# Boundedness ofn-Multiple Discrete Hardy Operators with Weights for 1<q<p<∞

Authors: Aigerim Kalybay; Ryskul Oinarov; Ainur Temirkhanova
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We find necessary and sufficient conditions on weighted sequencesωi, i=1,2,…,n-1, u, and v, for which the operator (Sn-1f)i=∑k1=1i‍ω1,k1⋯∑kn-1=1kn-2‍ωn-1,kn-1∑j=1kn-1‍fj,i≥1, is bounded from lp,v to lq,u for 1<q<p<∞.


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## Body

## 1. Introduction

Let1<p,q<∞ and 1/p+1/p′=1. Let v={vk}k=1∞ be positive, ωi={ωi,k}k=1∞, i=1,2,…,n-1, and u={uk}k=1∞ nonnegative number sequences, hereinafter referred to as weights.Letlp,v be the space of sequences f={fk}k=1∞, for which the following norm is finite:
(1)∥f∥p,v=(∑i=1∞‍|vifi|p)1/p.Consider the operatorSn-1 that acts on the sequence f={fk}k=1∞ as follows:
(2)(Sn-1f)i=∑k1=1i‍ω1,k1∑k2=1k1‍ω2,k2⋯∑kn-1=1kn-2‍ωn-1,kn-1∑j=1kn-1‍fj,i≥1,and call itn-multiple discrete Hardy operator with weights.By changing the order of summation in (2) we have
(3)(Sn-1f)i=∑j=1i‍An-1,1(i,j)fj,i≥1,
where An-1,1(i,j)≡1 for n=1 and
(4)An-1,1(i,j)=∑kn-1=ji‍ωn-1,kn-1∑kn-2=kn-1i‍ωn-2,kn-2⋯∑k1=k2i‍ω1,k1,i≥j≥1,forn>1.Together with operator (3) we consider its dual operator
(5)(Sn-1*f)j=∑i=j∞‍An-1,1(i,j)fi,j≥1.Whenn=1, the operators Sn-1 and Sn-1* are simple discrete Hardy operators (S0f)i=∑j=1ifj and (S0*f)i=∑j=i∞fj; the problem of boundedness from lp,v to lq,u is investigated in detail in [1–3] for all values of the parameters p and q.In paper [4] necessary and sufficient conditions of boundedness of operators (3) and (5) from lp,v to lq,u are obtained for the case 1<p≤q<∞. However, the problem of boundedness of these operators has not been studied for the case 1<q<p<∞. In this paper we consider this problem.Let us notice that the boundedness problem of continuous Hardy-type operators has been studied and developed in an extraordinary depth (see, e.g., [3]). The corresponding results for n-multiple integral Hardy operators were considered by Baiarystanov in [5]. Namely, he investigated the continuous analogue of operators (3) and (5) and found necessary and sufficient conditions of their boundedness from Lp to Lq for the case 1<p≤q<∞. Moreover, there he obtained sufficient conditions of the same problem for the case 1<q<p<∞. Later in [6] Sagindykov proved that sufficient conditions found by Baiarystanov for the case 1<q<p<∞ are also necessary. However, the method for the continuous case does not work in the discrete situation as it was mentioned in [4]. Thus, here we present a completely different method.In the sequel we suppose that the sum∑i=tk is equal to zero for t>k; the symbol F≪E means F≤cE, where a positive constant c does not depend on arguments of the expressions F and E but can depend on the parameters p and q. The relationship F≈E means F≪E≪F.


## 2. Auxiliary Statements

For alli≥j≥1 we assume that Al,m(i,j)≡1 if l<m and Al,m(i,j)=∑kl=jiωl,kl∑kl-1=kliωl-1,kl-1⋯∑km=km+1iωm,km if n-1≥l≥m≥1. Moreover, for all i<j we suppose that Al,m(i,j)=0 if l,m≥1.In paper [4] the following lemma is proved.Lemma 1 (see [4]).
