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volume 3, issue 1, article 2, 2002.

Received 14 May, 2001;

accepted 02 July, 2001.

Communicated by:R.P. Agarwal

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Journal of Inequalities in Pure and Applied Mathematics

THE DISCRETE VERSION OF OSTROWSKI’S INEQUALITY IN NORMED LINEAR SPACES

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

c

2000Victoria University ISSN (electronic): 1443-5756 042-01

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The Discrete Version of Ostrowski’s Inequality in

Normed Linear Spaces S.S. Dragomir

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J. Ineq. Pure and Appl. Math. 3(1) Art. 2, 2002

http://jipam.vu.edu.au

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.

1. Introduction

The following result is known in the literature as Ostrowski’s inequality [10].

Theorem 1.1. Letf : [a, b]→ Rbe a differentiable mapping on(a, b)with the property that|f⁰(t)| ≤M for allt∈(a, b). Then

(1.1)

f(x)− 1 b−a

Z b a

f(t)dt

≤

"

1

4+ x− ^a+b₂ 2

(b−a)²

#

(b−a)M

for allx ∈[a, b]. The constant ¹₄ 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:

(1.2) f(x) = 1 b−a

Z b a

f(t)dt+ 1 b−a

Z b a

p(x, t)f⁰(t)dt, x∈[a, b], where

p(x, t) :=







t−a if a≤t≤x t−b if x < t≤b

which also holds for absolutely continuous functionsf : [a, b]→R.

The following Ostrowski type result for absolutely continuous functions holds (see [6] – [8]).

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Theorem 1.2. Let f : [a, b] → Rbe absolutely continuous on [a, b]. Then, for allx∈[a, b], we have:

(1.3)

f(x)− 1 b−a

Z b a

f(t)dt

≤











1

4 +_x−a+b 2

b−a

²

(b−a)kf⁰k_∞ if f⁰ ∈L∞[a, b] ;

1 (p+1)¹^p

h x−a b−a

p+1

+ ^b−x_b−ap+1i¹_p

(b−a)¹^pkf⁰k_q if f⁰ ∈L_q[a, b],

1

p + ¹_q = 1, p > 1;

h1 2 +

x−^a+b

2

b−a

ikf⁰k₁;

wherek·k_r (r∈[1,∞]) are the usual Lebesgue norms onL_r[a, b], i.e., kgk_∞ :=ess sup

t∈[a,b]

|g(t)|

and

kgk_r :=

Z b a

|g(t)|^rdt ¹_r

, r∈[1,∞).

The constants ¹₄, ¹

(p+1)^p¹

and ¹₂ respectively are sharp in the sense presented in Theorem1.1.

The above inequalities can also be obtained from the Fink result in [9] on choosingn= 1and performing some appropriate computations.

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If one drops the condition of absolute continuity and assumes thatfis Hölder continuous, then one may state the result (see [5]):

Theorem 1.3. Letf : [a, b]→Rbe ofr−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:

(1.5)

f(x)− 1 b−a

Z b a

f(t)dt

≤ H r+ 1

"

b−x b−a

r+1

+

x−a b−a

r+1#

(b−a)^r.

The constant _r+1¹ is also sharp in the above sense.

Note that ifr = 1, i.e.,f is Lipschitz continuous, then we get the following version of Ostrowski’s inequality for Lipschitzian functions (with Linstead of H) (see [4])

(1.6)

f(x)− 1 b−a

Z b a

f(t)dt

≤



 1

4+ x− ^a+b₂ b−a

!2

(b−a)L.

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]).

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Theorem 1.4. Assume that f : [a, b] → Ris of bounded variation and denote byWb

a(f)its total variation. Then (1.7)

f(x)− 1 b−a

Z b a

f(t)dt

≤

"

1 2+

x−^a+b₂ b−a

# _b _

a

(f)

for allx∈[a, b]. The constant ¹₂ 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:

f(x)− 1 b−a

Z b a

f(t)dt (1.8)

≤ 1 b−a

[2x−(a+b)]f(x) + Z b

a

sgn(t−x)f(t)dt

≤ 1

b−a{(x−a) [f(x)−f(a)] + (b−x) [f(b)−f(x)]}

≤

"

1 2 +

x− ^a+b₂ b−a

#

[f(b)−f(a)].

