# Fractional Hadamard and Fejér-Hadamard Inequalities Associated with Exponentiallys,m-Convex Functions

Authors: Shuya Guo; Yu-Ming Chu; Ghulam Farid; Sajid Mehmood; Waqas Nazeer
Journal: Journal of Function Spaces
 (2020)

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## Abstract

The aim of this paper is to present the fractional Hadamard and Fejér-Hadamard inequalities for exponentiallys,m-convex functions. To establish these inequalities, we will utilize generalized fractional integral operators containing the Mittag-Leffler function in their kernels via a monotone function. The presented results in particular contain a number of fractional Hadamard and Fejér-Hadamard inequalities for s-convex, m-convex, s,m-convex, exponentially convex, exponentially s-convex, and convex functions.


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## Body

## 1. Introduction and Preliminaries

Integral operators play an important role in the subject of mathematical analysis. Fractional integral operators have been proven very useful in almost all fields of science and engineering. By using fractional integral operators, a lot of well-known inequalities have been studied, and in the consequent, they are generalized and extended in the subject of fractional calculus. Fractional integral inequalities provide support in the formation of modeling of physical phenomenons. They also provide their role in the uniqueness of solutions of fractional boundary value problems. A large number of fractional integral inequalities exist in literature due to fractional integral operators (see, [1–5]). The Riemann-Liouville fractional integral operators are the first formulation of fractional integral operators of nonintegral order.Definition 1 [6].
Letη∈L1a,b. The Riemann-Liouville fractional integral operators Ja+ση and Jb−ση of order σ∈ℂℛσ>0 are defined by
(1)Ja+σηu=1Γσ∫auu−zσ−1ηzdz,(2)Jb−σηu=1Γσ∫ubz−uσ−1ηzdz,where Γσ=∫0∞e−zzσ−1dz.Next, we give the definition of generalized fractional integral operators containing the Mittag-Leffler function in their kernels as follows.Definition 2 [7].
Letω,ν,τ,l,ρ,c∈ℂ, ℛν,ℛτ,ℛl>0, ℛc>ℛρ>0 with p≥0, r>0, and 0<q≤r+ℛν. Let η∈L1a,b and u∈a,b. Then, the generalized fractional integral operators Yν,τ,l,ω,a+ρ,r,q,cη and Yν,τ,l,ω,b−ρ,r,q,cη are defined by
(3)Yν,τ,l,ω,a+ρ,r,q,cηu;p=∫auu−zτ−1Eν,τ,lρ,r,q,cωu−zν;pηzdz,(4)Yν,τ,l,ω,b−ρ,r,q,cηu;p=∫ubz−uτ−1Eν,τ,lρ,r,q,cωz−uν;pηzdz,where Eν,τ,lρ,r,q,cz;p is the generalized Mittfag-Leffler function defined as follows:
(5)Eν,τ,lρ,r,q,cz;p=∑n=0∞βpρ+nq,c−ρβρ,c−ρcnqΓνn+τznlnr.Lemma 3 [7].
Ifω,ν,τ,l,ρ,c∈ℂ, ℛν,ℛτ,ℛl>0, ℛc>ℛρ>0 with p≥0, r>0 and 0<q≤r+ℛν, then
(6)1Γν∫01zν−11−zτ−1Eν,τ,lρ,r,q,cω1−zν;pdz=Eν,τ+ν,lρ,r,q,cω;p.Recently in [8], Farid defined the following unified integral operators.Definition 4.
