SOME COMPANIONS OF AN OSTROWSKI TYPE INEQUALITY AND APPLICATIONS

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Ostrowski Type Inequality Zheng Liu vol. 10, iss. 2, art. 52, 2009

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SOME COMPANIONS OF AN OSTROWSKI TYPE INEQUALITY AND APPLICATIONS

ZHENG LIU

Institute of Applied Mathematics, School of Science University of Science and Technology Liaoning Anshan 114051, Liaoning, China

EMail:lewzheng@163.net

Received: 11 February, 2009

Accepted: 12 May, 2009

Communicated by: S.S. Dragomir 2000 AMS Sub. Class.: 26D15.

Key words: Ostrowski type inequality, Perturbed trapezoid rule, Midpoint rule, Composite quadrature rule.

Abstract: We establish some companions of an Ostrowski type integral inequality for functions whose derivatives are absolutely continuous. Applications for composite quadrature rules are also given.

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1. Introduction

Motivated by [1], Dragomir in [2] has proved the following companion of the Os- trowski inequality:

(1.1) 1

2[f(x) +f(a+b−x)]− 1 b−a

Z b a

f(t) dt

≤











1

8 + 2_x−3a+b 4

b−a

²

(b−a)kf⁰k[a,b],∞ iff⁰ ∈L∞[a, b] ;

2¹^q (q+1)¹^q

x−a b−a

q+1

+a+b 2 −x b−a

q+1¹_q

(b−a)¹^q kf⁰k_[a,b],p, ifp > 1,¹_p +¹_q = 1, andf⁰ ∈L_p[a, b] ; h1

4 +

x−^3a+b

4

b−a

ikf⁰k_[a,b],1 iff⁰ ∈L₁[a, b],

for allx∈ a,^a+b₂

, wheref : [a, b]→Ris an absolutely continuous function.

In particular, the best result in (1.1) is obtained forx= ^a+3b₄ , giving the following trapezoid type inequalities:

(1.2) 1 2

f

3a+b 4

+f

a+ 3b 4

− 1 b−a

Z b a

f(t)dt

≤











1

8 (b−a)kf⁰k_[a,b],∞ if f⁰ ∈L_∞[a, b] ;

1 4 · ^(b−a)

1q

(q+1)¹^q

kf⁰k_[a,b],p, if f⁰ ∈L_p[a, b], p >1, ¹_p +¹_q = 1;

1

4kf⁰k_[a,b],1 if f⁰ ∈L₁[a, b]. Some natural applications of (1.1) and (1.2) are also provided in [2].

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In [3], Dedi´c et al. have derived the following trapezoid type inequality:

(1.3) 1 2

f

3a+b 4

+f

a+ 3b 4

− 1 b−a

Z b a

f(t)dt

≤ (b−a)²

32 kf⁰⁰k∞, for a function f : [a, b] → R whose derivative f⁰ is absolutely continuous and f⁰⁰ ∈L∞[a, b].

In [4], we find that for a functionf : [a, b]→Rwhose derivativef⁰ is absolutely continuous, the following perturbed trapezoid inequalities hold:

(1.4)

Z b a

f(t) dt− b−a

2 [f(a) +f(b)] + (b−a)²

8 [f⁰(b)−f⁰(a)]

≤











(b−a)³

24 kf⁰⁰k_∞ if f⁰⁰ ∈L_∞[a, b] ;

(b−a)^{2+ 1}^q 8(2q+1)¹^q

kf⁰⁰k_p, if f⁰⁰ ∈L_P[a, b], p >1, ¹_p +¹_q = 1;

(b−a)²

8 kf⁰⁰k₁ if f⁰⁰ ∈L₁[a, b].

In this paper, we provide some companions of Ostrowski type inequalities for functions whose first derivatives are absolutely continuous and whose second derivatives belong to the Lebesgue spacesL_p[a, b], 1≤p ≤ ∞. These improve (1.3) and recapture (1.4). Applications for composite quadrature rules are also given.

