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Baleanu et al. Advances in Difference Equations (2020) 2020:374 https://doi.org/10.1186/s13662-020-02837-0 RESEARCH Open Access Some modifications in conformable fractional integral inequalities Dumitru Baleanu 1,2,3 , Pshtiwan Othman Mohammed 4* , Miguel Vivas-Cortez 5 and Yenny Rangel-Oliveros 5 * Correspondence: [email protected] 4 Department of Mathematics, College of Education, University of Sulaimani, Sulaimani, Kurdistan Region, Iraq Full list of author information is available at the end of the article Abstract The prevalence of the use of integral inequalities has dramatically influenced the evolution of mathematical analysis. The use of these useful tools leads to faster advances in the presentation of fractional calculus. This article investigates the Hermite–Hadamard integral inequalities via the notion of -convexity. After that, we introduce the notion of μ -convexity in the context of conformable operators. In view of this, we establish some Hermite–Hadamard integral inequalities (both trapezoidal and midpoint types) and some special case of those inequalities as well. Finally, we present some examples on special means of real numbers. Furthermore, we offer three plot illustrations to clarify the results. MSC: Primary 26D07; secondary 26D10; 26D15; 26A33 Keywords: Integral inequality; Conformable operator; Convex functions 1 Introduction For any v 1 , v 2 [a, b] and [0, 1], the real-valued function g on an interval [a, b] is called a convex function if the following holds: g ( v 1 + (1 – )v 2 ) g (v 1 ) + (1 – )g (v 2 ). (1.1) The theory and application of convexity has a close relationship with theory and ap- plication of inequalities or integral inequalities. The convex function (1.1) has been ex- tended and generalized in several directions, such as pseudo-convex [1], quasi-convex [2], strongly convex [3], -convex [4], s-convex [5], h-convex [6, 7], (α, m)-convex [8, 9], invex and preinvex [1012], and other kinds of convex functions by a number of mathemati- cians; see [1321] for more details. Integral inequalities form an essential field of study among the field of mathematical analysis. They have been vital in providing bounds to solve some boundary value prob- lems in fractional calculus, and in establishing the existence and uniqueness of solutions for certain fractional differential equations. Convexity plays an important role in the field of integral inequality due to the behavior of its definition. Also, there is a strong connection © The Author(s) 2020. This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.

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  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 https://doi.org/10.1186/s13662-020-02837-0

    R E S E A R C H Open Access

    Some modifications in conformablefractional integral inequalitiesDumitru Baleanu1,2,3, Pshtiwan Othman Mohammed4* , Miguel Vivas-Cortez5 andYenny Rangel-Oliveros5

    *Correspondence:[email protected] of Mathematics,College of Education, University ofSulaimani, Sulaimani, KurdistanRegion, IraqFull list of author information isavailable at the end of the article

    AbstractThe prevalence of the use of integral inequalities has dramatically influenced theevolution of mathematical analysis. The use of these useful tools leads to fasteradvances in the presentation of fractional calculus. This article investigates theHermite–Hadamard integral inequalities via the notion of�-convexity. After that, weintroduce the notion of�μ-convexity in the context of conformable operators. Inview of this, we establish some Hermite–Hadamard integral inequalities (bothtrapezoidal and midpoint types) and some special case of those inequalities as well.Finally, we present some examples on special means of real numbers. Furthermore,we offer three plot illustrations to clarify the results.

    MSC: Primary 26D07; secondary 26D10; 26D15; 26A33

    Keywords: Integral inequality; Conformable operator; Convex functions

    1 IntroductionFor any v1, v2 ∈ [a, b] and � ∈ [0, 1], the real-valued function g on an interval [a, b] is calleda convex function if the following holds:

    g(�v1 + (1 – �)v2

    ) ≤ �g(v1) + (1 – �)g(v2). (1.1)

    The theory and application of convexity has a close relationship with theory and ap-plication of inequalities or integral inequalities. The convex function (1.1) has been ex-tended and generalized in several directions, such as pseudo-convex [1], quasi-convex [2],strongly convex [3], �-convex [4], s-convex [5], h-convex [6, 7], (α, m)-convex [8, 9], invexand preinvex [10–12], and other kinds of convex functions by a number of mathemati-cians; see [13–21] for more details.

    Integral inequalities form an essential field of study among the field of mathematicalanalysis. They have been vital in providing bounds to solve some boundary value prob-lems in fractional calculus, and in establishing the existence and uniqueness of solutionsfor certain fractional differential equations. Convexity plays an important role in the fieldof integral inequality due to the behavior of its definition. Also, there is a strong connection

    © The Author(s) 2020. This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use,sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the originalauthor(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or otherthird party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit lineto the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted bystatutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view acopy of this licence, visit http://creativecommons.org/licenses/by/4.0/.

    https://doi.org/10.1186/s13662-020-02837-0http://crossmark.crossref.org/dialog/?doi=10.1186/s13662-020-02837-0&domain=pdfhttp://orcid.org/0000-0001-6837-8075mailto:[email protected]

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 2 of 25

    between convexity and integral inequality. For this reason, many known integral inequal-ities have been established in the literature. The Hermite–Hadamard (HH) inequality isthe most well known one: for an L1 convex function g : I ⊆ R→ R with v1, v2 ∈ I , v1 < v2,the HH inequality is defined as follows:

    g(

    v1 + v22

    )≤ 1

    v2 – v1

    ∫ v2

    v1g(x) dx ≤ g(v1) + g(v2)

    2. (1.2)

    A huge number of researchers in the field of applied mathematics have dedicated theirinterest to generalize, improve, refine, counterpart, and extend HH inequality (1.2) forvarious types of convex functions; see e.g. [22–30].

    Recently, Samet [31] introduced a new notion of convexity for certain functions thatdepends on some axioms. This often generalizes various types of convexity e.g. �-convexfunctions, α-convex functions, h-convex functions, and so on. Also, for further details,visit [16, 32].

    Throughout our study, we suppose that I ⊆ R (R the set of real numbers), V ={(η1,η2);η� ∈ [v1, v2],� = 1, 2} and R̄ = {(η1,η2,η3);ηi ∈ R,� = 1, 2, 3}. Then the family ofF of functions � : R̄× [0, 1] → R satisfies the major axioms [31]:

    (Λ1) If y� ∈ L1(0, 1), � = 1, 2, 3, then for every γ ∈ [0, 1] we have∫ 1

    0�

    (y1(η), y2(η), y3(η),γ

    )dη

    = �(∫ 1

    0y1(η) dη,

    ∫ 1

    0y2(η) dη,

    ∫ 1

    0y3(η) dη,γ

    ).

    (Λ2) For every u ∈ L1(0, 1), w ∈ L∞(0, 1), and (z1, z2) ∈ R2, we have∫ 1

    0�

    (w(η)u(η), w(η)z1, w(η)z2,η

    )dη = T�,w

    (∫ 1

    0w(η)u(η) dη, z1, z2

    ),

    where T�,w : R̄→ R is a function depending on (F , w). Moreover, it is a nondecreas-ing function according to the first variable.

    (Λ3) For any (w, y1, y2, y3) ∈ R4, y4 ∈ [0, 1], we have

    w�(y1, y2, y3, y4) = �(wy1, wy2, wy3, y4) + Lw,

    where Lw ∈ R is a constant (depending on w).

    Definition 1.1 Let g : [v1, v2] ⊆ R→ R with v1 < v2 be a function, then we say that g is aconvex function according to � ∈F (or briefly �-convex function) iff

    �̄(g(ηx + (1 – η)y

    ), g(x), g(y),η

    ) ≤ 0, (x, y,η) ∈ V × [0, 1].

