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Hermite-Hadamard type inclusions via generalized Atangana-Baleanu fractional operator with application

  • Defining new fractional operators and employing them to establish well-known integral inequalities has been the recent trend in the theory of mathematical inequalities. To take a step forward, we present novel versions of Hermite-Hadamard type inequalities for a new fractional operator, which generalizes some well-known fractional integral operators. Moreover, a midpoint type fractional integral identity is derived for differentiable mappings, whose absolute value of the first-order derivatives are convex functions. Moreover, considering this identity as an auxiliary result, several improved inequalities are derived using some fundamental inequalities such as Hölder-İşcan, Jensen and Young inequality. Also, if we take the parameter ρ=1 in most of the results, we derive new results for Atangana-Baleanu equivalence. One example related to matrices is also given as an application.

    Citation: Soubhagya Kumar Sahoo, Fahd Jarad, Bibhakar Kodamasingh, Artion Kashuri. Hermite-Hadamard type inclusions via generalized Atangana-Baleanu fractional operator with application[J]. AIMS Mathematics, 2022, 7(7): 12303-12321. doi: 10.3934/math.2022683

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  • Defining new fractional operators and employing them to establish well-known integral inequalities has been the recent trend in the theory of mathematical inequalities. To take a step forward, we present novel versions of Hermite-Hadamard type inequalities for a new fractional operator, which generalizes some well-known fractional integral operators. Moreover, a midpoint type fractional integral identity is derived for differentiable mappings, whose absolute value of the first-order derivatives are convex functions. Moreover, considering this identity as an auxiliary result, several improved inequalities are derived using some fundamental inequalities such as Hölder-İşcan, Jensen and Young inequality. Also, if we take the parameter ρ=1 in most of the results, we derive new results for Atangana-Baleanu equivalence. One example related to matrices is also given as an application.



    H-H: Hermite-Hadamard; AB: Atangana-Baleanu; ABK: Atangana-Baleanu-Kashuri

    The concept of functions is one of the fundamental constructions of mathematics, and many researchers have concentrated on new functions and put forth attempts to order the space of functions. One such function characterized as a result of this extreme exertion is the convex function, which has a lot of applications in engineering, industrial optimization, probability theory, theory of mathematical inequalities (see [1,2,3]), etc. Starting now and into the foreseeable future various experts have inspected and applied different fractional operators for modelling of COVID-19 [4], modelling of Hepatitis-B epidemic [5], groundwater flow [6], RLC electric circuit [7], the hypothesis of viscoelasticity [8], fluid mechanics [9], etc.

    The theory of convex function was preceded by Jensen in [10] to the literature for the first time and since then attracted consideration to the fact that it seems to be the basis of the concept of incremental function. Many researchers have concentrated on the connections between convexity and the Hermite-Hadamard inequality.

    There are numerous inequalities in the literature for convexity theory. Hermite-Hadamard inequality may be the one that takes the most consideration of scientists and on which many investigations have been conducted. The concept of integral inequality is an intriguing topic of study in mathematical analysis. The fundamental integral inequalities can be used to cultivate convexity's subjective features. The convex function has a lovely representation based on an inequality shown when the functional value of a linear combination of two points in its domain does not exceed the linear combination of the functional values at those two locations.

    Definition 1.1. (see [11]) A real valued function Υ:KRR (set of real numbers) is said to be convex iff the following inequality satisfies

    Υ(uϱ1+(1u)ϱ2)uΥ(ϱ1)+(1u)Υ(ϱ2),

    for all ϱ1,ϱ2K,u[0,1].

    Let Υ:KRR be a convex function with ϱ1<ϱ2 and ϱ1,ϱ2K. Then the H-H inequality is expressed as follows (see [12]):

    Υ(ϱ1+ϱ22)1ϱ2ϱ1ϱ2ϱ1Υ(x)dxΥ(ϱ1)+Υ(ϱ2)2. (1.1)

    Numerous mathematicians have recently generalized and extended the standard H-H inequality (1.1) under the premise of certain interesting new definitions as a generalization of a convex function.

    It is known that fractional calculus aims at establishing mathematical models. As of late, it is seen that refining well-known integral inequalities utilizing fractional integral operators has become a surprising subject of exploration among mathematicians. Generalization of some existing integral inequalities such as Hermite-Hadamard, Simpson, Opial, Fejer and Ostrowski type inequalities using different types of fractional operators and innovative methodologies guided to a revolution in the inequality theory. Several researchers have included this concept in the premises of several types of convexities in recent decades. We refer the interested readers to check (see [13,14,15,16,17,18,19]) for collections of integral inequalities involving convexities.

    Mathematicians have recently been fascinated by the prospect of refining well-known integral inequalities using fractional integral operators. Sarikaya et al. [20] suggested the idea of presenting the Hermite-Hadamard inequality using the Rieman-Liouville fractional integral operator. Liu et al. [21] generalized the Hermite-Hadamard inequality for the ψ-Riemann-Liouville fractional operator, which was inspired by Sarikaya's research. Similarly, several researchers, including Mumcu et al. [22] employed a generalized Proportional fractional integral. Gürbüz et al. [23] worked on the Caputo-Fabrizio operator, whereas Fernandez et al. [24] combined the Atangana-Baleanu fractional operator with the Mittag-Leffler kernel. Mohammed et al. [25] used Tempered fractional integrals, Sahoo et al. [26] used kRiemann-Liouville fractional integrals and Khan et al. [27] worked on a generalised conformable operator.

