# Fractional Integral Inequalities via Atangana-Baleanu Operators for Convex and Concave Functions

Authors: Ahmet Ocak Akdemir; Ali Karaoğlan; Maria Alessandra Ragusa; Erhan Set
Journal: Journal of Function Spaces
 (2021)

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## Abstract

Recently, many fractional integral operators were introduced by different mathematicians. One of these fractional operators, Atangana-Baleanu fractional integral operator, was defined by Atangana and Baleanu (Atangana and Baleanu, 2016). In this study, firstly, a new identity by using Atangana-Baleanu fractional integral operators is proved. Then, new fractional integral inequalities have been obtained for convex and concave functions with the help of this identity and some certain integral inequalities.


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## Body

## 1. Introduction

Mathematics is a tool that serves pure and applied sciences with its deep-rooted history as old as human history and sheds light on how to express and then solve problems. Mathematics uses various concepts and their relations with each other while performing this task. By defining spaces and algebraic structures built on spaces, mathematics creates structures that contribute to human life and nature. The concept of function is one of the basic structures of mathematics, and many researchers have focused on new function classes and made efforts to classify the space of functions. One of the types of functions defined as a product of this intense effort is the convex function, which has applications in statistics, inequality theory, convex programming, and numerical analysis. This interesting class of functions is defined as follows.Definition 1.
The mappingf:θ1,θ2⊆ℝ⟶ℝ is said to be convex if
(1)fλx+1−λy≤λfx+1−λfy,is valid for all x,y∈θ1,θ2 and λ∈0,1.Many inequalities have been obtained by using this unique function type and varieties in inequality theory, which is one of the most used areas of convex functions. We will continue by introducing the Hermite-Hadamard inequality that generate limits on the mean value of a convex function and the famous Bullen inequality as follows.Assume thatf:I⊆ℝ⟶ℝ is a convex mapping defined on the interval I of ℝ, where θ1<b. The following statement:
(2)fθ1+θ22≤1κ2−θ1∫θ1θ2fxdx≤fθ1+fθ22,holds and known as Hermite-Hadamard inequality. Both inequalities hold in the reversed direction if f is concave.Bullen’s integral inequality can be presented as(3)1θ2−θ1∫θ1θ2fxdx≤12fθ1+θ22+fθ1+fθ22,where f:I⊂ℝ⟶ℝ is a convex mapping on the interval I of ℝ where κ1,θ2∈I with θ1<θ2.To provide detail information on convexity, let us consider some earlier studies that have been performed by many researchers. In [1], Jensen introduced the concept of convex function to the literature for the first time and drew attention to the fact that it seems to be the basis of the concept of incremental function. In [2], Beckenbach has mentioned about the concept of convexity and emphasized several features of this useful function class. In [3], the authors have focused the relations between convexity and Hermite-Hadamard’s inequality. This study has led many researchers to the link between convexity and integral inequalities, which has guided studies in this field. Based on these studies, many papers have been produced for different kinds of convex functions. In [4], Akdemir et al. have proved several new integral inequalities for geometric-arithmetic convex functions via a new integral identity. Several new Hadamard’s type integral inequalities have been established with applications to special means by Kavurmaci et al. in [5]. Therefore, a similar argument has been carried out by Zhang et al. but now for s−geometrically convex functions in [6]. On all of these, Xi et al. have extended the challenge to m− and α,m−convex functions by providing Hadamard type inequalities in [7].Although fractional analysis has been known since ancient times, it has recently become a more popular subject in mathematical analysis and applied mathematics. The adventure that started with the question of whether the solution will exist if the order is fractional in a differential equation has developed with many derivative and integral operators. By defining the derivative and integral operators in fractional order, the researchers who aimed to propose more effective solutions to the solution of physical phenomena have turned to new operators with general and strong kernels over time. This orientation has provided mathematics and applied sciences several operators with kernel structures that differ in terms of locality and singularity, as well as generalized operators with memory effect properties. The struggle that started with the question of how the order in the differential equation being a fraction would have consequences has now evolved into the problem of how to explain physical phenomena and find the most effective fractional operators that will provide effective solutions to real-world problems. Let us introduce some fractional derivative and integral operators that have broken ground in fractional analysis and have proven their effectiveness in different fields by using by many researchers.We will remember the Caputo-Fabrizio derivative operators. Also, we would like to note that the functions belong to Hilbert spaces denoted byH10,θ2.Definition 2.
