# Boundedness Criteria and Norm of Some Multilinear Hilbert-Type Operators

Authors: Justice S. Bansah; Benoît F. Sehba
Journal: Abstract and Applied Analysis
 (2017)

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## Abstract

We consider two families of multilinear Hilbert-type operators for which we give exact relations between the parameters so that they are bounded. We also find the exact norm of these operators.


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## Body

## 1. Introduction

Let1≤p<∞, and let t be a real number. We denote by Ltp((0,∞)) or simply Ltp the weighted Lebesgue space Lp(0,∞),ytdy. When t=0, we simply write Lp for the corresponding space. We will be using the notation (1)fp,t≔∫0∞fypytdy1/pand we recall that f∈Ltp((0,∞)) if and only if the above quantity is finite. For t=0, we simply write fp for fp,0. All over the text, for 1≤p≤∞ we denote by p′ its conjugate exponent, that is, the unique extended real number satisfying 1=1/p+1/p′.We recall that the Hilbert operator is defined by(2)Hfx≔∫0∞fyx+ydy.The boundedness of this operator has been heavily studied in the literature; in particular people have been very interested in the norm estimate of this operator and its siblings (see, e.g., the following and the references therein [1–6]). In [7], we considered a more general family of this operator for which we provided boundedness criteria and some sharp norm estimates. More precisely, for β, γ real parameters we considered the family of operators Hβ,γ defined for compactly supported functions by(3)Hβ,γfx≔∫0∞fyx+yγyβdy.This family as shown in [7] can be related to Bergman-type projections. This family can be extended in two different ways into m-linear operators on (0,∞)m. For the first family, we let β1,…,βm,γ1,…,γm be real parameters and put γ→=γ1,…,γm and β→=(β1,…,βm). Put f→=(f1,…,fm), where fj, j=1,…,m, are compactly supported functions on (0,∞). Consider the operators Tβ→,γ→ defined by (4)Tβ→,γ→f→x≔∫0,∞mf1y1f2y2⋯fmymx+y1γ1x+y2γ2⋯x+ymγmy1β1y2β2⋯ymβmdy1dy2⋯dym.Following the idea in [7], one can also relate this family to multilinear Bergman projections. To define the second family, we let β1,…,βm,γ be real parameters and put β→=(β1,…,βm). We define the operators Sβ→,γ as follows: (5)Sβ→,γf→x=∫0,∞mf1y1f2y2⋯fmymx+y1+⋯+ymγy1β1y2β2⋯ymβmdy1dy2⋯dym.This last family appears in [8] in the study of the Laplace representation of some mixed norm Bergman spaces in relation to the question of the boundedness of the Bergman projection in tube domains over symmetric cones. Many authors have provided the norm of the operators Sβ→,γ in some particular cases and some variations [9–14]. The fact is that, in these papers, the relations between the parameters are directly given without indicating how they are obtained and the authors are only interested in finding the exact norm of the operators or proving the corresponding Hardy-Hilbert inequality.We aim in this note to provide exact relations between the parameters so that the above operators are bounded fromLt1p1×⋯×Ltmpm to Ltp when 1≤p1,…,pm≤∞, t1,…,tm,t being real numbers with(6)1p1+⋯+1pm=1p,(7)t1p1+⋯+tmpm=tp.We also find the exact norm of these operators, extending the results in [9, 12]. Note that, in the above relations, we allow some (not all) of the exponents pjs to be infinite, a situation which has not been considered before as far as we know.


## 2. Statement of the Results

We give in this section all our results. We denote, byB(x,y), the Beta-function of x and y to be defined in the next section. We start by the following.Theorem 1.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm<∞, 1≤p<∞, let t1,…,tm,t be real numbers, and assume that (6) and (7) hold. Then the following conditions are equivalent. (i)
The operatorTβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)).(ii)
The parameters satisfy(8)∑j=1mγj=∑j=1mβj+m,(9)-pjγj-βj-1<tj+1<pjβj+1,j=1,…,m.In this case, if we denote by Tβ→,γ→∏j=1mLtjpj→Ltpm the operator norm of Tβ→,γ→, then(10)Tβ→,γ→∏j=1mLtjpj→Ltp=∏j=1mBβj+1-tj+1pj,γj-βj-1+tj+1pj.It is easy to see that condition (9) provides that, for any j∈{1,…,m},(11)γj>0;hence (12)∑j=1mγj>0.As -pj(γj-βj-1)<tj+1, j=1,…,m, we also have that (13)∑j=1mγj>∑j=1mβj+1-tj+1pj.Now observing that (8) can be written as (14)∑j=1mγj=∑j=1mβj+1-tj+1pj+t+1p,we conclude that t+1>0 or equivalently that t>-1.We also obtain the following result.Theorem 2.
Letβ1,…,βm,γ be real numbers. Let 1<p1,…,pm<∞, 1≤p<∞, let t1,…,tm,t be real numbers, and assume that (6) and (7) hold. Then the following conditions are equivalent. (i)
The operatorSβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)).(ii)
The parameters satisfy(15)0<γ=∑j=1mβj+m,t>-1,tj+1<pjβj+1,j=1,…,m.In this case, if we denote by Sβ→,γ∏j=1mLtjpj→Ltpm the operator norm of Sβ→,γ, then (16)Sβ→,γ∏j=1mLtjpj→Ltp=∏j=1mΓβj+1-tj+1/pjΓt+1/pΓγ.We note that the norm of the operatorSβ→,γ for β→=0→ and γ=m was computed in [9] for the unweighted case and [12] for the weighted case.If in relation (6) we allow only some (but not all) of the exponents to be finite while all the other exponents are equal to infinity, then we obtain a kind of mixed endpoints version of the previous results.Theorem 3.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pl<∞, 1≤l<m, 1≤p<∞, and assume that 1/p1+⋯+1/pl=1/p. Let t1,…,tl,t be real numbers and assume that t1/p1+⋯+tl/pl=t/p. Then the following conditions are equivalent. (i)
The operatorTβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltlpl((0,∞))×L∞((0,∞))m-l to Ltp((0,∞)).(ii)
The parameters satisfy(17)∑j=1mγj=∑j=1mβj+m;-pjγj-βj-1<tj+1<pjβj+1,forj=1,…,l,γj>βj+1>0,forj=l+1,…,m.In this case, if we denote by Tβ→,γ→∏j=1lLtjpj×(L∞)m-l→Ltp the operator norm of Tβ→,γ→, then (18)Tβ→,γ→∏j=1lLtjpj×L∞m-l→Ltp=∏j=1lBβj+1-tj+1pj,γj-βj-1+tj+1pj×∏j=l+1mBβj+1,γj-βj-1.We also obtain the following result.Theorem 4.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pl<∞, 1≤l<m, 1≤p<∞, and assume that 1/p1+⋯+1/pl=1/p. Let t1,…,tl,t be real numbers and assume that t1/p1+⋯+tl/pl=t/p. Then the following conditions are equivalent.(i)
The operatorSβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltlpl((0,∞))×L∞((0,∞))m-l to Ltp((0,∞)).(ii)
The parameters satisfy(19)0<γ=∑j=1mβj+m,t>-1;tj+1<pjβj+1,forj=1,…,l,βj>-1,forj=l+1,…,m.In this case, if we denote by Sβ→,γ∏j=1lLtjpj×(L∞)m-l→Ltp the operator norm of Sβ→,γ, then (20)Sβ→,γ∏j=1lLtjpj×L∞m-l→Ltp=∏j=1lΓβj+1-tj+1/pjΓt+1/pΓγ∏j=l+1mΓβj+1.We are essentially motivated by the need of explaining the right relations between the parameters that make these operators bounded and the generalization of the previous results on the norm estimates of these operators. In the proofs of the necessary parts, we appeal to duality and use appropriate local test functions. Note that as our operators arem-linear, each of them has m adjoints. For the computation of the norm of each operator, to simplify our presentation, we give an upper estimate and a lower estimate. The proof of the lower estimate appeals to a clever choice of test functions, a good decomposition of multiple integrals to find the right lower bound.


## 3. Some Useful Tools

We recall that the Beta-function of the cone(0,∞) is defined by (21)Bm,n=Bn,m=∫0∞um-11+um+ndu.We note that this integral converges if and only if m,n>0. Recall that (22)Bm,n=ΓmΓnΓn+m,where Γ is the classical Gamma-function. The following can be obtained by induction (see, e.g., [10, Lemma 5.1]):(23)∫0,∞s-1u1m1-1⋯us-1ms-1-11+u1+⋯+us-1∑j=1smjdu1⋯dus=∏j=1sΓmjΓ∑j=1smj,where mj>0, j=1,…,s.