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2. Some Integral Inequalities

Lemma 2.1. Letf : [a, b]→Rbe such that the derivativef⁰ is absolutely continu- ous on[a, b]. Then we have the equality

(2.1) 1 b−a

Z b a

f(t)dt− 1

2[f(x) +f(a+b−x)]

+ 1 2

x−3a+b 4

[f⁰(x)−f⁰(a+b−x)]

= 1

2 (b−a)

"

Z x a

(t−a)²f⁰⁰(t) dt+

Z a+b−x x

t− a+b 2

2

f⁰⁰(t)dt +

Z b a+b−x

(t−b)²f⁰⁰(t) dt

for anyx∈

a,^a+b₂ .

Proof. Using the integration by parts formula for Lebesgue integrals, we have Z x

a

(t−a)²f⁰⁰(t) dt= (x−a)²f⁰(x)−2 (x−a)f(x) + 2 Z x

a

f(t) dt,

Z a+b−x x

t− a+b 2

2

f⁰⁰(t)dt

=

x− a+b 2

2

[f⁰(a+b−x)−f⁰(x)]

+ 2

x−a+b 2

[f(x) +f(a+b−x)] + 2

Z a+b−x x

f(t)dt,

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and Z b

a+b−x

(t−b)²f⁰⁰(t)dt

=−(x−a)²f⁰(a+b−x)−2 (x−a)f(a+b−x) + 2 Z b

a+b−x

f(t)dt.

Summing the above equalities, we deduce the desired identity (2.1).

Theorem 2.2. Letf : [a, b]→Rbe such that the derivativef⁰ is absolutely contin- uous on[a, b]. Then we have the inequality

1 b−a

Z b a

f(t) dt− 1

2[f(x) +f(a+b−x)]

(2.2)

+ 1 2

x−3a+b 4

[f⁰(x)−f⁰(a+b−x)]

≤ 1

2 (b−a)

"

Z x a

(t−a)²|f⁰⁰(t)|dt+

Z a+b−x x

t− a+b 2

2

|f⁰⁰(t)|dt +

Z b a+b−x

(t−b)²|f⁰⁰(t)|dt

:=M(x) for anyx∈

a,^a+b₂ .

Iff⁰⁰ ∈L_∞[a, b], then we have the inequalities

M(x)≤ 1

2 (b−a)

"

(x−a)³

3 kf⁰⁰k[a,x],∞

(2.3)

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+ 2 3

a+b 2 −x

3

kf⁰⁰k[x,a+b−x],∞+(x−a)³

3 kf⁰⁰k[a+b−x,b]

#

≤











1

96+ ¹₂_x−3a+b 4

b−a

²

(b−a)²kf⁰⁰k[a,b],∞;

1 2^α−1

x−a b−a

3α

+_x−a+b 2

b−a

^3α_α¹

×h

kf⁰⁰k^β_[a,x],∞+kf⁰⁰k^β[x,a+b−x],∞+kf⁰⁰k^β[a+b−x,b],∞

i_β¹ _(b−a)2

3

if α >1,_α¹ +_β¹ = 1;

max

1 2

x−a b−a

3

,_x−a+b 2

b−a

3

×

kf⁰⁰k[a,x],∞+kf⁰⁰k[x,a+b−x],∞+kf⁰⁰k[a+b−x,b],∞

(b−a)² 3 ; for anyx∈

a,^a+b₂ .

The inequality (2.2), the first inequality in (2.3) and the constant ₉₆¹ are sharp.

Proof. The inequality (2.2) follows by Lemma2.1on taking the modulus and using its properties.

Iff⁰⁰ ∈L∞[a, b], then Z x

a

(t−a)²|f⁰⁰(t)|dt ≤ (x−a)³

3 kf⁰⁰k[a,x],∞, Z a+b−x

x

t− a+b 2

2

|f⁰⁰(t)|dt≤ 2 3

a+b 2 −x

3

kf⁰⁰k[x,a+b−x],∞,

Z b a+b−x

(t−b)²|f⁰⁰(t)|dt≤ (x−a)³

3 f⁰⁰k[a+b−x,b],∞

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and the first inequality in (2.3) is proved.

