IJANAL International Journal of Analysis 2314-4998 2314-498X Hindawi Publishing Corporation 10.1155/2016/6749213 6749213 Research Article Some New Ostrwoski’s Inequalities for Functions Whose nth Derivatives Are r-Convex http://orcid.org/0000-0002-0156-7864 Meftah Badreddine 1 Zayed Ahmed Laboratoire des Télécommunications Faculté des Sciences et de la Technologie Université 8 Mai 1945 de Guelma P.O. Box 401 24000 Guelma Algeria univ-guelma.dz 2016 28112016 2016 20 07 2016 14 11 2016 28112016 2016 Copyright © 2016 Badreddine Meftah. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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

1. Introduction

In 1938, A. M. Ostrowski proved an interesting integral inequality, given by the following theorem.

Theorem 1 ([<xref ref-type="bibr" rid="B2">1</xref>]).

Let f:IRR be a differentiable mapping on I (interior of I), and let a,bI with a<b. If fM for all xa,b, then(1)fx-1b-aabftdtMb-a14+x-a+b/22b-a2,xa,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 , Cerone et al. proved the following identity.

Lemma 2 ([<xref ref-type="bibr" rid="B1">2</xref>, Lemma <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M13"><mml:mrow><mml:mn>2.1</mml:mn></mml:mrow></mml:math></inline-formula>]).

Let f:a,bR be a mapping such that fn-1 is absolutely continuous on a,b. Then for all xa,b one has the identity(2)abftdt=k=0k=nb-xk+1+-1kx-ak+1k+1!fkx+-1nabPnx,tfntdt,where the kernel Pn:a,b2R is given by(3)Pnx,t=t-ann!ifta,xt-bnn!iftx,b,xa,band n is natural number, n1.

We also recall that a positive function f:IR is said to be r-convex on I, if the following inequality(4)ftx+1-tytfrx+1-tfry1/r,r0fx1-tfyt,r=0holds for all x,yI and t[0,1]; see .

In this paper we establish some new Ostrwoski’s inequalities for functions whose nth derivatives are r-convex.

2. Main Results

In order to establish our results, we need these lemmas.

Lemma 3 ([<xref ref-type="bibr" rid="B5">4</xref>]).

For α>0 and k>0, z>0,(5)Jα,k=011-tα-1ktdt=i=1lnki-1αi<(6)Hα,k,z=01tα-1ktdt=zαkzi=1-zlnki-1αi<,where αi=j=0i-1α+j.

Lemma 4 ([<xref ref-type="bibr" rid="B4">5</xref>]).

For a0 and b0, the following algebraic inequalities are true: (7)a+bλ2λ-1aλ+bλ,forλ1,a+bλaλ+bλ,for0λ1.

Theorem 5.

For nN, let f:IRR be n-time differentiable on a,bI such that fnLa,b. If fn is r-convex, then the following inequality (8)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxcrx-an+1n!fna+b-xn+1n!fnbB1r+1,n+1+x-an+1n!+b-xn+1n!rcr1+n+1rfnxifr0x-an+1n+1!fna+b-xn+1n+1!fnxifr=0,A=B=1x-an+1n+1!fna+b-xn+1n!fnxi=1lnBi-1n+1iifr=0,A=1Bb-xn+1n+1!fnx+x-an+1n!fnxi=1-lnAi-1n+1iifr=0,A1=Bfnxx-an+1n!i=1-lnAi-1n+1i+b-xn+1n!i=1lnBi-1n+1iifr=0,A1,B1holds for all xa,b, where A=fn(x)/fn(a), B=fn(b)/fn(x), and(9)cr=21/rif0<r<11ifr1.

Proof.

From Lemma 2 and properties of modulus, we have (10)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxaxu-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 r0, the use of Lemma 4 gives (11)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxx-an+1n!01tn1-tfnar+tfnxr1/rdt+b-xn+1n!011-tn1-tfnxr+tfnbr1/rdtx-an+1n!cr01tn1-t1/rfna+tn+1/rfnxdt+b-xn+1n!cr011-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 where r=0, (10) becomes(12)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxx-an+1n!01tnfna1-tfnxtdt+b-xn+1n!011-tnfnx1-tfnbtdt=x-an+1n!fna01tnAtdt+b-xn+1n!fnx011-tnBtdt,and we distinguish 4 cases.

If A=B=1, then (12) gives(13)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxx-an+1n!fna01tndt+b-xn+1n!fnx011-tndt=x-an+1n+1!fna+b-xn+1n+1!fnx.If A=1B, then (12) becomes(14)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxx-an+1n+1!fna+b-xn+1n!fnxi=1lnBi-1n+1i,where we have used (5).

If A1=B, then (12) becomes (15)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxb-xn+1n+1!fnx+x-an+1n!fnxi=1-lnAi-1n+1i,where we have used (6) with z=1.

In the case where A1 and B1, using Lemma 3, (12) gives(16)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxx-an+1n!fnxi=1-lnAi-1n+1i+b-xn+1n!fnxi=1lnBi-1n+1i.The desired result follows from (11) and (13)–(16).