## 4. The Norm ofTβ→,γ→ and Sβ→,γ

### 4.1. The Norm ofTβ→,γ→

We have the following upper estimate of the norm ofTβ→,γ→.Lemma 5.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm<∞, 1≤p<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold. Assume that (24)∑j=1mγj=∑j=1mβj+m,-pjγj-βj-1<tj+1<pjβj+1,j=1,…,m.Then the operator Tβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)). Moreover, (25)Tβ→,γ→∏j=1mLtjpj→Ltp≤∏j=1mBβj+1-tj+1pj,γj-βj-1+tj+1pj.Proof.
Forfi∈Ltipi, i=1,…,m, we write again f→=(f1,…,fm). Using an easy change of variables and Minkowski’s inequality, we obtain (26)Tβ→,γ→f→p,t=∫0∞Tβ→,γ→f→xpxtdx1/p≤∫0∞∫0,∞m∏i=1mfiyi∏i=1mx+yiγi∏i=1myiβidy1dy2⋯dympxtdx1/p=∫0∞∫0,∞m∏i=1mfiuix∏i=1m1+uiγi∏i=1muiβidu1du2⋯dumpxtdx1/p≤∫0,∞m∫0∞∏i=1mfiuixp∏i=1m1+uiγixtdx1/p∏i=1muiβidu1du2⋯dum=∫0,∞m∏i=1muiβi∏i=1m1+uiγi∫0∞∏i=1mfiuixpxtdx1/pdu1du2⋯dum.Using (6), (7), and Hölder’s inequality, we easily obtain that (27)∫0∞∏i=1mfiuixpxtdx≤∏i=1m∫0∞fiuixpixtidxp/pi=∏i=1mui-p/piti+1fipi,tip.It follows that (28)Tβ→,γ→f→p,t≤∏i=1mfipi,ti∫0,∞m∏i=1muiβi-ti+1/pi∏i=1m1+uiγidu1du2⋯dum=∏i=1mfipi,ti∏i=1m∫0∞uiβi-ti+1/pi1+uiγidui=∏i=1mfipi,ti∏i=1mBβi+1-ti+1pi,γi-βi-1+ti+1pi.The proof is complete.We also have the following result.Lemma 6.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pl<∞, 1≤l<m, 1≤p<∞, and assume that 1/p1+⋯+1/pl=1/p. Let t1,…,tl,t be real numbers and assume that t1/p1+⋯+tl/pl=t/p. Suppose that the parameters satisfy (29)∑j=1mγj=∑j=1mβj+m;-pjγj-βj-1<tj+1<pjβj+1,forj=1,…,l,γj>βj+1>0,forj=l+1,…,m.Then the operator Tβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltlpl((0,∞))×L∞((0,∞))m-l to Ltp((0,∞)). In this case, if we denote by Tβ→,γ→∏j=1lLtjpj×(L∞)m-l→Ltp the operator norm of Tβ→,γ→, then (30)Tβ→,γ→∏j=1lLtjpj×L∞m-l→Ltp≤∏j=1lBβj+1-tj+1pj,γj-βj-1+tj+1pj×∏j=l+1mBβj+1,γj-βj-1.Proof.
Forfi∈Ltipi, i=1,…,l, and fl+1,…,fm∈L∞, we write again f→=(f1,…,fm). Proceeding as at the beginning of the proof of the previous lemma, we obtain (31)Tβ→,γ→f→p,t≤∏j=l+1mfj∞∫0,∞m∏j=1mujβj∏j=1m1+ujγj∫0∞∏i=1lfiuixpxtdx1/pdu1du2⋯dum≤∏j=l+1mfj∞∏j=1lfjpj,tj∫0,∞m∏i=1luiβi-ti+1/pi∏i=l+1muiβi∏i=1m1+uiγidu1du2⋯dum=∏j=l+1mfj∞∏j=1lfjpj,tj∏i=1lBβi+1-ti+1pi,γi-βi-1+ti+1pi×∏i=l+1mBβi+1,γi-βi-1.The proof is complete.Let us now prove the lower bound for the norm ofTβ→,γ→.Lemma 7.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm≤∞, 1≤p<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold. Suppose that the parameters satisfy (32)∑j=1mγj=∑j=1mβj+m,-γj-βj-1<tj+1pj<βj+1,j=1,…,m.Then (33)Tβ→,γ→∏j=1mLtjpj→Ltp≥∏j=1mBβj+1-tj+1pj,γj-βj-1+tj+1pj.Proof.
For simplicity, let us put(34)K=Tβ→,γ→∏j=1mLtjpj→Ltp,Bm≔∏j=1mBβj+1-tj+1pj,γj-βj-1+tj+1pj.For any vector f→=(f1,…,fm), fj∈Ltjpj((0,∞)), j=1,…,m (for pj=∞, replace Ltjpj((0,∞)) by L∞((0,∞))), we have (35)K∏j=1mfjpj,tj≥Tβ→,γ→f→p,t;hence for any g∈Ltp′((0,∞)) (with Ltp′((0,∞)) replaced by L∞((0,∞)) if p=1), (36)K∏j=1mfjpj,tj≥∫0∞gxTβ→,γ→f→xxtdx.Let 0<ξ<min{pj(βj+1)-(tj+1):j=1,…,m} and define(37)fjx=0if0<x<1x-tj+1+ξ/pjifx≥1,gx=0if0<x<1x-t+1+ξ/p′ifx≥1.Then (38)fjpj,tj=1ξ1/pj,gp′,t=1ξ1/p′.Substituting these into (36), we obtain (39)Kξ≥∫1∞x-t+1+ξ/p′∫1,∞m∏j=1myjβj-tj+1+ξ/pj∏j=1mx+yjγjdy1⋯dymxtdx.Observing that ∫1∞=∫0∞-∫01, we obtain that (40)L≔∫1,∞m∏j=1myjβj-tj+1+ξ/pj∏j=1mx+yjγjdy1⋯dym=∫0,∞m∏j=1myjβj-tj+1+ξ/pj∏j=1mx+yjγjdy1⋯dym+∑k,l≠0,k+l=m-1lLkLl+-1m∫0,1m∏j=1myjβj-tj+1+ξ/pj∏j=1mx+yjγjdy1⋯dym=x-∑j=1mtj+1+ξ/pjBm+∑k,l≠0,k+l=m-1lLkLl+-1m∫0,1m∏j=1myjβj-tj+1+ξ/pj∏j=1mx+yjγjdy1⋯dym,where (41)Lk≔∫0,∞k∏s=1kyjsβjs-tjs+1+ξ/pjs∏s=1kx+yjsγjsdyj1⋯dyjk,Ll≔∫0,1l∏s=1lyjsβjs-tjs+1+ξ/pjs∏s=1lx+yjsγjsdyj1⋯dyjk.We have that(42)0<Lk=x-∑s=1kγjs+∑s=1kβjs+k-∑s=1ktjs+1+ξ/pjsBkξ,where (43)Bkξ≔∏s=1kBβjs+1-tjs+1+ξpjs,γjs-βjs-1+tjs+1pjs,(44)0<Ll≤x-∑s=1lγjs∫0,1l∏s=1lyjsβjs-tjs+1+ξ/pjsdyj1⋯dyjk=x-∑s=1lγjs∏s=1l1βjs+1-tjs+1+ξ/pjs≔x-∑s=1lγjsClξ.Let us write (45)L0≔∫0,1m∏j=1myjβj-tj+1+ξ/pj∏j=1mx+yjγjdy1⋯dym.Then(46)0<L0≤x-∑j=1mγjCmξwith (47)Cmξ≔∏j=1m1βj+1-tj+1+ξ/pj.It follows that if m is an even integer, then (48)L>x-∑j=1mtj+1+ξ/pjBm-∑k≠0,k+2l+1=mxαkBkξC2l+1ξ,where αk≔-∑j=1mγj+∑s=1kβjs+k-∑s=1k(tjs+1+ξ/pjs); and if m is an odd number, then (49)L>x-∑j=1mtj+1+ξ/pjBm-∑k≠0,k+2l+1=mxαkBkξC2l+1ξ-x-∑j=1mγjCmξ.Taking (48) into (39), we obtain that if m is even, then (50)Kξ≥Bm∫1∞xt-t+1+ξ/p′-∑j=1mtj+1+ξ/pjdx-∑k≠0,k+2l+1=mBkξC2l+1ξ∫1∞xt-t+1+ξ/p′+αkdx=Bm∫1∞x-1-ξdx-∑k≠0,k+2l+1=mBkξC2l+1ξ∫1∞x-∑s=12l+1βjs+1-tjs+1+ξ/pjs-1-ξdx=Bmξ-∑k≠0,k+2l+1=mBkξC2l+1ξ∑s=12l+1βjs+1-tjs+1+ξ/pjs+ξ.Thus (51)K≥Bm-∑k≠0,k+2l+1=mξBkξC2l+1ξ∑s=12l+1βjs+1-tjs+1+ξ/pjs+ξ.Taking (49) into (39) and doing the same type of calculations as above, we obtain that if m is an odd integer, then (52)K≥Bm-∑k≠0,k+2l+1=mξBkξC2l+1ξ∑s=12l+1βjs+1-tjs+1+ξ/pjs+ξ-ξCmξξ+∑j=1mβj+1-tj+1+ξ/pj.Letting ξ→0 in (51) or (52) we obtain that K≥Bm. That is, (53)Tβ→,γ→∏j=1mLtjpj→Ltp≥∏j=1mBβj+1-tj+1pj,γj-βj-1+tj+1pj.The proof is complete.


