# Some New Inequalities for p-Convex Functions via a K-Fractional Conformable Integral

Authors: Yan Dou; Muhammad Shoaib Saleem; Nimra Anwar; Haiping Gao
Journal: Journal of Mathematics
 (2022)

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## Abstract

The intention of this paper is to develop some new Hermite–Jensen–Mercer type inequalities forp−convex functions via k−fractional conformable integrals. Several existing results are also discussed which can be deduced from our results.


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## Body

## 1. Introduction

In the last few decades, the theory of convexity got a remarkable development due to its wide range of applications in optimization problems and other applied sciences [1, 2]. It has several important and interesting properties [3]. Because of its importance, several generalization and extensions have been established by a number of researchers [4]. The definition of a classical convex function is as given below:A functionμ : I ⊆ℝ⟶ℝ is said to be convex if(1)μτx+1−τy≤τμx+1−τy,holds for all x, y∈ℝ and τ∈0,1.Furthermore, the theory of convexity is interlinked to the theory of inequalities due to the nature of its definition having geometric interpretation. Several classical inequalities have been derived by many authors via convex functions, for example, see [5, 6]. One of them is Jensen’s inequality, see [7], which can be stated as follows.Definition 1 (See [26]).
Let 0<x1≤x2≤⋯≤xn and τ=τ1,τ2,…,τn be a nonnegative weight with ∑i=1nτi=1 and μ be a convex function on the interval ε,η. Then, the inequality(2)μ∑i=1nτixi≤∑i=1nτiμxi,holds for all xi∈ε,η and τi∈0,1i=1,2,…,n.Another important inequality for convex functions is the Hermite–Hadamard inequality [8] proposed by Hermite and Hadamard in 1883, which was published in 1893, and is considered as one of the fundamental tools in mathematical literature due to its vast applications.The Hermite–Hadamard inequality states that ifμ:I⊂ℝ⟶ℝ is a convex function defined on I and x,y∈I such that x<y, then(3)μx+y2≤1b−a∫xyμτdτ≤μx+μy2.It is also seldom called Hadamard’s inequality. A function is said to be convex if and only if it satisfies the Hermite–Hadamard inequality. Many new inequalities for convex functions have been established with the help of the Hermite–Hadamard inequality. These inequalities played an important role in both pure and applied mathematics. There are many recent studies which deal with several extensions, generalizations, and modifications of the Hermite–Hadamard inequality, for example, see [9, 10].In 2003, Mercer [11] proved the following variant of Jensen’s inequality.Theorem 1.
Ifμ:I⊂ℝ⟶ℝ is a convex function in ε,η, then(4)με+η−∑i=1nτixi≤με+μη−∑i=1nτiμxi,holds for all xi∈ε,η and μi∈0,1 with ∑i=1nxi=1i=1,2,…,n.It is known as the Jensen–Mercer inequality in literature. Recently, notable contributions have been made to the Jensen–Mercer inequality [12, 13].Fractional calculus is a branch of mathematical analysis which deals with the derivatives and integrals for a non-integer order and is as old as classical calculus. In the last forty years or so, it has become a rapidly growing area, mainly due to its extensively significant applications in widespread fields of engineering, biological, and physical sciences. Thousands of research papers have been published in this regard, for example, see [14, 15] and the references therein.Before we proceed further, let us have a brief review of previously known concepts which will be helpful in proving the results of this paper.


