Volume 8 (2007), Issue 1, Article 30, 11 pp.
ON NEW INEQUALITIES OF HADAMARD-TYPE FOR LIPSCHITZIAN
MAPPINGS AND THEIR APPLICATIONS
LIANG-CHENG WANG
S CHOOL OF M ATHEMATICA S CIENTIA ,
C HONGQING I NSTITUTE OF T ECHNOLOGY,
N O . 4 OF X INGSHENG L U , YANGJIA P ING 400050,
C HONGQING C ITY, CHINA.
wlc@cqit.edu.cn
Received 11 December, 2005; accepted 16 August, 2006
Communicated by F. Qi
A BSTRACT. In this paper, we study some new inequalities of Hadamard’s Type for Lipschitzian
mappings. some applications are also included.
Key words and phrases: Lipschitzian mappings, Hadamard inequality, Convex function.
2000 Mathematics Subject Classification. Primary 26D07; Secondary 26B25, 26D15.
1. I NTRODUCTION
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
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 1, 0, . . . , 0) (11 is
1
n
ith component, i = 1, 2, . . . , n) are special points in R ; G = 0, n × 0, n × · · · × 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), H = {t ∈ I|Tn ≤ 1} and L = {t ∈ I|Tn =
n
1} are subsets in Rn .
This author is partially supported by the Key Research Foundation of the Chongqing Institute of Technology under Grant 2004ZD94.
362-05
2
L IANG -C HENG WANG
We list these mappings by
Rn : I 7→ R,
4
Rn (t) =
Sn : H 7→ R,
n
b−a
4
Sn (t) =
!
n 1X
xi−1 + xi
dY,
ti yi + (1 − ti )
n i=1
2
n Z
f
D
1
(b − a)n
Z
f
V
n
X
ti yi + (1 − Tn )
i=1
a+b
2
!
dY
and
Pn : L 7→ R,
1
4
Pn (t) =
(b − a)n
Z
f
V
n
X
!
ti yi dY.
i=1
We write Pn+1 in the following equivalent form
! #
Z "Z b
n
X
1
4
Pn+1 : H 7→ R, Pn+1 (t) =
f
ti yi + (1 − Tn )x dx dY.
(b − a)n+1 V
a
i=1
(j)
(1)
(2)
(j)
Let g : A ⊆ Rn → R. For all t(j) = t1 , . . . , tn
∈ A(j = 1, 2) with 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
yi dY
=
f
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
for all t ∈ G,
Sn (t) ≤ Pn+1 (t)
(1.4)
for all t ∈ H, and
(1.5)
Z b
1
1
1
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.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 30, 11 pp.
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I NEQUALITIES OF H ADAMARD - TYPE FOR L IPSCHITZIAN M APPINGS
3
2. M AIN R ESULTS
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:
Theorem 2.1. Let f : [a, b] → R be an M −Lipschitzian mapping, then we have
n X
(2)
(1) Rn (t(2) ) − Rn (t(1) ) ≤ M (b − a)
ti − ti 4n2
i=1
(2.1)
(j)
(j)
for any t(j) = t1 , . . . , tn ∈ I(j = 1, 2),
a
+
b
f
≤ M (b − a)Tn
−
R
(t)
n
4n2
2
(2.2)
and
(2.3)
n Z
n
f
Rn (t) −
b−a
D
!
n
1X
M
yi dY ≤ 2 (b − a) (n − Tn )
4n
n i=1
for all t ∈ I, and
(2.4)
n Z
n
f
b−a
D
Proof. (1) For xi =
(b−a)i
n
!
Z b
n
M (n2 − 1)
X
1
1
yi dY −
f (x)dx ≤
(b − a).
n i=1
b−a a
3n2
(i = 0, 1, . . . , n; x0 = a, xn = b), from integral properties, we have
Rn t(2) − Rn t(1) n Z !
n X
n
1
x
+
x
i−1
i
(2)
(2)
≤
ti yi + (1 − ti )
f
b−a
n
2
D
i=1
!
n X
1
x
+
x
i−1
i
(1)
(1)
−f
t yi + (1 − ti )
dY
n i=1 i
2
n
Z X
n n
M
x
+
x
i−1
i (2)
(1)
≤
·
t − ti
yi −
dY
b−a
n D i=1 i
2
n
n
Z n
M X (2)
xi−1 + xi (1) ≤
·
t − ti dY
y i −
b−a
n i=1 i
2
D
n
n
b − a n−1 Z xi n
M X (2)
xi−1 + xi (1) =
·
t − ti y i −
dyi
b−a
n i=1 i
n
2
xi−1
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 30, 11 pp.
