# Time Scale Inequalities of the Ostrowski Type for Functions Differentiable on the Coordinates

Authors: Eze R. Nwaeze; Seth Kermausuor; Ana M. Tameru
Journal: Abstract and Applied Analysis
 (2018)

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## Abstract

In 2016, some inequalities of the Ostrowski type for functions (of two variables) differentiable on the coordinates were established. In this paper, we extend these results to an arbitrary time scale by means of a parameterλ∈0,1. The aforementioned results are regained for the case when the time scale T=R. Besides extension, our results are employed to the continuous and discrete calculus to get some new inequalities in this direction.


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## Body

## 1. Introduction

To find a bound for the difference of a function and its integral mean, the Ukraine-born mathematician Ostrowski [1], in 1938, established the subsequent result which is nowadays celebrated as the Ostrowski inequality.Theorem 1.
LetG:[α,β]→R be continuous on [α,β] and differentiable in (α,β) and its derivative G′:(α,β)→R is bounded in (α,β). If |G′(s)|≤M for all s∈[α,β], then we have(1)Gx-1β-α∫αβGsds≤14+x-α+β/22β-α2β-αM,for all x∈[α,β]. The inequality is sharp in the sense that the constant 1/4 cannot be replaced by a smaller one.In 2001, Cheng [2] gave the following improvement of the above inequality.Theorem 2.
LetG:[α,β]→R be continuous on [α,β] and differentiable in (α,β) such that there exist constants γ,Γ∈R with γ≤G′(s)≤Γ for all s∈[α,β]. Then for all x∈[α,β], one gets(2)12Gx-x-βGβ-x-αGα2β-α-1β-α∫αβGsds≤x-α2+β-x28β-αΓ-γ.Recently, Farid [3] extended Theorems 1 and 2 to functions of two variables that are differentiable on their coordinates. Specifically, he proved the following two theorems.Theorem 3.
LetG:I×J→R, where I, J are open intervals in R, be a mapping such that for α1,β1∈I,α2,β2∈J,α1<β1,α2<β2, the partial mappings(3)Gy:α1,β1⟶R,Gyξ≔Gξ,y,Gx:α2,β2⟶R,Gxζ≔Gx,ζ,defined for all y∈[α2,β2] and x∈[α1,β1], are differentiable, and Gy′s≤M,s∈α1,β1,Gx′s≤N,s∈[α2,β2]. Then we have(4)∫α1β1Gx,α2+Gx,β22dx+∫α2β2Gα1,y+Gβ1,y2dy-1β1-α1+1β2-α2∫α1β1∫α2β2Gx,ydydx≤M+N2β1-α1β2-α2,(5)∫α1β1Gx,α2+Gx,β24β1-α1dx+∫α2β2Gα1,y+Gβ1,y4β2-α2dy-1β1-α1β2-α2∫α1β1∫α2β2Gx,ydydx≤Mβ1-α1+Nβ2-α24,(6)∫α1β1Gx,α2+β22dx+∫α2β2Gα1+β12,ydy-1β1-α1+1β2-α2∫α1β1∫α2β2Gx,ydydx≤M+N4β1-α1β2-α2,(7)12β1-α1∫α1β1Gx,α2+β22dx+12β2-α2∫α2β2Gα1+β12,ydy-1β1-α1β2-α2∫α1β1∫α2β2Gx,ydydx≤Mβ1-α1+Nβ2-α28.Remark 4.
Inequality (7) is the correct version of inequality (2.18) as presented in [[3], Theorem 2.6].Theorem 5.
LetG:I×J→R, where I,J are open intervals in R, be a mapping such that for α1,β1∈I,α2,β2∈J,α1<β1,α2<β2, the partial mappings(8)Gy:α1,β1⟶R,Gyξ≔Gξ,y,Gx:α2,β2⟶R,Gxζ≔Gx,ζ,defined for all y∈[α2,β2] and x∈[α1,β1], are differentiable with γy≤Gy′s≤Γy,s∈[α1,β1], γx≤Gx′s≤Γx,s∈[α2,β2]. Then we have(9)∫α1β1Gx,α2+Gx,β22dx+∫α2β2Gα1,y+Gβ1,y2dy-1β1-α1+1β2-α2∫α1β1∫α2β2Gx,ydydx≤Γx+Γy-γx+γy8β1-α1β2-α2,∫α1β1Gx,α2+Gx,β24β1-α1dx+∫α2β2Gα1,y+Gβ1,y4β2-α2dy-1β1-α1β2-α2∫α1β1∫α2β2Gx,ydydx≤β1-α1Γy-γy+β2-α2Γx-γx16.In 1988, the idea of time scales [4] was initiated so as to bring together the continuous and discrete analysis into a unified fold. Since the introduction of this subject, many classical integral results have been extended to time scales. “A time scale T is an arbitrary nonempty closed subset of R.” We shall presume, all over this work, that the reader is familiar with the theory of time scale (see [5, 6] for more on this subject). We present here a result of Bohner and Matthews [7] which is embedded in Theorem 6 below. This result extends Theorem 1 to time scales. For more improvements and generalizations around this result, we refer the interested reader to see the papers [8–12] and the references therein.Theorem 6.
Letα,β,s,t∈T, α<β and G:[α,β]→R be differentiable. Then for all t∈[α,β], we have (10)Gt-1β-α∫αβGσsΔs≤Mβ-αh2t,α+h2t,β,where h2t,s=∫st(τ-s)Δτ for all s,t∈T and M=supα<t<β⁡GΔt<∞. This inequality is sharp in the sense that the right-hand side of (10) cannot be replaced by a smaller one.It is our purpose in this paper to extend inequalities (4), (5), (6), (7), and (9) to time scales by means of a parameter λ∈[0,1]. In Section 2, we frame and prove the main results followed by applications to the continuous and discrete calculus.


