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On Hermite-Hadamard type inequalities for co-ordinated convex function via conformable fractional integrals

  • In this study, some new Hermite-Hadamard type inequalities for co-ordinated convex functions were obtained with the help of conformable fractional integrals. We have presented some remarks to give the relation between our results and earlier obtained results. Moreover, an identity for partial differentiable functions has been established. By using this equality and concept of co-ordinated convexity, we have proven a trapezoid type inequality for conformable fractional integrals.

    Citation: Mehmet Eyüp Kiriş, Miguel Vivas-Cortez, Gözde Bayrak, Tuğba Çınar, Hüseyin Budak. On Hermite-Hadamard type inequalities for co-ordinated convex function via conformable fractional integrals[J]. AIMS Mathematics, 2024, 9(4): 10267-10288. doi: 10.3934/math.2024502

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  • In this study, some new Hermite-Hadamard type inequalities for co-ordinated convex functions were obtained with the help of conformable fractional integrals. We have presented some remarks to give the relation between our results and earlier obtained results. Moreover, an identity for partial differentiable functions has been established. By using this equality and concept of co-ordinated convexity, we have proven a trapezoid type inequality for conformable fractional integrals.



    The Hermite-Hadamard inequality stands as a cornerstone in the realm of convex functions, boasting a geometric interpretation and having broad applicability. Countless mathematicians have dedicated their endeavors to extending, refining, and providing counterparts for this inequality across various classes of functions, often involving convex mappings. The inequalities originally formulated by C. Hermite and J. Hadamard for convex functions hold significant importance in the existing literature. (see, e.g., [1], [2, p. 137]). The Hermite-Hadamard inequality is stated as follows:

    If F :IR is a convex function on the interval I of real numbers and η1,η2I with η1<η2 then,

    F(η1+η22)1η2η1η2η1F(δ)dδF(η1)+F(η2)2. (1.1)

    The Hermite-Hadamard inequality has attracted the attention of many mathematicians since the day it was proved. Especially in recent years, many generalizations and extensions of this inequality have been created. The definition of convexity is used a lot when creating new versions of the Hermite-Hadamard inequality. To define convexity on coordinates let us first consider a bidimensional interval Δ:=[η1,η2]×[υ1,υ2] in R2.

    Definition 1.1. [3] A function F:ΔR is called a co-ordinated convex on Δ, if it satifies the inequality

    F(μt+(1t)τ,sρ+(1s)σ)tsF(μ,ρ)+t(1s)F(μ,σ)+s(1t)F(τ,ρ)+(1t)(1s)F(τ,σ) (1.2)

    for all (μ,ρ),(τ,σ)Δ and t,s[0,1].

    In [3], Dragomir proved the Hermite-Hadamard inequality for co-ordinated convex functions on the rectangle from the plane R2. For several results concerning the Hermite-Hadamard type inequality for co-ordinated convex functions. Some papers devoted Hermite-Hadamard inequalities for co-ordinated convex functions [4,5,6]. Alomari and Darus presented some inequalities for s-convex function on co-ordinates.[7]. Vivas et al. proved some Hermite-Hadamard inequalities for co-ordinated convex interval valued functions [8].

    Fractional analysis is a current field of study with various uses in fields such as physics, engineering and biology. Fractional integral operators are also very important for mathematics because the generalization of many integral inequalities has been introduced to the literature thanks to the fractional integral operators For more information please refeer to the books [9,10,11]. Multiple fractional operators have been defined so far, i.e. Caputo, Riemann-Liouville, Hadamard, and Katugampola to name a few. The concept of conformable fractional integrals was given by Khalil et al. in 2014 [12]. The conformable fractional integral operator was used throughout this study. For all this, please see [13,14,15,16,17,18].

    Definition 1.2. [9] Let FL1[η1,η2]. The Riemann-Liouville fractional integrals Iβη+1F and  Iβη2F of order β>0 are given by

    Iαη+1F(x)=1Γ(β)xη1(xδ)β1F(δ)dδ,    x>η1, (1.3)
    Iαη2F(x)=1Γ(β)η2x(δx)β1F(δ)dδ,    x<η2. (1.4)

    The Riemann-Liouville fractional integrals will be provided for order β>0. The Riemann Liouville integrals will be equal to the classical Riemann integral for the condition β=1.

    Definition 1.3. [9] Let FL1(Δ). Riemann-Liouville fractional integrals Iα,βη+1,υ+1,F, Iα,βη+1,υ2,F, Iα,βη2,υ+1F and Iα,βη2,υ2F of orders α,β>0 with η1,υ10 are defined by

    Iα,βη+1,υ+1,F(δ,ξ)=1Γ(α)Γ(β)δη1ξυ1(δt)α1(ξs)β1F(t,s)dsdt,    δ>η1,ξ>υ1,
    Iα,βη+1,υ2,F(δ,ξ)=1Γ(α)Γ(β)δη1υ2ξ(δt)α1(sξ)β1F(t,s)dsdt,    δ>η1,ξ<υ2,
    Iα,βη2,υ+1F(δ,ξ)=1Γ(α)Γ(β)η2δξυ1(tδ)α1(ξs)β1F(t,s)dsdt,    δ<η2,ξ>υ1,

    and

    Iα,βη2,υ2F(δ,ξ)=1Γ(α)Γ(β)η2δυ2ξ(tδ)α1(sξ)β1F(t,s)dsdt,    δ<η2,ξ<υ2,

    respectively. Here, Γ is the gamma function.

    Definition 1.4. [19] For FL1[η1,η2], the conformable fractional integral operators  βIαη+1F and  βIαη2F of orders β>0 and α(0,1] are given by

     βJαη+1F(x)=1Γ(β)xη1((xη1)α(tη1)αα)β1F(t)(tη1)1αdt,    x>η1, (1.5)

    and

     βJαη2F(x)=1Γ(β)η2x((η2x)α(η2t)αα)β1F(t)(η2t)1αdt,    x<η2, (1.6)

    respectively.