All the inequalities in (1.8) are sharp and the constant ¹₂ is the best possible.

For other recent results including Ostrowski type inequalities forn-time differentiable functions, visit the RGMIA website at

http://rgmia.vu.edu.au/database.html.

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In this paper we point out some discrete Ostrowski type inequalities for vectors in normed linear spaces.

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2. Some Identities

The following lemma holds.

Lemma 2.1. Letx_i(i= 1, . . . , n)be vectors inX. Then we have the represen- tation

(2.1) x_i = 1 n

n

X

j=1

x_j+ 1 n

n

X

j=1

p(i, j) ∆x_j, i∈ {1, . . . , n},

where

(2.2) p(1, j) = j−n if 1≤j ≤n−1;

(2.3) p(n, j) = j if 1≤j ≤n−1;

and

(2.4) p(i, j) =







j if 1≤j ≤i−1, j−n if i≤j ≤n−1, where2≤i≤n−1and1≤j ≤n−1.

Proof. Fori= 1, we have to prove that

(2.5) x₁ = 1

n

X

j=1

x_j + 1 n

n

X

j=1

(j −n) ∆x_j.

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Using the summation by parts formula, we have

n

X

j=1

(j−n) ∆x_j = (j−n)x_j

n j=1−

n−1

X

j=1

∆ (j−n)x_j+1

= (n−1)x₁−

n−1

X

j=1

x_j+1

= nx₁−

n

X

j=1

x_j

and the formula (2.5) is proved.

Fori=n, we can prove similarly that

(2.6) x_n= 1

n

X

j=1

x_j+ 1 n

n−1

X

j=1

j∆x_j.

Let2≤i≤n−1. We have

n−1

X

j=1

p(i, j) ∆x_j =

i−1

X

j=1

p(i, j) ∆x_j +

n−1

X

j=i

p(i, j) ∆x_j (2.7)

=

i−1

X

j=1

i∆x_j+

n−1

X

j=i

(j −n) ∆x_j.

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Using the summation by parts formula, we have

i−1

X

j=1

i∆x_j = jx_j

n j=i−

i−1

X

j=1

∆ (i)x_j+1 (2.8)

= ix_i−x₁−

i−1

X

j=1

x_j+1 = (i−1)x_i−

i−1

X

j=1

x_j

and

n−1

X

j=i

(j−n) ∆xj = (j−n)xj

n j=i−

n−1

X

j=i

∆ (j−n)xj+1

(2.9)

= (n−i)x_i−

n−1

X

j=i

x_j+1

= (n−i+ 1)xi−

n

X

j=i

xj. Using (2.7) – (2.9), we deduce

n−1

X

j=1

p(i, j) ∆x_j = (i−1)x_i−

i−1

X

j=1

x_j+ (n−i+ 1)x_i−

n

X

j=i

x_j

= nx_i−

n

X

j=1

x_j

and the identity (2.1) is proved.

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The following corollaries hold.

Corollary 2.2. We have the identity

(2.10) x₁+x_n

2 = 1 n

n

X

j=1

x_j+ 1 n

n

X

j=1

j −n

2

∆x_j.

Corollary 2.3. Letn = 2m+ 1. Then we have

(2.11) xm+1 = 1

2m+ 1

2m+1

X

j=1

xj + 1 2m+ 1

2m

X

j=1

pm(j) ∆xj,

where

p_m(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 andx_i (i= 1, . . . , n)be vectors inX. Then we have the inequality

(3.1)

x_i− 1 n

n

X

k=1

x_k

≤ 1 n

"

i− n+ 1 2

2

+n²−1 4

#

k=1,...,n−1max k∆x_kk,

for alli ∈ {1, . . . , n}. The constantc = ¹₄ 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

x_i− 1 n

n

X

k=1

x_k

= 1 n

n−1

X

k=1

p(i, k) ∆x_k

≤ 1 n

n−1

X

k=1

|p(i, k)| k∆x_kk

≤ max

k=1,...,n−1k∆x_kk × 1 n

n−1

X

k=1

|p(i, k)|. Ifi= 1, then we have

n−1

X

k=1

|p(1, k)|=

n−1

X

k=1

|k−n|=

n−1

X

k=1

k = n(n−1) 2

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and as

1−n+ 1 2

2

+ n²−1

4 = n(n−1)

2 , for n≥1 the inequality (3.1) is valid fori= 1.

