THREE MAPPINGS RELATED TO
CHEBYSHEV-TYPE INEQUALITIES
LIANG-CHENG WANG
School of Mathematics Science
Chongqing Institute of Technology
No.4 of Xingsheng Lu, YangjiaPing 400050
Chongqing City, P.R. of China.
EMail: wlc@cqit.edu.cn
Mappings Related to
Chebyshev-type Inequalities
Liang-Cheng Wang, Li-Hong Liu
and Xiu-Fen Ma
vol. 9, iss. 4, art. 113, 2008
Title Page
LI-HONG LIU AND XIU-FEN MA
Chongqing Institute of Technology
P.R. of China.
EMail: llh-19831017@cqit.edu.cn
maxiufen86@cqit.edu.cn
Contents
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Received:
22 May, 2008
J
Accepted:
15 October, 2008
Page 1 of 23
Communicated by:
F. Qi
2000 AMS Sub. Class.:
26D15.
Key words:
Chebyshev-type inequality, Monotonicity, Refinement.
Abstract:
In this paper, by the Chebyshev-type inequalities we define three mappings, investigate their main properties, give some refinements for Chebyshev-type inequalities, obtain some applications.
Acknowledgements:
The first author is partially supported by the Key Research Foundation of the
Chongqing Institute of Technology under Grant 2004ZD94.
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Contents
1
Introduction
3
2
Main Results
8
3
Proof of Theorems
11
4
Applications
17
Mappings Related to
Chebyshev-type Inequalities
Liang-Cheng Wang, Li-Hong Liu
and Xiu-Fen Ma
vol. 9, iss. 4, art. 113, 2008
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1.
Introduction
Let n(≥ 2) be a given positive integer, A = (a1 , a2 , . . . , an ) and B = (b1 , b2 , . . . , bn )
be known as sequences of real numbers. Also, let pi > 0 and qi > 0 (i = 1, 2, . . . , n),
Pj = p1 + p2 + · · · + pj and Qj = q1 + q2 + · · · + qj (j = 1, 2, . . . , n).
If A and B are both increasing or both decreasing, then
! n !
n
n
X
X
X
(1.1)
ai
bi ≤ n
ai b i .
i=1
i=1
i=1
Mappings Related to
Chebyshev-type Inequalities
Liang-Cheng Wang, Li-Hong Liu
and Xiu-Fen Ma
vol. 9, iss. 4, art. 113, 2008
If one of the sequences A or B is increasing and the other decreasing, then the
inequality (1.1) is reversed.
The inequality (1.1) is called the Chebyshev’s inequality, see [1, 2].
For A and B both increasing or both decreasing, Behdzet in [3] extended inequality (1.1) to
! n
!
! n
!
n
n
X
X
X
X
(1.2)
p i ai
qi b i +
q i ai
p i bi
i=1
i=1
i=1
i=1
n
X
≤ Pn
i=1
i=1
i=1
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Page 3 of 23
qi ai bi + Qn
n
X
p i ai b i .
i=1
If one of the sequences A or B is increasing and the other decreasing, then the
inequality (1.2) is reversed.
For pi = qi , i = 1, 2, . . . , n, the inequality (1.2) reduces to
! n
!
n
n
X
X
X
(1.3)
p i ai
p i b i ≤ Pn
p i ai b i ,
i=1
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where, A and B are both increasing or both decreasing. If one of the sequences A or
B is increasing and the other decreasing, then the inequality (1.3) is reversed.
Let r, s : [a, b] → R be integrable functions, either both increasing or both decreasing. Furthermore, let p, q : [a, b] → [0, +∞) be the integrable functions. Then
Z
b
b
Z
Z
b
b
Z
p(t)r(t)dt
(1.4)
q(t)s(t)dt +
q(t)r(t)dt
p(t)s(t)dt
a
a
a
Z b
Z b
Z b
Z b
≤
p(t)dt
q(t)r(t)s(t)dt +
q(t)dt
p(t)r(t)s(t)dt.
a
a
a
a
a
If one of the functions r or s is increasing and the other decreasing, then the inequality (1.4) is reversed.
When p(t) = q(t), t ∈ [a, b], the inequality (1.4) reduces to
Z b
Z b
Z b
Z b
(1.5)
p(t)r(t)dt
p(t)s(t)dt ≤
p(t)dt
p(t)r(t)s(t)dt,
a
a
a
a
where r and s are both increasing or both decreasing. If one of the functions r or s
is increasing and the other decreasing, then the inequality (1.5) is reversed.
