# Hardy-Leindler-Type Inequalities via Conformable Delta Fractional Calculus

Authors: H. M. Rezk; Wedad Albalawi; H. A. Abd El-Hamid; Ahmed I. Saied; Omar Bazighifan; Mohamed S. Mohamed; M. Zakarya
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this article, some fractional Hardy-Leindler-type inequalities will be illustrated by utilizing the chain law, Hölder’s inequality, and integration by parts on fractional time scales. As a result of this, some classical integral inequalities will be obtained. Also, we would have a variety of well-known dynamic inequalities as special cases from our outcomes whenα=1.


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## Body

## 1. Introduction

The Hardy discrete inequality is known as (see [1])
(1)∑r=1∞1r∑j=1rljμ≤μμ−1μ∑r=1∞lμr,μ>1.where lr>0 for all r≥1.In [2], Hardy employed the calculus of variations and exemplified the continuous version for (1) as follows:
(2)∫0∞1y∫0ygsdsμdy≤μμ−1μ∫0∞gμydy,μ>1,where μ≥0 is integrable over any finite interval 0,y,gμ is convergent and integrable over 0,∞, and μ/μ−1μ is a sharp constant in (1) and (2).Leindler in [3] exemplified that if μ>1 and λr,fr>0, then
(3)∑r=1∞λr∑s=1rfsμ≤μμ∑r=1∞λ1−μr∑s=r∞λsμfμr,(4)∑r=1∞λr∑k=r∞fkμ≤μμ∑r=1∞λ1−μr∑s=1rλsμfμr.The converses of (3) and (4) are exemplified by Leindler in [4]. Precisely, he established that if 0<μ≤1, then
(5)∑r=1∞λr∑s=1rfsμ≥μμ∑r=1∞λ1−μr∑s=r∞λsμfμr,(6)∑r=1∞λr∑k=r∞fkμ≥μμ∑r=1∞λ1−μr∑s=1rλsμfμr.Saker [5] exemplified the time scale version of (3) and (4), respectively, as follows: suppose that T be a time scale and μ>1. If Λτ≔∫τ∞λsΔs,Φτ≔∫lτfsΔs, for any τ∈l,∞T, then
(7)∫l∞λτΦστμΔτ≤μμ∫l∞λ1−μτΛμτfμτΔτ.Also, ifΛ¯τ≔∫lτλsΔs and Φ¯τ≔∫τ∞fsΔs, for any τ∈l,∞T, then
(8)∫l∞λτΦ¯τμΔτ≤μμ∫l∞λ1−μτΛ¯στμfμτΔτ.The converses of (7) and (8) are established by Saker [5]. Precisely, he exemplified that, if T is a time scale, 0<μ≤1,Ωτ=∫τ∞λsΔs, and Ψτ=∫lτfsΔs, for any τ∈l,∞T, then
(9)∫l∞λτΨστμΔτ≥μμ∫l∞fμτΩμτλ1−μτΔτ.Also, ifΩ¯τ=∫lτλsΔs and Ψ¯τ=∫τ∞fsΔs, for any τ∈l,∞T, then
(10)∫l∞λτΨ¯τμΔτ≥μμ∫l∞fμτΩ¯στμλ1−μτΔτ,which are the time scale version for (5) and (6), respectively. For developing dynamic inequalities, see the papers ([6–11]).Our target in this article is proving some fractional dynamic inequalities for Hardy-Leindler’s type, and it is reversed with employing conformable calculus on time scales. This article is structured as follows: In Section2, we discuss the preliminaries of conformable fractional on time scale calculus which will be required in proving our main outcomes. In Section 3, we will exemplify the major consequences.


## 2. Basic Concepts

In this part, we introduce the essentials of conformable fractional integral and derivative of orderα∈0,1 on time scales that will be used in this article (see [12–15]). For a time scale T, we define the operator σ:T⟶T, as
(11)στ≔infs∈T:s>τ.Also, we define the functionμ:T⟶0,∞ by
(12)μτ≔στ−τ.Finally, for anyτ∈T, we refer to the notation ξστ by ξστ, i.e., ξσ=ξ∘σ. In the following, we define conformable α-fractional derivative and α-fractional integral on T.Definition 1 (see [16], Definition 3.1).
