# The Hermite–Hadamard–Jensen–Mercer Type Inequalities for Riemann–Liouville Fractional Integral

Authors: Hua Wang; Jamroz Khan; Muhammad Adil Khan; Sadia Khalid; Rewayat Khan
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we give Hermite–Hadamard type inequalities of the Jensen–Mercer type for Riemann–Liouville fractional integrals. We prove integral identities, and with the help of these identities and some other eminent inequalities, such as Jensen, Hölder, and power mean inequalities, we obtain bounds for the difference of the newly obtained inequalities.


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## Body

## 1. Introduction

The concept of convex functions plays a vital role in both pure and applied mathematics. Convex functions also have many applications in other branches of science such as finance, economics, and engineering.Definition 1.
(see [1]). A function ψ:m,M⟶ℝ is convex if(1)ψsx+1−sy≤sψx+1−sψy,for all x,y∈m,M and s∈0,1.
If the inequality in (1) is strict for x≠y, then ψ is said to be a strictly convex function, and if −ψ is convex, then ψ is said to be a concave function [2, 3].
Many important inequalities such as Jensen, Jensen–Mercer, Hermite–Hadamard, and support line inequalities hold for convex functions. The classical Jensen’s inequality is among the most prominent inequalities stated as follows [4, 5].
Ifψ:m,M⟶ℝ is convex, then(2)ψ∑i=1nwixi≤∑i=1nwiψxi,for all xi∈m,M and wi∈0,1i=1,2,…,n with ∑i=1nwi=1.
In [6], Mercer presented a type of Jensen’s inequality called Jensen–Mercer inequality.Theorem 1.
Ifψ:m,M⟶ℝ is convex, then(3)ψm+M−∑i=1nwixi≤ψm+ψM−∑i=1nwiψxi,for each xi∈m,M and wi∈0,1i=1,2,…,n with ∑i=1nwi=1.
For a convex function, there exist at least one line lies on or below the graph of the function.Definition 2.
(see [7]). A function ψ:I⟶ℝ has a support at x0∈I if(4)ψx0+cu−x0≤ψu,for all x0∈I and for each u∈m,M⊂I. Inequality (4) is said to be the support line inequality.
The following theorem connects the support line inequality with convex functions.Theorem 2.
(see [7]). The function ψ:m,M⟶ℝ is convex if and only if ψ has at least one line of support at each x0∈m,M.The Hermite–Hadamard inequality is one of the most investigated inequality in the theory of convex functions due to its geometrical significance and applications. Because of the importance of Hermite–Hadamard inequality, there is an ample amount of research work dedicated to the extensions, generalizations, refinements, and applications of the Hermite–Hadamard inequality. The Hermite–Hadamard inequality is given below [8].Letψ:I⟶ℝ be a convex function, where I is an interval and m,M∈I such that m<M. Then,(5)ψm+M2≤1M−m∫mMψxdx≤ψm+ψM2.Ifψ is concave, then (5) holds in the reversed direction. For more results associated with Hermite–Hadamard inequality, see [9–18].The Hermite–Hadamard inequality has been extended by means of fractional integral operators. Most popular of them is the Riemann–Liouville fractional operator given in the following definition [19–22].Definition 3.
(see [23, 24]). Let ψ be an integrable function defined on m,M. Then, the integrals Jm+αψx and JM−αψx defined by(6)Jm+αψx=1Γα∫mxx−sα−1ψsds,x>m,(7)JM−αψx=1Γα∫xMs−xα−1ψsds,x<M,are called the left and right Riemann–Liouville fractional integrals of order α>0 respectively. Here, Γ represents gamma function defined by Γα=∫0∞e−ssα−1ds.
In [25, 26], authors used the following lemmas to obtain trapezoidal and midpoint type inequalities.Lemma 1.
(see [25]). Let ψ:I∘⟶ℝ (where I∘ is the interior of I) be a differentiable function and m,M∈I∘ such that m<M. If ψ′∈Lm,M, then(8)ψm+ψM2−1M−m∫mMψudu=M−m2∫011−2sψ′sm+1−sMds.Lemma 2.
(see [26]). Let all the assumptions of Lemma 1 hold. Then,(9)1M−m∫mMψudu−ψm+M2=M−m∫01/2sψ′sm+1−sMds+∫1/21s−1ψ′sm+1−sMds.
In this article, we establish fractional Hermite–Hadamard–Jensen–Mercer type inequalities. We give identities involving fractional integrals, and from these identities, we derive trapezoidal and midpoint type inequalities.Throughout this article,α represents a positive real number.


