The Australian Journal of Mathematical Analysis and ...4 S. S. DRAGOMIR and then by (1.4), via the...

The Australian Journal of MathematicalAnalysis and Applications

AJMAA

Volume 14, Issue 1, Article 1, pp. 1-287, 2017

OSTROWSKI TYPE INEQUALITIES FOR LEBESGUE INTEGRAL:A SURVEY OF RECENT RESULTS

SILVESTRU SEVER DRAGOMIR1,2

Received 1 November, 2016; accepted 4 January, 2017; published 20 February, 2017.

1MATHEMATICS, COLLEGE OFENGINEERING & SCIENCE, V ICTORIA UNIVERSITY, PO BOX 14428,MELBOURNE CITY, MC 8001, AUSTRALIA .

2DST-NRF CENTRE OFEXCELLENCE, IN THE MATHEMATICAL AND STATISTICAL SCIENCES, SCHOOL OF

COMPUTERSCIENCE & A PPLIED MATHEMATICS, UNIVERSITY OF THE WITWATERSRAND, PRIVATE BAG 3,JOHANNESBURG2050, SOUTH AFRICA.

[email protected]: http://rgmia.org/dragomir

ABSTRACT. The main aim of this survey is to present recent results concerning Ostrowski typeinequalities for the Lebesgue integral of various classes of complex and real-valued functions.The survey is intended for use by both researchers in various fields of Classical and ModernAnalysis and Mathematical Inequalities and their Applications, domains which have grown ex-ponentially in the last decade, as well as by postgraduate students and scientists applying in-equalities in their specific areas.

Key words and phrases:Ostrowski inequality, Midpoint inequality, Lebesgue integral, Absolutely continuous functions, Func-tions of Bounded variation, Convex functions, Logarithmic convex functions, Monotonic functions,Mean value theorem.

2010Mathematics Subject Classification.26D15, 26D10

ISSN (electronic): 1449-5910

c© 2017 Austral Internet Publishing. All rights reserved.

http://ajmaa.org/

mailto: Sever S. Dragomir <[email protected]>

http://rgmia.org/dragomir

http://www.ams.org/msc/

Contents

1. Introduction 1

Chapter 1. Inequalities for Functions of Bounded Variation 31. Ostrowski Inequality for Functions of Bounded Variation 32. Refinements for Functions of Bounded Variation 43. Ostrowski for Cumulative Variation 84. Ostrowski Inequality for Monotonic Functions 165. Ostrowski Inequality for Convex Functions 18References 21

Chapter 2. Inequalities for Absolutely Continuous Functions 231. Ostrowski forL∞-norm 232. Ostrowski forL1-norm 263. Ostrowski forLp-norm 294. Ostrowski for Bounded Derivatives 325. Functional Ostrowski Inequality for Convex Mappings 36References 45

Chapter 3. Inequalities for Differentiable Functions 471. Ostrowski Via Cauchy Mean Value Theorem 472. Ostrowski Via General Cauchy Mean Value Theorem 523. Ostrowski Via Pompeiu Mean Value Theorem 584. Another Ostrowski Type Inequality Via Pompeiu’s Result 61References 63

Chapter 4. Inequalities of Pompeiu Type 651. Ostrowski Via a Generalized Pompeiu’s Inequality 652. Ostrowski Via Power Pompeiu’s Inequality 723. Ostrowski Via an Exponential Pompeiu’s Inequality 774. Ostrowski Via a Two Functions Pompeiu’s Inequality 87References 97

Chapter 5. Inequalities for Derivatives with Special Properties 991. Ostrowski for Convex Derivatives 992. Ostrowski for Derivatives that are Convex in Modulus 1023. Ostrowski for Derivatives that Have Certain Convexity Properties 1064. Ostrowski for Functions Whose Derivatives areh-Convex in Modulus 112References 121

Chapter 6. Other Ostrowski Type Inequalities 1231. Ostrowski for Products 1232. Ostrowski for S-Dominated Functions 1333. Ostrowski for H-Dominated Functions 138

3

4 S. S. DRAGOMIR

References 143

Chapter 7. Perturbed Ostrowski Type Inequalities 1451. Perturbed Ostrowski Type Inequalities for Functions of Bounded Variation 1452. Some Perturbed Ostrowski Type Inequalities for Absolutely Continuous Functions 1553. Other Perturbed Ostrowski Type Inequalities for Absolutely Continuous Function 1634. More Perturbed Ostrowski Type Inequalities for Absolutely Continuous Functions 172References 181

Chapter 8. Companions of Ostrowski’s Inequality 1831. A Companion of Ostrowski Inequality for Functions of Bounded Variation 1832. A Companion of Ostrowski Inequality for Absolutely Continuous Functions 1883. Perturbed Companions of Ostrowski Inequality for Functions of Bounded Variation 1954. Perturbed Companions of Ostrowski’s Inequality for Absolutely Continuous

Functions 2075. More Perturbed Companions of Ostrowski’s Inequality for Absolutely Continuous

Functions 216References 223

Chapter 9. Ostrowski’s Inequality for General Lebesgue Integral 2251. Ostrowski-Jensen Type Inequalities 2252. Other Ostrowski-Jensen Type Inequalities 2393. Inequalities for Functions Whose Derivatives in Absolute Value areh-Convex 249References 258

Chapter 10. List of Papers Related to Ostrowski’s Inequality 2611. Before 2000 2612. Period 2000-2010 2623. After 2010 272

AJMAA, Vol. 14, No. 1, Art. 1, pp. 1-287, 2017 AJMAA

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OSTROWSKITYPE INEQUALITIES FORLEBESGUEINTEGRAL 1

1. INTRODUCTION

The main aim of this survey is to present recent results concerning Ostrowski type inequal-ities for the Lebesgue integral of various classes of complex and real-valued functions. It isa natural continuation of the edited monograph from 2002 [Ostrowski Type Inequalities andApplications in Numerical Integration. Edited by Sever S. Dragomir and Themistocles M. Ras-sias. Kluwer Academic Publishers, Dordrecht, 2002. xx+481 pp. ISBN: 1-4020-0562-8 26-06]whose preprint version is available athttp://rgmia.org/monographs/Ostrowski.html .

As revealed by a simple search in the databaseMathSciNetwith the key words"Ostrowski"and"inequality" in the title, an exponential evolution of research papers devoted to this resulthas been registered in the last decade. There are now at least 470 papers that can be found byperforming the above search. Numerous extensions, generalizations in both the integral anddiscrete case have been discovered. More general versions forn-time differentiable functions,the corresponding versions on time scales, for vector valued functions or multiple integralshave been established as well. Numerous applications in Numerical Analysis, ApproximationTheory, Probability Theory & Statistics, Information Theory and other fields have been alsogiven.

In the first Chapter of the survey, some Ostrowski type inequalities for functions of boundedvariation, monotonic and convex functions are given. The case of absolutely continuous func-tions and bounds in terms of the Lebesguep-norms of the derivatives are presented in Chapter2 . Then, in Chapter 3 , by applying Cauchy Mean Value Theorem and Pompeiu Mean ValueTheorem some Ostrowski type inequalities for differentiable functions are provided. In Chapter4 , more Pompeiu type inequalities for absolutely continuous functions, powers and exponentialare given, while in Chapter 5 , Ostrowski type inequalities for convex derivatives, derivativesthat are convex orh-convex in modulus are presented. In Chapter 6 we introduce the class ofSandH-dominated functions and establish the corresponding Ostrowski type inequalities as wellas some inequalities for products of two functions. Further, in Chapter 7 , we established var-ious perturbed Ostrowski type inequalities while in Chapter 8 we present some companions ofOstrowski inequality for various classes of functions, including functions of bounded variation,Lipschitzian functions, convex functions and absolutely continuous functions whose derivativessatisfy certain properties. The last part, Chapter 9 , deals with Ostrowski-Jensen type inequal-ities in the general setting of abstract Lebesgue integral for composite functions in which onefunction is Lebesgue integrable on a measurable space while the second is either of boundedvariation or absolutely continuous and with the derivative satisfying some usual properties suchas boundedness or of Lipschitz type etc. . .

The survey also provides a chronological list of published papers, the corresponding ver-sions on time scales, for vector valued functions or multiple integrals devoted to Ostrowskiinequality in different settings including the integral and discrete case, the corresponding ver-sions on time scales, fractional integrals, for vector valued functions or multiple integrals.

The survey is intended for use by both researchers in various fields of Classical and ModernAnalysis and Mathematical Inequalities and their Applications, domains which have grownexponentially in the last decade, as well as by postgraduate students and scientists applyinginequalities in their specific areas.

Each chapter contains the necessary references and therefore can be read independently.


http://rgmia.org/monographs/Ostrowski.html

http://rgmia.org/monographs/Ostrowski.html

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CHAPTER 1

Inequalities for Functions of Bounded Variation

1. OSTROWSKI I NEQUALITY FOR FUNCTIONS OF BOUNDED VARIATION

The following inequality for functions of bounded variation holds:

THEOREM 1.1 (Dragomir, 1999 [4]). Letf : [a, b] → C be a function of bounded variationon [a, b] . Then for allx ∈ [a, b] , we have the inequality

(1.1)

∣∣∣∣∫ b

a

f (t) dt− (b− a) f (x)

∣∣∣∣ ≤ [1

2(b− a) +

∣∣∣∣x− a+ b

2

∣∣∣∣] b∨a

(f) ,

where∨b

a (f) denotes the total variation off. The constant12

is the best possible one.

PROOF. Using the integration by parts formula for Riemann-Stieltjes integrals we have∫ x

a

(t− a) df (t) = f (x) (x− a)−∫ x

a

f (t) dt

and ∫ b

x

(t− b)df (t) = f (x) (b− x)−∫ b

x

f (t) dt.

If we add the above two equalities, we obtain the following equality of interest, see [6]

(1.2) (b− a) f (x)−∫ b

a

f (t) dt =

∫ b

a

p (x, t) df (t) ,

where

p (x, t) :=

t− a if t ∈ [a, x)

t− b if x ∈ [x, b],

for all x, t ∈ [a, b] .It is well known [1, p. 177] that ifp : [a, b] → C is continuous on[a, b] andv : [a, b] → C is ofbounded variation on[a, b] , then

(1.3)

∣∣∣∣∫ b

a

p (x) dv (x)

∣∣∣∣ ≤ supx∈[a,b]

|p (x)|b∨a

(v).

Applying the inequality (1.3) forp (x, t) as above andv (x) = f (x) , x ∈ [a, b] , we get∣∣∣∣∫ b

a

p (x, t) df (t)

∣∣∣∣ ≤ supt∈[a,b]

|p (x, t)|b∨a

(f) = maxx− a, b− xb∨a

(f)(1.4)

=

[b− a

2+

∣∣∣∣x− a+ b

2

∣∣∣∣] b∨a

(f)

3

4 S. S. DRAGOMIR

and then by (1.4), via the identity (1.2), we deduce the desired inequality (1.1).Now to prove that1

2is the best possible constant assume that the inequality (1.1) holds with a

constantC > 0. That is,

(1.5)

∣∣∣∣∫ b

a

f (t) dt− f (x) (b− a)

∣∣∣∣ ≤ [C (b− a) +

∣∣∣∣x− a+ b

2

∣∣∣∣] b∨a

(f)

for all x ∈ [a, b] .Consider the functionf : [a, b] → R, given by

f (x) =

0 if x ∈ [a, b] \a+b

2

1 if x = a+b2

in (1.5). Thenf is of bounded variation on[a, b] , and∨b

a (f) = 2,∫ b

af (t) dt = 0 . Forx = a+b

2,

we get in (1.5)1 ≤ 2C , which implies thatC ≥ 12

and the theorem is completely proved.

COROLLARY 1.2 (Dragomir, 2000 [5]). Let f : [a, b] → C be of bounded variation. Thenwe have the inequality:

(1.6)

∣∣∣∣∫ b

a

f (t) dx− (b− a) f

(a+ b

2

)∣∣∣∣ ≤ 1

2(b− a)

b∨a

(f) .

The constant12

is best possible.

For other Ostrowski type inequalities, see [2].

2. REFINEMENTS FOR FUNCTIONS OF BOUNDED VARIATION

2.1. A Hölder Type Inequality for Total Variation. The following lemma is of interestin itself as well.

LEMMA 2.1 (Dragomir, 2008 [10]). Letf, u : [a, b] → C. If f is continuous on[a, b] anduis of bounded variation on[a, b] , then∣∣∣∣∫ b

a

f (t) du (t)

∣∣∣∣ ≤ ∫ b

a

|f (t)| d

(t∨a

(u)

)(2.1)

≤

[b∨a

(u)

] 1q∫ b

a

|f (t)|p d

(t∨a

(u)

) 1p

if p > 1,1

p+

1

q= 1,

≤ maxt∈[a,b]

|f (t)|b∨a

(u) .

PROOF. Since the Riemann-Stieltjes integral∫ b

af (t) du (t) exists, then for any divisionIn :

a = t0 < t1 < · · · < tn−1 < tn = b with the normv (In) := maxi∈0,...,n−1 (ti+1 − ti) → 0and for any intermediate pointsξi ∈ [ti, ti+1], i ∈ 0, . . . , n− 1 we have:∣∣∣∣∫ b

a

f (t) du (t)

∣∣∣∣ =

∣∣∣∣∣ limv(In)→0

n−1∑i=0

f (ξi) [u (ti+1)− u (ti)]

∣∣∣∣∣(2.2)

≤ limv(In)→0

n−1∑i=0

|f (ξi)| |u (ti+1)− u (ti)| .


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However,

(2.3) |u (ti+1)− u (ti)| ≤ti+1∨ti

(u) =

ti+1∨a

(u)−ti∨a

(u) ,

for anyi ∈ 0, . . . , n− 1 , and by (2.3) we have

∣∣∣∣∫ b

a

f (t) du (t)

∣∣∣∣ ≤ limv(In)→0

n−1∑i=0

|f (ξi)|

[ti+1∨

a

(u)−ti∨a

(u)

]=

∫ b

a

|f (t)| d

(t∨a

(u)

),

and the last Riemann-Stieltjes integral exists since|f | is continuous and∨·

a is monotonic non-decreasing on[a, b] .

The last part follows from the following well known Hölder type inequality for the Riemann-Stieltjes integral with monotonic integrator, namely:∣∣∣∣∫ b

a

g (t) dv (t)

∣∣∣∣(2.4)

≤ [v (b)− v (a)]1q

[∫ b

a

|g (t)|p dv (t)

] 1p

if p > 1,1

p+

1

q= 1,

≤ maxt∈[a,b]

|g (t)| [v (b)− v (a)] ,

holding for anyg continuous on[a, b] andv monotonic nondecreasing on[a, b] .The details are omitted.

2.2. Refinement of the Ostrowski Inequality.The following result that provides a refine-ments of Ostrowski inequality for functions of bounded variation may be stated:

THEOREM2.2 (Dragomir, 2008 [10]). Letf : [a, b] → R be a function of bounded variationon [a, b] . Then

(2.5)

∣∣∣∣∫ b

a

f (t) dt− f (x) (b− a)

∣∣∣∣ ≤ Q (x)

for anyx ∈ [a, b] , where

Q (x) := [2x− (a+ b)]x∨a

(f)−∫ b

a

sgn (x− t)t∨a

(f) dt(2.6)

≤ (b− x)b∨x

(f) + (x− a)x∨a

(f)

≤[1

2(b− a) +

∣∣∣∣x− a+ b

2

∣∣∣∣] b∨a

(f) .


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6 S. S. DRAGOMIR

We also have

Q (x) ≤

[b∨a

(f)

] 1q

[(x− a)p − (b− x)p]x∨a

(f)(2.7)

−p∫ b

a

rp (x, t) sgn (x− t)t∨a

(f) dt

1p

≤

[b∨a

(f)

] 1q

(b− x)pb∨x

(f) + (x− a)px∨a

(f)

1p

≤[1

2(b− a) +

∣∣∣∣x− a+ b

2

∣∣∣∣] b∨a

(f) ,

for anyx ∈ [a, b] , wherep > 1, 1p

+ 1q

= 1 andrp : [a, b]2 → R with

(2.8) rp (x, t) :=

(t− a)p−1 if t ∈ [a, x],

(b− t)p−1 if t ∈ (x, b].

PROOF. We use the identity obtained in [6]

(2.9) f (x) (b− a)−∫ b

a

f (t) dt =

∫ b

a

p (x, t) df (t) ,

wherep : [a, b]2 → R is given by

p (x, t) :=

t− a if t ∈ [a, x],

t− b if t ∈ (x, b],

andx ∈ [a, b].Now, if f is of bounded variation on[a, b] , then on applying the first inequality in (2.1) we

deduce ∣∣∣∣∫ b

a

p (x, t) df (t)

∣∣∣∣ ≤ ∫ b

a

|p (x, t)| d

(t∨a

(f)

)

=

∫ x

a

(t− a) d

(t∨a

(f)

)+

∫ b

x

(b− t) d

(t∨a

(f)

)

= (x− a)x∨a

(f)−∫ x

a

(t∨a

(f)

)dt

− (b− x)x∨a

(f) +

∫ b

x

(t∨a

(f)

)dt

= [2x− (a+ b)]x∨a

(f)−∫ b

a

sgn (x− t)t∨a

(f) dt

= Q (x) ,

for eachx ∈ [a, b] , and the inequality (2.5) is proved.


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Now, since∨·

a is monotonic nondecreasing on[a, b] , then

Q (x) ≤ [2x− (a+ b)]x∨a

(f) + (b− x)b∨a

(f)

= (b− x)b∨x

(f) + (x− a)x∨a

(f)

and the inequality (2.6) is proved.Utilising the second part of the second inequality in (2.1) we deduce that

(2.10) Q (x) ≤

[b∨a

(f)

] 1q∫ b

a

|p (x, t)|p d

(t∨a

(f)

) 1p

.

Now, observe that

R (x) :=

∫ b

a

|p (x, t)|p d

(t∨a

(f)

)(2.11)

=

∫ x

a

(t− a)p d

(t∨a

(f)

)+

∫ b

x

(b− t)p d

(t∨a

(f)

)

= (x− a)px∨a

(f)− p

∫ x

a

(t− a)p−1t∨a

(f) dt

− (b− x)px∨a

(f) + p

∫ b

x

(b− t)p−1t∨a

(f) dt

= [(x− a)p − (b− x)p]x∨a

(f)− p

∫ b

a

rp (x, t) sgn (x− t)t∨a

(f) dt

whererp is given in (2.8). Utilising (2.9), we deduce the first part of (2.7).Since

∨·a (t) is increasing, we also have

R (x) ≤ [(x− a)p − (b− x)p]x∨a

(f) + p

b∨a

(f) ·∫ b

x

(b− t)p−1 dt

= [(x− a)p − (b− x)p]x∨a

(f) + (b− x)p ·b∨a

(f)

= (x− a)p ·x∨a

(f) + (b− x)p ·b∨x

(f) ,

which proves the last part of (2.7).

The following particular case that provides a refinement of the midpoint inequality is ofinterest.


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8 S. S. DRAGOMIR

COROLLARY 2.3. Letf : [a, b] → C be a function of bounded variation on[a, b] . Then∣∣∣∣∫ b

a

f (t) dt− f

(a+ b

2

)(b− a)

∣∣∣∣(2.12)

≤∫ b

a

sgn

(t− a+ b

2

)·

t∨a

(f) dt

≤ p1/p

[b∨a

(f)

] 1q[∫ b

a

r (t) sgn

(t− a+ b

2

)·

t∨a

(f) dt

] 1p

≤ 1

2(b− a) ·

b∨a

(f) ,

where

r (t) =

(t− a)p−1 if t ∈[a, a+b

2

],

(b− t)p−1 if t ∈(

a+b2, b]

andp > 1, 1p

+ 1q

= 1.

REMARK 2.1. The inequalities (2.5)–(2.7) provide refinements for the Ostrowski inequality

(2.13)

∣∣∣∣∫ b

a

f (t) dt− f (x) (b− a)

∣∣∣∣ ≤ [1

2(b− a) +

∣∣∣∣x− a+ b

2

∣∣∣∣] b∨a

(f)

for anyx ∈ [a, b] , obtained by the author in [4] and [6]. The inequalities (2.12) are refinementsof the midpoint inequality for functions of bounded variation obtained in [5].

3. OSTROWSKI FOR CUMULATIVE VARIATION

3.1. Cumulative Variation Function. For a function of bounded variationv : [a, b] → Cwe define theCumulative Variation Function(CVF) V : [a, b] → [0,∞) by

V (t) :=t∨a

(v) ,

i.e., the total variation of the functionv on the interval[a, t] with t ∈ [a, b] .It is know that the CVF is monotonic nondecreasing on[a, b] and is continuous in a point

c ∈ [a, b] if and only if the generating functionv is continuing in that point. Ifv is Lipschitzianwith the constantL > 0, i.e.

|v (t)− v (s)| ≤ L |t− s| for anyt, s ∈ [a, b] ,

thenV is also Lipschitzian with the same constant.The following lemma is of interest in itself as well [1, p. 177], see also [10] for a general-

ization.

LEMMA 3.1. Letf, u : [a, b] → C. If f is continuous on[a, b] andu is of bounded variationon [a, b] , then the Riemann-Stieltjes integral

∫ b

af (t) du (t) exists and

(3.1)

∣∣∣∣∫ b

a

f (t) du (t)

∣∣∣∣ ≤ ∫ b

a

|f (t)| d (V (t)) ≤ maxt∈[a,b]

|f (t)|b∨a

(u) .


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3.2. Some Refinements.The following result may be stated.

THEOREM3.2 (Dragomir, 2013 [11]). Letf : [a, b] → C be a function of bounded variationon [a, b] . Then ∣∣∣∣∫ b

a

f (t) dt− f (x) (b− a)

∣∣∣∣(3.2)

≤∫ x

a

(x∨t

(f)

)dt+

∫ b

x

(t∨x

(f)

)dt

≤ (x− a)x∨a

(f) + (b− x)b∨x

(f)

≤

[

12(b− a) +

∣∣x− a+b2

∣∣]∨ba (f) ,[

12

∨ba (f) + 1

2

∣∣∣∨xa (f)−

∨bx (f)

∣∣∣] (b− a) ,

for anyx ∈ [a, b].

PROOF. We start with the equality

(3.3) f (x) (b− a)−∫ b

a

f (t) dt =

∫ x

a

(t− a) df (t) +

∫ b

x

(t− b) df (t)

that holds for anyx ∈ [a, b] andf : [a, b] → C a function of bounded variation on[a, b] (see [4]or [6]).

Taking the modulus in (3.3) and using the property (3.1) we have∣∣∣∣f (x) (b− a)−∫ b

a

f (t) dt

∣∣∣∣(3.4)

≤∣∣∣∣∫ x

a

(t− a) df (t)

∣∣∣∣+ ∣∣∣∣∫ b

x

(t− b) df (t)

∣∣∣∣≤∫ x

a

(t− a) d

(t∨a

(f)

)+

∫ b

x

(b− t) d

(t∨x

(f)

)

for anyx ∈ [a, b] .Utilising the integration by parts formula for the Riemann-Stieltjes integral, we have

∫ x

a

(t− a) d

(t∨a

(f)

)= (t− a)

t∨a

(f)

∣∣∣∣∣x

a

−∫ x

a

(t∨a

(f)

)dt(3.5)

= (x− a)x∨a

(f)−∫ x

a

(t∨a

(f)

)dt

=

∫ x

a

(x∨a

(f)−t∨a

(f)

)dt =

∫ x

a

(x∨t

(f)

)dt


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10 S. S. DRAGOMIR

and ∫ b

x

(b− t) d

(t∨x

(f)

)= (b− t)

t∨x

(f)

∣∣∣∣∣b

x

+

∫ b

x

(t∨x

(f)

)dt(3.6)

=

∫ b

x

(t∨x

(f)

)dt

for anyx ∈ [a, b] .Utilising (3.4)-(3.6) we deduce the first inequality in (3.2).Since

∨xt (f) ≤

∨xa (f) for t ∈ [a, x] and

∨tx (f) ≤

∨bx (f) for t ∈ [x, b] , then∫ x

a

(x∨t

(f)

)dt+

∫ b

x

(t∨x

(f)

)dt ≤ (x− a)

x∨a

(f) + (b− x)b∨x

(f)

for anyx ∈ [a, b] , which proves the second inequality in (3.2).The last part is obvious by the max properties and the fact that forc, d ∈ R we have

max c, d = c+d+|c−d|2

. The details are omitted.

The following midpoint inequality holds:

COROLLARY 3.3. Letf : [a, b] → C be a function of bounded variation on[a, b] . Then∣∣∣∣∫ b

a

f (t) dt− f

(a+ b

2

)(b− a)

∣∣∣∣(3.7)

≤∫ a+b

2

a

a+b2∨t

(f)

dt+

∫ b

a+b2

t∨a+b2

(f)

dt ≤ 1

2(b− a)

b∨a

(f) .

The first inequality in (3.7) is sharp and the constant12

in the second, is best possible.

PROOF. We must prove only the sharpness of the inequalities in (3.7).If we consider the functionf : [a, b] → R given by

f (t) :=

0, t ∈ [a, a+b2

)1, t = a+b

2

0 t ∈ (a+b2, b],

then this function is of bounded variation, we observe that∨t

a+b2

(f) = 1 for any t ∈ (a+b2, b]

and∨a+b

2t (f) = 1 for anyt ∈ [a, a+b

2). Also, we have

∨ba (f) = 2.

If we replace these values in (3.7) we obtain in all terms the same quantityb− a.

COROLLARY 3.4. Let f : [a, b] → C be a function of bounded variation on[a, b] . If p ∈(a, b) is a median point in variation, namely

∨pa (f) =

∨bp (f) , then we have the inequality∣∣∣∣∫ b

a

f (t) dt− f (p) (b− a)

∣∣∣∣(3.8)

≤∫ p

a

(p∨t

(f)

)dt+

∫ b

p

(t∨p

(f)

)dt ≤ 1

2(b− a)

b∨a

(f) .

The first inequality in (3.2) is useful when some properties for the CVF are available, likefor instance, below:


http://ajmaa.org


COROLLARY 3.5. Let f : [a, b] → C be a function of bounded variation on[a, b] . If forx ∈ (a, b) there existLx > 0 andα > −1 such that

(3.9)

∣∣∣∣∣t∨x

(f)

∣∣∣∣∣ ≤ Lx |t− x|α for anyt ∈ [a, b] \ x ,

then

(3.10)

∣∣∣∣∫ b

a

f (t) dt− f (x) (b− a)

∣∣∣∣ ≤ 1

α+ 1Lx

[(x− a)α+1 + (b− x)α+1] .

In particular, forα = 1 in (3.9) we get from (3.10) that

(3.11)

∣∣∣∣∫ b

a

f (t) dt− f (x) (b− a)

∣∣∣∣ ≤1

4+

(x− a+b

2

b− a

)2Lx (b− a)2 .

REMARK 3.1. If the CVF∨t

a (f) isK-Lipschitzian, i.e.,∣∣∣∣∣t∨s

(f)

∣∣∣∣∣ ≤ K |t− s| for anyt, s ∈ [a, b] ,

then

(3.12)

∣∣∣∣∫ b

a

f (t) dt− f (x) (b− a)

∣∣∣∣ ≤1

4+

(x− a+b

2

b− a

)2L (b− a)2 .


COROLLARY 3.6. If there exists a constantLa+b2> 0 andα > −1 such that

(3.13)

∣∣∣∣∣∣t∨

a+b2

(f)

∣∣∣∣∣∣ ≤ La+b2

∣∣∣∣t− a+ b

2

∣∣∣∣α for anyt ∈ [a, b] \a+ b

2

,

then we have the midpoint inequality

(3.14)

∣∣∣∣∫ b

a

f (t) dt− f

(a+ b

2

)(b− a)

∣∣∣∣ ≤ 1

2α (α+ 1)La+b

2(b− a)α+1 .

In particular, if we takeα = 1 in (3.13), then we get from (3.14)

(3.15)

∣∣∣∣∫ b

a

f (t) dt− f

(a+ b

2

)(b− a)

∣∣∣∣ ≤ 1

4La+b

2(b− a)2 .

The constant14

is best possible in (3.15).

PROOF. First, we notice that ifh : [a, b] → C is of bounded variation, then|h| : [a, b] →[0,∞) is of bounded variation and

(3.16)b∨a

(|h|) ≤b∨a

(h) .

Indeed, by the continuity property of the modulus, we have thatn−1∑j=0

||h (tj+1)| − |h (tj)|| ≤n−1∑j=0

|h (tj+1)− h (tj)|


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12 S. S. DRAGOMIR

for any divisiona = t0 < t1 < ... < tn−1 < tn = b, which, by taking the supremum over alldivisions of[a, b] , produces the desired inequality (3.16).

If we consider the functionf0 : [a, b]→ R, f0 (s) :=∣∣s− a+b

2

∣∣ then, by denoting withe theidentity function on[a, b] , i.e. e (t) = t, t ∈ [a, b] , we have∣∣∣∣∣∣

t∨a+b2

(f0)

∣∣∣∣∣∣ =

∣∣∣∣∣∣t∨

a+b2

(∣∣∣∣e−a+ b

2

∣∣∣∣)∣∣∣∣∣∣

≤

∣∣∣∣∣∣t∨

a+b2

(e−a+ b

2

)∣∣∣∣∣∣ =

∣∣∣∣∣∣t∨

a+b2

(e)

∣∣∣∣∣∣ =

∣∣∣∣t− a+ b

2

∣∣∣∣for anyt ∈ [a, b] .

Therefore the functionf0 satisfies the condition (3.13) forα = 1 and with the constantLa+b

2= 1. Since ∫ b

a

f0 (t) dt =

∫ b

a

∣∣∣∣t− a+ b

2

∣∣∣∣ dt =1

4(b− a)2 ,

then we obtain in both sides of the inequality (3.15) the same quantity14(b− a)2 .

REMARK 3.2. The inequalities (3.12) and (3.15) are known in the case of Lipschitzianfunctions with the constantL > 0. We obtained them here under weaker conditions for thefunctionf. This show that the refinement in terms of the CVF for the Ostrowski inequality (3.2)is also useful to extend known results to larger classes of functions.

3.3. Bounds for Shifted Function. We have the following result as well:

THEOREM3.7 (Dragomir, 2013 [11]). Letf : [a, b] → C be a function of bounded variationon [a, b] . Then ∣∣∣∣∫ b

a

f (t) dt− f (x) (b− a)

∣∣∣∣(3.17)

≤∫ x

a

(x∨t

(|f (x)− f |)

)dt+

∫ b

x

(t∨x

(|f (x)− f |)

)dt

≤ (x− a)x∨a

(|f (x)− f |) + (b− x)b∨x

(|f (x)− f |)

≤ (x− a)x∨a

(f) + (b− x)b∨x

(f)

≤

[

12(b− a) +

∣∣x− a+b2

∣∣]∨ba (f) ,[

12

∨ba (f) + 1

2

∣∣∣∨xa (f)−

∨bx (f)

∣∣∣] (b− a) ,


PROOF. Observe that∣∣∣∣f (x) (b− a)−∫ b

a

f (t) dt

∣∣∣∣ =

∣∣∣∣∫ b

a

[f (x)− f (t)] dt

∣∣∣∣(3.18)

≤∫ b

a

|f (x)− f (t)| dt


http://ajmaa.org


for anyx ∈ [a, b].For a fixedx ∈ [a, b] , define the functiongx : [a, b] → [0,∞) by gx (t) := |f (x)− f (t)| .

We observe thatgx is of bounded variation on[a, b] and

(3.19)

∣∣∣∣gx (x) (b− a)−∫ b

a

gx (t) dt

∣∣∣∣ =

∫ b

a

|f (x)− f (t)| dt.

Writing the inequality (3.2) for the functiongx we have∣∣∣∣gx (x) (b− a)−∫ b

a

gx (t) dt

∣∣∣∣(3.20)

≤∫ x

a

(x∨t

(|f (x)− f |)

)dt+

∫ b

x

(t∨x

(|f (x)− f |)

)dt

≤ (x− a)x∨a

(|f (x)− f |) + (b− x)b∨x

(|f (x)− f |) .

Utilising (3.18)-(3.20) we deduce the first two inequalities in (3.17).By the inequality (3.16) we have

x∨a

(|f (x)− f |) ≤x∨a

(f (x)− f) =x∨a

(f)

andb∨x

(|f (x)− f |) ≤b∨x

(f (x)− f) =b∨x

(f) ,

which proves the third inequality in (3.17).

COROLLARY 3.8. If f : [a, b] → C is a function of bounded variation on[a, b] , then∣∣∣∣∫ b

a

f (t) dt− f

(a+ b

2

)(b− a)

∣∣∣∣(3.21)

≤∫ a+b

2

a

a+b2∨t

(∣∣∣∣f (a+ b

2

)− f

∣∣∣∣) dt

+

∫ b

a+b2

t∨a+b2

(∣∣∣∣f (a+ b

2

)− f

∣∣∣∣) dt

≤ 1

2(b− a)

b∨a

(∣∣∣∣f (a+ b

2

)− f

∣∣∣∣) ≤ 1

2(b− a)

b∨a

(f) .

All inequalities in (3.21) are sharp.

PROOF. If we consider the functionf : [a, b] → R, with

f (t) :=

0, t ∈ [a, a+b2

)1, t = a+b

2

0 t ∈ (a+b2, b],


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14 S. S. DRAGOMIR

then this function is of bounded variation, we observe that∣∣∣∣f (a+ b

2

)− f (t)

∣∣∣∣ =

1, t ∈ [a, a+b2

)0, t = a+b

2

1 t ∈ (a+b2, b],

a+b2∨t

(∣∣∣∣f (a+ b

2

)− f

∣∣∣∣) = 1 for t ∈ [a,a+ b

2),

t∨a+b2

(∣∣∣∣f (a+ b

2

)− f

∣∣∣∣) = 1 for t ∈ (a+ b

2, b]

andb∨a

(∣∣∣∣f (a+ b

2

)− f

∣∣∣∣) = 2.

Replacing these values in (3.21) we get in all terms of this inequality the same quantityb− a.

COROLLARY 3.9. Let f : [a, b] → C be a function of bounded variation on[a, b] . If q ∈(a, b) is a point for which

q∨a

(|f (q)− f |) =b∨q

(|f (q)− f |) ,

then ∣∣∣∣∫ b

a

f (t) dt− f (q) (b− a)

∣∣∣∣(3.22)

≤∫ q

a

(q∨t

(|f (q)− f |)

)dt+

∫ b

q

(t∨q

(|f (q)− f |)

)dt

≤ 1

2

b∨a

(|f (q)− f |) ≤ 1

2(b− a)

b∨a

(f) .

REMARK 3.3. Since

(x− a)x∨a

(|f (x)− f |) + (b− x)b∨x

(|f (x)− f |)

≤

max x− a, b− x

∨ba |f (x)− f |

max∨x

a (|f (x)− f |) ,∨b

x (|f (x)− f |)

(b− a)

=

[

12(b− a) +

∣∣x− a+b2

∣∣]∨ba |f (x)− f |[

12

∨ba |f (x)− f |+ 1

2

∣∣∣∨xa (|f (x)− f |)−

∨bx (|f (x)− f |)

∣∣∣] ,AJMAA, Vol. 14, No. 1, Art. 1, pp. 1-287, 2017 AJMAA

http://ajmaa.org


then from (3.17) we also have the string of inequalities∣∣∣∣∫ b

a

f (t) dt− f (x) (b− a)

∣∣∣∣(3.23)

≤∫ x

a

(x∨t

(|f (x)− f |)

)dt+

∫ b

x

(t∨x

(|f (x)− f |)

)dt

≤ (x− a)x∨a

(|f (x)− f |) + (b− x)b∨x

(|f (x)− f |)

≤

[

12(b− a) +

∣∣x− a+b2

∣∣]∨ba |f (x)− f |[

12

∨ba |f (x)− f |+ 1

2

∣∣∣∨xa (|f (x)− f |)−

∨bx (|f (x)− f |)

∣∣∣]× (b− a) ,

for anyx ∈ [a, b] .

REMARK 3.4. If the functionf : [a, b] → R is monotonic nondecreasing on[a, b] , then∣∣∣∣∫ b

a

f (t) dt− f (x) (b− a)

∣∣∣∣(3.24)

≤∫ b

a

sgn (x− t) [f (x)− f (t)] dt

≤ (x− a) [f (x)− f (a)] + (b− x) [f (b)− f (x)]

≤

[

12(b− a) +

∣∣x− a+b2

∣∣] [f (b)− f (a)] ,[12[f (b)− f (a)] + 1

2

∣∣∣f (x)− f(a)+f(b)2

∣∣∣] (b− a) ,

for anyx ∈ [a, b] .In particular, forx = a+b

2, we get the midpoint inequality∣∣∣∣∫ b

a

f (t) dt− f

(a+ b

2

)(b− a)

∣∣∣∣(3.25)

≤∫ b

a

sgn

(a+ b

2− t

)[f

(a+ b

2

)− f (t)

]dt ≤ 1

2[f (b)− f (a)] (b− a) .

Moreover, ifp ∈ (a, b) is such that

f (p) =f (a) + f (b)

2,

then we have the trapezoid inequality∣∣∣∣f (a) + f (b)

2(b− a)−

∫ b

a

f (t) dt

∣∣∣∣(3.26)

≤∫ b

a

sgn (p− t)

[f (a) + f (b)

2− f (t)

]dt ≤ 1

2[f (b)− f (a)] (b− a) .


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16 S. S. DRAGOMIR

4. OSTROWSKI I NEQUALITY FOR M ONOTONIC FUNCTIONS

The following results of Ostrowski type holds.

THEOREM 4.1 (Dragomir, 1999 [3]). Let f : [a, b] → R be a monotonic nondecreasingfunction on[a, b] . Then for allx ∈ [a, b] , we have the inequality

∣∣∣∣∫ b

a

f (t) dt− (b− a) f (x)

∣∣∣∣(4.1)

≤ [2x− (a+ b)]f (x) +

∫ b

a

sgn(t− x)f (t) dt

≤ (x− a) [f (x)− f (a)] + (b− x) [f (b)− f (x)]

≤[b− a

2+

∣∣∣∣x− a+ b

2

∣∣∣∣] [f (b)− f (a)] .

All the inequalities in (4.1) are sharp and the constant12

is the best possible one.

PROOF. Using the integration by parts formula for Riemann-Stieltjes integrals we have

∫ x

a

(t− a) df (t) = f (x) (x− a)−∫ x

a

f (t) dt

and ∫ b

x

(t− b)df (t) = f (x) (b− x)−∫ b

x

f (t) dt.

If we add the above two equalities, we obtain the following equality of interest, see [6]

(4.2) (b− a) f (x)−∫ b

a

f (t) dt =

∫ b

a

p (x, t) df (t) ,

where

p (x, t) :=

t− a if t ∈ [a, x)

t− b if x ∈ [x, b],

for all x, t ∈ [a, b] .If p : [a, b] → C is continuous on[a, b] andv : [a, b] → R is monotonic nondecreasing on

[a, b] , then [1, p. 155]

(4.3)

∣∣∣∣∫ b

a

p (x) dv (x)

∣∣∣∣ ≤ ∫ b

a

|p (x)| dv (x) .


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Now, by the property (4.3) we have∣∣∣∣(b− a) f (x)−∫ b

a

f (t) dt

∣∣∣∣=

∣∣∣∣∫ x

a

(t− a) df (t) +

∫ b

x

(t− b) df (t)

∣∣∣∣≤∣∣∣∣∫ x

a

(t− a) df (t)

∣∣∣∣+ ∣∣∣∣∫ b

x

(t− b) df (t)

∣∣∣∣≤∫ x

a

|t− a| df (t) +

∫ b

x

|t− b| df (t) =

∫ x

a

(t− a)df (t) +

∫ b

x

(b− t)df (t)

= (t− a)f (t)|xa −∫ x

a

f (t) dt− (b− t)f (t)|bx +

∫ b

x

f (t) dt

= [2x− (a+ b)]f (x)−∫ x

a

f (t) dt+

∫ b

x

f (t) dt

= [2x− (a+ b)]f (x) +

∫ b

a

sgn(t− x)f (t) dt,

which proves the first part of (4.1).As f is monotonic nondecreasing on[a, b] , we can state that

∫ x

a

f (t) dt ≥ (x− a)f (a) and∫ b

x

f (t) dt ≤ (b− x)f (b)

so that ∫ b

a

sgn(t− x)f (t) dt ≤ (b− x)f (b)− (x− a)f (a) .

Consequently

[2x− (a+ b)]f (x) +

∫ b

a

sgn(t− x)f (t) dt

≤ [2x− (a+ b)]f (x) + (b− x)f (b)− (x− a)f (a)

= (b− x) [f (b)− f (x)] + (x− a) [f (x)− f (a)]

and the second part of (4.1) is proved.Finally, let us observe that

(b− x) [f (b)− f (x)] + (x− a) [f (x)− f (a)]

≤ maxb− x, x− a[f (b)− f (x) + f (x)− f (a)]

=

[b− a

2+

∣∣∣∣x− a+ b

2

∣∣∣∣] [f (b)− f (a)]

and the inequality (4.1) is thus proved.


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18 S. S. DRAGOMIR

Now for the sharpness of the inequalities, assume that (4.1) holds with a constantC > 0instead of1

2. That is, ∣∣∣∣(b− a) f (x)−

∫ b

a

f (t) dt

∣∣∣∣(4.4)

≤ [2x− (a+ b)] f (x) +

∫ b

a

sgn(t− x)f (t) dt

≤ (x− a) [f (x)− f (a)] + (b− x) [f (b)− f (x)]

≤[C (b− a) +

∣∣∣∣x− a+ b

2

∣∣∣∣] [f (b)− f (a)] .

Consider the mappingf0 : [a, b] → R given by

f0 (x) :=

−1 if x = a0 if x ∈ (a, b]

.

Putting in (4.4)f = f0 andx = a, we have∣∣∣∣(b− a) f (x)−∫ b

a

f (t) dt

∣∣∣∣= [2x− (a+ b)]f (x) +

∫ b

a

sgn(t− x)f (t) dt

= (x− a) [f (x)− f (a)] + (b− x) [f (b)− f (x)] = b− a

≤[C (b− a) +

∣∣∣∣x− a+ b

2

∣∣∣∣] [f (b)− f (a)] =

(C +

1

2

)(b− a) ,

which proves the sharpness of the first two inequalities and the fact thatC cannot not be lessthan 1

2.

The following corollary is interesting.

COROLLARY 4.2. Letf be as above. Then we have the midpoint inequality:∣∣∣∣(b− a) f

(a+ b

2

)−∫ b

a

f (t) dt

∣∣∣∣ ≤ ∫ b

a

sgn

(t− a+ b

2

)f (t) dt(4.5)

≤ 1

2(b− a) [f (b)− f (a)] .

For other Ostrowski type inequalities, see [2].

5. OSTROWSKI I NEQUALITY FOR CONVEX FUNCTIONS

The following results of Ostrowski type holds.

THEOREM 5.1 (Dragomir, 2002 [7]). Letf : [a, b] ⊂ R → R be a convex function on[a, b].Then for anyx ∈ [a, b] one has the inequality

1

2

[(b− x)2 f ′+ (x)− (x− a)2 f ′− (x)

](5.1)

≤∫ b

a

f (t) dt− (b− a) f (x)

≤ 1

2

[(b− x)2 f ′− (b)− (x− a)2 f ′+ (a)

].


http://ajmaa.org


The constant12

is sharp in both inequalities. The second inequality also holds forx = a orx = b.

PROOF. It is easy to see that for any locally absolutely continuous functionf : (a, b) → R,we have the identity

(5.2)∫ x

a

(t− a) f ′ (t) dt+

∫ b

x

(t− b) f ′ (t) dt = f (x)−∫ b

a

f (t) dt

for anyx ∈ (a, b) , wheref ′ is the derivative off which exists a.e. on(a, b) .Sincef is convex, then it is locally Lipschitzian and thus (5.2) holds. Moreover, for any

x ∈ (a, b), we have the inequalities

(5.3) f ′ (t) ≤ f ′− (x) for a.e.t ∈ [a, x]

and

(5.4) f ′ (t) ≥ f ′+ (x) for a.e.t ∈ [x, b] .

If we multiply (5.3) byt− a ≥ 0, t ∈ [a, x] and integrate on[a, x] , we get

(5.5)∫ x

a

(t− a) f ′ (t) dt ≤ 1

2(x− a)2 f ′− (x)

and if we multiply (5.4) byb− t ≥ 0, t ∈ [x, b], and integrate on[x, b] , we also have

(5.6)∫ b

x

(b− t) f ′ (t) dt ≥ 1

2(b− x)2 f ′+ (x) .

If we subtract (5.6) from (5.5) and use the representation (5.2), we deduce the first inequality in(5.1).

Now, assume that the first inequality (5.1) holds withC > 0 instead of12, i.e.,

(5.7) C[(b− x)2 f ′+ (x)− (x− a)2 f ′− (x)

]≤∫ b

a

f (t) dt− (b− a) f (x) .

Consider the convex functionf0 (t) := k∣∣t− a+b

2

∣∣, k > 0, t ∈ [a, b]. Then

f ′0+

(a+ b

2

)= k, f ′0−

(a+ b

2

)= −k, f0

(a+ b

2

)= 0

and ∫ b

a

f0 (t) dt =1

4k (b− a)2 .

If in (5.7) we choosef = f0, x = a+b2

, then we get

C

[1

4(b− a)2 k +

1

4(b− a)2 k

]≤ 1

4k (b− a)2

which givesC ≤ 12

and the sharpness of the constant in the first part of (5.1) is proved.If eitherf ′+ (a) = −∞ or f ′− (b) = −∞, then the second inequality in (5.1) holds true.Assume thatf ′+ (a) andf ′− (b) are finite. Sincef is convex on[a, b] , we have

(5.8) f ′ (t) ≥ f ′+ (a) for a.e.t ∈ [a, x] (x may be equal tob)

and

(5.9) f ′ (t) ≤ f ′− (b) for a.e.t ∈ [x, b] (x may be equal toa) .


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20 S. S. DRAGOMIR

If we multiply (5.8) byt− a ≥ 0, t ∈ [a, x] and integrate on[a, x] , then we deduce

(5.10)∫ x

a

(t− a) f ′ (t) dt ≥ 1

2(x− a)2 f ′+ (a)

and if we multiply (5.9) byb− t ≥ 0, t ∈ [x, b], and integrate on[x, b] , then we also have

(5.11)∫ b

x

(b− t) f ′ (t) dt ≤ 1

2(b− x)2 f ′− (b) .

Finally, if we subtract (5.10) from (5.11) and use the representation (5.2), we deduce the secondinequality in (5.1). Now, assume that the second inequality in (5.1) holds with a constantD > 0instead of1

2, i.e.,

(5.12)∫ b

a

f (t) dt− (b− a) f (x) ≤ D[(b− x)2 f ′− (b)− (x− a)2 f ′+ (a)

].

If we consider the convex functionf0 (t) = k∣∣t− a+b

2

∣∣, k > 0, t ∈ [a, b], then we havef ′− (b) = k, f ′+ (a) = −k and by (5.12) applied forf0 in x = a+b

2we get

1

4k (b− a)2 ≤ D

[1

4k (b− a)2 +

1

4k (b− a)2

],

givingD ≥ 12

which proves the sharpness of the constant12

in the second inequality in (5.1).

COROLLARY 5.2. With the assumptions of Theorem 5.1 and ifx ∈ (a, b) is a point ofdifferentiability forf , then

(5.13)

(a+ b

2− x

)f ′ (x) ≤ 1

b− a

∫ b

a

f (t) dt− f (x) .

Now, recall that the following inequality, which is well known in the literature as theHermite-Hadamard inequality for convex functions, holds

(5.14) f

(a+ b

2

)≤ 1

b− a

∫ b

a

f (t) dt ≤ f (a) + f (b)

2.

The following corollary provides both a sharper lower bound for the difference,

1

b− a

∫ b

a

f (t) dt− f

(a+ b

2

),

which we know is nonnegative, and an upper bound.

COROLLARY 5.3. Let f : [a, b] → R be a convex function on[a, b]. Then we have theinequality

0 ≤ 1

8

[f ′+

(a+ b

2

)− f ′−

(a+ b

2

)](b− a)(5.15)

≤ 1

b− a

∫ b

a

f (t) dt− f

(a+ b

2

)≤ 1

8

[f ′− (b)− f ′+ (a)

](b− a) .

The constant18

is sharp in both inequalities.


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EXAMPLE 5.1. Assume that−∞ < a < 0 < b < ∞ and consider the convex functionf : [a, b] → R, f (x) = exp |x| . We have

f ′ (x) =

−e−x if x < 0,

ex if x > 0;

andf ′− (0) = −1, f ′+ (0) = 1. Also,∫ b

a

f (t) dt =

∫ 0

a

e−xdx+

∫ b

0

exdx = exp (b) + exp (−a)− 2.

Now, if a+b26= 0, then by (5.15) we deduce the elementary inequality

(5.16) 0 ≤ exp (b) + exp (−a)− 2

b− a− exp

∣∣∣∣a+ b

2

∣∣∣∣ ≤ 1

8[exp (b) + exp (−a)] (b− a) .

If a+b2

= 0 and if we denoteb = u, u > 0, thusa = −u and by (5.15) we also have

(5.17)1

2u ≤ exp (u)− 1

u− 1 ≤ 1

2u exp (u) .

The reader may produce other elementary inequalities by choosing in an appropriate waythe convex functionf. We omit the details.

REFERENCES

[1] T. M. Apostol, Mathematical Analysis,Second Edition, Addison-Wesley Publishing Company,1975.

[2] P. Cerone and S. S. Dragomir, Midpoint-type rules from an inequalities point of view.Handbook ofanalytic-computational methods in applied mathematics, 135–200, Chapman & Hall/CRC, BocaRaton, FL, 2000.

[3] S. S. Dragomir, Ostrowski’s Inequality for monotonic mappings and applications,J. KSIAM,3(1999), 127-135.

[4] S. S. Dragomir, The Ostrowski integral inequality for mappings of bounded variation.Bull. Austral.Math. Soc.60 (1999), No. 3, 495–508.

[5] S. S. Dragomir, On the midpoint quadrature formula for mappings with bounded variation andapplications.Kragujevac J. Math.22 (2000), 13–19.

[6] S. S. Dragomir, On the Ostrowski’s integral inequality for mappings with bounded variation andapplications,Math. Ineq. Appl.4 (2001), No. 1, 59-66. Preprint:RGMIA Res. Rep. Coll.2 (1999),Art. 7, [Online: http://rgmia.org/papers/v2n1/v2n1-7.pdf].

[7] S. S. Dragomir, An inequality improving the first Hermite-Hadamard inequality forconvex functions defined on linear spaces and applications for semi-inner prod-ucts. J. Inequal. Pure Appl. Math.3 (2002), No. 2, Article 31, 8 pp.[Onlinehttp://www.emis.de/journals/JIPAM/article183.html?sid=183].

[8] S. S. Dragomir, An Ostrowski like inequality for convex functions and applications.Rev. Mat.Complut.16 (2003), No. 2, 373–382.

[9] S. S. Dragomir, An Ostrowski type inequality for convex functions.Univ. Beograd. Publ. Elek-trotehn. Fak. Ser. Mat.16 (2005), 12–25.

[10] S. S. Dragomir, Refinements of the generalised trapezoid and Ostrowski inequalities for functionsof bounded variation.Arch. Math.(Basel)91 (2008), No. 5, 450–460.


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22 S. S. DRAGOMIR

[11] S. S. Dragomir, Refinements of the Ostrowski inequality in terms of the cumulative variation andapplications,Analysis(Berlin) 34 (2014), No. 2, 223–240. Preprint:RGMIA Res. Rep. Coll.16(2013), Art. 29[Online:http://rgmia.org/papers/v16/v16a29.pdf] .


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CHAPTER 2

Inequalities for Absolutely Continuous Functions

1. OSTROWSKI FOR L∞-NORM

1.1. Ostrowski’s Inequality. In 1938, A. Ostrowski [10], proved the following inequalityconcerning the distance between the integral mean1

b−a

∫ b

af (t) dt and the valuef (x), x ∈ [a, b].

THEOREM1.1 (Ostrowski, 1938 [10]). Letf : [a, b] → R be continuous on[a, b] and differ-entiable on(a, b) such thatf ′ : (a, b) → R is bounded on(a, b), i.e.,‖f ′‖∞ := sup

t∈(a,b)

|f ′ (t)| <

∞. Then

(1.1)

∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤1

4+

(x− a+b

2

b− a

)2 ‖f ′‖∞ (b− a) ,

for all x ∈ [a, b] and the constant14

is the best possible.

In [7], S.S. Dragomir and S. Wang, by the use of the Montgomery integral identity [9, p.565],

f (x)− 1

b− a

∫ b

a

f (t) dt =1

b− a

∫ b

a

p (x, t) f ′ (t) dt , x ∈ [a, b] ,

wherep : [a, b]2 → R is given by

p (x, t) :=

t− a if t ∈ [a, x] ,

t− b if t ∈ (x, b],

gave a simple proof of Ostrowski’s inequality and applied it for special means (identric mean,logarithmic mean, etc.) and to the problem of estimating the error bound in approximating theRiemann integral

∫ b

af (t) dt by one arbitrary Riemann sum (see [7], Section 3).

1.2. A Refinement forL∞-norm. The following result, which is an improvement on Os-trowski’s inequality, holds.

23

24 S. S. DRAGOMIR

THEOREM 1.2 (Dragomir, 2002 [2]). Let f : [a, b] → C be an absolutely continuous func-tion on[a, b] whose derivativef ′ ∈ L∞ [a, b]. Then∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.2)

≤ 1

2 (b− a)

[‖f ′‖[a,x],∞ (x− a)2 + ‖f ′‖[x,b],∞ (b− x)2

]

≤

‖f ′‖[a,b],∞

[14

+(

x−a+b2

b−a

)2]

(b− a) ;

12

[‖f ′‖α

[a,x],∞ + ‖f ′‖α[x,b],∞

] 1α[(

x−ab−a

)2β+(

b−xb−a

)2β] 1

β(b− a) ,

where α > 1, 1α

+ 1β

= 1;

12

[‖f ′‖[a,x],∞ + ‖f ′‖[x,b],∞

] [12

+∣∣∣x−a+b

2

b−a

∣∣∣]2 (b− a)

for all x ∈ [a, b], where‖·‖[m,n],∞ denotes the usual norm onL∞ [m,n], i.e., we recall that

‖g‖[m,n],∞ = ess supt∈[m,n]

|g (t)| <∞.

PROOF. Using the integration by parts formula for absolutely continuous functions on[a, b],we have

(1.3)∫ x

a

(t− a) f ′ (t) dt = (x− a) f (x)−∫ x

a

f (t) dt

and

(1.4)∫ b

x

(t− b) f ′ (t) dt = (b− x) f (x)−∫ b

x

f (t) dt,

for all x ∈ [a, b].Adding these two equalities, we obtain the Montgomery identity (see for example [9, p.

565]):

(1.5) (b− a) f (x)−∫ b

a

f (t) dt =

∫ x

a

(t− a) f ′ (t) dt+

∫ b

x

(t− b) f ′ (t) dt

for all x ∈ [a, b].Taking the modulus, we deduce∣∣∣∣(b− a) f (x)−

∫ b

a

f (t) dt

∣∣∣∣(1.6)

≤∣∣∣∣∫ x

a

(t− a) f ′ (t) dt

∣∣∣∣+ ∣∣∣∣∫ b

x

(t− b) f ′ (t) dt

∣∣∣∣≤∫ x

a

(t− a) |f ′ (t)| dt+

∫ b

x

(b− t) |f ′ (t)| dt

≤ ‖f ′‖[a,x],∞

∫ x

a

(t− a) dt+ ‖f ′‖[x,b],∞

∫ b

x

(b− t) dt

=1

2

[‖f ′‖[a,x],∞ (x− a)2 + ‖f ′‖[x,b],∞ (b− x)2

]


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and the first inequality in (1.2) is proved.Now, let us observe that

‖f ′‖[a,x],∞ (x− a)2 + ‖f ′‖[x,b],∞ (b− x)2

≤ max‖f ′‖[a,x],∞ , ‖f ′‖[x,b],∞

[(x− a)2 + (b− x)2]

= max‖f ′‖[a,x],∞ , ‖f ′‖[x,b],∞

[1

2(b− a)2 + 2

(x− a+ b

2

)2]

= (b− a)2 max‖f ′‖[a,x],∞ , ‖f ′‖[x,b],∞

[1

2+ 2 ·

(x− a+b

2

)2(b− a)2

]

= (b− a)2 ‖f ′‖[a,b],∞

[1

2+ 2 ·

(x− a+b

2

)2(b− a)2

],

and the first part of the second inequality in (1.2) is proved.For the second inequality, we employ the elementary inequality for real numbers which can

be derived from Hölder’s discrete inequality

(1.7) 0 ≤ ms+ nt ≤ (mα + nα)1α ×

(sβ + tβ

) 1β ,

provided thatm, s, n, t ≥ 0, α > 1 and 1α

+ 1β

= 1.Using (1.7), we obtain

‖f ′‖[a,x],∞ (x− a)2 + ‖f ′‖[x,b],∞ (b− x)2

≤(‖f ′‖α

[a,x],∞ + ‖f ′‖α[x,b],∞

) 1α[(x− a)2β + (b− x)2β

] 1β

and the second part of the second inequality in (1.2) is also obtained.Finally, we observe that

‖f ′‖[a,x],∞ (x− a)2 + ‖f ′‖[x,b],∞ (b− x)2

≤ max(x− a)2 , (b− x)2 [‖f ′‖[a,x],∞ + ‖f ′‖[x,b],∞

]=

[b− a

2+

∣∣∣∣x− a+ b

2

∣∣∣∣]2 [‖f ′‖[a,x],∞ + ‖f ′‖[x,b],∞

]

and the last part of the second inequality in (1.2) is proved.

The following corollary is also natural.


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26 S. S. DRAGOMIR

COROLLARY 1.3. Under the above assumptions, we have the midpoint inequality

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.8)

≤ 1

8(b− a)

[‖f ′‖[a, a+b

2 ],∞ + ‖f ′‖[a+b2

,b],∞

]

≤

14(b− a) ‖f ′‖[a,b],∞ ;

1

23β−1

2

(b− a)[‖f ′‖α

[a, a+b2 ],∞ + ‖f ′‖α

[a+b2

,b],∞

] 1α,

where α > 1, 1α

+ 1β

= 1.

2. OSTROWSKI FOR L1-NORM

2.1. L1-norm Inequality. In 1997, Dragomir and Wang proved the following Ostrowskitype inequality [5].

THEOREM 2.1. Let f : [a, b] → R be an absolutely continuous function on[a, b]. Then wehave the inequality

(2.1)

∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤[

1

2+

∣∣x− a+b2

∣∣b− a

]‖f ′‖[a,b],1 ,

for all x ∈ [a, b], where‖·‖1 is the Lebesgue norm onL1 [a, b], i.e., we recall it

‖g‖[a,b],1 :=

∫ b

a

|g (t)| dt.

The constant12

is best possible.

Note that the fact that12

is the best constant for differentiable functions was proved in [11]and (2.1) can also be obtained from a more general result obtained by A. M. Fink in [8] choosingn = 1 and doing some appropriate computation. However the inequality (2.1) was not statedexplicitly in [8].

2.2. A Refinement forL1-norm. The following result, which is an improvement on theinequality (2.1), holds.


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THEOREM 2.2 (Dragomir, 2002 [1]). Let f : [a, b] → C be an absolutely continuous func-tion on[a, b]. Then∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.2)

≤ x− a

b− a‖f ′‖[a,x],1 +

b− x

b− a‖f ′‖[x,b],1

≤

12

[‖f ′‖[a,b],1 +

∣∣∣‖f ′‖[a,x],1 − ‖f ′‖[x,b],1

∣∣∣][(

x−ab−a

)β+(

b−xb−a

)β] 1β(∣∣∣‖f ′‖α

[a,x],1

∣∣∣+ ‖f ′‖α[x,b],1

) 1α

whereα > 1 and 1α

+ 1β

= 1,[12

+|x−a+b

2 |b−a

]‖f ′‖[a,b],1

for all x ∈ [a, b], where‖·‖[m,n],1 denotes the usual norm onL1 [m,n] with m < n, i.e., werecall that

‖g‖[m,n],1 :=

∫ n

m

|g (t)| dt <∞.


(2.3)∫ x

a

(t− a) f ′ (t) dt = (x− a) f (x)−∫ x

a

f (t) dt

and

(2.4)∫ b

x

(t− b) f ′ (t) dt = (b− x) f (x)−∫ b

x

f (t) dt

for all x ∈ [a, b].Adding the two inequalities, we obtain the Montgomery identity for absolutely continuous

functions (see for example [9, p. 565])

(2.5) (b− a) f (x)−∫ b

a

f (t) dt =

∫ x

a

(t− a) f ′ (t) dt+

∫ b

x

(t− b) f ′ (t) dt


∫ b

a

f (t) dt

∣∣∣∣(2.6)

≤∣∣∣∣∫ x

a

(t− a) f ′ (t) dt

∣∣∣∣+ ∣∣∣∣∫ b

x

(t− b) f ′ (t) dt

∣∣∣∣≤∫ x

a

(t− a) |f ′ (t)| dt+

∫ b

x

(b− t) |f ′ (t)| dt

≤ (x− a)

∫ x

a

|f ′ (t)| dt+ (b− x)

∫ b

x

|f ′ (t)| dt

= (x− a) ‖f ′‖[a,x],1 + (b− x) ‖f ′‖[x,b],1


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28 S. S. DRAGOMIR

and the first inequality in (2.2) is proved.Now, let us observe that

(x− a) ‖f ′‖[a,x],1 + (b− x) ‖f ′‖[x,b],1

≤ max‖f ′‖[a,x],1 , ‖f

′‖[x,b],1

(b− a)

=1

2

[‖f ′‖[a,x],1 + ‖f ′‖[x,b],1 +

∣∣∣‖f ′‖[a,x],1 − ‖f ′‖[x,b],1

∣∣∣] (b− a)

=1

2

[‖f ′‖[a,b],1 +

∣∣∣‖f ′‖[a,x],1 − ‖f ′‖[x,b],1

∣∣∣] (b− a)

and the first part of the second inequality is proved.For the second inequality, we employ the elementary inequality for real numbers which can

be derived from Hölder’s discrete inequality

(2.7) 0 ≤ ms+ nt ≤ (mα + nα)1α ×

(sβ + tβ

) 1β ,


+ 1β


(x− a) ‖f ′‖[a,x],1 + (b− x) ‖f ′‖[x,b],1

≤(‖f ′‖α

[a,x],1 + ‖f ′‖α[x,b],1

) 1α[(x− a)β + (b− x)β

] 1β


(x− a) ‖f ′‖[a,x],1 + (b− x) ‖f ′‖[x,b],1

≤ max x− a, b− x[‖f ′‖[a,x],1 + ‖f ′‖[x,b],1

]=

[b− a

2+

∣∣∣∣x− a+ b

2

∣∣∣∣] ‖f ′‖[a,b],1

and the last part of the second inequality in (2.2) is proved.


COROLLARY 2.3. Under the above assumptions, we have

(2.8)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

2‖f ′‖[a,b],1 .

Another interesting result is the following one.

COROLLARY 2.4. Under the above assumptions, and if there is anx0 ∈ [a, b] with

(2.9)∫ x0

a

|f ′ (t)| dt =

∫ b

x0

|f ′ (t)| dt

then we have the inequality

(2.10)

∣∣∣∣f (x0)−1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

2‖f ′‖[a,b],1 .


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3. OSTROWSKI FOR Lp-NORM

3.1. Lp-norm Inequality. In 1998, Dragomir and Wang proved the following Ostrowskitype inequality [6].

THEOREM 3.1. Let f : [a, b] → R be an absolutely continuous function on[a, b]. If f ′ ∈Lp [a, b] , then we have the inequality

∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.1)

≤ 1

(q + 1)1/q

[(x− a

b− a

)q+1

+

(b− x

b− a

)q+1]1/q

(b− a)1/q ‖f ′‖[a,b],p ,

for all x ∈ [a, b], wherep > 1, 1p

+ 1q

= 1 and‖·‖[a,b],p is thep-Lebesgue norm onLp [a, b], i.e.,we recall it

‖g‖[a,b],p :=

(∫ b

a

|g (t)|p dt)1/p

.

Note that the inequality (3.1) can also be obtained from a more general result obtained byA. M. Fink in [8] choosingn = 1 and doing some appropriate computation. However theinequality (3.1) was not stated explicitly in [8].

From (3.1) we get the following midpoint inequality

(3.2)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

2 (q + 1)1/q(b− a)1/q ‖f ′‖[a,b],p ,

and 12

is a best possible constant.Indeed, if we takef : [a, b] → R with f (t) =

∣∣t− a+b2

∣∣ , thenf is absolutely continuous∫ b

af (t) dt = (b−a)2

4, ‖f ′‖[a,b],p = (b− a)1/p and if we assume that (3.2) holds with a constant

C > 0 instead of12, then we get1

4(b− a) ≤ C

(q+1)1/q (b− a) for anyq > 1. Letting q → 1+,

we obtainC ≥ 12, which proves the sharpness of the constant.

In the following, we provide some refinements of (3.1) and (3.2).

3.2. A Refinement forLp-norm. The following new result, which is an improvement onthe inequality (3.1), holds.

THEOREM 3.2 (Dragomir, 2013 [4]). Let f : [a, b] → C be an absolutely continuous func-tion on[a, b]. If f ′ ∈ Lp [a, b] , then


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30 S. S. DRAGOMIR

∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.3)

≤ 1

(q + 1)1/q

[(x− a

b− a

) q+1q

‖f ′‖[a,x],p +

(b− x

b− a

) q+1q

‖f ′‖[x,b],p

](b− a)1/q

≤ 1

(q + 1)1/q

×

12

[‖f ′‖[a,x],p + ‖f ′‖[x,b],p +

∣∣∣‖f ′‖[a,x],p − ‖f ′‖[x,b],p

∣∣∣]×[(

x−ab−a

) q+1q +

(b−xb−a

) q+1q

](b− a)1/q

(‖f ′‖α

[a,x],p + ‖f ′‖α[x,b],p

) 1α[(

x−ab−a

) q+1q

β+(

b−xb−a

) q+1q

β] 1

β

(b− a)1/q

whereα > 1 and 1α

+ 1β

= 1,

[‖f ′‖[a,x],p + ‖f ′‖[x,b],p

] [12

+∣∣∣x−a+b

2

b−a

∣∣∣] q+1q

(b− a)1/q

for all x ∈ [a, b], wherep > 1, 1p

+ 1q

= 1 and‖·‖[m,n],p denotes the usualp-norm onLp [m,n]withm < n, i.e., we recall that

‖g‖[m,n],p :=

(∫ n

m

|g (t)| dt)1/p

<∞.


(3.4)∫ x

a

(t− a) f ′ (t) dt = (x− a) f (x)−∫ x

a

f (t) dt

and

(3.5)∫ b

x

(t− b) f ′ (t) dt = (b− x) f (x)−∫ b

x

f (t) dt

for all x ∈ [a, b].Adding the two inequalities, we obtain the Montgomery identity for absolutely continuous

functions (see for example [9, p. 565])

(3.6) (b− a) f (x)−∫ b

a

f (t) dt =

∫ x

a

(t− a) f ′ (t) dt+

∫ b

x

(t− b) f ′ (t) dt


∫ b

a

f (t) dt

∣∣∣∣(3.7)

≤∣∣∣∣∫ x

a

(t− a) f ′ (t) dt

∣∣∣∣+ ∣∣∣∣∫ b

x

(t− b) f ′ (t) dt

∣∣∣∣≤∫ x

a

(t− a) |f ′ (t)| dt+

∫ b

x

(b− t) |f ′ (t)| dt.


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Utilizing Hölder’s integral inequality we have∫ x

a

(t− a) |f ′ (t)| dt+

∫ b

x

(b− t) |f ′ (t)| dt

≤(∫ x

a

(t− a)q dt

)1/q (∫ x

a

|f ′ (t)|p dt)1/p

+

(∫ b

x

(b− t)q dt

)1/q (∫ b

x

|f ′ (t)|p dt)1/p

=1

(b− a) (q + 1)1/q

[(x− a)

q+1q ‖f ′‖[a,x],p + (b− x)

q+1q ‖f ′‖[x,b],p

]for all x ∈ [a, b], and the first inequality in (3.3) is proved.

Now, let us observe that

(x− a)q+1

q ‖f ′‖[a,x],p + (b− x)q+1

q ‖f ′‖[x,b],p

≤ max‖f ′‖[a,x],p , ‖f

′‖[x,b],p

[(x− a)

q+1q + (b− x)

q+1q

]=

1

2

[‖f ′‖[a,x],p + ‖f ′‖[x,b],p +

∣∣∣‖f ′‖[a,x],p − ‖f ′‖[x,b],p

∣∣∣]×[(x− a)

q+1q + (b− x)

q+1q

]and the first part of the second inequality is proved.

For the second inequality, we employ the elementary inequality for real numbers which canbe derived from Hölder’s discrete inequality

(3.8) 0 ≤ ms+ nt ≤ (mα + nα)1α ×

(sβ + tβ

) 1β ,


+ 1β


(x− a)q+1

q ‖f ′‖[a,x],p + (b− x)q+1

q ‖f ′‖[x,b],p

≤(‖f ′‖α

[a,x],p + ‖f ′‖α[x,b],p

) 1α[(x− a)

q+1q

β + (b− x)q+1

qβ] 1

β


(x− a)q+1

q ‖f ′‖[a,x],p + (b− x)q+1

q ‖f ′‖[x,b],p

≤ max

(x− a)q+1

q , (b− x)q+1

q

[‖f ′‖[a,x],p + ‖f ′‖[x,b],p

]=

[b− a

2+

∣∣∣∣x− a+ b

2

∣∣∣∣] q+1q [

‖f ′‖[a,x],p + ‖f ′‖[x,b],p

]and the last part of the second inequality in (3.3) is proved.



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32 S. S. DRAGOMIR

COROLLARY 3.3. Under the above assumptions, we have∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.9)

≤ 1

2(q+1)/q (q + 1)1/q

[‖f ′‖[a, a+b

2 ],p + ‖f ′‖[a+b2

,b],p

](b− a)1/q .

Another interesting result is the following one.

COROLLARY 3.4. Under the above assumptions, and if there is anx0 ∈ [a, b] with

(3.10)∫ x0

a

|f ′ (t)|p dt =

∫ b

x0

|f ′ (t)|p dt

then we have the inequality∣∣∣∣f (x0)−1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.11)

≤ 1

(q + 1)1/q

[(x0 − a

b− a

) q+1q

+

(b− x0

b− a

) q+1q

]‖f ′‖[a,x0],p (b− a)1/q .

REMARK 3.1. If we take in (3.3)α = p andβ = q, wherep > 1, 1p

+ 1q

= 1, then we getthe following refinement of (3.1)∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.12)

≤ 1

(q + 1)1/q

[(x− a

b− a

) q+1q

‖f ′‖[a,x],p +

(b− x

b− a

) q+1q

‖f ′‖[x,b],p

](b− a)1/q

≤ 1

(q + 1)1/q

[(x− a

b− a

)q+1

+

(b− x

b− a

)q+1]1/q

(b− a)1/q ‖f ′‖[a,b],p ,

for all x ∈ [a, b].This is true, since forα = p we have(

‖f ′‖α[a,x],p + ‖f ′‖α

[x,b],p

) 1α

=(‖f ′‖p

[a,x],p + ‖f ′‖p[x,b],p

) 1p

=

(∫ x

a

|f ′ (t)|p dt+

∫ b

x

|f ′ (t)|p)1/p

= ‖f ′‖[a,b],p .

4. OSTROWSKI FOR BOUNDED DERIVATIVES

We start with the following result.

THEOREM 4.1 (Dragomir, 2003 [3]). Let f : [a, b] → R be an absolutely continuous func-tion on[a, b] andx ∈ [a, b]. Suppose that there exist the functionsmi,Mi : [a, b] → R

(i = 1, 2

)with the properties:

(4.1) m1 (x) ≤ f ′ (t) ≤M1 (x) for a.e.t ∈ [a, x]

and

(4.2) m2 (x) ≤ f ′ (t) ≤M2 (x) for a.e.t ∈ (x, b] .


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Then we have the inequalities:1

2 (b− a)

[m1 (x) (x− a)2 −M2 (x) (b− x)2](4.3)

≤ f (x)− 1

b− a

∫ b

a

f (t) dt

≤ 1

2 (b− a)

[M1 (x) (x− a)2 −m2 (x) (b− x)2] .

The constant12

is sharp on both sides.

PROOF. As f : [a, b] → R is absolutely continuous on[a, b], it is differentiable a.e. on[a, b]and, by applying the integration by parts formula, we may write the Montgomery identity

(4.4) f (x) =1

b− a

∫ b

a

f (t) dt+1

b− a

∫ x

a

(t− a) f ′ (t) dt+1

b− a

∫ b

x

(t− b) f ′ (t) dt

for anyx ∈ [a, b].Using the assumption (4.1) and (4.2), we have:

(4.5) m1 (x) (t− a) ≤ (t− a) f ′ (t) ≤M1 (x) (t− a) for a.e.t ∈ [a, x]

and

(4.6) M2 (x) (t− b) ≤ f ′ (t) (t− b) ≤ m2 (x) (t− b) for a.e.t ∈ (x, b] .

Integrating (4.5) on[a, x] and (4.6) on[x, b] and summing the obtained inequalities, we have

1

2m1 (x) (x− a)2 − 1

2M2 (x) (b− x)2

≤∫ x

a

(t− a) f ′ (t) dt+

∫ b

x

(t− b) f ′ (t) dt

≤ 1

2M1 (x) (x− a)2 − 1

2m2 (x) (b− x)2 .

Using the representation (4.4), we deduce (4.3).Assume that the first inequality in (4.3) holds with a constantc > 0; that is,

(4.7)c

b− a

[m1 (x) (x− a)2 −M2 (x) (b− x)2] ≤ f (x)− 1

b− a

∫ b

a

f (t) dt.

Consider the functionf : [a, b] → R, f (t) = M |t− x|, M > 0. Thenf is absolutelycontinuous and

f ′ (t) =

−M if t ∈ [a, x]

M if t ∈ (x, b].

Thus, if we choosem1 = −M ,m2 = M in (4.7), we get

−M c

b− a

[(x− a)2 + (b− x)2] ≤ − M

b− a

∫ b

a

|t− x| dt

= − M

b− a

[(b− x)2 + (x− a)2

2

]for all x ∈ [a, b], implying thatc ≥ 1

2, that is,1

2is the best constant in the first member of (4.3).

Using a similar process, we may prove that12

is the best constant in the third member of(4.3) and the theorem is completely proved.


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34 S. S. DRAGOMIR

COROLLARY 4.2. If f : [a, b] → R is absolutely continuous on[a, b] and the derivativef ′ : [a, b] → R is bounded above and below, that is,

(4.8) −∞ < m ≤ f ′ (t) ≤M <∞ for a.e.t ∈ [a, b] ,


1

2 (b− a)

[m (x− a)2 −M (b− x)2](4.9)

≤ f (x)− 1

b− a

∫ b

a

f (t) dt

≤ 1

2 (b− a)

[M (x− a)2 −m (b− x)2]

for all x ∈ [a, b]. The constant12

is the best in both inequalities.

Applying Taylor’s formula

g (x) = g

(a+ b

2

)+

(x− a+ b

2

)g′(a+ b

2

)+

1

2

(x− a+ b

2

)2

g′′(a+ b

2

)for g (x) = M (x− a)2 −m (b− x)2, we obtain

g (x) =1

4(M −m) (b− a)2 + 2

(x− a+ b

2

)(M +m

2

)(b− a)

+ (M −m)

(x− a+ b

2

)2

.

The same formula applied forh (x) = m (x− a)2 −M (b− x)2, will reveal that

h (x) =1

4(M −m) (b− a)2 + 2

(x− a+ b

2

)(M +m

2

)(b− a)

− (M −m)

(x− a+ b

2

)2

.

Consequently, we may rewrite Corollary 4.2 in the following equivalent manner:

COROLLARY 4.3. With the assumptions on Corollary 4.2, we have:∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt−(x− a+ b

2

)(M +m

2

)∣∣∣∣(4.10)

≤ 1

2(M −m) (b− a)

(x− a+b2

b− a

)2

+1

4

for all x ∈ [a, b].


