/ / Article

Research Article | Open Access

Volume 2014 |Article ID 563096 | 11 pages | https://doi.org/10.1155/2014/563096

Fractional Integral Inequalities via Hadamard’s Fractional Integral

Academic Editor: Dumitru Baleanu
Accepted30 Mar 2014
Published22 Apr 2014

Abstract

We establish new fractional integral inequalities, via Hadamard’s fractional integral. Several new integral inequalities are obtained, including a Grüss type Hadamard fractional integral inequality, by using Young and weighted AM-GM inequalities. Many special cases are also discussed.

1. Introduction

Inequalities have proved to be one of the most powerful and far-reaching tools for the development of many branches of mathematics. The study of mathematical inequalities plays very important role in classical differential and integral equations which has applications in many fields. Fractional inequalities are important in studying the existence, uniqueness, and other properties of fractional differential equations. Recently many authors have studied integral inequalities on fractional calculus using Riemann-Liouville and Caputo derivative; see [17] and the references therein.

Another kind of fractional derivative that appears in the literature is the fractional derivative due to Hadamard introduced in 1892 [8], which differs from the Riemann-Liouville and Caputo derivatives in the sense that the kernel of the integral contains logarithmic function of arbitrary exponent. Details and properties of Hadamard fractional derivative and integral can be found in [914]. Recently in the literature, there appeared some results on fractional integral inequalities using Hadamard fractional integral; see [1517].

In this paper we present some new fractional integral inequalities using the Hadamard fractional integral. Several new integral inequalities are obtained by using Young and weighted AM-GM inequalities. Many special cases are also discussed. Moreover, a Grüss type Hadamard fractional integral inequality is obtained.

2. Preliminaries

In this section we give some preliminaries and basic proposition used in our subsequent discussion. The necessary background details are given in the book by Kilbas et al. [9].

Definition 1. The Hadamard fractional integral of order of a function , for all , is defined as where is the standard gamma function defined by , provided the integral exists, where .

Definition 2. The Hadamard fractional derivative of order , , of a function is given by

Next, we recall a proposition concerning a Hadamard integral and derivative.

Proposition 3 (see [9]). If , the following relations hold: respectively.

For the convenience of establishing our results, we give the semigroup property which implies the commutative property

3. Main Results

Now, we are in a position to give our main results.

Theorem 4. Let be an integrable function on . Assume the following.
() There exist two integrable functions on such that Then, for , , one has

Proof. From , for all , , we have Therefore, Multiplying both sides of (10) by , , we get Integrating both sides of (11) with respect to on , we obtain which yields Multiplying both sides of (13) by , , we have Integrating both sides of (14) with respect to on , we get Hence, we deduce inequality (8) as requested. This completes the proof.

As special cases of Theorems 4, we obtain the following results.

Corollary 5. Let be an integrable function on satisfying , for all and . Then for and , one has

Corollary 6. Let be an integrable function on . Assume that there exists an integrable function on and a constant such that for all . Then for and , one has

Example 7. Let be a function satisfying for . Then for and , we have

Theorem 8. Let be an integrable function on and satisfying . Suppose that holds. Then, for , , one has

Proof. According to the well-known Young’s inequality setting and , , we have Multiplying both sides of (22) by , , we get Integrating the above inequality with respect to and from to , we have which implies (20).

Corollary 9. Let be an integrable function on satisfying , for all and . Then for and , one has

Example 10. Let be a function satisfying for . Then for and , we have

Theorem 11. Let be an integrable function on and satisfying . In addition, suppose that holds. Then, for , , one has

Proof. From the well-known weighted AM-GM inequality by setting and , , we have Multiplying both sides of (29) by , , we get Integrating the above inequality with respect to and from to , we have Therefore, we deduce inequality (27).

Corollary 12. Let be an integrable function on satisfying , for all and . Then for and , one has

Example 13. Let be a function satisfying for . Then for and , we have

Lemma 14 (see [18]). Assume that , , and . Then

Theorem 15. Let be an integrable function on and constants , . In addition, assume that holds. Then for any , , , , the following two inequalities hold:

Proof. By condition and Lemma 14, for , , it follows that for any . Multiplying both sides of (36) by , , and integrating the resulting identity with respect to from to , one has which leads to inequality . Inequality is proved by similar arguments.

Corollary 16. Let be an integrable function on satisfying , for all and . Then for and , one has

Example 17. Let be a function satisfying for . Then for and , we have

Theorem 18. Let and be two integrable functions on . Suppose that holds and moreover one assumes the following.
There exist and integrable functions on such that Then, for , , , the following inequalities hold:

Proof. To prove , from and , we have for that Therefore, Multiplying both sides of (43) by , , we get Integrating both sides of (44) with respect to on , we obtain Then we have Multiplying both sides of (46) by , , we have Integrating both sides of (47) with respect to on , we get the desired inequality .
To prove , we use the following inequalities:

As a special case of Theorem 18, we have the following corollary.

Corollary 19. Let and be two integrable functions on . Assume the following.
There exist real constants such that Then, for , , , one has

Theorem 20. Let and be two integrable functions on and satisfying . Suppose that and hold. Then, for , , the following inequalities hold:

Proof. The inequalities can be proved by choosing of the parameters in the Young inequality

Theorem 21. Let and be two integrable functions on and satisfying . Suppose that and hold. Then, for , , the following inequalities hold:

Proof. The inequalities can be proved by choosing of the parameters in the weighted AM-GM:

Theorem 22. Let and be two integrable functions on and constants , . Assume that and hold. Then, for any , , , the following inequalities hold:

Proof. The inequalities can be proved by choosing of the parameters in Lemma 14:

Lemma 23. Let be an integrable function on and , are two integrable functions on . Assume that the condition holds. Then, for , , one has

Proof. For any and , we have Multiplying (58) by , , , and integrating the resulting identity with respect to from to , we get