ON NEW INEQUALITIES OF HADAMARD-TYPE
FOR LIPSCHITZIAN MAPPINGS AND THEIR
APPLICATIONS
Inequalities of Hadamard-type
for Lipschitzian Mappings
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
LIANG-CHENG WANG
School of Mathematica Scientia,
Chongqing Institute of Technology,
No. 4 of Xingsheng Lu, Yangjia Ping 400050,
Chongqing City, China.
EMail: wlc@cqit.edu.cn
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Received:
11 December, 2005
J
Accepted:
16 August, 2006
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Communicated by:
F. Qi
2000 AMS Sub. Class.:
Primary 26D07; Secondary 26B25, 26D15.
Key words:
Lipschitzian mappings, Hadamard inequality, Convex function.
Abstract:
In this paper, we study some new inequalities of Hadamard’s Type for Lipschitzian mappings. some applications are also included.
Acknowledgements:
This 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
6
3
Applications
18
Inequalities of Hadamard-type
for Lipschitzian Mappings
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
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1.
Introduction
Let f : [a, b] → R (a < b) be a continuous function.
If f is convex on [a, b], then
Z b
a+b
1
f (a) + f (b)
(1.1)
f
≤
f (x)dx ≤
.
2
b−a a
2
Inequalities of Hadamard-type
for Lipschitzian Mappings
The inequalities in (1.1) are known as the Hermite-Hadamard inequality [1].
For some recent results which generalize, improve, and extend this classic inequality, see references of [2] – [7]. In order to refine inequalities of (1.1), the author
of this paper in [2] defined the following some notations, symbols and mappings. we
list these notations and symbols by
Y = (y1 , y2 , . . . , yn ) ∈ Rn , t = (t1 , t2 , . . . , tn ) ∈ Rn , Tn = t1 +t2 +· · ·+tn ; 0 =
(0, 0, . . . , 0), 1 = (1, 1, . . . , 1), n1 = ( n1 , n1 , . . . , n1 ) and (1i , 0) = (0, . . . , 0,
1, 0, . 1. ., 0)
1
n
(1 is ith
i = 1, 2, . . . , n) are special points in R ; G = 0, n × 0, n ×
component,
1
· · · × 0, n , I = [0, 1] × [0, 1] × · · · × [0, 1], V = [a, b] × [a, b] × · · · × [a, b], D =
[a, x1 ] × [x1 , x2 ] × · · · × [xn−1 , b] (xi = a + (b−a)i
, i = 0, 1, . . . , n; x0 = a, xn = b),
n
H = {t ∈ I|Tn ≤ 1} and L = {t ∈ I|Tn = 1} are subsets in Rn .
We list these mappings by
n Z
!
n n
1X
xi−1 + xi
4
Rn : I 7→ R, Rn (t) =
f
ti yi + (1 − ti )
dY,
b−a
n i=1
2
D
Sn : H 7→ R,
4
Sn (t) =
1
(b − a)n
Z
f
V
n
X
i=1
ti yi + (1 − Tn )
a+b
2
!
dY
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
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and
Pn : L 7→ R,
1
Pn (t) =
(b − a)n
4
Z
f
V
We write Pn+1 in the following equivalent form
Z "Z b
1
4
Pn+1 : H 7→ R, Pn+1 (t) =
f
(b − a)n+1 V
a
Let g : A ⊆ Rn → R. For all t(j) =
(1)
(j)
n
X
!
ti yi dY.
i=1
n
X
!
#
ti yi + (1 − Tn )x dx dY.
i=1
Inequalities of Hadamard-type
for Lipschitzian Mappings
Liang-Cheng Wang
(j)
t1 , . . . , tn
∈ A(j = 1, 2) with
vol. 8, iss. 1, art. 30, 2007
(2)
ti ≤ ti (i = 1, 2, . . . , n), if g(t(1) ) ≤ g(t(2) ), then we call g increasing on A.
