# Some New Iterated Hardy-Type Inequalities

Authors: A. Gogatishvili; R. CH. Mustafayev; L.-E. Persson
Journal: Journal of Function Spaces and Applications
 (2012)

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## Abstract

We characterize the validity of the Hardy-type inequality∥∥∫s∞h(z)dz∥p,u,(0,t)∥q,w,(0,∞)≤c∥h∥θ,v(0,∞), where 0<p<∞, 0<q≤∞, 1<θ≤∞, u, w, and v are weight functions on (0,∞). Some fairly new discretizing and antidiscretizing techniques of independent interest are used.


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## Body

## 1. Introduction

Everywhere in the paper,u, v, and w are weights, that is, locally integrable nonnegative functions on (0,∞), and we denote
(1.1)U(s)=∫0s‍u(t)dt,Vθ(t)={∫t∞‍v(s)1-θ′dsfor1<θ<∞,∫t∞‍dsv(s)forθ=∞,
where
(1.2)θ':={∞ifθ=1,θθ-1if1<θ<∞,1ifθ=∞.
We assume that u is such that U(t)>0 for every t∈(0,∞).For0<p<∞ and w, a weight function on (a,b)⊆(0,∞), let us denote by Lp,w(a,b) the weighted Lebesgue space defined as the set of all measurable functions u on (a,b) for which the quantity
(1.3)∥u∥p,w,(a,b)={(∫ab‍|u(x)|pw(x)dx)1/pfor1≤p<∞,esssupa<x<b|u(x)|w(x)forp=∞
is finite.In this paper we characterize the validity of the inequality(1.4)∥∥∫s∞‍h(z)dz∥p,u,(0,t)∥q,w,(0,∞)≤c∥h∥θ,v,(0,∞),
where 0<p<∞, 0<q≤∞, 1<θ≤∞, u, w, and v are weight functions on (0,∞). Note that inequality (1.4) has been considered in the case p=1 in [1] (see also [2]), where the result is presented without proof, in the case p=∞ in [3] and in the case θ=1 in [4, 5], where weight functions v of special type were considered. For general weight functions v, the characterization of the inequality (1.4) in the case θ=1 does not follow directly by this method (there are some technical problems) and we are working on it.It is worth to mention that, by Fubini’s theorem,(1.5)∫0x‍u(t)(∫t∞‍g(s)ds)dt≈U(x)∫0∞‍g(s)U(s)dsU(x)+U(s).
Hence, we see that the inequality (1.4) (with p=1) is equivalent with the following inequality:
(1.6)∥Sh∥q,Uqw,(0,∞)≤c∥h∥θ,U-θv,(0,∞),
where the operator S defined by
(1.7)(Sh)(x)=∫0∞‍h(t)dtU(x)+U(t)
for all nonnegative measurable functions h on (0,∞). We call this operator the generalized Stieltjes transform; the usual Stieltjes transform is obtained on putting U(x)≡x.In the caseU(x)≡xλ, λ>0, the boundedness of the operator S between weighted Lp and Lq spaces was investigated in [6] (when 1≤p≤q≤∞) and in [7, 8] (when 1≤q<p≤∞).Our approach is based on discretization and antidiscretization methods developed in [4, 9, 10]. Some basic facts concerning these methods and other preliminaries are presented in Section 2. The main results (Theorems 3.1 and 3.2) are stated and proved in Section 3.Throughout the paper, we always denote byc or C a positive constant which is independent of the main parameters, but it may vary from line to line. However a constant with subscript such as c1 does not change in different occurrences. By a≲b, (b≳a), we mean that a≤λb, where λ>0 depends on inessential parameters. If a≲b and b≲a, we write a≈b and say that a and b are equivalent. We put 1/∞=0, 0·∞=0, 0/0=0, and ∞/∞=0.


## 2. Preliminaries

Let us now recall some definitions and basic facts concerning discretization and antidiscretization which can be found in [4, 9, 10].Definition 2.1.
Let{ak} be a sequence of positive real numbers. One says that {ak} is strongly increasing or strongly decreasing and write ak↑↑ or ak↓↓ when
(2.1)infk∈ℤak+1ak>1orsupk∈ℤak+1ak<1,
respectively.Definition 2.2.
LetU be a continuous strictly increasing function on [0,∞) such that U(0)=0 and limt→∞U(t)=∞. Then One says that U is admissible.
LetU be an admissible function. We say that a function φ is U-quasiconcave if φ is equivalent to an increasing function on (0,∞) and φ/U is equivalent to a decreasing function on (0,∞). We say that a U-quasiconcave function φ is nondegenerate if
(2.2)limt→0+φ(t)=limt→∞1φ(t)=limt→∞φ(t)U(t)=limt→0+U(t)φ(t)=0.
The family of nondegenerate U-quasiconcave functions will be denoted by ΩU. We say that φ is quasiconcave when φ∈ΩU with U(t)=t. A quasiconcave function is equivalent to a concave function. Such functions are very important in various parts of analysis. Let us just mention that, for example, the Hardy operator Hf(x)=∫0x‍f(t)dt of a decreasing function, the Peetre K-functional in interpolation theory, and the fundamental function ∥χE∥X, X is a rearrangement invariant space, all are quasiconcave.Definition 2.3.
Assume thatU is admissible and φ∈ΩU. One says that {xk}k∈ℤ is a discretizing sequence for φ with respect to U if(i)
x0=1 and U(xk)↑↑;(ii)
φ(xk)↑↑ and φ(xk)/U(xk)↓↓;(iii)
there is a decompositionℤ=ℤ1∪ℤ2 such that ℤ1∩ℤ2=∅ and for every t∈[xk,xk+1](2.3)φ(xk)≈φ(t)ifk∈ℤ1,φ(xk)U(xk)≈φ(t)U(t)ifk∈ℤ2.
Let us recall (see [9, Lemma 2.7]) that if φ∈ΩU, then there always exists a discretizing sequence for φ with respect to U.Definition 2.4.
LetU be an admissible function, and let ν be a nonnegative Borel measure on [0,∞). We say that the function φ defined by
(2.4)φ(t)=U(t)∫[0,∞)‍dν(s)U(s)+U(t),t∈(0,∞),
is the fundamental function of the measure ν with respect to U. One will also say that ν is a representation measure of φ with respect to U.
We say thatν is nondegenerate if the following conditions are satisfied for every t∈(0,∞):
(2.5)∫[0,∞)‍dν(s)U(s)+U(t)<∞,∫[0,1]‍dν(s)U(s)=∫[1,∞)‍dν(s)=∞.
We recall from [9, Remark 2.10] that
(2.6)φ(t)≈∫[0,t]‍dν(s)+U(t)∫[t,∞)‍U(s)-1dν(s),t∈(0,∞).Corollary 2.5 (see [10, Lemma 1.5]).
Letu, w be weights, and let φ be defined by
(2.7)φ(t)=esssups∈(0,t)U(s)1/pess supτ∈(s,∞)w(τ)U(τ)1/p,t∈(0,∞).
