# New Fractional Hermite–Hadamard–Mercer Inequalities for Harmonically Convex Function

Authors: Saad Ihsan Butt; Saba Yousaf; Atifa Asghar; Khuram Ali Khan; Hamid Reza Moradi
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In 2003, Mercer presented an interesting variation of Jensen’s inequality called Jensen–Mercer inequality for convex function. In the present paper, by employing harmonically convex function, we introduce analogous versions of Hermite–Hadamard inequalities of the Jensen–Mercer type via fractional integrals. As a result, we introduce several related fractional inequalities connected with the right and left differences of obtained new inequalities for differentiable harmonically convex mappings. As an application viewpoint, new estimates regarding hypergeometric functions and special means of real numbers are exemplified to determine the pertinence and validity of the suggested scheme. Our results presented here provide extensions of others given in the literature. The results proved in this paper may stimulate further research in this fascinating area.


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## Body

## 1. Introduction

The definition of convexity has been improved, generalized, and expanded in several directions in recent years. In the literature, Jensen’s inequality (J-I) and the Hermite–Hadamard’s (H-H) inequality are highly familiar results. Several new classes of convex functions along with their respective new variants of (J-I) and (H-H) inequalities are established. One of the well-known and most significant inequalities in mathematical analysis is (J-I) and its related variants. These inequalities are useful in Physics since they provide upper and lower limits for natural phenomena defined by integrals, such as mechanical work. The definition of a classical convex function is as follows:Definition 1.
A functionϕ:I⊂ℝ⟶ℝ is called a convex function on ϕ∈KI, if
(1)ϕζθ+1−ζΘ≤ζϕθ+1−ζϕΘ,holds provided that all θ,Θ∈I and ζ∈0,1.Jensen’s inequality is the key to success in extracting applications in information theory. It is effective in finding estimates for several quantitative measures in information theory about continuous random variables, see [1–3]. The (J-I) can be stated as a generalization of convex functions as follows:Theorem 2 (see [4]).
Ifϕ∈KI, then
(2)ϕ∑mj=1ωjaj≤∑mj=1ωjϕaj,for all aj∈I and ωj∈0,1, j=1,2,⋯,m with ∑mj=1ωj=1.The (H-H) inequality is another well-known inequality in the theory of convex analysis. Several notable results surrounding (H-H) inequality and its significance are compiled by Dragomir et al. in [5].Theorem 3.
Ifϕ∈KI on the interval I=θ,Θ with θ<Θ, then
(3)ϕθ+Θ2≤1Θ−θ∫θΘϕxdx≤ϕθ+ϕΘ2.In 2003, Mercer presents a variant of (J-I) which has a great impact on the theory of inequalities known as Jensen–Mercer (J-M) inequality.Theorem 4 (see [6]).
Ifϕ∈KI on the interval I=θ,Θ, then
(4)ϕθ+Θ−∑mj=1ωjaj≤ϕθ+ϕΘ−∑mj=1ωjϕaj,for all aj∈θ,Θ and ωj∈0,1, j=1,2,⋯,m with ∑mj=1ωj=1.In 2013 [7], Kian and Moslehian introduced new variant (H-H) type inequalities utilizing Mercer concept via convex functions. Recently in 2019, Moradi and Furuichi in [8] worked on some improvements and generalization of (J-M) type inequalities. Then in 2020 [9], Adil et al. gave applications of (J-M) inequality in information theory. He computed new estimates for Csiszár and related divergences. Also, he gave new bounds for Zipf-Mandelbrot Entropy via (J-M) inequality.Harmonic convex sets are introduced by investigating harmonic means. In 2003, the first harmonic convex set was introduced by Shi and Zhang [10]. The harmonic mean has been important in different fields of pure and applied sciences. Anderson et al. [11] and Íşcan [12] introduced a significant class of convex functions known as harmonic convex.Definition 5 (see [12]).
