Ostrowski Via a Two Functions Pompeiu's Inequality

(1)

This is the Published version of the following publication

Dragomir, Sever S (2016) Ostrowski via a two functions Pompeiu’s inequality.

Analele Stiintifice ale Universitatii Ovidius Constanta, Seria Matematica, 24 (3). 123 - 140. ISSN 1224-1784

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Downloaded from VU Research Repository https://vuir.vu.edu.au/34501/

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Ostrowski Via a Two Functions Pompeiu’s Inequality

S. S. Dragomir

Abstract

In this paper, some generalizations of Pompeiu’s inequality for two complex-valued absolutely continuous functions are provided. They are applied to obtain some new Ostrowski type results. Reverses for the integral Cauchy-Bunyakovsky-Schwarz inequality are provided as well.

1 Introduction

In 1946, Pompeiu [6] derived a variant ofLagrange’s mean value theorem, now known asPompeiu’s mean value theorem (see also [8, p. 83]).

Theorem 1 (Pompeiu, 1946 [6]). For every real valued function f differentiable on an interval[a, b] not containing0 and for all pairsx₁6=x₂ in [a, b], there exists a pointξbetween x₁ andx₂ such that

x1f(x2)−x2f(x1)

x₁−x₂ =f(ξ)−ξf⁰(ξ). (1)

The following inequality is useful to derive some Ostrowski type inequalities.

Key Words: Ostrowski’s inequality, Pompeiu’s mean value theorem, Integral inequalities, Schwarz inequality

2010 Mathematics Subject Classification: Primary 26D15; Secondary 26D10 Received: 23.03.2015

Revised: 10.07.2015 Accepted: 21.07.2015

123

(3)

Corollary 1(Pompeiu’s Inequality). With the assumptions of Theorem 1 and ifkf−`f⁰k_∞= sup_t∈(a,b)|f(t)−tf⁰(t)|<∞where `(t) =t, t∈[a, b],then

|tf(x)−xf(t)| ≤ kf−`f⁰k_∞|x−t| (2) for anyt, x∈[a, b].

The inequality (2) was stated by the author in [3].

In 1938, A. Ostrowski [4] proved the following result in the estimating the integral mean:

Theorem 2 (Ostrowski, 1938 [4]). Let f : [a, b]→R be continuous on[a, b]

and differentiable on(a, b)with |f⁰(t)| ≤M < ∞for all t∈(a, b). Then for anyx∈[a, b],we have the inequality

f(x)− 1 b−a

Z b a

f(t)dt

≤



 1

4+ x−^a+b₂ b−a

!²

M(b−a). (3)

The constant ¹₄ is best possible in the sense that it cannot be replaced by a smaller quantity.

In order to provide another approximation of the integral mean, by making use of the Pompeiu’s mean value theorem, the author proved the following result:

Theorem 3 (Dragomir, 2005 [3]). Let f : [a, b]→ Rbe continuous on [a, b]

and differentiable on(a, b)with[a, b]not containing0.Then for anyx∈[a, b], we have the inequality

a+b 2 ·f(x)

x − 1

b−a Z b

a

f(t)dt

(4)

≤ b−a

|x|



 1

4 + x−^a+b₂ b−a

!2

kf−`f⁰k_∞,

where`(t) =t, t∈[a, b].

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

In [7], E. C. Popa using a mean value theorem obtained a generalization of (4) as follows:

(4)

Theorem 4 (Popa, 2007 [7]). Let f : [a, b] →R be continuous on [a, b] and differentiable on (a, b). Assume that α /∈ [a, b]. Then for any x∈ [a, b], we have the inequality

a+b 2 −α

f(x) +α−x b−a

Z b a

f(t)dt

(5)

≤



 1

4 + x−^a+b₂ b−a

!²

(b−a)kf−`_αf⁰k_∞, where`_α(t) =t−α, t∈[a, b].

In [5], J. Peˇcari´c and S. Ungar have proved a general estimate with the p-norm, 1≤p≤ ∞which forp=∞gives Dragomir’s result.