    This article is arranged as follows: In Section 2, some basic notions and properties in the frame of fractional calculus are presented. Section 3 deals with presenting the main results of our article, where we have established some new H-H type inequalities involving ABK fractional operator. In Section 4, application related to matrices is given. Finally, in Section 5, conclusions and plans of this paper are hinted at.

    Now, we will present some basic definitions of different types of fractional operators that are exceptionally appealing in many parts of mathematics as follows:

    Definition 2.1. (see [20,28]) Let ΥL[ϱ1,ϱ2]. Then the left and right Riemann-Liouville fractional integrals Iαϱ+1 and Iαϱ2 of order α>0 are defined by

    Iαϱ+1Υ(x):=1Γ(α)xϱ1(xu)α1Υ(u)du,(0ϱ1<x<ϱ2)

    and

    Iαϱ2Υ(x):=1Γ(α)ϱ2x(ux)α1Υ(u)du,(0ϱ2<x<ϱ2),

    respectively.

    Theorem 2.1. (see [29]) Let Υ:[ϱ1,ϱ2]R be a convex function with 0ϱ1ϱ2. If ΥL[ϱ1,ϱ2], then the following inequality for Riemann-Liouville fractional integral operator holds true:

    Υ(ϱ1+ϱ22)2μ1Γ(μ+1)(ϱ2ϱ1)μ[Iμ(ϱ1+ϱ22)+Υ(ϱ2)+Iμ(ϱ1+ϱ22)Υ(ϱ1)]Υ(ϱ1)+Υ(ϱ2)2.

    Definition 2.2. (see [30]) Let [ϱ1,ϱ2]R be a finite interval. Then the left side and right side Katugampola fractional integral of order μ>0 of ΥXρc(ϱ1,ϱ2) are defined by:

    ρIμϱ+1{Υ(t)}=ρ1μΓ(μ)tϱ1Υ(x)(tρxρ)μ1xρ1dx,

    and

    ρIμϱ2{Υ(t)}=ρ1μΓ(μ)ϱ2tΥ(x)(xρtρ)μ1xρ1dx.

    With the use of fractional integral operators, this study was able to grow and obtain a variety of well-known integral inequalities. Several mathematicians have recently described distinct forms of the Riemann-Liouville fractional operator in order to propose novel generalisations and refinements of integral inequalities involving differentiable functions. In various domains of applied sciences, studies pertaining to fractional calculus have provided a new perspective and direction. With the use of recently defined fractional operators, it has provided insight into a variety of real-world challenges. A few fundamental rules have separated all of these new fractional operators, and some favourable applications have been contrasted with others.

    As of late, Atangana-Baleanu presented another fractional operator that involves a special function, i.e., the Mittag-Leffler function, which tackles the issue of recovering the original function. The Mittag-Leffler function is more reasonable than a power law in demonstrating the physical phenomenon around us. This made the AB fractional operator more powerful and accommodating. Thus, numerous researchers have shown a keen fascination for using this operator. Atangana-Baleanu presented the derivative both in Caputo and Reimann-Liouville sense.

    Note. From now on we will use B(μ)>0 as a normalization function satisfying B(0)=B(1)=1.

    Definition 2.3. Let p[1,) and [ϱ1,ϱ2] be an open subset of R, the Sobolev space Hp(ϱ1,ϱ2) is defined by

    Hp(ϱ1,ϱ2)={ΥL2(ϱ1,ϱ2):DμΥL2(ϱ1,ϱ2),forall|μ|p}.

    Definition 2.4. (see [31]) Let ϱ2>ϱ1, μ[0,1] and ΥH1(ϱ1,ϱ2). The new fractional derivative is given:

    ABCϱ1Dμt[Υ(t)]=B(μ)1μtϱ1Υ(x)Eμ[μ(tx)μ(1μ)]dx.

    Definition 2.5. (see[31]) Let ΥH1(ϱ1,ϱ2), ϱ1<ϱ2, μ[0,1]. The new fractional derivative is given:

    ABRϱ1Dμt[Υ(t)]=B(μ)1μddttϱ1Υ(x)Eμ[μ(tx)μ(1μ)]dx.

    However, in the same paper they have given the corresponding Atangana-Baleanu(AB)-fractional integral operator as:

    Definition 2.6. (see[31]) The fractional integral operator with non-local kernel of a function ΥH1(ϱ1,ϱ2) is defined as:

    ABϱ1Iμt{Υ(t)}=1μB(μ)Υ(t)+μB(μ)Γ(μ)tϱ1Υ(x)(tx)μ1dx,

    where ϱ2>ϱ1,μ[0,1].

    In [32], the right hand side of Atangana-Baleanu fractional integral operator as following;

    ABϱ2Iμt{Υ(t)}=1μB(μ)Υ(t)+μB(μ)Γ(μ)ϱ2tΥ(x)(xt)μ1dx,

    where, Γ(μ) is the Gamma function defined by Γ(μ)=0ettμ1dt.

