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Mathematical and Computer Modelling of Dynamical Systems
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Research Article

Harmonic conformable refinements of Hermite-Hadamard Mercer inequalities by support line and related applications

, &
Pages 385-416 | Received 07 Mar 2024, Accepted 18 Apr 2024, Published online: 15 May 2024

ABSTRACT

We establish new conformable fractional Hermite-Hadamard (H–H) Mercer type inequalities for harmonically convex functions using the concept of support line. We introduce two new conformable fractional auxiliary equalities in the Mercer sense and apply them to differentiable functions with harmonic convexity. We also use Power-mean, Hölder’s and improved Hölder inequality to derive new Mercer type inequalities via conformable fractional integrals. The accuracy and superiority of the offered technique are clearly depicted through impactful visual illustrations. We also use our technique to derive new estimates for hypergeometric functions and special means of real numbers that are more precise than existing ones. Some applications are provided as well. Our results generalize and extend some existing ones in the literature.

1 Introduction

Convexity holds significance, in both pure and applied mathematics as it allows for the generalization of existing results. For over 60 years researchers have extensively studied convex functions, which have applications across fields like mathematics, physics, economics and optimization. Research on functions and integral inequalities remains an area of interest. The study of convex functions and integral inequalities remains an area of research leading to significant advancements in convex analysis and fractional integral inequalities. Harmonic convex functions represent a development within the realm of convex functions with diverse applications in mathematics and physics. For instance, they find utility in theory, electrostatics and fluid mechanics. Numerous efforts have been dedicated to extending the concept of convexity and establishing (H–H) type inequalities, for them. A variety of unique and noteworthy inequalities can be obtained from harmonic convex functions. In the paper (ÍşCan Citation2014; Dragomir Citation2017; Latif et al. Citation2017; Gao et al. Citation2020; Butt et al. Citation2021) one can explore many excellent inequalities pertaining to harmonically convex functions. The classical definition of a convex mapping is as follows:

Definition 1.1.

Roberts and Varberg (Citation1973) A function ϕ:IRR is called convex, if

(1.1) ϕ(γω1+(1γ)ω2)γϕ(ω1)+(1γ)ϕ(ω2).(1.1)

For each ω1,ω2I and γ[0,1] holds.

The following convex mapping qualities are employed in the main outcomes.

Definition 1.2.

Roberts and Varberg (Citation1973) A mapping ϕ on interval I has a support at point k0I, if there exists an affine function A(k)=ϕ(k0)+c(kk0) such that A(k)ϕ(k) for all kI. The graph of support function A is support line for function ϕ at point k0.

Theorem 1.1.

Roberts and Varberg (Citation1973) ϕ:(ω1,ω2)R is a convex function iff there is at least one line of support for ϕ at each κ0(ω1,ω2).

Definition 1.3.

A mapping ϕ:IR{0}R is harmonically convex (ϕHC(I)), if

(1.2) ϕ(ω1ω2γω1+(1γ)ω2)γϕ(ω2)+(1γ)ϕ(ω1),(1.2)

for all ω1,ω2I and γ[0,1].

Remark 1.1. (Dragomir Citation2017) Let [ω1,ω2]I(0,), if function h:[1ω2,1ω1]R defined as g(κ)=ϕ(1κ), then ϕ is harmonically convex on [ω1,ω2] iff g is convex on [1ω2,1ω1].

Proposition 1.1.

ÍşCan (Citation2014) If ϕ : (0,α)R is convex and nondecreasing on (0,α), then ϕHC(I)

The classical (H–H) inequality, a well-known inequality in convex function theory, has a geometric meaning as well as various practical applications. It asserts that if ϕ is a convex mapping defined on the interval [ω1,ω2], then:

ϕω1+ω221ω2ω1ω1ω2ϕ(κ)ϕ(ω1)+ϕ(ω2)2.

The inequality holds in reversed direction if the function is concave on [ω1,ω2]. Several variants have been developed regarding Mercer (H–H) type inequalities (see (Bohner et al. Citation2023; Budak and Kara Citation2023; Sitthiwirattham et al. Citation2024)). (H-H) inequality is the utmost important and extensively used result involving harmonic convex functions. In (ÍşCan Citation2014), Iscan gave (H–H) type inequalities for harmonically convex mapping stated as

Theorem 1.2.

Let ϕHC(I) and ω1,ω2I with ω1<ω2. If ϕL[ω1,ω2], then

(1.3) ϕ2ω1ω2ω1+ω2ω1ω2ω2ω1ω1ω2ϕ(κ)κ2ϕ(ω1)+ϕ(ω2)2.(1.3)

The Jensen inequality is a generalization of the notion of convexity for n-convex combinations. Dragomir in (Dragomir Citation2015), gave the notable Jensen type inequality for harmonic convex functions in such a way that, for any ϕHC(I) and any non-negative weights γj[0,1], (j=1,2,3,.,k) with j=1kγj=1, we have:

(1.4) ϕ(1j=1kγjj)j=1kγjϕ(j).(1.4)

Jensen’s inequality is widely recognized as a fundamental component within the realm of mathematical inequalities. The concept has extensive application in the fields of mathematics, statistics, and information theory, enabling the derivation of several significant inequalities, including the arithmetic-geometric mean inequality, Minkowski inequality, Höder inequality, and Ky Fan’s inequality. In 2020, Baloch et al. presented a new variant of Jensen’s inequality. This variant is defined as

Theorem 1.3.

Let I=[ω1,ω2](0,) be an interval. If ϕHC(I), then

ϕ(11ω1+1ω2j=1kγjj)ϕ(ω1)+ϕ(ω2)j=1kγjϕ(j),

for all j[ω1,ω2] and γj[0,1],(j=1,2,3,.,k) with j=1kγj=1.

The field of fractional calculus gained significant attention from scholars across various academic areas due to its intriguing nature. This field of research explores the concepts of differentiation and integration of arbitrary order within the realm of mathematics. Academic researchers are attracted to this particular topic due to its ability to provide meaningful outcomes through the utilization of fractional operators for the purpose of modelling real-world problems. Fractional integral and derivative operators have developed over time (Khalil et al. Citation2014; Atangana and Baleanu Citation2016). Thus, these operators have been considered as one of the most powerful tools in the area of mathematical modelling. Many engineering, physical, chemical, and biological phenomena can be modelled by employing differential equations containing fractional derivatives (Atangana Citation2017; Gomez-Aguilar and Atangana Citation2022). Prominent academics like as R.P. Agrwal and D. Baleanu have made significant contributions to the enormous body of research in this particular topic. The publication entitled ‘Fractional calculus in the sky’ provides a comprehensive survey of the progress made in this particular field (Baleanu and Agarwal Citation2021). Furthermore, the field of inequalities has been enhanced by the discovery of several variants generated from distinct convexities, as demonstrated by their applications (Sarikaya et al. Citation2013; Chen and Katugampola Citation2013; Set et al. Citation2021; Sun Citation2021; Butt et al. Citation2021; Yuan et al. Citation2023). Over the course of several decades, there has been a notable emphasis on the investigation of diverse approaches aimed at establishing integral inequalities associated with harmonic convexity within the field of fractional calculus. In their publication (IşCan and Wu Citation2014), Iscan and S. Wu established a set of inequalities of the (H–H) type for a function ϕHC(I). These inequalities are expressed in the form of integrals, as demonstrated below.

