# Some Midpoint Inequalities forη-Convex Function via Weighted Fractional Integrals

Authors: Lei Chen; Waqas Nazeer; Farman Ali; Thongchai Botmart; Sarah Mehfooz
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this research, by using a weighted fractional integral, we establish a midpoint version of Hermite-Hadamrad Fejér type inequality forη-convex function on a specific interval. To confirm the validity, we considered some special cases of our results and relate them with already existing results. It can be observed that several existing results are special cases of our presented results.


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## Body

## 1. Introduction

In the last few decades, the classical convexity has a rapid development in fractional calculus [1]. We can say that convexity plays a vital role in fractional integral inequalities because of its geometric features [2–4].Take a functionf:I⟶ℝ be a continuous function. Then, this function is called convex if
(1)ftm+1−tn≤tfm+1−tfn,∀m,n∈I, and t∈0,1.There are many integral inequalities in the literature and one of the most common inequality is Hermite-Hadamarad or, shortly, the HH integral inequality, which is introduced by [5]:
(2)fm+n/2≤1/n−m∫mnfxdx≤fm+fn/2,m<n∈I.In the literature, we can notice that Hermite-Hadamarad inequality (2) has been applied to distinct convexities like exponential convexity [6, 7], s-convexity [8], quasiconvexity [9, 10], GA-convexity [11], α,m-convexity [12], MT-convexity [13], and also, other types of convexity (see [14, 15]). Different forms of fractional integrals like Riemann-Liouville (RL), Caputo Fabrizio, Hadamrad, Riesz, Prabhakar, Ψ-RL, and weighted integrals [16–20] have been established. A lot of integer-order integral inequalities like Simpson [21], Ostrowski [22], Rozanova [23], Gagliardo-Nirenberg [24], Olsen [25], Hardy [26], Opial [27, 28], and Akdemir et al. [29, 30] have been developed and generalized from fractional point of view.Definition 1.
LetI⊂ℝ be an interval and f:I⟶ℝ be a continuous function. Then, the function f is called η-convex if
(3)ftm+1−tn≤fn+tηfm,fn.Definition 2.
[18] Let f is positive convex function, continuous on closed interval m,n and x∈m,n when fx∈L1m,n with m<n, where left- and right-side RL fractional integrals are defined by
(4)RLIm+vfx=1Γv∫mxx−tv−1ftdt,RLIn−vfx=1Γv∫xnt−xv−1ftdt,where Γ is famous Gamma function and for any positive integer n,Γn=n−1!.Definition 3 (see [19]).
Letm,n⊆ℝ, f:m,n⟶ℝ and ϕ:m,n⟶ℝ be monotonically increasing positive function with a continuous derivative ϕ′x on m,n. Then, the left-sided and the right-sided weighted fractional integrals of f according to ϕ on m,n are defined by:
(5)Im+gv,ϕfx=gx−1Γv∫mxϕ′tϕx−ϕtv−1ftgtdt,In+gv,ϕfx=gx−1Γv∫xnϕ′tϕx−ϕtv−1ftgtdt,v>0.In this research, we denotegx−1=1/gx and the inverse of function ϕx by ϕ−1x.Remark 4.
From Definition3, we can see some special cases:
(i)
Ifϕx=x and gx=1, then weighted fractional integrals [14] deduce to the classical RL fractional integrals [9].(ii)
Ifgx=1, we get fractional integrals of function f with respect to function ϕx, which is defined by [16, 17]:
(6)Im+v:ϕfx=1Γv∫mxϕ′tϕx−ϕtv−1ftdt,n+Iv:ϕfx=1Γv∫xnϕ′tϕx−ϕtv−1ftdt,v>0.Lemma 5.
[31] Assume that g:m,n⟶0,∞ is integrable function and symmetric with respect to m+n/2,m<n. Then,
(i)
For eacht∈0,1, we have
(7)gt2m+2−t2n=g2−t2m+t2n.(ii)
Forv>0, we have
(8)Iϕ−1m+n/2+v:ϕgoϕϕ−1n=Iϕ−1m+n/2−goϕv:ϕϕ−1m=12Iϕ−1m+n/2+v:ϕgoϕϕ−1n=Iϕ−1m+n/2−goϕv:ϕϕ−1m.


## 2. Main Results

Theorem 6.
