Generalized perfect spaces

Indag. Mathern., N.S., 19 (3), 359-378 Generalized perfect spaces * by J.M. Calabuig, O. Delgado and E.A. Sanchez Perez September, 2008 Instituto Universitario de Matematlca Pura yAplicada, Universidad Politi!Cnica de Valencia, 46071 Valencia, Spain Communicated by Prof. H.W Broer ABSTRACT Given two Banach function spaces Xand Yrelated to ameasure JL, the Y-dual space XY of Xis defined as the space of the multipliers from Xto Y. The space XYis ageneralization of the classical Kothe dual space of X, which is obtained by taking Y=LJ(JL). Under minimal conditions, we can consider the Y-bidual space X YY of X(i.e. the Y-dual of XY ). As in the classical case, the containment X C X YY always holds. We give conditions guaranteeing that Xcomcides with X YY, in which case Xis said to be Y-perfect. We also study when Xis isometncally embedded in X YY. Properties involving p-convexity, p-concavity and the order of Xand Y, will have aspecial relevance. 1. INTRODUCTION In the theory of function spaces on ameasure space (Q, ~,/1), the classical Kothe dual (or associate) space X' of aBanach function space (briefly, B.f.s.) X, plays an important role due to the fact that it is identified with the elements x* of X*, MSC: Primary 46E30; Secondary 46B42 Key words and phrases: Banach function spaces, Kothe dual, Perfect spaces, Fatou property, p-Convexity, p-Concavity '" 1M. CalabUig was supported by Generalitat Valenciana (prQject GV/2007/191), MEC (project MTM2005-08350-C03-03) (Spain) and FEDER. O. Delgado was supported by Generalitat Valenciana (TSGD-07 and project GVPRE/2008/3l2), MEC (TSGD-08), D.G.I. #MTM2006-13000-C03-0l (Spain) and FEDER. E.A. Sanchez Perez was supported by MEC (project #MTM2006-ll690-C02-0l) (Spain) and FEDER. E-mails:[email protected](lM.Calabuig)[email protected] (0. Delgado), [email protected] (E.A. Sanchez Perez). 359 Indag. Mathern., N.S., 19 (3), 359-378 Generalized perfect spaces * by J.M. Calabuig, O. Delgado and E.A. Sanchez Perez September, 2008 Instituto Universitario de Matematlca Pura yAplicada, Universidad Politi!Cnica de Valencia, 46071 Valencia, Spain Communicated by Prof. H.W Broer ABSTRACT Given two Banach function spaces Xand Yrelated to ameasure JL, the Y-dual space XY of Xis defined as the space of the multipliers from Xto Y. The space XYis ageneralization of the classical Kothe dual space of X, which is obtained by taking Y=LJ(JL). Under minimal conditions, we can consider the Y-bidual space X YY of X(i.e. the Y-dual of XY ). As in the classical case, the containment X C X YY always holds. We give conditions guaranteeing that Xcomcides with X YY, in which case Xis said to be Y-perfect. We also study when Xis isometncally embedded in X YY. Properties involving p-convexity, p-concavity and the order of Xand Y, will have aspecial relevance. 1. INTRODUCTION In the theory of function spaces on ameasure space (Q, ~,/1), the classical Kothe dual (or associate) space X' of aBanach function space (briefly, B.f.s.) X, plays an important role due to the fact that it is identified with the elements x* of X*, MSC: Primary 46E30; Secondary 46B42 Key words and phrases: Banach function spaces, Kothe dual, Perfect spaces, Fatou property, p-Convexity, p-Concavity '" 1M. CalabUig was supported by Generalitat Valenciana (prQject GV/2007/191), MEC (project MTM2005-08350-C03-03) (Spain) and FEDER. O. Delgado was supported by Generalitat Valenciana (TSGD-07 and project GVPRE/2008/3l2), MEC (TSGD-08), D.G.I. #MTM2006-13000-C03-0l (Spain) and FEDER. E.A. Sanchez Perez was supported by MEC (project #MTM2006-ll690-C02-0l) (Spain) and FEDER. E-mails:[email protected](lM.Calabuig)[email protected] (0. Delgado), [email protected] (E.A. Sanchez Perez). 359 the topological dual space of X, acting on Xas an integral, i.e. there exists a measurable function gsuch that x* (f) =Jfgd J-L for all f EX. We can interpret X' in adifferent way, as the space of the multipliers from Xto L1(J-L), that is, the measurable functions gdefining amultiplication operator from X to LI (J-L). From this point of view, ageneralization of the Kothe dual is possible by taking any B.f.s. Yin the place of L1 (J-L). Namely, the Y-dual space of X, denoted by XY,is the space of multipliers from Xto Y, which under an elementary requirement becomes aB.f.s. when endowed with the usual operator norm. The Kothe dual of aB.f.s. takes acrucial part in the interplay between the order and the topology of X. For instance, Xis order continuous if and only if X* coincides with X', or Xsatisfies the Fatou property if and only if Xcoincides with the Kothe bidual X" of X. We pay attention to this second case, in which X is said to be aperfect space. Is there an analogous result for the general case? Or in other words, denoting by X YY the Y-bidual space of X(i.e. the Y-dual of XY ), when does Xcoincide with XY Y? When it does, it is called Y-perfect. We will study this question, which was already posed by Maligranda and Persson in [11, p. 337]. In this item they present some properties concerning the generalized duality and provides adescription for the space XYin some special cases, all of them of great usefulness in the development of this paper. As in the classical case, the containment XCX YY always holds and is continuous with IIf11 x Yy ~ IIf11x for all fE X. So, another weaker question will be dealt in this paper: When X is isometrically embedded in X YY ? In the following section we analyze in detail the difficulties of solving the above questions, by comparing with the classical case. Section 3collects some results for the Y-dual and the Y-bidual of aB.f.s. In particular, we prove that XYinherits some of the order properties of Yas the Fatou property (Proposition 3.3), and that the property of being an Y-perfect space is transitive (Proposition 3.4). This last fact will be of special relevance for the proof of our main result in Section 5, where conditions on Xand Y, involving p-convexity and q-concavity properties, are given for Xto be an Y-perfect space (Theorem 5.7). Another important tool used for proving this result is what we call the p-power of aB.f.s. In Section 4we study the Y-perfect property for different p-powers of the same B.f.s. Finally, in Section 6we exhibit some couples of particular rearrangement invariant B.f.s.' (namely: Lorentz, Marcinkiewicz and Orlicz spaces) satisfying the generalized perfectness property. 