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Journal of Functional Analysis 266 (2014) 616–659 www.elsevier.com/locate/jfa
Pointwise products of some Banach function spaces and factorization
Paweł Kolwicz a,∗,1, Karol Lesnik´ a,1, Lech Maligranda b a Institute of Mathematics of Electric Faculty, Pozna´n University of Technology, ul. Piotrowo 3a, 60-965 Pozna´n, Poland b Department of Engineering Sciences and Mathematics, Luleå University of Technology, SE-971 87 Luleå, Sweden
Received 31 October 2012; accepted 24 October 2013
Available online 19 November 2013
Communicated by K. Ball
Abstract The well-known factorization theorem of Lozanovski˘ı may be written in the form L1 ≡ E E ,where means the pointwise product of Banach ideal spaces. A natural generalization of this problem would be the question when one can factorize F through E, i.e., when F ≡ E M(E,F),whereM(E,F) is the space of pointwise multipliers from E to F . Properties of M(E,F) were investigated in our ear- lier paper [41] and here we collect and prove some properties of the construction E F . The formulas for pointwise product of Calderón–Lozanovski˘ı Eϕ-spaces, Lorentz spaces and Marcinkiewicz spaces are proved. These results are then used to prove factorization theorems for such spaces. Finally, it is proved in Theorem 11 that under some natural assumptions, a rearrangement invariant Banach function space may be factorized through a Marcinkiewicz space. © 2013 Elsevier Inc. All rights reserved.
Keywords: Banach ideal spaces; Banach function spaces; Calderón spaces; Calderón–Lozanovski˘ı spaces; Symmetric spaces; Orlicz spaces; Sequence spaces; Pointwise multipliers; Pointwise multiplication; Factorization
* Corresponding author. E-mail addresses: [email protected] (P. Kolwicz), [email protected] (K. Lesnik),´ [email protected] (L. Maligranda). 1 Research partially supported by the State Committee for Scientific Research, Poland, Grant N N201 362236.
0022-1236/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jfa.2013.10.028 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 617
1. Introduction and preliminaries
The well-known factorization theorem of Lozanovski˘ı says that for any ε>0 each z ∈ L1 can be factorized by x ∈ E and y ∈ E in such a way that
= + z xy and x E y E (1 ε) z L1 .
Moreover, if E has the Fatou property we may take ε = 0 in the above inequality. This theorem can be written in the form L1 ≡ E E, where
E F ={x · y: x ∈ E and y ∈ F }.
Then natural question arises: when is it possible to factorize F through E, i.e., when
F ≡ E M(E,F)?(1)
Here M(E,F) denotes the space of multipliers defined as M(E,F)= x ∈ L0: xy ∈ F for each y ∈ E with the operator norm
xM(E,F) = sup xyF . yE =1
The space of multipliers between function spaces was investigated by many authors, see for example [85,66,3,6] (see also [93,94,18,67,1,63,79,69,26,20,14,25,65,86,41]). In this paper we are going to investigate general properties of the product construction E F and calcu- late the product space E F for Calderón–Lozanovski˘ı, Lorentz and Marcinkiewicz spaces. This product space was of interest in [2,91,72,94,22,85,63,78,79,11,6,24,5,46,86]. The results on product construction will be used to give answers to the factorization question (1) in these special spaces. Let (Ω,Σ,μ) be a complete σ -finite measure space and L0 = L0(Ω) be the space of all classes of μ-measurable real-valued functions defined on Ω. A (quasi-)Banach space E = (E, ·E) is said to be a (quasi-)Banach ideal space on Ω if E is a linear subspace of L0(Ω) and satisfies the so-called ideal property, which means that if y ∈ E, x ∈ L0 and |x(t)| |y(t)| for μ-almost all t ∈ Ω, then x ∈ E and xE yE. We will also assume that a(quasi-)Banach ideal space on Ω is saturated, i.e. every A ∈ Σ with μ(A) > 0 has a subset B ∈ Σ of finite positive measure for which χB ∈ E. The last statement is equivalent with the existence of a weak unit, i.e., an element x ∈ E such that x(t) > 0 for each t ∈ Ω (see [39] and [63]). If the measure space (Ω,Σ,μ) is non-atomic we shall speak about (quasi-)Banach function space and if we replace the measure space (Ω,Σ,μ) by the counting measure space (N, 2N,m), then we say that E is a (quasi-)Banach sequence space (denoted by e). A point x ∈ E is said to have an order continuous norm (or to be order continuous ele- ment) if for each sequence (xn) ⊂ E satisfying 0 xn |x| and xn → 0 μ-a.e. on Ω, one has xnE → 0. By Ea we denote the subspace of all order continuous elements of E.Itisworth ∈ ↓ to notice that in case of Banach ideal spaces on Ω, x Ea if and only if xχAn E 0 for any 618 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 { } ∅ ⊃ ∞ = sequence An satisfying An (that is An An+1 and μ( n=1 An) 0). A Banach ideal space E is called order continuous if every element of E is order continuous, i.e., E = Ea. 0 A space E has the Fatou property if the conditions 0 xn ↑ x ∈ L with xn ∈ E and ∞ ∈ ↑ supn∈N xn E < imply that x E and xn E x E. We shall consider the pointwise product of Calderón–Lozanovski˘ı spaces Eϕ (which are gen- eralizations of Orlicz spaces Lϕ), but very useful will be also the general Calderón–Lozanovski˘ı construction ρ(E,F). A function ϕ :[0, ∞) →[0, ∞] is called a Young function (or an Orlicz function if it is finite-valued) if ϕ is convex, non-decreasing with ϕ(0) = 0; we assume also that ϕ is neither identically zero nor identically infinity on (0, ∞) and lim − ϕ(u) = ϕ(b ) if b < ∞, where u→bϕ ϕ ϕ bϕ = sup{u>0: ϕ(u) < ∞}. Note that from the convexity of ϕ and the equality ϕ(0) = 0 it follows that limu→0+ ϕ(u) = ϕ(0) = 0. Furthermore, from the convexity and ϕ ≡ 0 we obtain limu→∞ ϕ(u) =∞. If we denote aϕ = sup{u 0: ϕ(u) = 0}, then 0 aϕ bϕ ∞ and aϕ < ∞, bϕ > 0, since a Young function is neither identically zero nor identically infinity on (0, ∞). The function ϕ is continuous and non-decreasing on [0,bϕ) and is strictly increasing on [aϕ,bϕ).Ifaϕ = 0, then we write ϕ>0 and if bϕ =∞, then ϕ<∞. Each Young function ϕ defines the function ρϕ :[0, ∞) ×[0, ∞) →[0, ∞) in the following way −1 u vϕ ( v ), if u>0, ρϕ(u, v) = 0, if u = 0, where ϕ−1 denotes the right-continuous inverse of ϕ and is defined by − − − ϕ 1(v) = inf u 0: ϕ(u) > v for v ∈[0, ∞) with ϕ 1(∞) = lim ϕ 1(v). v→∞
If ρ = ρϕ and E, F are Banach ideal spaces over the same measure space (Ω,Σ,μ), then the Calderón–Lozanovski˘ı space ρ(E,F) is defined as all z ∈ L0(Ω) such that for some x ∈ E, y ∈ F with xE 1, yF 1 and for some λ>0wehave |z| λρ |x|, |y| μ-a.e. on Ω.
The norm zρ =zρ(E,F) of an element z ∈ ρ(E,F) is defined as the infimum values of λ for which the above inequality holds. It can be shown that 0 ρ(E,F)= z ∈ L (Ω): |z| ρ(x,y) μ-a.e. for some x ∈ E+,y∈ F+ with the norm zρ(E,F) = inf max xE, yF : |z| ρ(x,y), x ∈ E+,y∈ F+ . (2)
The Calderón–Lozanovski˘ı spaces, introduced by Calderón in [15] and developed by Lozanovski˘ı in [55,56,58,59], play a crucial role in the theory of interpolation since such construction is the in- terpolation functor for positive operators and under some additional assumptions on spaces E, F , like the Fatou property or separability, also for all linear operators (see [73,75,43,63]). If ρ(u,v) = uθ v1−θ with 0 <θ<1 we write Eθ F 1−θ instead of ρ(E,F) and these spaces are P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 619
Calderón spaces (cf. [15, p. 122]). Another important situation, investigated by Calderón (cf. [15, p. 121]) and independently by Lozanovski˘ı(cf.[51, Theorem 2], [52, Theorem 2]), appears ∞ when we put F ≡ L . In this case, in the definition of the norm, it is enough to take y = χΩ and then ∞ = | | z ρϕ (E,L ) inf λ>0: ϕ x /λ E 1 . (3) ∞ Thus the Calderón–Lozanovski˘ı space Eϕ = ρϕ(E, L ) for any Young function ϕ is defined by 0 Eϕ = x ∈ L : Iϕ(cx) < ∞ for some c = c(x) > 0 , and it is a Banach ideal space on Ω with the so-called Luxemburg–Nakano norm = x Eϕ inf λ>0: Iϕ(x/λ) 1 , where the convex semimodular Iϕ is defined as ϕ(|x|)E, if ϕ(|x|) ∈ E, Iϕ(x) := ∞, otherwise.
1 1 ϕ ϕ If E = L (E = l ), then Eϕ is the classical Orlicz function (sequence) space L (l ) equipped with the Luxemburg–Nakano norm (cf. [42,63]). If E is a Lorentz function (sequence) space, then Eϕ is the corresponding Orlicz–Lorentz function (sequence) space, equipped with the p Luxemburg–Nakano norm. On the other hand, if ϕ(u) = u ,1 p<∞, then Eϕ is the (p) =| |p1/p p-convexification E of E with the norm x E(p) x E . In the case 0
˜ := + t ˜ ∈ fE(t) infs∈(0,m(I))(1 s )fE(s). Then fE(t) fE(t) 2fE(t) for all t I . For any quasi- concave function φ on I the Marcinkiewicz function space Mφ is defined by the norm
t = ∗∗ ∗∗ = 1 ∗ x Mφ sup φ(t)x (t), x (t) x (s) ds. (4) t∈I t 0 = This is a symmetric Banach function space on I with the fundamental function fMφ (t) φ(t). →1 Moreover, E MfE for any symmetric space E because ∗∗ 1 ∗ 1 [ ] = ∈ x (t) x E χ 0,t E x E for any t I (5) t fE(t)
(see, for example, [43] or [7]). Although the fundamental function of a symmetric function space E need not to be concave, there always exists equivalent norm on E for which the new fundamental function is concave (cf. Zippin [95, Lemma 2.1]). Then for a symmetric function space E with the concave fundamental function fE there is also the smallest symmetric space with the same fundamental function. This space is the Lorentz function space ΛfE given by the norm ∗ + ∗ x = x (t) df (t) = f 0 x ∞ + x (t)f (t) dt. (6) ΛfE E E L (I) E I I
Then the inclusions
→1 →1 ΛfE E MfE (7) are satisfied, where fE is the fundamental function of E. More information about Banach ideal spaces, quasi-Banach ideal spaces, symmetric Banach and quasi-Banach spaces can be found, for example, in [39,48,35,33,43,7,63,4]. The paper is organized as follows. In Section 1 some necessary definitions and notations are collected including the Calderón–Lozanovski˘ı Eϕ-spaces. In Section 2 the product space E F is defined and some general results are presented. We prove important representation of 1 1/2 1/2 ≡ 1/2 1/2 (1/2) E F as 2 -concavification of the Calderón space E F , i.e., E F (E F ) . Such an equality was used by Schep in [86] but without any explanation, which seems to be not so evident. Then we present some properties of E F that follow from this representation. In particular, the symmetry is proved and formula for the fundamental function of the product space is given fEF (t) = fE(t)fF (t). We finish Section 2 with some sufficient conditions on E and F that E F is a Banach space (not only a quasi-Banach space). In Section 3 we collected properties connecting product spaces with the space of multipliers. There is a proof of the cancellation property of the product operation for multipliers M(E F, E G) ≡ M(F,G). The cancellation property for concrete sequence spaces was already ob- served by Bennett [6, pp. 72–73]. Section 4 is devoted to products of the Calderón–Lozanovski˘ı spaces of the type Eϕ as an im- provement of results on products, known for Orlicz spaces Lϕ which were proved by Ando [2], → Wang [91] and O’Neil [72]. The inclusion Eϕ1 Eϕ2 Eϕ follows from the results proved P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 621
→ in [41]. The reverse inclusion Eϕ Eϕ1 Eϕ2 is investigated here and we improve the suffi- cient and necessary conditions which were given in the case of Orlicz spaces by Zabre˘ıko and Ruticki˘ı [94], Dankert [21,22] and Maligranda [63]. Combining the above two inclusions we ob- = tain conditions on the equality Eϕ Eϕ1 Eϕ2 . For example, for two Young functions ϕ1,ϕ2 = = ⊕ we always have Eϕ1 Eϕ2 Eϕ, where ϕ ϕ1 ϕ2 is defined by (ϕ1 ⊕ ϕ2)(u) = inf ϕ1(v) + ϕ2(w) . u=vw
In Section 5 we deal with the product space of Lorentz and Marcinkiewicz spaces. The prod- ucts of those spaces are calculated. One of the main tools in the proof is the commutativity of Calderón construction with the symmetrizations (cf. Lemma 4). Section 6 starts with some general discussion about factorization. We prove that so-called E-perfectness of F is necessary for the factorization F ≡ E M(E,F). The rest of this sec- tion is divided into two parts. The first is devoted to the factorization of Calderón–Lozanovski˘ı Eϕ-spaces. Using the results from Section 4 and the paper [41] we examine when Eϕ can be fac- torized through the space Eϕ1 . In the second part, we investigate the possibility of factorization for Lorentz and Marcinkiewicz spaces. Finally, in Theorem 11 it is proved that under some nat- ural assumptions a rearrangement invariant Banach function space E may be factorized through Marcinkiewicz space and by duality, Lorentz space may be factorized through a rearrangement invariant Banach function space E.