For alli,j,τ:1≤j≤τ≤i one has
(6)maxm-1≤r≤lAr,m(i,τ)Al,r+1(τ,j)≤Al,m(i,j)≤∑r=m-1l‍Ar,m(i,τ)Al,r+1(τ,j),
when n-1≥l≥m≥1.Lemma 2.
Let(Bi,j)i,j=1∞ be a nonnegative matrix whose elements do not decrease in the first index and do not increase in the second index. Let γ>0. Then for 1≤t,k≤∞ one has
(7)Δj+(∑i=jk‍Bi,j)γ≈(∑i=jk‍Bi,j)γ-1Δj+(∑i=jk‍Bi,j),(8)Δj-(∑j=ti‍Bi,j)γ≈(∑j=ti‍Bi,j)γ-1Δi-(∑j=ti‍Bi,j),
whereΔj+Ei,j=Ei,j-Ei,j+1 and Δi-Ei,j=Ei,j-Ei-1,j.Generally speaking the statement of Lemma2 is known, but for more complete presentation let us give its proof. We prove only relation (7). Relation (8) can be proved similarly.Proof.
Letγ≥1. Then
(9)Δj+(∑i=jk‍Bi,j)γ=(∑i=jk‍Bi,j)γ-(∑i=j+1k‍Bi,j+1)γ≤γ(∑i=jk‍Bi,j)γ-1Δj+(∑i=jk‍Bi,j),Δj+(∑i=jk‍Bi,j)γ≥(∑i=jk‍Bi,j)γ-(∑i=jk‍Bi,j)γ-1∑i=j+1k‍Bi,j+1=(∑i=jk‍Bi,j)γ-1Δj+(∑i=jk‍Bi,j).
If0<γ<1, then these inequalities hold in opposite direction. Therefore (7) holds for all γ>0. The proof of Lemma 2 is complete.Moreover, we need the following obvious relations:(10)(∑i=tk‍Bi,t)γ=∑j=tk‍Δj+(∑i=jk‍Bi,j)γ,(∑j=tk‍Bk,j)γ=∑i=tk‍Δi-(∑j=ti‍Bi,j)γ.We also use the following lemma.Lemma 3 (see [1]).
Let1<p<q<∞. Then the operator S0* is bounded from lp,v to lq,u if and only if
(11)B0(1)=(∑i=1∞‍(∑j=1i‍ujq)p/(p-q)(∑k=i∞‍vk-p′)p(q-1)/(p-q)vi-p′)(p-q)/pq<∞;
moreover, B0(1)≈∥S0*∥.


## 3. Main and Associated Results

Forr=0,1,…,n-1, we assume that
(12)Ar(n)=(∑i=1∞‍(∑j=i∞‍Ar,1q(j,i)ujq)q/(p-q)nnnnnn×(∑k=1i‍An-1,r+1p′(i,k)vk-p′)q(p-1)/(p-q)nnnnnn×Δi+(∑j=i∞‍Ar,1q(j,i)ujq))(p-q)/pq,Br(n)=(∑i=1∞‍(∑j=i∞‍Ar,1p′(j,i)vj-p′)p(q-1)/(p-q)nnnnnn×(∑k=1i‍An-1,r+1q(i,k)ukq)p/(p-q)nnnnnn×Δi+(∑j=i∞‍Ar,1p′(j,i)vj-p′))(p-q)/pq.Theorem 4.
Let1<q<p<∞ and n≥1. Then operator (3) is bounded from lp,v to lq,u if and only if A(n)=max0≤r≤n-1Ar(n)<∞. Moreover, for the norm of operator (3) from lp,v to lq,u the following relation ∥Sn-1∥≈A(n) holds.Theorem 5.
