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volume 3, issue 1, article 12, 2002.

Received 22 June, 2001;

accepted 25 October, 2001.

Communicated by:Th. M. Rassias

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Journal of Inequalities in Pure and Applied Mathematics

ON WEIGHTED OSTROWSKI TYPE INEQUALITIES FOR OPERATORS AND VECTOR-VALUED FUNCTIONS

1N.S. BARNETT, ²C. BUSE,¹P. CERONE AND¹S.S. DRAGOMIR

1School of Communications and Informatics Victoria University of Technology

PO Box 14428 Melbourne City MC 8001 Victoria, Australia EMail:neil@matilda.vu.edu.au URL:http://sci.vu.edu.au/staff/neilb.html

2Department of Mathematics West University of Timi¸soara Bd. V. Pârvan 4

1900 Timi¸soara, România.

EMail:buse@hilbert.math.uvt.ro

URL:http://rgmia.vu.edu.au/BuseCVhtml/

EMail:pc@matilda.vu.edu.au URL:http://rgmia.vu.edu.au/cerone EMail:sever@matilda.vu.edu.au

URL:http://rgmia.vu.edu.au/SSDragomirWeb.html

c

2000School of Communications and Informatics,Victoria University of Technology ISSN (electronic): 1443-5756

050-01

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On Weighted Ostrowski Type Inequalities for Operators and

Vector-Valued Functions

N.S. Barnett,C. Bu¸se,P. Cerone andS.S. Dragomir

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J. Ineq. Pure and Appl. Math. 3(1) Art. 12, 2002

http://jipam.vu.edu.au

Abstract

Some weighted Ostrowski type integral inequalities for operators and vector- valued functions in Banach spaces are given. Applications for linear operators in Banach spaces and differential equations are also provided.

2000 Mathematics Subject Classification:Primary 26D15; Secondary 46B99 Key words: Ostrowski’s Inequality, Vector-Valued Functions, Bochner Integral.

1. Introduction

In [12], Peˇcari´c and Savi´c obtained the following Ostrowski type inequality for weighted integrals (see also [7, Theorem 3]):

Theorem 1.1. Let w : [a, b] → [0,∞)be a weight function on[a, b].Suppose thatf : [a, b]→Rsatisfies

(1.1) |f(t)−f(s)| ≤N|t−s|^α, for allt, s∈[a, b],

whereN >0and0< α≤1are some constants. Then for anyx∈[a, b]

(1.2)

f(x)− Rb

a w(t)f(t)dt Rb

aw(t)dt

≤N · Rb

a |t−x|^αw(t)dt Rb

a w(t)dt . Further, if for some constantscandλ

0< c≤w(t)≤λc, for all t∈[a, b], then for anyx∈[a, b], we have

(1.3)

f(x)− Rb

a w(t)f(t)dt Rb

a w(t)dt

≤N · λL(x)J(x) L(x)−J(x) +λJ(x), where

L(x) :=

1

2(b−a) +

x−a+b 2

α

and

J(x) := (x−a)^1+α+ (b−x)^1+α (1 +α) (b−a) .

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The inequality (1.2) was rediscovered in [4] where further applications for different weights and in Numerical Analysis were given.

For other results in connection to weighted Ostrowski inequalities, see [3], [8] and [10].

In the present paper we extend the weighted Ostrowski’s inequality for vector- valued functions and Bochner integrals and apply the obtained results to operatorial inequalities and linear differential equations in Banach spaces. Some numerical experiments are also conducted.

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2. Weighted Inequalities

Let X be a Banach space and −∞ < a < b < ∞. We denote by L(X) the Banach algebra of all bounded linear operators acting onX.The norms of vectors or operators acting onXwill be denoted byk·k.

A functionf : [a, b] → X is called measurable if there exists a sequence of simple functions f_n : [a, b] → X which converges punctually almost everywhere on [a, b] atf. We recall also that a measurable functionf : [a, b] → X is Bochner integrable if and only if its norm function (i.e. the function t 7→

kf(t)k: [a, b]→R+) is Lebesgue integrable on[a, b].