Inequalities (1.4) and (1.5) are the integral forms of inequalities (1.2) and (1.3),
respectively (see [1, 2]).
The results from other inequalities connected with (1.1) to (1.5) can be seen in
[1], [3] – [8] and [2, pp. 61–65].
e by c : N+ ×N+ →R,
We define three mappings c, C and C
(1.6) c(k, n; pi , qi ) = Pk
k
X
i=1
qi ai bi + Qk
k
X
i=1
p i ai b i
Mappings Related to
Chebyshev-type Inequalities
Liang-Cheng Wang, Li-Hong Liu
and Xiu-Fen Ma
vol. 9, iss. 4, art. 113, 2008
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+
!
n
X
p i ai
!
n
X
qi b i
k
X
+
i=1
i=k+1
n
X
+
!
n
X
p i ai
i=1
!
q i ai
n
X
k
X
+
i=1
i=k+1
qi b i
i=k+1
!
p i bi
!
!
n
X
q i ai
i=1
!
p i bi ,
i=k+1
Mappings Related to
Chebyshev-type Inequalities
Liang-Cheng Wang, Li-Hong Liu
and Xiu-Fen Ma
where k = 1, 2, . . . , n, and
n
X
q i ai =
i=n+1
n
X
p i bi =
i=n+1
n
X
i=n+1
n
X
p i ai =
qi b i = 0
vol. 9, iss. 4, art. 113, 2008
i=n+1
is assumed.
For C : [a, b] → R,
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Contents
(1.7) C(x; p, q; r, s)
Z x
Z x
Z x
Z x
=
p(t)dt
q(t)r(t)s(t)dt +
q(t)dt
p(t)r(t)s(t)dt
a
a
a
a
Z b
Z b
Z x
Z b
+
p(t)r(t)dt
q(t)s(t)dt +
p(t)r(t)dt
q(t)s(t)dt
x
a
a
x
Z b
Z b
Z x
Z b
+
q(t)r(t)dt
p(t)s(t)dt +
q(t)r(t)dt
p(t)s(t)dt
x
a
a
x
e : [a, b] → R,
and for C
Z
e p, q; r, s) =
(1.8) C(y;
b
Z
p(t)dt
y
b
Z
q(t)r(t)s(t)dt +
y
b
Z
q(t)dt
y
b
p(t)r(t)s(t)dt
y
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Z
y
+
Z
b
p(t)r(t)dt
q(t)s(t)dt +
a
a
Z
+
b
Z
q(t)s(t)dt
a
b
Z
q(t)r(t)dt
a
p(t)r(t)dt
y
y
y
Z
Z
p(t)s(t)dt +
b
y
Z
q(t)r(t)dt
a
y
p(t)s(t)dt.
a
We write
Mappings Related to
Chebyshev-type Inequalities
Liang-Cheng Wang, Li-Hong Liu
and Xiu-Fen Ma
(1.9) c1 (k, n; pi )
1
= c(k, n; pi , pi )
2
k
X
= Pk
p i ai b i +
i=1
vol. 9, iss. 4, art. 113, 2008
n
X
!
p i ai
n
X
!
p i bi
+
i=1
i=k+1
k
X
!
p i ai
i=1
n
X
!
p i bi ,
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i=k+1
Contents
(1.10)
c2 (k, n) = c1 (k, n; 1)
=k
k
X
ai b i +
i=1
n
X
!
ai
n
X
i=1
i=k+1
!
bi
+
k
X
i=1
!
ai
n
X
!
bi ,
i=k+1
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(1.11)
C0 (x; p; r, s)
1
= C(x; p, p; r, s)
2
Z x
Z x
Z b
Z b
=
p(t)dt
p(t)r(t)s(t)dt +
p(t)r(t)dt
p(t)s(t)dt
a
a
x
a
Z x
Z b
+
p(t)r(t)dt
p(t)s(t)dt
a
x
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and
(1.12)
e0 (y; p; r, s)
C
1e
= C(y;
p, p; r, s)
2
Z b
Z b
Z
=
p(t)dt
p(t)r(t)s(t)dt +
y
Z
y
b
+
Z
p(t)r(t)dt
a
Z
p(t)r(t)dt
y
y
b
p(t)s(t)dt
a
y
p(t)s(t)dt.
a
(1.10), (1.6), (1.9), (1.7) and (1.8), (1.11) and (1.12) are generated by the inequalities (1.1) to (1.5), respectively.
e and
The aim of this paper is to study the monotonicity properties of c, C and C,
obtain some refinements of (1.1) to (1.5) using these monotonicity properties. Some
applications are given.
