Volume 9 (2008), Issue 4, Article 106, 6 pp.
ON SOME q-INTEGRAL INEQUALITIES
KAMEL BRAHIM
I NSTITUT P RÉPARATOIRE AUX É TUDES D ’I NGÉNIEUR DE T UNIS
Kamel.Brahim@ipeit.rnu.tn
Received 24 June, 2008; accepted 10 November, 2008
Communicated by S.S. Dragomir
A BSTRACT. In this paper, we provide a q-analogue of an open problem posed by Q. A. Ngô et
al. in the paper, Note on an integral inequality, J. Inequal. Pure and Appl. Math., 7(4)(2006),
Art. 120, by using analytic and elementary methods in Quantum Calculus.
Key words and phrases: q-integral, Inequalities.
2000 Mathematics Subject Classification. 26D15.
1.
INTRODUCTION
In [9], Q.A. Ngô et al. studied an interesting integral inequality and proved the following
result:
Theorem 1.1. Let f (x) ≥ 0 be a continuous function on [0, 1] satisfying
Z 1
Z 1
(1.1)
f (t)dt ≥
tdt, ∀x ∈ [0, 1].
x
x
Then the inequalities
Z
1
f
(1.2)
α+1
Z
1
xα f (x)dx
(x)dx ≥
0
0
and
Z
1
f
(1.3)
α+1
Z
(x)dx ≥
0
1
xf α (x)dx
0
hold for every positive real number α > 0.
Then, they proposed the following open problem: Under what condition does the inequality
Z 1
Z 1
α+β
(1.4)
f
(x)dx ≥
xβ f α (x)dx
0
hold for α and β?
179-08
0
2
K AMEL B RAHIM
In view of the interest in this type of inequalities, much attention has been paid to the problem
and many authors have extended the inequality to more general cases (see [1, 3, 7, 8]). In this
paper, we shall discuss a q-analogue of Ngô’s problem.
This paper is organized as follows: In Section 2, we present definitions and facts from the
q-calculus necessary for understanding this paper. In Section 3, we discuss a q-analogue of the
inequalities given in [9] and [3].
2. N OTATIONS AND P RELIMINARIES
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
[4] and [6]). We write for a ∈ C,
1 − qa
.
[a]q =
1−q
The q-derivative Dq f of a function f is given by
(Dq f )(x) =
(2.1)
f (x) − f (qx)
, if x 6= 0,
(1 − q)x
(Dq f )(0) = f 0 (0), provided f 0 (0) exists.
The q-Jackson integral from 0 to a is defined by (see [5])
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 [5])
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 [6])
Z x
(2.4)
Dq
f (t)dq t = f (x).
a
If F is any anti q-derivative of the function f , namely Dq F = f , continuous at x = 0, then
Z a
(2.5)
f (x)dq x = F (a) − F (0).
0
A q-analogue of the integration by parts formula is given by
Z b
Z b
(2.6)
f (x)(Dq g(x))dq x = f (a)g(a) − f (b)g(b) −
(Dq f (x))g(qx)dq x.
a
a
Finally, we denote
[0, 1]q = {q k : k = 0, 1, 2, . . . , ∞}.
3. M AIN R ESULTS
Let us begin with the following useful result:
Lemma 3.1 ([9] General Cauchy inequality). Let α and β be positive real numbers satisfying
α + β = 1. Then for all positive real numbers x and y, we always have
(3.1)
αx + βy ≥ xα y β .
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 106, 6 pp.
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q-I NTEGRAL I NEQUALITIES
3
Theorem 3.2. Let f be a nonnegative function defined on [0, 1]q satisfying
Z 1
Z 1
β
(3.2)
f (t)dq t ≥
tβ dq t, ∀x ∈ [0, 1]q .
x
x
Then the inequality
Z
1
f
(3.3)
α+β
1
Z
xα f β (x)dq x,
(x)dq x ≥
0
0
holds for all positive real numbers α > 0 and β > 0.
To prove Theorem 3.2, we need the following lemma.
Lemma 3.3. Under the conditions of Theorem 3.2, we have
Z 1
1
.
(3.4)
xα f β (x)dq x ≥
[α + β + 1]q
0
Proof. By using a q-integration by parts, we obtain
Z 1
x=1
Z 1
Z 1
Z 1
qα
1
α
α−1
β
β
x
f (t)dq t dq x =
x
f (t)dq t
+
xα f β (x)dq x
[α]q
[α]
q
0
x
x
0
x=0
Z 1
qα
=
xα f β (x)dq x,
[α]q 0
which yields
Z
(3.5)
0
1
[α]q
x f (x)dq x = α
q
α β
Z
1
x
α−1
0
f (t)dq t dq x.
