# The Hermite–Hadamard–Mercer Type Inequalities via Generalized Proportional Fractional Integral Concerning Another Function

Authors: Tariq A. Aljaaidi; Deepak B. Pachpatte
Journal: International Journal of Mathematics and Mathematical Sciences
 (2022)

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## Abstract

In order to be able to study cosmic phenomena more accurately and broadly, it was necessary to expand the concept of calculus. In this study, we aim to introduce a new fractional Hermite–Hadamard–Mercer’s inequality and its fractional integral type inequalities. To facilitate that, we use the proportional fractional integral operators of integrable functions with respect to another continuous and strictly increasing function. Moreover, we establish some new fractional weightedφ-proportional fractional integral Hermite–Hadamard–Mercer type inequalities. Furthermore, in this article, we are keen to present some special cases related to our current study compared to the previous work of the inequality under study.


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## Body

## 1. Introduction

There is no doubt that a researcher in the field of calculus knows the significant importance that fractional calculus has acquired recently due to its multiple and important uses in many fields in the natural sciences and technology, especially in physics, fluid dynamics, biology, image processing, control theory, computer networking, and signal processing.Fractional calculus is the generalized form of classical integrals and derivatives for the order is a noninteger, which comes within the framework of mathematicians’ relentless pursuit of developing mathematics to make it more general and useable in most cases that may encounter when studying and analyzing natural phenomena. According to this, we can say fractional calculus has become the focus of a large number of researchers’ attention. As a result, a lot of extensions and generalizations have appeared especially on the classical fractional calculus like the definitions of Riemann–Liouville (RL) and Caputo. Actually, the derivative Riemann–Liouville is the most general concept and the most uniform and natural. In general, there are numerous other definitions of fractional operators such as Erdélyi–Kober, Hilfer, Katugampola, Hadamard, and Riesz which are just a few examples to make reference to [1, 2]. It should be noted that there are many modern fractional operators proposed by many researchers and perhaps the most prominent of them is the recently proposed ABC operator by Atangana and Baleanu [3, 4].Definition 1.
The functiong:a,z⊆ℝ⟶ℝ is said to be a convex function if the inequality(1)gηr+1−ηs≤ηgr+1−ηgs,holds for all r,s∈a,z and η∈0,1. We say that g is a concave function if inequality (1) is reversed. In general, the real-valued function g is said to be a convex function on a,z if and only if for all y1,y2,….,yn∈a,z and for any ηi∈a,z, i=1,2,….,n with ∑i=1i=nηi=1, we have(2)g∑i=1nηiyi≤∑i=1nηigyi.
This well-known inequality is called Jensen inequality [5].
Convexity of functions with their features is one of the most useful properties among other categories of functions in the important fields of applied sciences, especially statistics and mathematics, which according to its own useful definition has a geometric interpretation. Furthermore, it is a vital part of inequalities theory and has become the leading point for creating numerous inequalities such as Jensen’s inequality, Hadamard’s inequality with its type inequalities, and Steffensen’s inequality. One of these inequalities that are closely related to the convexity of functions is the Hermite–Hadamard inequality, which has a well-known area in the space of inequalities theory. This inequality was initially proposed by Hermite in 1881, but it did not come into prominence until it was enriched by Hadamard in 1893 [6] as follows:(3)ga+z2≤1a−z∫azgydy≤ga+gz2,a,z∈ℝ,a<z,where g is a convex function on a,z, which is called the Hermite–Hadamard (H-H) inequality. Significantly, H-H inequality recently has become the focus of attention of several mathematicians and researchers due to its remarkable applications and its spacious uses in several diverse areas. Concerning that, a large number of articles have appeared that contain extensions and generalizations of this inequality (see [7–12]).Lemma 1 (see [13]).