Letη,κ:a,b⟶ℝ, 0<a<b be the functions such that η be a positive and integrable and κ be a differentiable and strictly increasing. Also, let γ/u be an increasing function on a,∞ and τ,l,ρ,c∈ℂ, p,ν,r≥0, and 0<q≤r+ν. Then, for u∈a,b, the integral operators κYν,τ,l,a+γ,ρ,r,q,cη and κYν,τ,l,b−γ,ρ,r,q,cη are defined by
(7)κYν,τ,l,a+γ;ρ,r,q,cηu;p=∫auγκu−κzκu−κzEν,τ,lρ,r,q,cωκu−κzν;pηzdκz,(8)κYν,τ,l,b−γ;ρ,r,q,cηu;p=∫ubγκz−κuκz−κuEν,τ,lρ,r,q,cωκz−κuν;pηzdκz.If we takeγu=uτ in (7) and (9), then we get the following generalized fractional integral operators containing the Mittag-Leffler function.Definition 5.
Letη,κ:a,b⟶ℝ, 0<a<b be the functions such that η be a positive and integrable and κ be a differentiable and strictly increasing. Also, let ν,τ,l,ω,ρ,c∈ℂ, p,ν,r≥0, and 0<q≤r+ν. Then, for u∈a,b, the integral operators κYν,τ,l,ω,a+ρ,r,q,cη and κYν,τ,l,ω,b−ρ,r,q,cη are defined by
(9)κYν,τ,l,ω,a+ρ,r,q,cηu;p=∫auκu−κzτ−1Eν,τ,lρ,r,q,cωκu−κzν;pηzdκz,(10)κYν,τ,l,ω,b−ρ,r,q,cηu;p=∫ubκz−κuτ−1Eν,τ,lρ,r,q,cωκz−κuν;pηzdκz.Remark 6.
The operators (9) and (10) are the generalization of the following fractional integral operators:
(i)
By takingκu=u and ω=p=0, the Riemann-Liouville fractional integral operators (1) and (3) can be achieved(ii)
By takingκu=u, the fractional integral operators (5) and (6) can be achieved(iii)
By takingκu=u and p=0, the fractional integral operators defined by Salim-Faraj in [9] can be achieved(iv)
By takingκu=u and l=r=1, the fractional integral operators defined by Rahman et al. in [10] can be achieved(v)
By takingκu=u, p=0, and l=r=1, the fractional integral operators defined by Srivastava-Tomovski in [11] can be achieved(vi)
By takingκu=u, p=0, and l=r=q=1, the fractional integral operators defined by Prabhakar in [12] can be achievedFrom generalized fractional integral operator (9), we have
(11)κYν,τ,l,ω,a+ρ,r,q,c1u;p=∫auκu−κzτ−1Eν,τ,lρ,r,q,cωκu−κzν;pdκz=∫auκu−κzτ−1∑n=0∞βpρ+nq,c−ρβρ,c−ρcnqΓνn+τωnκu−κzνnlnrdκz=∑n=0∞βpρ+nq,c−ρβρ,c−ρcnqΓνn+τωnlnr∫auκu−κzνn+τ−1dκz=κu−κaτ∑n=0∞βpρ+nq,c−ρβρ,c−ρcnqΓνn+τωnlnrκu−κaνnνn+τ=κu−κaτ∑n=0∞βpρ+nq,c−ρβρ,c−ρcnqΓνn+τ+1ωnκu−κaνnlnr=κu−κaτEν,τ+1,lρ,r,q,cωκu−κaν;p.Hence,(12)κYν,τ,l,ω,a+ρ,r,q,c1u;p=κu−κaτEν,τ+1,lρ,r,q,cωκu−κaν;p.Similarly, from the generalized fractional integral operator (10), we have
(13)κYν,τ,l,ω,b−ρ,r,q,c1u;p=κb−κuτEν,τ+1,lρ,r,q,cωκb−κuν;p.In this paper, we will use the following notations frequently:(14)κζω,a+τu;p=κYν,τ,l,ω,a+ρ,r,q,c1u;p,(15)κζω,b−τu;p=κYν,τ,l,ω,b−ρ,r,q,c1u;p.Convex functions are also very important in the field of mathematical analysis (see, [13–15]).A real-valued functionη:I⟶ℝ is said to be convex on I (I is an interval in ℝ), if
