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Article

Innovative Interpolating Polynomial Approach to Fractional Integral Inequalities and Real-World Implementations

1
Department of Mathematics, University of Sargodha, Sargodha 40100, Pakistan
2
Department of Mathematics and Statistics, Hazara University, Mansehra 21300, Pakistan
3
Escuela de Ciencias Físicas y Matemáticas, Facultad de Ciencias Exactas y Naturales, Pontificia Universidad Católica del Ecuador, Av. 12 de Octubre 1076, Apartado, Quito 17-01-2184, Ecuador
*
Author to whom correspondence should be addressed.
Axioms 2023, 12(10), 914; https://doi.org/10.3390/axioms12100914
Submission received: 21 July 2023 / Revised: 25 August 2023 / Accepted: 8 September 2023 / Published: 26 September 2023
(This article belongs to the Special Issue Current Research on Mathematical Inequalities II)

Abstract

:
Our paper explores Hermite–Hadamard inequalities through the application of Abel–Gontscharoff Green’s function methodology, which involves interpolating polynomials and Riemann-type generalized fractional integrals. While establishing our main results, we explore new identities. These identities are used to estimate novel findings for functions, such that the second derivative of the functions is monotone, absolutely convex, and concave. A section relating the results of exploration to generalized means and trapezoid formulas is included in the applications. We anticipate that the method presented in this study will inspire further research in this field.

1. Introduction

The theory of inequality has evolved quickly in recent years. It is important to consider how closely related the convexity and inequality theories are to one another. Numerous new definitions, generalizations, and expansions of novel convexity have been provided in recent years, and related advancements in the theory of convexity inequality—particularly integral inequalities theory—have also been acknowledged. The Hermite–Hadamard inequality is one of the most significant causes of this development, among other significant inequalities. This inequality is a well known for convex functions that have been established in various ways and has a number of expansions and generalizations in the literature [1,2,3,4,5]. It defines upper and lower bounds for the integral mean of any convex function defined over a closed and bounded interval encompassing the function’s endpoints and midpoint. For a convex function ⊺, the Hermite–Hadamard inequality is stated as follows:
ς 1 + ς 2 2 1 ς 1 ς 2 ς 1 ς 2 ( ϱ ) d ϱ ( ς 1 ) + ( ς 2 ) 2 .
The above inequality will hold in reverse directions if ⊺ is a concave function. This inequality has many fractional extensions like [6,7,8]. We further refer the reader to [9,10].
Compared to other function classes, convex functions possess a geometric interpretation and find numerous applications in various fields, including mathematics, statistics, optimization theory, finance, decision making, and numerical analysis. They constitute not only fundamental components of inequality theory but also serve as key elements motivating several inequalities. These functions are associated with not just continuity and differentiability, but also inequalities. The exploration of integral inequalities is a captivating pursuit within mathematical analysis. Foundational integral inequalities can significantly contribute to the elucidation of subjective aspects of convexity. Convex functions find utility across multiple domains within mathematical analysis and statistics; however, their role within inequality theory is of paramount significance. In this context, a plethora of classical and analytical inequalities have been established, most notably the Hermite–Hadamard, Ostrowski, Simpson, Fejer, and Hardy-type inequalities [11,12]. The extensive body of literature concerning integral inequalities for convex functions underscores the immense importance of this subject [13,14,15,16,17]. The convex function is defined as follows:
Definition 1
([17]). A real valued function : [ ς 1 , ς 2 ] R is said to be convex (concave) if the inequality
( φ ς 1 + ( 1 φ ) ς 2 ) ( ) φ ( ς 1 ) + ( 1 φ ) ( ς 2 )
holds for all 0 φ 1 .
Fractional calculus has seen rapid progress in applied and pure mathematics because of its widespread usage in image processing, physics, and other fields. Experts from various fields have been quick to notice the fractional derivative. Classical derivations cannot be used to model the majority of practical problems. Fractional differential equations deal with the complexities of real-world problems [18,19,20,21,22]. For fractional integral operators, several definitions have been used, including the Hadamard integral, the k-Riemann–Liouville (RL) fractional integral, the Caputo–Fabrizio fractional integral, the RL fractional integral, and the conformable fractional integral [23,24,25]. One can extend such fractional integral operators by including additional parameters, leading to the fractional inequalities i.e., Minkowski, Hermite–Hadamard, Jensen, Ostrowski, and Grüss [23,26,27,28]. Future study is encouraged by these generalizations to provide additional innovative concepts using unified fractional operators, and to discover inequalities employing such generalised fractional operators.
We must keep in mind the preliminary equations and notations of a few well-known RL and k-RL fractional integral operators in order to derive certain results and corollaries. The RL fractional integrals, which are defined as follows, are the most traditional form of fractional integrals that have been described in the literature.
Definition 2
([29]). Let ς 1 , ς 2 R with ς 1 < ς 2 and L [ ς 1 , ς 2 ] . Then, the left-sided RL and right-sided RL fractional integrals ς 1 + ϖ and ς 2 ϖ of order ϖ > 0 on a finite interval [ ς 1 , ς 2 ] are defined by
ς 1 + ϖ ( χ ) = 1 Γ ( ϖ ) ς 1 χ ( χ φ ) ϖ 1 ( φ ) d φ , χ > ς 1
and
ς 2 ϖ ( χ ) = 1 Γ ( ϖ ) χ ς 2 ( φ χ ) ϖ 1 ( φ ) d φ , χ < ς 2 ,
respectively. The conventional Euler’s gamma function, denoted by Γ ( ϖ ) , is defined in the following manner:
Γ ( ϖ ) = 0 ϱ ϖ 1 e ϱ d ϱ , R e ( ϖ ) > 0 .
Definition 3
([30]). The definition of κ-fractional integrals with order ϖ and parameters κ > 0 and ς 0 is as follows:
ς 1 + ϖ , κ ( χ ) = 1 κ Γ κ ( ϖ ) ς 1 χ ( χ φ ) ϖ κ 1 ( φ ) d φ , χ > ς 1
and
ς 2 ϖ , κ ( χ ) = 1 κ Γ κ ( ϖ ) χ ς 2 ( φ χ ) ϖ κ 1 ( φ ) d φ , χ < ς 2 .
The κ-gamma function denoted as Γ κ , as defined by Diaz et al. [31], can be expressed as follows:
Γ κ ( ϖ ) = 0 φ ϖ 1 e φ κ κ d φ .
Corresponding to the choice of κ = 1 , the classical RL fractional integral is obtained as given in (1).
Definition 4
([28]). A real valued function : I R , where I is the range of continuous function g : [ ς 1 , ς 2 ] R . Then, the Jensen’s integral inequality,
ς 1 ς 2 g ( ϱ ) d λ ( ϱ ) ς 1 ς 2 d λ ( ϱ ) ς 1 ς 2 ( g ( ϱ ) ) d λ ( ϱ ) ς 1 ς 2 d λ ( ϱ ) ,
holds ifis continuous, provided that λ is nondecreasing, bounded, and λ ( ς 1 ) λ ( ς 2 ) ) .
The Abel–Gontscharoff polynomial and theorem for the two-point right focal" problem are referenced in [32]. The Abel–Gontscharoff polynomial for two-point right focal interpolating polynomial can be stated as a special choice for n = 2 .
( χ ) = ( ς 1 ) + ( χ ς 1 ) ( ς 2 ) + ς 1 ς 2 G ( χ , ψ ) ( ψ ) d ψ ,
where G ( χ , ψ ) is the Green’s function.
The following four functions were introduced by Mehmood et al. [33] based on the Abel–Gontscharoff Green’s function.
G 1 ( χ , ψ ) = ς 1 ψ , for ς 1 ψ χ ; ς 1 χ , for χ ψ ς 2 ;
G 2 ( χ , ψ ) = χ ς 2 , for ς 1 ψ χ ; ψ ς 2 , for χ ψ ς 2 ;
G 3 ( χ , ψ ) = χ ς 1 , for ς 1 ψ χ ; ψ ς 1 , for χ ψ ς 2 ;
G 4 ( χ , ψ ) = ς 2 ψ , for ς 1 ψ χ ; ς 2 χ , for χ ψ ς 2 .
Sarikaya et al. [34] derived the subsequent inequality of Hermite–Hadamard type for fractional integrals.
Theorem 1.
Let : [ ς 1 , ς 2 ] R be a positive function satisfying 0 ς 1 < ς 2 and L [ ς 1 , ς 2 ] . Ifis a convex function over [ ς 1 , ς 2 ] , then the following inequalities hold for fractional integrals.
ς 1 + ς 2 2 Γ ( ϖ + 1 ) 2 ( ς 2 ς 1 ) ϖ ς 1 + ϖ ( ς 2 ) + ς 2 ϖ ( ς 1 ) ( ς 1 ) + ( ς 2 ) 2 ,
where ϖ > 0 .
Inspired by the research conducted by Khan et al. as reported in [29], we intend to develop Hadamard-type fractional integral inequalities by utilizing appropriate Green’s functions. The findings we present in the following section extend the current body of knowledge and serve as a source of inspiration for researchers engaged in the study of mathematical inequalities. Finally, we refer the reader to learn about fractional calculus and its applications from the articles [35,36,37,38,39,40,41] and from the books [42,43]. The convex functions and related inequalities can also be studied in [44,45,46,47,48,49,50,51,52,53].

