# Fractional Minkowski-Type Integral Inequalities via the Unified Generalized Fractional Integral Operator

Authors: Tingmei Gao; Ghulam Farid; Ayyaz Ahmad; Waewta Luangboon; Kamsing Nonlaopon
Journal: Journal of Function Spaces
 (2022)

---

## Abstract

This paper is aimed at presenting the unified integral operator in its generalized form utilizing the unified Mittag-Leffler function in its kernel. We prove the boundedness of this newly defined operator. A fractional integral operator comprising a unified Mittag-Leffler function is used to establish further Minkowski-type integral inequalities. Several related fractional integral inequalities that have recently been published in various articles can be inferred.


---

## Body

## 1. Introduction

Integral operators are useful in the study of differential equations and in the formation of real-world problems in integral equations. They also behave like integral transformations in particular cases. In the past few decades, fractional integral operators have been defined extensively (see [1–4]). Recently, in [5] a unified integral operator is studied which has interesting consequences in the theory of fractional integral operators.This paper is aimed at presenting a unified integral operator in the more generalized form via the unified Mittag-Leffler function introduced in [6]. The boundedness of the newly defined integral operator is studied. By taking the power function ξβ;β>1, a unified generalized extended fractional integral operator is deduced and analyzed to construct Minkowski-type integral inequalities. This is the extension of our previous work on Minkowski-type integral inequalities [7]. The connection of the results of this paper is established with many published results of references [7–9]. We begin by reviewing several key Minkowski-type inequalities as well as some definitions that will be useful in our subsequent work.The well-known Minkowski inequality is given as follows:Theorem 1. Letϕ, ψ∈Lmu,v.
Then form≥1, we have
(1)∫uvϕξ+ψξmdξ1/m≤∫uvϕmξdξ1/m+∫uvψmξdξ1/m.Some more Minkowski-type inequalities are stated in the next results.Theorem 2. ([10]).
Letϕ, ψ∈Lmu,v. Also ϕ, ψ∈R+ such that 0<k1≤ϕξ/ψξ≤k2∀ξ∈u,v. Then for m≥1, the following inequality holds true
(2)∫uvϕmξdξ1/m+∫uvψmξdξ1/m≤1+k2−k1k1+1k2+1∫uvϕξ+ψξmdξ1/m.Theorem 3. ([11]).
Under the assumptions of Theorem2, we have the following inequality:
(3)∫uvϕmξdξ2/m+∫uvψmξdξ2/m≥2+k1−1k2+1k2∫uvϕmξdξ2/m∫uvψmξdξ2/m.Theorem 4. ([9]).
Letω∈ℝ,α,β,γ>0,θ>λ>0 with s≥0,r>0 and 0<k≤r+α. Let m≥1 and ϕ,ψ∈Lmu,v be positive functions satisfying
(4)0<k1≤ϕξψξ≤k2,ξ∈u,v.
Then the following inequality holds:(5)εϕmξ;s1/m+εψmξ;s1/m≤1+k2−k1k1+1k2+1εϕ+ψmξ;s1/m.Theorem 5. [9].
Letm,n>1 such that 1/m+1/n=1. Then under the assumptions of Theorem 4, we have
(6)εϕξ;s1/mεψξ;s1/n≤k2k11/mnεϕ1/mψ1/nξ;s.A special function known as the Mittag-Leffler function was introduced by a Swedish mathematician Gosta Mittag-Leffler [12] by the following series:
(7)Eαt=∑l=0∞tlΓαl+1,where t,α∈ℂ and Rα>0.This function is a direct extension of the exponential function that can be used to construct solutions of fractional differential equations. Due to its wide range of applications, this function has received considerable attention in recent decades. Many researchers provided its numerous generalized forms due to its intriguing results. We refer the readers to [2, 4, 13–16] for the study of generalized versions of the Mittag-Leffler function. In [17], Bhatnagar et al. introduced the generalization of Mittag-Leffler function in the form of generalized Q function as follows:Definition 6.
