Research article

Estimates of trapezium-type inequalities for h-convex functions with applications to quadrature formulae

  • Received: 19 December 2020 Accepted: 29 April 2021 Published: 10 May 2021
  • MSC : 26D15, 26D10, 26A33

  • In this article, we develop a new class of trapezium-type inequalities up to twice differentiable h-convex mappings for fractional integrals of Riemann-type. We conclude numerous existing results in literature from our general inequalities. Based on our consequences, we will obtain some quadrature formulas as applications.

    Citation: Muhammad Samraiz, Fakhra Nawaz, Bahaaeldin Abdalla, Thabet Abdeljawad, Gauhar Rahman, Sajid Iqbal. Estimates of trapezium-type inequalities for h-convex functions with applications to quadrature formulae[J]. AIMS Mathematics, 2021, 6(7): 7625-7648. doi: 10.3934/math.2021443

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  • In this article, we develop a new class of trapezium-type inequalities up to twice differentiable h-convex mappings for fractional integrals of Riemann-type. We conclude numerous existing results in literature from our general inequalities. Based on our consequences, we will obtain some quadrature formulas as applications.



    It is familiar that the convexity has an important and key part in many areas due to its broad applications. This idea has been broadened and summed up in different directions (see, e.g., [1,2,3,4,5,6,7,8]). Sanja Varošanec present a class of convex functions [9] which is very helpful for the mathematicians working in the field of mathematical inequalities. There is a famous inequality for convex functions known as Hermite-Hadamard inequality (see, e.g., [10,p. 137]). This inequality provides bounds of the mean value of a continuous convex function Ψ:(a,b)R.

    If Ψ:R on an interval of real numbers, such that ρ,ϱU with ρ<ϱ, we can write

    Ψ(ρ+ϱ2)1ϱρϱρΨ(ϖ)dϖΨ(ρ)+Ψ(ϱ)2. (1.1)

    Fractional calculus has vast applications in inequalities as well as in other physical sciences [11,12]. Recently Hermite-Hadamard inequalities have been reinvestigated and developed by many researchers by using several fractional calculus operators [13,14,15,16,17,18,19,20,21,22]. In [23] Sarikaya et al. studied the Hemite-Hadamard inequalities for Riemann-Liouville fractional integral (RLFI) using convex functions. Lio et al. in [24] investigated the said inequalities by involving RLFI operator for twice differentiable geometric-arithmetically s-convex functions. Such inequalities were examined by Wu et al. in [25] and Iqbal et al. in [26] for k-fractional operators via different convexities. Mevlut Tunc in [27] explored the inequality given by (1.1) for h-convex functions via RLFI operator. To enhance the flow of the work, we present some mathematical preliminaries of the theory of fractional calculus that are required to set up our results.

    Definition 1.1. ([24]) A function Ψ:DR+R+ is known as geometric-arithmetically s-convex on D if for s(0,1] and ,ρ,ϱD with ν[0,1], we have

    Ψ(ρνϱ1ν)νsΨ(ρ)+(1ν)sΨ(ϱ).

    We utilize the class of h-convex SX(h,D), h-concave SV(h,D), s-convex in second sense K2s and quasi convex P(I) functions.

    Sanja Varošanec present the following concept of convex and h-convex functions in [9] which are explained as follows:

    Definition 1.2. A function Ψ:[ρ,ϱ]R is said to be convex if the inequality

    Ψ(ωl+(1ω)r)ωΨ(l)+(1ω)Ψ(r)

    holds for all l,r[ρ,ϱ] with ω[0,1].

    Definition 1.3. Let h>0 be a function such that h:DRR. Then the function Ψ>0 is h-convex, if for Ψ:DRR, the inequality

    Ψ(ωρ+(1ω)ϱ)h(ω)Ψ(ρ)+h(1ω)Ψ(ϱ),ρ,ϱD (1.2)

    is true. The function ΨSV(h,D) is h-concave, if the inequality (1.2) is reversed.

    For more detail about the classes P(D) and K2s, we refer the reader to visit the article [28].

    Definition 1.4. ([29]) The left and right sided RLFI Iαρ+Ψ and IαϱΨ of order α>0 on an interval [ρ,ϱ] are defined by

    Iαρ+Ψ(ω)=1Γ(α)ωρ(ωt)α1Ψ(t)dt,ω>ρ

    and

    IαϱΨ(ω)=1Γ(α)ϱω(tω)α1Ψ(t)dt, ω<ϱ

    respectively. Here Γ(.) represents the Euler-Gamma function defined by

    Γ(t)=0ωt1eωdω,R(t)>0.

    The following results presented by Sarikaya et al. in [23] must be recalled in order to achieve our objectives.

    Theorem 1.5. If Ψ:[ρ,ϱ]R is a positive convex function on [ρ,ϱ] with 0ρ<ϱ and ΨL1[ρ,ϱ], then the inequality

    Ψ(ρ+ϱ2)Γ(α+1)2(ϱρ)ϱ[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]Ψ(ρ)+Ψ(ϱ)2

    holds.

    Lemma 1.6. Let Ψ:[ρ,ϱ]R be a differentiable mapping on (ρ,ϱ) with ρ<ϱ. If ΨL1[ρ,ϱ], then the identity

    Ψ(ρ)+Ψ(ϱ)2Γ(α+1)2(ρa)ϱ[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]=ϱρ210[(1ω)αωα]Ψ(ωρ+(1ω)ϱ)dω

    holds.

