EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 4, Article Number 6927 ISSN 1307-5543 – ejpam.com Published by New York Business Global Development of Quantum Hermite-Hadamard Type Inequalities Using Green’s Function Techniques Muhammad Adil Khan1, Tareq Saeed2,∗, Sajjad Ali1, Çetin Yildiz3, Mohammed Kbiri Alaoui4 1 Department of Mathematics, University of Peshawar, Peshawar 25000, Pakistan 2 Financial Mathematics and Actuarial Science (FMAS)-Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia 3 Department of Mathematics, K.K. Education Faculty, Ataturk University, 25240 Campus, Erzurum, Turkey 4 Department of Mathematics, College of Science, King Khalid University, P.O. Box 9004, 61413 Abha, Saudi Arabia Abstract. In this paper, we investigate the quantum Hermite-Hadamard inequality using the Green’s function. This process leads to the derivation of novel quantum identities, which are then employed to establish novel inequalities. Utilizing these identities, we establish novel inequalities. The main results of the paper are derived using various techniques such as q-identities, convex- ity and Jensen inequality. Furthermore, the study provides numerical validation and graphical representations to support the main results. 2020 Mathematics Subject Classifications: 26A51, 26D15, 68P30 Key Words and Phrases: Quantum integral, Green function, H-H-inequality 1. Introduction Scientists are very interested in the theory of convexity because of its many uses. Con- vexity is an important term in the extension and generalization of inequalities. As a result, convexity and inequality theory are closely related. Many inequalities have been motivated by convex functions, which are essential to inequality theory. Therefore, it is evident that the Hermite-Hadamard (H−H) inequality assumes particular significance in the context of convex functions. The integral mean of any convex function defined within a closed and bounded area, inclusive of the endpoints and midpoints of the function’s domain, can ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.6927 Email addresses: madilkhan@uop.edu.pk (M. Adil Khan), tsalmalki@kau.edu.sa (T. Saeed), sajjadbtk15302@gmail.com (S. Ali), cetin@atauni.edu.tr (Ç. Yildiz), mka la@yahoo.fr (M. K. Alaoui) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) M. Adil Khan et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6927 2 of 18 be estimated through the utilization of upper and lower bounds. This estimation is facil- itated by the H−H inequality, a geometric-based principle. The aforementioned double inequality can be articulated as follows: Let φ be a convex mapping on [ω1, ω2] ⊂ R, where ω1 ̸= ω2. Then φ ( ω1 + ω2 2 ) ≤ 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)dκ ≤ φ(ω1) + φ(ω2) 2 . One important finding in convexity theory is the H−H inequality. The field has advanced significantly as a result of the efforts of several mathematicians who have concentrated on enhancing and generalizing the inequality. We encourage interested readers to review some of the references and the papers [1–4] as it has been widely researched and used in a variety of settings. Definition 1. A function φ : I ⊆ R → R is said to be convex if φ (σω1 + (1− σ)ω2) ≤ σφ (ω1) + (1− σ)φ (ω2) holds for all ω1, ω2 ∈ I and σ ∈ [0, 1]. If −φ is convex, then φ is said to be concave. Convex function theory plays a crucial role in both pure and applied mathematics. Noteworthy inequalities have been derived using various types of convexity [5–9]. The study of integrals and derivatives requires a solid understanding of calculus, a fundamental branch of mathematics. A new mathematical framework called quantum cal- culus, or q-calculus, has emerged as a result of the evolution and adaptation of the classi- cal calculus concepts. Quantum calculus, which includes q-integral calculus, q-fractional calculus, and q-transform analysis, is the study of calculus without limits. Numerous mathematical and physical areas have shown how effective these techniques are. In the early 20th century, the first description of quantum calculus was provided by Jackson. The book [10] is recommended for those who wish to investigate deeper into this topic. q-deformation is a key idea in the field of quantum calculus. This procedure includes changing the characteristics of calculus operations, such as differentiation and integration, by adding a parameter q. Similar to ordinary differential equations in classical calculus, q-difference equations are used in q-calculus to define functions and their derivatives. q-integrals and q-derivatives, which are extensions of their classical counterparts, are in- troduced in quantum calculus. It is clear that these operators meet several q-analogues of the basic theorem of calculus and the Leibniz rule, which are characteristics of ordinary derivatives and integrals. This research paper’s main goal is to investigate the H−H inequality related to the quantum integral operator. Several features of the q-integral for a continuous function were defined and shown by Tariboon and Ntouyas in [11] in 2013. However, Kunt and Iscan showed in [12] that the H−H inequality derived in [11] is in- correct on the left. The following variation of the H−H inequality for the q-integral was then established by Alp et al. in [13]: φ ( qω1 + ω2 q + 1 ) ≤ 1 ω2 − ω1 ∫ ω2 ω1 φ(κ1)ω1dqκ1 ≤ qφ(ω1) + φ(ω2) q + 1 , (1) M. Adil Khan et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6927 3 of 18 where q ∈ (0, 1) and φ : [ω1, ω2] → R is a convex function. The aforementioned inequality (1) is referred to as the quantum H−H inequality. In recent years, this inequality has been the focus of research by numerous mathematicians. In [14], Ali et al. established an identity related to the quantum H−H inequality and the q-integral. Since integral identity and applications of power-mean inequality and Hölder inequality result in the q-integral form of the H−H inequality, it is clear that certain conclusions have been established. Previously, the findings were deduced for a certain value of q. Noor et al., in [15], established some novel quantum estimates for H−H inequalities via q-differentiable convex functions and q-differentiable quasi-convex functions. In the present study [16], the authors propose a novel definition of convexity (k-harmonically γ−convex function) and employ this definition to derive new H−H-type integral inequalities for quantum integrals. A number of authors engaged in research within this field have