    Remark 1.1 Suppose that (v1, v2) ∈ R2 with v1 < v2,(i) if g : [v1, v2] ⊂ R→ R is an ε-convex function, or equivalently [26]

    g(ηx + (1 – η)y

    ) ≤ ηg(x) + (1 – η)g(y) + ε, (x, y,η) ∈ V × [0, 1],

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 3 of 25

    then we define the functions � : R̄× [0, 1] → R as follows:

    �(y1, y2, y3, y4) = y1 – y4y2 – (1 – y4)y3 – ε, (1.3)

    and T�,w : R̄× [0, 1] → R as

    T�,w(y1, y2, y3) = y1 –(∫ 1

    0tw(η) dη

    )y2 –

    (∫ 1

    0(1 – η)w(η) dη

    )y3 – ε. (1.4)

    For

    Lw = (1 – w)ε, (1.5)

    we can observe that � ∈F and

    �(g(ηx + (1 – η)y

    ), g(x), g(y),η

    )= g

    (ηx + (1 – η)y

    )– ηg(x) – (1 – η)g(y) – ε ≤ 0,

    and this tells us g is an �-convex function. In a particular case, we take ε = 0 to show thatg is an �-convex function according to � when g is assumed to be a convex function.

    (ii) If g : [v1, v2] ⊂ R→ R is a μ-convex function with μ ∈ (0, 1], or equivalently

    g(ηx + (1 – η)y

    ) ≤ ημg(x) + (1 – ημ)g(y), (x, y,η) ∈ V × [0, 1].

    Then we define the function � : R̄× [0, 1] → R as follows:

    �(y1, y2, y3, y4) = y1 – yμ4 y2 –(1 – yμ4

    )y3, (1.6)

    and T�,w : R̄× [0, 1] → R as

    T�,w(y1, y2, y3) = y1 –(∫ 1

    0ημw(η) dη

    )y2 –

    (∫ 1

    0

    (1 – ημ

    )w(η) dη

    )y3. (1.7)

    For Lw = 0, we can observe that � ∈F and

    �(g(ηx + (1 – η)y

    ), g(x), g(y),η

    )= g

    (ηx + (1 – η)y

    )– ημg(x) –

    (1 – ημ

    )g(y) – ε ≤ 0,

    or g is an �-convex function.(iii) If h : I → R is a function and it is not identically 0, where (0, 1) ⊆ I . Also, suppose

    that g : [v1, v2] ⊂ I → [0,∞) is an h-convex function, that is,

    g(ηx + (1 – η)y

    ) ≤ h(η)g(x) + h(1 – η)g(y), (x, y,η) ∈ V × [0, 1].

    Then we define the functions � : R̄× [0, 1] → R as follows:

    �(y1, y2, y3, y4) = y1 – h(y4)y3 – h(1 – y4)y2, (1.8)

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 4 of 25

    and T�,w : R̄× [0, 1] → R as

    T�,w(y1, y2, y3) = y1 –(∫ 1

    0h(η)w(η) dη

    )y2 –

    (∫ 1

    0h(1 – η)w(η) dη

    )y3. (1.9)

    For Lw = (1 – w)ε, we can observe that � ∈F and

    �(g(ηx + (1 – η)y

    ), g(x), g(y),η

    )= g

    (ηx + (1 – η)y

    )– h(η)g(x) – h(1 – η)g(y) – ε ≤ 0,

    or we can say that g is an �-convex function.

    In recent years, many possible inequalities have been proposed in the context of frac-tional calculus including the midpoint and trapezoidal formula inequalities and inequal-ities for ε-convexity, α-convexity, (α, m)-convexity, and h-convexity; see [26, 31, 33] formore details.

    2 Conformable fractional operators and �μ-convexityIn the last fifteen years, the definition of fractional calculus has been more appropriateto describe historical dependence processes than the local limit definitions of integer or-dinary differential equations (ODEs) or partial differential equations (PDEs), and has re-ceived more and more attention in many mathematical and physical fields, see for de-tails [34–44]. Differential equations of fractional order are more accurate than differentialequations of integer order in describing the nature of things and objective laws. In 1695,Leibnitz discovered fractional derivatives, and after that more and more researchers havededicated themselves to the study of fractional calculus. The most commonly used frac-tional calculus definitions are Riemann–Liouville definition, Caputo definition, and con-formable fractional definition in basic mathematical and engineering application research.In the present paper, we deal with the conformable fractional definition [45–47] in orderto obtain our results.

    In this section, we recall some preliminaries and properties on conformable fractionalcalculus. For further details and applications, see the previously published articles [33, 45–54].

    Definition 2.1 ([47]) Let g : [0,∞) → R, then the μth order conformable derivative of gat η is defined by

    Dμ(g)(η) = lim�→0

    g(η + �η1–μ) – g(η)�

    , μ ∈ (0, 1),η > 0. (2.1)

    For μ-differentiable function g in some (0,μ),μ > 0, limt→0+ g(μ)(η) exist, define

    g(μ)(0) = limt→0+

    g(μ)(η).

    Furthermore, if g is differentiable, then we have

    Dμ(g)(η) = η1–μg ′(η), where g ′(η) = lim�→0

    g(η + �) – g(η)�

    . (2.2)

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 5 of 25

    Observe that we can write g(μ)(η) for dμdμη (g(η)) or simply Dμ(g)(η) to denote a μth orderconformable derivative of g at η. Furthermore, if the μth order conformable derivative ofg exists, then we can simply say g is μ-differentiable.

    Theorem 2.1 ([48]) Assume that μ ∈ (0, 1] and f , g are two μ-differentiable functions ata point η > 0. Then we have:

    1. Dμ(v1f + v2g) = v1Dμ(f ) + v2Dμ(g) for all v1, v2 ∈ R,2. Dμ(fg) = fDμ(g) + gDμ(f ),3. Dμ( fg ) =

    gDμ(f )–fDμ(g)g2 ,

    4. Dμ(c) = 0 for each constant function, namely g(η) = c,5. Dμ(1) = 0,6. Dμ( 1μη

    μ) = 1.

    Some basic properties of conformable operator are now stated, which are useful in whatfollows.

    Definition 2.2 ([47]) Assume that μ ∈ (0, 1], 0 ≤ v1 ≤ v2, and g : [v1, v2] ⊂ R → R, thenwe say that a function g is μ-fractional integrable on the interval [v1, v2] if the followingintegral

    ∫ v2

    v1g(η) dμη =

    ∫ v2

    v1g(η)ημ–1 dη (2.3)

    exists and is finite.

    Remark 2.1(a) We indicate by L1μ([v1, v2]) all μ-fractional integrable functions on an interval

    [v1, v2].(b) The usual Riemann improper integral has the form

    Iv1μ (g)(η) = Iv11

    (ημ–1g

    )=

    ∫ t

    v1xμ–1g(x) dx, μ ∈ (0, 1]. (2.4)

    Theorem 2.2 ([47, 48]) Let g : (v1, v2) → R be differentiable and μ ∈ (0, 1]. Then, for allη > v1, we have

    Iv1μ Dv1μ (g)(η) = g(η) – g(v1).

    Theorem 2.3 ([51]) Suppose that g : [v1,∞) → R such that g(n) is continuous. Then, foreach η > v1, we have

    Dv1μ Iv1μ (g)(η) = g(η), μ ∈ (n, n + 1],

    which is called the inverse property.

    Theorem 2.4 ([47, 48]) Let g : [v1, v2] ⊂ R→ R be two functions with fg is differentiable.Then

    ∫ v2

    v1g(x)Dv1μ (h)(x) dμx = gh|v2v1 –

    ∫ v2

    v1h(x)Dv1μ (g)(x) dμx.

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 6 of 25

    Theorem 2.5 ([47, 48]) Let f , g : [v1, v2] ⊂ R→ R be a continuous function on [v1, v2] andwith 0 ≤ v1 ≤ v2. Then

    ∣∣Iv1μ (g)(η)∣∣ ≤ Iv1μ |f |(η), μ ∈ (0, 1].

    It is time to define the concept of �μ-convexity on conformable integrals, namely thefamily of �μ.