The generalizedQ function denoted by Qα,β,γ,δ,μ,νλ,ρ,θ,k,n.;.,. is defined by the following series:
(8)Qα,β,γ,δ,μ,νλ,ρ,θ,k,nt;a¯,b¯=∑l=0∞Πi=1nBbi,lλρlθkltlΠi=1nBai,lγδlμνlΓαl+β,where a¯=a1,a2,⋯,an,b¯=b1,b2,⋯,bn,α,β,γ,δ,μ,ν,λ,ρ,θ,ai,bi∈ℂ,k∈0,1∪ℕ and minRα,Rβ,Rγ,Rθ,Rλ,Rδ,Rρ>0.Recently, in [7], we introduced the fractional integral operator associated with generalized Q function as follows:
(9)QIu+,α,β,γ,δ,μ,νω,λ,ρ,θ,k,nfξ;a¯,b¯=∫uξξ−tβ−1Qα,β,γ,δ,μ,νλ,ρ,θ,k,nωξ−tα;a¯,b¯ftdt,(10)QIv−,α,β,γ,δ,μ,νω,λ,ρ,θ,k,nfξ;a¯,b¯=∫ξvt−ξβ−1Qα,β,γ,δ,μ,νλ,ρ,θ,k,nωt−ξα;a¯,b¯ftdt.Andrić et al. in [2] introduced an extended and generalized Mittag-Leffler function along with the corresponding fractional integral operator as follows:Definition 7.
The extended and generalized Mittag-Leffler functionEα,β,γδ,μ,k,ν is defined by the following series:
(11)Eα,β,γλ,θ,k,rt;s=∑l=0∞Bsλ+lk,θ−λθlktlBλ,θ−λγlrΓαl+β,where t,α,β,γ,θ,λ∈ℂ,Rα,Rβ,Rγ,Rθ,Rλ>0, Rθ>Rλ with s≥0, r>0, 0<k≤r+Rα, and θlk=Γθ+lk/Γθ.Definition 8. Letf∈L1u,v.
Then forξ∈u,v, the fractional integral operator corresponding to (11) is defined by the following integrals:
(12)εu+,α,β,γω,λ,θ,k,rfξ;s=∫uξξ−tβ−1Eα,β,γ,λ,θ,k,rωξ−tα;sftdt,εv−,α,β,γω,λ,θ,k,rfξ;s=∫ξvt−ξβ−1Eα,β,γ,λ,θ,k,rωξ−tα;sftdt.In [5], Farid defined the unified integral operator based on the extended and generalized Mittag-Leffler function (11) as follows:Definition 9.
Letω,β,γ,λ,θ∈ℂ,Rβ,Rγ>0,Rθ>Rλ>0 with s≥0,α,r>0 and 0<k≤r+α. Let ϕ∈L1u,v,0<u<v<∞, be a positive function. Let g:u,v⟶ℝ be a differentiable function, strictly increasing. Also let ζξ/ξ be an increasing function on u,∞ and ξ∈u,v. Then the left-sided integral is defined by
(13)gζFu+,α,β,γω,λ,θ,k,rϕξ;s=∫uξζgξ−gtgξ−gtEα,β,γλ,θ,k,rωgξ−gtα;sg′tϕtdt.In [6], we have presented a further generalized unified Mittag-Leffler function and the associated fractional integral operator as follows:Definition 10.
Fora¯=a1,a2,⋯,an,b¯=b1,b2,⋯,bn,c¯=c1,c2,⋯,cn,whereai,bi,ci∈ℂ;i=1,⋯,n such that Rai,Rbi,Rci>0. Also let ,β,γ,δ,μ,ν,λ,ρ,θ,t∈ℂ, minRα,Rβ,Rγ,Rδ,Rλ,Rθ>0 and k∈0,1∪ℕ with s≥0. Letk+Rρ<Rδ+ν+α with Imρ=Imδ+ν+α, then the unified Mittag-Leffler function is defined as follows
(14)Mα,β,γ,δ,μ,νλ,ρ,θ,k,nt;a¯,b¯,c¯,s=∑l=0∞Πi=1nBsbi,aiλρlθklΠi=1nBci,aiγδlμνltlΓαl+β.Definition 11. Letϕ∈L1u,v.
Then forξ∈u,v, the fractional integral operators corresponding to (14) are defined by
(15)Iu+,α,β,γ,δ,μ,νω,λ,ρ,θ,k,nϕξ;a¯,b¯,c¯,s=∫uξξ−tβ−1Mα,β,γ,δ,μ,νλ,ρ,θ,k,nωξ−tα;a¯,b¯,c¯,sϕtdt,(16)Iv−,α,β,γ,δ,μ,νω,λ,ρ,θ,k,nϕξ;a¯,b¯,c¯,s=∫ξvt−ξβ−1Mα,β,γ,δ,μ,νλ,ρ,θ,k,nωt−ξα;a¯,b¯,c¯,sϕtdt.Fractional integral operators are used to extend different types of integral inequalities such as Opial-type inequalities [2, 18–22], Hadamard- and Fejér-Hadamard-type inequalities [23–32], Pólya-Szegö-, Chebyshev-, and Grüss-type inequalities [33–36] (see references therein), and Minkowski-type fractional inequalities [7–9]. In this paper we study Minkowski-type fractional inequalities via the unified Mittag-Leffler function.In Section2, we give the definition of further generalized integral operator containing the unified Mittag-Leffler function. The boundedness of this integral operator is proved under the conditions stated in the definition. In Section 3, by applying a particular fractional integral operator for the power function, Minkowski-type fractional integral inequalities are established. In Section 4, reverse Minkowski-type fractional integral inequalities are presented. The connection of these inequalities with previous work is stated in the form of remarks and corollaries.