(16)ηza+1−zb≤zηa+1−zηbholds, for all a,b∈I, and z∈0,1. The function η is said to be concave if the reverse of inequality (16) holds.Convex functions are further extended and generalized in various ways by using different techniques. For example in [16], Qiang et al. introduced exponentially s,m-convex function which is a generalization of s-convex, m-convex, s,m-convex, exponentially convex, exponentially s-convex, and convex functions.Recently,s is as follows.Definition 7.
Lets∈0,1 and I⊆0,∞ be an interval. A function η:I⟶ℝ is said to be exponentially s,m-convex on I, if
(17)ηza+m1−zb≤zsηaeαa+m1−zsηbeαbholds, for all a,b∈I, m,z∈0,1, and α∈ℝ.Remark 8.
(i)
If we takem=1 in (17), then exponentially s-convex function defined by Mehreen and Anwar in [17] can be achieved(ii)
If we takes=m=1 in (17), then exponentially convex function defined by Awan et al. in [18] can be achieved(iii)
If we takeα=0 in (17), then s,m-convex function defined by Efthekhari in [19] can be achieved(iv)
If we takeα=0 and m=1 in (17), then s-convex function defined by Hudzik and Maligranda in [20] can be achieved(v)
If we takeα=0 and s=1 in (17), then m-convex function defined by Toader in [21] can be achieved(vi)
If we takeα=0 and s=m=1 in (17), then convex function (16) can be achievedA convex function is also equally defined by the well-known Hadamard inequality stated as follows:(18)ηa+b2≤1b−a∫abηzdz≤ηa+ηb2,where η:a,b⟶ℝ is a convex function.In [22], Fejér gave a generalization of the Hadamard inequality stated as follows:
(19)ηa+b2∫abκzdz≤∫abηzκzdz≤ηa+ηb2∫abκzdz,where η:a,b⟶ℝ is a convex function and κ:a,b⟶ℝ is positive, integrable, and symmetric to a+b/2.The inequality (19) is well known as the Fejér-Hadamard inequality. The Hadamard and the Fejér-Hadamard inequalities have been analyzed by many authors and produced frequently their generalizations, refinements, and extensions (see, [23–37]).The main purpose of this article is to establish the Hadamard and the Fejér-Hadamard inequalities for exponentiallys,m-convex functions by utilizing the generalized fractional integral operators (9) and (10) containing the Mittag-Leffler function. These inequalities lead to produce results for several kinds of convexities and fractional integral operators composed in Remark 8 and Remark 6, respectively. In Section 2, we prove the Hadamard inequalities for generalized fractional integral operators (9) and (10) via exponentially s,m-convex functions. In Section 3, we prove the Fejér-Hadamard inequalities for these generalized fractional integral operators via exponentially s,m-convex functions. Also, it is shown that these inequalities are generalizations of the Hadamard and the Fejér-Hadamard inequalities proved in [27–30, 32, 33, 35–37]. In the whole paper, we will consider real parameters of the Mittag-Leffler function.