## 2. Preliminaries

A generalization of the convex function which is frequently used in convex analysis is thep−, p−convexity. In Reference [16], Zhang and Wan introduced the concept of p−, p− convexity as follows.Definition 2.
LetI∈0,∞ be a real interval and p∈ℝ∖0; then, I is said to be a p−, p−convex set, if(5)τxp+1−τyp1/p∈I,holds for all x, y∈I and τ∈0,1, where p=2k+1 or p=n/m, n=2r+1, m=2s+1, and k,r,s∈ℕ.Definition 3.
LetI∈0,∞ be a real interval and p∈ℝ∖0. A function μ:I⟶R is said to be a p−, p−convex function if(6)μτxp+1−τyp1/p≤1−τμx+τμy,holds for all x, y∈I and τ∈0,1.If the above inequality is reversed, thenμ is said to be p− concave. It can also be easily seen that for p=1, the p−, p−convexity reduces to ordinary convexity of functions.In References [17, 18], the Riemann–Liouville fractional integrals are defined as follows:Definition 4.
Letε,η be a finite interval on the real axis ℝ and μ∈Lε,η. The right- and left-sided Reimann–Liouville fractional integrals Iϵ+α and Iη−α of order α>0 with η>ε≥0 are defined by the following:(7)Iε+αμy=1Γα∫εyy−τα−1μτdτ,y>ε,and(8)Iη−αμy=1Γα∫yητ−yα−1μτdτ,y<η,respectively.Now, we define the Hadamard fractional integrals, which were introduced in [19].Definition 5.
Letα∈ℂ, Reα>0; then, the left- and right-sided Hadamard fractional integrals of order α are defined as follows:(9)Iε+αμy=1Γα∫εylny−lnτα−1μττdτ,and(10)Iη−αμy=1Γα∫yηlnτ−lnyα−1μττdτ,respectively.In [20], Katugampola introduced new fractional integral operators as follows.Definition 6.
Letε,η⊂ℝ be a finite interval. Then, the left- and right-sided Katugampola fractional integrals of order α>0 of μ∈XcPε,η are defined by the following:(11)Iβε+αμy=β1−αΓα∫εyτβ−1yβ−τβ1−αμτdτ,and(12)Iβη−αμy=β1−αΓα∫yητβ−1τβ−yβ1−αμτdτ,with ε<y<η and β>0.In [21], Jarad introduced the new fractional conformable integrals operators as follows,Definition 7.
Letα∈ℂ, Reα>0. Then, the left and right sided fractional conformable integral operators are defined by the following:(13)Iεβαμy=1Γα∫εyy−εα−τ−εααβ−1μττ−ε1−αdτ,and(14)Iβηαμy=1Γα∫yηη−yα−ε−τααβ−1μτη−τ1−αdτ,respectively.Remark 1.
(1)
The fractional integral (13) coincides with (7), (9), and (11), if we choose ε=0 , α=1, ε=0, α⟶0, and ε=0, respectively.(2)
The fractional integral (14) coincides with (8), (10), (12), if we choose η=0, α=1, η=0, α⟶0, and η=0, respectively.In [22], Qi introduced the generalized k-fractional conformable integrals which generalize the Riemann–Liouville fractional integrals [17], Hadamard fractional integrals [19], and Katugampola fractional integrals [20].Definition 8.
Letμ be a continuous function on a finite real interval ε,η. Then, the generalized left- and right-sided k−, k−fractional conformable integrals of order β∈ℂ, Reβ>0 are defined as follows:(15)Ikβε+α=1kΓkβ∫εyy−εα−τ−εααβ/k−1μττ−ε1−αdτ,and(16)Ikβη−α=1kΓkβ∫yηη−yα−η−τααβ/k−1μττ−ϵ1−αdτ,respectively.Diaz and Pariguan [23] established the k-generalized gamma function Γk, which is the generalization of the classical gamma function.Definition 9 (See [26]).
The notion of thek-gamma function is as given below:(17)Γkα=limt⟶∞τ!kττkα/k−1ατ,kk>0.
The integral from the abovementioned notion is given below as follows:(18)Γkα=∫0∞e−τk/kτα−1dτ,Reα>0,with αΓkk=Γkα+k, Γkk=1, ατ,k=Γkα+τk/Γkα.Definition 10.
The beta function is defined as follows:(19)βx,y=ΓxΓyΓx+y=∫01τx−11−τb−1dτ,x,y>0.Definition 11.
The hypergeometric function is defined as follows:(20)F21x,y;w;z=∫01τy−11−τw−y−11−zτ−xdτ,w>y>0,z<1.In this paper, by utilizing the Jensen–Mercer inequality, we derive the Hermite–Hadamard inequality forp−, p−convex functions via k−, k−fractional integral operators and establish some new inequalities related to the Hermite–Jensen–Mercer inequality.


