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Some Inequalities of Qi Type Using Fractional Integration
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International Journal of Nonlinear Science Vol.10(2010) No.4,pp.396-400
Some Inequalities of Qi Type Using Fractional Integration
Zoubir Dahmani
∗, Soumia Belarbi
Laboratory of Pure and Applied Mathematics, Faculty of SESNV, University Abdelhamid Ben Badis of Mostaganem, UMAB, Mostaganem, Algeria
(Received 17 April 2010 , accepted 1 September 2010)
Abstract: In the present paper, we use the Riemann-Liouville fractional integral to establish some integral results for certain classes of functions defined on some intervals of the real line. By introducing parameters 𝛼, 𝛽 and 𝛿, we give some sufficient conditions to generate some fractional inequalities of Qi type, and we give new generalizations for some results of [12,16].
Keywords:Fractional integration, Qi integral inequality
1
Introduction
Inequalities have proved to be one of the most powerful and far-reaching tools for the development of many branches of mathematics. In the last few decades, much significant development in the classical and new inequalities, particularly in analysis has been witnessed. As an example, let us cite the field of integration which is dominated by inequalities involving functions and their integrals [2,4,5,6,9,15]. One of the famous integral inequalities is Feng Qi inequality [14,15]. In [15], Qi proved that ∫ 𝑏 𝑎 [𝑓(𝜏)] 𝑛+2𝑑𝜏 ≥(∫ 𝑏 𝑎 𝑓(𝜏)𝑑𝜏 )𝑛+1 . (1)
In [13], the authors established the following inequality: ∫ 𝑏 𝑎 [𝑓(𝜏)] 𝛽𝑑𝜏 ≥(∫ 𝑏 𝑎 𝑓(𝜏)𝑑𝜏 )𝛽−1 , (2) where𝑓 ∈ 𝐶1([𝑎, 𝑏]), 𝑓(𝑎) ≥ 0 and 𝑓′(𝜏) > (𝛽 − 2)(𝜏 − 𝑎)𝛽−3, 𝜏 ∈ [𝑎, 𝑏] .
Many researchers have given considerable attention to (2) and a number of extensions, generalizations and variants have appeared in the literature, see [3,10,11,12,14].
In the case of fractional integral, in [7], the authors established some new fractional inequalities based on the paper [1].
The main purpose of this paper is to establish some fractional results of the inequality (2) using Riemann-Liouville fractional integral. Our results have some relationships with some inequalities obtained in [12,16].
2
Basic Definitions
In the following, we will give the necessary notations and basic definitions. For more details, one can consult [6,12]. Definition 1 A real valued function𝑓(𝑡), 𝑡 > 0 is said to be in the space 𝐶𝜇, 𝜇 ∈ ℝ if there exists a real number 𝑝 > 𝜇 such that𝑓(𝑡) = 𝑡𝑝𝑓1(𝑡), where 𝑓1(𝑡) ∈ 𝐶(]0, ∞[).
∗Corresponding author. E-mail address: [email protected]
Copyright c⃝World Academic Press, World Academic Union IJNS.2010.12.30/421
Z. Dahmani, S. Belarbi: Some Inequalities of Qi Type Using Fractional Integration 397
Definition 2 A function𝑓(𝑡), 𝑡 > 0 is said to be in the space 𝐶𝜇𝑛, 𝑛 ∈ ℕ, if 𝑓(𝑛)∈ 𝐶𝜇.
Definition 3 The Riemann-Liouville fractional integral operator of order𝛼 ≥ 0, for a function 𝑓 ∈ 𝐶𝜇, (𝜇 ≥ −1) is defined as 𝐽𝛼𝑓(𝑡) = 1 Γ(𝛼) ∫𝑡 0(𝑡 − 𝜏)𝛼−1𝑓(𝜏)𝑑𝜏; 𝛼 > 0, 𝑡 > 0, 𝐽0𝑓(𝑡) = 𝑓(𝑡), (3) whereΓ(𝛼) :=∫0∞𝑒−𝑢𝑢𝛼−1𝑑𝑢.
For the convenience of establishing the results, we give the semigroup property:
𝐽𝛼𝐽𝛽𝑓(𝑡) = 𝐽𝛼+𝛽𝑓(𝑡); 𝛼 ≥ 0, 𝛽 ≥ 0, (4)
which implies the commutative property
𝐽𝛼𝐽𝛽𝑓(𝑡) = 𝐽𝛽𝐽𝛼𝑓(𝑡). (5)
3
Main Results
Theorem 1 Suppose that 𝑓 ∈ 𝐶1([0, ∞[) satisfies 𝑓(0) ≥ 0 and 𝑓′(𝑥) ≥ (𝛽 − 2)(Γ(𝛿+1)𝑥𝛿 )𝛽−3 (
(𝑥−𝜏)𝛿−1
Γ(𝛿)
)𝛼−1
for 𝑥 ∈ [0, 𝑡]; 𝑡 > 0, 𝛽 ≥ 3. Then for all 𝛼 ≥ 1, the inequality
𝐽𝛼(𝑓𝛽(𝑡))≥ Γ𝛽−2(𝛼) (𝐽𝛼𝑓(𝑡))𝛽−1 (6)
is valid.
