# Hermite–Hadamard Inequalities for Harmonics,m-Convex Functions

Authors: Jian Zhong Xu; Umar Raza; Muhammad Waqas Javed; Zaryab Hussain
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

The objective of this article is to establish some Hermite–Hadamard-type inequalities via harmonics,m-convex functions in the framework of conformal fractional integral.


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## Body

## 1. Introduction and Preliminaries

LetC⊂R be an interval. Then, C is said to be convex, if(1)1−tu+tv∈C,holds ∀u,v∈C and t∈0,1.LetC⊂R be an interval. Then, a function f:C⟶R is said to be convex (concave), if(2)f1−tu+tv≤≥1−tfu+tfv,holds ∀u,v∈C and t∈0,1.It can be easily seen in [1–7] that the convex (concave) functions have extensive applications in pure and applied mathematics, and in the literature [8–15], many eminent inequalities and other properties can be found in the framework of convexity. One of the renowned inequalities in the literature of Hermite–Hadamard Integral Inequality is given below:(3)fu+v2≤1v−u∫uvfxx2dx≤fu+fv2.These both inequalities hold in reverse if the function is concave. Now, the harmonic convex set is defined as follows.Definition 1.
LetC⊂R be an interval. Then, C is said to be harmonic convex, if(4)uv1−tu+tv∈C,holds ∀u,v∈C with u,v≠0,0 and t∈0,1.
Iscan [8] introduced the concept of harmonic convex function.Definition 2.
(see [8]). Let C⊂R be an interval. Then, a function f:C⟶R is called harmonic convex (concave), if(5)fuv1−tu+tv≤≥tfu+1−tfv,holds for all u,v∈C with u,v≠0,0 and t∈0,1.
In [8], Iscan by using the concept of harmonic convex function gave a new refinement of Hermite–Hadamard inequality as(6)f2uvu+v≤uvv−u∫uvfxx2dx≤fu+fv2.Definition 3.
LetC⊂R be an interval and s,m∈0,1. Then, a function f:C⟶R is called harmonic s,m-convex (concave), if(7)fmuv1−tu+mtv≤≥tsfu+m1−tsfv,holds ∀u,v∈C with u,v≠0,0 and t∈0,1.
Ifs=m=1, then harmonic s,m-convex function becomes the classical harmonic convex function. So harmonic convex function is a special case of harmonic s,m-convex function.
The main purpose of this article is to establish some conformable fractional estimates of Hermite–Hadamard-type inequalities via harmonics,m-convex functions. Before going further towards our main results, let us have a brief review of the previously well known concepts and results. These preliminaries will be highly helpful in acquiring the main results.
The eminent gamma and beta functions are defined as(8)Γu=∫0∞e−ttu−1dt,foru>0,βu,v=∫01tu−11−tv−1dt=ΓuΓvΓu+v,foru,v>0.
The integral form of hypergeometric function is defined as(9)F21u,v;w;z=1βv,w−v∫01tv−11−tw−v−11−zt−udt,for z≥1.
Now, iff∈L1u,v with u≥0, then Riemann–Liouville integrals Iu+αf and Iv−αf of any positive order α are defined as(10)Iu+αfa=1Γα∫0aa−tα−1ftdt,a>u,Iv−αfa=1Γα∫avt−aα−1ftdt,a<v.
For more details, see [11].
Recently, Abdeljawad [16] introduced the notation of right and left conformable fractional integrals for any positive order α as follows.Definition 4.
(see [16]). Let α∈n,n+1. Then, the left and right conformable fractional integrals starting from u of any positive order α is given as(11)Iαuft=1n!∫utt−ana−uα−n−1fada,Iαvft=1n!∫tva−tnv−aα−n−1fada.Lemma 1.
(see [5]). Let f:C=u,v⊂R\0⟶R be differentiable on Co,u,v∈C, and f′∈Lu,v. Then,(12)fu+fv2−uvv−u∫uvfxx2dx≤uvv−u2∫011−2ttv+1−tu2f′uvtv+1−tudt.