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4
L IANG -C HENG WANG
"
n
Z xi−12+xi x + x
M X (2)
i−1
i
(1) =
− yi dyi
t − ti b − a i=1 i
2
xi−1
#
Z xi
xi−1 + xi
+ x +x yi −
dyi
i−1
i
2
2
n X
M
(2)
(1) = 2 (b − a)
ti − ti .
4n
i=1
This completes the proof of (2.1).
(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 xi
n Z
X
n
f (yi )dY.
(2.5)
f (x)dx =
f (yi )dyi =
b−a
a
i=1 xi−1
i=1 D
Using (2.5) and integral properties, we obtain
!
n Z
Z b
n
X
n
1
1
f
yi dY −
f (x)dx
b−a
n i=1
b−a a
D
!
n Z X
n
n
n
n
1X
1X
1
≤
f
yj −
f (yi ) dY
b−a
n j=1
n i=1
D n i=1
n
Z X
n X
n
M
n
1
·
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
n
x
x
i
j
n
MX X
=
·
yi dyi −
yj dyj
b − a n2 i=1 j=1
xi−1
xj−1
!#
Z xj
Z xi
n
X
+
yj dyj −
yi dyi
xj−1
j=i+1
n
n
MX
=
·
b − a n2 i=1
=
"
b−a
n
xi−1
2
i−1
X
j=1
(i − j) +
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.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 30, 11 pp.
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I NEQUALITIES OF H ADAMARD - TYPE FOR L IPSCHITZIAN M APPINGS
5
Corollary 2.2. Let f be convex on [a, b], with f+0 (a) and f−0 (b) existing. Then we obtain
0 ≤ Rn (t(2) ) − Rn (t(1) )
(2.6)
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
(2.7)
0 ≤ Rn (t) − f
≤
(b − a)Tn
2
4n2
and
n
b−a
0≤
(2.8)
≤
!
n
1X
yi dY − Rn (t)
n i=1
n Z
f
D
max{|f+0 (a)|, |f−0 (b)|}
(b − a)(n − Tn )
4n2
for all t ∈ I, and
1
0≤
b−a
(2.9)
Z
b
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
n X
(2)
(1) Sn (t(2) ) − Sn (t(1) ) ≤ M (b − a)
(2.11)
ti − ti 4
i=1
(j)
(j)
for any t(j) = (t1 , . . . , tn ) ∈ H(j = 1, 2),
M
a+b
(2.12)
− Sn (t) ≤
(b − a)Tn
f
2
4
and
(2.13)
Z
1
f
Sn (t) −
(b − a)n V
!
n
n
M
X
1X
yi dY ≤
(b − a)
|nti − 1|
4n
n i=1
i=1
for all t ∈ H, and
Z
a+b
1
(2.14)
−
f
f
2
(b − a)n V
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 30, 11 pp.
!
n
M
1X
yi dY ≤
(b − a).
n i=1
4
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6
L IANG -C HENG WANG
Proof. (1) From integral properties, we obtain
Sn t(2) − Sn t(1) !
!
Z n
n
X
X
1
a
+
b
(2)
(2)
≤
ti yi + 1 −
ti
f
n
(b − a) V
2
i=1
i=1
!
!
n
n
X
X
a + b (1)
(1)
−f
ti yi + 1 −
ti
dY
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
M
(2)
(1) yi − a + b dY
≤
t
−
t
i
i
(b − a)n i=1
2 V
n
Z b M X (2)
a + b (1) =
t − ti yi − 2 dyi
b − a i=1 i
a
"
#
Z b n
Z a+b
2
a
+
b
M X (2)
a
+
b
(1) − yi dyi +
dyi
=
yi −
t − ti a+b
2
2
b − a i=1 i
a
2
n X
M
(2)
(1) =
(b − a)
ti − ti .
4
i=1
This completes the proof of (2.11).
(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
(2.15)
0 ≤ Sn (t(2) ) − Sn (t(1) ) ≤
(j)
X (2)
max{|f+0 (a)|, |f−0 (b)|}
(1)
(b − a)
ti − ti
4
i=1
(j)
(2)
(1)
for any t(j) = (t1 , . . . , tn ) ∈ G(j = 1, 2) with ti ≥ ti (i = 1, 2, . . . , n),
max{|f+0 (a)|, |f−0 (b)|}
a+b
(2.16)
0 ≤ Sn (t) − f
≤
(b − a)Tn
2
4
and
!