Mappings Related to
Chebyshev-type Inequalities
Liang-Cheng Wang, Li-Hong Liu
and Xiu-Fen Ma
vol. 9, iss. 4, art. 113, 2008
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2.
Main Results
The monotonicity properties of the mapping c, c1 and c2 are embodied in the following theorem.
Theorem 2.1. Let c, c1 and c2 be defined as in the first section. If A and B are both
increasing or both decreasing, then we have the following refinements of (1.2), (1.3)
and (1.1)
! n
!
! n
!
n
n
X
X
X
X
(2.1)
p i ai
qi b i +
q i ai
p i bi
i=1
i=1
i=1
vol. 9, iss. 4, art. 113, 2008
i=1
= c(1, n; pi , qi ) ≤ · · · ≤ c(k, n; pi , qi ) ≤ c(k + 1, n; pi , qi ) ≤ · · ·
n
n
X
X
≤ c(n, n; pi , qi ) = Pn
qi ai bi + Qn
p i ai b i ,
i=1
(2.2)
n
X
!
pi ai
i=1
n
X
Contents
= c1 (1, n; pi ) ≤ · · · ≤ c1 (k, n; pi )
i=1
≤ c1 (k + 1, n; pi ) ≤ · · · ≤ c1 (n, n; pi )
n
X
= Pn
p i ai b i
i=1
(2.3)
i=1
!
ai
n
X
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!
bi
JJ
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and
n
X
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i=1
!
p i bi
Mappings Related to
Chebyshev-type Inequalities
Liang-Cheng Wang, Li-Hong Liu
and Xiu-Fen Ma
= c2 (1, n) ≤ · · · ≤ c2 (k, n)
i=1
≤ c2 (k + 1, n) ≤ · · · ≤ c2 (n, n) = n
n
X
i=1
ai b i ,
respectively. If one of the sequences A or B is increasing and the other decreasing,
then inequalities in (2.1)–(2.3) are reversed.
The monotonicity properties of the mappings C and C0 are given in the following
theorem.
Theorem 2.2. Let C and C0 be defined as in the first section. If r and s are both
increasing or both decreasing, then C(x; p, q; r, s) and C0 (x; p; r, s) are increasing
on [a, b] with x, and for x ∈ [a, b] we have the following refinements of (1.4) and
(1.5)
Z b
Z b
Z b
Z b
(2.4)
p(t)r(t)dt
q(t)s(t)dt +
q(t)r(t)dt
p(t)s(t)dt
a
a
a
a
a
a
and
Z
b
Z
p(t)r(t)dt
(2.5)
a
vol. 9, iss. 4, art. 113, 2008
a
= C(a; p, q; r, s) ≤ C(x; p, q; r, s) ≤ C(b; p, q; r, s)
Z b
Z b
Z b
Z b
=
p(t)dt
q(t)r(t)s(t)dt +
q(t)dt
p(t)r(t)s(t)dt
a
Mappings Related to
Chebyshev-type Inequalities
Liang-Cheng Wang, Li-Hong Liu
and Xiu-Fen Ma
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b
p(t)s(t)dt = C0 (a; p; r, s)
a
≤ C0 (x; p; r, s) ≤ C0 (b; p; r, s)
Z b
Z b
=
p(t)dt
p(t)r(t)s(t)dt,
a
a
respectively. If one of the functions r or s is increasing and the other decreasing,
then C(x; p, q; r, s) and C0 (x; p; r, s) are decreasing on [a, b] with x, and inequalities
in (2.4) and (2.5) are reversed.
e and C
f0 are given in the following theorem.