Let2≤i≤n−1. Then

n−1

X

k=1

|p(i, k)| =

i−1

X

k=1

|p(i, k)|+

n−1

X

k=i

|p(i, k)|

=

i−1

X

k=1

k+

n−1

X

k=i

(n−k)

= (i−1)i

2 +n(n−1−i+ 1)−

n−1

X

k=1

k−

i−1

X

k=1

k

!

= (i−1)i

2 +n(n−i)−

n(n−1)

2 − i(i−1) 2

= 1

2 2i²+n²−2ni+n

=

i− n+ 1 2

2

+n²−1 4

and the inequality (3.1) is also proved fori∈ {2, . . . , n−1}.

Fori=n, we havep(n, k) =k,k = 1, . . . , n−1giving

n−1

X

k=1

|p(n, k)|=

n−1

X

k=1

k= n(n−1) 2

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and as

n− n+ 1 2

2

+n² −1

4 = n(n−1) 2 the inequality (3.2) is also valid fori=n.

To prove the sharpness of the constantc= ¹₄, assume that (3.1) holds with a constantc >0, i.e.,

(3.2)

x_i− 1 n

n

X

k=1

x_k

≤ 1 n

"

i− n+ 1 2

2

+c n²−1

#

k=1,...,n−1max k∆x_kk

for anyx_k(k = 1, . . . , n)inX.

Letx_k = x₁+ (k−1)r, k = 1, . . . , n,r ∈ X, r 6= 0, x₁ 6= 0 andi = 1in (3.2). Then we get

(3.3)

x1− 1 n

n

X

k=1

(x1+ (k−1)r)

≤ 1 n

"

(n−1)²

4 +c n² −1

# krk and as

n

X

k=1

(x₁+ (k−1)r) =nx₁+n(n−1) 2 r, then from (3.3) we deduce

n−1 2

·r

≤ 1 n

"

(n−1)²

4 +c n²−1

# krk

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from where we get

1 2 ≤ 1

n

n−1

4 +c(n+ 1)

i.e.,

n+ 1 ≤4c(n+ 1), which implies thatc≥ ¹₄, and the theorem is proved.

Corollary 3.2. Under the above assumptions and ifn = 2m+ 1, then we have the inequality

(3.4)

x_m+1− 1 2m+ 1

2m+1

X

k=1

x_k

≤ m(m+ 1)

2m+ 1 max

k=1,...,2mk∆x_kk. The proof is obvious by the above Theorem3.1fori=m+ 1.

The following corollary also holds.

Corollary 3.3. Under the above assumptions, we have:

a) Ifn = 2k, then (3.5)

x1+x2k

2 − 1

2k

X

j=1

x_j

≤ 1

2(k−1) max

j=1,...,2k−1k∆x_jk. b) Ifn = 2k+ 1, then

(3.6)

x1+x2k+1

2 − 1

2k+ 1

2k+1

X

j=1

x_j

≤ 2k²+ 2k+ 1 2 (2k+ 1) max

j=1,...,2kk∆x_jk.

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Proof. The proof is as follows.

a) Ifn = 2k, then by Corollary2.2, we have

x₁+x_2k

2 − 1

2k

X

j=1

x_j

≤ 1 2k

2k−1

X

j=1

|j −k| k∆x_jk

≤ 1

2k max

j=1,...,2k−1k∆xjk

2k−1

X

j=1

|j−k|

= 1

2k max

j=1,...,2k−1k∆x_jk

k

X

j=1

(k−j) +

2k−1

X

j=k+1

(j−k)

!

