# Some New Ostrowski-Type Inequalities Involvingσ-Fractional Integrals

Authors: Bandar B. Mohsen; Muhammad Uzair Awan; Muhammad Zakria Javed; Muhammad Aslam Noor; Khalida Inayat Noor
Journal: Journal of Mathematics
 (2021)

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## Abstract

The aim of this paper is to derive some new fractional analogues of Ostrowski-type inequalities involving bounded functions using the concept ofσ-Riemann–Liouville fractional integrals.


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## Body

## 1. Introduction and Preliminaries

Inequalities play a pivotal role in modern analysis. Mathematical analysis depends upon many inequalities. In recent years, an extensive research has been carried out on obtaining various analogues of classical inequalities using different approaches, for details and applications, see [1–4]. A very interesting approach is to obtain fractional analogues of the inequalities. The fractional version of inequalities plays a significant role in the establishment of the uniqueness of solutions for certain fractional partial differential equations. Sarikaya et al. [5] were the first to introduce the concepts of fractional calculus in the theory of integral inequalities by obtaining the fractional analogues of classical Hermite–Hadamard’s inequality. Dragomir [6, 7] obtained fractional versions of Ostrowski-like inequalities. Erden et al. [8] recently obtained some more new fractional analogues of Ostrowski-type inequalities using bounded functions. Sarikaya [9] introduced the notion of two-dimensional Riemann–Liouville fractional integrals and obtained some new fractional variants of Hermite–Hadamard’s inequality on two dimensions. Having inspiration from the research work of Mubeen and Habibullah [10] and Sarikaya [9], Awan et al. [11] introduced the concepts of σ-Riemann–Liouville fractional integrals on two dimensions and obtained two-dimensional fractional integral inequalities. It is worth to mention here that if σ⟶1, then σ-Riemann–Liouville fractional integrals reduces to classical Riemann–Liouville fractional integral. Note that the concept of σ-Riemann–Liouville fractional integral is a significant generalization of classical Riemann–Liouville fractional integrals; as for σ≠1, the properties of σ–Riemann–Liouville fractional integrals are quite different from the classical Riemann–Liouville fractional integrals.The aim of this paper is to obtain some new fractional analogues the classical Ostrowski’s inequality using the concepts ofσ-fractional integrals. In order to obtain the main results of the paper, we first derive some new lemmas results, and then using these lemmas as auxiliary results, we derive our main results of the paper.Let us first recall some previously known concepts and results. The first one is the definition of the Riemann–Liouville fractional integrals.Definition 1.
(see [12]). Let F∈L1a,b. Then, the Riemann–Liouville integrals of order α1>o with a>0 are defined as follows:(1)Ja+α1Fx=1Γα1∫axx−t1α1−1Ft1dt1,x>a,Ja+α1Fx=1Γα1∫xbt1−xα1−1Ft1dt1,x<b,respectively. Here, Γα1 is the gamma function. These integrals are motivated by the well-known Cauchy formula:(2)∫axdt1∫at1dt2…∫atn−1ftndtn=1Γn∫axx−tn−1ftdt,n∈ℕ∗.
Mubeen and Habibullah [10] introduced the σ-Riemann–Liouville fractional integrals as follows.Definition 2.
(see [10]). Let Ξ∈L1a,b. The σ-Reimann–Liouville fractional integrals Jσa+α1Ξ and Jσb−α1Ξ of order α1>0 with a≥0 and σ>0 are defined as follows:(3)Jσa+α1Ξx=1σΓσα1∫abx−t1α1/σ−1Ξt1dt1,x>a,Jσb−α1Ξx=1σΓσα1∫xbt1−xα1/σ−1Ξt1dt1,x<b.
The above integrals for all functions are continuous and integrable on the interval0,∞. Note that if f∈L1a,b and a>0, then Jσaα1 exists almost everywhere on a,b. If α1≥1, σ>0, and f∈L1a,b, then Jσaα1f∈Ca,b. For more details, see [13].
Awan et al. [11] defined σ-Reimannn–Liouville fractional integrals of two variable functions as follows.Definition 3.
(see [11]). Let Ξ∈L1a,b×c,d. The Riemann–Liouville σ−fractional integrals Jσa+,c+α1,α2, Jσa+,d−α1,α2, Jσb−,c+α1,α2, and Jσb−,d−α1,α2 are defined as follows:(4)Jσa+,c+α1,α2Ξx,y=1σ2Γσα1Γσα2∫ax∫cyx−t1α1/σ−1y−t2α2/σ−1Ξt1,t2dt2dt1,x>a,y>c,Jσa+,d−α1,α2Ξx,y=1σ2Γσα1Γσα2∫ax∫ydx−t1α1/σ−1t2−yα2/σ−1Ξt1,t2dt2dt1,x>a,y<d,Jσb−,c+α1,α2Ξx,y=1σ2Γσα1Γσα2∫xb∫cyt1−xα1/σ−1y−t2α2/σ−1Ξt1,t2dt2dt1,x<b,y>c,Jσb−,c+α1,α2Ξx,y=1σ2Γσα1Γσα2∫xb∫ydt1−xα1/σ−1t2−yα2/σ−1Ξt1,t2dt2dt1,x<b,y<d,where α1,α2>0 and a,b,c,d≥0.
For the sake of simplicity, we define the following functions as(5)Mkα1a,b;x≔x−aα1/σ+b−xα1/σΓσα1+σ,Nkα1c,d;y≔y−cα2/σ+d−yα2/σΓσα2+σ,for x,y∈D:=a,b×c,d.


