# Some New Fractional Inequalities Involving Convex Functions and Generalized Fractional Integral Operator

Authors: Majid K. Neamah; Alawiah Ibrahim; Hala Shaker Mehdy; Saleh S. Redhwan; Mohammed S. Abdo
Journal: Journal of Function Spaces
 (2022)

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## Abstract

In this manuscript, we are getting some novel inequalities for convex functions by a new generalized fractional integral operator setting. Our results are established by merging thek,s-Riemann-Liouville fractional integral operator with the generalized Katugampola fractional integral operator. Certain special instances of our main results are considered. The detailed results extend and generalize some of the present results by applying some special values to the parameters.


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## Body

## 1. Introduction

Chebyshev [1] presented the celebrated functional described by
(1)Tf,g=1b−a∫abfxgxdx−1b−a∫abfxdx1b−a∫abgxdx,where the functions f and g are integrable on a,b. If f and g are synchronous, i.e.,
(2)fx1−fx2gx1−gx2≥0,for x1,x2∈a,b, then,Tf,g≥0.The relation (1) has stood out for some researchers because of the different applications in numerical quadrature, statistical problems, probability, and transform theory. Among those applications, the relation (1) was utilized to yield various integral inequalities (see, e.g., [2–6], for extremely late work, see additionally [7–10]).There is one more appealing and useful inequality, namely, the Pólya-Szegö inequality [11], which establishes the essential key of motivation in our study, which we can indicate as follows
(3)∫abf2xdx∫abg2xdx∫abfxgxdx2≤14MNmn+mnMN2,where m≤fx≤M and n≤gx≤N, for some m,M,n,N∈ℝ and for each x∈J≔a,b.In [12], Dragomir and Diamond introduced the following Grüss type integral inequality
(4)Tf,g≤M−mN−n4b−a2MmNn∫abfxdx∫abgxdx,by the Pólya-Szegö result, here, 0<m≤fx≤M<∞ and 0<n≤gx≤N<∞, for x∈J.The next integral inequalities(5)∫01vμ−1xdx≥∫01xμvxdx,(6)∫01vμ−1xdx≥∫01xvσxdx,have been proved by Ngo [13], where x>0 and v are a positive continuous on 0,1 such that
(7)∫h1vxdx≥∫h1xdx,h∈0,1.In this regard, Liu et al. [14] introduced the subsequent inequality
(8)∫abvμ+νxdx≥∫abx−aμvνxdx,with(9)∫abvςxdx≥∫abx−aςdx,where μ,ν>0 and v are a positive continuous on J, and ς=min1,ν, for x∈J.Since one of the primary inspiration points of fractional analysis is getting more general and valuable fundamental integral operators, the generalized fractional integral operator is a decent tool to sum up numerous past investigations and results, see [15–18]. Essentially, in inequality theory, scientists utilize such broad operators to generalize and extend their inequalities.Lately, these fractional integral operators have been considered and used to broaden particularly Grüss, Chebychev-Grüss, Pólya-Szegö, Gronwall, Minkowski, and Hermite-Hadamard type inequalities. For additional subtleties, Agarwal [19] proved some fractional integral inequalities with Hadamard’s fractional integral operators. Ntouyas et al. [20] established certain Chebyshev type inequalities involving Hadamard’s fractional integral operators. Some Grüss type inequalities under k-Riemann-Liouville fractional integral operators have been investigated by Set et al. [21]. In [22], the authors introduced some Pólya-Szegö type inequalities by Hadamard k-fractional integral operators. A new version for the Gronwall type inequality involving generalized proportional fractional integral operators was presented by Alzabut et al. [23]. Rahman et al. [24] established reverse Minkowski inequalities with generalized proportional fractional integral operators. Many Chebyshev type inequalities with generalized conformable fractional integral operators have been discussed by Nisar et al. [25].In this regard, Dahmani [26] established some new inequalities for convex functions involving Riemann-Liouville fractional integral operator. Jleli et al. [27] obtained new Hermite-Hadamard type inequalities for convex functions via generalized fractional integral operators. Some Hermite-Hadamard type inequalities for k,s-Riemann-Liouville fractional integral operators were obtained by Agarwal et al. [28]. The authors in [29] established certain Polya-Szego type inequalities involving generalized Katugampola fractional integral operator.Motivated by the above works and discussions, in this manuscript, we establish certain novel inequalities for convex functions under a new generalized fractional integral operator which integrates the two proposed fractional integral operators in [30, 31]. Moreover, we consider certain special cases of our main results. The results obtained extend and generalize some of the existing results by substituting some parameters.The manuscript is marshaled as follows: Section2 presents some main definitions and results. The acquired results are presented in Section 3. The last section concludes the manuscript.


