# Hadamard and Fejér–Hadamard Inequalities for Further Generalized Fractional Integrals Involving Mittag-Leffler Functions

Authors: M. Yussouf; G. Farid; K. A. Khan; Chahn Yong Jung
Journal: Journal of Mathematics
 (2021)

---

## Abstract

In this paper, generalized versions of Hadamard and Fejér–Hadamard type fractional integral inequalities are obtained. By using generalized fractional integrals containing Mittag-Leffler functions, some well-known results for convex and harmonically convex functions are generalized. The results of this paper are connected with various published fractional integral inequalities.


---

## Body

## 1. Introduction

First we give definitions of fractional integral operators which are useful in establishing the results of this paper. In the following, we give fractional integral operators defined by Andrić et al. in [1] via an extended generalized Mittag-Leffler function in their kernels.Definition 1 (see [1]).
Letω,τ,δ,ρ,c∈ℂ, ℜτ,ℜδ>0, ℜc>ℜρ>0 with p¯≥0, σ,r>0 and 0<k≤r+σ. Let φ∈L1ε1,ε2 and x∈ε1,ε2. Then, the generalized fractional integral operators ϵσ,τ,δ,ω,ε1+ρ,r,k,cφ and ϵσ,τ,δ,ω,ε2−ρ,r,k,cφ are defined by(1)εσ,τ,δ,ω,ε1+ρ,r,k,cφx;p¯=∫ε1xx−tτ−1Eσ,τ,δρ,r,k,cωx−tσ;p¯φtdt,εσ,τ,δ,ω,ε2−ρ,r,k,cφx;p¯=∫xε2t−xτ−1Eσ,τ,δρ,r,k,cωt−xσ;p¯φtdt,where(2)Eσ,τ,δρ,r,k,ct;p¯=∑n=0∞βp¯ρ+nk,c−ρcnktnβρ,c−ρΓσn+τδnr,is the extended generalized Mittag-Leffler function and βp¯ is the extension of beta function which is defined as follows:(3)βp¯x,y=∫01tx−11−ty−1e−p¯/t1−tdt,where x,y,p¯ are positive real numbers.
Recently, Farid defined elegantly a unified integral operator in [2] (see, also [3]) as follows.Definition 2.
Letφ,θ:ε1,ε2⟶ℝ, 0<ε1<ε2 be the functions such that φ be positive and φ∈L1ε1,ε2 and θ be a differentiable and strictly increasing function. Also, let χ/x be an increasing function on ε1,∞ and ω,τ,δ,ρ,c∈ℂ, ℜτ,ℜδ>0, ℜc>ℜρ>0 with p¯≥0, σ,r>0, and 0<k≤r+σ.
Then, forx∈ε1,ε2, the integral operators θΥσ,τ,δ,ε1+χ,ρ,r,k,cφ and θΥσ,τ,δ,ε1−χ,ρ,r,k,cφ are defined by(4)θΥσ,τ,δ,ε1+χ,ρ,r,k,cφx;p¯=∫ε1xχθx−θtθx−θtEσ,τ,δρ,r,k,cωθx−θtσ;p¯φtdθt,θΥσ,τ,δ,ε2−χ,ρ,r,k,cφx;p¯=∫xε2χθt−θxθt−θxEσ,τ,δρ,r,k,cωθt−θxσ;p¯φtdθt.
The following definition of generalized fractional integral operators containing extended Mittag-Leffler function in the kernel can be extracted from Definition2. It is generalization of Definition 1 by a monotonically increasing function.Definition 3.
Letφ,θ:ε1,ε2⟶ℝ, 0<ε1<ε2 be the functions such that φ be positive and φ∈L1ε1,ε2 and θ be a differentiable and strictly increasing function. Also, let ω,τ,δ,ρ,c∈ℂ, ℜτ,ℜδ>0, ℜc>ℜρ>0 with p¯≥0, σ,r>0 and 0<k≤r+σ. Then, for x∈ε1,ε2, fractional integral operators are defined by(5)θΥσ,τ,δ,ωε1+ρ,r,k,cφx;p¯=∫ε1xθx−θtτ−1Eσ,τ,δρ,r,k,cωθx−θtσ;p¯φtdθt,(6)θΥσ,τ,δ,ωε2−ρ,r,k,cφx;p¯=∫xε2θt−θxτ−1Eσ,τ,δρ,r,k,cωθt−θxσ;p¯φtdθt.
