# Hermite–Hadamard-Type Inequalities forF-Convex Functions via Katugampola Fractional Integral

Authors: Erhan Set; İlker Mumcu
Journal: Mathematical Problems in Engineering
 (2021)

---

## Abstract

This article is organized as follows: First, definitions, theorems, and other relevant information required to obtain the main results of the article are presented. Second, a new version of the Hermite–Hadamard inequality is proved for the F-convex function class using a fractional integral operator introduced by Katugampola. Finally, new fractional Hermite–Hadamard-type inequalities are given with the help of F-convexity.


---

## Body

## 1. Introduction and Preliminaries

First of all, let us recall the concept of convex function that is the fundamental notation of convex analysis.Definition 1.
The functionℏ:u,v⊆ℝ⟶ℝ is said to be convex if(1)ℏsx+1−sy≤sℏx+1−sℏy,for all x,y∈u,v and s∈0,1. We say that ℏ is concave if −ℏ is convex.
There are many inequalities in the literature for convex functions. But, among these inequalities, perhaps the one which takes the most attention of researchers is the Hermite–Hadamard inequality on which hundreds of studies have been conducted. The classical Hermite–Hadamard integral inequalities are as the follows.Theorem 1.
Assume thatℏ:I⊆ℝ⟶ℝ is a convex mapping defined on the interval I of ℝ where u<v. The statement(2)ℏu+v2≤1v−u∫uvℏxdx≤ℏu+ℏv2holds and known as Hermite–Hadamard inequality. Both inequalities hold in the reversed direction if f is concave.
Convex functions played a significant role in several areas such as engineering, finance, statistics, optimization, and mathematical inequalities. Convex functions have a paramount history and have been an intense study issue for over a century in mathematics. Various generalizations, extensions, and variants of the convex functions have been presented by many researchers. Recently, one of them has been introduced by Samet [1] as follows:
Letℱ be a family of a mappings ℱ:ℝ×ℝ×ℝ×0,1⟶ℝ that satisfy the following axioms:(A1)
Ifτi∈L10,1, i=1,2,3, then for every μ∈0,1, we have(3)∫01Fτ1s,τ2s,τ3sds=F∫01τ1sds,∫01τ2sds,∫01τ3sds,μ.(A2)
For everyτ∈L10,1, ψ∈L∞0,1 and z1,z2∈ℝ2, we have(4)∫01Fψsτs,ψsz1,ψsz2,sds,=TF,ψ∫01ψsτsds,z1,z2,whereTF,ψ:ℝ×ℝ×ℝ⟶ℝ is a function that depends on F,ψ, and it is nondecreasing with respect to the first variable.(A3)
For anyψ,τ1,τ2,τ3∈ℝ4, τ4∈0,1, we have(5)ψFτ1,τ2,τ3,τ4=Fψτ1,ψτ2,ψτ3,τ4+Lψ,where Lψ∈ℝ is a constant that depends only on ψ.Definition 2.
Letf:u,v⟶ℝ, u,v∈ℝ2, u<v, be a given function. We say that f is a convex function with respect to some F∈ℱ (or F-convex function) if(6)Ffsx+1−sy,fx,fy,s≤0,x,y,s∈u,v×u,v×0,1.Remark 1.
Letε≥0, and let f:u,v⟶ℝ, u,v∈ℝ2, u<v be an ε-convex function, that is,(7)fsx+1−sy≤sfx+1−sfy+ε,x,y,s∈u,v×u,v×0,1.
We define the functionsF:ℝ×ℝ×ℝ×0,1⟶ℝ by(8)Fτ1,τ2,τ3,τ4=τ1−τ4τ2−1−τ4τ3−εand TF,ψτ1,τ2,τ3ℝ×ℝ×ℝ⟶ℝ by(9)TF,ψτ1,τ2,τ3=τ1−∫01sψsdsτ2−∫011−sψsdsτ3−ε.
For(10)Lψ=1−ψε,it is explicit that F∈ℱ and(11)Ffsx+1−sy,fx,fy,s=fsx+1−sy−sfx−1−sfy−ε≤0,that is, f is an F-convex function. Particularly, taking ε=0, we show that if f is a convex function, then f is an F-convex function with respect to F defined above.Remark 2.
Letf:u,u⟶ℝ, u,v∈ℝ2, u<v be an α-convex function, α∈0,1, that is,(12)fsx+1−sy≤sαfx+1−sαfy,x,y,s∈u,v×u,v×0,1.
We define the functionsF:ℝ×ℝ×ℝ×0,1⟶ℝ by(13)Fτ1,τ2,τ3,τ4=τ1−τ4ατ2−1−τ4ατ3and TF,ψτ1,τ2,τ3:ℝ×ℝ×ℝ⟶ℝ by(14)TF,ψτ1,τ2,τ3=τ1−∫01sαψsdsτ2−∫011−sαψsdsτ3.