[8]. Let f∈H10,θ2, θ2>θ1, α∈0,1, then the definition of the new Caputo fractional derivative is
(4)DαCFfτ1=Mα1−α∫κ1τ1f′sexp−α1−ατ1−sds,where Mα is normalization function.Depending on this interesting fractional derivative operator, the authors have defined the Caputo-Fabrizio fractional integral operator as follows.Definition 3.
[9] Let f∈H10,θ2, θ2>θ1, α∈0,1, then the definition of the left and right side of Caputo-Fabrizio fractional integral is
(5):::θ1CFIατ1=1−αBαfτ1+αBα∫θ1τ1fydy,
and(6)CFIθ2ατ1=1−αBαfτ1+αBα∫τ1κ2fydy,where Bα is the normalization function.The Caputo-Fabrizio fractional derivative, which is used in dynamical systems, physical phenomena, disease models, and many other fields, is a highly functional operator by definition, but has a deficiency in terms of not meeting the initial conditions in the special caseα=1. The improvement to eliminate this deficiency has been provided by the new derivative operator developed by Atangana-Baleanu, which has versions in the sense of Caputo and Riemann. In the sequel of this paper, we will denote the normalization function with Bα with the same properties with the Mα which is defined in Caputo-Fabrizio definition.Definition 4.
[10] Let f∈H1θ1,θ2, θ2>κ1, α∈0,1, then the definition of the new fractional derivative is given:
(7)::::::θ1ABCDτ1αfτ1=Bα1−α∫aτ1f′xEα−ατ1−xα1−αdx.Definition 5.
[10] Let f∈H1θ1,θ2, θ2>κ1, α∈0,1, then the definition of the new fractional derivative is given:
(8):::::::θ1ABRDτ1αfτ1=Bα1−αddτ1∫θ1τ1fxEα−ατ1−xα1−αdx.Equations (7) and (8) have a nonlocal kernel. Also, in Equation (8), when the function is constant, we get zero.The associated fractional integral operator has been defined by Atangana-Baleanu as follows.Definition 6.
[10] The fractional integral associate to the new fractional derivative with nonlocal kernel of a function f∈H1κ1,θ2 as defined:
(9)::::θ1ABIαfτ1=1−αBαfτ1+αBαΓα∫θ1τ1fyτ1−yα−1dy,where θ2>θ1 and α∈0,1.In [11], Abdeljawad and Baleanu introduced the right hand side of integral operator as follows: the right fractional new integral with ML kernel of order α∈0,1 is defined by
(10)IABθ2αfτ1=1−αBαfτ1+αBαΓα∫τ1θ2fyy−τ1α−1dy.In [9], Abdeljawad and Baleanu has presented some new results based on fractional-order derivatives and their discrete versions. Conformable integral operators have been defined by Abdeljawad in [12]. This useful operator has been used to prove some new integral inequalities in [13]. Another important fractional operator—Riemann-Liouville fractional integral operators—have been used to provide some new Simpson type integral inequalities in [14]. Ekinci and Ozdemir have proved several generalizations by using Riemann-Liouville fractional integral operators in [15], and the authors have established some similar results with this operator in [16]. In [17], Akdemir et al. have presented some new variants of celebrated Chebyshev inequality via generalized fractional integral operators. The argument has been proceed with the study of Rashid et al. (see [18]) that involves new investigations related to generalized k-fractional integral operators. In [19], Rashid et al. have presented some motivated findings that extend the argument to the Hilbert spaces. For more information related to different kinds of fractional operators, we recommend to consider [20]. The applications of fractional operators have been demonstrated by several researchers; we suggest to see the papers [21–23].The main motivation of this paper is to prove an integral identity that includes the Atangana-Baleanu integral operator and to provide some new Bullen type integral inequalities for differentiable convex and concave functions with the help of this integral identity. Some special cases are also considered.


