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Research article

Generalized integral inequalities for ABK-fractional integral operators

  • Received: 05 March 2021 Accepted: 29 June 2021 Published: 09 July 2021
  • MSC : 26A33, 26A51, 26D10

  • In this paper, we employ new version of the Atangana-Baleanu integral operator namely ABK-fractional integrals to obtain two general integral identities complying second-order derivatives for a given function. Thus allowing us to derive new generalized Hermite-Hadamard type inequalities via ABK- fractional integrals. Moreover we give several new versions of Mid-point and Trapezoid type inequalities by employing Hölder, Young and Jensen inequalities.

    Citation: Saad Ihsan Butt, Erhan Set, Saba Yousaf, Thabet Abdeljawad, Wasfi Shatanawi. Generalized integral inequalities for ABK-fractional integral operators[J]. AIMS Mathematics, 2021, 6(9): 10164-10191. doi: 10.3934/math.2021589

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  • In this paper, we employ new version of the Atangana-Baleanu integral operator namely ABK-fractional integrals to obtain two general integral identities complying second-order derivatives for a given function. Thus allowing us to derive new generalized Hermite-Hadamard type inequalities via ABK- fractional integrals. Moreover we give several new versions of Mid-point and Trapezoid type inequalities by employing Hölder, Young and Jensen inequalities.



    Convexity, with a very old background, has more significance for the theory of inequalities and other areas of mathematics such as probability theory, graph theory, functional analysis and numerical computing. The definition of a classical convex function is:

    Definition 1. (see, e.g., [38]) The mapping f:[α,β]RR, is said to be convex if the following condition holds

    f(λx+(1λ)y)λf(x)+(1λ)f(y) (1.1)

    for all x,y[α,β] and λ[0,1].

    Integral inequalities have a significant role in the expansion of all branches of mathematics. One of the most powerful of these integral inequalities is the Hermite-Hadamard inequality obtained for convex function as follows:

    Theorem 1. (see, e.g., [37]) Suppose that f:IRR is a convex function on the interval I with α<β. Then

    f(α+β2)1βαβαf(x)dxf(α)+f(β)2. (1.2)

    There are many new results in the literature concerning these type of inequalities. For example, Hermite-Hadamard inequalities for m-convex and (α,m)-convex functions have been proved by Bakula et al. and similar results have been provided by Kırmacı et al. for sconvex functions in [1,2]. In [3], by using the classical definition of convexity, Kavurmacı et al. obtained new Hermite-Hadamard type inequalities. In [4], new Hermite-Hadamard type inequalities involving fractional operators were presented by Set et al. In [5], Bayraktar proved new results related to the inequality for concave and rconvex functions. Akdemir et al. [10], introduced new Hermite-Hadamard type inequalities for GGconvex functions. Also, in 2002-2005, Guessab and Schmeisser, worked on convexity and Hermite-Hadamard inequality in [31,32,33].

    One of the concepts that have played a significant role in the growth of inequality theory in recent years is fractional analysis. Since fractional calculus was presented toward the end of the nineteenth century, the subject has become a quickly developing area and has discovered numerous applications in different research fields. Fractional integrals are the most commonly used concept in calculus analysis to obtain new generalizations, extensions, and versions of classical integral inequalities. Until today, many definitions of fractional integrals have been presented such as Riemann-Liouville, Weyl, Erdelyi-Kober, Hadamard, Katugampola, conformable fractional integrals et. al. and fractional versions of many inequalities such as Hermite-Hadamard, Simpson, Ostrowski, Grüss, Chebyshev inequalities for these fractional integrals have been obtained. For example, Minkowski and Hermite-Hadamard integral inequalities for Riemann-Liouville fractional integrals were obtained by Z. Dahmani in [12]. In 2016 to 2020, R. Almeida worked on Caputo fractional derivative in [6,7,8]. Khalil et al. and Abdeljawad worked on conformable fractional calculus in [9,13], respectively and Akdemir et al. [11] obtained new integral inequalities by motivated from these studies. Also, Chebyshev type inequalities for Katugampola fractional integral operators and fractional integral inequalities for co-ordinated MTconvex functions were obtained, respectively, in [15,16]. Recently, some interesting and new results related to fractional analysis have been presented to the literature by the authors as Sene and Yavuz (see the papers [34,35,36]). Readers who want to learn more about fractional derivatives and integral operators can look to the book by Samko et al. in [14].

    Now let's give the concept of fractional integral, which has been recently defined and attracted many researchers.

    Definition 2. Xpc(α,β) (cR), 1p denotes the space of all complex-valued Lebesgue measurable functions f for which ||f||Xpc<, where the norm ||.||Xpc is defined by

    ||f||Xpc=(βαϰcf(ϰ)pdϰϰ)1p  (1p<)

    and for p=

    ||f||Xc=esssupαϰβϰcf(ϰ).

    Definition 3. [17] Let [α,β]R be a finite interval. Then, the left and right side Katugampola fractional integrals of order ζ(>0) of fXpc(α,β) are defined by

    τIζα+f(x)=τ1ζΓ(ζ)xαϰτ1(xτϰτ)1ζf(ϰ)dϰ,  x>α

    and

    τIζβf(x)=τ1ζΓ(ζ)βxϰτ1(ϰτxτ)1ζf(ϰ)dϰ,  x<β,

    where τ>0, if the integrals exits.

