Journal of Inequalities in Pure and
Applied Mathematics
THE DISCRETE VERSION OF OSTROWSKI’S INEQUALITY IN
NORMED LINEAR SPACES
volume 3, issue 1, article 2,
2002.
S.S. DRAGOMIR
School of Communications and Informatics
Victoria University of Technology
PO Box 14428
Melbourne City MC
8001 Victoria, Australia
EMail: sever@matilda.vu.edu.au
URL: http://rgmia.vu.edu.au/SSDragomirWeb.html
Received 14 May, 2001;
accepted 02 July, 2001.
Communicated by: R.P. Agarwal
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
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Abstract
Discrete versions of Ostrowski’s inequality for vectors in normed linear spaces
are given.
2000 Mathematics Subject Classification: Primary 26D15; Secondary 26D99.
Key words: Discrete Ostrowski’s Inequality.
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
Some Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3
Discrete Ostrowski’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . 12
4
Weighted Ostrowski Inequality . . . . . . . . . . . . . . . . . . . . . . . . . 23
References
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
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1.
Introduction
The following result is known in the literature as Ostrowski’s inequality [10].
Theorem 1.1. Let f : [a, b] → R be a differentiable mapping on (a, b) with the
property that |f 0 (t)| ≤ M for all t ∈ (a, b). Then
#
"
Z b
a+b 2
x
−
1
2
f (x) − 1
(1.1)
f (t) dt ≤
+
(b − a) M
b−a a
4
(b − a)2
for all x ∈ [a, b]. The constant 41 is the best possible in the sense that it cannot
be replaced by a smaller constant.
A simple proof of this fact can be done by using the identity:
Z b
Z b
1
1
(1.2) f (x) =
f (t) dt +
p (x, t) f 0 (t) dt, x ∈ [a, b] ,
b−a a
b−a a
where
p (x, t) :=

 t − a if a ≤ t ≤ x

t − b if x < t ≤ b
which also holds for absolutely continuous functions f : [a, b] → R.
The following Ostrowski type result for absolutely continuous functions holds
(see [6] – [8]).
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
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Theorem 1.2. Let f : [a, b] → R be absolutely continuous on [a, b]. Then, for
all x ∈ [a, b], we have:
Z b
1
f (x) −
f
(t)
dt
b−a a
a+b 2
x− 2
1
+
(b − a) kf 0 k∞
4
b−a
(1.3)



if f 0 ∈ L∞ [a, b] ;








h
i1
1
1
x−a p+1
b−x p+1 p
0
≤
p kf 0 k
+
(b
−
a)
1
q if f ∈ Lq [a, b] ,
b−a
b−a

p

(p+1)


1

+ 1q = 1, p > 1;

p

a+b i
h


 1 + x− 2 kf 0 k ;
1
2
b−a
where k·kr (r ∈ [1, ∞]) are the usual Lebesgue norms on Lr [a, b], i.e.,
kgk∞ := ess sup |g (t)|
t∈[a,b]
and
Z
kgkr :=
b
r1
|g (t)| dt , r ∈ [1, ∞).
r
a
The constants 41 ,
1
1
(p+1) p
and
1
2
respectively are sharp in the sense presented in
Theorem 1.1.
The above inequalities can also be obtained from the Fink result in [9] on
choosing n = 1 and performing some appropriate computations.
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
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If one drops the condition of absolute continuity and assumes that f is Hölder
continuous, then one may state the result (see [5]):
Theorem 1.3. Let f : [a, b] → R be of r − H−Hölder type, i.e.,
(1.4)
|f (x) − f (y)| ≤ H |x − y|r , for all x, y ∈ [a, b] ,
where r ∈ (0, 1] and H > 0 are fixed. Then, for all x ∈ [a, b] , we have the
inequality:
Z b
1
(1.5) f (x) −
f (t) dt
b−a a
"
r+1 r+1 #
H
b−x
x−a
≤
+
(b − a)r .
r+1
b−a
b−a
The constant
1
r+1
is also sharp in the above sense.
Note that if r = 1, i.e., f is Lipschitz continuous, then we get the following
version of Ostrowski’s inequality for Lipschitzian functions (with L instead of
H) (see [4])