## 3. New Hermite–Jensen–Mercer Type Inequalities viaK-Fractional Integrals

Theorem 2.
Letα,β>0 and μ:ε,η⟶ℝ be a p−, p−convex mapping. Then, the following fractional integral inequality holds(21)μεp+ηp−xp+yp21/p≤2αβ/k−1αβ/kΓkβ+kyp−xpαβ/kIkβεp+ηp−xp+yp/2+αμ°Ωεp+ηp−xp+Ikβεp+ηp−xp+yp/2−αμ°Ωεp+ηp−yp≤με+μη−μx+μy2,where Ωs=s1/p, s∈εp,ηp.Proof.
Sinceμ is a p−, p−convex function, we can write it as follows:(22)μεp+ηp−x1p+y1p21/p=μεp+ηp−x1p+εp+ηp−y1p21/p≤μεp+ηp−x1p1/p+μεp+ηp−y1p1/p2,for all x1,y1∈εp,ηp.
We considerx1p=τ/2xp+2−τ/2yp and y1p=2−τ/2xp+τ/2yp. Then, for x,y∈εp,ηp and τ∈0,1, we have the following:(23)2μεp+ηp−xp+yp21/p≤μεp+ηp−τ2xp+2−τ2yp1/p+μεp+ηp−2−τ2xp+τ2yp1/p.
Multiplying both sides of the abovementioned equation with1−1−τα/αβ/k−11−τα and then integrating w.r.t τ over the interval 0,1 and unliving k-fractional integral operators gives the following:(24)2μεp+ηp−xp+yp21/p∫011−1−τααβ/k−11−ταdτ≤∫011−1−τααβ/k−11−τα×μεp+ηp−τ2xp+2−τ2yp1/p+μεp+ηp−2−τ2xp+τ2yp1/pdτ.
By settingτ1=εp+ηp−τ/2xp+2−τ/2yp and τ2=εp+ηp−2−τ/2xp+τ/2yp, we have as follows:(25)=∫εp+ηp−ypxp+yp−xp+yp/21−εp+ηp−xp+yp/2−τ1/yp−xp/2ααβ/k−1εp+ηp−xp+yp/2−τ1yp−xp/2α−1μτ11/p2yp−xpdτ1+∫xp+yp−xp+yp/2εp+ηp−xp1−τ2−εp+ηp−xp+yp/2/yp−xp/2ααβ/k−1×τ2−εp+ηp−xp+yp/2yp−xp/2α−1μτ21/p2yp−xpdτ2.
Note that(26)∫011−1−τααβ/k−11−τα−1dτ=1β/kαβ/k.
Therefore,(27)2μεp+ηp−xp+yp21/p1β/kαβ/k≤2yp−xpαβ/kkΓkβIkβεp+ηp−xp+yp/2+αμ°Ωεp+ηp−xp+kΓkβIkβεp+ηp−xp+yp/2−αμ°Ωεp+ηp−yp.
So, the first inequality of (21) is proved.
Now, for the proof of the second inequality of (21), we first note that μ is a p−, p−convex function; then, for τ∈0,1, we can write as follows:(28)μεp+ηp−τ2xp+2−τ2yp1/p≤με+μη−τ2μx+2−τ2μy,and(29)μεp+ηp−2−τ2xp+τ2yp1/p≤με+μη−2−τ2μx+τ2μy.
By adding equations (28) and (29), we have as follows:(30)μεp+ηp−τ2xp+2−τ2yp1/p+μεp+ηp−2−τ2xp+τ2yp1/p≤2με+μη−μx+μy.
Multiplying both sides of the abovementioned equation with1−1−τα/αβ/k−11−τα and then integrating w.r.t τ over 0,1 and utilizing k-fractional integral operator gives the following:(31)∫011−1−τααβ/k−11−τα×μεp+ηp−τ2xp+2−τ2yp1/p+μεp+ηp−2−τ2xp+τ2yp1/pdτ≤2με+μη−μx+μy∫011−1−τααβ/k−11−ταdτ,2yp−xpαβ/kkΓkβkβIεp+ηp−xp+yp/2+αμ°Ωεp+ηp−xp+kΓkβIkβεp+ηp−xp+yp/2−αμ°Ωεp+ηp−yp≤1β/kαβ/k2με+μη−μx+μy,which completes the proof of (21).Remark 2.
Theorem2 leads to the following conclusions,(1)
If we setk = 1, x =  ε, y =  η, and p=1, then Theorem 2 coincides with Theorem 2.1 of [24].(2)
If we setα  = k = p = 1, x=ε, and y=η, then Theorem 2 coincides with Theorem 2 of [25].(3)
