# Hadamard Inequalities for Stronglyα,m-Convex Functions via Caputo Fractional Derivatives

Authors: Yanliang Dong; Muhammad Zeb; Ghulam Farid; Sidra Bibi
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we present two versions of the Hadamard inequality forα,m convex functions via Caputo fractional derivatives. Several related results are analyzed for convex and m-convex functions along with their refinements and generalizations. The error bounds of the Hadamard inequalities are established by applying some known identities.


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## Body

## 1. Introduction

Fractional calculus is a natural extension of classical calculus and the notions related to integer order derivatives and integrals have been extended to fractional order derivatives and integrals. Many classical inequalities related to integrals of real valued functions have been presented for fractional integrals. The Hadamard inequality is one of them which is studied extensively for different types of convex functions via fractional derivatives and integrals. For the detailed study of Hadamard fractional integral inequalities, we refer the readers to [1–13]. In 1967, Caputo made the most significant contributions to fractional calculus by reformulating the definition of the Riemann–Liouville fractional derivatives [14].In the following, we give the definition of Caputo fractional derivatives.Definition 1 [15].
Letψ∈ACna,b and n=ℜβ+1. Then, Caputo fractional derivatives of order β∈ℂ, ℜβ>0, of ψ are defined as follows:(1)Da+βCψx=1Γn−β∫axψnzx−zβ−n+1dz,x>a,Db−βCψx=−1nΓn−β∫xbψnzz−xβ−n+1dz,x<b.
Ifβ=n∈1,2,3,… and usual derivative of order n exists, then Caputo fractional derivative CDa+βψx coincides with ψnx, where as CDb−βψx coincides with ψnx with exactness to a constant multiplier −1n. In particular, we have(2)CDa+0ψx=CDb−0ψx=ψx,where n=1 and β=0.
Convex functions are represented in terms of different inequalities. Many of the well-known inequalities are consequences of convex functions. Strongly convexity is a strengthening of the notion of convexity; some properties of strongly convex functions are just “stronger versions” of known properties of convex functions. Strongly convex function was introduced by Polyak [16].Definition 2.
LetD be a convex subset of L and L,⋅ be a normed space. A function ψ:D⊂L⟶ℝ is called strongly convex function with modulus C if it satisfies(3)ψaz+1−zb≤zψa+1−zψb−Cz1−za−b2,∀a,b∈D,z∈0,1.
ForC=0, (3) gives the inequality satisfied by convex functions. Many authors have been inventing the properties and applications of strongly convex functions, see [17–21].
The concept ofm-convex functions and strongly m-convex functions are introduced in [22, 23], respectively. In [23], the definition of m-convex functions is given.Definition 3.
The functionf:0,b⟶ℝ is said to be m-convex, where m∈0,1, if for every x,y∈0,b and t∈0,1 we have(4)ftx+m1−ty≤tfx+m1−tfy.
In [22], strongly m-convex function is introduced as follows.Definition 4.
A functionψ:0,b⟶ℝ is called strongly m-convex function with modulus C if(5)ψzx+m1−zy≤zψx+m1−zψy−Cmz1−zx−y2,where x,y∈0,b and z∈0,1.
In [24], α,m-convex function is introduced as follows.Definition 5.
A functionf:0,b⟶ℝ is said to be α,m-convex function, where α,m∈0,12 and b>0, if for every x,y∈0,b and z∈0,1, we have(6)fzx+m1−zy≤zαfx+m1−zαfy.
Fromα,m-convex functions, one can obtain star-shaped, m-convex, convex, and α-convex functions. In literature, α,m-convex functions are considered by many researchers and mathematicians and many properties especially inequalities have been obtained for them, for example (see [25–30]).
The stronglyα,m-convex function is introduced as follows.Definition 6.
A functionf:0,+∞⟶ℝ is said to be strongly α,m-convex function with modulus C≥0, for α,m∈0,12, if(7)fzx+m1−zy≤zαfx+m1−zαfy−Cmzα1−zαy−x2,holds for all x,y∈0,+∞ and z∈0,1.