The monotonicity properties of C
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e and C
f0 be defined as in the first section. If r and s are both
Theorem 2.3. Let C
e p, q; r, s) and C
e0 (y; p; r, s) are decreasing
increasing or both decreasing, then C(y;
on [a, b] with y, and for y ∈ [a, b] we have the following refinements of (1.4) and
(1.5)
Z b
Z b
Z b
Z b
p(t)r(t)dt
(2.6)
q(t)s(t)dt +
q(t)r(t)dt
p(t)s(t)dt
a
a
a
a
e p, q; r, s) ≤ C(y;
e p, q; r, s) ≤ C(a;
e p, q; r, s)
= C(b;
Z b
Z b
Z b
Z b
=
p(t)dt
q(t)r(t)s(t)dt +
q(t)dt
p(t)r(t)s(t)dt
a
a
a
Mappings Related to
Chebyshev-type Inequalities
Liang-Cheng Wang, Li-Hong Liu
and Xiu-Fen Ma
vol. 9, iss. 4, art. 113, 2008
a
and
Title Page
Z
b
Z
p(t)r(t)dt
(2.7)
a
a
b
e0 (b; p; r, s) ≤ C
e0 (y; p; r, s) ≤ C
e0 (a; p; r, s)
p(t)s(t)dt = C
Z b
Z b
=
p(t)dt
p(t)r(t)s(t)dt,
a
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a
respectively. If one of the functions r or s is increasing and the other decreasing, then
e p, q; r, s) and C
f0 (y; p; r, s) are increasing on [a, b] with y, and the inequalities
C(y;
in (2.6) and (2.7) are reversed.
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3.
Proof of Theorems
Proof of Theorem 2.1. For k = 2, 3, . . . , n, we have
(3.1)
c(k, n; pi , qi ) − c(k − 1, n; pi , qi )
= (Pk−1 + pk )
k−1
X
!
q i ai b i + q k ak b k
Mappings Related to
Chebyshev-type Inequalities
Liang-Cheng Wang, Li-Hong Liu
and Xiu-Fen Ma
i=1
+ (Qk−1 + qk )
!
k−1
X
p i ai b i + p k ak b k
vol. 9, iss. 4, art. 113, 2008
i=1
"
− Pk−1
k−1
X
qi ai bi + Qk−1
k−1
X
i=1
p k ak +
+
k−1
X
−
!
p i ai
p i ai q k b k +
n
X
k−1
X
!
q i ai
i=1
"
= pk
k−1
X
p k ak +
q i ai p k b k +
n
X
i=1
i=k+1
k−1
X
q i ai b i + p k ak b k
i=1
n
X
qi b i
+
q i ai
Title Page
!
p i ai
n
X
Contents
qi b i
i=1
p i bi
q k ak +
n
X
!
q i ai
n
X
I
p i bi
qi b i − p k b k
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i=1
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p i bi
i=1
J
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!
q i − p k ak
II
Page 11 of 23
i=k+1
k−1
X
JJ
i=1
i=k+1
p i bi −
n
X
qi b i
i=1
i=k+1
n
X
i=k+1
k−1
X
i=1
qi b i −
!
n
X
p i ai
n
X
i=k+1
i=k+1
q k ak +
−
n
X
n
X
i=k+1
i=1
+
p i ai b i +
i=1
i=1
k−1
X
#
k−1
X
i=1
#
q i ai
"
+ qk
k−1
X
p i ai b i + q k ak b k
i=1
= pk
k−1
X
k−1
X
p i − q k ak
i=1
qi (ak − ai ) (bk − bi ) + qk
k−1
X
p i b i − qk b k
i=1
k−1
X
k−1
X
#
p i ai
i=1
pi (ak − ai ) (bk − bi ) .
i=1
i=1
Mappings Related to
Chebyshev-type Inequalities
Liang-Cheng Wang, Li-Hong Liu
and Xiu-Fen Ma
If A and B are both increasing or both decreasing, then
(3.2)
ak − ai bk − bi ≥ 0, (i = 1, 2, . . . , k − 1).
vol. 9, iss. 4, art. 113, 2008
Using (1.6), (3.1) and (3.2), we obtain (2.1).
If one of the sequences A or B is increasing and the other decreasing, then (3.2)
is reversed, which implies that the inequalities in (2.1) are reversed.
For i = 1, 2, . . . , n, replacing qi in (2.1) with pi and replacing pi in (2.2) with
1, we obtain (2.2) and (2.3), respectively. This completes the proof of Theorem
2.1.
