# Some New Inequalities Involvingκ-Fractional Integral for Certain Classes of Functions and Their Applications

Authors: Shuang-Shuang Zhou; Saima Rashid; Silvestru Sever Dragomir; Muhammad Amer Latif; Ahmet Ocak Akdemir; Jia-Bao Liu
Journal: Journal of Function Spaces
 (2020)

---

## Abstract

In this article, we present several new inequalities involving theκ-fractional integral for the integrable function ℱ which satisfies one of the following conditions: a ℱq is preinvex for some q>1; b ℱ′ is bounded; c ℱ′ is a Lipschitz function. As applications, we establish new inequalities for the weighted arithmetic and generalized logarithmic means.


---

## Body

## 1. Introduction

LetE⊆ℝ be a nonempty interval. Then, a real-valued function f:E⟶ℝ is said to be convex (concave) on E if the inequality(1)fλσ+1−λτ≤≥λfσ+1−λfτ,takes place for any σ,τ∈E and 0≤λ≤1.We all know that the convexity theory has penetrated into every branch of pure and applied mathematics [1–20], and it has more and more practical applications in physics, mechanics, statistics, operations research, and even in economics and meteorology [21–40]. Many remarkable inequalities in mathematics, control theory, and game theory can be found in the literature [41–60] by use of the convexity theory. In the past half century, to research the generalizations and variants for the convexity has always been a hot topic for mathematicians and physicists as well as engineers. Recently, a great deal of generalizations and variants has been made for the convexity, for example, the GA-convexity and GG-convexity [61], s-convexity [62, 63], preinvex convexity [64], strong convexity [65–68], and Schur convexity [69].When we talk about convex functions, we have to mention a classical and most important inequality, which is the well-known Hermite–Hadamard inequality [70] which states that the double inequality(2)ℱε+ζ2≤≥1ζ−ε∫εζℱxdx≤≥ℱε+ℱζ,2holds for all ε,ζ∈J with ε≠ζ if ℱ:J⟶ℝ is a convex (concave) function on J and J⊆ℝ is a nonempty interval. For a long time, numerous researchers have been devoted to the generalizations, improvements, refinements, and variations for inequality (2) [71–73].The aim of this article is to provide new Hermite–Hadamard-type inequalities for certain classes of functions via theκ-fractional integral and give their applications to the bivariate means.In order to clearly describe and prove our main results in the next sections, we have to recall some definitions which we present in this section.Definition 1.
LetΩ⊆ℝn be a nonempty set and η:Ω×Ω⟶ℝn be a mapping. Then, Ω is said to be an invex set with respect to the mapping η if(3)ε+θηζ,ε∈Ω,for all ε,ζ∈Ω and θ∈0,1.Definition 2.
LetΩ⊆ℝn be an invex set with respect to the mapping η:Ω×Ω⟶ℝn. Then, the mapping φ:Ω⟶ℝ is said to be preinvex with respect to the mapping η if the inequality(4)φε+θηζ,ε≤1−θφε+θφζ,holds for all ε,ζ∈Ω and θ∈0,1.Definition 3.
(see [74]). Let β,κ>0, ε,ζ∈ℝ with ε<ζ, and ℱ∈Lε,ζ. Then, the β order κ-fractional integral operators Jκε+β and Jκζ−β of ℱ are defined by(5)Tκε+βℱx=1κΓκβ∫εxx−τβ/κ−1ℱτdτ,Tκζ−βℱx=1κΓκβ∫xζτ−xβ/κ−1ℱτdτ,respectively, where(6)Γκx=∫0∞τx−1e−τκ/κdτ,is the κ-gamma function.