Suppose thatξ:T⟶ℝ and α∈0,1. Then, for τ>0, we define DαξΔτ to be the number with the property that, for any ε>0, there is a neighborhood V of τ s.t. ∀τ∈V, we have
(13)ξστ−ξsσ1−ατ−DαξΔτστ−s≤εστ−s.
The conformableα-fractional derivative on T at 0 is
(14)DαξΔ0=limτ⟶0DαξΔτ.Theorem 2 (see [16], Theorem 3.6).
Assume0<α≤1 and ν,ξ:T⟶ℝ are conformable α-fractional derivatives at τ∈Tk. Then, we have the following.
(i)
The sumν+ξ is a conformable α-fractional derivative and
(15)Dαν+ξΔτ=DανΔτ+DαξΔτ(ii)
The productνξ:T⟶ℝ is a conformable α-fractional derivative with
(16)DανξΔτ=DανΔτξτ+νστDαξΔτ=ντDαξΔτ+DανΔτξστ(iii)
Ifξτξστ≠0, then ν/ξ is a conformable α-fractional derivative with
(17)DανξΔτ=DανΔτξτ−ντDαξΔτξτξστLemma 3 (Chain rule).
Suppose thatξ:T⟶ℝ is continuous and α-fractional differentiable at τ∈T, for α∈0,1 and ν:ℝ⟶ℝ is continuously differentiable. Then, ν∘ξ:T⟶ℝ is α-fractional differentiable and
(18)Dαν∘ζΔτ=ν′ξdDαξΔτ,whered∈τ,στ.Definition 4 (see [16], Definition 4.1).
For0<α≤1, then the α-conformable Δ fractional integral of ξ is defined as
(19)IαξΔs=∫ξsΔαs=∫ξsσα−1sΔs,foralls∈Tk.Theorem 5 (see [16], Theorem 4.3).
Letl,m,n∈T,β∈ℝ,α∈0,1, and ν,ξ:T⟶ℝ be rd-continuous functions. Then,
(i)
∫lmνs+ξsΔαs=∫lmνsΔαs+∫lmξsΔαs(ii)
∫lmβνsΔαs=β∫lmνsΔαs(iii)
∫lmνsΔαs=−∫mlνsΔαs(iv)
∫lmνsΔαs=∫lnνsΔαs+∫nmνsΔαs(v)
∫llνsΔαs=0Lemma 6 (Integration by parts formula [16], Theorem 4.3).
Suppose thatl,m∈T where m>l. If ν,ξ are rd-continuous functions and α∈0,1, then
(20)i∫lmνsDαξΔsΔαs=νsξslm−∫lmξσsDαvΔsΔαs.(21)ii∫lmvσsDαξΔsΔαs=νsξslm−∫lmξsDαvΔsΔαs.Lemma 7 (Hölder’s inequality).
Letl,m∈T where m>l. If α∈0,1 and F,G:T⟶ℝ, then
(22)∫lmFsGsΔαs≤∫lmFsβΔαs1/β∫lmGsμΔαs1/μ,where β>1 and 1/β+1/μ=1.
Through our paper, we will consider the integrals are given exist (are finite, i.e., convergent).


## 3. Main Results

Here, we will exemplify our major results in this article. In the pursuing theorem, we will exemplify Leindler’s inequality (7) for fractional time scales as follows.Theorem 8.
Suppose thatT be a time scale and 0<α≤1. If μ>1,Λτ≔∫τ∞λsΔαs and Φτ≔∫lτfsΔαs, for any τ∈l,∞T, then
(23)∫l∞λτΦστμ−α+1Δατ≤μ−α+1∫l∞λ1−μτΛμτfμτΔατ1/μ×∫l∞λτΦστμμ−α/μ−1Δατμ−1/μ.Proof.
By utilizing (20) on
(24)∫l∞λτΦστμ−α+1Δατ,with ζστ=Φστμ−α+1 and DανΔτ=λτ, we have
(25)∫l∞λτΦστμ−α+1Δατ=ντΦμ−α+1τl∞+∫l∞−ντDαΦΔμ−α+1τΔατ,where
(26)ντ=−∫τ∞λsΔαs=−Λτ.