If we setp=1, then Theorem 2 coincides with Theorem 2.1 of [26].Theorem 3.
Letα,β>0 and μ:ε,η⟶R be a p-convex function. Then, the following fractional integral inequality holds,(32)μεp+ηp−xp+yp21/p≤με+μη−αβ/kΓkβ+k2yp−xpαβ/kIkβxp+αμ°Ωyp+Ikβyp−αμ°Ωxp≤με+μη−μx+y2,where Ωs=s1/p, s∈εp,ηp.Proof.
It follows from the Jensen–Mercer inequality that(33)μεP+ηp−x1p+y1p21/p≤με+μη−μx1+μy12,for all x1, y1∈εp,ηp. Now, by a change of variable x1p=τxp+1−τyp and y1=1−τxp+τyp for x, y∈εp,ηp and τ∈0,1 in the above inequality, we obtain the following:(34)μεP+ηp−xp+yp21/p≤με+μη−μτxp+1−τyp1/p+μ1−τxp+τyp1/p2.
Multiplying both sides of the abovementioned equation with1−1−τα/αβ/k−11−τα and then integrating w.r.t τ over the interval 0,1 and utilizing the fractional integral operator gives the following:(35)μεp+ηp−xp+yp21/p∫011−1−τααβ/k−11−ταdτ≤∫011−1−τααβ/k−11−τα×με+μη−μτxp+1−τyp1/p+μ1−τxp+τyp1/p2dτ.
By settingτ1=τxp+1−τyp and τ2=1−τxp+τyp, we have as follows:(36)μεp+ηp−xp+yp21/p∫011−1−τααβ/k−11−ταdτ≤με+μη∫011−1−τααβ/k−11−ταdτ×∫xpyp1−τ1−xp/yp−xpααβ/k−1τ1−xpyp−xpα−1μτ11/pdτ1+∫xpyp1−yp−τ2/yp−xpααβ/k−1yp−τ2yp−xpα−1μτ21/pdτ2.
By using fractional integral operators, we have as follows:(37)μεp+ηp−xp+yp21/p≤με+μη−αβ/kΓkβ+k2yp−xpαβ/k×Ikβxp+αμ°Ωyp+Ikβyp−αμ°Ωxp,which completes the proof of the first inequality of (32).
To prove the second inequality, we use thep-convexity of μ; for τ∈0,1, we obtain(38)μxp+yp21/p=μτxp+1−τyp+1−τxp+τyp21/p≤μτxp+1−τyp1/p+μ1−τxp+τyp1/p2.
Multiplying both sides of the abovementioned equation by1−1−τα/αβ/k−11−τα and then integrating with respect to τ over 0,1, we have the following:(39)μxp+yp21/p∫011−1−τααβ/k−11−ταdτ≤∫011−1−τααβ/k−11−ταμτxp+1−τyp1/p+μ1−τxp+τyp1/p2dτ.
By substitution of thek-fractional integral operators as discussed above, we have the following:(40)μxp+yp21/p≤αβ/kΓkβ+k2yp−xpαβ/kIkβxp+αμ°Ωyp+Ikβyp−αμ°Ωxp,−μxp+yp21/p≥−αβ/kΓkβ+k2yp−xpαβ/kIkβxp+αμ°Ωyp+Ikβyp−αμ°Ωxp.
Addingμε+μη on both sides of the abovementioned equation, we obtain as follows:(41)με+μη−μxp+yp21/p≥με+μη−αβ/kΓkβ+k2yp−xpαβ/k×Ikβxp+αμ°Ωyp+Ikβyp−αμ°Ωxp.
Combining (37) and (41), we obtain (32).Remark 3.
By assumingp=1, Theorem 3 reduces to the inequality (13) of Theorem 2.3 in [26].Corollary 1.
Settingα  =  β  = k = p = 1 in Theorem 3 gives the following result:(42)με+η−x+y2≤με+μη−∫01μτx+1−τydτ≤με+μη−μx+y2.Theorem 4.