## 2. Main Results

We begin this section with our first main result.Theorem 3.
Supposeψ:m,M⟶ℝ is a convex function and x,y∈m,M such that x<y. Then,(10)ψm+M−αx+yα+1≤ψm+ψM−Γα+1y−xαJx+αψy≤ψm+ψM−ψαx+yα+1,(11)ψm+M−αx+yα+1≤Γα+1y−xαJm+M−x−αψm+M−y≤αψm+M−x+ψm+M−yα+1≤ψm+ψM−αψx+ψyα+1.Proof.
Sinceψ is convex, it has support line at each point x0∈m,M, that is,(12)ψx0+cu−x0≤ψu,for each u∈m,M. Substituting x0=m+M−αx+y/α+1 and u=m+M−sx−1−sy, where s∈0,1, in inequality (12), we obtain(13)ψm+M−αx+yα+1+c−sx−1−sy+αx+yα+1≤ψm+M−sx−1−sy.
Multiplying (13) with αsα−1 and integrate with respect to s, we obtain(14)ψm+M−αx+yα+1+c−αx+yα+1+αx+yα+1≤α∫o1sα−1ψm+M−sx−1−syds⟹ψm+M−αx+yα+1≤α∫o1sα−1ψm+M−sx−1−syds.
Using Mercer’s inequality, we obtain(15)ψm+M−αx+yα+1≤α∫o1sα−1ψm+ψM−sψx+1−sψyds⟹ψm+M−αx+yα+1≤ψm+ψM−α∫o1sα−1sψx+1−sψyds.
Sinceψ is convex, we have −sψx+1−sψy≤−ψsx+y1−s and (15) becomes(16)ψm+M−αx+yα+1≤ψm+ψM−α∫o1sα−1ψsx+1−syds.
Substitutingsx+1−sy=w in (16), we obtain(17)ψm+M−αx+yα+1≤ψm+ψM−αy−xα∫xyy−wα−1ψwdw⟹ψm+M−αx+yα+1≤ψm+ψM−Γα+1y−xαJx+αψy.
Now, we prove the second inequality of (10). Put x0=αx+y/α+1 and u=sx+1−sy in (12), we obtain(18)ψαx+yα+1+csx+1−sy−αx+yα+1≤ψsx+1−sy.
Multiplying the above inequality withαsα−1 and integrating and using ∫01sα−1ds=1/α and ∫01sαds=1/α+1, we obtain(19)ψαx+yα+1≤α∫01sα−1ψsx+1−syds.
Putsx+1−sy=w, and we obtain(20)ψαx+yα+1≤αy−xα∫xyy−wα−1ψwdw⟹ψαx+yα+1≤Γα+1y−xαJx+αψy⟹−Γα+1y−xαJx+αψy≤−ψαx+yα+1.
Addingψm+ψM on both sides of (20), we obtain(21)ψm+ψM−Γα+1y−xαJx+αψy≤ψm+ψM−ψαx+yα+1,and on combining (17) and (21), we obtain (10).
Now, we prove the inequalities in (11). Let u=m+M−sx−1−sy⟹s=u−m−M+y/y−x and (14) become(22)ψm+M−αx+yα+1≤α∫m+M−ym+M−xu−m−M+yy−xα−1ψuduy−x=αΓαy−xα1Γα∫m+M−ym+M−xu−m+M−yα−1ψudu⟹ψm+M−αx+yα+1≤Γα+1y−xαJm+M−x−αψm+M−y.
Now, we prove the other two inequalities of (11). As ψ is a convex function, we have(23)ψm+M−sx−1−sy=ψsm+M−x+1−sm+M−y⟹ψm+M−sx−1−sy≤sψm+M−x+1−sψm+M−y≤ψm+ψM−sψx−1−sψy.
Multiplying withαsα−1 and integrating, we obtain(24)α∫01sα−1ψm+M−sx−1−syds≤αψm+M−x∫01sαds+αψm+M−y∫01sα−1−sαds≤ψm+ψM−αψx∫01sαds−αψy∫01sα−1−sαds⟹α∫01sα−1ψm+M−sx−1−syds≤αψm+M−x+ψm+M−yα+1≤ψm+ψM−αψx+ψyα+1.
By changing of variable, (24) becomes(25)α∫m+M−ym+M−xu−m−M+yy−xα−1ψuduy−x≤αψm+M−x+ψm+M−yα+1≤ψm+ψM−αψx+ψyα+1⟹αΓαy−xα1Γα∫m+M−ym+M−xu−m+M−yα−1ψudu≤αψm+M−x+ψm+M−yα+1≤ψm+ψM−αψx+ψyα+1⟹Γα+1y−xαJm+M−x−αψm+M−y≤αψm+M−x+ψm+M−yα+1≤ψm+ψM−αψx+ψyα+1,and on combining (22) and (25), we obtain (11).Remark 1.
If we putα=1 in Theorem 3 and in the obtained expressions substitute u  =  sx+1−sy and u=m+M−v, respectively, we obtain(26)ψm+M−x+y2≤ψm+ψM−∫01ψsx+1−syds≤ψm+ψM−ψx+y2,(27)ψm+M−x+y2≤1y−x∫xyψm+M−vdv≤ψm+M−x+ψm+M−y2≤ψm+ψM−ψx+ψy2,respectively. Inequalities (26) and (27) have been proved in [27].Remark 2.
Substitutingα=1, x=m, and y=M in (11), one can obtain Hermite–Hadamard inequality.