Suppose that the functiong:a,z⊆ℝ⟶ℝ is convex on a,z. Then, we have(4)ga+z−y≤ga+gz−gy.
In general, McD Mercer in [13] proved the following generalization of each of inequality (2) and inequality (4), which is called the well-known Jensen–Mercer inequality.Theorem 1.
Suppose that the functiong:a,z⊆ℝ⟶ℝ is convex on a,z. Then, for all y1,y2,….,yn∈a,z and for any ηi∈a,z, i=1,2,….,n, with ∑i=1i=nηi=1, we have(5)ga+z−∑i=1nηiyi≤ga+gz−∑i=1nηigyi.Inequality (5) is a matter of supreme interest due to much information and its explicit boundary conditions. In mathematics and engineering sciences, Jensen–Mercer’s inequality and associated inequalities have wonderful applications and their generalizations and extensions have been an excellent topic of research for mathematicians and authors in the past few years as seen through a variety of investigations on the subject. Moradi and Furuichi [14] (2020) presented some new generalizations and improvements of Jensen–Mercer’s type inequalities. Khan et al. [15] (2020) applied Jensen–Mercer’s inequality in information theory to compute new ratings for Csiszár and related divergence. For more generalizations and details of Jensen–Mercer’s type inequalities, see [16–18].Many researchers, motivated by all the above literature, did a lot of research and were able to derive a new inequality which is a mixture of H-H and Jensen–Mercer’s inequalities, which was named the Hermite–Hadamard–Mercer’s inequality which is our focus through introducing this article.Ogulmus and Sarikaya [19] (2019) established fractional integral Hermite–Hadamard–Mercer’s inequalities for RL operators.Theorem 2.
Letg:a,z⟶ℝ be a convex function. The following inequalities hold:(6)ga+z−x+y2≤ga+gz−Γβ+12y−xβℐx+βgy+ℐy−βgx≤ga+gz−gx+y2,(7)ga+z−x+y2≤Γβ+12y−xβℐa+z−y+βga+z−x+ℐa+z−x−βga+z−y≤ga+z−y+ga+z−x2≤ga+gz−gx+gy2.
The same authors, in the same work, presented the following inequality.Theorem 3.
Letg:a,z⟶ℝ be a convex function. The following inequalities hold:(8)ga+z−x+y2≤Γβ+12y−xβℐa+z−x+y2−βga+z−y+ℐa+z−x+y2+βga+z−x≤ga+gz−gx+gy2.Iscan [20] (2020) employed the RL fractional integral to investigate some weighted Hermite–Hadamard–Mercer’s type inequalities as follows.Theorem 4.
Letg:a,z⟶ℝ be a convex differentiable function on a,z and w:a,z⟶ℝ be a nonnegative integrable function. Then, the following inequalities hold:(9)ga+z−x+y2ℐa+z−x−βwa+z−y≤12φℐa+z−x−β,δgwa+z−y+ℐa+z−y+βgwa+z−x≤12ga+z−x+ga+z−yℐa+z−x−βwa+z−y.
Iscan also, in the same work, gave the following weighted Hermite–Hadamard–Mercer’s inequality.Theorem 5.
Letg:a,z⟶ℝ be a convex differentiable function on a,z and w:a,z⟶ℝ be a nonnegative integrable function. Then, the following inequalities hold:(10)ga+z−x+y2ℐa+z−x−βwa+z−y≤12ℐa+z−x−βgwa+z−y+ℐa+z−y+βgwa+z−x≤ga+gzℐx+βwy−12ℐx+βgwy+ℐy−βgwx≤ga+gz−gx+y2ℐx+βwy.Abdeljawad et al. [21] (2020) established some inequalities of Hermite–Hadamard–Mercer type inequalities employing RL fractional integral. Butt et al. [22] (2020) proved some Hermite–Hadamard–Mercer type inequalities for convex functions by using the conformable fractional integrals. Chu et al. [23] (2020) presented some generalizations of Hermite–Hadamard–Mercer type inequalities via Katugampola fractional integral. Recently, in 2021, Vivas-Cortez et al. [24] used generalized RL to present some Hermite–Hadamard–Mercer type inequalities involving convex functions. For more recent studies and generalizations of this inequality, please see [25–28].All of what we mentioned above prompts us to study Hermite–Hadamard–Mercer’s inequality via the recently generalized operators. Here, in this study, we aim to establish Hermite–Hadamard–Mercer’s inequality and its type inequalities for convex functions employing proportional fractional integral operators involving continuous strictly increasing functions. We also aim to present some fractional weighted Hermite–Hadamard–Mercer type inequalities via the current generalized integral operators. Along with this study, we are able to discuss some special cases and some relationships between our current study and previous studies.The organization of this research paper will be as follows: In Section 2, we will mention some notations, definitions, and preparatory acquaintance which are used in this work. Section 3 is devoted to the first part of our major results which contain Hermite–Hadamard–Mercer’s inequalities. Throughout Section 4, we provide the fractional weighted Hermite–Hadamard–Mercer’s type inequalities.


