ON SOME NEW FRACTIONAL INTEGRAL
INEQUALITIES
SOUMIA BELARBI AND ZOUBIR DAHMANI
vol. 10, iss. 3, art. 86, 2009
Department of Mathematics
University of Mostaganem
Algeria
EMail: soumia-math@hotmail.fr
Fractional Integral Inequalities
Soumia Belarbi and
Zoubir Dahmani
zzdahmani@yahoo.fr
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Received:
23 May, 2009
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Accepted:
24 June, 2009
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Communicated by:
G. Anastassiou
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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.
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Acknowledgements:
The authors would like to thank professor A. El Farissi for his helpful.
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Contents
1
Introduction
3
2
Description of Fractional Calculus
4
3
Main Results
5
Fractional Integral Inequalities
Soumia Belarbi and
Zoubir Dahmani
vol. 10, iss. 3, art. 86, 2009
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1.
Introduction
Let us consider the functional [1]:
1
(1.1) T (f, g) :=
b−a
Z
b
f (x) g (x) dx
a
1
−
b−a
Z
b
f (x) dx
a
1
b−a
Z
b
g (x) dx ,
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.
Fractional Integral Inequalities
Soumia Belarbi and
Zoubir Dahmani
vol. 10, iss. 3, art. 86, 2009
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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 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
f (n) ∈ Cµ .
Cµn , n
∈ R, if
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,
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:
(2.4)
J α tµ =
Γ(µ + 1) α+µ
t ,
Γ(α + µ + 1)
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which implies the commutative property:
(2.3)
Fractional Integral Inequalities
Soumia Belarbi and
Zoubir Dahmani
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:
Γ(α + 1) α
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.
(3.1)
J α (f g)(t) ≥
Therefore
(3.3)
Fractional Integral Inequalities
Soumia Belarbi and
Zoubir Dahmani
vol. 10, iss. 3, art. 86, 2009
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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
(3.4)
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
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Consequently,
Z t
1
(3.6) J (f g)(t) + f (ρ) g (ρ)
(t − τ )α−1 dτ
Γ (α) 0
Z t
Z
g (ρ)
f (ρ) t
α−1
≥
(t − τ )
f (τ ) dτ +
(t − τ )α−1 g (τ ) dτ.
Γ (α) 0
Γ (α) 0
α
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)α−1
,
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
and this ends the proof.
J α (f g)(t) ≥
1
J α (1)
vol. 10, iss. 3, art. 86, 2009
ρ ∈ (0, t), we obtain:
(t − ρ)α−1
(t − ρ)α−1 α
(3.8)
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)
(3.10)
Fractional Integral Inequalities
Soumia Belarbi and
Zoubir Dahmani
J α f (t)J α g(t),
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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)
tβ
tα
J β (f g)(t) +
J α (f g)(t)
Γ (α + 1)
Γ (β + 1)
≥ J α f (t)J β g(t) + J β f (t)J α g(t).
Proof. Using similar arguments as in the proof of Theorem 3.1, we can write
(t − ρ)β−1
(t − ρ)β−1 α
α
(3.12)
J (f g) (t) + J (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)
Fractional Integral Inequalities
Soumia Belarbi and
Zoubir Dahmani
vol. 10, iss. 3, art. 86, 2009
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By integrating (3.12) over (0, t) , we obtain
J
I
Z
J α (1) t
(t − ρ)β−1
(3.13) J (f g)(t)
dρ +
f (ρ) g (ρ) (t − ρ)β−1 dρ
Γ
(β)
Γ
(β)
0
0
Z
Z
J α f (t) 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
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α
Z
t
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.
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The third result is:
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
(3.14)
Jα
fi (t) ≥ (J α (1))1−n
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) ,
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t > 0, α > 0.
i=1
Qn−1 Since (fi )i=1,...,n are positive increasing functions, then
i=1 fi (t) is an increasQ
ing function. Hence we can apply Theorem 3.1 to the functions n−1
i=1 fi = g,
fn = f. We obtain:
!
!
n
n−1
Y
Y
(3.16) J α
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
n−1
Y
Y
−1
2−n
(3.17) J α
fi (t) ≥ (J α (1)) ((J α (1))
J α fi (t))J α (fn ) (t) ,
i=1
and this ends the proof.
vol. 10, iss. 3, art. 86, 2009
for all t > 0, α > 0.
Now, suppose that (induction hypothesis)
!
n−1
n−1
Y
Y
2−n
α
α
J α fi (t) ,
(3.15)
J
fi (t) ≥ (J (1))
i=1
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Zoubir Dahmani
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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, g is differentiable and there exists a real number m := inf t≥0 g 0 (t) . Then
the inequality
(3.18)
J α (f g)(t) ≥ (J α (1))−1 J α f (t)J α g(t) −
mt α
J f (t) + mJ α (tf (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:
α
(3.19)
J (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 f (t)
≥ (J (1)) J f (t)J g(t) −
Γ (α + 2)
mΓ (α + 1) t α
J f (t)
≥ (J α (1))−1 J α f (t)J α g(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)
mt α
−
J f (t) + mJ α (tf (t)) ,
α+1
Theorem 3.4 is thus proved.
Fractional Integral Inequalities
Soumia Belarbi and
Zoubir Dahmani
vol. 10, iss. 3, art. 86, 2009
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t > 0, α > 0.
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:
(3.21) J α (f g)(t) ≥ (J α (1))−1 J α f (t)J α g(t)−
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
(3.22) 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)
(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)
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− M2 J α tJ α f (t) + M1 M2 (J α t)2 .
Proof.
(A): Apply Theorem 3.1 to the functions f and G(t) := g(t) − m2 t.
vol. 10, iss. 3, art. 86, 2009
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− m2 J α tJ α f (t) + m1 m2 (J α t)2 .
is valid.
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Soumia Belarbi and
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(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.
Fractional Integral Inequalities
Soumia Belarbi and
Zoubir Dahmani
vol. 10, iss. 3, art. 86, 2009
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References
[1] P.L. CHEBYSHEV, Sur les expressions approximatives des integrales definies
par les autres prises entre les mêmes limites, Proc. Math. Soc. Charkov, 2 (1882),
93–98.
[2] R. GORENFLO AND F. MAINARDI, Fractional Calculus: Integral and Differential Equations of Fractional Order, Springer Verlag, Wien (1997), 223–276.
[3] S.M. MALAMUD, Some complements to the Jenson and Chebyshev inequalities and a problem of W. Walter, Proc. Amer. Math. Soc., 129(9) (2001), 2671–
2678.
[4] S. MARINKOVIC, P. RAJKOVIC AND M. STANKOVIC, The inequalities for
some types q-integrals, Comput. Math. Appl., 56 (2008), 2490–2498.
Fractional Integral Inequalities
Soumia Belarbi and
Zoubir Dahmani
vol. 10, iss. 3, art. 86, 2009
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Contents
[5] B.G. PACHPATTE, A note on Chebyshev-Grüss type inequalities for differential functions, Tamsui Oxford Journal of Mathematical Sciences, 22(1) (2006),
29–36.
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[6] I. PODLUBNI, Fractional Differential Equations, Academic Press, San Diego,
1999.
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