Theorem 1.4.

Let ϕ:I=[ω1,ω2](0,)R be a function such that ϕL[ω1,ω2] with 0<ω1<ω2. If ϕHC(I) on the interval I=[ω1,ω2], then

ϕ2ω1ω2ω1+ω2Γ(ϑ+1)2ω1ω2ω2ω1ϑJ1ω1ϑ(ϕg)(1ω2)+J1ω2+ϑ(ϕg)(1ω1)
ϕ(ω1)+ϕ(ω2)2.

For further recent developments one can (see (Rashid et al. Citation2020; Latif et al. Citation2022; Butt et al. Citation2022)). We need some important special functions used for our study.

Definition 1.4. (Tunc et al. Citation2015)

1. The Beta function has real number domain and defined as

(1.5) β(ω1,ω2)=Γ(ω1)Γ(ω2)Γ(ω1+ω2)=01γω11(1γ)ω21,ω1,ω2>0.(1.5)

2. The Hypergeometric function is defined as:

(1.6)  2F1(ω1,ω2;c;z)=1β(ω2,cω2)01γω211γcω21(1zγ)ω1,(1.6)

where c>ω2>0,|z|<1.

Lemma 1.1.

Prudnikov et al., (Citation1998) For 0<ξ1 and 0<ω1ω2 we have

ω1ξω2ξ(ω2ω1)ξ.

Conformable fractional operators are a new class of fractional operators that have been recently proposed to overcome some of the limitations and complications of the traditional fractional operators, such as the Riemann-Liouville and Caputo derivatives. Conformable fractional operators are based on the basic limit definition of the derivative and preserve some of the properties of the integer-order calculus, such as the product rule, the chain rule, and the Leibniz rule. Conformable fractional operators have been used to develop new results and applications in various fields of mathematics and science, such as differential equations, stability analysis, nonlinear observers, and approximation methods.

In their research (Nisar et al. Citation2019), Nisar et al. extended Minkowski and (H–H) inequalities using conformable fractional integral operator. The results are generalization of the integral inequalities obtained by Dahmani and Bougoffa cited in the literature. In their research (Adil Khan et al. Citation2018), Khan et al. presented an advanced version of the renowned (H–H) inequality, utilizing newly defined Generalized Conformable fractional operators. In (Abdeljawad Citation2015), left and right Conformable fractional integrals is defined as

Definition 1.5.

For ξ(,+1],=0,1,2,.,β=ξ,ω1,ω2R with ω1<ω2 and ϕL[ω1,ω2]. The left and right Conformable fractional integrals Iξω1 and  ω2Iξ of order ξ>0 are:

Iξω1=1!ω1γ(γκ)(κω1)β1ϕ(κ),γ>ω2,

and

 ω2Iξ=1!γω2(κγ)(ω2κ)β1ϕ(κ),γ<ω2,

respectively. If one takes ξ=+1 in this definition (for the left and right Conformable fractional integrals), one has the left and right Riemann–Liouville fractional integrals for ξN.

The primary objective of this analysis is to utilize fractional calculus to derive novel Mercer type inequalities for harmonically convex mappings. The study will introduce the concept of the line of support to obtain a new fractional (H–H) Mercer type inequality. Additionally, we aim to derive additional trapezoid-type inequalities by utilizing the right and left Conformable fractional integral in the Mercer sense. Furthermore, when ξ=+1 and ξ=1 in the obtained results, we will deduce inequalities of the classical Mercer type along with their different variants. Finally, to validate the reported results, we will compare them with diminished outcomes and conduct relevant implementations.

2. New Fractional H-H Mercer Type Inequalities for ϕHC(I)

In this section, we present new conformable fractional (H–H) Mercer type inequalities that are derived using only the right or left fractional integrals, separately, for harmonically convex functions.

Theorem 2.1.

Let ϕ:I=[1,2](0,)R be a function such that ϕL[1,2] with 0<1<2. If ϕHC(I) on the interval I=[1,2], then

ϕ(11ω1+1ω2(+1)1+(ξ)2(ξ+1)12)ϕ(ω1)+ϕ(ω2)Γ(ξ+1)Γ(ξ)1221ξIξ12(ϕg)(11)
(2.1) ϕ(ω1)+ϕ(ω2)ϕ(ξ+1)12(+1)1+(ξ)2,(2.1)

and

ϕ(11ω1+1ω2(+1)1+(ξ)212(ξ+1))(1221)ξΓ(ξ+1)Γ(ξ).1ω1+1ω212Iξ(ϕg)(1ω1+1ω211)
(+1)ϕ11ω1+1ω212+(ξ)ϕ11ω1+1ω211ξ+1
(2.2) ϕ(ω1)+ϕ(ω2)(+1)ϕ(2)+(ξ)ϕ(1)ξ+1,(2.2)

for all 1,2[ω1,ω2] and g(κ)=ϕ(1κ), κ[1ω2,1ω1].

Proof.

Since ϕ is harmonic convex on [ω1,ω2] by using Remark 1.1, g(κ)=ϕ(1κ) is convex on [1ω2,1ω1]. Hence, using Remark 1.1, there is at least one line of support.

(2.3) A(κ)=g(κ0)+cκκ0g(κ),(2.3)

we put κ0=(1ω1+1ω2(+1)1+(ξ)212(ξ+1)) and κ=1ω1+1ω2γ2(1γ)1 in (2.3).

A(κ)=g1ω1+1ω2(+1)1+(ξ)212(ξ+1)\break+cκ1ω11ω2+(+1)1+(ξ)212(ξ+1)g(κ),

for all κ[1ω2,1ω1] and c[g (1ω1+1ω2(+1)1+(ξ)2(ξ+1)12),g +(1ω1+1ω2(+1)1+(ξ)2(ξ+1)12)].

