# Boundedness of One Class of Integral Operators from Second Order Weighted Sobolev Space to Weighted Lebesgue Space

Authors: Aigerim Kalybay
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In the paper, for a certain class of Hardy operators with kernels, we consider the problem of their boundedness from a second order weighted Sobolev space to a weighted Lebesgue space.


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## Body

## 1. Introduction

LetI=0,∞ and 1<p,q<∞. Let u and v be positive functions locally integrable on the interval I. In addition, suppose that v−p′∈L1locI, where p′=p/p−1.LetWp,v2≡Wp2v,I be a set of functions f:I⟶ℝ having generalized derivatives up to the second order on I with the finite norm
(1)fWp,v2=vf′′p+f′1+f1,where ·p is the standard norm of the space LpI, 1<p<∞.In the paper, we consider the problem of boundedness of the integral operator(2)Kfx=∫0xKx,sfsds,x>0,with a kernel Kx,s≥0 from the weighted space Wp,v2 to the weighted space Lq,u≡Lqu,I with the norm fLq,u=ufq. This problem is equivalent to the validity of the following inequality
(3)uKfq≤CfWp,v2,f∈Wp,v2.LetC0∞I be the set of compactly supported functions infinitely time continuously differentiable on I. Due to the assumptions on v, we have that C0∞I⊂Wp,v2. Denote by Ẇp,v2≡Ẇp2v,I the closure of the set C0∞I with respect to norm defined by (1). Depending on the behaviour of the function v at zero and infinity, the set C0∞I can be dense or not dense in the space Wp,v2, i.e., Ẇp,v2=Wp,v2 or
(4)Ẇp,v2⊂Wp,v2andẆp,v2≠Wp,v2,respectively.In the paper, we study inequality (3) under condition (4) for a certain class of integral operators. Note that in the case when K is the identity operator I, inequalities of form (3) have been studied in many papers. Some results with proofs and a survey of other results with comments are given in Chapter 4 of the book [3]. Our work is related to the works [5, 6], in which inequality (3) with K≡I was studied under various zero boundary conditions for f∈Wp,v2.The boundedness of integral operators in form (2) from a first order weighted Sobolev space to a weighted Lebesgue space has been investigated in the series of papers (see, e.g., [1, 2] and references given therein).The paper is organized as follows. In Section2, we present definitions and statements required to prove the main results. In Section 3, we present and prove the main results, especially we obtain necessary and sufficient conditions for the validity of inequality (3). In Section 4, we present corollaries that follow from the results of Section 3.


## 2. Axillary Definitions and Statements

Let−∞≤a<b≤∞. In the paper, χa,b· is the characteristic function of the interval a,b. Moreover, the notation A≪B means A≤cB and A≈B means A≪B≪A.From the book [3], we have the following theorem.Theorem 1.
Let1<p≤q<∞.
(i)
The inequality(5)∫abut∫atfsdsqdt1/q≤C∫abvtftpdt1/p,holds if and only if
(6)A=supz∈a,b∫zbuqtdt1/q∫azv−p′sds1/p′<∞.
In addition,C≈A, where C is the best constant in (5).
(ii)
The inequality(7)∫abut∫tbfsdsqdt1/q≤C∫abvtftpdt1/p,holds if and only if
(8)A⋆=supz∈a,b∫azuqtdt1/q∫zbv−p′sds1/p′<∞.In addition,C≈A⋆, where C is the best constant in (7).The following definitions and statements are from the paper [7].Definition 2.
LetKx,s be a nonnegative function measurable on the set Ωx,s:a<s≤x<b and nonincreasing in the second argument. We say that the function Kx,s belongs to the class O1−Ω if there exist nonnegative functions wx and K0,1t,s measurable on Ω such that
(9)Kx,s≈Kx,t+wxK0,1t,s,for a<s≤t≤x<b; moreover, the equivalence coefficients in (9) do not depend on s, t, and x.Definition 3.
