Acta Mathematica Academiae Paedagogicae Nyı́regyháziensis
30 (2014), 53–65
www.emis.de/journals
ISSN 1786-0091
SOME GENERAL OSTROWSKI TYPE INEQUALITIES
ZHENG LIU
Abstract. A new general Ostrowski type inequality for functions whose
(n − 1)th derivatives are continuous functions of bounded variation is established. Some special cases are discussed.
1. Introduction
In 2001, J. Pečarić and S. Varošanec in [4] have proved the following Simpson
type inequality involving bounded variation:
Theorem 1. Let f : [a, b] → R be such that f (n−1) (n ≥ 1) is a continuous
function of bounded variation on [a, b]. Then
Z b
a+b
b−a
(1) [f (a) + 4f (
) + f (b)]
f (t) dt −
6
2
a
n
X
2
b
−
a
1
1
a
+
b
+
(−1)k
(
)k ( − )f (k−1) (
)
(k
−
1)!
2
k
3
2
k=5, k is odd
b
_
(f (n−1) ),
≤ Cn (b − a)
n
a
where C1 = 13 , C2 =
1
,
24
C3 =
1
324
and Cn =
1 n−3
,n
n! 3·2n
≥ 4.
In 2008, the author in [2] has proved the following Ostrowski type inequality
involving bounded variation:
2010 Mathematics Subject Classification. 26D15.
Key words and phrases. Ostrowski type inequality; Midpoint type inequality; Trapezoid type inequality; Simpson type inequality; Averaged midpoint-trapezoid type inequality;
Bounded variation.
53
54
ZHENG LIU
Theorem 2. Let f : [a, b] → R be such that f 0 is a continuous function of
bounded variation on [a, b]. Then for any x ∈ [a, b] and θ ∈ [0, 1] we have
Z b
(2) f (t) dt−
a
f (a) + f (b)
a+b 0
(b − a) (1 − θ)f (x) + θ
− (1 − θ)(x −
)f (x) 2
2
1
≤ [4(b − x)2 − 4θ(b − a)(b − x) + θ2 (b − a)2
16
b
_
2
2
2
+ |4(b − x) − 4θ(b − a)(b − x) − θ (b − a) |] (f 0 )
a
with θ ∈ [0, 1], and
for a ≤ x ≤
Z b
(3) f (t) dt − (b − a)[(1 − θ)f (x)
a+b
2
a+b 0
f (a) + f (b)
− (1 − θ)(x −
)f (x)]
+θ
2
2
1
≤ [4(x − a)2 − 4θ(b − a)(x − a) + θ2 (b − a)2
16
a
b
_
+ |4(x − a) − 4θ(b − a)(x − a) − θ (b − a) |] (f 0 )
2
2
2
a
for
a+b
2
< x ≤ b with θ ∈ [0, 1].
The purpose of this paper is to derive further generalization of the above
inequalities which will also lead to some interesting special cases.
2. The results