2. Generalized Fractional Integral Inequalities via Special Green’s Functions

In this section, we derive generalized fractional integral inequalities via Abel–Gontscharoff Green’s function given in (3).
Theorem 2.
Letbe a function that is twice differentiable and convex on the interval [ ς 1 , ς 2 ] . Then, the following double inequality holds for any positive values of ϖ and κ:
κ ς 1 + ϖ ς 2 ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ .
Proof. 
By substituting χ = κ ς 1 + ϖ ς 2 ϖ + κ into the Abel–Gontscharoff polynomial for the two-point right focal interpolating polynomial given by Equation (2), we derive
κ ς 1 + ϖ ς 2 ϖ + κ = ( ς 1 ) + κ ς 1 + ϖ ς 2 ϖ + κ ς 1 ( ς 2 ) + ς 1 ς 2 G κ ς 1 + ϖ ς 2 ϖ + κ , ψ ( ψ ) d ψ = ( ς 1 ) + ϖ ( ς 2 ς 1 ) ϖ + κ ( ς 2 ) + ς 1 ς 2 G κ ς 1 + ϖ ς 2 ϖ + κ , ψ ( ψ ) d ψ .
After multiplying both sides of Equation (2) by ϖ ( χ ς 1 ) ϖ κ 1 κ ( ς 2 ς 1 ) ϖ κ and integrating with respect to χ , we obtain
ϖ κ ( ς 2 ς 1 ) ϖ κ 1 ς 1 ς 2 ( χ ς 1 ) ϖ κ 1 ( χ ) d χ = ϖ κ ( ς 2 ς 1 ) ϖ κ 1 ς 1 ς 2 ( χ ς 1 ) ϖ κ 1 ( ς 1 ) d χ + ς 1 ς 2 ( χ ς 1 ) ϖ κ 1 ( χ ς 1 ) ( ς 2 ) d χ + ς 1 ς 2 ς 1 ς 2 G ( χ , ψ ) ( χ ς 1 ) ϖ κ 1 ( ψ ) d ψ d χ
= ϖ κ ( ς 2 ς 1 ) ϖ κ κ ( ς 1 ) ( ς 2 ς 1 ) ϖ κ ϖ + κ ( ς 2 ) ( ς 2 ς 1 ) ϖ κ + 1 ϖ + κ + ς 1 ς 2 ς 1 ς 2 G ( χ , ψ ) ( χ ς 1 ) ϖ κ 1 ( ψ ) d ψ d χ .
By using the relation Γ κ ( ϖ + κ ) = ϖ Γ κ ( κ ) , this can also be written as
Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) = ( ς 1 ) + ϖ ( ς 2 ) ( ς 2 ς 1 ) ( ϖ + κ ) + ϖ κ ( ς 2 ς 1 ) ϖ κ ς 1 ς 2 ς 1 ς 2 G ( χ , ψ ) ( χ ς 1 ) ϖ κ 1 ( ψ ) d ψ d χ .
From (9) and (10), we can write
κ ς 1 + ϖ ς 2 ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) = ς 1 ς 2 G κ ς 1 + ϖ ς 2 ϖ + κ , ψ ϖ κ ( ς 2 ς 1 ) ϖ κ ς 1 ς 2 G ( χ , ψ ) ( χ ς 1 ) ϖ κ 1 d χ ( ψ ) d ψ .
By employing the formula (3) and implementing certain simplifications, we arrive at the following:
ς 1 ς 2 G ( χ , ψ ) ( χ ς 1 ) ϖ κ 1 d χ = κ 2 ϖ ( ϖ + κ ) ( ψ ς 1 ) ϖ κ + 1 ϖ κ + 1 ( ς 2 ς 1 ) ϖ κ ( ψ ς 1 )
and
G κ ς 1 + ϖ ς 2 ϖ + κ , ψ = ς 1 ψ , for ς 1 ψ κ ς 1 + ϖ ς 2 ϖ + κ ; ϖ ( ς 1 ς 2 ) ϖ + κ , for κ ς 1 + ϖ ς 2 ϖ + κ ψ ς 2 .
If ς 1 ψ κ ς 1 + ϖ ς 2 ϖ + κ , then by utilizing (12) and (13) in (11), we have
G κ ς 1 + ϖ ς 2 ϖ + κ , ψ ϖ κ ( ς 2 ς 1 ) ϖ κ ς 1 ς 2 G ( χ , ψ ) ( χ ς 1 ) ϖ κ 1 d χ = ( ς 1 ψ ) κ ( ψ ς 1 ) ϖ κ + 1 ( ς 2 ς 1 ) ϖ κ ( ψ ς 1 ) ( ϖ κ + 1 ) ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ
= ( ς 1 ψ ) ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ κ ( ψ ς 1 ) ϖ κ + 1 + κ ( ψ ς 1 ) ( ϖ κ + 1 ) ( ς 2 ς 1 ) ϖ κ ( ς 2 ς 1 ) ϖ κ ( ϖ + κ ) = κ ( ψ ς 1 ) ϖ κ + 1 ( ς 2 ς 1 ) ϖ κ ( ϖ + κ ) 0 .
Again, for the choice κ ς 1 + ϖ ς 2 ϖ + κ ψ ς 2 , and making use of (12) and (13) in (11), we get
G κ ς 1 + ϖ ς 2 ϖ + κ , ψ ϖ κ ( ς 2 ς 1 ) ϖ κ ς 1 ς 2 G ( χ , ψ ) ( χ ς 1 ) ϖ κ 1 d χ = ϖ ( ς 1 ς 2 ) ϖ + κ κ ( ψ ς 1 ) ϖ κ + 1 ( ψ ς 1 ) ( ϖ κ + 1 ) ( ς 2 ς 1 ) ϖ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ = ϖ ( ς 1 ς 2 ) ( ς 2 ς 1 ) ϖ κ κ ( ψ ς 1 ) ϖ κ + 1 ( ψ ς 1 ) ( ϖ κ + 1 ) ( ς 2 ς 1 ) ϖ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ = ( ς 2 ς 1 ) ϖ κ ϖ ( ψ ς 2 ) + κ ( ψ ς 1 ) κ ( ψ ς 1 ) ϖ κ + 1 ( ς 2 ς 1 ) ϖ κ ( ϖ + κ ) .
Let
h ( ψ ) = ( ς 2 ς 1 ) ϖ κ ϖ ( ψ ς 2 ) + κ ( ψ ς 1 ) κ ( ψ ς 1 ) ϖ κ + 1 ( ς 2 ς 1 ) ϖ κ ( ϖ + κ ) .
It is notable that
h ( ς 2 ) = 0 and h ( ψ ) = 1 ( ψ ς 1 ) ϖ κ ( ς 2 ς 1 ) ϖ κ > 0 .
Therefore, followed by (15) to (16), we can write
G κ ς 1 + ϖ ς 2 ϖ + κ , ψ ϖ κ ( ς 2 ς 1 ) ϖ κ ς 1 ς 2 G ( χ , ψ ) ( χ ς 1 ) ϖ κ 1 d χ 0 .
Since ⊺ is convex and implies ( ψ ) 0 , and by making use of (14), (17) in (11), we can write
κ ς 1 + ϖ ς 2 ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) .
This establishes the left half of the inequality in Equation (8).
Now, we proceed to prove the right half inequality of Equation (8). To accomplish this, we select χ = ς 2 in Equation (2), yielding
( ς 2 ) = ( ς 1 ) + ( ς 2 ς 1 ) ( ς 2 ) + ς 1 ς 2 G ( ς 2 , ψ ) ( ψ ) d ψ .
Multiplying both sides by ϖ κ and then dividing by ( ϖ κ + 1 ) , Equation (19) is equivalent to
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ = ( ς 1 ) + ϖ ( ς 2 ς 1 ) ( ς 2 ) ϖ + κ + ϖ ϖ + κ ς 1 ς 2 G ( ς 2 , ψ ) ( ψ ) d ψ ;
by subtracting (10) from (20), we get the inequality
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) = ς 1 ς 2 ϖ ϖ + κ G ( ς 2 , ψ ) ϖ κ ( ς 2 ς 1 ) ϖ κ ς 1 ς 2 G ( χ , ψ ) ( χ ς 1 ) ϖ κ 1 d χ ( ψ ) d ψ .
By including the Green formula ( ς 1 ψ ) for ς 1 ψ ς 2 and Equation (12), we can express it as follows:
ϖ G ( ς 2 , ψ ) ϖ + κ ϖ κ ( ς 2 ς 1 ) ϖ κ ς 1 ς 2 G ( χ , ψ ) ( χ ς 1 ) ϖ κ 1 d χ = ϖ ( ς 1 ψ ) ϖ + κ κ ( ψ ς 1 ) ϖ κ + 1 ( ς 2 ς 1 ) ϖ κ ( ψ ς 1 ) ( ϖ κ + 1 ) ( ς 2 ς 1 ) ϖ κ ( ϖ + κ ) = κ ( ψ ς 1 ) ( ς 2 ς 1 ) ϖ κ ( ψ ς 1 ) ϖ κ ( ς 2 ς 1 ) ϖ κ ( ϖ + κ ) 0 .
By using the convexity of ⊺ and Equations (21) and (22), we obtain
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) .
By combining the inequalities (18) and (23), we obtained the required result. □
Remark 1.
By substituting κ = 1 into inequality (8), we obtain the following results as presented in ([29], Theorem 2.2)
ς 1 + ϖ ς 2 ϖ + 1 Γ ( ϖ + 1 ) ( ς 2 ς 1 ) ϖ ς 2 ϖ ( ς 1 ) ( ς 1 ) + ϖ ( ς 2 ) ϖ + 1 .
Theorem 3.
Letbe a function that is twice differentiable on the interval [ ς 1 , ς 2 ] , and let ϖ and κ be positive. Then, the following inequalities hold.
(i) 
If | | is an increasing function, then
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) ( ς 2 ) ϖ κ ς 2 ς 1 2 2 ( κ + ϖ ) ( ϖ + 2 κ ) ;
(ii) 