## 2. Main Results

### 2.1. Key Lemmas

In this section, we prove some lemmas which will help us in obtaining the main results of the paper.Lemma 1.
LetΞ:D⟶ℝ be an absolutely continuous, differentiable function such that ∂2Ξθ,μ/∂θ∂μ exists and is continuous on D⊆ℝ2. Then, for any x,y∈D, we have(6)1σ2Γσα1Γσα2∫ab∫cdGx,t1,y,t2∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1=Jσa+,c+α1,α2Ξx,y+Jσa+,d−α1,α2Ξx,y+Jσb−,c+α1,α2Ξx,y+Jσb−,d−α1,α2Ξx,y−Nkα2c,d:yJσa+α1Ξx,y+Jσb−α1Ξx,y−Mkα1a,b:xJσc+α1Ξx,y+Jσd−α1Ξx,y+Mkα1a,b:xNkα2c,d:yΞx,y=F1x,y;a,b,c,d,where(7)Gx,t1,y,t2=x−t1α1/σ−1y−t2α2/σ−1,a≤t1<xandc≤t2<y,x−t1α1/σ−1t2−yα2/σ−1,a≤t1<xandy≤t2<d,t1−xα1/σ−1y−t2α2/σ−1,x≤t1≤bandc≤t2<y,t1−xα1/σ−1t2−yα2/σ−1,x≤t1≤bandy≤t2≤d.Proof.
Now,(8)∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμ=Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,y=Lx,t1,y,t2.
This implies(9)I=1σ2Γσα1Γσα2∫ab∫cdGx,t1,y,t2∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1=1σ2Γσα1Γσα2∫ax∫cyx−t1α1/σ−1y−t2α2/σ−1Lx,t1,y,t2dt2dt1+1σ2Γσα1Γσα2∫ax∫ydx−t1α1/σ−1t2−yα2/σ−1Lx,t1,y,t2dt2dt1+1σ2Γσα1Γσα2∫xb∫cyt1−xα1/σ−1y−t2α2/σ−1Lx,t1,y,t2dt2dt1+1σ2Γσα1Γσα2∫xb∫ydt1−xα1/σ−1t2−yα2/σ−1Lx,t1,y,t2dt2dt1=I1+I2+I3+I4.
Now, consider(10)I1=1σ2Γσα1Γσα2∫ax∫cyx−t1α1/σ−1y−t2α2/σ−1Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,ydt2dt1=1σ2Γσα1Γσα2∫ax∫cyx−t1α1/σ−1y−t2α2/σ−1Ξt1,t2dt2dt1−1σ2Γσα1Γσα2∫ax∫cyx−t1α1/σ−1y−t2α2/σ−1Ξt1,ydt2dt1−1σ2Γσα1Γσα2∫ax∫cyx−t1α1/σ−1y−t2α2/σ−1Ξx,t2dt2dt1+1σ2Γσα1Γσα2∫ax∫cyx−t1α1/σ−1y−t2α2/σ−1Ξx,ydt2dt1=Jσa+,c+α1,α2Ξx,y−y−cα2/σΓσα2+σJσa+α1Ξx,y−x−aα1/σΓσα1+σJσc+α2Ξx,y+x−aα1/σy−cα2/σΓσα1+σΓσα2+σ.
Similarly,(11)I2=1σ2Γσα1Γσα2∫ax∫ydx−t1α1/σ−1t2−yα2/σ−1Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,ydt2dt1=Jσa+,d−α1,α2Ξx,y−d−yα2/σΓσα2+σJσa+α1Ξx,y−x−aα1/σΓσα1+σJσd−α2Ξx,y+x−aα1/σd−yα2/σΓσα1+σΓσα2+σ,(12)I3=1σ2Γσα1Γσα2∫xb∫cyt1−xα1/σ−1y−t2α2/σ−1Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,ydt2dt1=Jσb−,c+α1,α2Ξx,y−y−cα2/σΓσα2+σJσb−α1Ξx,y−b−xα1/σΓσα1+σJσc+α2Ξx,y+b−xα1/σy−cα2/σΓσα1+σΓσα2+σ,(13)I4=1σ2Γσα1Γσα2∫xb∫ydt1−xα1/σ−1y−t2α2/σ−1Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,ydt2dt1=Jσb−,d−α1,α2Ξx,y−d−yα2/σΓσα2+σJσb−α1Ξx,y−b−xα1/σΓσα1+σJσd−α2Ξx,y+b−xα1/σd−yα2/σΓσα1+σΓσα2+σ.
Using (10)–(13) in (9), we get the required result.Lemma 2.
LetΞ:D⟶ℝ be an absolutely continuous, differentiable function such that ∂2Ξθ,μ/∂θ∂μ exists and is continuous on D⊆ℝ2. Then, for any x,y∈D, we have(14)1σ2Γσα1Γσα2∫ab∫cdHt1,t2∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1=Jσx+,y+α1,α2Ξb,d+Jσx+,y−α1,α2Ξb,c+Jσx−,y+α1,α2Ξa,d+Jσx−,y−α1,α2Ξa,c−Nkα2c,d:yJσx+α1Ξb,y+Jσx−α1Ξa,y−Mkα1a,b:xJσy+α1Ξx,d+Jσy−α1Ξx,c+−Mkα1a,b:xNkα2c,d:yΞx,y=F2x,y;a,b,c,d,where(15)Ht1,t2=t1−aα1/σ−1t2−cα2/σ−1a≤t1<xandc≤t2<y,t1−aα1/σ−1d−t2α2/σ−1a≤t1<xandy≤t2≤d,b−t1α1/σ−1t2−cα2/σ−1x≤t1<bandc≤t2<y,b−t1α1/σ−1d−t2α2/σ−1x≤t1≤bandy≤t2<d.Proof.
The proof is same as the proof of Lemma1.Lemma 3.
LetΞ:D⟶ℝ be an absolutely continuous, differentiable function such that ∂2Ξθ,μ/∂θ∂μ exists and is continuous on D⊆ℝ2. Then, for any x,y∈D, we have(16)14σ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1+b−t1α1/σ−1×t2−cα2/σ−1+d−t2α2/σ−1∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1=Jσa+,c+α1,α2Ξb,d+Jσx+,d−α1,α2Ξb,c+Jσb−,c+α1,α2Ξa,d+Jσb−,d−α1,α2Ξa,c4−d−cα2/σ2Γσα2+σJσb−α1Ξa,y+Jσa+α1Ξb,y−b−aα1/σ2Γσα1+σJσd−α2Ξx,c+Jσc+α2Ξx,d+b−aα1/σd−cα2/σΓσα1+σΓσα2+σΞx,y=F3x,y;a,b,c,d.Proof.
Consider(17)I=14σ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1+b−t1α1/σ−1×t2−cα2/σ−1+d−t2α2/σ−1∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1=14σ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1t2−cα2/σ−1Lx,t1,y,t2dt2dt1+14σ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1d−t2α2/σ−1Lx,t1,y,t2dt2dt1+14σ2Γσα1Γσα2∫ab∫cdb−t1α1/σ−1t2−cα2/σ−1Lx,t1,y,t2dt2dt1+14σ2Γσα1Γσα2∫ab∫cdb−t1α1/σ−1d−t2α2/σ−1Lx,t1,y,t2dt2dt1=14I1+I2+I3+I4.
Now,(18)I1=1σ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1t2−cα2/σ−1Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,ydt2dt1=Jσb−,d−α1,α2Ξa,c−d−cα2/σΓσα2+σJσb−α1Ξa,y−b−aα1/σΓσα1+σJσd−α1Ξx,c+b−aα1/σd−cα2/σΓσα1+σΓσα2+σΞx,y.
Similarly,(19)I2=1σ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1d−t2α2/σ−1Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,ydt2dt1=Jσb−,c+α1,α2Ξa,d−d−cα2/σΓσα2+σJσb−α1Ξa,y−b−aα1/σΓσα1+σJσc+α1Ξx,d+b−aα1/σd−cα2/σΓσα1+σΓσα2+σΞx,y,I3=1σ2Γσα1Γσα2∫ab∫cdb−t1α1/σ−1t2−cα2/σ−1Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,ydt2dt1=Jσa+,d−α1,α2Ξb,c−d−cα2/σΓσα2+σJσa+α1Ξb,y−b−aα1/σΓσα1+σJσd−α1Ξx,c+b−aα1/σd−cα2/σΓσα1+σΓσα2+σΞx,y,I4=1σ2Γσα1Γσα2∫ab∫cdb−t1α1/σ−1d−t2α2/σ−1Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,ydt2dt1=Jσa+,c+α1,α2Ξb,d−d−cα2/σΓσα2+σJσa+α1Ξb,y−b−aα1/σΓσα1+σJσc+α1Ξx,d+b−aα1/σd−cα2/σΓσα1+σΓσα2+σΞx,y.
Using the values ofI1,I2,I3, and I4 in (17), we get the required result.