LetKx,s be a nonnegative function measurable on the set Ω and nonincreasing in the second argument. We say that the function Kx,s belongs to the class O2−Ω if there exist K1x,t∈O1−Ω and nonnegative functions wx, K0,2t,s, and K1,2t,s measurable on Ω such that
(10)Kx,s≈Kx,t+K1x,tK1,2t,s+wxK0,2t,s,for a<s≤t≤x<b; moreover, the equivalence coefficients in (10) do not depend on s, t, and x.Definition 4.
LetKx,s be a nonnegative function measurable on the set Ω and nonincreasing in the second argument. We say that the function Kx,s belongs to the class O3−Ω if there exist K1x,t∈O1−Ω, K2x,t∈O2−Ω, and nonnegative functions wx, K0,3t,s, K1,3t,s, and K2,3t,s measurable on Ω such that
(11)Kx,s≈Kx,t+K2x,tK2,3t,s+K1x,tK1,3t,s+wxK0,3t,s,for a<s≤t≤x<b; moreover, the equivalence coefficients in (11) do not depend on s, t, and x.
Let(12)B1,1a,b=supz∈a,b∫zbKqx,zuqxdx1/q∫azv−p′sds1/p′,(13)B1,0a,b=supz∈a,b∫zbuqxwqxdx1/q∫azK0,1p′z,sv−p′sds1/p′,(14)B1a,b=maxB1,1a,b,B1,0a,b.Theorem 5.
Let1<p≤q<∞. Let the kernel of operator (2) belong to the class O1−. Then, the inequality
(15)∫abux∫axKx,sfsdsqdx1/q≤C∫abvtftpdt1/p,holds if and only if B1a,b<∞. In addition, C≈B1a,b, where C is the best constant in (15).
Let(16)B2,2a,b=supz∈a,b∫zbKqx,zuqxdx1/q∫azv−p′sds1/p′,(17)B2,1a,b=supz∈a,b∫zbK1qx,zuqxdx1/q∫azK1,2p′z,sv−p′sds1/p′,(18)B2,0a,b=supz∈a,b∫zbuqxwqxdx1/q∫azK0,2p′z,sv−p′sds1/p′,(19)B2a,b=maxB2,2a,b,B2,1a,b,B2,0a,b.Theorem 6.
Let1<p≤q<∞. Let the kernel of operator (2) belong to the class O2−Ω. Then, inequality (15) holds if and only if B2a,b<∞. In addition, C≈B2a,b, where C is the best constant in (15).
Let(20)B3,3a,b=supz∈a,b∫zbKqx,zuqxdx1/q∫azv−p′sds1/p′,(21)B3,2a,b=supz∈a,b∫zbK2qx,zuqxdx1/q∫azK2,3p′z,sv−p′sds1/p′,(22)B3,1a,b=supz∈a,b∫zbK1qx,zuqxdx1/q∫azK1,3p′z,sv−p′sds1/p′,(23)B3,0a,b=supz∈a,b∫zbuqxwqxdx1/q∫azK0,3p′z,sv−p′sds1/p′,(24)B3a,b=maxB3,3a,b,B3,2a,b,B3,1a,b,B3,0a,b.Theorem 7.
Let1<p≤q<∞. Let the kernel of operator (2) belong to the class O3−Ω. Then, inequality (15) holds if and only if B3a,b<∞. In addition, C≈B3a,b, where C is the best constant in (15).Forf∈Wp,v2, we assume that limt⟶0+ft=f0, limt⟶0+f′t=f′0, limt⟶∞ft=f∞, and limt⟶∞f′t=f′∞ regardless of whether they are finite or infinite.The following statement is from the paper [4].Theorem 8.
Let1<p<∞. If the conditions
(25)∫0∞v−p′sds<∞and∫1∞tp′v−p′tdt=∞,hold; then, for f∈Wp,v2, there exist the finite values f0, f′0, and f′∞ such that
(26)Ẇp,v2=f∈Wp,v2:f0=f′0=f′∞=0.


## 3. Main Results

First, we state some necessary lemmas. Some of them are new and of independent interest, and therefore proved in detail.Lemma 9.