Z
n
1
1X
(2.17)
0≤
f
yi dY − Sn (t)
(b − a)n V
n i=1
max{|f+0 (a)|, |f−0 (b)|}
≤
(b − a)(1 − Tn )
4
for all t ∈ G, and
(2.18)
1
0≤
(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
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 30, 11 pp.
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I NEQUALITIES OF H ADAMARD - TYPE FOR L IPSCHITZIAN M APPINGS
7
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)
for any t
(2.20)
=
(j)
(j)
t1 , . . . , tn
Z
1
f
(b − a)n V
n−1 X
M
(2)
(1) (b − a)
ti − ti 3
i=1
∈ L(j = 1, 2),
!
n
n−1
M
X
1X
yi dY − pn (t) ≤
(b − a)
|nti − 1|
3n
n i=1
i=1
and
(2.21)
Z b
n−1
X
M
1
≤
pn (t) −
f
(x)dx
(b
−
a)
ti
b−a a
3
i=1
for all t ∈ L, and
Z
1
f
(2.22)
(b − a)n V
!
Z b
n
M (n − 1)
1X
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)
|Sn (t) − Pn+1 (t)| ≤
M
(b − a)(1 − Tn ).
4
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)
≤
ti yi + 1 −
ti
yn
f
n
(b − a) V
i=1
i=1
! !
n−1
n−1
X
X
(1)
(1)
−f
ti yi + 1 −
ti
yn dY
i=1
i=1
Z X
n−1 M
(2)
(1)
≤
t
−
t
(y
−
y
)
dY
i
n
i
i
n
(b − a) V i=1
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8
L IANG -C HENG WANG
Z bZ b
n−1 X
M
(2)
(1) n−2
≤
|yi − yn | dyi dyn
t − ti (b − a)
(b − a)n i=1 i
a
a
Z b
n−1 Z b Z x
X
M
(2)
(1) =
(x − y)dy +
(y − x)dy dx
t − ti (b − a)2 i=1 i
a
a
x
=
n−1 X
M
(2)
(1) (b − a)
ti − ti .
3
i=1
This completes the proof of (2.19).
(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
Using (2.25) and integral properties, we obtain
|Sn (t) − Pn+1 (t)|
!
! #
Z "Z b n
n
X
X
1
a+b
≤
f
t
y
+
(1
−
T
)
−
f
t
y
+
(1
−
T
)x
dx dY
i
i
n
i
i
n
n+1
(b − a)
2
V
a i=1
i=1
Z Z b a + b
M
dx dY
≤
(1
−
T
)
−
x
n
2
(b − a)n+1
V
a
"Z a+b #
Z b 2
a+b
M
a+b
− x dx +
dx
(1 − Tn )
x−
=
a+b
2
2
b−a
a
2
M
(b − a)(1 − Tn ).
4
This completes the proof of (2.23).
This completes the proof of Theorem 2.5.
=
Corollary 2.6. Let f be defined as in Corollary 2.2. For n ≥ 2, then we have
n−1 X
max{|f+0 (a)|, |f−0 (b)|}
(2)
(1) (2)
(1) (2.26)
Pn (t ) − Pn (t ) ≤
(b − a)
ti − ti 3
i=1
(j)
(j)
for any t(j) = (t1 , . . . , tn ) ∈ L(j = 1, 2) ,
1
0 ≤ Pn (t) −
(b − a)n
(2.27)
Z
f
V
!
n
1X
yi dY
n i=1
n−1
X
max{|f+0 (a)|, |f−0 (b)|}
≤
(b − a)
|nti − 1|
3n
i=1
and
(2.28)
1
0≤
b−a
Z
a
b
n−1
X
max{|f+0 (a)|, |f−0 (b)|}
f (x)dx − Pn (t) ≤
(b − a)
ti
3
i=1
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 30, 11 pp.
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I NEQUALITIES OF H ADAMARD - TYPE FOR L IPSCHITZIAN M APPINGS
9
for all t ∈ L, and
1
0≤
b−a
(2.29)
b
Z
a
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
≤
max{|f+0 (a)|, |f−0 (b)|}
(b − a)(1 − Tn ).
4
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).