### 4.2. The Norm ofSβ→,γ

Let us prove the following upper bound for the norm ofSβ→,γ.Lemma 8.
Letβ1,…,βm,γ be real numbers. Let 1<p1,…,pm<∞, 1≤p<∞, and let t1,…,tm,t be real numbers, and assume that relations (6) and (7) hold. Assume that (54)0<γ=∑j=1mβj+m,t>-1,tj+1<pjβj+1,j=1,…,m.Then the operator Sβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)). In this case, if we denote by Sβ→,γ∏j=1mLtjpj→Ltpm the operator norm of Sβ→,γ, then (55)Sβ→,γ∏j=1mLtjpj→Ltp≤∏j=1mΓβj+1-tj+1/pjΓt+1/pΓγ.Proof.
The proof is essentially the same as above. Let us give it here for completeness. Forfi∈Ltipi, i=1,…,m, we write again f→=(f1,…,fm). We remark that t>-1 and γ=∑i=1mβi+1-ti+1/pi+(t+1)/p. Using an easy change of variables and Minkowski’s inequality, Hölder’s inequality, and equality (23), we obtain (56)Sβ→,γf→p,t=∫0∞Sβ→,γf→xpxtdx1/p≤∫0∞∫0,∞m∏i=1mfiyix+y1+⋯+ymγ∏i=1myiβidy1dy2⋯dympxtdx1/p=∫0∞∫0,∞m∏i=1mfiuix1+u1+⋯+umγ∏i=1muiβidu1du2⋯dumpxtdx1/p≤∫0,∞m∫0∞∏i=1mfiuixp1+u1+⋯+umγxtdx1/p∏i=1muiβidu1du2⋯dum=∫0,∞m∏i=1muiβi1+u1+⋯+umγ∫0∞∏i=1mfiuixpxtdx1/pdu1du2⋯dum≤∏i=1mfipi,ti∫0,∞m∏i=1muiβi-ti+1/pi1+u1+⋯+umγdu1du2⋯dum=∏i=1mfipi,ti∏i=1mΓβi+1-ti+1/piΓt+1/pΓγ.
The proof is complete.Let us prove the following result.Lemma 9.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pl<∞, 1≤l<m, 1≤p<∞, and assume that 1/p1+⋯+1/pl=1/p. Let t1,…,tl,t be real numbers and assume that t1/p1+⋯+tl/pl=t/p. Suppose that the parameters satisfy (57)0<γ=∑j=1mβj+m,t>-1;tj+1<pjβj+1,forj=1,…,l,βj>-1,forj=l+1,…,m.Then the operator Sβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltlpl((0,∞))×L∞((0,∞))m-l to Ltp((0,∞)). In this case, if we denote by Sβ→,γ∏j=1lLtjpj×(L∞)m-l→Ltp the operator norm of Sβ→,γ, then (58)Sβ→,γ∏j=1lLtjpj×L∞m-l→Ltp≤∏j=1lΓβj+1-tj+1/pjΓt+1/pΓγ∏j=l+1mΓβj+1.Proof.
Forfi∈Ltipi, i=1,…,l, and fl+1,…,fm∈L∞, we write again f→=(f1,…,fm). We observe that t>-1 and γ=∑i=1l(βi+1-(ti+1)/pi)+∑i=l+1m(βi+1)+(t+1)/p. We proceed as in the proof of Lemma 8; we use a change of variables and Minkowski’s inequality, Hölder’s inequality, and (23) to obtain (59)Sβ→,γf→p,t≤∏j=l+1mfj∞∫0,∞m∏i=1muiβi1+u1+⋯+umγ×∫0∞∏i=1lfiuixpxtdx1/pdu1du2⋯dum≤∏j=l+1mfj∞∏i=1lfipi,ti×∫0,∞m∏i=1luiβi-ti+1/pi∏i=l+1muiβi1+u1+⋯+umγdu1du2⋯dum=∏i=1lΓβi+1-ti+1/pi∏i=l+1mΓβi+1Γt+1/pΓγ×∏j=l+1mfj∞∏i=1lfipi,ti.The proof is complete.We next obtain a lower bound of the norm ofSβ→,γf→.Lemma 10.
Letβ1,…,βm,γ be real numbers. Let 1<p1,…,pm≤∞, 1≤p<∞, and let t1,…,tm,t be real numbers with t>-1 and assume that relations (6) and (7) hold. Suppose that the parameters satisfy (60)0<γ=∑j=1mβj+m,t>-1,-γ-βj-1<tj+1pj<βj+1,j=1,…,m.Then (61)Sβ→,γ∏j=1mLtjpj→Ltp≥∏j=1mΓβj+1-tj+1/pjΓt+1/pΓγ.Proof.
Let us put(62)K=Sβ→,γ∏j=1mLtjpj→Ltp,Bm≔∏j=1mΓβj+1-tj+1/pjΓt+1/pΓγ.Then for any vector f→=(f1,…,fm), fj∈Ltjpj((0,∞)), j=1,…,m (for pj=∞, replace Ltjpj((0,∞)) by L∞((0,∞))), (63)K∏j=1mfjpj,tj≥Sβ→,γf→p,t;hence for any g∈Ltp′((0,∞)) (with Ltp′((0,∞)) replaced by L∞((0,∞)) if p=1), (64)K∏j=1mfjpj,tj≥∫0∞gxSβ→,γf→xxtdx.Let 0<ξ<min{pj(βj+1)-(tj+1):j=1,…,m} and define(65)fjx=0if0<x<1x-tj+1+ξ/pjifx≥1,gx=0if0<x<1x-t+1+ξ/p′ifx≥1.Then (66)fjpj,tj=1ξ1/pj,gp′,t=1ξ1/p′.Substituting these into (64), we obtain (67)Kξ≥∫1∞x-t+1+ξ/p′∫1,∞m∏j=1myjβj-tj+1+ξ/pjx+y1+⋯+ymγdy1⋯dymxtdx.Put (68)Lx≔∫1,∞m∏j=1myjβj-tj+1+ξ/pjx+y1+⋯+ymγdy1⋯dym.
Then(69)Lx=∫0,∞m∏j=1myjβj-tj+1+ξ/pjx+y1+⋯+ymγdy1⋯dym+∑k,l≠0,k+l=m-1lLk,lx+-1mL0x,where (70)L0x≔∫0,1m∏j=1myjβj-tj+1+ξ/pjx+y1+⋯+ymγdy1⋯dym,Lk,lx≔∫0,∞k∫0,1l∏s=1kyjsβjs-tjs+1+ξ/pjs∏s=1lyisβis-tis+1+ξ/pisx+y1+⋯+ymγdyj1⋯dyjkdyi1⋯dyil.We have (71)0<L0x≤x-γ∏j=1m1βj+1-tj+ξ+1/pj≔x-γAm,Lk,lx≤∫0,1l∏s=1lyisβis-tis+1+ξ/pisdyi1⋯dyil∫0,∞k∏s=1kyjsβjs-tjs+1+ξ/pjsx+yj1+⋯+yjkγdyj1⋯dyjk=∏s=1l1βis+1-tis+ξ+1/pis∫0,∞k∏s=1kyjsβjs-tjs+1+ξ/pjsx+yj1+⋯+yjkγdyj1⋯dyjk≤x-γ+∑s=1kβjs+1-tjs+1+ξ/pjsAlBk,l,where (72)Al≔∏s=1l1βis+1-tis+ξ+1/pis,Bk,l≔∏s=1kΓβjs+1-tjs+ξ+1/pjsΓt+1+ξ/p+∑s=1lβis+1-tis+ξ+1/pisΓγ.It follows that, for m even, (73)Lx>x-∑j=1mtj+1+ξ/pjBm-∑k≠0,k+2l+1=mx-γ+∑s=1kβjs+1-tjs+1+ξ/pjsA2l+1Bk,2l+1and, for m odd, (74)Lx>x-∑j=1mtj+1+ξ/pjBm-∑k≠0,k+2l+1=mx-γ+∑s=1kβjs+1-tjs+1+ξ/pjsA2l+1Bk,2l+1-x-γAm.Taking (73) into (67), we obtain that if m is even, then (75)Kξ≥Bm∫1∞x-1-ξdx-∑k≠0,k+2l+1=mA2l+1Bk,2l+1∫1∞x-1-ξ-∑s=1lβis+1-tis+ξ+1/pisdx=Bmξ-∑k≠0,k+2l+1=mA2l+1Bk,2l+11ξ+∑s=1lβis+1-tis+ξ+1/pis.Thus(76)K≥Bm-∑k≠0,k+2l+1=mA2l+1Bk,2l+1ξξ+∑s=1lβis+1-tis+ξ+1/pis.Taking (74) into (67), we obtain that if m is odd, then (77)Kξ≥Bm∫1∞x-1-ξdx-Am∫1∞x-γ-1-ξ+t+1+ξ/pdx-∑k≠0,k+2l+1=mA2l+1Bk,2l+1∫1∞x-1-ξ-∑s=1lβis+1-tis+ξ+1/pisdx=Bmξ-Amγ+ξ+t+1+ξ/p-∑k≠0,k+2l+1=mA2l+1Bk,2l+11ξ+∑s=1lβis+1-tis+ξ+1/pis.Thus for m odd, (78)K≥Bm-ξAmγ+ξ+t+1+ξ/p-∑k≠0,k+2l+1=mA2l+1Bk,2l+1ξξ+∑s=1lβis+1-tis+ξ+1/pis.Letting ξ→0 in (76) or (78) we obtain that K≥Bm; that is, (79)Sβ→,γ∏j=1mLtjpj→Ltp≥∏j=1mΓβj+1-tj+1/pjΓt+1/pΓγ.The proof is complete.


