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Article

Some New Estimates of Hermite–Hadamard Inequalities for Harmonical cr-h-Convex Functions via Generalized Fractional Integral Operator on Set-Valued Mappings

1
Department of Mathematics, College of Sciences, King Khalid University, Abha 61413, Saudi Arabia
2
Department of Mathematics, Government College University Lahore (GCUL), Lahore 54000, Pakistan
3
Department of Mathematics, University of Gujrat, Gujrat 50700, Pakistan
*
Author to whom correspondence should be addressed.
Mathematics 2023, 11(19), 4041; https://doi.org/10.3390/math11194041
Submission received: 1 September 2023 / Revised: 20 September 2023 / Accepted: 21 September 2023 / Published: 23 September 2023
(This article belongs to the Special Issue Variational Problems and Applications, 2nd Edition)

Abstract

:
The application of fractional calculus to interval analysis is vital for the precise derivation of integral inequalities on set-valued mappings. The objective of this article is to reformulated the well-known Hermite–Hadamard inequality into various new variants via fractional integral operator (Riemann–Liouville) and generalize the various previously published results on set-valued mappings via center and radius order relations using harmonical h-convex functions. First, using these notions, we developed the Hermite–Hadamard ( H H ) inequality, and then constructed some product form of these inequalities for harmonically convex functions. Moreover, to demonstrate the correctness of these results, we constructed some interesting non-trivial examples.

1. Introduction

Fractional calculus ( FC ) focuses on fractional order integrals and derivatives, as well as their applications, over real and complex domains. To generate more realistic results with fractional analysis, classical analysis arithmetic is essential. There are a wide variety of mathematical models that can be solved using fractional differential equations and integral equations. Mathematical models with fractional order have more broad and accurate conclusions than classical mathematical models because they are special instances of fractional order mathematical models. As opposed to integer orders, fractional theory permits handling of any number of orders, real or integer, so it is a more appropriate method. Almost no field of nonlinear disciplines or research in contemporary times is uninfluenced by ( FC ) methods and instruments. Many fields of engineering have numerous and fruitful applications, including electrical engineering, control theory, mechanical engineering, viscoelasticity, rheology, optics, and physics, see Refs. [1,2].
A mathematical technique called interval analysis limits errors caused by rounding and measurement in mathematical computations. The Japanese scientist Teruo Sunag has published the most leading paper on interval analysis, see Ref. [3]. There is not only a systematic investigation of the rules which govern basic operations with intervals in this publication, but also a mathematical analysis of the rules that govern them. Using only the terminal points of rational functions over intervals, the range of rational functions can be determined. The corresponding operations are also discussed for interval vectors as multidimensional intervals. Using interval arithmetic tools, a pointwise enclosure for the solution of an initial value problem is computed by enclosing the remainder term and bounding the value of a definite integral. During the last three decades, a compact interval has become an independent object in numerical analysis for verifying or enclosing solutions to a variety of mathematical problems or proving that such problems cannot be solved in a given domain. Further, here are some practical applications of interval analysis in various linear and nonlinear disciplines, see Refs. [4,5]. As we know, Hermite–Hadamard inequality is the first geometric interpretation of a convex function and it is widely used in several disciplines that involve convex optimization. We also know that in order to check the accuracy of various mathematical models based on real life, inequalities must be included to verify their existence, uniqueness, and stability. To visualize better accuracy of results, authors often use mathematical models based on nature as fractional forms, so to handle uncertainty and check the stability of differential models, authors have developed fractional forms of inequalities.
However, generalized convexity mapping is able to deal with a wide variety of problems in both applied and pure analysis. A number of well-known inequalities have been constructed using related classes of convexity, including Simpson, Ostrowski, Opial, Hardy, and the famous Hermite–Hadamard, which has been extended to interval-valued functions. To construct these inequalities, various authors employ different notions of convex classes and integral operators, including the standard Riemann integral, Caputo Fabrizo, Riemann–Liouville, and k-fractional operators. Wang et al. [6] investigated several identities for a differentiable functions involving Hadamard fractions and Riemann–Liouville fractions, and proved some inequalities based on s-convex, r-convex, m-convex, ( s , m ) -convex, etc. In addition, Işcan [7] utilized the notion of preinvexity to obtain various new forms of Hermite–Hadamard via fractional operators. Pachpette established the refined form of Hermite–Hadamard inequalities for the product of two convex mappings using fractional integral in [8]. Based on Riemann–Liouville fractional integrals, Chan established various Hermite–Hadamard inequalities for products of convex functions, see Ref. [9]. Using the Riemann–Liouville fractional integral operator, Khan et al. [10] developed ( H H ) inequalities in an expanded form for harmonically convex mappings in the context of fuzzy interval-valued setting. Shi et al. [11] began by developing Hermite–Hadamard inequality results using h-convex and harmonically h-convex functions, and extended their own work to coordinated convex interval-valued functions ( I . V . F . S ) through fractional integrals. With the idea of h-convexity, Dragomir developed Hermite–Hadamard inequalities using Riemann–Liouville fractional integrals, see Ref. [12]. Khan and his colleagues created Hermite–Hadamard inequalities using left-right set-valued functions via Riemann–Liouville integral operators, see Ref. [13]. Recently, Afzal et al. [14] employed the notions of Riemann integral operator via center-radius order relations to develop a more generalized form of double inequality for harmonically CR - ( h 1 , h 2 ) convex mappings. Sharma et al. [15] developed Hermite–Hadamard inequalities based on general h-harmonic preinvex functions for real-valued stochastic processes. Based on exponentially ( h , m ) -preinvexity, Chen et al. [16] developed various variants of Hermite–Hadamard inequality based on Riemann–Liouville fractional integrals. Awan et al. [17] established several new Hermite–Hadamard type inequalities by using n-polynomial mappings of harmonically convex functions. Viloria et al. [18] developed Hermite–Hadamard-type inequalities for harmonically convex functions based on n-coordinates. Kunt et al. [19] developed new fractional integral inequalities of Hermite–Hadamard–Fejer type for harmonically convex functions. Moshin et al. [20] developed different variants of Hermite–Hadamard inequalities with bounds through q-calculus. For some recent developments regarding developed inequalities, see Refs. [21,22,23,24,25,26].
Bhunia examined the optimality of multi-objective optimization problems using various interval metrics in 2014. He also introduced the concept of center-radius order using radius and interval midpoints, see Ref. [27]. A recent work by the following authors uses Bhunia’s notion to construct various types of inequalities more precisely based on different integral operators, see Refs. [28,29,30,31]. According to our literature review, most of these inequalities are caused by partial order relations. In the case of harmonically h-convex functions, the main advantage of center-radius order relations is that the inequality term can be predicted more accurately. This argument can be proved by valuable examples of proposed theorems. As a result, it is important to understand how total order relations can be used to analyze a variety of convex mapping classes.
This study is significant and original because it investigates the center-radius order relationship in relation to the harmonic convexity and fractional integral for the first time. It suggests a novel approach to inequalities research that includes set-valued functions. This order relation’s main benefit is that it has full order, which allows us to compare intervals while several other interval order relations have significant flaws that prevent us from doing so in set-valued mappings. Unlike the various partial order relations, this one can be calculated very differently, e.g.,: δ c = δ ̲ + δ ¯ 2 and δ r = δ ¯ δ ̲ 2 , are center and interval order representation of interval, respectively, where interval δ = [ δ ¯ , δ ̲ ] .
We are inspired by quality literature and particular articles for our research, see Refs. [13,14,28,29,30]. We developed some variants of Hermite–Hadamard inequalities on set-valued mappings via harmonically h-convex functions in conjunction with fractional integrals. Additionally, we created some examples to demonstrate their accuracy. To conclude, the article follows the following format. There is a brief background provided in Section 2. The main findings are discussed in Section 3. A brief conclusion is provided in Section 4.