### 2.2. Results and Discussion

In this section, we discuss our main results.Theorem 1.
Under the assumptions of Lemma1, if Ξ is bounded, that is,(20)Ξθ,μ∞=supθ,μ∈D∂2Ξθ,μ∂θ∂μ<∞,then(21)F1x,y:a,b,c,d≤Ξθ,μ∞Mσα1+1a,b:xNσα2+1c,d:y,∀x,y∈D.Proof.
Using Lemma1 and the fact that Ξθ,μ is bounded, we have(22)F1x,y:a,b,c,d≤Ξθ,μ∞1σ2Γσα1Γσα2∫ab∫cdGx,y:a,b,c,dt1−xt2−ydt2dt1=Ξθ,μ∞1σ2Γσα1Γσα2∫ax∫cyx−t1α1/σy−t2α2/σdt2dt1+1σ2Γσα1Γσα2∫ax∫ydx−t1α1/σt2−yα2/σdt2dt1+1σ2Γσα1Γσα2∫xb∫cyt1−xα1/σy−t2α2/σdt2dt1+1σ2Γσα1Γσα2∫xb∫ydt1−xα1/σt2−yα2/σdt2dt1=Ξθ,μ∞I1+I2+I3+I4.
Now,(23)I1=1σ2Γσα1Γσα2∫ax∫cyx−t1α1/σy−t2α2/σdt2dt1=x−aα1/σ+1y−cα2/σ+1Γσα1+2σΓσα2+2σ.
Similarly,(24)I2=1σ2Γσα1Γσα2∫ax∫ydx−t1α1/σt2−yα2/σdt2dt1=x−aα1/σ+1d−yα2/σ+1Γσα1+2σΓσα2+2σ,I3=1σ2Γσα1Γσα2∫xb∫cyt1−xα1/σy−t2α2/σdt2dt1=b−xα1/σ+1y−cα2/σ+1Γσα1+2σΓσα2+2σ,I4=1σ2Γσα1Γσα2∫xb∫ydt1−xα1/σt2−yα2/σdt2dt1=b−xα1/σ+1d−yα2/σ+1Γσα1+2σΓσα2+2σ.
Substituting the values ofI1,I2,I3, and I4 in (22), we get the required result.Corollary 1.
Consideringx=a+b/2 and y=c+d/2 in Theorem 1, we have(25)Jσa+,c+α1,α2Ξa+b2,c+d2+Jσa+,d−α1,α2Ξa+b2,c+d2+Jσb−,c+α1,α2Ξa+b2,c+d2+Jσb−,d−α1,α2Ξa+b2,c+d2−d−cα2/σ2α2/σ−1Γσα2+σJσa+α1Ξa+b2,c+d2+Jσb−α1Ξa+b2,c+d2−b−aα1/σ2α1/σ−1Γσα1+σJσc+α1Ξa+b2,c+d2+Jσd−α1Ξa+b2,c+d2+b−aα1/σd−cα2/σ2α1+α2/σ−2Γσα1+σΓσα2+σΞa+b2,c+d2≤b−aα1/σ+1d−cα1/σ+12α1+α2/σΓσα1+2σΓσα2+2σΞθ,μ∞.Theorem 2.
Under the assumptions of Lemma2, if Ξ is bounded, that is,(26)Ξθ,μ∞=supθ,μ∈D∂2Ξθ,μ∂θ∂μ<∞,then(27)F1x,y:a,b,c,d≤Ξθ,μ∞Mkα1+1a,b:xNkα2+1c,d:y,∀x,y∈D.Proof.
The proof of this theorem follows the same technique which was used in Theorem1 by considering Lemma 2.Corollary 2.
By takingx=a+b/2 and y=c+d/2 in Theorem 2, we have(28)|Jσa+b/2+,c+d/2+α1,α2Ξb,d+Jσa+b/2+,c+d/2−α1,α2Ξb,c+Jσa+b/2−,c+d/2+α1,α2Ξa,d+Jσa+b/2−,c+d/2−α1,α2Ξa,c−d−cα2/σ2α2/σ−1Γσα2+σJσa+b/2+α1Ξb,c+d2+Jσa+b/2−α1Ξa,c+d2−b−aα1/σ2α1/σ−1Γσα1+σJσc+d/2+α1Ξa+b2,d+Jσc+d/2−α1Ξa+b2,c+b−aα1/σd−cα2/σ2α1+α2/σ−2Γσα1+σΓσα2+σΞa+b2,c+d2|≤b−aα1/σ+1d−cα2/σ+12α1+α2/σΓσα1+2σΓσα2+2σΞθ,μ∞.
We now derive results for mappings whose elements are ofLp space.Theorem 3.
Under the assumptions of Lemma1, if ∂2Ξθ,μ/∂θ∂μ∈LpD for p>1 with 1/p+1/q=1 with(29)Ξθ,μp=∫ab∫cd∂2Ξθ,μ∂θ∂μp1/p<∞,then(30)F1x,y:a,b,c,d≤Ξθ,μpx−aα1/σ+1/q+b−xα1/σ+1/qy−cα2/σ+1/q+d−yα2/σ+1/qσ2Γσα1Γσα2α1/σ+1/qα2/σ+1/q,for all x,y∈D.Proof.
From Lemma1, property of the modulus, and applying definition of Ξ with the use of Holder’s inequality, we have(31)σ2Γσα1Γσα2F1x,y:a,b,c,d≤∫ab∫cdGx,t1,y,t2∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1≤∫ab∫cdGx,t1,y,t2t1−x1/qt2−y1/q∫xt1∫yt2∂2Ξθ,μ∂θ∂μpdθdμ1/pdt2dt1=Ξθ,μp∫ab∫cdGx,t1,y,t2t1−x1/qt2−y1/qdt2dt1=‖Ξθ,μ‖p∫ax∫cyx−t1α1/σ+1/q−1y−t2α2/σ+1/q−1dt2dt1+∫ax∫ydx−t1α1/σ+1/q−1t2−yα2/σ+1/q−1dt2dt1+∫xb∫cyt1−xα1/σ+1/q−1y−t2α2/σ+1/q−1dt2dt1+∫xb∫ydt1−xα1/σ+1/q−1t2−yα2/σ+1/q−1dt2dt1=Ξθ,μpI1+I2+I3+I4.
Now,(32)I1=∫ax∫cyx−t1α1/σ+1/q−1y−t2α2/σ+1/q−1dt2dt1=x−aα1/σ+1/qy−cα2/σ+1/qα1/σ+1/qα2/σ+1/q.
Similarly, we find the values ofI2,I3, and I4, and substituting their values in (31), we get the required result.Theorem 4.