β
x
On the other hand, from condition (3.2), we get
Z 1
Z 1
Z
α−1
β
x
f (t)dq t dq x ≥
0
1
Z
x
1
x
α−1
Z
0
1
t dq t dq x
β
x
Z
1
1
(xα−1 − xα+β )dq x
[β + 1]q 0
qα
=
.
[α]q [α + β + 1]q
=
Therefore, from (3.5), we obtain
Z 1
(3.6)
xα f β (x)dq x ≥
0
1
.
[α + β + 1]q
We now give the proof of Theorem 3.2.
Proof of Theorem 3.2. Using Lemma 3.1, we obtain
β
α
f α+β (x) +
xα+β ≥ xα f β (x),
α+β
α+β
(3.7)
which gives
Z
(3.8)
β
1
f
0
α+β
Z
1
(x)dq x + α
x
0
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 106, 6 pp.
α+β
Z
dq x ≥ (α + β)
1
xα f β (x)dq x.
0
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4
K AMEL B RAHIM
Moreover, by using Lemma 3.3, we get
Z 1
Z
α β
(α + β)
x f (x)dq x = α
(3.9)
1
Z
1
x f (x)dq x + β
xα f β (x)dq x
0
0
Z 1
α
+β
xα f β (x)dq x.
≥
[α + β + 1]q
0
0
α β
Then, from relation (3.8), we obtain
Z 1
Z 1
α
α
α+β
≥
+β
xα f β (x)dq x,
(3.10)
β
f
(x)dq x +
[α
+
β
+
1]
[α
+
β
+
1]
q
q
0
0
which completes the proof.
Taking β = 1 in Theorem 3.2, we obtain
Corollary 3.4. Let f be a nonnegative function defined on [0, 1]q satisfying
Z 1
Z 1
(3.11)
f (t)dq t ≥
tdq t, ∀x ∈ [0, 1]q .
x
x
Then the inequality
1
Z
f
(3.12)
α+1
1
Z
xα f (x)dq x
(x)dq x ≥
0
0
holds for every positive real number α > 0.
Theorem 3.5. Let f be a nonnegative function defined on [0, 1]q satisfying
Z 1
Z 1
(3.13)
f (t)dq t ≥
tdq t, ∀x ∈ [0, 1]q .
x
x
Then the inequality
1
Z
f
(3.14)
α+1
1
Z
xf α (x)dq x
(x)dq x ≥
0
0
holds for every positive real number α > 0.
Proof. We have
(3.15)
∀x ∈ [0, 1]q ,
(f α (x) − xα )(f (x) − x) ≥ 0,
so
(3.16)
f α+1 (x) + xα+1 ≥ xα f (x) + xf α (x).
By integrating with some simple calculations we deduce that
Z 1
Z 1
Z 1
1
α+1
α
(3.17)
f
(x)dq x +
≥
x f (x)dq x +
xf α (x)dq x.
[α
+
2]
q
0
0
0
Then, from Lemma 3.3 for β = 1, the result follows.
Theorem 3.6. Let f be a nonnegative function defined on [0, 1]q satisfying
Z 1
Z 1
(3.18)
f (t)dq t ≥
tdq t, ∀x ∈ [0, 1]q .
x
x
Then the inequality
Z
1
f
(3.19)
α+β
Z
(x)dq x ≥
0
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 106, 6 pp.
1
xα f β (x)dq x
0
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q-I NTEGRAL I NEQUALITIES
5
holds for all real numbers α > 0 and β ≥ 1.
Lemma 3.7. Under the conditions of Theorem 3.6, we have
Z 1
1
(3.20)
xα f β (x)dq x ≥
[α + β + 1]q
0
for all real numbers α > 0 and β ≥ 1.
Proof. Using Lemma 3.1, we obtain
1 β
β−1 β
(3.21)
f (x) +
x ≥ xβ−1 f (x),
β
β
which implies
Z 1
Z 1
Z 1
α β
α+β
(3.22)
x f (x)dq x + (β − 1)
x dq x ≥ β
xα+β−1 f (x)dq x.
0
0
Therefore, from Lemma 3.3, we get
Z 1
(3.23)
xα f β (x)dq x +
0
0
β
β−1
≥
.
[α + β + 1]q
[α + β + 1]q
Thus (3.20) is proved.
We now give the proof of Theorem 3.6.
Proof of Theorem 3.6. By using Lemma 3.1, we obtain
β
α
(3.24)
f α+β (x) +
xα+β ≥ xα f β (x),
α+β
α+β
which implies
Z 1
Z 1
α
α+β
(3.25)
β
f
(x)dq x +
≥ (α + β)
xα f β (x)dq x.
[α
+
β
+
1]
q
0
0
Then, from Lemma 3.7, we obtain
Z 1
Z 1
α
α
α+β
(3.26)
β
f
(x)dq x +
≥
+β
xα f β (x)dq x,
[α
+
β
+
1]
[α
+
β
+
1]
q
q
0
0
which completes the proof.
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http://jipam.vu.edu.au/