LetKx,s≡K1x,s∈O1−Ω, where K1x,s≈K1x,t+wxK0,1t,s for a<s≤t≤x<b. Then
(27)K2x,s≡∫sxK1x,tdt∈O2−Ω,(28)∫sτK1x,tt−sdt≈K1x,ττ−s2+wx∫sτK0,1τ,tt−sdt,for a<s≤τ≤x<b;
(29)K3x,s≡∫sxK1x,tt−sdt∈O3−Ω.Proof.
(i)
Fora<s≤τ≤x<b, we have(30)K2x,s=∫sxK1x,tdt=∫τxK1x,tdt+∫sτK1x,tdt≈K2x,τ+K1x,ττ−s+wx∫sτK0,1τ,tdt.
Therefore, by (10), we get that K2x,s∈O2−Ω.
(ii)
Fora<s≤τ≤x<b, it easily follows that(31)∫sτK1x,tt−sdt≈∫sτK1x,τt−sdt+∫sτwxK0,1τ,tt−sdt≈K1x,ττ−s2+wx∫sτK0,1τ,tt−sdt.(iii)
Using (28), for a<s≤τ≤x<b, we have(32)K3x,s=∫sxK1x,tt−sdt=∫τxK1x,tt−sdt+∫sτK1x,tt−sdt≈∫τxK1x,tt−τdt+∫τxK1x,tdtτ−s+K1x,ττ−s2+wx∫sτK0,1τ,tt−sdt=K3x,τ+K2x,ττ−s+K1x,ττ−s2+wx∫sτK0,1τ,tt−sdt.
Then, in view of (11), we obtain that K3x,s∈O3−Ω. The proof is complete.
Leta=0 and b=∞. Assume that
(33)B3,3−τ=sup0<z<τ∫zτ∫zxKx,tt−zdtquqxdx1/q∫0zv−p′sds1/p′,(34)B3,2−τ=sup0<z<τ∫zτ∫zxKx,tdtquqxdx1/q∫0zz−sp′v−p′sds1/p′,(35)B3,1−τ=sup0<z<τ∫zτKqx,zuqxdx1/q∫0zz−s2p′v−p′sds1/p′,(36)B3,0−τ=sup0<z<τ∫zτuqxwqxdx1/q∫0z∫szK0,1z,tt−sdtp′v−p′sds1/p′,(37)B3−τ=maxB3,3−τ,B3,2−τ,B3,1−τ,B3,0−τ.By using part (iii) of Lemma9 and Theorem 7, we have one more lemma.Lemma 10.
Let1<p≤q<∞ and Kx,s∈O1−Ω. Then, the inequality
(38)∫0τux∫0x∫sxKx,tt−sdtgsdsqdx1/q≤C1−∫0τvtgtpdt1/p,holds if and only if B3−τ<∞. In addition, C1−≈B3−τ, where C1− is the best constant in (38).
Let(39)B1,1−τ=∫τ∞Kqx,τuqxdx1/q∫0ττ−s2p′v−p′sds1/p′,(40)B1,0−τ=∫τ∞uqxwqxdx1/q∫0τ∫sτK0,1τ,tt−sdtp′v−p′sds1/p′,(41)B1−τ=maxB1,1−τ,B1,0−τ.Using (28) and the inverse Hölder’s inequality, by Theorem 5, we have the following lemma.Lemma 11.
Let1<p≤q<∞ and Kx,s∈O1−Ω. Then, the inequality
(42)∫τ∞ux∫0τ∫sτKx,tt−sdtgsdsqdx1/q≤C2−∫0τvtgtpdt1/p,holds if and only if B1−τ<∞. In addition, C2−≈B1−τ, where C2− is the best constant in (42).
Let(43)A−τ=∫τ∞∫τxKx,tdtquqxdx1/q∫0ττ−sp′v−p′sds1/p′.From part (i) of Theorem1, we can state the following lemma.Lemma 12.
Let1<p≤q<∞. Then, the inequality
(44)∫τ∞ux∫τxKx,tdt∫0ττ−sgsdsqdx1/q≤C3−∫0τvtgtpdt1/p,holds if and only if A−τ<∞. In addition, C3−≈A−τ, where C3− is the best constant in (44).
