# The Nonlocal Fractal Integral Reverse Minkowski’s and Other Related Inequalities on Fractal Sets

Authors: Gauhar Rahman; Kottakkaran Sooppy Nisar; Alireza Khalili Golamankaneh
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

In this paper, we study the generalized Riemann–Liouville fractional integral for the functions with fractal support. The aim of this article is to investigate reverse Minkowski’s inequalities and certain other related inequalities by employing the generalized Riemann–Liouville fractional integral for the functions with fractal support.


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## Body

## 1. Introduction

Fractional calculus involves integrals and derivatives of arbitrary order. The applications of fractional calculus have been found in the field of several sciences and engineering [1–8]. In [1, 2], the nonlocal fractional integrals and derivatives are utilized to model the processes with memory effect. In [9–11], the researchers used the nonlocal derivatives to model more appropriately the dynamics of the nonconservative systems in formulation of Hamilton and Lagrange.Fractal analysis has been studied by many researchers by using measure theory, harmonic analysis, stochastic process, fractional spaces, and other techniques [12–35].Recently, Parvate and Gangal [36–41] proposed the ℱα-calculus on the fractal subset of real line and fractal curves. The researchers have applied transport materials on disordered systems such as fractal sets and fractal curves [42–44]. In [45], the researchers established Schrödinger’s equation on a fractal curve.Such new developments in fractional calculus encourage future research to investigate new innovative ideas to unify the fractional operators and establish inequalities involving new fractional operators. The fractional integral inequalities (FII in short) and their applications play an important role in the field of applied mathematics. A wide number of integral inequalities and their extensions were built in the sense of classical fractional integral and fractional derivative operators (see, e.g., [46–50]).The inequalities, especially the Hölder, the reverse Minkowski, the arithmetic, and geometric inequalities, have played a key role in the field of both pure as well as applied mathematics. These and several other essential inequalities are now in common use and, therefore, it is not surprising that several studies associated to these areas have been made in order to accomplish a diversity of desired goals. In the past few decades, the theory of inequalities has established rapidly and unexpected results were investigated, along with simpler new proofs for existing results, and, accordingly, new direction for research has been opened up. Recently, the theory of inequalities has gained more considerable interest from many mathematicians, and a large number of new inequalities have been estimated in the literature. It is recognized that, in general, some specific inequalities provide a useful and important device in the development of different branches of mathematics. In [51], Dahmani has investigated the reverse Minkowski fractional integral inequalities. Sousa and Capelas de Oliveira [52] have investigated the reverse Minkowski inequalities and certain other related inequalities for Katugampola fractional Integral operators. In [53, 54], the authors have studied the reverse Minkowski inequalities by considering Hadamard fractional integral operators. In this present article, we study the said inequalities by considering the generalized Riemann–Liouville fractional integral for the functions with fractal support.The structure of the paper is follows.In Section2, we have given some known results and basic definitions. In Section 3, the nonlocal reverse Minkowski inequalities are presented for nonlocal fractal integrals on fractal subset of real line. In Section 4, some other related inequalities for nonlocal fractal integrals on fractal subset of real line are presented.


