Journal of Inequalities in Pure and
Applied Mathematics
NEW INEQUALITIES OF HADAMARD’S TYPE FOR
LIPSCHITZIAN MAPPINGS
volume 6, issue 2, article 37,
2005.
LIANG-CHENG WANG
School of Mathematics Scientia
Chongqing Institute of Technology
Xingsheng Lu 4, Yangjia Ping 400050
Chongqing City, China.
Received 16 September, 03;
accepted 06 February, 2005.
Communicated by: F. Qi
EMail: wlc@cqit.edu.cn
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
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Abstract
In this paper, we study several new inequalities of Hadamard’s type for Lipschitzian mappings.
2000 Mathematics Subject Classification: Primary 26D07; Secondary 26B25,
26D15.
Key words: Lipschitzian mappings, Hadamard inequality, Convex function.
New Inequalities of Hadamard’s
Type for Lipschitzian Mappings
This author is partially supported by the Key Research Foundation of the Chongqing
Institute of Technology under Grant 2004ZD94.
Liang-Cheng Wang
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
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1.
Introduction
Let us begin by defining some mappings that we shall need. If f is a continuous
function on a closed interval [a, b] (a < b), and for any t ∈ (0, 1), let u =
ta + (1 − t)b, then we can define
Z u Z b
1
4
A(t) =
f (tx + (1 − t)y) dy dx,
t(1 − t)(b − a)2 a
u
Z u Z b 1
(b − y)x + (y − u)u
4
B(t) =
f
dy dx
(1 − t)(b − a)2 a
t(b − a)
u
Z u Z b 1
(u − x)u + (x − a)y
+
f
dy dx,
t(b − a)2 a
(1 − t)(b − a)
u
Z u
Z b
t
1−t
4
C(t) =
f (x)dx +
f (y)dy.
(1 − t)(b − a) a
t(b − a) u
Let f be a continuous convex function on [a, b], then
Z b
1
f (a) + f (b)
a+b
≤
f (x)dx ≤
.
(1.1)
f
2
b−a a
2
The inequalities in (1.1) are called Hadamard inequalities [1] – [7]. In [2], the
author of this paper gave extensions and refinements of the inequalities in (1.1).
He obtained the following:
(1.2)
f (ta + (1 − t)b) ≤ A(t) ≤ B(t) ≤ C(t) ≤ tf (a) + (1 − t)f (b).
New Inequalities of Hadamard’s
Type for Lipschitzian Mappings
Liang-Cheng Wang
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Recently, Dragomir et al. [3], Yang and Tseng [4] and Matic and Pečarić [5]
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 A, B and C (or to (1.2)).
New Inequalities of Hadamard’s
Type for Lipschitzian Mappings
Liang-Cheng Wang
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2.
Main Results
For the mapping A(t), we have the following theorem:
Theorem 2.1. Let f : [a, b] → R be a M -Lipschitzian mapping. For any
t ∈ (0, 1), then
(2.1)
|A(t) − f (ta + (1 − t)b)| ≤
M
t(1 − t)(b − a).
3
Proof. Obviously, we have
(2.2)
b − u = t(b − a), u − a = (1 − t)(b − a).
From integral properties and (2.2), we have
Z u Z b
1
(2.3) f (ta + (1 − t)b) =
f (ta + (1 − t)b) dy dx.
t(1 − t)(b − a)2 a
u
t
u
x + 1−t
, then we have
For x ∈ [a, u], when y ∈ u, − 1−t
ta + (1 − t)b = u ≥ tx + (1 − t)y,
t
u
if y ∈ − 1−t
x + 1−t
, b , then the inequality (2.4) reverses.
