# Some Trapezium-Like Inequalities Involving Functions Having Stronglyn-Polynomial Preinvexity Property of Higher Order

Authors: Muhammad Uzair Awan; Sadia Talib; Muhammad Aslam Noor; Yu-Ming Chu; Khalida Inayat Noor
Journal: Journal of Function Spaces
 (2020)

---

## Abstract

The main objective of this paper is to introduce a new class of preinvex functions which is called asn-polynomial preinvex functions of a higher order. As applications of this class of functions, we discuss several new variants of trapezium-like inequalities. In order to obtain the main results of the paper, we use the concepts and techniques of k-fractional calculus. We also discuss some special cases of the obtained results which show that the main results of the paper are quite unifying one.


---

## Body

## 1. Introduction

Convexity is one of the most important and natural notions in mathematics; it plays a significant role in various branches of pure and applied sciences [1–15]. For example, the set of feasible points in optimization theory is convex; the loss function used to measure the quality of solution in statistics is convex. In particular, many remarkable inequalities have been established via the convexity theory [16–32].Recently, the classical concept of convexity has been extended and generalized in different directions. For instance, Hanson [33] introduced the notion of differentiable invex function but did not name it as invex; Craven [34] gave the term invex for this class of functions due to their property described as invariance by convexity. Mititelu [35] introduced the notion of invex set as follows.Definition 1 (see [35]).
LetX∈ℝ be a nonempty set and ζ:X×X↦ℝ be a real-valued function. Then X is said to be invex with respect to ζ if
(1)x+tζy,x∈Xfor all x,y∈X and t∈0,1.
We clearly see that the invexity reduces to the classical convexity ifζy,x=y−x. Therefore, every convex set is an invex with respect to ζy,x=y−x, but its converse is not true in general [35].
Weir and Mond [36] introduced the class of preinvex functions by use of the invex set.Definition 2 (see [36]).
LetX∈ℝ be a nonempty invex set with respect to ζ:X×X↦ℝ. Then the function F:X↦ℝ is said to be preinvex with respect to ζ if the inequality
(2)Fx+tζy,x≤1−tFx+tFyholds for all x,y∈X and t∈0,1.
Note that the preinvex function becomes the classical convex function ifζy,x=y−x; for more details regarding recent study on preinvexity property, we recommend the literature [37].
Very recently, Toplu et al. [38] introduced and investigated a new class of convexity named n-polynomial convexity, and Karamardian [39] and Polyak [40] independently introduced the class of strongly convex functions. Strong convexity is the strengthening of convexity.Definition 3 (see [39, 40]).
A functionF:X↦ℝ is said to be strongly convex with respect to modulus μ>0 if the inequality
(3)F1−tx+ty≤1−tFx+tFy−μt1−ty−x2holds for all x,y∈X and t∈0,1.
Lin et al. [41] introduced higher order strongly convex functions to simplify the study of mathematical programs with equilibrium constraints.Definition 4 (see [41]).
Letσ,μ>0. Then, the function F:X↦ℝ is said to be σ-order strongly convex with respect to modulus μ if
