Some New Riemann-Liouville Fractional Integral Inequalities

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 𝑓 and 𝑔 are two continuous functions on [𝑎,𝑏] satisfying 𝑚≤𝑓(𝑡)≤𝑀 and 𝑝≤𝑔(𝑡)≤𝑃 for all 𝑡∈[𝑎,𝑏], 𝑚,𝑀,𝑝,𝑃∈ℝ, then 1𝑏−𝑎^𝑏_𝑎𝑓(𝑡)𝑔(𝑡)1𝑑𝑡−(𝑏−𝑎)^2𝑏𝑎𝑓(𝑡)𝑑𝑡^𝑏_𝑎𝑔(𝑡)≤1𝑑𝑡4(𝑀−𝑚)𝑃−𝑝. (1)

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 𝑓 and 𝑔 be two integrable functions on [0,∞) satisfying the following conditions: (𝑡)(𝑡)[).𝑚≤𝑓≤𝑀,𝑝≤𝑔≤𝑃𝑚,𝑀,𝑝,𝑃∈ℝ,𝑡∈0,∞ (2) For all 𝑡>0, 𝛼>0, 𝛽>0, then 𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝑓𝑔(𝑡)−𝐽^𝛼𝑓(𝑡)𝐽^𝛼𝑔(𝑡≤𝑡_𝛼)Γ(𝛼+1)²(,𝑡𝑀−𝑚)𝑃−𝑝_𝛼Γ()𝐽𝛼+1^𝛽(𝑡)+𝑡𝑓𝑔^𝛽Γ𝐽𝛽+1^𝛼𝑓𝑔(𝑡)−𝐽^𝛼𝑓(𝑡)𝐽^𝛽𝑔(𝑡)−𝐽^𝛽𝑓(𝑡)𝐽^𝛼𝑔(𝑡)²≤𝑀𝑡^𝛼Γ(𝛼+1)−𝐽^𝛼𝑓(𝑡)𝐽^𝛽𝑓(𝑡)𝑡−𝑚^𝛽Γ+𝐽𝛽+1_𝛼𝑓(𝑡)𝑡−𝑚^𝛼Γ()×𝑀𝑡𝛼+1_𝛽Γ𝛽+1−𝐽^𝛽𝑓(𝑡)×𝑃𝑡^𝛼Γ(𝛼+1)−𝐽^𝛼𝑔(𝑡)𝐽^𝛽𝑔(𝑡)𝑡−𝑝^𝛽Γ+𝐽𝛽+1_𝛼𝑔(𝑡)𝑡−𝑝^𝛼Γ()×𝑃𝑡𝛼+1_𝛽Γ𝛽+1−𝐽^𝛽𝑔(𝑡). (3)

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 𝑓 and 𝑔, 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 𝛼≥0 of a function 𝑔∈𝐿¹((0,∞),ℝ) is defined by 𝐽^𝛼𝑓(𝑡)=^𝑡₀(𝑡−𝑠)^𝛼−1Γ(𝛼)𝑓(𝑠)𝐽𝑑𝑠,₀𝑓(𝑡)=𝑓(𝑡), (4) where Γ is the gamma function.

For the convenience of establishing our results, we give the semigroup property: 𝐽^𝛼𝐽^𝛽𝑓(𝑡)=𝐽^𝛼+𝛽𝑓(𝑡),𝛼≥0,𝛽≥0, (5) which implies the commutative property 𝐽^𝛼𝐽^𝛽𝑓(𝑡)=𝐽^𝛽𝐽^𝛼𝑓(𝑡). (6) From Definition 1, if 𝑓(𝑡)=𝑡^𝛾, then we have 𝐽^𝛼𝑡^𝛾=Γ𝛾+1Γ𝑡𝛼+𝛾+1^𝛼+𝛾,𝛼>0,𝛾>−1,𝑡>0. (7)