ϕ(11ω1+1ω2(+1)1+(ξ)2(ξ+1)12)
+c(1ω1+1ω2γ2(1γ)11ω11ω2+(+1)1+(ξ)2(ξ+1)12)
(2.4) ϕ11ω1+1ω2γ2(1γ)1.(2.4)

Multiplying all sides of (2.4) with 1!γ(1γ)ξ1 and integrating over [0,1] respect to γ, we have

011!γ(1γ)ξ1[ϕ(11ω1+1ω2(+1)1+(ξ)2(ξ+1)12)+c(γ2(1γ)1+(+1)1+(ξ)2(ξ+1)12)]+1![01γ(1γ)ξ1(γ2(1γ)1)+01γ(1γ)ξ1((+1)1+(ξ)2(ξ+1)12)]
=ϕ(11ω1+1ω2(+1)1+(ξ)2(ξ+1)12)β(+1,ξ)!+1![(+1)1+(ξ)2(ξ+1)12β(+1,ξ)!
+(+1)1+(ξ)2(ξ+1)12β(+1,ξ)!]
(2.5) ϕ(11ω1+1ω2(+1)1+(ξ)2(ξ+1)12)β(+1,ξ)!011!γ(1γ)ξ1ϕ(1ω1+1ω2γ2(1γ)1).(2.5)

Using Mercer’s inequality, we have

ϕ(11ω1+1ω2(+1)1+(ξ)2(ξ+1)12)β(+1,ξ)!
(2.6) 011!γ(1γ)ξ1ϕ(ω1)+ϕ(ω2)γϕ(2)(1γ)ϕ(1).(2.6)

Since ϕ is harmonic convex, we have (γϕ(2)+(1γ)ϕ(1)) ϕ12γ1+(1γ)2 and (2.6) becomes

ϕ(11ω1+1ω2(+1)1+(ξ)2(ξ+1)12)β(+1,ξ)!
(2.7) β(+1,ξ)![ϕ(ω1)+ϕ(ω2)]011!γ(1γ)ξ1ϕ12γ1+(1γ)2.(2.7)

Substitute 1u=12γ1+(1γ)2 in (2.7), we obtain

ϕ(11ω1+1ω2(+1)1+(ξ)2(ξ+1)12)β(+1,ξ)!
β(+1,ξ)![ϕ(ω1)+ϕ(ω2)]1221ξ1!121111uu12ξ1ϕ1udu
ϕ(11ω1+1ω2(+1)1+(ξ)2(ξ+1)12)β(+1,ξ)!
(2.8) β(+1,ξ)![ϕ(ω1)+ϕ(ω2)]1221ξIξ12(ϕg)(11).(2.8)

It means that

ϕ(11ω1+1ω2(+1)1+(ξ)2(ξ+1)12)ϕ(ω1)+ϕ(ω2)!β(+1,ξ)1221ξIξ12(ϕg)(11)
(2.9) =ϕ(ω1)+ϕ(ω2)Γ(ξ+1)Γ(ξ)1221ξIξ12(ϕg)(11).(2.9)

Now for second inequality of (2.1). Put κ0=(+1)1+(ξ)2(ξ+1)12 and κ=γ1+(1γ)212 in (2.3) we get:

(2.10) ϕ(ξ+1)12(+1)1+(ξ)2+cγ1+(1γ)212(+1)1+(ξ)2(ξ+1)12ϕ12γ1+(1γ)2,(2.10)

for all γ[0,1]. Multiplying all sides of (2.10) with 1!γ(1γ)ξ1 1!γ(1γ)ξ1 and integrating over [0,1] respect to γ, we have:

ϕ(ξ+1)12(+1)1+(ξ)2β(+1,ξ)!011!γ(1γ)ξ1ϕ12γ1+(1γ)2,

put 1u=12γ1+(1γ)2, we obtain

ϕ(ξ+1)12(+1)1+(ξ)2Γ(ξ+1)Γ(ξ)1221ξIξ12(ϕg)(11)
(2.11) Γ(ξ+1)Γ(ξ)1221ξIξ12(ϕg)(11)ϕ(ξ+1)12(+1)1+(ξ)2.(2.11)

Adding ϕ(ω1)+ϕ(ω2) on both sides of (2.11), we have

ϕ(ω1)+ϕ(ω2)Γ(ξ+1)Γ(ξ)1221ξIξ12(ϕg)(11)
(2.12) ϕ(ω1)+ϕ(ω2)ϕ(ξ+1)12(+1)1+(ξ)2,(2.12)

and on combining (2.9) and (2.12), we obtain (2.1). This completes the proof.

Now we prove (2.2). Let κ=1ω1+1ω2γ2(1γ)1γ=1221κ(1ω1+1ω211) and (2.5) becomes:

ϕ(11ω1+1ω2(+1)1+(ξ)2(ξ+1)12)β(+1,ξ)!
011!γ(1γ)ξ1ϕ(1ω1+1ω2γ2(1γ)1)
=(1221)ξ1ω1+1ω212Iξ(ϕg)(1ω1+1ω211)
ϕ(11ω1+1ω2(+1)1+(ξ)212(ξ+1))
(2.13) (1221)ξΓ(ξ+1)Γ(ξ)1ω1+1ω212Iξ(ϕg)(1ω1+1ω211).(2.13)

Now we prove other two inequalities of (2.2). By using the harmonic convexity of ϕ on [ω1,ω2], we have:

ϕ11ω1+1ω2γ2(1γ)1=ϕ1γ1ω1+1ω212+(1γ)1ω1+1ω211
ϕ11ω1+1ω2γ2(1γ)1γϕ11ω1+1ω212+(1γ)ϕ11ω1+1ω211
ϕ(ω1)+ϕ(ω2)γϕ(2)(1γ)ϕ(1).

Multiplying the above inequality with 1!γ(1γ)ξ1 and integrating over [0,1] respect to γ, we have

011!γ(1γ)ξ1ϕ11ω1+1ω2γ2(1γ)1
ϕ11ω1+1ω212011!γ+1(1γ)ξ1+ϕ11ω1+1ω211011!γ(1γ)ξ
011!γ(1γ)ξ1[ϕ(ω1)+ϕ(ω2)]ϕ(2)011!γ+1(1γ)ξ1
ϕ(1)011!γ(1γ)ξ
ϕ11ω1+1ω2γ2(1γ)1β(+1,ξ)!
(+1)ϕ11ω1+1ω212+(ξ)ϕ11ω1+1ω211ξ+1β(+1,ξ)!
(2.14) β(+1,ξ)!ϕ(ω1)+ϕ(ω2)(+1)ϕ(2)+(ξ)ϕ(1)ξ+1.(2.14)

By changing variable, (2.14) becomes

(1221)ξ1ω1+1ω212Iξ(ϕg)1ω1+1ω211
(+1)ϕ11ω1+1ω212+(ξ)ϕ11ω1+1ω211ξ+1
(2.15) ϕ(ω1)+ϕ(ω2)(+1)ϕ(2)+(ξ)ϕ(1)ξ+1.(2.15)

By combining (2.13) and (2.15), we obtain (2.2).□

Remark 2.1. If we put ξ=+1 after that ξ=1 in Theorem 2.1, we have

ϕ(11ω1+1ω21+2212)ϕ(ω1)+ϕ(ω2)01ϕ12γ1+(1γ)2
(2.16) ϕ(ω1)+ϕ(ω2)ϕ(2121+2),(2.16)

and

ϕ(11ω1+1ω21+2212)122111ω1+1ω21211ω1+1ω211ϕ(κ)κ2
(2.17) ϕ11ω1+1ω212+ϕ11ω1+1ω2112ϕ(ω1)+ϕ(ω2)ϕ(1)+ϕ(2)2,(2.17)

for all 1,2[ω1,ω2]. The proof of inequality (2.16) is given by Baloch et al. in [(Baloch et al. Citation2020), Theorem 3.5]. Inequality (2.17) is better than the inequality proved by Baloch et al. in [(Baloch et al. Citation2021), Theorem 2.1]. As (2.17) is refinement of that inequality.