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REMARK 4.1. If we assume that‖f ′‖∞ := ess supt∈[a,b]

|f ′ (t)|, then obviously we may choose

in (4.9)m = ‖f ′‖∞ andM = ‖f ′‖∞, obtaining Ostrowski’s inequality for absolutely continu-ous functions whose derivatives are essentially bounded:∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ ‖f ′‖∞2 (b− a)

[(x− a)2 + (b− x)2]

=

1

4+

(x− a+b

2

b− a

)2 (b− a) ‖f ′‖∞ ,

for all x ∈ [a, b]. The constant14

here is best.

REMARK 4.2. Ostrowski’s inequality for absolutely continuous mappings in terms of‖f ′‖∞basically states that

− ‖f ′‖∞2 (b− a)

[(x− a)2 + (b− x)2] ≤ f (x)− 1

b− a

∫ b

a

f (t) dt(4.11)

≤ ‖f ′‖∞2 (b− a)

[(x− a)2 + (b− x)2]

for all x ∈ [a, b].Now, if we assume that (4.1) and (4.2) hold, then−‖f ′‖∞ ≤ m1 (x), m2 (x) andM1 (x),

M2 (x) ≤ ‖f ′‖∞, which implies:

− ‖f ′‖∞2 (b− a)

[(x− a)2 + (b− x)2](4.12)

≤ 1

2 (b− a)

[m1 (x) (x− a)2 −M2 (x) (b− x)2] ≤ f (x)− 1

b− a

∫ b

a

f (t) dt

≤ 1

2 (b− a)

[M1 (x) (x− a)2 −m2 (x) (b− x)2]

≤ ‖f ′‖∞2 (b− a)

[(x− a)2 + (b− x)2] .

Thus, the inequality (4.3) may also be regarded as a refinement of the classical Ostrowski result.

An important particular case isx = a+b2

providing the following corollary.

COROLLARY 4.4. Assume that the derivativef ′ : [a, b] → R satisfy the conditions:

(4.13) −∞ < m1 ≤ f ′ (t) ≤M1 <∞ for a.e.t ∈[a,a+ b

2

]and

(4.14) −∞ < m2 ≤ f ′ (t) ≤M2 <∞ for a.e.t ∈(a+ b

2, b

].

Then we have the inequalities

1

8(m1 −M2) (b− a) ≤ f

(a+ b

2

)− 1

b− a

∫ b

a

f (t) dt(4.15)

≤ 1

8(M1 −m2) (b− a) .

The constant18

is the best in both inequalities.


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36 S. S. DRAGOMIR

Finally, if we know some global bounds for the derivativef ′ on [a, b], then we may state thefollowing corollary.

COROLLARY 4.5. Under the assumptions of Corollary 4.2, we have the midpoint inequality:

(4.16)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

8(M −m) (b− a) .

The constant18

is best.

PROOF. The inequality is obvious by Corollary 4.2 puttingx = a+b2

. We observe that thefunctionf : [a, b] → R, f (x) = k

∣∣x− a+b2

∣∣, k > 0 is absolutely continuous and−k ≤ f ′ (t) ≤k for all t ∈ [a, b]. Thus, we may chooseM = k,m = −k and as∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ =k (b− a)

4,

1

8(M −m) (b− a) =

k (b− a)

4

we conclude that the constant18

is best in (4.16).

5. FUNCTIONAL OSTROWSKI I NEQUALITY FOR CONVEX M APPINGS

5.1. A Generalization of Ostrowski’s Inequality. The following result holds:

THEOREM 5.1 (Dragomir, 2013 [4]). Letf : [a, b] → R be absolutely continuous on[a, b] .If Φ : R → R is convex (concave) onR then we have the inequalities

Φ

(f (x)− 1

b− a

∫ b

a

f (t) dt

)(5.1)

≤ (≥)1

b− a

[∫ x

a

Φ [(t− a) f ′ (t)] dt+

∫ b

x

Φ [(t− b) f ′ (t)] dt

]for anyx ∈ [a, b] .

PROOF. Utilising the Montgomery identity

f (x)− 1

b− a

∫ b

a

f (t) dt(5.2)

=1

b− a

[∫ x

a

(t− a) f ′ (t) dt+

∫ b

x

(t− b) f ′ (t) dt

]=x− a

b− a

(1

x− a

∫ x

a

(t− a) f ′ (t) dt

)+b− x

b− a

(1

b− x

∫ b

x

(t− b) f ′ (t) dt

),

which holds for anyx ∈ (a, b) and the convexity ofΦ : R → R, we have

Φ

(f (x)− 1

b− a

∫ b

a

f (t) dt

)(5.3)

≤ x− a

b− aΦ

(1

x− a

∫ x

a

(t− a) f ′ (t) dt

)+b− x

b− aΦ

(1

b− x

∫ b

x

(t− b) f ′ (t) dt

)


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for anyx ∈ (a, b), which is an inequality of interest in itself as well.If we use Jensen’s integral inequality

Φ

(1

d− c

∫ d

c

g (t) dt

)≤ 1

d− c

∫ d

c

Φ [g (t)] dt

we have

(5.4) Φ

(1

x− a

∫ x

a

(t− a) f ′ (t) dt

)≤ 1

x− a

∫ x

a

Φ [(t− a) f ′ (t)] dt

and

(5.5) Φ

(1

b− x

∫ b

x

(t− b) f ′ (t) dt

)≤ 1

b− x

∫ b

x

Φ [(t− b) f ′ (t)] dt

for anyx ∈ (a, b) .Making use of (5.3)-(5.5) we get the desired result (5.1) for the convex functions.If x = b, then

f (b)− 1

b− a

∫ b

a

f (t) dt =1

b− a

∫ b

a

(t− a) f ′ (t) dt

and by Jensen’s inequality we get

Φ

(f (b)− 1

b− a

∫ b

a

f (t) dt

)≤ 1

b− a

∫ b

a

Φ [(t− a) f ′ (t)] dt,

which proves the inequality (5.1) forx = b.The same argument can be applied forx = a.The case of concave functions goes likewise and the theorem is proved.

COROLLARY 5.2. With the assumptions of Theorem 5.1 we have

Φ (0) ≤ (≥)1

b− a

∫ b

a

Φ

(f (x)− 1

b− a

∫ b

a

f (t) dt

)dx(5.6)

≤ (≥)1

(b− a)2

[∫ b

a

(b− x) Φ [(x− a) f ′ (x)] dx

+

∫ b

a

(x− a) Φ [(x− b) f ′ (x)] dx

].

PROOF. By Jensen’s integral inequality we have

1

b− a

∫ b

a

Φ

(f (x)− 1

b− a

∫ b

a

f (t) dt

)dx

≥ (≤) Φ

[1

b− a

∫ b

a

(f (x)− 1

b− a

∫ b

a

f (t) dt

)dx

]= Φ (0) ,

which proves the first inequality in (5.6).Integrating the inequality (5.1) overx we have

1

b− a

∫ b

a

Φ

(f (x)− 1

b− a

∫ b

a

f (t) dt

)dx(5.7)

≤ (≥)1

(b− a)2

∫ b

a

[∫ x

a

Φ [(t− a) f ′ (t)] dt+

∫ b

x

Φ [(t− b) f ′ (t)] dt

]dx.


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38 S. S. DRAGOMIR

Integrating by parts we have

∫ b

a

(∫ x

a

Φ [(t− a) f ′ (t)] dt

)dx

= x

∫ x

a

Φ [(t− a) f ′ (t)] dt

∣∣∣∣ba

−∫ b

a

xd

(∫ x

a

Φ [(t− a) f ′ (t)] dt

)= b

∫ b

a

Φ [(t− a) f ′ (t)] dt−∫ b

a

xΦ [(x− a) f ′ (x)] dx

=

∫ b

a

(b− x) Φ [(x− a) f ′ (x)] dx

and

∫ b

a

(∫ b

x

Φ [(t− b) f ′ (t)] dt

)dx

= x

(∫ b

x

Φ [(t− b) f ′ (t)] dt

)∣∣∣∣ba

−∫ b

a

xd

(∫ b

x

Φ [(t− b) f ′ (t)] dt

)= −a

(∫ b

a

Φ [(t− b) f ′ (t)] dt

)+

∫ b

a

xΦ [(x− b) f ′ (x)] dx

=

∫ b

a

(x− a) Φ [(x− b) f ′ (x)] dx.

Utilising the inequality (5.7) we deduce the desired inequality (5.6).

REMARK 5.1. If we write the inequality (5.1) for the convex functionΦ (x) = |x|p , p ≥ 1then we get the inequality

∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣p(5.8)

≤ 1

b− a

[∫ x

a

(t− a)p |f ′ (t)|p dt+

∫ b

x

(b− t)p |f ′ (t)|p dt]

for x ∈ [a, b] .


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Utilising Hölder’s inequality we have

B (x) :=

∫ x

a

(t− a)p |f ′ (t)|p dt+

∫ b

x

(b− t)p |f ′ (t)|p dt(5.9)

≤

(x−a)p+1

p+1‖f ′‖p

[a,x],∞ if f ′ ∈ L∞ [a, x] ;

(x−a)p+1/α

(pα+1)1/α ‖f ′‖p[a,x],pβ

if f ′ ∈ Lpβ [a, x] ,α > 1, 1/α+ 1/β = 1;

(x− a)p ‖f ′‖p[a,x],p

+

(b−x)p+1

p+1‖f ′‖p

[x,b],∞ if f ′ ∈ L∞ [x, b] ;

(b−x)p+1/α

(pα+1)1/α ‖f ′‖p[x,b],pβ

if f ′ ∈ Lpβ [x, b] ,α > 1, 1/α+ 1/β = 1;

(b− x)p ‖f ′‖p[x,b],p

for x ∈ [a, b] .Utilising the inequalities (5.8) and (5.9) we have forx ∈ [a, b] that∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣p(5.10)

≤ 1

(b− a) (p+ 1)

[(x− a)p+1 ‖f ′‖p

[a,x],∞ + (b− x)p+1 ‖f ′‖p[x,b],∞

]≤ 1

(p+ 1)

[(x− a

b− a

)p+1

+

(b− x

b− a

)p+1]

(b− a)p ‖f ′‖p[a,b],∞

providedf ′ ∈ L∞ [a, b],∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣p(5.11)

≤ 1

(b− a) (pα + 1)1/α

[(x− a)p+1/α ‖f ′‖p

[a,x],pβ + (b− x)p+1/α ‖f ′‖p[x,b],pβ

]≤ 1

(pα + 1)1/α

[(x− a

b− a

)p+1/α

+

(b− x

b− a

)p+1/α]

(b− a)p−1/β ‖f ′‖p[a,b],pβ

providedf ′ ∈ Lpβ [a, b] , α > 1, 1/α+ 1/β = 1 and∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣p(5.12)

≤ 1

b− a

[(x− a)p ‖f ′‖p

[a,x],p + (b− x)p ‖f ′‖p[x,b],p

]≤ max

(x− a

b− a

)p

,

(b− x

b− a

)p(b− a)p−1 ‖f ′‖p

[a,b],p

=

1

2+

∣∣∣∣∣x− a+b2

b− a

∣∣∣∣∣p

(b− a)p−1 ‖f ′‖p[a,b],p

providedf ′ ∈ Lp [a, b].


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40 S. S. DRAGOMIR

REMARK 5.2. If we takep = 1 in the above inequalities (5.11)-(5.12), then we obtain∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(5.13)

≤ 1

2 (b− a)

[(x− a)2 ‖f ′‖[a,x],∞ + (b− x)2 ‖f ′‖[x,b],∞

]≤ 1

2

[(x− a

b− a

)2

+

(b− x

b− a

)2]

(b− a) ‖f ′‖[a,b],∞

=

1

4+

(x− a+b

2

b− a

)2 (b− a) ‖f ′‖[a,b],∞

for x ∈ [a, b] , providedf ′ ∈ L∞ [a, b],∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(5.14)

≤ 1

(b− a) (α+ 1)1/α

[(x− a)1+1/α ‖f ′‖[a,x],β + (b− x)1+1/α ‖f ′‖[x,b],β

]≤ 1

(α+ 1)1/α

[(x− a

b− a

)1+1/α

+

(b− x

b− a

)1+1/α]

(b− a)1/α ‖f ′‖[a,b],β

for x ∈ [a, b] , providedf ′ ∈ Lβ [a, b] , α > 1, 1/α+ 1/β = 1 and∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(5.15)

≤ 1

b− a

[(x− a) ‖f ′‖[a,x],1 + (b− x) ‖f ′‖[x,b],1

]=

1

2+

∣∣∣∣∣x− a+b2

b− a

∣∣∣∣∣‖f ′‖[a,b],1

for x ∈ [a, b].

5.2. Applications for p-Norms. We have the following inequalities for Lebesgue norms ofthe deviation of a function from its integral mean:

THEOREM 5.3 (Dragomir, 2013 [4]). Letf : [a, b] → R be absolutely continuous on[a, b] .

(i) If f ′ ∈ L∞ [a, b] , then

(5.16)

∥∥∥∥f − 1

b− a

∫ b

a

f (t) dt

∥∥∥∥[a,b],p

≤[

2

(p+ 1) (p+ 2)

]1/p

(b− a)1+ 1p ‖f ′‖[a,b],∞ .

(ii) If f ′ ∈ Lpβ [a, b] , with α > 1, 1/α+ 1/β = 1 then∥∥∥∥f − 1

b− a

∫ b

a

f (t) dt

∥∥∥∥[a,b],p

(5.17)

≤

[2

(pα + 1)1/α (p+ 1/α+ 1)

]1/p

‖f ′‖[a,b],pβ (b− a)1+ 1αp .


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(iii) We have

(5.18)

∥∥∥∥f − 1

b− a

∫ b

a

f (t) dt

∥∥∥∥[a,b],p

≤ 1

2

(2p+1 − 1

p+ 1

)1/p

(b− a) ‖f ′‖[a,b],p .

PROOF. Integrating on[a, b] the inequality (5.10) we have∫ b

a

∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣p dx(5.19)

≤ 1

(b− a) (p+ 1)‖f ′‖p

[a,b],∞

∫ b

a

[(x− a)p+1 + (b− x)p+1] dx

=1

(b− a) (p+ 1)‖f ′‖p

[a,b],∞

[2 (b− a)p+2

p+ 2

]=

2

(p+ 1) (p+ 2)‖f ′‖p

[a,b],∞ (b− a)p+1

which is equivalent with (5.16).Integrating the inequality (5.11)∫ b

a

∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣p dx(5.20)

≤ 1

(b− a) (pα + 1)1/α‖f ′‖p

[a,b],pβ

∫ b

a

[(x− a)p+1/α + (b− x)p+1/α

]dx

=1

(b− a) (pα + 1)1/α‖f ′‖p

[a,b],pβ

[2 (b− a)p+1/α+1

p+ 1/α+ 1

]=

2

(pα + 1)1/α (p+ 1/α+ 1)‖f ′‖p

[a,b],pβ (b− a)p+1/α

which is equivalent with (5.17).Integrating the inequality (5.12) we have∫ b

a

∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣p dx(5.21)

≤ 1

b− a‖f ′‖p

[a,b],p

∫ b

a

max (x− a)p , (b− x)p dx.

Since ∫ b

a

max (x− a)p , (b− x)p dx

=

∫ a+b2

a

(b− x)p dx+

∫ b

a+b2

(x− a)p dx

= −(

b−a2

)p+1

p+ 1+

(b− a)p+1

p+ 1+

(b− a)p+1

p+ 1−(

b−a2

)p+1

p+ 1

=1

p+ 1

(2p+1 − 1

2p

)(b− a)p+1


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42 S. S. DRAGOMIR

then from (5.21) we get∫ b

a

∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣p dx ≤ 1

p+ 1

(2p+1 − 1

2p

)(b− a)p ‖f ′‖p

[a,b],p ,

which is equivalent with (5.18).

5.3. Applications for the Exponential. If we write the inequality (5.1) for the convexfunctionΦ (x) = exp (x) then we get the inequality

exp

[f (x)− 1

b− a

∫ b

a

f (t) dt

](5.22)

≤ 1

b− a

[∫ x

a

exp [(t− a) f ′ (t)] dt+

∫ b

x

exp [(t− b) f ′ (t)] dt

]for x ∈ [a, b] .

If we write the inequality (5.1) for the convex functionΦ (x) = cosh (x) := ex+e−x

2then we

get the inequality

cosh

[f (x)− 1

b− a

∫ b

a

f (t) dt

](5.23)

≤ 1

b− a

[∫ x

a

cosh [(t− a) f ′ (t)] dt+

∫ b

x

cosh [(t− b) f ′ (t)] dt

]for x ∈ [a, b] .

Utilising the inequality (5.22) we have the following multiplicative version of Ostrowski’sinequality:

THEOREM 5.4 (Dragomir, 2013 [4]). Let f : [a, b] → (0,∞) be absolutely continuous on[a, b] . Then we have the inequalities

f (x)

exp[

1b−a

∫ b

aln f (t) dt

](5.24)

≤ 1

b− a

[∫ x

a

exp

[(t− a)

f ′ (t)

f (t)

]dt+

∫ b

x

exp

[(t− b)

f ′ (t)

f (t)

]dt

]for anyx ∈ [a, b] and ∫ b

af (x) dx

exp[

1b−a

∫ b

aln f (t) dt

](5.25)

≤ 1

b− a

[∫ b

a

(b− x) exp

[(x− a)

f ′ (x)

f (x)

]dx

+

∫ b

a

(x− a) exp

[(x− b)

f ′ (x)

f (x)

]dx

].

PROOF. If we replacef by ln f in (5.22) we get

exp

[ln f (x)− 1

b− a

∫ b

a

ln f (t) dt

](5.26)

≤ 1

b− a

[∫ x

a

exp

[(t− a)

f ′ (t)

f (t)

]dt+

∫ b

x

exp

[(t− b)

f ′ (t)

f (t)

]dt

]


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for anyx ∈ [a, b] .Since

exp

[ln f (x)− 1

b− a

∫ b

a

ln f (t) dt

]= exp

[ln f (x)− ln

exp

(1

b− a

∫ b

a

ln f (t) dt

)]

= exp

ln

f (x)

exp(

1b−a

∫ b

aln f (t) dt

)

=f (x)

exp(

1b−a

∫ b

aln f (t) dt

)for anyx ∈ [a, b] , then we get from (5.26) the desired inequality (5.24).

If we integrate the inequality (5.24) we get∫ b

af (x) dx

exp[

1b−a

∫ b

aln f (t) dt

](5.27)

≤ 1

b− a

∫ b

a

[∫ x

a

exp

[(t− a)

f ′ (t)

f (t)

]dt+

∫ b

x

exp

[(t− b)

f ′ (t)

f (t)

]dt

]dx.

Integrating by parts we have∫ b

a

(∫ x

a

exp

[(t− a)

f ′ (t)

f (t)

]dt

)dx

= x

∫ x

a

exp

[(t− a)

f ′ (t)

f (t)

]dt

∣∣∣∣ba

−∫ b

a

xd

(∫ x

a

exp

[(t− a)

f ′ (t)

f (t)

]dt

)= b

∫ b

a

exp

[(t− a)

f ′ (t)

f (t)

]dt−

∫ b

a

x exp

[(x− a)

f ′ (x)

f (x)

]dx

=

∫ b

a

(b− x) exp

[(x− a)

f ′ (x)

f (x)

]dx

and ∫ b

a

(∫ b

x

exp

[(t− b)

f ′ (t)

f (t)

]dt

)dx

= x

∫ b

x

exp

[(t− b)

f ′ (t)

f (t)

]dt

∣∣∣∣ba

−∫ b

a

xd

(∫ b

x

exp

[(t− b)

f ′ (t)

f (t)

]dt

)= −a

∫ b

a

exp

[(t− b)

f ′ (t)

f (t)

]dt+

∫ b

a

x exp

[(x− b)

f ′ (x)

f (x)

]dx

=

∫ b

a

(x− a) exp

[(x− b)

f ′ (x)

f (x)

]dx,

then by (5.27) we deduce the desired inequality (5.25).


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44 S. S. DRAGOMIR

5.4. Applications for Midpoint-Inequalities. We have from the inequality (5.1) writtenfor −f the following result:

PROPOSITION5.5. Let f : [a, b] → R be absolutely continuous on[a, b] . If Φ : R → R isconvex (concave) onR then from (5.1) we have the inequalities

Φ

(1

b− a

∫ b

a

f (t) dt− f

(a+ b

2

))(5.28)

≤ (≥)1

b− a

[∫ b

a+b2

Φ [(b− t) f ′ (t)] dt+

∫ a+b2

a

Φ [(a− t) f ′ (t)] dt

].

If f : [a, b] → R is convex on[a, b] , then by Hermite-Hadamard inequality we have

1

b− a

∫ b

a

f (t) dt ≥ f

(a+ b

2

).

We can state the following result in which the functionΦ is assumed be convex only on[0,∞)or (0,∞) .

PROPOSITION 5.6. If f : [a, b] → R is convex on[a, b] , monotonic nondecreasing on[a, a+b

2

]and monotonic nonincreasing

[a, a+b

2

].If Φ : [0,∞) , (0,∞)→ R is convex (concave)

on [0,∞) or (0,∞) , then (5.28) holds true.

If f : [a, b] → R is strictly convex on[a, b] , monotonic nondecreasing on[a, a+b

2

]and

monotonic nonincreasing[a, a+b

2

], then by takingΦ (x) = lnx, which is strictly concave on

(0,∞) , we get the logarithmic inequality

ln

(1

b− a

∫ b

a

f (t) dt− f

(a+ b

2

))(5.29)

≥ 1

b− a

[∫ b

a+b2

ln [(b− t) f ′ (t)] dt+

∫ a+b2

a

ln [(a− t) f ′ (t)] dt

].

If f : [a, b] → R is convex on[a, b] , monotonic nondecreasing on[a, a+b

2

]and monotonic

nonincreasing[a, a+b

2

], then by takingΦ (x) = xq, with q ∈ (0, 1) we also have(

1

b− a

∫ b

a

f (t) dt− f

(a+ b

2

))q

(5.30)

≥ 1

b− a

[∫ b

a+b2

[(b− t) f ′ (t)]qdt+

∫ a+b2

a

[(a− t) f ′ (t)]qdt

].

If Φ : [0,∞) , (0,∞)→ R is convex (concave) on[0,∞) or (0,∞) , and if we takef (t) :=∣∣t− a+b2

∣∣p , p ≥ 1, then we get from (5.28)

Φ

((b− a)p

2p (p+ 1)

)≤ (≥)

1

b− a

[∫ b

a+b2

Φ

[p (b− t)

(t− a+ b

2

)p−1]dt(5.31)

+

∫ a+b2

a

Φ

[(t− a)

(a+ b

2− t

)p−1]dt

].

Assume thatΦ : R → R is convex (concave) onR.


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Now, if we takef (t) = 1t

in (5.28), wheret ∈ [a, b] ⊂ (0,∞) , then we have

Φ

(A (a, b)− L (a, b)

A (a, b)L (a, b)

)(5.32)

≤ (≥)1

b− a

[∫ b

a+b2

Φ

(t− b

t2

)dt+

∫ a+b2

a

Φ

(t− a

t2

)dt

].

If we takef (t) = − ln t in (5.28), wheret ∈ [a, b] ⊂ (0,∞) , then we have

Φ

(ln

(A (a, b)

I (a, b)

))(5.33)

≤ (≥)1

b− a

[∫ b

a+b2

Φ

(t− b

t

)dt+

∫ a+b2

a

Φ

(t− a

t

)dt

].

If we takef (t) = tp, p ∈ R\ 0,−1 in (5.28), wheret ∈ [a, b] ⊂ (0,∞) , then we have

Φ(Lp

p (a, b)− Ap (a, b))

(5.34)

≤ (≥)1

b− a

[∫ b

a+b2

Φ[p (b− t) tp−1

]dt+

∫ a+b2

a

Φ[p (a− t) tp−1

]dt

].

REFERENCES

[1] S. S. Dragomir, A refinement of Ostrowski’s inequality for absolutely continuous functions andapplications.Acta Math. Vietnam.27 (2002), no. 2, 203–217.

[2] S. S. Dragomir, A refinement of Ostrowski’s inequality for absolutely continuous functions whosederivatives belong toL∞ and applications.Libertas Math.22 (2002), 49–63.

[3] S. S. Dragomir, Improvements of Ostrowski and generalised trapezoid inequality in terms of theupper and lower bounds of the first derivative.Tamkang J. Math.34 (2003), No. 3, 213–222.

[4] S. S. Dragomir, A functional generalization of Ostrowski inequality via Montgomery identity,ActaMath. Univ. Comenian.(N.S.) 84 (2015), no. 1, 63–78. PreprintRGMIA Res. Rep. Coll.16 (2013),Art. 65, pp. 15[Online http://rgmia.org/papers/v16/v16a65.pdf].

[5] S. S. Dragomir and S. Wang, A new inequality of Ostrowski’s type inL1 norm and applicationsto some special means and to some numerical quadrature rules,Tamkang J. of Math., 28 (1997),239-244.

[6] S. S. Dragomir and S. Wang, A new inequality of Ostrowski’s type inLp norm and applications tosome special means and to some numerical quadrature rules,Indian J. of Math., 40 (1998), No. 3,299-304.

[7] S. S. Dragomir and S. Wang, Applications of Ostrowski’s inequality to the estimation of error boundsfor some special means and for some numerical quadrature rules,Appl. Math. Lett.,11 (1) (1998),105-109.

[8] A. M. Fink, Bounds on the derivative of a function from its averages,Czechoslovak Math. J.,42(117) (1992), 289-310.

[9] D. S. Mitrinovic, J. E. Pecaric and A .M. Fink,Inequalities for Functions and Their Integrals andDerivatives, Kluwer Academic Publishers, Dordrecht, 1994.

[10] A. Ostrowski, Über die Absolutabweichung einer differentiierbaren Funktion von ihrem Inte-gralmittelwert,Comment. Math. Helv.,10 (1938), 226-227.


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46 S. S. DRAGOMIR

[11] T. C. Peachey, A. McAndrew and S. S. Dragomir, The best constant in an inequality of Ostrowskitype,Tamkang J. of Math.,30 (1999), No. 3, 219-222.


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CHAPTER 3

Inequalities for Differentiable Functions

1. OSTROWSKI V IA CAUCHY M EAN VALUE THEOREM

1.1. A Local Result. The following theorem holds.

THEOREM 1.1 (Dragomir, 2003 [1]). Let f : [a, b] → R be continuous on[a, b] and differ-entiable on(a, b). Letp ∈ (0,∞) and assume, for a givenx ∈ (a, b), we have that

(1.1) Mp (x) := supu∈(a,b)

|x− u|1−p |f ′ (u)|

<∞.

Then we have the inequality

(1.2)

∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ Mp (x)

p (p+ 1) (b− a)

[(x− a)p+1 + (b− x)p+1] .

PROOF. Letx ∈ (a, b) and define the mappingg1,x : (a, x) → R, g1,x (t) = (x− t)p.Applying the Cauchy mean value theorem, for anyt ∈ (a, x) there exists aη ∈ (t, x) such

that

[f (t)− f (x)] g′1,x (η) = [g1,x (t)− g1,x (x)] f ′ (η)

i.e.,

(−p) (f (t)− f (x)) (x− η)p−1 = (x− t)p f ′ (η)

from where we obtain

(1.3) |f (t)− f (x)| = (x− t)p |f ′ (η)|p (x− η)p−1 ≤ (x− t)p

pMp (x) , t ∈ (a, x) .

We define the mappingg2,x : (x, b) → R, g2,x (t) = (t− x)p. Applying the Cauchy mean valuetheorem, we can find aξ ∈ (x, t) such that

[f (t)− f (x)] p (ξ − x)p−1 = (t− x)p f ′ (ξ)

from where we get

(1.4) |f (t)− f (x)| = (t− x)p |f ′ (ξ)|p (ξ − x)p−1 ≤ (t− x)p

pMp (x) , t ∈ (x, b) .

In conclusion, by (1.3) and (1.4) we may write

(1.5) |f (t)− f (x)| ≤ 1

pMp (x) |t− x|p for all t ∈ (a, b) .

47

48 S. S. DRAGOMIR

Integrating (1.5) overt on [a, b], we get∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

b− a

∫ b

a

|f (t)− f (x)| dt ≤ 1

pMp (x)

1

b− a

∫ b

a

|t− x|p dt

=1

pMp (x)

1

b− a

[∫ x

a

(x− t)p dt+

∫ b

x

(t− x)p dt

]=

1

pMp (x)

(x− a)p+1 + (b− x)p+1

(p+ 1) (b− a),

and the inequality (1.2) is proved.

REMARK 1.1. Forp = 1, we obtain∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ (x− a)2 + (b− x)2

2 (b− a)‖f ′‖∞

=

1

4+

(x− a+b

2

b− a

)2 ‖f ′‖∞ (b− a) , x ∈ [a, b] ,

where‖f ′‖∞ := supt∈(a,b)

|f ′ (t)| < ∞, which is Ostrowski’s inequality [5]. It is obvious that

for p > 1, the accuracy order provided by (1.2) is higher than1, as provided by the classicalOstrowski’s inequality.

REMARK 1.2. If p ∈ (0, 1) andf ′ ∈ L∞ [a, b], then obviously

Mp (x) ≤ [max x− a, b− x]1−p ‖f ′‖∞ =

[a+ b

2+

∣∣∣∣x− a+ b

2

∣∣∣∣]1−p

‖f ′‖∞


The following midpoint formula holds.

COROLLARY 1.2. Letf andp be as in Theorem 1.1. Assume that

Mp

(a+ b

2

):= sup

u∈(a,b)

∣∣∣∣a+ b

2− u

∣∣∣∣1−p

|f ′ (u)|

<∞.

Then we have the midpoint inequality

(1.6)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ (b− a)p

p (p+ 1) 2pMp

(a+ b

2

).

1.2. Some Global Results.Before we continue our presentation, we recall the followingspecial means:

(a) Thearithmetic mean

A = A (a, b) :=a+ b

2, a, b ≥ 0;

(b) Thegeometric mean

G = G (a, b) :=√ab, a, b ≥ 0;


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(c) Theharmonic mean

H = H (a, b) :=2ab

a+ b, a, b > 0;

(d) Thelogarithmic mean

L = L (a, b) :=

a if a = b,

b−aln b−ln a

if a 6= b,a, b > 0;

(e) Theidentric mean

I = I (a, b) :=

a if a = b,

1e

(bb

aa

) 1b−a

if a 6= b,

a, b > 0;

(f) Thep-logarithmic mean

Lp = Lp (a, b) :=

[

bp+1−ap+1

(p+1)(b−a)

] 1p

if a 6= b,

a if a = b,

a, b > 0;

wherep ∈ R\ −1, 0 .The following result also holds.

THEOREM 1.3 (Dragomir, 2003 [1]). Letf : [a, b] → R be continuous on[a, b] with a > 0,and differentiable on(a, b). Letp ∈ R 0 and assume that

(1.7) Kp (f ′) := supu∈(a,b)

u1−p |f ′ (u)|

<∞.

Then we have the inequality:

(1.8)

∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ Kp (f ′)

|p| (b− a)

×

2xp (x− A) + (b− x)Lpp (b, x)− (x− a)Lp

p (x, a) if p ∈ (0,∞) ;

(x− a)Lpp (x, a)− (b− x)Lp

p (b, x)− 2xp (x− A) ifp ∈ (−∞,−1)∪ (−1, 0) ;

(x− a)L−1 (a, x)− (b− x)L−1 (b, x)− 2x

(x− A) if p = −1,


PROOF. Consider the mappingg : [a, b] → R, g (x) = xp. Applying the Cauchy meanvalue theorem, then for anyx andt ∈ [a, b], there exists aη betweenx andt such that

[f (t)− f (x)] g′ (η) = [g (t)− g (x)] f ′ (η)

i.e.,(f (t)− f (x)) pηp−1 = (tp − xp) f ′ (η)

from where we obtain:

|f (t)− f (x)| = |f ′ (η)| |tp − xp||p| ηp−1

≤ Kp (f ′)

|p||tp − xp| .


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50 S. S. DRAGOMIR

In conclusion, for anyt, x ∈ [a, b], we have the inequality

(1.9) |f (t)− f (x)| ≤ Kp (f ′)

|p||tp − xp| .


b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

b− a

∫ b

a

|f (t)− f (x)| dt

≤ Kp (f ′)

p

1

b− a

∫ b

a

|tp − xp| dt.

Forp > 0, we have∫ b

a

|tp − xp| dt =

∫ x

a

(xp − tp) dt+

∫ b

x

(tp − xp) dt

= 2xp (x− A) + (b− x)Lpp (b, x)− (x− a)Lp

p (x, a) .

Forp ∈ (−∞,−1) ∪ (−1, 0), we have∫ b

a

|xp − tp| dt =

∫ x

a

(tp − xp) dt+

∫ b

x

(xp − tp) dt

= (x− a)Lpp (x, a)− (b− x)Lp

p (b, x)− 2xp (x− A)

and, finally, forp = −1, we have∫ b

a

∣∣∣∣1x − 1

t

∣∣∣∣ dt =

∫ x

a

(1

t− 1

x

)dt+

∫ b

x

(1

x− 1

t

)dt

= (x− a)L−1 (a, x)− (b− x)L−1 (b, x)− 2

x(x− A)

and the theorem is proved.

The following corollary is natural.

COROLLARY 1.4. With the assumptions in Theorem 1.3, we have the midpoint inequality

(1.10)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ Kp (f ′)

|p|

×

12

(Lp

p (b, A)− Lpp (A, a)

)if p > 0;

12

(Lp

p (A, a)− Lpp (A, b)

)if p ∈ (−∞,−1) ∪ (−1, 0) ;

12(L−1 (a,A)− L−1 (A, b)) if p = −1.

The following theorem also holds.

THEOREM1.5 (Dragomir, 2003 [1]). Letf : [a, b] → R be continuous on[a, b] (with a > 0)and differentiable on(a, b). If

(1.11) P (f ′) := supu∈(a,b)

|uf ′ (u)| <∞


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then we have the inequality∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.12)

≤ P (f ′)

b− a

[ln

[[I (x, b)]b−x

[I (a, x)]x−a

]+ 2 (x− A) ln x

]


PROOF. Consider the mappingg : [a, b] → R, g (t) = ln t. Applying the Cauchy meanvalue theorem for anyx andt ∈ [a, b] , then there exists aη betweenx andt such that

(f (t)− f (x)) g′ (η) = (g (t)− g (x)) f ′ (η)

i.e.,

(f (t)− f (x))1

η= (ln t− lnx) f ′ (η)

from where we get

|f (t)− f (x)| = |ηf ′ (η)| |ln t− lnx| ≤ P (f ′) |ln t− lnx| .

In conclusion, for anyt, x ∈ [a, b], we have the inequality

(1.13) |f (t)− f (x)| ≤ P (f ′) |ln t− lnx| .


b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

b− a

∫ b

a

|f (t)− f (x)| dt ≤ P (f ′)1

b− a

∫ b

a

|ln t− lnx| dt

=P (f ′)

b− a

[∫ x

a

(lnx− ln t) dt+

∫ b

x

(ln t− lnx) dt

]=P (f ′)

b− a[(x− a) ln x− (x− a) ln I (a, x) + (b− x) ln I (b, x)− (b− x) ln x]

=P (f ′)

b− a[2 (x− A) ln x+ (b− x) ln I (x, b)− (x− a) ln I (a, x)]


The following corollary is natural.

COROLLARY 1.6. With the assumptions of Theorem 1.5, we have the inequality

(1.14)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

2P (f ′) ln

[I (A, b)

I (a,A)

],

whereA = A (a, b) = a+b2

.


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52 S. S. DRAGOMIR

2. OSTROWSKI V IA GENERAL CAUCHY M EAN VALUE THEOREM

2.1. A General Result.We may state the following theorem.

THEOREM 2.1 (Dragomir, 2005 [2]). Let f, g : [a, b] → R be continuous on[a, b] anddifferentiable on(a, b) . If g′ (t) 6= 0 for eacht ∈ (a, b) and

(2.1)

∥∥∥∥f ′g′∥∥∥∥∞

:= supt∈(a,b)

∣∣∣∣f ′ (t)g′ (t)

∣∣∣∣ <∞,

then for anyx ∈ [a, b] one has the inequality∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.2)

≤

∣∣∣∣∣2(x− a+b

2

b− a

)g (x) +

∫ b

xg (t) dt−

∫ x

ag (t) dt

b− a

∣∣∣∣∣ ·∥∥∥∥f ′g′∥∥∥∥∞.

PROOF. Letx, t ∈ [a, b] with t 6= x. Applying Cauchy’s mean value theorem, there exists aη betweent andx such that

(f (x)− f (t)) =f ′ (η)

g′ (η)(g (x)− g (t))

from where we get

(2.3) |f (x)− f (t)| =∣∣∣∣f ′ (η)g′ (η)

∣∣∣∣ |g (x)− g (t)| ≤∥∥∥∥f ′g′∥∥∥∥∞|g (x)− g (t)| ,

for anyt, x ∈ [a, b] .Using the properties of the integral, we deduce by (2.3), that∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

b− a

∫ b

a

|f (x)− f (t)| dt(2.4)

≤∥∥∥∥f ′g′∥∥∥∥∞

1

b− a

∫ b

a

|g (x)− g (t)| dt.

Sinceg′ (t) 6= 0 on (a, b) , it follows that eitherg′ (t) > 0 or g′ (t) < 0 for anyt ∈ (a, b) .If g′ (t) > 0 for all t ∈ (a, b) , theng is strictly monotonic increasing on(a, b) and∫ b

a

|g (x)− g (t)| dt =

∫ x

a

(g (x)− g (t)) dt+

∫ b

x

(g (t)− g (x)) dt

= 2

(x− a+ b

2

)g (x) +

∫ b

x

g (t) dt−∫ x

a

g (t) dt.

If g′ (t) < 0 for all t ∈ (a, b) , then∫ b

a

|g (x)− g (t)| dt = −[2

(x− a+ b

2

)g (x) +

∫ b

x

g (t) dt−∫ x

a

g (t) dt

]and the inequality (2.2) is proved.

The following midpoint inequality is a natural consequence of the above result.


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COROLLARY 2.2. With the above assumptions forf andg, one has the inequality∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.5)

≤ 1

b− a

∣∣∣∣∣∫ b

a+b2

g (t) dt−∫ a+b

2

a

g (t) dt

∣∣∣∣∣∥∥∥∥f ′g′∥∥∥∥∞.

REMARK 2.1. (1)(2) If in the above theorem, we chooseg (t) = t, then from (2.2) we recapture Ostrowski’s

inequality [5].(3) If in Theorem 2.1 we chooseg (t) = tp, p ∈ R\ 0 , or g (t) = ln t with t ∈ (a, b) ⊂

(0,∞) , then we obtain the results from [1].

One may obtain many inequalities from Theorem 2.1 on choosing different instances offunctionsg.

PROPOSITION2.3 (Dragomir, 2005 [2]). Let f : [a, b] ⊂ R → R be continuous on[a, b]and differentiable on(a, b) . If there exists a constantΓ <∞ such that

(2.6) |f ′ (t)| ≤ Γe−t for any t ∈ (a, b) ,

then one has the inequality:∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.7)

≤ Γ

[2

(x− A

b− a

)ex +

(b− x)E (x, b)− (x− a)E (a, x)

b− a

]for anyx ∈ (a, b) , whereA = A (a, b) = a+b

2andE is the exponential men, i.e.,

E (x, y) :=

ex − ey

x− yif x 6= y

ey if x = y

, x, y ∈ R.

In particular, we have

(2.8)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

2[E (A, b)− E (a,A)] Γ.

The proof is obvious by Theorem 2.1 on choosingg (t) = et and we omit the details.Another example is considered in the following proposition.

PROPOSITION2.4 (Dragomir, 2005 [2]). Let f : [a, b] ⊂(0, π

2

)→ R be continuous on

[a, b] and differentiable on(a, b) .

(i) If there exists a constantΓ1 <∞ such that

(2.9) |f ′ (t)| ≤ Γ1 cos t, t ∈ (a, b) ,

then one has the inequality∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.10)

≤ Γ1

[2

(x− A

b− a

)sin x+

(x− a)C (a, x)− (b− x)C (x, b)

b− a

]


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54 S. S. DRAGOMIR

for anyx ∈ (a, b) , whereC is the cos-mean value, i.e.,

C (x, y) :=

cosx− cos y

x− yif x 6= y

− sin y if x = y

.

In particular we have

(2.11)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

2[C (a,A)− C (A, b)] Γ1.

(ii) If there exists a constantΓ2 <∞ such that

(2.12) |f ′ (t)| ≤ Γ1 sin t, t ∈ (a, b) ,

then one has the inequality

∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ Γ2

[2

(x− A

b− a

)cosx+

(b− x)S (x, b)− (x− a)S (a, x)

b− a

],

for anyx ∈ (a, b) , whereS is the sin-mean value, i.e.,

S (x, y) :=

sin x− sin y

x− yif x 6= y

cos y if x = y

.


(2.13)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

2[S (A, b)− S (a,A)] Γ2.

2.2. A Result on Sub-intervals.The following result also holds.

THEOREM 2.5 (Dragomir, 2005 [2]). Let f, g : [a, b] → R be continuous on[a, b] anddifferentiable on(a, b) \ x , x ∈ (a, b) . If g′ (t) 6= 0 for t ∈ (a, x) ∪ (x, b) , then we have theinequality ∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.14)

≤ 1

b− a

∣∣∣∣g (x) (x− a)−∫ x

a

g (t) dt

∣∣∣∣ · ∥∥∥∥f ′g′∥∥∥∥

(a,x),∞

+1

b− a

∣∣∣∣g (x) (b− x)−∫ b

x

g (t) dt

∣∣∣∣ · ∥∥∥∥f ′g′∥∥∥∥

(x,b),∞.


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PROOF. We obviously have:∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.15)

=

∣∣∣∣ 1

b− a

∫ b

a

(f (x)− f (t)) dt

∣∣∣∣≤ 1

b− a

∫ b

a

|f (x)− f (t)| dt

=1

b− a

[∫ x

a

|f (x)− f (t)| dt+

∫ b

x

|f (x)− f (t)| dt].

Applying Cauchy’s mean value theorem on the interval(a, x) , we deduce (see the proof ofTheorem 2.1) that

(2.16) |f (x)− f (t)| ≤∥∥∥∥f ′g′∥∥∥∥

(a,x),∞|g (x)− g (t)|

for anyt ∈ (a, x) , and, similarly

(2.17) |f (x)− f (t)| ≤∥∥∥∥f ′g′∥∥∥∥

(x,b),∞|g (x)− g (t)|

for anyt ∈ (x, b) .Consequently∫ x

a

|f (x)− f (t)| dt ≤∥∥∥∥f ′g′∥∥∥∥

(a,x),∞

∫ x

a

|g (x)− g (t)| dt

and ∫ b

x

|f (x)− f (t)| dt ≤∥∥∥∥f ′g′∥∥∥∥

(x,b),∞

∫ b

x

|g (x)− g (t)| dt.

Sinceg′ has a constant sign in either(a, x) or (x, b) , it follows thatg is strictly increasing orstrictly decreasing in(a, x) and(x, b) .

Thus ∫ x

a

|g (x)− g (t)| dt

=

g (x) (x− a)−∫ x

ag (t) dt if g is increasing on[a, x]∫ x

ag (t) dt− g (x) (x− a) if g is decreasing

=

∣∣∣∣g (x) (x− a)−∫ x

a

g (t) dt

∣∣∣∣and, in a similar way∫ b

x

|g (x)− g (t)| dt =

∣∣∣∣g (x) (b− x)−∫ b

x

g (t) dt

∣∣∣∣ .Consequently, by the use of (2.15), we deduce the desired inequality (2.14).

The following particular case may be of interest.


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56 S. S. DRAGOMIR

COROLLARY 2.6. Let f, g : [a, b] → R be continuous on[a, b] and differentiable on(a, b) \

a+b2

. If g′ (t) 6= 0 on

(a, a+b

2

)∪(

a+b2, b), then we have the inequality

(2.18)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

2

∣∣∣∣∣g(a+ b

2

)− 2

b− a

∫ a+b2

a

g (t) dt

∣∣∣∣∣ ·∥∥∥∥f ′g′∥∥∥∥(a, a+b

2 ),∞

+

∣∣∣∣∣g(a+ b

2

)− 2

b− a

∫ b

a+b2

g (t) dt

∣∣∣∣∣ ·∥∥∥∥f ′g′∥∥∥∥(a+b

2,b),∞

.

The following result also holds.

PROPOSITION2.7 (Dragomir, 2005 [2]). Let f : [a, b] → R be continuous on[a, b] anddifferentiable on(a, b) \ x , x ∈ (a, b) . Assume that, forp > 0, we have

(2.19) |f ′ (t)| ≤

M1,p (x) (x− t)1−p for any t ∈ (a, x) ,

M2,p (x) (t− x)1−p for any t ∈ (x, b) .

whereM1,p (x) andM2,p (x) are positive constants depending onx. Then we have the inequality∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.20)

≤ 1

p (p+ 1)(b− a)

[M1,p (x) (x− a)p+1 +M2,p (x) (b− x)p+1] .

The proof follows by Theorem 2.5 applied forg (x) = |x− t|p , p > 0. We omit the details.

REMARK 2.2. If f is as in Proposition 2.7 and

(2.21) |f ′ (t)| ≤

M1

(a+b2

) (a+b2− t)1−p

for any t ∈(a, a+b

2

),

M2

(a+b2

) (t− a+b

2

)1−pfor any t ∈

(a+b2, b),

then, by (2.20), we get∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.22)

≤ (b− a)p+1

2p+1p (p+ 1)

[M1

(a+ b

2

)+M2

(a+ b

2

)].

REMARK 2.3. If f is as in Proposition 2.7 and

|f ′ (t)| ≤Mp (x) |x− t|1−p t ∈ (a, b) ,

then, by (2.20), we get∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.23)

≤ 1

p (p+ 1) (b− a)

[(x− a)p+1 + (b− x)p+1]Mp (x) ,

which is the result obtained in [1].


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2.3. Some Inequalities of Midpoint Type.(1) Let0 < a < b. Consider the functiong : [a, b] → R, g (t) = tp, t ∈ R\ 0,−1 . Then

g′ (t) = ptp−1, g(

a+b2

)= Ap (a, b) ,

2

b− a

∫ a+b2

a

g (t) dt = Lpp (a,A) ,

2

b− a

∫ b

a+b2

g (t) dt = Lpp (A, b) ,

and by Corollary 2.6, we may state the following proposition.

PROPOSITION 2.8. Let f : [a, b] ⊂ (0,∞) → R be continuous on[a, b] anddifferentiable on(a, b) \

a+b2

. If

(2.24) |f ′ (t)| ≤

M1

(a+b2

)tp, t ∈

(a, a+b

2

),

M2

(a+b2

)tp, t ∈

(a+b2, b),


(2.25)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

2p

M1

(a+ b

2

) ∣∣Ap (a, b)− Lpp (a,A)

∣∣+ M2

(a+ b

2

) ∣∣Lpp (A, b)− Ap (a, b)

∣∣ .The particular casep = 1 is of interest and so we may state the following corollary.

COROLLARY 2.9. Let f : [a, b] ⊂ (0,∞) → R be continuous on[a, b] and differ-entiable on(a, b) \

a+b2

. If

(2.26) |f ′ (t)| ≤

N1

(a+b2

)t, t ∈

(a, a+b

2

),

N2

(a+b2

)t, t ∈

(a+b2, b),

then we have the inequality:∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.27)

≤ 1

8

[N1

(a+ b

2

)+N2

(a+ b

2

)](b− a) .

(2) Let 0 < a < b. Consider the functiong : [a, b] → R, g (t) = 1t. Theng′ (t) = − 1

t2,

g(

a+b2

)= A−1 (a, b) ,

2

b− a

∫ a+b2

a

g (t) dt = L−1 (a,A) ,

2

b− a

∫ b

a+b2

g (t) dt = L−1 (A, b) ,

and by Corollary 2.6 we may state the following Proposition.


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58 S. S. DRAGOMIR

PROPOSITION2.10. Let f : [a, b] ⊂ (0,∞) → R be continuous on[a, b] anddifferentiable on(a, b) \

a+b2

. If

(2.28) |f ′ (t)| ≤

M1

(a+b2

)t−2, t ∈

(a, a+b

2

),

M2

(a+b2

)t−2, t ∈

(a+b2, b),

then we have the inequality:

(2.29)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

2

[M1

(a+ b

2

)· [A− L (a,A)]

L (a,A)A

+ M2

(a+ b

2

)· [L (A, b)− A]

L (A, b)A

].

(3) Let 0 < a < b. Consider the functiong : [a, b] → R, g (t) = ln t. Theng′ (t) = 1t,

g(

a+b2

)= lnA,

2

b− a

∫ a+b2

a

g (t) dt = ln I (a,A) ,

2

b− a

∫ b

a+b2

g (t) dt = ln I (A, b) ,

and by Corollary 2.6 we may state the following proposition.

PROPOSITION2.11. Let f : [a, b] ⊂ (0,∞) → R be continuous on[a, b] anddifferentiable on(a, b) \

a+b2

. If

(2.30) |f ′ (t)| ≤

M1

(a+b2

)t−1, t ∈

(a, a+b

2

),

M2

(a+b2

)t−1, t ∈

(a+b2, b),

then we have the inequality:∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.31)

≤ ln

G

([A

I (a,A)

]M1(a+b2 )

,

[I (A, b)

A

]M2(a+b2 ))

.

3. OSTROWSKI V IA POMPEIU M EAN VALUE THEOREM

3.1. Pompeiu Mean Value Theorem.In 1946, Pompeiu [6] derived a variant of Lagrange’smean value theorem, now known asPompeiu’s mean value theorem(see also [7, p. 83]).

LEMMA 3.1 (Pompeiu, 1946 [6]). For every real valued functionf differentiable on aninterval [a, b] not containing0 and for all pairsx1 6= x2 in [a, b] , there exists a pointξ betweenx1 andx2 such that

(3.1)x1f (x2)− x2f (x1)

x1 − x2

= f (ξ)− ξf ′ (ξ) .


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PROOF. Define a real valued functionF on the interval[

1b, 1

a

]by

(3.2) F (t) = tf

(1

t

).

Sincef is differentiable on(

1b, 1

a

)and

(3.3) F ′ (t) = f

(1

t

)− 1

tf ′(

1

t

),

then applying the mean value theorem toF on the interval[x, y] ⊂[

1b, 1

a

]we get

(3.4)F (x)− F (y)

x− y= F ′ (η)

for someη ∈ (x, y) .Let x2 = 1

x, x1 = 1

yandξ = 1

η. Then, sinceη ∈ (x, y) , we have

x1 < ξ < x2.

Now, using (3.2) and (3.3) on (3.4), we have

xf(

1x

)− yf

(1y

)x− y

= f

(1

η

)− 1

ηf ′(

1

η

),

that isx1f (x2)− x2f (x1)

x1 − x2

= f (ξ)− ξf ′ (ξ) .

This completes the proof of the theorem.

REMARK 3.1. Following [7, p. 84 – 85], we will mention here a geometrical interpreta-tion of Pompeiu’s theorem. The equation of the secant line joining the points(x1, f (x1)) and(x2, f (x2)) is given by

y = f (x1) +f (x2)− f (x1)

x2 − x1

(x− x1) .

This line intersects they−axis at the point(0, y) , wherey is

y = f (x1) +f (x2)− f (x1)

x2 − x1

(0− x1)

=x1f (x2)− x2f (x1)

x1 − x2

.

The equation of the tangent line at the point(ξ, f (ξ)) is

y = (x− ξ) f ′ (ξ) + f (ξ) .

The tangent line intersects they−axis at the point(0, y) , where

y = −ξf ′ (ξ) + f (ξ) .

Hence, the geometric meaning of Pompeiu’s mean value theorem is that the tangent of the point(ξ, f (ξ)) intersects on they−axis at the same point as the secant line connecting the points(x1, f (x1)) and(x2, f (x2)) .


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60 S. S. DRAGOMIR

3.2. Evaluating the Integral Mean. The following result holds.

THEOREM 3.2 (Dragomir, 2005 [3]). Let f : [a, b] → R be continuous on[a, b] and differ-entiable 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

∫ b

a

f (t) dt

∣∣∣∣(3.5)

≤ b− a

|x|

1

4+

(x− a+b

2

b− a

)2 ‖f − `f ′‖∞ ,

where` (t) = t, t ∈ [a, b] .The constant1

4is sharp in the sense that it cannot be replaced by a smaller constant.

PROOF. Applying Pompeiu’s mean value theorem, for anyx, t ∈ [a, b] , there is aξ betweenx andt such that

tf (x)− xf (t) = [f (ξ)− ξf ′ (ξ)] (t− x)

giving

|tf (x)− xf (t)| ≤ supξ∈[a,b]

|f (ξ)− ξf ′ (ξ)| |x− t|(3.6)

= ‖f − `f ′‖∞ |x− t|

for anyt, x ∈ [a, b] .Integrating overt ∈ [a, b] , we get∣∣∣∣f (x)

∫ b

a

tdt− x

∫ b

a

f (t) dt

∣∣∣∣(3.7)

≤∫ b

a

|tf (x)− xf (t)| dt

≤ ‖f − `f ′‖∞∫ b

a

|x− t| dt

= ‖f − `f ′‖∞

[(x− a)2 + (b− x)2

2

]

= ‖f − `f ′‖∞

[1

4(b− a)2 +

(x− a+ b

2

)2]

and since∫ b

atdt = b2−a2

2, we deduce from (3.7) the desired result (3.5).

Now, assume that (3.6) holds with a constantk > 0, i.e.,∣∣∣∣a+ b

2· f (x)

x− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.8)

≤ b− a

|x|

k +

(x− a+b

2

b− a

)2 ‖f − `f ′‖∞ ,



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Considerf : [a, b] → R, f (t) = αt+ β; α, β 6= 0. Then

‖f − `f ′‖∞ = |β| ,1

b− a

∫ b

a

f (t) dt =a+ b

2· α+ β,

and by (3.8) we deduce∣∣∣∣a+ b

2

(α+

β

x

)−(a+ b

2α+ β

)∣∣∣∣ ≤ b− a

|x|

k +

(x− a+b

2

b− a

)2 |β|

giving

(3.9)

∣∣∣∣a+ b

2− x

∣∣∣∣ ≤ (b− a) k +

(x− a+b

2

b− a

)2

for anyx ∈ [a, b] .If in (3.9) we letx = a or x = b, we deducek ≥ 1

4, and the sharpness of the constant is

proved.

The following interesting particular case holds.

COROLLARY 3.3. With the assumptions in Theorem 3.2, we have

(3.10)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ b− a

2 |a+ b|‖f − `f ′‖∞ .

4. ANOTHER OSTROWSKI TYPE I NEQUALITY V IA POMPEIU ’ S RESULT

4.1. Evaluating the Integral Mean. The following new result holds.

THEOREM 4.1 (Dragomir, 2013 [4]). Let f : [a, b] → R be continuous on[a, b] and differ-entiable on(a, b) with b > a > 0. Then for anyx ∈ [a, b] , we have the inequality∣∣∣∣f (x)

x− 1

b− a

∫ b

a

f (t)

tdt

∣∣∣∣(4.1)

≤ 2

b− a‖f − `f ′‖∞

(ln

x√ab

+a+b2− x

x

),

where` (t) = t, t ∈ [a, b] .The constant2 is best possible in (4.1).

PROOF. Applying Pompeiu’s mean value theorem [6] (see also [7, p. 83]), for anyx, t ∈[a, b] , there is aξ betweenx andt such that

tf (x)− xf (t) = [f (ξ)− ξf ′ (ξ)] (t− x)

giving|tf (x)− xf (t)| ≤ sup

ξ∈[a,b]

|f (ξ)− ξf ′ (ξ)| |x− t| = ‖f − `f ′‖∞ |x− t|

for anyt, x ∈ [a, b] , or, by dividing withx, t > 0, equivalently to

(4.2)

∣∣∣∣f (x)

x− f (t)

t

∣∣∣∣ ≤ ‖f − `f ′‖∞

∣∣∣∣1x − 1

t

∣∣∣∣for anyt, x ∈ [a, b] .


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62 S. S. DRAGOMIR

Integrating overt ∈ [a, b] , we get∣∣∣∣f (x)

x(b− a)−

∫ b

a

f (t)

tdt

∣∣∣∣ ≤ ∫ b

a

∣∣∣∣f (x)

x− f (t)

t

∣∣∣∣ dt(4.3)

≤ ‖f − `f ′‖∞∫ b

a

∣∣∣∣1x − 1

t

∣∣∣∣ dtand since ∫ b

a

∣∣∣∣1x − 1

t

∣∣∣∣ dt =

[∫ x

a

(1

t− 1

x

)dt+

∫ b

x

(1

x− 1

t

)dt

]=

(lnx

a− x− a

x+b− x

x− ln

b

x

)=

(lnx2

ab+a+ b− 2x

x

)for anyx ∈ [a, b] , then we deduce from (4.3) the desired result (4.1).

Now, assume that (4.1) holds with a constantk > 0, i.e.,∣∣∣∣f (x)

x− 1

b− a

∫ b

a

f (t)

tdt

∣∣∣∣(4.4)

≤ k

b− a‖f − `f ′‖∞

(ln

x√ab

+a+b2− x

x

),

for anyx ∈ [a, b] .Considerf : [a, b] → R, f (t) = 1. Then

‖f − `f ′‖∞ = 1,1

b− a

∫ b

a

f (t)

tdt =

1

b− alnb

a,

and by (4.4) we deduce∣∣∣∣1x − 1

b− alnb

a

∣∣∣∣ ≤ k

b− a

(ln

x√ab

+a+b2− x

x

)for anyx ∈ [a, b] .

If we take in this inequalityx = a, we get∣∣∣∣1a − 1

b− alnb

a

∣∣∣∣ ≤ k

b− a

(ln

a√ab

+b− a

2a

)(4.5)

=k

2 (b− a)

(lna2

ab+b− a

a

)=

k

2 (b− a)

(lna

b+b− a

a

).

In we multiply (4.5) with2 (b− a) we get

2

∣∣∣∣b− a

a− ln

b

a

∣∣∣∣ ≤ k

(b− a

a− ln

b

a

)which implies thatk ≥ 2.

The following interesting particular case holds.


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COROLLARY 4.2. With the assumptions in Theorem 4.1, we have

(4.6)

∣∣∣∣∣f(

a+b2

)a+b2

− 1

b− a

∫ b

a

f (t)

tdt

∣∣∣∣∣ ≤ 2

b− a‖f − `f ′‖∞ ln

(a+b2√ab

).

REMARK 4.1. If we consider the functionψ : [a, b] → R given by

ψ (x) := lnx√ab

+a+b2− x

x,

then we observe that

ψ′ (x) =x− a+b

2

x2,

which shows that

infx∈[a,b]

ψ (x) = ψ

(a+ b

2

)= ln

(a+b2√ab

),

meaning that the inequality (4.6) is the best possible one can get from (4.1).

REMARK 4.2. We can state from (4.1) the following inequality as well:

(4.7)

∣∣∣∣∣∣f(√

ab)

√ab

− 1

b− a

∫ b

a

f (t)

tdt

∣∣∣∣∣∣ ≤ 2

b− a‖f − `f ′‖∞

(a+b2−√ab

√ab

).

REFERENCES

[1] S. S. Dragomir, Some new inequalities of Ostrowski type,Austral. Math. Soc. Gaz. 30(2003), No. 1, 25–30. Preprint:RGMIA Res. Rep. Coll., 5 (2002), Supplement, Article 11,[ONLINE:http://rgmia.org/papers/v5e/IOT.pdf].

[2] S. S. Dragomir, Some Ostrowski type inequalities via Cauchy’s mean value theorem.New ZealandJ. Math.34 (2005), No. 1, 31–42.

[3] S. S. Dragomir, An inequality of Ostrowski type via Pompeiu’s mean value theo-rem. J. Inequal. Pure Appl. Math.6 (2005), No. 3, Article 83, 9 pp. [Onlinehttp://www.emis.de/journals/JIPAM/article556.html?sid=556].

[4] S. S. Dragomir, Another Ostrowski type inequality via Pompeiu’s mean value theorem,GeneralMathematics(Sibiu, Romania),21 (2013), No. 2, 3-16. PreprintRGMIA Res. Rep. Coll.16 (2013),Art. 64, pp. 9[Online http://rgmia.org/papers/v16/v16a64.pdf].

[5] A. Ostrowski, Uber die Absolutabweichung einer differentienbaren Funktionen von ihren Inte-gralmittelwert,Comment. Math. Hel,10 (1938), 226-227.

[6] D. Pompeiu, Sur une proposition analogue au théorème des accroissements finis,Mathematica(Cluj,Romania),22 (1946), 143-146.

[7] P. K. Sahoo and T. Riedel,Mean Value Theorems and Functional Equations, World Scientific, Sin-gapore, New Jersey, London, Hong Kong, 2000.


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CHAPTER 4

Inequalities of Pompeiu Type

1. OSTROWSKI V IA A GENERALIZED POMPEIU ’ S I NEQUALITY

1.1. Pompeiu’s Inequality for p-Norms. In 1946, Pompeiu [6] derived a variant of La-grange’s mean value theorem, now known asPompeiu’s mean value theorem(see also [8, p.83]).

LEMMA 1.1 (Pompeiu, 1946 [6]). For every real valued functionf differentiable on aninterval [a, b] not containing0 and for all pairsx1 6= x2 in [a, b] , there exists a pointξ betweenx1 andx2 such that

(1.1)x1f (x2)− x2f (x1)

x1 − x2

= f (ξ)− ξf ′ (ξ) .

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

COROLLARY 1.2 (Pompeiu’s Inequality).With the assumptions of Lemma 1.1 and if‖f − `f ′‖∞ =supt∈(a,b) |f (t)− tf ′ (t)| <∞ where` (t) = t, t ∈ [a, b] , then

(1.2) |tf (x)− xf (t)| ≤ ‖f − `f ′‖∞ |x− t|

for anyt, x ∈ [a, b] .

The inequality (1.2) was obtained by the author in [1].We can generalize the above inequality for the larger class of functions that are absolutely

continuous and complex valued as well as for other norms of the differencef − `f ′.

LEMMA 1.3 (Dragomir, 2013 [3]). Letf : [a, b] → C be an absolutely continuous functionon the interval[a, b] with b > a > 0. Then for anyt, x ∈ [a, b] we have

|tf (x)− xf (t)|(1.3)

≤

‖f − `f ′‖∞ |x− t| if f − `f ′ ∈ L∞ [a, b]

12q−1

‖f − `f ′‖p

∣∣ xq

tq−1 − tq

xq−1

∣∣1/qif f − `f ′ ∈ Lp [a, b]p > 1,

1p

+ 1q

= 1

‖f − `f ′‖1maxt,xmint,x

65

66 S. S. DRAGOMIR

or, equivalently∣∣∣∣f (x)

x− f (t)

t

∣∣∣∣(1.4)

≤

‖f − `f ′‖∞∣∣1

t− 1

x

∣∣ if f − `f ′ ∈ L∞ [a, b]

12q−1

‖f − `f ′‖p

∣∣ 1t2q−1 − 1

x2q−1

∣∣1/qif f − `f ′ ∈ Lp [a, b]

p > 1,1p

+ 1q

= 1

‖f − `f ′‖11

mint2,x2

PROOF. If f is absolutely continuous, thenf/` is absolutely continuous on the interval[a, b]that does not containing0 and∫ x

t

(f (s)

s

)′ds =

f (x)

x− f (t)

t

for anyt, x ∈ [a, b] with x 6= t.Since ∫ x

t

(f (s)

s

)′ds =

∫ x

t

f ′ (s) s− f (s)

s2ds

then we get the following identity

(1.5) tf (x)− xf (t) = xt

∫ x

t

f ′ (s) s− f (s)

s2ds

for anyt, x ∈ [a, b] .We notice that the equality (1.5) was proved for the smaller class of differentiable function

and in a different manner in [7].Taking the modulus in (1.5) we have

|tf (x)− xf (t)| =

∣∣∣∣xt∫ x

t

f ′ (s) s− f (s)

s2ds

∣∣∣∣(1.6)

≤ xt

∣∣∣∣∫ x

t

∣∣∣∣f ′ (s) s− f (s)

s2

∣∣∣∣ ds∣∣∣∣ := I

and utilizing Hölder’s integral inequality we deduce

I ≤ xt

sups∈[t,x]([x,t]) |f ′ (s) s− f (s)|∣∣∫ x

t1s2ds

∣∣∣∣∫ x

t|f ′ (s) s− f (s)|p ds

∣∣1/p ∣∣∫ x

t1

s2q ds∣∣1/q p > 1,

1p

+ 1q

= 1∣∣∫ x

t|f ′ (s) s− f (s)| ds

∣∣ sups∈[t,x]([x,t])

1s2

(1.7)

≤

‖f − `f ′‖∞ |x− t|

12q−1

‖f − `f ′‖p

∣∣ xq

tq−1 − tq

xq−1

∣∣1/q p > 1,1p

+ 1q

= 1

‖f − `f ′‖1maxt,xmint,x



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REMARK 1.1. The first inequality in (1.3) also holds in the same form for0 > b > a.

REMARK 1.2. If we take in (1.3)x = A = A (a, b) := a+b2

(the arithmetic mean) andt = G = G (a, b) :=

√ab (the geometric mean) then we get the simple inequality for functions

of means:

|Gf (A)− Af (G)|(1.8)

≤

‖f − `f ′‖∞ (A−G) if f − `f ′ ∈ L∞ [a, b]

12q−1

‖f − `f ′‖p

(A2q−1−G2q−1)1/q

A1/pG1/p

if f − `f ′ ∈ Lp [a, b]p > 1,

1p

+ 1q

= 1

‖f − `f ′‖1AG

1.2. Evaluating the Integral Mean. The following new result holds.

THEOREM 1.4 (Dragomir, 2013 [3]). Let f : [a, b] → C be an absolutely continuous func-tion on the interval[a, b] with b > a > 0. Then for anyx ∈ [a, b] we have∣∣∣∣a+ b

2· f (x)

x− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.9)

≤

b−ax

[14

+(

x−a+b2

b−a

)2]‖f − `f ′‖∞ if f − `f ′ ∈ L∞ [a, b]

1

(2q−1)x(b−a)1/q ‖f − `f ′‖p [Bq (a, b;x)]1/qif f − `f ′ ∈ Lp [a, b]p > 1,

1p

+ 1q

= 1

1b−a

‖f − `f ′‖1

(ln x

a+ b2−x2

2x2

),

,

where

(1.10) Bq (a, b;x) =

xq

2−q(2xq−2 − aq−2 − bq−2)

+ 1xq−1(q+1)

(bq+1 + aq+1 − 2xq+1) ,q 6= 2

x2 ln x2

ab+ b3+a3−2x3

3x, q = 2

PROOF. The first inequality can be proved in an identical way to the case of differentiablefunctions from [1] by utilizing the first inequality in (1.3).

Utilising the second inequality in (1.3) we have∣∣∣∣a+ b

2· f (x)− x

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.11)

≤ 1

b− a

∫ b

a


≤ 1

(2q − 1) (b− a)‖f − `f ′‖p

∫ b

a

∣∣∣∣ xq

tq−1− tq

xq−1

∣∣∣∣1/q

dt


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68 S. S. DRAGOMIR

Utilising Hölder’s integral inequality we have∫ b

a

∣∣∣∣ xq

tq−1− tq

xq−1

∣∣∣∣1/q

dt(1.12)

≤(∫ b

a

dt

)1/p(∫ b

a

[∣∣∣∣ xq

tq−1− tq

xq−1

∣∣∣∣1/q]q

dt

)1/q

= (b− a)1/p

(∫ b

a

∣∣∣∣ xq

tq−1− tq

xq−1

∣∣∣∣ dt)1/q

.

For q 6= 2 we have

∫ b

a

∣∣∣∣ xq

tq−1− tq

xq−1

∣∣∣∣ dt=

∫ x

a

(xq

tq−1− tq

xq−1

)dt+

∫ b

x

(tq

xq−1− xq

tq−1

)dt

= xq

∫ x

a

dt

tq−1− 1

xq−1

∫ x

a

tqdt+1

xq−1

∫ b

x

tqdt− xq

∫ b

x

1

tq−1dt

=xq

2− q

(1

x2−q− 1

a2−q

)− 1

xq−1 (q + 1)

(xq+1 − aq+1

)+

1

xq−1 (q + 1)

(bq+1 − xq+1

)− xq

2− q

(1

b2−q− 1

x2−q

)

=xq

2− q

(1

x2−q− 1

a2−q− 1

b2−q+

1

x2−q

)+

1

xq−1 (q + 1)

(bq+1 − xq+1 − xq+1 + aq+1

)=

xq

2− q

(2xq−2 − aq−2 − bq−2

)+

1

xq−1 (q + 1)

(bq+1 + aq+1 − 2xq+1

)= Bq (a, b;x) .

For q = 2 we have ∫ b

a

∣∣∣∣x2

t− t2

x

∣∣∣∣ dt=

∫ x

a

(x2

t− t2

x

)dt+

∫ b

x

(t2

x− x2

t

)dt

= x2

∫ x

a

dt

t− 1

x

∫ x

a

t2dt+1

x

∫ b

x

t2dt− x2

∫ b

x

1

tdt

= x2 lnx

a− 1

x

x3 − a3

3+

1

x

b3 − x3

3− x2 ln

b

x

= x2 lnx2

ab+

1

x

b3 + a3 − 2x3

3= B2 (a, b;x) .

Utilizing (1.11) and (1.12) we get the second inequality in (1.9).


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Utilising the third inequality in (1.3) we have∣∣∣∣a+ b

2· f (x)− x

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

b− a

∫ b

a

|tf (x)− xf (t)| dt(1.13)

≤ 1

b− a‖f − `f ′‖1

∫ b

a

max t, xmin t, x

dt.

Since ∫ b

a

max t, xmin t, x

dt =

∫ x

a

x

tdt+

∫ b

x

t

xdt = x ln

x

a+

1

x

b2 − x2

2,

then by (1.13) we have∣∣∣∣a+ b

2· f (x)− x

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

b− a

∫ b

a


≤ 1

b− a‖f − `f ′‖1

[x ln

x

a+

1

x

b2 − x2

2

],

and the last part of (1.9) is thus proved.


, then we get∣∣∣∣f (A)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.14)

≤

b−a4A‖f − `f ′‖∞ if f − `f ′ ∈ L∞ [a, b]

1

(2q−1)A(b−a)1/q ‖f − `f ′‖p [Bq (a, b;A)]1/qif f − `f ′ ∈ Lp [a, b]p > 1,

1p

+ 1q

= 1

1b−a

‖f − `f ′‖1

(ln A

a+ 1

2(b− a) a+3b

4A),

,

where

Bq (a, b;A) =

2Aq

2−q(Aq−2 − A (aq−2, bq−2))

+ 2(q+1)Aq−1 (A (bq+1, aq+1)− Aq+1) ,

q 6= 2

2A2 ln AG

+ 12(b− a)2 , q = 2

1.3. A Related Result.The following new result also holds.

THEOREM 1.5 (Dragomir, 2013 [3]). Let f : [a, b] → C be an absolutely continuous func-tion on the interval[a, b] with b > a > 0. Then for anyx ∈ [a, b] we have∣∣∣∣f (x)

x− 1

b− a

∫ b

a

f (t)

tdt

∣∣∣∣(1.15)

≤

2b−a

‖f − `f ′‖∞(ln x√

ab+

a+b2−x

x

)if f − `f ′ ∈ L∞ [a, b]

1

(2q−1)(b−a)1/q ‖f − `f ′‖p (Cq (a, b;x))1/qif f − `f ′ ∈ Lp [a, b]p > 1,

1p

+ 1q

= 1

1b−a

‖f − `f ′‖1x2+ab−2ax

x2a,

,


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70 S. S. DRAGOMIR

where

(1.16) Cq (a, b;x) =1

x2q−1(b+ a− 2x) +

a2−2q + b2−2q − 2x2−2q

2 (q − 1), q > 1.

PROOF. From the first inequality in (1.10) we have

∣∣∣∣f (x)

x− 1

b− a

∫ b

a

f (t)

tdt

∣∣∣∣ ≤ 1

b− a

∫ b

a

∣∣∣∣f (x)

x− f (t)

t

∣∣∣∣ dt(1.17)

≤ ‖f − `f ′‖∞1

b− a

∫ b

a

∣∣∣∣1t − 1

x

∣∣∣∣ dt.Since

∫ b

a

∣∣∣∣1x − 1

t

∣∣∣∣ dt =

[∫ x

a

(1

t− 1

x

)dt+

∫ b

x

(1

x− 1

t

)dt

]=

(lnx

a− x− a

x+b− x

x− ln

b

x

)=

(lnx2

ab+a+ b− 2x

x

)

for anyx ∈ [a, b] , then we deduce from (1.17) the first inequality in (1.15).From the second inequality in (1.10) we have

∣∣∣∣f (x)

x− 1

b− a

∫ b

a

f (t)

tdt

∣∣∣∣(1.18)

≤ 1

b− a

∫ b

a

∣∣∣∣f (x)

x− f (t)

t

∣∣∣∣ dt≤ 1

(2q − 1) (b− a)‖f − `f ′‖p

∫ b

a

∣∣∣∣ 1

t2q−1− 1

x2q−1

∣∣∣∣1/q

dt.

Utilising Hölder’s integral inequality we have

∫ b

a

∣∣∣∣ 1

t2q−1− 1

x2q−1

∣∣∣∣1/q

dt(1.19)

≤(∫ b

a

dt

)1/p(∫ b

a

[∣∣∣∣ 1

t2q−1− 1

x2q−1

∣∣∣∣1/q]q

dt

)1/q

= (b− a)1/p

(∫ b

a

∣∣∣∣ 1

t2q−1− 1

x2q−1

∣∣∣∣ dt)1/q

.


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Since ∫ b

a

∣∣∣∣ 1

t2q−1− 1

x2q−1

∣∣∣∣ dt=

∫ x

a

(1

t2q−1− 1

x2q−1

)dt+

∫ b

x

(1

x2q−1− 1

t2q−1

)dt

=x2−2q − a2−2q

2− 2q− 1

x2q−1(x− a) +

1

x2q−1(b− x)− b2−2q − x2−2q

2− 2q

=1

x2q−1(b+ a− 2x) +

2x2−2q − a2−2q − b2−2q

2− 2q

=1

x2q−1(b+ a− 2x) +

a2−2q + b2−2q − 2x2−2q

2 (q − 1)= Cq (a, b;x)

then by (1.18) and (1.19) we get∣∣∣∣f (x)

x− 1

b− a

∫ b

a

f (t)

tdt

∣∣∣∣≤ 1

(2q − 1) (b− a)‖f − `f ′‖p (b− a)1/p (Cq (a, b;x))1/q

and the second inequality in (1.15) is proved.From the third inequality in (1.10) we have∣∣∣∣f (x)

x− 1

b− a

∫ b

a

f (t)

tdt

∣∣∣∣ ≤ 1

b− a

∫ b

a

∣∣∣∣f (x)

x− f (t)

t

∣∣∣∣ dt(1.20)

≤ 1

b− a‖f − `f ′‖1

∫ b

a

1

min t2, x2dt.

Since ∫ b

a

1

min t2, x2dt =

∫ x

a

dt

t2+

∫ b

x

dt

x2=x− a

xa+b− x

x2

=x2 + ab− 2ax

x2a,

then by (1.20) we deduce the last part of (1.15).


, then we get∣∣∣∣f (A)

A− 1

b− a

∫ b

a

f (t)

tdt

∣∣∣∣(1.21)

≤

2b−a

‖f − `f ′‖∞ ln(

AG

)if f − `f ′ ∈ L∞ [a, b]

1

(2q−1)(b−a)1/q ‖f − `f ′‖p (Cq (a, b;A))1/qif f − `f ′ ∈ Lp [a, b]p > 1,

1p

+ 1q

= 1

12‖f − `f ′‖1

A+aA2a

,

,

where

Cq (a, b;A) =A (a2−2q, b2−2q)− A2−2q

q − 1, q > 1.


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72 S. S. DRAGOMIR

2. OSTROWSKI V IA POWER POMPEIU ’ S I NEQUALITY

2.1. Power Pompeiu’s Inequality. In 1946, Pompeiu [6] derived a variant of Lagrange’smean value theorem, now known asPompeiu’s mean value theorem(see also [8, p. 83]).

We can generalize the above Pompeiu’s inequality for the power function as follows.

