Volume 9 (2008), Issue 2, Article 43, 7 pp.
ON SOME FENG QI TYPE q-INTEGRAL INEQUALITIES
KAMEL BRAHIM, NÉJI BETTAIBI, AND MOUNA SELLAMI
I NSTITUT P RÉPARATOIRE AUX É TUDES D ’I NGÉNIEUR DE T UNIS
Kamel.Brahim@ipeit.rnu.tn
I NSTITUT P RÉPARATOIRE AUX É TUDES D ’I NGÉNIEUR DE M ONASTIR , 5000 M ONASTIR ,
T UNISIA .
Neji.Bettaibi@ipein.rnu.tn
I NSTITUT P RÉPARATOIRE AUX É TUDES D ’I NGÉNIEUR DE E L M ANAR , T UNIS , T UNISIA
sellami_mouna@yahoo.fr
Received 03 May, 2008; accepted 30 May, 2008
Communicated by F. Qi
A BSTRACT. In this paper, we provide some Feng Qi type q-Integral Inequalities, by using analytic and elementary methods in Quantum Calculus.
Key words and phrases: q-series, q-integral, Inequalities.
2000 Mathematics Subject Classification. 26D10, 26D15, 33D05, 33D15.
1. I NTRODUCTION
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
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 qcalculus necessary for understanding this paper. In Section 3, we discuss some generalizations
of the so-called Feng Qi inequality.
131-08
2
K AMEL B RAHIM , N ÉJI B ETTAIBI ,
AND
M OUNA S ELLAMI
2. N OTATIONS AND P RELIMINARIES
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
[0]q ! = 1,
(a; q)n =
n−1
Y
(1 − aq k ), n = 1, 2, . . . , ∞,
k=0
[n]q ! = [1]q [2]q ...[n]q ,
n = 1, 2, . . .
and
(x − a)nq =

 1
if n = 0
x ∈ C, n ∈ N.
 (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)
f (x) − f (qx)
,
(1 − q)x
if x 6= 0,
(Dq f )(0) = f 0 (0) provided f 0 (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
n=0
provided the sum converges absolutely.
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
n
Finally, for b > 0 and a = bq , n a positive integer, we write
[a, b]q = {bq k : 0 ≤ k ≤ n} and
(a, b]q = [q −1 a, b]q .
3. q-I NTEGRAL I NEQUALITIES OF F ENG Q I
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 .
Proof. We have
(3.1)
g p (x) − g p (qx)
1
Dq [g ](x) =
=
p
(1 − q)x
(1 − q)x
p
Z
g(x)
tp−1 dt.
g(qx)
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)] .
g(qx)
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F ENG Q I T YPE q-I NTEGRAL I NEQUALITIES
3
Therefore, according to the relation (3.1), we obtain
pg p−1 (qx)Dq g(x) ≤ Dq [g p ](x) ≤ pg p−1 (x)Dq g(x).
Proposition 3.2. Let f be a function defined on [a, b]q satisfying
f (a) ≥ 0 and
Dq f (x) ≥ (t − 2)(x − a)t−3
Then
b
Z
Z
t
Proof. Put g(x) =
Rx
a
and
t ≥ 3.
t−1
b
[f (x)] dq x ≥
a
x ∈ (a, b]q
for
f (qx)dq x
.
a
f (qu)dq u and
Z
Z x
t
F (x) =
[f (u)] dq u −
a
t−1
x
f (qu)dq u
.
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)
≥ f t (x) − (t − 1)g t−2 (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
Dq h(x) = Dq [f t−1 ](x) − (t − 1)Dq [g t−2 ](x).
By using Lemma 3.1, we obtain
(3.2)
(3.3)
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,
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
t−1
Z b
Z b
t
≥ 0.
f (qu)dq u
F (b) =
[f (u)] dq u −
a
a
Corollary 3.3. Let n be a positive integer and f be a function defined on [a, b]q satisfying
Dq f (x) ≥ n(x − a)n−1 ,
f (a) ≥ 0 and
x ∈ (a, b]q .
