# Multivariate Dynamic Sneak-Out Inequalities on Time Scales

Authors: Ammara Nosheen; Aneeqa Aslam; Khuram Ali Khan; Khalid Mahmood Awan; Hamid Reza Moradi
Journal: Journal of Mathematics
 (2021)

---

## Abstract

In this study, we extend some “sneak-out” inequalities on time scales for a function depending on more than one parameter. The results are proved by using the induction principle and time scale version of Minkowski inequalities. In seeking applications, these inequalities are discussed in classical, discrete, and quantum calculus.


---

## Body

## 1. Introduction

Bennett and Grosse-Erdmann [1] introduce the “sneak-out” principle concerned with the equivalence of two series. Bohner and Saker [2] extended the sneak-out principle on time scales and proved some new dynamic sneak-out inequalities and their converses on time scales which, as special cases, with T=ℕ, contain the discrete inequalities obtained by Bennett and Grosse-Erdmann (Section 6 in [1]). However, the sneak-out principle on time scales can be applied to formulate the corresponding integral inequalities by choosing T=ℝ. The paper aims to extend the work given by Bohner and Saker in [2] for functions depending on more than one parameter. Some other inequalities, such as Hardy-type, Hardy-Copson, and Copson-Leindler-type inequalities, are also studied for functions of more than one parameter [3–5] via time scales’ calculus. Some literature concerning with time scale can be seen in [6–13].The paper is organized as follows. Section2 provides some basics from time scales’ calculus. Section 3 features two dynamic inequalities of the Copson type, which are needed to prove further results. In Section 4, we present sneak-out inequalities on time scales for functions depending on more than one parameter.


## 2. Preliminaries

A time scaleT as well as close set in ℝ are nonempty [14, 15]. Some examples of time scales are ℤ, ℝ, and Cantor set. Assume that infT=ϕ, where ϕ is empty set and supT=∞. A time-scale interval is denoted by t0,∞T≔t0,∞∩T, for t0∈T.The operatorsσ:T⟶T defined by σl≔infb∈T;b>l and ρ:T⟶T defined by ρl≔supb∈T;b<l are forward as well as backward jump operators, respectively, for l∈T. The point l∈T is right-scattered if it satisfies σl>l, and left-scattered if ρl<l. The points which are at the same time left-scattered as well as right-scattered are called isolated. Furthermore, the point l∈T is right-dense if it satisfies l<SupT and σl=l, and left-dense if it satisfies l>infT and σl=l; furthermore, the point is called dense if it is left-dense as well as right-dense at the same time. A function μ:T⟶0,∞, defined by μl≔σl−l, is called the graininess function.If a functiong:T⟶ℝ is continuous at all right-dense points, the left-hand limits exist and are finite at left-dense points in T; then, it is right-dense continuous (rd-continuous) on T. The set denoted by CrdT contain all rd-continuous functions on T.Consider a functionβ:T⟶ℝ, and define the number βΔς if it exists with the property that, for given ε>0, there is a neighborhood U of ς which satisfies(1)βσς−βr−βΔςβς−r≤εσς−r,∀r∈U,then βΔς is delta derivative of function βς at ς∈T.Notation: ζσς=ζσς for any function ζ≔T⟶ℝ.(1)
Product and quotient rule for delta derivative (Theorem 1.20 in [14]): assume ζ,η:T⟶ℝ are differentiable; then,(2)iζηΔ=ζΔη+ζσηΔ=ζηΔ+ζΔησ,iiζηΔς=ζΔςης−ζςηΔςηςησς,ης≠0,ς∈T.(2)
Integration by parts formula (Theorem 1.77 in [14]): for two delta differentiable functions g,h:T⟶ℝ, and ς,a,m∈T, we have(3)∫amgςhΔςΔς=gςhς|am−∫amgΔςhσςΔς.(3)
Minkowski inequality (Theorem 6.16 in [14]): for three rd-continuous functions f:T⟶ℝ, g:T⟶ℝ,andh:T⟶ℝ, we have(4)∫abhtft+gtp1/p≤∫abhtftpΔt1/p+∫abhtgtpΔt1/p,wherep>1 and a,b,t∈T.(4)
Fubini’s theorem [16]: let there exist two time scales’ measure spaces ν,M,ϕΔ and υ,N,φΔ which have finite dimensions. If η:ν×υ⟶ℝ is a ϕΔ×φΔ which is an integrable function and the function ψ1m=∫νηl,mΔl exists for almost every m∈υ and ψ2l=∫υηl,mΔm exists for almost every l∈ν, then ψ1 is φΔ integrable on υ, ψ2 is ϕΔ integrable on ν, and(5)∫νΔl∫υηl,mΔm=∫υΔm∫vηl,mΔl.Notation:(6)∂Δτkgτ1,…,τk,…,τn=gΔkτ1,…,τk,…,τn,1≤k≤n.Some preliminary inequalities [2]: suppose g:T⟶ℝ is differentiable. Let β∈ℝ,ifgΔ is monotone, i.e., either always negative or always positive, then(7)βgΔgβ−1σ≤gβΔ≤βgΔgβ−1,if 0≤β≤1,(8)βgΔgβ−1≤gβΔ≤βgΔgβ−1σ,ifβ≥1,and if gΔ is positive, then(9)gβΔ≤gΔgβ−1σ,if 0≤β≤1,gβΔ≤gΔgβ−1σ,ifβ≥1.


