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Research article

Hermite-Jensen-Mercer type inequalities via Ψ-Riemann-Liouville k-fractional integrals

  • Received: 25 April 2020 Accepted: 11 June 2020 Published: 15 June 2020
  • MSC : 26A33, 26A51, 26D07, 26D10, 26D15

  • Integral inequalities involving various fractional integral operators are used to solve many fractional differential equations. In this paper, authors prove some Hermite-Jensen-Mercer type inequalities using Ψ-Riemann-Liouville k-Fractional integrals via convex functions. We established some new Ψ-Riemann-Liouville k-Fractional integral inequalities. We also give Ψ-Riemann-Liouville k-Fractional integrals identities for differentiable mapping, and these will be used to derive estimates for some fractional Hermite-Jensen-Mercer type inequalities. Some known results are recaptured from our results as special cases.

    Citation: Saad Ihsan Butt, Artion Kashuri, Muhammad Umar, Adnan Aslam, Wei Gao. Hermite-Jensen-Mercer type inequalities via Ψ-Riemann-Liouville k-fractional integrals[J]. AIMS Mathematics, 2020, 5(5): 5193-5220. doi: 10.3934/math.2020334

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  • Integral inequalities involving various fractional integral operators are used to solve many fractional differential equations. In this paper, authors prove some Hermite-Jensen-Mercer type inequalities using Ψ-Riemann-Liouville k-Fractional integrals via convex functions. We established some new Ψ-Riemann-Liouville k-Fractional integral inequalities. We also give Ψ-Riemann-Liouville k-Fractional integrals identities for differentiable mapping, and these will be used to derive estimates for some fractional Hermite-Jensen-Mercer type inequalities. Some known results are recaptured from our results as special cases.


    Inequalities have always proved to be useful in establishing mathematical models and their solutions in almost all branches of applied sciences, in particular, in physics and engineering. Convexity plays a very important role in the optimization of solutions of mathematical problems. Theory of convex functions has great importance in various fields of pure and applied sciences. It is known that theory of convex functions is closely related to theory of inequalities. It is now become a trending aspect of mathematical research to generalize classical known results via fractional integral operator.

    Fractional calculus which is calculus of integrals and derivatives of any arbitrary real or complex order has gained remarkable popularity and importance during the last four decades or so, due mainly to its demonstrated applications in diverse and widespread fields ranging from natural sciences to social sciences, see [4,5,6,7,15] and references therein. Many authors have established a variety of inequalities for those fractional integral and derivative operators, some of which have turned out to be useful in analyzing solutions of certain fractional integral and differential equations.

    Definition 1.1. The function Υ:[a,b]R is said to be convex, if we have

    Υ(λy1+(1λ)y2)λΥ(y1)+(1λ)Υ(y2)

    for all y1,y2[a,b] and λ[0,1].

    In 1883, Hermite–Hadamard's (H–H) inequality has been considered the most useful inequality in mathematical analysis. It is also known as classical H-H inequality.

    The Hermite–Hadamard inequality assert that, if Υ:JRR is a convex function in J and 1,2J, where 1<2, then

    Υ(1+22)12121Υ(λ)dλΥ(1)+Υ(2)2.

    Let 0<y1y2yn and let μ=(μ1,μ2,,μn) be non-negative weights such that ni=1 μi=1.

    The famous Jensen inequality, see [24] in the literature states that, if Υ is convex function on the interval [1,2], then

    Υ(ni=1μiyi)(ni=1μiΥ(yi)) (1.1)

    for all yi[1,2] and μi[0,1], (i=1,2,,n).

    In (2003) Mercer gave a variant of Jensen's inequality, see [19] as:

    Theorem 1.2. If Υ is a convex function on [1,2], then

    Υ(1+2ni=1μi yi)Υ(1)+Υ(2)ni=1μiΥ(yi) (1.2)

    for all yi[1,2] and μi[0,1], (i=1,2,,n).

    In [18], Matković et al. have proved Jensen's inequality of Mercer's type for operators with applications in 2006. Later, in 2009 Mercer's result was generalized to higher dimensions by M. Niezgoda [22]. In recent years, notable contributions have been made on Jensen–Mercer's type inequality. In 2014 M. Kian gave concept of Jensen inequality for superquadratic functions [13]. Further, in 2016 to 2018 E. Anjidani worked on Reverse Jensen–Mercer type operator inequalities and Jensen–Mercer operator inequalities for superquadratic functions [2,3]. M. M. Ali and A. R. Khan generalized integral Mercer's inequality and integral means in 2019 [1]. Some improvements have been made for Jensen–Mercer–Type inequalities by H. R. Moradi and S. Furuichi in 2019 [20].

    Now we give necessary definitions of fractional calculus theory which is used throughout this paper.

    Definition 1.3. [14] Let (1,2)(1<2) and α>0. Also let Ψ be an increasing and positive monotone function on (1,2], having a continuous derivative Ψ on (1,2). Then the left-sided and right-sided Ψ-Riemann-Liouville Fractional integrals of a function Υ with respect to another function Ψ on [1,2] are defined as follows:

    (Iα:Ψ1+)Υ(y1)=1Γ(α)y11Ψ(λ)(Ψ(y1)Ψ(λ))α1Υ(λ)dλ,          1<y1 (1.3)

    and

    (Iα:Ψ2)Υ(y1)=1Γ(α)2y1Ψ(λ)(Ψ(λ)Ψ(y1))α1Υ(λ)dλ,         y1<2, (1.4)

    respectively.

    Definition 1.4. [8] Diaz and Parigun have defined the k–gamma function Γk(), a generalization of the classical gamma function, which is given by the following formula

    Γk(y)=limn+n!kn(nk)yk1(y)n,k,k>0.

    It is shown that Mellin transform of the exponential function etkk is the k–gamma function clearly given by:

    Γk(α)=0etkktα1dt.

    Obviously, Γk(y+k)=yΓk(y), Γ(y)=limk1Γk(y) and Γk(y)=kyk1Γ(yk).

    Many researchers have generalized the classical and fractional operators by introducing a parameter k>0 about a decade ago. Mubeen and Habibullah [21] used special k-functions theory in fractional calculus for the first time in literature in the form of k–Riemann–Liouville integral. Recently, many researchers are presenting new fractional differential and integral operators and they generalized by iteration procedure and by introducing a new parameter k>0. They also found relationships of these generalized fractional operators with existing fractional and classical operators under the special values of the parameter k. Many k–fractional operators, their properties, related identities, and inequalities are proved during the past years.

    In [16], the author define a more general form of Riemann–Liouville k–fractional integrals with respect to an increasing function and use them to obtain Ostrowski-type inequalities. Utilizing a simple inequality via an increasing function and assumptions of Ostrowski inequality several fractional integral inequalities are obtained. These results provide the Ostrowski type inequalities for Riemann–Liouville fractional integrals which are published in [9].