## 2. Preliminaries

Let us first present the essential definitions and properties of fractional analysis that will be frequently used in this study.Definition 1 (see [15]).
The right and left-sided Riemann-Liouville fractional integral operators of orderα>0 of u∈L1a,b are defined by
(10)Iaαux=1Γα∫axx−yα−1uydy,a<x,(11)Ibαux=1Γα∫xby−xα−1uydy,x<b.Definition 2 (see [30]).
For>0,k>0, and s∈ℝ\−1, the left-sided k,s-Riemann-Liouville fractional integral operator is defined as
(12)Ia,s,kαux=s+11−α/kα/kkΓkα∫axxs+1−ys+1αk−1ysuydy,a<x.Definition 3 (see [31]).
Forα,ρ>0, and β,η,r∈ℝ, the left-sided generalized Katugampola fractional integral operator is defined as
(13)Iρa,η,rα,βux=ρ1−βxrΓα∫axxρ−yρα−1yρη+1−1uydy,a<x.
The subsequent important results have been established by Liu et al. [32].Theorem 4 (see [32]).
Letu and v be two positive continuous functions with u≤v on J. Assume that the functions u/v and u are decreasing and increasing, respectively. If the function Ω is a convex with Ω0=0, then
(14)∫abuxdx∫abvxdx≥∫abΩuxdx∫abΩvxdx.Theorem 5 (see [32]).
Letu, z, and v be positive continuous functions with u≤v on J. Let also u,z are increasing functions and u/v is a decreasing function. If the function Ω is a convex with Ω0=0, then
(15)∫abuxdx∫abvxdx≥∫abΩuxzxdx∫abΩvxzxdx.