## 4.1. The Norm ofTβ→,γ→

We have the following upper estimate of the norm ofTβ→,γ→.Lemma 5.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm<∞, 1≤p<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold. Assume that (24)∑j=1mγj=∑j=1mβj+m,-pjγj-βj-1<tj+1<pjβj+1,j=1,…,m.Then the operator Tβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)). Moreover, (25)Tβ→,γ→∏j=1mLtjpj→Ltp≤∏j=1mBβj+1-tj+1pj,γj-βj-1+tj+1pj.Proof.
Forfi∈Ltipi, i=1,…,m, we write again f→=(f1,…,fm). Using an easy change of variables and Minkowski’s inequality, we obtain (26)Tβ→,γ→f→p,t=∫0∞Tβ→,γ→f→xpxtdx1/p≤∫0∞∫0,∞m∏i=1mfiyi∏i=1mx+yiγi∏i=1myiβidy1dy2⋯dympxtdx1/p=∫0∞∫0,∞m∏i=1mfiuix∏i=1m1+uiγi∏i=1muiβidu1du2⋯dumpxtdx1/p≤∫0,∞m∫0∞∏i=1mfiuixp∏i=1m1+uiγixtdx1/p∏i=1muiβidu1du2⋯dum=∫0,∞m∏i=1muiβi∏i=1m1+uiγi∫0∞∏i=1mfiuixpxtdx1/pdu1du2⋯dum.Using (6), (7), and Hölder’s inequality, we easily obtain that (27)∫0∞∏i=1mfiuixpxtdx≤∏i=1m∫0∞fiuixpixtidxp/pi=∏i=1mui-p/piti+1fipi,tip.It follows that (28)Tβ→,γ→f→p,t≤∏i=1mfipi,ti∫0,∞m∏i=1muiβi-ti+1/pi∏i=1m1+uiγidu1du2⋯dum=∏i=1mfipi,ti∏i=1m∫0∞uiβi-ti+1/pi1+uiγidui=∏i=1mfipi,ti∏i=1mBβi+1-ti+1pi,γi-βi-1+ti+1pi.The proof is complete.We also have the following result.Lemma 6.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pl<∞, 1≤l<m, 1≤p<∞, and assume that 1/p1+⋯+1/pl=1/p. Let t1,…,tl,t be real numbers and assume that t1/p1+⋯+tl/pl=t/p. Suppose that the parameters satisfy (29)∑j=1mγj=∑j=1mβj+m;-pjγj-βj-1<tj+1<pjβj+1,forj=1,…,l,γj>βj+1>0,forj=l+1,…,m.Then the operator Tβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltlpl((0,∞))×L∞((0,∞))m-l to Ltp((0,∞)). In this case, if we denote by Tβ→,γ→∏j=1lLtjpj×(L∞)m-l→Ltp the operator norm of Tβ→,γ→, then (30)Tβ→,γ→∏j=1lLtjpj×L∞m-l→Ltp≤∏j=1lBβj+1-tj+1pj,γj-βj-1+tj+1pj×∏j=l+1mBβj+1,γj-βj-1.Proof.
Forfi∈Ltipi, i=1,…,l, and fl+1,…,fm∈L∞, we write again f→=(f1,…,fm). Proceeding as at the beginning of the proof of the previous lemma, we obtain (31)Tβ→,γ→f→p,t≤∏j=l+1mfj∞∫0,∞m∏j=1mujβj∏j=1m1+ujγj∫0∞∏i=1lfiuixpxtdx1/pdu1du2⋯dum≤∏j=l+1mfj∞∏j=1lfjpj,tj∫0,∞m∏i=1luiβi-ti+1/pi∏i=l+1muiβi∏i=1m1+uiγidu1du2⋯dum=∏j=l+1mfj∞∏j=1lfjpj,tj∏i=1lBβi+1-ti+1pi,γi-βi-1+ti+1pi×∏i=l+1mBβi+1,γi-βi-1.The proof is complete.Let us now prove the lower bound for the norm ofTβ→,γ→.Lemma 7.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm≤∞, 1≤p<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold. Suppose that the parameters satisfy (32)∑j=1mγj=∑j=1mβj+m,-γj-βj-1<tj+1pj<βj+1,j=1,…,m.Then (33)Tβ→,γ→∏j=1mLtjpj→Ltp≥∏j=1mBβj+1-tj+1pj,γj-βj-1+tj+1pj.Proof.
For simplicity, let us put(34)K=Tβ→,γ→∏j=1mLtjpj→Ltp,Bm≔∏j=1mBβj+1-tj+1pj,γj-βj-1+tj+1pj.For any vector f→=(f1,…,fm), fj∈Ltjpj((0,∞)), j=1,…,m (for pj=∞, replace Ltjpj((0,∞)) by L∞((0,∞))), we have (35)K∏j=1mfjpj,tj≥Tβ→,γ→f→p,t;hence for any g∈Ltp′((0,∞)) (with Ltp′((0,∞)) replaced by L∞((0,∞)) if p=1), (36)K∏j=1mfjpj,tj≥∫0∞gxTβ→,γ→f→xxtdx.Let 0<ξ<min{pj(βj+1)-(tj+1):j=1,…,m} and define(37)fjx=0if0<x<1x-tj+1+ξ/pjifx≥1,gx=0if0<x<1x-t+1+ξ/p′ifx≥1.Then (38)fjpj,tj=1ξ1/pj,gp′,t=1ξ1/p′.Substituting these into (36), we obtain (39)Kξ≥∫1∞x-t+1+ξ/p′∫1,∞m∏j=1myjβj-tj+1+ξ/pj∏j=1mx+yjγjdy1⋯dymxtdx.Observing that ∫1∞=∫0∞-∫01, we obtain that (40)L≔∫1,∞m∏j=1myjβj-tj+1+ξ/pj∏j=1mx+yjγjdy1⋯dym=∫0,∞m∏j=1myjβj-tj+1+ξ/pj∏j=1mx+yjγjdy1⋯dym+∑k,l≠0,k+l=m-1lLkLl+-1m∫0,1m∏j=1myjβj-tj+1+ξ/pj∏j=1mx+yjγjdy1⋯dym=x-∑j=1mtj+1+ξ/pjBm+∑k,l≠0,k+l=m-1lLkLl+-1m∫0,1m∏j=1myjβj-tj+1+ξ/pj∏j=1mx+yjγjdy1⋯dym,where (41)Lk≔∫0,∞k∏s=1kyjsβjs-tjs+1+ξ/pjs∏s=1kx+yjsγjsdyj1⋯dyjk,Ll≔∫0,1l∏s=1lyjsβjs-tjs+1+ξ/pjs∏s=1lx+yjsγjsdyj1⋯dyjk.We have that(42)0<Lk=x-∑s=1kγjs+∑s=1kβjs+k-∑s=1ktjs+1+ξ/pjsBkξ,where (43)Bkξ≔∏s=1kBβjs+1-tjs+1+ξpjs,γjs-βjs-1+tjs+1pjs,(44)0<Ll≤x-∑s=1lγjs∫0,1l∏s=1lyjsβjs-tjs+1+ξ/pjsdyj1⋯dyjk=x-∑s=1lγjs∏s=1l1βjs+1-tjs+1+ξ/pjs≔x-∑s=1lγjsClξ.Let us write (45)L0≔∫0,1m∏j=1myjβj-tj+1+ξ/pj∏j=1mx+yjγjdy1⋯dym.Then(46)0<L0≤x-∑j=1mγjCmξwith (47)Cmξ≔∏j=1m1βj+1-tj+1+ξ/pj.It follows that if m is an even integer, then (48)L>x-∑j=1mtj+1+ξ/pjBm-∑k≠0,k+2l+1=mxαkBkξC2l+1ξ,where αk≔-∑j=1mγj+∑s=1kβjs+k-∑s=1k(tjs+1+ξ/pjs); and if m is an odd number, then (49)L>x-∑j=1mtj+1+ξ/pjBm-∑k≠0,k+2l+1=mxαkBkξC2l+1ξ-x-∑j=1mγjCmξ.Taking (48) into (39), we obtain that if m is even, then (50)Kξ≥Bm∫1∞xt-t+1+ξ/p′-∑j=1mtj+1+ξ/pjdx-∑k≠0,k+2l+1=mBkξC2l+1ξ∫1∞xt-t+1+ξ/p′+αkdx=Bm∫1∞x-1-ξdx-∑k≠0,k+2l+1=mBkξC2l+1ξ∫1∞x-∑s=12l+1βjs+1-tjs+1+ξ/pjs-1-ξdx=Bmξ-∑k≠0,k+2l+1=mBkξC2l+1ξ∑s=12l+1βjs+1-tjs+1+ξ/pjs+ξ.Thus (51)K≥Bm-∑k≠0,k+2l+1=mξBkξC2l+1ξ∑s=12l+1βjs+1-tjs+1+ξ/pjs+ξ.Taking (49) into (39) and doing the same type of calculations as above, we obtain that if m is an odd integer, then (52)K≥Bm-∑k≠0,k+2l+1=mξBkξC2l+1ξ∑s=12l+1βjs+1-tjs+1+ξ/pjs+ξ-ξCmξξ+∑j=1mβj+1-tj+1+ξ/pj.Letting ξ→0 in (51) or (52) we obtain that K≥Bm. That is, (53)Tβ→,γ→∏j=1mLtjpj→Ltp≥∏j=1mBβj+1-tj+1pj,γj-βj-1+tj+1pj.The proof is complete.