Proof.Since𝑓′(𝑥) ≥ 0 for 𝑥 ∈ [0, ∞[ then 𝑓 is an increasing function on [0, ∞[. Hence for any 𝑡 > 0, we can write
𝑓 (𝜏) ≤ 𝑓 (𝑥) ; 𝜏 ∈ [0, 𝑥] , 𝑥 ≤ 𝑡. (7)
Multiplying both sides of by(𝑡 − 𝜏)𝛼−1, we get
(𝑡 − 𝜏)𝛼−1𝑓 (𝜏) ≤ (𝑡 − 𝜏)𝛼−1𝑓 (𝑥) . (8)
Integrating both sides of(8) over [0, 𝑥] , we obtain
𝑥 ∫ 0 (𝑡 − 𝜏)𝛼−1𝑓 (𝜏) 𝑑𝜏 ≤ 𝑡𝛼𝛼𝑓 (𝑥) . (9) Now we define 𝐹 (𝑥) := 𝑥 ∫ 0 (𝑡 − 𝜏)𝛼−1𝑓𝛽(𝜏) 𝑑𝜏 − ⎛ ⎝ 𝑥 ∫ 0 (𝑡 − 𝜏)𝛼−1𝑓 (𝜏) 𝑑𝜏 ⎞ ⎠ 𝛽−1 . Clearly𝐹 (0) = 0 and 𝐹′(𝑥) = (𝑡 − 𝑥)𝛼−1𝑓𝛽(𝑥) − (𝛽 − 1) ⎛ ⎝ 𝑥 ∫ 0 (𝑡 − 𝜏)𝛼−1𝑓 (𝜏) 𝑑𝜏 ⎞ ⎠ 𝛽−2 (𝑡 − 𝑥)𝛼−1𝑓 (𝑥) = (𝑡 − 𝑥)𝛼−1𝑓 (𝑥) ⎡ ⎢ ⎣𝑓𝛽−1(𝑥) − (𝛽 − 1) ⎛ ⎝ 𝑥 ∫ 0 (𝑡 − 𝜏)𝛼−1𝑓 (𝜏) 𝑑𝜏 ⎞ ⎠ 𝛽−2⎤ ⎥ ⎦ . Setting 𝐺(𝑥) = 𝑓𝛽−1(𝑥) − (𝛽 − 1) ⎛ ⎝ 𝑥 ∫ 0 (𝑡 − 𝜏)𝛼−1𝑓 (𝜏) 𝑑𝜏 ⎞ ⎠ 𝛽−2 .
Then we have𝐺(0) = 𝑓𝛽−1(0) ≥ 0 and 𝐺′(𝑥) = (𝛽 − 1) 𝑓𝛽−2(𝑥) 𝑓′ (𝑥) − (𝛽 − 1) (𝛽 − 2) ⎛ ⎝ 𝑥 ∫ 0 (𝑡 − 𝜏)𝛼−1𝑓 (𝜏) 𝑑𝜏 ⎞ ⎠ 𝛽−3 (𝑡 − 𝑥)𝛼−1𝑓 (𝑥) = (𝛽 − 1) 𝑓 (𝑥) ⎡ ⎢ ⎣𝑓𝛽−3(𝑥) 𝑓′(𝑥) − (𝛽 − 2) ⎛ ⎝ 𝑥 ∫ 0 (𝑡 − 𝜏)𝛼−1𝑓 (𝜏) 𝑑𝜏 ⎞ ⎠ 𝛽−3 (𝑡 − 𝑥)𝛼−1 ⎤ ⎥ ⎦ . From the conditions of Theorem 1 and inequality(9), we have
𝑓𝛽−3(𝑥) 𝑓′ (𝑥) ≥ (𝛽 − 2) ( 𝑡𝛼 𝛼𝑓 (𝑥) )𝛽−3 (𝑡 − 𝑥)𝛼−1 ≥ (𝛽 − 2) ⎛ ⎝ 𝑥 ∫ 0 (𝑡 − 𝜏)𝛼−1𝑓 (𝜏) 𝑑𝜏 ⎞ ⎠ 𝛽−3 (𝑡 − 𝑥)𝛼−1. Thus𝐺′(𝑥) ≥ 0 and 𝐺(0) ≥ 0, so we get 𝐺(𝑥) ≥ 0.
