Proceedings of The Fifth International Symposium ...depmath.ulbsibiu.ro/genmath/pdf/MathematicalInequalities.pdf5 Participants to The Fifth International Symposium "Mathematical Inequalities"

”Lucian Blaga” University of Sibiu

Department of Mathematics

and

Romanian Mathematical Scientific Society

Proceedings of The Fifth International

Symposium

”Mathematical Inequalities”

Sibiu, 25 - 27 September 2008

Editors:

Dumitru Acu

Emil C. Popa

Ana-Maria Acu

Florin Sofonea

Copyright c© 2009 Publishing of ”Lucian Blaga” University from Sibiu

ISBN 978-973-739-740-9

ISSN 2066-2386

COVER DESIGN AND EDITOR-IN-COMPUTER

Ana Maria Acu


Str. Dr. Ioan Ratiu, No. 5-7

550012-Sibiu, Romania

The Fifth International Symposium ”Mathematical Inequalities”

25 - 27 September 2008, Sibiu, Romania

ORGANIZING COMMITTEE

Professor Ph.D. Dumitru Acu - Head of the Department of Mathematics

Ph.D. Constantin Oprean - Rector of ”Lucian Blaga” University

Professor Ph.D. Dumitru Batar - Dean of the Faculty of Sciences

Professor Ph.D. Ilie Barza - Karlstad University, Sweden

Professor Ph.D. Josip E. Pecaric - University of Zagreb, Croatia

Professor Ph.D. Sever S. Dragomir - Victoria University of Technology, Australia

Assoc.Professor Ph.D. Mihai Damian - Strasbourg University, France

Assoc.Professor Ph.D. Sorina Barza - Karlstad University, Sweden

Professor Ph.D. Emil C. Popa - ”Lucian Blaga” University of Sibiu

Professor Ph.D. Vasile Berinde - North University of Baia Mare, Romania

Assoc.Professor Ph.D. Silviu Craciunas - ”Lucian Blaga” University of Sibiu

Assoc.Professor Ph.D. Florin Sofonea - ”Lucian Blaga” University of Sibiu

Lecturer Ph.D. Ana-Maria Acu - ”Lucian Blaga” University of Sibiu

Lecturer Lecturer Ph.D. Marian Olaru - ”Lucian Blaga” University of Sibiu

Lecturer Ph.D. Adrian Branga - ”Lucian Blaga” University of Sibiu

Lecturer Ph.D. Eugen Constantinescu - ”Lucian Blaga” University of Sibiu

Asist. Petrica Dicu - ”Lucian Blaga” University of Sibiu

5

Participants to The Fifth International Symposium

”Mathematical Inequalities” Sibiu, 25 - 27 September 2008

No.

Crt.Name/ e-mail Affiliation

1Ana Maria Acu

[email protected]

”Lucian Blaga” University

from Sibiu, Romania

2Mugur Acu

acu [email protected]


from Sibiu, Romania

3Dumitru Acu



of Sibiu, Romania

4Adrian Branga

adrian [email protected]


from Sibiu, Romania

5Daniel Breaz

[email protected]

”1 Decembrie 1918” University

of Alba-Iulia, Romania

6Nicoleta Breaz

[email protected]

”1 Decembrie 1918” University

of Alba-Iulia, Romania

7Amelia Bucur

[email protected]


from Sibiu, Romania

8Eugen Constantinescu

[email protected]


from Sibiu, Romania

9Daniela Dicu

[email protected]

Liceul Teoretic ” Gheorghe Lazar”

Avrig, Romania

10Gheorghe Dicu Gr. Sc. Forestier

Curtea de Arges, Romania

11Petrica Dicu

[email protected]


of Sibiu, Romania

12Irina Dorca

ira [email protected]


of Sibiu, Romania

13Eugen Draghici

[email protected]


from Sibiu, Romania

14Ali Ebadian

[email protected]

Urmia University,

Iran

6

15Bogdan Gavrea

[email protected]

”Tehnical University”

of Cluj-Napoca, Romania

16Ioan Gavrea

[email protected]



17Heiner Gonska

[email protected]

University of Duisburg-Essen

Germania

18Jose Luis Lopez-Bonilla

joseluis.lopezbonilla@gm

Escuela Superior de Ingenieria

Mecanica y Electrica Ins, Mexic

19Vasile Mihesan

[email protected]



20Nicusor Minculete

[email protected]

