International Journal of Analysis and Applications
ISSN 2291-8639
Volume 7, Number 1 (2015), 50-58
http://www.etamaths.com

NEW INTEGRAL INEQUALITIES IN QUANTUM CALCULUS

KAMEL BRAHIM1,∗, SABRINA TAF2 AND BOCHRA NEFZI1

Abstract. In this paper, we study the q-analogue of Klamkin-McLenaghan’s
and Grueb-Reinboldt’s inequalities then we use the Riemann-Liouville frac-

tional q-integral to get some new integral results.

1. introduction

Let us consider
(1.1)

T(f,g; a,b) =
1

b−a

∫ b
a

f(x)g(x)dx−

(
1

b−a

∫ b
a

f(x)dx

)(
1

b−a

∫ b
a

g(x)dx

)
where f and g are two integrable functions on [a,b],
and

Tq(f,g; a,b) =
1

b−a

∫ b
a

f(x)g(x)dqx−

(
1

b−a

∫ b
a

f(x)dqx

)(
1

b−a

∫ b
a

g(x)dqx

)
where f and g are two functions defined on [a,b]q.
The well-known Grüss integral inequality can be stated as follows (see [10, 15]):

(1.2) |T(f,g; a,b)| ≤
1

4
(M −m)(N −n),

provided that f and g are two integrable functions on [a,b] such that

(1.3) 0 < m ≤ f(x) ≤ M < ∞, 0 < n ≤ g(x) ≤ N < ∞, ,x ∈ [a,b].

The constant 1
4

is best possible.
Gauchman gave the q-integral Grüss inequality as follows (see [8]):

|Tq(f,g; a,b)| ≤
1

4
(M −m)(N −n),

In [12], the authors proved the following Klamkin-McLenaghan inequality
(1.4)

n∑
k=1

wka
2
k

n∑
k=1

wkb
2
k −

(
n∑
k=1

wkakbk

)2
≤

(√
M

n
−
√
m

N

)2 n∑
k=1

wkakbk

n∑
k=1

wkb
2
k,

2010 Mathematics Subject Classification. 35A23.
Key words and phrases. Grüss inequality, fractional q-integral, Klamkin-McLenaghan inequal-

ity, Grueb-Reinboldt inequality.

c©2015 Authors retain the copyrights of their papers, and all
open access articles are distributed under the terms of the Creative Commons Attribution License.

50



NEW INTEGRAL INEQUALITIES IN QUANTUM CALCULUS 51

where 0 < m
N
≤ ak

bk
≤ M

n
< ∞, wk > 0, k = 1, . . . ,n.

In [9], the authors proved the following Grueb-Reinboldt inequality

(1.5)

n∑
k=1

wka
2
k

n∑
k=1

wkb
2
k ≤

(MN + mn)2

4nmNM

(
n∑
k=1

wkakbk

)2
In [6], Dragomir and Diamond proved that

(1.6) |T(f,g; a,b)| ≤
1

4
·

(M −m)(N −n)
√
mMnN

·
1

b−a

∫ b
a

f(x)dx ·
1

b−a

∫ b
a

g(x)dx,

and
(1.7)

|T(f,g; a,b)| ≤
(√

M −
√
m

)(√
N −

√
n

)√
1

b−a

∫ b
a

f(x)dx ·
1

b−a

∫ b
a

g(x)dx.

In recent years, many researches have studies (1.1) and number of inequalities ap-
peared in literature (see [1, 2, 3, 4, 5, 14, 18]).

The main objective of this paper is to establish some new q-fractional integral
inequalities of Klamkin-McLenaghan and Grueb-Reinboldt type.

This paper is organized as follows: in section 2, we present some preliminary re-
sults and notation. In section 3, we state the q-analogue of Klamkin-McLenaghan
and Grueb-Reinboldt inequalities, then we establish some new q-fractional integral
inequalities.

