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Volume 3, Issue 2, Article 21, 2002

INTEGRAL INEQUALITIES OF THE OSTROWSKI TYPE

A. SOFO

SCHOOL OFCOMMUNICATIONS ANDINFORMATICS

VICTORIAUNIVERSITY OFTECHNOLOGY

P.O. BOX14428 MELBOURNECITYMC VICTORIA8001, AUSTRALIA. sofo@matilda.vu.edu.au URL:http://rgmia.vu.edu.au/sofo/

Received December, 2000; accepted 16 November, 2001.

Communicated by T. Mills

ABSTRACT. Integral inequalities of Ostrowski type are developed forn−times differentiable mappings, with multiple branches, on theLnorm. Some particular inequalities are also inves- tigated, which include explicit bounds for perturbed trapezoid, midpoint, Simpson’s, Newton- Cotes and left and right rectangle rules. The results obtained provide sharper bounds than those obtained by Dragomir [5] and Cerone, Dragomir and Roumeliotis [2].

Key words and phrases: Ostrowski Integral Inequality, Quadrature Formulae.

2000 Mathematics Subject Classification. 26D15, 41A55.

1. INTRODUCTION

In 1938 Ostrowski [18] obtained a bound for the absolute value of the difference of a function to its average over a finite interval. The theorem is as follows.

Theorem 1.1. Letf : [a, b] →Rbe a differentiable mapping on[a, b]and let|f0(t)| ≤ M for allt ∈(a, b), then the following bound is valid

(1.1)

f(x)− 1 b−a

Z b a

f(t)dt

≤(b−a)M

"

1

4+ x− a+b2 2

(b−a)2

#

for allx∈[a, b].

The constant 14 is sharp in the sense that it cannot be replaced by a smaller one.

Dragomir and Wang [12, 13] extended the result (1.1) and applied the extended result to numerical quadrature rules and to the estimation of error bounds for some special means.

ISSN (electronic): 1443-5756

c 2002 Victoria University. All rights reserved.

I wish to acknowledge the useful comments of an anonymous referee which led to an improvement of this work.

083-01

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Dragomir [8, 9, 10] further extended the result (1.1) to incorporate mappings of bounded varia- tion, Lipschitzian mappings and monotonic mappings.

Cerone, Dragomir and Roumeliotis [3] as well as Dedi´c, Mati´c and Peˇcari´c [4] and Pearce, Peˇcari´c, Ujevi´c and Varošanec [19] further extended the result (1.1) by considering n−times differentiable mappings on an interior pointx∈[a, b].

In particular, Cerone and Dragomir [1] proved the following result.

Theorem 1.2. Let f : [a, b] → Rbe a mapping such thatf(n−1) is absolutely continuous on [a, b], (AC[a, b]for short). Then for allx∈[a, b]the following bound is valid:

(1.2)

Z b a

f(t)dt−

n−1

X

j=0

(b−x)j+1+ (−1)j(x−a)j+1 (j+ 1)!

!

f(j)(x)

f(n)

(n+ 1)! (x−a)n+1+ (b−x)n+1

if f(n) ∈L[a, b], where

f(n)

:= sup

t∈[a,b]

f(n)(t) <∞.

Dragomir also generalised the Ostrowski inequality forkpoints,x1, . . . , xkand obtained the following theorem.

Theorem 1.3. Let Ik := a = x0 < x1 < ... < xk−1 < xk = b be a division of the interval [a, b], αi (i= 0, ..., k+ 1)be “k+ 2” points so thatα0 = a, αi ∈ [xi−1, xi] (i= 1, ..., k)and αk+1 =b. Iff : [a, b]→Ris AC[a, b], then we have the inequality

Z b a

f(x)dx−

k

X

i=0

i+1−αi)f(xi) (1.3)

"

1 4

k−1

X

i=0

h2i +

k−1

X

i=0

αi+1− xi+xi+1 2

2# kf0k

≤ 1 2kf0k

k−1

X

i=0

h2i

≤ 1

2(b−a)kf0kν(h) ,

wherehi :=xi+1−xi(i= 0, ..., k−1)andν(h) := max{hi|i= 0, ..., k−1}.

The constant 14 in the first inequality and the constant 12 in the second and third inequalities are the best possible.

The main aim of this paper is to develop the upcoming Theorem 3.1 (see page 10) of integral inequalities for n−times differentiable mappings. The motivation for this work has been to improve the order of accuracy of the results given by Dragomir [5], and Cerone and Dragomir [1]. In the case of Dragomir [5], the bound of (1.3) is of order 1, whilst in this paper we improve the bound of (1.3) to ordern.