## 4.2. The Norm ofSβ→,γ

Let us prove the following upper bound for the norm ofSβ→,γ.Lemma 8.
Letβ1,…,βm,γ be real numbers. Let 1<p1,…,pm<∞, 1≤p<∞, and let t1,…,tm,t be real numbers, and assume that relations (6) and (7) hold. Assume that (54)0<γ=∑j=1mβj+m,t>-1,tj+1<pjβj+1,j=1,…,m.Then the operator Sβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)). In this case, if we denote by Sβ→,γ∏j=1mLtjpj→Ltpm the operator norm of Sβ→,γ, then (55)Sβ→,γ∏j=1mLtjpj→Ltp≤∏j=1mΓβj+1-tj+1/pjΓt+1/pΓγ.Proof.
The proof is essentially the same as above. Let us give it here for completeness. Forfi∈Ltipi, i=1,…,m, we write again f→=(f1,…,fm). We remark that t>-1 and γ=∑i=1mβi+1-ti+1/pi+(t+1)/p. Using an easy change of variables and Minkowski’s inequality, Hölder’s inequality, and equality (23), we obtain (56)Sβ→,γf→p,t=∫0∞Sβ→,γf→xpxtdx1/p≤∫0∞∫0,∞m∏i=1mfiyix+y1+⋯+ymγ∏i=1myiβidy1dy2⋯dympxtdx1/p=∫0∞∫0,∞m∏i=1mfiuix1+u1+⋯+umγ∏i=1muiβidu1du2⋯dumpxtdx1/p≤∫0,∞m∫0∞∏i=1mfiuixp1+u1+⋯+umγxtdx1/p∏i=1muiβidu1du2⋯dum=∫0,∞m∏i=1muiβi1+u1+⋯+umγ∫0∞∏i=1mfiuixpxtdx1/pdu1du2⋯dum≤∏i=1mfipi,ti∫0,∞m∏i=1muiβi-ti+1/pi1+u1+⋯+umγdu1du2⋯dum=∏i=1mfipi,ti∏i=1mΓβi+1-ti+1/piΓt+1/pΓγ.
The proof is complete.Let us prove the following result.Lemma 9.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pl<∞, 1≤l<m, 1≤p<∞, and assume that 1/p1+⋯+1/pl=1/p. Let t1,…,tl,t be real numbers and assume that t1/p1+⋯+tl/pl=t/p. Suppose that the parameters satisfy (57)0<γ=∑j=1mβj+m,t>-1;tj+1<pjβj+1,forj=1,…,l,βj>-1,forj=l+1,…,m.Then the operator Sβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltlpl((0,∞))×L∞((0,∞))m-l to Ltp((0,∞)). In this case, if we denote by Sβ→,γ∏j=1lLtjpj×(L∞)m-l→Ltp the operator norm of Sβ→,γ, then (58)Sβ→,γ∏j=1lLtjpj×L∞m-l→Ltp≤∏j=1lΓβj+1-tj+1/pjΓt+1/pΓγ∏j=l+1mΓβj+1.Proof.
Forfi∈Ltipi, i=1,…,l, and fl+1,…,fm∈L∞, we write again f→=(f1,…,fm). We observe that t>-1 and γ=∑i=1l(βi+1-(ti+1)/pi)+∑i=l+1m(βi+1)+(t+1)/p. We proceed as in the proof of Lemma 8; we use a change of variables and Minkowski’s inequality, Hölder’s inequality, and (23) to obtain (59)Sβ→,γf→p,t≤∏j=l+1mfj∞∫0,∞m∏i=1muiβi1+u1+⋯+umγ×∫0∞∏i=1lfiuixpxtdx1/pdu1du2⋯dum≤∏j=l+1mfj∞∏i=1lfipi,ti×∫0,∞m∏i=1luiβi-ti+1/pi∏i=l+1muiβi1+u1+⋯+umγdu1du2⋯dum=∏i=1lΓβi+1-ti+1/pi∏i=l+1mΓβi+1Γt+1/pΓγ×∏j=l+1mfj∞∏i=1lfipi,ti.The proof is complete.We next obtain a lower bound of the norm ofSβ→,γf→.Lemma 10.
Letβ1,…,βm,γ be real numbers. Let 1<p1,…,pm≤∞, 1≤p<∞, and let t1,…,tm,t be real numbers with t>-1 and assume that relations (6) and (7) hold. Suppose that the parameters satisfy (60)0<γ=∑j=1mβj+m,t>-1,-γ-βj-1<tj+1pj<βj+1,j=1,…,m.Then (61)Sβ→,γ∏j=1mLtjpj→Ltp≥∏j=1mΓβj+1-tj+1/pjΓt+1/pΓγ.Proof.
Let us put(62)K=Sβ→,γ∏j=1mLtjpj→Ltp,Bm≔∏j=1mΓβj+1-tj+1/pjΓt+1/pΓγ.Then for any vector f→=(f1,…,fm), fj∈Ltjpj((0,∞)), j=1,…,m (for pj=∞, replace Ltjpj((0,∞)) by L∞((0,∞))), (63)K∏j=1mfjpj,tj≥Sβ→,γf→p,t;hence for any g∈Ltp′((0,∞)) (with Ltp′((0,∞)) replaced by L∞((0,∞)) if p=1), (64)K∏j=1mfjpj,tj≥∫0∞gxSβ→,γf→xxtdx.Let 0<ξ<min{pj(βj+1)-(tj+1):j=1,…,m} and define(65)fjx=0if0<x<1x-tj+1+ξ/pjifx≥1,gx=0if0<x<1x-t+1+ξ/p′ifx≥1.Then (66)fjpj,tj=1ξ1/pj,gp′,t=1ξ1/p′.Substituting these into (64), we obtain (67)Kξ≥∫1∞x-t+1+ξ/p′∫1,∞m∏j=1myjβj-tj+1+ξ/pjx+y1+⋯+ymγdy1⋯dymxtdx.Put (68)Lx≔∫1,∞m∏j=1myjβj-tj+1+ξ/pjx+y1+⋯+ymγdy1⋯dym.
