Volume 10 (2009), Issue 3, Article 86, 5 pp.
ON SOME NEW FRACTIONAL INTEGRAL INEQUALITIES
SOUMIA BELARBI AND ZOUBIR DAHMANI
D EPARTMENT OF M ATHEMATICS ,
U NIVERSITY OF M OSTAGANEM
soumia-math@hotmail.fr
zzdahmani@yahoo.fr
Received 23 May, 2009; accepted 24 June, 2009
Communicated by G. Anastassiou
A BSTRACT. 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.
Key words and phrases: Fractional integral inequalities, Riemann-Liouville fractional integral.
2000 Mathematics Subject Classification. 26D10, 26A33.
1. I NTRODUCTION
Let us consider the functional [1]:
Z b
Z b
Z b
1
1
1
(1.1) T (f, g) :=
f (x) g (x) dx −
f (x) dx
g (x) dx ,
b−a a
b−a
b−a a
a
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 any x, 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.
2. D ESCRIPTION OF F RACTIONAL C ALCULUS
We will give the necessary notation and basic definitions below. For more details, one can
consult [2, 6].
Definition 2.1. A real valued function f (t), t ≥ 0 is said to be in the space Cµ , µ ∈ R if there
exists a real number p > µ such that f (t) = tp f1 (t), where f1 (t) ∈ C([0, ∞[).
Definition 2.2. A function f (t), t ≥ 0 is said to be in the space Cµn , n ∈ R, if f (n) ∈ Cµ .
The authors would like to thank professor A. El Farissi for his helpful.
139-09
2
S OUMIA B ELARBI
AND
Z OUBIR DAHMANI
Definition 2.3. The Riemann-Liouville fractional integral operator of order α ≥ 0, for a function f ∈ Cµ , (µ ≥ −1) is defined as
Z t
1
α
(2.1)
J f (t) =
(t − τ )α−1 f (τ )dτ ;
α > 0, t > 0,
Γ(α) 0
J 0 f (t) = f (t),
R∞
where Γ(α) := 0 e−u uα−1 du.
For the convenience of establishing the results, we give the semigroup property:
(2.2)
J α J β f (t) = J α+β f (t),
α ≥ 0, β ≥ 0,
which implies the commutative property:
(2.3)
J α J β f (t) = J β J α f (t).
From (2.1), when f (t) = tµ we get another expression that will be used later:
Γ(µ + 1) α+µ
(2.4)
J α tµ =
t ,
α > 0; µ > −1, t > 0.
Γ(α + µ + 1)
3. M AIN R ESULTS
Theorem 3.1. Let f and g be two synchronous functions on [0, ∞[. Then for all t > 0, α > 0,
we have:
Γ(α + 1) α
(3.1)
J α (f g)(t) ≥
J f (t)J α g(t).
tα
Proof. Since the functions f and g are synchronous on [0, ∞[, then for all τ ≥ 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 )α−1
,
Γ(α)
τ ∈ (0, t), we get
(t − τ )α−1
(t − τ )α−1
f (τ )g(τ ) +
f (ρ)g(ρ)
Γ(α)
Γ(α)
(t − τ )α−1
(t − τ )α−1
≥
f (τ )g(ρ) +
f (ρ)g(τ ).
Γ(α)
Γ(α)
Then integrating (3.4) over (0, t), we obtain:
Z t
Z t
1
1
α−1
(3.5)
(t − τ ) f (τ )g(τ )dτ +
(t − τ )α−1 f (ρ)g(ρ)dτ
Γ(α) 0
Γ(α) 0
Z t
Z t
1
1
α−1
≥
(t − τ ) f (τ )g(ρ)dτ +
(t − τ )α−1 f (ρ)g(τ )dτ.
Γ(α) 0
Γ(α) 0
Consequently,
Z t
1
α
(3.6) J (f g)(t) + f (ρ) g (ρ)
(t − τ )α−1 dτ
Γ (α) 0
Z
Z
g (ρ) t
f (ρ) t
α−1
≥
(t − τ )
f (τ ) dτ +
(t − τ )α−1 g (τ ) dτ.
