THE EXTENSION OF MONTGOMERY IDENTITY VIA
FINK IDENTITY WITH APPLICATIONS
A. AGLIĆ ALJINOVIĆ, J. PEČARIĆ, AND A. VUKELIĆ
Received 30 January 2004
The new extension of the weighted Montgomery identity is given by using Fink identity
and is used to obtain some Ostrowski-type inequalities and estimations of the difference
of two integral means.
1. Introduction
The following Ostrowski inequality is well known [10]:
2 b
x − (a + b)/2
1
1
f (x) −
≤
f
(t)dt
(b − a)L,
+
b−a a
4
(b − a)2
x ∈ [a,b],
(1.1)
where f : [a,b] → R is a differentiable function such that | f (x)| ≤ L, for every x ∈ [a,b].
The Ostrowski inequality has been generalized over the last years in a number of ways.
Milovanović and Pečarić [8] and Fink [6] have considered generalizations of (1.1) in the
form
b
n
−1
1
1
≤ K(n, p,x)
f (n) p
f (x) + Fk (x) −
f (t)dt n
b−a a
(1.2)
k =1
which is obtained from the identity
n −1
1
1
f (x) + Fk (x) −
n
b
−
a
k =1
b
a
1
f (t)dt =
n!(b − a)
b
a
(x − t)n−1 k(t,x) f (n) (t)dt, (1.3)
where
Fk (x) =
n − k f (k−1) (a)(x − a)k − f (k−1) (b)(x − b)k
,
k!
b−a

t − a, a ≤ t ≤ x ≤ b,
k(t,x) = 
t − b, a ≤ x < t ≤ b.
Copyright © 2005 Hindawi Publishing Corporation
Journal of Inequalities and Applications 2005:1 (2005) 67–80
DOI: 10.1155/JIA.2005.67
(1.4)
68
The extension of Montgomery identity
In fact, Milovanović and Pečarić have proved that
K(n, ∞,x) =
(x − a)n+1 + (b − x)n+1
,
n(n + 1)!(b − a)
(1.5)
while Fink gave the following generalizations of this result.
Theorem 1.1. Let f (n−1) be absolutely continuous on [a,b] and let f (n) ∈ L p [a,b]. Then
inequality (1.2) holds with
K(n, p,x) =
(x − a)nq+1 + (b − x)nq+1
n!(b − a)
1/q
B (n − 1)q + 1, q + 1
1/q
,
(1.6)
where 1 < p ≤ ∞, 1/ p + 1/q = 1, B is the Beta function, and
K(n,1,x) =
(n − 1)n−1
max (x − a)n ,(b − x)n .
n
n n!(b − a)
(1.7)
Let f : [a,b] → R be differentiable on [a,b] and f : [a,b] → R integrable on [a,b].
Then the Montgomery identity holds [9]:
f (x) =
1
b−a
b
a
b
f (t)dt +
a
P(x,t) f (t)dt,
(1.8)
where P(x,t) is the Peano kernel defined by

t−a


,

b
P(x,t) =  − a

t−b ,
b−a
a ≤ t ≤ x,
(1.9)
x < t ≤ b.
Now, we suppose w : [a,b] → [0, ∞ is some probability density function, that is, an int
b
tegrable function satisfying a w(t)dt = 1, and W(t) = a w(x)dx for t ∈ [a,b], W(t) = 0
for t < a, and W(t) = 1 for t > b. The following identity (given by Pečarić in [12]) is the
weighted generalization of the Montgomery identity:
f (x) =
b
a
b
w(t) f (t)dt +
a
Pw (x,t) f (t)dt,
(1.10)
where the weighted Peano kernel is