## 2. Main Results

In this section, we will present our results involving double integrals. For some recent results in this regard, see [13–18]. The proofs of our findings shall be anchored on the subsequent lemmas.Lemma 7 (see [9]).
Letα,β,t,x∈T,α<β and G:[α,β]→R be differentiable. Then(11)1-λGx+λGα+Gβ2-1β-α∫αβGσtΔt≤Mβ-αh2α,α+λβ-α2+h2x,α+λβ-α2+h2x,β-λβ-α2+h2β,β-λβ-α2,for all λ∈0,1 such that α+λβ-α/2 and β-λβ-α/2 are in T and x∈α+λβ-α/2,β-λβ-α/2∩T, where M≔supα<x<β⁡|GΔ(x)|<∞. This is sharp provided that(12)λ2αβ-α+λ24β-α2≤∫αα+λβ-α/2tΔt.Lemma 8 (see [8]).
Letα,β,t,x∈T,α<β and G:[α,β]→R be differentiable. If GΔ∈Crd(T,R) and there exist γ,Γ∈R such that γ≤GΔ(t)≤Γ for all t∈[α,β], then for all x∈[α,β], we have(13)1-λ2Gx+λx-αGα+β-xGβ2β-α-1β-α∫αβGσtΔt-Γ+γ21β-αh2x,α-h2x,β-λx-α2-β-x22≤Γ-γ2β-αh2α,α+λx-α2+h2x,α+λx-α2+h2x,β-λβ-x2+h2β,β-λβ-x2,for all λ∈[0,1] such that α+λx-α/2 and β-λβ-x/2 are in T.We now formulate and prove our first result.Theorem 9.
Letα1,β1,x∈T1,α2,β2,y∈T2, with α1<β1,α2<β2 and G:[α1,β1]×[α2,β2]→R be such that the partial mappings(14)Gy:α1,β1⟶R,Gyξ≔Gξ,y,Gx:α2,β2⟶R,Gxζ≔Gx,ζ,defined for all y∈[α2,β2] and x∈[α1,β1], are differentiable. If M=supα1<t<β1⁡GyΔt and N=supα2<t<β2⁡GxΔt, then the succeeding inequalities(15)∫α1β1Gx,α2+Gx,β22Δx+∫α2β2Gα1,y+Gβ1,y2Δy-1β1-α1∫α1β1∫α2β2Gσx,yΔyΔx-1β2-α2∫α1β1∫α2β2Gx,σyΔyΔx≤Mβ2-α22β1-α13h2α1,α1+λβ1-α12+h2β1,α1+λβ1-α12+3h2β1,β1-λβ1-α12+h2α1,β1-λβ1-α12+Nβ1-α12β2-α23h2α2,α2+λβ2-α22+h2β2,α2+λβ2-α22+3h2β2,β2-λβ2-α22+h2α2,β2-λβ2-α22,(16)∫α1β1Gx,α2+Gx,β24β1-α1Δx+∫α2β2Gα1,y+Gβ1,y4β2-α2Δy-12β1-α1β2-α2∫α1β1∫α2β2Gx,σy+Gσx,yΔyΔx≤M4β1-α13h2α1,α1+λβ1-α12+h2β1,α1+λβ1-α12+3h2β1,β1-λβ1-α12+h2α1,β1-λβ1-α12+N4β2-α23h2α2,α2+λβ2-α22+h2β2,α2+λβ2-α22+3h2β2,β2-λβ2-α22+h2α2,β2-λβ2-α22,hold for all λ∈0,1 such that (α1+λβ1-α1/2,α2+λβ2-α2/2)∈T1×T2 and (β1-λβ1-α1/2,β2-λβ2-α2/2)∈T1×T2.Proof.