    The family of �μ of functions �μ : R̄× [0, 1] → R satisfies the major axioms:(Λ̄1) If y� ∈ L1(0, 1), � = 1, 2, 3, then for every γ ∈ [0, 1] we have

    ∫ 1

    0�μ

    (y1(η), y2(η), y3(η),γ

    )dη

    = �μ(∫ 1

    0y1(η) dη,

    ∫ 1

    0y2(η) dη,

    ∫ 1

    0y3(η) dη,γ

    ).

    (Λ̄2) For every u ∈ L1(0, 1), w ∈ L∞(0, 1), and (z1, z2) ∈ R2, we have∫ 1

    0�μ

    (w(η)u(η), w(η)z1, w(η)z2,η

    )dη = T�μ ,w

    (∫ 1

    0w(η)u(η) dη, z1, z2

    ),

    where T�μ ,w : R̄ → R is a nondecreasing function according to the first variablewhich depends on (�μ, w).

    (Λ̄3) For any (w, y1, y2, y3) ∈ R4, y4 ∈ [0, 1], we have

    w�μ(y1, y2, y3, y4) = �μ(wy1, wy2, wy3, y4) + Lw,

    where Lw ∈ R is a constant (depending on w).

    Definition 2.3 Let μ ∈ (0, 1] and g : [v1, v2] ⊂ R→ R with v1 < v2 be a function, then wesay g is a conformable convex function according to �μ ∈ F (or briefly �μ-conformableconvex function) if

    �μ

    (g(ημxμ +

    (1 – ημ

    )yμ

    ), g

    (xμ

    ), g

    (yμ

    ),ημ

    ) ≤ 0, (x, y,η) ∈ V × [0, 1].

    Remark 2.2 Suppose that (v1, v2) ∈ R2 with v1 < v2.(i) Let g : [v1, v2] ⊂ R→ R be an ε-conformable convex function, or equivalently

    g(ημxμ +

    (1 – ημ

    )yμ

    ) ≤ ημg(xμ) + (1 – ημ)g(yμ) + ε, (x, y,η) ∈ V × [0, 1].

    Then we define the function �μ : R̄× [0, 1] → R as follows:

    �μ(y1, y2, y3, y4) = y1 – yμ4 y2 –(1 – yμ4

    )y3 – ε, (2.5)

    and T�μ ,w : R̄× [0, 1] → R as

    T�μ ,w(y1, y2, y3) = y1 –(∫ 1

    0ημw

    (ημ

    )dμη

    )y2 –

    (∫ 1

    0

    (1 – ημ

    )w(η) dμη

    )y3 – ε. (2.6)

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 7 of 25

    For

    Lw = (1 – w)ε, (2.7)

    it can be observed that � ∈F and

    �μ

    (g(ημxμ +

    (1 – ημ

    )yμ

    ), g

    (xμ+

    ), g

    (yμ

    ),ημ

    )

    = g(ημxμ +

    (1 – ημ

    )yμ

    )– ημg

    (xμ

    )–

    (1 – ημ

    )g(yμ

    )– ε ≤ 0,

    or in another meaning g is an �-conformable convex function. In particular, g is an �-conformable convex function according to � for ε = 0 when g is a conformable convexfunction.

    (ii) Let g : [v1, v2] ⊂ I → R be a μ-conformable convex function μ ∈ (0, 1]; that is,

    g(ημxμ +

    (1 – ημ

    )yμ

    ) ≤ ηg(xμ) + (1 – η)g(yμ), (x, y,η) ∈ V × [0, 1].

    Then we define the functions �μ : R̄× [0, 1] → R as follows:

    �μ(y1, y2, y3, y4) = y1 – y4y2 – (1 – y4)y3, (2.8)

    and T�μ ,w : R̄× [0, 1] → R as

    T�μ ,w(y1, y2, y3) = y1 –(∫ 1

    0tw

    (ημ

    )dμη

    )y2 –

    (∫ 1

    0(1 – η)w

    (ημ

    )dμη

    )y3. (2.9)

    For Lw = 0, we can observe that �μ ∈F and

    �μ

    (g(ημxμ +

    (1 – ημ

    )yμ

    ), g

    (xμ

    ), g

    (yμ

    ),ημ

    )

    = g(ημxμ +

    (1 – ημ

    )yμ

    )– ηg

    (xμ

    )– (1 – η)g

    (yμ

    )– ε ≤ 0,

    or equivalently g is an �μ-conformable convex function.(iii) Let h : I → R be a function, which is not identically 0, where (0, 1) ⊆ I . Let g :

    [v1, v2] ⊂ I → [0,∞) be an h-conformable convex function, or let

    g(ημxμ +

    (1 – ημ

    )yμ

    ) ≤ h(ημ)g(xμ) + h(1 – ημ)g(yμ), (x, y,η) ∈ V × [0, 1].

    Then we define the functions �μ : R̄× [0, 1] → R as follows:

    �μ(y1, y2, y3, y4) = y1 – h(y4)y3 – h(1 – yμ4

    )y2, (2.10)

    and T�μ ,w : R̄× [0, 1] → R as

    T�μ ,w(y1, y2, y3) = y1 –(∫ 1

    0h(ημ

    )w

    (ημ

    )dμη

    )y2

    –(∫ 1

    0h(1 – ημ

    )w

    (ημ

    )dμη

    )y3. (2.11)

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 8 of 25

    For Lw = (1 – w)ε, we can observe that �μ ∈F and

    �μ

    (g(ημxμ +

    (1 – ημ

    )yμ

    ), g

    (xμ

    ), g

    (yμ

    ),ημ

    )

    = g(ημxμ +

    (1 – ημ

    )yμ

    )– h

    (ημ

    )g(xμ

    )– h

    (1 – ημ

    )g(yμ

    )– ε ≤ 0,

    or equivalently we can say g is an �μ-conformable convex function.

    For the conformable operators, we recall some early findings in the earlier literaturewhich may help us in finding our main results. For example in [55], Sarikaya et al. investi-gated new results for the conformable fractional operator, and their results are as follows.

    Theorem 2.6 ([55, Theorem 11]) Let μ ∈ (0, 1] and g : [v1, v2] ⊂ R→R be a μ-fractionaldifferentiable function on (v1, v2) with 0 ≤ v1 < v2. Then we have

    g(

    vμ1 + vμ2

    2

    )≤ μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx ≤ g(v

    μ1 ) + g(v

    μ2 )

    2. (2.12)

    Lemma 2.1 ([55, Lemma 3]) Let μ ∈ (0, 1] and g : [v1, v2] ⊂ R→R be a μ-fractional differ-entiable function on (v1, v2) with 0 ≤ v1 < v2. If Dμ(g) is a μ-fractional integrable functionon [v1, v2], then we have

    g(vμ1 ) + g(vμ2 )

    2–

    ∫ v2

    v1g(xμ

    )dμx

    =vμ2 – v

    μ1

    2

    ∫ 1

    0

    (1 – 2ημ

    )Dμ(g)

    (ημvμ1 +

    (1 – ημ

    )vμ2

    )dμη. (2.13)

    Lemma 2.2 ([55, Lemma 4]) Let μ ∈ (0, 1] and g : [v1, v2] ⊂ R→R be a μ-fractional differ-entiable function on (v1, v2) with 0 ≤ v1 < v2. If Dμ(g) is a μ-fractional integrable functionon [v1, v2], then we have

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx – g

    (vμ1 + v

    μ2

    2

    )

    =(vμ2 – v

    μ1)∫ 1

    0p(η)Dμ(g)

    (ημvμ1 +

    (1 – ημ

    )vμ2

    )dμη, (2.14)

    where

    P(η) =

    ⎧⎨

    ⎩ημ, 0 ≤ t ≤ 121/μ ,ημ – 1, 121/μ ≤ t ≤ 1.

    In view of these indices, we investigate some new inequalities of HH type for the � and�μ-convex functions involving conformable fractional operators in this attempt. Specifi-cally, we investigate some inequalities of trapezoidal and midpoint type.

    3 Hermite–Hadamard inequalities for �-convex functionsThis section deals with the investigation of HH-type inequalities for �-convex functions.

