Bulletin of Mathematical Analysis and Applications
ISSN: 1821-1291, URL: http://www.bmathaa.org
Volume 4 Issue 4 (2012), Pages 116-122
ON SOME QI TYPE INEQUALITIES USING FRACTIONAL
q-INTEGRAL
(COMMUNICATED BY NAIM BRAHA)
KAMEL BRAHIM & HEDI EL MONSER
Abstract. In this paper, we provide some Qi type inequalities using a fractional q-integral.
Key words and phrases: q-fractional integral, q-integral inequality.
2010 Mathematics Subject Classification: 26A33, 26A24, 26D15.
1. introduction
In [12], Feng Qi proposed the following problem:
Under what conditions does the inequality:
Z b
t−1
Z b
[f (x)]t dx ≥
f (x)dx
a
(1)
a
hold for t > 1?
In view of the interest in this type of inequalities, many attentions have been payed to
the problem and many authors have extended the inequality to more general cases (see
[1, 2, 3, 4, 9, 8, 11, 14]). In this paper, we establish some inequalities of F. Qi type by
using a fractional q-integral and we will generalize the inequalities given in [3].
This paper is organized as follows: In Section 2, we present some definitions and facts
from the q-calculus necessary for understanding this paper. In Section 3, we discuss some
generalizations of Qi’s inequality.
2. Notations and preliminaries
Throughout this paper, we will fix q ∈ (0, 1). For the convenience of the reader, we
provide in this section a summary of the mathematical notations and definitions used in
this paper (see [5, 7, 13]). We write for a ∈ C,
[a]q =
n−1
Y
1 − qa
(1 − aq k ), n = 1, 2, ...∞,
, (a; q)n =
1−q
k=0
[0]q ! = 1,
[n]q ! = [1]q [2]q ...[n]q ,
n = 1, 2, ...
and
(x − a)
(n)
=

 1,

n=0
(x − a)(x − qa)...(x − q n−1 a),
116
n 6= 0
ON SOME QI TYPE INEQUALITIES USING FRACTIONAL q-INTEGRAL
117
Their natural expansions to the reals are
(a − b)(α) = aα
( ab ; q)∞
(q α ab ; q)∞
, (a; q)α =
(a; q)∞
.
(aq α ; q)∞
Notice that
b
(a − b)(α) = aα ( ; q)α .
a
The q-derivative Dq f of a function f is given by
f (x) − f (qx)
, if x 6= 0,
(1 − q)x
(2)
Dq,t (b − t)(α) = −[α]q (b − qt)(α−1) .
(3)
(Dq f )(x) =
(Dq f )(0) = f ′ (0) provided f ′ (0) exists.
Clearly
The q-Jackson integrals from 0 to a is defined by (see [6])
Z a
∞
X
f (aq n )q n ,
f (x)dq x = (1 − q)a
0
(4)
n=0
provided the sum converges absolutely.
The q-Jackson integral in a generic interval [a, b] is given by (see [6])
Z b
Z b
Z a
f (x)dq x =
f (x)dq x −
f (x)dq x.
0
a
For any function f , we have ( see [7] )
Z
Dq
x
f (t)dq t
a
(5)
0
= f (x).
(6)
The fractional q-integral of the Riemann-Liouville type is[13]
Z x
1
α
α
(x − qt)(α−1) f (t)dq t,
Iq,a
(f )(x) = Iq,a
(f (t))(x) =
Γq (α) a
Finally, for b > 0 and a = bq n ,
α > 0.
(7)
n = 1, 2, . . . , ∞, we write
[a, b]q = {bq k : 0 ≤ k ≤ n}
and
(a, b]q = [q −1 a, b]q .
3. Fractional q-Integral Inequalities of Qi type
Let us begin by recalling the following useful result[3]:
Lemma 1. Let p ≥ 1 be a real number and g be a nonnegative and monotonous function
on [a, b]q . Then
pg p−1 (qx)Dq g(x) ≤ Dq [g p (x)] ≤ pg p−1 (x)Dq g(x),
x ∈ (a, b]q .
(8)
Proposition 1. Let f be a function defined on [a, b]q satisfying
f (a) ≥ 0
and
Dq f (x) ≥ (β − 2)Cαβ−2 (x − a)β−3
f or
x ∈ (a, b]q
and
β ≥ 3.
