# Hardy-Type Inequalities on Time Scale via Convexity in Several Variables

Authors: Tzanko Donchev; Ammara Nosheen; Josip Pečarić
Journal: ISRN Mathematical Analysis
 (2013)

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## Abstract

We extend some Hardy-type inequalities with general kernels to arbitrary time scales using multivariable convex functions. Some classical and new inequalities are deduced seeking applications.


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## Body

## 1. Introduction

The significant Hardy inequality is published in [1] (1952) (both in the continuous and discrete settings). More general Hardy integral inequalities have been studied in continuous cases. We notice only [2–6] and the references therein.Recently, this inequality is studied in discrete case, and some variants of it are proved in case of time scales [7–9].In [10], the authors study Hardy-type inequalities using convex functions of one variable with general kernels to arbitrary time scales.The aim of this paper is to provide Hardy-type inequalities using multivariable convex function with general kernels to arbitrary time scales. Notice that time scales include continuous and discrete time cases under unified approach.Firstly, we recall necessary preliminary facts needed afterward. The main results are given in Section3. Section 4 is devoted to some inequalities with certain kernels. In the last section, we discuss some particular cases of Hardy-type inequalities.


## 2. Preliminaries

First, we recall the basic concepts used in the paper and refer the interested reader to [11] for the theory of time scales. A time scale is any nonempty closed subset of the real line 𝕋⊂ℝ. On nondensity points we define forward, respectively, backward jump operators σ,ρ:𝕋→𝕋 as
(1)σ(t)=inf{s∈𝕋:s>t},ρ(t)=sup{s∈𝕋:s<t}.
The point t is said to be right-scattered if σ(t)>t and left-scattered if ρ(t)<t, respectively. Clearly, t is right-dense if σ(t)=t and left dense if ρ(t)=t, respectively.Letn∈ℕ; we define n-dimensional time scale by the Cartesian product of given time scales 𝕋i, i∈{1,…,n}, as
(2)Ωn={a→=(a1,a2,…,an):ai∈𝕋i,i∈{1,…,n}}.
Evidently, Ωn is a complete metric space with distance as follows:
(3)d(a→,b→)=(∑i=1n‍|bi-ai|2)1/2,a,b∈Ωn.
Now we are going to describe the construction of Lebesgue measure in Ωn. We refer to [12–14] for the theory of measure spaces and measurable functions on time scales.Letℱ be the family of all n-dimensional time scale intervals in Ωn; that is,
(4)V=[a→,b→)=[a1,b1)×[a2,b2)×⋯×[an,bn),
with a→=(a1,a2,…,an), b→=(b1,b2,…,bn)∈Ωn, and ai≤bi for all i∈{1,…,n}. Let m:ℱ→[0,∞) be the set function that assigns to each n-dimensional time scale interval V=[a,b) its volume as follows:
(5)m(V)=∏i=1n‍(bi-ai).
Let ∅≠E⊂Ωn. If there exists finite or countable system of pairwise disjoint n-dimensional time scale intervals Vk=[a→k,b→k) with E⊂⋃k≥1Vk, then the outer measure m* of E is defined by
(6)m*(E)=inf{∑k≥1‍m(Vk),E⊂⋃k≥1Vk}.