Then φ is the least U1/p-quasiconcave majorant of w, and
(2.8)supt∈(0,∞)φ(t)(1U(t)∫0t‍(∫s∞‍h(z)dz)pu(s)ds)1/p=esssupt∈(0,∞)w(t)(1U(t)∫0t‍(∫s∞‍h(z)dz)pu(s)ds)1/p
for any nonnegative measurable h on (0,∞). Further, for t∈(0,∞),
(2.9)φ(t)=ess supτ∈(0,∞)w(τ)min{1,(U(t)U(τ))1/p}=U(t)1/pess sups∈(t,∞)1U(s)1/pess supτ∈(0,s)w(τ),φ(t)≈ess sups∈(0,∞)w(s)(U(t)U(s)+U(t))1/p.Theorem 2.6 (see [9, Theorem 2.11]).
Letp,q,r∈(0,∞). Assume that U is an admissible function, ν is a nonnegative nondegenerate Borel measure on [0,∞), and φ is the fundamental function of ν with respect to Uq and σ∈ΩUp. If {xk} is a discretizing sequence for φ with respect to Uq, then
(2.10)∫[0,∞)‍φ(t)(r/q)-1σ(t)r/pdν(t)≈∑k∈ℤ‍φ(xk)r/qσ(xk)r/p.Lemma 2.7 (see [9, Corollary 2.13]).
Letq∈(0,∞). Assume that U is an admissible function, f∈ΩU, ν is a nonnegative nondegenerate Borel measure on [0,∞), and φ is the fundamental function of ν with respect to Uq. If {xk} is a discretizing sequence for φ with respect to Uq, then
(2.11)(∫[0,∞)‍(f(t)U(t))qdν(t))1/q≈(∑k∈ℤ‍(f(xk)U(xk))qφ(xk))1/q.Lemma 2.8 (see [9, Lemma 3.5]).
Letp,q,r∈(0,∞). Assume that U is an admissible function, φ∈ΩUq, and g∈ΩUp. If {xk} is a discretizing sequence for φ with respect to Uq and {λk} is a discretizing sequence of g with respect to Up, then
(2.12)∑k∈ℤ‍φ(xk)r/qg(xk)r/p≈∑ℓ∈ℤ‍φ(λℓ)r/qg(λℓ)r/p,supt∈(0,∞)φ(t)1/qg(t)1/p≈supk∈ℤφ(xk)1/qg(xk)1/p≈supℓ∈ℤφ(λℓ)1/qg(λℓ)1/p.Lemma 2.9 (see [4, Lemma 2.5]).
Ifτk↓↓, then
(2.13)∑k∈ℤ‍(∫0xk‍h)qτk≈∑k∈ℤ‍(∫xk-1xk‍h)qτk,supk∈ℤ(∫0xk‍h)qτk≈supk∈ℤ(∫xk-1xk‍h)qτk,∑k∈ℤ‍(∫xk∞‍h)qτk-1≈∑k∈ℤ‍(∫xkxk+1‍h)qτk-1,supk∈ℤ(∫xk∞‍h)qτk-1≈supk∈ℤ(∫xkxk+1‍h)qτk-1.Lemma 2.10 (see [9, Lemma 3.6]).
Letq∈(0,∞). Assume that U is an admissible function, ν is a nondegenerate nonnegative Borel measure on [0,∞), φ is the fundamental function of ν with respect to Uq, and f is a measurable function on [0,∞). If {xk} is a discretizing sequence for  φ with respect to  Uq, then
(2.14)∫0∞‍(∫0∞‍|f(t)|dtU(t)+U(x))qdν(x)≈∑k∈ℤ‍(∫0∞‍|f(t)|dtU(t)+U(xk))qφ(xk)≈∑k∈ℤ‍(U-1(xk)∫xk-1xk‍|f(y)|dy+∫xkxk+1‍|f(y)|U-1(y)dy)qφ(xk)≈∑k∈ℤ‍(∫xkxk+1‍|f(y)|U-1(y)φ(y)1/qdy)q.Lemma 2.11 (see [9, Lemma 3.7]).
Letq∈(0,∞). Assume that U is an admissible function, ν is a nondegenerate nonnegative Borel measure on [0,∞), φ is the fundamental function of ν with respect to Uq, and f is a measurable function on [0,∞). If {xk} is a discretizing sequence for φ with respect to Uq, then
(2.15)∫[0,∞)‍(ess supy∈(0,∞)|f(y)|U(x)+U(y))qdν(x)≈∑k∈ℤ‍(ess supy∈(0,∞)|f(y)|U(xk)+U(y))qφ(xk)≈∑k∈ℤ‍(U-1(xk)ess supxk-1≤y<xk|f(y)|+ess supxk≤y<xk+1|f(y)|U-1(y))qφ(xk)≈∑k∈ℤ‍ess supxk≤y<xk+1|f(y)|qU-q(y)φ(y).Lemma 2.12 (see [9, Lemma 3.8]).
Letq∈(0,∞). Assume that U is an admissible function, φ∈ΩUq, {xk} is a discretizing sequence for φ with respect to Uq, and f is a measurable function on [0,∞). Then
(2.16)supx∈(0,∞)(∫0∞‍|f(t)|dtU(t)+U(x))qφ(x)≈supk∈ℤ(∫0∞‍|f(t)|dtU(t)+U(xk))qφ(xk)≈supk∈ℤ(U-1(xk)∫xk-1xk‍|f(y)|dy+∫xkxk+1‍|f(y)|U-1(y)dy)qφ(xk)≈supk∈ℤ(∫xkxk+1‍|f(y)|U-1(y)φ(y)1/qdy)q.Lemma 2.13 (see [9, Lemma 3.9]).
LetU be an admissible function, φ∈ΩU, {xk} be a discretizing sequence for φ with respect to U, and f be a measurable function on [0,∞). Then
(2.17)supx∈(0,∞)φ(x)esssupy∈(0,∞)|f(y)|U(x)+U(y)≈supk∈ℤφ(xk)esssupy∈(0,∞)|f(y)|U(xk)+U(y)≈supk∈ℤφ(xk)U(xk)-1esssupxk-1≤y<xk|f(y)|+supk∈ℤφ(xk)esssupxk≤y<xk+1|f(y)|U(y)-1≈supk∈ℤesssupxk≤y<xk+1|f(y)|U(y)-1φ(y).Proposition 2.14 (see [9, Proposition 4.1]).
Let{ωk} and {υk}, k∈ℤ, be two sequences of positive real numbers. Let p,q∈(0,∞), and assume that the inequality
(2.18)(∑k∈ℤ‍akqυk)1/q≤c(∑k∈ℤ‍akpωk)1/p
is satisfied for every sequence {ak} of positive real numbers.(i)
Ifp≤q, then(2.19)supk∈ℤωk-q/pυk<∞.(ii)
Ifp>q and r=pq/(p-q), then(2.20)(∑k∈ℤ‍ωk-r/pυkr/q)1/r<∞.Lemma 2.15.