Γ (α) 0
Γ (α) 0
(3.4)
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 86, 5 pp.
http://jipam.vu.edu.au/
F RACTIONAL I NTEGRAL I NEQUALITIES
3
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
(3.8)
(t−ρ)α−1
,
Γ(α)
ρ ∈ (0, t), we obtain:
(t − ρ)α−1 α
(t − ρ)α−1
J (f g)(t) +
f (ρ) g (ρ) J α (1)
Γ (α)
Γ (α)
(t − ρ)α−1
(t − ρ)α−1
α
≥
g (ρ) J f (t) +
f (ρ) J α g(t).
Γ (α)
Γ (α)
Now integrating (3.8) over (0, t), we get:
Z
Z t
(t − ρ)α−1
J α (1) t
α
(3.9) J (f g)(t)
dρ +
f (ρ)g(ρ)(t − ρ)α−1 dρ
Γ(α)
Γ(α)
0
0
Z
Z
J α f (t) t
J α g(t) t
α−1
≥
(t − ρ) g(ρ)dρ +
(t − ρ)α−1 f (ρ)dρ.
Γ(α) 0
Γ(α) 0
Hence
J α (f g)(t) ≥
(3.10)
1
J α (1)
J α f (t)J α g(t),
and this ends the proof.
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)
tα
tβ
J β (f g)(t) +
J α (f g)(t) ≥ J α f (t)J β g(t) + J β f (t)J α g(t).
Γ (α + 1)
Γ (β + 1)
Proof. Using similar arguments as in the proof of Theorem 3.1, we can write
(3.12)
(t − ρ)β−1 α
(t − ρ)β−1
J (f g) (t) + J α (1)
f (ρ) g (ρ)
Γ (β)
Γ (β)
≥
(t − ρ)β−1
(t − ρ)β−1
g (ρ) J α f (t) +
f (ρ) J α g (t) .
Γ (β)
Γ (β)
By integrating (3.12) over (0, t) , we obtain
Z t
Z
(t − ρ)β−1
J α (1) t
α
(3.13) J (f g)(t)
dρ +
f (ρ) g (ρ) (t − ρ)β−1 dρ
Γ (β)
Γ (β) 0
0
Z t
Z
α
J f (t)
J α g (t) t
β−1
≥
(t − ρ)
g (ρ) dρ +
(t − ρ)β−1 f (ρ) dρ,
Γ (β) 0
Γ (β) 0
and this ends the proof.
Remark 1. The inequalities (3.1) and (3.11) are reversed if the functions are asynchronous on
[0, ∞[ (i.e. (f (x) − f (y))(g(x) − g(y)) ≤ 0, for any x, y ∈ [0, ∞[).
Remark 2. Applying Theorem 3.2 for α = β, we obtain Theorem 3.1.
The third result is:
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 86, 5 pp.
http://jipam.vu.edu.au/
4
S OUMIA B ELARBI
AND
Z OUBIR DAHMANI
Theorem 3.3. Let (fi )i=1,...,n be n positive increasing functions on [0, ∞[. Then for any t >
0, α > 0, we have
!
n
n
Y
Y
1−n
α
α
(3.14)
J
fi (t) ≥ (J (1))
J α fi (t) .
i=1
i=1
Proof. We prove this theorem by induction.
Clearly, for n = 1, we have J α (f1 ) (t) ≥ J α (f1 ) (t) , for all t > 0, α > 0.
For n = 2, applying (3.1), we obtain:
J α (f1 f2 ) (t) ≥ (J α (1))−1 J α (f1 ) (t) J α (f2 ) (t) ,
for all t > 0, α > 0.
Now, suppose that (induction hypothesis)
!
n−1
n−1
Y
Y
2−n
α
α
fi (t) ≥ (J (1))
J α fi (t) ,
(3.15)
J
i=1
t > 0, α > 0.
i=1
Qn−1 Since (fi )i=1,...,n are positive increasing functions, then
fi (t) is an increasing function.