    Definition 1.5. [16] Let (1,2)(1<2) and α,k>0. Also let Ψ be an increasing and positive monotone function on [1,2], having a continuous derivative Ψ on (1,2). Then the left-sided and right-sided Ψ–Riemann–Liouville k–Fractional integrals of a function Υ with respect to another function Ψ on [1,2] are defined as follows:

    (kIα:Ψ1+)Υ(y1)=1kΓk(α)y11Ψ(λ)(Ψ(y1)Ψ(λ))αk1Υ(λ)dλ,       1<y1 (1.5)

    and

    (kIα:Ψ2)Υ(y1)=1kΓk(α)2y1Ψ(λ)(Ψ(λ)Ψ(y1))αk1Υ(λ)dλ,       y1<2, (1.6)

    respectively.

    In this article, by using the Jensen–Mercer's inequality, we proved Hermite–Hadamard's inequalities for fractional integrals and we established some new Ψ–Riemann–Liouville k–Fractional integrals connected with the left and right sides of Hermite–Hadamard type inequalities for differentiable mappings whose derivatives in absolute value are convex. From our results some known results will be taken. At the end, a briefly conclusion is given as well.

    Throughout the paper, we need the following assumption:

    (A1): Let 01<2, Υ:[1,2] be a positive function and ΥL1[1,2]. Also suppose that Υ is a convex function on [1,2], Ψ() is an increasing and positive monotone function on (1,2], having a continuous derivative Ψ on (1,2) and α,k>0.

    Theorem 2.1. Let (A1) satisfied, then the following fractional integral inequalities hold:

    Υ(1+2y1+y22)[Υ(1)+Υ(2)]Γk(α+k)2(y2y1)αk×{(kIα:ΨΨ1(y1)+)(ΥΨ)(Ψ1(y2))+(kIα:ΨΨ1(y2))(ΥΨ)(Ψ1(y1))}[Υ(1)+Υ(2)]Υ(y1+y22) (2.1)

    and

    Υ(1+2y1+y22)Γk(α+k)2(y2y1)αk[(kIα:ΨΨ1(1+2y1)+)(ΥΨ)(Ψ1(1+2y2))+(kIα:ΨΨ1(1+2y2))(ΥΨ)(Ψ1(1+2y1))]Υ(1+2y1)+Υ(1+2y2)2Υ(1)+Υ(2)Υ(y1+y22) (2.2)

    for all y1,y2[1,2] and Γk() is the k-gamma function.

    Proof. Using the Jensen–Mercer inequality, we have

    Υ(1+2u+v2)Υ(1)+Υ(2)Υ(u)+Υ(v)2

    for all u,v[1,2].

    Now by change of variables u=λy1+(1λ)y2 and v=(1λ)y1+λy2,y1,y2[1,2] and λ[0,1], we get

    Υ(1+1y1+y22)Υ(1)+Υ(2)Υ(λy1+(1λ)y2)+Υ((1λ)y1+λy2)2.

    Multiplying both sides of above inequality by λαk1 and then integrating the resulting inequality with respect to λ over [0,1], we obtain

    kαΥ(1+2y1+y22)kα{Υ(1)+Υ(2)}            12{10λαk1(Υ(λy1+(1λ)y2)+Υ((1λ)y1+λy2))dλ},

    where

    α2k{10λαk1(Υ(λy1+(1λ)y2)+Υ((1λ)y1+λy2))dλ}=α2k10λαk1Υ(λy1+(1λ)y2)dλ+α2k10λαk1Υ((1λ)y1+λy2)dλ=α2kΨ1(y2)Ψ1(y1)(y2Ψ(γ)y2y1)αk1Υ(Ψ(γ))Ψ(γ)y2y1dγ+α2kΨ1(y2)Ψ1(y1)(Ψ(γ)y1y2y1)αk1Υ(Ψ(γ))Ψ(γ)y2y1dγ.

    So, final form will be of this type

    α2k{10λαk1(Υ(λy1+(1λ)y2)+Υ((1λ)y1+λy2))dλ}=Γk(α+k)2(y2y1)αk{(kIα:ΨΨ1(y1)+)(ΥΨ)(Ψ1(y2))+(kIα:ΨΨ1(y2))(ΥΨ)(Ψ1(y1))}

    and so the first inequality of (2.1) proved.

    Now for the proof of second inequality of (2.1), we first note that, if Υ is convex function, then for λ[0,1], it gives

    Υ(y1+y22)=Υ(λy1+(1λ)y2+(1λ)y1+λy22)Υ(λy1+(1λ)y2)+Υ((1λ)y1+λy2)2.

    Multiplying both sides of above inequality by λαk1 and then integrating the resulting inequality with respect to λ over [0,1], and let Ψ(γ)=λy1+(1λ)y2, and Ψ(β)=(1λ)y1+λy2, we have

    Υ(y1+y22)α2k{10λαk1(Υ(λy1+(1λ)y2)+Υ((1λ)y1+λy2))dλ}=Γk(α+k)2(y2y1)αk{(kIα:ΨΨ1(y1)+)(ΥΨ)(Ψ1(y2))+(kIα:ΨΨ1(y2))(ΥΨ)(Ψ1(y1))}.

    Multiplying by (1) and adding Υ(1)+Υ(2) both sides, we get the second inequality of (2.1).

    Now for the proof of inequality of (2.2), we first note that, if Υ is convex function, then for λ[0,1], it gives

    Υ(1+2u1+u22)=Υ(1+2u1+1+2u22)Υ(1+2u1)+Υ(1+2u2)2

    for all u1,u2[1,2].

    Now by change of variables u1=λ(1+2y1)+(1λ)(1+2y2) and u2=(1λ)(1+2y1)+λ(1+2y2), y1,y2[1,2] and λ[0,1], we have

    Υ(1+2y1+y22)  12[Υ(λ(1+2y1)+(1λ)(1+2y2))+Υ((1λ)(1+2y1)+λ(1+2y2))].

    Multiplying both sides of above inequality by λαk1 and then integrating the resulting inequality with respect to λ over [0,1], and let Ψ(γ)=λ(1+2y1)+(1λ)(1+2y2), and Ψ(β)=(1λ)(1+2y1)+λ(1+2y2), we get

    Υ(1+2y1+y22)α2k[10λαk1Υ(λ(1+2y1)+(1λ)(1+2y2))dλ+10λαk1Υ((1λ)(1+2y1)+λ(1+2y2))dλ]=α2kΨ1(1+2y2)Ψ1(1+2y1)((1+2y2)Ψ(γ)y2y1)αk1Υ(Ψ(γ))Ψ(γ)y2y1dγ+α2kΨ1(1+2y2)Ψ1(1+2y1)(Ψ(γ)(1+2y1)y2y1)αk1Υ(Ψ(γ))Ψ(γ)y2y1dγ.

    Hence

    Υ(1+2y1+y22)Γk(α+k)2(y2y1)αk[(kIα:ΨΨ1(1+2y1)+)(ΥΨ)(Ψ1(1+2y2))+(kIα:ΨΨ1(1+2y2))(ΥΨ)(Ψ1(1+2y1))]

    and so the first inequality of (2.2) proved.