## 2. Main Results

We will start with a new integral identity that will be used as proofs of our main findings.Lemma 7.
Letf:θ1,θ2⟶ℝ be differentiable function on θ1,θ2 with κ1<θ2. Then, we have the following identity for Atangana-Baleanu fractional integral operators:
(11)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1ABIθ1+θ2/2αfθ1+::::θ1ABIαfκ1+θ22+::θ1+θ2/2ABIαfθ2+ABIθ2αfθ1+θ22=∫011−τ1α−τ1αf′1+τ12θ1+1−τ12θ2dτ1+∫01τ1α−1−τ1αf′1+τ12θ2+1−τ12κ1dτ1,where α,τ1∈0,1, Γ. is the gamma function, and Bα is the normalization function.Proof.
By addingI1 and I2, we have
(12)I1+I2=∫011−τ1α−τ1αf′1+τ12θ1+1−τ12κ2dτ1+∫01τ1α−1−τ1αf′1+τ12θ2+1−τ12θ1dτ1.
By using integration, we have(13)I1=∫011−τ1α−τ1αf′1+τ12θ1+1−τ12κ2dτ1=1−τ1α−τ1αf1+τ1/2θ1+1−τ1/2θ2dτ1θ1−θ2/210−2ακ2−θ1∫011−τ1α−1+τ1α−1f1+τ12θ1+1−τ12κ2dτ1=−2θ1−θ2fθ1−2κ1−θ2fθ1+θ22−2ακ2−θ1∫011−τ1α−1f1+τ12θ1+1−τ12θ2dτ1−2αθ2−θ1∫01τ1α−1f1+τ12θ1+1−τ12θ2dτ1=2θ2−θ1fθ1+fκ1+θ22−2α+1ακ2−θ1α+1∫θ1θ1+κ2/2x−θ1α−1fxdx−2α+1αθ2−θ1α+1∫θ1θ1+θ2/2κ1+θ22−xα−1fxdx.
Multiplying both sides of (13) identity by κ2−θ1α+1/2α+1BαΓα, we have
(14)θ2−θ1α+12α+1BαΓαI1=θ2−θ1α2αBαΓαfθ1+fθ1+κ22−αBαΓα∫κ1θ1+θ2/2x−θ1α−1fxdx−αBαΓα∫θ1θ1+θ2/2θ1+θ22−xα−1fxdx.
Similarly, by using integration, we get(15)I2=∫01τ1α−1−τ1αf′1+τ12θ2+1−τ12κ1dτ1=τ1α−1−τ1αf1+τ1/2θ2+1−τ1/2θ1dτ1θ2−θ1/210−2ακ2−θ1∫01τ1α−1+1−τ1α−1f1+τ12θ2+1−τ12κ1dτ1=2θ2−θ1fθ2+fκ1+θ22−2α+1ακ2−θ1α+1∫θ1+θ2/2θ2x−θ1+θ22α−1fxdx−2α+1αθ2−θ1α+1∫θ1+θ2/2θ2θ2−xα−1fxdx.
Multiplying both sides of (13) identity by θ2−κ1α+1/2α+1BαΓα, we get
(16)θ2−θ1α+12α+1BαΓαI2=θ2−θ1α2αBαΓαfθ2+fθ1+κ22−αBαΓα∫θ1+θ2/2θ2x−θ1+κ22α−1fxdx−αBαΓα∫θ1+κ2/2θ2θ2−xα−1fxdx.