LEMMA 2.1 (Dragomir, 2013 [2]). Letf : [a, b] → C be an absolutely continuous functionon the interval[a, b] with b > a > 0. If r ∈ R, r 6= 0, then for anyt, x ∈ [a, b] we have

|trf (x)− xrf (t)|(2.1)

≤

1|r| ‖f

′`− rf‖∞ |tr − xr| , if f ′`− rf ∈ L∞ [a, b]

‖f ′`− rf‖p

1

|1−q(r+1)|

∣∣ tr

x1−q(r+1)−r − xr

t1−q(r+1)−r

∣∣ , for r 6= −1p

trxr |lnx− ln t| , for r = −1p

if f ′`− rf ∈ Lp [a, b]

‖f ′`− rf‖1trxr

minxr+1,tr+1

or, equivalently∣∣∣∣f (x)

xr− f (t)

tr

∣∣∣∣(2.2)

≤

1|r| ‖f

′`− rf‖∞∣∣ 1xr − 1

tr

∣∣ , if f ′`− rf ∈ L∞ [a, b]

‖f ′`− rf‖p

1

|1−q(r+1)|

∣∣ 1x1−q(r+1) − 1

t1−q(r+1)

∣∣ , for r 6= −1p

|lnx− ln t| , for r = −1p

if f ′`− rf ∈ Lp [a, b]

‖f ′`− rf‖11

minxr+1,tr+1

wherep > 1, 1p

+ 1q

= 1.

PROOF. If f is absolutely continuous, thenf/ (·)r is absolutely continuous on the interval[a, b] and ∫ x

t

(f (s)

sr

)′ds =

f (x)

xr− f (t)

tr

for anyt, x ∈ [a, b] with x 6= t.Since∫ x

t

(f (s)

sr

)′ds =

∫ x

t

f ′ (s) sr − rsr−1f (s)

s2rds =

∫ x

t

f ′ (s) s− rf (s)

sr+1ds


(2.3) trf (x)− xrf (t) = xrtr∫ x

t

f ′ (s) s− rf (s)

sr+1ds



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Taking the modulus in (2.3) we have

|trf (x)− xrf (t)| = xrtr∣∣∣∣∫ x

t

f ′ (s) s− rf (s)

sr+1ds

∣∣∣∣(2.4)

≤ xrtr∣∣∣∣∫ x

t

|f ′ (s) s− rf (s)|sr+1

ds

∣∣∣∣ := I


I ≤ xrtr

sups∈[t,x]([x,t]) |f ′ (s) s− rf (s)|

∣∣∫ x

t1

sr+1ds∣∣

∣∣∫ x

t|f ′ (s) s− rf (s)|p ds

∣∣1/p ∣∣∫ x

t1

sq(r+1)ds∣∣1/q

∣∣∫ x

t|f ′ (s) s− rf (s)| ds


1

sr+1

(2.5)

≤ xrtr

1|r| ‖f

′`− rf‖∞∣∣ 1xr − 1

tr

∣∣‖f ′`− rf‖p

1

|1−q(r+1)|

∣∣ 1x1−q(r+1) − 1

t1−q(r+1)

∣∣ , r 6= −1p

|lnx− ln t| , r = −1p

‖f ′`− rf‖11

minxr+1,tr+1 .

wherep > 1, 1p

+ 1q

= 1, and the inequality (2.1) is proved.

2.2. Some Ostrowski Type Results.The following new result also holds.

THEOREM 2.2 (Dragomir, 2013 [2]). Let f : [a, b] → C be an absolutely continuous func-tion on the interval[a, b] with b > a > 0. If r ∈ R, r 6= 0, andf ′` − rf ∈ L∞ [a, b] , then foranyx ∈ [a, b] we have∣∣∣∣br+1 − ar+1

r + 1f (x)− xr

∫ b

a

f (t) dt

∣∣∣∣(2.6)

≤ 1

|r|‖f ′`− rf‖∞

×

2rxr+1−xr(a+b)(r+1)+br+1+ar+1

r+1, if r > 0

xr(a+b)(r+1)−2rxr+1−br+1−ar+1

r+1, if r ∈ (−∞, 0) −1 .

Also, forr = −1, we have

(2.7)

∣∣∣∣f (x) lnb

a− 1

x

∫ b

a

f (t) dt

∣∣∣∣ ≤ 2 ‖f ′`+ f‖∞

(ln

x√ab

+a+b2− x

x

)for anyx ∈ [a, b] , providedf ′`+ f ∈ L∞ [a, b]

The constant2 in (2.7) is best possible.

PROOF. Utilising the first inequality in (2.1) forr 6= −1 we have∣∣∣∣br+1 − ar+1

r + 1f (x)− xr

∫ b

a

f (t) dt

∣∣∣∣ ≤∫ b

a

|trf (x)− xrf (t)| dt(2.8)

≤ 1

|r|‖f ′`− rf‖∞

∫ b

a

|tr − xr| dt.


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74 S. S. DRAGOMIR

Observe that∫ b

a

|tr − xr| dt

=

∫ x

a(xr − tr) dt+

∫ b

x(tr − xr) dt, if r > 0∫ x

a(tr − xr) dt+

∫ b

x(xr − tr) dt, if r ∈ (−∞, 0) −1 .

Then forr > 0 we have∫ x

a

(xr − tr) dt+

∫ b

x

(tr − xr) dt

= xr (x− a)− xr+1 − ar+1

r + 1+br+1 − xr+1

r + 1− xr (b− x)

= 2xr+1 − xr (a+ b) +br+1 + ar+1 − 2xr+1

r + 1

=2rxr+1 + 2xr+1 − xr (a+ b) (r + 1) + br+1 + ar+1 − 2xr+1

r + 1

=2rxr+1 − xr (a+ b) (r + 1) + br+1 + ar+1

r + 1

and forr ∈ (−∞, 0) −1 we have∫ x

a

(tr − xr) dt+

∫ b

x

(xr − tr) dt

= −2rxr+1 − xr (a+ b) (r + 1) + br+1 + ar+1

r + 1.

Making use of (2.8) we get (2.6).Utilizing the inequality (2.1) forr = −1 we have∣∣t−1f (x)− x−1f (t)

∣∣ ≤ ‖f ′`+ f‖∞∣∣t−1 − x−1

∣∣if f ′`+ f ∈ L∞ [a, b].

Integrating this inequality, we have∣∣∣∣f (x) lnb

a− x−1

∫ b

a

f (t) dt

∣∣∣∣ ≤ ∫ b

a

∣∣t−1f (x)− x−1f (t)∣∣ dt(2.9)

≤ ‖f ′`+ f‖∞∫ b

a

∣∣t−1 − x−1∣∣ dt.

Since ∫ b

a

∣∣∣∣1x − 1

t

∣∣∣∣ dt =

[∫ x

a

(1

t− 1

x

)dt+

∫ b

x

(1

x− 1

t

)dt

]=

(lnx

a− x− a

x+b− x

x− ln

b

x

)= ln

x2

ab+a+ b− 2x

x

= 2

(ln

x√ab

+a+b2− x

x

),

then by (2.9) we get the desired inequality (2.7).


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Now, assume that (2.7) holds with a constantC > 0, i.e.

(2.10)

∣∣∣∣f (x) lnb

a− x−1

∫ b

a

f (t) dt

∣∣∣∣ ≤ C ‖f ′`+ f‖∞

(ln

x√ab

+a+b2− x

x


If we take in (2.10)f (t) = 1, t ∈ [a, b] , then we get

(2.11)

∣∣∣∣ln ba − b− a

x

∣∣∣∣ ≤ C

(ln

x√ab

+a+b2− x

x

)for any for anyx ∈ [a, b] .

Makingx = a in (2.10) produces the inequality∣∣∣∣ln ba − b− a

a

∣∣∣∣ ≤ C

(b− a

2a− 1

2lnb

a

)which implies thatC ≥ 2.

This proves the sharpness of the constant2 in (2.7).

REMARK 2.1. Consider ther-Logarithmic mean

Lr = Lr (a, b) :=

[br+1 − ar+1

(r + 1) (b− a)

]1/r

defined forr ∈ R 0,−1 and the Logarithmic mean, defined as

L = L (a, b) :=b− a

ln b− ln a.

If A = A (a, b) := a+b2, then from (2.6) we get forx = A the inequality∣∣∣∣Lr

r (b− a) f (A)− Ar

∫ b

a

f (t) dt

∣∣∣∣(2.12)

≤ 2

|r|‖f ′`− rf‖∞

A(br+1,ar+1)−Ar+1

r+1, if r > 0

Ar+1−A(br+1,ar+1)r+1

, if r ∈ (−∞, 0) −1 ,while from (2.7) we get

(2.13)

∣∣∣∣L−1 (b− a) f (A)− A−1

∫ b

a

f (t) dt

∣∣∣∣ ≤ 2 ‖f ′`+ f‖∞ lnA

G

The following related result holds.

THEOREM 2.3 (Dragomir, 2013 [2]). Let f : [a, b] → C be an absolutely continuous func-tion on the interval[a, b] with b > a > 0. If r ∈ R, r 6= 0, then for anyx ∈ [a, b] we have∣∣∣∣f (x)

xr(b− a)−

∫ b

a

f (t)

trdt

∣∣∣∣(2.14)

≤ 1

|r|‖f ′`− rf‖∞

×

2x1−r−a1−r−b1−r

1−r+ 1

xr (b+ a− 2x) , r ∈ (0,∞) 1

a1−r+b1−r−2x1−r

1−r+ 1

xr (2x− a− b) , if r < 0.

,


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76 S. S. DRAGOMIR

Also, forr = 1, we have

(2.15)

∣∣∣∣f (x)

x(b− a)−

∫ b

a

f (t)

tdt

∣∣∣∣ ≤ 2 ‖f ′`− f‖∞

(ln

x√ab

+a+b2− x

x

)

for anyx ∈ [a, b] , providedf ′`− f ∈ L∞ [a, b] .The constant2 is best possible in (2.15).

PROOF. From the first inequality in (2.2) we have

(2.16)

∣∣∣∣f (x)

xr− f (t)

tr

∣∣∣∣ ≤ 1

|r|‖f ′`− rf‖∞

∣∣∣∣ 1

xr− 1

tr

∣∣∣∣ ,for anyt, x ∈ [a, b] , providedf ′`− rf ∈ L∞ [a, b] .

Integrating overt ∈ [a, b] we get∣∣∣∣f (x)

xr(b− a)−

∫ b

a

f (t)

trdt

∣∣∣∣ ≤ ∫ b

a

∣∣∣∣f (x)

xr− f (t)

tr

∣∣∣∣ dt(2.17)

≤ 1

|r|‖f ′`− rf‖∞

∫ b

a

∣∣∣∣ 1

xr− 1

tr

∣∣∣∣ dt,for r ∈ R, r 6= 0.

For r ∈ (0,∞) 1 we have∫ b

a

∣∣∣∣ 1

xr− 1

tr

∣∣∣∣ dt=

∫ x

a

(1

tr− 1

xr

)dt+

∫ b

x

(1

xr− 1

tr

)dt

=x1−r − a1−r

1− r− 1

xr(x− a) +

1

xr(b− x)− b1−r − x1−r

1− r

=2x1−r − a1−r − b1−r

1− r+

1

xr(b+ a− 2x)

for anyx ∈ [a, b] .For r < 0, we also have∫ b

a

∣∣∣∣ 1

xr− 1

tr

∣∣∣∣ dt =a1−r + b1−r − 2x1−r

1− r+

1

xr(2x− a− b)

for anyx ∈ [a, b] .For r = 1 we have ∫ b

a

∣∣∣∣1x − 1

t

∣∣∣∣ dt = 2

(ln

x√ab

+a+b2− x

x

)

for anyx ∈ [a, b] , and the inequality (2.15) is obtained.The sharpness of the constant2 follows as in the proof of Theorem 2.2 and the details are

omitted.


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REMARK 2.2. If we takex = A in Theorem 2.3, then we we have

∣∣∣∣f (A)

Ar(b− a)−

∫ b

a

f (t)

trdt

∣∣∣∣(2.18)

≤ 2

|r|‖f ′`− rf‖∞

A1−r−A(a1−r,b1−r)

1−r, r ∈ (0,∞) 1

A(a1−r,b1−r)−A1−r

1−r, if r < 0.

,

Also, for r = 1, we have

(2.19)

∣∣∣∣f (A)

A(b− a)−

∫ b

a

f (t)

tdt

∣∣∣∣ ≤ 2 ‖f ′`− f‖∞ lnA

G.

REMARK 2.3. The interested reader may obtain other similar results in terms of thep-norms‖f ′`− rf‖p with p ≥ 1. However, since some calculations are too complicated, the details arenot presented here.

3. OSTROWSKI V IA AN EXPONENTIAL POMPEIU ’ S I NEQUALITY

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

We can provide some similar results for complex-valued functions and the exponential asfollows.

LEMMA 3.1 (Dragomir, 2013 [4]). Letf : [a, b] → C be an absolutely continuous functionon the interval[a, b] andα ∈ C with Re (α) 6= 0. Then for anyt, x ∈ [a, b] we have

(3.1)

∣∣∣∣ f (x)

exp (αx)− f (t)

exp (αt)

∣∣∣∣

≤

|Re (α)| ‖f ′ − αf‖∞×∣∣∣ 1exp(t Re(α))

− 1exp(x Re(α))

∣∣∣ if f ′−αf∈ L∞ [a, b] ,

q1/q |Re (α)|1/q ‖f ′ − αf‖p

×∣∣∣ 1exp(tq Re(α))

− 1exp(xq Re(α))

∣∣∣1/q

if f ′−αf∈ Lp [a, b]p > 1,

1p+1

q= 1,

‖f ′ − αf‖11

minexp(t Re(α)),exp(x Re(α)) ,


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78 S. S. DRAGOMIR

or, equivalently

(3.2) |exp (αt) f (x)− f (t) exp (αx)|

≤

|Re (α)| ‖f ′ − αf‖∞× |exp (xRe (α))− exp (tRe (α))|

if f ′−αf∈ L∞ [a, b] ,

q1/q |Re (α)|1/q ‖f ′ − αf‖p

× |exp (xqRe (α))− exp (tqRe (α))|1/q

if f ′−αf∈ Lp [a, b]p > 1,

1p+1

q= 1,

‖f ′ − αf‖1 max exp (tRe (α)) , exp (xRe (α)) .

PROOF. If f is absolutely continuous, thenf/ exp (α·) is absolutely continuous on the in-terval [a, b] and

∫ x

t

(f (s)

exp (αs)

)′ds =

f (x)

exp (αx)− f (t)

exp (αt)

for anyt, x ∈ [a, b] with x 6= t.Since

∫ x

t

(f (s)

exp (αs)

)′ds =

∫ x

t

f ′ (s) exp (αs)− αf (s) exp (αs)

exp (2αs)ds

=

∫ x

t

f ′ (s)− αf (s)

exp (αs)ds,


(3.3)f (x)

exp (αx)− f (t)

exp (αt)=

∫ x

t

f ′ (s)− αf (s)

exp (αs)ds

for anyt, x ∈ [a, b] with x 6= t.Taking the modulus in (3.3) we have

∣∣∣∣ f (x)

exp (αx)− f (t)

exp (αt)

∣∣∣∣ =

∣∣∣∣∫ x

t

f ′ (s)− αf (s)

exp (αs)ds

∣∣∣∣(3.4)

≤∣∣∣∣∫ x

t

|f ′ (s)− αf (s)||exp (αs)|

ds

∣∣∣∣ := I


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80 S. S. DRAGOMIR

The inequality (3.2) follows by (3.1) on multiplying with|exp (αx) exp (αt)| and perform-ing the required calculation.

The following particular case is of interest.

COROLLARY 3.2. Let f : [a, b] → C be an absolutely continuous function on the interval[a, b] . Then for anyt, x ∈ [a, b] we have

(3.6)

∣∣∣∣ f (x)

exp (x)− f (t)

exp (t)

∣∣∣∣

≤

‖f ′ − f‖∞∣∣∣ 1exp(t)

− 1exp(x)

∣∣∣ if f ′ − f∈ L∞ [a, b] ,

q1/q ‖f ′ − f‖p

∣∣∣ 1exp(tq)

− 1exp(xq)

∣∣∣1/q

if f ′ − f∈ Lp [a, b]

p > 1,1p

+ 1q

= 1,

‖f ′ − f‖11

minexp(t),exp(x) ,

or, equivalently

(3.7) |exp (t) f (x)− f (t) exp (x)|

≤

‖f ′ − f‖∞ |exp (x)− exp (t)| if f ′ − f∈ L∞ [a, b] ,

q1/q ‖f ′ − f‖p |exp (xq)− exp (tq)|1/q

if f ′ − f∈ Lp [a, b]

p > 1,1p

+ 1q

= 1,

‖f ′ − f‖1 max exp (t) , exp (x) .


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82 S. S. DRAGOMIR

or, equivalently

(3.11) |exp (it) f (x)− f (t) exp (ix)| ≤

‖f ′ − if‖∞ |x− t| ,

‖f ′ − if‖p |x− t|1/q ,

‖f ′ − if‖1 ,


3.2. Inequalities of Ostrowski Type. The following result holds:

THEOREM 3.3 (Dragomir, 2013 [4]). Let f : [a, b] → C be an absolutely continuous func-tion on the interval[a, b] andα ∈ C with Re (α) > 0. Then for anyx ∈ [a, b] we have

(3.12)

∣∣∣∣f (x)exp (αb)− exp (αa)

α− exp (αx)

∫ b

a

f (t) dt

∣∣∣∣

≤

|Re (α)| ‖f ′ − αf‖∞B1(a, b, x, α)if f ′−αf∈ L∞ [a, b] ,

q1/q |Re (α)|1/q (b− a)1/p

×‖f ′ − αf‖p |Bq(a, b, x, α)|1/q

if f ′−αf∈ Lp [a, b]p > 1,

1p+1

q= 1,

‖f ′ − αf‖1B∞(a, b, x, α)

where

Bq(a, b, x, α) := 2

[exp (xqRe (α))

(x− a+ b

2

)+

1

qRe (α)

(exp (bqRe (α)) + exp (aqRe (α))

2− exp (xqRe (α))

)]for q ≥ 1 and

B∞(a, b, x, α) := exp (xRe (α)) (x− a) +exp (bRe (α))− exp (xRe (α))

Re (α).

PROOF. Utilising the first inequality in (3.2) we have∣∣∣∣f (x)

∫ b

a

exp (αt) dt− exp (αx)

∫ b

a

f (t) dt

∣∣∣∣(3.13)

≤∫ b

a

|exp (αt) f (x)− f (t) exp (αx)| dt

≤ |Re (α)| ‖f ′ − αf‖∞∫ b

a

|exp (xRe (α))− exp (tRe (α))| dt



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Observe that, sinceRe (α) > 0, then

∫ b

a

|exp (xRe (α))− exp (tRe (α))| dt

=

∫ x

a

(exp (xRe (α))− exp (tRe (α))) dt

+

∫ b

x

(exp (tRe (α))− exp (xRe (α))) dt

= exp (xRe (α)) (x− a)− exp (tRe (α))

Re (α)

∣∣∣∣xa

+exp (tRe (α))

Re (α)

∣∣∣∣bx

− (b− x) exp (xRe (α))

= exp (xRe (α)) (2x− a− b)− 1

Re (α)(exp (xRe (α))− exp (aRe (α)))

+1

Re (α)(exp (bRe (α))− exp (xRe (α)))

= exp (xRe (α)) (2x− a− b)

+1

Re (α)(exp (bRe (α)) + exp (aRe (α))− 2 exp (xRe (α)))

= 2

[exp (xRe (α))

(x− a+ b

2

)+

1

Re (α)

(exp (bRe (α)) + exp (aRe (α))

2− exp (xRe (α))

)]for anyx ∈ [a, b] .

Also

f (x)

∫ b

a


∫ b

a

f (t) dt

= f (x)exp (αb)− exp (αa)

α− exp (αx)

∫ b

a

f (t) dt

for anyx ∈ [a, b] and by (3.13) we get the first inequality in (3.12).Using the second inequality in (3.2) we have∣∣∣∣f (x)

∫ b

a


∫ b

a

f (t) dt

∣∣∣∣(3.14)

≤∫ b

a


≤ q1/q |Re (α)|1/q ‖f ′ − αf‖p

∫ b

a

|exp (xqRe (α))− exp (tqRe (α))|1/q dt



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84 S. S. DRAGOMIR

By Hölder’s integral inequality we also have∫ b

a

|exp (xqRe (α))− exp (tqRe (α))|1/q dt

≤(∫ b

a

dt

)1/p [∫ b

a

(|exp (xqRe (α))− exp (tqRe (α))|1/q

)q

dt

]1/q

= (b− a)1/p

[∫ b

a

|exp (xqRe (α))− exp (tqRe (α))| dt]1/q

,

for anyx ∈ [a, b] .Observe that, as above, we have∫ b

a

|exp (xqRe (α))− exp (tqRe (α))| dt

= 2

[exp (xqRe (α))

(x− a+ b

2

)+

1

qRe (α)

(exp (bqRe (α)) + exp (aqRe (α))

2− exp (xqRe (α))

)]= Bq(a, b, x, α)

for anyx ∈ [a, b] and by (3.14) we get the second part of (3.12).Using the third inequality in (3.2) we have∣∣∣∣f (x)

∫ b

a


∫ b

a

f (t) dt

∣∣∣∣(3.15)

≤∫ b

a


≤ ‖f ′ − αf‖1

∫ b

a

max exp (tRe (α)) , exp (xRe (α)) dt

for anyx ∈ [a, b] .Observe that,∫ b

a


=

∫ x

a


+

∫ b

x


=

∫ x

a

exp (xRe (α)) dt+

∫ b

x

exp (tRe (α)) dt =

= exp (xRe (α)) (x− a) +exp (bRe (α))− exp (xRe (α))

Re (α)

and by (3.15) we get the third part of (3.12).

REMARK 3.2. If Re (α) < 0, then a similar result may be stated. However the details areleft to the interested reader.


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COROLLARY 3.4. Let f : [a, b] → C be an absolutely continuous function on the interval[a, b]. Then for anyx ∈ [a, b] we have

(3.16)

∣∣∣∣f (x) [exp (b)− exp (a)]− exp (x)

∫ b

a

f (t) dt

∣∣∣∣

≤

‖f ′ − f‖∞B1(a, b, x) if f ′−f ∈ L∞ [a, b] ,

q1/q (b− a)1/p ‖f ′ − f‖p

× |Bq(a, b, x)|1/q

if f ′−f ∈ Lp [a, b]p > 1,

1p+1

q= 1,

‖f ′ − f‖1B∞(a, b, x)

where

Bq(a, b, x)

:= 2

[(x− a+ b

2

)exp (xq) +

1

q

(exp (bq) + exp (aq)

2− exp (xq)

)]for q ≥ 1 and

B∞(a, b, x) := (x− a) exp (x) + exp (b)− exp (x) .

REMARK 3.3. The midpoint case is as follows:

(3.17)

∣∣∣∣f (a+ b

2

)[exp (b)− exp (a)]− exp

(a+ b

2

)∫ b

a

f (t) dt

∣∣∣∣

≤

‖f ′ − f‖∞B1(a, b) if f ′−f ∈ L∞ [a, b] ,

q1/q (b− a)1/p ‖f ′ − f‖p

× |Bq(a, b)|1/q

if f ′−f ∈ Lp [a, b]p > 1,

1p+1

q= 1,

‖f ′ − f‖1B∞(a, b)

where

Bq(a, b, x) :=2

q

(exp (bq) + exp (aq)

2− exp

(a+ b

2q

))for q ≥ 1 and

B∞(a, b) :=b− a

2exp

(a+ b

2

)+ exp (b)− exp

(a+ b

2

).

The caseRe (α) = 0 is different and may be stated as follows.

THEOREM 3.5 (Dragomir, 2013 [4]). Let f : [a, b] → C be an absolutely continuous func-tion on the interval[a, b] andα ∈ C with Re (α) = 0 and Im (α) 6= 0. Then for anyx ∈ [a, b]


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86 S. S. DRAGOMIR

we have

(3.18)

∣∣∣∣f (x)exp (i Im (α) b)− exp (i Im (α) a)

i Im (α)− exp (i Im (α)x)

∫ b

a

f (t) dt

∣∣∣∣

≤

‖f ′ − i Im (α) f‖∞[14

+(

x−a+b2

b−a

)2]

(b− a)2if f ′−i Im (α) f∈ L∞ [a, b] ,

qq+1

‖f ′ − i Im (α) f‖p

×[(

b−xb−a

) q+1q +

(x−ab−a

) q+1q

](b− a)

q+1q

if f ′−i Im (α) f∈ Lp [a, b]p > 1,

1p+1

q= 1,

‖f ′ − i Im (α) f‖1 (b− a) .

PROOF. Utilizing the inequality (3.9) we have∣∣∣∣f (x)exp (i Im (α) b)− exp (i Im (α) a)

i Im (α)− exp (i Im (α)x)

∫ b

a

f (t) dt

∣∣∣∣(3.19)

≤∫ b

a

|exp (i Im (α) t) f (x)− f (t) exp (i Im (α)x)| dt

≤

‖f ′ − i Im (α) f‖∞

∫ b

a|x− t| dt,

‖f ′ − i Im (α) f‖p

∫ b

a|x− t|1/q dt,

‖f ′ − i Im (α) f‖1

∫ b

adt.

Since ∫ b

a

|x− t| dt =

1

4+

(x− a+b

2

b− a

)2 (b− a)2

and ∫ b

a

|x− t|1/q dt =(b− x)1+1/q + (x− a)1+1/q

1 + 1/q

=q

q + 1

[(b− x

b− a

) q+1q

+

(x− a

b− a

) q+1q

](b− a)

q+1q ,

then we get from (3.19) the desired result (3.18).


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COROLLARY 3.6. Let f : [a, b] → C be an absolutely continuous function on the interval[a, b] . Then for anyx ∈ [a, b] we have

(3.20)

∣∣∣∣f (x)exp (ib)− exp (ia)

i− exp (ix)

∫ b

a

f (t) dt

∣∣∣∣

≤

‖f ′ − if‖∞×[

14

+(

x−a+b2

b−a

)2]

(b− a)2if f ′−if∈ L∞ [a, b] ,

qq+1

‖f ′ − if‖p

×[(

b−xb−a

) q+1q +

(x−ab−a

) q+1q

](b− a)

q+1q

if f ′−if∈ Lp [a, b]p > 1,

1p+1

q= 1,

‖f ′ − if‖1 (b− a) .

REMARK 3.4. The midpoint case is as follows

∣∣∣∣f (a+ b

2

)exp (ib)− exp (ia)

i− exp

(ia+ b

2

)∫ b

a

f (t) dt

∣∣∣∣(3.21)

≤

14‖f ′ − if‖∞ (b− a)2 , if f ′−if ∈ L∞ [a, b] ,

q(q+1)21/q ‖f ′ − if‖p (b− a)

q+1q , if f ′−if ∈ Lp [a, b] .

Similar inequalities may be stated if one uses (3.1) and integrates overt on [a, b] . The detailsare left to the interested reader.

4. OSTROWSKI V IA A TWO FUNCTIONS POMPEIU ’ S I NEQUALITY

4.1. A General Pompeiu’s Inequality. We start with the following generalization of Pom-peiu’s inequality:

Assume thatf : [a, b] → R is differentiable on the interval(a, b) which is not containing0and if‖f − `f ′‖∞ = supt∈(a,b) |f (t)− tf ′ (t)| <∞ where` (t) = t, t ∈ [a, b] , then

(4.1) |tf (x)− xf (t)| ≤ ‖f − `f ′‖∞ |x− t|

for anyt, x ∈ [a, b] .The inequality (4.1) was stated by the author in [1].


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88 S. S. DRAGOMIR

THEOREM 4.1 (Dragomir, 2013 [5]). Let f, g : [a, b] → C be absolutely continuous func-tions on the interval[a, b] with g (t) 6= 0 for all t ∈ [a, b] . Then for anyt, x ∈ [a, b] we have

(4.2)

∣∣∣∣f (x)

g (x)− f (t)

g (t)

∣∣∣∣

≤

‖f ′g − fg′‖∞∣∣∣∫ x

t1

|g(s)|2ds∣∣∣ if f ′g − fg′ ∈ L∞ [a, b] ,

‖f ′g − fg′‖p

∣∣∣∫ x

t1

|g(s)|2q ds∣∣∣1/q

if f ′g − fg′ ∈ Lp [a, b]p > 1,

1p

+ 1q

= 1,

‖f ′g − fg′‖1 sups∈[t,x]([x,t])

1

|g(s)|2

or, equivalently

(4.3) |g (t) f (x)− f (t) g (x)|

≤

‖f ′g − fg′‖∞ |g (t) g (x)|∣∣∣∫ x

t1

|g(s)|2ds∣∣∣ if f ′g − fg′ ∈ L∞ [a, b] ,

‖f ′g − fg′‖p |g (t) g (x)|∣∣∣∫ x

t1

|g(s)|2q ds∣∣∣1/q

if f ′g − fg′ ∈ Lp [a, b]p > 1,

1p

+ 1q

= 1,

‖f ′g − fg′‖1 |g (t) g (x)| sups∈[t,x]([x,t])

1

|g(s)|2

.

PROOF. If f andg are absolutely continuous andg (t) 6= 0 for all t ∈ [a, b], thenf/g isabsolutely continuous on the interval[a, b] and∫ x

t

(f (s)

g (s)

)′ds =

f (x)

g (x)− f (t)

g (t)

for anyt, x ∈ [a, b] with x 6= t.Since ∫ x

t

(f (s)

g (s)

)′ds =

∫ x

t

f ′ (s) g (s)− f (s) g′ (s)

g2 (s)ds,


(4.4)f (x)

g (x)− f (t)

g (t)=

∫ x

t

f ′ (s) g (s)− f (s) g′ (s)

g2 (s)ds

for anyt, x ∈ [a, b] .Taking the modulus in (4.4) we have∣∣∣∣f (x)

g (x)− f (t)

g (t)

∣∣∣∣ =

∣∣∣∣∫ x

t

f ′ (s) g (s)− f (s) g′ (s)

g2 (s)ds

∣∣∣∣(4.5)

≤∣∣∣∣∫ x

t

|f ′ (s) g (s)− f (s) g′ (s)||g (s)|2

ds

∣∣∣∣ := I


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I ≤

sups∈[t,x]([x,t]) |f ′ (s) g (s)− f (s) g′ (s)|∣∣∣∫ x

t1

|g(s)|2ds∣∣∣ ,

∣∣∫ x

t|f ′ (s) g (s)− f (s) g′ (s)|p ds

∣∣1/p∣∣∣∫ x

t1

|g(s)|2q ds∣∣∣1/q p > 1,

1p

+ 1q

= 1,

∣∣∫ x

t|f ′ (s) g (s)− f (s) g′ (s)| ds


1

|g(s)|2

,

≤

‖f ′g − fg′‖∞∣∣∣∫ x

t1

|g(s)|2ds∣∣∣ ,

‖f ′g − fg′‖p

∣∣∣∫ x

t1

|g(s)|2q ds∣∣∣1/q p > 1,

1p

+ 1q

= 1,

‖f ′g − fg′‖1 sups∈[t,x]([x,t])

1

|g(s)|2


The following particular case extends Pompeiu’s inequality to otherp-norms thanp = ∞obtained in (4.2).

COROLLARY 4.2. 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

(4.6)

∣∣∣∣f (x)

x− f (t)

t

∣∣∣∣

≤

‖f − `f ′‖∞∣∣1

t− 1

x

∣∣ if f − `f ′ ∈ L∞ [a, b] ,

12q−1

‖f − `f ′‖p

∣∣ 1t2q−1 − 1

x2q−1

∣∣1/qif f − `f ′ ∈ Lp [a, b]

p > 1,1p

+ 1q

= 1,

‖f − `f ′‖11

mint2,x2

or, equivalently

(4.7) |tf (x)− xf (t)|

≤

‖f − `f ′‖∞ |x− t| if f − `f ′ ∈ L∞ [a, b] ,

12q−1

‖f − `f ′‖p

∣∣ xq

tq−1 − tq

xq−1

∣∣1/q if f − `f ′ ∈ Lp [a, b]p > 1, 1

p+ 1

q= 1,

‖f − `f ′‖1maxt,xmint,x ,

where` (t) = t, t ∈ [a, b] .

The proof follows by (4.2) forg (t) = ` (t) = t, t ∈ [a, b] .The general case for power functions is as follows.


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90 S. S. DRAGOMIR

COROLLARY 4.3. Let f : [a, b] → C be an absolutely continuous function on the interval[a, b] with b > a > 0. If r ∈ R, r 6= 0, then for anyt, x ∈ [a, b] we have

(4.8)

∣∣∣∣f (x)

xr− f (t)

tr

∣∣∣∣

≤

1|r| ‖f

′`− rf‖∞∣∣ 1xr − 1

tr

∣∣ , if f ′`− rf ∈ L∞ [a, b] ,

‖f ′`− rf‖p

×

1

|1−q(r+1)|

∣∣ 1x1−q(r+1) − 1

t1−q(r+1)

∣∣ , for r 6= −1p

|lnx− ln t| , for r = −1p

if f ′`− rf ∈ Lp [a, b] ,

‖f ′`− rf‖11

minxr+1,tr+1 ,

or, equivalently

(4.9) |trf (x)− xrf (t)|

≤

1|r| ‖f

′`− rf‖∞ |tr − xr| , if f ′`− rf ∈ L∞ [a, b] ,

‖f ′`− rf‖p

×

1

|1−q(r+1)|

∣∣ tr

x1−q(r+1)−r − xr

t1−q(r+1)−r

∣∣ , for r 6= −1p

trxr |lnx− ln t| , for r = −1p

if f ′`− rf ∈ Lp [a, b] ,

‖f ′`− rf‖1trxr

minxr+1,tr+1 ,

wherep > 1, 1p

+ 1q

= 1.

The proof follows by (4.2) forg (t) = tr, t ∈ [a, b] . The details for calculations are omitted.We have the following result for exponential.

COROLLARY 4.4. Let f : [a, b] → C be an absolutely continuous function on the interval[a, b] andα ∈ R, α 6= 0. Then for anyt, x ∈ [a, b] we have

(4.10)

∣∣∣∣ f (x)

exp (iαx)− f (t)

exp (iαt)

∣∣∣∣

≤

‖f ′ − iαf‖∞ |x− t| if f ′ − iαf ∈ L∞ [a, b] ,

‖f ′ − iαf‖p |x− t|1/qif f ′ − iαf ∈ Lp [a, b]

p > 1,1p

+ 1q

= 1,

‖f ′ − iαf‖1


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or, equivalently

(4.11) |exp (iαt) f (x)− f (t) exp (iαx)|

≤

‖f ′ − iαf‖∞ |x− t| if f ′ − iαf ∈ L∞ [a, b] ,

‖f ′ − iαf‖p |x− t|1/qif f ′ − iαf ∈ Lp [a, b]

p > 1,1p

+ 1q

= 1,

‖f ′ − iαf‖1 .

4.2. An Inequality Generalizing Ostrowski’s. The following result holds:

THEOREM 4.5 (Dragomir, 2013 [5]). Let f, g : [a, b] → C be absolutely continuous func-tions on the interval[a, b] . If 0 < m ≤ |g (t)| ≤M <∞ for anyt ∈ [a, b] , then

(4.12)

∣∣∣∣f (x)

∫ b

a

g (t) dt− g (x)

∫ b

a

f (t) dt

∣∣∣∣

≤(M

m

)2

‖f ′g − fg′‖∞ (b− a)2

[14

+(

t−a+b2

b−a

)2]

if f ′g − fg′ ∈ L∞ [a, b] ,

‖f ′g − fg′‖p

[(b−x)1+1/q+(x−a)1+1/q

1+1/q

] if f ′g − fg′ ∈ Lp [a, b]p > 1,

1p

+ 1q

= 1,

‖f ′g − fg′‖1 (b− a)


PROOF. Utilizing (4.3) we have

(4.13)

∣∣∣∣f (x)

∫ b

a

g (t) dt− g (x)

∫ b

a

f (t) dt

∣∣∣∣≤∫ b

a

|g (t) f (x)− f (t) g (x)| dt

≤

‖f ′g − fg′‖∞ |g (x)|∫ b

a

(|g (t)|

∣∣∣∫ x

t1

|g(s)|2ds∣∣∣) dt,

‖f ′g − fg′‖p |g (x)|∫ b

a

(|g (t)|

∣∣∣∫ x

t1

|g(s)|2q ds∣∣∣1/q)dt,

‖f ′g − fg′‖1 |g (x)|∫ b

a

(|g (t)| sups∈[t,x]([x,t])

1

|g(s)|2

)dt

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


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92 S. S. DRAGOMIR

Since0 < m ≤ |g (t)| ≤M <∞ for anyt ∈ [a, b] , then

|g (x)|∫ b

a

(|g (t)|

∣∣∣∣∫ x

t

1

|g (s)|2ds

∣∣∣∣) dt ≤ (Mm)2 ∫ b

a

|x− t| dt

=

(M

m

)21

4+

(t− a+b

2

b− a

)2 ,

|g (x)|∫ b

a

(|g (t)|

∣∣∣∣∫ x

t

1

|g (s)|2q ds

∣∣∣∣1/q)dt

≤(M

m

)2 ∫ b

a

|x− t|1/q dt =

(M

m

)2(b− x)1+1/q + (x− a)1+1/q

1 + 1/q

and

|g (x)|∫ b

a

(|g (t)| sup

s∈[t,x]([x,t])

1

|g (s)|2

)dt ≤

(M

m

)2 ∫ b

a

dt =

(M

m

)2

(b− a)

for anyx ∈ [a, b] and by (4.13) we get the desired result (4.12).

REMARK 4.1. If we takeg (t) = 1, t ∈ [a, b] in the first inequality (4.12) we recaptureOstrowski’s inequality.

COROLLARY 4.6. With the assumptions in Theorem 4.5 we have the midpoint inequalities

(4.14)

∣∣∣∣f (a+ b

2

)∫ b

a

g (t) dt− g

(a+ b

2

)∫ b

a

f (t) dt

∣∣∣∣

≤(M

m

)2

14(b− a)2 ‖f ′g − fg′‖∞ if f ′g − fg′ ∈ L∞ [a, b] ,

121/q(1+1/q)

(b− a)1+1/q ‖f ′g − fg′‖p

if f ′g − fg′ ∈ Lp [a, b]p > 1,

1p

+ 1q

= 1.

The following result also holds:

THEOREM 4.7 (Dragomir, 2013 [5]). Let f, g : [a, b] → C be absolutely continuous func-tions on the interval[a, b] , g (x) 6= 0 for x ∈ [a, b] andg−2 ∈ L∞ [a, b] . Then

(4.15)

∣∣∣∣f (x)

g (x)

∫ b

a

g (t) dt−∫ b

a

f (t) dt

∣∣∣∣

≤∥∥g−2

∥∥∞ ×

‖f ′g − fg′‖∞∫ b

a|g (t)| |x− t| dt, if f ′g − fg′ ∈ L∞ [a, b] ,

‖f ′g − fg′‖p

∫ b

a|g (t)| |x− t|1/q dt

if f ′g − fg′ ∈ Lp [a, b]p > 1,

1p

+ 1q

= 1,

‖f ′g − fg′‖1

∫ b

a|g (t)| dt


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PROOF. Utilizing (4.3) we have

(4.16)

∣∣∣∣f (x)

g (x)

∫ b

a

g (t) dt−∫ b

a

f (t) dt

∣∣∣∣

≤

‖f ′g − fg′‖∞∫ b

a

(|g (t)|

∣∣∣∫ x

t1

|g(s)|2ds∣∣∣) dt,

‖f ′g − fg′‖p

∫ b

a

(|g (t)|

∣∣∣∫ x

t1

|g(s)|2q ds∣∣∣1/q)dt,

‖f ′g − fg′‖1

∫ b

a

(|g (t)| sups∈[t,x]([x,t])

1

|g(s)|2

)dt

for anyx ∈ [a, b].Since ∣∣∣∣∫ x

t

1

|g (s)|2ds

∣∣∣∣ ≤ ∥∥g−2∥∥∞ |x− t| ,

∣∣∣∣∫ x

t

1

|g (s)|2q ds

∣∣∣∣1/q

≤∥∥g−2

∥∥∞ |x− t|

and

sups∈[t,x]([x,t])

1

|g (s)|2

≤∥∥g−2

∥∥∞

for anyx, t ∈ [a, b] , then on making use of (4.16) we get the desired result (4.15).

We have the midpoint inequalities:

COROLLARY 4.8. With the assumptions of Theorem 4.7 we have

(4.17)

∣∣∣∣∣f(

a+b2

)g(

a+b2

) ∫ b

a

g (t) dt−∫ b

a

f (t) dt

∣∣∣∣∣

≤∥∥g−2

∥∥∞ ×

‖f ′g − fg′‖∞∫ b

a|g (t)|

∣∣a+b2− t∣∣ dt, if f ′g − fg′ ∈ L∞ [a, b] ,

‖f ′g − fg′‖p

∫ b

a|g (t)|

∣∣a+b2− t∣∣1/q

dtif f ′g − fg′ ∈ Lp [a, b]

p > 1,1p

+ 1q

= 1.

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


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94 S. S. DRAGOMIR

THEOREM 4.9 (Dragomir, 2013 [5]). Let f : [a, b] → C be an absolutely continuous func-tion on the interval[a, b] andα ∈ R, α 6= 0. Then for anyx ∈ [a, b] we have

(4.18)

∣∣∣∣exp (iα (b− x))− exp (−iα (x− a))

iαf (x)−

∫ b

a

f (t) dt

∣∣∣∣

≤

‖f ′ − iαf‖∞ (b− a)2

[14

+(

t−a+b2

b−a

)2], if f ′ − iαf ∈ L∞ [a, b] ,

‖f ′ − iαf‖p(b−x)1+1/q+(x−a)1+1/q

1+1/q,

if f ′ − iαf∈ Lp [a, b]

p > 1,1p

+ 1q

= 1,

‖f ′ − iαf‖1 .

PROOF. If we write the inequality (4.13) forg (t) = exp (iαt) , t ∈ [a, b] , then we get

∣∣∣∣f (x)

∫ b

a

exp (iαt) dt− exp (iαx)

∫ b

a

f (t) dt

∣∣∣∣

≤

‖f ′ − iαf‖∞∫ b

a|x− t| dt, if f ′ − iαf ∈ L∞ [a, b]

‖f ′ − iαf‖p |g (x)|∫ b

a|x− t|1/q dt,

if f ′ − iαf∈ Lp [a, b]p > 1,

1p

+ 1q

= 1

‖f ′ − iαf‖1 ,

which, after simple calculation, is equivalent with (4.18).The details are omitted.

COROLLARY 4.10. With the assumptions of Theorem 4.9 we have the midpoint inequalities

(4.19)

∣∣∣∣∣exp(iα(

b−a2

))− exp

(−iα

(b−a2

))iα

f

(a+ b

2

)−∫ b

a

f (t) dt

∣∣∣∣∣

≤

14‖f ′ − iαf‖∞ (b− a)2 , if f ′ − iαf ∈ L∞ [a, b] ,

121/q(1+1/q)

(b− a)1+1/q ‖f ′ − iαf‖p ,if f ′ − iαf ∈ Lp [a, b]p > 1, 1

p+ 1

q= 1,


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or, equivalently

(4.20)

∣∣∣∣∣2 sin(α(

b−a2

))α

f

(a+ b

2

)−∫ b

a

f (t) dt

∣∣∣∣∣

≤

14‖f ′ − iαf‖∞ (b− a)2 , if f ′ − iαf ∈ L∞ [a, b] ,

121/q(1+1/q)

(b− a)1+1/q ‖f ′ − iαf‖p ,if f ′ − iαf ∈ Lp [a, b]p > 1, 1

p+ 1

q= 1.

4.3. An Application for CBS-Inequality. The following inequality is well known in theliterature as the Cauchy-Bunyakovsky-Schwarz inequality, or the CBS-inequality, for short:

(4.21)

∣∣∣∣∫ b

a

f (t) g (t) dt

∣∣∣∣2 ≤ ∫ b

a

|f (t)|2 dt∫ b

a

|g (t)|2 dt,

provided thatf, g ∈ L2 [a, b] .We have the following result concerning some reverses of the CBS-inequality:

THEOREM 4.11 (Dragomir, 2013 [5]). Let f, g : [a, b] → C be absolutely continuous func-tions on the interval[a, b] with g (t) 6= 0 for all t ∈ [a, b] . Then

(4.22) 0 ≤∫ b

a

|g (t)|2 dt∫ b

a

|f (t)|2 dt−∣∣∣∣∫ b

a

f (t) g (t) dt

∣∣∣∣2

≤ 1

2×

‖f ′g − fg′‖2∞

(∫ b

a|g (t)|2 dt

)2 (∫ b

a1

|g(t)|2dt)2

, iff ′g − fg′ ∈ L∞ [a, b] ,

1|g|2 ∈ L [a, b]

‖f ′g − fg′‖2p

(∫ b

a|g (t)|2 dt

)2 (∫ b

a1

|g(t)|2q dt)2/q

, if

f ′g − fg′ ∈ Lp [a, b] ,1

|g|2q ∈ L [a, b]

p > 1,1p

+ 1q

= 1,

‖f ′g − fg′‖21

(∫ b

a|g (t)|2 dt

)2

ess supt∈[a,b]

1

|g(t)|4

, if 1

|g| ∈ L∞ [a, b] .


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96 S. S. DRAGOMIR

PROOF. Utilising the inequality (4.3) we have

(4.23)∣∣∣g (t)f (x)− f (t) g (x)

∣∣∣

≤

‖f ′g − fg′‖∞ |g (t) g (x)|∣∣∣∫ x

t1

|g(s)|2ds∣∣∣ if f ′g − fg′

∈ L∞ [a, b] ,

‖f ′g − fg′‖p |g (t) g (x)|∣∣∣∫ x

t1

|g(s)|2q ds∣∣∣1/q

if f ′g − fg′

∈ Lp [a, b]p > 1,

1p

+ 1q

= 1,

‖f ′g − fg′‖1 |g (t) g (x)| sups∈[t,x]([x,t])

1

|g(s)|2

.

for anyt, x ∈ [a, b] .Taking the square in (4.23) and integrating over(t, x) ∈ [a, b]2 we have

(4.24)∫ b

a

∫ b

a

∣∣∣g (t)f (x)− f (t) g (x)∣∣∣2 dtdx

≤

‖f ′g − fg′‖2∞∫ b

a

∫ b

a|g (t) g (x)|2

∣∣∣∫ x

t1

|g(s)|2ds∣∣∣2 dtdx,

‖f ′g − fg′‖2p

∫ b

a

∫ b

a|g (t) g (x)|2

∣∣∣∫ x

t1

|g(s)|2q ds∣∣∣2/q

dtdx,

‖f ′g − fg′‖21

∫ b

a

∫ b

a|g (t) g (x)|2 sups∈[t,x]([x,t])

1

|g(s)|4

dtdx.

Observe that∫ b

a

∫ b

a

∣∣∣g (t)f (x)− f (t) g (x)∣∣∣2 dtdx

=

∫ b

a

∫ b

a

(|g (t)|2 |f (x)|2 − 2 Re

[g (t)f (x) f (t) g (x)

]+ |g (x)|2 |f (t)|2

)dtdx

=

∫ b

a

|g (t)|2 dt∫ b

a

|f (x)|2 dx− 2 Re

[∫ b

a

f (t) g (t)dt

∫ b

a

f (x) g (x) dx

]+

∫ b

a

|g (x)|2 dx∫ b

a

|f (t)|2 dt

= 2

[∫ b

a

|g (t)|2 dt∫ b

a

|f (t)|2 dt−∣∣∣∣∫ b

a

f (t) g (t) dt

∣∣∣∣2],

∫ b

a

∫ b

a

[|g (t) g (x)|2

∣∣∣∣∫ x

t

1

|g (s)|2ds

∣∣∣∣2]dtdx

≤(∫ b

a

|g (t)|2 dt)2(∫ b

a

1

|g (t)|2dt

)2

,


http://ajmaa.org


∫ b

a

∫ b

a

[|g (t) g (x)|2

∣∣∣∣∫ x

t

1

|g (s)|2q ds

∣∣∣∣2/q]dtdx

≤(∫ b

a

|g (t)|2 dt)2(∫ b

a

1

|g (t)|2q dt

)2/q

and ∫ b

a

∫ b

a

[|g (t) g (x)|2 sup

s∈[t,x]([x,t])

1

|g (s)|4

]dtdx

≤(∫ b

a

|g (t)|2 dt)2

ess supt∈[a,b]

1

|g (t)|4

,

then by (4.24) we get the desired result (4.22).

REFERENCES

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

[2] S. S. Dragomir, Power Pompeiu’s type inequalities for absolutely continuous functionswith applications to Ostrowski’s inequality,Stud. Univ. Babes-Bolyai Math.59 (2014),no. 2, 131–140. PreprintRGMIA Res. Rep. Coll.16 (2013), Art. 68, pp. 8 [Onlinehttp://rgmia.org/papers/v16/v16a68.pdf].

[3] S. S. Dragomir, Inequalities of Pompeiu’s type for absolutely continuous functions with ap-plications to Ostrowski’s inequality,Acta Math. Acad. Paedagog. Nyházi.(N.S.) 30 (2014),no. 1, 43–52. PreprintRGMIA Res. Rep. Coll.16 (2013), Art. 66, pp. 10 [Onlinehttp://rgmia.org/papers/v16/v16a66].

[4] S. S. Dragomir, Exponential Pompeiu’s type inequalities with applications to Ostrowski’s inequality,Acta Math. Univ. Comenian.(N.S.) 84 (2015), No. 1, 39–50. PreprintRGMIA Res. Rep. Coll.16(2013), Art. 70, pp. 12[Online http://rgmia.org/papers/v16/v16a70.pdf].

[5] S. S. Dragomir, Ostrowski via a two functions Pompeiu’s inequality,An. St. Univ. Ovidius Constanta,Vol. 24 (3), 2016, 123-139. PreprintRGMIA Res. Rep. Coll.16 (2013), Art. 90, 13pp.[Onlinehttp://rgmia.org/papers/v16/v16a90.pdf] .

[6] D. Pompeiu, Sur une proposition analogue au théorème des accroissements finis,Mathematica(Cluj,Romania),22(1946), 143-146.

[7] J. Pecaric and Š. Ungar, On an inequality of Ostrowski type,J. Ineq. Pure & Appl. Math.7 (2006).No. 4. Art. 151.

[8] P. K. Sahoo and T. Riedel,Mean Value Theorems and Functional Equations, World Scientific, Sin-gapore, New Jersey, London, Hong Kong, 2000.


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CHAPTER 5

Inequalities for Derivatives with Special Properties

1. OSTROWSKI FOR CONVEX DERIVATIVES

1.1. A Representation Result.In [8], the author pointed out the following identity in rep-resenting an absolutely continuous function.

LEMMA 1.1 (Dragomir, 2002 [8]). Letf : [a, b] → R be an absolutely continuous functionon [a, b] . Then for anyx ∈ [a, b] , one has the equality:

(1.1) f (x) =1

b− a

∫ b

a

f (t) dt+1

b− a

∫ b

a

(x− t)

(∫ 1

0

f ′ [(1− λ)x+ λt] dλ

)dt.

PROOF. For anyt, x ∈ [a, b] , x 6= t, one has

f (x)− f (t)

x− t=

1

x− t

∫ x

t

f ′ (u) du =

∫ 1

0

f ′ [(1− λ)x+ λt] dλ,

showing that

(1.2) f (x) = f (t) + (x− t)

∫ 1

0

f ′ [(1− λ)x+ λt] dλ

for anyt, x ∈ [a, b] .If we integrate (1.2) overt on [a, b] and divide by(b− a) , we deduce the desired identity

(1.1).

1.2. Ostrowski for Convex Derivatives.The following result holds.

THEOREM1.2 (Dragomir & Sofo, 2002 [17]). Letf : [a, b] → R be a differentiable functionsuch that its derivativef ′ is convex on(a, b) . Then for anyx ∈ (a, b) one has the inequality

1

2· b− x

b− a

[1

b− a

∫ b

x

f (u) du− f (b)− 1

2(b− x) f ′ (x)

](1.3)

+ 2(x− a)

b− a

[2

x− a

∫ x

a+x2

f (u) du− f

(a+ x

2

)]

≤ f (x)− 1

b− a

∫ b

a

f (t) dt

≤ 1

2· x− a

b− a

[1

x− a

∫ x

a

f (u) du− f (a)− 1

2(x− a) f ′ (x)

]+ 2

(b− x)

b− a

[2

b− x

∫ x+b2

x

f (u) du− f

(x+ b

2

)].

99

100 S. S. DRAGOMIR

PROOF. For anyx ∈ (a, b) , one has the identity (see Lemma 1.1)

(1.4) f (x)− 1

b− a

∫ b

a

f (t) dt

=1

b− a

[∫ x

a

(x− t)

(∫ 1

0

f ′ [(1− λ)x+ λt] dλ

)dt

+

∫ b

x

(x− t)

(∫ 1

0

f ′ [(1− λ)x+ λt] dλ

)dt

].

Sincef ′ is convex, then by the Hermite-Hadamard inequality forf ′ we have

(1.5) f ′(x+ t

2

)≤∫ 1

0

f ′ [(1− λ)x+ λt] dλ ≤ f ′ (x) + f ′ (t)

2

for anyt, x ∈ (a, b) .Assumea ≤ t ≤ x. Then by (1.5) we get∫ x

a

(x− t) f ′(x+ t

2

)dt ≤

∫ x

a

(x− t)

(∫ 1

0

f ′ [(1− λ)x+ λt] dλ

)dt(1.6)

≤∫ x

a

[f ′ (x) + f ′ (t)

2

](x− t) dt.

Assumex ≤ t ≤ b. Then by (1.5) we also get∫ b

x

(x− t)

[f ′ (x) + f ′ (t)

2

]dt ≤

∫ b

x

(x− t)

(∫ 1

0

f ′ [(1− λ)x+ λt] dλ

)dt(1.7)

≤∫ b

x

(x− t) f ′(x+ t

2

)dt.

Summing (1.6) with (1.7), dividing withb− a and using the identity (1.4) we deduce

1

b− a

[∫ x

a

(x− t) f ′(x+ t

2

)dt+

∫ b

x

(x− t)

[f ′ (x) + f ′ (t)

2

]dt

](1.8)

≤ f (x)− 1

b− a

∫ b

a

f (t) dt

≤ 1

b− a

[∫ x

a

[f ′ (x) + f ′ (t)

2

](x− t) dt+

∫ b

x

(x− t) f ′(x+ t

2

)dt

].

Since ∫ x

a

(x− t) f ′(x+ t

2

)dt = 2

∫ x

a

f

(x+ t

2

)dt− 2f

(x+ a

2

)(x− a)

= 4

∫ x

a+x2

f (u) du− 2f

(a+ x

2

)(x− a) ,

∫ b

x

(x− t)

[f ′ (x) + f ′ (t)

2

]dt =

1

2

∫ b

x

f (t) dt− 1

2(b− x) f (b)− 1

4(b− x)2 f ′ (x) ,

∫ x

a

[f ′ (x) + f ′ (t)

2

](x− t) dt =

1

2

∫ x

a

f (t) dt− 1

2(x− a) f (a) +

1

4(x− a)2 f ′ (x) ,


http://ajmaa.org


and ∫ b

x

(x− t) f ′(x+ t

2

)dt = 2

∫ b

x

f

(x+ t

2

)dt− 2f

(x+ b

2

)(b− x)

= 4

∫ x+b2

x

f (u) du− 2f

(x+ b

2

)(b− x) ,

then by (1.8) we deduce the desired result.

The following corollary is natural to be considered.

COROLLARY 1.3. Assume thatf : [a, b] → R is as in Theorem 1.2. Then

1

4

[2

b− a

∫ b

a+b2

f (u) du− f (b)− 1

4(b− a) f ′

(a+ b

2

)](1.9)

+4

b− a

∫ a+b2

3a+b4

f (u) du− f

(3a+ b

4

)≤ f

(a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

≤ 1

4

[2

b− a

∫ a+b2

a

f (u) du− f (a) +1

4(b− a) f ′

(a+ b

2

)]

+4

b− a

∫ a+3b4

a+b2

f (u) du− f

(a+ 3b

4

).

COROLLARY 1.4. Assume thatf : [a, b] → R is as in Theorem 1.2. Then

8

5 (b− a)

∫ b

a+b2

f (u) du− 1

5f (b)− 1

10(b− a) f ′ (a)− 4

5f

(a+ b

2

)(1.10)

≤ 4

5

[f (a) + f (b)

2

]− 1

b− a

∫ b

a

f (u) du

≤ 8

5 (b− a)

∫ a+b2

a

f (u) du− 1

5f (a) +

1

10(b− a) f ′ (b)− 4

5f

(a+ b

2

).

PROOF. At x = a, we have from (1.3)

1

2

[1

b− a

∫ b

a

f (u) du− f (b)− 1

2(b− a) f ′ (a)

](1.11)

≤ f (a)− 1

b− a

∫ b

a

f (u) du

≤ 2

[2

b− a

∫ a+b2

a

f (u) du− f

(a+ b

2

)].


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102 S. S. DRAGOMIR

At x = b,

2

[2

b− a

∫ b

a+b2

f (u) du− f

(a+ b

2

)](1.12)

≤ f (b)− 1

b− a

∫ b

a

f (u) du

≤ 1

2

[1

b− a

∫ b

a

f (u) du− f (a) +1

2(b− a) f ′ (b)

].

Manipulating (1.11) and (1.12) we arrive at (1.10).

2. OSTROWSKI FOR DERIVATIVES THAT ARE CONVEX IN M ODULUS

2.1. A Representation Result.In [8], the author pointed out the following identity in rep-resenting an absolutely continuous function.


(2.1) f (x) =1

b− a

∫ b

a

f (t) dt+1

b− a

∫ b

a

(x− t)

(∫ 1

0

f ′ [(1− λ)x+ λt] dλ

)dt.

We have the following results.

2.2. Some Refinements.Using the above lemma the following result can be pointed outimproving Ostrowski’s inequality.

THEOREM 2.2 (Barnett et al., 2001 [3]). Let f : [a, b] → C be an absolutely continuousfunction on[a, b] so that|f ′| is convex on(a, b). If f ′ ∈ L∞[a, b], then for anyx ∈ [a, b],∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(2.2)

≤ 1

2

1

4+

(x− a+b

2

b− a

)2 (b− a) [|f ′(x)|+ ‖f ′‖∞] .

The constant12

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

PROOF. Using (2.1) and taking the modulus, we have∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣ =1

b− a

∣∣∣∣∫ b

a

∫ 1

0

(x− t) f ′ [(1− λ)x+ λt] dλdt

∣∣∣∣≤ 1

b− a

∫ b

a

∫ 1

0

|x− t| |f ′ [(1− λ)x+ λt]| dλdt

:= K


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Utilizing the convexity of|f ′| we have

K ≤ 1

b− a

∫ b

a

∫ 1

0

|x− t| [(1− λ) |f ′(x)|+ λ |f ′(t)|] dλdt

=1

b− a

∫ b

a

|x− t|[|f ′(x)|

∫ 1

0

(1− λ) dλ+ |f ′(t)|∫ 1

0

λdλ

]dt

=1

b− a

∫ b

a

|x− t|[|f ′(x)|+ |f ′(t)|

2

]dt := M(x)

≤ 1

2

1

b− aess. sup

t∈[a,b]

[|f ′(x)|+ |f ′(t)|]∫ b

a

|x− t| dt

=1

2

[(x− a)2 + (b− x)2

2 (b− a)

][|f ′(x)|+ ‖f ′‖∞]

=1

2

1

4+

(x− a+b

2

b− a

)2 (b− a) [|f ′(x)|+ ‖f ′‖∞] ,

for anyx ∈ [a, b], and the inequality (2.2) is proved.Assume that (2.2) holds with a constantC > 0, that is,∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(2.3)

≤ C

1

4+

(x− a+b

2

b− a

)2 (b− a) [|f ′(x)|+ ‖f ′‖∞]

for anyx ∈ [a, b] with f as in the hypothesis of the theorem.Consider the function

f0 : [a, b] → R, f0(t) = k

∣∣∣∣t− a+ b

2

∣∣∣∣ , k > 0, t ∈ [a, b].

Since|f ′0(t)| = k, for anyt ∈ [a, b] and

1

b− a

∫ b

a

f0(t)dt =k

4(b− a) , ‖f ′0‖∞ = k

then choosingf = f0 andx = a+b2

in (2.3), we get

k

4(b− a) ≤ Ck (b− a)

2

givingC ≥ 12, which proves the sharpness of the constant1

2.

The following particular case is interesting.

COROLLARY 2.3. With the assumptions of Theorem 2.4, we have the inequality

(2.4)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣ ≤ 1

8(b− a)

[∣∣∣∣f ′(a+ b

2

)∣∣∣∣+ ‖f ′‖∞]

and the constant18


The following result in terms of thep-norms also holds:


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104 S. S. DRAGOMIR

THEOREM 2.4. Letf : [a, b] → C be as in Theorem 2.2. Iff ′ ∈ Lp[a, b], p > 1, 1p

+ 1q

= 1,

then for anyx ∈ [a, b],∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(2.5)

≤ 1

2 (q + 1)1q

[(b− x

b− a

)q+1

+

(x− a

b− a

)q+1] 1

q

(b− a)1q ‖|f ′(x)|+ |f ′|‖p .

The constant12

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

PROOF. According to the proof of Theorem 2.2, we have∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣ ≤ 1

b− a

∫ b

a

|x− t|[|f ′(x)|+ |f ′(t)|

2

]dt := M(x).

Using Hölder’s integral inequality forp > 1, 1p

+ 1q

= 1, we get that

M(x) ≤ 1

2 (b− a)

(∫ b

a

|x− t|q dt) 1

q(∫ b

a

(|f ′(x)|+ |f ′(t)|)pdt

) 1p

=1

2 (b− a)

[(b− x)q+1 + (x− a)q+1

q + 1

] 1q

‖|f ′(x)|+ |f ′|‖p

and the inequality (2.5) is proved.Reconsider the function utilised in Theorem 2.2,

f0 : [a, b] → R, f0(t) = k

∣∣∣∣t− a+ b

2

∣∣∣∣ , k > 0, t ∈ [a, b]

which has|f ′0(t)| (= k) convex in[a, b]. If we assume that (2.5) holds with a constantD > 0instead of1

2, so that∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣≤ D

(q + 1)1q

[(b− x

b− a

)q+1

+

(x− a

b− a

)q+1] 1

q

(b− a)1q ‖|f ′(x)|+ |f ′|‖p ,

then takingf = f0 overx = a+b2, we get,

k

4(b− a) ≤ D

(q + 1)1q

(1

2q

) 1q

(b− a)1q k (b− a)

1p ,

q > 1, p > 1, 1p

+ 1q

= 1 giving, on simplification,

D ≥ 1

2(q + 1)

1q , q > 1.

Taking the limit asq →∞ and since,

limq→∞

(q + 1)1q = exp

limq→∞

[ln(1 + q)

q

]= exp 0 = 1,

we deduce thatD ≥ 12, which proves the sharpness of the constant.

A particular case is the following mid-point inequality:


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COROLLARY 2.5. With the assumptions of Theorem 2.4, we have,∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(2.6)

≤ 1

4 (q − a)1q

(b− a)1q

(∫ b

a

[∣∣∣∣f ′(a+ b

2

)∣∣∣∣+ |f ′(t)|]p

dt

) 1p

(p > 1, 1

p+ 1

q= 1). The constant1

4is sharp in the previous sense.

Finally, the case involving the1-norm is embodied in the following theorem:

THEOREM 2.6. Let f : [a, b] → R be as in Theorem 2.2. Iff ′ ∈ L1[a, b], then, for anyx ∈ [a, b],

(2.7)

∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣ ≤ 1

2

[1

2+

∣∣∣∣∣x− a+b2

b− a

∣∣∣∣∣]

[(b− a) |f ′(x)|+ ‖f ′‖1] .

PROOF. We have, from the proof of Theorem 2.2, that

M(x) ≤ supt∈[a,b]

|x− t| 1

b− a

∫ b

a

[|f ′(x)|+ |f ′(t)|

2

]dt

=1

2(b− a)max (x− a, b− x)

[(b− a) |f ′(x)|+

∫ b

a

|f ′(t)| dt]

=1

2

[1

2+

∣∣∣∣∣x− a+b2

b− a

∣∣∣∣∣]

[(b− a) |f ′(x)|+ ‖f ′‖1]


In particular, we have the mid-point inequality:

COROLLARY 2.7. Assume thatf is as in Theorem 2.6. Then

(2.8)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣ ≤ 1

4

[(b− a)

∣∣∣∣f ′(a+ b

2

)∣∣∣∣+ ∫ b

a

|f ′(t)| dt].

Another way to estimate the difference∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣is presented in the following theorem.

THEOREM 2.8. Letf : [a, b] → R be an absolutely continuous function on[a, b] so that|f ′|is convex on(a, b). Then, for anyx ∈ [a, b],∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(2.9)

≤ 1

2

1

4+

(x− a+b

2

b− a

)2 |f ′(x)| (b− a)

+1

(q + 1)1q

[(b− x

b− a

)q+1

+

(x− a

b− a

)q+1] 1

q

(b− a)1q ‖f ′‖p

,

wherep > 1, 1p

+ 1q

= 1.


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106 S. S. DRAGOMIR

PROOF. With the notation of Theorem 2.2, we have,

M(x) =1

2 (b− a)

[|f ′(x)|

∫ b

a

|x− t| dt+

∫ b

a

|x− t| |f ′(t)| dt]

=1

2 (b− a)

[|f ′(x)| (x− a)2 + (b− x)2

2+

∫ b

a

|x− t| |f ′(t)| dt]

=1

2

|f ′(x)|1

4+

(x− a+b

2

b− a

)2 (b− a) +

1

b− a

∫ b

a

|x− t| |f ′(t)| dt

.

Using Hölder’s inequality,

1

b− a

∫ b

a

|x− t| |f ′(t)| dt

≤ 1

b− a

(∫ b

a

|x− t|q dt) 1

q(∫ b

a

|f ′(t)|p dt) 1

p

=1

b− a

[(b− x)q+1 + (x− a)q+1

q + 1

] 1q

‖f ′‖p

=1

(q + 1)1

q

[(b− x

b− a

)q+1

+

(x− a

b− a

)q+1] 1

q

(b− a)1q ‖f ′‖p ,


The following particular corollary is of interest providing a bound for the midpoint.

COROLLARY 2.9. Letf be as in the previous theorem. Then one has the inequality:∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(2.10)

≤ 1

4

1

2

∣∣∣∣f ′(a+ b

2

)∣∣∣∣ (b− a) +1

(q + 1)1q

(b− a)1q ‖f ′‖p

.

3. OSTROWSKI FOR DERIVATIVES THAT HAVE CERTAIN CONVEXITY PROPERTIES

3.1. Some Inequalities for|f ′| Convex. The following representation holds:

LEMMA 3.1 (Cerone & Dragomir, 2004 [7]). Letf : [a, b] → C be an absolutely continuousfunction on[a, b] . Then we have the representation

f (x) =1

b− a

∫ b

a

f (t) dt+ (x− a)2 · 1

b− a

∫ 1

0

λf ′ [(1− λ) a+ λx] dλ(3.1)

− (b− x)2 · 1

b− a

∫ 1

0

λf ′ [λx+ (1− λ) b] dλ



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PROOF. We start with the following Montgomery identity

f (x) =1

b− a

∫ b

a

f (t) dt+1

b− a

∫ x

a

(t− a) f ′ (t) dt(3.2)

+1

b− a

∫ b

x

(t− b) f ′ (t) dt

for anyx ∈ [a, b] .If we make the change of variablet = (1− λ) a+ λx, λ ∈ [0, 1] , then we get∫ x

a

(t− a) f ′ (t) dt = (x− a)2

∫ 1

0

λf ′ [(1− λ) a+ λx] dλ.

Also, the change of variablet = µx+ (1− µ) b, µ ∈ [0, 1] , will provide the equality∫ b

x

(t− b) f ′ (t) dt = − (b− x)2

∫ 1

0

µf ′ [µx+ (1− µ) b] dµ.

Using (3.2), we then deduce the desired identity (3.1).

The following Ostrowski-type inequality holds for|f ′| convex.

THEOREM 3.2 (Cerone & Dragomir, 2004 [7]). Let f : [a, b] → C be an absolutely con-tinuous function on[a, b] andx ∈ [a, b] . If |f ′| is convex on[a, x] and [x, b] , then one has theinequality:∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

6

[|f ′ (a)|

(x− a

b− a

)2

+ |f ′ (b)|(b− x

b− a

)2

(3.3)

+

1 + 2

(x− a+b

2

b− a

)2 |f ′ (x)|

(b− a) .

The constant16

is best possible in the sense that it cannot be replaced by a smaller value.

PROOF. Taking the modulus in (3.1) we have∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.4)

≤ (x− a)2 · 1

b− a

∫ 1

0

λ |f ′ [(1− λ) a+ λx]| dλ

+ (b− x)2 · 1

b− a

∫ 1

0

λ |f ′ [λx+ (1− λ) b]| dλ

:= M (x) .

Since|f ′| is convex on[a, x] and[x, b] , then obviously∫ 1

0

λ |f ′ [(1− λ) a+ λx]| dλ ≤ |f ′ (a)|∫ 1

0

λ (1− λ) dλ+ |f ′ (x)|∫ 1

0

λ2dλ

=1

6|f ′ (a)|+ 1

3|f ′ (x)|


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108 S. S. DRAGOMIR

and ∫ 1

0

λ |f ′ [λx+ (1− λ) b]| dλ ≤ |f ′ (x)|∫ 1

0

λ2dλ+ |f ′ (b)|∫ 1

0

λ (1− λ) dλ

=1

3|f ′ (x)|+ 1

6|f ′ (b)| .

Thus from (3.4)

M (x)

≤ 1

3

[1

2|f ′ (a)|+ |f ′ (x)|

](x− a)2 +

1

3

[|f ′ (x)|+ 1

2|f ′ (b)|

](b− x)2

=1

3

[|f ′ (a)| (x− a)2 + |f ′ (b)| (b− x)2

2+[(x− a)2 + (b− x)2] |f ′ (x)|]

=1

6

[|f ′ (a)| (x− a)2 + |f ′ (b)| (b− x)2 +

[(b− a)2 + 2

(x− a+ b

2

)2]|f ′ (x)|


To prove the sharpness of the constant16, assume that (3.3) holds with a constantC > 0.

Namely, ∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ C

[|f ′ (a)|

(x− a

b− a

)2

+ |f ′ (b)|(b− x

b− a

)2

(3.5)

+

1 + 2

(x− a+b

2

b− a

)2 |f ′ (x)|

(b− a) ,

provided that|f ′| is convex on[a, x] and[x, b] .If we choosex = a+b

2, then by (3.5) we deduce∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.6)

≤ C

[|f ′ (a)|+ |f ′ (b)|

4+

∣∣∣∣f ′(a+ b

2

)∣∣∣∣] (b− a) ,

where|f ′| is convex on[a, a+b

2

]and

[a+b2, b].

Consider the functionf0 : [a, b] → R, given by

f0 (x) =

a+b2− x, if x ∈

[a, a+b

2

],

x− a+b2, if x ∈

(a+b2, b].

The function is absolutely continuous on[a, b] and, obviously,|f ′| = 1 on[a, a+b

2

]and

[a+b2, b]

showing that it is convex on[a, a+b

2

]and

[a+b2, b].

On the other hand, we have

f0

(a+ b

2

)= 0,

1

b− a

∫ b

a

f0 (x) dx =b− a

4,

|f ′0 (a)| = |f ′0 (b)| =∣∣∣∣f ′0(a+ b

2

)∣∣∣∣ = 1


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and so from (3.6) we getb− a

4≤ C

(1

2+ 1

)(b− a)

givingC ≥ 16.

The following corollary is a natural consequence.

COROLLARY 3.3. Assume thatf : [a, b] → R is absolutely continuous such that|f ′| is aconvex function on

[a, a+b

2

]and

(a+b2, b]. Then we have the inequality∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.7)

≤ 1

6

[|f ′ (a)|+ |f ′ (b)|

4+

∣∣∣∣f ′(a+ b

2

)∣∣∣∣] (b− a) .

The 16

is best possible in (3.7) in the sense that it cannot be replaced by a smaller constant.

3.2. Inequalities for|f ′|Quasi-Convex. Firstly, let us recall the definition of quasi-convexfunctions.

DEFINITION 3.1. The functionh : [a, b] ⊂ R → R is said to bequasi-convex(QC) on theintervalI if

(3.8) h (λx+ (1− λ) y) ≤ max h (x) , h (y)for anyx, y ∈ I andλ ∈ [0, 1] .

Following [15], we say that for an intervalI ⊆ R, the mappingh : I → R is quasi-monotoneon I if it is either monotone onI = [c, d] or monotone nonincreasing on a propersubinterval[c, c′] ⊂ I and monotone nondecreasing on[c′, d] .

The classQM (I) of quasi-monotone functions onI provides an immediate characterizationof quasi-convex functions [15].

PROPOSITION3.4. SupposeI ⊆ R. Then the following statements are equivalent for afunctionh : I → R:

(a) h ∈ QM (I) ;(b) On any subinterval ofI, h achieves its supremum at an end point;(c) h ∈ QC (I) .

As examples of quasi-convex functions we may consider the class of monotonic functionson an intervalI for the class of convex functions on that interval.

The following Ostrowski type inequality for absolutely continuous functions for which|f ′|is quasi-convex holds.

THEOREM 3.5 (Cerone & Dragomir, 2004 [7]). Letf : [a, b] → R be an absolutely contin-uous function on[a, b] andx ∈ [a, b] . If |f ′| is quasi-convex on[a, x] and [x, b] , then one hasthe inequality ∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.9)

≤ 1

4

(x− a

b− a

)2

[|f ′ (a)|+ |f ′ (x)|+ ||f ′ (x)| − |f ′ (a)||]

+

(b− x

b− a

)2

[|f ′ (x)|+ |f ′ (b)|+ ||f ′ (x)| − |f ′ (b)||]

.


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110 S. S. DRAGOMIR

The constant14

is sharp in (3.9) in the sense that it cannot be replaced by a smaller value.

PROOF. Since|f ′| is quasi-convex on[a, x] and[x, b] , then from (3.8)∫ 1

0

λ |f ′ ((1− λ) a+ λx)| dλ ≤ max |f ′ (a)| , |f ′ (x)|∫ 1

0

λdλ

=1

2

[|f ′ (a)|+ |f ′ (x)|

2+

1

2||f ′ (x)| − |f ′ (a)||

]:= M1 (x)

and, similarly∫ 1

0

λ |f ′ (λx+ (1− λ) b)| dλ ≤ 1

2

[|f ′ (x)|+ |f ′ (b)|

2+

1

2||f ′ (x)| − |f ′ (b)||

]:= M2 (x) .

Using (3.4) and the notationM (x) for the right hand side of that inequality, we deduce that∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤M (x)

≤(x− a

b− a

)2

M1 (x) +

(b− x

b− a

)2

M2 (x)

and the result (3.9) is thus proved.The fact that1

4is the best possible constant will be shown in the following.

COROLLARY 3.6. Let f : [a, b] → R be an absolutely continuous function on[a, b] . If |f ′|is quasi-convex on

[a, a+b

2

]and

[a+b2, b], then one has the inequality:∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.10)

≤ 1

16

|f ′ (a)|+ 2

∣∣∣∣f ′(a+ b

2

)∣∣∣∣+ |f ′ (b)|+∣∣∣∣∣∣∣∣f ′(a+ b

2

)∣∣∣∣− |f ′ (a)|∣∣∣∣

+

∣∣∣∣|f ′ (b)| − ∣∣∣∣f ′(a+ b

2

)∣∣∣∣∣∣∣∣ (b− a) .

The constant116

is best possible.

PROOF. The inequality follows by (3.9) on choosingx = a+b2. To prove the sharpness of

the constant116, assume that (3.10) holds with a constantC > 0. That is,∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.11)

≤ C

|f ′ (a)|+ 2

∣∣∣∣f ′(a+ b

2

)∣∣∣∣+ |f ′ (b)|+∣∣∣∣∣∣∣∣f ′(a+ b

2

)∣∣∣∣− |f ′ (a)|∣∣∣∣

+

∣∣∣∣|f ′ (b)| − ∣∣∣∣f ′(a+ b

2

)∣∣∣∣∣∣∣∣ (b− a) .

Consider the functionf0 : [a, b] → R, f0 (t) =∣∣t− a+b

2

∣∣ . Thenf0 is absolutely continuous and|f ′0 (t)| = 1, t ∈ [a, b] . Thus, from (3.11), we deduce

b− a

4≤ C · 4 (b− a)


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givingC ≥ 12

and the corollary is proved.

