SOME COMPANIONS OF AN OSTROWSKI TYPE
INEQUALITY AND APPLICATIONS
Ostrowski Type Inequality
ZHENG LIU
Institute of Applied Mathematics, School of Science
University of Science and Technology Liaoning
Anshan 114051, Liaoning, China
Zheng Liu
vol. 10, iss. 2, art. 52, 2009
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EMail: lewzheng@163.net
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12 May, 2009
J
I
Communicated by:
S.S. Dragomir
Page 1 of 23
2000 AMS Sub. Class.:
26D15.
Key words:
Ostrowski type inequality, Perturbed trapezoid rule, Midpoint rule, Composite
quadrature rule.
Received:
11 February, 2009
Accepted:
Abstract:
We establish some companions of an Ostrowski type integral inequality for functions whose derivatives are absolutely continuous. Applications for composite
quadrature rules are also given.
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Contents
1
Introduction
3
2
Some Integral Inequalities
5
3
A Composite Quadrature Formula
19
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1.
Introduction
Motivated by [1], Dragomir in [2] has proved the following companion of the Ostrowski inequality:
Z b
1
1
(1.1) [f (x) + f (a + b − x)] −
f (t) dt
2
b−a a
 3a+b 2 
x− 4

 18 + 2
(b − a) kf 0 k[a,b],∞
if f 0 ∈ L∞ [a, b] ;

b−a




a+b q+1 1q

1
1
−x
x−a q+1
2q
+ 2b−a
(b − a) q kf 0 k[a,b],p , if p > 1, p1 + 1q = 1,
1
≤
b−a
(q+1) q




and f 0 ∈ Lp [a, b] ;

h
i

3a+b


 1 + x− 4 kf 0 k[a,b],1
if f 0 ∈ L1 [a, b] ,
4
b−a
, where f : [a, b] → R is an absolutely continuous function.
for all x ∈ a, a+b
2
In particular, the best result in (1.1) is obtained for x = a+3b
, giving the following
4
trapezoid type inequalities:
Z b
1
3a + b
a + 3b
1
(1.2) f
+f
−
f (t) dt
2
4
4
b−a a

1
0
0

 8 (b − a) kf k[a,b],∞ if f ∈ L∞ [a, b] ;



1
1 (b−a) q
0
≤
·
if f 0 ∈ Lp [a, b] , p > 1, p1 + 1q = 1;
1 kf k[a,b],p ,
4

(q+1) q



 1 kf 0 k
if f 0 ∈ L1 [a, b] .
[a,b],1
4
Some natural applications of (1.1) and (1.2) are also provided in [2].
Ostrowski Type Inequality
Zheng Liu
vol. 10, iss. 2, art. 52, 2009
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In [3], Dedić et al. have derived the following trapezoid type inequality:
Z b
1
(b − a)2 00
3a
+
b
a
+
3b
1
(1.3) f
+f
−
f (t) dt ≤
kf k∞ ,
2
4
4
b−a a
32
for a function f : [a, b] → R whose derivative f 0 is absolutely continuous and
f 00 ∈ L∞ [a, b].
In [4], we find that for a function f : [a, b] → R whose derivative f 0 is absolutely
continuous, the following perturbed trapezoid inequalities hold:
Z
2
b
(b
−
a)
b
−
a
(1.4) f (t) dt −
[f (a) + f (b)] +
[f 0 (b) − f 0 (a)]
a
2
8

(b−a)3

kf 00 k∞
if f 00 ∈ L∞ [a, b] ;


24



2+ 1
(b−a) q
00
≤
if f 00 ∈ LP [a, b] , p > 1, p1 + 1q = 1;
1 kf kp ,
q

8(2q+1)