Theorem 5(Peˇcari´c & Ungar, 2006 [5]). Let f : [a, b]→Rbe continuous on [a, b] and differentiable on (a, b) with 0< a < b. Then for 1≤p, q≤ ∞ with

1

p +¹_q = 1 we have the inequality

a+b 2 · f(x)

x − 1

b−a Z b

a

f(t)dt

≤P U(x, p)kf−`f⁰k_p, (6) forx∈[a, b], where

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# .

In the cases(p, q) = (1,∞),(∞,1) and(2,2)the quantityP U(x, p)has to be taken as the limit asp→1,∞and2,respectively.

For other inequalities in terms of the p-norm of the quantity f −`_αf⁰, where`_α(t) =t−α, t∈[a, b] andα /∈[a, b] see [1] and [2].

In this paper, some new Pompeiu’s type inequalities for two complex-valued absolutely continuous functions are provided. They are applied to obtain some Ostrowski type inequalities. Reverses for the integral Cauchy-Bunyakovsky- Schwarz inequality are provided as well.

2 A General Pompeiu’s Inequality

We start with the following generalization of (2).

(5)

Theorem 6. Let f, g : [a, b] →C be absolutely continuous functions on the interval[a, b] withg(t)6= 0for allt∈[a, b].Then for anyt, x∈[a, b] we have

f(x) g(x) −f(t)

g(t)

≤











kf⁰g−f g⁰k_∞

Rx t

1

|g(s)|²ds

iff⁰g−f g⁰∈L_∞[a, b],

kf⁰g−f g⁰k_p

Rx t

1

|g(s)|^2qds

1/q if f⁰g−f g⁰ ∈Lp[a, b]

p >1,

1

p+¹_q = 1, kf⁰g−f g⁰k₁sups∈[t,x]([x,t])

n 1

|g(s)|²

o

(7)

or, equivalently

|g(t)f(x)−f(t)g(x)|

≤











kf⁰g−f g⁰k_∞|g(t)g(x)|

Rx t

1

|g(s)|²ds

iff⁰g−f g⁰∈L_∞[a, b],

kf⁰g−f g⁰k_p|g(t)g(x)|

Rx t

1

|g(s)|^2qds

1/q if f⁰g−f g⁰∈L_p[a, b]

p >1,

1

p+¹_q = 1, kf⁰g−f g⁰k₁|g(t)g(x)|sups∈[t,x]([x,t])

n 1

|g(s)|²

o .

(8) Proof. Iff andgare absolutely continuous andg(t)6= 0 for allt∈[a, b], then f /gis absolutely continuous on the interval [a, b] and

Z x t

f(s) g(s)

0

ds=f(x) g(x) −f(t)

g(t) for anyt, x∈[a, b] withx6=t.

Since

Z x t

f(s) g(s)

⁰ ds=

Z x t

f⁰(s)g(s)−f(s)g⁰(s)

g²(s) ds,

then we get the following identity f(x)

g(x) −f(t) g(t) =

Z x t

g²(s) ds (9)

(6)

for anyt, x∈[a, b].

Taking the modulus in (9) we have

f(x) g(x) −f(t)

g(t)

=

Z x t

g²(s) ds

(10)

≤

Z x t

|f⁰(s)g(s)−f(s)g⁰(s)|

|g(s)|² ds

:=I and utilizing H¨older’s integral inequality we deduce

I≤











sups∈[t,x]([x,t])|f⁰(s)g(s)−f(s)g⁰(s)|

Rx t

1

|g(s)|²ds ,

Rx

t |f⁰(s)g(s)−f(s)g⁰(s)|^pds

1/p

Rx t

1

|g(s)|^2qds

1/q p >1,

1

p+¹_q = 1,

Rx

t |f⁰(s)g(s)−f(s)g⁰(s)|ds

sups∈[t,x]([x,t])

n 1

|g(s)|²

o,

≤











kf⁰g−f g⁰k_∞

Rx t

1

|g(s)|²ds ,

kf⁰g−f g⁰k_p

Rx t

1

|g(s)|^2qds

1/q p >1,

1

p+¹_q = 1, kf⁰g−f g⁰k₁sups∈[t,x]([x,t])

n 1

|g(s)|²

o

and the inequality (7) is proved.

The following particular case extends Pompeiu’s inequality to other p- norms thanp=∞obtained in (2).