We need the following result similar to that given by C. E. M. Pearce et al.
in [3]:
Lemma 1. Let {Pn }n∈N and {Qn }n∈N be two sequences of harmonic polynomials, i.e.,
Pn0 (t) = Pn−1 (t), P0 (t) = 1, t ∈ R,
and
Q0n (t) = Qn−1 (t), Q0 (t) = 1,
Set
t ∈ R.
(
Pn (t), t ∈ [a, x],
Sn (t, x) :=
Qn (t), t ∈ (x, b].
SOME GENERAL OSTROWSKI TYPE INEQUALITIES
55
Then we have the identity
Z b
Z b
n
(n−1)
(4) (−1)
Sn (t, x) df
(t) =
f (t) dt
+
n
X
a
a
(−1)k [Qk (b)f (k−1) (b) + (Pk (x) − Qk (x))f (k−1) (x) − Pk (a)f (k−1) (a)]
k=1
provided that f : [a, b] → R is such that f (n−1) is a continuous function of
bounded variation on [a, b].
Proof. Observe that Sn (t, x) is a piecewise continuous function of t on [a, b], the
Riemann-Stieltjes integral in the left-hand side of (4) is guaranteed to exist.
Then (4) is not difficult to find by using the integration by parts formula for
Riemann-Stieltjes integrals.
Lemma 2. Let f : [a, b] → R be such that f (n−1) is a continuous function of
bounded variation on [a, b]. Then for all x ∈ [a, b] and any θ ∈ [0, 1] we have
the identity
Z b
n
(5) (−1)
Kn (t, x, θ) df (n−1) (t) =
a
Z b
b−a
[θf (a) + 2(1 − θ)f (x) + θf (b)]
f (t) dt −
2
a
n−1 X
(−1)k (x − a)k+1 + (b − x)k+1
−
(k + 1)!
k=1
θ(b − a)[(−1)k (x − a)k + (b − x)k ]
−
f (k) (x),
2k!
where
( (t−a)n θ(b−a)(t−a)n−1
−
, t ∈ [a, x],
n!
2(n−1)!
(6)
Kn (t, x, θ) :=
(t−b)n
θ(b−a)(t−b)n−1
+
, t ∈ (x, b].
n!
2(n−1)!
Proof. The proof is immediate by using the identity (4) in Lemma 1.
Theorem 3. Let f : [a, b] → R be such that f
is a continuous function of
bounded variation on [a, b]. Then for any x ∈ [a, b] and θ ∈ [0, 1] we have
Z b
b−a
f (t) dt −
[θf (a) + 2(1 − θ)f (x) + θf (b)]
(7) 2
a
n−1 X
(−1)k (x − a)k+1 + (b − x)k+1
−
(k + 1)!
k=1
b
_
θ(b − a)[(−1)k (x − a)k + (b − x)k ]
(k)
−
f (x) ≤ In (θ, x) (f (n−1) ),
2k!
a
(n−1)
56
ZHENG LIU
where
(8) In (θ, x)
=
(
n
max{| (x−b)
+
n!
n
max{| (x−a)
−
n!
θ(b−a)(x−b)n−1 (n−1)n−1 θn (b−a)n
|,
},
2(n−1)!
n!2n
θ(b−a)(x−a)n−1 (n−1)n−1 θn (b−a)n
|,
},
2(n−1)!
n!2n
a≤x≤
a+b
2
a+b
,
2
< x ≤ b,
for 0 ≤ (n − 1)θ ≤ 1, and
 (n−1)n−1 θn (b−a)n