If there is no such covering of E, then m*(E)=∞.A subsetA of Ωn is said to be measurable (or Δ-measurable) if
(7)m*(E)=m*(E∩A)+m*(E∩Ac)
holds for all E⊂Ωn, where Ac=Ωn∖A. The family ℳ of all m*-measurable subsets of Ωn is a σ-algebra generated by ℱ. The restriction of m* to ℳ, which we denote by μΔ, is a σ-additive measure on ℳ. Clearly, ℱ⊂ℳ and μΔ(V)=m(V) for each V∈ℱ. The measure μΔ (called the Lebesgue Δ-measure on Ωn) is Carathéodory extension of the original measure m defined on ℱ. We call (Ωn,ℳ,μΔ) an n-dimensional time scale measure space.Denotea~i=inft∈𝕋it. Similarly, b~i=supt∈𝕋it. Let Ω∞n be the set of all points b→=(b1,b2,…,bn)∈Ωn for which there exists at least one bi such that bi=b~i. From [11, Theorem 3.1], we know that if t→=(t1,t2,…,tn)∈Ωn∖Ω∞n, then the single-point set {t} is Δ-measurable and
(8)μΔ({t})=∏i=1n‍(σi(ti)-ti).
Obviously, for all i∈{1,…,n}, the set Ω0n=Ωn∖Ω∞n can be represented as a finite or countable union of intervals of the family ℱ; hence it is Δ-measurable. Furthermore, the set Ω∞n=Ωn∖Ω0n is Δ-measurable being the difference of two Δ-measurable sets Ωn and Ω0n, but Ω∞n does not have a finite or countable covering intervals of ℱ, that is, for any Δ-measurable subset A of Ωn such that A∩Ω∞≠∅ has Δ-measure ∞. In particular, if a∈𝕋, where 𝕋 is an arbitrary time scale, then the set [a,∞)={t∈𝕋:a≤t} is Δ-measurable.The functionf:Ωn→ℝ-:=[-∞,∞] is said to be Δ-measurable if for every α∈ℝ, the set f-1([-∞,α))={t=(t1,…,tn)∈Ωn:f(t)<α} is Δ-measurable. It is easy to see that f is Δ-measurable if and only if for each open set G⊂ℝ, the set f-1(G)={t∈Ωn:f(t)∈G} is Δ-measurable. Moreover, if f:Ωn→ℝ is Δ-measurable and Φ:I⊂ℝ→ℝ is a continuous function, then Φ∘f:Ωn→ℝ is Δ-measurable.Having theσ-additive measure μΔ on Ωn, we possess the corresponding integration theory for functions f:E⊂Ωn→ℝ, according to the general Lebesgue integration theory (see, e.g., [15]). The Lebesgue integral associated with the measure μΔ on Ωn is called the Lebesgue Δ-integral. For a Δ-measurable set E⊂Ωn and a Δ-measurable function f:E→ℝ, the corresponding Δ-integral of f over E will be denoted by
(9)∫E‍f(t1,…,tn)Δt1⋯Δtnor∫E‍f(t)Δt.
Notice that all theorems of the general Lebesgue integration theory, including the Lebesgue dominated convergence theorem, hold also for Lebesgue Δ-integrals on Ωn.If𝕋 is a time scale and the interval [a,b)⊂𝕋 contains only isolated points, then
(10)∫ab‍f(t)Δt=∑t∈[a,b)‍(σ(t)-t)f(t).Let(Ω,ℳ,μΔ) and (Λ,ℒ,λΔ) be two finite dimensional time scale measure spaces. We define the product measure space (Ω×Λ,ℳ×ℒ,μΔ×λΔ), where ℳ×ℒ is the product σ-algebra generated by {E×F:E∈ℳ,F∈ℒ} and
(11)(μΔ×λΔ)(E×F)=μΔ(E)λΔ(F).Fubini's Theorem in Time Scales see ([16, Theorem 1.1]). If f:Ω×Λ→ℝ is a μΔ×λΔ-integrable function and if we define the function φ(y)=∫Ω‍f(x,y)Δx for a.e. y∈Λ and ψ(x)=∫Λ‍f(x,y)Δy for a.e. x∈Ω, then φ is λΔ-integrable on Λ, ψ is μΔ-integrable on Ω, and
(12)∫Ω‍Δx∫Λ‍f(x,y)Δy=∫Λ‍Δy∫Ω‍f(x,y)Δx.