Then
α
β−1
α
Iq,a
(f β (t))(b) ≥ Γβ−2
(α) Iq,a
(f (qt))(b)
,
q

 (b − qa)(α−1) , if α ≥ 1
where Cα = sup (b − qt)(α−1) =
 α−1
t∈[a,b]q
b
(q, q)α−1 , if 0 < α < 1.
(9)
118
KAMEL BRAHIM & HEDI EL MONSER
Proof. For x ∈ (a, b]q , we put
Z x
g(x) =
(b−qt)(α−1) f (qt)dq t
We have
and F (x) =
a
Z
x
(b−qt)(α−1) [f (t)]β dq t−
a
Z
x
(b − qt)(α−1) f (qt)dq t
a
Dq F (x) = (b − qx)(α−1) f β (x) − Dq [g β−1 ](x).
Since f and g increase on [a, b]q , we obtain from Lemma 1,
Dq F (x)
≥
(b − qx)(α−1) f β (x) − (β − 1)g β−2 (x)(b − qx)(α−1) f (qx)
≥
(b − qx)(α−1) f β (x) − (β − 1)g β−2 (x)(b − qx)(α−1) f (x) ≥ (b − qx)(α−1) f (x)h(x),
where h(x) = f β−1 (x) − (β − 1)g β−2 (x).
On the other hand, we have
Dq h(x) = Dq [f β−1 ](x) − (β − 1)Dq [g β−2 ](x).
Using Lemma 1 once again leads to
Dq h(x)
≥
(β − 1)f β−2 (qx)Dq f (x) − (β − 1)(β − 2)g β−3 (x)Dq g(x)
≥
(β − 1)f β−2 (qx)Dq f (x) − (β − 1)(β − 2)g β−3 (x)(b − qx)(α−1) f (qx)
h
i
(β − 1)f (qx) f β−3 (qx)Dq f (x) − (β − 2)(b − qx)(α−1) g β−3 (x) .
≥
The use of the inequality,
Z x
(b − qt)(α−1) f (qt)dq t
≤
Cα f (qx)(x − a)
a
gives
Dq h(x)
≥
≥
h
i
(β − 1)f β−2 (qx) Dq f (x) − (β − 2)(b − qx)(α−1) Cαβ−3 (x − a)β−3
h
i
(β − 1)f β−2 (qx) Dq f (x) − (β − 2)Cαβ−2 (x − a)β−3 ≥ 0.
From the fact h(a) = f β−1 (a) ≥ 0, we get h(x) ≥ 0, x ∈ [a, b]q .
Therefore Dq F (x) ≥ 0, and so F (x) ≥ 0 for all x ∈ [a, b]q , in particular
Z b
β−1
Z b
F (b) =
(b − qt)(α−1) [f (t)]β dq t −
(b − qt)(α−1) f (qt)dq t
≥ 0,
a
a
hence
β−1
Z b
Z b
1
1
(α−1)
(b − qt)(α−1) [f (t)]β dq t ≥ Γβ−2
(α)
(b
−
qt)
f
(qt)d
t
.
q
q
Γq (α) a
Γq (α) a
The proof is complete.
Remark: We can see that for α = 1, we obtain the proposition 3.2 of [3].
Corollary 1. Let n be a positive integer and f be a function defined on [a, b]q satisfying
f (a) ≥ 0
and
Dq f (x) ≥ nCαn (x − a)n−1 ,
x ∈ (a, b]q .
Then
α
n+1
α
Iq,a
(f n+2 (t))(b) ≥ Γn
.
q (α) Iq,a (f (qt))(b)
Proof. It suffices to take β = n + 2 in Proposition 1 and the result follows.
Corollary 2. Let n be a positive integer and f be a function defined on [a, b]q satisfying
Dqi f (a) ≥ 0,
0 ≤ i≤ n−1
and
Dqn f (x) ≥ n[n − 1]q !Cαn ;
Then
α
n+1
α
Iq,a
(f n+2 (t))(b) ≥ Γn
.
q (α) Iq,a (f (qt))(b)
x ∈ (a, b]q .
β−1
.