## 2. Hadamard Inequalities

Here, we will give two versions of the generalized fractional Hadamard inequality.Theorem 9.
Letη,κ:a,mb⊆0,∞⟶ℝ, 0<a<mb be the real-valued functions. If η be a integrable and exponentially s,m-convex and κ be a differentiable and strictly increasing. Then, for integral operators (9) and (10), the following inequalities hold:
(20)2sξαηκa+mκb2κζω¯,a+τκ−1mκb;p≤κYν,τ,l,ω¯,a+ρ,r,q,cη∘κκ−1mκb;p+mτ+1κYν,τ,l,ω¯mν,b−ρ,r,q,cη∘κκ−1κam;p≤mκb−κaτηκaeακa+mηκbeακbEν,τ+1,lρ,r,q,cω;p+mηκbeακb+mηκa/m2eακa/m2Γs+1Eν,τ+s+1,lρ,r,q,cz,where ξα=eακb for α<0, ξα=eακa for α≥0 and ω¯=ω/mκb−κaν.Proof.
Sinceη is exponentially s,m-convex function on a,mb, for z∈0,1, we have the following inequalities:
(21)2sηκa+mκb2≤ηzκa+m1−zκbeαzκa+m1−zκb+mη1−zκa/m+zκbeα1−zκa/m+zκb,(22)ηzκa+m1−zκb+mη1−zκam+zκb≤zsηκaeακa+mηκbeακb+m1−zsηκbeακb+mηκa/m2eακa/m2.Multiplying both sides of (21) with zτ−1Eν,τ,lρ,r,q,cωzν;p and integrating over 0,1, we have
(23)2sηκa+mκb2∫01zτ−1Eν,τ,lρ,r,q,cωzν;pdz≤∫01zτ−1Eν,τ,lρ,r,q,cωzν;pηzκa+m1−zκbeαzκa+m1−zκbdz+m∫01zτ−1Eν,τ,lρ,r,q,cωzν;pη1−zκa/m+zκbeα1−zκa/m+zκbdz.Puttingκu=zκa+m1−zκb and κv=1−zκa/m+zκb in (23), we get
(24)2sηκa+mκb2∫aκ−1mκbmκb−κuτ−1Eν,τ,lρ,r,q,cω¯mκb−κuν;pdκu≤∫aκ−1mκbmκb−κuτ−1Eν,τ,lρ,r,q,cω¯mκb−κuν;pηκueακudκu+mτ+1∫κ−1κa/mbκv−κamτ−1Eν,τ,lρ,r,q,cmνω¯κv−κamν;pηκveακvdκv.By using (9), (10), and (14), the first inequality of (20) is obtained.Now, multiplying both sides of (22) with zτ−1Eν,τ,lρ,r,q,cωzν;p and integrating over 0,1, we have
(25)∫01zτ−1Eν,τ,lρ,r,q,cωzν;pηzκa+m1−zκbdz+m∫01zτ−1Eν,τ,lρ,r,q,cωzν;pη1−zκam+zκbdz≤ηκaeακa+mηκbeακb∫01zs+τ−1Eν,τ,lρ,r,q,cωzν;pdz+mηκbeακb+mηκa/m2eακa/m22∫011−zszτ−1Eν,τ,lρ,r,q,cωzν;pdz.First, we calculate the following integrals of (25):
(26)∫01zs+τ−1Eν,τ,lρ,r,q,cωzν;pdz=∫01zs+τ−1∑n=0∞βpρ+nq,c−ρβρ,c−ρcnqΓνn+τωnzνnlnrdz=∑n=0∞βpρ+nq,c−ρβρ,c−ρcnqΓνn+τωnlnr∫01zs+τ−1+νndz=∑n=0∞βpρ+nq,c−ρβρ,c−ρcnqΓνn+τωnlnr1νn+τ+s,as νn+τ≤νn+τ+s for s∈0,1. Hence,
(27)∫01zs+τ−1Eν,τ,lρ,r,q,cωzν;pdz≤∑n=0∞βpρ+nq,c−ρβρ,c−ρcnqΓνn+τωnlnr1νn+τ=Eν,τ+1,lρ,r,q,cω;p.By using (9), we have
(28)∫011−zszτ−1Eν,τ,lρ,r,q,cωzν;pdz=Γs+1Eν,τ+s+1,lρ,r,q,cω;p.Now, puttingκu=zκa+m1−zκb, κv=1−zκa/m+zκb, and using (27), (28) in (25), the second inequality of (20) is obtained.Special cases of the above generalized fractional Hadamard inequality are highlighted in the following remark.Remark 10.
(i)
If we takem=1 in (20), then Hadamard inequality for the exponentially s-convex function of second kind is obtained(ii)
If we takem=s=1 in (20), then Hadamard inequality for the exponentially convex function is obtained(iii)
If we takeκu=u, α=0, and m=s=1 in (20), then [35] (Theorem 2.1) is obtained(iv)
If we takeκu=u, α=0, and s=1 in (20), then [35] (Theorem 3.1) is obtained(v)
If we takeκu=u, α=0, m=s=1, and p=0 in (20), then [27] (Theorem 2.1) is obtained(vi)
If we takeκu=u, α=0, s=1, and p=0 in (20), then [28] (Theorem 3) is obtained(vii)