## 2. Essential Preliminaries

Here, we characterize some of the basic properties and some definitions of several elementary fractional integral operators which include the final generalized fractional operator we used to obtain and discuss our new results.Definition 2 (see [1]).
Suppose that the functiong is integrable on a,z and a≥0. Then, for all β>0, we have(11)ℐa+βgy=1Γβ∫ayy−uβ−1gudu,u>a,(12)ℐz−βgy=1Γβ∫yzu−yβ−1gudu,y<z,where Γβ=∫0∞e−xxβ−1dx is the Gamma function and ℐa+0gy=ℐz−0gy=gy. The notations ℐa+βgy and ℐz−βgy are called, respectively, the left- and right-sided Riemann–Liouville fractional integrals of a function g for the order β.Definition 3 (see [1, 2]).
Suppose that the functiong is integrable on the interval Ϝ, and let φ be an increasing function, where φy∈C1Ϝ,ℝ such that φ′y≠0 and y∈Ϝ. Then, for all β>0, we have(13)φℐa+βgy=1Γβ∫ayφ′uφy−φuβ−1gudu,(14)φℐz−βgy=1Γβ∫yzφ′uφu−φyβ−1gudu.
The notationsφℐa+βgy and φℐz−βgy are, respectively, called the left- and right-sided φ-Riemann–Liouville fractional integrals of a function g for the order β.Definition 4 (see [29]).
For the functiong, let δ>0, and we have for all β∈ℂ and Reβ≥0,(15)Da+β,δgy=Dm,δℐa+m−β,δgy=Dym,δδm−βΓm−β∫ayexpδ−1δy−uy−um−β−1gudu,(16)Dz−β,δgy=γDm,δℐz−m−β,δgy=γDym,δδm−βΓm−β∫yzexpδ−1δu−yu−ym−β−1gudu,where(17)Dm,δ=DδDδ….Dδ︸m−times,m=Reβ+1,(18)γDδgy=1−δgy−δg′y,γDm,δ=γDγδDδ….γDδ︸m−times.
The notationsDa+β,δgy and Dz−β,δgy are, respectively, called the left- and right-sided proportional fractional derivatives of a function g for the order β.Definition 5 (see [29]).
For the integrable functiong, let δ>0, and we have for all β∈ℂ and Reβ≥0,(19)ℐa+β,δgy=1δβΓβ∫ayexpδ−1δy−uy−uβ−1gudu,(20)ℐz−β,δgy=1δβΓβ∫yzexpδ−1δu−yu−yβ−1gudu.
The notationsℐa+β,δgy and ℐz−β,δgy are, respectively, called the left- and right-sided proportional fractional integrals of a function g for the order β.Definition 6 (see [30]).
For the integrable functiong and for the strictly increasing continuous function φ on a,z, let δ∈(0,1], and we have for all β∈ℂ and Reβ≥0,(21)φDa+β,δgy=φDm,δφℐa+m−β,δgy=φDym,δδm−βΓm−β∫ayexpδ−1δφy−φuφy−φum−β−1φ′ugudu,(22)φDz−β,δgy=γφDm,δφℐz−m−β,δgy=φγDym,δδm−βΓm−β∫yzexpδ−1δφu−φyφu−φym−β−1φ′ugudu,where(23)φDm,δ=φDδφDδ…..φDδ︸m−times,m=Reβ+1,(24)φγDδgy=1−δgy−δg′yφ′y,φγDm,δ=φγDγδφDδ…..φγDδ︸m−times.
The notationsφDa+β,δgy and φDz−β,δgy are, respectively, called the left- and right-sided proportional fractional derivatives of a function g with respect to φ for the order β.Definition 7 (see [30]).
For the integrable functiong and for the continuous and strictly increasing function φ on a,z, let δ∈(0,1], and we have for all β∈ℂ and Reβ≥0,(25)φℐa+β,δgy=1δβΓβ∫ayexpδ−1δφy−φu(φy−φu)β−1φ′ugudu,(26)φℐz−β,δgy=1δβΓβ∫yzexpδ−1δφu−φy(φu−φy)β−1φ′ugudu.
The notationsφℐa+β,δgy and φℐz−β,δgy are, respectively, called the left- and right-sided proportional fractional integrals of a function g with respect to φ for the order β.Lemma 2 (see [30]).
Letφ be a continuous function on y≥a. If δ∈(0,1] and Reα,Reβ>0, we have(27)φℐa+β,δφℐa+α,δgy=φℐa+α,δφℐa+β,δgy=φℐa+β+α,δgy,(28)φℐz−β,δφℐz−α,δgy=φℐz−β,υφℐz−α,δgy=φℐz−β+α,δgy.Lemma 3 (see [30]).
Letφ be an integrable function defined on a,y and δ>a. If 0≤m<Reβ+1, then we have(29)φDm,δφℐa+β,δgy=φℐa+β−m,δgy,(30)φγDm,δφℐz−β,δgy=φℐz−β−m,δgy.
In this paper, we need the following identity as in [31].
Letδ∈(0,1], β∈ℂ, Reβ≥0, and φ be a strictly increasing continuous function. Then, for any constant k, we have(31)φℐx+β,δky=(φy−φx)βδβΓβ+1k.