    Now for the proof of second inequality of (2.2), we first note that, if Υ is convex function, then for λ[0,1], it gives

    Υ(λ(1+2y1)+(1λ)(1+2y2))λΥ(1+2y1)+(1λ)Υ(1+2y2);Υ((1λ)(1+2y1)+λ(1+2y2))(1λ)Υ(1+2y1)+λΥ(1+2y2).

    By adding these inequalities and using the Jensen–Mercer inequality, we have

    Υ(λ(1+2y1)+(1λ)(1+2y2))+Υ((1λ)(1+2y1)+λ(1+2y2))Υ(1+2y1)+Υ(1+2y2)2[Υ(1)+Υ(2)][Υ(y1)+Υ(y2)].

    Multiplying both sides of above inequality by λαk1 and then integrating the resulting inequality with respect to λ over [0,1], we obtain second and third inequalities of (2.2).

    Corollary 1. Under the assumption of Theorem 2.1, choosing Ψ(γ)=γ we get the following inequalities

    Υ(1+2y1+y22)[Υ(1)+Υ(2)]Γk(α+k)2(y2y1)αk{kJα(y2)Υ(y1) + kJα(y1)+Υ(y2)}[Υ(1)+Υ(2)]Υ(y1+y22)

    and

    Υ(1+2y1+y22)Γk(α+k)2(y2y1)αk[(kJα(1+2y1)+)Υ(1+2y2)+(kJα(1+2y2))Υ(1+2y1)]Υ(1+2y1)+Υ(1+2y2)2[Υ(1)+Υ(2)]Υ(y1+y22).

    Remark 1. For k=1 and taking Ψ(γ)=γ, we have the following inequalities proved in [23].

    Υ(1+2y1+y22)[Υ(1)+Υ(2)]Γ(α+1)2(y2y1)α{Jα(y2)Υ(y1)+Jα(y1)+Υ(y2)}[Υ(1)+Υ(2)]Υ(y1+y22)

    and

    Υ(1+2y1+y22)Γ(α+1)2(y2y1)α[(Jα(1+2y2)+)Υ(1+2y1)++(Jα(1+2y1))Υ(1+2y2)]Υ(1+2y1)+Υ(1+2y2)2[Υ(1)+Υ(2)]Υ(y1+y22).

    Remark 2. For Ψ(γ)=γ and α=k=1 in Theorem 2.1, we will get the following inequalities proved in [12] and [23].

    Υ(1+2y1+y22)[Υ(1)+Υ(2)]10Υ(λy1+(1λ)y2)dλ[Υ(1)+Υ(2)]Υ(y1+y22)

    and

    Υ(1+2y1+y22)1y2y1y2y1Υ(1+2λ)dλ[Υ(1)+Υ(2)]Υ(y1)+Υ(y2)2.

    Theorem 2.2. Let (A1) holds, then the following fractional integral inequalities will be of the form:

    Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk[(kIα:ΨΨ1(1+2y1+y22)+)(ΥΨ)(Ψ1(1+2y1))+(kIα:ΨΨ1(1+2y1+y22))(ΥΨ)(Ψ1(1+2y2))]Υ(1)+Υ(2)(Υ(y1) + Υ(y2)2) (2.3)

    for all y1,y2[1,2].

    Proof. To prove the first part of inequality (2.3), we have

    2Υ(1+2u+v2)Υ(1+2u)+Υ(1+2v)

    for all u,v[1,2].

    By change of variables u=λ2y1+2λ2y2 and v=2λ2y1+λ2y2, λ[0,1], we get

    2Υ(1+2y1+y22)Υ(1+2(λ2y1+2λ2y2  ))+Υ(1+2(2λ2y1+λ2y2 )).

    Multiplying both sides of above inequality by λαk1 and then integrating the resulting inequality with respect to λ over [0,1], and let Ψ(γ)=(1+2(λ2y1+2λ2y2  )) and Ψ(β)=(1+2(2λ2y1+t2y2 )), we have

    2Υ(1+2y1+y22)10λαk1dλ10λαk1×[Υ(1+2(λ2y1+2λ2y2  ))+Υ(1+2(2λ2y1+λ2y2 ))]dλ
    Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk{(kIα:ΨΨ1(1+2y1+y22)+)(ΥΨ)(Ψ1(1+2y1))+(kIα:ΨΨ1(1+2y1+y22))(ΥΨ)(Ψ1(1+2y2))}

    and so the first inequality of (2.3) is proved.

    Now for the proof of second inequality of (2.3), we first note that, if Υ is convex function, then for λ[0,1], it gives

    Υ(1+2(λ2y1+2λ2y2))Υ(1)+Υ(2)[λ2Υ(y1)+2λ2Υ(y2)]      (2.4)

    and

    Υ(1+2(2λ2y1+λ2y2))Υ(1)+Υ(2)[2λ2Υ(y1)+λ2Υ(y2)]. (2.5)

    By adding the inequalities of (2.4) and (2.5), we have

    Υ(1+2(λ2y1+2λ2y2  ))+Υ(1+2(2λ2y1+λ2y2 )) 2(Υ(1)+Υ(2))(Υ(y1)+Υ(y2)).

    Multiplying both sides of above inequality by λαk1 and then integrating the resulting inequality with respect to λ over [0,1], we get

    10λαk1[Υ(1+2(λ2y1+2λ2y2  ))+Υ(1+2(2λ2y1+λ2y2 ))]dλ {2(Υ(1)+Υ(2))(Υ(y1)+Υ(y2))} 10λαk1dλ.

    Hence, we obtain

    2αk kΓk(α)(y2y1)αk[(kIα:ΨΨ1(1+2y1+y22)+)(ΥΨ)(Ψ1(1+2y1))+(kIα:ΨΨ1(1+2y1+y22))(ΥΨ)(Ψ1(1+2y2))] {2(Υ(1)+Υ(2))(Υ(y1)+Υ(y2))}kα.

    Multiplying by α2k, we get

    2αk1 Γk(α+k)(y2y1)αk[(kIα:ΨΨ1(1+2y1+y22)+)(ΥΨ)(Ψ1(1+2y1))+(kIα:ΨΨ1(1+2y1+y22))(ΥΨ)(Ψ1(1+2y2))] (Υ(1)+Υ(2))Υ(y1)+Υ(y2)2

    and so the second inequality of (2.3) is proved.

    Corollary 2. Choosing Ψ(γ)=γ in Theorem 2.2, we will get the following inequalities

    Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk{kJα(1+2y1+y22)Υ(1+2y2)  +kJα(1+2y1+y22)+Υ(1+2y1)}Υ(1)+Υ(2)(Υ(y1) + Υ(y2)2).

    Remark 3. For k=1 and taking Ψ(γ)=γ, we will get the following inequalities proved in [23].

    Υ(1+2y1+y22)2α1Γ(α+1)(y2y1)α{Jα(1+2y1+y22)Υ(1+2y2)+Jα(1+2y1+y22)+Υ(1+2y1)}Υ(1)+Υ(2)(Υ(y1) + Υ(y2)2).

