Volume 10 (2009), Issue 2, Article 52, 12 pp.
SOME COMPANIONS OF AN OSTROWSKI TYPE INEQUALITY AND
APPLICATIONS
ZHENG LIU
I NSTITUTE OF A PPLIED M ATHEMATICS , S CHOOL OF S CIENCE
U NIVERSITY OF S CIENCE AND T ECHNOLOGY L IAONING
A NSHAN 114051, L IAONING , C HINA
lewzheng@163.net
Received 11 February, 2009; accepted 12 May, 2009
Communicated by S.S. Dragomir
A BSTRACT. 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.
Key words and phrases: Ostrowski type inequality, Perturbed trapezoid rule, Midpoint rule, Composite quadrature rule.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
Motivated by [1], Dragomir in [2] has proved the following companion of the Ostrowski
inequality:
(1.1)
Z b
1
1
[f (x) + f (a + b − x)] −
f (t) dt
2
b−a a
 3a+b 2 
x− 4
1


(b − a) kf 0 k[a,b],∞
+2

b−a
8




a+b q+1 1q

1
1
−x
x−a q+1
2q
(b − a) q kf 0 k[a,b],p ,
+ 2b−a
1
≤
b−a
(q+1) q





3a+b i
h



 1 + x− 4 kf 0 k[a,b],1
b−a
4
if f 0 ∈ L∞ [a, b] ;
if p > 1, p1 +
= 1,
and f 0 ∈ Lp [a, b] ;
if f 0 ∈ L1 [a, b] ,
for all x ∈ a, a+b
, where f : [a, b] → R is an absolutely continuous function.
2
041-09
1
q
2
Z HENG L IU
In particular, the best result in (1.1) is obtained for x = a+3b
, giving the following trapezoid
4
type inequalities:
Z b
1
3a
+
b
a
+
3b
1
(1.2) f
+f
−
f (t) dt
2
4
4
b−a a

1

(b − a) kf 0 k[a,b],∞ if f 0 ∈ L∞ [a, b] ;

8



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

(q+1)



 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].
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
b
(b − a)2 0
b−a
[f (a) + f (b)] +
[f (b) − f 0 (a)]
(1.4) f (t) dt −
a
2
8

(b−a)3

if f 00 ∈ L∞ [a, b] ;
 24 kf 00 k∞




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.
2. S OME I NTEGRAL I NEQUALITIES
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
1
3a + b
(2.1)
f (t) dt− [f (x)+f (a+b−x)]+
x−
[f 0 (x)−f 0 (a + b − x)]
b−a a
2
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
2 00
+
(t − b) f (t) dt
a+b−x
for any x ∈ a, a+b
.
2
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 52, 12 pp.
http://jipam.vu.edu.au/
O STROWSKI T YPE I NEQUALITY
3
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
a+b−x
x
and
Z
a
a+b
t−
2
2
2
a+b
f (t) dt = x −
[f 0 (a + b − x) − f 0 (x)]
2
Z a+b−x
a+b
+2 x−
[f (x) + f (a + b − x)] + 2
f (t) dt,
2
x
00
b
2
2
00
0
Z
b
(t − b) f (t) dt = − (x − a) f (a + b − x)−2 (x − a) f (a + b − x)+2
a+b−x
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
2
a
1
3a + b
0
0
+
x−
[f (x) − f (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
00
If f ∈ 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
#
3
2 a+b
(x − a)3 00
00
+
− x kf k[x,a+b−x],∞ +
kf k[a+b−x,b]
3
2
3
 3a+b 2 1
1 x− 4