Under the assumptions of Lemma2, if ∂2Ξθ,μ/∂θ∂μ∈LpD for p>1 with 1/p+1/q=1 with(33)Ξθ,μp=∫ab∫cd∂2Ξθ,μ∂θ∂μp1/p<∞,then(34)F2x,y:a,b,c,d≤Ξθ,μpΓσσ+σ/q2Γσα1+σ+σ/qΓσα2+σ+σ/q×x−aα1/σ+1/q+b−xα1/σ+1/qy−cα2/σ+1/q+d−yα2/σ+1/q,for any x,y∈D.Proof.
From Lemma2, using the definition of Ξ and Holder’s inequality, we have(35)F2x,y:a,b,c,d≤1σ2Γσα1Γσα2∫ab∫cdHt1,t2∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1≤1σ2Γσα1Γσα2∫ab∫cdHt1,t2t1−x1/qt2−y1/q∫xt1∫yt2∂2Ξθ,μ∂θ∂μpdθdμ1/pdt2dt1=Ξθ,μp∫ab∫cdHt1,t2t1−x1/qt2−y1/qdt2dt1=Ξθ,μpσ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1t2−cα2/σ−1t1−x1/qt2−y1/qdt2dt1+Ξθ,μpσ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1d−t2α2/σ−1t1−x1/qt2−y1/qdt2dt1+Ξθ,μpσ2Γσα1Γσα2∫ab∫cdb−t1α1/σ−1t2−cα2/σ−1t1−x1/qt2−y1/qdt2dt1+Ξθ,μpσ2Γσα1Γσα2∫ab∫cdb−t1α1/σ−1d−t2α2/σ−1t1−x1/qt2−y1/qdt2dt1=I1+I2+I3+I4.
Now, considerI1:(36)I1=Ξθ,μpσ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1t2−cα2/σ−1t1−x1/qt2−y1/qdt2dt1=Ξθ,μpσ2Γσα1Γσα2∫abt1−aα1/σ−1t1−x1/qσt2−cα2/σy−t21/qα2|cy+∫cyσt2−cα2/σy−t21/q−1qα2dt2dt1=Ξθ,μpσ2Γσα1Γσα2σt1−aα1/σx−t11/qα1|ax+∫axσt1−aα1/σx−t11/q−1qα1dt1+∫cyσt2−cα2/σy−t21/q−1qα2dt2=Ξθ,μpq2Γσα1+σΓσα2+σ∫axt1−aα1/σx−t11/q−1dt1∫cyt2−cα2/σy−t21/q−1dt2=Ξθ,μpx−aα1/σ+1/qx−aα1/σ+1/qq2Γσα1+σΓσα2+σ×∫01uα1/σ1−u1/q−1du×∫01vα1/σ1−v1/q−1dv.
Thus,(37)I1=Ξθ,μpΓσσ/q+σ2x−aα1/σ+1/qx−aα1/σ+1/qΓσα1+σ+σ/qΓσα2+σσ/q.
Similarly, we can find the valuesI2,I3, and I4, and substituting the values in (35), we get the required result.
We now obtain the results whenΞ is element of L1D.Theorem 5.
Under the assumptions of Lemma1, if ∂2Ξθ,μ/∂θ∂μ∈L1D for p>1 with 1/p+1/q=1 with(38)Ξθ,μ1=∫ab∫cd∂2Ξθ,μ∂θ∂μ<∞,then(39)F1x,y:a,b,c,d≤Ξθ,μ1Mkα1a,b:xNkα2c,d:y,for all x,y∈D. And(40)F2x,y:a,b,c,d≤Ξθ,μ1Mkα1a,b:xNkα2c,d:y,for any x,y∈D.Proof.
From Lemma1, the property of modulus, and using the definition of Ξ, we have(41)F1x,y:a,b,c,d≤1σ2Γσα1Γσα2∫ab∫cdGx,t1,y,t2∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1≤Ξθ,μ11σ2Γσα1Γσα2∫ab∫cdGx,t1,y,t2dt2dt1=Ξθ,μ11σ2Γσα1Γσα2∫ax∫cyx−t1α1/σ−1y−t2α2/σ−1dt2dt1+1σ2Γσα1Γσα2∫ax∫ydx−t1α1/σ−1t2−yα2/σ−1dt2dt1+1σ2Γσα1Γσα2∫xb∫cyt1−xα1/σ−1y−t2α2/σ−1dt2dt1+1σ2Γσα1Γσα2∫xb∫ydt1−xα1/σ−1t2−yα2/σ−1dt2dt1.
Integrating above inequality, we get the required result.
To prove the other part of the inequality, we use Lemma2 and the same technique as used in the above part.Theorem 6.
Under the assumptions of Lemma3, if ∂2Ξθ,μ/∂θ∂μ∈L1D for p>1 with 1/p+1/q=1 with(42)Ξθ,μ1=∫ab∫cd∂2Ξθ,μ∂θ∂μ<∞,then(43)F1x,y:a,b,c,d≤Ξθ,μ1b−aα1/Kd−cα2/KΓσα1+σΓσα1+σ,for all x,y∈D.Proof.
From Lemma3 and using modulus property with the definition of Ξ, we have(44)F3x,y:a,b,c,d≤14σ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1+b−t1α1/σ−1×t2−cα2/σ−1+d−t2α2/σ−1∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1≤Ξθ,μ114σ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1+b−t1α1/σ−1×t2−cα2/σ−1+d−t2α2/σ−1dt2dt1.
Calculating the above double integral, we get the required result.