For these mappings and if f is convex on [a, b], L.-C. Wang in [2] gave the following properties and inequalities:
Pn is convex on L; Rn and Sn are convex, increasing on I and G, respectively;
a+b
f
(1.2)
= Rn (0) ≤ Rn (t) ≤ Rn (1)
2
!
n Z
n
n
1X
=
f
yi dY
b−a
n i=1
D
Z b
1
≤
f (x)dx
b−a a
for any t ∈ I,
Z
a+b
1
1
(1.3) f
= Sn (0) ≤ Sn (t) ≤ Sn
=
f
2
n
(b − a)n V
!
n
1X
yi dY
n i=1
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for all t ∈ G,
(1.4)
Sn (t) ≤ Pn+1 (t)
for all t ∈ H, and
Z b
1
1
1
(1.5)
Sn
= Pn
≤ Pn (t) ≤ Pn (1i , 0) =
f (x)dx
n
n
b−a a
for all t ∈ L.
(1.2) – (1.5) are refinements of (1.1).
Recently, Dragomir et al. [3], Yang and Tseng [5], Matic and Pečarić [6] and
L.-C. Wang [7] proved some results for Lipschitzian mappings related to (1.1). In
this paper, we will prove some new inequalities for Lipschitzian mappings related to
the mappings Rn (or (1.2)), Sn (or (1.3)) and Pn (or (1.5) and (1.4)). Finally, some
applications are given.
Inequalities of Hadamard-type
for Lipschitzian Mappings
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
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2.
Main Results
A function f : [a, b] → R is called an M −Lipschitzian mapping, if for every two
elements x, y ∈ [a, b] and M > 0 we have
|f (x) − f (y)| ≤ M |x − y|.
For the mapping Rn (t), we have the following theorem:
Inequalities of Hadamard-type
for Lipschitzian Mappings
Theorem 2.1. Let f : [a, b] → R be an M −Lipschitzian mapping, then we have
n
(2.1)
(j)
for any t
(2.2)
X (2)
M
(1)
Rn (t ) − Rn (t ) ≤ 2 (b − a)
ti − ti
4n
i=1
(j)
(j)
= t1 , . . . , tn ∈ I(j = 1, 2),
M
a+b
− Rn (t) ≤ 2 (b − a)Tn
f
2
4n
(2)
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Rn (t) −
n
b−a
n Z
f
D
!
n
1X
M
yi dY ≤ 2 (b − a) (n − Tn )
n i=1
4n
for all t ∈ I, and
(2.4)
vol. 8, iss. 1, art. 30, 2007
(1)
and
(2.3)
Liang-Cheng Wang
n
b−a
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n Z
f
D
1
n
n
X
i=1
!
yi dY −
1
b−a
Z
b
f (x)dx
a
≤
M (n2 − 1)
(b − a).
3n2
Proof. (1) For xi = (b−a)i
(i = 0, 1, . . . , n; x0 = a, xn = b), from integral properties,
n
we have
Rn t(2) − Rn t(1)
n Z
!
n n
1 X (2)
(2) xi−1 + xi
≤
f
t yi + (1 − ti )
b−a
n i=1 i
2
D
!
n
1 X (1)
(1) xi−1 + xi
−f
t yi + (1 − ti )
dY
n i=1 i
2
n
Z X
n n
M
xi−1 + xi
(2)
(1)
≤
·
t − ti
yi −
dY
b−a
n D i=1 i
2
n
Z
n
xi−1 + xi
n
M X (2)
(1)
≤
·
ti − ti
yi −
dY
b−a
n i=1
2
D
n
n−1 Z xi
n
n
M X (2)
b−a
xi−1 + xi
(1)
=
·
ti − ti
yi −
dyi
b−a
n i=1
n
2
xi−1
"Z xi−1 +xi n
2
M X (2)
xi−1 + xi
(1)
=
t − ti
− yi dyi
b − a i=1 i
2
xi−1
#
Z xi
xi−1 + xi
+ x +x yi −
dyi
i−1
i
2
2
n
X (2)
M
(1)
= 2 (b − a)
ti − ti .
4n
i=1
This completes the proof of (2.1).
Inequalities of Hadamard-type
for Lipschitzian Mappings
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
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(2) The inequalities (2.2) and (2.3) follow from (2.1) by choosing t(1) = 0, t(2) = t
and t(1) = 1, t(2) = t, respectively. This completes the proof of (2.2) and (2.3).