Using (2.2) – (2.4) and integral properties, we obtain
(2.4)
|A(t) − f (ta + (1 − t)b)|
Z u Z b
1
≤
|f (tx + (1 − t)y) −f (u)| dy dx
t(1 − t)(b − a)2 a
u
New Inequalities of Hadamard’s
Type for Lipschitzian Mappings
Liang-Cheng Wang
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M
≤
t(1 − t)(b − a)2
Z
M
=
t(1 − t)(b − a)2
Z
u
b
Z
|tx + (1 − t)y − u| dy dx
a
a
u
u
"Z
t
u
− 1−t
x+ 1−t
#
(u − (tx + (1 − t)y)) dy dx
u
#
Z u "Z b
M
(tx + (1 − t)y − u) dy dx
+
u
t
t(1 − t)(b − a)2 a
x+ 1−t
− 1−t
Z u 2
M
t
1+t
2
=
x +t b−
u x
t(1 − t)(b − a)2 a 1 − t
1−t
1 + t2 2 1 − t 2
u +
b − bu dx
+
2(1 − t)
2
t2
t
1+t
M
2
2
(u + ua + a ) +
b−
u (u + a)
=
t(b − a) 3(1 − t)
2
1−t
1 + t2 2 1 − t 2
u +
b − bu
+
2(1 − t)
2
2
2t − 3t + 3
M
=
(ta + (1 − t)b)2
t(b − a)
6(1 − t)
−t(t + 3)a − 3(t2 − 3t + 2)b
+
(ta + (1 − t)b)
6(1 − t)
t2
t
1−t 2
M
+
a2 + ab +
b =
t(1 − t)(b − a).
3(1 − t)
2
2
3
This completes the proof of Theorem 2.1.
New Inequalities of Hadamard’s
Type for Lipschitzian Mappings
Liang-Cheng Wang
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For the mapping B(t), we have the following theorem:
Theorem 2.2. Let f be defined as in Theorem 2.1. For any t ∈ (0, 1), then
|B(t) − A(t)| ≤
(2.5)
M
t(1 − t)(b − a),
2
and
|B(t) − f (ta + (1 − t)b)| ≤
(2.6)
Proof. From
1
t(1−t)
(2.7)
A(t) =
=
1
t
+
1
,
1−t
we have
1
(b − a)2
1
1
+
t 1−t
M
t(1 − t)(b − a).
2
New Inequalities of Hadamard’s
Type for Lipschitzian Mappings
Liang-Cheng Wang
Z
u
Z
b
f (tx + (1 − t)y) dy dx.
a
u
Let F be defined by
F (y) = tx + (1 − t)y −
(b − y)x + (y − u)u
.
t(b − a)
Obviously, if F (y) is the function of first degree for y, then F (y) is monotone
with y on [u, b]. And for any x ∈ [a, u], F (u) = tx + (1 − t)u − x ≥ 0,
F (b) = tx + (1 − t)b − u ≥ 0. Hence for x ∈ [a, u] and y ∈ [u, b], we get
F (y) ≥ 0, i.e.,
(2.8)
tx + (1 − t)y ≥
(b − y)x + (y − u)u
.
t(b − a)
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Let G be defined by
G(x) =
(u − x)u + (x − a)y
− (tx + (1 − t)y) .
(1 − t)(b − a)
Using G(x) and the same method as in the proof of (2.8), we obtain
(2.9)
(u − x)u + (x − a)y
≥ tx + (1 − t)y,
(1 − t)(b − a)
where, x ∈ [a, u] and y ∈ [u, b].
Using (2.7) – (2.9) and (2.2), we obtain
New Inequalities of Hadamard’s
Type for Lipschitzian Mappings
Liang-Cheng Wang
|B(t) − A(t)|
Z uZ b (b − y)x + (y − u)u
1
≤
f
−f (tx + (1 − t)y) dy dx
2
(1 − t)(b − a) a
t(b − a)
u
Z u Z b (u − x)u + (x − a)y
1
f
dy dx
+
−f
(tx
+
(1
−
t)y)
t(b − a)2 a u (1 − t)(b − a)
Z u Z b (b − y)x + (y − u)u
M
≤
− (tx + (1 − t)y) dy dx
2
(1 − t)(b − a) a
t(b − a)
u
Z u Z b (u − x)u + (x − a)y
M
dy dx
+
−
(tx
+
(1
−
t)y)
(1 − t)(b − a)
t(b − a)2 a
u
Z u Z b M
(b − y)x + (y − u)u
=
tx + (1 − t)y −
dy dx
(1 − t)(b − a)2 a
t(b − a)
u
Z u Z b M
(u − x)u + (x − a)y
+
− (tx + (1 − t)y) dy dx
t(b − a)2 a
(1 − t)(b − a)
u
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M
=
t(1 − t)(b − a)3
Z
u
Z
b
(b − a)(2t − 1) (tx + (1 − t)y)
2
+ 2xy − (b + u)x − (a + u)y + 2u dy dx
Z u
M
1
=
((b − a)(2t − 1)(1 − t) − (a + u)) (b + u)
2
(1 − t)(b − a) a 2
2
+ (b − a)(2t − 1)tx + 2u dx
a
u
M
((b − a)(2t − 1)(1 − t) − (a + u)) (b + u)
2(b − a)
+ (b − a)(2t − 1)t(u + a) + 4u2
M
t(1 − t)(b − a).