## 2. Preliminaries

Some well-known basic definitions and results associated with classical fractional integrals and generalized fractional integrals are presented in this section. The reverse Minkowski’s integral inequalities can be found in the work of [27, 55]. The reverse Minkowski’s inequalities are the motivation of the work performed so far, involving the classical Riemann integrals which are presented by the following theorems.Theorem 1.
(see [55]). Let the two functions U1 and U2 be positive on 0,∞ and δ≥1. If 0<k≤U1ζ/h1ζ≤K, ζ∈x1,x2, then the following inequality holds:(1)∫x1x2U1δζdζ1/δ+∫x1x2U2δζdζ1/δ≤1+Kk+2k+1K+1∫x1x2U1+U2δζdζ1/δ.Theorem 2.
(see [55]). Let the two functions U1 and U2 be positive on 0,∞ and δ≥1. If 0<k≤U1ζ/h1ζ≤K, ζ∈x1,x2, then the following inequality holds:(2)∫x1x2U1δζdζ2/δ+∫x1x2U2δζdζ2/δ≥K+1k+1K−2∫x1x2U1δζdζ1/δ∫x1x2U2δζdζ1/δ.Definition 1.
(see [5, 6]). The well-known classical fractional integrals of order λ>0 are, respectively, defined by(3)Tλx1U1ζ=1Γλ∫x1ζζ−ρλ−1U1ρdρ,x1<ζ,and(4)Tx2λU1ζ=1Γλ∫ζx2ρ−ζλ−1gρdρ,ζ<x2,where λ∈ℂ with ℜλ>0.Dahmani [51] has investigated the following inequalities by using classical fractional integral.Theorem 3.
(see [51]). Let the two functions U1 and U2 be positive on 0,∞ such that, for all ζ>0, TτU1δζ<∞, TτU2δζ<∞. If 0<k≤U1ρ1/U2ρ1≤K, ρ1∈x1,ζ, then the following inequality holds:(5)TτU1δζ1/δ+TτU2δζ1/δ≤1+Kk+2k+1K+1TτU1+U2δζ1/δ,τ∈ℂ, ℜτ>0, δ≥1.Theorem 4.
(see [51]). Let the two functions U1, U2 be positive on 0,∞ such that, for all ζ>0, TτU1δζ<∞, TτU2δζ<∞. If 0<k≤U1ρ1/U2ρ1≤K, ρ1∈x1,ζ, then the following inequality holds:(6)TτU1δζ2/δ+TτU2δζ2/δ≥K+1k+1K−2TτU1δζ1/δTτU2δζ1/δ,τ∈ℂ, ℜτ>0, δ≥1.In [1], it is shown that the geometry of fractal is the geometry of real world. In [36, 37, 39], Parvate and Gangal proposed the calculus on fractals which is related to Riemann integrals.Definition 2.
(see [36, 37, 39]). For the thin Cantor set, the following integral staircase function is defined by(7)Sℱαρ=γαℱ,x1,x2,ifρ≥x0,−γαℱ,x1,x2,otherwise,where α is the γ-dimension of thin Cantor set.Definition 3.
(see [36, 37, 39]). The ℱα-derivative is defined by(8)DℱαUx=ℱ−limt⟶xUt−UxSℱαt−Sℱαx,ifx∈ℱ,0,otherwise,when the limit exists.Definition 4.
The Gamma function with the fractal support is defined by(9)Γℱατ=∫0∞e−SℱαxSℱαxSℱατ−1dℱαx,where e−Sℱαx=ℱ−limm⟶∞1−Sℱαt/mm.Here, we review the following nonlocal fractal integral operators for the functions with fractal support [56, 57].Definition 5.
IfU1x∈Cℱαx1,x2 (α-order differentiable function on x1,x2) and τ>0, then the left- and right-sided Riemann–Liouville fractal integral operators of order τ are, respectively, defined by [56, 57](10)Tτx1gζ=1Γℱατ∫x1ζSℱαζ−Sℱαρτ−αU1ρdℱαρ,Sℱαx1<Sℱαζand(11)Tx2τgζ=1Γℱατ∫ζx2Sℱαρ−Sℱαζτ−αU1ρdℱαρ,Sℱαζ<Sℱαx2,where Γℱατ is defined in (7), Sℱα is the staircase function, and ℱ is the fractal set with α-dimension (see, e.g., [56, 57]).Remark 1.
If we considerα=1 in (10) and (11), then we get (3) and (4), respectively.One can easily prove the following lemma [56, 57].Lemma 1.
(12)Tx1τSℱαζ−Sℱαx1λ=Γℱαλ+1Γℱαλ+τ+1Sℱαζ−Sℱαx1λ+τ,λ>−1.


