CT (g), which means that CT is not well deﬁned. In other words,6 the nonsingularity of T is a necessary condition in order to T induces a composition operator on F (X, ,µ). A Composition operators are relatively simple operators with a wide range of applications in areas such a partial diﬀerential equations, group repre- sentation theory, ergodic theory or dynamical systems, etc. For details on composition operator see [7, 10, 11, 12, 14, 15] and the references given therein. In what follows, we will consider the transformation CT from Lϕ,w into the space of all complex-valued measurable functions on X as

f (T (x)) , if x Y (CT f)(x)== ∈ ( 0, otherwise 100 RenéE.Castillo,Héctor C. Chaparro and Julio C. Ramos where Y is a measurable subset of X. Next, a necessary and sufficient condition for the boundedness of composition mapping CT is given. If (X, ,µ)isaσ-finite measure space and T : X X is a non-singular measurableA transformation and w is a weight function,→ define a measure ν on the σ-algebra as A µ(A) ν(A)= w(t) dt. Z0 Next, for A , ∈ A χA∗ (t) χA ϕ,w =inf k>0: ∞ ϕ w(t) dt 1 k k 0 k ≤ n R ³ ´ o χ (t) =inf k>0: ∞ ϕ (0,µ(A)) w(t) dt 1 0 k ≤ ( Z Ã ! ) µ(A) 1 =inf k>0: ϕ w(t) dt 1 . 0 k ≤ ( Z µ ¶ ) 1 Now, observe that if k = 1 1 ,then ϕ− ν(A) ¡ ¢ 1 1 1 1 ϕ 1 = ϕ ϕ− = , ⎛ 1 ⎞ ν(A) ν(A) ϕ− (ν(A)) µ µ ¶¶ ⎝ ⎠ thus

µ(A) 1 µ(A) 1 1 0 ϕ 1 w(t) dt = 0 ϕ ϕ− ν(A) w(t) dt 1 Ã ϕ− (ν(A)) ! R R ³ ³ ´´ µ(A) w(t) = dt ν(A) Z0 1 µ(A) = w(t) dt ν(A) Z0 1 = ν(A) ν(A) · =1. Therefore Orlicz-Lorentz Spaces and their Composition Operators 101

1 χA ϕ,w = . k k 1 1 ϕ− ν(A) ³ ´ Theorem 3.2. Let T : X X be a non-singular measurable transforma- → tion. Then CT induced by T is bounded on Lϕ,w if and only if there exists M 1 such that ≥ 1 (3.1) ν T − (A) Mν(A) A . ≤ ∀ ∈ A Moreover ³ ´ 1 ν(T − (A)) (3.2) CT (f) =sup . k k 0<ν(A)< Ã ν(A) ! ∞

Proof. Let CT be a bounded transformation on Lϕ,w,thenwecanﬁnd M 1 such that ≥

CT f ϕ,w M f ϕ,w f Lϕ,w. k k ≤ k k ∀ ∈ If A is such that ν(A)= , then (3.1) holds. Suppose A is such that∈νA(A) < ,thus ∞ ∈ A ∞ 0∞ ϕ (αχA∗ (t)) w(t) dt = 0∞ ϕ αχ0,µ(A)(t) w(t) dt R R ³ ´ µ(A) = ϕ(α)w(t) dt Z0 = ϕ(α)ν(A) < . ∞ Hence

(3.3) CT χA ϕ,w M χA ϕ,w. k k ≤ k k Note

1, if T(x) A (χA T )(x)=χA (T (x)) = ∈ ◦ 0, if T(x)/ A ( ∈ 1, if x T 1(A) = − 0, if x∈/ T 1(A) ( ∈ − = χ 1 (x). T − (A) Then 102 Ren´eE.Castillo,H´ector C. Chaparro and Julio C. Ramos

CT χA ϕ,w = χ 1 ϕ,w k k k T − (A)k 1 = , 1 1 ϕ 1 − ν(T − (A)) and ³ ´ 1 χA ϕ,w = . k k 1 1 ϕ− ν(A) ³ ´ Hence, we can write (3.3) as follows

1 M 1 1 1 1 ϕ 1 ≤ ϕ − ν(T − (A)) − ν(A) ³ ´ ³ ´ and so 1 1 ϕ 1 ϕ 1 . − ν(A) − ν(T 1(A)) µ ¶ ≤ µ − ¶ 1 1 1 1 Since ϕ− is concave and 0 = ϕ− (ϕ(0)) = ϕ− (0) thus ϕ− is increasing, then 1 1 M 1 ν(A) ≤ ν (T − (A)) 1 ν(T − (A)) Mν(A). ≤ Conversely, if inequality (3.1) holds for all A ,then Therefore ∈ A (f T )∗(t) Mf∗(t)a.e. ◦ ≤ Since ϕ(αt) αϕ(t)forα<1, then ≤

(f T ) (t) 1 f (t) ∞ ϕ ◦ ∗ w(t) dt ∞ ϕ ∗ w(t) dt 0 M f ϕ,w ≤ M 0 f ϕ,w Z Ã k k ! Z Ãk k ! f (t) ∞ ϕ ∗ w(t) dt 1. ≤ 0 f ϕ,w ≤ Z Ãk k ! Finally f T ϕ,w M f ϕ,w, k ◦ k ≤ k k Orlicz-Lorentz Spaces and their Composition Operators 103 that is CT f ϕ,w M f ϕ,w. k k ≤ k k On the one hand, let us prove (3.2). Indeed, let

ν T 1(A) N =sup − , 0<ν(A)< Ã ν(A) ! ∞ ¡ ¢ then 1 ν T − (A) Nν(A) ≤ and thus ³ ´ CT f ϕ,w N f ϕ,w, f Lϕ,w k k ≤ k k ∀ ∈ hence CT f ϕ,w k k N, for all 0 = f Lϕ,w. f ϕ,w ≤ 6 ∈ k k Therefore

CT (f) ϕ,w CT =supk k k k f=0 f ϕ,w 6 k k ν T 1(A)

CT (f) ϕ,w M = CT =supk k , k k f=0 f ϕ,w 6 k k then CT (f) ϕ,w k k M 0 = f Lϕ,w. f ϕ,w ≤ ∀ 6 ∈ k k In particular, if f = χA such that 0 <µ(A) < , A ,then ∞ ∈ A 1 CT (χA) ϕ,w ν T (A) k k = − M, χA ϕ,w ν(A) ≤ k k Ã ¡ ¢! therefore 1 ν T − (A) (3.5) sup M = CT . 0<ν(A)< Ã ν(A) ! ≤ k k ∞ ¡ ¢ 104 Ren´eE.Castillo,H´ector C. Chaparro and Julio C. Ramos

Combining 3.4 and 3.5 we have

1 ν T − (A) CT =sup . k k 0<ν(A)< Ã ν(A) ! ∞ ¡ ¢ 2

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René Erlin Castillo Departamento de Matemáticas, Universidad Nacional de Colombia, Bogotá, Colombia e-mail : [email protected]

Héctor Camilo Chaparro Departamento de Matemáticas, Universidad Nacional de Colombia, Bogotá, Colombia e-mail : [email protected]

and

Julio César Ramos Fernández Departamento de Matemáticas Universidad de Oriente, Cumaná, Estado Sucre, Venezuela e-mail : [email protected]