Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 2, Article 37, 2005
NEW INEQUALITIES OF HADAMARD’S TYPE FOR LIPSCHITZIAN MAPPINGS
LIANG-CHENG WANG
S CHOOL OF M ATHEMATICS S CIENTIA
C HONGQING I NSTITUTE OF T ECHNOLOGY
X INGSHENG L U 4, YANGJIA P ING 400050
C HONGQING C ITY, C HINA .
wlc@cqit.edu.cn
Received 16 September, 03; accepted 06 February, 2005
Communicated by F. Qi
A BSTRACT. In this paper, we study several new inequalities of Hadamard’s type for Lipschitzian
mappings.
Key words and phrases: Lipschitzian mappings, Hadamard inequality, Convex function.
2000 Mathematics Subject Classification. Primary 26D07; Secondary 26B25, 26D15.
1. I NTRODUCTION
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
a+b
1
f (a) + f (b)
(1.1)
f
≤
f (x)dx ≤
.
2
b−a a
2
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
This author is partially supported by the Key Research Foundation of the Chongqing Institute of Technology under Grant 2004ZD94.
123-03
2
L IANG -C HENG WANG
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).
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)).
2. M AIN R ESULTS
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
For x ∈ [a, u], when y ∈ u, − 1−t
x + 1−t
, then we have
(2.4)
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
|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
Z u Z b
M
≤
|tx + (1 − t)y − u| dy dx
t(1 − t)(b − a)2 a
u
#
Z u "Z − t x+ u
1−t
1−t
M
=
(u − (tx + (1 − t)y)) dy dx
t(1 − t)(b − a)2 a
u
#
Z u "Z b
M
+
(tx + (1 − t)y − u) dy dx
t
u
t(1 − t)(b − a)2 a
− 1−t
x+ 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
J. Inequal. Pure and Appl. Math., 6(2) Art. 37, 2005
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N EW I NEQUALITIES OF H ADAMARD ’ S T YPE FOR L IPSCHITZIAN M APPINGS
M
t2
t
1+t
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)
t
t2
1−t 2
2
+
a + ab +
b
3(1 − t)
2
2
M
=
t(1 − t)(b − a).
3
This completes the proof of Theorem 2.1.
3
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
(2.7)
1
t(1−t)
M
t(1 − t)(b − a).
2
1
+ 1−t
, we have
Z u Z b
1
1
1
A(t) =
+
f (tx + (1 − t)y) dy dx.
t 1 − t (b − a)2 a
u
=
1
t
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)
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].
J. Inequal. Pure and Appl. Math., 6(2) Art. 37, 2005
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4
L IANG -C HENG WANG
Using (2.7) – (2.9) and (2.2), we obtain
|B(t) − A(t)|
≤
≤
=
=
=
Z u Z b 1
(b
−
y)x
+
(y
−
u)u
f
dy dx
−
f
(tx
+
(1
−
t)y)
(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
(tx
+
(1
−
t)y)
t(b − a)2 a
(1
−
t)(b
−
a)
u
Z u Z b
(b − y)x + (y − u)u
M
dy dx
−
(tx
+
(1
−
t)y)
(1 − t)(b − a)2 a
t(b − a)
u
Z u Z b (u − x)u + (x − a)y
M
+
− (tx + (1 − t)y) dy dx
2
t(b − a) a
(1 − t)(b − 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
Z u Z b
M
(b − a)(2t − 1) (tx + (1 − t)y)
t(1 − t)(b − a)3 a
u
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
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) =
1
1
+
t 1−t
1
(b − a)2
Z
u
Z
b
f (ta + (1 − t)b) dy dx.
a
u
If x ∈ [a, u] and y ∈ [u, b], then we have
(2.11)
ta + (1 − t)b = u ≥
(b − y)x + (y − u)u
,
t(b − a)
(u − x)u + (x − a)y
≥ ta + (1 − t)b = u.