(3) From integral properties, we have
n−1 X
Z b
n Z
n Z xi
X
n
(2.5)
f (x)dx =
f (yi )dY.
f (yi )dyi =
b
−
a
D
a
x
i−1
i=1
i=1
Using (2.5) and integral properties, we obtain
!
n Z
Z b
n
1
n
1X
f
yi dY −
f (x)dx
b−a
n i=1
b−a a
D
!
n Z
n
n
n
X
X
1
1X
n
1
f
yj −
f (yi ) dY
≤
b−a
n j=1
n i=1
D n i=1
n
Z X
n
n
M
1X
n
·
yj − yi dY
≤
b−a
n D i=1 n j=1
n
n Z X
n
MX
n
≤
· 2
|yj − yi | dY
b−a
n i=1 D j=1
" i−1
#
n
n Z
n
X
X
n
MX
=
· 2
(yi − yj ) +
(yj − yi ) dY
b−a
n i=1 D j=1
j=i+1
!
" i−1 Z
Z xj
n
xi
n
MX X
=
·
yi dyi −
yj dyj
b − a n2 i=1 j=1
xi−1
xj−1
Inequalities of Hadamard-type
for Lipschitzian Mappings
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
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+
n
X
j=i+1
Z
xj
!#
xi
Z
yj dyj −
xj−1
yi dyi
xi−1
"
2
n
MX
b−a
n
=
·
b − a n2 i=1
n
i−1
X
(i − j) +
j=1
n
X
!#
(j − i)
j=i+1
M (n2 − 1)
(b − a).
3n2
This completes the proof of (2.4).
This completes the proof of Theorem 2.1.
Inequalities of Hadamard-type
for Lipschitzian Mappings
=
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
Corollary 2.2. Let f be convex on [a, b], with f+0 (a) and f−0 (b) existing. Then we
obtain
(2.6)
Contents
0 ≤ Rn (t(2) ) − Rn (t(1) )
n
X (2)
max{|f+0 (a)|, |f−0 (b)|}
(1)
≤
(b
−
a)
t
−
t
i
i
4n2
i=1
(j)
(j)
(2)
(1)
for any t(j) = (t1 , . . . , tn ) ∈ I(j = 1, 2) with ti ≥ ti (i = 1, 2, . . . , n),
max{|f+0 (a)|, |f−0 (b)|}
a+b
≤
(b − a)Tn
(2.7)
0 ≤ Rn (t) − f
2
4n2
and
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(2.8)
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0≤
n
b−a
n Z
f
D
1
n
n
X
!
yi dY − Rn (t)
i=1
max{|f+0 (a)|, |f−0 (b)|}
(b − a)(n − Tn )
≤
4n2
for all t ∈ I, and
1
0≤
b−a
(2.9)
b
Z
f (x)dx −
a
n
b−a
n Z
f
D
!
n
1X
yi dY
n i=1
max{|f+0 (a)|, |f−0 (b)|}(n2 − 1)
(b − a).
3n2
Proof. For any x, y ∈ [a, b], from properties of convex functions, we have the following max{|f+0 (a)|, |f−0 (b)|}−Lipschitzian inequality (see [8]):
≤
|f (x) − f (y)| ≤ max{|f+0 (a)|, |f−0 (b)|}|x − y|.
(2.10)
Since Rn is increasing on I, using (1.2), (2.10) and Theorem 2.1, we obtain (2.6)(2.9).
This completes the proof of Corollary (2.2).
For the mapping Sn (t), we have the following theorem:
Theorem 2.3. Let f be defined as in Theorem 2.1, then we obtain
Sn (t(2) ) − Sn (t(1) ) ≤
(2.11)
(j)
for any t
=
(2)
(1)
ti − ti
i=1
(j)
(j)
(t1 , . . . , tn )
(2.12)
M
(b − a)
4
n
X
f
∈ H(j = 1, 2),
a+b
M
− Sn (t) ≤
(b − a)Tn
2
4
and
(2.13)
1
Sn (t) −
(b − a)n
Z
f
V
!
n
n
X
1X
M
yi dY ≤
(b − a)
|nti − 1|
n i=1
4n
i=1
Inequalities of Hadamard-type
for Lipschitzian Mappings
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
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for all t ∈ H, and
Z
a+b
1
(2.14)
−
f
f
2
(b − a)n V
!
n
M
1X
yi dY ≤
(b − a).
n i=1
4
Proof. (1) From integral properties, we obtain
Sn t(2) − Sn t(1)
!