(4)F1−tx+ty≤1−tFx+tFy−μt1−ty−xσfor all x,y∈X and t∈0,1.
Ifσ=2, then Definition 4 becomes Definition 3. Therefore, higher order strong convexity is a generalization of strong convexity. Lin et al. [41] proved that the higher order strong convexity of a function is equivalent to higher order strong monotonicity of the gradient map of the function.
It is well known that the Hermite-Hadamard inequality [42–45] is one of the most important and classical inequalities in convex function theory, which can be stated as follows.Theorem 5.
LetF:a,b⊂ℝ↦ℝ be a convex function. Then
(5)Fa+b2≤1b−a∫abFxdx≤Fa+Fb2.
In recent decades, the fractional calculus has become a powerful tool in numerous branches of mathematics, physics, and engineering. The history of fractional calculus dates back to 1695 with the work of mathematicians such as L’Hospital and Leibniz, but the logical definitions were proposed by Liouville in 1834, Riemann in 1847, and Grünwald in 1867. Fractional calculus can be considered a super set of integer-order calculus, which has the potential to accomplish what integer-order calculus cannot. The classical form of the fractional calculus is given by the Riemann-Liouville integrals as follows.Definition 6 (see [46]).
Letα>0, 0≤a<b, and F∈L1a,b. Then, the α-order Riemann-Liouville integrals Ja+αF and Jb−αF are defined by
(6)Ja+αFx=1Γα∫axx−tα−1Ftdtx>a,Jb−αFx=1Γα∫xbt−xα−1Ftdtx<b,respectively, where
(7)Γα=∫0∞e−ttα−1dtis the Euler gamma function.
Sarikaya et al. [47] used the fractional calculus to obtain new variants of the Hermite-Hadamard inequality, which opened a new research area for the people who are working on the field of mathematical inequalities, particularly working on the inequalities involving convexity and its generalizations. For some more recent research work done in this direction, see [48–50].
Diaz et al. [51] introduced the generalized k-gamma function Γkx and k-beta function Bkx,y as follows.
(8)Γkx=∫0∞tx−1e−tk/kdt,Bkx,y=1k∫01tx/k−11−ty/k−1dt=ΓkxΓkyΓkx+y.
Making use of the generalizedk-gamma function, Sarikaya et al. [52] introduced the k-Riemann-Liouville fractional integrals and discussed their properties and applications.Definition 7 (see [52]).
Letα,k>0, 0≤a<b, and F∈L1a,b. Then, the α-order k-Riemann-Liouville integrals Jka+αF and Jkb−αF are defined by
(9)Jka+αFx=1kΓkα∫axx−ta/k−1Ftdtx>a,Jkb−αFx=1kΓkα∫xbt−xa/k−1Ftdtx<b.
In recent years, several authors have used the concepts ofk-fractional calculus in obtaining new variants of fractional analogues of classical inequalities. For example, Huang et al. [53] obtained k-fractional conformable analogues of Hermite-Hadamard’s inequality. Rahman et al. [54] obtained fractional analogues of the Gruss type inequalities using k-conformable fractional integral operators.
The aim of the article is to obtain some newk-analogues of trapezium-like inequalities involving a new class of functions called strongly n-polynomial preinvex function of higher order. We also discuss some special cases of the obtained results. We expect that the ideas and techniques of this article will inspire interested readers working in this field. This is the main motivation of the article.