3.3. Inequalities for |f ′| Log-Convex. In what follows,I will denote an interval of realnumbers. A functionf : I → (0,∞) is said to belog-convexor multiplicatively convexif log fis convex, or, equivalently, if for anyx, y ∈ I andt ∈ [0, 1] one has the inequality

(3.12) f (tx+ (1− t) y) ≤ [f (x)]t [f (y)]1−t .

We note that iff andg are convex andg is increasing, theng f is convex, moreover, sincef = exp [log f ] , it follows that a log-convex function is convex, but the converse may notnecessarily be true. This follows directly from (3.12) since, by the arithmetic-geometric meaninequality we have

(3.13) [f (x)]t [f (y)]1−t ≤ tf (x) + (1− t) f (y)

for all x, y ∈ I andt ∈ [0, 1] .The following result holds.

THEOREM 3.7 (Cerone & Dragomir, 2004 [7]). Letf : [a, b] → R be an absolutely contin-uous function on[a, b] andx ∈ [a, b] . If |f ′| is log-convex on[a, x] and [x, b] , then one has theinequality

∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.14)

≤

(x− a

b− a

)2

|f ′ (a)| A lnA+ 1− A

(lnA)2

+

(b− x

b− a

)2

|f ′ (b)| B lnB + 1−B

(lnB)2

(b− a) ,

where

A :=

∣∣∣∣f ′ (x)f ′ (a)

∣∣∣∣ , B :=

∣∣∣∣f ′ (x)f ′ (b)

∣∣∣∣ .AJMAA, Vol. 14, No. 1, Art. 1, pp. 1-287, 2017 AJMAA

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112 S. S. DRAGOMIR

PROOF. Using the representation (3.1) and the definition of log-convexity (3.12), we havesuccessively: ∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.15)

≤ (b− a)

(x− a

b− a

)2 ∫ 1

0

λ |f ′ ((1− λ) a+ λx)| dλ

+

(b− x

b− a

)2 ∫ 1

0

λ |f ′ (λx+ (1− λ) b)| dλ

≤ (b− a)

(x− a

b− a

)2 ∫ 1

0

λ |f ′ (a)|1−λ |f ′ (x)|λ dλ

+

(b− x

b− a

)2 ∫ 1

0

λ |f ′ (x)|λ |f ′ (b)|1−λdλ

= (b− a)

(x− a

b− a

)2

|f ′ (a)|∫ 1

0

λAλdλ

+

(b− x

b− a

)2

|f ′ (b)|∫ 1

0

λBλdλ

.

Since, a simple calculation shows that for anyC > 0, one has∫ 1

0

λCλdλ =C lnC + 1− C

(lnC)2 ,

then from (3.15) we deduce the desired result (3.14).

The following corollary holds.

COROLLARY 3.8. Let f : [a, b] → R be an absolutely continuous function on[a, b] . If |f ′|is log-convex on

[a, a+b

2

]and

[a+b2, b], then one has the inequality:∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.16)

≤ 1

4

[|f ′ (a)| α lnα+ 1− α

(lnα)2 + |f ′ (b)| β ln β + 1− β

(ln β)2

](b− a) ,

where

α :=

∣∣∣∣∣f ′(

a+b2

)f ′ (a)

∣∣∣∣∣ , β :=

∣∣∣∣∣f ′(

a+b2

)f ′ (b)

∣∣∣∣∣ .4. OSTROWSKI FOR FUNCTIONS WHOSE DERIVATIVES ARE h-CONVEX IN M ODULUS

4.1. Some Classes of Functions.We recall here some concepts of convexity that are wellknown in the literature. LetI be an interval inR.

DEFINITION 4.1 (Godunova-Levin, 1985 [18]). We say thatf : I → R is a Godunova-Levin function or thatf belongs to the classQ (I) if f is non-negative and for allx, y ∈ I andt ∈ (0, 1) we have

f (tx+ (1− t) y) ≤ 1

tf (x) +

1

1− tf (y) .


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Some further properties of this class of functions can be found in [12], [13], [16], [22],[24] and [25]. Among others, its has been noted that non-negative monotone and non-negativeconvex functions belong to this class of functions.

DEFINITION 4.2 (Dragomir et al., 1995 [16]). We say that a functionf : I → R belongs tothe classP (I) if it is nonnegative and for allx, y ∈ I andt ∈ [0, 1] we have

f (tx+ (1− t) y) ≤ f (x) + f (y) .

ObviouslyQ (I) containsP (I) and for applications it is important to note that alsoP (I)contain all nonnegative monotone, convex andquasi convex functions,i. e. nonnegative func-tions satisfying

f (tx+ (1− t) y) ≤ max f (x) , f (y)for all x, y ∈ I andt ∈ [0, 1] .

For some results onP -functions see [16] and [23] while for quasi convex functions, thereader can consult [14].

DEFINITION 4.3 (Breckner, 1978 [5]). Let s be a real number,s ∈ (0, 1]. A function f :[0,∞) → [0,∞) is said to bes-convex (in the second sense) or Breckners-convex if

f (tx+ (1− t) y) ≤ tsf (x) + (1− t)s f (y)

for all x, y ∈ [0,∞) andt ∈ [0, 1] .

For some properties of this class of functions see [1], [2], [5], [6], [10], [11], [19], [20] and[27].

In order to unify the above concepts, S. Varošanec introduced the concept ofh-convexfunctions as follows.

Assume thatI andJ are intervals inR, (0, 1) ⊆ J and functionsh andf are real non-negative functions defined inJ andI, respectively.

DEFINITION 4.4 (Varošanec, 2007 [30]). Let h : J → [0,∞) with h not identical to0. Wesay thatf : I → [0,∞) is anh-convex function if for allx, y ∈ I we have

f (tx+ (1− t) y) ≤ h (t) f (x) + h (1− t) f (y)

for all t ∈ (0, 1) .

For some results concerning this class of functions see [30], [4], [21], [28], [26] and [29].

4.2. Inequalities of Hermite-Hadamard Type. In [26] the authors proved the followingHermite-Hadamard type inequality for integrableh-convex functions.

THEOREM 4.1 (Sarikaya et al., 2008 [26]). Assume thatf : I → [0,∞) is anh-convexfunction,h ∈ L [0, 1] andf ∈ L [a, b] wherea, b ∈ I with a < b. Then

(HH)1

2h(

12

)f (a+ b

2

)≤ 1

b− a

∫ b

a

f (t) dt ≤ [f (a) + f (b)]

∫ 1

0

h (t) dt.

If we write (HH) for h (t) = t, then we get the classical Hermite-Hadamard inequality forconvex functions.

If we write it for the case ofP -type functions, i.e.,h (t) = 1, then we get the inequality

(4.1)1

2f

(a+ b

2

)≤ 1

b− a

∫ b

a

f (t) dt ≤ f (a) + f (b) ,

providedf ∈ L [a, b] , that has been obtained in [16].


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114 S. S. DRAGOMIR

If f is integrable on[a, b] and Breckners-convex on[a, b] , for s ∈ (0, 1) , then by takingh (t) = ts in (HH) we get

(4.2) 2s−1f

(a+ b

2

)≤ 1

b− a

∫ b

a

f (t) dt ≤ f (a) + f (b)

s+ 1

that was obtained in [10].Since for the case of Godunova-Levin class of function we haveh (t) = 1

t, which is not

Lebesgue integrable on(0, 1) , we cannot apply the left inequality in (HH).We can introduce now another class of functions.

DEFINITION 4.5 (Dragomir, 2013 [9]). We say that the functionf : I → [0,∞) is ofs-Godunova-Levin type, withs ∈ [0, 1] , if

(4.3) f (tx+ (1− t) y) ≤ 1

tsf (x) +

1

(1− t)sf (y) ,

for all t ∈ (0, 1) andx, y ∈ I.

We observe that fors = 0 we obtain the class ofP -functions while fors = 1 we obtainthe class of Godunova-Levin. If we denote byQs (I) the class ofs-Godunova-Levin functionsdefined onI, then we obviously have

P (I) = Q0 (I) ⊆ Qs1 (I) ⊆ Qs2 (I) ⊆ Q1 (I) = Q (I)

for 0 ≤ s1 ≤ s2 ≤ 1.We have the following Hermite-Hadamard type inequality.

THEOREM 4.2 (Dragomir, 2013 [9]). Assume that the functionf : I → [0,∞) is of s-Godunova-Levin type, withs ∈ [0, 1). If f ∈ L [a, b] wherea, b ∈ I anda < b, then

(4.4)1

2s+1f

(a+ b

2

)≤ 1

b− a

∫ b

a

f (t) dt ≤ f (a) + f (b)

1− s.

We notice that fors = 1 the first inequality in (4.4) still holds and was obtained for the firsttime in [16].

4.3. Inequalities for Functions Whose Derivatives areh-Convex in Modulus. In [8],the author pointed out the following identity in representing an absolutely continuous function.


(4.5) f (x) =1

b− a

∫ b

a

f (t) dt+1

b− a

∫ b

a

(x− t)

(∫ 1

0

f ′ [(1− λ)x+ λt] dλ

)dt.

The following result holds:

THEOREM 4.4 (Dragomir, 2013 [9]). Let f : [a, b] → C be an absolutely continuous func-tion on[a, b] so that|f ′| is h-convex on(a, b) with h ∈ L [0, 1].

(i) If f ′ ∈ L∞[a, b], then for anyx ∈ [a, b],∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(4.6)

≤

1

4+

(x− a+b

2

b− a

)2 (b− a) [|f ′(x)|+ ‖f ′‖∞]

∫ 1

0

h (t) dt.


http://ajmaa.org


(ii) If f ′ ∈ Lp[a, b], p > 1, 1p

+ 1q

= 1, then for anyx ∈ [a, b],∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(4.7)

≤ 1

(q + 1)1q

[(b− x

b− a

)q+1

+

(x− a

b− a

)q+1] 1

q

× (b− a)1q ‖|f ′(x)|+ |f ′|‖p

∫ 1

0

h (t) dt.

(iii) If f ′ ∈ L1[a, b], then for anyx ∈ [a, b],∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣ ≤[

1

2+

∣∣∣∣∣x− a+b2

b− a

∣∣∣∣∣]

(4.8)

× [(b− a) |f ′(x)|+ ‖f ′‖1]

∫ 1

0

h (t) dt.

PROOF. (i). Using (4.5) and taking the modulus, we have∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣ =1

b− a

∣∣∣∣∫ b

a

∫ 1

0

(x− t) f ′ [(1− λ)x+ λt] dλdt

∣∣∣∣≤ 1

b− a

∫ b

a

∫ 1

0

|x− t| |f ′ [(1− λ)x+ λt]| dλdt := K

Utilizing theh-convexity of|f ′| we have

K ≤ 1

b− a

∫ b

a

∫ 1

0

|x− t| [h (1− λ) |f ′(x)|+ h (λ) |f ′(t)|] dλdt

=1

b− a

∫ b

a

|x− t|[|f ′(x)|

∫ 1

0

h (1− λ) dλ+ |f ′(t)|∫ 1

0

h (λ) dλ

]dt

=1

b− a

∫ 1

0

h (λ) dλ

∫ b

a

|x− t| [|f ′(x)|+ |f ′(t)|] dt := M (x)

∫ 1

0

h (λ) dλ

≤ 1

b− a

∫ 1

0

h (λ) dλ ess supt∈[a,b]

[|f ′(x)|+ |f ′(t)|]∫ b

a

|x− t| dt

=

1

4+

(x− a+b

2

b− a

)2 (b− a) [|f ′(x)|+ ‖f ′‖∞]

∫ 1

0

h (λ) dλ,

for anyx ∈ [a, b], and the inequality (4.6) is proved.(ii). As above, we have∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣≤ 1

b− a

∫ b

a

|x− t| [|f ′(x)|+ |f ′(t)|] dt := M (x)

∫ 1

0

h (λ) dλ.


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116 S. S. DRAGOMIR

Using Hölder’s integral inequality forp > 1, 1p

+ 1q

= 1, we get that

M(x) ≤ 1

b− a

(∫ b

a

|x− t|q dt) 1

q(∫ b

a

(|f ′(x)|+ |f ′(t)|)pdt

) 1p

=1

b− a

[(b− x)q+1 + (x− a)q+1

q + 1

] 1q

‖|f ′(x)|+ |f ′|‖p

and the inequality (4.7) is proved.(iii). We also have that

M(x) ≤ supt∈[a,b]

|x− t| 1

b− a

∫ b

a

[|f ′(x)|+ |f ′(t)|] dt

=

[1

2+

∣∣∣∣∣x− a+b2

b− a

∣∣∣∣∣]

[(b− a) |f ′(x)|+ ‖f ′‖1]


The following particular case is interesting.

COROLLARY 4.5. With the assumptions of Theorem 4.4, we have the midpoint inequality∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(4.9)

≤ 1

4(b− a)

[∣∣∣∣f ′(a+ b

2

)∣∣∣∣+ ‖f ′‖∞] ∫ 1

0

h (t) dt,

providedf ′ ∈ L∞[a, b].If f ′ ∈ Lp[a, b], p > 1, 1

p+ 1

q= 1, then, we have,∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(4.10)

≤ 1

2(b− a)

1q

(∫ b

a

[∣∣∣∣f ′(a+ b

2

)∣∣∣∣+ |f ′(t)|]p

dt

) 1p∫ 1

0

h (t) dt.

If f ′ ∈ L1[a, b], then∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(4.11)

≤ 1

2

[(b− a)

∣∣∣∣f ′(a+ b

2

)∣∣∣∣+ ∫ b

a

|f ′(t)| dt] ∫ 1

0

h (t) dt.

REMARK 4.1. Assume that|f ′| is Breckners-convex on[a, b] , for s ∈ (0, 1) .

(a) If f ′ ∈ L∞[a, b], then for anyx ∈ [a, b],∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(4.12)

≤ 1

s+ 1

1

4+

(x− a+b

2

b− a

)2 (b− a) [|f ′(x)|+ ‖f ′‖∞] .


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(aa) Iff ′ ∈ Lp[a, b], p > 1, 1p

+ 1q

= 1, then for anyx ∈ [a, b],

∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(4.13)

≤ 1

(s+ 1) (q + 1)1q

[(b− x

b− a

)q+1

+

(x− a

b− a

)q+1] 1

q

× (b− a)1q ‖|f ′(x)|+ |f ′|‖p .

(aaa) Iff ′ ∈ L1[a, b], then for anyx ∈ [a, b],

∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣ ≤ 1

s+ 1

[1

2+

∣∣∣∣∣x− a+b2

b− a

∣∣∣∣∣]

(4.14)

× [(b− a) |f ′(x)|+ ‖f ′‖1] .

Assume that|f ′| is of s-Godunova-Levin type, withs ∈ [0, 1).

(b) If f ′ ∈ L∞[a, b], then for anyx ∈ [a, b],∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(4.15)

≤ 1

1− s

1

4+

(x− a+b

2

b− a

)2 (b− a) [|f ′(x)|+ ‖f ′‖∞] .

(bb) If f ′ ∈ Lp[a, b], p > 1, 1p

+ 1q

= 1, then for anyx ∈ [a, b],

∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(4.16)

≤ 1

(1− s) (q + 1)1q

[(b− x

b− a

)q+1

+

(x− a

b− a

)q+1] 1

q

× (b− a)1q ‖|f ′(x)|+ |f ′|‖p .

(bbb) If f ′ ∈ L1[a, b], then for anyx ∈ [a, b],

∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣ ≤ 1

1− s

[1

2+

∣∣∣∣∣x− a+b2

b− a

∣∣∣∣∣]

(4.17)

× [(b− a) |f ′(x)|+ ‖f ′‖1] .


THEOREM 4.6 (Dragomir, 2013 [9]). Let f : [a, b] → C be an absolutely continuous func-tion on[a, b] so that|f ′|p with p > 1 is h-convex on(a, b) andh ∈ L [0, 1].


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118 S. S. DRAGOMIR

(i) If f ′ ∈ L∞[a, b], then for anyx ∈ [a, b],∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(4.18)

≤

1

4+

(x− a+b

2

b− a

)2 (b− a)

×[|f ′(x)|p + ‖f ′‖p

∞]1/p

(∫ 1

0

h (t) dt

)1/p

.

(ii) If f ′ ∈ Lp[a, b], p > 1, 1p

+ 1q

= 1, then for anyx ∈ [a, b],∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(4.19)

≤ (b− a)1q

(q + 1)1/q

[(b− x

b− a

)q+1

+

(x− a

b− a

)q+1]1/q

×[(b− a) |f ′ (x)|p + ‖f ′‖p

p

]1/p(∫ 1

0

h (t) dt

)1/p

.

(iii) If f ′ ∈ Lp[a, b], then for anyx ∈ [a, b],∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣ ≤[

1

2+

∣∣∣∣∣x− a+b2

b− a

∣∣∣∣∣]

(4.20)

×∥∥|f ′ (x)|p + |f ′|p

∥∥p(∫ 1

0

h (t) dt

)1/p

≤

[1

2+

∣∣∣∣∣x− a+b2

b− a

∣∣∣∣∣]

×((b− a) |f ′ (x)|p + ‖f ′‖p

p

)1/p(∫ 1

0

h (t) dt

)1/p

.

PROOF. As in the proof of Theorem 4.4 we have∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣ =1

b− a

∣∣∣∣∫ b

a

∫ 1

0

(x− t) f ′ [(1− λ)x+ λt] dλdt

∣∣∣∣≤ 1

b− a

∫ b

a

|x− t|(∫ 1

0

|f ′ [(1− λ)x+ λt]| dλ)dt

:= K

for anyx ∈ [a, b].By Hölder’s integral inequality we have∫ 1

0

|f ′ [(1− λ)x+ λt]| dλ ≤(∫ 1

0

1qdλ

)1/q (∫ 1

0

|f ′ [(1− λ)x+ λt]|p dλ)1/p

=

(∫ 1

0

|f ′ [(1− λ)x+ λt]|p dλ)1/p

for anyx ∈ [a, b], where1p

+ 1q

= 1, p > 1.


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Since|f ′|p is h-convex on(a, b) with h ∈ L [0, 1] , then

∫ 1

0

|f ′ [(1− λ)x+ λt]|p dλ ≤[|f ′ (x)|p + |f ′ (t)|p

] ∫ 1

0

h (λ) dλ,

for anyx ∈ [a, b].Therefore

(4.21) K ≤ 1

b− a

(∫ 1

0

h (λ) dλ

)1/p ∫ b

a

|x− t|[|f ′ (x)|p + |f ′ (t)|p

]1/pdt

for anyx ∈ [a, b].(i). Now, if f ′ ∈ L∞ [a, b] then

∫ b

a

|x− t|[|f ′ (x)|p + |f ′ (t)|p

]1/pdt

≤ ess supt∈[a,b]

[|f ′ (x)|p + |f ′ (t)|p

]1/p∫ b

a

|x− t| dt

=[|f ′ (x)|p + ‖f ′‖p

∞]1/p 1

2

[(x− a)2 + (b− x)2]

for anyx ∈ [a, b], and utilizing (4.21), the inequality (4.18) is proved.(ii). If f ′ ∈ Lp[a, b], p > 1, 1

p+ 1

q= 1, then by Hölder’s inequality we have

∫ b

a

|x− t|[|f ′ (x)|p + |f ′ (t)|p

]1/pdt

≤(∫ b

a

|x− t|q dt)1/q (∫ b

a

([|f ′ (x)|p + |f ′ (t)|p

]1/p)p

dt

)1/p

=

[(b− x)q+1 + (x− a)q+1

q + 1

]1/q [(b− a) |f ′ (x)|p + ‖f ′‖p

p

]1/p

=(b− a)1+ 1

q

(q + 1)1/q

[(b− x

b− a

)q+1

+

(x− a

b− a

)q+1]1/q

×[(b− a) |f ′ (x)|p + ‖f ′‖p

1

]1/p

for anyx ∈ [a, b], and by (4.21) we deduce the desired inequality (4.19).


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120 S. S. DRAGOMIR

(iii). If f ′ ∈ Lp[a, b], then by Hölder’s inequality we also have∫ b

a

|x− t|[|f ′ (x)|p + |f ′ (t)|p

]1/pdt

≤ supt∈[a,b]

|x− t|∫ b

a

[|f ′ (x)|p + |f ′ (t)|p

]1/pdt

= max x− a, b− x∫ b

a

[|f ′ (x)|p + |f ′ (t)|p

]1/pdt

= (b− a)

[1

2+

∣∣∣∣∣x− a+b2

b− a

∣∣∣∣∣] ∥∥|f ′ (x)|p + |f ′|p

∥∥p

≤ (b− a)

[1

2+

∣∣∣∣∣x− a+b2

b− a

∣∣∣∣∣](∫ b

a

[|f ′ (x)|p + |f ′ (t)|p

]dt

)1/p

= (b− a)

[1

2+

∣∣∣∣∣x− a+b2

b− a

∣∣∣∣∣] (

(b− a) |f ′ (x)|p + ‖f ′‖pp

)1/p


The following midpoint type inequalities are of interest.

COROLLARY 4.7. With the assumptions of Theorem 4.6, we have the inequality∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(4.22)

≤ 1

4(b− a)

[∣∣∣∣f ′(a+ b

2

)∣∣∣∣p + ‖f ′‖p∞

]1/p(∫ 1

0

h (t) dt

)1/p

,

providedf ′ ∈ L∞[a, b].If f ′ ∈ Lp[a, b], p > 1, 1

p+ 1

q= 1, then we have∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(4.23)

≤ 1

2 (q + 1)1/q(b− a)

1q

×[(b− a)

∣∣∣∣f ′(a+ b

2

)∣∣∣∣p + ‖f ′‖pp

]1/p(∫ 1

0

h (t) dt

)1/p

.

If f ′ ∈ Lp[a, b], then∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣(4.24)

≤ 1

2

∥∥∥∥∣∣∣∣f ′(a+ b

2

)∣∣∣∣p + |f ′|p∥∥∥∥p(∫ 1

0

h (t) dt

)1/p

≤ 1

2

((b− a)

∣∣∣∣f ′(a+ b

2

)∣∣∣∣p + ‖f ′‖pp

)1/p(∫ 1

0

h (t) dt

)1/p

.


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REFERENCES

[1] M. Alomari and M. Darus, The Hadamard’s inequality fors-convex function.Int. J. Math. Anal.(Ruse)2 (2008), no. 13-16, 639–646.

[2] M. Alomari and M. Darus, Hadamard-type inequalities fors-convex functions.Int. Math. Forum3(2008), no. 37-40, 1965–1975.

[3] N. S. Barnett, P. Cerone, S. S. Dragomir, M. R. Pinheiro,and A. Sofo, Ostrowski type in-equalities for functions whose modulus of the derivatives are convex and applications.Inequal-ity Theory and Applications, Vol. 2 (Chinju/Masan, 2001), 19–32, Nova Sci. Publ., Haup-pauge, NY, 2003. Preprint: RGMIA Res. Rep. Coll.5 (2002), No. 2, Art. 1 [Onlinehttp://rgmia.org/papers/v5n2/Paperwapp2q.pdf] .

[4] M. Bombardelli and S. Varošanec, Properties ofh-convex functions related to the Hermite-Hadamard-Fejér inequalities.Comput. Math. Appl.58 (2009), no. 9, 1869–1877.

[5] W. W. Breckner, Stetigkeitsaussagen für eine Klasse verallgemeinerter konvexer Funktionen intopologischen linearen Räumen. (German)Publ. Inst. Math. (Beograd) (N.S.)23(37)(1978), 13–20.

[6] W. W. Breckner and G. Orbán, Continuity properties of rationallys-convex mappings with values inan ordered topological linear space. Universitatea "Babes-Bolyai”, Facultatea de Matematica, Cluj-Napoca, 1978. viii+92 pp.

[7] P. Cerone and S. S. Dragomir, Ostrowski type inequalities for functions whose derivatives satisfycertain convexity assumptions.Demonstratio Math.37 (2004), No. 2, 299–308.

[8] S. S. Dragomir, Ostrowski type inequalities for isotonic linear function-als, J. Ineq. Pure & Appl. Math., 3 (5) (2002), Art. 68. [Onlinehttp://www.emis.de/journals/JIPAM/article220.html?sid=220].

[9] S. S. Dragomir, Ostrowski type inequalities for functions whose derivatives areh-convex in absolutevalue, PreprintRGMIA Res. Rep. Coll.16 (2013), Art. 71.

[10] S.S. Dragomir and S. Fitzpatrick, The Hadamard inequalities fors-convex functions in the secondsense.Demonstratio Math.32 (1999), no. 4, 687–696.

[11] S.S. Dragomir and S. Fitzpatrick,The Jensen inequality fors-Breckner convex functions in linearspaces.Demonstratio Math.33 (2000), no. 1, 43–49.

[12] S. S. Dragomir and B. Mond, On Hadamard’s inequality for a class of functions of Godunova andLevin. Indian J. Math.39 (1997), no. 1, 1–9.

[13] S. S. Dragomir and C. E. M. Pearce, On Jensen’s inequality for a class of functions of Godunovaand Levin.Period. Math. Hungar.33 (1996), no. 2, 93–100.

[14] S. S. Dragomir and C. E. M. Pearce, Quasi-convex functions and Hadamard’s inequality,Bull.Austral. Math. Soc.57 (1998), 377-385.

[15] S. S. Dragomir and C. E. M. Pearce, Quasi-convex functions and Hadamard’s inequality,Bull.Australian Math. Soc., 57(1998), 377-385.

[16] S. S. Dragomir, J. Pecaric and L. Persson, Some inequalities of Hadamard type.Soochow J. Math.21 (1995), no. 3, 335–341.

[17] S. S. Dragomir and A. Sofo, Ostrowski type inequalities for functions whose derivatives are convex,Proc. 4th Int. Conf. on Modelling and Simulation, 2002, 369-374. Preprint:RGMIA Res. Rep. Coll.5(2002), Supplement[Online http://rgmia.org/papers/v5e/OTIDC2_col.pdf].

[18] E. K. Godunova and V. I. Levin, Inequalities for functions of a broad class that contains convex,monotone and some other forms of functions. (Russian)Numerical mathematics and mathematicalphysics(Russian), 138–142, 166, Moskov. Gos. Ped. Inst., Moscow, 1985


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122 S. S. DRAGOMIR

[19] H. Hudzik and L. Maligranda, Some remarks ons-convex functions.Aequationes Math. 48 (1994),no. 1, 100–111.

[20] U. S. Kirmaci, M. Klaricic Bakula, M. E Özdemir and J. Pecaric, Hadamard-type inequalities fors-convex functions.Appl. Math. Comput.193(2007), no. 1, 26–35.

[21] M. A. Latif, On some inequalities forh-convex functions.Int. J. Math. Anal.(Ruse)4 (2010), no.29-32, 1473–1482.

[22] D. S. Mitrinovic and J. E. Pecaric, Note on a class of functions of Godunova and Levin.C. R. Math.Rep. Acad. Sci. Canada12 (1990), no. 1, 33–36.

[23] C. E. M. Pearce and A. M. Rubinov,P -functions, quasi-convex functions, and Hadamard-typeinequalities.J. Math. Anal. Appl.240(1999), no. 1, 92–104.

[24] J. E. Pecaric and S. S. Dragomir, On an inequality of Godunova-Levin and some refinements ofJensen integral inequality.Itinerant Seminar on Functional Equations, Approximation and Convexity(Cluj-Napoca, 1989), 263–268, Preprint, 89-6, Univ. "Babes-Bolyai”, Cluj-Napoca, 1989.

[25] M. Radulescu, S. Radulescu and P. Alexandrescu, On the Godunova-Levin-Schur class of functions.Math. Inequal. Appl.12 (2009), no. 4, 853–862.

[26] M. Z. Sarikaya, A. Saglam, and H. Yildirim, On some Hadamard-type inequalities forh-convexfunctions.J. Math. Inequal.2 (2008), no. 3, 335–341.

[27] E. Set, M. E. Özdemir and M. Z. Sarıkaya, New inequalities of Ostrowski’s type fors-convexfunctions in the second sense with applications.Facta Univ. Ser. Math. Inform.27 (2012), no. 1,67–82.

[28] M. Z. Sarikaya, E. Set and M. E. Özdemir, On some new inequalities of Hadamard type involvingh-convex functions.Acta Math. Univ. Comenian.(N.S.)79 (2010), no. 2, 265–272.

[29] M. Tunç, Ostrowski-type inequalities viah-convex functions with applications to special means.J.Inequal. Appl.2013, 2013:326.

[30] S. Varošanec, Onh-convexity.J. Math. Anal. Appl.326(2007), no. 1, 303–311.


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CHAPTER 6

Other Ostrowski Type Inequalities

1. OSTROWSKI FOR PRODUCTS

1.1. An Identity for RS-Integral. The following identity is of interest in itself:

LEMMA 1.1 (Dragomir, 2013 [2]). Let f, g : [a, b] → C be two functions of boundedvariation and such that the Riemann-Stieltjes integral

∫ b

af (t) dg (t) exists. Ifh : [a, b] → C is

continuous, then the Riemann-Stieltjes integral∫ b

ah (t) d (f (t) g (t)) exists and

(1.1)∫ b

a

h (t) d (f (t) g (t)) =

∫ b

a

h (t) f (t) d (g (t)) +

∫ b

a

h (t) g (t) d (f (t)) .

PROOF. Sincef, g : [a, b] → C are of bounded variation, thenfg is of bounded vari-ation and sinceh : [a, b] → C is continuous, it follows that the Riemann-Stieltjes integral∫ b

ah (t) d (f (t) g (t)) exists.

Observe that, since the integral∫ b

af (t) dg (t) exists, then for anys ∈ [a, b] the integral

` (s) :=∫ s

af (t) dg (t) exists and the functionis of bounded variation on[a, b] .

Indeed, let

a = s0 < s1 < ... < sn−1 < sn = b

a division of the interval[a, b] . Then we have

n−1∑i=0

|` (si+1)− ` (si)| =n−1∑i=0

∣∣∣∣∫ si+1

a

f (t) dg (t)−∫ si

a

f (t) dg (t)

∣∣∣∣=

n−1∑i=0

∣∣∣∣∫ si+1

si

f (t) dg (t)

∣∣∣∣ ≤ n−1∑i=0

(sup

t∈[si,si+1]

|f (t)|si+1∨si

(g)

)

≤ supt∈[a,b]

|f (t)|n−1∑i=0

(si+1∨si

(g)

)= sup

t∈[a,b]

|f (t)|b∨a

(g) <∞,

which shows that is of bounded variation on[a, b] and

b∨a

(`) ≤ ‖f‖∞b∨a

(g) ,

where‖f‖∞ := supt∈[a,b] |f (t)| .Now, by the integration by parts theorem, since

∫ s

af (t) dg (t) exists for anys ∈ [a, b] , then∫ s

ag (t) df (t) also exists and we have the equality

(1.2) f (s) g (s) = f (a) g (a) +

∫ s

a

f (t) dg (t) +

∫ s

a

g (t) df (t)

for anys ∈ [a, b] .

123

124 S. S. DRAGOMIR

Since the functions∫ ·

af (t) dg (t) and

∫ ·ag (t) df (t) are of bounded variation, then the Riemann-

Stieltjes integrals∫ b

a

h (s) d

(∫ s

a

f (t) dg (t)

)and

∫ b

a

h (s) d

(∫ s

a

g (t) df (t)

)exist and

(1.3)∫ b

a

h (s) d

(∫ s

a

f (t) dg (t)

)=

∫ b

a

h (s) f (s) dg (s) ,

and

(1.4)∫ b

a

h (s) d

(∫ s

a

g (t) df (t)

)=

∫ b

a

h (s) g (s) df (s) .

Now, on utilizing (1.2), (1.3) and (1.4) we have∫ b

a

h (s) d (f (s) g (s)) =

∫ b

a

h (s) d (f (a) g (a))

+

∫ b

a

h (s) d

(∫ s

a

f (t) dg (t)

)+

∫ b

a

h (s)

(∫ s

a

g (t) df (t)

)=

∫ b

a

h (s) f (s) dg (s) +

∫ b

a

h (s) g (s) df (s)

and the equality (1.1) is proved.

REMARK 1.1. The dual case also holds, namely, when the functionsf, g : [a, b] → C arecontinuous and such that the Riemann-Stieltjes integral

∫ b

af (t) dg (t) exists, then for any func-

tion h : [a, b] → C of bounded variation the Riemann-Stieltjes integral∫ b

ah (t) d (f (t) g (t))

exists and the equality (1.1) is satisfied.

COROLLARY 1.2. Let f : [a, b] → C be a functions of bounded variation and such thatthe Riemann-Stieltjes integral

∫ b

af (t) df (t) exists. Ifh : [a, b] → C is continuous, then the

Riemann-Stieltjes integral∫ b

ah (t) d (f 2 (t)) exists and

(1.5)∫ b

a

h (t) d(f 2 (t)

)= 2

∫ b

a

h (t) f (t) d (f (t)) .

If∫ b

af (t) df (t) exists, then for any continuous functionh : [a, b] → C, the Riemann-

Stieltjes integral∫ b

ah (t) d

(|f (t)|2

)exists and

(1.6)∫ b

a

h (t) d(|f (t)|2

)=

∫ b

a

h (t) f (t) d(f (t)

)+

∫ b

a

h (t) f (t)d (f (t)) .

In particular, if h : [a, b] → R, then

(1.7)∫ b

a

h (t) d(|f (t)|2

)= 2 Re

(∫ b

a

h (t) f (t)d (f (t))

).


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1.2. Inequalities for Product Integrators. The first bound for the Riemann-Stieltjes inte-gral of product integrators is as follows:

THEOREM 1.3 (Dragomir, 2013 [2]). Let f, g : [a, b] → C be two functions of boundedvariation and such that the Riemann-Stieltjes integral

∫ b

af (t) dg (t) exists. Ifh : [a, b] → C is

continuous, then∣∣∣∣∫ b

a

h (t) d (f (t) g (t))

∣∣∣∣ ≤ ‖hf‖∞b∨a

(g) + ‖hg‖∞b∨a

(f)(1.8)

≤ ‖h‖∞

[‖f‖∞

b∨a

(g) + ‖g‖∞b∨a

(f)

].

Both inequalities in (1.8) are sharp.

PROOF. We know that ifp : [a, b] → C is bounded,v : [a, b] → C is of bounded variationand the Riemann-Stieltjes integral

∫ b

ap (s) dv (s) exists, then we have the inequality

(1.9)

∣∣∣∣∫ b

a

p (s) dv (s)

∣∣∣∣ ≤ ‖p‖∞b∨a

(v) ,

where‖p‖∞ = supt∈[a,b] |p (t)| <∞.Taking the modulus in (1.1) and using the property (1.9) we have∣∣∣∣∫ b

a

h (t) d (f (t) g (t))

∣∣∣∣ ≤ ∣∣∣∣∫ b

a

h (t) f (t) d (g (t))

∣∣∣∣+ ∣∣∣∣∫ b

a

h (t) g (t) d (f (t))

∣∣∣∣≤ ‖hf‖∞

b∨a

(g) + ‖hg‖∞b∨a

(f)

≤ ‖h‖∞ ‖f‖∞b∨a

(g) + ‖h‖∞ ‖g‖∞b∨a

(f)

= ‖h‖∞

[‖f‖∞

b∨a

(g) + ‖g‖∞b∨a

(f)


Now, to prove the sharpness of the inequalities (1.8) we consider the functionsf, g : [a, b] →R given by

f (t) :=

0 if t = a

1 if t ∈ (a, b],g (t) :=

1 if t ∈ [a, b)

0 if t = b.

The functionsf andg are of bounded variation,b∨a

(f) =b∨a

(g) = 1 and‖f‖∞ = ‖g‖∞ = 1.

We have

f (t) g (t) :=

0 if t = a

1 if t ∈ (a, b)

0 if t = b.


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126 S. S. DRAGOMIR

The functionfg is of bounded variation and for a continuous functionh : [a, b] → C theRiemann-Stieltjes integral

∫ b

ah (t) d (f (t) g (t)) exists and integrating by parts we have∫ b

a

h (t) d (f (t) g (t))(1.10)

= f (b) g (b)h (b)− f (a) g (a)h (a)−∫ b

a

f (t) g (t) d (h (t))

= −∫ b

a

f (t) g (t) d (h (t)) .

Consider the following sequence of divisions and intermediate points

a = x(n)0 < ξ

(n)0 < x

(n)1 < ... < x

(n)n−1 < ξ

(n)n−1 < x(n)

n = b

such that the norm of the division∆n := maxi∈0,...,n−1

(x

(n)i+1 − x

(n)i

)→ 0 asn→∞. By the

definition of the Riemann-Stieltjes integral we have∫ b

a

f (t) g (t) d (h (t)) = limn→∞

n−1∑i=0

f(ξ

(n)i

)g(ξ

(n)i

)(h(x

(n)i+1

)− h

(x

(n)i

))= lim

n→∞

n−1∑i=0

(h(x

(n)i+1

)− h

(x

(n)i

))= h (b)− h (a) ,

and then, by (1.10) we have∫ b

a

h (t) d (f (t) g (t)) = −h (b) + h (a) .

We also have

h (t) f (t) :=

0 if t = a

h (t) if t ∈ (a, b],h (t) g (t) :=

h (t) if t ∈ [a, b)

0 if t = b,

which implies that‖hf‖∞ = ‖hg‖∞ = ‖h‖∞ .

Therefore the inequality (1.8) reduces to

(1.11) |h (b)− h (a)| ≤ 2 ‖h‖∞ ≤ 2 ‖h‖∞ .

We observe that, this inequality is sharp since for continuous functionsh : [a, b] → R for which

0 < h (b) = −h (a) = supt∈[a,b]

h (t) ,

we get equality in (1.11).For instance, if we take

h (t) = t− a+ b

2, t ∈ [a, b] ,

then

|h (b)− h (a)| = b− a, ‖h‖∞ =b− a

2and the equality in (1.11) is realized.


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We say that the functionf : [a, b] → C is Lipschitzianwith the constantL > 0 if

|f (t)− f (s)| ≤ L |t− s|

for anyt, s ∈ [a, b] .

THEOREM1.4 (Dragomir, 2013 [2]). Assume that the functionf : [a, b] → C is Lipschitzianwith the constantL > 0, g : [a, b] → C is Lipschitzian with the constantK > 0 andh : [a, b] →C a continuous function on[a, b] . Then we have the inequality∣∣∣∣∫ b

a

h (t) d (f (t) g (t))

∣∣∣∣ ≤ K

∫ b

a

|h (t) f (t)| dt+ L

∫ b

a

|h (t) g (t)| dt(1.12)

≤ max K,L∫ b

a

|h (t)| (|f (t)|+ |g (t)|) dt.

The inequalities in (1.12) are sharp.

PROOF. It is known that, ifp : [a, b] → C is continuous andv : [a, b] → C is Lipschitzianwith the constantL > 0, then the Riemann-Stieltjes integral

∫ b

ap (s) dv (s) exists and we have

the inequality

(1.13)

∣∣∣∣∫ b

a

p (s) dv (s)

∣∣∣∣ ≤ L

∫ b

a

|p (s)| ds.

Taking the modulus in (1.1) and using the property (1.9) we have∣∣∣∣∫ b

a

h (t) d (f (t) g (t))

∣∣∣∣ ≤ ∣∣∣∣∫ b

a

h (t) f (t) d (g (t))

∣∣∣∣+ ∣∣∣∣∫ b

a

h (t) g (t) d (f (t))

∣∣∣∣≤ K

∫ b

a

|h (t) f (t)| dt+ L

∫ b

a

|h (t) g (t)| dt

≤ max K,L∫ b

a

|h (t)| (|f (t)|+ |g (t)|) dt,

and the inequality (1.12) is proved.Consider now the functionsf, g : [a, b] → R, f (t) = g (t) =

∣∣t− a+b2

∣∣ . We observe thatfandg are Lipschitzian with the constantL = 1.

Indeed, for anyt, s ∈ [a, b] we have

|f (t)− f (s)| =∣∣∣∣∣∣∣∣t− a+ b

2

∣∣∣∣− ∣∣∣∣s− a+ b

2

∣∣∣∣∣∣∣∣≤ |t− s| ,

which shows that the functionf is Lipschitzian with the constantL = 1.Now ∣∣∣∣∫ b

a

h (t) d (f (t) g (t))

∣∣∣∣ =

∫ b

a

h (t) d

((t− a+ b

2

)2)

= 2

∣∣∣∣∫ b

a

h (t)

(t− a+ b

2

)dt

∣∣∣∣and

K

∫ b

a

|h (t) f (t)| dt+ L

∫ b

a

|h (t) g (t)| dt = 2

∫ b

a

|h (t)|∣∣∣∣t− a+ b

2

∣∣∣∣ dtAJMAA, Vol. 14, No. 1, Art. 1, pp. 1-287, 2017 AJMAA

http://ajmaa.org

128 S. S. DRAGOMIR

and the first inequality in (1.12) becomes∣∣∣∣∫ b

a

h (t)

(t− a+ b

2

)dt

∣∣∣∣ ≤ ∫ b

a

|h (t)|∣∣∣∣t− a+ b

2

∣∣∣∣ dt.We observe that the equality case holds if we takeh : [a, b] → R, h (t) = t− a+b

2.

THEOREM 1.5 (Dragomir, 2013 [2]). Assume thatf, g : [a, b] → R are monotonic non-decreasing on[a, b] and such that the Riemann-Stieltjes integral

∫ b

af (t) dg (t) exists, andh :

[a, b] → C is continuous on[a, b] . Then we have the inequality

(1.14)

∣∣∣∣∫ b

a

h (t) d (f (t) g (t))

∣∣∣∣ ≤ ∫ b

a

|f (t)h (t)| dg (t) +

∫ b

a

|g (t)h (t)| df (t) .

The inequality (1.14) is sharp.

PROOF. It is well known that ifp : [a, b] → C is continuous andv : [a, b] → R is monotonicnondecreasing, then the Riemann-Stieltjes integral

∫ b

ap (t) dv (t) exists and∣∣∣∣∫ b

a

p (t) dv (t)

∣∣∣∣ ≤ ∫ b

a

|p (t)| dv (t) .

Taking the modulus in (1.1) we have∣∣∣∣∫ b

a

h (t) d (f (t) g (t))

∣∣∣∣ ≤ ∣∣∣∣∫ b

a

f (t)h (t) dg (t)

∣∣∣∣+ ∣∣∣∣∫ b

a

g (t)h (t) df (t)

∣∣∣∣≤∫ b

a

|f (t)h (t)| dg (t) +

∫ b

a

|g (t)h (t)| df (t) .

Consider the functionsf, g : [a, b] → R defined by

f (t) :=

0 if t = a

1 if t ∈ (a, b],g (t) :=

−1 if t ∈ [a, b)

0 if t = b.

The functionsf andg are monotonic nondecreasing. We will show that the Riemann-Stieltjesintegral

∫ b

af (t) dg (t) exists.

Take the sequence of divisions and intermediate points

dn : a = x(n)0 < ξ

(n)0 < x

(n)1 < ... < x

(n)n−1 < ξ

(n)n−1 < x(n)

n = b

such that∆ (dn) := maxi∈0,...,n−1

x

(n)i+1 − x

(n)i

→ 0 asn→∞.

By the definition of the Riemann-Stieltjes integral∫ b

af (t) dg (t) we have∫ b

a

f (t) dg (t) = limn→∞

n−1∑i=0

f(ξ

(n)i

) [g(x

(n)i+1

)− g

(x

(n)i

)]= lim

n→∞

n−2∑i=0

f(ξ

(n)i

) [g(x

(n)i+1

)− g

(x

(n)i

)]+ lim

n→∞f(ξ

(n)n−1

) [g (b)− g

(x

(n)n−1

)]= 0 + 1 = 1.


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Now, define the function : [a, b] → R by

` (t) := f (t) g (t) =

0 if t = a

−1 if t ∈ (a, b)

0 if t = a.

For a continuous functionh : [a, b] → R, since` is of bounded variation, then the integral∫ b

ah (t) d` (t) exists.Take the sequence of divisions and intermediate points

dn : a = x(n)0 < ξ

(n)0 < x

(n)1 < ... < x

(n)n−1 < ξ

(n)n−1 < x(n)

n = b


x

(n)i+1 − x

(n)i

→ 0 asn→∞.Then we have∫ b

a

h (t) d` (t) = limn→∞

n−1∑i=0

h(ξ

(n)i

) [`(x

(n)i+1

)− `(x

(n)i

)]= lim

n→∞h(ξ

(n)0

) [`(x

(n)1

)− ` (a)

]+ lim

n→∞

n−2∑i=1

h(ξ

(n)i

) [`(x

(n)i+1

)− `(x

(n)i

)]+ lim

n→∞h(ξ

(n)n−1

) [` (b)− `

(x

(n)n−1

)]= lim

n→∞h(ξ

(n)0

)(−1− 0) + 0 + lim

n→∞h(ξ

(n)n−1

)[0− (−1)]

= h (b)− h (a) .

Consider the functionsu, v : [a, b] → R given by

u (t) := |f (t)h (t)| =

0 if t = a

|h (t)| if t ∈ (a, b],

and

v (t) := |g (t)h (t)| =

|h (t)| if t ∈ [a, b)

0 if t = b.

Take the sequence of divisions and intermediate points

dn : a = x(n)0 < ξ

(n)0 < x

(n)1 < ... < x

(n)n−1 < ξ

(n)n−1 < x(n)

n = b


x

(n)i+1 − x

(n)i

→ 0 asn→∞.Then we have∫ b

a

u (t) dg (t) = limn→∞

n−1∑i=0

u(ξ

(n)i

) [g(x

(n)i+1

)− g

(x

(n)i

)]= lim

n→∞u(ξ

(n)n−1

) [g (b)− g

(x

(n)n−1

)]= lim

n→∞

∣∣∣h(ξ(n)n−1

)∣∣∣ [0− (−1)] = |h (b)|


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130 S. S. DRAGOMIR

and ∫ b

a

v (t) df (t) = limn→∞

n−1∑i=0

v(ξ

(n)i

) [f(x

(n)i+1

)− f

(x

(n)i

)]= lim

n→∞v(ξ

(n)0

) [f(x

(n)1

)− f (a)

]= lim

n→∞

∣∣∣h(ξ(n)0

)∣∣∣ (1− 0) = |h (a)| .

Replacing these values in (1.14) we have

(1.15) |h (b)− h (a)| ≤ |h (b)|+ |h (a)| .

This inequality reduces to an equality if we choose a continuous functionh : [a, b] → R suchthath (b) = −h (a) . For instance, forh : [a, b] → R, h (t) = t − a+b

2, we get in both sides of

(1.15) the same quantityb− a.

1.3. Ostrowski Type Inequalities for Products. The following result holds:

THEOREM1.6 (Dragomir, 2013 [2]). Letf, g : [a, b] → C be two functions of bounded vari-ation and such that forx ∈ (a, b) the Riemann-Stieltjes integrals

∫ x

af (t) dg (t) and

∫ b

xf (t) dg (t)

exist. Then we have∣∣∣∣f (x) g (x) (b− a)−∫ b

a

f (t) g (t) dt

∣∣∣∣(1.16)

≤ (x− a) supt∈[a,x]

|f (t)|x∨a

(g) + (x− a) supt∈[a,x]

|g (t)|x∨a

(f)

+ (b− x) supt∈[x,b]

|f (t)|b∨x

(g) + (b− x) supt∈[x,b]

|g (t)|b∨x

(f)

≤[1

2(b− a) +

∣∣∣∣x− a+ b

2

∣∣∣∣][‖g‖∞

b∨a

(f) + ‖f‖∞b∨a

(g)

].

In particular, if the Riemann-Stieltjes integrals∫ a+b

2

af (t) dg (t) and

∫ ba+b2f (t) dg (t) exist, then

we have ∣∣∣∣f (a+ b

2

)g

(a+ b

2

)(b− a)−

∫ b

a

f (t) g (t) dt

∣∣∣∣(1.17)

≤ b− a

2

supt∈[a, a+b

2 ]|f (t)|

a+b2∨a

(g) + supt∈[a, a+b

2 ]|g (t)|

a+b2∨a

(f)

+ supt∈[a+b

2,b]|f (t)|

b∨a+b2

(g) + supt∈[a+b

2,b]|g (t)|

b∨a+b2

(f)

≤ 1

2(b− a)

[‖g‖∞

b∨a

(f) + ‖f‖∞b∨a

(g)

].

The inequalities in (1.17) are sharp.


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PROOF. We use the following identity (see for instance [1])

(1.18) F (x) (b− a)−∫ b

a

F (t) dt =

∫ x

a

(t− a) dF (t) +

∫ b

x

(t− b) dF (t)

that holds for any function of bounded variationF : [a, b] → C and anyx ∈ [a, b] .If we write the equality (1.18) forF = fg we get

f (x) g (x) (b− a)−∫ b

a

f (t) g (t) dt(1.19)

=

∫ x

a

(t− a) d (f (t) g (t)) +

∫ b

x

(t− b) d (f (t) g (t)) ,

for any functionsf, g : [a, b] → C of bounded variation and anyx ∈ [a, b] .Taking the modulus on (1.19) and utilizing Theorem 1.3 on the intervals[a, x] and[x, b] we

have successively that∣∣∣∣f (x) g (x) (b− a)−∫ b

a

f (t) g (t) dt

∣∣∣∣(1.20)

≤∣∣∣∣∫ x

a

(t− a) d (f (t) g (t))

∣∣∣∣+ ∣∣∣∣∫ b

x

(t− b) d (f (t) g (t))

∣∣∣∣≤ sup

t∈[a,x]

(t− a) |f (t)|x∨a

(g) + supt∈[a,x]

(t− a) |g (t)|x∨a

(f)

+ supt∈[x,b]

(b− t) |f (t)|b∨x

(g) + supt∈[x,b]

(b− t) |g (t)|b∨x

(f)

≤ (x− a) supt∈[a,x]

|f (t)|x∨a

(g) + (x− a) supt∈[a,x]

|g (t)|x∨a

(f)

+ (b− x) supt∈[x,b]

|f (t)|b∨x

(g) + (b− x) supt∈[x,b]

|g (t)|b∨x

(f)

≤ max x− a, b− x supt∈[a,b]

|f (t)|b∨a

(g)

+ max x− a, b− x supt∈[a,b]

|g (t)|b∨a

(f)

=

[1

2(b− a) +

∣∣∣∣x− a+ b

2

∣∣∣∣][‖g‖∞

b∨a

(f) + ‖f‖∞b∨a

(g)

],

which proves the desired result (1.16).The inequality (1.17) is obvious from (1.16).Consider now the functionsf, g : [a, b] → R defined by

f (t) :=

0 if t ∈[a, a+b

2

)1 if t ∈

[a+b2, b] g (t) :=

1 if t ∈[a, a+b

2

]0 if t ∈

(a+b2, b].


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132 S. S. DRAGOMIR

We observe thatf andg are of bounded variation and

b∨a

(f) =b∨a

(g) = 1.

The Riemann-Stieltjes integrals∫ a+b

2

af (t) dg (t) and

∫ ba+b2f (t) dg (t) exist since one function

is continuous while the other is of bounded variation on those intervals.We observe that for these functions we have

f

(a+ b

2

)g

(a+ b

2

)(b− a)−

∫ b

a

f (t) g (t) dt = b− a,

supt∈[a, a+b

2 ]|f (t)|

a+b2∨a

(g) + supt∈[a, a+b

2 ]|g (t)|

a+b2∨a

(f)

+ supt∈[a+b

2,b]|f (t)|

b∨a+b2

(g) + supt∈[a+b

2,b]|g (t)|

b∨a+b2

(f)

= 2

and

‖g‖∞b∨a

(f) + ‖f‖∞b∨a

(g) = 2.

Replacing these values in (1.17) we obtain in all terms the same quantityb − a, which provesthe sharpness of the inequalities.

In a similar way we can prove the following results as well:

THEOREM 1.7 (Dragomir, 2013 [2]). Let f : [a, b] → C be Lipschitzian with the constantL > 0 andg : [a, b] → C be Lipschitzian with the constantK > 0. Then for anyx ∈ [a, b] wehave ∣∣∣∣f (x) g (x) (b− a)−

∫ b

a

f (t) g (t) dt

∣∣∣∣(1.21)

≤ L

(∫ x

a

(t− a) |g (t)| dt+

∫ b

x

(b− t) |g (t)| dt)

+K

(∫ x

a

(t− a) |f (t)| dt+

∫ b

x

(b− t) |f (t)| dt)

≤

[1

4(b− a)2 +

(x− a+ b

2

)2]

(L ‖g‖∞ +K ‖f‖∞) .


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In particular, we have∣∣∣∣f (a+ b

2

)g

(a+ b

2

)(b− a)−

∫ b

a

f (t) g (t) dt

∣∣∣∣(1.22)

≤ L

(∫ a+b2

a

(t− a) |g (t)| dt+

∫ b

a+b2

(b− t) |g (t)| dt

)

+K

(∫ a+b2

a

(t− a) |f (t)| dt+

∫ b

a+b2

(b− t) |f (t)| dt

)≤ 1

4(b− a)2 (L ‖g‖∞ +K ‖f‖∞) .

We also have:

THEOREM 1.8 (Dragomir, 2013 [2]). Assume thatf, g : [a, b] → R are monotonic nonde-creasing on[a, b] and such that forx ∈ (a, b) the Riemann-Stieltjes integrals

∫ x

af (t) dg (t) and∫ b

xf (t) dg (t) exist. Then we have∣∣∣∣f (x) g (x) (b− a)−

∫ b

a

f (t) g (t) dt

∣∣∣∣(1.23)

≤∫ x

a

(t− a) |g (t)| df (t) +

∫ b

x

(b− t) |g (t)| df (t)

+

∫ x

a

(t− a) |f (t)| dg (t) +

∫ b

x

(b− t) |f (t)| dg (t)

≤[1

2(b− a) +

∣∣∣∣x− a+ b

2

∣∣∣∣](∫ b

a

|g (t)| df (t) +

∫ b

a

|f (t)| dg (t)

).

In particular, if the Riemann-Stieltjes integrals∫ a+b

2

af (t) dg (t) and

∫ ba+b2f (t) dg (t) exist, then

we have ∣∣∣∣f (a+ b

2

)g

(a+ b

2

)(b− a)−

∫ b

a

f (t) g (t) dt

∣∣∣∣(1.24)

≤∫ a+b

2

a

(t− a) |g (t)| df (t) +

∫ b

a+b2

(b− t) |g (t)| df (t)

+

∫ a+b2

a

(t− a) |f (t)| dg (t) +

∫ b

a+b2

(b− t) |f (t)| dg (t)

≤ 1

2(b− a)

(∫ b

a

|g (t)| df (t) +

∫ b

a

|f (t)| dg (t)

).

2. OSTROWSKI FOR S-DOMINATED FUNCTIONS

2.1. S-Dominated Functions.We start with the following definition:

DEFINITION 2.1 (Dragomir, 2013 [4]). Assume thatu, v : [a, b] → R aremonotonic non-decreasingon the interval[a, b] . We say that the complex-valued functionh : [a, b] → C isS-dominatedby the pair(u, v) if

(S) |h (y)− h (x)|2 ≤ [u (y)− u (x)] [v (y)− v (x)]

for anyx, y ∈ [a, b] .


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134 S. S. DRAGOMIR

We observe that by the monotonicity of the functionsu andv and by the symmetry of theinequality (S) overx andy we can assume that (S) is satisfied only fory > x with x, y ∈ [a, b] .

We can give numerous examples of such functions.For instance, if we takef, g ∈ L2 [a, b] the Hilbert space of all complex-valued functions

that are square-Lebesgue integrable and denote

h (x) :=

∫ x

a

f (t) g (t) dt, u (x) :=

∫ x

a

|f (t)|2 dt andv (x) :=

∫ x

a

|g (t)|2 dt,

then we observe thatu andv are monotonic nondecreasing on[a, b] and by Cauchy-Bunyakovsky-Schwarz integral inequality we have for anyy > x with x, y ∈ [a, b] that

|h (y)− h (x)|2 =

∣∣∣∣∫ y

x

f (t) g (t) dt

∣∣∣∣2 ≤ ∫ y

x

|f (t)|2 dt∫ y

x

|g (t)|2 dt

≤ [u (y)− u (x)] [v (y)− v (x)] .

Now, for p, q > 0 if we considerf (t) := tp andg (t) := tq for t ≥ 0, then

hp,q (x) :=

∫ x

0

tp+qdt =1

p+ q + 1xp+q+1

and

up (x) :=

∫ x

0

t2pdt =1

2p+ 1x2p+1, vq (x) :=

∫ x

0

t2qdt =1

2q + 1x2q+1.

Taking into account the above comments we observe that the functionhp,q is S-dominatedbythe pair(up, vq) on any subinterval of[0,∞) .

PROPOSITION2.1 (Dragomir, 2013 [4]). If h : [a, b] → C is S-dominated by the pair(u, v) ,thenh is of bounded variation on any subinterval[c, d] ⊂ [a, b] and

(2.1)

[d∨c

(h)

]2

≤ [u (d)− u (c)] [v (d)− v (c)] .

PROOF. Consider a divisionδ of the interval[c, d] given by

δ : c = x0 < x1 < ... < xn−1 < xn = b.

Sinceh : [a, b] → C is S-dominated by the pair(u, v) then we have

|h (xi+1)− h (xi)| ≤ [u (xi+1)− u (xi)]1/2 [v (xi+1)− v (xi)]

1/2

for anyi ∈ 0, ..., n− 1 .Summing this inequality overi from 0 to n − 1 and utilizing the Cauchy-Bunyakovsky-

Schwarz discrete inequality we haven−1∑i=1

|h (xi+1)− h (xi)|(2.2)

≤n−1∑i=1

[u (xi+1)− u (xi)]1/2 [v (xi+1)− v (xi)]

1/2

≤

(n−1∑i=1

[u (xi+1)− u (xi)]

)1/2(n−1∑i=1

[v (xi+1)− v (xi)]

)1/2

= [u (d)− u (c)]1/2 [v (d)− v (c)]1/2 .

Taking the supremum overδ we deduce the desired result (2.1).


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COROLLARY 2.2. If h : [a, b] → C is S-dominated by the pair(u, v) , then the cumulativevariation functionV : [a, b] → [0,∞) defined by

V (x) :=x∨a

(h)

is also S-dominated by the pair(u, v) .

THEOREM 2.3 (Dragomir, 2013 [4]). Assume thatu, v : [a, b] → R are monotonic non-decreasing on the interval[a, b] . If h : [a, b] → C is S-dominated by the pair(u, v) and f :

[a, b] → C is a continuous function on[a, b] , then the Riemann-Stieltjes integral∫ b

af (t) dh (t)

exists and

(2.3)

∣∣∣∣∫ b

a

f (t) dh (t)

∣∣∣∣2 ≤ ∫ b

a

|f (t)| du (t)

∫ b

a

|f (t)| dv (t) .


af (t) dh (t) exists, then for any sequence of

partitionsI(n)n : a = t

(n)0 < t

(n)1 < · · · < t

(n)n−1 < t(n)

n = b

with the normv(I(n)n

):= max

i∈0,...,n−1

(t(n)i+1 − t

(n)i

)→ 0

asn→∞, and for any intermediate pointsξ(n)i ∈ [t

(n)i , t

(n)i+1], i ∈ 0, . . . , n− 1 we have:∣∣∣∣∫ b

a

f (t) dh (t)

∣∣∣∣ =

∣∣∣∣∣∣ limvI(n)n

→0

n−1∑i=0

f(ξ

(n)i

) [h(t(n)i+1

)− h

(t(n)i

)]∣∣∣∣∣∣(2.4)

≤ limvI(n)n

→0

n−1∑i=0

∣∣∣f (ξ(n)i

)∣∣∣ ∣∣∣h(t(n)i+1

)− h

(t(n)i

)∣∣∣≤ lim

vI(n)n

→0

n−1∑i=0

∣∣∣f (ξ(n)i

)∣∣∣ [u(t(n)i+1

)− u

(t(n)i

)]1/2

×[v(t(n)i+1

)− v

(t(n)i

)]1/2

≤ limvI(n)n

→0

(n−1∑i=0

∣∣∣f (ξ(n)i

)∣∣∣ [u(t(n)i+1

)− u

(t(n)i

)])1/2

× limvI(n)n

→0

(n−1∑i=0

∣∣∣f (ξ(n)i

)∣∣∣ [v (t(n)i+1

)− v

(t(n)i

)])1/2

=

∫ b

a

|f (t)| du (t)

∫ b

a

|f (t)| dv (t) ,

where for the last inequality we employed the Cauchy-Bunyakovsky-Schwarz weighted discreteinequality

n∑k=1

mkakbk ≤

(n∑

k=1

mka2k

)1/2( n∑k=1

mkb2k

)1/2

,

wheremk, ak, bk ≥ 0 for k ∈ 1, ..., n .


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136 S. S. DRAGOMIR

2.2. Ostrowski Type Inequality. We can state the following new result.

THEOREM 2.4 (Dragomir, 2013 [4]). Assume thatu, v : [a, b] → R are monotonic nonde-creasing on the interval[a, b] . If h : [a, b] → C is S-dominated by the pair(u, v) , then

∣∣∣∣h (x) (b− a)−∫ b

a

h (t) dt

∣∣∣∣2(2.5)

≤[[2x− (a+ b)]u (x) +

∫ b

a

sgn(t− x)u (t) dt

]×[[2x− (a+ b)]v (x) +

∫ b

a

sgn(t− x)v (t) dt

]≤ [(b− x) [u (b)− u (x)] + (x− a) [u (x)− u (a)]]

× [(b− x) [v (b)− v (x)] + (x− a) [v (x)− v (a)]]

≤[b− a

2+

∣∣∣∣x− a+ b

2

∣∣∣∣]2

[u (b)− u (a)] [v (b)− v (a)] ,


PROOF. Integrating by parts on the Riemann-Stieltjes integral we have [1]

(2.6) h (x) (b− a)−∫ b

a

h (t) dt =

∫ x

a

(t− a) dh (t) +

∫ b

x

(b− t) dh (t) ,

for anyx ∈ [a, b] .Taking the modulus in (2.6) we have∣∣∣∣h (x) (b− a)−

∫ b

a

h (t) dt

∣∣∣∣(2.7)

≤∣∣∣∣∫ x

a

(t− a) dh (t)

∣∣∣∣+ ∣∣∣∣∫ b

x

(b− t) dh (t)

∣∣∣∣ .Applying the inequality (2.3) twice, we have

∣∣∣∣∫ x

a

(t− a) dh (t)

∣∣∣∣ ≤ (∫ x

a

(t− a) du (t)

)1/2(∫ x

a

(t− a) dv (t)

)1/2

and ∣∣∣∣∫ b

x

(b− t) dh (t)

∣∣∣∣ ≤ (∫ b

x

(b− t) du (t)

)1/2(∫ b

x

(b− t) dv (t)

)1/2

.

Summing these inequalities and utilizing the elementary result

αβ + λδ ≤(α2 + λ2

)1/2 (β2 + δ2

)1/2


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whereα, β, λ, δ ≥ 0, we have∣∣∣∣∫ x

a

(t− a) dh (t)

∣∣∣∣+ ∣∣∣∣∫ b

x

(b− t) dh (t)

∣∣∣∣(2.8)

≤(∫ x

a

(t− a) du (t)

)1/2(∫ x

a

(t− a) dv (t)

)1/2

+

(∫ b

x

(b− t) du (t)

)1/2(∫ b

x

(b− t) dv (t)

)1/2

≤(∫ x

a

(t− a) du (t) +

∫ b

x

(b− t) du (t)

)1/2

+

(∫ x

a

(t− a) dv (t) +

∫ b

x

(b− t) dv (t)

)1/2

.

Integrating by parts in the Riemann-Stieltjes integral we have∫ x

a

(t− a)du (t) +

∫ b

x

(b− t)du (t)(2.9)

= (t− a)u (t)|xa −∫ x

a

u (t) dt− (b− t)u (t)|bx +

∫ b

x

u (t) dt

= [2x− (a+ b)]u (x)−∫ x

a

u (t) dt+

∫ b

x

u (t) dt

= [2x− (a+ b)]u (x) +

∫ b

a

sgn(t− x)u (t) dt.

The same equality holds forv as well.Making use of (2.7) and (2.8) we deduce the first inequality in (2.5).As u is monotonic nondecreasing on[a, b] , we can state that∫ x

a

u (t) dt ≥ (x− a)u (a) and∫ b

x

u (t) dt ≤ (b− x)u (b)

so that ∫ b

a

sgn(t− x)u (t) dt ≤ (b− x)u (b)− (x− a)u (a) .

Consequently

[2x− (a+ b)]u (x) +

∫ b

a

sgn(t− x)u (t) dt

≤ [2x− (a+ b)]u (x) + (b− x)u (b)− (x− a)u (a)

= (b− x) [u (b)− u (x)] + (x− a) [u (x)− u (a)]

and a similar inequality forv.Finally, let us observe that

(b− x) [u (b)− u (x)] + (x− a) [u (x)− u (a)]

≤ maxb− x, x− a[u (b)− u (x) + u (x)− u (a)]

=

[b− a

2+

∣∣∣∣x− a+ b

2

∣∣∣∣] [u (b)− u (a)]

and a similar inequality forv.


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138 S. S. DRAGOMIR

Utilising these inequalities we deduce the last part of (2.5).

As a particular case of interest, we can state the following midpoint inequality obtained in[4].

COROLLARY 2.5 (Dragomir, 2013 [4]). With the assumptions of Theorem 2.4 we have themidpoint inequality∣∣∣∣h(a+ b

2

)(b− a)−

∫ b

a

h (t) dt

∣∣∣∣2(2.10)

≤∫ b

a

sgn(t− a+ b

2)u (t) dt

∫ b

a

sgn(t− a+ b

2)v (t) dt

≤ 1

4(b− a)2 [u (b)− u (a)] [v (b)− v (a)] .

3. OSTROWSKI FOR H-DOMINATED FUNCTIONS

3.1. H-Dominated Functions.Assume thatu, v : [a, b] → R aremonotonic nondecreas-ing on the interval[a, b] . Assume everywhere in what follows thatp, q > 1 with 1

p+ 1

q= 1.

We can introduce the following concept:

DEFINITION 3.1 (Dragomir, 2013 [3]). We say that the complex-valued functionh : [a, b] →C is (p, q)-H-dominatedby the pair(u, v) if

(H) |h (y)− h (x)| ≤ [u (y)− u (x)]1/p [v (y)− v (x)]1/q

for anyx, y ∈ [a, b] with y ≥ x.

We can give numerous examples of such functions.For instance, if we takef, g two measurable complex-valued functions such that|f |p and

|g|q are Lebesgue integrable and denote

h (x) :=

∫ x

a

f (t) g (t) dt, u (x) :=

∫ x

a

|f (t)|p dt andv (x) :=

∫ x

a

|g (t)|q dt,

then we observe thatu andv are monotonic nondecreasing on[a, b] and byHölder integralinequalitywe have for anyy ≥ x with x, y ∈ [a, b] that

|h (y)− h (x)| =∣∣∣∣∫ y

x

f (t) g (t) dt

∣∣∣∣ ≤ (∫ y

x

|f (t)|p dt)1/p(∫ y

x

|g (t)|q dt)1/q

≤ [u (y)− u (x)]1/p [v (y)− v (x)]1/q .

Now, form,n > 0 if we considerf (t) := tm andg (t) := tn for t ≥ 0, then

hm,n (x) :=

∫ x

0

tm+ndt =1

m+ n+ 1xm+n+1

and

um,p (x) : =

∫ x

0

tpmdt =1

2pm+ 1x2pm+1,

vn,q (x) : =

∫ x

0

tqndt =1

2qn+ 1x2qn+1,

for p, q > 1 with 1p

+ 1q

= 1.

Taking into account the above comments we observe that the functionhm,n is (p, q)-H-dominatedby the pair(um,p, vn,q) on any subinterval of[0,∞) .


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PROPOSITION3.1 (Dragomir, 2013 [3]). If h : [a, b] → C is (p, q)-H-dominated by the pair(u, v) , thenh is of bounded variation on any subinterval[c, d] ⊂ [a, b] and

(3.1)d∨c

(h) ≤ [u (d)− u (c)]1/p [v (d)− v (c)]1/q .

for p, q > 1 with 1p

+ 1q

= 1.

PROOF. Consider a divisionδ of the interval[c, d] given by

δ : c = x0 < x1 < ... < xn−1 < xn = b.

Sinceh : [a, b] → C is (p, q)-H-dominatedby the pair(u, v) then we have

|h (xi+1)− h (xi)| ≤ [u (xi+1)− u (xi)]1/p [v (xi+1)− v (xi)]

1/q

for anyi ∈ 0, ..., n− 1 .Summing this inequality overi from 0 to n− 1 and utilizing the Hölder discrete inequality

we haven−1∑i=1

|h (xi+1)− h (xi)|(3.2)

≤n−1∑i=1

[u (xi+1)− u (xi)]1/p [v (xi+1)− v (xi)]

1/q

≤

(n−1∑i=1

[u (xi+1)− u (xi)]

)1/p(n−1∑i=1

[v (xi+1)− v (xi)]

)1/q

= [u (d)− u (c)]1/p [v (d)− v (c)]1/q .

Taking the supremum overδ we deduce the desired result (3.1).

COROLLARY 3.2. If h : [a, b] → C is (p, q)-H-dominated by the pair(u, v) , then thecumulative variation functionV : [a, b] → [0,∞) defined by

V (x) :=x∨a

(h)

is also(p, q)-H-dominated by the pair(u, v) .

The following result is a kind of Hölder integral inequality for the Riemann-Stieltjes inte-gral:

THEOREM 3.3 (Dragomir, 2013 [3]). Assume thatu, v : [a, b] → R are monotonic non-decreasing on the interval[a, b] . If h : [a, b] → C is (p, q)-H-dominated by the pair(u, v)and f : [a, b] → C is a continuous function on[a, b] , then the Riemann-Stieltjes integral∫ b

af (t) dh (t) exists and

(3.3)

∣∣∣∣∫ b

a

f (t) dh (t)

∣∣∣∣ ≤ (∫ b

a

|f (t)| du (t)

)1/p(∫ b

a

|f (t)| dv (t)

)1/q

.


af (t) dh (t) exists, then for any sequence of

partitions

I(n)n : a = t

(n)0 < t

(n)1 < · · · < t

(n)n−1 < t(n)

n = b


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140 S. S. DRAGOMIR

with the norm

v(I(n)n

):= max

i∈0,...,n−1

(t(n)i+1 − t

(n)i

)→ 0

asn→∞, and for any intermediate pointsξ(n)i ∈ [t

(n)i , t

(n)i+1], i ∈ 0, . . . , n− 1 we have:

∣∣∣∣∫ b

a

f (t) dh (t)

∣∣∣∣ =

∣∣∣∣∣∣ limvI(n)n

→0

n−1∑i=0

f(ξ

(n)i

) [h(t(n)i+1

)− h

(t(n)i

)]∣∣∣∣∣∣(3.4)

≤ limvI(n)n

→0

n−1∑i=0

∣∣∣f (ξ(n)i

)∣∣∣ ∣∣∣h(t(n)i+1

)− h

(t(n)i

)∣∣∣≤ lim

vI(n)n

→0

n−1∑i=0

∣∣∣f (ξ(n)i

)∣∣∣ [u(t(n)i+1

)− u

(t(n)i

)]1/p

×[v(t(n)i+1

)− v

(t(n)i

)]1/q

≤ limvI(n)n

→0

(n−1∑i=0

∣∣∣f (ξ(n)i

)∣∣∣ [u(t(n)i+1

)− u

(t(n)i

)])1/p

× limvI(n)n

→0

(n−1∑i=0

∣∣∣f (ξ(n)i

)∣∣∣ [v (t(n)i+1

)− v

(t(n)i

)])1/q

=

(∫ b

a

|f (t)| du (t)

)1/p(∫ b

a

|f (t)| dv (t)

)1/q

,

where for the last inequality we employed the Hölder weighted discrete inequality

n∑k=1

mkakbk ≤

(n∑

k=1

mkapk

)1/p( n∑k=1

mkbqk

)1/q

,

wheremk, ak, bk ≥ 0 for k ∈ 1, ..., n .

3.2. Ostrowski Type Inequality. We have the following new result:

THEOREM 3.4 (Dragomir, 2013 [3]). Assume thatu, v : [a, b] → R are monotonic nonde-creasing on the interval[a, b] . If h : [a, b] → C is (p, q)-H-dominated by the pair(u, v) for


http://ajmaa.org


p, q > 1 with 1p

+ 1q

= 1, then∣∣∣∣h (x) (b− a)−∫ b

a

h (t) dt

∣∣∣∣(3.5)

≤[[2x− (a+ b)]u (x) +

∫ b

a

sgn(t− x)u (t) dt

]1/p

×[[2x− (a+ b)]v (x) +

∫ b

a

sgn(t− x)v (t) dt

]1/q

≤ [(b− x) [u (b)− u (x)] + (x− a) [u (x)− u (a)]]1/p

× [(b− x) [v (b)− v (x)] + (x− a) [v (x)− v (a)]]1/q

≤[b− a

2+

∣∣∣∣x− a+ b

2

∣∣∣∣] [u (b)− u (a)]1/p [v (b)− v (a)]1/q


PROOF. Integrating by parts on the Riemann-Stieltjes integral we have [1]

(3.6) h (x) (b− a)−∫ b

a

h (t) dt =

∫ x

a

(t− a) dh (t) +

∫ b

x

(b− t) dh (t) .

Taking the modulus in (3.6) we have

(3.7)

∣∣∣∣h (x) (b− a)−∫ b

a

h (t) dt

∣∣∣∣ ≤ ∣∣∣∣∫ x

a

(t− a) dh (t)

∣∣∣∣+ ∣∣∣∣∫ b

x

(b− t) dh (t)

∣∣∣∣ .Applying the inequality (3.3) twice, we have∣∣∣∣∫ x

a

(t− a) dh (t)

∣∣∣∣ ≤ (∫ x

a

(t− a) du (t)

)1/p(∫ x

a

(t− a) dv (t)

)1/q

and ∣∣∣∣∫ b

x

(b− t) dh (t)

∣∣∣∣ ≤ (∫ b

x

(b− t) du (t)

)1/p(∫ b

x

(b− t) dv (t)

)1/q

.

Summing these inequalities and utilizing the elementary result

αβ + λδ ≤ (αp + λp)1/p (βq + δq)1/q

for α, β, λ, δ ≥ 0 andp, q > 1 with 1p

+ 1q

= 1, we have∣∣∣∣∫ x

a

(t− a) dh (t)

∣∣∣∣+ ∣∣∣∣∫ b

x

(b− t) dh (t)

∣∣∣∣(3.8)

≤(∫ x

a

(t− a) du (t)

)1/p(∫ x

a

(t− a) dv (t)

)1/q

+

(∫ b

x

(b− t) du (t)

)1/p(∫ b

x

(b− t) dv (t)

)1/q

≤(∫ x

a

(t− a) du (t) +

∫ b

x

(b− t) du (t)

)1/p

+

(∫ x

a

(t− a) dv (t) +

∫ b

x

(b− t) dv (t)

)1/q

.


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142 S. S. DRAGOMIR


a

(t− a)du (t) +

∫ b

x

(b− t)du (t)(3.9)

= (t− a)u (t)|xa −∫ x

a

u (t) dt− (b− t)u (t)|bx +

∫ b

x

u (t) dt

= [2x− (a+ b)]u (x)−∫ x

a

u (t) dt+

∫ b

x

u (t) dt

= [2x− (a+ b)]u (x) +

∫ b

a

sgn(t− x)u (t) dt.

The same equality holds forv as well.Making use of (3.7) and (3.8) we deduce the first inequality in (3.5).As u is monotonic nondecreasing on[a, b] , we can state that∫ x

a

u (t) dt ≥ (x− a)u (a) and∫ b

x

u (t) dt ≤ (b− x)u (b)

so that ∫ b

a

sgn(t− x)u (t) dt ≤ (b− x)u (b)− (x− a)u (a) .

Consequently

[2x− (a+ b)]u (x) +

∫ b

a

sgn(t− x)u (t) dt

≤ [2x− (a+ b)]u (x) + (b− x)u (b)− (x− a)u (a)

= (b− x) [u (b)− u (x)] + (x− a) [u (x)− u (a)]

and a similar inequality forv.Finally, let us observe that

(b− x) [u (b)− u (x)] + (x− a) [u (x)− u (a)]

≤ maxb− x, x− a[u (b)− u (x) + u (x)− u (a)]

=

[b− a

2+

∣∣∣∣x− a+ b

2

∣∣∣∣] [u (b)− u (a)]

and a similar inequality forv.Utilising these inequalities we deduce the last part of (3.5).