## 3. Inequalities with General Kernels

Letf(y)=(f1(y),…,fm(y)) be m-tuple of functions such that fj(y) are λΔ-integrable for all j∈{1,…,m}. Then ∫Λf(y)Δ(y) denotes the m-tuple (∫Λf1(y)Δ(y),…,∫Λfm(y)Δ(y)); that is, Δ-integral acts on each component of f(y).The following Jensen's inequality on time scales is given in [17, Theorem 2.1].Theorem 1.
Let(Ω,ℳ,μΔ) and (Λ,ℒ,λΔ) be two time scale measure spaces. Suppose U⊂ℝm is a closed convex set, Φ∈C(U,ℝ) is convex, and f(Λ)⊂U. Moreover, let k:Ω×Λ→ℝ be nonnegative such that k(x,·) is λΔ-integrable. Then one has
(13)Φ(∫Λk(x,y)f(y)Δy∫Λk(x,y)Δy)≤∫Λk(x,y)Φ(f(y))Δy∫Λk(x,y)Δy.Throughout this section, we assume that the following hypotheses hold.H1
(Ω,ℳ,μΔ) and (Λ,ℒ,λΔ) are two time scale measure spaces.H2
k:Ω×Λ→ℝ+ is such that K(x):=∫Λ‍k(x,y)Δy<∞,x∈Ω.H3
ξ:Ω→ℝ+ is such that w(y):=∫Ω‍(k(x,y)ξ(x)/K(x))Δx,y∈Λ.Theorem 2.
IfU⊂ℝm is a closed convex set such that Φ:U→ℝ is convex and continuous, then
(14)∫Ω‍ξ(x)Φ(1K(x)∫Λ‍k(x,y)f(y)Δy)Δx≤∫Λ‍w(y)Φ(f(y))Δy
holds for all λΔ-integrable functions f:Λ→ℝm such that f(Λ)⊂U.Proof.
Using Jensen's inequality (13) for several variables and the Fubini theorem on time scales, we find that
(15)∫Ω‍ξ(x)Φ(1K(x)∫Λ‍k(x,y)f(y)Δy)Δx=∫Ω‍ξ(x)Φ(1K(x)∫Λ‍k(x,y)f1(y)Δy,…,1K(x)×∫Λ‍k(x,y)fm(y)Δy)Δx≤∫Ω‍ξ(x)K(x)(∫Λ‍k(x,y)Φ(f(y))Δy)Δx=∫Λ‍Φ(f(y))(∫Ω‍k(x,y)ξ(x)K(x)Δx)Δy=∫Λ‍w(y)Φ(f(y))Δy.
The proof is therefore complete.Remark 3.
IfΦ is concave, then (14) holds in reverse direction.Corollary 4.
LetΨ:[l1,l´1)×⋯×[lm,l´m)→ℝ+ be continuous function and define
(16)L~j(fj,Λ)=Lj-1(∫Λ‍k(x,y)Lj(fj(y))ΔyK(x))
for all j∈{1,…,m}. If Φ(s1,…,sm)=Ψ(L1-1(s1),…,Lm-1(sm)) is convex, then
(17)∫Ω‍ξ(x)Ψ(L~1(f1,Λ),…,L~m(fm,Λ))Δx≤∫Λ‍w(y)Ψ(f1(y),…,fm(y))Δy
holds for all fj(Λ)⊂[lj,l´j) and continuous monotone functions Lj:[lj,l´j)→ℝ such that Lj∘fj are λΔ-integrable for all j∈{1,…,m}.Proof.
Replace in Theorem2  fj(y) with Lj(fj(y)) for all j∈{1,…,m} and Φ(s1,…,sm) with Ψ(L1-1(s1),…,Lm-1(sm)).Remark 5.
In case that𝕋=ℕ and m=2, Corollary 4 is as [18, Corollary 1.2].Remark 6.
In case thatm=2, we can use the results of Beck [19] (see also [20], page 194) as applications of Corollary 4, which corresponds to the generalizations of Hölder’s and Minkoski's inequalities. In classical case, many authors have studied these types of generalizations; see, for example, [21–23].Furthermore, in the paper, we usep>1 with (1/p)+(1/p´)=1.Corollary 7.
Iff(x)=(f1(x),f2(x)), then
(18)∫Ω‍ξ(x)(1K(x)∫Λ‍k(x,y)(f1(y))pΔy)1/p×(1K(x)∫Λ‍k(x,y)(f2(y))p´Δy)1/p´Δx≥∫Λ‍w(y)f1(y)f2(y)Δy
holds for all λΔ-integrable fj:Λ→ℝ+, where j∈{1,2}.Proof.
Usem=2,Ψ(s1,s2)=s1s2,L1(t1)=(t1)p,L2(t2)=(t2)p´ in Corollary 4; then Φ(s1,s2)=s11/ps21/p´ is concave in Theorem 2.Corollary 8.
Iff(x)=(f1(x),f2(x)), then
(19)∫Ω‍ξ(x)((1K(x)∫Λ‍k(x,y)(f1(y))pΔy)1/p+(1K(x)∫Λ‍k(x,y)(f2(y))pΔy)1/p)pΔx≥∫Λ‍w(y)(f1(y)+f2(y))pΔy
holds for all λΔ-integrable fj:Λ→ℝ+, where j∈{1,2}.Proof.
Usem=2, Ψ(s1,s2)=(s1+s2)p, L1(t1)=(t1)p, L2(t2)=(t2)p in Corollary 4; then Φ(s1,s2)=(s11/p+s21/p)p is concave in Theorem 2.Remark 9.
Ifp<1, then reverse inequalities hold in Corollaries 4, 7, and 8.