Substituting (26) into (25), we get
(27)∫l∞λτΦστμ−α+1Δατ=ντΦμ−α+1τl∞+∫l∞ΛτDαΦΔμ−α+1τΔατ.
UsingΦl=0 and Λ∞=0 in (27), we have
(28)∫l∞λτΦστμ−α+1Δατ=∫l∞ΛτDαΦΔμ−α+1τΔατ.
Utilizing the chain rule (18), we get
(29)DαΦΔμ−α+1τ=μ−α+1Φμ−αdDαΦΔτ≤μ−α+1DαΦΔτΦστμ−α.
SinceDαΦΔτ=fτ, we get
(30)DαΦΔμ−α+1τ≤μ−α+1fτΦστμ−α.
Substituting (30) into (28) yields
(31)∫l∞λτΦστμ−α+1Δατ≤μ−α+1∫l∞ΛτfτΦστμ−αΔατ.Inequality (31) can be written as
(32)∫l∞λτΦστμ−α+1Δατ≤μ−α+1∫l∞Λτfτλμ−1/μτλμ−1/μτΦστμ−αΔατ.
Implementing Hölder’s inequality on the R.H.S of (32) with indices μ,μ/μ−1, we get
(33)∫l∞Λτfτλμ−1/μτλμ−1/μτΦστμ−αΔατ≤∫l∞Λτfτλμ−1/μτμΔατ1/μ∫l∞λμ−1/μτΦστμ−αμ/μ−1Δατμ−1/μ=∫l∞Λμτfμτλμ−1τΔατ1/μ∫l∞λτΦστμμ−α/μ−1Δατμ−1/μ.By substituting (33) into (32), we get
(34)∫l∞λτΦστμ−α+1Δατ≤μ−α+1∫l∞λ1−μτΛμτfμτΔατ1/μ×∫l∞λτΦστμμ−α/μ−1Δατμ−1/μ,which is (23).Corollary 9.
Atα=1 in Theorem 8, then
(35)∫l∞λτΦστμΔτ≤μ∫l∞λ1−μτΛμτfμτΔτ1/μ×∫l∞λτΦστμΔτμ−1/μ,where >1,Λτ≔∫τ∞λsΔs, and Φτ≔∫lτfsΔs, for any τ∈l,∞T.Remark 10.
In Corollary9, if we divide both sides of (35) by the factor
(36)∫l∞λτΦστμΔτμ−1/μ,and using the fact that 1−μ−1/μ=1/μ, then
(37)∫l∞λτΦστμΔτ1/μ≤μ∫l∞λ1−μτΛμτfμτΔτ1/μ.
Elevating the last inequality to theμth power, we get
(38)∫l∞λτΦστμΔτ≤μμ∫l∞λ1−μτΛμτfμτΔτ,which is (7) in Introduction.Remark 11.
If we putT=ℝ (i.e., στ=τ) in Theorem 5, then
(39)∫l∞λτΦμ−α+1ττα−1dτ≤μ−α+1∫l∞λ1−μτΛμτfμττα−1dτ1/μ×∫l∞λτΦτμμ−α/μ−1τα−1dτμ−1/μ,where α∈0,1,μ>1,Λτ≔∫τ∞λssα−1ds, and Φτ≔∫lτfssα−1ds, for any τ∈l,∞.Remark 12.
Clearly, forα=1 and l=1, Remark 12 coincides with Remark 10 in [5].Remark 13.
WhenT=ℤ (i.e., στ=τ+1),μ>1, and l=1 in (23), then we get
(40)∑r=1∞λr∑s=1σr−1fss+1α−1μ−α+1r+1α−1≤μ−α+1∑r=1∞λ1−μr∑s=r∞λss+1α−1μfμrr+1α−11/μ×∑r=1∞λr∑s=1σr−1fss+1α−1μμ−α/μ−1r+1α−1μ−1/μ.
Ifα=1, then (40) becomes
(41)∑r=1∞λr∑s=1σr−1fsμ≤μ∑r=1∞λ1−μr∑s=r∞λsμfμr1/μ×∑r=1∞λr∑s=1σr−1fsμμ−1/μ,which is Remark 11 in [5].In the pursuing theorem, we will exemplify Leindler’s inequality (8) on fractional time scales as follows.Theorem 14.