!2 
Z b
a+b
x− 2
1
f (x) − 1
 (b − a) L.
(1.6)
f (t) dt ≤  +
b−a a
4
b−a
Here the constant 14 is also best.
Moreover, if one drops the condition of the continuity of the function, and
assumes that it is of bounded variation, then the following result may be stated
(see [2]).
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
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Theorem
1.4. Assume that f : [a, b] → R is of bounded variation and denote
Wb
by a (f ) its total variation. Then
# b
"
Z b
x − a+b _
1
1
2 f (x) −
+
(f )
(1.7)
f (t) dt ≤
b−a a
2 b−a a
for all x ∈ [a, b]. The constant
1
2
is the best possible.
If we assume more about f , i.e., f is monotonically increasing, then the
inequality (1.7) may be improved in the following manner [3] (see also [1]).
Theorem 1.5. Let f : [a, b] → R be monotonic nondecreasing. Then for all
x ∈ [a, b], we have the inequality:
Z b
1
(1.8) f (x) −
f (t) dt
b−a a
Z b
1
≤
[2x − (a + b)] f (x) +
sgn (t − x) f (t) dt
b−a
a
1
≤
{(x − a) [f (x) − f (a)] + (b − x) [f (b) − f (x)]}
#
"b − a 1 x − a+b
2 ≤
+
[f (b) − f (a)] .
2 b−a All the inequalities in (1.8) are sharp and the constant
1
2
is the best possible.
For other recent results including Ostrowski type inequalities for n-time differentiable functions, visit the RGMIA website at
http://rgmia.vu.edu.au/database.html.
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
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In this paper we point out some discrete Ostrowski type inequalities for vectors in normed linear spaces.
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
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2.
Some Identities
The following lemma holds.
Lemma 2.1. Let xi (i = 1, . . . , n) be vectors in X. Then we have the representation
n
(2.1)
n
1X
1X
xi =
xj +
p (i, j) ∆xj , i ∈ {1, . . . , n} ,
n j=1
n j=1
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
where
(2.2)
p (1, j) = j − n if 1 ≤ j ≤ n − 1;
(2.3)
p (n, j) = j if 1 ≤ j ≤ n − 1;
S.S. Dragomir
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and
(2.4)
p (i, j) =

 j

if 1 ≤ j ≤ i − 1,
j − n if i ≤ j ≤ n − 1,
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Proof. For i = 1, we have to prove that
(2.5)
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where 2 ≤ i ≤ n − 1 and 1 ≤ j ≤ n − 1.
n
JJ
J
n
1X
1X
xj +
(j − n) ∆xj .
x1 =
n j=1
n j=1
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Using the summation by parts formula, we have
n
X
n−1
X
n
(j − n) ∆xj = (j − n) xj j=1 −
∆ (j − n) xj+1
j=1
j=1
= (n − 1) x1 −
n−1
X
xj+1
j=1
= nx1 −
n
X
xj
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
j=1
and the formula (2.5) is proved.
For i = n, we can prove similarly that
n
S.S. Dragomir
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n−1
1X
1X
xn =
xj +
j∆xj .
n j=1
n j=1
(2.6)
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Let 2 ≤ i ≤ n − 1. We have
(2.7)
n−1
X
p (i, j) ∆xj =
j=1
i−1
X
p (i, j) ∆xj +
j=1
=
i−1
X
j=1
n−1
X
p (i, j) ∆xj
j=i
i∆xj +
n−1
X
j=i
(j − n) ∆xj .
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Using the summation by parts formula, we have
i−1
X
(2.8)
i−1
X
n
i∆xj = jxj j=i −
∆ (i) xj+1
j=1
j=1
= ixi − x1 −
i−1
X
xj+1 = (i − 1) xi −
j=1
i−1
X
xj
j=1
and
n−1
X
(2.9)
n
(j − n) ∆xj = (j − n) xj j=i −
j=i
n−1
X
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
∆ (j − n) xj+1
j=i
= (n − i) xi −
n−1
X
S.S. Dragomir
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j=i
= (n − i + 1) xi −
n
X
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xj .
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j=i
Using (2.7) – (2.9), we deduce
n−1
X
p (i, j) ∆xj = (i − 1) xi −
i−1
X
j=1
j=1
= nxi −
n
X
xj + (n − i + 1) xi −
n
X
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xj
Page 10 of 34
j=1
and the identity (2.1) is proved.
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The following corollaries hold.
Corollary 2.2. We have the identity
(2.10)
n
n
x1 + xn
1X
n
1 X
=
xj +
j−
∆xj .
2
n j=1
n j=1
2
Corollary 2.3. Let n = 2m + 1. Then we have
(2.11)
xm+1
2m+1
2m
X
X
1
1
=
xj +
pm (j) ∆xj ,
2m + 1 j=1
2m + 1 j=1
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
where
pm (j) =