    The identity proved by Wang et al. in [30] stated as follows:

    Lemma 1.7. Let Ψ:[ρ,ϱ]R be a twice differentiable mapping on (ρ,ϱ) with ρ<ϱ. If ΨL1[ρ,ϱ], then the equality

    Ψ(ρ)+Ψ(ϱ)2Γ(α+1)2(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]=(ϱρ)22101(1ω)α+1ωα+1α+1Ψ(ωρ+(1ω)ϱ)dω

    is true.

    Lemma 1.8. ([31]) For ω[0,1], we have

    (1ω)α21αωαα[0,1]

    and

    (1ω)α21αωαα[1,].

    Lemma 1.9. ([19]) Assume that Ψ:[ρ,ϱ]R is twice differentiable on (ρ,ϱ) with ρ<ϱ. If ΨL1[ρ,ϱ], then the identity

    Γ(α+1)2(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]Ψ(ρ+ϱ2)=(ϱρ)2210m(ω)Ψ(ωρ+(1ω)ϱ)dω

    is true, where

    m(ω)={ω1(1ω)α+1ωα+1α+1,ω[0,12);(1ω)1(1ω)α+1ωα+1α+1,ω[12,1).

    The k-RLFI introduced by Mubeen et al. in [32] is given in the following definition.

    Definition 1.10. The k-RLFI Iαρ+,kΨ and Iαϱ,kΨ of order α>0, k>0 and ρ0 for ΨL1[ρ,ϱ] are defined by

    Iαρ+,kΨ(ω)=1kΓk(α)ωρ(ωμ)αk1Ψ(μ)dμ,ω>ρ

    and

    Iαϱ,kΨ(ω)=1kΓk(α)ϱω(μω)αk1Ψ(μ)dμ,ω<ϱ

    respectively. Here Γk(.) is the k-Gamma function and I0ρ+,kΨ = I0ϱ,kΨ = Ψ.

    The following results are given in [25].

    Theorem 1.11. Let Ψ:[ρ,ϱ]R be a positive convex mapping for 0ρ<ϱ and ΨL1[ρ,ϱ], then the below inequalities

    Ψ(ρ+ϱ2)Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]Ψ(ρ)+Ψ(ϱ)2

    holds.

    Lemma 1.12. Let Ψ:[ρ,ϱ]R be a differentiable mapping on (ρ,ϱ) with ρ<ϱ. If ΨL1[ρ,ϱ], then the identity for generalized RLFI

    Ψ(ρ)+Ψ(ϱ)2Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]=ϱρ210[(1ω)αkωαk]Ψ(ωρ+(1ω)ϱ)dω

    is true.

    Recently, Iqbal et al. prove the following lemma's in [26].

    Lemma 1.13. Let Ψ:[ρ,ϱ]R be a twice differentiable mapping on (ρ,ϱ) with ρ<ϱ. If ΨL1[ρ,ϱ], then the upcoming identity for generalized RLFI

    Ψ(ρ)+Ψ(ϱ)2Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]=(ϱρ)22101(1ω)αk+1ωαk+1(αk+1)Ψ(ωρ+(1ω)ϱ)dω

    holds.

    Lemma 1.14. Assume that Ψ:[ρ,ϱ]R is twice differentiable mapping on (ρ,ϱ) with ρ<ϱ. Let k>0 and ΨL1[ρ,ϱ], then

    Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]Ψ(ρ+ϱ2)=(ϱρ)2210m(ω)Ψ(ωρ+(1ω)ϱ)dω,

    where

    m(ω)={ω1(1ω)αk+1ωαk+1αk+1,ω[0,12);(1ω)1(1ω)αk+1ωαk+1αk+1,ω[12,1).

    Motivated by a certain class of h-convex functions presented by Sanja Varošanec in [9] and the fractional Hermite-Hadamard inequalities [23,24,25,26,27], we will study such inequalities for h-convex functions via Riemann-type integrals. Our main results are stated in sections below.

    This section focuses on trapezoid-type inequalities for twice-differentiable h-convex functions. In approximating the Riemann integral by a trapezoidal formula, such a type of inequality offers a priori error bounds. In general, they also show that the mid-point rule provides the best approximation of all Riemann sums sampled at the inside points of a given partition in the class.

    Theorem 2.1. Let Iαρ+Ψ and IαϱΨ be the left and right sided RLFI of order α>0. Let Ψ:[ρ,ϱ]R be a positive mapping with ΨL1[ρ,ϱ] and 0ρ<ϱ. If Ψ is h-convex on [ρ,ϱ], then we have the inequality

    Ψ(ρ+ϱ2)h(12)Γ(α+1)(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]αh(12)[Ψ(ρ)+Ψ(ϱ)]10ωα1[h(ω)+h(1ω)]dω. (2.1)

    Proof. Since Ψ is an h-convex function, so we can write

    Ψ(ωx+(1ω)y)h(ω)Ψ(x)+h(1ω)Ψ(y), (2.2)
    Ψ((1ω)x+ωy)h(1ω)Ψ(x)+h(ω)Ψ(y). (2.3)

    For ω=12, we get

    Ψ(x+y2)h(12)Ψ(x)+h(12)Ψ(y)=h(12)[Ψ(x)+Ψ(y)].

    If we choose x=ωρ+(1ω)ϱ and y=ρ(1ω)+ϱω, then we have

    Ψ(ρ+ϱ2)h(12)[Ψ(ωρ+(1ω)ϱ)+Ψ(ρ(1ω)+ωϱ)].

