ON SOME NEW FRACTIONAL INTEGRAL INEQUALITIES

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Fractional Integral Inequalities Soumia Belarbi and

Zoubir Dahmani vol. 10, iss. 3, art. 86, 2009

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ON SOME NEW FRACTIONAL INTEGRAL INEQUALITIES

SOUMIA BELARBI AND ZOUBIR DAHMANI

Department of Mathematics University of Mostaganem Algeria

EMail:soumia-math@hotmail.fr zzdahmani@yahoo.fr

Received: 23 May, 2009

Accepted: 24 June, 2009

Communicated by: G. Anastassiou 2000 AMS Sub. Class.: 26D10, 26A33.

Key words: Fractional integral inequalities, Riemann-Liouville fractional integral.

Abstract: In this paper, using the Riemann-Liouville fractional integral, we establish some new integral inequalities for the Chebyshev functional in the case of two synchronous functions.

Acknowledgements: The authors would like to thank professor A. El Farissi for his helpful.

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1. Introduction

Let us consider the functional [1]:

(1.1) T(f, g) := 1 b−a

Z b

a

f(x)g(x)dx

− 1 b−a

Z b

a

f(x)dx 1 b−a

Z b

a

g(x)dx

,

where f and g are two integrable functions which are synchronous on [a, b]

i.e.

(f(x)−f(y))(g(x)−g(y))≥0,for anyx, y ∈[a, b]

.

Many researchers have given considerable attention to (1.1) and a number of inequalities have appeared in the literature, see [3,4,5].

The main purpose of this paper is to establish some inequalities for the functional (1.1) using fractional integrals.

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2. Description of Fractional Calculus

We will give the necessary notation and basic definitions below. For more details, one can consult [2,6].

Definition 2.1. A real valued functionf(t), t≥0is said to be in the spaceC_µ, µ ∈ R if there exists a real number p > µ such that f(t) = t^pf1(t), where f1(t) ∈ C([0,∞[).

Definition 2.2. A function f(t), t ≥ 0 is said to be in the space C_µⁿ, n ∈ R, if f⁽ⁿ⁾ ∈C_µ.

Definition 2.3. The Riemann-Liouville fractional integral operator of orderα ≥0, for a functionf ∈C_µ,(µ≥ −1)is defined as

J^αf(t) = 1 Γ(α)

Z t

0

(t−τ)^α−1f(τ)dτ; α >0, t >0, (2.1)

J⁰f(t) = f(t), whereΓ(α) :=R∞

0 e^−uu^α−1du.

For the convenience of establishing the results, we give the semigroup property:

(2.2) J^αJ^βf(t) =J^α+βf(t), α ≥0, β ≥0, which implies the commutative property:

(2.3) J^αJ^βf(t) = J^βJ^αf(t).

From (2.1), whenf(t) =t^µwe get another expression that will be used later:

(2.4) J^αt^µ= Γ(µ+ 1)

Γ(α+µ+ 1)t^α+µ, α >0; µ >−1, t >0.

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3. Main Results

Theorem 3.1. Let f and g be two synchronous functions on [0,∞[. Then for all t >0, α >0,we have:

(3.1) J^α(f g)(t)≥ Γ(α+ 1)

t^α J^αf(t)J^αg(t).

Proof. Since the functionsfandgare synchronous on[0,∞[,then for allτ ≥0, ρ≥ 0,we have

(3.2)

f(τ)−f(ρ)

g(τ)−g(ρ)

≥0.

Therefore

(3.3) f(τ)g(τ) +f(ρ)g(ρ)≥f(τ)g(ρ) +f(ρ)g(τ).

Now, multiplying both sides of (3.3) by ^(t−τ)_Γ(α)^α−1, τ ∈(0, t),we get (3.4) (t−τ)^α−1

Γ(α) f(τ)g(τ) + (t−τ)^α−1

Γ(α) f(ρ)g(ρ)

≥ (t−τ)^α−1

Γ(α) f(τ)g(ρ) + (t−τ)^α−1

Γ(α) f(ρ)g(τ).

Then integrating (3.4) over(0, t), we obtain:

(3.5) 1 Γ(α)

Z t

0

(t−τ)^α−1f(τ)g(τ)dτ+ 1 Γ(α)

Z t

0

(t−τ)^α−1f(ρ)g(ρ)dτ

≥ 1 Γ(α)

Z t

0

(t−τ)^α−1f(τ)g(ρ)dτ+ 1 Γ(α)

Z t

0

(t−τ)^α−1f(ρ)g(τ)dτ.