## 4. Inequalities with Special Kernels

Throughout this section and in the next section, we assume that the following hypothesis holds.H4
Ω=Λ=[a1,b1)𝕋×[a2,b2)𝕋×⋯×[an,bn)𝕋, 0≤ai<bi≤∞ for all i∈{1,…,n}, where 𝕋 is an arbitrary time scale.Corollary 10.
Assume that(20)ξ:Ω→ℝ+is  such  thatw(y)=∫y1b1⋯∫ynbnk(x1,…,xn,y)ξ(x1,…,xn)K(x1,…,xn)×Δx1⋯Δxn,y∈Λ.
If U⊂ℝm is a closed convex set such that Φ:U→ℝ is convex and continuous; then
(21)∫a1b1‍⋯∫anbn‍ξ(x1,…,xn)Φ((Akf)(x1,…,xn))Δx1⋯Δxn≤∫a1b1‍⋯∫anbn‍w(y1,…,yn)Φ(f(y1,…,yn))×Δy1⋯Δyn
holds for all λΔ-integrable f:Λ→ℝm such that f(Λ)⊂U, where
(22)(Akf)(x)=1K(x)∫a1σ(x1)‍⋯∫anσ(xn)‍k(x,y1,…,yn)×f(y1,…,yn)Δy1⋯Δyn.Proof.
Statement follows from Theorem2 using
(23)k(x1,…,xn,y1,…,yn)=0,ifai≤yi≤σ(xi)for  alli∈{1,…,n},
since in this case
(24)K(x)=∫a1σ(x1)‍⋯∫anσ(xn)‍k(x,y1,…,yn)Δy1⋯Δyn.Corollary 11.
Assume that(25)ξ:Ω⟶ℝ+is  such  thatw(y)=∫a1y1‍⋯∫anyn‍k(x1,…,xn,y)ξ(x1,…,xn)K(x1,…,xn)×Δx1⋯Δxn,y∈Λ.
If U⊂ℝm is a closed convex set such that Φ:U→ℝ is convex and continuous, then (21) holds for all λΔ-integrable f:Λ→ℝm such that f(Λ)⊂U, and
(26)(Akf)(x)=1K(x)∫σ(x1)b1⋯∫σ(xn)bnk(x,y1,…,yn)×f(y1,…,yn)Δy1⋯Δyn.Proof.
Statement follows from Theorem2 using
(27)k(x1,…,xn,y1,…,yn)=0,ifai≤σ(xi)≤yifor  alli∈{1,…,n},
since in this case
(28)K(x)=∫σ(x1)b1⋯∫σ(xn)bnk(x,y1,…,yn)Δy1⋯Δyn.Theorem 12.
Assume that(29)ξ:Ω→ℝ+is  such  thatw~(y)=∫y1b1‍⋯∫ynbn‍ξ(x1,…,xn)∏i=1n(σ(xi)-ai)Δx1⋯Δxn,y∈Λ.
If U⊂ℝm is a closed convex set such that Φ:U→ℝ is convex and continuous, then
(30)∫a1b1‍⋯∫anbn‍ξ(x1,…,xn)Φ((Ak~f)(x1,…,xn))Δx1⋯Δxn≤∫a1b1‍⋯∫anbn‍w~(y1,…,yn)Φ(f(y1,…,yn))×Δy1⋯Δyn
holds for all λΔ-integrable f:Λ→ℝm such that f(Λ)⊂U, where
(31)(Ak~f)(x1,…,xn)=1∏i=1n(σ(xi)-ai)×∫a1σ(x1)‍⋯∫anσ(xn)‍f~(y1,…,yn)×Δy1⋯Δyn.Proof.
Statement follows from Theorem2 using
(32)k(x1,…,xn,y1,…,yn)={1ifai≤yi<σ(xi)≤bi,i∈{1,…,n}0otherwise,},
since in this case
(33)K(x1,…,xn)=∫a1σ(x1)‍⋯∫anσ(xn)‍Δy1⋯Δyn=∏i=1n‍(σ(xi)-ai).
Thus Ak=A~k,w=w~.Corollary 13.
Ifai=0 for all i∈{1,…,n} in H4 and U⊂ℝm is a closed convex set such that Φ:U→ℝ is convex and continuous, then
(34)∫0b1‍⋯∫0bn‍Φ((Akf)(x1,…,xn))Δx1⋯Δxnx1⋯xn≤∫0b1‍⋯∫0bn‍∏i=1n‍(1yi-1bi)Φ(f(y1,…,yn))Δy1⋯Δyn
holds for all λΔ-integrable f:Λ→ℝm such that f(Λ)⊂U, where
(35)(Akf)(x1,…,xn)≔1∏i=1nσ(xi)×∫0σ(x1)‍⋯∫0σ(xn)‍f(y1,…,yn)×Δy1⋯Δyn.Proof.
The statement follows from Theorem12 using ξ(x1,…,xn)=1/x1⋯xn, since in this case
(36)w(y1,…,yn)=∫y1b1‍⋯∫ynbn‍1∏i=1nxiσ(xi)Δx1⋯Δxn=∏i=1n‍(1yi-1bi).Example 14.
Ifbi=∞ for all i∈{1,…,n} in addition to the assumptions of Corollary 13, then (34) takes the form
(37)∫0∞‍⋯∫0∞‍Φ((Akf)(x1,…,xn))Δx1⋯Δxnx1⋯xn≤∫0∞‍⋯∫0∞‍Φ(f(y1,…,yn))Δy1⋯Δyny1⋯yn.Remark 15.
Clearly, if the left-hand side is∞ in (34), then right-hand side is also ∞.Remark 16.
For𝕋=ℝ and m=1, inequality (37) is proved in [4, 24].