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 9 of 25

    Theorem 3.1 Let g : [v1, v2] ⊂ R → R be a μ-fractional differentiable function on (v1, v2)with 0 ≤ v1 < v2. If g is an �-convex function on [v1, v2] for some � ∈F , then

    (g(

    vμ1 + vμ2

    2

    ),

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx,

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx,

    12

    )+

    ∫ 1

    0Lw(η) dη

    ≤ 0, (3.1)

    T�,w(

    2μvμ2 – v

    μ1

    ∫ v2

    v1g(xμ

    )dμx, g

    (vμ1

    )+ g

    (vμ2

    ), g

    (vμ1

    )+ g

    (vμ2

    ))

    +∫ 1

    0Lw(η) dη ≤ 0. (3.2)

    Proof The �-convexity of g leads to

    (g(

    x + y2

    ), g(x), g(y),

    12

    ), x, y ∈ [v1, v2].

    For the values x = ημvμ1 + (1 – ημ)vμ2 and y = (1 – ημ)v

    μ1 + ημv

    μ2 , where η ∈ [0, 1], we obtain

    (g(

    vμ1 + vμ2

    2

    ), g

    (ημvμ1 +

    (1 – ημ

    )vμ2

    ), g

    ((1 – ημ

    )vμ1 + η

    μvμ2),

    12

    )≤ 0.

    Multiplying this inequality w(η) = 1 and making use of axiom (Λ3), we get

    (g(

    vμ1 + vμ2

    2

    ), g

    (ημvμ1 +

    (1 – ημ

    )vμ2

    ), g

    ((1 – ημ

    )vμ1 + η

    μvμ2),

    12

    )+ Lw(η) ≤ 0.

    Integrating over [0, 1] according to η and making use of axiom (Λ1), we get

    (∫ 1

    0g(

    vμ1 + vμ2

    2

    )dμη,

    ∫ 1

    0g(ημvμ1 +

    (1 – ημ

    )vμ2

    )dμη,

    ∫ 1

    0g((

    1 – ημ)vμ1 + η

    μvμ2)

    dμη,12

    )+

    ∫ 1

    0Lw(η) dμη ≤ 0,

    that is,

    (g(

    vμ1 + vμ2

    2

    ),

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx,

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx,

    12

    )+

    ∫ 1

    0Lw(η) dη

    ≤ 0,

    where we have used

    ∫ 1

    0g(ημvμ1 +

    (1 – ημ

    )vμ2

    )dμη =

    ∫ 1

    0g((

    1 – ημ)vμ1 + η

    μvμ2)

    dμη

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx.

    This completely gives the proof of (3.1). On the other hand, since g is �-convex, we have

    �(g(ημvμ1 +

    (1 – ημ

    )vμ2

    ), g

    (vμ1

    ), g

    (vμ2

    ),η

    ) ≤ 0,�

    (g((

    1 – ημ)vμ1 + η

    μvμ2), g

    (vμ2

    ), g

    (vμ1

    ),η

    ) ≤ 0.

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 10 of 25

    We make use of the linearity of � to get

    �(g(ημvμ1 +

    (1 – ημ

    )vμ2

    )+ g

    ((1 – ημ

    )vμ1 + η

    μvμ2), g

    (vμ1

    )+ g

    (vμ2

    ), g

    (vμ1

    )+ g

    (vμ2

    ),η

    ) ≤ 0.

    Applying the axiom (Λ3) for w(η) = 1, we get

    �(g(ημvμ1 +

    (1 – ημ

    )vμ2

    )+ g

    ((1 – ημ

    )vμ1 + η

    μvμ2), g

    (vμ1

    )+ g

    (vμ2

    ), g

    (vμ1

    )+ g

    (vμ2

    ),η

    )

    + Lw(η) ≤ 0.

    Integrating over [0, 1] according to η and making use of axiom (Λ2) we get

    T�,w(

    2μvμ2 – v

    μ1

    ∫ v2

    v1g(xμ

    )dμx, g

    (vμ1

    )+ g

    (vμ2

    ), g

    (vμ1

    )+ g

    (vμ2

    ))+

    ∫ 1

    0Lw(η) dη ≤ 0.

    This completes the proof of (3.2). Thus, the proof of Theorem 3.1 is completed. �

    Corollary 3.1 Theorem 3.1 with g to be ε-convex leads to

    g(

    vμ1 + vμ2

    2

    )– ε ≤ μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx ≤ g(v

    μ1 ) + g(v

    μ2 )

    2+

    ε

    2. (3.3)

    Proof By making use of w(η) = 1 in (1.5), we get

    ∫ 1

    0Lw(η) dη = 0. (3.4)

    Making use of (1.3), (3.1), and (3.4), we get

    0 ≥ F(

    g(

    vμ1 + vμ2

    2

    ),

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx,

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx,

    12

    )+

    ∫ 1

    0Lw(η) dη

    = g(

    vμ1 + vμ2

    2

    )–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx – ε,

    or equivalently,

    g(

    vμ1 + vμ2

    2

    )– ε ≤ μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx.

    Making use of w(η) = 1 in (1.4), we have

    T�,w(y1, y2, y3) = y1 –(∫ 1

    0t dη

    )y2 –

    (∫ 1

    0(1 – η) dη

    )y3 – ε

    = y1 –y2 + y3

    2– ε, y1, y2, y3 ∈ R. (3.5)

    Now, from (3.2) and (3.5), we can deduce

    0 ≥ T�,w(

    2μvμ2 – v

    μ1

    ∫ v2

    v1g(xμ

    )dμx, g

    (vμ1

    )+ g

    (vμ2

    ), g

    (vμ1

    )+ g

    (vμ2

    ))+

    ∫ 1

    0Lw(η) dη

    =2μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx –

    (g(vμ1

    )+ g

    (vμ2

    ))– ε.

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 11 of 25

    This gives

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx ≤ g(v

    μ1 ) + g(v

    μ2 )

    2+

    ε

    2.

    This ends the proof of (3.3). �

    Remark 3.1 Inequality (3.3) with ε = 0 becomes inequality (2.12).

    Corollary 3.2 Theorem 3.1 with g to be h-convex leads to

    12h( 12 )

    g(

    vμ1 + vμ2

    2

    )≤ μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx ≤ g(v

    μ1 ) + g(v

    μ2 )

    2

    ∫ 1

    0h(η) dη. (3.6)

    Proof Making use (1.5) and (3.1) with Lw(η) = 0, we have

    0 ≥ F(

    g(

    vμ1 + vμ2

    2

    ),

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx,

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx,

    12

    )+

    ∫ 1

    0Lw(η) dη

    = g(

    vμ1 + vμ2

    2

    )– h

    (12

    )2μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx,

    or equivalently,

    12h( 12 )

    g(

    vμ1 + vμ2

    2

    )≤ μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx.

    Now, by making use of w(η) = 1 in (1.4) and (3.2), we get

    0 ≥ T�,w(

    2μvμ2 – v

    μ1

    ∫ v2

    v1g(xμ

    )dμx, g

    (vμ1

    )+ g

    (vμ2

    ), g

    (vμ1

    )+ g

    (vμ2

    ))+

    ∫ 1

    0Lw(η) dη

    =2μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx –

    (g(vμ1

    )+ g

    (vμ2

    ))(∫ 1

    0h(η) dη

    ).

    This gives

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx ≤

    (g(vμ1 ) + g(v

    μ2 )

    2

    )(∫ 1

    0h(η) dη

    ).

    Thus, the proof of (3.6) is completed. �

    4 Hermite–Hadamard inequalities for �μ-convex functionsHere, we deal with the investigation of HH-type inequalities for�μ-convex functions. Thissection is separated into two subsections: a section for the trapezoidal formula inequalityand the other one for the midpoint formula inequality of HH type, respectively.