=
2
=
This completes the proof of inequality (2.5).
1
1
From (2.3) and t(1−t)
= 1t + 1−t
, we have
(2.10) f (ta + (1 − t)b)
Z u Z b
1
1
1
=
+
f (ta + (1 − t)b) dy dx.
t 1 − t (b − a)2 a
u
If x ∈ [a, u] and y ∈ [u, b], then we have
(2.11)
New Inequalities of Hadamard’s
Type for Lipschitzian Mappings
ta + (1 − t)b = u ≥
(b − y)x + (y − u)u
,
t(b − a)
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(u − x)u + (x − a)y
≥ ta + (1 − t)b = u.
(1 − t)(b − a)
Using (2.10)-(2.11) and (2.2), we obtain
|B(t) − f (ta + (1 − t)b)|
Z u Z b 1
(b − y)x + (y − u)u
≤
− f (u) dy dx
f
2
(1 − t)(b − a) a
t(b − a)
u
Z u Z b 1
(u
−
x)u
+
(x
−
a)y
f
dy dx
−
f
(u)
+
t(b − a)2 a
(1 − t)(b − a)
u
Z u Z b (b − y)x + (y − u)u
M
− u dy dx
≤
2
(1 − t)(b − a) a
t(b − a)
u
Z u Z b (u − x)u + (x − a)y
M
dy dx
−
u
+
(1 − t)(b − a)
t(b − a)2 a
u
Z u Z b (b − y)x + (y − u)u
M
=
u−
dy dx
(1 − t)(b − a)2 a
t(b − a)
u
Z u Z b M
(u − x)u + (x − a)y
+
− u dy dx
t(b − a)2 a
(1 − t)(b − a)
u
Z uZ b
M
=
(b − a)(2t − 1)u
t(1 − t)(b − a)3 a
u
2
+ (2x − u − a)y − (b + u)x + 2u dy dx
Z u
M
1
2
=
− (u + a)(b + u) + (b − a)(2t − 1)u + 2u dx
(1 − t)(b − a)2 a
2
New Inequalities of Hadamard’s
Type for Lipschitzian Mappings
Liang-Cheng Wang
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M
1
2
=
− (u + a)(b + u) + (b − a)(2t − 1)u + 2u
b−a
2
M
=
t(1 − t)(b − a).
2
This completes the proof of inequality (2.6).
This completes the proof of Theorem (2.2).
For the mapping C(t), we have the following theorem:
Theorem 2.3. Let f be defined as in Theorem 2.1. For any t ∈ (0, 1), then
|C(t) − B(t)| ≤
(2.12)
M
t(1 − t)(b − a),
2
and
(2.13)
Liang-Cheng Wang
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|C(t) − (tf (a) + (1 − t)f (b))| ≤ M t(1 − t)(b − a).
Proof. From the integral property and (2.2), we have
Z u Z b
1
(2.14) C(t) =
f (x)dy dx
(1 − t)(b − a)2 a
u
Z u Z b
1
+
f (y)dy dx.
t(b − a)2 a
u
x≤
(b − y)x + (y − u)u
,
t(b − a)
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If x ∈ [a, u] and y ∈ [u, b], then we have
(2.15)
New Inequalities of Hadamard’s
Type for Lipschitzian Mappings
(u − x)u + (x − a)y
≤ y.