## 5. Some Particular Cases

In this section, firstly, we give Hilbert-type inequality on time scales.Theorem 17.
Ifn=1, a1=0,andb1=∞ in H4 and for q>1 one defines
(38)K1(x)=∫0∞‍(y/x)-1/qx+yΔy,K2(y)=∫0∞‍(y/x)1-1/qx+yΔx,
then
(39)∫0∞(∫0∞(y/x)1-1/q(g1(y))px+yΔy)1/p×(∫0∞(y/x)1-1/q(g2(y))p′x+yΔy)1/p′Δx≥∫0∞‍K2(y)g1(y)g2(y)Δy
holds for all λΔ-integrable gj:Λ→ℝ+, where j∈{1,2}.Proof.
Useξ(x)=K1(x)/x and k(x,y)={(y/x)-1/q/(x+y),x≠0,y≠0,x+y≠0,0,otherwise} in Corollary 7 to obtain
(40)∫0∞K1(x)x(1K1(x)∫0∞(y/x)-1/q(f1(y))px+yΔy)1/p×(1K1(x)∫0∞(y/x)-1/q(f2(y))p′x+yΔy)1/p′Δx≥∫0∞w(y)f1(y)f2(y)Δy,
where
(41)w(y)=∫0∞‍k(x,y)ξ(x)K1(x)Δx=∫0∞‍k(x,y)xΔx=1y∫0∞‍(y/x)1-1/qx+yΔx=K2(y)y.
Using this value in (40), we obtain
(42)∫0∞(∫0∞(y/x)-1/q(f1(y))px+yΔy)1/p×(∫0∞(y/x)-1/q(f2(y))p′x+yΔy)1/p′Δxx≥∫0∞‍K2(y)f1(y)f2(y)Δyy.
Now, if we replace f1(y) with g1(y)y1/p and f2(y) with g2(y)y1/p′, we obtain (39).Another Proof of (42). Consider L.H.S of (40) and applying Hölder inequality on time scale [25, Theorem 6.2] and Fubini theorem on time scale [16, Theorem 1.1], we have
(43)∫0∞K1(x)x(1K1(x)∫0∞(y/x)-1/q(f1(y))px+yΔy)1/p×(1K1(x)∫0∞(y/x)-1/q(f2(y))p′x+yΔy)1/p′Δx≥∫0∞(∫0∞(y/x)-1/qf1(y)f2(y)x+yΔy)Δxx=∫0∞f1(y)f2(y)(∫0∞(y/x)-1/qx+yΔxx)Δy=∫0∞f1(y)f2(y)y(∫0∞(y/x)1-1/qx+yΔx)Δy=∫0∞K2(y)f1(y)f2(y)Δyy.Example 18.
It is known that(44)∫0∞‍(y/x)-1/qx+ydy=∫0∞‍(y/x)1-1/qx+ydx=πsin(π/q),
for all x,y∈ℝ+=(0,∞) with q>1. If 𝕋=ℝ; then from (39), we obtain
(45)∫0∞(∫0∞(y/x)1-1/q(f1(y))px+ydy)1/p×(∫0∞(y/x)1-1/q(f2(y))p′x+ydy)1/p′dx≥πsin(π/q)∫0∞f1(y)f2(y)dy.
In the rest of paper, we taken=1, a≥0, b=∞ in H4.Theorem 19.
If (38) is satisfied, then
(46)∫0∞((∫0∞(y/x)1-1/q(f1(y))px+yΔy)1/p+(∫0∞(y/x)1-1/q(f2(y))px+yΔy)1/p)pΔx≥∫0∞K2(y)(f1(y)+f2(y))pΔy
holds for all λΔ-integrable fj:Λ→ℝ+, where j∈{1,2}.Proof.
Useξ(x)=K1(x)/x and k(x,y)={(y/x)-1/q/(x+y),x≠0,y≠0,x+y≠0,0,otherwise} in Corollary 7 to obtain
(47)∫0∞((∫0∞(y/x)-1/q(f1(y))px+yΔy)1/p+(∫0∞(y/x)-1/q(f2(y))px+yΔy)1/p)pΔxx≥∫0∞K2(y)(f1(y)+f2(y))pΔyy.