On the other hand𝐹 (0) = 0 and
𝐹′(𝑥) = (𝑡 − 𝑥)𝛼−1𝑓 (𝑥) 𝐺(𝑥) ≥ 0 for all 𝑥 ∈ [0, 𝑡] . In particular 𝐹 (𝑡) = 𝑡 ∫ 0 (𝑡 − 𝜏)𝛼−1𝑓𝛽(𝜏) 𝑑𝜏 − ⎛ ⎝ 𝑡 ∫ 0 (𝑡 − 𝜏)𝛼−1𝑓 (𝜏) 𝑑𝜏 ⎞ ⎠ 𝛽−1 ≥ 0, and then 1 Γ (𝛼) 𝑡 ∫ 0 (𝑡 − 𝜏)𝛼−1𝑓𝛽(𝜏) 𝑑𝜏 ≥ Γ𝛽−2(𝛼) Γ𝛽−1(𝛼) ⎛ ⎝ 𝑥 ∫ 0 (𝑡 − 𝜏)𝛼−1𝑓 (𝜏) 𝑑𝜏 ⎞ ⎠ 𝛽−1 . Theorem 1 is thus proved.
Remark 2 In Theorem 1, if we take𝛼 = 1, we obtain the inequality (2) on [0, 𝑡] .
Theorem 3 Suppose that𝛼 and 𝛽 are two positive real numbers such that 𝛼 > 𝛽 ≥ 2, 𝑚 = [𝛽] and let 𝑓(𝑥) ∈ 𝐶1[0, ∞[ satisfying𝑓′(𝑥) ≥ 𝑓(𝑥) ≥ 0 and[𝑓𝛼−𝛽(𝑥)]′ ≥ (𝛼 − 𝛽)(𝛼−1)(𝛼−2)...(𝛼−𝑚+1)𝛽(𝛽−1)...(𝛽−𝑚+1) (Γ(𝛿+1)𝑥𝛿 )𝛽−𝑚 ( (𝑥−𝜏)𝛿−1 Γ(𝛿) )𝑚−1 . Then for any𝑡 > 1 and 𝛿 > 1, we have
𝐽𝛿𝑓𝛼(𝑡) ≥(𝐽𝛿𝑓(𝑡))𝛽. (1.2)
Proof.Using the fact that𝑓′(𝑥) ≥ 𝑓(𝑥) ≥ 0, [0, 𝑥] ⊂ [0, 𝑡] , we get 𝑓(𝑥) ∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑑𝜏 ≥ ∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏, 𝜏 ∈ [0, 𝑥] , that is 𝑓(𝑥)Γ(𝛿 + 1)𝑥𝛿−1 ≥ ∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏. Now we define: 𝐹 (𝑥) := ∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓𝛼(𝜏)𝑑𝜏 − (∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏 )𝛽 , 𝑥 ∈ [0, 𝑡] . We have: 𝐹′(𝑥) = 𝑓(𝑥)(𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝐺1(𝑥),
Z. Dahmani, S. Belarbi: Some Inequalities of Qi Type Using Fractional Integration 399 where: 𝐺1(𝑥) = 𝑓𝛼−1(𝑥) − 𝛽 (∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏 )𝛽−1 . The derivative of the function𝐺1gives
𝐺′ 1(𝑥) = (𝛼 − 1) 𝑓𝛼−2(𝑥)𝑓′(𝑥) − 𝛽 (𝛽 − 1) (∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏 )𝛽−2 𝑓(𝑥)(𝑥 − 𝜏)Γ(𝛿)𝛿−1 ≥ (𝛼 − 1) 𝑓𝛼−1(𝑥) − 𝛽 (𝛽 − 1)(∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏 )𝛽−2 𝑓(𝑥)(𝑥 − 𝜏)Γ(𝛿)𝛿−1 = 𝑓(𝑥) ⎛ ⎝(𝛼 − 1) 𝑓𝛼−2(𝑥) − 𝛽 (𝛽 − 1)(∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏 )𝛽−2 (𝑥 − 𝜏)𝛿−1 Γ(𝛿) ⎞ ⎠ = 𝑓(𝑥)𝐺2(𝑥), where 𝐺2(𝑥) := ⎛ ⎝(𝛼 − 1) 𝑓𝛼−2(𝑥) − 𝛽 (𝛽 − 1)(∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏 )𝛽−2 (𝑥 − 𝜏)𝛿−1 Γ(𝛿) ⎞ ⎠ . It follows that 𝐺′ 2(𝑥) = (𝛼 − 1) (𝛼 − 2) 𝑓𝛼−3(𝑥)𝑓′(𝑥) − 𝛽 (𝛽 − 1) ((𝛽 − 2) (∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏 )𝛽−3 𝑓(𝑥) ( (𝑥 − 𝜏)𝛿−1 Γ(𝛿) )2 + (∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏 )𝛽−2( (𝑥 − 𝜏)𝛿−1 Γ(𝛿) )′ ) = (𝛼 − 1) (𝛼 − 2) 𝑓𝛼−3(𝑥)𝑓′(𝑥) − 𝛽 (𝛽 − 1) ((𝛽 − 2)(∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏 )𝛽−3 ×𝑓(𝑥) ( (𝑥 − 𝜏)𝛿−1 Γ(𝛿) )2 + (∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏 )𝛽−2( (𝑥 − 𝜏)𝛿−2 Γ(𝛿 − 1) ) ≥ (𝛼 − 1) (𝛼 − 2) 𝑓𝛼−2(𝑥) − 𝛽 (𝛽 − 1) (𝛽 − 2)(∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏 )𝛽−3 ×𝑓(𝑥) ( (𝑥 − 𝜏)𝛿−1 Γ(𝛿) )2 = 𝑓(𝑥) (𝛼 − 1) (𝛼 − 2) 𝑓𝛼−3(𝑥) − 𝛽 (𝛽 − 1) (𝛽 − 2)(∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏 )𝛽−3 × ( (𝑥 − 𝜏)𝛿−1 Γ(𝛿) )2 = 𝑓(𝑥)𝐺3(𝑥).