,

n!2n



if a ≤ x ≤ b − (n−1)θ
(b − a),

2


max{| (x−a)n − θ(b−a)(x−a)n−1 |, | (x−b)n + θ(b−a)(x−b)n−1 |},
n!
2(n−1)!
n!
2(n−1)!
(9) In (θ, x) =
(n−1)θ
(n−1)θ

if b − 2 (b − a) < x < a + 2 (b − a),



(n−1)n−1 θn (b−a)n


,

n!2n


(n−1)θ
if a + 2 (b − a) ≤ x ≤ b
for 1 < (n − 1)θ ≤ 2, and
(10) In (θ, x)
(x − a)n θ(b − a)(x − a)n−1 (x − b)n θ(b − a)(x − b)n−1 ,
−
+
= max , n!
n!
2(n − 1)!
2(n − 1)!
for a ≤ x ≤ b with (n − 1)θ > 2.
Proof. Using the identity (5) in Lemma 2, we can easily derive that
Z b
b−a
(11) [θf (a) + 2(1 − θ)f (x) + θf (b)]
f (t) dt −
2
a
n−1 X
(−1)k (x − a)k+1 + (b − x)k+1
−
(k + 1)!
k=1
b
_
θ(b − a)[(−1)k (x − a)k + (b − x)k ]
(k)
−
f (x) ≤ In (θ, x) (f (n−1) ).
2k!
a
where
In (θ, x) = max |Kn (t, x, θ)|.
(12)
t∈[a,b]
For brevity, we put
nθ
(b − a)], t ∈ [a, b],
2
nθ
Qn (t) := (t − b)n−1 [t − b + (b − a)], t ∈ [a, b],
2
where θ ∈ [0, 1]. It is clear that both Pn (t) and Qn (t) have only one zero in
(a, b) for 0 < nθ < 2. Denote t1 = a + nθ
(b − a) and t2 = b − nθ
(b − a). It is
2
2
Pn (t) := (t − a)n−1 [t − a −
SOME GENERAL OSTROWSKI TYPE INEQUALITIES
57
easy to find that a ≤ t1 < a+b
< t2 ≤ b if and only if 0 ≤ nθ < 1 as well as
2
a+b
a ≤ t2 ≤ 2 ≤ t1 ≤ b if and only if 1 ≤ nθ ≤ 2.
By differentiation, we get
(n − 1)θ
Pn0 (t) := (t − a)n−2 [t − a −
(b − a)], t ∈ [a, b],
2
(n − 1)θ
Q0n (t) := (t − b)n−2 [t − b +
(b − a)], t ∈ [a, b],
2
where θ ∈ [0, 1]. It is clear that both Pn0 (t) and Q0n (t) have only one zero in
(a, b) for 0 < (n−1)θ < 2. Denote t∗1 = a+ (n−1)θ
(b−a) and t∗2 = b− (n−1)θ
(b−a).
2
2
∗
It is easy to find that a ≤ t∗1 < a+b
<
t
≤
b
if
and
only
if
0
≤
(n
−
1)θ
< 1 as
2
2
a+b
∗
∗
well as a ≤ t2 ≤ 2 ≤ t1 ≤ b if and only if 1 ≤ (n − 1)θ ≤ 2.
By differential calculus, it is easy to find that t∗1 is a minimum point of Pn (t)
with
(n − 1)n−1 θn (b − a)n
(13)
Pn (t∗1 ) = −
,
n!2n
as well as t∗2 is a minimum point of Qn (t) for an even n and a maximum point
of Qn (t) for an odd n > 1 with
(n − 1)n−1 θn (b − a)n
.
n!2n
From (6), (13) and (14) it is not difficult to find that
Qn (t∗2 ) = (−1)n−1
(14)
(15)
max |Kn (t, x, θ)|
(
n
n−1
n−1 n
n
max{| (x−b)
+ θ(b−a)(x−b)
|, (n−1) n!2θn (b−a) }, a ≤ x ≤ a+b
,
n!
2(n−1)!
2
=
θ(b−a)(x−a)n−1 (n−1)n−1 θn (b−a)n
(x−a)n
|,
}, a+b
< x ≤ b,
max{| n! −
2(n−1)!
n!2n
2
t∈[a,b]
for 0 ≤ (n − 1)θ ≤ 1, and
(16)
max |Kn (t, x, θ)|
 (n−1)n−1 θn (b−a)n

,

n!2n



if a ≤ x ≤ b − (n−1)θ
(b − a),

2


max{| (x−a)n − θ(b−a)(x−a)n−1 |, | (x−b)n + θ(b−a)(x−b)n−1 |},
n!
2(n−1)!
n!
2(n−1)!
=
(n−1)θ
(n−1)θ

