# Some New Tempered Fractional Pólya-Szegö and Chebyshev-Type Inequalities with Respect to Another Function

Authors: Gauhar Rahman; Kottakkaran Sooppy Nisar; Thabet Abdeljawad; Muhammad Samraiz
Journal: Journal of Mathematics
 (2020)

---

## Abstract

In this present article, we establish certain new Pólya–Szegö-type tempered fractional integral inequalities by considering the generalized tempered fractional integral concerning another functionΨ in the kernel. We then prove certain new Chebyshev-type tempered fractional integral inequalities for the said operator with the help of newly established Pólya–Szegö-type tempered fractional integral inequalities. Also, some new particular cases in the sense of classical tempered fractional integrals are discussed. Additionally, examples of constructing bounded functions are considered. Furthermore, one can easily form new inequalities for Katugampola fractional integrals, generalized Riemann–Liouville fractional integral concerning another function Ψ in the kernel, and generalized fractional conformable integral by applying different conditions.


---

## Body

## 1. Introduction

The well-known Chebyshev functional [1] is defined by(1)Tf1,f2=1x2−x1∫x1x2f1θf2θdθ−1x2−x1∫x1x2f1θdθ1x2−x1∫x1x2f2θdθ,where the functions f1 and f2 are integrable on x1,x2. If the functions f1 and f2 are synchronous on x1,x2, i.e.,(2)f1ϑ−f1ζf2ϑ−f2ζ≥0,for any ϑ,ζ∈x1,x2, then Tf1,f2≥0. Functional (1) has gained more recognition due to its diverse applications in the fields of transform theory, numerical quadrature, probability, and statistical problems. Additionally, the researchers have established a large number of integral inequalities by utilizing functional (1). The interesting readers may consult [2–5]. In [6], Tassaddiq et al. recently established certain inequalities via fractional conformable integrals by considering functional (1).In [7], Grüss introduced the following inequality:(3)Tf1,f2≤M1−m1N1−n14,where the functions f1 and f2 are integrable on x1,x2 such that f1 and f2 satisfy the inequalities m1≤f1ϑ≤M1 and n1≤f2ζ≤N1, for all ϑ,ζ∈x1,x2 and for some constant m1,n1,M1,N1∈ℝ.In [8], Pólya–Szegö presented the following inequality:(4)∫x1x2f12θdθ∫x1x2f22θdθ∫x1x2f1θf2θdθ2≤14M1N1m1n1+m1n1M1N12.In [9], Dragomir and Diamond presented the following inequality with the help of Pólya–Szegö inequality:(5)Tf1,f2≤M1−m1N1−n14x2−x12m1M1n1N1∫x1x2f1θdθ∫x1x2f2θdθ,where the functions f1 and f2 are positive and integrable on x1,x2 such that f1 and f2 satisfy the inequalities m1≤f1θ≤M1 and n1≤f2ζ≤N1, for all θ,ζ∈x1,x2 and for some constant m1,n1,M1,N1∈ℝ.In the last few decades, the researchers have considered that fractional integral inequalities are the most powerful tools for the development of both applied and pure mathematics. In [10], the authors presented some Grüss-type integral inequalities by considering fractional integrals. Some new integral inequalities in sense of Riemann–Liouville fractional integrals can be found in the work of Dahmani [11].In [12], Sarikaya et al. gave the idea of generalized k,s-fractional integrals with applications. Set et al. [13] investigated some Grüss-type inequalities by considering generalized k-fractional integrals.Very recently, the idea of fractional conformable and proportional fractional integral operators was proposed by Jarad et al. [14, 15]. Later on, Huang et al. [16] presented generalized Hermite–Hadamard-type inequalities by considering generalized fractional conformable integrals. In [17], Qi et al. established Chebyshev-type inequalities for generalized fractional conformable integrals.In [18], Ntouyas et al. investigated some new Pólya–Szegö- and Chebyshev-type inequalities by considering Riemann–Liouville fractional integrals. The tempered fractional integral was first studied by Buschman [19], but Li et al. [20] and Meerschaert et al. [21] have described the associated tempered fractional calculus more explicitly. Fernandez and Ustaoǧlu [22] investigated several analytic properties of the tempered fractional integral. In [23], Fahad et al. proposed the general form of the generalized tempered fractional integral concerning another function. In this paper, we investigate the said inequalities for the so-called tempered fractional integrals containing another function in the kernel.The structure of the paper as follows.In Section2, some basic definitions are presented. Some new Pólya–Szegö-type for the so-called generalized tempered fractional integral in the sense of another function is presented in Section 3. In Section 4, we present some new generalized Chebyshev-type tempered fractional integral inequalities. In Section 5, certain new particular cases in terms of classical tempered fractional integrals are discussed. An example of constructing bounding functions is considered in Section 6. Finally, the concluding remarks are discussed in Section 7.