W(t),
a ≤ t ≤ x,
Pw (x,t) = 
W(t) − 1, x < t ≤ b.
(1.11)
A. Aglić Aljinović et al.
69
The aim of this paper is to give the extension of the weighted Montgomery identity
(1.10) using identity (1.2), and further, obtain some new Ostrowski-type inequalities, as
well as the generalizations of the estimations of the difference of two weighted integral
means (generalizations of the results from [1, 3, 7, 11]).
2. The extension of Montgomery identity via Fink identity
Theorem 2.1. Let f : [a,b] → R be such that f (n−1) is an absolutely continuous function
on [a,b] for some n ≥ 1. If w : [a,b] → [0, ∞ is some probability density function, then the
following identity holds:
f (x) =
b
a
w(t) f (t)dt −
n
−1
Fk (x) +
k =1
1
+
(n − 1)!(b − a)
b
1
−
(n − 1)!(b − a)
a
n
−1 b
k =1 a
w(t)Fk (t)dt
(x − y)n−1 k(y,x) f (n) (y)d y
b b
a
a
(2.1)
w(t)(t − y)
n −1
k(y,t)dt f (n) (y)d y.
Proof. We apply identity (1.3) with f (t):
f (t) = −
+
n
−1
f (b) − f (a)
n − k f (k) (a)(t − a)k − f (k) (b)(t − b)k
+n
k!
b−a
b−a
k =1
1
(n − 1)!(b − a)
b
a
(t − y)n−1 k(y,t) f (n+1) (y)d y
(2.2)
n
−1
n − k f (k) (a)(t − a)k − f (k) (b)(t − b)k
=−
k!
b−a
k =0
1
+
(n − 1)!(b − a)
b
a
(t − y)n−1 k(y,t) f (n+1) (y)d y.
Now, by putting this formula in the weighted Montgomery identity (1.10), we obtain
f (x) =
b
a
−
w(t) f (t)dt
n
−1
n−k
k!
k =0
b
a
Pw (x,t)
1
+
(n − 1)!(b − a)
f (k) (a)(t − a)k − f (k) (b)(t − b)k
dt
b−a
b
a
Pw (x,t)
b
a
(t − y)
n −1
k(y,t) f
(n+1)
(2.3)
(y)d y dt.
70
The extension of Montgomery identity
Further,
b
a
Pw (x,t)
=
f (k) (a)(x − a)k+1 − f (k) (b)(x − b)k+1
(b − a)(k + 1)
−
b
a
f (k) (a)(t − a)k − f (k) (b)(t − b)k
dt
b−a
b
w(t)
a
f (k) (a)(t − a)k+1 − f (k) (b)(t − b)k+1
dt,
(b − a)(k + 1)
(2.4)
Pw (x,t)(t − y)n−1 k(y,t)dt
=
1
1
(x − y)n k(y,x) −
n
n
b
a
w(t)(t − y)n k(y,t)dt.
Now, if we replace n with n − 1, we will get (2.1). This identity is valid for n − 1 ≥ 1, that
is, n > 1.
Remark 2.2. We could also obtain identity (2.1) by applying identity (1.3) such that we
multiply this identity by w(x) and than integrate it to obtain
b
a
w(x) f (x)dx = −
n
−1 b
k =1 a
w(x)Fk (x)dx +
1
+
(n − 1)!(b − a)
b b
a
a
b
a
w(x)dx
w(x)(x − t)
n
b
b−a
n −1
a
f (t)dt
(2.5)
k(t,x)dx f
(n)
(t)dt.
If we subtract this identity from (1.3) we will obtain (2.1).
Remark 2.3. In the special case, if we take w(t) = 1/(b − a), t ∈ [a,b], we will have
1
n
−1 b
b − a k =1
a
=
1
b−a
b
a
n
−1
n−k
Fk (t)dt =
b − a k=1 k!
1
b
a
f (k−1) (a)(t − a)k − f (k−1) (b)(t − b)k
dt
b−a
n
−1
n − k (k−1)
f
(a)(b − a)k−1 + f (k−1) (b)(a − b)k−1 ,
(k
+
1)!
k =1
(b − y)n
(a − y)n
− k(y,a)
n(b − a)
n(b − a)
(y − a)(b − y)n (y − b)(a − y)n
=
−
.
n(b − a)
n(b − a)
(t − y)n−1 k(y,t)dt = k(y,b)
(2.6)
A. Aglić Aljinović et al.
71
We denote
In =
b
1
n!(b − a)2
a
(y − a)(b − y)n − (y − b)(a − y)n f (n) (y)d y.
(2.7)
Then we have
In =
b
1
n!(b − a)2
(a − y)n − (b − y)n f (n−1) (y)d y + In−1 = Jn + In−1 ,
a
(2.8)
where
I0 =
b
1
(b − a)2
a
(b − a) f (y)d y =
1
b−a
b
a
f (y)d y.
(2.9)
Further,
1 (n−2)
f
(a)(b − a)n−2 + f (n−2) (b)(a − b)n−2 + Jn−1 ,
n!
b
b
1
1
(a
−
b)
f
(y)d
y
=
−
f (y)d y.
J1 =
(b − a)2 a
b−a a
Jn =
(2.10)
So,
Jn =
n
−1
1 (k−1)
f
(a)(b − a)k−1 + f (k−1) (b)(a − b)k−1 + J1 ,
