# k -Fractional Variants of Hermite-Mercer-Type Inequalities via s-Convexity with Applications

Authors: Saad Ihsan Butt; Jamshed Nasir; Shahid Qaisar; Khadijah M. Abualnaja
Journal: Journal of Function Spaces
 (2021)

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## Abstract

This article is aimed at studying novel generalizations of Hermite-Mercer-type inequalities within the Riemann-Liouvillek-fractional integral operators by employing s-convex functions. Two new auxiliary results are derived to govern the novel fractional variants of Hadamard-Mercer-type inequalities for differentiable mapping Ψ whose derivatives in the absolute values are convex. Moreover, the results also indicate new lemmas for Ψ′, Ψ′′, and Ψ′′′ and new bounds for the Hadamard-Mercer-type inequalities via the well-known Hölder’s inequality. As an application viewpoint, certain estimates in respect of special functions and special means of real numbers are also illustrated to demonstrate the applicability and effectiveness of the suggested scheme.


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## Body

## 1. Introduction

Recently, two fundamental notions have been introduced in pure and applied analysis having potential utilities in every field and are known as convexity and concavity. Interestingly, the convexity theory is attributed to Jensen. Several monographs and articles have played a prominent role in the developments, speculations, and modifications of convexity in different directions such asη-convexity, harmonic convexity, h-convexity, and strong convexity. Moreover, a strong connection has been developed between diverse kinds of convex functions and inequality theory. Their fertile applications in optimization theory, functional analysis, physics, and statistical theory have made it a much fascinating subject, and hence, it is assumed as an incorporative subject between combinatorics, orthogonal polynomials, hypergeometric functions, quantum theory, and linear programming. This is the major motivation behind the keen investigation and progress of the integral inequalities in the literature [1, 2].Let0<ζ1≤ζ2≤⋯≤ζn and let ρ=ρ1,ρ2,⋯,ρn be nonnegative weights such that ∑j=1nρj=1. The famous Jensen’s inequality (see [1]) in the literature states that if Ψ is a convex function on the interval θ1,θ2, then
(1)Ψ∑j=1nρjζj≤∑j=1nρjΨζj,for all ζj∈θ1,θ2, ρj∈0,1, and j=1,2,⋯,n. It is one of the key inequalities that helps to extract bounds for useful distances in information theory (see [3, 4]).In 2003, a new variant of Jensen’s inequality was introduced by Mercer [5].IfΨ is a convex function on θ1,θ2, then
(2)Ψθ1+θ2−∑j=1nρjζj≤Ψθ1+Ψθ2−∑j=1nρjΨζj,for all ζj∈θ1,θ2, ρj∈0,1, and j=1,2,⋯,n.Matkovic and Pećarić proposed several generalizations on Jensen-Mercer operator inequalities [6]. Later on, Niezgoda [7] has provided several extensions to higher dimensions for Mercer-type inequalities. Recently, the Jensen-Mercer-type inequality has made a significant contribution to inequality theory due to its prominent characterizations.In the present study, we considers∈0,∞, the class of Breckner s-convex functions (which for 0<s<1 were called in [8, 9]), s-convex in the second sense. In [10], Dragomir and Fitzpatrick introduce the concept of a real-valued Breckner s-convex function Ψ on a convex subset C of a linear space V as
(3)Ψρ1ζ1+ρ2ζ2≤ρsψζ1+1−ρsΨζ2,whenever ρ1,ρ2≥0 with ρ1+ρ2=1 and ζ1,ζ2∈C. For s=1, it reduces to the usual notion of convexity. As a result, he generalizes Jensen’s inequality (1) as
(4)Ψ∑nj=1ρjζj≤∑nj=1ρjsΨζj,whenever ρi≥0, ζi∈C, and ∑i=1nρi=1.In [9], the class of s-convex functions in the first and second senses is introduced along with their significant properties.Definition 1.
Lets∈0,1, a real-valued function Ψ on an interval I=0,∞ is s-convex in the second sense provided that (3) holds for all ζ1,ζ2∈I and ρ1,ρ2≥0 with ρ1+ρ2=1.They denote this class of function byΨ∈Ks2. Moreover, they proved that the class Ψ∈Ks2 is stronger than convexity in the first and original sense for 0<s<1. Several properties of s-convex functions in both senses are presented in a comprehensive manner with supporting examples. It is interesting to see that if 0<s<1 and Ψ∈Ks2, then ψ is nonnegative. This result may not hold in general when the function is convex (i.e., s=1). Also, the situation is more interesting when f0=0.Viewing this literature, we intend to extend the Jensen-Mercer inequality for Breckners-convex functions. For this, we use the ideology of Mercer’s concept [5] and give the following important lemma.Lemma 2.
IfΨ is a real-valued Breckner s-convex function on the interval θ1,θ2⊂ℝ+ and s>0 such that ρ1,ρ2≥0, ρ1+ρ2=1, and ρ1s+ρ2s≤1, then for any finite positive increasing sequence ζnk=1n∈θ1,θ2, we have
(5)Ψθ1+θ2−ζk≤Ψθ1+Ψθ2−Ψζk,for all 1≤k≤n.Proof.
Letyk=θ1+θ2−ζk. Then, θ1+θ2=yk+ζk, so the pairs θ1,θ2 and yk,ζk possess the same midpoint. Since that is the case, there exist ρ1,ρ2∈0,1, with ρ1+ρ2=1 such that ζk=ρ1θ1+ρ2θ2 and yk=ρ2θ1+ρ1θ2. Therefore, employing (3) and the assumed condition, we get
(6)Ψyk=Ψρ2θ1+ρ1θ2≤ρ2sΨθ1+ρ1sΨθ2≤1−ρ1sΨθ1+1−ρ2sΨθ2=Ψθ1+Ψθ2−ρ1sΨθ1−ρ2sΨθ2=Ψθ1+Ψθ2−ρ1sΨθ1+ρ2sΨθ2≤Ψθ1+Ψθ2−Ψρ1θ1+ρ2θ2=Ψθ1+Ψθ2−Ψζk,which completes the proof.Now, we give the result for the Jensen-Mercer inequality in the Breckners-sense.Theorem 3.
Letρ1,ρ2,⋯,ρn be positive real numbers n≥2 such that ∑k=1nρk=1 and ∑k=1nρks≤1. If ψ is a real-valued Breckner s-convex function on θ1,θ2⊂ℝ+, then for any finite positive increasing sequence ζnk=1n∈θ1,θ2, we have
(7)Ψθ1+θ2−∑k=1nρkζk≤Ψθ1+Ψθ2−∑k=1nρksΨζk.Proof.
One can prove it by following a similar idea as in [5]; however, we need to employ Lemma 2 and generalized Jensen’s inequality (4) for Breckner s-convex functions.For further properties and applications ofs-convex functions, see [9, 11] and references therein. The following lemma is of great interest for applications.Lemma 4 (see [11]).
LetΨ:θ1,θ2⟶ℝ be a convex function. Then, the following results hold:
(i)
IfΨ is nonnegative, then it is s-convex for s∈0,1(ii)
IfΨ is nonpositive, then it is s-convex for s∈1,∞
One of the famous integral inequalities for convex functions is the Hermite-Hadamard inequality (see [2]):
(8)Ψθ1+θ22≤1θ2−θ1∫θ1θ2Ψλdλ≤Ψθ1+Ψθ22,provided that if a mapping Ψ:I⊆ℝ⟶ℝ is a convex function on I and θ1,θ2∈I.Fractional-order calculus deals with more general behavior than integer-order calculus, and it not only provides new mathematical methods for practical systems but also has been applied into various fields due to its accurate description in many active fields, such as fraction-order memristive chaotic circuit, fractional-orderrelaxation-oscillation model, mathematical biology, and economics (see [12]).The theory of Riemann-Liouvillek-fractional integrals is a pertinent extension of Riemann-Liouville fractional integrals. It is important to note that if k≠1, the properties of Riemann-Liouville k-fractional integrals are quite dissimilar from those of general fractional integrals. For this, the Riemann-Liouville k-fractional integrals have agitated the interest of many researchers. Now, we demonstrate some essential concepts of k-fractional calculus for the investigation of our results.Definition 5 (see [13]).
Diaz and Pariguan have defined thek-gamma function Γk·, a generalization of the classical gamma function, which is given by the following formula:
(9)Γkα=limm⟶∞m!kmmkα/k−1αm,k,k>0,α∈C\kZ−,where αm,k is the Pochhammer k-symbol given by
(10)αm,k=αα+kα+2k,⋯,α+m−1k.
It is shown that the Mellin transform of the exponential functione−μk/k is the k-gamma function clearly given by
(11)Γkα=∫0∞e−μk/kμα−1dμ,for Reα>0 with αΓkα=Γkα+k, where Γk. stands for the k-gamma function.Many researchers have generalized the classical and fractional operators by introducing a parameterk>0 about a decade ago. Mubeen et al. [14] used special k-function theory in fractional calculus for the first time in the literature in the form of Riemann-Liouville k-fractional integrals.Definition 6 (see [15]).
LetΨ∈L1θ1,θ2. The Riemann-Liouville k-fractional integrals of Ψ with k>0,kJθ1+αΨ and Jkθ2−αΨ of order α>0 with θ1≥0 are defined by
(12)Jkθ1+αΨζ=1kΓα∫θ1ζζ−λα/k−1Ψλdλ,ζ>θ1,kJθ2−αΨζ=1kΓα∫ζθ2λ−ζα/k−1Ψλdλ,ζ<θ2,respectively.Remark 7.
Ifk=1, then Riemann-Liouville k-fractional integrals reduce to classical Riemann-Liouville fractional integrals. And if α=1 and k=1, the fractional integral reduces to the classical integral.Recently, many researchers are presenting new fractional differential and integral operators and they generalized by using the iteration procedure and by introducing a new parameterk>0. They also found relationships of these generalized fractional operators with existing fractional and classical operators under the special values of the parameter k. Many k-fractional operators, their properties, related identities, and inequalities are proved during the past years. For instance, see [16, 17] and references therein.