The following theorem holds.

Theorem 2.1. Assume that B : [a, b] → L(X)is Hölder continuous on[a, b], i.e.,

(2.1) kB(t)−B(s)k ≤H|t−s|^α for all t, s ∈[a, b], whereH >0andα ∈(0,1].

Iff : [a, b]→Xis Bochner integrable on[a, b], then we have the inequality:

(2.2)

B(t) Z b

a

f(s)ds− Z b

a

B(s)f(s)ds

≤H Z b

a

|t−s|^αkf(s)kds

≤H×











(b−t)^α+1+(t−a)^α+1

α+1 k|f|k_[a,b],∞ if f ∈L∞([a, b] ;X) ; h_(b−t)qα+1+(t−a)^qα+1

qα+1

i¹_q

k|f|k_[a,b],p if p >1, ¹_p +¹_q = 1 and f ∈L_p([a, b] ;X) ; ₁

2(b−a) +

t− ^a+b₂

α

k|f|k_[a,b],1

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for anyt ∈[a, b], where

|kfk|_[a,b],∞ :=ess sup

t∈[a,b]

kf(t)k and

|kfk|_[a,b],p:=

Z b a

kf(t)k^pdt

1 p

, p≥1.

Proof. Firstly, we prove that theX−valued functions7→B(s)f(s)is Bochner integrable on [a, b]. Indeed, let (f_n)be a sequence ofX−valued, simple functions which converge almost everywhere on [a, b]at the function f. The maps s 7→ B(s)f_n(s)are measurable (because they are continuous with the excep- tion of a finite number of pointssin[a, b]). Then

kB(s)fn(s)−B(s)f(s)k ≤ kB(s)k kfn(s)−f(s)k →0 a.e. on [a, b]

whenn → ∞so that the functions 7→B(s)f(s) : [a, b]→ X is measurable.

Now, using the estimate

kB(s)f(s)k ≤ sup

ξ∈[a,b]

kB(ξ)k · kf(s)k, for alls∈[a, b],

it is easy to see that the functions 7→B(s)f(s)is Bochner integrable on[a, b].

We have successively

B(t) Z b

a

f(s)ds− Z b

a

B(s)f(s)ds

=

Z b a

(B(t)−B(s))f(s)ds

≤ Z b

a

k(B(t)−B(s))f(s)kds

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≤ Z b

a

k(B(t)−B(s))k kf(s)kds≤H Z b

a

|t−s|^αkf(s)kds=:M(t) for anyt∈[a, b], proving the first inequality in (2.2).

Now, observe that

M(t) ≤ Hk|f|k_[a,b],∞

Z b a

|t−s|^αds

= Hk|f|k_[a,b],∞· (b−t)^α+1+ (t−a)^α+1 α+ 1

and the first part of the second inequality is proved.

Using Hölder’s integral inequality, we may state that M(t) ≤ H

Z b a

|t−s|^qαds

¹q Z b a

kf(s)k^pds ¹p

= H

"

(b−t)^qα+1+ (t−a)^qα+1 qα+ 1

#¹_q

k|f|k_[a,b],p, proving the second part of the second inequality.

Finally, we observe that M(t) ≤ H sup

s∈[a,b]

|t−s|^α Z b

a

kf(s)kds

= Hmax{(b−t)^α,(t−a)^α} k|f|k_[a,b],1

= H

1

2(b−a) +

t− a+b 2

α

k|f|k_[a,b],1

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and the theorem is proved.

The following corollary holds.

Corollary 2.2. Assume thatB : [a, b]→ L(X)is Lipschitzian with the constant L >0.Then we have the inequality

(2.3)

B(t) Z b

a

f(s)ds− Z b

a

B(s)f(s)ds

≤L Z b

a

|t−s| kf(s)kds

≤L×









 h1

4(b−a)²+ t− ^a+b₂ 2i

k|f|k_[a,b],∞ if f ∈L∞([a, b] ;X) ; h_(b−t)q+1+(t−a)^q+1

q+1

i¹_q

k|f|k_[a,b],p if p > 1, ¹_p + ¹_q = 1 and f ∈L_p([a, b] ;X) ; ₁

2(b−a) +

t−^a+b₂

k|f|k_[a,b],1 for anyt ∈[a, b].

