# Properties of a Generalized Class of Weights Satisfying Reverse Hölder’s Inequality

Authors: S. H. Saker; S. S. Rabie; R. P. Agarwal
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we will prove some fundamental properties of the discrete power mean operatorMpun=1/n∑k=1nupk1/p,forn∈I⊆ℤ+, of order p, where u is a nonnegative discrete weight defined on I⊆ℤ+ the set of the nonnegative integers. We also establish some lower and upper bounds of the composition of different operators with different powers. Next, we will study the structure of the generalized discrete class BpqB of weights that satisfy the reverse Hölder inequalityMqu≤BMpu, for positive real numbers p, q, and B such that 0<p<q and B>1. For applications, we will prove some self-improving properties of weights from BpqB and derive the self improving properties of the discrete Gehring weights as a special case. The paper ends by a conjecture with an illustrative sharp example.


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## Body

## 1. Introduction

In [1], Muckenhoupt introduced a full characterization of the Ap−class of weights in connection with the boundedness of the Hardy-Littlewood maximal operator in the space Lwpℝ+ with a weight w. Another important class of weights, the Gehring class Gq, for 1<q<∞, was introduced by Gehring [2, 3] in connection with local integrability properties of the gradient of quasiconformal mappings. Due to the importance of these two classes in mathematical and harmonic analysis, the structure of them has been studied by several authors, and various results regarding the relation between them and their applications have been established. We refer the reader to the papers [1–23] and the references cited therein.In recent years, the study of the discrete analogues in harmonic analysis becomes an active field of research. For example, the study of regularity and boundedness of discrete operator onlp analogues for Lp−regularity, higher summability, and structure of discrete Muckenhoupt and Gehring weights has been considered by some authors, and we refer the reader to the papers [24–34] and the references they are cited.We confine ourselves, in this paper in proving some new fundamental properties of a generalized discrete space of weights that satisfy reverse Hölder’s inequality and prove some self-improving properties. As special cases, we will derive the self-improving properties of the discrete Gehring weights.In the following, for the sake of completeness, we present the background and the basic definitions that will be used in this paper. Throughout this paper,ℤ+ stands for the set of nonnegative integers, i.e., ℤ+=1,2,⋯. By an interval J, we mean a finite subset of ℤ+ consisting of consecutive integers, i.e., J=a,a+1,⋯,a+n, a,n∈ℤ+, and ∣J∣ stands for its cardinality. We assume that 1<p<∞ and fix an interval I⊂ℤ+ of the form I=1,2,⋯,N, where N a nonnegative integer (or 1,N⊂ℤ+). A discrete weight v defined on I is a sequence of nonnegative real numbers.A discrete weightu defined on I⊆ℤ+ belongs to the discrete Muckenhoupt class A2A for p>1 and A>1, if the inequality
(1)1J∑k∈Juk≤Auk,forallk∈Jholds for every subinterval J⊂I, where J is the cardinality of the set J. A discrete weight u defined on I⊂ℤ+ is said to belong to the discrete Muckenhoupt class ApA for p>1 and A>1 if the inequality
(2)1J∑k∈Juk1J∑k∈Ju−1p−1kp−1≤Aholds for every subinterval interval J⊂I. For a given exponent p>1, we define the Ap−norm by the following quantity
(3)Apu≔supJ⊂I1J∑k∈Juk1I∑k∈Ju−1p−1kp−1,where the supremum is taken over all intervals J⊂I. Note that by Hölder’s inequality Apu≥1 for all 1<p<∞ and the following inclusions are true:
(4)if1<p≤q<∞,thenA1⊂Ap⊂AqandAqu≤Apu.For a given exponentq>1 and a constant K>1, a discrete nonnegative weight u defined on I belongs to the discrete Gehring class GqK (or satisfies the reverse Hölder inequality) if for every subinterval J⊆I, we have
(5)1Jk∈Juqk1/q≤K1J∑k∈Juk.For a given exponentq>1, we define the Gq−norm by
(6)Gqu≔supJ⊂I1J∑k∈Juk−11J∑k∈Juqk1qqq−1,where the supremum is taken over all intervals J⊆I and represents the best constant for which the Gq−condition holds true independently on the interval J⊆I. Note that by Hölder’s inequality Gqu≥1 for all 1<q<∞ and the following inclusion is true:
(7)if1<p≤q<∞,thenGq⊂Gpand1≤Gpu≤Gqu.By the generalized power mean operatorMqu of order q≠0 and nonnegative weight u defined on I, we mean an operator of the form
(8)Mqu≔1J∑k∈Juqk1/q,forJ⊂I.In [27], Böttcher and Seybold considered the operator (8) and proved that if