We begin the process by obtaining the following integral equalities.

2. INTEGRAL IDENTITIES

Theorem 2.1. LetIk : a=x0 < x1 <· · ·< xk−1 < xk =bbe a division of the interval[a, b]

andαi (i= 0, . . . , k+ 1)be ‘k+ 2’ points so that α0 = a, αi ∈ [xi−1, xi] (i= 1, . . . , k)and

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αk+1 =b. Iff : [a, b] →Ris a mapping such thatf(n−1) is AC[a, b], then for allxi ∈[a, b]we have the identity:

(2.1) Z b

a

f(t)dt+

n

X

j=1

(−1)j j!

k−1

X

i=0

n

(xi+1−αi+1)jf(j−1)(xi+1)

−(xi−αi+1)jf(j−1)(xi)o

= (−1)n Z b

a

Kn,k(t)f(n)(t)dt, where the Peano kernel

(2.2) Kn,k(t) :=





























(t−α1)n

n! , t∈[a, x1) (t−α2)n

n! , t∈[x1, x2) ...

(t−αk−1)n

n! , t∈[xk−2, xk−1) (t−αk)n

n! , t∈[xk−1, b], nandk are natural numbers,n≥1,k ≥1andf(0)(x) =f(x).

Proof. The proof is by mathematical induction. Forn = 1, from (2.1) we have the equality Z b

a

f(t)dt =

k−1

X

i=0

h

(xi+1−αi+1)jf(xi+1)−(xi−αi+1)jf(xi) i (2.3)

− Z b

a

K1,k(t)f0(t)dt, where

K1,k(t) :=

























(t−α1), t ∈[a, x1) (t−α2), t ∈[x1, x2)

...

(t−αk−1), t ∈[xk−2, xk−1) (t−αk), t ∈[xk−1, b]. To prove (2.3), we integrate by parts as follows

Z b a

K1,k(t)f0(t)dt

=

k−1

X

i=0

Z xi+1

xi

(t−αi+1)f0(t)dt

=

k−1

X

i=0

(t−αi+1)f(t)

xi+1

xi − Z xi+1

xi

f(t)dt

(4)

=

k−1

X

i=0

[(xi+1−αi+1)f(xi+1)−(xi−αi+1)f(xi)]−

k−1

X

i=0

Z xi+1

xi

f(t)dt.

Z b a

f(t)dt + Z b

a

K1,k(t)f0(t)dt =

k−1

X

i=0

[(xi+1−αi+1)f(xi+1)−(xi−αi+1)f(xi)]. Hence (2.3) is proved.

Assume that (2.1) holds for ‘n’ and let us prove it for ‘n+ 1’. We need to prove the equality (2.4)

Z b a

f(t)dt+

k−1

X

i=0 n+1

X

j=1

(−1)j j!

n

(xi+1−αi+1)jf(j−1)(xi+1) −(xi−αi+1)jf(j−1)(xi)o

= (−1)n+1 Z b

a

Kn+1,k(t)f(n+1)(t)dt, where from (2.2)

Kn+1,k(t) :=

























(t−α1)n+1

(n+1)! , t∈[a, x1)

(t−α2)n+1

(n+1)! , t∈[x1, x2) ...

(t−αk−1)n+1

(n+1)! , t∈[xk−2, xk−1)

(t−αk)n+1

(n+1)! , t∈[xk−1, b]. Consider

Z b a

Kn+1,k(t)f(n+1)(t)dt=

k−1

X

i=0

Z xi+1

xi

(t−αi+1)n+1

(n+ 1)! f(n+1)(t)dt and upon integrating by parts we have

Z b a

Kn+1,k(t)f(n+1)(t)dt

=

k−1

X

i=0

"

(t−αi+1)n+1

(n+ 1)! f(n)(t)

xi+1

xi

− Z xi+1

xi

(t−αi+1)n

n! f(n)(t)dt

#

=

k−1

X

i=0

((xi+1−αi+1)n+1

(n+ 1)! f(n)(xi+1)−(xi−αi+1)n+1

(n+ 1)! f(n)(xi) )

− Z b

a

Kn,k(t)f(n)(t)dt.

Upon rearrangement we may write Z b

a

Kn,k(t)f(n)(t)dt=

k−1

X

i=0

((xi+1−αi+1)n+1

(n+ 1)! f(n)(xi+1)−(xi−αi+1)n+1

(n+ 1)! f(n)(xi) )

− Z b

a

Kn+1,k(t)f(n+1)(t)dt.

