Strong Factorizations between Couples of Operators on Banach Function Spaces

Strong Factorizations between Couples of Operators on Banach Function Spaces O. Delgado Departamento de Matem´atica Aplicada I, Universidad de Sevilla, Avenida Reina Mercedes s/n, 41012 Sevilla, Spain [email protected] E. A. S´anchez Pe´rez Instituto Universitario de Matem´atica Pura y Aplicada, Universidad Polite´cnica de Valencia, Camino de Vera s/n, 46022 Valencia, Spain [email protected] Let T : X1 → Y1 and S : X2 → Y2 be two continuous linear operators between Banach function spaces related to a finite measure space. Under some lattice requirements on the spaces involved, we give characterizations by means of inequalities of when T can be strongly factorized through S, that is, T = Mg ◦ S ◦Mf with Mf : X1 → X2 and Mg : Y2 → Y1 being multiplication operators defined by some measurable functions f and g. In particular, we study the cases when S is a composition operator or a kernel operator. Keywords: Banach function spaces, factorization of operators, multiplication operators, product spaces, vector measures 2000 Mathematics Subject Classification: Primary 46E30, 47B38; Secondary 46B42 1. Introduction Let (Ω, Σ, µ) be a finite measure space and consider two continuous linear operators T : X1 → Y1 and S : X2 → Y2 between Banach function spaces related to µ. The aim of this paper is to study when it is possible to factorize T through S as T = Mg ◦ S ◦ Mf where Mf : X1 → X2 and Mg : Y2 → Y1 are multiplication operators defined by measurable functions f and g. This type of factorization is called strong factorization for T , see for instance [16, Section III.H.§9]. There are many classical results relating inequalities for operators and factorizations, as the well known ones coming from the Maurey-Rosenthal cycle of ideas which allow to associate some convexity/concavity inequalities for operators with strong factorizations through Lp-spaces, see for instance [16, Proposition III.H.10] and The research of the first author has been partially supported by MTM2012-36732-C03-03 (Ministerio de Econom´ıa y Competitividad, Spain), FQM-262, FQM-4643, FQM-7276 (Junta de Andaluc´ıa) and Feder Funds (European Union). The research of the second author has been supported by MTM2012-36740-C02-02 (Ministerio de Econom´ıa y Competitividad, Spain). [5, 6, 7]. In these results, the operator S appearing in the factorization of T is constructed from the inequality satisfied by T . In the present paper we are interested in analyzing the factorization of T through a specifically given operator S. Our main results establish that, under some lattice type requirements on the spaces involved, mainly order continuity, a factorization for Tas Mg◦S◦Mfis equivalent to a domination of Tby S(Theorem 4.1) or by its transpose operator S′(Theorem 4.2). Section 5 is devoted to one of these requirements, namely we give conditions under which a product space is order continuous (Proposition 5.2). The proof of Theorem 4.2 uses a Radon-Nikod´ym Theorem for vector measures which characterizes when, for