    4.1 Trapezoidal inequalities for �μ-convex functionsTheorem 4.1 Let g : [v1, v2] ⊂ R → R be a μ-fractional differentiable function on (v1, v2)and Dμ(g) be a μ-fractional integrable function on [v1, v2] with 0 ≤ v1 < v2. If |Dμ(g)| is

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 12 of 25

    an �μ-convex function on [v1, v2] for some �μ ∈ F and the function η ∈ [0, 1] → Lw(ημ)belongs to L1[0, 1], where w(ημ) = |1 – 2ημ|, then we have the inequality

    T�μ ,w(

    2vμ2 – v

    μ1

    ∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣,

    ∣∣Dμ(g)(vμ1

    )∣∣,∣∣Dμ(g)

    (vμ2

    )∣∣,η)

    +∫ 1

    0Lw(η) dμη ≤ 0. (4.1)

    Proof The �μ-convexity of |Dμ(g)| leads to

    �μ

    (∣∣Dμ(g)(ημvμ1 +

    (1 – ημ

    )vμ2

    )∣∣,∣∣Dμ(g)

    (vμ1

    )∣∣,∣∣Dμ(g)

    (vμ2

    )∣∣,η) ≤ 0.

    By applying axiom (Λ̄3) for w(ημ) = |1 – 2ημ|, η ∈ [0, 1], we can deduce

    �μ

    (w

    (ημ

    )∣∣Dμ(g)(ημvμ1 +

    (1 – ημ

    )vμ2

    )∣∣, w(ημ

    )∣∣Dμ(g)(vμ1

    )∣∣, w(ημ

    )∣∣Dμ(g)(vμ2

    )∣∣,η) ≤ 0.

    Integrating over [0, 1] according to η and by making use of axiom (Λ̄2), we obtain

    T�μ ,w(∫ 1

    0w

    (ημ

    )∣∣Dμ(g)(ημvμ1 +

    (1 – ημ

    )vμ2

    )∣∣dμη,∣∣Dμ(g)

    (vμ1

    )∣∣,∣∣Dμ(g)

    (vμ2

    )∣∣,η)

    +∫ 1

    0Lw(ημ) dμt ≤ 0.

    From Lemma 2.1, we have

    ∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣

    ≤ vμ2 – v

    μ1

    2

    ∫ 1

    0w

    (ημ

    )∣∣Dμ(g)(ημvμ1 +

    (1 – ημ

    )vμ2

    )∣∣dμη.

    Since T�μ ,w is nondecreasing according to the first variable, then we can deduce

    T�μ ,w(

    2vμ2 – v

    μ1

    ∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣,

    ∣∣Dμ(g)

    (vμ1

    )∣∣,∣∣Dμ(g)

    (vμ2

    )∣∣,η)

    +∫ 1

    0Lw(ημ) dμη ≤ 0,

    which ends the proof of (4.1). �

    Corollary 4.1 Theorem 4.1 with |Dμ(g)| to be ε conformable convex leads to∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣

    ≤ vμ2 – v

    μ1

    2

    (23μ2 + 6 × 2μ2 – 8

    6μ × 23μ2)(∣∣Dμ(g)

    (vμ1

    )∣∣ +∣∣Dμ(g)

    (vμ2

    )∣∣ +2μ – 1

    2με

    ). (4.2)

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 13 of 25

    Proof We know that any ε-convex is �μ-convex. So, by making use of w(ημ) = |1 – 2ημ|in (2.7) and by using Definition 2.2, we get

    ∫ 1

    0Lw(η) dμη = ε

    ∫ 1

    0

    (1 – w(η)

    )dμη =

    12μ

    ε.

    By making use of w(ημ) = |1 – 2ημ| in (2.6), we get

    T�μ ,w(y1, y2, y3) = y1 –(∫ 1

    0ημ

    ∣∣1 – 2ημ∣∣dμη

    )y2 –

    (∫ 1

    0

    (1 – ημ

    )∣∣1 – 2ημ∣∣dμη

    )y3 – ε

    = y1 –(

    23μ2 + 6 × 2μ2 – 86μ × 23μ2

    )(y2 + y3) – ε

    for y1, y2, y3 ∈ R. By making use of Theorem 4.1, we get

    0 ≥ T�μ ,w(

    2vμ2 – v

    μ1

    ∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣,

    ∣∣Dμ(g)

    (vμ1

    )∣∣,∣∣Dμ(g)

    (vμ2

    )∣∣,η)

    +∫ 1

    0Lw(ημ) dμη

    =2

    vμ2 – vμ1

    ∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣

    –(

    23μ2 + 6 × 2μ2 – 86μ × 23μ2

    )(∣∣Dμ(g)(vμ1

    )∣∣ +∣∣Dμ(g)

    (vμ2

    )∣∣) – ε +1

    2με.

    This rearranges to the required inequality (4.2). �

    Remark 4.1 Corollary 4.1 with ε = 0 becomes Theorem 13 in [55].

    Corollary 4.2 Theorem 4.1 with |Dμ(g)| to be μ-conformable convex leads to∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣

    ≤ vμ2 – v

    μ1

    2(μ + 1)(2μ + 1)

    (1 +

    μ

    21/μ

    )(∣∣Dμ(g)

    (vμ1

    )∣∣ +∣∣Dμ(g)

    (vμ2

    )∣∣). (4.3)

    Proof We know that any μ-convex is �μ-convex. So, by making use of w(ημ) = |1 – 2ημ|in (2.9), we get

    T�μ ,w(y1, y2, y3) = y1 –(∫ 1

    0t∣∣1 – 2ημ

    ∣∣dμη)

    y2 –(∫ 1

    0(1 – η)

    ∣∣1 – 2ημ∣∣dμη

    )y3

    = y1 –1

    (μ + 1)(2μ + 1)

    (1 +

    μ

    21/μ

    )(y2 + y3)

    for y1, y2, y3 ∈ R. Then, by applying Theorem 4.1, we have

    0 ≥ T�μ ,w(

    2vμ2 – v

    μ1

    ∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣,

    ∣∣Dμ(g)

    (vμ1

    )∣∣,∣∣Dμ(g)

    (vμ2

    )∣∣,η)

    +∫ 1

    0Lw(ημ) dμη

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 14 of 25

    =2

    vμ2 – vμ1

    ∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣

    –1

    (μ + 1)(2μ + 1)

    (1 +

    μ

    21/μ

    )(∣∣Dμ(g)(vμ1

    )∣∣ +∣∣Dμ(g)

    (vμ2

    )∣∣).

    This rearranges to the required inequality (4.3). �

    Corollary 4.3 Theorem 4.1 with |Dμ(g)| to be h-conformable convex leads to

    ∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣

    ≤ vμ2 – v

    μ1

    2

    (∫ 1

    0h(ημ

    )∣∣1 – 2ημ∣∣dμη

    )(∣∣Dμ(g)(vμ1

    )∣∣ +∣∣Dμ(g)

    (vμ2

    )∣∣). (4.4)

    Proof It is known that every μ-convex is�μ-convex. So, by making use of w(ημ) = |1–2ημ|in (2.11), we get

    T�μ ,w(y1, y2, y3) = y1 –(∫ 1

    0h(ημ

    )∣∣1 – 2ημ∣∣dμη

    )y2

    –(∫ 1

    0h(1 – ημ

    )∣∣1 – 2ημ∣∣dμη

    )y3

    = y1 –(∫ 1

    0h(ημ

    )∣∣1 – 2ημ∣∣dμη

    )(y2 + y3)

    for y1, y2, y3 ∈ R. Then, by using Theorem 4.1, we get

    0 ≥ T�μ ,w(

    2vμ2 – v

    μ1

    ∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣,

    ∣∣Dμ(g)

    (vμ1

    )∣∣,∣∣Dμ(g)

    (vμ2

    )∣∣,η)

    +∫ 1

    0Lw(ημ) dμη

    =2

    vμ2 – vμ1

    ∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣

    –(∫ 1

    0h(ημ

    )∣∣1 – 2ημ∣∣dμη

    )(∣∣Dμ(g)

    (vμ1

    )∣∣ +∣∣Dμ(g)

    (vμ2

    )∣∣).