(b − a)2 kf 00 k[a,b],∞ ;
 96 + 2
b−a




3α α1

3α x− a+b

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 .
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 52, 12 pp.
http://jipam.vu.edu.au/
4
Z HENG L IU
The inequality (2.2), the first inequality in (2.3) and the constant
1
96
are sharp.
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
and the first inequality in (2.3) is proved.
Denote
3
(x − a)3 00
1 a+b
(x − a)3 00
00
M̄ (x) :=
kf k[a,x],∞ +
− x kf k[x,a+b−x],∞ +
kf k[a+b−x,b]
6
3
2
6
for x ∈ a, a+b
.
2
Firstly, observe that
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 1 a + b
(x − a)3
×
+
−x +
6
3
2
6
"
#
2
2
(b − a)
1
3a + b
= kf 00 k[a,b],∞
+
x−
(b − a)
96
2
4
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−
3α
"
3
#α ) α1
a+b
(x − a)
+
2
2
h
i β1
× 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)
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 function
f : [a, b] → R, f (t) = t2 for any x ∈ a, a+b
to
obtain
the corresponding equalities.
2
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
0
≤ 1 (b − a)2 kf 00 k∞
f
(t)
dt
−
[f
(a)
+
f
(b)]
+
[f
(b)
−
f
(a)]
b − a
24
2
8
a
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 52, 12 pp.
http://jipam.vu.edu.au/
O STROWSKI T YPE I NEQUALITY
with
1
24
5
as a sharp constant. If we choose x = a+b
, then we get
2
Z b
i
1
a + b 1 h 00
00
+
kf
k
f
(t)
dt
−
f
≤
kf
k
a+b
a+b
b − a
48
[a, 2 ],∞
[ 2 ,b],∞
2
a
1
≤
(b − a)2 kf 00 k[a,b],∞
24
with the constants
1
48
1
24
and
being sharp.
Corollary 2.3. With the assumptions in Theorem 2.2, one has the inequality
1 Z b
a+3b 3a+b
+
f
f
1
4
4
(b − a)2 kf 00 k[a,b],∞ .
(2.4)
f (t) dt −
≤
96
b − a a
2
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:
"
1
(2.5) M (x) ≤
2 (2q + 1)
a+b
2
−x
b−a
1
+2 q
≤
1
1
1
q
×
2 (2q + 1) q





































2
x−a
b−a
!2+ 1q
2+ 1q
kf 00 k[a,x],p
kf 00 k[x,a+b−x],p
1
x−a 2+ q
b−a
+2
1
q
a+b
−x
2
b−a
x−a
b−a
2+ 1q

1
kf 00 k[a+b−x,b],p  (b − a)1+ q
2+ 1q 1
× max kf 00 k[a,x],p , kf 00 k[x,a+b−x],p , kf 00 k[a+b−x,b],p (b − a)1+ q ;
a+b 2α+ αq α1
α
α
−x
x−a 2α+ q
2
q
+2
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
1 1 a+b
−x
x−a 2+ q
2
q
,2
max
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
Proof. Using Hölder’s integral inequality for p > 1,
Z
x
2
(t − a) |f 00 (t) | dt ≤
a
Z
1
p
+
1
q
= 1, we have
x
2q
(t − a)
1q
dt kf 00 k[a,x],p
a
1
=
(x − a)2+ q
(2q + 1)
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 52, 12 pp.
1
q
kf 00 k[a,x],p ,
http://jipam.vu.edu.au/
6
Z HENG L IU
Z
a+b−x
x
a+b
t−
2
2
1q
a + b 2q
|f (t) | dt ≤
|t −
| dt kf 00 k[x,a+b−x],p
2
x
2+ 1q
1
2 q a+b
−
x
2
=
kf 00 k[x,a+b−x],p ,
1
q
(2q + 1)
Z
00
a+b−x
and
Z
b
2
Z
00
b
(t − b) |f (t) | dt ≤
a+b−x
2q
(b − t)
1q
dt kf 00 k[a+b−x,b],p
a+b−x
1
=
(x − a)2+ q
1
q
kf 00 k[a+b−x,b],p .
(2q + 1)
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, p1 + 1q = 1, p > 1, then we get the inequality