## 2. Main Results

This section contains several new generalizations of Hermite-Hadamard-Mercer-type inequalities fors-convex functions in the second sense (Breckner sense) via k-fractional calculus theory.Throughout the paper, we assumed the following assumptions:A1: let ζ1,ζ2∈θ1,θ2⊆ℝ+ with ζ1<ζ2, α,k>0, and for some fixed s∈0,1, λ∈0,1 and Γk. is the k-gamma function.A2: for λ∈0,1, 1+λ/2s+1−λ/2s≤1, whenever we use the definition of the Jensen-Mercer inequality for the s-convex function.Theorem 8.
LetΨ:θ1,θ2⟶ℝ be the s-convex function such that Ψ∈L1θ1,θ2 along with assumptions A1 and A2. Then, the following Riemann-Liouville k-fractional integral inequalities hold:
(13)12α/k−sΨθ1+θ2−ζ1+ζ22≤Γkα+kζ2−ζ1α/k×Jkθ1+θ2−ζ1−αΨθ1+θ2−ζ1+ζ22+Jkθ1+θ2−ζ2+αΨθ1+θ2−ζ1+ζ22≤Ψθ1+Ψθ22α/k−1−αk2α/k+sUα,k,s,λ+Bαk,s+1Ψζ1+Ψζ2,(14)Ψθ1+θ2−ζ1+ζ22≤Ψθ1+Ψθ2−2α/k−sΓkα+kζ2−ζ1α/k×Jkζ1+αΨζ1+ζ22+Jkζ2−αΨζ1+ζ22≤Ψθ1+Ψθ2−Ψζ1+ζ22,where Uα,k,s,λ=∫01λα/k−11+λsdλ and B−,− is the beta function.Proof.
By employing the definition of thes-convex function Ψ, we get
(15)Ψθ1+θ2−u+v2=Ψθ1+θ2−u+θ1+θ2−v2≤12sΨθ1+θ2−u+Ψθ1+θ2−v∀u,v∈θ1,θ2.
By change of variablesu=1+λ/2ζ1+1−λ/2ζ2 and v=1−λ/2ζ1+1+λ/2ζ2, λ∈0,1, we get
(16)2sΨθ1+θ2−ζ1+ζ22≤Ψθ1+θ2−1+λ2ζ1+1−λ2ζ2+Ψθ1+θ2−1−λ2ζ1+1+λ2ζ2.
Now, multiplying the above inequality byλα/k−1 and then integrating w.r.t. λ over 0,1 yield
(17)2sΨθ1+θ2−ζ1+ζ22∫01λα/k−1dλ≤∫01λα/k−1Ψθ1+θ2−1+λ2ζ1+1−λ2ζ2+Ψθ1+θ2−1−λ2ζ1+1+λ2ζ2dλ.
By change of variable, we have(18)12α/k−sΨθ1+θ2−ζ1+ζ22≤Γkα+kζ2−ζ1α/k×Jkθ1+θ2−ζ1−αΨθ1+θ2−ζ1+ζ22+Jkθ1+θ2−ζ2+αΨθ1+θ2−ζ1+ζ22,and so the first inequality of (13) is proved.
Now, for the proof of the second inequality of (13), we first note that if Ψ is an s-convex function, then for λ∈0,1, it gives
(19)Ψθ1+θ2−1+λ2ζ1+1−λ2ζ2≤Ψθ1+Ψθ2−1+λ2sΨζ1+1−λ2sΨζ2,(20)Ψθ1+θ2−1−λ2ζ1+1+λ2ζ2≤Ψθ1+Ψθ2−1−λ2sΨζ1+1+λ2sΨζ2.
By adding the inequalities of (19) and (20), we have
(21)Ψθ1+θ2−1+λ2ζ1+1−λ2ζ2+Ψθ1+θ2−1−λ2ζ1+1+λ2ζ2≤2Ψθ1+Ψθ2−1−λ2s+1+λ2sΨζ1+Ψζ2.
Now, multiplying the above inequality byλα/k−1 and then integrating w.r.t. λ over 0,1, we get
(22)∫01λα/k−1Ψθ1+θ2−1+λ2ζ1+1−λ2ζ2+Ψθ1+θ2−1−λ2ζ1+1+λ2ζ2dλ≤2α/kΨθ1+Ψθ2×12s∫01λα/k−11+λsdλ+∫01λα/k−11−λsdλΨζ1+Ψζ2.
Consequently, we get(23)2α/kkΓkαζ2−ζ1α/k×Jkθ1+θ2−ζ1−αΨθ1+θ2−ζ1+ζ22+Jkθ1+θ2−ζ2+αΨθ1+θ2−ζ1+ζ22≤2α/kΨθ1+Ψθ2−12sUα,k,s,λ+Bαk,s+1Ψζ1+Ψζ2,(24)Γkα+kζ2−ζ1α/k×Jkθ1+θ2−ζ1−αΨθ1+θ2−ζ1+ζ22+Jkθ1+θ2−ζ2+αΨθ1+θ2−ζ1+ζ22≤Ψθ1+Ψθ22α/k−1−αk2α/k+sUα,k,s,λ+Bαk,s+1Ψζ1+Ψζ2.
Combining (18) and (23), one can get (13). In order to prove (14), we employ the Jensen-Mercer inequality for the s-convex function Ψ in the second sense; then, for λ∈0,1, it yields
(25)Ψθ1+θ2−u+v2≤Ψθ1+Ψθ2−Ψu+Ψv2s,u,v∈θ1,θ2.
Now, by change of variablesu=1+λ/2ζ1+1−λ/2ζ2 and v=1−λ/2ζ1+1+λ/2ζ2, ∀ζ1,ζ2∈θ1,θ2 and λ∈0,1 in (25), we have
(26)Ψθ1+θ2−ζ1+ζ22≤Ψθ1+Ψθ2−Ψ1+λ/2ζ1+1−λ/2ζ2+Ψ1−λ/2ζ1+1+λ/2ζ22s.
Multiplying byλα/k−1 and then integrating w.r.t. λ over 0,1 give
(27)2skαΨθ1+θ2−ζ1+ζ22≤2skαΨθ1+Ψθ2−∫01λα/k−1Ψ1+λ2ζ1+1−λ2ζ2+Ψ1−λ2ζ1+1+λ2ζ2dλ.
It follows that(28)2skαΨθ1+θ2−ζ1+ζ22≤2skαΨθ1+Ψθ2−2α/k−1Γkα+kζ2−ζ1α/kJkζ1−αΨζ1+ζ22+Jkζ2+αΨζ1+ζ22Ψθ1+θ2−ζ1+ζ22≤Ψθ1+Ψθ2−2α/k−sΓkα+kθ2−θ1α/k×Jkζ1+αΨζ1+ζ22+Jkζ2−αΨζ1+ζ22,and so the first inequality of (14) is proved.
Now, for the proof of the second inequality of (14), we first note that if Ψ is an s-convex function, then for λ∈0,1, it gives
(29)Ψζ1+ζ22=Ψ1+λ/2ζ1+1−λ/2ζ2+1−λ/2ζ1+1+λ/2ζ22≤Ψ1+λ/2ζ1+1−λ/2ζ2+Ψ1−λ/2ζ1+1+λ/2ζ22s.
Multiplying byλα/k−1 and then integrating w.r.t. λ over 0,1 give
(30)2skαΨζ1+ζ22≤∫01λα/k−1Ψ1+λ2ζ1+1−λ2ζ2+Ψ1−λ2ζ1+1+λ2ζ2dλΨζ1+ζ22≤2α/k−sΓkα+kζ2−ζ1α/kJkζ1+αΨζ1+ζ22+Jkζ2−αΨζ1+ζ22.
Therefore, we have(31)−2α/k−sΓkα+kζ2−ζ1αJkζ1+αΨζ1+ζ22+Jkζ2−αΨζ1+ζ22≤−Ψζ1+ζ22.
AddingΨθ1+Ψθ2 to both sides in (31), we get the second inequality of (14).Remark 9.
Under the assumption of Theorem8 for inequality (13) with s=α=k=1, one has
(32)Ψθ1+θ2−ζ1+ζ22≤1ζ2−ζ1∫ζ1ζ2Ψθ1+θ2−λdλ≤Ψθ1+Ψθ2−Ψζ1+Ψζ22.
The inequality (32) is proposed by Kian and Moslehian in [18].Remark 10.
If we chooses=k=1 in Theorem 8, we get Theorem 2 in [19].


