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volume 7, issue 4, article 151, 2006.

Received 22 August, 2006;

accepted 28 August, 2006.

Communicated by:W.-S. Cheung

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Journal of Inequalities in Pure and Applied Mathematics

ON AN INEQUALITY OF OSTROWSKI TYPE

JOSIP PE ˇCARI ´C AND SIME UNGAR

Faculty of Textile Technology University of Zagreb Pierottijeva 6

10000 Zagreb, Croatia EMail:pecaric@hazu.hr Department of Mathematics University of Zagreb Pierottijeva 6

10000 Zagreb, Croatia EMail:ungar@math.hr

c

2000Victoria University ISSN (electronic): 1443-5756 225-06

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On an Inequality of Ostrowski type

Josip Peˇcari´c and Sime Ungar

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J. Ineq. Pure and Appl. Math. 7(4) Art. 151, 2006

http://jipam.vu.edu.au

Abstract

We prove an inequality of Ostrowski type forp-norm, generalizing a result of Dragomir [1].

2000 Mathematics Subject Classification:26D15, 26D10.

Key words: Ostrowski’s inequality,p-norm.

1. Introduction

For a differentiable function f: [a, b] → R, a · b > 0, Dragomir has in [1]

proved, using Pompeiu’s mean value theorem [3], the following Ostrowski type inequality:

(1.1)

a+b

2 · f(x)

x − 1

b−a Z b

a

f(t) dt

≤D(x)· kf −ι f⁰k∞,

whereι(t) = t,t∈[a, b], and

(1.2) D(x) = b−a

|x|



 1

4 + x− ^a+b₂ b−a

!2

.

We are going to prove a general estimate with the p-norm, 1 ≤ p ≤ ∞, which will forp=∞give the Dragomir result.

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2. The Main Result

Theorem 2.1. Let the functionf: [a, b]→Rbe continuous on[a, b]and differ- entiable on(a, b)with0< a < b. Then for¹_p +¹_q = 1, with1≤p, q ≤ ∞, and allx∈[a, b], the following inequality holds:

(2.1)

a+b

2 ·f(x)

x − 1

b−a Z b

a

f(t) dt

≤P U(x, p)· kf−ι f⁰kp

whereι(t) = t,t∈[a, b], and

(2.2) P U(x, p) = (b−a)¹^p⁻¹·

"

a^2−q−x^2−q

(1−2q)(2−q)+ x^2−q−a^1+qx^1−2q (1−2q)(1 +q)

¹_q

+

b^2−q−x^2−q

(1−2q)(2−q) +x^2−q−b^1+qx^1−2q (1−2q)(1 +q)

¹_q# .

Note that in cases(p, q) = (1,∞),(∞,1), and(2,2)the constant P U(x, p) has to be taken as the limit asp→1,∞, and2, respectively.

Proof. Define F: [1/b,1/a] → R byF(t) := t f(¹_t). The function F is continuous and is differentiable on (1/b,1/a), and for all x1, x2 ∈ [1/b,1/a]we have

F(x₁)−F(x₂) = Z x1

x2

F⁰(t) dt

= Z x1

x2

f 1

t

− 1 t f⁰

1 t

dt

u:= 1

t (2.3)

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=− Z 1/x1

1/x2

(f(u)−uf⁰(u)) 1 u² du . (2.4)

Denotex₁ =: 1/xandx₂ =: 1/t. Then for allx, t∈[a, b]from(2.4)we get 1

xf(x)− 1

t f(t) = Z t

x

(f(u)−uf⁰(u)) 1 u² du, i.e.,

(2.5) tf(x)−xf(t) =x t Z t

x

(f(u)−uf⁰(u)) 1 u² du . Integrating ontand dividing byx, we obtain

b²−a²

2 · f(x)

x −

Z b a

f(t) dt= Z b

a

t Z t

x

(f(u)−uf⁰(u)) 1 u² du

dt ,

and therefore

b²−a²

2 · f(x)

x −

Z b a

f(t) dt (2.6)

≤ Z b

a

Z t x

(f(u)−uf⁰(u)) t u²

du

dt

= Z x

a

Z x t

(f(u)−uf⁰(u)) t u²

du

dt

+ Z b

x

Z t x

(f(u)−uf⁰(u)) t u²

du

dt .

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First, consider the case 1 < p, q < ∞. Applying Hölder’s inequality, the sum in the last line of (2.5) is

≤ Z x

a

Z x t

|f(u)−uf⁰(u)|^pdu

dt ¹_p

· Z x

a

Z x t

t^q u^2qdu

dt

¹_q (2.7)

+ Z b

x

Z t x

|f(u)−uf⁰(u)|^pdu

dt ¹p

· Z b

x

Z t x

t^q u^2qdu

dt

¹q

≤ Z b

a

Z b a

|f(u)−uf⁰(u)|^p du

dt ¹p

×

"

Z x a

Z x t

t^q u^2q du

dt

¹_q +

Z b x

Z t x

t^q u^2q du

dt

¹q# .

The first factor in(2.7)equals (2.8)

Z b a

|f(u)−uf⁰(u)|^p du

dt ¹_p

= (b−a)¹^pkf −ι f⁰k_p,

and for the second factor, forp, q 6= 2, we get (2.9)

Z x a

Z x t

t^q u^2qdu

dt

¹_q +

Z b x

Z t x

t^q u^2q du

dt

¹_q

=

a^2−q−x^2−q

(1−2q)(2−q) +x^2−q−a^1+qx^1−2q (1−2q)(1 +q)

¹_q

+

b^2−q−x^2−q

(1−2q)(2−q)+ x^2−q−b^1+qx^1−2q (1−2q)(1 +q)

¹_q .