If we takeκu=u, α=0, m=s=1, and p=ω=0 in (20), then [36] (Theorem 2) is obtained(viii)
If we takeκu=u, α=0, s=1, and p=ω=0 in (20), then [30] (Theorem 2.1) is obtained(ix)
If we takeκu=u, α=0, m=s=1, τ=1, and p=ω=0 in (20), then classical Hadamard inequality is obtainedIn the following, we give another version of the Hadamard inequality for generalized fractional integral operators.Theorem 11.
Letη,κ:a,mb⊆0,∞⟶ℝ, 0<a<mb be the real-valued functions. If η be a integrable and s,m-convex and κ be a differentiable and strictly increasing. Then, for integral operators (9) and (10), the following inequalities hold:
(29)2sξαηκa+mκb2κζω¯2ν,κ−1κa+mκb/2+τκ−1mκb;p≤κYν,τ,l,ω¯2ν,κ−1κa+mκb/2+ρ,r,q,cηκ−1mκb;p+mτ+1κYν,τ,l,ω¯2mν,κ−1κa+mκb/2m−ρ,r,q,cηκ−1κam;p≤mκb−κaτ2τ+sηκaeακa+mηκbeακbEν,τ+1,lρ,r,q,cω;p+mηκbeακb+mηκa/m2eακa/m2∫01∞2−zszτ−1Eν,τ,lρ,r,q,cωzν;pdz,where ξα=eακb for α<0, ξα=eακa for α≥0 and ω¯=ω/mκb−κaν.Proof.
Sinceη is exponentially s,m-convex function on a,mb, for z∈0,1, we have the following inequalities:
(30)2sηκa+mκb2≤ηz/2κa+m2−z/2κbeαz/2κa+m2−z/2κb+mηz/2κb+2−z/2κa/meαz/2κb+2−z/2κa/m,(31)ηz2κa+m2−z2κb+mηz2κb+2−z2κam≤z2sηκaeακa+mηκbeακb+m2−z2sηκbeακb+mηκa/m2eακa/m2.Multiplying both sides of (30) with zτ−1Eν,τ,lρ,r,q,cωzν;p and integrating over 0,1, we have
(32)2sηκa+mκb2∫01zτ−1Eν,τ,lρ,r,q,cωzν;pdz≤∫01zτ−1Eν,τ,lρ,r,q,cωzν;pηz/2κa+m2−z/2κbeαz/2κa+m2−z/2κbdz+m∫01zτ−1Eν,τ,lρ,r,q,cωzν;pηz/2κb+2−z/2κa/meαz/2κb+2−z/2κa/mdz.Puttingκu=z/2κa+m2−z/2κb and κv=z/2κb+2−z/2κa/m in (32), we get
(33)2sηκa+mκb2∫κ−1κa+mκb/2κ−1mκbmκb−κuτ−1Eν,τ,lρ,r,q,c2νω¯mκb−κuν;pdκu≤∫κ−1κa+mκb/2κ−1mκbmκb−κuτ−1Eν,τ,lρ,r,q,c2νω¯mκb−κuν;pηκueακudκu+mτ+1∫κ−1κa/mκ−1κa+mκb/2mκv−κamτ−1Eν,τ,lρ,r,q,c2mνω¯κv−κamν;pηκveακvdκv.By using (9), (10), and (14), the first inequality of (29) is obtained.Now, multiplying both sides of (31) with zτ−1Eν,τ,lρ,r,q,cωzν;p and integrating over 0,1, we have
(34)∫01zτ−1Eν,τ,lρ,r,q,cωzν;pηz2κa+m2−z2κbdz+m∫01zτ−1Eν,τ,lρ,r,q,cωzν;pηz2κb+2−z2κamdz≤12sηκaeακa+mηκbeακb∫01zs+τ−1Eν,τ,lρ,r,q,cωzν;pdz+mηκbeακb+mηκa/m2eακa/m2∫012−zszτ−1Eν,τ,lρ,r,q,cωzν;pdzPuttingκu=z/2κa+m2−z/2κb and κv=z/2κb+2−z/2κa/m in (34), then by using (9), (10), and (27), the second inequality of (29) is obtained.The following remark establishes the connection with already published results.Remark 12.
(i)
If we takem=1 in (29), then Hadamard inequality for the exponentially s-convex function of second kind is obtained(ii)
If we takes=m=1 in (29), then Hadamard inequality for the exponentially convex function is obtained(iii)
If we takeκu=u, α=0, and s=1 in (29), then [32] (Theorem 2.11) is obtained(iv)
If we takeκu=u, α=0, s=1, and p=0 in (29), then [29] (Theorem 3.10) is obtained(v)
If we takeκu=u, α=0, m=s=1, and p=ω=0 in (29), then [37] (Theorem 4) is obtained(vi)
If we takeκu=u, α=0, m=s=1, τ=1, and p=ω=0 in (29), then classical Hadamard inequality is obtained