2. Preliminaries

There are a few terms used in the paper without being defined, see Ref. [28]. You will find it very helpful throughout the paper to be familiar with a few basic arithmetic principles related to interval analysis. As a result, all the intervals and positive intervals can be represented as R I and R I + of set of real R . The following are the definitions for scalar multiplication and interval addition:
[ δ ] = [ δ ̲ , δ ¯ ] ( k R , δ ̲ x δ ¯ ; k R )
[ ρ ] = [ ρ ̲ , ρ ¯ ] ( k R , ρ ̲ x ρ ¯ ; k R )
[ δ ] + [ ρ ] = [ δ ̲ , δ ¯ ] + [ ρ ̲ , ρ ¯ ] = [ δ ̲ + ρ ̲ , δ ¯ + ρ ¯ ]
ϵ δ = ϵ [ δ ̲ , δ ¯ ] = ϵ δ ̲ , ϵ δ ¯ ( ϵ > 0 ) { 0 } ( ϵ = 0 ) ϵ δ ¯ , ϵ δ ̲ ( ϵ < 0 ) ,
where ϵ R .
Let δ = [ δ ̲ , δ ¯ ] R I , then δ c = δ ¯ + δ ̲ 2 and δ r = δ ¯ δ ̲ 2 are the representation of center- radius form of interval δ , respectively. The CR representation of interval δ can be defined as:
δ = δ c , δ r = δ ¯ + δ ̲ 2 , δ ¯ δ ̲ 2 .
Definition 1
(See [28]). The CR -order relations for δ = [ δ ̲ , δ ¯ ] = δ c , δ r , ρ = [ ρ ̲ , ρ ¯ ] = ρ c , ρ r R I represented as (Figure 1):
δ c r ρ δ c < ρ c , i f δ c ρ c ; δ r ρ r , i f δ c = ρ c .
Theorem 1
(See [28]). Let Ψ : α 1 , α 2 R I be I . V . F defined as Ψ ( ϵ ) = [ Ψ ̲ ( ϵ ) , Ψ ¯ ( ϵ ) ] for every ϵ α 1 , α 2 and Ψ ̲ , Ψ ¯ are Riemann integrable ( IR ) over interval α 1 , α 2 . Then, we said that Ψ is IR over α 1 , α 2 , and
α 1 α 2 Ψ ( ϵ ) d ϵ = α 1 α 2 Ψ ̲ ( ϵ ) d ϵ , α 1 α 2 Ψ ¯ ( ϵ ) d ϵ .
It is most efficient to carry integrals and derivatives to fractional orders or orders other than integers and natural numbers. We describe the idea of the ( R L ) integral operator to facilitate discussions of fractional integrals.
Definition 2.
Let χ L α 1 , α 2 . The ( R L ) integrals J α 1 + ϵ χ ( s ) and J α 2 ϵ χ ( s ) of order ϵ > 0 and α 1 0 are defined as:
J α 1 + ϵ χ ( s ) = 1 Γ ( ϵ ) α 1 s ( s ϵ ) ϵ 1 χ ( ϵ ) d ϵ , s > α 1 ,
and
J α 2 ϵ χ ( s ) = 1 Γ ( ϵ ) s α 2 ( ϵ s ) ϵ 1 χ ( ϵ ) d ϵ , s < α 2 .
Corollary 1
(See [14]). Let χ : α 1 , α 2 R I + be I . V . F given by χ = [ χ ̲ , χ ¯ ] , with χ IR α 1 , α 2 , then
J α 1 + ϵ χ ( s ) = J α 1 + ϵ χ ̲ ( s ) , J α 1 + ϵ χ ¯ ( s )
and
J α 2 + ϵ χ ( s ) = J α 2 + ϵ χ ̲ ( s ) , J α 2 + ϵ χ ¯ ( s ) .
Theorem 2
(See [28]). Let Ψ , χ : α 1 , α 2 R I + be I . V . F . S given by Ψ = [ Ψ ̲ , Ψ ¯ ] , and χ = [ χ ̲ , χ ¯ ] (Figure 2). If Ψ , χ IR α 1 , α 2 , and Ψ ( ϵ ) CR χ ( ϵ )  ∀  ϵ α 1 , α 2 , then
α 1 α 2 Ψ ( ϵ ) d ϵ CR α 1 α 2 χ ( ϵ ) d ϵ .
We will use some interesting examples to demonstrate the above theorem.
Example 1.
Consider Ψ = [ z , z + 1 ] and χ = [ z 2 + 1 , 2 z + 1 ] , ∀  z [ 0 , 1 ] (Figure 3).
χ r = z z 2 2 , Ψ r = 1 2 , χ c = z 2 2 + z + 1 a n d Ψ c = 2 z + 1 2 .
From Definition 1, one has Ψ ( z ) CR χ ( z ) , ∀  z [ 0 , 1 ] .
Since,
0 1 [ z , z + 1 ] d z = 1 2 , 3 2 .
and
0 1 [ z 2 + 1 , 2 z + 1 ] d z = 4 3 , 2
From Theorem 2, we have
0 1 Ψ ( z ) d z CR 0 1 χ ( z ) d z .
Definition 3
([28]). Let h : [ 0 , 1 ] R + and a function Ψ : α 1 , α 2 R + is said to be a harmonically h-convex function, or that Ψ S H X ( h , α 1 , α 2 , R + ) , if ∀  α 1 , α 2 α 1 , α 2 and ϵ [ 0 , 1 ] , one has
Ψ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 h ( ϵ ) Ψ ( α 1 ) + h ( 1 ϵ ) Ψ ( α 2 ) .
If in (1) ≤ is altered with ≥, then it is called harmonical h-concave function or Ψ S H V ( h , α 1 , α 2 , R + ) .
The following is the newly introduced class of harmonic-convex ( I . V . F . S ) for CR order relations:
Definition 4
([22]). Let h : [ 0 , 1 ] R + and a function Ψ : α 1 , α 2 R + is harmonically CR -h-convex function, or that Ψ S H X ( CR h , α 1 , α 2 , R + ) , if ∀  α 1 , α 2 α 1 , α 2 and ϵ [ 0 , 1 ] , one has
Ψ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 CR h ( ϵ ) Ψ ( α 1 ) + h ( 1 ϵ ) Ψ ( α 2 ) .
If in (2)≤ changed with ≥, it is known as harmonical CR -h-concave function or Ψ S H V ( CR h , α 1 , α 2 , R + ) .
Example 2.
Consider α 1 , α 2 = [ 1 , 2 ] , h ( ϵ ) = ϵ and for all ϵ [ 0 , 1 ] . If Ψ : α 1 , α 2 R I + is defined as (Figure 4).
Ψ ( z ) = 1 z 2 + 3 , 1 z 2 + 4 , z [ 1 , 2 ] .
Then,
Ψ C ( σ ) = 7 2 , Ψ R ( σ ) = 1 z 2 + 1 2 , ϵ [ 0 , 1 ] .
It is obvious Ψ C ( z ) , Ψ R ( z ) are harmonically h-convex functions over [ 1 , 2 ] .
Remark 1.
  • If h = 1 , Definition assimilates harmonic CR -P-function.
  • If h ( ϵ ) = 1 h ( ϵ ) , Definition assimilates harmonic CR -h-Godunova–Levin function.
  • If h ( ϵ ) = ϵ s , Definition assimilates harmonic CR -s-convex function.
  • If h ( ϵ ) = 1 ϵ s , Definition assimilates harmonic CR -s-Godunova–Levin function.