## 2. Main Results

In this section, we will present our main results.Theorem 1.
Letf:C=a,b⊂R\0⟶R be a harmonic (s, m)-convex function such that f∈L1u,v and s,m∈0,1. Then,(13)f2uvu+v≤12suvv−u∫uvfxx2dx+m∫uvfx/mx2dx≤12sfx+mfym.Proof.
By applying the definition of harmonic (s, m)-convex function for t=1/2, we have(14)fxyx+y≤12sfx+mfym.
Putx=uv/tu+1−tv and y=uv/tv+1−tu. Then,(15)f2uvu+v≤12sfuvtu+1−tv+mfuvmtv+1−tu,≤12s∫01fuvtu+1−tvdt+m∫01fuvmtv+1−tudt.
We know that(16)∫0112sfuvtu+1−tvdt=uvv−u∫uvfxx2dx.⇒f2uvu+v≤12suvv−u∫uvfxx2dx+m∫uvfx/mx2dx.
Now, consider a functionf:R⟶R such that fx=0. Then,(17)fmxymty+1−tx=0.
Also,(18)tsfx+m1−tsfy=0.
Sof is harmonic (s, m)-convex, and also, we have(19)f2uvu+v=0,12suvv−u∫uvfxx2dx+m∫uvfx/mx2dx=0,which implies that the inequality holds.Theorem 2.
Letf:C=u,v⊂R\0⟶R be a harmonic (s, m)-convex function such that f∈L1u,v, where s,m∈0,1. Then,(20)Γα−nΓα+1f2uvu+v≤12suvv−uαIα1/vfog1u+mIα1/ufog1v.Here, fog is the composition function.Proof.
From inequality 1, we have(21)f2xyx+y≤12sfx+mfym.
Putx=uv/tu+1−tv and y=uv/tv+1−tu. Then,(22)f2uvu+v≤12sfuvtu+1−tv+mfuvmtv+1−tu,(23)1n!f2uvu+v∫01tn1−tα−n−1dt≤1n!∫01tn1−tα−n−1fuvtu+1−tvdt,+mn!∫01tn1−tα−n−1fuvmtv+1−tudt=J1+J2.
By using change of variable technique of integration, we have(24)J1=1n!∫01tn1−tα−n−1fuvtu+1−tvdt=uvv−uαIα1/vfog1u,where gx=1/x, and(25)J2=mn!∫01tn1−tα−n−1fuvmtv+1−tudt=muvv−uαIα1/ufog1v,where gx=1/x.
Now, from (23), we have(26)Γα−nΓα+1f2uvu+v≤12suvv−uαIα1/vfog1u+mIα1/ufog1v.
So the inequality holds.Theorem 3.
Letf:C=u,v⊂R\0⟶R be differentiable on Co,u,v/m∈C,m∈0,1, and f′∈Lu,v. If f′q for q≥1 is harmonic s,m-convex on u,v/m, then(27)fu+fv2−uvv−u∫uvfxx2dx≤uvv−u22−1/qΩ1f′uq+mΩ2f′v/mq1/q,where(28)Ω1=∫01tstv+1−tu2dt=u−2s+1F212,s+1;s+2;1−vu,Ω2=∫011−tstv+1−tu2dt=−u−2s+1F212,1;s+2;1−vu.Proof.
Ho..lder’s inequality and Lemma 1 implies that(29)fu+fv2−uvv−u∫uvfxx2dx≤uvv−u2∫011−2ttv+1−tu2f′uvtv+1−tudt≤uvv−u2∫011−2tdt1−1/q×∫011tv+1−tu2f′uvtv+1−tuqdt1/q.
Sincef′q is harmonically s,m-convex on u,v/m, we have(30)≤uvv−u2121−1/q∫011tv+1−tu2tsf′uq+mf′vmq1/qdt,≤uv22−1/q∫01tstv+1−tu2dtf′aq+m∫011−tstv+1−tu2dtf′vmq1/q,=uvv−u22−1/qΩ1f′uq+mΩ2f′vmq1/q.Here,(31)Ω1=∫01tstv+1−tu2dt=u−2s+1F122,s+1;s+2;1−v/u,Ω2=∫011−tstv+1−tu2dt=−u−2s+1F122,1;s+2;1−v/u.Theorem 4.
Letf:C=u,v⊂R\0⟶R be differentiable on Co,u,v/m∈C,m∈0,1, and f′∈Lu,v. If f′q for q≥1 is harmonic s,m-convex on u,v/m, then(32)fu+fv2−uvv−u∫uvfxx2dx≤uvv−u2u−2F212,1;2;1−vu1−1/q×−Ψf′uq+mΨf′vmq1/q,where Ψ=s/s+1s+2.Proof.
Ho..lder’s inequality and Lemma 1 implies that(33)fu+fv2−uvv−u∫uvfxx2dx≤uvv−u2∫011−2ttv+1−tu2f′uvtv+1−tudt≤uvv−u2∫01dttv+1−tu21−1/q×∫011−2tf′uvtv+1−tuqdt1/q.
Sincef′q is harmonically s,m-convex on u,v/m, we have(34)≤uvv−u2u−2F212,1;2;1−vu1−1/q×∫011−2ttsf′uq+m1−tsf′vmqdt1/q,≤uvv−u2v−2F212,1;2;1−vu1−1/q×∫011−2ttsdtf′uq+m∫011−2t1−tsdtf′vmq1/q,≤uvv−u2u−2F212,1;2;1−vu1−1/q×−ss+1s+2f′uq+mss+1s+2f′vmq1/q,≤uvv−u2u−2F212,1;2;1−vu1−1/q×−Ψf′aq+mΨf′vmq1/q,where Ψ=s/s+1s+2.Theorem 5.
Letf:I=u,v⊂R∖0⟶R be differentiable on Co,u,v/m∈C,m∈0,1, and f′∈Lu,v. If f′q for q≥1 is harmonic s,m-convex on u,v/m, then(35)fu+fv2−uvv−u∫uvfxx2dx≤uvv−u21p+11/p×Φ1f′uq+mΦ2f′vmq1/q,where(36)Φ1=∫01tstv+1−tu2qdt=u−2qs+1F212q,s+1;s+2;1−vu,Φ2=∫011−tstv+1−tu2qdt=−u−2qs+1F212q,1;s+2;1−vu.Proof.
Ho..lder’s inequality and Lemma 1 implies that(37)fu+fv2−uvv−u∫uvfxx2dx≤uvv−u2∫011−2ttv+1−tu2f′uvtv+1−tudt,≤uvv−u2∫011−2tp1/p×∫011tv+1−tu2qf′uvtv+1−tuqdt1/q.
Sincef′q is harmonically s,m-convex on u,v/m, we have(38)≤uvv−u21p+11/p×tsf′uq+m1−tsf′vmqdt1/q,≤uvv−u21p+11/p×∫01tstv+1−tu2qdtf′uq+m∫011−tstv+1−tu2qdtf′vmq1/p,≤uvv−u21p+11/p×Φ1f′uq+mΦ2f′vmq1/p,where(39)Φ1=∫01tstv+1−tu2qdt=u−2qs+1F212q,s+1;s+2;1−vu,Φ2=∫011−tstv+1−tu2qdt=−u−2qs+1F212q,1;s+2;1−vu.
This completes our arguments.


## 3. Conclusion

In this article, we have established new Hermite–Hadamard-type inequalities via harmonics,m-convex functions in the framework of fractional integrals. This article may be useful for the researchers to derive some other type of inequalities.



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