If | | is decreasing function, then
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) ( ς 1 ) ϖ κ ς 2 ς 1 2 2 ( κ + ϖ ) ( ϖ + 2 κ ) ;
(iii) 
If | | is a convex function, then
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) max ( ς 1 ) , ( ς 2 ) ϖ κ ς 2 ς 1 2 2 ( κ + ϖ ) ( ϖ + κ ) .
Proof. 
(i) From (21) and (22), the increasing monotonicity of | | and
( ψ ς 1 ) ( ς 2 ς 1 ) ϖ κ ( ψ ς 1 ) ϖ κ + 1 0 ,
for ς 1 ψ ς 2 . Consider
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) κ ( ς 2 ) ( ς 2 ς 1 ) ϖ κ ( ϖ + κ ) ς 1 ς 2 ( ψ ς 1 ) ( ς 2 ς 1 ) ϖ κ ( ψ ς 1 ) ϖ κ + 1 d ψ κ ( ς 2 ) ( ς 2 ς 1 ) ϖ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ( ς 2 ς 1 ) 2 2 κ ( ς 2 ς 1 ) ϖ κ + 2 ϖ + 2 κ = κ ( ς 2 ) ϖ ( ς 2 ς 1 ) 2 2 ( ϖ + κ ) ( ϖ + 2 κ ) .
This corresponds to inequality (24).
(ii) By employing (21) and (22) and following the same procedure as in case (i), we derive
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) κ ( ς 2 ς 1 ) ϖ κ ( ϖ + κ ) ς 1 ς 2 ( ψ ς 1 ) ( ς 2 ς 1 ) ϖ κ ( ψ ς 1 ) ϖ κ + 1 | ( ψ ) | d ψ κ ( ς 1 ) ( ς 2 ς 1 ) ϖ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ( ς 2 ς 1 ) 2 2 κ ( ς 2 ς 1 ) ϖ κ + 2 ϖ + 2 κ = κ ( ς 1 ) ϖ ( ς 2 ς 1 ) 2 2 ( ϖ + κ ) ( ϖ + 2 κ ) .
(iii) From (21) and (22),
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) κ ( ς 2 ς 1 ) ϖ κ ( ϖ + κ ) ς 1 ς 2 ( ψ ς 1 ) ( ς 2 ς 1 ) ϖ κ ( ψ ς 1 ) ϖ κ + 1 ( ψ ) d ψ .
By using (27) and the fact that | | is a convex function on the interval [ ς 1 , ς 2 ] , we get
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) κ max ( ς 1 ) , ( ς 2 ) ( ϖ + κ ) ( ( ς 2 ς 1 ) ϖ κ ) ς 1 ς 2 ( ψ ς 1 ) ( ς 2 ς 1 ) ϖ κ ( ψ ς 1 ) ϖ κ + 1 d ψ = max | ( ς 1 ) | , | ( ς 2 ) | κ ϖ ς 2 ς 1 2 2 ( ϖ + κ ) ( ϖ + 2 κ ) .
This corresponds to inequality (26). □
Remark 2.
By setting κ = 1 in inequalities (24)–(26), we derive the following results as presented in ([29], Theorem 2.6)
( ς 1 ) + ϖ ( ς 2 ) ϖ + 1 Γ ( ϖ + 1 ) ( ς 2 ς 1 ) ϖ ς 2 ϖ ( ς 1 ) ( ς 2 ) ϖ ς 2 ς 1 2 2 ( ϖ + 1 ) ( ϖ + 2 ) , ( ς 1 ) + ϖ ( ς 2 ) ϖ + 1 Γ ( ϖ + 1 ) ( ς 2 ς 1 ) ϖ ς 2 ϖ ( ς 1 ) ( ς 1 ) ϖ ς 2 ς 1 2 2 ( ϖ + 1 ) ( ϖ + 2 )
and
( ς 1 ) + ϖ ( ς 2 ) ϖ + 1 Γ ( ϖ + 1 ) ( ς 2 ς 1 ) ϖ ς 2 ϖ ( ς 1 ) max ( ς 1 ) , ( ς 2 ) ϖ ς 2 ς 1 2 2 ( ϖ + 1 ) ( ϖ + 2 ) .
Theorem 4.
Letbe a function that is twice differentiable on the interval [ ς 1 , ς 2 ] , and let ϖ and κ be positive. The following statements hold.
(i) 
If | | is an increasing function, then
κ ς 1 + ϖ ς 2 ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) ϖ κ 2 ( ς 2 ς 1 ) 2 ( ϖ + κ ) ϖ κ + 3 ( ϖ + 2 κ ) ( κ ς 1 + ϖ ς 2 ϖ + κ ) ( ϖ ) ϖ κ + 1 + ( ς 2 ) ( ϖ + κ ) ϖ κ + 1 2 ( ϖ ) ϖ κ + 1 2 ;
(ii) 
If | | is a decreasing function, then
κ ς 1 + ϖ ς 2 ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) ϖ κ 2 ( ς 2 ς 1 ) 2 ( ϖ + κ ) ϖ κ + 3 ( ϖ + 2 κ ) ( ς 1 ) ( ϖ ) ϖ κ + 1 + κ ς 1 + ϖ ς 2 ϖ + κ ( ϖ + κ ) ϖ κ + 1 2 ( ϖ ) ϖ κ + 1 2 ;
(iii) 
If | | is a convex function, then
κ ς 1 + ϖ ς 2 ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) ϖ κ 2 ( ς 2 ς 1 ) 2 ( ϖ + κ ) ϖ κ + 3 ( ϖ + 2 κ ) max ( ς 1 ) , κ ς 1 + ϖ ς 2 ϖ + κ ( ϖ ) ϖ κ + 1 + max κ ς 1 + ϖ ς 2 ϖ + κ , ( ς 2 ) ( ϖ + κ ) ϖ κ + 1 2 ( ϖ ) ϖ κ + 1 2 .
Proof. 
(i) By using (11), (14) and (15), we get
κ ς 1 + ϖ ς 2 ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) = ς 1 κ ς 1 + ϖ ς 2 ϖ + κ G κ ς 1 + ϖ ς 2 ϖ + κ , ψ ϖ κ ( ς 2 ς 1 ) ϖ κ ς 1 ς 2 G ( χ , ψ ) ( χ ς 1 ) ϖ κ 1 d χ ( ψ ) d ψ + κ ς 1 + ϖ ς 2 ϖ + κ ς 2 G κ ς 1 + ϖ ς 2 ϖ + κ , ψ ϖ κ ( ς 2 ς 1 ) ϖ κ ς 1 ς 2 G ( χ , ψ ) ( χ ς 1 ) ϖ κ 1 d χ ( ψ ) d ψ = κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 1 κ ς 1 + ϖ ς 2 ϖ + κ ( ψ ς 1 ) ϖ κ + 1 ( ψ ) d ψ + κ ς 1 + ϖ ς 2 ϖ + κ ς 2 ( ψ ς 1 ) ϖ κ + 1 + ϖ κ ( ς 2 ψ ) ( ς 2 ς 1 ) ϖ κ ( ς 2 ς 1 ) ϖ κ ( ψ ς 1 ) ( ψ ) d ψ .
The triangle inequality, and the increasing monotonicity of | ( ψ ) | , enabled us to obtain the following from Equation (29):
κ ς 1 + ϖ ς 2 ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ κ ς 1 + ϖ ς 2 ϖ + κ ς 1 κ ς 1 + ϖ ς 2 ϖ + κ ( ψ ς 1 ) ϖ κ + 1 d ψ + ( ς 2 ) κ ς 1 + ϖ ς 2 ϖ + κ ς 2 ( ψ ς 1 ) ϖ κ + 1 + ϖ κ ( ς 2 ψ ) ( ς 2 ς 1 ) ϖ κ ( ς 2 ς 1 ) ϖ κ ( ψ ς 1 ) d ψ = ϖ κ 2 ( ς 2 ς 1 ) 2 ( ϖ + κ ) ϖ κ + 3 ( ϖ + 2 κ ) ( κ ς 1 + ϖ ς 2 ϖ + κ ) ϖ ϖ κ + 1 + ( ς 2 ) ( ϖ + κ ) ϖ κ + 1 2 ( ϖ ) ϖ κ + 1 2 .
This completes part (i) of the result.
(ii) The proof of this part can be carried out using the same procedure as described above.
(iii) Given that every convex function ⊺ defined on an interval [ ς 1 , ς 2 ] is bounded above by max ( ς 1 ) , ( ς 2 ) , we have the following inequality:
κ ς 1 + ϖ ς 2 ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 )
κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ( max ( ς 1 ) , κ ς 1 + ϖ ς 2 ϖ + κ × ς 1 κ ς 1 + ϖ ς 2 ϖ + κ ( ψ ς 1 ) ϖ κ + 1 d ψ + max κ ς 1 + ϖ ς 2 ϖ + κ , ( ς 2 ) × κ ς 1 + ϖ ς 2 ϖ + κ ς 2 ( ψ ς 1 ) ϖ κ + 1 + ϖ κ ( ς 2 ψ ) ( ς 2 ς 1 ) ϖ κ ( ψ ς 1 ) ( ς 2 ς 1 ) ϖ κ d ψ ) .
By calculation, we get inequality (28). □
Remark 3.
By choosing κ = 1 , we get the results presented in ([29], Theorem 2.4).
Theorem 5.
Suppose thatis a function that is twice differentiable and | | is convex on the interval [ ς 1 , ς 2 ] . Then, for ϖ , κ > 0 , the following inequality holds:
κ ς 1 + ϖ ς 2 ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) κ 3 ( ς 2 ς 1 ) 2 6 ( ϖ + κ ) 3 ( ϖ + 3 κ ) ( ς 1 ) ( 7 ϖ 2 + 17 ϖ κ + 12 κ 2 ) ϖ + 2 κ + ( ς 2 ) 9 ϖ 2 κ + 23 ϖ + 12 κ .
Proof. 
By substituting ψ = ϱ ς 1 + ( 1 ϱ ) ς 2 with ϱ [ 0 , 1 ] , Equation (29) can be written as