!
Z
n
n
X
X
1
a+b
(2)
(2)
≤
f
ti yi + 1 −
ti
(b − a)n V
2
i=1
i=1
!
!
n
n
X
X
a+b
(1)
(1)
dY
−f
ti yi + 1 −
ti
2
i=1
i=1
Z X
n M
a+b
(2)
(1)
≤
t − ti
yi −
dY
(b − a)n V i=1 i
2
Z
n
X
a+b
M
(2)
(1)
≤
ti − ti
yi −
dY
n
(b − a) i=1
2
V
Z b
n
M X (2)
a+b
(1)
=
ti − ti
yi −
dyi
b − a i=1
2
a
"Z a+b #
Z b n
2
a+b
a+b
M X (2)
(1)
t − ti
− yi dyi +
dyi
=
yi −
a+b
b − a i=1 i
2
2
a
2
n
X (2)
M
(1)
=
(b − a)
ti − ti .
4
i=1
This completes the proof of (2.11).
Inequalities of Hadamard-type
for Lipschitzian Mappings
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
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(2) The inequalities (2.12) and (2.13) follow from (2.11) by choosing t(1) = 0,
t(2) = t and t(1) = n1 , t(2) = t, respectively. The inequalities (2.14) follow from
(2.12) by choosing t(1) = n1 . This completes the proof of (2.12)-(2.14).
This completes the proof of Theorem 2.3.
Corollary 2.4. Let f be defined as in Corollary 2.2, then we have
n
X (2)
max{|f+0 (a)|, |f−0 (b)|}
(1)
(2.15) 0 ≤ Sn (t ) − Sn (t ) ≤
(b − a)
ti − ti
4
i=1
(2)
(1)
Inequalities of Hadamard-type
for Lipschitzian Mappings
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
(j)
for any t
(2.16)
=
(j)
(j)
(t1 , . . . , tn )
(2)
ti
(1)
ti
∈ G(j = 1, 2) with
≥
(i = 1, 2, . . . , n),
max{|f+0 (a)|, |f−0 (b)|}
a+b
0 ≤ Sn (t) − f
≤
(b − a)Tn
2
4
and
0≤
(2.17)
≤
1
(b − a)n
n
Z
f
V
1X
yi dY − Sn (t)
n i=1
max{|f+0 (a)|, |f−0 (b)|}
4
!
(b − a)(1 − Tn )
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for all t ∈ G, and
(2.18)
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0≤
≤
1
(b − a)n
Z
f
V
!
n
1X
a+b
yi dY − f
n i=1
2
max{|f+0 (a)|, |f−0 (b)|}
(b − a).
4
Close
Proof. Since Sn is increasing on G, using (1.3), (2.10) and Theorem 2.3, we obtain
(2.15) – (2.18).
This completes the proof of Corollary (2.4).
For the mapping Pn (t), we have the following theorem:
Theorem 2.5. Let f be defined as in Theorem 2.1. For n ≥ 2, then we obtain
|pn (t(2) ) − pn (t(1) )| ≤
(2.19)
(j)
(j)
for any t(j) = t1 , . . . , tn
M
(b − a)
3
n−1
X
(2)
(1)
ti − ti
Inequalities of Hadamard-type
for Lipschitzian Mappings
Liang-Cheng Wang
i=1
vol. 8, iss. 1, art. 30, 2007
∈ L(j = 1, 2),
Title Page
(2.20)
1
(b − a)n
Z
f
V
!
n
n−1
X
1X
M
yi dY − pn (t) ≤
(b − a)
|nti − 1|
n i=1
3n
i=1
and
pn (t) −
(2.21)
1
b−a
Z
b
f (x)dx ≤
a
M
(b − a)
3
n−1
X
ti
i=1
for all t ∈ L, and
(2.22)
1
(b − a)n
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Z
f
V
!