## 3. The Nonlocal Reverse Minkowski Inequalities on Fractal Sets

In this section, we present the nonlocal fractal reverse Minkowski integral inequalities in the fractal support by using the generalized nonlocal fractal integral operator. The nonlocal reverse Minkowski fractal integral inequalities in fractal support are presented in the following theorems.Theorem 5.
Letτ∈ℂ, ℜτ>0, δ≥1, and U1 and U2∈Cℱαx1,x2, (α-order differentiable functions on x1,x2) be two positive functions on 0,∞ such that, for all theta>0, Tτx1g1δζ<∞ and Tτx2U2δζ<∞. If 0<k≤U1ρ1/h1ρ1≤K, ρ1∈x1,ζ, then the following inequality holds:(13)Tτx1U1δζ1/δ+Tτx1U2δζ1/δ≤1+Kk+2k+1K+1Tτx1U1+U2δζ1/δ.Proof.
Under the given hypothesis of Theorem5, U1ρ1/U2ρ1≤K, ρ1∈x1,ζ, ζ>0, we have(14)K+1δU1δρ1≤KδU1+U2δρ1.
Consider a function:(15)Λζ,ρ1=1ΓℱατSℱαζ−Sℱαρ1τ−α.
We conclude that the functionΛζ,ρ1 is positive for all ρ1∈x1,ζ, 0≤x1<ζ≤x2, as each term of Λζ,ρ1 defined in (15) is positive in view of hypothesis of Theorem 5.
Therefore, conducting product on both sides of (14) by Λζ,ρ1 and integrating the estimated inequality with respect to ρ1 from Sℱαa to Sℱαζ, we have(16)K+1δΓℱατ∫x1ζSℱαζ−Sℱαρ1τ−αU1δρ1dℱαρ1≤KδΓℱατ∫x1ζSℱαζ−Sℱαρ1τ−αU1+U2δρ1dℱαρ1,which can be written as(17)Tτx1U1δζ≤KδK+1δTτ,ηx1U1+U2δζ.
Hence, it follows that(18)Tτx1U1δζ1/δ≤KK+1Tτx1U1+U2δζ1/δ.
Now, utilizing the conditionkU1ρ1≤U2ρ1, we have(19)1+1kU2ρ1≤1kU1ρ1+U2ρ1.
It follows that(20)1+1kδU2δρ1≤1kδU1ρ1+U2ρ1δ.
Again, conducting product on both sides of (20) by Λζ,ρ1 and integrating the estimated inequality with respect to ρ1 from Sℱαx1 to Sℱαζ, we obtain(21)Tτx1U2δζ1/δ≤1k+1Tτx1U1+U2δζ1/δ.
Thus, by adding inequalities (18) and (21) yields the desired inequality.Theorem 6.
Letτ∈ℂ, ℜτ>0, δ≥1 and let U1 and U2∈Cℱαx1,x2 (α-order differentiable functions on x1,x2) be two positive functions on 0,∞ such that, for all ζ>0, ℑτx1g1δζ<∞ and ℑτx1hδζ<∞. If 0<k≤g1ρ1/hρ1≤K, ρ1∈x1,ζ, then the following inequality holds:(22)Tτx1U1δζ2/δ+Tτx1U2δζ2/δ≥K+1k+1K−2Tτx1U1δζ1/δTτx1U2δζ1/δ.Proof.
The multiplication of inequalities (18) and (21) yields(23)K+1k+1MTτx1U1δζ1/δTτ,ηx1U2δζ1/δ≤Tτx1U1ζ+U2ζδ1/δ2
By utilizing the Minkowski inequality to the right-hand side of [17], we have(24)Tτx1gζ+U2ζδ1/δ2≤Tτx1U1δζ1/δ+Tτx1U2δζ1/δ2≤Tτx1U1δζ2/δ+Tτx1U2δζ2/δ+2Tτx1U1δζ1/δTτx1U2δζ1/δ.
Thus, from inequalities (23) and (24), we get the desired inequality (22).