Let1<q<p<∞ and n≥1. Then operator (5) is bounded from lp,v to lq,u if and only if B(n)=max0≤r≤n-1Br(n)<∞. Moreover, for the norm of operator (5) from lp,v to lq,u the following relation ∥Sn-1*∥≈B(n) holds.For the proofs of Theorems4 and 5 we need to establish several statements.Assume that(13)u~iq=Δi-(∑j=1i‍An-1,r+1q(i,j)ujq),(Srf)i=∑j=1i‍Ar,1(i,j)fj,(Sr*g)j=∑i=j∞‍Ar,1(i,j)gi,i,j≥1,0≤r≤n-1.Consider the inequality(14)∥Sr*g∥q,u~≤C∥g∥p,v.Lemma 6.
Let1<p,q<∞. Then the inequality
(15)∥Srfu~q∥p′,v-1≤C∥f∥q′,u~q-1,f≥0,
is dual to (14) with respect to the linear form ∑i=1∞figiu~iq.Proof of Lemma6.
IfC>0 is the best constant in (14), then on the basis of principle of duality in lq we have
(16)C=supg≥0∥Sr*g∥q,u~∥g∥p,v=supg≥0supf≥0∑k=1∞fk(Sr*g)ku~kq∥f∥q′,u~q-1∥g∥p,v=supg≥0supf≥0∑i=1∞gi(Srfu~q)i∥g∥p,v∥f∥q′,u~q-1=supf≥0∥Srfu~q∥p′,v-1∥f∥q′,u~q-1.Lemma 7.
Let1<q<p<∞ and 0≤r≤n-1. If inequality (14) holds with the least constant C>0, then
(17)Br(n)≤C.Proof of Lemma7.
Suppose that inequality (14) holds. Then due to Lemma 6 inequality (15) also holds with the same best constant as in (14).
Lett≥x≥1. Introduce the test sequence f={fj}j=1∞ as follows: fj=0 for j>x and fj=(∑i=1ju~iq)(q-1)/(p-q)(∑k=jtAr,1p′(k,j)vk-p′)(p-1)(q-1)/(p-q) for 1≤j≤x.
If we substitute this sequence in the right side of (15), we have
(18)∥f∥q′,u~q-1=(∑j=1x‍(∑i=1j‍u~iq)q/(p-q)nnnnnn×(∑k=jt‍Ar,1p′(k,j)vk-p′)q(p-1)/(p-q)u~jq)1/q′.
Now we will work with the left side of (15) and the test sequence:
(19)∥Srfu~q∥p′,v-1p′=∑i=1∞‍vi-p′(∑j=1i‍Ar,1(i,j)fju~jq)(∑k=1i‍Ar,1(i,k)fku~kq)p′-1
(reduce the sum in the second brackets)
(20)≥∑i=1∞‍vi-p′∑j=1i‍Ar,1(i,j)fju~jq(∑k=1j‍Ar,1(i,k)fku~kq)p′-1
(change order of summation)
(21)=∑j=1∞‍fj∑i=j∞‍Ar,1(i,j)vi-p′(∑k=1j‍Ar,1(i,k)fku~kq)p′-1u~jq
(use nonincreasing of Ar,1(i,k) in the second argument)
(22)≥∑j=1∞‍fj∑i=j∞‍Ar,1p′(i,j)vi-p′(∑k=1j‍fku~kq)p′-1u~jq
(substitute the test sequence)
(23)≥∑j=1x‍(∑i=1j‍u~iq)(q-1)/(p-q)(∑k=jt‍Ar,1p′(k,j)vk-p′)(p-1)(q-1)/(p-q)×∑i=jt‍Ar,1p′(i,j)vi-p′×(∑k=1j‍(∑i=1k‍u~iq)(q-1)/(p-q)nnnnnnnn×(∑i=kt‍Ar,1p′(i,k)vi-p′)(p-1)(q-1)/(p-q)u~kq)p′-1u~jq
(join together the similar cofactors)
(24)≥∑j=1x‍(∑i=1j‍u~iq)(q-1)/(p-q)(∑k=jt‍Ar,1p′(k,j)vk-p′)q(p-1)/(p-q)×(∑k=1j‍(∑i=1k‍u~iq)(q-1)/(p-q)u~kq)p′-1u~jq
(use (8))
(25)≫∑j=1x‍(∑i=1j‍u~iq)(q-1)/(p-q)(∑k=jt‍Ar,1p′(k,j)vk-p′)q(p-1)/(p-q)×(∑k=1j‍Δk-(∑i=1k‍u~iq)(p-1)/(p-q))p′-1u~jq
(simplify)
(26)=∑j=1x‍(∑k=jt‍Ar,1p′(k,j)vk-p′)q(p-1)/(p-q)×(∑i=1j‍u~iq)q/(p-q)u~jq.
