# Caputo Fractional Derivative Hadamard Inequalities for Stronglym-Convex Functions

Authors: Xue Feng; Baolin Feng; Ghulam Farid; Sidra Bibi; Qi Xiaoyan; Ze Wu
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, two versions of the Hadamard inequality are obtained by using Caputo fractional derivatives and stronglym-convex functions. The established results will provide refinements of well-known Caputo fractional derivative Hadamard inequalities for m-convex and convex functions. Also, error estimations of Caputo fractional derivative Hadamard inequalities are proved and show that these are better than error estimations already existing in literature.


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## Body

## 1. Introduction

Strongly convex function was introduced by Polyak in [1]. Strong 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 functions have been used for proving the convergence of a gradient-type algorithm for minimizing a function. They play an important role in the optimization theory and mathematical economics.Definition 1.
LetD be a convex subset of X,X,. be a normed space. A function ψ:D⊂X⟶R is called strongly convex function with modulus C if it satisfies
(1)ψaz+1−zb≤zψa+1−zψb−Cz1−za−b2,∀a,b∈D,z∈0,1 and C≥0.
Many authors have been inventing the properties and applications of strongly convex function, for detailed information, see [2–6].The concepts ofm-convex functions and strongly m-convex functions were introduced in [7, 8], respectively. Toader [7] gave the idea of m-convex functions as follows.Definition 2.
A functionψ:0,b⟶ℝ is said to be m-convex, where m∈0,1, if for every x,y∈0,b and z∈0,1, we have
(2)ψzx+m1−zy≤zψx+m1−zψy.Lara et al. introduced stronglym-convex functions as follows:Definition 3 (see [8]).
A functionψ:I⟶ℝ is called strongly m-convex function with modulus C≥0 if
(3)ψza+m1−zb≤zψa+m1−zψb−Cmz1−za−b2,for a,b∈I and z∈0,1.Nowadays, fractional integral inequalities are in the study of several researchers (see, [9–17] and references therein), where they have used different kinds of well-known functions and fractional integral operators. Here, in this paper, we are interested to produce fractional integral inequalities for Caputo fractional derivatives of strongly m-convex functions. The Caputo fractional derivative operators are defined as follows.Definition 4 (see [18]).
Letψ∈ACna,b. The Caputo fractional derivatives of order β>0 of ψ are defined as follows:
(4)CDa+βψx=1Γn−β∫axψnzx−zβ−n+1dz,x>a,CDb−βψx=−1nΓn−β∫xbψnzz−xβ−n+1dz,x<b,where n=β+1. If β=n∈1,2,3,⋯ and usual derivative of order n exists, then Caputo fractional derivative DCa+βψx coincides with ψnx, whereas DCa−βψx coincides with ψnx with exactness to a constant multiplier −1n. In particular, we have
(5)DCa+0ψx=CDb−0ψx=ψx,where n=1 and β=0.The Hadamard inequality is another interpretation of convex function. It is stated as follows.Definition 5 (see [19]).
Letψ:I⟶ℝ be a convex function on interval I⊂ℝ and a,b∈I where a<b. Then, the following inequality holds:
(6)ψa+b2≤1b−a∫abψxdx≤ψa+ψb2.
If order in (6) is reversed, then it holds for concave function.Farid et al. [20] have proved the following Hadamard inequality for Caputo fractional derivatives of convex functions:Theorem 6.
Letψ:a,b⟶ℝ be the function with ψ∈ACna,b and 0≤a<b. Also, let ψn be positive and convex function on a,b. Then, the following inequality holds for Caputo fractional derivatives:
(7)ψna+b2≤Γn−β+12b−an−βCDa+βψb+−1nCDb−βψa≤ψna+ψnb2.They also established the following identity.Lemma 7 (see [20]).
Letψ:a,b⟶ℝ,0≤a<b, be the function such that ψ∈ACna,b. Then, the following equality for Caputo fractional derivatives holds:
(8)ψna+ψnb2−Γn−β+12b−an−βDCa+βψb+−1nDCb−βψa=b−a2∫011−zn−β−zn−βψn+1za+1−zbdz.Farid et al. [20] also proved the following inequality for Caputo fractional derivatives.Theorem 8.