 (b−a)2 kf 00 k
if f 00 ∈ L1 [a, b] .
1
8
In this paper, we provide some companions of Ostrowski type inequalities for
functions whose first derivatives are absolutely continuous and whose second derivatives belong to the Lebesgue spaces Lp [a, b], 1 ≤ p ≤ ∞. These improve (1.3) and
recapture (1.4). Applications for composite quadrature rules are also given.
Ostrowski Type Inequality
Zheng Liu
vol. 10, iss. 2, art. 52, 2009
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2.
Some Integral Inequalities
Lemma 2.1. Let f : [a, b] → R be such that the derivative f 0 is absolutely continuous on [a, b]. Then we have the equality
Z b
1
1
(2.1)
f (t) dt − [f (x) + f (a + b − x)]
b−a a
2
1
3a + b
+
x−
[f 0 (x) − f 0 (a + b − x)]
2
4
"Z
2
Z a+b−x x
a+b
1
2 00
(t − a) f (t) dt +
t−
f 00 (t) dt
=
2 (b − a) a
2
x
Z b
2 00
+
(t − b) f (t) dt
a+b−x
for any x ∈ a, a+b
.
2
Proof. Using the integration by parts formula for Lebesgue integrals, we have
Z x
Z x
2 00
2 0
(t − a) f (t) dt = (x − a) f (x) − 2 (x − a) f (x) + 2
f (t) dt,
a
Z
x
a+b−x
a
2
a+b
f 00 (t) dt
t−
2
2
a+b
[f 0 (a + b − x) − f 0 (x)]
= x−
2
Z a+b−x
a+b
+2 x−
[f (x) + f (a + b − x)] + 2
f (t) dt,
2
x
Ostrowski Type Inequality
Zheng Liu
vol. 10, iss. 2, art. 52, 2009
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and
Z
b
(t − b)2 f 00 (t) dt
a+b−x
2
0
= − (x − a) f (a + b − x) − 2 (x − a) f (a + b − x) + 2
Z
b
f (t) dt.
a+b−x
Summing the above equalities, we deduce the desired identity (2.1).
Theorem 2.2. Let f : [a, b] → R be such that the derivative f 0 is absolutely continuous on [a, b]. Then we have the inequality
Z b
1
1
(2.2) f (t) dt − [f (x) + f (a + b − x)]
b−a a
2
3a + b
1
x−
[f 0 (x) − f 0 (a + b − x)]
+
2
4
"Z
2
Z a+b−x x
1
a+b
2
00
≤
(t − a) |f (t) | dt +
t−
|f 00 (t) | dt
2 (b − a) a
2
x
Z b
+
(t − b)2 |f 00 (t) | dt
a+b−x
:= M (x)
for any x ∈ a, a+b
.
2
If f 00 ∈ L∞ [a, b], then we have the inequalities
"
1
(x − a)3 00
(2.3) M (x) ≤
kf k[a,x],∞
2 (b − a)
3
Ostrowski Type Inequality
Zheng Liu
vol. 10, iss. 2, art. 52, 2009
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#
3
3
a+b
(x
−
a)
− x kf 00 k[x,a+b−x],∞ +
kf 00 k[a+b−x,b]
2
3
 3a+b 2 1
1 x− 4


+2
(b − a)2 kf 00 k[a,b],∞ ;

96
b−a



1


3α x− a+b 3α α

1
x−a

2

+ b−a

2α−1 b−a



h
i β1
(b−a)2
00 β
00 β
00 β
≤
×
kf
k
+
kf
k
+
kf
k
[a,x],∞
[x,a+b−x],∞
[a+b−x,b],∞
3


1
1


if α > 1, α + β = 1;