## 2. Preliminaries

In this section, we consider some well-known definitions and mathematical preliminaries.Definition 1.
(see [7]). Suppose that the functions f1,f2:x1,x1⟶ℝ are positive with A≤f1ϑ≤ℬ and C≤f2ϑ≤D, for all ϑ∈x1,x1, then the following inequality holds:(6)1x2−x1∫x1x2f1ϑf2ϑdϑ−1x2−x1∫x1x2f1ϑdϑ1x2−x1∫x1x2f2ϑdϑ≤14ℬ−AD−C,where the constants ℬ, A, C,D∈ℝ and 1/4 is the sharp of inequality (6).Definition 2.
(see [24, 25]). The function f1 will be in the space Lp,r0,∞ if(7)Lp,r0,∞=f1:f1Lp,r0,∞=∫rsf1ϑpϑrdϑ1/p<∞,1≤p<∞,r≥0.
If we applyr=0, then (7) gives(8)Lp0,∞=f1:f1Lp0,∞=∫rsf1ϑpdϑ1/p<∞,1≤p<∞.Definition 3.
(see [26]). Suppose that the function f1∈L10,∞ and assume that the function Ψ is positive, monotone, and increasing on 0,∞ and having continuous derivative Ψ′ on 0,∞ with Ψ0=0. Then, the Lebesgue real-valued measurable function f1 defined on 0,∞ is said to be in the space XΨp0,∞, 1≤p<∞ if(9)f1XΨp=∫rsf1ϑpΨ′ϑdϑ1/p<∞,1≤p<∞.
Whenp=∞, then(10)f1XΨ∞=esssup0≤ϑ<∞Ψ′ϑf1ϱ.
Note that, the spaceXΨp0,∞ coincides with the space Lp0,∞ if Ψϑ=ϑ for 1≤p<∞ and similarly with the space Lp,r1,∞ if Ψϑ=lnϑ for 1≤p<∞.Definition 4.
(see [20–22]). The left-sided tempered fractional integral of order κ>0 and τ≥0 with ℜκ>0 and ℜτ≥0 is defined by(11)ℛκ,τx1f1θ=1Γκ∫x1θe−τθ−ϑϱ−tκ−1f1ϑdϑ,x1<θ.Remark 1.
By settingτ=0 in (11) yields the following Riemann–Liouville fractional integral, which is defined by(12)ℛκx1f1θ=1Γκ∫x1θθ−ϑκ−1f1ϑdϑ,x1<θ.
The following results for (11) hold:(13)ℛκ,τx1ℛλ,τx1f1θ=ℛκ+λ,τx1f1θ.
We define the following one-sided tempered fractional integral.Definition 5.
The one-sided tempered fractional integral of orderκ>0, τ≥0 is defined by(14)ℛ0κ,τf1θ=ℛκ,τf1θ=1Γκ∫0θe−τθ−ϑθ−ϑκ−1f1ϑdϑ.Definition 6.
(see [23]). Let the function f1 be an integrable in the space XΨp0,∞ and assume that the function Ψ is positive, monotone, and increasing on 0,∞, and its derivative Ψ′ is continuous on 0,∞ with Ψ0=0. Then, the left-sided generalized tempered fractional integral of the function f1 concerning another function Ψ in the kernel is defined by(15)ℛκ,τx1Ψf1θ=1Γκ∫x1θe−τΨθ−ΨθΨθ−Ψθκ−1Ψ′θf1θdϑ,x1<θ,where τ≥0, κ,∈ℂ with ℜκ>0, and Γ. is the well-known gamma function.Remark 2.
The following results can be obtained:(i)
Applying Definition6 for Ψθ=θ, we get (11).(ii)
Applying Definition6 for τ=0, then it will reduce to the left-sided generalized Riemann–Liouville fractional integral operator [27].(iii)
Applying Definition6 for Ψθ=lnθ, then it will reduce to the following left-sided Hadamard tempered integral defined by [23](16)ℛκ,τx1Ψf1θ=1Γκ∫x1θexp−τlnθ−lnϑlnθ−lnϑκ−1f1ϑϑdϑ,x1<θ.(iv)
Applying Definition6 for Ψθ=θη/η, η>0 and τ=0, then it will reduce to the left-sided Katugampola [24] fractional integral.(v)
Applying Definition6 for Ψθ=θ and τ=0, the left Riemann–Liouville fractional integral (12) will be obtained(vi)
Applying Definition6 for Ψθ=θα+s/α+s and τ=0 (where α∈0,1, s∈ℝ and α+s≠0), then it reduces to the left-sided generalized fractional conformable integral given by [28](vii)
Applying Definition6 for Ψθ=θ−x1α/α, α>0, then it reduces to the fractional conformable integral defined by Jarad et al. [14].
In this article, we consider the following one-sided GTF-integral.Definition 7.
Let the functionf1 be integrable in the space XΨp0,∞ and assume that the function Ψ is positive, monotone, and increasing on 0,∞, and its derivative Ψ′ is continuous on 0,∞ with Ψ0=0. Then, the one-sided generalized tempered fractional integral of the function f1 concerning another function Ψ in the kernel is defined by(17)ℛ0κ,τΨf1θ=1Γκ∫0θe−τΨθ−ΨθΨθ−Ψθκ−1Ψ′θf1θdϑ.Definition 8.
For0=x0<x1<⋯<xp<xp+1=X, we define the following subintegrals for the generalized tempered integral(18)ℛxi,xi+1κ,τΨf1X=1Γκ∫xixi+1e−τΨX−ΨϑΨX−Ψϑκ−1Ψ′ϑf1ϑdϑ.
Note that(19)ℛ0κ,τΨf1X=∑i=0pℛti,ti+1κ,τΨf1X=1Γκ∫0x1e−τΨX−ΨϑΨX−Ψϑκ−1Ψ′ϑf1ϑdϑ+1Γκ∫x1x2e−τΨX−ΨϑΨX−Ψϑκ−1Ψ′ϑf1ϑdϑ+⋯+1Γκ∫xpXe−τΨX−ΨϑΨX−Ψϑκ−1Ψ′ϑf1ϑdϑ.Remark 3.
If we setΨX=X and τ=0, then (18) will reduce to the subintegrals of Riemann–Liouville fractional integral defined by [18].