## 3. Fractional Hermite–Hadamard–Mercer Inequalities Involvingφ-Proportional Fractional Integrals

This section is the first part of our main contributions. Here, we present basic generalization in Hermite–Hadamard–Mercer’s inequalities which involve convex functions for generalized proportional fractional integral operators concerning another strictly increasing continuous function.Theorem 6.
Letφ:Ϝ⟶a,z⊆ℝ, with 0≤a<z, be a continuous strictly increasing function and g:a,z⟶ℝ be a convex differentiable function on a,z satisfying that g°φ:Ϝ⟶ℝ is an integrable mapping on Ϝ. Then, we have(32)ga+z−x+y2≤ga+gz−δβΓβ+12y−xβφℐφ−1x+β,δg°φφ−1y+φℐφ−1y−β,δg°φφ−1x≤ga+gz−gx+y2,(33)ga+z−x+y2≤δβΓβ+12y−xβφℐφ−1a+z−y+β,δg°φφ−1a+z−x+φℐφ−1a+z−x−β,δg°φφ−1a+z−y≤ga+z−y+ga+z−x2≤ga+gz−gx+gy2.Proof.
According to the Jensen–Mercer inequality and forr,s∈a,z, we have(34)ga+z−r+s2≤ga+gz−gr+gs2.
Now, we change the variablesrands with r=ηx+1−ηy and s=1−ηx+ηy, and we get(35)ga+z−ηx+1−ηy+1−ηx+ηy2≤ga+gz−gηx+1−ηy+g1−ηx+ηy2,which leads to(36)ga+z−x+y2≤ga+gz−gηx+1−ηy+g1−ηx+ηy2.
On both sides of (36), taking product by expδ−1/δηy−xηβ−1 and then integrating the estimating inequality with respect to η over 0,1, we obtain(37)∫01ga+z−x+y2expδ−1δηy−xηβ−1dη≤∫01ga+gzexpδ−1δηy−xηβ−1dη−∫01expδ−1δηy−xgηx+1−ηy+g1−ηx+ηy2ηβ−1dη.
Using identity (31) on both sides of (37), we obtain(38)2βga+z−x+y2≤2βga+gz−∫01expδ−1δηy−xηβ−1gηx+1−ηydη+∫01expδ−1δηy−xηβ−1g1−ηx+ηydη.
Next,(39)δβΓβ+12y−xβφℐφ−1x+β,δg°φφ−1y+φℐφ−1y−β,δg°φφ−1x=β2y−xβ∫φ−1xφ−1yexpδ−1δy−φuy−φuβ−1g°φuφ′udu+∫φ−1xφ−1yexpδ−1δφv−xφv−xβ−1g°φvφ′vdv=β2∫φ−1xφ−1yexpδ−1δy−φuy−φuy−xβ−1g°φuφ′uy−xdu+∫φ−1xφ−1yexpδ−1δφv−xφv−xy−xβ−1g°φvφ′vy−xdv.
Puttingφu=ηx+1−ηy and φv=1−ηx+ηy, we get(40)ga+gz−δβΓβ+12y−xβφℐφ−1x+β,δg°φφ−1y+φℐφ−1y−β,δg°φφ−1x=ga+gz−β2∫01expδ−1δηy−xηβ−1gηx+1−ηydη+∫01expδ−1δηy−xηβ−1g1−ηx+ηydη≥ga+z−x+y2.
This proves the first inequality in (32). To prove the second inequality and by using the convexity of g, we can be certain that(41)gφu+φv2≤g°φu+g°φv2.Then, for φu=ηx+1−ηy and φv=1−ηx+ηy, we have(42)gx+y2≤gηx+1−ηy+g1−ηx+ηy2,η∈0,1.
On both sides of (42), taking product by expδ−1/δηy−xηβ−1 and then integrating the estimating inequality with respect to η over 0,1, we get(43)gx+y2≤β2∫01expδ−1δηy−xηβ−1gηx+1−ηydη+∫01expδ−1δηy−xηβ−1g1−ηx+ηydη=δβΓβ+12y−xβφℐφ−1x+β,δg°φφ−1y+φℐφ−1y−β,δg°φφ−1x.
Therefore, we have(44)−gx+y2≥−δβΓβ+12y−xβφℐφ−1x+β,δg°φφ−1y+φℐφ−1y−β,δg°φφ−1x.
On both sides of inequality (44), adding ga+gz, we get the second inequality in (32). Hence, desired inequality (32) is thus proved. We now give the proof of inequalities in (33). We have, according to the convexity of the function g, for all r,s∈a,z, that(45)ga+z−r+s2=ga+z−r+a+z−s2≤12ga+z−r+ga+z−s.
By applying the change of variablesa+z−r=ηa+z−x+1−ηa+z−y and a+z−s=ηa+z−y+1−ηa+z−x, we obtain(46)ga+z−x+y2≤12gηa+z−x+1−ηa+z−y+gηa+z−y+1−ηa+z−x.
On both sides of (46), taking product by expδ−1/δηy−xηβ−1 and then integrating the estimating inequality with respect to η over 0,1, we obtain(47)ga+z−x+y2≤β2∫01expδ−1δηy−xηβ−1gηa+z−x+1−ηa+z−ydη+∫01expδ−1δηy−xηβ−1gηa+z−y+1−ηa+z−xdη.
Next,(48)δβΓβ+12y−xβφℐφ−1a+z−y+β,δg°φφ−1a+z−x+φℐφ−1a+z−x−β,δg°φφ−1a+z−y=β2y−xβ∫φ−1a+z−yφ−1a+z−xexpδ−1δa+z−x−φua+z−x−φuβ−1g°φuφ′udu+∫φ−1a+z−yφ−1a+z−xexpδ−1δφv−a+z−yφv−a+z−yβ−1g°φvφ′vdv=β2∫φ−1a+z−yφ−1a+z−xexpδ−1δa+z−x−φua+z−x−φuy−xβ−1g°φuφ′uy−xdu+∫φ−1a+z−yφ−1a+z−xexpδ−1δφv−a+z−yφv−a+z−yy−xβ−1g°φvφ′vy−xdv.
Puttingφu=ηa+z−y+1−ηa+z−x and φv=ηa+z−x+1−ηa+z−y, we get(49)δβΓβ+12y−xβφℐφ−1a+z−y+β,δg°φφ−1a+z−x+φℐφ−1a+z−x−β,δg°φφ−1a+z−y=β2∫01expδ−1δηy−xηβ−1gηa+z−x+1−ηa+z−ydη+∫01expδ−1δηy−xηβ−1gηa+z−y+1−ηa+z−xdη≥ga+z−x+y2.