2. On the product space E F
Given two Banach ideal spaces (real or complex) E and F on (Ω,Σ,μ) let us define the pointwise product space E F as
E F ={x · y: x ∈ E and y ∈ F }, with a functional ·EF defined by the formula zEF = inf xEyF : z = xy, x ∈ E, y ∈ F . (8)
We will show in the sequel that E F is, in general, a quasi-Banach ideal space even if both E and F are Banach ideal spaces. Let us collect some general properties of the product space and its norm.
Proposition 1. If E and F are Banach ideal spaces on (Ω,Σ,μ), then E F has an ideal property. Moreover, = | | z E F z EF = inf xEyF : |z|=xy, x ∈ E+,y∈ F+ = inf xEyF : |z| xy, x ∈ E+,y∈ F+ . 622 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659
iθ Proof. We show first that zEF =|z|EF .Ifz = xy with x ∈ E, y ∈ F , then |z|=ze = xyeiθ, where θ : Ω → R, and | | iθ = z EF xe E y F x E y F .
−iθ −iθ Hence, |z|EF zEF . Similarly, if |z|=xy with x ∈ E, y ∈ F , then z =|z|e = xye and −iθ = z E F xe E y F x E y F , from which we obtain the estimate zEF |z|EF . Combining the above estimates we get zEF =|z|EF . To show the ideal property of E F assume that z ∈ E F and |w| |z|. By the definition for any ε>0 we can find x ∈ E, y ∈ F such that z = xy and xEyF zEF + ε.Weset = w(t) = = = = = | | | | h(t) z(t) if z(t) 0 and h(t) 0ifz(t) 0. Then w hz hxy and since hx x we have w = hxy ∈ E F with
wEF hxEyF xEyF zEF + ε.
Since ε>0 was arbitrary we obtain wEF zEF . Note that in the above proofs we needed only the ideal property of one of the spaces E or F . Next, if |z(t)|=x(t)y(t), t ∈ Ω, then, taking x0 =|x|, y0 =|y|, we obtain x0 0, y0 0, x0y0 =|x||y|=|xy|=|z|, which gives the proof of the second equality. The proof of the third equality follows from the fact that if 0 x ∈ E,0 y ∈ F and |z| xy, then |z|=uxy = x0y0, = = = |z| = where x0 ux, y0 y and u xy on the support of xy and u 0 elsewhere. Since 0 u 1, it follows that x0 x, y0 y and this proves (the non-trivial part of) the last equality. 2
Proposition 1 shows that in the investigation of product spaces it is enough to consider real spaces, therefore from now on we shall consider only real Banach ideal spaces. The product space can be described with the help of the Calderón construction. To come to this result we first prove some description of E1/pF 1−1/p spaces and p-convexification. Let us start in Theorem 1(i) below with a reformulation of Lemma 31 from [40].
Theorem 1. Let E and F be a couple of Banach ideal spaces on (Ω,Σ,μ).
(i)<∞If 1
z E1/pF 1−1/p 1/p 1−1/p = inf max xE, yF : |z|=x y ,x∈ E+,y∈ F+ 1/p 1−1/p = inf max xE, yF : |z|=x y , xE =yF ,x∈ E+,y∈ F+ .
(ii) If 1 − E(p) F (p ) ≡ E1/pF 1 1/p, where 1/p + 1/p = 1. P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 623 (iii) For 0 that is, = 2 2 | |= = ∈ ∈ z EF inf max x E, y F : z xy, x E y F ,x E+,y F+ . (10) Proof. (i) Let z ∈ E1/pF 1−1/p and z = x1/py1−1/p, where 0 x ∈ E,0 y ∈ F . Suppose x E = a>1 (for 0 −(1−1/p) 1 x1 = a x, y1 = a p y. = 1/p 1−1/p = = 1/p 1−1/p Then x1 E x E y F y1 F and z x1 y1 . Of course, max x1E, y1F max xE, yF , which ends the proof. (ii) If z ∈ E(p) F (p ), then using Proposition 1 and the definition of p-convexification we obtain = | |= ∈ (p) ∈ (p ) z E(p)F (p ) inf g E(p) h F (p ) : z gh, 0 g E , 0 h F = 1/p 1−1/p | |= 1/p 1−1/p ∈ ∈ inf x E y F : z x y ,x E+,y F+ , and, using Theorem 1(i) to the last expression, we get 1/p 1−1/p | |= 1/p 1−1/p x E = ∈ ∈ inf inf x E y F : z x y , a, x E+,y F+ a>0 yF 1/p 1/p 1−1/p 1/p 1/p 1−1/p = inf inf a u y : |z|=a u y , uE =yF ,u∈ E+,y∈ F+ a>0 E F | | 1/p z 1/p 1−1/p = inf a inf uE: = u y , uE =yF ,u∈ E+,y∈ F+ a>0 a1/p 1/p z = a =z 1/p 1−1/p . inf 1/p E F a>0 a E1/pF 1−1/p (iii) One has p 1/p 1/p 1/p p = | | = | | = ∈ + ∈ + z (EF)(p) z EF inf x E y F : z xy, x E ,y F = p 1/p p 1/p | |p = p p ∈ (p) ∈ (p) inf u E v F : z u v ,u E+ ,v F+ = | |= ∈ (p) ∈ (p) = inf u E(p) v F (p) : z uv, u E+ ,v F+ z E(p)F (p) . 624 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 (iv) The proof is an immediate consequence of Theorem 1(ii) and (iii) since E F ≡ (E F)(2) (1/2) ≡ E1/2F 1/2 (1/2). Moreover, 2 = = | | z E F z (E1/2F 1/2)(1/2) z E1/2F 1/2 √ 2 = inf max xE, yF : |z|= xy, xE =yF ,x∈ E+,y∈ F+ = 2 2 | |= = ∈ ∈ inf max x E, y F : z xy, x E y F ,x E+,y F+ , and the proof is complete. 2 As a consequence of the representation (9) we obtain the following results: Corollary 1. Let E and F be a couple of Banach ideal spaces on (Ω,Σ,μ). (i) Then E F is a quasi-Banach ideal space and the triangle inequality is satisfied with constant 2, i.e., x + yEF 2 xEF +yEF . (ii) If both E and F satisfy the Fatou property, then E F has the Fatou property. (iii) The space E F has order continuous norm if and only if the couple (E, F ) is not jointly order discontinuous, i.e., (E, F )∈ / (JOD) (for the definition, see [40]). Proof. (i) It is a consequence of the representation E F ≡ (E1/2F 1/2)(1/2) and the fact that for 1/2-concavification of a Banach ideal space G = E1/2F 1/2 we have | + |1/2 2 | |1/2 + | |1/2 2 x y G x G y G | |1/2 2 + | |1/2 2 2 x G y G . (ii) It is again a consequence of (9) and the fact that E1/2F 1/2 has the Fatou property when E and F have the Fatou property (see [54, p. 595] and [82]). (iii) Using (9) and Theorem 13 in [40] we can show that E1/2F 1/2 has order continuous norm if and only if (E, F )∈ / (JOD), which gives the statement. 2 Lozanovski˘ı [52, Theorem 4] formulated result on the Köthe dual of p-convexification E(p) with no proof. The proof can be found in paper by Schep [86, Theorem 2.9] and we present another proof which follows from the Lozanovski˘ı duality result and our Theorem 1(ii). Corollary 2. Let E be a Banach ideal space and 1 Proof. Using the Lozanovski˘ı theorem on the duality of Calderón spaces (see [54, Theorem 2]) and our Theorem 1(ii) we obtain ∞ − − E(p) ≡ E1/p L 1 1/p ≡ E 1/p L1 1 1/p ≡ E (p) L1 (p ) ≡ E (p) Lp . 2 Remark 1. In general, [E(p)] = (E)(p). In fact, for the classical Lorentz space E = Lr,1 with 1 Example 1. (a) If 1 p, q ∞,1/p + 1/q = 1/r, then Lp Lq ≡ Lr . In particular, Lp Lp ≡ Lp/2. p q In fact, by the Hölder–Rogers inequality xyr xpyq for x ∈ L , y ∈ L (see p q r r [63, p. 69]) we obtain L L ⊂ L and zr zLpLq . On the other hand, if z ∈ L , then x =|z|r/p sgn z ∈ Lp, y =|z|r/q sgn z ∈ Lq and xy = z with r/p r/q r/p+r/q xpyq =zr zr =zr =zr , r p q which shows that L ⊂ L L and zLpLq zr . More general, if 1 p,q < ∞, 1/p +1/q = 1/r and E is a Banach ideal space, then E(p) E(q) ≡ E(r) (cf. [66, Lemma 1] and [71, Lemma 2.21(i)]). 1 ∞ 1 1 ∞ ∞ ∞ ∞ (b) We have c0 l ≡ l l ≡ l and c0 ≡ c0 l = l l ≡ l . This example shows that for the Fatou property of E F it is not necessary that both E and F do have the Fatou property. The next interesting question about the product space is its symmetry. Theorem 2. Let E and F be symmetric Banach function spaces on I = (0, 1) or I = (0, ∞) with the fundamental functions fE and fF , respectively. Then E F is a symmetric quasi-Banach function space on I and its fundamental function fEF is given by the formula fEF (t) = fE(t)fF (t) for t ∈ I. (11) Proof. Using Lemma 4.3 from [43, p. 93] we can easily show that E1/2F 1/2 (even ρ(E,F))is a Banach symmetric space and (9) gives the symmetry of E F . The inequality fEF (t) fE(t)fF (t) for t ∈ I is clear. We prove the reverse inequality using →1 the fact that each symmetric Banach function space E satisfies E MfE , where MfE is the Marcinkiewicz space (see (5)), some classical inequality on rearrangement (see, for example, [29, p. 277] or [7,p.44]or [43, p. 64]) and the reverse Chebyshev inequality (see Lemma 1 below). For any 0 x ∈ E,0 y ∈ F such that xy = χ[0,t] we have x y x y E F MfE MfF u v f (u) ∗ f (v) ∗ sup E x (s) ds sup F y (s) ds 0 t t f (t)f (t) 1 ∗ ∗ E F x (s) ds y (s) ds t t 0 0 t t f (t)f (t) 1 E F x(s)ds y(s)ds t t 0 0 t f (t)f (t) E F x(s)y(s)ds t 0 t fE(t)fF (t) = χ[ ](s) ds = f (t)f (t), t 0,t E F 0 and so χ[0,t]EF fE(t)fF (t). 