3. Main Results

Our goal in this section is to prove some inequalities of Hermite–Hadamard type for set-valued harmonically CR -h-convex mappings involving fractional integrals.
Theorem 3.
Let Ψ : α 1 , α 2 R I + be ( I . V . F ) such that Ψ ( ϵ ) = [ Ψ ( ϵ ) ̲ , Ψ ( ϵ ) ¯ ] , and h : 0 , 1 R + with h 1 2 0 . If Ψ S H X ( CR h , α 1 , α 2 , R I + ) , then the following inequality holds:
1 2 h ( 1 2 ) Ψ 2 α 1 α 2 α 1 + α 2 CR Γ ( β + 1 ) 2 α 1 α 2 α 2 α 1 β J 1 α 2 + β ( Ψ η ) 1 α 1 + J 1 α 1 β ( Ψ η ) 1 α 2 CR β Ψ α 1 + Ψ α 2 2 0 1 ϵ β 1 [ h ( ϵ ) + h ( 1 ϵ ) ] d ϵ .
Proof. 
Since Ψ is a harmonically CR -h-convex function, one has
Ψ 2 f 1 g 1 f 1 + g 1 CR h 1 2 Ψ ( f 1 ) + Ψ ( g 1 ) .
Substitute f 1 = α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 and g 1 = α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 in (4); we have
1 h ( 1 / 2 ) Ψ 2 α 1 α 2 α 1 + α 2 CR Ψ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 + Ψ α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 .
Multiplying by ϵ β 1 in (5) and integrating over [ 0 , 1 ] , one has
1 h ( 1 / 2 ) Ψ 2 α 1 α 2 α 1 + α 2 0 1 ϵ β 1 d ϵ CR ( IR ) 0 1 ϵ β 1 Ψ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 d ϵ + ( IR ) 0 1 ϵ β 1 Ψ α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 d ϵ .
From Theorem 3, we have
( IR ) 0 1 ϵ β 1 Ψ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 d ϵ = ( R ) 0 1 ϵ β 1 Ψ ̲ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 d ϵ , ( R ) 0 1 ϵ β 1 Ψ ¯ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 d ϵ = α 1 α 2 α 2 α 1 β ( R ) 1 α 2 1 α 1 1 α 1 x Ψ ̲ 1 x d x , α 1 α 2 α 2 α 1 β ( R ) 1 α 2 1 α 1 1 α 1 x Ψ ¯ 1 x d x = α 1 α 2 α 2 α 1 β Γ ( β ) J 1 α 2 + β ( Ψ ̲ η ) 1 α 1 , Γ ( β ) J 1 α 2 + β ( Ψ ¯ η ) 1 α 1 = Γ ( β ) α 1 α 2 α 2 α 1 β J 1 α 2 + β ( Ψ η ) 1 α 1 .
Similarly, one has
( IR ) 0 1 ϵ β 1 Ψ α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 d ϵ = ( R ) 0 1 ϵ β 1 Ψ ̲ α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 d ϵ , ( R ) 0 1 ϵ β 1 Ψ ¯ α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 d ϵ = Γ ( β ) α 1 α 2 α 2 α 1 β J 1 α 1 β ( Ψ η ) 1 α 2 .
Hence, by the inequality (6), we obtain
1 h ( 1 / 2 ) β Ψ 2 α 1 α 2 α 1 + α 2 CR Γ ( β ) α 1 α 2 α 2 α 1 β J 1 α 2 + β ( Ψ η ) 1 α 1 + J 1 α 1 β ( Ψ η ) 1 α 2 ,
As a result, first inequality (12) is proven. In order to prove its other side from definition, we have
Ψ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 CR h ( ϵ ) Ψ α 2 + h ( 1 ϵ ) Ψ α 1
and
Ψ α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 CR h ( ϵ ) Ψ α 1 + h ( 1 ϵ ) Ψ α 2 .
Adding (7) and (8), we have
Ψ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 + Ψ α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 CR [ h ( ϵ ) + h ( 1 ϵ ) ] [ Ψ α 1 + Ψ α 2 ] .
Multiplying by ϵ β 1 in (9) and integrating over [ 0 , 1 ] , one has
( IR ) 0 1 ϵ β 1 Ψ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 d ϵ + ( IR ) 0 1 ϵ β 1 Ψ α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 d ϵ CR [ Ψ α 1 + Ψ α 2 ] 0 1 ϵ β 1 [ h ( ϵ ) + h ( 1 ϵ ) ] d ϵ .
As a result, the proof has been completed. □
Remark 2.
(i) 
If h ( ϵ ) = ϵ and Ψ ̲ = Ψ ¯ with β = 1 , then we obtain the following output, see Ref. [32].
Ψ 2 α 1 α 2 α 1 + α 2 α 1 α 2 α 2 α 1 α 1 α 2 Ψ ( ϵ ) ϵ 2 Ψ α 1 + Ψ α 2 2 .
(ii) 
If h ( ϵ ) = ϵ and Ψ ̲ = Ψ ¯ , then we obtain the following output, see Ref. [33].
Ψ 2 α 1 α 2 α 1 + α 2 Γ ( β + 1 ) ( α 2 α 1 ) β J 1 α 2 + β ( Ψ η ) 1 α 1 + J 1 α 1 β ( Ψ η ) 1 α 2 Ψ α 1 + Ψ α 2 2 .
Example 3.
Let [ α 1 , α 2 ] = [ 1 , 2 ] , h ( ϵ ) = ϵ , β = 1 and ∀  ϵ ( 0 , 1 ) . If Ψ : [ α 1 , α 2 ] R I + is defined as
Ψ ( ϵ ) = 1 ϵ 2 + 3 , 1 ϵ 2 + 4 ,
then, we have
1 2 h ( 1 2 ) Ψ 2 α 1 α 2 α 1 + α 2 = 39 16 , 73 16 , Γ ( β + 1 ) 2 α 1 α 2 α 2 α 1 β J 1 α 2 + β ( Ψ η ) 1 α 1 + J 1 α 1 β ( Ψ η ) 1 α 2 = 29 12 , 55 12
and
β Ψ α 1 + Ψ α 2 2 0 1 ϵ β 1 [ h ( ϵ ) + h ( 1 ϵ ) ] d ϵ = 19 8 , 37 8
Consequently, Theorem is verified.
39 16 , 73 16 CR 29 12 , 55 12 CR 19 8 , 37 8 .
Theorem 4.
Let Ψ , χ : α 1 , α 2 R I + be ( I . V . F . S ) such that Ψ ( ϵ ) = [ Ψ ( ϵ ) ̲ , Ψ ( ϵ ) ¯ ] , χ ( ϵ ) = [ χ ( ϵ ) ̲ , χ ( ϵ ) ¯ ] , and h : 0 , 1 R + with h 1 2 0 . If Ψ , χ S H X ( CR h , α 1 , α 2 , R I + ) , then the following inequality holds:
Γ ( β + 1 ) 2 α 1 α 2 α 2 α 1 β J 1 α 2 + β ( Ψ η ) 1 α 1 ( χ η ) 1 α 1 + J 1 α 1 β ( Ψ η ) 1 α 1 ( χ η ) 1 α 2 CR β M α 1 , α 2 2 0 1 ϵ β 1 [ h 2 ( ϵ ) + h 2 ( 1 ϵ ) ] d ϵ + N α 1 , α 2 0 1 ϵ β 1 h ( ϵ ) h ( 1 ϵ ) d ϵ
where
M ( α 1 , α 2 ) = Ψ ( α 1 ) χ ( α 1 ) + Ψ ( α 2 ) χ ( α 2 )
and
N ( α 1 , α 2 ) = Ψ ( α 1 ) χ ( α 2 ) + Ψ ( α 2 ) χ ( α 1 ) .
Proof. 
Since Ψ , χ S H X ( CR h , α 1 , α 2 , R I + ) , one has