## 2. Generalized Version of a Unified Integral Operator

In this section, we introduce a generalized version of a unified integral operator containing a unified Mittag-Leffler function in its kernel and also discuss its boundedness.Definition 12.
Letω,a¯=a1,a2,⋯,an,b¯=b1,b2,⋯,bn,c¯=c1,c2,⋯,cn,whereai,bi,ci∈ℂ;i=1,⋯,n such that Rai,Rbi,Rci>0∀i. Also let ,γ,μ,λ,θ,t∈ℂ, minRβ,Rγ,Rμ,Rλ,Rθ>0, ρ,δ,ν,α>0 and k∈0,1∪ℕ. Let k+ρ<δ+ν+α with s≥0. Let ϕ∈L1u,v, 0<u<v<∞ be a positive function, and let g:u,v⟶ℝ be a differentiable and strictly increasing function. Also let ζξ/ξ be an increasing function on u,∞ for ξ∈u,v. Then the unified integral operator in its generalized form is defined by the following integral:
(17)Ωgζu+,α,β,γ,δ,μ,νω,λ,ρ,θ,k,nϕξ;s=∫uξζgξ−gtgξ−gtMα,β,γ,δ,μ,νλ,ρ,θ,k,nωgξ−gtα;sg′tϕtdt.
On a particular case, by takingζξ=ξβ;β>1 and replacing ℂ by ℝ, the above operator takes the following form:
(18)gΩu+,α,β,γ,δ,μ,νω,λ,ρ,θ,k,nϕξ;s=∫uξgξ−gtβ−1Mα,β,γ,δ,u,νλ,ρ,θ,k,nωgξ−gtα;sg′tϕtdt,where ω,a¯=a1,a2,⋯,an,b¯=b1,b2,⋯,bn,c¯=c1,c2,⋯,cn,ai,bi,ci∈ℝ;i=1,⋯,n such that ai,bi,ci>0∀i. Also α,γ,δ,μ,ν,λ,ρ,θ>0,β>1 and k∈0,1∪ℕ with k+ρ<δ+ν+α and s≥0.Definition 13.
By settingai=l,s=0 and ρ>0 in (18), we will get the following integral operator:
(19)QgΩu+,α,β,γ,δ,μ,νω,λ,ρ,θ,k,nϕξ;s=∫uξgξ−gtβ−1Qα,β,γ,δ,u,νλ,ρ,θ,k,nωgξ−gtα;sg′tϕtdt.Remark 14.
(i)
By consideringn=1, b1=λ+lk, a1=θ−λ, c1=λ, ρ=ν=0, δ>0 in (17), the unified integral operator given in (13) is deduced(ii)
By considering the functiong to be an identity function in (19), the fractional integral operator given in (9) is deduced(iii)
By consideringn=1, b1=λ+lk, a1=θ−λ, c1=λ, ρ=ν=0, and δ>0 in (18), the generalized fractional integral operator ([8], Definition 1.4) is deduced(iv)
By considerings=0=ω in (18), then the left-sided Riemann-Liouville fractional integral operator of a function ϕ with respect to another function g of order β given in [1, 3] is deduced(v)
By consideringg to be an identity function, (18) is deduced to (15)(vi)
By consideringg as identity function and setting n=1, b1=λ+lk, a1=θ−λ, c1=λ, ρ=ν=0, and δ>0, in (18), the generalized fractional integral operator (21) is deduced
For simplicity, we will use the following notations throughout this paper:Mα,β,γ,δ,μ,νλ,ρ,θ,k,n≔M, Iu+,α,β,γ,δ,μ,νω,λ,ρ,θ,k,n=I, QIu+,α,β,γ,δ,μ,νω,λ,ρ,θ,k,n=QI, gζΩu+,α,β,γ,δ,μ,νω,λ,ρ,θ,k,n:=ζΩ, gΩu+,α,β,γ,δ,μ,νω,λ,ρ,θ,k,n≔Ω, QgΩu+,α,β,γ,δ,μ,νω,λ,ρ,θ,k,n:=QΩ.
Next, we discuss the boundedness of the newly defined generalized form of unified fractional integral operator.Theorem 15.
Letω∈ℝ,a¯=a1,a2,⋯,an,b¯=b1,b2,⋯,bn,c¯=c1,c2,⋯,cn,ai,bi,ci∈ℝ;i=1,⋯,n such that ai,bi,ci>0. Also α,β,γ,δ,μ,ν,λ,ρ,θ,t∈ℝ, k∈0,1∪ℕ and minα,β,γ,δ,λ,θ>0 with k+ρ<δ+ν+α with s≥0. Let ϕ∈L1u,v, 0<u<v<∞ be a positive function, and let g:u,v⟶ℝ be a differentiable and strictly increasing function. Also let ζ/ξ be an increasing function on u,∞. Then for ξ∈u,v, we get
(20)Ωζϕξ;s≤ζgξ−guMωgξ−guα;sϕu,ξ,(21)Ωζϕξ;s≤ζgv−gξMωgξ−guα;sϕξ,v,where ϕu,ξ=supt∈u,ξϕt and ϕξ,v=supt∈ξ,vϕt.Proof.
According to the statement,ζ/ξ is an increasing function; therefore, the following inequality prevails:
(22)ζgξ−gtgξ−gt≤ζgξ−gugξ−gu.
Sinceg is differentiable and increasing and ϕ is a positive function, so the above inequality remains preserved by multiplying it with g′tϕt. Therefore, we obtain the following inequality:
(23)ζgξ−gtgξ−gtg′tϕt≤ζgξ−gugξ−gug′tϕt.
Multiplying (23) by Mωgξ−gtα;s and integrating over u,ξ one can get
(24)ζΩϕξ;s≤ζgξ−gugξ−guϕu,ξ∫uξMωgξ−gtα;sg′tdt.
Solving the above definite integral, we get(25)ζΩϕξ;s≤ζgξ−guMωgξ−guα;sϕu,ξ.
Similarly, one can easily prove (21).