 j

if 1 ≤ j ≤ m,
j − 2m − 1 if m + 1 ≤ j ≤ 2m.
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3.
Discrete Ostrowski’s Inequality
The following discrete inequality of Ostrowski type holds.
Theorem 3.1. Let (X, k·k) be a normed linear space and xi (i = 1, . . . , n) be
vectors in X. Then we have the inequality
"
#
2
n
n2 − 1
1X n+1
1
xk ≤
(3.1) xi −
i−
+
max k∆xk k ,
k=1,...,n−1
n k=1 n
2
4
for all i ∈ {1, . . . , n}. The constant c = 41 in the right hand side is best in the
sense that it cannot be replaced by a smaller one.
Proof. We use the representation (2.1) and the generalised triangle inequality to
obtain
n
n−1
1
1X X
xk =
p (i, k) ∆xk xi −
n
n
k=1
1
≤
n
k=1
n−1
X
|p (i, k)| k∆xk k
k=1
n−1
≤
max
k=1,...,n−1
k∆xk k ×
1X
|p (i, k)| .
n k=1
S.S. Dragomir
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If i = 1, then we have
n−1
X
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
n−1
X
n−1
X
n (n − 1)
|p (1, k)| =
|k − n| =
k=
2
k=1
k=1
k=1
Page 12 of 34
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and as
2
n2 − 1
n+1
n (n − 1)
+
=
, for n ≥ 1
1−
2
4
2
the inequality (3.1) is valid for i = 1.
Let 2 ≤ i ≤ n − 1. Then
n−1
X
|p (i, k)| =
k=1
=
i−1
X
k=1
i−1
X
k=1
|p (i, k)| +
n−1
X
|p (i, k)|
k=i
k+
n−1
X
(n − k)
k=i
!
n−1
i−1
X
X
(i − 1) i
+ n (n − 1 − i + 1) −
k−
k
=
2
k=1
k=1
n (n − 1) i (i − 1)
(i − 1) i
=
+ n (n − i) −
−
2
2
2
1
=
2i2 + n2 − 2ni + n
2
2
n+1
n2 − 1
=
i−
+
2
4
and the inequality (3.1) is also proved for i ∈ {2, . . . , n − 1}.
For i = n, we have p (n, k) = k, k = 1, . . . , n − 1 giving
n−1
X
n−1
X
n (n − 1)
|p (n, k)| =
k=
2
k=1
k=1
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
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and as
n+1
n−
2
2
+
n2 − 1
n (n − 1)
=
4
2
the inequality (3.2) is also valid for i = n.
To prove the sharpness of the constant c = 14 , assume that (3.1) holds with a
constant c > 0, i.e.,
"
#
2
n
1 X
n
+
1
1
xk ≤
+ c n2 − 1
(3.2) xi −
i−
max k∆xk k
k=1,...,n−1
n k=1 n
2
for any xk (k = 1, . . . , n) in X.
Let xk = x1 + (k − 1) r, k = 1, . . . , n, r ∈ X, r 6= 0, x1 6= 0 and i = 1 in
(3.2). Then we get
"
#
n
1 (n − 1)2
X
1
(3.3)
(x1 + (k − 1) r) ≤
+ c n2 − 1 krk
x1 −
n
n k=1
4
and as
n
X
(x1 + (k − 1) r) = nx1 +
k=1
n (n − 1)
r,
2
then from (3.3) we deduce
"
#
1 (n − 1)2
n−1
· r
+ c n2 − 1 krk
≤ n
2
4
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
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from where we get
1
1 n−1
≤
+ c (n + 1)
2
n
4
i.e.,
n + 1 ≤ 4c (n + 1) ,
which implies that c ≥ 14 , and the theorem is proved.
Corollary 3.2. Under the above assumptions and if n = 2m + 1, then we have
the inequality
2m+1
m (m + 1)
X
1
xk ≤
(3.4)
max k∆xk k .
xm+1 −
2m + 1
2m + 1 k=1,...,2m
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