3. Main Results

Theorem 2. Let 𝑓 be an integrable function on [0,∞). Assume that (𝐻₁)there exist two integrable functions 𝜑₁, 𝜑₂ on [0,∞) such that 𝜑₁(𝑡)≤𝑓(𝑡)≤𝜑₂(𝑡)∀𝑡∈[0,∞). (8)Then, for 𝑡>0, 𝛼,𝛽>0, one has 𝐽^𝛽𝜑₁(𝑡)𝐽^𝛼𝑓(𝑡)+𝐽^𝛼𝜑₂(𝑡)𝐽^𝛽𝑓(𝑡)≥𝐽^𝛼𝜑₂(𝑡)𝐽^𝛽𝜑₁(𝑡)+𝐽^𝛼𝑓(𝑡)𝐽^𝛽𝑓(𝑡). (9)

Proof. From (𝐻₁), for all 𝜏≥0, 𝜌≥0, we have 𝜑₂(𝜏)−𝑓(𝜏)𝑓𝜌−𝜑₁𝜌≥0. (10) Therefore 𝜑₂(𝜏)𝑓𝜌+𝜑₁𝜌𝑓(𝜏)≥𝜑₁𝜌𝜑₂(𝜏)+𝑓(𝜏)𝑓𝜌. (11) Multiplying both sides of (11) by (𝑡−𝜏)^𝛼−1/Γ(𝛼), 𝜏∈(0,𝑡), we get 𝑓𝜌(𝑡−𝜏)^𝛼−1Γ(𝛼)𝜑₂(𝜏)+𝜑₁𝜌(𝑡−𝜏)^𝛼−1Γ(𝛼)𝑓(𝜏)≥𝜑₁𝜌(𝑡−𝜏)^𝛼−1Γ(𝛼)𝜑₂(𝜏)𝜌()+𝑓𝑡−𝜏^𝛼−1Γ(𝛼)𝑓(𝜏). (12) Integrating both sides of (12) with respect to 𝜏 on (0,𝑡), we obtain 𝑓𝜌^𝑡₀(𝑡−𝜏)^𝛼−1Γ(𝛼)𝜑₂(𝜏)𝑑𝜏+𝜑₁𝜌^𝑡₀(𝑡−𝜏)^𝛼−1Γ(𝛼)𝑓(𝜏)𝑑𝜏≥𝜑₁𝜌^𝑡₀(𝑡−𝜏)^𝛼−1Γ(𝛼)𝜑₂(𝜏)𝜌𝑑𝜏+𝑓^𝑡₀(𝑡−𝜏)^𝛼−1Γ(𝛼)𝑓(𝜏)𝑑𝜏, (13) which yields 𝑓𝜌𝐽^𝛼𝜑₂(𝑡)+𝜑₁𝜌𝐽^𝛼𝑓(𝑡)≥𝜑₁𝜌𝐽^𝛼𝜑₂(𝑡)+𝑓𝜌𝐽^𝛼𝑓(𝑡). (14) Multiplying both sides of (14) by (𝑡−𝜌)^𝛽−1/Γ(𝛽), 𝜌∈(0,𝑡), we have 𝐽^𝛼𝜑₂(𝑡)𝑡−𝜌^𝛽−1Γ𝛽𝑓𝜌+𝐽^𝛼𝑓(𝑡)𝑡−𝜌^𝛽−1Γ𝛽𝜑₁𝜌≥𝐽^𝛼𝜑₂(𝑡)𝑡−𝜌^𝛽−1Γ𝛽𝜑₁𝜌+𝐽^𝛼𝑓(𝑡)𝑡−𝜌^𝛽−1Γ𝛽𝑓𝜌. (15) Integrating both sides of (15) with respect to 𝜌 on (0,𝑡), we get 𝐽^𝛼𝜑₂(𝑡)^𝑡₀𝑡−𝜌^𝛽−1Γ𝛽𝑓𝜌𝑑𝜌+𝐽^𝛼𝑓(𝑡)^𝑡₀𝑡−𝜌^𝛽−1Γ𝛽𝜑₁𝜌𝑑𝜌≥𝐽^𝛼𝜑₂(𝑡)^𝑡₀𝑡−𝜌^𝛽−1Γ𝛽𝜑₁𝜌𝑑𝜌+𝐽^𝛼𝑓(𝑡)^𝑡₀𝑡−𝜌^𝛽−1Γ𝛽𝑓𝜌𝑑𝜌. (16) 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 𝑓 be an integrable function on [0,∞) satisfying 𝑚≤𝑓(𝑡)≤𝑀, for all 𝑡∈[0,∞) and 𝑚,𝑀∈ℝ. Then, for 𝑡>0 and 𝛼,𝛽>0, one has 𝑚𝑡^𝛽Γ𝐽𝛽+1^𝛼𝑓(𝑡)𝑡+𝑀^𝛼Γ()𝐽𝛼+1^𝛽𝑓(𝑡)𝑡≥𝑚𝑀^𝛼+𝛽Γ(𝛼+1)Γ𝛽+1+𝐽^𝛼𝑓(𝑡)𝐽^𝛽𝑓(𝑡). (17)