    Multiplying above inequality by ωα1 and integrating with respect to ω over [0,1], we get

    Ψ(ρ+ϱ2)10ωα1dωh(12)[10ωα1Ψ(ωρ+(1ω)ϱ)dω+10ωα1Ψ((1ω)ρ+ωϱ)dω],

    which can also be written as

    1αΨ(ρ+ϱ2)h(12)[10ωα1Ψ(ωρ+(1ω)ϱ)dω+10ωα1Ψ((1ω)ρ+ωϱ)dω]. (2.4)

    Now, replacing ωρ+(1ω)ϱ=u and (1ω)ρ+ωϱ=v, the inequality (2.4) becomes

    1αΨ(ρ+ϱ2)h(12)Γ(α)(ϱρ)α[1Γ(α)ϱρ(ϱu)α1Ψ(u)du+1Γ(α)ϱρ(vρ)α1Ψ(v)dv].

    By managing the terms, we get

    Ψ(ρ+ϱ2)h(12)Γ(α+1)(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]. (2.5)

    By replacing x=ρ, y=ϱ in (2.2) and (2.3) respectively, then adding, we have

    Ψ(ωρ+(1ω)ϱ)+Ψ((1ω)ρ+ωϱ)[h(ω)+h(1ω)][Ψ(ρ)+Ψ(ϱ)]. (2.6)

    Multiplying (2.6) by ωα1, then integrating over [0,1], we get

    h(12)Γ(α+1)(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]αh(12)[Ψ(ρ)+Ψ(ϱ)]10ωα1[h(ω)+h(1ω)]dω. (2.7)

    Combining (2.5) and (2.7), we get the desired result.

    Remark 2.2. If we choose h(ω)=ω in Theorem 2.1, then we get the result of Sarikaya et al. [23,Theorem 2].

    Example 2.3. By plotting graphs of double inequality (2.1) corresponding to the choice Ψ()=e and h(ω)=ω, we prove that both inequalities are correct. Obviously

    Iαρ+eϱ=1Γ(α)ϱρ(ϱ)α1ed (2.8)

    and

    Iαϱeρ=1Γ(α)ϱρ(ρ)α1ed. (2.9)

    By utilizing these expressions into (2.1), we get

    2eρ+ϱ2α(ϱρ)α10[(ϱ)α1+(ρ)α1]edeρ+eϱ. (2.10)

    The three functions given by the double inequality on the left, middle and right sides (2.10) are plotted in Figure 1 against α(0,1]. The functions curves indicate that dual inequality is correct.

    Figure 1.  For the case ρ=0, ϱ=1 the graphs illustrate that the double inequality (2.10) is correct.

    Theorem 2.4. Let Iαρ+Ψ and IαϱΨ be the left and right sided RLFI of order α>0. Let ΨSX(h,D), ρ,ϱD with ρ<ϱ and ΨL1[ρ,ϱ]. Let |Ψ|b be h-convex function, then one has the following inequality

    |Ψ(ρ)+Ψ(ϱ)2Γ(α+1)2(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]|(ϱρ)22(α+1)(12(α+1)g+1)1g(|Ψ(ρ)|b10h(ω)dω+|Ψ(ϱ)|b10h(1ω)dω)1b, (2.11)

    where 1g+1b=1.

    Proof. By using Lemma 1.7, Definition 1.3 and Hölder's inequality respectively, we have

    |Ψ(ρ)+Ψ(ϱ)2Γ(α+1)2(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]|(ϱρ)2210|1(1ω)α+1ωα+1α+1||Ψ(ωρ+(1ω)ϱ)|dω(ϱρ)22(α+1)(10(1(1ω)α+1ωα+1)gdω)1g(10|Ψ(ωρ+(1ω)ϱ)|bdω)1b.

    By utilizing the fact (CD)gCgDg for any C>D0 and g1, we can write

    |Ψ(ρ)+Ψ(ϱ)2Γ(α+1)2(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]|(ϱρ)22(α+1)(10(1(1ω)(α+1)gω(α+1)g)dω)1g(10|Ψ(ωρ+(1ω)ϱ)|bdω)1b(ϱρ)22(α+1)(11g(α+1)+11g(α+1)+1)1g(10(h(ω)|Ψ(ρ)|b+h(1ω)|Ψ(ϱ)|b)dω)1b=(ϱρ)22(α+1)(12g(α+1)+1)1g(|Ψ(ρ)|b10h(ω)dω+|Ψ(ϱ)|b10h(1ω)dω)1b.

    This completes the proof of inequality (2.11).

    Corollary 2.5. If we choose h(ω)=ω, then we get the result for convex function i.e.,

    |Ψ(ρ)+Ψ(ϱ)2Γ(α+1)2(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]|(ϱρ)22(α+1)(12(α+1)g+1)1g(|Ψ(ρ)|b+|Ψ(ϱ)|b2)1b.

    Corollary 2.6. Corresponding to the choice h(ω)=ωs, where s(0,1), we get the result for geometric-arithmetically s-convex function.

    |Ψ(ρ)+Ψ(ϱ)2Γ(α+1)2(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]|(ϱρ)22(α+1)(12(α+1)g+1)1g(|Ψ(ρ)|b+|Ψ(ϱ)|bs+1)1b.

    Example 2.7. By plotting graphs of dual inequality in Theorem 2.4 for a convex function Ψ()=e, corresponding to b=2, g=2 and h()=, we prove the validity of the results.