!2q+1  1q
1
2q+1
a+b
1
−x
2q
2
 x−a
 (b − a)1+ q kf 00 k[a,b],p ,
(2.6) M (x) ≤
+
1
b−a
b−a
2 (2q + 1) q
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
2
8
a
1
1+
1 (b − a) q kf 00 k[a,b],p
≤ ·
.
1
8
(2q + 1) q
The constant
1
8
is best possible in the sense that it cannot be replaced by a smaller constant.
Proof. Indeed, if we assume that (2.7) holds with a constant C > 0, instead of 81 , 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
2
8
a
1
≤C·
(b − a)1+ q kf 00 k[a,b],p
1
,
(2q + 1) q
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
1
kf 00 k[a,b],p = 2k (b − a) p ;
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 52, 12 pp.
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O STROWSKI T YPE I NEQUALITY
7
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
giving C ≥
proved.
(2q+1) q
24
. Letting q → 1+, we deduce C ≥ 18 , and the sharpness of the constant is
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
In both inequalities the constant
constant.
1
8
p > 1,
1 1
+ = 1.
p q
is sharp in the sense that it cannot be replaced by a smaller
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
≤C· 1
kf
k
+
kf
k
a+b
a+b
1
[a, 2 ],p
[ 2 ,b],p
2 q (2q + 1) q
1
≤D·
(b − a)1+ q
(2q + 1)
1
q
kf 00 k[a,b],p .
2
, k > 0, we have
For the function f : [a, b] → R, f (x) = k x − a+b
2
Z b
a+b
1
k (b − a)2
f
= 0,
f (t) dt =
,
2
b−a a
12
1
1
1
b−a p
00
00
kf k[a, a+b ],p + kf k[ a+b ,b],p = 4k
= 21+ q (b − a) p k,
2
2
2
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
≤
D
·
≤C·
1 ,
1
12
(2q + 1) q
(2q + 1) q
1
q
for any q > 1. Letting q → 1+, we deduce C, D ≥ 18 and the sharpness
giving C, D ≥ (2q+1)
24
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].
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 52, 12 pp.
http://jipam.vu.edu.au/
8
Z HENG L IU
Corollary 2.5. With the assumptions in Theorem 2.4, one has the inequality
(2.11)
Z b
1
3a
+
b
a
+
3b
1
f
+
f
−
f
(t)
dt
2
4
4
b−a a
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+ 1q
1
3a
+
b
a
+
3b
1
(b
−
a)
00
(2.12) f
+f
−
f (t) dt ≤ E ·
1 kf k[a,b],p .
2
4
4
b−a a
(2q + 1) q
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

 −2 x − 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
.
We have
  x −
0
f (x) =
 x−
3a+b 4
if
x ∈ a, a+b
,
2
if
x∈
a+3b 4
a+b
,b
2
.
Then f 0 is absolutely continuous and f 00 ∈ Lp [a, b], p > 1. We also have
1
3a + b
a + 3b
f
+f
= 0,
2
4
4
Z b
1
(b − a)2
f (t) dt =
,
b−a a
96
1
kf 00 k[a,b],p = (b − a) p ,
and then, by (2.12), we obtain
(b − a)2
(b − a)2
≤E·
1 ,
96
(2q + 1) q
1
giving E ≥
(2q+1) q
96
for any q > 1, i.e., E ≥
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 52, 12 pp.
1
,
32
and the corollary is proved.
http://jipam.vu.edu.au/
O STROWSKI T YPE I NEQUALITY
9
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
x−a
b−a
kf 00 k[a,x],1
(2.13)
M (x) ≤
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

2 2 a+b
−x
b−a
x−a

2

2 b−a + b−a

2





× max kf 00 k[a,x],1 , kf 00 k[x,a+b−x],1 , kf 00 k[a+b−x,b],1 ;



