# New Variant of Hermite–Jensen–Mercer Inequalities via Riemann–Liouville Fractional Integral Operators

Authors: Qiong Kang; Saad Ihsan Butt; Waqas Nazeer; Mehroz Nadeem; Jamshed Nasir; Hong Yang
Journal: Journal of Mathematics
 (2020)

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## Abstract

In this paper, certain Hermite–Hadamard–Mercer-type inequalities are proved via Riemann–-Liouville fractional integral operators. We established several new variants of Hermite–Hadamard’s inequalities for Riemann–Liouville fractional integral operators by utilizing Jensen–Mercer inequality for differentiable mappingϒ whose derivatives in the absolute values are convex. Moreover, we construct new lemmas for differentiable functions ϒ′, ϒ″, and ϒ‴ and formulate related inequalities for these differentiable functions using variants of Hölder’s inequality.


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## Body

## 1. Introduction

The study of convex functions started over the period 1905 to 1906 by thought provoking ideas and fascinating work of Jensen. It is used as a major tool to solve optimization problems in analysis. However, inequalities involving convex functions are quite stimulating in the expansion of numerous sections of mathematics, for example, mathematical finance, economics, management sciences, and optimization theory. That is why the study of such inequalities have been given great importance in the literature [1–4].Let0<μ1≤μ2≤⋯≤μn and let ξ=ξ1,ξ2,…,ξn be nonnegative weights such that ∑i=1nξi=1. The famous Jensen inequality (see [5]) in the literature states that if Υ is a convex function on the interval θ1,θ2, then(1)ϒ∑i=1nξiμi≤∑i=1nξiϒμi,for all μi∈θ1,θ2 and all ξi∈0,1,i=1,2,…,n.Mercer gave a variant of Jensen’s inequality [6] as follows.Theorem 1.
IfΥ is a convex function on θ1,θ2, then(2)ϒθ1+θ2−∑i=1nξiμi≤ϒθ1+ϒθ2−∑i=1nξiϒμi,∀μi∈θ1,θ2 and ξi∈0,1,i=1,2,…,n.Matkovic and Pecaric worked on Jensen’s inequality of Mercer’s type for operators with applications [7]. Later, Mercer’s result was generalized to higher dimensions by Niezgoda [8]. In recent years, notable contributions have been made on Jensen–Mercer’s type inequality. Kian gave concept of Jensen inequality for superquadratic functions [9]. Furthermore, Anjidani and Changalvaiy worked on reverse Jensen–Mercer-type operator inequalities and Jensen–Mercer operator inequalities for superquadratic functions (see [10, 11]). Ali and Khan generalized integral Mercer’s inequality and integral means in [12].Another important inequality that characterize convex function is Hermite–Hadamard inequality, that is, if a mappingϒ:J⊆ℝ⟶ℝ is a convex function on J and θ1,θ2∈J, θ1<θ2, then(3)ϒθ1+θ22≤1θ2−θ1∫θ1θ2ϒλdλ≤ϒθ1+ϒθ22.Fractional calculus is the study of fractional order derivatives and integrals. Fractional calculus has found applications in diverse fields ranging from physical sciences and engineering to biological sciences and economics (see [13–15]). Now, we give necessary definition of fractional calculus theory, which is used throughout this paper.Definition 1.
Letϒ∈θ1,θ2. The Riemann–Liouville integrals Jθ1+αϒ and Jθ2−αϒ of order α>0 with θ1≥0 are defined by(4)Jθ1+αϒμ=1Γα∫θ1μμ−λα−1ϒλdλ,μ>θ1,Jθ2−αϒμ=1Γα∫μθ2λ−μα−1ϒλdλ,μ<θ2,Respectively, where Γ⋅ is Euler Gamma function and Jθ1+0ϒμ=Jθ2−0ϒμ=ϒμ.
In the case ofα=1, the fractional integral reduces to classical integral. Several researchers have focused on new integral inequalities involving Riemann–Liouville fractional integrals in recent years (see [16–18]) and references therein.