Assume that(45)B2,2+τ=supz>τ∫z∞∫zxKx,tdtquqxdx1/q∫τzs−τp′v−p′sds1/p′,(46)B2,1+τ=supz>τ∫z∞Kqx,zuqxdx1/q∫τzz−sp′s−τp′v−p′sds1/p′,(47)B2,0+τ=supz>τ∫z∞uqxwqxdx1/q∫τz∫szK0,1z,tdtp′s−τp′v−p′sds1/p′,(48)B2+τ=maxB2,2+τ,B2,1+τ,B2,0+τ.By using part (i) of Lemma9 and Theorem 6, we get the following statement.Lemma 13.
Let1<p≤q<∞ and Kx,s∈O1−Ω. Then, the inequality
(49)∫τ∞ux∫τx∫sxKx,tdts−τgsdsqdx1/q≤C1+∫τ∞vtgtpdt1/p,holds if and only if B2+τ<∞. In addition, C1+≈B2+τ, where C1+ is the best constant in (49).
Let(50)B1,1+τ=supz>τ∫z∞Kqx,zuqxdx1/q∫τzs−τ2p′v−p′sds1/p′,(51)B1,0+τ=supz>τ∫z∞uqxwqxdx1/q∫τzK0,1p′z,ss−τ2p′v−p′sds1/p′,(52)A+τ=supz>τ∫z∞uqxwqxdx1/q∫τz∫τsK0,1s,tt−τdtp′v−p′sds1/p′,(53)B1A+τ=maxB1,1+τ,B1,0+τ,A+τ.Lemma 14.
Let1<p≤q<∞ and Kx,s∈O1−Ω. Then, the inequality
(54)∫τ∞ux∫τx∫τsKx,tt−τdtgsdsqdx1/q≤C2+∫τ∞vtgtpdt1/p,holds if and only if B1A+τ<∞. In addition, C2+≈B1A+τ, where C2+ is the best constant in (54).Proof.
SinceKx,s∈O1−Ω, by Lemma 9, we have that
(55)∫τsKx,tt−τdt≈Kx,ss−τ2+wx∫τsK0,1s,tt−τdt.
Hence, inequality (54) is equivalent to simultaneous fulfilment of the following inequalities:
(56)∫τ∞ux∫τxKx,ss−τ2gsdsqdx1/q≤C2,2+∫τ∞vtgtpdt1/p,(57)∫τ∞uxwx∫τx∫τsK0,1s,tt−τdtgsdsqdx1/q≤C2,1+∫τ∞vtgtpdt1/p.
In addition,maxC2,1+,C2,2+≈C2+, where C2,1+ and C2,2+ are the best constants in (56) and (57), respectively. By Theorem 5, inequality (56) holds if and only if maxB1,1+τ,B1,0+τ<∞, and in addition, C2,2+≈maxB1,1+τ,B1,0+τ. By part (i) of Theorem 1, inequality (57) holds if and only if A+τ<∞, and in addition, C2,1+≈A+τ. Then, inequality (54) holds if and only if B1A+τ<∞ and C2+≈B1A+τ. The proof is complete.Assume that(58)A∗+τ=supz>τ∫τz∫τxKx,tt−τdtquqxdx1/q∫x∞v−p′sds1/p′.By using part (ii) of Theorem1, we have the following lemma.Lemma 15.