## 3. Dynamic Copson-Type Inequalities for Finite Numbers of Parameters

We assume throughout that all the functions are nonnegative and the integrals considered exist. Forh∈ℕ, ι∈1,2,…,h, let Tι be time scales.Presume 1:(10)H1=SupTι=∞,bι∈0,∞Tι,υι:Tι⟶ℝ+is rd−continuous,Aιτι≔∫bιτιυιsιΔsι,forτι∈Tι.Theorem 1.
AssumeH1. Suppose g:T1×⋯×Th⟶ℝ+ is such that(11)ϕτ1,…,τh≔∫τ1∞⋯∫τh∞∏ι=1hυιsιAισιsιgs1,…,shΔsh⋯Δs1,is well defined and m≥1. Then,(12)∫b1∞⋯∫bh∞∏ι=1hυιτιϕmτ1,…,τhΔτh⋯Δτ1⩽mhm∫b1∞⋯∫bh∞∏ι=1hυιτιgmτ1,…,τhΔτh⋯Δτ1.Theorem 2.
AssumeH1. Suppose g:T1×⋯×Th⟶ℝ is such that(13)ϕς1,…,ςh≔∫ς1∞⋯∫ςh∞∏ι=1hυιsιgs1,…,shΔsh⋯Δs1,ς1,…,ςh∈T1×⋯×Th,is well defined. Let m≥1 and 0≤cι<1. Then,(14)∫a1∞⋯∫ah∞∏ι=1hυιςιAιcισιςιϕmς1,…,ςhΔςh⋯Δς1≤∏ι=1hm1−cιm∫a1∞⋯∫ah∞∏ι=1hυιςιAιm−cισιςιgmς1,…,ςhΔςh⋯Δς1.Proofs of Theorem1 and Theorem 2 are after sneak-out inequalities.