In this article, by using the Jensen–Mercer inequality, we proved Hermite–Hadamard’s inequalities for fractional integrals and established some new Riemann–Liouville fractional integrals connected with the left and right sides of Hermite–Hadamard-type inequalities for differentiable mappings whose derivatives in the absolute value are convex. Moreover, there will be further equalities forϒ′, ϒ″, and ϒ‴ and related inequalities for these differentiable functions using Hölder’s inequality.


## 2. Variant of Hermite–Hadamard–Mercer-Type Inequalities

By using Jensen–Mercer inequalities, Hermite–Hadamard-type inequalities can be expressed in Riemann–Liouville fractional integral form as follows:Theorem 2.
Letϒ:θ1,θ2⟶ℝ be a convex function. Then, the following fractional integral inequalities hold:(5)ϒθ1+θ2−z1+z22≤2α−1Γα+1z2−z1αJθ1+θ2−z1α−ϒθ1+θ2−z1+z22+Jθ1+θ2−z2α+ϒθ1+θ2−z1+z22≤ϒθ1+ϒθ2−ϒz1+ϒz22,(6)ϒθ1+θ2−z1+z22≤ϒθ1+ϒθ2−Γα+12z2−z1αJz1+αϒz1+z22+Jz2−αϒz1+z22≤ϒθ1+ϒθ2−ϒz1+z22,where ∀z1,z2∈θ1,θ2,α>0 and Γ⋅ is the Gamma function.Proof.
Using convexity ofϒ, we have(7)ϒθ1+θ2−u+v2=ϒθ1+θ2−u+θ1+θ2−v2≤12ϒθ1+θ2−u+ϒθ1+θ2−v,∀u,v∈θ1,θ2.
By change of variablesu=1+λ/2z1+1−λ/2z2 and v=1−λ/2z1+1+λ/2z2,λ∈0,1, we obtain(8)2ϒθ1+θ2−z1+z22≤ϒθ1+θ2−1+λ2z1+1−λ2z2+ϒθ1+θ2−1−λ2z1+1+λ2z2.
Multiply both sides byλα−1 and then integrate the resulting inequality with respect to λ over 0,1. Let ν=θ1+θ2−1+λ/2z1+1−λ/2z2 and w=θ1+θ2−1−λ/2z1+1+λ/2z2, and we have(9)1αϒθ1+θ2−z1+z22∫01λα−1dλ≤∫01λα−1ϒθ1+θ2−1+λ2z1+1−λ2z2+ϒθ1+θ2−1−λ2z1+1+λ2z2dλ,(10)ϒθ1+θ2−z1+z22≤2α−1Γα+1z2−z1α×Jθ1+θ2−z2α+ϒθ1+θ2−z1+z22+Jθ1+θ2−z1α−ϒθ1+θ2−z1+z22,so the first inequality of (5) is proved.
Now, for the proof of second inequality of (5), we first note that if ϒ is convex function, then for λ∈0,1, it gives(11)ϒθ1+θ2−1+λ2z1+1−λ2z2≤ϒθ1+ϒθ2−1+λ2ϒz1+1−λ2ϒz2,(12)ϒθ1+θ2−1−λ2z1+1+λ2z2≤ϒθ1+ϒθ2−1−λ2ϒz1+1+λ2ϒz2.
By adding the inequalities of (11) and (12), we have(13)ϒθ1+θ2−1+λ2z1+1−λ2z2+ϒθ1+θ2−1−λ2z1+1+λ2z2≤2ϒθ1+ϒθ2−ϒz1+ϒz2.
Multiplying both sides byλα−1 and then integrating the resulting inequality with respect to λ over 0,1, we have(14)∫01λα−1ϒθ1+θ2−1+λ2z1+1−λ2z2+ϒθ1+θ2−1−λ2z1+1+λ2z2dλ≤2ϒθ1+ϒθ2−ϒz1+ϒz2∫01λα−1dλ.
Simplifying the integrals gives(15)2α−1Γα+1z2−z1αJθ1+θ2−z2+αϒθ1+θ2−z1+z22+Jθ1+θ2−z1−αϒθ1+θ2−z1+z22≤ϒθ1+ϒθ2−ϒz1+ϒz22.
Combining (10) and (15), we will get (5). Now, in order to prove (6), we employ Jensen–Mercer inequality as(16)ϒθ1+θ2−u+v2≤ϒθ1+ϒθ2−ϒu+ϒv2,where ∀u,v∈θ1,θ2.
