# Inequalities for a Unified Integral Operator viaα,m-Convex Functions

Authors: Baizhu Ni; Ghulam Farid; Kahkashan Mahreen
Journal: Journal of Mathematics
 (2020)

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## Abstract

Recently, a unified integral operator has been introduced by Farid, 2020, which produces several kinds of known fractional and conformable integral operators defined in recent decades (Kwun, 2019, Remarks 6 and 7). The aim of this paper is to establish bounds of this unified integral operator by means ofα,m-convex functions. The resulting inequalities provide the bounds of all associated fractional and conformable integral operators in a compact form. Also, the results of this paper hold for different kinds of convex functions connected with α,m-convex functions.


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## Body

## 1. Introduction

To prove the mathematical inequalities, fractional integral operators play an important role in the field of different branches of mathematics and engineering. Many mathematicians have used fractional integrals and conformable fractional integrals to develop integral inequalities [1–14]. We start from definitions of fractional integral operators which are direct consequences of unified integral operators given in (8) and (9).Definition 1.
(see [15]). Let η1:a,b⟶ℝ be an integrable function. Also, let η2 be an increasing and positive function on a,b, having a continuous derivative η2′ on a,b. The left-sided and right-sided fractional integrals of a function η1 with respect to another function η2 on a,b of order μ, where ℜμ>0, are defined by(1)Ia+η2μη1x=1Γμ∫axη2x−η2tμ−1η2′tη1tdt,x>a,Ib−η2μη1x=1Γμ∫xbη2t−η2xμ−1η2′tη1tdt,x<b,where Γ⋅ is the gamma function.
Ak-analogue of the above definition is defined as follows.Definition 2.
(see [16]). Let η1:a,b⟶ℝ be an integrable function. Also, let η2 be an increasing and positive function on a,b, having a continuous derivative η2′ on a,b. The left-sided and right-sided fractional integrals of a function η1 with respect to another function η2 on a,b of order μ;ℜμ,k>0 are defined by(2)Ia+kη2μη1x=1kΓkμ∫axη2x−η2tμ/k−1η2′tη1tdt,x>a,(3)Ib−kη2μη1x=1kΓkμ∫xbη2t−η2xμ/k−1η2′tη1tdt,x<b,where Γk⋅ is defined by [17](4)Γkx=∫0∞tx−1e−tk/kdt,ℜx>0.
A well-known function named Mittag-Leffler function is defined by [18](5)Eαz=∑n=0∞znΓαn+1,where α,z∈ℂ and ℜα>0.
One can see [19–22] to study the Mittag-Leffler function and its generalizations. Also, (2) and (3) produce many types of fractional integral operators (see [10], Remark 6).
A generalized fractional integral operator containing an extended generalized Mittag-Leffler function is defined as follows.Definition 3.