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Consequently,

(3.6) J^α(f g)(t) +f(ρ)g(ρ) 1 Γ (α)

Z t

0

(t−τ)^α−1dτ

≥ g(ρ) Γ (α)

Z t

0

(t−τ)^α−1f(τ)dτ + f(ρ) Γ (α)

Z t

0

(t−τ)^α−1g(τ)dτ.

So we have

(3.7) J^α(f g)(t) +f(ρ)g(ρ)J^α(1)≥g(ρ)J^α(f)(t) +f(ρ)J^α(g)(t).

Multiplying both sides of (3.7) by ^(t−ρ)_Γ(α)^α−1, ρ∈(0, t),we obtain:

(3.8) (t−ρ)^α−1

Γ (α) J^α(f g)(t) + (t−ρ)^α−1

Γ (α) f(ρ)g(ρ)J^α(1)

≥ (t−ρ)^α−1

Γ (α) g(ρ)J^αf(t) + (t−ρ)^α−1

Γ (α) f(ρ)J^αg(t).

Now integrating (3.8) over(0, t),we get:

(3.9) J^α(f g)(t) Z t

0

(t−ρ)^α−1

Γ(α) dρ+J^α(1) Γ(α)

Z t

0

f(ρ)g(ρ)(t−ρ)^α−1dρ

≥ J^αf(t) Γ(α)

Z t

0

(t−ρ)^α−1g(ρ)dρ+ J^αg(t) Γ(α)

Z t

0

(t−ρ)^α−1f(ρ)dρ.

Hence

(3.10) J^α(f g)(t)≥ 1

J^α(1)J^αf(t)J^αg(t), and this ends the proof.

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The second result is:

Theorem 3.2. Let f and g be two synchronous functions on [0,∞[. Then for all t >0, α >0, β >0,we have:

(3.11) t^α

Γ (α+ 1)J^β(f g)(t) + t^β

Γ (β+ 1)J^α(f g)(t)

≥J^αf(t)J^βg(t) +J^βf(t)J^αg(t).

Proof. Using similar arguments as in the proof of Theorem3.1, we can write

(3.12) (t−ρ)^β−1

Γ (β) J^α(f g) (t) +J^α(1)(t−ρ)^β−1

Γ (β) f(ρ)g(ρ)

≥ (t−ρ)^β−1

Γ (β) g(ρ)J^αf(t) + (t−ρ)^β−1

Γ (β) f(ρ)J^αg(t).

By integrating (3.12) over(0, t),we obtain (3.13) J^α(f g)(t)

Z t

0

(t−ρ)^β−1

Γ (β) dρ+ J^α(1) Γ (β)

Z t

0

f(ρ)g(ρ) (t−ρ)^β−1dρ

≥ J^αf(t) Γ (β)

Z t

0

(t−ρ)^β−1g(ρ)dρ+J^αg(t) Γ (β)

Z t

0

(t−ρ)^β−1f(ρ)dρ, and this ends the proof.

Remark 1. The inequalities (3.1) and (3.11) are reversed if the functions are asyn- chronous on[0,∞[(i.e. (f(x)−f(y))(g(x)−g(y))≤0,for anyx, y ∈[0,∞[).

Remark 2. Applying Theorem3.2forα=β,we obtain Theorem3.1.

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The third result is:

Theorem 3.3. Let(f_i)_i=1,...,n benpositive increasing functions on[0,∞[.Then for anyt >0, α >0,we have

(3.14) J^α

n

Y

i=1

f_i

!

(t)≥(J^α(1))¹⁻ⁿ

n

Y

i=1

J^αf_i(t).

Proof. We prove this theorem by induction.

Clearly, forn = 1,we haveJ^α(f1) (t)≥J^α(f1) (t),for allt >0, α >0.

Forn = 2,applying (3.1), we obtain:

J^α(f₁f₂) (t)≥(J^α(1))⁻¹J^α(f₁) (t)J^α(f₂) (t), for allt >0, α >0.

Now, suppose that (induction hypothesis) (3.15) J^α

n−1

Y

i=1

f_i

!