κ ς 1 + ϖ ς 2 ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) = κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ( 1 κ ϖ + κ ϱ ς 1 + ς 2 ϱ ς 2 ς 1 ϖ κ + 1 ϱ ς 1 + ( 1 ϱ ) ς 2 × ς 2 ς 1 d ϱ κ ϖ + κ 0 ( ϱ ς 1 + ς 2 ϱ ς 2 ς 1 ϖ κ + 1 + ϖ κ ( ς 2 ϱ ς 1 ς 2 + ϱ ς 2 ) ς 2 ς 1 ϖ κ ϱ ς 1 + ς 2 ϱ ς 2 ς 1 ς 2 ς 1 ϖ κ ) × ( ϱ ς 1 + ( 1 ϱ ) ς 2 ) ( ς 2 ς 1 ) d ϱ ) = κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ( κ ϖ + κ 1 1 ϱ ϖ κ + 1 ς 2 ς 1 ϖ κ + 1 × ϱ ς 1 + ( 1 ϱ ) ς 2 ς 2 ς 1 d ϱ + 0 κ ϖ + κ ( 1 ϱ ϖ κ + 1 ς 2 ς 1 ϖ κ + 1
+ ϖ κ ϱ ς 2 ς 1 ϖ κ + 1 1 ϱ ς 2 ς 1 ϖ κ + 1 ) × ( ϱ ς 1 + ( 1 ϱ ) ς 2 ) ( ς 2 ς 1 ) d ϱ ) = κ ( ς 2 ς 1 ) ϖ κ + 2 ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ κ ϖ + κ 1 1 ϱ ϖ κ + 1 ϱ ς 1 + ( 1 ϱ ) ς 2 d ϱ + 0 κ ϖ + κ 1 ϱ ϖ κ + 1 ϱ ς 1 + ( 1 ϱ ) ς 2 d ϱ + 0 κ ϖ + κ ϱ + ϖ κ ϱ 1 ϖ κ + 1 ϱ ς 1 + ( 1 ϱ ) ς 2 d ϱ = κ ( ς 2 ς 1 ) 2 ϖ + κ 0 1 1 ϱ ϖ κ + 1 ϱ ς 1 + ( 1 ϱ ) ς 2 d ϱ 0 κ ϖ + κ 1 ϱ ϖ κ ϱ ϱ ς 1 + ( 1 ϱ ) ς 2 d ϱ .
Taking absolute value on both sides, using the fact 1 ϱ ϖ κ ϱ < 1 ϱ and applying triangular inequality, we get
κ ς 1 + ϖ ς 2 ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) κ ( ς 2 ς 1 ) 2 ϖ + κ 0 1 1 ϱ ϖ κ + 1 ϱ ς 1 + ( 1 ϱ ) ς 2 d ϱ
+ 0 κ ϖ + κ 1 ϱ ϖ κ ϱ ϱ ς 1 + ( 1 ϱ ) ς 2 d ϱ κ ( ς 2 ς 1 ) 2 ( ϖ + κ ) 0 κ ϖ + κ 1 ϱ ϖ κ ϱ ϱ | ( ς 1 ) | + ( 1 ϱ ) | ( ς 2 ) | d ϱ + 0 1 1 ϱ ϖ κ + 1 ϱ ( ς 1 ) + ( 1 ϱ ) ( ς 2 ) d ϱ κ 3 ( ς 2 ς 1 ) 2 6 ( ϖ + κ ) 3 ( ϖ + 3 κ ) ( ς 1 ) ( 7 ( ϖ 2 ) + 17 ϖ κ + 12 κ 2 ) ( ϖ + 2 κ ) + ( ς 2 ) 9 ϖ 2 κ + 23 ϖ + 12 κ ,
This proved the desired result. □
Remark 4.
By setting κ = 1 , we obtain Theorem 2.8 presented in [29].
Theorem 6.
Supposeis a function that is twice differentiable and | | is convex on the interval [ ς 1 , ς 2 ] . Then, we formulate the following inequality:
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) κ ϖ ( ς 2 ς 1 ) 2 3 ( ϖ + κ ) ( ϖ + 3 κ ) ( ς 1 ) ( ϖ + 5 κ ) 2 ( ϖ + 2 κ ) + ( ς 2 ) .
Proof. 
Let ϱ [ 0 , 1 ] and ψ = ϱ ς 1 + ( 1 ϱ ) ς 2 . Then, from (21) and (22), we obtain
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) = κ ( ς 2 ς 1 ) ϖ κ ( ϖ + κ ) 1 0 ( ς 2 ς 1 ϖ κ ϱ ς 1 + ( 1 ϱ ) ς 2 ς 1 ϱ ς 1 + ( 1 ϱ ) ς 2 ς 1 ϖ κ + 1 ) ( ϱ ς 1 + ( 1 ϱ ) ς 2 ) ( ς 2 ς 1 ) d ϱ = κ ( ς 2 ς 1 ) 2 ϖ + κ 0 1 1 ϱ ( 1 ϱ ) ϖ κ + 1 ϱ ς 1 + ( 1 ϱ ) ς 2 d ϱ .
By considering the relation of absolute value in an inequality and using the convexity of | | , we get
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) κ ( ς 2 ς 1 ) 2 ϖ + κ 0 1 1 ϱ ( 1 ϱ ) ϖ κ + 1 ϱ ς 1 + ( 1 ϱ ) ς 2 d ϱ κ ( ς 2 ς 1 ) 2 ϖ + κ 0 1 1 ϱ ( 1 ϱ ) ϖ κ + 1 ϱ | ( ς 1 ) | + ( 1 ϱ ) ( ς 2 ) d ϱ = ϖ κ ( ς 2 ς 1 ) 2 3 ( ϖ + κ ) ( ϖ + 3 κ ) ( ς 1 ) ( ϖ + 5 κ ) 2 ( ϖ + 2 κ ) + ( ς 2 ) .
This completes the proof. □
Remark 5.
Corresponding to the choice κ = 1 in Theorem 6, we obtain the result explored by Muhammad Adil Khan et al. in ([29], Theorem 2.10).
Theorem 7.
Suppose thatis a function that is twice differentiable and that | | is concave on the interval [ ς 1 , ς 2 ] . Then, the following inequality,
κ ς 1 + ϖ ς 2 ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) κ ( ς 2 ς 1 ) 2 ϖ + κ κ ϖ + 2 κ κ ς 1 + 2 ς 2 + ϖ κ ς 2 ϖ + 3 κ + κ 2 ( ϖ + κ ) κ ς 1 + 2 ς 2 + 3 ϖ κ ς 2 3 ( ϖ + κ ) ,
holds.
Proof. 
It follows from (31) and using Jensen’s integral inequality that
κ ς 1 + ϖ ς 2 ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) κ ( ς 2 ς 1 ) 2 ϖ + κ 0 1 ( 1 ϱ ) ϖ κ + 1 d ϱ 0 1 ( 1 ϱ ) ϖ κ + 1 ( ϱ ς 1 + ( 1 ϱ ) ς 2 ) d ϱ 0 1 ( 1 ϱ ) ϖ κ + 1 d ϱ
+ 0 κ ϖ + κ 1 ϱ ϖ κ ϱ d ϱ 0 1 1 ϱ ϖ κ ϱ ( ϱ ς 1 + ( 1 ϱ ) ς 2 ) d ϱ 0 1 1 ϱ ϖ κ ϱ d ϱ κ ( ς 2 ς 1 ) 2 ϖ + κ κ ϖ + 2 κ κ 2 ( ς 1 + 2 ς 2 + ϖ κ ς 2 ) ( ϖ + 2 κ ) κ ( ϖ + 2 κ ) ( ϖ + 3 κ ) + κ 2 ( ϖ + κ ) 2 κ 2 ( ς 1 + 2 ς 2 + 3 ϖ κ ς 2 ) ( ϖ + κ ) 6 κ ( ϖ + κ ) 2 = κ ( ς 2 ς 1 ) 2 ϖ + κ κ 2 ( ϖ + κ ) κ ς 1 + 3 ϖ ς 2 + 2 κ ς 2 3 ( ϖ + κ ) + κ ϖ + 2 κ κ ς 1 + ϖ ς 2 + 2 κ ς 2 ϖ + 3 κ .
Hence, the required result is proved. □
Remark 6.
By setting κ = 1 in Theorem 7, we arrive at the result presented in ([29], Theorem 2.12).
Theorem 8.
Suppose thatis a function that is twice differentiable and that | | is a concave function on the interval [ ς 1 , ς 2 ] . Then, for any positive values of ϖ and κ, the following inequality holds:
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) κ ϖ ( ς 2 ς 1 ) 2 2 ( ϖ + κ ) ( ϖ + 2 κ ) ς 1 ( ϖ + 5 κ ) + 2 ς 2 ( ϖ + 2 κ ) 3 ( ϖ + 3 κ ) .
Proof. 
By using Jensen’s inequality on (32), we get
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ Γ κ ( ϖ + κ ) ( ς 2 ς 1 ) ϖ κ ς 2 ϖ , κ ( ς 1 ) κ ( ς 2 ς 1 ) 2 ( ϖ + κ ) 0 1 ϱ + 1 ( 1 ϱ ) ϖ κ + 1 d ϱ × 0 1 1 ϱ ( 1 ϱ ) ϖ κ + 1 ϱ ς 1 + ( 1 ϱ ) ς 2 d ϱ 0 1 ϱ + 1 ( 1 ϱ ) ϖ κ + 1 d ϱ = κ ϖ ( ς 2 ς 1 ) 2 2 ( ϖ + κ ) ( ϖ + 2 κ ) ( ϖ + 5 κ ) ς 1 + 2 ς 2 ( ϖ + 2 κ ) 3 ( ϖ + 3 κ ) .
Remark 7.
The choice of κ = 1 in Theorem 8 gives the result presented in ([29], Theorem 2.14)
Remark 8.
If an alternative approach is taken using Green’s function G 2 , as defined in Equation (4), analogous results to those expounded in this article can be obtained. Nevertheless, the utilization of the method proposed in this article in conjunction with Green’s functions G 3 (described by Equation (5)) or G 4 (described by Equation (6)) exclusively reproduces previously established findings. This strategic choice substantiates the exclusive focus on Green’s function G 1 within the discourse of this article. As an intellectual exercise, interested scholars are encouraged to explore the remaining three Green functions to elicit their respective outcomes.