(k + 1)!
k =1
(2.11)
and then
In =
n
Jm + nJ1 + I0
m=2
=
n
−1
n − k (k−1)
f
(a)(b − a)k−1 + f (k−1) (b)(a − b)k−1
(k + 1)!
k =1
−
n−1
b−a
(2.12)
b
a
f (y)d y.
Consequently, identity (2.1) reduces to identity (1.3). So we may regard it as a weighted
Fink identity.
72
The extension of Montgomery identity
Remark 2.4. Applying identity (2.1) with x = a and x = b, we get
f (a) =
b
a
n − k (k−1)
w(t) f (t)dt −
(b)(a − b)k−1 +
f
k!
k =1
k =1
1
+
(n − 1)!(b − a)
1
−
(n − 1)!(b − a)
f (b) =
b
a
n −1 b
n
−1
b
a
b b
a
a
w(t)(t − y)
n −1
k(y,t)dt f (n) (y)d y,
1
−
(n − 1)!(b − a)
a
(2.13)
n −1 b
n − k (k−1)
w(t) f (t)dt −
(a)(b − a)k−1 +
f
k!
k =1
k =1
b
w(t)Fk (t)dt
(a − y)n−1 (y − b) f (n) (y)d y
n
−1
1
+
(n − 1)!(b − a)
a
a
w(t)Fk (t)dt
(b − y)n−1 (y − a) f (n) (y)d y
b b
a
a
w(t)(t − y)
n −1
k(y,t)dt f (n) (y)d y.
So, we get the generalized trapezoid identity
1
f (a) + f (b) =
2
b
a
w(t) f (t)dt +
n
−1 b
k =1 a
w(t)Fk (t)dt
n −1
−
1 n − k (k−1)
f
(a)(b − a)k−1 + f (k−1) (b)(a − b)k−1
2 k=1 k!
+
1
2(n − 1)!(b − a)
1
−
(n − 1)!(b − a)
b
a
(a − y)n−1 (y − b) + (b − y)n−1 (y − a) f (n) (y)d y
b b
a
a
w(t)(t − y)
n −1
k(y,t)dt f (n) (y)d y.
(2.14)
Similarly, applying identity (2.1) with x = (a + b)/2, we get
f
a+b
=
2
b
a
−
w(t) f (t)dt +
n
−1 b
k =1 a
n
−1
n − k (k−1)
f
(a)(b − a)k−1 + f (k−1) (b)(a − b)k−1
k
2 k!
k =1
1
+
(n − 1)!(b − a)
−
w(t)Fk (t)dt
1
(n − 1)!(b − a)
b
a
a+b
−y
2
b b
a
a
n−1 (2.15)
a + b (n)
k y,
f (y)d y
2
w(t)(t − y)n−1 k(y,t)dt f (n) (y)d y.
We can regard this as the second Euler-Maclaurin formula (the generalized midpoint
identity).
A. Aglić Aljinović et al.
73
3. Ostrowski-type inequalities
We denote, for n ≥ 2,
Tw,n (x) =
n
−1
Fk (x) −
k =1
n
−1 b
k =1 a
w(t)Fk (t)dt.
(3.1)
Theorem 3.1. Assume (p, q) is a pair of conjugate exponents, that is, 1 ≤ p, q ≤ ∞, 1/ p +
1/q = 1. Let | f (n) | p : [a,b] → R be an R-integrable function for some n > 1. Then, for x ∈
[a,b], the following inequality holds:
b
f (x) − w(t) f (t)dt + Tw,n (x)
a
 q 1/q
b b
1
 Pw (x,t)(t − y)n−2 k(y,t)dt d y  f (n) p .
≤
(n − 2)!(b − a) a a
(3.2)
b b
The constant (1/(n − 2)!(b − a))( a | a Pw (x,t)(t − y)n−2 k(y,t)dt |q d y)1/q is sharp for 1 <
p ≤ ∞ and is the best possible for p = 1.
Proof. From Theorem 2.1 we have
(x − y)n−1 k(y,x) −
b
a
w(t)(t − y)n−1 k(y,t)dt = (n − 1)
b
a
Pw (x,t)(t − y)n−2 k(y,t)dt.
(3.3)
b
We denote C1 (y) = (1/(n − 2)!(b − a)) a Pw (x,t)(t − y)n−2 k(y,t)dt. We use identity (2.1)
and apply the Hölder inequality to obtain
b
f (x) − w(t) f (t)dt + Tw,n (x)
a
1/q
b
b
q
(n) (n)
f .
C1 (y) d y
= C1 (y) f (y)d y ≤
p
a
a
(3.4)
b
For the proof of the sharpness of the constant ( a |C1 (y)|q d y)1/q , we will find a function
f for which the equality in (3.2) is obtained.
For 1 < p < ∞, take f to be such that
1/(p−1)
f (n) (y) = sgn C1 (y) · C1 (y)
.
(3.5)
For p = ∞, take
f (n) (y) = sgn C1 (y).
(3.6)
74
The extension of Montgomery identity
For p = 1, we will prove that
b
b
(n)
(n)
C1 (y) f (y)dt ≤ max C1 (y)
f (y) d y
a
y ∈[a,b]
a
(3.7)
is the best possible inequality. Suppose that |C1 (y)| attains its maximum at y0 ∈ [a,b].
First we assume that C1 (y0 ) > 0. For ε small enough, define fε (y) by