(1 − t)(b − a)
J. Ineq. Pure and Appl. Math. 6(2) Art. 37, 2005
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Using (2.14) – (2.15) and (2.2), we obtain
|C(t) − B(t)|
≤
≤
=
=
=
Z u Z b 1
(b
−
y)x
+
(y
−
u)u
f
dy dx
−
f
(x)
(1 − t)(b − a)2 a
t(b − a)
u
Z u Z b (u − x)u + (x − a)y
1
dy dx
+
f
−
f
(y)
t(b − a)2 a
(1
−
t)(b
−
a)
u
Z u Z b (b − y)x + (y − u)u
M
dy dx
−
x
(1 − t)(b − a)2 a
t(b − a)
u
Z u Z b (u − x)u + (x − a)y
1
dy dx
+
−
y
(1 − t)(b − a)
t(b − a)2 a
u
Z u Z b M
(b − y)x + (y − u)u
− x dy dx
(1 − t)(b − a)2 a
t(b − a)
u
Z u Z b 1
(u − x)u + (x − a)y
+
y−
dy dx
t(b − a)2 a
(1 − t)(b − a)
u
Z uZ b
M
(b − a) ((1 − t)y − tx)
t(1 − t)(b − a)3 a
u
2
− 2xy + (b + u)x + (a + u)y − 2u dy dx
Z u
1
M
((b − a)(1 − t) + (a + u)) (b + u)
2
(1 − t)(b − a) a 2
2
− (b − a)tx − 2u dx
New Inequalities of Hadamard’s
Type for Lipschitzian Mappings
Liang-Cheng Wang
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M
1
2
=
(((b − a)(1 − t) + (a + u)) (b + u) − (b − a)t(u + a)) − 2u
b−a 2
M
=
t(1 − t)(b − a).
2
This completes the proof of inequality (2.12).
From integral property and (2.2), we have
Z u
Z b
t
1−t
(2.16) tf (a)+(1−t)f (b) =
f (a)dx+
f (b)dy.
(1 − t)(b − a) a
t(b − a) u
Using (2.16) and (2.2), we obtain
|C(t) − (tf (a) + (1 − t)f (b))|
Z u
Z
1
t
1−t b
≤
|f (x) − f (a)| dx +
|f (y) − f (b)| dy
b−a 1−t a
t
u
Z u
Z
M
t
1−t b
≤
|x − a|dx +
|y − b|dy
b−a 1−t a
t
u
Z u
Z
M
t
1−t b
=
(x − a)dx +
(b − y)dy
b−a 1−t a
t
u
M
t
1 2
1−t
1 2
2
2
=
(u − a ) − a(u − a) +
b(b − u) − (b − u )
b−a 1−t 2
t
2
= M t(1 − t)(b − a).
This completes the proof of inequality (2.13).
This completes the proof of Theorem 2.3.
New Inequalities of Hadamard’s
Type for Lipschitzian Mappings
Liang-Cheng Wang
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Corollary 2.4. Let f be defined as in Theorem 2.1. For any t ∈ (0, 1), then
(2.17)
|C(t) − f (ta + (1 − t)b)| ≤ M t(1 − t)(b − a),
(2.18)
|A(t) − C(t)| ≤ M t(1 − t)(b − a),
(2.19)
|A(t) − (tf (a) + (1 − t)f (b))| ≤ 2M t(1 − t)(b − a),
New Inequalities of Hadamard’s
Type for Lipschitzian Mappings
Liang-Cheng Wang
(2.20)
|B(t) − (tf (a) + (1 − t)f (b))| ≤
3M
t(1 − t)(b − a),
2
and
(2.21)
|f (ta + (1 − t)b) − (tf (a) + (1 − t)f (b))| ≤ 2M t(1 − t)(b − a).
Proof. Using (2.5) and (2.12), we get inequality (2.18):
|A(t) − C(t)| ≤ |A(t) − B(t)| + |B(t) − C(t)| ≤ M t(1 − t)(b − a).
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Using the same method as the proof of (2.18), from (2.6) and (2.12), (2.13) and
(2.18), (2.12) and (2.13), and (2.13) and (2.17), we get (2.17), (2.19), (2.20) and
(2.21), respectively.
This completes the proof of Corollary 2.4.
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Remark 1. If we let t = 21 , then (2.13), (2.17) and (2.21) reduce to
(2.22)
Z b
1
f (a) + f (b) M
f (x)dx −
b − a
≤ 4 (b − a),
2
a
(2.23)
Z b
M
1
a
+
b
≤
f
(x)dx
−
f
(b − a),
b − a
2
4
a
New Inequalities of Hadamard’s
Type for Lipschitzian Mappings
(2.24)
M
a
+
b
f
(a)
+
f
(b)
f
≤
−
(b − a).