2 In the fifth inequality above we used the reverse Chebyshev inequality which we will prove in the lemma below. On the classical Chebyshev inequality for decreasing functions (see, for example, [68, p. 39] or [30, p. 213]). Lemma 1. Let 0 <μ(A)<∞ and x(s)y(s) = a>0 for all s ∈ A with 0 x, y ∈ L1(A). Then μ(A) xy dμ xdμ ydμ. (12) A A A Proof. For any s,t ∈ A we have x(s)y(s) − x(s)y(t) − x(t)y(s) + x(t)y(t) = 2a − x(s)y(t) − x(t)y(s) x(s)a x(t)a x(s) x(t) = 2a − − = a 2 − − x(t) x(s) x(t) x(s) 2x(s)x(t) − x(s)2 − x(t)2 [x(s) − x(t)]2 = a =−a 0. x(s)x(t) x(s)x(t) Now integrating over A with respect to s and over A with respect to t we obtain the desired inequality (12). 2 Remark 2. Formula (11) is a generalization of the well-known equality on fundamental functions = = ∈ fE(t)fE (t) t fL1 (t) for t I and it is also true for symmetric sequence spaces with the same proof. Example 1(a) shows that E F is, in general, a quasi-Banach ideal space even if both E and F are Banach ideal spaces. We can ask under which additional conditions on E and F the P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 627 product space E F is a Banach ideal space. Before formulation the theorem we need a notion of p-convexity. A Banach lattice E is said to be p-convex (1 p<∞) with a constant K 1if 1/p 1/p n n | |p p xk K xk E , k=1 E k=1 n ⊂ ∈ N for any sequence (xk)k=1 X and any n . If a Banach lattice E is p-convex with a con- stant K 1 and 1 q Theorem 3. Suppose that E, F are Banach ideal spaces such that E is p0-convex with con- stant 1, F is p -convex with constant 1 and 1 + 1 1. Then E F is a Banach space which 1 p0 p1 is even p -convex, where 1 = 1 ( 1 + 1 ). 2 p 2 p0 p1 Before the proof of Theorem 3 let us present the following lemma, which was mentioned in [80] and proved in [90, p. 219]. Lemma 2. If E is p0-convex with constant K0 and F is p1-convex with constant K1, then − E1−θ F θ is p-convex with constant K K1 θ Kθ , where 1 = 1−θ + θ . 0 1 p p0 p1 = 1/2 1/2 1 = Proof of Theorem 3. By Lemma 2, Z E F is p-convex with constant 1, where p 1 + 1 . The assumption on p ,p gives that p 2 and since 1/2-concavification of Z is 2p0 2p1 0 1 p/2-convex with constant 1 (p/2 1) it follows that it is 1-convex with constant 1 which gives that the norm of E F = Z(1/2) satisfies the triangle inequality, and consequently is a Banach space. This completes the proof. 2 Remark 3. By duality arguments, Theorem 3 can be also formulated in the terms of q-concavity of the Köthe dual spaces. A Banach lattice E is q-concave (1 Remark 4. Since inclusion G → E F means also factorization z = xy, where x ∈ E and y ∈ F , therefore sometimes these inclusions or identifications of the product spaces E F = G are called factorizations of concrete spaces as, for example, lp and Cesàro sequence spaces or Lp and Cesàro function spaces (cf. [6,5,86]), factorization of tent spaces or other spaces (cf. [16,17,31,81,8]). 3. The product spaces and multipliers Let us collect properties connecting the product space with the space of multipliers. We start with the Cwikel and Nilsson result [20, Theorem 3.5]. They proved that if a Banach ideal space E has the Fatou property and 0 <θ<1, then − − E ≡ M F (1/θ),E1 θ F θ (1 θ). 628 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 We will prove a generalization of this equality, which in the case of G = L∞ coincides with their result. Proposition 2. Let E, F , G be Banach ideal spaces. Suppose that E has the Fatou property and 0 <θ<1. Then − − − M(G,E)≡ M G1 θ F θ ,E1 θ F θ (1 θ). 1 Proof. First, let us prove the inclusion →.Letx ∈ M(G,E). We want to show that x ∈ M(G1−θ F θ ,E1−θ F θ )(1−θ), that is, |x|1−θ ∈ M(G1−θ F θ ,E1−θ F θ ) or equivalently x1−θ |y|∈ E1−θ F θ for any y ∈ G1−θ F θ . Take arbitrary y ∈ G1−θ F θ with the norm < 1. Then there are 1−θ θ w ∈ G, v ∈ F satisfying wG 1, vF 1 and |y| |w| |v| . Clearly, − − − |x|1 θ |y| |xw|1 θ |v|θ ∈ E1 θ F θ since x ∈ M(G,E), w ∈ G gives xw ∈ E. This proves the inclusion. Moreover, | |1−θ | | | |1−θ | |θ x y E1−θ F θ xw v E1−θ F θ 1−θ θ xw E v F 1−θ 1−θ θ 1−θ x M(G,E) w G v F x M(G,E). Thus, | |1−θ 1/(1−θ) x M(G1−θ F θ ,E1−θ F θ ) x M(G,E). ←1 = | |1−θ = The inclusion .Let x M(G1−θ F θ ,E1−θ F θ )(1−θ) 1, i.e. x M(G1−θ F θ ,E1−θ F θ ) 1. We need to show that for any w ∈ G we have xw ∈ M(F(1/θ),E1−θ F θ )(1−θ), that is, |xw|1−θ ∈ M(F(1/θ),E1−θ F θ ). Really, by the Cwikel–Nilsson result, we obtain xw ∈ E for any w ∈ G. Let w ∈ G and v ∈ F . Since the norm of x is 1 it follows that − z |x|1 θ 1 z G1−θ F θ E1−θ F θ for each 0 = z ∈ G1−θ F θ . Consequently, for z =|w|1−θ |v|θ , we obtain 1−θ θ 1−θ 1−θ θ |xw| |v| 1−θ θ = |x| |w| |v| 1−θ θ E F E F | |1−θ | |θ 1−θ θ w v G1−θ F θ w G v F . This proves our inclusion because, from the assumption on x and the fact that |w|1−θ |v|θ ∈ G1−θ F θ ,wehave|xw|1−θ |v|θ =|x|1−θ |w|1−θ |v|θ ∈ E1−θ F θ . Moreover, by the Cwikel– Nilsson result and the last estimate with v =|m|1/θ , we conclude P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 629 = = x M(G,E) sup xw E sup xw M(F(1/θ),E1−θ F θ )(1−θ) wG1 wG1 = | |1−θ 1/(1−θ) sup sup xw m 1−θ θ E F w G 1 m F (1/θ) 1 1/(1−θ) sup sup w G m (1/θ) 1, F w G 1 m F (1/θ) 1 and the theorem is proved with the equality of norms. 2 Proposition 2 together with the representation of the product space as the 1/2-concavification of the Calderón space will give the “cancellation” property for multipliers of products. Let us observe that Bennett in [6] proved this property for some concrete sequence spaces. Our result shows that the cancellation property is true in general. Theorem 4. Let E, F , G be Banach ideal spaces. If G has the Fatou property, then M(E F,E G) ≡ M(F,G). (13) Proof. Applying Theorem 1(iv), property (g) from [66] and Proposition 2 we obtain M(E F,E G) ≡ M E1/2F 1/2 (1/2), E1/2G1/2 (1/2) ≡ M E1/2F 1/2, E1/2G1/2 (1/2) ≡ M(F,G), and (13) is proved. 2 Remark 5. Note that Proposition 2 and Theorem 4 are equivalent. Proposition 2 can be written in the following form: if F has the Fatou property, then − − M Eθ G1 θ ,Fθ G1 θ ≡ M(E,F)(1/θ), and it can be proved using Theorem 4. In fact, applying Theorem 1(ii), the cancellation property from Theorem 4 and property (g) from [66] we obtain − − − − M Eθ G1 θ ,Fθ G1 θ ≡ M E(1/θ) G(1/(1 θ)),F(1/θ) G(1/(1 θ)) ≡ M E(1/θ),F(1/θ) ≡ M(E,F)(1/θ). From Theorem 4 we can also get the equality mentioned by Raynaud [79] which can be proved also directly (cf. also [86, Proposition 1.4]). Corollary 3. Let E, F be Banach ideal spaces. Then (E F) ≡ M F,E ≡ M E,F . (14) 630 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 Proof. Using the Lozanovski˘ı factorization theorem (for more discussion see Section 6) and the cancellation property (13) we obtain (E F) ≡ M E F,L1 ≡ M E F,E E ≡ M F,E , and (E F) ≡ M E F,L1 ≡ M E F,F F ≡ M E,F . Note that the second identity in (14) follows also from the general properties of multipli- ers (see [66, property (e)] or [41, property (vii)]) because we have M(F,E) ≡ M(E,F) ≡ M(E,F). 2 Corollary 4. Let E, F be Banach ideal spaces. If F has the Fatou property and xyEF 1 for all x ∈ E with xE 1, then yF 1. Proof. Since by the assumption yM(E,EF) 1, then using Theorem 4, together with the facts that M(L∞,F)≡ F , E L∞ ≡ E, we obtain yF =yM(L∞,F ) =yM(EL∞,EF) =yM(E,EF) 1. 2 Corollary 5. Let E, F , G be Banach ideal spaces. If F and G have the Fatou property, then M(E F,G)≡ M E,M(F,G) . Proof. Using Theorem 4, the Lozanovski˘ı factorization theorem, Corollary 3 with the fact that the Fatou property of F gives by Corollary 1(ii) that F G has the Fatou property, again Corol- lary 3 and the Fatou property of G we obtain M(E F,G)≡ M E F G ,G G ≡ M E F G ,L1 ≡ E F G ≡ F G E ≡ M E, F G ≡ M E,M F,G ≡ M E,M(F,G) , which establishes the formula. 