By the same argument as before, we obtain
𝐺𝑚−1(𝑥) = (𝛼 − 1) (𝛼 − 2) ... (𝛼 − 𝑚 + 2) 𝑓𝛼−𝑚+1(𝑥) − 𝛽 (𝛽 − 1) (𝛽 − 2) ... (𝛽 − 𝑚 + 2)(∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏 )𝛽−𝑚+1( (𝑥 − 𝜏)𝛿−1 Γ(𝛿) )𝑚−2
Obviously 𝐺′ 𝑚−1(𝑥) = (𝛼 − 1) (𝛼 − 2) ... (𝛼 − 𝑚 + 1) 𝑓𝛼−𝑚(𝑥)𝑓′(𝑥) − 𝛽 (𝛽 − 1) (𝛽 − 2) ...(𝛽 − 𝑚 + 2) (𝛽 − 𝑚 + 1) (∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏 )𝛽−𝑚 𝑓(𝑥) ( (𝑥 − 𝜏)𝛿−1 Γ(𝛿) )𝑚−1 + (𝛽 − 𝑚 + 1) (∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏 )𝛽−𝑚+1( (𝑥 − 𝜏)𝛿−1 Γ(𝛿) )𝑚−3 ≥ (𝛼 − 1) (𝛼 − 2) ... (𝛼 − 𝑚 + 1) 𝑓𝛼−𝑚(𝑥)𝑓′(𝑥) − 𝛽 (𝛽 − 1) (𝛽 − 2) ... (𝛽 − 𝑚 + 2) (𝛽 − 𝑚 + 1) 𝑓𝛽−𝑚(𝑥)( 𝑥𝛿 Γ(𝛿 + 1) )𝛽−𝑚 𝑓(𝑥) ( (𝑥 − 𝜏)𝛿−1 Γ(𝛿) )𝑚−1 = 𝑓𝛽−𝑚+1(𝑥) (𝛼 − 1) (𝛼 − 2) ... (𝛼 − 𝑚 + 1) 𝑓𝛼−𝛽−1(𝑥)𝑓′(𝑥) − 𝛽 (𝛽 − 1) (𝛽 − 2) ... (𝛽 − 𝑚 + 2) (𝛽 − 𝑚 + 1) ( 𝑥𝛿 Γ(𝛿 + 1) )𝛽−𝑚((𝑥 − 𝜏)𝛿−1 Γ(𝛿) )𝑚−1 = 𝑓𝛽−𝑚+1(𝑥)[(𝛼 − 1) (𝛼 − 2) ... (𝛼 − 𝑚 + 1) 1 𝛼−𝛽 ( 𝑓𝛼−𝛽(𝑥))′− 𝛽 (𝛽 − 1) (𝛽 − 2) ... (𝛽 − 𝑚 + 2) (𝛽 − 𝑚 + 1) ( 𝑥𝛿 Γ(𝛿 + 1) )𝛽−𝑚((𝑥 − 𝜏)𝛿−1 Γ(𝛿) )𝑚−1] .
We have𝐺′𝑚−1(𝑥) ≥ 0, it follows that 𝐺𝑚−1(𝑥) is increasing on [0, 𝑡] . Hence 𝐹′(𝑥) ≥ 0. And then 𝐹 (𝑥) is increasing on[0, 𝑡] . Finally we can obtain
∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓𝛼(𝜏)𝑑𝜏 − (∫ 𝑥 0 (𝑡 − 𝜏)𝛿−1 Γ(𝛿) 𝑓(𝜏)𝑑𝜏 )𝛽 ≥ 0. In particular, for𝑥 = 𝑡, we get (10).
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