Example 4. Let 𝑓 be a function satisfying 𝑡≤𝑓(𝑡)≤𝑡+1 for 𝑡∈[0,∞). Then, for 𝑡>0 and 𝛼>0, we have 2𝑡^𝛼+1Γ()+𝑡𝛼+2Γ(^𝛼)𝐽𝛼+1^𝛼𝑓(𝑡)≥𝑡^𝛼+1Γ()+𝑡𝛼+2Γ(^𝛼)𝑡𝛼+1Γ(^𝛼+1)+𝐽𝛼+2^𝛼𝑓(𝑡)². (18)

Theorem 5. Let 𝑓 and 𝑔 be two integrable functions on [0,∞). Suppose that (𝐻₁) holds and moreover one assumes that (𝐻₂) there exist 𝜓₁ and 𝜓₂ integrable functions on [0,∞) such that 𝜓₁(𝑡)≤𝑔(𝑡)≤𝜓₂(𝑡),∀𝑡∈[0,∞). (19)Then, for 𝑡>0, 𝛼,𝛽>0, the following inequalities hold: (𝑎)𝐽^𝛽𝜓₁(𝑡)𝐽^𝛼𝑓(𝑡)+𝐽^𝛼𝜑₂(𝑡)𝐽^𝛽𝑔(𝑡)≥𝐽^𝛽𝜓₁(𝑡)𝐽^𝛼𝜑₂(𝑡)+𝐽^𝛼𝑓(𝑡)𝐽^𝛽𝑔(𝑡),(𝑏)𝐽^𝛽𝜑₁(𝑡)𝐽^𝛼𝑔(𝑡)+𝐽^𝛼𝜓₂(𝑡)𝐽^𝛽𝑓(𝑡)≥𝐽^𝛽𝜑₁(𝑡)𝐽^𝛼𝜓₂(𝑡)+𝐽^𝛽𝑓(𝑡)𝐽^𝛼𝑔(𝑡),(𝑐)𝐽^𝛼𝜑₂(𝑡)𝐽^𝛽𝜓₂(𝑡)+𝐽^𝛼𝑓(𝑡)𝐽^𝛽𝑔(𝑡)≥𝐽^𝛼𝜑₂(𝑡)𝐽^𝛽𝑔(𝑡)+𝐽^𝛽𝜓₂(𝑡)𝐽^𝛼𝑓(𝑡),(𝑑)𝐽^𝛼𝜑₁(𝑡)𝐽^𝛽𝜓₁(𝑡)+𝐽^𝛼𝑓(𝑡)𝐽^𝛽𝑔(𝑡)≥𝐽^𝛼𝜑₁(𝑡)𝐽^𝛽𝑔(𝑡)+𝐽^𝛽𝜓₁(𝑡)𝐽^𝛼𝑓(𝑡). (20)