A functionϕ:I⊂ℝ\0⟶ℝ is said to be harmonically convex on ϕ∈HKI, if
(5)ϕθΘζθ+1−ζΘ≤ζϕΘ+1−ζϕθ,
for allθ,Θ∈I and ζ∈0,1 holds.The harmonic mean is useful in electrical circuit theory and different fields of research. It is well known that the total resistance of a set of parallel resistors can be calculated by adding the reciprocals of each individual resistance value and then taking the reciprocal of the total resistance. For example, ift1 and t2 are the resistance of two parallel resistors, the total resistance is
(6)T=t1t2t1+t2=12Ht1,t2,which is the half of the harmonic mean [13]. The harmonic mean is also important in the development of parallel algorithms for solving nonlinear problems [14]. The harmonic mean of the effective masses, as well as the three crystallographic directions, is often used to describe a semiconductor’s “conductivity effective mass” [15]; see also [16].Dragomir is the first to introduce (J-I) forϕ∈HKI as:Theorem 6 (see [17]).
Ifϕ∈HKI on the interval I⊆0,∞, then
(7)ϕ1∑mj=1ωj/aj≤∑mj=1ωjϕaj,
for allaj∈I and ωj∈0,1, j=1,2,⋯,m with ∑mj=1ωj=1.In [12], Íşcan proved the (H-H) inequality for ϕ∈HKI as:Theorem 7 (see [12]).
LetI⊆0,∞ be an interval. If ϕ∈HKI and ϕ∈Lθ,Θ and for all θ, Θ∈I with θ<Θ then
(8)ϕ2θΘθ+Θ≤θΘΘ−θ∫01ϕxx2dx≤ϕθ+ϕΘ2.Very recently, Baloch et al. [18] present a variant of (J-I) which has a great impact on the theory of inequalities known as (J-M) inequality for ϕ∈HKI:Theorem 8 (see [18]).
LetI=θ,Θ⊆0,∞ be an interval. If ϕ∈HKI, then inequality
(9)ϕ11/θ+1/Θ−∑mj=1ωj/aj≤ϕθ+ϕΘ−∑mj=1ωjϕaj,
for allaj∈θ,Θ and ωj∈0,1, j=1,2,⋯,m with ∑mj=1ωj=1.For some recent results connected with (J-M) inequality forϕ∈HKI, see [18, 19].Let us recall some important functions and inequality.(i) Beta function(10)βξ1,ξ2=Γξ1Γξ2Γξ1+ξ2=∫01ζξ1−11−ζξ2−1dζ,ξ1,ξ2>0.(ii) Hypergeometric function: [20]
(11)F12x,y;z;k=1βy,z−y∫01ζy−11−ζz−y−11−kζ−xdζ,z>y>0,k<1.Lemma 9 (see [21, 22]).
For0<α≤1 and 0≤x<y, we have
(12)xα−yα≤y−xα.One of the concepts that have played a significant role in the growth of inequality theory in recent years is fractional analysis. Fractional integrals are the most commonly used concept in calculus analysis to obtain new generalizations, extensions, and versions of classical integral inequalities. Since fractional calculus was presented toward the end of the nineteenth century, the subject has become a quickly developing area and has discovered numerous applications in different research fields. Fractional calculus is now concerned with fractional-order integral and derivative operators in real and complex analysis and their applications. Fractional calculus is used in several fields of engineering and science worldwide, including fluid dynamics, electrochemistry, electromagnetics, viscoelasticity, biological population models, optics, and signal processing. It has been used to model physical and engineering processes that are best represented by fractional differential equations.Now, we give the definition of Riemann-Liouville (RL) integrals which we will use in this paper.Definition 10.
Letϕ∈Lθ,Θ. The left and right sided (RL) fractional integrals of order α>0 with θ≥0 are stated as:
(13)Jθ+αϕr=1Γα∫θrr−uα−1ϕudu,r>θ,JΘ−αϕr=1Γα∫rΘu−rα−1ϕudu,r<Θ,
respectively, withΓα=∫0∞e−uuα−1du and Jθ+0ϕr= JΘ−0ϕr=ϕr.In recent times, the topic of investigating fractional (H-H) inequalities by employing the Mercer concept along with its applications is worth study, as evident from several publications in this direction (see [23–26]). This study is done by utilizing convex functions. But in this paper, we first time introduce and analyze this concept for harmonic convex functions. In this paper, by using (J-M) inequality, we derive Hermite-Hadamard–Mercer’s (H-H-M) inequalities for ϕ∈HKI via (RL) fractional integral, and we established several new fractional inequalities pertaining (H-H-M) type inequalities for differentiable harmonically convex mappings. Some applications to special means of positive real numbers will also be provided in Section 4. We hope that the new idea and techniques formulated in the present paper are more invigorating than the accessible ones.