Applying Lemma7 to Gy at x=β1, we get(17)1-λ2Gβ1,y+λ2Gα1,y-1β1-α1∫α1β1Gσx,yΔx≤Mβ1-α1h2α1,α1+λβ1-α12+h2β1,α1+λβ1-α12+2h2β1,β1-λβ1-α12.Integrating (17) over [α2,β2] gives(18)1-λ2∫α2β2Gβ1,yΔy+λ2∫α2β2Gα1,yΔy-1β1-α1∫α1β1∫α2β2Gσx,yΔyΔx≤Mβ2-α2β1-α1h2α1,α1+λβ1-α12+h2β1,α1+λβ1-α12+2h2β1,β1-λβ1-α12.
Applying, again, Lemma7 to Gy at x=α1, and integrating over [α2,β2] give(19)1-λ2∫α2β2Gα1,yΔy+λ2∫α2β2Gβ1,yΔy-1β1-α1∫α1β1∫α2β2Gσx,yΔyΔx≤Mβ2-α2β1-α12h2α1,α1+λβ1-α12+h2α1,β1-λβ1-α12+h2β1,β1-λβ1-α12.Using (18) and (19), we have(20)∫α2β2Gα1,y+Gβ1,y2Δy-1β1-α1∫α1β1∫α2β2Gσx,yΔyΔx≤Mβ2-α22β1-α13h2α1,α1+λβ1-α12+h2β1,α1+λβ1-α12+3h2β1,β1-λβ1-α12+h2α1,β1-λβ1-α12.Similarly, doing the same thing for Gx at y=α2 and y=β2, and then integrating the resultant inequality over [α1,β1], we get(21)∫α1β1Gx,α2+Gx,β22Δx-1β2-α2∫α1β1∫α2β2Gx,σyΔyΔx≤Nβ1-α12β2-α23h2α2,α2+λβ2-α22+h2β2,α2+λβ2-α22+3h2β2,β2-λβ2-α22+h2α2,β2-λβ2-α22.Using (20) and (21) amounts to (15). Also, from (20) and (21), we get(22)∫α2β2Gα1,y+Gβ1,y2β2-α2Δy-1β1-α1β2-α2∫α1β1∫α2β2Gσx,yΔyΔx≤M2β1-α13h2α1,α1+λβ1-α12+h2β1,α1+λβ1-α12+3h2β1,β1-λβ1-α12+h2α1,β1-λβ1-α12,(23)∫α1β1Gx,α2+Gx,β22β1-α1Δx-1β1-α1β2-α2∫α1β1∫α2β2Gx,σyΔyΔx≤N2β2-α23h2α2,α2+λβ2-α22+h2β2,α2+λβ2-α22+3h2β2,β2-λβ2-α22+h2α2,β2-λβ2-α22.Combining (22) and (23), one gets (16).Theorem 10.
Under the assumptions of Theorem9 and suppose also the intervals contain the mid points, then we have(24)
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Next, we now apply Lemma7 to Gy at x=α1+β1/2 and thereafter integrate the resulting inequality over [α2,β2] to get(26)1-λ∫α2β2Gα1+β12,yΔy+λ∫α2β2Gα1,yΔy+∫α2β2Gβ1,yΔy2-1β1-α1∫α1β1∫α2β2Gσx,yΔyΔx≤Mβ2-α2β1-α1h2α1,α1+λβ1-α12+h2α1+β12,α1+λβ1-α12+h2α1+β12,β1-λβ1-α12+h2β1,β1-λβ1-α12.Similarly, if one