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In this paper, we establish general form of an inequality of Ostrowski type for twice differentiable mappings in terms ofLp−norm, with first derivative absolutely continuous

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http://jipam.vu.edu.au/

Volume 7, Issue 3, Article 112, 2006

AN OSTROWSKI TYPE INEQUALITY FOR p−NORMS

A. RAFIQ AND NAZIR AHMAD MIR

CENTER FORADVANCEDSTUDIES INPURE ANDAPPLIEDMATHEMATICS

BAHAUDDINZAKARIYAUNIVERSITY

MULTAN, PAKISTAN

[email protected]

Received 22 February, 2005; accepted 04 May, 2005 Communicated by A. Sofo

ABSTRACT. In this paper, we establish general form of an inequality of Ostrowski type for twice differentiable mappings in terms ofLp−norm, with first derivative absolutely continuous.

The integral inequality of similar type already pointed out in literature is a special case of ours.

The already established inequality contains a mistake and as a result incorrect consequences and applications. The corrected version of the inequality is pointed out and the inequality is also applied to special means and numerical integration.

Key words and phrases: Ostrowski inequality, Numerical integration, Special means.

2000 Mathematics Subject Classification. 26D15.

1. INTRODUCTION

We establish here the general form of an inequality of Ostrowski type, different to that of Cerone, Dragomir and Roumeliotis [1], for twice differentiable mappings in terms ofLp−norm.

The integral inequality of similar type already pointed out by N.S. Barnett, P. Cerone, S.S.

Dragomir, J. Roumeliotis and A. Sofo [2], contains a mistake which has already been reported by N.A. Mir and A. Rafiq in their research work [3]. The same mistake has been carried out in their other research article, namely Theorem 20 of [2] and as a result incorrect consequences and applications of this theorem. The corrected form of the theorem is as follows:

Theorem 1.1. Letg : [a, b]−→Rbe a mapping whose first derivative is absolutely continuous on[a, b].If we assume that the second derivativeg00 ∈ Lp(a, b),1< p < ∞,then we have the

ISSN (electronic): 1443-5756

c 2006 Victoria University. All rights reserved.

050-05

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inequality

(1.1)

Z b a

g(t)dt− 1 2

g(x) + g(a) +g(b) 2

(b−a) + 1

2(b−a)

x− a+b 2

g0(x)

≤ 1 2

b−a 2

2+1q

kg00kp

× (

[B(q+ 1, q+ 1) +Bx1(q+ 1, q+ 1) + Ψx2(q+ 1, q+ 1)]1q forx∈ a,a+b2

, [B(q+ 1, q+ 1) +Bx3(q+ 1, q+ 1) +Bx4(q+ 1, q+ 1)]1q forx∈(a+b2 , b], where 1p +1q = 1, p >1, q > 1,andB(·,·)is the Beta function of Euler given by

B(l, s) = Z 1

0

tl−1(1−t)s−1dt, l, s > 0.

Further

Br(l, s) = Z r

0

tl−1(1−t)s−1dt is the incomplete Beta function,

Ψr(l, s) = Z r

0

tl−1(1 +t)s−1dt is the real positive valued integral,

x1 = 2(x−a)

b−a , x2 = 1−x1, x3 =x1−1, x4 = 2−x1 and

kg00kp :=

Z b a

|g00(t)|pdt 1p

. If we assume thatg00∈L1(a, b),then we have

(1.2)

Z b a

g(t)dt− 1 2

g(x) + g(a) +g(b) 2

(b−a) + 1

2(b−a)

x− a+b 2

g0(x)

≤ kg00k1

8 (b−a)2, where

kg00k1 :=

Z b a

|g00(t)|dt.

2. MAINRESULTS

The following theorem is now proved and subsequently applied to numerical integration and special means.

Theorem 2.1. Letg : [a, b]−→Rbe a mapping whose first derivative is absolutely continuous on[a, b]. If we assume that the second derivativeg00 ∈ Lp(a, b),1 < p <∞,then we have the

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inequality (2.1)

1 α+β

α x−a

Z x a

g(t)dt+ β b−x

Z b x

g(t)dt

− 1

2g(x)− 1 2(α+β)

x−a+b 2

g(x)

α

x−a − β b−x

+ (b−a) 2

α

x−ag(a) + β b−xg(b)

−(α+β)

x−a+b 2

g0(x)

b−a 2

2+1q

kg00kp



















 h β

α+β 1 b−x

q

B(q+ 1, q+ 1) + α

α+β 1 x−a

q

Bx1(q+ 1, q+ 1) +

β α+β

1 b−x

q

Ψx2(q+ 1, q+ 1)i1q

forx∈ a,a+b2

, h α

α+β 1 x−a

q

B(q+ 1, q+ 1) +

α α+β

1 x−a

q

Bx3(q+ 1, q+ 1) +

β α+β

1 b−x

q

Bx4(q+ 1, q+ 1)i1q

forx∈ a+b2 , b , where 1p +1q = 1, p >1, q > 1,andB(·,·)is the Beta function of Euler given by

B(l, s) = Z 1

0

tl−1(1−t)s−1dt, l, s >0.

