# Some New Ostrwoski’s Inequalities for Functions Whosenth Derivatives Are r-Convex

Authors: Badreddine Meftah
Journal: International Journal of Analysis
 (2016)

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## Abstract

We prove some new Ostrowski’s inequalities for functions whosenth derivatives are r-convex.


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## Body

## 1. Introduction

In 1938, A. M. Ostrowski proved an interesting integral inequality, given by the following theorem.Theorem 1 ([1]).
Letf:I⊆R→R be a differentiable mapping on I∘ (interior of I), and let a,b∈I∘ with a<b. If f′≤M for all x∈a,b, then(1)fx-1b-a∫abftdt≤Mb-a14+x-a+b/22b-a2,∀x∈a,b.Inequality (1) has attracted much interest due to its diversity of applications in numerical analysis, probability theory, and other areas. We note that a numerous variants, extensions, and generalizations of inequality (1) have been discovered.In [2], Cerone et al. proved the following identity.Lemma 2 ([2, Lemma 2.1]).
Letf:a,b→R be a mapping such that fn-1 is absolutely continuous on a,b. Then for all x∈a,b one has the identity(2)∫abftdt=∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx+-1n∫abPnx,tfntdt,where the kernel Pn:a,b2→R is given by(3)Pnx,t=t-ann!ift∈a,xt-bnn!ift∈x,b,x∈a,band n is natural number, n≥1.We also recall that a positive functionf:I→R is said to be r-convex on I, if the following inequality(4)ftx+1-ty≤tfrx+1-tfry1/r,r≠0fx1-tfyt,r=0holds for all x,y∈I and t∈[0,1]; see [3].In this paper we establish some new Ostrwoski’s inequalities for functions whosenth derivatives are r-convex.


