FURTHER DEVELOPMENT OF AN OPEN PROBLEM

FURTHER DEVELOPMENT OF AN OPEN PROBLEM

YU MIAO AND JUNFEN LI

COLLEGE OFMATHEMATICS ANDINFORMATIONSCIENCE

HENANNORMALUNIVERSITY

HENANPROVINCE, 453007, CHINA

yumiao728@yahoo.com.cn junfen_li@yahoo.com.cn

Received 06 March, 2008; accepted 5 October, 2008 Communicated by N.S. Barnett

ABSTRACT. In this paper, we generalize an open problem posed by Ngô et al. in [Notes on an Integral Inequality, JIPAM, 7(4) (2006), Art.120] and give some answers which extend the results of Boukerrioua-Guezane-Lakoud [On an open question regarding an integral inequality, JIPAM, 8(3) (2007), Art. 77.] and Liu-Li-Dong [On an open problem concerning an integral inequality, JIPAM, 8 (3) (2007), Art. 74.].

Key words and phrases: Integral inequality, Cauchy inequality, Open problem.

2000 Mathematics Subject Classification. 26D15; 60E15.

1. INTRODUCTION

In [4], the following inequality is found.

Theorem A. 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

(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.

The authors next proposed the following open problem:

The authors are indebted to the referees and the editor for their many helpful and valuable comments.

069-08

Open Problem 1. Letf(x)≥0be a continuous function on[0,1]satisfying

(1.4)

Z 1

x

f(t)dt≥ Z 1

x

tdt, ∀x∈[0,1].

Under what conditions does the inequality

(1.5)

Z 1

0

f^α+β(x)≥ Z 1

0

x^αf^β(x)dx hold forαandβ?

Several answers and extension results have been given to this open problem [1, 3]. In the present paper, we obtain a generalization of the above Open Problem 1 and provide some res- olutions to it. Here, and in what follows, we useX to denote a non-negative random variable (r.v.) on[0,∞)with probability density functionp(x). E(X)denotes the mathematical expec- tation ofX and1_A := 1_A(X)denotes the indicator function of the eventA. LetA_t = [t,∞).

We now consider the following generalization of Open Problem 1.

Open Problem 2. Letf(x) ≥ 0be a continuous function on[0,∞). Under what conditions does the inequality

(1.6) E(f^α+β(X))≥E(X^αf^β(X))

hold forαandβ?

Remark 1. LetX possess a uniform distribution on the support interval[0,1], i.e., the proba- bility density function ofX is equal to1, x ∈ [0,1] and zero elsewhere, then the above open problem becomes Open Problem 1.

For convenience, we assume that all the necessary functions in the following are integrable.

2. MAINRESULTS

Theorem 2.1. Letf(x)≥0be a continuous function on[0,∞)satisfying (2.1) E(f^β(X)1_At)≥E(X^β1_At), ∀t ∈[0,∞).

Then

(2.2) E(f^α+β(X))≥E(X^αf^β(X))

holds for every positive real numberαandβ.

Proof. By Fubini’s Theorem and (2.1), we have

E(X^αf^β(X)) = Z ∞

0

x^αf^β(x)p(x)dx (2.3)

= 1 α

Z ∞

0

Z x

0

t^α−1dt

f^β(x)p(x)dx

= 1 α

Z ∞

0

t^α−1 Z ∞

t

f^β(x)p(x)dx

dt

= 1 α

Z ∞

0

t^α−1E(f^β(X)1At)dt

≥1 α

Z ∞

0

t^α−1E(X^β1_At)dt

= 1 α

Z ∞

0

t^α−1 Z ∞

t

x^βp(x)dx

dt

= 1 α

Z ∞

0

Z x

0

t^α−1dt

x^βp(x)dx

= Z ∞

0

x^β+αp(x)dx=E(X^β+α).

Using Cauchy’s inequality, we have

(2.4) β

α+βf^α+β(X) + α

α+βX^α+β ≥X^αf^β(X), which, by (2.3), yields

(2.5) β

α+βE(f^α+β(X)) + α

α+βE(X^α+β)≥E(X^αf^β(X))≥E(X^β+α).

Remark 2. If we assume thatX possesses a uniform distribution on the support interval[0,1]

(or[0, b]), then the above theorem is Theorem 2.1 (or Theorem 2.4) of Liu-Li-Dong in [3].

Theorem 2.2. Letf(x)≥0be a continuous function on[0,∞)satisfying (2.6) E(f(X)1At)≥E(X1At), ∀t∈[0,∞).

Then

(2.7) E(f^α+β(X))≥E(X^βf^α(X))

holds for every positive real numberα≥1andβ.