Proof. To prove (𝑎), from (𝐻₁) and (𝐻₂), we have for 𝑡∈[0,∞) that 𝜑₂(𝜏)−𝑓(𝜏)𝑔𝜌−𝜓₁𝜌≥0. (21) Therefore 𝜑₂(𝜏)𝑔𝜌+𝜓₁𝜌𝑓(𝜏)≥𝜓₁𝜌𝜑₂(𝜏)+𝑓(𝜏)𝑔𝜌. (22) Multiplying both sides of (22) by (𝑡−𝜏)^𝛼−1/Γ(𝛼), 𝜏∈(0,𝑡), we get 𝑔𝜌(𝑡−𝜏)^𝛼−1Γ(𝛼)𝜑₂(𝜏)+𝜓₁𝜌(𝑡−𝜏)^𝛼−1Γ(𝛼)𝑓(𝜏)≥𝜓₁𝜌(𝑡−𝜏)^𝛼−1Γ(𝛼)𝜑₂(𝜏)𝜌()+𝑔𝑡−𝜏^𝛼−1Γ(𝛼)𝑓(𝜏). (23) Integrating both sides of (23) with respect to 𝜏 on (0,𝑡), we obtain 𝑔𝜌^𝑡₀(𝑡−𝜏)^𝛼−1Γ(𝛼)𝜑₂(𝜏)𝑑𝜏+𝜓₁𝜌^𝑡₀(𝑡−𝜏)^𝛼−1Γ(𝛼)𝑓(𝜏)𝑑𝜏≥𝜓₁𝜌^𝑡₀(𝑡−𝜏)^𝛼−1Γ(𝛼)𝜑₂(𝜏)𝜌𝑑𝜏+𝑔^𝑡₀(𝑡−𝜏)^𝛼−1Γ(𝛼)𝑓(𝜏)𝑑𝜏. (24) Then we have 𝑔𝜌𝐽^𝛼𝜑₂(𝑡)+𝜓₁𝜌𝐽^𝛼𝑓(𝑡)≥𝜓₁𝜌𝐽^𝛼𝜑₂(𝑡)+𝑔𝜌𝐽^𝛼𝑓(𝑡). (25) Multiplying both sides of (25) by (𝑡−𝜌)^𝛽−1/Γ(𝛽), 𝜌∈(0,𝑡), we have 𝐽^𝛼𝜑₂(𝑡)𝑡−𝜌^𝛽−1Γ𝛽𝑔𝜌+𝐽^𝛼𝑓(𝑡)𝑡−𝜌^𝛽−1Γ𝛽𝜓₁𝜌≥𝐽^𝛼𝜑₂(𝑡)𝑡−𝜌^𝛽−1Γ𝛽𝜓₁𝜌+𝐽^𝛼𝑓(𝑡)𝑡−𝜌^𝛽−1Γ𝛽𝑔𝜌. (26) Integrating both sides of (26) with respect to 𝜌 on (0,𝑡), we get the desired inequality (𝑎).
To prove (𝑏)--(𝑑), we use the following inequalities: (𝑏)𝜓₂(𝜏)−𝑔(𝜏)𝑓𝜌−𝜑₁𝜌(𝑐)𝜑≥0,₂(𝜏)−𝑓(𝜏)𝑔𝜌−𝜓₂𝜌(𝑑)𝜑≤0,₁(𝜏)−𝑓(𝜏)𝑔𝜌−𝜓₁𝜌≤0. (27)

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

Corollary 6. Let 𝑓 and 𝑔 be two integrable functions on [0,∞). Assume that (𝐻₃) there exist real constants 𝑚,𝑀,𝑛,𝑁 such that 𝑚≤𝑓(𝑡)≤𝑀,𝑛≤𝑔(𝑡)≤𝑁∀𝑡∈[0,∞). (28)Then, for 𝑡>0, 𝛼,𝛽>0, we have 𝑎₁𝑛𝑡^𝛽Γ𝐽𝛽+1^𝛼𝑓(𝑡)+𝑀𝑡^𝛼Γ()𝐽𝛼+1^𝛽𝑔(𝑡)≥𝑛𝑀𝑡^𝛼+𝛽Γ(𝛼+1)Γ𝛽+1+𝐽^𝛼𝑓(𝑡)𝐽^𝛽𝑔(𝑡),𝑏₁𝑚𝑡^𝛽Γ𝐽𝛽+1^𝛼𝑔(𝑡)+𝑁𝑡^𝛼Γ()𝐽𝛼+1^𝛽𝑓(𝑡)≥𝑚𝑁𝑡^𝛼+𝛽Γ(𝛼+1)Γ𝛽+1+𝐽^𝛽𝑓(𝑡)𝐽^𝛼𝑔(𝑡),𝑐₁𝑀𝑁𝑡^𝛼+𝛽Γ(𝛼+1)Γ𝛽+1+𝐽^𝛼𝑓(𝑡)𝐽^𝛽𝑔(𝑡)≥𝑀𝑡^𝛼Γ()𝐽𝛼+1^𝛽𝑔(𝑡)+𝑁𝑡^𝛽Γ𝐽𝛽+1^𝛼𝑓(𝑡),𝑑₁𝑚𝑛𝑡^𝛼+𝛽Γ(𝛼+1)Γ𝛽+1+𝐽^𝛼𝑓(𝑡)𝐽^𝛽𝑔(𝑡)≥𝑚𝑡^𝛼Γ()𝐽𝛼+1^𝛽𝑔(𝑡)+𝑛𝑡^𝛽Γ𝐽𝛽+1^𝛼𝑓(𝑡). (29)