(5)

Now substituteRb

aKn,k(t)f(n)(t)dtfrom the induction hypothesis (2.1) such that (−1)n

Z b a

f(t)dt+ (−1)n

k−1

X

i=0

" n X

j=1

(−1)j j!

n

(xi+1−αi+1)jf(j−1)(xi+1)

− (xi−αi+1)jf(j−1)(xi)oi

=

k−1

X

i=0

((xi+1−αi+1)n+1

(n+ 1)! f(n)(xi+1)− (xi−αi+1)n+1

(n+ 1)! f(n)(xi) )

− Z b

a

Kn+1,k(t)f(n+1)(t)dt.

Collecting the second and third terms and rearranging, we can state

Z b a

f(t)dt+

k−1

X

i=0 n+1

X

j=1

(−1)j j!

n

(xi+1−αi+1)jf(j−1)(xi+1) −(xi−αi+1)jf(j−1)(xi)o

= (−1)n+1 Z b

a

Kn+1,k(t)f(n+1)(t)dt,

which is identical to (2.4), hence Theorem 2.1 is proved.

The following corollary gives a slightly different representation of Theorem 2.1, which will be useful in the following work.

Corollary 2.2. From Theorem 2.1, the equality (2.1) may be represented as

(2.5) Z b

a

f(t)dt+

n

X

j=1

(−1)j j!

" k X

i=0

n

(xi−αi)j−(xi−αi+1)jo

f(j−1)(xi)

#

= (−1)n Z b

a

Kn,k(t)f(n)(t)dt.

Proof. From (2.1) consider the second term and rewrite it as

(2.6) S1+S2 :=

k−1

X

i=0

{−(xi+1−αi+1)f(xi+1) + (xi−αi+1)f(xi)}

+

k−1

X

i=0

" n X

j=2

(−1)j j!

n

(xi+1−αi+1)jf(j−1)(xi+1)

−(xi−αi+1)jf(j−1)(xi)o .

(6)

Now

S1 = (a−α1)f(a) +

k−1

X

i=1

(xi−αi+1)f(xi)

+

k−2

X

i=0

{−(xi+1−αi+1)f(xi+1)} −(b−αk)f(b)

= (a−α1)f(a) +

k−1

X

i=1

(xi−αi+1)f(xi)

+

k−1

X

i=1

{−(xi−αi)f(xi)} −(b−αk)f(b)

= −(α1−a)f(a)−

k−1

X

i=1

i+1−αi)f(xi)−(b−αk)f(b).

Also,

S2 =

k−2

X

i=0

" n X

j=2

(−1)j j!

n

(xi+1−αi+1)jf(j−1)(xi+1)o

#

+

n

X

j=2

(−1)j j!

n

(xk−αk)jf(j−1)(xk)o

n

X

j=2

(−1)j j!

n

(x0−α1)jf(j−1)(x0)o

k−1

X

i=1

" n X

j=2

(−1)j j!

n

(xi−αi+1)jf(j−1)(xi)o

#

=

n

X

j=2

(−1)j

j! (b−αk)jf(j−1)(b)−

n

X

j=2

(−1)j

j! (a−α1)jf(j−1)(a) +

k−1

X

i=1

" n X

j=2

(−1)j j!

n

(xi−αi)j −(xi−αi+1)jo

f(j−1)(xi)

# .

From (2.6)

S1+S2 := − (

(b−αk)f(b) + (α1−a)f(a) +

k−1

X

i=1

i+1−αi)f(xi) )

+

n

X

j=2

(−1)j j!

n

(b−αk)jf(j−1)(b)−(a−α1)jf(j−1)(a)o

+

k−1

X

i=1

" n X

j=2

(−1)j j!

n

(xi−αi)j −(xi−αi+1)jo

f(j−1)(xi)

#

(7)

= − (

1−a)f(a) +

k−1

X

i=1

i+1−αi)f(xi) + (b−αk)f(b) )

+

n

X

j=2

(−1)j j!

"

−(a−α1)jf(j−1)(a)

+

k−1

X

i=1

n

(xi−αi)j−(xi−αi+1)j o

f(j−1)(xi) + (b−αk)jf(j−1)(b)

# . Keeping in mind thatx0 =a,α0 = 0,xk =bandαk+1 =bwe may write

S1+S2 = −

k

X

i=0

i+1−αi)f(xi)

+

n

X

j=2

(−1)j j!