## 4. Dynamic Sneak-Out Inequalities for Finite Numbers of Parameters

Leti,j,r∈1,…,h and i1,…,ih=j1,…,jh=1,…,h.Presume 2:(15)H2=x:T1×⋯×Th⟶ℝ+is rd−continuous,yτi1,…,τih≔∫τi1∞⋯∫τih∞xsi1,…,sihΔsih⋯Δsi1,τi1,…,τih∈T1×⋯×Th,ψτi1,…,τih≔∫τi1∞⋯∫τih∞∏ik=1hAikαikσiksikxsi1,…,sihΔsih⋯Δsi1.Lemma 1.
LetTι be the time scales for ι∈1,2,…,h, under H1 and H2, and we have(16)ψτi1,…,τih≤∏k=1hAikαikσikτikyτi1,…,τih+∑1≤j1<⋯<jr≤h∏m=1rαjm∏m=1h−rAimαimσimτim×∫∏1≤j1<⋯<jr≤hτjm∞∏m=1rAjmαjm−1σjmsjmυjmsjmyτi1,…,τih−r,sj1,…,sjrΔ∏1≤j1<⋯<jr≤hjmsjm.Proof.
Forh=1, (16) is true by Theorems 4.1 in [2], i.e.,(17)ψτ1≤A1α1σ1τ1yτ1+α1∫τ1∞υ1s1A1α1−1σ1s1ys1Δs1.
Suppose (16) is true for 1≤h≤p. To prove for h=p+1, by using H2, we have defined as(18)ψτi1,…,τip+1=∫τi1∞⋯∫τip∞∏k=1pAikαikσiksik×∫τip+1∞Aip+1αip+1σip+1sip+1yΔ1⋯Δp+1si1,…,sip+1Δsip+1Δsip⋯Δsi1.
Denote(19)Zip+1=∫τip+1∞Aip+1αip+1σip+1sip+1yΔ1⋯Δp+1si1,…,sip+1Δsip+1.
Use integration by parts’ formula (3) in (19) to obtain(20)Zip+1=−Aip+1αip+1sp+1yΔ1⋯Δpsi1,…,sip+1τp+1∞−∫τip+1∞−∂Δτip+1Aip+1αip+1sip+1yΔ1⋯Δpsi1,…,sip+1Δsip+1.
Use the right-hand side part of inequality (8) with Aip+1≤Aip+1σip+1 in (20)(21)Zip+1≤Aip+1αip+1σip+1τip+1yΔ1⋯Δpsi1,…,sip,τip+1+αip+1∫τip+1∞υip+1sip+1Aip+1αip+1−1σip+1sip+1yΔ1⋯Δpsi1,…,sip+1Δsip+1.
Substitute (21) in (18):(22)ψτi1,…,τip+1=∫τi1∞⋯∫τip∞∏k=1pAikαikσiksikAip+1αip+1σip+1τip+1yΔ1⋯Δpsi1,…,sip,τip+1Δsip⋯Δsi1+αip+1∫τi1∞⋯∫τip∞∏k=1pAikαikσiksik×∫τip+1∞υip+1sip+1Aip+1αip+1−1σip+1sip+1yΔ1⋯Δpsi1,…,sip+1Δsip+1Δsip⋯Δsi1.
Use (5) “p times” in second term of (22):(23)ψτi1…τip+1≤Aip+1αip+1σip+1τip+1∫τi1∞⋯∫τip∞∏k=1pAikαikσiksikyΔ1⋯Δpsi1,…,sip,τip+1Δsip⋯Δsi1+αip+1∫τip+1∞υip+1sip+1Aip+1αip+1−1σip+1sip+1(24)×∫τi1∞⋯∫τip∞∏k=1pAikαikσiksikyΔ1⋯Δpsi1,…,sip+1Δsip⋯Δsi1Δsip+1.
Use induction hypothesis forψτi1,…,τip+1 with fix τip+1,sip+1∈Tip+1 (instead for ψτi1,…,τip) in (23) and (25) to obtain(25)ψτi1…τip+1≤Aip+1αip+1σip+1τip+1×∏k=1pAikαikσikτikyτi1,…,τip+1+∑1≤j1<⋯<jr≤p∏m=1rαjm∏im=1p−rAimαimσimτim×∫∏1≤j1<⋯<jr≤pτjm∞∏m=1rAjmαjm−1σjmsjmυjmsjmyτi1,…,τip+1−r,sj1,…,sjrΔ∏1≤j1<⋯<jr≤pjmsjm+αip+1∫τip+1∞υip+1sip+1Aip+1αip+1−1σip+1sip+1×∏k=1pAikαikσikτikyτi1,…,τip,sip+1+∑1≤j1<⋯<jr≤p∏m=1rαjm∏m=1p−rAimαimσimτim×∫∏1≤j1<⋯<jr≤pτjm∞∏m=1rAjmαjm−1σjmsjmυjmsjmyτi1,…,τip−r−1,sj1,…,sjrΔ∏1≤j1<⋯<jr≤pjmsjmΔsip+1.
By applying (5) “p times” on (25) and making simplification, we obtain(26)=∏k=1pAikαikσikτik×Aip+1αip+1σip+1τip+1yτi1,…,τip+1+αip+1∫τip+1∞yτi1,…,τip,sip+1Δsip+1yτi1,…,τip,sip+1Δsip+1+∑1≤j1<⋯<jr≤p∏m=1rαjm∏m=1ip−rAimαimσimτim∫∏1≤j1<⋯<jr≤pτjm∞∏m=1rAjmαjm−1σjmsjmυjmsjm×Aip+1αip+1σip+1τip+1yτi1,…,τip+1−r,sj1,…,sjr+αip+1∫τip+1∞υip+1sip+1Aip+1αip+1−1σip+1sip+1yτi1,…,τip−r−1,sj1,…,sjrΔsip+1Δ∏1≤j1<⋯<jr≤pjmsjm.