(see [1]). Let ω,μ,α,l,γ,c∈ℂ, ℜμ,ℜα,ℜl>0, and ℜc>ℜγ>0 with p≥0, δ>0, and 0<k≤δ+ℜμ. Let η1∈L1a,b and x∈a,b. Then, the generalized fractional integral operators ϵμ,α,l,ω,a+γ,δ,k,cη1 and ϵμ,α,l,ω,b−γ,δ,k,cη1 are defined by(6)ϵμ,α,l,ω,a+γ,δ,k,cη1x;p=∫axx−tα−1Eμ,α,lγ,δ,k,cωx−tμ;pη1tdt,ϵμ,α,l,ω,b−γ,δ,k,cη1x;p=∫xbt−xα−1Eμ,α,lγ,δ,k,cωt−xμ;pη1tdt,where(7)Eμ,α,lγ,δ,k,ct;p=∑n=0∞βpγ+nk,c−γβγ,c−γcnkΓμn+αtnlnδ,is the extended generalized Mittag-Leffler function.
Recently, Farid defined a unified integral operator which unifies several kinds of fractional and conformable integrals in a compact formula which is defined as follows.Definition 4.
(see [23]). Let η1,η2:a,b⟶ℝ, 0<a<b, be the functions such that η1 be positive and η1∈L1a,b and η2 be differentiable and strictly increasing. Also, let ϕ/x be an increasing function on a,∞ and α,l,γ,c∈ℂ, ℜα,ℜl>0, ℜc>ℜγ>0, p,μ,δ≥0, and 0<k≤δ+μ. Then, for x∈a,b, the left and right integral operators are defined by(8)Fη2μ,α,l,a+ϕ,γ,δ,k,cη1x,ω;p=∫axKxyEμ,α,lγ,δ,k,c,η2;ϕη1ydη2y,(9)Fη2μ,α,l,b−ϕ,γ,δ,k,cη1x,ω;p=∫xbKyxEμ,α,lγ,δ,k,c,η2;ϕη1ydη2y,where KxyEμ,α,lγ,δ,k,c,η2;ϕ=ϕη2x−η2y/η2x−η2yEμ,α,lγ,δ,k,cωη2x−η2yμ;p.
For suitable settings of functionϕ, η2, and certain values of parameters included in Mittag-Leffler function (7), very interesting consequences are obtained which are comprised in Remarks 6 and 7 of [10].
The objective of this paper is to obtain bounds of unified integral operators explicitly which are directly linked with various fractional and conformable integrals. Theα,m-convexity has been used for establishing these bounds. The notion of α,m-convexity is defined by Mihesan in [24].Definition 5.
A functionη1:0,b⟶ℝ,b>0, is said to be α,m-convex, where α,m∈0,12, if(10)η1tx+m1−ty≤tαη1x+m1−tαη1y,holds for all x,y∈0,bandt∈0,1.Remark 1.
(i)
If we putα,m = 1,m, then (10) gives the definition of m-convex function(ii)
If we putα,m = 1,1, then (10) gives the definition of convex function(iii)
If we putα,m = 1,0, then (10) gives the definition of star-shaped function
For some recent citations and utilizations ofα,m-convex functions, one can see [9, 25–28] and references therein. In the upcoming section, bounds of unified integral operators are established by using α,m-convexity. These bounds provide general formulas to obtain bounds of fractional and conformable integral operators described in Remarks 6 and 7 of [10]. Among the well-known inequalities which are related to the integral mean of a convex function, the Hadamard inequality is of great importance. Many mathematicians worked on new types of Hadamard inequalities using convex functions, see [8, 29–31]. We also established the general Hadamard-type inequality by applying Lemma 1 which further produces various inequalities of Hadamard type for fractional and conformable integrals. At the end, by using α,m-convexity of η1′, a modulus inequality is obtained.