a+b 3 

3
x−


max 12 x−a
, b−a2

b−a



2

× kf 00 k[a,x],∞ + kf 00 k[x,a+b−x],∞ + kf 00 k[a+b−x,b],∞ (b−a)
;
3
a+b for any x ∈ a, 2 .
1
The inequality (2.2), the first inequality in (2.3) and the constant 96
are sharp.
2
+
3
Proof. The inequality (2.2) follows by Lemma 2.1 on taking the modulus and using
its properties.
If f 00 ∈ L∞ [a, b], then
Z x
(x − a)3 00
2
00
(t − a) |f (t) | dt ≤
kf k[a,x],∞ ,
3
a
2
3
Z a+b−x 2 a+b
a+b
00
t−
|f (t) | dt ≤
− x kf 00 k[x,a+b−x],∞ ,
2
3
2
x
Z b
(x − a)3 00
2
00
(t − b) |f (t) | dt ≤
f k[a+b−x,b],∞
3
a+b−x
Ostrowski Type Inequality
Zheng Liu
vol. 10, iss. 2, art. 52, 2009
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and the first inequality in (2.3) is proved.
Denote
3
(x − a)3 00
1 a+b
M̄ (x) :=
kf k[a,x],∞ +
− x kf 00 k[x,a+b−x],∞
6
3
2
+
(x − a)3 00
kf k[a+b−x,b]
6
Ostrowski Type Inequality
Zheng Liu
for x ∈ a, a+b
.
2
Firstly, observe that
vol. 10, iss. 2, art. 52, 2009
M̄ (x) ≤ max kf 00 k[a,x],∞ , kf 00 k[x,a+b−x],∞ , kf 00 k[a+b−x,b],∞
#
"
3
(x − a)3
(x − a)3 1 a + b
×
+
−x +
6
3
2
6
"
2 #
2
(b
−
a)
1
3a
+
b
= kf 00 k[a,b],∞
+
x−
(b − a)
96
2
4
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and the first part of the second inequality in (2.3) is proved.
Using the Hölder inequality for α > 1, α1 + β1 = 1, we also have
M̄ (x) ≤
1
3
("
3
(x − a)
2
#α
+ x−
a+b
2
"
3α
+
3
(x − a)
2
#α ) α1
i β1
h
× kf 00 kβ[a,x],∞ + kf 00 kβ[x,a+b−x],∞ + kf 00 kβ[a+b−x,b],∞
giving the second part of the second inequality in (2.3)
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Finally, we also observe that
(
3 )
1
(x − a)3
a+b
M̄ (x) ≤ max
, x−
3
2
2
00
× kf k[a,x],∞ + kf 00 k[x,a+b−x],∞ + kf 00 k[a+b−x,b],∞ .
The sharpness of the inequalities mentioned follows
from
the fact that we can choose
a+b
2
a function f : [a, b] → R, f (t) = t for any x ∈ a, 2 to obtain the corresponding
equalities.
Remark 1. If in Theorem 2.2 we choose x = a, then we recapture the first part of
the inequality (1.4), i.e.,
Z b
1
1
b−a 0
1
0
f (t) dt − [f (a) + f (b)] +
[f (b) − f (a)] ≤
(b − a)2 kf 00 k∞
b − a
2
8
24
a
with
1
24
as a sharp constant. If we choose x = a+b
, then we get
2
Z b
i
1
a + b 1 h 00
00
f (t) dt − f
,∞ + kf k[ a+b
,b],∞
≤ 48 kf k[a, a+b
b − a
2 ]
2
2
a
1
≤
(b − a)2 kf 00 k[a,b],∞
24
with the constants
1
48
and
1
24
being sharp.
Corollary 2.3. With the assumptions in Theorem 2.2, one has the inequality
1 Z b
a+3b f 3a+b
+
f
1
4
4
(2.4)
f (t) dt −
(b − a)2 kf 00 k[a,b],∞ .
≤
b − a a
96
2
Ostrowski Type Inequality
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vol. 10, iss. 2, art. 52, 2009
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1
The constant 96
is best possible in the sense that it cannot be replaced by a smaller
constant. Clearly (2.4) is an improvement of (1.3).
Theorem 2.4. Let f : [a, b] → R be such that the derivative f 0 is absolutely continuous on [a, b] and f 00 ∈ Lp [a, b], p > 1. If M (x) is as defined in (2.2), then we have
the bounds:
"
2+ 1q
x−a
1
(2.5) M (x) ≤
kf 00 k[a,x],p
1
b
−
a
2 (2q + 1) q

!2+ 1q
2+ 1q
a+b
1
1
−x
x−a
2
kf 00 k[x,a+b−x],p
kf 00 k[a+b−x,b],p  (b − a)1+ q
+2 q
b−a
b−a



