This completes the proof of the first inequality in (33). To prove the second inequality and by using the convexity of g, we can be certain that(50)gηa+z−x+1−ηa+z−y≤ηga+z−x+1−ηga+z−y,(51)gηa+z−y+1−ηa+z−x≤ηga+z−y+1−ηga+z−x.
Adding inequalities (50) and (51), we obtain(52)gηa+z−x+1−ηa+z−y+gηa+z−y+1−ηa+z−x≤ga+z−y+ga+z−x≤2ga+gz−gx+gy.
On both sides of (52), taking product by expδ−1/δηy−xηβ−1 and then integrating the estimating inequality with respect to η over 0,1, we obtain(53)∫01expδ−1δηy−xηβ−1gηa+z−x+1−ηa+z−ydη+∫01expδ−1δηy−xηβ−1gηa+z−y+1−ηa+z−xdη≤1βga+z−y+1βga+z−x≤2βga+gz−1βgx+gy.
On the left-hand side in (53), applying the same arguments as above, we obtain(54)δβΓβ+12y−xβφℐφ−1a+z−y+β,δg°φφ−1a+z−x+φℐφ−1a+z−x−β,δg°φφ−1a+z−y≤ga+z−y+ga+z−x2≤ga+gz−gx+gy2,which is the second and third inequalities in (33). Hence, the desired inequalities in (33) are thus proved.Remark 1.
The proportional fractional integral version of Theorem6 was provided by K. Yildirim and S. Yildirim in [32].If we putδ=1, in Theorem 6, we obtain its φ-Riemann–Liouville fractional integral version, which was proved by Butt et al. in [33].Fixingδ=1 and φx=x in Theorem 6 for all x∈a,z, it gives(55)ga+z−x+y2≤ga+gz−Γβ+12y−xβℐx+βgy+ℐy−βgx≤ga+gz−gx+y2,(56)ga+z−x+y2≤Γβ+12y−xβℐa+z−y+βga+z−x+ℐa+z−x−βga+z−y≤ga+z−y+ga+z−x2≤ga+gz−gx+gy2.This was proved by Ogulmus and Sarikaya in [19].Fixingδ=1, β=1, and φx=x in Theorem 6 for all x∈a,z, it gives(57)ga+z−x+y2≤ga+gz−∫01gηx+1−ηydη≤ga+gz−gx+y2,(58)ga+z−x+y2≤1y−x∫xyga+z−ηdη≤ga+gz−gx+gy2.This was given by Kian and Moslehian in [34].Theorem 7.
Letφ:Ϝ⟶a,z⊆ℝ, with 0≤a<z, be a continuous strictly increasing function and g:a,z⟶ℝ be a convex differentiable function on a,z satisfying that g°φ:Ϝ⟶ℝ is an integrable mapping on Ϝ. Then, we have(59)ga+z−x+y2≤δβΓβ+12y−xβφℐφ−1a+z−x+y2−β,δg°φφ−1a+z−y+φℐφ−1a+z−x+y2+β,δg°φφ−1a+z−x≤ga+gz−gx+gy2.Proof.
According to the convexity of the functiong for all r,s∈a,z, we have(60)ga+z−r+s2=ga+z−r+a+z−s2≤12ga+z−r+ga+z−s.
Puttingr=η/2x+2−η/2y and s=2−η/2x+η/2y, it follows, for all r,s∈a,z and η∈0,1, that(61)ga+z−x+y2≤12ga+z−η2x+2−η2y+ga+z−2−η2x+η2y.
On both sides of (61), taking product by expδ−1/δη/2y−xη/2β−1 and then integrating the estimating inequality with respect to η over 0,1, we obtain(62)1βga+z−x+y2≤∫01expδ−1δη2y−xη2β−1ga+z−η2x+2−η2ydη+∫01expδ−1δη2y−xη2β−1ga+z−2−η2x+η2ydη.
Next,(63)δβΓβ+12y−xβφℐφ−1a+z−x+y2−β,δg°φφ−1a+z−y+φℐφ−1a+z−x+y2+β,δg°φφ−1a+z−x=β2y−xβ∫φ−1a+z−yφ−1a+z−x+y2expδ−1δφu−a+z−yφu−a+z−yβ−1g°φuφ′udu+∫φ−1a+z−x+y2φ−1a+z−xexpδ−1δa+z−x−φva+z−x−φvβ−1g°φvφ′vdv=β∫φ−1a+z−yφ−1a+z−x+y2expδ−1δφu−a+z−yφu−a+z−yy−xβ−1g°φuφ′uy−xdu+∫φ−1a+z−x+y2φ−1a+z−xexpδ−1δa+z−x−φva+z−x−φvy−xβ−1g°φvφ′vy−xdv.
Puttingφu=a+z−η/2x+2−η/2y and φv=a+z−2−η/2x+η/2y, we obtain(64)δβΓβ+12y−xβφℐφ−1a+z−x+y2−β,δg°φφ−1a+z−y+φℐφ−1a+z−x+y2+β,δg°φφ−1a+z−x=β∫01expδ−1δη2y−xη2β−1ga+z−η2x+2−η2ydη+∫01expδ−1δη2y−xη2β−1ga+z−2−η2x+η2ydη≥ga+z−x+y2.
This proves the first inequality in (59). To prove the second inequality and by using the Jensen–Mercer inequality, we can be certain that(65)ga+z−η2x+2−η2y≤ga+gz−η2gx+2−η2gy,(66)ga+z−2−η2x+η2y≤ga+gz−2−η2gx+η2gy.
Adding inequalities (65) and (66), we get(67)ga+z−η2x+2−η2y+ga+z−2−η2x+η2y≤2ga+gz−gx+gy.
On both sides of (67), taking product by expδ−1/δη/2y−xη/2β−1 and then integrating the estimating inequality with respect to η over 0,1, we obtain(68)β∫01expδ−1δη2y−xη2β−1ga+z−η2x+2−η2ydη+∫01expδ−1δη2y−xη2β−1ga+z−2−η2x+η2ydη≤ga+gz−12gx+gy.
By comparing the left-hand side of inequality (64) with the left-hand side of inequality (68), we can deduce(69)δβΓβ+12y−xβφℐφ−1a+z−x+y/2−β,δg°φφ−1a+z−y+φℐφ−1a+z−x+y/2+β,δg°φφ−1a+z−x≤ga+gz−gx+gy2,which is the second inequality in (59). The proof is thus completed.