## 3. Pólya–Szegö-Type Tempered Fractional Integral Inequalities

In this section, we provide some new Pólya–Szegö-type tempered fractional integral inequalities for positive and integrable functions via tempered fractional integral (17) containing another function Ψ in the kernel.Lemma 1.
Let the functionsf1 and f2 be positive and integrable on 0,∞ and assume that the function Ψ is positive, monotone, and increasing on 0,∞, and its derivative Ψ′ is continuous on 0,∞ with Ψ0=0. Suppose that U1,U2,V1 and V2 are four positive and integrable functions on 0,∞ such that(20)H10<U1ϑ≤f1ϑ≤U2ϑ,0<V1ϑ≤f2ϑ≤V2ϑ,ϑ∈0,θ,θ>0.
Then, forκ>0, τ≥0, and θ>0, the following tempered fractional integral inequality holds:(21)ℛ0κ,τΨV1V2f12θΨℛ0κ,τU1U2f22θℛ0κ,τΨU1V1+U2V2f1f2θ2≤14.Proof.
From the given hypothesis, we have(22)U2ϑV1ϑ−f1ϑf2ϑ≥0.
Similarly, we have(23)f1ϑf2ϑ−U1ϑV2ϑ≥0.
Taking product of (22) and (23), we get(24)U2ϑV1ϑ−f1ϑf2ϑf1ϑf2ϑ−U1ϑV2ϑ≥0.
From (24), it can be written as(25)U1ϑV1ϑ+U2ϑV2ϑf1ϑf2ϑ≥V1ϑV2ϑf12ϑ+U1ϑU2ϑf22ϑ.
Now, taking product of (25) with e−τΨθ−ΨθΨθ−Ψθκ−1Ψ′θ/Γκ and integrating the resultant identity with respect to ϑ over 0,θ, we have(26)1Γκ∫0θe−τΨθ−ΨϑΨθ−Ψϑκ−1Ψ′ϑU1ϑV1ϑ+U2ϑV2ϑf1ϑf2ϑdϑ≥1Γκ∫0θe−τΨθ−ΨϑΨθ−Ψϑκ−1Ψ′ϑV1ϑV2ϑf12ϑdϑ+1Γκ∫0θe−τΨθ−ΨϑΨθ−Ψϑκ−1Ψ′ϑU1ϑU2ϑf22ϑdϑ.
With the aid of Definition8, we can write(27)ℛ0κ,τΨU1V1+U2V2f1f2θ≥ℛ0κ,τΨV1V2f12θ+ℛ0κ,τΨU1U2f22θ.
By applying AM-GM inequality, i.e.,x1+x2≥2x1x2,x1,x2∈ℝ+, we get(28)ℛ0κ,τΨU1V1+U2V2f1f2θ≥2ℛ0κ,τΨV1V2f12θℛ0κ,τΨU1U2f22θ.
It follows that(29)ℛ0κ,τΨV1V2f12θℛ0κ,τΨU1U2f22θ≤14ℛ0κ,τΨU1V1+U2V2f1f2θ2,which gives the desired assertion (21).Corollary 1.
Let the functionsf1 and f2 be positive and integrable on 0,∞ and assume that the function Ψ is positive, monotone, and increasing on 0,∞, and its derivative Ψ′ is continuous on 0,∞ with Ψ0=0. Suppose that U1,U2,V1, and V2 are four positive and integrable functions on 0,∞ such that(30)H20<m1≤f1ϑ≤M1<∞,0<n1≤f2ϑ≤N1<∞,ϑ∈0,θ,θ>0.
Then, forκ>0, τ≥0, and θ>0, the following tempered fractional integral inequality holds:(31)ℛ0κ,τΨV1V2f12θℛ0κ,τΨU1U2f22θℛ0κ,τΨf1f2θ2≤14m1n1M1N1+M1N1m1n12.Lemma 2.
Let all the conditions of Lemma1 hold. Then, for κ,λ>0, τ≥0, and θ>0, the following tempered fractional integral inequality holds:(32)ℛ0κ,τΨf12θℛ0κ,τΨf22θℛ0κ,τΨU1U2θℛ0κ,τΨV1V2θℛ0κ,τΨU1f1θℛ0κ,τΨV1f2θ+ℛ0κ,τΨU2f1θℛ0κ,τΨV2f2θ2≤14.Proof.
From the hypothesisH1 defined by (20), we have(33)U2ϑV1ζ−f1ϑf2ζ≥0,(34)f1ϑf2ζ−U1ϑV2ζ≥0,which follows that(35)U1ϑV2ζ+U2ϑV1ζf1ϑf2ζ≥f12ϑf22ζ+U1ϑU2ϑU1ζU2ζ.
Taking product on both sides of (18) by V1ζV2ζf22ζ, we obtain(36)U1ϑf1ϑV1ζf2ζ+U2ϑf1ϑV2ζf2ζ≥V1ζV2ζf12ϑ+U1ϑU2ϑf22ζ.