    Now, we present the definition of integral operator that generalizes Atangana-Baleanu integral operator and Katugampola integral operator.

    Definition 2.7. (see [33]) Let [ϱ1,ϱ2]R be a finite interval. Then the left side and right side ABK-fractional integral of order μ>0 of ΥXρc(ϱ1,ϱ2) are defined as follows:

    ABKϱ+1ρIμt{Υ(t)}=1μB(μ)Υ(t)+ρ1μμB(μ)Γ(μ)tϱ1Υ(x)(tρxρ)μ1xρ1dx,

    and

    ABKϱ2ρIμt{Υ(t)}=1μB(μ)Υ(t)+ρ1μμB(μ)Γ(μ)ϱ2tΥ(x)(xρtρ)μ1xρ1dx.

    For other generalizations and detailed knowledge about fractional-calculus operators, the interested readers can see [34,35,36,37,38,39,40,41,42,43,44,45].

    The purpose of this investigation is to get new estimations of Hermite-Hadamard type inequalities employing a generalized Atangana-Baleanu fractional integral operator, i.e., ABK for convex functions. The principal motivation for applying ABK fractional operators is that the consequences demonstrated by this operator permit us to generate unique and comprehensive inequalities of the Hermite-Hadamard type. A novel version of Hermite-Hadamard inequality and an integral identity including generalized Atangana-Baleanu fractional operator is established. Next, some refinements of H-H type inequalities in the setting of (ABK) fractional operator are discussed as well. It is apparent that if one selects ρ = 1, and μ=1 in the main results, multiple alternatives of the classical H-H inequality will be recaptured.

    Definition 2.8. (Hölder's inequality [46]) Let p>1 and 1p+1q=1. If Υ and Ψ are real functions defined on [ϱ1,ϱ2] and if |Υ|p and |Ψ|q are integrable on [ϱ1,ϱ2], then the following inequality holds true:

    ϱ2ϱ1|Υ(x)Ψ(x)|dx(ϱ2ϱ1|Υ(x)|pdx)1p(ϱ2ϱ1|Ψ(x)|qdx)1q.

    Definition 2.9. (Power-mean inequality [46]) Let q1. If Υ and Ψ are real functions defined on [ϱ1,ϱ2] and if |Υ|, |Υ||Ψ|q are integrable on [ϱ1,ϱ2], then the following inequality holds true:

    ϱ2ϱ1|Υ(x)Ψ(x)|dx(ϱ2ϱ1|Υ(x)|dx)11q(ϱ2ϱ1|Υ(x)||Ψ(x)|qdx)1q.

    Definition 2.10. (Hölder-İşcan integral inequality [46]) Let p>1 and 1p+1q=1. If Υ and Ψ are real functions defined on [ϱ1,ϱ2] and if |Υ|p and |Ψ|q are integrable on [ϱ1,ϱ2], then the following inequality holds true:

    ϱ2ϱ1|Υ(x)Ψ(x)|dx1ϱ2ϱ1(ϱ2ϱ1(ϱ2x)|Υ(x)|pdx)1p(ϱ2ϱ1(ϱ2x)|Ψ(x)|qdx)1q+1ϱ2ϱ1(ϱ2ϱ1(xϱ1)|Υ(x)|pdx)1p(ϱ2ϱ1(xϱ1)|Ψ(x)|qdx)1q.

    The main objective of this section is to establish several novel refinements of H-H type inequalities via ABK fractional operator.

    Theorem 3.1. Let Υ:X=[ϱρ1,ϱρ2]R be a function with ϱ2>ϱ10, ρ>0 and ΥXρc(ϱρ1,ϱρ2). If Υ is a convex function on X, then the following ABK fractional integral inequalities hold true:

    1μB(μ)[Υ(ϱρ1)+Υ(ϱρ2)]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)[ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)][Υ(ϱρ1+Υ(ϱρ2))]2μ[(ϱρ2ϱρ1)μρ1μ+2μΓ(μ)(1μ)B(μ)Γ(μ)],

    where μ(0,1).

    Proof. From the convexity theory, we have

    Υ(xρ+yρ2)Υ(xρ)+Υ(yρ)2.

    Choosing

    xρ=uρ2ϱρ1+2uρ2ϱρ2

    and

    yρ=uρ2ϱρ2+2uρ2ϱρ1.

    Consequently, we get

    2Υ(ϱρ1+ϱρ22)Υ(uρ2ϱρ1+2uρ2ϱρ2)+Υ(uρ2ϱρ2+2uρ2ϱρ1). (3.1)

    Multiplying both sides of (3.1) by μB(μ)Γ(μ)uρμ1(μ>0) and integrating over the closed inteval [0, 1], we obtain

    2B(μ)Γ(μ)Υ(ϱρ1+ϱρ22)μ2μB(μ)Γ(μ)ϱ2(ϱρ1+ϱρ22)1ρ(ϱρ2xρϱρ2ϱρ1)μ1Υ(xρ)xρ1ϱρ2ϱρ1dx+μ2μB(μ)Γ(μ)(ϱρ1+ϱρ22)1ρϱ1(xρϱρ1ϱρ2ϱρ1)μ1Υ(xρ)xρ1ϱρ2ϱρ1dx.