One has the following Hardy-type inequalities.(a)
Let1<θ≤p<∞. Then the inequality
(2.21)∥∫sxk‍h(z)dz∥p,u,(xk-1,xk)≤c∥h∥θ,v,(xk-1,xk)holds for all nonnegative measurableh if and only if
(2.22)A:=supxk-1<t<xk(∫xk-1t‍u(s)ds)1/p(∫txk‍v(s)1-θ′ds)1/θ′<∞,and the best constant in (2.21) satisfies c≈A.(b)
Let1<θ<∞, p=∞. Then the inequality (2.21) holds if and only if
(2.23)B:=supxk-1<t<xk(esssupxk-1<s<tu(s))(∫txk‍v(s)1-θ′ds)1/θ′<∞,and the best constant in (2.21) satisfies c≈B.(c)
Letθ=p=∞. Then the inequality (2.21) holds if and only if
(2.24)C∶=supxk-1<t<xk(esssupxk-1<s<tu(s))∫txk‍dsv(s)<∞,and the best constant in (2.21) satisfies c≈C.(d)
Let1<θ<∞, 0<p<θ, and 1/r=1/p-1/θ. Then the inequality (2.21) holds if and only if
(2.25)D:=(∫xk-1xk‍(∫xk-1t‍u(s)ds)r/p(∫txk‍v(s)1-θ′ds)r/p'v(t)1-θ′dt)1/r<∞,and the best constant in (2.21) satisfies c≈D.(e)
Letθ=∞, 0<p<∞. Then the inequality (2.21) holds if and only if
(2.26)E:=(∫xk-1xk‍(∫xk-1t‍u(s)ds)(∫txk‍dsv(s))p-1dtv(t))1/p<∞,and the best constant in (2.21) satisfies c≈E.These results are just classical results of Maz’ja [11] and Sinnamon [12] (cf. also [13, 14]).


## 3. The Main Results

In this section we characterize the validity of the inequalities(3.1)(∫0∞‍(1U(t)∫0t‍(∫s∞‍h(z)dz)pu(s)ds)q/pw(t)dt)1/q≤c∥h∥θ,v,(0,∞),(3.2)esssupt∈(0,∞)w(t)(1U(t)∫0t‍(∫s∞‍h(z)dz)pu(s)ds)1/p≤c∥h∥θ,v,(0,∞).Denote(3.3)𝒰(x,t):=U(x)U(t)+U(x).First we characterize (3.1) as follows.Theorem 3.1.
Let0<q<∞, 0<p<∞, 1<θ≤∞, and let u,v,w be weights. Assume that u is such that Uq/p is admissible and the measure w(t)dt is nondenerate with respect to Uq/p. Then the inequality (3.1) holds for every measurable function f on (0,∞) if and only if(i)
1<θ≤p<∞, θ≤q(3.4)A1:=supx∈(0,∞)(∫0∞‍𝒰(x,t)q/pw(t)dt)1/qU(x)-1/psupt∈(0,x)U(t)1/pVθ(t)1/θ′<∞.Moreover, the best constantc in (3.1) satisfies c≈A1.(ii)
1<θ≤p<∞, 0<q<θ<∞,  l=θq/(θ-q)(3.5)A2:=(∫0∞‍(∫0∞‍𝒰(x,t)q/pw(t)dt)(l-q)/qw(x)U(x)-l/psupt∈(0,x)U(t)l/pVθ(t)l/θ′dx)1/l<∞.Moreover, the best constantc in (3.1) satisfies c≈A2.(iii)
0<p<θ<∞, 1<θ≤q<∞, r=θp/(θ-p)(3.6)A3:=supx∈(0,∞)(∫0∞‍𝒰(x,t)q/pw(t)dt)1/q(∫0∞‍𝒰(t,x)r/pVθ(t)r/p'v(t)1-θ′dt)1/r<∞.Moreover, the best constantc in (3.1) satisfies c≈A3.(iv)
0<p<θ<∞, 1<θ<∞, q<θ,  r=θp/(θ-p),  l=θq/(θ-q),
(3.7)A4:=(∫0∞‍(∫0∞‍𝒰(x,t)q/pw(t)dt)(l-q)/qw(x)(∫0∞‍𝒰(t,x)r/pVθ(t)r/p'v(t)1-θ′dt)l/rdx)1/l<∞.Moreover, the best constantc in (3.1) satisfies c≈A4.(v)
Letθ=∞, 0<p<∞, 0<q<∞,
(3.8)A5:=(∫0∞‍(𝒰(t,x)Vθ(t)p-1dtv(t))q/pw(x)dx)1/q<∞.Moreover, the best constantc in (3.1) satisfies c≈A5.Proof.
Define(3.9)φ(x)=∫0∞‍𝒰(x,s)q/pw(s)ds.
Then φ∈ΩUq/p, and therefore there exists a discretizing sequence for φ with respect to Uq/p. Let {xk} be one such sequence. Then φ(xk)↑↑ and φ(xk)U-q/p↓↓. Furthermore, there is a decomposition ℤ=ℤ1∪ℤ2, ℤ1∩ℤ2=∅ such that for every k∈ℤ1 and t∈[xk,xk+1], φ(xk)≈φ(t) and for every k∈ℤ2 and t∈[xk,xk+1], φ(xk)U(xk)-q/p≈φ(t)U(t)-q/p.
For the left-hand side of (3.1), by using Lemma 2.7 with
(3.10)dν(t)=w(t)dt,f(t)=∫0t‍(∫s∞‍h(z)dz)pu(s)ds,
we get that
(3.11)J:=(∫0∞‍(1U(t)∫0t‍(∫s∞‍h(z)dz)pu(s)ds)q/pw(t)dt)1/q≈(∑k∈ℤ‍(∫0xk‍(∫s∞‍h(z)dz)pu(s)ds)q/pφ(xk)Uq/p(xk))1/q.
Moreover, by using Lemma 2.9, we get that
(3.12)J≈(∑k∈ℤ‍(∫xk-1xk‍(∫s∞‍h(z)dz)pu(s)ds)q/pφ(xk)Uq/p(xk))1/q=(∑k∈ℤ‍(∫xk-1xk‍(∫sxk‍h(z)dz+∫xk∞‍h(z)dz)pu(s)ds)q/pφ(xk)Uq/p(xk))1/q≈(∑k∈ℤ‍(∫xk-1xk‍(∫sxk‍h(z)dz)pu(s)ds)q/pφ(xk)Uq/p(xk))1/q+(∑k∈ℤ‍(∫xk-1xk‍(∫xk∞‍h(z)dz)pu(s)ds)q/pφ(xk)Uq/p(xk))1/q≈(∑k∈ℤ‍(∫xk-1xk‍(∫sxk‍h(z)dz)pu(s)ds)q/pφ(xk)Uq/p(xk))1/q+(∑k∈ℤ‍(∫xk∞‍h(z)dz)q(∫xk-1xk‍u(s)ds)q/pφ(xk)Uq/p(xk))1/q.
By now using the fact that∫xk-1xk‍u(s)ds=U(xk)-U(xk-1)≈U(xk), we find that
(3.13)J≈(∑k∈ℤ‍(∫xk-1xk‍(∫sxk‍h(z)dz)pu(s)ds)q/pφ(xk)Uq/p(xk))1/q+(∑k∈ℤ‍(∫xk∞‍h(z)dz)qφ(xk))1/q,
is, by using Lemma 2.9 on the second term,
(3.14)J≈(∑k∈ℤ‍(∫xk-1xk‍(∫sxk‍h(z)dz)pu(s)ds)q/pφ(xk)Uq/p(xk))1/q+(∑k∈ℤ‍(∫xkxk+1‍h(z)dz)qφ(xk))1/q:=I+II.
Now we will distinguish several cases. We start with the case 1<θ≤p<∞. Then, by using Lemma 2.15, we get that
(3.15)I=(∑k∈ℤ‍(∫xk-1xk‍(∫sxk‍h(z)dz)pu(s)ds)q/pφ(xk)Uq/p(xk))1/q≤(∑k∈ℤ‍φ(xk)Uq/p(xk)supxk-1<t<xkU(t)q/pVθ(t)q/θ′(∫xk-1xk‍h(z)θv(z)dz)q/θ)1/q.