Proof of Theorem 2.2. For any x1 , x2 ∈ [a, b], x1 < x2 , we write
Z x2
Z x2
Z x2
Z x2
I1 =
p(t)dt
q(t)r(t)s(t)dt +
q(t)dt
p(t)r(t)s(t)dt
x1
x1
x1
x1
Z x2
Z x2
Z x2
Z x2
−
p(t)r(t)dt
q(t)s(t)dt −
q(t)r(t)dt
p(t)s(t)dt.
x1
x1
x1
x1
For t ∈ [a, x1 ], u ∈ [x1 , x2 ], using the properties of double integrals, we get
ZZ
I2 =
p(t)q(u) r(t) − r(u) s(t) − s(u) dtdu
[a,x1 ]×[x1 ,x2 ]
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Z
x1
=
a
Z
p(t)dt
Z
−
x2
Z
x1
Z
x2
q(t)r(t)s(t)dt +
q(t)dt
p(t)r(t)s(t)dt
x1
a
x1
Z x2
Z x1
Z x2
x1
p(t)r(t)dt
q(t)s(t)dt −
p(t)s(t)dt
q(t)r(t)dt
a
x1
a
x1
and
ZZ
I3 =
p(u)q(t) r(t) − r(u) s(t) − s(u) dtdu
[a,x ]×[x ,x ]
Z x1 1 1 Z2 x2
Z x1
Z x2
=
q(t)dt
p(t)r(t)s(t)dt +
q(t)r(t)s(t)dt
p(t)dt
a
x1
a
x1
Z x1
Z x2
Z x1
Z x2
−
q(t)r(t)dt
p(t)s(t)dt −
q(t)s(t)dt
p(t)r(t)dt.
a
x1
a
C(x2 ; p, q; r, s) − C(x1 ; p, q; r, s)
Z x2
Z x2
Z x2
Z x2
=
p(t)dt
q(t)r(t)s(t)dt +
q(t)dt
p(t)r(t)s(t)dt
x1
x1
x1
x1
Z x2
Z x2
Z x2
Z x2
−
p(t)r(t)dt
q(t)s(t)dt −
q(t)r(t)dt
p(t)s(t)dt
x1
x1
= I1 .
When x1 > a, from (1.7), we have
(3.4)
C(x2 ; p, q; r, s) − C(x1 ; p, q; r, s)
x1
vol. 9, iss. 4, art. 113, 2008
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x1
When x1 = a, from (1.7), we get
(3.3)
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Liang-Cheng Wang, Li-Hong Liu
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x1
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Z
=
x1
Z
x2 Z
x1
Z
x2 +
+
p(t)dt
q(t)r(t)s(t)dt
a
x1
x1
Z x1 Z x2 Z x1 Z x2 +
+
q(t)dt
+
p(t)r(t)s(t)dt
a
x1
a
x1
Z x1
Z x1
Z x1
Z x1
−
p(t)dt
q(t)r(t)s(t)dt −
q(t)dt
p(t)r(t)s(t)dt
a
a
a
a
Z x1 Z x2 Z b Z b
+
p(t)r(t)dt
+
+
q(t)s(t)dt
a
x2
x1
x2
Z x1 Z x2 Z b
+
+
p(t)r(t)dt
q(t)s(t)dt
a
x1
x2
Z x2 Z b Z x1 Z x2 Z b −
+
p(t)r(t)dt
+
+
q(t)s(t)dt
x1
x2
a
x1
x2
Z x2 Z b Z x1
−
p(t)r(t)dt
+
q(t)s(t)dt
a
x1
x2
Z x1 Z x2 Z b Z b
+
q(t)r(t)dt
+
+
p(t)s(t)dt
x2
a
x1
x2
Z x1 Z x2 Z b
+
+
q(t)r(t)dt
p(t)s(t)dt
a
x1
x2
Z x2 Z b Z x1 Z x2 Z b +
−
+
q(t)r(t)dt
+
p(t)s(t)dt
x1
x2
a
x1
x2
Z x2 Z b Z x1
−
q(t)r(t)dt
+
p(t)s(t)dt
a
a
x1
x2
Mappings Related to
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Liang-Cheng Wang, Li-Hong Liu
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vol. 9, iss. 4, art. 113, 2008
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Z
=
x2
Z
x2
Z
x2
Z
x2
p(t)dt
q(t)r(t)s(t)dt +
q(t)dt
p(t)r(t)s(t)dt
x1
x1
x1
x1
Z x2
Z x2
Z x2
Z x2
−
p(t)r(t)dt
q(t)s(t)dt −
q(t)r(t)dt
p(t)s(t)dt
x1
x1
x1
x1
Z x1
Z x2
Z x1
Z x2
+
p(t)dt
q(t)r(t)s(t)dt +
p(t)r(t)s(t)dt
q(t)dt
a
x1
a
x1
Z x1
Z x2
Z x1
Z x2
−
p(t)r(t)dt
q(t)s(t)dt −
p(t)s(t)dt
q(t)r(t)dt
a
x1
a
x1
Z x1
Z x2
Z x1
Z x2
+
q(t)dt
p(t)r(t)s(t)dt +
q(t)r(t)s(t)dt
p(t)dt
a
x1
a
x1
Z x1
Z x2
Z x1
Z x2
−
q(t)r(t)dt
p(t)s(t)dt −
q(t)s(t)dt
p(t)r(t)dt
a
x1
a
Mappings Related to
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Liang-Cheng Wang, Li-Hong Liu
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vol. 9, iss. 4, art. 113, 2008
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x1
= I1 + I2 + I3 .