## 3. New Identities and Related Results via Riemann-Liouvillek-Fractional Integrals

Lemma 11.
LetΨ:θ1,θ2⟶ℝ be a differentiable mapping on θ1,θ2 with θ1<θ2 along with assumption A1. If Ψ′∈Lθ1,θ2, then the following equality for Riemann-Liouville k-fractional integrals holds:
(33)Ψθ1+θ2−ζ1+Ψθ1+θ2−ζ22−2α/k−1Γkα+kζ2−ζ1α/k×Jkθ1+θ2−ζ2+αΨθ1+θ2−ζ1+ζ22+Jkθ1+θ2−ζ1−αΨθ1+θ2−ζ1+ζ22=ζ2−ζ14∫01λα/kΨ′θ1+θ2−1+λ2ζ1+1−λ2ζ2−Ψ′θ1+θ2−1−λ2ζ1+1+λ2ζ2dλ.Proof.
It suffices to write that(34)I=ζ2−ζ14I1−I2,where
(35)I1=∫01λα/kΨ′θ1+θ2−1+λ2ζ1+1−λ2ζ2dλ=2Ψθ1+θ2−ζ1ζ2−ζ1−2α/k+1α/kζ2−ζ1α/k+1∫θ1+θ2−ζ1+ζ2/2θ1+θ2−ζ1w−θ1+θ2−ζ1+ζ22α/k−1Ψwdw=2Ψθ1+θ2−ζ1ζ2−ζ1−2α/k+1Γkα+kζ2−ζ1α/k+1Jkθ1+θ2−ζ1−αΨθ1+θ2−ζ1+ζ22.
Analogously,(36)I2=∫01λα/kΨ′θ1+θ2−1−λ2ζ1+1+λ2ζ2dλ=−2Ψθ1+θ2−ζ2ζ2−ζ1+2α/k+1α/kζ2−ζ1α/k+1×∫θ1+θ2−ζ2θ1+θ2−ζ1+ζ2/2θ1+θ2−ζ1+ζ22−wα/k−1Ψwdw=−2Ψθ1+θ2−ζ2ζ2−ζ1+2α/k+1Γkα+kζ2−ζ1α/k+1Jkθ1+θ2−ζ2+αΨθ1+θ2−ζ1+ζ22.
Combining (35) and (36) with (34), we get (33).Corollary 12.
Forα=k=1 in Lemma 11, we acquire
(37)Ψθ1+θ2−ζ1+iθ1+θ2−ζ22−1ζ2−ζ1∫θ1+θ2−ζ2θ1+θ2−ζ1Ψudu=ζ2−ζ14∫01λΨ′θ1+θ2−1+λ2ζ1+1−λ2ζ2−Ψ′θ1+θ2−1−λ2ζ1+1+λ2ζ2dλ.Remark 13.
Takingα=k=1 with ζ1=θ1 and ζ2=θ2 in Lemma 11, we get Lemma 2.1 in [20] and the following equality holds:
(38)Ψθ1+Ψθ22−1θ2−θ1∫θ1θ2Ψudu=θ2−θ14∫01λΨ′1−λ2θ1+1+λ2θ2−Ψ′1+λ2θ1+1−λ2θ2dλ.Theorem 14.
Suppose thatΨ:θ1,θ2⟶ℝ is a differentiable mapping on θ1,θ2 with θ1<θ2 and Ψ′∈Lθ1,θ2 along with assumptions A1 and A2. If Ψ′ is an s-convex function on θ1,θ2, then the following inequality for Riemann-Liouville k-fractional integrals holds:
(39)Ψθ1+θ2−ζ1+Ψθ1+θ2−ζ22−2α/k−1Γkα+kζ2−ζ1α/k×kJθ1+θ2−ζ1−αΨθ1+θ2−ζ1+ζ22+Jkθ1+θ2−ζ2+αΨθ1+θ2−ζ1+ζ22≤ζ2−ζ142α/k+1Ψ′θ1+Ψ′θ2−12sUα,k,s,λ+Bαk,s+1Ψζ1+Ψζ2,where Uα,k,s,λ=∫01λα/k1+λsdλ.Proof.
By using Lemma11 and the Jensen-Mercer inequality and the s-convexity of ∣Ψ′∣, we have
(40)Ψθ1+θ2−ζ1+Ψθ1+θ2−ζ22−2α/k−1Γkα+kζ2−ζ1α/k×Jkθ1+θ2−ζ2+αΨθ1+θ2−ζ1+ζ22+Jkθ1+θ2−ζ1−αΨθ1+θ2−ζ1+ζ22≤ζ2−ζ14∫01λα/kΨ′θ1+θ2−1+λ2ζ1+1−λ2ζ2+Ψ′θ1+θ2−1−λ2ζ1+1+λ2ζ2dλ≤ζ2−ζ14∫01λα/kΨ′θ1+Ψ′θ2−1+λ2sΨ′ζ1+1−λ2sΨ′ζ2dλ+∫01λα/kΨ′θ1+Ψ′θ2−1−λ2sΨ′ζ1+1+λ2sΨ′ζ2dλ.
After simple computations, we get the required result of (39).Corollary 15.
Fors=α=k=1 in Theorem 14, we get
(41)Ψθ1+θ2−ζ1+Ψθ1+θ2−ζ22−1ζ2−ζ1∫θ1+θ2−ζ2θ1+θ2−ζ1Ψudu≤ζ2−ζ14Ψ′θ1+Ψ′θ2−Ψ′ζ1+Ψ′ζ22.Remark 16.
Takings=α=k=1 with ζ1=θ1 and ζ2=θ2 in Theorem 14, we recapture Theorem 2.2 in [21]:
(42)Ψθ1+Ψθ22−1θ2−θ1∫θ1θ2Ψudu≤θ2−θ14Ψ′θ1+Ψ′θ22.Remark 17.
If we chooses=k=1 in Theorem 14, we get Theorem 4 in [19].Theorem 18.
Suppose thatΨ:θ1,θ2⟶ℝ is a differentiable function on θ1,θ2 with θ1<θ2 along with assumptions A1 and A2. If Ψ′∈Lθ1,θ2 and Ψ′q is an s-convex function on θ1,θ2, where 1/r+1/q=1, r>1, with q=r/r−1, then the following inequality for Riemann-Liouville k-fractional integrals holds:
(43)Ψθ1+θ2−ζ1+Ψθ1+θ2−ζ22−2α/k−1Γkα+kζ2−ζ1α/k×Jkθ1+θ2−ζ2+αΨθ1+θ2−ζ1+ζ22+Jkθ1+θ2−ζ1−αΨθ1+θ2−ζ1+ζ22≤ζ2−ζ14×1α/kr+1rΨ′θ1q+Ψ′θ2q−2S+1−12ss+1Ψ′ζ1q+12ss+1Ψ′ζ2q1/q+Ψ′θ1q+Ψ′θ2q−12ss+1Ψ′ζ1q+2S+1−12ss+1Ψ′ζ2q1/q.Proof.
Employing Lemma11 and the Jensen-Mercer inequality with noted Hölder’s inequality and utilizing the s-convexity of Ψ′q, we have