Remark 2.2. If we put ξ=1,ω1=1 and ω2=2 in (2.2), we have (1.3) (H-H inequality) appeared in (Butt et al. Citation2022).

3. Novel Mercer type inequalities for Ф є HC(I) via conformable fractional integrals

We make the following assumptions throughout the rest of this paper:

Ω= Let ϕ:I=[ω1,ω2](0,)R be a differentiable function on [ω1,ω2] with 0<ω1ω2.

Iϕ1(g;ξ,1,2)=(ξ)ϕ(11ω1+1ω212)+(+1)ϕ(11ω1+1ω211)ξ+1(1221)ξΓ(ξ+1)Γ(ξ)Iξ1ω1+1ω211(ϕg)(1ω1+1ω212),
Iϕ2(g;ξ,1,2)=(ξ)ϕ(11ω1+1ω211)+(+1)ϕ(11ω1+1ω212)ξ+11221ξ\breakΓ(ξ+1)Γ(ξ)1ω1+1ω212Iξ(ϕg)(1ω1+1ω211),

with =0,1,2,3, and ξ=(,+1].

In this section, we prove two new Conformable fractional equalities used in forward results.

Lemma 3.1.

If ϕ L[1,2] along with assumption Ω, then following identity for fractional integral holds:

Iϕ1(g;ξ,1,2)=2112(ξ+1)β(+1,ξ)01(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)
(3.1) ×1(1ω1+1ω2γ1(1γ)2)2ϕ (11ω1+1ω2γ1(1γ)2),(3.1)

with =0,1,2, and ξ=(,+1].

Proof.

Let Aγ=1ω1+1ω2γ1(1γ)2. Applying partial integration to the right-hand side of equation (3.1) as follows:

=2112(ξ+1)01(ξ+1)βγ(+1,ξ)β(+1,ξ)(ξ)×1Aγ2ϕ (1Aγ)
=1β(+1,ξ)010γκ(1κ)ξ12112Aγ2ϕ (1Aγ)ξξ+1ϕ(1Aγ)|01
=[1β(+1,ξ)0γκ(1κ)ξ1)ϕ1Aγ|0101γ(1γ)ξ1ϕ(1Aγ
ξξ+1ϕ11ω1+1ω211)ϕ11ω1+1ω212
=1β(+1,ξ)(β(+1,ξ)ϕ(11ω1+1ω211)01γ(1γ)ξ1ϕ(1Aγ)
+ξξ+1ϕ11ω1+1ω212ϕ11ω1+1ω211)
=(ξ)ϕ(11ω1+1ω212)+(+1)ϕ(11ω1+1ω211)ξ+1(1221)ξΓ(ξ+1)Γ(ξ)Iξ1ω1+1ω211(ϕg)(1ω1+1ω212).

This completes the proof.□

Remark 3.1. Another approach to proof Lemma 3.1 is by selecting a=11ω1+1ω211 and b=11ω1+1ω212 in Lemma 3.2 of (Sanli et al. Citation2018). This choice directly aligns Lemma 3.2 of (Sanli et al. Citation2018) with (3.2).

Remark 3.2. If we take ξ=+1 in Lemma 3.1, then we have

ϕ(11ω1+1ω212)+ξϕ(11ω1+1ω211)ξ+11β(ξ,1)01γξ1ϕ(1Aγ)=2112(ξ+1)β(ξ,1)
01(ξ+1)βγ(ξ,1)β(ξ,1)×1Aγ2ϕ (1Aγ),

which is new in the literature.

Remark 3.3. If we take ω1=1 and ω2=2 in Lemma 3.1, then we have

(ξ)ϕ(1)+(+1)ϕ(2)ξ+1(1221)ξΓ(ξ+1)Γ(ξ)Iξ12(ϕg)(11)

=12(21)(ξ+1)β(+1,ξ)01(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)×1γ1+(1γ)22ϕ(12γ1+(1γ)2),which is appeared in (Sanli et al. Citation2018).

Remark 3.4. If we put ω1=1,ω2=2 and ξ=+1 in Lemma 3.1, then we have

ϕ(1)+ξϕ(2)ξ+1Γ(ξ+1)1221ξI12+ξ(ϕg)(11)
=12(21)ξ+101(ξ+1)(1γ)ξ1(γ2+(1γ)1)2ϕ12γ2+(1γ)1,

which is appeared in (Sanli et al. Citation2020).

Remark 3.5. If we put ω1=1,ω2=2,ξ=+1 after that ξ=1 in Lemma 3.1, then we have

ϕ(1)+ϕ(2)2122112ϕ(κ)κ2\break=12(21)20112γ(γ2+(1γ)1)2ϕ12γ2+(1γ)1,

which is appeared in (ͺͺCan Citation2014).

Lemma 3.2.

If ϕ L[1,2] along with assumption Ω, then following identity for fractional integral holds:

(ξ)ϕ(11ω1+1ω211)+(+1)ϕ(11ω1+1ω212)ξ+11221ξΓ(ξ+1)Γ(ξ)1ω1+1ω212Iξ(ϕg)(1ω1+1ω211)
(3.2) =2112(ξ+1)β(+1,ξ)01(ξ)β(+1,ξ)(ξ+1)β1γ(+1,ξ)×1Aγ2ϕ (1Aγ),(3.2)

with =0,1,2,3, and ξ=(,+1].

Proof.

In Lemma 3.2 of Sanli and Köroğlu (Citation2018), if we choose

a=11ω1+1ω211,

and

b=11ω1+1ω212,

then Lemma 3.2 of (Sanli et al. Citation2018) reduces to (3.2).□

Remark 3.6. If we take ξ=+1 in Lemma 3.2, then we have

ϕ(11ω1+1ω211)+ξϕ(11ω1+1ω212)ξ+11β(ξ,1)01(1γ)ξ1ϕ(1Aγ)=2112(ξ+1)β(ξ,1)
01β(ξ,1)(ξ+1)β1γ(ξ,1)×1Aγ2ϕ (1Aγ),

which is new in literature.

Remark 3.7. If we take ω1=1 and ω2=2 in Lemma 3.2, then we have the equality proved in (Sanli et al. Citation2018).

Remark 3.8. If we put ω1=1,ω2=2 and ξ=+1 in Lemma 3.2, then we have the equality proved in (Sanli et al. Citation2020).