## 2. Main Results

In order to establish our results, we need these lemmas.Lemma 3 ([4]).
Forα>0 and k>0, z>0,(5)Jα,k=∫011-tα-1ktdt=∑i=1∞ln⁡ki-1αi<∞(6)Hα,k,z=∫01tα-1ktdt=zαkz∑i=1∞-zln⁡ki-1αi<∞,where αi=∏j=0i-1α+j.Lemma 4 ([5]).
Fora≥0 and b≥0, the following algebraic inequalities are true: (7)a+bλ≤2λ-1aλ+bλ,forλ≥1,a+bλ≤aλ+bλ,for0≤λ≤1.Theorem 5.
Forn∈N, let f:I⊂R→R be n-time differentiable on a,b⊂I such that fn∈La,b. If fn is r-convex, then the following inequality (8)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤crx-an+1n!fna+b-xn+1n!fnbB1r+1,n+1+x-an+1n!+b-xn+1n!rcr1+n+1rfnxifr≠0x-an+1n+1!fna+b-xn+1n+1!fnxifr=0,A=B=1x-an+1n+1!fna+b-xn+1n!fnx∑i=1∞ln⁡Bi-1n+1iifr=0,A=1≠Bb-xn+1n+1!fnx+x-an+1n!fnx∑i=1∞-ln⁡Ai-1n+1iifr=0,A≠1=Bfnxx-an+1n!∑i=1∞-ln⁡Ai-1n+1i+b-xn+1n!∑i=1∞ln⁡Bi-1n+1iifr=0,A≠1,B≠1holds for all x∈a,b, where A=fn(x)/fn(a), B=fn(b)/fn(x), and(9)cr=21/rif0<r<11ifr≥1.Proof.
From Lemma2 and properties of modulus, we have (10)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤∫axu-ann!fnudu+∫xbb-unn!fnudu=x-an+1n!∫01tnfn1-ta+txdt+b-xn+1n!∫011-tnfn1-tx+tbdt.Since fn is r-convex function, for r≠0, the use of Lemma 4 gives (11)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤x-an+1n!∫01tn1-tfnar+tfnxr1/rdt+b-xn+1n!∫011-tn1-tfnxr+tfnbr1/rdt≤x-an+1n!cr∫01tn1-t1/rfna+tn+1/rfnxdt+b-xn+1n!cr∫011-tn+1/rfnx+1-tnt1/rfnbdt=x-an+1n!crBn+1,1r+1fna+r1+n+1rfnx+b-xn+1n!crr1+n+1rfnx+B1r+1,n+1fnb=crx-an+1n!fna+b-xn+1n!fnbB1r+1,n+1+x-an+1n!+b-xn+1n!rcr1+n+1rfnx,where cr is defined as in (9).
In the case wherer=0, (10) becomes(12)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤x-an+1n!∫01tnfna1-tfnxtdt+b-xn+1n!∫011-tnfnx1-tfnbtdt=x-an+1n!fna∫01tnAtdt+b-xn+1n!fnx∫011-tnBtdt,and we distinguish 4 cases.
IfA=B=1, then (12) gives(13)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤x-an+1n!fna∫01tndt+b-xn+1n!fnx∫011-tndt=x-an+1n+1!fna+b-xn+1n+1!fnx.If A=1≠B, then (12) becomes(14)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤x-an+1n+1!fna+b-xn+1n!fnx∑i=1∞ln⁡Bi-1n+1i,where we have used (5).
IfA≠1=B, then (12) becomes (15)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤b-xn+1n+1!fnx+x-an+1n!fnx∑i=1∞-ln⁡Ai-1n+1i,where we have used (6) with z=1.
In the case whereA≠1 and B≠1, using Lemma 3, (12) gives(16)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤x-an+1n!fnx∑i=1∞-ln⁡Ai-1n+1i+b-xn+1n!fnx∑i=1∞ln⁡Bi-1n+1i.The desired result follows from (11) and (13)–(16).Theorem 6.
Forn∈N, let f:I⊂R→R be n-time differentiable on a,b⊂I such that fn∈La,b and let q>1. If fnq is r-convex, then the following inequality(17)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤rcr1+r1/qx-an+1np+11/pn!fnaq+fnxq1/q+b-xn+1np+11/pn!fnxq+fnbq1/qifr≠0x-an+1np+11/pn!fna+b-xn+1np+11/pn!fnxifr=0,A=B=1x-an+1np+11/pn!fna+b-xn+1np+11/pn!Bq-1qln⁡B1/qfnxifr=0,A=1≠Bb-xn+1np+11/pn!fnx+x-an+1np+11/pn!Aq-1qln⁡A1/qfnaifr=0,A≠1=Bx-an+1np+11/pn!Aq-1qln⁡A1/qfna+b-xn+1np+11/pn!Bq-1qln⁡B1/qfnxifr=0,A≠1,B≠1holds for all x∈a,b, where A=fn(x)/fn(a), B=fn(b)/fn(x), 1/p+1/q=1, and cr is defined as in (9).Proof.