Now, by change of variablesu=1+λ/2z1+1−λ/2z2 and v=1−λ/2z1+1+λ/2z2, ∀z1,z2∈θ1,θ2 and λ∈0,1, in (16), we have(17)ϒθ1+θ2−z1+z22≤ϒθ1+ϒθ2−ϒ1+λ/2z1+1−λ/2z2+ϒ1−λ/2z1+1+λ/2z22.
Multiplying both sides byλα−1 and then integrating the resulting inequality with respect to λ over 0,1, we have(18)1αϒθ1+θ2−z1+z22≤1αϒθ1+ϒθ2−∫01λα−1ϒ1+λ/2z1+1−λ/2z2+ϒ1−λ/2z1+1+λ/2z2dλ2,ϒθ1+θ2−z1+z22≤ϒθ1+ϒθ2−2α−1Γα+1z2−z1αJz1+αϒz1+z22+Jz2−αϒz1+z22,so the first inequality of (6) is proved.
Now, for the proof of second inequality of (6), we first note that if ϒ is convex function, then for λ∈0,1,(19)ϒz1+z22=ϒ1+λ/2z1+1−λ/2z2+1−λ/2z1+1+λ/2z22≤ϒ1+λ/2z1+1−λ/2z2+ϒ1−λ/2z1+1+λ/2z22.
Multiplying both sides byλα−1 and then integrating the resulting inequality with respect to λ over 0,1, we have(20)1αϒz1+z22≤12∫01λα−1ϒ1+λ2z1+1−λ2z2+ϒ1−λ2z1+1+λ2z2ϒz1+z22≤2α−1Γα+1z2−z1αJz1+αϒz1+z22+Jz2−αϒz1+z22.
Multiplying by−1,(21)−2α−1Γα+1z2−z1αJz1+αϒz1+z22+Jz2−αϒz1+z22≤−ϒz1+z22.
Addingϒθ1+ϒθ2 both sides in (21), we get the second inequality of (6).Remark 1.
Under the assumption of Theorem2 for inequality (5) with α=1, one has(22)ϒθ1+θ2−z1+z22≤1z2−z1∫z1z2ϒθ1+θ2−λdλ≤ϒθ1+ϒθ2−ϒz1+ϒz22.
Inequality (22) is proved by Kian and Moslehian in [19].


## 3. New Identities and Related Results

In this section, we prove new lemmas forϒ′, ϒ′′, and ϒ′′′, which plays a key part in proving our inequalities. Throughout the rest of the paper, we need the following assumption.(A1) Let z1,z2∈θ1,θ2, α>0, λ∈0,1, and Γ. is the Gamma function.Lemma 1.
LetΥ:θ1,θ2⟶ℝ be a differentiable mapping on θ1,θ2 with θ1<θ2 along with assumption A1. If Υ′∈Lθ1,θ2, then the following equality holds:(23)Υθ1+θ2−z1+Υθ1+θ2−z22−2αΓα+12z2−z1α×Jθ1+θ2−z2+αΥθ1+θ2−z1+z22+Jθ1+θ2−z1−αΥθ1+θ2−z1+z22=z2−z14∫01λαΥ′θ1+θ2−1+λ2z1+1−λ2z2−Υ′θ1+θ2−1−λ2z1+1+λ2z2dλ.Proof.
It suffices to note that(24)I=z2−z14I1−I2,where(25)I1=∫01λαϒ′θ1+θ2−1+λ2z1+1−λ2z2dλ=2ϒθ1+θ2−z1z2−z1−2α+1αz2−z1α+1∫θ1+θ2−z1+z2/2θ1+θ2−z1w−θ1+θ2−z1+z22α−1ϒwdw=2ϒθ1+θ2−z1z2−z1−2α+1Γα+1z2−z1α+1Jθ1+θ2−z1−αϒθ1+θ2−z1+z22,(26)I2=∫01λαϒ′θ1+θ2−1−λ2z1+1+λ2z2dλ=−2ϒθ1+θ2−z2z2−z1+2α+1αz2−z1α+1∫θ1+θ2−z2θ1+θ2−z1+z2/2θ1+θ2−z1+z22−wα−1ϒwdw=−2ϒθ1+θ2−z2z2−z1+2α+1Γα+1z2−z1α+1Jθ1+θ2−z2+αϒθ1+θ2−z1+z22.