also studied the symmetric quantum calculus of the H−H inequality. Researchers interested in further works may refer to studies [17] and [18]. In [19], Budak and colleagues took into account the class of coordinated convex functions in order to derive the extended form of the quantum H−H inequality. In order to further generalize the quantum H−H inequality, the double integral identity has been developed. Furthermore, as mentioned in [20] and [21], it has been shown that the above inequality holds for the class of s-convex and r-convex functions, respectively. This study’s main goal is to analyze the quantum H−H inequality using a Green function method. Several novel quantum identities were inferred throughout this partic- ular technique. New inequalities have been made possible by the use of these identities. Convexity, Jensen’s inequality for convex mappings, and q-identities are among the meth- ods used in the study to arrive at the main results of the work. To support the primary findings, the study offers graphical representations and numerical confirmation. 2. Preliminaries and Definitions of q-calculus The following discussion will commence with a presentation of these fundamental def- initions. Definition 2. [11] Let φ : [ω1, ω2] → R be a continuous function, and let c ∈ [ω1, ω2]. Then the expression ω1Dqφ(c) = φ(c)− φ(qc+ (1− q)ω1) (1− q)(c− ω1) , c ̸= ω1,ω1 Dqφ(ω1) = lim c→b1 ω1Dqφ(c) is called the q-derivative on [ω1, ω2] of the function at c. We call φ q-differentiable on [ω1, ω2] if ω1Dqφ(c) exists for all c ∈ [ω1, ω2]. The q-Jackson integral, or q-integral ([22]), was found by Jackson in 1910. Definition 3. [11] If φ : [ω1, ω2] → R is a continuous function, then the q-integral of φ on [ω1, ω2] is defined as:∫ c ω1 φ(κ) ω1dqκ = (c− ω1)(1− q) ∞∑ k=0 qkφ ( qkc+ (1− qk)ω1 ) , M. Adil Khan et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6927 4 of 18 where 0 < q < 1 and c ∈ [ω1, ω2]. There are several important properties of q-integral, for example interval addition, linearity, triangular, and monotonicity property. Theorem 1. [11] If φ : [ω1, ω2] → R is a continuous function. Then∫ c ξ ω1Dqφ(c)ω1dqκ = φ(c)− φ(ξ) where ξ ∈ (ω1, c). Theorem 2. [23] If φ,Ω : [ω1, ω2] → R are two continuous functions and suppose φ(κ) ≤ Ω(κ), ∀κ ∈ [ω1, ω2]. Then we have∫ c ω1 φ(κ)ω1dqκ ≤ ∫ c ω1 Ω(κ) ω1dqκ. Theorem 3. [11] If φ : [ω1, ω2] → R is a continuous function. Then we have ω1Dq ∫ c ω1 φ(κ)ω1dqκ = φ(c);∫ c ξ ω1Dqφ(κ)ω1dqκ = φ(c)− φ(ξ) where ξ ∈ (ω1, c). Theorem 4. [11] If φ,Ω : [ω1, ω2] → R are two continuous functions and suppose κ ∈ R, c ∈ [ω1, ω2], and ξ ∈ (ω1, c). Then we have∫ c ω1 [φ(κ) + Ω(κ)]ω1dqκ = ∫ c ω1 φ(κ)ω1dqκ + ∫ c ω1 Ω(κ)ω1dqκ;∫ c ω1 κφ(κ)ω1dqκ = κ ∫ c ω1 φ(κ)ω1dqκ;∫ c ξ φ(κ)ω1DqΩ(κ)ω1dqκ =φ(c)Ω(c)− φ(ξ)Ω(ξ)− ∫ c ξ Ω(qκ + (1− q)ω1)ω1Dqφ(κ)ω1dqκ. 3. Main Results The fundamental results will be established through the utilization of the following lemma. M. Adil Khan et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6927 5 of 18 Lemma 1. [24, 25] Let G be the Green function defined on [ω1, ω2]× [ω1, ω2] by G(κ, ℓ) = { ω1 − ℓ, ω1 ≤ ℓ ≤ κ, ω1 − κ, κ ≤ ℓ ≤ ω2. Then any φ ∈ C2([ω1, ω2]) can be expressed as φ(κ) = φ(ω1) + (κ − ω1)φ ′(ω2) + ∫ ω2 ω1 G(κ, ℓ)φ′′(ℓ)dℓ. (2) Theorem 5. Let φ ∈ C2[ω1, ω2] such that φ′′ is convex and 0 < q < 1. Then 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)ω1dqκ − φ ( qω1 + ω2 q + 1 ) ≤ 1 ω2 − ω1 [ 1 6 ( φ′′(ω2)− φ′′(ω1) )( (ω2 − ω1) 3 (1 + q)(1 + q2) ) − ( φ′′(ω2) + φ′′(ω1) 2 )( (ω2 − ω1) 1 + q ) ω2 1 − ( 2φ′′(ω2) + φ′′(ω1) 6 ) ω3 1 − ( φ′′(ω2) + φ′′(ω1) 6 )( (qω1 + ω2) (1 + q) )3 + ( φ′′(ω2)ω1 − φ′′(ω1)ω2 2 ) ( (qω1 + ω2) (1 + q) )2 + ω1ω2φ ′′(ω1)(ω2 − ω1) (1 + q) + φ′′(ω1)ω 2 1ω2 + 1 2 ( (ω2 − ω1) 3 1 + q + q2 ) φ′′(ω1) ] . (3) Proof. If we set κ = qω1+ω2 1+q in (2), then we get φ ( qω1 + ω2 q + 1 ) =φ(ω1) + ( qω1 + ω2 q + 1 − ω1 ) φ ′ (ω2) + ∫ ω2 ω1 G ( qω1 + ω2 q + 1 , ℓ ) φ ′′ dℓ =φ(ω1) + ω2 − ω1 q + 1 φ′(ω2) + ∫ ω2 ω1 G ( qω1 + ω2 q + 1 , ℓ ) φ′′(ℓ)dℓ. (4) Also from (2), we obtain that 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)ω1dqκ = 1 ω2 − ω1 ∫ ω2 ω1 { φ(ω1) + (κ − ω1)φ ′(ω2) + ∫ ω2 ω1 G(κ, ℓ)φ′′(ℓ)dℓ } ω1 dqκ =φ(ω1) + ω2 − ω1 q + 1 φ′(ω2) + 1 ω2 − ω1 ∫ ω2 ω1 ∫ ω2 ω1 G(κ, ℓ)φ′′(ℓ)dℓω1dqκ. (5) M. Adil Khan et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6927 6 of 18 Subtracting (4) from (5), we get: 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)ω1dqκ − φ ( qω1 + ω2 q + 1 ) =φ(ω1) + ω2 − ω1 q + 1 φ′(ω2) + 1 ω2 − ω1 ∫ ω2 ω1 ∫ ω2 ω1 G(κ, ℓ)φ′′(ℓ)dℓω1dqκ − φ(ω1)− ω2 − ω1 q + 1 φ′(ω2)− ∫ ω2 ω1 G ( qω1 + ω2 q + 1 , ℓ ) φ′′(ℓ)dℓ. = ∫ ω2 ω1 { 1 ω2 − ω1 ∫ ω2 ω1 G(κ, ℓ)ω1dqκ − G ( qω1 + ω2 q + 1 , ℓ )} φ′′(ℓ)dℓ. (6) Let γ(ℓ) = 1 ω2 − ω1 ∫ ω2 ω1 G(κ, ℓ)ω1dqκ − G ( qω1 + ω2 q + 1 , ℓ ) . Clearly γ(ℓ) is the difference of middle and left side of (1), for the Green function therefore γ(ℓ) is non-negative. Let ℓ = ℓ− ω1 ω2 − ω1 ω2 + ω2 − ℓ ω2 − ω1 ω1, then from (6) we have 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)ω1dqκ − φ ( qω1 + ω2 q + 1 ) = ∫ ω2 ω1 γ(ℓ)φ′′ ( ℓ− ω1 ω2 − ω1 ω2 + ω2 − ℓ ω2 − ω1 ω1 ) dℓ. By convexity of φ′′, we obtain 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)ω1dqκ − φ ( qω1 + ω2 q + 1 ) ≤ ∫ ω2 ω1 γ(ℓ) { ℓ− ω1 ω2 − ω1 φ′′(ω2) + ( ω2 − ℓ ω2 − ω1 ) φ′′(ω1) } dℓ ⇒ 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)ω1dqκ − φ ( qω1 + ω2 q + 1 ) ≤ 1 (ω2 − ω1) { φ′′(ω2) ∫ ω2 ω1 γ(ℓ)(ℓ− ω1)dℓ+ φ′′(ω1)∫ ω2 ω1 γ(ℓ)(ω2 − ℓ)dℓ } . (7) Now we find the integral ∫ ω2 ω1 γ(ℓ)(ℓ− ω1)dℓ. If φ(ℓ) = 1 6ℓ 3 − 1 2ω1ℓ 2, then φ ′′ (ℓ) = ℓ− ω1, using these functions in (6) we obtain.∫ ω2 ω1 γ(ℓ)(ℓ− ω1)dℓ = 1 ω2 − ω1 ∫ ω2 ω1 ( κ3 6 − b1 κ2 2 ) dqκ − 1 6 ( qω1 + ω2 1 + q )3 + ω1 2 ( qω1 + ω2 1 + q )2 . Finding the above integrals, we deduce.∫ ω2 ω1 γ(ℓ)(ℓ− ω1)dℓ = 1 6 (ω2 − ω1) 3 (1 + q)(1 + q2) − 1 2 ω2 1(ω2 − ω1) 1 + q − 1 3 ω3 1 − 1 6 ( qω1 + ω2 1 + q )3 + ω1 2 ( qω1 + ω2 1 + q )2 . (8) M. Adil Khan et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6927 7 of 18 Now we find the integral ∫ ω2 ω1 γ(ℓ)(ω2 − ℓ)dℓ. If φ(ℓ) = 1 2ω2ℓ 2 − 1 6ℓ 3, then φ ′′ (ℓ) = ω2 − ℓ, using these functions in (6) we obtain.∫ ω2 ω1 γ(ℓ)(ω2 − ℓ)dℓ = 1 ω2 − ω1 ∫ ω2 ω1 ( ω2 2 κ2 − 1 6 κ3 ) ω1dqκ − ω2 2 ( qω1 + ω2 1 + q )2 − 1 6 ( qω1 + ω2 1 + q )3 . Finding the above integrals, we deduce.