Ψ α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 CR h ( ϵ ) Ψ α 1 + h ( 1 ϵ ) Ψ α 2
and
χ α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 CR h ( ϵ ) χ α 1 + h ( 1 ϵ ) χ α 2 .
Multiplying (14) and (15), we obtain
Ψ α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 χ α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 CR h 2 ( ϵ ) Ψ α 1 χ α 1 + h 2 ( ϵ ) Ψ α 2 χ α 2 + h ( ϵ ) h ( 1 ϵ ) [ Ψ α 1 χ α 2 + Ψ α 2 χ α 1 ] .
Similarly, we have
Ψ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 χ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 CR h 2 ( ϵ ) Ψ α 1 χ α 1 + h 2 ( ϵ ) Ψ α 2 χ α 2 + h ( ϵ ) h ( 1 ϵ ) [ Ψ α 1 χ α 2 + Ψ α 2 χ α 1 ]
Adding (16) and (17), we have the following relation
Ψ α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 χ α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 + Ψ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 χ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 CR [ h 2 ( ϵ ) + h 2 ( 1 ϵ ) ] M α 1 , α 2 + 2 h ( ϵ ) h ( 1 ϵ ) N α 1 , α 2 .
Multiplying by ϵ β 1 in (18) and integrating over ( 0 , 1 ) , one has
( IR ) 0 1 ϵ β 1 Ψ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 χ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 d ϵ + ( IR ) 0 1 ϵ β 1 Ψ α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 χ α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 d ϵ CR M α 1 , α 2 0 1 ϵ β 1 [ h 2 ( ϵ ) + h 2 ( 1 ϵ ) ] d ϵ + 2 N α 1 , α 2 0 1 ϵ β 1 h ( ϵ ) h ( 1 ϵ ) d ϵ .
Using Theorem 3 in relation (19), we have
( IR ) 0 1 ϵ β 1 Ψ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 χ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 d ϵ = Γ ( β ) α 1 α 2 α 2 α 1 β J 1 α 2 + β ( Ψ η ) 1 α 1 ( χ η ) 1 α 1
and
( IR ) 0 1 ϵ β 1 Ψ α 1 α 2 ϵ α 2 + ( 1 ϵ ) α 1 χ α 1 α 2 ϵ α 1 + ( 1 ϵ ) α 2 d ϵ = Γ ( β ) α 1 α 2 α 2 α 1 β J 1 α 1 β ( Ψ η ) 1 α 2 ( χ η ) 1 α 2 .
Putting (20) and (21) in (19), our desired outcome has been achieved.
Γ ( β + 1 ) 2 α 1 α 2 α 2 α 1 β J 1 α 2 + β ( Ψ η ) 1 α 1 ( χ η ) 1 α 1 + J 1 α 1 β ( Ψ η ) 1 α 1 ( χ η ) 1 α 2 CR β M α 1 , α 2 2 0 1 ϵ β 1 [ h 2 ( ϵ ) + h 2 ( 1 ϵ ) ] d ϵ + N α 1 , α 2 0 1 ϵ β 1 h ( ϵ ) h ( 1 ϵ ) d ϵ .
Example 4.
Let [ α 1 , α 2 ] = [ 1 , 2 ] , h ( ϵ ) = ϵ , β = 1 and ∀  ϵ ( 0 , 1 ) . If Ψ , χ : [ α 1 , α 2 ] R I + are defined as
Ψ ( ϵ ) = 1 + 2 ϵ 2 ϵ 2 , 1 + 3 ϵ 2 ϵ 2 and χ ( ϵ ) = 1 + ϵ ϵ , 1 + 2 ϵ ϵ .
then, we have
Γ ( β + 1 ) 2 α 1 α 2 α 2 α 1 β J 1 α 2 + β ( Ψ η ) 1 α 1 ( χ η ) 1 α 1 + J 1 α 1 β ( Ψ η ) 1 α 1 ( χ η ) 1 α 2
0.385 , 9.885
and
β M α 1 , α 2 2 0 1 ϵ β 1 [ h 2 ( ϵ ) + h 2 ( 1 ϵ ) ] d ϵ + N α 1 , α 2 0 1 ϵ β 1 h ( ϵ ) h ( 1 ϵ ) d ϵ 0.375 , 10
Consequently, Theorem is verified.
0.385 , 9.885 CR 0.375 , 10 .
Theorem 5.
Let Ψ , χ : α 1 , α 2 R I + be ( I . V . F . S ) such that Ψ ( ϵ ) = [ Ψ ( ϵ ) ̲ , Ψ ( ϵ ) ¯ ] , χ ( ϵ ) = [ χ ( ϵ ) ̲ , χ ( ϵ ) ¯ ] , and h : [ 0 , 1 ] R + with h 1 2 0 . If Ψ , χ S H X ( CR h , α 1 , α 2 , R I + ) , then the following inequality holds:
1 2 h 2 ( 1 / 2 ) Ψ 2 α 1 α 2 α 1 + α 2 χ 2 α 1 α 2 α 1 + α 2 CR Γ ( β ) α 1 α 2 α 2 α 1 β J 1 α 2 + β ( Ψ η ) 1 α 1 ( χ η ) 1 α 1 + J 1 α 1 β ( Ψ η ) 1 α 2 ( χ η ) 1 α 2 + β N α 1 , α 2 2 0 1 ϵ β 1 [ h 2 ( ϵ ) + h 2 ( 1 ϵ ) ] d ϵ + M α 1 , α 2 0 1 ϵ β 1 [ h ( ϵ ) + h ( 1 ϵ ) ] d ϵ .
Proof. 
For ϵ [ 0 , 1 ] , one has
2 α 1 α 2 α 1 + α 2 = 2 α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 + α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 .
Since Ψ , χ S H X ( CR h , α 1 , α 2 , R I + ) , one has
1 2 h 2 ( 1 / 2 ) Ψ 2 α 1 α 2 α 1 + α 2 χ 2 α 1 α 2 α 1 + α 2 = 1 2 h 2 ( 1 / 2 ) Ψ 2 α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 + α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 χ 2 α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 + α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 CR Ψ α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 + Ψ α 1 α 2 ( 1 ϵ ) α 2 + ϵ α 1 × χ α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 + χ α 1 α 2 ( 1 ϵ ) α 2 + ϵ α 1 = Ψ α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 χ α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 + Ψ α 1 α 2 ( 1 ϵ ) α 2 + ϵ α 1 χ α 1 α 2 ( 1 ϵ ) α 2 + ϵ α 1 + Ψ α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 χ α 1 α 2 ( 1 ϵ ) α 2 + ϵ α 1 + Ψ α 1 α 2 ( 1 ϵ ) α 2 + ϵ α 1 χ α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 CR Ψ α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 χ α 1 α 2 ( 1 ϵ ) α 2 + ϵ α 1 + Ψ α 1 α 2 ( 1 ϵ ) α 2 + ϵ α 1 χ α 1 α 2 ( 1 ϵ ) α 2 + ϵ α 1 + [ h 2 ( ϵ ) + h 2 ( 1 ϵ ) ] N α 1 , α 2 + 2 h ( ϵ ) h ( 1 ϵ ) M α 1 , α 2 .
Multiplying by ϵ β 1 in (24) and integrating over [ 0 , 1 ] , one has