## 3. Main Results

In this section, we will establish some new tolerances for the convex functions under a new fractional integral operator which combine together the two operators proposed in [30, 31].Definition 6.
Letα>0,k>0,s>−1, and β,η,r∈ℝ. Then, the generalized fractional integral operator of order α for a continuous function u is defined as
(16)Iksa,η,rα,βux=s+11−β/kβ/kxrkΓkα∫axxs+1−ys+1α/kαk−1ys+1η+suydy,a<x.Remark 7.
(1)
Settingk=1,s=ρ−1 in Eq. (16), the fractional integral operator Eq. (16) reduces to the generalized Katugampola fractional integral operator defined by Eq. (13)(2)
Settingα=β,r=0, and η=0 in Eq. (16), the fractional integral operator Eq. (16) reduces to the generalized fractional integral operator defined by Eq. (12)(3)
Settingk=1,α=β,r=0, η=0, and s⟶−1 in Eq. (16), with L’Hôpital’s rule, the fractional integral operator Eq. (16) reduces to the Hadamard fractional integral operator, namely,(17)IHaαux=1Γα∫axlogxyα−1uyydy,a<x.(4)
Settingk=1,β=0,s=ρ−1, and r=−ρη+α, in Eq. (16), the fractional integral operator Eq. (16) reduces to the Erdélyi-Kober fractional integral operator (see [15])(5)
Settingα=β,r=0, η=0, and s=0 in Eq. (16), the fractional integral operator Eq. (16) reduces to the k-Riemann-Liouville fractional integral operator, i.e.,(18)Iksa,η,rα,βux⟶Ia,kαux=1kΓkα∫axx−yα/kαk−1uydy,a<x.(6)
Settingk=1,α=β,r=0, η=0, and s=0 in Eq. (16), the fractional integral operator Eq. (16) reduces to the Riemann-Liouville fractional integral operator defined by(19)Iaαux=1Γα∫axx−yα−1uydy,a<x.
Now, we are ready to provide the inequalities for convex functions by using the considered fractional integral operator defined by Eq. (16).Theorem 8.
Letu and v be positive continuous functions with u≤v on J. If u and u/v are increasing and decreasing on J, respectively, then for any convex function Ω with Ω0=0, we have
(20)Iksa,η,rα,βuxIksa,η,rα,βvx≥Iksa,η,rα,βΩuxIksa,η,rα,βΩvx,where α,β,a,η,r,s,k are as in Definition 6.Proof.
By the hypotheses of theorem,Ω is convex with Ω0=0. Then the function Ωx/x is increasing. Since u is an increasing function, thus, Ωux/ux is an increasing function too.
Obviously,ux/vx is a decreasing function. Therefore, for each ζ,ξ∈J, we have
(21)Ωuζuζ−Ωuξuξuξvξ−uζvζ≥0.
It follows that(22)Ωuζuζuξvξ+Ωuξuξuζvζ−Ωuξuξuξvξ−Ωuζuζuζvζ≥0.
Multiplying Eq. (22) by vζvξ, we obtain
(23)Ωuζuζuξvζ+Ωuξuξuζvξ−Ωuξuξuξvζ−Ωuζuζuζvξ≥0.
Multiplying Eq. (23) by s+11−β/kxr/kΓkαxs+1−ζs+1α/k−1ζs+1η+s and integrating Eq. (23) with respect to ζ over a,x,a<x≤b, we get
(24)s+11−β/kxrkΓkα∫axxs+1−ζs+1α/k−1ζs+1η+sΩuζuζuξvζdζ+s+11−β/kxrkΓkα∫axxs+1−ζs+1α/k−1ζs+1η+sΩuξuξuζvξdζ−s+11−β/kxrkΓkα∫axxs+1−ζs+1α/k−1ζs+1η+sΩuξuξuξvζdζ−s+11−β/kxrkΓkα∫axxs+1−ζs+1α/k−1ζs+1η+sΩuζuζuζvξdζ≥0.
Hence(25)uξIksa,η,rα,βΩuxuxvx+ΩuξuξvξIksa,η,rα,βux−ΩuξuξuξIksa,η,rα,βvx−vξIksa,η,rα,βΩuxuxux≥0.
Again, multiplying Eq. (25) by s+11−β/kxr/kΓkαxs+1−ξs+1α/k−1ξs+1η+s and integrating Eq. (25) with respect to ξ over a,x,a<x≤b, we obtain
(26)Iksa,η,rα,βuxIksa,η,rα,βΩuxuxvx+Iksa,η,rα,βΩuxuxvxIksa,η,rα,βux≥Iksa,η,rα,βΩuxIksa,η,rα,βvx+Iksa,η,rα,βvxIksa,η,rα,βΩux.