## 3. Fejér-Hadamard Inequalities

Here, we give two versions of the Fejér-Hadamard inequality.Theorem 13.
Letη,κ:a,mb⊆0,∞⟶ℝ, 0<a<mb be the real-valued functions. If η be a integrable, exponentially s,m-convex and ηκv/eακv=ηκa+mκb−mκv/eακa+mκb−mκv and κ be a differentiable and strictly increasing. Also, let γ:a,mb⟶ℝ be a function which is nonnegative and integrable. Then, for integral operators (9) and (10), the following inequalities hold:
(35)2sξαηκa+mκb2κYν,τ,l,ω¯mν,b−ρ,r,q,cγ∘κκ−1κam;p≤1+mκYν,τ,l,ω¯mν,b−ρ,r,q,cη∘κγ∘κκ−1κam;p≤mκb−κamτηκaeακa+mηκbeακb×∫01zs+τ−1Eν,τ,lρ,r,q,cωzν;pγ1−zκam+zκbdz+mηκbeακb+mηκa/m2eακa/m2∫011−zszτ−1Eν,τ,lρ,r,q,cωzν;pγ1−zκam+zκbdzwhere ξα=eακb for α<0, ξα=eακa for α≥0 and ω¯=ω/mκb−κaν.Proof.
Multiplying both sides of (21) with zτ−1Eν,τ,lρ,r,q,cωzν;pγ1−zκa/m+zκb and integrating over 0,1, we have
(36)2sηκa+mκb2∫01zτ−1Eν,τ,lρ,r,q,cωzν;pγ1−zκam+zκbdz≤∫01zτ−1Eν,τ,lρ,r,q,cωzν;pηzκa+m1−zκbeαzκa+m1−zκbγ1−zκam+zκbdz+m∫01zτ−1Eν,τ,lρ,r,q,cωzν;pη1−zκa/m+zκbeα1−zκa/m+zκbγ1−zκam+zκbdz.Puttingκv=1−zκa/m+zκb in (36), we get
(37)2sηκa+mκb2∫κ−1κa/mbκv−κamτ−1×Eν,τ,lρ,r,q,cmνω¯κv−κamν;pγκvdκv≤∫κ−1κa/mbκv−κamτ−1Eν,τ,lρ,r,q,cmνω¯κv−κamν;p×ηκa+mκb−mκveακa+mκb−mκvγκvdκv+m∫κ−1κa/mbκv−κamτ−1×Eν,τ,lρ,r,q,cmνω¯κv−κamν;pηκveακvγκvdκv.By using (10) and given condition ηκv/eακv=ηκa+mκb−mκv/eακa+mκb−mκv, the first inequality of (35) is obtained.Now, multiplying both sides of (22) with zτ−1Eν,τ,lρ,r,q,cωzν;pγ1−zκa/m+zκb and integrating over 0,1, we have
(38)∫01zτ−1Eν,τ,lρ,r,q,cωzν;pηzκa+m1−zκbγ1−zκam+zκbdz+m∫01zτ−1Eν,τ,lρ,r,q,cωzν;pη1−zκam+zκbγ1−zκam+zκbdz.≤ηκaeακa+mηκbeακb∫01zs+τ−1Eν,τ,lρ,r,q,cωzν;pγ1−zκam+zκbdz+mηκbeακb+mηκa/m2eακa/m2∫011−zszτ−1Eν,τ,lρ,r,q,cωzν;pγ1−zκam+zκbdz.From the above inequality, the second inequality of (35) can be achieved.The following remark establishes the connection with already published results.Remark 14.
(i)
If we takem=1 in (35), then Fejér-Hadamard inequality for the exponentially s-convex function of second kind is obtained(ii)
If we takem=s=1 in (35), then Fejér-Hadamard inequality for the exponentially convex function is obtained(iii)
If we takeκu=u, α=0, and m=s=1 in (35), then [35] (Theorem 2.2) is obtained(iv)
If we takeκu=u, α=0, and s=1 in (35), then [35] (Theorem 3.2) is obtained(v)
If we takeκu=u, α=0, m=s=1, and p=0 in (35), then [27] (Theorem 2.2) is obtained(vi)
If we takeκu=u, α=0, m=s=1, and p=ω=0 in (35), then [33] (Theorem 4) is obtained(vii)
If we takeκu=u, α=0, m=s=1, τ=1, and p=ω=0 in (35), then classical Fejér-Hadamard inequality is obtainedIn the following, we give another generalized fractional version of the Fejér-Hadamard inequality.Theorem 15.