Qn−1 i=1
Hence we can apply Theorem 3.1 to the functions i=1 fi = g, fn = f. We obtain:
!
!
n
n−1
Y
Y
(3.16)
Jα
fi (t) = J α (f g) (t) ≥ (J α (1))−1 J α
fi (t) J α (fn ) (t) .
i=1
i=1
Taking into account the hypothesis (3.15), we obtain:
!
n
Y
(3.17)
Jα
fi (t) ≥ (J α (1))−1 ((J α (1))2−n
i=1
n−1
Y
!
J α fi (t))J α (fn ) (t) ,
i=1
and this ends the proof.
We further have:
Theorem 3.4. Let f and g be two functions defined on [0, +∞[ , such that f is increasing, g is
differentiable and there exists a real number m := inf t≥0 g 0 (t) . Then the inequality
mt α
J f (t) + mJ α (tf (t))
(3.18)
J α (f g)(t) ≥ (J α (1))−1 J α f (t)J α g(t) −
α+1
is valid for all t > 0, α > 0.
Proof. We consider the function h (t) := g (t) − mt. It is clear that h is differentiable and it is
increasing on [0, +∞[. Then using Theorem 3.1, we can write:
(3.19)
J α (g − mt) f (t) ≥ (J α (1))−1 J α f (t) J α g(t) − mJ α (t)
m (J α (1))−1 tα+1 α
≥ (J (1)) J f (t)J g(t) −
J f (t)
Γ (α + 2)
mΓ (α + 1) t α
≥ (J α (1))−1 J α f (t)J α g(t) −
J f (t)
Γ (α + 2)
mt α
≥ (J α (1))−1 J α f (t)J α g(t) −
J f (t).
α+1
α
−1
α
α
Hence
(3.20) J α (f g)(t) ≥ (J α (1))−1 J α f (t)J α g(t) −
Theorem 3.4 is thus proved.
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 86, 5 pp.
mt α
J f (t) + mJ α (tf (t)) ,
α+1
t > 0, α > 0.
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F RACTIONAL I NTEGRAL I NEQUALITIES
5
Corollary 3.5. Let f and g be two functions defined on [0, +∞[.
(A) Suppose that f is decreasing, g is differentiable and there exists a real number M :=
supt≥0 g 0 (t). Then for all t > 0, α > 0, we have:
Mt α
J f (t) + M J α (tf (t)) .
α+1
(B) Suppose that f and g are differentiable and there exist m1 := inf t≥0 f 0 (x) , m2 :=
inf t≥0 g 0 (t). Then we have
(3.21)
J α (f g)(t) ≥ (J α (1))−1 J α f (t)J α g(t) −
(3.22) J α (f g)(t) − m1 J α tg(t) − m2 J α tf (t) + m1 m2 J α t2
≥ (J α (1))−1 J α f (t)J α g(t) − m1 J α tJ α g(t) − m2 J α tJ α f (t) + m1 m2 (J α t)2 .
(C) Suppose that f and g are differentiable and there exist M1 := supt≥0 f 0 (t) , M2 :=
supt≥0 g 0 (t) . Then the inequality
(3.23) J α (f g)(t) − M1 J α tg(t) − M2 J α tf (t) + M1 M2 J α t2
≥ (J α (1))−1 J α f (t)J α g(t) − M1 J α tJ α g(t) − M2 J α tJ α f (t) + M1 M2 (J α t)2 .
is valid.
Proof.
(A): Apply Theorem 3.1 to the functions f and G(t) := g(t) − m2 t.
(B): Apply Theorem 3.1 to the functions F and G, where: F (t) := f (t) − m1 t, G(t) :=
g(t) − m2 t.
To prove (C), we apply Theorem 3.1 to the functions
F (t) := f (t) − M1 t, G(t) := g(t) − M2 t.
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J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 86, 5 pp.
http://jipam.vu.edu.au/