ON SOME FENG QI TYPE q-INTEGRAL
INEQUALITIES
KAMEL BRAHIM
Institut Préparatoire aux Études d’Ingénieur de Tunis
EMail: Kamel.Brahim@ipeit.rnu.tn
Feng Qi Type q-Integral
Inequalities
Kamel Brahim, Néji Bettaibi
and Mouna Sellami
vol. 9, iss. 2, art. 43, 2008
NÉJI BETTAIBI
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Institut Préparatoire aux Études d’Ingénieur de Monastir,
5000 Monastir, Tunisia.
Contents
EMail: Neji.Bettaibi@ipein.rnu.tn
MOUNA SELLAMI
Institut Préparatoire aux Études d’Ingénieur de El Manar,
Tunis, Tunisia
EMail: sellami_mouna@yahoo.fr
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Received:
03 May, 2008
Accepted:
30 May, 2008
Communicated by:
F. Qi
2000 AMS Sub. Class.:
26D10, 26D15, 33D05, 33D15.
Key words:
q-series, q-integral, Inequalities.
Abstract:
In this paper, we provide some Feng Qi type q-Integral Inequalities, by using
analytic and elementary methods in Quantum Calculus.
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Contents
1
Introduction
3
2
Notations and Preliminaries
4
3
q-Integral Inequalities of Feng Qi type
6
Feng Qi Type q-Integral
Inequalities
Kamel Brahim, Néji Bettaibi
and Mouna Sellami
vol. 9, iss. 2, art. 43, 2008
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1.
Introduction
In [9], F. Qi studied an interesting integral inequality and proved the following result:
Theorem 1.1. For a positive integer n and an nth order continuous derivative function f on an interval [a, b] such that f (i) (a) ≥ 0, 0 ≤ i ≤ n − 1 and f (n) (a) ≥ n!,
we have
Z b
n+1
Z b
n+2
(1.1)
[f (t)] dt ≥
f (t)dt
.
a
a
Feng Qi Type q-Integral
Inequalities
Kamel Brahim, Néji Bettaibi
and Mouna Sellami
vol. 9, iss. 2, art. 43, 2008
Then, he proposed the following open problem:
Under what condition is the inequality (1.1) still true if n is replaced by any
positive real number p?
In view of the interest in this type of inequality, much attention has been paid to
the problem and many authors have extended the inequality to more general cases
(see [1, 8]). In this paper, we shall discuss a q-analogue of the Feng Qi problem and
we will generalize the inequalities given in [1], [7] and [8].
This paper is organized as follows: In Section 2, we present definitions and facts
from q-calculus necessary for understanding this paper. In Section 3, we discuss
some generalizations of the so-called Feng Qi inequality.
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2.
Notations and Preliminaries
Throughout this paper, we will fix q ∈ (0, 1). For the convenience of the reader, we
provide a summary of the mathematical notations and definitions used in this paper
(see [3] and [5]). We write for a ∈ C,
1 − qa
[a]q =
,
1−q
(a; q)n =
n−1
Y
(1 − aq k ), n = 1, 2, . . . , ∞,
k=0
Feng Qi Type q-Integral
Inequalities
Kamel Brahim, Néji Bettaibi
and Mouna Sellami
[0]q ! = 1,
[n]q ! = [1]q [2]q ...[n]q ,
n = 1, 2, . . .
vol. 9, iss. 2, art. 43, 2008
and
(x −
a)nq
=

 1
 (x − a)(x − qa) · · · (x − q n−1 a) if n 6= 0
The q-derivative Dq f of a function f is given by
(Dq f )(x) =
(2.1)
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if n = 0
f (x) − f (qx)
,
(1 − q)x
if x 6= 0,
x ∈ C, n ∈ N.
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0
0
(Dq f )(0) = f (0) provided f (0) exists.