ON SOME QI TYPE INEQUALITIES USING FRACTIONAL q-INTEGRAL
119
Proof. Since Dqn f (x) ≥ n[n − 1]q !Cαn , then by q-integrating n − 1 times over [a, x], we get
Dq f (x) ≥ nCαn (x − a)(n−1) ≥ nCαn (x − a)n−1 . The result turns out from Corollary 1. Proposition 2. Let p ≥ 1 be a real number and f be a function defined on [a, b]q satisfying
f (a) ≥ 0,
Dq f (x) ≥ pCαp , x ∈ (a, b]q .
(10)
p+1
[Γq (α)]p α
Iq,a (f (qt))(b)
.
(b − a)p−1
(11)
Then
α
Iq,a
(f p+2 (t))(b) ≥
Proof. Define g(x) =
H(x) =
Z
x
Z
x
(b − qt)(α−1) f (qt)dq t
and
a
(b−qt)(α−1) f p+2 (t)dq t−
a
1
(b − a)p−1
Z
x
(b − qt)(α−1) f (qt)dq t
a
p+1
,
x ∈ [a, b]q .
We have
=
Dq H(x)
(b − qx)(α−1) f p+2 (x) −
1
Dq [g p+1 ](x),
(b − a)p−1
x ∈ (a, b]q .
Since f and g increase on [a, b]q , we obtain according to Lemma 1,
Dq H(x)
≥
≥
≥
1
(p + 1)g p (x)Dq g(x)
(b − a)p−1
1
(b − qx)(α−1) f p+2 (x) −
(p + 1)g p (x)(b − qx)(α−1) f (qx)
(b − a)p−1
1
p
(p
+
1)g
(x)
(b − qx)(α−1) f (x) = (b − qx)(α−1) f (x)h(x),
f p+1 (x) −
(b − a)p−1
(b − qx)(α−1) f p+2 (x) −
1
(p + 1)g p (x).
(b − a)p−1
On the other hand, we have
where h(x) = f p+1 (x) −
Dq h(x) = Dq [f p+1 ](x) −
1
(p + 1)Dq [g p ](x).
(b − a)p−1
In virtue of Lemma 1, it follows that
Dq h(x)
≥
≥
(p + 1)p p−1
g
(x)(b − qx)(α−1) f (qx)
(p + 1)f p (qx)Dq f (x) −
(b − a)p−1
p
p−1
(α−1)
g
(x)(b
−
qx)
.
(p + 1)f (qx) f p−1 (qx)Dq f (x) −
(b − a)p−1
Since f is non-negative and increasing function, then
Z x
g(x) =
(b − qt)(α−1) f (qt)dq t ≤
f (qx)Cα (x − a),
a
hence
Dq h(x)
≥
(p + 1)f p (qx) [Dq f (x) − pCαp ] ≥ 0,
which implies that h increases on [a, b]q .
Finally, since h(a) = f p+1 (a) ≥ 0, then H increases and H(b) ≥ H(a) ≥ 0.
The proof is complete.
In what follows, we will adopt the terminology of the following definition.
120
KAMEL BRAHIM & HEDI EL MONSER
Definition 1. Let b > 0 and a = bq n , n be a positive integer. For each real number r, we
α
denote by Eq,r
([a, b]) the set of functions defined on [a, b]q such that
f (a) ≥ 0
and
where Kα = sup (b − q 2 t)(α−1) =
t∈[a,b]q
Proposition 3. Let f ∈
α
Iq,a
Dq f (x) ≥ [r]q Kα , ∀x ∈ (a, b]q ,

 (b − q 2 a)(α−1) if α ≥ 1

bα−1 (q 2 , q)α−1
if 0 < α < 1.
α
Eq,2
([a, b]).
f
2p+1
Then for all p > 0, we have
α
2
(t) (b) > Γq (α) Iq,a
(f p (t)) (b) .
(12)
Proof. For x ∈ [a, b]q , let
Z x
2
Z x
F (x) =
(b − qt)(α−1) f 2p+1 (t)dq t −
(b − qt)(α−1) f p (t)dq t
a Z
a
x
(α−1) p
and g(x) =
(b − qt)
f (t)dq t.
a
We have for x ∈ [a, b]q ,
Dq F (x)
=
(b − qx)(α−1) f 2p+1 (x) − (b − qx)(α−1) f p (x)(g(x) + g(qx))
=
(b − qx)(α−1) f p (x)G(x),
where G(x) = f p+1 (x) − [g(x) + g(qx)] .