Hence, by using (17) for τip+1∈Tp+1, we obtain(27)ψτi1,…,τip+1≤∏k=1p+1Aikαikσikτikyτi1,…,τip+1+∑1≤j1<⋯<jr≤p+1∏m=1rαjm∏m=1p+1−rAimαimσimτim×∫∏1⩽j1<⋯<jr⩽p+1τjm∞∏m=1rAjmαjm−1σjmsjmυjmsjmyτi1,…,τip+1−r,sj1,…,sjrΔ∏1⩽j1<⋯<jr⩽p+1jmsjm.
Thus, by mathematical induction, (16) holds for all h∈ℕ, which completes the proof.Remark 1.
If we choseh=3 in Lemma 1, then (16) becomes the following inequality:(28)ψτ1,τ2,τ3≤A1α1σ1τ1A2α2σ2τ2A3α3σ3τ3yτ1,τ2,τ3+α1A2α2σ2τ2A3α3σ3τ3∫τ1∞A1α1−1σ1s1a1s1ys1,τ2,τ3Δs1+α2A1α1σ1τ1A3α3σ3τ3∫τ2∞A2α2−1σ2s2a2s2yτ1,s2,τ3Δs2+α3A1α1σ1τ1A2α2σ2τ2∫τ3∞A3α3−1σ3s3a3s3yτ1,τ2,s3Δs3+α1α2A3α3σ3τ3∫τ2∞A2α2−1σ2s2a2s2∫τ1∞A1α1−1σ1s1a1s1ys1,s2,τ3Δs1Δs2+α1α3A2α2σ2τ2∫τ1∞A1α1−1σ1s1a1s1∫τ3∞A3α3−1σ3s3a3s3ys1,τ2,s3Δs3Δs1+α2α3A1α1σ1τ1∫τ2∞A2α2−1σ2s2a2s2∫τ3∞A3α3−1σ3s3a3s3yτ1,s2,s3Δs3Δs2+α1α2α3∫τ3∞A3α3−1σ3s3a3s3∫τ2∞A2α2−1σ2s2a2s2∫τ1∞A1α1−1σ1s1a1s1ys1,s2,s3Δs1Δs2Δs3.Theorem 3.
AssumeH1, H2, and l,αι≥1,∀ι∈1,2,…,h,h∈ℕ. Then,(29)∫∏m=1hbjm∞∏jm=1hυjmτjmψlτj1,…,τjhΔ∏m=1hjmτjm≤1+∑1≤j1<⋯<jr≤hlr∏m=1rαjml×∫∏m=1hbjm∞∏jm=1hυjmτjmAjmlαjmσjmτjmylτj1,…,τjhΔ∏m=1hjmτjm.Proof.
We prove the result by using mathematical induction. Forh=1, statement is true by Theorems 4.1 in [2]. Assume for 1≤h≤p, (29) holds. To prove the result for h=p+1, take L.H.S of (29) in the following form:(30)∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmψlτj1,…,τjp+1Δ∏m=1p+1jmτjm.
Using (27) in (30) for h=p+1,(31)∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmψlτj1,…,τjp+1Δ∏m=1p+1jmτjm1/l≤∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmIp+1lΔ∏m=1p+1jmτjm.1/l,where(32)Ip+1=∏k=1p+1Aikαikσikτikyτi1,…,τip+1+∑1≤j1<⋯<jr≤p+1∏m=1rαjm∏m=1p+1−rAimαimσimτim×∫∏1≤j1<⋯<jr≤p+1τjm∞∏m=1rAjmαjm−1σjmsjmυjmsjmyτi1,…,τip+1−r,sj1,…,sjrΔsΠ1≤j1<⋯<jr≤p+1.
Apply Minkowski’s inequality (4) on (31) to obtain(33)∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmψlτj1,…,τjp+1Δ∏m=1p+1jmτjm1/l≤∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmAjmlαjmσjmτjmylτj1,…,τjp+1Δ∏m=1p+1jmτjm1/l+∑1≤j1<⋯<jr≤p+1∏m=1rαjm∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmI˜p+1lΔ∏m=1p+1jmτjm1/l,where(34)I˜p+1=∏m=1p+1−rAimαimσimτim×∫∏1≤j1<⋯<jr≤p+1τjm∞∏jm=1rAjmαjm−1σjmsjmυjmsjmyτi1,…,τip+1−r,sj1,…,sjrΔsΠ1≤j1<⋯<jr≤p+1.