A well-known inequality named Hadamard inequality is another interpretation of convex function. It is stated as follows [31].Theorem 1.
Letψ:I⟶ℝ be a convex function on interval I⊂ℝ and a,b∈I, where a<b. Then, the following inequality holds:(8)ψa+b2≤1b−a∫abψxdx≤ψa+ψb2.If order in (8) is reversed, then it holds for concave function.Fractional integral inequalities are useful in establishing the uniqueness of solutions for certain fractional partial differential equations. These inequalities also provide upper as well as lower bounds for solutions of the fractional boundary value problems. Fractional integral inequalities are in the study of several mathematicians. For fractional versions of Hadamard inequality, we refer the researchers and references [1–5, 11, 12, 32].Farid et al. [33] established the following identity for Caputo fractional derivatives.Lemma 1.
Letψ:a,b⟶ℝ, 0≤a<b, be the function such that ψ∈Cna,b. Also, let ψn+1 be positive and convex function on a,b. Then, the following equality for Caputo fractional derivatives holds:(9)ψna+ψnb2−Γn−β+12b−an−βCDa+βψb+−1nCDb−βψa=b−a2∫011−zn−β−zn−βψn+1za+1−zbdz.The following identity is established in [34].Lemma 2.
Letψ:a,b⟶ℝ be a differentiable mapping on a,b with a<b. If ψ∈Cn+1a,b, then the following equality for Caputo fractional derivatives holds:(10)2n−β/k−1kΓkn−β/k+kmb−an−β/kDa+bm/2+β,kCψmb+mn−β/k+1−1nDa+mb/2m−β,kCψam−12ψna+mb2+mψna+mb2m=mb−a4∫01zn−β/kψn+1z2a+m2−z2b−∫01zn−β/kψn+12−z2ma+z2bdz,with β>0.The Hadamard inequality for Caputo fractional derivatives of convex functions is studied in [7, 33, 34]; also, the error estimations are established by using aforementioned identities. The aim of this paper is to prove the Hadamard inequality for Caputo fractional derivatives of strongly α,m-convex functions. We have obtained refinements of various inequalities proved for convex and m-convex functions.In Section2, we will give two versions of the Hadamard inequality for Caputo fractional derivatives using strongly α,m-convex functions. Also, we connect the particular cases with some classical results. In Section 3, by applying known identities, we will derive refinements of some well-known inequalities.


## 2. Main Results

The following results give the Hadamard inequality for Caputo fractional derivatives of stronglyα,m-convex functions.Theorem 2.
Letψ:a,b⟶ℝ be a positive function with ψ∈Cna,b, α,m∈0,12, and 0≤a<mb,m≠0. If ψn is a strongly α,m-convex function with modulus C, then the following inequality for Caputo fractional derivatives holds:(11)ψnbm+a2+mCn−β2αn−β+21−12αb−a2+2b−aa/m−mbn−β+1+2a/m−mb2n−βn−β+1≤Γn−β+1bm−an−β1−12αmn−β+1−1nDb−βCψam+12αDa+βCψmb≤n−βn−β+α1−12αm2ψna/m2+mψnbn−β+mψnb+ψna2α−Cmαm2α−1b−a/m22+b−a22αn−β+2α,with β>0 and n=β+1.Proof.
Sinceψn is strongly α,m-convex function with modulus C, for x,y∈a,b, we have(12)ψnmx+y2≤1−12αmψnx+ψny2α−mC2α1−12αx−y2.
Letx=1−za/m+zb≤b and y=m1−zb+za≥a, z∈0,1. Then, we have(13)ψnbm+a2≤1−12αmψn1−zam+zb+ψnm1−zb+za2α−mC2α1−12α1−zam+zb−m1−zb+za2.