Denote

M¯ (x) := (x−a)³

6 kf⁰⁰k_[a,x],∞+1 3

a+b 2 −x

3

kf⁰⁰k[x,a+b−x],∞

+(x−a)³

6 kf⁰⁰k[a+b−x,b]

forx∈ a,^a+b₂

. Firstly, observe that

M¯ (x)≤max

kf⁰⁰k[a,x],∞,kf⁰⁰k[x,a+b−x],∞,kf⁰⁰k[a+b−x,b],∞

×

"

(x−a)³

6 + 1

3

a+b 2 −x

3

+(x−a)³ 6

#

=kf⁰⁰k_[a,b],∞

"

(b−a)² 96 +1

2

x−3a+b 4

2#

(b−a) and the first part of the second inequality in (2.3) is proved.

Using the Hölder inequality forα >1, _α¹ + ¹_β = 1, we also have

M¯ (x)≤ 1 3

("

(x−a)³ 2

#α

+

x− a+b 2

3α

+

"

(x−a)³ 2

#α)_α¹

×h

kf⁰⁰k^β_[a,x],∞+kf⁰⁰k^β[x,a+b−x],∞+kf⁰⁰k^β[a+b−x,b],∞

i_β¹

giving the second part of the second inequality in (2.3)

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Finally, we also observe that M¯ (x)≤ 1

3max

((x−a)³

2 ,

x− a+b 2

3)

×

kf⁰⁰k[a,x],∞+kf⁰⁰k[x,a+b−x],∞+kf⁰⁰k[a+b−x,b],∞

. The sharpness of the inequalities mentioned follows from the fact that we can choose a functionf : [a, b]→R,f(t) =t²for anyx∈

a,^a+b₂

to obtain the corresponding equalities.

Remark 1. If in Theorem2.2 we choose x = a, then we recapture the first part of the inequality (1.4), i.e.,

1 b−a

Z b a

f(t) dt− 1

2[f(a) +f(b)] + b−a

8 [f⁰(b)−f⁰(a)]

≤ 1

24(b−a)²kf⁰⁰k∞

with ₂₄¹ as a sharp constant. If we choosex= ^a+b₂ , then we get

1 b−a

Z b a

f(t) dt−f

a+b 2

≤ 1 48

hkf⁰⁰k[â,â+b2 ]^,∞+kf⁰⁰k[â+b2 ,b]^,∞

i

≤ 1

24(b−a)²kf⁰⁰k[a,b],∞

with the constants ₄₈¹ and ₂₄¹ being sharp.

Corollary 2.3. With the assumptions in Theorem2.2, one has the inequality (2.4)

1 b−a

Z b a

f(t)dt− f ^3a+b₄

+f ^a+3b₄ 2

≤ 1

96(b−a)²kf⁰⁰k[a,b],∞.

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The constant ₉₆¹ is best possible in the sense that it cannot be replaced by a smaller constant. Clearly (2.4) is an improvement of (1.3).

Theorem 2.4. Letf : [a, b]→Rbe such that the derivativef⁰ is absolutely contin- uous on[a, b]andf⁰⁰ ∈L_p[a, b],p >1. IfM(x)is as defined in (2.2), then we have the bounds:

(2.5) M(x)≤ 1 2 (2q+ 1)¹^q

"

x−a b−a

2+¹_q

kf⁰⁰k_[a,x],p

+2¹^q

a+b 2 −x b−a

!2+¹_q

kf⁰⁰k[x,a+b−x],p

x−a b−a

2+¹_q

kf⁰⁰k[a+b−x,b],p



(b−a)¹⁺¹^q

≤ 1

2 (2q+ 1)¹^q

×











2 ^x−a_b−a2+¹_q

+ 2¹^q a+b

2 −x b−a

2+¹_q

×max

kf⁰⁰k_[a,x],p,kf⁰⁰k[x,a+b−x],p,kf⁰⁰k[a+b−x,b],p (b−a)¹⁺¹^q ;

2 ^x−a_b−a2α+^α_q

+ 2^α^q a+b 2 −x b−a

^2α+^α_q¹_α

×h

kf⁰⁰k^β_[a,x],p+kf⁰⁰k^β[x,a+b−x],p+kf⁰⁰k^β[a+b−x,b],p

i_β¹

(b−a)¹⁺¹^q ifα >1,_α¹ +_β¹ = 1, max

x−a b−a

2+¹_q

,2¹^q a+b 2 −x b−a

²⁺¹_q

×

kf⁰⁰k[a,x],p+kf⁰⁰k[x,a+b−x],p+kf⁰⁰k[a+b−x,b],p

(b−a)¹⁺¹^q ; for anyx∈

a,^a+b₂ .