Moreover, by applying Hölder’s inequality for II, we find that
(3.16)II=(∑k∈ℤ‍φ(xk)(∫xkxk+1‍h(z)dz)q)1/q≤(∑k∈ℤ‍φ(xk)(∫xkxk+1‍v(z)1-θ'dz)q/θ'(∫xkxk+1‍h(z)θv(z)dz)q/θ)1/q≤(∑k∈ℤ‍φ(xk)Vθ(xk)q/θ'(∫xkxk+1‍h(z)θv(z)dz)q/θ)1/q.(i)
In the caseq/θ≥1, according to (3.15), we have that
(3.17)I≤supk∈ℤφ(xk)1/qU1/p(xk)supxk-1<t<xkU(t)1/pVθ(t)1/θ′∥h∥θ,v,(0,∞).Similarly, ifq/θ≥1, then, according to (3.16), we obtain that
(3.18)II≤supk∈ℤφ(xk)1/qVθ(xk)1/θ′∥h∥θ,v,(0,∞),and, finally, by using (3.9), Lemma 2.13, and (3.14), we get that
(3.19)J≲supk∈ℤφ(xk)1/q(U-1/p(xk)supxk-1<t<xkU(t)1/pVθ(t)1/θ′+Vθ(xk)1/θ′)∥h∥θ,v,(0,∞)≈supx∈(0,∞)φ(x)1/qsupt∈(0,∞)𝒰(t,x)1/pVθ(t)1/θ′∥h∥θ,v,(0,∞)=supx∈(0,∞)(∫0∞‍𝒰(x,t)q/pw(t)dt)1/qsupt∈(0,∞)𝒰(t,x)1/pVθ(t)1/θ′∥h∥θ,v,(0,∞)≈supx∈(0,∞)(∫0∞‍𝒰(x,t)q/pw(t)dt)1/qU(x)-1/psupt∈(0,x)U(t)1/pVθ(t)1/θ′∥h∥θ,v,(0,∞).(ii)
For the case0<q<θ<∞, l=θq/(θ-q), by applying Hölder’s inequality for sums to the right-hand side of (3.15) and (3.16) with exponents θ/q and l/q, we find that
(3.20)I≤(∑k∈ℤ‍φ(xk)l/qUl/p(xk)supxk-1<t<xkU(t)l/pVθ(t)l/θ′)1/l∥h∥θ,v,(0,∞),II≤(∑k∈ℤ‍φ(xk)l/qVθ(xk)l/θ′)1/l∥h∥θ,v,(0,∞).Therefore, we get that(3.21)I+II≤(∑k∈ℤ‍φ(xk)l/q(U(xk)-l/psupxk-1<t<xkU(t)l/pVθ(t)l/θ′+Vθ(xk)l/θ′))1/l∥h∥θ,v,(0,∞),so that, in view of Lemma2.11, Theorem 2.6, and (3.14),
(3.22)J≲(∑k∈ℤ‍φ(xk)l/qsupt∈(0,∞)𝒰(t,xk)l/pVθ(t)l/θ′)1/l∥h∥θ,v,(0,∞)≈(∫0∞‍φ(x)(l/q)-1supt∈(0,∞)𝒰(t,x)l/pVθ(t)l/θ′w(x)dx)1/l∥h∥θ,v,(0,∞)≈(∫0∞‍(∫0∞‍𝒰(x,t)q/pw(t)dt)(l-q)/qw(x)U(x)-l/psupt∈(0,x)U(t)l/pVθ(t)l/θ′dx)1/l∥h∥θ,v,(0,∞).Now let us assume that0<p<θ<∞, 1<θ<∞, 1/r=1/p-1/θ. By Lemma 2.15, we have that
(3.23)I≤(∑k∈ℤ‍φ(xk)Uq/p(xk)(∫xk-1xk‍(∫xk-1t‍u(s)ds)r/p(∫txk‍v(s)1-θ′ds)r/p'v(t)1-θ′dt)q/r×(∫xk-1xk‍h(z)θv(z)dz)q/θ)1/q.Moreover, by applying Hölder’s inequality forII, we find that
(3.24)II≤(∑k∈ℤ‍φ(xk)Vθ(xk)q/θ′(∫xkxk+1‍h(z)θv(z)dz)q/θ)1/q.(iii)
Now, we assume thatq/θ≥1. Then, according to (3.7) and (3.24), we obtain that
(3.25)I≲supk∈ℤφ(xk)1/qU1/p(xk)(∫xk-1xk‍U(t)r/pVθ(t)r/p'v(t)1-θ′dt)1/r∥h∥θ,v,(0,∞),II≤supk∈ℤφ(xk)1/qVθ(xk)1/θ′∥h∥θ,v,(0,∞).Hence, using Lemmas2.9 and 2.12, and (3.14), we get that
(3.26)J≲supk∈ℤφ(xk)1/q(U-r/p(xk)∫xk-1xk‍U(t)r/pVθ(t)r/p′v(t)1-θ′dt+∫xk∞‍Vθ(t)r/p′v(t)1-θ′dt)1/r×∥h∥θ,v,(0,∞)≈supk∈ℤφ(xk)1/q(U-r/p(xk)∫xk-1xk‍U(t)r/pVθ(t)r/p′v(t)1-θ′dt+∫xkxk+1‍Vθ(t)r/p′v(t)1-θ′dt)1/r×∥h∥θ,v,(0,∞)≈supk∈ℤφ(xk)1/q(∫0∞‍𝒰(t,xk)r/pVθ(t)r/p'v(t)1-θ′dt)1/r∥h∥θ,v,(0,∞)≈supx∈(0,∞)φ(x)1/q(∫0∞‍𝒰(t,x)r/pVθ(t)r/p'v(t)1-θ′dt)1/r∥h∥θ,v,(0,∞)=supx∈(0,∞)(∫0∞‍𝒰(x,t)q/pw(t)dt)1/q(∫0∞‍𝒰(t,x)r/pVθ(t)r/p'v(t)1-θ′dt)1/r×∥h∥θ,v,(0,∞).(iv)
Next, we consider the case0<q<θ, 1/l=1/q-1/θ. By using Hölder’s inequality for sums to the right-hand side of (3.7) and (3.24) with exponents θ/q and l/q, we get that
(3.27)I≤(∑k∈ℤ‍φ(xk)l/qUl/p(xk)(∫xk-1xk‍U(t)r/pVθ(t)r/p′v(t)1-θ′dt)l/r)1/l∥h∥θ,v,(0,∞),II≤(∑k∈ℤ‍φ(xk)l/qVθ(xk)l/θ′)1/l∥h∥θ,v,(0,∞).Therefore, using Lemmas2.9 and 2.10, Theorem 2.6, and (3.14), we find that