Further,

Br(l, s) = Z r

0

tl−1(1−t)s−1dt is the incomplete Beta function,

Ψr(l, s) = Z r

0

tl−1(1 +t)s−1dt is a real positive valued integral,

x1 = 2(x−a)

b−a , x2 = 1−x1, x3 =x1−1, x4 = 2−x1 and

kg00kp :=

Z b a

|g00(t)|pdt

1 p

. If we assume thatg00∈L1(a, b),then we have

(2.2)

1 α+β

α x−a

Z x a

g(t)dt+ β b−x

Z b x

g(t)dt

− 1 2g(x)

− 1 2(α+β)

x− a+b 2

g(x)

α

x−a − β b−x

+(b−a) 2

α

x−ag(a) + β b−xg(b)

−(α+β)

x− a+b 2

g0(x)

≤ 1

2kg00k1kK(x, t)k, where

kg00k1 = Z b

a

|g00(t)|dt,

(4)

and

kK(x, t)k = 1

α+β max α

x−a, β b−x

(b−a)2

4 forx∈[a, b].

Proof. We begin by recalling the following integral equality proved by N.A. Mir and A. Rafiq [3] which is generalization of an integral equality proved by Dragomir and Wang [4].

(2.3)

1 α+β

α x−a

Z x a

g(t)dt+ β b−x

Z b x

g(t)dt

− 1 2g(x)

− 1 2(α+β)

x− a+b 2

g(x)

α

x−a − β b−x

+(b−a) 2

α

x−ag(a) + β b−xg(b)

−(α+β)

x− a+b 2

g0(x)

= 1 2

Z b a

p(x, t)

t− a+b 2

g00(t)dt whose left hand side is equivalent to that of (2.1). From the right hand side of (2.3) we have, by Hölder’s inequality, that

Z b a

p(x, t)

t−a+b 2

g00(t)dt

≤ Z b

a

|g00(t)|pdt

1pZ b a

|p(x, t)|q

t− a+b 2

q

dt 1q

=kg00kp Z b

a

|p(x, t)|q

t− a+b 2

q

dt

1 q

, and from (2.3) we obtain the inequality

(2.4)

1 α+β

α x−a

Z x a

g(t)dt+ β b−x

Z b x

g(t)dt

− 1

2g(x)− 1 2(α+β)

x−a+b 2

g(x)

α

x−a − β b−x

+ (b−a) 2

α

x−ag(a) + β b−xg(b)

−(α+β)

x−a+b 2

g0(x)

≤ 1 2kg00kp

Z b a

|p(x, t)|q

t− a+b 2

q

dt 1q

. From the right hand side of (2.4) we may define

I :=

Z b a

|p(x, t)|q

t−a+b 2

q

dt

= α

α+β · 1 x−a

qZ x a

(t−a)q

t− a+b 2

q

dt +

β

α+β · 1 b−x

qZ b x

|t−b|q

t− a+b 2

q

dt (2.5)

such that we can identify two distinct cases.

(5)

(a) Forx∈ a,a+b2 IA=

α α+β

1 x−a

qZ x a

(t−a)q

a+b 2 −t

q

dt +

β α+β

1 b−x

qZ a+b2

x

(b−t)q

a+b 2 −t

q

dt +

β α+β

1 b−x

qZ b

a+b 2

(b−t)q

t− a+b 2

q

dt.

Investigating the three separate integrals, we may evaluate as follows:

I1 = Z x

a

(t−a)q

a+b 2 −t

q

dt, making the change of variablet=a+ b−a2

w,we arrive at I1 =

b−a 2

2q+1Z x1

0

wq(1−w)qdw,

=

b−a 2

2q+1

Bx1(q+ 1, q+ 1), whereBx1(·,·)is the incomplete Beta function andx1 = 2(x−a)b−a .

I2 = Z a+b2

x

(b−t)q

a+b 2 −t

q

dt, making the change of variablet= a+b2b−a2

w,we obtain I2 =

b−a 2

2q+1Z x2

0

wq(1 +w)qdw=

b−a 2

2q+1

Ψx2(q+ 1, q+ 1), where

Ψx2 :=

Z x2

0

wq(1 +w)qdw andx2 = a+b−2xb−a = 1−x1.

I3 = Z b

a+b 2

(b−t)q

t− a+b 2

q

dt, making the change of variablet= a+b2 + b−a2

w,we get I3 =

b−a 2

2q+1Z 1 0

wq(1−w)qdw =

b−a 2

2q+1

B(q+ 1, q+ 1), whereB(·,·)is the Beta function.