Remark 3.9. If we put ω1=1,ω2=2,ξ=+1 after that ξ=1 in Lemma 3.1, then we have the equality proved in (ÍşCan Citation2014).

Theorem 3.1.

If |ϕ |q is harmonically convex function for some fixed q1 and ϕ L[ω1,ω2] along with assumption Ω and 0<ξ<1, then the following inequality for fractional integral holds:

|Iϕ1(g;ξ,1,2)|
(3.3) 2112(ξ+1)β(+1,ξ)K111q(1,2,ξ)K1(1,2,ξ)|ϕ (ω1)|q+K1(1,2,ξ)|ϕ (ω2)|qK2(1,2,ξ)|ϕ (1)|qK3(1,2,ξ)|ϕ (2)|q1q,(3.3)

where

K1(1,2,ξ)=01(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)×1Aγ2
K2(1,2,ξ)=01(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)×1Aγ2γdγ
K3(1,2,ξ)=K1(1,2,ξ)K2(1,2,ξ),

with =0,1,2,3, and ξ=(,+1].

Proof.

Using Lemma 3.1, Power-mean inequality and since |ϕ |q is harmonically convex, by using Mercer inequality, we have

|Iϕ1(g;ξ,1,2)|
2112(ξ+1)β(+1,ξ)01(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)
×1Aγ2ϕ (1Aγ)
2112(ξ+1)β(+1,ξ)01(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)×1Aγ211q
×01(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)×1Aγ2ϕ (1Aγ)q1q
2112(ξ+1)β(+1,ξ)01(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)×1Aγ211q
×01(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)×1Aγ2|ϕ(ω1)|q+|ϕ(ω2)|qγ|ϕ(1)|q(1γ)|ϕ(2)|q1q
2112(ξ+1)β(+1,ξ)01(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)×1Aγ211q

×|ϕ(ω1)|q+|ϕ(ω2)|q01(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)×1Aγ2|ϕ(1)|q01(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)×1Aγ2γdγ|ϕ(2)|q01(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)×1Aγ2(1γ)1q
2112(ξ+1)β(+1,ξ)K111q(1,2,ξ)K1(1,2,ξ)|ϕ (ω1)|q+K1(1,2,ξ)|ϕ (ω2)|qK2(1,2,ξ)|ϕ (1)|qK3(1,2,ξ)|ϕ (2)|q1q.

This completes the proof.□

Remark 3.10. If we take ξ=+1 in Theorem 3.1, then we have

ϕ(11ω1+1ω212)+ξϕ(11ω1+1ω211)ξ+11β(ξ,1)01γξ1ϕ(1Aγ)
2112(ξ+1)β(+1,ξ)K111q(1,2,ξ)K1(1,2,ξ)|ϕ (ω1)|q+K1(1,2,ξ)|ϕ (ω2)|qK2(1,2,ξ)|ϕ (1)|qK3(1,2,ξ)|ϕ (2)|q1q,

where

K1(1,2,ξ)=01(ξ+1)βγ(ξ,1)β(ξ,1)×1Aγ2
K2(1,2,ξ)=01(ξ+1)βγ(ξ,1)β(ξ,1)×1Aγ2γdγ
K3(1,2,ξ)=K1(1,2,ξ)K2(1,2,ξ),

which is new in the literature.

Remark 3.11. If we take ξ=+1 after that ξ=1 in Theorem 3.1, then we have

ϕ(11ω1+1ω212)+ϕ(11ω1+1ω211)201ϕ(1Aγ)
22112K111q(1,2)K1(1,2)|ϕ (ω1)|q+K1(1,2)|ϕ (ω2)|qK2(1,2)|ϕ (1)|qK3(1,2)|ϕ (2)|q1q,

where

K1(1,2)=01×|12γ|Aγ2
K2(1,2)=01×|12γ|Aγ2γdγ
K3(1,2)=K1(1,2)K2(1,2),

which is new in the literature.

Remark 3.12. If we take ω1=1 and ω2=2 in Theorem 3.1, then we have the inequality proved in (Sanli et al. Citation2018).

Remark 3.13. If we take ω1=1 and ω2=2 and ξ=+1 in Theorem 3.1, then we have the inequality proved in (Sanli et al. Citation2020), Theorem 9].

Remark 3.14. If we put ω1=1,ω2=2,ξ=+1 after that ξ=1 in Theorem 3.1, then we have then we have the inequality proved in (ÍşCan Citation2014).

Example 3.1. Case 1: Let ϕ(κ)=κ44,κ>0. If we set ξ=+1 after that ξ=1, ω1=1, ω1=2, 1=2, 2=5 and q[1.1,100], then by Proposition 1.1 mapping, |ϕ (κ)|q=κ3q is harmonic convex, then we can say that the inequality (3.3) will deduce to

15[0.055]q1q×{(0.016)×[2]3q+(0.039)×[5]3q}1q
24+5485625u2du47.625
15[0.055]q1q×{(0.016)×[2]3q+(0.039)×[5]3q}1q.

Case 2: Let ϕ(κ)=κ44,κ>0. Suppose we set ξ=+1 after that ξ=1, ω1=2, ω2=5, q=2 and 1=[2,3], 2=[4,5] in (3.3).

Figure 1. Case 1 visual illustration for ξ=+1 after that ξ=1, ω1=1, ω2=2, 1=2, 2=5 and q[1.1,100].

Figure 1. Case 1 visual illustration for ξ=ℵ+1 after that ξ=1, ω1=ℓ1, ω2=ℓ2, ℓ1=2, ℓ2=5 and q∈[1.1,100].

Figure 2. Case 2 visual illustration for ξ=+1 after that ξ=1, ω1=2, ω2=5, q=2 and 1=[2,3], 2=[4,5].

Figure 2. Case 2 visual illustration for ξ=ℵ+1 after that ξ=1, ω1=2, ω2=5, q=2 and ℓ1=[2,3], ℓ2=[4,5].

Left, middle, and right mappings from the inequalities (3.3) are plotted against q[1.1,100] in . Left, middle, and right mappings from the inequalities (3.3) are plotted against 1=[1,1.5], 2=[4,5] in . The graphs of the functions prove the correctness of the Theorem 3.1 with ξ=+1 after that ξ=1.

Theorem 3.2.

If |ϕ |q is harmonically convex function for some fixed q1 and ϕ L[ω1,ω2] along with assumption Ω and 0<ξ<1, then the following inequality for fractional integral holds:

|Iϕ1(g;ξ,1,2)|
(3.4) 2112(ξ+1)β(+1,ξ)K41p(ξ,)K5(ξ,)|ϕ (ω1)|q+K5(ξ,)|ϕ (ω2)|qK6(1,2,ξ)|ϕ (2)|qK7(1,2,ξ)|ϕ (1)|q1q,(3.4)

where

K4(ξ,)=01(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)p,
K5(1,2,ξ)=[1ω1+1ω212]2q2F12q,1;2,11ω1+1ω2111ω1+1ω212,
K6(1,2)=(1ω1+1ω212)2q22F12q,2;3;11ω1+1ω2111ω1+1ω212,
K7(1,2)=(1ω1+1ω212)2q22F12q,1;3;11ω1+1ω2111ω1+1ω212.

