in the paper, Note on an integral inequality, J

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ON SOME q-INTEGRAL INEQUALITIES

KAMEL BRAHIM

INSTITUTPRÉPARATOIRE AUXÉTUDES D’INGÉNIEUR DETUNIS

Received 24 June, 2008; accepted 10 November, 2008 Communicated by S.S. Dragomir

ABSTRACT. In this paper, we provide aq-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. Letf(x)≥0be a continuous function on[0,1]satisfying

(1.1)

Z 1

x

f(t)dt≥ Z 1

x

tdt, ∀x∈[0,1].

Then the inequalities

(1.2)

Z 1

0

f^α+1(x)dx≥ Z 1

0

x^αf(x)dx

and

(1.3)

Z 1

0

f^α+1(x)dx≥ Z 1

0

xf^α(x)dx

hold for every positive real numberα >0.

Then, they proposed the following open problem: Under what condition does the inequality (1.4)

Z 1

0

f^α+β(x)dx≥ Z 1

0

x^βf^α(x)dx hold forαandβ?

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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 aq-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 aq-analogue of the inequalities given in [9] and [3].

2. NOTATIONS AND PRELIMINARIES

Throughout this paper, we will fixq ∈(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 fora ∈C,

[a]q = 1−q^a 1−q . Theq-derivativeD_qf of a functionf is given by

(2.1) (D_qf)(x) = f(x)−f(qx)

(1−q)x , if x6= 0, (D_qf)(0) =f⁰(0),providedf⁰(0)exists.

Theq-Jackson integral from0toais defined by (see [5]) (2.2)

Z a

0

f(x)d_qx= (1−q)a

∞

X

n=0

f(aqⁿ)qⁿ, provided the sum converges absolutely.

Theq-Jackson integral in a generic interval[a, b]is given by (see [5]) (2.3)

Z b

a

f(x)dqx= Z b

0

f(x)dqx− Z a

0

f(x)dqx.

We recall that for any functionf, we have (see [6])

(2.4) Dq

Z x

a

f(t)dqt

=f(x).

IfF is any antiq-derivative of the functionf, namelyD_qF =f, continuous atx= 0, then (2.5)

Z a

0

f(x)d_qx=F(a)−F(0).

Aq-analogue of the integration by parts formula is given by (2.6)

Z b

a

f(x)(D_qg(x))d_qx=f(a)g(a)−f(b)g(b)− Z b

a

(D_qf(x))g(qx)d_qx.

Finally, we denote

[0,1]_q ={q^k : k = 0,1,2, . . . ,∞}.

3. MAINRESULTS

Let us begin with the following useful result:

Lemma 3.1 ([9] General Cauchy inequality). Let αandβ be positive real numbers satisfying α+β = 1.Then for all positive real numbersxandy, we always have

(3.1) αx+βy ≥x^αy^β.

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Theorem 3.2. Letf be a nonnegative function defined on[0,1]_qsatisfying

(3.2)

Z 1

x

f^β(t)d_qt≥ Z 1

x

t^βd_qt, ∀x∈[0,1]_q. Then the inequality

(3.3)

Z 1

0

f^α+β(x)d_qx≥ Z 1

0

x^αf^β(x)d_qx,

holds for all positive real numbersα >0andβ >0.

To prove Theorem 3.2, we need the following lemma.

Lemma 3.3. Under the conditions of Theorem 3.2, we have (3.4)

Z 1

0

x^αf^β(x)d_qx≥ 1 [α+β+ 1]_q. Proof. By using aq-integration by parts, we obtain

Z 1

0

x^α−1 Z 1

x

f^β(t)d_qt

d_qx= 1 [α]q

x^α

Z 1

x

f^β(t)d_qt ^x=1

x=0

+ q^α [α]q

Z 1

0

x^αf^β(x)d_qx

= q^α [α]_q

Z 1

0

x^αf^β(x)d_qx,

which yields (3.5)

Z 1

0

x^αf^β(x)d_qx= [α]_q q^α

Z 1

0

x^α−1 Z 1

x

f^β(t)d_qt

d_qx.