Consequently, we have(27)Iksa,η,rα,βuxIksa,η,rα,βvx≥Iksa,η,rα,βΩuxIksa,η,rα,βΩux/uxvx.
Sinceux≤vx for all x∈J and the function Ωx/x is an increasing, thus, for ζ∈a,x,a<x≤b, we have
(28)Ωuζuζ≤Ωvζvζ.
Multiplying both sides of Eq. (28) by s+11−β/kxr/kΓkαxs+1−ζs+1α/k−1ζs+1η+svζ then integrating with respect to ζ over a,x,a<x≤b, we get
(29)s+11−β/kxrkΓkα∫axxs+1−ζs+1α/k−1ζs+1η+sΩuζuζvζdζ≤s+11−β/kxrkΓkα∫axxs+1−ζs+1α/k−1ζs+1η+sΩvζdζ.
As per Eq. (16) can be written Eq. (29) as follows
(30)Iksa,η,rα,βΩuxuxvx≤Iksa,η,rα,βΩvx.
Hence, from Eq. (27) and Eq. (30), we obtain the desired result Eq. (20).Remark 9.
(i)
When=1,α=β,r=0, η=0, and s=0 in Theorem 8, we get the result (Theorem 3.1) proved by Dahmani [26].(ii)
Whenα=β=1,k=1,r=η=s=0, and x=b in Theorem 8, we recapture Theorem 4(iii)
In Theorem8, if we replace the operator ksIa,η,rα,β with the generalized proportional fractional integral operator, then, we obtain the result (Theorem 3.1) proved by Neamah and Ibrahim [33].Theorem 10.
Letu and v be positive continuous functions with u≤v on J. If u is increasing and u/v is decreasing on J, then for any convex function Ω with Ω0=0, we have
(31)Iksa,η,rα,βuxIksa,η,rγ,βΩvx+Iksa,η,rγ,βuxIksa,η,rα,βΩvxIksa,η,rα,βvxIksa,η,rγ,βΩux+Iksa,η,rγ,βvxIksa,η,rα,βΩux≥1,where γ>0 and α,β,a,η,r,s,k are as in Definition 6.Proof.
Thanks to the hypotheses of theorem,Ω is convex with Ω0=0. Thus, Ωx/x is an increasing function. Furthermore, since u is increasing function, the function Ωux/ux is increasing. Distinctly, ux/vx is decreasing function. Thus, by multiplying Eq. (25) by s+11−β/kxr/kΓkγxs+1−ξs+1γ/k−1ξs+1η+s and integrating the nascent identity with respect to ξ over a,x,a<x≤b, we obtain
(32)Iksa,η,rγ,βuxIksa,η,rα,βΩuxuxvx+Iksa,η,rγ,βΩuxuxvxIksa,η,rα,βux≥Iksa,η,rγ,βΩuxuxuxIksa,η,rα,βvx+Iksa,η,rγ,βvxIksa,η,rα,βΩuxuxux.
Consequently, the inequalities Eq. (30) and Eq. (32) give the inequality Eq. (31).Remark 11.
(i)
Applying Theorem10 for α=γ, we get Theorem 8(ii)
Applying Theorem10 for =1,α=β,r=0, η=0, and s=0, we get the result (Theorem 8) proved by Dahmani [26].(iii)
Applying Theorem10 for α=β=γ=1,k=1,r=η=s=0, and x=b, we get Theorem 4Theorem 12.
Letu, z, and v be three positive continuous functions and u≤v on J. If u/v is decreasing, u and z are increasing, and Ω is convex function with Ω0=0. Then
(33)Iksa,η,rα,βuxIksa,η,rα,βvx≥Iksa,η,rα,βΩuxzxIksa,η,rα,βΩvxzx.Proof.
In view of conditions of theorem,Ω is convex with Ω0=0. Thus, Ωx/x is increasing. Besides, from the increasing of u,Ωux/ux is increasing. Obviously, the function ux/vx is decreasing. So, for all ζ,ξ∈a,x and a<x≤b, we have
(34)Ωuζuζzζ−Ωuξuξzξuξvζ−uζvξ≥0.
Then,(35)Ωuζzζuζuξvζ+Ωuξzξuξuζvξ−Ωuξzξuξuξvζ−Ωuζzζuζuζvξ≥0.
Multiplying Eq. (35) by s+11−β/kxr/kΓkαxs+1−ζ1s+1α/k−1ζs+1η+s then integrating the resulting inequality with respect to ζ over a,x,a<x≤b, we get
(36)s+11−β/kxrkΓkα∫axxs+1−ζs+1α/k−1ζs+1η+sΩuζuζuξvζzζdζ+s+11−β/kxrkΓkα∫axxs+1−ζs+1α/k−1ζs+1η+sΩuξuξuζvξzξdζ−s+11−β/kxrkΓkα∫axxs+1−ζs+1α/k−1ζs+1η+sΩuξuξuξvζzξdζ−s+11−β/kxrkΓkα∫axxs+1−ζs+11α/k−1ζs+1η+sΩuζuζuζvξzζdζ≥0.