    This fractional integrals are generalized the Riemann-Liouville and Hadamard fractional integrals which are defined as follows:

    Definition 4. Let fL[α,β]. The Riemann-Liouville integrals Jαα+f and Jαβf of order ζ>0 with α0 are defined by

    Jζα+f(x)=1Γ(ζ)xα(xt)ζ1f(t)dt,  x>α

    and

    Jζβf(x)=1Γ(ζ)βx(tx)ζ1f(t)dt,  x<β

    respectively where Γ(ζ)=0etuα1du. Here is J0α+f(x)=J0βf(x)=f(x).

    Recently, the Caputo-Fabrizio (CF) fractional integral has become a term that many researchers use as follows.

    Definition 5. [18] Let fH1(0,β), β>α, ζ[0,1] then, the definition of the new Caputo-Fabrizio fractional derivative is:

    CFDζf(ϰ)=B(ζ)1ζϰαf(s)exp[ζ(1ζ)(ϰs)]ds (2.1)

    where B(ζ) is a normalization function such that B(0)=B(1)=1.

    The related CF fractional derivative integral formula can be given as follows.

    Definition 6. [19] Suppose that fH1(0,β), β>α, ζ[0,1] then, the definition of the left and right side of CF fractional integral is:

    (CFαIζ)f(ϰ)=1ζB(ζ)f(ϰ)+ζB(ζ)ϰαf(y)dy,

    and

    (CFIζβ)f(ϰ)=1ζB(ζ)f(ϰ)+ζB(ζ)βϰf(y)dy

    where B(ζ) is normalization function.

    For recent results concerning these operators, we refer to [20,21].

    Fractional calculus is a branch of mathematical analysis that studies the various unique possibilities of describing real or complex numbers powers of the differentiation and integration operators. The fractional calculus has a significant part in different scientific fields because of its few applications in dynamic problems including hydrodynamics, signals, dynamics, fluid, viscoelastic theory, control theory, biology, computer networking, image processing, and lots of others. Fractional calculus is an important technique with significant applications in science and engineering. Atangana and Baleanu have recently introduced AB-fractional calculus and have drawn a large number of scientists in various scientific fields to investigate various topics. Atangana and Baleanu recommended a greatly improved variant of a derivative with no singular kernel. Their derivative depends on the notable generalized Mittag-Leffler function. It very well may be reviewed that, the Mittag-Leffler function has been acquainted giving a reaction to a traditional inquiry of complex analysis.

    The Atangana-Baleanu fractional integral operator is defined as follows.

    Definition 7. [22] The AB-fractional integral of a function fH1(α,β) is given by

    ABαIζϰ{f(ϰ)}=1ζB(ζ)f(ϰ)+ζB(ζ)Γ(ζ)ϰαf(y)(ϰy)ζ1dy

    where β>α,ζ[0,1].

    Similarly, [23], the authors gave opposite side of the AB-fractional integral operator is given by

    ABIζβ{f(ϰ)}=1ζB(ζ)f(ϰ)+ζB(ζ)Γ(ζ)βϰf(y)(yϰ)ζ1dy.

    For recent results related to AB-fractional operators see [24,25,26,27,40,41].

    In [30], a connection between the Atangana-Baleanu and the Riemann-Liouville fractional integrals of a function with respect to a monotone function with nonsingular kernel is given as follows:

    Lemma 1. For any L1 function f(x) on an interval [α,β] with x[α,β], the ζth Atangana-Baleanu (AB) fractional integral of a function f(x) with respect to another function can be represented as follows:

    ABψ(x)Iζα{f(ϰ)}=ζB(ζ)ψ(x)Jζα{f(ϰ)}+1ζB(ζ)f(ϰ),0<ζ<1 (2.2)

    where B(ζ) is as before and ψ(x) is an increasing positive monotone function on the interval [α,β] with ψ(x)L1(α,β).

    The left and right side of ABK-fractional integrals, the special case of (2.2), was given by Kashuri as:

    Definition 8. [28] Let [α,β]R be a finite interval. Then, the left and right side ABK-fractional integrals of order ζ(0,1) of fXpc(α,β) are defined by

    ABKτα+Iζϰ{f(ϰ)}=1ζB(ζ)f(ϰ)+τ1ζζB(ζ)Γ(ζ)ϰαyτ1(ϰτyτ)1ζf(y)dy,  ϰ>α0 (2.3)

    and

    ABKτβIζϰ{f(ϰ)}=1ζB(ζ)f(ϰ)+τ1ζζB(ζ)Γ(ζ)βϰyτ1(yτϰτ)1ζf(y)dy,  ϰ<β, (2.4)

    where τ>0 and B(ζ)>0 satisfies the property B(0)=B(1)=1.

    Since the normalization function B(ζ)>0 is positive, it directly implies that the fractional ABK-integral of a positive function is positive. Note that, if τ1 then we recover the AB-fractional integral. The Atangana-Baleanu fractional integral operator differs positively from other fractional integral operators with its time memory effect feature in the solutions of the real world problems and pysical phenomenon with the superior features of its kernel structure. In addition to preserving the features of this unique operator, ABK- fractional integral operators have started to be used widely in various disciplines in many application areas in the literature as they form a general form.

    Here, we aim to establish certain new generalizations of Hermite-Hadamard type integral inequalities for twice differentiable convex functions via the fractional integral operators (2.3) and (2.4). The inequalities presented here, being very general, are pointed out to be specialized to yield many new and known inequalities associated with some known fractional integrals.