Here, 1p+1q=1, =0,1,2,3, and ξ=(,+1].

Proof.

By using Lemma 3.1, Hölder’s inequality and since |ϕ |q is harmonically convex, by using Mercer inequality, we have

|Iϕ1(g;ξ,1,2)|
2112(ξ+1)β(+1,ξ)01(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)
×1Aγ2ϕ (1Aγ)
2112(ξ+1)β(+1,ξ)01(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)p1p
(011Aγ2qϕ (1Aγ)q)1q
2112(ξ+1)β(+1,ξ)01(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)p1p
(011Aγ2q|ϕ (ω1)|q+|ϕ (ω2)|qγ|ϕ (1)|q(1γ)|ϕ (2)|q)1q
(3.5) 2112(ξ+1)β(+1,ξ)K41p(1,2,ξ)[K5(ξ,)|ϕ (ω1)|q+K5(ξ,)|ϕ (ω2)|qK6(1,2,ξ)|ϕ (2)|qK7(1,2,ξ)|ϕ (1)|q]1q.(3.5)

Calculating the appearing integrals in (3.5), we have

011Aγ2q=011γ1ω1+1ω212+(1γ)1ω1+1ω2112q
=011γ1ω1+1ω211+(1γ)1ω1+1ω2122q
=[1ω1+1ω212]2q01(1γ11ω1+1ω2111ω1+1ω212)2q
=[1ω1+1ω212]2q2F12q,1;2,11ω1+1ω2111ω1+1ω212
=K5(1,2,ξ)
01γAγ2q=(1ω1+1ω212)2q201γ1γ1ω1+1ω2111ω1+1ω212
=(1ω1+1ω212)2q22F12q,2;3;11ω1+1ω2111ω1+1ω212
=K6(1,2),

and

011γAγ2q=(1ω1+1ω212)2q201(1γ)1γ1ω1+1ω2111ω1+1ω212
=(1ω1+1ω212)2q22F12q,1;3;11ω1+1ω2111ω1+1ω212
=K7(1,2).

This ends the proof.□

Remark 3.15. If we take ξ=+1 in Theorem 3.2, then we have

ϕ(11ω1+1ω212)+ξϕ(11ω1+1ω211)ξ+11β(ξ,1)01γξ1ϕ(1Aγ)2112(ξ+1)β(ξ,1)
012112(ξ+1)β(ξ,1)K41p(ξ)K5(ξ)|ϕ (ω1)|q+K5(ξ)|ϕ (ω2)|qK6(1,2)|ϕ (2)|qK7(1,2)|ϕ (1)|q1q,

where

K4(ξ)=01(ξ+1)βγ(ξ,1)β(ξ,1)p,

and K5K7 are defined above. This is new inequality in the literature.

Remark 3.16. If we take ω1=1 and ω2=2 in Theorem 3.2, then we have the inequality proved in (Sanli et al. Citation2018).

Remark 3.17. If we put ω1=1,ω2=2 and ξ=+1 in Theorem 3.2, then we have the inequality proved in (Sanli et al. Citation2020), Theorem 10].

Remark 3.18. If we put ω1=1,ω2=2,ξ=+1 after that ξ=1 in Theorem 3.2, then we have then we have the inequality proved in (ÍşCan Citation2014).

Example 3.2. Case 1: Let ϕ(κ)=κ44,κ>0. If we set ξ=+1 after that ξ=1, ω1=1, ω1=2, 1=1, 2=5 and q[1.1,100] in (3.4), then by Proposition 1.1 mapping, |ϕ (κ)|q=κ3q is harmonic convex.

10[q12q1]q1q×μ3(1)3q+μ4(5)3q1q
215415κ365.33
(3.6) 10[q12q1]q1q×μ3(1)3q+μ4(5)3q1q,(3.6)

where

μ3=122q+512q[4(12q)1]2(4)2(1q)(12q)
μ4=522q+112q[4(12q)1]2(4)2(1q)(12q).

Case 2: Let ϕ(κ)=κ44,κ>0. Suppose we set ξ=1,ω1=1, ω2=5, q=2 and 1=[1,1.5], 2=[4,5] in (3.4).

Figure 3. Case 1 visual illustration for ξ=+1 after that ξ=1, ω1=1, ω2=2, 1=1, 2=5 and q[1.1,100].

Figure 3. Case 1 visual illustration for ξ=ℵ+1 after that ξ=1, ω1=ℓ1, ω2=ℓ2, ℓ1=1, ℓ2=5 and q∈[1.1,100].

Figure 4. Case 2 visual illustration for ξ=+1 after that ξ=1,ω1=1, ω2=5, q=2 and 1=[1,1.5], 2=[4,5].

Figure 4. Case 2 visual illustration for ξ=ℵ+1 after that ξ=1,ω1=1, ω2=5, q=2 and ℓ1=[1,1.5], ℓ2=[4,5].

Left, middle, and right mappings from the inequalities (3.4) are plotted against q[1.1,100] in . Left, middle, and right mappings from the inequalities (3.4) are plotted against 1=[1,1.5], 2=[4,5] in . The graphs of the functions prove the correctness of the Theorem 3.2 with ξ=+1 after that ξ=1.

Theorem 3.3.

If |ϕ |q is harmonically convex function for some fixed q1 and ϕ L[ω1,ω2] along with assumption Ω and 0<ξ<1, then the following inequality for fractional integral holds:

|Iϕ1(g;ξ,1,2)|
2112(ξ+1)β(+1,ξ)
Υ11pΛ1|ϕ (ω1)|q+Λ1|ϕ (ω2)|qΛ2|ϕ (1)|qΛ3|ϕ (2)|q1q
+Υ21pΛ4|ϕ (ω1)|q+Λ4|ϕ (ω2)|qΛ5|ϕ (1)|qΛ2|ϕ (2)|q1q},

where

Υ1=01(1γ)(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)p1p,
Υ2=01γ(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)p1p,
Λ1=12[1ω1+1ω212]2q2F12q,1;3,11ω1+1ω2111ω1+1ω212,
Λ2=16[1ω1+1ω212]2q2F12q,2;4,11ω1+1ω2111ω1+1ω212,
Λ3=13[1ω1+1ω212]2q2F12q,1;4,11ω1+1ω2111ω1+1ω212,
Λ4=12[1ω1+1ω212]2q2F12q,2;3,11ω1+1ω2111ω1+1ω212,
Λ5=13[1ω1+1ω212]2q2F12q,3;4,11ω1+1ω2111ω1+1ω212.