2. PRELIMINARIES Let (Q, ~,J-L) be aa-finite measure space. We denote by LO(J-L) the space of all measurable finite real functions on Q, where functions which are equal J-L-a.e. are identified. ABanach function space related to J-L (briefly B.f.s.) is aBanach space X ~ LO(J-L) with norm II· IIx, such that if fE LO(J-L), gEXand If I ~ Igi f-L-a.e. then fEXand II f II x ~ II g II x. In particular, X is aBanach lattice with the pointwise J-L-a.e. order. AB.f.s. Xhas the Fatou property if for every sequence (fn) ~ Xsuch that 0 ~ fn tf f-L-a.e. and SUPn II fn II x< 00, we have that fEX and II fn II xt II f II x· AB.f.s. Xis order semi-continuous if for every f, fn EX, such 360 the topological dual space of X, acting on Xas an integral, i.e. there exists a measurable function gsuch that x* (f) =Jigd J1 for all I EX. We can interpret Xl in adifferent way, as the space of the multipliers from Xto L1 (J1), that is, the measurable functions gdefining amultiplication operator from Xto L1(J1). From this point of view, ageneralization of the Kothe dual is possible by taking any B.f.s. Yin the place of L1 (IJ-). Namely, the Y-dual space of X, denoted by X Y, is the space of multipliers from Xto Y, which under an elementary requirement becomes aB.f.s. when endowed with the usual operator norm. The Kothe dual of aB.f.s. takes acrucial part in the interplay between the order and the topology of X. For instance, Xis order continuous if and only if X* coincides with XI, or Xsatisfies the Fatou property if and only if Xcoincides with the Kothe bidual X" of X. We pay attention to this second case, in which X is said to be aperfect space. Is there an analogous result for the general case? Or in other words, denoting by X YY the Y-bidual space of X(i.e. the Y-dual of XY ), when does Xcoincide with X YY? When it does, it is called Y-perfect. We will study this question, which was already posed by Maligranda and Persson in [II, p. 337]. In this item they present some properties concerning the generalized duality and provides adescription for the space XYin some special cases, all of them of great usefulness in the development of this paper. As in the classical case, the containment XCX YY always holds and is continuous with IlfllxYy ~ 1I/IIx for all lEX. So, another weaker question will be dealt in this paper: When X is isometrically embedded in X YY ? In the following section we analyze in detail the difficulties of solving the above questions, by comparing with the classical case. Section 3collects some results for the Y-dual and the Y-bidual of aB.f.s. In particular, we prove that XYinherits some of the order properties of Yas the Fatou property (Proposition 3.3), and that the property of being an Y-perfect space is transitive (Proposition 3.4). This last fact will be of special relevance for the proof of our main result in Section 5, where conditions on Xand Y, involving p-convexity and q-concavity properties, are given for Xto be an Y-perfect space (Theorem 5.7). Another important tool used for proving this result is what we call the p-power of aB.f.s. In Section 4we study the Y-perfect property for different p-powers of the same B.f.s. Finally, in Section 6we exhibit some couples of particularrearrangement invariant B.f.s.' (namely: Lorentz, Marcinkiewicz and Orlicz spaces) satisfying the generalized perfectness property. 2. PRELIMINARIES Let (Q, ~,IJ-) be aa-finite measure space. We denote by LO(IJ-) the space of all measurable finite real functions on Q, where functions which are equal IJ--a.e. are identified. ABanach function space related to J1 (briefly B.f.s.) is aBanach space X ~ LO(J1) with norm II . IIx, such that if IE LO(J1), gEXand III ~ Igi J1-a.e. then IEXand 1I/IIx ~ Ilglix. In particular, Xis aBanach lattice with the pointwise J1-a.e. order. AB.f.s. Xhas the Fatou property if for every sequence (fn) ~ Xsuch that 0 ~ In tI J1-a.e. and sUP n II/nllx < 00, we have that IEX and II In II Xt II I II x· AB.f.s. Xis ordersemi-continuous if for every I, In EX, such 360 that 0 ~ In tI /L- a.e., we have that II/nllx t IIf11x. Note that if aB.f.s. Xhas the Fatou property, then Xis order semi-continuous. For issues related to B.f.s.', see [17, Ch. 15] considering the function norm pdefined as p(f) = IIf11x if IEXand p (f) = 00 in other case. Given two B.f.s.' Xand Y, the Y-dual space of Xis defined as XY := {h ELO(/L): hi EY foraB I EX}, i.e. the space of multipliers from Xto Y. The map II . II xygiven by IIhll x Y:= sup IIhIIlY, for hE XY, IEBX defines anatural seminorm on XY.Note that the supremum above is finite. Indeed, if 0 ~ hEX Y, then it defines apositive multiplication operator between two Banach lattices and so it is continuous, see [9, p. 2]. The same holds for ageneral hEXY by taking positive and negative parts. In order to II . II xYbe anorm, it is necessary and sufficient to require that Xto be saturated, that is, there exists no AE ~ with /L(A) >0such that I XA =0/L-a.e. for all I EX. Note that Xis saturated if and only if Xhas a weak unit (i.e. gE X such that g>0/L-a.e.). In particular, X =1= {OJ. Let X, Y be B.f.s.' with Xsaturated. Then, XYis aB.f.s. endowed with the norm II . II x Y, see [11, Proposition 2]. The space XYgeneralizes the classical Kothe dual space X' of aB.f.s. X, which is obtained taking Yas the space L1(/L). In this classical case, Xsaturated always implies that X' is saturated, [17, Ch. 15, §71, Theorem 4]. This fails for the general case. In order to obtain asecond Y-dual space of Xwith structure ofB.f.s., XYis needed to be saturated. This fact justifies the following comments about saturation for this space. First of all, we note that XY may be trivial. Example 2.1. Suppose (Q,~, /L) is non-atomic. Then, LP(/L)U(/-L) = {OJ whenever 1 ~ p<q ~ 00. See [11, Theorem 2]. Also, even if XYis non-trivial, it may be non-saturated. For instance, every B.f.s. Ysatisfies that is, both spaces coincide with equal norms. This fact, mentioned in [11, §2.(f)], can be directly proved. Then, when Yis non-trivial and non-saturated, UXJ(/L)Y is so. Note that XYsaturated implies that Yis saturated. Indeed, if not, then there exists AE ~ such that /L(A) >0and gXA =0/L-a.e. for all gEY. Sin~e Xis saturated, we can take IEXsuch that I>0/L-a.e. Then, for every hEX Y, we have that hi EYand so hlxA =0/L-a.e. Hence, hXA =0/L-a.e., contradicting the fact that XYis saturated. The converse does not hold, that is, there exist saturated B.f.s.' X and Ysuch that XYis non-trivial and non-saturated. 361 that 0:::; fn tfJL-a.e., we have that IIfnllx t IIf11x. Note that if aB.f.s. Xhas the Fatou property, then Xis order semi-continuous. For issues related to B.f.s.', see [17, Ch. 15] considering the function norm pdefined as p(f) =Ilflix if fEXand p(f) = 00 in other case. Given two B.f.s.' Xand Y, the Y-dual space of Xis defined as XY := {h E LO(JL): hf EYfor all fE Xl, i.e. the space of multipliers from Xto Y. The map II· II xYgiven by IIhll x Y:= sup IIhflIY, for hE XY, IEBX defines anatural seminorm on XY.Note that the supremum above is finite. Indeed, if 0 :::; hEXY,then it defines apositive multiplication operator between two Banach lattices and so it is continuous, see [9, p. 2]. The same holds for ageneral hEXY by taking positive and negative parts. In order to II . IIxY be anorm, it is necessary and sufficient to require that Xto be saturated, that is, there exists no AE 1: with JL(A) >0such that f XA =0 JL-a.e. for all f EX. Note that Xis saturated if and only if Xhas a weak unit (i.e. gE X such that g>0JL-a.e.). In particular, Xi- {O}. Let X, Ybe B.f.s.' with Xsaturated. Then, XYis aB.f.s. endowed with the norm II . II x Y, see [11, Proposition 2]. The space XYgeneralizes the classical Kothe dual space X' of aB.f.s. X, which is obtained taking Yas the space L1 (JL). In this classical case, Xsaturated always implies that X' is saturated, [17, Ch. 15, §71, Theorem 4]. This fails for the general case. In order to obtain asecond Y-dual space of Xwith structure ofB.f.s., XYis needed to be saturated. This fact justifies the following comments about saturation for this space. First of all, we note that XY may be trivial. Example 2.1. Suppose (Q, 1:, JL) is non-atomic. Then, LP(JL)Lq(J.L) = {O} whenever 1 :::; p<q :::; 00. See [11, Theorem 2]. Also, even if XYis non-trivial, it may be non-saturated. For instance, every B.f.s. Ysatisfies (1) LOO(JL)Y=Y, that is, both spaces coincide with equal norms. This fact, mentioned in [11, §2.