The following remark provides connection of Definition3 with existing fractional integral operators.Remark 1.
(i)
If we setp¯=0 and θx=x in equations (5) and (6), then these reduce to fractional integral operators defined by Salim and Faraj in [4].(ii)
If we setδ=r=1 and θx=x in equations (5) and (6), then these reduce to the fractional integral operators Υσ,τ,1,ω,ε1+ρ,1,k,cθ and Υσ,τ,1,ω,ε1−ρ,1,k,cθ containing generalized Mittag-Leffler function Eσ,τ,1ρ,1,k,ct;p¯ defined by Rahman et al. in [5].(iii)
If we takep¯=0,δ=r=1 and θx=x in equations (5) and (6), then these reduce to fractional integral operators containing extended generalized Mittag-Leffler function introduced by Srivastava and Tomovski in [6].(iv)
If we setp¯=0,δ=r=k=1 and θx=x in equations (5) and (6), then these reduce to fractional integral operators defined by Prabhaker in [7].(v)
Forp¯=ω=0 and θx=x in equations (5) and (6), these reduce to renowned Riemann–Liouville fractional integral operators [8].
The Riemann–Liouville fractional integrals for a functionφ∈L1ε1,ε2 of order τ∈ℝτ>0 are defined by(7)Iε1+τφx=1Γτ∫ε1xx−tτ−1φtdt,x>ε1,Iε2−τφx=1Γτ∫xε2t−xτ−1φtdt,x<ε2.
After introducing generalized fractional integral operators, now we define notions of functions for which generalized fractional integral operators are utilized to get main results of this paper.Definition 4 (see [9]).
A functionφ:ε1,ε2⟶ℝ is said to be convex if(8)φtx1+1−tx2≤tφx1+1−tφx2,holds for all x1,x2∈ε1,ε2 and t∈0,1.Definition 5 (see [10]).
LetI be an interval such that I⊆ℝ+. Then, a function φ:I⟶ℝ is said to be harmonically convex, if(9)φabta+1−tb≤tφb+1−tφa,holds for all a,b∈I and t∈0,1.Definition 6 (see [11]).
LetJ⊂0,∞ be a real interval and p∈ℝ\0. Then, a function φ:J⟶ℝ is said to be p-convex, if(10)φtε1p+1−tε2p1/p≤tφε1+1−tφε2,holds for ε1,ε2∈J and t∈0,1.
It is easy to see that forp=1 and p=−1, the p-convexity reduces to convexity and harmonical convexity, respectively.Definition 7 (see [11]).
Letp∈ℝ\0. Then, a function φ:ε1,ε2⊂0,∞⟶ℝ is said to be p-symmetric with respect to ε1p+ε2p/21/p if(11)φt1/p=φε1p+ε2p−t1/p,holds, for t∈ε1,ε2.
Convex functions are equivalently studied by the Hadamard inequality.Theorem 1.
Letφ:ε1,ε2⟶ℝ be a convex function such that ε1<ε2. Then, the following inequality holds:(12)φε1+ε22≤1ε2−ε1∫ε1ε2φxdx≤φε1+φε22.The Fejér–Hadamard inequality is a weighted version of the Hadamard inequality given by Fejér in [12].Theorem 2.
Letφ:ε1,ε2⟶ℝ be a convex function and g:ε1,ε2⟶ℝ be non-negative, integrable, and symmetric about ε1+ε2/2. Then, the following inequality holds:(13)φε1+ε22∫ε1ε2gxdx≤∫ε1ε2φxgxdx≤φε1+φε22∫ε1ε2gxdx.In recent decades, the Hadamard and the Fejér–Hadamard fractional integral inequalities have been studied extensively for different kinds of convex functions (see [1, 3, 13–21]). In this paper, we find Hadamard and Fejér–Hadamard inequalities for a generalized fractional integral operator involving an extended generalized Mittag-Leffler function.In the upcoming section, we give two versions of the Hadamard inequality as well as two versions of the Fejér–Hadamard inequality. Their special cases are also discussed along with noticing connections with published results.


## 2. Main Results

First we give the following version of the Hadamard inequality.Theorem 3.