2 4. The product of Calderón–Lozanovski˘ı Eϕ -spaces The pointwise product of Orlicz spaces was investigated already by Krasnoselski˘ı and Ruticki˘ı in their book, where sufficient conditions on inclusion Lϕ1 Lϕ2 ⊂ Lϕ are given in the case when Ω is bounded closed subset of Rn (cf. [42, Theorems 13.7 and 13.8]). For the same set Ω, ϕ ϕ ϕ Ando [2] proved that L 1 L 2 ⊂ L if and only if there exist C>0, u0 > 0 such that ϕ(Cuv) ϕ1(u) + ϕ2(v) for u, v u0. O’Neil [72] presented necessary and sufficient conditions for the inclusion Lϕ1 Lϕ2 ⊂ Lϕ in the case when the measure space is either non-atomic and infinite or non-atomic and finite or P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 631 counting measure on N. Moreover, he observed that condition ϕ(Cuv) ϕ1(u) + ϕ2(v) for all −1 −1 −1 [large, small] u, v > 0 is equivalent to condition on inverse functions C1ϕ1 (u)ϕ2 (u) ϕ (u) for all [large, small] u>0. O’Neil’s results were also presented, with his proofs, in the books [63, pp. 71–75] and [78, pp. 179–184]. The reverse inclusion Lϕ ⊂ Lϕ1 Lϕ2 and the equality Lϕ1 Lϕ2 = Lϕ were considered by Zabre˘ıko and Ruticki˘ı [94, Theorem 8], Dankert [22, pp. 63–68] and Maligranda [63, pp. 69–71]. We will prove the above results for more general spaces, that is, for the Calderón–Lozanovski˘ı → Eϕ-spaces. Results on the inclusion Eϕ1 Eϕ2 Eϕ need the following relations between −1 −1 ≺ −1 Young functions (cf. [72]): we say ϕ1 ϕ2 ϕ for all arguments [for large arguments] (for small arguments) if there is a constant C>0 [there are constants C,u0 > 0] (there are con- stants C, u0 > 0) such that the inequality −1 −1 −1 Cϕ1 (u)ϕ2 (u) ϕ (u) (15) holds for all u>0 [for all u u0] (for all u u0), respectively. Remark 6. The inequality (15) implies the generalized Young inequality: ϕ(Cuv) ϕ1(u) + ϕ2(v) for all u, v > 0 such that ϕ1(u), ϕ2(v) < ∞. (16) + −1 −1 2 −1 On the other hand, if ϕ(Cuv) ϕ1(u) ϕ2(v) for all u, v > 0, then ϕ1 (w)ϕ2 (w) C ϕ (w) for each w>0(see[72] and [41]). Similar equivalences hold for large and small arguments. ∈ ∈ ∈ In [41] the question when the product xy Eϕ provided x Eϕ1 and y Eϕ2 was investi- gated, as a generalization of O’Neil’s theorems [72], and the following results were proved (see [41, Theorems 4.1, 4.2 and 4.5]): Theorem A. Let ϕ1, ϕ2 and ϕ be three Young functions. (a) If E is a Banach ideal space with the Fatou property and one of the following conditions holds: −1 −1 ≺ −1 (a1) ϕ1 ϕ2 ϕ for all arguments, −1 −1 ≺ −1 ∞ → (a2) ϕ1 ϕ2 ϕ for large arguments and L E, −1 −1 ≺ −1 → ∞ (a3) ϕ1 ϕ2 ϕ for small arguments and E L , ∈ ∈ ∈ then, for every x Eϕ1 and y Eϕ2 the product xy Eϕ, which means that → Eϕ1 Eϕ2 Eϕ. = { } → (b) If a Banach ideal space E with the Fatou property is such that Ea 0 and Eϕ1 Eϕ2 −1 −1 −1 Eϕ, then ϕ ϕ ≺ ϕ for large arguments. 1 2 ∞ (c) If a Banach ideal space E has the Fatou property, supp Ea = Ω, L → E and → −1 −1 ≺ −1 Eϕ1 Eϕ2 Eϕ, then ϕ1 ϕ2 ϕ for small arguments. ∞ ∞ → (d) If e is a Banach sequence space with the Fatou property, supk∈N ek e < , l e and → −1 −1 ≺ −1 eϕ1 eϕ2 eϕ, then ϕ1 ϕ2 ϕ for small arguments. −1 −1 ≺ −1 Note that in case (c) we can even conclude the relation ϕ1 ϕ2 ϕ for all arguments, using (b) and (c). 632 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 → The sufficient and necessary conditions on the reverse inclusion Eϕ Eϕ1 Eϕ2 need also the reverse relations between Young functions, the same as in [41]. −1 ≺ −1 −1 The symbol ϕ ϕ1 ϕ2 for all arguments [for large arguments] (for small arguments) means that there is a constant D>0 [there are constants D, u0 > 0] (there are constants D,u0 > 0) such that the inequality −1 −1 −1 ϕ (u) Dϕ1 (u)ϕ2 (u) (17) holds for all u>0 [for all u u0] (for all 0 Theorem 5. Let ϕ1,ϕ2 and ϕ be three Young functions. (a) If E is a Banach ideal space with the Fatou property and one of the following conditions holds: −1 ≺ −1 −1 (a1) ϕ ϕ1 ϕ2 for all arguments, −1 ≺ −1 −1 ∞ → (a2) ϕ ϕ1 ϕ2 for large arguments and L E, −1 ≺ −1 −1 → ∞ (a3) ϕ ϕ1 ϕ2 for small arguments and E L , → then Eϕ Eϕ1 Eϕ2 . (b) If E is a symmetric Banach function space on I with the Fatou property, Ea = {0} and → −1 ≺ −1 −1 Eϕ Eϕ1 Eϕ2 , then ϕ ϕ1 ϕ2 for large arguments. (c) If E is a symmetric Banach function space on I with the Fatou property, supp Ea = Ω, ∞ → → −1 ≺ −1 −1 L E and Eϕ Eϕ1 Eϕ2 , then ϕ ϕ1 ϕ2 for small arguments. (d) Let e be a symmetric Banach sequence space with the Fatou property and order continuous → −1 ≺ −1 −1 norm. If eϕ eϕ1 eϕ2 , then ϕ ϕ1 ϕ2 for small arguments. Proof. (a1) The idea of the proof is taken from [63, Theorem 10.1(b)].Forz ∈ Eϕ\{0} let y = |z| ϕ( ) and z Eϕ ⎧ ⎨ |z(t)| −1 − − ϕ (y(t)), if t ∈ supp z, = ϕ 1(y(t))ϕ 1(y(t)) i zi(t) ⎩ 1 2 0, otherwise, for i = 1, 2. The elements zi are well defined. Indeed, if aϕ = 0, then y(t) > 0forμ-a.e. ∈ t supp z.Ifaϕ > 0, then the assumption on functions implies that aϕ1 > 0 and aϕ2 > 0. Conse- −1 = −1 = quently, ϕ1 (0) aϕ1 and ϕ2 (0) aϕ2 . Now we will prove the inequality z ϕ i y, i = 1, 2. (18) i D z Eϕ → = If aϕ > 0, taking u 0in(17) we obtain aϕ Daϕ1 aϕ2 .Ify(t) 0, then | | z a = z(t) −1 Eϕ ϕ −1 −1 zi(t) ϕi (0) ϕi (0) D z Eϕ ϕi (0) aϕ1 aϕ2 aϕ1 aϕ2 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 633 and consequently ϕ ( zi (t) ) = 0 = y(t).Ify(t) > 0, then i D z Eϕ |z(t)| − D|z(t)| − − z (t) = ϕ 1 y(t) ϕ 1 y(t) = Dz ϕ 1 y(t) . i −1 −1 i ϕ−1(y(t)) i ϕ i ϕ1 (y(t))ϕ2 (y(t)) This proves (18) and consequently z |z| I 1 y = ϕ 1. ϕ1 E D z Eϕ z Eϕ E | |= Thus z1 Eϕ D z Eϕ and similarly z2 Eϕ D z Eϕ . Since z z1z2, it follows that ∈ 1 2 z Eϕ1 Eϕ2 and z Eϕ Eϕ D z Eϕ . ∞1 2 ∞ ∞ (a2) If bϕ < ∞ and L → E, then Eϕ = L with equivalent norms and clearly Eϕ = L → =∞ = −1 Eϕ1 Eϕ2 . Suppose bϕ . Set v0 ϕ (u0), where u0 is from (17) and let v>0 be such that [ ] = = | | max ϕ1(v), ϕ2(v) χΩ E 1/2. For z Eϕ 1lety ϕ( z ) and A = t ∈ supp z: z(t) v0 ,B= supp z\A = t ∈ supp z: z(t) Define ⎧ ⎪ |z(t)| −1 ∈ ⎨⎪ −1 −1 ϕi (y(t)), if t A, ϕ1 (y(t))ϕ2 (y(t)) zi(t) = √ ⎪ |z(t)|,t∈ B, ⎩⎪ 0, otherwise, for i = 1, 2. Since ϕ(v0)>0, the functions zi are well defined. If t ∈ A, then √ |z(t)| − D|z(t)| − − z (t) = ϕ 1 y(t) ϕ 1 y(t) Dϕ 1 y(t) , i −1 −1 i ϕ−1(y(t)) i i ϕ1 (y(t))ϕ2 (y(t)) whence z1 1 z1 1 1 Iϕ √ χA Iϕ √ χA yE , 1 2 D 2 1 D 2 2 and vz vz 1 √ 1 = √ 1 Iϕ1 χB ϕ1 χB ϕ1(v) χΩ E . v0 v0 E 2 √ √ = { v0 } Then, for λ max v , 2 D , we obtain z z z z vz 1 1 + 1 √1 + √ 1 Iϕ1 Iϕ1 χA Iϕ1 χB Iϕ1 χA Iϕ1 χB 1. λ λ λ 2 D v0 634 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 Thus z λ and similarly z λ. Since |z|=z z , it follows that z ∈ E E and 1 Eϕ1 2 Eϕ2 1 2 ϕ1 ϕ2 z λ2. Consequently, z λ2z for each z ∈ E . Eϕ1 Eϕ2 Eϕ1 Eϕ2 Eϕ ϕ −1 ≺ −1 −1 (a3) Since ϕ ϕ1 ϕ2 for small arguments, it follows that for any u1 >u0 there is a constant D1 D such that −1 −1 −1 ϕ (u) D1ϕ1 (u)ϕ2 (u) (19) for any u u1. We follow the same way as in the proof of (a1) replacing D by D1 from (19) for M ∞ u1 = M, where M is the constant of the inclusion E → L . −1 ≺ −1 −1 (b) Suppose that the condition ϕ ϕ1 ϕ2 is not satisfied for large arguments. Then there ∞ n −1 −1 −1 ∈ N is a sequence (un) with un such that 2 ϕ1 (un)ϕ2 (un) ϕ (un) for all n . We repeat the construction of sequence (zn), as it was given in [41] in the proof of Theo- z n Eϕ1 Eϕ2 rem 4.2(i), showing that →∞as n →∞. zn Eϕ Since Ea = {0}, it follows that there is a nonzero 0 x ∈ Ea and so there is a set A of positive measure such that χA ∈ Ea. Of course, for large enough n one has unχAE 1. Applying Dobrakov’s result from [27] we conclude that the submeasure ω(B) =unχB E has the Darboux ∈ N = property. Consequently, for each n there exists a set An such that unχAn E 1. Define = −1 = −1 = xn ϕ1 (un)χAn ,yn ϕ2 (un)χAn and zn xnyn. Let us consider two cases: 0 =∞ ∞ =∞ −1 = 1 . Suppose either bϕ1 or bϕ1 < and ϕ1(bϕ1 ) . Then ϕ1(ϕ1 (u)) u for u 0 and for 0 <λ<1, by the convexity of ϕ1, we obtain − ϕ 1(u ) xn = 1 n 1 −1 = 1 Iϕ1 ϕ1 χAn ϕ1 ϕ1 (un) χAn E un χAn E > 1. λ λ E λ λ 0 ∞ ∞ 2 .Letbϕ1 < and ϕ1(bϕ1 )< . Then, for sufficiently large n and 0 <λ<1, we have xn =∞ Iϕ1 ( λ ) . In both cases x 1 and similarly y 1. Applying Theorem 2, we get n Eϕ1 n Eϕ2 −1 −1 z = ϕ (u )ϕ (u )f m(A ) n Eϕ1 Eϕ2 1 n 2 n Eϕ1 Eϕ2 n − − = ϕ 1(u )ϕ 1(u )f m(A ) f m(A ) 1 n 2 n Eϕ1 n Eϕ2 n =x y 1. n Eϕ1 n Eϕ2 On the other hand, using the relation between functions on a sequence (un) and the fact that ϕ(ϕ−1(u)) u for u>0, we obtain n = n −1 −1 −1 = Iϕ 2 zn ϕ 2 ϕ1 (un)ϕ2 (un) χAn E ϕ ϕ (un) χAn E unχAn E 1, z n n Eϕ1 Eϕ2 n →∞ →∞ i.e., zn Eϕ 1/2 which gives 2 as n . zn Eϕ P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 635 (c) It can be done by combining methods from the proof of Theorem 5(b) and Theorem A(c). −1 ≺ −1 −1 (d) Suppose condition ϕ ϕ1 ϕ2 is not satisfied for small arguments. Then we can find → n −1 −1 −1 ∈ N a sequence un 0 such that 2 ϕ1 (un)ϕ2 (un) ϕ (u n) for all n . m =∞ ∈ N Assumption on a sequence space e gives limm→∞ k=0 ek e . Thus, for each n there is a number mn such that + mn m n 1 un ek 1 mn mn = −1 = −1 = xn ϕ1 (un) ek,yn ϕ2 (un) ek and zn xnyn. k=1 k=1 mn Then Iϕ1 (xn) un k=1 ek e 1 and mn mn − I (2x ) = ϕ 2ϕ 1(u ) e 2u e → 2asn →∞. ϕ1 n 1 1 n k n k k=1 e k=1 e Therefore, for n large enough 1 x 1/2aswellas1 y 1/2. Consequently, n eϕ1 n eϕ2 z n n Eϕ1 Eϕ2 explaining like in (b) one has zn Eϕ Eϕ 1/4 and Iϕ(2 zn) 1, which gives 1 2 zn Eϕ 2n−2 →∞as n →∞and the proof of Theorem 5 is complete. 