(2.30)
0 ≤ Pn+1 (t) − Sn (t) ≤
Remark 2.7. 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 2.8. 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.
3. A PPLICATIONS
In this section, we agree that when ti = 0,
" ti
#
t
1
b 2n2 a 2ni2
ln b − ln a
−
=
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




(2)
(1)
t
t
(2)
(1)
i
i
t
t
n
n
Y
1  b 2n2 a 2ni 2  Y 1  b 2n2 a 2ni 2 
(3.1)
0≤
−
−
−


(2)  a
(1)  a
b
b
t
t
i=1 i
i=1 i
1
≤
4
ln b − ln a
n2
n+1 12 X
n b
(2)
(1)
ti − ti ,
a
i=1
" ti
#
n Y
t
n
n2
1
b 2n2 a 2ni2
0≤
−
−1
ln b − ln a
t
a
b
i
i=1
12
ln b − ln a b
≤
Tn
4n2
a
(3.2)
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=1
i=1 i
n+1 12
1 ln b − ln a
b
≤
(n − Tn ) .
2
4
n
a
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 30, 11 pp.
http://jipam.vu.edu.au/
10
L IANG -C HENG WANG
For b > a > 0,
(3.4)
1
n
0 ≤ (ab)
(2)
(1)
≥ ti ≥ ti ≥ 0 and
n
Y
1 P
(2)
1− n
i=1 ti
2
(2)
i=1
n+1
≤
b (ln b − ln a)
4
1
n
≥ ti ≥ 0 (i = 1, 2, . . . , n), we have
(2)
(2)
bti − ati
ti
n X
(2)
(1)
ti − ti
n
Y
1 (1)
(1)
(1)
bti − ati
ti
,
i=1
Pn
i=1 ti
2
≤
P
(1)
1− n
i=1 ti
2
i=1
1−
1
0≤
n (ab)
(ln b − ln a)
(3.5)
− (ab)
n
Y
1
1 ti
b − ati − (ab) 2
t
i=1 i
b (ln b − ln a)
Tn
4
and
P
n
1
n
Y
1− n
i=1 ti
1
1 ti
0 ≤ nb n − na n − (ab) 2
b − ati
t
i=1 i
(3.6)
≤
b (ln b − ln a)n+1
(1 − Tn ) .
4
(j)
(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
n
n
n−1 Y 1 (2)
Y
(1)
b (ln b − ln a)n+1 X
(2)
(1) 1
(2)
(1) ti
ti
ti
ti
b −a
−
b −a
(3.7) ≤
ti − ti .
(2)
(1)
3
i=1 ti
i=1 ti
i=1
For b > a > 0, 1 ≥ ti ≥ 0 (i = 1, 2, . . . , n; n ≥ 2) and Tn = 1, we have
(3.8)
n
n−1
n b (ln b − ln a)n+1 X
Y
1
1
1 ti
ti
n
n
b − a − nb − na
|nti − 1|
0≤
≤
t
3n
i=1 i
i=1
and
(3.9)
n
n−1
Y
b (ln b − ln a) X
b−a
1
1 ti
ti
−
b −a ≤
ti .
0≤
ln b − ln a (ln b − ln a)n i=1 ti
3
i=1
For b > a > 0, we have
(3.10)
b−a
0≤
−
ln b − ln a
≤
n
" 1
#
n
Y
b 2n2 a 2n12
(ab)
−
a
b
i=1
1
2
b(n2 − 1)
(ln b − ln a) ,
3n2
1
(3.11)
n2
ln b − ln a
0≤
1
nb n − na n
ln b − ln a
!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
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 30, 11 pp.
http://jipam.vu.edu.au/
I NEQUALITIES OF H ADAMARD - TYPE FOR L IPSCHITZIAN M APPINGS
11
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.
R EFERENCES
[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.
[5] G.-S. YANG AND K.-L. TSENG, Inequalities of Hadamard’s type for Lipschitzian mappings, J.
Math. Anal. Appl., 260 (2001), 230–238.
[6] M. MATIĆ AND J. PEČARIĆ, Note on inequalities of Hadamard’s type for Lipschitzian mappings,
Tamkang J. Math., 32(2) (2001), 127–130.
[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].
[8] L.-C. WANG, Convex Functions and Their Inequalities, Sichuan University Press, Chengdu, China,
2001. (Chinese).
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 30, 11 pp.
http://jipam.vu.edu.au/