" k X

i=0

n

(xi −αi)j−(xi−αi+1)jo

f(j−1)(xi)

#

=

n

X

j=1

(−1)j j!

" k X

i=0

n

(xi−αi)j−(xi−αi+1)jo

f(j−1)(xi)

# .

And substitutingS1+S2into the second term of (2.1) we obtain the identity (2.5).

If we now assume that the points of the divisionIkare fixed, we obtain the following corol- lary.

Corollary 2.3. LetIk :a=x0 < x1 <· · ·< xk−1 < xk =bbe a division of the interval[a, b].

Iff : [a, b]→Ris as defined in Theorem 2.1, then we have the equality (2.7)

Z b a

f(t)dt+

n

X

j=1

1 2jj!

" k X

i=0

n

−hji + (−1)jhji−1 o

f(j−1)(xi)

#

= (−1)n Z b

a

Kn,k(t)f(n)(t)dt, wherehi :=xi+1−xi,h−1 := 0andhk := 0.

Proof. Choose

α0 = a, α1 = a+x1

2 , α2 = x1+x2 2 , . . . , αk−1 = xk−2+xk−1

2 ,αk= xk−1 +xk

2 and αk+1 =b.

From Corollary 2.2, the term

(b−αk)f(b) + (α1−a)f(a) +

k−1

X

i=1

i+1−αi)f(xi)

= 1 2

(

h0f(a) +

k−1

X

i=1

(hi+hi−1)f(xi) +hk−1f(b) )

,

(8)

the term

n

X

j=2

(−1)j j!

n

(b−αk)jf(j−1)(b)−(a−α1)jf(j−1)(a)o

=

n

X

j=2

(−1)j j!2j

n

hjk−1f(j−1)(b)−(−1)jhj0f(j−1)(a)o

and the term

k−1

X

i=1

" n X

j=2

(−1)j j!

n

(xi −αi)j−(xi−αi+1)jo

f(j−1)(xi)

#

=

k−1

X

i=1

" n X

j=2

(−1)j j!2j

n

hji−1−(−1)jhjio

f(j−1)(xi)

# .

Putting the last three terms in (2.5) we obtain

Z b a

f(t)dt− 1 2

(

h0f(a) +

k−1

X

i=1

(hi+hi−1)f(xi) +hk−1f(b) )

+

n

X

j=2

(−1)j j!2j

n

hjk−1f(j−1)(b)−(−1)jhj0f(j−1)(a)o

+

k−1

X

i=1

" n X

j=2

(−1)j j!2j

n

hji−1−(−1)jhjio

f(j−1)(xi)

#

= (−1)n Z b

a

Kn,k(t)f(n)(t)dt.

Collecting the inner three terms of the last expression, we have

Z b a

f(t)dt+

n

X

j=1

(−1)j j!2j

k

X

i=0

n

hji−1−(−1)jhjio

f(j−1)(xi) = (−1)n Z b

a

Kn,k(t)f(n)(t)dt,

which is equivalent to the identity (2.7).

The case of equidistant partitioning is important in practice, and with this in mind we obtain the following corollary.

Corollary 2.4. Let

(2.8) Ik :xi =a+i

b−a k

, i= 0, . . . , k

(9)

be an equidistant partitioning of [a, b], and f : [a, b] → R be a mapping such thatf(n−1) is AC[a, b],then we have the equality

(2.9) Z b

a

f(t)dt+

n

X

j=1

b−a 2k

j

1 j!

"

−f(j−1)(a)

+

k−1

X

i=1

n

(−1)j −1 o

f(j−1)(xi) + (−1)jf(j−1)(b)

#

= (−1)n Z b

a

Kn,k(t)f(n)(t)dt.

It is of some interest to note that the second term of (2.9) involves only even derivatives at all interior pointsxi,i= 1, . . . , k−1.

Proof. Using (2.8) we note that h0 = x1−x0 = b−a

k , hk−1 = (xk−xk−1) = b−a k , hi = xi+1−xi = b−a

k and hi−1 =xi−xi−1 = b−a

k , (i= 1, . . . , k−1) and substituting into (2.7) we have

Z b a

f(t)dt+

n

X

j=1

1 j!2j

"

b−a 2k

j

f(j−1)(a) +

k−1

X

i=0

(

b−a k

j

+ (−1)j

b−a k

j)

f(j−1)(xi) + (−1)j

b−a k

j

f(j−1)(b)

#

= (−1)n Z b

a

Kn,k(t)f(n)(t)dt,

which simplifies to (2.9) after some minor manipulation.