(9)MpuM−qu≤C,where p>1 and 1/p+1/q=1, then there exists a constant δ>0 and C1<∞ depending only on p and u such that
(10)Mp1+εu≤C1Mpu,for all ε∈0,δ] and all J of the form J=2r with r∈ℕ (the set of natural numbers). In [34], the authors proved that M−1u≤M1u,Mqu≥M1u, for all q≥1,Mqu≤M1u, for all q<1 and if p≤q, then
(11)Mpu≤Mqu,for any nonnegative weight u defined on I. In the present paper, we consider the class BpqB of all nonnegative weights u that satisfy the reverse Hölder inequality
(12)Mqu≤BMpu,forJ⊂I,where the constant B>1 is independent of p,q, and J and q>p. The smallest constant B independent on the interval J and satisfies the inequality (12) is called the Bpq−norm which is given by
(13)Bpqu≔supJ⊂IMpu−1pMqu1q,forJ⊂I.We say thatu is a Bpq−weight if its Bpq−norm is finite, i.e.,
(14)v∈Bpq⇐Bpqu<+∞.When we fix a constantC>1, the triple of real numbers p,q,C defines the Bpq−discrete class:
(15)u∈BpqC⇐Bpqu≤C,and we will refer to C as the Bpq−constant of the class. It is immediate to observe that the classes Ap and Gq are special cases of the discrete class Bpq of weights as follows:
(16)Ap≔B1/p−11,andGq≔B1q.In this paper, we aim to study the structure of the general classBpq and use the new properties to prove some self-improving properties. The paper is organized as follows: In Section 2, we state and prove some basic lemmas concerning the bounds of the generalized power mean operator Mp. In Section 3, we will establish some lower and upper bounds of the composition of operators by using two special functions ρp and ρq (will be defined later) and prove some inclusion properties. For example, we prove that if u∈BpqB, then Mqu∈BpδB1 with exact values of δ and B1. In Section 4, we present some applications of the main results and prove the self-improving property of a monotone weights from Bpq, i.e., we will prove that if u∈BpqB, then u∈BpλB1 with exact values of λ and B1. For illustrations, we will derive the self improving property of the discrete Gehring weights as special cases. The paper ends by a conjecture with the self-improving of the Muckenhoupt weights with an illustrative example.


## 2. Basic Lemmas

In this section, we state and prove the basic lemmas and establish some properties of the power mean operators that will be used to prove the main results later. We will assume thatI≪1,2,⋯,N is a fixed finite subset of ℤ+, and we recall the power mean operator Mqu that we will consider in this paper is given by
(17)Mqun=1n∑k=1nuqk1/q,foralln∈I,for any nonnegative weight u:I⟶ℝ+ and q∈ℝ\0 and by Mqpuk, we mean that Mqukp. For the sake of conventions, we assume that 0·∞=0 and 0/0=0 and ∑k=abyk=0, whenever a>b, and
(18)Δ∑k=1n−1yk=yn,∑k=1n−1Δyk=yn−y1.The product rule in the discrete form is given by(19)Δukvk=ukΔvk+vk+1Δuk,where Δuk=uk+1−uk. The summation by parts now is given by
(20)∑k=1nΔukvk+1=ukvkk=1n+1−∑k=1nukΔvk.Lemma 1.
Letp<q and p.q≠0, and u:I⟶ℝ+ is a nonnegative weight. Then, the following
(21)∑k=1nΔk−1Mqpuk−1=nMqpun,(22)∑k=1nΔk−1Mpquk−1=nMpqunhold for all n∈I.Proof.
By applying the second relation in (18) with uk=k−1Mqpuk−1, we have
(23)∑k=1nΔk−1Mqpuk−1=∑k=1nkMqpuk−k−1Mqpuk−1=nMqpun,which is the desired equations (21). Similarly, by applying the second relation in (18) with uk=k−1Mpquk−1, we have
(24)∑k=1nΔk−1Mpquk−1=∑k=1nkMpquk−k−1Mpquk−1=nMpqun,which is the equality (22). The proof is complete.Lemma 2.
Assume thatu:I→ℝ+ be any nonnegative weight and q∈ℝ\0. Then, following properties hold:
(1)
Ifu is nonincreasing, then Mqu is nonincreasing and Mqun≥un, for all n∈I(2)
Ifu is nondecreasing, then Mqu is nondecreasing and Mqun≤un, for all n∈IProof.
(1)). From the definition ofMqu and the fact that u is nonincreasing, we get for q=1 that
(25)M1un=1n∑k=1nuk≥1n∑k=1nun=un.
For the general case whenq≠1, we have also for all n∈I that
(26)Mqun=1n∑k=1nuqk1/q≥1n∑k=1nuqn1/q=un.
From this inequality, we get that(27)nuqn≤∑k=1nuqk,foralln∈I.
Now, by using (27) and the fact that u is nonincreasing, we obtain that
(28)ΔMqun=1n+11/q∑k=1n+1uqk1/q−1n1/q∑k=1nuqk1/q=n∑k=1n+1uqk1/q−n+1∑k=1nuqk1/qnn+11/q=nuqn+1+n∑k=1nuqk1/q−n∑k=1nuqk+∑k=1nuqk1/qnn+11/q≤nuqn+n∑k=1nuqk1/q−n∑k=1nuqk+∑k=1nuqk1/qnn+11/q≤∑k=1nuqk+n∑k=1nuqk1/q−n∑k=1nuqk+∑k=1nuqk1/qnn+11/q=0,and thus Mqun is nonincreasing.
2). From the definition ofMqun and the fact that un is nondecreasing, we have for q=1 that
(29)M1un=1n∑k=1nuk≤1n∑k=1nun=un.
For the general case whenq≠1, we have also for all n∈I that
(30)Mqun=1n∑k=1nuqk1/q≤1n∑k=1nuqn1/q=un.