    Remark 1.1. If we consider that α=1 in Definition 1.4, then the fractional integrals (1.5) and (1.6) reduce to the Riemann-Liouville fractional integrals (1.3) and (1.4), respectively.

    Definition 1.5. [20] Let FL1([η1,η2]×[υ1,υ2]), γ1, γ2(0,1], α>0 and β>0. The conformable fractional integrals of orders α, β of F(δ,ξ) are defined by

    (γ1γ2Jα,βη+1,υ+1F)(δ,ξ)=[1Γ(α)Γ(β)δη1ξυ1((δη1)γ1(tη1)γ1γ1)α1×((ξυ1)γ2(sυ1)γ2γ2)β1F(t,s)(tη1)1γ1(sυ1)1γ2dsdt],   (1.7)
    (γ1γ2Jα,βη2,υ+1F)(δ,ξ)=[1Γ(α)Γ(β)η2δξυ1((η2δ)γ1(η2t)γ1γ1)α1×((ξυ1)γ2(sυ1)γ2γ2)β1F(t,s)(η2t)1γ1(sυ1)1γ2dsdt], (1.8)
    (γ1γ2Jα,βη+1,υ2F)(δ,ξ)=[1Γ(α)Γ(β)δη1υ2ξ((δη1)γ1(tη1)γ1γ1)α1×((υ2ξ)γ2(υ2s)γ2γ2)β1F(t,s)(tη1)1γ1(υ2s)1γ2dsdt], (1.9)

    and

    (γ1γ2Jα,βη2,υ2F)(δ,ξ)=[1Γ(α)Γ(β)η2δυ2ξ((η2δ)γ1(η2t)γ1γ1)α1×((υ2ξ)γ2(υ2s)γ2γ2)β1F(t,s)(η2t)1γ1(υ2s)1γ2dsdt]. (1.10)

    Remark 1.2. If we consider that γ1=γ2=1 in Definition 1.5, then Definition 1. reduces to Definition 1.3.

    By Definition 1.5, we can write the following conformable fractional integrals:

    Definition 1.6. Let FL1([η1,η2]×[υ1,υ2]), γ1, γ2(0,1], α>0 and β>0. In this case, the following equations can be written:

    (γ1Jαη+1F)(δ,υ1+υ22)=1Γ(α)δη1((δη1)γ1(tη1)γ1γ1)α1F(t,υ1+υ22)(tη1)1γ1dt, δ>η1, (1.11)
    (γ1Jαη2F)(δ,υ1+υ22)=1Γ(α)η2δ((η2δ)γ1(η2t)γ1γ1)α1F(t,υ1+υ22)(η2t)1γ1dt, δ<η2, (1.12)
    (γ2Jβυ+1F)(η1+η22,ξ)=1Γ(β)ξυ1((ξυ1)γ2(sυ1)γ2γ2)β1F(η1+η22,s)(sυ1)1γ2ds, ξ>υ1, (1.13)

    and

    (γ2Jβυ2F)(η1+η22,ξ)=1Γ(β)υ2ξ((υ2ξ)γ2(υ2s)γ2γ2)β1F(η1+η22,s)(υ2s)1γ2ds, ξ<υ2. (1.14)

    Theorem 1.1. [21] Assume that F:[η1,η2]R is a convex function. Then, for β>0 and α(0,1], the following inequalities for the fractional conformable integrals hold:

    F(η1+η22)Γ(β+1)αβ2(η2η1)αβ[βJαη+1F(η2)+βJαη2F(η1)]F(η1)+F(η2)2. (1.15)

    For some results connected with fractional integral inequalities, see [22,23,24,25].

    The purpose of this article is to establish the Hermite-Hadamard-type inequality for co-ordinated convex mappings by using the conformable fractional integral operators.

    In this part, we obtain new versions of the Hermite-Hadamard inequality for co-ordinated convex functions involving conformable fractional integrals.

    Theorem 2.1. Let F:ΔR be co-ordinated convex on Δ and FL1(Δ). Then, we have the following Hermite-Hadamard inequality for conformable fractional integrals

    F(η1+η22,υ1+υ22)Γ(α+1)Γ(β+1)γα1γβ24(η2η1)γ1α(υ2υ1)γ2β[γ1γ2Jα,βη+1,υ+1F(η2,υ2)+γ1γ2Jα,βη+1,υ2F(η1,υ1)+γ1γ2Jα,βη+2,υ+1F(η1,υ2)+γ1γ2Jα,βη2,υ2F(η2,υ1)]F(η1,υ1)+F(η1,υ2)+F(η2,υ1)+F(η2,υ2)4. (2.1)

    Proof. For t, s[0,1], we can write

    F(η1+η22,υ1+υ22)=F(14(tη1+(1t)η2,sυ1+(1s)υ2)+14(tη1+(1t)η2+(1s)υ1+sυ2)+14(1t)η1+tη2,sυ1+(1s)υ2)+14(1t)η1+tη2+(1s)υ1+sυ2).

    With the help of the co-ordinated convexity of F, we have

    F(η1+η22,υ1+υ22)14(F(tη1+(1t)η2,sυ1+(1s)υ2)+F(tη1+(1t)η2+(1s)υ1+sυ2)+F(1t)η1+tη2,sυ1+(1s)υ2)+F(1t)η1+tη2+(1s)υ1+sυ2)F(η1,υ1)+F(η1,υ2)+F(η2,υ1)+F(η2,υ2)4. (2.2)

    If we multiply the inequality (2.2) by (1(1t)γ1γ1)α1(1t)γ11(1(1s)γ2γ2)β1(1s)γ21 and integrate the resulting inequality on [0,1]×[0,1],we have