Letφ,θ:ε1,ε2⊂0,∞⟶ℝ, Range θ⊂ε1,ε2 be the functions such that φ be positive and φ∈L1ε1,ε2, and θ be a differentiable and strictly increasing function. If φ is p-convex, p∈ℝ\0, then the following inequalities for fractional integral operators (5) and (6) hold:(i)
Ifp>0, then(14)φθpε1+θpε221/pΥσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,cθ1θ−1θpε2;p¯≤12Υσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,cθφ∘ψθ−1θpε2;p¯+Υσ,τ,δ,ω′,θ−1θpε2−ρ,r,k,cθφ∘ψθ−1θpε1;p¯≤φθε1+φθε22Υσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,cθ1θ−1θpε2;p¯,whereω′=ω/θpε2−θpε1σ and ψt=θ1/pt for all t∈ε1p,ε2p.(ii)
Ifp<0, then(15)φθpε1+θpε221/pεσ,τ,δ,ω′,θ−1θpε1−ρ,r,k,c1θ−1θpε2;p¯≤12Υσ,τ,δ,ω′,θ−1θpε2+ρ,r,k,cθφ∘ψθ−1θpε1;p¯+Υσ,τ,δ,ω′,θ−1θpε1−ρ,r,k,cθφ∘ψθ−1θpε2;p¯≤φθε1+φθε22Υσ,τ,δ,ω′,θ−1θpε1−ρ,r,k,cθ1θ−1θpε2;p¯,whereω′=ω/θpε1−θpε2σ and ψt=θ1/pt for all t∈ε2p,ε1p.Proof.
(i) Sinceφ is p-convex over ε1,ε2, for all x,y∈I, we have(16)φθpx+θpy21/p≤φθx+φθy2.
Settingθx=tθpε1+1−tθpε21/p and θy=tθpε2+1−tθpε11/p in above inequality, we have(17)φθpε1+θpε221/p≤φtθpε1+1−tθpε21/p+φtθpε2+1−tθpε11/p2.
Multiplying both sides of (17) by 2tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯ and then integrating over 0,1, we have(18)2φθpε1+θpε221/p∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯dt≤∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯φtθpε1+1−tθpε21/pdt+∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯φtθpε2+1−tθpε11/pdt.
By choosingθx=tθpε1+1−tθpε2 and θy=tθpε2+1−tθpε1 in (18), we have(19)2φθpε1+θpε221/p∫θ−1θpε1θ−1θpε2θpε2−θxτ−1Eσ,τ,δρ,r,k,cω′θpε2−θxσ;p¯1dθx≤∫θ−1θpε1θ−1θpε2θpε2−θxτ−1Eσ,τ,δρ,r,k,cω′θpε2−θxσ;p¯φθx1/pdθx+∫θ−1θpε1θ−1θpε2θy−θpε1τ−1Eσ,τ,δρ,r,k,cω′θy−θpε1σ;p¯φθy1/pdθy,where ω′=ω/θpε2−θpε1σ.
This implies(20)2φθpε1+θpε221/pεσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,c1θ−1θpε2;p¯≤Υσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,cθφ∘ψθ−1θpε2;p¯+Υσ,τ,δ,ω′,θ−1θpε2−ρ,r,k,cθφ∘ψθ−1θpε1;p¯.
To prove the second inequality of (14), again from p-convexity of φ over ε1,ε2 and for t∈0,1, we have(21)φtθpε1+1−tθpε21/p+φtθpε2+1−tθpε11/p≤φθε1+φθε2.
Multiplying both sides of (21) by tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯ and then integrating over 0,1, we have(22)∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯φtθpε1+1−tθpε21/pdt+∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯φtθpε2+1−tθpε11/pdt≤φθε1+φθε2∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯dt.
Settingθx=tθpε1+1−tθpε2 and θy=tθpε2+1−tθpε1 in (22), we have(23)θΥσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,cφ∘ψθ−1θpε2;p¯+Υσ,τ,δ,ω′,θ−1θpε2−ρ,r,k,cθφ∘ψθ−1θpε1;p¯≤φθε1+φθε2Υσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,cθ1θ−1θpε2;p¯.
By combining (20) and (23), we get (14).
(ii) Proof is similar to the proof of (i).Remark 2.