Remark 2.1. If we chooset = ^a+b₂ in (2.2) and (2.3), then we get the following midpoint inequalities:

(2.4)

B

a+b 2

Z b a

f(s)ds− Z b

a

B(s)f(s)ds

≤H Z b

a

s− a+b 2

α

kf(s)kds

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≤H×











1

2^α(α+1)(b−a)^α+1k|f|k_[a,b],∞ if f ∈L_∞([a, b] ;X) ;

1 2^α(qα+1)¹^q

(b−a)^α+¹^q k|f|k_[a,b],p if p >1, ¹_p +¹_q = 1 and f ∈L_p([a, b] ;X) ;

1

2^α (b−a)^αk|f|k_[a,b],1 and

B

a+b 2

Z b a

f(s)ds− Z b

a

B(s)f(s)ds (2.5)

≤L Z b

a

s− a+b 2

kf(s)kds

≤L×











1

4(b−a)²k|f|k_[a,b],∞ if f ∈L∞([a, b] ;X) ;

1 2(q+1)

1

q (b−a)¹⁺¹^q k|f|k_[a,b],p if p > 1, ¹_p + ¹_q = 1 and f ∈Lp([a, b] ;X) ;

1

2(b−a)k|f|k_[a,b],1 respectively.

Remark 2.2. Consider the functionΨ_α : [a, b]→R,Ψ_α(t) :=Rb

a|t−s|^αkf(s)kds, α ∈(0,1). Iff is continuous on[a, b], thenΨ_αis differentiable and

dΨ_α(t)

dt = d

dt Z t

a

(t−s)^αkf(s)kds+ Z b

t

(s−t)^αkf(s)kds

= α

Z t a

kf(s)k (t−s)^1−αds−

Z b t

kf(s)k (s−t)^1−αds

.

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Ift₀ ∈(a, b)is such that Z t0

a

kf(s)k

(t₀−s)^1−αds = Z b

t0

kf(s)k (s−t₀)^1−αds

andΨ⁰(·)is negative on(a, t0)and positive on(t0, b),then the best inequality we can get in the first part of (2.2) is the following one

(2.6)

B(t0) Z b

a

f(s)ds− Z b

a

B(s)f(s)ds

≤H Z b

a

|t0−s|^αkf(s)kds.

Ifα= 1, then, for

Ψ (t) :=

Z b a

|t−s| kf(s)kds, we have

dΨ (t)

dt =

Z t a

kf(s)kds− Z b

t

kf(s)kds, t∈(a, b), d²Ψ (t)

dt² = 2kf(t)k ≥0, t∈(a, b), which shows thatΨis convex on(a, b).

Ift_m ∈(a, b)is such that Z tm

a

kf(s)kds = Z b

tm

kf(s)kds,

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then the best inequality we can get from the first part of (2.3) is (2.7)

B(t_m) Z b

a

f(s)ds− Z b

a

B(s)f(s)ds

≤L Z b

a

sgn(s−t_m)skf(s)kds.

Indeed, as inf

t∈[a,b]

Z b a

|t−s| kf(s)kds

= Z b

a

|t_m−s| kf(s)kds

= Z tm

a

(t_m−s)kf(s)kds+ Z b

tm

(s−t_m)kf(s)kds

=t_m Z tm

a

kf(s)kds− Z b

tm

kf(s)kds

+ Z b

tm

skf(s)kds− Z tm

a

skf(s)kds

= Z b

tm

skf(s)kds− Z tm

a

skf(s)kds

= Z b

a

sgn(s−t_m)skf(s)kds,

then the best inequality we can get from the first part of (2.3) is obtained for t =tm ∈(a, b).