1 h 2 ( 1 / 2 ) ( IR ) 0 1 ϵ β 1 Ψ 2 α 1 α 2 α 1 + α 2 χ 2 α 1 α 2 α 1 + α 2 d ϵ CR ( IR ) 0 1 ϵ β 1 Ψ α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 χ α 1 α 2 ( 1 ϵ ) α 1 + ϵ α 2 d ϵ + ( IR ) 0 1 ϵ β 1 Ψ α 1 α 2 ( 1 ϵ ) α 2 + ϵ α 1 χ α 1 α 2 ( 1 ϵ ) α 2 + ϵ α 1 d ϵ + N α 1 , α 2 0 1 ϵ β 1 [ h 2 ( ϵ ) + h 2 ( 1 ϵ ) ] d ϵ + 2 M α 1 , α 2 0 1 ϵ β 1 h ( ϵ ) h ( 1 ϵ ) d ϵ .
Inequality (25) was achieved by changing the integrating variable:
1 2 h 2 ( 1 / 2 ) Ψ 2 α 1 α 2 α 1 + α 2 χ 2 α 1 α 2 α 1 + α 2 CR Γ ( β ) α 1 α 2 α 2 α 1 β J 1 α 2 + β ( Ψ η ) 1 α 1 ( χ η ) 1 α 1 + J 1 α 1 β ( Ψ η ) 1 α 2 ( χ η ) 1 α 2 + β N α 1 , α 2 2 0 1 ϵ β 1 [ h 2 ( ϵ ) + h 2 ( 1 ϵ ) ] d ϵ + M α 1 , α 2 0 1 ϵ β 1 [ h ( ϵ ) + h ( 1 ϵ ) ] d ϵ .
Example 5.
Let [ α 1 , α 2 ] = [ 1 , 2 ] , h ( ϵ ) = ϵ , β = 1 and ∀  ϵ ( 0 , 1 ) . If Ψ , χ : [ α 1 , α 2 ] R I + are defined as
Ψ ( ϵ ) = 1 + 2 ϵ 2 ϵ 2 , 1 + 3 ϵ 2 ϵ 2 and χ ( ϵ ) = 1 + ϵ ϵ , 1 + 2 ϵ ϵ .
then, we have
1 2 h 2 ( 1 / 2 ) Ψ 2 α 1 α 2 α 1 + α 2 χ 2 α 1 α 2 α 1 + α 2 = 23 128 , 627 128 ,
and
Γ ( β ) α 1 α 2 α 2 α 1 β J 1 α 2 + β ( Ψ η ) 1 α 1 ( χ η ) 1 α 1 + J 1 α 1 β ( Ψ η ) 1 α 2 ( χ η ) 1 α 2 + β N α 1 , α 2 2 0 1 ϵ β 1 [ h 2 ( ϵ ) + h 2 ( 1 ϵ ) ] d ϵ + M α 1 , α 2 0 1 ϵ β 1 [ h ( ϵ ) + h ( 1 ϵ ) ] d ϵ
= 137 96 , 1171 32 .
Consequently, Theorem is verified.
23 128 , 627 128 CR 137 96 , 1171 32 .
Theorem 6.
Let Ψ , χ : α 1 , α 2 R I + be ( I . V . F . S ) such that Ψ ( ϵ ) = [ Ψ ( ϵ ) ̲ , Ψ ( ϵ ) ¯ ] , χ ( ϵ ) = [ χ ( ϵ ) ̲ , χ ( ϵ ) ¯ ] , and h : [ 0 , 1 ] R + with h 1 2 0 . If Ψ , χ S H X ( CR h , α 1 , α 2 , R I + ) , then the following inequality holds:
1 2 h 2 ( 1 / 2 ) Ψ 2 α 1 α 2 α 1 + α 2 χ 2 α 1 α 2 α 1 + α 2 CR Γ ( β + 1 ) 2 1 β α 1 α 2 α 1 + α 2 β J α 1 + α 2 2 α 1 α 2 + β ( Ψ η ) 1 α 1 + J α 1 + α 2 2 α 1 α 2 β ( Ψ η ) 1 α 2 + β M α 1 , α 2 0 1 ϵ β 1 h 2 ϵ 2 h ϵ 2 d ϵ + N α 1 , α 2 2 0 1 ϵ β 1 h 2 2 ϵ 2 + h 2 ϵ 2 d ϵ .
Proof. 
Since Ψ , χ S H X ( CR h , α 1 , α 2 , R I + ) , one has
Ψ 2 f 1 g 1 f 1 + g 1 CR h ϵ 2 [ Ψ ( f 1 ) + Ψ ( g 1 ) ] .
For f 1 = 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 and g 1 = 2 α 1 α 2 ( 2 ϵ ) α 1 + ϵ α 2 , we obtain
1 h 1 2 Ψ 2 α 1 α 2 α 1 + α 2 CR Ψ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 + Ψ 2 α 1 α 2 ϵ α 2 + ( 2 ϵ ) α 1 .
Similarly, we obtain
1 h 1 2 Ψ 2 α 1 α 2 α 1 + α 2 CR χ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 + χ 2 α 1 α 2 ϵ α 2 + ( 2 ϵ ) α 1 .
Multiplying the inequality (28) and (29), we have
1 h 2 1 2 Ψ 2 α 1 α 2 α 1 + α 2 χ 2 α 1 α 2 α 1 + α 2 CR Ψ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 χ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 + Ψ 2 α 1 α 2 ϵ α 2 + ( 2 ϵ ) α 1 χ 2 α 1 α 2 ϵ α 2 + ( 2 ϵ ) α 1 + Ψ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 χ 2 α 1 α 2 ϵ α 2 + ( 2 ϵ ) α 1 + Ψ 2 α 1 α 2 ϵ α 2 + ( 2 ϵ ) α 1 χ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 CR Ψ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 χ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 + Ψ 2 α 1 α 2 ϵ α 2 + ( 2 ϵ ) α 1 χ 2 α 1 α 2 ϵ α 2 + ( 2 ϵ ) α 1 + h 2 ϵ 2 Ψ α 1 + h ϵ 2 Ψ α 2 × h 2 ϵ 2 χ α 2 + h ϵ 2 χ α 1 + h 2 ϵ 2 Ψ α 2 + h ϵ 2 Ψ α 1 × h 2 ϵ 2 χ α 1 + h ϵ 2 χ α 2 = Ψ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 χ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 + Ψ 2 α 1 α 2 ϵ α 2 + ( 2 ϵ ) α 1 χ 2 α 1 α 2 ϵ α 2 + ( 2 ϵ ) α 1 + 2 M α 1 , α 2 h 2 ϵ 2 h ϵ 2 + h 2 2 ϵ 2 h 2 ϵ 2 N α 1 , α 2 .
Multiplying by ϵ β 1 in (30) and integrating over [ 0 , 1 ] , we obtain our desired result (26). □
Example 6.
Let [ α 1 , α 2 ] = [ 1 , 2 ] , h ( ϵ ) = ϵ , β = 1 and ∀  ϵ ( 0 , 1 ) . If Ψ , χ : [ α 1 , α 2 ] R I + are defined as
Ψ ( ϵ ) = 1 + 2 ϵ 2 ϵ 2 , 1 + 3 ϵ 2 ϵ 2 and χ ( ϵ ) = 1 + ϵ ϵ , 1 + 2 ϵ ϵ .
then, we have
1 2 h 2 ( 1 / 2 ) Ψ 2 α 1 α 2 α 1 + α 2 χ 2 α 1 α 2 α 1 + α 2 = 23 128 , 627 128 ,
and
Γ ( β + 1 ) 2 1 β α 1 α 2 α 1 + α 2 β J α 1 + α 2 2 α 1 α 2 + β ( Ψ η ) 1 α 1 + J α 1 + α 2 2 α 1 α 2 β ( Ψ η ) 1 α 2 + β M α 1 , α 2 0 1 ϵ β 1 h 2 ϵ 2 h ϵ 2 d ϵ + N α 1 , α 2 2 0 1 ϵ β 1 h 2 2 ϵ 2 + h 2 ϵ 2 d ϵ
= 37 240 , 5 .
Consequently, Theorem is verified.
23 128 , 627 128 CR 37 240 , 5 .
Theorem 7.
Let Ψ , χ : α 1 , α 2 R I + be ( I . V . F . S ) such that Ψ ( ϵ ) = [ Ψ ( ϵ ) ̲ , Ψ ( ϵ ) ¯ ] , χ ( ϵ ) = [ χ ( ϵ ) ̲ , χ ( ϵ ) ¯ ] , and h : [ 0 , 1 ] R + with h 1 2 0 . If Ψ , χ S H X ( CR h , α 1 , α 2 , R I + ) , then the following inequality holds:
Γ ( β + 1 ) 2 1 β α 1 α 2 α 1 + α 2 β J α 1 + α 2 2 α 1 α 2 + β ( Ψ η ) 1 α 1 + J α 1 + α 2 2 α 1 α 2 β ( Ψ η ) 1 α 2 CR β M α 1 , α 2 2 0 1 ϵ β 1 h 2 2 ϵ 2 + h 2 ϵ 2 d ϵ + N α 1 , α 2 0 1 ϵ β 1 h ϵ 2 h 2 ϵ 2 d ϵ .
Proof. 
Since Ψ , χ S H X ( CR h , α 1 , α 2 , R I + ) , one has
Ψ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 CR h 2 ϵ 2 Ψ α 1 + h ϵ 2 Ψ α 2
and