## 2. (H-H-M) Inequalities forϕ∈HKI via (RL) Fractional Integrals

By using (J-M) inequality, we give the following (H-H-M) inequalities forϕ∈HKI.Theorem 11.
Letϕ:I=θ,Θ⊆0,∞⟶ℝ be a function such that ϕ∈Lθ,Θ with 0<θ<Θ. If ϕ∈HKI on the interval I=θ,Θ, then
(14)ϕ11/θ+1/Θ−x+y/2xy≤ϕθ+ϕΘ−Γα+12xyy−xαJ1/y+αϕ∘ℏ1x+J1/x−αϕ∘ℏ1y≤ϕθ+ϕΘ−ϕ2xyx+y,(15)ϕ11/θ+1/Θ−x+y/2xy≤Γα+12xyy−xαJ1/θ+1/Θ−1/x+αϕ∘ℏ1θ+1Θ−1y+J1/θ+1/Θ−1/y−αϕ∘ℏ1θ+1Θ−1x≤12ϕ11/θ+1/Θ−1/x+ϕ11/θ+1/Θ−1/y≤ϕθ+ϕΘ−ϕx+ϕy2,for all x, y∈θ,Θ, α>0, and ℏr=1/r, r∈1/Θ,1/θ.Proof.
By employing (J-M) inequality forϕ∈HKI, we have
(16)ϕ11/θ+1/Θ−x1+y1/2x1y1≤ϕθ+ϕΘ−ϕx1+ϕy12,for all x1, y1∈θ,Θ. By changing of the variables x1=xy/ζx+1−ζy, y1=xy/ζy+1−ζa for all x, y∈θ,Θ, and ζ∈0,1 in (16), we obtain
(17)ϕ11/θ+1/Θ−x+y/2xy≤ϕθ+ϕΘ−ϕxy/ζx+1−ζy+ϕxy/ζy+1−ζa2.
Conducting product on both sides of (17) by ζα−1 and then integrating the obtained inequality w.r.t ζ over 0,1, we have
(18)1αϕ11/θ+1/Θ−x+y/2xy≤1αϕθ+ϕΘ−12∫01ζα−1ϕxyζx+1−ζy+ϕxyζy+1−ζadζ=1αϕθ+ϕΘ−12xyy−xα×∫1/y1/x1x−uα−1ϕ1udu+∫1/y1/xu−1yα−1ϕ1udu=1αϕθ+ϕΘ−Γα2xyy−xαJ1/y+αϕ∘ℏ1x+J1/x−αϕ∘ℏ1y.
That is(19)ϕ11/θ+1/Θ−x+y/2xy≤ϕθ+ϕΘ−Γα+12xyy−xαJ1/y+αϕ∘ℏ1x+J1/x−αϕ∘ℏ1y,whereℏr=1r.
Thus, the first inequality of (14) is proved. Now, we prove the second inequality in (14), since ϕ∈HKI; then, for ζ∈0,1, it yields
(20)ϕ2xyx+y=ϕ2ζ/x+1−ζ/y+1−ζ/x+ζ/y≤12ϕxyζy+1−ζx+ϕxyζx+1−ζy.
Conducting product on both sides of (20) by ζα−1 and then integrating the obtained inequality w.r.t ζ over 0,1, we obtain
(21)1αϕ2xyx+y≤12∫01ζα−1ϕxyζy+1−ζx+ϕxyζx+1−ζydζ=Γα2xyy−xαJ1/y+αϕ∘ℏ1x+J1/x−αϕ∘ℏ1y,and then
(22)−ϕ2xyx+y≥−Γα+12xyy−xαJ1/y+αϕ∘ℏ1x+J1/x−αϕ∘ℏ1y.
Addingϕθ+ϕΘ to both sides of (22), we find the second inequality of (14).
Next for the proof of the inequality (15), take ϕ∈HKI; we have for any x,y∈θ,Θ(23)ϕ2xyx+y=ϕ11/2ζ/x+1−ζ/y+1−ζ/x+ζ/y≤12ϕ1ζ/x+1−ζ/y+ϕ11−ζ/x+ζ/y≤ϕx+ϕy2.