two vector measures n, m : Σ →Ewith values in a Banach space E, there exists a bounded function hsuch that n(A) = RAh dm for all A∈Σ. This Radon-Nikod´ym Theorem was proved by Musia l in [13, Theorem 1] by using a Ryll-Nardzewski lemma. In Section 3 we provide a simple proof of this characterization which is based in a completely different fundamental tool, a separation argument based on Ky Fan’s Lemma. We end the paper by studying in Section 6 the particular cases when Sis a composition operator or a kernel operator. In the first case, we added a new equivalent condition for Tto be strongly factorized through S(Theorem 6.1), which involves a generalization of the concept of local operator (see for instance [1, Definition 2.2]). 2. Basic concepts and notation We give in this section some basic definitions and results on Banach function spaces and vector measures which will be used along this paper. Let (Ω,Σ, µ) be a finite measure space and L0(µ) denote the space of all measurable real functions on Ω, where functions which are equal µ-a.e. are identified. As usual, L∞(µ) will denote the space of functions in L0(µ) which are bounded µ-a.e. By Banach function space (briefly B.f.s.) we will mean a Banach space X⊂L0(µ) with norm k · kX such that if f∈L0(µ), g∈Xand |f| ≤ |g|µ-a.e. then f∈Xand kfkX≤ kgkX. Note that every B.f.s. is a Banach lattice with the pointwise µ-a.e. order. A B.f.s. Xhas the Fatou property if for every sequence (fn)⊂Xsuch that 0 ≤fn↑f µ-a.e. and supnkfnkX<∞, it follows that f∈Xand kfnkX↑ kfkX. We will say that Xis order continuous if for every f, fn∈Xsuch that 0 ≤fn↑f µ-a.e., we have that fn→fin norm. For issues related to B.f.s.’, see [17, Ch. 15] considering the function norm ρdefined as ρ(f) = kfkXif f∈Xand ρ(f) = ∞in other case. Given two B.f.s.’ Xand Y, the space of multipliers from Xto Yis defined as XY=h∈L0(µ) : hf ∈Yfor all f∈X. The map k · kXYgiven by khkXY= supf∈BXkhfkYfor all h∈XY, where BXis the open unit ball of X, defines a natural seminorm on XYwhich becomes a norm only in the case when Xis saturated, i.e. there is no A∈Σ with µ(A)>0 such that fχA= 0 µ-a.e. for all f∈X. Note that Xis saturated if and only if it has a weak unit (i.e. g∈Xsuch that g > 0µ-a.e.). In this case, XYis a B.f.s. The particular multiplier space XL1is just the classical K¨othe dual of Xwhich will be denoted by X′. For each h∈XY, we denote by Mh:X→Ythe multiplication operator defined as Mh(f) = hf for all f∈X. For more information on these spaces see [12] and also [2]. The product space XπY of two B.f.s.’ Xand Yis defined as the space of functions f∈L0(µ) such that |f| ≤ Pi≥1|xiyi|µ-a.e. for some sequences (xi)⊂Xand (yi)⊂Ysatisfying Pi≥1kxikXkyikY<∞. For f∈XπY , denote kfkXπY = inf (X i≥1 kxikXkyikY), where the infimum is taken over all sequences (xi)⊂Xand (yi)⊂Ysuch that |f| ≤ Pi≥1|xiyi|µ-a.e. and Pi≥1kxikXkyikY<∞. If X,Yand XY′are saturated then XπY is a saturated B.f.s. with norm k · kXπY . An study of these product spaces can be found in [8]. Let m: Σ →Ebe a vector measure, that is, a countably additive set function, where