Lemma 7. Let 𝑓 be an integrable function on [0,∞) and let 𝜑₁, 𝜑₂ be two integrable functions on [0,∞). Assume that the condition (𝐻₁) holds. Then, for 𝑡>0, 𝛼>0, we have 𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝑓²(𝑡)−𝐽^𝛼𝑓(𝑡)²=𝐽^𝛼𝜑₂(𝑡)−𝐽^𝛼𝑓(𝑡)𝐽^𝛼𝑓(𝑡)−𝐽^𝛼𝜑₁(𝑡)−𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝜑₂(𝑡)−𝑓(𝑡)𝑓(𝑡)−𝜑₁(𝑡+𝑡_𝛼)Γ()𝐽𝛼+1^𝛼𝜑₁𝑓(𝑡)−𝐽^𝛼𝜑₁(𝑡)𝐽^𝛼𝑓(𝑡)+𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝜑₂𝑓(𝑡)−𝐽^𝛼𝜑₂(𝑡)𝐽^𝛼𝑓(𝑡)+𝐽^𝛼𝜑₁(𝑡)𝐽^𝛼𝜑₂(𝑡)−𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝜑₁𝜑₂(𝑡). (30)

Proof. For any 𝜏>0 and 𝜌>0, we have 𝜑₂𝜌−𝑓𝜌𝑓(𝜏)−𝜑₁(𝜏)+𝜑₂(𝜏)−𝑓(𝜏)𝑓𝜌−𝜑₁𝜌−𝜑₂(𝜏)−𝑓(𝜏)𝑓(𝜏)−𝜑₁(𝜏)−𝜑₂𝜌−𝑓𝜌𝑓𝜌−𝜑₁𝜌=𝑓²(𝜏)+𝑓²𝜌−2𝑓(𝜏)𝑓𝜌+𝜑₂𝜌𝑓(𝜏)+𝜑₁(𝜏)𝑓𝜌−𝜑₁(𝜏)𝜑₂𝜌+𝜑₂(𝜏)𝑓𝜌+𝜑₁𝜌𝑓(𝜏)−𝜑₁𝜌𝜑₂(𝜏)−𝜑₂(𝜏)𝑓(𝜏)+𝜑₁(𝜏)𝜑₂(𝜏)−𝜑₁(𝜏)𝑓(𝜏)−𝜑₂𝜌𝑓𝜌+𝜑₁𝜌𝜑₂𝜌−𝜑₁𝜌𝑓𝜌. (31) Multiplying (31) by (𝑡−𝜏)^𝛼−1/Γ(𝛼), 𝜏∈(0,𝑡), 𝑡>0 and integrating the resulting identity with respect to 𝜏, from 0 to 𝑡, we get 𝜑₂𝜌−𝑓𝜌𝐽^𝛼𝑓(𝑡)−𝐽^𝛼𝜑₁(𝑡+𝐽_𝛼)𝜑₂(𝑡)−𝐽^𝛼𝑓(𝑡)𝑓𝜌−𝜑₁𝜌−𝐽^𝛼𝜑₂(𝑡)−𝑓(𝑡)𝑓(𝑡)−𝜑₁(𝑡)−𝜑₂𝜌−𝑓𝜌𝑓𝜌−𝜑₁𝜌𝑡^𝛼Γ(𝛼+1)=𝐽^𝛼𝑓²(𝑡)+𝑓²𝜌𝑡^𝛼Γ()𝜌𝐽𝛼+1−2𝑓^𝛼𝑓(𝑡)+𝜑₂𝜌𝐽^𝛼𝑓(𝑡)+𝑓𝜌𝐽^𝛼𝜑₁(𝑡)−𝜑₂𝜌𝐽^𝛼𝜑₁(𝑡)+𝑓𝜌𝐽^𝛼𝜑₂(𝑡)+𝜑₁𝜌𝐽^𝛼𝑓(𝑡)−𝜑₁𝜌𝐽^𝛼𝜑₂(𝑡)−𝐽^𝛼𝜑₂𝑓(𝑡)+𝐽^𝛼𝜑₁𝜑₂(𝑡)−𝐽^𝛼𝜑₁𝑓(𝑡)−𝜑₂𝜌𝑓𝜌𝑡^𝛼Γ(𝛼+1)+𝜑₁𝜌𝜑₂𝜌𝑡^𝛼Γ(𝛼+1)−𝜑₁𝜌𝑓𝜌𝑡^𝛼Γ().