    This completes the proof of (4.4). �

    Theorem 4.2 Let g : [v1, v2] ⊂ R → R be a μ-fractional differentiable function on (v1, v2)and Dμ(g) be a μ-fractional integrable function on [v1, v2] with 0 ≤ v1 < v2. If |Dμ(g)|

    pp–1 is

    an �μ-convex function on [v1, v2] for some �μ ∈F , then we have

    T�μ ,1(v1(g, p),

    ∣∣Dμ(g)(vμ1

    )∣∣p

    p–1 ,∣∣Dμ(g)

    (vμ2

    )∣∣p

    p–1) ≤ 0, (4.5)

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 15 of 25

    where

    v1(g, p) =(

    2vμ2 – v

    μ1

    ) pp–1

    (1

    2μ(p + 1)

    {2 –

    (1 –

    12μ2–1

    )p+1–

    (1

    2μ2–1– 1

    )p+1}) –1p–1

    ×∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣

    pp–1

    .

    Proof By using the �μ-convexity of |Dμ(g)|p

    p–1 , we have

    �μ

    (∣∣Dμ(g)(ημvμ1 +

    (1 – ημ

    )vμ2

    )∣∣p

    p–1 ,∣∣Dμ(g)

    (vμ1

    )∣∣p

    p–1 ,∣∣Dμ(g)

    (vμ2

    )∣∣p

    p–1 ,η) ≤ 0.

    By making use of w(ημ) = 1 in axiom (Λ̄3), we obtain

    T�μ ,1(∫ 1

    0

    ∣∣Dμ(g)(ημvμ1 +

    (1 – ημ

    )vμ2

    )∣∣p

    p–1 dμη,∣∣Dμ(g)

    (vμ1

    )∣∣p

    p–1 ,∣∣Dμ(g)

    (vμ2

    )∣∣p

    p–1

    )≤ 0.

    Then, by making use Lemma of 2.2, we have

    ∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣

    ≤ vμ2 – v

    μ1

    2

    (1

    2μ(p + 1)

    {2 –

    (1 –

    12μ2–1

    )p+1–

    (1

    2μ2–1– 1

    )p+1}) 1p

    ×(∫ 1

    0

    ∣∣Dμ(g)(ημvμ1 +

    (1 – ημ

    )vμ2

    )∣∣p

    p–1 dμη) p–1

    p.

    Since T�μ ,w is nondecreasing according to the first variable, then we can deduce

    T�μ ,1(v1(g, p),

    ∣∣Dμ(g)(vμ1

    )∣∣p

    p–1 ,∣∣Dμ(g)

    (vμ2

    )∣∣p

    p–1) ≤ 0.

    This completes the proof of (4.5). �

    Corollary 4.4 Theorem 4.2 with |Dμ(g)|p

    p–1 to be ε-conformable convex leads to

    ∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣

    ≤ vμ2 – v

    μ1

    2

    (1

    2μ(p + 1)

    {2 –

    (1 –

    12μ2–1

    )p+1–

    (1

    2μ2–1– 1

    )p+1}) 1p

    ×( |Dμ(g)(vμ1 )|

    pp–1 + |Dμ(g)(vμ2 )|

    pp–1

    2μ+ ε

    ) p–1p

    . (4.6)

    Proof By making use of w(ημ) = |1 – 2ημ| in (2.6) and by Definition 2.3, we get

    T�μ ,w(y1, y2, y3) = y1 –(∫ 1

    0ημ

    ∣∣1 – 2ημ

    ∣∣dμη

    )y2 –

    (∫ 1

    0

    (1 – ημ

    )∣∣1 – 2ημ∣∣dμη

    )y3 – ε

    = y1 –y2 + y3

    2μ– ε

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 16 of 25

    for y1, y2, y3 ∈ R. Then, by using Theorem 4.2, we have

    0 ≥ T�μ ,1(v1(g, p),

    ∣∣Dμ(g)(vμ1

    )∣∣p

    p–1 ,∣∣Dμ(g)

    (vμ2

    )∣∣p

    p–1)

    – �

    = v1(g, p) –|Dμ(g)(vμ1 )|

    pp–1 + |Dμ(g)(vμ2 )|

    pp–1

    2μ– ε

    =(

    2vμ2 – v

    μ1

    ) pp–1

    (1

    2μ(p + 1)

    {2 –

    (1 –

    12μ2–1

    )p+1–

    (1

    2μ2–1– 1

    )p+1}) –1p–1

    ×∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣

    pp–1

    –|Dμ(g)(vμ1 )|

    pp–1 + |Dμ(g)(vμ2 )|

    pp–1

    2μ– ε.

    This completes our proof. �

    Remark 4.2 Corollary 4.4 with ε = 0 becomes Theorem 13 in [55].

    Corollary 4.5 Theorem 4.2 with |Dμ(g)| to be μ-convex leads to

    ∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣

    ≤ vμ2 – v

    μ1

    2

    (1

    2μ(p + 1)

    {2 –

    (1 –

    12μ2–1

    )p+1–

    (1

    2μ2–1– 1

    )p+1}) 1p

    ×(

    μ|Dμ(g)(vμ1 )|p

    p–1 + |Dμ(g)(vμ2 )|p

    p–1

    μ(μ + 1)

    ) p–1p

    . (4.7)

    Proof By making use of w(ημ) = 1 in (2.9), we get

    T�μ ,1(y1, y2, y3) = y1 –(∫ 1

    0η dμη

    )y2 –

    (∫ 1

    0(1 – η) dμη

    )y3

    = y1 –μy2 + y3μ(μ + 1)

    for y1, y2, y3 ∈ R. Then, by using Theorem 4.2, we have

    0 ≥ T�μ ,1(v1(g, p),

    ∣∣Dμ(g)(vμ1

    )∣∣p

    p–1 ,∣∣Dμ(g)

    (vμ2

    )∣∣p

    p–1)

    = v1(g, p) –μ|Dμ(g)(vμ1 )|

    pp–1 + |Dμ(g)(vμ2 )|

    pp–1

    μ(μ + 1).

    This rearranges to the required inequality (4.7). �

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 17 of 25

    Corollary 4.6 Theorem 4.2 with |Dμ(g)| to be h-convex leads to∣∣∣∣g(vμ1 ) + g(v

    μ2 )

    2–

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx

    ∣∣∣∣

    ≤ vμ2 – v

    μ1

    2

    (1

    2μ(p + 1)

    {2 –

    (1 –

    12μ2–1

    )p+1–

    (1

    2μ2–1– 1

    )p+1}) 1p

    ×(∫ 1

    0h(ημ

    )dμη

    ) p–1p (∣∣Dμ(g)

    (vμ1

    )∣∣p

    p–1 +∣∣Dμ(g)

    (vμ2

    )∣∣p

    p–1) p–1

    p . (4.8)

    Proof By making use of (2.11) with w(ημ) = 1, we have

    T�μ ,1(y1, y2, y3) = y1 –(∫ 1

    0h(ημ

    )dμη

    )y2 –

    (∫ 1

    0h(1 – ημ

    )dμη

    )y3

    = y1 –(∫ 1

    0h(ημ

    )dμη

    )(y2 + y3)

    for y1, y2, y3 ∈ R. Then, by making use of Theorem 4.2, we get

    0 ≥ T�μ ,1(v1(g, p),

    ∣∣Dμ(g)

    (vμ1

    )∣∣p

    p–1 ,∣∣Dμ(g)

    (vμ2

    )∣∣p

    p–1)

    = v1(g, p) –(∫ 1

    0h(ημ

    )dμη

    )(∣∣Dμ(g)(vμ1

    )∣∣p

    p–1 +∣∣Dμ(g)

    (vμ2

    )∣∣p

    p–1).