(t)], can be directly proved. Then, when Yis non-trivial and non-saturated, L OO(JLl is so. Note that XYsaturated implies that Yis saturated. Indeed, if not, then there exists AE 1: such that JL(A) >0and gXA =0 JL-a.e. for all gEY. Sin~e Xis saturated, we can take fEXsuch that f>0 JL-a.e. Then, for every hEX Y, we have that hf EYand so hf XA =0 JL-a.e. Hence, h XA =0JL-a.e., contradicting the fact that XYis saturated. The converse does not hold, that is, there exist saturated B.f.s.' X and Ysuch that XYis non-trivial and non-saturated. 361 Example 2.2. Consider the measure space ([0,2], /3([0, 2]), A), where /3([0,2]) is the a-algebra of all Borel sets of [0,2] and A is the Lebesgue measure on [0,2]. Let us define the saturated B.f.s.' X := {J EL D(A): fX[O,I] ELI (A) and fX[l,2] EL2(A)} with norm IIfllx := IIfX[D,l]lll +IIfX[l,2]1I2, and let Y=L2(A). Note that Xy is non-trivial, since for instance X[I,2] EXY.Let us see that the space XYis non-saturated. Let hEXY.For every gE LI(A), we have that gX[O,I] EXand so hgX[o,I]EY=L2(A). Then, hX[o,I] ELI(A)L2(Al = {OJ (see Example 2.1), that is, h X[O, I] =0 A-a.e. Remark 2.3. Let X, Ybe B.f.s.' with Xsaturated. Acondition guaranteeing that XYis saturated, is that X s; Y. Indeed, this containment holds if and only if L 00 (M) S; XYand, since L 00 (M) is saturated, XYis so. As an immediate consequence of Remark 2.3, YYis saturated for every saturated B.f.s. Y. Moreover, from [11, Theorem 1], we have that Let X, Ybe B.f.s.' with Xsaturated. Whenever XYis saturated, we can consider the Y-bidual space of X, that is, the Y-dual space of X Y, which will be denoted by X YY .Then, that is the space of multipliers from XYto Y. The space X YY is aB.f.s. endowed with the norm IIH x YY:= sup lI~hlly. hEBxY Taking Yas the space LI(M), we obtain that X YY is just the classical Kothe bidual space X" of X. Analogously to X", the space X YY always contains Xand (3) IIfllx Yy :::; IIflix for all fE X. In particular, X YY is saturated. At this point natural questions arise: Questions 2.4. (i) When is Xisometrically embedded in X YY (i.e. (3) is an equality)? (ii) When is X a Y-perfect space (i.e. x:;X YY )? Note that in the expression "X is Y-perfect" or when X YY appears without any specification, it must be understood that the minimal requirements which allow to consider the Y-bidual space of Xare satisfied, namely Xand XYsaturated. 362 Example 2.2. Consider the measure space ([0,2], /3([0, 2]), A), where /3([0,2]) is the a-algebra of all Borel sets of [0,2] and Ais the Lebesgue measure on [0,2]. Let us define the saturated B.f.s.' X := {J EL D(A): fX[O,I] ELI (A) and fX[l,2] EL2(A)} with norm IIfllx := IIfX[D,l]lll +IIfX[l,2]1I2, and let Y=L2(A). Note that Xy is non-trivial, since for instance X[I,2] EXy.Let us see that the space Xyis non-saturated. Let hEXY.For every gE LI(A), we have that gX[O,I] EXand so hgX[o,I]EY=L2(A). Then, hX[o,I] ELI(A)L2(Al = {OJ (see Example 2.1), that is, h X[O, I] =0 A-a.e. Remark 2.3. Let X, Ybe B.f.s.' with Xsaturated. Acondition guaranteeing that XYis saturated, is that X s; Y. Indeed, this containment holds if and only if L 00 (JL) S; XYand, since L 00 (JL) is saturated, XYis so. As an immediate consequence of Remark 2.3, YYis saturated for every saturated B.f.s. Y. Moreover, from [11, Theorem 1], we have that (2) YY-==LOO(JL). Let X, Ybe B.f.s.' with Xsaturated. Whenever XYis saturated, we can consider the Y-bidual space of X, that is, the Y-dual space of X Y, which will be denoted by X YY .Then, X YY := {~ EL o(JL): ~h EYfor all hE XY}, that is the space of multipliers from XYto Y. The space X YY is aB.f.s. endowed with the norm IIH x YY:= sup lI~hlly. hEBxY Taking Yas the space LI (JL), we obtain that X YY is just the classical Kothe bidual space X" of X. Analogously to X", the space X YY always contains Xand (3) IIfllx Yy :::; IIflix for all fEX. In particular, X YY is saturated. At this point natural questions arise: Questions 2.4. (i) When is Xisometrically embedded in X YY (i.e. (3) is an equality)? (ii) When is X a Y-perfect space (i.e. X -== X YY )? Note that in the expression "X is Y-perfect" or when X YY appears without any specification, it must be understood that the minimal requirements which allow to consider the Y-bidual space of Xare satisfied, namely Xand XYsaturated. 362 For asaturated B.f.s. Y, simple examples of Y-perfect spaces follow from (l) and (2). Namely, L 00(11-) and Yare always Y-perfect. Questions 2.4 are solved for the classical case X". In this case, (3) is an equality (i.e. X' is anorming subspace of X*) if and only if Xis order semi-continuous. Moreover, X == X" if and only if Xhas the Fatou property (see for instance [9, Proposition l.b.18] and [17, Ch. 15, §71, Theorem 1]). In the general case, Xhaving the Fatou property is neither anecessary nor sufficient condition for Xto be Y-perfect. Indeed, Xis X-perfect, even if Xdoes not have the Fatouproperty. For the converse implication, we can consider the following counterexample. Example 2.5. If 1 ~ p<q ~ 00, then (£P/ q == £00 (see [11, Theorem 2]). SO, £P has the Fatou property but is not £q -perfect, since Similarly, the order semi-continuity of Xis neither anecessary nor sufficient condition for (3) to be an equality. Indeed, (3) is an equality whenever Xand Y coincide, even if Xis not order semi-continuous. Conversely, in Example 2.5, £P is order semi-continuous but (3) is not an equality for Y= £q. Trying to solve Questions 2.4 in the general case by giving equivalent conditions turns out to be avery difficult (if not impossible) task, due to the fact already shown that the general duality includes plenty of cases totally different to the classical one. We only need to notice that every saturated B.f.s. Yis Y-perfect, without any kind of requirement on Y. Therefore, we will not look for equivalent conditions but conditions guaranteeing that X is Yperfect or that Xis isometrically embedded in X YY. Before that, we will exhibit some properties of the generalized dual spaces which will be used throughout this paper. 