Letψ:a,b⟶ℝ,0≤a<b be the function with ψ∈ACn+1a,b and also let ψn+1 be convex on a,b. Then, the following inequality for Caputo fractional derivatives holds:
(9)ψna+ψnb2−Γn−β+12b−an−βCDa+βψb+−1nCDb−βψa≤b−a2n−β+11−12n−βψn+1a+ψn+1b.Kang et al. [21] proved the following version of the Hadamard inequality for Caputo fractional derivatives.Theorem 9.
Letψ:a,b⟶ℝ be a positive function with ψ∈ACna,b and 0≤a<b. If ψn is convex function on a,b, then the following inequality for Caputo fractional derivatives holds:
(10)ψna+b2≤2n−β−1Γn−β+1b−an−βDCa+b/2+βψb+−1nCDa+b/2−βψa≤ψna+ψnb2.
Farid et al. [22] established the following identity.Lemma 10.
Letψ:a,b⟶ℝ be a differentiable mapping on a,b with a<mb,m∈0,1. If ψ∈ACn+1a,b, then the following equality for Caputo fractional derivatives holds:
(11)2n−β/k−1kΓkn−β/k+kmb−an−β/kCDa+bm/2+β,kψmb+mn−β/k+1−1nCDa+mb/2m−β,kψam−12ψna+mb2+mψna+mb2m=mb−a4∫01zn−β/kψn+1z2a+m2−z2b−∫01zn−β/kψn+12−z2ma+z2bdz,β>0.Kang et al. [21] also proved the following inequalities for Caputo fractional derivatives.Theorem 11.
Letψ:a,b⟶ℝ be a differentiable mapping on a,b with ψ∈ACn+1a,b and a<b. If ψn+1q is convex on a,b for q≥1, then the following inequality for Caputo fractional derivatives holds:
(12)2n−β−1Γn−β+1b−an−βDCa+b/2+βψb+−1nDCa+b/2−βψa−ψna+b2≤b−a4n−β+11/p12n−β+1n−β+21/qn−β+1ψn+1aq+n−β+3ψn+1bq1/q+n−β+3ψn+1aq+β+1ψn+1bq1/q,β>0.Theorem 12 (see [21]).
Letψ:a,b⟶ℝ be a function with ψ∈ACn+1a,b and a<b. If ψn+1q is convex on a,b for q>1, then the following inequality for Caputo fractional derivatives holds:
(13)2n−β−1Γn−β+1b−an−βDCa+b/2+βψb+−1nDCa+b/2−βψa−ψna+b2≤b−a41np−βp+11/pψn+1aq+3ψn+1bq41/q+3ψn+1aq+ψn+1bq41/q≤b−a443np−βp+11/pψn+1a+ψn+1b,β>0,1p+1q=1.
We will study all of the above fractional inequalities for stronglym-convex functions and at the same time will obtain their generalizations and refinements. In Section 2, we will give refinements of two versions of the Hadamard inequality for Caputo fractional derivatives. We will connect their particular cases with some well-known results. In Section 3, by applying known identities, we will give refinements of error estimations of the Hadamard inequalities.


## 2. Main Results

The following result is the generalization of Theorem6 which in a particular case also provides its refinement.Theorem 13.
Letψ∈ACna,b, 0≤a<mb be a positive function. If ψn is a strongly m-convex function with modulus C≥0,m∈0,1, then the following inequality for Caputo fractional derivatives holds:
(14)ψnbm+a2+mCn−β4n−β+2b−a2+2b−aa/m−mbn−β+1+2a/m−mb2n−βn−β+1≤Γn−β+12bm−an−βmn−β+1−1nDCb−βψam+DCa+βψmb≤n−β2n−β+1m2ψna/m2+mψnbn−β+mψnb+ψna−Cmb−a2+b−a/m22n−β+2,with β>0.Proof.
Sinceψn is strongly m-convex function with modulus C, for x,y∈a,b, we have
(15)ψnmx+y2≤mψnx+ψny2−mC4x−y2.
Letx=1−za/m+zb≤b and y=m1−zb+za≥a, z∈0,1. Then, we have
(16)2ψnbm+a2≤mψn1−zam+zb+ψnm1−zb+za−mC21−zam+zb−m1−zb+za2.