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We recall that a functionF : [a, b]→ L(X)is said to be strongly continuous if for allx ∈ X,the mapss 7→ F (s)x : [a, b] → X are continuous on [a, b].

In this case the function s 7→ kB(s)k : [a, b] → R+is (Lebesgue) measurable and bounded ([6]). The linear operator L = Rb

a F(s)ds (defined by Lx :=

Rb

a F(s)xdsfor allx∈X) is bounded, because kLxk ≤

Z b a

kF (s)kds

· kxk for all x∈X.

In a similar manner to Theorem2.1, we may prove the following result as well.

Theorem 2.3. Assume thatf : [a, b]→Xis Hölder continuous, i.e., (2.8) kf(t)−f(s)k ≤K|t−s|^β for all t, s∈[a, b], whereK >0andβ∈(0,1].

If B : [a, b] → L(X) is strongly continuous on [a, b], then we have the inequality:

(2.9)

Z b a

B(s)ds

f(t)− Z b

a

B(s)f(s)ds

≤K Z b

a

|t−s|^βkB(s)kds

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≤K×











(b−t)^β+1+(t−a)^β+1

β+1 k|B|k_[a,b],∞ if kB(·)k ∈L∞([a, b] ;R+) ; h_(b−t)qβ+1+(t−a)^qβ+1

qβ+1

i¹_q

k|B|k_[a,b],p if p > 1, ¹_p + ¹_q = 1

and kB(·)k ∈L_p([a, b] ;R+) ; ₁

2(b−a) +

t−^a+b₂

β

k|B|k_[a,b],1 for anyt ∈[a, b].

Corollary 2.4. Assume that f and B are as in Theorem 2.3. If, in addition, Rb

a B(s)dsis invertible inL(X),then we have the inequality:

(2.10)

f(t)− Z b

a

B(s)ds

⁻¹Z b a

B(s)f(s)ds

≤K

Z b a

B(s)ds ⁻¹

Z b a

|t−s|^βkB(s)kds

for anyt ∈[a, b].

Remark 2.3. It is obvious that the inequality (2.10) contains as a particular case what is the so called Ostrowski’s inequality for weighted integrals (see (1.2)).

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3. Inequalities for Linear Operators

Let0 ≤ a < b < ∞andA ∈ L(X). We recall that the operatorial norm ofA is given by

kAk= sup{kAxk:kxk ≤1}.

The resolvent set ofA(denoted byρ(A)) is the set of all complex scalarsλfor whichλI−Ais an invertible operator. HereI is the identity operator inL(X).

The complementary set of ρ(A) in the complex plane, denoted by σ(A), is the spectrum of A. It is known that σ(A) is a compact set in C. The series P

n≥0 (tA)ⁿ

n!

converges absolutely and locally uniformly for t ∈ R. If we denote bye^tA its sum, then

e^tA

≤e^|t|kAk, t∈R.

Proposition 3.1. LetXbe a real or complex Banach space,A ∈ L(X)andβ be a non-null real number such that−β ∈ρ(A). Then for all0 ≤a < b <∞ and eachs∈[a, b], we have

(3.1)

e^βb−e^βa

β ·e^sA−(βI+A)⁻¹

e^b(βI+A)−e^a(βI+A)

≤ kAke^bkAk·

"

1

4(b−a)²+

s− a+b 2

2#

·max

e^βb, e^βa . Proof. We apply the second inequality from Corollary2.2in the following particular case.

B(τ) :=e^{τ A}, f(τ) = e^βτx, τ ∈[a, b], x∈X.

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For allξ, η ∈[a, b]there exists anαbetweenξandηsuch that kB(ξ)−B(η)k =

∞

X

n=1

(ξⁿ−ηⁿ) n! Aⁿ

=

(ξ−η)A

∞

X

n=0

(αA)ⁿ n!

≤ kAk e^αA

· |ξ−η|

≤ kAke^bkAk· |ξ−η|.