Replacingx and y by 1/1/θ+1/Θ−1/x and 1/1/θ+1/Θ−1/y, respectively, in (23), we get
(24)ϕ11/θ+1/Θ−x+y/2xy≤12ϕ11/θ+1/Θ−ζ/x+1−ζ/y+ϕ11/θ+1/Θ−1−ζ/x+ζ/y≤12ϕ11/θ+1/Θ−1/x+ϕ11/θ+1/Θ−1/y.
Conducting product on both sides of (24) by ζα−1 and then integrating the obtained inequality w.r.t ζ over 0,1, we have
(25)1αϕ11/θ+1/Θ−x+y/2xy≤12∫01ζα−1ϕ11/θ+1/Θ−ζ/x+1−ζ/ydζ+∫01ζα−1ϕ11/θ+1/Θ−1−ζ/x+ζ/ydζ≤12αϕ11/θ+1/Θ−1/x+ϕ11/θ+1/Θ−1/y.
It is obvious that(26)12∫01ζα−1ϕ11/θ+1/Θ−ζ/x+1−ζ/ydζ+∫01ζα−1ϕ11/θ+1/Θ−1−ζ/x+ζ/ydζ=Γα2xyy−xα∫1/θ+1/Θ−1/x1/θ+1/Θ−1/y1θ+1Θ−1y−uα−1ϕ1udu+∫1/θ+1/Θ−1/x1/θ+1/Θ−1/yu−1θ+1Θ−1xα−1ϕ1udu=Γα2xyy−xαJ1/θ+1/Θ−1/x+αϕ∘ℏ1θ+1Θ−1y+J1/θ+1/Θ−1/y−αϕ∘ℏ1θ+1Θ−1x.
Using (J-M) inequality forϕ∈HKI, we conclude that
(27)ϕ11/θ+1/Θ−x+y/2xy≤Γα+12xyy−xαJ1/θ+1/Θ−1/x+αϕ∘ℏ1θ+1Θ−1y+J1/θ+1/Θ−1/y−αϕ∘ℏ1θ+1Θ−1x≤12ϕ11/θ+1/Θ−1/x+ϕ11/θ+1/Θ−1/y≤ϕθ+ϕΘ−ϕx+ϕy2.
So, the inequality (15) is proved.☐Remark 12.
Under the assumptions of Theorem11 with α=1, one has
(28)ϕ11/θ+1/Θ−x+y/2xy≤ϕθ+ϕΘ−∫01ϕxyζx+1−ζydζ≤ϕθ+ϕΘ−ϕ2xyx+y,ϕ11/θ+1/Θ−x+y/2xy≤xyy−x∫xy1ζ2ϕ11/θ+1/Θ−1/ζdζ≤ϕθ+ϕΘ−ϕx+ϕy2,for all x, y∈θ,Θ. The proof of remark is proved by Baloch et al. in [18], Theorem 3.5, and [19], Theorem 2.1.


## 3. Related Variants of (H-H-M) Type Inequalities forϕ∈HKI via (RL) Fractional Integrals

Throughout the paper, we assumed the following assumption.A1=Let ϕ:I=θ,Θ⊆0,∞⟶ℝ be a differentiable function on θ,Θ with 0<θ<Θ,
(29)Iϕℏ;α,x,y=12ϕ11/θ+1/Θ−1/x+ϕ11/θ+1/Θ−1/y−Γα+12xyy−xαJ1/θ+1/Θ−1/x+αϕ∘ℏ1θ+1Θ−1y+J1/θ+1/Θ−1/y−αϕ∘ℏ1θ+1Θ−1x,where x,y∈I with x<y, α>0, and ℏr=1/r, r∈1/Θ,1/θ.We give the new following lemma for our results.Lemma 13.
Ifϕ′∈Lθ,Θ along with assumption A1, then the following equality for fractional integrals holds:
(30)Iϕℏ;α,x,y=y−x2xy∫01ζα−1−ζα1/θ+1/Θ−ζ/x+1−ζ/y2ϕ′11/θ+1/Θ−ζ/x+1−ζ/ydζ.Proof.
LetAζ=1/θ+1/Θ−ζ/x+1−ζ/y. It suffices to note that
(31)Iϕℏ;α,x,y=y−x2xy∫01ζα−1−ζαAζ2ϕ′1Aζdζ=12y−xxy∫01ζαAζ2ϕ′1Aζdζ−y−xxy∫011−ζαAζ2ϕ′1Aζdζ=12I1−I2.