From this inequality, we see that(31)nuqn≥∑k=1nuqk,foralln∈I.
Then, by using inequality (31) and the fact that u is nondecreasing and proceeding as in the first case, we obtain that
(32)ΔMqun=1n+1∑k=1n+1uqk1/q−1n∑k=1nuqk1/q=n∑k=1n+1uqk1/q−n+1∑k=1nuqk1/qnn+11/q.
We proceed as in the proof of the nonincreasing case to get thatΔMqun≥0, and thus Mqun is nondecreasing. The proof is complete.The following lemma will play an important rule in proving the main results.Lemma 3.
Letα and β be positive numbers and g:I→ℝ+ be any nonnegative weight such that
(33)α1k∑τ=1kgτ≤gk≤β1k∑τ=1kgτ,forallk∈I.
Then, for everyr, s∈0,N] such that r<s, we have that
(34)srα−11r∑k=1rgk≤1s∑k=1sgk≤srβ−11r∑k=1rgk.Proof.
The left-side of inequality (33) writes
(35)1k∑τ=1kgτ≤1αgk,and by multiplying both sides by k−αand summating from k=m>1 to s, we have
(36)∑k=msk−α−1∑τ=1kgτ≤∑k=ms1αk−αgk.
The left-side of inequality (36) can be written in the form
(37)∑k=msk−α−1∑τ=1kgτ=∑k=1sk−α−1∑τ=1kgτ−∑k=1m−1k−α−1∑τ=1kgτ,≥∑k=1sk−α−1∑τ=1kgτ−∑k=1m−1k−α−1∑τ=1k+1gτ.
By applying Fubini’s Theorem on the right-hand side, we have that(38)∑k=msk−α−1∑τ=1kgτ≥∑k=1sgk∑τ=ksτ−α−1−∑k=1m−1gk∑τ=k+1m−1τ−α−1.
By using the inequality,(39)γxγ−1x−y≥xγ−yγ≥γyγ−1x−y,forx≥y>0,γ>1orγ<0,with γ=−α<0, and we have
(40)∑τ=ksτ−α−1≥∑τ=ks−1αΔτ−α=k−αα−s+1−αα≥k−αα−s−αα,∑τ=k+1m−1τ−α−1≤∑τ=k+1m−1−1αΔτ−1−α=k−αα−m−1−αα.
Then, (38) becomes
(41)∑k=msk−α−1∑τ=1kgτ≥∑k=1sk−αα−s−ααgk−∑k=1m−1k−αα−m−1−ααgk=∑k=msk−ααgk−s−αα∑k=1sgk+m−1−αα∑k=1m−1gk.
By substituting (41) into (36), we have
(42)∑k=ms1αk−αgk≥∑k=msk−α−1∑τ=1kgτ≥∑k=msk−ααgk−s−αα∑k=1sgk+m−1−αα∑k=1m−1gk,which implies that
(43)−s−αα∑k=1sgk+m−1−αα∑k=1m−1gk≤0,which is equivalent to
(44)1/r∑k=1rgk1/s∑k=1sgk≤srrsα=rsα−1,where r=m−1>0. This proves the left-side of inequality (34). Now, the right-side of inequality (33) writes
(45)1k∑τ=1kgτ≥1βgk,and by multiplying both sides by k−β, and summating from k=m>1 to s, we have
(46)∑k=msk−β−1∑τ=1kgτ≥∑k=ms1βk−βgk.
The left-side of inequality (46) can be written in the form
(47)∑k=msk−β−1∑τ=1kgτ=∑k=1sk−β−1∑τ=1kgτ−∑k=1m−1k−β−1∑τ=1kgτ≤∑k=1sk−β−1∑τ=1k+1gτ−∑k=1m−1k−β−1∑τ=1kgτ.
This implies by applying Fubini’s Theorem on the left-hand side that(48)∑k=msk−β−1∑τ=1kgτ≤∑k=1sgk∑τ=k+1sτ−β−1−∑k=1m−1gk∑τ=km−1τ−β−1.
By using the inequality,(49)γxγ−1x−y≥xγ−yγ≥γyγ−1x−y,forx≥y>0,γ>1orγ<0,with γ=−β<0, and we have
(50)∑τ=k+1sτ−β−1≤∑τ=k+1s−1βΔτ−1−β=k−ββ−s−ββ,∑τ=km−1τ−β−1≥∑τ=km−1−1βΔτ−β=k−ββ−m−ββ≥k−ββ−m−1−ββ.
Then, (48) becomes
(51)∑k=msk−β−1∑τ=1kgτ≤∑k=1sk−ββ−s−ββgk−∑k=1m−1k−ββ−m−1−ββgk=∑k=msk−ββgk−s−ββ∑k=1sgk+m−1−ββ∑k=1m−1gk.
By substituting (51) into (46), we have
(52)∑k=ms1βk−βgk≤∑k=msk−β−1∑τ=1kgτ≤∑k=msk−ββgk−s−ββ∑k=1sgk+m−1−ββ∑k=1m−1gk,which implies that
(53)−s−ββ∑k=1sgk+m−1−ββ∑k=1m−1gk≥0,which is equivalent to
(54)1/r∑k=1rgk1/s∑k=1sgk≥srrsβ=rsβ−1,where r=m−1>0. This proves the right-side of inequality (34). The proof is complete.