    Now, by using Definition 2.7 of ABK Fractional operator, we get

    2ρ1μ(ϱρ2ϱρ1)μ2μρB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)μρ1μB(μ)Γ(μ)ϱ2(ϱρ1+ϱρ22)1ρ(ϱρ2xρ)μ1Υ(xρ)xρ1dx+μρ1μB(μ)Γ(μ)(ϱρ1+ϱρ22)1ρϱ1(xρϱρ1)μ1Υ(xρ)xρ1dx.

    This proves the first inequality asserted by Theorem 3.1.

    1μB(μ)[Υ(ϱρ1)+Υ(ϱρ2)]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)[ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)].

    For the proof of the second inequality in Theorem 3.1, we need the following results:

    Υ(uρ2ϱρ1+2uρ2ϱρ2)+Υ(uρ2ϱρ2+2uρ2ϱρ1)uρ2Υ(ϱρ1)+2uρ2Υ(ϱρ2)+uρ2Υ(ϱρ2)+2uρ2Υ(ϱρ1)=Υ(ϱρ1)+Υ(ϱρ2). (3.2)

    Multiplying both sides of (3.2) by μB(μ)Γ(μ)uρμ1(μ>0) and integrating over the closed inteval [0, 1], we obtain

    μB(μ)Γ(μ)10Υ(uρ2ϱρ1+2uρ2ϱρ2)uρμ1du+μB(μ)Γ(μ)10Υ(uρ2ϱρ2+2uρ2ϱρ1)uρμ1duμ[Υ(ϱρ1)+Υ(ϱρ2)]B(μ)Γ(μ)10uρμ1du.

    It follows

    [ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)]Υ(ϱρ1)+Υ(ϱρ2)2μ[(ϱρ2ϱρ1)μρ1μ+2μΓ(μ)(1μ)B(μ)Γ(μ)].

    This leads us to the proof of the Theorem 3.1.

    Corollary 3.1. When we choose ρ=1, Theorem 3.1 yields the following result for Atangana-Baleanu fractional operator.

    1μB(μ)[Υ(ϱ1)+Υ(ϱ2)]+(ϱ2ϱ1)μ2μ1B(μ)Γ(μ)Υ(ϱ1+ϱ22)[AB(ϱ1+ϱ22)+Iμϱ2Υ(ϱ2)+AB(ϱ1+ϱ22)Iμϱ1Υ(ϱ1)][Υ(ϱ1+Υ(ϱ2))]2μ[(ϱ2ϱ1)μ+2μΓ(μ)(1μ)B(μ)Γ(μ)].

    Lemma 3.1. Let Υ:X=[ϱρ1,ϱρ2]R be a differentiable function on X, with 0ϱ1<ϱ2, ρ>0 and μ(0,1). Then the following ABK fractional integral equality holds true:

    [ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)]1μB(μ)[Υ(ϱρ1)+Υ(ϱρ2)]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)=(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)[10Υ(uρ2ϱρ1+2uρ2ϱρ2)uρμuρ1du+10Υ(uρ2ϱρ2+2uρ2ϱρ1)uρμuρ1du].

    Proof. The proof directly follows by using integration by parts technique

    10Υ(uρ2ϱρ1+2uρ2ϱρ2)uρμuρ1du+10Υ(uρ2ϱρ2+2uρ2ϱρ1)uρμuρ1du=I1+I2.I1=10Υ(uρ2ϱρ1+2uρ2ϱρ2)uρμuρ1du=10uρμ[ddu(Υ(uρ2ϱρ1+2uρ2ϱρ2))2ρ(ϱρ1ϱρ2)]du.Integrating by partsuρμΥ(uρ2ϱρ1+2uρ2ϱρ2)2ρ(ϱρ1ϱρ2)|1010μuρμ1Υ(uρ2ϱρ1+2uρ2ϱρ2)2(ϱρ1ϱρ2)du=2ρ(ϱρ2ϱρ1)Υ(ϱρ1+ϱρ22)+2μ+1μ(ϱρ2ϱρ1)μ+1ϱ2(ϱρ1+ϱρ22)1ρ(ϱρ2xρ)μ1Υ(xρ)xρ1dx.Similarly,I2=10Υ(uρ2ϱρ2+2uρ2ϱρ1)uρμuρ1du=2ρ(ϱρ2ϱρ1)Υ(ϱρ1+ϱρ22)2μ+1μ(ϱρ2ϱρ1)μ+1(ϱρ1+ϱρ22)1ρϱ1(xρϱρ1)μ1Υ(xρ)xρ1dx.

    Multiplying both I1 and I2 by ρ1μB(μ)Γ(μ)(ϱρ1+ϱρ22)μ+1 and then subtracting the resultant I2 from I1, we have the desired proof.

    Theorem 3.2. Let Υ:X=[ϱρ1,ϱρ2]R be a differentiable function on X, with 0ϱ1<ϱ2, ρ>0 and μ(0,1). If |Υ| is a convex function, then the following ABK fractional integral inequality holds true:

    |[ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)]1μB(μ)[Υ(ϱρ1)+Υ(ϱρ2)]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)|(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)[|Υ(ϱρ1)|+|Υ(ϱρ2)ρ(μ+1)].