## 3. Unified Versions of Minkowski-Type Fractional Integral Inequalities

In this section, we giveproof of unified versions of generalized Minkowski-type integral inequalities.Theorem 16.
Letω∈ℝ,a¯=a1,a2,⋯,an,b¯=b1,b2,⋯,bn,c¯=c1,c2,⋯,cn,ai,bi,ci∈ℝ;i=1,⋯,n such that ai,bi,ci>0. Also α,γ,δ,μ,ν,λ,ρ,θ>0,β>1 and k∈0,1∪ℕ with k+ρ<δ+ν+α with s≥0. Let g:u,v⟶ℝ be a differentiable and strictly increasing function, and let ϕ,ψ,ζ1,ζ2 be m−power integrable and positive functions on u,v such that the ratio ϕξ/ψξ is bounded above by ζ2 and bounded below by ζ1∀ξ∈u,v. Let m,n>1 such that 1/m+1/n=1; then
(26)Ωϕξ;s1/mΩψξ;s1/n≤Ωζ21/mnϕ1/mψ1/nξ;s1/mΩζ1−1/mnϕ1/mψ1/nξ;s1/n.Proof.
According to the statement of the theorem, we have(27)0<ζ1t≤ϕtψt≤ζ2t,t∈u,v.
By considering the lower bound, the above inequality produces(28)ψt≤1ζ11/mϕ1/mtψ1/nt.
By multiplying both sides of the above inequality withgξ−gtβ−1Mωgξ−gtα;sg′t and integrating on u,ξ, we get
(29)Ωψξ;s1/n≤Ωζ1−1/mnϕ1/mψ1/nξ;s1/n.
Also, by considering the upper bound of inequality (27), the following inequality holds:
(30)ϕt≤ζ21/nϕ1/mtψ1/nt.
Multiplying both sides of the above inequality withgξ−gtβ−1Mωgξ−gtα;sg′t and integrating on u,ξ, we get the following inequality:
(31)Ωϕξ;s1/m≤Ωζ21/mnϕ1/mψ1/nξ;s1/m.
The product of (29) and (31) results in inequality (26).Corollary 17.
Under the assumptions of Theorem16 along with the condition that ζ1ξ=k1andζ2ξ=k2with0<k1<k2∀ξ∈u,v, (26) takes the following form:
(32)Ωϕξ;s1/mΩψξ;s1/n≤k2k11/mnΩϕ1/mψ1/nξ;s.Corollary 18.
Under the assumptions of above theorem and substitutingai=l,s=0 and ρ>0 in (26), we get the following inequality:
(33)QΩϕξ1/mQΩψξ1/n≤QΩζ21/mnϕ1/mψ1/nξ1/mQΩζ1−1/mnϕ1/mψ1/nξ1/n.Remark 19.
(i)
By settingn=1, b1=λ+lk, a1=θ−λ, c1=λ, ρ=ν=0, and δ>0 in (32), the inequality given in [8], Theorem 5, is deduced(ii)
By settingg, the identity function in (32) the inequality [7], Theorem 4, is deduced(34)Iϕξ;s1/mIψξ;s1/n≤k2k11/mnIϕ1/mψ1/nξ;s(iii)
Under the assumptions of the Corollary21 along with the condition that ζ1ξ=k1andζ2ξ=k2with0<k1<k2∀ξ∈u,v and setting the function g to be an identity function, the inequality given in [7], Corollary 1, is deduced(iv)
By settingn=1, b1=λ+lk, a1=θ−λ, c1=λ, ρ=ν=0, and δ>0 in (34), the Minkowski-type inequality (6) is deduced
In the proof of our next result, we will use Young’s inequality forx,y≥0 with m,n>1 satisfying m−1+n−1=1:
(35)xy≤m−1xm+n−1yn.
Also, the following inequality will be required:(36)x+ym≤2m−1xm+ym;x,y≥0andm>1.Theorem 20.
Under the assumptions of Theorem16 the following inequality holds:
(37)Ωϕψξ;s≤m−12m−1Ωζ2ζ2+1mϕm+ψmξ;s+n−12n−1Ω1ζ1+1nϕn+ψnξ;s.Proof.
Taking the left side of inequality (27), we obtain the following form:
(38)ψnt≤ζ1+1−nϕt+ψtn.
Multiplying both sides of inequality bygξ−gtβ−1Mωgξ−gtα;sg′t and integrating over u,ξ, the above inequality gives