∫ ω2 ω1 γ(ℓ)(ω2 − ℓ)dℓ = ω2 2 ( (ω2 − ω1) 2 1 + q + q2 ) + ω1ω2(ω2 − ω1) 1 + q + ω1 2ω2 − 1 6 (ω2 − ω1) 3 (1 + q)(1 + q2) − 1 2 ω1(ω2 − ω1) 2 1 + q + q2 − 1 2 ω2 1(ω2 − ω1) 1 + q − 1 6 ω3 1 − ω2 2 ( qω1 + ω2 1 + q )2 − 1 6 ( qω1 + ω2 1 + q )3 . (9) using (8) and (9) in (7), we get ≤ 1 (ω2 − ω1) [ φ′′(ω2) {1 6 (ω2 − ω1) 3 (1 + q)(1 + q2) − 1 2 ω2 1(ω2 − ω1) 1 + q − 1 3 ω3 1 − 1 6 ( qω1 + ω2 1 + q )3 + ω1 2 ( qω1 + ω2 1 + q )2 } + φ′′(ω1) {ω2 2 ( (ω2 − ω1) 2 1 + q + q2 ) + ω1ω2(ω2 − ω1) 1 + q + ω1 2ω2 − 1 6 (ω2 − ω1) 3 (1 + q)(1 + q2) − 1 2 ω1(ω2 − ω1) 2 1 + q + q2 − 1 2 ω2 1(ω2 − ω1) 1 + q − 1 6 ω3 1 − ω2 2 ( qω1 + ω2 1 + q )2 − 1 6 ( qω1 + ω2 1 + q )3 }] . = 1 ω2 − ω1 [ 1 6 ( φ′′(ω2)− φ′′(ω1) )( (ω2 − ω1) 3 (1 + q)(1 + q2) ) − ( φ′′(ω2) 2 + φ′′(ω1) 2 )( (ω2 − ω1) 1 + q ) ω2 1 − ( φ′′(ω2) 3 + φ′′(ω1) 6 ) ω3 1 − ( φ′′(ω2) 6 + φ′′(ω1) 6 )( (qω1 + ω2) (1 + q) )3 + ( φ′′(ω2)ω1 2 − φ′′(ω1)ω2 2 )( (qω1 + ω2 (1 + q) )2 + φ′′(ω1)ω2 2 ( (ω2 − ω1) 2 (1 + q + q2) ) + ω1ω2φ ′′(ω1)(ω2 − ω1) (1 + q) + φ′′(ω1)ω 2 1ω2 − ω1φ ′′(ω1)(ω2 − ω1) 2 2(1 + q + q2) ] . = 1 ω2 − ω1 [ 1 6 ( φ′′(ω2)− φ′′(ω1) )( (ω2 − ω1) 3 (1 + q)(1 + q2) ) M. Adil Khan et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6927 8 of 18 − ( φ′′(ω2) + φ′′(ω1) 2 )( (ω2 − ω1) 1 + q ) ω2 1 − ( 2φ′′(ω2) + φ′′(ω1) 6 ) ω3 1 − ( φ′′(ω2) + φ′′(ω1) 6 )( (qω1 + ω2) (1 + q) )3 + ( φ′′(ω2)ω1 − φ′′(ω1)ω2 2 )( (qω1 + ω2) (1 + q) )2 + ω1ω2φ ′′(ω1)(ω2 − ω1) (1 + q) + φ′′(ω1)ω 2 1ω2 + 1 2 ( (ω2 − ω1) 3 1 + q + q2 ) φ′′(ω1) ] . (10) (10) is equivalent to (3). Remark 1. Under the assumptions of Theorem 5 with the limit as q → 1, we have the following H−H inequality: 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)dκ − φ ( ω1 + ω2 2 ) ≤ 1 ω2 − ω1 [ 1 24 ( φ′′(ω2)− φ′′(ω1) ) (ω2 − ω1) 3 − 1 4 ( φ′′(ω2) + φ′′(ω1) ) (ω2 − ω1)ω 2 1 − 1 6 ( 2φ′′(ω2) + φ′′(ω1) ) ω3 1 − 1 48 ( φ′′(ω2) + φ′′(ω1) ) (ω1 + ω2) 3 + 1 8 ( φ′′(ω2)ω1 − φ′′(ω1)ω2 ) (ω1 + ω2) 2 + 1 2 ( ω1ω2φ ′′(ω1)(ω2 − ω1) ) + φ′′(ω1)ω 2 1ω2 + 1 6 (ω2 − ω1) 3φ′′(ω1) ] . Theorem 6. Let φ ∈ C2[ω1, ω2] such that φ′′ is convex and 0 < q < 1. Then qφ(ω1) + φ(ω2) q + 1 − 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)ω1dqκ ≤ 1 ω2 − ω1 [ − 1 6 ( φ′′(ω2)− φ′′(ω1) )( (ω2 − ω1) 3 (1 + q)(1 + q2) ) − ( φ′′(ω2)− φ′′(ω1) 2 )( (ω2 − ω1) 1 + q ) ω2 1 + ( 2φ′′(ω2) + φ′′(ω1) 6 ) ω3 1 − 1 2 ( (ω2 − ω1) 3 1 + q + q2 ) φ′′(ω1)− ω1ω2φ ′′(ω1)(ω2 − ω1) (1 + q) − 1 2 φ′′(ω1)ω 2 1ω2 − ( 2φ′′(ω2) + φ′′(ω1) 6 )( qω3 1 1 + q ) + ( φ′′(ω2) + 2φ′′(ω1) 6 )( ω3 2 1 + q ) M. Adil Khan et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6927 9 of 18 − ( ω2φ ′′(ω2)− qω1φ ′′(ω1) 2 )( ω1ω2 1 + q )] . (11) Proof. If we set κ = ω2 in (2), then we get φ(ω2) = φ(ω1) + (ω2 − ω1)φ ′(ω2) + ∫ ω2 ω1 G(ω2, ℓ)φ ′′(ℓ)dℓ. Adding qφ(ω1) and divide by (q + 1) both sides we get qφ(ω1) + φ(ω2) q + 1 = φ(ω1) + ω2 − ω1 q + 1 φ′(ω2) + 1 q + 1 ∫ ω2 ω1 G(ω2, ℓ)φ ′′(ℓ)dℓ. (12) Subtracting (5) from (12), we get: qφ(ω1) + φ(ω2) q + 1 − 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)ω1dqκ = ∫ ω2 ω1 {G(ω2, ℓ) q + 1 − 1 ω2 − ω1 ∫ ω2 ω1 G(κ, ℓ)ω1dqκ } φ′′(ℓ)dℓ. (13) Let γ(ℓ) = G(ω2, ℓ) q + 1 − 1 ω2 − ω1 ∫ ω2 ω1 G(κ, ℓ)ω1dqκ. Clearly γ(ℓ) is the difference of right and middle side of (1), for the Green function therefore by γ(ℓ) is non-negative. Let ℓ = ℓ− ω1 ω2 − ω1 ω2 + ω2 − ℓ ω2 − ω1 ω1. Then from (13) we have qφ(ω1) + φ(ω2) q + 1 − 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)ω1dqκ = ∫ ω2 ω1 γ(ℓ)φ′′ ( ℓ− ω1 ω2 − ω1 ω2 + ω2 − ℓ ω2 − ω1 ω1 ) dℓ. By convexity of φ′′, we obtain qφ(ω1) + φ(ω2) q + 1 − 