The following Taylor-like formula with integral remainder also holds.

Corollary 2.5. Let g : [a, y] → R be a mapping such that g(n) is AC[a, y]. Then for all xi ∈[a, y]we have the identity

(2.10) g(y) = g(a)−

k−1

X

i=0

" n X

j=1

(−1)j j!

n

(xi+1−αi+1)jg(j)(xi+1)

− (xi−αi+1)jg(j)(xi)o

+ (−1)n Z y

a

Kn,k(y, t)g(n+1)(t)dt or

(2.11) g(y) = g(a)−

n

X

j=1

(−1)j j!

" k X

i=0

n

(xi−αi)j −(xi−αi+1)jo

g(j)(xi)

#

+ (−1)n Z y

a

Kn,k(y, t)g(n+1)(t)dt.

The proof of (2.10) and (2.11) follows directly from (2.1) and (2.5) respectively upon choos- ingb =yandf =g0.

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3. INTEGRALINEQUALITIES

In this section we utilise the equalities of Section 2 and develop inequalities for the represen- tation of the integral of a function with respect to its derivatives at a multiple number of points within some interval.

Theorem 3.1. LetIk : a=x0 < x1 <· · ·< xk−1 < xk =bbe a division of the interval[a, b]

andαi (i= 0, . . . , k+ 1)be ‘k+ 2’ points so that α0 = a, αi ∈ [xi−1, xi] (i= 1, . . . , k)and αk+1 =b. Iff : [a, b] →Ris a mapping such thatf(n−1) is AC[a, b], then for allxi ∈[a, b]we have the inequality:

Z b a

f(t)dt+

n

X

j=1

(−1)j j!

" k X

i=0

n

(xi−αi)j−(xi−αi+1)jo

f(j−1)(xi)

# (3.1)

f(n) (n+ 1)!

k−1

X

i=0

i+1−xi)n+1+ (xi+1−αi+1)n+1

f(n) (n+ 1)!

k−1

X

i=0

hn+1i

f(n)

(n+ 1)! (b−a)νn(h) if f(n) ∈L[a, b], where

f(n)

:= sup

t∈[a,b]

f(n)(t) <∞, hi := xi+1−xi and

ν(h) := max{hi|i= 0, . . . , k−1}. Proof. From Corollary 2.2 we may write

(3.2)

Z b a

f(t)dt+

n

X

j=1

(−1)j j!

" k X

i=0

n

(xi−αi)j−(xi−αi+1)jo

f(j−1)(xi)

#

=

(−1)n Z b

a

Kn,k(t)f(n)(t)dt , and

(−1)n Z b

a

Kn,k(t)f(n)(t)dt

≤ f(n)

Z b

a

|Kn,k(t)|dt,

Z b a

|Kn,k(t)|dt =

k−1

X

i=0

Z xi+1

xi

|t−αi+1|n

n! dt

=

k−1

X

i=0

Z αi+1

xi

i+1−t)n n! dt+

Z xi+1

αi+1

(t−αi+1)n

n! dt

= 1

(n+ 1)!

k−1

X

i=0

i+1−xi)n+1+ (xi+1−αi+1)n+1

(11)

and thus

(−1)n Z b

a

Kn,k(t)f(n)(t)dt

f(n) (n+ 1)!

k−1

X

i=0

i+1−xi)n+1+ (xi+1−αi+1)n+1 . Hence, from (3.2), the first part of the inequality (3.1) is proved. The second and third lines follow by noting that

f(n) (n+ 1)!

k−1

X

i=0

i+1−xi)n+1+ (xi+1−αi+1)n+1

f(n) (n+ 1)!

k−1

X

i=0

hn+1i , since for0< B < A < C it is well known that

(3.3) (A−B)n+1+ (C−A)n+1 ≤(C−B)n+1. Also

f(n) (n+ 1)!

k−1

X

i=0

hn+1i

f(n) (n+ 1)!νn(h)

k−1

X

i=0

hi =

f(n)

(n+ 1)! (b−a)νn(h),

where ν(h) := max{hi|i= 0, . . . , k−1}and therefore the third line of the inequality (3.1)

follows, hence Theorem 3.1 is proved.

When the points of the divisionIk are fixed, we obtain the following inequality.

Corollary 3.2. Letf : [a, b] →Rbe a mapping such thatf(n−1) is AC[a, b],andIkbe defined as in Corollary 2.3, then

(3.4)

Z b a

f(t)dt+

n

X

j=1

(−1)j 2jj!