Here, 1p+1q=1, =0,1,2,3, and ξ=(,+1].

Proof.

By taking modulus in Lemma 3.1, using modified Hölder’s inequality [35] and since |ϕ |q is harmonically convex, by using Mercer inequality, we have

|Iϕ1(g;ξ,1,2)|
2112(ξ+1)β(+1,ξ){01(1γ)(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)p1p
(011γAγ2qϕ (1Aγ)q)1q
+01γ(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)p1p
(01γAγ2qϕ (1Aγ)q)1q}
2112(ξ+1)β(+1,ξ){01(1γ)(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)p1p
(011γAγ2q|ϕ (ω1)|q+|ϕ (ω2)|qγ|ϕ (1)|q(1γ)|ϕ (2)|q)1q
+2112(ξ+1)β(+1,ξ)01γ(ξ+1)βγ(+1,ξ)(ξ)β(+1,ξ)p1p

(3.7) (01γAγ2q|ϕ (ω1)|q+|ϕ (ω2)|qγ|ϕ (1)|q(1γ)|ϕ (2)|q)1q}.(3.7)

Calculating the appearing integrals in (3.7), we have

Λ1=011γAγ2q
=011γγ1ω1+1ω211+(1γ)1ω1+1ω2122q
=[1ω1+1ω212]2q01(1γ)(1γ11ω1+1ω2111ω1+1ω212)2q
=12[1ω1+1ω212]2q2F12q,1;3,11ω1+1ω2111ω1+1ω212
Λ2=01γ(1γ)γ1ω1+1ω211+(1γ)1ω1+1ω2122q
=[1ω1+1ω212]2q01γ(1γ)(1γ11ω1+1ω2111ω1+1ω212)2q
=16[1ω1+1ω212]2q2F12q,2;4,11ω1+1ω2111ω1+1ω212
Λ3=01(1γ)2γ1ω1+1ω211+(1γ)1ω1+1ω2122q
=[1ω1+1ω212]2q01(1γ)2(1γ11ω1+1ω2111ω1+1ω212)2q
=13[1ω1+1ω212]2q2F12q,1;4,11ω1+1ω2111ω1+1ω212
Λ4=01γγ1ω1+1ω211+(1γ)1ω1+1ω2122q
=[1ω1+1ω212]2q01γ(1γ11ω1+1ω2111ω1+1ω212)2q
=12[1ω1+1ω212]2q2F12q,2;3,11ω1+1ω2111ω1+1ω212
Λ5=01γ2γ1ω1+1ω211+(1γ)1ω1+1ω2122q
=[1ω1+1ω212]2q01γ2(1γ11ω1+1ω2111ω1+1ω212)2q
=13[1ω1+1ω212]2q2F12q,3;4,11ω1+1ω2111ω1+1ω212.

This ends the proof.□

Remark 3.19. If we put ω1=1,ω2=2,ξ=+1 after that ξ=1 in Theorem 3.3, then we have

|ϕ(1)+ϕ(2)2122112ϕ(κ)κ2|12(21)212(p+1)1p
{Λ2|ϕ (2)|q+Λ3|ϕ (1)|q1q+Λ2|ϕ (1)|q+Λ5|ϕ (2)|q1q},

where

Λ2=01γ(1γ)γ2+(1γ)12q
=1(21)2[2212q112q(12q)222q122q(22q)]
1(21)3[22212q112q(12q)+232q132q(32q)22222q122q(22q)]
Λ3=01(1γ)2γ2+(1γ)12q
=1(21)3[22212q112q(12q)+232q132q(32q)22222q122q(22q)],

and Λ5 is calculated above. This is new in literature.

4. Comparison between Hölder and improved Hölder fractional integral inequalities

Here, we provide a comparative analysis of Hölder’s integral inequality and its improved version. In Theorem 3.3, by utilizing the enhanced form of H”older’s inequality, we derive a superior lower upper bound compared to the original form in Theorem 3.2.

Example 4.1.

If one chooses ϕ(κ)=13κ3,κ>0, then by Proposition 1.1, mapping |ϕ (κ)|q=κ2q for q>1 and κ>0 is ϕHC(I). In the instance of 1=2,2=3 and q=2, let us find the right side of the inequalities in Theorem 3.2 and Theorem 3.3 with ξ=+1 after that ξ=1 and ω1=1, ω2=2 i.e. Remark 3.18 and Remark 3.19, respectively. After reducing common factor 12(21)2, the right side of Remark 3.18 is

=[1p+1]1p[μ1|ϕ (1)|q+μ2|ϕ (2)|q]1q
=1312×7648(81)+154(16)12
0.624846.

The right side of Remark 3.19, is

=[12(p+1)]1p×{Λ2|ϕ (2)|q+Λ3|ϕ (1)|q1q
+Λ2|ϕ (1)|q+Λ5|ϕ (2)|q1q}
=1612×{1216(81)+172(16)12+1162(81)+1216(16)12}
0.624815.

Clearly,

0.624846>0.624815,

which verifies that the Remark 3.19 error estimate is more accurate than the one in the Remark 3.18. We obtain the lower upper bound better than that of the original form of Hölder inequality.

Case 1: Furthermore, let us set 1=2, 2=5 and q[1.1,50] as the unknown variables. The right-hand sides of Remark 3.18 and Remark 3.19 are denoted by F1(q) and F2(q), respectively.

Figure 5. For the case 1=2, 2=5 and q[1.1,50] the diagram for example 4.1.

Figure 5. For the case ℓ1=2, ℓ2=5 and q∈[1.1,50] the diagram for example 4.1.

Plotting F1(q) and F2(q) in , which illustrates the upper bound derived in Remark 3.19 is better than that of Remark 3.18.

Case 2: Now, we set ξ=1, 1=2, 2=4, q=2 and λ1=[2,2.1], λ2=[3.5,4] as the unknown. The right-hand sides of Theorem 3.2 and Theorem 3.3 are denoted by F3(λ1,λ2) and F4(1,2), respectively.

Figure 6. For the case 1=2, 2=4, q=2 and 1=[2,2.1], 2=[3.5,4] the diagram for example 4.1.

Figure 6. For the case ℓ1=2, ℓ2=4, q=2 and ℓ1=[2,2.1], ℓ2=[3.5,4] the diagram for example 4.1.

Plotting F3(1,2) and F4(1,2) in , which illustrates the upper bound derived in Theorem 3.3 is better than that of Theorem 3.2.

Theorem 4.1.