Let1<p≤q<∞. Then, the inequality
(59)∫τ∞ux∫τxKx,tt−τdt∫x∞gsdsqdx1/q≤C3+∫τ∞vtgtpdt1/p,holds if and only if A∗+τ<∞. In addition, C3+≈A∗+τ, where C3+ is the best constant in (59).Let infinitely differentiable functionsφ and ψ be such that 1≥φ≥0, 1≥ψ≥0, φt=1 for 0<t<1/2, φt=0 for t≥3/2, ψt=1 for t>3/2 and ψt=0 for 0<t≤1/2. Moreover, φt>0, ψt>0 for 1/2<t<3/2 and φt+ψt=1 for all t∈I.Assume thatP− and P+ are polynomials such that P−t=c0+c1t and P+t=c2t, where ci∈ℝ, i=0,1,2. Denote by P− and P+ the sets of polynomials in the form P− and P+, respectively.Let the conditions of Theorem8 hold. Then, from (26), we have
(60)Wp,v2=Ẇp,v2⊕φχ0,1P−⊕ψχ1,∞P+,where ⊕ means the direct sum.From Theorem8, it follows that (1) is equivalent to the norm
(61)fWp,v2=vf′′p+f′0+f0.Therefore, forf∈Wp,v2 we have
(62)fWp,v2=vf′′p.First, using (60), we establish inequality (3) on the set Wp,v2, which, due to (62), has the form:
(63)uKfq≤Cvf′′p,f∈Ẇp,v2.Assume that(64)Eτ=maxB3−τ,B1−τ,A−τ,Fτ=maxB2+τ,B1A+τ,A∗+τ,(65)EF=infτ∈ImaxEτ,Fτ.Our first main result reads.Theorem 16.
Let1<p≤q<∞ and Kx,s∈O1−Ω. Let condition (25) hold. Then, inequality (63) holds if and only if EF<∞. In addition, C≈EF, where C is the best constant in (63).Proof.
Sufficiency. From (25) by Theorem 8, it follows the validity of (26). As in Theorem 2.1 of [5], using (26), for f∈Wp,v2, we get
(66)ft=χ0,τt∫0tt−sf′′sds+χτ,∞t∫0ττ−sf′′sds−∫τts−τf′′sds−t−τ∫t∞f′′sds,where τ∈I. Assuming f′′=g in (66), we have that g∈Lp,vI. Moreover, from (26), it follows that ∫0∞gsds=0.
Assume thatL~p,vI=g∈Lp,vI:∫0∞gsds=0. Then, in (66), the condition f∈Ẇp,v2 is equivalent to the condition g∈L~p,vI. Replacing (66) into the left-hand side of (63), we find that
(67)F≡uKfqq=∫0τux∫0xKx,tftdtqdx+∫τ∞ux∫0xKx,tftdtqdx=∫0τux∫0xKx,t∫0tt−sgsdsqdx+∫τ∞ux∫0τKx,t∫0tt−sgsdsdt+ux∫τxKx,t∫0ττ−sgsdsdt−ux∫τxKx,t∫τts−τgsdsdt−ux∫τxKx,tt−τ∫t∞gsdsdtqdx=∫0τux∫0x∫sxKx,tt−sdtgsdsqdx+∫τ∞ux∫0τ∫sτKx,tt−sdtgsds+ux∫τxKx,tdt∫0ττ−sgsds−ux∫τx∫sxKx,tdts−τgsds−ux∫τx∫τsKx,tt−τdtgsds−ux∫τxKx,tt−τdt∫x∞gsdsqdx.
Therefore, inequality (63) has the form
(68)∫0τux∫0x∫sxKx,tt−sdtgsdsqdx+∫τ∞ux∫0τ∫sτKx,tt−sdtgsds+ux∫τxKx,tdt∫0ττ−sgsds−ux∫τx∫sxKx,tdts−τgsds−ux∫τx∫τsKx,tt−τdtgsds−ux∫τxKx,tt−τdt∫x∞gsdsqdx1/q≤C∫0∞vsgspds1/p,g∈L~p,vI.
In the left-hand side of (68), using the Minkowski’s inequality for sums, then, applying Lemmas 10, 11, 12, 13, and 14 to each term, we get
(69)F≪Eτ∫0τvsgspds1/p+Fτ∫τ∞vsgspds1/p≪maxEτ,Fτ∫0∞vsgspds1/p.
Since the left-hand side of inequality (63) does not depend on τ∈I, then, taking in the right-hand side of (69) infimum with respect to τ∈I, we can conclude that
(70)C≪EF,where C is the best constant in (63).