The q-Jackson integral from 0 to a is defined by (see [4])
Z a
∞
X
(2.2)
f (x)dq x = (1 − q)a
f (aq n )q n ,
0
provided the sum converges absolutely.
n=0
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The q-Jackson integral in a generic interval [a, b] is given by (see [4])
Z b
Z b
Z a
(2.3)
f (x)dq x =
f (x)dq x −
f (x)dq x.
a
0
0
We recall that for any function f , we have (see [5])
Z x
(2.4)
Dq
f (t)dq t = f (x).
a
Feng Qi Type q-Integral
Inequalities
Kamel Brahim, Néji Bettaibi
and Mouna Sellami
n
Finally, for b > 0 and a = bq , n a positive integer, we write
[a, b]q = {bq k : 0 ≤ k ≤ n} and
vol. 9, iss. 2, art. 43, 2008
(a, b]q = [q −1 a, b]q .
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3.
q-Integral Inequalities of Feng Qi type
Let us begin with the following useful result:
Lemma 3.1. Let p ≥ 1 be a real number and g be a nonnegative and monotone
function on [a, b]q . Then
pg p−1 (qx)Dq g(x) ≤ Dq [g(x)]p ≤ pg p−1 (x)Dq g(x),
x ∈ (a, b]q .
Feng Qi Type q-Integral
Inequalities
Kamel Brahim, Néji Bettaibi
Proof. We have
(3.1)
and Mouna Sellami
g p (x) − g p (qx)
1
=
p
(1 − q)x
(1 − q)x
Dq [g p ](x) =
Z
vol. 9, iss. 2, art. 43, 2008
g(x)
tp−1 dt.
g(qx)
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Since g is a nonnegative and monotone function, we have
Z g(x)
p−1
g (qx) [g(x) − g(qx)] ≤
tp−1 dt ≤ g p−1 (x) [g(x) − g(qx)] .
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g(qx)
Therefore, according to the relation (3.1), we obtain
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pg p−1 (qx)Dq g(x) ≤ Dq [g p ](x) ≤ pg p−1 (x)Dq g(x).
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Proposition 3.2. Let f be a function defined on [a, b]q satisfying
f (a) ≥ 0 and
t−3
Dq f (x) ≥ (t − 2)(x − a)
for x ∈ (a, b]q
Then
Z
b
t
Z
[f (x)] dq x ≥
a
t−1
b
f (qx)dq x
a
.
Close
and
t ≥ 3.
Proof. Put g(x) =
Rx
a
f (qu)dq u and
x
Z
t
Z
[f (u)] dq u −
F (x) =
t−1
x
f (qu)dq u
a
.
a
We have
Dq F (x) = f t (x) − Dq [g t−1 ](x).
Since f and g increase on [a, b]q , we obtain from Lemma 3.1,
Dq F (x) ≥ f t (x) − (t − 1)g t−2 (x)f (qx)
t
≥ f (x) − (t − 1)g
t−2
Dq h(x) = Dq [f t−1 ](x) − (t − 1)Dq [g t−2 ](x).
By using Lemma 3.1, we obtain
(3.3)
and Mouna Sellami
vol. 9, iss. 2, art. 43, 2008
(x)f (x) = f (x)h(x),
where h(x) = f t−1 (x) − (t − 1)g t−2 (x).
On the other hand, we have
(3.2)
Feng Qi Type q-Integral
Inequalities
Kamel Brahim, Néji Bettaibi
Dq h(x) ≥ (t − 1)f t−2 (qx)Dq f (x) − (t − 1)(t − 2)g t−3 (x)Dq g(x)
≥ (t − 1)f (qx) f t−3 (qx)Dq f (x) − (t − 2)g t−3 (x) .
Since the function f increases, we have
Z x
f (qu)dq u ≤ f (qx)(x − a).
a
Then, from the conditions of the proposition and inequalities (3.2) and (3.3), we get
Dq h(x) ≥ (t − 1)f t−2 (qx) Dq f (x) − (t − 2)(x − a)t−3 ≥ 0,
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and from the fact h(a) = f t−1 (a) ≥ 0, we get h(x) ≥ 0, x ∈ [a, b]q .