    (ϱρ)2(α+1)(2α+12α+3)12(e2ρ+e2ϱ2)12eρ+eϱα(ϱρ)αϱρ[(ϱ)α1+(ρ)α1]ed(ϱρ)2(α+1)(2α+12α+3)12(e2ρ+e2ϱ2)12. (2.12)

    The three functions given by the double inequality on the left, middle and right sides (2.12) are plotted in Figure 2 against α(0,1]. The graphs of the functions prove the validity of dual inequality.

    Figure 2.  For the case ρ=0, ϱ=1 the graphs illustrate that the double inequality (2.12) is correct.

    Theorem 2.8. Let Iαρ+Ψ and IαϱΨ be the left and right sided RLFI of order ϱ>0. Let ΨSX(h,D), ρ,ϱD with ρ<ϱ and ΨL1[ρ,ϱ]. Let |Ψ|b be h-convex, then one has inequality via fractional integrals

    |Ψ(ρ)+Ψ(ϱ)2Γ(α+1)2(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]|(ϱρ)2max{21α1,121α}2(α+1)(|Ψ(ρ)|b10h(ω)dω+|Ψ(ϱ)|b10h(1ω)dω)1b,

    where 1g+1b=1.

    Proof. Using Lemma 1.7, Definition 1.3, Lemma 1.8 and Hölder's inequality respectively, we have

    |Ψ(ρ)+Ψ(ϱ)2Γ(α+1)2(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]|(ϱρ)22(α+1)10|1(1ω)α+1ωα+1||Ψ(ωρ+(1ω)ϱ)|dω(ϱρ)22(α+1)(10|1(1ω)α+1ωα+1|gdω)1g(10|Ψ(ωρ+(1ω)ϱ)|bdω)1b. (2.13)

    Now, 2-cases arise.

    Case 1: For ϱ[0,1], the inequality (2.13) becomes

    |Ψ(ρ)+Ψ(ϱ)2Γ(α+1)2(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]|(ϱρ)22(α+1)(10(21α1)gdω)1g(10|Ψ(ωρ+(1ω)ϱ)|bdω)1b(ϱρ)22(α+1)(21α1)(10[h(ω)|Ψ(ρ)|b+h(1ω)|Ψ(ϱ)|b]dω)1b=(ϱρ)2(21α1)2(α+1)(|Ψ(ρ)|b10h(ω)dω+|Ψ(ϱ)|b10h(1ω)dω)1b. (2.14)

    Case-2: For α[1,), the inequality (2.13) can be written as:

    |Ψ(ρ)+Ψ(ϱ)2Γ(α+1)2(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]|(ϱρ)22(α+1)(10(121α)gdω)1g(10|Ψ(ωρ+(1ω)ϱ)|bdω)1b(ϱρ)22(α+1)(121α)(10[h(ω)|Ψ(ρ)|b+h(1ω)|Ψ(ϱ)|b]dω)1b=(ϱρ)2(121α)2(α+1)(|Ψ(ρ)|b10h(ω)dω+|Ψ(ϱ)|b10h(1ω)dω)1b. (2.15)

    By combining (2.14) and (2.15), we get the required result.

    Corollary 2.9. If we choose h(ω)=ω in Theorem 2.8, then we get the result for the classical convex function.

    |Ψ(ρ)+Ψ(ϱ)2Γ(α+1)2(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]|(ϱρ)2max{21α1,121α}2(α+1)(|Ψ(ρ)|b+|Ψ(ϱ)|b2)1b.

    Remark 2.10. If we choos h(ω)=ωs in Theorem 2.8, where s(0,1), then we get the result of Liao et al. [24,Theorem 3.2].

    Theorem 2.11. Let Iαρ+Ψ and IαϱΨ be the left and right sided RLFI of order α>0. Let ΨSX(h,D), ρ,ϱD with ρ<ϱ and ΨL1[ρ,ϱ]. Let |Ψ| be h-convex, then one has the following inequality

    |Ψ(ρ)+Ψ(ϱ)2Γ(α+1)2(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]|(ϱρ)22(α+1)[10[h(ω)|Ψ(ρ)|+h(1ω)|Ψ(ϱ)|]dω10[h(ω)(1ω)α+1|Ψ(ρ)|+h(1ω)(1ω)α+1|Ψ(ϱ)|]dω10[h(ω)ωα+1|Ψ(ρ)|+h(1ω)ωα+1|Ψ(ϱ)|]dω]. (2.16)

    Proof. By using Lemma 1.7 and Definition 1.3 respectively, we have

    |Ψ(ρ)+Ψ(ϱ)2Γ(α+1)2(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]|(ϱρ)2210|1(1ω)α+1ωα+1α+1||Ψ(ωa+(1ω)ϱ)|dω=(ϱρ)22(α+1)10(1(1ω)α+1ωα+1)|Ψ(ωρ+(1ω)ϱ)|dω
    (ϱρ)22(α+1)10(1(1ω)α+1ωα+1)[h(ω)|Ψ(ρ)|+h(1ω)|Ψ(ϱ)|]dω=(ϱρ)22(α+1)[10[h(ω)|Ψ(ρ)|+h(1ω)|Ψ(ϱ)|]dω10[h(ω)(1ω)α+1|Ψ(ρ)|+h(1ω)(1ω)α+1|Ψ(ϱ)|]dω10[h(ω)ωα+1|Ψ(ρ)|+h(1ω)ωα+1|Ψ(ϱ)|]dω].