3. SOME PROPERTIES RELATED TO GENERALIZED DUALITY Given two B.f.s.' Xand Y, we will use the expression "X ~ c Y" to mean that X is continuously contained in Ywith II f II y ~ c II f II xfor all f EX. The expression "X ~ i Y" will mean that Xis continuously contained in Ywith II fIIx= II f II y for all f EX. In the case when X=Yas sets, the identity map between Xand Yis an order isomorphism, since it is apositive linear operator between Banach lattices, see [9, p. 2]. As in the previous section, we will write X == Ywhenever the isomorphism is an isometry, that is, the norms coincide. Lemma 3.1. Let X, Y, Z be Bfs.' with Xsaturated. Then, (a) Y ~c(i) Z => Xy ~c(i) XZ, (b) X ~c Z => zY ~c Xy. Moreover, ifY is order semi-continuous, 363 For asaturated B.f.s. Y, simple examples of Y-perfect spaces follow from (l) and (2). Namely, L 00(11-) and Yare always Y-perfect. Questions 2.4 are solved for the classical case X". In this case, (3) is an equality (i.e. X' is anorming subspace of X*) if and only if Xis order semi-continuous. Moreover, X == X" if and only if Xhas the Fatou property (see for instance [9, Proposition l.b.18] and [17, Ch. 15, §71, Theorem 1]). In the general case, Xhaving the Fatou property is neither anecessary nor sufficient condition for Xto be Y-perfect. Indeed, Xis X-perfect, even if Xdoes not have the Fatouproperty. For the converse implication, we can consider the following counterexample. Example 2.5. If 1 ~ p<q ~ 00, then (£P/q == £00 (see [11, Theorem 2]). SO, £P has the Fatou property but is not £q -perfect, since (£P(e q ==(£oo( ==£q 12£P. Similarly, the order semi-continuity of Xis neither anecessary nor sufficient condition for (3) to be an equality. Indeed, (3) is an equality whenever Xand Y coincide, even if Xis not order semi-continuous. Conversely, in Example 2.5, £P is order semi-continuous but (3) is not an equality for Y= £q. Trying to solve Questions 2.4 in the general case by giving equivalent conditions turns out to be avery difficult (if not impossible) task, due to the fact already shown that the general duality includes plenty of cases totally different to the classical one. We only need to notice that every saturated B.f.s. Yis Y-perfect, without any kind of requirement on Y. Therefore, we will not look for equivalent conditions but conditions guaranteeing that X is Yperfect or that Xis isometrically embedded in X YY. Before that, we will exhibit some properties of the generalized dual spaces which will be used throughout this paper. 3. SOME PROPERTIES RELATED TO GENERALIZED DUALITY Given two B.f.s.' Xand Y, we will use the expression "X ~ c Y" to mean that X is continuously contained in Ywith II f II y ~ c II f II xfor all f EX. The expression "X ~ i Y" will mean that Xis continuously contained in Ywith II fIIx= II f II y for all f EX. In the case when X=Yas sets, the identity map between Xand Yis an order isomorphism, since it is apositive linear operator between Banach lattices, see [9, p. 2]. As in the previous section, we will write X == Ywhenever the isomorphism is an isometry, that is, the norms coincide. Lemma 3.1. Let X, Y, Z be Bfs.' with Xsaturated. Then, (a) Y ~c(i) Z => Xy ~c(i) XZ, (b) X ~c Z => zY ~c Xy. Moreover, ifY is order semi-continuous, 363 Proof. Parts (a) and (b) can be directly proved. Suppose Yis order semicontinuous. If X "--+ i Z, from (b) we have that Zy "--+ 1X y. Let hEZ Y. Given fEB z, consider asequence of simple functions (0/ n) satisfying that 0 :::::; 0/ ntIf I· Note that since Xis saturated, it has aweak unit, which is equivalent to the existence of asequence (An) C ~ such that n= Un An and XA nEX. Taking CPn := o/nXUn AEX, we have that 0:::::; CPn tIII. Then, 0 ~ CPnlhl t Ilhl EY. Since 1] Yis order semi-continuous, II/hlly =lim IICPnhily ~ IlhllxY lim IICPnlix n----'J-oo n----+oo = IlhllxY lim IICPnliz ~ Ilhllxrllfllz:::::; II h ll x Y' n~oo Then, IIhllzy ~ IlhllxY. So, (c) holds. 0 Lemma 3.1(b) and otherproperties related to generalized duality can be found in [11,§2]. Remark 3.2. As aconsequence of (3) and Lemma 3.1(b), we obtain that Xyis always Y-perfect, for each couple of Rf.s.' X, Y with X and XYbeing saturated. An special feature of the Kothe dual space X' of asaturated B.f.s. Xfollows by taking Y= L1(J-L) in Remark 3.2: X' always has the Fatou property (see also [9, p. 30]). This does not hold in the general case. For instance, we can take a Rf.s. Y which does not satisfy the Fatou property and then L 00 (J-L) Y == Yfails in having this property. However, if Y has the Fatou property (as L1 (J-L)), this property is transferred to XYfor every saturated Rf.s. X. The analogous for Ybeing order semi-continuous also holds. Proposition 3.3. Let X, Ybe Bfs.' with Xbeing saturated. (a) /fY has the Fatou property then XYalso does. (b) /fY is order semi-continuous then XYis so. Proof. Let us prove (a), and asimilar argument works for (b). Let (hn) be a sequence in XYsuch that 0 ~ hnth J-L-a.e. and sUPn IIh n II xY< 00. Then, for every fE X, we have that 0:::::; Iflh nt I/lh J-L-a.e., where (fh n)CYwith sup II fh n lIy :::::; IIflix .sup IIh n II xY< 00. n n Hence, if Yhas the Fatou property, it follows that f hEY and IIfhll y=lim IIfhn llY ~ IIf11x ·limllhnllxY' n n Therefore, hE XY and IIhlixY ~ limn IlhnllxY. This fact, together with IlhnllxY ~ IlhllxY for all n, implies that IIh n IlxY t IIhll x Y. So, XY has the Fatou property. 0 364 (c) X ~i z::::} Zy ~i XY. Proof. Parts (a) and (b) can be directly proved. Suppose Yis order semicontinuous. If X ~ i Z, from (b) we have that Zy ~ 1X Y. Let hEZ Y. Given fEB z, consider asequence of simple functions (0/ n) satisfying that 0 :::::; 0/ ntIf I· Note that since Xis saturated, it has aweak unit, which is equivalent to the existence of asequence (An) C ~ such that n=Un An and XA nEX. Taking CPn := o/nXUn AEX, we have that 0:::::; CPn tIII. Then, 0 ~ CPnlhl t Ilhl EY. Since 1] Yis order semi-continuous, II/hlly=lim IICPnhlly:::::; IlhllxY lim IICPnlix n----'J-oo n----+oo = IlhllxY lim IICPnllz:::::; Ilhllxrllfllz:::::; II h ll x Y' n~oo Then, IIhllzy :::::; IlhllxY. So, (c) holds. 0 Lemma 3.1(b) and otherproperties related to generalized duality can be found in [11,§2]. Remark 3.2. As aconsequence of (3) and Lemma 3.1(b), we obtain that Xyis always Y-perfect, for each couple of Rf.s.' X, Y with X and XYbeing saturated. An special feature of the Kothe dual space X' of asaturated B.f.s. Xfollows by taking Y= L1(J-L) in Remark 3.2: X' always has the Fatou property (see also [9, p. 30]). This does not hold in the general case. For instance, we can take a Rf.s. Y which does not satisfy the Fatou property and then L 00 (J-L) Y == Yfails in having this property. However, if Y has the Fatou property (as L1 (J-L)), this property is transferred to XYfor every saturated Rf.s. X. The analogous for Ybeing order semi-continuous also holds. Proposition 3.3. Let X, Ybe Bfs.' with Xbeing saturated. (a) lfY has the Fatou property then XYalso does. (b) lfY is order semi-continuous then XY is so. Proof. Let us prove (a), and asimilar argument works for (b). Let (hn) be a sequence in XYsuch that 0:::::; hnth J-L-a.e. and sUPn IIh n II xY< 00. Then, for every fE X, we have that 0:::::; Iflh nt I/lh J-L-a.e., where (fh n)CYwith sup II fh n lIy :::::; IIflix .sup IIh n II xY< 00. n n Hence, if Yhas the Fatou property, it follows that f hEY and IIfhlly=lim IIfh n lly :::::; IIf11x ·limllhnllxY' n n Therefore, hE XY and IIhlixY :::::; limn Ilhnllxy. This fact, together with Ilhnllxy :::::; IlhllxY for all n, implies that IIh n IlxY t IIhll x Y. So, XY has the Fatou property. 