University ”Dimitrie Cantemir”

of Brasov, Romania

21Shahram Najafzadeh

[email protected]

University of Maragheh,

Iran

22Marian Olaru

[email protected]


from Sibiu, Romania

23Stefan Poka Gheorghe Sincai High Scholl

Cluj-Napoca, Romania

24Emil C. Popa

[email protected]


from Sibiu, Romania

25Ioan Popa

ioanpopa [email protected]

Edmond Nicolau College


26Luminita Preoteasa Gr. Sc. Forestier

Curtea de Arges, Romania

27Arif Rafiq

[email protected]

COMSATS Institute of

Information Technology

Lahore, Pakistan

28Sofonea Florin

[email protected]


from Sibiu, Romania

29Gheorghe Sandru Scoala Generala Vistea de Jos

Brasov, Romania

30Doru Stefanescu

[email protected]

University of Bucharest

Romania

31Ioan Tincu

[email protected]


from Sibiu, Romania

32Andrei Vernescu

[email protected]

University Valahia

of Targoviste, Romania

7

Contents

A. M. Acu, M. Acu, A. Rafiq – Some inequalities of Ostrowski type in the case of

weighted integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

D. Acu – Some interesting elementary inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .23

A. Branga – An inequality for generalized spline functions . . . . . . . . . . . . . . . . . . . . . . . . . 33

D. Breaz, N. Breaz – Some starlikeness conditions proved by inequalities . . . . . . . . . 40

I. Dorca – Note on subclass of β-starlike and β-convex functions with negative

coefficients associated with some hyperbola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

B. Gavrea – On some inequality for convex functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

I. Gavrea – On some inequalities for convex functions of higher order . . . . . . . . . . . . . . 67

H. Gonska, I. Rasa – A Voronovskaya estimate with second order modulus of

smoothness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

V. Mihesan – Popoviciu type inequalities for pseudo arithmetic and

geometric means . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

V. Mihesan – Rado type inequalities for weighted power pseudo means . . . . . . . . . . . .98

N. Minculete – Several inequalities about arithmetic functions which use the

e-divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

N. Minculete, P.Dicu – Inequalities between some arithmetic functions . . . . . . . . . . 116

I. M. Olaru – An integral inequality for convex functions of three order . . . . . . . . . . . 126

E.C. Popa – On a problem of A. Shafie . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

F. Sofonea, A.M. Acu, A. Rafiq – An error analysis for a family of four-point

quadrature formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

I. Tincu, G. Sandru – A proof of an inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

8

Proceedings of the Fifth International Symposium


SOME INEQUALITIES OF OSTROWSKI TYPE IN THE

CASE OF WEIGHTED INTEGRALS

Ana Maria Acu, Mugur Acu, Arif Rafiq

Abstract

Some new inequalities of Ostrowski type are established. In this

paper we considered the weighted integral case. Some of this inequal-

ities are obtained using the mean value theorems.

2000 Mathematics Subject Classification: 65D30 , 65D32

Key words and phrases: quadrature rule, Ostrowski inequality

1. INTRODUCTION

In 1938, A. M. Ostrowski proved the following classical inequality [5]:

Theorem 1. Let f : [a, b] → R be continuous on [a, b] and differentiable

on (a, b), whose first derivative f ′ : (a, b) → R is bounded on (a, b), i.e.,

|f ′(x)| ≤ M < ∞. Then,

∣∣∣∣f(x)− 1

b− a

∫ b

a

f(t)dt

∣∣∣∣ ≤

1

4+

(x− a + b

2

)2

(b− a)2

(b− a)M,

for all x ∈ [a, b], where M is a constant.

9

A.M. Acu, M. Acu, A. Rafiq - Some inequalities of Ostrowski type...

In [10], N. Ujevic proved the following generalization of Ostrowski ′s

inequality:

Theorem 2.[10] Let I ⊂ R be a open interval and a, b ∈ I, a < b. If

f : I → R is a differentiable function such that γ ≤ f ′(t) ≤ Γ, for all

t ∈ [a, b], for some constants Γ, γ ∈ R, then we have

∣∣∣∣f(x)−Γ+γ

2

(x− a + b

2

)− 1

b−a

∫ b

a

f(t)dt

∣∣∣∣ ≤1

2(Γ−γ)(b−a)

[1

4+

(x− a+b

2

)2

(b−a)2

].