(b
−
a)
<
x
<
a
+
(b
− a),
if
b
−

2
2


(n−1)n−1 θn (b−a)n


,

n!2n


(n−1)θ
if a + 2 (b − a) ≤ x ≤ b
t∈[a,b]
for 1 < (n − 1)θ ≤ 2, and
(17)
max |Kn (t, x, θ)|
t∈[a,b]
58
ZHENG LIU
(x − a)n θ(b − a)(x − a)n−1 (x − b)n θ(b − a)(x − b)n−1 ,
= max −
+
, n!
n!
2(n − 1)!
2(n − 1)!
for a ≤ x ≤ b with (n − 1)θ > 2.
Consequently, the inequality (7) with (8), (9) and (10) follows from (11),
(12) and (15)-(17).
Remark 1. It is clear that Theorem 2 is just the special case n = 2 of Theorem
3.
Corollary 1. Let the assumptions of Theorem 3 hold. Then for all n ≥ 1 and
x ∈ [a, b] we have midpoint type inequalities
Z
n−1 b
k+1
k
k+1
X
(b − x)
+ (−1) (x − a)
(k)
(18) f (t) dt − (b − a)f (x) −
f (x)
a
(k + 1)!
k=1
(
b
(b−x)n
_
, a ≤ x ≤ a+b
,
(n−1)
n!
2
≤ (f
) × (x−a)
n
a+b
,
≤ x ≤ b.
n!
2
a
Proof. Letting θ = 0 in (7) with (8) readily produces the result (18).
Remark 2. For n = 1, it is clear that (18) can be written as
Z b
b
1
a + b _
f (t) dt − (b − a)f (x) ≤ (b − a) + x −
(f )
2
2 a
a
which was first appeared in [1].
Corollary 2. Let the assumptions of Theorem 3 hold. Then for n = 1, 2 we
have trapezoid inequalities
Z b
b
b−a_
b
−
a
≤
(19)
[f
(a)
+
f
(b)]
(f )
f
(t)
dt
−
2
2 a
a
and
(20)
Z b
b
(b − a)2 _
b
−
a
(f 0 ),
f (t) dt −
[f (a) + f (b)] ≤
2
8
a
a
and for n ≥ 3 we have trapezoid type inequalities
Z
n−1 b
X
b−a
(−1)k (x − a)k+1 + (b − x)k+1
f (t) dt −
[f (a) + f (b)] −
(21) a
2
(k + 1)!
k=1
b
_
(b − a)[(−1)k (x − a)k + (b − x)k ]
(k)
−
f (x) ≤ (f (n−1) )×
2k!
a
(x − a)n (b − a)(x − a)n−1 (x − b)n (b − a)(x − b)n−1 ,
.
× max −
+
n!
n!
2(n − 1)!
2(n − 1)!
SOME GENERAL OSTROWSKI TYPE INEQUALITIES
59
Proof. Letting θ = 1 in (7) with (8), (9) and (10) readily produces the results
(19)-(21).
Corollary 3. Let the assumptions of Theorem 3 hold. Then for n = 1, 2, 3, 4, 5, 6,
we have Simpson type inequalities
Z b
b
−
a
f (t) dt −
(22) [f (a) + 4f (x) + f (b)]
6
a
(
b
a+5b
_
− x, a ≤ x ≤ a+b
,
6
2
≤ (f ) ×
a+b
5a+b
≤ x ≤ b,
x− 6 ,
2
a
Z b
b−a
2(b − a)
a + b 0 (23) f (t) dt −
[f (a) + 4f (x) + f (b)] +
(x −
)f (x)
6
3
2
a
(
b
(b−a)2
1 a+b
b−a a+b
2
_
(
−
x)
+
(
−
x)
+
, a ≤ x ≤ a+b
,
0
3
2
24
2
≤ (f ) × 12 2 a+b 2 b−a
2
(b−a)
(x − 2 ) + 3 (x − a+b
) + 24 , a+b
≤ x ≤ b,
2
2
2
a
Z b
b−a
(24) [f (a) + 4f (x) + f (b)]
f (t) dt −
6
a
2
2(b − a)
a+b
b−a
a+b
0
00
+
x−
f (x) −
x−
f (x)
3
2
3
2

1 a+b

( 2 − x)3 + b−a
( a+b
− x)2

12
12
2

3

(b−a)2 a+b


( 2 − x) + (b−a)

48
648
 1 a+b


+| 12 ( 2 − x)3 + b−a
( a+b
− x)2

12
2


2
3
b
+ (b−a) ( a+b − x) − (b−a) |,
_
if a ≤ x ≤ a+b
,
2
≤ (f 00 ) × 1 48 a+b2 3 b−a 648 a+b 2

(x − 2 ) + 12 (x − 2 )

12
a

3

(b−a)2


+ 48 (x − a+b
) + (b−a)
+

2
648


a+b 3
b−a
a+b 2
1

| 12 (x − 2 ) + 12 (x − 2 )



+ (b−a)2 (x − a+b ) − (b−a)3 |,
if a+b
≤ x ≤ b,
48
2
648
2
Z b
b−a
2(b − a)
a+b
f (t) dt −
(25) [f (a) + 4f (x) + f (b)] +
x−
f 0 (x)
6
3
2
a
2
3
b
−
a
b−a
a+b
a
+
b
f 00 (x) +
f 000 (x)
−
x−
x−
3
2
9
2
≤
b
(b − a)4 _ 000
(f ),
1152 a
a ≤ x ≤ b,
60
ZHENG LIU
Z b
2(b − a)
a+b 0
b−a
(26) [f (a) + 4f (x) + f (b)] +
(x −
)f (x)
f (t) dt −
6
3
2
a
b−a
a + b 2 00
b−a
a + b 3 000
−
(x −
) f (x) +
(x −
) f (x)
3
2
9
2
(b − a)5 b − a
a + b 4 (4) +[
−
(x −
) ]f (x)
2880
36
2
 (b−a)5