ForLψ=0, it is clear that F∈ℱ and(15)Ffsx+1−sy,fx,fy,s=fsx+1−sy−sαfx−1−sαfy≤0,that is, f is an F-convex function.Remark 3.
Leth:J⟶0,∞ be a given function which is not identical to 0, where J is a interval in ℝ such that 0,1⊆J. Let f:u,v⟶0,∞, u,v∈ℝ2, u<v be an h-convex function, that is,(16)fsx+1−sy≤hsfx+h1−sfy,x,y,s∈u,v×u,v×0,1.
We define the functionsF:ℝ×ℝ×ℝ×0,1⟶ℝ by(17)Fτ1,τ2,τ3,τ4=τ1−hτ4τ2−h1−τ4τ3and TF,ψτ1,τ2,τ3ℝ×ℝ×ℝ⟶ℝ by(18)TF,ψτ1,τ2,τ3=τ1−∫01hsψsdsτ2−∫01h1−sψsdsτ3.
ForLψ=0, it is clear that F∈ℱ and(19)Ffsx+1−sy,fx,fy,s=fsx+1−sy−hsfx−h1−sfy≤0,that is, f is an F-convex function.
We define the beta function as ([2], p18)(20)Bu,v=ΓuΓvΓu+v=∫01su−11−sv−1ds,u,v>0,where Γ is the gamma function.
The incomplete Beta function is defined by(21)Bxu,v=∫0xsu−11−sv−1ds,u,v>0,0≤x≤1.Definition 3.
Letf∈L1a,b. The Riemann–Liouville integrals Ja+μf and Jb−μf of order α>0 are defined by(22)Ja+μfx=1Γμ∫axx−sμ−1fsds,x>aJb−μfx=1Γμ∫xbs−xμ−1fsds,x<b,respectively, where Γμ=∫0∞e−suμ−1du. Here, Ja+0fx=Jb−0fx=fx.
Forμ=1, the fractional integral reduces to a classical integral.
Fractional calculus has the great impact in pure and applied sciences. In recent years, fractional analysis has become one of the most frequently used methods to obtain new and different versions of the results available in the literature. In [3], by using Riemann–Liouville fractional integrals, a new version of Hermite–Hadamard’s inequalities was proved by Budak et al. for F-convex function classes as follows.Theorem 2.
LetI⊆ℝ be an interval and f:I⟶ℝ be a differentiable mapping on I,a,b∈Io,a<b. If it is F−convex on a,b, for some F∈ℱ, then we have(23)Ffa+b2Γα+1b−aαJa+αfb,Γα+1b−aαJb−αfa,12+∫01Lψsds≤0and(24)TF,ψΓα+1b−aαJa+αfb+Jb−αfa,fa+fb,fa+fb+∫01Lψsds≤0,where ψs=αsα−1.
For results related to F-convex functions, one can see [1, 3–5].
Here, we present some definitions of fractional integrals.Definition 4 (see [6]).
Leta,b0≤a<b<∞ be a finite or infinite interval on the half-axis ℝ+. The Hadamard fractional integrals (left sided and right sided) of order α∈ℂ, Reα>0 of a real function f∈La,b are defined by(25)Ha+αfx≔1Γα∫axlogxsα−1fssds,a<x<band(26)Hb−αfx≔1Γα∫xblogsxα−1fssds,a<x<b.
We consider the spaceXcpa,bc∈ℝ,1≤p≤∞, 1≤p≤∞ consists of those complex-valued Lebesque measurable functions φ on a,b for which φXcp<∞, with(27)φXcp=∫abxcφxpdxx1/p,1≤p<∞,φXcp=esssupx∈a,bxcφx.
A new fractional integral operator, which is a generalization of Riemann–Liouville and Hadamard fractional integrals to a single form, is introduced by Katugampola as follows.Definition 5 (see [7]).
Leta,b⊂ℝ be a finite interval. Then, the left- and right-side Katugampola fractional integrals of order α>0 of f∈Xcpa,b are defined:(28)Ia+αρfx=ρ1−αΓα∫axsρ−1xρ−sρ1−αfsds,Ib−αρfx=ρ1−αΓα∫xbsρ−1sρ−xρ1−αfsds,with a<x<b and ρ>0, if the integral exists.
If we take asq = 1 in this definition, the Riemann–Liouville fractional integral operator that is well known in the literature used to describe Riemann–Liouville and Caputo fractional derivatives is obtained [6, 8, 9]. Using L’Hospital rule, when ρ⟶0+, we have Hadamard fractional integrals of (25) and (26).
For results associated with Katugampola fractional operators, we refer the reader to the some recent papers (see [7, 10, 11]).
Motivated from the studies presenting Hermite–Hadamard-type inequalities obtained with the help of fractional integral operators for the F-convexity class, to obtain more general and new versions of Hermite–Hadamard-type inequalities by using Katugampola fractional integrals is the main purpose of this article.