## 3. Fundamental Properties of Power Mean Operators

In this section, we will prove some fundamental properties of the generalized power mean operatorMpun which is given by
(55)Mpun=1n∑k=1nupk1/p,foralln∈I,where p is assumed to be positive for the rest of the paper. In order to prove the main results, we will use the properties of the function
(56)ρpλ=1−pλ1/p,p>0of the variable λ∈−∞,0∪p,∞). It is clear that the function ρpλ is continuous and increases from 1 to +∞ on −∞,0 and from 0 to 1 on p,∞ and for λ≠p, we have that
(57)ρpλρpp−λ=1−pλ1/p1−pp−λ1/p=λ−pλ1/pλλ−p1/p=1.To understand the importance of the functionρpλ, we consider the sequence u0n=n−1/λ. Then, we have
(58)Mpu0n=1n∑k=1nk−p/λ1/p.We consider the different cases of the power−p/λ+1. First, assume that 0<−p/λ+1<1, and by employing the inequality
(59)γxγ−1x−y≤xγ−yγ≤γyγ−1x−y,forx≥y>0,0≤γ≤1,with γ=−p/λ+1, we have
(60)Δk−11−p/λ=k1−p/λ−k−11−p/λ≥1−p/λk−p/λ.Then, we have(61)Mpu0n=1n∑k=1nk−p/λ1/p≤1n∑k=1n11−p/λΔk−11−p/λ1/p=1n11−p/λn1−p/λ1/p=n−1/λ1−p/λ1/p=u0nρpλ,foralln∈I.Next, we consider the case when−p/λ+1>1, and by employing inequality (39) with γ=1−p/λ>1, we have
(62)Δk−11−p/λ=k1−p/λ−k−11−pλ≤1−pλk−pλ.Then, we have that(63)Mpu0n=1n∑k=1nk−p/λ1/p≥1n∑k=1n11−p/λΔk−11−p/λ1/p=11−p/λ1/pn1−p/λn1/p=u0nρpλ,foralln∈I.The meaning ofρpλ now arises from the fact that the sequence u0n=n−1/λ satisfies the equivalence between Mpu0n and the fraction u0n/ρpλ, for all n∈I. Let 0<p<q and define the function Sp,qλ by
(64)Sp,qλ≔ρpλρqλ,forλ∈−∞,0∪q,∞.The functionCp,qλ is continuous and increases on the interval −∞,0 and decreases on the interval q,∞ with Cp,qβ,∞=1,∞. Therefore, for any B>1, the equation
(65)Sp,qλ=1−p/λ1/p1−q/λ1/q=Bhas exactly two roots: a positive root λ+∈q,∞ and a negative root λ−∈−∞,0. The nonnegative weight v:I⟶ℝ+ is said to be belong to BpqB if v satisfies the reverse Hölder inequality
(66)Mqvn≤BMpvn,foralln∈I;that is,
(67)1n∑k=1nvqk1/q≤B1n∑k=1nvpk1/p,for all n∈I, where the constant B>1 is independent of p,q, and q>p. Now, we are ready to state and prove the main properties of the operator (55) and the composition of different operators with different powers.Theorem 4.
Let0<p<q, and u:I⟶ℝ+ be any nonnegative, monotone weight. If u∈BpqBforB>1, then
(68)ρpλ+≤MqunMpMqun≤ρpλ−,foralln∈I,where λ+ and λ− are the roots of (65).Proof.
By applying the product rule (19) on the term Δk−1Mqpuk−1 with u=k−1 and v=Mqpuk−1, we obtain that
(69)Δk−1Mqpuk−1=Δk−11k−1∑s=1k−1uqsp/q=1k∑s=1kuqsp/q+k−1Δ1k−1∑s=1k−1uqsp/q.
Now, we find the estimate of the second term in (69) and consider two cases of the behavior of the monotone weight u. First, we assume that u is nondecreasing. Then, by Lemma 2, we have that
(70)Mquk=1k∑s=1kuqs1/qis also nondecreasing and by applying the elementary inequality (59), for γ=p/q<1, we obtain
(71)k−1Δ∑s=1k−1uqsk−1p/q=k−1∑s=1kuqskp/q−∑s=1k−1uqsk−1p/q≥pqk−11k∑s=1kuqsp/q−1k−1∑s=1kuqs−k∑s=1k−1uqskk−1=pqk−11k∑s=1kuqsp/q−1kuqk−∑s=1kuqskk−1=pq1k∑s=1kuqsp/q−1uqk−pq1k∑s=1kuqsp/q.
By combining (71) and (69), we obtain
(72)Δk−1Mqpuk−1≥q−pq∑s=1kuqskp/q+pq∑s=1kuqskp/q−1uqk.
Next, we assume thatu is nonincreasing. Then, by Lemma 2, we see that Mquk is nonincreasing and by employing the inequality (59) again, we have that
(73)k−1Δ∑s=1k−1uqsk−1p/q=−k−1∑s=1k−1uqsk−1p/q−∑s=1kuqskp/q≥−pqk−1∑s=1kuqskp/q−1k∑s=1k−1uqs−k−1∑s=1kuqskk−1=−pqk−11k∑s=1kuqsp/q−1∑s=1kuqskk−1−uqkk−1=−pq1k∑s=1kuqsp/q+pq1k∑s=1kuqsp/q−1uqk.