(i)
By settingp¯=ω=0 and θ=I, Theorem 9 of [11] is obtained.(ii)
By settingp¯=0, p=−1, and θ=I, Theorem 2.1 of [22] is obtained.(iii)
By settingθ=I and p=−1, Theorem 2.1 of [23] is obtained.(iv)
By settingω=p¯=0, p=−1, and θ=I, Theorem 4 of [18] is obtained.(v)
By settingp=−1, Theorem 2.1 of [24] is obtained.(vi)
By settingp=−1 and ψx=x, Corollary 2.3 of [24] is obtained.Corollary 1.
In (15), if we take ω=p¯=0, p=−1, and θ=I, then we get the following Hadamard inequality for the RL fractional integrals:(24)φ2ε1ε2ε1+ε2≤Γτ+12ε1ε2ε2−ε1τI1\ε1−τφ∘ψ1ε2+I1\ε1+τφ∘ψ1ε1≤φε1+φε22.Now we obtain Fejér–Hadamard type fractional integral inequalities forp-convex function via generalized fractional integral operators; for this, first we prove the following lemma.Lemma 1.
Letφ,θ:ε1,ε2⊂0,∞⟶ℝ, Range θ⊂ε1,ε2 be the functions such that φ be positive and φ∈L1ε1,ε2, ε1<ε2, and θ be a differentiable and strictly increasing function. If φ is p-convex, p∈ℝ\0, and φθ1/px=φθpε1+θpε2−θx1/p, then for generalized fractional integral operators (5) and (6), we have(i)
Ifp>0, then(25)θΥσ,τ,δ,ω,θ−1θpε1+ρ,r,k,cφ∘ψθ−1θpε2;p¯=Υσ,τ,δ,ω,θ−1θpε2−ρ,r,k,cθφ∘ψθ−1θpε1;p¯=12Υσ,τ,δ,ω,θ−1θpε1+ρ,r,k,cθφ∘ψθ−1θpε2;p¯+Υσ,τ,δ,ω,θ−1θpε2−ρ,r,k,cθφ∘ψθ−1θpε1;p¯,withψt=θ1/pt, t∈ε1p,ε2p.(ii)
Ifp<0, then(26)θΥσ,τ,δ,ω,θ−1θpε2+ρ,r,k,cφ∘ψθ−1θpε1;p¯=Υσ,τ,δ,ω,θ−1θpε1−ρ,r,k,cθφ∘ψθ−1θpε2;p¯=12Υσ,τ,δ,ω,θ−1θpε2+ρ,r,k,cθφ∘ψθ−1θpε1;p¯+Υσ,τ,δ,ω′,θ−1θpε1−ρ,r,k,cθφ∘ψθ−1θpε2;p¯,withψt=θ1/pt, t∈ε2p,ε1p.Proof.
(i) By definition of generalized fractional integral operators (5) and (6), we have(27)θΥσ,τ,δ,ω,θ−1θpε1+ρ,r,k,cφ∘ψθ−1θpε2;p¯=∫θ−1θpε1θ−1θpε2θpε2−θxτ−1Eσ,τ,δρ,r,k,cωθpε2−θxσ;p¯φ∘ψxdθx=∫θ−1θpε1θ−1θpε2θpε2−θxτ−1Eσ,τ,δρ,r,k,cωθpε2−θxσ;p¯φθx1/pdθx=∫θ−1θpε1θ−1θpε2θpε2−θxτ−1Eσ,τ,δρ,r,k,cωθpε2−θxσ;p¯φθpε1+θpε2−θx1/pdθx.
Settingθt=θpε1+θpε2−θx in the above equation and using φθ1/px=φθpε1+θpε2−θx1/p, we have(28)θΥσ,τ,δ,ω,θ−1θpε1+ρ,r,k,cφ∘ψθ−1θpε2;p¯=∫θ−1θpε1θ−1θpε2θt−θpε1τ−1Eσ,τ,δρ,r,k,cωθt−θpε1σ;p¯φθt1/pdθt=∫θ−1θpε1θ−1θpε2θt−θpε1τ−1Eσ,τ,δρ,r,k,cωθt−θpε1σ;p¯φ∘ψtdθt.