(1) If r and s are both increasing or both decreasing, then we have
(3.5)
r(t) − r(u) s(t) − s(u) ≥ 0,
i.e., I2 ≥ 0 and I3 ≥ 0. By the inequality (1.4), I1 ≥ 0 holds . Using (3.3) and (3.4),
we obtain that C(x; p, q; r, s) is increasing on [a, b] with x. Further, from (1.11), we
get that C0 (x; p; r, s) is increasing on [a, b] with x.
From (1.7) and (1.11), using the increasing properties of C(x; p, q; r, s) and C0 (x; p; r, s),
we obtain (2.4) and (2.5), respectively.
(2) If one of the functions r or s is increasing and the other decreasing, then the
inequality in (3.5) is reversed, which implies that I2 ≤ 0 and I3 ≤ 0. By the reverse
of (1.4), I1 ≤ 0 holds. From (3.3) and (3.4), (1.11), we obtain that C(x; p, q; r, s),
C0 (x; p; r, s) are decreasing on [a, b] with x, respectively.
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From (1.7) and (1.11), using the decreasing properties of C(x; p, q; r, s) and C0 (x; p; r, s),
we obtain the reverse of (2.4) and (2.5), respectively.
This completes the proof of Theorem 2.2.
Proof of Theorem 2.3. Using the same arguments as those in the proof of Theorem
2.2, we can prove Theorem 2.3.
Mappings Related to
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vol. 9, iss. 4, art. 113, 2008
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4.
Applications
Let I be a real interval and u, v, w : I → [0, +∞). For any α, β ∈ R and any xi ∈ I
(i = 1, 2, . . . , n, n ≥ 2) satisfying x1 ≤ x2 ≤ · · · ≤ xn , we define
K(k, n) =
k
X
v(xi )wβ (xi )
i=1
+
k
X
+
u(xi )w−β (xi )
i=1
v(xi )w−α (xi )
i=1
k
X
k
X
k
X
n
X
u(xi )wα (xi ) +
i=1
α
v(xi )w (xi )
i=1
n
X
v(xi )wα (xi )
n
X
u(xi )w
n
X
(xi ) +
i=k+1
v(xi )w
−β
(xi )
+
n
X
u(xi )wβ (xi )
v(xi )w−β (xi )
i=1
n
X
Contents
u(xi )wβ (xi ),
i=k+1
and
L(k, n) =
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i=1
i=k+1
k
X
vol. 9, iss. 4, art. 113, 2008
i=1
i=k+1
−α
u(xi )w−α (xi )
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Liang-Cheng Wang, Li-Hong Liu
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k
X
v(xi )wβ (xi )
i=1
+
k
X
i=1
n
X
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v(xi )wα (xi )
i=k+1
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u(xi )wα (xi )
n
X
Close
u(xi )wβ (xi )
i=1
+
k
X
i=1
α
v(xi )w (xi )
n
X
i=k+1
u(xi )wβ (xi ),
where, k = 1, 2, . . . , n,
n
X
β
u(xi )w (xi ) =
i=n+1
n
X
u(xi )w−α (xi ) =
i=n+1
n
X
i=n+1
n
X
v(xi )wα (xi ) = 0,
v(xi )w−β (xi ) = 0.