Letα,β>0 and μ:ε,η⟶R be a p-convex function. Then, the following fractional integral inequality holds,(43)μεp+ηp−xp+yp21/p≤αβ/kΓkβ+k2yp−xpαβ/kIkβεp+ηp−xp+αμ°Ωεp+ηp−yp+Ikβεp+ηp−yp−αμ°Ωεp+ηp−xp≤με+μη−μx+μy2,where Ωs=s1/p, s. ∈εp,ηp.Proof.
Sinceμ is a p-convex function, for τ∈0,1, we have the following:(44)μεp+ηp−x1p+y1p21/p=μεp+ηp−x1p+εp+ηp−y1p21/p≤μεp+ηp−x1p1/p+μεp+ηp−y1p1/p2,for all x1,y1∈εp,ηp.
Letx1p=τxp+1−τyp and y1=1−τxp+τyp. Then,(45)2μεp+ηp−xp+yp21/p∫011−1−τααβ/k−11−ταdτ≤∫011−1−τααβ/k−11−τα×μεp+ηp−τxp+1−τyp1/p+μεp+ηp−1−τxp+τyp1/pdτ.
Puttingτ1=εp+ηp−τxp+1−τyp and τ2=εp+ηp−1−τxp+τyp and solving the above inequality, we obtain the following:(46)=∫εp+ηp−ypxp+yp−xp1−εp+ηp−xp−τ1/yp−xp/2ααβ/k−1⋅εp+ηp−xp−τ1yp−xp/2α−1μτ11/p2yp−xpdτ1+∫εp+ηp−ypεp+ηp−xp1−τ2−εp+ηp−yp/yp−xp/2ααβ/k−1⋅τ2−εp+ηp−ypyp−xp/2α−1μτ21/p2yp−xpdτ2.
By usingk-fractional integral operators and (26), we have the following:(47)μεp+ηp−xp+yp21/p≤αβ/kΓkβ+k2yp−xpαβ/kIkβεp+ηp−yp+αμ°Ωεp+ηp−xp+Ikβεp+ηp−xp−αμ°Ωεp+ηp−yp.
This completes the first part of the inequality (43). To prove the second inequality, we use the p-convexity of μ.(48)μεp+ηp−τxp+1−τyp1/p≤τμεp+ηp−xp1/p+1−τμεp+ηp−yp1/p,(49)μεp+ηp−1−τxp+τyp1/p≤1−τμεp+ηp−xp1/p+τμεp+ηp−yp1/p.
Adding (48) and (49), and then using the Jensen–Mercer inequality, we obtain the following:(50)μεp+ηp−τxp+1−τyp1/p+μεp+ηp−1−τxp+τyp1/p≤μεp+ηp−xp1/p+μεp+ηp−yp1/p≤2με+μη−μx+μy.
Multiplying both sides of the abovementioned equation by1−1−τα/αβ/k−11−τα and then integrating with respect to τ over 0,1, we have the following:(51)∫011−1−τααβ/k−11−τα×μεp+ηp−τxp+1−τyp1/p+μεp+ηp−1−τxp+τyp1/pdτ≤με+μη−μx+μy∫011−1−τααβ/k−11−ταdτ,αβ/kΓkβ+k2yp−xpαβ/kIkβεp+ηp−yp+αμ°Ωεp+ηp−xp+Ikβεp+ηp−xp−αμ°Ωεp+ηp−yp≤με+μη−μx+μy2.
Combining (48) and (51) leads to (43).Remark 4.
By assumingp=1, Theorem 4 reduces to the inequality (14) of Theorem 2.3 in [26].Corollary 2.
By assumingp =  α  =  β  = k = 1, Theorem 4 reduces to the following inequality:(52)μμ+η−x+y2≤1y−x−∫xyμε+η−τdτ≤με+μη−μx+μy2.Lemma 1.
Letα,β>0, ε<η and μ:ε,η⟶ℝ be a differential mapping, such that μ′∈ε,η. Then, we have the following:(53)2αβ/k−1αβ/kΓkβ+kyp−xpαβ/k×Ikβεp+ηp−xp+yp/2+αμ°Ωεp+ηp−xp+Ikβεp+ηp−xp+yp/2−αμ°Ωεp+ηp−yp−μεp+ηp−xp+yp21/p=yp−xpαβ/k4p∫011−1−τααβ/kAτ1/p−1μ′εp+ηp−2−τ2xp+τ2yp1/p−Bτ1/p−1μ′εp+ηp−τ2xp+2−τ2yp1/pdτ,where Aτ=εp+ηp−2−τ/2xp+τ/2yp, Bτ=εp+ηp−τ/2xp+2−τ/2yp, and Ωs=s1/p, s∈εp,ηp.Proof.
Let(54)I=yp−xpαβ/k4I1−I2,where(55)I1=∫011−1−τααβ/kAτ1/p−1μ′εp+ηp−2−τ2xp+τ2yp1/pdτ,I2=∫011−1−τααβ/kBτ1/p−1μ′εp+ηp−τ2xp+2−τ2yp1/pdτ.