    Proof. Taking the equality given in Lemma 3.1 into consideration and convexity of |Υ|, we have

    |[ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)]1μB(μ)[Υ(ϱρ1)+Υ(ϱρ2)]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)|(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)[10(|Υ(ϱρ1)|uρ2+2uρ2|Υ(ϱρ2)|)uρμuρ1du+10(|Υ(ϱρ2)|uρ2+2uρ2|Υ(ϱρ1)|)uρμuρ1du](ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)[|Υ(ϱρ1)|+|Υ(ϱρ2)10uρμ+ρ1du]=(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)[|Υ(ϱρ1)|+|Υ(ϱρ2)ρ(μ+1)].

    This completes the proof.

    Corollary 3.2. Using the same notations as in Theorem 3.2 and choosing |Υ|M, we have the following inequality:

    |[ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)]1μB(μ)[Υ(ϱρ1)+Υ(ϱρ2)]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)|(ϱρ1+ϱρ22)μ+1Mρμ1B(μ)Γ(μ).

    Corollary 3.3. Using the same notations as in Theorem 3.2 and choosing ρ=1, we have the following inequality involving Atangana-Baleanu fractional operators:

    |[AB(ϱ1+ϱ22)+Iμϱ2Υ(ϱ2)+AB(ϱ1+ϱ22)Iμϱ1Υ(ϱ1)]1μB(μ)[Υ(ϱ1)+Υ(ϱ2)]+(ϱ2ϱ1)μ2μ1B(μ)Γ(μ)Υ(ϱ1+ϱ22)|(ϱ1+ϱ22)μ+11B(μ)Γ(μ)[|Υ(ϱ1)|+|Υ(ϱ2)(μ+1)].

    Theorem 3.3. Let Υ:X=[ϱρ1,ϱρ2]R be a differentiable function on X, with 0ϱ1<ϱ2, ρ>0 and μ(0,1). If |Υ|q is a convex function, then the following ABK fractional integral inequality holds true:

    |[ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)]1μB(μ)[Υ(ϱρ1)+Υ(ϱρ2)]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)|(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)[(1ρ(μ+1))11q{12ρ(μ+2)|Υ(ϱρ1)|q+μ+32ρ(μ+1)(μ+2)|Υ(ϱρ2)|q}1q+(1ρ(μ+1))11q{12ρ(μ+2)|Υ(ϱρ2)|q+μ+32ρ(μ+1)(μ+2)|Υ(ϱρ1)|q}1q],

    where q1.

    Proof. Taking the equality given in Lemma 3.1 into consideration and power-mean inequality, we have

    |[ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)]1μB(μ)[Υ(ϱρ1)+Υ(ϱρ2)]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)|(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)[(10uρμuρ1du)11q(10|Υ(uρ2ϱρ1+2uρ2ϱρ2)|quρμuρ1du)1q+(10uρμuρ1du)11q(10|Υ(uρ2ϱρ2+2uρ2ϱρ1)|quρμuρ1du)1q]. (3.3)

    Using the convexity of |Υ|q on X, we have

    10|Υ(uρ2ϱρ2+2uρ2ϱρ1)|quρμuρ1du=10(uρ2|Υ(ϱρ1)|+2uρ2|Υ(ϱρ2)|)uρμuρ1du=12ρ(μ+2)|Υ(ϱρ1)|q+μ+32ρ(μ+1)(μ+2)|Υ(ϱρ2)|q,

    and

    10|Υ(uρ2ϱρ2+2uρ2ϱρ1)|quρμuρ1du=10(uρ2|Υ(ϱρ2)|+2uρ2|Υ(ϱρ1)|)uρμuρ1du=12ρ(μ+2)|Υ(ϱρ2)|q+μ+32ρ(μ+1)(μ+2)|Υ(ϱρ1)|q,

    and also

    10uρμuρ1du=1ρ(μ+1).

    Substitution of the above computations in (3.3), completes rest of the proof.

    Corollary 3.4. Using the same notations as the above Theorem 3.3 and if we take |Υ|M, then we get:

    |[ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)]1μB(μ)[Υ(ϱρ1)+Υ(ϱρ2)]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)|(ϱρ1+ϱρ22)μ+12Mρμ1B(μ)Γ(μ)(1ρ(μ+1))11q.

    Corollary 3.5. Using the same notations as in Theorem 3.3 and choosing ρ=1, we have the following inequality involving Atangana-Baleanu fractional operators:

    |[AB(ϱ1+ϱ22)+Iμϱ2Υ(ϱ2)+AB(ϱ1+ϱ22)Iμϱ1Υ(ϱ1)]1μB(μ)[Υ(ϱ1)+Υ(ϱ2)]+(ϱ2ϱ1)μ2μ1B(μ)Γ(μ)Υ(ϱ1+ϱ22)|(ϱ1+ϱ22)μ+11B(μ)Γ(μ)[(1(μ+1))11q{12(μ+2)|Υ(ϱ1)|q+μ+32(μ+1)(μ+2)|Υ(ϱ2)|q}1q+(1(μ+1))11q{12(μ+2)|Υ(ϱ2)|q+μ+32(μ+1)(μ+2)|Υ(ϱ1)|q}1q].