0,





 1 a ≤ y ≤ y0 ,
n
fε (y) =  εn! y − y0 ,



n −1

1


y − y0
,
n!
y0 ≤ y ≤ y0 + ε,
(3.8)
y0 + ε ≤ y ≤ b.
Then, for ε small enough,
y +ε
b
y0 +ε
1 1 0
(n)
C1 (y) f (y)d y = C1 (y) d y C1 (y)d y.
=
a
y0
ε ε y0
(3.9)
Now, from inequality (3.7) we have
1
ε
y0 +ε
y0
C1 (y)d y ≤ C1 y0
y0 +ε
y0
1
d y = C1 y0 .
ε
(3.10)
Since
1
ε→0 ε
lim
y0 +ε
y0
C1 (y)d y = C1 y0 ,
(3.11)
the statement follows. In case C1 (y0 ) < 0, we take
 n−1
1


y − y0 − ε
,



 n!
fε (y) = − 1 y − y0 − εn ,


εn!


0,
and the rest of the proof is the same as above.
a ≤ y ≤ y0 ,
y0 ≤ y ≤ y0 + ε,
(3.12)
y0 + ε ≤ y ≤ b,
A. Aglić Aljinović et al.
75
Remark 3.2. For w(t) = 1/(b − a), n = 2, and q = 1 in Theorem 3.1, we get
b
1
a+b f (x) −
f (t)dt − x −
f (b) − f (a) b−a a
2
b
b
1
1
(x − y)k(y,x) −
(t
− y)k(y,t)dt d y f ∞
≤
b−a a
b−a a
b
(y
−
a)(b
−
y)
1
(x − y)k(y,x) −
d y f =
∞
b−a a
2
b
x
1
(y − a)(2x − y − b)d y + (b − y)(−2x + y + a)d y f =
∞
2(b − a) a
x
1 4 3
1
f ,
=
δ (x) − δ 2 (x) +
∞
3
2
24
(3.13)
where δ(x) = |x − (a + b)/2|.
If instead of q = 1 (p = ∞) we put p = 1, then, similarly we have
b
1
a+b f (x) −
f (t)dt − x −
f (b) − f (a) b−a a
2
1
≤
max max (y − a)(2x − y − b), max (b − y)(−2x + y + a) f 1
2(b − a)
y ∈[a,x]
y ∈[x,b]
1 1 a + b 2
1
f 1 .
=
−2 x−
+
4 4
4
2
(3.14)
These two inequalities are proved in [5].
Corollary 3.3. Suppose that all the assumptions of Theorem 3.1 hold. Then the following
inequality holds:
b
f (x) − w(t) f (t)dt + Tw,n (x)
a
1/q
b
(n) q
1
f .
≤
(b − y)(y − a)n−1 + (y − a)(b − y)n−1 d y
p
(n − 1)!(b − a) a
(3.15)
76
The extension of Montgomery identity
Proof. Since 0 ≤ W(t) ≤ 1, t ∈ [a,b], so |Pw (x,t)| ≤ 1. Then, for every y ∈ [a,b], we have
b
Pw (x,t)(t − y)n−2 k(y,t)dt a
b
Pw (x,t)(t − y)n−2 k(y,t)dt
≤
a
≤
b
(t − y)n−2 k(y,t)dt
a
=
=
y
a
(y − t)
n −2
(b − y)dt +
(3.16)
b
y
(t − y)
n −2
(y − a)dt
1 (b − y)(y − a)n−1 + (y − a)(b − y)n−1 .
n−1
So,