## 2.1. Key Lemmas

In this section, we prove some lemmas which will help us in obtaining the main results of the paper.Lemma 1.
LetΞ:D⟶ℝ be an absolutely continuous, differentiable function such that ∂2Ξθ,μ/∂θ∂μ exists and is continuous on D⊆ℝ2. Then, for any x,y∈D, we have(6)1σ2Γσα1Γσα2∫ab∫cdGx,t1,y,t2∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1=Jσa+,c+α1,α2Ξx,y+Jσa+,d−α1,α2Ξx,y+Jσb−,c+α1,α2Ξx,y+Jσb−,d−α1,α2Ξx,y−Nkα2c,d:yJσa+α1Ξx,y+Jσb−α1Ξx,y−Mkα1a,b:xJσc+α1Ξx,y+Jσd−α1Ξx,y+Mkα1a,b:xNkα2c,d:yΞx,y=F1x,y;a,b,c,d,where(7)Gx,t1,y,t2=x−t1α1/σ−1y−t2α2/σ−1,a≤t1<xandc≤t2<y,x−t1α1/σ−1t2−yα2/σ−1,a≤t1<xandy≤t2<d,t1−xα1/σ−1y−t2α2/σ−1,x≤t1≤bandc≤t2<y,t1−xα1/σ−1t2−yα2/σ−1,x≤t1≤bandy≤t2≤d.Proof.
Now,(8)∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμ=Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,y=Lx,t1,y,t2.
This implies(9)I=1σ2Γσα1Γσα2∫ab∫cdGx,t1,y,t2∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1=1σ2Γσα1Γσα2∫ax∫cyx−t1α1/σ−1y−t2α2/σ−1Lx,t1,y,t2dt2dt1+1σ2Γσα1Γσα2∫ax∫ydx−t1α1/σ−1t2−yα2/σ−1Lx,t1,y,t2dt2dt1+1σ2Γσα1Γσα2∫xb∫cyt1−xα1/σ−1y−t2α2/σ−1Lx,t1,y,t2dt2dt1+1σ2Γσα1Γσα2∫xb∫ydt1−xα1/σ−1t2−yα2/σ−1Lx,t1,y,t2dt2dt1=I1+I2+I3+I4.
Now, consider(10)I1=1σ2Γσα1Γσα2∫ax∫cyx−t1α1/σ−1y−t2α2/σ−1Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,ydt2dt1=1σ2Γσα1Γσα2∫ax∫cyx−t1α1/σ−1y−t2α2/σ−1Ξt1,t2dt2dt1−1σ2Γσα1Γσα2∫ax∫cyx−t1α1/σ−1y−t2α2/σ−1Ξt1,ydt2dt1−1σ2Γσα1Γσα2∫ax∫cyx−t1α1/σ−1y−t2α2/σ−1Ξx,t2dt2dt1+1σ2Γσα1Γσα2∫ax∫cyx−t1α1/σ−1y−t2α2/σ−1Ξx,ydt2dt1=Jσa+,c+α1,α2Ξx,y−y−cα2/σΓσα2+σJσa+α1Ξx,y−x−aα1/σΓσα1+σJσc+α2Ξx,y+x−aα1/σy−cα2/σΓσα1+σΓσα2+σ.
Similarly,(11)I2=1σ2Γσα1Γσα2∫ax∫ydx−t1α1/σ−1t2−yα2/σ−1Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,ydt2dt1=Jσa+,d−α1,α2Ξx,y−d−yα2/σΓσα2+σJσa+α1Ξx,y−x−aα1/σΓσα1+σJσd−α2Ξx,y+x−aα1/σd−yα2/σΓσα1+σΓσα2+σ,(12)I3=1σ2Γσα1Γσα2∫xb∫cyt1−xα1/σ−1y−t2α2/σ−1Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,ydt2dt1=Jσb−,c+α1,α2Ξx,y−y−cα2/σΓσα2+σJσb−α1Ξx,y−b−xα1/σΓσα1+σJσc+α2Ξx,y+b−xα1/σy−cα2/σΓσα1+σΓσα2+σ,(13)I4=1σ2Γσα1Γσα2∫xb∫ydt1−xα1/σ−1y−t2α2/σ−1Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,ydt2dt1=Jσb−,d−α1,α2Ξx,y−d−yα2/σΓσα2+σJσb−α1Ξx,y−b−xα1/σΓσα1+σJσd−α2Ξx,y+b−xα1/σd−yα2/σΓσα1+σΓσα2+σ.
Using (10)–(13) in (9), we get the required result.Lemma 2.
