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`International Journal of AnalysisVolume 2013 (2013), Article ID 127061, 5 pageshttp://dx.doi.org/10.1155/2013/127061`
Research Article

## On Right Caputo Fractional Ostrowski Inequalities Involving Three Functions

Department of Mathematics, Dr. B.A.M. University, Aurangabad, Maharashtra 431004, India

Received 28 August 2012; Accepted 27 December 2012

Copyright © 2013 Deepak B. Pachpatte. 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.

#### Abstract

We establish Ostrowski inequalities involving three functions in right Caputo fractional derivative in spaces.

#### 1. Introduction

In 1938, Ostrowski proved the following useful inequality.

Let be continuous on and differentiable on whose derivative is bounded on , that is, . Then for any . The constant is best possible.

In [1, 2] Pachpatte has proved Ostrowski inequality in three independent variables.

In past few years many authors have obtained various generalisation and variant of the above type of inequality and other on fractional as well as time scale calculus see [36].

Here we give some basic definition from fractional calculus used in [79].

Definition 1. Let , . The right and left Riemann-Liouville integrals and of order with are defined by respectively, where and .

Definition 2 (see [10, page 2]). Let ( be in ), , , ( the ceiling of the number). We define the right Caputo fractional derivative of order by
If , then
If , we define .

Definition 3 (see [9, page 74]). Let , with , and then the fractional integral is defined by
We give here the theorems proved in [10].

Theorem 4. Let , , . Assume that , and . Then

Theorem 5. Let , , . Assume that , and . Then

Theorem 6. Let ; , , , . Assume that , and . Then

#### 2. Main Results

Our main results are given in the following theorems.

Theorem 7. Let , , . Assume that , and . Then

Proof. Let we have Multiplying (10), (11), and (12) by , , and , respectively, and adding them, we have Integrating both sides of (13) with respect to and rewriting above equation we have From (14) and using the properties of modulus we have It is easy to observe that The proof of the theorem is complete.

Remark 8. If we take , and hence , in Theorem 7, then we get Theorem 4.

Theorem 9. Let , , . Assume that , , and . Then

Proof. From (15) we have
This proves the theorem.

Remark 10. If we take , , and hence , in Theorem 9, then we get Theorem 5.

Theorem 11. Let ; , , , . Assume that , , and . Then

Proof. From (15) we have
Applying Holder’s inequality to (21), we get
We have
Substituting (22) into (21), we get the required inequality.

Remark 12. If we take and hence , in Theorem 11, then we get Theorem 6.

#### References

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