    We start with the following lemma for ABK-fractional integral operators, which is the main motivation of this paper:

    Lemma 2. Let ζ(0,1) and τ>0 and f:[ατ,βτ]R be a twice differentiable mapping on (ατ,βτ) with 0α<β. Then the following equality is valid for ABK-fractional integral operators

    ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)=(ϰτατ)ζ+1τζ(ζ+1)B(ζ)Γ(ζ)f(ατ)+(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(1kτ)ζ+1kτ1f(kτϰτ+(1kτ)ατ)dk(βτϰτ)ζ+1τζ(ζ+1)B(ζ)Γ(ζ)f(βτ)+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(kτ)ζ+1kτ1f(kτβτ+(1kτ)ϰτ)dk

    where g(u)=uτ, ϰτ[ατ,βτ].

    Proof. Employing integration by parts gives

    (ϰτατ)ζ+1τζ1B(ζ)Γ(ζ)10(1kτ)ζkτ1f(kτϰτ+(1kτ)ατ)dk=(ϰτατ)ζ+1τζ1B(ζ)Γ(ζ)[(1kτ)ζ+1τ(ζ+1)f(kτϰτ+(1kτ)ατ)|10  +(ϰτατ)10(1kτ)ζ+1ζ+1kτ1f(kτϰτ+(1kτ)ατ)dk]=(ϰτατ)ζ+1τζ1B(ζ)Γ(ζ)[1τ(ζ+1)f(ατ)+(ϰτατ)ζ+110(1kτ)ζ+1kτ1f(kτϰτ+(1kτ)ατ)dk]. (3.1)

    Similarly

    (βτϰτ)ζ+1τζ1B(ζ)Γ(ζ)10(kτ)ζkτ1f(kτβτ+(1kτ)ϰτ)dk=(βτϰτ)ζ+1τζ1B(ζ)Γ(ζ)[(kτ)ζ+1τ(ζ+1)f(kτβτ+(1kτ)ϰτ)|10  (βτϰτ)ζ+110(kτ)ζ+1kτ1f(kτβτ+(1kτ)ϰτ)dk]=(βτϰτ)ζ+1τζ1B(ζ)Γ(ζ)[1τ(ζ+1)f(βτ)(βτϰτ)(ζ+1)10(kτ)ζ+1kτ1f(kτβτ+(1kτ)ϰτ)dk]. (3.2)

    By adding (3.1) and (3.2), we obtain

    ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)=(ϰτατ)ζ+1τζ(ζ+1)B(ζ)Γ(ζ)f(ατ)+(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(1kτ)ζ+1kτ1f(kτϰτ+(1kτ)ατ)dk(βτϰτ)ζ+1τζ(ζ+1)B(ζ)Γ(ζ)f(βτ)+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(kτ)ζ+1kτ1f(kτβτ+(1kτ)ϰτ)dk.

    So, the proof is completed.

    Remark 1. For τ1 in Lemma 2, we will get Lemma 2.1 in [29].

    Now, by using this important equality, we will establish the generalizations of the Hermite-Hadamard type inequalities via ABK-fractional integral operators.

    Theorem 2. Let ζ(0,1) and τ>0 and f:[ατ,βτ]R be a twice differentiable mapping on (ατ,βτ) with 0α<β and fXτc(ατ,βτ). If |f| is a convex on [ατ,βτ], then following inequality for ABK-fractional integral operators

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[|f(ϰτ)|τ(ζ+2)(ζ+3)+|f(ατ)|τ(ζ+3)]+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[|f(βτ)|τ(ζ+3)+|f(ϰτ)|τ(ζ+2)(ζ+3)]

    where g(u)=uτ, ϰτ[ατ,βτ], ζ(0,1), B(ζ) is normalization function.

    Proof. Taking into account the identity obtained in Lemma 2, we may write

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|=|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(1kτ)ζ+1kτ1f(kτϰτ+(1kτ)ατ)dk+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(kτ)ζ+1kτ1f(kτβτ+(1kτ)ϰτ)dk|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(1kτ)ζ+1kτ1|f(kτϰτ+(1kτ)ατ)|dk+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(kτ)ζ+1kτ1|f(kτβτ+(1kτ)ϰτ)|dk.

    By using convexity of |f|, it yields

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(1kτ)ζ+1kτ1|f(kτϰτ+(1kτ)ατ)|dk+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(kτ)ζ+1kτ1|f(kτβτ+(1kτ)ϰτ)|dk(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(1kτ)ζ+1kτ1[kτ|f(ϰτ)|+(1kτ)|f(ατ)|]dk+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(kτ)ζ+1kτ1[kτ|f(βτ)|+(1kτ)|f(ϰτ)|]dk=(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[|f(ϰτ)|τ(ζ+2)(ζ+3)+|f(ατ)|τ(ζ+3)]+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[|f(βτ)|τ(ζ+3)+|f(ϰτ)|τ(ζ+2)(ζ+3)]

    and the proof is completed.

    Remark 2. For τ1 in Theorem 2, we will get Theorem 2.1 in [29].

    Corollary 1. Set ϰτ=ατ+βτ2 in Theorem 2, we obtain

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)](βτατ)ζ+12ζ+1τζ(ζ+1)B(ζ)Γ(ζ)[f(ατ)f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[2|f(ατ+βτ2)|τ(ζ+2)(ζ+3)+|f(ατ)|τ(ζ+3)+|f(βτ)|τ(ζ+3)].

    Remark 3. For τ1 in Corollary 1, we will get Corollary 2.1 in [29].