Let0≤m<n and f:m,n⟶ℝ be an L1η-convex function and g:m,n⟶ℝ be an integrable, positive and weighted symmetric function with respect to m+n/2. If, in addition, ϕ is an increasing and positive function from m,n onto itself such that its derivative ϕ′x is continues on m,n, then for v>0, the following inequalities are valid:
(9)fm+n2×Iϕϕ−1m+n/2+v:ϕgoϕϕ−1n+Iϕϕ−1m+n/2−v:ϕgoϕϕ−1m≤gnIϕ−1m+n/2+goϕv:ϕfoϕϕ−1n+η2gmft2m+2−t2n,f2−t2m+t2nϕ−1m.≤gnIϕ−1m+n/2+goϕv:ϕfoϕϕ−1n+η2gmft2m+2−t2n,f2−t2m+t2nϕ−1m.Proof.
Theη-convexity of f on m,n, for all x,y∈m,n gives
(10)fx+y2≤fy+ηfx,fy2,setting x=t/2m+2−t/2n and y=2−t/2m+t/2n(11)2fm+n2≤f2−t2m+t2n+ηft2m+2−t2n,f2−t2m+t2n.
Multiplying both sides of inequality (11) by tv−1gt/2m+2−t/2n and integrating over 0,1, we get
(12)2fm+n2∫01tv−1gt2m+2−t2ndt≤∫01tv−1gt2m+2−t2nf2−t2m+t2ndt+∫01ηtv−1gt2m+2−t2nft2m+2−t2n,f2−t2m+t2n.
From the left side of inequality (12), we use
(13)2v−1Γvn−mvIϕ−1m+n/2+v:ϕgoϕϕ−1n+Iϕϕ−1m+n/2−v:ϕgoϕϕ−1m=2vΓvn−mvIϕϕ−1m+n/2+v:ϕgoϕϕ−1n=2vn−mv∫ϕ−1m+n/2ϕ−1nn−ϕxv−1goϕxϕ′xdx=∫ϕ−1m+n/2ϕ−1n2n−ϕxn−mv−1goϕxϕ′x2dxn−m=∫01tv−1gt2m+2−t2ndt,where t=2n−ϕx/n−m. It follows that
(14)2fm+n2∫01tv−1gt2m+2−t2ndt=2vΓvn−mvfm+n2×Iϕϕ−1m+n/2+v:ϕgoϕϕ−1n+Iϕϕ−1m+n/2−v:ϕgoϕϕ−1m.
By evaluating the weighted fractional operators, we see that(15)gnIϕϕ−1m+n/2+goϕv:ϕfoϕϕ−1n+gmIgoϕϕϕ−1m+n/2−v:ϕfoϕϕ−1m=gngoϕ−1ϕ−1nΓv∫ϕ−1m+n/2ϕ−1nn−ϕxv−1foϕxgoϕxϕ′xdx+gmgoϕ−1ϕ−1nΓv∫ϕ−1mϕ−1m+n/2ϕx−mv−1foϕxgoϕxϕ′xdx=n−mv2vΓv∫ϕ−1m+n/2ϕ−1n2n−ϕxn−mv−1foϕxgoϕxϕ′x2dxn−m+n−mv2vΓv∫ϕ−1mϕ−1m+n/22ϕx−mn−mv−1foϕxgoϕxϕ′x2dxn−m,where
(16)goϕϕ−1y−1=1goϕϕ−1y=1gy,for y=m,n.
Settingu1=2n−ϕx/n−m and u2=2ϕx−m/n−m, one can deduce that
(17)gnIϕ−1m+n/2+goϕv:ϕfoϕϕ−1n+gmIgoϕϕ−1m+n/2−v:ϕfoϕϕ−1m=n−mv2vΓv∫01u1v−1fu12m+2−u12ngu12m+2−u12ndu1+∫01u2v−1f2−u22m+u22ng2−u22m+u22ndu2=n−mv2vΓv∫01tv−1ft2m+2−t2ngt2m+2−t2ndt+∫01tv−1f2−t2m+t2ngt2m+2−t2ndt,(18)∫01tv−1ft2m+2−t2ngt2m+2−t2ndt+∫01ηtv−1gt2m+2−t2nft2m+2−t2n,f2−t2m+t2n=2vΓvn−mvgnIϕ−1m+n/2+goϕv:ϕfoϕϕ−1n+ηgmft2m+2−t2n,f2−t2m+t2nϕ−1m.