## 2. Results and Discussions

In this section, we discuss our main results. First of all, let us give the definition of stronglyn-polynomial preinvex function of higher order.Definition 8.
Letn∈ℕ and X∈ℝ be a nonempty invex set with respect to ζ:X×X↦ℝ. A nonnegative function F:X↦ℝ is said to be strongly n-polynomial preinvex function of higher order, if
(10)Fa+tζb,a≤1n∑i=1n1−tiFa+1n∑i=1n1−1−tiFb−μtσ1−t+t1−tσζb,aσ,∀a,b∈X,t∈0,1,μ,σ>0,
Note that ifσ=2, then the class of strongly n-polynomial preinvex function of higher order reduces to the class of strongly n-polynomial preinvex function which is also new class. If we consider μ=0, then the class of strongly n-polynomial preinvex function of higher order reduces to the class of n-polynomial preinvex functions which is also new in the literature. Similarly, if we take ζb,a=b−a, then we have new class of strongly n-polynomial convex function of higher order. If we take n=1, then the above class reduces to simple strongly preinvex function of higher order. If we take n=2, then we have strongly 2-polynomial preinvex function of higher order:
(11)Fa+tζb,a≤2−t−t22Fa+3t−t22Fb−μtσ1−t+t1−tσζb,aσ,∀a,b∈X,t∈0,1,μ,σ>0,
In order to obtain following result, we need the famous ConditionC which was introduced by Mohan and Neogy [55].Condition C.
LetX⊂ℝ be an invex set with respect to bifunction ζ⋅,⋅. Then, for any x,y∈X and t∈0,1,
(12)ζy,y+tζx,y=−tζx,y,ζx,y+tζx,y=1−tζx,y.
Note that for everyx,y∈X, t1,t2∈0,1, and from Condition C, we have
(13)ζy+t2ζx,y,y+t1ζx,y=t2−t1ζx,y.
We now give our first main result.Theorem 9.
LetF:a,a+ζb,a↦ℝ be an strongly n-polynomial preinvex function with ζb,a>0 and F∈La,a+ζb,a. If ζ⋅,⋅ satisfies Condition C, then
(14)nn+2−n−1F2a+ζb,a2+μα2−αk+2k4α+kα+2kζb,aσ≤Γkα+kζα/kb,aJka+αFa+ζb,a+Jka+ζb,a−αFa≤Fa+Fbn∑i=1n2skα+sk−2kαμζb,aσα+kα+2k.Proof.
Since it is given thatF is an an strongly n-polynomial preinvex function and ζ⋅,⋅ satisfies Condition C, then
(15)F2a+ζb,a2≤1n∑i=1n1−12iFa+tζb,a+Fa+1−tζb,a−μ4ζ2a+1−tζb,a,a+tζb,a=1n∑i=1n1−12iFa+tζb,a+Fa+1−tζb,a−μ41−2t2ζ2b,a.
Multiplying the above inequality bytα/k−1 and then integrating the above inequality with respect to t on 0,1 yields
(16)nn+2−n−1kαF2a+ζb,a2+μkα2−αk+2k4αα+kα+2kζ2b,a≤∫01ta/k−1Fa+tζb,adt+∫01ta/k−1Fa+1−tζb,adt=I1+I2.
Now(17)I1=∫aa+ζb,au−aζb,aa/k−1Fuduζb,a=kΓkαζα/kb,aJka+ζb,a−αFa.
Similarly(18)I2=kΓkαζα/kb,aJka+αFa+ζb,a.
This imples(19)nn+2−n−1F2a+ζb,a2+μα2−αk+2k4α+kα+2kζ2b,a≤Γkα+kζα/kb,aJka+αFa+ζb,a+Jka+ζb,a−αFa.