On the other hand, we have
f p+1 (x) − f p+1 (qx) − (b − qx)(α−1) f p (x) + q(b − q 2 x)(α−1) f p (qx)
Dq G(x) =
(1 − q)x
=
f p (x)
f (x) − (1 − q)x(b − qx)(α−1)
f (qx) + q(b − q 2 x)(α−1) (1 − q)x
− f p (qx)
.
(1 − q)x
(1 − q)x
From the relation Dq f (x) ≥ Kα [2]q ≥ (b − q 2 x)(α−1) [2]q , we obtain
f (x) ≥ f (qx) + (1 − q 2 )x(b − q 2 x)(α−1) ,
hence
f p (x) − f p (qx)
[f (qx) + q(b − q 2 x)(α−1) (1 − q)x] > 0
Dq G(x) ≥
(1 − q)x
x ∈ (a, b]q .
(13)
Therefore G is strictly increasing on [a, b]q . Moreover, we have
G(a) = [f (a)]p+1 + (b − qa)(α−1) (f (a))p ≥ 0,
then for all x ∈ (a, b]q , G(x) > G(a) ≥ 0,which proves that Dq F (x) > 0, and so F is
strictly increasing on [a, b]q . In particular, F (b) > F (a) = 0.
α
Corollary 3. Let β > 0 and f ∈ Eq,2
([a, b]). Then for all positive integers m, we have
h
i2m
m
m
α
α
Iq,a
f (β+1)2 −1 (t) (b) > Γq2 −1 (α) Iq,a
.
(14)
f β (t) (b)
Proof. We suggest here a proof by induction. For this purpose, we note
pm (β) = (β + 1)2m − 1.
We have
pm (β) > 0 and pm+1 (β) = 2pm (β) + 1.
From Proposition 3, we deduce that the inequality (14) is true for m = 1.
Suppose that (14) holds for an integer m and let us prove it for m + 1.
By using the relation (15) and Proposition 3, we obtain
h
i2
m+1
m
α
−1
α
Iq,a
(f (β+1)2
(t))(b) > Γq (α) Iq,a
(f (β+1)2 −1 (t))(b) .
(15)
(16)
ON SOME QI TYPE INEQUALITIES USING FRACTIONAL q-INTEGRAL
And by assumption, we have
h
i2m
m
m
α
α
Iq,a
f (β+1)2 −1 (t) (b) > Γq2 −1 (α) Iq,a
.
f β (t) (b)
121
(17)
Finally, the relations (16) and (17) imply that the inequality (14) is true for m + 1. This
completes the proof.
α
Corollary 4. Let f ∈ Eq,2
([a, b]) and β > 0. For m ∈ N, we have
1
h
i 1m
h
i m+1
1
m
m+1
2
α
α
−1
f (β+1)2 −1 (t) (b)
> [Γq (α)] 2m+1 Iq,a
.
Iq,a
f (β+1)2
(t) (b) 2
(18)
Proof. Since, from Proposition 3,
m+1
α
Iq,a
(f (β+1)2
−1
h
i2
m
α
(t))(b) > Γq (α) Iq,a
(f (β+1)2 −1 (t))(b) ,
then
1
h
i 1m
h
i m+1
1
m
m+1
2
α
α
−1
.
f (β+1)2 −1 (t) (b)
> [Γq (α)] 2m+1 Iq,a
Iq,a
f (β+1)2
(t) (b) 2
(19)
(20)
α
Corollary 5. Let f ∈ Eq,2
([a, b]). For all integers m ≥ 2, we have
α
2m
m
α
2m+1 −1
Iq,a (f
(t))(b) > [Γq (α)]2 −1 Iq,a
(f (t)) (b)
.
Proof. By using Proposition 3 for β = 1, we obtain the result.
(21)
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K. Brahim. Institut Supérieur des Études Préparatoires en Biologie Géologie Soukra
Tunisia.
E-mail address:
kamel.brahim@ipeit.rnu.tn*
H. Elmonser. Institut National des sciences appliques et de technology, Tunis,
Tunisia.
E-mail address:
monseur2004@yahoo.fr