## 3. Bounds for the Difference of Hermite–Hadamard–Jensen–Mercer Type Inequalities

Throughout this section, we considerψ:m,M⟶ℝ is a differentiable function. To give the bounds for the difference of Hermite–Hadamard–Jensen–Mercer type inequalities, first, we present the following lemmas.Lemma 3.
Letx,y∈m,M such that x<y and let ψ′∈Lm,M. Then,(28)αψm+M−x+ψm+M−yα+1−Γα+1y−xαJm+M−x−αψm+M−y=y−xα+1∫01α+1sα−1ψ′m+M−sx−1−syds.Proof.
Using the techniques of integration, we have(29)y−xα+1∫01α+1sα−1ψ′m+M−sx−1−syds=y−x∫01sαψ′m+M−sx−1−syds−y−xα+1∫01ψ′m+M−sx−1−syds=sαψm+M−sx−1−sy|01−α∫01sα−1ψm+M−sx−1−syds−1α+1ψm+M−sx−1−sy|01,(30)=αψm+M−x+ψm+M−yα+1−α∫01sα−1ψm+M−sx−1−syds=αψm+M−x+ψm+M−yα+1−αy−xα∫m+M−ym+M−xu−m+M−yα−1ψudu=αψm+M−x+ψm+M−yα+1−Γα+1y−xαJm+M−x−αψm+M−y.Remark 3.
Substitutingα=1, x=m, and y=M in (28), we obtain (8).Lemma 4.
Let all the assumptions of Lemma3 hold. Then,(31)Γα+1y−xαJm+M−x−αψm+M−y−ψm+M−αx+yα+1=y−x∫α/α+111−sαψ′m+M−sx−1−syds−∫0α/α+1sαψ′m+M−sx−1−syds.Proof.
Using techniques of integration, we have(32)y−x∫α/α+111−sαψ′m+M−sx−1−syds−y−x∫0α/α+1sαψ′m+M−sx−1−syds=y−x∫α/α+11ψ′m+M−sx−1−syds−y−x∫01sαψ′m+M−sx−1−syds=ψm+M−sx−1−sy|α/α+11−sαψm+M−sx−1−sy|01+α∫01sα−1ψm+M−sx−1−syds=−ψm+M−αx+yα+1+α∫01sα−1ψm+M−sx−1−syds=Γα+1y−xαJm+M−x−αψm+M−y−ψm+M−αx+yα+1.Remark 4.
If we putα=1, x=m, and y=M in (31), we obtain (9).
We use Lemmas3 and 4 and obtain bounds for the difference of the inequalities in (11).Theorem 4.
Letψ′ be a convex function defined on m,M and let x,y∈m,M such that x<y. Then,(33)αψm+M−x+ψm+M−yα+1−Γα+1y−xαJm+M−x−αψm+M−y≤y−xα+1P1αψ′m+P1αψ′M−P2αψ′x−P3αψ′y,where(34)P1α=2αα+1α+1/α,P2α=α2+α+12/α2α+2α+12/α,P3α=α4α+2α+11/α−1−α+12/α−22α+2α+12/α.Proof.
From Lemma3, we have(35)αψm+M−x+ψm+M−yα+1−Γα+1y−xαJm+M−x−αψm+M−y≤y−xα+1∫01α+1sα−1ψ′m+M−sx−1−syds.
Sinceψ′ is convex, using Mercer’s inequality, we obtain(36)αψm+M−x+ψm+M−yα+1−Γα+1y−xαJm+M−x−αψm+M−y≤y−xα+1∫01α+1sα−1ψ′m+ψ′M−sψ′x−1−sψ′yds=y−xα+1∫01/α+11/α1−α+1sαψ′m+ψ′M−sψ′x−1−sψ′yds+∫1/α+11/α1α+1sα−1ψ′m+ψ′M−sψ′x−1−sψ′yds,equivalent to(37)αψm+M−x+ψm+M−yα+1−Γα+1y−xαJm+M−x−αψm+M−y≤y−xα+1L1α+N1αψ′m+L1α+N1αψ′M−L2α+N2αψ′x−L3α+N3αψ′y,where(38)L1α=∫01/α+11/α1−α+1sαds=αα+1α+1/α,L2α=∫01/α+11/α1−α+1sαsds=α2α+2α+12/α,L3α=∫01/α+11/α1−α+1sα1−sds=α2α+2α+11/α−1−12α+2α+12/α,N1α=∫1/α+11/α1α+1sα−1ds=αα+1α+1/α,N2α=∫1/α+11/α1α+1sα−1sds=αα+12/α+12α+2α+12/α,N3α=∫1/α+11/α1α+1sα−11−sds=α2α+2α+11/α−1−α+12/α−12α+2α+12/α.