    F(η1+η22,υ1+υ22)×1010(1(1t)γ1γ1)α1(1t)γ11(1(1s)γ2γ2)β1(1s)γ21dsdt14[1010(1(1t)γ1γ1)α1(1t)γ11(1(1s)γ2γ2)β1(1s)γ21×(F(tη1+(1t)η2,sυ1+(1s)υ2)+F(tη1+(1t)η2,(1s)υ1+sυ2))dsdt+1010(1(1t)γ1γ1)α1(1t)γ11(1(1s)γ2γ2)β1(1s)γ21×(F((1t)η1+tη2,sυ1+(1s)υ2)+F((1t)η1+tη2,(1s)υ1+sυ2))dsdt]F(η1,υ1)+F(η1,υ2)+F(η2,υ1)+F(η2,υ2)4×1010(1(1t)γ1γ1)α1(1t)γ11(1(1s)γ2γ2)β1(1s)γ21dsdt. (2.3)

    By applying the change of variables technique, we get

    1010(1(1t)γ1γ1)α1(1t)γ11(1(1s)γ2γ2)β1(1s)γ21×F(tη1+(1t)η2,sυ1+(1s)υ2)dsdt=1(η2η1)(υ2υ1)η2η1υ2υ1(1(δη1η2η1)γ1γ1)α1(δη1η2η1)γ11×(1(ξυ1υ2υ1)γ2γ2)β1(ξυ1υ2υ1)γ21dξdδ=(1η2η1)γ1α(1υ2υ1)γ2βη2η1υ2υ1((η2η1)γ1(δη1)γ1γ1)α1×((υ2υ1)γ2(ξυ1)γ2γ2)β1dξ(ξυ1)1γ2dδ(δη1)1γ1=Γ(α)Γ(β)(η2η1)γ1α(υ2υ1)γ2β(γ1γ2Iα,βη+1,υ+1F)(η2,υ2). (2.4)

    Similarly we have

    1010(1(1t)γ1γ1)α1(1t)γ11(1(1s)γ2γ2)β1(1s)γ21×F(tη1+(1t)η2,(1s)υ1+sυ2)dsdt=Γ(α)Γ(β)(η2η1)γ1α(υ2υ1)γ2β(γ1γ2Iα,βη+1,υ2F)(η2,υ1), (2.5)
    1010(1(1t)γ1γ1)α1(1t)γ11(1(1s)γ2γ2)β1(1s)γ21×F((1t)η1+tη2,sυ1+(1s)υ2)dsdt=Γ(α)Γ(β)(η2η1)γ1α(υ2υ1)γ2β(γ1γ2Iα,βη2,υ+1F)(η1,υ2), (2.6)

    and

    1010(1(1t)γ1γ1)α1(1t)γ11(1(1s)γ2γ2)β1(1s)γ21×F((1t)η1+tη2,(1s)υ1+sυ2)dsdt=Γ(α)Γ(β)(η2η1)γ1α(υ2υ1)γ2β(γ1γ2Iα,βη2υ2F)(η1,υ1). (2.7)

    On the other side, we have

    1010(1(1t)γ1γ1)α1(1t)γ11(1(1s)γ2γ2)β1(1s)γ21dsdt=1γα1γβ2αβ. (2.8)

    If we substitute the Eqs (2.4)–(2.8) in (2.3), then we get

    F(η1+η22,υ1+υ22)1γα1γβ2αβ14[Γ(α)Γ(β)(η2η1)γ1α(υ2υ1)γ2β(γ1γ2Jα,βη+1,υ+1F)(η2,υ2)+Γ(α)Γ(β)(η2η1)γ1α(υ2υ1)γ2β(γ1γ2Jα,βη+1,υ2F)(η2,υ1)+Γ(α)Γ(β)(η2η1)γ1α(υ2υ1)γ2β(γ1γ2Jα,βη2,υ+1F)(η1,υ2)+Γ(α)Γ(β)(η2η1)γ1α(υ2υ1)γ2β(γ1γ2Jα,βη2,υ2F)(η1,υ1)]F(η1,υ1)+F(η1,υ2)+F(η2,υ1)+F(η2,υ2)41γα1γβ2αβ, (2.9)

    which concludes the proof.

    Remark 2.1. In Theorem 2.1, if we choose γ1=1 and γ2=1, then we have the following inequalities for Riemann-Liouville fractional integrals

    F(η1+η22,υ1+υ22)Γ(α+1)Γ(β+1)4(η2η1)α(υ2υ1)β[Iα,βη+1,υ+1F(η2,υ2)+Iα,βη+1,υ2F(η2,υ1) +Iα,βη2,υ+1F(η1,υ2)+Iα,βη2,υ2F(η1,υ1)]F(η1,υ1)+F(η1,υ2)+F(η2,υ1)+F(η2,υ2)4, (2.10)

    which is proved by Sarikaya in [23, Theorem 3].

    Remark 2.2. In Theorem 2.1, if we choose γ1=1, γ2=1,α=1 and β=1, then we have the following inequalities

    F(η1+η22,υ1+υ22)1(η2η1)(υ2υ1)η2η1υ2υ1F(t,s)dsdtF(η1,υ1)+F(η1,υ2)+F(η2,υ1)+F(η2,υ2)4, (2.11)

    which is given by Alomari and Darus [26, Theorem 1.1].

    Theorem 2.2. Let F:ΔR be co-ordinated convex on Δ and FL1(Δ). Then the following Hermite-Hadamard inequality for conformable fractional integrals holds:

    F(η1+η22,υ1+υ22)Γ(α+1)γα14(η2η1)γ1α[γ1Jαη+1F(η2,υ1+υ22)+γ1Jαη2F(η1,υ1+υ22)]+Γ(β+1)γβ24(υ2υ1)γ2β[γ2Jβυ+1F(η1+η22,υ2)+γ2Jβυ2F(η1+η22,υ1)]Γ(α+1)Γ(β+1)γα1γβ24(η2η1)γ1α(υ2υ1)γ2β[γ1γ2Jα,βη+1,υ+1F(η2,υ2)+γ1γ2Jα,βη+1,υ2F(η2,υ1)+γ1γ2Jα,βη2,υ+1F(η1,υ2)+γ1γ2Jα,βη2,υ2F(η1,υ1)]Γ(α+1)γα18(η2η1)γ1α[γ1Jαη+1F(η2,υ1)+γ1Jαη+1F(η2,υ2)+γ1Jαη2F(η1,υ1)+γ1Jαη2F(η1,υ2)]+Γ(β+1)γβ28(υ2υ1)γ2β[γ2Jβυ+1F(η1,υ2)+γ2Jβυ+1F(η2,υ2)+γ2Jβυ2F(η1,υ1)+γ2Jβυ2F(η2,υ1)]F(η1,υ1)+F(η1,υ2)+F(η2,υ1)+F(η2,υ2)4 (2.12)

    for γ1,γ2(0,1], α>0, β>0.