𝛼+1 (32) Multiplying (32) by (𝑡−𝜌)^𝛼−1/Γ(𝛼), 𝜌∈(0,𝑡), 𝑡>0 and integrating the resulting identity with respect to 𝜌, from 0 to 𝑡, we have 𝐽^𝛼𝜑₂(𝑡)−𝐽^𝛼𝑓(𝑡)𝐽^𝛼𝑓(𝑡)−𝐽^𝛼𝜑₁(𝑡+𝐽_𝛼)𝜑₂(𝑡)−𝐽^𝛼𝑓(𝑡)𝐽^𝛼𝑓(𝑡)−𝐽^𝛼𝜑₁(𝑡)−𝐽^𝛼𝜑₂(𝑡)−𝑓(𝑡)𝑓(𝑡)−𝜑₁(𝑡)𝑡^𝛼Γ(𝛼+1)−𝐽^𝛼𝜑₂(𝑡)−𝑓(𝑡)𝑓(𝑡)−𝜑₁(𝑡)𝑡^𝛼Γ()=𝑡𝛼+1_𝛼Γ()𝐽𝛼+1^𝛼𝑓²(𝑡)+𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝑓²(𝑡)−2𝐽^𝛼𝑓(𝑡)𝐽^𝛼𝑓(𝑡)+𝐽^𝛼𝜑₂(𝑡)𝐽^𝛼𝑓(𝑡)+𝐽^𝛼𝜑₁(𝑡)𝐽^𝛼𝑓(𝑡)−𝐽^𝛼𝜑₁(𝑡)𝐽^𝛼𝜑₂(𝑡)+𝐽^𝛼𝜑₂(𝑡)𝐽^𝛼𝑓(𝑡)+𝐽^𝛼𝜑₁(𝑡)𝐽^𝛼𝑓(𝑡)−𝐽^𝛼𝜑₁(𝑡)𝐽^𝛼𝜑₂(𝑡)−𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝜑₂𝑓(𝑡)+𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝜑₁𝜑₂(𝑡)−𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝜑₁𝑓(𝑡)−𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝜑₂𝑓(𝑡)+𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝜑₁𝜑₂(𝑡)−𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝜑₁𝑓(𝑡), (33) which implies (30).

If 𝜑₁(𝑡)≡𝑚 and 𝜑₂(𝑡)≡𝑀, 𝑚,𝑀∈ℝ, for all 𝑡∈[0,∞), then inequality (30) reduces to the following corollary [10, Lemma 3.2].

Corollary 8. Let 𝑓 be an integrable function on [0,∞) satisfying 𝑚≤𝑓(𝑡)≤𝑀, for all 𝑡∈[0,∞). Then, for all 𝑡>0, 𝛼>0, one has 𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝑓²(𝑡)−𝐽^𝛼𝑓(𝑡)²=𝑀𝑡^𝛼Γ(𝛼+1)−𝐽^𝛼𝑓(𝑡)𝐽^𝛼𝑓(𝑡)𝑡−𝑚^𝛼Γ()−𝑡𝛼+1_𝛼Γ()𝐽𝛼+1^𝛼𝑀−𝑓(𝑡)𝑓(𝑡)−𝑚. (34)