    Hence the inequality (2.16) holds.

    Remark 2.12. By choosing h(ω)=ωs, where s(0,1), we get the result of Liao et al. [24,Theorem 3.1].

    Example 2.13. By plotting the graphs of (2.16) for a convex function Ψ()=e corresponding to h()=, we prove the validity of the results.

    α(ϱρ)2(eρ+eϱ)(α+1)(α+2)eρ+eϱα(ϱρ)αϱρ[(ϱ)α1+(ρ)α1]edα(ϱρ)2(eρ+eϱ)(α+1)(α+2). (2.17)

    The graphs of the functions given by the left, middle and right sides of the double inequality (2.17) are plotted in Figure 3 against α(0,1]. The functions curves indicate that dual inequality is correct.

    Figure 3.  The graphs illustrate the validity of dual inequality (2.17) for the case ρ=0 and ϱ=1.

    Theorem 2.14. Let Iαρ+Ψ and IαϱΨ be the left and right sided RLFI of order α>0. Let ΨSX(h,D), ρ,ϱD with ρ<ϱ and ΨL1[ρ,ϱ]. Let |Ψ|b be h-convex on [ρ,ϱ] for fixed α(0,), then one has the inequality

    |Γ(α+1)2(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]Ψ(ρ+ϱ2)|(ϱρ)22(α+1)(10[h(ω)|Ψ(ρ)|b+h(1ω)|Ψ(ρ)|b]dω)1b×((α+1)2g1+(ϱ+0.5)g+1ϱg+1g+1)1g, (2.18)

    where 1g+1b=1.

    Proof. By using Lemma 1.9, Hölder's inequality and Definition 1.3 respectively, we have

    |Γ(α+1)2(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]Ψ(ρ+ϱ2)|(ϱρ)2210|m(ω)||Ψ(ωρ+(1ω)ϱ)|dω(ϱρ)22(10|m(ω)|gdω)1g(10|Ψ(ωρ+(1ω)ϱ)|bdω)1b(ϱρ)22(10|m(ω)|gdω)1g(10[h(ω)|Ψ(ρ)|b+h(1ω)|Ψ(ϱ)|b]dω)1b=(ϱρ)22(10[h(ω)|Ψ(ρ)|b+h(1ω)|Ψ(ϱ)|b]dω)1b×[(120|(ω1(1ω)α+1ωα+1α+1)|gdω)+(112|(1ω1(1ω)α+1ωα+1α+1)|gdω)]1g.

    By simple calculation and using the fact (1ω)α+1+ωα+11, we get

    |Γ(α+1)2(ϱρ)α[Iαρ+Ψ(ϱ)+IαϱΨ(ρ)]Ψ(ρ+ϱ2)|(ϱρ)22(α+1)(10[h(ω)|Ψ(ρ)|b+h(1ω)|Ψ(ϱ)|b]dω)1b
    [(α+1)120ωgdω+112(α+1ω)gdω]1g=(ϱρ)22(α+1)(10[h(ω)|Ψ(ρ)|b+h(1ω)|Ψ(ϱ)|b]dω)1b((α+1)2g1+(ϱ+0.5)g+1ϱg+1g+1)1g.

    This completes the proof of Theorem 2.14.

    Remark 2.15. If we choose h(ω)=ωs in Theorem 2.14, where s(0,1), then we get the result of Liao et al. [24,Theorem 4.2].

    In this section, we present the estimates of mid-point type inequalities for k-RLFI via h-convex functions.

    Theorem 3.1. Let Iαρ+,kΨ and Iαϱ,kΨ be the left and right sided k-RLFI of order α>0. Let Ψ:[ρ,ϱ]R be a positive function with 0ρ<ϱ and ΨL1[ρ,ϱ]. If Ψ is h-convex function on [ρ,ϱ], then

    Ψ(ρ+ϱ2)h(12)Γk(α+k)(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]αh(12)k[Ψ(ρ)+Ψ(ϱ)]10ωαk1[h(ω)+h(1ω)]dω.

    Proof. Since Ψ is h-convex, then we can write

    Ψ(ωx+(1ω)y)h(ω)Ψ(x)+h(1ω)Ψ(y), (3.1)
    Ψ((1ω)x+ωy)h(1ω)Ψ(x)+h(ω)Ψ(y). (3.2)

    Corresponding to the choice ω=12, we can write

    Ψ(x+y2)h(12)Ψ(x)+h(12)Ψ(y)=h(12)[Ψ(x)+Ψ(y)].

    By replacing x=ωa+(1ω)ϱ and y=a(1ω)+ϱω, we get

    Ψ(ρ+ϱ2)h(12)[Ψ(ωρ+(1ω)ϱ)+Ψ(a(1ω)+ωϱ)].

    Multiplying by ωαk1 and integrating over [0,1], we get

    Ψ(ρ+ϱ2)10ωαk1dωh(12)[10ωαk1Ψ(ωρ+(1ω)ϱ)dω+10ωαk1Ψ((1ω)ρ+ωϱ)dω].