2. Basic Definitions

For the convenience of the reader, we provide in this section a summary of the
mathematical notations and definitions used in this paper (see [7, 13, 16]). We
write for a,b ∈ C and q ∈ (0, 1),

(a; q)∞ =

∞∏
k=0

(1 −aqk), (a− b)(α) = aα
( b
a
; q)∞

(qα b
a
; q)∞

.

The q-Jackson integral from 0 to a is defined by (see [11])

(2.1)

∫ a
0

f(x)dqx = (1 −q)a
∞∑
n=0

f(aqn)qn,

provided the sum converges absolutely.
The q-Jackson integral in a generic interval [a,b] is given by (see [11])

(2.2)

∫ b
a

f(x)dqx =

∫ b
0

f(x)dqx−
∫ a

0

f(x)dqx.

In the case a = bqn, we can write

(2.3)

∫ b
a

f(x)dqx = (1 −q)b
n−1∑
k=0

f(bqk)qk.



52 BRAHIM, TAF AND NEFZI

The fractional q-integral of the Riemann-Liouville type is (see [16])(
Jαq f

)
(x) =

1

Γq(α)

∫ x
0

(x−qt)(α−1)f(t)dqt; α > 0

=
xα

Γq(α)
(1 −q)

∞∑
n=0

(1 −qn+1)(α−1)f(xqn)qn.(2.4)

where

Γq(α) =
1

1 −q

∫ 1
0

(
u

1 −q

)α−1
eq(qu)dqu, and eq(t) =

∞∏
k=0

(1 −qkt).

The q-fractional integration has the following semi-group property for α,β ∈ R+

(Jβq J
α
q f)(x) = (J

α+β
q f)(x).

For the expression (2.4), when f(x) = xλ, we get another expression that will be
used later:

Jαq (x
λ) =

Γq(λ + 1)

Γ(α + λ + 1)
xα+λ.

Finally, for b > 0 and a = bqn,n = 1, 2, ...,∞, we write

(2.5) [a,b]q = {bqk : 0 ≤ k ≤ n}, [0,b]q = {bqk : k ∈ N}.

3. Main results

Theorem 1. Let f and g be two functions defined on [a,b]q satisfying the condition

0 < m ≤ f(t) ≤ M < ∞, 0 < n ≤ g(t) ≤ N < ∞, t ∈ [a,b]q(3.1)

Then one has the inequality
(3.2)

|Tq(f,g; a,b)| ≤
(√

M−
√
m

)(√
N−
√
n

)√
1

b−a

∫ b
a

f(x)dqx ·
1

b−a

∫ b
a

g(x)dqx.

The following Lemma is used to prove Theorem 1:

Lemma 1. Let h and l are two functions defined on [a,b]q such that

0 < m1 ≤ h(t) ≤ M1 < ∞, 0 < n1 ≤ l(t) ≤ N1 < ∞, t ∈ [a,b]q.(3.3)

Then, we have

(3.4)∫ b
a

h2(x)dqx

∫ b
a

l2(x)dqx −
(∫ b

a

h(x)l(x)dqx

)2
≤

(√
M1
n1
−
√
m1
N1

)2 ∫ b
a

h(x)l(x)dqx

∫ b
a

l2(x)dqx

Proof. From the condition (3.8), we have

m1
√
qk ≤ h(bqk)

√
qk ≤ M1

√
qk,

and

n1
√
qk ≤ l(bqk)

√
qk ≤ N1

√
qk.



NEW INTEGRAL INEQUALITIES IN QUANTUM CALCULUS 53

Using the Klamkin-McLenaghan inequality (1.4), we obtain

n−1∑
k=0

h2(bqk)qk
n−1∑
k=0

l2(bqk)qk −

(
n−1∑
k=0

h(bqk)l(bqk)qk

)2

≤

(√
M1
n1
−
√
m1
N1

)2 n−1∑
k=0

h(bqk)l(bqk)qk
n−1∑
k=0

l2(bqk)qk.