Now, if we replace f1(y) with f1(y)y1/p and f2(y) with f2(y)y1/p, we obtain (46).Remark 20.
(a) We can give another proof of (47) using Minkowski's inequality on time scale [25, Theorem  7.2].
(b) Ifp<1, then we have reverse inequalities.
Now we consider some generalizations of the Pólya-Knopp type inequalities.Corollary 21.
If (29) holds, and furthermore, if U⊂ℝm is a closed convex set such that Φ:U→ℝ is convex and continuous, then
(48)∫a∞‍ξ(x)Φ(1σ(x)-a∫aσ(x)‍f(y)Δy)Δx≤∫a∞‍w(y)Φ(f(y))Δy
holds for all λΔ-integrable f:Λ→ℝm such that f(Λ)⊂U.Proof.
The statement follows from Theorem12 using n=1.Corollary 22.
Assume (29). Then
(49)∫a∞‍ξ(x)(1σ(x)-a∫aσ(x)‍(f1(y))pΔy)1/p×(1σ(x)-a∫aσ(x)‍(f1(y))p′Δy)1/p′Δx≥∫a∞‍w(y)f1(y)f2(y)Δy
holds for all λΔ-integrable fj:Λ→ℝ+, where j∈{1,2}.Proof.
Statement follows from Corollary7 using m=2.Corollary 23.
Assume (29). Then
(50)∫a∞ξ(x)((1σ(x)-a∫aσ(x)(f1(y))pΔy)1/p+(1σ(x)-a∫aσ(x)(f2(y))pΔy)1/p)pΔx≥∫a∞w(y)(f1(y)+f2(y))pΔy
holds for all λΔ-integrable fj:Λ→ℝ+, where j∈{1,2}.Proof.
Statement follows from Corollary8 using m=2.Example 24.
When𝕋 consists of isolated points, then from Corollary 22, we have
(51)∑x∈[a,∞)‍ξ(x)(σ(x)-x)σ(x)-a×(∑y∈[a,σ(x))‍(f1(y))p(σ(y)-y))1/p×(∑y∈[a,σ(x))‍(f2(y))p′(σ(y)-y))1/p′≥∑y∈[a,∞)‍w(y)f1(y)f2(y)(σ(y)-y),
where w(y)=∑x∈[y,∞)(ξ(x)(σ(x)-x)/(σ(x)-a)).Example 25.
When𝕋 consists of isolated points, then from Corollary 23, we have
(52)∑x∈[a,∞)‍ξ(x)(σ(x)-x)σ(x)-a×((∑y∈[a,σ(x))‍(f1(y))p(σ(y)-y))1/p+(∑y∈[a,σ(x))‍(f2(y))p(σ(y)-y))1/p)p≥∑y∈[a,∞)‍w(y)(f1(y)+f2(y))p(σ(y)-y),
where w(y) is the same as in Example 24.Example 26.
For𝕋=hℕ={hn:n∈ℕ} with  h>0, a=1, and ξ(x)=1/σ(x),  (51) takes the form
(53)∑n=1∞‍1n(n+1)(∑k=1n‍(f1(kh))p)1/p×(∑k=1n‍(f2(kh))p′)1/p′≥∑n=1∞‍1nf1(nh)f2(nh).
Inequality (52) takes the form
(54)∑n=1∞‍1n(n+1)((∑k=1n‍(f1(kh))p)1/p+(∑k=1n‍(f2(kh))p)1/p)p≥∑n=1∞‍1n(f1(nh)+f2(nh))p.Example 27.