    Theorem 3. Let ζ(0,1) and τ>0 and f:[ατ,βτ]R be a twice differentiable mapping on (ατ,βτ) with 0α<β and fXτc(ατ,βτ). If |f|s is a convex on [ατ,βτ], then the following inequality for ABK-fractional integral operators

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(1τ(ζr+r+1))1r(|f(ϰτ)|s+|f(ατ)|s2τ)1s+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(1τ(ζr+r+1))1r(|f(βτ)|s+|f(ϰτ)|s2τ)1s

    where g(u)=uτ, r1+s1=1, ϰτ[ατ,βτ], ζ(0,1), s>1, B(ζ) is normalization function.

    Proof. By using Lemma 2, we get

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(1kτ)ζ+1kτ1|f(kτϰτ+(1kτ)ατ)|dk+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(kτ)ζ+1kτ1|f(kτβτ+(1kτ)ϰτ)|dk.

    By applying Hölder inequality, we get

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(10(1kτ)(ζ+1)rkτ1dk)1r(10kτ1|f(kτϰτ+(1kτ)ατ)|sdk)1s]+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(10(kτ)(ζ+1)rkτ1dk)1r(10kτ1|f(kτβτ+(1kτ)ϰτ)|sdk)1s].

    Using convexity of |f|s, we get

    10kτ1|f(kτϰτ+(1kτ)ατ)|sdk10kτ1[kτ|f(ϰτ)|s+(1kτ)|f(ατ)|s]dk,10kτ1|f(kτβτ+(1kτ)ϰτ)|sdk10kτ1[kτ|f(βτ)|s+(1kτ)|f(ϰτ)|s]dk.

    After simplification, we get desired result.

    Remark 4. For τ1 in Theorem 3, we will get Theorem 2.2 in [29].

    Corollary 2. Set ϰτ=ατ+βτ2 in Theorem 3, we obtain

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)](βτατ)ζ+12ζ+1τζ(ζ+1)B(ζ)Γ(ζ)[f(ατ)f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(1τ(ζr+r+1))1r[(|f(ατ+βτ2)|s+|f(ατ)|s2τ)1s+(|f(βτ)|s+|f(ατ+βτ2)|s2τ)1s].

    Remark 5. For τ1 in Corollary 2, we will get Corollary 2.2 in [29].

    Theorem 4. Let ζ(0,1) and τ>0 and f:[ατ,βτ]R be a twice differentiable mapping on (ατ,βτ) with 0α<β and fXτc(ατ,βτ). If |f|s is a convex on [ατ,βτ], then the following inequality for ABK-fractional integral operators

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(1τr(ζr+r+1)+|f(ϰτ)|s+|f(ατ)|s2τs)+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(1τr(ζr+r+1)+|f(βτ)|s+|f(ϰτ)|s2τs)

    where g(u)=uτ, r1+s1=1, ϰτ[ατ,βτ], ζ(0,1), s>1, B(ζ) is normalization function.

    Proof. By using Lemma 2, we have

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(1kτ)ζ+1kτ1|f(kτϰτ+(1kτ)ατ)|dk+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(kτ)ζ+1kτ1|f(kτβτ+(1kτ)ϰτ)|dk.

    By using the Young inequality as xy1rxr+1sys

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[1r10(1kτ)(ζ+1)rkτ1dk+1s10kτ1|f(kτϰτ+(1kτ)ατ)|sdk]+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[1r10(kτ)(ζ+1)rkτ1dk+1s10kτ1|f(kτβτ+(1kτ)ϰτ)|sdk].

    Now by employing convexity of |f|s along with simple computations give the required result.

    Remark 6. For τ1 in Theorem 4, we will get Theorem 2.3 in [29].

    Corollary 3. Set ϰτ=ατ+βτ2 in Theorem 4, we obtain

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)](βτατ)ζ+12ζ+1τζ(ζ+1)B(ζ)Γ(ζ)[f(ατ)f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(2τr(ζr+r+1)+2|f(ατ+βτ2)|s+|f(ατ)|s+|f(βτ)|s2τs).

    Remark 7. For τ1 in Corollary 3, we will get Corollary 2.3 in [29].

    Theorem 5. Let ζ(0,1) and τ>0 and f:[ατ,βτ]R be a twice differentiable mapping on (ατ,βτ) with 0α<β and fXτc(ατ,βτ). If |f|s is a convex on [ατ,βτ], then the following inequality for ABK-fractional integral operators

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(1τ(ζ+2))11s(|f(ϰτ)|sτ(ζ+2)(ζ+3)+|f(ατ)|sτ(ζ+3))1s]+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(1τ(ζ+2))11s(|f(βτ)|sτ(ζ+3)+|f(ϰτ)|sτ(ζ+2)(ζ+3))1s]

    where g(u)=uτ, ϰτ[ατ,βτ], ζ(0,1), s1, B(ζ) is normalization function.

    Proof. By using Lemma 2, we get

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(1kτ)ζ+1kτ1|f(kτϰτ+(1kτ)ατ)|dk+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(kτ)ζ+1kτ1|f(kτβτ+(1kτ)ϰτ)|dk.

    By applying power mean inequality, we get

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(10(1kτ)ζ+1kτ1dk)11s(10(1kτ)ζ+1kτ1|f(kτϰτ+(1kτ)ατ)|sdk)1s]+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(10(kτ)ζ+1kτ1dk)11s(10(kτ)ζ+1kτ1|f(kτβτ+(1kτ)ϰτ)|sdk)1s].