This implies(29)θΥσ,τ,δ,ω,θ−1θpε1+ρ,r,k,cφ∘ψθ−1θpε2;p¯=Υσ,τ,δ,ω,θ−1θpε2−ρ,r,k,cθφ∘ψθ−1θpε1;p¯.
By addingθΥσ,τ,δ,ω,θ−1θpε1+ρ,r,k,cφ∘ψθ−1θpε2;p¯ on both sides of (29), we have(30)2θΥσ,τ,δ,ω,θ−1θpε1+ρ,r,k,cφ∘ψθ−1θpε2;p¯=Υσ,τ,δ,ω,θ−1θpε1+ρ,r,k,cθφ∘ψθ−1θpε2;p¯+Υσ,τ,δ,ω,θ−1θpε2−ρ,r,k,cθφ∘ψθ−1θpε1;p¯.
From equations (29) and (30), the required result can be obtained.
(ii) Proof is on the same lines as the proof of (i).Theorem 4.
Letφ,θ,h:ε1,ε2⊂0,∞⟶ℝ, Range θ, Range h⊂ε1,ε2 be the functions such that φ be positive and φ∈L1ε1,ε2, ε1<ε2, and θ be a differentiable and strictly increasing function where h is a non-negative and integrable function. If φ is p-convex, p∈ℝ\0, and φθ1/px=φθpε1+θpε2−θx1/p, then the following inequalities for generalized fractional integral operators (5) and (6) hold:(i)
Ifp>0, then(31)φθpε1+θpε221/pΥσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,cθh∘ψθ−1θpε2;p¯+Υσ,τ,δ,ω′,θ−1θpε2−ρ,r,k,cθh∘ψθ−1θpε1;p¯≤Υσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,cθφh∘ψθ−1θpε2;p¯+Υσ,τ,δ,ω′,θ−1θpε2−ρ,r,k,cθφh∘ψθ−1θpε1;p¯≤φθε1+φθε22Υσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,cθh∘ψθ−1θpε2;p¯+Υσ,τ,δ,ω′,θ−1θpε2−ρ,r,k,cθh∘ψθ−1θpε1;p¯,whereω′=ω/θpε2−θpε1σ and ψt=θ1/pt for all t∈ε1p,ε2p.(ii)
Ifp<0, then(32)φθpε1+θpε221/pΥσ,τ,δ,ω′,θ−1θpε2+ρ,r,k,cθh∘ψθ−1θpε1;p¯+Υσ,τ,δ,ω′,θ−1θpε1−ρ,r,k,cθh∘ψθ−1θpε2;p¯≤Υσ,τ,δ,ω′,θ−1θpε2+ρ,r,k,cθφh∘ψθ−1θpε1;p¯+Υσ,τ,δ,ω′,θ−1θpε1−ρ,r,k,cθφh∘ψθ−1θpε2;p¯≤φθε1+φθε22Υσ,τ,δ,ω′,θ−1θpε2+ρ,r,k,cθh∘ψθ−1θpε1;p¯+Υσ,τ,δ,ω′,θ−1θpε1−ρ,r,k,cθh∘ψθ−1θpε2;p¯,whereω′=ω/θpε1−θpε2σ and ψt=θ1/pt for all t∈ε2p,ε1p.Proof.
(i) Multiplying both sides of (17) by 2tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯htθpε1+1−tθpε21/p and then integrating over 0,1, we have(33)2φθpε1+θpε221/p∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯htθpε1+1−tθpε21/pdt≤∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯φtθpε1+1−tθpε21/phtθpε1+1−tθpε21/pdt+∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯φtθpε2+1−tθpε11/phtθpε1+1−tθpε21/pdt.
By choosingθx=tθpε1+1−tθpε2, that is, θpε1+θpε2−θx=tθpε2+1−tθpε1, in (33) and using φθ1/px=φθpε1+θpε2−θx1/p, we have(34)2φθpε1+θpε221/p∫θ−1θpε1θ−1θpε2θpε2−θxτ−1Eσ,τ,δρ,r,k,cω′θpε2−θxσ;p¯h∘ψxdθx≤∫θ−1θpε1θ−1θpε2θpε2−θxτ−1Eσ,τ,δρ,r,k,cω′θpε2−θxσ;p¯φh∘ψxdθx+∫θ−1θpε1θ−1θpε2θx−θpε1τ−1Eσ,τ,δρ,r,k,cω′θx−θpε1σ;p¯φh∘ψxdθx.