From F (a) = 0 and Dq F (x) = f (x)h(x) ≥ 0, it follows that F (x) ≥ 0 for all
x ∈ [a, b]q , in particular
Z b
t−1
Z b
t
F (b) =
≥ 0.
[f (u)] dq u −
f (qu)dq u
a
a
Corollary 3.3. Let n be a positive integer and f be a function defined on [a, b]q
satisfying
f (a) ≥ 0 and
n−1
Dq f (x) ≥ n(x − a)
,
Then,
Z
b
(f (x))n+2 dq x ≥
a
Z
f (qx)dq x
.
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a
Corollary 3.4. Let n be a positive integer and f be a function defined on [a, b]q
satisfying
≥ 0,
0 ≤ i ≤ n − 1 and
Dqn f (x)
≥ n[n − 1]q !
Then,
Z
a
b
(f (x))n+2 dq x ≥
Z
x ∈ (a, b]q .
f (qx)dq x
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n+1
b
vol. 9, iss. 2, art. 43, 2008
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Proof. It suffices to take t = n + 2 in Proposition 3.2 and the result follows.
Dqi f (a)
and Mouna Sellami
x ∈ (a, b]q .
n+1
b
Feng Qi Type q-Integral
Inequalities
Kamel Brahim, Néji Bettaibi
.
a
Proof. Since Dqn f (x) ≥ n[n − 1]q !, then by q-integrating n − 1 times over [a, x], we
get
Dq f (x) ≥ n(x − a)n−1
≥ n(x − a)n−1 .
q
The result follows from Corollary 3.3.
Close
Proposition 3.5. Let p ≥ 1 be a real number and f be a function defined on [a, b]q
satisfying
f (a) ≥ 0,
(3.4)
Dq f (x) ≥ p,
∀ x ∈ (a, b]q .
Then we have
b
Z
[f (x)]p+2 dq x ≥
(3.5)
a
1
(b − a)p−1
Feng Qi Type q-Integral
Inequalities
Kamel Brahim, Néji Bettaibi
p+1
b
Z
f (qx)dq x
.
and Mouna Sellami
a
vol. 9, iss. 2, art. 43, 2008
Proof. Put g(t) =
Rt
Z
(3.6)
H(t) =
a
f (qx)dq x and
t
p+2
[f (x)]
a
1
dq x −
(b − a)p−1
Z
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p+1
t
f (qx)dq x
,
t ∈ [a, b]q .
We have
Dq H(t) = [f (t)]p+2 −
1
Dq [g p+1 ](t),
(b − a)p−1
t ∈ (a, b]q .
Since f and g increase on [a, b]q , we obtain, according to Lemma 3.1, for t ∈ (a, b]q ,
1
(p + 1)g p (t)f (qt)
(b − a)p−1
1
(p + 1)g p (t)f (t)
≥ [f (t)]p+2 −
(b − a)p−1
1
p+1
p
≥ [f (t)]
−
(p + 1)g (t) f (t) = h(t)f (t),
(b − a)p−1
Dq H(t) ≥ [f (t)]p+2 −
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a
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where
h(t) = [f (t)]p+1 −
1
(p + 1)g p (t).
(b − a)p−1
On the other hand, we have
Dq h(t) = Dq [f p+1 ](t) −
1
(p + 1)Dq [g p ](t).
(b − a)p−1
By using Lemma 3.1, we obtain
(p + 1)p p−1
Dq h(t) ≥ (p + 1)f p (qt)Dq f (t) −
g (t)f (qt)
(b − a)p−1
p
p−1
p−1
≥ (p + 1)f (qt) f (qt)Dq f (t) −
g (t) .
(b − a)p−1
Since f increases, then for t ∈ [a, b]q ,
Z t
(3.7)
f (qx)dq x ≤ (b − a)f (qt),
a
therefore,
(3.8)
Dq h(t) ≥ (p + 1)f p (qt)[Dq f (t) − p].
We deduce, from the relation (3.4), 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,
which completes the proof.