0 364 The converse of Proposition 3.3 does not hold. For instance, consider asaturated B.f.s. Ywhich does not satisfy the Fatou property (or is not order semi-continuous) and take the space Yy == L 00 (Il) which has the Fatou property. Note that in the case when XYis saturated, Proposition 3.3 can be obtained as a particular case of the following more general result by taking Z=LI (Il). Proposition 3.4. Let X, Y, Zbe Bfs.' with X, XY and X YZ saturated. (a) IfY is Z-perfect then XYis so. (b) IfY "-+, y ZZ then XY "-+, XYZz. Proof. For (a) see [11, §2(h»). Suppose that yZis saturated. Let us show that The first inclusion follows from (3). Let us prove the second one. Consider TJ E X Yzz .For every IEXand ~ EyZ,we have that I~ EX YZ ,since nh EZfor all hE XY. Then TJ/~ EZand so TJI Ey Zz .Hence, TJ EX Yzz .Moreover, IITJllxyzz =sup IITJfII y zz =sup sup IITJnllz JEBx jEBx~EByZ :::;; IITJllxYzz' sup sup IInllx Yz JEBx ~EByZ = IITJll x Yzz' sup sup sup II/~hllz JEBx ~EByZ hEBxY :::;; IITJll x Yzz' sup sup II/hlly:::;; IITJll x Yzz. JEBx hEBxY If Y "-+, y ZZ ,from Lemma 3.1(a) we have that XY "-+i X Yzz .This fact together with (4) gives XY "-+i XYZz. So (b) holds. D Let us see that being ageneralized perfect space is atransitive property. In Section 5, this fact will allow us to recognize B.f.s.' X, Ysuch that Xis Y-perfect passing through an LP -space. As can be expected, p-convexity and p-concavity properties must be required for the spaces X and Y. Proposition 3.5. Let X, Y, Zbe Bfs.' with X and XZsaturated. (a) If Xis Y-perfectand Yis Z-perfect, then Xis Z-perfect. (b) If X "-+i X YY and Y "-+i y ZZ , th~n X "-+, X ZZ Proof. Suppose that XY and YZare saturated. Then, This can be directly checked in the same line of the proof of (4). 365 The converse of Proposition 3.3 does not hold. For instance, consider asaturated B.f.s. Ywhich does not satisfy the Fatou property (or is not order semi-continuous) and take the space Yy == L 00 «(.1.) which has the Fatou property. Note that in the case when XYis saturated, Proposition 3.3 can be obtained as a particular case of the following more general result by taking Z=L1 «(.1.). Proposition 3.4. Let X, Y, Zbe Bfs.' with X, XY and X YZ saturated. (a) IfY is Z-perfect then XYis so. (b) IfY "-+, y ZZ then XY "-+, X Yzz . Proof. For (a) see [II, §2(h»). Suppose that yZis saturated. Let us show that (4) XY "-41 X YZZ "-+1 X yZZ . The first inclusion follows from (3). Let us prove the second one. Consider 17 E XYZ z. For every IEXand ~ EY z, we have that I ~ EXYZ, since I ~ hEZfor all hE XY. Then 17/~ EZand so 171 E yZz. Hence, 17 EX Yzz .Moreover, 1I17ll xy zz =sup lI17fII y zz =sup sup II17/Hz IEBx IEBx~EByZ ~ 1l17llxYzz' sup sup 1IJ;llx yz IEBx ~EByZ = 1I17ll xYzz' sup sup sup II/~hllz IEBx ~EByZ hEBxY ~ 1I17ll xYzz' sup sup IIfhlly ~ 1I17ll x Yzz. IEBx hEBxY If Y "-+, yZZ, from Lemma 3.l(a) we have that XY "-+i X Yzz .This fact together with (4) gives XY "-+i X Yzz .So (b) holds. D Let us see that being ageneralized perfect space is atransitive property. In Section 5, this fact will allow us to recognize B.f.s.' X, Y such that Xis Y-perfect passing through an LP -space. As can be expected, p-convexity and p-concavity properties must be required for the spaces Xand Y. Proposition 3.5. Let X, Y, Z be Bfs.' with X and XZsaturated. (a) If Xis Y-perfect and Yis Z-perfect, then Xis Z-perfect. (b) If X "-+i XYY and Y "-+i y ZZ , th~n X "-+, X ZZ Proof. Suppose that XYand YZare saturated. Then, (5) X "-+1 X ZZ "-+( x yyZZ This can be directly checked in the same line of the proof of (4). 365 If Xis Y-perfect and Yis Z-perfect, then XYY == Xand y ZZ == Y, and so x yyZZ == X. This, together with (5), gives X == X ZZ ,that is Xis Z-perfect. So (a) holds. Suppose that X Y-i XYY and Y Y-i y Zz .Then, from Lemma 3.1(a), This fact, together with (5), gives X Y-i X ZZ . SO, (b) holds. 0 Remark 3.6. Taking Z=LI (JL) in Proposition 3.5, it follows: (i) In the case when Yhas the Fatou property, Xbeing Y-perfect implies that Xhas the Fatou property. (ii) In the case when Yis order semi-continuous, X Y-i XYY implies that Xis order semi-continuous. The converse of (i) and (ii) does not hold as Example 2.5 shows. The following useful result, mentioned in [11, §2(b)], can be directly proved. Lemma 3.7. Let X, Y, Z be Bfs.' with X and Ysaturated. Then, As aconsequence of Lemma 3.7, for B.f.s.' X, Zsuch that Xand XZare d, h h xZz -ZxZ- 00 h fi f saturate we ave tat X= (X) =L (JL). Ten, or every B. .s. Y satisfying X Y-l Y Y-l X ZZ ,we have that and so In particular, Xis Y-perfect in the only case of X=Y, since X YY == LOO(JL)Y == Y. For instance, taking Z=L1 (JL) and Y= X", if Xdoes not have the Fatou property then Xis not X"_perfect. Note that if any of the inclusions in X ~ Y ~ X ZZ is continuous with other constant different of 1, then XY=L 00 (JL) with equivalent norms. 4. PERFECT SPACES INVOLVING P-POWERS OF A B.F.S. Let Xbe asaturated B.f.s. The maximal normed extension of Xis the B.f.s. defined as [X]:= {g E LO(JL): sup{lIf11x: fEX, 0 ~ f ~ Igil <oo}, 366 If Xis Y-perfect and Yis Z-perfect, then x yy == Xand y ZZ == Y, and so x yyZZ == X. This, together with (5), gives X == X ZZ ,that is Xis Z-perfect. So (a) holds. Suppose that X Y-i X YY and Y Y-i y Zz .Then, from Lemma 3.1(a), X Y-, X YY Y-, x yyzz . This fact, together with (5), gives X Y-i X ZZ . SO, (b) holds. 0 Remark 3.6. Taking Z=LI (JL) in Proposition 3.5, it follows: (i) In the case when Yhas the Fatou property, Xbeing Y-perfect implies that Xhas the Fatou property. (ii) In the case when Yis order semi-continuous, X Y-i X YY implies that Xis order semi-continuous. The converse of (i) and (ii) does not hold as Example 2.5 shows. The following useful result, mentioned in [11, §2(b)], can be directly proved. Lemma 3.7. Let X, Y, Z be Bfs.' with X and Ysaturated. Then, X yZ == y Xz As aconsequence of Lemma 3.7, for B.f.s.' X, Zsuch that Xand XZare saturated, we have that Xx zz == (Xz)x z == L'x'(JL). Then, for every B.f.s. Y satisfying X Y-l Y Y-l X ZZ ,we have that L OO(JL) == XX Y-I XY Y-I Xx zz == L 00 (JL) , and so (6) XY == L 00 (JL). In particular, Xis Y-perfect in the only case of X= Y, since X YY == L 00 (JL) Y == Y. For instance, taking Z=L1 (JL) and Y= X", if Xdoes not have the Fatou property then Xis not X"_perfect. Note that if any of the inclusions in X ~ Y ~ X ZZ is continuous with other constant different of 1, then XY=L 00 (JL) with equivalent norms. 