LetΞ:D⟶ℝ be an absolutely continuous, differentiable function such that ∂2Ξθ,μ/∂θ∂μ exists and is continuous on D⊆ℝ2. Then, for any x,y∈D, we have(14)1σ2Γσα1Γσα2∫ab∫cdHt1,t2∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1=Jσx+,y+α1,α2Ξb,d+Jσx+,y−α1,α2Ξb,c+Jσx−,y+α1,α2Ξa,d+Jσx−,y−α1,α2Ξa,c−Nkα2c,d:yJσx+α1Ξb,y+Jσx−α1Ξa,y−Mkα1a,b:xJσy+α1Ξx,d+Jσy−α1Ξx,c+−Mkα1a,b:xNkα2c,d:yΞx,y=F2x,y;a,b,c,d,where(15)Ht1,t2=t1−aα1/σ−1t2−cα2/σ−1a≤t1<xandc≤t2<y,t1−aα1/σ−1d−t2α2/σ−1a≤t1<xandy≤t2≤d,b−t1α1/σ−1t2−cα2/σ−1x≤t1<bandc≤t2<y,b−t1α1/σ−1d−t2α2/σ−1x≤t1≤bandy≤t2<d.Proof.
The proof is same as the proof of Lemma1.Lemma 3.
LetΞ:D⟶ℝ be an absolutely continuous, differentiable function such that ∂2Ξθ,μ/∂θ∂μ exists and is continuous on D⊆ℝ2. Then, for any x,y∈D, we have(16)14σ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1+b−t1α1/σ−1×t2−cα2/σ−1+d−t2α2/σ−1∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1=Jσa+,c+α1,α2Ξb,d+Jσx+,d−α1,α2Ξb,c+Jσb−,c+α1,α2Ξa,d+Jσb−,d−α1,α2Ξa,c4−d−cα2/σ2Γσα2+σJσb−α1Ξa,y+Jσa+α1Ξb,y−b−aα1/σ2Γσα1+σJσd−α2Ξx,c+Jσc+α2Ξx,d+b−aα1/σd−cα2/σΓσα1+σΓσα2+σΞx,y=F3x,y;a,b,c,d.Proof.
Consider(17)I=14σ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1+b−t1α1/σ−1×t2−cα2/σ−1+d−t2α2/σ−1∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1=14σ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1t2−cα2/σ−1Lx,t1,y,t2dt2dt1+14σ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1d−t2α2/σ−1Lx,t1,y,t2dt2dt1+14σ2Γσα1Γσα2∫ab∫cdb−t1α1/σ−1t2−cα2/σ−1Lx,t1,y,t2dt2dt1+14σ2Γσα1Γσα2∫ab∫cdb−t1α1/σ−1d−t2α2/σ−1Lx,t1,y,t2dt2dt1=14I1+I2+I3+I4.
Now,(18)I1=1σ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1t2−cα2/σ−1Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,ydt2dt1=Jσb−,d−α1,α2Ξa,c−d−cα2/σΓσα2+σJσb−α1Ξa,y−b−aα1/σΓσα1+σJσd−α1Ξx,c+b−aα1/σd−cα2/σΓσα1+σΓσα2+σΞx,y.
Similarly,(19)I2=1σ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1d−t2α2/σ−1Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,ydt2dt1=Jσb−,c+α1,α2Ξa,d−d−cα2/σΓσα2+σJσb−α1Ξa,y−b−aα1/σΓσα1+σJσc+α1Ξx,d+b−aα1/σd−cα2/σΓσα1+σΓσα2+σΞx,y,I3=1σ2Γσα1Γσα2∫ab∫cdb−t1α1/σ−1t2−cα2/σ−1Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,ydt2dt1=Jσa+,d−α1,α2Ξb,c−d−cα2/σΓσα2+σJσa+α1Ξb,y−b−aα1/σΓσα1+σJσd−α1Ξx,c+b−aα1/σd−cα2/σΓσα1+σΓσα2+σΞx,y,I4=1σ2Γσα1Γσα2∫ab∫cdb−t1α1/σ−1d−t2α2/σ−1Ξt1,t2−Ξt1,y−Ξx,t2+Ξx,ydt2dt1=Jσa+,c+α1,α2Ξb,d−d−cα2/σΓσα2+σJσa+α1Ξb,y−b−aα1/σΓσα1+σJσc+α1Ξx,d+b−aα1/σd−cα2/σΓσα1+σΓσα2+σΞx,y.
Using the values ofI1,I2,I3, and I4 in (17), we get the required result.