Proof. By Fubini’s Theorem and (2.6), we have

E(X^α+1f(X)) = Z ∞

0

x^α+1f(x)p(x)dx (2.8)

= 1

α+ 1 Z ∞

0

Z x

0

t^αdt

f(x)p(x)dx

= 1

α+ 1 Z ∞

0

t^α Z ∞

t

f(x)p(x)dx

dt

≥ 1 α+ 1

Z ∞

0

t^α Z ∞

t

xp(x)dx

dt

= 1

α+ 1 Z ∞

0

xp(x) Z x

0

t^αdt

dx=E(X^α+2).

Applying Cauchy’s inequality, we get

(2.9) 1

αf^α(X) + α−1

α X^α ≥f(X)X^α−1. Multiplying both sides of (2.9) byX^β, we have

(2.10) 1

αE(f^α(X)X^β) + α−1

α E(X^α+β)≥E(f(X)X^α+β−1)≥E(X^α+β),

which implies

(2.11) E(f^α(X)X^β)≥E(X^α+β).

The remainder of the proof is similar to that of Theorem 2.1.

Remark 3. If we assume thatX possesses a uniform distribution on the support interval[0,1]

then the above theorem is Theorem 2.3 of Boukerrioua and Guezane-Lakoud in [1].

Next we consider the case “α >0, β >2” by using the ideas of Dragomir-Ngô in [2].

Lemma 2.3 ([2]). Letf : [a, b] → [0,∞)be a continuous function andg : [a, b] → [0,∞)be non-decreasing, differentiable on(a, b)satisfying

Z b

x

f(t)dt ≥ Z b

x

g(t)dt.

Then

Z b

x

f^β(t)dt ≥ Z b

x

g^β(t)dt holds forβ >1.

Lemma 2.4. Letf(x) ≥ 0be a continuous function on[0,∞)withf⁰(x) ≥ 0on(0,∞)and satisfying

(2.12) E(f(X)1_At)≥E(X1_At), ∀t∈[0,∞).

Then

(2.13) E(f^β(X)1_At))≥E(X^β1_At) holds for every positive real numberβ ≥1.

Proof. The proof is a direct extension of Theorem 3 in [2].

Theorem 2.5. Letf(x)≥0be a continuous function on[0,∞)withf⁰(x)≥0on(0,∞)and satisfying

(2.14) E(f(X)1_At)≥E(X1_At), ∀t∈[0,∞).

In addition, for every positive real numberα >0, β >2satisfying

(2.15) lim

x→∞f^α(x)x^β−1E[(f(X)−X)1_Ax] = 0, and

(2.16) lim

x→∞f^α(x)xE[(f^β−1(X)−X^β−1)1Ax] = 0, then

(2.17) E(f^α+β(X))≥E(X^βf^α(X)).

Proof. It is obvious that

(f(x)−x)(f^β−1(x)−x^β−1)≥0, which implies that

(2.18) f^β+α(x)≥x^β−1f^1+α(x) +xf^β−1+α(x)−x^βf^α(x).

Integrating by parts and using (2.14) and (2.15), we have E(f^α(X)X^β−1(f(X)−X))

= Z ∞

0

f^α(x)x^β−1(f(x)−x)p(x)dx

=− Z ∞

0

f^α(x)x^β−1d Z ∞

x

(f(t)−t)p(t)dt

dx

=−f^α(x)x^β−1 Z ∞

x

(f(t)−t)p(t)dt

∞ 0

+ Z ∞

0

αf⁰(x)f^α−1(x)x^β−1+ (β−1)x^β−2f^α(x)Z ∞ x

(f(t)−t)p(t)dt

dx

= Z ∞

0

αf⁰(x)f^α−1(x)x^β−1+ (β−1)x^β−2f^α(x)Z ∞ x

(f(t)−t)p(t)dt

dx

= Z ∞

0

αf⁰(x)f^α−1(x)x^β−1+ (β−1)x^β−2f^α(x)

E((f(X)−X)1_Ax) dx≥0, which yields

(2.19) E(f^α+1(X)X^β−1)≥E(f^α(X)X^β).

Furthermore, by Lemma 2.4 and the condition (2.16), we have E(f^α(X)X(f^β−1(X)−X^β−1))

= Z ∞

0

f^α(x)x(f^β−1(x)−x^β−1)p(x)dx

=− Z ∞

0

f^α(x)xd Z ∞

x

(f^β−1(t)−t^β−1)p(t)dt

dx

=−f^α(x)x Z ∞

x

(f^β−1(t)−t^β−1)p(t)dt

∞ 0

+ Z ∞

0

(αxf⁰(x)f^α−1(x) +f^α(x)) Z ∞

x

(f^β−1(t)−t^β−1)p(t)dt

dx

= Z ∞

0

(αxf⁰(x)f^α−1(x) +f^α(x))E (f^β−1(X)−X^β−1)1_Ax

dx≥0, which yields

(2.20) E(f^α+β−1(X)X)≥E(f^α(X)X^β).