Taking product of both sides of (36) with e−τΨθ−ΨθΨθ−Ψθκ−1Ψ′θ/Γκ and integrating the resultant identity with respect to ϑ over 0,θ, we have(37)V1ζf2ζ1Γκ∫0θe−τΨθ−ΨϑΨθ−Ψϑκ−1Ψ′ϑU1ϑf1ϑdϑ+V2ζf2ζ1Γκ∫0θe−τΨθ−ΨϑΨθ−Ψϑκ−1Ψ′ϑU2ϑf1ϑdϑ≥V1ζV2ζ1Γκ∫0θe−τΨθ−ΨϑΨθ−Ψϑκ−1Ψ′ϑf12ϑdϑ+f22ζ1Γκ∫0θe−τΨθ−ΨϑΨθ−Ψϑκ−1Ψ′ϑU1ϑU2ϑdϑ,which by applying (8) becomes(38)V1ζf2ζℛ0κ,τΨU1f1θ+V2ζf2ζℛ0κ,τΨU2f1θ≥V1ζV2ζℛ0κ,τΨf12θ+f22ζℛ0κ,τΨU1U2θ.
Again, taking product of both sides of (38) with e−τΨθ−ΨζΨθ−Ψζλ−1Ψ′ζ/Γλ and integrating the resultant identity with respect to ζ over 0,θ and then applying (17), we get(39)ℛ0λ,τΨV1f2θℛ0λ,τΨU1f1θ+ℛ0λ,τΨV2f2θℛ0λ,τΨU2f1θ≥ℛ0λ,τΨV1V2θℛ0λ,τΨf12θ+ℛ0λ,τΨf22θℛ0λ,τΨU1U2θ.
By using AM-GM inequality, we get(40)ℛ0λ,τΨV1f2θℛ0λ,τΨU1f1θ+ℛ0λ,τΨV2f2θℛ0λ,τΨU2f1θ≥2ℛ0λ,τΨV1V2θℛ0λ,τΨf12θℛ0λ,τΨf22θℛ0λ,τΨU1U2θ.
It follows that(41)ℛ0λ,τΨV1V2θℛ0λ,τΨf12θℛ0λ,τΨf22θℛ0λ,τΨU1U2θ≤14ℛ0λ,τΨV1f2θℛ0λ,τΨU1f1θ+ℛ0λ,τΨV2f2θℛ0λ,τΨU2f1θ2,which completes the desired assertion (17).Corollary 2.
Let the functionsf1 and f2 be positive and integrable on 0,∞ satisfying the hypothesis H2 defined by (16) and assume that the function Ψ is positive, monotone, and increasing on 0,∞, and its derivative Ψ′ is continuous on 0,∞ with Ψ0=0. Then, for κ,λ>0, τ≥0, and θ>0, the following tempered fractional integral inequality holds:(42)γκ,τΨθγκ,τΨθτκ+λΓκΓλℛ0κ,τΨf12θℛ0κ,τΨf22θℛ0κ,τΨf1θℛ0κ,τΨf2θ2≤14m1n1M1N1+M1N1m1n12,where(43)ℛ0κ,τΨ1=1Γκ∫0θexp−τΨθ−ΨϑΨθ−Ψϑκ−1Ψ′ϑdϑ=1Γκ∫0Ψθe−τzzκ−1dz,Ψ0=0=1τκΓκ∫0τΨθe−uuκ−1du=γκ,τΨθτκΓκ,and γκ,t=∫0te−uuκ−1du is the well-known incomplete gamma function (see [22]).Lemma 3.
Suppose that all the conditions of Lemma1 hold and assume that the function Ψ is positive, monotone, and increasing on 0,∞, and its derivative Ψ′ is continuous on 0,∞ with Ψ0=0. Then, for κ,λ>0, τ≥0, and θ>0, the following tempered fractional integral inequality holds:(44)ℛ0κ,τΨf12θℛ0κ,τΨf22θ≤ℛ0κ,τΨU2f1f2V1θℛ0κ,τΨV2f1f2U1θ.Proof.
From the hypothesisH1 defined by (20), we have(45)1Γκ∫0θe−τΨθ−ΨϑΨθ−Ψϑκ−1Ψ′ϑf12ϑdϑ≤1Γκ∫0θe−τΨθ−ΨϑΨθ−Ψϑκ−1Ψ′ϑU2ϑV1ϑf1ϑf2ϑdϑ,which in view of (17) yields(46)ℛ0κ,τΨf12θ≤ℛ0κ,τΨU2f1f2V1θ.
Similarly, one can obtain(47)ℛ0λ,τΨf22θ≤ℛ0λ,τΨV2f1f2U1θ.
Hence, the product of (46) and (47) yields the desired assertion (44).Corollary 3.
Let the functionsf1 and f2 be positive and integrable on 0,∞ satisfying the hypothesis H2 defined by (30) and assume that the function Ψ is positive, monotone, and increasing on 0,∞, and its derivative Ψ′ is continuous on 0,∞ with Ψ0=0. Then, for κ,λ>0, τ≥0, and θ>0, the following tempered fractional integral inequality holds:(48)ℛ0κ,τΨf12θℛ0κ,τΨf22θℛ0κ,τΨf1f2θℛ0κ,τΨf1f2θ≤M1N1m1n1.