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)ω1dqκ ≤ ∫ ω2 ω1 γ(ℓ) { ℓ− ω1 ω2 − ω1 φ′′(ω2) + ( ω2 − ℓ ω2 − ω1 ) φ′′(ω1) } dℓ. ⇒ qφ(ω1) + φ(ω2) q + 1 − 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)ω1dqκ ≤ 1 (ω2 − ω1) { φ′′(ω2) ∫ ω2 ω1 γ(ℓ)(ℓ− ω1)dℓ+ φ′′(ω1)∫ ω2 ω1 γ(ℓ)(ω2 − ℓ)dℓ } . (14) Now we find the integral ∫ ω2 ω1 γ(ℓ)(ℓ− ω1)dℓ. If φ(ℓ) = 1 6ℓ 3 − 1 2ω1ℓ 2, then φ ′′ (ℓ) = ℓ− ω1, using these functions in (13) we obtain. ∫ ω2 ω1 γ(ℓ)(ℓ− ω1)dℓ = q ( ω3 1 6 − ω1ω2 1 2 ) + ω3 2 6 − ω1ω2 2 2 1 + q − 1 ω2 − ω1 ∫ ω2 ω1 ( κ3 6 − b1 κ2 2 ) ω1 dqκ. M. Adil Khan et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6927 10 of 18 Finding the above integrals, we deduce.∫ ω2 ω1 γ(ℓ)(ℓ− ω1)dℓ = −1 6 (ω2 − ω1) 3 (1 + q)(1 + q2) − 1 2 ω2 1(ω2 − ω1) 1 + q + 1 3 ω3 1 + − qω3 1 3 + ω3 2 6 − ω1ω2 2 2 1 + q . (15) Now we find the integral ∫ ω2 ω1 γ(ℓ)(ω2 − ℓ)dℓ. If φ(ℓ) = 1 2ω2ℓ 2 − 1 6ℓ 3, then φ ′′ (ℓ) = ω2 − ℓ. using these functions in (13) we obtain. ∫ ω2 ω1 γ(ℓ)(ω2 − ℓ)dℓ = q ( ω2ω2 1 2 − ω3 1 6 ) + ω2ω2 2 2 − ω3 2 6 1 + q − 1 ω2 − ω1 ∫ ω2 ω1 ( ω2 2 κ2 − 1 6 κ3 ) ω1dqκ. Finding the above integrals, we deduce.∫ ω2 ω1 γ(ℓ)(ω2 − ℓ)dℓ =− ω2 2 ( (ω2 − ω1) 2 1 + q + q2 ) − ω1ω2(ω2 − ω1) 1 + q − ω1 2ω2 2 + 1 6 (ω2 − ω1) 3 (1 + q)(1 + q2) + 1 2 ω1(ω2 − ω1) 2 1 + q + q2 + 1 2 ω2 1(ω2 − ω1) 1 + q + 1 6 ω3 1 + qω2 1ω2 2 − qω3 1 6 + ω3 2 3 1 + q . (16) using (15) and (16) in (14), we get ≤ 1 (ω2 − ω1) [ φ′′(ω2) { − 1 6 (ω2 − ω1) 3 (1 + q)(1 + q2) − 1 2 ω2 1(ω2 − ω1) 1 + q + 1 3 ω3 1 + − qω3 1 3 + ω3 2 6 − ω1ω2 2 2 1 + q } + φ′′(ω1) { − ω2 2 ( (ω2 − ω1) 2 1 + q + q2 ) − ω1ω2(ω2 − ω1) 1 + q − ω1 2ω2 2 + 1 6 (ω2 − ω1) 3 (1 + q)(1 + q2) + 1 2 ω1(ω2 − ω1) 2 1 + q + q2 + 1 2 ω2 1(ω2 − ω1) 1 + q + 1 6 ω3 1 + qω2 1ω2 2 − qω3 1 6 + ω3 2 3 1 + q }] . = 1 ω2 − ω1 [ − 1 6 ( φ′′(ω2)− φ′′(ω1) )( (ω2 − ω1) 3 (1 + q)(1 + q2) ) − ( φ′′(ω2)− φ′′(ω1) 2 )( (ω2 − ω1) 1 + q ) ω2 1 + ( 2φ′′(ω2) + φ′′(ω1) 6 ) ω3 1 − 1 2 ( (ω2 − ω1) 3 1 + q + q2 ) φ′′(ω1)− ω1ω2φ ′′(ω1)(ω2 − ω1) (1 + q) − 1 2 φ′′(ω1)ω 2 1ω2 − ( 2φ′′(ω2) + φ′′(ω1) 6 )( qω3 1 1 + q ) + ( φ′′(ω2) + 2φ′′(ω1) 6 )( ω3 2 1 + q ) − ( ω2φ ′′(ω2)− qω1φ ′′(ω1) 2 )( ω1ω2 1 + q )] . (17) (17) is equivalent to (11). M. Adil Khan et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6927 11 of 18 Remark 2. Under the assumptions of Theorem 6 with the limit as q → 1, we have the following H−H inequality: φ(ω1) + φ(ω2) 2 − 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)dκ ≤ 1 ω2 − ω1 [ − 1 24 ( φ′′(ω2)− φ′′(ω1) ) (ω2 − ω1) 3 − 1 4 ( φ′′(ω2)− φ′′(ω1) ) (ω2 − ω1)ω 2 1 + 1 6 ( 2φ′′(ω2) + φ′′(ω1) ) ω3 1 − 1 6 (ω2 − ω1) 3 φ′′(ω1)− 1 2 ( ω1ω2φ ′′(ω1)(ω2 − ω1) ) − 1 2 φ′′(ω1)ω 2 1ω2 − 1 12 ( 2φ′′(ω2) + φ′′(ω1) ) ω3 1 + 1 12 ( φ′′(ω2) + 2φ′′(ω1) ) ω3 2 − 1 4 ( ω2φ ′′(ω2)− ω1φ ′′(ω1) ) ω1ω2 ] . A wide range of inequalities for convex functions have been published in the literature, and Jensen’s inequality has a special place among them. The following is the presentation of Jensen’s inequality: Lemma 2. [26] Let p : [ω1, ω2] → I be integrable functions with p(κ) ≥ 0, ∀ κ ∈ [ω1, ω2], and ∫ ω2 ω1 p(κ)dκ > 0. If φ : I → R is convex function, then φ (∫ ω2 ω1 p(κ)κdκ∫ ω2 ω1 p(κ)dκ ) ≤ ∫ ω2 ω1 p(κ)φ (κ) dκ∫ ω2 ω1 p(κ)dκ . (18) Theorem 7. Let φ ∈ C2[ω1, ω2] such that φ′′ is a convex and 0 < q < 1. Then 