4. PERFECT SPACES INVOLVING P-POWERS OF A B.F.S. Let Xbe asaturated B.f.s. The maximal normed extension of Xis the B.f.s. defined as [X]:= {g ELO(JL): sup{lIf11x: fEX, 0 ~ f ~ Igl} <oo}, 366 endowed with the norm IIgll[Xj := sup{lIf11x: fE X, 0 (f (igil. for gE [Xl We always have that X ~, [X] ~ I X". In particular, [X] is saturated. Note that [X] is the largest B.f.s. having Xas aclosed subspace. Then, [X] == X" if and only if X is order semi-continuous. These topics appear in [II, §I], where [1] is referred as original source. From (6), we have that (7) XX == X[Xj == XX" == L 00 (J1-). Let 0<p< 00. The p-power of X is the space defined as endowed with the quasi-norm IIf11xp = IIlfl P II~P, for fEXP. For 1 (p, 11·llxp is anorm, [11, Proposition 1]. In the case when 0<p< 1, if Xis (1/ p)-convex with constant 1(see Section 5for this concept), then II . IIxp is also anorm. This follows from [5, Lemma 3] noting that there our space XP is denoted by XI/p. We will only consider these cases, for which XP is aB.f.s. Note that XP is saturated, since Xis so, and [XP] == [X]P. Moreover, XP has the Fatou property ifand only if Xhas the Fatou property. Similarly, XPis order semi-continuous if and only if Xis so. Let 1( q<p< 00. Then, where ~ = ~ - i, see [11, Theorem 5]. In particular, since p and rplay the same role, we have that That is, [XP] and XP are Xq-perfect. Example 4.1. For every 1( q(p( 00, we have that with ~ = ~ - i. In the case q<p< 00, this follows from (8) by taking X=LI (J1-). Note that, LI (J1-) == [LI(/L)] == L1(/Lt, since L1(/L) has the Fatou property. The remaining cases follow from (I) and (2). See also [11,Proposition 3]. Then, LP(J1-) is U(/L)-perfect. There is another possibility left of generalized dual space as combination of p-powers and maximal normed extension of X, namely (XP)[xqj. We give a description for this space when Xsatisfies certain conditions. 367 endowed with the nonn IlglI[x] := suplllfllx: fEX, 0 ~ f ~ IglL for gE [X]. We always have that X ~ I [X] ~ I X". In particular, [X] is saturated. Note that [X] is the largest B.f.s. having Xas aclosed subspace. Then, [X] == X" if and only if Xis order semi-continuous. These topics appear in [11, §1], where [1] is referred as original source. From (6), we have that (7) XX == X[X] == XX" == LOO (J.1). Let 0<p< 00. The p-power of Xis the space defined as XP={JELO(J.1): IfI PEX }, endowed with the quasi-norm IIf11xp = IIlfl P II~P, for fEXP. For 1 ~ p, II ·lIxp is a nonn, [11, Proposition 1]. In the case when 0<p< 1, if Xis (1/ P)-convex with constant 1(see Section 5for this concept), then II . IIxp is also anonn. This follows from [5, Lemma 3] noting that there our space XP is denoted by XI/p. We will only consider these cases, for which XP is aB.f.s. Note that XP is saturated, since Xis so, and [X P] == [X]p. Moreover, XPhas the Fatou property if and only if Xhas the Fatou property. Similarly, XPis order semi-continuous if and only if Xis so. Let 1 ~ q<p< 00. Then, (8) [XP]x q == xr and (XP)x q == [XP][x qj == [X r ], where ~ = ~ - i, see [11, Theorem 5]. In particular, since pand rplay the same role, we have that [Xpr qXq == (Xr(q == [XP] and (XP)xqx q == [xrrq ==XP. That is, [XP] and XP are XLperfect. Example 4.1. For every 1 ~ q ~ p ~ 00, we have that LP (J.1)U (11-) == Lr (J.1) with ~ = ~ - i. In the case q<p< 00, this follows from (8) by taking X=LI (J.1). Note that, LI (J.1) == [L I (J.1)] == LI (J.1'>;', since LI (J.1) has the Fatou property. The remaining cases follow from (1) and (2). See also [II, Proposition 3]. Then, U(J.1) is U(J.1)-perfect. There is another possibility left of generalized dual space as combination of p-powers and maximal nonned extension of X, namely (XP)[x q ]. We give a description for this space when Xsatisfies certain conditions. 367 Every r.i. Rf.s. Xis an intermediate space between aLorentz and aMarcinkiewicz space. Namely, where CPx is thefundamentalfunction of X(i.e. cpx(t):= IIX[o.t]lIx for tEl, which can be assumed to be concave). The spaces Arp and Mrp have fundamental function cp and t/ cp respectively. An r.i. Rf.s. Xhas the majorant property if whenever gE LOU ... ), fEXand f~ g*(s)ds ~ f~ f*(s)ds for all °< tEl, it follows that gEXand IIglix ~ IIfllx· For instance, this property is satisfied if Xis separable (which is equivalent to being order continuous in the setting of r.i. Rf.s.') or has the Fatou property. The Fatou property holds for Arp and Mrp and so does the majorant property. Proposition 6.4. Let Xbe an r.i. Bfs. such that (Mrp)F <; X. Then, M;x = Mrp with equivalent norms if and only if there exists k>°satisfying (14) t -~kcp xx (t), forallO<tEI. cp(t) Mq; Moreover, Mrp is X-perfect if and only if (14) holdsfor k= 1. Proof. Suppose that M;x = Mrp. Then, the fundamental functions of both spaces are equivalent and in particular, (14) holds. Note that if M;x == Mrp, then (14) holds for k= l. Conversely, suppose that (14) holds. Then, Since we always have Mrp ~ 1 M;x, it follows that M;x = Mrp with equivalent norms. If k=1then, M;x == Mrp. 0 Note that, under conditions of Proposition 6.4, if Xhas the majorant property, cp (0+) =0, cp (00) =00 (in the case of a=00) and k>°satisfies (15) , cp(t) Ilcp X[O,t] IIx ~ k CPx (t)-t-' for all °< tEl, then (14) holds for k. Indeed, since X[O,t] EM; and IIX[o,t] 11 M,}' = Ilcp' X[O,t] Ilx, where cp' denotes the derivative function of cp (see [11, Theorem 3]), then IIx[o,t]lIx cp xx(t)= sup IIhX[o,t]lIx~ , . Mq; hEBMx IIcp X[O,t] Ilx q; Example 6.5. Let cp, 1/J : 1---+ [0,00) be increasing concave maps vanishing only at zero with (fi(0+) =0, cp(oo) =00 (if a=00) and satisfying that *is increasing. 374 Every r.i. Rf.s. Xis an intermediate space between aLorentz and aMarcinkiewicz space. Namely, Acpx "--+1 X"--+1 M tNx ' where CPx is thefundamentalfunction of X(i.e. cpx(t):= IIX[o.tJllx for tEl, which can be assumed to be concave). The spaces Acp and Mcp have fundamental function cp and t/ cp respectively. An r.i. Rf.s. Xhas the majorant property if whenever gE LOU ... ), fEXand f~ g*(s)ds ~ f~ f*(s)ds for all °< tEl, it follows that gEXand IIglix ~ IIfllx· For instance, this property is satisfied if Xis separable (which is equivalent to being order continuous in the setting of r.i. Rf.s.') or has the Fatou property. The Fatou property holds for Acp and Mcp and so does the majorant property. Proposition 6.4. Let Xbe an r.i. Bfs. such that (Mcp)F <; X. Then, M;x = Mcp with equivalent norms if and only if there exists k>°satisfying (14) t cp(t) ~ k cp M$X (t), for all 0< tEl. Moreover, Mcp is X-perfect if and only if (14) holdsfor k= 1. Proof. Suppose that M;x = Mcp. Then, the fundamental functions of both spaces are equivalent and in particular, (14) holds. Note that if M;x == Mcp, then (14) holds for k= l. Conversely, suppose that (14) holds. Then, M XX "--+1 MtN "--+k Mcp. cp M$X Since we always have Mcp "--+ 1 M;x, it follows that M;x = Mcp with equivalent norms. If k=1then, M;x == Mcp. 0 Note that, under conditions of Proposition 6.4, if Xhas the majorant property, cp (0+) =0, cp (00) =00 (in the case of a=00) and k>°satisfies (15) , cp(t) Ilcp X[O,t] IIx ~ kCPx(t)-, for all °< tEl, t then (14) holds for k. Indeed, since X[O,t] EM; and IIX[o,tJ II M$= Ilcp' X[O,t] Ilx, where cp' denotes the derivative function of cp (see [11, Theorem 3]), then CPMXX(t)= sup IIhX[otJllx>- IIX[o,t]lIx 'P hEB M$, :;-- IIcp'X[o,tJllx' Example 6.5. Let cp, 1/J : 1---+ [0,00) be increasing concave maps vanishing only at zero with (fi(0+) =0, cp(oo) =00 (if a=00) and satisfying that *is increasing. 