    Remark 4. For Ψ(γ)=γ and α=k=1 in Theorem 2.2, we obtain

    Υ(1+2y1+y22)1y2y1y2y1Υ(1+2λ)dλΥ(1)+Υ(2)(Υ(y1) + Υ(y2)2)

    for all y1,y2[1,2]. This inequality was proved by Kian and Moslehian in [12].

    Theorem 2.3. Let (A1) holds, then the following fractional integral inequalities will be of the form:

    Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk[(kIα:ΨΨ1(1+2y2)+)(ΥΨ)(Ψ1(1+2y1+y22))+(kIα:ΨΨ1(1+2y1))(ΥΨ)(Ψ1(1+2y1+y22))]Υ(1)+Υ(2)(Υ(y1) + Υ(y2)2) (2.6)

    for all y1,y2[1,2].

    Proof. To prove the first part of inequality (2.6), we have

    2Υ(1+2u+v2)Υ(1+2u)+Υ(1+2v)

    u,v[1,2].

    By change of variables u=1+λ2y1+1λ2y2 and v=1λ2y1+1+λ2y2, λ[0,1], we get

    2Υ(1+2y1+y22)Υ(1+2(1+λ2y1+1λ2y2))+Υ(1+2(1λ2y1+1+λ2y2)).

    Multiplying both sides of above inequality by λαk1 and then integrating the resulting inequality with respect to λ over [0,1], and let Ψ(γ)=(1+2(1+λ2y1+1λ2y2  )) and Ψ(β)=(1+2(1λ2y1+1+λ2y2)), we obtain

    2Υ(1+2y1+y22)10λαk1dλ10λαk1[Υ(1+2(1+λ2y1+1λ2y2))+Υ(1+2(1λ2y1+1+λ2y2 ))]dλ.

    So, we have

    Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk[(kIα:ΨΨ1(1+2y2)+)((ΥΨ)(Ψ1(1+2y1+y22)))+(kIα:ΨΨ1(1+2y1))((ΥΨ)(Ψ1(1+1y1+y22)))] (2.7)

    and so the first inequality of (2.6) is proved.

    Now for the proof of second inequality of (2.6), we first note that, if Υ is convex function, then for λ[0,1], it gives

    Υ(1+2(1+λ2y1+1λ2y2))Υ(1)+Υ(2)[1+λ2Υ(y1)+1λ2Υ(y2)] (2.8)

    and

    Υ(1+2(1λ2y1+1+λ2y2 ))Υ(1)+Υ(2)[1λ2Υ(y1)+1+λ2Υ(y2)]. (2.9)

    By adding the inequalities of (2.8) and (2.9), we have

    Υ(1+2(1+λ2y1+1λ2y2  ))+Υ(1+2(1λ2y1+1+λ2y2 )) 2(Υ(1)+Υ(2))(Υ(y1)+Υ(y2)).

    Multiplying both sides by λαk1 and then integrating the resulting inequality with respect to λ over [0,1], we have

    10λαk1[Υ(1+2(1+λ2y1+1λ2y2  ))+Υ(1+2(1λ2y1+1+λ2y2 ))]dλ (2(Υ(1)+Υ(2))(Υ(y1)+Υ(y2))) 10λαk1dλ.

    Then, we obtain

    2αk kΓk(α)(y2y1)αk[(kIα:ΨΨ1(1+2y2)+)((ΥΨ)(Ψ1(1+2y1+y22)))+(kIα:ΨΨ1(1+2y1))((ΥΨ)(Ψ1(1+2y1+y22)))] (2(Υ(1)+Υ(2))(Υ(y1)+Υ(y2)))kα.

    Multiplying by α2k, we get

    2αk1 Γk(α+k)(y2y1)αk[(kIα:ΨΨ1(1+2y2)+)((ΥΨ)(Ψ1(1+2y1+y22)))+(kIα:ΨΨ1(1+2y1))((ΥΨ)(Ψ1(1+2y1+y22)))] (Υ(1)+Υ(2))Υ(y1)+Υ(y2)2. (2.10)

    From (2.7) and (2.10), we obtain (2.6).

    Corollary 3. Choosing Ψ(γ)=γ in Theorem 2.3, we get the following inequalities

    Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk{kJα(1+2y1)Υ(1+2y1+y22)+kJα(1+2y2)+Υ(1+2y1+y22)}Υ(1)+Υ(2)(Υ(y1) + Υ(y2)2).

    Remark 5. For k=1 and taking Ψ(γ)=γ, we have

    Υ(1+2y1+y22)2α1Γ(α+1)(y2y1)α{Jα(1+2y1)Υ(1+2y1+y22)+Jα(1+2y2)+Υ(1+2y1+y22)}Υ(1)+Υ(2)(Υ(y1) + Υ(y2)2).

    We start this section with the following important lemma:

    Lemma 3.1. Let 01<2, Υ:[1,2] be a positive function and ΥL1[1,2]. Also suppose that Υ is a convex function on [1,2], Ψ(γ) is an increasing and positive monotone function on (1,2], having a continuous derivative Ψ on (1,2) and α,k>0. Then the following identity holds:

    Υ(1+2y1)+Υ (1+2y2)2 Γk(α+k)2(y2y1)αk[(kIα:ΨΨ1(1+2y2)+)(ΥΨ)(Ψ1(1+1y1))+(kIα:ΨΨ1(1+2y1))(ΥΨ)(Ψ1(1+2y2))]=12(y2y1)αk×Ψ1(1+2y1)Ψ1(1+2y2)((Ψ(γ)(1+2y2))αk((1+2y1)Ψ(γ))αk)(ΥΨ)(γ)Ψ(γ)dγ (3.1)

    for all y1,y2[1,2].

    Proof. It suffices to note that

    I=Υ(1+2y1)Υ(1+2y2)2{I1+I2}, (3.2)

    where

    I1=Γk(α+k)2(y2y1)αk[kIα:ΨΨ1(1+2y2)+(ΥΨ)(Ψ1(1+2y1))]=α2k(y2y1)αkΨ1(1+2y1)Ψ1(1+2y2)Ψ(γ)((1+2y1)Ψ(γ))αk1(ΥΨ)(γ)dγ=12(y2y1)αk[Υ(1+2y2)(y2y1)αk]+Ψ1(1+2y1)Ψ1(1+2y2)Ψ(γ)((1+2y1)Ψ(γ))αk(ΥΨ)(γ)dγ (3.3)

    and

    I2= Γk(α+k)2(y2y1)αk[kIα:ΨΨ1(1+2y1)((ΥΨ)(Ψ1(1+2y2)))]=α2k(y2y1)αkΨ1(1+2y1)Ψ1(1+2y2)Ψ(γ)((1+2y2))+Ψ(γ))αk1(ΥΨ)(γ)dγ=12(y2y1)αk[Υ(1+2y1)(y2y1)αk]Ψ1(1+2y1)Ψ1(1+2y2)Ψ(γ)((1+2y2)+Ψ(γ))αk(ΥΨ)(γ)dγ. (3.4)

    Substituting (3.3) and (3.4) in (3.2), we get (3.1).