Eis a real Banach space. A set A∈Σ is m-null if m(B) = 0 for every B∈Σ with B⊂A. For each x∗in the topological dual E∗of E, we denote by |x∗m|the variation of the real measure x∗mgiven by the composition of mwith x∗. There exists x∗ 0∈E∗such that |x∗ 0m|has the same null sets as m, see [9, Ch. IX. 2]. We will call |x∗ 0m|aRybakov control measure for m. A measurable function f: Ω →Ris integrable with respect to mif: (i) R|f|d|x∗m|<∞for all x∗∈E∗. (ii) For each A∈Σ, there exists xA∈Esuch that x∗(xA) = ZA f dx∗m, for all x∗∈E. The element xAwill be written as RAf dm. Denote by L1(m) the space of integrable functions with respect to m, where functions which are equal m-a.e. are identified. The space L1(m) is a Banach space endowed with the norm kfkm= sup x∗∈BE∗Z|f|d|x∗m|. Note that L∞(|x∗ 0m|)⊂L1(m). In particular every measure of the type |x∗m|is finite as |x∗m|(Ω) ≤ kx∗k · kχΩkm. Given f∈L1(m), the set function mf: Σ →Egiven by mf(A) = RAf dm for all A∈Σ is a vector measure. Moreover, g∈L1(mf) if and only if gf ∈L1(m) and in this case Rg dmf=Rgf dm. For a complete overview about integration with respect to vector measures we refer to [4], [14, Ch. 3] and the references therein. 3. The Radon-Nikod´ym theorem for vector measures In the next section we will use the following Radon-Nikod´ym theorem for vector measures proved by Musia l in [13, Theorem 1]. We include here a totally different 0 0 proof for the real Banach space case based on Ky Fan’s lemma (see for instance [15, E. 4]). Theorem 3.1. Let m, n: Σ → E be two vector measures and take |x∗m| a Rybakov control measure for m. The following statements are equivalent. (1) There exists a positive constant K such that x∗n(A) ≤ K|x∗m|(A) for all A ∈ Σ and x∗ ∈ E∗. (2) There exists a function g ∈ L∞(|x∗m|) such that n(A) = ZA g dm for all A∈Σ. Proof. Let us show that (1) implies (2). Denote ξ=|x∗ 0m|. For each x∗∈E∗, since the measure x∗mis absolutely continuous with respect to ξ, there exists hx∗∈L1(ξ) such that x∗m(A) = ZA hx∗dξ for all A∈Σ, i.e. hx∗is the Radon-Nikod´ym derivative of x∗mwith respect to ξ. For every finite measurable partition {A1, ..., An}of Ω and every finite set of vectors x∗ 1, ..., x∗ n∈E∗, we define the function φ:BL∞(ξ)→Rby φ(f) = n X i=1 x∗ in(Ai)−K n X i=1 ZAi fhx∗ idξ for all f∈BL∞(ξ). Note that φis convex. Considering the weak* topology on BL∞(ξ), we have that φis a continuous function on a compact convex set. Let Fdenote the family of the functions φdefined in this way. If φ1and φ2are two functions in Fand 0 < α < 1, since αφ1(f) + (1 −α)φ2(f) = n X i=1 αx∗ 1,in(A1 i)−K n X i=1 ZA1 i fhαx∗ 1,i dξ + k X j=1 (1 −α)x∗ 2,jn(A2 j)−K k X j=1 ZA2 j fh(1−α)x∗ 2,j dξ = n X i=1 k X j=1 (αx∗ 1,i + (1 −α)x∗ 2,j)n(A1 i∩A2 j) −K n X i=1 k X j=1 ZA1 i∩A2 j f hαx∗ 1,i+(1−α)x∗ 2,j dξ, we have that αφ1+ (1 −α)φ2∈ F as {A1 i∩A2 j:i= 1, ..., n, j = 1, ..., k}is a partition of Ω. Then, it follows that Fis a concave family of real functions. If (1) holds, for each φ∈ F, taking fφ=Pn i=1 sign(hx∗ i)χAi∈BL∞(ξ), we have that n X i=1 x∗ in(Ai)≤K n X i=1 |x∗ im|(Ai) = K n X i=1 ZAi |hx∗ i|dξ =K n X i=1 ZAi fφhx∗ idξ, that is φ(fφ)≤0. Therefore, by applying Ky Fan’s lemma, there exists a function f0∈BL∞(ξ)such that φ(f0)≤0 for all φ∈ F. Consequently, for every A∈Σ and x∗∈E∗, if we consider the partition {A, Ω\A}and the vectors x∗and 0, we obtain x∗n(A)≤KZA f0hx∗dξ and −x∗n(A)≤KZA f0h−x∗dξ =−KZA f0hx∗dξ, that is x∗n(A) = KRAf0hx∗dξ. Since this happens for every x∗∈E∗, taking into account that RAf0hx∗dξ =RAf0dx∗m=x∗RAf0dm, we have that n(A) = KRAf0dm for every A∈Σ and so (2) holds for g=Kf0. The converse implication is obtained easily as follows. If (2) holds, for every A∈Σ and x∗∈E∗we have that x∗n(A)≤ |x∗n(A)|=x∗ZA g dm=ZA g dx∗m ≤ZA |g|d|x∗m| ≤ kgk∞|x∗m|(A). 4. Strongly factorized operators between B.f.s.’ Let (Ω,Σ, µ) be a fixed finite measure space and X1, X2, Y1, Y2B.f.s.’ related to µ such that L∞(µ)⊂X1⊂X2and L∞(µ)⊂Y2⊂Y1. This guarantees that XX2 1 and YY1 2are B.f.s.’ containing L∞(µ). Consider two continuous linear operators T:X1→Y1and S:X2→Y2. This section is devoted to characterize when Tfactorizes strongly through S, that is, the following diagram commutes X1 T// Mf  Y1 X2 S//Y2 Mg OO(1) for some f∈XX2 1and g∈YY1 2. We will give two different characterizations. In the first one we will use the Ky Fan’s lemma ([15, E. 4]) and the product space Y2πY ′ 1which under our conditions is a saturated B.f.s. (see [8, Proposition 2.2]). Theorem 4.1. Suppose that Y1, Y2πY ′ 1are order continuous and moreover Y1has the Fatou property. The following statements are equivalent: (i) There exists a function h ∈ X1 X2 such that n X i=1 ZT(xi)y′ idµ ≤     n X i=1 S(hxi)y′ i    Y2πY ′ 1 for every x1, ..., xn∈X1and y′ 1, ..., y′ n∈Y′ 1. (ii)There exist functions f∈XX2 1and g∈YY1 2such that T(x) = gS(fx)for all x∈X1, i.e. Tfactorizes strongly through Sas in (1). Proof. First note that since Y2πY ′ 1is order continuous, the K¨othe dual (Y2πY ′ 1)′ can be identified with the topological dual (Y2πY ′ 1)∗(see [10, p. 29]). On the other hand, Y1=Y′′ 1as Y1has the Fatou property (see [10, p. 30]), and so by [8, Proposition 2.2] we have that (Y2πY ′ 1)′=YY′′ 1 2=YY1 2. Then, applying the Hahn-Banach theorem, for each ξ∈Y2πY ′ 1there exists ξ′∈BY2Y1such that kξkY2πY ′ 1=hξ′, ξi=Rξ ξ′dµ. Suppose (i) holds for some h∈XX2 1. For every x1, ...xn∈X1and y′ 1, ..., y′ n∈Y′ 1, we define the convex function φ:BY2Y1→Rby φ(g) = n X i=1 ZT(xi)y′ idµ − n X i=1 ZgS(hxi)y′ idµ, for all g∈BY2Y1. Since (Y2πY ′ 1)∗=Y2Y1, considering the weak* topology we have that BY2Y1is a compact convex set and φis continuous. Denote by Fthe family of all functions φdefined in this way, which is obviously concave. Let φ∈ F defined through x1, ...xn∈X1and y′ 1, ..., y′ n∈Y′ 1. As we have noted before, for Pn i=1 S(hxi)y′ i∈Y2πY ′ 1there exists gφ∈BY2Y1such that      n X i=1 S(hxi)y′ i    Y2πY ′ 1 =Zgφ n X i=1 S(hxi)y′ idµ. Then, from (i) we have that φ(gφ)≤0. By Ky Fan’s lemma, there exists a function g∈BY2Y1such that φ(g)≤0 for all φ∈ F. In particular, ZT(x)y′dµ ≤ZgS(hx)y′dµ (2) for every x∈X1and y′∈Y′ 1. Since this holds for every y′∈Y′ 1, by taking −y′we obtain also −ZT(x)y′dµ ≤ − ZgS(hx)y′dµ. Then ZT(x)y′dµ =ZgS(hx)y′dµ for every x∈X1and y′∈Y′ 1. Since Y1is order continuous it follows that T(x) = gS(hx) for all x∈X1, that is, (ii) holds. Suppose now that (ii) holds for f∈XX2 1and g∈YY1 2. For every x1, ..., xn∈X1 and y′ 1, ..., y′ n∈Y′ 1we have that n X i=1 ZT(xi)y′ idµ = n X i=1 ZgS(fxi)y′ idµ ≤ kgkYY1 2 ·     n X i=1 S(fxi)y′ i    Y2πY ′ 1 , that is, (i) holds for h=kgkYY1 2 f. Note that the previous theorem requires conditions on the B.f.s.’ Y1,Y2and the operator S. The second characterization of when Tfactorizes strongly through S is a dual result of Theorem 4.1 in the sense that the conditions are required on X1,X2and the transpose operator S′of S. For the proof we will use also the Radon-Nikod´ym theorem proved in Section 3. Theorem 4.2. Suppose that X1,Y1,X2and X1πX′ 2are order continuous and X2 has the Fatou property. The following statements are equivalent. (i)There exists a function h∈YY1 2such that n X i=1 ZT(xi)y′ idµ ≤     n X i=1 |S′(hy′ i)xi|    X1πX′ 2 for every x1, ..., xn∈X1and y′ 1, ..., y′ n∈Y′ 1. (ii)There exist functions f∈XX2 1and g∈YY1 2such that T(x) = gS(fx)for all x∈X1. Proof. Since X1πX′ 2is order continuous and X2has the Fatou property it follows that (X1πX′ 2)∗=XX2 1. Suppose (i) holds for some h∈YY1 2. It is direct to check that YY1 2⊂Y′Y′ 2 1, then since X2is order continuous, we have that S′(hy′)∈X′ 2for all y′∈Y′ 1. For every x1, ...xn∈X1and y′ 1, ..., y′ n∈Y′ 1, we define the convex function φ:BX1X2→Rby φ(f) = n X i=1 ZT(xi)y′ idµ − n X i=1 Zf|S′(hy′ i)xi|dµ, for all f∈BX1X2. Considering the weak* topology, φis continuous on a compact convex set. Let Fbe the concave family of all functions φdefined in this way. If φ∈ F is defined through x1, ...xn∈X1and y′ 1, ..., y′ n∈Y′ 1, taking fφ∈BX1X2such that      n X i=1 |S′(hy′ i)xi|    X1πX′ 2 =Zfφ n X i=1 |S′(hy′ i)xi|dµ, from (i) we have that φ(fφ) ≤ 0. By Ky Fan’s lemma, there exists a function f ∈ BX1X2 such that φ(f) ≤ 0 for all φ ∈ F. In particular, ZT(x)y′dµ ≤Zf|S′(hy′)x|dµ (3) for every x∈X1and y′∈Y′ 1. We can assume that f≥0. On the other hand, consider the continuous linear operator R:X1→Y1given by R=Mh◦S◦Mfand the set function mR: Σ →Y1defined as mR(A) = R(χA) for all A∈Σ. Since X1is order continuous we have that mRis a vector measure and for every x∈X1it follows that x∈L1(mR) with Rx dmR=R(x), see for instance [3, Section 3]. For each y′∈Y′ 1and A∈Σ, we have that y′mR(A) = hy′, R(χA)i=Zy′hS(fχA)dµ =hy′h, S(fχA)i=hS′(y′h), fχAi=ZA S′(y′h)f dµ, where in the last inequality we have used that X2is order continuous and so S′:Y∗ 2→X∗ 2=X′ 2. Then, |y′mR|(A) = ZA |S′(y′h)|f dµ. (4) Now, considering the vector measure mT: Σ →Y1given by mT(A) = T(χA) for all A∈Σ and applying (3) for x=χAand (4), we have that y′mT(A) = ZT(χA)y′dµ ≤ZA f|S′(hy′)|dµ =|y′mR|(A). Hence, since Y1is order continuous, Theorem 3.1 provides a function f1∈L∞(ξ), where ξis a Rybakov control measure for mR, such that mT(A) = ZA f1dmR for all A∈Σ. Since X1⊂L1(mT) and the integration operator with respect to mTrestricted to X1coincides with