Theorem 6.

For nN, let f:IRR be n-time differentiable on a,bI such that fnLa,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!fkxrcr1+r1/qx-an+1np+11/pn!fnaq+fnxq1/q+b-xn+1np+11/pn!fnxq+fnbq1/qifr0x-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-1qlnB1/qfnxifr=0,A=1Bb-xn+1np+11/pn!fnx+x-an+1np+11/pn!Aq-1qlnA1/qfnaifr=0,A1=Bx-an+1np+11/pn!Aq-1qlnA1/qfna+b-xn+1np+11/pn!Bq-1qlnB1/qfnxifr=0,A1,B1holds for all xa,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 Lemma 2, properties of modulus, and Hölder’s inequality, we have(18)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxaxu-ann!fnudu+xbb-unn!fnudu=x-an+1n!01tnfn1-ta+txdt+b-xn+1n!011-tnfn1-tx+tbdtx-an+1n!01tnpdt1/p01fn1-ta+txqdt1/q+b-xn+1n!011-tnpdt1/p01fn1-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 r0, the use of r-convexity of fnq and Lemma 4 gives(19)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxx-an+1np+11/pn!011-tfnaqr+tfnxqr1/rdt1/q+b-xn+1np+11/pn!011-tfnxqr+tfnbqr1/rdt1/qx-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!fkxx-an+1np+11/pn!fna01Aqtdt1/q+b-xn+1np+11/pn!fnx01Bqtdt1/q.Analogously to Theorem 5, we will treat the 4 cases.

If A=B=1, then (20) gives(21)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxx-an+1np+11/pn!fna+b-xn+1np+11/pn!fnx.If A=1B, then (20) becomes (22)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxx-an+1np+11/pn!fna+b-xn+1np+11/pn!fnxq01Bqtdt1/qx-an+1np+11/pn!fna+b-xn+1np+11/pn!Bq-1qlnB1/qfnx.If A1=B, then (20) becomes(23)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxb-xn+1np+11/pn!fnx+x-an+1np+11/pn!fnaq01Aqtdt1/qb-xn+1np+11/pn!fnx+x-an+1np+11/pn!Aq-1qlnA1/qfna.In the case where A1 and B1, (20) gives(24)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxx-an+1np+11/pn!Aq-1qlnA1/qfna+b-xn+1np+11/pn!Bq-1qlnB1/qfnx.The desired result follows from (19) and (21)–(24).

Theorem 7.

For nN, let f:IRR be n-time differentiable on a,bI such that fnLa,b and let q>1. If fnq is convex, then the following inequality (25)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxn+11/qn+1!cr1/qx-an+1Bn+1,1r+1fnaq+r1+n+1rfnxq1/q+b-xn+1r1+n+1rfnxq+B1r+1,n+1fnbq1/qifr0x-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-1qlnB1/qfnxifr=0,A=1Bb-xn+1np+11/pn!fnx+x-an+1np+11/pn!Aq-1qlnA1/qfnaifr=0,A1=Bfnxn+11/qx-an+1n+1!i=1-lnAqi-1n+1i1/q+n+11/qb-xn+1n+1!i=1lnBqi-1n+1i1/qifr=0,A1,B1holds for all xa,b, where A=fn(x)/fn(a), B=fn(b)/fn(x), and cr is defined as in (9).

Proof.

From Lemma 2, properties of modulus, and power mean inequality, we have (26)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxaxu-ann!fnudu+xbb-unn!fnudu=x-an+1n!01tnfn1-ta+txdt+b-xn+1n!011-tnfn1-tx+tbdtx-an+1n!01tndt1-1/q01tnfn1-ta+txqdt1/q+b-xn+1n!011-tndt1-1/q011-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 r0, using r-convexity of fnq and Lemma 4, we get(27)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxn+11/qx-an+1n+1!cr1/q01tn1-t1/rfnaq+tn+1/rfnxqdx1/q+n+11/qb-xn+1n+1!cr1/q01tn1-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!fkxn+11/qx-an+1n+1!fnaq01tnAqtdt1/q+n+11/qb-xn+1n+1!fnxq011-tnBqtdt1/q.

If A=B=1, then (27) gives(29)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxx-an+1n+1!fna+b-xn+1n+1!fnx.If A=1B, then (27) becomes(30)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxx-an+1n+1!fna+n+11/qb-xn+1n+1!fnxi=1lnBqi-1n+1i1/q.If A1=B, then (27) becomes(31)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxb-xn+1n+1!fnx+n+11/qx-an+1n+1!fnxi=1-lnAqi-1n+1i1/q.In the case where A1 and B1, (27) gives(32)abftdt-k=0k=nb-xk+1+-1kx-ak+1k+1!fkxn+11/qx-an+1n+1!fnxi=1-lnAqi-1n+1i1/q+n+11/qb-xn+1n+1!fnxi=1lnBqi-1n+1i1/q,where we have used Lemma 3. The desired result follows from (26) and (28)–(31).

Competing Interests

The author declares that there are no competing interests regarding the publication of this paper.

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