The function τ 7→ e^{τ A} is thus Lipschitzian on [a, b] with the constant L :=

kAke^b·kAk. On the other hand we have Z b

a

e^{τ A} e^βτx

dτ = Z b

a

e^{τ A} e^βτIx dτ

= Z b

a

e^τ(A+βI)xdτ

= (A+βI)⁻¹

e^b(A+βI)−e^a(A+βI) x, and

|kfk|_[a,b],∞ = sup

τ∈[a,b]

e^{τ β}x

= max

e^βb, e^βa · kxk.

Placing all the above results in the second inequality from (2.3) and taking the supremum for allx∈X,we will obtain the desired inequality (3.1).

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Remark 3.1. LetA ∈ L(X)such that0∈ρ(A). Taking the limit asβ → 0in (3.1), we get the inequality

(b−a)e^sA−A⁻¹

e^bA−e^aA

≤ kAke^bkAk·

"

1

4(b−a)²+

s− a+b 2

2# ,

wherea, bandsare as in Proposition3.1.

Proposition 3.2. LetA∈ L(X)be an invertible operator,t ≥0and0≤s ≤t.

Then the following inequality holds:

(3.2)

t²

2 sin (sA)−A⁻²[sin (tA)−tAcos (tA)]

≤ 2s³+ 2t³−3st²

6 kAk.

In particular, ifX =R,A= 1ands= 0it follows the scalar inequality

|sint−tcost| ≤ t³

3, for allt ≥0.

Proof. We apply the inequality from (2.3) in the following particular case:

B(τ) = sin (τ A) :=

∞

X

n=0

(−1)ⁿ (τ A)²ⁿ⁺¹

(2n+ 1)!, τ ≥0,

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and

(3.3) f(τ) =τ ·x, for fixedx∈X.

For eachξ, η ∈[0, t],we have kB(ξ)−B(η)k =

A

∞

X

n=0

(−1)ⁿ(ξ−η)α²ⁿ (2n)! A²ⁿ

!

≤ kAk |ξ−η| · kcos (αA)k

≤ kAk |ξ−η|,

where α is a real number between ξ and η, i.e., the function τ 7−→ B(τ) : R⁺ → L(X)iskAk −Lipschitzian.

Moreover, it is easy to see that Z t

0

B(τ)f(τ)dτ =A⁻²[sin (tA)−tAcos (tA)]x and

(3.4)

Z t 0

|s−τ| |f(τ)|dτ = 2s³+ 2t³−3st²

6 kxk.

Applying the first inequality from (2.3) and taking the supremum for x ∈ X withkxk ≤1, we get (3.2).

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4. Quadrature Formulae

Consider the division of the interval[a, b]given by

(4.1) In :a =t0 < t1 <· · ·< tn−1 < tn=b and h_i := t_i+1 − t_i, ν(h) := max

i=0,n−1

h_i. For the intermediate points ξ :=

(ξ₀, . . . , ξn−1)withξ_i ∈[t_i, t_i+1], i= 0, n−1, define the sum (4.2) S_n⁽¹⁾(B, f;I_n, ξ) :=

n−1

X

i=0

B(ξ_i) Z ti+1

ti

f(s)ds.

Then we may state the following result in approximating the integral Z b

a

B(s)f(s)ds, based on Theorem2.1.

Theorem 4.1. Assume that B : [a, b] → L(X)is Hölder continuous on[a, b], i.e., it satisfies the condition (2.1) andf : [a, b]→ X is Bochner integrable on [a, b]. Then we have the representation

(4.3)

Z b a

B(s)f(s)ds=S_n⁽¹⁾(B, f;I_n, ξ) +R⁽¹⁾_n (B, f;I_n, ξ),

whereSn⁽¹⁾(B, f;I_n, ξ)is as given by (4.2) and the remainderRn⁽¹⁾(B, f;I_n, ξ)

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satisfies the estimate R⁽¹⁾_n (B, f;I_n, ξ)