(39)n−1Ωψnξ;s≤n−1Ωζ1+1−nϕ+ψnξ;s.
Also, by considering right side of inequality (27), we have the following inequality:
(40)ϕmt≤ζ2ζ2+1mψt+ϕtm.
Multiplying both sides of the above inequality bygξ−gtβ−1Mωgξ−gtα;sg′t, integrating over u,ξ and multiplying resulting inequality by m−1, we get
(41)m−1Ωϕmξ;s≤m−1Ωζ2ζ2+1mψ+ϕmξ;s.
By Young’s inequality, we have(42)ϕtψt≤m−1ϕmt+n−1ψnt.
Multiplying both sides of the above inequality bygξ−gtβ−1Mωgξ−gtα;sg′t and integrating over u,ξ, the above inequality takes the form
(43)Ωϕψξ;s≤m−1Ωϕmξ;s+n−1Ωψnξ;s.
Applying (43) to the sum of (39) and (41), we get the following inequality:
(44)Ωϕψξ;s≤m−1Ωζ2ζ2+1mϕ+ψmξ;s+n−1Ω1ζ1+1nϕ+ψnξ;s.
Inequality (37) follows by using (36) in (44).Corollary 21.
Under the assumptions of Theorem20 together with the condition that ζ1ξ=k1andζ2ξ=k2with0<k1<k2∀ξ∈u,v, (37) becomes
(45)Ωϕψξ;s≤m−12m−1k2k2+1mΩϕm+ψmξ;s+n−12n−11k1+1nΩϕn+ψnξ;s.Corollary 22.
Under the assumptions of the above theorem and settingai=l,s=0 and ρ>0 in (37), the following inequality holds true:
(46)QΩϕψξ≤m−12m−1QΩζ2ζ2+1mϕm+ψmξ+n−12n−1QΩ1ζ1+1nϕn+ψnξ.Remark 23.
(i)
By settingn=1, b1=λ+lk, a1=θ−λ, c1=λ, ρ=ν=0, and δ>0 in (45), the inequality given in [8], Theorem 6, is deduced(ii)
By settingg the identity function, (45) is deduced to the following inequality given in [7]:(47)Iϕψξ;s≤m−12m−1k2k2+1mIϕm+ψmξ;s+n−12n−11k1+1nIϕn+ψnξ;s(iii)
Under the assumptions of Corollary25 along with the condition that ζ1ξ=k1andζ2ξ=k2with0<k1<k2∀ξ∈u,v and setting the function g the identity function, the inequality given in [7], Corollary 2, is deduced(iv)
By settingn=1, b1=λ+lk, a1=θ−λ, c1=λ, ρ=ν=0, and δ>0 in (47), the Minkowski-type inequality given in [9], Theorem 3.2, is deducedTheorem 24.
Suppose the assumptions of Theorem16 hold; then for m≥1, the following inequalities hold:
(48)Ω1ζ2ϕψξ;s≤Ω1ζ1+1ζ2+1ϕ+ψ2ξ;s≤Ω1ζ1ϕψξ;s.Proof.
Considering right side of inequality (27), we get the following inequalities:
(49)ϕt+ψt≤ζ2t+1ψt,(50)ζ2−1tζ2t+1ϕt≤ϕt+ψt.
Also, from the left side of inequality (27), we have the following inequalities:
(51)ϕt+ψt≥ζ1t+1ψt,(52)ζ1−1tζ1t+1ϕt≥ϕt+ψt.
Combining the inequalities (49) and (51), the following inequality holds
(53)ζ1t+1ψt≤ϕt+ψt≤ζ2t+1ψt.
By the combining the inequalities (50) and (52), we get
(54)ζ2−1tζ2t+1ϕt≤ϕt+ψt≤ζ1−1tζ1t+1ϕt.
The product of the above two inequalities yields(55)ζ2−1tϕtψt≤1ζ1t+1ζ2t+1ϕt+ψt2≤ζ1−1tϕtψt.
Now, multiplyinggξ−gtβ−1Mωgξ−gtα;sg′t with the above inequality and integrating over u,ξ, we get the required inequality (48).Corollary 25.
Under the assumptions of Theorem24 and taking ζ1ξ=k1andζ2ξ=k2with0<k1<k2∀ξ∈u,v, (48) takes the following form:
(56)1k2Ωϕψξ;s≤1k1+1k2+1Ωϕ+ψ2ξ;s≤1k1Ωϕψξ;s.Corollary 26.
Under the assumptions of the above theorem and consideringai=l,s=0 and ρ>0 in (48), the following inequality holds:
(57)QΩ1ζ2ϕψξ≤QΩ1ζ1+1ζ2+1ϕ+ψ2ξ≤QΩ1ζ1ϕψξ.Remark 27.
(i)
By settingn=1, b1=λ+lk, a1=θ−λ, c1=λ, ρ=ν=0, and δ>0 in (56), the inequality given in [8], Theorem 16 is deduced(ii)
By settingg the identity function, (56) gives the following inequality [7]:(58)1k2Iϕψξ;s≤1k1+1k2+1Iϕ+ψ2ξ;s≤1k1Iϕψξ;s(iii)
Under the assumptions of Corollary29 along with the condition that ζ1ξ=k1 and ζ2ξ=k2with0<k1<k2∀ξ∈u,v and setting the function g to be the identity function, the inequality given in [7], Corollary 3, is deduced(iv)
By settingn=1, b1=λ+lk, a1=θ−λ, c1=λ, ρ=ν=0, and δ>0 in (58), the Minkowski-type inequality given in [9], Theorem 3.3, is deducedTheorem 28.
Let the assumptions of Theorem16 hold true. Also, let ϕ,ψ,ζ1,ζ2,f be m−power integrable and positive functions on u,v such that 0<ft<ζ1t≤ϕt/ψt≤ζ2t∀t∈u,v; then the following inequalities hold for m≥1:
(59)Ωϕ−fψζ2−fmξ;s1/m+Ωζ2ϕ−fψζ2−fmξ;s1/m≤Ωϕmξ;s1/m+Ωψmξ;s1/m≤Ωϕ−fψζ1−fmξ;s1/m+Ωζ1ϕ−fψζ1−fmξ;s1/m.Proof.
By the assumption of the theorem, we have(60)0<ft<ζ1t≤ϕtψt≤ζ2t,t∈u,v.
The above inequality can be arranged as follows:(61)ζ1t−ft≤ϕt−ftψtψt≤ζ2t−ft.
From which we can write(62)ϕt−ftψtmζ2t−ftm≤ψmt≤ϕt−ftψtmζ1t−ftm.
The following inequality follows by multiplyinggξ−gtβ−1Mωgξ−gtα;sg′t throughout the above inequality and integrating over u,ξ:
(63)Ωϕ−fψζ2−fmξ;s1/m≤Ωψmξ;s1/m≤Ωϕ−fψζ1−fmξ;s1/m.
Also, from (60), one can have
(64)ζ1t−ftζ1t≤ϕt−ftψtϕt≤ζ2t−ftζ2t,which can also be written as
(65)ζ2tϕt−ftψtζ2t−ft≤ϕt≤ζ1tϕt−ftψtζ1t−ft.
Taking the powerm, after multiplying by gξ−gtβ−1Mωgξ−gtα;sg′t throughout the above inequality and integrating over u,ξ, one can get the following inequality:
(66)Ωζ2ϕ−fψζ2−fmξ;s1/m≤Ωϕmξ;s1/m≤Ωζ1ϕ−fψζ1−fmξ;s1/m.
The sum of (63) and (66) produces the required inequality (59).Corollary 29.
Under the assumptions of Theorem28 along with the condition that fξ=m,ζ1ξ=k1andζ2ξ=k2with0<k1<k2∀ξ∈u,v in (59), the following inequality holds:
(67)k2+1k2−mΩϕ−mψmξ;s1/m≤Ωϕmξ;s1/m+Ωψmξ;s1/m≤k1+1k1−mΩϕ−mψmξ;s1/m.Corollary 30.
Under the assumptions of the above theorem with the condition thatai=l,s=0 and ρ>0 in (59), the following inequality holds:
(68)QΩϕ−fψζ2−fmξ1/m+QΩζ2ϕ−fψζ2−fmξ1/m≤QΩϕmξ;s1/m+QΩψmξ1/m≤Ωϕ−fψζ1−fmξ1/m+Ωζ1ϕ−fψζ1−fmξ1/m.Remark 31.
(i)
By settingn=1, b1=λ+lk, a1=θ−λ, c1=λ, ρ=ν=0, and δ>0 in (67), the inequality given in [8], Theorem 8, is deduced(ii)
By settingg the identity function, (67) is deduced to the following inequality [7]:(69)k2+1k2−mIϕ−mψmξ;s1/m≤Iϕmξ;s1/m+Iψmξ;s1/m≤k1+1k1−mIϕ−mψmξ;s1/m(iii)
Under the assumptions of Corollary33 along with the condition that ζ1ξ=k1 and ζ2ξ=k2with0<k1<k2∀ξ∈u,v and setting the function g to be the identity function, the inequality given in [7], Corollary 4, is deduced(iv)
By settingn=1, b1=λ+lk, a1=θ−λ, c1=λ, ρ=ν=0, and δ>0 in (69), the Minkowski-type inequality given [9], Theorem 3.4, is deduced