## 2.2. Results and Discussion

In this section, we discuss our main results.Theorem 1.
Under the assumptions of Lemma1, if Ξ is bounded, that is,(20)Ξθ,μ∞=supθ,μ∈D∂2Ξθ,μ∂θ∂μ<∞,then(21)F1x,y:a,b,c,d≤Ξθ,μ∞Mσα1+1a,b:xNσα2+1c,d:y,∀x,y∈D.Proof.
Using Lemma1 and the fact that Ξθ,μ is bounded, we have(22)F1x,y:a,b,c,d≤Ξθ,μ∞1σ2Γσα1Γσα2∫ab∫cdGx,y:a,b,c,dt1−xt2−ydt2dt1=Ξθ,μ∞1σ2Γσα1Γσα2∫ax∫cyx−t1α1/σy−t2α2/σdt2dt1+1σ2Γσα1Γσα2∫ax∫ydx−t1α1/σt2−yα2/σdt2dt1+1σ2Γσα1Γσα2∫xb∫cyt1−xα1/σy−t2α2/σdt2dt1+1σ2Γσα1Γσα2∫xb∫ydt1−xα1/σt2−yα2/σdt2dt1=Ξθ,μ∞I1+I2+I3+I4.
Now,(23)I1=1σ2Γσα1Γσα2∫ax∫cyx−t1α1/σy−t2α2/σdt2dt1=x−aα1/σ+1y−cα2/σ+1Γσα1+2σΓσα2+2σ.
Similarly,(24)I2=1σ2Γσα1Γσα2∫ax∫ydx−t1α1/σt2−yα2/σdt2dt1=x−aα1/σ+1d−yα2/σ+1Γσα1+2σΓσα2+2σ,I3=1σ2Γσα1Γσα2∫xb∫cyt1−xα1/σy−t2α2/σdt2dt1=b−xα1/σ+1y−cα2/σ+1Γσα1+2σΓσα2+2σ,I4=1σ2Γσα1Γσα2∫xb∫ydt1−xα1/σt2−yα2/σdt2dt1=b−xα1/σ+1d−yα2/σ+1Γσα1+2σΓσα2+2σ.
Substituting the values ofI1,I2,I3, and I4 in (22), we get the required result.Corollary 1.
Consideringx=a+b/2 and y=c+d/2 in Theorem 1, we have(25)Jσa+,c+α1,α2Ξa+b2,c+d2+Jσa+,d−α1,α2Ξa+b2,c+d2+Jσb−,c+α1,α2Ξa+b2,c+d2+Jσb−,d−α1,α2Ξa+b2,c+d2−d−cα2/σ2α2/σ−1Γσα2+σJσa+α1Ξa+b2,c+d2+Jσb−α1Ξa+b2,c+d2−b−aα1/σ2α1/σ−1Γσα1+σJσc+α1Ξa+b2,c+d2+Jσd−α1Ξa+b2,c+d2+b−aα1/σd−cα2/σ2α1+α2/σ−2Γσα1+σΓσα2+σΞa+b2,c+d2≤b−aα1/σ+1d−cα1/σ+12α1+α2/σΓσα1+2σΓσα2+2σΞθ,μ∞.Theorem 2.
Under the assumptions of Lemma2, if Ξ is bounded, that is,(26)Ξθ,μ∞=supθ,μ∈D∂2Ξθ,μ∂θ∂μ<∞,then(27)F1x,y:a,b,c,d≤Ξθ,μ∞Mkα1+1a,b:xNkα2+1c,d:y,∀x,y∈D.Proof.
The proof of this theorem follows the same technique which was used in Theorem1 by considering Lemma 2.Corollary 2.
By takingx=a+b/2 and y=c+d/2 in Theorem 2, we have(28)|Jσa+b/2+,c+d/2+α1,α2Ξb,d+Jσa+b/2+,c+d/2−α1,α2Ξb,c+Jσa+b/2−,c+d/2+α1,α2Ξa,d+Jσa+b/2−,c+d/2−α1,α2Ξa,c−d−cα2/σ2α2/σ−1Γσα2+σJσa+b/2+α1Ξb,c+d2+Jσa+b/2−α1Ξa,c+d2−b−aα1/σ2α1/σ−1Γσα1+σJσc+d/2+α1Ξa+b2,d+Jσc+d/2−α1Ξa+b2,c+b−aα1/σd−cα2/σ2α1+α2/σ−2Γσα1+σΓσα2+σΞa+b2,c+d2|≤b−aα1/σ+1d−cα2/σ+12α1+α2/σΓσα1+2σΓσα2+2σΞθ,μ∞.
We now derive results for mappings whose elements are ofLp space.Theorem 3.
Under the assumptions of Lemma1, if ∂2Ξθ,μ/∂θ∂μ∈LpD for p>1 with 1/p+1/q=1 with(29)Ξθ,μp=∫ab∫cd∂2Ξθ,μ∂θ∂μp1/p<∞,then(30)F1x,y:a,b,c,d≤Ξθ,μpx−aα1/σ+1/q+b−xα1/σ+1/qy−cα2/σ+1/q+d−yα2/σ+1/qσ2Γσα1Γσα2α1/σ+1/qα2/σ+1/q,for all x,y∈D.Proof.
From Lemma1, property of the modulus, and applying definition of Ξ with the use of Holder’s inequality, we have(31)σ2Γσα1Γσα2F1x,y:a,b,c,d≤∫ab∫cdGx,t1,y,t2∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1≤∫ab∫cdGx,t1,y,t2t1−x1/qt2−y1/q∫xt1∫yt2∂2Ξθ,μ∂θ∂μpdθdμ1/pdt2dt1=Ξθ,μp∫ab∫cdGx,t1,y,t2t1−x1/qt2−y1/qdt2dt1=‖Ξθ,μ‖p∫ax∫cyx−t1α1/σ+1/q−1y−t2α2/σ+1/q−1dt2dt1+∫ax∫ydx−t1α1/σ+1/q−1t2−yα2/σ+1/q−1dt2dt1+∫xb∫cyt1−xα1/σ+1/q−1y−t2α2/σ+1/q−1dt2dt1+∫xb∫ydt1−xα1/σ+1/q−1t2−yα2/σ+1/q−1dt2dt1=Ξθ,μpI1+I2+I3+I4.
Now,(32)I1=∫ax∫cyx−t1α1/σ+1/q−1y−t2α2/σ+1/q−1dt2dt1=x−aα1/σ+1/qy−cα2/σ+1/qα1/σ+1/qα2/σ+1/q.
Similarly, we find the values ofI2,I3, and I4, and substituting their values in (31), we get the required result.Theorem 4.
Under the assumptions of Lemma2, if ∂2Ξθ,μ/∂θ∂μ∈LpD for p>1 with 1/p+1/q=1 with(33)Ξθ,μp=∫ab∫cd∂2Ξθ,μ∂θ∂μp1/p<∞,then(34)F2x,y:a,b,c,d≤Ξθ,μpΓσσ+σ/q2Γσα1+σ+σ/qΓσα2+σ+σ/q×x−aα1/σ+1/q+b−xα1/σ+1/qy−cα2/σ+1/q+d−yα2/σ+1/q,for any x,y∈D.Proof.