By combining (69) and (73), we again obtain the inequality (72). Now, by summing (72) from 1 to n, and applying (21), we obtain
(74)Mqunp≥1−pq1n∑k=1n1k∑s=1kuqsp/q+pq1n∑k=1n1k∑s=1kuqsp/q−1uqk.
From the definition ofMqu, we see that the first term in (74) is given by
(75)1n∑k=1n1k∑s=1kuqsp/q=MpMqunp.
Now, we simplify the term(76)1n∑k=1n1k∑s=1kuqsp/q−1uqk.
By applying reverse Hölder’s inequality forp/q<1 and p/p−q, we obtain that
(77)1n∑k=1n1k∑s=1kuqsp/q−1uqk≥1n∑k=1n1k∑s=1kuqsp/qp−q/p1n∑k=1nupkq/p=MpMqunp−qMpunq.
By substituting (75) and (77) into (74), dividing by pMpMqunp and then applying (66), we obtain
(78)1pMqunpMpMqunp≥1p−1q+1qMpunqMpMqunq≥1p−1q+B−q1qMqunqMpMqunq.
By setting(79)λ≔pMpMqunpMpMqunp−Mqunp,we see that the inequality (78) can be written in the form
(80)B−q1−MpMqunp−MqunpMpMqunpq/p=B−qMqunpMpMqunpq/p≤1−qp+qpMqunpMpMqunp≤1−qp1−MqunpMpMqunp=1−qpMpMqunp−MqunpMpMqunp.
This inequality can be written now as(81)1−pλ1/p≤B1−qλ1/q,or equivalently
(82)Sp,qλ=1−p/λ1/p1−q/λ1/q≤B,forallp<q.
This means thatλ∈−∞,λ−∪λ+,+∞). The properties of ρp imply that
(83)ρpλ+≤ρpλ≤ρpλ−,and since
(84)ρpλ=1−pλ1/p=1−MpMqunp−MqunpMpMqunp1/p=1−1+MqunpMpMqunp1/p=MqunMpMqun,we obtain that
(85)ρpλ+≤MqunMpMqun≤ρpλ−,which is the desired inequality (68). The proof is complete.Theorem 5.
Let0<p<q, and u:I→ℝ+ be any nonnegative, monotone weight. If u∈BpqB for B>1, then
(86)ρqλ+≤MpunMqMpun≤ρqλ−,foralln∈I,where λ+ and λ− are the roots of (65).Proof.
By applying the product rule (19) on the term Δk−1Mpquk−1, with u=k−1 and v=Mpquk−1, we obtain that
(87)Δk−1Mpquk−1=Δk−11k−1∑s=1k−1upsq/p=1k∑s=1kupsq/p+k−1Δ1k−1∑s=1k−1upsq/p.
First, we assume thatu is nonincreasing. Then, by Lemma 2, we have then Mpuk that is nonincreasing. By employing inequality (39) with γ=q/p>1, we obtain
(88)k−1Δ∑s=1k−1upsk−1q/p=−k−1∑s=1k−1upsk−1q/p−∑s=1kupskq/p≤−qp1k∑s=1kupsq/p−1∑s=1kups−kupsk=−qp1k∑s=1kupsq/p+qp1k∑s=1kupsq/p−1ups.
By substituting (88) into (87), we obtain
(89)Δk−1Mpquk−1≤p−qp∑s=1kupskq/p+qp∑s=1kupskq/p−1upk.
Next, we assume thatu is nondecreasing. Then, by Lemma 2, we have that Mpuk is nondecreasing and by applying the inequality (39) with γ=q/p>1, we get
(90)k−1Δ∑s=1k−1upsk−1q/p=k−1∑s=1kupskq/p−∑s=1k−1upsk−1q/p≤qpk−11k∑s=1kupsq/p−1−∑s=1kups+kupskk−1=−qp1k∑s=1kupsq/p+qp1k∑s=1kupsq/p−1upk.
Now, by combining (87) and (90), we obtain again (89). Now, by summing (89) from 1 to n and applying (22), we obtain that
(91)Mpunq≤1−qp1n∑k=1n1k∑s=1kupsq/p+qp1n∑k=1n1k∑s=1kupsq/p−1upk.
The first term in (91) is given by
(92)1n∑k=1n1k∑s=1kupsq/p=MqMpunq.
Now, we simplify the second term(93)1n∑k=1n1k∑s=1kupsq/p−1upk,in (91), by applying Hölder’s inequality for q/p>1 and q/q−p, to obtain
(94)1n∑k=1n1k∑s=1kupsq/p−1upk≤1n∑k=1n1k∑s=1kupsq/pq−p/q1n∑k=1nuqkp/q=MqMpunq−pMqunp.
By substituting (92) and (94) into (91), dividing by qMqMpunq and applying (66), we obtain
(95)MpunqqMqMpunq≤1q−1p+1pMpunpMqMpunp≤1q−1p+BppMpunpMqMpunp.