From Lemma2, properties of modulus, and Hölder’s inequality, we have(18)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤∫axu-ann!fnudu+∫xbb-unn!fnudu=x-an+1n!∫01tnfn1-ta+txdt+b-xn+1n!∫011-tnfn1-tx+tbdt≤x-an+1n!∫01tnpdt1/p∫01fn1-ta+txqdt1/q+b-xn+1n!∫011-tnpdt1/p∫01fn1-tx+tbqdt1/q=x-an+1np+11/pn!∫01fn1-ta+txqdt1/q+b-xn+1np+11/pn!∫01fn1-tx+tbqdt1/q.In the case where r≠0, the use of r-convexity of fnq and Lemma 4 gives(19)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤x-an+1np+11/pn!∫011-tfnaqr+tfnxqr1/rdt1/q+b-xn+1np+11/pn!∫011-tfnxqr+tfnbqr1/rdt1/q≤x-an+1cr1/qnp+11/pn!∫011-t1/rfnaq+t1/rfnxqdt1/q+b-xn+1cr1/qnp+11/pn!∫011-t1/rfnxq+t1/rfnbqdt1/q=x-an+1np+11/pn!rcr1+r1/qfnaq+fnxq1/q+b-xn+1np+11/pn!rcr1+r1/qfnxq+fnbq1/q.For r=0, using the r-convexity of fnq, (18) gives (20)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤x-an+1np+11/pn!fna∫01Aqtdt1/q+b-xn+1np+11/pn!fnx∫01Bqtdt1/q.Analogously to Theorem 5, we will treat the 4 cases.
IfA=B=1, then (20) gives(21)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤x-an+1np+11/pn!fna+b-xn+1np+11/pn!fnx.If A=1≠B, then (20) becomes (22)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤x-an+1np+11/pn!fna+b-xn+1np+11/pn!fnxq∫01Bqtdt1/q≤x-an+1np+11/pn!fna+b-xn+1np+11/pn!Bq-1qln⁡B1/qfnx.If A≠1=B, then (20) becomes(23)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤b-xn+1np+11/pn!fnx+x-an+1np+11/pn!fnaq∫01Aqtdt1/q≤b-xn+1np+11/pn!fnx+x-an+1np+11/pn!Aq-1qln⁡A1/qfna.In the case where A≠1 and B≠1, (20) gives(24)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤x-an+1np+11/pn!Aq-1qln⁡A1/qfna+b-xn+1np+11/pn!Bq-1qln⁡B1/qfnx.The desired result follows from (19) and (21)–(24).Theorem 7.
Forn∈N, let f:I⊂R→R be n-time differentiable on a,b⊂I such that fn∈La,b and let q>1. If fnq is convex, then the following inequality (25)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤n+11/qn+1!cr1/qx-an+1Bn+1,1r+1fnaq+r1+n+1rfnxq1/q+b-xn+1r1+n+1rfnxq+B1r+1,n+1fnbq1/qifr≠0x-an+1np+11/pn!fna+b-xn+1np+11/pn!fnxifr=0,A=B=1x-an+1np+11/pn!fna+b-xn+1np+11/pn!Bq-1qln⁡B1/qfnxifr=0,A=1≠Bb-xn+1np+11/pn!fnx+x-an+1np+11/pn!Aq-1qln⁡A1/qfnaifr=0,A≠1=Bfnxn+11/qx-an+1n+1!∑i=1∞-ln⁡Aqi-1n+1i1/q+n+11/qb-xn+1n+1!∑i=1∞ln⁡Bqi-1n+1i1/qifr=0,A≠1,B≠1holds for all x∈a,b, where A=fn(x)/fn(a), B=fn(b)/fn(x), and cr is defined as in (9).Proof.
From Lemma2, properties of modulus, and power mean inequality, we have (26)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤∫axu-ann!fnudu+∫xbb-unn!fnudu=x-an+1n!∫01tnfn1-ta+txdt+b-xn+1n!∫011-tnfn1-tx+tbdt≤x-an+1n!∫01tndt1-1/q∫01tnfn1-ta+txqdt1/q+b-xn+1n!∫011-tndt1-1/q∫011-tnfn1-tx+tbqdt1/q=n+11/qx-an+1n+1!∫01tnfn1-ta+txqdt1/q+n+11/qb-xn+1n+1!∫011-tnfn1-tx+tbqdt1/q.If r≠0, using r-convexity of fnq and Lemma 4, we get(27)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤n+11/qx-an+1n+1!cr1/q∫01tn1-t1/rfnaq+tn+1/rfnxqdx1/q+n+11/qb-xn+1n+1!cr1/q∫01tn1-t1/rfnxq+tn+1/rfnbqdt1/q=n+11/qx-an+1n+1!cr1/qBn+1,1r+1fnaq+r1+n+1rfnxq1/q+n+11/qb-xn+1n+1!cr1/qr1+n+1rfnxq+B1r+1,n+1fnbq1/q.Now, suppose that r=0, and from the convexity of fnq, (25) becomes(28)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤n+11/qx-an+1n+1!fnaq∫01tnAqtdt1/q+n+11/qb-xn+1n+1!fnxq∫011-tnBqtdt1/q.
IfA=B=1, then (27) gives(29)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤x-an+1n+1!fna+b-xn+1n+1!fnx.If A=1≠B, then (27) becomes(30)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤x-an+1n+1!fna+n+11/qb-xn+1n+1!fnx∑i=1∞ln⁡Bqi-1n+1i1/q.If A≠1=B, then (27) becomes(31)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤b-xn+1n+1!fnx+n+11/qx-an+1n+1!fnx∑i=1∞-ln⁡Aqi-1n+1i1/q.In the case where A≠1 and B≠1, (27) gives(32)∫abftdt-∑k=0k=nb-xk+1+-1kx-ak+1k+1!fkx≤n+11/qx-an+1n+1!fnx∑i=1∞-ln⁡Aqi-1n+1i1/q+n+11/qb-xn+1n+1!fnx∑i=1∞ln⁡Bqi-1n+1i1/q,where we have used Lemma 3. The desired result follows from (26) and (28)–(31).



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