,
if a ≤ x ≤ a + 2a+b
,

36455
3


(b−a)
b−a
a+b
4

b
 5760 − 72 (x − 2 )
_
(4)
1
≤ (f ) ×
+ b−a
|x − a+b
|| 15
(x − a+b
)4
8
2
2

4

a

+ b−a
(x − a+b
)2 − (b−a)
|,
if 2a+b
< x < a+2b
,

18 5
2
144
3
3

 (b−a)
a+2b
,
if 3 ≤ x ≤ b,
3645
Z b
b−a
(27) f (t) dt −
[f (a) + 4f (x) + f (b)]
6
a
2
a+b
b−a
a+b
2(b − a)
0
x−
f (x) −
x−
f 00 (x)
+
3
2
3
2
3
4
b−a
a+b
(b − a)5 b − a
a+b
000
+
x−
f (x) + [
−
x−
]f (4) (x)
9
2
2880
36
2
"
5 #
(b − a)5
a+b
b−a
a+b
−
x−
−
x−
f (5) (x)
2880
2
180
2

625(b−a)6

if a ≤ x ≤ a + 5a+b
,
 6718464 ,
6

4
2
6

(b−a)
(b−a)
(b−a)
a+b
a+b

2
4

b
 46080 + 2304 (x − 2 ) − 576 (x − 2 )
_
4
(5)
1
≤ (f )× − 720
(x − a+b
)6 + b−a
[ (b−a)
2
360
16


a

−(x − a+b
)4 ]|x − a+b
|,
if 5a+b
< x < a+5b
,

2
2
6
6


6
 625(b−a) ,
a+5b
if 6 ≤ x ≤ b
6718464
and for n ≥ 7 with a ≤ x ≤ b we have Simpson type inequalities
Z b
b−a
f (t) dt −
[f (a) + 4f (x) + f (b)]
(28) 6
a
n−1 X
(−1)k (x − a)k+1 + (b − x)k+1
−
(k + 1)!
k=1
b
_
(b − a)[(−1)k (x − a)k + (b − x)k ]
(k)
−
f (x) ≤ (f (n−1) )×
6k!
a
(x − a)n (b − a)(x − a)n−1 (x − b)n (b − a)(x − b)n−1 ,
.
× max −
+
n!
n!
6(n − 1)!
6(n − 1)!
SOME GENERAL OSTROWSKI TYPE INEQUALITIES
Proof. Letting θ =
(22)-(28).
1
3
61
in (7) with (8), (9) and (10) readily produces the results
Corollary 4. Let the assumptions of Theorem 3 hold. Then for n = 1, 2, 3, 4,
we have averaged midpoint-trapezoid type inequalities
Z b
b−a
(29) f (t) dt −
[f (a) + 2f (x) + f (b)]
4
a
(
b
a+3b
_
− x,
4
≤ (f ) ×
3a+b
x− 4 ,
a
if a ≤ x ≤ a+b
,
2
a+b
if 2 ≤ x ≤ b,
Z b
b−a
b−a
a+b
0
f (t) dt −
(30) [f (a) + 2f (x) + f (b)] +
x−
f (x)
4
2
2
a
 1 a+b
2
( 2 − x)2 + b−a
( a+b
− x) + (b−a)
+

4
8
2
64


2
b

(b−a)
1
a+b
b−a
a+b
2
_
| 4 ( 2 − x) + 8 ( 2 − x) − 64 |, a ≤ x ≤ a+b
,
0
2
≤ (f ) × 1
(b−a)2
a+b 2
b−a
a+b

(x − 2 ) + 8 (x − 2 ) + 64 +

a

2
 41
| 4 (x − a+b
)2 + b−a
(x − a+b
) − (b−a)
|, a+b
≤ x ≤ b,
2
8
2
64
2
Z b
b
−
a
b
−
a
a
+
b
(31) f (t) dt −
[f (a) + 2f (x) + f (b)] +
x−
f 0 (x)
4
2
2
a
"
2 #
b
(b − a)2 _
(b − a)3 b − a
a+b
00
+
−
x−
f (x) ≤
(f 00 ), a ≤ x ≤ b
48
4
2
96
a
and
Z b
b−a
a+b
b−a
(32) f (t) dt −
[f (a) + 2f (x) + f (b)] +
x−
f 0 (x)
4
2
2
a
"
2 #
(b − a)3 b − a
(b − a)3
a+b
a+b
00
+
−
f (x) −
x−
x−
48
4
2
48
2
#
3
a+b
b−a
f 000 (x)
x−
−
12
2
 27(b−a)4
,
a ≤ x ≤ a + 3a+b
,