Corollary 2. Let f : [a, b]→C be an absolutely continuous function on the interval[a, b]with b > a >0. Then for anyt, x∈[a, b] we have

(7)

f(x) x −f(t)

t

≤











kf−`f⁰k_∞ ¹_t −_x¹

iff−`f⁰∈L_∞[a, b],

1

(2q−1)^1/q kf−`f⁰k_p

_t2q−1¹ −_x2q−1¹

1/q if f−`f⁰∈Lp[a, b]

p >1,

1

p +¹_q = 1, kf−`f⁰k₁_min{t¹2,x²}

(11)

or, equivalently

|tf(x)−xf(t)|

≤











kf−`f⁰k_∞|x−t| iff−`f⁰∈L_∞[a, b],

1

(2q−1)^1/qkf−`f⁰k_p ^x

q

t^q−1 −_xq−1^t^q

1/q if f−`f⁰∈Lp[a, b]

p >1,¹_p+¹_q = 1, kf−`f⁰k₁^max{t,x}_min{t,x},

(12)

where`(t) =t, t∈[a, b].

The proof follows by (7) forg(t) =`(t) =t, t∈[a, b]. The general case for power functions is as follows.

Corollary 3. Let f : [a, b]→C be an absolutely continuous function on the interval[a, b]withb > a >0.Ifr∈R,r6= 0,then for anyt, x∈[a, b]we have

(8)

f(x) x^r −f(t)

t^r

≤











1

|r|kf⁰`−rfk_∞

_x¹_r −_t¹_r

, iff⁰`−rf ∈L_∞[a, b], kf⁰`−rfk_p

×







1

|1−q(r+1)|^1/q

_x1−q(r+1)¹ −_t1−q(r+1)¹

1/q, forr6=−¹_p

|lnx−lnt|^1/q, forr=−¹_p if f⁰`−rf ∈L_p[a, b],

kf⁰`−rfk₁_min{x_r+1¹ _,t_r+1_},

(13)

or, equivalently

|t^rf(x)−x^rf(t)|

≤











1

|r|kf⁰`−rfk_∞|t^r−x^r|,iff⁰`−rf ∈L∞[a, b], kf⁰`−rfk_p

×







t^rx^r

|1−q(r+1)|^1/q

_x_1−q(r+1)¹ −_t_1−q(r+1)¹

1/q, forr6=−¹_p

t^rx^r|lnx−lnt|^1/q, forr=−¹_p if f⁰`−rf ∈Lp[a, b],

kf⁰`−rfk₁_min{x^tr+1^r^x^r,t^r+1},

(14)

wherep >1,¹_p+¹_q = 1.

The proof follows by (7) forg(t) =t^r, t∈[a, b].The details for calculations are omitted.

We have the following result for exponential.

Corollary 4. Let f : [a, b]→C be an absolutely continuous function on the

(9)

interval[a, b]andα∈R,α6= 0.Then for any t, x∈[a, b]we have

f(x)

exp (iαx)− f(t) exp (iαt)

≤











kf⁰−iαfk_∞|x−t| if f⁰−iαf ∈L_∞[a, b],

kf⁰−iαfk_p|x−t|^1/q

iff⁰−iαf∈Lp[a, b]

p >1,

1

p+¹_q = 1, kf⁰−iαfk₁

(15)

or, equivalently

|exp (iαt)f(x)−f(t) exp (iαx)|

≤











kf⁰−iαfk_∞|x−t| if f⁰−iαf ∈L_∞[a, b],

kf⁰−iαfk_p|x−t|^1/q

iff⁰−iαf∈Lp[a, b]

p >1,

1

p+¹_q = 1, kf⁰−iαfk₁.

(16)

3 An Inequality Generalizing Ostrowski’s

The following result holds:

Theorem 7. Let f, g : [a, b] →C be absolutely continuous functions on the interval[a, b]. If0< m≤ |g(t)| ≤M <∞ for anyt∈[a, b], then

(10)

f(x) Z b

a

g(t)dt−g(x) Z b

a

f(t)dt

≤ M

m ²











kf⁰g−f g⁰k_∞(b−a)²

1

4+_x−a+b 2

b−a

²

if f⁰g−f g⁰∈L_∞[a, b],

kf⁰g−f g⁰k_ph_(b−x)1+1/q+(x−a)^1+1/q 1+1/q

i iff⁰g−f g⁰∈Lp[a, b]

p >1,

1

p+¹_q = 1, kf⁰g−f g⁰k₁(b−a)

(17) for anyx∈[a, b].