Corollary 3.6. Let p ≥ 1 be a real number and f be a nonnegative function on [0, 1]
such that Dq f (x) ≥ 1. Then
Z 1
p+1
Z 1
1
p+2
(3.9)
[f (x)] dq x ≥
f (qx)dq x
.
p 0
0
Feng Qi Type q-Integral
Inequalities
Kamel Brahim, Néji Bettaibi
and Mouna Sellami
vol. 9, iss. 2, art. 43, 2008
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Proof. Replacing, in the previous proposition, f (x) by pf (x), b by 1 and a by q N
(N = 1, 2, . . . ), we obtain then the result by tending N to ∞.
In what follows, we will adopt the terminology of the following definition.
Definition 3.7. Let b > 0 and a = bq n , where n is a positive integer. For each real
number r, we denote by Eq,r ([a, b]) the set of functions defined on [a, b]q such that
f (a) ≥ 0
Dq f (x) ≥ [r]q ,
and
Feng Qi Type q-Integral
Inequalities
Kamel Brahim, Néji Bettaibi
∀x ∈ (a, b]q .
and Mouna Sellami
Proposition 3.8. Let f ∈ Eq,2 ([a, b]). Then for all p > 0, we have
Z
b
2p+1
[f (x)]
(3.10)
Z
dq x >
a
b
p
(f (x)) dq x
vol. 9, iss. 2, art. 43, 2008
2
.
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a
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Proof. For t ∈ [a, b]q , we put
Z
F (t) =
t
[f (x)]2p+1 dq x −
a
Z
t
(f (x))p dq x
2
Z
and g(t) =
a
Then, we have for t ∈ [a, b]q ,
Dq F (t) = [f (t)]2p+1 − [f (t)]p (g(t) + g(qt))
= [f (t)]p [f (t)]p+1 − [g(t) + g(qt)]
= [f (t)]p G(t),
where G(t) = [f (t)]p+1 − [g(t) + g(qt)] .
a
t
[f (x)]p dq x.
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On the other hand, we have
f p+1 (t) − f p+1 (qt)
− f p (t) − qf p (qt)
(1 − q)t
f (qt) + q(1 − q)t
f (t) − (1 − q)t
− f p (qt)
.
= f p (t)
(1 − q)t
(1 − q)t
Dq G(t) =
By using the relation Dq f (t) ≥ [2]q , we obtain f (t) ≥ f (qt) + (1 − q 2 )t, therefore
(3.11)
Dq G(t) ≥
f p (t) − f p (qt)
[f (qt) + q(1 − q)t] > 0,
(1 − q)t
Feng Qi Type q-Integral
Inequalities
Kamel Brahim, Néji Bettaibi
and Mouna Sellami
t ∈ (a, b]q .
Hence, G is strictly increasing on [a, b]q . Moreover, we have
vol. 9, iss. 2, art. 43, 2008
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G(a) = [f (a)]p+1 + (1 − q)af (a) ≥ 0,
for all t ∈ (a, b]q , G(t) > G(a) ≥ 0, which proves that Dq F (t) > 0, for all t ∈
(a, b]q . Thus, F is strictly increasing on [a, b]q . In particular, F (b) > F (a) = 0.
Corollary 3.9. Let α > 0 and f ∈ Eq,2 ([a, b]). Then for all positive integers m, we
have
Z b
2m
Z b
m
(3.12)
[f (x)](α+1)2 −1 dq x >
[f (x)]α dq x
.
a
a
Proof. We suggest here a proof by induction. For this purpose, we put:
pm (α) = (α + 1)2m − 1.
We have
(3.13)
pm (α) > 0
and
pm+1 (α) = 2pm (α) + 1.
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From Proposition 3.8, we deduce that the inequality (3.12) is true for m = 1.
Suppose that (3.12) holds for an integer m and let us prove it for m + 1.