By using (14) and (18) in (12), we get
(19)fm+n2×Iϕϕ−1m+n/2+v:ϕgoϕϕ−1n+Iϕϕ−1m+n/2−v:ϕgoϕϕ−1m≤gnIϕ−1m+n/2+goϕv:ϕfoϕϕ−1n+ηtgmft2m+2−t2n,f2−t2m+t2nϕ−1m.
The left side of Theorem6 is completed.
Now, we will prove right side of inequality (9) by using η-convexity.
(20)f2−t2m+t2n+tηft2m+2−t2n,f2−t2m+t2n≤fn+2−t2ηfm,fn+ηft2m+2−t2n,f2−t2m+t2n.
Multiply Equation (20) by tv−1gt/2m+2−t/2n and integrate over 0,1 leads us to
(21)∫01tv−1ft2m+2−t2ngt2m+2−t2ndt+∫01tv−1f2−t2m+t2ngt2m+2−t2ndt.≤fn+2−t2ηfm,fn+ηft2m+2−t2n,f2−t2m+t2n.×∫01tv−1gt2m+2−t2ndt.
By using (7) and (14) in (21), we get
(22)gnIϕ−1m+n/2+goϕv:ϕfoϕϕ−1n+ηtgmft2m+2−t2n,f2−t2m+t2nϕ−1m≤gnIϕ−1m+n/2+goϕv:ϕfoϕϕ−1n+η2gmft2m+2−t2n,f2−t2m+t2nϕ−1m.≤gnIϕ−1m+n/2+goϕv:ϕfoϕϕ−1n+η2gmft2m+2−t2n,f2−t2m+t2nϕ−1m.
This completes our proof.Remark 7.
From Theorem6, we can get following special case:
Ifϕx=x, then inequality (9) becomes
(23)fm+n2Im+n/2+RLvgn+IRLm+n/2−vgm≤gnIm+n/2+RLgvfn+gmη2IgRLm+n/2−vfm+η2ft2m+2−t2n,f2−t2m+t2n≤gnIm+n/2+RLgvfn+gmη2IgRLm+n/2−vfm+η2ft2m+2−t2n,f2−t2m+t2n.Lemma 8.
[31] Let 0≤m<n and f:m,n⟶ℝ be a continuous with a derivative f′∈L1m,n such that fx=fm+∫mxf′tdt and let g:m,n⟶ℝ be an integrable, positive, and weighted symmetric function with respect to m+n/2. If ϕ is a continuous increasing mapping form the interval m,n onto itself with a derivative ϕ′x which is continuous on m,n, then for v>0, the following equality is valid:
(24)fm+n2Iϕ−1m+n/2+v:ϕgoϕϕ−1n+Iϕ−1m+n/2−v:ϕgoϕϕ−1m−gnIϕ−1m+n/2+goϕv:ϕfoϕϕ−1n+gmIgoϕϕ−1m+n/2−v:ϕfoϕϕ−1m=1Γv∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕ′xϕx−mv−1goϕxdxf′oϕtϕ′tdt−1Γv∫ϕ−1m+n/2ϕ−1n∫tϕ−1nϕ′xn−ϕxv−1goϕxdxf′oϕtϕ′tdt.Remark 9.
From Lemma8, we obtain the following special case:
Ifϕx=x, then equality (24) becomes
(25)fm+n2Im+n/2+RLvgn+IRlm+n/2−vgm−gnIm+n/2+RLgvfn+gmIgRLm+n/2−vfm=1Γv∫mm+n/2∫mtx−mv−1gxdxf′tdt−1Γv∫m+n/2n∫tnn−xv−1gxdxf′tdt.Theorem 10.