2 To formulate the results on equality of product spaces we need to introduce the equivalences −1 −1 ≈ −1 of inverses of Young functions ϕ1, ϕ2 and ϕ. The symbol ϕ1 ϕ2 ϕ for all arguments [for −1 −1 ≺ −1 −1 ≺ −1 −1 large arguments] (for small arguments) means that ϕ1 ϕ2 ϕ and ϕ ϕ1 ϕ2 , that is provided there are constants C,D > 0 [there are constants C,D,u0 > 0] (there are constants C,D,u0 > 0) such that the inequalities −1 −1 −1 −1 −1 Cϕ1 (u)ϕ2 (u) ϕ (u) Dϕ1 (u)ϕ2 (u) (20) hold for all u>0 [for all u u0] (for all 0 Corollary 6. Let ϕ1,ϕ2 and ϕ be three Young functions. (a) Suppose E is a symmetric Banach function space with the Fatou property, L∞ → E and = = −1 −1 ≈ −1 supp Ea Ω. Then Eϕ1 Eϕ2 Eϕ if and only if ϕ1 ϕ2 ϕ for all arguments. (b) Suppose E is a symmetric Banach function space with the Fatou property, L∞ → E and = { } = −1 −1 ≈ −1 Ea 0 . Then Eϕ1 Eϕ2 Eϕ if and only if ϕ1 ϕ2 ϕ for large arguments. (c) Suppose e is a symmetric Banach sequence space with the Fatou property and order contin- = −1 −1 ≈ −1 uous norm. Then eϕ1 eϕ2 eϕ if and only if ϕ1 ϕ2 ϕ for small arguments. 636 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 The following construction appeared in [94] and in [23]: for two Young functions ϕ1,ϕ2 (or even for only the so-called ϕ-functions, cf. the definition below) one can define a new function ϕ1 ⊕ ϕ2 by the formula u (ϕ1 ⊕ ϕ2)(u) = inf ϕ1(v) + ϕ2(w) = inf ϕ1(v) + ϕ2 , (21) u=vw v>0 v for u 0. This operation was investigated in [94,13,63,66,88,89]. Note that ϕ1 ⊕ ϕ2 is non- decreasing, left-continuous function and is 0 at u = 0. Moreover, as it was proved in [94, pp. 267, 271] and [88, Theorem 1]:ifϕ1,ϕ2 are nondegenerate Orlicz functions and ϕ = ϕ1 ⊕ ϕ2 then −1 −1 −1 −1 ϕ (t) ϕ1 (t)ϕ2 (t) ϕ (2t) for all t>0. The function ϕ need not to be convex even if both ϕ1 and ϕ2 are convex functions. However, −1 −1 −1 −1 if ϕ is a convex function, then ϕ ϕ1 ϕ2 2ϕ and, by Theorems A(a1) and 5(a1),we = obtain Eϕ1 Eϕ2 Eϕ. We will prove the last result without the explicit assumption that ϕ is convex, but to do this we need to extend definition of the Calderón–Lozanovski˘ı Eϕ-space to this case (cf. [37] for the definition and some results). For a non-decreasing and left-continuous function ϕ :[0, ∞) →[0, ∞) with ϕ(0) = 0 (such a function is called ϕ-function) assume that there exist C,α > 0 such that ϕ(st) Ctαϕ(s) for all s>0 and 0 Then the Calderón–Lozanovski˘ı Eϕ-space is a quasi-Banach ideal space (since ϕ need not to be convex) with the quasi-norm = | | x Eϕ inf λ>0: ϕ x /λ E 1 . Note that for a convex function ϕ the condition (22) holds with C = α = 1 and the space Eϕ is normable when (22) holds with α 1. We know that if ϕ1 and ϕ2 are convex functions, then ϕ = ϕ1 ⊕ ϕ2 is not necessary a convex function but (22) holds with C = 1 and α = 1/2 and the space Eϕ is a quasi-Banach ideal space. In fact, for any s>0 and 0 := ⊕ = Theorem 6. Let ϕ1, ϕ2 be Young functions. If ϕ ϕ1 ϕ2, then Eϕ1 Eϕ2 Eϕ, where E is a Banach ideal space with the Fatou property. → ⊕ Proof. We prove that Eϕ1 Eϕ2 Eϕ. By the definition of ϕ1 ϕ2 one has ϕ(uv) ϕ1(u) + ϕ2(v) P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 637 ∞ ∞ ∈ = for each u, v > 0 with ϕ1(u) < , ϕ2(v) < .Letz Eϕ1 Eϕ2 , z 0, and take arbitrary ∈ ∈ | |= = = 0 x Eϕ1 ,0 y Eϕ2 with z xy. Since (22) holds with C 1 and α 1/2, it follows that for 0 1 1 1/t Thus z x y and consequently z z . Finally, E E → Eϕ t Eϕ1 Eϕ2 Eϕ t Eϕ1 Eϕ2 ϕ1 ϕ2 Eϕ. → The proof of the inclusion Eϕ Eϕ1 Eϕ2 is exactly the same as the proof of Theorem 5(a1) since the convexity of ϕ has not been used there, which proves the theorem. 2 5. The product of Lorentz and Marcinkiewicz spaces Before proving results on the product of Lorentz and Marcinkiewicz spaces on I = (0, 1) or I = (0, ∞) we need some auxiliary lemmas on the Calderón construction and the notion of dilation operator. The dilation operator Ds , s>0, defined by Dsx(t) = x(t/s)χI (t/s), t ∈ I is bounded in any symmetric space E on I and DsE→E max(1,s) (see [87, Lemma 1] in the case I = (0, 1), [43, pp. 96–98] for I = (0, ∞) and [48, p. 130] for both cases). Moreover, the Boyd indices of E are defined by ln DsE→E ln DsE→E αE = lim ,βE = lim , s→0+ ln s s→∞ ln s and we have 0 αE βE 1. Lemma 3. Let E, F be symmetric function spaces on I and 0 <θ<1. Then 1 ∗ ∗ ∗ ∗ ∗ z − z − z − , 2 Eθ F 1 θ Eθ F 1 θ Eθ F 1 θ ∗ ∗ ∗ ∗ ∗ ∗ ∗ − where z := inf{max(x , y ): z (x )θ (y )1 θ ,x∈ E+,y∈ F+}. Eθ F 1−θ E F Proof. Since ∗ ∗ ∗ ∗ x = D2D 1 x D2E→E D 1 x 2 D 1 x E 2 E 2 E 2 E and (|x|θ |y|1−θ )∗(t) x∗(t/2)θ y∗(t/2)1−θ for any t ∈ I (cf. [43, p. 67]), it follows that 638 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 ∗ = ∗ θ 1−θ ∈ ∈ z θ 1−θ inf max x E, y F : z x y ,x E+,y F+ E F ∗ ∗ θ ∗ 1−θ inf max xE, yF : z (t) x (t/2) y (t/2) ,x∈ E+,y∈ F+ ∗ ∗ ∗ ∗ θ ∗ 1−θ = inf max D 1 x , D 1 y : z (t) x (t) y (t) ,x∈ E+,y∈ F+ 2 E 2 F 1 ∗ ∗ ∗ ∗ θ ∗ 1−θ inf max x , y : z (t) x (t) y (t) ,x∈ E+,y∈ F+ 2 E F 1 ∗ ∗ = z − . 2 Eθ F 1 θ The other estimate is clear and the lemma follows. 2 As a consequence of representation (10) and the above lemma with θ = 1/2 we obtain Corollary 7. Let E, F be symmetric function spaces on I . Then ∗ ∗ ∗ ∗ ∗ ∈ + ∈ + z EF inf x E y F : z x y ,x E ,y F 2 z EF . The idea of proof of the next result is coming from the Calderón paper [15, Part 13.5].Fora Banach function space E on I = (0, 1) or (0, ∞) define new spaces E(∗) and E(∗∗) (symmetriza- tions of E)as ∗ ∗ ∗∗ ∗∗ E( ) = x ∈ L0(I): x ∈ E ,E( ) = x ∈ L0(I): x ∈ E = ∗ = ∗∗ ∞ with the functionals x E(∗) x E and x E(∗∗) x E. If there is a constant 1 C< such that ∗ ∗ ∗ ∈ D2x E C x E for all x E, (23) (∗) then E is a quasi-Banach symmetric space. Then by CE we denote the smallest constant C satisfying the above inequality. The space E(∗∗) is always a Banach symmetric space. Consider the Hardy operator H and its dual H ∗ defined by t l 1 ∗ x(s) Hx(t)= x(s)ds, H x(t) = ds with l = m(I), t ∈ I. (24) t s 0 t Remark 7. If E is a Banach function space on I and the operator H is bounded in E, then (23) holds with CE 2H E→E. This follows directly from the estimates 1 1/2 ∗ ∗ ∗ 1 ∗ Hx = x (st) ds x (st) ds x (t/2) . E 2 E 0 E 0 E As we already mentioned before the Calderón spaces Eθ F 1−θ can be also defined for quasi- Banach spaces E, F (cf. [74,70,37]). P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 639 Lemma 4. Let E and F be Banach function spaces on I and 0 <θ<1. Suppose that both operators H , H ∗ are bounded in E and F . Then ∗ ∗ − C1 − ∗ C2 ∗ ∗ − E( ) θ F ( ) 1 θ → Eθ F 1 θ ( ) → E( ) θ F ( ) 1 θ , (25) = θ 1−θ = ∗θ ∗1−θ where C1 CECF , C2 HH E→E HH F →F and ∗∗ ∗∗ − 1 − ∗∗ C3 ∗∗ ∗∗ − E( ) θ F ( ) 1 θ → Eθ F 1 θ ( ) → E( ) θ F ( ) 1 θ , (26) ∗ θ ∗ 1−θ where C3 =[H E→EHHH E→E] [HF →F HHH F →F ] . Proof. Inclusions (25).Letz ∈ (E(∗))θ (F (∗))1−θ . Then |z| λ|x|θ |y|1−θ for some λ>0 and ∗ ∗ x E 1, y F 1. Thus ∗ θ ∗ 1−θ ∗ θ 1−θ ∗ ∗ θ ∗ 1−θ x (t/2) y (t/2) z (t) λ |x| |y| (t) λx (t/2) y (t/2) = λC1 CE CF θ 1−θ (∗) for any t ∈ I , which means that z ∈ (E F ) with the norm λC1. On the other hand, if z ∈ (Eθ F 1−θ )(∗), then z∗ ∈ Eθ F 1−θ and so ∗ θ 1−θ z λ|x| |y| with some λ>0, xE 1, yF 1. The following equality is true ∗ ∗ HH x(t) = Hx(t)+ H x(t), t ∈ I. (27) In fact, using the Fubini theorem, we obtain for x 0 t l t r l t ∗ 1 x(r) 1 x(r) 1 x(r) HH x(t) = dr ds = ds dr + ds dr t r t r t r 0 s 0 0 t 0 ∗ = Hx(t)+ H x(t). Using the equality (27) and twice the Hölder–Rogers inequality we obtain ∗ ∗ ∗ ∗ ∗ ∗ ∗ z H z H z + H z = HH z ∗ − ∗ ∗ − λHH |x|θ |y|1 θ λH H |x| θ H |y| 1 θ ∗ ∗ − λ HH |x| θ HH |y| 1 θ . By the Ryff theorem there exists a measure-preserving transformation ω : I → I such that |z|=z∗(ω) a.e. (cf. [7, Theorem 7.5] for I = (0, 1) or Corollary 7.6 for I = (0, ∞) under the additional assumption that z∗(∞) = 0). Thus ∗ ∗ ∗ − − |z|=z (ω) λ HH |x| (ω) θ HH |y| (ω) 1 θ = λuθ v1 θ . 