From Lemma2, using the definition of Ξ and Holder’s inequality, we have(35)F2x,y:a,b,c,d≤1σ2Γσα1Γσα2∫ab∫cdHt1,t2∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1≤1σ2Γσα1Γσα2∫ab∫cdHt1,t2t1−x1/qt2−y1/q∫xt1∫yt2∂2Ξθ,μ∂θ∂μpdθdμ1/pdt2dt1=Ξθ,μp∫ab∫cdHt1,t2t1−x1/qt2−y1/qdt2dt1=Ξθ,μpσ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1t2−cα2/σ−1t1−x1/qt2−y1/qdt2dt1+Ξθ,μpσ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1d−t2α2/σ−1t1−x1/qt2−y1/qdt2dt1+Ξθ,μpσ2Γσα1Γσα2∫ab∫cdb−t1α1/σ−1t2−cα2/σ−1t1−x1/qt2−y1/qdt2dt1+Ξθ,μpσ2Γσα1Γσα2∫ab∫cdb−t1α1/σ−1d−t2α2/σ−1t1−x1/qt2−y1/qdt2dt1=I1+I2+I3+I4.
Now, considerI1:(36)I1=Ξθ,μpσ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1t2−cα2/σ−1t1−x1/qt2−y1/qdt2dt1=Ξθ,μpσ2Γσα1Γσα2∫abt1−aα1/σ−1t1−x1/qσt2−cα2/σy−t21/qα2|cy+∫cyσt2−cα2/σy−t21/q−1qα2dt2dt1=Ξθ,μpσ2Γσα1Γσα2σt1−aα1/σx−t11/qα1|ax+∫axσt1−aα1/σx−t11/q−1qα1dt1+∫cyσt2−cα2/σy−t21/q−1qα2dt2=Ξθ,μpq2Γσα1+σΓσα2+σ∫axt1−aα1/σx−t11/q−1dt1∫cyt2−cα2/σy−t21/q−1dt2=Ξθ,μpx−aα1/σ+1/qx−aα1/σ+1/qq2Γσα1+σΓσα2+σ×∫01uα1/σ1−u1/q−1du×∫01vα1/σ1−v1/q−1dv.
Thus,(37)I1=Ξθ,μpΓσσ/q+σ2x−aα1/σ+1/qx−aα1/σ+1/qΓσα1+σ+σ/qΓσα2+σσ/q.
Similarly, we can find the valuesI2,I3, and I4, and substituting the values in (35), we get the required result.
We now obtain the results whenΞ is element of L1D.Theorem 5.
Under the assumptions of Lemma1, if ∂2Ξθ,μ/∂θ∂μ∈L1D for p>1 with 1/p+1/q=1 with(38)Ξθ,μ1=∫ab∫cd∂2Ξθ,μ∂θ∂μ<∞,then(39)F1x,y:a,b,c,d≤Ξθ,μ1Mkα1a,b:xNkα2c,d:y,for all x,y∈D. And(40)F2x,y:a,b,c,d≤Ξθ,μ1Mkα1a,b:xNkα2c,d:y,for any x,y∈D.Proof.
From Lemma1, the property of modulus, and using the definition of Ξ, we have(41)F1x,y:a,b,c,d≤1σ2Γσα1Γσα2∫ab∫cdGx,t1,y,t2∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1≤Ξθ,μ11σ2Γσα1Γσα2∫ab∫cdGx,t1,y,t2dt2dt1=Ξθ,μ11σ2Γσα1Γσα2∫ax∫cyx−t1α1/σ−1y−t2α2/σ−1dt2dt1+1σ2Γσα1Γσα2∫ax∫ydx−t1α1/σ−1t2−yα2/σ−1dt2dt1+1σ2Γσα1Γσα2∫xb∫cyt1−xα1/σ−1y−t2α2/σ−1dt2dt1+1σ2Γσα1Γσα2∫xb∫ydt1−xα1/σ−1t2−yα2/σ−1dt2dt1.
Integrating above inequality, we get the required result.
To prove the other part of the inequality, we use Lemma2 and the same technique as used in the above part.Theorem 6.
Under the assumptions of Lemma3, if ∂2Ξθ,μ/∂θ∂μ∈L1D for p>1 with 1/p+1/q=1 with(42)Ξθ,μ1=∫ab∫cd∂2Ξθ,μ∂θ∂μ<∞,then(43)F1x,y:a,b,c,d≤Ξθ,μ1b−aα1/Kd−cα2/KΓσα1+σΓσα1+σ,for all x,y∈D.Proof.
From Lemma3 and using modulus property with the definition of Ξ, we have(44)F3x,y:a,b,c,d≤14σ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1+b−t1α1/σ−1×t2−cα2/σ−1+d−t2α2/σ−1∫xt1∫yt2∂2Ξθ,μ∂θ∂μdθdμdt2dt1≤Ξθ,μ114σ2Γσα1Γσα2∫ab∫cdt1−aα1/σ−1+b−t1α1/σ−1×t2−cα2/σ−1+d−t2α2/σ−1dt2dt1.
Calculating the above double integral, we get the required result.


## 3. Conclusion

We have derived three new auxiliary results. Using these new auxiliary results, we have derived some newσ-fractional analogues of Ostrowski-type inequalities involving bounded functions in Lp,L∞, and L1 spaces. We have also discussed some new special cases in which we have obtained some midpoint-type inequalities. We hope that the techniques used in this paper will inspire interested readers.



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