(44)Ψθ1+θ2−ζ1+Ψθ1+θ2−ζ22−2α/k−1Γkα+kζ2−ζ1α/k×Jkθ1+θ2−ζ2+αΨθ1+θ2−ζ1+ζ22+Jkθ1+θ2−ζ1−αΨθ1+θ2−ζ1+ζ22≤ζ2−ζ14∫01λα/kΨ′θ1+θ2−1+λ2ζ1+1−λ2ζ2+Ψ′θ1+θ2−1−λ2ζ1+1+λ2ζ2dλ≤ζ2−ζ14∫01λα/krdλ1/r∫01Ψ′θ1+θ2−1+λ2ζ1+1−λ2ζ2qdλ1/q+∫01Ψ′θ1+θ2−1−λ2ζ1+1+λ2ζ2qdλ1/q≤ζ2−ζ14∫01λα/krdλ1/r×∫01Ψ′θ1q+Ψ′θ2q−1+λ2sΨ′ζ1q+1−λ2sΨ′ζ2qdλ1/q+∫01Ψ′θ1q+Ψ′θ2q−1−λ2sΨ′ζ1q+1+λ2sΨ′ζ2qdλ1/q≤ζ2−ζ141α/kr+11/rΨ′θ1q+Ψ′θ2q−2S+1−12ss+1Ψ′ζ1q+12ss+1Ψ′ζ2q1/q+Ψ′θ1q+Ψ′θ2q−12ss+1Ψ′ζ1q+2S+1−12ss+1Ψ′ζ2q1/q.
This completes the proof.Theorem 19.
Suppose thatΨ:θ1,θ2⟶ℝ is a differentiable function on θ1,θ2 with θ1<θ2 along with assumptions A1 and A2. If Ψ′∈Lθ1,θ2 and Ψ′q is an s-convex function on θ1,θ2, where 1/r+1/q=1, r≥1, with q=r/r−1, then the following inequality for Riemann-Liouville k-fractional integrals holds:
(45)Ψθ1+θ2-ζ1+Ψθ1+θ2-ζ22-2α/k-1Γkα+kζ2-ζ1α/k×Jkθ1+θ2-ζ2+αΨθ1+θ2-ζ1+ζ22+Jkθ1+θ2-ζ1-αΨθ1+θ2-ζ1+ζ22≤ζ2-ζ141α/k+11-1/q×Ψ′θ1q+Ψ′θ2qα/k+1-Uα,k,s,λ2sΨ′ζ1q+Bα/k+1,s+12sΨ′ζ2q1/q+Ψ′θ1q+Ψ′θ2qα/k+1-Bα/k+1,s+12sΨ′ζ1q+Uα,k,s,λ2sΨ′ζ2q1/q,where Uα,k,s,λ=∫01λα/k1+λsdλ.Proof.
Forr≥1, taking into account Lemma 11 and the Jensen-Mercer inequality with the noted power-mean inequality and utilizing the s-convexity of Ψ′q, we have
(46)Ψθ1+θ2-ζ1+Ψθ1+θ2-ζ22-2α/k-1Γkα+kζ2-ζ1α/k×Jkθ1+θ2-ζ2+αΨθ1+θ2-ζ1+ζ22+Jkθ1+θ2-ζ1-αΨθ1+θ2-ζ1+ζ22≤ζ2-ζ14∫01λα/k×Ψ′θ1+θ2-1+λ2ζ1+1-λ2ζ2+Ψ′θ1+θ2-1-λ2ζ1+1+λ2ζ2dλ≤ζ2-ζ14∫01λα/kdλ1-1/q×∫01λα/kΨ′θ1+θ2-1+λ2ζ1+1-λ2ζ2qdλ1/q+∫01λα/kΨ′θ1+θ2-1-λ2ζ1+1+λ2ζ2qdλ1/q≤ζ2-ζ14∫01λα/kdλ1-1/q×∫01λα/k1+λ2sΨ′θ1q+Ψ′θ2q-1+λ2sΨ′ζ1q+1-λ2sΨ′ζ2qdλ1/q+∫01λα/k1+λ2sΨ′θ1q+Ψ′θ2q-1-λ2sΨ′ζ1q+1+λ2sΨ′ζ2qdλ1/q.
Simple computations yield the desired inequality (45).Next, we demonstrate results for twice differentiable functionsΨ′′. For that, we give the following new lemma.Lemma 20.
LetΨ:θ1,θ2⟶ℝ be a differentiable mapping on θ1,θ2 with θ1<θ2 along with assumption A1. If Ψ′′∈Lθ1,θ2, then the following equality for Riemann-Liouville k-fractional integrals holds:
(47)2α/k−1Γkα+kζ2−ζ1α/kJkθ1+θ2−ζ1+ζ2/2+αΨθ1+θ2−ζ1+Jkθ1+θ2−ζ1+ζ2/2−αΨθ1+θ2−ζ2−Ψθ1+θ2−ζ1+ζ22=ζ2−ζ128α/k+1∫011−λα/k+1Ψ″θ1+θ2−1+λ2ζ1+1−λ2ζ2dλ+∫011−λα/k+1Ψ″θ1+θ2−1−λ2ζ1+1+λ2ζ2dλ.Proof.
It suffices to write that(48)I=ζ2−ζ128α/k+1I1+I2,where
(49)I1=∫011−λα/k+1Ψ″θ1+θ2−1+λ2ζ1+1−λ2ζ2dλ=−2ζ2−ζ1Ψ′θ1+θ2−ζ1+ζ22+2α/k+1θ2−θ1×∫011−λα/kΨ′θ1+θ2−1+λ2ζ1+1−λ2ζ2dλ=−2ζ2−ζ1Ψ′θ1+θ2−ζ1+ζ22−4α/k+1ζ2−ζ12Ψθ1+θ2−ζ1+ζ22+4α/k+1ζ2−ζ12∫011−λα/k−1Ψθ1+θ2−1+λ2ζ1+1−λ2ζ2dλ=−2ζ2−ζ1Ψ′θ1+θ2−ζ1+ζ22−4α/k+1ζ2−ζ12Ψθ1+θ2−ζ1+ζ22+2α/k+2α/k+1Γkα+kζ2−ζ1α/k+2Jkθ1+θ2−ζ1+ζ2/2+αΨθ1+θ2−ζ1,and similarly, we can find
(50)I2=∫011−λα/k+1Ψ″θ1+θ2−1−λ2ζ1+1+λ2ζ2dλ=2ζ2−ζ1Ψ′θ1+θ2−ζ1+ζ22−4α/k+1ζ2−ζ12Ψθ1+θ2−ζ1+ζ22+2α/k+2α/k+1Γkα+kζ2−ζ1α/k+2kJθ1+θ2−ζ1+ζ2/2−αΨθ1+θ2−ζ2.
Combining (49) and (50) with (48), we get the identity (47).Remark 21.
In Lemma20, taking k=1, with ζ1=θ1 and ζ2=θ2, recaptures Lemma 1 in [22].Remark 22.
Forα=k=1, with ζ1=θ1 and ζ2=θ2, in Lemma 20, it reduces to Lemma 2 proved in [22].Theorem 23.