q 1/q
b
b
 Pw (x,t)(t − y)n−1 k(y,t)dt d y
a a
≤
1
n−1
b
a
(b − y)(y − a)n−1 + (y − a)(b − y)
n −1 q
1/q
(3.17)
dt
and, by applying (3.2), the inequality is proved.
Remark 3.4. Inequality (3.15) reduces to the following: for n = 2,
b
f (x) − w(t) f (t)dt + Tw,2 (x)
a
≤
2
b
b−a
a
q
1/q
q
(b − y) (y − a) d y
= 2(b − a)(q+1)/q
1
0
f 1/q
(1 − s)q sq ds
p
f (3.18)
p
= 2(b − a)(q+1)/q B(q + 1, q + 1)1/q f p .
For n = 3,
b
f (x) − w(t) f (t)dt + Tw,3 (x)
a
≤
=
1
b
2(b − a)
a
q
1/q
q
q
(b − y) (y − a) (b − a) d y
1
(b − a)(2q+1)/q B(q + 1, q + 1)1/q f p .
2
f p
(3.19)
A. Aglić Aljinović et al.
77
Remark 3.5. If we use the identities (2.14) and (2.15) for n = 2 and w(t) = 1/(b − a),
t ∈ [a,b], and then apply the Hölder inequality with p = ∞, q = 1, we will obtain
b
f (a) + f (b)
(b − a)2 1
f ,
≤
−
f
(t)dt
∞
2
b−a a
12
a+b
(b − a)2 b
1
f .
f
≤
f
(t)dt
−
∞
2
b−a a
24
(3.20)
By doing the same for n = 3, we have
b
f (a) + f (b)
1
b −a (b − a)3 f ,
≤
−
f (t)dt −
f (b) − f (a) ∞
2
b−a a
12
192
a+b
b
b −a 1
(b − a)3 f .
f
−
f
(t)dt
+
f
(b)
− f (a) ≤
∞
2
b−a a
24
192
(3.21)
The first two inequalities were obtained in [4] and the last two in [2].
4. Estimations of the difference of two weighted integral means
In this section, we will denote, for n > 1,
[a,b]
(x) =
Tw,n
n
−1
k =1
Fk[a,b] (x) −
n
−1 b
k =1 a
w(t)Fk[a,b] (t)dt,
(4.1)
for a function f : [a,b] → R such that f (n−1) is an absolutely continuous function on
[a,b].
The following results are generalizations of the results from [3] in two cases. The first
case is when [c,d] ⊆ [a,b] and the second is when [a,b] ∩ [c,d] = [c,b]. Other two possible cases, when [a,b] ∩ [c,d] = ∅ ([a,b] ⊂ [c,d] and [a,b] ∩ [c,d] = [a,d]) are simply
got by change a ↔ c, b ↔ d.
Theorem 4.1. Let f : [a,b] ∪ [c,d] → R be such that f (n−1) is an absolutely continuous
function on [a,b] for some n > 1, and let w : [a,b] → [0, ∞ and u : [c,d] → [0, ∞ be some
probability density functions. Then, if [a,b] ∩ [c,d] = ∅ and x ∈ [a,b] ∩ [c,d],
b
a
w(t) f (t)dt −
d
c
[a,b]
[c,d]
u(t) f (t)dt − Tw,n
(x) + Tu,n
(x) =
max{b,d}
min{a,c}
Kn (x, y) f (n) (y)d y,
(4.2)
78
The extension of Montgomery identity
where, in case [c,d] ⊆ [a,b],

b

−1

n−2 [a,b]

Pw (x,t)(t − y) k (y,t)dt ,



(n − 2)!(b − a) a





b


−1

n−2 [a,b]


P
(x,t)(t
−
y)
k
(y,t)dt

 (n − 2)!(b − a) a w
Kn (x, y) = 
d

1

n
−2 [c,d]


+
Pu (x,t)(t − y) k (y,t)dt ,


(n − 2)!(d − c) c






b

−1

n
−
2
[a,b]