## 4. Reverse Minkowski-Type Fractional Integral Inequalities

In this section, we state and prove some reverse versions of Minkowski-type inequalities that are the generalizations of (2), (3), and (5).Theorem 32.
Under the assumptions of Theorem16, the following inequality holds for m≥1:
(70)Ωϕmξ;s1/m+Ωψmξ;s1/m≤Ωζ21+ζ2mϕ+ψmξ;s1/m+Ω11+ζ1mϕ+ψmξ;s1/m.Proof.
From (27), one can obtain the following inequality:
(71)ψmt≤11+ζ1tmϕt+ψtm.
Multiplying both sides of inequality bygξ−gtβ−1Mωgξ−gtα;sg′t and integrating on u,ξ, the above inequality can take the form as follows:
(72)Ωψmξ;s1/m≤Ω11+ζ1tmψ+ϕmξ;s1/m.
Also, by considering inequality (27), one can have the following inequality:
(73)ϕmt≤ζ21+ζ2tmψt+ϕtm.
Multiplying both sides of the above inequality bygξ−gtβ−1Mωgξ−gtα;sg′t and integrating over u,ξ, we can get
(74)Ωϕmξ;s1/m≤Ωζ21+ζ2tmψ+ϕmξ;s1/m.
Adding (72) and (74), inequality (70) can be obtained.Corollary 33.
Under the assumptions of Theorem32 along with the condition that ζ1ξ=k1andζ2ξ=k2with0<k1<k2∀ξ∈u,v, (70) takes the following form:
(75)Ωϕmξ;s1/m+Ωψmξ;s1/m≤1+k2−k1k1+1k2+1Ωϕ+ψmξ;s1/m.Corollary 34.
Under the assumptions of the above theorem and takingai=l,s=0, and ρ>0 in (70), the following inequality is obtained:
(76)QΩϕmξ1/m+QΩψmξ1/m≤QΩζ21+ζ2mϕ+ψmξ1/m+QΩ11+ζ1mϕ+ψmξ1/m.Remark 35.
(i)
By settingn=1, b1=λ+lk, a1=θ−λ, c1=λ, ρ=ν=0, and δ>0 in (70), the inequality introduced by Andric et. al [8] (Theorem 3) is generated(ii)
Takingg:u,v⟶ℝ to be an identity function, (75) gives the following inequality [7]:(77)Iϕmξ;s1/m+Iψmξ;s1/m≤1+k2−k1k1+1k2+1Iϕ+ψmξ;s1/m(iii)
Under the assumptions of Corollary37 along with the condition that ζ1ξ=k1 and ζ2ξ=k2 with 0<k1<k2∀ξ∈u,v and setting the function g to be an identity function, we obtain the inequality as in [7], Corollary 5(iv)
By settingn=1, b1=λ+lk, a1=θ−λ, c1=λ, ρ=ν=0, and δ>0 in (77), Minkowski-type inequality (5) is deducedTheorem 36.
Under the assumptions of Theorem16, for m≥1, we have
(78)Ωϕmξ;s2/m+Ωψmξ;s2/m≥Ω1+ζ2ζ2mϕmξ;s1/mΩ1+ζ1mψmξ;s1/m−2Ωϕmξ;s1/mΩψmξ;s1/m.Proof.
Inequalities (71) and (73) from the previous theorem can be arranged in the following forms:
(79)1+ζ1tmψmt≤ϕt+ψtm,1+ζ2tζ2tmϕmt≤ψt+ϕtm.