1
×
≤
1

2 (2q + 1) q 
















2
1
x−a 2+ q
b−a
00
+2
1
q
a+b
−x
b−a
2
00
Ostrowski Type Inequality
Zheng Liu
vol. 10, iss. 2, art. 52, 2009
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2+ 1q 00
1+ 1q
× max kf k[a,x],p , kf k[x,a+b−x],p , kf k[a+b−x,b],p (b − a)
;
1
α
2α+ q α
a+b
2α+ αq
α
−x
2
q
2 x−a
+
2
b−a
b−a
h
i β1
1
× kf 00 kβ[a,x],p + kf 00 kβ[x,a+b−x],p + kf 00 kβ[a+b−x,b],p (b − a)1+ q
if α > 1, α1 + β1 = 1,
2+ 1q
2+ 1q 1 a+b
−x
2
q
max x−a
,
2
b−a
b−a
00
1
× kf k[a,x],p + kf 00 k[x,a+b−x],p + kf 00 k[a+b−x,b],p (b − a)1+ q ;
for any x ∈ a, a+b
.
2
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1
p
+
2q
1q
dt kf 00 k[a,x],p
Proof. Using Hölder’s integral inequality for p > 1,
x
Z
2
x
Z
00
(t − a) |f (t) | dt ≤
(t − a)
a
1
q
= 1, we have
a
1
=
(x − a)2+ q
(2q + 1)
1
q
kf 00 k[a,x],p ,
Ostrowski Type Inequality
Zheng Liu
Z
a+b−x
t−
x
a+b
2
2
1q
a
+
b
|f 00 (t) | dt ≤
|t −
|2q dt kf 00 k[x,a+b−x],p
2
x
2+ 1
1
2 q a+b
− x q 00
2
kf k[x,a+b−x],p ,
=
1
(2q + 1) q
Z
a+b−x
and
Z
b
(t − b)2 |f 00 (t) | dt ≤
a+b−x
Z
b
(b − t)2q
a+b−x
2+ 1q
=
(x − a)
1
(2q + 1) q
1q
dt kf 00 k[a+b−x,b],p
kf 00 k[a+b−x,b],p .
Summing the above inequalities, we deduce the first bound in (2.5).
The last part may be proved in a similar fashion to the one in Theorem 2.2, and
we omit the details.
Remark 2. If in (2.5) we choose α = q, β = p,
1
p
+
1
q
= 1, p > 1, then we get the
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inequality

2q+1
2
x−a

(2.6) M (x) ≤
+
1
b−a
2 (2q + 1) q
1
q
a+b
2
−x
b−a
!2q+1  1q

1
× (b − a)1+ q kf 00 k[a,b],p ,
Ostrowski Type Inequality
.
for any x ∈ a, a+b
2
Remark 3. If in Theorem 2.4 we choose x = a, then we recapture the second part of
the inequality (1.4), i.e.,
Z b
1
1
b
−
a
0
0
(2.7) f (t) dt − [f (a) + f (b)] +
[f (b) − f (a)]
b−a a
2
8
1+ 1q
≤
The constant
constant.
1
8
kf 00 k[a,b],p
1 (b − a)
.
·
1
8
(2q + 1) q
is best possible in the sense that it cannot be replaced by a smaller
Zheng Liu
vol. 10, iss. 2, art. 52, 2009
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,
8
Proof. Indeed, if we assume that (2.7) holds with a constant C > 0, instead of
i.e.,
Z b
1
1
b
−
a
0
0
(2.8) f (t) dt − [f (a) + f (b)] +
[f (b) − f (a)]
b−a a
2
8
Close
1
≤C·
(b − a)1+ q kf 00 k[a,b],p
1
(2q + 1) q
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,
then for the function f : [a, b] → R, f (x) = k x −
a+b 2
,
2
k > 0, we have
f (a) + f (b)
(b − a)2
=k·
,
2
4
f 0 (b) − f 0 (a) = 2k (b − a) ,
Z b
1
(b − a)2
f (t) dt = k ·
,
b−a a
12
Ostrowski Type Inequality
Zheng Liu
1
kf 00 k[a,b],p = 2k (b − a) p ;
vol. 10, iss. 2, art. 52, 2009
and by (2.8) we deduce
k (b − a)2 k (b − a)2 k (b − a)2 2C · k (b − a)2
−
+
,
≤
1
12
4
4
(2q + 1) q
1
(2q+1) q
24
giving C ≥
. Letting q → 1+, we deduce C ≥
constant is proved.
1
,
8
and the sharpness of the
Remark 4. If in Theorem 2.4 we choose x = a+b
, then we get the midpoint inequality
2
Z b
1
a
+
b
(2.9)
f
(t)
dt
−
f
b − a
2
a
1
i
1 (b − a)1+ q h 00
00
≤ · 1
kf k[a, a+b ],p + kf k[ a+b ,b],p
2
2
8 2 q (2q + 1) 1q
1
1 (b − a)1+ q 00
≤ ·
kf k[a,b],p ,
8 (2q + 1) 1q
p > 1,
1 1
+ = 1.
p q
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In both inequalities the constant 18 is sharp in the sense that it cannot be replaced by
a smaller constant.
To show this fact, assume that (2.9) holds with C, D > 0, i.e.,
Z b
1
a
+
b
(2.10)
f (t) dt − f
b − a
2
a
1
i
(b − a)1+ q h 00
00
+
kf
k
≤C· 1
kf
k
1
,p
,b],p
[a, a+b
[ a+b
2 ]
2
2 q (2q + 1) q
1
≤D·
(b − a)1+ q
(2q + 1)
1
q
Ostrowski Type Inequality
Zheng Liu
vol. 10, iss. 2, art. 52, 2009
kf 00 k[a,b],p .
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a+b 2
For the function f : [a, b] → R, f (x) = k x − 2 , k > 0, we have
Z b
a+b
1
k (b − a)2
f
= 0,
f (t) dt =
,
2
b−a a
12
kf 00 k[a, a+b ],p + kf 00 k[ a+b ,b],p = 4k
2
2
b−a
2
1
p1
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= 21+ q (b − a) p k,
1
kf 00 k[a,b],p = 2 (b − a) p k;
and then by (2.10) we deduce
k (b − a)2
2k (b − a)2
2k (b − a)2
≤C·
≤D·
1 ,
1
12
(2q + 1) q
(2q + 1) q
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1
q
giving C, D ≥ (2q+1)
for any q > 1. Letting q → 1+, we deduce C, D ≥ 18 and the
24
sharpness of the constants in (2.9) is proved.
The following result is useful in providing the best quadrature rule in the class
for approximating the integral of a function f : [a, b] → R whose first derivative is
absolutely continuous on [a, b] and whose second derivative is in Lp [a, b].
Corollary 2.5. With the assumptions in Theorem 2.4, one has the inequality
Z b
1
3a
+
b
a
+
3b
1
f
(2.11)
+
f
−
f
(t)
dt
2
4
4
b−a a
Ostrowski Type Inequality
Zheng Liu
vol. 10, iss. 2, art. 52, 2009
1
1 (b − a)1+ q 00
kf k[a,b],p ,
·
≤
32 (2q + 1) 1q
where p1 + 1q = 1.
1
The constant 32
is the best possible in the sense that it cannot be replaced by a
smaller constant.
Proof. The inequality follows by Theorem 2.4 and (2.6) on choosing x = 3a+b
.
4
To prove the sharpness of the constant, assume that (2.11) holds with a constant
E > 0, i.e.,
Z b
1
3a + b
a + 3b
1
(2.12) f
+f
−
f (t) dt
2
4
4
b−a a
1
≤E·
(b − a)1+ q
(2q + 1)
1
q
kf 00 k[a,b],p .
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Consider the function f : [a, b] → R,
 1
2
− 2 x − 3a+b