By using the relation (3.13) and Proposition 3.8, we obtain
Z b
2
Z b
(α+1)2m+1 −1
(α+1)2m −1
(3.14)
[f (x)]
dq x >
[f (x)]
dq x .
a
a
Feng Qi Type q-Integral
Inequalities
Kamel Brahim, Néji Bettaibi
And, by assumption, we have
Z b
2m
Z b
(α+1)2m −1
α
(3.15)
[f (x)]
>
[f (x)] dq x
.
a
and Mouna Sellami
vol. 9, iss. 2, art. 43, 2008
a
Finally, the relations (3.14) and (3.15) imply that the inequality (3.12) is true for
m + 1. This completes the proof.
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Corollary 3.10. Let f ∈ Eq,2 ([a, b]) and α > 0. For m ∈ N, we have
b
Z
[f (x)](α+1)2
(3.16)
m+1 −1
1
2m+1
Z
b
[f (x)](α+1)2
>
dq x
m −1
21m
dq x
.
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Proof. Since, from Proposition 3.8,
Z b
2
Z b
(α+1)2m+1 −1
(α+1)2m −1
(3.17)
[f (x)]
dq x >
[f (x)]
dq x ,
a
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then
Z
b
(α+1)2m+1 −1
[f (x)]
(3.18)
a
1
2m+1
dq x
Z
>
b
(α+1)2m −1
[f (x)]
a
21m
dq x
.
Corollary 3.11. Let f ∈ Eq,2 ([a, b]). For all integers m ≥ 2, we have
Z
b
2m+1 −1
[f (x)]
(3.19)
Z
[f (x)]3 dq x
dq x >
a
2m−1
a
Z
(3.20)
b
>
2m
b
f (x)dq x
.
a
Proof. By using Proposition 3.8 and the two previous corollaries for α = 1, we
obtain the required result.
Feng Qi Type q-Integral
Inequalities
Kamel Brahim, Néji Bettaibi
and Mouna Sellami
vol. 9, iss. 2, art. 43, 2008
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References
[1] L. BOUGOFFA, Notes on Qi type inequalities, J. Inequal. Pure and Appl. Math.,
4(4) (2003), Art. 77. [ONLINE: http://jipam.vu.edu.au/article.
php?sid=318].
[2] W.S. CHEUNG AND Z̆. HANJS̆ AND J. PEČARIĆ, Some Hardy-type inequalities, J. Math. Anal. Applics., 250 (2000), 621–634.
[3] G. GASPER AND M. RAHMAN, Basic Hypergeometric Series, 2nd Edition,
(2004), Encyclopedia of Mathematics and Its Applications, 96, Cambridge University Press, Cambridge.
[4] F.H. JACKSON, On a q-definite integrals, Quarterly Journal of Pure and Applied
Mathematics, 41 (1910), 193–203.
[5] V.G. KAC AND P. CHEUNG, Quantum Calculus, Universitext, Springer-Verlag,
New York, (2002).
[6] T.H. KOORNWINDER, q-Special Functions, a Tutorial, in Deformation Theory
and Quantum Groups with Applications to Mathematical Physics, M. Gerstenhaber and J. Stasheff (eds.), Contemp. Math., 134, Amer. Math. Soc., (1992).
[7] S. MAZOUZI AND F. QI, On an open problem regarding an integral inequality, J.
Inequal. Pure Appl. Math., 4(2) (2003), Art. 31. [ONLINE: http://jipam.
vu.edu.au/article.php?sid=73].
[8] T.K. POGÁNY, On an open problem of F. Qi, J. Inequal Pure Appl. Math., 3(4)
(2002), Art. 54. [ONLINE: http://jipam.vu.edu.au/article.php?
sid=206].
[9] F. QI, Several integral inequalities, J. Inequal. Pure Appl. Math., 1(2) (2002),
Art. 19. [ONLINE: http://jipam.vu.edu.au/article.php?sid=
113].
Feng Qi Type q-Integral
Inequalities
Kamel Brahim, Néji Bettaibi
and Mouna Sellami
vol. 9, iss. 2, art. 43, 2008
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