## 2. Hermite–Hadamard Inequalities for F-Convexity via Katugampola Fractional Integrals

Now, let us give the Hermite–Hadamard inequality for F-convex functions via Katugampola fractional integrals as follows.Theorem 3.
Letα>0 and ρ>0. Let f:aρ,bρ⟶ℝ be a positive function with 0<a<b and f∈Xcpa,b. If f is F-convex on a,b, for some F∈ℱ, then the following inequalities hold:(29)Ffaρ+bρ2,ραΓα+12bρ−aραIa+αρfogb,ραΓα+12bρ−aραIb−αρfoga,12+∫01Lψtdt≤0and(30)TF,ψραΓα+1bρ−aραIa+αρfogb+Ib−αρfoga,faρ+fbρ,faρ+fbρ+∫01Lψtdt≤0,where gx=xρ.Proof.
Lett∈0,1, x,y∈a,b, α≥0. Since f is F-convex, we can write(31)Ffxρ+yρ2,fxρ,fyρ,12≤0.
Forxρ=tρaρ+1−tρbρ and yρ=1−tρaρ+tρbρ, we have(32)Ffaρ+bρ2,ftρaρ+1−tρbρ,f1−tρaρ+tρbρ,12≤0.
Multiplying both sides of the last inequality byαρtαρ−1, α>0, and using axiom (A3), we have(33)Fαρtαρ−1faρ+bρ2,αρtαρ−1ftρaρ+1−tρbρ,αρtαρ−1f1−tρaρ+tρbρ,12+Lψt≤0.
Integrating the last inequality with respect tot over 0,1 and using axiom (A1), we get(34)Fαρfaρ+bρ2∫01tαρ−1dt,αρ∫01tαρ−1ftρaρ+1−tρbρdt,αρ∫01tαρ−1f1−tρaρ+tρbρdt,12+∫01Lψtdt=Ffaρ+bρ2,ραΓα+12bρ−aραIa+αρfogb,ραΓα+12bρ−aραIb−αρfoga,12+∫01Lψtdt≤0,where(35)∫01tαρ−1ftρaρ+1−tρbρdt=∫abbρ−xρbρ−aρα−1fxρxρ−1aρ−bρdx=ρα−1Γαbρ−aραIa+αρfogb,∫01tαρ−1f1−tρaρ+tρbρdt=∫abxρ−aρbρ−aρα−1fxρxρ−1bρ−aρdx=ρα−1Γαbρ−aραIb−αρfoga.
This establishes the first inequality. For the proof of the second inequality, sincef is F-convex, we have(36)Fftρaρ+1−tρbρ,faρ,fbρ,t≤0Ff1−tρaρ+tρbρ,fbρ,faρ,t≤0.
By adding these inequalities, we get(37)Fftρaρ+1−tρbρ+f1−tρaρ+tρbρ,faρ+fbρ,faρ+fbρ,t≤0.