Letη,κ:a,mb⊆0,∞⟶ℝ, 0<a<mb be the real-valued functions. If η be a integrable, exponentially s,m-convex and ηκv/eακv=ηκa+mκb−mκv/eακa+mκb−mκv and κ be a differentiable and strictly increasing. Also, let γ:a,mb⟶ℝ be a function which is nonnegative and integrable. Then, for integral operators (9) and (10), the following inequalities hold:
(39)2sξαηκa+mκb2κYν,τ,l,ω¯2mν,κ−1κa+mκb/2m−ρ,r,q,cγ∘κκ−1κam;p≤1+mκYν,τ,l,ω¯2mν,κ−1κa+mκb/2m−ρ,r,q,cη∘κγ∘κκ−1κam;p≤12smκb−κa2mτηκaeακa+mηκbeακb×∫01zs+τ−1Eν,τ,lρ,r,q,cωzν;pγz2κb+2−z2κamdz+mηκbeακb+mηκa/m2eακa/m2∫012−zszτ−1Eν,τ,lρ,r,q,cωzν;pγz2κb+2−z2κamdzwhere ξα=eακb for α<0, ξα=eακa for α≥0 and ω¯=ω/mκb−κaν.Proof.
Multiplying both sides of (30) with zτ−1Eν,τ,lρ,r,q,cωzν;pγz/2κb+2−z/2κa/m and integrating over 0,1, we have
(40)2sηκa+mκb2∫01zτ−1Eν,τ,lρ,r,q,cωzν;pγz2κb+2−z2κamdz≤∫01zτ−1Eν,τ,lρ,r,q,cωzν;pηz/2κa+m2−z/2κbeαz/2κa+m2−z/2κbγz2κb+2−z2κamdz+m∫01zτ−1Eν,τ,lρ,r,q,cωzν;pηz/2κb+2−z/2κa/meαz/2κb+2−z/2κa/mγz2κb+2−z2κamdz.
Puttingκv=z/2κb+2−z/2κa/m in (40), we get
(41)2sηκa+mκb2∫κ−1κa/mκ−1κa+mκb/2mκv−κamτ−1×Eν,τ,lρ,r,q,c2mνω¯κv−κamν;pγκvdκv≤∫κ−1κa/mκ−1κa+mκb/2mκv−κamτ−1Eν,τ,lρ,r,q,c2mνω¯κv−κamν;p×ηκa+mκb−mκveακa+mκb−mκvγκvdκv+m∫κ−1κa/mκ−1κa+mκb/2mκv−κamτ−1×Eν,τ,lρ,r,q,c2mνω¯κv−κamν;pηκveακvγκvdκv.
By using (10) and the given condition ηκv/eακv=ηκa+mκb−mκv/eακa+mκb−mκv, the first inequality of (39) is obtained.
Now, multiplying both sides of (31) with zτ−1Eν,τ,lρ,r,q,cωzν;pγz/2κb+2−z/2κa/m and integrating over 0,1, we have
(42)∫01zτ−1Eν,τ,lρ,r,q,cωzν;pηz2κa+m2−z2κbγz2κb+2−z2κamdz+m∫01zτ−1Eν,τ,lρ,r,q,cωzν;pηz2κb+2−z2κamγz2κb+2−z2κamdz.≤ηκaeακa+mηκbeακb∫01zs+τ−1Eν,τ,lρ,r,q,cωzν;pγz2κb+2−z2κamdz+mηκbeακb+mηκa/m2eακa/m2∫012−zszτ−1Eν,τ,lρ,r,q,cωzν;pγz2κb+2−z2κamdz.
From the above inequality, the second inequality of (39) can be achieved.Remark 16.
(i)
If we takem=1 in (39), then Fejér-Hadamard inequality for the exponentially s-convex function of second kind is obtained(ii)
If we takem=s=1 in (39), then Fejér-Hadamard inequality for the exponentially convex function is obtained


## 4. Concluding Remarks

We have established two generalized fractional versions of the Hadamard inequality. Also, two generalized fractional versions of the Fejér-Hadamard inequality are proved. For proving these fractional inequalities, we have utilized exponentiallys,m-convex functions as well as generalized integral operators containing Mittag-Leffler functions in their kernels. The presented results hold for various kinds of convexities and well-known fractional integral operators given in Remarks 6 and 8. Therefore, the reader can get a lot of desired fractional Hadamard and fractional Fejér-Hadamard inequalities from this paper.



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