Then(69)Lx=∫0,∞m∏j=1myjβj-tj+1+ξ/pjx+y1+⋯+ymγdy1⋯dym+∑k,l≠0,k+l=m-1lLk,lx+-1mL0x,where (70)L0x≔∫0,1m∏j=1myjβj-tj+1+ξ/pjx+y1+⋯+ymγdy1⋯dym,Lk,lx≔∫0,∞k∫0,1l∏s=1kyjsβjs-tjs+1+ξ/pjs∏s=1lyisβis-tis+1+ξ/pisx+y1+⋯+ymγdyj1⋯dyjkdyi1⋯dyil.We have (71)0<L0x≤x-γ∏j=1m1βj+1-tj+ξ+1/pj≔x-γAm,Lk,lx≤∫0,1l∏s=1lyisβis-tis+1+ξ/pisdyi1⋯dyil∫0,∞k∏s=1kyjsβjs-tjs+1+ξ/pjsx+yj1+⋯+yjkγdyj1⋯dyjk=∏s=1l1βis+1-tis+ξ+1/pis∫0,∞k∏s=1kyjsβjs-tjs+1+ξ/pjsx+yj1+⋯+yjkγdyj1⋯dyjk≤x-γ+∑s=1kβjs+1-tjs+1+ξ/pjsAlBk,l,where (72)Al≔∏s=1l1βis+1-tis+ξ+1/pis,Bk,l≔∏s=1kΓβjs+1-tjs+ξ+1/pjsΓt+1+ξ/p+∑s=1lβis+1-tis+ξ+1/pisΓγ.It follows that, for m even, (73)Lx>x-∑j=1mtj+1+ξ/pjBm-∑k≠0,k+2l+1=mx-γ+∑s=1kβjs+1-tjs+1+ξ/pjsA2l+1Bk,2l+1and, for m odd, (74)Lx>x-∑j=1mtj+1+ξ/pjBm-∑k≠0,k+2l+1=mx-γ+∑s=1kβjs+1-tjs+1+ξ/pjsA2l+1Bk,2l+1-x-γAm.Taking (73) into (67), we obtain that if m is even, then (75)Kξ≥Bm∫1∞x-1-ξdx-∑k≠0,k+2l+1=mA2l+1Bk,2l+1∫1∞x-1-ξ-∑s=1lβis+1-tis+ξ+1/pisdx=Bmξ-∑k≠0,k+2l+1=mA2l+1Bk,2l+11ξ+∑s=1lβis+1-tis+ξ+1/pis.Thus(76)K≥Bm-∑k≠0,k+2l+1=mA2l+1Bk,2l+1ξξ+∑s=1lβis+1-tis+ξ+1/pis.Taking (74) into (67), we obtain that if m is odd, then (77)Kξ≥Bm∫1∞x-1-ξdx-Am∫1∞x-γ-1-ξ+t+1+ξ/pdx-∑k≠0,k+2l+1=mA2l+1Bk,2l+1∫1∞x-1-ξ-∑s=1lβis+1-tis+ξ+1/pisdx=Bmξ-Amγ+ξ+t+1+ξ/p-∑k≠0,k+2l+1=mA2l+1Bk,2l+11ξ+∑s=1lβis+1-tis+ξ+1/pis.Thus for m odd, (78)K≥Bm-ξAmγ+ξ+t+1+ξ/p-∑k≠0,k+2l+1=mA2l+1Bk,2l+1ξξ+∑s=1lβis+1-tis+ξ+1/pis.Letting ξ→0 in (76) or (78) we obtain that K≥Bm; that is, (79)Sβ→,γ∏j=1mLtjpj→Ltp≥∏j=1mΓβj+1-tj+1/pjΓt+1/pΓγ.The proof is complete.


## 5. Necessity for Boundedness ofTβ→,γ→ and Sβ→,γ

### 5.1. Necessity for Boundedness ofTβ→,γ→

Lemma 11.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm≤∞, 1≤p<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold. If the operator Tβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)), then the parameters satisfy (80)∑j=1mγj=∑j=1mβj+m.Proof.
For simplicity, we putγ=∑j=1mγj and β=∑j=1mβj. Let R>0 be a real number. Given a function f, we denote, by fR, the function defined by fR(x)=f(Rx). One easily checks that, for f∈Ltjpj, (81)fjRpj,tj=R-tj+1/pjfjpj,tj.For a vector f→=(f1,…,fm), we write f→R=((f1)R,…,(fm)R). Using some easy changes of variables, we obtain (82)Tβ→,γ→f→Rx=Rγ-β-mTβ→,γ→f→Rx.It comes that (83)Tβ→,γ→f→Rp,t=Rγ-β-m-t+1/pTβ→,γ→f→p,t.Recall that the boundedness of Tβ→,γ→ means that there exists a constant C>0 such that, for any fj∈Ltjpj, j=1,…,m, (84)Tβ→,γ→f→Rp,t≤C∏j=1mfjRpj,tj.It follows using (81) and (83) that (85)Rγ-β-m-t+1/pTβ→,γ→f→p,t≤CR-∑j=1mtj+1/pj∏j=1mfjpj,tjor equivalently, (86)Rγ-β-m-t+1/p+∑j=1mtj+1/pjTβ→,γ→f→p,t≤C∏j=1mfjpj,tj.As (86) holds for any f1∈Lt1p1,…,fm∈Ltmpm and any real number R>0, we necessarily have that (87)γ-β-m-t+1p+∑j=1mtj+1pj=0which combined with (7) gives that (88)∑j=1mγj=∑j=1mβj+m.Lemma 12.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm<∞, 1<p<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold. Then if the operator Tβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)), then the parameters satisfy (89)-pjγj-βj-1<tj+1<pjβj+1,j=1,…,m.Proof.
Assume thatTβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)). Then its j-th adjoint Tβ→,γ→∗,j defined by (90)Tβ→,γ→∗,jf→jyj=yjβj-tj∫0,∞mgx∏1=k≠jmfkyk∏k=1mx+ykγkxt∏1=k≠jmykβkdykdx,where f→j=(f1,…,fj-1,g,fj+1,…,fm), is bounded from Lt1p1((0,∞))×⋯×Ltj-1pj-1((0,∞))×Ltp′((0,∞))×Ltj+1pj+10,∞×⋯×Ltmpm((0,∞)) to Ltjpj′((0,∞)). Let us take g(x)=χ[1,2](x) and fk(y)=χ[1,2](y), j≠k∈{1,…,m}. We observe that, for x,yk∈[1,2], (x+yk)γk≃1 while (x+yj)γj≃(1+yj)γj. It follows for this choice of functions that, for any yj∈(0,∞), (91)Tβ→,γ→∗,jf→jyj≃yjβj-tj1+yjγj.It follows from the boundedness of Tβ→,γ→∗,j that there is a constant C>0 such that (92)∫0∞yjpj′βj-tj1+yjγjpj′yjtjdyj≃Tβ→,γ→∗,jf→jpj′,tj≤C.From the properties of the Beta-function, we know that this implies that pj′(βj-tj)+tj+1>0 and γjpj′-pj′(βj-tj)-tj-1>0. These two inequalities are equivalent to (93)-pjγj-βj-1<tj+1<pjβj+1.The proof is complete.The proof of the following result follows the same steps as in the proof of the above lemma.Lemma 13.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold with p=1. If the operator Tβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Lt1((0,∞)), then the parameters satisfy (94)-pjγj-βj-1<tj+1<pjβj+1,j=1,…,m.