On the other hand, from condition (3.2), we get Z 1

0

x^α−1 Z 1

x

f^β(t)dqt

dqx≥ Z 1

0

x^α−1 Z 1

x

t^βdqt

dqx

= 1

[β+ 1]_q Z 1

0

(x^α−1 −x^α+β)d_qx

= q^α

[α]_q[α+β+ 1]_q. Therefore, from (3.5), we obtain

(3.6)

Z 1

0

x^αf^β(x)d_qx≥ 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

(3.7) β

α+βf^α+β(x) + α

α+βx^α+β ≥x^αf^β(x), which gives

(3.8) β

Z 1

0

f^α+β(x)dqx+α Z 1

0

x^α+βdqx≥(α+β) Z 1

0

x^αf^β(x)dqx.

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Moreover, by using Lemma 3.3, we get (α+β)

Z 1

0

x^αf^β(x)d_qx=α Z 1

0

x^αf^β(x)d_qx+β Z 1

0

x^αf^β(x)d_qx (3.9)

≥ α

[α+β+ 1]_q +β Z 1

0

x^αf^β(x)d_qx.

Then, from relation (3.8), we obtain

(3.10) β

Z 1

0

f^α+β(x)d_qx+ α

[α+β+ 1]_q ≥ α

[α+β+ 1]_q +β Z 1

0

x^αf^β(x)d_qx,

which completes the proof.

Takingβ = 1in Theorem 3.2, we obtain

Corollary 3.4. Letf be a nonnegative function defined on[0,1]_qsatisfying

(3.11)

Z 1

x

f(t)dqt≥ Z 1

x

tdqt, ∀x∈[0,1]q. Then the inequality

(3.12)

Z 1

0

f^α+1(x)d_qx≥ Z 1

0

x^αf(x)d_qx

holds for every positive real numberα >0.

(3.13)

Z 1

x

f(t)d_qt≥ Z 1

x

td_qt, ∀x∈[0,1]_q. Then the inequality

(3.14)

Z 1

0

f^α+1(x)d_qx≥ Z 1

0

xf^α(x)d_qx

holds for every positive real numberα >0.

Proof. We have

(3.15) ∀x∈[0,1]_q, (f^α(x)−x^α)(f(x)−x)≥0, so

(3.16) f^α+1(x) +x^α+1 ≥x^αf(x) +xf^α(x).

By integrating with some simple calculations we deduce that (3.17)

Z 1

0

f^α+1(x)d_qx+ 1 [α+ 2]_q ≥

Z 1

0

x^αf(x)d_qx+ Z 1

0

xf^α(x)d_qx.

Then, from Lemma 3.3 forβ = 1, the result follows.

(3.18)

Z 1

x

f(t)d_qt≥ Z 1

x

td_qt, ∀x∈[0,1]_q. Then the inequality

(3.19)

Z 1

0

f^α+β(x)dqx≥ Z 1

0

x^αf^β(x)dqx

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holds for all real numbersα >0andβ ≥1.

Lemma 3.7. Under the conditions of Theorem 3.6, we have (3.20)

Z 1

0

x^αf^β(x)dqx≥ 1 [α+β+ 1]_q for all real numbersα >0andβ ≥1.

Proof. Using Lemma 3.1, we obtain

(3.21) 1

βf^β(x) + β−1

β x^β ≥x^β−1f(x), which implies

(3.22)

Z 1

0

x^αf^β(x)d_qx+ (β−1) Z 1

0

x^α+βd_qx≥β Z 1

0

x^α+β−1f(x)d_qx.

Therefore, from Lemma 3.3, we get (3.23)

Z 1

0

x^αf^β(x)dqx+ β−1

[α+β+ 1]_q ≥ β [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

(3.24) β

α+βf^α+β(x) + α

α+βx^α+β ≥x^αf^β(x), which implies

(3.25) β

Z 1

0

f^α+β(x)d_qx+ α

[α+β+ 1]_q ≥(α+β) Z 1

0

x^αf^β(x)d_qx.

Then, from Lemma 3.7, we obtain

(3.26) β

Z 1

0

f^α+β(x)d_qx+ α

[α+β+ 1]_q ≥ α

[α+β+ 1]_q +β Z 1

0

x^αf^β(x)d_qx,

which completes the proof.

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