By adding identities (14) and (16), we obtain
(17)θ2−θ1α+12α+1BαΓαI1+I2=θ2−θ1α+1−α2αΓα2αBαΓαfθ1+fθ1+θ22−1−αBαfθ1−αBαΓα∫θ1θ1+θ2/2x−θ1α−1fxdx−1−αBαfθ1+θ22−αBαΓα∫θ1θ1+θ2/2θ1+θ22−xα−1fxdx+θ2−θ1α+1−α2αΓα2αBαΓαfθ2+fθ1+θ22−1−αBαfθ2−αBαΓα∫κ1+θ2/2θ2θ2−xα−1fxdx−1−αBαfθ1+θ22−αBαΓα∫κ1+θ2/2θ2x−θ1+θ22α−1fxdx.
Using the definition of Atangana-Baleanu fractional integral operators, we get(18)θ2−θ1α+12α+1BαΓα∫011−τ1α−τ1αf′1+τ12θ1+1−τ12θ2dτ1+∫01τ1α−1−τ1αf′1+τ12θ2+1−τ12θ1dτ1=θ2−θ1α+1−α2αΓα2αBαΓαfκ1+fθ2+2fθ1+θ22−ABIθ1+θ2/2αfκ1+::::θ1ABIαfθ1+θ22+::θ1+θ2/2ABIαfθ2+IABθ2αfθ1+θ22.
☐☐Theorem 8.
Letf:θ1,θ2⟶ℝ be differentiable function on θ1,θ2 with κ1<θ2 and f′∈L1θ1,θ2. If f′ is a convex function, we have the following inequality for Atangana-Baleanu fractional integral operators
(19)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+∷∷θ1ABIαfκ1+θ22+::θ1+θ2/2ABIαfθ2+IABθ2αfθ1+θ22≤2f′θ1+f′θ2α+1,where α∈0,1 and Bα is the normalization function.Proof.
By using Lemma7, we can write
(20)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+::::θ1ABIαfκ1+θ22+::θ1+θ2/2ABIαfθ2+IABθ2αfθ1+θ22≤∫011−τ1αf′1+τ12θ1+1−τ12θ2dτ1+∫01τ1αf′1+τ12θ1+1−τ12θ2dτ1+∫01τ1αf′1+τ12θ2+1−τ12θ1dτ1+∫011−τ1αf′1+τ12θ2+1−τ12θ1dτ1.
By using convexity off′, we get
(21)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+::::θ1ABIαfκ1+θ22+I::θ1+θ2/2ABαfθ2+IABθ2αfθ1+θ22≤∫011−τ1α1+τ12f′θ1+1−τ12f′θ2dτ1+∫01τ1α1+τ12f′θ1+1−τ12f′θ2dτ1+∫01τ1α1+τ12f′θ2+1−τ12f′κ1dτ1+∫011−τ1α1+τ12f′θ2+1−τ12f′θ1dτ1.
By computing the above integral, we obtain(22)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+I::::θ1ABαfκ1+θ22+I∷θ1+θ2/2ABαfθ2+ABIθ2αfθ1+θ22≤2f′θ1+f′θ2α+1,and the proof is completed.☐☐Corollary 9.
In Theorem8, if we choose α=1, we obtain
(23)fθ1+fθ2+2fθ1+κ2/2θ2−θ1−4θ2−κ12∫θ1θ2fxdx≤f′θ1+f′κ22.Theorem 10.
Letf:θ1,θ2⟶ℝ be differentiable function on θ1,θ2 with κ1<θ2 and f′∈L1θ1,θ2. If f′q is a convex function, then we have the following inequality for Atangana-Baleanu fractional integral operators:
(24)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+I::::θ1ABαfκ1+θ22+I::θ1+θ2/2ABαfθ2+IABθ2αfθ1+θ22≤2αp+11/p3f′θ1q+f′κ2q41/q+3f′θ2q+f′κ1q41/q,where p−1+q−1=1, α∈0,1, q>1, and Bα is the normalization function.Proof.
By using the identity that is given in Lemma7, we have
(25)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+I::::θ1ABαfκ1+θ22+I::θ1+θ2/2ABαfθ2+IABθ2αfθ1+θ22≤∫011−τ1αf′1+τ12θ1+1−τ12θ2dτ1+∫01τ1αf′1+τ12θ1+1−τ12θ2dτ1+∫01τ1αf′1+τ12θ2+1−τ12θ1dτ1+∫011−τ1αf′1+τ12θ2+1−τ12θ1dτ1.