Denote(35)Wp+1=∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmI˜p+1lΔ∏m=1p+1jmτjm,and one has that(36)Wp+1=∫∏m=1p+1bjm∞∏jm=1p+1υjmτjm∏m=1p+1−rAimlαimσimτim×∫∏1≤j1<⋯<jr≤p+1τjm∞∏jm=1rυjmsjmAjmσjmsjmAjmαjmσjmsjmyτi1,…,τip+1−r,sj1,…,sjrΔsΠ1≤j1<⋯<jr≤p+1l×Δ∏m=1p+1jmτjm.
Use Theorem1 in (36) by taking gsj1,…,sjr=∏jm=1rAjmαjmσjmsjmyτi1,…,τip+1−r,sj1,…,sjr to obtain(37)Wp+1≤l∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmAjmlαjmσjmτjmylτj1,…,τjp+1Δ∏m=1p+1jmτjm.
Substitute (37) in (33) and take power l on both sides to obtain(38)∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmψlτj1,…,τjp+1Δ∏m=1p+1jmτjm≤1+∑1≤j1<⋯<jr≤p+1lr∏m=1rαjml×∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmAjmlαjmσjmτjmylτj1,…,τjp+1Δ∏m=1p+1jmτjm.
Thus, by mathematical induction, (29) holds for all h∈ℕ.Example 1.
LetTι=ℕ and bjm=1,∀j,m∈1,…,h and njm,kjm,h∈ℕ∀jm. In this case, (29) in Theorem 3 takes the form(39)∑kj1=1∞⋯∑kjh=1∞∏m=1hυjmkjm∑nj1=kj1∞⋯∑njh=kjh∞∏m=1hAjmαjmnjm+1xnj1,…,njhl≤1+∑1≤j1<⋯<jr≤p+1lr∏m=1rαjml∑kj1=1∞⋯∑kjh=1∞∏m=1hυjmkjmAjmαjmlkjm+1×∑nj1=kj1∞⋯∑njh=kjh∞xnj1,…,njhl,where(40)Ajmkjm=∑njm=1kjm−1υjmnjm,kjm∈ℕ.
Note that (39) is extension of Example 4.4 in [2].Example 2.
LetTι=ℝ∀j,m∈1,…,h, in Theorem 3. In this case, (29) takes the form(41)∫∏m=1hbjm∞∏jm=1hυjmτjm∫∏m=1hbjm∞∏m=1hAjmαjmsjmxsj1,…,sjhd∏m=1hjmsjmld∏m=1hjmτjm≤1+∑1≤j1<⋯<jr≤p+1lr∏m=1rαjml∫∏m=1hbjm∞∏jm=1hυjmτjmAjmαjmlτjm×∫∏m=1hτjm∞xsj1,…,sjhd∏m=1hjmsjmld∏m=1hjmτjm,where(42)Ajmτjm=∫bjmτjmυjmsjmd∏m=1hjmsjm,τjm∈ℝ.
Note that (41) is extension of Example 4.3 in [2].Example 3.
LetTι=qιℕ0, qjm>1, and ∀j,m∈1,…,h, in Theorem 3. In this case, (29) takes the form(43)∑kj1=1∞⋯∑kjh=1∞∏jm=1hυjmqjmkjmqjmkjml∑nj1=kj1∞⋯∑njh=kjh∞∏m=1hqjmnjmAjmαjmqjmnjm+1xqj1nj1,…,qjhnjhl≤1+∑1≤j1<⋯<jr≤p+1lr∏m=1rαjml×∑kj1=1∞⋯∑kjh=1∞∏jm=1hυjmqjmkjmAjmαjmlqjmkjm+1qjmkjml∑nj1=kj1∞⋯∑njh=kjh∞∏m=1hqjmnjmxqj1nj1,…,qjhnjhl,where(44)Ajmqjmkjm=∑njm=1kjm−1υjmqjmnjmqjmnjmqjm−1,kjm∈ℕ0.Lemma 2.
LetTι be the time scales for ι,j,r∈1,2,…,h, under H1 and H2, we have(45)yτi1,…,τih≤∏k=1hAik−αikσikτikψτi1,…,τih+∑1≤j1<…<jr≤h∏m=1h−rAim−αimσimτim×∫∏1≤j1<⋯<jr≤hτjm∞∏m=1rAjm−αjm−1σjmsjmυjmsjmψτi1,…,τih−r,sj1,…,sjrΔ∏1≤j1<⋯<jr≤hjmsjm.Proof.
Forh=1, (45) is true by Theorems 4.6 in [2], i.e.,(46)yτ1≤A1−α1σ1τ1ψτ1+∫τ1∞υ1s1A1α1−1σ1s1ψs1Δs1.
Suppose (45) is true for 1≤h≤p. To prove for h=p+1, by using H2, we have(47)yτi1,…,τip+1=∫τi1∞⋯∫τip∞∏k=1pAik−αikσiksik×∫τip+1∞Aip+1−αip+1σip+1sip+1ψΔ1⋯Δp+1si1,…,sip+1Δsip+1Δsip⋯Δsi1.
Denote(48)Zip+1=∫τip+1∞Aip+1−αip+1σip+1sip+1ψΔ1⋯Δp+1si1,…,sip+1Δsip+1.