= 1

k max

j=1,...,2k−1k∆x_jk(k−1)k 2

= 1

2(k−1) max

j=1,...,2k−1k∆x_jk, and the inequality (3.5) is proved.

b) Ifn = 2k+ 1, then by Corollary2.2, we have

x₁+x_2k+1

2 − 1

2k+ 1

2k+1

X

j=1

x_j

≤ 1 2k+ 1

2k+1

X

j=1

j− 2k+ 1 2

k∆x_jk

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≤ 1

2k+ 1 max

j=1,...,2kk∆x_jk

2k+1

X

j=1

j−k−1 2

= 1

2k+ 1 max

j=1,...,2kk∆x_jk

" _k X

j=1

k+ 1

2−j

+

2k+1

X

j=k+1

j −k− 1 2

#

= 1

2k+ 1 max

j=1,...,2kk∆xjk

"

1 2k+

k

X

j=1

(k−j)−1

2(k+ 1) +

2k+1

X

j=k+1

(j−k)

#

= 1

2k+ 1 max

j=1,...,2kk∆x_jk

k²−k+k²+ 3k+ 2−1 2

= max

j=1,...,2kk∆xjk2k²+ 2k+ 1 2 (2k+ 1) and the inequality (3.6) is proved.

The following result including a version of a discrete Ostrowski inequality forl_p−norms of{∆x_i}_i=1,n−1 also holds.

(3.7)

x_i− 1 n

n

X

k=1

x_k

≤ 1

n [s_α(i−1) +s_α(n−i)]¹^α

"_n−1 X

k=1

k∆x_kk^β

#¹_β

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for allα >1, _α¹ +_β¹ = 1, wheres_α(·)denotes the sum:

s_α(m) :=

m

X

j=1

j^α.

Whenm= 0, the sum is assumed to be zero.

Proof. Using representation (2.2) and the generalised triangle inequality, we have:

x_i− 1 n

n

X

k=1

x_k

= 1 n

n−1

X

k=1

p(i, k) ∆x_k (3.8)

≤ 1 n

n−1

X

k=1

|p(i, k)| k∆x_kk=:B.

Using Hölder’s discrete inequality, we have

(3.9) B ≤ 1

n

n−1

X

k=1

|p(i, k)|^α

!_α¹ _n−1 X

k=1

k∆x_kk^β

!¹_β .

However,

n−1

X

k=1

|p(i, k)|^α =

i−1

X

k=1

|p(i, k)|^α+

n−1

X

k=i

|p(i, k)|^α

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=

i−1

X

k=1

k^α+

n−1

X

k=i

(n−k)^α

= 1^α+· · ·+ (i−1)^α+ (n−i)^α+· · ·+ 1^α

= s_α(i−1) +s_α(n−i)

and the inequality (3.7) then follows by (3.8) and (3.9).

The case ofα=β = 2can be useful in practical applications.

Corollary 3.5. With the assumptions of Theorem3.4, we have

(3.10)

x_i− 1 n

n

X

k=1

x_k

≤ 1

√n

"

i− n+ 1 2

2

+n²−1 12

#¹₂ "_n−1 X

k=1

k∆x_kk²

#¹₂ .

Proof. Forα = 2, we have

s₂(i−1) =

i−1

X

k=1

k² = i(i−1) (2i−1) 6

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

"

i− n+ 1 2

2

+n²−1 12

# , then, by (3.7) we deduce the desired inequality (3.10).

Corollary 3.6. Under the above assumptions and ifn = 2m+ 1, then we have the inequality:

(3.11)

x_m+1− 1 2m+ 1

2m+1

X

k=1

x_k

≤ 2^α¹

2m+ 1[s_α(m)]^α¹

"_2m X

k=1

k∆x_kk^β

#_β¹

forα >1, _α¹ + ¹_β = 1.

In particular, forα=β = 2, we have

(3.12)

x_m+1− 1 2m+ 1

2m+1

X

k=1

x_k

≤ s

m(m+ 1) 3 (2m+ 1)

"_2m X

k=1

k∆x_kk²

#¹₂ . The following result providing an upper bound in terms of thel₁−norm of (∆x_k)_k=1,n−1also holds.