    Theorem 3.4. Let Υ:X=[ϱρ1,ϱρ2]R be a differentiable function on X with 0ϱ1<ϱ2, ρ>0 and μ(0,1). If |Υ|n is a convex function, then the following ABK fractional integral inequality holds true:

    |[ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)]1μB(μ)[Υ(ϱρ1)+Υ(ϱρ2)]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)|(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)[2mρ(mμ+1)+|Υ(ϱρ1)|n+|Υ(ϱρ2)|nnρ],

    where 1m+1n=1.

    Proof. Taking the equality given in Lemma 3.1 into consideration, Young's inequality and convexity of |Υ|n, we have

    |[ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)]1μB(μ)[Υ(ϱρ1)+Υ(ϱρ2)]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)|(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)[10|Υ(uρ2ϱρ1+2uρ2ϱρ2)|uρμuρ1du+10|Υ(uρ2ϱρ2+2uρ2ϱρ1)|uρμuρ1du].(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)[1m10umρμuρ1du+1n10|Υ(uρ2ϱρ1+2uρ2ϱρ2)|nuρ1du+1m10umρμuρ1du+1n10|Υ(uρ2ϱρ2+2uρ2ϱρ1)|nuρ1du].(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)[1m10umρμuρ1du+10(|Υ(ϱρ1)|nuρ2+2uρ2|Υ(ϱρ2)|n)uρ1du1m10umρμuρ1du+10(|Υ(ϱρ2)|nuρ2+2uρ2|Υ(ϱρ1)|n)uρ1du].(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)[2mρ(mμ+1)+|Υ(ϱρ1)|n+|Υ(ϱρ2)|nnρ],

    where

    1m10umρμuρ1du=1m[1ρ(mμ+1)],1n10uρuρ12du=14nρ,

    and

    1n10(2uρ)uρ12du=34nρ.

    This completes the proof.

    Corollary 3.6. Using the same notations as in Theorem 3.4 and choosing ρ=1, we have the following inequality involving Atangana-Baleanu fractional operators:

    |[AB(ϱ1+ϱ22)+Iμϱ2Υ(ϱ2)+AB(ϱ1+ϱ22)Iμϱ1Υ(ϱ1)]1μB(μ)[Υ(ϱ1)+Υ(ϱ2)]+(ϱ2ϱ1)μ2μ1B(μ)Γ(μ)Υ(ϱ1+ϱ22)|(ϱ1+ϱ22)μ+11B(μ)Γ(μ)[2m(mμ+1)+|Υ(ϱ1)|n+|Υ(ϱ2)|nn].

    Theorem 3.5. Let Υ:X=[ϱρ1,ϱρ2]R be a differentiable function on X with 0ϱ1<ϱ2, ρ>0 and μ(0,1). If |Υ| is a concave function on X, then the following ABK fractional integral inequality holds true:

    |[ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)]1μB(μ)[Υ(ϱρ1)+Υ(ϱρ2)]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)|(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)(1ρ(μ+1))[|Υ((μ+1)ϱρ2+(μ+3)ϱρ12(μ+2))|+|Υ((μ+1)ϱρ1+(μ+3)ϱρ22(μ+2))|].

    Proof. Taking the equality given in Lemma 3.1 into consideration and Jensen's inequality, we have

    |[ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)]1μB(μ)[Υ(ϱρ1)+Υ(ϱρ2)]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)|(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)[10|Υ(uρ2ϱρ1+2uρ2ϱρ2)|uρμuρ1du+10|Υ(uρ2ϱρ2+2uρ2ϱρ1)|uρμuρ1du].(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)[(10uρμuρ1du)|Υ(10uρμuρ1(uρ2ϱρ1+2uρ2ϱρ2)du10uρμuρ1du)|+(10uρμuρ1du)|Υ(10uρμuρ1(uρ2ϱρ2+2uρ2ϱρ1)du10uρμuρ1du)|](ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)[1ρ(μ+1)|Υ(ϱρ12ρ(μ+2)+(μ+3)ϱρ22ρ(μ+1)(μ+2)1ρ(μ+1))|+1ρ(μ+1)|Υ(ϱρ22ρ(μ+2)+(μ+3)ϱρ12ρ(μ+1)(μ+2)1ρ(μ+1))|](ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)(1ρ(μ+1))[|Υ((μ+1)ϱρ2+(μ+3)ϱρ12(μ+2))|+|Υ((μ+1)ϱρ1+(μ+3)ϱρ22(μ+2))|].

    This led us to the proof of the desired Theorem 3.5.

    Corollary 3.7. Using the same notations as in Theorem 3.5 and choosing ρ=1, we have the following inequality involving Atangana-Baleanu fractional operators:

    |[AB(ϱ1+ϱ22)+Iμϱ2Υ(ϱ2)+AB(ϱ1+ϱ22)Iμϱ1Υ(ϱ1)]1μB(μ)[Υ(ϱ1)+Υ(ϱ2)]+(ϱ2ϱ1)μ2μ1B(μ)Γ(μ)Υ(ϱ1+ϱ22)|(ϱ1+ϱ22)μ+11B(μ)Γ(μ)(1(μ+1))[|Υ((μ+1)ϱ2+(μ+3)ϱ12(μ+2))|+|Υ((μ+1)ϱ1+(μ+3)ϱ22(μ+2))|].