By multiplying (16) with zn−β−1 on both sides and making integration over 0,1, we get
(17)2ψnbm+a2∫01zn−β−1dz≤∫01mψn1−zam+zbzn−β−1dz+∫01ψnm1−zb+zazn−β−1dz−mC2∫01zn−β−1×1−zam+zb−m1−zb+za2dz.
By using change of variables and computing the last integral, from (17), we get
(18)2n−βψnbm+a2≤1bm−a∫ambm2ψnxmx−abm−an−β−1dx+∫ambψnybm−ybm−an−β−1dy−mC2b−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(19)ψnbm+a2≤Γn−β+12bm−an−βmn−β+1−1nDCb−βψam+CDa+βψmb−mCn−β4n−β+2b−a2+2b−aa/m−mbn−β+1+2a/m−mb2n−βn−β+1.
Sinceψn is strongly m-convex function with modulus C, for z∈0,1, then one has
(20)mψn1−zam+zb+ψnm1−zb+za≤mzψnb+m21−zψnam2+zψna+m1−zψnb−Cmz1−zb−a2−Cm2z1−zb−am22.
By multiplying (20) with zn−β−1 on both sides and making integration over 0,1, we get
(21)∫01mψn1−zam+zbzn−β−1dz+∫01ψnm1−zb+zazn−β−1dz≤∫01mψnbzn−βdz+∫01m2ψnam21−zzn−β−1dz+∫01ψnazn−βdz+∫01mψnb1−zzn−β−1dz−Cmb−a2+b−am2∫011−zzn−βdz.
By using change of variables and computing the last integral, from (21), we get
(22)1bm−a∫a/mbm2ψnxmx−abm−an−β−1dx+1bm−a∫ambψnybm−ybm−an−β−1dy≤m2ψna/m2+mψnbn−βn−β+1+mψnb+ψnan−β+1−Cmb−a2+b−a/m22n−β+1n−β+2.
Further, it takes the following form(23)Γn−β+12bm−an−βmn−β+1−1nDCb−βψam+CDa+βψmb≤n−β2n−β+1m2ψna/m2+mψnbn−β+mψnb+ψna−Cmb−a2+b−a/m22n−β+2.
Inequalities (19) and (11) constituted the required inequality.The consequences of Theorem13 are stated in the following corollary and remark:Corollary 14.
By settingm=1 in inequality (14), we will get ([23], Theorem 6)
(24)ψnb+a2+Cb−a2β−n+2+n−β24n−β+1n−β+2≤Γn−β+12b−an−β−1nCDb−βψa+CDa+βψb≤ψna+ψnb2−Cn−βb−a2n−β+1n−β+2.Remark 15.
IfC=0 and m=1 in (14), then we will get the fractional Hadamard inequality stated in Theorem 6.
The upcoming result is the refinement of another version of the Hadamard inequality for Caputo fractional derivatives stated in a theorem in [19].Theorem 16.
Under the assumptions of Theorem13, the following inequality for Caputo fractional derivatives holds:
(25)ψnbm+a2+mCn−β8n−β+2b−a22+b−aa/m−mbn−β+3n−β+1+a/m−mb2n−β2+5n−5β+82n−βn−β+1≤2n−β−1Γn−β+1bm−an−βmn−β+1−1nCDa+bm/2−βψam+CDa+bm/2+βψmb≤n−β4n−β+1n−β+2mψnb+m2ψna/m2n−β+mψnb+ψna−mCn−β+3b−a2+mb−a/m222n−β+2,with β>0.Proof.
Letx=a/m2−z/2+bz/2 and y=az/2+m2−z/2b, z∈0,1 in (15), then we have
(26)2ψnbm+a2≤mψnam2−z2+bz2+ψnaz2+m2−z2b−mC2am2−z2+bz2−az2−m2−z2b2.
By multiplying (26) with zn−β−1 on both sides and making integration over 0,1, we get
(27)2ψnbm+a2∫01zn−β−1dz≤∫01mψnam2−z2+bz2zn−β−1dz+∫01ψnaz2+m2−z2bzn−β−1dz−mC2∫01am2−z2+bz2−az2−m2−z2b2zn−β−1dz.