Inequality (95) now takes the form
(96)B1−MqMpunq−MpunqMqMpunq1/qp=BMpunqMqMpunq1/qp=BpMpunpMqMpunp≥1−pq+pqMpunqMqMpunq=1−pq1−MpunqMqMpunq=1−pqMqMpunq−MpunqMqMpunq.
By setting(97)λ≔qMqMpunqMqMpunq−Mpunq,we see that inequality (96) becomes
(98)1−pλ1/p≤B1−qλ1/q,or equivalently,
(99)Sp,qλ=1−p/λ1/p1−q/λ1/q≤B.
This means thatλ∈−∞,λ−∪λ+,+∞). The properties of ρq implies that
(100)ρqλ+≤ρqλ≤ρqλ−,and since
(101)ρqλ=1−qλ1/q=1−MqMpunq−MpunqMqMpunq1/q=1−1+MpunqMqMpunq1/q=MpunMqMpun,we obtain that
(102)ρqλ+≤MpunMqMpun≤ρqλ−,which is the required inequality (86). The proof is complete.The assumptions and the conclusions of Theorems4 and 5 will be used in proving the following theorems.Theorem 6.
Assume that the conditions in Theorems4 and 5 hold. Then, the compositions
(103)n1/λ−MqMpun,andn1/λ−MpMqunare nonincreasing for all n∈I, and the compositions
(104)n1/λ+MpMqun,andn1/λ+MqMpunare nondecreasing for all n∈I.Proof.
Raise the inequality (68) to the power p, we get that
(105)1−pλ+≤1/k∑τ=1kuqτp/q1/s∑k=1s1/k∑τ=1kuqτp/q≤1−pλ−.
Setting(106)gk=Mquk=1k∑s=1kuqsp/q,we see that gk satisfies the inequality (34) in Lemma 3 with
(107)α=1−pλ+andβ=1−pλ−.
Therefore, we see that(108)srα−1≤1/s∑k=1s1/k∑τ=1kuqτp/q1/r∑k=1r1/k∑s=1kuqτp/q≤srβ−1,and so we have that
(109)sr−pλ+≤1/s∑k=1s1/k∑τ=1kuqτp/q1/r∑k=1r1/k∑s=1kuqτp/q≤sr−pλ−.
If raised the last inequality to the power−1/p, we see that this inequality is equivalent to the monotonicity of the compositions
(110)n1/λ±MpMqun.
Analogously, the inequality (86) and the same proof imply the monotonicity of the compositions
(111)n1/λ±MqMpun.
The proof is complete.Theorem 7.
Let0<p<q and u be any nonnegative weight, and λ+ and λ− are the roots of equation (65).
(i)
Ifu∈BpqB for B>1 and λ<λ+ and λ≠0, then(112)Mpu∈Bqλρqλ−ρλλ+,andMqu∈Bpλρpλ−ρλλ+.(ii)
Ifu∈BpqB for B>1 and λ>λ−, and λ≠0, then(113)Mpu∈Bλqρλλ−ρqλ+,andMqu∈Bλpρλλ−ρpλ+.Proof.
(i). Sinceλ is either positive or negative, we will discuss the two cases:
(1) Assume thatλ>0. By raising (86) to the power λ, we obtain for m<n that
(114)m−λ/λ−ρqλλ+m1/λ−MqMpumλ≤Mpumλ≤m−λ/λ+ρqλλ−m1/λ+MqMpumλ.
By using the monotonicity of(115)n1/λ−MqMpun,andn1/λ+MqMpun,
(see Theorem6), we have that
(116)m−λ/λ−ρqλλ+n1/λ−MqMpunλ≤Mpumλ≤m−λ/λ+ρqλλ−n1/λ+MqMpunλ.
Sinceλ<λ+, by summing (116) from m=1 to n, dividing by n, and raising it to the power 1/λ>0, we get
(117)ρqλλ+n1/λ−MqMpunλn∑m=1nm−λ/λ−1/λ≤1n∑m=1nMpumλ1/λ≤ρqλλ−n1/λ+MqMpunλn∑m=1nm−λ/λ+1/λ.
Since−1<−λ/λ+<0, we have that 0<−λ/λ++1<1. By applying (59) with γ=−λ/λ++1<1, we have
(118)Δm−1−λ/λ++1=m−λ/λ++1−m−1−λ/λ++1≥−λ/λ++1m−λ/λ+,and then
(119)∑m=1nm−λ/λ+≤1−λ/λ++1∑m=1nΔm−1−λ/λ++1=n−λ/λ++1−λ/λ++1.
So, we have(120)ρqλλ−n1/λ+MqMpunλn∑m=1nm−λ/λ+1/λ≤ρqλλ−n1/λ+MqMpunλnn−λ/λ++1−λ/λ++11/λ=ρqλλ−MqMpunλ11−λ/λ+1/λ=ρqλ−ρλλ+MqMpun.
Similarly, sinceλ−<0, then −λ/λ−>0 and hence −λ/λ−+1>1. By applying (39) with γ=−λ/λ−+1>1, we have
(121)Δm−1−λ/λ−+1=m−λ/λ−+1−m−1−λ/λ−+1≤−λ/λ−+1m−λ/λ−,and then
(122)∑m=1nm−λ/λ−≥∑m=1nΔm−1−λ/λ−+1−λ/λ−+1=n−λ/λ−+1−λ/λ−+1.