    Theorem 3.6. Let Υ:X=[ϱρ1,ϱρ2]R be a differentiable function on [ϱρ1,ϱρ2] with 0ϱ1<ϱ2, ρ>0 and μ(0,1). If |Υ|q is a convex function on X, then the following inequality ABK fractional integral inequality holds true:

    |[ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)]1μB(μ)[Υ(ϱρ1)+Υ(ϱρ2)]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)|(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)×[(1ρμs+ρ1ρμs+ρ+1)1s(|Υ(ϱρ1)|q2[12ρ(2ρ+1)]+|Υ(ϱρ2)|q2[32ρ+12ρ+12ρ+1])+(1ρμs+ρ+1)1s(|Υ(ϱρ1)|q2(2ρ+1)+|Υ(ϱρ2)|q(3ρ+1)2(ρ+1)(2ρ+1))1q+(1ρμs+ρ+1)1s(|Υ(ϱρ2)|q2(2ρ+1)+|Υ(ϱρ1)|q(3ρ+1)2(ρ+1)(2ρ+1))1q+(1ρμs+ρ1ρμs+ρ+1)1s(|Υ(ϱρ2)|q2[12ρ(2ρ+1)]+|Υ(ϱρ1)|q2[32ρ+12ρ+12ρ+1])],

    where 1s+1q=1.

    Proof. Taking the equality given in Lemma 3.1 into consideration and using Hölder-İşcan inequality

    |[ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)]1μB(μ)[Υ(ϱρ1)+Υ(ϱρ2)]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)|(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)×[10|Υ(uρ2ϱρ1+2uρ2ϱρ2)|uρμuρ1du+10|Υ(uρ2ϱρ2+2uρ2ϱρ1)|uρμuρ1du].(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)×[(10(1u)uρμsuρ1du)1s(10(1u)uρ1|Υ(uρ2ϱρ1+2uρ2ϱρ2)|q)1q+(10uuρμsuρ1du)1s(10uuρ1|Υ(uρ2ϱρ1+2uρ2ϱρ2)|q)1q+(10(1u)uρμsuρ1du)1s(10(1u)uρ1|Υ(uρ2ϱρ2+2uρ2ϱρ1)|q)1q+(10uuρμsuρ1du)1s(10uuρ1|Υ(uρ2ϱρ2+2uρ2ϱρ1)|q)1q]. (3.4)

    By the convexity of |Υ|q, we have

    10(u)uρ1|Υ(uρ2ϱρ1+2uρ2ϱρ2)|qdu=|Υ(ϱρ1)|q2(2ρ+1)+|Υ(ϱρ2)|q2[2ρ+112ρ+1],10(1u)uρ1|Υ(uρ2ϱρ1+2uρ2ϱρ2)|qdu=|Υ(ϱρ1)|q2[12ρ12ρ+1]+|Υ(ϱρ2)|q2[32ρ2ρ+1+12ρ+1],10(u)uρ1|Υ(uρ2ϱρ2+2uρ2ϱρ1)|qdu=|Υ(ϱρ2)|q2(2ρ+1)+|Υ(ϱρ1)|q2[2ρ+112ρ+1],10(1u)uρ1|Υ(uρ2ϱρ2+2uρ2ϱρ1)|qdu=|Υ(ϱρ2)|q2[12ρ12ρ+1]+|Υ(ϱρ1)|q2[32ρ2ρ+1+12ρ+1].

    Also,

    10(1u)uρμsuρ1du=1ρμs+ρ1ρμs+ρ+1,10(u)uρμsuρ1du=1ρμs+ρ+1.

    Upon, substituting the above computations in (3.4), we have the required result.

    Corollary 3.8. Using the same notations as in Theorem 3.6 and choosing ρ=1, we have the following inequality involving Atangana-Baleanu fractional operators:

    |[AB(ϱ1+ϱ22)+Iμϱ2Υ(ϱ2)+AB(ϱ1+ϱ22)Iμϱ1Υ(ϱ1)]1μB(μ)[Υ(ϱ1)+Υ(ϱ2)]+(ϱ2ϱ1)μ2μ1B(μ)Γ(μ)Υ(ϱ1+ϱ22)|(ϱ1+ϱ22)μ+11B(μ)Γ(μ)[(1μs+11μs+2)1s(|Υ(ϱ1)|q12+5|Υ(ϱ2)|q12)+(1μs+2)1s(|Υ(ϱ1)|q6+|Υ(ϱ2)|q3)1q+(1μs+2)1s(|Υ(ϱ2)|q6+7|Υ(ϱ1)|q3)1q+(1μs+11μs+2)1s(|Υ(ϱ2)|q12+5|Υ(ϱ1)|q12)].

    Theorem 3.7. Let Υ:X=[ϱρ1,ϱρ2]R be a differentiable function with 0ϱ1<ϱ2 ρ>0, μ(0,1). If |Υ|q is a convex function on X, for q>1 and 1s+1q=1, then the following ABK fractional integral inequality holds true:

    |[ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)]1μB(μ)[Υ(ϱρ1)+Υ(ϱρ2)]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)|(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)(1s(ρμ)+ρ)1s{(|Υ(ϱρ1)|q4ρ+3|Υ(ϱρ2)|q4ρ)1q+(|Υ(ϱρ2)|q4ρ+3|Υ(ϱρ1)|q4ρ)1q}.