3. Some Relations of Means with Trapezoid Formulae

In this section, we will present some propositions that highlight both the practical use and immense significance of our outcomes. At the same time, we will verify the validity of the conclusions by using special means of real numbers ς 1 and ς 2 , where ς 1 ς 2 . These propositions present the estimates of differences of generalized means that are concluded from our main results. First, we need to recall the following relations. For ς 1 , ς 2 > 0 , one of the important average values is
(i)
The Arithmetic mean:
A = A ( ς 1 , ς 2 ) = ς 1 + ς 2 2 ;
(ii)
The Harmonic mean:
H = H ( ς 1 , ς 2 ) = 2 1 ς 1 + 1 ς 2 ;
(iii)
The logarithmic mean:
L ( ς 1 , ς 2 ) = ς 2 ς 1 ln ς 2 ln ς 1 , ς 1 ς 2 ;
(iv)
The generalized logarithmic mean presented in [54] is defined as follows:
L m m ( ς 1 , ς 2 ) = ς 2 m + 1 ς 1 m + 1 ( m + 1 ) ( ς 2 ς 1 ) , m Z { 1 , 0 } , ς 1 ς 2 .
Proposition 1.
Let ς 1 , ς 2 R , ς 1 < ς 2 , then we can derive the following inequalities:
A ( e ς 1 , e ς 2 ) L ( e ς 1 , e ς 2 ) e ς 2 ( ς 2 ς 1 ) 2 12 , A ( e ς 1 , e ς 2 ) L ( e ς 1 , e ς 2 ) e ς 1 ( ς 2 ς 1 ) 2 12 , A ( e ς 1 , e ς 2 ) L ( e ς 1 , e ς 2 ) max e ς 1 , e ς 2 ( ς 2 ς 1 ) 2 12
and
A ( e ς 1 , e ς 2 ) L ( e ς 1 , e ς 2 ) ( ς 2 ς 1 ) 2 e ς 1 + e ς 2 24 .
Proof. 
By utilizing Theorem 3 and performing some simplifications, we can express them:
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ ϖ κ ( ς 2 ς 1 ) ϖ κ ς 1 ς 2 ( χ ς 1 ) ϖ κ 1 ( χ ) d χ ϖ κ ( ς 2 ) ς 2 ς 1 2 2 ( ϖ + κ ) ( ϖ + 2 κ ) .
By substituting κ = ϖ , ( ς ) = e ς and using simple calculation, we obtain
e ς 1 + e ς 2 2 e ς 2 e ς 1 ( ς 2 ς 1 ) e ς 2 ς 2 ς 1 2 12 .
Now, making use of (34) and (37), we arrive at the result
A ( e ς 1 , e ς 2 ) L ( e ς 1 , e ς 2 ) | e ς 2 | ( ς 2 ς 1 ) 2 12 .
The remaining inequalities can be obtained by applying the same procedure as in parts (ii) and (iii) of Theorem 3.
Theorem 6 can also be proved by using same function as in Theorem 3, after making some simplifications to Theorem 6
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ ϖ κ ( ς 2 ς 1 ) ϖ κ ς 1 ς 2 ( χ ς 1 ) ϖ κ 1 ( χ ) d χ κ ϖ ( ς 2 ς 1 ) 2 3 ( ϖ + κ ) ( ϖ + 3 κ ) ( ς 1 ) | ( ϖ + 5 κ ) 2 ( ϖ + 2 κ ) + ( ς 2 ) .
By substituting κ = ϖ , ( ς ) = e ς and using simple calculation, we obtain
e ς 1 + e ς 2 2 e ς 2 e ς 1 ( ς 2 ς 1 ) ( ς 2 ς 1 ) 2 ( e ς 1 + e ς 2 ) 24 .
By employing (34) and (37), we obtain the following result
A ( e ς 1 , e ς 2 ) L ( e ς 1 , e ς 2 ) ( ς 2 ς 1 ) 2 e ς 1 + e ς 2 24 .
Proposition 2.
Let ς 1 , ς 2 R + with ς 1 < ς 2 , then the inequalities hold:
H 1 ( ς 1 , ς 2 ) L 1 ( ς 1 , ς 2 ) ( ς 2 ς 1 ) 2 | ς 2 3 | 6 , H 1 ( ς 1 , ς 2 ) L 1 ( ς 1 , ς 2 ) ( ς 2 ς 1 ) 2 | ς 1 3 | 6 , H 1 ( ς 1 , ς 2 ) L 1 ( ς 1 , ς 2 ) max ς 1 3 , ς 2 3 ( ς 2 ς 1 ) 2 6
and
H 1 ( ς 1 , ς 2 ) L 1 ( ς 1 , ς 2 ) ( ς 2 ς 1 ) 2 ς 1 3 + ς 2 3 12 .
Proof. 
By utilizing Theorem 3 and simplifying the expressions, we derive
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ ϖ κ ( ς 2 ς 1 ) ϖ κ ς 1 ς 2 ( χ ς 1 ) ϖ κ 1 ( χ ) d χ ϖ κ ( ς 2 ) ς 2 ς 1 2 2 ( ϖ + κ ) ( ϖ + 2 κ ) .
By substituting κ = ϖ and ( ς ) = 1 ς , where ς > 0 , we obtain
ς 2 + ς 1 2 ς 1 ς 2 ln ς 2 ln ς 1 ( ς 2 ς 1 ) ς 2 3 ς 2 ς 1 2 6
Now, by using Equations (35) and (36), we get
H 1 ( ς 1 , ς 2 ) L 1 ( ς 1 , ς 2 ) ( ς 2 ς 1 ) 2 ς 2 3 6 .
Similarly, by some simple calculations in Theorem 6,
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ ϖ κ ( ς 2 ς 1 ) ϖ κ ς 1 ς 2 ( χ ς 1 ) ϖ κ 1 ( χ ) d χ κ ϖ ( ς 2 ς 1 ) 2 3 ( ϖ + κ ) ( ϖ + 3 κ ) ( ς 1 ) ( ϖ + 5 κ ) 2 ( ϖ + 2 κ ) + ( ς 2 ) ;
using the same polynomial function, we obtain
ς 2 + ς 1 2 ς 1 ς 2 ln ς 2 ln ς 1 ( ς 2 ς 1 ) ( ς 2 ς 1 ) 2 ς 1 3 + ς 2 3 12 ,
which is the required result. □
Proposition 3.
Let ς 1 , ς 2 R + with ς 1 < ς 2 , then the inequalities
A ( ς 1 m , ς 2 m ) L m m ( ς 1 , ς 2 ) ( ς 2 ς 1 ) 2 m ( m 1 ) ς 2 m 2 12 , A ( ς 1 m , ς 2 m ) L m m ( ς 1 , ς 2 ) ( ς 2 ς 1 ) 2 m ( m 1 ) ς 1 m 2 12 , A ( ς 1 m , ς 2 m ) L m m ( ς 1 , ς 2 ) max m ( m 1 ) ς 1 m 2 , m ( m 1 ) ς 2 m 2 ( ς 2 ς 1 ) 2 12
and
A ( ς 1 m , ς 2 m ) L m m ( ς 1 , ς 2 ) ( ς 2 ς 1 ) 2 24 m ( m 1 ) ς 1 m 2 + ς 2 m 2
are true for m Z with | m ( m 1 ) | 2 .
Proof. 
Using Theorem 3 and making some simplification, we get
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ ϖ κ ( ς 2 ς 1 ) ϖ κ ς 1 ς 2 ( χ ς 1 ) ϖ κ 1 ( χ ) d χ ϖ κ ( ς 2 ) ς 2 ς 1 2 2 ( ϖ + κ ) ( ϖ + 2 κ ) .
By substituting κ = ϖ and ( ς ) = ς m , where ς > 0 and m ( m 1 ) 2 , we obtain
ς 1 m + ς 2 m 2 ς 2 m + 1 ς 1 m + 1 ( m + 1 ) ( ς 2 ς 1 ) m ( m 1 ) ς 2 m 2 ς 2 ς 1 2 12 .
Now, by using Equations (34) and (37), we get
A ( ς 1 m , ς 2 m ) L m m ( ς 1 , ς 2 ) ( ς 2 ς 1 ) 2 m ( m 1 ) ς 2 m 2 12 .
Similarly, by some simple calculations in Theorem 6,
κ ( ς 1 ) + ϖ ( ς 2 ) ϖ + κ ϖ κ ( ς 2 ς 1 ) ϖ κ ς 1 ς 2 ( χ ς 1 ) ϖ κ 1 ( χ ) d χ κ ϖ ( ς 2 ς 1 ) 2 3 ( ϖ + κ ) ( ϖ + 3 κ ) ( ς 1 ) ( ϖ + 5 κ ) 2 ( ϖ + 2 κ ) + ( ς 2 ) ,
using the same polynomial function, we obtain
ς 1 m + ς 2 m 2 ς 2 m + 1 ς 1 m + 1 ( m + 1 ) ( ς 2 ς 1 ) ( ς 2 ς 1 ) 2 24 m ( m 1 ) ς 1 m 2 + ς 2 m 2 .