Combining (25) and (26) with (24), we get (22).Corollary 1.
Forα=1 in Lemma 1, we will obtain(27)ϒθ1+θ2−z1+ϒθ1+θ2−z22−1z2−z1∫θ1+θ2−z2θ1+θ2−z1ϒudu=z2−z14∫01λϒ′θ1+θ2−1+λ2z1+1−λ2z2−ϒ′θ1+θ2−1−λ2z1+1+λ2z2dλ.Remark 2.
Takingα=1, z1=θ1, and z2=θ2 in Lemma 1, we get Lemma 2.1 in [20] and the following inequality holds:(28)ϒθ1+ϒθ22−1θ2−θ1∫θ1θ2ϒudu=θ2−θ14∫01λϒ′1−λ2θ1+1+λ2θ2−ϒ′1+λ2θ1+1−λ2θ2dλ.Theorem 3.
Suppose thatϒ:θ1,θ2⟶ℝ is a differentiable mapping on θ1,θ2 with 0≤θ1<θ2 and ϒ∈Lθ1,θ2 along with assumption A1. If ϒ′ is differentiable function on θ1,θ2, then the following inequality for Riemann–Liouville fractional integral holds:(29)ϒθ1+θ2−z1+ϒθ1+θ2−z22−2αΓα+12z2−z1αJθ1+θ2−z2+αϒθ1+θ2−z1+z22+Jθ1+θ2−z1−αϒθ1+θ2−z1+z22≤z2−z124α+2supξ∈θ1,θ2ϒ″ξ.Proof.
By using Lemma1 and applying mean value theorem for the function ϒ′, we have(30)ϒθ1+θ2−z1+ϒθ1+θ2−z22−2αΓα+12z2−z1αJθ1+θ2−z2+αϒθ1+θ2−z1+z22+Jθ1+θ2−z1−αϒθ1+θ2−z1+z22=z2−z122∫01λαλϒ″ξdλ,where ξ∈θ1,θ2. This leads us to(31)ϒθ1+θ2−z1+ϒθ1+θ2−z22−2αΓα+12z2−z1α×Jθ1+θ2−z2+αϒθ1+θ2−z1+z22+Jθ1+θ2−z1−αϒθ1+θ2−z1+z22≤z2−z124∫01λαλϒ″ξdλ≤z2−z124supξ∈θ1,θ2ϒ″ξ∫01λα+1dλ=z2−z124α+2supξ∈θ1,θ2ϒ″ξ.Theorem 4.
Suppose thatϒ:θ1,θ2⟶ℝ is a differentiable mapping on θ1,θ2 with θ1<θ2 and ϒ∈Lθ1,θ2 along with assumption A1. If ϒ′ is convex function on θ1,θ2, then the following inequality for Riemann–Liouville fractional integral holds:(32)ϒθ1+θ2−z1+ϒθ1+θ2−z22−2αΓα+12z2−z1α×Jθ1+θ2−z2+αϒθ1+θ2−z1+z22+Jθ1+θ2−z1−αϒθ1+θ2−z1+z22≤z2−z12α+1ϒ′θ1+ϒ′θ2−ϒ′z1+ϒ′z22.Proof.
By using Lemma1 and Jensen–Mercer inequality, we have(33)ϒθ1+θ2−z1+ϒθ1+θ2−z22−2αΓα+12z2−z1αJθ1+θ2−z2+αϒθ1+θ2−z1+z22+Jθ1+θ2−z1−αϒθ1+θ2−z1+z22≤z2−z14∫01λαϒ′θ1+θ2−1+λ2z1+1−λ2z2dλ+∫01λα−ϒ′θ1+θ2−1−λ2z1+1+λ2z2dλ,(34)≤z2−z14∫01λαϒ′θ1+ϒ′θ2−1+λ2Υ′z1+1−λ2ϒ′z2dλ+∫01λαΥ′θ1+ϒ′θ2−1−λ2ϒ′z1+1+λ2ϒ′z2dλ.
Solving the integrals gives the required result.Corollary 2.