    Corollary 4. Choosing Ψ(γ)=γ in Lemma 3.1, we get the following equality

    Υ(1+2y1)+Υ (1+2y2)2Γk(α+k)2(y2y1)αk[ kJα(1+2y2)+Υ(1+2y1)+ kJα(1+2y1)Υ(1+2y2)]= 12(y2y1)αk(1+2y1)(1+2y2)[(γ(1+2y2))αk((1+2y1)γ)αk]Υ(γ)dγ.

    Remark 6. For k=1 and taking y1=1 and y2=2, we get the following Lemma by O'Regan et al. in [17]:

    Υ(1)+Υ (2)2 Γ(α+1)2(21)α{(Iα:ΨΨ1(1)+)(ΥΨ)(Ψ1(2))+(Iα:ΨΨ1(2))(ΥΨ)(Ψ1(1))}= 12(21)α Ψ1(2)Ψ1(1)((Ψ(γ)(1))α((2)Ψ(γ))α) (ΥΨ)(γ)Ψ(γ)dγ.

    Theorem 3.2. Let (A1) holds. Also suppose that |Υ| is a convex function on [1,2], then the following fractional integral inequality holds:

    |Υ(1+2y1)+Υ(1+2y2)2Γk(α+k)2(y2y1)αk[(kIα:ΨΨ1(1+2y2)+ )Υ(1+2y1)+(kIα:ΨΨ1(1+2y1) )Υ(1+2y2)]|(y2y1αk+1)(112αk){|Υ(1)|+|Υ(2)|(|Υ(y1)|+|Υ(y2)|2)} (3.5)

    for all y1,y2[1,2].

    Proof. By using Lemma 3.1, Jensen–Mercer inequality and properties of modulus, we have

    |Υ(1+2y1)+Υ(1+2y2)2Γk(α+k)2(y2y1)αk[(kIα:Ψ(1+2y2)+ )Υ(1+2y1)+(kIα:Ψ(1+2y1) )Υ(1+2y2)]|Ψ1(1+2y1)Ψ1(1+2y2)|((Ψ(γ)(1+2y2))αk((1+2y1)Ψ(γ))αk)|×12(y2y1)αk|(ΥΨ)(γ)|Ψ(γ)dγ=(y2y1)2 10|λαk(1λ)αk||Υ((1+2(λy1+(1λ)y2))|dλ(y2y1)2 10|λαk(1λ)αk|{|Υ(1)|+|Υ(2)|(λ|Υ(y1)|+(1λ)|Υ(y2)|)}dλ=(y2y1)2[I1+I2],

    where

    I1=120((1λ)αkλαk){|Υ(1)|+|Υ(2)|(λ|Υ(y1)|+(1λ)|Υ(y2)|)}=(|Υ(1)|+|Υ(2)|)(1(αk+1)2αk(αk+1)){|Υ(y1)|(1(αk+1)(α+2)2αk1(αk+1))+|Υ(y2)|(1(αk+2)2αk1(αk+1))} (3.6)

    and

    I2=112(λαk(1λ)αk){|Υ(1)|+|Υ(2)|(λ|Υ(y1)|+(1λ)|Υ(y2)|)}=(|Υ(1)|+|Υ(2)|)(1(αk+1)2αk(αk+1)){|Υ(y1)|(1(αk+2)2αk1(αk+1))+|Υ(y2)|(1(αk+1)(αk+2)2αk1(αk+1))}. (3.7)

    Combining inequalities (3.6) and (3.7), we get the desired inequality (3.5).

    Corollary 5. Choosing Ψ(γ)=γ in Theorem 3.2, we get the following inequality

    |Υ(1+2y1)+Υ(1+2y2)2Γk(α+k)2(y2y1)αk[(kJα(1+2y2)+ )Υ(1+2y1)+(kJα(1+2y1) )Υ(1+2y2)]|(y2y1αk+1)(112αk){|Υ(1)|+|Υ(2)|(|Υ(y1)|+|Υ(y2)|2)}.

    Remark 7. For y1=1 and y2=2 and k=1 in Theorem 3.2, we will get inequality proved in [17]:

    |Υ(1)+Υ(2)2Γ(α+1)2(21)α[(Iα:ΨΨ1(1)+ )Υ(2)+(Iα:ΨΨ1(2) )Υ(1)]|(21)2(α+1)(112α)[|Υ(1)|+|Υ(2)|].

    Remark 8. For k=1 and taking Ψ(γ)=γ, we obtain Theorem 4 in [23].

    Lemma 3.3. Let 01<2, Υ:[1,2] be a positive function and ΥL1[1,2]. Also suppose that Υ is a convex function on [1,2], Ψ(γ) is an increasing and positive monotone function on (1,2], having a continuous derivative Ψ on (1,2) and α,k>0. Then the following identity holds:

    Υ(1+2y1)+Υ(1+2y2)22αk1Γk(α+k)(y2y1)αk×[(kIα:ΨΨ1(1+2y2)+)((ΥΨ)(Ψ1(1+2y1+y22)))+(kIα:ΨΨ1(1+2y1))((ΥΨ)(Ψ1(1+2y1+y22)))]=(y2y1)4[10λαk Υ(1+2(1+λ2y1+1λ2y2))dλ10λαk Υ(1+2(1λ2y1+1+λ2y2 ))dλ]. (3.8)

    Proof. It suffices to note that

    I=(y2y1)4{I1I2}, (3.9)

    where

    I1=10λαk Υ(1+2(1+λ2y1+1λ2y2))dλ= 2(y2y1) Υ(1+2y1)2(αk)y2y110λαk1 Υ(1+2(1+λ2y1+1λ2y2))dλ=2(y2y1) Υ(1+2y1)2αk+1Γk(α+k)(y2y1)αk+1(kIα:ΨΨ1(1+2y1))((ΥΨ)(Ψ1(1+2y1+y22))) (3.10)

    and

    I2=10λαkΥ(1+2(1λ2y1+1+λ2y2 ))dλ= 2(y2y1) Υ(1+2y2)2(αk)y2y110λαk1 Υ(1+2(1λ2y1+1+λ2y2))dλ=2(y2y1) Υ(1+2y2)2αk+1Γk(α+k)(y2y1)αk+1(kIα:ΨΨ1(1+2y2)+)((ΥΨ)(Ψ1(1+2y1+y22))) (3.11)

    Substituting (3.10) and (3.11) in (3.9), we get (3.8).

    Corollary 6. Choosing Ψ(γ)=γ in Lemma 3.3, we get the following equality

    Υ(1+2y1)+Υ (1+2y2)2 2αk1Γk(α+k)(y2y1)αk×{(kJα(1+2y2)+)Υ(1+2y1+y22)+(kJα(1+2y1))Υ(1+2y1+y22)}= (y2y1)4 10λαk[Υ(1+2(1+λ2y1+1λ2y2))                                           Υ(1+2(1λ2y1+1+λ2y2))]dλ.

