FURTHER DEVELOPMENT OF AN OPEN PROBLEM

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Further Development of an Open Problem Yu Miao and Junfen Li vol. 9, iss. 4, art. 108, 2008

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FURTHER DEVELOPMENT OF AN OPEN PROBLEM

YU MIAO AND JUNFEN LI

College of Mathematics and Information Science Henan Normal University

Henan Province, 453007, China

EMail:yumiao728@yahoo.com.cn junfen_li@yahoo.com.cn

Received: 06 March, 2008

Accepted: 5 October, 2008

Communicated by: N.S. Barnett 2000 AMS Sub. Class.: 26D15; 60E15.

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

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.].

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

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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:

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?

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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 Problem1 and provide some resolutions to it. Here, and in what follows, we useXto denote a non- negative random variable (r.v.) on [0,∞) with probability density function p(x).

E(X) denotes the mathematical expectation of X and 1_A := 1_A(X) denotes the indicator function of the eventA. LetA_t = [t,∞). We now consider the following generalization of Open Problem1.

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. Let X possess a uniform distribution on the support interval [0,1], i.e., the probability density function of X is equal to 1, x ∈ [0,1] and zero elsewhere, then the above open problem becomes Open Problem1.

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

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2. Main Results

Theorem 2.1. Letf(x)≥0be a continuous function on[0,∞)satisfying (2.1) E(f^β(X)1_A_t)≥E(X^β1_A_t), ∀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)1_A_t)dt

≥1 α

Z ∞

0

t^α−1E(X^β1_A_t)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

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= 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 that X 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)1_A_t)≥E(X1_A_t), ∀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

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= 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 Theorem2.1.

Remark 3. If we assume that X 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].

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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. Let f(x) ≥ 0be a continuous function on[0,∞)with f⁰(x) ≥ 0 on (0,∞)and satisfying

(2.12) E(f(X)1_A_t)≥E(X1_A_t), ∀t∈[0,∞).

Then

(2.13) E(f^β(X)1At))≥E(X^β1At) 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_A_t)≥E(X1_A_t), ∀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)1Ax] = 0,

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and

(2.16) lim

x→∞f^α(x)xE[(f^β−1(X)−X^β−1)1_A_x] = 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_A_x) dx≥0,

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which yields

(2.19) E(f^α+1(X)X^β−1)≥E(f^α(X)X^β).

Furthermore, by Lemma2.4and 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

∞ 0

+ Z ∞

0

(αxf⁰(x)f^α−1(x) +f^α(x)) Z ∞

x

dx

= Z ∞

0

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

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.

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3. Further Discussion

Letg(x) ≥ 0, 0 < R∞

0 g(x)dx < ∞. If p(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 ≥ 1and 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.

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Theorem 3.3. Let f(x) ≥ 0, g(x) ≥ 0 be 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 numbersα andβ, let g(x)be differentiable with [g^α(x)]⁰ ≥0andg(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),

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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. Let f(x) ≥ 0, g(x) ≥ 0 be 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.

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References

[1] K. BOUKERRIOUA AND A. GUEZANE-LAKOUD, On an open question re- garding 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. LIANDJ.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. DAT AND D.A. TUAN, Note on an integral inequality, J. Inequal. Pure Appl. Math., 7(4) (2006), Art. 120. [ONLINE:

http://jipam.vu.edu.au/article.php?sid=737].