From (2.3), we get∫ b
a

h2(x)dqx

∫ b
a

l2(x)dqx −
(∫ b

a

h(x)l(x)dqx

)2
≤

(√
M1
n1
−
√
m1
N1

)2 ∫ b
a

h(x)l(x)dqx

∫ b
a

l2(x)dqx

Lemma 1 is thus proved. �

Proof of Theorem 1:
Using the Cauchy-Schwartz inequality for double integrals, we have

|Tq(f,g; a,b)| =

∣∣∣∣∣ 12(b−a)2
∫ b
a

∫ b
a

(f(x) −f(y))(g(x) −g(y))dqxdqy

∣∣∣∣∣
≤

1

2(b−a)2

[∫ b
a

∫ b
a

(f(x) −f(y))2dqxdqy ·
∫ b
a

∫ b
a

(g(x) −g(y))2dqxdqy
]1

2

=
1

2(b−a)2

[
4

[
(b−a)

∫ b
a

f2(x)dqx−
(∫ b

a

f(x)dqx

)2]
×

[
(b−a)

∫ b
a

g2(x)dqx−
(∫ b

a

g(x)dqx

)2]]1
2

=

[
1

b−a

∫ b
a

f2(x)dqx−
(

1

b−a

∫ b
a

f(x)dqx

)2]1
2

×
[

1

b−a

∫ b
a

g2(x)dqx−
(

1

b−a

∫ b
a

g(x)dqx

)2]1
2

.(3.5)

By Lemma 1, we get

1

b−a

∫ b
a

f2(x)dqx −

(
1

b−a

∫ b
a

f(x)dqx

)2

≤
(√

M −
√
m

)2
1

b−a

∫ b
a

f(x)dqx(3.6)

and

1

b−a

∫ b
a

g2(x)dqx −

(
1

b−a

∫ b
a

g(x)dqx

)2

≤
(√

N −
√
n

)2
1

b−a

∫ b
a

g(x)dqx(3.7)



54 BRAHIM, TAF AND NEFZI

From (3.5), (3.6) and (3.7), we deduce the desired inequality (3.9).
Theorem 1 is thus proved.

Corollary 1. Let f and g be two functions defined on [0,b]q satisfying the condition

0 < m ≤ f(t) ≤ M < ∞, 0 < n ≤ g(t) ≤ N < ∞, t ∈ [0,b]q(3.8)

Then one has the inequality

(3.9) |Tq(f,g; 0,b)| ≤
1

b

(√
M −

√
m

)(√
N −

√
n

)√∫ b
0

f(x)dqx ·
∫ b

0

g(x)dqx.

Proof. By taking a = bqn in the previous theorem and by tending n to ∞ we obtain
the result. �

Theorem 2. Let f and g be two functions defined on [a,b]q satisfying the condition
(3.1). Then one has the inequality

(3.10) |Tq(f,g; a,b)| ≤
(M −m)(N −n)

4
√
nmNM

·
1

b−a

∫ b
a

f(x)dqx ·
1

b−a

∫ b
a

g(x)dqx

Lemma 2. [17] Let f and g are two functions defined on [a,b]q satisfying the
condition (3.8). Then we have

(3.11)

∫ b
a

f2(x)dqx

∫ b
a

g2(x)dqx ≤
(MN + mn)2

4nmNM

(∫ b
a

f(x)g(x)dqx

)2
Proof of Theorem 2:

Using Lemma 2, we get

(b−a)
∫ b
a

f2(x)dqx ≤
(M + m)2

4mM

(∫ b
a

f(x)dqx

)2
,

Hence
(3.12)

1

b−a

∫ b
a

f2(x)dqx−
(

1

b−a

∫ b
a

f(x)dqx

)2
≤

(M −m)2

4mM

(
1

b−a

∫ b
a

f(x)dqx

)2
.