4




 1 x − 3a+b 2
2
4
f (x) =

1
a+3b 2

−
x
−

2
4



2
 1
x − a+3b
2
4
if
x ∈ a, 3a+b
,
4
if
x∈
3a+b a+b
, 2
4
,
if
x∈
a+b a+3b
, 4
2
,
if
x∈
a+3b
,b
4
.
Zheng Liu
vol. 10, iss. 2, art. 52, 2009
We have
  x −
0
f (x) =
 x−
,
if x ∈ a, a+b
2
if
3a+b 4
a+3b 4
00
x∈
a+b
,b
2
.
Then f 0 is absolutely continuous and f ∈ Lp [a, b], p > 1. We also have
1
3a + b
a + 3b
f
+f
= 0,
2
4
4
Z b
(b − a)2
1
f (t) dt =
,
b−a a
96
1
p
00
kf k[a,b],p = (b − a) ,
and then, by (2.12), we obtain
2
(b − a)
(b − a)
≤E·
1 ,
96
(2q + 1) q
1
(2q+1) q
96
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2
giving E ≥
Ostrowski Type Inequality
for any q > 1, i.e., E ≥
1
,
32
and the corollary is proved.
Theorem 2.6. Let f : [a, b] → R be such that the derivative f 0 is absolutely continuous on [a, b] and f 00 ∈ L1 [a, b]. If M (x) is as defined in (2.2), then we have the
bounds:
"
2
b−a
x−a
(2.13) M (x) ≤
kf 00 k[a,x],1
2
b−a