## 2. Main Results

Bounds of unified integral operators (8) and (9) using α,m-convexity are studied in the following result:Theorem 1.
Letη1:a,b⟶ℝ be a positive integrable α,m-convex function with m∈0,1. Let η2:a,b⟶ℝ be differentiable and strictly increasing function, and also, let ϕ/x be an increasing function on a,b. If α,l,γ,c∈ℂ, ℜα,ℜl>0, ℜc>ℜγ>0, p,μ,δ≥0, and 0<k≤δ+μ, then for x∈a,b, we have(11)Fη2μ,α,l,a+ϕ,γ,δ,k,cη1x,ω;p≤KxaEμ,α,lγ,δ,k,c,η2;ϕmη1xmη2x−η1aη2a−Γα+1x−aαmη1xm−η1aIαa+η2x,Fη2μ,α,l,b−ϕ,γ,δ,k,cη1x,ω;p≤KbxEμ,α,lγ,δ,k,c,η2;ϕη1bη2b−mη1xmη2xΓα+1b−xαη1b−mη1xmIαb−η2x,and hence,(12)Fη2μ,α,l,a+ϕ,γ,δ,k,cη1x,ω;p+Fη2μ,α,l,b−ϕ,γ,δ,k,cη1x,ω;p≤KxaEμ,α,lγ,δ,k,c,η2;ϕmη1xmη2x−η1aη2a−Γα+1x−aαmη1xm−η1aαIa+η2x+KbxEμ,α,lγ,δ,k,c,η2;ϕη1bη2b−mη1xmη2x−Γα+1b−xαη1b−mη1xmαIb−η2x.Proof.
Under the assumptions ofϕ and η2, one can write the following inequality:(13)ϕη2x−η2tη2x−η2t≤ϕη2x−η2aη2x−η2a;t∈a,x,x∈a,b.
Multiplying withEμ,α,lγ,δ,k,cωη2x−η2tμ;pη2′t, we can obtain(14)KxtEμ,α,lγ,δ,k,c,η2;ϕη2′t≤ϕη2x−η2aη2x−η2aEμ,α,lγ,δ,k,cωη2x−η2tμ;pη2′t.
By usingEμ,α,lγ,δ,k,cωη2x−η2tμ;p≤Eμ,α,lγ,δ,k,cωη2x−η2aμ;p, the following inequality is obtained:(15)KxtEμ,α,lγ,δ,k,c,η2;ϕη2′t≤KxaEμ,α,lγ,δ,k,c,η2;ϕη2′t.
Using the definition ofα,m-convexity for η1, the following inequality is valid:(16)η1t≤x−tx−aαη1a+m1−x−tx−aαη1xm.
Multiplying (15) with (16) and integrating over a,x, one can obtain(17)∫axKxtEμ,α,lγ,δ,k,c,η2;ϕη1tdη2t≤η1aKxaEμ,α,lγ,δ,k,c,η2;ϕ∫axx−tx−aαdη2t+mη1xmKxaEμ,α,lγ,δ,k,c,η2;ϕ∫ax1−x−tx−aαdη2t.
By using (8) of Definition 4 and integrating by parts, the following inequality is obtained:(18)Fη2μ,α,l,a+ϕ,γ,δ,k,cη1x,ω;p≤KxaEμ,α,lγ,δ,k,c,η2;ϕx−aαmη1xmη2x−η1aη2aΓα+1x−aαmη1xm−η1aαIa+η2x.
Now, on the other side, fort∈x,b and x∈a,b, the following inequality holds true:(19)KxtEμ,α,lγ,δ,k,c,η2;ϕη2′t≤KbxEμ,α,lγ,δ,k,c,η2;ϕη2′t.
Usingα,m-convexity of η1, we have(20)η1t≤t−xb−xαη1b+m1−t−xb−xαη1xm.
Adopting the same procedure as we did for (15) and (16), the following inequality from (19) and (20) can be obtained:(21)Fη2μ,α,l,b−ϕ,γ,δ,k,cη1x,ω;p≤KbxEμ,α,lγ,δ,k,c,η2;ϕη1bη2b−mη1xmη2x−Γα+1b−xαη1b−mη1xmαIb−η2x.
By adding (18) and (21), (12) can be obtained.Remark 2.
(i)
If we considerα,m = (1,1) in (12), Theorem 8 in [10] is obtained(ii)
If we considerϕt=Γμtμ/k/kΓkμ for the left-hand integral and ϕt=Γνtν/k/kΓkν for the right-hand integral and p=ω=0 in (12), then Theorem 1 in [9] can be obtained(iii)
If we considerμ=ν in the result of (ii), then Corollary 1 in [9] can be obtained(iv)
If we considerϕt=Γμtμ, p=ω=0, and α,m = (1,1) in (12), Theorem 1 in [6] is obtained(v)
If we considerμ=ν in the result of (iv), Corollary 1 in [6] is obtained(vi)