This implies(35)2φθpε1+θpε221/pΥσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,cθh∘ψθ−1θpε2;p¯≤Υσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,cθφh∘ψθ−1θpε2;p¯+Υσ,τ,δ,ω′,θ−1θpε2−ρ,r,k,cθφh∘ψθ−1θpε1;p¯.
Using Lemma1 (i) in above inequality, we have(36)φθpε1+θpε221/pΥσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,cθh∘ψθ−1θpε2;p¯+Υσ,τ,δ,ω′,θ−1θpε2−ρ,r,k,cθh∘ψθ−1θpε1;p¯≤Υσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,cθφh∘ψθ−1θpε2;p¯+Υσ,τ,δ,ω′,θ−1θpε2−ρ,r,k,cθφh∘ψθ−1θpε1;p¯.
To prove the second inequality of (31), multiplying both sides of (21) by tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯htθpε1+1−tθpε21/p and then integrating over 0,1, we have(37)∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯φtθpε1+1−tθpε21/phtθpε1+1−tθpε21/pdt+∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯φtθpε2+1−tθpε11/phtθpε1+1−tθpε21/pdt≤φθε1+φθε2∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯htθpε1+1−tθpε21/pdt.
Settingθx=tθpε1+1−tθpε2 and using φθ1/px=φθpε1+θpε2−θx1/p in (37) and after simplification, we have(38)θΥσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,cφh∘ψθ−1θpε2;p¯+Υσ,τ,δ,ω′,θ−1θpε2−ρ,r,k,cθφh∘ψθ−1θpε1;p¯≤φθε1+φθε2Υσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,cθh∘ψθ−1θpε2;p¯.
Using Lemma1 (i), inequality (38) becomes(39)θΥσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,cφh∘ψθ−1θpε2;p¯+Υσ,τ,δ,ω′,θ−1θpε2−ρ,r,k,cθφh∘ψθ−1θpε1;p¯≤φθε1+φθε22Υσ,τ,δ,ω′,θ−1θpε1+ρ,r,k,cθh∘ψθ−1θpε2;p¯+Υσ,τ,δ,ω′,θ−1θpε2−ρ,r,k,cθh∘ψθ−1θpε1;p¯.
By combining (36) and (39), we get (31).
(ii) Proof is similar to the proof of (i) by using (ii) of Lemma1.Remark 3.
(i)
By settingp¯=0 and θ=I, Theorem 2.2 of [25] is obtained.(ii)
By settingp¯=ω=0 and θ=I, Theorem 9 of [11] is obtained.(iii)
By settingp¯=0, hx=1, p=−1, and θ=I, Theorem 2.1 of [22] is obtained.(iv)
By settingθ=I, hx=1, and p=−1, Theorem 2.1 of [23] is obtained.(v)
By settingω=p¯=0, hx=1, p=−1, and θ=I, Theorem 4 of [18] is obtained.(vi)
By settingp=−1, Theorem 2.5 of [24] is obtained.Corollary 2.
If we putp=−1, p¯=0, and θ=I in Theorem 4 (ii), we get the following Fejér–Hadamard inequalities for harmonically convex function via generalized fractional integral operators:(40)φ2ε1ε2ε1+ε2Υσ,τ,δ,ω′,1/ε2+ρ,r,k,cθh∘ψ1ε1+Υσ,τ,δ,ω′,1/ε1−ρ,r,k,cθh∘ψ1ε2≤Υσ,τ,δ,ω′,1/ε2+ρ,r,k,cθφh∘ψ1ε1+Υσ,τ,δ,ω′,1/ε1−ρ,r,k,cθφh∘ψ1ε2≤φε1+φε22Υσ,τ,δ,ω′,1/ε2+ρ,r,k,cθh∘ψ1ε1+Υσ,τ,δ,ω′,1/ε1−ρ,r,k,cθh∘ψ1ε2.Now we give another version of the Hadamard inequality.Theorem 5.