Substituting these values in (37), we get (33).Remark 5.
If we putα=1, x=m, and y=M in (33), we get the inequality given in Theorem 2.2 of [25].Theorem 5.
Letψ′q be a convex function for q≥1 and let x,y∈m,M such that x<y. Then,(39)αψm+M−x+ψm+M−yα+1−Γα+1y−xαJm+M−x−αψm+M−y≤y−xα+1P1α1−1/q×P1αψ′mq+P1αψ′Mq−P2αψ′xq−P3αψ′yq1/q,where P1α, P2α, and P3α are the same as defined in Theorem 4.Proof.
From Lemma3, we have (35). Applying power mean inequality, we obtain(40)αψm+M−x+ψm+M−yα+1−Γα+1y−xαJm+M−x−αψm+M−y≤y−xα+1∫01α+1sα−1ds1−1/q×∫01α+1sα−1ψ′m+M−sx−1−syqds1/q.
Sinceψ′q is convex, using Mercer’s inequality, we have(41)αψm+M−x+ψm+M−yα+1−Γα+1y−xαJm+M−x−αψm+M−y≤y−xα+1∫01/α+11/α1−α+1sαds+∫1/α+11/α1α+1sα−1ds1−1/q×∫01α+1sα−1ψ′mq+ψ′Mq−sψ′xq−1−sψ′yqds1/q=y−xα+12αα+1α+1/α1−1/q×∫01/α+11/α1−α+1sαψ′mq+ψ′Mq−sψ′xq−1−sψ′yqds+∫1/α+11/α1α+1sα−1ψ′mq+ψ′Mq−sψ′xq−1−sψ′yqds1/q.
This implies that(42)αψm+M−x+ψm+M−yα+1−Γα+1y−xαJm+M−x−αψm+M−y≤y−xα+12αα+1α+1/α1−1/qL1α+N1αψ′mq+L1α+N1αψ′Mq−L2α+N2αψ′xq−L3α+N3αψ′yq1/q.
Substituting the values ofL1, L2, L3, N1, N2, and N3 as given in the proof of Theorem 4 in (42), we get (39).Remark 6.
If we putα=1, x=m, and y=M in (39), we obtain the inequality proved in Theorem 1 of [28].
In the following theorem, we derive trapezoidal type inequality.Theorem 6.
Letp, q>1 and ψ′q be a convex function, and let x,y∈m,M such that x<y. Then,(43)αψm+M−x+ψm+M−yα+1−Γα+1y−xαJm+M−x−αψm+M−y≤y−xα+1L4α,p1/pψ′mq+ψ′Mq−ψ′xq+ψ′yq21/q,where L4α,p=∫01α+1sα−1pds such that 1/p+1/q=1.Proof.
Using Lemma3, we have (35). Applying Hölder’s inequality, we obtain(44)αψm+M−x+ψm+M−yα+1−Γα+1y−xαJm+M−x−αψm+M−y≤y−xα+1∫01α+1sα−1pds1/p∫01ψ′m+M−sx−1−syqds1/q≤y−xα+1L4α,p1/p∫01ψ′mq+ψ′Mq−sψ′xq−1−sψ′yqds1/q=y−xα+1L4α,p1/pψ′mq+ψ′Mq−ψ′xq+ψ′yq21/q.Remark 7.
Substitutingα=1, x=m, and y=M in Theorem 6, we obtain Theorem 2.3 of [25].Theorem 7.
Letψ′ be a convex function and let x,y∈m,M such that x<y. Then,(45)Γα+1y−xαJm+M−x−αψm+M−y−ψm+M−αx+yα+1≤y−xP5αψ′x+P6αψ′y,where(46)P5α=−α2α+2α+12,P6α=α2α+2α+12.Proof.