Forα=1 in Theorem 4, we obtain(35)ϒθ1+θ2−z1+ϒθ1+θ2−z22−1z2−z1∫θ1+θ2−z2θ1+θ2−z1ϒudu≤z2−z14ϒ′θ1+ϒ′θ2−f′z1+f′z22.Remark 3.
Takingα=1, z1=θ1, and z2=θ2 in Theorem 4, we get (Theorem 2.2, [21]), and following inequality holds:(36)ϒθ1+ϒθ22−1θ2−θ1∫θ1θ2ϒudu≤θ2−θ14ϒ′θ1+ϒ′θ22.
Now we state our results for twice differentiable functionsϒ″:Lemma 2.
LetΥ:θ1,θ2⟶ℝ be a differentiable mapping on θ1,θ2 with θ1<θ2 along with assumption A1. If ϒ″∈Lθ1,θ2, then the following equation holds:(37)2α−1Γα+1z2−z1αJθ1+θ2−z1+z2/2+αϒθ1+θ2−z1+Jθ1+θ2−z1+z2/2−αϒθ1+θ2−z2−ϒθ1+θ2−z1+z22=z2−z128α+1∫011−λα+1ϒ″θ1+θ2−1+λ2z1+1−λ2z2dλ+∫011−λα+1ϒ″θ1+θ2−1−λ2z1+1+λ2z2dλ.Proof.
It suffices to note that(38)I=z2−z128α+1I1+I2,where(39)I1=∫011−λα+1ϒ″θ1+θ2−1+λ2z1+1−λ2z2dλ=−2z2−z1ϒ′θ1+θ2−z1+z22+2α+1z2−z1∫011−λαϒ′θ1+θ2−1+λ2z1+1−λ2z2dλ=−2z2−z1ϒ′θ1+θ2−z1+z22−4α+1z2−z12ϒθ1+θ2−z1+z22+4αα+1z2−z12∫011−λα−1ϒθ1+θ2−1+λ2z1+1−λ2z2dλ=−2z2−z1ϒ′θ1+θ2−z1+z22−4α+1z2−z12ϒθ1+θ2−z1+z22+2α+2Γα+2z2−z1α+2Jθ1+θ2−z1+z2/2−αϒθ1+θ2−z2,and(40)I2=∫011−λα+1ϒ″θ1+θ2−1−λ2z1+1+λ2z2dλ=2z2−z1ϒ′θ1+θ2−z1+z22−2α+1z2−z1∫011−λαϒ′θ1+θ2−1−λ2z1+1+λ2z2dλ=2z2−z1ϒ′θ1+θ1−z1+z22−4α+1z2−z12ϒθ1+θ2−z1+z22+4αα+1z2−z12∫011−λα−1ϒθ1+θ2−1−λ2z1+1+λ2z2dλ=2z2−z1ϒ′θ1+θ2−z1+z22−4α+1z2−z12ϒθ1+θ2−z1+z22+2α+2Γα+2z2−z1α+2Jθ1+θ2−z1+z2/2+αϒθ1+θ2−z1.
Combining (39) and (40) with (38), we get (33).Remark 4.
Takingz1=θ1 and z2=θ2 in Lemma 2 gives Lemma 1 of [22].Remark 5.
Forα=1, z1=θ1, and z2=θ2 in Lemma 2, it reduces to Lemma 2 proved in [22].Theorem 5.
Suppose thatϒ:θ1,θ2⟶ℝ is a differentiable mapping on θ1,θ2 with θ1<θ2 and ϒ∈Lθ1,θ2 along with assumption A1. If ϒ″ is convex function on θ1,θ2, then the following inequality for Riemann–Liouville fractional integral holds:(41)2α−1Γα+1z2−z1αJθ1+θ2−z1+z2/2+αϒθ1+θ2−z1+Jθ1+θ2−z1+z2/2−αϒθ1+θ2−z2−ϒθ1+θ2−z1+z22≤z2−z124α+1α+2ϒ″θ1+ϒ″θ2−ϒ″z1+ϒ″z22.Proof.