Therefore,
(27)∥Srfu~q∥p′,v-1p′≫(∑j=1x‍(∑k=jt‍Ar,1p′(k,j)vk-p′)q(p-1)/(p-q)nnnnnn×(∑i=1j‍u~iq)q/(p-q)u~jq)1/p′,
which, together with (15) and (18), gives
(28)C≫(∑j=1x‍(∑k=jt‍Ar,1p′(k,j)vk-p′)q(p-1)/(p-q)nnnnnn×(∑i=1j‍u~iq)q/(p-q)u~jq)(p-q)/pq,
for allt≥x≥1.
Now in (28) we approach t→∞ and use (8)
(29)C≫(∑j=1x‍(∑k=j∞‍Ar,1p′(k,j)vk-p′)q(p-1)/(p-q)nnnnn×Δj-(∑i=1j‍u~iq)p/(p-q))(p-q)/pq
(then we use the Abelian transformation)
(30)=(∑j=1x‍(∑i=1j‍u~iq)p/(p-q)Δj+(∑k=j∞‍Ar,1p′(k,j)vk-p′)q(p-1)/(p-q)nnnnn+(∑i=1x‍u~iq)p/(p-q)nnnnn×(∑k=x+1∞‍Ar,1p′(k,j)vk-p′)q(p-1)/(p-q))(p-q)/pq
(and use (7))
(31)≫(∑j=1x‍(∑i=1j‍u~iq)p/(p-q)(∑k=j∞‍Ar,1p′(k,j)vk-p′)p(q-1)/(p-q)nnnnn×Δj+(∑k=j∞‍Ar,1p′(k,j)vk-p′))(p-q)/pq.
Substituting u~iq and approaching x→∞, we obtain (17).Lemma 8.
Let1<q<∞ and 0≤r≤n-1. Then
(32)∥Sr*f∥q,u~≪∥Sn-1*f∥q,u,f≥0.Proof of Lemma8.
The expressionAn-1,1(i,j) can be transformed as follows:
(33)An-1,1(i,j)=∑kn-1=ji‍ωn-1,kn-1…∑kr+1=kr+2i‍ωr+1,kr+1×∑kr=kr+1i‍ωr,kr⋯∑k1=k2i‍ω1,k1=∑kn-1=ji‍ωn-1,kn-1⋯∑kr+2=kr+3i‍ωr+2,kr+2×∑kr+1=kr+2i‍ωr+1,kr+1Ar,1(i,kr+1)
(change of order of summation)
(34)=∑kr+1=ji‍Ar,1(i,kr+1)ωr+1,kr+1×∑kn-1=jkr+1‍ωn-1,kn-1⋯∑kr+2=kr+3kr+1‍ωr+2,kr+2=∑t=ji‍Ar,1(i,t)ωr+1,tAn-1,r+2(t,j).
Since Δt-An-1,r+1(t,j)=ωr+1,tAn-1,r+2(t,j), then
(35)An-1(i,j)=∑t=ji‍Ar,1(i,t)Δt-An-1,r+1(t,j).