We may now write IA=I1+I2+I3

=

b−a 2

2q+1 α α+β

1 x−a

q

Bx1(q+ 1, q+ 1) +

β α+β

1 b−x

q

Ψx2(q+ 1, q+ 1) + β

α+β 1 b−x

q

B(q+ 1, q+ 1)

(6)

forx∈ a,a+b2

.

(b) Forx∈ a,a+b2 IB =

α α+β

1 x−a

qZ a+b2

a

(t−a)q

a+b 2 −t

q

dt +

α α+β

1 x−a

qZ x

a+b 2

(t−a)q

t− a+b 2

q

dt +

β α+β

1 b−x

qZ b x

(b−t)q

t− a+b 2

q

dt.

In a similar fashion to the previous case, we have I4 =

Z a+b2

a

(t−a)q

a+b 2 −t

q

dt.

Lettingt =a+ b−a2

w,we obtain I4 =

b−a 2

2q+1Z 1 0

wq(1−w)qdw =

b−a 2

2q+1

B(q+ 1, q+ 1), whereB(·,·)is the Beta function.

I5 = Z x

a+b 2

(t−a)q

t− a+b 2

q

dt, making the change of variablet= a+b2 + b−a2

w,we arrive at I5 =

b−a 2

2q+1Z x3

0

wq(1−w)qdw=

b−a 2

2q+1

Bx3(q+ 1, q+ 1), whereBx3(·,·)is the incomplete Beta function andx3 =x1−1.

I6 = Z b

x

(b−t)q

t− a+b 2

q

dt, making the change of variablet=b− b−a2

w,we get I6 =

b−a 2

2q+1Z x4

0

wq(1−w)qdw=

b−a 2

2q+1

Bx4(q+ 1, q+ 1), whereBx4(·,·)is the incomplete Beta function andx4 = 2−x1.

IB =I4+I5+I6

=

b−a 2

2q+1 α α+β

1 x−a

q

B(q+ 1, q+ 1) + α

α+β 1 x−a

q

Bx3(q+ 1, q+ 1) +

β α+β

1 b−x

q

Bx4(q+ 1, q+ 1)

forx∈(a+b2 , b].

(7)

Also from (2.5) I =IA+IB

=

b−a 2

2q+1

















α

α+β 1 x−a

q

Bx1(q+ 1, q+ 1) +

β α+β

1 b−x

q

Ψx2(q+ 1, q+ 1) +

β α+β

1 b−x

q

B(q+ 1, q+ 1)forx∈

a,a+b2 , α

α+β 1 x−a

q

B(q+ 1, q+ 1) + α

α+β 1 x−a

q

Bx3(q+ 1, q+ 1) +

β α+β

1 b−x

q

Bx4(q+ 1, q+ 1)forx∈a+b

2 , b . Using (2.4), we obtain the result (2.1). Using the inequality (2.3), we can also state that

1 α+β

α x−a

Z x a

g(t)dt+ β b−x

Z b x

g(t)dt

− 1

2g(x)− 1 2(α+β)

x−a+b 2

g(x)

α

x−a − β b−x

+ (b−a) 2

α

x−ag(a) + β b−xg(b)

−(α+β)

x−a+b 2

g0(x)

≤ 1

2kg00k1kK(x, t)k, where

kK(x, t)k=p(x, t)

t− a+b 2

. As it is easy to see that

kK(x, t)k= 1

α+β ·max α

x−a, β b−x

· (b−a)2

4 forx∈[a, b],

we deduce (2.2).

Remark 2.2. Puttingα =x−aandβ =b−xin (2.1) and (2.2), we get the inequalities (1.1) and (1.2).

Remark 2.3. Simple manipulation of (2.1) will allow for the corrected result of (1.1) and (1.2), owing to a missing factor of12 in the third term of the original result (1.1) of the Barnett, Cerone, Dragomir, Roumeliotis and Sofo, this will not be done here.

REFERENCES

[1] P. CERONE, S.S. DRAGOMIRANDJ. ROUMELIOTIS, An inequality of Ostrowski-Grüss type for twice differentiable mappings and applicatios in numerical integration, Kyungpook Mathematical Journal, 39(2) (1999), 331–341.

[2] N.S. BARNETT, P. CERONE, S.S. DRAGOMIR, J. ROUMELIOTISANDA. SOFO, A survey on Ostrowski type inequalities for twice differentiable mappings and applications, Inequality Theory and Applications, 1 (2001), 33–86.

[3] N.A. MIRANDA. RAFIQ, An integral inequality for twice differentiable bounded mappings with first derivative absolutely continuous and applications, submitted.

[4] S.S. DRAGOMIRANDS. WANG, Applications of Ostrowski’s inequality for the estimation of error bounds for some special means and some numerical quadrature rules, Appl. Math. Lett., 11 (1998), 105–109.

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