640 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 Since H ∗|x| is a non-increasing function, it follows that HH∗|x| is also a non-increasing func- tion and HH∗|x|=[HH∗|x|]∗ =[(H H ∗|x|)(ω)]∗. Similarly for HH∗|y|. Hence, ∗ ∗ ∗ u (∗) = u = HH |x| (ω) E E E = ∗| | ∗ ∗ HH x E HH E→E x E HH E→E and ∗ ∗ ∗ v (∗) = v = HH |y| (ω) F F F = ∗| | ∗ ∗ HH y F HH F →F y F HH F →F , (∗) θ (∗) 1−θ which means that z ∈ (E ) (F ) with the norm λC2. To finish the proof in the case I = (0, ∞) we need to show that z∗(∞) = 0. If we will have z∗(∞) = a>0, then λ|x(t)|θ |y(t)|1−θ a for almost all t>0 and considering the sets A = {t>0: |x(t)| a/λ}, B ={t>0: |y(t)| a/λ} we obtain A ∪ B = (0, ∞) up to a set of measure zero. Then ∞ ∗ |x(s)| a H |x|(t) = ds ds s λs t A∩(t,∞) and ∞ ∗ |y(s)| a H |y|(t) = ds ds, s λs t B∩(t,∞) which means that H ∗|x|(t) + H ∗|y|(t) =+∞for all t>0. Since ∗ ∗ (0, ∞) = t>0: H |x|(t) =∞ ∪ t>0: H |y|(t) =∞ (maybe except a set of measure zero), it follows that H ∗|x| ∈/ E or H ∗|y| ∈/ F , which is a con- tradiction. Inclusions (26).Letz ∈ (E(∗∗))θ (F (∗∗))1−θ . Then |z| λ|x|θ |y|1−θ for some λ>0 and ∗∗ ∗∗ x E 1, y F 1. Thus ∗∗ − ∗∗ ∗∗ ∗∗ − z (t) λ |x|θ |y|1 θ (t) λx (t)θ y (t)1 θ for any t ∈ I , which means that z ∈ (Eθ F 1−θ )(∗∗) with the norm λ. On the other hand, if z ∈ (Eθ F 1−θ )(∗∗), then z∗∗ ∈ Eθ F 1−θ and repeating the above argu- ments we obtain ∗ ∗∗ ∗ |z|=z (ω) z (ω) = Hz (ω) ∗| | θ ∗ | | 1−θ = θ 1−θ λ HHH x (ω) HHH y (ω) λu1v1 . P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 641 Since HHH∗|x|=[HHH∗|x|]∗ =[(HHH∗|x|)(ω)]∗, it follows that = ∗∗ = ∗ ∗ u1 E(∗∗) u1 Hu1 H E→E u1 E E E ∗ ∗ ∗ =HE→E HHH |x| (ω) =H E→E HHH |x| E E ∗ ∗ → → H E E HHH E→E x E H E E HHH E→E and = ∗∗ = ∗ ∗ v1 F (∗∗) v1 Hv1 H F →F v1 F F F ∗ ∗ ∗ =HF →F HHH |y| (ω) =H F →F HHH |y| F F ∗ ∗ → → H F F HHH F →F y F H F F HHH F →F , (∗∗) θ (∗∗) 1−θ which implies that z ∈ (E ) (F ) with the norm λC3, and the lemma follows. 2 Note that our proofs are working for both cases I = (0, 1) and I = (0, ∞). The inclu- sions (25) were proved by Calderón but his result is true only in the case when I = (0, ∞) (cf. [15, pp. 167–169]). Since he was working with the other composition H ∗H , which in the = ∞ ∗ = + ∗ = case I (0, ) gives the equality H H H H .ForI (0, 1) one gets another formula ∗ = + ∗ − 1 H Hx(t) Hx(t) H x(t) 0 x(s)ds, which does not allow then to prove the same result in this case. For the identification of product spaces we will need a result on the Calderón construction p 0 p for weighted Lebesgue spaces L (w) ={x ∈ L (μ): xw ∈ L (μ)} with the norm xLp(w) = xwLp , where 1 p ∞ and w 0. Such a result for p0 = p1 was given in [75, p. 459], and for general Banach ideal spaces in [44, Theorem 2] (for 1 p0, p1 < ∞ it also follows implicitly from [9, Theorem 5.5.3] and results on relation between the complex method and the Calderón construction). We present here a direct proof. Lemma 5. Let 1 p0,p1 ∞ and 0 <θ<1. Then p0 1−θ p1 θ p L (w0) L (w1) ≡ L (w), (28) − where 1 = 1−θ + θ and w = w1 θ wθ . p p0 p1 0 1 p 1−θ p θ 1−θ θ Proof. Suppose 1 p0,p1 < ∞.Ifx ∈ L 0 (w0) L 1 (w1) , then |x| λ|x0| |x1| with p p x0 L 0 (w0) 1 and x1 L 1 (w1) 1. Using the Hölder–Rogers inequality we obtain p p (1−θ)p θp |xw| dμ λ |x0w0| |x1w1| dμ (1−θ)p/p0 θp/p1 p p0 p1 λ |x0w0| dμ |x1w1| dμ = p (1−θ)p θp λ x0w0 Lp0 x1w1 Lp1 , that is 642 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 1−θ θ p x L (w) λ x0w0 Lp0 x1w1 Lp1 λ, and so x ∈ Lp(w) with the norm λ. p |x(t)w(t)|p/pi 1 On the other hand, if 0 = x ∈ L (w) then, considering xi(t) = on the support p/pi wi (t) x Lp(w) 1−θ θ of wi and 0 otherwise (i = 0, 1), we obtain |x(t)|=xLp(w)x0(t) x1(t) and | |p pi pi x(t)w(t) xi p = xi(t)wi(t) dμ = dμ = 1. L i (wi ) p x Lp(w) p 1−θ p θ Therefore, x ∈ L 0 (w0) L 1 (w1) with the norm xLp(w) and (28) is proved. The proof for the case when one or both p0,p1 are equal ∞ is even simpler. 2 We want to calculate the product spaces of Lorentz space Λφ and Marcinkiewicz space Mφ on I , where φ is a quasi-concave function on I with φ(0+) = 0. We will do this, in fact, for some other closely related spaces. Consider the Lorentz space Λφ,p for 0 1 a The space Λφ,1 is a Banach space and if φ(t) atφ (t) for all t ∈ I , then Λφ,1 → Λφ → Λφ,1 ∗ (space Λφ was defined in (6)). Consider also another Marcinkiewicz space Mφ than the space Mφ defined in (4), that is ∗ ∗ ∗ M = M (I) = x ∈ L0(I): x ∗ = sup φ(t)x (t) < ∞ . φ φ Mφ t∈I →1 ∗ This Marcinkiewicz space need not to be a Banach space and always we have Mφ Mφ. More- ∗ →C over, Mφ Mφ if and only if t 1 t ds C for all t ∈ I. (29) φ(s) φ(t) 0 1 ∈ ∗ In fact, since φ Mφ,so(29) is necessary for this inclusion. On the other hand, if (29) holds and ∈ ∗ x Mφ, then t = ∗∗ = φ(t) 1 ∗ x Mφ sup φ(t)x (t) sup φ(s)x (s) ds t∈I t∈I t φ(s) 0 t ∗ φ(t) 1 sup φ(s)x (s) sup ds Cx ∗ . Mφ s∈I t∈I t φ(s) 0 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 643 ∗ We can consider spaces Λw,p and Mw for more general weights w 0, but then the problem of being quasi-Banach space or Banach space will appear. Such investigations can be found in [19] and [36]. Since the indices of the quasi-concave function on I are useful in the formulation of further results let us define them. The lower index pφ,I and upper index qφ,I of a function φ on I are numbers defined as ln m (t) ln m (t) φ(st) p = lim φ,I ,q= lim φ,I , where m (t) = sup . φ,I + φ,I →∞ φ,I t→0 ln t t ln t s∈I,st∈I φ(s) It is known (see, for example, [43,62,63]) that for a quasi-concave function φ on [0, ∞) we have 0 pφ,[0,∞) pφ,[0,1] qφ,[0,1] qφ,[0,∞) 1. Moreover, (29) is equivalent to qφ,I < 1. We also need for a differentiable increasing function φ on I with φ(0+) = 0theSimonenko indices defined by tφ(t) tφ(t) sφ,I = inf ,σφ,I = sup . t∈I φ(t) t∈I φ(t) They satisfy 0 sφ,I pφ,I qφ,I σφ,I (cf. [62, p. 22] and [63, Theorem 11.11]). Theorem 7. ∗ →1 ∗ ∗ →2 ∗ (i) If φ, ψ are quasi-concave functions on I , then Mφψ Mφ Mψ Mφψ. + + (ii) Let φ, ψ and φψ be increasing concave functions on I with φ(0 ) = ψ(0 ) = 0.Ifsφ,I + ∗ 4 →4/a 2→/b ∗ a>0 and sφψ,I b>0, then Λφ Mψ Λφψ Λφ Mψ . (iii) Let φ, ψ be quasi-concave functions on I such that 0 = ∗ = ∗ ∗ = ∗ Λφ,1 Λψ,1 Λφψ,1/2,Λφ,1 Mψ Λφψ,1,Mφ Mψ Mφψ with equivalent quasi-norms. z ∗ z ∗ ∈ ∗ ∗ Mφψ Mφψ ∈ ∗ ∗ Proof. (i) For each z Mφψ one has z φψ , but since φψ Mφ Mψ , it follows that ∗ ∗ z ∈ M M and z ∗ ∗ z ∗ . φ ψ Mφ Mψ Mφψ ∈ ∗ ∗ ∗ ∗ ∗ ∗ ∈ ∗ ∗ ∈ ∗ If z Mφ Mψ then, by Corollary 7,wehavez x y for some x Mφ, y Mψ and x ∗ y ∗ ∗ ∗ ∗ ∗ ∗ Mφ Mψ inf{x ∗ y ∗ : z x y } 2z ∗ ∗ .Butx y and so Mφ Mψ Mφ Mψ φ ψ ∗ ∗ ∗ z ∗ = sup φ(t)ψ(t)z (t) sup φ(t)ψ(t)x (t)y (t) x ∗ y ∗ . Mφψ Mφ Mψ t∈I t∈I Therefore, ∗ ∗ ∗ z ∗ inf x ∗ y ∗ : z x y 2z ∗ ∗ . Mφψ Mφ Mψ Mφ Mψ 644 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 ∈ ∗ ∈ ∈ ∗ = (ii) Let z Λφ Mψ . Then for any ε>0 we can find x Λφ, y Mψ such that z xy and 1 x y ∗ (1 + ε)z ∗ . Since ψ (t) ψ(t)/t and φ(t) tφ (t), it follows that Λφ Mψ Λφ Mψ a ∗ ∗ ∗ z (t) d(φψ)(t) x (t/2)y (t/2) φ (t)ψ(t) + φ(t)ψ (t) dt I I ∗ ∗ 1 ∗ ∗ ψ(t) 2 x (t/2)y (t/2)ψ(t/2)φ (t) dt + x (t/2)y (t/2)tφ (t) dt a t I I ∗ ∗ 2 ∗ ∗ 2supψ(s)y (s) x (t/2)φ (t) dt + sup ψ(s)y (s) x (t/2)φ (t) dt s∈I a s∈I I I (2 + 2/a)y ∗ D x Mψ 2 Λφ (4 + 4/a)y ∗ x (4 + 4/a)(1 + ε)z ∗ . Mψ Λφ Λφ Mψ Since ε>0 is arbitrary the first inclusion of (ii) is proved. To prove the second one assume that ∗ z = z ∈ Λφψ. Then φ(t)ψ(t) ∗ w(t) = z(t) ∈ L1(I) and w = w . t Moreover, ∗ φ(t)ψ(t) 1 ∗ 1 w 1 = z (t) dt z (t) d(φψ)(t) = z . L t b b Λφψ I I Using the Lorentz result on the duality (Λφ) ≡ Mt/φ(t) (see [49, Theorem 6], [50, Theo- rem 3.6.1]; see also [43, Theorem 5.2] for separable Λφ, [30, Proposition 2.5(a)] with p = q = 1, 1 [38, Theorem 2.2]) and Lozanovski˘ı’s factorization theorem L ≡ Λφ (Λφ) ≡ Λφ Mt/φ(t) we can find u ∈ Λφ, v ∈ Mt/φ(t) such that ∗ = w uv and u Λφ v Mt/φ(t) w L1 . ∈ ∈ →1 ∗ By Corollary 7, there are u0 Λφ, v0 Mt/φ(t) Mt/φ(t) with ∗ ∗ ∗ w u0v0 and u0 Λφ v0 Mt/φ(t) 2 u Λφ v Mt/φ(t). Let ∗ ∗ = t w (t) = w x(t) and y ∗ . φ(t) v0 Mt/φ(t) u0 w∗(t) ∗ 1 ∗ ∗ φ(t)ψ(t) Then x(t) ∗ u (t) because