Suppose thatΨ:θ1,θ2⟶ℝ is a differentiable mapping on θ1,θ2 with θ1<θ2 and Ψ∈Lθ1,θ2 along with assumptions A1 and A2. If Ψ′′ is an s-convex function on θ1,θ2, then the following inequality holds:
(51)2α/k−1Γkα+kζ2−ζ1α/kkJθ1+θ2−ζ1+ζ2/2+αΨθ1+θ2−ζ1+kJθ1+θ2−ζ1+ζ2/2−αΨθ1+θ2−ζ2−Ψθ1+θ2−ζ1+ζ22≤ζ2−ζ128α/k+12Ψ″θ1+Ψ″θ2α/k+2−U1α,k,s,λ2s+Γα/k+s+22sΓα/k+s+3Ψ″ζ1+Ψ″ζ2,where U1α,k,s,λ=∫011−λα/k+11+λsdλ.Proof.
By using Lemma20 with the Jensen-Mercer inequality and the s-convexity of Ψ′′, we have
(52)2α/k−1Γkα+kζ2−ζ1α/kkJθ1+θ2−ζ1+ζ2/2+αΨθ1+θ2−ζ1+kJθ1+θ2−ζ1+ζ2/2−αΨθ1+θ2−ζ2−Ψθ1+θ2−ζ1+ζ22≤ζ2−ζ128α/k+1∫011−λα/k+1Ψ″θ1+θ2−1+λ2ζ1+1−λ2ζ2dλ+∫011−λα/k+1Ψ″θ1+θ2−1−λ2ζ1+1+λ2ζ2dλ≤ζ2−ζ128α/k+1×∫011−λα/k+1Ψ″θ1+Ψ″θ2−1+λ2sΨ″ζ1+1−λ2sΨ″ζ2dλ+∫011−λα/k+1Ψ″θ1+Ψ″θ2−1−λ2sΨ″ζ1+1+λ2sΨ″ζ2dλ.
After simplifications, we get the required result.Remark 24.
For choosingk=1 with ζ1=θ1 and ζ2=θ2 in Theorem 23, we will get Theorem 2 proved in [22].Theorem 25.
Suppose thatΨ:θ1,θ2⟶ℝ is a differentiable function on I along with assumptions A1 and A2. If Ψ′′∈Lθ1,θ2 and Ψ′′q is an s-convex function, where 1/r+1/q=1, q>1, then the following inequality holds:
(53)2α/k−1Γkα+kζ2−ζ1α/kJkθ1+θ2−ζ1+ζ2/2+αΨθ1+θ2−ζ1+Jkθ1+θ2−ζ1+ζ2/2−αΨθ1+θ2−ζ2−Ψθ1+θ2−ζ1+ζ22≤ζ2−ζ128α/k+11α/k+1r+11/r×Ψ″θ1q+Ψ″θ2q−2S+1−12ss+1Ψ″ζ1q+12ss+1Ψ″ζ2q1/q+Ψ″θ1q+Ψ″θ2q−12ss+1Ψ″ζ1q+2S+1−12ss+1Ψ″ζ2q1/q.Proof.
By using Lemma20 and the well-known Hölder’s inequality and the Jensen-Mercer inequality along with the fact that Ψ′′q is an s-convex function, we have
(54)2α/k-1Γkα+kζ2-ζ1α/k×kJθ1+θ2-ζ1+ζ2/2+αΨθ1+θ2-ζ1+kJθ1+θ2-ζ1+ζ2/2-αΨθ1+θ2-ζ2-Ψθ1+θ2-ζ1+ζ22≤ζ2-ζ128α/k+1∫011-λα/k+1×Ψ′θ1+θ2-1+λ2ζ1+1-λ2ζ2dλ+∫011-λα/k+1×Ψ′θ1+θ2-1-λ2ζ1+1+λ2ζ2dλ≤ζ2-ζ128α/k+1×∫011-λrα/k+1dλ×∫01Ψ′θ1+Ψ′θ2-1+λ2Ψ′ζ1+1-λ2Ψ′ζ2qdλ1/q+∫011-λrα/k+1dλ1/r×∫01Ψ′θ1+Ψ′θ2-1-λ2Ψ′ζ1+1+λ2Ψ′ζ2qdλ1/q≤ζ2-ζ128α/k+1∫011-λα/k+1rdλ1/r×∫011-λ2sΨ′θ1q+Ψ′θ2q-1+λ2sΨ′ζ1q+1-λ2sΨ′ζ2q1/q+∫011-λ2sΨ′θ1q+Ψ′θ2q-1-λ2sΨ′ζ1q+1+λ2sΨ′ζ2q1/q.
By direct computations, we get the required result.Remark 26.
For choosingk=1 with ζ1=θ1 and ζ2=θ2 in Theorem 25, we will get Theorem 3 proved in [22].Corollary 27.
Forα=k=1 in Theorem 25, we get
(55)1ζ2−ζ1∫θ1+θ2−ζ2θ1+θ2−ζ1Ψudu−Ψθ1+θ2−ζ1+ζ22≤ζ2−ζ121612r+11/r×Ψ′′θ1q+Ψ′′θ2q−2s+1−12ss+1Ψ′′ζ1q+12ss+1Ψ′′ζ2q1/q+Ψ′′θ1q+Ψ′′θ2q−12ss+1Ψ′′ζ1q+2s+1−12ss+1Ψ′′ζ2q1/q.
Finally, we state our results for third-order differentiable functionsΨ′′′.Lemma 28.
LetΨ:θ1,θ2⟶ℝ be a differentiable mapping on θ1,θ2 with θ1<θ2 along with assumption A1. If Ψ′′′∈Lθ1,θ2, then the following equality for Riemann-Liouville k-fractional integrals holds:
(56)2α/k−1Γkα+kζ2−ζ1α/k×kJθ1+θ2−ζ1+ζ2/2+αΨθ1+θ2−ζ1+kJθ1+θ2−ζ1+ζ2/2−αΨθ1+θ2−ζ2−ζ2−ζ124α/k+1α/k+2Ψ″θ1+θ2−ζ1+ζ22−Ψθ1+θ2−ζ1+ζ22=ζ2−ζ1316α/k+1α/k+2∫011−λα/k+2Ψ‴θ1+θ2−1+λ2ζ1+1−λ2ζ2dλ−∫011−λα/k+2Ψ‴θ1+θ2−1−λ2ζ1+1+λ2ζ2dλ.Proof.
It suffices to write that(57)I=ζ2−ζ1316α/k+1α/k+2I1−I2,where
(58)I1=∫011−λα/k+2Ψ‴θ1+θ2−1+λ2ζ1+1−λ2ζ2dλ=−2ζ2−ζ1Ψ″θ1+θ2−ζ1+ζ22+2α/k+2ζ2−ζ1∫011−λα/k+1Ψ′′θ1+θ2−1+λ2ζ1+1−λ2ζ2dλ=−2ζ2−ζ1Ψ″θ1+θ2−ζ1+ζ22−4α/k+2ζ2−ζ12Ψ′θ1+θ2−ζ1+ζ22+4α/k+1α/k+2ζ2−ζ12∫011−λα/kΨ′θ1+θ2−1+λ2ζ1+1−λ2ζ2dλ=−2ζ2−ζ1Ψ″θ1+θ2−ζ1+ζ22−4α/k+2ζ2−ζ12Ψ′θ1+θ2−ζ1+ζ22−8α/k+2α/k+1ζ2−ζ13Ψθ1+θ2−ζ1+ζ22+2α/k+3α/k+1α/k+2Γkα+kζ2−ζ1α/k+3kJθ1+θ2−ζ1+ζ2/2+αΨθ1+θ2−ζ1=−2ζ2−ζ1Ψ″θ1+θ2−ζ1+ζ22−4α/k+2ζ2−ζ12Ψ′θ1+θ2−ζ1+ζ22−8α/k+2α/k+1ζ2−ζ13Ψθ1+θ2−ζ1+ζ22+2α/k+3Γkα+3kζ2−ζ1α/k+3Jθ1+θ2−ζ1+ζ2/2+αΨθ1+θ2−ζ1,and similarly, we can find