Integrating by parts, we get(32)I1=y−xxy∫01ζαAζ2ϕ′1Aζdζ=ζαϕ1Aζ01−α∫01ζα−1ϕ1Aζdζ=ϕ11/θ+1/Θ−1/x−αxyy−xα∫1/θ+1/Θ−1/x1/θ+1/Θ−1/y1θ+1Θ−1y−uα−1ϕ1udu=ϕ11/θ+1/Θ−1/x−Γα+1xyy−xαJ1/θ+1/Θ−1/x+αϕ∘ℏ1θ+1Θ−1y.
Similarly, we get(33)I2=y−xxy∫011−ζαAζ2ϕ′1Aζdζ=1−ζαϕ1Aζ01+α∫011−ζα−1ϕ1Aζdζ=−ϕ11/θ+1/Θ−1/y+αxyy−xα∫1/θ+1/Θ−1/x1/θ+1/Θ−1/yu−1θ+1Θ−1xα−1ϕ1udu=−ϕ11/θ+1/Θ−1/y+Γα+1xyy−xαJ1/θ+1/Θ−1/y−αϕ∘ℏ1θ+1Θ−1x.
Using (32) and (33) in (31), we get inequality (30).☐Corollary 14.
If we chooseα=1 in Lemma 13, then we have the following equality:
(34)12ϕ11/θ+1/Θ−1/x+ϕ11/θ+1/Θ−1/y−xyy−x∫1/θ+1/Θ−1/x1/θ+1/Θ−1/yϕ1udu=y−x2xy∫012ζ−11/θ+1/Θ−ζ/x+1−ζ/y2ϕ′11/θ+1/Θ−ζ/x+1−ζ/ydζ.Remark 15.
If we takex=θ and y=Θ in Corollary 14, then the equality (34) reduces to the equality
(35)ϕθ+ϕΘ2−θΘΘ−θ∫θΘϕxx2dx=θΘΘ−θ2∫012ζ−1ζθ+1−ζΘ2ϕ′θΘζθ+1−ζΘdζ,which is proved by Íşcan in [12].Using Lemma13, we present the following fractional integral inequality for ϕ′q∈HKI as follows.Theorem 16.
Ifϕ′q∈HKθ,Θ for some fixed q≥1 and ϕ′∈Lθ,Θ along with assumption A1, then the following inequality for fractional integrals holds:
(36)Iϕℏ;α,x,y≤y−x2xyK11−1/qα;x,yK1α;x,yϕ′θq+ϕ′Θq−K2α;x,yϕ′xq+K3α;x,yϕ′yq1/q,where
(37)K1α;x,y=1/θ+1/Θ−1/y−2α+1F122,1;α+2;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y+F122,α+1;α+2;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y,K2α;x,y=1/θ+1/Θ−1/y−2α+21α+1F122,2;α+3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y+F122,α+2;α+3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y,K3α;x,y=1/θ+1/Θ−1/y−2α+2F122,1;α+3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y+1α+1F122,α+1;α+3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y.Proof.
LetAζ=1/θ+1/Θ−ζ/x+1−ζ/y=ζ1/θ+1/Θ−1/x+1−ζ1/θ+1/Θ−1/y. From Lemma 13 and Lemma 9, using the properties of the modulus, the power mean inequality, and ϕ′q∈HKI, we find that
(38)Iϕℏ;α,x,y≤y−x2xy∫011−ζα−ζαAζ2ϕ′1Aζdζ≤y−x2xy∫011−ζα−ζαAζ2dζ1−1/q∫011−ζα−ζαAζ2ϕ′1Aζqdζ1/q≤y−x2xy∫011−ζα+ζαAζ2dζ1−1/q∫011−ζα+ζαAζ2ϕ′θq+ϕ′Θq−ζϕ′xq+1−ζϕ′yqdζ1/q≤y−x2xyK11−1/qα;x,yK1α;x,yϕ′θq+ϕ′Θq−K2α;x,yϕ′xq+K3α;x,yϕ′yq1/q.