If |ϕ |q is harmonically convex function for some fixed q1 and ϕ L[ω1,ω2] along with assumption Ω and 0<ξ<1, then the following inequality for fractional integral holds:

|Iϕ2(g;ξ,1,2)|
(4.1) 2112(ξ+1)β(+1,ξ)K711q(1,2,ξ)K7(1,2,ξ)|ϕ (ω1)|q+K7(1,2,ξ)|ϕ (ω2)|qK8(1,2,ξ)|ϕ (1)|qK9(1,2,ξ)|ϕ (2)|q1q,(4.1)

where

K7(1,2,ξ)=01(ξ)β(+1,ξ)(ξ+1)β1γ(+1,ξ)×1Aγ2
K8(1,2,ξ)=01(ξ)β(+1,ξ)(ξ+1)β1γ(+1,ξ)×1Aγ2γdγ
K9(1,2,ξ)=K7(1,2,ξ)K8(1,2,ξ),

with =0,1,2,3, and ξ=(,+1]

Proof.

Similarly the proof of Theorem 3.1, by using Lemma 3.2, Power mean inequality and since |ϕ |q is harmonically convex, by using Mercer inequality, we have (4.1).□

Remark 4.1. If we take ω1=1 and ω2=2 in Theorem 4.1, then we have the inequality proved in (Sanli et al. Citation2018).

Remark 4.2. If we take ω1=1 and ω2=2 and ξ=+1 in Theorem 4.1, then we have the inequality proved in [(Sanli et al. Citation2020), Theorem 7].

Remark 4.3. If we put ω1=1,ω2=2, ξ=+1 after that ξ=1 in Theorem 4.1, then we have the inequality proved in (ÍşCan Citation2014).

Theorem 4.2.

If |ϕ |q is harmonically convex function for some fixed q1 and ϕ L[ω1,ω2] along with assumption Ω and 0<ξ<1, then the following inequality for fractional integral holds

Iϕ2(g;ξ,1,2)
((4.(4. 2112(ξ+1)β(+1,ξ)K41p(ξ,)[K4(ξ,)|ϕ (ω1)|q+K4(ξ,)|ϕ (ω2)|qK5(1,2,ξ)|ϕ (2)|qK6(1,2,ξ)|ϕ (1)|q]1q,((4.(4.

where K4(ξ,),K5(1,2) and K6(1,2) are the same as Theorem 3.2 and ξ=(,+1].

Proof.

Similarly the proof of Theorem 3.2, by using Lemma 3.2, Hölder inequality and since |ϕ |q is harmonically convex, by using Mercer inequality, we have (4.2).□

Remark 4.4. If we take ω1=1 and ω2=2 in Theorem 4.2, then we have the inequality proved in (Sanli et al. Citation2018).

Remark 4.5. If we put ω1=1,ω2=2 and ξ=+1 in Theorem 3.2, then we have the inequality proved in [(Sanli et al. Citation2020), Theorem 10].

Remark 4.6. If we put ω1=1,ω2=2,ξ=+1 after that ξ=1 in Theorem 4.2, then we have the inequality proved in (ÍşCan Citation2014).

5 Applications

5.1 Special means

Now, for numbers 0<1<2, we consider special means and j-logarithmic means:

A(1,2)=1+22,
G(1,2)=(12)12,

and

Lj(1,2)=2(j+1)1(j+1)(j+1)(21)1j,jZ{1,0}.

Proposition 5.1.

Using the same assumptions as in Theorem 3.3, if we take ψ(ω)=ω(k+1)k+1 with ω>0, k1 and ξ=+1 after that ξ=1, then we get

|A11ω1+1ω212k+1,11ω1+1ω211k+1
(k+1)G2(1,2)L(k1)k111ω1+1ω212,11ω1+1ω211|
(1+k)(21)2G2(1,2)12(p+1){Λ1(ω1)kq+Λ1(ω2)kqΛ2(1)kqΛ3(2)kq1q
+Λ4(ω1)kq+Λ4(ω2)kqΛ5(1)kqΛ2(2)kq1q},

where Λ1Λ5 are same as in Theorem 3.3.

Corollary 5.1.

Under the same assumptions as in Proposition 5.1, if we take ω1=1 and ω2=2, then we get

|A(1k+1,2k+1)G2(1,2)L(k1)k1(1,2)|
(1+k)G2(1,2)(21)2[12(p+1)]1p{[Λ3(1)kq+Λ2(2)kq]1q
+[Λ2(1)kq+Λ5(2)kq]1q},

where Λ2Λ5 are same as in Remark 3.19.

5.2 Applications to Numerical Quadrature Rule

In this section, we will extend the idea by giving applications to numerical quadrature rule for harmonic convex inequality.

Theorem 5.1.

Under the assumption of Theorem 3.2 for ξ=+1 after that ξ=1, let In:ω1=κ0<κ1<κ2<κn1<κn=ω2 is a partition of [ω1,ω2], 1,2[κi,κi+1] and gi=(κi+1κi), i=0,1,,n1, then we have:

11ω1+1ω21111ω1+1ω212ϕ(κ)κ2=B(In,ϕ)+R(In,ϕ),

where

(5.1) B(In,ϕ)=i=0n1ϕ11κi+1κi+11κi,2+ϕ11κi+1κi+11κi,12κi,1κi,2gi,(5.1)

and the remainder term satisfy the estimation:

|R(In,ϕ)|
i=0n1gi22(κi,1κi,2)2K41pK5|ϕ (ω1)|q+K5|ϕ (ω2)|qK6|ϕ (2)|qK7|ϕ (1)|q1q,

where K4K6 are same as calculated in Remark 3.18.

Proof.

Applying Theorem 3.2 with ξ=+1 after that ξ=1 on interval [κi,κi+1], i=0,1,,n1, we get

|ϕ11κi+1κi+11κi,2+ϕ11κi+1κi+11κi,12κi,1κi,2gi11κi+1κi+11κi,111κi+1κi+11κi,2ϕ(κ)κ2|
gi22(κi,1κi,2)2K41pK5|ϕ (ω1)|q+K5|ϕ (ω2)|qK6|ϕ (2)|qK7|ϕ (1)|q1q,

for all i=0,1,,n1. Summing over 0 to n1 and using the triangular inequality we obtain the above estimation.□

6 Conclusion

The study dealt with the analysis of novel (H–H) Mercer type inequalities in fractional calculus for harmonic convex functions. We presented two new Conformable fractional trapezoidal type auxiliary equalities in Mercer sense. We enhanced the study of Mercer type integral inequalities using Power-mean, Hölder’s and modified Hölder integral inequalities using the novel approach. We showed by example and graph that the modified Hölder inequality offer lower upper bound than the original form. Similarly, modified power inequality offer lower upper bound than the original form. The remarkable techniques and ideas presented in this article may be developed to coordinates and fractional integral calculus. Our goal for the future is to continue and expand our research in this regard.

Acknowledgments

This study was supported via funding from the Pontifical Catholic University of Ecuador Project No. (070-UIO-2022).

Disclosure statement

No potential conflict of interest was reported by the author(s).

Additional information

Funding

The work was supported by the Pontificia Universidad Católica del Ecuador [Project No. 070-UIO-2022].

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