k=1
The proof is obvious by the above Theorem 3.1 for i = m + 1.
The following corollary also holds.
Corollary 3.3. Under the above assumptions, we have:
a) If n = 2k, then
2k
x + x
1 X 1
1
2k
−
xj ≤ (k − 1) max k∆xj k .
(3.5)
j=1,...,2k−1
2
2k j=1 2
b) If n = 2k + 1, then
2k+1
x + x
1 X 1
2k 2 + 2k + 1
2k+1
−
xj ≤
max k∆xj k .
(3.6) 2
2k + 1 j=1 2 (2k + 1) j=1,...,2k
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Proof. The proof is as follows.
a) If n = 2k, then by Corollary 2.2, we have
2k
x + x
X
1
1
2k
−
xj 2
2k
j=1
≤
2k−1
1 X
|j − k| k∆xj k
2k j=1
1
≤
2k
=
1
2k
max
j=1,...,2k−1
max
j=1,...,2k−1
k∆xj k
2k−1
X
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
|j − k|
S.S. Dragomir
j=1
k∆xj k
k
X
(k − j) +
j=1
2k−1
X
!
(j − k)
j=k+1
1
(k − 1) k
=
max k∆xj k
k j=1,...,2k−1
2
1
= (k − 1) max k∆xj k ,
j=1,...,2k−1
2
and the inequality (3.5) is proved.
b) If n = 2k + 1, then by Corollary 2.2, we have
2k+1
2k+1 x + x
X
1
1 X 2k + 1 1
2k+1
−
xj ≤
j−
k∆xj k
2
2k + 1 j=1 2k + 1 j=1 2 Title Page
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2k+1
X
1
1 ≤
max k∆xj k
j − k − 2 2k + 1 j=1,...,2k
j=1
" k 2k+1
#
X
X
1
1
1
=
max k∆xj k
k+ −j +
j−k−
j=1,...,2k
2k + 1
2
2
j=1
j=k+1
"
#
k
2k+1
X
X
1
1
1
=
max k∆xj k k +
(k − j) − (k + 1) +
(j − k)
2k + 1 j=1,...,2k
2
2
j=1
j=k+1
2
2
k − k + k + 3k + 2 − 1
1
=
max k∆xj k
2k + 1 j=1,...,2k
2
2
2k + 2k + 1
= max k∆xj k
j=1,...,2k
2 (2k + 1)
and the inequality (3.6) is proved.
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
Title Page
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The following result including a version of a discrete Ostrowski inequality
for lp −norms of {∆xi }i=1,n−1 also holds.
Theorem 3.4. Let (X, k·k) be a normed linear space and xi (i = 1, . . . , n) be
vectors in X. Then we have the inequality
(3.7)
" n−1
# β1
n
1
X
X
1
1
xk ≤ [sα (i − 1) + sα (n − i)] α
k∆xk kβ
xi −
n k=1 n
k=1
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for all α > 1,
1
α
+
1
β
= 1, where sα (·) denotes the sum:
sα (m) :=
m
X
j α.
j=1
When m = 0, the sum is assumed to be zero.
Proof. Using representation (2.2) and the generalised triangle inequality, we
have:
n−1
n
1
1X X
(3.8)
xk =
p (i, k) ∆xk xi −
n k=1 n k=1
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
n−1
1X
|p (i, k)| k∆xk k =: B.
≤
n k=1
Using Hölder’s discrete inequality, we have
(3.9)
B≤
1
n
n−1
X
! α1
|p (i, k)|α
k=1
n−1
X
Contents
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J
! β1
k∆xk kβ
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.
k=1
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However,
n−1
X
k=1
α
|p (i, k)| =
i−1
X
k=1
α
|p (i, k)| +
n−1
X
k=i
|p (i, k)|α
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=
i−1
X
α
k +
k=1
α
n−1
X
(n − k)α
k=i
