IJANAL International Journal of Analysis 2314-4998 2314-498X Hindawi Publishing Corporation 10.1155/2016/3918483 3918483 Research Article Straightforward Proofs of Ostrowski Inequality and Some Related Results http://orcid.org/0000-0002-4103-7745 Farid Ghulam 1 Wang Lianwen COMSATS Institute of Information Technology Attock Campus Attock Pakistan comsats.edu.pk 2016 9112016 2016 19 07 2016 25 10 2016 9112016 2016 Copyright © 2016 Ghulam Farid. 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 give proofs of some known results in very simple and antique way. Also we find some general bounds of a nonnegative difference of the Hadamard inequality and an Ostrowski-Grüss type inequality is proved.

1. Introduction

A function f is convex on an interval [a,b] if for all x1, x2[a,b] where 0λ1,(1)fλx1+1-λx2λfx1+1-λfx2.The most classical result for convex functions can be seen in the following theorem.

Theorem 1.

Let f:IR be a convex function defined on the interval I of real numbers and a,bI with a<b. Then following double inequality (2)fa+b21b-aabfxdxfa+fb2holds.

It is known as the Hadamard inequality. History of this inequality begins with the papers of Hermite  and Hadamard  in the years 1883–1893 (see ). A rich literature of mathematical inequalities is due to convex functions equally determined by the Hadamard inequality, inspired from this inequality many closely related results have been established which have their applications in approximately all fields of mathematical analysis up to some extent (see, ).

In 1935 Grüss proved an inequality well known as the Grüss inequality stated in the following theorem .

Theorem 2.

Let f,g:[a,b]R be integrable functions such that ϕf(x)Φ and γg(x)Γ for all x[a,b], where ϕ, Φ, γ, Γ are constants. Then we have (3) 1 b - a a b f x g x d x - 1 b - a a b f x d x · 1 b - a a b g x d x 1 4 Φ - ϕ Γ - γ , where the constant 1/4 is sharp.

In 1938, Ostrowski established the following inequality known as the Ostrowski inequality stated.

Theorem 3.

Let f:IR, where I is an interval in R, be a mapping differentiable in Io the interior of I and a,bIo, a<b. If f(t)M, for all t[a,b], then we have(4)fx-1b-aabftdt14+x-a+b/22b-a2b-aM,for all x[a,b].

By using Grüss inequality some results have been established which are well known as Ostrowski-Grüss or Ostrowski-Grüss type inequalities (see  and references therein). Several quadrature rules of numerical integration have been estimated using Ostrowski and Ostrowski-Grüss type inequalities (see [4, 5, 8, 10, 12]).

In  Cerone and Dragomir have estimated differences of the Hadamard inequality as follows.

Theorem 4.

Suppose that f:[a,b]R be a twice differentiable function on (a,b) and suppose that γf(t)Γ for all t(a,b). Then we have the double inequality:(5)γb-a2241b-aabftdt-fa+b2Γb-a224.

Theorem 5.

Under the assumptions of Theorem 4 we have(6)γb-a212fa+fb2-1b-aabftdtΓb-a212.

Ujević in  also estimated differences of the Hadamard inequality.

Theorem 6.

Suppose that f:[a,b]R be a twice differentiable function on (a,b) and suppose that γf(t)Γ for all t(a,b). Then we have (7)3S-2Γ24b-a21b-aabftdt-fa+b23S-2γ24b-a2,where S=(f(b)-f(a))/(b-a).

Theorem 7.

Under the assumptions of Theorem 6 we have(8)3S-Γ24b-a2fa+fb2-1b-aabftdt3S-γ24b-a2,where S=(fb-fa)/(b-a).

The aim of this paper is in fact to establish proof of well known Ostrowski inequality in a very straightforward way, and to establish bounds of a difference of the Hadamard inequality given in [4, 8] in very simple way, here there is no need to define a two variable kernel. In the last by involving a parameter a similar but general result have been found and some particular bounds of a difference of the Hadamard inequality are calculated, also an Ostrowski-Grüss type inequality is obtained by elementary calculation.

2. Some Alternative Proofs

First we give a proof of well-known Ostrowski inequality, and then proofs of Theorems 4 and 7 are given.

2.1. Proof of Theorem <xref ref-type="statement" rid="thm1.3">3</xref> Proof.

It is clear that (9)t-aft+M0,axt-aft+Mdt0.Integrating by parts we have(10)x-afx-axftdt+M2x-a20.Also (11)b-tM-ft0,xbb-tM-ftdt0.Integrating by parts we have(12)b-xfx-xbftdt+M2b-x20.By adding (10) and (12) one has(13)fx-1b-aabftdt-M2b-ax-a2+x-b2.On the other hand using positivity of (b-t)(f(t)+M) and (t-a)(M-f(t)) we have (14)axt-aM-ftdt+xbb-tM+ftdt0.From which one can have(15)fx-1b-aabftdtM2b-ax-a2+x-b2.From inequalities in (13) and (15) we have (16)fx-1b-aabftdtM2b-ax-a2+x-b2.Using the following identity one can get inequality in (4) (17)x-a2+x-b2=2b-a214+x-a+b/22b-a2.