## 4. Certain Related Nonlocal Fractal Integral Inequalities on Fractal Sets

This section is devoted to deriving certain related nonlocal fractal integral inequalities on the fractal set.Theorem 7.
Letτ∈ℂ, ℜτ>0, r>1,1/δ+1/σ=1 and let U1 and U2∈Cℱαx1,x2 (α-order differentiable functions on x1,x2) be two positive functions on 0,∞ such that Tτx1U1ζ<∞ and Tτx1U2ζ<∞. If 0<k≤gρ1/hρ1≤K<∞, ρ1∈x1,ζ,ζ>x1, we have(25)Tτx1U1ζ1/δTτx1U2ζ1/σ≤Kk1/rsTτx1U1ϑ1/δU2ϑ1/σ.Proof.
SinceU1ρ1/U2ρ1≤K<∞, ρ1∈x1,ζ,ζ>a, therefore, we have(26)U2ρ11/σ≥K−1/σU1ρ11/σ.
It follows that(27)U1ρ11/δU2ρ11/σ≥K−1/σU1ρ11/δU1ρ11/σ≥K−1/σU1ρ11/δ+1/σ≥K−1/σU1ρ1.
Conducting multiplication on both sides of (27) by Λζ,ρ1 where Λζ,ρ1 is defined by (15) and integrating the estimated inequality with respect to ρ1 from Sℱαx1 to Sℱαζ, we have(28)1Γℱατ∫x1ζSℱαζ−Sℱαρ1τ−αU1ρ11/δU2ρ11/σdℱαρ1≥K−1/σΓℱατ∫x1ζSℱαζ−Sℱαρ1τ−αgρ1dℱαρ1.
It follows that(29)Tτx1U1ζ1/δhζ1/σ≥K−1/δTτx1U1ζ.
Consequently, we have(30)Tτx1U1ζ1/δhζ1/σ≥K−1/rsTτx1U1ζ1/δ.
On the contrary,kU1ρ1≤U2ρ1, ρ1∈x1,ζ,ζ>x1; therefore, we have(31)U1ρ11/δ≥k1/δU2ρ11/δ.
It follows that(32)U1ρ11/δhρ11/σ≥k1/δU1ρ11/δU2ρ11/σ≥k1/δU2ρ11/δ+1/σ≥k1/δU2ρ1.
Again, conducting multiplication on both sides of (32) by Λζ,ρ1where Λζ,ρ1 is defined by (15) and integrating the estimated inequality with respect to ρ1 from Sℱαx1 to Sℱαζ, we have(33)1Γℱατ∫x1ζSℱαζ−Sℱαρ1τ−αU1ρ11/δU2ρ11/σdℱαρ1≥k1/δΓℱατ∫x1ζSℱαζ−Sℱαρ1τ−αU2ρ1dℱαρ1.
Hence, we can write(34)Tτx1U1ζ1/δU2ϑ1/σ1/δ≥k1/rsTτx1U1ζ1/σ.
Multiplying (30) and (34), we get the desired inequality.Theorem 8.
Letτ∈ℂ, ℜτ>0, δ>1,1/δ+1/σ=1 and let U1 and U2∈Cℱαx1,x2 (α-order differentiable functions on x1,x2) be two positive functions on 0,∞ such that Tτx1U1δζ<∞ and Tτx1U2σζ<∞. If 0<k≤U1ρ1δ/U2ρ1σ≤K<∞, ρ1∈a,ϑ,ζ>x1, we have(35)Tτx1U1δζ1/δTτx1U2σζ1/σ≤Kk1/rsTτx1U1ζ1/δU2ζ1/σ.Proof.
ReplacingU1ζ and U2ζ by U1δζ and h1σζ, x1<ζ≤x2 in Theorem 7, and we get the desired inequality (35).Theorem 9.
Letτ∈ℂ, ℜτ>0, δ>1,1/δ+1/σ=1 and let U1 and U2∈Cℱαx1,x2 (α-order differentiable functions on x1,x2) be two positive functions on 0,∞ such that Tτx1U1δζ<∞ and Tτx1U2σζ<∞. If 0<k≤U1ρ1δ/U2ρ1σ≤K<∞ where k,K∈ℝ, ρ1∈x1,ζ,ζ>x1, then the following inequality for left nonlocal fractal integral with fractal support holds:(36)Tτx1U1ζU2ζ≤2δ−1KδδK+1δTτx1U1δ+hδζ+2σ−1σk+1σTτx1U1σ+U2σϑ.Proof.