    Proof. Since F:ΔR is a co-ordinated convex function, then the function hδ:[υ1,υ2]R, hδ(ξ)=F(δ,ξ) is convex on [υ1,υ2] for all δ[η1,η2]. Then, by applying (1.15), we can write

    hδ(υ1+υ22)Γ(β+1)γβ22(υ2υ1)γ2β[ βJγ2υ+1hδ(υ2)+ βJγ2υ2hδ(υ1)]hδ(υ2)+hδ(υ1)2, δ[η1,η2]. (2.13)

    That is,

    F(δ,υ1+υ22)βγβ22(υ2υ1)γ2β[υ2υ1((υ2υ1)γ2(ξυ1)γ2γ2)β1F(δ,ξ)(ξυ1)1γ2dξ+υ2υ1((υ2υ1)γ2(υ2ξ)γ2γ2)β1F(δ,ξ)(υ2ξ)1γ2dξ]F(δ,υ1)+F(δ,υ2)2 (2.14)

    for all δ[η1,η2]. Then multiplying both sides of (2.14) by

    αγα12(η2η1)γ1α((η2η1)γ1(δη1)γ1γ1)α11(δη1)1γ1

    and integrating with respect to δ over [η1,η2], we have

    αγα12(η2η1)γ1αη2η1((η2η1)γ1(δη1)γ1γ1)α1F(δ,υ1+υ22)(δη1)1γ1dδαβγα1γβ24(η2η1)γ1α(υ2υ1)γ2β×[η2η1υ2υ1((η2η1)γ1(δη1)γ1γ1)α1((υ2υ1)γ2(ξυ1)γ2γ2)β1F(δ,ξ)(δη1)1γ1(ξυ1)1γ2dξdδ+η2η1υ2υ1((η2η1)γ1(δη1)γ1γ1)α1((υ2υ1)γ2(υ2ξ)γ2γ2)β1F(δ,ξ)(δη1)1γ1(υ2ξ)1γ2dξdδ]αγα14(η2η1)γ1α[η2η1((η2η1)γ1(δη1)γ1γ1)α1F(δ,υ1)(δη1)1γ1dδ +η2η1((η2η1)γ1(δη1)γ1γ1)α1F(δ,υ2)(δη1)1γ1dδ]. (2.15)

    Similarly, let us multiply both sides of (2.14) by

    αγα12(η2η1)γ1α((η2η1)γ1(η2δ)γ1γ1)α11(η2δ)1γ1

    and integrate with respect to δ on the interval [η1,η2]; then, we have

    αγα12(η2η1)γ1αη2η1((η2η1)γ1(η2δ)γ1γ1)α1F(δ,υ1+υ22)(η2η1)1γ1dδαβγα1γβ24(η2η1)γ1α(υ2υ1)γ2β×[η2η1υ2υ1((η2η1)γ1(η2δ)γ1γ1)α1((υ2υ1)γ2(ξυ1)γ2γ2)β1F(δ,ξ)(η2δ)1γ1(ξυ1)1γ2dξdδ+η2η1υ2υ1((η2η1)γ1(η2δ)γ1γ1)α1((υ2υ1)γ2(υ2ξ)γ2γ2)β1F(δ,ξ)(η2δ)1γ1(υ2ξ)1γ2dξdδ]αγα14(η2η1)γ1α[η2η1((η2η1)γ1(η2δ)γ1γ1)α1F(δ,υ1)(η2δ)1γ1dδ +η2η1((η2η1)γ1(η2δ)γ1γ1)α1F(δ,υ2)(η2δ)1γ1dδ]. (2.16)

    In the same way, since F:ΔR is a co-ordinated convex function, the function gξ:[η1,η1]R, gξ(δ)=F(δ,ξ) is convex on [η1,η1] for all ξ[υ1,υ2]. Then, by applying (1.15), we can write,

    gξ(η1+η22)Γ(α+1)γα12(η2η1)γ1α[ αJγ1η+1gξ(η2)+ αJγ1η2gξ(η1)]gξ(η1)+gξ(η2)2. (2.17)

    That is,

    F(η1+η22,ξ)αγα12(η2η1)γ1α[η2η1((η2η1)γ1(δη1)γ1γ1)α1F(δ,ξ)(δη1)1γ1dδ+η2η1((η2η1)γ1(η2δ)γ1γ1)α1F(δ,ξ)(η2δ)1γ1dδ]F(η1,ξ)+F(η2,ξ)2 (2.18)

    for all ξ[υ1,υ2]. Then multiplying both sides of (2.18) by

    βγβ22(υ2υ1)γ2β((υ2υ1)γ2(ξυ1)γ2γ2)β11(ξυ1)1γ2

    and integrating with respect to ξ over [υ1,υ2], we have,

    βγβ22(υ2υ1)γ2βυ2υ1((υ2υ1)γ2(ξυ1)γ2γ2)β1F(η1+η22,ξ)(ξυ1)1γ2dξαβγα1γβ24(η2η1)γ1α(υ2υ1)γ2β×[η2η1υ2υ1((η2η1)γ1(δη1)γ1γ1)α1((υ2υ1)γ2(ξυ1)γ2γ2)β1F(δ,ξ)(δη1)1γ1(ξυ1)1γ2dξdδ+η2η1υ2υ1((η2η1)γ1(η2δ)γ1γ1)α1((υ2υ1)γ2(ξυ1)γ2γ2)β1F(δ,ξ)(η2δ)1γ1(ξυ1)1γ2dξdδ]βγβ24(υ2υ1)γ2β[υ2υ1((υ2υ1)γ2(ξυ1)γ2γ2)β1F(η1,ξ)(ξυ1)1γ2dξ +υ2υ1((υ2υ1)γ2(ξυ1)γ2γ2)β1F(η2,ξ)(ξυ1)1γ2dξ]. (2.19)