Theorem 9. Let 𝑓 and 𝑔 be two integrable functions on [0,∞) and let 𝜑₁, 𝜑₂, 𝜓₁, and 𝜓₂ be four integrable functions on [0,∞) satisfying the conditions (𝐻₁) and (𝐻₂) on [0,∞). Then, for all 𝑡>0, 𝛼>0, one has 𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝑓𝑔(𝑡)−𝐽^𝛼𝑓(𝑡)𝐽^𝛼𝑔(𝑡≤)𝑇𝑓,𝜑₁,𝜑₂𝑇𝑔,𝜓₁,𝜓₂, (35) where 𝑇(𝑢,𝑣,𝑤) is defined by 𝑇(𝑢,𝑣,𝑤)=𝐽^𝛼𝑤(𝑡)−𝐽^𝛼𝑢(𝑡)𝐽^𝛼𝑢(𝑡)−𝐽^𝛼𝑣(𝑡+𝑡_𝛼)Γ()𝐽𝛼+1^𝛼𝑣𝑢(𝑡)−𝐽^𝛼𝑣(𝑡)𝐽^𝛼𝑢(𝑡)+𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝑤𝑢(𝑡)−𝐽^𝛼𝑤(𝑡)𝐽^𝛼𝑢(𝑡)+𝐽^𝛼𝑣(𝑡)𝐽^𝛼𝑤(𝑡)−𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝑣𝑤(𝑡). (36)

Proof. Let 𝑓 and 𝑔 be two integrable functions defined on [0,∞) satisfying (𝐻₁) and (𝐻₂). Define 𝐻𝑓(𝜏)𝜌𝑔(𝜏)𝜌,()𝜏,𝜌∶=−𝑓−𝑔𝜏,𝜌∈0,𝑡,𝑡>0. (37) Multiplying both sides of (37) by (𝑡−𝜏)^𝛼−1(𝑡−𝜌)^𝛼−1/Γ²(𝛼), 𝜏,𝜌∈(0,𝑡) and integrating the resulting identity with respect to 𝜏 and 𝜌, from 0 to 𝑡, we can state that 12Γ²(𝛼)^𝑡₀(𝑡−𝜏)^𝛼−1𝑡−𝜌^𝛼−1𝐻=𝑡𝜏,𝜌𝑑𝜏𝑑𝜌_𝛼Γ()𝐽𝛼+1^𝛼𝑓𝑔(𝑡)−𝐽^𝛼𝑓(𝑡)𝐽^𝛼𝑔(𝑡). (38) Applying the Cauchy-Schwarz inequality to (38), we have 𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝑓𝑔(𝑡)−𝐽^𝛼𝑓(𝑡)𝐽^𝛼𝑔(𝑡)²≤𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝑓²(𝑡)−𝐽^𝛼𝑓(𝑡)²×𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝑔²(𝑡)−𝐽^𝛼𝑔(𝑡)². (39) Since (𝜑₂(𝑡)−𝑓(𝑡))(𝑓(𝑡)−𝜑₁(𝑡))≥0 and (𝜓₂(𝑡)−𝑔(𝑡))(𝑔(𝑡)−𝜓₁(𝑡))≥0, for 𝑡∈[0,∞), we have 𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝜑₂(𝑡)−𝑓(𝑡)𝑓(𝑡)−𝜑₁(𝑡)𝑡≥0,_𝛼Γ()𝐽𝛼+1^𝛼𝜓₂(𝑡)−𝑔(𝑡)𝑔(𝑡)−𝜓₁(𝑡)≥0. (40) Thus, from Lemma 7, we get 𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝑓²(𝑡)−𝐽^𝛼𝑓(𝑡)²≤𝐽^𝛼𝜑₂(𝑡)−𝐽^𝛼𝑓(𝑡)𝐽^𝛼𝑓(𝑡)−𝐽^𝛼𝜑₁(𝑡)+𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝜑₁𝑓(𝑡)−𝐽^𝛼𝜑₁(𝑡)𝐽^𝛼𝑓(𝑡)+𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝜑₂𝑓(𝑡)−𝐽^𝛼𝜑₂(𝑡)𝐽^𝛼𝑓(𝑡)+𝐽^𝛼𝜑₁(𝑡)𝐽^𝛼𝜑₂(𝑡)−𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝜑₁𝜑₂(𝑡)=𝑇𝑓,𝜑₁,𝜑₂,𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝑔²(𝑡)−𝐽^𝛼𝑔(𝑡)²≤𝐽^𝛼𝜓₂(𝑡)−𝐽^𝛼𝑔(𝑡)𝐽^𝛼𝑔(𝑡)−𝐽^𝛼𝜓₁(𝑡)+𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝜓₁𝑔(𝑡)−𝐽^𝛼𝜓₁(𝑡)𝐽^𝛼𝑔(𝑡)+𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝜓₂𝑔(𝑡)−𝐽^𝛼𝜓₂(𝑡)𝐽^𝛼𝑔(𝑡)+𝐽^𝛼𝜓₁(𝑡)𝐽^𝛼𝜓₂(𝑡)−𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝜓₁𝜓₂(𝑡)=𝑇𝑔,𝜓₁,𝜓₂. (41)(42) From (39), (41), and (42), we obtain (35).