## 4. Chebyshev-Type Tempered Fractional Integral Inequalities

In this section, certain Chebyshev-type inequalities via tempered fractional integral (20) are presented with the help of Pólya–Szegö integral inequality given by Lemma 1.Theorem 1.
Let the functionsf1 and f2 be positive and integrable on 0,∞ and assume that the function Ψ is positive, monotone, and increasing on 0,∞, and its derivative Ψ′ is continuous on 0,∞ with Ψ0=0. Suppose that U1,U2,V1, and V2 are four positive and integrable functions on 0,∞ satisfying the hypothesis H1 defined by (10). Then, for κ,λ>0, τ≥0, and θ>0, the following tempered fractional integral inequality holds:(49)γκ,τΨθτκΓκℛ0κ,τΨf1f2θ+γλ,τΨθτλΓλΨℛ0λ,τf1f2θ−ℛ0κ,τΨf1θℛ0κ,τΨf2θ−ℛ0κ,τΨf2θℛ0κ,τΨf1θ≤F1f1,U1,U2θ+F2f1,U1,U2θ1/2×F1f2,V1,V2θ+F2f1,V1,V2θ1/2,where(50)F1f1,U1,U2θ=γλ,τΨθ4τλΓλℛ0κ,τΨU1+U2f12ℛ0κ,τΨU1U2θ−ℛ0κ,τΨf1θℛ0κ,τΨf1θ,(51)F2f1,U1,U2θ=γκ,τΨθ4τκΓκℛ0λ,τΨU1+U2f12ℛ0λ,τΨU1U2θ−ℛ0λ,τΨf1θℛ0λ,τΨf1θ.Proof.
By the given hypothesis, both the functionsf1 and f2 are positive and integrable functions on 0,∞. Therefore, for ϑ,ζ∈0,θ with θ>0, we define Aϑ,ζ by(52)Aϑ,ζ=f1ϑ−f1ζf2ϑ−f2ζ=f1ϑf2ϑ+f1ζf2ζ−f1ϑf2ζ−f1ζf2ϑ.
Multiplying (52) by 1/ΓκΓλexp−τΨθ−Ψϑexp−τΨθ−ΨζΨθ−Ψϑκ−1Ψθ−Ψζλ−1Ψ′ϑΨ′ζ and double integrating the resultant identity with respect to ϑ and ζ over 0,θ and then using (17), we obtain(53)1ΓκΓλ∫0θ∫0θexp−τΨθ−Ψϑexp−τΨθ−Ψζ×Ψθ−Ψϑκ−1Ψθ−Ψζλ−1Ψ′ϑΨ′ζAϑ,ζdϑdζ=γκ,τΨθτκΓκℛ0λ,τΨf1f2θ+γλ,τΨθτλΓλℛ0λ,τΨf1f2θ−ℛ0λ,τΨf1θℛ0λ,τΨf2θ−ℛ0λ,τΨf1θℛ0λ,τΨf2θ.
By applying Cauchy–Schwartz inequality for double integrals, we have(54)1ΓκΓλ∫0θ∫0θexp−τΨθ−Ψϑexp−τΨθ−Ψζ×Ψθ−Ψϑκ−1Ψθ−Ψζλ−1Ψ′ϑΨ′ζAϑ,ζdϑdζ≤1ΓκΓλ∫0θ∫0θexp−τΨθ−Ψϑexp−τΨθ−Ψζ×Ψθ−Ψϑκ−1Ψθ−Ψζλ−1Ψ′ϑΨ′ζf12ϑdϑdζ+1ΓκΓλ∫0θ∫0θexp−τΨθ−Ψϑexp−τΨθ−Ψζ×Ψθ−Ψϑκ−1Ψθ−Ψζλ−1Ψ′ϑΨ′ζf12ζdϑdζ−21ΓκΓλ∫0θ∫0θexp−τΨθ−Ψϑexp−τΨθ−Ψζ×Ψθ−Ψϑκ−1Ψθ−Ψζλ−1Ψ′ϑΨ′ζf1ϑf1ζdϑdζ1/2×1ΓκΓλ∫0θ∫0θexp−τΨθ−Ψϑexp−τΨθ−Ψζ×Ψθ−Ψϑκ−1Ψθ−Ψζλ−1Ψ′ϑΨ′ζf22ϑdϑdζ+1ΓκΓλ∫0θ∫0θexp−τΨθ−Ψϑexp−τΨθ−Ψζ×Ψθ−Ψϑκ−1Ψθ−Ψζλ−1Ψ′ϑΨ′ζf22ζdϑdζ−21ΓκΓλ∫0θ∫0θexp−τΨθ−Ψϑexp−τΨθ−Ψζ×Ψθ−Ψϑκ−1Ψθ−Ψζλ−1Ψ′ϑΨ′ζf2ϑf2ζdϑdζ1/2.