    This can also be written as:

    kαΨ(ρ+ϱ2)h(12)[10ωαk1Ψ(ωρ+(1ω)ϱ)dω+10tαk1Ψ((1ω)ρ+ωϱ)dω]. (3.3)

    Substituting ωρ+(1ω)ϱ=u and (1ω)ρ+ωϱ=v, the inequality (3.3) can be written as:

    1αΨ(ρ+ϱ2)h(12)Γk(α)(ϱρ)αk[1kΓk(α)ϱρ(ϱu)αk1Ψ(u)du+1kΓk(α)ϱρ(vρ)αk1Ψ(v)dv]

    which implies

    Ψ(ρ+ϱ2)h(12)Γk(α+k)(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]. (3.4)

    Choose x=ρ and y=ϱ in (3.1) and (3.2), then adding, we get

    Ψ(ωρ+(1ω)ϱ)+Ψ((1ω)ρ+ωϱ)[h(ω)+h(1ω)][Ψ(ρ)+Ψ(ϱ)]. (3.5)

    Multiplying (3.5) by tαk1 and integrating over [0,1], we obtain

    h(12)Γk(α+k)(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]αh(12)k[Ψ(ρ)+Ψ(ϱ)]10ωαk1[h(ω)+h(1ω)]dω. (3.6)

    Combining (3.4) and (3.6), we get the required result.

    Remark 3.2. If we choose h(ω)=ω in Theorem 3.1, then we get the result of Iqbal et al. [25,Theorem 2.2].

    Theorem 3.3. Let Iαρ+,kΨ and Iαϱ,kΨ be the left and right sided k-RLFI of order α>0. Let ΨSX(h,D), ρ,ϱD with ρ<ϱ and ΨL1[ρ,ϱ]. Let |Ψ|b be h-convex, then the inequality

    |Ψ(ρ)+Ψ(ϱ)2Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]|(ϱρ)22(αk+1)(12(αk+1)g+1)1g(|Ψ(ρ)|b10h(ω)dω+|Ψ(ϱ)|b10h(1ω)dω)1b

    holds, where 1g+1b=1.

    Proof. By using Lemma 1.13, Definition 1.3 and Hölder's inequality respectively, we can have

    |Ψ(ρ)+Ψ(ϱ)2Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]|(ϱρ)2210|1(1ω)αk+1ωαk+1(αk+1)||Ψ(ωρ+(1ω)ϱ)|dω(ϱρ)22(αk+1)(10(1(1ω)αk+1ωαk+1)gdω)1g(10|Ψ(ωρ+(1ω)ϱ)|bdω)1b(ϱρ)22(αk+1)(11g(αk+1)+11g(αk+1)+1)1g(10(h(ω)|Ψ(ρ)|b+h(1ω)|Ψ(ϱ)|b)dω)1b=(ϱρ)22(αk+1)(12g(αk+1)+1)1g(|Ψ(ρ)|b10h(ω)dω+|Ψ(ϱ)|b10h(1ω)dω)1b.

    Using the fact (1(1ω)αk+1tαk+1)g1(1ω)g(αk+1)ωg(αk+1) for any ω[0,1], which follows from (CD)gCgDg, for any C>D0 and g1.

    Corollary 3.4. If we choose h(ω)=ω in Theorem 3.3, then the result for a simple convex function is given by

    |Ψ(ρ)+Ψ(ϱ)2Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]|(ϱρ)22(αk+1)(12(αk+1)g+1)1g(|Ψ(ρ)|b+|Ψ(ϱ)|b2)1b.

    Corollary 3.5. If we choose h(ω)=ωs corresponding to s(0,1) in Theorem 3.3, then we get the result for geometrically arithmetically s-convex function i.e.,

    |Ψ(ρ)+Ψ(ϱ)2Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]|(ϱρ)22(αk+1)(12(αk+1)g+1)1g(|Ψ(ρ)|b+|Ψ(ϱ)|bs+1)1b.

    Theorem 3.6. Let Iαρ+,kΨ and Iαϱ,kΨ be the left and right sided k-RLFI of order α>0. Let ΨSX(h,D), ρ,ϱI with ρ<ϱ and ΨL1[ρ,ϱ]. Let |Ψ|b be h-convex, then one has the inequality via fractional integrals.

    |Ψ(ρ)+Ψ(ϱ)2Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]|(ϱρ)2max{21αk1,121αk}2(αk+1)(|Ψ(ρ)|b10h(ω)dω+|Ψ(ϱ)|b10h(1ω)dω)1b,

    where 1g+1b=1.

    Proof. By using Lemma 1.13, Hölder's inequality, Lemma 1.8 and Definition 1.3 respectively, we can write

    |Ψ(ρ)+Ψ(ϱ)2Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(a)]|(ϱρ)22(αk+1)10|1(1ω)αk+1ωαk+1||Ψ(ωρ+(1ω)ϱ)|dω(ϱρ)22(αk+1)(10|1(1ω)αk+1ωαk+1|gdω)1g(10|Ψ(ωρ+(1ω)ϱ)|bdω)1b. (3.7)

    Now, we have the following two cases:

    Case 1: For α[0,1], the inequality (3.7) can be written as:

    |Ψ(ρ)+Ψ(ϱ)2Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]|(ϱρ)22(αk+1)(10(21αk1)gdω)1g(10|Ψ(ωρ+(1ω)ϱ)|bdω)1b(ϱρ)22(αk+1)(21αk1)(10[h(ω)|Ψ(ρ)|b+h(1ω)|Ψ(ϱ)|b]dω)1b=(ϱρ)2(21αk1)2(αk+1)(|Ψ(ρ)|b10h(ω)dω+|Ψ(ϱ)|b10h(1ω)dω)1b. (3.8)

    Case 2: For the choice of α[1,), the inequality (3.7) can be written as:

    |Ψ(ρ)+Ψ(ϱ)2Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]|(ϱρ)22(αk+1)(10(121αk)gdω)1g(10|Ψ(ωρ+(1ω)ϱ)|bdω)1b(ϱρ)22(αk+1)(121αk)(10[h(ω)|Ψ(ρ)|b+h(1ω)|Ψ(ϱ)|b]dω)1b=(ϱρ)2(121αk)2(αk+1)(|Ψ(ρ)|b10h(ω)dω+|Ψ(ϱ)|b10h(1ω)dω)1b. (3.9)

    Combining inequalities (3.8) and (3.9), we get the inequality (3.7).