Mt/φ(t) → M .Alsoy(t) v (t) and z(t) = v0 (t) 0 t/φ(t) 0 t = φ(t) w(t) x(t) t v0 Mt/φ(t), hence P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 645 1 z(t) = x(t) v ψ(t) 0 Mt/φ(t) v ∈ 0 Mt/φ(t) ∈ ∗ with x Λφ and ψ(t) Mψ . Moreover, 1 ∗ z Λ M x Λφ v0 Mt/φ(t) u0 Λφ v0 Mt/φ(t) φ ψ ψ ∗ Mψ 2 2u v 2w 1 z Λφ Mt/φ(t) L b Λφψ and the proof of (ii) is complete. ∗ 1 φ(t) (iii) If 0 6. Factorization of some Banach ideal spaces The factorization theorem of Lozanovski˘ı states that for any Banach ideal space E the space L1 has a factorization L1 ≡ E E. The natural generalization of the type F ≡ E M(E,F) (30) is not true without additional assumptions on the spaces, as we can see in the example below. = p,1 = 1 1/p−1 ∗ ∞ Example 2. If E L with the norm x E p I t x (t) dt for 1 Therefore, the factorization (30) is not true and we even do not have the factorization p p p,∞ L = E M(E,L ) with equivalent norms. Similarly, if F = L with the norm xF = 1/p ∗∗ ∞ p p,∞ ≡ p,1 p ≡ ∞ supt∈I t x (t) for 1 Consequently, again the factorization (30) is not true and we do not even have the equality F = Lp M(Lp,F)with equivalent norms. Let us collect some factorization results of type (30). First of all the Lozanovski˘ı factorization theorem was announced in 1967 (cf. [53, Theorem 4]) and published with detailed proof in 1969 (cf. [54, Theorem 6]; see also [57]). His proof uses the Calderón space F = E1/2(E)1/2 and the result about its dual F ≡ F ≡ L2 (cf. [54, Theorem 5]; see also [63, p. 185] and [82,83]). The Lozanovski˘ı factorization theorem was new even for finite dimensional spaces. In 1976 Jamison and Ruckle [32] proved that l1 factors through every normal Banach sequence space and its Köthe dual. The proof even in the finite dimensional case is indirect and it uses the Brouwer fixed point theorem. Later on Lozanovski˘ı’s factorization result was proved by Gillespie [28] and the method was inspired by the theory of reflexive algebras of operators on Hilbert space. If E, F are finite dimensional ideal spaces and BE, BF denote their unit balls, then Bol- lobás and Leader [12], with help of the Jamison–Ruckle method, proved the factorization BE BM(E,F) ≡ BF under assumptions that BF is a strictly unconditional body and BM(E,F) is smooth. Nilsson [70, Lemma 2.5], using the Maurey factorization theorem (cf. [67, Theorem 8];see also [92, pp. 264–266] and [84]), proved the following result of the type (30):ifE is a Banach ideal space which is p-convex with constant 1, then E ≡ Lp M E,Lp ≡ Lp M Lp ,E . (31) P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 647 By the duality result and the equality (31) we conclude that if F is a Banach ideal space with the Fatou property which is q-concave with constant 1 for 1 The factorization (32) was proved and used by Nilsson [70, Theorem 2.4] in a new proof of the Pisier theorem (cf. [76, Theorem 2.10], [77, Theorem 2.2]; see also [90, Theorem 28.1]): if a Banach ideal space E with the Fatou property is p-convex and q-concave with constants 1, ∞ 1 = θ + − 1 = θ ≡ q (1−θ) 1 1 − − s s | |1 θ 1 θ ≡ | |1/s ≡ | |1/s x M(Lq ,E) x M(Ls ,E(1/p)) x M((E(1/p)),Ls ), (1/p) (1/p) s = r E is a Banach space and M((E ) ,L ) is s -convex with constant 1, where s θp . Second part uses (32) and by Theorem 1(ii) 1 1 − θ − r − 1 θ r ≡ 1 θ θ ≡ 1 θ q ≡ q q ≡ E0 L E0 L E0 L M L ,E L E. Schep proved the factorization (32) and also the reverse implication, that is, if (32) holds, then the space F is q-concave with constant 1 (cf. [86, Theorem 3.9]). He has also proved another factorization result (even equivalence – see [86, Theorem 3.3]): if Banach ideal space E with the Fatou property is p-convex with constant 1 (1 Note that the factorization theorem of type (32): F = lq M(lq ,F) for any q-concave Banach space F with a monotone unconditional basis was proved already in 1980 (cf. [47, Corol- lary 3.2]). If a space E has the Fatou property, then in the definition of the norm of E E we may take “minimum” instead of “infimum”. It is known that the Fatou property of E is equivalent with the isometric equality E ≡ E. Then E is called perfect. This notion can be generalized to F -perfectness. We say that E is F -perfect if M(M(E,F),F) ≡ E (see [66,14,86] for more in- formation about F -perfectness). Is there any connection between factorization (30) and E being F -perfect? Theorem 8. Let E, F be Banach ideal spaces with the Fatou property. Then the factorization E M(E,F)≡ F implies F -perfectness of E, i.e., M(M(E,F),F) ≡ E. Proof. Schep [86, Theorem 2.8] proved that if E F is a Banach ideal space, then M(E,E F) ≡ F (see also Theorem 4 above). Since E M(E,F) ≡ F is a Banach ideal space by the assumption, therefore from the above Schep result we obtain M M(E,F),F ≡ M M(E,F),E M(E,F) ≡ E, which is the F -perfectness of E. 2 The example of Bollobás and Brightwell [10], presented in [86, Example 3.6], shows that the reverse implication is not true, even for three-dimensional spaces. 648 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 Almost all proofs in factorization theorems are tricky or use powerful theorems and, in fact, equality E M(E,F) ≡ F is proved without calculating M(E,F) directly. Except for some special cases it seems to be the only way to prove the equality of norms in (30). However, it seemstobealsousefultohave(30) with just the equivalence of norms, that is F = E M(E,F). (34) This can be done by finding M(E,F) and E M(E,F) separately and we will do so. Notice that similarly was done in [6], where the author has proved a number of factorization theorems for lp spaces, Cesàro sequence spaces, their duals and convexifications. Observe also that if a Banach ideal space E is p-convex (1 (1/p) −p 0 = (1/p) ·0 defines the norm on E with K x E(1/p) x x E(1/p) . Thus E0 (E , ) (p) = 1 = | |p0 1/p is a Banach ideal space and its p-convexification E0 E with the norm x ( x ) is p-convex with constant 1 (cf. [48, Lemma 1.f.11] and [71, Proposition 2.23]), and we can use the result from (31) to obtain E = Lp M(Lp ,E). As a straightforward conclusion from Corollary 6.1 in [41] and Theorem A(a) with Theo- rem 5(a) we get the following factorization theorem for Calderón–Lozanovski˘ı Eϕ-spaces. Theorem 9. Let E be a Banach ideal space with the Fatou property and supp E = Ω. Suppose that for two Young functions ϕ, ϕ1 there is a Young function ϕ2 such that one of the following conditions holds: −1 −1 ≈ −1 (i) ϕ1 ϕ2 ϕ for all arguments, −1 −1 ≈ −1 ∞ → (ii) ϕ1 ϕ2 ϕ for large arguments and L E, −1 −1 ≈ −1 → ∞ (iii) ϕ1 ϕ2 ϕ for small arguments and E L . = Then the factorization Eϕ1 M(Eϕ1 ,Eϕ) Eϕ with equivalent norms is valid and, in conse- quence, the space Eϕ1 is Eϕ-perfect up to the equivalence of norms. Moreover, applying Lemma 7.4 from [41] to Theorem 9(i) one has the following special case. Corollary 8. Let ϕ,ϕ1 be two Orlicz functions, and let E be a Banach ideal space with the Fatou ϕ(uv) property and supp E = Ω. If the function fv(u) = is non-increasing on (0, ∞) for any ϕ1(u) = v>0, then the factorization Eϕ1 M(Eϕ1 ,Eϕ) Eϕ is valid with equivalent norms and, in consequence, the space Eϕ1 is Eϕ-perfect up to the equivalence of norms. Proof. It is enough to take as ϕ2 the function defined by ϕ2(u) = (ϕ ϕ1)(u) = sup ϕ(uv) − ϕ1(v) v>0 −1 −1 ≈ −1 2 and use the fact proved in [41, Lemma 7.4] showing that ϕ1 ϕ2 ϕ for all arguments. P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 649 Before we consider the factorization of Lorentz and Marcinkiewicz spaces, let us calculate “missing” spaces of multipliers. Proposition 3. Suppose φ, ψ are non-decreasing, concave functions on I with φ(0+) = + ψ(0 ) = 0. Let E and F be symmetric spaces on I with fundamental functions fE(t) = φ(t) = = ψ(s) ∈ and fF (t) ψ(t).Ifω(t) sup0 →1 →1 ∗ →1 M(Λφ,F) Mω,M(E,Mψ ) Mω, and Mω M(Λφ,1,Λψ,1). ∗ 1→/a If, moreover, sφ,I a>0, then Mω M(Λφ,Λψ ). Proof. By Theorem 2.2(iv) in [41] the function ω is the fundamental function of M(Λφ,F) and, by the maximality of the space Mω and Theorem 2.2(i) in [41], we obtain inclusion 1 M(Λφ,F) → Mω. On the other hand, using the property (e) from [66, p. 326], the duality (Mψ ) ≡ Λt/ψ(t) and the above result we conclude 1 M(E,Mψ ) ≡ M (Mψ ) ,E ≡ M Λt/ψ(t),E → Mω. Two other inclusions will be proved if we show that 1 belongs to the corresponding spaces. ω ∗ ∗ = ∗ Since, by Theorem 2.2(ii) in [41],wehave y M(E,F) supx E 1 x y F , it follows that ∗ 1 ∗ 1 ψ(t) ∗ 1 ψ(t) = sup x (t) dt = sup x (t) dt ω ω t ω(t) t M(Λφ,1,Λψ,1) x Λφ,1 1 x Λφ,1 1 I I ∗ φ(s) ψ(t) ∗ φ(t) = sup x (t) inf dt sup x (t) dt 1, xΛ 1 0 Putting together previous results on the products and multipliers of Lorentz and Marcinkie- wicz spaces we are ready to prove the factorization of these spaces. Theorem 10. Let φ, ψ be non-decreasing, concave functions on I with φ(0+) = ψ(0+) = 0. ψ(t) Suppose φ(t) is a non-decreasing function on I . (a) If sφ,I > 0 and sψ,I > 0, then Λφ M(Λφ,Λψ ) = Λψ . 