For the proof of right-hand side inequality,(20)Fa+tζb,a+Fa+1−tζb,a≤Fa+Fb1n∑i=1n1−ti+1n∑i=1n1−1−ti−2μt1−tζ2b,a.
Multiplying the above inequality bytα/k−1, then integrating the resulting inequality with respect to t over 0,1, we obtain
(21)kΓkαζα/kb,aJka+αFa+ζb,a+Jka+ζb,a−αFa≤Fa+Fbn∑i=1n∫01ta/k−12−ti−1−tidt−2k2μζ2b,aα+kα+2k.
This implies that(22)Γkα+kζα/kb,aJka+αFa+ζb,a+Jka+ζb,a−αFa≤Fa+Fbn∑i=1n2skα+sk−2kαμζ2b,aα+kα+2k.
Combining (19) and (22) completes the proof.Note that if we takeα=k=n=1 in Theorem 9, then we get the Hermite-Hadamard like inequality involving strongly preinvex functions.We now derive a new auxiliary result which will be helpful in obtaining the next results of the article.Lemma 10.
LetF:a,a+ζb,a↦ℝ be a differentiable function and F′∈a,a+ζb,a. Then, for any 0<α≤1,0<λ≤1, the following equality for k-fractional integrals holds:
(23)k+α1−λFa+ζb,a+k−α1−λFa2k−Γkα+k2ζα/kb,aJka+αFa+ζb,a+Jka+ζb,a−αFa=ζb,a2∫01ta/k+αk1−λ−1−ta/kF′a+tζb,adt.Proof.
It suffices to show that(24)J=ζb,a2∫01ta/k+αk1−λ−1−ta/kF′a+tζb,adt=ζb,a2∫01ta/k+αk1−λF′a+tζb,adt−∫011−ta/kF′a+tζb,adt=ζb,a2J1+J2.
Integrating by parts(25)J1=∫01ta/k+αk1−λF′a+tζb,adt=α1−λ+kFa+ζb,a−α1−λFakζb,a−αkζb,a∫01ta/kFa+tζb,adt=α1−λ+kFa+ζb,a−α1−λFakζb,a−αkζα/k+1b,a∫aa+ζb,au−aa/k−1Fudu=α1−λ+kFa+ζb,a−α1−λFakζb,a−Γkα+kζa/k+1b,akJa+ζb,a−αFa.
Similarly,(26)J2=−∫011−ta/kF′a+tζb,adt=Faζb,a−αkζb,a∫011−ta/kFa+tζb,adt=Faζb,a−αkζa/k+1b,a∫aa+ζb,aa+ζb,a−ua/k−1Fudu=Faζb,a−Γkα+kζa/k+1b,akJa+αFa+ζb,a.
Using (25) and (26) in (24) completes the proof.Theorem 11.
LetF:a,a+ζb,a↦ℝ be a differentiable function on a,a+ζb,a with ζb,a>0 and F′∈La,a+ζb,a. If F′ is a strongly n-polynomial preinvex function of higher order, then
(27)k+α1−λFa+ζb,a+k−α1−λFa2k−Γkα+k2ζα/kb,aJka+αFa+ζb,a+Jka+ζb,a−αFa≤ζb,a2nF′a∑i=1nM1+F′b∑i=1nM2−μζb,aσM3,where
(28)M1=αi1−λki+1+kBkα+k,sk+k−kα+sk+k,M2=kα+sk+k−kBksk+k,α+k+iα1−λki+1,M3=2α1−λkσ+1σ+2−kBkσk+k,α+k+kBkα+2k,σk+k+kBkσk+2k,α+k.Proof.
Using Lemma10 and the fact that F′ is a strongly n-polynomial preinvex function of higher order, we have
(29)k+α1−λFa+ζb,a+k−α1−λFa2k−Γkα+k2ζa/kb,aJka+αFa+ζb,a+Jka+ζb,a−αFa=ζb,a2∫01ta/k+αk1−λ−1−ta/kF′a+tζb,adt≤ζb,a2∫01ta/k+αk1−λ−1−ta/kF′a+tζb,adt≤ζb,a2∫01ta/k+αk1−λ−1−ta/k1n∑i=1n1−tiF′a+1n∑i=1n1−1−tiF′b−μζb,aσtσ1−t+t1−tσdt=ζb,a2nF′a∫01ta/k+αk1−λ−1−ta/k∑i=1n1−tidt+F′b∫01ta/k+αk1−λ−1−ta/k∑i=1n1−1−tidt−μζb,aσ∫01ta/k+αk1−λ−1−ta/ktσ1−t+t1−tσdt=ζb,a2nF′a∑i=1n∫01ta/k+αk1−λ−1−ta/k1−tidt+F′b∑i=1n∫01ta/k+αk1−λ−1−ta/k1−1−tidt−μζb,aσ∫01ta/k+αk1−λ−1−ta/ktσ1−t+t1−tσdt=ζb,a2nF′a∑i=1nM1+F′b∑i=1nM2−μζb,aσM3.