## 2. Main Results

Throughout this section, we always assume thatℤ∗ is the set of positive integers, β,κ>0, θ∈0,1, Ω⊆ℝ is an open invex set with respect to the mapping η:Ω×Ω⟶ℝ\0, ε,ζ∈Ω with ε<ζ, and ℱ:Ω⟶ℝ is a differentiable mapping such that ℱ′ is integrable on ε,ε+ηζ,ε for ηζ,ε>0, and(7)Ληβ,κ,θ;x=1−θηx,εβ/κℱε+ℱε+ηx,ε+ηζ,xβ/κℱx+ℱx+ηζ,xηζ,ε+θηζ,εηx,εβ/κℱε+1n+1ηx,ε+ℱε+nn+1ηx,ε+ηζ,xβ/κℱx+nn+1ηζ,x+ℱx+1n+1ηζ,x−n+1β/κΓκβ+κηζ,εTκε+ηx,ε−βℱε+nn+1ηx,ε+Tκε+βℱε+1n+1ηx,ε+Tκx+βℱx+1n+1ηζ,x+Tκx+ηζ,x−βℱx+nn+1ηζ,x.Lemma 1.
LetΛηβ,κ,θ;x be defined by (7). Then, we have the identity(8)Ληβ,κ,θ;x=ηx,εβ/κn+1ηζ,ε∫01τβ/κ−θℱ′ε+n+τn+1ηx,εdτ−∫01τβ/κ−θℱ′ε+1−τn+1ηx,εdτ−ηζ,xβ/κn+1ηζ,ε∫01τβ/κ−θℱ′x+n+τn+1ηζ,xdτ−∫01τβ/κ−θℱ′x+1−τn+1ηζ,xdτ.Proof.
Making use of integration by parts and variable transformation, one has(9)I1=∫01τβ/κ−θℱ′ε+n+τn+1ηx,εdτ=n+1ηx,ετβ/κ−θℱε+n+τn+1ηx,ε01−βn+1κηx,ε∫01τβ/κℱε+n+τn+1ηx,εdτ=n+1ηx,ε1−θℱε+ηx,ε+θℱε+nn+1ηx,ε−n+1β/κ+1Γκβ+κηx,εβ/κ+1Tκε+ηx,ε−βℱε+nn+1ηx,ε.
Analogously, we also have(10)I2=−n+1ηx,ε1−θℱε+θℱε+1n+1ηx,ε−n+1β/κ+Γκβ+κηx,εβ/κ+1Tκε+βℱε+1n+1ηx,ε,(11)I3=n+1ηζ,x1−θℱx+ηζ,x+θℱx+nn+1ηζ,x−n+1β/κ+1Γκβ+κηζ,xβ/κ+1Tκx+ηζ,x−βℱε+nn+1ηζ,x,(12)I4=−n+1ηζ,x1−θℱx+ηζ,x+θℱx+nn+1ηζ,x−n+1β/κ+1Γκβ+κηζ,xβ/κ+1Tκx+βℱε+1n+1ηζ,x.
Therefore, identity (8) follows from multiplying (9) and (10) by ηx,εβ/κ+2/n+1ηζ,ε, multiplying (11) and (12) by ηζ,xβ/κ+2/n+1ηζ,ε, and then adding them.Remark 1.
Lemma1 leads to the conclusions as follows:(i)
Letn=1. Then, Lemma 1 leads to Lemma 2.1 of [75].(ii)
Letηω,υ=ω−υ. Then, one has(13)Λβ,κ,θ;x=1−θx−εβ/κℱε+ℱx+ζ−xβ/κℱx+ℱζζ−ε+θζ−εx−εβ/κℱnε+xn+1+ℱnx+εn+1+ζ−xβ/κℱx+nζn+1+ℱnx+ζn+1−n+1β/κΓκβ+κζ−εTκx−βℱnx+εn+1+Tκε+βℱnε+xn+1Tκx+βℱnx+ζn+1+Tκζ−βℱx+nζn+1.(iii)
Ifθ=0 and β=κ=1, then (13) reduces to(14)Λ1,1,0;x=ℱx+ζ−xℱζ+x−εℱεζ−ε−n+1ζ−ε∫εζℱzdz.(iv)
Ifθ=1 and β=κ=1, then (13) becomes(15)Λ1,1,1;x=x−εℱnε+x/n+1+ℱε+nx/n+1+ζ−xℱnζ+x/n+1+ℱζ+nx/n+1ζ−ε−n+1ζ−ε∫εζℱzdz.(v)
Ifηω,υ=ω−υ and n=1, then Corollary 2.1 of [75] can be derived from Lemma 1.Theorem 1.