applies Lemma 7 to Gx at y=α2+β2/2 and thereafter integrate the resulting inequality over [α1,β1], one gets(27)1-λ∫α1β1Gx,α2+β22Δx+λ∫α1β1Gx,α2Δx+∫α1β1Gx,β2Δx2-1β2-α2∫α1β1∫α2β2Gx,σyΔyΔx≤Nβ1-α1β2-α2h2α2,α2+λβ2-α22+h2α2+β22,α2+λβ2-α22+h2α2+β22,β2-λβ2-α22+h2β2,β2-λβ2-α22.Combining (26) and (27), we get (24). Finally, from (26) and (27), we get(28)1-λβ2-α2∫α2β2Gα1+β12,yΔy+λ2β2-α2∫α2β2Gα1,y+Gβ1,yΔy-1β1-α1β2-α2∫α1β1∫α2β2Gσx,yΔyΔx≤Mβ1-α1h2α1,α1+λβ1-α12+h2α1+β12,α1+λβ1-α12+h2α1+β12,β1-λβ1-α12+h2β1,β1-λβ1-α12,1-λβ1-α1∫α1β1Gx,α2+β22Δx+λ2β1-α1∫α1β1Gx,α2+Gx,β2Δx-1β1-α1β2-α2∫α1β1∫α2β2Gx,σyΔyΔx≤Nβ2-α2h2α2,α2+λβ2-α22+h2α2+β22,α2+λβ2-α22+h2α2+β22,β2-λβ2-α22+h2β2,β2-λβ2-α22.Using (28) amounts to (25). Thus, the proof of Theorem 9 is complete.Corollary 11.
If we letT1=T2=R in Theorems 9 and 10, then we obtain the inequality(29)∫α1β1Gx,α2+Gx,β22dx+∫α2β2Gα1,y+Gβ1,y2dy-1β1-α1+1β2-α2∫α1β1∫α2β2Gx,ydydx≤λ2-λ+12β1-α1β2-α2M+N,∫α1β1Gx,α2+Gx,β24β1-α1dx+∫α2β2Gα1,y+Gβ1,y4β2-α2dy-1β1-α1β2-α2∫α1β1∫α2β2Gx,ydydx≤λ2-λ+14Mβ1-α1+Nβ2-α2,1-λ∫α1β1Gx,α2+β22dx+∫α2β2Gα1+β12,ydy+λ2∫α1β1Gx,α2+Gx,β2dx+λ2∫α2β2Gα1,y+Gβ1,ydy-1β1-α1+1β2-α2∫α1β1∫α2β2Gx,ydydx≤2λ2-2λ+14β1-α1β2-α2M+N,1-λ2β1-α1∫α2β2Gx,α2+β22dx+1-λ2β2-α2∫α2β2Gα1+β12,ydy+λ4β1-α1∫α1β1Gx,α2+Gx,β2dx+λ4β2-α2∫α2β2Gα1,y+Gβ1,ydy-1β1-α1β2-α2∫α1β1∫α2β2Gx,ydydx≤2λ2-2λ+18Mβ1-α1+Nβ2-α2.Remark 12.
Corollary11 becomes Theorem 3 if λ=0.Corollary 13.
If we letT1=T2=Z in Theorems 9 and 10, then we get the succeeding inequalities (30)∑x=α1β1-1Gx,α2+Gx,β22+∑y=α2β2-1Gα1,y+Gβ1,y2-1β1-α1∑x=α1β1-1∑y=α2β2-1Gx+1,y-1β2-α2∑x=α1β1-1∑y=α2β2-1Gx,y+1≤λ2-λ+12β1-α1β2-α2M+N,∑x=α1β1-1Gx,α2+Gx,β24β1-α1+∑y=α2β2-1Gα1,y+Gβ1,y4β2-α2-12β1-α1β2-α2∑x=α1β1-1∑y=α2β2-1Gx,y+1+Gx+1,y≤λ2-λ+14Mβ1-α1+Nβ2-α2,1-λ∑x=α1β1-1Gx,α2+β22+∑y=α2β2-1Gα1+β12,y+λ2∑x=α1β1-1Gx,α2+Gx,β2+λ2∑y=α2β2-1Gα1,y+Gβ1,y-