    Theorem 3.4. Let (A1) holds. Also suppose that Υ is a differentiable and Υ is bounded function on [1,2], then the following fractional integral inequality holds:

    |Υ(1+2y1)+Υ(1+2y2)22αk1Γk(α+k)(y2y1)αk×[(kIα:ΨΨ1(1+2y2)+)((ΥΨ)(Ψ1(1+2y1+y22)))+(kIα:ΨΨ1(1+2y1))((ΥΨ)(Ψ1(1+2y1+y22)))]| (y2y1)24(αk+2)supξ[1,2]|Υ(ξ)| (3.12)

    for all y1,y2[1,2].

    Proof. By using Lemma 3.3 and applying mean value theorem for the function Υ, we have

    Υ(1+2y1)+Υ(1+2y2)22αk1Γk(α+k)(y2y1)αk×[(kIα:ΨΨ1(1+2y2)+)((ΥΨ)(Ψ1(1+2y1+y22)))+(kIα:ΨΨ1(1+2y1))((ΥΨ)(Ψ1(1+2y1+y22)))]=(y2y1)2410(λ)αk λΥ(ξ)dλ, (3.13)

    where ξ(λ)[1,2]. This leads us to

    |Υ(1+2y1)+Υ(1+2y2)22αk1Γk(α+k)(y2y1)αk×[(kIα:ΨΨ1(1+2y2)+)((ΥΨ)(Ψ1(1+2y1+y22)))+(kIα:ΨΨ1(1+2y1))((ΥΨ)(Ψ1(1+2y1+y22)))]|(y2y1)2410(λ)αkλ|Υ(ξ)|dλ(y2y1)24supξ[1,2]|Υ(ξ)|×10(λ)αk+1dλ=(y2y1)24(αk+2)supξ[1,2]|Υ(ξ)|,

    which completes the proof.

    Corollary 7. Choosing Ψ(γ)=γ in Theorem 3.4, we get the following inequality

    |Υ(1+2y1)+Υ (1+2y2)2 2αk1Γk(α+k)(y2y1)αk×{(kJα(1+2y2)+)(Υ(1+2y1+y22))+(kJα(1+2y1))(Υ(1+2y1+y22))}|(y2y1)24(αk+2)supξ[1,2]|Υ(ξ)|.

    Theorem 3.5. Let (A1) holds. Also suppose that |Υ| is a convex function on [1,2], then the following fractional integral inequality holds:

    |Υ(1+2y2)+Υ(1+2y1)22αk1Γk(α+k)(y2y1)αk×[(kIα:Ψ(Ψ1(1+2y2))+ )Υ(1+2y1+y22)+(kIα:Ψ(Ψ1(1+2y1)) )Υ(1+2y1+y22)]| (y2y1)2(αk+1){|Υ(1)|+|Υ(2)|(|Υ(y1)|+|Υ(y2)|2)} (3.14)

    for all y1,y2[1,2].

    Proof. By using Lemma 3.3, Jensen–Mercer inequality and properties of modulus, we have

    |Υ(1+2y2)+Υ(1+2y1)22αk1Γk(α+k)(y2y1)αk×[(kIα:Ψ(Ψ1(1+2y2))+ )Υ(1+2y1+y22)+(kIα:Ψ(Ψ1(1+2y1)) )Υ(1+2y1+y22)]|(y2y1)4 [10λαk |Υ((1+2(1+λ2y1+1λ2y2))|dλ+10λαk |Υ((1+2(1λ2y1+1+λ2y2 ))|dλ](y2y1)4[10λαk{|Υ(1)|+|Υ(2)|(1+λ2|Υ(y1)|+(1λ)2|Υ(y2)|)}dλ+10λαk{|Υ(1)|+|Υ(2)|((1λ)2|Υ(y1)|+1+λ2|Υ(y2)|)}dλ].

    After integration, we obtain the required result.

    Corollary 8. Choosing Ψ(γ)=γ in Theorem 3.5, we get the following inequality

    |Υ(1+2y1)+Υ (1+2y2)2 2αk1Γk(α+k)(y2y1)αk×{(kJα(1+2y2)+)(Υ(1+2y1+y22))+(kJα(1+2y1))(Υ(1+2y1+y22))}| (y2y1)2(αk+1){|Υ(1)|+|Υ(2)|(|Υ(y1)|+|Υ(y2)|2)}.

    Lemma 3.6. Let 01<2, Υ:[1,2] be a positive function and ΥL1[1,2]. Also suppose that Υ is a convex function on [1,2], Ψ(γ) is an increasing and positive monotone function on (1,2], having a continuous derivative Ψ on (1,2) and α,k>0. Then the following identity holds:

    Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk[(kIα:ΨΨ1(1+2y1+y22)+)(ΥΨ)(Ψ1(1+2y1))+(kIα:ΨΨ1(1+2y1+y22))(ΥΨ)(Ψ1(1+2y2))]=(y2y1)4 [10λαk Υ(1+2(λ2y1+2λ2y2))dλ10λαk Υ(1+2(2λ2y1+λ2y2 ))dλ].

    Proof. The proof of this Lemma is similar to the proof of Lemma 3.3.

    Corollary 9. Choosing Ψ(γ)=γ in Lemma 3.6, we get the following identity

    Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk×[(kJα(1+2y1+y22)+)Υ(1+2y1)+(kJα(1+2y1+y22))Υ(1+2y2)]=(y2y1)4 [10λαk Υ(1+2(λ2y1+2λ2y2))dλ10λαk Υ(1+2(2λ2y1+λ2y2 ))dλ].

    Remark 9. For k=1 and taking Ψ(γ)=γ in Lemma 3.6, we obtain Lemma 2 in [23].

    Remark 10. For Ψ(γ)=γ, y1=1 and y2=2, we have Lemma 3 in [25].

    Theorem 3.7. Let (A1) holds. Also suppose that |Υ| is a convex function on [1,2], then the following fractional integral inequality holds:

    |Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk×[(kIα:Ψ(Ψ1(1+2y1+y22))+ )Υ(1+2y1)+(kIα:Ψ(Ψ1(1+2y1+y22)) )Υ(1+2y2)]|(y2y1)2(αk+1){|Υ(1)|+|Υ(2)|(|Υ(y1)|+|Υ(y2)|2)} (3.15)

    for all y1,y2[1,2].

    Proof. By using Lemma 3.6, Jensen–Mercer inequality and properties of modulus, we have

    |Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk×[(kIα:Ψ(Ψ1(1+2y1+y22))+ )Υ(1+2y1)+(kIα:Ψ(Ψ1(1+2y1+y22)) )Υ(1+2y2)]|(y2y1)4 [10λαk|Υ(1+2(λ2y1+2λ2y2))|dλ+10λαk|Υ(1+2(2λ2y1+λ2y2 ))|dλ](y2y1)4[10λαk{|Υ(1)|+|Υ(2)|(λ2|Υ(y1)|+(2λ)2|Υ(y2)|)}dλ+10λαk{|Υ(1)|+|Υ(2)|((2λ)2|Υ(y1)|+λ2|Υ(y2)|)}dλ] (y2y1)2(αk+1){|Υ(1)|+|Υ(2)|(|Υ(y1)|+|Υ(y2)|2)},

    which completes the proof.

