#### Abstract

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

#### 1. Introduction

In [1] (see also [2]), 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 [3–9] and the references cited therein. Dahmani et al. [10] 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 [10] 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 [10]. 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.