χ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 CR h 2 ϵ 2 χ α 1 + h ϵ 2 χ α 2
Multiplying (32) and (33), we have
Ψ 2 α 1 α 2 ( 2 ϵ ) α 1 + ϵ α 2 χ 2 α 1 α 2 ( 2 ϵ ) α 1 + ϵ α 2 CR h 2 2 ϵ 2 Ψ α 1 χ α 1 + h 2 ϵ 2 Ψ α 2 χ α 2 + h 2 ϵ 2 h ϵ 2 [ Ψ α 1 χ α 2 + Ψ α 2 χ α 1 ] .
Similarly, we obtain
Ψ 2 α 1 α 2 ( 2 ϵ ) α 1 + ϵ α 2 χ 2 α 1 α 2 ( 2 ϵ ) α 1 + ϵ α 2 CR h 2 2 ϵ 2 Ψ α 1 χ α 1 + h 2 ϵ 2 Ψ α 1 χ α 1 + h 2 ϵ 2 h ϵ 2 [ Ψ α 1 χ α 2 + Ψ α 2 χ α 1 ] .
Adding (34) and (35), we obtain the following relation:
Ψ 2 α 1 α 2 ( 2 ϵ ) α 1 + ϵ α 2 χ 2 α 1 α 2 ( 2 ϵ ) α 1 + ϵ α 2 + Ψ 2 α 1 α 2 ( 2 ϵ ) α 2 + ϵ α 1 χ 2 α 1 α 2 ( 2 ϵ ) α 2 + ϵ α 1 CR h 2 2 ϵ 2 [ Ψ α 1 χ α 1 + Ψ α 2 χ α 2 ] + h 2 ϵ 2 [ Ψ α 1 χ α 1 + Ψ α 2 χ α 2 ] + 2 h ϵ 2 h 2 ϵ 2 [ Ψ α 1 χ α 2 + Ψ α 2 χ α 1 ] = h 2 2 ϵ 2 + h 2 ϵ 2 M α 1 , α 2 + 2 h ϵ 2 h 2 ϵ 2 N α 1 , α 2 .
Multiplying by ϵ β 1 in (36) and integrating over ( 0 , 1 ) , one has
( IR ) 0 1 ϵ β 1 Ψ 2 α 1 α 2 ( 2 ϵ ) α 1 + ϵ α 2 χ 2 α 1 α 2 ( 2 ϵ ) α 1 + ϵ α 2 d ϵ + ( IR ) 0 1 ϵ β 1 Ψ 2 α 1 α 2 ( 2 ϵ ) α 2 + ϵ α 1 χ 2 α 1 α 2 ( 2 ϵ ) α 2 + ϵ α 1 d ϵ CR M α 1 , α 2 0 1 ϵ β 1 h 2 2 ϵ 2 + h 2 ϵ 2 d ϵ + 2 N α 1 , α 2 0 1 ϵ β 1 h ϵ 2 h 2 ϵ 2 d ϵ .
By using Theorem 3 in relation (37), we obtain our desired inequality. □
Example 7.
Let [ α 1 , α 2 ] = [ 1 , 2 ] , h ( ϵ ) = ϵ , β = 1 and ∀  ϵ ( 0 , 1 ) . If Ψ , η : [ α 1 , α 2 ] R I + are defined as
Ψ ( ϵ ) = 1 + 2 ϵ 2 ϵ 2 , 1 + 3 ϵ 2 ϵ 2 and η ( ϵ ) = 1 + ϵ ϵ , 1 + 2 ϵ ϵ .
then, we have
Γ ( β + 1 ) 2 1 β α 1 α 2 α 1 + α 2 β J α 1 + α 2 2 α 1 α 2 + β ( Ψ η ) 1 α 1 + J α 1 + α 2 2 α 1 α 2 β ( Ψ η ) 1 α 2
= 37 96 , 949 96
and
β M α 1 , α 2 2 0 1 ϵ β 1 h 2 2 ϵ 2 + h 2 ϵ 2 + N α 1 , α 2 0 1 ϵ β 1 h ϵ 2 h 2 ϵ 2 d ϵ = 3 8 , 10
Consequently, Theorem 7 is verified.
37 96 , 949 96 CR 3 8 , 10 .
Theorem 8.
Let Ψ : α 1 , α 2 R I + be ( I . V . F ) such that Ψ ( ϵ ) = [ Ψ ( ϵ ) ̲ , Ψ ( ϵ ) ¯ ] , and h : [ 0 , 1 ] R + with h 1 2 0 . If Ψ S H X ( CR h , α 1 , α 2 , R I + ) , then the following inequality holds:
1 2 h ( 1 / 2 ) Ψ α 1 + α 2 2 CR Γ ( β + 1 ) 2 1 β J α 1 + α 2 2 α 1 α 2 + β ( Ψ η ) 1 α 1 + J α 1 + α 2 2 α 1 α 2 β ( Ψ η ) 1 α 2 CR β Ψ α 1 + Ψ α 2 2 0 1 ϵ β 1 h 2 ϵ 2 + h ϵ 2 d ϵ .
Proof. 
Since Ψ S H X ( CR h , α 1 , α 2 , R I + ) , one has
Ψ 2 f 1 g 1 f 1 + g 1 CR h 1 2 [ Ψ ( f 1 ) + Ψ ( g 1 ) ]
For f 1 = 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 and g 1 = 2 α 1 α 2 ( 2 ϵ ) α 1 + ϵ α 2 , we obtain
1 h 1 2 Ψ α 1 + α 2 2 CR Ψ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 + Ψ 2 α 1 α 2 ϵ α 2 + ( 2 ϵ ) α 1 .
Multiplying by ϵ β 1 in (39) and integrating over [ 0 , 1 ] , one has
1 h 1 2 Ψ α 1 + α 2 2 0 1 ϵ β 1 d ϵ CR ( IR ) 0 1 ϵ β 1 Ψ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 d ϵ + ( IR ) 0 1 ϵ β 1 Ψ 2 α 1 α 2 ϵ α 2 + ( 2 ϵ ) α 1 d ϵ
Using Theorem 3 in the relation (40), we have
( IR ) 0 1 ϵ β 1 Ψ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 d ϵ = ( R ) 0 1 ϵ β 1 Ψ ̲ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 d ϵ , ( R ) 0 1 ϵ β 1 Ψ ¯ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 d ϵ = 2 α 1 α 2 α 1 + α 2 β ( R ) α 1 + α 2 2 α 1 α 2 1 α 1 1 α 1 u ( 1 α 1 ) ̲ d u , 2 α 1 α 2 α 1 + α 2 β ( R ) α 1 + α 2 2 α 1 α 2 1 1 α 1 u Ψ ¯ ( 1 α 1 ) d u = 2 α 1 α 2 α 1 + α 2 β Γ ( β ) J 2 α 1 α 2 α 1 + α 2 + β ( Ψ ̲ η ) 1 α 1 , 2 α 1 α 2 α 1 + α 2 β Γ ( β ) J 2 α 1 α 2 α 1 + α 2 + β ( Ψ ¯ η ) 1 α 1 = Γ ( β ) 2 α 1 α 2 α 1 + α 2 β J α 1 + α 2 2 α 1 α 2 + β ( Ψ η ) 1 α 1 .
Similarly, we obtain
( IR ) 0 1 ϵ β 1 Ψ 2 α 1 α 2 ϵ α 2 + ( 2 ϵ ) α 1 d ϵ = 2 α 1 α 2 α 1 + α 2 β Γ ( β ) J 2 α 1 α 2 α 1 + α 2 + β ( Ψ ̲ η ) 1 α 2 , 2 α 1 α 2 α 1 + α 2 β Γ ( β ) J 2 α 1 α 2 α 1 + α 2 + β ( Ψ ¯ η ) 1 α 2 = Γ ( β ) 2 α 1 α 2 α 1 + α 2 β J α 1 + α 2 2 α 1 α 2 + β ( Ψ η ) 1 α 2 .
As a result, the first inequality (38) is proven. In order to prove the second inequality, since Ψ is a harmonically CR -h-convex function, one has
Ψ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 CR h 2 ϵ 2 Ψ α 1 + h ϵ 2 Ψ α 2
and
Ψ 2 α 1 α 2 ϵ α 2 + ( 2 ϵ ) α 1 CR h ϵ 2 Ψ α 1 + h 2 ϵ 2 Ψ α 2
Adding (41) and (42), we obtain
Ψ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 + Ψ 2 α 1 α 2 ϵ α 2 + ( 2 ϵ ) α 1 CR [ Ψ α 1 + Ψ α 2 ] h 2 ϵ 2 + h ϵ 2 .
Multiplying by σ 1 ϵ 1 in (43) and integrating over [ 0 , 1 ] , one has
( IR ) 0 1 ϵ β 1 Ψ 2 α 1 α 2 ϵ α 1 + ( 2 ϵ ) α 2 d ϵ + ( IR ) 0 1 ϵ β 1 Ψ 2 α 1 α 2 ϵ α 2 + ( 2 ϵ ) α 1 d ϵ CR ( IR ) 0 1 ϵ β 1 h 2 ϵ 2 + h ϵ 2 [ Ψ α 1 + Ψ α 2 ] d ϵ .
Inequality (38) was achieved by changing the integrating variable. □
Example 8.
Let [ α 1 , α 2 ] = [ 1 , 2 ] , h ( ϵ ) = ϵ , β = 1 and ∀  ϵ ( 0 , 1 ) . If Ψ , η : [ α 1 , α 2 ] R I + are defined as
Ψ ( ϵ ) = 1 ϵ 2 + 5 , 1 ϵ 2 + 6 and η ( ϵ ) = 1 ϵ + 2 , 1 ϵ + 3 .