CalculatingK1α;x,y, K2α;x,y, and K3α;x,y, we have
(39)K1α;x,y=∫011−ζα+ζαAζ2dζ=1θ+1Θ−1y−2∫011−ζα1−ζ1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2dζ+∫01ζα1−ζ1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2dζ=1/θ+1/Θ−1/y−2α+1F122,1;α+2;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y+F122,α+1;α+2;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y,(40)K2α;x,y=∫011−ζα+ζαAζ2ζdζ=1θ+1Θ−1y−2∫01ζ1−ζα1−ζ1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2dζ+∫01ζα+11−ζ1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2dζ=1/θ+1/Θ−1/y−2α+21α+1F122,2;α+3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y+F122,α+2;α+3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y,(41)K3α;x,y=∫011−ζα+ζαAζ21−ζdζ=1θ+1Θ−1y−2∫011−ζα+11−ζ1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2dζ+∫01ζα1−ζ1−ζ1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2dζ=1/θ+1/Θ−1/y−2α+2F122,1;α+3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y+1α+1F122,α+1;α+3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y.
Using (39), (40), and (41) in (38), we get the inequality of (36).☐Remark 17.
If we takex=θ and y=Θ in Theorem 16, then it becomes Theorem 5 proved by Íşcan and Whu in [27].When0<α≤1, using Lemma 9 and Lemma 13, we can obtain another results for ϕ′q∈HKI via fractional integral as follows.Theorem 18.
Ifϕ′q∈HKθ,Θ for some fixed q≥1 and ϕ′∈Lθ,Θ along with assumption A1 and 0<α≤1, then the following inequality for fractional integrals holds:
(42)Iϕℏ;α,x,y≤y−x2xyK11−1/qα;x,yK1α;x,yϕ′θq+ϕ′Θq−K2α;x,yϕ′xq+K3α;x,yϕ′yq1/q,
where(43)K1α;x,y=1/θ+1/Θ−1/y−2α+1F122,α+1;α+2;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−F122,1;α+2;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y+F122,1;α+2;121−1/θ+1/Θ−1/x1/θ+1/Θ−1/y,K2α;x,y=1/θ+1/Θ−1/y−2α+2F122,α+2;α+3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−1α+1F122,2;α+3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y+12α+1F122,2;α+3;121−1/θ+1/Θ−1/x1/θ+1/Θ−1/y,K3α;x,y=1/θ+1/Θ−1/y−2α+21α+1F122,α+1;α+3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−F122,1;α+3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y+F122,1;α+3;121−1/θ+1/Θ−1/x1/θ+1/Θ−1/y+12α+1F122,2;α+3;121−1/θ+1/Θ−1/x1/θ+1/Θ−1/y.Proof.
LetAζ=1/θ+1/Θ−ζ/x+1−ζ/y=ζ1/θ+1/Θ−1/x+1−ζ1/θ+1/Θ−1/y. From Lemma 13 and Lemma 9, using the properties of the modulus, the power mean inequality, and ϕ′q∈HKθ,Θ, we find that
(44)Iϕℏ;α,x,y≤y−x2xy∫011−ζα−ζαAζ2ϕ′1Aζdζ≤y−x2xy∫011−ζα−ζαAζ2dζ1−1/q∫011−ζα−ζαAζ2ϕ′1Aζqdζ1/q≤y−x2xy∫011−ζα−ζαAζ2dζ1−1/q∫011−ζα−ζαAζ2ϕ′θq+ϕ′Θq−ζϕ′xq+1−ζϕ′yqdζ1/q≤y−x2xyZ11−1/qZ1ϕ′θq+ϕ′Θq−Z2ϕ′xq+Z3ϕ′yq1/q,where
(45)Z1=∫011−ζα−ζαAζ2dζ,Z2=∫011−ζα−ζαAζ2ζdζ,Z3=∫011−ζα−ζαAζ21−ζdζ.
CalculatingZ1, Z2, and Z3, by lemma, we have
(46)Z1=∫011−ζα−ζαAζ2dζ=∫01/21−ζα−ζαAζ2dζ+∫1/21ζα−1−ζαAζ2dζ=∫01ζα−1−ζαAζ2dζ+2∫01/21−ζα−ζαAζ2dζ≤∫01ζαAζ2dζ−∫011−ζαAζ2dζ+2∫01/21−2ζαAζ2dζ=1θ+1Θ−1y−2∫01ζα1−ζ1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2dζ−∫011−ζα×1−ζ1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2dζ+∫011−uα1−u21−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2du=1/θ+1/Θ−1/y−2α+1F122,α+1;α+2;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−F122,1;α+2;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y+F122,1;α+2;121−1/θ+1/Θ−1/x1/θ+1/Θ−1/y=K1α;x,y.