By using change of variables and computing the last integral, from (27), we get
(28)2n−βψnbm+a2≤m∫a/ma+bm/2m2my−a/mbm−an−β−1ψny2mdybm−a+∫a+mb/2mb2mb−xbm−an−β−1ψnx2dxbm−a−mC2b−a24n−β+2+b−aa/m−mbn−β+32n−β+1n−β+2+a/m−mb2n−β2+5n−5β+84n−βn−β+1n−β+2.
Further, it takes the following form:(29)ψnbm+a2≤2n−β−1Γn−β+1bm−an−βmn−β+1−1nCDa+bm/2m−βψam+CDa+bm/2+βψmb−mCn−β8n−β+2b−a22+b−aa/m−mbn−β+3n−β+1+a/m−mb2n−β2+5n−5β+82n−βn−β+1.
Sinceψn is strongly m-convex function and z∈0,1, we have the following inequality:
(30)mψnam2−z2+bz2+ψnaz2+m2−z2b≤m22−z2ψnam2+mz2ψnb+z2ψna+m2−z2ψnb−m2Cz2−z4b−am22−mCz2−z4b−a2.
By multiplying (30) with zn−β−1 on both sides and making integration over 0,1, we get
(31)∫01mψnam2−z2+bz2zn−β−1dz+∫01ψnaz2+m2−z2bzn−β−1dz≤∫01m22−z2ψnam2zn−β−1dz+∫01mz2ψnbzn−β−1dz+∫01z2ψnazn−β−1dz+∫01m2−z2ψnbzn−β−1dz−∫01mCz2−z4mb−am22+b−a2zn−β−1dz.
By using change of variables and computing the last integral, from (31), we get
(32)m∫a/ma+bm/2m2my−a/mbm−an−β−1ψny2mdybm−a+∫a+mb/2mb2mb−xbm−an−β−1ψnx2dxbm−a≤n−β+2mψnb+m2ψna/m22n−βn−β+1+mψnb+ψna2n−β+1−mCn−β+3b−a2+mb−a/m224n−β+1n−β+2.
Further, it takes the following form(33)2n−β−1Γn−β+1bm−an−βmn−β+1−1nCDa+bm/2m−βψam+DCa+bm/2+βψmb≤n−β4n−β+2mψnb+m2ψna/m2n−βn−β+1+mψnb+ψnan−β+1−mCn−β+3b−a2+mb−a/m222n−β+1n−β+2.
From (29) and (33), (25) can be obtained.Corollary 17.
By settingm=1 in inequality (25), we will get ([23], Theorem 7)
(34)ψnb+a2+Cb−a22n−β+1n−β+2≤2n−β−1Γn−β+1b−an−β−1nCDa+b/2−βψa+CDa+b/2+βψb≤ψnb+ψna2−Cb−a2n−βn−β+34n−β+1n−β+2.Remark 18.
IfC=0 and m=1 in (25), then we will get the fractional Hadamard inequality stated in Theorem 9.


## 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 19.
Letψ∈ACn+1a,b,a<b be a differentiable mapping on a,b. If ψn+1 is a strongly m-convex function on a,mb,m∈0,1, then the following inequality for Caputo fractional derivatives holds:
(35)ψna+ψnb2−Γn−β+12b−an−βDCa+βψb+−1nDCb−βψa≤b−a2n−β+11−12n−βψn+1a+mψn+1bm−Cmb−ab/m−a2n−β+2n−β+31−n−β+42n−β+2,with β>0.Proof.
Since∣ψ′∣ is strongly m-convex function on a,b, for z∈0,1, we have
(36)f′za+1−zb=f′za+m1−zbm≤zf′a+m1−zf′b/m−Cmz1−zbm−a2.
By applying Lemma7 and the strongly m-convexity of ψn+1, we find
(37)ψna+ψnb2−Γn−β+12b−an−βDCa+βψb+−1nDb−βψa≤b−a2∫011−zn−β−zn−βψn+1za+m1−zbmdz≤b−a2∫011−zn−β−zn−βzψn+1a+m1−zψn+1bm−Cmz1−zbm−a2dz≤b−a2∫01/21−zn−β−zn−βzψn+1a+m1−zψn+1bm−Cmz1−zbm−a2dz+∫1/21zn−β−1−zn−βzψn+1a+m1−zψn+1bm−Cmz1−zbm−a2dz.