### 5.2. Necessity for Boundedness ofSβ→,γ

Lemma 14.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm≤∞, 1≤p<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold. If the operator Sβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)), then the parameters satisfy (95)γ=∑j=1mβj+m.Proof.
This is obtained exactly the same way as in the proof of Lemma11. We leave it to the interested reader.Lemma 15.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm≤∞, 1≤p<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold. If the operator Sβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)), then t>-1 and γ>0.Proof.
Let us takefk(y)=χ[1,2](y), k∈{1,…,m}. We observe that, for y1,…,ym∈[1,2], (x+y1+⋯+ym)γ≃1+xγ. It follows for this choice of functions that, for any x∈(0,∞), (96)Sβ→,γf→x≃11+xγ.It follows from the boundedness of Sβ→,γ that there is a constant C>0 such that (97)∫0∞xt1+xγpdx≃Sβ→,γf→p,t≤C.Clearly, this is only possible if t>-1 and γ>0. The proof is complete.Lemma 16.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm<∞, 1<p<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold. If the operator Sβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)), then the parameters satisfy (98)-pjγ-βj-1<tj+1<pjβj+1,j=1,…,m.Proof.
Assume thatSβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)), Then its j-th adjoint Sβ→,γ∗,j is defined by (99)Sβ→,γ∗,jf→jyj=∫0,∞myjβj-tjgxxtx+∑i=1myiγ∏1=k≠jmykβkfkykdy1⋯dyj-1dyj+1⋯dymdx,where again f→j=(f1,…,fj-1,g,fj+1,…,fm) is bounded from Lt1p1((0,∞))×⋯×Ltj-1pj-1((0,∞))×Ltp′((0,∞))×Ltj+1pj+1((0,∞))×⋯×Ltmpm((0,∞)) to Ltjpj′((0,∞)). Let us take g(x)=χ[1,2](x) and fk(y)=χ[1,2](y), j≠k∈{1,…,m}. We observe that, for x,yk∈[1,2], j≠k∈{1,…,m}, (x+y1+⋯+ym)γ≃(1+yj)γ. It follows for this choice of functions that, for any yj∈(0,∞), (100)Sβ→,γ∗,jf→jyj≃yjβj-tj1+yjγ.It follows from the boundedness of Sβ→,γ∗,j that there is a constant C>0 such that (101)∫0∞yjpj′βj-tj1+yjγpj′yjtjdyj≃Sβ→,γ∗,jf→jpj′,tj≤C.This as in the proof of Lemma 12 implies that (102)-pjγ-βj-1<tj+1<pjβj+1.The proof is complete.The following is obtained as above.Lemma 17.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold with p=1. If the operator Sβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Lt1((0,∞)), then the parameters satisfy and (103)-pjγ-βj-1<tj+1<pjβj+1,j=1,…,m.


### 5.3. Necessity for the Other Cases

Let us start by proving the followingLemma 18.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pl<∞, 1≤l<m, 1≤p<∞, and assume that 1/p1+⋯+1/pl=1/p. Let t1,…,tl,t be real numbers and assume that t1/p1+⋯+tl/pl=t/p. If the operator Tβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltlpl((0,∞))×L∞((0,∞))m-l to Ltp((0,∞)), then the parameters satisfy (104)-pjγj-βj-1<tj+1<pjβj+1,j=1,…,l,γj>βj+1>0,j=l+1,…,m.Proof.
Suppose thatTβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltlpl((0,∞))×L∞((0,∞))m-l to Ltp((0,∞)). Then for 1≤j≤l, the j-th adjoint, Tβ→,γ→∗j, of Tβ→,γ→ is defined as in the proof of Lemmas 12 and 13. Hence the proof of the first inequalities follows from the proof of Lemmas 12 and 13. To prove the second inequalities, we observe that, for j>l, j-th adjoint, Tβ→,γ→∗j, of Tβ→,γ→ is defined by(105)Tβ→,γ→∗,jf→jyj=yjβj∫0,∞mgx∏1=k≠jmfkyk∏k=1mx+ykγk∏1=k≠jmykβkdykxtdx,where f→j=(f1,…,fj-1,g,fj+1,…,fm) is bounded from (106)∏k=1lLtkpk0,∞×L∞0,∞j-l-1×Ltp′0,∞×L∞0,∞m-jto L1((0,∞)). Hence the second inequality can now be established similarly as in the proof of Lemma 12.Let us finish with the following.Lemma 19.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pl<∞, 1≤l<m, 1≤p<∞, and assume that 1/p1+⋯+1/pl=1/p. Let t1,…,tl,t be real numbers and assume that t1/p1+⋯+tl/pl=t/p. If the operator Sβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltlpl((0,∞))×L∞((0,∞))m-l to Ltp((0,∞)), then the parameters satisfy (107)tj+1<pjβj+1,j=1,…,m,βj>-1,j=l+1,…,m.Proof.
Suppose thatSβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltlpl((0,∞))×L∞((0,∞))m-l to Ltp((0,∞)). Then for 1≤j≤l, the j-th adjoint, Sβ→,γ∗j, of Sβ→,γ is defined as in the proof of Lemmas 16 and 17. Hence the proof of the first inequality follows from the proof of Lemmas 16 and 17. To prove the second inequality, we observe that, for j>l, Sβ→,γ∗j, of Sβ→,γ, is given by (108)Sβ→,γ∗,jf→jyj=yjβj∫0,∞mxtgxx+∑i=1myiγ∏1=k≠jmykβkfkykdy1⋯dyj-1dyj+1⋯dymdx,where again f→j=(f1,…,fj-1,g,fj+1,…,fm) is bounded from (109)∏k=1lLtkpk0,∞×L∞0,∞j-l-1×Ltp′0,∞×L∞0,∞m-jto L1((0,∞)). Hence the second inequality can now be established similarly as in the proof of Lemma 16.