Let f≥0; then
(36)∥Sn-1*f∥q,uq=∑j=1∞‍ujq(∑i=j∞‍An-1,1(i,j)fi)q-1×∑k=j∞‍An-1,1(k,j)fk≥∑j=1∞‍ujq∑k=j∞‍An-1,1(k,j)fk×(∑i=k∞‍An-1,1(i,j)fi)q-1
(use (35))
(37)=∑j=1∞‍ujq∑k=j∞‍fk∑t=jk‍Ar,1(k,t)Δt-An-1,r+1(t,j)×(∑i=k∞‍An-1,1(i,j)fi)q-1
(change order of summation)
(38)=∑j=1∞‍ujq∑t=j∞‍Δt-An-1,r+1(t,j)×∑k=t∞‍Ar,1(k,t)fk(∑i=k∞‍An-1,1(i,j)fi)q-1
(use An-1,1(i,j)≥Ar,1(i,t)An-1,r+1(t,j), which follows from (6))
(39)≥∑j=1∞‍ujq∑t=j∞‍(Δt-An-1,r+1(t,j))An-1,r+1q-1(t,j)×∑k=t∞‍Ar,1(k,t)fk(∑i=k∞‍Ar,1(i,t)fi)q-1
(use (7) and (8))
(40)≫∑j=1∞‍ujq∑t=j∞‍Δt-An-1,r+1q(t,j)∑k=t∞‍Δk+(∑i=k∞‍Ar,1(i,t)fi)q
(use the relation Δk+(∑i=k∞Ar,1(i,t)fi)=Ar,1(k,t)fk)
(41)=∑j=1∞‍ujq∑t=j∞‍Δt-An-1,r+1q(t,j)(∑i=t∞‍Ar,1(i,t)fi)q
(rearrange the cofactors)
(42)=∑t=1∞‍(∑i=t∞‍Ar,1(i,t)fi)q∑j=1t‍Δt-An-1,r+1q(t,j)ujq=
(take into account that An-1,r+1(t-1,t)=0)
(43)=∑t=1∞‍(∑i=t∞‍Ar,1(i,t)fi)qΔt-(∑j=1t‍An-1,r+1q(t,j)ujq)=∥Sr*f∥q,u~;
that is, we have (32).


## 4. Proofs of Theorems4 and 5

First we prove Theorem5.Proof of Theorem5.
Necessity. Suppose that operator (5) is bounded from lp,v to lq,u that equivalently means the validity of the following inequality:
(44)∥Sn-1*f∥q,u≤C∥f∥p,v,f≥0.
Then due to (32) inequality (14) holds for any r=0,1,…,n-1. Therefore, by Lemma 7 we have Br(n)≪C,  0≤r≤n-1, that means B(n)≪C, where C is the best constant in (44); that is, C=∥Sn-1*∥.
Sufficiency. will be proved by the induction method.
Forn=1 the operator Sn-1*=S0* is the Hardy operator. Thus by Lemma 3 if B0(1)<∞, then the operator S0* is bounded from lp,v to lq,u with the estimate ∥S0*∥≪B0(1).
Next we assume that forn=1,2,…,l, 1≤l<n-1, if B(n)<∞, then the operator Sn-1*, n=1,2,…,l, is bounded from lp,v to lq,u with the estimate ∥Sn-1*∥≪B(n).
Now we need to prove that forn=l+1 if B(l+1)=max0≤r≤lBr(l+1)<∞, then the operator Sl* is bounded from lp,v to lq,u with the estimate ∥Sl*∥≪B(l+1).
Letf≥0 and Tj={k∈ℤ:2-k≤(Sl*f)j}, j∈ℕ. Suppose that kj=infTj for Tj≠∅ and kj=∞ for Tj=∅. When kj<∞, then 2-kj≤(Sl*f)j<2-kj+1. Further, for exception of the trivial case we suppose that (Sl*f)1>0.