Similarly, we have
(3.13)

1

b−a

∫ b
a

g2(x)dqx−
(

1

b−a

∫ b
a

g(x)dqx

)2
≤

(N −n)2

4nN

(
1

b−a

∫ b
a

g(x)dqx

)2
.

From (3.5), (3.12) and (3.13), we deduce the desired inequality (3.14).
Theorem 2 is thus proved.

Corollary 2. Let f and g be two functions defined on [0,b]q satisfying the condition
(3.8). Then one has the inequality

(3.14) |Tq(f,g; 0,b)| ≤
(M −m)(N −n)

4b2
√
nmNM

·
∫ b

0

f(x)dqx ·
∫ b

0

g(x)dqx

Theorem 3. Let f and g be two positive functions defined on [0,∞) satisfying the
condition

(3.15) 0 < m ≤ f(τ) ≤ M < ∞, 0 < n ≤ g(τ) ≤ N < ∞, τ ∈ [0, t], t > 0.



NEW INTEGRAL INEQUALITIES IN QUANTUM CALCULUS 55

Then for all α > 0, b > 0 we have

(3.16)

∣∣∣∣ bαΓq(α + 1)Jαq (f(b)g(b)) −Jαq f(b)Jαq g(b)
∣∣∣∣

≤
(√

M −
√
m

)(√
N −

√
n

)
bα

Γq(α + 1)

√
Jαq f(b)J

α
q g(b)

The following Lemma is used to prove Theorem 3:

Lemma 3. Let h and l be two positive functions on [0,∞) such that

(3.17) 0 < m1 ≤ h(τ) ≤ M1 < ∞, 0 < n1 ≤ l(τ) ≤ N1 < ∞, τ ∈ [0, t], t > 0.

Then for all α > 0, b > 0, we have
(3.18)

Jαq h
2(b)Jαq l

2(b) −
(
Jαq (h(b)l(b))

)2
≤

(√
M1
n1
−
√
m1
N1

)2
Jαq (h(b)l(b))J

α
q l

2(b).

Proof. Using the Klamkin-McLenaghan inequality (1.4), we obtain

∞∑
k=0

h2(bqk)(1 −qk+1)(α−1)qk
∞∑
k=0

l2(bqk)(1 −qk+1)(α−1)qk

−

(
∞∑
k=0

h(bqk)l(bqk)(1 −qk+1)(α−1)qk
)2

≤

(√
M1
n1
−
√
m1
N1

)2 ∞∑
k=0

h(bqk)l(bqk)(1 −qk+1)(α−1)qk
∞∑
k=0

l2(bqk)(1 −qk+1)(α−1)qk.

From (2.4), we obtain

Jαq h
2(b)Jαq l

2(b) −
(
Jαq (h(b)l(b))

)2
≤

(√
M1
n1
−
√
m1
N1

)2
Jαq (h(b)l(b))J

α
q l

2(b).

Lemma 3 is thus proved. �

Proof of Theorem 3:
Define

(3.19) Q(τ,ρ) = (f(τ) −f(ρ))(g(τ) −g(ρ))

Multiplying (3.19) by
(b−qτ)(α−1)(b−qρ)(α−1)

Γ2q(α)
and double integrating with respect to

τ and ρ from 0 to b, we get

1

Γ2q(α)

∫ b
0

∫ b
0

(b−qτ)(α−1)(b−qρ)(α−1)Q(τ,ρ)dqτdqρ(3.20)

= 2
bα

Γq(α + 1)
Jαq (f(b)g(b)) − 2J

α
q f(b)J

α
q g(b).