By the given hypothesisU1ρ1/U2ρ1≤K, we have(37)K+1δU1δρ1≤KδU1+U2δρ1.
Conducting multiplication on both sides of inequality (37) by Λζ,ρ1 where Λζ,ρ1 id defined by (15) and integrating the estimated inequality with respect to ρ1 over Sℱαx1,Sℱαζ, we obtain(38)K+1δΓℱατ∫x1ζSℱαζ−Sℱαρ1τ−αU1δρ1dℱαρ1≤KδΓℱατ∫x1ζSℱαζ−Sℱαρ1τ−αU1+U2δρ1dℱαρ1.
It follows that(39)Tτx1U1δζ≤KδK+1δTτx1U1+U2δζ.
On the contrary, usingk≤U1ρ1/U2ρ1, a≤ρ1≤ζ, we have(40)k+1σU2σρ1≤U1+U2σρ1.
Again, conducting multiplication on both sides of inequality (40) by Λζ,ρ1 where Λζ,ρ1 is defined by (15) and integrating the estimated inequality with respect to ρ1 over Sℱαx1,Sℱαζ, we obtain(41)Tτx1U2σζ≤1k+1σTτx1U1+U2σζ.
Now, using Young’s inequality, we have(42)U1ρ1U2ρ1≤U1δρ1δ+U1σρ1σ.
Taking product on both sides of inequality (40) by Λζ,ρ1 where Λζ,ρ1 id defined by (15) and integrating the resultant identity with respect to ρ1 over Sℱαx1 to Sℱαζ, we obtain(43)Tτx1U1ζU2ζ≤1δTτx1U1δζ+1σTτx1U1σζ.
With the aid of (39) and (41), (43) can be written as(44)Tτx1U1ζU2ζ≤1δTτx1U1δζ+1σTτx1U1σζ≤KδδK+1δTτx1U1+U2δζ+1σk+1σTτx1U1+U2σζ.
Now, using the inequalityρ1+ωδ≤2σ−1ρ1δ+ωδ,δ>1,ρ1,ω>0, one can obtain(45)Tτx1U1+U2δζ≤Tτx1U1δ+U2δζand(46)Tτx1U1+U2σζ≤Tτx1U1σ+U2σζ.
Hence, the proof of (36) can be followed from (44)–(46).Theorem 10.
Letτ∈ℂ, ℜτ>0, δ≥1 and let U1 and U2∈Cℱαx1,x2 (α-order differentiable functions on x1,x2) be two positive functions on 0,∞ such that Tτx1U2δζ<∞, Tτx1U2σζ<∞. If 0<l<k≤U1ρ1/U2ρ1≤K<∞ where l,k,K∈ℝ, ζ>a, then we have(47)K+1K−kTτx1U1ζ−lU2ζ≤Tτx1U1δζ1/δ+Tτx1U2δζ1/δ≤k+1k−lTτx1U1ζ−lU2ζ1/δ.Proof.
Under the given hypothesis0<l<k≤U1ρ1/U2ρ1≤K<∞, we have(48)kl≤Kl⇒kl+m≤kl+K≤Kl+K⇒K+1k−l≤k+1k−l.
It can be written as(49)K+1K−l≤k+1k−l.
Also, we have(50)k−l≤U1ρ1−lU2ρ1U2ρ1≤K−l.
It follows that(51)U1ρ1−lU2ρ1δK−lδ≤U2δρ1≤U1ρ1−lU2ρ1δk−lδ.
Also, we have(52)1K≤U2ρ1U1ρ1≤1K⇒k−lkl≤U1ρ1−lU2ρ1lU1ρ1≤k−llK.
It follows that(53)KK−lδ≤U1ρ1−lU2ρ1δ≤U1δρ1≤kk−lδ≤U1ρ1−lU2ρ1δ.