Now, using integration by parts,(56)I1=−2pyp−xp1−1−τααβ/kμεp+ηp−2−τ2xp+τ2yp1/p01+2pβ/kyp−xp∫011−1−τααβ/k−11−τα−1μεp+ηp−2−τ2xp+τ2yp1/pdτ,I1=−2pαβ/kyp−xpμεp+ηp−xp+yp21/p+2β/k+1pΓkβ+kyp−xpαβ/k+1Ikβεp+ηp−xp+yp/2+αμ°Ωεp+ηp−xp.
Similarly, we have the following:(57)I2=∫011−1−τααβ/kBτ1/p−1μ′εp+ηp−τ2x+2−τ21/pdτ=2pαβ/kyp−xpμεp+ηp−xp+yp21/p−2β/k+1pΓkβ+kyp−xpαβ/k+1Ikβεp+ηp−xp+yp/2−αμ°Ωεp+ηp−yp.
Therefore, by substituting values ofI1 and I2 in (54), we will obtain inequality (53).Remark 5.
We can deduce the following conclusions from Lemma1:(1)
If we substitutek = p = 1, x =  ϵ, and y =  η, then Lemma 1 converts into Lemma 3.1 of [24].(2)
If we substitutek = p =  α  = 1, x =  ϵ, and y =  η, then Lemma 1 converts into Lemma 3 of [27].(3)
If we takep = 1, then Lemma 1 coincides with Lemma 2.5 of [26].Theorem 5.
Assume thatq>1 and α, β>0 and ε<η, and μ:ε,η⟶ℝ is a differential function, where μ′∈Lε,η and μ′q is p-convex, then(58)2αβ/k−1αβ/kΓkβ+kyp−xpαβ/k×Ikβεp+ηp−xp+yp/2+αμ°Ωεp+ηp−xp+Ikβεp+ηp−xp+yp/2−αμ°Ωεp+ηp−yp−μεp+ηp−xp+yp/21/p≤yp−xpαβ/k4p+εp+ηp−xp1/p−1F11−1p,1;2;yp−xp2εp+ηp−xp1−1/p×μ′εq+μ′ηq1αβ/k+1Bβk+1,1α−μ′xq12αβ/k+1Bβk+1,1α+Bβk+1,2α−μ′yq12αβ/k+1Bβk+1,2α−Bβk+1,1α1/q+εp+ηp−yp1/p−1F11−1p,1;2;xp−yp2εp+ηp−yp1−1/q×μ′εq+μ′ηq1αβ/k+1Bβk+1,1α−μ′xq12αβ/k+1Bβk+1,2α−Bβk+1,1α−μ′yq12αβ/k+1Bβk+1,1α+Bβk+1,2α1/q,holds for all x, y∈εp,ηp.Proof.
By using Lemma1, the Jensen–Mercer inequality, modulus property, Holder’s inequality, and p-convexity of μ′q, we have the following:(59)2αβ/k−1αβ/kΓkβ+kyp−xpαβ/k×Ikβεp+ηp−xp+yp/2+αμ°Ωεp+ηp−xp+Ikβεp+ηp−xp+yp/2−αμ°Ωεp+ηp−yp−μεp+ηp−xp+yp21/p=yp−xpαβ/k4p∫011−1−τααβ/k×Aτ1/p−1μ′εp+ηp−2−τ2xp+τ2yp1/p−Bτ1/p−1μ′εp+ηp−τ2xp+2−τ2yp1/pdτ≤yp−xpαβ/k4p∫01Aτ1/p−1dτ1−1/q×∫011−1−τααβ/kμ′εp+ηp−2−τ2xp+τ2yp1/pqdτ1/q+∫01Aτ1/p−1dτ1−1/q×∫011−1−τααβ/kμ′εp+ηp−τ2xp+2−τ2yp1/pqdτ1/q.
Therefore, after some simple calculations, we obtain (58).


## 4. Conclusion

One of the most important tools to deal with the problems of pure and applied mathematics, particularly, the optimization theory is convex functions. In this paper, we studiedp−, p−convex functions and established several inequalities in the setting of k−, k−fractional integral for p− convex functions. The inequalities established in this paper can be considered as generalizations and extensions of several existing results. It is worthy to mention here that by taking k=1, the results of this paper reduce to the similar results for the usual fractional integral.



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