Applying axiom (A3) forψt=αρtαρ−1, we obtain(38)Fαρtαρ−1ftρaρ+1−tρbρ+f1−tρaρ+tρbρ,αρtαρ−1faρ+fbρ,αρtαρ−1faρ+fbρ,t+Lψt≤0.
Integrating over0,1 the last inequality and using axiom (A2), we have(39)TF,ψ∫01αρtαρ−1ftρaρ+1−tρbρ+f1−tρaρ+tρbρdt,faρ+fbρ,faρ+fbρ+∫01Lwtdt=TF,ψραΓα+1bρ−aραIa+αρfogb+ραΓα+1bρ−aραIb−αρfoga,faρ+fbρ,faρ+fbρ+∫01Lψtdt≤0,and thus, the proof is completed.Remark 4.
In Theorem3, taking ρ=1, we obtain inequalities of (23) and (24).Remark 5.
In Theorem3, taking limit ρ⟶0+, we get(40)Ff1,Γα+12lnb/aαHa+αf1,Γα+12lnb/aαHb−αf1,12+∫01Lψtdt≤0TF,ψΓα+1lnb/aαIa+αf1+Hb−αf1,2f1,2f1+∫01Lψtdt≤0.Corollary 1.
Iff is ε-convex on [a, b] with 0<a<b, ε≥0, then we have(41)faρ+bρ2−ε≤ραΓα+12bρ−aραIa+αρfogb+Ib−αρfoga≤faρ+fbρ2+ε2,where gx=xρ.Proof.
It is known that anε-convex is an F-convex. Using (10) with ψt=αρtαρ−1, we have(42)∫01Lψtdt=ε∫011−αρtαρ−1dt=0.
Using (8), (29), and (42), we can write(43)0≥Ffaρ+bρ2,ραΓα+12bρ−aραIa+αρfogb,ραΓα+12bρ−aραIb−αρfoga,12+∫01Lψtdt=faρ+bρ2−ραΓα+12bρ−aραIa+αρfogb+Ib−αρfoga−ε.
So, we get(44)faρ+bρ2−ε≤ραΓα+12bρ−aραIa+αρfogb+Ib−αρfoga.
On the other hand, using (9) with ψt=αρtαρ−1, we have(45)TF,ψτ1,τ2,τ3=τ1−αρ∫01tαρdtτ2−αρ∫011−ttαρ−1dtτ3−ε=τ1−αρτ2+τ3αρ+1−ε,for τ1,τ2,τ3∈ℝ. So, from (30) and (45), we obtain(46)0≥TF,ψραΓα+1bρ−aραIa+αρfogb+ραΓα+1bρ−aραIb−αρfoga,faρ+fbρ,faρ+fbρ+∫01Lψtdt=ραΓα+1bρ−aραIa+αρfogb+Ib−αρfoga−1αρ+1αρfaρ+fbρ+faρ+fbρ−ε=ραΓα+1bρ−aραIa+αρfogb+Ib−αρfoga−faρ+fbρ−ε.
This implies that(47)ραΓα+1bρ−aραIa+αρfogb+Ib−αρfoga≤faρ+fbρ+ε.
The proof is completed.Remark 6.
If we takeε=0 in Corollary 1, then f′ is convex and we have Theorem 2.1 in [12].