Multiplying (13) with zn−β−1 on both sides and making integration over 0,1, we get(14)ψnbm+a2∫01zn−β−1dz≤1−12α∫01mψn1−zam+zbzn−β−1dz+12α∫01ψnm1−zb+zazn−β−1dz−mC2α1−12α∫01zn−β−1×1−zam+zb−m1−zb+za2dz.
By using change of the variables and computing the last integral, from (14), we get(15)1n−βψnbm+a2≤1bm−a1−12α∫a/mbm2ψnxmx−abm−an−β−1dx+12α∫ambψnybm−ybm−an−β−1dy−mC2α1−12α×b−a2n−β+2+2b−aa/m−mbn−β+1n−β+2+2a/m−mb2n−βn−β+1n−β+2.
Further, it takes the following form:(16)ψnbm+a2≤Γn−β+1bm−an−β1−12αmn−β+1−1nDb−βCψam+12αDa+βCψmb−mCn−β2α1−12α+b−a2n−β+2+2b−aa/m−mbn−β+1n−β+2+2a/m−mb2n−βn−β+1n−β+2.
Sinceψn is strongly α,m-convex function with modulus C, for z∈0,1, then one has(17)mψn1−zam+zb≤mzαψnb+m21−zαψnam2−Cm2zα1−zαb−am22.
By multiplying (17) with 1−1/2αzn−β−1 on both sides and making integration over 0,1, we get(18)1−12α∫01mψn1−zam+zbzn−β−1dz≤1−12α∫01mψnbzn−β+α−1dz+∫01m2ψnam21−zαzn−β−1dz−Cm2b−am22∫011−zαzn−β+α−1dz.
By using change of the variables and computing the last integral, from (18), we get(19)1−12α1bm−a∫a/mbm2ψnxmx−abm−an−β−1dx≤1−12αm2ψna/m2+mψnbn−βn−β+α−Cm2αb−a/m22n−β+αn−β+2α.
Further, it takes the following form:(20)Γn−β+1bm−an−β1−12αmn−β+1−1nDb−βCψam≤1−12αn−βm2ψna/m2+mψnbn−βn−β+α−Cm2αb−a/m22n−β+αn−β+2α.
Sinceψn is strongly α,m-convex function with modulus C, for z∈0,1, then one has(21)ψnm1−zb+za≤zαψna+m1−zαψnb−Cmzα1−zαb−a2.
By multiplying (21) with 1/2αzn−β−1 on both sides and making integration over 0,1, we get(22)12α∫01ψnm1−zb+zazn−β−1dz≤12α∫01ψnazn−β+α−1dz+∫01mψnb1−zαzn−β−1dz−Cmb−a2∫011−zαzn−β+α−1dz.
By using change of the variables and computing the last integral, from (22), we get(23)12αbm−a∫ambψnybm−ybm−an−β−1dy≤mψnb+ψna2αn−β+α−Cmαb−a22αn−β+αn−β+2α.
Further, it takes the following form:(24)Γn−β+1bm−an−β12αDa+βCψmb≤n−β2αmψnb+ψnan−β+α−Cmαb−a2n−β+αn−β+2α.
By adding (20) and (24), we have(25)Γn−β+1bm−an−β1−12αmn−β+1−1nDb−βCψam+12αDa+βCψmb≤n−β1−12αm2ψna/m2+mψnbn−βn−β+α+mψnb+ψna2αn−β+α−Cmαm2α−1b−a/m22+b−a22αn−β+αn−β+2α.
Inequalities (16) and (25) constituted the required inequality.
The consequences of Theorem2 are stated in the following remark.Remark 1.
IfC=0, m=1, and α=1 in (11), then we get the fractional Hadamard inequality for convex function given in [33], Theorem 3.
The upcoming result is the refinement of another version of the Hadamard inequality for Caputo fractional integrals stated in [7], Theorem 2.Theorem 3.