= 1 + · · · + (i − 1)α + (n − i)α + · · · + 1α
= sα (i − 1) + sα (n − i)
and the inequality (3.7) then follows by (3.8) and (3.9).
The case of α = β = 2 can be useful in practical applications.
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
Corollary 3.5. With the assumptions of Theorem 3.4, we have
(3.10)
n
X
1
xk xi −
n k=1 S.S. Dragomir
1
≤√
n
"
n+1
i−
2
2
n2 − 1
+
12
# 12 " n−1
X
k∆xk k2
k=1
Proof. For α = 2, we have
s2 (i − 1) =
i−1
X
k=1
k2 =
Title Page
# 12
i (i − 1) (2i − 1)
6
.
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and
s2 (n − i) =
n−i
X
k=1
k2 =
(n − i) (n − i + 1) [2 (n − i) + 1]
.
6
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As simple algebra proves that
"
s2 (i − 1) + s2 (n − i) = n
n+1
i−
2
2
#
n2 − 1
+
,
12
then, by (3.7) we deduce the desired inequality (3.10).
Corollary 3.6. Under the above assumptions and if n = 2m + 1, then we have
the inequality:
" 2m
# β1
1
2m+1
X
X 1
2α
1
β
xk ≤
[sα (m)] α
k∆xk k
(3.11) xm+1 −
2m + 1
2m + 1
k=1
k=1
for α > 1, α1 + β1 = 1.
In particular, for α = β = 2, we have
s
" 2m
# 12
2m+1
X
X
1
m (m + 1)
(3.12)
xk ≤
k∆xk k2 .
xm+1 −
2m + 1 k=1 3 (2m + 1) k=1
The following result providing an upper bound in terms of the l1 −norm of
(∆xk )k=1,n−1 also holds.
Theorem 3.7. Let (X, k·k) be a normed linear space and xi (i = 1, . . . , n) be
vectors in X. Then we have the inequality
n−1
n
1X n + 1 X
1 1
(3.13)
xk ≤
(n − 1) + i −
k∆xk k
xi −
n k=1 n 2
2 k=1
for all i ∈ {1, . . . , n}.
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
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Proof. As in Theorem 3.4, we have
n
1X (3.14)
xk ≤ B,
xi −
n k=1 where
n−1
1X
B :=
|p (i, k)| k∆xk k .
n k=1
It is obvious that
" i−1
#
n−1
X
1 X
B =
k k∆xk k +
(n − k) k∆xk k
n k=1
k=i
"
#
i−1
n−1
X
X
1
(i − 1)
k∆xk k + (n − i)
k∆xk k
≤
n
k=1
k=i
" i−1
#
n−1
X
X
1
=
max {i − 1, n − i}
k∆xk k +
k∆xk k
n
k=1
k=i
X
n−1
1 1
1
(n − 1) + |n − i − i + 1|
k∆xk k
=
n 2
2
k=1
n−1
X
n
+
1
1 1
(n − 1) + i −
k∆xk k
=
n 2
2 k=1
and the inequality (3.13) is proved.
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
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The following corollary contains the best inequality we can get from (3.13).
Corollary 3.8. Let (X, k·k) be as above and n = 2m + 1. Then we have the
inequality
2m+1
2m
X
1
m X
(3.15)
x
≤
k∆xk k .
x
−
m+1
k
2m + 1
2m + 1
k=1
k=1
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
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4.
Weighted Ostrowski Inequality
We start with the following theorem.
Theorem 4.1. Let (X, k·k) be P
a normed linear space, xi ∈ X (i = 1, . . . , n)
and pi ≥ 0 (i = 1, . . . , n) with ni=1 pi = 1. Then we have the inequality:
n
n
X
X
(4.1) xi −
pj xj ≤
pj |j − i| · max k∆xk k
k=1,n−1
j=1
j=1
 n−1 ,
+ i − n+1