(59)I2=∫011−λα/k+2Ψ′′′θ1+θ2−1−λ2ζ1+1+λ2ζ2dλ=2ζ2−ζ1Ψ′′θ1+θ2−ζ1+ζ22−4α/k+2ζ2−ζ12Ψ′θ1+θ2−ζ1+ζ22+8α/k+2α/k+1ζ2−ζ13Ψθ1+θ2−ζ1+ζ22−2α/k+3Γkα+3kζ2−ζ1α/k+3Jθ1+θ2−ζ1+ζ2/2−αΨθ1+θ2−ζ2.
Replacing the values of the integralsI1 and I2 in (57), we get the identity (56).Remark 29.
In Lemma28, choosing k=1 with ζ1=θ1 and ζ2=θ2, it recaptures Lemma 3.1 proved in [23].Remark 30.
Fork=α=1 with ζ1=θ1 and ζ2=θ2 in Lemma 28, it reduces to Lemma 2.1 proved in [24].Theorem 31.
Suppose thatΨ:θ1,θ2⟶ℝ is a three times differentiable mapping on θ1,θ2 with θ1<θ2 and Ψ∈Lθ1,θ2 along with assumptions A1 and A2. If Ψ′′′ is an s-convex function on θ1,θ2, then the following inequality for k-fractional integrals holds:
(60)2α/k−1Γkα+kζ2−ζ1α/k×kJθ1+θ2−ζ1+ζ2/2+αΨθ1+θ2−ζ1+kJθ1+θ2−ζ1+ζ2/2−αΨθ1+θ2−ζ2−ζ2−ζ124α/k+1α/k+2Ψ″θ1+θ2−ζ1+ζ22−Ψθ1+θ2−ζ1+ζ22≤ζ2−ζ1316α/k+1α/k+22Ψ‴θ1+Ψ‴θ2α/k+3−U2α,k,s,λ2s+12sα/k+s+3Ψ‴ζ1+Ψ‴ζ2,where U2α,k,s,λ=∫011−λα/k+21+λsdλ.Proof.
By using Lemma28 and the Jensen-Mercer inequality and the s-convexity of Ψ′′′, we have
(61)2α/k−1Γkα+kζ2−ζ1α/k×kJθ1+θ2−ζ1+ζ2/2+αΨθ1+θ2−ζ1+kJθ1+θ2−ζ1+ζ2/2−αΨθ1+θ2−ζ2−ζ2−ζ124α/k+1α/k+2Ψ″θ1+θ2−ζ1+ζ22−Ψθ1+θ2−ζ1+ζ22≤ζ2−ζ1316α/k+1α/k+2∫011−λα/k+2Ψ‴θ1+θ2−1+λ2ζ1+1−λ2ζ2dλ−∫011−λα/k+2Ψ‴θ1+θ2−1−λ2ζ1+1+λ2ζ2dλ≤ζ2−ζ1316α/k+1α/k+2∫011−λα/k+2Ψ‴θ1+Ψ‴θ2−1+λ2sΨ‴ζ1+1−λ2sΨ‴ζ2dλ+∫011−λα/k+2Ψ‴θ1+Ψ‴θ2−1−λ2sΨ‴ζ1+1+λ2sΨ‴ζ2dλ≤ζ2−ζ1316α/k+1α/k+22Ψ‴θ1+Ψ‴θ2α/k+3−U2α,k,s,λ2s+12sα/k+s+3Ψ‴ζ1+Ψ‴ζ2.Remark 32.
Choosingk=s=1 with ζ1=θ1 and ζ2=θ2 in Theorem 31, we will get Theorem 18 proved in [23].Theorem 33.
Suppose thatΨ:θ1,θ2⟶ℝ is a three times differentiable function on I along with assumptions A1 and A2. If Ψ′′′∈Lθ1,θ2 and Ψ′′′q is an s-convex function, where 1/r+1/q=1, q>1, then the following k-fractional integral inequality holds:
(62)2α/k−1Γkα+kζ2−ζ1α/k×kJθ1+θ2−ζ1+ζ2/2+αΨθ1+θ2−ζ1+kJθ1+θ2−ζ1+ζ2/2−αΨθ1+θ2−ζ2−ζ2−ζ124α/k+1α/k+2Ψ″θ1+θ2−ζ1+ζ22−Ψθ1+θ2−ζ1+ζ22≤ζ2−ζ1316α/k+1α/k+21α/k+2r+11/r×Ψ‴θ1q+Ψ‴θ2q−2S+1−12ss+1Ψ‴ζ1q+12ss+1Ψ‴ζ2q1/q+Ψ‴θ1q+Ψ‴θ2q−12ss+1Ψ‴ζ1q+2S+1−12ss+1Ψ‴ζ2q1/q.Proof.
By using Lemma28 with the Jensen-Mercer inequality and the well-known Hölder’s inequality on the fact that Ψ′′′q is s-convexity, we have
(63)2α/k-1Γkα+kζ2-ζ1α/k×kJθ1+θ2-ζ1+ζ2/2+αΨθ1+θ2-ζ1+kJθ1+θ2-ζ1+ζ2/2-αΨθ1+θ2-ζ2-ζ2-ζ124α/k+1α/k+2×Ψ′′θ1+θ2-ζ1+ζ22-Ψθ1+θ2-ζ1+ζ22≤ζ2-ζ1316α/k+1α/k+2×∫011-λα/k+2×Ψ′′′θ1+θ2-1+λ2ζ1+1-λ2ζ2dλ-∫011-λα/k+2×Ψ′′′θ1+θ2-1-λ2ζ1+1+λ2ζ2dλ≤ζ2-ζ1316α/k+1α/k+2×∫011-λα/k+2rdλ1/r×∫01Ψ′′′θ1+Ψ′′′θ2-1+λ2Ψ′′′ζ1+1-λ2Ψ′′′ζ2qdλ1/q+∫011-λα/k+2rdλ1/r×∫01Ψ′′′θ1+Ψ′′′θ2-1-λ2Ψ′′′ζ1+1+λ2Ψ′′′ζ2qdλ1/q≤ζ2-ζ1316α/k+1α/k+2×∫011-λα/k+2rdλ1/r×∫01Ψ′′′θ1q+Ψ′′′θ2q-1+λ2sΨ′′′ζ1q+1-λ2sΨ′′′ζ2q1/q+∫01Ψ′′′θ1q+Ψ′′′θ2q-1-λ2sΨ′′′ζ1q+1+λ2sΨ′′′ζ2q1/q.
After some simplifications, we get the required results.Remark 34.
Choosingk=s=1 with ζ1=θ1 and ζ2=θ2 in Theorem 33, we will get Theorem 19 proved in [23].