Letψ:a,b⟶ℝ be a positive function with ψ∈Cna,b, α,m∈0,12, and 0≤a<mb,m≠0. If ψn is a strongly α,m-convex function with modulus C, then the following inequality for Caputo fractional derivatives holds:(26)ψnbm+a2+mCn−β1−1/2α2α+1n−β+2b−a22+b−aa/m−mbn−β+3n−β+1+a/m−mb2n−β2+5n−5β+82n−βn−β+1≤2n−βΓn−β+1bm−an−β×1−12αmn−β+1−1nDa+bm/2m−βCψam+12αDa+bm/2+βCψmb≤n−β2αn+α−β1−12αmψnb+m2ψna/m22αn+α−β−n−βn−β+mψna+ψna2α−mC2α2α+n−β−α+n−β2α2α+n−β×1−12αmb−am22+b−a22α.with β>0 and n=β+1.Proof.
Letx=a/m2−z/2+bz/2 and y=az/2+m2−z/2b, z∈0,1 in (12), then we have(27)ψnbm+a2≤1−12αmψnam2−z2+bz2+12αψnaz2+m2−z2b−mC2α1−12αam2−z2+bz2−az2−m2−z2b2.
By multiplying (27) with zn−β−1 on both sides and making integration over 0,1, we get(28)ψnbm+a2∫01zn−β−1dz≤∫011−12αmψnam2−z2+bz2+12αψnaz2+m2−z2bzn−β−1dz−mC2α1−12α∫01am2−z2+bz2−az2−m2−z2b2zn−β−1dz.
By using change of variables and computing the last integral, from (14), we get(29)1n−βψnbm+a2≤1−12αm∫a/ma+bm/2m2my−a/mbm−an−β−1ψny2mdybm−a+12α∫a+mb/2mb2mb−xbm−an−β−1ψnx2dxbm−a−mC2αn−β+21−12αb−a24+b−aa/m−mbn−β+32n−β+1+a/m−mb2n−β2+5n−5β+84n−βn−β+1.
Further, it takes the following form:(30)ψnbm+a2≤2n−βΓn−β+1bm−an−β1−12αmn−β+1−1nDa+bm/2m−βCψam+12αDa+bm/2+βCψmb−mCn−β2α+11−12αb−a22n−β+2+b−aa/m−mbn−β+3n−β+1n−β+2+a/m−mb2n−β2+5n−5β+82n−βn−β+1n−β+2.
Sinceψn is strongly α,m-convex function and z∈0,1, we have the following inequality:(31)mψnam2−z2+bz2≤m21−z2αψnam2+mz2αψnb−m2Cz2α1−z2αb−am22.
By multiplying (31) with 1−1/2αzn−β−1 on both sides and making integration over 0,1, we get(32)1−12α∫01mψnam2−z2+bz2zn−β−1dz≤1−12α×∫01m21−z2αψnam2zn−β−1dz+∫01mz2αψnbzn−β−1dz−1−12α∫01mCz2α1−z2αmb−am22zn−β−1dz.
By using change of variables and computing the last integral, from (32), we get(33)21−1/2αbm−a∫a/ma+bm/2m2my−a/mbm−an−β−1m2ψnydy≤1−1/2αmψnb+m2ψna/m22αn+α−β−n−β2αn−βn+α−β−mC1−1/2α2α2α+n−β−α+n−βmb−a/m2222αα+n−β2α+n−β.
Further, it takes the following form:(34)2n−βΓn−β+1bm−an−β1−12αmn−β+1−1nDa+bm/2m−βCψam≤n−β2αn+α−β1−12αmψnb+m2ψna/m22αn+α−β−n−βn−β−mC2α2α+n−β−α+n−βmb−a/m222α2α+n−β.
Again by using stronglyα,m-convexity on the function ψn for z∈0,1, we have the following inequality:(35)ψnaz2+m2−z2b≤z2αψna+m1−z2αψnb−mCz2α1−z2αb−a2.
By multiplying (35) with 1/2αzn−β−1 on both sides and making integration over 0,1, we get(36)12α∫01ψnaz2+m2−z2bzn−β−1dz≤12α×∫01z2αψnazn−β−1dz+∫01m1−z2αψnbzn−β−1dz−12α∫01mCz2α1−z2αb−a2zn−β−1dz.