2
2






! p1
! 1q


n
n
 P
P
p
q
|j − i|
pj
if p > 1,
≤ max k∆xk k ×
j=1
j=1

k=1,n−1





h 2

i

n −1
n+1 2

max {pj }
+ i− 2

4
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
1
p
+
1
q
= 1,
j=1,n
for all i ∈ {1, . . . , n}.
Proof. Using the properties of the norm, we have
n
n
X
X
(4.2)
pj kxi − xj k ≥ pj (xi − xj )
j=1
j=1
n
n
n
X
X
X
= xi
pj −
pj xj = xi −
pj xj ,
j=1
j=1
j=1
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for all i ∈ {1, . . . , n}.
On the other hand,
(4.3)
n
X
pj kxi − xj k
j=1
=
i−1
X
pj kxi − xj k +
j=1
n
X
pj kxi − xj k
j=i+1
j−1
n
i−1
X
X
X
=
pj (xk+1 − xk ) +
pj (xl+1 − xl )
j=1
j=i+1
k=j
l=i
!
!
j−1
i−1
i−1
n
X
X
X
X
≤
pj
k∆xk k +
pj
k∆xl k
i−1
X
j=1
j=i+1
k=j
l=i
=: A.
k=j
S.S. Dragomir
Title Page
Contents
JJ
J
Now, as
i−1
X
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
k∆xk k ≤ (i − j) max k∆xk k (where j ≤ i − 1)
k=j,i−1
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and
s−1
X
l=i
Quit
k∆xl k ≤ (s − i) max k∆xl k (where i ≤ s − 1),
l=i,n−1
Page 24 of 34
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then we deduce that
A ≤
i−1
X
pj (i − j) · max k∆xk k +
k=j,i−1
j=1
≤
max k∆xk k
" i−1
X
k=1,n−1
=
n
X
max k∆xk k ·
k=1,n−1
j=1
n
X
pj (j − i) · max k∆xl k
l=i,n−1
j=i+1
pj (i − j) +
#
n
X
pj (j − i)
j=i+1
pj |i − j|
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
j=1
and the first inequality in (4.1) is proved.
Now, we observe that
n
X
j=1
S.S. Dragomir
pj |i − j| ≤ max |i − j|
n
X
j=1,n
pj
j=1
= max |i − j|
j=1,n
= max {i − 1, n − i}
n − 1 n + 1 =
+ i −
,
2
2 which proves the first part of the second inequality in (4.1).
By Hölder’s discrete inequality, we also have
! p1
! 1q
n
n
n
X
X
X
|i − j|p
,
pj |i − j| ≤
pqj
j=1
j=1
j=1
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where p > q and p1 + 1q = 1, and the second part of the second inequality in
(4.1) holds.
Finally, we also have
n
X
pj |i − j| ≤ max |pj |
j=1,n
j=1
n
X
|i − j| .
j=1
Now, let us observe that
n
X
|i − j| =
j=1
=
i
X
|i − j| +
n
X
j=1
j=i+1
i
X
n
X
j=1
(i − j) +
|i − j|
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
(j − i)
j=i+1
n
i
X
i (i + 1) X
+
j−
j − i (n − i)
= i2 −
2
j=1
j=1
2
2
n −1
n+1
=
+ i−
4
2
and the last part of the second inequality in (4.1) is proved.
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Remark 4.1. In some practical applications the case p = q = 2 in the second
part of the second inequality may be useful. As
"
2 #
n
n
n
2
X
X
X
n
+
1
n
−
1
(j − i)2 =
j 2 − 2i
j + ni2 = n
+ i−
,
12
2
j=1
j=1
j=1
Quit
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then we may state the inequality
n
X
(4.4) xi −
pj xj j=1
"
2 # 21
√ n2 − 1
n+1
≤ n
+ i−
12
2
n
X
! 12
p2j
max k∆xk k
k=1,n−1
j=1
for all i ∈ {1, . . . , n}.
The following particular case was proved in a different manner in Theorem
3.1.
Corollary 4.2. If xi (i = 1, . . . , n) are vectors in the normed linear space (X, k·k),
then we have
"
2 #
n
2
X
1
n+1
1 n −1
(4.5)
xj ≤
+ i−
max k∆xk k .
xi −
n
n
4
2
k=1,n−1
j=1
The following result also holds.
Theorem 4.3. Let (X, k·k) be aPnormed linear space, xi ∈ X (i = 1, . . . , n)
and pi ≥ 0 (i = 1, . . . , n) with ni=1 pi = 1. Then, for α > 1, α1 + β1 = 1, we
have the inequality:
(4.6)
n
n
X
X
1
pj xj ≤
|i − j| β pj
xi −
j=1
j=1
n−1
X
k=1
! α1
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
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α
k∆xk k
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≤
 1
1
β,

(n − 1) + i − n+1

2
2






! 1δ
! γ1
1

!