650 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 Moreover, for any symmetric space F on I with the fundamental function fF (t) = ψ(t) and under the above assumptions on φ and ψ we have Λφ M(Λφ,F)= F if and only if F = Λψ . (35) (b) If σφ,I < 1 and σψ,I < 1, then Mφ M(Mφ,Mψ ) = Mψ . Moreover, for any symmetric space E on I having Fatou property, with the fundamental function fE(t) = φ(t) and under the above assumptions on φ and ψ we have E M(E,Mψ ) = Mψ if and only if E = Mφ. (36) (c) If σφ,I < 1, sψ,I > 0 and sψ/φ,I > 0, then Mφ M(Mφ,Λψ ) = Λψ . (37) Proof. (a) Using Proposition 3 we have →1 →1 ∗ 1→/a = ψ(s) M(Λφ,Λψ ) Mω Mω M(Λφ,Λψ ), where ω(t) sup . 0 Since ψ/φ is a non-decreasing function on I , it follows that φω = ψ,sφω;I = sψ,I > 0 and, by Theorem 7(ii), = ∗ = Λφ M(Λφ,Λψ ) Λφ Mω Λψ with equivalent norms. Under the assumptions on F we obtain from Proposition 3 the inclusion 1 M(Λφ,F) → Mω and then, by Theorem 7(ii), = →1 →1 ∗ = F Λφ M(Λφ,F) Λφ Mω Λφ Mω Λψ . Minimality of Λψ gives F = Λψ . (b) Applying the fact that (Mφ) ≡ Λt/φ(t), the general duality property of multipliers (see [66, property (e)]) and using Proposition 3 we obtain ≡ ≡ = ∗ M(Mφ,Mψ ) M Mψ ,Mφ M(Λt/ψ(t),Λt/φ(t)) Mω because st/ψ(t),I = 1 − σψ,I > 0. Since ψ/φ is a non-decreasing function on I , it follows that φω = ψ and σφω,I = σψ,I < 1. By Theorem 7(i) and the estimate σφ,I < 1wehave = = ∗ ∗ = ∗ = ∗ = Mφ M(Mφ,Mψ ) Mφ Mω Mφ Mω Mφω Mψ Mψ . 1 Under the assumptions on E we obtain from Proposition 3 that M(E,Mψ ) → Mω and 1 Mψ = E M(E,Mψ ) → E Mω. P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 651 On the other hand, by Theorem 7(i) and the assumption qψ;I < 1 →1 ∗ ∗ →2 ∗ ≡ ∗ = Mφ Mω Mφ Mω Mφω Mψ Mψ . C C Therefore, Mφ Mω → E Mω. Using now Schep’s theorem, saying that if E F → E G, C C then F → G (see [86, Theorem 2.5]), we obtain Mφ → E. Maximality of the Marcinkiewicz space Mφ implies that E = Mφ since the fundamental function of E is fE(t) = φ(t) for all t ∈ I . ≡ = (c) Using Theorem 2.2(v) from [41] we obtain M(Mφ,Λψ ) Λη, where η(t) t s ∞ 0 ( φ(s)) ψ (s) ds < . Since t t φ(s)ψ(s) − φ(s)sψ(s) ψ(s) η(t) = ds ds φ(s)2 φ(s) 0 0 t 1 ψ 1 ψ(t) (s) ds = , sψ/φ φ sψ/φ φ(t) 0 and t t φ(s)ψ(s) − φ(s)sψ(s) φ(s)ψ(s) − φ(s)ψ(s) η(t) = ds ds φ(s)2 φ(s)2 0 0 t ψ ψ(t) = (s) ds = , φ φ(t) 0 it follows that M(Mφ,Λψ ) = Λψ/φ. Using the assumption σφ,I < 1 to this equality and Theo- rem 7(ii) we obtain = = ∗ = Mφ M(Mφ,Λψ ) Mφ Λψ/φ Mφ Λψ/φ Λψ , and the theorem is proved. 2 Applying the above theorem to classical Lorentz Lp,1 and Marcinkiewicz Lp,∞ spaces we obtain the following factorization results: Example 3. (a) If 1 p q<∞, then Lp,1 = Lq,1 M(Lq,1,Lp,1). (b) If 1 Example 4. If either 1 r p In fact, if 1 r p Finally, by Theorem 1(iii) and Theorem 7(ii) with φ(t)= tr/q and ψ(t)= tr/p−r/q, we get [ − ] ∞ Lq,r M Lq,r,Lp,r = Lq/r,1 (r) Lpq/ r(q p) , (r) [ − ] ∞ = Lq/r,1 Lpq/ r(q p) , (r) = Lp/r,1 (r) = Lp,r. The case 1 Theorem 11. Let φ be an increasing, concave function on I with 0 (a) Suppose F is a symmetric space on I with the lower Boyd index αF >qφ,I and such that ∗ = { } M(Mφ,F) 0 . Then = ∗ ∗ = F Mφ M Mφ,F Mφ M(Mφ,F). (b) Suppose E is a symmetric space on I with the Fatou property, which Boyd indices satisfy 0 <αE βE Λφ,1 = E M(E,Λφ). Let us start with the following identifications. Lemma 6. Under assumptions on φ from Theorem 11 we have ∞ ∗ ∞ ∗ ∗ M L (φ)( ),F = M L (φ), F ( ) ≡ F(1/φ)( ). (38) Proof. Since we have the equivalences ∗ ∞ ∗ ∗ ∞ z z ∈ M L (φ), F ( ) ⇔ z ∈ M L (φ), F ⇔ ∈ F φ ∗ ∗ ⇔ z ∈ F(1/φ) ⇔ z ∈ F(1/φ)( ) with the equalities of norms, the equality M(L∞(φ), F )(∗) ≡ F(1/φ)(∗) follows. Let I = (0, 1). We prove the equality M(L∞(φ)(∗),F)= F(1/φ)(∗). Clearly, it is enough to show that the fol- lowing conditions are equivalent: P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 653 10. z ∈ M(L∞(φ)(∗),F), 0 z ∈ : → 2 . φ(ω) F for every measure-preserving transformation (mpt) ω I I , ∗ 0 z ∈ 3 . φ F . Moreover, we shall prove that ∗ ∗ z z = z sup z M[L∞(φ)(∗),F ] D2 F →F . (39) φ F ω-mpt φ(ω) F φ F 0 ⇒ 0 ∈ ∞ ∗ : → 1 ∗ = 1 = 1 2 .Letz M(L (φ) ,F)and take arbitrary mpt ω I I . Since ( φ(ω)) φ φ φ 1 ∞ ∗ 1 ∈ ( ) z ∈ z ∞ ∗ it follows that φ(ω) L (φ) . Therefore, φ(ω) F and φ(ω) F z M[L (φ)( ),F ]. Conse- quently, z sup z M(L∞(φ)(∗),F ). ω-mpt φ(ω) F 0 0 ∞ (∗) ∗ 2 ⇒ 1 .Letx ∈ L (φ) with the norm 1. Take an mpt ω0 such that |x|=x (ω0). Then ∗ |z| |zx|=|z|x (ω0) ∈ F, φ(ω0) ∗ ∗ ∞ (∗) because 1 x φL∞ =(x φ)(ω0)L∞ . Thus z ∈ M(L (φ) ,F)and z z M(L∞(φ)(∗),F ) sup . ω-mpt φ(ω) F 0 0 ∗ 2 ⇒ 3 .Takeanmptω0 such that |z|=z (ω0). Then ∗ ∗ ∗ z z |z| z |z| z ∼ (ω0) = ∈ F and = sup . φ φ φ(ω0) φ F φ(ω0) F ω-mpt φ(ω) F 30 ⇒ 20. For each mpt ω : I → I we have ∗ ∗ ∗ ∗ z z ∗ 1 z (t/2) z (t) ∼ (t) z (t/2) (t/2) = = D (t). φ(ω) φ(ω) φ(ω) φ(t/2) 2 φ z ∈ By the symmetry of F we obtain φ(ω) F and ∗ z z sup D2F →F . ω-mpt φ(ω) F φ F The proof of (39) and also (38) is finished for I = (0, 1).IfI = (0, ∞), then the existence of a measure-preserving transformation ω0 : I → I requires the additional assumption. In the first case we need φ(∞) =∞, which is satisfied because pφ,I > 0. In the second case, we need ∗ ∞ = ∈ ∗ = { } ∗ ∞ = to have z ( ) 0 when z M(Mφ,F) 0 . Suppose, on the contrary, that z ( ) a>0. 654 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 ∗ ∈ ∗ a z∗ ∈ ∈ Since z M(Mφ,F), it follows that φ φ F and 1/φ F gives, by the maximality of ∈ ∗ = Marcinkiewicz space, that 1/φ Mψ , where the fundamental function of F is fF (t) ψ(t). ψ(t) ∞ − Thus supt>0 φ(t) < . On the other hand, since pψ αF >qφ and pψ/φ pψ qφ > 0, then for 0 <ε<(pψ − qφ)/2 and for large t ψ(t) ψ(1) ψ(1) − − + ψ(1) − − tpψ ε (qφ ε) = tpψ qφ 2ε →∞ as t →∞, φ(t) φ(1)mφ(t)mψ (1/t) φ(1) φ(1) which gives a contradiction. 2 Proof of Theorem 11. (a) We have ∗ ∗ ≡ ∞ (∗) ∞ (∗) Mφ M Mφ,F L (φ) M L (φ) ,F (using Lemma 6) ∞ ∗ ∗ = L (φ)( ) F(1/φ)( ) by Theorem 1(iv) ∞ ∗ ∗ ≡ L (φ)( ) 1/2 F(1/φ)( ) 1/2 (1/2) using Lemma 4 ∞ ∗ = L (φ)1/2F(1/φ)1/2 ( ) (1/2) by Theorem 2 in [44] ∞ ∗ ∗ ∗ = L 1/2F 1/2 ( ) (1/2) ≡ F (2) ( ) (1/2) ≡ F ( ) ≡ F. Note that Lemma 4 can be used in the above equality since 0 We must only control if the assumptions of Lemma 4 are satisfied here, that is, if operators H , H ∗ φ(t) are bounded in E and in E ( t ). Since 0 <αE βE t φ(t) s x(s) ds C x for all x ∈ E , 2 1 E t φ(s) 0 E whichistrueifβE l φ(t) 1 ∈ x(s) ds C2 x E for all x E , t φ(s) t E which is true if αE > 1−pφ,I (cf. [60, Theorem 1] or [61, Theorem 1]). The last strict inequality means that αE = 1 − βE > 1 − pφ,I or βE Examples 5. (a) If E = Lq , F = Lp and φ(t)= t1/r, where 1 p ∞ ∞ Lr, M Lr, ,Lp = Lp and Lq M Lq ,Lr,1 = Lr,1. (40) Equalities (40) we can also get from (33) and (32). In fact, the space Lr,∞ satisfies upper r-estimate (cf. [64, Theorem 5.4(a)] and [34, Theorem 3.1 and Corollary 3.9]). Therefore, for p ∞ ∞ Ls, M Ls, ,Lp,r = Lp,r and Lq,r M Lq,r,Ls,1 = Ls,1. (41) (c) If F = Λψ and αF = pψ,I >qφ,I, then from Theorem 11(a) we also obtain the factoriza- tion (37) since pψ/φ,I pψ,I − qφ,I > 0. 656 P. Kolwicz et al. / Journal of Functional Analysis 266 (2014) 616–659 = ∗ → Remark 8. In the case I (0, 1) the assumption αF >qφ,I implies the inclusion Mφ F ,even ∗ → the inclusion Mφ Λψ , where ψ is a fundamental function of F because pψ,1 αF >qφ,1 and pψ/φ,1 pψ,1 − qφ,1 > 0gives 1 1 1 ψ(t) dt ψ (t) dt < ∞. φ(t) φ(t) t 0 0 ∗ = { } Consequently, M(Mφ,F) 0 . References [1] J.M. Anderson, A.L. 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