By applying Hölder inequality, we have(26)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+I::::θ1ABαfκ1+θ22+I::θ1+θ2/2ABαfθ2+IABθ2αfθ1+θ22≤∫011−τ1αpdτ11/p∫01f′1+τ12κ1+1−τ12θ2qdτ11/q+∫01τ1αpdτ11/p∫01f′1+τ12θ1+1−τ12θ2qdτ11/q+∫01τ1αpdτ11/p∫01f′1+τ12θ2+1−τ12θ1qdτ11/q+∫011−τ1αpdτ11/p∫01f′1+τ12θ2+1−τ12θ1qdτ11/q.
By using convexity off′q, we obtain
(27)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+I::::θ1ABαfκ1+θ22+I::θ1+θ2/2ABαfθ2+IABθ2αfθ1+θ22≤∫011−τ1αpdτ11/p∫011+τ12f′κ1q+1−τ12f′κ2qdv1/q+∫01τ1αpdτ11/p∫011+τ12f′κ1q+1−τ12f′κ2qdτ11/q+∫01τ1αpdτ11/p∫011+τ12f′κ2q+1−τ12f′κ1qdτ11/q+∫011−τ1αpdτ11/p∫011+τ12f′κ2q+1−τ12f′κ1qdτ11/q.
By calculating the integrals that is in the above inequalities, we get desired result.☐☐Corollary 11.
In Theorem10, if we choose α=1, we obtain
(28)fθ1+fθ2+2fκ1+θ2/2θ2−θ1−4κ2−θ12∫θ1θ2fxdx≤1p+11/p3f′θ1q+f′κ2q41/q+3f′θ2q+f′θ1q41/q.Theorem 12.
Letf:θ1,θ2⟶ℝ be differentiable function on θ1,θ2 with κ1<θ2 and f′∈L1θ1,θ2. If f′q is a convex function, then we have the following inequality for Atangana-Baleanu fractional integral operators:
(29)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+I::::θ1ABαfκ1+θ22+I::θ1+θ2/2ABαfθ2+ABIθ2αfθ1+θ22≤1α+11−1/qα+32α+1α+2f′κ1q+12α+2f′κ2q1/q+2α+32α+1α+2f′θ1q+12α+1α+2f′θ2q1/q+2α+32α+1α+2f′θ2q+12α+1α+2f′θ1q1/q+α+32α+1α+2f′θ2q+12α+2f′θ1q1/q,where α∈0,1, q≥1, and Bα is the normalization function.Proof.
By Lemma7, we get
(30)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+I::::θ1ABαfκ1+θ22+I::θ1+θ2/2ABαfθ2+IABθ2αfθ1+θ22≤∫011−τ1αf′1+τ12θ1+1−τ12θ2dτ1+∫01tαf′1+τ12κ1+1−τ12θ2dτ1+∫01τ1αf′1+τ12θ2+1−τ12θ1dτ1+∫011−τ1αf′1+τ12θ2+1−τ12θ1dτ1.
By applying power mean inequality, we get(31)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+::::θ1ABIαfκ1+θ22+::θ1+θ2/2ABIαfθ2+IABθ2αfθ1+θ22≤∫011−τ1αdτ11−1/q∫011−τ1αf′1+τ12θ1+1−τ12θ2qdτ11/q+∫01τ1αdτ11−1/q∫01τ1αf′1+τ12θ1+1−τ12θ2qdτ11/q+∫01τ1αdτ11−1/q∫01τ1αf′1+τ12θ2+1−τ12θ1qdτ11/q+∫011−τ1αdτ11−1/q∫011−τ1αf′1+τ12θ2+1−τ12θ1qdτ11/q.