Letm1=1M1={j∈ℕ:kj=km1}. Assume that supM1+1=m2. If supM1<∞, then m2<∞ and supM1=maxM1=m2-1. It is obvious that m1<m2. Suppose that we have found m1<m2<⋯<ms<∞, s>1. Then we determine ms+1 as ms+1=supMs+1, where Ms={j∈ℕ:kj=kms}. By the definition 2-kms≤(Sl*f)j<2-kms+1, ms≤j≤ms+1-1, and kms<kms+1, s≥1.
Further, for convenience letkms=ns and ms+1-1=ms+1′; then
(45)2-ns≤(Sl*f)j<2-ns+1,ms≤j≤ms+1′,
andns<ns+1, s≥1.
Letℕ0={s∈ℕ:ms<∞}. Then from (45) it follows
(46)ℕ=⋃s∈ℕ0‍[ms,ms+1).
There are two possible cases:(1)ℕ0=ℕ and (2)ℕ0={1,2,…,s0}, s0>1.
Case(1)ℕ0=ℕ. Since ns<ns+1, s≥1, then -ns-1≤-ns+2+1. Therefore, using (45) and (6), we have
(47)2-ns-1=2-ns-2-ns-1≤2-ns-2-ns+2+1≤(Sl*f)ms+1′-(Sl*f)ms+2=∑i=ms+1′ms+2′‍Al,1(i,ms+1′)fi+∑i=ms+2∞‍[Al,1(i,ms+1′)-Al,1(i,ms+2)]fi≤∑i=ms+1′ms+2′‍Al,1(i,ms+1′)fi+∑i=0l-1‍Al,r+1(ms+2,ms+1′)×∑i=ms+2∞‍Ar,1(i,ms+2)fi.
Using (45), (46), and (47), we get
(48)∥Sl*f∥q,uq=∑s≥1‍∑i=msms+1′‍uiq(Sl*f)iq<∑s‍(2-ns+1)q∑i=msms+1′‍uiq=22q∑s‍(2-ns-1)q∑i=msms+1′‍ui≪∑s‍(∑i=ms+1′ms+2′‍Al,1(i,ms+1′)fi)q∑i=msms+1′‍uiq+∑r=0l-1‍∑s‍Al,r+1q(ms+2,ms+1′)×(∑i=ms+2∞‍Ar,1(i,ms+2)fi)q∑i=msms+1′‍uiq=Il+∑i=0l-1‍Ir.
First we estimate Il. Using Hölder’s inequality twice, we have
(49)Il≤∑s‍(∑i=ms+1′ms+2′‍Al,1p′(i,ms+1′)vi-p′)q/p′×∑i=msms+1′‍uiq(∑i=ms+1′ms+2′‍(vifi)p)q/p≤(∑s‍(∑i=ms+1′ms+2′‍Al,1p′(i,ms+1′)vi-p′)q(p-1)/(p-q)nnnnn×(∑i=msms+1′‍uiq)p/(p-q))(p-q)/p×(∑s‍∑i=ms+1′ms+2′‍(vifi)p)q/p≪B~l(p-q)/p∥f∥p,vq,
where
(50)B~l=∑s‍(∑i=ms+1′ms+2′‍Al,1p′(i,ms+1′)vi-p′)q(p-1)/(p-q)×(∑i=msms+1′‍uiq)p/(p-q).
Next we need the following obvious inequality:(51)Δi+(∑j=i∞‍Al,1p′(j,i)vj-p′)=Al,1p′(i,i)+∑j=i+1∞‍Δi+Al,1p′(j,i)vj-p′≥Al,1p′(i,i)+∑j=i+1ms+2′‍Δi+Al,1p′(j,i)vj-p′=Δi+(∑j=ims+2′‍Al,1p′(j,i)vj-p′),ms+1′≤i≤ms+2′.