In view of (17) and (43), we get(55)1ΓκΓλ∫0θ∫0θexp−τΨθ−Ψϑexp−τΨθ−Ψζ×Ψθ−Ψϑκ−1Ψθ−Ψζλ−1Ψ′ϑΨ′ϑAϑ,ζdϑdζ≤γκ,τΨθτκΓκℛ0λ,τΨf12θ+γλ,τΨθτλΓλℛ0λ,τΨf12θ−2ℛ0λ,τΨf1θℛ0λ,τΨf1θ1/2×γκ,τΨθτκΓκℛ0λ,τΨf22θ+γλ,τΨθτλΓλℛ0λ,τΨf22θ−2ℛ0λ,τΨf2θℛ0λ,τΨf2θ1/2.
Applying Lemma1 for V1θ=V2θ=f2θ=1, we get(56)γλ,τΨθτλΓλℛ0κ,τΨf12θ≤γλ,τΨθ4τλΓλℛ0κ,τΨU1+U2f12ℛ0κ,τΨU1U2θ.
It follows that(57)γλ,τΨθτλΓλℛ0κ,τΨf12θ−ℛ0κ,τΨf1θℛ0κ,τΨf1θ≤γλ,τΨθ4τλΓλℛ0κ,τΨU1+U2f12ℛ0κ,τΨU1U2θ−ℛ0κ,τΨf1θℛ0κ,τΨf1θ=F1f1,U1,U2θ.
Similarly, one can get(58)γκ,τΨθτκΓκℛ0λ,τΨf12θ−ℛ0λ,τΨf1θℛ0λ,τΨf1θ≤γκ,τΨθ4τκΓκℛ0λ,τΨU1+U2f12ℛ0λ,τΨU1U2θ−ℛ0λ,τΨf1θℛ0λ,τΨf1θ=F2f1,U1,U2θ.
Again applying Lemma1 for U1θ=U2θ=f1θ=1, we get(59)γλ,τΨθτλΓλℛ0κ,τΨf22θ−ℛ0κ,τΨf2θℛ0κ,τΨf2θ≤γλ,τΨθ4τλΓλℛ0κ,τΨV1+V2f22ℛ0κ,τΨV1V2θ−ℛ0κ,τΨf2θℛ0κ,τΨf2θ=F1f2,U1,U2θ,(60)γκ,τΨθτκΓκℛ0λ,τΨf22θ−ℛ0λ,τΨf2θℛ0λ,τΨf2θ≤γκ,τΨθ4τκΓκℛ0λ,τΨV1+V2f22ℛ0λ,τΨV1V2θ−ℛ0λ,τΨf2θℛ0λ,τΨf2θ=F2f2,U1,U2θ.
Thus, by considering (53) to (60), we arrive at the desired assertion (49). This is the desired proof of Theorem 1.Theorem 2.
Suppose that all the conditions of Theorem1 are satisfied. Then, for κ>0, τ≥0, and θ>0, the following tempered fractional integral inequality holds:(61)γκ,τΨθτκΓκℛ0κ,τΨf1f2θ−ℛ0κ,τΨf1θℛ0κ,τΨf2θ≤Ff1,U1,U2θFf2,V1,V2θ1/2,where(62)Ff1,U1,U2θ=γκ,τΨθ4τκΓκℛ0κ,τΨU1+U2f12ℛ0κ,τΨU1U2θ−ℛ0κ,τΨf1θ2.Proof.
Applying Theorem1 for κ=λ, we get the desired result in (61).Remark 4.
If we considerU1=m1, U2=M1, V1=n1, and V2=N1, then we have(63)Ff1,m1,M1θ=M1−m124M1m1ℛ0κ,τΨf1θ2,(64)Ff2,m1,M1θ=N1−n124N1n1ℛ0κ,τΨf2θ2.Corollary 4.
Let the functionsf1 and f2 be positive and integrable on 0,∞ and satisfying the hypothesis H2 given by (30). Then, for κ>0, τ≥0, and θ>0, the following tempered fractional integral inequality holds:(65)γκ,τΨθτκΓκℛ0κ,τΨf1f2θ−ℛ0κ,τΨf1θℛ0κ,τΨf2θ≤M1−m1N1−n14m1M1n1N1ℛ0κ,τΨf1θℛ0κ,τΨf2θ.