Letp,q>1 such that 1/p+1/q=1 and ℱq be preinvex on Ω. Then, the inequality(16)Ληβ,κ,θ;x≤Ψ1/pβ,κ,θ,pηx,εβ/κ+1n+1ηζ,εℱ′εq+2n+1ℱ′xq2n+11/q+2n+1ℱ′εq+ℱ′xq2n+11/q+ηζ,xβ/κ+1n+1ηζ,εℱ′xq+2n+1ℱ′ζq2n+11/q+2n+1ℱ′xq+ℱ′ζq2n+11/q,holds for all x∈ε,ε+ηζ,ε, where(17)Ψβ,κ,θ,p=κβp+κ,θ=0,κ1−2θβp+κ/ββp+κ+2θβp+κ/β−θp,0<θ<1,Γp+1Γβ/κ+1Γβ/κ+1+p,θ=1.Proof.
It follows from Lemma1 and the preinvexity of ℱq together with the Hölder’s inequality that(18)ληβ,κ,θ;x≤ηx,εβ/κ+1n+1ηζ,ε∫01τβ/κ−θpdτ1/p∫01ℱ′ε+n+τn+1ηx,εqdτ+∫01ℱ′ε+1−τn+1ηx,εqdτ1/q−ηζ,xβ/κ+1n+1ηζ,ε∫01τβ/κ−θpdτ1/p×∫01ℱ′x+n+τn+1ηζ,xqdτ+∫01ℱ′x+1−τn+1ηζ,xqdτ1/q.
Note that(19)∫01ℱ′ε+n+τn+1ηx,εqdτ≤∫011−τn+1ℱ′εq+n+τn+1ℱ′xqdτ=ℱ′εq+2n+1ℱ′xq2n+1.
Analogously, we have(20)∫01ℱ′ε+1−τn+1ηx,εqdτ≤2n+1ℱ′εq+ℱ′xq2n+1,∫01ℱ′x+n+τn+1ηζ,xqdτ≤ℱ′xq+2n+1ℱ′ζq2n+1,∫01ℱ′x+1−τn+1ηζ,xqdτ≤2n+1ℱ′xq+ℱ′ζq2n+1.
We clearly see that(21)∫01τβ/κ−θpdτ=κβp+κ,for θ=0,(22)∫01τβ/κ−θpdτ=Γp+1Γβ/κ+1Γβ/κ+1+p,for θ=1, and(23)∫01τβ/κ−θpdτ=κ1−2θβp+κ/ββp+κ+2θβp+κ/β−θp,for 0<θ<1.
Therefore, inequality (16) can be derived from the above inequalities and identities.Remark 2.
Theorem1 leads to the conclusion as follows:(i)
Theorem 2.1 of [75] can be obtained from Theorem 1 if we take n=1.(ii)
Ifθ=κ=1 and ηω,υ=ω−υ, Theorem 1 reduces to(24)Λβ,1,1:x=x−εβℱnε+x/n+1+ℱnε+x/n+1+ζ−xβℱnζ+x/n+1+ℱnx+ζ/n+1ζ−ε−n+1βΓβ+1ζ−εTκx−βℱnx+εn+1+Tκε+βℱnε+xn+1+Tκx+βℱnζ+xn+1+Tκζ−βℱnx+ζn+1≤Γp+1Γβ+1Γp+β+1n+11/q+121/qx−εβ+1ζ−εℱεq+2n+1ℱxq1/q+2n+1ℱεq+ℱxq1/q+ζ−xβ+1ζ−εℱζq+2n+1ℱxq1/q+2n+1ℱxq+ℱζq1/q.(iii)
Letn=1. Then, (24) leads to Corollary 2.2 of [75].Theorem 2.
Letq>1 and ℱq be preinvex on Ω. Then, the inequality(25)Ληβ,κ,θ;x≤Φ11−1/qηx,εβ/κ+1n+1ηζ,ε1n+1Φ1−Φ2ℱ′εq+Φ1+nΦ2ℱ′xq1/q+1n+1nΦ1+Φ2ℱ′εq+Φ1−Φ2ℱ′xq1/q+ηx,εβ/κ+1n+1ηζ,ε1n+1Φ1−Φ2ℱ′xq+Φ1+nΦ2ℱ′ζq1/q+1n+1nΦ1+Φ2ℱ′xq+Φ1−Φ2ℱ′ζq1/qholds for x∈ε,ε+ηζ,ε, where(26)Φ1=∫01τβ/κ−θdτ=κ+2βθβ+κ/ββ+κ−θ,Φ2=∫01ττβ/κ−θdτ=κ+βθβ+2κ/ββ+2κ−θ2.Proof.