## 3. Hermite–Hadamard-Type Inequalities for F-Convexity via Katugampola Fractional Integrals

To prove our main results in this section, let us consider the following lemma.Lemma 1 (see [12]).
Letf:aρ,bρ⟶ℝ be a differentiable mapping on aρ,bρ with 0<a<b. Then, the following equality holds if the fractional integrals exist:(48)faρ+fbρ2−αραΓα+12b−aαIa+αρfogb+Ib−αρfoga=bρ−aρ2∫011−tρα−tραtρ−1f′tρaρ+1−tρbρdt,where gx=xρ.Theorem 4.
LetI⊆ℝ be an interval and f:I⟶ℝ be a differentiable mapping on I,a,b∈I,0<a<b. Suppose that f′ is F-convex on a,b, for some F∈ℱ, and the function t∈0,1⟶Lψt belongs to L10,1. Then, we have the inequality(49)TF,ψ2bρ−aρfaρ+fbρ2−αραΓα+12b−aαIa+αρfogb+Ib−αρfoga,f′aρ,f′bρ+∫01Lψtdt≤0,where gx=xρ and ψt=1−tρα−tρα.Proof.
By using F-convexity off′, we can write(50)Ff′tρaρ+1−tρbρ,f′aρ,f′bρ,t≤0,t∈0,1.
Using axiom (A3) withψt=1−tρα−tρα, we get(51)Fψtf′tρaρ+1−tρbρ,ψtf′aρ,ψtf′bρ,t+Lψt≤0,t∈0,1.
Integrating over0,1 and using axiom (A2), we obtain(52)TF,ψ∫01ψtf′tρaρ+1−tρbρdt,f′aρ,f′bρ,t+∫01Lψtdt≤0,t∈0,1.
From Lemma1, we get(53)2bρ−aρfaρ+fbρ2−αραΓα+12b−aαIa+αρfogb+Ib−αρfoga≤∫011−tρα−tραtρ−1f′tρaρ+1−tρbρdt.
SinceTF,ψ is nondecreasing with respect to the first variable, we establish(54)TF,ψ2bρ−aρfaρ+fbρ2−αραΓα+12b−aαIa+αρfogb+Ib−αρfoga,f′aρ,f′bρ,t+∫01Lψtdt≤0.
The proof is completed.Remark 7.
In Theorem4, taking ρ=1, we obtain Theorem 2.5 in [3].Remark 8.
In Theorem4, taking limit ρ⟶0+, we get(55)TF,ψ2lnb/af1−αΓα+12lnb/aαHa+αf1+Hb−αf1,f′1,f′1+∫01Lψtdt≤0.Corollary 2.
Iff′ is ε-convex on a,b with 0<a<b, ε≥0, then we have(56)faρ+fbρ2−αραΓα+12b−aαIa+αρfogb+Ib−αρfoga≤bρ−aρ2ρα+11−12αf′aρ+f′bρ+ε,where gx=xρ.Proof.
From (10), with ψt=1−tρα−tραtρ−1, we have(57)∫01Lψtdt=ε∫011−1−tρα−tραtρ−1dt=ε∫01dt−∫01/21−tρα−tαρtρ−1dt−∫1/21tαρ−1−tραtρ−1dt=ε1−2ρα+11−12α.
Using (9) with ψt=1−tρα−tραtρ−1,(58)TF,ψτ1,τ2,τ3=τ1−∫01t1−tρα−tραtρ−1τ2−∫011−t1−tρα−tραtρ−1τ3−ε=τ1−1ρB1/21/ρ+1,α+1−B1/2α+1,1/ρ+1+1α+1/ρ+11−12α+1/ρ×τ2−τ3−2ρα+11−12ατ3−ε.
Then, by Theorem4, we get(59)0≥TF,ψ2bρ−aρfaρ+fbρ2−αραΓα+12b−aαIa+αρfogb+Ib−αρfoga,f′aρ+f′bρ2,f′aρ+f′bρ2+∫01Lψtdt=2bρ−aρfaρ+fbρ2−αραΓα+12b−aαIa+αρfogb+Ib−αρfoga−2αρα+11−12αf′aρ+f′bρ2−ε+ε1−2ρα+11−12α.
So, we get the desired result.Remark 9.
If we takeε=0 in Corollary 2, then f′ is convex and we have Theorem 2.5 in [12].



---

*Source: 5549258-2021-05-08.xml*