Volume 9 (2008), Issue 4, Article 108, 7 pp.
FURTHER DEVELOPMENT OF AN OPEN PROBLEM
YU MIAO AND JUNFEN LI
C OLLEGE OF M ATHEMATICS AND I NFORMATION S CIENCE
H ENAN N ORMAL U NIVERSITY
H ENAN P ROVINCE , 453007, C HINA
yumiao728@yahoo.com.cn
junfen_li@yahoo.com.cn
Received 06 March, 2008; accepted 5 October, 2008
Communicated by N.S. Barnett
A BSTRACT. 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. I NTRODUCTION
In [4], the following inequality is found.
Theorem A. 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
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.
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
2
Y. M IAO AND J. F. L I
Open Problem 1. Let f (x) ≥ 0 be a continuous function on [0, 1] satisfying
Z 1
Z 1
(1.4)
f (t)dt ≥
tdt,
∀ x ∈ [0, 1].
x
x
Under what conditions does the inequality
Z 1
Z
α+β
(1.5)
f
(x) ≥
0
1
xα f β (x)dx
0
hold for α and β?
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 resolutions to it. Here, and in what follows, we use X to 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 1A := 1A (X) denotes the indicator function of the event A. Let At = [t, ∞).
We now consider the following generalization of Open Problem 1.
Open Problem 2. Let f (x) ≥ 0 be a continuous function on [0, ∞). Under what conditions
does the inequality
(1.6)
E(f α+β (X)) ≥ E(X α f β (X))
hold for α and β?
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 Problem 1.
For convenience, we assume that all the necessary functions in the following are integrable.
2. M AIN R ESULTS
Theorem 2.1. Let f (x) ≥ 0 be a continuous function on [0, ∞) satisfying
(2.1)
E(f β (X)1At ) ≥ E(X β 1At ),
∀ t ∈ [0, ∞).
Then
(2.2)
E(f α+β (X)) ≥ E(X α f β (X))
holds for every positive real number α and β.
Proof. By Fubini’s Theorem and (2.1), we have
Z ∞
α β
(2.3)
E(X f (X)) =
xα f β (x)p(x)dx
0
Z Z x
1 ∞
α−1
=
t dt f β (x)p(x)dx
α 0
0
Z ∞
Z ∞
1
α−1
β
=
t
f (x)p(x)dx dt
α 0
t
Z
1 ∞ α−1
=
t E(f β (X)1At )dt
α 0
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 108, 7 pp.
http://jipam.vu.edu.au/
F URTHER D EVELOPMENT OF AN O PEN P ROBLEM
3
Z
1 ∞ α−1
≥
t E(X β 1At )dt
α 0
Z ∞
Z
1 ∞ α−1
β
=
t
x p(x)dx dt
α 0
t
Z Z x
1 ∞
α−1
=
t dt xβ p(x)dx
α 0
0
Z ∞
=
xβ+α p(x)dx = E(X β+α ).
0
Using Cauchy’s inequality, we have
β
α
f α+β (X) +
X α+β ≥ X α f β (X),
α+β
α+β
(2.4)
which, by (2.3), yields
(2.5)
β
α
E(f α+β (X)) +
E(X α+β ) ≥ E(X α f β (X)) ≥ E(X β+α ).
α+β
α+β
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. Let f (x) ≥ 0 be 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 β f α (X))
holds for every positive real number α ≥ 1 and β.
Proof. By Fubini’s Theorem and (2.6), we have
Z ∞
α+1
(2.8)
E(X f (X)) =
xα+1 f (x)p(x)dx
0
Z ∞ Z x
1
α
=
t dt f (x)p(x)dx
α+1 0
0
Z ∞ Z ∞
1
α
=
t
f (x)p(x)dx dt
α+1 0
t
Z ∞ Z ∞
1
α
≥
t
xp(x)dx dt
α+1 0
t
Z x
Z ∞
1
α
=
xp(x)
t dt dx = E(X α+2 ).
α+1 0
0
Applying Cauchy’s inequality, we get
1 α
α−1 α
f (X) +
X ≥ f (X)X α−1 .