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Putting (2.8) and (2.9) into (2.7) and dividing (2.6) and (2.7) by (b −a) gives the required inequality(2.1)in the case1< p, q < ∞,p, q 6= 2.

Forp=q = 2, instead of(2.9)we obtain

(2.10)

Z x a

Z x t

t² u⁴ du

dt

¹₂ +

Z b x

Z t x

t² u⁴du

dt

¹₂

= 1 3

lnx

a 3

+ a³ x³ −1

¹₂ +

lnx

b 3

+ b³ x³ −1

¹₂!

which is easily shown to be equal to the limit of the right hand side in(2.9)for q →2, i. e.

(2.11) lim

p→2P U(x, p)

= 1

3(b−a)¹²

lnx a

3

+ a³ x³ −1

¹₂ +

lnx

b 3

+ b³ x³ −1

¹₂! .

Now consider the casep=∞,q= 1. The last line in(2.6)is

≤ sup

a≤u≤b

|f(u)−u f⁰(u)|

(2.12)

× Z x

a

Z x t

t u² du

dt+

Z b x

Z t x

t u² du

dt

=kf−ι f⁰k∞·

a²+b²

2x +x−a−b

.

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Putting(2.12)into(2.6)and dividing by(b−a)gives

a+b

2 · f(x)

x − 1

b−a Z b

a

f(t) dt

≤ 1 b−a

a²+b²

2x +x−a−b

· kf−ι f⁰k∞

which reproves the Dragomir’s result [1].

It is easy to see that 1 b−a

a²+b²

2x +x−a−b

= lim

p→∞P U(x, p)

proving(2.1)in the casep=∞,q = 1.

Finally consider the casep= 1,q =∞. In this case the last line of(2.6)is

≤ Z x

a

Z x t

|f(u)−uf⁰(u)| max

t≤u≤x a≤t≤x

t u² du

(2.13) dt

+ Z b

x

Z t x

|f(u)−uf⁰(u)| max

x≤u≤t x≤t≤b

t u² du

dt

≤ Z b

a

Z b a

|f(u)−u f⁰(u)| dudt· 1

a + b x²

= (b−a) 1

a + b x²

· kf−ι f⁰k₁.

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Appending(2.13)to(2.6)and dividing by(b−a)gives (2.14)

a+b

2 · f(x)

x − 1

b−a Z b

a

f(t) dt

≤ 1

a + b x²

· kf −ι f⁰k₁.

It is not too difficult to show that

(2.15) lim

p→1P U(x, p) = 1 a + b

x²,

so(2.14)proves formula(2.1)forp= 1,q =∞, proving the theorem.

Remark 1. We have considered only the positive case,0 < a < b. In the case a < b < 0,a similar but more cumbersome formula holds, where mosta’s,b’s andx’s have to be replaced by their absolute values.

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3. The Weighted Case

In the weighted case we have the following result:

Theorem 3.1. Let the function f: [a, b] → R be continuous on[a, b] and dif- ferentiable on (a, b)with0 < a < b, and letw: [a, b] → R be a nonnegative integrable function. Then for ¹_p +¹_q = 1, with1≤p, q ≤ ∞, and allx∈ [a, b], the following inequality holds:

(3.1)

f(x) x

Z b a

t w(t) dt− Z b

a

f(t)w(t) dt

≤ kf−ιf⁰k_p· (b−a)¹^p (1−2q)¹^q

·

"

x^1−2q

Z x a

t^q (w(t))^q dt− Z x

a

t^1−q (w(t))^q dt ¹_q

+ Z b

x

t^1−q(w(t))^q dt−x^1−2q Z b

x

t^q(w(t))^q dt ¹q#

.

Proof. Multiplying(2.5)byw(t)/xand integrating ont, we get f(x)

x Z b

a

t w(t) dt− Z b

a

f(t)w(t) dt

= Z b

a

t w(t) Z t

x

(f(u)−u f⁰(u)) 1 u² du

dt ,

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and as in the proof of Theorem2.1we have

f(x) x

Z b a

t w(t) dt− Z b

a

f(t)w(t) dt

≤ Z x

a

Z x t

(f(u)−uf⁰(u))t w(t) u²

du

dt

+ Z b

x

Z t x

(f(u)−uf⁰(u))t w(t) u²

du

dt

≤ Z x

a

Z x t

|f(u)−u f⁰(u)|^p du

dt ¹_p

· Z x

a

Z x t

t^q (w(t))^q u^2q du

dt

¹_q

+ Z b

x

Z t x

dt

1 p

· Z b

x

Z t x

dt

1 q

≤ Z b

a

Z b a

dt ¹_p

·

"

Z x a

Z x t

dt

¹_q +

Z b x

Z t x

dt

¹_q#

which gives(3.1).

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References

[1] S.S. DRAGOMIR, An inequality of Ostrowski type via Pompeiu’s mean value theorem, J. of Inequal. in Pure and Appl. Math., 6(3) (2005), Art. 83. [ONLINE: http://jipam.vu.edu.au/article.php?

sid=532].

[2] A. OSTROWSKI, Über die Absolutabweichung einer differentienbaren Funktionen von ihren Integralmittelwert, Comment. Math. Hel., 10 (1938), 226–227.

[3] D. POMPEIU, Sur une proposition analogue au théorème des accroisse- ments finis, Mathematica (Cluj, Romania), 22 (1946), 143–146.