## 4. Applications

### 4.1. Some Applications of the Means

Let us consider the following special means for different values ofa1 and a2:
(1)
The arithmetic mean:(64)Aa1,a2=a1+a22.(2)
The geometric mean:(65)Ga1,a2=a1a21/2.(3)
The harmonic mean:(66)H=2a1a2a1+a2.Proposition 35.
Supposea1,a2∈ℝ,0<a1<a2,0∉a1,a2. Then, for all r>1, the following inequality holds:
(67)1ζ2−ζ1s+12Aa1,a2−a1s+1−2Aa1,a2−a2s+1−2Aa1,a2s≤ζ2−ζ1216ss−112r+11/r×2Aa1s−2q,a2s−2q−2s+1−12ss+1a1s−2q+12ss+1a2s−2q1/q+2Aa1s−2q,a2s−2q−12ss+1a1s−2q+2s+1−12ss+1a2s−2q1/q.Proof.
The proof is an immediate consequence from Corollary27 by selecting Ψx=xs and s∈0,1.Proposition 36.
Supposea1,a2∈ℝ,s∈0,1,0<a1<a2,0∉a1,a2. Then, for all r>1, the following inequality holds:
(68)1ζ2−ζ11−s2Aa1,a2−a2s−1−2Aa1,a2−a1s−1G22Aa1,a2−a1,2Aa1,a2−a2s−1−Ha1,a2s≤ζ2−ζ1216ss+112r+11/r×2Aa1−s+2q,a2−s+2q−2s+1−12ss+1a1−s+2q+12ss+1a2−s+2q1/q+2Aa1−s+2q,a2−s+2q−12ss+1a1−s+2q+2s+1−12ss+1a2−s+2q1/q.Proof.
The proof is an immediate consequence from Corollary27 by taking into account Lemma 4 by selecting Ψx=1/xs and s∈0,1.


### 4.2.q-Digamma Function

Suppose0<q<1; the q-digamma function φq is the q-analogue of the digamma function φ (see [25]) given as
(69)φq=−ln1−q+lnq∑k=0∞qk+ζ1−qk+ζ=−ln1−q+lnq∑k=0∞qkζ1−qkζ.Forq>1 and ζ>0, the q-digamma function φq can be given as
(70)φq=−lnq−1+lnqζ−12−∑k=0∞q−k+ζ1−q−k+ζ=−lnq−1+lnqζ−12−∑k=0∞q−kζ1−q−kζ.Proposition 37.
Leta1, a2, q, and q be real numbers such that 0<a1<a2, q>1, and 0<q<1. Then, the following inequality holds:
(71)φq2Aa1,a2−a1−φq2Aa1,a2−a2ζ2−ζ1−φq′Aa1,a2≤ζ2−ζ121612r+11/r×φq3a1q+φq3a2q−2s+1−12ss+1φq3a1q+12ss+1φq3a2q1/q+φq3a1q+φq3a2q−12ss+1φq3a1q+2s+1−12ss+1φq3a2q1/q.Proof.
By employing the definition of theq-digamma function φqζ, it is easy to notice that the q-trigamma function ζ⟶φq′ζ is completely monotonic on 0,∞. This ensures that the function φq3 is again completely monotonic on 0,∞ for each q∈0,1 and consequently is convex and nonnegative (see [26], p. 167). Now, by applying Corollary 27, we extract that the inequality (71) is valid for q∈0,1.
As another application of inequality (71), we can deliver the following inequalities for the q-trigamma and q-polygamma functions and the analogue of harmonic numbers Hnq defined by
(72)Hnq=∑qk1−qk,n∈N.
So, from inequality (71), we use the equation
(73)φqn+1=φq1−logqHnq,n∈N.
Consequently, we obtain the following result.Corollary 38.
Supposen∈N, q>1, and 0<q<1. Then, the following inequality holds:
(74)−logqHnqn−φq′A1,n+1≤n21612r+11/r×φq31q+φq3n+1q−2s+1−12ss+1φq31q+12ss+1φq3n+1q1/q+φq31q+φq3n+1q−12ss+1φq31q+2s+1−12ss+1φq3n+1q1/q.Proposition 39.
Supposen is an integer and q>1. Then, the following inequality holds:
(75)Hnn−φ′A1,n+1≤n21612r+11/r×φ31q+φ3n+1q−2s+1−12ss+1φ31q+12ss+1φ3n+1q1/q+φ31q+φ3n+1q−12ss+1φ31q+2s+1−12ss+1φ3n+1q1/q.Proof.
From inequality (74), when q⟶1, we use the relation
(76)limq⟶1logqHnq=limq⟶1logqq−1.q−1Hnq=−limq⟶1∑k=1n1−q1−qk=−Hn.
We obtain the required result.Remark 40.
By using the equation(77)Hn=γ+φn+1,where γ is the Euler-Mascheroni constant, the inequality (75) becomes
(78)γ+φn+1n−φ′A1,n+1≤n21612r+11/r×φ31q+φ3n+1q−2s+1−12ss+1φ31q+12ss+1φ3n+1q1/q+φ31q+φ3n+1q−12ss+1φ31q+2s+1−12ss+1φ3n+1q1/q.