" k X

i=0

n−hji + (−1)jhji+1o

f(j−1)(xi)

#

f(n) (n+ 1)!2n

k−1

X

i=0

hn+1i forf(n)∈L[a, b].

Proof. From Corollary 2.3 we choose

α0 = a, α1 = a+x1

2 , . . . , αk−1 = xk−2+xk−1

2 , αk = xk−1+xk

2 andαk+1 =b.

Now utilising the first line of the inequality (3.1), we may evaluate

k−1

X

i=0

i+1−xi)n+1+ (xi+1−αi+1)n+1 =

k−1

X

i=0

2 hi

2 n+1

and therefore the inequality (3.4) follows.

For the equidistant partitioning case we have the following inequality.

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Corollary 3.3. Letf : [a, b]→Rbe a mapping such thatf(n−1)is AC[a, b]and letIkbe defined by (2.8). Then

(3.5)

Z b a

f(t)dt+

n

X

j=1

b−a 2k

j

1 j!

×

"

−f(j−1)(a) +

k−1

X

i=1

n

(−1)j−1 o

f(j−1)(xi) + (−1)jf(j−1)(b)

#

f(n)

(n+ 1)! (2k)n(b−a)n+1 forf(n)∈L[a, b].

Proof. We may utilise (2.9) and from (3.1), note that h0 =x1−x0 = b−a

k and hi =xi+1−xi = b−a

k , i= 1, . . . , k−1

in which case (3.5) follows.

The following inequalities for Taylor-like expansions also hold.

Corollary 3.4. Letgbe defined as in Corollary 2.5. Then we have the inequality (3.6)

g(y)−g(a) +

n

X

j=1

(−1)j j!

" k X

i=0

n

(xi−αi)j−(xi−αi+1)jo

g(j)(xi)

#

g(n+1) (n+ 1)!

k−1

X

i=0

i+1−xi)n+1+ (xi+1−αi+1)n+1 ifg(n+1) ∈L[a, b]

for allxi ∈[a, y]where

g(n+1)

:= sup

t∈[a,y]

g(n+1)(t) <∞.

Proof. Follows directly from (2.11) and using the norm as in (3.1).

When the points of the divisionIk are fixed we obtain the following.

Corollary 3.5. Let g be defined as in Corollary 2.5 andIk : a = x0 < x1 < · · · < xk−1 <

xk =ybe a division of the interval[a, y]. Then we have the inequality (3.7)

g(y)−g(a) +

n

X

j=1

1 2jj!

" k X

i=0

n−hji + (−1)jhji+1o

g(j)(xi)

#

g(n+1) (n+ 1)!2n

k−1

X

i=0

hn+1i ifg(n+1) ∈L[a, y].

Proof. The proof follows directly from using (2.7).

For the equidistant partitioning case we have:

Corollary 3.6. Letgbe defined as in Corollary 2.5 and Ik :xi =a+i·

y−a k

, i= 0, . . . , k

(13)

be an equidistant partitioning of[a, y], then we have the inequality:

(3.8)

g(y)−g(a) +

n

X

j=1

y−a 2k

j

1 j!

×

"

−g(j)(a) +

k−1

X

i=1

n(−1)j−1o

g(j)(xi) + (−1)jg(j)(y)

#

g(n+1)

(n+ 1)! (2k)n(y−a)n+1 ifg(n+1) ∈L[a, y]. Proof. The proof follows directly upon using (2.9) withf0 =gandb=y.

4. THECONVERGENCE OF AGENERALQUADRATUREFORMULA

Let

m :a =x(m)0 < x(m)1 <· · ·< x(m)m−1 < x(m)m =b

be a sequence of division of[a, b]and consider the sequence of real numerical integration for- mula

(4.1) Im f, f0, . . . , f(n),∆m, wm :=

m

X

j=0

w(m)j f x(m)j

n

X

r=2

(−1)r r!

" m X

j=0

(

x(m)j −a−

j−1

X

s=0

w(m)s

!r

− x(m)j −a−

j

X

s=0

ws(m)

!r)

f(r−1) x(m)j

# ,

wherewj (j = 0, . . . , m)are the quadrature weights and assume thatPm

j=0w(m)j =b−a. The following theorem contains a sufficient condition for the weights w(m)j so that Im f, f0, . . . , f(n),∆m, wm

approximates the integral Rb

a f(x)dx with an error expressed in terms of

f(n) .

Theorem 4.1. Letf : [a, b]→Rbe a continuous mapping on[a, b]such thatf(n−1)is AC[a, b].