Letφ,θ:ε1,ε2⊂0,∞⟶ℝ, Range θ⊂ε1,ε2 be the functions such that φ be positive and φ∈L1ε1,ε2, ε1<ε2, and θ be a differentiable and strictly increasing function. If φ is p-convex, p∈ℝ\0, then for generalized fractional integral operators (5) and (6), we have(i)
Ifp>0, then(41)φθpε1+θpε221/pΥσ,τ,δ,2σω′,θ−1θpε1+θpε2/2+ρ,r,k,cθ1θ−1θpε2;p¯≤Υσ,τ,δ,2σω′,θ−1θpε1+θpε2/2+ρ,r,k,cθφ∘ψθ−1θpε2;p¯+Υσ,τ,δ,2σω′,θ−1θpε1+θpε2/2−ρ,r,k,cθφ∘ψθ−1θpε1;p¯≤φθε1+φθε22Υσ,τ,δ,2σω′,θ−1θpε1+θpε2/2+ρ,r,k,cθ1θ−1θpε2;p¯,whereω′=ω/θpε2−θpε1σ and ψt=θ1/pt for all t∈ε1p,ε2p.(ii)
Ifp<0, then(42)φθpε1+θpε221/pΥσ,τ,δ,2σω′,θ−1θpε1+θpε2/2−ρ,r,k,cθ1θ−1θpε2;p¯≤Υσ,τ,δ,2σω′,θ−1θpε1+θpε2/2+ρ,r,k,cθφ∘ψθ−1θpε1;p¯+Υσ,τ,δ,2σω′,θ−1θpε1+θpε2/2−ρ,r,k,cθφ∘ψθ−1θpε2;p¯≤φθε1+φθε22Υσ,τ,δ,2σω′,θ−1θpε1+θpε2/2−ρ,r,k,cθ1θ−1θpε2;p¯,whereω′=ω/θpε1−θpε2σ and ψt=θ1/pt for all t∈ε2p,ε1p.Proof.
(i) Settingθx=t/2θpε1+2−t/2θpε21/p and θy=t/2θpε2+2−t/2θpε11/p in (12), we have(43)φθpε1+θpε221/p≤φt/2θpε1+2−t/2θpε21/p+φt/2θpε2+2−t/2θpε11/p2.
Multiplying both sides of (43) by 2tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯ and then integrating over 0,1, we have(44)2φθpε1+θpε221/p∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯dt≤∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯φt2θpε1+2−t/2θpε21/pdt+∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯φt2θpε2+2−t/2θpε11/pdt.
By choosingθx=t/2θpε1+2−t/2θpε2 and θy=t/2θpε2+2−t/2θpε1 in (44) and by (5) and (6), we get first inequality of (41).
To prove the second inequality of (41), again from p-convexity of φ over ε1,ε2 and for t∈0,1, we have(45)φt2θpε1+2−t2θpε21/p+φt2θpε2+2−t2θpε11/p≤φθε1+φθε2.
Multiplying both sides of (45) by tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯ and then integrating over 0,1, we have(46)∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯φt2θpε1+2−t2θpε21/pdt+∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯φt2θpε2+2−t2θpε11/pdt≤φθε1+φθε2∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯dt.
Settingθx=t/2θpε1+2−t/2θpε2 and θy=t/2θpε2+2−t/2θpε1 in (46) and using (5) and (6), we get second inequality of (41).
(ii) Proof is similar to the proof of (i).Remark 4.
(i)
By settingp¯=0, p=−1, and θ=I, Theorem 2.3 of [22] is obtained.(ii)
By settingθ=I and p=−1, Theorem 2.3 of [23] is obtained.(iii)
By settingp=−1, Theorem 2.7 of [24] is obtained.
Now we obtain another Fejér–Hadamard type fractional integral inequality forp-convex function via generalized fractional integral operators (5) and (6).Theorem 6.