In this case, we have(123)ρqλλ+n1/λ−MqMpunλn∑m=1nm−λ/λ−1/λ≥ρqλλ+n1/λ−MqMpunλnn−λ/λ−+1−λ/λ−+11/λ=ρqλ+ρλλ−MqMpun.
By substituting (120) and (123) into (117), we obtain that
(124)ρqλ+ρλλ−MqMpun≤MλMpun≤ρqλ−ρλλ+MqMpun,which implies that
(125)Mpu∈Bqλρqλ−ρλλ+,that is the first relation in (112) holds for λ>0.
(2). Assume thatλ<0. By raising (86) to the power λ, we obtain for m<n that
(126)m−λ/λ−ρqλλ+m1/λ−MqMpumλ≥Mpumλ≥m−λ/λ+ρqλλ−m1/λ+MqMpumλ.
By using the monotonicity of(127)n1/λ−MqMpun,andn1/λ+MqMpun,
(see Theorem6), we have that
(128)m−λ/λ−ρqλλ+n1/λ−MqMpunλ≥Mpumλ≥m−λ/λ+ρqλλ−n1/λ+MqMpunλ.
Sinceλ<λ+, by summing (128) from m=1 to n, dividing by n, and raising it to the power 1/λ>0, we get
(129)ρqλλ+n1/λ−MqMpunλ1n∑m=1nm−λ/λ−1/λ≤1n∑m=1nMpumλ1λ,≤ρqλλ−n1λ+MqMpunλ1n∑m=1nm−λλ+1λ.
Since0<−λ/λ+, then −λ/λ++1>1 and by applying (39) with γ=−λ/λ++1, we have
(130)Δm−1−λ/λ++1=m−λ/λ++1−m−1−λ/λ++1≤−λ/λ++1m−λ/λ+,and then
(131)∑m=1nm−λ/λ+≥1−λ/λ++1∑m=1nΔm−1−λ/λ++1=n−λ/λ++1−λ/λ++1.
Sinceλ<0, we have that
(132)ρqλλ−n1/λ+MqMpunλn∑m=1nm−λ/λ+1/λ≤ρqλλ−n1/λ+MqMpunλnn−λ/λ++1−λ/λ++11/λ=ρqλ−ρλλ+MqMpunn−λ/λ++1n−λ/λ++11/λ=ρqλ−ρλλ+MqMpun.
Similarly, sinceλ−<0 and λ<0, then −λ/λ−<0 and −λ/λ−+1<1. If 0<−λ/λ−+1<1, we apply inequality (59) to obtain
(133)Δm−1−λ/λ−+1=m−λ/λ−+1−m−1−λ/λ−+1≥−λ/λ−+1m−λ/λ−,and then
(134)∑m=1nm−λ/λ−≤1−λ/λ−+1∑m=1nΔm−1−λ/λ−+1=n−λ/λ−+1−λ/λ−+1.
Sinceλ<0, we have that
(135)ρqλλ+n1/λ−MqMpunλn∑m=1nm−λ/λ−1/λ≥ρqλλ+n1/λ−MqMpunλnn−λ/λ−+1−λ/λ−+11/λ=ρqλ+ρλλ−MqMpunn−λ/λ−+1n−λ/λ−+11/λ=ρqλ+ρλλ−MqMpun.
If−λ/λ−+1<0, we apply (39) to obtain
(136)Δm−1−λ/λ−+1=m−λ/λ−+1−m−1−λ/λ−+1≤−λ/λ−+1m−λ/λ−,and then
(137)∑m=1nm−λ/λ−≤1−λ/λ−+1∑m=1nΔm−1−λ/λ−+1≥n−λ/λ−+1−λ/λ−+1.
In this case, we have(138)ρqλλ+n1/λ−MqMpunλn∑m=1nm−λ/λ−1/λ≥ρqλλ+n1/λ−MqMpunλnn−λ/λ−+1−λ/λ−+11/λ=ρqλ+ρλλ−MqMpun.
From the two cases, we conclude that for allλ−, we have
(139)ρqλλ+n1/λ−MqMpunλn∑m=1nm−λ/λ−1/λ≥ρqλ+ρλλ−MqMpun.
By substituting (132) and (139) into (117), we have
(140)ρqλ+ρλλ−MqMpun≤MλMpun≤ρqλ−ρλλ+MqMpun.
This gives us that(141)Mpu∈Bqλρqλ−ρλλ+,which is the first relation in (112) in the case λ<0. Similarly, we can prove the first relation in (112) by using relation (68). Analogously, we prove the two relations in (113) by using the same technique and inequalities (68) and (86). The proof is complete.


## 4. Self-Improving Properties

In Theorem7, we proved that the power mean operators Mpu and Mqu of the weight u∈Bpq satisfy the reverse Hölder inequality with some better exponents. However, the fact that the mean Mpu or Mqu belongs to some class Bp′q′ does not imply that the weight u itself belongs to Bp′q′. Thus, Theorem 7 does not guarantee the self-improvement of the summability exponents of the weight u∈Bpq. But if we additionally assume the condition of the monotonicity of the weight u, then we can obtain the following results for self-improving of exponents.Theorem 8.
Let0<p<q and u be any nonnegative weight belongs to BpqB for B>1, and λ+ and λ− are the roots of equation (65).