Conducting product on both sides of inequality (51) by Λζ,ρ1 where Λζ,ρ1 is defined by (15) and integrating the resultant identity with respect to ρ1 over Sℱαx1,Sℱαζ, we obtain(54)1K−lδΓℱατ∫x1ζSℱαζ−Sℱαρ1τ−αU1ρ1−lU2ρ1δdℱαρ1≤1Γℱατ∫x1ζSℱαζ−Sℱαρ1τ−αU2δρ1dℱαρ1≤1k−lδΓℱατ∫x1ζSℱαζ−Sℱαρ1τ−αU1ρ1−lU2ρ1δdℱαρ1.
It follows that(55)1K−lTτx1U1ζlU2ζδ1/δ≤Tτx1U2δζ1/δ≤1k−lTτx1U1ζ−lU2ζδ1/δ.
Again, conducting product on both sides of inequality (53) by Λζ,ρ1 where Λζ,ρ1 is defined by (15) and integrating the resultant identity with respect to ρ1 over Sℱαx1,Sℱαζ, we obtain(56)KK−lTτx1U1ζ−lU2ζδ1/δ≤Tτx1U1δζ1/δ≤kk−lTτx1U1ζ−lU2ζδ1/δ.
Hence, by adding inequalities (55) and (56), we get the desired inequality (47).Theorem 11.
Letτ∈ℂ, ℜτ>0, δ≥1 and let U1 and U2∈Cℱαx1,x2 (α-order differentiable functions on x1,x2) be two positive functions on 0,∞ such that Tτx1U1δζ<∞, Tτx1U2δζ<∞. If 0≤α≤U1ρ1≤A and 0≤σ≤h1ρ1≤ℬ for all ρ1∈x1,ζ,ζ>x1, then we have(57)Tτx1U1δζ1/δ+Tτx1U2δζ1/δ≤Aα+ℬ+ℬσ+AA+σℬ+αTτx1U1+U2δζ1/δ.Proof.
Under the given hypothesis, we have(58)1ℬ≤1U2ρ1≤1σ.
The product of inequality (58) with 0≤α≤U1ρ1≤A gives(59)αℬ≤U1ρ1U2ρ1≤Aσ.
From (59), we obtain(60)U2δρ1≤ℬα+ℬδU1ρ1+U2ρ1δand(61)U1δρ1≤Aσ+AδU1ρ1+U2ρ1δ.
Now, conducting product on both sides of (60) and (61), respectively, by Λζ,ρ1 where Λζ,ρ1 id is defined by (15) and integrating the estimated identity with respect to ρ1 over Sℱαx1,Sℱαζ, we obtain(62)Tτx1U2δζ1/δ≤ℬα+ℬTτx1U1ζ+U2ζδ1/δand(63)Tτx1U1δζ1/δ≤Aσ+ATτx1U1ζ+U2ζδ1/δ.
Hence, by adding (62) and (63), we get the desired proof.Theorem 12.
Letτ∈ℂ, ℜτ>0, δ≥1 and let U1 and U2∈Cℱαx1,x2 (α-order differentiable functions on x1,x2) be two positive functions on 0,∞ such that TTx1U1ζ<∞, TTx1U2ζ<∞. If 0<k≤U1ρ1/h1ρ1≤K where k,K∈ℝ for all ρ1∈x1,ζ,ζ>x1, then we have(64)1KTτx1U1ζU2ζ≤1k+1K+1Tτx1U1ζ+U2ζ2≤1kTτx1U1ζU2ζ.Proof.
Under the given hypothesis,0<k≤U1ρ1/U2ρ1≤K, we have(65)U2ρ1k+1≤U2ρ1+U1ρ1≤U2ρ1K+1.
Also, we have1/K≤U2ρ1/U1ρ1≤1/K, which gives(66)U1ρ1K+1K≤U1ρ1+U2ρ1≤U1ρ1k+1k.
The multiplication of (65) and (66) yields(67)U1ρ1U2ρ1K≤U1ρ1+U2ρ12k+1K+1≤U1ρ1U2ρ1k.