Similarly, we get(47)Z2=∫011−ζα−ζαAζ2ζdζ≤1θ+1Θ−1y−2∫01ζα+11−ζ1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2dζ−∫01ζ1−ζα×1−ζ1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2dζ+12∫01u1−uα1−u21−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2du=1/θ+1/Θ−1/y−2α+2F122,α+2;α+3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−1α+1×F122,2;α+3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y+12α+1F122,2;α+3;121−1/θ+1/Θ−1/x1/θ+1/Θ−1/y=K2α;x,y,(48)Z3=∫011−ζα−ζαAζ21−ζdζ≤1θ+1Θ−1y−2∫01ζα1−ζ1−ζ1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2dζ−∫011−ζα+1×1−ζ1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2dζ+∫011−uα1−u21−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2du+12∫01u1−uα1−u21−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2du=1/θ+1/Θ−1/y−2α+21α+1F122,α+1;α+3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−F122,1;α+3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y+F122,1;α+3;121−1/θ+1/Θ−1/x1/θ+1/Θ−1/y+12α+1F122,2;α+3;121−1/θ+1/Θ−1/x1/θ+1/Θ−1/y=K3α;x,y.
Using (46), (47), and (48) in (44), we get the inequality of (42).☐Remark 19.
If we takex=θ and y=Θ in Theorem 18, then it becomes Theorem 6 proved by Íşcan and Whu in [27].Remark 20.
If we takeα=1, x=θ, and y=Θ in Theorem 18, then it becomes Theorem 2.6 proved by Íşcan in [12].Theorem 21.
Ifϕ′q∈HKθ,Θ for some fixed q>1 and ϕ′∈Lθ,Θ along with assumption A1 and 0<α≤1, then the following inequality for fractional integrals holds:
(49)Iϕℏ;α,x,y≤y−x2xy1αp+11/p1θ+1Θ−1y−2ϕ′θq+ϕ′Θq−ϕ′xq+ϕ′yq21/q×2F11/p2p,1;αp+2;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y+2F11/p2p,αp+1;αp+2;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y,where 1/p+1/q=1.Proof.
LetAζ=1/θ+1/Θ−ζ/x+1−ζ/y=ζ1/θ+1/Θ−1/x+1−ζ1/θ+1/Θ−1/y. From Lemma 13 and Lemma 9, using the Hölder inequality and ϕ′q∈HKθ,Θ, we find
(50)Iϕℏ;α,x,y≤y−x2xy∫011−ζαAζ2ϕ′1Aζdζ+∫01ζαAζ2ϕ′1Aζdζ≤y−x2xy∫011−ζαpAζ2pdζ1/p∫01ϕ′1Aζqdζ1/q+∫01ζαpAζ2pdζ1/p∫01ϕ′1Aζqdζ1/q≤y−x2xyZ41/p+Z51/p∫01ϕ′θq+ϕ′Θq−ζϕ′xq+1−ζϕ′yqdζ1/q≤y−x2xyZ41/p+Z51/pϕ′θq+ϕ′Θq−ϕ′xq+ϕ′yq21/q.
CalculatingZ4 and Z5, we have
(51)Z4=∫011−ζαpAζ2pdζ=1θ+1Θ−1y−2p∫011−ζαp1−ζ1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2pdζ=1/θ+1/Θ−1/y−2pαp+1F122p,1;αp+2;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y,(52)Z5=∫01ζαpAζ2pdζ=1θ+1Θ−1y−2p∫01ζαp1−ζ1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2pdζ=1/θ+1/Θ−1/y−2pαp+1F122p,αp+1;αp+2;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y.
Using (51) and (52) in (50), we get the inequality of (49). This completes the proof.☐Remark 22.
If we takex=θ and y=Θ in Theorem 21, then it becomes Theorem 7 proved by Íşcan and Whu in [27].Theorem 23.