Z b
n
1X
1
M (n − 1)
yi dY −
f (x)dx ≤
(b − a).
n i=1
b−a a
3n
For n ≥ 1 and all t ∈ H, then we have
(2.23)
Contents
|Sn (t) − Pn+1 (t)| ≤
M
(b − a)(1 − Tn ).
4
Close
Proof. (1) Since n ≥ 2 and Tn = t1 + · · · + tn−1 + tn = 1, we can write Pn (t) in the
following equivalent form
! !
Z
n−1
n−1
X
X
1
(2.24)
Pn (t) =
f
ti yi + 1 −
ti yn dY.
(b − a)n V
i=1
i=1
Using (2.24) and integral properties, we obtain
Pn t(2) − Pn t(1)
! !
Z
n−1
n−1
X
X
1
(2)
(2)
≤
f
ti yi + 1 −
ti
yn
(b − a)n V
i=1
i=1
! !
n−1
n−1
X
X
(1)
(1)
−f
ti yi + 1 −
ti
yn dY
M
≤
(b − a)n
i=1
n−1
X
Z
V
i=1
(2)
ti
−
(1)
ti
(yi − yn ) dY
i=1
Z bZ b
n−1
X
M
(2)
(1)
n−2
≤
t − ti (b − a)
|yi − yn | dyi dyn
(b − a)n i=1 i
a
a
Z b Z x
Z b
n−1
X
M
(2)
(1)
=
t − ti
(x − y)dy +
(y − x)dy dx
(b − a)2 i=1 i
a
a
x
n−1
X (2)
M
(1)
=
(b − a)
ti − ti .
3
i=1
This completes the proof of (2.19).
Inequalities of Hadamard-type
for Lipschitzian Mappings
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
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(2) The inequalities (2.20) and (2.21) follow from (2.19) by choosing t(1) = n1 ,
t(2) = t and t(1) = (1n , 0) = (0, . . . , 0, 1), t(2) = t, respectively. The inequalities
(2.22) follow from (2.21) by choosing t = n1 . This completes the proof of (2.20) –
(2.22).
(3) Using integral properties, we write Sn (t) in the following equivalent form
! #
Z "Z b
n
X
1
a+b
(2.25)
Sn (t) =
f
ti yi + (1 − Tn )
dx dY.
(b − a)n+1 V
2
a
i=1
Inequalities of Hadamard-type
for Lipschitzian Mappings
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
Using (2.25) and integral properties, we obtain
|Sn (t) − Pn+1 (t)|
Z "Z b
1
≤
f
(b − a)n+1 V
a
−f
n
X
Title Page
n
X
ti yi + (1 − Tn )
i=1
!
ti yi + (1 − Tn )x
a+b
2
!
dx dY
Z Z b
M
a+b
≤
(1 − Tn )
− x dx dY
(b − a)n+1
2
V
a
"Z a+b #
Z b 2
M
a+b
a+b
=
(1 − Tn )
− x dx +
x−
dx
a+b
b−a
2
2
a
2
M
(b − a)(1 − Tn ).
4
This completes the proof of (2.23).
This completes the proof of Theorem 2.5.
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#
i=1
=
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Corollary 2.6. Let f be defined as in Corollary 2.2. For n ≥ 2, then we have
n−1
(2.26)
X (2)
max{|f+0 (a)|, |f−0 (b)|}
(1)
Pn (t ) − Pn (t ) ≤
(b − a)
ti − ti
3
i=1
(2)
(1)
(j)
(j)
for any t(j) = (t1 , . . . , tn ) ∈ L(j = 1, 2) ,
Z
1
(2.27)
f
0 ≤ Pn (t) −
(b − a)n V
max{|f+0 (a)|, |f−0 (b)|}
≤
3n
Inequalities of Hadamard-type
for Lipschitzian Mappings
!
n
1X
yi dY
n i=1
(b − a)
n−1
X
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
|nti − 1|
Title Page
i=1
and
(2.28)
Contents
1
0≤
b−a
Z
n−1
b
f (x)dx − Pn (t) ≤
a
X
max{|f+0 (a)|, |f−0 (b)|}
(b − a)
ti
3
i=1
for all t ∈ L, and
(2.29)
Z
a
b
1
f (x)dx −
(b − a)n
Z
f
V
!
n
1X
yi dY
n i=1
max{|f+0 (a)|, |f−0 (b)|}(n − 1)
(b − a).