## 4.1. Some Applications of the Means

Let us consider the following special means for different values ofa1 and a2:
(1)
The arithmetic mean:(64)Aa1,a2=a1+a22.(2)
The geometric mean:(65)Ga1,a2=a1a21/2.(3)
The harmonic mean:(66)H=2a1a2a1+a2.Proposition 35.
Supposea1,a2∈ℝ,0<a1<a2,0∉a1,a2. Then, for all r>1, the following inequality holds:
(67)1ζ2−ζ1s+12Aa1,a2−a1s+1−2Aa1,a2−a2s+1−2Aa1,a2s≤ζ2−ζ1216ss−112r+11/r×2Aa1s−2q,a2s−2q−2s+1−12ss+1a1s−2q+12ss+1a2s−2q1/q+2Aa1s−2q,a2s−2q−12ss+1a1s−2q+2s+1−12ss+1a2s−2q1/q.Proof.
The proof is an immediate consequence from Corollary27 by selecting Ψx=xs and s∈0,1.Proposition 36.
Supposea1,a2∈ℝ,s∈0,1,0<a1<a2,0∉a1,a2. Then, for all r>1, the following inequality holds:
(68)1ζ2−ζ11−s2Aa1,a2−a2s−1−2Aa1,a2−a1s−1G22Aa1,a2−a1,2Aa1,a2−a2s−1−Ha1,a2s≤ζ2−ζ1216ss+112r+11/r×2Aa1−s+2q,a2−s+2q−2s+1−12ss+1a1−s+2q+12ss+1a2−s+2q1/q+2Aa1−s+2q,a2−s+2q−12ss+1a1−s+2q+2s+1−12ss+1a2−s+2q1/q.Proof.
The proof is an immediate consequence from Corollary27 by taking into account Lemma 4 by selecting Ψx=1/xs and s∈0,1.


## 4.2.q-Digamma Function

Suppose0<q<1; the q-digamma function φq is the q-analogue of the digamma function φ (see [25]) given as
(69)φq=−ln1−q+lnq∑k=0∞qk+ζ1−qk+ζ=−ln1−q+lnq∑k=0∞qkζ1−qkζ.Forq>1 and ζ>0, the q-digamma function φq can be given as
(70)φq=−lnq−1+lnqζ−12−∑k=0∞q−k+ζ1−q−k+ζ=−lnq−1+lnqζ−12−∑k=0∞q−kζ1−q−kζ.Proposition 37.
Leta1, a2, q, and q be real numbers such that 0<a1<a2, q>1, and 0<q<1. Then, the following inequality holds:
(71)φq2Aa1,a2−a1−φq2Aa1,a2−a2ζ2−ζ1−φq′Aa1,a2≤ζ2−ζ121612r+11/r×φq3a1q+φq3a2q−2s+1−12ss+1φq3a1q+12ss+1φq3a2q1/q+φq3a1q+φq3a2q−12ss+1φq3a1q+2s+1−12ss+1φq3a2q1/q.Proof.
By employing the definition of theq-digamma function φqζ, it is easy to notice that the q-trigamma function ζ⟶φq′ζ is completely monotonic on 0,∞. This ensures that the function φq3 is again completely monotonic on 0,∞ for each q∈0,1 and consequently is convex and nonnegative (see [26], p. 167). Now, by applying Corollary 27, we extract that the inequality (71) is valid for q∈0,1.
As another application of inequality (71), we can deliver the following inequalities for the q-trigamma and q-polygamma functions and the analogue of harmonic numbers Hnq defined by
(72)Hnq=∑qk1−qk,n∈N.
So, from inequality (71), we use the equation
(73)φqn+1=φq1−logqHnq,n∈N.
Consequently, we obtain the following result.Corollary 38.
Supposen∈N, q>1, and 0<q<1. Then, the following inequality holds:
(74)−logqHnqn−φq′A1,n+1≤n21612r+11/r×φq31q+φq3n+1q−2s+1−12ss+1φq31q+12ss+1φq3n+1q1/q+φq31q+φq3n+1q−12ss+1φq31q+2s+1−12ss+1φq3n+1q1/q.Proposition 39.
Supposen is an integer and q>1. Then, the following inequality holds:
(75)Hnn−φ′A1,n+1≤n21612r+11/r×φ31q+φ3n+1q−2s+1−12ss+1φ31q+12ss+1φ3n+1q1/q+φ31q+φ3n+1q−12ss+1φ31q+2s+1−12ss+1φ3n+1q1/q.Proof.
From inequality (74), when q⟶1, we use the relation
(76)limq⟶1logqHnq=limq⟶1logqq−1.q−1Hnq=−limq⟶1∑k=1n1−q1−qk=−Hn.
We obtain the required result.Remark 40.
By using the equation(77)Hn=γ+φn+1,where γ is the Euler-Mascheroni constant, the inequality (75) becomes
(78)γ+φn+1n−φ′A1,n+1≤n21612r+11/r×φ31q+φ3n+1q−2s+1−12ss+1φ31q+12ss+1φ3n+1q1/q+φ31q+φ3n+1q−12ss+1φ31q+2s+1−12ss+1φ3n+1q1/q.


## 5. Conclusion

In this paper, we have explored newk-fractional variants of Hermite-Mercer-type integral inequalities for s-convex functions. New results and novel connections are built for the left and right sides of Hermite-Hadamard-type inequalities for differentiable mappings whose derivatives in absolute values at certain powers are s-convex in the second sense. New integral identities for differentiable mappings are obtained, and related results are established. In the application viewpoint, our findings illustrate new generalizations with the connection of special function theory (special means of real numbers and q-digamma function) and harmonic numbers. It is quite open to think about Jensen-Hermite-Mercer variants for generalized integral operators having nonlocal and nonsingular kernels by applying generalized convexities. However, it is not easy to extend such inequalities for other existing types of convexities. The suggested scheme is viable, effective, and computationally appealing in fractional differential equations, optimization theory, and other related areas of convexity.



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