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Proof. Using Hölder’s integral inequality forp >1, ¹_p +¹_q = 1, we have Z x

a

(t−a)²|f⁰⁰(t)|dt≤ Z x

a

(t−a)^2q dt ¹_q

kf⁰⁰k_[a,x],p

= (x−a)²⁺¹^q (2q+ 1)¹^q

kf⁰⁰k_[a,x],p,

Z a+b−x x

t− a+b 2

2

|f⁰⁰(t)|dt ≤

Z a+b−x

x

|t− a+b 2 |^2qdt

¹_q

kf⁰⁰k[x,a+b−x],p

= 2¹^q ^a+b₂ −x2+¹_q

(2q+ 1)¹^q

kf⁰⁰k[x,a+b−x],p,

and

Z b a+b−x

(t−b)²|f⁰⁰(t)|dt ≤ Z b

a+b−x

(b−t)^2q dt ¹q

kf⁰⁰k[a+b−x,b],p

= (x−a)²⁺¹^q (2q+ 1)¹^q

kf⁰⁰k[a+b−x,b],p.

Summing the above inequalities, we deduce the first bound in (2.5).

The last part may be proved in a similar fashion to the one in Theorem 2.2, and we omit the details.

Remark 2. If in (2.5) we chooseα = q, β = p, ¹_p + ¹_q = 1,p > 1, then we get the

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inequality

(2.6) M(x)≤ 2¹^q 2 (2q+ 1)¹^q





x−a b−a

2q+1

+

a+b 2 −x b−a

!2q+1



1 q

×(b−a)¹⁺¹^q kf⁰⁰k_[a,b],p, for anyx∈

a,^a+b₂ .

Remark 3. If in Theorem2.4we choosex=a, then we recapture the second part of the inequality (1.4), i.e.,

(2.7)

1 b−a

Z b a

f(t) dt− 1

2[f(a) +f(b)] + b−a

8 [f⁰(b)−f⁰(a)]

≤ 1

8· (b−a)¹⁺¹^qkf⁰⁰k[a,b],p

(2q+ 1)¹^q .

The constant ¹₈ is best possible in the sense that it cannot be replaced by a smaller constant.

Proof. Indeed, if we assume that (2.7) holds with a constantC > 0, instead of ¹₈, i.e.,

(2.8)

1 b−a

Z b a

f(t)dt− 1

2[f(a) +f(b)] + b−a

8 [f⁰(b)−f⁰(a)]

≤C· (b−a)¹⁺¹^qkf⁰⁰k_[a,b],p (2q+ 1)¹^q

,

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then for the functionf : [a, b]→R,f(x) = k x− ^a+b₂ 2

,k > 0, we have f(a) +f(b)

2 =k·(b−a)² 4 , f⁰(b)−f⁰(a) = 2k(b−a), 1

b−a Z b

a

f(t)dt =k· (b−a)² 12 , kf⁰⁰k_[a,b],p= 2k(b−a)¹^p; and by (2.8) we deduce

k(b−a)²

12 −k(b−a)²

4 + k(b−a)² 4

≤ 2C·k(b−a)² (2q+ 1)¹^q

,

givingC ≥ ^(2q+1)

1q

24 . Lettingq → 1+, we deduceC ≥ ¹₈, and the sharpness of the constant is proved.

Remark 4. If in Theorem2.4we choosex= ^a+b₂ , then we get the midpoint inequality

1 b−a

Z b a

f(t) dt−f

a+b 2

(2.9)

≤ 1

8 · (b−a)¹⁺¹^q 2¹^q (2q+ 1)¹^q

hkf⁰⁰k[â,â+b2 ]^,p+kf⁰⁰k[â+b2 ,b]^,p i

≤ 1

8 ·(b−a)¹⁺¹^q (2q+ 1)¹^q

kf⁰⁰k_[a,b],p, p >1,1 p+ 1

q = 1.

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In both inequalities the constant ¹₈ is sharp in the sense that it cannot be replaced by a smaller constant.