Now, taking into account the above inequality (51) and using (10) and (7), we estimate B~l:
(52)B~l≪∑s‍∑i=ms+1′ms+2′‍(∑j=ims+2′‍Al,1p′(j,i)vj-p′)p(q-1)/(p-q)×(∑k=msms+1′‍ukq)p/(p-q)Δi+(∑j=ims+2′‍Al,1p′(j,i)vj-p′)≤∑s‍∑i=ms+1′ms+2′‍(∑j=i∞‍Al,1p′(j,i)vj-p′)p(q-1)/(p-q)×(∑k=1i‍ukq)p/(p-q)Δi+(∑j=i∞‍Al,1p′(j,i)vj-p′)≪∑i=1∞‍(∑j=i∞‍Al,1p′(j,i)vj-p′)p(q-1)/(p-q)×(∑k=1i‍ukq)p/(p-q)Δi+(∑j=i∞‍Al,1p′(j,i)vj-p′)=(Bl(l+1))pq/(p-q).
From (49) and (52) we have
(53)Il≪Blq(l+1)∥f∥p,vq≤Bq(l+1)∥f∥p,vq.
Now we estimateIr, r=0,1,…,l-1. Let ℕ1={k=ms+2:s∈ℕ}.
Assume thatΔkq=Al,r+1q(ms+2,ms+1′)∑i=msms+1′uiq if k=ms+2∈ℕ1 and Δkq=0 if k∉ℕ1. Then
(54)Ir=∑k=1∞‍Δkq(∑i=k∞‍Ar,1(i,k)fi)q=∥Sr*f∥q,Δq,r=0,1,…,l-1.
The operatorSr* is the operator Sn-1* for n=r+1 and 1≤r+1≤l. Therefore, by our assumption we have ∥Sr*∥≪B~(r+1)=max0≤t≤rB~t(r+1). Hence,
(55)Ir≪max0≤t≤rB~t(r+1),
where
(56)(B~t(r+1))pq/(p-q)=∑i=1∞‍(∑j=i∞‍At,1p′(j,i)vj-p′)p(q-1)/(p-q)×(∑k=1i‍Ar,t+1q(i,k)Δkq)p/(p-q)×Δi+(∑j=i∞‍At,1p′(j,i)vj-p′).
We estimate the expression ∑k=1iAr,t+1q(i,k)Δkq:
(57)∑k=1i‍Ar,t+1q(i,k)Δkq=∑ms+2≤i‍Ar,t+1q(i,ms+2)Al,r+1q(ms+2,ms+1′)∑j=msms+1′‍ujq
(use the left side of (6) and nonincreasing of Al,t+1q(i,j) in the second argument)
(58)≤∑ms+2≤i‍Al,t+1q(i,ms+1′)∑j=msms+1′‍ujq≤∑ms+2≤i‍∑j=msms+1′‍Al,t+1q(i,j)ujq≤∑j=1i‍Al,t+1q(i,j)ujq.
Substituting the obtained estimate in (56), we have B~t(r+1)≤Bt(l+1). Therefore, from (55) we get
(59)Ir≪max0≤t≤rBtq(l+1)∥f∥p,vq≤Bq(l+1)∥f∥p,vq,lr=0,1,…,l-1.
Then from (48), (53), and (59) we obtain
(60)∥Sl*f∥q,u≪B(l+1)∥f∥p,v,(61)∥Sl*∥≪B(l+1).
Now we turn to case(2)ℕ0={1,2,…,s0}, 1≤s0<∞. In this case we have ms0<∞ and ms0+1=∞. Here there are two possible cases: ns0<∞ and ns0=∞.
Below we suppose that∑s=kt=∑s=1t if k<0.
Letns0<∞. Then
(62)∥Sl*f∥q,uq=∑s=1s0‍∑j=msms+1‍(Sl*f)jquj=∑s=1s0-2‍∑j=msms+1′‍(Sl*f)jqujq+∑j=ms0-1ms0′‍(Sl*f)jqujq+∑j=ms0∞‍(Sl*f)jqujq=J1+J2+J3.
If J1≠0, then s0>2, and as in the case ℕ0=ℕ using (47), we estimate J1 and as a result we get
(63)J1≪Bq(l+1)∥f∥p,vq.