Pw (x,t)(t − y) k (y,t)dt ,

(n − 2)!(b − a) a
y ∈ [a,c],
(4.3)
y ∈ c,d],
y ∈ d,b],
and in case [a,b] ∩ [c,d] = [c,b],

b

−1

n−2 [a,b]

P
(x,t)(t
−
y)
k
(y,t)dt
,

w


(n − 2)!(b − a) a





b


−1

n−2 [a,b]


P (t)(t − y) k (y,t)dt

 (n − 2)!(b − a) a w
Kn (x, y) = 
d

1

n
−2 [c,d]

+
(y,t)dt ,

 (n − 2)!(d − c) c Pu (t)(t − y) k






d

1

n−2 [c,d]


P
(t)(t
−
y)
k
(y,t)dt
,
u

(n − 2)!(d − c) c
y ∈ [a,c],
(4.4)
y ∈ c,b],
y ∈ b,d].
Proof. We subtract identity (2.1) for intervals [a,b] and [c,d] to get formula (4.2).
Theorem 4.2. Assume (p, q) is a pair of conjugate exponents, that is, 1 ≤ p, q ≤ ∞, 1/ p +
1/q = 1. Let | f (n) | p : [a,b] → R be an R-integrable function for some n > 1. Then
d
b
[a,b]
[c,d]
w(t) f (t)dt −
u(t)
f
(t)dt
− Tw,n (x) + Tu,n (x)
a
c
1/q
max{b,d} (n) Kn (x, y)q d y
f ≤
(4.5)
p
min{a,c}
max{b,d}
for every x ∈ [a,b] ∩ [c,d]. The constant ( min{a,c} |Kn (x, y)|q d y)1/q in inequality (4.5) is
sharp for 1 < p ≤ ∞ and is the best possible for p = 1.
Proof. Use identity (4.2) and apply the Hölder inequality to obtain
d
b
[a,b]
[c,d]
w(t) f (t)dt −
u(t) f (t)dt − Tw,n (x) + Tu,n (x)
a
c
(4.6)
1/q
max{b,d}
max{b,d} (n) q
(n) Kn (x, y) f (y)d y ≤
Kn (x, y) d y
f ,
≤
p
min{a,c}
min{a,c}
A. Aglić Aljinović et al.
79
which proves inequality (4.5). The proofs for sharpness and best possibility are as in
Theorem 3.1.
Corollary 4.3. Suppose that all the assumptions of Theorem 4.2 hold. Then, for x ∈ [a,b] ∩
[c,d],
d
b
[a,b]
[c,d]
w(t) f (t)dt −
u(t)
f
(t)dt
− Tw,n
(x)
+
T
(x)
u,n
a
c
1/q
max{b,d} (n) n−1 2
q
n
−1
f .
(y − a)
dy
≤
+ max{b,d} − y
p
(n − 1)! a
(4.7)
Proof. We have
−1
Kn (x, y) =
(n − 2)!
max{b,d} min{a,c}
k[a,b] (y,t)
k[c,d] (y,t)
Pw (x,t)
− Pu (x,t)
(t − y)n−2 dt
b−a
d−c
(4.8)
because Pw (x,t) = 0, for x ∈
/ [a,b] and Pu (x,t) = 0, for x ∈
/ [c,d]. Since
−1 ≤ Pw (x,t),Pu (x,t),
k[a,b] (y,t) k[c,d] (y,t)
≤ 1,
,
b−a
d−c
(4.9)
we get
k[a,b] (y,t)
k[c,d] (y,t) Pw (x,t)
≤ 2,
− Pu (x,t)
b−a
d−c (4.10)
and then we have
Kn (x, y) ≤
2
(n − 2)!
max{b,d}
a
|t − y |n−2 dt =
2 (y − a)n−1 + max{b,d} − y
(n − 1)!
n−1 .
(4.11)
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[2]
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[5]
[6]
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A. Aglić Aljinović: Department of Applied Mathematics, Faculty of Electrical Engineering and
Computing, University of Zagreb, Unska 3, 10000 Zagreb, Croatia
E-mail address: andrea@zpm.fer.hr
J. Pečarić: Department of Mathematics, Faculty of Textile Technology, University of Zagreb, Pierottijeva 6, 10000 Zagreb, Croatia
E-mail address: pecaric@hazu.hr
A. Vukelić: Mathematics Department, Faculty of Food Technology and Biotechnology, University
of Zagreb, Pierottijeva 6, 10000 Zagreb, Croatia
E-mail address: avukelic@pbf.hr