Use integration by parts formula (3) in (48) to obtain(49)Zip+1=−Aip+1−αip+1sp+1ψΔ1⋯Δpsi1,…,sip+1τp+1∞−∫τip+1∞−∂Δτip+1Aip+1−αip+1sip+1ψΔ1⋯Δpsi1,…,sip+1Δsip+1.
Useψsi1,…,sip,∞=0 and the right-hand side part of inequality (8) with Aip+1≤Aip+1σip+1 in (49)(50)Zip+1≤Aip+1−αip+1σip+1τip+1ψΔ1⋯Δpsi1,…,sip,τip+1+∫τip+1∞υip+1sip+1Aip+1−αip+1−1σip+1sip+1ψΔ1⋯Δpsi1,…,sip+1Δsip+1.
Substitute (50) in (47)(51)yτi1,…,τip+1=∫τi1∞⋯∫τip∞∏k=1pAik−αikσiksikAip+1−αip+1σip+1τip+1ψΔ1⋯Δpsi1,…,sip,τip+1×Δsip⋯Δsi1+∫τi1∞⋯∫τip∞∏k=1pAik−αikσiksik×∫τip+1∞υip+1sip+1Aip+1−αip+1−1σip+1sip+1ψΔ1⋯Δpsi1,…,Δsip+1Δsip+1Δsip⋯Δsi1.
Use (5) “p times” on (51):(52)yτi1,…,τip+1≤Aip+1−αip+1σip+1τip+1∫τi1∞⋯∫τip∞∏k=1pAik−αikσiksikψΔ1⋯Δpsi1,…,sip,τip+1Δsip⋯Δsi1+∫τip+1∞υip+1sip+1Aip+1−αip+1−1σip+1sip+1∫τi1∞⋯∫τip∞∏k=1pAik−αikσiksikψΔ1⋯Δpsi1,…,sip+1Δsip⋯Δsi1×Δsip+1.
Use induction hypothesis foryτi1,…,τip+1 with fix τip+1,sip+1∈Tip+1 (instead for yτi1,…,τip) in (52) to obtain(53)yτi1,…,τip+1≤Aip+1−αip+1σip+1τip+1×∏k=1pAik−αikσikτikψτi1,…,τip+1+∑1≤j1<⋯<jr≤p∏m=1p−rAim−αimσimτim×∫∏1≤j1<⋯<jr≤pτjm∞∏m=1rAjm−αjm−1σjmsjmυjmsjmψτi1,…,τip+1−r,sj1,…,sjrΔ∏1≤j1<⋯<jr≤pjmsjm+∫τip+1∞υip+1sip+1Aip+1−αip+1−1σip+1sip+1×∏k=1pAik−αikσikτikψτi1,…,τip,sip+1+∑1≤j1<⋯<jr≤p∏m=1p−rAim−αimσimτim×∫∏1≤j1<⋯<jr≤pτjm∞∏m=1rAjm−αjm−1σjmsjmυjmsjmψτi1,…,τip−r−1,sj1,…,sjrΔ∏1≤j1<⋯<jr≤pjmsjmΔsip+1,=∏k=1pAik−αikσikτik×Aip+1−αip+1σip+1τip+1ψτi1,…,τip+1+∫τip+1∞υip+1sip+1Aip+1−αip+1−1σip+1sip+1ψτi1,…,τip,sip+1Δsip+1+∑1≤j1<⋯<jr≤p∏m=1p−rAim−αimσimτim∫∏1≤j1<⋯<jr≤pτjm∞∏m=1rAjm−αjm−1σjmsjmυjmsjm×Aip+1αip+1σip+1τip+1ψτi1,…,τip+1−r,sj1,…,sjp+∫τip+1∞υip+1sip+1Aip+1−αip+1−1σip+1sip+1ψτi1,…,τip−r−1,sj1,…,sjpΔsip+1Δ∏1≤j1<⋯<jr≤pjmsjm.
Hence, by using (46) for τip+1, we obtain(54)yτi1,…,τip+1≤∏k=1p+1Aik−αikσikτikψτi1,…,τip+1+∑1≤j1<⋯<jr≤p+1∏m=1ip+1−rAim−αimσimτim×∫∏1≤j1<⋯<jr≤p+1τjm∞∏m=1rAjm−αjm−1σjmsjmυjmsjmψτi1,…,τp+1−r,sj1,…,sjrΔ∏1≤j1<⋯<jr≤p+1jmsjm.
Thus, by mathematical induction, (54) holds for all h∈ℕ, which completes the proof.Theorem 4.
AssumeH1, H2, and l,αι≥1. for ι∈1,2,…,h, h∈ℕ. Then,(55)∫∏m=1hbjm∞∏jm=1hυjmτjmψlτj1,…,τjhΔ∏m=1hjmτjm≥∑1≤j1<⋯<jr≤p+11+lr∏m=1rαjm1+lr∏m=1rαjm+lrl×∫∏m=1hbjm∞∏jm=1hυjmτjmAjmlαjmσjmτjmylτj1,…,τjhΔ∏m=1hjmτjm.Proof.
We prove the result by using mathematical induction. Forh=1, statement is true by Theorems 4.6 in [2]. Assume for 1≤h≤p, (55) holds. To prove the result for h=p+1, take L.H.S of (55) with h=p+1 in the following form:(56)∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmAjmlαjmσjmτjmylτj1,…,τjp+1Δ∏m=1p+1jmτjm.