Then,
Z
b
n+2
(f (x))
Z
dq x ≥
a
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 43, 7 pp.
n+1
b
f (qx)dq x
.
a
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4
K AMEL B RAHIM , N ÉJI B ETTAIBI ,
AND
M OUNA S ELLAMI
Proof. It suffices to take t = n + 2 in Proposition 3.2 and the result follows.
Corollary 3.4. 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 ! x ∈ (a, b]q .
Then,
b
Z
n+2
(f (x))
dq x ≥
a
n+1
b
Z
f (qx)dq x
.
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.
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
Z
b
p+2
[f (x)]
(3.5)
a
Proof. Put g(t) =
Rt
Z
p+1
b
f (qx)dq x
.
a
f (qx)dq x and
a
Z
(3.6)
1
dq x ≥
(b − a)p−1
H(t) =
t
p+2
[f (x)]
a
1
dq x −
(b − a)p−1
Z
p+1
t
f (qx)dq x
,
t ∈ [a, b]q .
a
We have
1
Dq [g p+1 ](t), t ∈ (a, b]q .
p−1
(b − a)
Since f and g increase on [a, b]q , we obtain, according to Lemma 3.1, for t ∈ (a, b]q ,
Dq H(t) = [f (t)]p+2 −
1
(p + 1)g p (t)f (qt)
(b − a)p−1
1
≥ [f (t)]p+2 −
(p + 1)g p (t)f (t)
(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 −
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
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F ENG Q I T YPE q-I NTEGRAL I NEQUALITIES
5
Since f increases, then for t ∈ [a, b]q ,
Z t
(3.7)
f (qx)dq x ≤ (b − a)f (qt),
a
therefore,
Dq h(t) ≥ (p + 1)f p (qt)[Dq f (t) − p].
(3.8)
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
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 3.1. 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
∀x ∈ (a, b]q .
Proposition 3.7. Let f ∈ Eq,2 ([a, b]). Then for all p > 0, we have
Z b
2
Z b
2p+1
p
(3.10)
[f (x)]
dq x >
(f (x)) dq x .
a
a
Proof. For t ∈ [a, b]q , we put
Z t
2
Z t
2p+1
p
F (t) =
[f (x)]
dq x −
(f (x)) dq x
a
Z
and
g(t) =
a
t
[f (x)]p dq x.
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)] .
On the other hand, we have
f p+1 (t) − f p+1 (qt)
− f p (t) − qf p (qt)
(1 − q)t
f (t) − (1 − q)t
f (qt) + q(1 − q)t
= f p (t)
− f p (qt)
.
(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
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 43, 7 pp.
t ∈ (a, b]q .
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K AMEL B RAHIM , N ÉJI B ETTAIBI ,
AND
M OUNA S ELLAMI
Hence, G is strictly increasing on [a, b]q . Moreover, we have
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 3.8. 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
pm (α) > 0
(3.13)
and
pm+1 (α) = 2pm (α) + 1.
From Proposition 3.7, 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 3.7, 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
And, by assumption, we have
Z b
2m
Z b
m
(3.15)
[f (x)](α+1)2 −1 >
[f (x)]α dq x
.
a
a
Finally, the relations (3.14) and (3.15) imply that the inequality (3.12) is true for m + 1. This
completes the proof.
Corollary 3.9. Let f ∈ Eq,2 ([a, b]) and α > 0. For m ∈ N, we have
1
Z b
2m+1
Z b
21m
(α+1)2m −1
(α+1)2m+1 −1
(3.16)
[f (x)]
dq x
>
[f (x)]
dq x
.
a
a
Proof. Since, from Proposition 3.7,
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
a
then
Z
b
(α+1)2m+1 −1
[f (x)]
(3.18)
1
2m+1
dq x
b
Z
>
(α+1)2m −1
[f (x)]
a
21m
dq x
.
a
Corollary 3.10. Let f ∈ Eq,2 ([a, b]). For all integers m ≥ 2, we have
Z b
2m−1
Z b
2m+1 −1
3
(3.19)
[f (x)]
dq x >
[f (x)] dq x
a
a
Z
(3.20)
>
2m
b
f (x)dq x
.
a
Proof. By using Proposition 3.7 and the two previous corollaries for α = 1, we obtain the
required result.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 43, 7 pp.
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F ENG Q I T YPE q-I NTEGRAL I NEQUALITIES
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