To show this fact, assume that (2.9) holds withC, D > 0, i.e.,

1 b−a

Z b a

f(t)dt−f

a+b 2

(2.10)

≤C· (b−a)¹⁺¹^q 2¹^q (2q+ 1)¹^q

hkf⁰⁰k[â,â+b2 ]^,p+kf⁰⁰k[â+b2 ,b]^,p i

≤D· (b−a)¹⁺¹^q (2q+ 1)¹^q

kf⁰⁰k_[a,b],p.

For the functionf : [a, b]→R,f(x) = k x− ^a+b₂ 2

,k >0, we have f

a+b 2

= 0, 1

b−a Z b

a

f(t)dt = k(b−a)² 12 ,

kf⁰⁰k[â,â+b2 ]^,p+kf⁰⁰k[â+b2 ,b]^,p= 4k

b−a 2

¹_p

= 2¹⁺¹^q (b−a)¹^pk, kf⁰⁰k_[a,b],p= 2 (b−a)¹^pk;

and then by (2.10) we deduce k(b−a)²

12 ≤C·2k(b−a)² (2q+ 1)¹^q

≤D· 2k(b−a)² (2q+ 1)¹^q

,

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givingC, D ≥ ^(2q+1)

1 q

24 for anyq >1. Lettingq →1+, we deduceC, D ≥ ¹₈ and the sharpness of the constants in (2.9) is proved.

The following result is useful in providing the best quadrature rule in the class for approximating the integral of a functionf : [a, b] → Rwhose first derivative is absolutely continuous on[a, b]and whose second derivative is inL_p[a, b].

Corollary 2.5. With the assumptions in Theorem2.4, one has the inequality

(2.11) 1 2

f

3a+b 4

+f

a+ 3b 4

− 1 b−a

Z b a

f(t) dt

≤ 1

32· (b−a)¹⁺¹^q (2q+ 1)¹^q

kf⁰⁰k_[a,b],p,

where ¹_p + ¹_q = 1.

The constant ₃₂¹ is the best possible in the sense that it cannot be replaced by a smaller constant.

Proof. The inequality follows by Theorem2.4and (2.6) on choosingx= ^3a+b₄ . To prove the sharpness of the constant, assume that (2.11) holds with a constant E >0, i.e.,

(2.12) 1 2

f

3a+b 4

+f

a+ 3b 4

− 1 b−a

Z b a

f(t) dt

≤E· (b−a)¹⁺¹^q (2q+ 1)¹^q

kf⁰⁰k_[a,b],p.

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Consider the functionf : [a, b]→R,

f(x) =











−¹₂ x− ^3a+b₄ 2

if x∈

a,^3a+b₄ ,

1

2 x− ^3a+b₄ 2

if x∈ ^3a+b₄ ,^a+b₂ ,

−¹₂ x− ^a+3b₄ 2

if x∈ ^a+b₂ ,^a+3b₄ ,

1

2 x− ^a+3b₄ 2

if x∈ ^a+3b₄ , b . We have

f⁰(x) =







x− ^3a+b₄

if x∈ a,^a+b₂

, x− ^a+3b₄

if x∈ ^a+b₂ , b .

Thenf⁰is absolutely continuous andf⁰⁰ ∈L_p[a, b],p > 1. We also have 1

2

f

3a+b 4

+f

a+ 3b 4

= 0, 1

b−a Z b

a

f(t) dt= (b−a)² 96 , kf⁰⁰k_[a,b],p = (b−a)¹^p, and then, by (2.12), we obtain

(b−a)²

96 ≤E· (b−a)² (2q+ 1)¹^q

,

givingE ≥ ^(2q+1)

1q

96 for anyq >1, i.e.,E ≥ ₃₂¹, and the corollary is proved.