!2
2
a+b
−x
x−a
2
+
kf 00 k[x,a+b−x],1 +
kf 00 k[a+b−x,b],1 
b−a
b−a

a+b 2
−x
b−a
x−a 2

2

2
+

b−a
2
b−a


00

00
00


×
max
kf
k
,
kf
k
,
kf
k
;
[a,x],1
[x,a+b−x],1
[a+b−x,b],1


α1


2α

−x
 b−a 2 x−a 2α + a+b
2
2
b−a
b−a
≤
h
i β1



00 β
00 β
00 β

×
kf
k
+
kf
k
+
kf
k

[a,x],1
[x,a+b−x],1
[a+b−x,b],1




if α > 1, α1 + β1 = 1;


h
i

2

 b−a | x− 3a+b
4
| + 1 kf 00 k
;
2
b−a
4
[a,b],1
for any x ∈ a, a+b
.
2
The proof is as in Theorem 2.2 and we need only to prove the third inequality of
the last part as

!2 
2

a+b
−x 
b−a
x−a
2
M (x) ≤
max
,
 b−a

2
b−a
× kf 00 k[a,x],1 + kf 00 k[x,a+b−x],1 + kf 00 k[a+b−x,b],1
Ostrowski Type Inequality
Zheng Liu
vol. 10, iss. 2, art. 52, 2009
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b−a
=
2
"
#
x − 3a+b 1 2
4
kf 00 k[a,b],1 .
+
b−a 4
Remark 5. By the use of Theorem 2.6, for x = a, we recapture the third part of the
inequality (1.4), i.e.,
Z b
1
b
−
a
1
0
0
f
(t)
dt
−
[f
(a)
+
f
(b)]
+
[f
(b)
−
f
(a)]
b − a
2
8
a
1
≤ (b − a) kf 00 k[a,b],1 .
8
If in (2.13) we choose x = a+b
, then we get the mid-point inequality
2
Z b
1
a + b 1
00
f (t) dt − f
b − a
≤ 8 (b − a) kf k[a,b],1 .
2
a
Corollary 2.7. With the assumptions in Theorem 2.6, one has the inequality
1 Z b
3a+b
a+3b +
f
f
1
4
4
f (t) dt −
(b − a) kf 00 k[a,b],1 .
≤
b − a a
32
2
Ostrowski Type Inequality
Zheng Liu
vol. 10, iss. 2, art. 52, 2009
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3.
A Composite Quadrature Formula
We use the following inequalities obtained in the previous section:
Z b
1
3a + b
a + 3b
1
(3.1) f
+f
−
f (t) dt
2
4
4
b−a a

2
1

(b − a) kf 00 k[a,b],∞ if f 00 ∈ L∞ [a, b] ;

96



1+ 1
(b−a) q
1
00
≤
if f 00 ∈ Lp [a, b] , p > 1,
·
1 kf k[a,b],p
32

(2q+1) q



 1
(b − a) kf 00 k[a,b],1
if f 00 ∈ L1 [a, b] .
32
Ostrowski Type Inequality
Zheng Liu
1
p
+
1
q
= 1;
Let In : a = x0 < x1 < · · · < xn−1 < xn = b be a division of the interval [a, b] and
hi := xi+1 − xi (i = 0, . . . , n − 1) and ν (In ) := max {hi |i = 0, . . . , n − 1}.
Consider the composite quadrature rule
(3.2)
n−1 1X
3xi + xi+1
xi + 3xi+1
Qn (In , f ) :=
f
+f
hi .
2 i=0
4
4
The following result holds.
Theorem 3.1. Let f : [a, b] → R be such that the derivative f 0 is absolutely continuous on [a, b]. Then we have
Z b
f (t) dt = Qn (In , f ) + Rn (In , f ) ,
a
where Qn (In , f ) is defined by the formula (3.2), and the remainder satisfies the
vol. 10, iss. 2, art. 52, 2009
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estimates
(3.3)
|Rn (In , f ) | ≤









1
kf 00 k[a,b],∞
96
00
1
1








32(2q+1) q
n−1
P
i=0
h3i
n−1
P
kf k[a,b],p
1
kf 00 k[a,b],1
32
i=0
[ν (In )]2
if f 00 ∈ L∞ [a, b] ;
h2q+1
i
1q
if f 00 ∈ Lp [a, b] ,
p > 1, p1 + 1q = 1;
if f 00 ∈ L1 [a, b] .
Proof. Applying inequality (3.1) on the interval [xi , xi+1 ], we may state that
Z xi+1
3x
+
x
x
+
3x
1
i
i+1
i
i+1
(3.4) f (t) dt −
f
+f
hi 2
4
4
xi
 1 3 00
h kf k[xi ,xi+1 ],∞ ;