    By using convexity of |f|s, we obtain

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(10(1kτ)ζ+1kτ1dk)11s(10(1kτ)ζ+1kτ1[kτ|f(ϰτ)|s+(1kτ)|f(ατ)|s]dk)1s]+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(10(kτ)ζ+1kτ1dk)11s(10(kτ)ζ+1kτ1[kτ|f(βτ)|s+(1kτ)|f(ϰτ)|s]dk)1s]=(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(1τ(ζ+2))11s(|f(ϰτ)|sτ(ζ+2)(ζ+3)+|f(ατ)|sτ(ζ+3))1s]+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(1τ(ζ+2))11s(|f(βτ)|sτ(ζ+3)+|f(ϰτ)|sτ(ζ+2)(ζ+3))1s].

    So, the proof is completed.

    Remark 8. For τ1 in Theorem 5, we will get Theorem 2.4 in [29].

    Corollary 4. Set ϰτ=ατ+βτ2 in Theorem 5, we obtain

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)](βτατ)ζ+12ζ+1τζ(ζ+1)B(ζ)Γ(ζ)[f(ατ)f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(1τ(ζ+2))11s[(|f(ατ+βτ2)|sτ(ζ+2)(ζ+3)+|f(ατ)|sτ(ζ+3))1s+(|f(βτ)|sτ(ζ+3)+|f(ατ+βτ2)|sτ(ζ+2)(ζ+3))1s].

    Remark 9. For τ1 in Corollary 4, we will get Corollary 2.4 in [29].

    Theorem 6. Let ζ(0,1) and τ>0 and f:[ατ,βτ]R be twice differentiable mapping on (ατ,βτ) with 0α<β and fXτc(ατ,βτ). If |f|s is a concave for s>1, then we have

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(1τ(ζ+2))|f(1ζ+3ϰτ+ζ+2ζ+3ατ)|+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(1τ(ζ+2))|f(ζ+2ζ+3βτ+1ζ+3ϰτ)|

    where g(u)=uτ, ϰτ[ατ,βτ], ζ(0,1).

    Proof. Using Jensen integral inequality and Lemma 2, we have

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(1kτ)ζ+1kτ1|f(kτϰτ+(1kτ)ατ)|dk+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)10(kτ)ζ+1kτ1|f(kτβτ+(1kτ)ϰτ)|dk(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(10(1kτ)ζ+1kτ1dk)|f(10(1kτ)ζ+1kτ1(kτϰτ+(1kτ)ατ)dk10(1kτ)ζ+1kτ1dk)|+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(10(kτ)ζ+1kτ1dk)|f(10(kτ)ζ+1kτ1(kτβτ+(1kτ)ϰτ)dk10(kτ)ζ+1kτ1dk)|.

    By computing the above integrals we have

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(1τ(ζ+2))|f(1ζ+3ϰτ+ζ+2ζ+3ατ)|+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(1τ(ζ+2))|f(ζ+2ζ+3βτ+1ζ+3ϰτ)|.

    Remark 10. For τ1 in Theorem 6, we will get Theorem 2.5 in [29].

    Corollary 5. Set ϰτ=ατ+βτ2 in Theorem 6, we obtain

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)](βτατ)ζ+12ζ+1τζ(ζ+1)B(ζ)Γ(ζ)[f(ατ)f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(1τ(ζ+2))[|f(ατ+βτ2(ζ+3)+ζ+2ζ+3ατ)|+|f(ζ+2ζ+3βτ+ατ+βτ2(ζ+3))|].

    Remark 11. For τ1 in Corollary 5, we will get Corollary 2.5 in [29].

    Theorem 7. Let ζ(0,1) and τ>0 and f:[ατ,βτ]R be a twice differentiable mapping on (ατ,βτ) with 0α<β and fXτc(ατ,βτ). If |f|s is a concave mapping, we have

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(1τ(ζr+r+1))1r|f(ατ+ϰτ2τ)|+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(1τ(ζr+r+1))1r|f(βτ+ϰτ2τ)|

    where g(u)=uτ, r1+s1=1, ϰτ[ατ,βτ], ζ(0,1), s>1.

    Proof. Taking into account Lemma 2 along with Hölder's inequality yields

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(10(1kτ)(ζ+1)rkτ1dk)1r(10kτ1|f(kτϰτ+(1kτ)ατ)|sdk)1s+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(10(kτ)(ζ+1)rkτ1dk)1r(10kτ1|f(kτβτ+(1kτ)ϰτ)|sdk)1s.

    Now by employing concavity of |f|s under the essence of Jensen's integral inequality gives

    10kτ1|f(kτϰτ+(1kτ)ατ)|sdk=10kokτ1|f(kτϰτ+(1kτ)ατ)|sdk(10k0dk)|f(10kτ1(kτϰτ+(1kτ)ατ)dk10k0dk)|s=|f(ατ+ϰτ2τ)|s

    Similarly

    10kτ1|f(kτβτ+(1kτ)ϰτ)|sdk|f(βτ+ϰτ2τ)|s.

    So, we obtain

    |ABKτα+Iζϰ{(fg)(ϰ)}+ABKτβIζϰ{(fg)(ϰ)}(ϰτατ)ζf(ατ)+(βτϰτ)ζf(βτ)τζB(ζ)Γ(ζ)(ϰτατ)ζ+1f(ατ)(βτϰτ)ζ+1f(βτ)τζ(ζ+1)B(ζ)Γ(ζ)2(1ζ)f(ϰτ)B(ζ)|(ϰτατ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(1τ(ζr+r+1))1r|f(ατ+ϰτ2τ)|+(βτϰτ)ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(1τ(ζr+r+1))1r|f(βτ+ϰτ2τ)|.

    Remark 12. For τ1 in Theorem 7, we will get Theorem 2.6 in [29].