Necessity. By the conditions of Theorem 16, we have that v−1∈Lp′I. Then, for any τ∈I, there exists kτ such that
(71)∫0τv−p′tdt=kτ∫τ∞v−p′tdt,in addition, kτ increases in τ, limτ⟶0+kτ=0 and limτ⟶∞kτ=∞.
Let us use the ideas in the paper [5]. For τ∈I, we consider two sets L1=g∈Lp,v0,τ:g≥0 and L2=g∈Lp,vτ,∞:g≤0. For each g1∈L1 and g2∈L2, we, respectively, construct the functions g2∈L2 and g1∈L1 so that gt=g1t for 0<t≤τ and gt=g2t for t>τ belongs to the set L~p,vI.
We define a strictly increasing functionρ:0,τ⟶τ,∞ from the relation
(72)∫0sv−p′tdt=kτ∫ρs∞v−p′tdt,s∈0,τ;(73)∫0ρ−1sv−p′tdt=kτ∫s∞v−p′tdt,s∈τ,∞,where ρ−1 is inverse to ρ. From (73), it follows that the functions ρ and ρ−1 are locally absolutely continuous, ρτ=τ and lims⟶0+ρs=∞.Differentiating both relations in (73), we have
(74)1kτ=v−p′ρsv−p′sρ′s,s∈0,τ;kτ=v−p′ρ−1sv−p′sρ−1s′,s∈τ,∞.
Then, forg1∈L1, we construct
(75)g2t=−kτg1ρ−1tv−p′tv−p′ρ−1t,t>τ,while for g2∈L2, we construct
(76)g1t=−1kτg2ρtv−p′tv−p′ρt,0<t≤τ.
Changing the variablesρ−1t=s and using the first equality in (74), we find that
(77)∫τ∞vtg2tpdt=kτp∫τ∞vtg1ρ−1tv−p′tv−p′ρ−1tpdt=kτp∫0τvpρsg1psv−pp′ρsv−pp′sρ′sds=kτp∫0τvsg1spv−p′ρsv−p′sρ′sds=kτp−1∫0τvsg1spds<∞.
Similarly, using the second equality in (74), we get
(78)∫0τvtg1tpdt=kτ1−p∫τ∞vsg2spds<∞.
From (77) and (78), assuming that gt=g1t for 0<t≤τ and gt=g2t for t>τ, we have
(79)∫0∞vtgtpdt=1+kτp−1∫0τvtg1tpdt=1+kτ1−p∫τ∞vtg2tpdt<∞,
i.e.,g∈Lp,vI. For any τ∈I integrating both sides of (75) from τ to ∞ and (76) from 0 to τ, we find that
(80)∫τ∞gtdt=−∫0τgtdt,i.e.,∫0∞gtdt=0.
Hence, constructed from the functionsg1∈L1 and g2∈L2, the function g belongs to L~p,vI. Replacing it into (68), we get
(81)∫0τux∫0x∫sxKx,tt−sdtg1sdsqdx+∫τ∞ux∫0τ∫sτKx,tt−sdtg1sds+ux∫τxKx,tdt∫0ττ−sg1sds+ux∫τx∫sxKx,tdts−τg2sds+ux∫τx∫τsKx,tτ−sdtg2sds+ux∫τxKx,tt−τdt∫x∞g2sdsqdx1/q≤C∫0∞vsgspds1/p,where all terms in the left-hand side are nonnegative.
Let the functiong∈L~p,vI constructed from the function g1∈L1. Then, from (81) and (79), we have
(82)∫0τux∫0x∫sxKx,tt−sdtg1sdsqdx+∫τ∞ux∫0τ∫sτKx,tt−sdtg1sds+ux∫τxKx,tdt∫0ττ−sg1sqdx1/q≤C1+kτp−11/p∫0τvsg1spds1/p.
Due to arbitrariness ofg1∈Lp,v0,τ, by Lemmas 10, 11, and 12, the latter gives that
(83)Eτ≪1+kτp−11/pC.