Using Lemma4, we have(47)Γα+1y−xαJm+M−x−αψm+M−y−ψm+M−αx+yα+1≤y−x∫α/α+111−sαψ′m+M−sx−1−syds−y−x∫0α/α+1sαψ′m+M−sx−1−syds.
Sinceψ′ is convex, using Mercer’s inequality, we obtain(48)Γα+1y−xαJm+M−x−αψm+M−y−ψm+M−αx+yα+1≤y−x∫α/α+111−sαψ′m+ψ′M−sψ′x−1−sψ′yds−y−x∫0α/α+1sαψ′m+ψ′M−sψ′x−1−sψ′yds,which implies that(49)Γα+1y−xαJm+M−x−αψm+M−y−ψm+M−αx+yα+1≤y−xN5α−L5αψ′x+N6α−L6αψ′y,where(50)L5α=∫α/α+111−sαsds=αα+1α+2αα+22α+2α+1α+2,L6α=∫α/α+111−sα1−sds=4αα+1−αα+1α2α+2α+1α+2,N5α=∫0α/α+1sα+1ds=αα+2α+2α+1α+2,N6α=∫0α/α+1sα1−sds=2αα+1α+2α+1α+2.
Substituting these values in (49), we get (45).
In next theorem, we use power mean inequality and derive midpoint type inequality.Theorem 8.
Letψ′q be a convex function for q≥1 and let x,y∈m,M such that x<y. Then,(51)Γα+1y−xαJm+M−x−αψm+M−y−ψm+M−αx+yα+1≤y−xL8α1−1/q×L8αψ′mq+L8αψ′Mq−L5αψ′xq−L6αψ′yq1/q−L8αψ′mq+L8αψ′Mq−N5αψ′xq−N6αψ′yq1/q,where(52)L8α=∫α/α+111−sαds=∫0α/α+1sαds=αα+1α+1α+2,and L5α, L6α, N5α, and N6α are given in the proof of Theorem 7.Proof.
Using power mean inequality in (47), we obtain(53)Γα+1y−xαJm+M−x−αψm+M−y−ψm+M−αx+yα+1≤y−x∫α/α+111−sαds1−1/q∫α/α+111−sαψ′m+M−sx−1−syqds1/q−y−x∫0α/α+1sαds1−1/q∫0α/α+1sαψ′m+M−sx−1−syqds1/q.
Sinceψ′q is convex, using Mercer’s inequality, we obtain(54)Γα+1y−xαJm+M−x−αψm+M−y−ψm+M−αx+yα+1≤y−xαα+1α+1α+21−1/q×∫α/α+111−sαψ′mq+ψ′Mq−sψ′xq−1−sψ′yqds1/q−∫0α/α+1sαψ′mq+ψ′Mq−sψ′xq−1−sψ′yqds1/q=y−xL8α1−1/q×L8αψ′mq+L8αψ′Mq−L5αψ′xq−L6αψ′yq1/q−L8αψ′mq+L8αψ′Mq−N5αψ′xq−N6αψ′yq1/q.Remark 8.
Substitutingα=1, x=m, and y=M in Theorem 8, we obtain the following midpoint type inequality:(55)1M−m∫mMfudu−ψm+M2≤M−m8ψ′mq+2ψ′Mq31/q−2ψ′mq+ψ′Mq31/q.
Another midpoint type inequality is presented in the following theorem.Theorem 9.
Letp, q>1 and ψ′q be convex, and let x,y∈m,M such that x<y. Then,(56)Γα+1y−xαJm+M−x−αψm+M−y−ψm+M−αx+yα+1≤y−xL9α,p1/p×1α+1ψ′mq+1α+1ψ′Mq−2α+12α+12ψ′xq−12α+12ψ′yq1/q−y−xL10α,p1/p×αα+1ψ′mq+αα+1ψ′Mq−α22α+12ψ′xq−αα+22α+12ψ′yq1/q,where L9α,p=∫α/α+111−sαpds and L10α,p=∫0α/α+1sαpds such that 1/p+1/q=1.Proof.
Applying Hölder’s inequality in (47), we have(57)Γα+1y−xαJm+M−x−αψm+M−y−ψm+M−αx+yα+1≤y−x∫α/α+111−sαpds1/p∫α/α+11ψ′m+M−sx−1−syqds1/q−y−x∫0α/α+1sαpdt1/p∫0α/α+1ψ′m+M−sx−1−syqds1/q.