α
α
β
Multiplying both sides of (2.9) by X , we have
(2.9)
(2.10)
1
α−1
E(f α (X)X β ) +
E(X α+β ) ≥ E(f (X)X α+β−1 ) ≥ E(X α+β ),
α
α
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 108, 7 pp.
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4
Y. M IAO AND J. F. L I
which implies
E(f α (X)X β ) ≥ E(X α+β ).
(2.11)
The remainder of the proof is similar to that of Theorem 2.1.
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].
Lemma 2.3 ([2]). Let f : [a, b] → [0, ∞) be a continuous function and g : [a, b] → [0, ∞) be
non-decreasing, differentiable on (a, b) satisfying
Z b
Z b
f (t)dt ≥
g(t)dt.
x
x
Then
Z
b
β
Z
f (t)dt ≥
x
b
g β (t)dt
x
holds for β > 1.
0
Lemma 2.4. Let f (x) ≥ 0 be a continuous function on [0, ∞) with f (x) ≥ 0 on (0, ∞) and
satisfying
E(f (X)1At ) ≥ E(X1At ),
(2.12)
∀ t ∈ [0, ∞).
Then
E(f β (X)1At )) ≥ E(X β 1At )
(2.13)
holds for every positive real number β ≥ 1.
Proof. The proof is a direct extension of Theorem 3 in [2].
0
Theorem 2.5. Let f (x) ≥ 0 be a continuous function on [0, ∞) with f (x) ≥ 0 on (0, ∞) and
satisfying
E(f (X)1At ) ≥ E(X1At ),
(2.14)
∀ t ∈ [0, ∞).
In addition, for every positive real number α > 0, β > 2 satisfying
lim f α (x)xβ−1 E[(f (X) − X)1Ax ] = 0,
(2.15)
x→∞
and
lim f α (x)xE[(f β−1 (X) − X β−1 )1Ax ] = 0,
(2.16)
x→∞
then
E(f α+β (X)) ≥ E(X β f α (X)).
(2.17)
Proof. It is obvious that
(f (x) − x)(f β−1 (x) − xβ−1 ) ≥ 0,
which implies that
(2.18)
f β+α (x) ≥ xβ−1 f 1+α (x) + xf β−1+α (x) − xβ f α (x).
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 108, 7 pp.
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F URTHER D EVELOPMENT OF AN O PEN P ROBLEM
5
Integrating by parts and using (2.14) and (2.15), we have
E(f α (X)X β−1 (f (X) − X))
Z ∞
=
f α (x)xβ−1 (f (x) − x)p(x)dx
0
Z ∞
Z ∞
α
β−1
=−
f (x)x d
(f (t) − t)p(t)dt dx
0
x
Z ∞
∞
α
β−1
= −f (x)x
(f (t) − t)p(t)dt 0
x
Z ∞
Z ∞
0
α−1
β−1
β−2 α
+
αf (x)f
(x)x
+ (β − 1)x f (x)
(f (t) − t)p(t)dt dx
0
x
Z ∞
Z ∞
0
α−1
β−1
β−2 α
=
αf (x)f
(x)x
+ (β − 1)x f (x)
(f (t) − t)p(t)dt dx
0
x
Z ∞
0
=
αf (x)f α−1 (x)xβ−1 + (β − 1)xβ−2 f α (x) E ((f (X) − X)1Ax ) dx ≥ 0,
0
which yields
E(f α+1 (X)X β−1 ) ≥ E(f α (X)X β ).
(2.19)
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 ∞
f α (x)x(f β−1 (x) − xβ−1 )p(x)dx
=
0
Z ∞
Z ∞
β−1
β−1
α
(f (t) − t )p(t)dt dx
=−
f (x)xd
x
0
Z ∞
∞
β−1
β−1
α
(f (t) − t )p(t)dt = − f (x)x
0
x
Z ∞
Z ∞
0
β−1
β−1
α−1
α
+
(αxf (x)f
(x) + f (x))
(f (t) − t )p(t)dt dx
x
0
Z ∞
0
=
(αxf (x)f α−1 (x) + f α (x))E (f β−1 (X) − X β−1 )1Ax dx ≥ 0,
0
which yields
E(f α+β−1 (X)X) ≥ E(f α (X)X β ).