    Corollary 6. Set ϰτ=ατ+βτ2 in Theorem 7, we obtain

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)](βτατ)ζ+12ζ+1τζ(ζ+1)B(ζ)Γ(ζ)[f(ατ)f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)(1τ(ζr+r+1))1r[|f(3ατ+βτ4τ)|+|f(3βτ+ατ4τ)|].

    Remark 13. For τ1 in Corollary 6, we will get Corollary 2.6 in [29].

    Now, we will utilize another change of Lemma 2 by considering an alternate technique to eliminate the first derivative of the function. Substituting ϰτ=ατ+βτ2 in Lemma 2 lead us to new results.

    Lemma 3. Let ζ(0,1) and τ>0 and f:[ατ,βτ]R be a twice differentiable mapping on (ατ,βτ) with 0α<β. Then the following equality for ABK-fractional integral operators

    ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)=(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[10((1kτ)ζ+11)kτ1f(kτατ+βτ2+(1kτ)ατ)dk+10((kτ)ζ+11)kτ1f(kτβτ+(1kτ)ατ+βτ2)dk].

    where ζ(0,1),ϰτ[ατ,βτ], and Γ(.) is Gamma function.

    Proof. By setting ϰτ=ατ+βτ2 in Lemma 2, we get

    ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)=(βτατ)ζ+12ζ+1τζ(ζ+1)B(ζ)Γ(ζ)[f(ατ)f(βτ)]+(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[10(1kτ)ζ+1kτ1f(kτατ+βτ2+(1kτ)ατ)dk+10(kτ)ζ+1kτ1f(kτβτ+(1kτ)ατ+βτ2)dk]. (4.1)

    Then, we can write

    [f(ατ)f(βτ)]=βτατf(x)dx=[βτατ210τkτ1f(kτατ+βτ2+(1kτ)ατ)dk+βτατ210τkτ1f(kτβτ+(1kτ)ατ+βτ2)dk]. (4.2)

    Putting equality (4.2) in (4.1), we get

    ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)=(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[10((1kτ)ζ+11)kτ1f(kτατ+βτ2+(1kτ)ατ)dk+10((kτ)ζ+11)kτ1f(kτβτ+(1kτ)ατ+βτ2)dk].

    Remark 14. For τ1 in Lemma 3, we will get Lemma 3.1 in [29].

    Theorem 8. Let ζ(0,1) and τ>0 and f:[ατ,βτ]R be a twice differentiable mapping on (ατ,βτ) with 0α<β and fXτc(ατ,βτ). If |f| is a convex on [ατ,βτ], then the following inequality for ABK-fractional integral operators

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(|f(ατ)|+|f(βτ)|)(1τ(ζ+3)12τ)+2|f(ατ+βτ2)|(1τ(ζ+2)(ζ+3)12τ)]

    where ϰτ[ατ,βτ], ζ(0,1).

    Proof. Employing Lemma 3 gives

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|=|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[10((1kτ)ζ+11)kτ1f(kτατ+βτ2+(1kτ)ατ)dk+10((kτ)ζ+11)kτ1f(kτβτ+(1kτ)ατ+βτ2)dk]|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[10((1kτ)ζ+11)kτ1|f(kτατ+βτ2+(1kτ)ατ)|dk+10((kτ)ζ+11)kτ1|f(kτβτ+(1kτ)ατ+βτ2)|dk].

    By using convexity of |f|, we get

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[10((1kτ)ζ+11)kτ1(kτ|f(ατ+βτ2)|+(1kτ)|f(ατ)|)dk+10((kτ)ζ+11)kτ1(kτ|f(βτ)|+(1kτ)|f(ατ+βτ2)|)dk]=(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(|f(ατ)|+|f(βτ)|)(1τ(ζ+3)12τ)+2|f(ατ+βτ2)|(1τ(ζ+2)(ζ+3)12τ)]

    and the proof is completed.

    Remark 15. For τ1 in Theorem 8, we will get Theorem 3.1 in [29].

    Theorem 9. Let ζ(0,1) and τ>0 and f:[ατ,βτ]R be a twice differentiable mapping on (ατ,βτ) with 0α<β and fXτc(ατ,βτ). If |f|s is a convex on [ατ,βτ], then the following inequality for ABK-fractional integral operators

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)((1τ(ζr+r+1))1r1)×[(|f(ατ+βτ2)|s+|f(ατ)|s2τ)1s+(|f(βτ)|s+|f(ατ+βτ2)|s2τ)1s]

    where r1+s1=1, ϰτ[ατ,βτ], ζ(0,1), s>1.

    Proof. Utilizing Lemma 3, we have

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[10((1kτ)ζ+11)kτ1|f(kτατ+βτ2+(1kτ)ατ)|dk+10((kτ)ζ+11)kτ1|f(kτβτ+(1kτ)ατ+βτ2)|dk]=(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[10(1kτ)ζ+1kτ1|f(kτατ+βτ2+(1kτ)ατ)|dk10kτ1|f(kτατ+βτ2+(1kτ)ατ)|dk+10(kτ)ζ+1kτ1|f(kτβτ+(1kτ)ατ+βτ2)|dk10kτ1|f(kτβτ+(1kτ)ατ+βτ2)|dk].

    By applying Hölder inequality, we get

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(10(1kτ)(ζ+1)rkτ1dk)1r(10kτ1|f(kτατ+βτ2+(1kτ)ατ)|sdk)1s(10(k0)rdk)1r(10kτ1|f(kτατ+βτ2+(1kτ)ατ)|sdk)1s+(10(kτ)(ζ+1)rkτ1dk)1r(10kτ1|f(kτβτ+(1kτ)ατ+βτ2)|sdk)1s(10(k0)rdk)1r(10kτ1|f(kτβτ+(1kτ)ατ+βτ2)|sdk)1s].