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Theorem 2.6. Letf : [a, b]→Rbe such that the derivativef⁰ is absolutely contin- uous on [a, b] andf⁰⁰ ∈ L₁[a, b]. If M(x) is as defined in (2.2), then we have the bounds:

M(x)≤ b−a 2

"

x−a b−a

2

kf⁰⁰k_[a,x],1 (2.13)

+

a+b 2 −x b−a

!2

kf⁰⁰k[x,a+b−x],1+

x−a b−a

2

kf⁰⁰k[a+b−x,b],1





≤











b−a 2

2 ^x−a_b−a2

+a+b 2 −x b−a

²

×max

kf⁰⁰k_[a,x],1,kf⁰⁰k[x,a+b−x],1,kf⁰⁰k[a+b−x,b],1

;

b−a 2

2 ^x−a_b−a2α

+ a+b

2 −x b−a

2α_α¹

×h

kf⁰⁰k^β_[a,x],1+kf⁰⁰k^β[x,a+b−x],1+kf⁰⁰k^β[a+b−x,b],1

i_β¹ if α >1,_α¹ +_β¹ = 1;

b−a 2

h

|^x−

3a+b 4

b−a |+ ¹₄ i2

kf⁰⁰k[a,b],1; for anyx∈

a,^a+b₂ .

The proof is as in Theorem2.2 and we need only to prove the third inequality of the last part as

M(x)≤ b−a 2 max







x−a b−a

2

,

a+b 2 −x b−a

!2





×

kf⁰⁰k_[a,x],1+kf⁰⁰k[x,a+b−x],1+kf⁰⁰k[a+b−x,b],1

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= b−a 2

"

x−^3a+b₄ b−a

+ 1 4

#2

kf⁰⁰k_[a,b],1.

Remark 5. By the use of Theorem2.6, forx =a, we recapture the third part of the inequality (1.4), i.e.,

1 b−a

Z b a

f(t) dt− 1

2[f(a) +f(b)] + b−a

8 [f⁰(b)−f⁰(a)]

≤ 1

8(b−a)kf⁰⁰k_[a,b],1. If in (2.13) we choosex= ^a+b₂ , then we get the mid-point inequality

1 b−a

Z b a

f(t) dt−f

a+b 2

≤ 1

8(b−a)kf⁰⁰k_[a,b],1. Corollary 2.7. With the assumptions in Theorem2.6, one has the inequality

1 b−a

Z b a

f(t) dt− f ^3a+b₄

+f ^a+3b₄ 2

≤ 1

32(b−a)kf⁰⁰k_[a,b],1.

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3. A Composite Quadrature Formula

We use the following inequalities obtained in the previous section:

(3.1) 1 2

f

3a+b 4

+f

a+ 3b 4

− 1 b−a

Z b a

f(t) dt

≤











1

96(b−a)²kf⁰⁰k[a,b],∞ if f⁰⁰∈L∞[a, b] ;

1

32 ·^(b−a)^{1+ 1}^q

(2q+1)

1

q kf⁰⁰k_[a,b],p if f⁰⁰∈L_p[a, b], p >1, ¹_p +¹_q = 1;

1

32(b−a)kf⁰⁰k_[a,b],1 if f⁰⁰∈L₁[a, b].

LetI_n :a=x₀ < x₁ <· · ·< xn−1 < x_n =b be a division of the interval[a, b]and h_i :=x_i+1−x_i (i= 0, . . . , n−1)andν(I_n) := max{h_i|i= 0, . . . , n−1}.

Consider the composite quadrature rule (3.2) Q_n(I_n, f) := 1

2

n−1

X

i=0

f

3x_i+x_i+1 4

+f

x_i+ 3x_i+1 4

h_i.

The following result holds.

Theorem 3.1. Letf : [a, b]→Rbe such that the derivativef⁰ is absolutely contin- uous on[a, b]. Then we have

Z b a

f(t) dt=Q_n(I_n, f) +R_n(I_n, f),

where Q_n(I_n, f) is defined by the formula (3.2), and the remainder satisfies the

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estimates

(3.3) |R_n(I_n, f)| ≤











1

96kf⁰⁰k[a,b],∞

n−1

P

i=0

h³_i if f⁰⁰ ∈L∞[a, b] ;

1 32(2q+1)¹^q

kf⁰⁰k_[a,b],p _n−1

P

i=0

h^2q+1_i ¹_q

if f⁰⁰ ∈L_p[a, b], p > 1, ¹_p + ¹_q = 1;

1

32kf⁰⁰k_[a,b],1[ν(I_n)]² if f⁰⁰ ∈L₁[a, b]. Proof. Applying inequality (3.1) on the interval[x_i, x_i+1], we may state that (3.4)

Z xi+1

xi

f(t) dt− 1 2

f

3x_i+x_i+1 4

+f

x_i+ 3x_i+1 4

h_i

≤











1

96h³_ikf⁰⁰k_[x_i_,x_i+1],∞;

1 32(2q+1)¹^q

h²⁺

1 q

i kf⁰⁰k_[x_i_,x_i+1_],p, p >1, ¹_p +¹_q = 1;

1

32h²_ikf⁰⁰k_[x_i_,x_i+1_],1; for eachi∈ {0, . . . , n−1}.