Proof. Utilizing (8) we have

f(x) Z b

a

g(t)dt−g(x) Z b

a

f(t)dt

≤ Z b

a

|g(t)f(x)−f(t)g(x)|dt

≤











kf⁰g−f g⁰k_∞|g(x)|Rb a

|g(t)|

Rx t

1

|g(s)|²ds

dt,

kf⁰g−f g⁰k_p|g(x)|Rb a

|g(t)|

Rx t

1

|g(s)|^2qds

1/q dt,

kf⁰g−f g⁰k₁|g(x)|Rb a

|g(t)|sups∈[t,x]([x,t])

n 1

|g(s)|²

odt

(18)

for anyx∈[a, b],which is of interest in itself.

(11)

Since 0< m≤ |g(t)| ≤M <∞for anyt∈[a, b], then

|g(x)|

Z b a

|g(t)|

Z x t

1

|g(s)|²ds

! dt≤

M m

2Z b a

|x−t|dt

= M

m ²



 1

4 + x−^a+b₂ b−a

!2

,

|g(x)|

Z b a



|g(t)|

Z x t

1

|g(s)|^2qds

1/q

dt

≤ M

m ²Z b

a

|x−t|^1/qdt= M

m

²(b−x)^1+1/q+ (x−a)^1+1/q 1 + 1/q

and

|g(x)|

Z b a

|g(t)| sup

s∈[t,x]([x,t])

( 1

|g(s)|² )!

dt≤ M

m ²Z b

a

dt= M

m ²

(b−a) for anyx∈[a, b] and by (18) we get the desired result (17).

Remark1.If we takeg(t) = 1, t∈[a, b] in the first inequality (17) we recapture Ostrowski’s inequality.

Corollary 5. With the assumptions in Theorem 7 we have the midpoint inequalities

f

a+b 2

Z b a

g(t)dt−g a+b

2 Z b

a

f(t)dt

≤ M

m ²











1

4(b−a)²kf⁰g−f g⁰k_∞ iff⁰g−f g⁰∈L_∞[a, b],

1

2^1/q(1+1/q)(b−a)^1+1/qkf⁰g−f g⁰k_p

if f⁰g−f g⁰∈Lp[a, b]

p >1,

1

p+¹_q = 1.

(19) The following result also holds:

(12)

Theorem 8. Let f, g : [a, b] →C be absolutely continuous functions on the interval[a, b], g(x)6= 0 forx∈[a, b]andg⁻²∈L∞[a, b]. Then

f(x) g(x)

Z b a

g(t)dt− Z b

a

f(t)dt

≤ g⁻²

∞×











kf⁰g−f g⁰k_∞Rb

a|g(t)| |x−t|dt, if f⁰g−f g⁰ ∈L_∞[a, b], kf⁰g−f g⁰k_pRb

a |g(t)| |x−t|^1/qdt

iff⁰g−f g⁰∈Lp[a, b]

p >1,

1

p+¹_q = 1, kf⁰g−f g⁰k₁Rb

a|g(t)|dt

(20) for anyx∈[a, b].

Proof. Utilizing (8) we have

f(x) g(x)

Z b a

g(t)dt− Z b

a

f(t)dt

≤











kf⁰g−f g⁰k_∞Rb a

|g(t)|

Rx t

1

|g(s)|²ds

dt,

kf⁰g−f g⁰k_pRb a

|g(t)|

Rx t

1

|g(s)|^2qds

1/q dt,

kf⁰g−f g⁰k₁Rb a

|g(t)|sups∈[t,x]([x,t])

n 1

|g(s)|²

o dt

(21)

for anyx∈[a, b].

Since

Z x t

1

|g(s)|²ds

≤ g⁻²

∞|x−t|,

Z x t

1

|g(s)|^2qds

1/q

≤ g⁻²

_∞|x−t|^1/q and

sup

s∈[t,x]([x,t])

( 1

|g(s)|² )

≤ g⁻²

_∞

(13)

for any x, t ∈ [a, b], then on making use of (21) we get the desired result (20).