This completes the proof.Theorem 12.
LetF:a,a+ζb,a↦ℝ be a differentiable function on a,a+ζb,a with ζb,a>0, p−1+q−1=1 and F′∈La,a+ζb,a. If F′q is strongly n-polynomial preinvex function of higher order, then
(30)k+α1−λFa+ζb,a+k−α1−λFa2k−Γkα+k2ζa/kb,aJka+αFa+ζb,a+Jka+ζb,a−αFa≤ζb,a2M42n∑i=1nii+1AF′aq,F′bq−2μζb,aσσ+1σ+21/q,where
(31)M4=α1−λk+2kαp+k1/p,and A⋅,⋅ is the arithmetic mean.Proof.
Using Lemma10, Hölder’s integral inequality and F′q is strongly n-polynomial preinvex function of higher order, we have
(32)k+α1−λFa+ζb,a+k−α1−λFa2k−Γkα+k2ζa/kb,aJka+αFa+ζb,a+Jka+ζb,a−αFa=ζb,a2∫01ta/k+αk1−λ−1−ta/kF′a+tζb,adt=ζb,a2∫01ta/k+αk1−λF′a+tζb,adt+∫011−ta/kF′a+tζb,adt≤ζb,a2∫01ta/k+αk1−λpdt1/p∫01F′a+tζb,aqdt1/q+∫011−tap/kdt1/p∫01F′a+tζb,aqdt1/q≤ζb,a2∫01tap/kdt1/p+∫01+αpkp1−λpdt1/p+∫011−tap/kdt1/p×F′aqn∑i=1n∫011−tidt+F′bqn∑i=1n∫011−1−tidt−μζb,aσ∫01tσ1−t+t1−tσdt=ζb,a2M4F′aqn∑i=1nii+1+F′bqn∑i=1nii+1−2μζb,aσσ+1σ+2=ζb,a2M42n∑i=1nii+1AF′aq,F′bq−2μζb,aσσ+1σ+21/q.
This completes the proof.Theorem 13.
LetF:a,a+ζb,a↦ℝ be a differentiable function on a,a+ζb,a with ζb,a>0, q>1 and F′∈La,a+ζb,a. If F′q is strongly n-polynomial preinvex function of higher order, then
(33)k+α1−λFa+ζb,a+k−α1−λFa2k−Γkα+k2ζα/kb,aJka+αFa+ζb,a+Jka+ζb,a−αFa≤ζb,a2α1−λk1−1/qF′aqn∑i=1nM1+F′bqn∑i=1nM2−μζb,aσM3,where M1,M2, and M3 are given in Theorem 11.Proof.
Using Lemma10, the power mean integral inequality and F′q is strongly n-polynomial preinvex function of higher order, we have
(34)k+α1−λFa+ζb,a+k−α1−λFa2k−Γkα+k2ζα/kb,aJka+αFa+ζb,a+Jka+ζb,a−αFa=ζb,a2∫01ta/k+αk1−λ−1−ta/kF′a+tζb,adt≤ζb,a2∫01ta/k+αk1−λ−1−ta/kF′a+tζb,adt≤ζb,a2∫01ta/k+αk1−λ−1−ta/kdt1−1/q×∫01ta/k+αk1−λ−1−ta/k1n∑i=1n1−tiF′aq+1n∑i=1n1−1−tiF′bq−μζb,aσtσ1−t+t1−tσdt1/q=ζb,a2α1−λk1−1/qF′aqn∫01ta/k+αk1−λ−1−ta/k∑i=1n1−tidt+F′bqn∫01ta/k+αk1−λ−1−ta/k∑i=1n1−1−tidt−μζb,aσ∫01ta/k+αk1−λ−1−ta/ktσ1−t+t1−tσdt1/q=ζb,a2α1−λk1−1/qF′aqn∑i=1n∫01ta/k+αk1−λ−1−ta/k1−tidt+F′bqn∑i=1n∫01ta/k+αk1−λ−1−ta/k1−1−tidt−μζb,aσ∫01ta/k+αk1−λ−1−ta/ktσ1−t+t1−tσdt1/q=ζb,a2α1−λk1−a/qF′aqn∑i=1nM1+F′bqn∑i=1nM2−μζb,aσM31/q.
This completes the proof.Theorem 14.
LetF:a,a+ζb,a↦ℝ be a differentiable function on a,a+ζb,a with ζb,a>0,q>1, and F′∈La,a+ζb,a. If F′q is strongly n-polynomial preinvex function of higher order, then
(35)k+α1−λFa+ζb,a+k−α1−λFa2k−Γkα+k2ζα/kb,aJka+αFa+ζb,a+Jka+ζb,a−αFa≤ζb,a2α2k1−λ−αkα+kα+2k1−1/qF′aqn∑i=1nM5+F′bqn∑i=1nM6−μζb,aσM71/q+ζb,a2αkα+kα+2k+α2k1−λ1−1/qF′aqn∑i=1nM8+F′bqn∑i=1nM9−μζb,aσM101/q,where