4. Concluding Remarks

In mathematics, inequalities play a crucial role as they are widely employed in diverse fields of study. They enable us to compare and contrast the relative magnitudes of distinct mathematical expressions, thereby facilitating a deeper comprehension of their interrelationships. Inequalities are not only critical for theoretical purposes but also have significant practical applications in optimization problems and statistical data analysis. The comprehension of inequalities is a fundamental aspect of mathematical literacy as it enables individuals to evaluate and interpret quantitative information and make well-informed decisions in various aspects of their lives. The inequality theory makes use of well-known results such as the Hölder’s and Jensen’s inequalities to derive compelling consequences. In the work presented, we used some special Green’s functions and utilized the Jensen’s inequality to establish the fractional Hermite–Hadamard type inequalities. We have adopted an innovative approach to obtain the novel findings. Specifically, we have utilized quadrature formulae to estimate differences between particular average values. The third section of our research deals with practical applications to real-world problems. In this section, we estimate the errors of generalized means differences that are very important in real life problems. Going forward, our objective is to explore additional inequalities by integrating Green’s functions G 2 , G 3 , and G 4 . We hope that our research will encourage other scholars who are working on fractional integral inequalities using different types of convex functions.

Author Contributions

Conceptualization, M.S., A.G. and M.V.-C.; methodology, S.N. and G.R.; software, M.S., A.G. and M.V.-C.; validation, M.V.-C. and G.R.; formal analysis, M.S. and S.N.; investigation, M.V.-C., M.S., S.N. and M.V.-C.; data duration, G.R., M.V.-C. and S.N.; writing—original draft preparation, M.S. and G.R.; writing—review and editing, M.S., G.R. and S.N.; visualization, M.V.-C.; supervision, A.G., G.R. and M.S.; project administration, A.G. and M.S.; funding acquisition, M.V.-C. All authors have read and agreed to the published version of the manuscript.

Funding

This research has received funding support from the National Science, Research and Innovation Fund (NSRF), Thailand.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