We have the midpoint inequalities:

Corollary 6. With the assumptions of Theorem 8 we have

f ^a+b₂ g ^a+b₂

Z b a

g(t)dt− Z b

a

f(t)dt

≤ g⁻²

∞×











kf⁰g−f g⁰k_∞Rb a|g(t)|

^a+b₂ −t

dt, iff⁰g−f g⁰∈L∞[a, b],

kf⁰g−f g⁰k_pRb a |g(t)|

^a+b₂ −t

1/qdt

if f⁰g−f g⁰∈L_p[a, b]

p >1,

1

p+¹_q = 1.

(22) We have the following exponential version of Ostrowski’s inequality as well:

Theorem 9. Let f : [a, b] →C be an absolutely continuous function on the interval[a, b]andα∈R,α6= 0.Then for any x∈[a, b]we have

exp (iα(b−x))−exp (−iα(x−a))

iα f(x)−

Z b a

f(t)dt

≤











kf⁰−iαfk_∞(b−a)²

1

4+_t−a+b 2

b−a

²

, if f⁰−iαf ∈L∞[a, b],

kf⁰−iαfk_p^(b−x)^1+1/q_1+1/q^+(x−a)^1+1/q,

iff⁰−iαf

∈L_p[a, b]

p >1,

1

p+¹_q = 1, kf⁰−iαfk₁.

(23)

(14)

Proof. If we write the inequality (18) for g(t) = exp (iαt), t∈[a, b],then we get

f(x) Z b

a

exp (iαt)dt−exp (iαx) Z b

a

f(t)dt

≤











kf⁰−iαfk_∞Rb

a |x−t|dt, iff⁰−iαf∈L_∞[a, b]

kf⁰−iαfk_p|g(x)|Rb

a|x−t|^1/qdt,

iff⁰−iαf

∈Lp[a, b]

p >1,

1

p +¹_q = 1 kf⁰−iαfk₁,

which, after simple calculation, is equivalent with (23).

The details are omitted.

Corollary 7. With the assumptions of Theorem 9 we have the midpoint inequalities

exp iα ^b−a₂

−exp −iα ^b−a₂

iα f

a+b 2

− Z b

a

f(t)dt

≤











1

4kf⁰−iαfk_∞(b−a)², iff⁰−iαf ∈L_∞[a, b],

1

2^1/q(1+1/q)(b−a)^1+1/qkf⁰−iαfk_p, iff⁰−iαf∈Lp[a, b]

p >1,¹_p+¹_q = 1,

(24)

or, equivalently

(15)

2 sin α ^b−a₂

α f

a+b 2

− Z b

a

f(t)dt

≤











1

4kf⁰−iαfk_∞(b−a)², iff⁰−iαf ∈L∞[a, b],

1

2^1/q(1+1/q)(b−a)^1+1/qkf⁰−iαfk_p, iff⁰−iαf∈L_p[a, b]

p >1,¹_p+¹_q = 1.

(25)

4 An Application for CBS-Inequality

The following inequality is well known in the literature as the Cauchy-Bunyakovsky- Schwarz inequality, or the CBS-inequality, for short:

Z b a

f(t)g(t)dt

2

≤ Z b

a

|f(t)|²dt Z b

a

|g(t)|²dt, (26) provided thatf, g∈L2[a, b].

We have the following result concerning some reverses of the CBS-inequality:

Theorem 10. Let f, g: [a, b]→Cbe absolutely continuous functions on the interval[a, b]with g(t)6= 0for all t∈[a, b]. Then

0≤ Z b

a

|g(t)|²dt Z b

a

|f(t)|²dt−

Z b a

f(t)g(t)dt

2

≤1 2×











kf⁰g−f g⁰k²_∞Rb

a|g(t)|²dt²Rb a

1

|g(t)|²dt²

, if f⁰g−f g⁰∈L_∞[a, b],

1

|g|² ∈L[a, b]

kf⁰g−f g⁰k²_pRb

a |g(t)|²dt²Rb a

1

|g(t)|^2qdt^2/q , if

f⁰g−f g⁰∈Lp[a, b],

1

|g|^2q ∈L[a, b]

p >1,

1

p+¹_q = 1, kf⁰g−f g⁰k²₁Rb

a|g(t)|²dt²

esssup_t∈[a,b]n

1

|g(t)|⁴

o

, if _|g|¹ ∈L∞[a, b].