Ifϕ′q∈HKθ,Θ for some fixed q>1 and ϕ′∈Lθ,Θ along with assumption A1 and 0<α≤1, then following inequality for fractional integrals holds:
(53)Iϕℏ;α,x,y≤y−x2xy1αp+11/p1θ+1Θ−1y−22F12q,1;2;3−1/θ+1/Θ−1/x1/θ+1/Θ−1/y,ϕ′θq+ϕ′Θq−F122q,2;3;1−1/θ+1/Θ−1/x/1/θ+1/Θ−1/yϕ′xq+F122q,1;3;1−1/θ+1/Θ−1/x/1/θ+1/Θ−1/yϕ′yq21/q.Proof.
LetAζ=1/θ+1/Θ−ζ/x+1−ζ/y=ζ1/θ+1/Θ−1/x+1−ζ1/θ+1/Θ−1/y. From Lemma 13 and Lemma 9, using the Hölder inequality and ϕ′q∈HKθ,Θ, we find
(54)Iϕℏ;α,x,y≤y−x2xy∫011−ζα−ζαAζ2ϕ′1Aζdζ≤y−x2xy∫011−ζα−ζαpdζ1/p∫011Aζ2qϕ′1Aζqdζ1/q≤y−x2xy∫011−2ζαpdζ1/p∫011Aζ2qϕ′θq+ϕ′Θq−ζϕ′xq+1−ζϕ′yqdζ1/q≤y−x2xyZ61/pZ7ϕ′θq+ϕ′Θq−Z8ϕ′xq+Z9ϕ′yq1/q.
CalculatingZ6, Z7, Z8, and Z9, we have
(55)Z6=∫011−2ζαpdζ=1αp+1,(56)Z7=∫011Aζ2qdζ=1θ+1Θ−1y−2q∫011−ζ1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2qdζ=1θ+1Θ−1y−2qF122q,1;2;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y,(57)Z8=∫01ζAζ2qdζ=1θ+1Θ−1y−2q∫01ζ1−ζ1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2qdζ=1/θ+1/Θ−1/y−2q2F122q,2;3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y,(58)Z9=∫011−ζAζ2qdζ=1θ+1Θ−1y−2q∫011−ζ1−ζ1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y−2qdζ=1/θ+1/Θ−1/y−2q2F122q,1;3;1−1/θ+1/Θ−1/x1/θ+1/Θ−1/y.
Using (55), (56), (57), and (58) in (54), we get the inequality of (53). This completes the proof.☐Remark 24.
If we takex=θ and y=Θ in Theorem 23, then it becomes Theorem 9 proved by Íşcan and Whu in [27].Remark 25.
If we takeα=1, x=θ, and y=Θ in Theorem 23, then it becomes Theorem 2.7 proved by Íşcan in [12].


## 4. Application to Special Means

For positive numbersθ>0 and Θ>0 with θ≠Θ.(1)
The arithmetic mean(59)Aθ,Θ=θ+Θ2.(2)
The geometric mean(60)Gθ,Θ=θΘ.(3)
The harmonic mean(61)Hθ,Θ=2θΘθ+Θ.(4)
Thep-logarithmic mean
(62)Lpθ,Θ=Θp+1−θp+1p+1Θ−θ1/p,wherep∈ℝ\−1,0.Proposition 26.
Let0<θ<Θ. Then, the following inequalities holds
(63)12Aθ−1,Θ−1−H−1x,yr+2≤2Aθr+2,Θr+2−G2x,yLrrx,y≤2Aθr+2,Θr+2−Hr+2x,y,(64)12Aθ−1,Θ−1−H−1x,yr+2≤L−r+2−r+22Aθ−1,Θ−1−x−1,2Aθ−1,Θ−1−y−1≤2Aθr+2,Θr+2−Axr+2,yr+2.Proof.
Letα=1 and ϕx=xr+2 where x>0, r∈−1,∞\0 in Theorem 11 leads to the desired inequality (63) and (64) immediately, respectively.☐


## 5. Conclusion

In this paper, we present the (H-H-M) inequalities involving (RL) fractional integrals for the class of harmonic convex function (instead of convex function) and established some integral inequalities connected with the right and left sides of fractional (H-H-M) type inequalities for differentiable mappings whose derivatives in absolute value are harmonically convex. Some applications to special means have also been presented. Our obtained results are an extension of previously known results. An interesting topic is whether we can use the techniques in this paper to establish the (H-H-M) inequalities for other kinds of convex functions via (RL) fractional integrals. Our ideas and approach may stimulate further research for the researchers working in this field.



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