If we considerϕt=Γμtμ/k/kΓkμ for the left-hand integral and ϕt=Γνtν/k/kΓkνfor the right-hand integral, α,m =(1,1), η2x=x, and p=ω=0, then Theorem 1 in [4] can be obtained(vii)
If we considerμ=ν in the result of (vi), then Corollary 1 in [4] can be obtained(viii)
If we considerϕt=Γμtμ for the left-hand integral and ϕt=Γνtν for the right-hand integral, η2x=x, and p=ω=0 and α,m =(1,1) in (12), then Theorem 1 in [5] is obtained(ix)
By settingμ=ν in the result of (viii), Corollary 1 in [5] can be obtained(x)
By settingμ=ν=1 and x=aorx=b in the result of (ix), Corollary 2 in [5] can be obtained(xi)
By settingμ=ν=1 and x=a+b/2 in the result of (ix), Corollary 3 in [5] can be obtained
To prove the next result, we need the following lemma [9].Lemma 1.
Letη1:0,∞⟶ℝ be an α,m-convex function with m∈0,1. If η1x=η1a+b−x/m,0<a<b, then the following inequality holds:(22)η1a+b2≤12α1+m2α−1η1x,for all x∈a,b and m∈0,1.The following result provides upper and lower bounds of the sum of operators (8) and (9) in the form of a Hadamard inequality.Theorem 2.
With the assumptions of Theorem1 in addition, if η1x=η1a+b−x/m, then we have(23)2αη1a+b/21+m2α−1Fη2μ,α,l,b−ϕ,γ,δ,k,c1a,ω;p+Fη2μ,α,l,a+ϕ,γ,δ,k,c1b,ω;p≤Fη2μ,α,l,b−ϕ,γ,δ,k,cη1a,ω;p+Fη2μ,α,l,a+ϕ,γ,δ,k,cη1b,ω;p≤2KbaEμ,α,lγ,δ,k,c,η2;ϕη1bη2b−mη1amη2a−Γα+1b−aαη1b−mη1amαIb−η2a.Proof.
Under the assumptions ofϕ and η2, we have(24)ϕη2x−η2aη2x−η2a≤ϕη2b−η2aη2b−η2a.
Multiplying withEμ,α,lγ,δ,k,cωη2x−η2aμ;pη2′x, we can obtain from (24) the following inequality:(25)KxaEμ,α,lγ,δ,k,c,η2;ϕη2′x≤ϕη2b−η2aη2b−η2aEμ,α,lγ,δ,k,cωη2x−η2aμ;pη2′x.
By usingEμ,α,lγ,δ,k,cωη2x−η2aμ;p≤Eμ,α,lγ,δ,k,cωη2b−η2aμ;p, the following inequality is obtained:(26)KxaEμ,α,lγ,δ,k,c,η2;ϕη2′x≤KbaEμ,α,lγ,δ,k,c,η2;ϕη2′x.
Usingα,m-convexity of η1 for x∈a,b, we have(27)η1x≤x−ab−aαη1b+m1−x−ab−aαη1am.
Multiplying (26) and (27) and integrating the resulting inequality over a,b, one can obtain(28)∫abKxaEμ,α,lγ,δ,k,c,η2;ϕη1xdη2x≤mη1amKbaEμ,α,lγ,δ,k,c,η2;ϕ∫ab1−x−ab−aαdη2x+η1bϕη2b−η2aη2b−η2aEμ,α,lγ,δ,k,cωη2b−η2aμ;p∫abx−ab−aαdη2x.
By using Definition4 and integrating by parts, the following inequality is obtained:(29)Fη2μ,α,l,a+ϕ,γ,δ,k,cη1b,ω;p≤KbaEμ,α,lγ,δ,k,c,η2;ϕη1bη2b−mη1amη2a−Γα+1b−aαη1b−mη1amαIb−η2a.
On the other hand, forx∈a,b, the following inequality holds true:(30)KbxEμ,α,lγ,δ,k,c,η2;ϕη2′x≤KbaEμ,α,lγ,δ,k,c,η2;ϕη2′x.
Adopting the same pattern of simplification as we did for (26) and (27), the following inequality can be observed from (27) and (30):(31)Fη2μ,α,l,b−ϕ,γ,δ,k,cη1a;p≤KbaEμ,α,lγ,δ,k,c,η2;ϕη1bη2b−mη1amη2a−Γα+1b−aαη1b−mη1amαIb−η2a.
By adding (29) and (31), the following inequality can be obtained:(32)Fη2μ,α,l,b−ϕ,γ,δ,k,cη1a,ω;p+Fη2μ,α,l,a+ϕ,γ,δ,k,cη1b,ω;p≤KbaEμ,α,lγ,δ,k,c,η2;ϕη1bη2b−mη1amη2a−Γα+1b−aαη1b−mη1amαIb−η2a.