    This rearranges to the required inequality (4.8). �

    4.2 Midpoint formula inequalities for �μ-convex functionsTheorem 4.3 Let g : [v1, v2] ⊂ R → R be a μ-fractional differentiable function on (v1, v2)and Dμ(g) be a μ-fractional integrable function on [v1, v2] with 0 ≤ v1 < v2. If |Dμ(g)| is an�μ-convex function on [v1, v2] for some �μ ∈F and the function η ∈ [0, 1] → Lw(η) belongsto L1[0, 1], where w(η) = |P(η)| (P(η) is given in Lemma 2.2, then we have the inequality

    T�μ ,w(

    1vμ2 – v

    μ1

    ∣∣∣∣

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx – g

    (vμ1 + v

    μ2

    2

    )∣∣∣∣,

    ∣∣Dμ(g)

    (vμ1

    )∣∣,∣∣Dμ(g)

    (vμ2

    )∣∣,η)

    +∫ 1

    0Lw(η) dμη ≤ 0. (4.9)

    Proof By using the �μ-convexity of |Dμ(g)|, we have

    �μ

    (∣∣Dμ(g)(ημvμ1 +

    (1 – ημ

    )vμ2

    )∣∣,∣∣Dμ(g)

    (vμ1

    )∣∣,∣∣Dμ(g)

    (vμ2

    )∣∣,η) ≤ 0.

    Making use of axiom (Λ̄3) for w(η) = |P(η)|, η ∈ [0, 1], we obtain

    �μ

    (w(η)

    ∣∣Dμ(g)(ημvμ1 +

    (1 – ημ

    )vμ2

    )∣∣, w(η)∣∣Dμ(g)

    (vμ1

    )∣∣, w(η)∣∣Dμ(g)

    (vμ2

    )∣∣,η) ≤ 0.

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 18 of 25

    Integrating over [0, 1] according to η and by making use of axiom (Λ̄2), we can obtain

    T�μ ,w(∫ 1

    0w(η)

    ∣∣Dμ(g)

    (ημvμ1 +

    (1 – ημ

    )vμ2

    )∣∣dμη,∣∣Dμ(g)

    (vμ1

    )∣∣,∣∣Dμ(g)

    (vμ2

    )∣∣,η)

    +∫ 1

    0Lw(η) dμη ≤ 0.

    From Lemma 2.2, we have

    ∣∣∣∣

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx – g

    (vμ1 + v

    μ2

    2

    )∣∣∣∣

    ≤ (vμ2 – vμ1)∫ 1

    0w(η)

    ∣∣Dμ(g)(ημvμ1 +

    (1 – ημ

    )vμ2

    )∣∣dμη.

    Since T�μ ,w is nondecreasing according to the first variable, then we can deduce

    T�,w(

    1vμ2 – v

    μ1

    ∣∣∣∣

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx – g

    (vμ1 + v

    μ2

    2

    )∣∣∣∣,

    ∣∣Dμ(g)

    (vμ1

    )∣∣,∣∣Dμ(g)

    (vμ2

    )∣∣,η)

    +∫ 1

    0Lw(η) dμη ≤ 0.

    This completes the proof of (4.9). �

    Corollary 4.7 Theorem 4.3 with |Dμ(g)| to be ε-convex leads to∣∣∣∣

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx – g

    (vμ1 + v

    μ2

    2

    )∣∣∣∣

    ≤ (vμ2 – vμ1)( |Dμ(g)(vμ1 )| + |Dμ(g)(vμ2 )|

    8μ+

    4μ – 3μ

    ε

    ). (4.10)

    Proof By making use of w(ημ) = |P(η)| in (2.7) as well as Definition 2.3, we get∫ 1

    0Lw(η) dμη = ε

    ∫ 1

    0

    (1 –

    ∣∣P(η)

    ∣∣)dμη =

    34μ

    ε.

    Then, by making use (2.6) with w(ημ) = |P(η)|, we get

    T�μ ,w(y1, y2, y3) = y1 –(∫ 1

    0ημ

    ∣∣P(η)∣∣dμη

    )y2 –

    (∫ 1

    0

    (1 – ημ

    )∣∣P(η)∣∣dμη

    )y3 – ε

    = y1 –y2 + y3

    8μ– ε,

    for y1, y2, y3 ∈ R. Thus, by using Theorem 4.3, we have

    0 ≥ T�,w(

    1vμ2 – v

    μ1

    ∣∣∣∣

    μ

    vμ2 – vμ1

    ∫ v2

    ag(xμ

    )dμx – g

    (vμ1 + v

    μ2

    2

    )∣∣∣∣,

    ∣∣Dμ(g)

    (vμ1

    )∣∣,∣∣Dμ(g)

    (vμ2

    )∣∣,η)

    +∫ 1

    0Lw(η) dμη

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 19 of 25

    =1

    vμ2 – vμ1

    ∣∣∣∣

    μ

    vμ2 – vμ1

    ∫ v2

    ag(xμ

    )dμx – g

    (vμ1 + v

    μ2

    2

    )∣∣∣∣

    –|Dμ(g)(vμ1 )| + |Dμ(g)(vμ2 )|

    8μ– ε +

    34μ

    ε.

    This completes the proof of (4.10). �

    Remark 4.3 Corollary 4.7 with ε = 0 becomes Theorem 14 in [55].

    Corollary 4.8 Theorem 4.3 with |Dμ(g)| to be μ-convex leads to∣∣∣∣

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx – g

    (vμ1 + v

    μ2

    2

    )∣∣∣∣

    ≤ (vμ2 – vμ1){ μ

    (μ + 1)(2μ + 1)

    (1 –

    121/μ + 1

    )∣∣Dμ(g)

    (vμ1

    )∣∣

    +[

    14μ

    –μ

    (μ + 1)(2μ + 1)

    (1 –

    121/μ + 1

    )]∣∣Dμ(g)

    (vμ2

    )∣∣}

    . (4.11)

    Proof By making use of w(ημ) = |P(η)| in (2.9), we get

    T�μ ,w(y1, y2, y3) = y1 –(∫ 1

    0t∣∣P(η)

    ∣∣dμη

    )y2 –

    (∫ 1

    0(1 – η)

    ∣∣P(η)

    ∣∣dμη

    )y3

    = y1 –μ

    (μ + 1)(2μ + 1)

    (1 –

    121/μ + 1

    )y2

    –[

    14μ

    –μ

    (μ + 1)(2μ + 1)

    (1 –

    121/μ + 1

    )]y3

    for y1, y2, y3 ∈ R. It follows from Theorem 4.3 that

    0 ≥ T�μ ,w(

    1vμ2 – v

    μ1

    ∣∣∣∣

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx – g

    (vμ1 + v

    μ2

    2

    )∣∣∣∣,

    ∣∣Dμ(g)(vμ1

    )∣∣,∣∣Dμ(g)

    (vμ2

    )∣∣,η)

    +∫ 1

    0Lw(ημ) dμη

    =1

    vμ2 – vμ1

    ∣∣∣∣

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx – g

    (vμ1 + v

    μ2

    2

    )∣∣∣∣

    –μ

    (μ + 1)(2μ + 1)

    (1 –

    121/μ + 1

    )∣∣Dμ(g)(vμ1

    )∣∣

    –[

    14μ

    –μ

    (μ + 1)(2μ + 1)

    (1 –

    121/μ + 1

    )]∣∣Dμ(g)

    (vμ2

    )∣∣.