Use (27) in (56) for h=p+1:(57)∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmAjmlαjmσjmτjmylτj1,…,τjp+1Δ∏m=1p+1jmτjm.1/l≤∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmAjmlαjmσjmτjmIp+1lΔ∏m=1p+1jmτjm1/l,where(58)Ip+1=∏k=1p+1Aik−αikσikτikψτi1,…,τip+1+∑1≤j1<⋯<jr≤p+1∏m=1p+1−rAim−αimσimτim×∫∏1≤j1<⋯<jr≤p+1τjm∞∏m=1rAjm−αjm−1σjmsjmυjmsjmψτi1,…,τip+1−r,sj1,…,sjrΔsΠ1≤j1<⋯<jr≤p+1.
Apply Minkowski’s inequality (4) on (57) to obtain(59)∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmAjmlαjmσjmτjmylτj1,…,τjp+1Δ∏m=1p+1jmτjm1/l≤∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmAjm−lαjmσjmτjmAjmlαjmσjmτjmψlτj1,…,τjp+1Δ∏m=1p+1jmτjm1/l+∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmAjmlαjmσjmτjmI˜p+1lΔ∏m=1p+1jmτjm1/l,where(60)I˜p+1=∏m=1p+1−rAim−αjmσimτim×∫1≤j1<⋯<jr≤p+1∞σjmsjmυjmsjmψτi1,…,τip+1−r,sj1,…,sjrΔsΠ1≤j1<⋯<jr≤p+1.
Denote(61)Wp+1=∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmAjmlαjmσjmτjmI˜p+1lΔ∏m=1p+1jmτjm,and one has that(62)Wp+1=∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmAim−lαimσimτim×∫∏1≤j1<⋯<jr≤p+1τjm∞∏jm=1rυjmsjmAjm−αjm−1σjmsjmψτi1,…,τip+1−r,sj1,…,sjrΔsΠ1≤j1<⋯<jr≤p+1l×Δ∏m=1p+1jmτjm.
Use Theorem2 in (62) by taking gsj1,…,sjr=∏jm=1rAjm−αjm−1σjmsjmψτi1,…,τip+1−r,sj1,…,sjr and c=∏m=1rαjml:(63)wp+1≤lrl1+lrl∏m=1rαjml∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmAjmαjmlσjmτjmψlτj1,…,τjp+1Δ∏m=1p+1jmτjm.
Substitute (63) in (59) and take power l on both sides to obtain(64)∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmψlτj1,…,τjp+1Δ∏m=1p+1jmτjm≥∑1≤j1<⋯<jr≤p+11+lr∏m=1rαjm1+lr∏m=1rαjm+lrl×∫∏m=1p+1bjm∞∏jm=1p+1υjmτjmAjmlαjmσjmτjmylτj1,…,τjp+1Δ∏m=1p+1jmτjm.
Thus, by mathematical induction, (55) holds for all h, which completes the proof.Example 4.
LetTι=ℕ and bjm=1,∀j,m∈1,…,h and njm,kjm,h∈ℕ∀jm, in Theorem 4. In this case, (55) takes the form(65)∑kj1=1∞⋯∑kjh=1∞∏m=1hυjmkjm∑nj1=kj1∞⋯∑njh=kjh∞∏m=1hAjmαjmnjm+1xnj1,…,njhl≥∑1≤j1<⋯<jr≤p+11+lr∏m=1rαjm1+lr∏m=1rαjm+lrl∑kj1=1∞⋯∑kjh=1∞∏m=1hυjmkjmAjmαjmlkjm+1×∑nj1=kj1∞⋯∑njh=kjh∞xnj1,…,njhl,where(66)Ajmkjm=∑njm=1kjm−1υjmnjm,kjm∈ℕ.
Note that (65) is extension of Example 4.7 in [2].Example 5.
LetTι=ℝ∀j,m∈1,…,h, in Theorem 4. In this case, (55) takes the form(67)∫∏m=1hbjm∞∏jm=1hυjmτjm∫∏m=1hτjm∞∏m=1hAjmαjmsjmxsj1,…,sjhd∏m=1hjmsjmld∏m=1hjmτjm≥∑1≤j1<⋯<jr≤p+11+lr∏m=1rαjm1+lr∏m=1rαjm+lrl∫∏m=1hbjm∞∏m=1hυjmτjmAjmαjmlτjm×∫∏m=1hτjm∞xsj1,…,sjhd∏m=1hjmsjmld∏m=1hjmτjm,where(68)Ajmτjm=∫bjmτjmυjmsjmd∏m=1hjmsjm,τjm∈ℝ.
Note that (67) is extension of Example 4.8 in [2].Example 6.