For𝕋=ℕ2={n2:n∈ℕ},a=1, and
(55)ξ(x)=2(σ(x)-1)(σ(x)-x)2(2x+3),
inequality (51) takes the form
(56)∑n=1∞‍2(2n+1)(2n+3)×(∑k=1n‍(2k+1)(f1(k2))p)1/p×(∑k=1n‍(2k+1)(f2(k2))p′)1/p′≥∑k=1∞‍f1(k2)f2(k2).
Inequality (52) takes the form
(57)∑n=1∞‍2(2n+1)(2n+3)×((∑k=1n‍(2k+1)(f1(k2))p)1/p+(∑k=1n‍(2k+1)(f2(k2))p)1/p)p≥∑k=1∞‍(f1(k2)+f2(k2))p.
If we take(58)ξ(x)=σ(x)-1xσ(x),
Inequality (51) takes the form
(59)∑n=1∞‍(2n+1)n2(n+1)2(∑k=1n‍(2k+1)(f1(k2))p)1/p×(∑k=1n‍(2k+1)(f2(k2))p′)1/p′≥∑k=1∞‍(2k+1)k2f1(k2)f2(k2).
Inequality (52) takes the form
(60)∑n=1∞‍(2n+1)n2(n+1)2×((∑k=1n‍(2k+1)(f1(k2))p)1/p+(∑k=1n‍(2k+1)(f2(k2))p)1/p)p≥∑k=1∞‍(2k+1)k2(f1(k2)+f2(k2))p.
By replacingf1(k2)=(k2/(2k+1))1/pf1(k2),f2(k2)=(k2/(2k+1))1/p´f2(k2) in (59) and in (60), we have
(61)∑n=1∞‍(2n+1)n2(n+1)2(∑k=1n‍k2(f1(k2))p)1/p×(∑k=1n‍k2(f2(k2))p′)1/p′≥∑k=1∞‍f1(k2)f2(k2),∑n=1∞‍(2n+1)n2(n+1)2((∑k=1n‍k2(f1(k2))p)1/p+(∑k=1n‍k2(f2(k2))p)1/p)p≥∑k=1∞‍(f1(k2)+f2(k2))p,
respectively.Example 28.
For𝕋=qℕ={qn:n∈{ℕ}, q>1}, a=q and
(62)ξ(x)=σ(x)-aσ(x)(σ(x)-x),
The inequality (51) takes the form
(63)∑n=1∞‍(q-1)qn(∑k=1n‍qk-1(f1(qk))p)1/p×(∑k=1n‍qk-1(f2(qk))p′)1/p′≥∑n=1∞‍f1(qn)f2(qn).
Inequality (52) takes the form
(64)∑n=1∞‍(q-1)qn×((∑k=1n‍qk-1(f1(qk))p)1/p+(∑k=1n‍qk-1(f2(qk))p)1/p)p≥∑n=1∞‍(f1(qn)+f2(qn))p.Remark 29.
(a) In classical case forh=1, inequalities (53) and (54) are the same as (1.7) and (1.9), also (63) and (64) are the same as (1.6) and (1.8) in [18, Corollary 1.3], respectively, while according to authors knowledge, (56), (57), (59), (60), and (61) are not existing in the literature.
(b) Forp<1, we get the reverse inequalities.Remark 30.
The results given in Section5 can analogously be proved for Ω=Λ=[a1,∞)𝕋×⋯×[an,∞)𝕋.Remark 31.
The results given in Corollaries7 and 8 and in their given applications can also be obtained for finite value of m>2 analogously.



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*Source: 903196-2013-09-03.xml*