(3.28)J≲(∑k∈ℤ‍φ(xk)l/q(U-l/p(xk)(∫xk-1xk‍U(t)r/pVθ(t)r/p′v(t)1-θ′dt)l/r+V(xk)l/θ′))1/l×∥h∥θ,v,(0,∞)≈(∑k∈ℤ‍φ(xk)l/q(U-r/p(xk)∫xk-1xk‍U(t)r/pVθ(t)r/p′v(t)1-θ′dt+∫xk∞‍Vθ(t)r/p′v(t)1-θ′dt)l/r)1/l×∥h∥θ,v,(0,∞)≈(∑k∈ℤ‍φ(xk)l/q(U-r/p(xk)∫xk-1xk‍U(t)r/pVθ(t)r/p′v(t)1-θ′dt+∫xkxk+1‍Vθ(t)r/p′v(t)1-θ'dt)l/r)1/l×∥h∥θ,v,(0,∞)≈(∑k∈ℤ‍φ(xk)l/q(∫0∞‍𝒰(t,xk)r/pVθ(t)r/p′v(t)1-θ'dt)l/r)l/l∥h∥θ,v,(0,∞)≈(∫0∞‍φ(t)l/q-1(∫0∞‍𝒰(t,x)r/pVθ(t)r/p′v(t)1-θ'dt)l/rw(x)dx)l/l∥h∥θ,v,(0,∞)=(∫0∞‍(∫0∞‍𝒰(x,t)q/pw(t)dt)(l-q)/qw(x)(∫0∞‍𝒰(t,x)r/pVθ(t)r/p′v(t)1-θ'dt)l/rdx)1/l×∥h∥θ,v,(0,∞).(v)
Letθ=∞, 0<p<∞, 0<q<∞. According to Lemma 2.15, we have that
(3.29)I≤(∑k∈ℤ‍φ(xk)Uq/p(xk)(∫xk-1xk‍(∫xk-1t‍u(s)ds)(∫txk‍dzv(z))p-1dtv(t))q/p)1/q∥h∥θ,v,(0,∞).Moreover, it yields that(3.30)II≤(∑k∈ℤ‍φ(xk)(∫xkxk+1‍dzv(z))q)1/q∥h∥θ,v,(0,∞).Hence, by integrating by parts, using Lemmas2.9 and 2.10, and (3.14), we get that
(3.31)J≲(∑k∈ℤ‍φ(xk)(U(xk)-1∫xk-1xk‍U(s)(∫s∞‍dzv(z))p-1dsv(s)+∫xk∞‍(∫s∞‍dzv(z))p-1dsv(s))q/p)1/q×∥h∥∞,v,(0,∞)≈(∑k∈ℤ‍φ(xk)(U(xk)-1∫xk-1xk‍U(s)(∫s∞‍dzv(z))p-1dsv(s)+∫xkxk+1‍(∫s∞‍dzv(z))p-1dsv(s))q/p)1/q×∥h∥∞,v,(0,∞)≈(∫0∞‍(𝒰(t,x)(∫t∞‍dzv(z))p-1dtv(t))q/pω(x)dx)1/q∥h∥∞,v,(0,∞).
Now we prove the lower bounds (necessity). Let0<q<∞ and {xk} be a discretizing sequence for φ from (3.9). Then, by (3.14), we find that
(3.32)(∑k∈ℤ‍(∫xk-1xk‍(∫sxk‍h(z)dz)pu(s)ds)q/pφ(xk)Uq/p(xk))1/q+(∑k∈ℤ‍(∫xkxk+1‍h(z)dz)qφ(xk))1/q≲(∑k∈ℤ‍∫xk-1xk‍h(z)θv(z)dz)1/θ.
Let 1<θ≤p<∞. For k∈ℤ, let hk be functions that saturate the Hardy inequality (2.21) and Hölder’s inequality, that is, functions hk satisfying
(3.33)supphk⊂[xk-1,xk],∫xk-1xk‍hk(t)θv(t)dt=1,∥∫sxk‍hk(z)dz∥p,u,(xk-1,xk)≳supxk-1<t<xk(∫xk-1t‍u(s)ds)1/p(∫txk‍v(s)1-θ′ds)1/θ′,∫xk-1xk‍hk(t)dt≳(∫xk-1xk‍v(t)1-θ′dt)1/θ′.
Now we define the test function
(3.34)h(t)=∑k∈ℤ‍akhk(t),
where {ak} is a sequence of positive real numbers. Thus, using test function (3.34) in (3.32), we get that
(3.35)(∑k∈ℤakqφ(xk)Uq/p(xk)supxk-1<t<xk(∫xk-1tu(s)ds)q/p(∫txk‍v(s)1-θ′ds)q/θ′)1/q+(∑k∈ℤakqφ(xk)(∫xkxk+1v(t)1-θ′dt)q/θ′)1/q≲(∑k∈ℤakθ)1/θ.
Now using Proposition 2.14 for the case θ≤q, we obtain that
(3.36)supk∈ℤφ(xk)1/q(∫xkxk+1‍v(t)1-θ′dt)1/θ′+supk∈ℤφ(xk)1/qU1/p(xk)supxk-1<t<xk(∫xk-1t‍u(s)ds)1/p(∫txk‍v(s)1-θ′ds)1/θ′<∞.
On the other hand, using Lemma 2.9, we get that
(3.37)A1≲supk∈ℤφ(xk)1/qU1/p(xk)supxk-1<t<xkU(t)1/pVθ(t)1/θ′+supk∈ℤφ(xk)1/qVθ(xk)1/θ′≲supk∈ℤφ(xk)1/qU1/p(xk)supxk-1<t<xkU(t)1/p(∫txk‍v(s)1-θ′ds)1/θ′+supk∈ℤφ(xk)1/qVθ(xk)1/θ′≲supk∈ℤφ(xk)1/qU1/p(xk)supxk-1<t<xk(∫xk-1t‍u(s)ds)1/p(∫txk‍v(s)1-θ′ds)1/θ′+supk∈ℤφ(xk)1/qVθ(xk)1/θ′≲supk∈ℤφ(xk)1/qU1/p(xk)supxk-1<t<xk(∫xk-1t‍u(s)ds)1/p(∫txk‍v(s)1-θ′ds)1/θ′+supk∈ℤφ(xk)1/q(∫xkxk+1‍v(s)1-θ′ds)1/θ′<∞.
Let 0<q<θ<∞. From (3.35) and Proposition 2.14, we obtain that
(3.38)(∑k∈ℤ‍φ(xk)l/q(∫xkxk+1‍v(t)1-θ′dt)l/θ′)1/l+(∑k∈ℤ‍φ(xk)l/qUl/p(xk)supxk-1<t<xk(∫xk-1t‍u(s)ds)l/p(∫txk‍v(s)1-θ′ds)l/θ′)1/l<∞.
Since
(3.39)A2≈(∑k∈ℤ‍φ(xk)l/qVθ(xk)l/θ′)1/l+(∑k∈ℤ‍φ(xk)l/qU(xk)l/psupxk-1<t<xkU(t)l/pVθ(t)l/θ′)1/l≲(∑k∈ℤ‍φ(xk)l/qVθ(xk)l/θ′)1/l+(∑k∈ℤ‍φ(xk)l/qU(xk)l/psupxk-1<t<xkU(t)l/p(∫txk‍v(s)1-θ′ds)l/θ′)1/l≲(∑k∈ℤ‍φ(xk)l/qVθ(xk)l/θ′)1/l+(∑k∈ℤ‍φ(xk)l/qU(xk)l/psupxk-1<t<xk(∫xk-1t‍u(s)ds)l/p(∫txk‍v(s)1-θ′ds)l/θ′)1/l,
by Lemma 2.9, we arrive at
(3.40)A2≲(∑k∈ℤ‍φ(xk)l/q(∫txk‍v(s)1-θ′ds)l/θ′)1/l+(φ(xk)l/qU(xk)l/psupxk-1<t<xk(∫xk-1t‍u(s)ds)l/p(∫txk‍v(s)1-θ′ds)l/θ′)1/l<∞.