Multiplying both sides of (22) by KxaEμ,α,lγ,δ,k,c,η2;ϕdη2x and integrating over a,b, we have(33)η1a+b2∫abKxaEμ,α,lγ,δ,k,c,η2;ϕdη2x≤12α1+m2α−1×∫abKxaEμ,α,lγ,δ,k,c,η2;ϕη1xdη2x.
From Definition4, the following inequality is obtained:(34)η1a+b22α1+m2α−1Fη2μ,α,l,b−ϕ,γ,δ,k,c1a;p≤Fη2μ,α,l,b−ϕ,γ,δ,k,cη1a;p.
Similarly, multiplying both sides of (22) by KbxEμ,α,lγ,δ,k,c,η2;ϕdη2x and integrating over a,b, we have(35)η1a+b22α1+m2α−1η2Fμ,α,l,a+ϕ,γ,δ,k,c1b;p≤η2Fμ,α,l,a+ϕ,γ,δ,k,cη1b;p.
By adding (34) and (35), the following inequality is obtained:(36)η1a+b22α1+m2α−1Fη2μ,α,l,b−ϕ,γ,δ,k,c1a,ω;p+Fη2μ,α,l,a+ϕ,γ,δ,k,c1b,ω;p≤Fη2μ,α,l,b−ϕ,γ,δ,k,cη1a,ω;p+Fη2μ,α,l,a+ϕ,γ,δ,k,cη1b,ω;p.
Using (32) and (36), inequality (23) can be achieved.Remark 3.
(i)
If we considerα,m = (1,1) in (23), Theorem 22 in [10] is obtained(ii)
If we considerϕt=Γμtμ/k+1 for the left-hand integral and ϕt=Γνtν/k+1 and p=ω=0 in (23), then Theorem 3 in [9] can be obtained(iii)
If we considerμ=ν in the result of (ii), then Corollary 3 in [9] can be obtained(iv)
If we considerϕt=Γμtμ+1 for the left-hand integral and ϕt=Γνtν+1for the right-hand integral in (23), p=ω=0, and α,m = (1,1) in (23), Theorem 3 in [6] is obtained(v)
If we considerμ=ν in the result of (iv), Corollary 3 in [6] is obtained(vi)
If we considerϕt=Γμtμ/k+1 for the left-hand integral and ϕt=Γνtν/k+1for the right-hand integral, α,m = (1,1), η2x=x, and p=ω=0 in (23), then Theorem 3 in [4] can be obtained(vii)
If we considerμ=ν in the result of (vi), then Corollary 6 in [4] can be obtained(viii)
By settingϕt=Γμtμ+1 for the left-hand integral and ϕt=Γνtν+1 for the right-hand integral, p=ω=0, α,m=1, and gt=t in (23), Theorem 3 in [5] can be obtained(ix)
By settingμ=ν in the result of (viii), Corollary 6 in [5] can be obtainedTheorem 3.
Letη1:a,b⟶ℝ be a differentiable function. η1′ is α,m-convex with m∈0,1, and let η2:a,b⟶ℝ be differentiable and strictly increasing function; also, let ϕ/x be an increasing function on a,b. If α,l,γ,c∈ℂ, ℜα,ℜl>0, ℜc>ℜγ>0, p,μ,δ≥0, and 0<k≤δ+μ, then for x∈a,b, we have(37)Fη2μ,α,l,a+ϕ,γ,δ,k,cη1x,ω;p+Fη2μ,α,l,b−ϕ,γ,δ,k,cη1x,ω;p≤KxaEμ,α,lγ,δ,k,c,η2;ϕmη1′xmη2x−η1′aη2a−Γα+1x−aαmη1′xm−η1′aαIa+η2x+KbxEμ,α,lγ,δ,k,c,η2;ϕη1′bη2b−mη1′xmη2x−Γα+1b−xαη1′b−mη1′xmαIb−η2x,where(38)Fη2μ,α,l,a+ϕ,γ,δ,k,cη1∗η2x,ω;p≔∫axKxtEμ,α,lγ,δ,k,c,η2;ϕη1′tdη2t,Fη2μ,α,l,b−ϕ,γ,δ,k,cη1∗η2x,ω;p≔∫xbKtxEμ,α,lγ,δ,k,c,η2;ϕη1′tdη2t.Proof.
Letx∈a,b and t∈a,x. Then, using α,m-convexity of η1′, we have(39)η1′t≤x−tx−aαη1′a+m1−x−tx−aαη1′xm.
Inequality (39) can be written as follows:(40)−x−tx−aαη1′a+m1−x−tx−aαη1′xm≤η1′t≤x−tx−aαη1′a+m1−x−tx−aαη1′xm.
Let us consider the second inequality of (40):(41)η1′t≤x−tx−aαη1′a+m1−x−tx−aαη1′xm.
Multiplying (15) and (41) and integrating over a,x, we can obtain(42)∫axKxtEμ,α,lγ,δ,k,c,η2;ϕdη2t≤η1aKxaEμ,α,lγ,δ,k,c,η2;ϕ∫axx−tx−aαdη2t+mη1xmKxaEμ,α,lγ,δ,k,c,η2;ϕ∫ax1−x−tx−aαdη2t.