Now, conducting multiplication on both sides of inequality (67) by Λζ,ρ1 where Λζ,ρ1 id is defined by (15) and integrating the resultant identity with respect to ρ1 over Sℱαx1,Sℱαζ, we have(68)1KΓℱατ∫x1ζSℱαζ−Sℱαρ1τ−αU1ρ1U2ρ1dℱαρ1≤1k+1K+1Γℱατ∫x1ζSℱαζ−Sℱαρ1τ−αU1ρ1+U2ρ12dℱαρ1≤1kΓℱατ∫x1ζSℱαζ−Sℱαρ1τ−αU1ρ1U2ρ1dℱαρ1.
It follows that(69)1KTτx1U1ζU2ζ≤1k+1K+1Tτx1U1ζ+U2ζ2≤1kTτx1U1ζU2ζ,which completes the desired proof.Theorem 13.
Letτ∈ℂ, ℜτ>0, δ≥1 and let U1 and U2∈Cℱαx1,x2 (α-order differentiable functions on x1,x2) be two positive functions on 0,∞ such that Tτx1U1ζ<∞ and Tτx1U2ζ<∞. If 0<k≤U1ρ1/h1ρ1≤K where k,K∈ℝ for all ρ1∈x1,ζ,ζ>x1, then the following inequality for the left nonlocal fractal integral on fractal set holds:(70)Tτx1U1δζ1/δ+Tτx1U2δζ1/δ≤2Tτx1hδU1ζ,U2ζ,where hU1ζ,U2ζ=maxKK/k+1U1ρ1−KU2ρ1,k+KU2ρ1−U1ρ1/k.Proof.
Under the given hypothesis0<k≤U1ρ1/U2ρ1≤K where ρ1∈x1,ζ,ζ>x1, we have(71)0<k≤K+k−U1ρ1U2ρ1and(72)K+k−U1ρ1U2ρ1≤K.
From (71) and (72), we have(73)U2ρ1<K+kU2ρ1−U1ρ1k≤hU1ρ1,U2ρ1,where hU1ρ1,U2ρ1=maxKK/k+1U1ρ1−KU2ρ1,k+KU2ρ1−U1ρ1/k. Also, from the given hypothesis 0<1/K≤U2ρ1/U1ρ1≤1/K, we have(74)1K≤1K+1k−U2ρ1U1ρ1and(75)1K+1k−U2ρ1U1ρ1≤1k.
From (74) and (75), we obtain(76)1K≤1/K+kU1ρ1−U2ρ1U1ρ1≤1k.
It follows that(77)U1ρ1=K1K+1kU1ρ1−KU2ρ1=KK+kgρ1−K2kU2ρ1kK=Kk+1U1ρ1−KU2ρ1=KKk+1U1ρ1−KU2ρ1≤hU1ρ1,U2ρ1.
From (73) and (77), we can write(78)U1δρ1≤hδgρ1,hρ1and(79)hδρ1≤hδgρ1,hρ1.
Now, conducting multiplication on both sides of (78) and (79), respectively, by Λζ,ρ1 where Λζ,ρ1 id is defined by (15) and integrating the resultant identity with respect to ρ1 over Sℱαx1,Sℱαζ, we obtain(80)1Γℱατ∫x1ζSℱαζ−Sℱαρ1τ−αU1δρ1dℱαρ1≤1Γℱατ∫x1ζSℱαζ−Sℱαρ1τ−αhU1ρ1,U2ρ1dℱαρ1.
It follows that(81)Tτx1U1δζ1/δ≤Tτx1hU1ζ,U2ζ1/δ.
Similarly, from (79), we obtain(82)Tτx1U2δζ1/δ≤Tτx1hδU1ζ,U2ζ1/δ.
Hence, by summing (81) and (82), we obtain the required proof.Remark 2.
We note that all results lead to standard fractional calculus by settingα=1, that is, Sℱαx=x.


## 5. Concluding Remarks

In this present investigation, we presented the nonlocal reverse Minkowski’s inequalities and some other inequalities for nonlocal fractal integral operator on fractal sets. The special cases of this work can be found in the work of [51, 58, 59].



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