Let0≤m<n and f:m,n⊆0,∞⟶ℝ be a differentiable function on the interval m,n such that fx=fm+∫mxf′tdt and let g:m,n⟶ℝ be an integrable, positive, and weighted symmetric function with respect to m+n/2. If, in addition, f′ is convex on m,n, and ϕ is an increasing and positive function from m,n onto itself such that its derivative ϕ′x is continuous on m,n, then for v>0, the following inequalities hold:
(26)∣σ1+σ2∣=1Γv∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕ′xϕx−mv−1goϕxdxf′oϕtϕ′tdt−1Γv∫ϕ−1m+n/2ϕ−1n∫tϕ−1nϕ′xn−ϕxv−1goϕxdxf′oϕtϕ′tdt≤n−mv+1Γv2v+1v+1gm,m+n/2,∞f′n+gm,m+n/2,∞f′n+n−mv+1Γv2v+2v+2gm,m+n/2,∞ηf′m,f′n+n−mv+1Γv2v+2v+1v+2gm,m+n/2,∞ηf′m,f′n≤2n−mv+1gm,n,∞Γv2v+1v+1f′n+n−mv+1gm,n,∞Γv2v+2v+1v+2+n−mv+1Γv2v+2v+2gm,n,∞ηf′m,f′n.Proof.
By using Lemma8 and properties of modulus, we get
(27)σ1+σ2=1Γv∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕ′xϕx−mv−1goϕxdxf′oϕtϕ′tdt+1Γv∫ϕ−1m+n/2ϕ−1n∫tϕ−1nϕ′xn−ϕxv−1goϕxdxf′oϕtϕ′tdt≤1Γv∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕ′xϕx−mv−1goϕxdxf′oϕtϕ′tdt+1Γv∫ϕ−1m+n/2ϕ−1n∫tϕ−1nϕ′xn−ϕxv−1goϕxdxf′oϕtϕ′tdt.
Sincef′ is η-convex on m,n for t∈ϕ−1m,ϕ−1n, so
(28)f′oϕt=f′n−ϕtn−mm+ϕt−mn−mn≤f′n+n−ϕtn−mηf′m,f′n.
So, using (28), we obtain
(29)σ1+σ1≤gm,m+n/2,∞Γv∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕ′xϕx−mv−1dx×f′n+n−ϕtn−mηf′m,f′nϕ′tdt+gm+n/2,n,∞Γv∫ϕ−1m+n/2ϕ−1n∫tϕ−1nϕ′xn−ϕxv−1dx×f′n+n−ϕtn−mηf′m,f′nϕ′tdt≤n−mv+1Γv+12v+1gm,m+n/2,∞f′n+gm,m+n/2,∞f′n+n−mv+1Γv+32v+2gm,m+n/2,∞ηf′m,f′n+n−mv+1Γv+22v+2v+1gm,m+n/2,∞ηf′m,f′n≤2n−mv+1gm,n,∞Γv+12v+1f′n+n−mv+1gm,n,∞Γv+22v+2v+1+n−mv+1gm,n,∞Γv+32v+2ηf′m,f′n,where
(30)∫ϕ−1mtϕ′xϕx−mv−1dx=ϕt−mvv,∫tϕ−1nϕ′xn−ϕxv−1dx=n−ϕtvv,∫ϕ−1mϕ−1m+n/2ϕt−mvn−ϕtϕ′tdt=n−mv+2v+32v+2v+1v+2,∫ϕ−1m+n/2ϕ−1nn−ϕtvn−ϕtϕ′tdt=n−mv+22v+2v+2,∫m+n/2ϕ−1nn−ϕtvϕ′tdt=∫ϕ−1mϕ−1m+n/2ϕt−mvϕ′tdt=n−mv+12v+1v+1.
This completes our proof.Remark 11.
From Theorem10, we can get following inequalities:
(1)
Ifϕx=x, then inequality (26) becomes
(31)fm+n2Im+n/2+RLvgn+RLIm+n/2−vgm−gnIm+n/2+RLgvfn+gmIgRLm+n/2−vfm≤n−mv+1Γv2v+1v+1gm,m+n/2,∞f′n+gm,m+n/2,∞f′n+n−mv+1Γv2v+2v+2gm,m+n/2,∞ηf′m,f′n+n−mv+1Γv2v+2v+1v+2gm,m+n/2,∞ηf′m,f′n≤2n−mv+1gm,n,∞Γv2v+1v+1f′n+n−mv+1gm,n,∞Γv2v+2v+1v+2+n−mv+1gm,n,∞Γv2v+2v+2ηf′m,f′n.(2)
Ifϕx=x and gx=1, then inequality (26) becomes
(32)2v−1Γv+1n−mvIm+n/2+RLvfn+RLIm+n/2−vfm−fm+n2≤n−mv+1Γv2v+1v+1gm,m+n/2,∞f′n+gm,m+n/2,∞f′n+n−mv+1Γv2v+2v+2gm,m+n/2,∞ηf′m,f′n+n−mv+1Γv2v+2v+1v+2gm,m+n/2,∞ηf′m,f′n≤2n−mv+1gm,n,∞Γv2v+1v+1f′n+n−mv+1gm,n,∞Γv2v+2v+1v+2+n−mv+1gm,n,∞Γv2v+2v+2ηf′m,f′n.(3)
Ifϕx=x, gx=1 and v=1, then inequality (26) becomes
(33)1n−m∫mnfxdx−fm+n2≤n−m24f′n+n−m2+2n−m248ηf′m,f′n.Theorem 12.