α
n−1
n
n
 P
δ
P
X
|i − j| β
pγj
k∆xk kα ×
j=1
j=1


k=1





n

1
P


|i − j| β max {pj }

if γ > 1,
1
γ
+
1
δ
= 1,
j=1,n
j=1
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
for all i ∈ {1, . . . , n}.
Proof. Using Hölder’s discrete inequality, we may write that
! α1
i−1
i−1
X
X
1
k∆xk k ≤ (i − j) β
k∆xk kα
k=j
k=j
s−1
X
s−1
X
Title Page
and
1
k∆xl k ≤ (s − i) β
l=i
S.S. Dragomir
Contents
! α1
k∆xl kα
JJ
J
,
l=i
which implies for A, as defined in the proof of Theorem 4.1, that
! α1
! α1
i−1
i−1
n
s−1
X
X
X
X
1
1
A≤
(i − j) β
k∆xk kα
pj +
(s − i) β
k∆xl kα
ps
j=1
≤
i−1
X
k=1
s=i+1
k=j
! α1
k∆xk k
α
i−1
X
j=1
1
β
(i − j) pj +
n−1
X
l=i
k∆xl k
α
n
X
s=i+1
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l=i
! α1
II
I
1
Page 28 of 34
(s − i) β ps
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≤
n−1
X
k∆xk k
α
! α1 " i−1
X
=
n−1
X
(i − j) pj +
j=1
k=1
! α1
k∆xk kα
n
X
n
X
1
β
#
1
β
(s − i) ps
s=i+1
1
|i − j| β pj ,
j=1
k=1
which proves the first inequality in (4.6).
Now it is obvious that
n
n
X
1
1 X
|i − j| β pj ≤ max |i − j| β
pj
j=1,n
j=1
j=1
n
o
1
1
= max (i − 1) β , (n − i) β
1
1
n + 1 β
,
(n − 1) + i −
=
2
2 proving the first part of the second inequality in (4.6).
For γ, δ > 1 with γ1 + 1δ = 1, we have
! γ1
! 1δ
n
n
n
X
X
X
1
δ
|i − j| β pj ≤
pγj
|i − j| β
j=1
j=1
j=1
obtaining the second part of the second inequality in (4.6).
Finally, we observe that
n
n
X
X
1
1
|i − j| β pj ≤ max {pj }
|i − j| β ,
j=1
j=1,n
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
j=1
S.S. Dragomir
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and the theorem is proved.
Corollary 4.4. If xi (i = 1, . . . , n) are vectors in the normed space (X, k·k),
then for all i ∈ {1, . . . , n} we have:
n
X
1
(4.7) xi −
xj n j=1 ! α1
n
n−1
X
1
1
1
1X
≤
|i − j| β
k∆xk kα
, α > 1, + = 1.
n j=1
α β
k=1
Finally, we may state the following result as well.
S.S. Dragomir
Theorem 4.5. Let X, xi and pi (i = 1, . . . , n) be as in Theorem 4.3. Then we
have the inequality:

n−1
P

max
{P
,
1
−
P
}
k∆xk k


i−1
i
n


k=1
X
(4.8)
pj xj ≤
xi −
i−1

n−1

P
P
j=1


k∆xk k ,
k∆xk k
 (1 − pi ) max
k=1
≤ (1 − pi )
n−1
X
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
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for all i ∈ {1, . . . , n}, where
Pm :=
m
X
i=1
Page 30 of 34
pi , m = 1, . . . , n
J. Ineq. Pure and Appl. Math. 3(1) Art. 2, 2002
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and P0 := 0.
Proof. It is obvious that
i−1
X
and
i−1
X
k∆xk k ≤
k=j
k=1
s−1
X
n−1
X
k∆xl k ≤
k∆xk k
k∆xl k ,
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
l=i
l=i
Then, for A as defined in the proof of Theorem 4.1, we have that
A≤
i−1
X
k∆xk k
i−1
X
pj +
j=1
k=1
n−1
X
n
X
k∆xl k
S.S. Dragomir
pj
Title Page
j=i+1
l=i
Contents
=: B
≤ max {Pi−1 , 1 − Pi }
" i−1
X
k∆xj k +
j=1
= max {Pi−1 , 1 − Pi }
n−1
X
n−1
X
#
k∆xj k
j=i+1
k∆xk k .
k=1
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Also, we observe that
B ≤ max
JJ
J
( i−1
X
j=1
k∆xj k ,
n−1
X
j=i+1
)
Page 31 of 34
k∆xj k (Pi−1 + 1 − Pi )
J. Ineq. Pure and Appl. Math. 3(1) Art. 2, 2002
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= (1 − pi ) max
( i−1
X
k=1
k∆xk k ,
n−1
X
)
k∆xk k
k=i
and the theorem is thus proved.
Corollary 4.6. Let X and xi (i = 1, . . . , n) be as in Corollary 4.4. Then

n−1
P
1 1

i − n+1 
k∆xk k ,
(n
−
1)
+

2
n

 n 2
k=1
X
1
(4.9)
xj ≤
xi −
i−1
n j=1 
n−1

P
P
n
−
1


k∆xk k
max
k∆xk k ,

n
k=1
k=i
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
for all i ∈ {1, . . . , n}.
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References
[1] P. CERONE AND S.S. DRAGOMIR, Midpoint type rules from an inequalities point of view, in Analytic-Computational Methods in Applied Mathematics, G.A. Anastassiou (Ed), CRC Press, New York, 2000, 135–200.
[2] S.S. DRAGOMIR, On the Ostrowski’s inequality for mappings of bounded
variation and applications, Math. Ineq. & Appl., 4(1) (2001), 59–66.
Preprint on line: RGMIA Res. Rep. Coll., 2(1) (1999), Article 7,
http://rgmia.vu.edu.au/v2n1.html
[3] S.S. DRAGOMIR, Ostrowski’s inequality for monotonous mappings and
applications, J. KSIAM, 3(1) (1999), 127–135.
[4] S.S. DRAGOMIR, The Ostrowski’s integral inequality for Lipschitzian
mappings and applications, Comp. and Math. with Appl., 38 (1999), 33–
37.
[5] S.S. DRAGOMIR, P. CERONE, J. ROUMELIOTIS AND S. WANG, A
weighted version of Ostrowski inequality for mappings of Hölder type and
applications in numerical analysis, Bull. Math. Soc. Sci. Math. Roumanie,
42(90)(4) (1992), 301–314.
[6] S.S. DRAGOMIR AND S. WANG, A new inequality of Ostrowski’s type in
L1 −norm and applications to some special means and to some numerical
quadrature rules, Tamkang J. of Math., 28 (1997), 239–244.
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
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Page 33 of 34
[7] S.S. DRAGOMIR AND S. WANG, A new inequality of Ostrowski’s type
in Lp −norm, Indian J. of Math., 40(3) (1998), 245–304.
J. Ineq. Pure and Appl. Math. 3(1) Art. 2, 2002
http://jipam.vu.edu.au
[8] S.S. DRAGOMIR AND S. WANG, Applications of Ostrowski’s inequality to the estimation of error bounds for some special means and some
numerical quadrature rules, Appl. Math. Lett., 11 (1998), 105–109.
[9] A.M. FINK, Bounds on the derivation of a function from its averages,
Czech. Math. J., 42(117) (1992), 289–310.
[10] A. OSTROWSKI, Uber die Absolutabweichung einer differentienbaren
Funktionen von ihren Integralmittelwert, Comment. Math. Hel, 10 (1938),
226–227.
The Discrete Version of
Ostrowski’s Inequality in
Normed Linear Spaces
S.S. Dragomir
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