By using change of variables and computing the last integral, from (36), we get(37)22αbm−a∫a+mb/2mb2mb−xbm−an−β−1ψnxdx≤mψnb+ψna22αα+n−β−mC2α2α+n−β−α+n−βb−a223αα+n−β2α+n−β.
Further, it takes the following form:(38)2n−βΓn−β+1bm−an−β12αDa+bm/2+βCψmb≤n−β22αn+α−βmψnb+ψna−mC2α2α+n−β−α+n−βb−a22α2α+n−β.
By adding (34) and (38), we get(39)2n−βΓn−β+1bm−an−β1−12αmn−β+1−1nDa+bm/2m−βCψam+12αDa+bm/2+βCψmb≤n−β2αn+α−β1−12αmψnb+m2ψna/m22αn+α−β−n−βn−β+mψna+ψna2α−mC2α2α+n−β−α+n−β2α2α+n−β×1−12αmb−am22+b−a22α.
From (30) and (39), (26) can be obtained.Remark 2.
IfC=0, m=1, and α=1 in (26), then we get the fractional Hadamard inequality for convex function given in [7], Theorem 2.2.


## 3. Error Bounds of Fractional Hadamard Inequalities

In this section, we give refinements of the error bounds of fractional Hadamard inequalities for Caputo fractional derivatives.Theorem 4.
Letψ:a,b⟶ℝ be a differentiable mapping on a,b with ψ∈Cn+1a,b, α,m∈0,12, and 0≤a<mb,m≠0. If ψn+1 is a strongly α,m-convex function on a,b, then the following inequality for Caputo fractional derivatives holds:(40)ψna+ψnb2−Γn−β+12b−an−βDa+βCψb+−1nDb−βCψa≤ψn+1a1−1/2np−βp+1np−βp+11/p12qα+1qα+11/q+2n−β+α+1−22n−β+α+1n−β+α+1−1/2np−βp+1np−βp+11/p2qα+1−12qα+1qα+11/q+mψn+1bm22α+12n−β−1n−β+α+1+n−β+11−2n−β+α2n−β+α+1n−β+1n−β+α+1−1−1/2np−βp+1np−βp+11/p12qα+1qα+11/q+1/2np−βp+1np−βp+11/p2qα+1−12qα+1qα+11/q−Cmbm−a21−1/2np−βp+1np−βp+11/p1/2qα+1qα+11/q−1/22qα+12qα+11/q+1/2np−βp+1np−βp+11/p22qα+1−122qα+12qα+11/q−2qα+1−12qα+1qα+11/q+2n−β+α+11−2n−β+2α+2αn−β+2α+12n−β+α−12n−β+2α+1n−β+α+1n−β+2α+1,with β>0 and n=β+1.Proof.
Sinceψn+1 is strongly α,m-convex function on a,b and z∈0,1, we have(41)ψn+1za+1−zb=ψn+1za+m1−zbm≤zαψn+1a+m1−zαψn+1bm−Cmzα1−zαbm−a2.
By using Lemma1 and inequality (41), we have(42)ψna+ψnb2−Γn−β+12b−an−βDa+βCψb+−1nDb−βCψa≤b−a2∫011−zn−β−zn−βψn+1za+m1−zbmdz≤b−a2∫011−zn−β−zn−βzαψn+1a+m1−zαψn+1bm−Cmzα1−zαbm−a2dz≤b−a2∫01/21−zn−β−zn−βzαψn+1a+m1−zαψn+1bm−Cmzα1−zαbm−a2dz+∫1/21zn−β−1−zn−βzαψn+1a+m1−zαψn+1bm−Cmzα1−zαbm−a2dz.