Let0≤m≤n and f:m,n⊆0,∞⟶ℝ be a continuously differentiable function on the interval m,n such that fx=fm+∫mxf′tdt, and let g:m,m⟶ℝ be integrable, positive, and weighted symmetric function with respect to m+n/2. If, in addition, f′q is convex on m,n,q≤1, and ϕ is increasing and positive function from m,n onto itself such that its derivative ϕ′x is continuous on m,m, then for v>0, we have:
(34)σ1+σ2≤n−mv+1Γv+12v+1+1/qgm,m+n/2,∞f′nq1/q+gm,m+n/2,∞f′nq1/q+n−mv+1Γv+32v+2+1/qgm,m+n/2,∞ηf′mq,f′nq1/q+n−mv+1Γv+22v+2+1/qv+11/qgm,m+n/2,∞ηf′mq,f′nq1/q≤2n−mv+1gm,n,∞Γv+12v+1+1/qf′nq1/q+n−mv+1gm,n,∞Γv+22v+2+1/qv+11/q+n−mv+1gm,n,∞Γv+32v+2+1/qηf′m,f′n1/q.Proof.
Sincef′q is η-convex on m,n for t∈ϕ−1m,ϕ−1n, so
(35)f′oϕtq=f′n−ϕtn−mm+ϕt−mn−mnq≤f′nq+n−ϕtn−mηf′mq,f′nq.
By using power mean integral, Lemma8, and η-convexity of f′q, we have
(36)σ1+σ2≤1Γv∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕ′xϕx−mv−1goϕxdxf′oϕtϕ′tdt+1Γv∫ϕ−1m+n/2ϕ−1n∫tϕ−1nϕ′xn−ϕxv−1goϕxdxf′oϕtϕ′tdt≤1Γv∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕ′xϕx−mv−1goϕxdxϕ′tdt1−1/q×∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕ′xϕx−mv−1goϕxdxf′oϕtqϕ′tdt1/q+1Γv∫ϕ−1m+n/2ϕ−1n∫tϕ−1nϕ′xn−ϕxv−1goϕxdxϕ′tdt1−1/q×∫ϕ−1m+n/2ϕ−1n∫tϕ−1nϕ′xn−ϕxv−1goϕxdxf′oϕtqϕ′tdt1/q≤gm,m+n/2,∞Γv∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕ′xϕx−mv−1dxϕ′tdt1−1/q×∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕ′xϕx−mv−1dxf′oϕtqϕ′tdt1/q+gm,m+n/2,∞Γv∫ϕ−1m+n/2ϕ−1n∫tϕ−1mϕ′xn−ϕxv−1dxϕ′tdt1−1/q×∫ϕ−1m+n/2ϕ−1n∫tϕ−1nϕ′xn−ϕxv−1dxf′oϕtqϕ′tdt1/q≤gm,m+n/2,∞Γv∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕ′xϕx−mv−1dxϕ′tdt1−1/q×∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕ′xϕx−mv−1dx×f′nq+n−ϕtn−mηf′mq,f′nqϕ′tdt1/q+gm,m+n/2,∞Γv×∫ϕ−1m+n/2ϕ−1n∫tϕ−1mϕ′xn−ϕxv−1dxϕ′tdt1−1/q×∫ϕ−1m+n/2ϕ−1n∫tϕ−1nϕ′xn−ϕxv−1dx×f′nq+n−ϕtn−mηf′mq,f′nqϕ′tdt1/q=n−mv+1Γv+12v+1+1/qgm,m+n/2,∞f′nq1/q+gm,m+n/2,∞f′nq1/q+n−mv+1Γv+32v+2+1/qgm,m+n/2,∞ηf′mq,f′nq1/q+n−mv+1Γv+22v+2+1/qv+11/qgm,m+n/2,∞ηf′mq,f′nq1/q≤2n−mv+1gm,n,∞Γv+12v+1+1/qf′nq1/q+n−mv+1gm,n,∞Γv+22v+2+1/qv+11/q+n−mv+1gm,n,∞Γv+32v+2+1/qηf′m,f′n1/q,where
(37)∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕt−mv−1n−ϕtϕ′tdt=n−mv+2v+32v+2v+1v+2,∫ϕ−1m+n/2ϕ−1n∣∫tϕ−1nn−ϕtvn−ϕt∣ϕ′tdt=n−mv+22v+2v+2.Remark 13.