≤H×









 1

α+ 1k|f|k_[a,b],∞

n−1

P

i=0

(t_i+1−ξ_i)^α+1+ (ξ_i−t_i)^α+1 1

(qα+ 1)¹^q

k|f|k_[a,b],p _n−1

P

i=0

(t_i+1−ξ_i)^qα+1+ (ξ_i−t_i)^qα+1 ¹_q

, p > 1, ¹_p +¹_q = 1

1

2ν(h) + max

i=0,n−1

ξ_i− t_i+1+t_i 2

α

k|f|k_[a,b],1

≤H×









 1

α+ 1k|f|k_[a,b],∞

n−1

P

i=0

h^α+1_i 1

(qα+ 1)¹^q

k|f|k_[a,b],p _n−1

P

i=0

h^qα+1_i ¹_q

, p >1, ¹_p +¹_q = 1 1

2ν(h) + max

i=0,n−1

ξ_i− ti+1+ti

2

α

k|f|k_[a,b],1

≤H×









 1

α+ 1k|f|k_[a,b],∞[ν(h)]^α (b−a)¹^q

(qα+ 1)¹^q

k|f|k_[a,b],p[ν(h)]^α k|f|k_[a,b],1[ν(h)]^α.

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Proof. Applying Theorem4.1on[x_i, x_i+1] i= 0, n−1

,we may write that

Z ti+1

ti

B(s)f(s)ds−B(ξ_i) Z ti+1

ti

f(s)ds

≤H×









 h_(t

i+1−ξi)^α+1+(ξi−ti)^α+1 α+1

ik|f|k_[t

i,ti+1],∞

h(ti+1−ξ_i)^qα+1+(ξi−t_i)^qα+1 qα+1

i¹_q

k|f|k_[t

i,ti+1],p

₁

2(t_i+1−t_i) +

ξ_i− ^tⁱ⁺¹₂^+tⁱ

^α

k|f|k_[t

i,ti+1],1. Summing overi from0to n−1and using the generalised triangle inequality we get

R⁽¹⁾_n (B, f;I_n, ξ)

≤

n−1

X

i=0

Z ti+1

ti

B(s)f(s)ds−B(ξi) Z ti+1

ti

f(s)ds

≤H×











1 α+1

n−1

P

i=0

(ti+1−ξi)^α+1+ (ξi−ti)^α+1 k|f|k_[t

i,ti+1],∞

1 (qα+1)¹^q

_n−1 P

i=0

(t_i+1−ξ_i)^qα+1+ (ξ_i−t_i)^qα+1 ¹_q

k|f|k_[t

i,ti+1],p n−1

P

i=0

₁

2h_i+

^α

k|f|k_[t

i,ti+1],1.

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Now, observe that

n−1

X

i=0

(ti+1−ξi)^α+1+ (ξi−ti)^α+1 k|f|k_[t

i,ti+1],∞

≤ k|f|k_[a,b],∞

n−1

X

i=0

(t_i+1−ξ_i)^α+1+ (ξ_i−t_i)^α+1

≤ k|f|k_[a,b],∞

n−1

X

i=0

h^α+1_i

≤ k|f|k_[a,b],∞(b−a) [ν(h)]^α. Using the discrete Hölder inequality, we may write that

"_n−1 X

i=0

(t_i+1−ξ_i)^qα+1+ (ξ_i−t_i)^qα+1

#¹_q

k|f|k_[t

i,ti+1],p

≤

"_n−1 X

i=0

(t_i+1−ξ_i)^qα+1+ (ξ_i−t_i)^qα+1¹_q^q

#¹_q

×

"_n−1 X

i=0

k|f|k^p_[t

i,ti+1],p

#¹_p

= (_n−1

X

i=0

(t_i+1−ξ_i)^qα+1+ (ξ_i−t_i)^qα+1 )¹_q

Z b a

kf(t)k^pds ¹_p

≤

n−1

X

i=0

h^qα+1_i

!¹_q

k|f|k^p_[a,b],p

≤(b−a)¹^q k|f|k_[a,b],p[ν(h)]^α.

