Some New Riemann-Liouville Fractional Integral Inequalities

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