    By using convexity of |f|s, we obtain

    10kτ1|f(kτατ+βτ2+(1kτ)ατ)|sdk10kτ1[kτ|f(ατ+βτ2)|s+(1kτ)|f(ατ)|s]dk,10kτ1|f(kτβτ+(1kτ)ατ+βτ2)|sdk10kτ1[kτ|f(βτ)|q+(1kτ)|f(ατ+βτ2)|s]dk.

    By computation the integrals that is in the above inequalities, we get desired result.

    Remark 16. For τ1 in Theorem 9, we will get Theorem 3.2 in [29].

    Theorem 10. Let ζ(0,1) and τ>0 and f:[ατ,βτ]R be a twice differentiable mapping on (ατ,βτ) with 0α<β and fXτc(ατ,βτ). If |f|s is a convex on [ατ,βτ], then the following inequality for ABK-fractional integral operators

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[2τr(ζr+r+1)2τr]

    where r1+s1=1, ϰτ[ατ,βτ], ζ(0,1), s>1, B(ζ) is normalization function.

    Proof. By using Lemma 3, we get

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[10((1kτ)ζ+11)kτ1|f(kτατ+βτ2+(1kτ)ατ)|dk+10((kτ)ζ+11)kτ1|f(kτβτ+(1kτ)ατ+βτ2)|dk]=(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[10(1kτ)ζ+1kτ1|f(kτατ+βτ2+(1kτ)ατ)|dk10kτ1|f(kτατ+βτ2+(1kτ)ατ)|dk+10(kτ)ζ+1kτ1|f(kτβτ+(1kτ)ατ+βτ2)|dk10kτ1|f(kτβτ+(1kτ)ατ+βτ2)|dk].

    By using the Young inequality as xy1rxr+1sys

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[1r10(1kτ)(ζ+1)rkτ1dk+1s10kτ1|f(kτατ+βτ2+(1kτ)ατ)|sdk1r10(k0)rdk1s10kτ1|f(kτατ+βτ2+(1kτ)ατ)|sdk+1r10(kτ)(ζ+1)rkτ1dk+1s10kτ1|f(kτβτ+(1kτ)ατ+βτ2)|sdk1r10(k0)rdk1s10kτ1|f(kτβτ+(1kτ)ατ+βτ2)|sdk].

    By using convexity of |f|s and by a simple computation, we have the desired result.

    Remark 17. For τ1 in Theorem 10, we will get Theorem 3.3 in [29].

    Theorem 11. Let ζ(0,1) and τ>0 and f:[ατ,βτ]R be a twice differentiable mapping on (ατ,βτ) with 0α<β and fXτc(ατ,βτ). If |f|s is a convex on [ατ,βτ], then the following inequality for ABK-fractional integral operators

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(1τ(ζ+2))11s(|f(ατ+βτ2)|sτ(ζ+2)(ζ+3)+|f(ατ)|sτ(ζ+3))1s+(1τ(ζ+2))11s(|f(βτ)|sτ(ζ+3)+|f(ατ+βτ2)|sτ(ζ+2)(ζ+3))1s(|f(ατ+βτ2)|s+|f(ατ)|s2τ)1s(|f(βτ)|s+|f(ατ+βτ2)|s2τ)1s]

    where ϰτ[ατ,βτ], ζ(0,1), s1, B(ζ) is normalization function.

    Proof. By using Lemma 3, we get

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[10((1kτ)ζ+11)kτ1|f(kτατ+βτ2+(1kτ)ατ)|dk+10((kτ)ζ+11)kτ1|f(kτβτ+(1kτ)ατ+βτ2)|dk]=(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[10(1kτ)ζ+1kτ1|f(kτατ+βτ2+(1kτ)ατ)|dk10kτ1|f(kτατ+βτ2+(1kτ)ατ)|dk+10(kτ)ζ+1kτ1|f(kτβτ+(1kτ)ατ+βτ2)|dk10kτ1|f(kτβτ+(1kτ)ατ+βτ2)|dk].

    By applying power mean inequality, we get

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(10(1kτ)ζ+1kτ1dk)11s(10(1kτ)ζ+1kτ1|f(kτατ+βτ2+(1kτ)ατ)|sdk)1s(10k0dk)11s(10kτ1|f(kτατ+βτ2+(1kτ)ατ)|sdk)1s+(10(kτ)ζ+1kτ1dk)11s(10(kτ)ζ+1kτ1|f(kτβτ+(1kτ)ατ+βτ2)|sdk)1s(10k0dk)11s(10kτ1|f(kτβτ+(1kτ)ατ+βτ2)|sdk)1s].

    By using convexity of |f|s and after further simplifications, we have the desired result.

    Remark 18. For τ1 in Theorem 11, we will get Theorem 3.4 in [29].

    Theorem 12. Let ζ(0,1) and τ>0 and f:[ατ,βτ]R be a twice differentiable mapping on (ατ,βτ) with 0α<β and fXτc(ατ,βτ). If |f|s is a concave for s>1, then we have

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(1τ(ζ+2))|f(ατ+βτ2(ζ+3)+ζ+2ζ+3ατ)|+(1τ(ζ+2))|f(ζ+2ζ+3βτ+ατ+βτ2(ζ+3))||f(3ατ+βτ4τ)||f(3βτ+ατ4τ)|]

    where ϰτ[ατ,βτ], ζ(0,1).