## 4. Weighted Fractional Hermite–Hadamard–Mercer Inequalities Involvingφ-Proportional Fractional Integrals

This section is the second part of our main contributions, within which we give the fractional weighted Hermite–Hadamar–Mercer’s inequalities which involve convex functions for generalized proportional fractional integral operators concerning another strictly increasing continuous function.Theorem 8.
Letφ:Ϝ⟶a,z⊆ℝ, with 0≤a<z, be a continuous strictly increasing function, g:a,z⟶ℝ be a convex differentiable function on a,z, and w:a,z⟶ℝ be a nonnegative integrable function satisfying that g°φ,w°φ:Ϝ⟶ℝ are integrable mappings on Ϝ. Then, we have(70)ga+z−x+y2φℐφ−1a+z−x−β,δw°φφ−1a+z−y≤12φℐφ−1a+z−x−β,δg°φw°φφ−1a+z−y+φℐφ−1a+z−y+β,δg°φw°φφ−1a+z−x≤12ga+z−x+ga+z−yφℐφ−1a+z−x−β,δw°φφ−1a+z−y≤ga+gz−gx+gy2φℐφ−1a+z−x−β,δw°φφ−1a+z−y.Proof.
According to the convexity of the functiong on a,z, we have(71)ga+z−x+y2=g12a+z−ηx+1−ηy+a+z−ηy+1−ηx≤ga+z−ηx+1−ηy+ga+z−ηy+1−ηx2.
On both sides of (71), taking product by ha+z−ηx+1−ηy and then integrating the estimating inequality with respect to η over 0,1, we obtain(72)ga+z−x+y2∫01ha+z−ηx+1−ηydη≤12∫01ga+z−ηx+1−ηyha+z−ηx+1−ηydη+∫01ga+z−ηy+1−ηxha+z−ηx+1−ηydη.
Applying the change of variablesφu=a+z−ηx+1−ηy, to each of the left-hand side and the first integration in (72) and applying the changeof variables φu=a+z−ηy+1−ηx, to the second integration on the right-hand side, we get(73)ga+z−x+y2∫φ−1a+z−yφ−1a+z−xh°φuφ′uy−xdu≤12∫φ−1a+z−yφ−1a+z−xg°φuh°φuφ′uy−xdu+∫φ−1a+z−yφ−1a+z−xg°φuh2a+z−x+y+φuφ′uy−xdu.
Choosing(74)h°φu=1δβΓβexpδ−1δ(φu−a+z−y)φu−a+z−yβ−1w°φu,then (73) leads to(75)1y−xga+z−x+y2φℐφ−1a+z−x−β,δw°φφ−1a+z−y≤12y−xφℐφ−1a+z−x−β,δg°φw°φφ−1a+z−y+φℐφ−1a+z−y+β,δg°φw°φφ−1a+z−x,which proves the first inequality in (70). To prove the second inequality, by using the convexity of g, we have(76)ga+z−ηx+1−ηy=gηa+z−x+1−ηa+z−y≤ηga+z−x+1−ηga+z−y,(77)ga+z−ηy+1−ηx=gηa+z−y+1−ηa+z−x≤ηga+z−y+1−ηga+z−x.
Adding inequalities (76) and (77), we obtain(78)ga+z−ηx+1−ηy+ga+z−ηy+1−ηx≤ga+z−x+ga+z−y.
On both sides of (78), taking product by 1/2ha+z−ηx+1−ηy and then integrating the estimating inequality with respect to η over 0,1, we obtain(79)12∫01ga+z−ηx+1−ηyha+z−ηx+1−ηydη+∫01ga+z−ηy+1−ηxha+z−ηx+1−ηydη≤12∫01ga+z−x+ga+z−yha+z−ηx+1−ηydη.
Applying the change of variablesφu=a+z−ηy+1−ηx, to the second integration in the left-hand side of (79) and applying the change of variables φu=a+z−ηx+1−ηy, to the first integration on left-hand side and to the right-hand side, then we obtain(80)12y−xφℐφ−1a+z−x−β,δg°φw°φφ−1a+z−y+φℐφ−1a+z−y+β,δg°φw°φφ−1a+z−x≤12y−xga+z−x+ga+z−yφℐφ−1a+z−x−β,δw°φφ−1a+z−y,and the second inequality is thus proved. To prove the third inequality in (70), by using the convexity of g and Lemma 1, we have(81)ga+z−x≤ga+gz−gx,(82)ga+z−y≤ga+gz−gy.
Adding inequalities (81) and (82), we get(83)ga+z−x+ga+z−y≤2ga+gz−gx−gy.
On both sides of (83), taking product by 1/2ha+z−ηx+1−ηy, integrating the estimating inequality with respect to η over 0,1, and then applying the changing of variables φu=a+z−ηx+1−ηy, we get(84)ga+z−x+ga+z−y2∫φ−1a+z−yφ−1a+z−xh°φuφ′uy−xdu≤ga+gz−gx+gy2∫φ−1a+z−yφ−1a+z−xh°φuφ′uy−xdu.
By choosing(85)h°φu=1δβΓβexpδ−1δ(φu−a+z−y)φu−a+z−yβ−1w°φu,we can reach the third desired inequality in (70). Hence, the proof is thus completed. □Corollary 1.