In the following, we compute integrals appearing on the right hand side of inequality (42) by using Holder inequality:(43)∫01/21−zn−β−zn−βzαψn+1a+m1−zαψn+1bm−Cmzα1−zαbm−a2dz=ψn+1a∫01/2zα1−zn−β−zn−β+αdz+mψn+1bm∫01/21−zn−β1−zα−1−zαzn−βdz−Cmbm−a2∫01/2zα−z2α1−zn−βdz−∫01/2zn−β+α1−zαdz=ψn+1a1−1/2np−βp+1np−βp+11/p12qα+1qα+11/q−1/2n−β+α+1n−β+α+1+mψn+1bm2α+12n−β−1n−β+α+1+n−β+12n−β+α+1n−β+1n−β+α+1−1−1/2np−βp+1np−βp+11/p12qα+1qα+11/q−Cmbm−a2n−β+α+1−2αn−β+2α+12n−β+2α+1n−β+α+1n−β+2α+1+1−1/2np−βp+1np−βp+11/p1/2qα+1qα+11/q−1/22qα+12qα+11/q,(44)∫1/21zn−β−1−zn−βzαψn+1a+m1−zαψn+1bm−Cmzα1−zαbm−a2dz=ψn+1a∫1/21zn−β+α−zα1−zn−βdzψn+1bm∫1/211−zαzn−β−1−zα1−zn−βdz−Cmbm−a2∫1/211−zαzn−β+αdz−∫1/21zα−z2α1−zn−βdz=ψn+1a2n−β+α+1−12n−β+α+1n−β+α+1−1/2np−βp+1np−βp+11/p2qα+1−12qα+1qα+11/q+mψn+1bm1/2np−βp+1np−βp+11/p2qα+1−12qα+1qα+11/q+2α+12n−β−1n−β+α+1−n−β+12n−β+α+1−12n−β+α+1n−β+1n−β+α+1−Cmbm−a22αn−β+2α+12n−β+α+1−1−n−β+α+12n−β+2α+1−12n−β+2α+1n−β+α+1n−β+2α+1+1/2np−βp+1np−βp+11/p22qα+1−122qα+12qα+11/q−2qα+1−12qα+1qα+11/q.
By putting the values of (43) and (44) in (42), we get (40).
By using Lemma2, we give the following error bounds of Caputo fractional derivative inequality (26).Theorem 5.
Letψ:a,b⟶ℝ be a differentiable mapping on a,b with ψ∈Cn+1a,b, α,m∈0,12, and 0≤a<mb,m≠0. If ψn+1q is strongly α,m-convex function on a,b for q≥1, then the following inequality for Caputo fractional derivatives holds:(45)2n−β−1Γn−β+1mb−an−βDa+bm/2+βCψmb+mn−β+1−1nDa+mb/2m−βCψam−12ψna+mb2+mψna+mb2m≤mb−a4n−β+11/p1n−β+α+11/q×ψn+1aq2α+mψn+1bq2αn−β+α+1−n−β+12αn−β+1−Cmb−a22αn−β+2α+1−n−β+α+14n−β+2α+11/q+mψn+1a/m2q2αn−β+α+1−n−β+12αn−β+1+ψn+1bq2α−Cmb−a/m222αn−β+2α+1−n−β+α+14n−β+2α+11/q,with β>0 and n=β+1.Proof.
By takingk=1 in Lemma 2 and using power mean inequality, we have(46)2n−β−1Γn−β+1mb−an−βDa+bm/2+βCψmb+mn−β+1−1nDa+bm/2m−βCψam−12ψna+mb2+mψna+mb2m≤mb−a4∫01zn−βψn+1z2a+m2−z2b+ψn+12−z2ma+z2bdz≤mb−a4∫01zn−βdz1−1/q∫01zn−βψn+1az2+m2−z2bqdz1/q+∫01zn−βψn+1a2−z2m+bz2qdz1/q.