96 i



2+ 1q
1
≤
kf 00 k[xi ,xi+1 ],p , p > 1, p1 + 1q = 1;
1 hi
q

32(2q+1)


 1 2 00
h kf k[xi ,xi+1 ],1 ;
32 i
for each i ∈ {0, . . . , n − 1}.
Summing the inequality (3.4) over i from 0 to n − 1 and using the generalized
triangle inequality, we get
 Pn−1
1
3
00


i=0 hi kf k[xi ,xi+1 ],∞ ;
96



Pn−1 2+ 1q 00
1
(3.5) |Rn (In , f ) | ≤
kf k[xi ,xi+1 ],p , p > 1, p1 + 1q = 1;
1
i=0 hi
q

32(2q+1)



 1 Pn−1 2 00
i=0 hi kf k[xi ,xi+1 ],1 .
32
Ostrowski Type Inequality
Zheng Liu
vol. 10, iss. 2, art. 52, 2009
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Now, we observe that
n−1
X
h3i kf 00 k[xi ,xi+1 ],∞
00
≤ kf k[a,b],∞
i=0
n−1
X
h3i .
i=0
Using Hölder’s discrete inequality, we may write that
n−1
X
2+ 1
hi q kf 00 k[xi ,xi+1 ],p
≤
n−1
X
(2+ 1 )q
hi q
! 1q
n−1
X
i=0
i=0
=
kf 00 kp[xi ,xi+1 ],p
n−1
X
n−1 Z
X
h2q+1
i
i=0
n−1
X
xi+1
Ostrowski Type Inequality
Zheng Liu
vol. 10, iss. 2, art. 52, 2009
i=0
! 1q
i=0
=
! p1
! p1
|f 00 (t) |p dt
Title Page
xi
Contents
! 1q
h2q+1
i
00
kf k[a,b],p .
i=0
Also, we note that
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n−1
X
i=0
h2i kf 00 k[xi ,xi+1 ],1
n−1
X
≤ max h2i
kf 00 k[xi ,xi+1 ],1
0≤i≤n−1
2
i=0
00
= [ν (In )] kf k[a,b],1 .
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Consequently, by the use of (3.5), we deduce the desired result (3.3).
For the particular case where the division In is equidistant, i.e.,
In := xi = a + i ·
b−a
,
n
i = 0, . . . , n,
we may consider the quadrature rule:
n−1 b−aX
4i + 1
(3.6) Qn (f ) :=
f a+
(b − a)
2n i=0
4n
4i + 3
+f a +
(b − a) .
4n
Ostrowski Type Inequality
The following corollary will be more useful in practice.
Corollary 3.2. With the assumption of Theorem 3.1, we have
Z b
f (t) dt = Qn (f ) + Rn (f ) ,
Zheng Liu
vol. 10, iss. 2, art. 52, 2009
Title Page
a
where Qn (f ) is defined by (3.6) and the remainder Rn (f ) satisfies the estimate:

(b−a)3
1
00

;
kf
k
[a,b],∞
 96
n2



2+ 1
(b−a) q
1
00
|Rn (In , f ) | ≤
, p > 1, p1 + 1q = 1;
1 kf k[a,b],p
2
n
q

32(2q+1)



2
 1 00
kf k[a,b],1 (b−a)
.
32
n2
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References
[1] A. GUESSAB AND G. SCHMEISSER, Sharp integral inequalities of the
Hermite-Hadamard type, J. Approx. Theory, 115 (2002), 260–288.
[2] S. S. DRAGOMIR, Some companions of Ostrowski’s inequality for absolutely
continuous functions and applications, Bull. Korean Math. Soc., 42(2) (2005),
213–230.
[3] Lj. DEDIĆ, M.MATIĆ AND J.PEČARIĆ, On generalizations of Ostrowski inequality via some Euler-type identities, Math. Inequal. Appl., 3(3) (2000), 337–
353.
[4] P. CERONE AND S.S. DRAGOMIR, Trapezoidal type rules from an inequalities point of view, Handbook of Analytic-Computational Methods in Applied
Mathematics, CRC Press N.Y. (2000), 65–134.
Ostrowski Type Inequality
Zheng Liu
vol. 10, iss. 2, art. 52, 2009
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