(1 − t)(b − a)
J. Inequal. Pure and Appl. Math., 6(2) Art. 37, 2005
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N EW I NEQUALITIES OF H ADAMARD ’ S T YPE FOR L IPSCHITZIAN M APPINGS
5
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
dy dx
f
≤
−
f
(u)
(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
(u)
t(b − a)2 a
(1
−
t)(b
−
a)
u
Z u Z b
(b − y)x + (y − u)u
M
dy dx
≤
−
u
(1 − t)(b − a)2 a
t(b − a)
u
Z u Z b (u − x)u + (x − a)y
M
+
− u dy dx
2
t(b − a) a
(1 − t)(b − a)
u
Z u Z b M
(b − y)x + (y − u)u
=
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 u Z b
M
2
(b − a)(2t − 1)u + (2x − u − a)y − (b + u)x + 2u dy dx
=
t(1 − t)(b − a)3 a
u
Z u
M
1
2
=
− (u + a)(b + u) + (b − a)(2t − 1)u + 2u dx
(1 − t)(b − a)2 a
2
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)
|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
Z u Z b
1
1
(2.14)
C(t) =
f (x)dy dx +
f (y)dy dx.
(1 − t)(b − a)2 a
t(b − a)2 a
u
u
If x ∈ [a, u] and y ∈ [u, b], then we have
(2.15)
x≤
(b − y)x + (y − u)u
,
t(b − a)
J. Inequal. Pure and Appl. Math., 6(2) Art. 37, 2005
(u − x)u + (x − a)y
≤ y.
(1 − t)(b − a)
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6
L IANG -C HENG WANG
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
dy dx
≤
f
−
f
(x)
(1 − t)(b − a)2 a
t(b
−
a)
u
Z u Z b
1
(u
−
x)u
+
(x
−
a)y
f
+
− f (y) dy dx
t(b − a)2 a
(1 − t)(b − a)
u
Z u Z b (b − y)x + (y − u)u
M
≤
− x dy dx
2
(1 − t)(b − a) 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
M
=
t(1 − t)(b − a)3
Z u Z b
2
×
(b − a) ((1 − t)y − tx) − 2xy + (b + u)x + (a + u)y − 2u dy dx
a
u
Z u
M
1
2
=
((b − a)(1 − t) + (a + u)) (b + u) − (b − a)tx − 2u dx
(1 − t)(b − a)2 a 2
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).
J. Inequal. Pure and Appl. Math., 6(2) Art. 37, 2005
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N EW I NEQUALITIES OF H ADAMARD ’ S T YPE FOR L IPSCHITZIAN M APPINGS
7
This completes the proof of Theorem 2.3.
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),
(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).
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.
Remark 2.5. If we let t = 12 , then (2.13), (2.17) and (2.21) reduce to
Z b
M
1
f
(a)
+
f
(b)
≤
(2.22)
f
(x)dx
−
(b − a),
b − a
2
4
a
Z b
1
M
a
+
b
≤
(2.23)
f (x)dx − f
(b − a),
b − a
2
4
a
f (a) + f (b) M
a+b
−
(2.24)
≤ 2 (b − a).
f
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 2.6. 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
max{|f+0 (a)|, |f−0 (b)|}
(2.25)
0 ≤ A(t) − f (ta + (1 − t)b) ≤
t(1 − t)(b − a),
3
(2.26)
(2.27)
(2.28)
(2.29)
0 ≤ B(t) − A(t) ≤
max{|f+0 (a)|, |f−0 (b)|}
t(1 − t)(b − a),
2
0 ≤ B(t) − f (ta + (1 − t)b) ≤
0 ≤ C(t) − B(t) ≤
max{|f+0 (a)|, |f−0 (b)|}
t(1 − t)(b − a),
2
max{|f+0 (a)|, |f−0 (b)|}
t(1 − t)(b − a),
2
0 ≤ tf (a) + (1 − t)f (b) − C(t) ≤ max{|f+0 (a)|, |f−0 (b)|}t(1 − t)(b − a),
J. Inequal. Pure and Appl. Math., 6(2) Art. 37, 2005
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8
L IANG -C HENG WANG
(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),
(2.32)
0 ≤ tf (a) + (1 − t)f (b) − A(t) ≤ 2 max{|f+0 (a)|, |f−0 (b)|}t(1 − t)(b − a),
(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)
|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 (2.6).
Remark 2.7. The condition in Corollary 2.6 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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J. Inequal. Pure and Appl. Math., 6(2) Art. 37, 2005
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