If the quadrature weights,wj(m)satisfy the condition

(4.2) x(m)i −a ≤

i

X

j=0

w(m)j ≤x(m)i+1 −a for alli= 0, . . . , m−1

then we have the estimation

(4.3)

Im f, f0, . . . , f(n),∆m, wm

− Z b

a

f(t)dt

(14)

f(n)

(n+ 1)!

m−1

X

i=0

 a+

i

X

j=0

w(m)j −x(m)i

!n+1

− x(m)i+1 −a−

i

X

j=0

wj(m)

!n+1

f(n) (n+ 1)!

m−1

X

i=0

h(m)i

n+1

f(n) (n+ 1)!

ν h(m)

(b−a), where f(n) ∈L[a, b]. Also

ν h(m)

:= max

i=0,...,m−1

n h(m)i

o

h(m)i := x(m)i+1−x(m)i . In particular, if

f(n)

<∞, then lim

ν(h(m))→0

Im f, f0, . . . , f(n),∆m, wm

= Z b

a

f(t)dt uniformly by the influence of the weightswm.

Proof. Define the sequence of real numbers

α(m)i+1 :=a+

i

X

j=0

wj(m), i= 0, . . . , m.

Note that

α(m)i+1 =a+

m

X

j=0

w(m)j =a+b−a=b.

By the assumption (4.2), we have α(m)i+1 ∈h

x(m)i , x(m)i+1i

for all i= 0, . . . , m−1.

Defineα0(m)=aand compute α(m)1 −α(m)0 = w0(m) α(m)i+1−α(m)i = a+

i

X

j=0

wj(m)−a−

i−1

X

j=0

w(m)j =wi(m) (i= 0, . . . , m−1) and

α(m)m+1−α(m)m =a+

m

X

j=0

wj(m)−a−

m−1

X

j=0

w(m)j =wm(m). Consequently

m

X

i=0

α(m)i+1 −α(m)i f

x(m)i

=

m

X

i=0

wi(m)f x(m)i

,

(15)

and let

m

X

j=0

wj(m)f

x(m)j

n

X

r=2

(−1)r r!

" m X

j=0

(

x(m)j −a−

j−1

X

s=0

ws(m)

!r

− x(m)j −a−

j

X

s=0

w(m)s

!r)

f(r−1) x(m)j

#

:=Im f, f0, . . . , f(n),∆m, wm .

Applying the inequality (3.1) we obtain the estimate (4.3).

The case when the partitioning is equidistant is important in practice. Consider the equidistant partition

Em :=x(m)i :=a+ib−a

m , (i= 0, . . . , m) and define the sequence of numerical quadrature formulae

Im f, f0, . . . , f(n),∆m, wm :=

m

X

j=0

w(m)j f

a+ j(b−a) n

n

X

r=2

(−1)r r!

" m X

j=0

( j(b−a)

n −

j−1

X

s=0

ws(m)

!r

− j(b−a)

n −

j

X

s=0

ws(m)

!r) f(r−1)

a+ j(b−a) n

# . The following corollary holds.

Corollary 4.2. Let f : [a, b] → R be AC[a, b]. If the quadrature weights w(m)j satisfy the condition

i

m ≤ 1

b−a

i

X

j=0

w(m)j ≤ i+ 1

m , i= 0,1, . . . , n−1, then the following bound holds:

Im f, f0, . . . , f(n),∆m, wm

− Z b

a

f(t)dt

f(n) (n+ 1)!

m−1

X

i=0

i

X

j=0

wj(m)−i

b−a m

!n+1

− (i+ 1)

b−a m

i

X

j=0

w(m)j

!n+1

f(n) (n+ 1)!

b−a m

n+1

, where f(n) ∈L[a, b]. In particular, if

f(n)

<∞, then

m→∞lim Im f, f0, . . . , f(n), wm

= Z b

a

f(t)dt uniformly by the influence of the weightswm.

The proof of Corollary 4.2 follows directly from Theorem 4.1.

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5. GRÜSSTYPE INEQUALITIES

The Grüss inequality [15], is well known in the literature. It is an integral inequality which establishes a connection between the integral of a product of two functions and the product of the integrals of the two functions.