    Remark 3.7. If we choose h(ω)=ωs corresponding to s(0,1) in Theorem 3.6, then we get the result of Iqbal et al. [26,Theorem 2.4].

    Theorem 3.8. Let Iαρ+,kΨ and Iαϱ,kΨ be the left and right sided k-RLFI of order α>0. Let ΨSX(h,D), ρ,ϱD with ρ<ϱ and ΨL1[ρ,ϱ]. Let |Ψ| be h-convex, then the inequality

    |Ψ(ρ)+Ψ(ϱ)2Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]|(ϱρ)22(αk+1)[10[h(ω)|Ψ(ρ)|+h(1ω)|Ψ(ϱ)|]dω10[h(ω)(1ω)αk+1|Ψ(ρ)|+h(1ω)(1ω)αk+1|Ψ(ϱ)|]dω10[h(ω)ωαk+1|Ψ(ρ)|+h(1ω)ωαk+1|Ψ(ϱ)|]dω]

    holds.

    Proof. Using Lemma 1.13 and Definition 1.3 respectively, we get

    |Ψ(ρ)+Ψ(ϱ)2Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]|(ϱρ)22(αk+1)10(1(1ω)αk+1ωαk+1)|Ψ(ωρ+(1ω)ϱ)|dω(ϱρ)22(αk+1)10(1(1ω)αk+1ωαk+1)[h(ω)|Ψ(ρ)|+h(1ω)|Ψ(ϱ)|]dω=(ϱρ)22(αk+1)[10[h(ω)|Ψ(ρ)|+h(1ω)|Ψ(ϱ)|]dω10[h(ω)(1ω)αk+1|Ψ(ρ)|+h(1ω)(1ω)αk+1|Ψ(ϱ)|]dω10[h(ω)ωαk+1|Ψ(ρ)|+h(1ω)ωαk+1|Ψ(ϱ)|]dω].

    This completes the proof of Theorem 3.8.

    Remark 3.9. By choosing h(ω)=ωs, where s(0,1), we get the result of Iqbal et al. [26,Theorem 2.3].

    Theorem 3.10. Let Iαρ+,kΨ and Iαϱ,kΨ be the left and right sided k-RLFI of order α>0. Let ΨSX(h,D), ρ,ϱD with ρ<ϱ and ΨL1[ρ,ϱ]. If |Ψ|b is h-convex on [ρ,ϱ] for α(0,) and ϱ(0,), we get the inequality

    |Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]Ψ(ρ+ϱ2)|(ϱρ)22(αk+1)(10[h(ω)|Ψ(ρ)|b+h(1ω)|Ψ(ρ)|b]dω)1b×((αk+1)2g1+(αk+0.5)g+1(αk)g+1g+1)1g,

    where 1g+1b=1.

    Proof. Using Lemma 1.14, Hölder's inequality and Definition 1.3 respectively, we have

    |Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]Ψ(a+ϱ2)|(ϱρ)22(10|m(ω)|gdω)1g(10|Ψ(ωρ+(1ω)ϱ)|bdω)1b(ϱρ)22(10|m(ω)|gdω)1g(10[h(ω)|Ψ(ρ)|b+h(1ω)|Ψ(ϱ)|b]dω)1b=(ϱρ)22(10[h(ω)|Ψ(ρ)|b+h(1ω)|Ψ(ϱ)|b]dω)1b×[(120|(ω1(1ω)αk+1ωαk+1αk+1)|gdω)+(112|(1ω1(1ω)αk+1ωαk+1αk+1)|gdω)]1g=(ϱρ)22(αk+1)(10[h(ω)|Ψ(ρ)|b+h(1ω)|Ψ(ϱ)|b]dω)1b×[(120|((αk+1)ω1+(1ω)αk+1+ωαk+1)|gdω)+(112|(αk+1(αk+1)ω1+(1ω)αk+1+ωαk+1)|gdω)]1g

    Since (1ω)αk+1+ωαk+11, therefore we can write

    |Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]Ψ(ρ+ϱ2)|(ϱρ)22(αk+1)(10[h(ω)|Ψ(ρ)|b+h(1ω)|Ψ(ϱ)|b]dω)1b×[(αk+1)120ωgdω+112(αk+1ω)gdω]1g=(ϱρ)22(αk+1)(10[h(ω)|Ψ(ρ)|b+h(1ω)|Ψ(ϱ)|b]dω)1b[(αk+1)2g1+(ϱ+0.5)g+1ϱg+1g+1]1g,

    The proof of Theorem 3.10 is done.

    Remark 3.11. If we choose h(ω)=ωs,s(0,1] in Theorem 3.10, then we get the result of Iqbal et al. [26,Theorem 2.6].