    Similarly, let us multiply both sides of (2.18) by

    βγβ22(υ2υ1)γ2β((υ2υ1)γ2(υ2ξ)γ2γ2)β11(υ2ξ)1γ2

    and integrate with respect to ξ in the interval [υ1,υ2]; then, we have

    βγβ22(υ2υ1)γ2βυ2υ1((υ2υ1)γ2(υ2ξ)γ2γ2)β1F(η1+η22,ξ)(υ2ξ)1γ2dξαβγα1γβ24(υ2υ1)γ1α(υ2υ1)γ2β×[η2η1υ2υ1((η2η1)γ1(δη1)γ1γ1)α1((υ2υ1)γ2(υ2ξ)γ2γ2)β1F(δ,ξ)(δη1)1γ1(υ2ξ)1γ2dξdδ+η2η1υ2υ1((η2η1)γ1(η2δ)γ1γ1)α1((υ2υ1)γ2(υ2ξ)γ2γ2)β1F(δ,ξ)(η2δ)1γ1(υ2ξ)1γ2dξdδ]βγβ24(υ2υ1)γ2β[υ2υ1((υ2υ1)γ2(υ2ξ)γ2γ2)β1F(η1,ξ)(υ2ξ)1γ2dξ +υ2υ1((υ2υ1)γ2(υ2ξ)γ2γ2)β1F(η2,ξ)(υ2ξ)1γ2dξ]. (2.20)

    If we add (2.15), (2.16), (2.19) and (2.20) and divide by 2, we get,

    Γ(α+1)γα14(η2η1)γ1α[γ1Jαη+1F(η2,υ1+υ22)+ γ1Jαη2F(η1,υ1+υ22)]+Γ(β+1)γβ22(υ2υ1)γ2β[γ2Jβυ+1F(η1+η22,υ2)+ γ2Jβυ2F(η1+η22,υ1)]Γ(α+1)Γ(β+1)γα1γβ24(η2η1)γ1α(υ2υ1)γ2β[γ1γ2Jα,βη+1,υ+1F(η2,υ2)+ γ1γ2Jα,βη+1,υ2F(η2,υ1) + γ1γ2Jα,βη2,υ1+F(η1,υ2)+ γ1γ2Jα,βη2,υ2F(η1,υ1)]Γ(α+1)γα18(η2η1)γ1α[γ1Jαη+1F(η2,υ1)+ γ1Jαη+1F(η2,υ2)+ γ1Jαη2F(η1,υ1)+ γ1Jαη2F(η1,υ2)]+Γ(β+1)γβ28(η2η1)γ2β[γ2Jβυ+1F(η1,υ2)+ γ2Jβυ+1F(η2,d)+ γ2Jβυ2F(η1,υ1)+ γ2Jβυ2F(η2,υ1)], (2.21)

    which give the second and the third inequalities in (2.12).

    Now, let us write δ=η1+η22 on the left side of the inequality (2.14); we have

    F(η1+η22,υ1+υ22)βγβ22(υ2υ1)γ2β[υ2υ1((υ2υ1)γ2(ξυ1)γ2γ2)β1F(η1+η22,ξ)(ξυ1)1γ2dξ+υ2υ1((υ2υ1)γ2(υ2ξ)γ2γ2)β1F(η1+η22,ξ)(υ2ξ)1γ2dξ], (2.22)

    and then, incorporating ξ=υ1+υ22 on the left side of the inequality (2.18), we have

    F(η1+η22,υ1+υ22)αγα12(η2η1)γ1α[η2η1((η2η1)γ1(δη1)γ1γ1)α1F(δ,υ1+υ22)(δη1)1γ1dδ+η2η1((η2η1)γ1(η2δ)γ1γ1)α1F(δ,υ1+υ22)(η2δ)1γ1dδ]. (2.23)

    If we add the inequalities (2.22) and (2.23) and divide by 2, then we get the following inequality

    F(η1+η22,υ1+υ22)Γ(α+1)γα14(η2η1)γ1α[γ1Jαη1+F(η2,υ1+υ22)+ γ1Jαη2F(η1,υ1+υ22)+Γ(β+1)γβ24(υ2υ1)γ2β[γ1Jαυ+1F(η1+η22,υ2)+ γ1Jαυ2F(η1+η22,υ1)]]. (2.24)

    The inequality in (2.24) is the first inequality of (2.12).

    Finally, assuming that ξ=υ1 and ξ=υ2 on the right-hand side of (2.18), we have

    αγα12(η2η1)γ1α[η2η1((η2η1)γ1(δη1)γ1γ1)α1F(δ,υ1)(δη1)1γ1dδ+η2η1((η2η1)γ1(η2δ)γ1γ1)α1F(δ,υ1)(η2δ)1γ1dδ]F(η1,υ1)+F(η2,υ1)2 (2.25)

    and

    αγα12(η2η1)γ1α[η2η1((η2η1)γ1(δη1)γ1γ1)α1F(δ,υ2)(δη1)1γ1dδ+η2η1((η2η1)γ1(η2δ)γ1γ1)α1F(δ,υ2)(η2δ)1γ1dδ]F(η1,υ2)+F(η2,υ2)2, (2.26)

    respectively. Likewise, assuming that δ=η1 and δ=η2 on the right-hand side of (2.14), we have