Similarly, due to (81) and (79), for the function g∈L~p,vI constructed from the function g2∈L2, we obtain
(84)Fτ≪1+kτ1−p1/pC.
From (83) and (84), we find that
(85)EF=infτ∈ImaxEτ,Fτ≪Cinfτ∈Imax1+kτp−11+kτ1−p1/p≤41/pC.
Therefore,EF≪C, which, together with (70), yields that EF≈C, where C is the best constant in (63). The proof is complete.Let(86)G=max∫0∞ux∫0xKx,tdtqdx1/q,∫0∞ux∫0xKx,ttdtqdx1/q.Our main result concerning Hardy-type inequality (3) reads.Theorem 17.
Let1<p≤q<∞ and Kx,s∈O1−Ω. Let conditions in (25) hold. Then, inequality (3) holds if and only if maxG,EF<∞. In addition, C≈maxG,EF, where C is the best constant in (3).Proof.
Due to (60), we consider inequality (3) on the set
(87)φχ0,1P−+ψχ1,∞P+=H.
The functionf∈H has the form
(88)ft=φtχ0,1tc0+c1t+ψtχ1,∞tc2t.
Hence,f′′t=0 almost everywhere on t∈I. Therefore, on the basis of (61), we have
(89)uKfq≤Cf′0+f0,f∈H.
Letf0∈H be such that
(90)f0t=φtχ0,1t+ψtχ1,∞tt.
Then, from (89), we obtain
(91)C≥∫0∞ux∫0xKx,tf0tdtqdx1/q≥minφ1,ψ1×∫01ux∫0xKx,tdtqdx+∫1∞ux∫01Kx,tdt+ux∫1xKx,ttdtqdx1/q,which implies that C≫G.
Letmaxc0+c1,c2=L. Then, ft≤Lχ0,1t+χ1,∞tt. Replacing the function f into the left-hand side of (89), we get
(92)∫0∞ux∫0xKx,tftdtqdx1/q≤L∫01ux∫0xKx,tdtqdx+∫1∞ux∫01Kx,tdt+ux∫1xKx,ttdtqdx1/q≪LG.
The latter, together withC≫G, gives that C≈G. Then, by Theorem 16, it follows that C≈maxG,EF, where C is the best constant in (3). The proof is complete.


## 4. Corollaries

Assume that the kernelKx,s of operator (2) satisfies the Oinarov condition
(93)Kx,s≈Kx,t+Kt,s,0<s≤t≤x<∞,which is often applied for integral operators. Then, in B3−τ, the expression B3,0−τ turns to the expression
(94)B3,0−τ=sup0<z<τ∫zτuqxdx1/q∫0z∫szKz,tt−sdtp′v−p′sds1/p′,in B1−τ the expression B1,0−τ turns to the expression
(95)B1,0−τ=∫τ∞uqxdx1/q∫0τ∫sτKτ,tt−sdtp′v−p′sds1/p′,in B2+τ the expression B2,0+τ turns to the expression
(96)B2,0+τ=supz>τ∫z∞uqxdx1/q∫τz∫szKz,tdtp′s−τp′v−p′sds1/p′,and in B1A+τ, the expressions B1,0+τ, A+τ, respectively, turn to the expressions
(97)B1,0+τ=supz>τ∫z∞uqxdx1/q∫τzKp′z,ss−τ2p′v−p′sds1/p′,(98)A+τ=supz>τ∫z∞uqxdx1/q∫τz∫τsKs,tt−τdtp′v−p′sds1/p′.After these changes, we denoteEτ by E~τ, Fτ by F~τ, and EF by E~F~ and get the following statement.Corollary 18.
Let1<p≤q<∞ and the kernel of (2) satisfy condition (93). Let conditions (25) hold.
(i)
Inequality (63) holds if and only if E~F~<∞. In addition, C≈E~F~, where C is the best constant in (63)(ii)
Inequality (3) holds if and only if maxG,E~F~<∞. In addition, C≈maxG,E~F~, where C is the best constant in (3)
Letn≥3. Instead of operator (2), we consider the operator of Riemann-Liouville In−2, defined by
(99)In−2fx=1n−3!∫0xx−tn−3ftdt.