    Corollary 10. Choosing Ψ(γ)=γ in Theorem 3.7, we get the following inequality

    |Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk×[(kJα(1+2y1+y22)+)Υ(1+2y1)+(kJα(1+2y1+y22))Υ(1+2y2)]|(y2y1)2(αk+1){|Υ(1)|+|Υ(2)|(|Υ(y1)|+|Υ(y2)|2)}.

    Remark 11. For k=1 and taking Ψ(γ)=γ in Theorem 3.7, we obtain Theorem 5 in [23].

    Theorem 3.8. Let (A1) holds. If |Υ|q is convex function, then for q1 and 1p+1q=1, the following fractional integral inequality holds:

    |Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk[(kIα:Ψ(Ψ1(1+2y1+y22))+ )Υ(1+2y1)+(kIα:Ψ(Ψ1(1+2y1+y22)) )Υ(1+2y2)]|(y2y1)4(kpα+k)1p[((|Υ(1)|q+|Υ(2)|q(14|Υ(y1))|q+34|Υ(y2)|q))1q+(|Υ(1)|q+|Υ(2)|q(34|Υ(y1)|q+14|Υ(y2)|q))1q] (3.16)

    for all y1,y2[1,2].

    Proof. By using Lemma 3.6, Holder's inequality, Jensen-Mercer inequality, the fact that |Υ|q is convex function and properties of modulus, we have

    |Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk[(kIα:Ψ(Ψ1(1+2y1+y22))+ )Υ(1+2y1)+(kIα:Ψ(Ψ1(1+2y1+y22)) )Υ(1+2y2)]|(y2y1)4 [10λαk|Υ(1+2(λ2y1+2λ2y2))|dλ+10λαk|Υ(1+2(2λ2y1+λ2y2 ))|dλ](y2y14)10λαk{|Υ(1)+Υ(2)(λ2Υ(y1)+(2λ)2Υ(y2))|}dλ+((y2y1)4)10λαk{|Υ(1)+Υ(2)((2λ)2Υ(y1)+λ2Υ(y2))|}dλ(y2y1)4(10λp(αk)dλ)1p(10|Υ(1)+Υ(1)(λ2Υ(y1)+(2λ)2Υ(y2))|qdλ)1q+(y2y1)4(10λp(αk)dλ)1p(10|Υ(1)+Υ(2)((2λ)2Υ(y1)+λ2Υ(y2))|qdλ)1q(y2y1)4(kpα+k)1p(|Υ(1)|q+|Υ(2)|q(14|Υ(y1)|q+34|Υ(y2)|q))1q+(y2y1)4(kpα+k)1p(|Υ(1)|q+|Υ(2)|q(34|Υ(y2)|q+14|Υ(y1)|q))1q=(y2y1)4(kpα+k)1p[(|Υ(1)|q+|Υ(2)|q(14|Υ(y1))|q+34|Υ(y2)|q))1q+(|Υ(1)|q+|Υ(2)|q(34|Υ(y1)|q+14|Υ(y2)|q))1q],

    which completes the proof.

    Corollary 11. Choosing Ψ(γ)=γ in Theorem 3.8, we get the following inequality

    |Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk×[(kJα(1+2y1+y22)+)Υ(1+2y1)+(kJα(1+2y1+y22))Υ(1+2y2)]|(y2y1)4(kp(α)+k)1p[((|Υ(1)|q+|Υ(2)|q(14|Υ(y1))|q+34|Υ(y2)|q))1q+(|Υ(1)|q+|Υ(2)|q(34|Υ(y2)|q+14|Υ(y1)|q))1q].

    Remark 12. For k=1 and taking Ψ(γ)=γ in Theorem 3.8, we have Theorem 6 in [23].

    Theorem 4.1. Let (A1) holds. Also suppose that |Υ|q is a convex function on [1,2], then for q1, the following fractional integral inequality holds:

    |Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk                                               ×[(kIα:ΨΨ1(1+2y1+y22)+)((ΥΨ)(Ψ1(1+2y1)))                            +(kIα:ΨΨ1(1+2y1+y22))((ΥΨ)(Ψ1(1+2y2)))]|
    (y2y1)4[{(1(αk+1)(αk+2))11q(|Υ(1)|q +|Υ(2)|q(αk+1)(αk+2)(|Υ(y1)|q2(αk+2)(αk+3) +(αk+5)|Υ(y2)|q2(αk+1)(αk+2)(αk+3)))1q+(1(αk+2))11q(|Υ(1)|q +|Υ(2)|q(αk+2)(|Υ(y1)|q2(αk+3) +|Υ(y2)|q2(αk+2)(αk+3)))1q}+ {(1(αk+1)(αk+2))11q(|Υ(1)|q +|Υ(2)|q(αk+1)(αk+2)((αk+5)|Υ(y1)|q2(αk+1)(αk+2)(αk+3) +|Υ(y2)|q2(αk+2)(αk+3)))1q+(1(αk+2))11q×(|Υ(1)|q +|Υ(2)|q(αk+2)((αk+4)|Υ(y1)|q2(αk+2)(αk+3) +|Υ(y2)|q2(αk+3)))1q}] (4.1)

    for all y1,y2[1,2].

    Proof. By using Lemma 3.6, Jensen–Mercer inequality for |Υ|q, applying the Improved power-mean integral inequality (see [11]) and properties of modulus, we have

    |Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk                                               ×[(kIα:ΨΨ1(1+2y1+y22)+)((ΥΨ)(Ψ1(1+2y1)))                            +(kIα:ΨΨ1(1+2y1+y22))((ΥΨ)(Ψ1(1+2y2)))]|
    (y2y1)4{(10(1λ)λαk dλ)11q(10(1λ)λαk|Υ(1+2(λ2y1+2λ2y2))|q dλ)1q+(10λαk+1 dλ)11q(10λαk+1|Υ(1+2(λ2y1+2λ2y2))|q dλ)1q}+ {(10(1λ)λαk dλ)11q(10(1λ)λαk|Υ(1+2(λ2y2+2λ2y1))|q dλ)1q+(10λαk+1 dλ)11q(10λαk+1|Υ(1+2(λ2y2+2λ2y1))|q dλ)1q}. (4.2)

    It is easy to see that

    10(1λ) λαk dλ= 1(αk+1)(αk+2) (4.3)
    10(1λ)λαk|Υ(1+2(λ2y1+2λ2y2))|q dλ(|Υ(1)|q +|Υ(2)|q)(αk+1)(αk+2)(|Υ(y1)|q2(αk+2)(αk+3) +|Υ(y2)|q2(αk+1)(αk+2)(αk+3)) (4.4)
    10λαk+1 dλ= 1(αk+2) (4.5)
    10λαk+1 |Υ(1+2(λ2y1+2λ2y2))|q dλ|Υ(1)|q +|Υ(2)|q(αk+2)(|Υ(y1)|q2(αk+3) +|Υ(y2)|q2(αk+2)(αk+3)) (4.6)
    10(1λ)λαk|Υ(1+2(2λ2y1+λ2y2))|q dλ |Υ(1)|q +|Υ(2)|q(αk+1)(αk+2)((αk+5)|Υ(y1)|q2(αk+1)(αk+2)(αk+3) +|Υ(y2)|q2(αk+2)(αk+3)) (4.7)
    10λαk+1|Υ(1+2(2λ2y1+λ2y2))|q dλ  |Υ(1)|q +|Υ(2)|q(αk+2)((αk+4)|Υ(y1)|q2(αk+2)(αk+3) +|Υ(y2)|q2(αk+3)). (4.8)

    By substituting (4.3)(4.8) in (4.2), we obtain the desired inequality (4.1).