Suppose thatT be a time scale and 0<α≤1. If >1,Λ¯τ≔∫lτλsΔαs, and Ψτ≔∫τ∞fsΔαs, for any τ∈l,∞T, then
(42)∫l∞λτΨτμ−α+1Δατ≤μ−α+1∫l∞λ1−μτΛ¯στμfμτΔατ1/μ×∫l∞λτΨτμμ−α/μ−1Δατμ−1/μ.Proof.
By utilizing (20) on
(43)∫l∞λτΨτμ−α+1Δατ,with ντ=Ψτμ−α+1 and DαζΔτ=λτ, we have
(44)∫l∞λτΨτμ−α+1Δατ=ζτΨτμ−α+1l∞+∫l∞ζστ−DαΨΔμ−α+1τΔατ,where
(45)ζτ=∫lτλsΔαs=Λ¯τ.
Substituting (45) into (44), we get
(46)∫l∞λτΨτμ−α+1Δατ=Λ¯τΨτμ−α+1l∞+∫l∞Λ¯τσ−DαΨΔμ−α+1τΔατ.
Using the fact thatΨ∞=0 and Λ¯l=0, (46) became
(47)∫l∞λτΨτμ−α+1Δατ=∫l∞Λ¯τσ−DαΨΔμ−α+1τΔατ.
Utilizing chain rule (18), we get
(48)−DαΨΔμ−α+1τ=−μ−α+1Ψμ−αdDαΨΔτ=−μ−α+1−fτΨμ−αd≤μ−α+1fτΨμ−ατ.
By substituting (48) into (47), we get
(49)∫l∞λτΨτμ−α+1Δατ≤μ−α+1∫l∞Λ¯τσfτΨμ−ατΔατ.Inequality (49) can be written as
(50)∫l∞λτΨτμ−α+1Δατ≤μ−α+1∫l∞Λ¯τσfτλμ−1/μτλμ−1/μτΨμ−ατΔατ.
Implementing Hölder’s inequality on the R.H.S of (50) with indices μ,μ/μ−1, we get
(51)∫l∞Λ¯τσfτλμ−1/μτλμ−1/μτΨμ−ατΔατ≤∫l∞Λ¯τσfτλμ−1/μτμΔατ1/μ∫l∞λμ−1/μτΨμ−ατμ/μ−1Δατμ−1/μ=∫l∞Λ¯στμfμτλμ−1τΔατ1/μ∫l∞λτΨτμμ−α/μ−1Δατμ−1/μ.By substituting (51) into (50), we get
(52)∫l∞λτΨμ−α+1τΔατ≤μ−α+1∫l∞λ1−μτΛ¯στμfμτΔατ1/μ×∫l∞λτΨτμμ−α/μ−1Δατμ−1/μ,
which is (42).Corollary 15.
Atα=1 in Theorem 14, then
(53)∫l∞λτΨτμΔτ≤μ∫l∞λ1−μτΛ¯στμfμτΔτ1/μ×∫l∞λτΦ¯τμΔτμ−1/μ,where μ>1,Λ¯τ≔∫lτλsΔs, and Ψτ≔∫τ∞fsΔs, for any τ∈l,∞T.Remark 16.
In Corollary15, if we divide both sides of (53) by the factor
(54)∫l∞λτΨτμΔτμ−1/μ,and using the fact that 1−μ−1/μ=1/μ, then
(55)∫l∞λτΨτμΔτ1/μ≤μ∫l∞λ1−μτΛ¯στμfμτΔτ1/μ.
Elevating the last inequality to theμth power, we get
(56)∫l∞λτΨτμΔτ≤μμ∫l∞λ1−μτΛ¯στμfμτΔτ,which is (8) in Introduction.Remark 17.
As a result, ifT=ℝ (i.e., στ=τ) in Theorem 14, then
(57)∫l∞λτΨμ−α+1ττα−1dτ≤μ−α+1∫l∞λ1−μτΛ¯τμfμττα−1dτ1/μ×∫l∞λτΨτμμ−α/μ−1τα−1dτμ−1/μ,where α∈0,1,μ>1,Λ¯τ≔∫lτλssα−1ds, and Ψτ≔∫τ∞fssα−1ds, for any τ∈l,∞.Remark 18.