(t)≥(J^α(1))²⁻ⁿ

n−1

Y

i=1

J^αf_i(t), t >0, α >0.

Since(f_i)_i=1,...,n are positive increasing functions, then Qn−1 i=1 f_i

(t)is an increasing function. Hence we can apply Theorem 3.1 to the functions Qn−1

i=1 f_i = g, f_n =f.We obtain:

(3.16) J^α

n

Y

i=1

f_i

!

(t) = J^α(f g) (t)≥(J^α(1))⁻¹J^α

n−1

Y

i=1

f_i

!

(t)J^α(f_n) (t).

Taking into account the hypothesis (3.15), we obtain:

(3.17) J^α

n

Y

i=1

f_i

!

(t)≥(J^α(1))⁻¹((J^α(1))²⁻ⁿ

n−1

Y

i=1

J^αf_i

!

(t))J^α(f_n) (t), and this ends the proof.

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We further have:

Theorem 3.4. Let f and g be two functions defined on [0,+∞[, such that f is increasing,gis differentiable and there exists a real numberm:= inf_t≥0g⁰(t).Then the inequality

(3.18) J^α(f g)(t)≥(J^α(1))⁻¹J^αf(t)J^αg(t)− mt

α+ 1J^αf(t) +mJ^α(tf(t)) is valid for allt >0, α >0.

Proof. We consider the functionh(t) := g(t)−mt.It is clear thathis differentiable and it is increasing on[0,+∞[. Then using Theorem3.1, we can write:

J^α

(g−mt)f(t) (3.19)

≥(J^α(1))⁻¹J^αf(t)

J^αg(t)−mJ^α(t)

≥(J^α(1))⁻¹J^αf(t)J^αg(t)− m(J^α(1))⁻¹t^α+1

Γ (α+ 2) J^αf(t)

≥(J^α(1))⁻¹J^αf(t)J^αg(t)− mΓ (α+ 1)t

Γ (α+ 2) J^αf(t)

≥(J^α(1))⁻¹J^αf(t)J^αg(t)− mt

α+ 1J^αf(t).

Hence

(3.20) J^α(f g)(t)≥(J^α(1))⁻¹J^αf(t)J^αg(t)

− mt

α+ 1J^αf(t) +mJ^α(tf(t)), t >0, α >0.

Theorem3.4is thus proved.

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Corollary 3.5. Letf andgbe two functions defined on[0,+∞[.

(A) Suppose thatf is decreasing,gis differentiable and there exists a real number M := sup_t≥0g⁰(t). Then for allt >0,α >0, we have:

(3.21) J^α(f g)(t)≥(J^α(1))⁻¹J^αf(t)J^αg(t)− M t

α+ 1J^αf(t)+M J^α(tf(t)). (B) Suppose that f and g are differentiable and there exist m₁ := inft≥0f⁰(x),

m₂ := inft≥0g⁰(t). Then we have

(3.22) J^α(f g)(t)−m₁J^αtg(t)−m₂J^αtf(t) +m₁m₂J^αt²

≥(J^α(1))⁻¹

J^αf(t)J^αg(t)−m₁J^αtJ^αg(t)

−m₂J^αtJ^αf(t) +m₁m₂(J^αt)² .

(C) Suppose that f and g are differentiable and there exist M₁ := sup_t≥0f⁰(t), M2 := sup_t≥0g⁰(t).Then the inequality

(3.23) J^α(f g)(t)−M₁J^αtg(t)−M₂J^αtf(t) +M₁M₂J^αt²

≥(J^α(1))⁻¹

J^αf(t)J^αg(t)−M₁J^αtJ^αg(t)

−M₂J^αtJ^αf(t) +M₁M₂(J^αt)² .

is valid.

Proof.

(A): Apply Theorem3.1to the functionsf andG(t) :=g(t)−m₂t.

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(B): Apply Theorem3.1to the functionsF andG, where:F(t) :=f(t)−m₁t, G(t) :=

g(t)−m₂t.

To prove(C),we apply Theorem3.1to the functions

F(t) :=f(t)−M₁t, G(t) := g(t)−M₂t.

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References

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[3] S.M. MALAMUD, Some complements to the Jenson and Chebyshev inequali- ties and a problem of W. Walter, Proc. Amer. Math. Soc., 129(9) (2001), 2671–

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