## 5. Particular Cases

The following new Pólya–Szegö- and Chebyshev-type inequalities for classical tempered fractional integral (14) can be easily established.Lemma 4.
Let the functionsf1 and f2 be positive and integrable on 0,∞. Suppose that U1,U2,V1, and V2 are four positive and integrable functions on 0,∞ satisfying the hypothesis H1 defined by (20). Then, for κ>0, τ≥0, and θ>0, the following tempered fractional integral inequality holds:(66)ℛ0κ,τV1V2f12θℛ0κ,τU1U2f22θℛ0κ,τU1V1+U2V2f1f2θ2≤14.Proof.
Applying Lemma1 for Ψθ=θ, we get Lemma 4.Lemma 5.
Let all the conditions of Lemma4 are satisfied. Then, for κ,λ>0, τ≥0, and θ>0, the following tempered fractional integral inequality holds:(67)ℛ0κ,τf12θℛ0λ,τf22θℛ0κ,τU1U2θℛ0λ,τV1V2θℛ0κ,τU1f1θℛ0λ,τV1f2θ+ℛ0κ,τU2f1θℛ0λ,τV2f2θ2≤14.Proof.
Applying Lemma2 for Ψθ=θ, we get Lemma 5.Similarly, we can derive the particular case of Lemma3. The following theorem represents the particular case of Theorem 1 in terms of classical tempered fractional integral.Theorem 3.
Let the functionsf1 and f2 be positive and integrable on 0,∞. Suppose that U1,U2,V1, and V2 are four positive and integrable functions on 0,∞ satisfying the hypothesis H1 defined by (20). Then, for κ,λ>0, τ≥0, and θ>0, the following tempered fractional integral inequality holds:(68)γκ,τθτκΓκℛ0κ,τf1f2θ+γλ,τθτλΓλℛ0λ,τf1f2θ−ℛ0κ,τf1θℛ0λ,τf2θ−ℛ0κ,τf2θℛ0λ,τf1θ≤F1f1,U1,U2θ+F2f1,U1,U2θ1/2×F1f2,V1,V2θ+F2f1,V1,V2θ1/2,where(69)F1f1,U1,U2θ=γλ,τθ4τλΓλℛ0κ,τU1+U2f12ℛ0κ,τU1U2θ−ℛ0κ,τf1θℛ0λ,τf1θ,(70)F2f1,U1,U2θ=γκ,τθ4τκΓκℛ0λ,τU1+U2f12ℛ0λ,τU1U2θ−ℛ0κ,τf1θℛ0λ,τf1θ.Proof.
Applying Theorem1 for Ψθ=θ, then we get the desired Theorem 3.Similarly, we can derive particular result of Theorem2.