Clearly, forα=1 and l=1, Remark 17 coincides with Remark 12 in [5].Remark 19.
WhenT=ℤ (i.e., στ=τ+1), μ>1, and l=1 in (42), we get
(58)∑r=1∞λr∑s=r∞fss+1α−1μ−α+1r+1α−1≤μ−α+1∑r=1∞λ1−μr∑k=1r−1λkk+1α−1μfμrr+1α−11/μ×∑r=1∞λr∑s=r∞fss+1α−1μμ−α/μ−1r+1α−1μ−1/μ.
Ifα=1, then (58) becomes
(59)∑r=1∞λr∑s=r∞fsμ≤μ∑r=1∞λ1−μr∑k=1rλkμfμr1/μ×∑r=1∞λr∑s=r∞fsμμ−1/μ,
which is Remark13 in [5].In the pursuing theorem, we will exemplify Leindler’s inequality (9) for fractional time scales as follows.Theorem 20.
Suppose thatT be a time scale and α∈0,1. If 0<μ≤1,Ωτ=∫τ∞λsΔαs and Ϝτ=∫lτfsΔαs, for any τ∈l,∞T, then
(60)∫l∞λτϜστμ−α+1Δατμ≥μ−α+1μ∫l∞fμτΩμτλ1−μτΔατ×∫l∞λτϜστμα−μ/1−μΔατμ−1.Proof.
By applying (20) on
(61)∫l∞λτϜστμ−α+1Δατ,with ζστ=Ϝστμ−α+1 and DανΔτ=λτ, we have
(62)∫l∞λτϜστμ−α+1Δατ=ντϜμ−α+1τ∞l+∫l∞−ντDαϜΔμ−α+1τΔατ,where
(63)ντ=−∫τ∞λsΔαs=−Ωτ.
Substituting (63) into (62) yields
(64)∫l∞λτϜστμ−α+1Δατ=−ΩτϜμ−α+1τ∞l+∫l∞ΩτDαϜΔμ−α+1τΔατ.
Using the fact thatν∞=0 and Ϝl=0, (64) became
(65)∫l∞λτϜστμ−α+1Δατ=∫l∞ΩτDαϜΔμ−α+1τΔατ.
Utilizing chain rule (18), we get
(66)DαϜΔμ−α+1τ=μ−α+1Ϝμ−αdDαϜΔτ≥μ−α+1DαϜΔτϜστμ−α.
SinceDαϜΔτ=fτ, we obtain
(67)DαϜΔμ−α+1τ≥μ−α+1fτϜστμ−α.By substituting (67) into (65), we have
(68)∫l∞λτϜστμ−α+1Δατ≥μ−α+1∫l∞ΩτfτϜστμ−αΔατ=μ−α+1∫l∞ΩμτfμτϜστμμ−α1/μΔατ.
Raises (68) to the factor μ, we get
(69)∫l∞λτϜστμ−α+1Δατμ≥μ−α+1μ∫l∞λΩμτfμτϜστμμ−α1/μΔατμ.By applying Hölder’s inequality on
(70)∫l∞ΩμτfμτϜστμμ−α1/μΔατμ,with indices 1/μ,1/1−μ, and
(71)Fτ=fμτΩμτϜστμα−μ,Gτ=λ1‐μτϜστμα‐μ,we see that
(72)∫l∞F1/μτΔατμ=∫l∞fμτΩμτϜστμα−μ1/μΔατμ≥∫l∞FτGτΔατ∫l∞G1/1−μτΔατ1−μ=∫l∞fμτΩμτλ1−μτϜστμα−μϜστμα−μΔατ×∫l∞λ1−μτϜστμα−μ1/1−μΔατμ−1=∫l∞fμτΩμτλ1−μτΔατ∫l∞λτϜστμα−μ/1−μΔατμ−1.