2
2
2
(2.22) is better than (2.2) in [3]. (2.23) and (2.24) are (2.1) and (2.4) in [3],
respectively.
Corollary 2.5. Let f be a convex mapping on [a, b], with f+0 (a) and f−0 (b) existing. For any t ∈ (0, 1), then we have
(2.25)
0 ≤ A(t) − f (ta + (1 − t)b)
max{|f+0 (a)|, |f−0 (b)|}
≤
t(1 − t)(b − a),
3
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(2.26)
0 ≤ B(t) − A(t)
max{|f+0 (a)|, |f−0 (b)|}
≤
t(1 − t)(b − a),
2
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(2.27)
0 ≤ B(t) − f (ta + (1 − t)b)
max{|f+0 (a)|, |f−0 (b)|}
t(1 − t)(b − a),
≤
2
(2.28)
0 ≤ C(t) − B(t)
max{|f+0 (a)|, |f−0 (b)|}
≤
t(1 − t)(b − a),
2
New Inequalities of Hadamard’s
Type for Lipschitzian Mappings
(2.29)
0 ≤ tf (a) + (1 − t)f (b) − C(t)
≤ max{|f+0 (a)|, |f−0 (b)|}t(1 − t)(b − a),
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(2.30)
(2.31)
0 ≤ C(t) − f (ta + (1 − t)b)
≤ max{|f+0 (a)|, |f−0 (b)|}t(1 − t)(b − a),
0 ≤ C(t) − A(t)
≤ max{|f+0 (a)|, |f−0 (b)|}t(1 − t)(b − a),
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(2.32)
0 ≤ tf (a) + (1 − t)f (b) − A(t)
≤ 2 max{|f+0 (a)|, |f−0 (b)|}t(1 − t)(b − a),
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(2.33)
0 ≤ tf (a) + (1 − t)f (b) − B(t)
3
≤ max{|f+0 (a)|, |f−0 (b)|}t(1 − t)(b − a),
2
and
(2.34)
0 ≤ tf (a) + (1 − t)f (b) − f (ta + (1 − t)b)
≤ 2 max{|f+0 (a)|, |f−0 (b)|}t(1 − t)(b − a).
Proof. For any x, y ∈ [a, b], from the properties of convex functions, we have
the following max{|f+0 (a)|, |f−0 (b)|}-Lipschitzian inequality (see [7]):
(2.35)
New Inequalities of Hadamard’s
Type for Lipschitzian Mappings
Liang-Cheng Wang
|f (x) − f (y)| ≤ max{|f+0 (a)|, |f−0 (b)|}|x − y|.
Using (1.2), (2.35), Theorem 2.1 – 2.2 and Corollary 2.4, we obtain (2.25) –
(2.34).
This completes the proof of Corollary (2.5).
Remark 2. The condition in Corollary 2.5 is better than that in Corollary 2.2,
Corollary 4.2 and Theorem 3.3 of [3]. Actually, it is that f is a differentiable
convex mapping on [a, b] with M = sup |f 0 (t)| < ∞.
t∈[a,b]
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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, On extensions and refinements of Hermite-Hadamard inequalities for convex functions, Math. Ineq. Appl., 6(4) (2003), 659–666.
[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] G.-S. YANG AND K.-L. TSENG, Inequalities of Hadamard’s type for Lipschitzian mappings, J. Math. Anal. Appl., 260 (2001), 230–238.
[5] M. MATIC AND J. PEČARIĆ, Note on inequalities of Hadamard’s type for
Lipschitzian mappings, Tamkang J. Math., 32(2) (2001), 127–130.
[6] L.-C. WANG, On some extentions of Hadamard inequalities of convex
function, Math. in Practice and Theory, 32 (2002), 1027–1030. (In Chinese).
[7] L.-C. WANG, Convex Functions and Their Inequalities, Sichuan University
Press, Chengdu, China, 2001, pp.55. (In Chinese).
New Inequalities of Hadamard’s
Type for Lipschitzian Mappings
Liang-Cheng Wang
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