3n
For n ≥ 1 and all t ∈ H , we have
(2.30)
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Page 16 of 22
1
0≤
b−a
≤
JJ
0 ≤ Pn+1 (t) − Sn (t) ≤
max{|f+0 (a)|, |f−0 (b)|}
(b − a)(1 − Tn ).
4
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Proof. Using (1.5), (1.4), (2.10) and Theorem 2.5, we obtain (2.26) – (2.30).
This completes the proof of Corollary (2.6).
Remark 1. The condition in Corollary 2.2 (or Corollary 2.4 or 2.6) is better than the
condition in Corollary 2.2 (or Corollary 4.2 or Theorem 3.3) of [3]. This is due to the
fact that f is a differentiable convex function on [a, b] with M = sup |f 0 (t)| < ∞.
t∈[a,b]
Remark 2. When n = 1, (2.1) and (2.11), (2.2) and (2.12), (2.3) and (2.13) and
(2.23) reduce to (3.4), (3.2), (3.1) and (4.3) of [3], respectively. When n = 2, (2.19),
(2.20), and (2.21) reduce to (4.6), (4.1) and (4.2) of [3], respectively.
Inequalities of Hadamard-type
for Lipschitzian Mappings
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
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3.
Applications
In this section, we agree that when ti = 0,
" ti
#
t
ln b − ln a
1
b 2n2 a 2ni2
−
=
b
n2
ti
a
(2)
and
bti − ati
= ln b − ln a.
ti
(1)
For b > a > 0, 1 ≥ ti ≥ ti ≥ 0 and 1 ≥ ti ≥ 0 (i = 1, 2, . . . , n), we have




(1)
(2)
t
t
(1)
(2)
i
i
t
t
n
n
i
i
Y 1  b 2n2
a 2n2 
a 2n2  Y 1  b 2n2
(3.1)
−
−
0≤
−




(1)
(2)
b
a
a
b
i=1 ti
i=1 ti
1
≤
4
ln b − ln a
n2
n Y
n
#
" ti
t
1
b 2n2 a 2ni2
−1
b
ti
a
n2
ln b − ln a
i=1
12
ln b − ln a b
≤
Tn
4n2
a
(3.2)
n+1 12 X
n b
(2)
(1)
ti − ti ,
a
i=1
0≤
and
(3.3)
" 1
#
" ti
#
t
n
n
Y
Y
b 2n2 a 2n12
1
b 2n2 a 2ni2
0≤
−
−
−
a
b
t
a
b
i
i=1
i=1
n+1 12
1 ln b − ln a
b
(n − Tn ) .
≤
4
n2
a
Inequalities of Hadamard-type
for Lipschitzian Mappings
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
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Contents
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For b > a > 0,
(3.4)
1
n
(2)
(1)
≥ ti ≥ ti ≥ 0 and
0 ≤ (ab)
P
(2)
1− n
i=1 ti
2
1
n
n
Y
1 (2)
(2)
(2)
bti − ati
ti
i=1
− (ab)
≥ ti ≥ 0 (i = 1, 2, . . . , n), we have
P
(1)
1− n
i=1 ti
2
n
Y
1 (1)
i=1
(1)
(1)
bti − ati
ti
Inequalities of Hadamard-type
for Lipschitzian Mappings
n
b (ln b − ln a)n+1 X (2)
(1)
≤
ti − ti ,
4
i=1
(3.5)
Pn
i=1 ti
2
1−
1
0≤
(ab)
(ln b − ln a)n
≤
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
n
Y
1
1 ti
b − ati − (ab) 2
t
i=1 i
Title Page
Contents
b (ln b − ln a)
Tn
4
and
(3.6)
1
n
0 ≤ nb − na
1
n
n
− (ab)
P
1− n
i=1 ti
2
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Page 19 of 22
n
Y
1 ti
b − ati
t
i=1 i
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b (ln b − ln a)n+1
≤
(1 − Tn ) .