Summing the inequality (3.4) over ifrom 0to n−1and using the generalized triangle inequality, we get

(3.5) |Rn(In, f)| ≤











1 96

Pn−1

i=0 h³_ikf⁰⁰k_[x_i_,x_i+1_],∞;

1 32(2q+1)

1 q

Pn−1 i=0 h²⁺

1 q

i kf⁰⁰k_[x_i_,x_i+1_],p, p > 1, ¹_p + ¹_q = 1;

1 32

Pn−1

i=0 h²_ikf⁰⁰k_[x_i_,x_i+1_],1.

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Now, we observe that

n−1

X

i=0

h³_ikf⁰⁰k_[x_i_,x_i+1_],∞≤ kf⁰⁰k_[a,b],∞

n−1

X

i=0

h³_i.

Using Hölder’s discrete inequality, we may write that

n−1

X

i=0

h²⁺

1 q

i kf⁰⁰k_[x_i_,x_i+1_],p≤

n−1

X

i=0

h(²⁺¹q)^q

i

!¹_q _n−1 X

i=0

kf⁰⁰k^p_[x

i,xi+1],p

!_p¹

=

n−1

X

i=0

h^2q+1_i

!¹_q _n−1 X

i=0

Z xi+1

xi

|f⁰⁰(t)|^pdt

!¹_p

=

n−1

X

i=0

h^2q+1_i

!¹_q

kf⁰⁰k_[a,b],p.

Also, we note that

n−1

X

i=0

h²_ikf⁰⁰k_[x_i_,x_i+1_],1 ≤ max

0≤i≤n−1

h²_i

n−1

X

i=0

kf⁰⁰k_[x_i_,x_i+1_],1

= [ν(I_n)]²kf⁰⁰k_[a,b],1.

Consequently, by the use of (3.5), we deduce the desired result (3.3).

For the particular case where the divisionI_nis equidistant, i.e., I_n:=x_i =a+i· b−a

n , i= 0, . . . , n,

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we may consider the quadrature rule:

(3.6) Q_n(f) := b−a 2n

n−1

X

i=0

f

a+

4i+ 1 4n

(b−a)

+f

a+

4i+ 3 4n

(b−a)

.

The following corollary will be more useful in practice.

Corollary 3.2. With the assumption of Theorem3.1, we have

Z b a

f(t) dt=Q_n(f) +R_n(f),

whereQ_n(f)is defined by (3.6) and the remainderR_n(f)satisfies the estimate:

|R_n(I_n, f)| ≤











1

96kf⁰⁰k[a,b],∞(b−a)³ n² ;

1 32(2q+1)

1

qkf⁰⁰k_[a,b],p^(b−a)_n2^{2+ 1}^q, p >1, ¹_p +¹_q = 1;

1

32kf⁰⁰k_[a,b],1^(b−a)_n2 ².

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References

[1] A. GUESSAB AND G. SCHMEISSER, Sharp integral inequalities of the Hermite-Hadamard type, J. Approx. Theory, 115 (2002), 260–288.

[2] S. S. DRAGOMIR, Some companions of Ostrowski’s inequality for absolutely continuous functions and applications, Bull. Korean Math. Soc., 42(2) (2005), 213–230.

[3] Lj. DEDI Ć, M.MATI Ć AND J.PE ˇCARI Ć, On generalizations of Ostrowski in- equality via some Euler-type identities, Math. Inequal. Appl., 3(3) (2000), 337–

353.

[4] P. CERONE AND S.S. DRAGOMIR, Trapezoidal type rules from an inequal- ities point of view, Handbook of Analytic-Computational Methods in Applied Mathematics, CRC Press N.Y. (2000), 65–134.