Theorem 5.1. Leth, g : [a, b] → Rbe two integrable functions such thatφ ≤ h(x) ≤ Φand γ ≤g(x)≤Γfor allx∈[a, b],φ,Φ, γandΓare constants. Then we have

(5.1) |T (h, g)| ≤ 1

4(Φ−φ) (Γ−γ), where

(5.2) T (h, g) := 1 b−a

Z b a

h(x)g(x)dx− 1 b−a

Z b a

h(x)dx· 1 b−a

Z b a

g(x)dx and the inequality (5.1) is sharp, in the sense that the constant 14 cannot be replaced by a smaller one.

For a simple proof of this fact as well as generalisations, discrete variants, extensions and associated material, see [17]. The Grüss inequality is also utilised in the papers [6, 7, 14] and the references contained therein.

A premature Grüss inequality is the following.

Theorem 5.2. Letf andg be integrable functions defined on[a, b]and letγ ≤ g(x) ≤ Γfor allx∈[a, b]. Then

(5.3) |T (h, g)| ≤ Γ−γ

2 (T(f, f))12 , whereT(f, f)is as defined in (5.2).

Theorem 5.2 was proved in 1999 by Mati´c, Peˇcari´c and Ujevi´c [16] and it provides a sharper bound than the Grüss inequality (5.1). The term premature is used to highlight the fact that the result (5.3) is obtained by not fully completing the proof of the Grüss inequality. The premature Grüss inequality is completed if one of the functions,f org, is explicitly known.

We now give the following theorem based on the premature Grüss inequality (5.3).

Theorem 5.3. Let Ik : a = x0 < x1 < · · · < xk−1 < xk = b be a division of the interval [a, b], αi (i= 0, . . . , k+ 1) be ‘k+ 1’ points such that α0 = a, αi ∈ [xi−1, xi] (i= 1, . . . , k) andαk =b. Iff : [a, b] →Ris AC[a, b]andntime differentiable on[a, b], then assuming that thenthderivativef(n): (a, b)→Rsatisfies the condition

m≤f(n) ≤M for allx∈(a, b), we have the inequality

(5.4)

(−1)n Z b

a

f(t)dt+ (−1)n

n

X

j=1

(−1)j j!

×

"k−1 X

i=0

( hi

2 −δi j

f(j−1)(xi+1)− hi

2 +δi j

f(j−1)(xi) )#

f(n−1)(b)−f(n−1)(a) (b−a) (n+ 1)!

k−1 X

i=0

hi 2

n+1

×

"n+1 X

r=0

n+ 1 r

i hi

r

{1 + (−1)r}

#

(17)

≤ M−m 2

"

b−a (2n+ 1) (n!)2

k−1

X

i=0

hi 2

2n+1

×

"2n+1 X

r=0

2n+ 1 r

i hi

r

{1 + (−1)r}

#

− 1

(n+ 1)!

k−1

X

i=0

hi

2

n+1"n+1 X

r=0

n+ 1 r

i

hi r

{1 + (−1)r}

#!2

1 2

where

hi := xi+1−xi and δi := αi+1− xi+1+xi

2 , i= 0, . . . , k−1.

Proof. We utilise (5.2) and (5.3), multiply through by(b−a) and chooseh(t) := Kn,k(t)as defined by (2.2) andg(t) :=f(n)(t),t ∈[a, b]such that

(5.5)

Z b a

Kn,k(t)f(n)(t)dt− 1 b−a

Z b a

f(n)(t)dt· Z b

a

Kn,k(t)dt

≤ Γ−γ 2

"

(b−a) Z b

a

Kn,k2 (t)dt− Z b

a

Kn,k(t)dt 2#12

. Now we may evaluate

Z b a

f(n)(t)dt=f(n−1)(b)−f(n−1)(a) and

G1 :=

Z b a

Kn,k(t)dt

=

k−1

X

i=0

Z xi+1

xi

1

n!(t−αi+1)ndt

= 1

(n+ 1)!

k−1

X

i=0

(xi+1−αi+1)n+1+ (αi+1−xi)n+1 . Using the definitions ofhi andδi we have

xi+1−αi+1 = hi

2 −δi and αi+1−xi = hi 2 +δi such that

G1 = 1 (n+ 1)!

k−1

X

i=0

( hi

2 −δi n+1

+ hi

2 +δi

n+1)

= 1

(n+ 1)!

k−1

X

i=0

"n+1 X

r=0

n+ 1 r

(−δi)r

hi 2

n+1−r

+

n+1

X

r=0

n+ 1 r

δir

hi 2

n+1−r#

= 1

(n+ 1)!

k−1

X

i=0

hi 2

n+1"n+1 X

r=0

n+ 1 r

i hi

r

{1 + (−1)r}

# .

Hivatkozások

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