i=n+1
Proposition 4.1. Let w and u/v be both increasing or both decreasing. If α > β,
then we have
(4.1)
n
X
v(xi )wα (xi )
i=1
n
X
u(xi )w−α (xi ) +
i=1
n
X
v(xi )w−β (xi )
i=1
n
X
u(xi )wβ (xi )
i=1
= K(1, n) ≤ · · · ≤ K(k, n) ≤ K(k + 1, n) ≤ · · · ≤ K(n, n)
n
n
n
n
X
X
X
X
β
−β
−α
=
v(xi )w (xi )
u(xi )w (xi ) +
v(xi )w (xi )
u(xi )wα (xi )
i=1
i=1
i=1
i=1
and
(4.2)
Mappings Related to
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Liang-Cheng Wang, Li-Hong Liu
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vol. 9, iss. 4, art. 113, 2008
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n
X
v(xi )wα (xi )
i=1
n
X
u(xi )wβ (xi )
i=1
= L(1, n) ≤ · · · ≤ L(k, n) ≤ L(k + 1, n) ≤ · · · ≤ L(n, n)
n
n
X
X
=
v(xi )wβ (xi )
u(xi )wα (xi ).
i=1
i=1
If α < β, then the inequalities in (4.1) and (4.2) are reversed.
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Proof. Replacing pi , qi , ai and bi in (2.1) (or the reverse of (2.1)) with v(xi )wβ (xi ),
v(xi )w−α (xi ), wα−β (xi ) and u(xi )/v(xi ), respectively, we obtain (4.1) (or the reverse of (4.1)). Replacing pi , ai and bi in (2.2) (or the reverse of (2.2)) with v(xi )wβ (xi ),
wα−β (xi ) and u(xi )/v(xi ), respectively, we obtain (4.2) (or the reverse of (4.2)).
This completes the proof of Proposition 4.1.
Remark 1. (4.1) and (4.2) are generated by Proposition 1 in [4].
Let f : [a, b] → R be a continuous convex function with f+0 (a) (= f−0 (a) is
assumed) and f−0 (b), {f (x)|x ∈ [a, b]} = [d, e]. Also, let h : [d, e] → (0, +∞) be an
integrable function, and g : [d, e] → R be a strict monotonic function. We define
" Z
#
−1 Z b
b
(4.3) E(g; f, h) = g −1
h(f (t))f−0 (t)dt
h(f (t))g(f (t))f−0 (t)dt ,
a
a
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and Xiu-Fen Ma
vol. 9, iss. 4, art. 113, 2008
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(4.4)
M (g; f, h) = g −1
" Z
b
#
−1 Z b
h(f (t))dt
h(f (t))g(f (t))dt ,
a
a
(4.5) R(x; g; f, h)
#
" Z
−1 Z b
b
C0 x; h(f ); g(f ), f−0
= g −1
h(f (t))dt
h(f (t))f−0 (t)dt
a
a
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and
e g; f, h)
(4.6) R(y;
#
" Z
−1 Z b
b
0
−1
0
e0 y; h(f ); g(f ), f
C
=g
h(f (t))dt
h(f (t))f− (t)dt
.
−
a
a
Proposition 4.2. If f is monotone, Then we have
1. R(x; g; f, h) is increasing on [a, b] with x. For x ∈ [a, b] we have
(4.7)
M (g; f, h) = R(a; g; f, h)
≤ R(x; g; f, h) ≤ R(b; g; f, h) = E(g; f, h).
2. R̃(y; g; f, h) is decreasing on [a, b] with y. For y ∈ [a, b] we have
(4.8)
M (g; f, h) = R̃(b; g; f, h)
≤ R̃(y; g; f, h) ≤ R̃(a; g; f, h) = E(g; f, h).
Proof. From the convexity of f , we get that f−0 (t) is increasing on [a, b] and the intee g; f, h) are valid (see [5]). From h(x) > 0,
grals in E(g; f, h), R(x; g; f, h) and R(y;
x ∈ [d, e], we have
Z b
(4.9)
h(f (t))dt > 0.
a
From the convexity of f , when f (a) < f (b) or f (a) > f (b), Wang in [5] proved that
h(f (t))f−0 (t)dt > 0
(4.10)
vol. 9, iss. 4, art. 113, 2008
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Z
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or
Z
(4.11)
b
h(f (t))f−0 (t)dt < 0.
a
(1) Let us first assume that g is a strictly increasing function.