This implies that(73)∫l∞ΩτfτϜστμ−αΔατμ≥∫l∞fμτΩμτλ1−μτΔατ×∫l∞λτϜστμα−μ/1−μΔατμ−1.By substituting (73) into (69), we get
(74)∫l∞λτϜστμ−α+1Δατμ≥μ−α+1μ∫l∞fμτΩμτλ1−μτΔατ×∫l∞λτϜστμα−μ/1−μΔατμ−1,which is (60).Corollary 21.
Atα=1 in Theorem 20, then
(75)∫l∞λτϜστμΔτμ≥μμ∫l∞fμτΩμτλ1−μτΔτ×∫l∞λτϜστμΔτμ−1,where 0<μ≤1,Ωτ=∫τ∞λsΔs, and Ϝτ=∫lτfsΔs, for any τ∈l,∞.Remark 22.
In Corollary21, if we divide both sides of (75) by the factor
(76)∫l∞λτϜστμτΔτμ−1,then (75) can be written as
(77)∫l∞λτϜστμΔτ≥μμ∫l∞fμτΩμτλ1−μτΔτ,which is (9) in Introduction.Remark 23.
As a result, ifT=ℝ (i.e., στ=τ) in Theorem 20, then
(78)∫l∞λτϜτμ−α+1τα−1dτμ≥μ−α+1μ∫l∞fμτΩμτλ1−μττα−1dτ×∫l∞λτϜτμα−μ/1−μτα−1dτμ−1,where α∈0,1,0<μ≤1,Ωτ=∫τ∞λssα−1ds, and Ϝτ=∫lτfssα−1ds, for any τ∈l,∞.Remark 24.
Clearly, forα=1 and l=1, Remark 23 coincides with Remark 16 in [5].Remark 25.
WhenT=ℤ (i.e., στ=τ+1), μ≤1, and l=1 in (60), then we get
(79)∑r=1∞λr∑s=1σr−1fss+1α−1μ−α+1r+1α−1μ≥μ−α+1μ∑r=1∞fμr∑s=r∞λss+1α−1μλ1−μrr+1α−1×∑r=1∞λr∑s=1σr−1fss+1α−1μ−α+1/1−μr+1α−1μ−1.
Ifα=1, then (79) becomes
(80)∑r=1∞λr∑s=1σr−1fsμμ≥μμ∑r=1∞fμr∑s=r∞λsμλ1−μr×∑r=1∞λr∑s=1σr−1fsμμ−1,which is Remark 17 in [5].In the pursuing theorem, we will exemplify Leindler’s inequality (10) for fractional time scales as follows.Theorem 26.
Suppose thatT be a time scale and α∈0,1. If 0<μ≤1,Ω¯τ=∫lτλsΔαs, and Γτ=∫τ∞fsΔαs, for any τ∈l,∞T, then
(81)∫l∞λτΓτμ−α+1Δατμ≥μ−α+1μ∫l∞fμτΩ¯στμλ1−μτΔατ×∫l∞λτΓτμα−1/1−μΔατμ−1.Proof.
By applying (20) on
(82)∫l∞λτΓτμ−α+1Δατ,with ντ=Γτμ−α+1 and DαζΔτ=λτ, we have
(83)∫l∞λτΓτμ−α+1Δατ=ζτΓτμ−α+1l∞+∫l∞ζστ−DαΓΔμ−α+1τΔατ,where
(84)ζτ=∫lτλsΔαs=Ω¯τ.
Substituting (84) into (83), we get
(85)∫l∞λτΓτμ−α+1Δατ=Ω¯τΓτμ−α+1l∞+∫l∞Ω¯στ−DαΓΔμ−α+1τΔατ.
Using the fact thatΓ∞=0 and Ω¯l=0, (85) became
(86)∫l∞λτΓτμ−α+1Δατ=∫l∞Ω¯στ−DαΓΔμ−α+1τΔατ.
Utilizing chain rule (18), we have
(87)−DαΓΔμ−α+1τ=−μ−α+1Γμ−αdDαΓΔτ≥−μ−α+1Γμ−ατDαΓΔτ.
SinceDαΓΔτ=−fτ, we get
(88)−DαΓΔμ−α+1τ≥μ−α+1fτΓμ−ατ.By substituting (88) into (86), we get
(89)∫l∞λτΓτμ−α+1Δατ≥μ−α+1∫l∞Ω¯στfτΓμ−ατΔατ,=∫l∞Ω¯στμfμτΓμμ−ατ1/μΔατ.