The kernelx−tn−3 of the operator In−2 satisfies condition (93), and therefore, it belongs to the class O1−Ω. In this case, we replace Kx,t and K0,1x,t by x−tn−3 and assume that wx≡1. For the kernel x−tn−3 inequality, (38) has the form
(100)∫0τux∫0xx−sn−1gsdsqdx1/q≤C1−∫0τvsgspds1/p.
Then, according to Theorem5, we have C1−≈B1−τ=maxB1,1−τ,B1,2−τ, where
(101)B1,1−τ=sup0<z<τ∫zτx−zqn−1uqxdx1/q∫0zv−p′sds1/p′,(102)B1,0−τ=sup0<z<τ∫zτuqxdx1/q∫0zz−sp′n−1v−p′sds1/p′.
For the sum of kernels of the operators in (42) and (44), we deduce that
(103)∫sτx−tn−3t−sdt+∫τxx−tn−3dtτ−s≈x−τn−3τ−s2+τ−sn−1+x−τn−2τ−s≈τ−sn−1+x−τn−2τ−s.
Then instead of inequalities (42) and (44), we, respectively, have
(104)∫τ∞ux∫0ττ−sn−1gsdsqdx1/q≤C2−∫0τvsgspds1/p,(105)∫τ∞uxx−τn−2∫0ττ−sgsdsqdx1/q≤C3−∫0τvsgspds1/p.
By part (i) of Theorem1, this yields that
(106)C2−≈∫τ∞uqxdx1/q∫0ττ−sp′n−1v−p′sds1/p′=A1−τ,(107)C3−≈∫τ∞x−τqn−2uqxdx1/q∫0ττ−sp′v−p′sds1/p′=A2−τ.
Assume thatA−τ=maxA1−τ,A2−τ. Now, for the sum of kernels of the operators in (49) and (54), we get
(108)∫sxx−tn−3dts−τ+∫τsx−tn−3t−τdt≈x−sn−2s−τ+x−sn−3s−τ2+s−τn−1≈x−sn−2s−τ+s−τn−1≈x−τn−2s−τ.
Then instead of inequalities (49) and (54), we obtain
(109)∫τ∞uxx−τn−2∫τxs−τgsdsqdx1/q≤C1+∫τ∞vsgspds1/p.
Hence, by part (i) of Theorem1 we have
(110)C1+≈supz>τ∫z∞x−τqn−2uqxdx1/q∫τzs−τp′v−p′sds1/p′=A+τ.
For the kernelx−tn−3 inequality (59) can be written as follows:
(111)∫τ∞uxx−τn−1∫x∞gsdsqdx1/q≤C2+∫τ∞vsgspds1/p.
Therefore, by using part (ii) of Theorem1, we find
(112)C2+≈∫τzx−τqn−1uqxdx1/q∫z∞v−p′sds1/p′=A∗+τ.
Assume that(113)Eτ=maxB1−τ,A−τ,Fτ=maxA+τ,A∗+τ,(114)EF=infτ∈ImaxEτ,Fτ.
Thus, for inequality (63) with operator (99)
(115)uIn−2fq≤Cvf′′p,f∈Wp,v2,we can conclude the following statement.Corollary 19.
Let1<p≤q<∞ and conditions (25) hold. Then, inequality (115) holds if and only if EF<∞. In addition, C≈EF, where C is the best constant in (115).Assume thatIn−2ft=gt in (115). Then, gn−2t=ft and gi0=0, i=0,1,⋯,n−3. Moreover, gnt=f′′t and inequality (115) turns to the inequality
(116)ugq≤Cvgnp,with conditions
(117)gi0=0,i=0,1,⋯,n−1,gn−1∞=0.From Corollary19, we get one more corollary.Corollary 20.
Let1<p≤q<∞ and conditions (25) hold. Then, inequality (116) with conditions (117) holds if and only if EF<∞. In addition, C≈EF, where C is the best constant in (116).The statement of Corollary20 gives one of the results of the work [6].



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