(3.13)

x_i− 1 n

n

X

k=1

x_k

≤ 1 n

1

2(n−1) +

i− n+ 1 2

ⁿ⁻¹ X

k=1

k∆x_kk for alli∈ {1, . . . , n}.

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Proof. As in Theorem3.4, we have

(3.14)

x_i− 1 n

n

X

k=1

x_k

≤B,

where

B := 1 n

n−1

X

k=1

|p(i, k)| k∆x_kk. It is obvious that

B = 1

n

"_i−1 X

k=1

kk∆xkk+

n−1

X

k=i

(n−k)k∆xkk

#

≤ 1 n

"

(i−1)

i−1

X

k=1

k∆xkk+ (n−i)

n−1

X

k=i

k∆xkk

#

= 1

nmax{i−1, n−i}

"_i−1 X

k=1

k∆xkk+

n−1

X

k=i

k∆xkk

#

= 1 n

1

2(n−1) + 1

2|n−i−i+ 1|

ⁿ⁻¹ X

k=1

k∆x_kk

= 1 n

1

2(n−1) +

i− n+ 1 2

ⁿ⁻¹ X

k=1

k∆x_kk and the inequality (3.13) is proved.

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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

(3.15)

x_m+1− 1 2m+ 1

2m+1

X

k=1

x_k

≤ m

2m+ 1

2m

X

k=1

k∆x_kk.

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4. Weighted Ostrowski Inequality

We start with the following theorem.

Theorem 4.1. Let (X,k·k)be a normed linear space, x_i ∈ X (i= 1, . . . , n) andp_i ≥0 (i= 1, . . . , n)withPn

i=1p_i = 1. Then we have the inequality:

(4.1)

x_i−

n

X

j=1

p_jx_j

≤

n

X

j=1

p_j|j−i| · max

k=1,n−1

k∆x_kk

≤ max

k=1,n−1

k∆x_kk ×











n−1 2 +

i− ⁿ⁺¹₂ ,

n

P

j=1

|j−i|^p

!¹_p

n

P

j=1

p^q_j

!¹_q

if p > 1, ¹_p + ¹_q = 1,

hn²−1

4 + i−ⁿ⁺¹₂ 2i max

j=1,n

{p_j}

for alli∈ {1, . . . , n}.

Proof. Using the properties of the norm, we have

n

X

j=1

p_jkx_i−x_jk ≥

n

X

j=1

p_j(x_i−x_j) (4.2)

=

x_i

n

X

j=1

p_j −

n

X

j=1

p_jx_j

=

x_i−

n

X

j=1

p_jx_j ,

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for alli∈ {1, . . . , n}.

On the other hand,

n

X

j=1

p_jkx_i−x_jk (4.3)

=

i−1

X

j=1

p_jkx_i−x_jk+

n

X

j=i+1

p_jkx_i−x_jk

=

i−1

X

j=1

p_j

i−1

X

k=j

(x_k+1−x_k)

+

n

X

j=i+1

p_j

j−1

X

l=i

(x_l+1−x_l)

≤

i−1

X

j=1

p_j

i−1

X

k=j

k∆x_kk

! +

n

X

j=i+1

p_j

j−1

X

l=i

k∆x_lk

!

=:A.

Now, as

i−1

X

k=j

k∆x_kk ≤(i−j) max

k=j,i−1

k∆x_kk (wherej ≤i−1)

and s−1

X

l=i

k∆x_lk ≤(s−i) max

l=i,n−1

k∆x_lk (wherei≤s−1),

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then we deduce that

A ≤

i−1

X

j=1

pj(i−j)· max

k=j,i−1

k∆xkk+

n

X

j=i+1

pj(j−i)· max

l=i,n−1

k∆xlk

≤ max

k=1,n−1k∆xkk

"_i−1 X

j=1

pj(i−j) +

n

X

j=i+1

pj(j−i)

#

= max

k=1,n−1

k∆x_kk ·

n

X

j=1

p_j|i−j| and the first inequality in (4.1) is proved.