(36)M5=iα1−λi+32ki+1i+2−kBksk+2k,α+k+kBksk+k,α+k−k2α+sk+kα+sk+2k−αkα+kα+2k,M6=iα1−λ2ki+2−kBkα+k,sk+k+kBkα+2k,sk+k+kα+sk+2kM7=2k3kσ+α+kkσ+α+2kkσ+α+3k+αk1−λ2σ+1σ+2σ+3+kBk2k,kσ+k−kBkα+2k,kσ+k+2kBkkσ+2k,α+k−kBkkσ+k,α+k+kBkα+2k,kσ+k−kBk2k,kσ+α+k−kBkkσ+3k,α+k−kBkα+3k,kσ+k+kBk3k,kσ+α+k,M8=αkα+kα+2k+iα1−λ2ki+2+kBksk+k,α+k−kα+sk+2kM9=αkα+kα+2k−kBksk+2k,α+k+iα1−λi+32ki+1i+2+k2α+sk+kα+sk+2k,M10=k2kσ+α+2kkσ+α+3k+αk1−λ1σ+2σ+3+kBk3k,kσ+k−kBkkσ+2k,α+k+kBkkσ+3k,α+k+kBkα+3k,kσ+k−kBk3k,kσ+α+k.Proof.
Using Lemma10. improved power mean integral inequality and F′q is strongly n-polynomial preinvex function of higher order, we have
(37)k+α1−λFa+ζb,a+k−α1−λFa2k−Γkα+k2ζa/kb,aJka+αFa+ζb,a+Jka+ζb,a−αFa=ζb,a2∫01ta/k+αk1−λ−1−ta/kF′a+tζb,adt≤ζb,a2∫011−tta/k+αk1−λ−1−ta/kdt1−1/q×∫011−tta/k+αk1−λ−1−ta/kF′a+tζb,aqdt1/q+ζb,a2∫01tta/k+αk1−λ−1−ta/kdt1−1/q×∫01tta/k+αk1−λ−1−ta/kF′a+tζb,aqdt1/q≤ζb,a2α2k1−λ−αkα+kα+2k1−1/q×F′aqn∑i=1n∫011−t1−tita/k+αk1−λ−1−ta/kdt+F′bqn∑i=1n∫011−t1−1−tita/k+αk1−λ−1−ta/kdt−μζb,aσ∫011−ttσ1−t+t1−tσta/k+αk1−λ−1−ta/kdt1/q+ζb,a2αkα+kα+2k+α2k1−λ1−1/q×F′aqn∑i=1n∫01t1−tita/k+αk1−λ−1−ta/kdt+F′bqn∑i=1n∫01t1−1−tita/k+αk1−λ−1−ta/kdt−μζb,aσ∫01ttσ1−t+t1−tσta/k+αk1−λ−1−ta/kdt1/q=ζb,a2α2k1−λ−αkα+kα+2k1−1/qF′aqn∑i=1nM5+F′bqn∑i=1nM6−μζb,aσM71/q+ζb,a2αkα+kα+2k+α2k1−λ1−1/qF′aqn∑i=1nM8+F′bqn∑i=1nM9−μζb,aσM101/q.
This completes the proof.


## 3. Conclusion

In this article, we have introduced the notion of stronglyn-polynomial preinvex function of higher order. We have derived a new k-fractional analogue of classical Hermite-Hadamard’s integral inequality utilizing the class of strongly n-polynomial preinvex functions. We established a new auxiliary result pertaining to k-fractional integrals, and utilizing this new result, we obtained several new variants of trapezium-like inequalities using the concept of strongly n-polynomial preinvex functions of higher order. We would like to emphasize here that we can recapture some other new results from the main results of this article under some suitable conditions. For example, if we take σ=2, then all the results reduce to the results for strongly n-polynomial preinvex functions. If ζb,a=b−a, then we have results for n-polynomial convex functions of higher order. Similarly for other suitable choices, other new and known results, we left the details to interested readers. This shows that the results obtained in this article are quite a unifying one. We hope that the ideas and techniques of this article will inspire interested readers.



---

*Source: 9154139-2020-08-13.xml*