Asψ′q is convex, applying Mercer’s inequality, we obtain(58)Γα+1y−xαJm+M−x−αψm+M−y−ψm+M−αx+yα+1≤y−xL9α,p1/p×∫α/α+11ψ′mq+ψ′Mq−sψ′xq−1−sψ′yqds1/q−y−xL10α,p1/p×∫0α/α+1ψ′mq+ψ′Mq−sψ′xq−1−sψ′yqds1/q=y−xL9α,p1/p×1α+1ψ′mq+1α+1ψ′Mq−2α+12α+12ψ′xq−12α+12ψ′yq1/q−y−xL10α,p1/p×αα+1ψ′mq+αα+1ψ′Mq−α22α+12ψ′xq−αα+22α+12ψ′yq1/q.Remark 9.
Substitutingα=1, x=m, and y=M in Theorem 9, we obtain(59)1M−m∫mMfudu−ψm+M2≤M−m4p+11/pψ′mq+3ψ′Mq41/q−3ψ′mq+ψ′Mq41/q.Theorem 10.
Letx,y∈m,M such that x<y. If ψ′ is concave on m,M, then(60)αψm+M−x+ψm+M−yα+1−Γα+1y−xαJm+M−x−αψm+M−y≤y−xα+1L1αψ′m+M−L11αx−L12αy+N1αψ′m+M−L13αx−L14αy,where L1 and N1 are given in the proof of Theorem 4 and(61)L11α=α+11−1/α2α+2,L12α=α+11−1/α2α+2α+11/α−1−12α+2,L13α=α+11−1/αα+12/α+12α+2,L14α=α+11−1/α2α+2α+11/α−1−α+12/α−12α+2.Proof.
Lemma3 implies that(62)αψm+M−x+ψm+M−yα+1−Γα+1y−xαJm+M−x−αψm+M−y≤y−xα+1∫01α+1sα−1ψ′m+M−sx−1−syds=y−xα+1∫01/α+11/α1−α+1sαψ′m+M−sx−1−syds+∫1/α+11/α1α+1sα−1ψ′m+M−sx−1−syds.
Sinceψ′ is concave, using Jensen’s inequality, we obtain(63)αψm+M−x+ψm+M−yα+1−Γα+1y−xαJm+M−x−αψm+M−y≤y−xα+1∫01/α+11/α1−α+1sαds×ψ′∫01/α+11/α1−α+1sαm+M−sx−1−syds∫01/α+11/α1−α+1sαds+∫1/α+11/α1α+1sα−1ds×ψ′∫1/α+11/α1α+1sα−1m+M−sx−1−syds∫1/α+11/α1α+1sα−1ds.=y−xα+1L1αψ′m+M−L11αx−L12αy+N1αψ′m+M−L13αx−L14αy.Theorem 11.
Letψ′ be a concave function, and let x,y∈m,M such that x<y. Then,(64)Γα+1y−xαJm+M−x−αψm+M−y−ψm+M−αx+yα+1≤y−xαα+1α+1α+2ψ′m+M−L15αx−L16αy−ψ′m+M−L17αx−L18αy,where(65)L15α=α+1α+2αα+12ααα+2,L16α=4αα−α+1α2ααα+2,L17α=αα+2,L18α=2α+2.Proof.
From Lemma4, we have (47). As ψ′ is concave, using Jensen’s inequality, we obtain(66)Γα+1y−xαJm+M−x−αψm+M−y−ψm+M−αx+yα+1≤y−x∫α/α+111−sαdsψ′∫α/α+111−sαm+M−sx−1−syds∫α/α+111−sαds−y−x∫0α/α+1sαdsψ′∫0α/α+1sαm+M−sx−1−syds∫0α/α+1sαds=y−xαα+1α+1α+2ψ′m+M−L15αx−L16αy−ψ′m+M−L17αx−L18αy.


## 4. Conclusion

In this paper, we establish the fractional Hermite–Hadamard type inequalities of Mercer type by using support line inequality. We expect that this work will lead to the new fractional integral studies for Hermite–Hadamard inequality. It is an open problem to prove inequalities (10) and (11) by any other method.



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