Raises (89) to the factor μ, we have
(90)∫l∞λτΓτμ−α+1Δατμ≥μ−α+1μ∫l∞Ω¯στμfμτΓμμ−ατ1/μΔατμ.By applying Hölder’s inequality on
(91)∫l∞Ω¯στμfμτΓμμ−ατ1/μΔατμ,with indices 1/μ,1/1−μ, and
(92)Fτ=fμτΩ¯στμΓμα−μτ,Gτ=λ1−μτΓτμα−μ,we see that
(93)∫l∞F1/μτΔατμ=∫l∞fμτΩ¯στμΓμα−μτ1/μΔατμ≥∫l∞FτGτΔατ∫l∞G1/1−μτΔατ1−μ=∫l∞fμτλ1−μτΩ¯στμΓτμα−μΓμα−μτΔατ×∫l∞λ1−μτΓτμα−μ1/1−μΔατμ−1=∫l∞fμτΩ¯στμλ1−μτΔατ×∫l∞λτΓτμα−μ/1−μΔατμ−1.
This implies that(94)∫l∞Ω¯στμfμτΓμμ−ατ1/μΔατμ≥∫l∞fμτΩ¯στμλ1−μτΔατ×∫l∞λτΓτμα−μ/1−μΔατμ−1.By substituting (94) into (90), we get
(95)∫l∞λτΓτμ−α+1Δατμ≥μ−α+1μ∫l∞fμτΩ¯στμλ1−μτΔατ×∫l∞λτΓτμα−μ/1−μΔατμ−1,which is (81).Corollary 27.
Atα=1 in Theorem 26, then
(96)∫l∞λτΓτμΔτμ≥μμ∫l∞fμτΩ¯στμλ1−μτΔτ×∫l∞λτΓτμΔτμ−1,where 0<μ≤1,Ω¯τ=∫lτλsΔs, and Γτ=∫τ∞fsΔs, for any τ∈l,∞T.Remark 28.
In Corollary27, if we divide both sides of (96) by the factor
(97)∫l∞λτΓτμΔτμ−1,then (96) can be written as
(98)∫l∞λτΓτμΔτ≥μμ∫l∞fμτΩ¯στμλ1−μτΔτ,which is (10) in Introduction.Remark 29.
As a result, ifT=ℝ (i.e., στ=τ) in Theorem 26, then
(99)∫l∞λτΓτμ−α+1τα−1dτμ≥μ−α+1μ∫l∞fμτΩ¯τμλ1−μττα−1dτ×∫l∞λτΓτμα−μ/1−μτα−1dτμ−1,where α∈0,1,0<μ≤1,Ω¯τ=∫lτλssα−1ds, and Γτ=∫τ∞fssα−1ds, for any τ∈l,∞.Remark 30.
Clearly, forα=1 and l=1, Remark 29 coincides with Remark 18 in [5].Remark 31.
WhenT=ℤ (i.e., στ=τ+1), μ≤1, and l=1 in (81), then we get
(100)∑r=1∞λr∑k=r∞fkk+1α−1μ−α+1r+1α−1μ≥μ−α+1μ∑r=1∞λ1−μr∑k=1r−1λkk+1α−1μfμrr+1α−1×∑r=1∞λr∑k=r∞fkk+1α−1μα−μ/1−μr+1α−1μ−1.
Ifα=1, then (100) becomes
(101)∑r=1∞λr∑k=r∞fkμμ≥μμ∑r=1∞λ1−μr∑k=1r−1λkμfμr×∑r=1∞λr∑k=r∞fkμμ−1,which is Remark 19 in [5].


## 4. Conclusions and Future Work

In this article, we explore new generalizations of the integral Hardy-Leindler-type inequalities by the utilization of the delta conformable calculus on time scales which are used in various problems involving symmetry. We generalize a number of those inequalities to a general time scale measure space. In addition to this, in order to obtain some new inequalities as special cases, we also extend our inequalities to a discrete and continuous calculus. In future work, we will continue to generalize more fractional dynamic inequalities by using Specht’s ratio, Kantorovich’s ratio, and n-tuple fractional integral.



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