    Proof. Taking the equality given in Lemma 3.1 into consideration and Hölder inequality, we have

    |[ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2Υ(ϱρ2)+ABK(ϱρ1+ϱρ22)1ρρIμϱρ1Υ(ϱρ1)]1μB(μ)[Υ(ϱρ1)+Υ(ϱρ2)]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)Υ(ϱρ1+ϱρ22)|(ϱρ1+ϱρ22)μ+1ρμ1B(μ)Γ(μ)[(10uρμsuρ1du)1s(10|Υ(uρ2ϱρ1+2uρ2ϱρ2)|quρ1du)1q+(10uρμsuρ1du)1s(10|Υ(uρ2ϱρ2+2uρ2ϱρ1)|quρ1du)1q]. (3.5)

    Using the convexity of |Υ|q on [ϱρ1,ϱρ2], we have

    10|Υ(uρ2ϱρ2+2uρ2ϱρ1)|quρμuρ1du=10(uρ2|Υ(ϱρ1)|+2uρ2|Υ(ϱρ2)|)uρ1du=|Υ(ϱρ1)|q4ρ+3|Υ(ϱρ2)|q4ρ,

    and

    10|Υ(uρ2ϱρ2+2uρ2ϱρ1)|quρμuρ1du=10(uρ2|Υ(ϱρ2)|+2uρ2|Υ(ϱρ1)|)uρ1du=|Υ(ϱρ2)|q4ρ+3|Υ(ϱρ1)|q4ρ.

    Substituting the above computations in (3.5), completes rest of the proof.

    Corollary 3.9. Using the same notations as in Theorem 3.7 and choosing ρ=1, we have the following inequality involving Atangana-Baleanu fractional operators:

    |[AB(ϱ1+ϱ22)+Iμϱ2Υ(ϱ2)+AB(ϱ1+ϱ22)Iμϱ1Υ(ϱ1)]1μB(μ)[Υ(ϱ1)+Υ(ϱ2)]+(ϱ2ϱ1)μ2μ1B(μ)Γ(μ)Υ(ϱ1+ϱ22)|(ϱ1+ϱ22)μ+11B(μ)Γ(μ)(1sμ)1s{(|Υ(ϱ1)|q4+3|Υ(ϱ2)|q4)1q+(|Υ(ϱ2)|q4+3|Υ(ϱ1)|q4)1q}.

    We represent Cn as the set of n×n complex matrices and Mn as the algebra of n×n complex matrices and M+n represents as the strictly positive matrices in M. That is, AM+n, if Au,u>0 for all nonzero uCn.

    Sababheh [47], proved that Υ(u)=∥AuXB1u+A1uXBu, A,BM+n,XMn is convex for all u[0,1]. Then, by using Theorem 3.1, we have

    1μB(μ)[||Aϱρ1XB1ϱρ1+A1ϱρ1XBϱρ1||+||Aϱρ2XB1ϱρ2+A1ϱρ2XBϱρ2||]+(ϱρ2ϱρ1)μ2μ1ρμB(μ)Γ(μ)||A(ϱρ1+ϱρ22)XB1(ϱρ1+ϱρ22)+A1(ϱρ1+ϱρ22)XB(ϱρ1+ϱρ22)||2αsΓ(α+1)(ϱρ2ϱρ1)α[ABK(ϱρ1+ϱρ22)1ρ+ρIμϱρ2||Aϱρ2XB1ϱρ2+A1ϱρ2XBϱρ2||+ABK(ϱρ1+ϱρ22)1ρρIμϱρ2||Aϱρ1XB1ϱρ1+A1ϱρ1XBϱρ1||]||Aϱρ1XB1ϱρ1+A1ϱρ1XBϱρ1||+||Aϱρ2XB1ϱρ2+A1ϱρ2XBϱρ2||2μ[(ϱρ2ϱρ1)μρ1μ+2μΓ(μ)(1μ)B(μ)Γ(μ)].

    Recently, fractional calculus has been one of the often utilized concepts to acquire new variants of some well-known integral inequalities. This improvement in the field of mathematical inequality dealing with fractional calculus has prompted another direction in different areas of mathematics and applied sciences. In this article, we have investigated ABK fractional integrals and, by using these fractional integrals, we have derived H-H type inequality (see Theorem 3.1) and then we presented an identity (see Lemma 3.1). Using the identity, some refinements of H-H type inequalities (see Theorems 3.2–3.7) are discussed for convex functions. If we choose ρ=1, our presented results give some new inequalities of (ϱ1+ϱ22) type for AB fractional operator. The ABK fractional operator being a new and unified operator, it will be quite interesting to check whether we can apply this integral operator to establish Ostrowski type inequality, Simpson type inequality. For future work, we will apply the ABK operator using Minkowski and Markov inequalities. Also, we will check its applicability on interval valued analysis and on coordinates for integral inequalities.

    The authors declare no conflict of interest.



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