By using Lemma2 and Jensen–Mercer inequality, we have(42)2α−1Γα+1z2−z1αJθ1+θ2−z1+z2/2+αϒθ1+θ2−z1+Jθ1+θ2−z1+z2/2−αϒθ1+θ2−z2−ϒθ1+θ2−z1+z22≤z2−z128α+1∫011−λα+1ϒ″θ1+θ2−1+λ2z1+1−λ2z2dλ+∫011−λα+1ϒ″θ1+θ2−1−λ2z1+1+λ2z2dλ≤z2−z128α+1∫011−λα+1ϒ″θ1+ϒ″θ2−1+λ2Υ″z1+1−λ2Υ″z2dλ+∫011−λα+1ϒ″θ1+ϒ″θ2−1−λ2ϒ″z1+1+λ2ϒ″z2dλ.
By simplifying the integrals, we get the required result.Remark 6.
Forz1=θ1 and z2=θ2 in Theorem 5, we will get Theorem 5 proved in [22] given as follows:(43)2α−1Γα+1θ2−θ1αJθ1+θ2/2+αϒθ2+Jθ1+θ2/2−αϒθ1−ϒθ1+θ22≤θ2−θ128α+1α+2ϒ″θ1+ϒ″θ2.Theorem 6.
Suppose thatϒ:θ1,θ2⟶ℝ is a three times differentiable function on I along with assumption A1. If Υ″∈Lθ1,θ2 and Υ″q is convex function, where 1/r+1/q=1, q≥1, then(44)2α−1Γα+1z2−z1αJθ1+θ2−z1+z2/2+αϒθ1+θ2−z1+Jθ1+θ2−z1+z2/2−αϒθ1+θ2−z2−ϒθ1+θ2−z1+z22≤z2−z128α+11rα+1+11/rϒ″θ1q+ϒ″θ2q−34ϒ″z1q+12ϒ″z2q1/q+ϒ″θ1q+ϒ″θ2q−34ϒ″yq+12ϒ″xq1/q.Proof.
By using Lemma2 and well-known Hölder’s inequality with Jensen–Mercer inequality on the fact that ϒ″q is convex function, we have(45)2α−1Γα+1z2−z1αJθ1+θ2−z1+z2/2+αϒθ1+θ2−z1+Jθ1+θ2−z1+z2/2−αϒθ1+θ2−z2−ϒθ1+θ2−z1+z22≤z2−z128α+1∫011−λα+1ϒ″θ1+θ2−1+λ2z1+1−λ2z2dλ+∫011−λα+1ϒ″θ1+θ2−1−λ2z1+1+λ2z2dλ≤z2−z128α+1∫011−λα+1ϒ″θ1+Υ″θ2−1+λ2ϒ″z1+1−λ2ϒ″z2dλ+∫011−λα+1ϒ″θ1+ϒ″θ2−1−λ2ϒ″z1+1+λ2ϒ″z2dλ≤z2−z128α+1∫011−λrα+1dλ1/r×∫01ϒ″θ1+ϒ″θ2−1+λ2ϒ″z1+1−λ2ϒ′′z2qdλ1/q+∫011−λrα+1dλ1/r×∫01ϒ″θ1+ϒ″θ2−1−λ2ϒ″z1+1+λ2ϒ″z2qdλ1/q≤z2−z128α+11rα+1+11/rϒ″θ1q+ϒ″θ2q−34ϒ″z1q+12ϒ″z2q1/q+1rα+1+11/rϒ″θ1q+ϒ″θ2q−34ϒ″z2q+12ϒ″z1q1/q=z2−z128α+11rα+1+11/r×ϒ″θ1q+ϒ″θ2q−34ϒ″z1q+12ϒ″z2q1/q+ϒ″θ1q+Υ″θ2q−34ϒ″z2q+12ϒ″z1q1/q.
Finally, we state our results for three times differentiable functionsϒ‴:Lemma 3.
Letϒ:θ1,θ2⟶ℝ be a differentiable mapping on θ1,θ2 with θ1<θ2 along with assumption A1. If ϒ‴∈Lθ1,θ2, then the following equation holds:(46)2α−1Γα+1z2−z1αJθ1+θ2−z1+z2/2+αϒθ1+θ2−z1+Jθ1+θ2−z1+z2/2−αϒθ1+θ2−z2−z2−z134α+1α+2ϒ″θ1+θ2−z1+z22−ϒθ1+θ2−z1+z22=z2−z1316α+1α+2∫011−λα+2ϒ‴θ1+θ2−1+λ2z1+1−λ2z2dλ−∫011−λα+2ϒ‴θ1+θ2−1−λ2z1+1+λ2z2dλ.Proof.