It follows from Lemma1 and the preinvexity of ℱq together with the power-mean inequality that(27)Ληβ,κ,θ;x≤ηx,εβ/κ+1n+1ηζ,ε∫01τβ/κ−θdτ1−1/qχ11/q+χ21/q−ηζ,xβ/κ+1n+1ηζ,ε∫01τβ/κ−θdτ1−1/qχ31/q+χ41/q,where(28)χ1=∫01τβ/κ−θℱ′ε+n+τn+1ηx,εqdτ≤∫01τβ/κ−θ1−τn+1ℱ′εq+n+τn+1ℱ′xqdτ=1n+1∫01τβ/κ−θdτ−∫01ττβ/κ−θdτℱ′εq+∫01τβ/κ−θdτ+n∫01ττβ/κ−θdτℱ′xq,χ2=∫01τβ/κ−θℱ′ε+1−τn+1ηx,εqdτ≤1n+1n∫01τβ/κ−θdτ+∫01ττβ/κ−θdτℱ′εq+∫01τβ/κ−θdτ−∫01ττβ/κ−θdτℱ′xq,χ3=∫01τβ/κ−θℱ′x+n+τn+1ηζ,xqdτ≤1n+1∫01τβ/κ−θdτ−∫01ττβ/κ−θdτℱ′xq+∫01τβ/κ−θdτ+n∫01ττβ/κ−θdτℱ′ζq,χ4=∫01τβ/κ−θℱ′x+1−τn+1ηζ,xqdτ≤1n+1n∫01τβ/κ−θdτ+∫01ττβ/κ−θdτℱ′xq+∫01τβ/κ−θdτ−∫01ττβ/κ−θdτℱ′ζq.
Substituting the above inequalities in (27), we get inequality (25).Remark 3.
From Theorem2, we have two conclusions as follows:(i)
Ifn=1, then we get Theorem 2.2 of [76].(ii)
Ifθ=κ=1 and ηω,υ=ω−υ, then(29)Λβ,1,1;x≤12β+21/qζ−εβn+1β+11+1/q×x−εββ+2ℱ′εq+βn+2+n+4ℱ′xq1/q+β+2ℱ′xq+βn+2+n+4ℱ′εq1/q+ζ−xββ+2ℱ′ζq+βn+2+n+4ℱ′xq1/q+β+2ℱ′xq+βn+2+n+4ℱ′ζq1/q.Theorem 3.
Ifr,ℛ∈0,∞ with r<ℛ, and r≤ℱ′y≤ℛ for all y∈ε,ε+ηζ,ε, then we have(30)Ληβ,κ,θ;x≤ηx,εβ/κ+ηζ,xβ/κℛ−rn+1ηζ,εκ+2βθβ+κ/ββ+κ−θ.Proof.
It follows from Lemma1 that(31)Ληβ,κ,θ;x=ηx,εβ/κn+1ηζ,ε∫01τβ/κ−θℱ′ε+n+τn+1ηx,ε−r+ℛ2dτ−∫01τβ/κ−θℱ′ε+1−τn+1ηx,ε−r+ℛ2dτ−ηζ,xβ/κn+1ηζ,ε∫01τβ/κ−θℱ′x+n+τn+1ηζ,x−r+ℛ2dτ−∫01τβ/κ−θℱ′x+1−τn+1ηζ,x−r+ℛ2dτ.
Making use of the fact that(32)r−r+ℛ2≤ℱ′ε+1−τn+1ηε,x−r+ℛ2≤ℛ−r+ℛ2,one has(33)ℱ′ε+n+τn+1ηx,ε−r+ℛ2≤ℛ−r2.
Similarly, we have(34)ℱ′ε+1−τn+1ηx,ε−r+ℛ2≤ℛ−r2,ℱ′x+n+τn+1ηζ,x−r+ℛ2≤ℛ−r2,ℱ′x+1−τn+1ηζ,x−r+ℛ2≤ℛ−r2.
Therefore,(35)Ληβ,κ,θ;x≤ηx,εβ/κn+1ηζ,ε∫01τβ/κ−θℱ′ε+n+τn+1ηx,ε−r+ℛ2dτ−∫01τβ/κ−θℱ′ε+1−τn+1ηx,ε−r+ℛ2dτ−ηζ,xβ/κn+1ηζ,ε∫01τβ/κ−θℱ′x+n+τn+1ηζ,x−r+ℛ2dτ−∫01τβ/κ−θℱ′x+1−τn+1ηζ,x−r+ℛ2dτ≤ηx,εβ/κn+1ηζ,εℛ−r2∫01τβ/κ−θdτ+ηζ,xβ/κn+1ηζ,εℛ−r2∫01τβ/κ−θdτ=ηx,εβ/κ+ηζ,xβ/κℛ−rn+1ηζ,εΦ1,which completes the proof of Theorem 3.Remark 4.
From Theorem3, we get two conclusions as follows:(i)
Letn=1. Then, Theorem 3.1 of [75] can be derived from Theorem 3.(ii)
Ifβ=κ=1 and ηω,υ=ω−υ, then Theorem 3 leads to(36)Λη1,1,θ;x=1−θx−εℱε+ℱx+ζ−xℱζ+ℱxζ−ε+θx−εℱnε+x/n+1+ℱε+nx/n+1+ζ−xℱnζ+x/n+1+ℱζ+nx/n+1ζ−ε−n+1ζ−ε∫εζℱzdz≤ℛ+rx−ε2+ζ−x2n+1ζ−ε2θ2−2θ+12.(iii)