(27)

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Proof. Utilising the inequality (8) we have

g(t)f(x)−f(t)g(x)

≤











kf⁰g−f g⁰k_∞|g(t)g(x)|

Rx t

1

|g(s)|²ds

iff⁰g−f g⁰

∈L∞[a, b],

kf⁰g−f g⁰k_p|g(t)g(x)|

Rx t

1

|g(s)|^2qds

1/q

iff⁰g−f g⁰

∈Lp[a, b]

p >1,

1

p+¹_q = 1, kf⁰g−f g⁰k₁|g(t)g(x)|sups∈[t,x]([x,t])

n 1

|g(s)|²

o .

(28) for anyt, x∈[a, b].

Taking the square in (28) and integrating over (t, x)∈[a, b]² we have Z b

a

Z b a

g(t)f(x)−f(t)g(x)

2

dtdx

≤











kf⁰g−f g⁰k²_∞Rb a

Rb

a |g(t)g(x)|²

Rx t

1

|g(s)|²ds

2

dtdx,

kf⁰g−f g⁰k²_pRb a

Rb

a |g(t)g(x)|²

Rx t

1

|g(s)|^2qds

2/q

dtdx,

kf⁰g−f g⁰k²₁Rb a

Rb

a |g(t)g(x)|²sups∈[t,x]([x,t])

n 1

|g(s)|⁴

o dtdx.

(29)

Observe that

(17)

Z b a

g(t)f(x)−f(t)g(x)

2

dtdx

= Z b

a

Z b a

|g(t)|²|f(x)|²−2Reh

g(t)f(x)f(t)g(x)i

+|g(x)|²|f(t)|² dtdx

= Z b

a

|g(t)|²dt Z b

a

|f(x)|²dx−2Re

"

Z b a

f(t)g(t)dt Z b

a

f(x)g(x)dx

#

+ Z b

a

|g(x)|²dx Z b

a

|f(t)|²dt

= 2



 Z b

a

|g(t)|²dt Z b

a

|f(t)|²dt−

Z b a

f(t)g(t)dt

2

,

Z b a



|g(t)g(x)|²

Z x t

1

|g(s)|²ds

2

dtdx

≤ Z b

a

|g(t)|²dt

!2

Z b a

1

|g(t)|²dt

!2

,

Z b a



|g(t)g(x)|²

Z x t

1

|g(s)|^2qds

2/q

dtdx

≤ Z b

a

|g(t)|²dt

!2

Z b a

1

|g(t)|^2qdt

!2/q

and

Z b a

"

|g(t)g(x)|² sup

s∈[t,x]([x,t])

( 1

|g(s)|⁴ )#

dtdx

≤ Z b

a

|g(t)|²dt

!2

ess sup

t∈[a,b]

( 1

|g(t)|⁴ )

,

then by (29) we get the desired result (27).

Acknowledgement. The authors would like to thank the anonymous referee for valuable comments that have been implemented in the final version of the paper.

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References

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Arts 2011, no. 3(16), 281–287.

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[3] S. S. Dragomir, An inequality of Ostrowski type via Pompeiu’s mean value theorem. J. Inequal. Pure Appl. Math.6 (2005), no. 3, Article 83, 9 pp.

[4] A. Ostrowski, ¨Uber die Absolutabweichung einer differentienbaren Funk- tionen von ihren Integralmittelwert, Comment. Math. Hel, 10 (1938), 226-227.

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[7] E. C. Popa, An inequality of Ostrowski type via a mean value theorem, General Mathematics Vol.15, No. 1, 2007, 93-100.

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Silvestru Sever DRAGOMIR,

Mathematics, College of Engineering & Science Victoria University, PO Box 14428

Melbourne City, MC 8001, Australia.

DST-NRF Center of Excellence in the Mathematical and Statistical Sciences, School of Computer Science & Applied Mathematics,

University of the Witwatersrand,

Private Bag 3, Johannesburg 2050, South Africa.

Email: [email protected]

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