Let 1<θ<∞, 0<p<θ. For k∈ℤ, let hk be functions that saturate the Hardy inequality (2.21) and Hölder’s inequality, that is, functions hk satisfying
(3.41)supphk⊂[xk-1,xk],∫xk-1xk‍hk(t)θv(t)dt=1,∥∫sxk‍hk(z)dz∥p,u,(xk-1,xk)≳(∫xk-1xk‍(∫xk-1t‍u(s)ds)r/p(∫txk‍v(s)1-θ′ds)r/p′v(t)1-θ′dt)1/r,∫xk-1xk‍hk(t)dt≳(∫xk-1xk‍v(t)1-θ′dt)1/θ′.
Now we define the test function
(3.42)h(t)=∑k∈ℤ‍akhk(t),
where {ak} is a sequence of positive real numbers. Thus, using test function (3.42) in (3.32), we get that
(3.43)(∑k∈ℤ‍akqφ(xk)Uq/p(xk)(∫xk-1xk‍(∫xk-1t‍u(s)ds)r/p(∫txk‍v(s)1-θ′ds)r/p′v(t)1-θ′dt)q/r)1/q+(∑k∈ℤ‍akqφ(xk)(∫xkxk+1‍v(t)1-θ′dt)q/θ′)1/q≲(∑k∈ℤ‍akθ)1/θ.
Now using Proposition 2.14 for the case θ≤q, we obtain that
(3.44)supk∈ℤφ(xk)1/qU1/p(xk)(∫xk-1xk‍(∫xk-1t‍u(s)ds)r/p(∫txk‍v(s)1-θ′ds)r/p′v(t)1-θ′dt)1/r+supk∈ℤφ(xk)1/q(∫xkxk+1‍v(t)1-θ′dt)1/θ′<∞.
Since
(3.45)A3≈supk∈ℤφ(xk)1/q(U-r/p(xk)∫xk-1xk‍U(t)r/pVθ(t)r/p'v(t)1-θ′dt+∫xk∞‍Vθ(t)r/p'v(t)1-θ′dt)1/r≲supk∈ℤφ(xk)1/qU1/p(xk)(∫xk-1xk‍U(t)r/pVθ(t)r/p'v(t)1-θ′dt)1/r+supk∈ℤφ(xk)1/qVθ(xk)1/θ'≲supk∈ℤφ(xk)1/qU1/p(xk)(∫xk-1xk‍(∫xk-1t‍u(s)ds)r/pVθ(t)r/p'v(t)1-θ'dt)1/r+supk∈ℤφ(xk)1/qVθ(xk)1/θ′,
by integrating by parts, we get that
(3.46)A3≲supk∈ℤφ(xk)1/qU1/p(xk)(∫xk-1xk‍(∫t∞‍v(s)1-θ′ds)(r/p′)+1u(t)(∫xk-1t‍u(s)ds)(r/p)-1dt)1/r+supk∈ℤφ(xk)1/qVθ(xk)1/θ′≲supk∈ℤφ(xk)1/qU1/p(xk)(∫xk-1xk‍(∫txk‍v(s)1-θ′ds)(r/p′)+1u(t)(∫xk-1t‍u(s)ds)(r/p)-1dt)1/r+supk∈ℤφ(xk)1/qVθ(xk)1/θ′.
Again integrating by parts, we arrive at
(3.47)A3≲supk∈ℤφ(xk)1/qU1/p(xk)(∫xk-1xk‍(∫xk-1t‍u(s)ds)r/p(∫txk‍v(s)1-θ′ds)r/p'v(t)1-θ′dt)1/r+supk∈ℤφ(xk)1/qVθ(xk)1/θ′<∞.
Now let 1<θ<∞, 0<p<θ, q<θ. By using (3.43) and Proposition 2.14, we obtain
(3.48)(∑k∈ℤ‍φ(xk)l/pU1/p(xk)(∫xk-1xk‍(∫xk-1t‍u(s)ds)r/p(∫txk‍v(s)1-θ′ds)r/p′v(t)1-θ′dt)l/r)1/l+(∑k∈ℤ‍φ(xk)l/p(∫xkxk+1‍v(t)1-θ′dt)l/θ′)1/l<∞.
Since
(3.49)A4≈(∑k∈ℤ‍φ(xk)l/q(U-l/p(xk)(∫xk-1xk‍U(t)r/pVθ(t)r/p′v(t)1-θ′dt)l/r+Vθ(xk)l/θ′))1/l≲(∑k∈ℤ‍φ(xk)l/qUl/p(xk)(∫xk-1xk‍U(t)r/pVθ(t)r/p′v(t)1-θ′dt)l/r)1/l+(∑k∈ℤ‍φ(xk)l/qVθ(xk)l/θ′)1/l≲(∑k∈ℤ‍φ(xk)l/qUl/p(xk)(∫xk-1xk‍(∫xk-1t‍u(s)ds)r/pVθ(t)r/p′v(t)1-θ′dt)l/r)1/l+(∑k∈ℤ‍φ(xk)l/qVθ(xk)l/θ′)1/l,
integrating by parts, we find that
(3.50)A4≲(∑k∈ℤ‍φ(xk)l/qUl/p(xk)(∫xk-1xk‍(∫t∞‍v(s)1-θ′ds)(r/p′)+1u(t)(∫xk-1t‍u(s)ds)(r/p)-1dt)l/r)1/l+(∑k∈ℤ‍φ(xk)l/qVθ(xk)l/θ′)1/l≲(∑k∈ℤ‍φ(xk)l/qUl/p(xk)(∫xk-1xk‍(∫txk‍v(s)1-θ′ds)(r/p′)+1u(t)(∫xk-1t‍u(s)ds)(r/p)-1dt)l/r)1/l+(∑k∈ℤ‍φ(xk)l/qVθ(xk)l/θ′)1/l.
Again integrating by parts, we arrive at
(3.51)A4≲(∑k∈ℤ‍φ(xk)l/qUl/p(xk)(∫xk-1xk‍(∫xk-1t‍u(s)ds)r/p(∫txk‍v(s)1-θ′ds)r/p′v(t)1-θ′dt)l/r)1/l+(∑k∈ℤ‍φ(xk)l/qVθ(xk)l/θ′)1/l<∞.
Now let0<p<∞, θ=∞, 0<q<∞. For k∈ℤ, let hk be functions that saturate the Hardy inequality (2.21) and Hölder’s inequality for θ=∞, that is, functions hk satisfying
(3.52)supphk⊂[xk-1,xk],∥hk∥∞,v,(0,∞)=1,∥∫sxk‍hk(z)dz∥p,u,(xk-1,xk)≳(∫xk-1xk‍(∫sxk‍dzv(z))pu(s)ds)1/p,∫xk-1xk‍hk(t)dt≳∫xkxk+1‍dzv(z).
Now we define the test function
(3.53)h(t)=∑k∈ℤ‍akhk(t),
where {ak} is a sequence of positive real numbers. Thus, using test function (3.53) in (3.32), we get
(3.54)(∑k∈ℤ‍akqφ(xk)U(xk)q/p(∫xk-1xk‍(∫sxk‍dzv(z))pu(s)ds)q/p)1/q+(∑k∈ℤ‍akqφ(xk)(∫xkxk+1‍dzv(z))q)1/q≲supk∈ℤak.