From Theorem12, we can get following special cases:
(1)
Ifϕx=x, then inequality (34) becomes
(38)fm+n2Im+n/2+RLvgn+IRLm+n/2−vgm−gnIm+n/2+RLgvfn+gmIgRLm+n/2−vfm≤n−mv+1Γv+12v+1+1/qgm,m+n/2,∞f′nq1/q+gm,m+n/2,∞f′nq1/q+n−mv+1Γv+32v+2+1/qgm,m+n/2,∞ηf′mq,f′nq1/qn−mv+1Γv+22v+2+1/qv+11/qgm,m+n/2,∞ηf′mq,f′nq1/q≤2n−mv+1gm,n,∞Γv+12v+1+1/qf′nq1/qn−mv+1gm,n,∞Γv+22v+2+1/qv+11/q+n−mv+1gm,n,∞Γv+32v+2+1/qηf′m,f′n1/q.(2)
Ifϕx=x and gx=1, then inequality (34) becomes
(39)2v−1Γv+1n−mvIm+n/2+RLvfn+IRLm+n/2−vfm−fm+n2≤2n−mv+1gm,n,∞Γv+12v+1+1/qf′nq1/q+n−mv+1gm,n,∞Γv+22v+2+1/qv+11/qn−mv+1gm,n,∞Γv+32v+2+1/qηf′m,f′n1/q.(3)
Ifϕx=x, gx=1 and v=1, then inequality (34) becomes
(40)1n−m∫mnfxdx−fm+n2≤2n−m222+1/qf′nq1/q+n−m2+21/qn−m221/q33+1/qηf′m,f′n1/q.Theorem 14.
Let0≤m≤n and f:m,n⊆0,∞⟶ℝ be a continuously differentiable function on the interval m,n such that fx=fm+∫mxf′tdt, and let g:m,m⟶ℝ be integrable, positive and weighted symmetric function with respect to m+n/2. If, in addition, f′q is convex on m,n,q≤1, and ϕ is increasing and positive function from m,n onto itself such that its derivative ϕ′x is continuous on m,m, then for v>0, we have
(41)σ1+σ2≤n−mv+1Γv+12v+1+2/qgm,m+n/2,∞f′nq1/q+gm,m+n/2,∞f′nq1/q+n−mv+1Γv+32v+2+2/qgm,m+n/2,∞ηf′mq,f′nq1/q+n−mv+1Γv+22v+2+2/qpv+11/pgm,m+n/2,∞ηf′mq,f′nq1/q≤2n−mv+1gm,n,∞Γv+12v+1+2/qf′nq1/q+n−mv+1gm,n,∞Γv+22v+2+2/qpv+11/p+n−mv+1gm,n,∞Γv+32v+2+2/qηf′m,f′n1/q.Proof.
Sincef′q is η-convex on m,n, for t∈ϕ−1m,ϕ−1n, we get
(42)f′oϕtq=f′n−ϕtn−mm+ϕt−mn−mnq≤f′nq+n−ϕtn−mηf′mq,f′nq.