T, considering a mR-null set Zsuch that f2=f1χΩ\Zis bounded in all Ω, for each x∈X1it follows that T(x) = Zx dmT=Zxf1dmR=Zxf2dmR. Noting that xf2∈X1, we have that T(x) = R(xf2) = hS(fxf2) where ff2∈XX2 1. So (ii) holds. Suppose now that (ii) holds for f∈XX2 1and g∈YY1 2. For every x1, ..., xn∈X1 and y′ 1, ..., y′ n∈Y′ 1we have that n X i=1 ZT(xi)y′ idµ = n X i=1 ZgS(fxi)y′ idµ = n X i=1 ZfxiS′(gy′ i)dµ ≤Z|f| n X i=1 |xiS′(gy′ i)|dµ ≤ kfkXX2 1     n X i=1 |xiS′(gy′ i)|    X1πX′ 2 , that is, (i) holds for h=gkfkXX2 1 . An easy example is given by the case when X1, X2, Y1, Y2all coincide with an order continuous B.f.s. Xhaving the Fatou property and containing L∞(µ). By a classical Lozanovskii’s result ([11, Theorem 6]) we have that XπX′=L1(µ) (with equal norms) is order continuous. Also note that XX=L∞(µ) (with equal norms), see [12, Theorem 1]. Then, Theorems 4.1 and 4.2 produce the following result. Corollary 4.3. The following statements are equivalent: (i)Tfactorizes strongly through S, i.e. XT// Mf  X XS//X Mg OO for some f, g ∈L∞(µ). (ii)There exists a function h∈L∞(µ)such that n X i=1 ZT(xi)x′ idµ ≤Z n X i=1 S(hxi)x′ idµ for every x1, ..., xn∈Xand x′ 1, ..., x′ n∈X′. (iii)There exists a function h∈L∞(µ)such that ZT(x)x′dµ ≤Z|S′(hx′)x|dµ for every x∈Xand x′∈X′. Example 6.5. Consider the measure space given by the interval [0, 1], its Borel σ-algebra and the Lebesgue measure. Let K be the kernel given by K(s, t) = χ[0,s](t) for all s, t ∈[0,1]. Then, SKis just the Volterra operator. Suppose that SK:X→Xis well defined and continuous (e.g. X=Lp[0,1] with 1 ≤p < ∞). The following statements are equivalent: (i) There exist g, f ∈L∞[0,1] such that T(x)(s) = g(s)Zs 0 f(t)x(t)dt a.e.(s) for all x∈X. (ii) There exists h∈L∞[0,1] such that Z1 0 T(x)(t)x′(t)dt ≤Z1 0x(t)Z1 t h(s)x′(s)dsdt for all x∈Xand x′∈X′. (iii) There exists h∈L∞[0,1] such that |T′(x′)(t)| ≤ Z1 t h(s)x′(s)dsa.e.(t) for all x′∈X′. Note that the equivalence between (i) and (ii) follows from applying (5). If (ii) holds, given x′∈X′we consider the sets B+={t∈[0,1] : T′(x′)≥0}and B−= {t∈[0,1] : T′(x′)<0}. For each measurable set A, taking x=χA(χB+−χB−), we have that ZA |T′(x′)(t)|dt =Z1 0 T′(x′)(t)x(t)dt =hT′(x′), xi=hx′, T (x)i =Z1 0 T(x)(t)x′(t)dt ≤Z1 0x(t)Z1 t h(s)x′(s)dsdt ≤ZAZ1 t h(s)x′(s)dsdt. so (iii) holds. Conversely, if (iii) holds, for every x∈Xand x′∈X′, it follows that Z1 0 T(x)(t)x′(t)dt =Z1 0 T′(x′)(t)x(t)dt ≤Z1 0 |T′(x′)(t)x(t)|dt ≤Z1 0x(t)Z1 t h(s)x′(s)dsdt. If we consider now the Hardy operator which is given by the kernel K(s, t) = 1 sχ[0,s](t) for all s, t ∈[0,1] and suppose that SK:X→Xis well defined and continuous (e.g. X=Lp[0,1] with 1 < p < ∞), in a similar way we obtain that the following statements are equivalent: (i) There exist g, f ∈L∞[0,1] such that T(x)(s) = g(s) sZs 0 f(t)x(t)dt a.e.(s) for all x∈X. (ii) There exists h∈L∞[0,1] such that Z1 0 T(x)(t)x′(t)dt ≤Z1 0x(t)Z1 t h(s)x′(s) sdsdt for all x∈Xand x′∈X′. (iii) There exists h∈L∞[0,1] such that |T′(x′)(t)| ≤ Z1 t h(s)x′(s) sdsa.e.(t) for all x′∈X′. 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