Using (45) and Hölder’s inequality, we estimate the value J2:
(64)J2<2q(2-ns0-1)q∑j=ms0-1ms0′‍ujq≪(∑i=ms0′∞‍Al,1(i,ms0′)fi)q∑j=ms0-1ms0′‍ujq≤(∑i=ms0′∞‍Al,1p′(i,ms0′)vi-p′)q/p′×∑j=ms0-1ms0′‍ujq(∑i=ms0′∞‍|vifi|p)q/p≤[(∑i=ms0′∞‍Al,1p′(i,ms0′)vi-p′)q(p-1)/(p-q)nnnnnn×(∑j=ms0-1ms0′‍ujq)p/(p-q)](p-q)/p∥f∥p,vq.
Using (7) and (10), we estimate the expression in the square brackets:
(65)(∑i=ms0′∞‍Al,1p′(i,ms0′)vi-p′)q(p-1)/(p-q)(∑j=ms0-1ms0′‍ujq)p/(p-q)≪∑i=ms0′∞‍(∑j=i∞‍Al,1(j,i)vj-p′)p(q-1)/(p-q)×Δi+(∑j=i∞‍Al,1(j,i)vj-p′)(∑j=ms0-1ms0′‍ujq)p/(p-q)≤∑i=ms0′∞‍(∑j=i∞‍Al,1(j,i)vj-p′)p(q-1)/(p-q)×(∑j=1i‍ujq)p/(p-q)Δi+(∑j=i∞‍Al,1(j,i)vj-p′)≤(Bl(l+1))pq/(p-q).
Therefore,
(66)J2≪(Bl(l+1))q∥f∥p,vq≤Bq(l+1)∥f∥p,vq.
Now we estimate J3. Since 2-ns0≤(Sl*f)j<2-ns0+1 for all j≥ms0, then (Sl*f)j≤2(Sl*f)t for all j,t≥ms0. Hence,
(67)J3≤supt≥ms0∑j=ms0t‍(Sl*f)jqujq≤2qsupt≥ms0(∑i=t∞‍An-1,1(i,t)fi)q∑j=ms0t‍ujq.
Further, as for the estimate of J2 by Hölder’s inequality, we have
(68)J3≪supt≥ms0[(∑i=t∞‍Al,1q(i,t)vi-p′)q(p-1)/(p-q)nnnnnnnnn×(∑j=ms0t‍ujq)p/(p-q)](p-q)/p∥f∥p,vq≪Blq(l+1)∥f∥p,vq≤Bq(l+1)∥f∥p,vq.
From (62), (63), (66), and (68) we have (60) and (61). If ns0=∞ that means kms=∞, then Tj=∅ for j≥ms0; that is, (Sl*f)j=0 for j≥ms0. By the assumption (Sl*f)1>0, therefore, s0>1. Then m2<∞ and s0≥2. Hence,
(69)∥Sl*f∥q,uq=∑s=1s0-2‍∑j=msms+1′‍(Sl*f)jqujq+∑j=ms0-1ms0′‍(Sn*f)jqujq=J1+J2.
This, together with estimates (63) and (66), gives (60) and (61). Thus, if B(l+1)<∞, then the operator Sl* is bounded from lp,v to lq,u and estimate (61) holds. Consequently, for any n≥1 from B(n)<∞ we have that the operator Sn-1* is bounded from lp,v to lq,u and the estimate ∥Sn-1*∥≪B(n) holds. The last estimate and the estimate ∥Sn-1*∥≫B(n) obtained in the necessity part yield ∥Sn-1*∥≈B(n).The proof of Theorem4 follows from Theorem 5, because boundedness of the operator Sn-1 from lp,v to lq,u is equivalent to boundedness of the operator Sn-1* from lq′,u-1 to lp′,v-1, and the respective norms of the operators Sn-1* and Sn-1 coincide.



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*Source: 121767-2013-06-06.xml*