Letφ:Ϝ⟶a,z⊆ℝ, with 0≤a<z, be a continuous strictly increasing function, g:a,z⟶ℝ be a convex differentiable function on a,z, and w:a,z⟶ℝ be a nonnegative integrable function satisfying that g°φ,w°φ:Ϝ⟶ℝ are integrable mappings on Ϝ. Then, we have(86)ga+z−x+y2φℐφ−1a+z−y+β,δw°φφ−1a+z−x≤12φℐφ−1a+z−y+β,δg°φw°φφ−1a+z−x+φℐφ−1a+z−x−β,δg°φw°φφ−1a+z−y≤12ga+z−x+ga+z−yφℐφ−1a+z−y+β,δw°φφ−1a+z−x≤ga+gz−gx+gy2φℐφ−1a+z−y+β,δw°φφ−1a+z−x.Proof.
This corollary can be easily demonstrated by following the proof of Theorem8, taking(87)h°φu=1δβΓβexpδ−1δa+z−x−φu(a+z−x−φu)β−1w°φu.Remark 2.
By adding inequalities (70) and (86), we can derive the following inequality:(88)ga+z−x+y2φℐφ−1a+z−x−β,δw°φφ−1a+z−y+φℐφ−1a+z−y+β,δw°φφ−1a+z−x≤φℐφ−1a+z−x−β,δg°φw°φφ−1a+z−y+φℐφ−1a+z−y+β,δg°φw°φφ−1a+z−x≤12ga+z−x+ga+z−yφℐφ−1a+z−x−β,δw°φφ−1a+z−y+φℐφ−1a+z−y+β,δw°φφ−1a+z−x≤ga+gz−gx+gy2φℐφ−1a+z−x−β,δw°φφ−1a+z−y+φℐφ−1a+z−y+β,δw°φφ−1a+z−x.Remark 3.
Puttingδ=1, we obtain the proportional fractional integral version of each of Theorem 8, Corollary 1, and inequality (88) with respect to the positive increasing function φBy takingφx=x, for all x∈a,z, we obtain the proportional fractional integral version for each of Theorem 8, Corollary 1, and inequality (88)If we putδ=1 and φx=x, for all x∈a,z, we obtain inequality (9), for Riemann–Liouville fractional integral introduced by Iscan [20]Puttingδ=1 and φx=x, for all x∈a,z, and if we choose x=a and y=z, we get the following inequality:(89)gx+y2ℐz−βwa+ℐa+βwz≤ℐz−βgwa+ℐa+βgwz≤ga+gz2ℐz−βwa+ℐa+βwz,which was introduced by Iscan [35]
The next result is as follows.Theorem 9.
Letφ:Ϝ⟶a,z⊆ℝ, with 0≤a<z, be a continuous strictly increasing function, g:a,z⟶ℝ be a convex differentiable function on a,z, and w:a,z⟶ℝ be a nonnegative integrable function satisfying that g°φ,w°φ:Ϝ⟶ℝ are integrable mappings on Ϝ. Then, we have(90)ga+z−x+y2φℐφ−1a+z−x−β,δw°φφ−1a+z−y≤12φℐφ−1a+z−x−β,δg°φw°φφ−1a+z−y+φℐφ−1a+z−y+β,δg°φw°φφ−1a+z−x≤ga+gzφℐφ−1x+β,δw°φφ−1y−12φℐφ−1x+β,δg°φw°φφ−1y+φℐφ−1y−β,δg°φw°φφ−1x≤ga+gz−gx+y2φℐφ−1x+β,δw°φφ−1y.Proof.
The first inequality in (90) is already proved in Theorem 8. To prove the second inequality, we have by Lemma (1) according to the convexity of the function g on a,z what follows(91)ga+z−ηx+1−ηy≤ga+gz−gηx+1−ηy,(92)ga+z−ηy+1−ηx≤ga+gz−gηy+1−ηx.
Adding inequalities (91) and (92), we get(93)ga+z−ηx+1−ηy+ga+z−ηy+1−ηx≤2ga+2gz−gηx+1−ηy+gηy+1−ηx.
On both sides of (93), taking product by 1/2ha+z−ηx+1−ηy, integrating the estimating inequality with respect to η over 0,1, and then applying the change of variables φu=a+z−ηx+1−ηy, to the left-hand side, and applying the change of variables φu=ηx+1−ηy, to the right-hand side, we get(94)12∫φ−1a+z−yφ−1a+z−xg°φuh°φuφ′uy−xdu+∫φ−1a+z−yφ−1a+z−xg°φuh2a+z−x+y+φuφ′uy−xdu≤ga+gz∫φ−1xφ−1yha+z−φuφ′uy−xdu−12∫φ−1xφ−1yg°φuha+z−φuφ′uy−xdu+∫φ−1xφ−1yg°φuha+z−x−y+φuφ′uy−xdu.
Choosing(95)h°φu=1δβΓβexpδ−1δ(φu−a+z−y)φu−a+z−yβ−1w°φu,then (94) leads to(96)12y−xφℐφ−1a+z−x−β,δg°φw°φφ−1a+z−y+φℐφ−1a+z−y+β,δg°φw°φφ−1a+z−x≤ga+gzy−xφℐφ−1x+β,δw°φφ−1y−12y−xφℐφ−1x+β,δg°φw°φφ−1y+φℐφ−1y−β,δg°φw°φφ−1x,which proves the second inequality in (90). To prove the last inequality, by using the convexity of g, we have, for all x,y∈a,z,(97)gx+y2=gηx+1−ηy+ηy+1−ηx2≤12gηx+1−ηy+gηy+1−ηx,which can be rewritten as(98)ga+gz−12gηx+1−ηy+gηy+1−ηx≤ga+gz−gx+y2.