As a particular case of interest, we can state the following midpoint inequality obtained in[3].

COROLLARY 3.5 (Dragomir, 2013 [3]). With the assumptions of Theorem 3.4 we have themidpoint inequality∣∣∣∣h(a+ b

2

)(b− a)−

∫ b

a

h (t) dt

∣∣∣∣(3.10)

≤[∫ b

a

sgn(t− a+ b

2)u (t) dt

]1/p [∫ b

a

sgn(t− a+ b

2)v (t) dt

]1/q

≤ 1

2(b− a) [u (b)− u (a)]1/p [v (b)− v (a)]1/q .


http://ajmaa.org


REFERENCES

[1] S. S. Dragomir, On the Ostrowski’s integral inequality for mappings with bounded variation andapplications,Math. Ineq. Appl.4 (2001), No. 1, 59-66. Preprint:RGMIA Res. Rep. Coll.2 (1999),Art. 7, [Online: http://rgmia.org/papers/v2n1/v2n1-7.pdf].

[2] S. S. Dragomir, Inequalities for the Riemann-Stieltjes integral of product integrators with applica-tions,J. Korean Math. Soc.51 (2014), no. 4, 791–815. Preprint,RGMIA Res. Rep. Coll.16 (2013),Article 37.[Online http://rgmia.org/papers/v16/v16a37.pdf] .

[3] S. S. Dragomir, Inequalities for the Riemann-Stieltjes Integral of(p; q)-H-Dominated Integratorswith Applications,Appl. Math. E-Notes15 (2015), 243–260. PreprintRGMIA Res. Rep. Coll.16(2013), Art. 33[Online http://rgmia.org/papers/v16/v16a33.pdf].

[4] S. S. Dragomir, Inequalities for the Riemann-Stieltjes integral ofS-dominated integrators with ap-plications (I),Probl. Anal. Issues Anal.4 (22) (2015), no. 1, 11–37. PreprintRGMIA Res. Rep. Coll.16 (2013), Art. 31[Online http://rgmia.org/papers/v16/v16a31.pdf].


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CHAPTER 7

Perturbed Ostrowski Type Inequalities

1. PERTURBED OSTROWSKI TYPE I NEQUALITIES FOR FUNCTIONS OF BOUNDED

VARIATION

1.1. Some Identities.We start with the following identity that will play an important rolein the following:

LEMMA 1.1 (Dragomir, 2013 [10]). Let f : [a, b] → C be a function of bounded variationon [a, b] andx ∈ [a, b] . Then for anyλ1 (x) andλ2 (x) complex numbers, we have

f (x) +1

2 (b− a)

[(b− x)2 λ2 (x)− (x− a)2 λ1 (x)

]− 1

b− a

∫ b

a

f (t) dt(1.1)

=1

b− a

∫ x

a

(t− a) d [f (t)− λ1 (x) t] +1

b− a

∫ b

x

(t− b) d [f (t)− λ2 (x) t] ,

where the integrals in the right hand side are taken in the Riemann-Stieltjes sense.

PROOF. Utilising the integration by parts formula in the Riemann-Stieltjes integral, we have∫ x

a

(t− a) d [f (t)− λ1 (x) t](1.2)

= (t− a) [f (t)− λ1 (x) t]|xa −∫ x

a

[f (t)− λ1 (x) t] dt

= (x− a) [f (x)− λ1 (x)x]−∫ x

a

f (t) dt+1

2λ1 (x)

(x2 − a2

)= (x− a) f (x)− λ1 (x)x (x− a)−

∫ x

a

f (t) dt+1

2λ1 (x)

(x2 − a2

)= (x− a) f (x)−

∫ x

a

f (t) dt− 1

2(x− a)2 λ1 (x)

and ∫ b

x

(t− b) d [f (t)− λ2 (x) t](1.3)

= (t− b) [f (t)− λ2 (x) t]|bx −∫ b

x

[f (t)− λ2 (x) t] dt

= (b− x) [f (x)− λ2 (x)x]−∫ b

x

f (t) dt+1

2λ2 (x)

(b2 − x2

)= (b− x) f (x)−

∫ b

x

f (t) dt− (b− x)λ2 (x)x+1

2λ2 (x)

(b2 − x2

)= (b− x) f (x)−

∫ b

x

f (t) dt+1

2(b− x)2 λ2 (x) .

145

146 S. S. DRAGOMIR

By adding the equalities (1.2) and (1.3) and dividing byb− a we get the desired representation(1.1).

COROLLARY 1.2. With the assumption in Lemma 1.1, we have for anyλ (x) ∈ C that

f (x) +

(a+ b

2− x

)λ (x)− 1

b− a

∫ b

a

f (t) dt(1.4)

=1

b− a

∫ x

a

(t− a) d [f (t)− λ (x) t] +1

b− a

∫ b

x

(t− b) d [f (t)− λ (x) t] .

We have the following midpoint representation:

COROLLARY 1.3. With the assumption in Lemma 1.1, we have for anyλ1, λ2 ∈ C that

f

(a+ b

2

)+

1

8(b− a) (λ2 − λ1)−

1

b− a

∫ b

a

f (t) dt(1.5)

=1

b− a

∫ a+b2

a

(t− a) d [f (t)− λ1t]

+1

b− a

∫ b

a+b2

(t− b) d [f (t)− λ2t] .

In particular, if λ1 = λ2 = λ, then we have the equality

f

(a+ b

2

)− 1

b− a

∫ b

a

f (t) dt(1.6)

=1

b− a

∫ a+b2

a

(t− a) d [f (t)− λt] +1

b− a

∫ b

a+b2

(t− b) d [f (t)− λt] .

REMARK 1.1. If we takeλ (x) = 0 in (1.4) we recapture the Montgomery type identityestablished in [2].

1.2. Inequalities for Functions of Bounded Variation. The following lemma will be usedin the sequel and is of interest in itself as well [1, p. 177]. For a simple proof see also [6].


∫ b


(1.7)

∣∣∣∣∫ b

a

f (t) du (t)

∣∣∣∣ ≤ ∫ b

a

|f (t)| d

(t∨a

(u)

)≤ max

t∈[a,b]|f (t)|

b∨a

(u) .

We denote by : [a, b] → [a, b] the identity function, namely(t) = t for anyt ∈ [a, b] .We have the following result:


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THEOREM1.5 (Dragomir, 2013 [10]). Letf : [a, b] → C be a function of bounded variationon [a, b] andx ∈ [a, b] . Then for anyλ1 (x) andλ2 (x) complex numbers, we have the inequality

∣∣∣∣f (x) +1

2 (b− a)

[(b− x)2 λ2 (x)− (x− a)2 λ1 (x)

]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.8)

≤ 1

b− a

[∫ x

a

(x∨t

(f − λ1 (x) `)

)dt+

∫ b

x

(t∨x

(f − λ2 (x) `)

)dt

]

≤ 1

b− a

[(x− a)

x∨a

(f − λ1 (x) `) + (b− x)b∨x

(f − λ2 (x) `)

]

≤

max

x∨a

(f − λ1 (x) `) ,b∨x

(f − λ2 (x) `)

[12

+∣∣∣x−a+b

2

b−a

∣∣∣]( x∨a

(f − λ1 (x) `) +b∨x

(f − λ2 (x) `)

),

whered∨c

(g) denotes the total variation ofg on the interval[c, d] .

PROOF. Taking the modulus in (1.1) and using the property (1.7) we have

∣∣∣∣f (x) +1

2 (b− a)

[(b− x)2 λ2 (x)− (x− a)2 λ1 (x)

]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.9)

≤ 1

b− a

∣∣∣∣∫ x

a

(t− a) d [f (t)− λ1 (x) t]

∣∣∣∣+

1

b− a

∣∣∣∣∫ b

x

(t− b) d [f (t)− λ2 (x) t]

∣∣∣∣≤ 1

b− a

∫ x

a

(t− a) d

(t∨a

(f − λ1 (x) `)

)

+1

b− a

∫ b

x

(b− t) d

(t∨a

(f − λ2 (x) `)

).


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148 S. S. DRAGOMIR

Integrating by parts in the Riemann-Stieltjes integral, we have

∫ x

a

(t− a) d

(t∨a

(f − λ1 (x) `)

)

= (t− a)t∨a

(f − λ1 (x) `)

∣∣∣∣∣x

a

−∫ x

a

(t∨a

(f − λ1 (x) `)

)dt

= (x− a)x∨a

(f − λ1 (x) `)−∫ x

a

(t∨a

(f − λ1 (x) `)

)dt

=

∫ x

a

(x∨t

(f − λ1 (x) `)

)dt

and

∫ b

x

(b− t) d

(t∨a

(f − λ2 (x) `)

)

= (b− t)t∨a

(f − λ2 (x) `)

∣∣∣∣∣b

x

+

∫ b

x

(t∨a

(f − λ2 (x) `)

)dt

=

∫ b

x

(t∨a

(f − λ2 (x) `)

)dt− (b− x)

x∨a

(f − λ2 (x) `)

=

∫ b

x

(t∨x

(f − λ2 (x) `)

)dt.

Using (1.9) we deduce the first inequality in (1.8).We also have

∫ x

a

(x∨t

(f − λ1 (x) `)

)dt ≤ (x− a)

x∨a

(f − λ1 (x) `)

and ∫ b

x

(t∨x

(f − λ2 (x) `)

)dt ≤ (b− x)

b∨x

(f − λ2 (x) `) ,

which prove the second inequality in (1.8).The last part is obvious.



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COROLLARY 1.6. Let f : [a, b] → C be a function of bounded variation on[a, b] andx ∈ [a, b] . Then for anyλ (x) a complex number, we have the inequality

∣∣∣∣f (x) +

(a+ b

2− x

)λ (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.10)

≤ 1

b− a

[∫ x

a

(x∨t

(f − λ (x) `)

)dt+

∫ b

x

(t∨x

(f − λ (x) `)

)dt

]

≤ 1

b− a

[(x− a)

x∨a

(f − λ (x) `) + (b− x)b∨x

(f − λ (x) `)

]

≤

12

b∨a

(f − λ (x) `) + 12

∣∣∣∣∣b∨x

(f − λ (x) `)−x∨a

(f − λ (x) `)

∣∣∣∣∣[

12

+∣∣∣x−a+b

2

b−a

∣∣∣] b∨a

(f − λ (x) `) .

REMARK 1.2. Letf : [a, b] → C be a function of bounded variation on[a, b]. Then for anyλ ∈ C we have the inequalities

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.11)

≤ 1

b− a

∫ a+b2

a

a+b2∨t

(f − λ`)

dt+

∫ b

a+b2

t∨a+b2

(f − λ`)

dt

≤ 1

2

b∨a

(f − λ`) .

This is equivalent to

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.12)

≤ 1

b− ainfλ∈C

∫ a+b2

a

a+b2∨t

(f − λ`)

dt+

∫ b

a+b2

t∨a+b2

(f − λ`)

dt

≤ 1

2infλ∈C

[b∨a

(f − λ`)

].

1.3. Inequalities for Lipshitzian Functions. We can state the following result:

THEOREM 1.7 (Dragomir, 2013 [10]). Let f : [a, b] → C be a bounded function on[a, b]andx ∈ (a, b) . If λ1 (x) andλ2 (x) are complex numbers and there exist the positive numbersL1 (x) andL2 (x) such thatf − λ1 (x) ` is Lipschitzian with the constantL1 (x) on the interval


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150 S. S. DRAGOMIR

[a, x] andf − λ2 (x) ` is Lipschitzian with the constantL2 (x) on the interval[x, b] , then

∣∣∣∣f (x) +1

2 (b− a)

[(b− x)2 λ2 (x)− (x− a)2 λ1 (x)

]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.13)

≤ 1

2

[(x− a

b− a

)2

L1 (x) +

(b− x

b− a

)2

L2 (x)

](b− a)

≤

[14

+(

x−a+b2

b−a

)2]

max L1 (x) , L2 (x) (b− a) ,

12

[(x−ab−a

)2q+(

b−xb−a

)2q]1/q

(Lp1 (x) + Lp

2 (x))1/p (b− a) ,

p > 1, 1p

+ 1q

= 1,[12

+∣∣∣x−a+b

2

b−a

∣∣∣] L1(x)+L2(x)2

(b− a) .

PROOF. It is known that, ifg : [c, d] → C is Riemann integrable andu : [c, d] → C isLipschitzian with the constantL > 0, then the Riemann-Stieltjes integral

∫ d

cg (t) du (t) exists

and

(1.14)

∣∣∣∣∫ d

c

g (t) du (t)

∣∣∣∣ ≤ L

∫ d

c

|g (t)| dt.

Taking the modulus in (1.1) and using the property (1.14) we have

∣∣∣∣f (x) +1

2 (b− a)

[(b− x)2 λ2 (x)− (x− a)2 λ1 (x)

]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

b− a

∣∣∣∣∫ x

a

(t− a) d [f (t)− λ1 (x) t]

∣∣∣∣+

1

b− a

∣∣∣∣∫ b

x

(t− b) d [f (t)− λ2 (x) t]

∣∣∣∣≤ 1

b− a

[L1 (x)

∫ x

a

(t− a) dt+ L2 (x)

∫ b

x

(b− t) dt

]=L1 (x) (x− a)2 + L2 (x) (b− x)2

2 (b− a)

=1

2

[L1 (x)

(x− a

b− a

)2

+ L2 (x)

(b− x

b− a

)2]

(b− a) ,

and the first inequality in (1.13) is proved.


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By Hölder’s inequality we have

L1 (x)

(x− a

b− a

)2

+ L2 (x)

(b− x

b− a

)2

≤

[(x−ab−a

)2+(

b−xb−a

)2]max L1 (x) , L2 (x)

[(x−ab−a

)2q+(

b−xb−a

)2q]1/q

(Lp1 (x) + Lp

2 (x))1/p ,

p > 1, 1p

+ 1q

= 1

max(

x−ab−a

)2,(

b−xb−a

)2[L1 (x) + L2 (x)] ,

which proves, upon simple calculations, the last part of the inequality (1.13).

COROLLARY 1.8. Let f : [a, b] → C be a bounded function on[a, b] andx ∈ (a, b) . Ifλ (x) is a complex number and there exist the positive numberL (x) such thatf − λ (x) ` isLipschitzian with the constantL (x) on the interval[a, b], then

∣∣∣∣f (x) +

(a+ b

2− x

)λ (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.15)

≤

1

4+

(x− a+b

2

b− a

)2L (x) (b− a) .

REMARK 1.3. If λ is a complex number and there exist the positive numberL such thatf − λ` is Lipschitzian with the constantL on the interval[a, b], then

(1.16)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

4L (b− a) .

1.4. Inequalities for Monotonic Functions. Now, the case of monotonic integrators is asfollows:

THEOREM 1.9 (Dragomir, 2013 [10]). Let f : [a, b] → R be a bounded function on[a, b]and x ∈ (a, b) . If λ1 (x) and λ2 (x) are real numbers such thatf − λ1 (x) ` is monotonicnondecreasing on the interval[a, x] andf−λ2 (x) ` is monotonic nondecreasing on the interval


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152 S. S. DRAGOMIR

[x, b] , then

∣∣∣∣f (x) +1

2 (b− a)

[(b− x)2 λ2 (x)− (x− a)2 λ1 (x)

]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.17)

≤ 1

b− a

[(2x− a− b) f (x) +

∫ b

a

sgn (t− x) f (t) dt

−1

2

[λ1 (x) (x− a)2 + λ2 (x) (b− x)2]]

≤ 1

b− a(x− a) [f (x)− f (a)− λ1 (x) (x− a)]

+ (b− x) [f (b)− f (x)− λ2 (x) (b− x)]

≤

12[f (b)− f (a)− λ1 (x) (x− a)− λ2 (x) (b− x)]

+∣∣∣f (x)− f(a)+f(b)

2− 1

2λ1 (x) (x− a) + 1

2λ2 (x) (b− x)

∣∣∣ ,[12

+∣∣∣x−a+b

2

b−a

∣∣∣]× [f (b)− f (a)− λ1 (x) (x− a)− λ2 (x) (b− x)] .

PROOF. It is known that, ifg : [c, d] → C is continuous andu : [c, d] → C is monotonicnondecreasing, then the Riemann-Stieltjes integral

∫ d

cg (t) du (t) exists and

(1.18)

∣∣∣∣∫ d

c

g (t) du (t)

∣∣∣∣ ≤ ∫ d

c

|g (t)| du (t) .

Taking the modulus in (1.1) and using the property (1.18) we have

∣∣∣∣f (x) +1

2 (b− a)

[(b− x)2 λ2 (x)− (x− a)2 λ1 (x)

]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.19)

≤ 1

b− a

∣∣∣∣∫ x

a

(t− a) d [f (t)− λ1 (x) t]

∣∣∣∣+

1

b− a

∣∣∣∣∫ b

x

(t− b) d [f (t)− λ2 (x) t]

∣∣∣∣≤ 1

b− a

∫ x

a

(t− a) d [f (t)− λ1 (x) t]

+1

b− a

∫ b

x

(b− t) d [f (t)− λ2 (x) t] .


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a

(t− a) d [f (t)− λ1 (x) t]

= (t− a) [f (t)− λ1 (x) t]|xa −∫ x

a

[f (t)− λ1 (x) t] dt

= (x− a) [f (x)− λ1 (x)x]−∫ x

a

[f (t)− λ1 (x) t] dt

= (x− a) f (x)− λ1 (x)x (x− a)−∫ x

a

f (t) dt+ λ1 (x)x2 − a2

2

= (x− a) f (x)−∫ x

a

f (t) dt− 1

2λ1 (x) (x− a)2

and ∫ b

x

(b− t) d [f (t)− λ2 (x) t]

= (b− t) [f (t)− λ2 (x) t]|bx +

∫ b

x

[f (t)− λ2 (x) t] dt

=

∫ b

x

f (t) dt− λ2 (x)

∫ b

x

tdt− (b− x) [f (x)− λ2 (x)x]

=

∫ b

x

f (t) dt− λ2 (x)b2 − x2

2− (b− x) f (x) + (b− x)λ2 (x)x

=

∫ b

x

f (t) dt− (b− x) f (x)− 1

2λ2 (x) (b− x)2 .

If we add these equalities, we get∫ x

a

(t− a) d [f (t)− λ1 (x) t] +

∫ b

x

(b− t) d [f (t)− λ2 (x) t]

= (x− a) f (x)−∫ x

a

f (t) dt− 1

2λ1 (x) (x− a)2

+

∫ b

x

f (t) dt− (b− x) f (x)− 1

2λ2 (x) (b− x)2

= (2x− a− b) f (x) +

∫ b

a


− 1

2

[λ1 (x) (x− a)2 + λ2 (x) (b− x)2]

and by (1.19) we get the first inequality in (1.17).Now, sincef − λ1 (x) ` is monotonic nondecreasing on the interval[a, x], then∫ x

a

(t− a) d [f (t)− λ1 (x) t]

≤ (x− a) [f (x)− λ1 (x)x− f (a) + λ1 (x) a]

= (x− a) [f (x)− f (a)− λ1 (x) (x− a)]


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154 S. S. DRAGOMIR

and, sincef − λ2 (x) ` is monotonic nondecreasing on the interval[x, b] , then also∫ b

x

(b− t) d [f (t)− λ2 (x) t]

≤ (b− x) [f (b)− λ2 (x) b− f (x) + λ2 (x)x]

= (b− x) [f (b)− f (x)− λ2 (x) (b− x)] .

These prove the second inequality in (1.17).The last part follows by the properties of maximum and the details are omitted.

COROLLARY 1.10. Let f : [a, b] → R be a bounded function on[a, b] andx ∈ (a, b) . Ifλ (x) is a real number such thatf − λ (x) ` is monotonic nondecreasing on the interval[a, b],then ∣∣∣∣f (x) +

(a+ b

2− x

)λ (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.20)

≤ 1

b− a

[(2x− a− b) f (x) +

∫ b

a


−

1

4+

(x− a+b

2

b− a

)2 (b− a)2 λ (x)

≤ 1

b− a(x− a) [f (x)− f (a)− λ (x) (x− a)]

+ (b− x) [f (b)− f (x)− λ (x) (b− x)]

≤

f(b)−f(a)2

− 12λ (x) (b− a)

+∣∣∣f (x)− f(a)+f(b)

2− 1

2λ (x) (2x− a− b)

∣∣∣ ,[12

+∣∣∣x−a+b

2

b−a

∣∣∣]× [f (b)− f (a)− λ (x) (b− a)] .

REMARK 1.4. If λ is a real number such thatf − λ` is monotonic nondecreasing on theinterval[a, b], then ∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.21)

≤ 1

b− a

[∫ b

a

sgn

(t− a+ b

2

)f (t) dt− 1

4λ (b− a)2

]≤ 1

2[f (b)− f (a)− λ (b− a)] .


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2. SOME PERTURBED OSTROWSKI TYPE I NEQUALITIES FOR ABSOLUTELY

CONTINUOUS FUNCTIONS

2.1. Some Identities.We start with the following identity that will play an important rolein the following:

LEMMA 2.1 (Dragomir, 2013 [7]). Let f : [a, b] → C be an absolutely continuous on[a, b]andx ∈ [a, b] . Then for anyλ1 (x) andλ2 (x) complex numbers, we have

f (x) +1

2 (b− a)

[(b− x)2 λ2 (x)− (x− a)2 λ1 (x)

]− 1

b− a

∫ b

a

f (t) dt(2.1)

=1

b− a

∫ x

a

(t− a) [f ′ (t)− λ1 (x)] dt+1

b− a

∫ b

x

(t− b) [f ′ (t)− λ2 (x)] dt,

where the integrals in the right hand side are taken in the Lebesgue sense.

PROOF. Utilising the integration by parts formula in the Lebesgue integral, we have∫ x

a

(t− a) [f ′ (t)− λ1 (x)] dt(2.2)

= (t− a) [f (t)− λ1 (x) t]|xa −∫ x

a

[f (t)− λ1 (x) t] dt

= (x− a) [f (x)− λ1 (x)x]−∫ x

a

f (t) dt+1

2λ1 (x)

(x2 − a2

)= (x− a) f (x)− λ1 (x)x (x− a)−

∫ x

a

f (t) dt+1

2λ1 (x)

(x2 − a2

)= (x− a) f (x)−

∫ x

a

f (t) dt− 1

2(x− a)2 λ1 (x)

and ∫ b

x

(t− b) [f ′ (t)− λ2 (x)] dt(2.3)

= (t− b) [f (t)− λ2 (x) t]|bx −∫ b

x

[f (t)− λ2 (x) t] dt

= (b− x) [f (x)− λ2 (x)x]−∫ b

x

f (t) dt+1

2λ2 (x)

(b2 − x2

)= (b− x) f (x)−

∫ b

x

f (t) dt− (b− x)λ2 (x)x+1

2λ2 (x)

(b2 − x2

)= (b− x) f (x)−

∫ b

x

f (t) dt+1

2(b− x)2 λ2 (x) .

If we add the identities (2.2) and (2.3) and divide byb−awe deduce the desired identity (2.1).

COROLLARY 2.2. With the assumption in Lemma 2.1, we have for anyλ (x) ∈ C that

f (x) +

(a+ b

2− x

)λ (x)− 1

b− a

∫ b

a

f (t) dt(2.4)

=1

b− a

∫ x

a

(t− a) [f ′ (t)− λ (x)] dt+1

b− a

∫ b

x

(t− b) [f ′ (t)− λ (x)] dt.


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156 S. S. DRAGOMIR

REMARK 2.1. If we takeλ (x) = 0 in (2.4), then we get Montgomery’s identity for ab-solutely continuous functions, i.e.

f (x)− 1

b− a

∫ b

a

f (t) dt(2.5)

=1

b− a

∫ x

a

(t− a) f ′ (t) dt+1

b− a

∫ b

x

(t− b) f ′ (t) dt,

for x ∈ [a, b] .

We have the following midpoint representation:

COROLLARY 2.3. With the assumption in Lemma 2.1, we have for anyλ1, λ2 ∈ C that

f

(a+ b

2

)+

1

8(b− a) (λ2 − λ1)−

1

b− a

∫ b

a

f (t) dt(2.6)

=1

b− a

∫ a+b2

a

(t− a) [f ′ (t)− λ1] dt+1

b− a

∫ b

a+b2

(t− b) [f ′ (t)− λ2] dt.

In particular, if λ1 = λ2 = λ, then we have the equality

f

(a+ b

2

)− 1

b− a

∫ b

a

f (t) dt(2.7)

=1

b− a

∫ a+b2

a

(t− a) [f ′ (t)− λ] dt+1

b− a

∫ b

a+b2

(t− b) [f ′ (t)− λ] dt.

REMARK 2.2. The identity (2.1) has many particular cases of interest.If we assume that the derivativesf ′+ (a), f ′− (b) andf ′ (x) exist and are finite, then by taking

λ1 (x) =f ′+ (a) + f ′ (x)

2andλ2 (x) =

f ′ (x) + f ′− (b)

2in (2.1) we get

f (x) +1

2

(a+ b

2− x

)f ′ (x)− 1

b− a

∫ b

a

f (t) dt(2.8)

+1

4 (b− a)

[(b− x)2 f ′− (b)− (x− a)2 f ′+ (a)

]=

1

b− a

∫ x

a

(t− a)

[f ′ (t)−

f ′+ (a) + f ′ (x)

2

]dt

+1

b− a

∫ b

x

(t− b)

[f ′ (t)−

f ′ (x) + f ′− (b)

2

]dt.


f

(a+ b

2

)+

1

16(b− a)

[f ′− (b)− f ′+ (a)

]− 1

b− a

∫ b

a

f (t) dt(2.9)

=1

b− a

∫ a+b2

a

(t− a)

[f ′ (t)−

f ′+ (a) + f ′(

a+b2

)2

]dt

+1

b− a

∫ b

a+b2

(t− b)

[f ′ (t)−

f ′(

a+b2

)+ f ′− (b)

2

]dt.


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2.2. Inequalities for Bounded Derivatives.Now, for γ,Γ ∈ C and [a, b] an interval ofreal numbers, define the sets of complex-valued functions

U[a,b] (γ,Γ)

:=f : [a, b] → C|Re

[(Γ− f (t))

(f (t)− γ

)]≥ 0 for almost everyt ∈ [a, b]

and

∆[a,b] (γ,Γ) :=

f : [a, b] → C|

∣∣∣∣f (t)− γ + Γ

2

∣∣∣∣ ≤ 1

2|Γ− γ| for a.e.t ∈ [a, b]

.

The following representation result may be stated.

PROPOSITION2.4 (Dragomir, 2013 [7]). For anyγ,Γ ∈ C, γ 6= Γ,we have thatU[a,b] (γ,Γ)and∆[a,b] (γ,Γ) are nonempty, convex and closed sets and

(2.10) U[a,b] (γ,Γ) = ∆[a,b] (γ,Γ) .

PROOF. We observe that for anyz ∈ C we have the equivalence∣∣∣∣z − γ + Γ

2

∣∣∣∣ ≤ 1

2|Γ− γ|

if and only if

Re [(Γ− z) (z − γ)] ≥ 0.

This follows by the equality

1

4|Γ− γ|2 −

∣∣∣∣z − γ + Γ

2

∣∣∣∣2 = Re [(Γ− z) (z − γ)]

that holds for anyz ∈ C.The equality (2.10) is thus a simple consequence of this fact.

On making use of the complex numbers field properties we can also state that:

COROLLARY 2.5. For anyγ,Γ ∈ C, γ 6= Γ,we have that

U[a,b] (γ,Γ) = f : [a, b] → C | (Re Γ− Re f (t)) (Re f (t)− Re γ)(2.11)

+ (Im Γ− Im f (t)) (Im f (t)− Im γ) ≥ 0 for a.e.t ∈ [a, b] .

Now, if we assume thatRe (Γ) ≥ Re (γ) and Im (Γ) ≥ Im (γ) , then we can define thefollowing set of functions as well:

S[a,b] (γ,Γ) := f : [a, b] → C | Re (Γ) ≥ Re f (t) ≥ Re (γ)(2.12)

andIm (Γ) ≥ Im f (t) ≥ Im (γ) for a.e.t ∈ [a, b] .

One can easily observe thatS[a,b] (γ,Γ) is closed, convex and

(2.13) ∅ 6= S[a,b] (γ,Γ) ⊆ U[a,b] (γ,Γ) .

THEOREM 2.6 (Dragomir, 2013 [7]). Let f : [a, b] → C be an absolutely continuous on[a, b] andx ∈ (a, b) . Suppose thatγi,Γi ∈ C with γi 6= Γi, i = 1, 2 andf ′ ∈ U[a,x] (γ1,Γ1) ∩


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158 S. S. DRAGOMIR

U[x,b] (γ2,Γ2), then we have∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt(2.14)

+1

2 (b− a)

[(b− x)2 Γ2 + γ2

2− (x− a)2 Γ1 + γ1

2

]∣∣∣∣≤ 1

4

[|Γ1 − γ1|

(x− a

b− a

)2

+ |Γ2 − γ2|(b− x

b− a

)2]

(b− a)

≤ 1

4(b− a)

×

[14

+(

x−a+b2

b−a

)2]

max |Γ1 − γ1| , |Γ2 − γ2| .

[(x−ab−a

)2p+(

b−xb−a

)2p]1/p

[|Γ1 − γ1|q + |Γ2 − γ2|

q]1/q

p > 1, 1p

+ 1q

= 1,[12

+∣∣∣x−a+b

2

b−a

∣∣∣] [|Γ1 − γ1|+ |Γ2 − γ2|] .

PROOF. Sincef ′ ∈ U[a,x] (γ1,Γ1) ∩ U[x,b] (γ2,Γ2) , then by taking the modulus in (2.1) forλ1 (x) = Γ1+γ1

2andλ2 (x) = Γ2+γ2

2we get∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt

+1

2 (b− a)

[(b− x)2 Γ2 + γ2

2− (x− a)2 Γ1 + γ1

2

]∣∣∣∣≤ 1

b− a

∣∣∣∣∫ x

a

(t− a)

[f ′ (t)− Γ1 + γ1

2

]dt

∣∣∣∣+

1

b− a

∣∣∣∣∫ b

x

(t− b)

[f ′ (t)− Γ2 + γ2

2

]dt

∣∣∣∣≤ 1

b− a

∫ x

a

(t− a)

∣∣∣∣f ′ (t)− Γ1 + γ1

2

∣∣∣∣ dt+

1

b− a

∫ b

x

(t− b)

∣∣∣∣f ′ (t)− Γ2 + γ2

2

∣∣∣∣ dt≤ 1

b− a

|Γ1 − γ1|2

∫ x

a

(t− a) dt+1

b− a

|Γ2 − γ2|2

∫ b

x

(b− t) dt

=1

4

[|Γ1 − γ1|

(x− a

b− a

)2

+ |Γ2 − γ2|(b− x

b− a

)2]

(b− a)

and the first inequality in (2.14) is proved.The last part follows by Hölder’s inequality

mn+ pq ≤ (mα + pα)1/α (nβ + qβ)1/β

,

wherem,n, p, q ≥ 0 andα > 1 with 1α

+ 1β

= 1.


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COROLLARY 2.7. Let f : [a, b] → C be an absolutely continuous on[a, b] andx ∈ (a, b) .Suppose thatγ,Γ ∈ C with γ 6= Γ, andf ′ ∈ U[a,b] (γ,Γ), then we have∣∣∣∣f (x) +

(a+ b

2− x

)Γ + γ

2− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.15)

≤ 1

2|Γ− γ|

1

4+

(x− a+b

2

b− a

)2 (b− a) .


(2.16)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

8|Γ− γ| (b− a) .

REMARK 2.3. If the derivativef ′ : [a, b] → R is bounded above and below, that is, thereexists the constantsM > m such that

−∞ < m ≤ f ′ (t) ≤M <∞ for a.e.t ∈ [a, b] ,

then we recapture from (2.15) the inequality [5]∣∣∣∣f (x) +

(a+ b

2− x

)M +m

2− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

2(M −m)

1

4+

(x− a+b

2

b− a

)2 (b− a) .

REMARK 2.4. Let f : [a, b] → C be an absolutely continuous on[a, b] . Suppose thatγi,Γi ∈ C with γi 6= Γi, i = 1, 2 andf ′ ∈ U[a, a+b

2 ] (γ1,Γ1) ∩ U[a+b2

,b] (γ2,Γ2), then we have

from (2.14) that∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt+1

8(b− a)

(Γ2 + γ2

2− Γ1 + γ1

2

)∣∣∣∣(2.17)

≤ 1

16[|Γ1 − γ1|+ |Γ2 − γ2|] (b− a) .

2.3. Inequalities for Derivatives of Bounded Variation. Assume that the functionf :I → C is differentiable on the interior ofI, denotedI, and[a, b] ⊂ I . Then, as in (2.8), we havethe equality

f (x) +1

2

(a+ b

2− x

)f ′ (x)− 1

b− a

∫ b

a

f (t) dt(2.18)

+1

4 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]=

1

b− a

∫ x

a

(t− a)

[f ′ (t)− f ′ (a) + f ′ (x)

2

]dt

+1

b− a

∫ b

x

(t− b)

[f ′ (t)− f ′ (x) + f ′ (b)

2

]dt,



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160 S. S. DRAGOMIR

THEOREM 2.8 (Dragomir, 2013 [7]). Let f : I → C be a differentiable function onI and[a, b] ⊂ I . If the derivativef ′ : I → C is of bounded variation on[a, b] , then

∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt+1

2

(a+ b

2− x

)f ′ (x)(2.19)

+1

4 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]∣∣∣∣

≤ 1

4

[(x− a

b− a

)2 x∨a

(f ′) +

(b− x

b− a

)2 b∨x

(f ′)

](b− a)

≤ 1

4(b− a)

×

[14

+(

x−a+b2

b−a

)2][

12

b∨a

(f ′) + 12

∣∣∣∣∣x∨a

(f ′)−b∨x

(f ′)

∣∣∣∣∣],

[(x−ab−a

)2p+(

b−xb−a

)2p]1/p

[[x∨a

(f ′)

]q

+

[b∨x

(f ′)

]q]1/q

p > 1, 1p

+ 1q

= 1,

[12

+∣∣∣x−a+b

2

b−a

∣∣∣] b∨a

(f ′) ,


PROOF. Taking the modulus in (2.18) we have

∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt+1

2

(a+ b

2− x

)f ′ (x)(2.20)

+1

4 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]∣∣∣∣≤ 1

b− a

∣∣∣∣∫ x

a

(t− a)

[f ′ (t)− f ′ (a) + f ′ (x)

2

]dt

∣∣∣∣+

1

b− a

∣∣∣∣∫ b

x

(t− b)

[f ′ (t)− f ′ (x) + f ′ (b)

2

]dt

∣∣∣∣≤ 1

b− a

∫ x

a

(t− a)

∣∣∣∣f ′ (t)− f ′ (a) + f ′ (x)

2

∣∣∣∣ dt+

1

b− a

∫ b

x

(b− t)

∣∣∣∣f ′ (t)− f ′ (x) + f ′ (b)

2

∣∣∣∣ dt.AJMAA, Vol. 14, No. 1, Art. 1, pp. 1-287, 2017 AJMAA

http://ajmaa.org


Sincef ′ : I → C is of bounded variation on[a, x] and[x, b] , then∣∣∣∣f ′ (t)− f ′ (a) + f ′ (x)

2

∣∣∣∣ =|f ′ (t)− f ′ (a) + f ′ (t)− f ′ (x)|

2

≤ 1

2[|f ′ (t)− f ′ (a)|+ |f ′ (x)− f ′ (t)|]

≤ 1

2

x∨a

(f ′)

for anyt ∈ [a, x] and, similarly,∣∣∣∣f ′ (t)− f ′ (x) + f ′ (b)

2

∣∣∣∣ ≤ 1

2

b∨x

(f ′)

for anyt ∈ [x, b] .Then ∫ x

a

(t− a)

∣∣∣∣f ′ (t)− f ′ (a) + f ′ (x)

2

∣∣∣∣ dt ≤ 1

2

x∨a

(f ′)

∫ x

a

(t− a) dt

=1

4(x− a)2

x∨a

(f ′)

and ∫ b

x

(b− t)

∣∣∣∣f ′ (t)− f ′ (x) + f ′ (b)

2

∣∣∣∣ dt ≤ 1

2

b∨x

(f ′)

∫ b

x

(b− t) dt

=1

4(b− x)2

b∨x

(f ′)

and by (2.20) we get the desired inequality (2.19).The last part follows by Hölder’s inequality


,


+ 1β

= 1.

COROLLARY 2.9. Let f : I → C be a differentiable function onI and [a, b] ⊂ I . If thederivativef ′ : I → C is of bounded variation on[a, b] , then∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt+1

16(b− a) [f ′ (b)− f ′ (a)]

∣∣∣∣(2.21)

≤ 1

16(b− a)

b∨a

(f ′) .


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162 S. S. DRAGOMIR

REMARK 2.5. If p ∈ (a, b) is a median point in bounded variation for the derivative, i.e.p∨a

(f ′) =b∨p

(f ′) , then under the assumptions of Theorem 2.8, we have

∣∣∣∣f (p)− 1

b− a

∫ b

a

f (t) dt+1

2

(a+ b

2− p

)f ′ (p)(2.22)

+1

4 (b− a)

[(b− p)2 f ′ (b)− (p− a)2 f ′ (a)

]∣∣∣∣≤ 1

8(b− a)

1

4+

(p− a+b

2

b− a

)2 b∨

a

(f ′) .

2.4. Inequalities for Lipschitzian Derivatives. We say thatv : [a, b] → C is Lipschitzianwith the constantL > 0, if it satisfies the condition

|v (t)− v (s)| ≤ L |t− s| for anyt, s ∈ [a, b] .

THEOREM 2.10 (Dragomir, 2013 [7]). Let f : I → C be a differentiable function onI and[a, b] ⊂ I . Letx ∈ (a, b) . If the derivativef ′ : I → C is Lipschitzian with the constantK1 (x)on [a, x] and constantK2 (x) on [x, b] , then∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt+1

2

(a+ b

2− x

)f ′ (x)(2.23)

+1

4 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]∣∣∣∣≤ 1

8

[(x− a

b− a

)3

K1 (x) +

(b− x

b− a

)3

K2 (x)

](b− a)2

≤ 1

8(b− a)2

×

[(x−ab−a

)3+(

b−xb−a

)3]max K1 (x) , K2 (x) ,

[(x−ab−a

)2p+(

b−xb−a

)2p]1/p

[Kq1 (x) +Kq

2 (x)]1/q

p > 1, 1p

+ 1q

= 1,

[12

+∣∣∣x−a+b

2

b−a

∣∣∣]3 [K1 (x) +K2 (x)] .

PROOF. Sincef ′ : I → C is Lipschitzian with the constantK1 (x) on [a, x] and constantK2 (x) on [x, b] , then∣∣∣∣f ′ (t)− f ′ (a) + f ′ (x)

2

∣∣∣∣ =|f ′ (t)− f ′ (a) + f ′ (t)− f ′ (x)|

2

≤ 1

2[|f ′ (t)− f ′ (a)|+ |f ′ (x)− f ′ (t)|]

≤ 1

2K1 (x) [|t− a|+ |x− t|]

=1

2K1 (x) (x− a)


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for anyt ∈ [a, x] and, similarly,∣∣∣∣f ′ (t)− f ′ (x) + f ′ (b)

2

∣∣∣∣ ≤ 1

2K2 (x) [|t− x|+ |b− t|]

=1

2K2 (x) (b− x)

for anyt ∈ [x, b] .Then∫ x

a

(t− a)

∣∣∣∣f ′ (t)− f ′ (a) + f ′ (x)

2

∣∣∣∣ dt ≤ 1

2K1 (x) (x− a)

∫ x

a

(t− a) dt

=1

8(x− a)3K1 (x)

and ∫ b

x

(b− t)

∣∣∣∣f ′ (t)− f ′ (x) + f ′ (b)

2

∣∣∣∣ dt ≤ 1

2K2 (x) (b− x)

∫ b

x

(b− t) dt

=1

8(b− x)3K2 (x) .

Making use of the inequality (2.20) we deduce the first bound in (2.23).The second part is obvious.

COROLLARY 2.11. Let f : I → C be a differentiable function onI and [a, b] ⊂ I . If thederivativef ′ : I → C is Lipschitzian with the constantK on [a, b] then∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt+1

2

(a+ b

2− x

)f ′ (x)(2.24)

+1

4 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]∣∣∣∣≤ 1

8

[(x− a

b− a

)3

+

(b− x

b− a

)3]K (b− a)2

for anyx ∈ [a, b] .In particular, we have∣∣∣∣f (a+ b

2

)+

1

16(b− a) [f ′ (b)− f ′ (a)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.25)

≤ 1

32K (b− a)2 .

3. OTHER PERTURBED OSTROWSKI TYPE I NEQUALITIES FOR ABSOLUTELY

CONTINUOUS FUNCTION

3.1. Inequalities for Derivatives of Bounded Variation. Assume that the functionf :I → C is differentiable on the interior ofI, denotedI, and [a, b] ⊂ I . Then, we have theequality [8]

f (x) +

(a+ b

2− x

)f ′ (x)− 1

b− a

∫ b

a

f (t) dt(3.1)

=1

b− a

∫ x

a

(t− a) [f ′ (t)− f ′ (x)] dt+1

b− a

∫ b

x

(t− b) [f ′ (t)− f ′ (x)] dt


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164 S. S. DRAGOMIR

for anyx ∈ [a, b] .We have the following result:

THEOREM 3.1 (Dragomir, 2013 [8]). Let f : I → C be a differentiable function onI and[a, b] ⊂ I . If the derivativef ′ : I → C is of bounded variation on[a, b] , then∣∣∣∣f (x) +

(a+ b

2− x

)f ′ (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.2)

≤ 1

b− a

[∫ x

a

(t− a)x∨t

(f ′) dt+

∫ b

x

(b− t)t∨x

(f ′) dt

]

≤ 1

2(b− a)

[(x− a

b− a

)2 x∨a

(f ′) dt+

(b− x

b− a

)2 b∨x

(f ′)

]

≤ 1

2(b− a)

×

[14

+(

x−a+b2

b−a

)2][

12

b∨a

(f ′) + 12

∣∣∣∣∣x∨a

(f ′)−b∨x

(f ′)

∣∣∣∣∣],

[(x−ab−a

)2p+(

b−xb−a

)2p]1/p

[[x∨a

(f ′)

]q

+

[b∨x

(f ′)

]q]1/q

p > 1, 1p

+ 1q

= 1,

[12

+∣∣∣x−a+b

2

b−a

∣∣∣] b∨a

(f ′) ,


PROOF. Taking the modulus in (3.1) we have∣∣∣∣f (x) +

(a+ b

2− x

)f ′ (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.3)

≤ 1

b− a

∣∣∣∣∫ x

a

(t− a) [f ′ (t)− f ′ (x)] dt

∣∣∣∣+

1

b− a

∣∣∣∣∫ b

x

(t− b) [f ′ (t)− f ′ (x)] dt

∣∣∣∣≤ 1

b− a

∫ x

a

(t− a) |f ′ (t)− f ′ (x)| dt

+1

b− a

∫ b

x

(b− t) |f ′ (t)− f ′ (x)| dt.

Since the derivativef ′ : I → C is of bounded variation on[a, x] and[x, b] , then

|f ′ (t)− f ′ (x)| ≤x∨t

(f ′) for t ∈ [a, x]


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and

|f ′ (t)− f ′ (x)| ≤t∨x

(f ′) for t ∈ [x, b] .

Therefore ∫ x

a

(t− a) |f ′ (t)− f ′ (x)| dt ≤∫ x

a

(t− a)x∨t

(f ′) dt

≤ 1

2(x− a)2

x∨a

(f ′) dt

and ∫ b

x

(b− t) |f ′ (t)− f ′ (x)| dt ≤∫ b

x

(b− t)t∨x

(f ′) dt

≤ 1

2(b− x)2

b∨x

(f ′) ,

which, by (3.3) produce the first two inequalities in (3.2).The last part follows by Hölder’s inequality


,


+ 1β

= 1.

COROLLARY 3.2. With the assumptions of Theorem 3.1, we have∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.4)

≤ 1

b− a

∫ a+b2

a

(t− a)

a+b2∨t

(f ′) dt+

∫ b

a+b2

(b− t)t∨

a+b2

(f ′) dt

≤ 1

8(b− a)

b∨a

(f ′) dt.

REMARK 3.1. If p ∈ (a, b) is a median point in bounded variation for the derivative, i.e.p∨a

(f ′) =b∨p

(f ′) , then under the assumptions of Theorem 3.1 we have

∣∣∣∣f (p) +

(a+ b

2− p

)f ′ (p)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.5)

≤ 1

b− a

[∫ p

a

(t− a)

p∨t

(f ′) dt+

∫ b

p

(b− t)t∨p

(f ′) dt

]

≤ 1

4(b− a)

1

4+

(p− a+b

2

b− a

)2 b∨

a

(f ′) .


http://ajmaa.org

166 S. S. DRAGOMIR

3.2. Inequalities for Lipschitzian Derivatives. We start with the following result.

THEOREM 3.3 (Dragomir, 2013 [8]). Let f : I → C be a differentiable function onI and[a, b] ⊂ I . Letx ∈ (a, b) . If αi > −1 andLαi

> 0 with i = 1, 2 are such that

(3.6) |f ′ (t)− f ′ (x)| ≤ Lα1 (x− t)α1 for anyt ∈ [a, x)

and

(3.7) |f ′ (t)− f ′ (x)| ≤ Lα2 (t− x)α2 for anyt ∈ (x, b],

then we have∣∣∣∣f (x) +

(a+ b

2− x

)f ′ (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.8)

≤ 1

b− a

[Lα1

(α1 + 1) (α1 + 2)(x− a)α1+2 +

Lα2

(α2 + 1) (α2 + 2)(b− x)α2+2

].

PROOF. Taking the modulus in (3.1) we have∣∣∣∣f (x) +

(a+ b

2− x

)f ′ (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.9)

≤ 1

b− a

∫ x

a

(t− a) |f ′ (t)− f ′ (x)| dt

+1

b− a

∫ b

x

(b− t) |f ′ (t)− f ′ (x)| dt.

Using the properties (3.6) and (3.7) we have∫ x

a

(t− a) |f ′ (t)− f ′ (x)| dt ≤ Lα1

∫ x

a

(t− a) (x− t)α1 dt

= Lα1 (x− a)α1+2

∫ 1

0

u (1− u)α1 du

= Lα1 (x− a)α1+2

∫ 1

0

uα1 (1− u) du

=1

(α1 + 1) (α1 + 2)Lα1 (x− a)α1+2

and ∫ b

x

(b− t) |f ′ (t)− f ′ (x)| dt ≤ Lα2

∫ b

x

(b− t) (t− x)α2 dt

=1

(α2 + 1) (α2 + 2)Lα2 (b− x)α2+2 .

Utilising (3.9) we get the desired result (3.8).

COROLLARY 3.4. Let f : I → C be a differentiable function onI and [a, b] ⊂ I . If thederivative isf ′ of r-H-Hölder type on[a, b] , i.e. we have the condition

|f ′ (t)− f ′ (s)| ≤ H |t− s|r


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for anyt, s ∈ [a, b] , wherer ∈ (0, 1] andH > 0 are given, then∣∣∣∣f (x) +

(a+ b

2− x

)f ′ (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.10)

≤ H

(r + 1) (r + 2)

[(x− a

b− a

)r+2

+

(b− x

b− a

)r+2]

(b− a)r+1 ,

for anyx ∈ [a, b] .In particular, if f ′ is Lipschitzian with the constantL > 0, then∣∣∣∣f (x) +

(a+ b

2− x

)f ′ (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.11)

≤ 1

6L

[(x− a

b− a

)3

+

(b− x

b− a

)3]

(b− a)2 ,


3.3. Inequalities for Differentiable Convex Functions.The case of convex functions isas follows:

THEOREM 3.5 (Dragomir, 2013 [8]). Let f : I → C be a differentiable convex function onI and[a, b] ⊂ I . Then for anyx ∈ [a, b] we have

(3.12) 0 ≤ 1

b− a

∫ b

a

f (t) dt− f (x)−(a+ b

2− x

)f ′ (x) ≤

I1 (x)

I2 (x)

I3 (x)

where

I1 (x) :=(b− x) f (b) + (x− a) f (a)

b− a− f (x)− 2f ′ (x)

(a+ b

2− x

),

I2 (x) :=1

2

f ′ (b) (b− x)2 − f ′ (a) (x− a)2

b− a− f ′ (x)

(a+ b

2− x

)and

I3 (x) :=1

2

[f (b) (b− x) + f (a) (x− a)

b− a− f (x)

]− f ′ (x)

(a+ b

2− x

)PROOF. We have the equality

1

b− a

∫ b

a

f (t) dt− f (x)−(a+ b

2− x

)f ′ (x)(3.13)

=1

b− a

∫ x

a

(t− a) [f ′ (x)− f ′ (t)] dt+1

b− a

∫ b

x

(b− t) [f ′ (t)− f ′ (x)] dt


Sincef is a differentiable convex function onI, thenf ′ is monotonic nondecreasing onIand then ∫ x

a

(t− a) [f ′ (x)− f ′ (t)] dt ≥ 0


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168 S. S. DRAGOMIR

and ∫ b

x

(b− t) [f ′ (t)− f ′ (x)] dt ≥ 0,

which proves the first inequality in (3.12).We have∫ x

a

(t− a) [f ′ (x)− f ′ (t)] dt ≤ (x− a)

∫ x

a

[f ′ (x)− f ′ (t)] dt

= (x− a) [f ′ (x) (x− a)− f (x) + f (a)]

and ∫ b

x

(b− t) [f ′ (t)− f ′ (x)] dt ≤ (b− x)

∫ b

x

[f ′ (t)− f ′ (x)] dt

= (b− x) [f (b)− f (x)− f ′ (x) (b− x)] .

Adding these inequalities we get∫ x

a

(t− a) [f ′ (x)− f ′ (t)] dt+

∫ b

x

(b− t) [f ′ (t)− f ′ (x)] dt

≤ (x− a) [f ′ (x) (x− a)− f (x) + f (a)]

+ (b− x) [f (b)− f (x)− f ′ (x) (b− x)]

= (b− x) f (b) + (x− a) f (a)− (b− a) f (x)

+ f ′ (x) [2x− (a+ b)] (b− a)

and by (3.13) we get the second inequality forI1 (x) .We also have∫ x

a

(t− a) [f ′ (x)− f ′ (t)] dt ≤∫ x

a

(t− a) [f ′ (x)− f ′ (a)] dt

=1

2[f ′ (x)− f ′ (a)] (x− a)2

and ∫ b

x

(b− t) [f ′ (t)− f ′ (x)] dt ≤∫ b

x

(b− t) [f ′ (b)− f ′ (x)] dt

=1

2[f ′ (b)− f ′ (x)] (b− x)2 .

Adding these inequalities we get∫ x

a

(t− a) [f ′ (x)− f ′ (t)] dt+

∫ b

x

(b− t) [f ′ (t)− f ′ (x)] dt

≤ 1

2[f ′ (x)− f ′ (a)] (x− a)2 +

1

2[f ′ (b)− f ′ (x)] (b− x)2

=1

2

[f ′ (b) (b− x)2 − f ′ (a) (x− a)2 + f ′ (x) (b− a) [2x− (a+ b)]

]and by (3.13) we get the second inequality forI2 (x) .

Further, we use theCebyšev inequality for asynchronous functions (functions of oppositemonotonicity), namely

1

d− c

∫ d

c

g (t)h (t) dt ≤ 1

d− c

∫ d

c

g (t) dt · 1

d− c

∫ d

c

h (t) dt.


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Therefore

1

x− a

∫ x

a

(t− a) [f ′ (x)− f ′ (t)] dt

≤ 1

x− a

∫ x

a

(t− a) dt · 1

x− a

∫ x

a

[f ′ (x)− f ′ (t)] dt

=(x− a)2

2 (x− a)· f

′ (x) (x− a)− f (x) + f (a)

x− a

=1

2[f ′ (x) (x− a)− f (x) + f (a)]

and

1

b− x

∫ b

x

(b− t) [f ′ (t)− f ′ (x)] dt

≤ 1

b− x

∫ b

x

(b− t) dt · 1

b− x

∫ b

x

[f ′ (t)− f ′ (x)] dt

=(b− x)2

2 (b− x)· f (b)− f (x)− f ′ (x) (b− x)

b− x

=1

2[f (b)− f (x)− f ′ (x) (b− x)] .

Adding these inequalities, we have

1

b− a

∫ x

a

(t− a) [f ′ (x)− f ′ (t)] dt+1

b− a

∫ b

x

(b− t) [f ′ (t)− f ′ (x)] dt

≤ 1

2

[f ′ (x) (x− a)− f (x) + f (a)] (x− a)

b− a

+1

2

[f (b)− f (x)− f ′ (x) (b− x)] (b− x)

b− a

=1

2 (b− a)[[f ′ (x) (x− a)− f (x) + f (a)] (x− a)]

+1

2 (b− a)[[f (b)− f (x)− f ′ (x) (b− x)] (b− x)]

=1

2

[f (b) (b− x) + f (a) (x− a)

b− a− f (x)

]+ f ′ (x)

(x− a+ b

2

)which proves the inequality forI3 (x) .

REMARK 3.2. From the first inequality in (3.12) we have

(3.14)1

b− a

∫ b

a

f (t) dt ≤ (b− x) f (b) + (x− a) f (a)

b− a− f ′ (x)

(a+ b

2− x


From the second inequality in (3.12) we have

(3.15)1

b− a

∫ b

a

f (t) dt− f (x) ≤ 1

2· f

′ (b) (b− x)2 − f ′ (a) (x− a)2

b− a



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170 S. S. DRAGOMIR

From the third inequality in (3.12) we have

(3.16)1

b− a

∫ b

a

f (t) dt ≤ 1

2

[f (b) (b− x) + f (a) (x− a)

b− a+ f (x)

]for anyx ∈ [a, b] .

3.4. Inequalities for Absolutely Continuous Derivatives.We use theLebesguep-normsdefined as follows:

‖g‖[c,d],p :=

(∫ d

c

|g (s)|p dt)1/p

, g ∈ Lp [c, d] , p ≥ 1

and‖g‖[c,d],∞ := ess sup

s∈[c,d]

|g (s)| , g ∈ L∞ [c, d] .

The case of absolutely continuous derivatives is as follows:

THEOREM 3.6 (Dragomir, 2013 [8]). Let f : I → C be a differentiable function onI and[a, b] ⊂ I . If the derivativef ′ is absolutely continuous on[a, b] , then for anyx ∈ [a, b]

(3.17)

∣∣∣∣f (x) +

(a+ b

2− x

)f ′ (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣

≤ 1

b− a×

16(x− a)3 ‖f ′′‖[a,x],∞ ,

q(q+1)(q+2)

(x− a)1/q+2 ‖f ′′‖[a,x],p ,

12(x− a)2 ‖f ′′‖[a,x],1 ,

+1

b− a×

16(b− x)3 ‖f ′′‖[x,b],∞ ,

q(q+1)(q+2)

(b− x)1/q+2 ‖f ′′‖[x,b],p ,

12(b− x)2 ‖f ′′‖[x,b],1 ,

wherep > 1, 1p

+ 1q

= 1.

PROOF. Taking the modulus in (3.1) we have∣∣∣∣f (x)− 1

b− a

∫ b

a

f (t) dt+

(a+ b

2− x

)f ′ (x)

∣∣∣∣(3.18)

≤ 1

b− a

∫ x

a

(t− a) |f ′ (t)− f ′ (x)| dt+1

b− a

∫ b

x

(b− t) |f ′ (t)− f ′ (x)| dt

=1

b− a

∫ x

a

(t− a)

∣∣∣∣∫ t

x

f ′′ (s) ds

∣∣∣∣+ 1

b− a

∫ b

x

(b− t)

∣∣∣∣∫ t

x

f ′′ (s) ds

∣∣∣∣≤ 1

b− a

∫ x

a

(t− a)

∫ x

t

|f ′′ (s)| ds+1

b− a

∫ b

x

(b− t)

∫ t

x

|f ′′ (s)| ds.


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Using Hölder’s integral inequality we have forp > 1, 1p

+ 1q

= 1,

∫ x

a

(t− a)

∫ x

t

|f ′′ (s)| ds ≤

∫ x

a(t− a) (x− t) ‖f ′′‖[t,x],∞ dt∫ x

a(t− a) (x− t)1/q ‖f ′′‖[t,x],p dt∫ x

a(t− a) ‖f ′′‖[t,x],1 dt

≤

‖f ′′‖[a,x],∞

∫ x

a(t− a) (x− t) dt

‖f ′′‖[a,x],p

∫ x

a(t− a) (x− t)1/q dt

‖f ′′‖[a,x],1

∫ x

a(t− a) dt

=

16(x− a)3 ‖f ′′‖[a,x],∞

q(q+1)(q+2)

(x− a)1/q+2 ‖f ′′‖[a,x],p

12(x− a)2 ‖f ′′‖[a,x],1

and, similarly

∫ b

x

(b− t)

∫ t

x

|f ′′ (s)| ds ≤

16(b− x)3 ‖f ′′‖[x,b],∞

q(q+1)(q+2)

(b− x)1/q+2 ‖f ′′‖[x,b],p

12(b− x)2 ‖f ′′‖[x,b],1 .

Utilizing the inequality (3.18) we get the desired result (3.17).

REMARK 3.3. Since

1

6(x− a)3 ‖f ′′‖[a,x],∞ +

1

6(b− x)3 ‖f ′′‖[x,b],∞

≤ 1

6

[(x− a)3 + (b− x)3]max

‖f ′′‖[a,x],∞ , ‖f ′′‖[x,b],∞

=

1

6(b− a)

[(x− a)2 − (x− a) (b− x) + (b− x)2] ‖f ′′‖[a,b],∞ ,

then by (3.17) we get∣∣∣∣f (x) +

(a+ b

2− x

)f ′ (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.19)

≤ 1

6

[(x− a

b− a

)2

−(x− a

b− a

)(b− x

b− a

)+

(b− x

b− a

)2]

× (b− a)2 ‖f ′′‖[a,b],∞ ,



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172 S. S. DRAGOMIR

Since

(x− a)1/q+2 ‖f ′′‖[a,x],p + (b− x)1/q+2 ‖f ′′‖[x,b],p

≤[(x− a)2q+1 + (b− x)2q+1]1/q

[‖f ′′‖p

[a,x],p + ‖f ′′‖[x,b],p

]1/p

=[(x− a)2q+1 + (b− x)2q+1]1/q ‖f ′′‖[a,b],p ,


(a+ b

2− x

)f ′ (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.20)

≤ q

(q + 1) (q + 2)

[(x− a

b− a

)2q+1

+

(b− x

b− a

)2q+1]1/q

× (b− a)1+1/q ‖f ′′‖[a,b],p ,

for anyx ∈ [a, b] .Since

(x− a)2 ‖f ′′‖[a,x],1 + (b− x)2 ‖f ′′‖[x,b],1

≤ max(x− a)2 , (b− x)2 [‖f ′′‖[a,x],1 + ‖f ′′‖[x,b],1

]=

[1

2(b− a) +

∣∣∣∣x− a+ b

2

∣∣∣∣]2

‖f ′′‖[a,b],1 ,


(a+ b

2− x

)f ′ (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

2

[1

2+

∣∣∣∣∣x− a+b2

b− a

∣∣∣∣∣]2

(b− a) ‖f ′′‖[a,b],1


4. MORE PERTURBED OSTROWSKI TYPE I NEQUALITIES FOR ABSOLUTELY


4.1. Inequalities for Derivatives of Bounded Variation. Assume that the functionf :I → C is differentiable on the interior ofI, denotedI, and [a, b] ⊂ I . Then, we have theequality

f (x) +1

2 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]− 1

b− a

∫ b

a

f (t) dt(4.1)

=1

b− a

∫ x

a

(t− a) [f ′ (t)− f ′ (a)] dt+1

b− a

∫ b

x

(b− t) [f ′ (b)− f ′ (t)] dt,



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In particular, forx = a+b2, we have

f

(a+ b

2

)+

1

8(b− a) [f ′ (b)− f ′ (a)]− 1

b− a

∫ b

a

f (t) dt(4.2)

=1

b− a

∫ a+b2

a

(t− a) [f ′ (t)− f ′ (a)] dt

+1

b− a

∫ b

a+b2

(b− t) [f ′ (b)− f ′ (t)] dt.

THEOREM 4.1. Let f : I → C be a differentiable function onI and [a, b] ⊂ I . If thederivativef ′ : I → C is of bounded variation on[a, b] , then for anyx ∈ [a, b]∣∣∣∣f (x) +

1

2 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.3)

≤ 1

b− a

[∫ x

a

(t− a)t∨a

(f ′) dt+

∫ b

x

(b− t)b∨t

(f ′) dt

]

≤ 1

b− a

12(x− a)2

x∨a

(f ′) ,

1

(q+1)1/q (x− a)1+1/q

(∫ x

a

(t∨a

(f ′)

)p

dt

)1/p

,

(x− a)∫ x

a

(t∨a

(f ′)

)dt

+1

b− a

12(b− x)2

b∨x

(f ′) ,

1

(q+1)1/q (b− x)1+1/q

(∫ b

x

(b∨t

(f ′)

)p

dt

)1/p

,

(b− x)∫ b

x

(b∨t

(f ′)

)dt.

wherep > 1, 1p

+ 1q

= 1.

PROOF. Taking the modulus in (4.1) we have∣∣∣∣f (x) +1

2 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.4)

≤ 1

b− a

∫ x

a

(t− a) |f ′ (t)− f ′ (a)| dt+1

b− a

∫ b

x

(b− t) |f ′ (b)− f ′ (t)| dt,



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174 S. S. DRAGOMIR

Since the derivativef ′ : I → C is of bounded variation on[a, b] , then

|f ′ (t)− f ′ (a)| ≤t∨a

(f ′) for anyt ∈ [a, x]

and

|f ′ (b)− f ′ (t)| ≤b∨t

(f ′) for anyt ∈ [x, b] .

Therefore ∫ x

a

(t− a) |f ′ (t)− f ′ (a)| dt ≤∫ x

a

(t− a)t∨a

(f ′) dt

and

∫ b

x

(b− t) |f ′ (b)− f ′ (t)| dt ≤∫ b

x

(b− t)b∨t

(f ′) dt

for anyx ∈ [a, b] .Adding these two inequalities and dividing byb− a we get the first inequality in (4.3).Using Hölder’s integral inequality we have

∫ x

a

(t− a)t∨a

(f ′) dt ≤

x∨a

(f ′)∫ x

a(t− a) dt,

(∫ x

a(t− a)q dt

)1/q

(∫ x

a

(t∨a

(f ′)

)p

dt

)1/p

,

(x− a)∫ x

a

(t∨a

(f ′)

)dt,

=

12(x− a)2

x∨a

(f ′) ,

1

(q+1)1/q (x− a)1+1/q

(∫ x

a

(t∨a

(f ′)

)p

dt

)1/p

,

(x− a)∫ x

a

(t∨a

(f ′)

)dt


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and

∫ b

x

(b− t)b∨t

(f ′) dt ≤

12(b− x)2

b∨x

(f ′) ,

1

(q+1)1/q (b− x)1+1/q

(∫ b

x

(b∨x

(f ′)

)p

dt

)1/p

,

(b− x)∫ b

x

(b∨x

(f ′)

)dt.

REMARK 4.1. From the first branch in (4.3) we have the sequence of inequalities

∣∣∣∣f (x) +1

2 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.5)

≤ 1

b− a

[∫ x

a

(t− a)t∨a

(f ′) dt+

∫ b

x

(b− t)b∨t

(f ′) dt

]

≤ 1

2(b− a)

[(x− a

b− a

)2 x∨a

(f ′) +

(b− x

b− a

)2 b∨x

(f ′)

]

≤ 1

2(b− a)

×

[14

+(

x−a+b2

b−a

)2][

12

b∨a

(f ′) + 12

∣∣∣∣∣x∨a

(f ′)−b∨x

(f ′)

∣∣∣∣∣],

[(x−ab−a

)2p+(

b−xb−a

)2p]1/p

[[x∨a

(f ′)

]q

+

[b∨x

(f ′)

]q]1/q

p > 1, 1p

+ 1q

= 1,

[12

+∣∣∣x−a+b

2

b−a

∣∣∣] b∨a

(f ′) ,



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176 S. S. DRAGOMIR

From the second branch in (4.3) we have∣∣∣∣f (x) +1

2 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.6)

≤ 1

b− a

[∫ x

a

(t− a)t∨a

(f ′) dt+

∫ b

x

(b− t)b∨t

(f ′) dt

]

≤ 1

(q + 1)1/q

(x− a

b− a

)1+1/q(∫ x

a

(t∨a

(f ′)

)p

dt

)1/p

+

(b− x

b− a

)1+1/q(∫ b

x

(b∨t

(f ′)

)p

dt

)1/p (b− a)1/q

≤ 1

(q + 1)1/q

[(x− a

b− a

)q+1

+

(b− x

b− a

)q+1]1/p

×

[∫ x

a

(t∨a

(f ′)

)p

dt+

∫ b

x

(b∨t

(f ′)

)p

dt

]1/p

(b− a)1/q

≤ 1

(q + 1)1/q

[(x− a

b− a

)q+1

+

(b− x

b− a

)q+1]1/p

×

[(x− a)

(x∨a

(f ′)

)p

+ (b− x)

(b∨x

(f ′)

)p]1/p

(b− a)1/q

for anyx ∈ [a, b] andp > 1, 1p

+ 1q

= 1.From the third branch in (4.3) we have∣∣∣∣f (x) +

1

2 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.7)

≤ 1

b− a

[∫ x

a

(t− a)t∨a

(f ′) dt+

∫ b

x

(b− t)b∨t

(f ′) dt

]

≤(x− a

b− a

)∫ x

a

(t∨a

(f ′)

)dt+

(b− x

b− a

)∫ b

x

(b∨t

(f ′)

)dt

≤

[12

+∣∣∣x−a+b

2

b−a

∣∣∣] [∫ x

a

(t∨a

(f ′)

)dt+

∫ b

x

(b∨t

(f ′)

)dt

][(

x−ab−a

)q+(

b−xb−a

)q]1/q

×

[[∫ x

a

(t∨a

(f ′)

)dt

]p

+

[∫ b

x

(b∨t

(f ′)

)dt

]p]1/p

max

∫ x

a

(t∨a

(f ′)

)dt,∫ b

x

(b∨t

(f ′)

)dt


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for anyx ∈ [a, b] andp > 1, 1p

+ 1q

= 1.

REMARK 4.2. We observe that, if we takex = a+b2

in (4.5) then we get the perturbedmidpoint inequality∣∣∣∣f (a+ b

2

)+

1

8(b− a) [f ′ (b)− f ′ (a)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.8)

≤ 1

b− a

[∫ a+b2

a

(t− a)t∨a

(f ′) dt+

∫ b

a+b2

(b− t)b∨t

(f ′) dt

]

≤ 1

8(b− a)

b∨a

(f ′) .

4.2. Inequalities for Lipschitzian Derivatives. We start with the following result.

THEOREM 4.2. Let f : I → C be a differentiable function onI and [a, b] ⊂ I . Let x ∈(a, b) . If αi > −1 andLαi

> 0 with i = 1, 2 are such that

(4.9) |f ′ (t)− f ′ (a)| ≤ Lα1 (t− a)α1 for anyt ∈ [a, x)

and

(4.10) |f ′ (b)− f ′ (t)| ≤ Lα2 (b− t)α2 for anyt ∈ (x, b],

then we have∣∣∣∣f (x) +1

2 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.11)

≤ 1

b− a

[Lα1

α1 + 2(x− a)α1+2 +

Lα2

α2 + 2(b− x)α2+2

].

PROOF. Using the conditions (4.9) and (4.10) we have∣∣∣∣f (x) +1

2 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

b− a

∫ x

a

(t− a) |f ′ (t)− f ′ (a)| dt+1

b− a

∫ b

x

(b− t) |f ′ (b)− f ′ (t)| dt

≤ 1

b− aLα1

∫ x

a

(t− a)α1+1 dt+1

b− aLα2

∫ b

x

(b− t)α2+1 dt

=1

b− aLα1

(x− a)α1+2

α1 + 2+

1

b− aLα2

(b− x)α2+2

α2 + 2

=1

b− a

[Lα1

α1 + 2(x− a)α1+2 +

Lα2

α2 + 2(b− x)α2+2

]and the inequality (4.11) is obtained.

COROLLARY 4.3. Let f : I → C be a differentiable function onI and [a, b] ⊂ I . If thederivative isf ′ of r-H-Hölder type on[a, b] , i.e. we have the condition

|f ′ (t)− f ′ (s)| ≤ H |t− s|r


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178 S. S. DRAGOMIR

for anyt, s ∈ [a, b] , wherer ∈ (0, 1] andH > 0 are given, then∣∣∣∣f (x) +1

2 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.12)

≤ H

r + 2

[(x− a

b− a

)r+2

+

(b− x

b− a

)r+2]

(b− a)r+1 ,

for anyx ∈ [a, b] .In particular, if f ′ is Lipschitzian with the constantL > 0, then∣∣∣∣f (x) +

1

2 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.13)

≤ 1

3L

[(x− a

b− a

)3

+

(b− x

b− a

)3]

(b− a)2 ,


REMARK 4.3. With the assumptions of Corollary 4.3 we have the midpoint inequality∣∣∣∣f (a+ b

2

)+

1

8(b− a) [f ′ (b)− f ′ (a)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.14)

≤ H

2r+1 (r + 2)(b− a)r+1 .

If if f ′ is Lipschitzian with the constantL > 0, then∣∣∣∣f (a+ b

2

)+

1

8(b− a) [f ′ (b)− f ′ (a)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.15)

≤ 1

12L (b− a)2 .

4.3. Inequalities for Differentiable Functions with the Property (S). Let f : I → C bea differentiable convex function onI and[a, b] ⊂ I . Thenf ′ is monotonic nondecreasing andby the equality (4.1) we have

(4.16) f (x) +1

2 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]− 1

b− a

∫ b

a

f (t) dt ≥ 0

or, equivalently

(4.17)1

2 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]≥ 1

b− a

∫ b

a

f (t) dt− f (x)

for anyx ∈ [a, b] .We observe that the inequalities (4.16) and (4.17) remain valid for the larger class of differ-

entiable functionsf that that satisfy theproperty(S) on the interval[a, b] , namely

(S) f ′ (a) ≤ f ′ (t) ≤ f ′ (b)

for anyt ∈ [a, b] .

THEOREM 4.4. Letf : I → C be a differentiable function onI and [a, b] ⊂ I .(i) Let x ∈ [a, b] . If f satisfies the property(S) on the interval[a, x] and [x, b] , then

(4.18) f ′ (x)

(a+ b

2− x

)≤ 1

b− a

∫ b

a

f (t) dt− f (x) .


http://ajmaa.org


(ii) If f satisfies the property (S) on the interval[a, b] , then for anyx ∈ [a, b]

f (a) (x− a) + f (b) (b− x)

b− a− 1

b− a

∫ b

a

f (t) dt(4.19)

≤ 1

2 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

].

PROOF. (i) Sincef satisfies the property(S) on the interval[a, x] and[x, b] , then

f (x) +1

2 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]− 1

b− a

∫ b

a

f (t) dt

=1

b− a

∫ x

a

(t− a) [f ′ (t)− f ′ (a)] dt+1

b− a

∫ b

x

(b− t) [f ′ (b)− f ′ (t)] dt

≤ 1

b− a

∫ x

a

(t− a) [f ′ (x)− f ′ (a)] dt+1

b− a

∫ b

x

(b− t) [f ′ (b)− f ′ (x)] dt

=f ′ (x)− f ′ (a)

b− a

∫ x

a

(t− a) dt+f ′ (b)− f ′ (x)

b− a

∫ b

x

(b− t) dt

=f ′ (x)− f ′ (a)

b− a· (x− a)2

2+f ′ (b)− f ′ (x)

b− a· (b− x)2

2

=1

2 (b− a)

[(f ′ (x)− f ′ (a)) (x− a)2 + (f ′ (b)− f ′ (x)) (b− x)2]

=1

2 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]− f ′ (x)

(a+ b

2− x

),

which proves the inequality (4.18).(ii) If f satisfies the property(S) on the interval[a, b] , then for anyx ∈ [a, b]

1

b− a

∫ x

a

(t− a) [f ′ (t)− f ′ (a)] dt+1

b− a

∫ b

x

(b− t) [f ′ (b)− f ′ (t)] dt

≤ x− a

b− a

∫ x

a

[f ′ (t)− f ′ (a)] dt+b− x

b− a

∫ b

x

[f ′ (b)− f ′ (t)] dt

=1

b− a(x− a) [f (x)− f (a)− f ′ (a) (x− a)]

+1

b− a(b− x) [f ′ (b) (b− x)− f (b) + f (x)]

=1

b− a

[f (x) (x− a)− f (a) (x− a)− f ′ (a) (x− a)2]

+1

b− a

[f ′ (b) (b− x)2 − f (b) (b− x) + f (x) (b− x)

]=

1

b− a

f ′ (b) (b− x)2 − f ′ (a) (x− a)2 − f (a) (x− a)− f (b) (b− x)

+f (x) (b− a)

=f ′ (b) (b− x)2 − f ′ (a) (x− a)2

b− a+ f (x)− f (a) (x− a) + f (b) (b− x)

b− a,

which proves the inequality (4.19).

REMARK 4.4. The inequality (4.18) was obtained for the case of convex functions in [3]while (4.19) was established for convex functions in [4] with different proofs.


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180 S. S. DRAGOMIR

Further, we use theCebyšev inequality for synchronous functions (functions with samemonotonicity), namely

(4.20)1

d− c

∫ d

c

g (t)h (t) dt ≥ 1

d− c

∫ d

c

g (t) dt · 1

d− c

∫ d

c

h (t) dt.

THEOREM 4.5. Let f : I → C be a differentiable function onI and [a, b] ⊂ I . Let x ∈[a, b] . If f is convex on the interval[a, x] and [x, b] , then

(4.21)1

2

[f (x) +

f (a) (x− a) + f (b) (b− x)

b− a

]≥ 1

b− a

∫ b

a

f (t) dt.

PROOF. We have

f (x) +1

2 (b− a)

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]− 1

b− a

∫ b

a

f (t) dt(4.22)

=1

b− a

∫ x

a

(t− a) [f ′ (t)− f ′ (a)] dt+1

b− a

∫ b

x

(b− t) [f ′ (b)− f ′ (t)] dt

for anyx ∈ [a, b] .Sincef ′ is monotonic nondecreasing on[a, x] , then byCebyšev inequality (4.12) we have∫ x

a

(t− a) [f ′ (t)− f ′ (a)] dt ≥ 1

x− a

∫ x

a

(t− a) dt ·∫ x

a

[f ′ (t)− f ′ (a)] dt

=1

2(x− a) [f (x)− f (a)− f ′ (a) (x− a)]

=1

2

[f (x) (x− a)− f (a) (x− a)− f ′ (a) (x− a)2]

and, by the same inequality,∫ b

x

(b− t) [f ′ (b)− f ′ (t)] dt ≥ 1

b− x

∫ b

x

(b− t) dt ·∫ b

x

[f ′ (b)− f ′ (t)] dt

=1

2(b− x) [f ′ (b) (b− x)− f (b) + f (x)]

=1

2

[f ′ (b) (b− x)2 − f (b) (b− x) + f (x) (b− x)

].

If we add these two inequalities, then we get∫ x

a

(t− a) [f ′ (t)− f ′ (a)] dt+

∫ b

x

(b− t) [f ′ (b)− f ′ (t)] dt

≥ 1

2

[f (x) (x− a)− f (a) (x− a)− f ′ (a) (x− a)2]

+1

2

[f ′ (b) (b− x)2 − f (b) (b− x) + f (x) (b− x)

]=

1

2

[(b− x)2 f ′ (b)− (x− a)2 f ′ (a)

]+

1

2f (x) (b− a)

− 1

2[f (a) (x− a) + f (b) (b− x)] .

Dividing by b− a and utilizing the equality (4.22) we deduce the inequality (4.21).

REMARK 4.5. If the function is convex on the whole interval[a, b], then the inequality (4.21)is true for anyx ∈ [a, b] .


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REFERENCES

[1] T. M. Apostol,Mathematical Analysis,Second Edition, Addison-Wesley Publishing Company, 1975

[2] S. S. Dragomir, On the Ostrowski’s integral inequality for mappings with bounded variation andapplications,Math. Ineq. & Appl., 4(1) (2001), 33-40. Preprint,RGMIA Res. Rep. Coll.2 (1999), No.1, Article 7. [http://rgmia.org/papers/v2n1/v2n1-7.pdf].