    This rearranges to the required inequality (4.11). �

    Corollary 4.9 Theorem 4.3 with |Dμ(g)| to be h-convex leads to∣∣∣∣

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx – g

    (vμ1 + v

    μ2

    2

    )∣∣∣∣

    ≤ (vμ2 – vμ1)(∫ 1

    0h(ημ

    )∣∣1 – 2ημ∣∣dμη

    )(∣∣Dμ(g)

    (vμ1

    )∣∣ +∣∣Dμ(g)

    (vμ2

    )∣∣). (4.12)

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 20 of 25

    Proof By making use of w(ημ) = |P(η)| in (2.11), we get

    T�μ ,w(y1, y2, y3) = y1 –(∫ 1

    0h(ημ

    )∣∣P(η)∣∣dμη

    )y2 –

    (∫ 1

    0h(1 – ημ

    )∣∣P(η)∣∣dμη

    )y3

    = y1 –(∫ 1

    0h(ημ

    )∣∣P(η)∣∣dμη

    )(y2 + y3)

    for y1, y2, y3 ∈ R. Then, by using Theorem 4.3, we get

    0 ≥ T�μ ,w(

    1vμ2 – v

    μ1

    ∣∣∣∣

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx – g

    (vμ1 + v

    μ2

    2

    )∣∣∣∣,

    ∣∣Dμ(g)

    (vμ1

    )∣∣,∣∣Dμ(g)

    (vμ2

    )∣∣,η)

    +∫ 1

    0Lw(ημ) dμη

    =1

    vμ2 – vμ1

    ∣∣∣∣

    μ

    vμ2 – vμ1

    ∫ v2

    v1g(xμ

    )dμx – g

    (vμ1 + v

    μ2

    2

    )∣∣∣∣

    –(∫ 1

    0h(ημ

    )∣∣P(η)∣∣dμη

    )(∣∣Dμ(g)

    (vμ1

    )∣∣ +∣∣Dμ(g)

    (vμ2

    )∣∣),

    which rearranges to the required inequality (4.12). �

    5 Application testIn this section we give some applications of our theorems to the special means for thepositive numbers v1 > 0 and v2 > 0:

    • Arithmetic mean:

    A(v1, v2) =v1 + v2

    2.

    • Harmonic mean:

    H = H(v1, v2) =2v1v2

    v1 + v2, v1, v2 > 0.

    • Logarithmic mean:

    L(v1, v2) =v2 – v1

    ln |v2| – ln |v1| , |v1| = |v2|, v1, v2 = 0, v1, v2 ∈ R.

    • Generalized log-mean:

    Lp(v1, v2) =[

    v2p+1 – v1p+1

    (p + 1)(v2 – v1)

    ] 1p

    , p ∈ Z \ {–1, 0}, v1, v2 ∈ R, v1 = v2.

    Proposition 5.1 Let μ ∈ (0, 1], v1, v2 ∈ R with 0 < v1 < v2. Then we have

    (A–1

    (vμ1 , v

    μ2),L–1

    (vμ1 , v

    μ2),L–1

    (vμ1 , v

    μ2),

    12

    )≤ 0, (5.1)

    T�,w(

    2L–1(vμ1 , v

    μ2),

    12H–1

    (vμ1 , v

    μ2),

    12H–1

    (vμ1 , v

    μ2))

    ≤ 0. (5.2)

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 21 of 25

    Proof The assertion follows from Theorem 3.1 and a simple computation applied to g(x) =1x , x ∈ [v1, v2], where g is convex and therefore �-convex function on [v1, v2] according to� defined in (1.3) with ε = 0. �

    Proposition 5.2 Let μ ∈ (0, 1], v1, v2 ∈ R with 0 < v1 < v2. Then we have

    A–1(vμ1 , v

    μ2) ≤L–1(vμ1 , vμ2

    ) ≤H–1(vμ1 , vμ2). (5.3)

    Proof The assertion follows from Corollary 3.1 and a simple computation applied to g(x) =1x , x ∈ [v1, v2], where it is easy to check that g is convex and therefore ε-convex with ε = 0.�

    Proposition 5.3 Let μ ∈ (0, 1], v1, v2 ∈ R with 0 < v1 < v2. Then we have

    (An

    (vμ1 , v

    μ2),Lnn

    (vμ1 , v

    μ2),Lnn

    (vμ1 , v

    μ2),

    12

    )≤ 0, (5.4)

    T�,w(2Lnn

    (vμ1 , v

    μ2), vnμ1 + v

    nμ2 , v

    nμ1 + v

    nμ2

    ). (5.5)

    Proof The assertion follows from Theorem 3.1 and a simple computation applied to g(x) =xn, x ∈ [v1, v2] with n ≥ 2, where g is convex and therefore �-convex function on [v1, v2]according to � defined in (1.3) with ε = 0. �

    Proposition 5.4 Let μ ∈ (0, 1], v1, v2 ∈ R with 0 < v1 < v2. Then we have

    ∣∣H–1

    (vμ1 , v

    μ2)

    –L–1(vμ1 , v

    μ2)∣∣ ≤ v

    μ2 – v

    μ1

    2

    (23μ2 + 6 × 2μ2 – 8

    6μ × 23μ2)(

    v–μ(1+μ)1 + v–μ(1+μ)2

    ). (5.6)

    Proof The assertion follows from Corollary 4.1 and a simple computation applied to g(x) =– 1x , x ∈ [v1, v2], where it is easy to check that |Dμ(g)| is convex and therefore ε-convex withε = 0. �

    Proposition 5.5 Let μ ∈ (0, 1], v1, v2 ∈ R with 0 < v1 < v2. Then we have

    ∣∣L–1

    (vμ1 , v

    μ2)

    – A–1(vμ1 , v

    μ2)∣∣ ≤ (vμ2 – vμ1

    )(v–μ(1+μ)1 + v–μ(1+μ)2

    ). (5.7)

    Proof The assertion follows from Corollary 4.7 and a simple computation applied to g(x) =– 1x , x ∈ [v1, v2], where it is easy to check that |Dμ(g)| is convex and therefore ε-convex withε = 0. �

    6 Three illustrative plotsIn this section, we give three plots of three dimensions to the above propositions in theprevious section.

    • Fig. 1 represents Proposition 5.2 with μ = 12 , v1 = x, v2 = y.• Fig. 2 represents Proposition 5.4 μ = 12 , v1 = x, v2 = y.• Fig. 3 represents Proposition 5.5 μ = 12 , v1 = x, v2 = y.

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 22 of 25

    Figure 1 Figure representation for Proposition 5.2

    Figure 2 Figure representation for Proposition 5.4

    Figure 3 Figure representation for Proposition 5.5

  • Baleanu et al. Advances in Difference Equations (2020) 2020:374 Page 23 of 25

    7 ConclusionIntroducing new definitions in the calculus will always open new doors in the field of sci-ence and technology. The use of these new definitions in mathematical analysis alwaysrequires the presentation of integral inequalities related to them in order to find the ex-istence and uniqueness of such problems. One of the new definitions presented for localfractional calculus is conformable fractional operator. In this study, we have consideredthe Hermite–Hadamard integral inequalities in the context of conformable fractional cal-culus. Also, we have introduced the notion of �μ-convexity. For this, we have establishedsome Hermite–Hadamard inequalities and related results in the contexts of fractional cal-culus and conformable fractional calculus.

    AcknowledgementsWe express our special thanks to the editor and referees of this manuscript.

    DeclarationsNot applicable.

    FundingNot applicable.

    Availability of data and materialsNot applicable.

    Competing interestsThe authors declare no conflict of interests.

    Authors’ contributionsThe four authors have contributed equally to the attempt. All four authors have read carefully and approved the finalversion of the study.

    Author details1Institute of Space Sciences, P.O. Box, MG-23, Magurele-Bucharest, R 76900, Romania. 2Department of Mathematics,Çankaya University, Ankara, Turkey. 3Department of Medical Research, China Medical University Hospital, China MedicalUniversity, Taichung, Taiwan. 4Department of Mathematics, College of Education, University of Sulaimani, Sulaimani,Kurdistan Region, Iraq. 5Facultad de Ciencias Exactas y Naturales, Escuela de Ciencias Fisicas y Matematica, PontificiaUniversidad Católica del Ecuador, Quito, Ecuador.

    Publisher’s NoteSpringer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

    Received: 23 April 2020 Accepted: 13 July 2020

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    Some modifications in conformable fractional integral inequalitiesAbstractMSCKeywords

    IntroductionConformable fractional operators and Fµ-convexityHermite-Hadamard inequalities for F-convex functionsHermite-Hadamard inequalities for Fµ-convex functionsTrapezoidal inequalities for Fµ-convex functionsMidpoint formula inequalities for Fµ-convex functions

    Application testThree illustrative plotsConclusionAcknowledgementsDeclarationsFundingAvailability of data and materialsCompeting interestsAuthors' contributionsAuthor detailsPublisher's NoteReferences