By using (8) of Definition 4 and integrating by parts, the following inequality is obtained:(43)Fη2μ,α,l,a+ϕ,γ,δ,k,cη1x,ω;p≤KxaEμ,α,lγ,δ,k,c,η2;ϕ×mη1′xmη2x−η1′aη2a−Γα+1x−aαmη1′xm−η1′aαIa+η2x.
If we consider the left-hand side from inequality (40) and adopt the same pattern as we did for the right-hand side inequality, then(44)Fη2μ,α,l,a+ϕ,γ,δ,k,cη1∗η2x,ω;p≥−KxaEμ,α,lγ,δ,k,c,η2;ϕ×mη1′xmη2x−η1′aη2a−Γα+1x−aαmη1′xm−η1′aαIa+η2x.
From (43) and (44), the following inequality is observed:(45)Fη2μ,α,l,a+ϕ,γ,δ,k,cη1∗η2x,ω;p≤KxaEμ,α,lγ,δ,k,c,η2;ϕ×mη1′xmη2x−η1′aη2a−Γα+1x−aαmη1′xm−η1′aαIa+η2x.
Now, usingα,m-convexity of η1′ on x,b for x∈a,b, we have(46)η1′t≤t−xb−xαη1′b+m1−t−xb−xαη1′xm.
On the same procedure as we did for (15) and (39), one can obtain the following inequality from (19) and (46):(47)Fη2μ,α,l,b−ϕ,γ,δ,k,cη1∗η2x,ω;p≤KbxEμ,α,lγ,δ,k,c,η2;ϕ×η1′bη2b−mη1′xmη2x−Γα+1b−xαη1′b−mη1′xmαIb−η2x.
By adding (45) and (47), inequality (37) can be achieved.Remark 4.
(i)
If we considerα,m = (1,1) in (37), then Theorem 25 in [10] is obtained(ii)
If we considerϕt=Γμtμ/k+1 for the left-hand integral and ϕt=Γνtν/k+1 for the right-hand integral and p=ω=0 in (37), then Theorem 2 in [9] can be obtained(iii)
If we considerμ=ν in the result of (ii), then Corollary 2 in [9] can be obtained(iv)
If we considerϕt=Γμtμ+1 for the left-hand integral and ϕt=Γνtν+1 for the right-hand integral, p=ω=0, and α,m = (1,1) in (37), then Theorem 2 in [6] is obtained(v)
If we considerμ=ν in the result of (iv), then Corollary 2 in [6] is obtained(vi)
If we considerϕt=Γμtμ/k+1 for the left-hand integral and ϕt=Γνtν/k+1 for the right-hand integral, α,m =(1,1), η2x=x, and p=ω=0 in (37), then Theorem 2 in [4] can be obtained(vii)
If we considerμ=ν in the result of (vi), then Corollary 4 in [4] can be obtained(viii)
If we considerμ=ν=k=1 and x=a+b/2 in the result of (vii), then Corollary 5 in [4] can be obtained(ix)
If we considerϕt=Γμtμ+1 for the left-hand integral and ϕt=Γνtν+1 for the right-hand integral, η2x=x, p=ω=0, and α,m =(1,1) in (37), then Theorem 2 in [5] is obtained(x)
By settingμ=ν in the result of (ix), then Corollary 5 in [5] can be obtained


## 3. Concluding Remarks

This research paper explores fractional and conformable fractional integral inequalities in a unified form, which provide the bounds of conformable fractional integral operators and fractional integral operators containing Mittag-Leffler functions in their kernels. The results of this paper hold for fractional and conformable integral operators and convex,m-convex, and star-shaped functions (see Remarks 6 and 7 of [10] and Remark 1) simultaneously.



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