56 BRAHIM, TAF AND NEFZI

On the other hand, using the Cauchy-Schwartz inequality we get∣∣∣∣∣ 1Γ2q(α)
∫ b

0

∫ b
0

(b−qτ)(α−1)(b−qρ)(α−1)Q(τ,ρ)dqτdqρ

∣∣∣∣∣
≤

[
1

Γ2q(α)

∫ b
0

∫ b
0

(b−qτ)(α−1)(b−qρ)(α−1)(f(τ) −f(ρ))2dqτdqρ
]1

2

×
[

1

Γ2q(α)

∫ b
0

∫ b
0

(b−qτ)(α−1)(b−qρ)(α−1)(g(τ) −g(ρ))2dqτdqρ
]1

2

.

Then, ∣∣∣∣∣ 1Γ2q(α)
∫ b

0

∫ b
0

(b−qτ)(α−1)(b−qρ)(α−1)Q(τ,ρ)dqτdqρ

∣∣∣∣∣(3.21)
≤ 2

[
bα

Γq(α + 1)
Jαq f

2(b) −
(
Jαq f(b)

)2]12 ×[ bα
Γq(α + 1)

Jαq g
2(b) −

(
Jαq g(b)

)2]12
Using Lemma 3, we get

(3.22)
bα

Γq(α + 1)
Jαq f

2(b) −
(
Jαq f(b)

)2 ≤ (√M −√m)2 bα
Γq(α + 1)

(
Jαq f(b)

)
and

(3.23)
bα

Γq(α + 1)
Jαq g

2(b) −
(
Jαq g(b)

)2 ≤ (√N −√n)2 bα
Γq(α + 1)

(
Jαq g(b)

)
.

Using (3.20), (3.22) and (3.23), we deduce the desired inequality (3.16).
Theorem 3 is thus proved.

Theorem 4. Let f and g be two positive functions defined on [0,∞) satisfying the
condition (3.15). Then for all α > 0, b > 0, we have
(3.24)∣∣∣∣ bαΓq(α + 1)Jαq (f(b)g(b)) −Jαq f(b)Jαq g(b)

∣∣∣∣ ≤ (M −m)(N −n)
4
√
mMnN

Jαq f(b)J
α
q g(b).

To prove Theorem 4 we need the following result:

Lemma 4. Let h and l be two positive functions on [0,∞) such that the condition
(3.17). Then for all α > 0, b > 0, we have

(3.25) Jαq h
2(b)Jαq l

2(b) ≤
(MN + mn)2

4nmNM
(Jαq h(b)l(b))

2

Proof. Using Grueb-Reinboldt inequality (1.5), we obtain(
∞∑
k=0

h2(bqk)(1 −qk+1)α−1qk
)(

∞∑
k=0

l2(bqk)(1 −qk+1)α−1qk
)

≤
(MN + mn)2

4nmNM

(
∞∑
k=0

h(bqk)l(bqk)(1 −qk+1)α−1qk
)2

(3.26)



NEW INTEGRAL INEQUALITIES IN QUANTUM CALCULUS 57

Which implies that

(3.27)
(Jαq h

2(b))(Jαq l
2(b))

(Jαq h(b)l(b))
2

≤
(MN + mn)2

4nmNM

Lemma 4 is thus proved. �

Proof of Theorem 4:
Using Lemma 4, we get

bα

Γq(α + 1)
Jαq f

2(b) ≤
(M + m)2

4mM
(Jαq f(b))

2

thus,

(3.28)
bα

Γq(α + 1)
Jαq f

2(b) − (Jαq f(b))
2 ≤

(M −m)2

4mM
(Jαq f(b))

2

By the same, we obtain

(3.29)
bα

Γq(α + 1)
Jαq g

2(b) − (Jαq g(b))
2 ≤

(N −n)2

4nN
(Jαq g(b))

2

Using (3.20), (3.28) and (3.29), we deduce the desired inequality (3.24).
Theorem 4 is thus proved.

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1Departement of Mathematics, Faculty of Science of Tunis, Tunisia

2Department of Mathematics, Faculty SEI, UMAB University of Mostaganem, Alge-

ria

∗Corresponding author