On both sides of (98), taking product by ha+z−ηx+1−ηy and integrating the estimating inequality with respect to η over 0,1,(99)ga+gz∫01ha+z−ηx+1−ηydη−12∫01gηx+1−ηyha+z−ηx+1−ηydη+∫01gηy+1−ηxha+z−ηx+1−ηydη≤ga+gz−gx+y2∫01ha+z−ηx+1−ηydη.
Applying the change of variablesφu=a+z−ηx+1−ηy, to the right-hand side and applying the same process to the left-hand side as above, we get(100)ga+gzy−xφℐφ−1x+β,δw°φφ−1y−12y−xφℐφ−1x+β,δg°φw°φφ−1y+φℐφ−1y−β,δg°φw°φφ−1x≤ga+gz−gx+y2φℐφ−1x+β,δw°φφ−1y,which proves the last inequality in (90). Hence, the proof is thus completed.Corollary 2.
Letφ:Ϝ⟶a,z⊆ℝ, with 0≤a<z, be a continuous strictly increasing function, g:a,z⟶ℝ be a convex differentiable function on a,z, and w:a,z⟶ℝ be a nonnegative integrable function satisfying that g°φ,w°φ:Ϝ⟶ℝ are integrable mappings on Ϝ. Then, we have(101)ga+z−x+y2φℐφ−1a+z−y+β,δw°φφ−1a+z−x≤12φℐφ−1a+z−x−β,δg°φw°φφ−1a+z−y+φℐφ−1a+z−y+β,δg°φw°φφ−1a+z−x≤ga+gzφℐφ−1y−β,δw°φφ−1x−12φℐφ−1x+β,δg°φw°φφ−1y+φℐφ−1y−β,δg°φw°φφ−1x≤ga+gz−gx+y2φℐφ−1y−β,δw°φφ−1x.Proof.
This corollary can be easily demonstrated by following the proof of Theorem9, taking(102)h°φu=1δβΓβexpδ−1δa+z−x−φu(a+z−x−φu)β−1w°φu.Remark 4.
By adding inequalities (90) and (101), we can derive the following inequality:(103)ga+z−x+y2φℐφ−1a+z−x−β,δw°φφ−1a+z−y+φℐφ−1a+z−y+β,δw°φφ−1a+z−x≤φℐφ−1a+z−x−β,δg°φw°φφ−1a+z−y+φℐφ−1a+z−y+β,δg°φw°φφ−1a+z−x≤ga+gzφℐφ−1x+β,δw°φφ−1y+φℐφ−1y−β,δw°φφ−1x−φℐφ−1x+β,δg°φw°φφ−1y+φℐφ−1y−β,δg°φw°φφ−1x≤ga+gz−gx+y2φℐφ−1x+β,δw°φφ−1y+φℐφ−1y−β,δw°φφ−1x.Remark 5.
Puttingδ=1, we obtain the proportional fractional integral version of each of Theorem 9, Corollary 2, and inequality (103) with respect to the positive increasing function φBy takingφx=x, for all x∈a,z, we obtain the proportional fractional integral version for each of Theorem 9, Corollary 2, and inequality (103).If we putδ=1 and φx=x, for all x∈a,z, we obtain inequality (10), for Riemann–Liouville fractional integral introduced by Iscan [20].


## 5. Conclusion

In view of the significant importance recently achieved by fractional calculus and its very important applications in the interpretation and modeling of natural phenomena, it has become necessary to develop and refine our capabilities to generalize some of the recent results related to this topic. We achieved our goals of introducing a new fractional Hermite–Hadamard–Mercer’s inequality and its fractional integral type inequalities by employing the proportional fractional operators of integrable functions with respect to another continuous and strictly increasing function. We enhanced our work by establishing some new fractional weightedφ-proportional fractional integral Hermite–Hadamard–Mercer type inequalities. Also, in this article, we were keen to present some special cases related to our current study compared to the previous work of the inequality under study. In future work, we recommend researchers study the current inequality via recent fractional operators such as the Atangana + Baleanu operator or Caputo + Fabrizio operator.



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