## 6. Applications

In this section, we define a way for constructing four bounded functions and then utilize them to present certain estimates of Chebyshev-type tempered fractional integral inequalities of two unknown functions.Let the unit functionℏθ be defined by(71)ℏθ=1,θ>0,0,θ≤0,and let the Heaviside unit step function ℏaθ be defined by(72)ℏaθ=1,θ>a,0,θ≤a.Suppose that the functionU1 is piecewise continuous function on 0,X defined by(73)U1x=m11ℏ0x−ℏx1x+m12ℏx1x−ℏx2x+⋯+m1p+1ℏxpx=m11ℏ0x+m12−m11ℏx1x+m13−m12ℏx2x+⋯+m1p+1−m1pℏxpx=∑i=0pm1i+1−m1iℏxix,where m10=0 and 0=x0<x1<x2<⋯<xp<xp+1=X. Similarly, we define(74)U2x=∑i=0pM1i+1−M1iℏxix,(75)V1x=∑i=0pn1i+1−n1iℏxix,(76)V2x=∑i=0pN1i+1−N1iℏxix,where the constants n10=N10=M10=0. If there exists an integrable function f1 on 0,X satisfying the hypothesis H1, then we have m1i+1≤f1x≤M1i+1 for each x∈xi,xi+1, i=0,1,2,…,p.Proposition 1.
Let the functionsf1 and f2 be two positive and integrable on 0,X. Assume that the functions U1, U2, V1, and V2 are defined by (73)–(76), respectively, and satisfying the hypothesis H1 defined by (30). Then, for κ>0, the following inequality for tempered fractional integral holds:(77)∑i=0pn1i+1N1i+1ℛxi,xi+1κ,τΨf12X2∑i=0pm1i+1M1i+1ℛxi,xi+1κ,τΨf22X2≤14∑i=0pn1i+1N1i+1+m1i+1M1i+1ℛxi,xi+1κ,τΨf1f2X.Proof.
By applying the Definition8, we have(78)ℛ0,Xκ,τΨV1V2f12X=∑i=0pn1i+1N1i+1ℛxi,xi+1κ,τΨf12X,(79)ℛ0,Xκ,τΨU1U2f22X=∑i=0pm1i+1M1i+1ℛxi,xi+1κ,τΨf22X,(80)ℛ0,Xκ,τΨU1V1+U2V2f1f2X=∑i=0pm1i+1n1i+1+M1i+1N1i+1ℛxi,xi+1κ,τΨf1f2X.
Hence, by applying Lemma1, we get the desired assertion (77).Proposition 2.
By settingΨθ=θ in Proposition 1, then we arrive to the following result in terms of classical tempered fractional integral:(81)∑i=0pn1i+1N1i+1ℛxi,xi+1κ,τf12X2∑i=0pm1i+1M1i+1ℛxi,xi+1κ,τf22X2≤14∑i=0pn1i+1N1i+1+m1i+1M1i+1ℛxi,xi+1κ,τf1f2X.Remark 5.
Throughout in the paper, if we applyΨθ=θ and τ=0, then all the newly presented inequalities will be reduced to the work derived earlier by Ntouyas et al. [18].


## 7. Concluding Remarks

Certain new Pólya–Szegö- and Chebyshev-type inequalities by utilizing tempered fractional integral are presented in this paper. These inequalities generalized the existing inequalities. We can easily get the said Pólya–Szegö- and Chebyshev-type inequalities for Katugampola, generalized Riemann–Liouville, classical Riemann–Liouville, generalized conformable, and conformable fractional integrals by applying different conditions on functionΨ given in Remark 2.



---

*Source: 9858671-2020-11-02.xml*