    βγβ22(υ2υ1)γ2β[υ2υ1((υ2υ1)γ2(ξυ1)γ2γ2)β1F(η1,ξ)(ξυ1)1γ2dξ+υ2υ1((υ2υ1)γ2(υ2ξ)γ2γ2)β1F(η1,ξ)(υ2ξ)1γ2dξ]F(η1,υ1)+F(η1,υ2)2 (2.27)

    and

    βγβ22(υ2υ1)γ2β[υ2υ1((υ2υ1)γ2(ξυ1)γ2γ2)β1F(η2,ξ)(ξυ1)1γ2dξ+υ2υ1((υ2υ1)γ2(υ2ξ)γ2γ2)β1F(η2,ξ)(υ2ξ)1γ2dξ]F(η2,υ1)+F(η2,υ2)2, (2.28)

    respectively. If we add the inequalites (2.25)–(2.28) and divide by 4, then we get the following inequality

    Γ(α+1)γα18(η2η1)γ1α[γ1Jαη+1F(η2,υ1)+ γ1Jαη+1F(η2,υ2)+ γ1Jαη2F(η1,υ1)+ γ1Jαη2F(η1,υ2)]+2γ1β3Γ(β+1)γβ28(υ2υ1)γ2β[γ2Jβυ+1F(η1,υ2)+ γ2Jβυ+1F(η2,υ2)+ γ2Jβυ2F(η1,υ1)+ γ2Jβυ2F(η2,υ1)]F(η1,υ1)+F(η1,υ2)+F(η2,υ1)+F(η2,υ2)4,

    which gives the last inequality in (2.12). This completes the proof.

    Remark 2.3. In Theorem 2.2, if we choose γ1=1 and γ2=1, then we have the following Hermite-Hadamard inequalities for Riemann-Liouville fractional integrals

    F(η1+η22,υ1+υ22)Γ(α+1)4(η2η1)α[Iαη+1F(η2,υ1+υ22)+Iαη2F(η1,υ1+υ22)]+Γ(β+1)4(υ2υ1)β[Iβυ+1F(η1+η22,υ2)+Iβυ2F(η1+η22,υ1)]Γ(α+1)Γ(β+1)4(η2η1)α(υ2υ1)β[Iα,βη+1,υ+1F(η2,υ2)+Iα,βη+1,υ2F(η2,υ1) +Iα,βη2,υ+1F(η1,υ2)+Iα,βη2,υ2F(η1,υ1)]Γ(α+1)8(η2η1)α[Iαη+1F(η2,υ1)+Iαη+1F(η2,υ2)+Iαη2F(η1,υ1)+Iαη2F(η1,υ2)]+Γ(β+1)8(υ2υ1)β[Iβυ+1F(η1,υ2)+Iβυ+1F(η2,υ2)+Iβυ2F(η1,υ1)+Iβυ2F(η2,υ1)]F(η1,υ1)+F(η1,υ2)+F(η2,υ1)+F(η2,υ2)4 (2.29)

    which is proved by Sarikaya in [23, Theorem 4].

    Remark 2.4. In Theorem 2.2, if we choose γ1=1, γ2=1,α=1 and β=1,we have the following Hermite-Hadamard inequalities

    F(η1+η22,υ1+υ22)12(η2η1)η2η1F(t,υ1+υ22)dt+12(υ2υ1)υ2υ1F(η1+η22,s)ds1(η2η1)(υ2υ1)η2η1υ2υ1F(t,s)dsdt14(η2η1)[η2η1F(t,υ1)dt+η2η1F(t,υ2)dt]+14(υ2υ1)[υ2υ1F(η1,s)ds+υ2υ1F(η2,s)ds]F(η1,υ1)+F(η1,υ2)+F(η2,υ1)+F(η2,υ2)4, (2.30)

    which is proved by Dragomir in [3, Theorem 1].

    In this section, we prove a trapezoid type inequality by using conformable fractional integrals. First, we need the following lemma.

    Lemma 3.1. Let F:ΔR2R be a partial differentiable mapping. If 2F/tsL1(Δ), then the following equality holds:

    F(η1,υ1)+F(η1,υ2)+F(η2,υ1)+F(η2,υ2)4+Γ(α+1)Γ(β+1)γα1γβ24(η2η1)γ1α(υ2υ1)γ2β×[γ1γ2Jα,βη+1,υ+1F(η2,υ2)+γ1γ2Jα,βη+1,υ2F(η2,υ1)+γ1γ2Jα,βη2,υ+1F(η1,υ2)+γ1γ2Jα,βη2,υ2F(η1,υ1)]A=(η2η1)(υ2υ1)γα1γβ241010[(1(1t)γ1γ1)α(1tγ1γ1)α]×[(1(1s)γ2γ2)β(1sγ2γ2)β]2Fts(tη1+(1t)η2,sυ1+(1s)υ2)dsdt, (3.1)

    where

    A=Γ(β+1)γβ24(υ2υ1)γ2β[γ2Jβυ+1F(η1,υ2)+γ2Jβυ+1F(η2,υ2)+γ2Jβυ2F(η1,υ1)+γ2Jβυ2F(η2,υ1)]+Γ(α+1)γα14(η2η1)γ1α[γ1Jαη+1F(η2,υ1)+γ1Jαη+1F(η2,υ2)+γ1Jαη2F(η1,υ1)+γ1Jαη2F(η1,υ2)]. (3.2)

    Proof. By integration by parts, we get

    I1=1010(1(1t)γ1γ1)α(1(1s)γ2γ2)β2Fts(tη1+(1t)η2,sυ1+(1s)υ2)dsdt=10(1(1s)γ2γ2)β{1η2η1(1(1t)γ1γ1)αFs(tη1+(1t)η2,sυ1+(1s)υ2)|10+αη2η110(1(1t)γ1γ1)α1(1t)γ11Fs(tη1+(1t)η2,sυ1+(1s)υ2)dt}ds=10(1(1s)γ2γ2)β{1γα1(η2η1)Fs(η1,sυ1+(1s)υ2)+αη2η110(1(1t)γ1γ1)α1(1t)γ11Fs(tη1+(1t)η2,sυ1+(1s)υ2)dt}ds=1γα1(η2η1)10(1(1s)γ2γ2)βFs(η1,sυ1+(1s)υ2)ds+αη2η110(1(1t)γ1γ1)α1(1t)γ11[10(1(1s)γ2γ2)β×Fs(tη1+(1t)η2,sυ1+(1s)υ2)ds]dt=1γα1γβ2(η2η1)(υ2υ1)F(η1,υ1)βγα1(η2η1)(υ2υ1)10(1(1s)γ2γ2)β1(1s)γ21F(η1,sυ1+(1s)υ2)dsαγβ2(η2η1)(υ2υ1)10(1(1t)γ1γ1)α1(1t)γ11F(tη1+(1t)η2,υ1)dt+αβ(υ2υ1)(η2η1)1010(1(1t)γ1γ1)α1(1t)γ11×(1(1s)γ2γ2)β1(1s)γ21F(tη1+(1t)η2,sυ1+(1s)υ2)dsdt. (3.3)