Ifn=1, then (36) becomes Corollary 3.1 of [75].(iv)
Ifθ=0, then (36) leads to(37)x−εℱε+ℱx+ζ−xℱζ+ℱxζ−ε−n+1ζ−ε∫εζℱzdz≤ℛ+rx−ε2+ζ−x22n+1ζ−ε.(v)
Ifθ=1/2, then (36) leads to the conclusion that(38)x−εℱε+ℱx+ζ−xℱζ+ℱx2ζ−ε+x−εℱnε+x/n+1+ℱε+nx/n+1+ζ−xℱnζ+x/n+1+ℱζ+nx/n+12ζ−ε−n+1ζ−ε∫εζℱzdz≤ℛ+rx−ε2+ζ−x24n+1ζ−ε.(vi)
Ifθ=1, then (36) gives(39)x−εℱnε+x/n+1+ℱε+nx/n+1+ζ−xℱnζ+x/n+1+ℱζ+nx/n+1ζ−ε−n+1ζ−ε∫εζℱzdz≤ℛ+rx−ε2+ζ−x22n+1ζ−ε.Theorem 4.
Ifℱ′ is a Lipschitz function on Ω with the Lipschitz constant L>0, then the fractional integral inequality(40)Ληβ,κ,θ;x≤Lηx,εβ/κ+2+Lηζ,xβ/κ+2n+1ηζ,ε×n−1n+1κ+2βθβ+κ/ββ+κ−θ+2n+1κ+βθβ+2κ/ββ+2κ−θ2,holds for x∈ε,ε+ηζ,ε.Proof.
It follows from Lemma1 that(41)Ληβ,κ,θ;x=ηx,εβ/κ+1n+1ηζ,ε∫01τβ/κ−θℱ′ε+n+τn+1ηx,ε−ℱ′ε+1−τn+1ηx,εdτ−ηζ,xβ/κ+1n+1ηζ,ε∫01τβ/κ−θℱ′x+n+τn+1ηζ,x−ℱ′x+1−τn+1ηζ,xdτ.
Sinceℱ′ is a Lipschitz function on ε,ε+ηζ,ε with Lipschitz constant L>0, we get(42)ℱ′ε+n+τn+1ηx,ε−ℱ′ε+1−τn+1ηx,ε≤L2τ+n−1n+1ηx,ε.
Similarly, we have(43)ℱ′x+n+τn+1ηζ,x−ℱ′x+1−τn+1ηζ,x≤L2τ+n−1n+1ηζ,x.
Therefore, one has(44)Ληβ,κ,θ;x≤ηx,εβ/κ+1n+1ηζ,ε∫01τβ/κ−θℱ′ε+n+τn+1ηx,ε−ℱ′ε+1−τn+1ηx,εdτ−ηζ,xβ/κ+1n+1ηζ,ε∫01τβ/κ−θℱ′x+n+τn+1ηζ,x−ℱ′x+1−τn+1ηζ,xdτ≤L2τ+n−1n+1ηx,εβ/κ+2+ηζ,xβ/κ+2n+1ηζ,εn−1n+1∫01τβ/κ−θdτ+2n+1∫01ττβ/κ−θdτ=Lηx,εβ/κ+2+ηζ,xβ/κ+2n+1ηζ,εn−1n+1κ+2βθβ+κ/ββ+κ−θ+2n+1κ+βθβ+2κ/ββ+2κ−θ2.Remark 5.
From Theorem4, we clearly see that the following six results are true:(i)
Ifn=1, then we get(45)Ληβ,κ,θ;x≤Lηx,εβ/κ+2+ηζ,xβ/κ+22ηζ,εκ+βθβ+2κ/ββ+2κ−θ2.(ii)
Ifηω,υ=ω−υ, then (45) becomes(46)Λβ,κ,θ;x≤Lx−εβ/κ+2+ζ−xβ/κ+22ζ−εκ+βθβ+2κ/ββ+2κ−θ2.(iii)
Ifβ=κ=1, then (46) reduces to(47)Λ1,1,θ;x≤Lx−ε3+ζ−x3ζ−ε2θ3−3θ+26.(iv)
Ifθ=0, then inequality (47) leads to(48)Λ1,1,0;x=ζ−xℱζ+x−εℱεζ−ε−2ζ−ε∫εζℱzdz≤Lx−ε3+ζ−x33ζ−ε.(v)
Ifθ=1/2, then inequality (47) gives(49)Λ1,1,12;x=x−εℱε+ℱx+ζ−xℱζ+ℱx2ζ−ε+x−εℱε+x/2+ζ−xℱζ+x/22ζ−ε−2ζ−ε∫εζℱzdz≤Lx−ε3+ζ−x36ζ−ε.(vi)
Ifθ=1, then inequality (47) becomes(50)Λ1,1,1;x=x−εℱε+x/2ζ−xℱζ+x/2ζ−ε−1ζ−ε∫εζℱzdz≤Lx−ε3+ζ−x312ζ−ε.