Hence, by Proposition 2.14, we have that
(3.55)(∑k∈ℤ‍φ(xk)U(xk)q/p(∫xk-1xk‍(∫sxk‍dzv(z))pu(s)ds)q/p)1/q+(∑k∈ℤ‍φ(xk)(∫xkxk+1‍dzv(z))q)1/q<∞.
On the other hand,
(3.56)A5≈(∑k∈ℤ‍φ(xk)(U(xk)-1∫xk-1xk‍U(s)(∫s∞‍dzv(z))p-1dsv(s)+∫xk∞‍(∫s∞‍dzv(z))p-1dsv(s))q/p)1/q≈(∑k∈ℤ‍φ(xk)U(xk)q/p(∫xk-1xk‍U(s)(∫s∞‍dzv(z))p-1dsv(s))q/p)1/q+(∑k∈ℤ‍φ(xk)(∫xk∞‍dzv(z))q)1/q.
Integrating by part and using Lemma 2.9, we get that
(3.57)A5≲(∑k∈ℤ‍φ(xk)U(xk)q/p(∫xk-1xk‍(∫s∞‍dzv(z))pu(s)ds)q/p)1/q+(∑k∈ℤ‍φ(xk)(∫xk∞‍dzv(z))q)1/q≲(∑k∈ℤ‍φ(xk)U(xk)q/p(∫xk-1xk‍(∫sxk‍dzv(z))pu(s)ds)q/p)1/q+(∑k∈ℤ‍φ(xk)(∫xkxk+1‍dzv(z))q)1/q<∞.
The proof is complete.We now state the announced characterization of (3.2).Theorem 3.2.
Let0<p<∞, 1<θ≤∞, and let u,v,w be weights. Assume that u is such that U1/p is admissible and the measure w(t)dt is nondenerate with respect to U1/p. Then the inequality (3.2) holds for every measurable function f on (0,∞) if and only if(i)
1<θ≤p<∞ and
(3.58)B1:=supx∈(0,∞)esssups∈(0,∞)w(s)𝒰(x,s)1/psupt∈(0,∞)Vθ(t)1/θ′𝒰(t,x)1/p<∞.Moreover, the best constantc in (3.2) satisfies c≈B1.(ii)
0<p<θ<∞, 1<θ<∞,  r=θp/(θ-p) and
(3.59)B2:=supx∈(0,∞)esssups∈(0,∞)w(s)𝒰(x,s)1/p(∫0∞‍𝒰(t,x)r/pVθ(t)r/p′v(t)1-θ′dt)1/r<∞.Moreover, the best constantc in (3.2) satisfies c≈B2.(iii)
0<p<∞, θ=∞ and
(3.60)B3:=supx∈(0,∞)(∫0∞‍𝒰(s,t)(∫s∞‍dzv(z))p-1dsv(s))1/pesssups∈(0,∞)w(s)𝒰(x,s)1/p<∞.Moreover, the best constantc in (3.2) satisfies c≈B3.Proof.
Using Corollary2.5, Lemmas 2.8 and 2.9, we obtain for the left-hand side J0 of (3.2) that (φ is defined by (2.7))
(3.61)J0≈supt∈(0,∞)φ(t)U(t)1/p(∫0t‍(∫s∞‍h(z)dz)pu(s)ds)1/p≈supk∈ℤφ(xk)U(xk)1/p(∫0xk‍(∫s∞‍h(z)dz)pu(s)ds)1/p≈supk∈ℤφ(xk)U(xk)1/p(∫xk-1xk‍(∫sxk‍h(z)dz)pu(s)ds)1/p+supk∈ℤφ(xk)∫xkxk+1‍h(z)dz:=III+IV.(i)
For the case1<θ≤p<∞, by using Lemma 2.15 for III and applying Hölder’s inequality for IV, we arrive at
(3.62)III≲supk∈ℤφ(xk)U(xk)1/psupxk-1<t<xkU(t)1/pVθ(t)1/θ′(∫xk-1xk‍h(z)θv(z)dz)1/θ,IV≲supk∈ℤφ(xk)Vθ(xk)1/θ′(∫xk-1xk‍h(z)θv(z)dz)1/θ,so that, by Lemma2.13 and (3.61), we obtain that
(3.63)J0≲(supk∈ℤφ(xk)U(xk)1/psupxk-1<t<xkU(t)1/pVθ(t)1/θ′+supk∈ℤφ(xk)Vθ(xk)1/θ′)∥h∥θ,v,(0,∞)≈supx∈(0,∞)φ(x)sup0<t<∞𝒰(t,x)1/pVθ(t)1/θ′∥h∥θ,v,(0,∞)=supx∈(0,∞)esssups∈(0,∞)w(s)𝒰(x,s)1/psup0<t<∞Vθ(t)1/θ′𝒰(t,x)1/p∥h∥θ,v,(0,∞).(ii)
Let now0<p<θ<∞, 1<θ<∞, r=θp/(θ-p). By using Lemma 2.15 for III and applying Hölder’s inequality for IV, we find that
(3.64)III≲supk∈ℤφ(xk)U(xk)1/p(∫xk-1xk‍(∫xk-1t‍u(s)ds)r/p(∫txk‍v(s)1-θ′ds)r/p′v(t)1-θ′dt)1/r×(∫xk-1xk‍h(z)θv(z)dz)1/θ,IV≲supk∈ℤφ(xk)Vθ(xk)1/θ′(∫xk-1xk‍h(z)θv(z)dz)1/θand, by Lemmas2.9 and 2.12, and (3.61), we get that
(3.65)J0≲supk∈ℤφ(xk)(U(xk)-r/p∫xk-1xk‍U(t)r/pVθ(t)r/p′v(t)1-θ′dt+∫xk∞‍Vθ(t)r/p′v(t)1-θ′dt)1/r×∥h∥θ,v,(0,∞)≈supk∈ℤφ(xk)(U(xk)-r/p∫xk-1xk‍U(t)r/pVθ(t)r/p'v(t)1-θ′dt+∫xkxk+1‍Vθ(t)r/p'v(t)1-θ′dt)1/r×∥h∥θ,v,(0,∞)≈supx∈(0,∞)φ(x)(∫0∞‍𝒰(t,x)r/pVθ(t)r/p'v(t)1-θ′dt)1/r∥h∥θ,v,(0,∞)=supx∈(0,∞)esssups∈(0,∞)w(s)𝒰(x,s)1/p(∫0∞‍𝒰(t,x)r/pVθ(t)r/p'v(t)1-θ′dt)1/r∥h∥θ,v,(0,∞).(iii)
Now let0<p<∞, θ=∞. By using Lemma 2.15 for III, we deduce that
(3.66)III≲supk∈ℤφ(xk)U(xk)1/p(∫xk-1xk‍(∫sxk‍dzv(z))pu(s)ds)1/p∥h∥∞,v,(0,∞).Moreover, forIV, it yields that
(3.67)IV≲supk∈ℤφ(xk)∫xkxk+1‍dzv(z)∥h∥∞,v,(0,∞).Therefore, by using integration by parts, Lemma2.12, and (3.61), we get that
(3.68)J0≲supk∈ℤφ(xk)(U(xk)-1∫xk-1xk‍U(s)(∫s∞‍dzv(z))p-1dsv(s)+∫xk∞‍(∫s∞‍dzv(z))p-1dsv(s))1/p×∥h∥∞,v,(0,∞)≈supx∈(0,∞)(∫0∞‍𝒰(s,t)(∫s∞‍dzv(z))p-1dsv(s))1/pφ(x)∥h∥∞,v,(0,∞)=supx∈(0,∞)(∫0∞‍𝒰(s,t)(∫s∞‍dzv(z))p-1dsv(s))1/pesssups∈(0,∞)w(s)𝒰(x,s)1/p∥h∥∞,v,(0,∞).