By using convexity off′q, we obtain
(32)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1ABIθ1+θ2/2αfθ1+∷∷θ1ABIαfκ1+θ22+∷θ1+θ2/2ABIαfθ2+ABIθ2αfθ1+θ22≤∫011−τ1αdτ11−1/q∫011−τ1α1+τ12f′θ1q+1−τ12f′θ2qdτ11/q+∫01τ1αdτ11−1/q∫01τ1α1+τ12f′θ1q+1−τ12f′θ2qdτ11/q+∫01τ1αdτ11−1/q∫01τ1α1+τ12f′θ2q+1−τ12f′θ1qdτ11/q+∫011−τ1αdτ11−1/q∫011−τ1α1+τ12f′θ2q+1−τ12f′θ1qdτ11/q.
By computing the above integrals, the proof is completed.☐☐Corollary 13.
In Theorem12, if we choose α=1, we obtain
(33)2fθ1+fθ2+2fκ1+θ2/2θ2−θ1−8θ2−θ12∫θ1θ2fxdx≤121−1/q2f′θ1q+f′θ2q61/q+5f′θ1q+f′θ2q121/q+5f′θ2q+f′θ1q121/q+2f′θ2q+f′θ1q61/q.Theorem 14.
Letf:θ1,θ2⟶ℝ be differentiable function on θ1,θ2 with κ1<θ2 and f′∈L1θ1,θ2. If f′q is a convex function, then we have the following inequality for Atangana-Baleanu fractional integral operators:
(34)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+I::::θ1ABαfκ1+θ22+I∷θ1+θ2/2ABαfθ2+ABIθ2αfθ1+θ22≤4pαp+1+2f′κ1q+f′θ2qq,where p−1+q−1=1, α∈0,1, q>1, and Bα is the normalization function.Proof.
By using identity that is given in Lemma7, we get
(35)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+I::::θ1ABαfκ1+θ22+I::θ1+θ2/2ABαfθ2+IABθ2αfθ1+θ22≤∫011−τ1αf′1+τ12θ1+1−τ12θ2dτ1+∫01τ1αf′1+τ12θ1+1−τ12θ2dτ1+∫01τ1αf′1+τ12θ2+1−τ12θ1dτ1+∫011−τ1αf′1+τ12θ2+1−τ12θ1dτ1.
By using the Young inequality asxy≤1/pxp+1/qyq(36)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+I::::θ1ABαfκ1+θ22+I::θ1+θ2/2ABαfθ2+ABIθ2αfθ1+θ22≤1p∫011−τ1αpdτ1+1q∫01f′1+τ12θ1+1−τ12θ2qdτ1+1p∫01τ1αpdτ1+1q∫01f′1+τ12θ1+1−τ12θ2qdτ1+1p∫01τ1αpdτ1+1q∫01f′1+τ12θ2+1−τ12θ1qdτ1+1p∫011−τ1αpdτ1+1q∫01f′1+τ12θ2+1−τ12θ1qdτ1.
By using convexity off′q and by a simple computation, we have the desired result.☐☐Corollary 15.
In Theorem14, if we choose α=1, we obtain
(37)fθ1+fθ2+2fθ1+κ2/2θ2−θ1−4θ2−κ12∫θ1θ2fxdx≤2p2+p+f′θ1q+f′θ2qq.Theorem 16.
Letf:θ1,θ2⟶ℝ be differentiable function on θ1,θ2 with κ1<θ2 and f′∈L1θ1,θ2. If f′ is a concave for q>1, then we have
(38)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+I::::θ1ABαfκ1+θ22+I∷θ1+θ2/2ABαfθ2+ABIθ2αfθ1+θ22≤1α+1f′θ1α+3+θ2α+12α+2+f′θ12α+3+θ22α+2+f′θ22α+3+θ12α+2+f′κ2α+3+θ1α+12α+2,where α∈0,1 and Bα is the normalization function.Proof.
From Lemma7 and the Jensen integral inequality, we have
(39)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+I::::θ1ABαfκ1+θ22+I::θ1+θ2/2ABαfθ2+IABθ2αfθ1+θ22≤∫011−τ1αdτ1f′∫011−τ1α1+τ1/2θ1+1−τ1/2θ2dτ1∫011−τ1αdτ1+∫01τ1αdτ1f′∫01τ1α1+τ1/2θ1+1−τ1/2θ2dτ1∫01τ1αdτ1+∫01τ1αdτ1f′∫01τ1α1+τ1/2θ2+1−τ1/2θ1dτ1∫01τ1αdτ1+∫011−τ1αdτ1f′∫011−τ1α1+τ1/2θ2+1−τ1/2θ1dτ1∫011−τ1αdτ1.