Now, by using the stronglyα,m-convexity of ψn+1q, we have the last expression(47)≤mb−a4n−β+11/pψn+1aq∫01zn−β+α2αdz+mψn+1bq∫01zn−β2α−zα2αdz−Cmb−a24∫012αzn−β+α−zn−β+2αdz1/q+mψn+1am2q∫01zn−β2α−zα2αdz+ψn+1bq∫01zn−β+α2αdz−Cmb−a/m224∫012αzn−β+α−zn−β+2αdz1/q=mb−a4n−β+11/pψn+1aq2αn−β+α+1+mψn+1bq2αn−β+α+1−n−β+12αn−β+1n−β+α+1−Cmb−a22αn−β+2α+1−n−β+α+14n−β+α+1n−β+2α+11/q+mψn+1a/m2q2αn−β+α+1−n−β+12αn−β+1n−β+α+1+ψn+1bq2αn−β+α+1−Cmb−a/m222αn−β+2α+1−n−β+α+14n−β+α+1n−β+2α+11/q,and inequality (45) is obtained.Remark 3.
IfC=0, m=1, and α=1 in (45), then we get the fractional Hadamard inequality for convex function given in [7], Theorem 3.2.Theorem 6.
Letψ:a,b⟶ℝ be a differentiable mapping on a,b with ψ∈Cn+1a,b, α,m∈0,12, and 0≤a<mb,m≠0. If ψn+1q is a strongly α,m-convex function on a,b for q>1, then the following inequality for Caputo fractional derivatives holds:(48)2n−β−1Γn−β+1mb−an−βDa+bm/2+βCψmb+mn−β+1−1nDa+mb/2m−βCψam−12ψna+mb2+mψna+mb2m≤bm−a4np−βp+11/p2αα+11/q×ψn+1a+m1/qψn+1b2αα+1−11/qq−Cmb−a22α2α+1−α+12α2α+11/q+m1/qψn+1am22αα+1−11/q+ψn+1bq−Cmb−a/m222α2α+1−α+12α2α+11/q,with β>0 and n=β+1.Proof.
By takingk=1 in Lemma 2 and with the help of modulus property, we get(49)2n−β−1Γn−β+1mb−an−βDa+bm/2+βCψmb+mn−β+1−1nDa+mb/2−βCψam−12ψna+mb2+mψna+mb2m≤mb−a4∫01zn−βψn+1z2a+m2−z2bdz+∫01zn−βψn+12−z2ma+z2bdz.
Now, applying Holder’s inequality, we get(50)2n−β−1Γn−β+1mb−an−βDa+bm/2+βCψmb+mn−β+1−1nDa+mb/2−βCψam−12ψna+mb2+mψna+mb2m≤mb−a41np−βp+11/p∫01ψn+1z2a+m2−zb2dz1/q+∫01ψn+12−z2ma+z2bdz1/q.
Using strongα,m-convexity of ψn+1q, we get(51)2n−β−1Γn−β+1mb−an−βDa+bm/2+βCψmb+mn−β+1−1nDa+mb/2m−βCψam−12ψna+mb2+mψna+mb2m≤mb−a41np−βp+11/pψn+1aq∫01z2αdz+mψn+1bq∫011−z2αdz−Cmb−a222α∫012αzα−z2αdz1/q+mψn+1am2q∫011−z2αdz+ψn+1bq∫01z2αdz−Cmb−a/m2222α∫012αzα−z2αdz1/q=mb−a41np−βp+11/pψn+1aq+mψn+1bq2αα+1−12αα+1−Cmb−a22α2α+1−α+122αα+12α+11/q+mψn+1a/m2q2αα+1−1+ψn+1bq2αα+1−Cmb−a/m222α2α+1−α+122αα+12α+11/q≤bm−a41np−βp+11/pψn+1a+m1/qψn+1b2αα+1−11/qq2αα+1−Cmb−a22α2α+1−α+122αα+12α+11/q+m1/qψn+1a/m22αα+1−11/q+ψn+1bq2αα+1−Cmb−a/m222α2α+1−α+122αα+12α+11/q.
Here, we have used the fact thata1+b1q≥a1q+b1q, where q>1, a1,b1≥0. This completes the proof.Remark 4.
IfC=0, m=1, and α=1 in (48), then we get the fractional Hadamard inequality for convex function given in [7], Theorem 3.3.



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