2α α1


−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
i2


 b−a | x− 3a+b
4
| + 1 kf 00 k
;
2
b−a
[a,b],1
4
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
a+b
2
!2 
−x 
x−a
b−a
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
"
#
1 2
b − a x − 3a+b
4
=
kf 00 k[a,b],1 .
+
b−a 4
2
M (x) ≤
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
1
1
b
−
a
0
0
f (t) dt − [f (a) + f (b)] +
[f (b) − f (a)] ≤ (b − a) kf 00 k[a,b],1 .
b − a
2
8
8
a
If in (2.13) we choose x =
a+b
,
2
then we get the mid-point inequality
Z b
1
1
a
+
b
≤ (b − a) kf 00 k[a,b],1 .
f
(t)
dt
−
f
b − a
8
2
a
Corollary 2.7. With the assumptions in Theorem 2.6, one has the inequality
1 Z b
a+3b f 3a+b
+
f
1
4
4
f (t) dt −
(b − a) kf 00 k[a,b],1 .
≤
b − a a
32
2
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 52, 12 pp.
http://jipam.vu.edu.au/
10
Z HENG L IU
3. A C OMPOSITE Q UADRATURE F ORMULA
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
q

(2q+1)



 1
(b − a) kf 00 k[a,b],1
if f 00 ∈ L1 [a, b] .
32
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
n−1 1X
3xi + xi+1
xi + 3xi+1
(3.2)
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 estimates

n−1
P 3

1
00

kf
k
hi
if f 00 ∈ L∞ [a, b] ;

[a,b],∞

96

i=0


n−1
1q

P
1
(3.3)
|Rn (In , f ) | ≤
kf 00 k[a,b],p
h2q+1
if f 00 ∈ Lp [a, b] ,
i
 32(2q+1) 1q

i=0



p > 1, p1 + 1q = 1;


 1 00
kf k[a,b],1 [ν (In )]2
if f 00 ∈ L1 [a, b] .
32
Proof. Applying inequality (3.1) on the interval [xi , xi+1 ], we may state that
Z xi+1
1
3x
+
x
x
+
3x
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,
h
1
i
q

32(2q+1)


 1 2 00
h kf k[xi ,xi+1 ],1 ;
32 i
1
p
+
1
q
= 1;
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
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 52, 12 pp.
http://jipam.vu.edu.au/
O STROWSKI T YPE I NEQUALITY
11
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
≤
i=0
n−1
X
(2+ 1 )q
hi q
! 1q
n−1
X
i=0
=
kf 00 kp[xi ,xi+1 ],p
i=0
n−1
X
! 1q
n−1 Z
X
h2q+1
i
i=0
=
! p1
i=0
n−1
X
xi+1
! p1
|f 00 (t) |p dt
xi
! 1q
h2q+1
i
kf 00 k[a,b],p .
i=0
Also, we note that
n−1
X
h2i kf 00 k[xi ,xi+1 ],1
n−1
2 X
≤ max hi
kf 00 k[xi ,xi+1 ],1
0≤i≤n−1
i=0
2
i=0
00
= [ν (In )] kf k[a,b],1 .
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:
(3.6)
n−1 b−aX
4i + 1
4i + 3
Qn (f ) :=
f a+
(b − a) + f a +
(b − a) .
2n i=0
4n
4n
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 ) ,
a
where Qn (f ) is defined by (3.6) and the remainder Rn (f ) satisfies the estimate:

3
1

kf 00 k[a,b],∞ (b−a)
;
2

96
n



2+ 1
(b−a) q
1
00
|Rn (In , f ) | ≤
kf
k
, p > 1, p1 + 1q = 1;
1
[a,b],p
n2
q

32(2q+1)




(b−a)2
1
00
kf
k
.
[a,b],1
32
n2
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 52, 12 pp.
http://jipam.vu.edu.au/
12
Z HENG L IU
R EFERENCES
[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.
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 52, 12 pp.
http://jipam.vu.edu.au/