Then, all the assumptions of Theorem (8) are satisfied.
Theorem 9.
Let Ψ : α 1 , α 2 R I + be ( I . V . F ) such that Ψ ( ϵ ) = [ Ψ ( ϵ ) ̲ , Ψ ( ϵ ) ¯ ] , and h : [ 0 , 1 ] R + with h 1 2 0 . If Ψ S H X ( CR h , α 1 , α 2 , R I + ) , then the following inequality holds:
1 2 h 2 ( 1 / 2 ) Ψ 2 α 1 α 2 α 1 + α 2 CR Γ ( β + 1 ) 2 1 β α 1 α 2 α 2 α 1 β J 1 α 1 β ( Ψ η ) 2 α 1 α 2 α 1 + α 2 + J 1 α 2 + β ( Ψ η ) 2 α 1 α 2 α 1 + α 2 CR β Ψ α 1 + Ψ α 2 2 0 1 ϵ β 1 h 1 + ϵ 2 + h 1 ϵ 2 d ϵ .
Proof. 
Since Ψ S H X ( CR h , α 1 , α 2 , R I + ) , one has
Ψ 2 f 1 g 1 f 1 + g 1 CR h 1 2 [ Ψ ( f 1 ) + Ψ ( g 1 ) ] .
For f 1 = 2 α 1 α 2 ( 1 ϵ ) α 1 + ( 1 + ϵ ) α 2 and g 1 = 2 α 1 α 2 ( 1 + ϵ ) α 1 + ( 1 ϵ ) α 2 , we have
1 h 1 2 Ψ 2 α 1 α 2 α 1 + α 2 CR χ 2 α 1 α 2 ( 1 ϵ ) α 1 + ( 1 + ϵ ) α 2 + χ 2 α 1 α 2 ( 1 + ϵ ) α 1 + ( 1 ϵ ) α 2 .
Multiplying by ϵ β 1 in (45) and integrating over [ 0 , 1 ] , we have
1 h 1 2 Ψ 2 α 1 α 2 α 1 + α 2 0 1 ϵ β 1 d ϵ CR ( IR ) 0 1 ϵ β 1 Ψ 2 α 1 α 2 ( 1 ϵ ) α 1 + ( 1 + ϵ ) α 2 d ϵ + ( IR ) 0 1 ϵ β 1 Ψ 2 α 1 α 2 ( 1 + ϵ ) α 1 + ( 1 ϵ ) α 2 d ϵ
By using Theorem 3 in the relation (46), we have
( IR ) 0 1 ϵ β 1 Ψ 2 α 1 α 2 ( 1 ϵ ) α 1 + ( 1 + ϵ ) α 2 d ϵ = ( R ) 0 1 ϵ β 1 Ψ ̲ 2 α 1 α 2 ( 1 ϵ ) α 1 + ( 1 ϵ ) α 2 d ϵ , ( R ) 0 1 ϵ β 1 Ψ ¯ 2 α 1 α 2 ( 1 ϵ ) α 1 + ( 1 + ϵ ) α 2 d ϵ = 2 α 1 α 2 α 2 α 1 β ( R ) α 1 + α 2 2 α 1 α 2 1 α 1 u α 1 + α 2 2 α 1 α 2 Ψ ̲ ( 1 α 1 ) d u , 2 α 1 α 2 α 2 α 1 β ( R ) α 1 + α 2 2 α 1 α 2 1 α 1 u α 1 + α 2 2 α 1 α 2 Ψ ¯ ( 1 α 1 ) d u = Γ ( β ) 2 α 1 α 2 α 2 α 1 β J 1 α 1 β Ψ 2 α 1 α 2 α 1 + α 2 .
Similarly, we obtain
( IR ) 0 1 ϵ β 1 Ψ 2 α 1 α 2 ( 1 + ϵ ) α 1 + ( 1 ϵ ) α 2 d ϵ = Γ ( β ) 2 α 1 α 2 α 2 α 1 β J 1 α 2 β Ψ 2 α 1 α 2 α 1 + α 2 .
As a result, the first inequality (44) is proven. In order to prove the second inequality, since Ψ is a harmonically CR -h-convex function, one has
Ψ 2 α 1 α 2 ( 1 ϵ ) α 1 + ( 1 + ϵ ) α 2 CR h 1 + ϵ 2 Ψ α 1 + h 1 ϵ 2 Ψ α 2
and
Ψ 2 α 1 α 2 ( 1 + ϵ ) α 1 + ( 1 ϵ ) α 2 CR h 1 ϵ 2 Ψ α 1 + h 1 + ϵ 2 Ψ α 2 .
Adding (47) and (48), we obtain
Ψ 2 α 1 α 2 ( 1 ϵ ) α 1 + ( 1 + ϵ ) α 2 + Ψ 2 α 1 α 2 ( 1 + ϵ ) α 1 + ( 1 ϵ ) α 2 CR [ Ψ α 1 + Ψ α 2 ] h 1 ϵ 2 + h 1 ϵ 2 .
Multiplying by ϵ β 1 in (49) and integrating over [ 0 , 1 ] , one has
( IR ) 0 1 ϵ β 1 Ψ 2 α 1 α 2 ( 1 ϵ ) α 1 + ( 1 + ϵ ) α 2 d ϵ + ( IR ) 0 1 ϵ β 1 Ψ 2 α 1 α 2 ( 1 + ϵ ) α 1 + ( 1 ϵ ) α 2 d ϵ CR [ Ψ α 1 + Ψ α 2 ] 0 1 ϵ β 1 h 1 ϵ 2 + h 1 + ϵ 2 d ϵ .
As a result, the proof has been completed. □
Example 9.
Let [ α 1 , α 2 ] = [ 1 , 2 ] , h ( ϵ ) = ϵ , β = 1 and ∀  ϵ ( 0 , 1 ) . If Ψ , η : [ α 1 , α 2 ] R I + are defined as
Ψ ( ϵ ) = 1 ϵ 2 + 2 , 1 ϵ 2 + 3 and η ( ϵ ) = 1 ϵ + 1 , 1 ϵ + 2 .
Then, all the assumptions of Theorem (9) are satisfied.
Theorem 10.
Let Ψ , χ : α 1 , α 2 R I + be ( I . V . F . S ) such that Ψ ( ϵ ) = [ Ψ ( ϵ ) ̲ , Ψ ( ϵ ) ¯ ] , χ ( ϵ ) = [ χ ( ϵ ) ̲ , χ ( ϵ ) ¯ ] , and h : [ 0 , 1 ] R + with h 1 2 0 . If Ψ , χ S H X ( CR h , α 1 , α 2 , R I + ) , then the following inequality holds:
Γ ( β + 1 ) 2 1 β α 1 α 2 α 2 α 1 β J 1 α 1 β ( Ψ η ) 2 α 1 α 2 α 1 + α 2 ( χ η ) 2 α 1 α 2 α 1 + α 2 + J 1 α 2 + β ( Ψ η ) 2 α 1 α 2 α 1 + α 2 ( χ η ) 2 α 1 α 2 α 1 + α 2 CR β M α 1 , α 2 2 0 1 ϵ β 1 h 2 1 ϵ 2 + h 2 1 + ϵ 2 d ϵ + N α 1 , α 2 0 1 ϵ β 1 h 1 + ϵ 2 h 1 ϵ 2 d ϵ .
Proof. 
Since Ψ , χ S H X ( CR h , α 1 , α 2 , R I + ) , one has
Ψ 2 α 1 α 2 ( 1 ϵ ) α 2 + ( 1 + ϵ ) α 1 CR h 1 ϵ 2 Ψ α 1 + h 1 + ϵ 2 Ψ α 2
and
χ 2 α 1 α 2 ( 1 ϵ ) α 2 + ( 1 + ϵ ) α 1 CR h 1 ϵ 2 χ α 1 + h 1 + ϵ 2 χ α 2 .
Multiplying (51) and (52), we have
Ψ 2 α 1 α 2 ( 1 ϵ ) α 2 + ( 1 + ϵ ) α 1 χ 2 α 1 α 2 ( 1 ϵ ) α 2 + ( 1 + ϵ ) α 1 CR h 2 1 ϵ 2 Ψ α 1 χ α 1 + h 2 1 ϵ 2 Ψ α 2 χ α 2 + h 1 + ϵ 2 h 1 ϵ 2 [ Ψ α 1 χ α 2 + Ψ α 2 χ α 1 ]
Similarly, we obtain
Ψ 2 α 1 α 2 ( 1 ϵ ) α 2 + ( 1 + ϵ ) α 1 χ 2 α 1 α 2 ( 1 ϵ ) α 2 + ( 1 + ϵ ) α 1 CR h 2 1 + ϵ 2 Ψ α 1 χ α 1 + h 2 1 + ϵ 2 Ψ α 2 χ α 2 + h 1 + ϵ 2 h 1 ϵ 2 [ Ψ α 1 χ α 2 + Ψ α 2 χ α 1 ]
Adding (53) and (54), one has
Ψ 2 α 1 α 2 ( 1 + ϵ ) α 2 + ( 1 ϵ ) α 1 χ 2 α 1 α 2 ( 1 + ϵ ) α 2 + ( 1 ϵ ) α 1 + Ψ 2 α 1 α 2 ( 1 + ϵ ) α 2 + ( 1 ϵ ) α 1 χ 2 α 1 α 2 ( 1 + ϵ ) α 2 + ( 1 ϵ ) α 1 CR h 2 1 + ϵ 2 [ Ψ α 1 χ α 1 + Ψ α 2 χ α 2 ] + h 2 1 ϵ 2 [ Ψ α 1 χ α 1 + Ψ α 2 χ α 2 ] + 2 h 1 + ϵ 2 h 1 ϵ 2 [ Ψ α 1 χ α 1 + Ψ α 2 χ α 2 ] .
Multiplying by ϵ β 1 in (53) and integrating over [ 0 , 1 ] , one has
( IR ) 0 1 ϵ β 1 Ψ 2 α 1 α 2 ( 1 ϵ ) α 2 + ( 1 + ϵ ) α 1 χ 2 α 1 α 2 ( 1 ϵ ) α 2 + ( 1 + ϵ ) α 1 d ϵ + ( IR ) 0 1 ϵ β 1 Ψ 2 α 1 α 2 ( 1 ϵ ) α 2 + ( 1 + ϵ ) α 1 χ 2 α 1 α 2 ( 1 ϵ ) α 2 + ( 1 + ϵ ) α 1 d ϵ CR M α 1 , α 2 0 1 ϵ β 1 h 2 1 ϵ 2 + h 2 1 + ϵ 2 d ϵ + 2 N α 1 , α 2 0 1 ϵ β 1 h 1 + ϵ 2 h 1 ϵ 2 d ϵ .
By using Theorem 3 in relation (56), we obtain the desired output. □