(2.20)
From (2.18)-(2.20), inequality (2.17) holds.
3. F URTHER D ISCUSSION
R∞
Let g(x) ≥ 0, 0 < 0 g(x)dx < ∞. If p(x) := R ∞g(x)
, then it is easy to check that p(x) is
0 g(x)dx
a probability density function on the interval [0, ∞). Thus we have the following:
Theorem 3.1. Let f (x) ≥ 0 be a continuous function on [0, ∞) satisfying
Z ∞
Z ∞
β
(3.1)
f (t)g(t)dt ≥
tβ g(t)dt
∀ t ∈ [0, ∞).
t
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 108, 7 pp.
t
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6
Y. M IAO AND J. F. L I
Then
Z
∞
f
(3.2)
α+β
Z
∞
xβ f α (x)g(x)dx
(x)g(x)dx ≥
0
0
holds for every positive real number α and β.
Theorem 3.2. Let f (x) ≥ 0 be a continuous function on [0, ∞) satisfying
Z ∞
Z ∞
(3.3)
f (t)g(t)dt ≥
tg(t)dt,
∀ t ∈ [0, ∞).
t
t
Then
Z
∞
f
(3.4)
α+β
Z
(x)g(x)dx ≥
0
∞
xβ f α (x)g(x)dx
0
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. Let f (x) ≥ 0, g(x) ≥ 0 be two continuous functions on [0, ∞) satisfying
Z ∞
Z ∞
β
(3.5)
f (t)dt ≥
g β (t)dt
∀ t ∈ [0, ∞).
t
t
0
Furthermore, for any positive real numbers α and β, let g(x) be differentiable with [g α (x)] ≥ 0
and g(0) = 0, then
Z ∞
Z ∞
α+β
(3.6)
f
(x)dx ≥
g β (x)f α (x)dx.
0
0
α
Proof. Denoting the derivative of g (x) by G(x), we obtain,
Z ∞
Z ∞ Z x
α
β
(3.7)
g (x)f (x)dx =
G(t)dt f β (x)dx
0
0
0
Z ∞
Z ∞
β
=
G(t)
f (x)dx dt
0
t
Z ∞
Z ∞
β
≥
G(t)
g (x)dx dt
t
0
Z x
Z ∞
β
=
g (x)
G(t)dt dx
0
0
Z ∞
=
g β+α (x)dx.
0
Using Cauchy’s inequality, we have
β
α
(3.8)
f α+β (x) +
g α+β (x) ≥ g α (x)f β (x),
α+β
α+β
which, by (3.7), yields
Z ∞
Z ∞
Z ∞
β
α
α+β
α+β
(3.9)
f
(x)dx +
g
(x)dx ≥
g α (x)f β (x)dx
α+β 0
α+β 0
Z0 ∞
≥
g β+α (x)dx.
0
The desired result then follows.
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 108, 7 pp.
http://jipam.vu.edu.au/
F URTHER D EVELOPMENT OF AN O PEN P ROBLEM
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A similar proof yields the following:
Theorem 3.4. Let f (x) ≥ 0, g(x) ≥ 0 be two continuous functions on [0, ∞) satisfying
Z ∞
Z ∞
(3.10)
f (t)dt ≥
g(t)dt,
∀ t ∈ [0, ∞).
t
t
Furthermore, for every pair of positive real numbers satisfying “α ≥ 1 and β > 0”, let g(x) be
0
differentiable with [g α (x)] ≥ 0 and g(0) = 0, then
Z ∞
Z ∞
α+β
(3.11)
f
(x)dx ≥
g β (x)f α (x)dx.
0
0
Additionally, for every positive real number “α > 0, β > 2” satisfying (2.15) and (2.16),
inequality (3.11) holds.
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edu.au/article.php?sid=885].
[2] S.S. DRAGOMIR AND Q.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,
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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.
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J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 108, 7 pp.
http://jipam.vu.edu.au/