By using Hölder’s inequality, Lemma8, η-convexity of f′q, and properties of modulus, we get
(43)σ1+σ2≤1Γv∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕ′xϕx−mv−1goϕxdxf′oϕtϕ′tdt+1Γv∫ϕ−1m+n/2ϕ−1n∫tϕ−1nϕ′xn−ϕxv−1goϕxdxf′oϕtϕ′tdt≤1Γv∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕ′xϕx−mv−1goϕxdxpϕ′tdt1/p×∫ϕ−1mϕ−1m+n/2f′oϕtqϕ′tdt1/q+1Γv∫ϕ−1m+n/2ϕ−1n∫tϕ−1nϕ′xn−ϕxv−1goϕxdxpϕ′tdt1/p×∫ϕ−1m+n/2ϕ−1nf′oϕtqϕ′tdt1/q≤gm,m+n/2,∞Γv×∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕ′xϕx−mv−1dxpϕ′tdt1/p×∫ϕ−1mϕ−1m+n/2f′oϕtqϕ′tdt1/q+gm,m+n/2,∞Γv×∫ϕ−1m+n/2ϕ−1n∫tϕ−1mϕ′xn−ϕxv−1dxpϕ′tdt1/p×∫ϕ−1m+n/2ϕ−1nf′oϕtqϕ′tdt1/q≤gm,m+n/2,∞Γv∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕ′xϕx−mv−1dxpϕ′tdt1/p×∫ϕ−1mϕ−1m+n/2f′nq+n−ϕtn−mηf′mq,f′nqϕ′tdt1/q+gm,m+n/2,∞Γv×∫ϕ−1m+n/2ϕ−1n∫tϕ−1mϕ′xn−ϕxv−1dxpϕ′tdt1/p×∫ϕ−1m+n/2ϕ−1nf′nq+n−ϕtn−mηf′mq,f′nqϕ′tdt1/q=n−mv+1Γv+12v+1+2/qgm,m+n/2,∞f′nq1/q+gm,m+n/2,∞f′nq1/q+n−mv+1Γv+32v+2+2/qgm,m+n/2,∞ηf′mq,f′nq1/q+n−mv+1Γv+22v+2+2/qpv+11/pgm,m+n/2,∞ηf′mq,f′nq1/q≤2n−mv+1gm,n,∞Γv+12v+1+2/qf′nq1/q+n−mv+1gm,n,∞Γv+22v+2+2/qpv+11/p+n−mv+1gm,n,∞Γv+32v+2+2/qηf′m,f′n1/q,where
(44)∫ϕ−1mϕ−1m+n/2∫ϕ−1mtϕt−mv−1n−ϕtpϕ′tdt=n−mpv+2pv+32pv+2pv+1pv+2,∫ϕ−1m+n/2ϕ−1n∫tϕ−1nn−ϕtvn−ϕtpϕ′tdt=n−mpv+22pv+2pv+2.
This completes the proof.Remark 15.
From Theorem14, we can obtain following special cases:
(1)
Ifϕx=x, then inequality (41) becomes
(45)fm+n2Im+n/2+RLvgn+IRLm+n/2−vgm−gnIm+n/2+RLgvfn+gmIgRLm+n/2−vfm≤n−mv+1Γv+12v+1+2/qgm,m+n/2,∞f′nq1/q+gm,m+n/2,∞f′nq1/q+n−mv+1Γv+32v+2+2/qgm,m+n/2,∞ηf′mq,f′nq1/q+n−mv+1Γv+22v+2+2/qpv+11/pgm,m+n/2,∞ηf′mq,f′nq1/q≤2n−mv+1gm,n,∞Γv+12v+1+2/qf′nq1/q+n−mv+1gm,n,∞Γv+22v+2+2/qpv+11/p+n−mv+1gm,n,∞Γv+32v+2+2/qηf′m,f′n1/q.(2)
Ifϕx=x and gx=1, then inequality (41) becomes
(46)2v−1Γv+1n−mvIm+n/2+RLvfn+IRLm+n/2−vfm−fm+n2≤2n−mv+1gm,n,∞Γv+12v+1+2/qf′nq1/q+n−mv+1gm,n,∞Γv+22v+2+2/qpv+11/p+n−mv+1gm,n,∞Γv+32v+2+2/qηf′m,f′n1/q.(3)
Ifϕx=x, gx=1 and v=1, then inequality (41) becomes
(47)1n−m∫mnfxdxs−fm+n2≤2n−m223+2/qf′nq1/q+n−m2+p+11/pn−m2p+11/p23+2/qηf′m,f′n1/q.


## 3. Conclusion

In this paper, we established Hermite-Hadamarad Fejér type inequalities forη-convex function by using weighted fractional integrals. Our results are extensions and generalizations of many existing results in the literature.



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*Source: 1652888-2022-01-12.xml*