    Proof. Using Lemma 3 and the Jensen integral inequality, we get

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[10((1kτ)ζ+11)kτ1|f(kτατ+βτ2+(1kτ)ατ)|dk+10((kτ)ζ+11)kτ1|f(kτβτ+(1kτ)ατ+βτ2)|dk]=(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[10(1kτ)ζ+1kτ1|f(kτατ+βτ2+(1kτ)ατ)|dk10kτ1|f(kτατ+βτ2+(1kτ)ατ)|dk+10(kτ)ζ+1kτ1|f(kτβτ+(1kτ)ατ+βτ2)|dk10kτ1|f(kτβτ+(1kτ)ατ+βτ2)|dk](βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(10(1kτ)ζ+1kτ1dk)|f(10(1kτ)ζ+1kτ1(kτατ+βτ2+(1kτ)ατ)dk10(1kτ)ζ+1kτ1dk)|(10k0dk)|f(10k0kτ1(kτατ+βτ2+(1kτ)ατ)dk10k0dk)|+(10(kτ)ζ+1kτ1dk)|f(10(kτ)ζ+1kτ1(kτβτ+(1kτ)ατ+βτ2)dk10(kτ)ζ+1kτ1dk)|(10k0dk)|f(10k0kτ1(kτβτ+(1kτ)ατ+βτ2)dk10k0dk)|].

    By computing the above integrals we have

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(1τ(ζ+2))|f(ατ+βτ2(ζ+3)+ζ+2ζ+3ατ)|+(1τ(ζ+2))|f(ζ+2ζ+3βτ+ατ+βτ2(ζ+3))||f(3ατ+βτ4τ)||f(3βτ+ατ4τ)|].

    Remark 19. For τ1 in Theorem 12, we will get Theorem 3.5 in [29].

    Theorem 13. Let ζ(0,1) and τ>0 and f:[ατ,βτ]R be a twice differentiable mapping on (ατ,βτ) with 0α<β and fXτc(ατ,βτ). If |f|s is a concave mapping, then

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(1τ(ζr+r+1))1r|f(3ατ+βτ4τ)||f(3ατ+βτ4τ)|+(1τ(ζr+r+1))1r|f(3βτ+ατ4τ)||f(3βτ+ατ4τ)|].

    where r1+s1=1, ϰτ[ατ,βτ], ζ(0,1), s>1.

    Proof. From Lemma 3 and Hölder integral inequality, we get

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(10(1kτ)(ζ+1)rkτ1dk)1r(10kτ1|f(kτατ+βτ2+(1kτ)ατ)|sdk)1s(10(k0)rdk)1r(10kτ1|f(kτατ+βτ2+(1kτ)ατ)|sdk)1s+(10(kτ)(ζ+1)rkτ1dk)1r(10kτ1|f(kτβτ+(1kτ)ατ+βτ2)|sdk)1s(10(k0)rdk)1r(10kτ1|f(kτβτ+(1kτ)ατ+βτ2)|sdk)1s].

    By using concavity of |f|s and Jensen integral inequality, we get

    10kτ1|f(kτατ+βτ2+(1kτ)ατ)|sdk=10k0kτ1|f(kτατ+βτ2+(1kτ)ατ)|sdk(10k0dk)|f(10kτ1(kτατ+βτ2+(1kτ)ατ)dk10k0dk)|s=|f(3ατ+βτ4τ)|s.

    Similarly

    10kτ1|f(kτβτ+(1kτ)ϰτ)|sdk|f(3βτ+ατ4τ)|s.

    So, we obtain

    |ABKτα+Iζ(ατ+βτ2)1/τf(ατ+βτ2)+ABKτβIζ(ατ+βτ2)1/τf(ατ+βτ2)(βτατ)ζ2ζτζB(ζ)Γ(ζ)[f(ατ)+f(βτ)]2(1ζ)f(ατ+βτ2)B(ζ)|(βτατ)ζ+22ζ+2τζ1(ζ+1)B(ζ)Γ(ζ)[(1τ(ζr+r+1))1r|f(3ατ+βτ4τ)||f(3ατ+βτ4τ)|+(1τ(ζr+r+1))1r|f(3βτ+ατ4τ)||f(3βτ+ατ4τ)|].

    Remark 20. For τ1 in Theorem 13, we will get Theorem 3.6 in [29].

    In this paper, we formulate two new integral identities with the second-order derivatives involving ABK-fractional integrals. We obtain generalized fractional Mid-point and Trapezoid type inequalities via ABK-fractional integrals with the help of these identities. Some related bounds are presented by using H¨older's, Young, power-mean, and Jensen inequalities. Several applications of main findings can be provided for the special means of real numbers by choosing α=1 and ρ=1.

    Recently, researchers working in the field of fractional analysis have defined some integral operators that include new and general forms. The usefulness and effectiveness of these new operators in the solutions of the real world problems have been tried to be demonstrated with the help of various applications and simulations. The interested readers can find several illustrative simulations related to the ABKfractional integral operators in [39]. Also, in that study, the authors have established some new integral inequalities via ABK fractional integral operators that have been provided general forms of the earlier studies.

    Our main findings will be motivated to obtain new extensions by using different kinds of convex functions for the interested researchers.

    We would like to express our gratitude to the editor and reviewers who helped to improve this article and reach its final form. Also, we would like to thank to Prof. Dr. Ahmet Ocak Akdemir for his contributions in the revision process of this article. The authors T. Abdeljawad and W. Shatanawi would like to thank Prince Sultan University for funding this work through the TAS Lab.



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