In the following, we compute integrals appearing on the right side of inequality (37):
(38)∫01/21−zn−β−zn−βzψn+1a+m1−zψn+1bm−Cmz1−zbm−a2dz=ψn+1a∫01/2z1−zn−β−zn−β+1dz+mψn+1bm∫01/21−zn−β+1−1−zzn−βdz−Cmbm−a2∫01/2z1−zn−β+1dz−∫01/2zn−β+11−zdz=ψn+1a1n−β+1n−β+2−1/2n−β+1n−β+1+mψn+1bm1n−β+2−1/2n−β+1n−β+1−Cmb/m−a2n−β+2n−β+31−n−β+42n−β+2,(39)∫1/21zn−β−1−zn−βzψn+1a+m1−zψn+1bm−Cmz1−zbm−a2dz=ψn+1a∫1/21zn−β+1−z1−zn−βdz+mψn+1bm∫1/211−zzn−β−1−zn−β+1dz−Cmbm−a2∫1/211−zzn−β+1dz−∫1/21z1−zn−β+1dz=ψn+1a1n−β+2−1/2n−β+1n−β+1+mψn+1bm1n−β+1n−β+2−1/2n−β+1n−β+1−Cmb/m−a2n−β+2n−β+31−n−β+42n−β+2.
By putting the values of (38) and (39) in (37), we get (35).Corollary 20.
By settingm=1 in inequality (35), we will get ([23], Theorem 8)
(40)ψna+ψnb2−Γn−β+12b−an−βDCa+βψb+−1nDCb−βψa≤b−a2n−β+11−12n−βψn+1a+ψn+1b−Cb−a3n−β+2n−β+31−n−β+42n−β+2.Remark 21.
IfC=0 and m=1 in (35), we get the fractional Hadamard inequality which is stated in Theorem 8.
By using Lemma (11), we give the following Caputo fractional derivative inequality.Theorem 22.
Letψ∈ACn+1a,b, a<b, be a differentiable mapping on a,b. If ψn+1q is strongly m-convex function on a,mb,m∈0,1, for q≥1, then the following inequality for Caputo fractional derivatives holds:
(41)2n−β−1Γn−β+1mb−an−βDCa+bm/2+βψmb+mn−β+1−1nDCa+mb/2m−βψam−12ψna+mb2+mψna+mb2m≤mb−a4n−β+11/p12n−β+1n−β+21/qn−β+1ψn+1aq+n−β+3mψn+1bq−Cmb−a2n−β+1n−β+42n−β+31/q+n−β+3mψn+1am2q+n−β+1ψn+1bq−Cmb−a/m22n−β+1n−β+42n−β+31/q.Proof.
By takingk=1 in Lemma 10 and using the strongly m-convexity of ∣ψn+1∣, we have
(42)2n−β−1Γn−β+1mb−an−βDCa+bm/2+βψmb+mn−β+1−1nDCa+mb/2m−βψam−12ψna+mb2+mψna+mb2m≤mb−a4∫01zn−βψn+1z2a+m2−z2b+ψn+12−z2ma+z2bdz≤mb−a4ψn+1a+ψn+1b∫01zn−β+12dz+ψn+1am2+ψn+1b∫012−zzn−β2dz−Cm4b−a2−b−am2∫012−zzn−β+1dz≤mb−a4n−β+2ψn+1a+ψn+1b+mψn+1a/m2+ψn+1b2n−β+1−Cm4b−a2−b−am2n−β+4n−β+3.
Now, applying Lemma10, using power mean inequality, we have
(43)2n−β−1Γn−β+1mb−an−βDCa+bm/2+βψmb+mn−β+1−1nDCa+mb/2m−βψam−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.