Now we prove the lower bounds (necessity). Let{xk} be a discretizing sequence for φ defined by (2.7). Then, by (3.61), we have
(3.69)supk∈ℤφ(xk)U(xk)1/p(∫xk-1xk‍(∫sxk‍h(z)dz)pu(s)ds)1/p+supk∈ℤφ(xk)∫xkxk+1‍h(z)dz≲(∑k∈ℤ‍∫xk-1xk‍h(z)θv(z)dz)1/θ.
Let 1<θ≤p<∞. If we use in (3.69) the test function defined by (3.34), we obtain that
(3.70)supk∈ℤakφ(xk)U(xk)1/psupxk-1<t<xk(∫xk-1t‍u(s)ds)1/p(∫txk‍v(s)1-θ′ds)1/θ′+supk∈ℤakφ(xk)(∫xkxk+1‍v(t)1-θ′dt)1/θ′≲(∑k∈ℤ‍akθ)1/θ.
Therefore, by Proposition 2.14, we have that
(3.71)supk∈ℤφ(xk)U(xk)1/psupxk-1<t<xk(∫xk-1t‍u(s)ds)1/p(∫txk‍v(s)1-θ′ds)1/θ′+supk∈ℤφ(xk)(∫xkxk+1‍v(t)1-θ′dt)1/θ′<∞.
Since
(3.72)B1≈supk∈ℤφ(xk)U(xk)1/psupxk-1<t<xkU(t)1/pVθ(t)1/θ′+supk∈ℤφ(xk)Vθ(xk)1/θ′≲supk∈ℤφ(xk)U(xk)1/psupxk-1<t<xk(∫xk-1t‍u(s)ds)1/p(∫txk‍v(s)1-θ′ds)1/θ′+supk∈ℤφ(xk)Vθ(xk)1/θ′,
by Lemma 2.9, we get that
(3.73)B1≲supk∈ℤφ(xk)U(xk)1/psupxk-1<t<xk(∫xk-1t‍u(s)ds)1/p(∫txk‍v(s)1-θ′ds)1/θ′+supk∈ℤφ(xk)(∫xkxk+1‍v(t)1-θ′dt)1/θ′<∞.
Now let 0<p<θ<∞, 1<θ<∞, r=θp/(θ-p). By using in (3.69) the test function defined by (3.42), we obtain that
(3.74)supk∈ℤakφ(xk)U(xk)1/p(∫xk-1xk‍(∫xk-1t‍u(s)ds)r/p(∫txk‍v(s)1-θ′ds)r/p′v(t)1-θ′dt)1/r+supk∈ℤakφ(xk)(∫xkxk+1‍v(t)1-θ′dt)1/θ′≲(∑k∈ℤ‍akθ)1/θ.
Then, by Proposition 2.14, we get that
(3.75)supk∈ℤφ(xk)U(xk)1/p(∫xk-1xk‍(∫xk-1t‍u(s)ds)r/p(∫txk‍v(s)1-θ′ds)r/p′v(t)1-θ′dt)1/r+supk∈ℤφ(xk)(∫xkxk+1‍v(t)1-θ′dt)1/θ′<∞.
Since
(3.76)B2≈supk∈ℤφ(xk)(U(xk)-r/p∫xk-1xk‍U(t)r/pVθ(t)r/p'v(t)1-θ′dt+∫xk∞‍Vθ(t)r/p'v(t)1-θ′dt)1/θ′≲supk∈ℤφ(xk)U(xk)1/p(∫xk-1xk‍U(t)r/pVθ(t)r/p'v(t)1-θ′dt)1/r+supk∈ℤφ(xk)Vθ(xk)1/θ′≲supk∈ℤφ(xk)U(xk)1/p(∫xk-1xk‍(∫xk-1t‍u(s)ds)r/pVθ(t)r/p′v(t)1-θ′dt)1/r+supk∈ℤφ(xk)Vθ(xk)1/θ′,
by integrating by parts, we find that
(3.77)B2≲supk∈ℤφ(xk)U(xk)1/p(∫xk-1xk‍(∫t∞‍v(s)1-θ′ds)(r/p′)+1u(t)(∫xk-1t‍u(s)ds)(r/p)-1dt)1/r+supk∈ℤφ(xk)Vθ(xk)1/θ′≲supk∈ℤφ(xk)U(xk)1/p(∫xk-1xk‍(∫txk‍v(s)1-θ′ds)(r/p′)+1u(t)(∫xk-1t‍u(s)ds)(r/p)-1dt)1/r+supk∈ℤφ(xk)Vθ(xk)1/θ′.
Moreover, by again integrating by parts, we arrive at
(3.78)B2≲supk∈ℤφ(xk)U(xk)1/p(∫xk-1xk‍(∫xk-1t‍u(s)ds)r/p(∫txk‍v(s)1-θ′ds)r/p′v(t)1-θ′dt)1/r+supk∈ℤφ(xk)Vθ(xk)1/θ′<∞.
Finally, let0<p<∞, θ=∞. By using the test function defined by (3.53) in (3.69), we get that
(3.79)supk∈ℤakφ(xk)U(xk)1/p(∫xk-1xk‍(∫sxk‍dzv(z))pu(s)ds)1/p+supk∈ℤakφ(xk)∫xkxk+1‍dzv(z)≲supk∈ℤak.
Hence, by Proposition 2.14, we have that
(3.80)supk∈ℤφ(xk)U(xk)1/p(∫xk-1xk‍(∫sxk‍dzv(z))pu(s)ds)1/p+supk∈ℤφ(xk)∫xkxk+1‍dzv(z)<∞.
Since
(3.81)B3≈supk∈ℤφ(xk)(U(xk)-1∫xk-1xk‍U(s)(∫s∞‍dzv(z))p-1dsv(s)+∫xk∞‍(∫s∞‍dzv(z))p-1dsv(s))1/p≲supk∈ℤφ(xk)U(xk)1/p(∫xk-1xk‍U(s)(∫s∞‍dzv(z))p-1dsv(s))1/p+supk∈ℤφ(xk)∫xk∞‍dzv(z)≲supk∈ℤφ(xk)U(xk)1/p(∫xk-1xk‍U(s)(∫sxk‍dzv(z))p-1dsv(s))1/p+supk∈ℤφ(xk)∫xk∞‍dzv(z),
by integrating by parts and using Lemma 2.9, we obtain that
(3.82)B3≲supk∈ℤφ(xk)U(xk)1/p(∫xk-1xk‍(∫sxk‍dzv(z))pu(s)ds)1/p+supk∈ℤφ(xk)∫xkxk+1‍dzv(z)<∞.
The proof is complete.



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*Source: 734194-2012-12-10.xml*