2.2. Proof of Theorem <xref ref-type="statement" rid="thm1.4">4</xref> Proof.

It is clear that (18)t-a2ft-γ0,aa+b/2t-a2ft-γdt0.Integrating by parts we have(19)b-a22fa+b2-b-afa+b2+2aa+b/2ftdt-b-a3γ240.Now using (t-b)2(f(t)-γ)0 and integrating on [(a+b)/2,b] we have(20)-b-a22fa+b2-b-afa+b2+2a+b/2bftdt-b-a3γ240.Adding (19) and (20) one can have(21)1b-aabftdt-fa+b2b-a2γ24.Similarly using (t-a)2(Γ-f(t))0 and integrating on [a,(a+b)/2] we have(22)-b-a22fa+b2+b-afa+b2-2aa+b/2ftdt+b-a3Γ240.Now using (t-b)2(Γ-f(t))0, integrating on [(a+b)/2,b] we have(23)b-a22fa+b2+b-afa+b2-2a+b/2bftdt+b-a3Γ240.Adding (22) and (23) we have(24)1b-aabftdt-fa+b2b-a2Γ24.From (21) and (24) we have (5).

2.3. Proof of Theorem <xref ref-type="statement" rid="thm1.7">7</xref> Proof.

It is clear that (25)t-a+b22ft-γ0,abt-a+b22ft-γ0.Integrating by parts we have(26)fa+fb2-1b-aabftdtb-a23S-γ24.Similarly using (t-a+b/2)2(Γ-f(t))0 and integrating on [a,b] we have(27)fa+fb2-1b-aabftdtb-a23S-Γ24.From (26) and (27) we have (8).

3. Some Related Results

In this section we give some more results in a very simple way. First by involving a parameter, we prove a general result that provides bounds of a nonnegative difference of the Hadamard inequality and gives particular bounds, and then an Ostrowski-Grüss type inequality is proved.

Theorem 8.

Suppose that f:[a,b]R be a twice differentiable function on (a,b) and suppose that γf(t)Γ for all t(a,b). Then we have (28)24+αγ-24S24αb-a21b-aabftdt-fa+b224+αΓ-24S24αb-a2,where S=(f(b)-fa)/(b-a).

Proof.

It is clear that (29)abft-γdt0.From which one has (30)fb-fab-ab-a-γb-a0.Further we can say for some α>0(31)S-γαb-a20.Adding (21) and (31) one has(32)1b-aabftdt-fa+b224+αγ-24S24αb-a2.On the other hand abΓ-ftdt0, which gives for α>0(33)S-Γαb-a20.Adding (24) and (33) one has(34)1b-aabftdt-fa+b224+αΓ-24S24αb-a2.From (32) and (34) we have the required inequality.

Corollary 9.

If one selects, for example, α=24 and α=48 in Theorem 8, then (35)2γ-S24b-a21b-aabftdt-fa+b22Γ-S24b-a2,3γ-S48b-a21b-aabftdt-fa+b23Γ-S48b-a2.

In the following, adopting the pattern of proofs we give the following Ostrowski-Grüss type inequality. It is remarkable to mention here that in  Cheng has proved an improved result adopting a comparatively different method.

Theorem 10.

Let f:[a,b]R, where I is an interval in R be a mapping differentiable in Io, the interior of I and a,bIo, a<b. If γf(t)Γ for all t(a,b), then we have (36) 1 2 f x - x - b f b - x - a f a 2 b - a - 1 b - a a b f t d t x - a 2 + b - x 2 4 b - a Γ - γ , for all x[a,b].

Proof.

It is clear that (x-t)(f(t)-γ)0 and (t-a)(Γ-f(t))0 for all t[a,x]. Therefore (37)axx-tft-γdt+axt-aΓ-ftdt0.After some computation we have(38)x-afx+x-afa-2axftdt12x-a2Γ-γ.Also it is easy to see (t-x)(Γ-f(t))0 and (b-t)(f(t)-γ)0 for all t[x,b]. Therefore (39)xbt-xΓ-ftdt+xbb-tft-γdt0.After some computation it can be seen(40)b-xfx-x-bfb-2xbftdt12b-x2Γ-γ.By adding (38) and (40) one has(41)12fx-x-bfb-x-afa2b-a-1b-aabftdtx-a2+b-x24b-aΓ-γ.On the other hand we have (42)axt-aft-γdt+axx-tΓ-ftdt0;ta,x.This gives(43)x-afx+x-afa-2axftdt-12x-a2Γ-γ.Also we have (44)xbb-tΓ-ftdt+xbt-xft-γdt0;tx,b,which gives(45)b-xfx-x-bfb-2xbftdt-12b-x2Γ-γ.Adding (43) and (45), then combining with (41) one can get (36).

Competing Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

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