    Similarly, by integration by parts, it follows that

    I2=1010(1(1t)γ1γ1)α(1sγ2γ2)β2Fts(tη1+(1t)η2,sυ1+(1s)υ2)dsdt=1γα1γβ2(η2η1)(υ2υ1)F(η1,υ2)+βγα1(η2η1)(υ2υ1)10(1sγ2γ2)β1sγ21F(η1,sυ1+(1s)υ2)ds+αγβ2(η2η1)(υ2υ1)10(1(1t)γ1γ1)α1(1t)γ11F(tη1+(1t)η2,υ2)dtαβ(υ2υ1)(η2η1)1010(1(1t)γ1γ1)α1(1t)γ11(1sγ2γ2)β1sγ21×F(tη1+(1t)η2,sυ1+(1s)υ2)dsdt, (3.4)
    I3=1010(1tγ1γ1)α(1(1s)γ2γ2)β2Fts(tη1+(1t)η2,sυ1+(1s)υ2)dsdt=1γα1γβ2(η2η1)(υ2υ1)F(η2,υ1)+βγα1(η2η1(υ2υ1)10(1(1s)γ2γ2)β1(1s)γ21F(η2,sυ1+(1s)υ2)ds+αγβ2(η2η1)(υ2υ1)10(1tγ1γ1)α1tγ11F(tη1+(1t)η2,υ1)dtαβ(υ2υ1)(η2η1)1010(1tγ1γ1)α1tγ11(1(1s)γ2γ2)β1×(1s)γ21F(tη1+(1t)η2,sυ1+(1s)υ2)dsdt, (3.5)

    and

    I4=1010(1tγ1γ1)α(1sγ2γ2)β2Fts(tη1+(1t)η2,sυ1+(1s)υ2)dsdt=1γα1γβ2(η2η1)(υ2υ1)F(η2,υ2)βγα1(η2η1)(υ2υ1)10(1sγ2γ2)β1sγ21F(η2,sυ1+(1s)υ2)dsαγβ2(η2η1)(υ2υ1)10(1tγ1γ1)α1tγ11F(tη1+(1t)η2,υ2)dt+αβ(υ2υ1)(η2η1)1010(1tγ1γ1)α1tγ11(1sγ2γ2)β1sγ21×F(tη1+(1t)η2,sυ1+(1s)υ2)dsdt. (3.6)

    By using the inequalities comprising (3.3)–(3.6) and applying the change of variables technique to δ=tη1+(1t)η2 and ξ=sυ1+(1s)υ2 for (t,s)[0,1], we can write

    II2I3+I4=F(η1,υ1)+F(η2,υ1)+F(η1,υ1)+F(η2,υ2)γα1γβ2(η2η1)(υ2υ1)Γ(β+1)γα1(η2η1)γ1α+1(υ2υ1)[γ2Jβυ+1F(η1,υ2)+ γ2Jβυ+1F(η2,υ2)+ γ2Jβυ2F(η1,υ1)+ γ2Jβυ2F(η2,υ1)]Γ(α+1)γβ2(η2η1)(υ2υ1)γ2β+1[γ1Jαη+1F(η2,υ1)+ γ1Jαη+1F(η2,υ2)+ γ1Jαη2F(η1,υ1)+ γ1Jαη2F(η1,υ2)]+Γ(α+1)Γ(β+1)(η2η1)γ1α+1(υ2υ1)γ2β+1×[γ1γ2Jα,βη+1,υ+1F(η2,υ2)+ γ1γ2Jα,βη+1,υ2F(η2,υ1)+ γ1γ2Jα,βη2,υ+1F(η1,υ2)+ γ1γ2Jα,βη2,υ2F(η1,υ1)]. (3.7)

    Multiplying the both sides of (3.7) by (η2η1)(υ2υ1)γα1γβ24, we obtain the required result (3.1).

    Now, we can present the following trapezoid-type inequality.

    Theorem 3.1. Let F:ΔR2R be a partial differentiable mapping on Δ:=[η1,η2]×[υ1,υ2] in R2 with 0η1η2,0υ1υ2,γ1,γ20, and α,β(0,1]. If |2F/ts| is a convex function on the Δ, then the following inequality holds:

    |F(η1,υ1)+F(η1,υ2)+F(η2,υ1)+F(η2,υ2)4+Γ(α+1)Γ(β+1)γα1γβ24(η2η1)γ1α(υ2υ1)γ2β×[γ1γ2Jα,βη+1,υ+1F(η2,υ2)+γ1γ2Jα,βη+1,υ2F(η2,υ1)+γ1γ2Jα,βη2,υ+1F(η1,υ2)+γ1γ2Jα,βη2,υ2F(η1,υ1)]A|(η2η1)(υ2υ1)4γ1γ2[2B(1γ1,α+1,(12)γ1)B(1γ1,α+1)]×[2B(1γ2,β+1,(12)γ2)B(1γ2,β+1)]×[|2Fts(η1,υ1)|+|2Fts(η2,υ1)|+|2Fts(η1,υ2)|+|2Fts(η2,υ2)|], (3.8)

    where A is defined as in (3.2), and B and B are the beta function and the incomplete beta function, respectively, defined by

    B(μ,ν)=10ζμ1(1ζ)ν1dζ,B(μ,ν,r)=r0ζμ1(1ζ)ν1dζ.