[3] S. S. Dragomir, An inequality improving the first Hermite-Hadamard inequality for convex functionsdefined on linear spaces and applications for semi-inner products.J. Inequal. Pure Appl. Math.3(2002), no. 2, Article 31, 8 pp.

[4] S. S. Dragomir, An inequality improving the second Hermite-Hadamard inequality for convex func-tions defined on linear spaces and applications for semi-inner products.J. Inequal. Pure Appl. Math.3 (2002), no. 3, Article 35, 8 pp.

[5] S. S. Dragomir, Improvements of Ostrowski and generalised trapezoid inequality in terms of theupper and lower bounds of the first derivative.Tamkang J. Math.34 (2003), no. 3, 213–222.

[6] S. S. Dragomir, Refinements of the generalised trapezoid and Ostrowski inequalities for functions ofbounded variation.Arch. Math.(Basel)91 (2008), no. 5, 450–460.

[7] S. S. Dragomir, Some perturbed Ostrowski type inequalities for absolutely continuous functions(I), Acta Univ. M. Belii Ser. Math.2015, 1–16. PreprintRGMIA Res. Rep. Coll.16 (2013), Art. 94.[Online http://rgmia.org/papers/v16/v16a94.pdf].

[8] S. S. Dragomir, Some perturbed Ostrowski type inequalities for absolutely continuous functions (II),Acta Univ. Apulensis Math. Inform.No. 43 (2015), 209–228. PreprintRGMIA Res. Rep. Coll.16(2013), Art. 95.[Online http://rgmia.org/papers/v16/v16a95.pdf].

[9] S. S. Dragomir, Some perturbed Ostrowski type inequalities for absolutely continuous functions(III), Transylv. J. Math. Mech.7 (2015), no. 1, 31–43. PreprintRGMIA Res. Rep. Coll.16 (2013), Art.96. [Online http://rgmia.org/papers/v16/v16a96.pdf].

[10] S. S. Dragomir, Some perturbed Ostrowski type inequalities for functions of bounded variation,Asian-Eur. J. Math.8 (2015), no. 4, 1550069, 14 pp. PreprintRGMIA Res. Rep. Coll.16 (2013), Art.93. [Online http://rgmia.org/papers/v16/v16a93.pdf].


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CHAPTER 8

Companions of Ostrowski’s Inequality

1. A COMPANION OF OSTROWSKI I NEQUALITY FOR FUNCTIONS OF BOUNDED

VARIATION

1.1. Introduction. In [13], Guessab and Schmeisser have proved among others, the fol-lowing companion of Ostrowski’s inequality.

THEOREM 1.1 (Guessab & Schmeisser, 2002 [13]). Letf : [a, b] → R be such that

(1.1) |f (t)− f (s)| ≤ H |t− s|k , for anyt, s ∈ [a, b]

with k ∈ (0, 1], i.e.,f ∈ LipH (k) . Then, for eachx ∈[a, a+b

2

], we have the inequality∣∣∣∣f (x) + f (a+ b− x)

2− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.2)

≤

[2k+1 (x− a)k+1 + (a+ b− 2x)k+1

2k (k + 1) (b− a)

]H.

This inequality is sharp for each admissiblex. Equality is obtained if and only iff = ±Hf∗+cwith c ∈ R and

(1.3) f∗ (t) =

(x− t)k for a ≤ t ≤ x

(t− x)k for x ≤ t ≤ 12(a+ b)

f∗ (a+ b− t) for 12(a+ b) ≤ t ≤ b.

We remark that fork = 1, i.e.,f ∈ LipH , since

4 (x− a)2 + (a+ b− 2x)2

4 (b− a)=

1

8+ 2

(x− 3a+b

4

b− a

)2 (b− a)

then we have the inequality∣∣∣∣f (x) + f (a+ b− x)

2− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.4)

≤

1

8+ 2

(x− 3a+b

4

b− a

)2 (b− a)H

for anyx ∈[a, a+b

2

].

The inequality18

is best possible in (1.4) in the sense that it cannot be replaced by a smallerconstant.

183

184 S. S. DRAGOMIR

We must also observe that the best inequality in (1.4) is obtained forx = a+3b4, giving the

trapezoid type inequality

(1.5)

∣∣∣∣∣f(

3a+b4

)+ f

(a+3b

4

)2

− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣∣ ≤ 1

8(b− a)H.

The constant18

is sharp in (1.5) in the sense mentioned above.

1.2. Some Integral Inequalities.The following identity holds.

LEMMA 1.2 (Dragomir, 2002 [5]). Assume that the functionf : [a, b] → R is of boundedvariation on[a, b] . Then we have the equality

1

2[f (x) + f (a+ b− x)]− 1

b− a

∫ b

a

f (t) dt(1.6)

=1

b− a

[∫ x

a

(t− a) df (t) +

∫ a+b−x

x

(t− a+ b

2

)df (t) +

∫ b

a+b−x

(t− b) df (t)

]for anyx ∈

[a, a+b

2

].

PROOF. Obviously, all the Riemann-Stieltjes integrals from the right hand side of (1.6) existbecause the functions(· − a) ,

(· − a+b

2

)and(· − b) are continuous on those intervals andf is

of bounded variation.Using the integration by parts formula for Riemann-Stieltjes integrals, we have, for any

x ∈[a, a+b

2

], that ∫ x

a

(t− a) df (t) = f (x) (x− a)−∫ x

a

f (t) dt,

∫ a+b−x

x

(t− a+ b

2

)df (t)

= f (a+ b− x)

(a+ b

2− x

)− f (x)

(x− a+ b

2

)−∫ a+b−x

x

f (t) dt

and ∫ b

a+b−x

(t− b) df (t) = (x− a) f (a+ b− x)−∫ b

a+b−x

f (t) dt.

Summing the above equalities we deduce (1.6).

REMARK 1.1. A version of this identity for piecewise continuously differentiable functionshas been obtained in [13, Lemma 3.2].

The following companion of Ostrowski’s inequality holds.

THEOREM 1.3 (Dragomir, 2002 [5]). Assume that the functionf : [a, b] → R is of boundedvariation on[a, b] . Then we have the inequalities:∣∣∣∣12 [f (x) + f (a+ b− x)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(1.7)

≤ 1

b− a

[(x− a)

x∨a

(f) +

(a+ b

2− x

) a+b−x∨x

(f) + (x− a)b∨

a+b−x

(f)

]


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≤

[1

4+

∣∣∣∣∣x− 3a+b4

b− a

∣∣∣∣∣]

b∨a

(f)

[2

(x− a

b− a

)α

+

(a+b2− x

b− a

)α] 1α

×

[[x∨a

(f)

]β

+

[a+b−x∨

x

(f)

]β

+

[b∨

a+b−x

(f)

]β] 1

β

, if α > 1, 1α

+ 1β

= 1

(x+ b−3a

2

b− a

)max

x∨a

(f) ,a+b−x∨

x

(f) ,b∨

a+b−x

(f)

for anyx ∈

[a, a+b

2

], where

∨dc (f) denotes the total variation off on [c, d] . The constant1

4is

best possible in the first branch of the second inequality in (1.7).

PROOF. We use the fact that for a continuous functionp : [c, d] → R and a functionv :[a, b] → R of bounded variation, one has the inequality

(1.8)

∣∣∣∣∫ d

c

p (t) dv (t)

∣∣∣∣ ≤ supt∈[c,d]

|p (t)|d∨c

(v) .

Taking the modulus in (1.6) we have∣∣∣∣12 [f (x) + f (a+ b− x)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

b− a

[∣∣∣∣∫ x

a

(t− a) df (t)

∣∣∣∣+ ∣∣∣∣∫ a+b−x

x

(t− a+ b

2

)df (t)

∣∣∣∣+ ∣∣∣∣∫ b

a+b−x

(t− b) df (t)

∣∣∣∣]≤ 1

b− a

[(x− a)

x∨a

(f) +

(a+ b

2− x

) a+b−x∨x

(f) + (x− a)b∨

a+b−x

(f)

]=: M (x)

and the first inequality in (1.7) is obtained.Now, observe that

M (x) ≤ 1

b− amax

x− a,

a+ b

2− x

[ x∨a

(f) +a+b−x∨

x

(f) +b∨

a+b−x

(f)

]

=1

b− a

[1

4(b− a) +

∣∣∣∣x− 3a+ b

4

∣∣∣∣] b∨a

(f)

and the first branch in the second inequality in (1.7) is proved.Using Hölder’s discrete inequality we have (forα > 1, 1

α+ 1

β= 1) that

M (x) ≤ 1

b− a

[(x− a)α +

(a+ b

2− x

)α

+ (x− a)α

] 1α

×

[ x∨a

(f)

]β

+

[a+b−x∨

x

(f)

]β

+

[b∨

a+b−x

(f)

]β 1

β

giving the second branch in the second inequality.


http://ajmaa.org

186 S. S. DRAGOMIR

Finally, we have

M (x) ≤ 1

b− amax

x∨a

(f) ,a+b−x∨

x

(f) ,b∨

a+b−x

(f)

×[(x− a) +

(a+ b

2− x

)+ (x− a)

],

which is equivalent with the last inequality in (1.7).The sharpness of the constant1

4in the first branch of the second inequality in (1.7) will be

proved in a particular case later.

COROLLARY 1.4. With the assumptions in Theorem 1.3, one has the trapezoid inequality

(1.9)

∣∣∣∣f (a) + f (b)

2− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

2

b∨a

(f) .

The constant12


PROOF. Follows from the first inequality in (1.7) on choosingx = a.For the sharpness of the constant, assume that (1.9) holds with a constantA > 0, i.e.,

(1.10)

∣∣∣∣f (a) + f (b)

2− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ Ab∨a

(f) .

If we choosef : [a, b] → R with

f (x) =

1 if x = a,

0 if x ∈ (a, b) ,

1 if x = b,

thenf is of bounded variation on[a, b] and

f (a) + f (b)

2= 1,

∫ b

a

f (t) dt = 0, andb∨a

(f) = 2,

giving in (1.10)1 ≤ 2A, thusA ≥ 12

and the corollary is proved.

REMARK 1.2. The inequality (1.9) was first proved in a different manner in [2].

COROLLARY 1.5. With the assumptions in Theorem 1.3, one has the midpoint inequality

(1.11)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

2

b∨a

(f) .

The constant12


PROOF. Follows from the first inequality in (1.7) on choosingx = a+b2.

For the sharpness of the constant, assume that (1.11) holds with a constantB > 0, i.e.,

(1.12)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ Bb∨a

(f) .


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If we choosef : [a, b] → R with

f (x) =

0 if x ∈

[a, a+b

2

),

1 if x = a+b2,

0 if x ∈(

a+b2, b],

thenf is of bounded variation on[a, b] , and

f

(a+ b

2

)= 1,

∫ b

a

f (t) dt = 0, andb∨a

(f) = 2,

giving in (1.12),1 ≤ 2B, thusB ≥ 12.

REMARK 1.3. The inequality (1.11) was firstly proved in a different manner in [3].

The best inequality we may get from Theorem 1.3 on using the bound provided by the firstbranch in the second inequality in (1.7) is incorporated in the following corollary.

COROLLARY 1.6 (Dragomir, 2002 [5]). With the assumptions in Theorem 1.3 , one has theinequality:

(1.13)

∣∣∣∣∣f(

3a+b4

)+ f

(a+3b

4

)2

− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣∣ ≤ 1

4

b∨a

(f) .

The constant14

is best possible.

PROOF. Follows by Theorem 1.3 on choosingx = 3a+b4.

To prove the sharpness of the constant14, assume that (1.13) holds with a constantC > 0,

i.e.,

(1.14)

∣∣∣∣∣f(

3a+b4

)+ f

(a+3b

4

)2

− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣∣ ≤ Cb∨a

(f) .

Consider the functionf : [a, b] → R, given by

f (x) =

1 if x ∈

3a+b4, a+3b

4

,

0 if x ∈ [a, b] \

3a+b4, a+3b

4

.

Thenf is of bounded variation on[a, b] ,

f(

3a+b4

)+ f

(a+3b

4

)2

= 1,

∫ b

a

f (t) dt = 0

andb∨a

(f) = 4,

giving in (1.14)4C ≥ 1, thusC ≥ 14.

This example can be used to prove the sharpness of the constant14

in (1.7) as well.


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188 S. S. DRAGOMIR

2. A COMPANION OF OSTROWSKI I NEQUALITY FOR ABSOLUTELY CONTINUOUS

FUNCTIONS

2.1. An Identity. The following identity holds.

LEMMA 2.1 (Dragomir, 2002 [6]). Assume thatf : [a, b] → R is an absolutely continuousfunction on[a, b]. Then we have the equality

1

2[f (x) + f (a+ b− x)]− 1

b− a

∫ b

a

f (t) dt(2.1)

=1

b− a

∫ x

a

(t− a) f ′ (t) dt+1

b− a

∫ a+b−x

x

(t− a+ b

2

)f ′ (t) dt

+1

b− a

∫ b

a+b−x

(t− b) f ′ (t) dt,

for anyx ∈[a, a+b

2

].

PROOF. Using the integration by parts formula for Lebesgue integrals, we have∫ x

a

(t− a) f ′ (t) dt = f (x) (x− a)−∫ x

a

f (t) dt,

∫ a+b−x

x

(t− a+ b

2

)f ′ (t) dt

= f (a+ b− x)

(a+ b

2− x

)− f (x)

(x− a+ b

2

)−∫ a+b−x

x

f (t) dt

and ∫ b

a+b−x

(t− b) f ′ (t) dt = (x− a) f (a+ b− x)−∫ b

a+b−x

f (t) dt.

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

REMARK 2.1. The identity (2.1) was obtained in [13, Lemma 3.2] for the case of piecewisecontinuously differentiable functions on[a, b] .

2.2. The case of Sup-Norm.The following result holds.

THEOREM 2.2 (Dragomir, 2002 [6]). Let f : [a, b] → R be an absolutely continuous func-tion on[a, b] . Then we have the inequality∣∣∣∣12 [f (x) + f (a+ b− x)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.2)

≤ 1

b− a

[∫ x

a

(t− a) |f ′ (t)| dt+

∫ a+b−x

x

∣∣∣∣t− a+ b

2

∣∣∣∣ |f ′ (t)| dt+

∫ b

a+b−x

(b− t) |f ′ (t)| dt]

:= M (x)

for anyx ∈[a, a+b

2

].


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If f ′ ∈ L∞ [a, b] , then we have the inequalities

(2.3) M (x) ≤ 1

b− a

[(x− a)2

2‖f ′‖[a,x],∞ +

(a+ b

2− x

)2

‖f ′‖[x,a+b−x],∞

+(x− a)2

2‖f ′‖[a+b−x,b],∞

]

≤

1

8+ 2

(x− 3a+b

4

b− a

)2 (b− a) ‖f ′‖[a,b],∞

1

2α−1

(x− a

b− a

)2α

+

(x− a+b

2

b− a

)2α 1

α

×[‖f ′‖β

[a,x],∞ + ‖f ′‖β[x,a+b−x],∞ + ‖f ′‖β

[a+b−x,b],∞

] 1β

(b− a)

if α > 1, 1α

+ 1β

= 1,

max

1

2

(x− a

b− a

)2

,

(x− a+b

2

b− a

)2

×[‖f ′‖[a,x],∞ + ‖f ′‖[x,a+b−x],∞ + ‖f ′‖[a+b−x,b],∞

](b− a)

for anyx ∈[a, a+b

2

].

The inequality (2.2), the first inequality in (2.3) and the constant18

are sharp.

PROOF. The inequality (2.2) follows by Lemma 2.1 on taking the modulus and using itproperties.

If f ′ ∈ L∞ [a, b] , then∫ x

a

(t− a) |f ′ (t)| dt ≤ (x− a)2

2‖f ′‖[a,x],∞ ,

∫ a+b−x

x

∣∣∣∣t− a+ b

2

∣∣∣∣ |f ′ (t)| dt ≤ (a+ b

2− x

)2

‖f ′‖[x,a+b−x],∞ ,

∫ b

a+b−x

(b− t) |f ′ (t)| dt ≤ (x− a)2

2‖f ′‖[a+b−x,b],∞

and the first inequality in (2.3) is proved.Denote

M (x) :=(x− a)2

2‖f ′‖[a,x],∞ +

(a+ b

2− x

)2

‖f ′‖[x,a+b−x],∞

+(x− a)2

2‖f ′‖[a+b−x,b],∞

for x ∈[a, a+b

2

].


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190 S. S. DRAGOMIR

Firstly, observe that

M (x) ≤ max‖f ′‖[a,x],∞ , ‖f ′‖[x,a+b−x],∞ , ‖f ′‖[a+b−x,b],∞

×

[(x− a)2

2+

(a+ b

2− x

)2

+(x− a)2

2

]

= ‖f ′‖[a,b],∞

[1

8(b− a)2 + 2

(x− 3a+ b

4

)2]

and the first inequality in (2.3) is proved.Using Hölder’s inequality forα > 1, 1

α+ 1

β= 1, we also have

M (x) ≤

[(x− a)2

2

]α

+

(x− a+ b

2

)2α

+

[(x− a)2

2

]α 1α

×[‖f ′‖β

[a,x],∞ + ‖f ′‖β[x,a+b−x],∞ + ‖f ′‖β

[a+b−x,b],∞

] 1β

giving the second inequality in (2.3).Finally, we also observe that

M (x) ≤ max

(x− a)2

2,

(x− a+ b

2

)2

×[‖f ′‖[a,x],∞ + ‖f ′‖[x,a+b−x],∞ + ‖f ′‖[a+b−x,b],∞

].

The sharpness of the inequalities follows easily. The details are omitted.

REMARK 2.2. If in Theorem 2.2 we choosex = a, then we get

(2.4)

∣∣∣∣f (a) + f (b)

2− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

4(b− a) ‖f ′‖[a,b],∞

with 14

as a sharp constant (see for example [11, p. 25]).If in the same theorem we now choosex = a+b

2, then we get∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.5)

≤ 1

8(b− a)

[‖f ′‖[a, a+b

2 ],∞ + ‖f ′‖[a+b2

,b],∞

]≤ 1

4(b− a) ‖f ′‖[a,b],∞

with the constants18

and 14

being sharp. This result was obtained in [4].

It is natural to consider the following corollary.

COROLLARY 2.3. With the assumptions in Theorem 2.2, one has the inequality:

(2.6)

∣∣∣∣∣f(

3a+b4

)+ f

(a+3b

4

)2

− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣∣ ≤ 1

8(b− a) ‖f ′‖[a,b],∞ .

The constant18

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


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2.3. The Case ofp-Norm. The case whenf ′ ∈ Lp [a, b] , p > 1 is embodied in the follow-ing theorem.

THEOREM 2.4 (Dragomir, 2002 [6]). Let f : [a, b] → R be an absolutely continuous func-tion on [a, b] so thatf ′ ∈ Lp [a, b] , p > 1. If M (x) is as defined in (2.2), then we have thebounds:

(2.7) M (x) ≤ 1

(q + 1)1q

[(x− a

b− a

)1+ 1q

‖f ′‖[a,x],p

+21q

(a+b2− x

b− a

)1+ 1q

‖f ′‖[x,a+b−x],p +

(x− a

b− a

)1+ 1q

‖f ′‖[a+b−x,b],p

(b− a)1q

≤ 1

(q + 1)1q

[2(

x−ab−a

)1+ 1q + 2

1q

(a+b2−x

b−a

)1+ 1q

]×max

‖f ′‖[a,x],p , ‖f ′‖[x,a+b−x],p , ‖f ′‖[a+b−x,b],p

(b− a)

1q

[2(

x−ab−a

)α+αq + 2

αq

(a+b2−x

b−a

)α+αq

] 1α

×[‖f ′‖β

[a,x],p + ‖f ′‖β[x,a+b−x],p + ‖f ′‖β

[a+b−x,b],p

] 1β

(b− a)1q

if α > 1, 1α

+ 1β

= 1,

max

(x−ab−a

)1+ 1q , 2

1q

(a+b2−x

b−a

)1+ 1q

×[‖f ′‖[a,x],p + ‖f ′‖[x,a+b−x],p + ‖f ′‖[a+b−x,b],p

](b− a)

1q

for anyx ∈[a, a+b

2

].

PROOF. Using Hölder’s integral inequality forp > 1, 1p

+ 1q

= 1, we have∫ x

a

(t− a) |f ′ (t)| dt ≤(∫ x

a

(t− a)q dt

) 1q

‖f ′‖[a,x],p =(x− a)1+ 1

q

(q + 1)1q

‖f ′‖[a,x],p ,

∫ a+b−x

x

∣∣∣∣t− a+ b

2

∣∣∣∣ |f ′ (t)| dt ≤ (∫ a+b−x

x

∣∣∣∣t− a+ b

2

∣∣∣∣q dt)1q

‖f ′‖[x,a+b−x],p

=2

1q(

a+b2− x)1+ 1

q

(q + 1)1q

‖f ′‖[x,a+b−x],p

and ∫ b

a+b−x

(b− t) |f ′ (t)| dt ≤(∫ b

a+b−x

(b− t)q dt

) 1q

‖f ′‖[a+b−x,b],p

=(x− a)1+ 1

q

(q + 1)1q

‖f ′‖[a+b−x,b],p .

Summing the above inequalities, we deduce the first bound in (2.7).The last part may be proved in a similar fashion to the one in Theorem 2.2, and we omit the

details.


http://ajmaa.org

192 S. S. DRAGOMIR

REMARK 2.3. If in (2.7) we chooseα = q, β = p, 1p

+ 1q

= 1, p > 1, then we get theinequality

(2.8) M (x) ≤ 21q

(q + 1)1q

(x− a

b− a

)q+1

+

(a+b2− x

b− a

)q+1 1

q

(b− a)1q ‖f ′‖[a,b],p

for anyx ∈[a, a+b

2

].

REMARK 2.4. If in Theorem 2.4 we choosex = a, then we get the trapezoid inequality

(2.9)

∣∣∣∣f (a) + f (b)

2− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

2·(b− a)

1q ‖f ′‖[a,b],p

(q + 1)1q

,

The constant12

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

Indeed, if we assume that (2.9) holds with a constantC > 0, instead of12, i.e.,

(2.10)

∣∣∣∣f (a) + f (b)

2− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ C ·(b− a)

1q ‖f ′‖[a,b],p

(q + 1)1q

,

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

2

∣∣ , k > 0, we have

f (a) + f (b)

2= k · b− a

2,

1

b− a

∫ b

a

f (t) dt = k · b− a

4,

‖f ′‖[a,b],p = k (b− a)1p ;

and by (2.10) we deduce∣∣∣∣k (b− a)

2− k (b− a)

4

∣∣∣∣ ≤ C · k (b− a)

(q + 1)1q

,

giving C ≥ (q+1)1q

4. Letting q → 1+, we deduceC ≥ 1

2, and the sharpness of the constant is

proved.

REMARK 2.5. If in Theorem 2.4 we choosex = a+b2, then we get the midpoint inequality∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.11)

≤ 1

2· (b− a)

1q

21q (q + 1)

1q

[‖f ′‖[a, a+b

2 ],p + ‖f ′‖[a+b2

,b],p

]≤ 1

2· (b− a)

1q

(q + 1)1q

‖f ′‖[a,b],p , p > 1,1

p+

1

q= 1.

In both inequalities the constant12

is sharp in the sense that it cannot be replaced by a smallerconstant.


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To show this fact, assume that (2.11) holds withC,D > 0, i.e.,∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(2.12)

≤ C · (b− a)1q

21q (q + 1)

1q

[‖f ′‖[a, a+b

2 ],p + ‖f ′‖[a+b2

,b],p

]≤ D · (b− a)

1q

(q + 1)1q

‖f ′‖[a,b],p .

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

2

∣∣ , k > 0, we have

f

(a+ b

2

)= 0,

1

b− a

∫ b

a

f (t) dt =k (b− a)

4,

‖f ′‖[a, a+b2 ],p + ‖f ′‖[a+b

2,b],p = 2

(b− a

2

) 1p

k = 21q (b− a)

1p k,

‖f ′‖[a,b],p = (b− a)1p k;

and then by (2.12) we deduce

k (b− a)

4≤ C · k (b− a)

(q + 1)1q

≤ D · k (b− a)

(q + 1)1q

,

giving C,D ≥ (q+1)1q

4for anyq > 1. Letting q → 1+, we deduceC,D ≥ 1

2and the sharpness

of the constants in (2.11) are proved.The following result is useful in providing the best quadrature rule in the class for approxi-

mating the integral of an absolutely continuous function whose derivative is inLp [a, b] .

COROLLARY 2.5. Assume thatf : [a, b] → R is an absolutely continuous function so thatf ′ ∈ Lp [a, b] , p > 1. Then one has the inequality

(2.13)

∣∣∣∣∣f(

3a+b4

)+ f

(a+3b

4

)2

− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣∣ ≤ 1

4

(b− a)1q

(q + 1)1q

‖f ′‖[a,b],p ,

where1p

+ 1q

= 1.

The constant14

is the best possible in the sense that it cannot be replaced by a smallerconstant.

PROOF. The inequality follows by Theorem 2.4 and Remark 2.3 on choosingx = 3a+b4.

To prove the sharpness of the constant, assume that (2.13) holds with a constantE > 0, i.e.,

(2.14)

∣∣∣∣∣f(

3a+b4

)+ f

(a+3b

4

)2

− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣∣ ≤ E · (b− a)1q

(q + 1)1q

‖f ′‖[a,b],p .

Consider the functionf : [a, b] → R,

f (x) =

∣∣∣∣x− 3a+ b

4

∣∣∣∣ if x ∈[a, a+b

2

]∣∣∣∣x− a+ 3b

4

∣∣∣∣ if x ∈(

a+b2, b].


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194 S. S. DRAGOMIR

Thenf is absolutely continuous andf ′ ∈ Lp [a, b] , p > 1. We also have

1

2

[f

(3a+ b

4

)+ f

(a+ 3b

4

)]= 0,

1

b− a

∫ b

a

f (t) dt =b− a

8

‖f ′‖[a,b],p = (b− a)1p ,

and then, by (2.14), we obtain:b− a

8≤ E

(b− a)

(q + 1)1q

givingE ≥ (q+1)1q

8for anyq > 1, i.e.,E ≥ 1

4, and the corollary is proved.

2.4. The Case of1-Norm. If one is interested in obtaining bounds in terms of the1-normfor the derivative, then the following result may be useful.

THEOREM 2.6 (Dragomir, 2002 [6]). Assume that the functionf : [a, b] → R is absolutelycontinuous on[a, b] . If M (x) is as in equation (2.2), then we have the bounds

M (x)(2.15)

≤(x− a

b− a

)‖f ′‖[a,x],1 +

(a+b2− x

b− a

)‖f ′‖[x,a+b−x],1 +

(x− a

b− a

)‖f ′‖[a+b−x,b],1

≤

[1

4+

∣∣∣∣∣x− 3a+b4

b− a

∣∣∣∣∣]‖f ′‖[a,b],1

[2

(x− a

b− a

)α

+

(a+b2− x

b− a

)α] 1α

×[‖f ′‖β

[a,x],1 + ‖f ′‖β[x,a+b−x],1 + ‖f ′‖β

[a+b−x,b],1

] 1β

if α > 1, 1α

+ 1β

= 1,[x+ b−3a

2

b− a

]max

[‖f ′‖[a,x],1 , ‖f ′‖[x,a+b−x],1 , ‖f ′‖[a+b−x,b],1

].

The proof is as in Theorem 2.2 and we omit it.

REMARK 2.6. By the use of Theorem 2.4, forx = a, we get the trapezoid inequality (seefor example [11, p. 55])

(2.16)

∣∣∣∣f (a) + f (b)

2− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

2‖f ′‖[a,b],1 .

If in (2.15) we also choosex = a+b2, then we get the mid point inequality (see for example [11,

p. 56])

(2.17)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

2‖f ′‖[a,b],1 .

The following corollary also holds.


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COROLLARY 2.7. With the assumption in Theorem 2.4, one has the inequality:

(2.18)

∣∣∣∣∣f(

3a+b4

)+ f

(a+3b

4

)2

− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣∣ ≤ 1

4‖f ′‖[a,b],1 .

3. PERTURBED COMPANIONS OF OSTROWSKI I NEQUALITY FOR FUNCTIONS OF

BOUNDED VARIATION

3.1. Some Identities.The following identity holds.

LEMMA 3.1 (Dragomir, 2014 [9]). Assume thatf : [a, b] → C is a function of boundedvariation on[a, b]. Then we have the equality

1

2[f (x) + f (a+ b− x)] +

1

2(x− a)2 λ3 (x)− λ1 (x)

b− a− 1

b− a

∫ b

a

f (t) dt(3.1)

=1

b− a

∫ x

a

(t− a) d [f (t)− λ1 (x) t]

+1

b− a

∫ a+b−x

x

(t− a+ b

2

)d [f (t)− λ2 (x) t]

+1

b− a

∫ b

a+b−x

(t− b) d [f (t)− λ3 (x) t] ,

for anyx ∈[a, a+b

2

]andλi (x) , i = 1, 2, 3 complex numbers.

PROOF. Using the integration by parts formula for Riemann-Stieltjes integrals, we have∫ x

a

(t− a) d [f (t)− λ1 (x) t]

=

∫ x

a

(t− a) df (t)− λ1 (x)

∫ x

a

(t− a) dt

= (x− a) f (x)−∫ x

a

f (t) dt− 1

2λ1 (x) (x− a)2 ,

∫ a+b−x

x

(t− a+ b

2

)d [f (t)− λ2 (x) t]

=

∫ a+b−x

x

(t− a+ b

2

)df (t)− λ2 (x)

∫ a+b−x

x

(t− a+ b

2

)dt

= f (a+ b− x)

(a+ b

2− x

)− f (x)

(x− a+ b

2

)−∫ a+b−x

x

f (t) dt− λ2 (x)

∫ a+b−x

x

(t− a+ b

2

)dt

= f (a+ b− x)

(a+ b

2− x

)− f (x)

(x− a+ b

2

)−∫ a+b−x

x

f (t) dt

since, by symmetry ∫ a+b−x

x

(t− a+ b

2

)dt = 0


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196 S. S. DRAGOMIR

and ∫ b

a+b−x

(t− b) d [f (t)− λ3 (x) t]

=

∫ b

a+b−x

(t− b) df (t)− λ3 (x)

∫ b

a+b−x

(t− b) dt

= (x− a) f (a+ b− x)−∫ b

a+b−x

f (t) dt+1

2λ3 (x) (x− a)2 .

Summing the above equalities, we deduce∫ x

a

(t− a) d [f (t)− λ1 (x) t] +

∫ a+b−x

x

(t− a+ b

2

)d [f (t)− λ2 (x) t]

+

∫ b

a+b−x

(t− b) d [f (t)− λ3 (x) t]

= (b− a)f (x) + f (a+ b− x)

2−∫ b

a

f (t) dt+1

2[λ3 (x)− λ1 (x)] (x− a)2 ,

which is equivalent with the desired identity (3.1).

The following particular cases are of interest:

COROLLARY 3.2 (Dragomir, 2014 [9]). With the assumption of Lemma 3.1 we have theequalities

(3.2)f (a) + f (b)

2− 1

b− a

∫ b

a

f (t) dt =1

b− a

∫ b

a

(t− a+ b

2

)d [f (t)− λ2t] ,

f

(a+ b

2

)+

1

8(b− a) (λ3 − λ1)−

1

b− a

∫ b

a

f (t) dt(3.3)

=1

b− a

∫ a+b2

a

(t− a) d [f (t)− λ1t] +1

b− a

∫ b

a+b2

(t− b) d [f (t)− λ3t] ,

and

1

2

[f

(3a+ b

4

)+ f

(a+ 3b

4

)]+

1

32(b− a) (λ3 − λ1)−

1

b− a

∫ b

a

f (t) dt(3.4)

=1

b− a

∫ 3a+b4

a

(t− a) d [f (t)− λ1t]

+1

b− a

∫ a+3b4

3a+b4

(t− a+ b

2

)d [f (t)− λ2t]

+1

b− a

∫ b

a+3b4

(t− b) d [f (t)− λ3t] ,

for anyλ1, λ2, λ3 ∈ C.

The following particular result with no parameter in the left hand term holds:


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COROLLARY 3.3 (Dragomir, 2014 [9]). Assume thatf : [a, b] → C is a function of boundedvariation on[a, b]. Then we have the equality

1

2[f (x) + f (a+ b− x)]− 1

b− a

∫ b

a

f (t) dt(3.5)

=1

b− a

∫ x

a

(t− a) d [f (t)− λ1 (x) t]

+1

b− a

∫ a+b−x

x

(t− a+ b

2

)d [f (t)− λ2 (x) t]

+1

b− a

∫ b

a+b−x

(t− b) d [f (t)− λ1 (x) t] ,

for anyx ∈[a, a+b

2

]andλi (x) , i = 1, 2 complex numbers.

REMARK 3.1. We get from (3.3) the following particular case:

f

(a+ b

2

)− 1

b− a

∫ b

a

f (t) dt(3.6)

=1

b− a

∫ a+b2

a

(t− a) d [f (t)− λ1t] +1

b− a

∫ b

a+b2

(t− b) d [f (t)− λ1t] ,

for anyλ1 ∈ C, while from (3.4) we get

1

2

[f

(3a+ b

4

)+ f

(a+ 3b

4

)]− 1

b− a

∫ b

a

f (t) dt(3.7)

=1

b− a

∫ 3a+b4

a

(t− a) d [f (t)− λ1t]

+1

b− a

∫ a+3b4

3a+b4

(t− a+ b

2

)d [f (t)− λ2t]

+1

b− a

∫ b

a+3b4

(t− b) d [f (t)− λ1t] ,

for anyλ1, λ2 ∈ C.

3.2. Inequalities for Functions of Bounded Variation. The following lemma will be usedin the sequel and is of interest in itself as well [1, p. 177]. For a simple proof see [7].


∫ b


(3.8)

∣∣∣∣∫ b

a

f (t) du (t)

∣∣∣∣ ≤ ∫ b

a

|f (t)| d

(t∨a

(u)

)≤ max

t∈[a,b]|f (t)|

b∨a

(u) .

We denote by : [a, b] → [a, b] the identity function, namely` (t) = t for anyt ∈ [a, b] .We have the following result:


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198 S. S. DRAGOMIR

THEOREM 3.5 (Dragomir, 2014 [9]). Letf : [a, b] → C be a function of bounded variationon [a, b] . Then we have the inequalities

∣∣∣∣12 [f (x) + f (a+ b− x)] +1

2(x− a)2 λ3 (x)− λ1 (x)

b− a− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.9)

≤ 1

b− a

[∫ x

a

(x∨t

(f − λ1 (x) `)

)dt+

∫ a+b2

x

(t∨x

(f − λ2 (x) `)

)dt

+

∫ a+b−x

a+b2

(a+b−x∨

t

(f − λ2 (x) `)

)dt+

∫ b

a+b−x

(t∨

a+b−x

(f − λ3 (x) `)

)dt

]

≤ 1

b− a

[(x− a)

x∨a

(f − λ1 (x) `) +

(a+ b

2− x

) a+b−x∨x

(f − λ2 (x) `)

+ (x− a)b∨

a+b−x

(f − λ3 (x) `)

]

≤

[1

4+

∣∣∣∣∣x− 3a+b4

b− a

∣∣∣∣∣]

×[∨x

a (f − λ1 (x) `) +∨a+b−x

x (f − λ2 (x) `) +∨b

a+b−x (f − λ3 (x) `)]

x+ b−3a2

b− a×max

∨xa (f − λ1 (x) `) ,

∨a+b−xx (f − λ2 (x) `) ,

∨ba+b−x (f − λ3 (x) `)

for anyx ∈[a, a+b

2

]andλi (x) , i = 1, 2, 3 complex numbers.

PROOF. Taking the modulus on (3.1) and making use of (3.8), we have

∣∣∣∣12 [f (x) + f (a+ b− x)] +1

2(x− a)2 λ3 (x)− λ1 (x)

b− a− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.10)

≤ 1

b− a

∫ x

a

(t− a) d

(t∨a

(f − λ1 (x) `)

)

+1

b− a

∫ a+b−x

x

∣∣∣∣t− a+ b

2

∣∣∣∣ d(

t∨a

(f − λ2 (x) `)

)

+1

b− a

∫ b

a+b−x

(b− t) d

(t∨a

(f − λ3 (x) `)

).


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Integrating by parts in the Riemann-Stieltjes integral, we have

∫ x

a

(t− a) d

(t∨a

(f − λ1 (x) `)

)

= (t− a)t∨a

(f − λ1 (x) `)

∣∣∣∣∣x

a

−∫ x

a

t∨a

(f − λ1 (x) `) dt

= (x− a)x∨a

(f − λ1 (x) `)−∫ x

a

t∨a

(f − λ1 (x) `) dt

=

∫ x

a

(x∨t

(f − λ1 (x) `)

)dt.

Also

∫ a+b−x

x

∣∣∣∣t− a+ b

2

∣∣∣∣ d(

t∨a

(f − λ2 (x) `)

)

=

∫ a+b2

x

(a+ b

2− t

)d

(t∨a

(f − λ2 (x) `)

)

+

∫ a+b−x

a+b2

(t− a+ b

2

)d

(t∨a

(f − λ2 (x) `)

)

=

(a+ b

2− t

)( t∨a

(f − λ2 (x) `)

)∣∣∣∣∣a+b2

x

+

∫ a+b2

x

(t∨a

(f − λ2 (x) `)

)dt

+

(t− a+ b

2

)( t∨a

(f − λ2 (x) `)

)∣∣∣∣∣a+b−x

a+b2

−∫ a+b−x

a+b2

(t∨a

(f − λ2 (x) `)

)dt

=

∫ a+b2

x

(t∨a

(f − λ2 (x) `)

)dt−

(a+ b

2− x

)( x∨a

(f − λ2 (x) `)

)

+

(a+ b

2− x

)(a+b−x∨a

(f − λ2 (x) `)

)−∫ a+b−x

a+b2

(t∨a

(f − λ2 (x) `)

)dt

=

∫ a+b−x

a+b2

(a+b−x∨

t

(f − λ2 (x) `)

)dt+

∫ a+b2

x

(t∨x

(f − λ2 (x) `)

)dt


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200 S. S. DRAGOMIR

and ∫ b

a+b−x

(b− t) d

(t∨a

(f − λ3 (x) `)

)

= (b− t)

(t∨a

(f − λ3 (x) `)

)∣∣∣∣∣b

a+b−x

+

∫ b

a+b−x

(t∨a

(f − λ3 (x) `)

)dt

=

∫ b

a+b−x

(t∨a

(f − λ3 (x) `)

)dt− (b− (a+ b− x))

(a+b−x∨

a

(f − λ3 (x) `)

)

=

∫ b

a+b−x

(t∨

a+b−x

(f − λ3 (x) `)

)dt.

Making use of (3.10) we deduce the first inequality in (3.9).Since ∫ x

a

(x∨t

(f − λ1 (x) `)

)dt ≤ (x− a)

x∨a

(f − λ1 (x) `) ,

∫ a+b−x

a+b2

(a+b−x∨

t

(f − λ2 (x) `)

)dt+

∫ a+b2

x

(t∨x

(f − λ2 (x) `)

)dt

≤(a+ b

2− x

) a+b−x∨a+b2

(f − λ2 (x) `) +

(a+ b

2− x

) a+b2∨x

(f − λ2 (x) `)

=

(a+ b

2− x

) a+b−x∨x

(f − λ2 (x) `)

and ∫ b

a+b−x

(t∨

a+b−x

(f − λ3 (x) `)

)dt ≤ (x− a)

b∨a+b−x

(f − λ3 (x) `) ,

the second inequality is also proved.The last inequality is obvious by the maximum properties.

The following midpoint and trapezoid type inequalities hold:

COROLLARY 3.6. Letf : [a, b] → C be a function of bounded variation on[a, b] . Then wehave the inequalities∣∣∣∣f (a) + f (b)

2− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.11)

≤ 1

b− a

[∫ a+b2

a

(t∨a

(f − λ2`)

)dt+

∫ b

a+b2

(b∨t

(f − λ2`)

)dt

]

≤ 1

2

b∨a

(f − λ2`) ,


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∣∣∣∣f (a+ b

2

)+

1

8(b− a) (λ3 − λ1)−

1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.12)

≤ 1

b− a

∫ a+b2

a

a+b2∨t

(f − λ1`)

dt+

∫ b

a+b2

t∨a+b2

(f − λ3`)

dt

≤ 1

2

a+b2∨a

(f − λ1`) +b∨

a+b2

(f − λ3`)

and

∣∣∣∣12[f

(3a+ b

4

)+ f

(a+ 3b

4

)]+

1

32(b− a) (λ3 − λ1)−

1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.13)

≤ 1

b− a

∫ 3a+b4

a

3a+b4∨t

(f − λ1`)

dt+

∫ a+b2

3a+b4

t∨3a+b

4

(f − λ2`)

dt

+

∫ a+3b4

a+b2

a+3b4∨t

(f − λ2`)

dt+

∫ b

a+3b4

t∨a+3b

4

(f − λ3`)

dt

≤ 1

4

3a+b4∨a

(f − λ1`) +

a+3b4∨

3a+b4

(f − λ2`) +b∨

a+3b4

(f − λ3`)


COROLLARY 3.7. Assume thatf : [a, b] → C is a function of bounded variation on[a, b].Then we have the inequalities

∣∣∣∣12 [f (x) + f (a+ b− x)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.14)

≤ 1

b− a

[∫ x

a

(x∨t

(f − λ1 (x) `)

)dt+

∫ a+b2

x

(t∨x

(f − λ2 (x) `)

)dt

+

∫ a+b−x

a+b2

(a+b−x∨

t

(f − λ2 (x) `)

)dt+

∫ b

a+b−x

(t∨

a+b−x

(f − λ1 (x) `)

)dt

]

≤ 1

b− a

[(x− a)

x∨a

(f − λ1 (x) `) +

(a+ b

2− x

) a+b−x∨x

(f − λ2 (x) `)

+ (x− a)b∨

a+b−x

(f − λ1 (x) `)

]


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202 S. S. DRAGOMIR

≤

[1

4+

∣∣∣∣∣x− 3a+b4

b− a

∣∣∣∣∣]

×[∨x

a (f − λ1 (x) `) +∨a+b−x

x (f − λ2 (x) `) +∨b

a+b−x (f − λ1 (x) `)]

x+ b−3a2

b− a×max

∨xa (f − λ1 (x) `) ,

∨a+b−xx (f − λ2 (x) `) ,

∨ba+b−x (f − λ1 (x) `)

for anyx ∈[a, a+b

2

]andλi (x) , i = 1, 2, complex numbers.

REMARK 3.2. We have the particular inequalities of interest

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.15)

≤ 1

b− a

∫ a+b2

a

a+b2∨t

(f − λ1`)

dt+

∫ b

a+b2

t∨a+b2

(f − λ1`)

dt

≤ 1

2

b∨a

(f − λ1`)

and

∣∣∣∣12[f

(3a+ b

4

)+ f

(a+ 3b

4

)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.16)

≤ 1

b− a

∫ 3a+b4

a

3a+b4∨t

(f − λ1`)

dt+

∫ a+b2

3a+b4

t∨3a+b

4

(f − λ2`)

dt

+

∫ a+3b4

a+b2

a+3b4∨t

(f − λ2`)

dt+

∫ b

a+3b4

t∨a+3b

4

(f − λ1`)

dt

≤ 1

4

3a+b4∨a

(f − λ1`) +

a+3b4∨

3a+b4

(f − λ2`) +b∨

a+3b4

(f − λ1`)



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If we takeλ1 = λ2 = λ, then we get∣∣∣∣12[f

(3a+ b

4

)+ f

(a+ 3b

4

)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.17)

≤ 1

b− a

∫ 3a+b4

a

3a+b4∨t

(f − λ`)

dt+

∫ a+b2

3a+b4

t∨3a+b

4

(f − λ`)

dt

+

∫ a+3b4

a+b2

a+3b4∨t

(f − λ`)

dt+

∫ b

a+3b4

t∨a+3b

4

(f − λ`)

dt

≤ 1

4

[b∨a

(f − λ`)

]for anyλ1, λ2 ∈ C.

From (3.15) we deduce the simpler inequality∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.18)

≤ 1

b− a

∫ b

a

∣∣∣∣∣∣a+b2∨t

(f − λ`)

∣∣∣∣∣∣ dt ≤ 1

2

b∨a

(f − λ`)

while from (3.17) we get∣∣∣∣12[f

(3a+ b

4

)+ f

(a+ 3b

4

)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.19)

≤ 1

b− a

∫ a+b2

a

∣∣∣∣∣∣3a+b

4∨t

(f − λ`)

∣∣∣∣∣∣ dt+

∫ b

a+b2

∣∣∣∣∣∣a+3b

4∨t

(f − λ`)

∣∣∣∣∣∣ dt

≤ 1

4

b∨a

(f − λ`)

for anyλ ∈ C.

We can state the following result.

PROPOSITION 3.8 (Dragomir, 2014 [9]). Assume thatf : [a, b] → C is a function ofbounded variation on[a, b]. If there exists the constantsγ,Γ ∈ C such that

b∨a

(f − γ + Γ

2`

)≤ 1

2|Γ− γ| ,

then ∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

4|Γ− γ|

and ∣∣∣∣12[f

(3a+ b

4

)+ f

(a+ 3b

4

)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

8|Γ− γ| .

The inequalities follow by (3.18) and (3.19).


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204 S. S. DRAGOMIR

PROPOSITION 3.9 (Dragomir, 2014 [9]). Assume thatf : [a, b] → C is a function ofbounded variation on[a, b]. If for someλ ∈ C the cumulative variation functionVλ : [a, b] →[0,∞),

Vλ (t) :=t∨a

(f − λ`)

is Lipschitzian with the constantLγ > 0, then

(3.20)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

4Lγ (b− a)

and

(3.21)

∣∣∣∣12[f

(3a+ b

4

)+ f

(a+ 3b

4

)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

8Lγ (b− a) .

PROOF. From (3.18) we have∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

b− a

∫ b

a

∣∣∣∣∣∣a+b2∨t

(f − λ`)

∣∣∣∣∣∣ dt =1

b− a

∫ b

a

∣∣∣∣Vλ

(a+ b

2

)− Vλ (t)

∣∣∣∣ dt≤ Lγ

b− a

∫ b

a

∣∣∣∣a+ b

2− t

∣∣∣∣ dt =1

4Lγ (b− a)

and the inequality (3.20) is proved.From (3.19) we have∣∣∣∣12

[f

(3a+ b

4

)+ f

(a+ 3b

4

)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

b− a

∫ a+b2

a

∣∣∣∣∣∣3a+b

4∨t

(f − λ`)

∣∣∣∣∣∣ dt+

∫ b

a+b2

∣∣∣∣∣∣a+3b

4∨t

(f − λ`)

∣∣∣∣∣∣ dt

=

1

b− a

[∫ a+b2

a

∣∣∣∣Vλ

(3a+ b

4

)− Vλ (t)

∣∣∣∣ dt+

∫ b

a+b2

(∣∣∣∣Vλ

(a+ 3b

4

)− Vλ (t)

∣∣∣∣) dt]

≤ Lγ

b− a

[∫ a+b2

a

∣∣∣∣3a+ b

4− t

∣∣∣∣ dt+

∫ b

a+b2

(∣∣∣∣a+ 3b

4− t

∣∣∣∣) dt]

=Lγ

b− a

[∫ a+b2

a

∣∣∣∣3a+ b

4− t

∣∣∣∣ dt+

∫ b

a+b2

(∣∣∣∣a+ 3b

4− t

∣∣∣∣) dt]

=Lγ

b− a

[1

16(b− a)2 +

1

16(b− a)2

]=Lγ

8(b− a)



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3.3. Inequalities for Lipschitzian Functions. We say that a functionf : [c, d] → C isLipschitzianwith a constantL > 0 on the interval[c, d] if

|f (t)− f (s)| ≤ L |t− s|

for anyt, s ∈ [c, d] .

THEOREM 3.10 (Dragomir, 2014 [9]). Let f : [a, b] → C be a bounded function on[a, b].For x ∈

[a, a+b

2

]and λi (x) , i = 1, 2, 3 complex numbers, assume thatf − λ1 (x) ` is Lip-

schitzian with the constantL1 (x) > 0 on [a, x] , f − λ2 (x) ` with the constantL2 (x) > 0 on[x, a+ b− x] andf − λ3 (x) ` with the constantL3 (x) > 0 on [a+ b− x, b] , then∣∣∣∣12 [f (x) + f (a+ b− x)] +

1

2(x− a)2 λ3 (x)− λ1 (x)

b− a− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.22)

≤ 1

(b− a)

[1

2(x− a)2 L1 (x) +

(x− a+ b

2

)2

L2 (x) +1

2(x− a)2 L3 (x)

]

≤

1

8+ 2

(x− 3a+b

4

b− a

)2 (b− a) max L1 (x) , L2 (x) , L3 (x) .

PROOF. It is known that ifg : [c, d] → C is Riemann integrable on[c, d] andu : [c, d] → Cis Lipschitzian with the constantL > 0, then the Riemann-Stieltjes integral

∫ b

af (t) du (t) exists

and we have the inequality

(3.23)

∣∣∣∣∫ b

a

f (t) du (t)

∣∣∣∣ ≤ L

∫ b

a

|f (t)| dt.

Taking the modulus in (3.1) and using the property (3.23) we have∣∣∣∣12 [f (x) + f (a+ b− x)] +1

2(x− a)2 λ3 (x)− λ1 (x)

b− a− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

b− a

∣∣∣∣∫ x

a

(t− a) d [f (t)− λ1 (x) t]

∣∣∣∣+

1

b− a

∣∣∣∣∫ a+b−x

x

(t− a+ b

2

)d [f (t)− λ2 (x) t]

∣∣∣∣+

1

b− a

∣∣∣∣∫ b

a+b−x

(t− b) d [f (t)− λ3 (x) t]

∣∣∣∣≤ 1

b− aL1 (x)

∫ x

a

(t− a) dt+1

b− aL2 (x)

∫ a+b−x

x

∣∣∣∣t− a+ b

2

∣∣∣∣ dt+

1

b− aL3 (x)

∫ b

a+b−x

(b− t) dt

=1

(b− a)

[1

2(x− a)2 L1 (x) +

(x− a+ b

2

)2

L2 (x) +1

2(x− a)2 L3 (x)

],

which proves the first inequality in (3.22).


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206 S. S. DRAGOMIR

Since

1

2(x− a)2 L1 (x) +

(x− a+ b

2

)2

L2 (x) +1

2(x− a)2 L3 (x)

≤

[1

2(x− a)2 +

(x− a+ b

2

)2

+1

2(x− a)2

]max L1 (x) , L2 (x) , L3 (x)

=

1

8+ 2

(x− 3a+b

4

b− a

)2 (b− a) max L1 (x) , L2 (x) , L3 (x) ,

the last part of (3.22) is also proved.

COROLLARY 3.11 (Dragomir, 2014 [9]). Letf : [a, b] → C be a bounded function on[a, b].(i) If forλ2 ∈ C the functionf − λ2` is Lipschitzian with the constantL2 > 0 on [a, b] , then

(3.24)

∣∣∣∣f (a) + f (b)

2− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

4(b− a)L2.

(ii) If for λ1, λ2 ∈ C the functionf − λ1` is Lipschitzian with the constantL1 > 0 on[a, a+b

2

]andf − λ3` is Lipschitzian with the constantL3 > 0 on

[a+b2, b], then∣∣∣∣f (a+ b

2

)+

1

8(b− a) (λ3 − λ1)−

1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.25)

≤ 1

4

(L1 + L3

2

)(b− a) .

(iii) If for λ1, λ2, λ3 ∈ C the functionf − λ1` is Lipschitzian with the constantL1 > 0on[a, 3a+b

4

], f − λ2` is Lipschitzian with the constantL2 > 0 on

[3a+b

4, a+3b

4

]andf − λ3` is

Lipschitzian with the constantL3 > 0 on[

a+3b4, b], then∣∣∣∣12

[f

(3a+ b

4

)+ f

(a+ 3b

4

)]+

1

32(b− a) (λ3 − λ1)−

1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.26)

≤ 1

16

(1

2L1 + L2 +

1

2L3

)(b− a) .

REMARK 3.3. We have the following particular cases of interest.If for someλ ∈ C the functionf − λ` is Lipschitzian with the constantLλ > 0 on [a, b] ,

then

(3.27)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

4Lλ (b− a)

and ∣∣∣∣12[f

(3a+ b

4

)+ f

(a+ 3b

4

)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.28)

≤ 1

8Lλ (b− a) .

The following lemma may be stated:

LEMMA 3.12. Let u : [a, b] → R and l, L ∈ R with L > l. The following statements areequivalent:

(i) The functionu− l+L2· e, wheree (t) = t, t ∈ [a, b] is 1

2(L− l)-Lipschitzian;


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(ii) We have the inequalities

(3.29) l ≤ u (t)− u (s)

t− s≤ L for each t, s ∈ [a, b] with t 6= s;

(iii) We have the inequalities

(3.30) l (t− s) ≤ u (t)− u (s) ≤ L (t− s) for each t, s ∈ [a, b] with t > s.

Following [12], we can introduce the definition of(l, L)-Lipschitzian functions:

DEFINITION 3.1. The functionu : [a, b] → R which satisfies one of the equivalent condi-tions (i) – (iii) from Lemma 3.12 is said to be(l, L)-Lipschitzian on[a, b] .

If L > 0 and l = −L, then(−L,L)−Lipschitzian meansL-Lipschitzian in the classicalsense.

Utilising Lagrange’s mean value theorem, we can state the following result that providesexamples of(l, L)-Lipschitzian functions.

PROPOSITION3.13. Letu : [a, b] → R be continuous on[a, b] and differentiable on(a, b) .If −∞ < l = inft∈(a,b) u

′ (t) and supt∈(a,b) u′ (t) = L < ∞, thenu is (l, L)-Lipschitzian on

[a, b] .

As consequences of the inequalities (3.27) and (3.28) for real valued functions we can statethe following result.

PROPOSITION3.14 (Dragomir, 2014 [9]). Let l, L ∈ R with L > l andf : [a, b] → R an(l, L)-Lipschitzian function on[a, b] , then

(3.31)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

8(L− l) (b− a)

and ∣∣∣∣12[f

(3a+ b

4

)+ f

(a+ 3b

4

)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(3.32)

≤ 1

16(L− l) (b− a) .

4. PERTURBED COMPANIONS OF OSTROWSKI ’ S I NEQUALITY FOR ABSOLUTELY


4.1. Some Identities.The following identity holds.

LEMMA 4.1 (Dragomir, 2014 [10]). Assume thatf : [a, b] → C is an absolutely continuousfunction on[a, b]. Then we have the equality

1

2[f (x) + f (a+ b− x)] +

1

2(x− a)2 λ3 (x)− λ1 (x)

b− a− 1

b− a

∫ b

a

f (t) dt(4.1)

=1

b− a

∫ x

a

(t− a) [f ′ (t)− λ1 (x)] dt

+1

b− a

∫ a+b−x

x

(t− a+ b

2

)[f ′ (t)− λ2 (x)] dt

+1

b− a

∫ b

a+b−x

(t− b) [f ′ (t)− λ3 (x)] dt,

for anyx ∈[a, a+b

2

]andλj (x) , j = 1, 2, 3 complex numbers.


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208 S. S. DRAGOMIR

PROOF. Using the integration by parts formula for Lebesgue integral, we have∫ x

a

(t− a) [f ′ (t)− λ1 (x)] dt

=

∫ x

a

(t− a) f ′ (t) dt− λ1 (x)

∫ x

a

(t− a) dt

= (x− a) f (x)−∫ x

a

f (t) dt− 1

2λ1 (x) (x− a)2 ,

∫ a+b−x

x

(t− a+ b

2

)[f ′ (t)− λ2 (x)] dt

=

∫ a+b−x

x

(t− a+ b

2

)f ′ (t) dt− λ2 (x)

∫ a+b−x

x

(t− a+ b

2

)dt

= f (a+ b− x)

(a+ b

2− x

)− f (x)

(x− a+ b

2

)−∫ a+b−x

x

f (t) dt

− λ2 (x)

∫ a+b−x

x

(t− a+ b

2

)dt

= f (a+ b− x)

(a+ b

2− x

)− f (x)

(x− a+ b

2

)−∫ a+b−x

x

f (t) dt,

since, by symmetry ∫ a+b−x

x

(t− a+ b

2

)dt = 0

and ∫ b

a+b−x

(t− b) [f ′ (t)− λ3 (x)] dt

=

∫ b

a+b−x

(t− b) f ′ (t) dt− λ3 (x)

∫ b

a+b−x

(t− b) dt

= (x− a) f (a+ b− x)−∫ b

a+b−x

f (t) dt+1

2λ3 (x) (x− a)2 .

Summing the above equalities, we deduce∫ x

a

(t− a) [f ′ (t)− λ1 (x)] dt+

∫ a+b−x

x

(t− a+ b

2

)[f ′ (t)− λ2 (x)] dt

+

∫ b

a+b−x

(t− b) [f ′ (t)− λ3 (x)] dt

= (b− a)f (x) + f (a+ b− x)

2−∫ b

a

f (t) dt+1

2[λ3 (x)− λ1 (x)] (x− a)2 ,

which is equivalent with the desired identity (4.1).



(4.2)f (a) + f (b)

2− 1

b− a

∫ b

a

f (t) dt =1

b− a

∫ b

a

(t− a+ b

2

)[f ′ (t)− λ2] dt,


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f

(a+ b

2

)+

1

8(b− a) (λ3 − λ1)−

1

b− a

∫ b

a

f (t) dt(4.3)

=1

b− a

∫ a+b2

a

(t− a) [f ′ (t)− λ1] dt+1

b− a

∫ b

a+b2

(t− b) [f ′ (t)− λ3] dt,

and

1

2

[f

(3a+ b

4

)+ f

(a+ 3b

4

)]+

1

32(b− a) (λ3 − λ1)−

1

b− a

∫ b

a

f (t) dt(4.4)

=1

b− a

∫ 3a+b4

a

(t− a) [f ′ (t)− λ1] dt

+1

b− a

∫ a+3b4

3a+b4

(t− a+ b

2

)[f ′ (t)− λ2] dt

+1

b− a

∫ b

a+3b4

(t− b) [f ′ (t)− λ3] dt



COROLLARY 4.3 (Dragomir, 2014 [10]). Assume thatf : [a, b] → C is absolutely continu-ous on[a, b]. Then we have the equality

1

2[f (x) + f (a+ b− x)]− 1

b− a

∫ b

a

f (t) dt(4.5)

=1

b− a

∫ x

a

(t− a) [f ′ (t)− λ1 (x)] dt

+1

b− a

∫ a+b−x

x

(t− a+ b

2

)[f ′ (t)− λ2 (x)] dt

+1

b− a

∫ b

a+b−x

(t− b) [f ′ (t)− λ1 (x)] dt,

for anyx ∈[a, a+b

2



f

(a+ b

2

)− 1

b− a

∫ b

a

f (t) dt(4.6)

=1

b− a

∫ a+b2

a

(t− a) [f ′ (t)− λ1] dt+1

b− a

∫ b

a+b2

(t− b) [f ′ (t)− λ1] dt,


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210 S. S. DRAGOMIR


1

2

[f

(3a+ b

4

)+ f

(a+ 3b

4

)]− 1

b− a

∫ b

a

f (t) dt(4.7)

=1

b− a

∫ 3a+b4

a

(t− a) [f ′ (t)− λ1] dt

+1

b− a

∫ a+3b4

3a+b4

(t− a+ b

2

)[f ′ (t)− λ2] dt

+1

b− a

∫ b

a+3b4

(t− b) [f ′ (t)− λ1] dt



U[a,b] (γ,Γ)

:=f : [a, b] → C|Re

[(Γ− f (t))

(f (t)− γ


and

∆[a,b] (γ,Γ) :=

f : [a, b] → C|

∣∣∣∣f (t)− γ + Γ

2

∣∣∣∣ ≤ 1

2|Γ− γ| for a.e.t ∈ [a, b]

.


PROPOSITION4.4. For anyγ,Γ ∈ C, γ 6= Γ, we have thatU[a,b] (γ,Γ) and∆[a,b] (γ,Γ) arenonempty, convex and closed sets and

(4.8) U[a,b] (γ,Γ) = ∆[a,b] (γ,Γ) .


2

∣∣∣∣ ≤ 1

2|Γ− γ|

if and only if

Re [(Γ− z) (z − γ)] ≥ 0.


1

4|Γ− γ|2 −

∣∣∣∣z − γ + Γ

2

∣∣∣∣2 = Re [(Γ− z) (z − γ)]







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(4.11) ∅ 6= S[a,b] (γ,Γ) ⊆ U[a,b] (γ,Γ) .

THEOREM 4.6 (Dragomir, 2014 [10]). Assume thatf : [a, b] → C is an absolutely contin-uous function on[a, b] andx ∈

[a, a+b

2

]. If there exists the complex numbersγj (x) 6= Γj (x) ,

j = 1, 2, 3 such that

f ′ ∈ ∆[a,x] (γ1 (x) ,Γ1 (x)) ∩ ∆[x,a+b−x] (γ2 (x) ,Γ2 (x))(4.12)

∩ ∆[a+b−x,b] (γ3 (x) ,Γ3 (x)) ,


∣∣∣∣12 [f (x) + f (a+ b− x)] +1

4(x− a)2 γ3 (x) + Γ3 (x)− γ1 (x)− Γ1 (x)

b− a(4.13)

− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

4 (b− a)

[|Γ1 (x)− γ1 (x)| (x− a)2 + 2 |Γ2 (x)− γ2 (x)|

(a+ b

2− x

)2

+ |Γ3 (x)− γ3 (x)| (x− a)2] .PROOF. Taking the modulus in the equality (4.1) written for

λ1 (x) =γ1 (x) + Γ1 (x)

2, λ2 (x) =

γ2 (x) + Γ2 (x)

2,

λ3 (x) =γ3 (x) + Γ3 (x)

2


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212 S. S. DRAGOMIR

and utilizing the condition (4.12) we have∣∣∣∣12 [f (x) + f (a+ b− x)] +1

4(x− a)2 γ3 (x) + Γ3 (x)− γ1 (x)− Γ1 (x)

b− a

− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

b− a

∫ x

a

(t− a)

∣∣∣∣f ′ (t)− γ1 (x) + Γ1 (x)

2

∣∣∣∣ dt+

1

b− a

∫ a+b−x

x

∣∣∣∣t− a+ b

2

∣∣∣∣ ∣∣∣∣f ′ (t)− γ2 (x) + Γ2 (x)

2

∣∣∣∣ dt+

1

b− a

∫ b

a+b−x

(b− t)

∣∣∣∣f ′ (t)− γ3 (x) + Γ3 (x)

2

∣∣∣∣ dt≤ 1

4 (b− a)|Γ1 (x)− γ1 (x)| (x− a)2

+2

4 (b− a)|Γ2 (x)− γ2 (x)|

(a+ b

2− x

)2

+1

4 (b− a)|Γ3 (x)− γ3 (x)| (x− a)2


COROLLARY 4.7. Assume thatf : [a, b] → C is an absolutely continuous function on[a, b]andx ∈

[a, a+b

2

]. If there exists the complex numbersγj (x) 6= Γj (x) , j = 1, 2 such that

f ′ ∈ ∆[a,x] (γ1 (x) ,Γ1 (x)) ∩ ∆[x,a+b−x] (γ2 (x) ,Γ2 (x))(4.14)

∩ ∆[a+b−x,b] (γ1 (x) ,Γ1 (x)) ,

then we have the inequality∣∣∣∣12 [f (x) + f (a+ b− x)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.15)

≤ 1

2 (b− a)

[|Γ1 (x)− γ1 (x)| (x− a)2 + |Γ2 (x)− γ2 (x)|

(a+ b

2− x

)2].

REMARK 4.2 (Dragomir, 2014 [10]). Assume thatf : [a, b] → C is an absolutely continu-ous function on[a, b] .

If there exists the complex numbersγ2 6= Γ2 such thatf ′ ∈ ∆[a,b] (γ2,Γ2) , then

(4.16)

∣∣∣∣12 [f (a) + f (b)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

8(b− a) |Γ2 − γ2| .

If there exists the complex numbersγj 6= Γj, j = 1, 3 such that

f ′ ∈ ∆[a, a+b2 ] (γ1,Γ1) ∩ ∆[a+b

2,b] (γ3,Γ3) ,

then we have the inequality∣∣∣∣f (a+ b

2

)+

1

16(b− a) (Γ3 + γ3 − Γ1 − γ1)−

1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.17)

≤ 1

16(b− a) [|Γ1 − γ1|+ |Γ3 − γ3|] .


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In particular, iff ′ ∈ ∆[a,b] (γ1,Γ1) then

(4.18)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

8(b− a) |Γ1 − γ1| .

If there exists the complex numbersγj 6= Γj, j = 1, 2, 3 such that

f ′ ∈ ∆[a, 3a+b4 ] (γ1,Γ1) ∩ ∆[ 3a+b

4, a+3b

4 ] (γ2,Γ2) ∩ ∆[a+3b4

,b] (γ3,Γ3) ,

then ∣∣∣∣12[f

(3a+ b

4

)+ f

(a+ 3b

4

)]+

1

64(b− a) (Γ3 + γ3 − Γ1 − γ1)(4.19)

− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

64(b− a) [|Γ1 − γ1|+ 2 |Γ2 − γ2|+ |Γ3 − γ3|] .

In particular, ifγ3 = γ1 andΓ3 = Γ1, then∣∣∣∣12[f

(3a+ b

4

)+ f

(a+ 3b

4

)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.20)

≤ 1

32(b− a) [|Γ1 − γ1|+ |Γ2 − γ2|] ,

providedf ′ ∈ ∆[a, 3a+b

4 ] (γ1,Γ1) ∩ ∆[ 3a+b4

, a+3b4 ] (γ2,Γ2) ∩ ∆[a+3b

4,b] (γ1,Γ1) .

Moreover, iff ′ ∈ ∆[a,b] (γ1,Γ1) then∣∣∣∣12[f

(3a+ b

4

)+ f

(a+ 3b

4

)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.21)

≤ 1

16(b− a) |Γ1 − γ1| .

The case of real-valued functions is of interest.

REMARK 4.3. If the functionf : [a, b] → R is absolutely continuous and if there existsthe constantsl < L such thatl ≤ f ′ (t) ≤ L for almost everyt ∈ [a, b] , then we have theinequalities

(4.22)

∣∣∣∣12 [f (a) + f (b)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

8(b− a) (L− l) ,

(4.23)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

8(b− a) (L− l)

and ∣∣∣∣12[f

(3a+ b

4

)+ f

(a+ 3b

4

)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.24)

≤ 1

16(b− a) (L− l) .

These results improve the corresponding inequalities from Introduction.


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214 S. S. DRAGOMIR

4.3. Inequalities for Derivatives of Bounded Variation. Assume thatf : I → C is anabsolutely continuous function on[a, b] ⊂ I , the interior ofI. Then from (4.1) we have forλ1 (x) = f ′ (a) , λ2 (x) = f ′(x)+f ′(a+b−x)

2andλ3 (x) = f ′ (b) the equality

1

2[f (x) + f (a+ b− x)] +

1

2(x− a)2 f

′ (b)− f ′ (a)

b− a− 1

b− a

∫ b

a

f (t) dt(4.25)

=1

b− a

∫ x

a

(t− a) [f ′ (t)− f ′ (a)] dt

+1

b− a

∫ a+b−x

x

(t− a+ b

2

)[f ′ (t)− f ′ (x) + f ′ (a+ b− x)

2

]dt

+1

b− a

∫ b

a+b−x

(t− b) [f ′ (t)− f ′ (b)] dt,

for anyx ∈[a, a+b

2

].

We can state the following result.

THEOREM 4.8 (Dragomir, 2014 [10]). Assume thatf : I → C is an absolutely continuousfunction on[a, b] ⊂ I . If the derivativef ′ is of bounded variation on[a, b] , then

∣∣∣∣12 [f (x) + f (a+ b− x)] +1

2(x− a)2 f

′ (b)− f ′ (a)

b− a− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.26)

≤ 1

b− a

∫ x

a

(t− a)t∨a

(f ′) dt+1

2 (b− a)

(x− a+ b

2

)2 a+b−x∨x

(f ′)

+1

b− a

∫ b

a+b−x

(b− t)b∨t

(f ′) dt

≤ 1

2 (b− a)

[(x− a)2

x∨a

(f ′) +

(x− a+ b

2

)2 a+b−x∨x

(f ′) + (x− a)2b∨

a+b−x

(f ′)

]

≤

12(b−a)

max

(x− a)2 ,(x− a+b

2

)2 b∨a

(f ′)

12(b−a)

max

x∨a

(f ′) ,a+b−x∨

x

(f ′) ,b∨

a+b−x

(f ′)

[2 (x− a)2 +

(x− a+b

2

)2]

for anyx ∈[a, a+b

2

].


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PROOF. If we take the modulus in (4.25) we get∣∣∣∣12 [f (x) + f (a+ b− x)] +1

2(x− a)2 f

′ (b)− f ′ (a)

b− a− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.27)

≤ 1

b− a

∫ x

a

(t− a) |f ′ (t)− f ′ (a)| dt

+1

b− a

∫ a+b−x

x

∣∣∣∣t− a+ b

2

∣∣∣∣ ∣∣∣∣f ′ (t)− f ′ (x) + f ′ (a+ b− x)

2

∣∣∣∣ dt+

1

b− a

∫ b

a+b−x

(b− t) |f ′ (t)− f ′ (b)| dt := K,

for anyx ∈[a, a+b

2

].

Let x ∈(a, a+b

2

). Sincef ′ is of bounded variation on[a, b] , then

|f ′ (t)− f ′ (a)| ≤t∨a

(f ′) ,

for anyt ∈ [a, x] and∣∣∣∣f ′ (t)− f ′ (x) + f ′ (a+ b− x)

2

∣∣∣∣=

∣∣∣∣f ′ (t)− f ′ (x) + f ′ (t)− f ′ (a+ b− x)

2

∣∣∣∣≤ 1

2[|f ′ (t)− f ′ (x)|+ |f ′ (a+ b− x)− f ′ (t)|] ≤ 1

2

a+b−x∨x

(f ′)

for anyt ∈ [x, a+ b− x] .We also have

|f ′ (t)− f ′ (b)| ≤b∨t

(f ′) , t ∈ [a+ b− x, b] .

Then we get

K ≤ 1

b− a

∫ x

a

(t− a)t∨a

(f ′) dt+1

2 (b− a)

a+b−x∨x

(f ′)

∫ a+b−x

x

∣∣∣∣t− a+ b

2

∣∣∣∣ dt+

1

b− a

∫ b

a+b−x

(b− t)b∨t

(f ′) dt

≤ 1

b− a

x∨a

(f ′)

∫ x

a

(t− a) dt+1

2 (b− a)

a+b−x∨x

(f ′)

∫ a+b−x

x

∣∣∣∣t− a+ b

2

∣∣∣∣ dt+

1

b− a

b∨a+b−x

(f ′)

∫ b

a+b−x

(b− t) dt


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216 S. S. DRAGOMIR

=1

2 (b− a)(x− a)2

x∨a

(f ′) +1

2 (b− a)

(x− a+ b

2

)2 a+b−x∨x

(f ′)

+1

2 (b− a)(x− a)2

b∨a+b−x

(f ′) ,

which proves the first two inequalities in (4.26).The last part is obvious by the maximum properties.

COROLLARY 4.9 (Dragomir, 2014 [10]). With the assumptions of Theorem 4.8 we have

(4.28)

∣∣∣∣12 [f (a) + f (b)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

8(b− a)

b∨a

(f ′) ,

∣∣∣∣f (a+ b

2

)+

1

8(b− a) [f ′ (b)− f ′ (a)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(4.29)

≤ 1

b− a

∫ a+b2

a

(t− a)t∨a

(f ′) dt+1

b− a

∫ b

a+b2

(b− t)b∨t

(f ′) dt

≤ 1

8(b− a)

b∨a

(f ′)

and ∣∣∣∣12[f

(3a+ b

4

)+ f

(a+ 3b

4

)]+

1

32(b− a) [f ′ (b)− f ′ (a)](4.30)

− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

b− a

∫ 3a+b4

a

(t− a)t∨a

(f ′) dt+1

32(b− a)

a+3b4∨

3a+b4

(f ′)

+1

b− a

∫ b

a+3b4

(b− t)b∨t

(f ′) dt

≤ 1

32(b− a)

b∨a

(f ′) .

5. MORE PERTURBED COMPANIONS OF OSTROWSKI ’ S I NEQUALITY FOR

ABSOLUTELY CONTINUOUS FUNCTIONS

5.1. Some Identities.In the recent paper [10] we established the following identity:


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LEMMA 5.1 (Dragomir, 2014 [10]). Assume thatf : [a, b] → C is an absolutely continuousfunction on[a, b]. Then we have the equality

1

2[f (x) + f (a+ b− x)] +

1

2(x− a)2 λ3 (x)− λ1 (x)

b− a− 1

b− a

∫ b

a

f (t) dt(5.1)

=1

b− a

∫ x

a

(t− a) [f ′ (t)− λ1 (x)] dt

+1

b− a

∫ a+b−x

x

(t− a+ b

2

)[f ′ (t)− λ2 (x)] dt

+1

b− a

∫ b

a+b−x

(t− b) [f ′ (t)− λ3 (x)] dt,

for anyx ∈[a, a+b

2

]andλj (x) , j = 1, 2, 3 complex numbers.



(5.2)f (a) + f (b)

2− 1

b− a

∫ b

a

f (t) dt =1

b− a

∫ b

a

(t− a+ b

2

)[f ′ (t)− λ2] dt,

f

(a+ b

2

)+

1

8(b− a) (λ3 − λ1)−

1

b− a

∫ b

a

f (t) dt(5.3)

=1

b− a

∫ a+b2

a

(t− a) [f ′ (t)− λ1] dt+1

b− a

∫ b

a+b2

(t− b) [f ′ (t)− λ3] dt,

and

1

2

[f

(3a+ b

4

)+ f

(a+ 3b

4

)]+

1

32(b− a) (λ3 − λ1)−

1

b− a

∫ b

a

f (t) dt(5.4)

=1

b− a

∫ 3a+b4

a

(t− a) [f ′ (t)− λ1] dt

+1

b− a

∫ a+3b4

3a+b4

(t− a+ b

2

)[f ′ (t)− λ2] dt

+1

b− a

∫ b

a+3b4

(t− b) [f ′ (t)− λ3] dt




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218 S. S. DRAGOMIR

COROLLARY 5.3 (Dragomir, 2014 [10]). Assume thatf : [a, b] → C is absolutely continu-ous on[a, b]. Then we have the equality

1

2[f (x) + f (a+ b− x)]− 1

b− a

∫ b

a

f (t) dt(5.5)

=1

b− a

∫ x

a

(t− a) [f ′ (t)− λ1 (x)] dt

+1

b− a

∫ a+b−x

x

(t− a+ b

2

)[f ′ (t)− λ2 (x)] dt

+1

b− a

∫ b

a+b−x

(t− b) [f ′ (t)− λ1 (x)] dt,

for anyx ∈[a, a+b

2



f

(a+ b

2

)− 1

b− a

∫ b

a

f (t) dt(5.6)

=1

b− a

∫ a+b2

a

(t− a) [f ′ (t)− λ1] dt+1

b− a

∫ b

a+b2

(t− b) [f ′ (t)− λ1] dt,


1

2

[f

(3a+ b

4

)+ f

(a+ 3b

4

)]− 1

b− a

∫ b

a

f (t) dt(5.7)

=1

b− a

∫ 3a+b4

a

(t− a) [f ′ (t)− λ1] dt

+1

b− a

∫ a+3b4

3a+b4

(t− a+ b

2

)[f ′ (t)− λ2] dt

+1

b− a

∫ b

a+3b4

(t− b) [f ′ (t)− λ1] dt


5.2. Inequalities for Lipschitzian Derivatives. We say that the functiong : [a, b] → C isLipschitzianwith the constantL > 0 if

|g (t)− g (s)| ≤ L |t− s|for anyt, s ∈ [a, b] .