Proof. It suffices to note that(47)I=z2−z2316α+1α+2I1−I2,where(48)I1=∫011−λα+2ϒ‴θ1+θ2−1+λ2z1+1−λ2z2dλ=−2z2−z1ϒ″θ1+θ2−z1+z22+2α+2z2−z1∫011−λα+1ϒ″θ1+θ2−1+λ2z1+1−λ2z2dλ=−2z2−z1ϒ″θ1+θ2−z1+z22−4α+2z2−z12ϒ′θ1+θ2−z1+z22+4α+1α+2z2−z12∫011−λαϒ′θ1+θ2−1+λ2z1+1−λ2z2dλ=−2z2−z1ϒ″θ1+θ2−z1+z22−4α+2z2−z12ϒ′θ1+θ2−z1+z22−8α+2α+1z2−z13ϒθ1+θ2−z1+z22+2α+3α+1α+2Γα+1z2−z1α+3Jθ1+θ2−z1+z2/2+αϒθ1+θ2−z1=−2z2−z1ϒ″θ1+θ2−z1+z22−4α+2z2−z12ϒ′θ1+θ2−z1+z22−8α+2α+1z2−z13ϒθ1+θ2−z1+z22+2α+3Γα+3z2−z1α+3Jθ1+θ2−z1+z2/2+αϒθ1+θ2−z1,I2=∫011−λα+2ϒ‴θ1+θ2−1−λ2z1+1+λ2z2dλ=2z2−z1ϒ″θ1+θ2−z1+z22−2α+2z2−z1∫011−λα+1ϒ″θ1+θ2−1−λ2z1+1+λ2z2dλ=2z2−z1ϒ″θ1+θ2−z1+z22−4α+2z2−z12ϒ′θ1+θ2−z1+z22+4α+1α+2z2−z12∫011−λαϒ′θ1+θ2−1−λ2z1+1+λ2z2dλ=2z2−z1ϒ″θ1+θ2−z1+z22−4α+2z2−z12ϒ′θ1+θ2−z1+z22+8α+2α+1z2−z13ϒθ1+θ2−z1+z22−2α+3α+1α+2Γα+1z2−z1α+3Jθ1+θ2−z1+z2/2−αϒθ1+θ2−z2=2z2−z1ϒ″θ1+θ2−z1+z22−4α+2z2−z12ϒ′θ1+θ2−z1+z22+8α+2α+1z2−z13ϒθ1+θ2−z1+z22−2α+3Γα+3z2−z1α+3Jθ1+θ2−z1+z2/2−αϒθ1+θ2−z2.
Replacing the values of the integralsI1 and I2 in (47), we get (46).Remark 7.
Forz1=θ1 and z2=θ2 in Lemma 3, it reduces to Lemma 1 proved in [23].Remark 8.
Forα=1, z1=θ1, and z2=θ2 in Lemma 3, it reduces to Lemma 2.1 proved in [24].Theorem 7.
Suppose thatϒ:θ1,θ2⟶ℝ is a differentiable mapping on θ1,θ2 with θ1<θ2 and ϒ∈Lθ1,θ2 along with assumption A1. If ϒ‴ is convex function on θ1,θ2, then the following inequality for Riemann–Liouville fractional integral holds:(49)2α−1Γα+1z2−z1αJθ1+θ2−z1+z2/2+αϒθ1+θ2−z1+Jθ1+θ2−z1+z2/2−αϒθ1+θ2−z2−z2−z134α+1α+2ϒ″θ1+θ2−z1+z22−ϒθ1+θ2−z1+z22≤z2−z138α+1α+2α+3ϒ‴θ1+ϒ‴θ2−ϒ‴z1+ϒ‴z22.Proof.