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Finally, we have

n−1

X

i=0

1 2h_i+

ξ_i −t_i+1+t_i 2

α

k|f|k_[t

i,ti+1],1

≤ 1

2 max

i=0,n−1

h_i+ max

i=0,n−1

ξ_i− t_i+1+t_i 2

α

k|f|k_[a,b],1

≤[ν(h)]^αk|f|k_[a,b],1 and the theorem is proved.

Corollary 4.2. If B is Lipschitzian with the constantL, then we have the rep- resentation (4.3) and the remainderRn⁽¹⁾(B, f;I_n, ξ)satisfies the estimates:

(4.4)

R⁽¹⁾_n (B, f;I_n, ξ)

≤L×











k|f|k_[a,b],∞

1 4

n−1

P

i=0

h²_i +

n−1

P

i=0

ξ_i− ^tⁱ⁺¹₂^+tⁱ2

1 (q+1)¹^q

k|f|k_[a,b],p _n−1

P

i=0

(t_i+1−ξ_i)^q+1+ (ξ_i−t_i)^q+1 ¹_q

, p >1, ¹_p +¹_q = 1

1

2ν(h) + max

i=0,n−1

k|f|k_[a,b],1

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≤L×











1

2k|f|k_[a,b],∞

n−1

P

i=0

h²_i

1 (q+1)¹^q

k|f|k_[a,b],p _n−1

P

i=0

h^q+1_i ¹_q

1

2ν(h) + max

i=0,n−1

k|f|k_[a,b],1

≤L×











1

2k|f|k_[a,b],∞(b−a)ν(h)

(b−a)¹^q (q+1)

1

q k|f|k_[a,b],pν(h) k|f|k_[a,b],1ν(h).

The second possibility we have for approximating the integralRb

a B(s)f(s)ds is embodied in the following theorem based on Theorem2.3.

Theorem 4.3. Assume thatf : [a, b]→Xis Hölder continuous, i.e., the condi- tion (2.8) holds. IfB : [a, b] → L(X)is strongly continuous on[a, b], then we have the representation:

(4.5)

Z b a

B(s)f(s)ds=S_n⁽²⁾(B, f;I_n, ξ) +R⁽²⁾_n (B, f;I_n, ξ), where

(4.6) S_n⁽²⁾(B, f;I_n, ξ) :=

n−1

X

i=0

Z ti+1

ti

B(s)ds

f(ξ_i)

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and the remainderR⁽²⁾n (B, f;I_n, ξ)satisfies the estimate:

(4.7)

R⁽²⁾_n (B, f;I_n, ξ)

≤K ×











1

β+1k|B|k_[a,b],∞

n−1

P

i=0

h

(t_i+1−ξ_i)^β+1+ (ξ_i −t_i)^β+1i

1 (qβ+1)¹^q

k|B|k_[a,b],p _n−1

P

i=0

h

(t_i+1−ξ_i)^qβ+1+ (ξ_i−t_i)^qβ+1i¹_q , p > 1, ¹_p + ¹_q = 1

1

2ν(h) + max

i=0,n−1

ξi− ^tⁱ⁺¹₂^+tⁱ

β

k|B|k_[a,b],1

≤K ×











1

β+1k|B|k_[a,b],∞

n−1

P

i=0

h^β+1_i

1 (qβ+1)¹^q

k|B|k_[a,b],p _n−1

P

i=0

h^qβ+1_i ¹_q

, p >1, ¹_p + ¹_q = 1;

1

2ν(h) + max

i=0,n−1

β

k|B|k_[a,b],1

≤K ×











1

β+1k|B|k_[a,b],∞(b−a) [ν(h)]^β

(b−a)¹^q (qβ+1)

1

q k|B|k_[a,b],p[ν(h)]^β, p > 1, ¹_p + ¹_q = 1 k|B|k_[a,b],1[ν(h)]^β .