1β1-α1∑x=α1β1-1∑y=α2β2-1Gx+1,y-1β2-α2∑x=α1β1-1∑y=α2β2-1Gx,y+1≤2λ2-2λ+14β1-α1β2-α2M+N,1-λ2β1-α1∑y=α2β2-1Gx,α2+β22+1-λ2β2-α2∑y=α2β2-1Gα1+β12,y+λ4β1-α1∑x=α1β1-1Gx,α2+Gx,β2+λ4β2-α2∑y=α2β2-1Gα1,y+Gβ1,y-12β1-α1β2-α2∑x=α1β1-1∑y=α2β2-1Gx+1,y+Gx,y+1≤2λ2-2λ+18Mβ1-α1+Nβ2-α2.Theorem 14.
Letα1,β1,x∈T1,α2,β2,y∈T2, with α1<β1,α2<β2 and G:[α1,β1]×[α2,β2]→R be such that the partial mappings(31)Gy:α1,β1⟶R,Gyξ≔Gξ,y,Gx:α2,β2⟶R,Gxζ≔Gx,ζ,defined for all y∈[α2,β2] and x∈[α1,β1], are differentiable. If GxΔ,GyΔ∈Crd(T,R) and there exist γx,γy,Γx,Γy∈R such that γy≤GyΔt≤Γy,t∈α1,β1,γx≤GxΔt≤Γx,t∈[α2,β2], then the succeeding inequalities(32)1-λ2∫α1β1Gx,α2Δx+∫α2β2Gα1,yΔy+λ2∫α1β1Gx,β2Δx+∫α2β2Gβ1,yΔy-1β1-α1∫α1β1∫α2β2Gσx,yΔyΔx-1β2-α2∫α1β1∫α2β2Gx,σyΔyΔx+Γx+γx2β1-α1β2-α2h2α2,β2-λ2β2-α22+Γy+γy2β2-α2β1-α1h2α1,β1-λ2β1-α12≤Γx-γx2β1-α1β2-α2h2α2,β2-λβ2-α22+h2β2,β2-λβ2-α22+Γy-γy2β2-α2β1-α1h2α1,β1-λβ1-α12+h2β1,β1-λβ1-α12,(33)14β1-α1∫α1β12-λGx,α2+λGx,β2Δx+14β2-α2∫α2β22-λGα1,y+λGβ1,yΔy-12β1-α1β2-α2∫α1β1∫α2β2Gx,σy+Gσx,yΔyΔx+Γy+γy41β1-α1h2α1,β1-λ2β1-α12+Γx+γx41β2-α2h2α2,β2-λ2β2-α22≤Γy-γy41β1-α1h2α1,β1-λβ1-α12+h2β1,β1-λβ1-α12+Γx-γx41β2-α2h2α2,β2-λβ2-α22+h2β2,β2-λβ2-α22,hold for all λ∈[0,1] such that β1-λβ1-α1/2∈T1 and β2-λβ2-α2/2∈T2.Proof.
Applying Lemma8 to the mapping Gy at x=α1 gives(34)1-λ2Gα1,y+λ2Gβ1,y-1β1-α1∫α1bfσt,yΔy+Γy+γy21β1-α1h2α1,β1-λ2β1-α12≤Γy-γy21β1-α1h2α1,β1-λβ1-α12+h2β1,β1-λβ1-α12.Integrating (34) over [α2,β2] yields(35)1-λ2∫α2dfα1,yΔy+λ2∫α2dfβ1,yΔy-1β1-α1∫α1β1∫α2dfσx,yΔyΔx+Γy+γy2β2-α2β1-α1h2α1,β1-λ2β1-α12≤Γy-γy2β2-α2β1-α1h2α1,β1-λβ1-α12+h2β1,β1-λβ1-α12.Similarly, applying Lemma 8 to the mapping Gx at y=α2 and then integrating the resulting inequality over [α1,β1] give(36)1-λ2∫α1bfx,α2Δx+λ2∫α1bfx,β2Δx-1β2-α2∫α1β1∫α2dfx,σyΔyΔx+Γx+γx2β1-α1β2-α2h2α2,β2-λ2β2-α22≤Γx-γx2β1-α1β2-α2h2α2,β2-λβ2-α22+h2β2,β2-λβ2-α22.Using (35) and (36), we get (32). Also, we obtain from (35) and (36) the following inequalities:(37)1-λ/2β2-α2∫α2dfα1,yΔy+λ2β2-α2∫α2dfβ1,yΔy-1β1-α1β2-α2∫α1β1∫α2dfσx,yΔyΔx+Γy+γy21β1-α1h2α1,β1-λ2β1-α12≤Γy-γy21β1-α1h2α1,β1-λβ1-α12+h2β1,β1-λβ1-α12,1-λ/2β1-α1∫α1bfx,α2Δx+λ2β1-α1∫α1bfx,β2Δx-1β1-α1β2-α2∫α1β1∫α2dfx,σyΔyΔx+Γx+γx21β2-α2h2α2,β2-λ2β2-α22≤Γx-γx21β2-α2h2α2,β2-λβ2-α22+h2β2,β2-λβ2-α22.Using (37) amounts to (33). That completes the proof of Theorem 14.Corollary 15.
If we letT1=T2=R in Theorem 10, then the succeeding inequalities hold:(38)1-λ2∫α1β1Gx,α2dx+∫α2β2Gα1,ydy+λ2∫α1β1Gx,β2dx+∫α2β2Gβ1,ydy-1β1-α1+1β2-α2∫α1β1∫α2β2Gx,ydydx+β1-α1β2-α241-λΓx+γx+Γy+γy≤β1-α1β2-α28λ-12+1Γx-γx+Γy-γy,14β1-α1∫α1β12-λGx,α2+λGx,β2dx+14β2-α2∫α2β22-λGα1,y+λGβ1,ydy-1β1-α1β2-α2∫α1β1∫α2β2Gx,ydydx+1-λ8Γy+γyβ1-α1+Γx+γxβ2-α2≤λ-12+116Γy-γyβ1-α1+Γx-γxβ2-α2.Remark 16.
Corollary15 becomes Theorem 5 if λ=1.Corollary 17.
If we letT1=T2=Z in Theorem 10, then the succeeding inequalities hold:(39)1-λ2∑x=α1β1-1Gx,α2+∑y=α2β2-1Gα1,y+λ2∑x=α1β1-1Gx,β2+∑y=α2β2-1Gβ1,y-1β1-α1∑x=α1β1-1∑y=α2β2-1Gx+1,y-1β2-α2∑x=α1β1-1∑y=α2β2-1Gx,y+1+Γx+γx4β1-α1β2-α2+1-λβ2-α2+Γy+γy4β2-α2β1-α1+1-λβ1-α1≤Γx-γx8β1-α1β2-α2λ2-2λ+2-2λ+2+Γy-γy8β2-α2β1-α1λ2-2λ+2-2λ+2,14β1-α1∑x=α1β1-12-λGx,α2+λGx,β2+14β2-α2∑y=α2β2-12-λGα1,y+λGβ1,y-12β1-α1β2-α2∑x=α1β1-1∑y=α2β2-1Gx,y+1+Gx+1,y+Γy+γy8β1-α1+1-λβ1-α1+Γx+γx8β2-α2+1-λβ2-α2≤Γy-γy16β1-α1λ2-2λ+2-2λ+2+Γx-γx16β2-α2λ2-2λ+2-2λ+2.


## 3. Conclusion

Three main theorems are hereby established. The results of Farid [3] are obtained as special cases of our results. Loads of interesting new inequalities can be obtained by choosing different values of λ∈[0,1], and considering a different time scale different from R and Z.



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