By using Lemma3 and Jensen–Mercer inequality, we have(50)2α−1Γα+1z2−z1αJθ1+θ2−z1+z2/2+αϒθ1+θ2−z1+Jθ1+θ2−z1+z2/2−αϒθ1+θ2−z2−z2−z134α+1α+2ϒ″θ1+θ2−z1+z22−ϒθ1+θ2−z1+z22≤z2−z1316α+1α+2∫011−λα+2ϒ‴θ1+θ2−1+λ2z1+1−λ2z2dλ+∫011−λα+2ϒ‴θ1+θ2−1−λ2z1+1+λ2z2dλ≤z2−z1316α+1α+2∫011−λα+2ϒ‴θ1+ϒ‴θ2−1+λ2ϒ‴z1+1−λ2ϒ‴z2dλ+∫011−λα+2ϒ‴θ1+ϒ‴θ2−1−λ2ϒ‴z1+1+λ2ϒ‴z2dλ≤z2−z138α+1α+2α+3ϒ‴θ1+ϒ‴θ2−ϒ‴z1+ϒ‴z22.Remark 9.
Forz1=θ1 and z2=θ2 in Theorem 7, we will get Theorem 4 proved in [23].Theorem 8.
Suppose thatϒ:θ1,θ2⟶ℝ is a three times differentiable function on I along with assumption A1. If ϒ‴∈Lθ1,θ2 and ϒ‴q is a convex function, where 1/r+1/q=1, q≥1, then(51)2α−1Γα+1z2−z1αJθ1+θ2−z1+z2/2+αϒθ1+θ2−z1+Jθ1+θ2−z1+z2/2−αϒθ1+θ2−z2−z2−z134α+1α+2ϒ″θ1+θ2−z1+z22−ϒθ1+θ2−z1+z22≤z2−z1316α+1α+21rα+2+11/r×ϒ‴θ1q+ϒ‴θ2q−34ϒ‴z1q+12ϒ‴z2q1/q+ϒ‴θ1q+ϒ‴θ2q−34ϒ‴z2q+12ϒ‴z1q1/q.Proof.
By using Lemma3 with Jensen–Mercer inequality and well-known Hölder’s inequality on the fact that ϒ‴q is a convex function, we have(52)2α−1Γα+1z2−z1αJθ1+θ2−z1+z2/2+αϒθ1+θ2−z1+Jθ1+θ2−z1+z2/2−αϒθ1+θ2−z2−z2−z134α+1α+2ϒ′′θ1+θ2−z1+z22−ϒθ1+θ2−z1+z22≤z2−z1316α+1α+2∫011−λα+2ϒ‴θ1+θ2−1+λ2z1+1−λ2z2dλ+∫011−λα+2ϒ‴θ1+θ2−1−λ2z1+1+λ2z2dλ≤z2−z1316α+1α+2∫011−λα+2ϒ‴θ1+ϒ‴θ2−1+λ2ϒ‴z1+1−λ2ϒ‴z2dλ+∫011−λα+2ϒ‴θ1+ϒ‴θ2−1−λ2ϒ‴z1+1+λ2ϒ‴z2dλ≤z2−z1316α+1α+2∫011−λrα+2dλ1/r×∫01ϒ‴θ1+ϒ‴θ2−1+λ2ϒ‴z1+1−λ2ϒ‴z2qdλ1/q+z2−z1316α+1α+2∫011−λrα+2dλ1/r×∫01ϒ‴θ1+ϒ‴θ2−1−λ2ϒ‴z1+1+λ2ϒ‴z2qdt1/q≤z2−z1316α+1α+21rα+2+11/r×ϒ‴θ1q+ϒ‴θ2q−34ϒ‴z1q+12ϒ‴z2q1/q+z2−z1316α+1α+21rα+2+11/rϒ‴θ1q+ϒ‴θ2q−34ϒ‴z2q+12ϒ‴z1q1/q.
After some further simplification of integrals, we get the required results.


## 4. Conclusion

In this paper, we have obtained some new Hermite–Jensen–Mercer-type integral inequalities for Riemann–Liouville Fractional integral operators. The results presented here would provide elegant connections and generalizations of well-known results established recently. These results can be applied in convex analysis, optimization, and different areas of pure and applied sciences. The authors hope that these results will serve as a motivation for future work in this fascinating area. This research is open for further work by investigating such results for other classes of convex functions. However, it is not easy to give Jensen–Mercer inequality for other classes of convex functions.



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