Remark 10. If 𝑇(𝑓,𝜑₁,𝜑₂)=𝑇(𝑓,𝑚,𝑀) and 𝑇(𝑔,𝜓₁,𝜓₂)=𝑇(𝑔,𝑝,𝑃), 𝑚,𝑀,𝑝,𝑃∈ℝ, then inequality (35) reduces to 𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝑓𝑔(𝑡)−𝐽^𝛼𝑓(𝑡)𝐽^𝛼𝑔(𝑡≤𝑡_𝛼)2Γ(𝛼+1)²(𝑀−𝑚)𝑃−𝑝. (43) See [10, Theorem 3.1].

Example 11. Let 𝑓 and 𝑔 be two functions satisfying 𝑡≤𝑓(𝑡)≤𝑡+1 and 𝑡−1≤𝑔(𝑡)≤𝑡 for 𝑡∈[0,∞). Then, for 𝑡>0 and 𝛼>0, we have 𝑡^𝛼Γ()𝐽𝛼+1^𝛼𝑓𝑔(𝑡)−𝐽^𝛼𝑓(𝑡)𝐽^𝛼𝑔(𝑡≤)𝑇𝑓,𝑡,𝑡+1𝑇𝑔,𝑡−1,𝑡, (44) where 𝑇=𝑡𝑓,𝑡,𝑡+1_𝛼+1Γ()+𝑡𝛼+2Γ(𝛼^𝛼+1)−𝐽^𝛼𝑓(𝑡)×𝐽^𝛼𝑓(𝑡)−𝑡^𝛼+1Γ()+𝑡𝛼+2_𝛼Γ()𝐽𝛼+1^𝛼(𝑡)−𝑡𝑡𝑓^𝛼+1Γ()𝐽𝛼+2^𝛼𝑓(𝑡)+𝑡^𝛼Γ()𝐽𝛼+1^𝛼()𝑓(𝑡)−𝑡𝑡+1_𝛼+1Γ()+𝑡𝛼+2Γ(^𝛼)𝐽𝛼+1^𝛼𝑓(𝑡)+𝑡^𝛼+1Γ()𝑡𝛼+2Γ(^𝛼+1)+𝑡𝛼+2Γ(^𝛼)−𝑡𝛼+1_𝛼Γ(𝛼+1)2𝑡^𝛼+2Γ()+𝑡𝛼+3Γ(^𝛼+1),𝑇=𝑡𝛼+2𝑔,𝑡−1,𝑡_𝛼+1Γ(𝛼+2)−𝐽^𝛼𝑔(𝑡)×𝐽^𝛼𝑔(𝑡)−𝑡^𝛼+1Γ()+𝑡𝛼+2Γ(^𝛼)+𝑡𝛼+1_𝛼Γ()𝐽𝛼+1^𝛼()𝑔(𝑡)−𝑡𝑡−1_𝛼+1Γ()−𝑡𝛼+2Γ(^𝛼)𝐽𝛼+1^𝛼𝑔(𝑡)+𝑡^𝛼Γ()𝐽𝛼+1^𝛼(𝑡)−𝑡𝑡𝑔^𝛼+1Γ()𝐽𝛼+2^𝛼𝑔(𝑡)+𝑡^𝛼+1Γ()−𝑡𝛼+2Γ(^𝛼)𝑡𝛼+1Γ(^𝛼+1)−𝑡𝛼+2_𝛼Γ(𝛼+1)2𝑡^𝛼+2Γ()−𝑡𝛼+3Γ(^𝛼+1).𝛼+2 (45)(46)

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.