374 Then, every fE Mcp with ex := A(SUpp f) < 00 belongs to M1fr with IlfIIM1{f ~ ~i~i IIfliMcp for ex> O. That is, (Mcp)F S; M1fr. Moreover, U ,If' 1I<p x[o.tJIIM1{f =sup ,I,() <p (S)X[O,/](s)ds O<uEI 'f/ Uo <p(t) <pet) =ljr(t) = <PM1{f (t)-t-' for every 0< tEl. That is, (15) holds for X=M 1fr and k= 1. Since M 1fr has the majorant property, then (14) holds for X= M1fr and k= 1. Therefore, from Proposition 6.4, it follows that Mcp is Mrperfect. Functions satisfYing the above required conditions are for instance <p(t) =t1/p and ljr(t)=t1/qwith I ~ p ~ q ~ 00. Note that in this case, Mcp and M1fr are just the Lorentz spaces LP,OO(A) and Lq,OO(A) respectively, with 1 +..\. =1+1=I(see p p q q [2, Theorem IY.4.6]). Example 6.6. Let <P, ljr: 1-+ [0,00) be increasing concave maps vanishing only at zero with <p(0+) =0, <p(oo) = 00 (if a= 00) and satisfYing: (i) Ii Pip dljr(s)< 00 for all (some) 0<b<a. (ii) There exists k>0such that I~ <p'(s)dljr(s) ~ k 1fr(t~cp(t) for all 0< tEl. Condition (i) implies that (Mcp)F S; 1I.1fr. Indeed, given fE Mcp with ex '- A(SUpp f) < 00 we have that a a s f f*(s)dljr(s) =f f*(s)dljr(s) ~ f ~ f f*(u)dudljr(s) o0 0 a f <pes) ~ IlfIIMcp -s- dljr(s). o Note that, since ~ is decreasing, if (i) holds for some b< 00 then it holds for all b< 00. Condition (ii) is just (15) for X= 1I.1fr which has the majorant property, so (14) holds for X= 1I.1fr. Then, from1>roposition 6.4·we have that M:1{f,A1{f = Mcp with equivalent norms. It can be proved that (ii) holds for k= ~ only in the case <pet) =t, for which Mcp == L 00 (A) and so Mcp is obviously 1I.rperfect. The functions <pet) =t1/pand ljr(t)=t1/qwith 1<p< 00 and 1 ~ q< /-1' satisfY the conditions (i) and (ii) with constant k=(q +pqp)-l. In this case, 1I.1fr is just the Lorentz space U,l(A) (see [2, Theorem 1Y.4.3]). 375 Then, every fE Mcp with ex := A(SUpp f) < 00 belongs to M1fr with IlfIIM1{f ~ ~i~i IIfliMcp for ex> O. That is, (Mcp)F S; M1fr. Moreover, U II cpl X[o.t) II M1{f =sup .1. (1 )f cpl (s)X[O,/](s) ds O<uEI 'f/ Uo _ cp(t) -ljr(t) = CPM". (t) cp(t) 'I' t' for every 0< tEl. That is, (15) holds for X=M 1fr and k= 1. Since M 1fr has the majorant property, then (14) holds for X= M1fr and k= 1. Therefore, from Proposition 6.4, it follows that Mcp is Mrperfect. Functions satisfYing the above required conditions are for instance cp(t) =t1/p and ljr(t) =t1/qwith I ~ p ~ q ~ 00. Note that in this case, Mcp and M1fr are just the Lorentz spaces LP,OO(A) and Lq,OO(A) respectively, with 1 +..\. =1+1=I(see p p q q [2, Theorem IY.4.6]). Example 6.6. Let cp, ljr: 1-+ [0,00) be increasing concave maps vanishing only at zero with cp(O+) =0, cp(oo) = 00 (if a= 00) and satisfYing: (i) Ii ~ dljr(s) < 00 for all (some) 0<b<a. (ii) There exists k>0such that I~ cpl(s)dljr(s) ~ k 1fr(t~cp(t) for all 0< tEl. Condition (i) implies that (Mcp)F S; 1I.1fr. Indeed, given fE Mcp with ex '- A(SUpp f) < 00 we have that a a s f f*(s)dljr(s) =f f*(s)dljr(s) ~ f ~ f f*(u)dudljr(s) o0 0 a fcp(s) ~ IlfIIMcp -s- dljr(s). o Note that, since ~ is decreasing, if (i) holds for some b< 00 then it holds for all b< 00. Condition (ii) is just (15) for X= 1I.1fr which has the majorant property, so (14) holds for X= 1I.1fr. Then, from1>roposition 6.4·we have that M:1{f·A1{f = Mcp with equivalent norms. It can be proved that (ii) holds for k= ~ only in the case cp(t) = t, for which Mcp == L 00 (A) and so Mcp is obviously 1I.rperfect. The functions cp(t) =t1/pand ljr(t) =t1/qwith 1<p< 00 and 1 ~ q< /-1' satisfY the conditions (i) and (ii) with constant k=(q +pqp)-l. In this case, 1I.1fr is just the Lorentz space U,l(A) (see [2, Theorem 1Y.4.3]). 375 6.2. Orlicz spaces Let <P be aYoungfunction, i.e. <P: [0, 00) ~ [0,00) is continuous, convex, increasing, <p(0) = 0 and liml--+oo <p(t) =00. The Orlicz space related to <P is the r.i. B.f.s. defined as endowed with the Luxemburg norm The following result about generalized duality ofOrlicz spaces is alittle variation of [II, Theorem 4] (the constant! in (i» and the proof is almost the same, so we omit it. Note that the reason of this variation is for the representation obtained to be isometric. Theorem 6.7. Let <P, <Po, <PI be Youngfunctions satisfYing: (i) <p(st) ~ !(<Po(s) + <PI (t»,for all s, t? 0, (ii) <p- I(t) ~ <POl (t)<pjI (t),for all t? O. The hypothesis of Theorem 6.7 are satisfied for instance if <P, <Po are Young functions such that ~o is increasing and, for any s> 0, (16) . <p(st). <po(t) hmsup-- =hmsup-- =0, 1--+00 <po(t) 1--+0 <p(st) by taking <PI(t):= suPs>0(2<p(st) - <po(s» for all t? O. Again, the reason of this fact can be found in [II, Example 2], noting that in the notation there, our <PI is just (2<p) G <Po. Remark 6.8. Let <P be aYoung function such that <Pit) is increasing and (17) . t . <P (t) hmsup-- =hmsup-- =0. t~oo <p(t) t~O t Then, taking 4>(t) =suPs>0(2st -<p(s» for all t?0, we obtain that (L <p)' =L cP. Note that this improves the representation obtained for (L <p)' via the complementary Young function W of <P. The Young function tP (1 <p<00) whose Orlicz space is just the classical LP (order isometrically), and the Young functions tlog+(t) and exp(t) -I whose Orlicz 376 6.2. Orlicz spaces Let <P be a Youngfunction, i.e. <P: [0, 00) ~ [0,00) is continuous, convex, increasing, <p(0) = 0 and liml--+oo <p(t) =00. The Orlicz spacerelated to <P is the r.i. B.f.s. defined as L <P := {fEL0 () .. ): J <P C ~ I) dA <00 for some c> O}, endowed with the Luxemburg norm Ilfllvp :=inf{C>O: J <pC~I)dA~ IJ. The following result about generalized duality ofOrlicz spaces is alittle variation of [11, Theorem 4] (the constant! in (i» and the proof is almost the same, so we omit it. Note that the reason of this variation is for the representation obtained to be isometric. Theorem 6.7. Let <P, <Po, <PI be Youngfunctions satisfYing: (i) <p(st) ~ !(<Po(s) + <PI (t»,for all s, t) 0, (ii) <P-I(t) ~ <POl (t)<pjI(t),forall t )0. Then, (L <PO)L <I> ;: L <Pl. The hypothesis of Theorem 6.7 are satisfied for instance if <P, <Po are Young functions such that ~o is increasing and, for any s > 0, (16) . <P(st). <Po(t) hmsup-- =hmsup-- =0, 1--+00 <Po(t) 1--+0 <P(st) by taking <PI(t):= suPs>0(2<P(st)- <Po(s» for all t) O. Again, the reason of this fact can be found in [11, Example 2], noting that in the notation there, our <PI is just (2<P) G <Po. Remark 6.8. Let <P be aYoung function such that <Pit) is increasing and (17) . t .<P(t) hmsup-- =hmsup-- =0. t~oo <P(t) t~O t Then, taking 4>(t) =suPs>0(2st-<P(s» for all t)0, we obtain that (L <P)';: L cP. Note that this improves the representation obtained for (L <P)' via the complementary Young function W of <P. The Young function tP (1 <p<00) whose Orlicz space is just the classical LP (order isometrically), and the Young functions tlog+(t) and exp(t) -1 whose Orlicz 376 spaces, in the case of finite measure, are (orderisomorphically) the Zygmund spaces Llog Land Lexp respectively (see [2, Definition IV6.l D, satisfy the conditions of Remark 6.8. As acorollary of Theorem 6.7 we obtain that L <1>0 is L <I> -perfect. Corollary 6.9. Let <1>, <1>0 be Young functions satisfYing the conditions (i) and (ii) of Theorem 6.7 for some other Youngfunction <1> 1. Then, Proof. Note that the Young functions <1>0, <1>1 play the same role in the hypothesis of Theorem 6.7. Hence, it follows that (L <l>O)L ¢> == L <1>] and (L <l>1)L ¢> == L <1>0. So, (L <l>O)L¢>L¢> == (L <PI )L¢> == L <1>0. 