References

  1. Azpeitia, A.G. Convex functions and the Hadamard inequality. Rev. Colomb. Mat. 1994, 28, 7–12. [Google Scholar]
  2. Sarikaya, M.Z.; Saglam, A.; Yildirim, H. On some Hadamard-type inequalities for h-convex functions. J. Math. Inequal. 2008, 2, 335–341. [Google Scholar] [CrossRef]
  3. Dragomir, S.S.; Pearce, C. Selected Topics on Hermite–Hadamard Inequalities and Applications. Sci. Direct Work. Pap. 2003, 4, S1574-0358. [Google Scholar]
  4. Set, E.; Özdemir, M.; Dragomir, S.S. On Hadamard-type inequalities involving several kinds of convexity. J. Inequal. Appl. 2010, 2010, 286845. [Google Scholar] [CrossRef]
  5. Vivas-Cortez, M.; Ali, M.A.; Kashuri, A.; Budak, H. Generalizations of fractional Hermite–Hadamard-mercer like inequalities for convex functions. AIMS Math. 2021, 6, 9397–9421. [Google Scholar] [CrossRef]
  6. Nápoles Valdés, J.E.; Rodríguez, J.M.; Sigarreta, J.M. New Hermite–Hadamard type inequalities involving non-comformable integral operators. Symmetry 2019, 11, 1108. [Google Scholar] [CrossRef]
  7. Bosch, P.; Carmenate, H.J.; Rodríguez, J.M.; Sigarreta, J.M. Generalized inequalities involving fractional operators of Riemann-Liouville type. AIMS Math. 2021, 7, 1470–1485. [Google Scholar] [CrossRef]
  8. Guzmán, P.M.; Valdés, J.E.N.; Stojiljkovic, V. New extensions of Hermite–Hadamard inequality. Contrib. Math. 2023, 7, 60–66. [Google Scholar] [CrossRef]
  9. Vivas-Cortez, M.; García, C. Ostrowski type inequalities for functions whose derivatives are (m,h1,h2)-convex. Appl. Math. Inf. Sci. 2017, 11, 79–86. [Google Scholar] [CrossRef]
  10. Vivas-Cortez, M.; Saleem, M.S.; Sajid, S. Fractional version of Hermite–Hadamard-Mercer inequalities for convex stochastic processes via Ψk-Riemann-Liouville fractional integrals and its applications. Appl. Math. Inf. Sci. 2022, 16, 695–709. [Google Scholar]
  11. Bayraktar, B.; Népoles, J.E.; Rabossi, F. On generalizations of integral inequalities. Probl. Anal. Issues Anal. 2022, 11, 3–23. [Google Scholar] [CrossRef]
  12. Mehmood, S.; Valdes, J.E.N.; Fatima, N.; Aslam, W. Some integral inequalities via fractional derivatives. Euro-Tbil. Math. J. 2022, 15, 31–44. [Google Scholar] [CrossRef]
  13. Sahoo, S.K.; Ahmad, H.; Tariq, M.; Kodamasingh, B.; Aydi, H.; Sen, M.D.L. Hermite–Hadamard type inequalities involving k-fractional operator for (h, m)-convex functions. Symmetry 2021, 13, 1686. [Google Scholar] [CrossRef]
  14. Han, J.; Mohammed, P.O.; Zeng, H. Generalized fractional integral inequalities of Hermite–Hadamard-type for a convex function. Open Math. 2020, 18, 794–806. [Google Scholar] [CrossRef]
  15. Ivastava, H.M.; Sahoo, S.K.; Mohammed, P.O.; Kodamasingh, B.; Nonlaopon, K.; Abualnaja, K.M. Interval valued Hadamard-Fejer and Pachpatte type inequalities pertaining to a new fractional integral operator with exponential kernel. AIMS Math. 2022, 7, 15041–15063. [Google Scholar] [CrossRef]
  16. Baleanu, D.; Samraiz, M.; Perveen, Z.; Iqbal, S.; Nisar, K.S.; Rahman, G. Hermite–Hadamard-Fejer type inequalities via fractional integrals of a function concerning another function. AIMS Math. 2021, 6, 4280–4295. [Google Scholar] [CrossRef]
  17. Niculescu, C.P.; Persson, L.E. Convex Functions and Their Applications; Springer: New York, NY, USA, 2006. [Google Scholar]
  18. Belarbi, S.; Dahmani, Z. On some new fractional integral inequalities. Int. J. Math. Ana. 2010, 4, 185–191. [Google Scholar]
  19. Podlubny, I.; Chechkin, A.; Skovranek, T.; Chen, Y.; Jara, B.M.V. Matrix approach to discrete fractional calculus II: Partial fractional differential equations. J. Comput. Phy. 2009, 228, 3137–3153. [Google Scholar] [CrossRef]
  20. Samko, S.G.; Kilbas, A.A.; Marichev, O.I. Fractional Integrals and Derivatives: Theory and Applications; Gordon and Breach Science Publishers: Zurich, Switzerland; Philadelphia, PA, USA, 1993. [Google Scholar]
  21. Dahmani, Z.; Tabharit, L. On weighted Grüss type inequalities via fractional integration. J. Adv. Res. Pure Math. 2010, 2, 31–38. [Google Scholar] [CrossRef]
  22. Samraiz, M.; Mehmood, A.; Iqbal, S.; Naheed, S.; Chu, Y.-M. Generalized fractional operator with applications in mathematical physics. Chaos Solitons Fractals 2022, 165, 112830. [Google Scholar] [CrossRef]
  23. Budak, H.; Hezenci, F.; Kara, H. On parameterized inequalities of Ostrowski and Simpson type for convex functions via generalized fractional integrals. Math. Methods Appl. Sci. 2021, 44, 12522–12536. [Google Scholar] [CrossRef]
  24. Li, Y.; Samraiz, M.; Gul, A.; Vivas-Cortez, M.; Rahman, G. Hermite–Hadamard fractional integral inequalities via Abel–Gontscharoff Green’s function. Fractal Fract. 2022, 6, 126. [Google Scholar] [CrossRef]
  25. Yildiz, C.; Cotirla, L.I. Examining the Hermite–Hadamard Inequalities for k-Fractional Operators Using the Green Function. Fractal Fract. 2023, 7, 161. [Google Scholar] [CrossRef]
  26. Wang, H.; Khan, J.; Khan, M.A.; Khalid, S.; Khan, R. The Hermite–Hadamard-Jensen-Mercer type inequalities for Rieman-Liouville fractional integral. J. Math. 2021, 2021, 5516987. [Google Scholar]
  27. Desta, H.D.; Pachpatte, D.B.; Mijena, J.B.; Abdi, T. Univariate and multivariate Ostrowski-type inequalities using Atangana-Baleanu Caputo fractional derivative. Axioms 2022, 11, 482. [Google Scholar] [CrossRef]
  28. Pachpatte, B.G. Mathematical Inequalities; Elsevier: Amsterdam, The Netherlands, 2005. [Google Scholar]
  29. Iqbal, A.; Khan, M.A.; Suleman, M.; Chu, Y.-M. The right Riemann-Liouville fractional Hermite–Hadamard type inequalities derived from Green’s function. AIP Adv. 2020, 10, 045032. [Google Scholar] [CrossRef]