6144
4


3
4
b
 (b−a)
(b−a)
a+b
_
+ 96 |x − 2 |
384
≤ (f 000 ) ×
b−a
1

− 24 |x − a+b
|3 − 24
(x − a+b
)4 , 3a+b
< x < a+3b
,

2
2
4
4
a

 27(b−a)4
a+3b
,
≤ x ≤ b,
6144
4
62
ZHENG LIU
and for n ≥ 5 with a ≤ x ≤ b we have averaged midpoint-trapezoid type
inequalities
Z b
b−a
(33) f (t) dt −
[f (a) + 2f (x) + f (b)]
4
a
n−1 X
(−1)k (x − a)k+1 + (b − x)k+1
−
(k + 1)!
k=1
b
_
(b − a)[(−1)k (x − a)k + (b − x)k ]
(k)
−
f (x) ≤ (f (n−1) )×
4k!
a
(x − a)n (b − a)(x − a)n−1 (x − b)n (b − a)(x − b)n−1 ,
.
× max −
+
n!
n!
4(n − 1)!
4(n − 1)!
Proof. Letting θ =
(29)-(33).
1
2
in (7) with (8), (9) and (10) readily produces the results
Corollary 5. Let the assumptions of Theorem 3 hold. Then for any θ ∈ [0, 1]
we have
Z b
a
+
b
b
−
a
(34) θf (a) + 2(1 − θ)f
+ θf (b)
f (x) dx −
2
2
a
]
[ n−1
2
2k+1
X
[1 − (n + 1)θ](b − a)
a + b (2k)
−
f
(2k + 1)!22k
2
k=1
(
b
max{|1−nθ|,(n−1)n−1 θn }(b−a)n
_
, n<
n
(n−1)
≤ (f
) × (nθ−1)(b−a)n(n!)2
,
n≥
(n!)2n
a
where [ n−1
] denotes the integer part of
2
Proof. Letting x =
(34).
a+b
2
1
θ
1
θ
+ 1,
+ 1.
n−1
.
2
in (7) with (8), (9) and (10) readily produces the result
Remark 3. If we take θ = 0 in (34), we get the midpoint type inequality
n−1
Z b
]
[X
2
2k+1
a+b
(b − a)
a + b f (x) dx − (b − a)f
(35) f (2k)
−
2k
2
(2k + 1)!2
2
a
k=1
≤
b
(b − a)n _ (n−1)
(f
).
n!2n a
SOME GENERAL OSTROWSKI TYPE INEQUALITIES
63
If we take θ = 1 in (34), we get the trapezoid type inequality
Z b
[ n−1
]
2
2k+1
X
b−a
k(b − a)
a + b (2k)
(36) f (x) dx −
[f (a) + f (b)] +
f
2
(2k + 1)!22k−1
2
a
k=1
b−a
b
_
,
n = 1,
(n−1)
2
n
≤ (f
) × (n−2)(b−a)
, n ≥ 2.
n!2n+1
a
If we take θ =
1
3
in (34), we get the Simpson type inequality
Z b
a+b
b−a
(37) f (a) + 4f
+ f (b)
f (x) dx −
6
2
a
[ n−1
]
2
2k+1
X (k − 1)(b − a)
a + b (2k)
f
+
2k−1
3(2k
+
1)!2
2
k=1
 b−a
,

3


2
b
(b−a)