0 Note that under the assumptions of Corollary 6.9, L <1>1 is also L <I> -perfect, since it can be interchanged with L <1>0. Finally let us show an example of functions satisfying the hypothesis of Corollary 6.9. Consider cP(t) := tP log-.B (2 + t) and cPo(t) := exp(ta) - I which are Young functions for 2<p< 00, 0< f3 ::::;; p-2and I ::::;; ex < 00. In the case when p< ex, the function cPo/¢ is increasing and (16) is satisfied. So, 1'1>0 is 1'1> -perfect. Note that if the measure is finite, for these Young functions, L¢and L ¢o are order isomorphic to the Zygmund spaces LP(logL)-.B/p and L:~~ respectively (see [2, Definition IV6.11 D. Note. Just before submission for publication of this paper we became aware of the preprint by A.R. Schep [15] in which some of our results concerning p-convexity and p-concavity, have been independently obtained in the setting of products of Banach function spaces. REFERENCES [I] AbramOVIC Iu.A. -The maximal normed extension of partially ordered normed spaces, Vestnik Leningrad Univ. 25 (1970) 7-17 (in Russian). [2] Bennett C., Sharpley R. -Interpolation of Operators, Academic Press, 1988. [3] Curbera G.P. -Operators into L] of avector measure and applications to Banach lattices, Math. Ann. 293 (1992) 317-330. [4] Curbera G.P., Ricker WJ. -Banach lattices with the Fatou property and optimal domains of kernel operators, Indag. Mathern. (N.S.) 17 (2006) 187-204. [5] Defant A. -Variants of the Maurey-Rosenthal theorem for quasi Kothe function spaces, Positivity 5(2001) 153-175. [6] del Campo R., Fernandez A., Ferrando I~, Mayoral E, Naranjo E - Multiplication operators on spaces of integrable functions with respect to avector measure, J. Math. Anal. App!. 343 (2008) 514-524. [7] Fernandez A., Mayoral E, Naranjo E, Saez C., Sanchez Perez E.A. -Spaces of p-integrable functions with respect to avector measure, Positivity 10 (2006) 1-16. [8] Krein S.G., Petunin Iu.I., Semenov E.M. -Interpolation of Linear Operators, Trans!. Math. Monogr., vol. 54, Amer. Math. Soc., Providence, RI, 1982. [9] Lindenstrauss J., Tzafriri L. -Classical Banach Spaces II, Springer-Verlag, Berlin, 1979. 377 spaces, in the case of finite measure, are (orderisomorphically) the Zygmund spaces Llog Land Lexp respectively (see [2, Definition IV6.l D, satisfy the conditions of Remark 6.8. As acorollary of Theorem 6.7 we obtain that L <1>0 is L <I> -perfect. Corollary 6.9. Let <1>, <1>0 be Young functions satisfYing the conditions (i) and (ii) of Theorem 6.7 for some other Young function <1> 1. Then, (L <1>0) L <Jl L <Jl == L <1>0 • Proof. Note that the Young functions <1>0, <1>1 play the same role in the hypothesis of Theorem 6.7. Hence, it follows that (L <l>O)L <Jl == L <1>] and (L <l>1)L <Jl == L <1>0. So, (L <l>O)L<JlL<Jl == (L <PI )L<Jl == L <1>0. 0 Note that under the assumptions of Corollary 6.9, L <1>1 is also L <I> -perfect, since it can be interchanged with L <1>0. Finally let us show an example of functions satisfying the hypothesis of Corollary 6.9. Consider ¢(t) := tP log-.B (2 +t) and ¢o(t) := exp(ta) - I which are Young functions for 2<p< 00, 0< f3 ::::;; p-2and I ::::;; ex < 00. In the case when p< ex, the function ¢o/¢ is increasing and (16) is satisfied. So, 1'1>0 is 1'1> -perfect. Note that if the measure is finite, for these Young functions, L¢and L ¢o are order isomorphic to the Zygmund spaces LP(logL)-.B/p and L:~~ respectively (see [2, Definition IV6.11 D. Note. Just before submission for publication of this paper we became aware of the preprint by A.R. Schep [15] in which some of our results concerning p-convexity and p-concavity, have been independently obtained in the setting of products of Banach function spaces. REFERENCES [I] AbramOVIC Iu.A. -The maximal normed extension of partially ordered normed spaces, Vestnik Leningrad Univ. 25 (1970) 7-17 (in Russian). [2] Bennett C., Sharpley R. -Interpolation of Operators, Academic Press, 1988. [3] Curbera G.P. -Operators into L] of avector measure and applications to Banach lattices, Math. Ann. 293 (1992) 317-330. [4] Curbera G.P., Ricker WJ. -Banach lattices with the Fatou property and optimal domains of kernel operators, Indag. Mathern. (N.S.) 17 (2006) 187-204. [5] Defant A. -Variants of the Maurey-Rosenthal theorem for quasi Kothe function spaces, Positivity 5(2001) 153-175. [6] del Campo R., Fernandez A., Ferrando I~, Mayoral E, Naranjo E - Multiplication operators on spaces of integrable functions with respect to avector measure, J. Math. Anal. App!. 343 (2008) 514-524. [7] Fernandez A., Mayoral E, Naranjo E, Saez C., Sanchez Perez E.A. -Spaces of p-integrable functions with respect to avector measure, Positivity 10 (2006) 1-16. [8] Krein S.G., Petunin Iu.I., Semenov E.M. -Interpolation of Linear Operators, Trans!. Math. Monogr., vol. 54, Amer. Math. Soc., Providence, RI, 1982. [9] Lindenstrauss J., Tzafriri L. -Classical Banach Spaces II, Springer-Verlag, Berlin, 1979. 377 [10] Lozanovskii G.Ya. -On some Banach lattices, Siberian Math. 1. 10 (1969) 419--430. [II] Maligranda L., Persson L.E. -Generalized duality of some Banach function spaces, Indag. Mathern. 51 (1989) 323-338. [12] Reisner S. - A factonzatlOn theorem in Banach lattices and its apphcation to Lorentz spaces, Ann. Inst. Fourier, Grenoble 31 (1981) 239-255. [13] Rutlckii Ja.B. -On properties of certain operations over spaces, Operator methods in differential equations, Voronez (1979) 79-84 (in Russian). [14] Sanchez Perez E.A. ~ Compactness arguments for spaces of p-integrable functions with respect to avector measure and factorization of operators through Lebesgue-Bochner spaces, Illinois 1. Math. 45 (2001) 907-923. [15] Schep A.R. ~ Products and factors of Banach function spaces. Preprint. [16] Stefansson G.P. - L[ ofa vector measure, Le Matematiche 48 (1993) 219-234. [17] Zaanen A.C. -Integration, 2nd rev. ed., North-Holland, Amsterdam, 1967. (Received ... 2008) 378 [10] Lozanovskii G.Ya. -On some Banach lattices, Siberian Math. 1. 10 (1969) 419--430. [II] Maligranda L., Persson L.E. -Generalized duality of some Banach function spaces, Indag. Mathern. 51 (1989) 323-338. [12] Reisner S. - A factonzatlOn theorem in Banach lattices and its apphcation to Lorentz spaces, Ann. Inst. Fourier, Grenoble 31 (1981) 239-255. [13] Rutlckii Ja.B. -On properties of certain operations over spaces, Operator methods in differential equations, Voronez (1979) 79-84 (in Russian). [14] Sanchez Perez E.A. ~ Compactness arguments for spaces of p-integrable functions with respect to avector measure and factorization of operators through Lebesgue-Bochner spaces, Illinois 1. Math. 45 (2001) 907-923. [15] Schep A.R. ~ Products and factors of Banach function spaces. Preprint. [16] Stefansson G.P. - L[ ofa vector measure, Le Matematiche 48 (1993) 219-234. [17] Zaanen A.C. -Integration, 2nd rev. ed., North-Holland, Amsterdam, 1967. (Received ... 2008) 378