THEOREM 5.4 (Dragomir, 2014 [10]). Assume thatf : I → C is an absolutely continuousfunction on[a, b] ⊂ I , the interior ofI. If the derivativef ′ is Lipschitzian with the constantK > 0, then∣∣∣∣12 [f (x) + f (a+ b− x)] +

1

2(x− a)2 f

′ (b)− f ′ (a)

b− a− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣(5.8)

≤ 2K

3 (b− a)

[(x− a)3 +

(a+ b

2− x

)3]

for anyx ∈[a, a+b

2

].


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PROOF. If we take in the equality (5.1)λ1 (x) = f ′ (a) , λ2 (x) = f ′(

a+b2

)andλ3 (x) =

f ′ (b) then we have

1

2[f (x) + f (a+ b− x)] +

1

2(x− a)2 f

′ (b)− f ′ (a)

b− a− 1

b− a

∫ b

a

f (t) dt(5.9)

=1

b− a

∫ x

a

(t− a) [f ′ (t)− f ′ (a)] dt

+1

b− a

∫ a+b−x

x

(t− a+ b

2

)[f ′ (t)− f ′

(a+ b

2

)]dt

+1

b− a

∫ b

a+b−x

(t− b) [f ′ (t)− f ′ (b)] dt,

for anyx ∈[a, a+b

2

].

Taking the modulus in (5.9) we have∣∣∣∣12 [f (x) + f (a+ b− x)] +1

2(x− a)2 f

′ (b)− f ′ (a)

b− a− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

b− a

∫ x

a

(t− a) |f ′ (t)− f ′ (a)| dt

+1

b− a

∫ a+b−x

x

∣∣∣∣t− a+ b

2

∣∣∣∣ ∣∣∣∣f ′ (t)− f ′(a+ b

2

)∣∣∣∣ dt+

1

b− a

∫ b

a+b−x

(b− t) |f ′ (b)− f ′ (t)| dt

≤ K

b− a

[∫ x

a

(t− a)2 dt+

∫ a+b−x

x

(t− a+ b

2

)2

dt+

∫ b

a+b−x

(b− t)2 dt

]

=2

3

K

b− a

[(x− a)3 +

(a+ b

2− x

)3]

for anyx ∈[a, a+b

2


COROLLARY 5.5 (Dragomir, 2014 [10]). With the assumptions of Theorem 5.4 we have theinequalities

(5.10)

∣∣∣∣12 [f (a) + f (b)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

12K (b− a)2 ,

(5.11)

∣∣∣∣f (a+ b

2

)+

1

8(b− a) [f ′ (b)− f ′ (a)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

12K (b− a)2

and ∣∣∣∣12[f

(3a+ b

4

)+ f

(a+ 3b

4

)]+

1

32(b− a) [f ′ (b)− f ′ (a)](5.12)

− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

48K (b− a)2

The following dual result also holds:


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220 S. S. DRAGOMIR

THEOREM 5.6 (Dragomir, 2014 [10]). With the assumptions of Theorem 5.4 we have∣∣∣∣12 [f (x) + f (a+ b− x)] +1

2(x− a)2 f

′ (a+ b− x)− f ′ (x)

b− a(5.13)

− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ K

3 (b− a)

[(x− a)3 + 2

(a+ b

2− x

)3]

for anyx ∈[a, a+b

2

].

PROOF. If we take in the equality (5.1)λ1 (x) = f ′ (x) , λ2 (x) = f ′(

a+b2

)andλ3 (x) =

f ′ (a+ b− x) then we have

1

2[f (x) + f (a+ b− x)] +

1

2(x− a)2 f

′ (a+ b− x)− f ′ (x)

b− a(5.14)

− 1

b− a

∫ b

a

f (t) dt

=1

b− a

∫ x

a

(t− a) [f ′ (t)− f ′ (x)] dt

+1

b− a

∫ a+b−x

x

(t− a+ b

2

)[f ′ (t)− f ′

(a+ b

2

)]dt

+1

b− a

∫ b

a+b−x

(t− b) [f ′ (t)− f ′ (a+ b− x)] dt,

for anyx ∈[a, a+b

2

].

Taking the modulus in (5.14) we have∣∣∣∣12 [f (x) + f (a+ b− x)] +1

2(x− a)2 f

′ (a+ b− x)− f ′ (x)

b− a(5.15)

− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

b− a

∫ x

a

(t− a) |f ′ (t)− f ′ (x)| dt

+1

b− a

∫ a+b−x

x

∣∣∣∣t− a+ b

2

∣∣∣∣ ∣∣∣∣f ′ (t)− f ′(a+ b

2

)∣∣∣∣ dt+

1

b− a

∫ b

a+b−x

(b− t) |f ′ (t)− f ′ (a+ b− x)| dt

≤ K

b− a

[∫ x

a

(t− a) (x− t) dt+

∫ a+b−x

x

(t− a+ b

2

)2

dt

+

∫ b

a+b−x

(b− t) (t− a− b+ x) dt

]:= J,

for anyx ∈[a, a+b

2

].


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However ∫ x

a

(t− a) (x− t) dt =1

6(x− a)3 ,∫ b

a+b−x

(b− t) (t− a− b+ x) dt =1

6(x− a)3

and ∫ a+b−x

x

(t− a+ b

2

)2

dt =2

3

(a+ b

2− x

)3

for anyx ∈[a, a+b

2

].

Then

J =K

3 (b− a)

[(x− a)3 + 2

(a+ b

2− x

)3]

and by (5.15) we get the desired result (5.13).

COROLLARY 5.7 (Dragomir, 2014 [10]). With the assumptions of Theorem 5.4 we have

(5.16)

∣∣∣∣f (a+ b

2

)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤ 1

24K (b− a)2

and ∣∣∣∣12[f

(3a+ b

4

)+ f

(a+ 3b

4

)](5.17)

+1

32(b− a)

[f ′(a+ 3b

4

)− f ′

(3a+ b

4

)]− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣≤ 1

64K (b− a)2 .

5.3. Inequalities for Convex Functions.Suppose thatI is an interval of real numberswith interior I and f : I → R is a convex function onI. Thenf is continuous onI andhas finite left and right derivatives at each point ofI. Moreover, if x, y ∈I and x < y, thenf ′− (x) ≤ f ′+ (x) ≤ f ′− (y) ≤ f ′+ (y) which shows that bothf ′− and f ′+ are nondecreasingfunction onI. It is also known that a convex function must be differentiable except for at mostcountably many points.

For a convex functionf : I → R, the subdifferential off denoted by∂f is the set of all

functionsϕ : I → [−∞,∞] such thatϕ(

I)⊂ R and

f (x) ≥ f (a) + (x− a)ϕ (a) for anyx, a ∈ I.It is also well known that iff is convex onI, then∂f is nonempty,f ′−, f ′+ ∈ ∂f and if

ϕ ∈ ∂f , thenf ′− (x) ≤ ϕ (x) ≤ f ′+ (x) for anyx ∈ I.

In particular,ϕ is a nondecreasing function.If f is differentiable and convex onI, then∂f = f ′ .

THEOREM 5.8 (Dragomir, 2014 [10]). Let f : [a, b] → R be a convex function with thelateral derivativesf ′− (b) andf ′+ (a) finite. Then we have the inequality

(5.18)1

b− a

∫ b

a

f (t) dt ≤ 1

2[f (x) + f (a+ b− x)] +

1

2(x− a)2 f

′− (b)− f ′+ (a)

b− a

for anyx ∈[a, a+b

2

].


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222 S. S. DRAGOMIR

PROOF. If we take in the equality (5.1)λ1 (x) = f ′+ (a) , λ2 (x) = f ′+(

a+b2

)andλ3 (x) =

f ′− (b) then we have

1

2[f (x) + f (a+ b− x)] +

1

2(x− a)2 f

′− (b)− f ′+ (a)

b− a− 1

b− a

∫ b

a

f (t) dt(5.19)

=1

b− a

∫ x

a

(t− a)[ϕ (t)− f ′+ (a)

]dt

+1

b− a

∫ a+b−x

x

(t− a+ b

2

)[ϕ (t)− f ′+

(a+ b

2

)]dt

+1

b− a

∫ b

a+b−x

(t− b)[ϕ (t)− f ′− (b)

]dt

for anyx ∈[a, a+b

2

]andϕ ∈ ∂f, sincef ′ = ϕ almost everywhere on[a, b] .

Let x ∈(a, a+b

2

). We have

(t− a)[ϕ (t)− f ′+ (a)

]≥ 0 for anyt ∈ [a, x]

and

(t− b)[ϕ (t)− f ′− (b)

]= (b− t)

[f ′− (b)− ϕ (t)

]≥ 0 for anyt ∈ [a+ b− x, b] .

Also (t− a+ b

2

)[ϕ (t)− f ′+

(a+ b

2

)]≥ 0 for anyt ∈ [a, a+ b− x] .

Therefore the right hand side of (5.19) is nonnegative and the inequality (5.18) is proved.

COROLLARY 5.9 (Dragomir, 2014 [10]). With the assumptions in Theorem 5.8 we have

(5.20) 0 ≤ 1

b− a

∫ b

a

f (t) dt− f

(a+ b

2

)≤ 1

8(b− a)

[f ′− (b)− f ′+ (a)

]and

1

b− a

∫ b

a

f (t) dt ≤ 1

2

[f

(3a+ b

4

)+ f

(a+ 3b

4

)](5.21)

+1

32(b− a)

[f ′− (b)− f ′+ (a)

].

REMARK 5.2. If thef : [a, b] → R is a twice differentiable convex function with the lateralderivativesf ′− (b) andf ′+ (a) finite, then we have the inequality

0 ≤ 1

2[f (x) + f (a+ b− x)] +

1

2(x− a)2 f

′− (b)− f ′+ (a)

b− a(5.22)

− 1

b− a

∫ b

a

f (t) dt

≤ 2S

3 (b− a)

[(x− a)3 +

(a+ b

2− x

)3]

for anyx ∈[a, a+b

2

], provided that0 ≤ f ′′ (t) ≤ S for anyt ∈ (a, b) .


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In particular we have

0 ≤ f

(a+ b

2

)+

1

8(b− a) [f ′ (b)− f ′ (a)]− 1

b− a

∫ b

a

f (t) dt(5.23)

≤ 1

12S (b− a)2

and

0 ≤ 1

2

[f

(3a+ b

4

)+ f

(a+ 3b

4

)]+

1

32(b− a) [f ′ (b)− f ′ (a)](5.24)

− 1

b− a

∫ b

a

f (t) dt

≤ 1

48S (b− a)2 .

REFERENCES

[1] T. M. Apostol,Mathematical Analysis,Second Edition, Addison-Wesley Publishing Company, 1975.

[2] S. S. Dragomir, The Ostrowski integral inequality for mappings of bounded variation,Bull. Austral.Math. Soc.,60(1) (1999), 495-508.

[3] S. S. Dragomir, On the midpoint quadrature formula for mappings with bounded variation and ap-plications,Kragujevac J. Math.,22 (2000), 13-18.

[4] S. S. Dragomir, A refinement of Ostrowski’s inequality for absolutely continuous functions whosederivatives belong toL∞ and applications,Libertas Math.22 (2002), 49–63.

[5] S. S. Dragomir, A companion of Ostrowski’s inequality for functions of bounded varia-tion and applications, PreprintRGMIA Res. Rep. Coll.5 (2002), Suppl. Art. 28.[Onlinehttp://rgmia.org/papers/v5e/COIFBVApp.pdf].

[6] S. S. Dragomir, Some companions of Ostrowski’s inequality for absolutely continuous functions andapplications.Bull. Korean Math. Soc.42 (2005), no. 2, 213–230. PreprintRGMIA Res. Rep. Coll.5(2002), Suppl. Art. 29.[Online http://rgmia.org/papers/v5e/COIACFApp.pdf].

[7] S. S. Dragomir, Refinements of the generalised trapezoid and Ostrowski inequalities for functions ofbounded variation.Arch. Math.(Basel)91 (2008), no. 5, 450–460.

[8] S. S. Dragomir, Perturbed companions of Ostrowski’s inequality for absolutely contin-uous functions (II), PreprintRGMIA Res. Rep. Coll.17 (2014), Art 19. [Onlinehttp://rgmia.org/papers/v17/v17a19.pdf].

[9] S. S. Dragomir, Perturbed companions of Ostrowski’s inequality for functions of bounded variation,Asian-Eur. J. Math.8 (2015), no. 4, 155-169, 14 pp. PreprintRGMIA Res. Rep. Coll.17 (2014), Art.1. [Online http://rgmia.org/papers/v17/v17a01.pdf].

[10] S. S. Dragomir, Perturbed companions of Ostrowski’s inequality for absolutely continuous func-tions (I),An. Univ. Vest Timis. Ser. Mat.-Inform.54 (2016), no. 1, 119–138. PreprintRGMIA Res. Rep.Coll. 17 (2014), Art. 7.[Online http://rgmia.org/papers/v17/v17a07.pdf].

[11] S. S. Dragomir and Th.M. Rassias (Ed.),Ostrowski Type Inequalities and Applications in Numeri-cal Integration,Kluwer Academic Publishers, Dordrecht/Boston/London, 2002.

[12] Z. Liu, Refinement of an inequality of Grüss type for Riemann-Stieltjes integral,Soochow J. Math.,30(4) (2004), 483-489.

[13] A. Guessab and G. Schmeisser, Sharp integral inequalities of the Hermite-Hadamard type,J. Ap-prox. Th.,115(2002), 260-288.


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CHAPTER 9

Ostrowski’s Inequality for General Lebesgue Integral

1. OSTROWSKI -JENSEN TYPE I NEQUALITIES

1.1. Introduction. Let (Ω,A, µ) be a measurable space consisting of a setΩ, aσ – algebraA of parts ofΩ and a countably additive and positive measureµ onA with values inR∪∞ .Assume, for simplicity, that

∫Ωdµ = 1. Consider the Lebesgue space

L (Ω, µ) := f : Ω → R, f is µ-measurable and∫

Ω

|f (t)| dµ (t) <∞.

For simplicity of notation we write everywhere in the sequel∫

Ωwdµ instead of

∫Ωw (t) dµ (t) .

In order to provide a reverse of the celebrated Jensen’s integral inequality for convex func-tions, S.S. Dragomir obtained in 2002 [1] the following result:

THEOREM 1.1 (Dragomir 2002, [1]). Let Φ : [m,M ] ⊂ R → R be a differentiable convexfunction on(m,M) andf : Ω → [m,M ] so thatΦ f, f, Φ′ f, (Φ′ f) · f ∈ L (Ω, µ) . Thenwe have the inequality:

0 ≤∫

Ω

Φ fdµ− Φ

(∫Ω

fdµ

)(1.1)

≤∫

Ω

f · (Φ′ f) dµ−∫

Ω

Φ′ fdµ∫

Ω

fdµ

≤ 1

2[Φ′ (M)− Φ′ (m)]

∫Ω

∣∣∣∣f − ∫Ω

fdµ

∣∣∣∣ dµ.In the case of discrete measure, we have:

COROLLARY 1.2. Let Φ : [m,M ] → R be a differentiable convex function on(m,M) . Ifxi ∈ [m,M ] andwi ≥ 0 (i = 1, . . . , n) withWn :=

∑ni=1wi = 1, then one has the counterpart

of Jensen’s weighted discrete inequality:

0 ≤n∑

i=1

wiΦ (xi)− Φ

(n∑

i=1

wixi

)(1.2)

≤n∑

i=1

wiΦ′ (xi)xi −

n∑i=1

wiΦ′ (xi)

n∑i=1

wixi

≤ 1

2[Φ′ (M)− Φ′ (m)]

n∑i=1

wi

∣∣∣∣∣xi −n∑

j=1

wjxj

∣∣∣∣∣ .REMARK 1.1. We notice that the inequality between the first and the second term in (1.2)

was proved in 1994 by Dragomir & Ionescu, see [6].

If f, g : Ω → R areµ−measurable functions andf, g, fg ∈ L (Ω, µ) , then we may considertheCebyšev functional

(1.3) T (f, g) :=

∫Ω

fgdµ−∫

Ω

fdµ

∫Ω

gdµ.

225

226 S. S. DRAGOMIR

The following result is known in the literature as theGrüss inequality

(1.4) |T (f, g)| ≤ 1

4(Γ− γ) (∆− δ) ,

provided

(1.5) −∞ < γ ≤ f (t) ≤ Γ <∞, −∞ < δ ≤ g (t) ≤ ∆ <∞for µ – a.e.t ∈ Ω.

The constant14

is sharp in the sense that it cannot be replaced by a smaller quantity.If we assume that−∞ < γ ≤ f (x) ≤ Γ < ∞ for µ – a.e. x ∈ Ω, then by the Grüss

inequality forg = f and by the Schwarz’s integral inequality, we have

(1.6)∫

Ω

∣∣∣∣f − ∫Ω

fdµ

∣∣∣∣ dµ ≤[∫

Ω

f 2dµ−(∫

Ω

fdµ

)2] 1

2

≤ 1

2(Γ− γ) .

On making use of the results (1.1) and (1.6), we can state the following string of reverse in-equalities

0 ≤∫

Ω

Φ fdµ− Φ

(∫Ω

fdµ

)(1.7)

≤∫

Ω

f · (Φ′ f) dµ−∫

Ω

Φ′ fdµ∫

Ω

fdµ

≤ 1

2[Φ′ (M)− Φ′ (m)]

∫Ω

∣∣∣∣f − ∫Ω

fdµ

∣∣∣∣ dµ≤ 1

2[Φ′ (M)− Φ′ (m)]

[∫Ω

f 2dµ−(∫

Ω

fdµ

)2] 1

2

≤ 1

4[Φ′ (M)− Φ′ (m)] (M −m) ,

provided thatΦ : [m,M ] ⊂ R → R is a differentiable convex function on(m,M) andf : Ω →[m,M ] so thatΦ f, f, Φ′ f, f · (Φ′ f) ∈ L (Ω, µ) , with

∫Ωdµ = 1.

The following reverse of the Jensen’s inequality also holds [2]:

THEOREM 1.3. Let Φ : I → R be a continuous convex function on the interval of realnumbersI andm,M ∈ R,m < M with [m,M ] ⊂ I, whereI is the interior ofI. If f : Ω → Ris µ-measurable, satisfies the bounds

−∞ < m ≤ f (t) ≤M <∞ for µ-a.e.t ∈ Ω

and such thatf, Φ f ∈ L (Ω, µ) , then

0 ≤∫

Ω

Φ fdµ− Φ

(∫Ω

fdµ

)(1.8)

≤(M −

∫Ω

fdµ

)(∫Ω

fdµ−m

)Φ′− (M)− Φ′

+ (m)

M −m

≤ 1

4(M −m)

[Φ′− (M)− Φ′

+ (m)],

whereΦ′− is the left andΦ′

+ is the right derivative of the convex functionΦ.

For other reverse of Jensen inequality and applications to divergence measures see [2].In 1938, A. Ostrowski [8], proved the following inequality concerning the distance between

the integral mean1b−a

∫ b

aΦ (t) dt and the valueΦ (x), x ∈ [a, b].


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THEOREM 1.4. Let Φ : [a, b] → R be continuous on[a, b] and differentiable on(a, b) suchthatΦ′ : (a, b) → R is bounded on(a, b), i.e.,‖Φ′‖∞ := sup

t∈(a,b)

|Φ′ (t)| <∞. Then

(1.9)

∣∣∣∣Φ (x)− 1

b− a

∫ b

a

f (t) dt

∣∣∣∣ ≤1

4+

(x− a+b

2

b− a

)2 ‖Φ′‖∞ (b− a) ,

for all x ∈ [a, b] and the constant14


Motivated by the above results, in this section we investigate the magnitude of the quantity∫Ω

Φ gdµ− Φ (x)− λ

(∫Ω

gdµ− x

), x ∈ [a, b] ,

for various assumptions on the absolutely continuous functionΦ, which in the particular case ofx =

∫Ωgdµ provides some results connected with Jensen’s inequality while in the caseλ = 0

provides some generalizations of Ostrowski’s inequality. Applications for divergence measuresare provided as well.

1.2. Some Identities.The following result holds:

LEMMA 1.5 (Dragomir, 2014 [4]). Let Φ : I → C be an absolutely continuous functionson [a, b] ⊂ I, the interior ofI. If g : Ω → [a, b] is Lebesgueµ-measurable onΩ and such thatΦ g, g ∈ L (Ω, µ) , then we have the equality∫

Ω


(∫Ω

gdµ− x

)(1.10)

=

∫Ω

[(g − x)

∫ 1

0

(Φ′ ((1− s)x+ sg)− λ) ds

]dµ

for anyλ ∈ C andx ∈ [a, b].In particular, we have

(1.11)∫

Ω

Φ gdµ− Φ (x) =

∫Ω

[(g − x)

∫ 1

0

Φ′ ((1− s)x+ sg) ds

]dµ,


PROOF. SinceΦ is absolutely continuous on[a, b] , then for anyu, v ∈ [a, b] we have

(1.12) Φ (u)− Φ (v) = (u− v)

∫ 1

0

Φ′ ((1− s) v + su) ds.

This implies that

Φ (g (t))− Φ (x) = (g (t)− x)

∫ 1

0

Φ′ ((1− s)x+ sg (t)) ds

for anyt ∈ Ω, or equivalently,

(1.13) Φ g − Φ (x) = (g − x)

∫ 1

0

Φ′ ((1− s)x+ sg) ds.

SinceΦ : I → C is an absolutely continuous functions on[a, b] the Lebesgue integral overµ inthe right side of (1.10) exists for anyλ ∈ C andx ∈ [a, b] .

Integrating (1.13) over the measureµ onΩ and since∫

Ωdµ = 1, then we have

(1.14)∫

Ω

Φ gdµ− Φ (x) =

∫Ω

[(g − x)

∫ 1

0

Φ′ ((1− s)x+ sg) ds

]dµ.


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228 S. S. DRAGOMIR

Now, observe that forλ ∈ C we have∫Ω

[(g − x)

∫ 1

0

(Φ′ ((1− s)x+ sg)− λ) ds

]dµ(1.15)

=

∫Ω

[(g − x)

(∫ 1

0

Φ′ ((1− s)x+ sg) ds− λ

)]dµ

=

∫Ω

[(g − x)

∫ 1

0

Φ′ ((1− s)x+ sg) ds

]dµ− λ

∫Ω

(g − x) dµ

=

∫Ω

[(g − x)

∫ 1

0

Φ′ ((1− s)x+ sg) ds

]dµ− λ

(∫Ω

gdµ− x

).

Making use of (1.14) and (1.15) we deduce the desired result (1.10).

REMARK 1.2. With the assumptions of Lemma 1.5 we have∫Ω

Φ gdµ− Φ

(a+ b

2

)(1.16)

=

∫Ω

[(g − a+ b

2

)∫ 1

0

Φ′(

(1− s)a+ b

2+ sg

)ds

]dµ.

COROLLARY 1.6 (Dragomir, 2014 [4]). With the assumptions of Lemma 1.5 we have∫Ω

Φ gdµ− Φ

(∫Ω

gdµ

)(1.17)

=

∫Ω

[(g −

∫Ω

gdµ

)∫ 1

0

Φ′(

(1− s)

∫Ω

gdµ+ sg

)ds

]dµ.

PROOF. We observe that sinceg : Ω → [a, b] and∫

Ωdµ = 1 then

∫Ωgdµ ∈ [a, b] and by

takingx =∫

Ωgdµ in (1.11) we get (1.17).


Φ gdµ− 1

b− a

∫ b

a

Φ (x) dx− λ

(∫Ω

gdµ− a+ b

2

)(1.18)

=

∫Ω

1

b− a

∫ b

a

[(g − x)

∫ 1

0

(Φ′ ((1− s)x+ sg)− λ) ds

]dx

dµ.

PROOF. Follows by integrating the identity (1.10) overx ∈ [a, b] , dividing by b − a > 0and using Fubini’s theorem.

COROLLARY 1.8 (Dragomir, 2014 [4]). Let Φ : I → C be an absolutely continuous func-tions on[a, b] ⊂ I, the interior ofI. If g, h : Ω → [a, b] are Lebesgueµ-measurable onΩ andsuch thatΦ g, Φ h, g, h ∈ L (Ω, µ) , then we have the equality∫

Ω

Φ gdµ−∫

Ω

Φ hdµ− λ

(∫Ω

gdµ−∫

Ω

hdµ

)(1.19)

=

∫Ω

∫Ω

[(g (t)− h (τ))

∫ 1

0

(Φ′ ((1− s)h (τ) + sg (t))− λ) ds

]× dµ (t) dµ (τ)

for anyλ ∈ C andx ∈ [a, b].


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In particular, we have∫Ω

Φ gdµ−∫

Ω

Φ hdµ(1.20)

=

∫Ω

∫Ω

[(g (t)− h (τ))

∫ 1

0

Φ′ ((1− s)h (τ) + sg (t)) ds

]dµ (t) dµ (τ) ,


PROOF. From (1.10) we have for anyτ ∈ Ω that∫Ω

Φ gdµ− Φ (h (τ))− λ

(∫Ω

gdµ− Φ (h (τ))

)=

∫Ω

[(g − Φ (h (τ)))

∫ 1

0

(Φ′ ((1− s) Φ (h (τ)) + sg)− λ) ds

]dµ

for anyλ ∈ C andx ∈ [a, b].Integrating onΩ overdµ (τ) and using Fubini’s theorem we get the desired result (1.19).

REMARK 1.3. The above inequality (1.19) can be extended for two measures as follows∫Ω1

Φ gdµ1 −∫

Ω2

Φ hdµ2 − λ

(∫Ω1

gdµ1 −∫

Ω2

hdµ2

)(1.21)

=

∫Ω1

∫Ω2

[(g (t)− h (τ))

∫ 1

0

(Φ′ ((1− s)h (τ) + sg (t))− λ) ds

]× dµ1 (t) dµ2 (τ) ,

for anyλ ∈ C andx ∈ [a, b] and provided thatΦg, g ∈ L (Ω1, µ1) while Φh, h ∈ L (Ω2, µ2) .

REMARK 1.4. If w ≥ 0 µ-almost everywhere(µ-a.e.) on Ω with∫

Ωwdµ > 0, then by

replacingdµ with wdµRΩ wdµ

in (1.10) we have the weighted equality

1∫Ωwdµ

∫Ω

w (Φ g) dµ− Φ (x)− λ

(1∫

Ωwdµ

∫Ω

wgdµ− x

)(1.22)

=1∫

Ωwdµ

∫Ω

w ·[(g − x)

∫ 1

0

(Φ′ ((1− s)x+ sg)− λ) ds

]dµ

for anyλ ∈ C andx ∈ [a, b], providedΦ g, g ∈ Lw (Ω, µ) where

Lw (Ω, µ) :=

g|∫

Ω

w |g| dµ <∞.

The other equalities have similar weighted versions. However the details are omitted.

If we use the discrete measure, then forΦ : I → C an absolutely continuous functions on[a, b] ⊂ I, the interior ofI, xj ∈ [a, b] andpj ≥ 0 with

∑nj=1 pj = 1 we can state the following

identityn∑

j=1

pjΦ (xj)− Φ (x)− λ

(n∑

j=1

pjxj − x

)(1.23)

=n∑

j=1

pj

[(xj − x)

∫ 1

0

(Φ′ ((1− s)x+ sxj)− λ) ds

]for anyλ ∈ C andx ∈ [a, b].


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230 S. S. DRAGOMIR


(1.24)n∑

j=1

pjΦ (xj)− Φ (x) =n∑

j=1

pj

[(xj − x)

∫ 1

0

Φ′ ((1− s)x+ sxj) ds

]for anyx ∈ [a, b] and

n∑j=1

pjΦ (xj)− Φ

(a+ b

2

)(1.25)

=n∑

j=1

pj

[(xj −

a+ b

2

)∫ 1

0

Φ′(

(1− s)a+ b

2+ sxj

)ds

]and

n∑j=1

pjΦ (xj)− Φ

(n∑

k=1

pkxk

)(1.26)

=n∑

j=1

pj

[(xj −

n∑k=1

pkxk

)∫ 1

0

Φ′

((1− s)

n∑k=1

pkxk + sxj

)ds

].

If xj ∈ [a, b] andpj ≥ 0, j ∈ 1, ..., n with∑n

j=1 pj = 1 and if yk ∈ [a, b] andqk ≥ 0,

k ∈ 1, ...,m with∑m

k=1 qk = 1, then we can state the following identity as well:

n∑j=1

pjΦ (xj)−m∑

k=1

qkΦ (yk)− λ

(n∑

j=1

pjxj −m∑

k=1

qkyk

)(1.27)

=n∑

j=1

pj

m∑k=1

qk

[(xj − yk)

∫ 1

0

(Φ′ ((1− s) yk + sxj)− λ) ds

].

In particular, we haven∑

j=1

pjΦ (xj)−m∑

k=1

qkΦ (yk)(1.28)

=n∑

j=1

pj

m∑k=1

qk

[(xj − yk)

∫ 1

0

Φ′ ((1− s) yk + sxj) ds

].

1.3. Bounds in Terms ofp-Norms. We use the notations

‖k‖Ω,p :=

∫Ω

|k (t)|p dµ (t)

1/p

<∞, p ≥ 1, k ∈ Lp (Ω, µ) ;

ess supt∈Ω |k (t)| <∞, p = ∞, k ∈ L∞ (Ω, µ) ;

and

‖Φ‖[0,1],p :=

(∫ 1

0|Φ (s)|p ds

)1/p

<∞, p ≥ 1, Φ ∈ Lp (0, 1) ;

ess sups∈[0,1] |Φ (s)| <∞, p = ∞, Φ ∈ L∞ (0, 1) .


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If we consider the identity function : [0, 1] → [0, 1], ` (s) = s we have

∫ 1

0

|Φ′ ((1− s)x+ sg (t))− λ|p ds = ‖Φ′ ((1− `)x+ `g (t))− λ‖p[0,1],p

and

ess sups∈[0,1]

|Φ′ ((1− s)x+ sg (t))− λ| = ‖Φ′ ((1− `)x+ `g (t))− λ‖[0,1],∞

for t ∈ Ω.

THEOREM 1.9 (Dragomir, 2014 [4]). LetΦ : I → C be an absolutely continuous functionson [a, b] ⊂ I, the interior ofI. If g : Ω → [a, b] is Lebesgueµ-measurable onΩ and such thatΦ g, g ∈ L (Ω, µ) , then∣∣∣∣∫

Ω


(∫Ω

gdµ− x

)∣∣∣∣(1.29)

≤∫

Ω

|g − x| ‖Φ′ ((1− `)x+ `g)− λ‖[0,1],1 dµ

≤

‖g − x‖Ω,∞

∥∥∥‖Φ′ ((1− `)x+ `g)− λ‖[0,1],1

∥∥∥Ω,1

;

‖g − x‖Ω,p

∥∥∥‖Φ′ ((1− `)x+ `g)− λ‖[0,1],1

∥∥∥Ω,q

p > 1, 1p

+ 1q

= 1;

‖g − x‖Ω,1

∥∥∥‖Φ′ ((1− `)x+ `g)− λ‖[0,1],1

∥∥∥Ω,∞

;

for anyλ ∈ C andx ∈ [a, b] .In particular, we have∣∣∣∣∫

Ω

Φ gdµ− Φ (x)

∣∣∣∣(1.30)

≤∫

Ω

|g − x| ‖Φ′ ((1− `)x+ `g)‖[0,1],1 dµ

≤

‖g − x‖Ω,∞

∥∥∥‖Φ′ ((1− `)x+ `g)‖[0,1],1

∥∥∥Ω,1

;

‖g − x‖Ω,p

∥∥∥‖Φ′ ((1− `)x+ `g)‖[0,1],1

∥∥∥Ω,q

p > 1, 1p

+ 1q

= 1;

‖g − x‖Ω,1

∥∥∥‖Φ′ ((1− `)x+ `g)‖[0,1],1

∥∥∥Ω,∞

;



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232 S. S. DRAGOMIR

PROOF. Taking the modulus in the equality (1.10), we have

∣∣∣∣∫Ω


(∫Ω

gdµ− x

)∣∣∣∣(1.31)

≤∫

Ω

∣∣∣∣(g − x)

∫ 1

0

(Φ′ ((1− s)x+ sg)− λ) ds

∣∣∣∣ dµ≤∫

Ω

|g − x|∫ 1

0

|Φ′ ((1− s)x+ sg)− λ| dsdµ

=

∫Ω

|g − x| ‖Φ′ ((1− `)x+ `g)− λ‖[0,1],1 dµ

for anyλ ∈ C andx ∈ [a, b] .Utilising Hölder’s inequality for theµ-measurable functionsF,G : Ω → C,

∣∣∣∣∫Ω

FGdµ

∣∣∣∣ ≤ (∫Ω

|F |p dµ)1/p(∫

Ω

|G|q dµ)1/q

, p > 1,1

p+

1

q= 1

and

∣∣∣∣∫Ω

FGdµ

∣∣∣∣ ≤ ess supt∈Ω

|F (t)|∫

Ω

|G| dµ

we get from (1.31) the desired result (1.29).

REMARK 1.5. If we takex = a+b2

in (1.29), then we get

∣∣∣∣∫Ω

Φ gdµ− Φ

(a+ b

2

)− λ

(∫Ω

gdµ− a+ b

2

)∣∣∣∣(1.32)

≤∫

Ω

∣∣∣∣g − a+ b

2

∣∣∣∣ ∥∥∥∥Φ′(

(1− `)a+ b

2+ `g

)− λ

∥∥∥∥[0,1],1

dµ

≤

∥∥g − a+b2

∥∥Ω,∞

∥∥∥∥∥Φ′ ((1− `) a+b2

+ `g)− λ∥∥

[0,1],1

∥∥∥Ω,1

;

∥∥g − a+b2

∥∥Ω,p

∥∥∥∥∥Φ′ ((1− `) a+b2

+ `g)− λ∥∥

[0,1],1

∥∥∥Ω,q

p > 1, 1p

+ 1q

= 1;

∥∥g − a+b2

∥∥Ω,1

∥∥∥∥∥Φ′ ((1− `) a+b2

+ `g)− λ∥∥

[0,1],1

∥∥∥Ω,∞

;


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for anyλ ∈ C and, in particular, forλ = 0 we have∣∣∣∣∫Ω

Φ gdµ− Φ

(a+ b

2

)∣∣∣∣(1.33)

≤∫

Ω

∣∣∣∣g − a+ b

2

∣∣∣∣ ∥∥∥∥Φ′(

(1− `)a+ b

2+ `g

)∥∥∥∥[0,1],1

dµ

≤

∥∥g − a+b2

∥∥Ω,∞

∥∥∥∥∥Φ′ ((1− `) a+b2

+ `g)∥∥

[0,1],1

∥∥∥Ω,1

;

∥∥g − a+b2

∥∥Ω,p

∥∥∥∥∥Φ′ ((1− `) a+b2

+ `g)∥∥

[0,1],1

∥∥∥Ω,q

p > 1, 1p

+ 1q

= 1;

∥∥g − a+b2

∥∥Ω,1

∥∥∥∥∥Φ′ ((1− `) a+b2

+ `g)∥∥

[0,1],1

∥∥∥Ω,∞

;

If we takex =∫

Ωgdµ in (1.29), then we get∣∣∣∣∫

Ω

Φ gdµ− Φ

(∫Ω

gdµ

)∣∣∣∣(1.34)

≤∫

Ω

∣∣∣∣g − ∫Ω

gdµ

∣∣∣∣ ∥∥∥∥Φ′(

(1− `)

∫Ω

gdµ+ `g

)− λ

∥∥∥∥[0,1],1

dµ

≤

∥∥g − ∫Ωgdµ

∥∥Ω,∞

∥∥∥∥∥Φ′ ((1− `)∫

Ωgdµ+ `g

)− λ∥∥

[0,1],1

∥∥∥Ω,1

;


∥∥Ω,p

∥∥∥∥∥Φ′ ((1− `)∫

Ωgdµ+ `g

)− λ∥∥

[0,1],1

∥∥∥Ω,q

p > 1, 1p

+ 1q

= 1;


∥∥Ω,1

∥∥∥∥∥Φ′ ((1− `)∫

Ωgdµ+ `g

)− λ∥∥

[0,1],1

∥∥∥Ω,∞

;

for anyλ ∈ C and, in particular, forλ = 0 we have∣∣∣∣∫Ω

Φ gdµ− Φ

(∫Ω

gdµ

)∣∣∣∣(1.35)

≤∫

Ω

∣∣∣∣g − ∫Ω

gdµ

∣∣∣∣ ∥∥∥∥Φ′(

(1− `)

∫Ω

gdµ+ `g

)∥∥∥∥[0,1],1

dµ

≤


∥∥Ω,∞

∥∥∥∥∥Φ′ ((1− `)∫

Ωgdµ+ `g

)∥∥[0,1],1

∥∥∥Ω,1

;


∥∥Ω,p

∥∥∥∥∥Φ′ ((1− `)∫

Ωgdµ+ `g

)∥∥[0,1],1

∥∥∥Ω,q

p > 1, 1p

+ 1q

= 1;


∥∥Ω,1

∥∥∥∥∥Φ′ ((1− `)∫

Ωgdµ+ `g

)∥∥[0,1],1

∥∥∥Ω,∞

.

COROLLARY 1.10 (Dragomir, 2014 [4]). Let Φ : I → C be an absolutely continuousfunctions on[a, b] ⊂ I, the interior ofI. If g : Ω → [a, b] is Lebesgueµ-measurable onΩ and


http://ajmaa.org

234 S. S. DRAGOMIR

such thatΦ g, g ∈ L (Ω, µ) , then

(1.36)

∣∣∣∣∫Ω

Φ gdµ− Φ (x)

∣∣∣∣ ≤ ‖Φ′‖[a,b],∞

∫Ω

|g − x| dµ

for anyx ∈ [a, b] .In particular, we have

(1.37)

∣∣∣∣∫Ω

Φ gdµ− Φ

(a+ b

2

)∣∣∣∣ ≤ ‖Φ′‖[a,b],∞

∫Ω

∣∣∣∣g − a+ b

2

∣∣∣∣ dµand

(1.38)

∣∣∣∣∫Ω

Φ gdµ− Φ

(∫Ω

gdµ

)∣∣∣∣ ≤ ‖Φ′‖[a,b],∞

∫Ω

∣∣∣∣g − ∫Ω

gdµ

∣∣∣∣ dµ.PROOF. We have from (1.29) that

(1.39)

∣∣∣∣∫Ω

Φ gdµ− Φ (x)

∣∣∣∣ ≤ ∫Ω

|g − x|(∫ 1

0

|Φ′ ((1− s)x+ sg)| ds)dµ

for anyx ∈ [a, b] .However, for anyt ∈ Ω and almost everys ∈ [0, 1] we have

|Φ′ ((1− s)x+ sg (t))| ≤ ess supu∈[a,b]

|Φ′ (u)| = ‖Φ′‖[a,b],∞ ,

for anyx ∈ [a, b] .Making use of (1.39) we get (1.36).

REMARK 1.6. We remark that, the quantity from Corollary 1.10

δµ (g, x) :=

∫Ω

|g − x| dµ

cannot be computed in general.However, in the case whenΩ = [a, b] , g : [a, b] → [a, b] , g (t) = t andµ (t) = 1

b−adt, we

have

δµ (g, x) :=1

b− a

∫ b

a

|t− x| dt =1

b− a

[∫ x

a

(x− t) dt+

∫ b

x

(t− x) dt

]=

1

b− a

[(x− a)2 + (b− x)2]

=

1

4+

(x− a+b

2

b− a

)2 (b− a) ,

wherex ∈ [a, b] .Utilising the inequality (1.36) we get Ostrowski’s inequality

(1.40)

∣∣∣∣ 1

b− a

∫ b

a

Φ (t) dt− Φ (x)

∣∣∣∣ ≤ ‖Φ′‖[a,b],∞

1

4+

(x− a+b

2

b− a

)2 (b− a)

for anyx ∈ [a, b] .From the inequalities (1.37) and (1.38) we get the midpoint inequality

(1.41)

∣∣∣∣ 1

b− a

∫ b

a

Φ (t) dt− Φ

(a+ b

2

)∣∣∣∣ ≤ 1

4‖Φ′‖[a,b],∞ (b− a) .


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REMARK 1.7. If we consider the dispersion or the standard variation

σµ (g) :=

(∫Ω

(g −

∫Ω

gdµ

)2

dµ

)1/2

=

(∫Ω

g2dµ−(∫

Ω

gdµ

)2)1/2

,

then by (1.38) we have the inequalities∣∣∣∣∫Ω

Φ gdµ− Φ

(∫Ω

gdµ

)∣∣∣∣ ≤ ‖Φ′‖[a,b],∞ δµ

(g,

∫Ω

gdµ

)(1.42)

≤ ‖Φ′‖[a,b],∞ σµ (g) .

In general, we have by Cauchy-Bunyakovsky-Schwarz’s inequality that

(1.43) δµ (g, x) :=

∫Ω

|g − x| dµ ≤(∫

Ω

(g − x)2 dµ

)1/2

.

Since ∫Ω

(g − x)2 dµ

=

∫Ω

(g −

∫Ω

gdµ+

∫Ω

gdµ− x

)2

dµ

=

∫Ω

(g −

∫Ω

gdµ

)2

dµ+ 2

∫Ω

(g −

∫Ω

gdµ

)(∫Ω

gdµ− x

)dµ

+

∫Ω

(∫Ω

gdµ− x

)2

dµ

=

∫Ω

(g −

∫Ω

gdµ

)2

dµ+

(∫Ω

gdµ− x

)2

for anyx ∈ [a, b] , then by (1.36) and (1.43) we get the inequalities∣∣∣∣∫Ω

Φ gdµ− Φ (x)

∣∣∣∣ ≤ ‖Φ′‖[a,b],∞ δµ (g, x)(1.44)

≤ ‖Φ′‖[a,b],∞

[σ2

µ (g) +

(∫Ω

gdµ− x

)2]1/2


If we use the discrete measure, then from (1.44) we have∣∣∣∣∣n∑

j=1

pjΦ (xj)− Φ (x)

∣∣∣∣∣(1.45)

≤ ‖Φ′‖[a,b],∞

n∑j=1

pj |xj − x|

≤ ‖Φ′‖[a,b],∞

n∑j=1

pjx2j −

(n∑

j=1

pjxj

)2

+

(n∑

j=1

pjxj − x

)21/2

,

for anyx ∈ [a, b] , wherexj ∈ [a, b] andpj ≥ 0 with∑n

j=1 pj = 1.


http://ajmaa.org

236 S. S. DRAGOMIR

In particular, we have∣∣∣∣∣n∑

j=1

pjΦ (xj)− Φ

(a+ b

2

)∣∣∣∣∣ ≤ ‖Φ′‖[a,b],∞

n∑j=1

pj

∣∣∣∣xj −a+ b

2

∣∣∣∣(1.46)

≤ 1

2(b− a) ‖Φ′‖[a,b],∞

and ∣∣∣∣∣n∑

j=1

pjΦ (xj)− Φ

(n∑

k=1

pkxk

)∣∣∣∣∣(1.47)

≤ ‖Φ′‖[a,b],∞

n∑j=1

pj

∣∣∣∣∣xj −n∑

k=1

pkxk

∣∣∣∣∣≤ ‖Φ′‖[a,b],∞

n∑j=1

pjx2j −

(n∑

j=1

pjxj

)21/2

≤ 1

2(b− a) ‖Φ′‖[a,b],∞ .


U[a,b] (γ,Γ)

:=f : [a, b] → C|Re

[(Γ− f (t))

(f (t)− γ


and

∆[a,b] (γ,Γ) :=

f : [a, b] → C|

∣∣∣∣f (t)− γ + Γ

2

∣∣∣∣ ≤ 1

2|Γ− γ| for a.e.t ∈ [a, b]

.


PROPOSITION1.11. For anyγ,Γ ∈ C, γ 6= Γ, we have thatU[a,b] (γ,Γ) and ∆[a,b] (γ,Γ)are nonempty, convex and closed sets and

(1.48) U[a,b] (γ,Γ) = ∆[a,b] (γ,Γ) .


2

∣∣∣∣ ≤ 1

2|Γ− γ|

if and only ifRe [(Γ− z) (z − γ)] ≥ 0.


1

4|Γ− γ|2 −

∣∣∣∣z − γ + Γ

2

∣∣∣∣2 = Re [(Γ− z) (z − γ)]







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(1.51) ∅ 6= S[a,b] (γ,Γ) ⊆ U[a,b] (γ,Γ) .


THEOREM1.13 (Dragomir, 2014 [4]). LetΦ : I → C be an absolutely continuous functionson [a, b] ⊂ I, the interior ofI. For someγ,Γ ∈ C, γ 6= Γ, assume thatΦ′ ∈ U[a,b] (γ,Γ) =∆[a,b] (γ,Γ) . If g : Ω → [a, b] is Lebesgueµ-measurable onΩ and such thatΦg, g ∈ L (Ω, µ) ,then we have the inequality

(1.52)

∣∣∣∣∫Ω

Φ gdµ− Φ (x)− γ + Γ

2

(∫Ω

gdµ− x

)∣∣∣∣ ≤ 1

2|Γ− γ|

∫Ω

|g − x| dµ

for anyx ∈ [a, b].In particular, we have∣∣∣∣∫

Ω

Φ gdµ− Φ

(a+ b

2

)− γ + Γ

2

(∫Ω

gdµ− a+ b

2

)∣∣∣∣(1.53)

≤ 1

2|Γ− γ|

∫Ω

∣∣∣∣g − a+ b

2

∣∣∣∣ dµ ≤ 1

4(b− a) |Γ− γ|

and ∣∣∣∣∫Ω

Φ gdµ− Φ

(∫Ω

gdµ

)∣∣∣∣ ≤ 1

2|Γ− γ|

∫Ω

∣∣∣∣g − ∫Ω

gdµ

∣∣∣∣ dµ(1.54)

≤ 1

2|Γ− γ|

(∫Ω

g2dµ−(∫

Ω

gdµ

)2)1/2

≤ 1

4(b− a) |Γ− γ| .

PROOF. By the equality (1.10) forλ = γ+Γ2

we have∫Ω


2

(∫Ω

gdµ− x

)(1.55)

=

∫Ω

[(g − x)

∫ 1

0

(Φ′ ((1− s)x+ sg)− γ + Γ

2

)ds

]dµ.

SinceΦ′ ∈ ∆[a,b] (γ,Γ) we have

(1.56)

∣∣∣∣Φ′ ((1− s)x+ sg (t))− γ + Γ

2

∣∣∣∣ ≤ 1

2|Γ− γ|

for a.e.s ∈ [0, 1] and for anyx ∈ [a, b] and anyt ∈ Ω.Integrating (1.56) overs on [0, 1] we get

(1.57)∫ 1

0

∣∣∣∣Φ′ ((1− s)x+ sg (t))− γ + Γ

2

∣∣∣∣ ds ≤ 1

2|Γ− γ|

for anyx ∈ [a, b] and anyt ∈ Ω.


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238 S. S. DRAGOMIR

Taking the modulus in (1.55) we get via (1.57) that∣∣∣∣∫Ω


2

(∫Ω

gdµ− x

)∣∣∣∣(1.58)

≤∫

Ω

[|g − x|

∣∣∣∣∫ 1

0

(Φ′ ((1− s)x+ sg)− γ + Γ

2

)ds

∣∣∣∣] dµ≤∫

Ω

[|g − x|

∫ 1

0

∣∣∣∣Φ′ ((1− s)x+ sg (t))− γ + Γ

2

∣∣∣∣ ds] dµ≤ 1

2|Γ− γ|

∫Ω

|g − x| dµ

and the proof of (1.52) is completed.

COROLLARY 1.14 (Dragomir, 2014 [4]). LetΦ : I → R be a convex function on[a, b] ⊂ I,the interior ofI. If g : Ω → [a, b] is Lebesgueµ-measurable onΩ and such thatΦ g, g ∈L (Ω, µ) , then we have the inequality∣∣∣∣∫

Ω

Φ gdµ− Φ (x)−Φ′

+ (a) + Φ′− (b)

2

(∫Ω

gdµ− x

)∣∣∣∣(1.59)

≤ 1

2

[Φ′− (b)− Φ′

+ (a)] ∫

Ω

|g − x| dµ


Ω

Φ gdµ− Φ

(a+ b

2

)−

Φ′+ (a) + Φ′

− (b)

2

(∫Ω

gdµ− a+ b

2

)∣∣∣∣(1.60)

≤ 1

2

[Φ′− (b)− Φ′

+ (a)] ∫

Ω

∣∣∣∣g − a+ b

2

∣∣∣∣ dµ ≤ 1

2(b− a)

[Φ′− (b)− Φ′

+ (a)]

and

(1.61) 0 ≤∫

Ω

Φ gdµ− Φ

(∫Ω

gdµ

)≤ 1

2

[Φ′− (b)− Φ′

+ (a)] ∫

Ω

∣∣∣∣g − ∫Ω

gdµ

∣∣∣∣ dµ.The inequality (1.61) is not as good as (1.1).The discrete case is as follows:

REMARK 1.8. LetΦ : I → C be an absolutely continuous functions on[a, b] ⊂ I, theinterior of I. For someγ,Γ ∈ C, γ 6= Γ, assume thatΦ′ ∈ ∆[a,b] (γ,Γ) . If xj ∈ [a, b] andpj ≥ 0 with

∑nj=1 pj = 1 then we have the inequality

(1.62)

∣∣∣∣∣n∑

j=1

pjΦ (xj)− Φ (x)− γ + Γ

2

(n∑

k=1

pkxk − x

)∣∣∣∣∣ ≤ 1

2|Γ− γ|

n∑j=1

pj |xj − x|

for anyx ∈ [a, b].In particular, we have∣∣∣∣∣

n∑j=1

pjΦ (xj)− Φ

(a+ b

2

)− γ + Γ

2

(n∑

k=1

pkxk −a+ b

2

)∣∣∣∣∣(1.63)

≤ 1

2|Γ− γ|

n∑j=1

pj

∣∣∣∣xj −a+ b

2

∣∣∣∣ ≤ 1

4(b− a) |Γ− γ|


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and ∣∣∣∣∣n∑

j=1

pjΦ (xj)− Φ

(n∑

k=1

pkxk

)∣∣∣∣∣(1.64)

≤ 1

2|Γ− γ|

n∑j=1

pj

∣∣∣∣∣xj −n∑

k=1

pkxk

∣∣∣∣∣≤ 1

2|Γ− γ|

n∑j=1

pjx2j −

(n∑

k=1

pkxk

)21/2

≤ 1

4(b− a) |Γ− γ| .

If Φ : I → R is a convex function on[a, b] , then we have∣∣∣∣∣n∑

j=1

pjΦ (xj)− Φ (x)−Φ′

+ (a) + Φ′− (b)

2

(n∑

k=1

pkxk − x

)∣∣∣∣∣(1.65)

≤ 1

2

[Φ′− (b)− Φ′

+ (a)] n∑

j=1

pj |xj − x|

for anyx ∈ [a, b].In particular, we have∣∣∣∣∣

n∑j=1

pjΦ (xj)− Φ

(a+ b

2

)−

Φ′+ (a) + Φ′

− (b)

2

(n∑

k=1

pkxk −a+ b

2

)∣∣∣∣∣(1.66)

≤ 1

2

[Φ′− (b)− Φ′

+ (a)] n∑

j=1

pj

∣∣∣∣xj −a+ b

2

∣∣∣∣ ≤ 1

4(b− a)

[Φ′− (b)− Φ′

+ (a)].

2. OTHER OSTROWSKI -JENSEN TYPE I NEQUALITIES

2.1. Some Identities.The following result holds [4]:

LEMMA 2.1 (Dragomir, 2014 [4]). Let Φ : I → C be an absolutely continuous functionson [a, b] ⊂ I, the interior ofI. If g : Ω → [a, b] is Lebesgueµ-measurable onΩ and such thatΦ g, g ∈ L (Ω, µ) , then we have the equality∫

Ω


(∫Ω

gdµ− x

)(2.1)

=

∫Ω

[(g − x)

∫ 1

0

(Φ′ ((1− s)x+ sg)− λ) ds

]dµ

for anyλ ∈ C andx ∈ [a, b].In particular, we have

(2.2)∫

Ω

Φ gdµ− Φ (x) =

∫Ω

[(g − x)

∫ 1

0

Φ′ ((1− s)x+ sg) ds

]dµ,



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240 S. S. DRAGOMIR

REMARK 2.1. With the assumptions of Lemma 2.1 we have∫Ω

Φ gdµ− Φ

(a+ b

2

)(2.3)

=

∫Ω

[(g − a+ b

2

)∫ 1

0

Φ′(

(1− s)a+ b

2+ sg

)ds

]dµ.


Φ gdµ− Φ

(∫Ω

gdµ

)(2.4)

=

∫Ω

[(g −

∫Ω

gdµ

)∫ 1

0

Φ′(

(1− s)

∫Ω

gdµ+ sg

)ds

]dµ.

PROOF. We observe that sinceg : Ω → [a, b] and∫

Ωdµ = 1 then

∫Ωgdµ ∈ [a, b] and by

takingx =∫

Ωgdµ in (2.2) we get (2.4).


Φ gdµ− 1

b− a

∫ b

a

Φ (x) dx− λ

(∫Ω

gdµ− a+ b

2

)(2.5)

=

∫Ω

1

b− a

∫ b

a

[(g − x)

∫ 1

0

(Φ′ ((1− s)x+ sg)− λ) ds

]dx

dµ.

PROOF. Follows by integrating the identity (2.1) overx ∈ [a, b] , dividing by b− a > 0 andusing Fubini’s theorem.

COROLLARY 2.4 (Dragomir, 2014 [4]). Let Φ : I → C be an absolutely continuous func-tions on[a, b] ⊂ I, the interior ofI. If g, h : Ω → [a, b] are Lebesgueµ-measurable onΩ andsuch thatΦ g, Φ h, g, h ∈ L (Ω, µ) , then we have the equality∫

Ω

Φ gdµ−∫

Ω

Φ hdµ− λ

(∫Ω

gdµ−∫

Ω

hdµ

)(2.6)

=

∫Ω

∫Ω

[(g (t)− h (τ))

∫ 1

0

(Φ′ ((1− s)h (τ) + sg (t))− λ) ds

]× dµ (t) dµ (τ)

for anyλ ∈ C andx ∈ [a, b].In particular, we have∫

Ω

Φ gdµ−∫

Ω

Φ hdµ(2.7)

=

∫Ω

∫Ω

[(g (t)− h (τ))

∫ 1

0

Φ′ ((1− s)h (τ) + sg (t)) ds

]dµ (t) dµ (τ) ,


REMARK 2.2. The above inequality (2.6) can be extended for two measures as follows∫Ω1

Φ gdµ1 −∫

Ω2

Φ hdµ2 − λ

(∫Ω1

gdµ1 −∫

Ω2

hdµ2

)(2.8)

=

∫Ω1

∫Ω2

[(g (t)− h (τ))

∫ 1

0

(Φ′ ((1− s)h (τ) + sg (t))− λ) ds

]× dµ1 (t) dµ2 (τ) ,


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for anyλ ∈ C andx ∈ [a, b] and provided thatΦg, g ∈ L (Ω1, µ1) while Φh, h ∈ L (Ω2, µ2) .

REMARK 2.3. If w ≥ 0 µ-almost everywhere(µ-a.e.) on Ω with∫

Ωwdµ > 0, then by

replacingdµ with wdµRΩ wdµ

in (2.1) we have the weighted equality

1∫Ωwdµ

∫Ω

w (Φ g) dµ− Φ (x)− λ

(1∫

Ωwdµ

∫Ω

wgdµ− x

)(2.9)

=1∫

Ωwdµ

∫Ω

w ·[(g − x)

∫ 1

0

(Φ′ ((1− s)x+ sg)− λ) ds

]dµ

for anyλ ∈ C andx ∈ [a, b], providedΦ g, g ∈ Lw (Ω, µ) where

Lw (Ω, µ) :=

g|∫

Ω

w |g| dµ <∞.

The other equalities have similar weighted versions. However the details are omitted.

2.2. Inequalities for Derivatives of Bounded Variation. The following result holds:

THEOREM 2.5 (Dragomir, 2014 [5]). LetΦ : I → C be an absolutely continuous functionson[a, b] ⊂ I, the interior ofI and with the property that the derivativeΦ′ is of bounded variationon [a, b] . If g : Ω → [a, b] is Lebesgueµ-measurable onΩ and such thatΦ g, g ∈ L (Ω, µ) ,then we have ∣∣∣∣∫

Ω

Φ gdµ− Φ (x)− Φ′ (a) + Φ′ (b)

2

(∫Ω

gdµ− x

)∣∣∣∣(2.10)

≤ 1

2

b∨a

(Φ′)

∫Ω

|g − x| dµ


Ω

Φ gdµ− Φ

(a+ b

2

)− Φ′ (a) + Φ′ (b)

2

(∫Ω

gdµ− a+ b

2

)∣∣∣∣(2.11)

≤ 1

2

b∨a

(Φ′)

∫Ω

∣∣∣∣g − a+ b

2

∣∣∣∣ dµ ≤ 1

2(b− a)

b∨a

(Φ′)


Φ gdµ− Φ

(∫Ω

gdµ

)∣∣∣∣ ≤ 1

2

b∨a

(Φ′)

∫Ω

∣∣∣∣g − ∫Ω

gdµ

∣∣∣∣ dµ(2.12)

≤ 1

2

b∨a

(Φ′)

(∫Ω

g2dµ−(∫

Ω

gdµ

)2)1/2

≤ 1

4(b− a)

b∨a

(Φ′) .

PROOF. From the identity (2.1) we have∫Ω

Φ gdµ− Φ (x)− Φ′ (a) + Φ′ (b)

2

(∫Ω

gdµ− x

)(2.13)

=

∫Ω

[(g − x)

∫ 1

0

(Φ′ ((1− s)x+ sg)− Φ′ (a) + Φ′ (b)

2

)ds

]dµ


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242 S. S. DRAGOMIR

for anyx ∈ [a, b].Taking the modulus in (2.13) we get∣∣∣∣∫

Ω

Φ gdµ− Φ (x)− Φ′ (a) + Φ′ (b)

2

(∫Ω

gdµ− x

)∣∣∣∣(2.14)

≤∫

Ω

∣∣∣∣(g − x)

∫ 1

0

(Φ′ ((1− s)x+ sg)− Φ′ (a) + Φ′ (b)

2

)∣∣∣∣ dsdµ≤∫

Ω

|g − x|∫ 1

0

∣∣∣∣Φ′ ((1− s)x+ sg)− Φ′ (a) + Φ′ (b)

2

∣∣∣∣ dsdµfor anyx ∈ [a, b].

SinceΦ′ is of bounded variation on[a, b] , then for anys ∈ [0, 1] , x ∈ [a, b] andt ∈ Ω wehave ∣∣∣∣Φ′ ((1− s)x+ sg (t))− Φ′ (a) + Φ′ (b)

2

∣∣∣∣=

1

2|Φ′ ((1− s)x+ sg (t))− Φ′ (a) + Φ′ ((1− s)x+ sg (t))− Φ′ (b)|

≤ 1

2[|Φ′ ((1− s)x+ sg (t))− Φ′ (a)|+ |Φ′ (b)− Φ′ ((1− s)x+ sg (t))|]

≤ 1

2

b∨a

(Φ′) .

Then we have ∫Ω

|g − x|∫ 1

0

∣∣∣∣Φ′ ((1− s)x+ sg)− Φ′ (a) + Φ′ (b)

2

∣∣∣∣ dsdµ(2.15)

≤ 1

2

b∨a

(Φ′)

∫Ω

|g − x| dµ

for anyx ∈ [a, b].Making use of (2.14) and (2.15) we deduce the desired result (2.10).

REMARK 2.4. LetΦ : I → C be an absolutely continuous functions on[a, b] ⊂ I, theinterior of I and with the property that the derivativeΦ′ is of bounded variation on[a, b] . Ifxi ∈ [m,M ] andwi ≥ 0 (i = 1, . . . , n) with Wn :=

∑ni=1wi = 1, then one has the weighted

discrete inequality:∣∣∣∣∣n∑

i=1

wiΦ (xi)− Φ (x)− Φ′ (a) + Φ′ (b)

2

(n∑

i=1

wixi − x

)∣∣∣∣∣(2.16)

≤ 1

2

b∨a

(Φ′)n∑

i=1

wi |xi − x|



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In particular, we have∣∣∣∣∣n∑

i=1

wiΦ (xi)− Φ

(a+ b

2

)− Φ′ (a) + Φ′ (b)

2

(n∑

i=1

wixi −a+ b

2

)∣∣∣∣∣(2.17)

≤ 1

2

b∨a

(Φ′)n∑

i=1

wi

∣∣∣∣xi −a+ b

2

∣∣∣∣ ≤ 1

4(b− a)

b∨a

(Φ′)

and ∣∣∣∣∣n∑

i=1

wiΦ (xi)− Φ

(n∑

i=1

wixi

)∣∣∣∣∣ ≤ 1

2

b∨a

(Φ′)n∑

i=1

wi

∣∣∣∣∣xi −n∑

i=1

wixi

∣∣∣∣∣(2.18)

≤ 1

2

b∨a

(Φ′)

n∑j=1

wix2j −

(n∑

k=1

wkxk

)21/2

≤ 1

4(b− a)

b∨a

(Φ′) .

2.3. Inequalities for Lipschitzian Derivatives. The following result holds:

THEOREM 2.6 (Dragomir, 2014 [5]). LetΦ : I → C be an absolutely continuous functionson [a, b] ⊂ I, the interior ofI and with the property that the derivativeΦ′ is Lipschitzian withthe constantK > 0 on [a, b] . If g : Ω → [a, b] is Lebesgueµ-measurable onΩ and such thatΦ g, g ∈ L (Ω, µ) , then we have∣∣∣∣∫

Ω

Φ gdµ− Φ (x)− Φ′ (x)

(∫Ω

gdµ− x

)∣∣∣∣(2.19)

≤ 1

2K

[σ2

µ (g) +

(∫Ω

gdµ− x

)2]

for anyx ∈ [a, b], whereσµ (g) is the dispersion or the standard variation, namely

σµ (g) :=

(∫Ω

(g −

∫Ω

gdµ

)2

dµ

)1/2

=

(∫Ω

g2dµ−(∫

Ω

gdµ

)2)1/2

.

In particular, we have∣∣∣∣∫Ω

Φ gdµ− Φ

(a+ b

2

)− Φ′

(a+ b

2

)(∫Ω

gdµ− a+ b

2

)∣∣∣∣(2.20)

≤ 1

2K

[σ2

µ (g) +

(∫Ω

gdµ− a+ b

2

)2]

and

(2.21)

∣∣∣∣∫Ω

Φ gdµ− Φ

(∫Ω

gdµ

)∣∣∣∣ ≤ 1

2Kσ2

µ (g) ≤ 1

8K (b− a)2 .


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244 S. S. DRAGOMIR

PROOF. From the identity (2.1) we have forλ = Φ′ (x) that∫Ω


(∫Ω

gdµ− x

)(2.22)

=

∫Ω

[(g − x)

∫ 1

0

(Φ′ ((1− s)x+ sg)− Φ′ (x)) ds

]dµ


Ω


(∫Ω

gdµ− x

)∣∣∣∣(2.23)

≤∫

Ω

|g − x|∣∣∣∣∫ 1

0

(Φ′ ((1− s)x+ sg)− Φ′ (x)) ds

∣∣∣∣ dµ≤∫

Ω

[|g − x|

∫ 1

0

|(Φ′ ((1− s)x+ sg)− Φ′ (x))| ds]dµ

≤ K

∫Ω

[|g − x|

∫ 1

0

s |g − x| ds]dµ =

1

2K

∫Ω

(g − x)2 dµ

for anyx ∈ [a, b].However,∫

Ω

(g − x)2 dµ

=

∫Ω

(g −

∫Ω

gdµ+

∫Ω

gdµ− x

)2

dµ

=

∫Ω

(g −

∫Ω

gdµ

)2

dµ+ 2

∫Ω

(g −

∫Ω

gdµ

)(∫Ω

gdµ− x

)dµ

+

∫Ω

(∫Ω

gdµ− x

)2

dµ

=

∫Ω

(g −

∫Ω

gdµ

)2

dµ+

(∫Ω

gdµ− x

)2

for anyx ∈ [a, b], and by (2.23) we get the desired result (2.19).

COROLLARY 2.7. Let Φ : I → C be a twice differentiable functions on[a, b] ⊂ I with‖Φ′′‖[a,b],∞ := ess supt∈[a,b] |Φ′′ (t)| < ∞. Then the inequalities (2.19)-(2.21) hold forK =

‖Φ′′‖[a,b],∞ .

REMARK 2.5. LetΦ : I → C be an absolutely continuous functions on[a, b] ⊂ I andwith the property that the derivativeΦ′ is Lipschitzian with the constantK > 0 on [a, b] . Ifxi ∈ [m,M ] andwi ≥ 0 (i = 1, . . . , n) with Wn :=

∑ni=1wi = 1, then one has the weighted

discrete inequality: ∣∣∣∣∣n∑

i=1

wiΦ (xi)− Φ (x)− Φ′ (x)

(n∑

i=1

wixi − x

)∣∣∣∣∣(2.24)

≤ 1

2K

σ2w (x) +

(n∑

i=1

wixi − x

)2


http://ajmaa.org


for anyx ∈ [a, b], where

σw (x) :=

n∑i=1

wi

(xi −

n∑k=1

wkxk

)21/2

=

n∑i=1

wix2i −

(n∑

k=1

wkxk

)21/2

.

The following lemma may be stated:

LEMMA 2.8. Let u : [a, b] → R and l, L ∈ R with L > l. The following statements areequivalent:

(i) The functionu− l+L2· e, wheree (t) = t, t ∈ [a, b] is 1

2(L− l)−Lipschitzian;

(ii) We have the inequalities

(2.25) l ≤ u (t)− u (s)

t− s≤ L for each t, s ∈ [a, b] with t 6= s;

(iii) We have the inequalities

(2.26) l (t− s) ≤ u (t)− u (s) ≤ L (t− s) for each t, s ∈ [a, b] with t > s.

DEFINITION 2.1. The functionu : [a, b] → R which satisfies one of the equivalent condi-tions (i) – (iii) from Lemma 2.8 is said to be(l, L)-Lipschitzian on[a, b] .

If L > 0 and l = −L, then(−L,L)−Lipschitzian meansL-Lipschitzian in the classicalsense.

Utilising Lagrange’s mean value theorem, we can state the following result that providesexamples of(l, L)-Lipschitzian functions.

PROPOSITION2.9. Letu : [a, b] → R be continuous on[a, b] and differentiable on(a, b) . If−∞ < l = inft∈[a,b] u

′ (t) andsupt∈[a,b] u′ (t) = L <∞, thenu is (l, L)-Lipschitzian on[a, b] .

The following result holds.

COROLLARY 2.10 (Dragomir, 2014 [5]). Let Φ : I → R be an absolutely continuousfunctions on[a, b] ⊂ I, with the property that the derivativeΦ′ is (l, L)-Lipschitzian on[a, b] ,wherel, L ∈ R with L > l. If g : Ω → [a, b] is Lebesgueµ-measurable onΩ and such thatΦ g, g ∈ L (Ω, µ) , then we have∣∣∣∣∫

Ω


(∫Ω

gdµ− x

)(2.27)

−1

4(L+ l)

[σ2

µ (g) +

(∫Ω

gdµ− x

)2]∣∣∣∣∣

≤ 1

4(L− l)

[σ2

µ (g) +

(∫Ω

gdµ− x

)2]


Ω

Φ gdµ− Φ

(a+ b

2

)− Φ′

(a+ b

2

)(∫Ω

gdµ− a+ b

2

)(2.28)

−1

4(L+ l)

[σ2

µ (g) +

(∫Ω

gdµ− a+ b

2

)2]∣∣∣∣∣

≤ 1

4(L− l)

[σ2

µ (g) +

(∫Ω

gdµ− a+ b

2

)2]


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246 S. S. DRAGOMIR


Φ gdµ− Φ

(∫Ω

gdµ

)− 1

4(L+ l)σ2

µ (g)

∣∣∣∣ ≤ 1

4(L− l)σ2

µ (g)(2.29)

≤ 1

16(L− l) (b− a)2 .

PROOF. Consider the auxiliary functionΨ : [a, b] → R given by

Ψ (x) = Φ (x)− 1

4(L+ l)x2.

We observe thatΨ is differentiable and

Ψ′ (x) = Φ′ (x)− 1

2(L+ l)x.

SinceΦ′ is (l, L)-Lipschitzian on[a, b] it follows that Ψ′ is Lipschitzian with the constant12(L− l) , so we can apply Theorem 2.6 forΨ, i.e. we have the inequality∣∣∣∣∫

Ω

Ψ gdµ−Ψ (x)−Ψ′ (x)

(∫Ω

gdµ− x

)∣∣∣∣(2.30)

≤ 1

4(L− l)

[σ2

µ (g) +

(∫Ω

gdµ− x

)2].

However ∫Ω

Ψ gdµ−Ψ (x)−Ψ′ (x)

(∫Ω

gdµ− x

)=

∫Ω


(∫Ω

gdµ− x

)− 1

4(L+ l)

[∫Ω

g2dµ− x2 − 2x

(∫Ω

gdµ− x

)]=

∫Ω


(∫Ω

gdµ− x

)− 1

4(L+ l)

[σ2

µ (g) +

(∫Ω

gdµ− x

)2]

and by (2.30) we get the desired result (2.27).

REMARK 2.6. We observe that if the functionΦ is twice differentiable onI and for[a, b] ⊂ Iwe have

−∞ < l ≤ Φ′′ (x) ≤ L <∞ for anyx ∈ [a, b] ,

thenΦ′ is (l, L)-Lipschitzian on[a, b] and the inequalities (2.27)-(2.29) hold true.


THEOREM2.11 (Dragomir, 2014 [5]). LetΦ : I → C be an absolutely continuous functionson [a, b] ⊂ I, the interior ofI and with the property that the derivativeΦ′ is Lipschitzian withthe constantK > 0 on [a, b] . If g : Ω → [a, b] is Lebesgueµ-measurable onΩ and such that


http://ajmaa.org


Φ g, g ∈ L (Ω, µ) , then we have∣∣∣∣∫Ω

Φ gdµ− Φ (x)− Φ′(∫

Ω

gdµ

)(∫Ω

gdµ− x

)∣∣∣∣(2.31)

≤ 1

2K

[∣∣∣∣x− ∫Ω

gdµ

∣∣∣∣ ∫Ω

|g − x| dµ+

∫Ω

|g − x|∣∣∣∣g − ∫

Ω

gdµ

∣∣∣∣ dµ]≤ 1

2K

[∣∣∣∣x− ∫Ω

gdµ

∣∣∣∣+ ∥∥∥∥g − ∫Ω

gdµ

∥∥∥∥Ω,∞

]∫Ω

|g − x| dµ

for anyx ∈ [a, b], where∥∥∥∥g − ∫Ω

gdµ

∥∥∥∥Ω,∞

:= ess supt∈Ω

∣∣∣∣g (t)−∫

Ω

gdµ

∣∣∣∣ <∞.


Φ gdµ− Φ

(a+ b

2

)− Φ′

(∫Ω

gdµ

)(∫Ω

gdµ− a+ b

2

)∣∣∣∣(2.32)

≤ 1

2K

[∣∣∣∣a+ b

2−∫

Ω

gdµ

∣∣∣∣ ∫Ω

∣∣∣∣g − a+ b

2

∣∣∣∣ dµ+

∫Ω

∣∣∣∣g − a+ b

2

∣∣∣∣ ∣∣∣∣g − ∫Ω

gdµ

∣∣∣∣ dµ]≤ 1

2K

[∣∣∣∣a+ b

2−∫

Ω

gdµ

∣∣∣∣+ ∥∥∥∥g − ∫Ω

gdµ

∥∥∥∥Ω,∞

]∫Ω

∣∣∣∣g − a+ b

2

∣∣∣∣ dµ.PROOF. From the identity (2.1) we have forλ = Φ′ (∫

Ωgdµ

)that∫

Ω

Φ gdµ− Φ (x)− Φ′(∫

Ω

gdµ

)(∫Ω

gdµ− x

)(2.33)

=

∫Ω

[(g − x)

∫ 1

0

(Φ′ ((1− s)x+ sg)− Φ′

(∫Ω

gdµ

))ds

]dµ


Ω

Φ gdµ− Φ (x)− Φ′(∫

Ω

gdµ

)(∫Ω

gdµ− x

)∣∣∣∣(2.34)

≤∫

Ω

|g − x|∣∣∣∣∫ 1

0

(Φ′ ((1− s)x+ sg)− Φ′

(∫Ω

gdµ

))ds

∣∣∣∣ dµ≤∫

Ω

[|g − x|

∫ 1

0

∣∣∣∣(Φ′ ((1− s)x+ sg)− Φ′(∫

Ω

gdµ

))∣∣∣∣ ds] dµ≤ K

∫Ω

[|g − x|

∫ 1

0

∣∣∣∣(1− s)x+ sg −∫

Ω

gdµ

∣∣∣∣ ds] dµ= K

∫Ω

[|g − x|

∫ 1

0

∣∣∣∣(1− s)x+ sg − (1− s)

∫Ω

gdµ− s

∫Ω

gdµ

∣∣∣∣ ds] dµ:= B.


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248 S. S. DRAGOMIR

Using the triangle inequality we have for anyt ∈ Ω∫ 1

0

∣∣∣∣(1− s)x+ sg (t)− (1− s)

∫Ω

gdµ− s

∫Ω

gdµ

∣∣∣∣ ds≤∫ 1

0

(1− s)

∣∣∣∣x− ∫Ω

gdµ

∣∣∣∣ ds+

∫ 1

0

s

∣∣∣∣g (t)−∫

Ω

gdµ

∣∣∣∣ ds=

1

2

[∣∣∣∣x− ∫Ω

gdµ

∣∣∣∣+ ∣∣∣∣g (t)−∫

Ω

gdµ

∣∣∣∣]and then

B ≤ 1

2K

∫Ω

|g − x|[∣∣∣∣x− ∫

Ω

gdµ

∣∣∣∣+ ∣∣∣∣g (t)−∫

Ω

gdµ

∣∣∣∣] dµ(2.35)

=1

2K

[∣∣∣∣x− ∫Ω

gdµ

∣∣∣∣ ∫Ω

|g − x| dµ+

∫Ω

|g − x|∣∣∣∣g − ∫

Ω

gdµ

∣∣∣∣ dµ] .Making use of (2.34) and (2.35) we deduce the desired result (2.31).

COROLLARY 2.12 (Dragomir, 2014 [5]). Let Φ : I → R be an absolutely continuousfunctions on[a, b] ⊂ I, with the property that the derivativeΦ′ is (l, L)-Lipschitzian on[a, b] ,wherel, L ∈ R with L > l. If g : Ω → [a, b] is Lebesgueµ-measurable onΩ and such thatΦ g, g ∈ L (Ω, µ) , then we have∣∣∣∣∫

Ω

Φ gdµ− Φ (x)− Φ′(∫

Ω

gdµ

)(∫Ω

gdµ− x

)(2.36)

−1

4(L+ l)

[σ2

µ (g)−(x−

∫Ω

gdµ

)2]∣∣∣∣∣

≤ 1

4(L− l)

[∣∣∣∣x− ∫Ω

gdµ

∣∣∣∣ ∫Ω

|g − x| dµ+

∫Ω

|g − x|∣∣∣∣g − ∫

Ω

gdµ

∣∣∣∣ dµ]≤ 1

4(L− l)

[∣∣∣∣x− ∫Ω

gdµ

∣∣∣∣+ ∥∥∥∥g − ∫Ω

gdµ

∥∥∥∥Ω,∞

]∫Ω

|g − x| dµ


Ω

Φ gdµ− Φ

(a+ b

2

)− Φ′

(∫Ω

gdµ

)(∫Ω

gdµ− a+ b

2

)(2.37)

−1

4(L+ l)

[σ2

µ (g)−(a+ b

2−∫

Ω

gdµ

)2]∣∣∣∣∣

≤ 1

4(L− l)

[∣∣∣∣a+ b

2−∫

Ω

gdµ

∣∣∣∣ ∫Ω

∣∣∣∣g − a+ b

2

∣∣∣∣ dµ+

∫Ω

∣∣∣∣g − a+ b

2

∣∣∣∣ ∣∣∣∣g − ∫Ω

gdµ

∣∣∣∣ dµ]≤ 1

4(L− l)

[∣∣∣∣a+ b

2−∫

Ω

gdµ

∣∣∣∣+ ∥∥∥∥g − ∫Ω

gdµ

∥∥∥∥Ω,∞

]∫Ω

∣∣∣∣g − a+ b

2

∣∣∣∣ dµ.