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)ω1dqκ − φ ( qω1 + ω2 q + 1 ) ≥ ( 1 2 ( (ω2 − ω1) 2 1 + q + q2 ) + ω1(ω2 − ω1) (1 + q) + 1 2 ω2 1 − 1 2 ( qω1 + ω2 1 + q )2 ) φ′′  1 6 (ω2−ω1)3 (1+q)(1+q2) + 1 2 ω1(ω2−ω1)2 1+q+q2 + 1 2 ω2 1(ω2−ω1) 1+q + 1 6ω 3 1 − 1 6 ( qω1+ω2 1+q )3 1 2 (ω2−ω1)2 1+q+q2 + ω1(ω2−ω1) 1+q + 1 2ω 2 1 − 1 2 ( qω1+ω2 1+q )2  . (19) Proof. From (18), we have∫ ω2 ω1 p(κ)dκφ (∫ ω2 ω1 p(κ)κdκ∫ ω2 ω1 p(κ)dκ ) ≤ ∫ ω2 ω1 p(κ)φ (κ) dκ. (20) M. Adil Khan et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6927 12 of 18 Comparing (20) with (6), we have p(κ) = γ(ℓ) and φ(κ) = φ′′(ℓ). (20), becomes ∫ ω2 ω1 γ(ℓ)φ′′ (ℓ) dℓ ≥ ∫ ω2 ω1 γ(ℓ)dℓ.φ′′ (∫ ω2 ω1 γ(ℓ)ℓdℓ∫ ω2 ω1 γ(ℓ)dℓ ) . By (6), we have 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)ω1dqκ − φ ( qω1 + ω2 q + 1 ) ≥ ∫ ω2 ω1 γ(ℓ)dℓ.φ′′ (∫ ω2 ω1 γ(ℓ)ℓdℓ∫ ω2 ω1 γ(ℓ)dℓ ) . (21) Now we solve the integral, ∫ ω2 ω1 γ(ℓ)dℓ. If φ(ℓ) = 1 2ℓ 2, then φ′′(ℓ) = 1, using these functions in (6), we obtain.∫ ω2 ω1 γ(ℓ)dℓ = 1 ω2 − ω1 ∫ ω2 ω1 1 2 (ℓ2)ω1dqℓ− 1 2 ( qω1 + ω2 1 + q )2 . Finding the above integrals, we get.∫ ω2 ω1 γ(ℓ)dℓ = 1 2 ( (ω2 − ω1) 2 1 + q + q2 ) + ω1(ω2 − ω1) (1 + q) + 1 2 ω2 1 − 1 2 ( qω1 + ω2 1 + q )2 . (22) Now we solve the integral, ∫ ω2 ω1 γ(ℓ)ℓdℓ. If φ(ℓ) = 1 6ℓ 3, then φ′′(ℓ) = ℓ, using these functions in (6), we obtain.∫ ω2 ω1 γ(ℓ)ℓdℓ = 1 ω2 − ω1 ∫ ω2 ω1 1 6 (ℓ3)ω1dqℓ− 1 6 ( qω1 + ω2 1 + q )3 . Finding the above integrals, we get.∫ ω2 ω1 γ(ℓ)ℓdℓ = 1 6 (ω2 − ω1) 3 (1 + q)(1 + q2) + 1 2 ω1(ω2 − ω1) 2 1 + q + q2 + 1 2 ω2 1(ω2 − ω1) 1 + q + 1 6 ω3 1 − 1 6 ( qω1 + ω2 1 + q )3 . (23) Using (22) and (23) in (21), we get 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)ω1dqκ − φ ( qω1 + ω2 q + 1 ) ≥ ( 1 2 ( (ω2 − ω1) 2 1 + q + q2 ) + ω1(ω2 − ω1) (1 + q) + 1 2 ω2 1 − 1 2 ( qω1 + ω2 1 + q )2 ) φ′′  1 6 (ω2−ω1)3 (1+q)(1+q2) + 1 2 ω1(ω2−ω1)2 1+q+q2 + 1 2 ω2 1(ω2−ω1) 1+q + 1 6ω 3 1 − 1 6 ( qω1+ω2 1+q )3 1 2 (ω2−ω1)2 1+q+q2 + ω1(ω2−ω1) 1+q + 1 2ω 2 1 − 1 2 ( qω1+ω2 1+q )2  . (24) (24) is equivalent to (19). M. Adil Khan et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6927 13 of 18 Remark 3. Under the assumptions of Theorem 7 with the limit as q → 1, we have the following H−H inequality: 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)dκ − φ ( ω1 + ω2 2 ) ≥ ( 1 6 (ω2 − ω1) 2 + 1 2 ω1(ω2 − ω1) + 1 2 ω2 1 − 1 8 (ω1 + ω2) 2 ) φ′′ ( 1 24(ω2 − ω1) 3 + 1 6(ω2 − ω1) 2ω1 + 1 4(ω2 − ω1)ω 2 1 + 1 6ω 3 1 − 1 48(ω1 + ω2) 3 1 6(ω2 − ω1)2 + 1 2(ω2 − ω1)ω1 + 1 2ω 2 1 − 1 8(ω1 + ω2)2 ) . Theorem 8. Let φ ∈ C2[ω1, ω2] such that φ′′ is convex and 0 < q < 1. Then qφ(ω1) + φ(ω2) q + 1 − 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)ω1dqκ ≥ ( 1 2 ( qω2 1 + ω2 2 1 + q ) − 1 2 ( (ω2 − ω1) 2 1 + q + q2 ) − ω1(ω2 − ω1) (1 + q) − 1 2 ω2 1 ) φ′′  1 6 ( qω3 1+ω3 2 1+q ) − 1 6 (ω2−ω1)2 (1+q)(1+q2) − 1 2 ω1(ω2−ω1)2 1+q+q2 − 1 2 ω2 1(ω2−ω1) 1+q − 1 6ω 3 1 1 2 ( qω2 1+ω2 2 1+q ) − 1 2 (ω2−ω1)2 1+q+q2 − ω1(ω2−ω1) 1+q − 1 2ω 2 1  . (25) Proof. From (18), we have∫ ω2 ω1 p(κ)dκφ (∫ ω2 ω1 p(κ)κdκ∫ ω2 ω1 p(κ)dκ ) ≤ ∫ ω2 ω1 p(κ)φ (κ) dκ. (26) Comparing (26) with (13), we have p(κ) = γ(ℓ) and φ(κ) = φ′′(ℓ). (26), becomes ∫ ω2 ω1 γ(ℓ)φ′′ (ℓ) dℓ ≥ ∫ ω2 ω1 γ(ℓ)dℓ.φ (∫ ω2 ω1 γ(ℓ)ℓdℓ∫ ω2 ω1 γ(ℓ)dℓ ) . By (13), we have qφ(ω1) + φ(ω2) q + 1 − 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)ω1dqκ ≥ ∫ ω2 ω1 γ(ℓ)dℓ.