Hence,(37)uξIksa,η,rα,βΩuxuxvxzx+ΩuξuξvξzξIksa,η,rα,βux−ΩuξuξuξzξIksa,η,rα,βvx−vξIksa,η,rα,βΩuxuxuxzx≥0.
With the same arguments as above for Eq. (37), we obtain
(38)Iksa,η,rα,βuxIksa,η,rα,βΩuxuxvxzx+Iksa,η,rα,βΩuxuxvxzxIksa,η,rα,βux≥Iksa,η,rα,βΩuxzxIksa,η,rα,βvx+Iksa,η,rα,βvxIksa,η,rα,βΩuxzx.
It follows that(39)Iksa,η,rα,βuxIksa,η,rα,βvx≥Iksa,η,rα,βΩuxzxIksa,η,rα,βΩux/uxvxzx.
Further, sinceu≤v on J then using fact that the function Ωx/x is increasing, we can write
(40)Ωuζuζ≤Ωvζvζ,forζ∈a,x.
By some previously repeated procedure, the inequality Eq. (40) leads to
(41)s+11−β/kxrkΓkα∫axxs+1−ζs+1α/k−1ζs+1η+sΩuζuζvζzζdζ≤s+11−β/kxrkΓkα∫axxs+1−ζs+1α/k−1ζs+1η+sΩvζzζdζ.
Given Eq. (16), the inequality Eq. (41) can be written as
(42)Iksa,η,rα,βΩuxuxvxzx≤Iksa,η,rα,βΩvxzx.
Therefore, from Eq. (42) and Eq. (39), we get Eq. (33), which completes the proof.Remark 13.
(1)
Applying Theorem12 for k=1,α=β, and r=η=s=0, we get the result (Theorem 10) proved by Dahmani [26](2)
It is noteworthy that Theorem5 is a special case of Theorem 12 when α=β=1,k=1,r=η=s=0, and x=bTheorem 14.
Letu, z, and v be three positive continuous functions and u≤v on J. If u/v is decreasing, u and z are increasing on J, and Ω is convex function such that Ω0=0. Then, we have
(43)Iksa,η,rα,βuxIksa,η,rγ,βΩvxzx+Iksa,η,rγ,βuxIksa,η,rα,βΩvxzxIksa,η,rα,βvxIksa,η,rγ,βΩuxzx+Iksa,η,rγ,βvxIksa,η,rα,βΩuxzx≥1.Proof.
Applyings+11−β/kxr/kΓkγxs+1−ξs+1γ/k−1ξs+1η+s on both sides of Eq. (37), then integrating the resulting inequality with respect to ξ over a,x,a<x≤b, we get
(44)Iksa,η,rγ,βuxIksa,η,rα,βΩuxuxvxzx+Iksa,η,rγ,βΩuxuxvxzxIksa,η,rα,βux≥Iksa,η,rγ,βΩuxuxuxzxIksa,η,rα,βvx+Iksa,η,rγ,βvxIksa,η,rα,βΩuxuxuxzx.
Sinceu≤v on J, then using fact that the function Ωx/x is increasing, we have
(45)Ωuζuζ≤Ωvζvζ,forζ∈a,ξandξ∈J.
Multiplying Eq. (45) by
(46)s+11−β/kxrkΓkαxs+1−ζs+1α/k−1ζs+1η+svζzζ,then integration with respect to ζ over a,x,a<x≤b, we obtain
(47)Iksa,η,rα,βΩuxuxvxzx≤Iksa,η,rα,βΩvxzx.
Following similar arguments as mentioned earlier, we conclude that(48)Iksa,η,rγ,βΩuxuxvxzx≤Iksa,η,rγ,βzΩvxzx.
Hence, by virtue of Eq. (44), Eq. (47), and Eq. (48), we obtain Eq. (43). Thus, the proof is completed.Remark 15.
(i)
Applying Theorem14 for α=γ, we obtain Theorem 12(ii)
Applying Theorem14 for k=1,α=β=γ, and r=η=s=0, we obtain Theorem 12 proved by Dahmani [26]


## 4. Conclusions

In this work, we have established certain Pólya-Szegö inequalities by using convex functions under a new generalized fractional integral operator. More precisely, some new results have been established by merging thek,s-Riemann-Liouville fractional integral operator with the generalized Katugampola fractional integral operator. Moreover, we have introduced several new special results that cover many classical fractional integral operators.In future work, it will be very interesting to study the inequalities considered in this work under a more general fractional integral operator in terms of another functionψ, precisely, we hint to ψx=xs+1, and this is what we will think about in the next work.



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