_
,
(n−1)
24
≤ (f
) × (b−a)3

,

324
a

n
 (n−3)(b−a)
3(n!)2n
n = 1,
n = 2,
n = 3,
, n≥4
which is just equal to (1) in Theorem 1.
If we take θ = 12 in (34), we get the averaged midpoint-trapezoid type
inequality
Z b
b
−
a
a
+
b
f (x) dx −
(38) f (a) + 2f
+ f (b)
4
2
a
]
[ n−1
2
2k+1
X
(2k − 1)(b − a)
a + b (2k)
f
+
(2k + 1)!22k+1
2
k=1

b−a

b
 4 ,2
_
≤ (f (n−1) ) × (b−a)
,
32

 (n−2)(b−a)n
a
n!2n+1
n = 1,
n = 2,
, n ≥ 3.
64
ZHENG LIU
Theorem 4. Let f : [a, b] → R be such that f (n−1) is absolutely continuous on
[a, b]. Then for any x ∈ [a, b] and θ ∈ [0, 1] we have
Z b
b−a
f (t) dt −
[θf (a) + 2(1 − θ)f (x) + θf (b)]
2
a
n−1 X
(−1)k (x − a)k+1 + (b − x)k+1
−
(k + 1)!
k=1
θ(b − a)[(−1)k (x − a)k + (b − x)k ]
(k)
−
f (x) ≤ In (θ, x)kf (n) k1 ,
2k!
where In (θ, x) is as given in (8)-(10) and kf (n) k1 =
Lebesgue norm on L1 [a, b].
Rb
a
|f (n) (t)| dt is the usual
Proof. It is immediate from Theorem 3, since if f (n−1) is absolutely continuous
on [a, b] then we certainly have
b
_
(f (n−1) ) = kf (n) k1 .
a
Theorem 5. Let f : [a, b] → R be such that f (n−1) is L-Lipschitzian on [a, b].
Then for any x ∈ [a, b] and θ ∈ [0, 1] we have
Z b
b−a
[θf (a) + 2(1 − θ)f (x) + θf (b)]
(39) f (t) dt −
2
a
n−1 X
(−1)k (x − a)k+1 + (b − x)k+1
−
(k + 1)!
k=1
θ(b − a)[(−1)k (x − a)k + (b − x)k ]
(k)
−
f (x) ≤ In (θ, x)L(b − a),
2k!
where In (θ, x) is as given in (8)-(10).
Proof. It is immediate from Theorem 3, since if f (n−1) is L-Lipschitzian on
[a, b] then we certainly have
b
_
(f (n−1) ) = L(b − a).
a
SOME GENERAL OSTROWSKI TYPE INEQUALITIES
65
Theorem 6. Let f : [a, b] → R be such that f (n−1) is monotonic on [a, b].
Then for any x ∈ [a, b] and θ ∈ [0, 1] we have
Z b
b−a
[θf (a) + 2(1 − θ)f (x) + θf (b)]
(40) f (t) dt −
2
a
n−1 X
(−1)k (x − a)k+1 + (b − x)k+1
−
(k + 1)!
k=1
θ(b − a)[(−1)k (x − a)k + (b − x)k ]
(k)
−
f (x)
2k!
≤ In (θ, x)|f (n−1) (b) − f (n−1) (a)|,
where In (θ, x) is as given in (8)-(10).
Proof. It is immediate from Theorem 3, since if f (n−1) is monotonic on [a, b]
then we certainly have
b
_
(f (n−1) ) = |f (n−1) (b) − f (n−1) (a)|.
a
Remark 4. It should be noticed that we can also get some further results
similar to Corollaries 1-5 for Theorem 4-6 and so are omitted.
References
[1] S. S. Dragomir. On the Ostrowski’s integral inequality for mappings with bounded variation and applications. Math. Inequal. Appl., 4(1):59–66, 2001.
[2] Z. Liu. Some Ostrowski type inequalities. Math. Comput. Modelling, 48(5-6):949–960,
2008.
[3] C. E. M. Pearce, J. Pečarić, N. Ujević, and S. Varošanec. Generalizations of some inequalities of Ostrowski-Grüss type. Math. Inequal. Appl., 3(1):25–34, 2000.
[4] J. Pečarić and S. Varošanec. Harmonic polynomials and generalization of Ostrowski
inequality with applications in numerical integration. In Proceedings of the Third World
Congress of Nonlinear Analysts, Part 4 (Catania, 2000), volume 47, pages 2365–2374,
2001.
Received December 20, 2011.
Institute of Applied Mathematics, School of Science,
University of Science and Technology Liaoning,
Anshan 114051, Liaoning,
China
E-mail address: lewzheng@163.net