By computing the above integrals, we have(40)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1ABIθ1+θ2/2αfθ1+∷∷θ1ABIαfκ1+θ22+I::θ1+θ2/2ABαfθ2+IABθ2αfθ1+θ22≤1α+1f′θ1α+3+θ2α+12α+2+f′θ12α+3+θ22α+2+f′θ22α+3+θ12α+2+f′κ2α+3+θ1α+12α+2.
So, the proof is completed.☐☐Corollary 17.
In Theorem16, if we choose α=1, we obtain
(41)fθ1+fθ2+2fκ1+θ2/2θ2−θ1−4κ2−θ12∫θ1θ2fxdx≤14f′2θ1+θ23+f′5θ1+θ26+f′5θ2+θ16+f′2θ2+θ13.Theorem 18.
Letf:θ1,θ2⟶ℝ be differentiable function on θ1,θ2 with κ1<θ2 and f′∈L1θ1,θ2. If f′q is a concave function, we have
(42)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+I::::θ1ABαfκ1+θ22+I::θ1+θ2/2ABαfθ2+IABθ2αfθ1+θ22≤2αp+11/pf′3θ1+θ24+f′3θ2+θ14,where p−1+q−1=1, α∈0,1, and q>1.Proof.
By using the Lemma7 and Hölder integral inequality, we can write
(43)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+I::::θ1ABαfκ1+θ22+I::θ1+θ2/2ABαfθ2+IABθ2αfθ1+θ22≤∫011−τ1αpdτ11/p∫01f′1+τ12κ1+1−τ12θ2qdτ11/q+∫01τ1αpdτ11/p∫01f′1+τ12θ1+1−τ12θ2qdτ11/q+∫01τ1αpdτ11/p∫01f′1+τ12θ2+1−τ12θ1qdτ11/q+∫011−τ1αpdτ11/p∫01f′1+τ12θ2+1−τ12θ1qdτ11/q.
By using concavity off′q and Jensen integral inequality, we get
(44)∫01f′1+τ12θ1+1−τ12θ2qdτ1=∫01τ10f′1+τ12θ1+1−τ12θ2qdτ1≤∫01τ10dτ1f′∫01τ101+τ1/2θ1+1−τ1/2θ2dτ1∫01τ10dτ1q=f′3θ1+θ24q.
Similarly,(45)∫01f′1+τ12θ2+1−τ12θ1qdτ1≤f′3θ2+θ14q,so we obtain
(46)2θ2−θ1α+1−α2α+1Γαθ2−θ1α+1fκ1+fθ2+2fθ1+θ22−2α+1BαΓαθ2−κ1α+1IABθ1+θ2/2αfθ1+∷∷θ1ABIαfκ1+θ22+I∷θ1+θ2/2ABαfθ2+ABIθ2αfθ1+θ22≤2αp+11/pf′3θ1+θ24+f′3θ2+θ14.
☐☐Corollary 19.
In Theorem18, if we choose α=1, we obtain
(47)fθ1+fθ2+2fκ1+θ2/2θ2−θ1−4κ2−θ12∫θ1θ2fxdx≤1p+11/pf′3θ1+θ24+f′3θ2+θ14.


## 3. Conclusion

In this study, an integral identity including Atangana-Baleanu integral operators has been proved. Some integral inequalities are established by using Hölder inequality, power-mean inequality, Young inequality, and convex functions with the help of Lemma7 which has the potential to produce Bullen type inequalities. Some special cases of the results in this general form have been pointed out. Researchers can establish new equations such as the integral identity in the study and reach similar inequalities of these equality-based inequalities.



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*Source: 1055434-2021-09-30.xml*