LetTι=qιℕ0, qjm>1, and ∀j,m∈1,…,h., in Theorem 4. In this case, (55) takes the form(69)∑kj1=1∞⋯∑kjh=1∞∏jm=1hυjmqjmkjmqjmkjml∑nj1=kj1∞⋯∑njh=kjh∞∏m=1hqjmnjmAjmαjmqjmnjm+1xqj1nj1,…,qjhnjhl≥∑1≤j1<⋯<jr≤p+11+lr∏m=1rαjm1+lr∏m=1rαjm+lrl×∑kj1=1∞⋯∑kjh=1∞∏jm=1hυjmqjmkjmAjmαjmlqjmkjm+1qjmkjml∑nj1=kj1∞⋯∑njh=kjh∞∏m=1hqjmnjmxqj1nj1,…,qjhnjhl,where(70)Ajmqjmkjm=∑njm=1kjm−1υjmqjmnjmqjmnjmqjm−1,kjm∈ℕ0.Proof.
We use mathematical induction to prove the result. Forh=1, (12) is true by Theorem 3.1 in [2]. Assume for 1≤h≤p, and (12) holds. To prove the result for h=p+1, take L.H.S of (12) in the following form:(71)∫b1∞⋯∫bp∞∏ι=1pυιτι×∫bp+1∞υp+1τp+1ϕmτ1,…,τp+1Δτp+1Δτp⋯Δτ1.
Denote(72)Ip+1=∫bp+1∞υp+1τp+1ϕmτ1,…,τp+1Δτp+1.
Use Theorem 3.1 in [2] in (72) with respect to τp+1∈Tp+1 for fix τ1,…,τp∈T1×⋯×Tp to obtain(73)⇒Ip+1m≤mm∫bp+1∞υp+1τp+1ϕpmτ1,…,τp+1Δτp+1,where(74)ϕpτ1,…,τp+1=∫τ1∞⋯∫τp∞∏ι=1pυιsιAισιsιgs1,…,sp,τp+1Δsp⋯Δs1.
Substitute (73) in (71) and use (5) “p times” in resultant inequality to obtain(75)∫b1∞⋯∫bp+1∞∏ι=1p+1υιτιϕmτ1,…,τp+1Δτp+1⋯Δτ1≤mm∫b1∞⋯∫bp∞∏ι=1pυιτι∫bp+1∞υp+1τp+1ϕpmτ1,…,τp+1Δτp+1Δτp⋯Δτ1.
Use induction hypothesis forϕpτ1…τp+1 with fix τp+1∈Tp+1, instead for ϕpτ1…τp to obtain(76)∫b1∞⋯∫bp+1∞∏ι=1p+1υιτιϕmτ1,…,τp+1Δτp+1⋯Δτ1≤mp+1m∫b1∞⋯∫bp+1∞∏ι=1p+1υιτιgmτ1,…,τp+1Δτp+1⋯Δτ1.
Thus, by mathematical induction, (12) holds for all h∈ℕ.Proof.
We prove the result by using mathematical induction. Forh=1, statement is true by Theorem 3.3 in [2]. Assume for 1≤h≤p, (14) holds. To prove the result for h=p+1, left-hand side of (14) can be written as(77)∫a1∞⋯∫ap∞∏ι=1pυιςιAιcισιςι∫ap+1∞υp+1ςp+1Ap+1cp+1σp+1ςp+1ϕmς1,…,ςp+1Δςp+1Δςp⋯Δς1.
Denote(78)Ip+1=∫ap+1∞υp+1ςp+1Ap+1cp+1σp+1ςp+1ϕmς1,…,ςp+1Δςp+1.
Use Theorem 3.3 in [2] in (72) with respect to ςp+1∈Tp+1 for fix ς1,…,ςp∈T1×⋯×Tp to obtain(79)Ip+1m≤m1−cp+1m∫ap+1∞υp+1ςp+1Ap+1m−cp+1σp+1ςp+1ϕpmς1,…,ςp+1Δςp+1,where(80)ϕpς1,…,ςp+1≐∫ς1∞⋯∫ςp∞∏ι=1pυιsιgs1,…,sp,ςp+1Δsp⋯Δs1.
Substitute (79) in (77) and use fd5(5) “p times” in resultant inequality to obtain(81)∫a1∞⋯∫ap∞∏ι=1pυιςιAιcισιςι∫ap+1∞υp+1ςp+1Ap+1cp+1σp+1ςp+1ϕmς1,…,ςp+1Δςp+1Δςp⋯Δς1≤m1−cp+1m∫ap+1∞υp+1ςp+1Ap+1m−cp+1σp+1ςp+1Δςp+1×∫a1∞⋯∫ap∞∏ι=1pυιςιAιcισιςιϕpmς1,…,ςp+1Δςp⋯Δς1.
Use induction hypothesis forϕpς1,…,ςp+1 with fix ςp+1∈Tp+1, instead for ϕpς1,…,ςp, to obtain(82)∫a1∞⋯∫ap+1∞∏ι=1p+1υιςιAιcισιςιϕmς1,…,ςp+1Δςp+1⋯Δς1≤∏ι=1p+1m1−cιp+1m∫a1∞⋯∫ap+1∞∏ι=1p+1υιςιAιm−cισιςιgmς1,…,ςp+1Δςp+1⋯Δς1.
Hence, by mathematical induction, (14) is true for all h∈ℕ.



---

*Source: 9978050-2021-05-27.xml*