(1)
Ifu is nonincreasing, and λ<λ+, then(142)u∈Bqλρpλ−ρλλ+ρpλ+,u∈Bpλρpλ−ρpλ+θλλ+.(2)
Ifu is nondecreasing, and λ>λ−, then(143)u∈Bλqρqλ−ρλλ−ρqλ+,u∈Bλpρpλ−ρλλ−ρpλ+.Proof.
(1). Sinceu is nonincreasing, then Lemma 2 implies that Mqun is also nonincreasing and Mqun≥un, and hence
(144)MλMqun≥Mλun.
By applying the second relation in (112), we obtain that
(145)MλMqun≤ρpλ−ρλλ+MpMqun.
By applying the left-inequality in (68) and since Mqun is nonincreasing (see Lemma 2), we from (144) and (145) that
(146)Mλun≤MλMqun≤ρpλ−ρλλ+MpMqun≤ρpλ−ρλλ+ρpλ+Mqun.
That is,(147)u∈Bqλρpλ−ρλλ+ρpλ+,which is the first relation in (142). Similarly, since u is nonincreasing, we have Mpun≥un, and so
(148)MλMpun≥Mλun.
By applying the first relation in (112), we obtain that
(149)MλMpun≤ρqλ−ρλλ+MqMpun.
By applying the left-inequality in (86), and using (148), (149), and the fact that Mpun is nonincreasing (see Lemma 2), we have that
(150)Mλun≤MλMpun≤ρqλ−ρλλ+MqMpun≤ρqλ−ρλλ+ρqλ+Mpun.
That is,(151)u∈Bpλρqλ−ρλλ+ρqλ+,which is the second relation in (142).
(2). Since u is nondecreasing, then Lemma 2 implies that Mpun is nondecreasing and Mpun≤un, and we have that
(152)MλMpun≤Mλun.
By applying the first relation in (113), we obtain that
(153)MλMpun≥ρqλ+ρλλ−MqMpun.
By applying the right-inequality in (86), we have that
(154)MλMpun≥ρqλ+ρλλ−MqMpun≥ρqλ+Mpunρqλ−ρλλ−.
By combining (152) and (154), we have that
(155)Mpun≤ρqλ−ρλλ−ρqλ+Mλun;that is,
(156)u∈Bλpρqλ−ρλλ−ρqλ+,which is the second relation in (143). Again, since u is nondecreasing, then Lemma 2 implies that Mqun is nondecreasing and Mqun≤un, and so
(157)MλMqun≤Mλun.
By applying the second relation in (113), we obtain that
(158)MλMqun≥ρpλ+ρλλ−MpMqun.
By applying the right-inequality in (68), we have that
(159)Mλun≥MλMqun≥ρpλ+ρλλ−MpMqun≥ρpλ+ρpλ−ρλλ−Mqun.
By combining (157) and (159), we have that
(160)Mqun≤ρpλ−ρλλ−ρpλ+Mλun.
That is,(161)u∈Bλqρpλ−ρλλ−ρpλ+,which is the first relation in (143). The proof is complete.Now, we derive the self-improving properties of the classGq≔B1q.Theorem 9.
Letq>1 and u be any nondecreasing weight, add weight after nondecreasing GqB for B>1. Then, u∈GλB1″ for λ∈q,λ+), where λ+ is the root of the equation
(162)x−1xxx−q1q=B.Proof.
SinceGq≔B1q, then equation (65) becomes C1,qx=B1′ which writes
(163)xx−1−1xx−q1q=B1′,which is the desired equation (162), and the constant B1″ is obtained from (142) and given by
(164)B1″=ρ1λ−ρλλ−ρ1λ+.
The proof is complete.Remark 10.
Our technique is only applicable in the case when0<p<q, and it remains an open problem to prove the main results for all p and q such that p<q and p.q≠0 to be able to get the main results of Muckenhoupt weights similar to the Gehring weights. This leads to the following conjecture.Theorem 11.
Letp<q such that p.q≠0 and u be any nonnegative weight that belongs to BpqB for B>1, and λ+ and λ− are the roots of equation (65).
(1)
Ifu is nonincreasing, and λ<λ+, then(165)u∈Bqλρpλ−ρλλ+ρpλ+,u∈Bpλρpλ−ρpλ+θλλ+.(2)
Ifu is nondecreasing, and λ>λ− then(166)u∈Bλqρqλ−ρλλ−ρqλ+,u∈Bλpρpλ−ρλλ−ρpλ+.From this theorem, we can obtain the following sharp result.Theorem 12.
Letp>1 and ube any nonnegative and nondecreasing belong toApB for B>1. Then, u∈AλB2″ for λ∈(x−,p], where x− is the root of the equation
(167)p−xp−1Bx1p−1=1.Proof.
SinceAp≔B1/1−p1, then equation (65) becomes Cp,qx=B2′ which is written by
(168)xx−1p−1xp−1x+1p−1=B.
By applying the transformλ→1/1−x, we see that λ− is determined from the equation
(169)p−λp−1Bλ1p−1=1,which is the desired equation (167), and the constant B2″ is obtained from (143) and given by
(170)B2″=ρ1λ−ρλλ−ρ1λ+.
The proof is complete.



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