Now, we observe that

n

X

j=1

p_j|i−j| ≤ max

j=1,n

|i−j|

n

X

j=1

p_j

= max

j=1,n

|i−j|

= max{i−1, n−i}

= n−1 2 +

i−n+ 1 2

, which proves the first part of the second inequality in (4.1).

By Hölder’s discrete inequality, we also have

n

X

j=1

p_j|i−j| ≤

n

X

j=1

p^q_j

!¹_q _n X

j=1

|i−j|^p

!¹_p ,

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where p > q and ¹_p + ¹_q = 1, and the second part of the second inequality in (4.1) holds.

Finally, we also have

n

X

j=1

p_j|i−j| ≤max

j=1,n

|p_j|

n

X

j=1

|i−j|.

Now, let us observe that

n

X

j=1

|i−j| =

i

X

j=1

|i−j|+

n

X

j=i+1

|i−j|

=

i

X

j=1

(i−j) +

n

X

j=i+1

(j−i)

= i² −i(i+ 1)

2 +

n

X

j=1

j−

i

X

j=1

j−i(n−i)

= n²−1

4 +

i− n+ 1 2

2

and the last part of the second inequality in (4.1) is proved.

Remark 4.1. In some practical applications the case p= q = 2in the second part of the second inequality may be useful. As

n

X

j=1

(j−i)² =

n

X

j=1

j²−2i

n

X

j=1

j+ni² =n

"

n²−1 12 +

i− n+ 1 2

2# ,

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then we may state the inequality

(4.4)

x_i−

n

X

j=1

p_jx_j

≤√ n

"

n²−1 12 +

i−n+ 1 2

2#¹₂ _n X

j=1

p²_j

!¹₂ max

k=1,n−1

k∆x_kk

for alli∈ {1, . . . , n}.

The following particular case was proved in a different manner in Theorem 3.1.

Corollary 4.2. Ifx_i(i= 1, . . . , n)are vectors in the normed linear space(X,k·k), then we have

(4.5)

x_i− 1 n

n

X

j=1

x_j

≤ 1 n

"

n² −1

4 +

i− n+ 1 2

2# max

k=1,n−1

k∆x_kk.

The following result also holds.

Theorem 4.3. Let (X,k·k)be a normed linear space, x_i ∈ X (i= 1, . . . , n) andpi ≥0 (i= 1, . . . , n)withPn

i=1pi = 1. Then, forα >1, _α¹ + ¹_β = 1,we have the inequality:

(4.6)

x_i−

n

X

j=1

p_jx_j

≤

n

X

j=1

|i−j|^β¹ p_j

n−1

X

k=1

k∆x_kk^α

!_α¹

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≤

n−1

X

k=1

k∆xkk^α

!¹_α

×









 ₁

2(n−1) +

i−ⁿ⁺¹₂

_β¹ ,

n

P

j=1

|i−j|^β^δ

!¹_δ

n

P

j=1

p^γ_j

!_γ¹

if γ >1, _γ¹ + ¹_δ = 1,

n

P

j=1

|i−j|¹^β max

j=1,n

{p_j}

for alli∈ {1, . . . , n}.

Proof. Using Hölder’s discrete inequality, we may write that

i−1

X

k=j

k∆x_kk ≤(i−j)^β¹

i−1

X

k=j

k∆x_kk^α

!¹_α

and

s−1

X

l=i

k∆x_lk ≤(s−i)^β¹

s−1

X

l=i

k∆x_lk^α

!_α¹ , which implies forA, as defined in the proof of Theorem4.1, that

A≤

i−1

X

j=1

(i−j)^β¹

i−1

X

k=j

k∆x_kk^α

!_α¹ p_j +

n

X

s=i+1

(s−i)^β¹

s−1

X

l=i

k∆x_lk^α

!_α¹ p_s

≤

i−1

X

k=1

k∆x_kk^α

!_α¹ _i−1 X

j=1

(i−j)^β¹ p_j +

n−1

X

l=i

k∆x_lk^α

!_α¹ _n X

s=i+1

(s−i)^β¹ p_s

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≤

n−1

X

k=1

k∆x_kk^α

!_α¹ "_i−1 X

j=1

(i−j)¹^β p_j+

n

X

s=i+1

(s−i)¹^β p_s

#

=

n−1

X

k=1

k∆x_kk^α

!_α¹ _n X

j=1

|i−j|^β¹ p_j, which proves the first inequality in (4.6).