4
(j)
JJ
Close
(j)
(j)
(j)
For b > a > 0, 1 ≥ ti ≥ 0 (i = 1, 2, . . . , n; n ≥ 2) and t1 +t2 +· · ·+tn = 1
(j = 1, 2), we have
(3.7)
n
Y
1 (2)
i=1
b
(2)
ti
−a
(2)
ti
−
ti
n
Y
1 (1)
i=1
(1)
(1)
bti − ati
ti
n−1
b (ln b − ln a)n+1 X (2)
(1)
≤
ti − ti .
3
i=1
For b > a > 0, 1 ≥ ti ≥ 0 (i = 1, 2, . . . , n; n ≥ 2) and Tn = 1, we have
Inequalities of Hadamard-type
for Lipschitzian Mappings
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
(3.8)
n−1
n
n b (ln b − ln a)n+1 X
Y
1
1
1 ti
ti
0≤
b − a − nb n − na n ≤
|nti − 1|
3n
t
i
i=1
i=1
Title Page
and
(3.9)
Contents
0≤
n
n−1
Y
b (ln b − ln a) X
1
1 ti
b−a
ti
−
b
−
a
≤
ti .
ln b − ln a (ln b − ln a)n i=1 ti
3
i=1
b−a
0≤
−
ln b − ln a
≤
n2
ln b − ln a
n
0≤
J
I
#
" 1
n
Y
b 2n2 a 2n12
(ab)
−
a
b
i=1
1
nb n − na n
ln b − ln a
Go Back
1
2
b(n2 − 1)
(ln b − ln a) ,
3n2
1
(3.11)
II
Page 20 of 22
For b > a > 0, we have
(3.10)
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!n
1
− (ab) 2 ≤
b (ln b − ln a)
4
and
(3.12)
n
 1
1
n − an
n
b
b−a
 ≤ b(n − 1) (ln b − ln a) .
0≤
−
ln b − ln a
ln b − ln a
3n
Indeed, (3.1) – (3.12) follow from (2.6) – (2.8), (2.15) – (2.17), (2.26) – (2.28),
(2.9), (2.18) and (2.29) applied to the convex function f : [ln a, ln b] 7→ [a, b], f (x) =
ex , with some simple manipulations, respectively.
Inequalities of Hadamard-type
for Lipschitzian Mappings
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
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Contents
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References
[1] J. HADAMARD, Etude sur les propriétés des fonctions entières et en particulier
d’une fonction considérée par Riemann, J. Math. Pures Appl., 58 (1893), 171–
215.
[2] L.-C. WANG, Three mapping related of Hermite-Hadamard inequalities, J.
Sichuan Univ., 39 (2002), 652–656. (Chinese).
[3] S.S. DRAGOMIR, Y.J. CHO AND S.S. KIM, Inequalities of Hadamard’s type for
Lipschitzian mappings and their applications, J. Math. Anal. Appl., 245 (2000),
489–501.
[4] L.-C. WANG, On extensions and refinements of Hermite-Hadamard inequalities
for convex functions, Math. Inequal. & Applics., 6(4) (2003), 659–666.
Inequalities of Hadamard-type
for Lipschitzian Mappings
Liang-Cheng Wang
vol. 8, iss. 1, art. 30, 2007
Title Page
Contents
[5] G.-S. YANG AND K.-L. TSENG, Inequalities of Hadamard’s type for Lipschitzian mappings, J. Math. Anal. Appl., 260 (2001), 230–238.
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[6] M. MATIĆ AND J. PEČARIĆ, Note on inequalities of Hadamard’s type for Lipschitzian mappings, Tamkang J. Math., 32(2) (2001), 127–130.
Page 22 of 22
[7] L.-C. WANG, New inequalities of Hadamard’s type for Lipschitzian mappings, J. Inequal. Pure Appl. Math., 6(2) (2005), Art. 37. [ONLINE: http:
//jipam.vu.edu.au/article.php?sid=506].
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[8] L.-C. WANG, Convex Functions and Their Inequalities, Sichuan University
Press, Chengdu, China, 2001. (Chinese).