Case 1. From the increasing properties of f , we have f (a) < f (b). Further, (4.10)
holds. To prove that R(x; g; f, h) is increasing, from (4.5), (4.9) and (4.10), we only
need to prove that
(4.12) C0 x; h(f ); g(f ), f−0
Z b
Z b
0
=
h(f (t))dt
h(f (t))f− (t)dt g R(x; g; f, h)
a
a
is increasing on [a, b] with x.
Indeed, since f is increasing on [a, b], we have that g(f (t)) is increasing on [a, b].
By Theorem 2.2, C0 (x; h(f ); g(f ), f−0 ) is monotonically increasing with x ∈ [a, b].
For x ∈ [a, b], from (4.3), (4.4), (4.5), (4.9) and (4.10), then (4.7) is equivalent to
Z b
Z b
(4.13)
h(f (t))g(f (t))dt
h(f (t))f−0 (t)dt
a
=
a
C0 (a; h(f ); g(f ), f−0 )
Z b
Z b
=
≤
C0 (b; h(f ); g(f ), f−0 )
h(f (t))g(f (t))f−0 (t)dt.
h(f (t))dt
a
≤
C0 (x; h(f ); g(f ), f−0 )
Mappings Related to
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vol. 9, iss. 4, art. 113, 2008
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a
Replacing p(t), r(t) and s(t) in (2.5) with h(f (t)), g(f (t)) and f−0 (t), respectively,
we obtain (4.13).
Case 2. If f is decreasing on [a, b], then we have f (a) > f (b), i.e. (4.11) holds. To
prove that R(x; g; f, h) is increasing, from (4.5), (4.9) and (4.11), we only need to
prove that C0 (x; h(f ); g(f ), f−0 ) (see (4.12)) is decreasing on [a, b] with x.
Indeed, since f is decreasing on [a, b], then g(f (t)) is decreasing on [a, b]. By Theorem 2.2, C0 (x; h(f ); g(f ), f−0 ) is decreasing with x ∈ [a, b].
For x ∈ [a, b], from (4.3), (4.4), (4.5), (4.9) and (4.11), then (4.7) is equivalent to the
reverse of (4.13). Replacing p(t), r(t) and s(t) in the reverse of (2.5) with h(f (t)),
g(f (t)) and f−0 (t), respectively, we obtain the reverse of (4.13).
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The second case: g is a strictly decreasing function. Using the same arguments for g
as a strictly increasing function, we can also prove (1).
(2) Using the same arguments as those for (1), with (2.6) and (2.7), we can prove
that R̃(x; g; f, h) is decreasing on [a, b] with x, and (4.8) holds.
This completes the proof of Proposition 4.2.
Remark 2. (4.7)–(4.8) can be generated by (∗) in [6] or Proposition 8.1 in [5].
Mappings Related to
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Liang-Cheng Wang, Li-Hong Liu
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vol. 9, iss. 4, art. 113, 2008
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References
[1] D.S. MITRINOVIĆ, Analytic Inequalities, Springer-Verlag. Berlin, 1970.
[2] J.-C. KUANG, Applied Inequalities, Shandong Science and Technology Press,
2004. (Chinese).
[3] M. BEHDZET, Some results connected with Cebysev’s inequality, Rad. Mat.,
1(2) (1985), 185–190.
Mappings Related to
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Liang-Cheng Wang, Li-Hong Liu
and Xiu-Fen Ma
[4] L.-C. WANG, On the monotonicity of difference generated by the inequalities
of Chebyshev type, J. Sichuan Univ., 39(3) (2002), 398–403. (Chinese).
vol. 9, iss. 4, art. 113, 2008
[5] L.-C. WANG, Convex Functions and Their Inequalities, Sichuan University
Press, Chengdu, China, 2001. (Chinese).
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[6] B.-N. GUO AND FENG QI, Inequalities for generalized weighted mean values
of convex function, Math. Inequalities Appl., 4(2) (2001), 195–202.
[7] F. QI, L.-H. CUI AND S.-L. XU, Some inequalities constructed by Tchebysheff’s
integral inequality, Math. Inequalities Appl., 2(4) (1999), 517–528.
[8] F. QI, S.-L. XU AND L. DEBNATH, A new proof of monotonicity for extended
mean values, J. Math. Math. Sci., 22(2) (1999), 417–421.
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