4. Conclusions

In this work, we construct some Hermite–Hadamard inequalities in a novel manner using harmonical convex functions via Riemann–Liouville integral operator. We do this by using the set-valued mappings for center and radius order. We can evaluate Hermite–Hadamard inequalities from a new angle by combining these ideas. Since it is well known that this fractional integral generalizes the conventional Riemann integral, in this study we generalize various previous findings. A few striking examples are also provided to demonstrate the validity of the conclusions that have been proven. More investigation into equivalent inequalities using different integral operators and convexity types will be fascinating. Additionally, readers will have to construct these findings using fuzzy environments, time scale calculus, and coordinates.

Author Contributions

Conceptualization, W.A. and Y.A.; investigation, W.A.; methodology, W.A. and Y.A.; validation, W.A.; visualization, Y.A.; writing—original draft, W.A.; writing review and editing, W.A. and Y.A. All authors have read and agreed to the published version of the manuscript.

Funding

The authors extend their appreciation to the Deanship of Scientific Research at King Khalid University for funding this work through large group Research Project under grant number RGP2/366/44.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

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Figure 1. The figure illustrates the correctness of CR -order relationships defined in Definition 1.
Figure 1. The figure illustrates the correctness of CR -order relationships defined in Definition 1.
Mathematics 11 04041 g001
Figure 2. The figure illustrates the validity of Theorem 2.
Figure 2. The figure illustrates the validity of Theorem 2.
Mathematics 11 04041 g002
Figure 3. χ r = z z 2 2 , Ψ r = 1 2 , χ c = z 2 2 + z + 1 and Ψ c = 2 z + 1 2 .
Figure 3. χ r = z z 2 2 , Ψ r = 1 2 , χ c = z 2 2 + z + 1 and Ψ c = 2 z + 1 2 .
Mathematics 11 04041 g003
Figure 4. In the above figure, Ψ ̲ is shown as blue and Ψ ¯ as orange, respectively.
Figure 4. In the above figure, Ψ ̲ is shown as blue and Ψ ¯ as orange, respectively.
Mathematics 11 04041 g004
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Almalki, Y.; Afzal, W. Some New Estimates of Hermite–Hadamard Inequalities for Harmonical cr-h-Convex Functions via Generalized Fractional Integral Operator on Set-Valued Mappings. Mathematics 2023, 11, 4041. https://doi.org/10.3390/math11194041

AMA Style

Almalki Y, Afzal W. Some New Estimates of Hermite–Hadamard Inequalities for Harmonical cr-h-Convex Functions via Generalized Fractional Integral Operator on Set-Valued Mappings. Mathematics. 2023; 11(19):4041. https://doi.org/10.3390/math11194041

Chicago/Turabian Style

Almalki, Yahya, and Waqar Afzal. 2023. "Some New Estimates of Hermite–Hadamard Inequalities for Harmonical cr-h-Convex Functions via Generalized Fractional Integral Operator on Set-Valued Mappings" Mathematics 11, no. 19: 4041. https://doi.org/10.3390/math11194041

APA Style

Almalki, Y., & Afzal, W. (2023). Some New Estimates of Hermite–Hadamard Inequalities for Harmonical cr-h-Convex Functions via Generalized Fractional Integral Operator on Set-Valued Mappings. Mathematics, 11(19), 4041. https://doi.org/10.3390/math11194041

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