## 5.1. Necessity for Boundedness ofTβ→,γ→

Lemma 11.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm≤∞, 1≤p<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold. If the operator Tβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)), then the parameters satisfy (80)∑j=1mγj=∑j=1mβj+m.Proof.
For simplicity, we putγ=∑j=1mγj and β=∑j=1mβj. Let R>0 be a real number. Given a function f, we denote, by fR, the function defined by fR(x)=f(Rx). One easily checks that, for f∈Ltjpj, (81)fjRpj,tj=R-tj+1/pjfjpj,tj.For a vector f→=(f1,…,fm), we write f→R=((f1)R,…,(fm)R). Using some easy changes of variables, we obtain (82)Tβ→,γ→f→Rx=Rγ-β-mTβ→,γ→f→Rx.It comes that (83)Tβ→,γ→f→Rp,t=Rγ-β-m-t+1/pTβ→,γ→f→p,t.Recall that the boundedness of Tβ→,γ→ means that there exists a constant C>0 such that, for any fj∈Ltjpj, j=1,…,m, (84)Tβ→,γ→f→Rp,t≤C∏j=1mfjRpj,tj.It follows using (81) and (83) that (85)Rγ-β-m-t+1/pTβ→,γ→f→p,t≤CR-∑j=1mtj+1/pj∏j=1mfjpj,tjor equivalently, (86)Rγ-β-m-t+1/p+∑j=1mtj+1/pjTβ→,γ→f→p,t≤C∏j=1mfjpj,tj.As (86) holds for any f1∈Lt1p1,…,fm∈Ltmpm and any real number R>0, we necessarily have that (87)γ-β-m-t+1p+∑j=1mtj+1pj=0which combined with (7) gives that (88)∑j=1mγj=∑j=1mβj+m.Lemma 12.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm<∞, 1<p<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold. Then if the operator Tβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)), then the parameters satisfy (89)-pjγj-βj-1<tj+1<pjβj+1,j=1,…,m.Proof.
Assume thatTβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)). Then its j-th adjoint Tβ→,γ→∗,j defined by (90)Tβ→,γ→∗,jf→jyj=yjβj-tj∫0,∞mgx∏1=k≠jmfkyk∏k=1mx+ykγkxt∏1=k≠jmykβkdykdx,where f→j=(f1,…,fj-1,g,fj+1,…,fm), is bounded from Lt1p1((0,∞))×⋯×Ltj-1pj-1((0,∞))×Ltp′((0,∞))×Ltj+1pj+10,∞×⋯×Ltmpm((0,∞)) to Ltjpj′((0,∞)). Let us take g(x)=χ[1,2](x) and fk(y)=χ[1,2](y), j≠k∈{1,…,m}. We observe that, for x,yk∈[1,2], (x+yk)γk≃1 while (x+yj)γj≃(1+yj)γj. It follows for this choice of functions that, for any yj∈(0,∞), (91)Tβ→,γ→∗,jf→jyj≃yjβj-tj1+yjγj.It follows from the boundedness of Tβ→,γ→∗,j that there is a constant C>0 such that (92)∫0∞yjpj′βj-tj1+yjγjpj′yjtjdyj≃Tβ→,γ→∗,jf→jpj′,tj≤C.From the properties of the Beta-function, we know that this implies that pj′(βj-tj)+tj+1>0 and γjpj′-pj′(βj-tj)-tj-1>0. These two inequalities are equivalent to (93)-pjγj-βj-1<tj+1<pjβj+1.The proof is complete.The proof of the following result follows the same steps as in the proof of the above lemma.Lemma 13.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold with p=1. If the operator Tβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Lt1((0,∞)), then the parameters satisfy (94)-pjγj-βj-1<tj+1<pjβj+1,j=1,…,m.


## 5.2. Necessity for Boundedness ofSβ→,γ

Lemma 14.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm≤∞, 1≤p<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold. If the operator Sβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)), then the parameters satisfy (95)γ=∑j=1mβj+m.Proof.
This is obtained exactly the same way as in the proof of Lemma11. We leave it to the interested reader.Lemma 15.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm≤∞, 1≤p<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold. If the operator Sβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)), then t>-1 and γ>0.Proof.
Let us takefk(y)=χ[1,2](y), k∈{1,…,m}. We observe that, for y1,…,ym∈[1,2], (x+y1+⋯+ym)γ≃1+xγ. It follows for this choice of functions that, for any x∈(0,∞), (96)Sβ→,γf→x≃11+xγ.It follows from the boundedness of Sβ→,γ that there is a constant C>0 such that (97)∫0∞xt1+xγpdx≃Sβ→,γf→p,t≤C.Clearly, this is only possible if t>-1 and γ>0. The proof is complete.Lemma 16.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm<∞, 1<p<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold. If the operator Sβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)), then the parameters satisfy (98)-pjγ-βj-1<tj+1<pjβj+1,j=1,…,m.Proof.
Assume thatSβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Ltp((0,∞)), Then its j-th adjoint Sβ→,γ∗,j is defined by (99)Sβ→,γ∗,jf→jyj=∫0,∞myjβj-tjgxxtx+∑i=1myiγ∏1=k≠jmykβkfkykdy1⋯dyj-1dyj+1⋯dymdx,where again f→j=(f1,…,fj-1,g,fj+1,…,fm) is bounded from Lt1p1((0,∞))×⋯×Ltj-1pj-1((0,∞))×Ltp′((0,∞))×Ltj+1pj+1((0,∞))×⋯×Ltmpm((0,∞)) to Ltjpj′((0,∞)). Let us take g(x)=χ[1,2](x) and fk(y)=χ[1,2](y), j≠k∈{1,…,m}. We observe that, for x,yk∈[1,2], j≠k∈{1,…,m}, (x+y1+⋯+ym)γ≃(1+yj)γ. It follows for this choice of functions that, for any yj∈(0,∞), (100)Sβ→,γ∗,jf→jyj≃yjβj-tj1+yjγ.It follows from the boundedness of Sβ→,γ∗,j that there is a constant C>0 such that (101)∫0∞yjpj′βj-tj1+yjγpj′yjtjdyj≃Sβ→,γ∗,jf→jpj′,tj≤C.This as in the proof of Lemma 12 implies that (102)-pjγ-βj-1<tj+1<pjβj+1.The proof is complete.The following is obtained as above.Lemma 17.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pm<∞, and let t1,…,tm,t be real numbers and assume that relations (6) and (7) hold with p=1. If the operator Sβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltmpm((0,∞)) to Lt1((0,∞)), then the parameters satisfy and (103)-pjγ-βj-1<tj+1<pjβj+1,j=1,…,m.


## 5.3. Necessity for the Other Cases

Let us start by proving the followingLemma 18.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pl<∞, 1≤l<m, 1≤p<∞, and assume that 1/p1+⋯+1/pl=1/p. Let t1,…,tl,t be real numbers and assume that t1/p1+⋯+tl/pl=t/p. If the operator Tβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltlpl((0,∞))×L∞((0,∞))m-l to Ltp((0,∞)), then the parameters satisfy (104)-pjγj-βj-1<tj+1<pjβj+1,j=1,…,l,γj>βj+1>0,j=l+1,…,m.Proof.
Suppose thatTβ→,γ→ is bounded from Lt1p1((0,∞))×⋯×Ltlpl((0,∞))×L∞((0,∞))m-l to Ltp((0,∞)). Then for 1≤j≤l, the j-th adjoint, Tβ→,γ→∗j, of Tβ→,γ→ is defined as in the proof of Lemmas 12 and 13. Hence the proof of the first inequalities follows from the proof of Lemmas 12 and 13. To prove the second inequalities, we observe that, for j>l, j-th adjoint, Tβ→,γ→∗j, of Tβ→,γ→ is defined by(105)Tβ→,γ→∗,jf→jyj=yjβj∫0,∞mgx∏1=k≠jmfkyk∏k=1mx+ykγk∏1=k≠jmykβkdykxtdx,where f→j=(f1,…,fj-1,g,fj+1,…,fm) is bounded from (106)∏k=1lLtkpk0,∞×L∞0,∞j-l-1×Ltp′0,∞×L∞0,∞m-jto L1((0,∞)). Hence the second inequality can now be established similarly as in the proof of Lemma 12.Let us finish with the following.Lemma 19.
Letβ1,…,βm,γ1,…,γm be real numbers. Let 1<p1,…,pl<∞, 1≤l<m, 1≤p<∞, and assume that 1/p1+⋯+1/pl=1/p. Let t1,…,tl,t be real numbers and assume that t1/p1+⋯+tl/pl=t/p. If the operator Sβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltlpl((0,∞))×L∞((0,∞))m-l to Ltp((0,∞)), then the parameters satisfy (107)tj+1<pjβj+1,j=1,…,m,βj>-1,j=l+1,…,m.Proof.
Suppose thatSβ→,γ is bounded from Lt1p1((0,∞))×⋯×Ltlpl((0,∞))×L∞((0,∞))m-l to Ltp((0,∞)). Then for 1≤j≤l, the j-th adjoint, Sβ→,γ∗j, of Sβ→,γ is defined as in the proof of Lemmas 16 and 17. Hence the proof of the first inequality follows from the proof of Lemmas 16 and 17. To prove the second inequality, we observe that, for j>l, Sβ→,γ∗j, of Sβ→,γ, is given by (108)Sβ→,γ∗,jf→jyj=yjβj∫0,∞mxtgxx+∑i=1myiγ∏1=k≠jmykβkfkykdy1⋯dyj-1dyj+1⋯dymdx,where again f→j=(f1,…,fj-1,g,fj+1,…,fm) is bounded from (109)∏k=1lLtkpk0,∞×L∞0,∞j-l-1×Ltp′0,∞×L∞0,∞m-jto L1((0,∞)). Hence the second inequality can now be established similarly as in the proof of Lemma 16.


## 6. Proof of the Results

Theorem1 follows from Lemmas 5, 7, 11, 12, and 13. Theorem 2 follows from Lemmas 8, 10, 14, 15, 16, and 17. Theorem 3 follows from Lemmas 6, 7, 11, and 18. Theorem 4 follows from Lemmas 9, 10, 14, 15, and 19. The proof is complete.



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