From (2.18)-(2.20), inequality (2.17) holds.

3. FURTHERDISCUSSION

Letg(x)≥0,0<R∞

0 g(x)dx <∞. Ifp(x) := ^R∞g(x)

0 g(x)dx, then it is easy to check thatp(x)is a probability density function on the interval[0,∞). Thus we have the following:

Theorem 3.1. Letf(x)≥0be a continuous function on[0,∞)satisfying (3.1)

Z ∞

t

f^β(t)g(t)dt≥ Z ∞

t

t^βg(t)dt ∀t∈[0,∞).

Then (3.2)

Z ∞

0

f^α+β(x)g(x)dx≥ Z ∞

0

x^βf^α(x)g(x)dx holds for every positive real numberαandβ.

Theorem 3.2. Letf(x)≥0be a continuous function on[0,∞)satisfying (3.3)

Z ∞

t

f(t)g(t)dt≥ Z ∞

t

tg(t)dt, ∀t ∈[0,∞).

Then (3.4)

Z ∞

0

f^α+β(x)g(x)dx≥ Z ∞

0

x^βf^α(x)g(x)dx

holds for every pair of positive real numbers “α ≥ 1 and β > 0”. Furthermore, for every positive real number “α >0, β >2” satisfying (2.15) and (2.16), the inequality (3.4) holds.

Two more general results follow.

Theorem 3.3. Letf(x)≥0,g(x)≥0be two continuous functions on[0,∞)satisfying (3.5)

Z ∞

t

f^β(t)dt≥ Z ∞

t

g^β(t)dt ∀t ∈[0,∞).

Furthermore, for any positive real numbersαandβ, letg(x)be differentiable with[g^α(x)]⁰ ≥0 andg(0) = 0, then

(3.6)

Z ∞

0

f^α+β(x)dx≥ Z ∞

0

g^β(x)f^α(x)dx.

Proof. Denoting the derivative ofg^α(x)byG(x), we obtain, Z ∞

0

g^α(x)f^β(x)dx= Z ∞

0

Z x

0

G(t)dt

f^β(x)dx (3.7)

= Z ∞

0

G(t) Z ∞

t

f^β(x)dx

dt

≥ Z ∞

0

G(t) Z ∞

t

g^β(x)dx

dt

= Z ∞

0

g^β(x) Z x

0

G(t)dt

dx

= Z ∞

0

g^β+α(x)dx.

Using Cauchy’s inequality, we have

(3.8) β

α+βf^α+β(x) + α

α+βg^α+β(x)≥g^α(x)f^β(x), which, by (3.7), yields

β α+β

Z ∞

0

f^α+β(x)dx+ α α+β

Z ∞

0

g^α+β(x)dx≥ Z ∞

0

g^α(x)f^β(x)dx (3.9)

≥ Z ∞

0

g^β+α(x)dx.

The desired result then follows.

A similar proof yields the following:

Theorem 3.4. Letf(x)≥0,g(x)≥0be two continuous functions on[0,∞)satisfying (3.10)

Z ∞

t

f(t)dt≥ Z ∞

t

g(t)dt, ∀t∈[0,∞).

Furthermore, for every pair of positive real numbers satisfying “α≥1andβ >0”, letg(x)be differentiable with[g^α(x)]⁰ ≥0andg(0) = 0, then

(3.11)

Z ∞

0

f^α+β(x)dx≥ Z ∞

0

g^β(x)f^α(x)dx.

Additionally, for every positive real number “α > 0, β > 2” satisfying (2.15) and (2.16), inequality (3.11) holds.

REFERENCES

[1] K. BOUKERRIOUA AND A. GUEZANE-LAKOUD, On an open question regarding an integral inequality, J. Inequal. Pure Appl. Math., 8(3) (2007), Art. 77. [ONLINE:http://jipam.vu.

edu.au/article.php?sid=885].

[2] S.S. DRAGOMIRANDQ.A. NGÔ, On an integral inequality, RGMIA Research Report Collection, 11(1) (2008), Art. 13. [ONLINE:http://www.staff.vu.edu.au/RGMIA/v11n1.asp].

[3] W.J. LIU, C.C. LI AND J.W. DONG, On an open problem concerning an integral inequality, J. Inequal. Pure Appl. Math., 8(3) (2007), Art. 74. [ONLINE:http://jipam.vu.edu.au/

article.php?sid=882].

[4] Q.A. NGÔ, D.D. THANG, T.T. DATANDD.A. TUAN, Note on an integral inequality, J. Inequal.

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