Letφ,θ,h:ε1,ε2⊂0,∞⟶ℝ, Range θ, Range h⊂ε1,ε2 be the functions such that φ be positive and φ∈L1ε1,ε2, ε1<ε2, and θ be a differentiable and strictly increasing function where h is a non-negative and integrable function. If φ is p-convex, p∈ℝ\0, and φθ1/px=φθpε1+θpε2−θx1/p, then the following inequalities for generalized fractional integral operators (5) and (6) hold:(i)
Ifp>0, then(47)φθpε1+θpε221/pΥσ,τ,δ,2σω′,θ−1θpε1+θpε2/2+ρ,r,k,cθh∘ψθ−1θpε2;p¯≤Υσ,τ,δ,2σω′,θ−1θpε1+θpε2/2+ρ,r,k,cθφh∘ψθ−1θpε2;p¯+Υσ,τ,δ,2σω′,θ−1θpε1+θpε2/2−ρ,r,k,cθφh∘ψθ−1θpε1;p¯≤φθε1+φθε22Υσ,τ,δ,2σω′,θ−1θpε1+θpε2/2+ρ,r,k,cθh∘ψθ−1θpε2;p¯,whereω′=ω/θpε2−θpε1σ and ψt=θ1/pt for all t∈ε1p,ε2p.(ii)
Ifp<0, then(48)φθpε1+θpε221/pΥσ,τ,δ,2σω′,θ−1θpε1+θpε2/2−ρ,r,k,cθh∘ψθ−1θpε2;p¯≤Υσ,τ,δ,2σω′,θ−1θpε1+θpε2/2+ρ,r,k,cθφh∘ψθ−1θpε1;p¯+Υσ,τ,δ,2σω′,θ−1θpε1+θpε2/2−ρ,r,k,cθφh∘ψθ−1θpε2;p¯≤φθε1+φθε22Υσ,τ,δ,2σω′,θ−1θpε1+θpε2/2−ρ,r,k,cθh∘ψθ−1θpε2;p¯,whereω′=ω/θpε1−θpε2σ and ψt=θ1/pt for all t∈ε2p,ε1p.Proof.
(i)
Multiplying (43) by 2tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯ht/2θpε1+2−t/2θpε21/p and then integrating over 0,1, we have(49)2φθpε1+θpε221/p∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯ht2θpε1+2−t2θpε21/pdt≤∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯φt2θpε1+2−t2θpε21/pht2θpε1+2−t2θpε21/pdt+∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯φt2θpε2+2−t2θpε11/pht2θpε1+2−t2θpε21/pdt.
By choosingθx=t/2θpε1+2−t/2θpε2, that is, θpε1+θpε2−θx=t/2θpε2+2−t/2θpε1, in (49) and using the condition φθ1/px=φθpε1+θpε2−θx1/p, one can get first inequality of (47).
To prove the second inequality of (47), multiplying both sides of (45) by tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯ht/2θpε1+2−t/2θpε21/p and then integrating over 0,1, we have(50)∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯φt2θpε1+2−t2θpε21/pht2θpε1+2−t2θpε21/pdt+∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯φt2θpε2+2−t2θpε11/pht2θpε1+2−t2θpε21/pdt≤φθε1+φθε2∫01tτ−1Eσ,τ,δρ,r,k,cωtσ;p¯ht2θpε1+2−t2θpε21/pdt.
Settingθx=t/2θpε1+2−t/2θpε2 in (38) and using the condition φθ1/px=φθpε1+θpε2−θx1/p, one can get second inequality of (47).
(ii) Proof is similar to the proof of (i).Remark 5.
(i)
By settingp=−1, p¯=0, and θ=I, Theorem 2.6 of [22] is obtained.(ii)
By settingθ=I and p=−1, Theorem 2.6 of [23] is obtained.(iii)
By settingp=−1, Theorem 2.10 of [24] is obtained.Corollary 3.
When we setp=−1, ω=p¯=0, and θ=I in Theorem 6, then we get the following inequalities via RL fractional integrals.(51)φ2ε1ε2ε1+ε2Iε1+ε2/2ε1ε2+τh∘ψ1ε1+Iε1+ε2/2ε1ε2−τh∘ψ1ε2≤Iε1+ε2/2ε1ε2+τφh∘ψ1ε1+Iε1+ε2/2ε1ε2−τφh∘ψ1ε2≤φε1+φε22Iε1+ε2/2ε1ε2+τh∘ψ1ε1+Iε1+ε2/2ε1ε2−τh∘ψ1ε2.


## 3. Conclusion

We have established Hadamard and Fejér–Hadamard fractional integral inequalities for generalized fractional integrals ofp-convex functions. The results of this paper hold simultaneously for convex and harmonically convex functions for different fractional integral operators containing Mittag-Leffler functions.



---

*Source: 5589405-2021-03-16.xml*