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3. INEQUALITIES FOR FUNCTIONS WHOSE DERIVATIVES IN ABSOLUTE VALUE ARE

h-CONVEX

3.1. Inequalities for |Φ′| is h-Convex, Quasi-convex or Log-convex.We use the nota-tions

‖k‖Ω,p :=

∫Ω

|k (t)|p dµ (t)

1/p

<∞, p ≥ 1, k ∈ Lp (Ω, µ) ;

ess supt∈Ω |k (t)| <∞, p = ∞, k ∈ L∞ (Ω, µ) ;

and

‖Φ‖[0,1],p :=

(∫ 1

0|Φ (s)|p ds

)1/p

<∞, p ≥ 1, Φ ∈ Lp (0, 1) ;

ess sups∈[0,1] |Φ (s)| <∞, p = ∞, Φ ∈ L∞ (0, 1) .


THEOREM 3.1 (Dragomir, 2014 [5]). Let Φ : I → C be a differentiable function onI,the interior ofI and such that|Φ′| is h-convex on the interval[a, b] ⊂ I . If g : Ω → [a, b] isLebesgueµ-measurable onΩ and such thatΦ g, g ∈ L (Ω, µ) , then we have the inequality∣∣∣∣∫

Ω

Φ gdµ− Φ (x)

∣∣∣∣(3.1)

≤∫ 1

0

h (s) ds

‖g − x‖Ω,∞

[|Φ′ (x)|+ ‖Φ′ g‖Ω,1

]if Φ′ g ∈ L (Ω, µ) ;

‖g − x‖Ω,p ‖|Φ′ (x)|+ |Φ′ g|‖Ω,q

if Φ′ g ∈ Lq (Ω, µ) , p > 1, 1p

+ 1q

= 1;

‖g − x‖Ω,1

[|Φ′ (x)|+ ‖Φ′ g‖Ω,∞

]if Φ′ g ∈ L∞ (Ω, µ)


Ω

Φ gdµ− Φ

(∫Ω

gdµ

)∣∣∣∣(3.2)

≤∫ 1

0

h (s) ds


∥∥Ω,∞

[∣∣Φ′ (∫Ωgdµ

)∣∣+ ‖Φ′ g‖Ω,1

],

if Φ′ g ∈ L (Ω, µ) ;∥∥g − ∫Ωgdµ

∥∥Ω,p

∥∥∣∣Φ′ (∫Ωgdµ

)∣∣+ |Φ′ g|∥∥

Ω,q

if Φ′ g ∈ Lq (Ω, µ) , p > 1, 1p

+ 1q

= 1;


∥∥Ω,1

[∣∣Φ′ (∫Ωgdµ

)∣∣+ ‖Φ′ g‖Ω,∞

]if Φ′ g ∈ L∞ (Ω, µ)


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250 S. S. DRAGOMIR


Φ gdµ− Φ

(a+ b

2

)∣∣∣∣(3.3)

≤∫ 1

0

h (s) ds

∥∥g − a+b2

∥∥Ω,∞

[∣∣Φ′ (a+b2

)∣∣+ ‖Φ′ g‖Ω,1

]if Φ′ g ∈ L (Ω, µ) ; ;∥∥g − a+b

2

∥∥Ω,p

∥∥∣∣Φ′ (a+b2

)∣∣+ |Φ′ g|∥∥

Ω,q

if Φ′ g ∈ Lq (Ω, µ) , p > 1, 1p

+ 1q

= 1;

∥∥g − a+b2

∥∥Ω,1

[∣∣Φ′ (a+b2

)∣∣+ ‖Φ′ g‖Ω,∞

]if Φ′ g ∈ L∞ (Ω, µ)

≤ 1

2(b− a)

∫ 1

0

h (s) ds

[∣∣Φ′ (a+b2

)∣∣+ ‖Φ′ g‖Ω,1

];∥∥∣∣Φ′ (a+b

2

)∣∣+ |Φ′ g|∥∥

Ω,q

p > 1, 1p

+ 1q

= 1;[∣∣Φ′ (a+b2

)∣∣+ ‖Φ′ g‖Ω,∞

].

PROOF. We have

(3.4)

∣∣∣∣∫Ω

Φ gdµ− Φ (x)

∣∣∣∣ ≤ ∫Ω

|g − x|∣∣∣∣∫ 1

0

Φ′ ((1− s)x+ sg) ds

∣∣∣∣ dµ,for anyx ∈ [a, b].

Utilising Hölder’s inequality for theµ-measurable functionsF,G : Ω → C,∣∣∣∣∫Ω

FGdµ

∣∣∣∣ ≤ (∫Ω

|F |p dµ)1/p(∫

Ω

|G|q dµ)1/q

, p > 1,1

p+

1

q= 1


FGdµ

∣∣∣∣ ≤ ess supt∈Ω

|F (t)|∫

Ω

|G| dµ,

we have

B :=

∫Ω

|g − x|∣∣∣∣∫ 1

0

Φ′ ((1− s)x+ sg) ds

∣∣∣∣ dµ(3.5)

≤

ess supt∈Ω |g (t)− x|∫

Ω

∣∣∣∫ 1

0Φ′ ((1− s)x+ sg) ds

∣∣∣ dµ;

(∫Ω|g − x|p dµ

)1/p(∫

Ω

∣∣∣∫ 1

0Φ′ ((1− s)x+ sg) ds

∣∣∣q dµ)1/q

p > 1, 1p

+ 1q

= 1;

∫Ω|g − x| dµ ess supt∈Ω

∣∣∣∫ 1

0Φ′ ((1− s)x+ sg) ds

∣∣∣ ,for anyx ∈ [a, b].


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Since|Φ′| is h-convex on the interval[a, b] , then we have for anyt ∈ Ω that∣∣∣∣∫ 1

0

Φ′ ((1− s)x+ sg (t)) ds

∣∣∣∣ ≤ ∫ 1

0

|Φ′ ((1− s)x+ sg (t))| ds

≤ |Φ′ (x)|∫ 1

0

h (1− s) ds+ |Φ′ (g (t))|∫ 1

0

h (s) ds

= [|Φ′ (x)|+ |Φ′ (g (t))|]∫ 1

0

h (s) ds,

for anyx ∈ [a, b].This implies that

(3.6)∫

Ω

∣∣∣∣∫ 1

0

Φ′ ((1− s)x+ sg) ds

∣∣∣∣ dµ ≤ ∫ 1

0

h (s) ds

[|Φ′ (x)|+

∫Ω

|Φ′ g| dµ]

for anyx ∈ [a, b].We have for anyt ∈ Ω that∣∣∣∣∫ 1

0

Φ′ ((1− s)x+ sg (t)) ds

∣∣∣∣q ≤[∫ 1

0

|Φ′ ((1− s)x+ sg (t))| ds]q

≤[[|Φ′ (x)|+ |Φ′ (g (t))|]

∫ 1

0

h (s) ds

]q

=

[∫ 1

0

h (s) ds

]q

[|Φ′ (x)|+ |Φ′ (g (t))|]q

for anyx ∈ [a, b].This implies (∫

Ω

∣∣∣∣∫ 1

0

Φ′ ((1− s)x+ sg) ds

∣∣∣∣q dµ)1/q

(3.7)

≤∫ 1

0

h (s) ds

[∫Ω

[|Φ′ (x)|+ |Φ′ (g (t))|]q dµ]1/q

=

∫ 1

0

h (s) ds

[∫Ω

[|Φ′ (x)|+ |Φ′ g|]q dµ]1/q

.

Also

ess supt∈Ω

∣∣∣∣∫ 1

0

Φ′ ((1− s)x+ sg) ds

∣∣∣∣(3.8)

≤[|Φ′ (x)|+ ess sup

t∈Ω|Φ′ (g (t))|

] ∫ 1

0

h (s) ds

=

[|Φ′ (x)|+ ess sup

t∈Ω|Φ′ g|

] ∫ 1

0

h (s) ds

for anyx ∈ [a, b].Making use of (3.6)-(3.8) we get the desired result (3.1).

REMARK 3.1. With the assumptions of Theorem 3.1 and if|Φ′| is convex on the interval[a, b] , then

∫ 1

0h (s) ds = 1

2and the inequalities (3.1)-(3.3) hold with1

2instead of

∫ 1

0h (s) ds. If

|Φ′| is of s-Godunova-Levin type, withs ∈ [0, 1) on the interval[a, b] , then∫ 1

01tsdt = 1

1−sand

the inequalities (3.1)-(3.3) hold with11−s

instead of∫ 1

0h (s) ds.


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252 S. S. DRAGOMIR

Firstly, let us recall the definition of quasi-convex functions.

DEFINITION 3.1. The functionh : [a, b] ⊂ R → R is said to bequasi-convex(QC) on theintervalI if

(3.9) h (λx+ (1− λ) y) ≤ max h (x) , h (y)

for anyx, y ∈ I andλ ∈ [0, 1] .

Following [7], we say that for an intervalI ⊆ R, the mappingh : I → R is quasi-monotoneon I if it is either monotone onI = [c, d] or monotone nonincreasing on a proper subinterval[c, c′] ⊂ I and monotone nondecreasing on[c′, d] .

The classQM (I) of quasi-monotone functions onI provides an immediate characterizationof quasi-convex functions [7].

PROPOSITION3.2. SupposeI ⊆ R. Then the following statements are equivalent for afunctionh : I → R:

(a) h ∈ QM (I) ;(b) On any subinterval ofI, h achieves its supremum at an end point;(c) h ∈ QC (I) .

As examples of quasi-convex functions we may consider the class of monotonic functionson an intervalI for the class of convex functions on that interval.

THEOREM 3.3 (Dragomir, 2014 [5]). Let Φ : I → C be a differentiable function onI, theinterior of I and such that|Φ′| is quasi-convex on the interval[a, b] ⊂ I . If g : Ω → [a, b] isLebesgueµ-measurable onΩ and such thatΦ g, g ∈ L (Ω, µ) andΦ′ g ∈ L∞ (Ω, µ) , thenwe have the inequality∣∣∣∣∫

Ω

Φ gdµ− Φ (x)

∣∣∣∣ ≤∫

Ω

|g − x|max |Φ′ (x)| , |Φ′ g| dµ(3.10)

≤ max|Φ′ (x)| , ‖Φ′ g‖Ω,∞

‖g − x‖Ω,1


Ω

Φ gdµ− Φ

(∫Ω

gdµ

)∣∣∣∣(3.11)

≤∫

Ω

∣∣∣∣g − ∫Ω

gdµ

∣∣∣∣max

∣∣∣∣Φ′(∫

Ω

gdµ

)∣∣∣∣ , |Φ′ g|dµ

≤ max|Φ′ (x)| , ‖Φ′ g‖Ω,∞

∥∥∥∥g − ∫Ω

gdµ

∥∥∥∥Ω,1


Φ gdµ− Φ

(a+ b

2

)∣∣∣∣(3.12)

≤∫

Ω

∣∣∣∣g − a+ b

2

∣∣∣∣max

∣∣∣∣Φ′(a+ b

2

)∣∣∣∣ , |Φ′ g|dµ

≤ max

∣∣∣∣Φ′(a+ b

2

)∣∣∣∣ , ‖Φ′ g‖Ω,∞

∥∥∥∥g − a+ b

2

∥∥∥∥Ω,1

.


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PROOF. From (3.4) we have∣∣∣∣∫Ω

Φ gdµ− Φ (x)

∣∣∣∣ ≤ ∫Ω

|g − x|(∫ 1

0

|Φ′ ((1− s)x+ sg)| ds)dµ(3.13)

≤∫

Ω

|g − x|max |Φ′ (x)| , |Φ′ g| dµ,

for anyx ∈ [a, b].Observe that

|(Φ′ g) (t)| ≤ ‖Φ′ g‖Ω,∞ for almost everyt ∈ Ω

and then ∫Ω

|g − x|max |Φ′ (x)| , |Φ′ g| dµ(3.14)

≤∫

Ω

|g − x|max|Φ′ (x)| , ‖Φ′ g‖Ω,∞

dµ

= max|Φ′ (x)| , ‖Φ′ g‖Ω,∞

∫Ω

|g − x| dµ,

for anyx ∈ [a, b].Using (3.13) and (3.14) we get the desired result (3.10).

In what follows,I will denote an interval of real numbers. A functionf : I → (0,∞) issaid to belog-convexor multiplicatively convexif log f is convex, or, equivalently, if for anyx, y ∈ I andt ∈ [0, 1] one has the inequality

(3.15) f (tx+ (1− t) y) ≤ [f (x)]t [f (y)]1−t .

We note that iff andg are convex andg is increasing, theng f is convex, moreover, sincef = exp [log f ] , it follows that a log-convex function is convex, but the converse may notnecessarily be true. This follows directly from (3.15) since, by the arithmetic-geometric meaninequality we have

(3.16) [f (x)]t [f (y)]1−t ≤ tf (x) + (1− t) f (y)

for all x, y ∈ I andt ∈ [0, 1] .

THEOREM 3.4 (Dragomir, 2014 [5]). Let Φ : I → C be a differentiable function onI,the interior ofI and such that|Φ′| is log-convex on the interval[a, b] ⊂ I . If g : Ω → [a, b]is Lebesgueµ-measurable onΩ and such thatΦ g, Φ′ g, g ∈ L (Ω, µ) then we have theinequality ∣∣∣∣∫

Ω

Φ gdµ− Φ (x)

∣∣∣∣(3.17)

≤∫

Ω

|g − x|L (|Φ′ g| , |Φ′ (x)|) dµ

≤ 1

2

[|Φ′ (x)|

∫Ω

|g − x| dµ+

∫Ω

|g − x| |Φ′ g| dµ]

(≤ 1

2

[|Φ′ (x)|+ ‖Φ′ g‖Ω,∞

]‖g − x‖Ω,1 if Φ′ g ∈ L∞ (Ω, µ)

)


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254 S. S. DRAGOMIR

for anyx ∈ [a, b], whereL (·, ·) is the logarithmic mean, namely forα, β > 0

L (α, β) :=

α−β

ln α−ln β, α 6= β

α, α = β.


Φ gdµ− Φ

(∫Ω

gdµ

)∣∣∣∣(3.18)

≤∫

Ω

∣∣∣∣g − ∫Ω

gdµ

∣∣∣∣L(|Φ′ g| ,∣∣∣∣Φ′(∫

Ω

gdµ

)∣∣∣∣) dµ≤ 1

2

[∣∣∣∣Φ′(∫

Ω

gdµ

)∣∣∣∣ ∫Ω

∣∣∣∣g − ∫Ω

gdµ

∣∣∣∣ dµ+

∫Ω

∣∣∣∣g − ∫Ω

gdµ

∣∣∣∣ |Φ′ g| dµ]

(≤ 1

2

[∣∣∣∣Φ′(∫

Ω

gdµ

)∣∣∣∣+ ‖Φ′ g‖Ω,∞

] ∥∥∥∥g − ∫Ω

gdµ

∥∥∥∥Ω,1

if Φ′ g ∈ L∞ (Ω, µ))


Φ gdµ− Φ

(a+ b

2

)∣∣∣∣(3.19)

≤∫

Ω

∣∣∣∣g − a+ b

2

∣∣∣∣L(|Φ′ g| ,∣∣∣∣Φ′(a+ b

2

)∣∣∣∣) dµ≤ 1

2

[∣∣∣∣Φ′(a+ b

2

)∣∣∣∣ ∫Ω

∣∣∣∣g − a+ b

2

∣∣∣∣ dµ+

∫Ω

∣∣∣∣g − a+ b

2

∣∣∣∣ |Φ′ g| dµ]

(≤ 1

2

[∣∣∣∣Φ′(a+ b

2

)∣∣∣∣+ ‖Φ′ g‖Ω,∞

] ∥∥∥∥g − a+ b

2

∥∥∥∥Ω,1

if Φ′ g ∈ L∞ (Ω, µ)).

PROOF. From (3.4) we have∣∣∣∣∫Ω

Φ gdµ− Φ (x)

∣∣∣∣ ≤ ∫Ω

|g − x|(∫ 1

0

|Φ′ ((1− s)x+ sg)| ds)dµ(3.20)

≤∫

Ω

|g − x|(∫ 1

0

|Φ′ (x)|1−s |Φ′ g|s ds)dµ,

for anyx ∈ [a, b].Since, for anyC > 0, one has ∫ 1

0

Cλdλ =C − 1

lnC,


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then for anyt ∈ Ω we have∫ 1

0

|Φ′ (x)|1−s |Φ′ (g (t))|s ds = |Φ′ (x)|∫ 1

0

∣∣∣∣Φ′ (g (t))

Φ′ (x)

∣∣∣∣s ds(3.21)

= |Φ′ (x)|

∣∣∣Φ′(g(t))Φ′(x)

∣∣∣− 1

ln∣∣∣Φ′(g(t))

Φ′(x)

∣∣∣=

|Φ′ (g (t))| − |Φ′ (x)|ln |Φ′ (g (t))| − ln |Φ′ (x)|

= L (|Φ′ (g (t))| , |Φ′ (x)|) ,

for anyx ∈ [a, b].Making use of (3.20) and (3.21) we get the first inequality in (3.17).The second inequality in (3.17) follows by the fact that

L (α, β) ≤ α+ β

2for anyα, β > 0.

The last inequality in (3.17) is obvious.

3.2. Inequalities for |Φ′|q is h-Convex or Log-convex.We have:

THEOREM 3.5 (Dragomir, 2014 [5]). Let Φ : I → C be a differentiable function onI, theinterior of I and such that forp > 1, q > 1 with 1

p+ 1

q= 1, |Φ′|q is h-convex on the interval

[a, b] ⊂ I . If g : Ω → [a, b] is Lebesgueµ-measurable onΩ and such thatΦ g, g ∈ L (Ω, µ)andΦ′ g ∈ Lq (Ω, µ) then we have the inequality∣∣∣∣∫

Ω

Φ gdµ− Φ (x)

∣∣∣∣(3.22)

≤(∫ 1

0

h (s) ds

)1/q

‖g − x‖Ω,p

(|Φ′ (x)|q +

∫Ω

|Φ′ g|q dµ)1/q

≤(∫ 1

0

h (s) ds

)1/q

‖g − x‖Ω,p

(|Φ′ (x)|+ ‖Φ′ g‖Ω,q

)for anyx ∈ [a, b].


Φ gdµ− Φ

(∫Ω

gdµ

)∣∣∣∣(3.23)

≤(∫ 1

0

h (s) ds

)1/q

×∥∥∥∥g − ∫

Ω

gdµ

∥∥∥∥Ω,p

(∣∣∣∣Φ′(∫

Ω

gdµ

)∣∣∣∣q +

∫Ω

|Φ′ g|q dµ)1/q

≤(∫ 1

0

h (s) ds

)1/q

×∥∥∥∥g − ∫

Ω

gdµ

∥∥∥∥Ω,p

(∣∣∣∣Φ′(∫

Ω

gdµ

)∣∣∣∣+ ‖Φ′ g‖Ω,q

)


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256 S. S. DRAGOMIR


Φ gdµ− Φ

(a+ b

2

)∣∣∣∣(3.24)

≤(∫ 1

0

h (s) ds

)1/q

×∥∥∥∥g − a+ b

2

∥∥∥∥Ω,p

(∣∣∣∣Φ′(a+ b

2

)∣∣∣∣q +

∫Ω

|Φ′ g|q dµ)1/q

≤(∫ 1

0

h (s) ds

)1/q

×∥∥∥∥g − a+ b

2

∥∥∥∥Ω,p

(∣∣∣∣Φ′(a+ b

2

)∣∣∣∣+ ‖Φ′ g‖Ω,q

).

PROOF. From the proof of Theorem 3.1 we have∣∣∣∣∫Ω

Φ gdµ− Φ (x)

∣∣∣∣(3.25)

≤∫

Ω

|g − x|∣∣∣∣∫ 1

0

Φ′ ((1− s)x+ sg) ds

∣∣∣∣ dµ≤(∫

Ω

|g − x|p dµ)1/p(∫

Ω

∣∣∣∣∫ 1

0

Φ′ ((1− s)x+ sg) ds

∣∣∣∣q dµ)1/q

≤(∫

Ω

|g − x|p dµ)1/p(∫

Ω

(∫ 1

0

|Φ′ ((1− s)x+ sg)|q ds)dµ

)1/q

for p > 1, q > 1 with 1p

+ 1q

= 1 andx ∈ [a, b].Since|Φ′|q is h-convex on the interval[a, b], then∫ 1

0

|Φ′ ((1− s)x+ sg (t))|q ds ≤ |Φ′ (x)|q∫ 1

0

h (1− s) ds+ |Φ′ (g (t))|q∫ 1

0

h (s) ds

=[|Φ′ (x)|q + |Φ′ (g (t))|q

] ∫ 1

0

h (s) ds

for anyx ∈ [a, b] andt ∈ Ω.Therefore (∫

Ω

(∫ 1

0


)1/q

(3.26)

≤(∫

Ω

([|Φ′ (x)|q + |Φ′ (g (t))|q

] ∫ 1

0

h (s) ds

)dµ

)1/q

=

(∫ 1

0

h (s) ds

)1/q (|Φ′ (x)|q +

∫Ω

|Φ′ g|q dµ)1/q

for anyx ∈ [a, b] .This proves the first inequality in (3.22).Now, we observe that the following elementary inequality holds:

(3.27) (α+ β)r ≥ (≤)αr + βr

for anyα, β ≥ 0 andr ≥ 1 (0 < r < 1) .


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Indeed, if we consider the functionfr : [0,∞) → R, fr (t) = (t+ 1)r − tr we havef ′r (t) = r

[(t+ 1)r−1 − tr−1

]. Observe that forr > 1 and t > 0 we have thatf ′r (t) > 0

showing thatfr is strictly increasing on the interval[0,∞). Now for t = αβ

(β > 0, α ≥ 0) we

havefr (t) > fr (0) giving that(

αβ

+ 1)r

−(

αβ

)r

> 1, i.e., the desired inequality (3.27).

For r ∈ (0, 1) we have thatfr is strictly decreasing on[0,∞) which proves the second casein (3.27).

Making use of (3.27) forr = 1/q ∈ (0, 1) we have

(|Φ′ (x)|q +

∫Ω

|Φ′ g|q dµ)1/q

≤ |Φ′ (x)|+(∫

Ω

|Φ′ g|q dµ)1/q

and then we get the second part of (3.22).

Finally, we have:

THEOREM 3.6 (Dragomir, 2014 [5]). Let Φ : I → C be a differentiable function onI, theinterior of I and such that forp > 1, q > 1 with 1

p+ 1

q= 1, |Φ′|q is log-convex on the interval

[a, b] ⊂ I . If g : Ω → [a, b] is Lebesgueµ-measurable onΩ and such thatΦ g, g ∈ L (Ω, µ)andΦ′ g ∈ Lq (Ω, µ) then we have the inequality

∣∣∣∣∫Ω

Φ gdµ− Φ (x)

∣∣∣∣(3.28)

≤ ‖g − x‖Ω,p

(∫Ω

L(|Φ′ g|q , |Φ′ (x)|q

)dµ

)1/q

≤ 1

21/q‖g − x‖Ω,p

[|Φ′ (x)|q +

∫Ω

|Φ′ g|q dµ]1/q

≤ 1

21/q‖g − x‖Ω,p

[|Φ′ (x)|+ ‖Φ′ g‖Ω,q

]for anyx ∈ [a, b].


∣∣∣∣∫Ω

Φ gdµ− Φ

(∫Ω

gdµ

)∣∣∣∣(3.29)

≤∥∥∥∥g − ∫

Ω

gdµ

∥∥∥∥Ω,p

(∫Ω

L

(|Φ′ g|q ,

∣∣∣∣Φ′(∫

Ω

gdµ

)∣∣∣∣q) dµ)1/q

≤ 1

21/q

∥∥∥∥g − ∫Ω

gdµ

∥∥∥∥Ω,p

[∣∣∣∣Φ′(∫

Ω

gdµ

)∣∣∣∣q +

∫Ω

|Φ′ g|q dµ]1/q

≤ 1

21/q

∥∥∥∥g − ∫Ω

gdµ

∥∥∥∥Ω,p

[∣∣∣∣Φ′(∫

Ω

gdµ

)∣∣∣∣+ ‖Φ′ g‖Ω,q

]


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258 S. S. DRAGOMIR


Φ gdµ− Φ

(a+ b

2

)∣∣∣∣(3.30)

≤∥∥∥∥g − a+ b

2

∥∥∥∥Ω,p

(∫Ω

L

(|Φ′ g|q ,

∣∣∣∣Φ′(a+ b

2

)∣∣∣∣q) dµ)1/q

≤ 1

21/q

∥∥∥∥g − a+ b

2

∥∥∥∥Ω,p

[∣∣∣∣Φ′(a+ b

2

)∣∣∣∣q +

∫Ω

|Φ′ g|q dµ]1/q

≤ 1

21/q

∥∥∥∥g − a+ b

2

∥∥∥∥Ω,p

[∣∣∣∣Φ′(a+ b

2

)∣∣∣∣+ ‖Φ′ g‖Ω,q

].

PROOF. Since|Φ′|q is log-convex on the interval[a, b], then∫ 1

0

|Φ′ ((1− s)x+ sg (t))|q ds ≤∫ 1

0

|Φ′ (x)|q(1−s) |g (t)|sq ds

= |Φ′ (x)|q∫ 1

0

∣∣∣∣ g (t)

Φ′ (x)

∣∣∣∣sq ds= L

(|Φ′ (g (t))|q , |Φ′ (x)|q

)for anyx ∈ [a, b] andt ∈ Ω.

Then(∫Ω

(∫ 1

0


)1/q

≤(∫

Ω

L(|Φ′ g|q , |Φ′ (x)|q

)dµ

)1/q

and by (3.25) we get the first inequality in (3.28).Since, in general

L (α, β) ≤ α+ β

2for anyα, β > 0,

then ∫Ω

L(|Φ′ g|q , |Φ′ (x)|q

)dµ ≤ 1

2

∫Ω

[|Φ′ g|q + |Φ′ (x)|q

]dµ

=1

2

[|Φ′ (x)|q +

∫Ω

|Φ′ g|q dµ]

and we get the second inequality in (3.28).The last part is obvious.

REFERENCES

[1] S. S. Dragomir, A Grüss type inequality for isotonic linear functionals and applications.Demonstra-tio Math. 36 (2003), no. 3, 551–562. Preprint RGMIARes. Rep. Coll.5(2002), Suplement, Art. 12.[Online http://rgmia.org/v5(E).php] .

[2] S. S. Dragomir, Reverses of the Jensen inequality in terms of first derivative and applications,ActaMath. Vietnam.38 (2013), no. 3, 429–446. PreprintRGMIA Res. Rep. Coll.14 (2011), Art. 71.[http://rgmia.org/papers/v14/v14a71.pdf].

[3] S. S. Dragomir. General Lebesgue integral inequalities of Jensen and Ostrowski type for differen-tiable functions whose derivatives in absolute value areh-convex and applications,Ann. Univ. MariaeCurie-Skłodowska Sect. A69 (2015), no. 2, 17–45. PreprintRGMIA Res. Rep. Coll.17 (2014), Art.38. [http://rgmia.org/papers/v17/v17a38.pdf].


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[4] S. S. Dragomir, Jensen and Ostrowski type inequalities for general Lebesgue integral with applica-tions,Ann. Univ. Mariae Curie-Skłodowska Sect. A70 (2016) , No.2, 29–49. PreprintRGMIA Res.Rep. Coll.17 (2014), Art. 25. [http://rgmia.org/papers/v17/v17a25.pdf].

[5] S. S. Dragomir, New Jensen and Ostrowski type inequalities for general Lebesgue integral withapplications,Iranian Journal of Mathematical Sciences and Informatics, Vol. 11, No. 2 (2016),pp. 1-22. DOI: 10.7508/ijmsi.2016.02.001. PreprintRGMIA Res. Rep. Coll.17 (2014), Art. 27.[http://rgmia.org/papers/v17/v17a27.pdf].

[6] S. S. Dragomir and N. M. Ionescu, Some converse of Jensen’s inequality and applications.Rev. Anal.Numér. Théor. Approx.23 (1994), no. 1, 71–78.

[7] S. S. Dragomir and C. E. M. Pearce, Quasi-convex functions and Hadamard’s inequality,Bull. Aus-tralian Math. Soc., 57(1998), 377-385.

[8] A. Ostrowski, Über die Absolutabweichung einer differentienbaren Funktionen von ihren Inte-gralmittelwert,Comment. Math. Hel.,10 (1938), 226-227.


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CHAPTER 10

List of Papers Related to Ostrowski’s Inequality

1. BEFORE 2000

1. Milovanovic, Gradimir V.; Pecaric, Josip E. On generalization of the inequality of A.Ostrowski and some related applications. Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat.Fiz. No. 544-576 (1976), 155–158.

2. Fink, A. M. Bounds on the deviation of a function from its averages. Czechoslovak Math.J. 42(117) (1992), no. 2, 289–310.

3. Anastassiou, George A. Ostrowski type inequalities. Proc. Amer. Math. Soc. 123(1995), no. 12, 3775–3781.

4. Dragomir, S. S.; Wang, S. An inequality of Ostrowski-Grüss type and its applications tothe estimation of error bounds for some special means and for some numerical quadrature rules.Comput. Math. Appl. 33 (1997), no. 11, 15–20.

5. Anastassiou, G. A. Multivariate Ostrowski type inequalities. Acta Math. Hungar. 76(1997), no. 4, 267–278.

6. Dragomir, Sever S.; Wang, Song A new inequality of Ostrowski’s type inL1 norm andapplications to some special means and to some numerical quadrature rules. Tamkang J. Math.28 (1997), no. 3, 239–244.

7. Dragomir, S. S.; Wang, S. Applications of Ostrowski’s inequality to the estimation oferror bounds for some special means and for some numerical quadrature rules. Appl. Math.Lett. 11 (1998), no. 1, 105–109.

8. Dragomir, Sever S.; Wang, Song A new inequality of Ostrowski’s type inLp-norm.Indian J. Math. 40 (1998), no. 3, 299–304.

9. Cerone, P.; Dragomir, S. S.; Roumeliotis, J. An inequality of Ostrowski type for mappingswhose second derivatives belong toL1(a, b) and applications. Honam Math. J. 21 (1999), no.1, 127–137.

10. Merkle, Milan Remarks on Ostrowski’s and Hadamard’s inequality. Univ. Beograd.Publ. Elektrotehn. Fak. Ser. Mat. 10 (1999), 113–117.

11. Peachey, T. C.; Mcandrew, A.; Dragomir, S. S. The best constant in an inequality ofOstrowski type. Tamkang J. Math. 30 (1999), no. 3, 219–222.

12. Fedotov, I.; Dragomir, S. S. An inequality of Ostrowski type and its applications forSimpson’s rule and special means. Math. Inequal. Appl. 2 (1999), no. 4, 491–499.

13. Dragomir, S. S.; Barnett, N. S.; Wang, S. An Ostrowski type inequality for a randomvariable whose probability density function belongs toLp[a, b], p > 1. Math. Inequal. Appl. 2(1999), no. 4, 501–508.

14. Dragomir, S. S. The Ostrowski integral inequality for mappings of bounded variation.Bull. Austral. Math. Soc. 60 (1999), no. 3, 495–508.

15. Barnett, N. S.; Dragomir, S. S. An inequality of Ostrowski’s type for cumulative distri-bution functions. Kyungpook Math. J. 39 (1999), no. 2, 303–311.

16. Cerone, P.; Dragomir, S. S.; Roumeliotis, J. An inequality of Ostrowski-Grüss type fortwice differentiable mappings and applications in numerical integration. Kyungpook Math. J.39 (1999), no. 2, 333–341.

261

262 S. S. DRAGOMIR

17. Dragomir, S. S. The Ostrowski’s integral inequality for Lipschitzian mappings andapplications. Comput. Math. Appl. 38 (1999), no. 11-12, 33–37.

18. Cerone, P.; Dragomir, S. S.; Roumeliotis, J. Some Ostrowski type inequalities forn-timedifferentiable mappings and applications. Demonstratio Math. 32 (1999), no. 4, 697–712.

19. Dragomir, S. S.; Barnett, N. S. An Ostrowski type inequality for mappings whosesecond derivatives are bounded and applications. J. Indian Math. Soc. (N.S.) 66 (1999), no.1-4, 237–245.

20. Dragomir, S. S.; Cerone, P.; Roumeliotis, J.; Wang, S. A weighted version of Ostrowskiinequality for mappings of Hölder type and applications in numerical analysis. Bull. Math.Soc. Sci. Math. Roumanie (N.S.) 42(90) (1999), no. 4, 301–314.

2. PERIOD 2000-2010

21. Dedic, Lj.; Matic, M.; Pecaric, J. On some generalizations of Ostrowski inequality forLipschitz functions and functions of bounded variation. Math. Inequal. Appl. 3 (2000), no. 1,1–14.

22. Pearce, C. E. M.; Pecaric, J.; Ujevic, N.; Varošanec, S. Generalizations of some inequal-ities of Ostrowski-Grüss type. Math. Inequal. Appl. 3 (2000), no. 1, 25–34.

23. Brnetic, I.; Pecaric, J. On an Ostrowski type inequality for a random variable. Math.Inequal. Appl. 3 (2000), no. 1, 143–145.

24. Matic, M.; Pecaric, J.; Ujevic, N. Improvement and further generalization of inequalitiesof Ostrowski-Grüss type. Comput. Math. Appl. 39 (2000), no. 3-4, 161–175.

25. Dragomir, S. S.; Cerone, P.; Roumeliotis, J. A new generalization of Ostrowski’s in-tegral inequality for mappings whose derivatives are bounded and applications in numericalintegration and for special means. Appl. Math. Lett. 13 (2000), no. 1, 19–25.

26. Dragomir, S. S.; Pecaric, J.; Wang, S. The unified treatment of trapezoid, Simpson, andOstrowski type inequality for monotonic mappings and applications. Math. Comput. Modelling31 (2000), no. 6-7, 61–70.

27. Dragomir, S. S.; Barnett, N. S.; Cerone, P. Ann-dimensional version of Ostrowski’sinequality for mappings of the Hölder type. Kyungpook Math. J. 40 (2000), no. 1, 65–75.

28. Dedic, Lj.; Matic, M.; Pecaric, J. On generalizations of Ostrowski inequality via someEuler-type identities. Math. Inequal. Appl. 3 (2000), no. 3, 337–353.

29. Dragomir, S. S.; Cerone, P.; Barnett, N. S.; Roumeliotis, J. An inequality of the Os-trowski type for double integrals and applications for cubature formulae. Tamsui Oxf. J. Math.Sci. 16 (2000), no. 1, 1–16.

30. Dragomir, S. S. On the Ostrowski’s inequality for Riemann-Stieltjes integral and appli-cations. Korean J. Comput. Appl. Math. 7 (2000), no. 3, 611–627.

31. Pearce, C. E. M.; Pecaric, J. On Anastassiou’s generalizations of the Ostrowski inequal-ity and related results. J. Comput. Anal. Appl. 2 (2000), no. 3, 215–235.

32. Pachpatte, B. G. On an inequality of Ostrowski type in three independent variables. J.Math. Anal. Appl. 249 (2000), no. 2, 583–591.

33. Gavrea, B.; Gavrea, I. Ostrowski type inequalities from a linear functional point of view.JIPAM. J. Inequal. Pure Appl. Math. 1 (2000), no. 2, Article 11, 1–9.

34. Barnett, N. S.; Dragomir, S. S. An Ostrowski’s type inequality for a random variablewhose probability density function belongs toL∞[a, b]. Nonlinear Anal. Forum 5 (2000), 125–135.

35. Culjak, V.; Pearce, C. E. M.; Pecaric, J. The unified treatment of some inequalities ofOstrowski and Simpson types. Soochow J. Math. 26 (2000), no. 4, 377–390.


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36. Matic, M.; Pecaric, J.; Ujevic, N. Generalizations of weighted version of Ostrowski’sinequality and some related results. J. Inequal. Appl. 5 (2000), no. 6, 639–666.

37. Hanna, G.; Cerone, P.; Roumeliotis, J. An Ostrowski type inequality in two dimensionsusing the three point rule. Proceedings of the 1999 International Conference on ComputationalTechniques and Applications (Canberra). ANZIAM J. 42 (2000), (C), C671–C689.

38. Cerone, P.; Dragomir, S. S.; Roumeliotis, J. An Ostrowski type inequality for mappingswhose second derivatives belong toLp(a, b)and applications. J. Indian Math. Soc. (N.S.) 67(2000), no. 1-4, 59–67.

39. Dragomir, S. S. On the Ostrowski’s integral inequality for mappings with boundedvariation and applications. Math. Inequal. Appl. 4 (2001), no. 1, 59–66.

40. Pachpatte, B. G. On a new Ostrowski type inequality in two independent variables.Tamkang J. Math. 32 (2001), no. 1, 45–49.

41. Dragomir, S. S. A generalization of the Ostrowski integral inequality for mappingswhose derivatives belong toLp[a, b] and applications in numerical integration. J. Math. Anal.Appl. 255 (2001), no. 2, 605–626.

42. Barnett, N. S.; Dragomir, S. S. An Ostrowski type inequality for double integrals andapplications for cubature formulae. Soochow J. Math. 27 (2001), no. 1, 1–10.

43. Sofo, A.; Dragomir, S. S. An inequality of Ostrowski type for twice differentiablemappings in terms of theLp norm and applications. Soochow J. Math. 27 (2001), no. 1,97–111.

44. Matic, M.; Pecaric, J. Two-point Ostrowski inequality. Math. Inequal. Appl. 4 (2001),no. 2, 215–221.

45. Cheng, Xiao-Liang Improvement of some OstrowskiGrüss type inequalities. Comput.Math. Appl. 42 (2001), no. 1-2, 109–114.

46. Matic, M.; Pecaric, J.; Ujevic, N. Weighted version of multivariate Ostrowski typeinequalities. Rocky Mountain J. Math. 31 (2001), no. 2, 511–538.

47. Dragomir, S. S. Better bounds in some OstrowskiGrüss type inequalities. Tamkang J.Math. 32 (2001), no. 3, 211–216.

48. Barnett, N. S.; Dragomir, S. S.; Sofo, A. Better bounds for an inequality of the Ostrowskitype with applications. Demonstratio Math. 34 (2001), no. 3, 533–542.

49. Dragomir, S. S. A generalization of Ostrowski integral inequality for mappings whosederivatives belong toL1[a, b] and applications in numerical integration. J. Comput. Anal. Appl.3 (2001), no. 4, 343–360.

50. Sofo, A.; Dragomir, S. S. A perturbed version of the Ostrowski inequality for twicedifferentiable mappings. Turkish J. Math. 25 (2001), no. 3, 379–412.

51. Cerone, P.; Dragomir, S. S. Some bounds in terms of∆-seminorms for OstrowskiGrüsstype inequalities. Soochow J. Math. 27 (2001), no. 4, 423–434.

52. Dragomir, S. S.; Fang, M. L. Improvement of an Ostrowski type inequality for monotonicmappings and its application for some special means. JIPAM. J. Inequal. Pure Appl. Math. 2(2001), no. 3, Article 31, 6 pp.

53. Dragomir, S. S. On the Ostrowski’s integral inequality for Lipschitzian mappings andapplications. Studia Univ. Babes-Bolyai Math. 46 (2001), no. 1, 33–40.

54. Sofo, A. A perturbed version of the Ostrowski integral inequality. Proceedings of theThird World Congress of Nonlinear Analysts, Part 4 (Catania, 2000). Nonlinear Anal. 47(2001), no. 4, 2353–2364.

55. Pecaric, Josip; Varošanec, Sanja Harmonic polynomials and generalization of Ostrowskiinequality with applications in numerical integration. Proceedings of the Third World Congressof Nonlinear Analysts, Part 4 (Catania, 2000). Nonlinear Anal. 47 (2001), no. 4, 2365–2374.


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264 S. S. DRAGOMIR

56. Dragomir, S. S.; Glušcevic, V.; Pearce, C. E. M. Csiszár f-divergence, Ostrowski’sinequality and mutual information. Proceedings of the Third World Congress of NonlinearAnalysts, Part 4 (Catania, 2000). Nonlinear Anal. 47 (2001), no. 4, 2375–2386.

57. Pachpatte, B. G. An inequality of Ostrowski type in n independent variables. FactaUniv. Ser. Math. Inform. No. 16 (2001), 21–24.

58. Barnett, N. S.; Cerone, P.; Dragomir, S. S.; Roumeliotis, J.; Sofo, A. A survey on Os-trowski type inequalities for twice differentiable mappings and applications. Inequality theoryand applications. Vol. I, 33–85, Nova Sci. Publ., Huntington, NY, 2001.

59. Sofo, A.; Dragomir, S. S. An OstrowskiGrüss type inequality for twice differentiablemappings using a multi-branch Peano kernel. Inequality theory and applications. Vol. I, 281–294, Nova Sci. Publ., Huntington, NY, 2001.

60. Pachpatte, B. G. A note on Ostrowski type inequalities. Demonstratio Math. 35 (2002),no. 1, 27–30.

61. Dragomir, S. S. The discrete version of Ostrowski’s inequality in normed linear spaces.JIPAM. J. Inequal. Pure Appl. Math. 3 (2002), no. 1, Article 2, 16 pp.

62. Barnett, N. S.; Buse, C.; Cerone, P.; Dragomir, S. S. On weighted Ostrowski typeinequalities for operators and vector-valued functions. JIPAM. J. Inequal. Pure Appl. Math. 3(2002), no. 1, Article 12, 21 pp.

63. Cerone, P. A new Ostrowski type inequality involving integral means over end intervals.Tamkang J. Math. 33 (2002), no. 2, 109–118.

64. Anastassiou, George A. Univariate Ostrowski inequalities, revisited. Monatsh. Math.135 (2002), no. 3, 175–189.

65. Sofo, A. Integral inequalities of the Ostrowski type. JIPAM. J. Inequal. Pure Appl.Math. 3 (2002), no. 2, Article 21, 39 pp.

66. Anastassiou, George A. Multivariate Montgomery identities and Ostrowski inequalities.Numer. Funct. Anal. Optim. 23 (2002), no. 3-4, 247–263.

67. Dragomir, S. S. Approximating the finite Hilbert transform via an Ostrowski type in-equality for functions of bounded variation. JIPAM. J. Inequal. Pure Appl. Math. 3 (2002), no.4, Article 51, 19 pp.

68. Pachpatte, B. G. On multivariate Ostrowski type inequalities. JIPAM. J. Inequal. PureAppl. Math. 3 (2002), no. 4, Article 58, 5 pp.

69. Barnett, N. S.; Buse, C.; Cerone, P.; Dragomir, S. S. Ostrowski’s inequality for vector-valued functions and applications. Comput. Math. Appl. 44 (2002), no. 5-6, 559–572.

70. Cerone, P. On relationships between Ostrowski, trapezoidal and Chebychev identitiesand inequalities. Soochow J. Math. 28 (2002), no. 3, 311–328.

71. Dragomir, S. S. A refinement of Ostrowski’s inequality for absolutely continuous func-tions whose derivatives belong toL∞ and applications. Libertas Math. 22 (2002), 49–63.

72. Lü, Zhong Xue On a generalized Ostrowski integral inequality. (Chinese) J. XuzhouNorm. Univ. Nat. Sci. Ed. 20 (2002), no. 3, 19–22.

73. Dragomir, Sever S.; Rassias, Themistocles M. Generalisations of the Ostrowski in-equality and applications. Ostrowski type inequalities and applications in numerical integration,1–63, Kluwer Acad. Publ., Dordrecht, 2002.

74. Barnett, Neil S.; Cerone, Pietro; Dragomir, Sever S. Ostrowski type inequalities formultiple integrals. Ostrowski type inequalities and applications in numerical integration, 285–329, Kluwer Acad. Publ., Dordrecht, 2002.

75. Hanna, George Some results for double integrals based on an Ostrowski type inequality.Ostrowski type inequalities and applications in numerical integration, 331–371, Kluwer Acad.Publ., Dordrecht, 2002.


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76. Dedic, Lj.; Matic, M.; Pecaric, J. Some further generalizations of Ostrowski inequalityfor Hölder functions and functions with bounded derivatives. J. Comput. Anal. Appl. 4 (2002),no. 4, 313–337.

77. Hanna, G.; Dragomir, S. S.; Cerone, P. A general Ostrowski type inequality for doubleintegrals. Tamkang J. Math. 33 (2002), no. 4, 319–333.

78. Zhongxue, Lü; Hongzheng, Xie On some new generalizations of Ostrowski integralinequality. Int. J. Math. Math. Sci. 32 (2002), no. 2, 93–101.

79. Dragomir, S. S. Approximating the finite Hilbert transform via Ostrowski type inequal-ities for absolutely continuous functions. Bull. Korean Math. Soc. 39 (2002), no. 4, 543–559.

80. Dragomir, S. S. A refinement of Ostrowski’s inequality for absolutely continuous func-tions and applications. Acta Math. Vietnam. 27 (2002), no. 2, 203–217.

81. Anastassiou, G. A. Multidimensional Ostrowski inequalities, revisited. Acta Math.Hungar. 97 (2002), no. 4, 339–353.

82. Matic, M.; Pecaric, J.; Ujevic, N. Generalization of an inequality of Ostrowski type andsome related results. Indian J. Math. 44 (2002), no. 2, 189–209.

83. Anastassiou, George A. Probabilistic Ostrowski type inequalities. Stochastic Anal.Appl. 20 (2002), no. 6, 1177–1189.

84. Ujevic, Nenad Inequalities of Ostrowski-Grüss type and applications. Appl. Math.(Warsaw) 29 (2002), no. 4, 465–479.

85. Dedic, L. J.; Matic, M.; Pecaric, J.; Vukelic, A. On generalizations of Ostrowski in-equality via Euler harmonic identities. J. Inequal. Appl. 7 (2002), no. 6, 787–805.

86. Buse, C.; Dragomir, S. S.; Sofo, A. Ostrowski’s inequality for vector-valued functionsof bounded semivariation and applications. New Zealand J. Math. 31 (2002), no. 2, 137–152.

87. Ujevic, Nenad Perturbations of an Ostrowski type inequality and applications. Int. J.Math. Math. Sci. 32 (2002), no. 8, 491–500.

88. Dragomir, S. S. Ostrowski type inequalities for isotonic linear functionals. JIPAM. J.Inequal. Pure Appl. Math. 3 (2002), no. 5, Article 68, 13 pp.

89. Yang, Xiaojing Refinement of Hölder inequality and application to Ostrowski inequality.Appl. Math. Comput. 138 (2003), no. 2-3, 455–461.

90. Dragomir, S. S. A weighted Ostrowski type inequality for functions with values inHilbert spaces and applications. J. Korean Math. Soc. 40 (2003), no. 2, 207–224.

91. Dedic, Lj.; Pecaric, J.; Ujevic, N. On generalizations of Ostrowski inequality and somerelated results. Czechoslovak Math. J. 53(128) (2003), no. 1, 173–189

92. Pachpatte, B. G. On multidimensional Ostrowski and Grüss type finite difference in-equalities. JIPAM. J. Inequal. Pure Appl. Math. 4 (2003), no. 1, Article 7, 8 pp.

93. Ujevic, Nenad Inequalities of Ostrowski type and applications in numerical integration.Appl. Math. E-Notes 3 (2003), 71–79

94. Dragomir, S. S. Some new inequalities of Ostrowski type. Austral. Math. Soc. Gaz. 30(2003), no. 1, 25–30

95. Anastassiou, George A. Fractional Ostrowski type inequalities. Commun. Appl. Anal.7 (2003), no. 2-3, 203–208.

96. Dragomir, S. S. Improvements of Ostrowski and generalised trapezoid inequality interms of the upper and lower bounds of the first derivative. Tamkang J. Math. 34 (2003), no. 3,213–222.

97. Ujevic, N. New bounds for the first inequality of Ostrowski-Grüss type and applications.Comput. Math. Appl. 46 (2003), no. 2-3, 421–427.

98. Anastassiou, George A. Fuzzy Ostrowski type inequalities. Comput. Appl. Math. 22(2003), no. 2, 279–292.


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266 S. S. DRAGOMIR

99. Dragomir, Sever S. An Ostrowski like inequality for convex functions and applications.Rev. Mat. Complut. 16 (2003), no. 2, 373–382.

100. Barnett, N. S.; Cerone, P.; Dragomir, S. S.; Pinheiro, M. R.; Sofo, A. Ostrowskitype inequalities for functions whose modulus of the derivatives are convex and applications.Inequality theory and applications, Vol. 2 (Chinju/Masan, 2001), 19–32, Nova Sci. Publ.,Hauppauge, NY, 2003

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102. Hanna, G.; Dragomir, S. S. Some Ostrowski type inequalities for double integrals offunctions whose partial derivatives satisfy certain convexity properties. Inequality theory andapplications, Vol. 2 (Chinju/Masan, 2001), 113–126, Nova Sci. Publ., Hauppauge, NY, 2003.

103. Ujevic, Nenad Ostrowski-Grüss type inequalities in two dimensions. JIPAM. J. In-equal. Pure Appl. Math. 4 (2003), no. 5, Article 101, 12 pp.

104. Lü, Zhongxue On Ostrowski integral inequality in three independent variables andapplications. Int. J. Math. Sci. 2 (2003), no. 1, 185–203.

105. Ujevic, Nenad Two sharp Ostrowskilike inequalities and applications. Methods Appl.Anal. 10 (2003), no. 3, 477–486.

106. Roumeliotis, John Improved weighted OstrowskiGrüss type inequalities. Inequalitytheory and applications. Vol. 3, 153–160, Nova Sci. Publ., Hauppauge, NY, 2003.

107. Ujevic, Nenad Generalized perturbed inequalities of Ostrowski type and applications.Inequality theory and applications. Vol. 3, 197–206, Nova Sci. Publ., Hauppauge, NY, 2003.

108. Ujevic, N. A generalization of Ostrowski’s inequality and applications in numericalintegration. Appl. Math. Lett. 17 (2004), no. 2, 133–137.

109. Pachpatte, B. G. A note on Ostrowski and Grüss type discrete inequalities. TamkangJ. Math. 35 (2004), no. 1, 61–65.

110. Hwang, Dah-Yan; Yang, Gou-Sheng On Ostrowski and Grüss type discrete inequalitiesfor second forward differences. Tamkang J. Math. 35 (2004), no. 1, 87–93.

111. Pachpatte, B. G. New Ostrowski type inequalities for mappings whose derivativesbelong to Lp spaces. Demonstratio Math. 37 (2004), no. 2, 293–298.

112. Cerone, Pietro; Dragomir, Sever S. Ostrowski type inequalities for functions whosederivatives satisfy certain convexity assumptions. Demonstratio Math. 37 (2004), no. 2, 299–308.

113. Pachpatte, B. G. New inequalities of Ostrowski type for twice differentiable mappings.Tamkang J. Math. 35 (2004), no. 3, 219–226.

114. Pachpatte, B. G. On a new generalisation of Ostrowski’s inequality. JIPAM. J. Inequal.Pure Appl. Math. 5 (2004), no. 2, Article 36, 4 pp.

115. Ujevic, N. Sharp inequalities of Simpson type and Ostrowski type. Comput. Math.Appl. 48 (2004), no. 1-2, 145–151.

116. Pachpatte, B. G. New inequalities of Ostrowski and trapezoid type forn-time differ-entiable functions. Bull. Korean Math. Soc. 41 (2004), no. 4, 633–639.

117. Aljinovic, A. A.; Pecaric, J. Discrete weighted Montgomery identity and discreteOstrowski type inequalities. Comput. Math. Appl. 48 (2004), no. 5-6, 731–745.

118. Dragomir, S. S. Some companions of Ostrowski’s inequality for absolutely continuousfunctions and applications. Facta Univ. Ser. Math. Inform. No. 19 (2004), 1–16.

119. Ujevic, Nenad Inequalities of Ostrowski type in two dimensions. Rocky Mountain J.Math. 35 (2005), no. 1, 331–347.

120. Wu, Qingbiao; Yang, Shijun A note to Ujevic’s generalization of Ostrowski’s inequal-ity. Appl. Math. Lett. 18 (2005), no. 6, 657–665.


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121. de la Cal, J.; Cárcamo, J. A general Ostrowskitype inequality. Statist. Probab. Lett. 72(2005), no. 2, 145–152.

122. Dragomir, S. S. Some Ostrowski type inequalities via Cauchy’s mean value theorem.New Zealand J. Math. 34 (2005), no. 1, 31–42.

123. Dragomir, S. S. Some companions of Ostrowski’s inequality for absolutely continuousfunctions and applications. Bull. Korean Math. Soc. 42 (2005), no. 2, 213–230.

124. Kumar, P. The Ostrowski type moment integral inequalities and moment-bounds forcontinuous random variables. Comput. Math. Appl. 49 (2005), no. 11-12, 1929–1940.

125. Dragomir, S. S. An Ostrowski type inequality for convex functions. Univ. Beograd.Publ. Elektrotehn. Fak. Ser. Mat. 16 (2005), 12–25.

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

127. Pachpatte, B. G. A note on Ostrowski like inequalities. JIPAM. J. Inequal. Pure Appl.Math. 6 (2005), no. 4, Article 114, 4 pp.

128. Pachpatte, B. G. On OstrowskiGrüss-Cebyšev type inequalities for functions whosemodulus of derivatives are convex. JIPAM. J. Inequal. Pure Appl. Math. 6 (2005), no. 4,Article 128, 15 pp.

129. Aglic Aljinovi c, A.; Pecaric, J. On some Ostrowski type inequalities via Montgomeryidentity and Taylor’s formula. Tamkang J. Math. 36 (2005), no. 3, 199–218.

130.Culjak, V.; Pecaric, J.; Persson, L. E. A generalized Simpson, trapezoid and Ostrowskitype inequalities for convex functions. Soochow J. Math. 31 (2005), no. 4, 617–627.

131. Aglic Aljinovi c, A.; Marangunic, Lj.; Pecaric, J. On Landau type inequalities viaOstrowski inequalities. Nonlinear Funct. Anal. Appl. 10 (2005), no. 4, 565–579.

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134. Dragomir, S. S. Ostrowski type inequalities for functions defined on linear spaces andapplications for semi-inner products. J. Concr. Appl. Math. 3 (2005), no. 1, 91–103

135. Chen, Ying Shu; Yu, Yuan Hong An inequality of Ostrowski type for twice differen-tiable functions. (Chinese) Math. Practice Theory 35 (2005), no. 12, 188–192.

136. Aglic Aljinovi c, A.; Matic, M.; Pecaric, J. Improvements of some Ostrowski typeinequalities. J. Comput. Anal. Appl. 7 (2005), no. 3, 289–304.

137. Pachpatte, B. G. New Ostrowski and Grüss type inequalities. An. Stiint. Univ. Al. I.Cuza Iasi. Mat. (N.S.) 51 (2005), no. 2, 377–386 (2006).

138. Gavrea, Ioan On an Ostrowski type inequality. Automat. Comput. Appl. Math. 14(2005), no. 1, 57–61.

139. Dragomir, S. S.; Sofo, A. An inequality for monotonic functions generalizing Os-trowski and related results. Comput. Math. Appl. 51 (2006), no. 3-4, 497–506.

140. Anastassiou, George A. Csiszar and Ostrowski type inequalities via Euler-type andFink identities. Panamer. Math. J. 16 (2006), no. 2, 77–91

141. Anastasiou, K. S.; Kechriniotis, Aristides I.; Kotsos, B. A. Generalizations of theOstrowski’s inequality. J. Interdiscip. Math. 9 (2006), no. 1, 49–60.

142. Yang, Shijun A unified approach to some inequalities of Ostrowski-Grüss type. Com-put. Math. Appl. 51 (2006), no. 6-7, 1047–1056.

143. Liu, Zheng A sharp integral inequality of Ostrowski-Grüss type. Soochow J. Math. 32(2006), no. 2, 223–231.

144. Pachpatte, B. G. A new Ostrowski type inequality for double integrals. Soochow J.Math. 32 (2006), no. 2, 317–322.


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145. Mir, Nazir Ahmad; Rafiq, Arif A note on Ostrowski like inequalities inL1(a, b) spaces.Gen. Math. 14 (2006), no. 1, 23–30.

146. Barnett, N. S.; Dragomir, S. S. Applications of Ostrowski’s version of the Grüssinequality for trapezoid type rules. Tamkang J. Math. 37 (2006), no. 2, 163–173.

147. Li, Xiu Qin; Song, Guo Hua; Yu, Yuan Hong Some generalizations of Ostrowski’sinequality. (Chinese) Heilongjiang Daxue Ziran Kexue Xuebao 23 (2006), no. 4, 483–485.

148. Pachpatte, B. G. New Ostrowski type inequalities involving the product of two func-tions. JIPAM. J. Inequal. Pure Appl. Math. 7 (2006), no. 3, Article 104, 5 pp.

149. Rafiq, A.; Mir, Nazir Ahmad An Ostrowski type inequality forp-norms. JIPAM. J.Inequal. Pure Appl. Math. 7 (2006), no. 3, Article 112, 7 pp.

150. Rafiq, A.; Mir, N. A.; Zafar, Fiza A generalized OstrowskiGrüss type inequality fortwice differentiable mappings and applications. JIPAM. J. Inequal. Pure Appl. Math. 7 (2006),no. 4, Article 124, 7 pp.

151. Pachpatte, B. G. New Ostrowski type inequalities via mean value theorems. JIPAM. J.Inequal. Pure Appl. Math. 7 (2006), no. 4, Article 137, 5 pp.

152. Pecaric, Josip; Ungar, Šime On an inequality of Ostrowski type. JIPAM. J. Inequal.Pure Appl. Math. 7 (2006), no. 4, Article 151, 5 pp.

153. Jia, Baoguo A generalization for Ostrowski’s inequality in R2. JIPAM. J. Inequal.Pure Appl. Math. 7 (2006), no. 5, Article 182, 4 pp. (electronic).

154. Rafiq, A.; Barnett, N. S. A note on a weighted Ostrowski type inequality for cumulativedistribution functions. JIPAM. J. Inequal. Pure Appl. Math. 7 (2006), no. 5, Article 188, 5 pp.

155. Liu, Zheng A sharp inequality of Ostrowski-Grüss type. JIPAM. J. Inequal. Pure Appl.Math. 7 (2006), no. 5, Article 192, 10 pp.

156. Pachpatte, B. G. New Ostrowski and Grüss type inequalities. Tamsui Oxf. J. Math.Sci. 22 (2006), no. 2, 159–166.

157. Wang, Chung-Shin; Ho, Ming-In; Hwang, Shiow-Ru Ostrowski and Grüss type in-equalities involving several functions. Tamsui Oxf. J. Math. Sci. 22 (2006), no. 2, 231–241.

158. Anastassiou, George A. Multivariate Euler type identity and Ostrowski type inequali-ties. Numerical analysis and approximation theory, 27–54, Casa Cartii de stiinta, Cluj-Napoca,2006.

159. Pachpatte, B. G. New Ostrowski and Grüss type inequalities for double integrals.Demonstratio Math. 39 (2006), no. 4, 783–791

160. Rafiq, Arif; Mir, Nazir Ahmad; Ahmad, Farooq Weighted Ostrowski type inequalityfor differentiable mappings whose first derivatives belong to Lp(a,b). Gen. Math. 14 (2006),no. 3, 91–102.

161. Ahmad, Farooq; Rafiq, Arif; Mir, Nazir Ahmad Weighted OstrowskiGrüss type in-equality for differentiable mappings. Glob. J. Pure Appl. Math. 2 (2006), no. 2, 147–154.

162. Pachpatte, B. G. On generalizations of Ostrowski type inequalities and some relatedresults. An. Stiint. Univ. Al. I. Cuza Iasi. Mat. (N.S.) 52 (2006), no. 2, 365–375.

163. Liu, Zheng Some inequalities of Ostrowski type and applications. Appl. Math. E-Notes 7 (2007), 93–101

164. Cheung, W.-S.; Dragomir, S. S. Two Ostrowski type inequalities for the Stieltjes inte-gral of monotonic functions. Bull. Austral. Math. Soc. 75 (2007), no. 2, 299–311.

165. Liu, Zheng Some sharp OstrowskiGrüss type inequalities. Univ. Beograd. Publ.Elektrotehn. Fak. Ser. Mat. 18 (2007), 14–21.

166. Anastassiou, George A. High order Ostrowski type inequalities. Appl. Math. Lett. 20(2007), no. 6, 616–621.


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167. Liu, Zheng Note on a paper by N. Ujevic: "Sharp inequalities of Simpson type andOstrowski type” [Comput. Math. Appl. 48 (2004), no. 1-2, 145–151; MR2086792]. Appl.Math. Lett. 20 (2007), no. 6, 659–663.

168. Hwang, Shiow-Ru Ostrowski-Grüss-Cebyšev type inequalities involving several func-tions. Tamsui Oxf. J. Math. Sci. 23 (2007), no. 1, 105–125.

169. Liu, Zheng Some OstrowskiGrüss type inequalities and applications. Comput. Math.Appl. 53 (2007), no. 1, 73–79.

170. Aglic Aljinovi c, A.; Hoxha, R.; Pecaric, J. On some Ostrowski type integral inequali-ties. Sarajevo J. Math. 3(15) (2007), no. 1, 29–40.

171. Rafiq, Arif; Ahmad, Farooq; Ahmad Mir, Nazir New OstrowskiGrüss type inequalities.Int. J. Appl. Math. 20 (2007), no. 4, 483–493.

172. Anastassiou, George A. Multivariate Fink type identity and multivariate Ostrowski,comparison of means and Grüss type inequalities. Math. Comput. Modelling 46 (2007), no.3-4, 351–374.

173. Anastassiou, George A. Optimal multivariate Ostrowski Euler type inequalities. Stud.Univ. Babes-Bolyai Math. 52 (2007), no. 1, 25–61.

174. Anastassiou, George A.; Goldstein, Jerome A. Ostrowski type inequalities over Euclid-ean domains. Atti Accad. Naz. Lincei Cl. Sci. Fis. Mat. Natur. Rend. Lincei (9) Mat. Appl.18 (2007), no. 3, 305–310.

175. Anastassiou, George A. Taylor-Widder representation formulae and Ostrowski, Grüss,integral means and Csiszar type inequalities. Comput. Math. Appl. 54 (2007), no. 1, 9–23.

176. Cerone, P.; Cheung, W. S.; Dragomir, S. S. On Ostrowski type inequalities for Stieltjesintegrals with absolutely continuous integrands and integrators of bounded variation. Comput.Math. Appl. 54 (2007), no. 2, 183–191.

177. Anastassiou, George Multivariate Euler type identity and optimal multivariate Os-trowski type inequalities. Adv. Nonlinear Var. Inequal. 10 (2007), no. 2, 51–104.

178. Anastassiou, George A. Multivariate fractional Ostrowski type inequalities. Comput.Math. Appl. 54 (2007), no. 3, 434–447.

179. Rafiq, Arif; Mir, Nazir Ahmad; Ahmad, Farooq WeightedCebyšev-Ostrowski typeinequalities. Appl. Math. Mech. (English Ed.) 28 (2007), no. 7, 901–906.

180. Aljinovic, A. Aglic; Pecaric, J. Generalizations of weighted Euler identity and Os-trowski type inequalities. Adv. Stud. Contemp. Math. (Kyungshang) 14 (2007), no. 1,141–151.

181. Pachpatte, B. G. Some new Ostrowski and Grüss type inequalities. Tamkang J. Math.38 (2007), no. 2, 111–120.

182. Civljak, A.; Dedic, Lj.; Matic, M. On Ostrowski and Euler-Grüss type inequalitiesinvolving measures. J. Math. Inequal. 1 (2007), no. 1, 65–81.

183. Rafiq, Arif; Ahmad, Farooq Ostrowski, Grüss,Cebyšev type inequalities for func-tions whose second derivatives belong to Lp(a,b) and whose modulus of second derivatives areconvex. Rev. Colombiana Mat. 41 (2007), no. 1, 1–13.

184. Pachpatte, B. G. New Ostrowski type inequality for triple integrals. Tamkang J. Math.38 (2007), no. 3, 253–259.

185. Pachpatte, B. G. New inequalities of Ostrowski-Grüss type. Fasc. Math. No. 38(2007), 97–104.

186. Pachpatte, B. G. New discrete OstrowskiGrüss like inequalities. Facta Univ. Ser.Math. Inform. 22 (2007), no. 1, 15–20.

187. Popa, Emil C. An inequality of Ostrowski type via a mean value theorem. Gen. Math.15 (2007), no. 1, 93–100.


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188. Barnett, N. S.; Dragomir, S. S. On the weighted Ostrowski inequality. JIPAM. J.Inequal. Pure Appl. Math. 8 (2007), no. 4, Article 96, 10 pp.

189. Anastassiou, George A. Ostrowski type inequalities over balls and shells via a Taylor-Widder formula. JIPAM. J. Inequal. Pure Appl. Math. 8 (2007), no. 4, Article 106, 13 pp.

190. Liu, Wen-jun; Xue, Qiao-ling; Wang, Shun-Feng Several new perturbed Ostrowskiliketype inequalities. JIPAM. J. Inequal. Pure Appl. Math. 8 (2007), no. 4, Article 110, 6 pp.

191. Pachpatte, B. G. On a new generalization of Ostrowski type inequality. Tamkang J.Math. 38 (2007), no. 4, 335–339.

192. Rafiq, Arif; Zafar, Fiza New bounds for the first inequality of Ostrowski-Grüss typeand applications in numerical integration. Nonlinear Funct. Anal. Appl. 12 (2007), no. 1,75–85.

193. Pachpatte, B. G. New generalization of certain Ostrowski type inequalities. TamsuiOxf. J. Math. Sci. 23 (2007), no. 4, 393–403.

194. Dinu, Cristian Ostrowski type inequalities on time scales. An. Univ. Craiova Ser. Mat.Inform. 34 (2007), 43–58.

195. Kechriniotis, Aristides I. Complements of Ostrowski type inequalities. Appl. Math.E-Notes 8 (2008), 89–97.

196. Anastassiou, George A.; Goldstein, Jerome A. Higher order Ostrowski type inequali-ties over Euclidean domains. J. Math. Anal. Appl. 337 (2008), no. 2, 962–968.

197. Bohner, Martin; Matthews, Thomas Ostrowski inequalities on time scales. JIPAM. J.Inequal. Pure Appl. Math. 9 (2008), no. 1, Article 6, 8 pp.

198. Tseng, Kuei-Lin; Hwang, Shiow Ru; Dragomir, S. S. Generalizations of weightedOstrowski type inequalities for mappings of bounded variation and their applications. Comput.Math. Appl. 55 (2008), no. 8, 1785–1793.

199. Yeh, Cheh-Chih Ostrowski’s inequality on time scales. Appl. Math. Lett. 21 (2008),no. 4, 404–409.

200. Liu, Wenjun; Dong, Jianwei On new Ostrowski type inequalities. Demonstratio Math.41 (2008), no. 2, 317–322.

201. Liu, Zheng A note on an Ostrowski type inequality. Tamsui Oxf. J. Math. Sci. 24(2008), no. 1, 7–10.

202. Hwang, Shiow-Ru; Cheng, Kuang-Tsan; Wang, Chung-Shin A note on multivariateOstrowski type inequalities. Tamsui Oxf. J. Math. Sci. 24 (2008), no. 1, 97–108.

203. Liu, Zheng A sharp generalized OstrowskiGrüss inequality. Tamsui Oxf. J. Math. Sci.24 (2008), no. 2, 175–184.

204. Rafiq, Arif; Mir, Nazir Ahmad; Zafar, Fiza A generalized Ostrowski type inequalityfor a random variable whose probability density function belongs toL∞[a, b]. DemonstratioMath. 41 (2008), no. 3, 723–732.

205. Liu, Wen-Jun; Huang, Yu; Pan, Xing-xia New weighted OstrowskiGrüss-Cebyšev typeinequalities. Bull. Korean Math. Soc. 45 (2008), no. 3, 477–483.

206. Liu, Zheng A sharpL2 inequality of Ostrowski type. ANZIAM J. 49 (2008), no. 3,423–429.

207. Rafiq, Arif; Mir, Nazir Ahmad; Zafar, Fiza A generalized Ostrowski type inequality fora random variable whose probability density function belongs to Lp[a,b]. Appl. Math. E-Notes8 (2008), 246–253.

208. Liu, Zheng Some Ostrowski type inequalities. Math. Comput. Modelling 48 (2008),no. 5-6, 949–960.

209. Anastassiou, George A. Ostrowski type inequalities over spherical shells. SerdicaMath. J. 34 (2008), no. 3, 629–650.


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210. Liu, Wenjun; Ngô, Quõc-Anh A generalization of Ostrowski inequality on time scalesfor k points. Appl. Math. Comput. 203 (2008), no. 2, 754–760.

211. Dragomir, Sever S. Refinements of the generalised trapezoid and Ostrowski inequali-ties for functions of bounded variation. Arch. Math. (Basel) 91 (2008), no. 5, 450–460.

212. Lü, Zhongxue On sharp inequalities of Simpson type and Ostrowski type in two inde-pendent variables. Comput. Math. Appl. 56 (2008), no. 8, 2043–2047.

213. Miao, Yu; Shen, Luming Some general OstrowskiGrüss type inequalities. Acta Math.Acad. Paedagog. Nyházi. (N.S.) 24 (2008), no. 2, 243–248.

214. Kikianty, Eder; Dragomir, S. S.; Cerone, P. Sharp inequalities of Ostrowski type forconvex functions defined on linear spaces and application. Comput. Math. Appl. 56 (2008),no. 9, 2235–2246.

215. Qayyum, A. A weighted OstrowskiGrüss type inequality for twice differentiable map-pings and applications. Int. J. Math. Comput. 1 (2008), N08, 63–71.

216. Kikianty, Eder; Dragomir, Sever S.; Cerone, Pietro Ostrowski type inequality forabsolutely continuous functions on segments in linear spaces. Bull. Korean Math. Soc. 45(2008), no. 4, 763–780.

217. Liu, Wenjun; Quoc Anh Ngô; Chen, Wenbing A perturbed Ostrowskitype inequalityon time scales fork points for functions whose second derivatives are bounded. J. Inequal.Appl. 2008, Art. ID 597241, 12 pp.

218. Karpuz, Basak; Özkan, Umut Mutlu Generalized Ostrowski’s inequality on timescales. JIPAM. J. Inequal. Pure Appl. Math. 9 (2008), no. 4, Article 112, 7 pp.

219. Rafiq, Arif; Mir, Nazir Ahmad; Zafar, Fiza A generalization of OstrowskiGrüss typeinequality for twice differentiable mappings in Euclidean norm. Gen. Math. 16 (2008), no. 3,51–72.

220. Mir, Nazir Ahmad; Rafiq, Arif; Rizwan, Muhammad Ostrowski GrüssCebyšev typeinequalities for functions whose modulus of second derivatives are convex. Gen. Math. 16(2008), no. 3, 73–97.

221. Rafiq, Arif; Ahmad, Farooq Another weighted OstrowskiGrüss type inequality fortwice differentiable mappings. Kragujevac J. Math. 31 (2008), 43–51.

222. Tseng, Kuei-Lin Improvements of some inequalities of Ostrowski type and their ap-plications. Taiwanese J. Math. 12 (2008), no. 9, 2427–2441.

223. Tong, Fengru; Guan, Linlin A simple proof of the generalized OstrowskiGrüss typeintegral inequality. Int. J. Math. Anal. (Ruse) 2 (2008), no. 17-20, 889–892.

224. Liu, Zheng Another sharp L2 inequality of Ostrowski type. ANZIAM J. 50 (2008),no. 1, 129–136.

225. Ngô, Quoc Anh; Liu, Wenjun A sharp Grüss type inequality on time scales and ap-plication to the sharp OstrowskiGrüss inequality. Commun. Math. Anal. 6 (2009), no. 2,33–41.

226. Barnett, N. S.; Cheung, W.-S.; Dragomir, S. S.; Sofo, A. Ostrowski and trapezoid typeinequalities for the Stieltjes integral with Lipschitzian integrands or integrators. Comput. Math.Appl. 57 (2009), no. 2, 195–201.

227. Miao, Yu Note on the discrete OstrowskiGrüss type inequality. J. Math. Inequal. 3(2009), no. 1, 93–98.

228. Liu, Zheng Some companions of an Ostrowski type inequality and applications. JI-PAM. J. Inequal. Pure Appl. Math. 10 (2009), no. 2, Article 52, 12 pp.

229. Huy, Vu Nhat; Ngô, Quoc-Anh New inequalities of Ostrowski-like type involving nknots and theLp-norm of them-th derivative. Appl. Math. Lett. 22 (2009), no. 9, 1345–1350.

230. Acu, Ana Maria; Gonska, Heiner Ostrowski inequalities and moduli of smoothness.Results Math. 53 (2009), no. 3-4, 217–228.


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231. Liu, Zheng Some Ostrowski type inequalities and applications. Vietnam J. Math. 37(2009), no. 1, 15–22.

232. Barnett, N. S.; Dragomir, S. S.; Gomm, I. A companion for the Ostrowski and thegeneralised trapezoid inequalities. Math. Comput. Modelling 50 (2009), no. 1-2, 179–187.

233. Niezgoda, Marek A new inequality of Ostrowski-Grüss type and applications to somenumerical quadrature rules. Comput. Math. Appl. 58 (2009), no. 3, 589–596.

234. Wang, Mingjin; Zhao, Xilai Ostrowski type inequalities for higher-order derivatives.J. Inequal. Appl. 2009, Art. ID 162689, 8 pp.

235. Pachpatte, B. G. New inequalities of Ostrowski and Grüss type for triple integrals.Tamkang J. Math. 40 (2009), no. 2, 117–127.

236. Liu, Wenjun; Ngô, Quoc-Anh An OstrowskiGrüss type inequality on time scales.Comput. Math. Appl. 58 (2009), no. 6, 1207–1210.

237. Anastassiou, George A. Caputo fractional Ostrowski type inequalities. J. Appl. Funct.Anal. 4 (2009), no. 2, 218–236.

238. Liu, Wenjun J.; Quoc-Anh Ngô; Chen, Wenbing B. A new generalization of Ostrowskitype inequality on time scales. An. Stiint. Univ. "Ovidius” Constanta Ser. Mat. 17 (2009), no.2, 101–114.

239. Aglic Aljinovi c, A. A note on generalization of weightednnCebyšev and Ostrowskiinequalities. J. Math. Inequal. 3 (2009), no. 3, 409–416.

240. Matic, Marko; Ungar, Šime More on the two-point Ostrowski inequality. J. Math.Inequal. 3 (2009), no. 3, 417–426.

241. Ahmad, Farooq; Barnett, N. S.; Dragomir, S. S. New weighted Ostrowski andCebyševtype inequalities. Nonlinear Anal. 71 (2009), no. 12, e1408–e1412.

242. Rafiq, Arif; Ahmad, Farooq On weighted Ostrowski type inequalities inL1(a, b)spaces. Math. Commun. 14 (2009), no. 1, 27–33.

3. AFTER 2010

243. Liu, Wen-Jun; Xue, Qiao-Ling; Wang, Shun-Feng New generalization of perturbedOstrowski type inequalities and applications. J. Appl. Math. Comput. 32 (2010), no. 1,157–169.

244. Lian, Bao-Sheng; Yang, Qiao-Hua Ostrowski type inequalities on H-type groups. J.Math. Anal. Appl. 365 (2010), no. 1, 158–166.

245. Zhao, Changjian; Cheung, Wing-Sum On Ostrowskitype inequalities for higher-orderpartial derivatives. J. Inequal. Appl. 2010, Art. ID 960672, 8 pp.

246. Özkan, Umut Mutlu; Yildirim, Hüseyin Ostrowski type inequality for double integralson time scales. Acta Appl. Math. 110 (2010), no. 1, 283–288.

247. Liu, Wenjun; Ngô, Quoc Anh; Chen, Wenbing Ostrowski type inequalities on timescales for double integrals. Acta Appl. Math. 110 (2010), no. 1, 477–497.

248. Liu, Heng-Xing; Luan, Jing-Wen Ostrowski type inequalities in the Grushin plane. J.Inequal. Appl. 2010, Art. ID 987484, 9 pp.

249. Huy, Vu Nhat; Ngô, Quoc-Anh A new way to think about Ostrowskilike type inequal-ities. Comput. Math. Appl. 59 (2010), no. 9, 3045–3052.

250. Civljak, Ambroz; Dedic, Ljuban Generalizations of Ostrowski inequality via bipara-metric Euler harmonic identities for measures. Banach J. Math. Anal. 4 (2010), no. 1, 170–184.

251. Liu, Zheng A sharp OstrowskiGrüss type inequality. Math. Inequal. Appl. 13 (2010),no. 2, 243–253.

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