φ′′ (∫ ω2 ω1 γ(ℓ)ℓdℓ∫ ω2 ω1 γ(ℓ)dℓ ) . (27) Now we solve the integral, ∫ ω2 ω1 γ(ℓ)dℓ. If φ(ℓ) = 1 2ℓ 2, then φ′′(ℓ) = 1, using these functions in (13), we obtain.∫ ω2 ω1 γ(ℓ)dℓ = 1 2qω 2 1 + 1 2ω 2 2 1 + q − 1 ω2 − ω1 ∫ ω2 ω1 1 2 (ℓ2)ω1dqℓ. M. Adil Khan et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6927 14 of 18 Finding the above integrals, we get.∫ ω2 ω1 γ(ℓ)dℓ = 1 2 ( qω2 1 + ω2 2 1 + q ) − 1 2 ( (ω2 − ω1) 2 1 + q + q2 ) − ω1(ω2 − ω1) (1 + q) − 1 2 ω2 1. (28) Now we solve the integral, ∫ ω2 ω1 γ(ℓ)ℓdℓ. If φ(ℓ) = 1 6ℓ 3, then φ′′(ℓ) = ℓ, using these functions in (13), we obtain.∫ ω2 ω1 γ(ℓ)ℓdℓ = 1 6qω 3 1 + 1 6ω 3 2 1 + q − 1 ω2 − ω1 ∫ ω2 ω1 1 6 (ℓ3)ω1dqℓ. Finding the above integrals, we get.∫ ω2 ω1 γ(ℓ)ℓdℓ = 1 6 ( qω3 1 + ω3 2 1 + q ) − 1 6 (ω2 − ω1) 3 (1 + q)(1 + q2) − 1 2 ω1(ω2 − ω1) 2 1 + q + q2 −1 2 ω2 1(ω2 − ω1) 1 + q − 1 6 ω3 1. (29) Using (28) and (29) in (27), we get qφ(ω1) + φ(ω2) q + 1 − 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)ω1dqκ ≥ ( 1 2 ( qω2 1 + ω2 2 1 + q ) − 1 2 ( (ω2 − ω1) 2 1 + q + q2 ) − ω1(ω2 − ω1) (1 + q) − 1 2 ω2 1 ) φ  1 6 ( qω3 1+ω3 2 1+q ) − 1 6 (ω2−ω1)2 (1+q)(1+q2) − 1 2 ω1(ω2−ω1)2 1+q+q2 − 1 2 ω2 1(ω2−ω1) 1+q − 1 6ω 3 1 1 2 ( qω2 1+ω2 2 1+q ) − 1 2 (ω2−ω1)2 1+q+q2 − ω1(ω2−ω1) 1+q − 1 2ω 2 1  . (30) (30) is equivalent to (25). Remark 4. Under the assumptions of Theorem 8 with the limit as q → 1, we have the following H−H inequality: φ(ω1) + φ(ω2) 2 − 1 ω2 − ω1 ∫ ω2 ω1 φ(κ)dκ ≥ ( 1 4 (ω2 1 + ω2 2)− 1 6 (ω2 − ω1) 2 − 1 2 (ω2 − ω1)ω1 − 1 2 ω2 1 ) φ′′ ( 1 12(ω 3 1 + ω3 2)− 1 24(ω2 − ω1) 2 − 1 6(ω2 − ω1) 2ω1 − 1 4(ω2 − ω1)ω 2 1 − 1 6ω 3 1 1 4(ω 2 1 + ω2 2)− 1 6(ω2 − ω1)2 − 1 2(ω2 − ω1)ω1 − 1 2ω 2 1 ) . M. Adil Khan et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6927 15 of 18 4. Numerical Examples and Graphical Analysis In this section, we present our primary results through numerical examples and graph- ical representations. Under the assumption of Theorem 5, we take q ∈ (0, 1), φ(κ) = κ2, and [ω1, ω2] = [6, 8], as a variable to illustrate a Figure 1 between the left and right-hand sides of Theorem 5. Figure 1: LHS vs RHS of Theorem 5 for q ∈ (0, 1), φ(κ) = κ2, on [ω1, ω2] = [6, 8]. Under the assumption of Theorem 6, we take q ∈ (0, 1), φ(κ) = κ2, and [ω1, ω2] = [2, 2.4], as a variable to illustrate a Figure 2 between the left and right-hand sides of Theorem 6. Figure 2: LHS vs RHS of Theorem 6 for q ∈ (0, 1), φ(κ) = κ2, on [ω1, ω2] = [2, 2.4]. M. Adil Khan et al. / Eur. J. Pure Appl. Math, 18 (4) (2025), 6927 16 of 18 Under the assumption of Theorem 7, we take q ∈ (0, 1), φ(κ) = κ2, and [ω1, ω2] = [0, 1], as a variable to illustrate a Figure 3 between the left and right-hand sided of Theorem 7. Figure 3: LHS vs RHS of Theorem 7 for q ∈ (0, 1), φ(κ) = κ2, on [ω1, ω2] = [0, 1]. Under the assumption of Theorem 8, we take q ∈ (0, 1), φ(κ) = κ4, and [ω1, ω2] = [7, 8] as a variable to illustrate a Figure 4 between the left and right hand side of Theorem 8. Figure 4: LHS vs RHS of Theorem 8 for q ∈ (0, 1), φ(κ) = κ4, on [ω1, ω2] = [7, 8]. 5. Conclusion The present study has employed a Green function technique to analyze the quantum H−H inequality. 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