By using stronglym-convexity of ψn+1q, we have
(44)≤mb−a4n−β+11/pψn+1aq∫01zn−β+12dz+mψn+1bq∫012zn−β−zn−β+12dz−Cmb−a24∫012zn−β+1−zn−β+2dz1/q+mψn+1am2q∫012zn−β−zn−β+12dz+ψn+1bq∫01zn−β+12dz−Cmb−a/m224∫012zn−β+1−zn−β+2dz1/q=mb−a4n−β+11/p12n−β+1n−β+21/qn−β+1ψn+1aq+n−β+3mψn+1bq−Cmb−a2n−β+1n−β+42n−β+31/q+n−β+3mψn+1am2q+n−β+1ψn+1bq−Cmb−a/m22n−β+1n−β+42n−β+31/q,inequality (41) is obtained.Corollary 23.
By settingm=1 in inequality (41), we will get ([23], Theorem 9)
(45)2n−β−1Γn−β+1b−an−βDCa+b/2+βψb+−1nCDa+b/2−βψa−ψna+b2≤b−a4n−β+11/p12n−β+1n−β+21/qn−β+1ψn+1aq+n−β+3ψn+1bq−Cb−a2n−β+1n−β+42n−β+31/q+n−β+3ψn+1aq+n−β+1ψn+1bq−Cb−a2n−β+1n−β+42n−β+31/q.Remark 24.
(i) IfC=0 and m=1 in (41), then we will get inequality stated in Theorem 11.Theorem 25.
Letψ∈ACn+1a,b,a<b, be a differentiable mapping on a,b. If ψn+1q is a strongly m-convex function on a,mb,m∈0,1, for q>1, then the following inequality for Caputo fractional derivatives holds:
(46)2n−β−1Γn−β+1mb−an−βCDa+bm/2+βψmb+mn−β+1−1nCDa+mb/2m−βψam−12ψna+mb2+mψna+mb2m≤bm−a164np−βp+11/pψn+1a+3m1/qψn+1bq−2Cmb−a231/q+3m1/qψn+1am2+ψn+1bq−2Cmb−a/m2231/q.Proof.
By takingk=1 in Lemma 10 and with the help of modulus property, we get
(47)2n−β−1Γn−β+1mb−an−βDCa+bm/2+βψmb+mn−β+1−1nDCa+mb/2m−βψam−12ψna+mb2+mψna+mb2m≤mb−a4∫01zn−βψn+1z2a+m2−z2bdz+∫01zn−βψn+12−z2ma+z2bdz.
Now applying Holder’s inequality, we get(48)2n−β−1Γn−β+1mb−an−βDa+bm/2+βψmb+mn−β+1−1nDa+mb/2m−βψam−12ψna+mb2+mψna+mb2m≤mb−a41np−βp+11/p∫01ψn+1z2a+m2−zb2dz1/q+∫01zn−αψn+12−z2ma+z2bdz1/q.
Using stronglym-convexity of ψn+1q, we get
(49)2n−β−1Γn−β+1mb−an−βDCa+bm/2+βψmb+mn−β+1−1nDCa+mb/2m−βψam−12ψna+mb2+mψna+mb2m≤mb−a41np−βp+11/pψn+1aq∫01z2dz+mψn+1bq∫012−z2dz−Cmb−a24∫012z−z2dz1/q+mψn+1am2q∫012−z2dz+ψn+1bq∫01z2dz−Cmb−a/m224∫012z−z2dz1/q=mb−a164np−βp+11/pψn+1aq+3mψn+1bq−2Cmb−a231/q+3mψn+1am2q+ψn+1bq−2Cmb−a/m2231/q≤bm−a164np−βp+11/pψn+1a+3m1/qψn+1bq−2Cmb−a231/q+3m1/qψn+1am2+ψn+1bq−2Cmb−a/m2231/q.
Here, we have used the fact thata1+b1q≥a1q+b1q, where q>1, a1,b1≥0. This completes the proof.Corollary 26.
By settingm=1 in inequality (46), we will get ([23], Theorem 10)
(50)2n−β−1Γn−β+1b−an−βDCa+b/2+βψb+−1nDCa+b/2−βψa−ψna+b2≤b−a164np−βp+11/pψn+1a+31/qψn+1bq−2Cb−a231/q+31/qψn+1a+ψn+1bq−2Cb−a231/q.Remark 27.
IfC=0 and m=1 in (46), then we will get inequality stated in Theorem 12.



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*Source: 6642655-2021-04-22.xml*