By multiplying withgξ−gtβ−1Mωgξ−gtα;sg′t and integrating over u,ξ and taking the power 1/m of the resulting inequalities, the above inequalities further take the following forms:
(80)Ω1+ζ1mψmξ;s1/m≤Ωϕ+ψmξ;s1/m,(81)Ω1+ζ2ζ2mϕmξ;s1/m≤Ωψ+ϕmξ;s1/m.
By multiplying (80) and (81), we get the following inequality:
(82)Ω1+ζ1mψmξ;s1/mΩ1+ζ2ζ2mϕmξ;s1/m≤Ωψ+ϕmξ;s1/m2.
Applying Minkowski’s inequality on the term within the square brackets at the right side of the above inequality and then usinga+b2=a2+2ab+b2, the above inequality gives the required inequality (78).Corollary 37.
Under the assumptions of Theorem36 along with the condition that ζ1ξ=k1andζ2ξ=k2with0<k1<k2∀ξ∈u,v, (78) takes the following form:
(83)Ωϕmξ;s2/m+Ωψmξ;s2/m≥2+k1−1k2+1k2Ωϕmξ;s1/mΩψmξ;s1/m.Corollary 38.
Under the assumptions of above theorem together with the conditionai=l,s=0, and ρ>0, (78) results in the following inequality:
(84)QΩϕmξ2/m+QΩψmξ2/m≥QΩ1+ζss2ζ2mϕmξ1/mQΩ1+ζ1mψmξ1/m−2QΩϕmξ1/mQΩψmξ1/m.Remark 39.
(i)
By settingn=1, b1=λ+lk, a1=θ−λ, c1=λ, ρ=ν=0, and δ>0 in (78), the inequality given in [8], Theorem 4, is deduced(ii)
By settingg the identity function in (83), the following inequality is deduced [7]:(85)Iϕmξ;s2/m+Iψmξ;s2/m≥2+k1−1k2+1k2Iϕmξ;s1/mIψmξ;s1/m(iii)
Under the assumptions of Corollary 13 along with the condition thatζ1ξ=k1andζ2ξ=k2with0<k1<k2∀ξ∈u,v and setting g the identity function, the inequality given in [7], Corollary 6, is deduced(iv)
By settingn=1, b1=λ+lk, a1=θ−λ, c1=λ, ρ=ν=0, and δ>0 in (85), the Minkowski-type inequality given in [9], Theorem 2.2, is deducedRemark 40.
All results of this paper hold for the right-sided integral operator:(86)gΩv−,α,β,γ,δ,μ,νω,λ,ρ,θ,k,nϕξ;s=∫ξvgt−gξβ−1Mα,β,γ,δ,u,νλ,ρ,θ,k,nωgt−gξα;sg′tϕtdt.


## 5. Conclusion

A generalized integral operator with the help of a unified Mittag-Leffler function is defined, and its boundedness is proved. By giving specific values to parameters and considering suitable functions involved in the kernel of this operator, various kinds of well-known integral and fractional integral operators can be reproduced. For a fractional integral operator, we have constructed several Minkowski- and reverse Minkowski-type inequalities. The particular cases of the results of this paper are connected with many already published results.



---

*Source: 2890981-2022-02-27.xml*