#### Abstract

In this paper, some new fractional integral inequalities are established.

#### 1. Introduction

In  (see also ), the Grüss inequality is defined as the integral inequality that establishes a connection between the integral of the product of two functions and the product of the integrals. The inequality is as follows.

If and are two continuous functions on satisfying and for all , , then

The literature on Grüss type inequalities is now vast, and many extensions of the classical inequality were intensively studied by many authors. In the past several years, by using the Riemann-Liouville fractional integrals, the fractional integral inequalities and applications have been addressed extensively by several researchers. For example, we refer the reader to  and the references cited therein. Dahmani et al.  gave the following fractional integral inequalities by using the Riemann-Liouville fractional integrals. Let and be two integrable functions on satisfying the following conditions: For all , , , then

In this paper, we use the Riemann-Liouville fractional integrals to establish some new fractional integral inequalities of Grüss type. We replace the constants appeared as bounds of the functions and , by four integrable functions. From our results, the above inequalities of  and the classical Grüss inequalities can be deduced as some special cases.

In Section 2 we briefly review the necessary definitions. Our results are given in Section 3. The proof technique is close to that presented in . But the obtained results are new and also can be applied to unbounded functions as shown in examples.

#### 2. Preliminaries

Definition 1. The Riemann-Liouville fractional integral of order of a function is defined by where is the gamma function.

For the convenience of establishing our results, we give the semigroup property: which implies the commutative property From Definition 1, if , then we have

#### 3. Main Results

Theorem 2. Let be an integrable function on . Assume that there exist two integrable functions , on such that Then, for , , one has

Proof. From , for all , , we have Therefore Multiplying both sides of (11) by , , we get Integrating both sides of (12) with respect to on , we obtain which yields Multiplying both sides of (14) by , , we have Integrating both sides of (15) with respect to on , we get Hence, we deduce inequality (9) as requested. This completes the proof.

As a special case of Theorem 2, we obtain the following result.

Corollary 3. Let be an integrable function on satisfying , for all and . Then, for and , one has

Example 4. Let be a function satisfying for . Then, for and , we have

Theorem 5. Let and be two integrable functions on . Suppose that holds and moreover one assumes that 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 (22) by , , we get Integrating both sides of (23) with respect to on , we obtain Then we have Multiplying both sides of (25) by , , we have Integrating both sides of (26) with respect to on , we get the desired inequality .
To prove , we use the following inequalities:

As a special case of Theorem 5, we have the following corollary.

Corollary 6. Let and be two integrable functions on . Assume that there exist real constants such that Then, for , , we have

Lemma 7. Let be an integrable function on and let , be two integrable functions on . Assume that the condition holds. Then, for , , we have

Proof. For any and , we have Multiplying (31) by , , and integrating the resulting identity with respect to , from to , we get Multiplying (32) by , , and integrating the resulting identity with respect to , from to , we have which implies (30).

If and , , for all , then inequality (30) reduces to the following corollary [10, Lemma 3.2].

Corollary 8. Let be an integrable function on satisfying , for all . Then, for all , , one has

Theorem 9. Let and be two integrable functions on and let , , , and be four integrable functions on satisfying the conditions and on . Then, for all , , one has where is defined by

Proof. Let and be two integrable functions defined on satisfying and . Define Multiplying both sides of (37) by , and integrating the resulting identity with respect to and , from to , we can state that Applying the Cauchy-Schwarz inequality to (38), we have Since and , for , we have Thus, from Lemma 7, we get From (39), (41), and (42), we obtain (35).

Remark 10. If and , , then inequality (35) reduces to See [10, Theorem 3.1].

Example 11. Let and be two functions satisfying and for . Then, for and , we have where

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

This research was funded by King Mongkut’s University of Technology North Bangkok, Thailand. Project code: KMUTNB-GRAD-56-02. Sotiris K. Ntouyas is a member of Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group at King Abdulaziz University, Jeddah, Saudi Arabia.