## 3. Applications to Special Bivariate Means

A bivariate functionϒ:0,∞×0,∞⟶0,∞ is said to be a mean if minε,ζ≤ϒε,ζ≤maxε,ζ for all ε,ζ∈0,∞. It is well known that the bivariate means are closely related to the special functions. Recently, the inequalities between different bivariate means have attracted the attention of many researchers [77–82].In this section, we use our obtained results in Section2 to provide two new inequalities between the weighted arithmetic mean(51)Aε,ζ;w1,w2=w1ε+w2ζw1+w2ε,ζ,w1,w2>0and sth generalized logarithmic mean(52)Λsε,ζ=ζs+1−εs+1s+1ζ−ε1/sε,ζ>0,ε≠ζ,s∈ℤ\−1,0.Theorem 5.
Letp,q>1 with 1/p+1/q=1, ε,ζ>0 with ε≠ζ, and s≥2 be a positive integer. Then, one has(53)ΛAsε,ζ,3,1,Asε,ζ,1,3−Λssε,ζ≤sζ−ε8121+2/q11+p1/p×A1/qεqs−1,ε+ζ2qs−1;1,3+A1/qεqs−1,ε+ζ2qs−1;3,1+A1/qε+ζ2qs−1,ζqs−1;1,3+A1/qε+ζ2qs−1,ζqs−1;3,1.Proof.
Letβ=κ=θ=1, ηω,υ=ω−υ, and ℱx=xs. Then, Theorem 5 follows from Theorem 1.Theorem 6.
Letp,q>1 with 1/p+1/q=1, ε,ζ>0 with ε≠ζ, and s≥2 be a positive integer. Then, one has(54)ΛAsε,ζ,3,1,Asε,ζ,1,3−Λssε,ζ≤1411241/q×A1/qεqs−1,ε+ζ2qs−1;3,8+A1/qεqs−1,ε+ζ2qs−1;8,3+A1/qε+ζ2qs−1,ζqs−1;3,8+A1/qε+ζ2qs−1,ζqs−1;8,3.Proof.
Theorem6 follows directly from Theorem 2 if we take β=κ=θ=1 and ηω,υ=ω−υ together with ℱx=xs.



---

*Source: 5285147-2020-04-29.xml*