  30. Wu, S.; Iqbal, S.; Aamir, M.; Samraiz, M. On some Hermite–Hadamard inequalities involving k-fractional operator. J. Inequal. Appl. 2021, 2021, 32. [Google Scholar] [CrossRef]
  31. Diaz, R.; Pariguan, E. On hypergeometric functions and Pochhammer k-symbol. Divulg. Mat. 2007, 15, 179–192. [Google Scholar]
  32. Agarwal, R.P.; Wong, P.-J.-Y. Error Inequalities in Polynomial Interpolation and Their Applications; Kluwer Academic Publishers: Dordrecht, The Netherlands, 1993. [Google Scholar]
  33. Mehmood, N.; Agarwal, R.P.; Butt, S.I.; Pecaric, J.E. New generalizations of Popoviciu-type inequalities via new Green’s functions and Montgomery identity. J. Inequal. Appl. 2017, 2017, 108. [Google Scholar] [CrossRef]
  34. Roman, H.E.; Porto, M. Fractional derivatives of random walks: Time series with long-time memory. Phys. Rev. E 2008, 78, 031127. [Google Scholar] [CrossRef]
  35. Rossikhin, Y.A.; Shitikova, M.V. Analysis of two colliding fractionally damped spherical shells in modelling blunt human head impacts. Cent. Eur. J. Phys. 2013, 11, 760–778. [Google Scholar] [CrossRef]
  36. Balci, M.A. Fractional virus epidemic model on financial networks. Open Math. 2016, 14, 1074–1086. [Google Scholar] [CrossRef]
  37. Moreles, M.A.; Lainez, R. Mathematical modelling of fractional order circuit elements and bioimpedance applications. Commun. Nonlinear Sci. Numer. Simul. 2017, 46, 81–88. [Google Scholar] [CrossRef]
  38. Reyes-Melo, E.; Martinez-Vega, J.; Guerrero-Salazar, C.; Ortiz-Mendez, U. Application of fractional calculus to the modeling of dielectric relaxation phenomena in polymeric materials. J. Appl. Polym. Sci. 2005, 98, 923–935. [Google Scholar] [CrossRef]
  39. Carcione, J.M. Theory and modeling of constant-Q P- and S-waves using fractional time derivatives. Geophysics 2009, 74, 1–11. [Google Scholar] [CrossRef]
  40. Khan, T.U.; Khan, M.A. Generalized conformable fractional operators. J. Comput. Appl. Math. 2019, 346, 378–389. [Google Scholar] [CrossRef]
  41. Jarad, F.; Ugurlu, E.; Abdeljawad, T.; Baleanu, D. On a new class of fractional operators. Adv. Differ. Equ. 2017, 2017, 247. [Google Scholar] [CrossRef]
  42. Kilbas, A.A.; Marichev, O.I.; Samko, S.G. Fractional Integrals and Derivatives: Theory and Applications; Gordonand Breach: Zurich, Switzerland, 1993. [Google Scholar]
  43. Kilbas, A.A.; Srivastava, H.M.; Trujillo, J.J. Theory and Applications of Fractional Differential Equations; North-Holland Mathematics Studies: New York, NY, USA; London, UK, 2006. [Google Scholar]
  44. Sarikaya, M.Z.; Set, E.; Yaldiz, H.; Basak, N. Hermite–Hadamard inequalities for fractional integrals and related fractional inequalities. Math. Comput. Model. 2013, 57, 2403–2407. [Google Scholar] [CrossRef]
  45. Vivas-Cortez, M.; Saleem, M.S.; Sajid, S.; Zahoor, M.S.; Kashuri, A. Hermite-Jensen-Mercer-Type Inequalities via Caputo-Fabrizio Fractional Integral for h-Convex Function. Fractal Fract. 2021, 5, 269. [Google Scholar] [CrossRef]
  46. Jensen, J.L.W.V. Sur les fonctions convexes et les inegalites entre les valeurs moyennes. Acta Math. 1905, 30, 175–193. [Google Scholar] [CrossRef]
  47. Hudzik, H.; Maligranda, L. Some remarks on s-convex functions. Aequationes Math. 1994, 48, 100–111. [Google Scholar] [CrossRef]
  48. Tunç, T. Hermite–Hadamard type inequalities for interval-valued fractional integrals with respect to another function. Math. Slovaca 1994, 72, 1501–1512. [Google Scholar] [CrossRef]
  49. Chen, H.; Katugampola, U.N. Hermite–Hadamard and Hermite–Hadamard-Fejer type inequalities for generalized fractional integrals. J. Math. Anal. Appl. 2017, 446, 1274–1291. [Google Scholar] [CrossRef]
  50. Sarikaya, M.Z.; Hüseyin, Y. On Hermite–Hadamard type inequalities for Riemann-Liouville fractional integrals. Miskolc Math. Notes 2017, 17, 1049–1059. [Google Scholar] [CrossRef]
  51. Ödemir, M.E.; Avci, M.; Set, E. On some inequalities of Hermite–Hadamard-type via m-convexity. Appl. Math. Lett. 2010, 23, 1065–1070. [Google Scholar] [CrossRef]
  52. Pečarić, J.; Proschan, F.; Tong, Y.L. Convex Functions, Partial Orderings and Statistical Application; Acadmic Press: Cambridge, MA, USA, 1992. [Google Scholar]
  53. Set, E.; Iscan, I.; Sarikaya, M.Z.; Ozdemir, M.E. On new inequalities of Hermite–Hadamard-Fejér type for convex functions via fractional integrals. Appl. Math. Comput. 2015, 259, 875–881. [Google Scholar] [CrossRef]
  54. Gürbüz, M.; Akdemir, A.O.; Rashid, S.; Set, E. Hermite–Hadamard inequality for fractional integrals of Caputo-Fabrizio type and related inequalities. J. Inequal. Appl. 2020, 2020, 172. [Google Scholar] [CrossRef]
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Samraiz, M.; Naheed, S.; Gul, A.; Rahman, G.; Vivas-Cortez, M. Innovative Interpolating Polynomial Approach to Fractional Integral Inequalities and Real-World Implementations. Axioms 2023, 12, 914. https://doi.org/10.3390/axioms12100914

AMA Style

Samraiz M, Naheed S, Gul A, Rahman G, Vivas-Cortez M. Innovative Interpolating Polynomial Approach to Fractional Integral Inequalities and Real-World Implementations. Axioms. 2023; 12(10):914. https://doi.org/10.3390/axioms12100914

Chicago/Turabian Style

Samraiz, Muhammad, Saima Naheed, Ayesha Gul, Gauhar Rahman, and Miguel Vivas-Cortez. 2023. "Innovative Interpolating Polynomial Approach to Fractional Integral Inequalities and Real-World Implementations" Axioms 12, no. 10: 914. https://doi.org/10.3390/axioms12100914

APA Style

Samraiz, M., Naheed, S., Gul, A., Rahman, G., & Vivas-Cortez, M. (2023). Innovative Interpolating Polynomial Approach to Fractional Integral Inequalities and Real-World Implementations. Axioms, 12(10), 914. https://doi.org/10.3390/axioms12100914

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