Abstract

The main aim of this paper is to establish some new fractional integral inequalities of Grüss-type via the Saigo fractional integral operator.

1. Introduction

Grüss inequality is an inequality which establishes a connection between the integral of the product of two functions and the product of the integrals of the two functions. In 1935, Grüss proved the following well-known classical integral inequality; see [1, 2].

Theorem 1 (see [2]). Let be two integrable functions such that and for all ; , and are constant; then where the constant is sharp.

In the past decade, many researchers had numerous generalizations, variants, and extensions of Grüss inequality done in literature; to mention a few, see [1–12] and the references cited therein.

Recently, many authors have studied the fractional integral inequalities via Caputo, Riemann-Liouville, and -fractional integral; see [6, 13–21]. Some authors have studied the Saigo fractional integral operator; for example, we refer the reader to [22–29] and references cited therein. In [19], Dahmani et al. gave the following fractional integral inequality using Riemann-Liouville fractional integral.

Theorem 2 (see [19]). Let and be two integrable functions on satisfying the condition on . Then, for all and , one has

In literature few results have been obtained on some fractional integral inequalities using Hadamard fractional integral and Saigo fractional integral operator in [25, 26, 30–32]. Our purpose in this paper is to establish some new results using Saigo fractional integral.

2. Preliminaries

We give some definitions and properties which will be used later. For more details, see [28, 33].

Definition 3. A real-valued function () is said to be in space , if there exists a real number such that , where .

Definition 4 (see [27, 28]). Let , ; then the Saigo fractional integral of order for a real-valued continuous function is defined by where the function in the right-hand side of (4) is the Gaussian hypergeometric function defined by And is the Pochhammer symbol For in (4), we have the known result [28] as .

3. Grüss-Type Fractional Integral Inequality

In this section, we establish Grüss-type fractional integral inequality involving Saigo fractional integer operator (4), for which we assume the following.

There exist two integrable functions and on , such that

There exist two integrable functions and on , such that

Lemma 5. Suppose that is an integrable function on and and are two integrable functions on . Assume that the condition holds.
Then, for all , , , , one has

Proof. Let be an integrable function on . For all , we have Consider Clearly, function positive since for all . Multiplying both sides of (11) by , then integrating resulting identity with respect to from to , we get
Multiplying both sides of (13) by and is positive since for all  , and defined as in (12). Then integrating resulting identity with respect to from to , we have It follows that which gives (10).

If and for all , then inequality (10) reduces to following lemma.

Lemma 6. Let , and be an integrable function on satisfying the condition . Then for all , , , , one has

Theorem 7. Let and be two integrable functions on , and , , , and are four integrable functions on satisfying the conditions and on . Then for all , , , , one has where is defined by

Proof. Let and be two functions defined on satisfying the conditions and . Define It follows that
Then, multiplying both sides of (20) by which is defined by (12) and is positive because for all . Then integrating resulting identity with respect to from to , we have
Again, multiplying both sides of (21) by which is defined by (12) and is positive because for all since each term of (12) is positive. Then integrating resulting identity with respect to from to , we have Applying the Cauchy-Schwarz inequality to (22), we have Since and , we have Thus we have Combining (23) and (25) we get the required inequality (17).

Theorem 8. Suppose that is an integrable function defined on and condition holds. Then for all , , , , , , ,

Proof. From the condition , for all , we have which implies that Multiplying both sides of (28) by which is defined by (12) and remain positive because for all since each term of (12) is positive. Then integrating resulting identity with respect to from to , we get Therefore,
Multiplying (30) by , , which remains positive. Then integrating the resulting identity with respect to from to , we have which gives (26) and proves the theorem.

Remark 9. Suppose that is an integrable function defined on , such that , for all and . Then for all ,  ,  ,  ,  ,  ,  , we have

Theorem 10. Suppose that and are two integrable functions defined on and suppose that and hold. Then for all , ,  , , , , , the following inequalities are satisfied:

Proof. To show , we use conditions and , for all ; we have which implies that Multiplying both sides of (35) by , which is defined by (12) and remain positive because for all , since each term of (12) is positive. Then integrating the resulting identity with respect to from to , we get Therefore, Multiplying (37) by , ), which remains positive. Then by integrating the resulting identity with respect to from to , we have
This gives the desired inequality . Using the following condition we prove ()–():

Remark 11. Suppose that and are two integrable functions defined on . Assume that there exist real constants , and such that Then for all , , , , , , , the following inequalities are satisfied: It is noted that the results give some contributions to the theory of integral inequalities and fractional calculus. Moreover, they are expected to lead to some applications for establishing uniqueness of solutions in fractional differential equations.

Conflict of Interests

The authors declare that they have no conflict of interests.

Authors’ Contribution

All authors have equal contributions. All authors read and approved the final paper.