    Proof. From Lemma 3.1, we have

    |F(η1,υ1)+F(η1,υ2)+F(η2,υ1)+F(η2,υ2)4+Γ(α+1)Γ(β+1)γα1γβ24(η2η1)γ1α(υ2υ1)γ2β×[γ1γ2Jα,βη+1,υ+1F(η2,υ2)+ γ1γ2Jα,βη+1,υ2F(η2,υ1)+ γ1γ2Jα,βη2,υ+1F(η1,υ2)+ γ1γ2Jα,βη2,υ2F(η1,υ1)]A|(η2η1)(υ2υ1)γα1γβ241010|(1(1t)γ1γ1)α(1tγ1γ1)α||(1(1s)γ2γ2)β(1sγ2γ2)β|×|2Fts(tη1+(1t)η2,sυ1+(1s)υ2)|dsdt. (3.9)

    Since |2Fts| is a co-ordinated convex function on Δ, then one has

    |F(η1,υ1)+F(η1,υ2)+F(η2,υ1)+F(η2,υ2)4+Γ(α+1)Γ(β+1)γα1γβ24(η2η1)γ1α(υ2υ1)γ2β×[γ1γ2Jα,βη+1,υ+1F(η2,υ2)+ γ1γ2Jα,βη+1,υ2F(η2,υ1)+ γ1γ2Jα,βη2,υ+1F(η1,υ2)+ γ1γ2Jα,βη2,υ2F(η1,υ1)]A|(η2η1)(υ2υ1)γα1γβ241010|(1(1t)γ1γ1)α(1tγ1γ1)α||(1(1s)γ2γ2)β(1sγ2γ2)β|×[ts|2Fts(η1,υ1)|+s(1t)|2Fts(η2,υ1)|+t(1s)|2Fts(η1,υ2)|+(1s)(1t)|2Fts(η2,υ2)|]dsdt=(η2η1)(υ2υ1)γα1γβ24(1010ts|(1(1t)γ1γ1)α(1tγ1γ1)α||(1(1s)γ2γ2)β(1sγ2γ2)β|dsdt)×[|2Fts(η1,υ1)|+|2Fts(η2,υ1)|+|2Fts(η1,υ2)|+|2Fts(η2,υ2)|]. (3.10)

    Here, we have

    10t|(1(1t)γ1γ1)α(1tγ1γ1)α|dt=1γα1[120t[(1tγ1)α(1(1t)γ1)α]dt+112t[(1(1t)γ1)α(1tγ1)α]dt]=1γα1[120t[(1tγ1)α(1(1t)γ1)α]dt+120(1t)[(1tγ1)α(1(1t)γ1)α]dt]=1γα1120[(1tγ1)α(1(1t)γ1)α]dt=1γα+11[2B(1γ1,α+1,(12)γ1)B(1γ1,α+1)]

    and similarly

    10s|(1(1s)γ2γ2)β(1sγ2γ2)β|ds=1γβ+12[2B(1γ2,β+1,(12)γ2)B(1γ2,β+1)].

    This completes the proof.

    Corollary 3.1. In Theorem 3.1, if we choose γ1=1 and γ2=1, then we have the following trapezoid type inequalities for Riemann-Liouville fractional integrals

    |F(η1,υ1)+F(η1,υ2)+F(η2,υ1)+F(η2,υ2)4+Γ(α+1)Γ(β+1)4(η2η1)α(υ2υ1)β×[Iα,βη+1,υ+1F(η2,υ2)+Iα,βη+1,υ2F(η2,υ1)+Iα,βη2,υ+1F(η1,υ2)+Iα,βη2,υ2F(η1,υ1)]B|(η2η1)(υ2υ1)4(α+1)(β+1)(112α)(112β)×[|2Fts(η1,υ1)|+|2Fts(η2,υ1)|+|2Fts(η1,υ2)|+|2Fts(η2,υ2)|],

    where

    B=Γ(β+1)4(υ2υ1)β[Iβυ+1F(η1,υ2)+Iβυ+1F(η2,υ2)+Iβυ2F(η1,υ1)+Iβυ2F(η2,υ1)]+Γ(α+1)4(η2η1)α[Iαη+1F(η2,υ1)+Iαη+1F(η2,υ2)+Iαη2F(η1,υ1)+Iαη2F(η1,υ2)].

    Remark 3.1. In Theorem 3.1, if we choose γ1=1, γ2=1,α=1 and β=1,we have the following trapezoid type inequality

    |F(η1,υ1)+F(η1,υ2)+F(η2,υ1)+F(η2,υ2)4+1(η2η1)(υ2υ1)η2η1υ2υ1F(t,s)dsdt12(η2η1)η2η1[F(t,υ1)+F(t,υ2)]dt12(υ2υ1)υ2υ1[F(η1,s)+F(η2,s)]ds|(η2η1)(υ2υ1)64[|2Fts(η1,υ1)|+|2Fts(η2,υ1)|+|2Fts(η1,υ2)|+|2Fts(η2,υ2)|],

    which is proved by Sarikaya et al. in [27, Theorem 2].

    In this study, new Hermite-Hadamard type inequalities for coordinated convex functions were obtained through the use of conformable fractional integrals. Some remarks have been presented to show the relationship between our results and earlier obtained results. Furthermore, an identity has been established for partially differentiable functions. By using this equality and the concept of coordinated convexity, a trapezoid type inequality for conformable fractional integrals has been proved. This study demonstrates how conformable fractional integrals can be used in Hermite-Hadamard type inequalities for coordinated convex functions. It also introduces a new identity for partially differentiable functions. These results indicate that such inequalities and identities can be applied to a wide range of studies. For researchers, the findings of this study can provide a basis for further studies in this field.

    The authors declare that they have not used artificial intelligence tools in the creation of this article.

    The authors declare that there are no conflicts of interest regarding the publication of this article.



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