    Theorem 3.12. Let Iαρ+,kΨ and Iαϱ,kΨ be the left and right sided k-RLFI of order α>0. Let h:JRR and Ψ:[ρ,ϱ]R be positive functions with 0ρ<ϱ and hb(ω)L1[0,1], ΨL1[ρ,ϱ]. If |Ψ| is h-convex mapping on [ρ,ϱ], then the inequality

    |Ψ(ρ)+Ψ(ϱ)2Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]|(ϱρ)[|Ψ(ρ)|+|Ψ(ϱ)|]2[(2gαk+112pαk+1(gαk+1))1g(12gαk+1(gαk+1))1g]×[(120(h(ω))bdω)1b+(112(h(ω))bdω)1b]

    holds, where α>0, g>0 and 1g+1b=1.

    Proof. Since |Ψ| is h-convex, then by Lemma 1.12, we can write

    |Ψ(ρ)+Ψ(ϱ)2Γk(α+k)2(ϱρ)αk[Iαρ+,kΨ(ϱ)+Iαϱ,kΨ(ρ)]|ϱρ210|(1ω)αkωαk||Ψ(ωρ+(1ω)ϱ)|dωϱρ2[120[(1ω)αkωαk][h(ω)|Ψ(ρ)|+h(1ω)|Ψ(ϱ)|]dω+112[ωαk(1ω)αk][h(ω)|Ψ(ρ)|+h(1ω)|Ψ(ϱ)|]dω]=ϱρ2[|Ψ(ρ)|120(1ω)αkh(ω)dω|Ψ(ρ)|120ωαkh(ω)dω+|Ψ(ϱ)|120(1ω)αkh(1ω)dω|Ψ(ϱ)|120ωαkh(1ω)dω+|Ψ(ρ)|112ωαkh(ω)dω|Ψ(ρ)|112(1ω)αkh(ω)dω+|Ψ(ϱ)|112ωαkh(1ω)dω|Ψ(ϱ)|112(1ω)αkh(1ω)dω]. (3.10)

    Using Hölder's inequality on the right side of (3.10), we get

    |Ψ(ρ)+Ψ(ϱ)2Γk(α+k)2(ϱρ)αk[Iαρ+Ψ(ϱ)+IϱbΨ(ρ)]|=(ϱρ)[|Ψ(ρ)|+|Ψ(ϱ)|]2[(2gαk+112gαk+1(gαk+1))1g(12gαk+1(gαk+1))1g]×[(120(h(ω))bdω)1b+(112(h(ω))bdω)1b].

    This completes the proof of the result.

    Remark 3.13. If we choose k=1 in Theorem 3.12, we get the result of Tunc. [27,Theorem 5].

    This section is devoted to some particular inequalities which generalize some classical results such as trapezoid-type, mid-point-type and Hadamard's inequality.

    Proposition 4.1. (Hadamard's inequality). Under the assumptions of Theorem 2.1 with α=1 and h(ω)=ω, we get the "Hadamard's inequality"

    Ψ(ρ+ϱ2)1ϱρϱρΨ(ω)dωΨ(ρ)+Ψ(ϱ)2.

    Proposition 4.2. Let the assumptions of Theorem 2.4 be fulfilled with α=1 and h(ω)=ω, then we obtain the "Trapezoid inequality"

    |(ϱρ)Ψ(ρ)+Ψ(ϱ)2ϱρΨ(ω)dω|(ϱρ)34(2g12g+1)1g(|Ψ(ρ)|b+|Ψ(ϱ)|b2)1b.

    Proposition 4.3. Let the assumptions of Theorem 2.11 be fulfilled with α=1 and h(ω)=ω, then we obtain the "Trapezoid inequality"

    |(ϱρ)Ψ(ρ)+Ψ(ϱ)2ϱρΨ(ω)dω|(ϱρ)324[|Ψ(ρ)|+|Ψ(ϱ)|].

    Proposition 4.4. Under the assumptions of Theorem 2.14 with α=1 and h(ω)=ω, we get the following "Mid-point inequality"

    |(ϱρ)Ψ(ρ+ϱ2)ϱρΨ(ω)dω|(ϱρ)34(1+3g+12g2g(g+1))1g(|Ψ(ρ)|b+|Ψ(ϱ)|b2)1b.

    Proposition 4.5. Let the assumptions of Theorem 3.12 be fulfilled with α=1 and h(ω)=ω, then we get the following "Trapezoid inequality"

    |(ϱρ)Ψ(ρ)+Ψ(ϱ)2ϱρΨ(ω)dω|(ϱρ)2[|Ψ(ρ)|+|Ψ(ϱ)|]2×[(2g+112g+1(g+1))1g(12g+1(g+1))1g][(2b+112b+1(b+1))1b+(12b+1(b+1))1b].

    We aim at designing the generalizations of some Trapezoid-type inequalities for Riemann-type fractional integrals in the present article. We use h-convex function to this ends and develops numerous inequalities. We prove the validity of our results by using the examples and their graphs. The work presented consists of the formulas of quadrature as a boundary of the new inequalities. The results of these studies complement those of earlier studies. The earlier findings are supported by simple, recent studies and play an additional role in generalizations.

    The authors Bahaaeldin Abdalla and Thabet Abdeljawad would like to thank Prince Sultan University for funding this research through the group: Nonlinear Analysis Methods in Applied Mathematics (NAMAM), group number RG-DES-2017-01-17.

    The authors declares that there is no conflict of interests regarding the publication of this paper.



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