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If we consider the quadrature (4.8) M_n⁽¹⁾(B, f;I_n) :=

n−1

X

i=0

B

t_i+t_i+1 2

Z ti+1

ti

f(s)ds, then we have the representation

(4.9)

Z b a

B(s)f(s)ds=M_n⁽¹⁾(B, f;I_n) +R⁽¹⁾_n (B, f;I_n), and the remainderR⁽¹⁾n (B, f;I_n)satisfies the estimate:

R⁽¹⁾_n (B, f;In) (4.10)

≤H×











1

2^α(α+1)k|f|k_[a,b],∞

n−1

P

i=0

h^α+1_i

1 2^α(qα+1)¹^q

k|f|k_[a,b],p _n−1

P

i=0

h^qα+1_i ¹_q

, p >1, ¹_p + ¹_q = 1;

1

2^α [ν(h)]^αk|f|k_[a,b],1

≤H×











1

2^α(α+1)(b−a)k|f|k_[a,b],∞[ν(h)]^α

(b−a)¹^q 2^α(qα+1)

1

q k|f|k_[a,b],p[ν(h)]^α, p >1, _p¹ +¹_q = 1;

1

2^α k|f|k_[a,b],1[ν(h)]^α,

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provided thatBandf are as in Theorem4.1.

Now, if we consider the quadrature (4.11) M_n⁽²⁾(B, f;I_n) :=

n−1

X

i=0

Z ti+1

ti

B(s)ds

f

t_i+t_i+1 2

, then we also have

(4.12)

Z b a

B(s)f(s)ds=M_n⁽²⁾(B, f;I_n) +R⁽²⁾_n (B, f;I_n), and in this case the remainder satisfies the bound

(4.13)

R⁽²⁾_n (B, f;I_n)

≤K×











1

2^β(β+1)k|B|k_[a,b],∞

n−1

P

i=0

h^β+1_i

1 2^β(qβ+1)¹^q

k|B|k_[a,b],p _n−1

P

i=0

h^qβ+1_i ¹_q

, p >1, ¹_p +¹_q = 1;

1

2^β [ν(h)]^βk|B|k_[a,b],1

≤K×











1

2^β(β+1)(b−a)k|B|k_[a,b],∞[ν(h)]^β

(b−a)¹^q 2^β(qβ+1)

1

q k|B|k_[a,b],p[ν(h)]^β, p >1, ¹_p +¹_q = 1;

1

2^β k|B|k_[a,b],1[ν(h)]^β,

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providedB andf satisfy the hypothesis of Theorem4.3.

Now, if we consider the equidistant partitioning of[a, b], E_n :t_i :=a+

b−a n

·i, i= 0, n,

thenMn⁽¹⁾(B, f;E_n)becomes (4.14) M_n⁽¹⁾(B, f) :=

n−1

X

i=0

B

a+

i+ 1 2

· b−a n

Z a+^b−a_n ·(i+1) a+^b−a_n ·i

f(s)ds

and then (4.15)

Z b a

B(s)f(s)ds=M_n⁽¹⁾(B, f) +R⁽¹⁾_n (B, f), where the remainder satisfies the bound

(4.16)

R⁽¹⁾_n (B, f)

≤H×











(b−a)^α+1

2^α(α+1)n^α k|f|k_[a,b],∞

(b−a)^{α+ 1}^q

2^α(α+1)n^α k|f|k_[a,b],p, p > 1, ¹_p + ¹_q = 1;

(b−a)^α

2^αn^α k|f|k_[a,b],1. Also, we have

(4.17)

Z b a

B(s)f(s)ds=M_n⁽²⁾(B, f) +R⁽²⁾_n (B, f),

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where

M_n⁽²⁾(B, f) :=

n−1

X

i=0

Z a+^b−a_n ·(i+1) a+^b−a_n ·i

B(s)ds

! f

a+

i+1

2

· b−a n

,

and the remainderR⁽²⁾n (B, f)satisfies the estimate

(4.18)

R⁽²⁾_n (B, f)

≤K×











(b−a)^β+1

2^β(β+1)n^β kBk_[a,b],∞

(b−a)^{β+ 1}^q

2^β(β+1)n^β k|B|k_[a,b],p, p > 1, ¹_p +¹_q = 1;

(b−a)^β

2^βn^β k|B|k_[a,b],1.