    Corollary 12. Choosing Ψ(γ)=γ in Theorem 4.1, we get the following inequality

    |Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk×[(kJα(1+2y1+y22)+)Υ(1+2y1)+(kJα(1+2y1+y22))Υ(1+2y2)]|
    (y2y1)4[{(1(αk+1)(αk+2))11q(|Υ(1)|q +|Υ(2)|q(αk+1)(αk+2)(|Υ(y1)|q2(αk+2)(αk+3) +(αk+5)|Υ(y2)|q2(αk+1)(αk+2)(αk+3)))1q+(1(αk+2))11q(|Υ(1)|q +|Υ(2)|q(αk+2)(|Υ(y1)|q2(αk+3) +|Υ(y2)|q2(αk+2)(αk+3)))1q}+ {(1(αk+1)(αk+2))11q(|Υ(1)|q +|Υ(2)|q(αk+1)(αk+2)((αk+5)|Υ(y1)|q2(αk+1)(αk+2)(αk+3) +|Υ(y2)|q2(αk+2)(αk+3)))1q+(1(αk+2))11q(|Υ(1)|q +|Υ(2)|q(αk+2)((αk+4)|Υ(y1)|q2(αk+2)(αk+3) +|Υ(y2)|q2(αk+3)))1q}].

    Theorem 4.2. Let (A1) holds. Also suppose that |Υ|q is a convex function on [1,2], then for q>1 and 1p+1q=1, the following fractional integral inequalities holds:

    |Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk[(kIα:ΨΨ1(1+2y1+y22)+)(ΥΨ)(Ψ1(1+2y1))                                  +(kIα:ΨΨ1(1+2y1+y22))(ΥΨ)(Ψ1(1+2y2))]|
    (y2y1)4[{(1(αpk+1)(αpk+2))1p(12(|Υ(1)|q +|Υ(2)|q)(112|Υ(y1)|q +512|Υ(y2)|q))1q+(1(αpk+2))1p(12(|Υ(1)|q +|Υ(2)|q)(16|Υ(y1)|q +13|Υ(y2)|q))1q}+ {(1(αpk+1)(αpk+2))1p(12(|Υ(1)|q +|Υ(2)|q)(512|Υ(y1)|q +112|Υ(y2)|q))1q+(1αpk+2)1p×(12(|Υ(1)|q +|Υ(2)|q)(13|Υ(y1)|q +16|Υ(y2)|q))1q}] (4.9)

    for all y1,y2[1,2].

    Proof. By using Lemma 3.6, Jensen–Mercer inequality, applying the Hölder- İşcan integral inequality (see [10]) and properties of modulus, we have

    |Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk                                           ×[(kIα:ΨΨ1(1+2y1+y22)+)(ΥΨ)(Ψ1(1+2y1))                                          +(kIα:ΨΨ1(1+2y1+y22))(ΥΨ)(Ψ1(1+2y2))]|
    (y2y1)4{(10(1λ)λαkp dλ)1p(10(1λ)|Υ(1+2(λ2y1+2λ2y2))|q dλ)1q+(10λαkp+1 dλ)1p(10λ|Υ(1+2(λ2y1+2λ2y2))|q dλ)1q}+ {(10(1λ)λαpk dλ)1p(10(1λ)|Υ(1+2(2λ2y1+t2y2))|q dλ)1q+(10λαpk+1 dλ)1p(10λ|Υ(1+2(2λ2y1+λ2y2))|q dλ)1q}. (4.10)

    By the convexity of |Υ|q, we get

    |Υ(1+2(λ2y1+2λ2y2))|q
    |Υ(1)|q+|Υ(2)|q(λ2|Υ(y1)|q+2λ2|Υ(y2)|q). (4.11)

    It is easy to see that

    10(1λ) λαpk dλ= 1(αpk+1)(αpk+2); (4.12)
    10(1λ)|Υ(1+2(λ2y1+2λ2y2))|q dλ                                     12(|Υ(1)|q +|Υ(2)|q)(112|Υ(y1)|q +512|Υ(y2)|q); (4.13)
    10λαpk+1 dλ= 1(αpk+2); (4.14)
    10λ|Υ(1+2(λ2y1+2λ2y2))|q dλ                   12(|Υ(1)|q +|Υ(2)|q)(16|Υ(y1)|q +13|Υ(y2)|q); (4.15)
    10(1λ)|Υ(1+2(2λ2y1+λ2y2))|q dλ                        12(|Υ(1)|q +|Υ(2)|q)(512|Υ(y1)|q +112|Υ(y2)|q) (4.16)

    and

    10λ|Υ(1+2(2λ2y1+λ2y2))|q dλ                                        12(|Υ(1)|q +|Υ(2)|q)(13|Υ(y1)|q +16|Υ(y2)|q). (4.17)

    By substituting (4.11)(4.17) in (4.10), we obtain the required inequality (4.9).

    Corollary 13. Choosing Ψ(γ)=γ in Theorem 4.2, we obtain the following inequality

    |Υ(1+2y1+y22)2αk1Γk(α+k)(y2y1)αk×[(kJα(1+2y1+y22)+)Υ(1+2y1)+(kJα(1+2y1+y22))Υ(1+2y2)]|
    (y2y1)4[{(1(αpk+1)(αpk+2))1p(12(|Υ(1)|q +|Υ(2)|q)(112|Υ(y1)|q +512|Υ(y2)|q))1q+(1(αpk+2))1p(12(|Υ(1)|q +|Υ(2)|q)(16|Υ(y1)|q +13|Υ(y2)|q))1q}+ {(1(αpk+1)(αpk+2))1p(12(|Υ(1)|q +|Υ(2)|q)(512|Υ(y1)|q +112|Υ(y2)|q))1q+(1αpk+2)1p(12(|Υ(1)|q +|Υ(2)|q)(13|Υ(y1)|q +16|Υ(y2)|q))1q}].

    In this article, authors obtained some Hermite–Jensen–Mercer type inequalities using ψ–Riemann–Liouville k–Fractional integrals and several ψ–Riemann–Liouville k–Fractional integral inequalities are provided as well. Some known results are recaptured as special cases of our results. We hope that our new idea and technique may inspired many researcher in this fascinating field.

    The research of the first author has been supported by H.E.C. Pakistan under NRPU project 7906.

    The authors declare no conflict of interest.



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