Now it is obvious that

n

X

j=1

|i−j|¹^β pj ≤ max

j=1,n

|i−j|^β¹

n

X

j=1

pj

= maxn

(i−1)¹^β ,(n−i)¹^βo

= 1

2(n−1) +

i− n+ 1 2

_β¹ , proving the first part of the second inequality in (4.6).

Forγ, δ >1with _γ¹ + ¹_δ = 1, we have

n

X

j=1

|i−j|^β¹ p_j ≤

n

X

j=1

p^γ_j

!¹_γ _n X

j=1

|i−j|^β^δ

!¹_δ

obtaining the second part of the second inequality in (4.6).

Finally, we observe that

n

X

j=1

|i−j|^β¹ p_j ≤max

j=1,n

{p_j}

n

X

j=1

|i−j|^β¹ ,

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and the theorem is proved.

Corollary 4.4. If x_i (i= 1, . . . , n) are vectors in the normed space(X,k·k), then for alli∈ {1, . . . , n}we have:

(4.7)

x_i− 1 n

n

X

j=1

x_j

≤ 1 n

n

X

j=1

|i−j|¹^β

n−1

X

k=1

k∆x_kk^α

!_α¹

, α >1, 1 α + 1

β = 1.

Finally, we may state the following result as well.

Theorem 4.5. Let X,x_i andp_i (i= 1, . . . , n)be as in Theorem4.3. Then we have the inequality:

x_i−

n

X

j=1

p_jx_j

≤











max{Pi−1,1−P_i}

n−1

P

k=1

k∆x_kk

(1−p_i) max _i−1

P

k=1

k∆x_kk,

n−1

P

k=i

k∆x_kk (4.8)

≤(1−p_i)

n−1

X

j=1

k∆x_kk

for alli∈ {1, . . . , n}, where P_m :=

m

X

i=1

p_i, m = 1, . . . , n

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andP₀ := 0.

Proof. It is obvious that

i−1

X

k=j

k∆x_kk ≤

i−1

X

k=1

k∆x_kk

and s−1

X

l=i

k∆xlk ≤

n−1

X

l=i

k∆xlk,

Then, forAas defined in the proof of Theorem4.1, we have that A≤

i−1

X

k=1

k∆x_kk

i−1

X

j=1

p_j+

n−1

X

l=i

k∆x_lk

n

X

j=i+1

p_j

=:B

≤max{Pi−1,1−P_i}

"_i−1 X

j=1

k∆x_jk+

n−1

X

j=i+1

k∆x_jk

#

= max{Pi−1,1−P_i}

n−1

X

k=1

k∆x_kk.

Also, we observe that B ≤max

(_i−1 X

j=1

k∆x_jk,

n−1

X

j=i+1

k∆x_jk )

(Pi−1+ 1−P_i)

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= (1−p_i) max (_i−1

X

k=1

k∆x_kk,

n−1

X

k=i

k∆x_kk )

and the theorem is thus proved.

Corollary 4.6. LetX andx_i(i= 1, . . . , n)be as in Corollary4.4. Then

(4.9)

x_i− 1 n

n

X

j=1

x_j

≤









 1 n

₁

2(n−1) +

i− ⁿ⁺¹₂

ⁿ⁻¹P

k=1

k∆x_kk,

n−1

n max

_i−1 P

k=1

k∆x_kk,

n−1

P

k=i

k∆x_kk

for alli∈ {1, . . . , n}.

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References

[1] P. CERONEANDS.S. DRAGOMIR, Midpoint type rules from an inequal- ities point of view, in Analytic-Computational Methods in Applied Mathe- matics, 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. DRAGOMIRANDS. 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.

[7] S.S. DRAGOMIR AND S. WANG, A new inequality of Ostrowski’s type inL_p−norm, Indian J. of Math., 40(3) (1998), 245–304.