2. THE CASE OF WEIGHTED INTEGRALS

In this section we obtain some inequalities of Ostrowski type in the case

of weighted integrals. Let w : [a, b] → R be a nonnegative and integrable

function on [a, b] defined by w(t) = (b− t)(t− a).

Theorem 3. Let f : [a, b] → R be a differentiable mapping on (a, b) and

suppose that γ ≤ f ′(t) ≤ Γ for all t ∈ (a, b). Then we have

(1)

∣∣∣∣∫ b

a

w(t)f(t)dt− (b− a)3

12(f(a) + f(b))

∣∣∣∣ ≤5

192(b− a)4(Γ− γ),

(2)1

12(b−a)4(γ−S) ≤

∫ b

a

w(t)f(t)dt−(b− a)3

12(f(a) + f(b)) ≤ 1

12(b−a)4(Γ−S),

where S =f(b)− f(a)

b− a.

Proof. We define

P (t) =

(b− a

2

)2 (t− a + b

2

)− 1

3

(t− a + b

2

)3

.

10


Integrating by parts, we have

∫ b

a

P (t)f ′(t)dt = −∫ b

a

(b− t)(t− a)f(t)dt +1

12(b− a)3 [f(a) + f(b)] .

We also have

∫ b

a

P (t)dt =

[1

2

(b− a

2

)2 (t− a + b

2

)2

− 1

12

(t− a + b

2

)4]∣∣∣∣∣

b

a

= 0

Using the relations

∫ b

a

|P (t)| dt =

∫ a+b2

a

[−

(b− a

2

)2 (t− a + b

2

)+

1

3

(t− a + b

2

)3]

dt

+

∫ b

a+b2

[(b−a

2

)2 (t− a+b

2

)− 1

3

(t− a+b

2

)3]

dt=5

96(b−a)4,

∣∣∣∣f ′(t)−Γ + γ

2

∣∣∣∣ ≤Γ− γ

2,

and

∫ b

a

P (t)

(f ′(t)− Γ + γ

2

)dt = −

∫ b

a

(b−t)(t−a)f(t)dt+1

12(b−a)3 (f(a) + f(b))

we obtain

∣∣∣∣∫ b

a


12(f(a) + f(b))

∣∣∣∣ ≤∫ b

a

|P (t)| ·∣∣∣∣f ′(t)−

Γ + γ

2

∣∣∣∣ dt

≤ Γ− γ

2

∫ b

a

|P (t)| dt =5

192(b− a)4(Γ− γ).

We have

∫ b

a

P (t) (f ′(t)− γ) dt = −∫ b

a

(b− t)(t− a)f(t)dt+1

12(b− a)3 [f(a) + f(b)] .

Since

11


∫ b

a

P (t)(f ′(t)−γ)dt≤max |P (t)|∫ b

a

(f ′(t)−γ)dt

= max |P (t)| [f(b)−f(a)−γ(b−a)]=max |P (t)| (S−γ)(b−a),

max |P (t)| = 1

12(b− a)3

we obtain

(3)

∫ b

a


12(f(a) + f(b)) ≥ 1

12(b− a)4(γ − S).

In a similar way we can prove that

(4)

∫ b

a


12(f(a) + f(b)) ≤ 1

12(b− a)4(Γ− S).

From (3) and (4) we get the desired inequality (2).

Theorem 4. Let f : [a, b] → R be a differentiable mapping on (a, b) and

suppose that γ ≤ f ′(t) ≤ Γ for all t ∈ (a, b). Then we have

∣∣∣∣∫ b

a

w(t)f(t)dt− 1

6(b− a)3f(x) +

Γ + γ

12(b− a)3

(x− a + b

2

)∣∣∣∣(5)

≤ 1

12(Γ− γ)(b− a)4

[3

16+

3

2

(x− a+b

2

)2

(b− a)2−

(x− a+b

2

)4

(b− a)4

].

(b−a)4

12

[γ

(M−2

x− a+b2

b−a

)−MS

]≤

∫ b

a

w(t)f(t)dt− (b−a)3

6f(x)(6)

≤(b−a)4

12

[Γ

(M−2

x− a+b2

b−a

)−MS

],

for all x ∈ [a, b], where S =f(b)−f(a)

b−aand M =1+

∣∣∣∣∣3x− a+b

2

b−a−4

(x− a+b

2

)3

(b−a)3

∣∣∣∣∣.

12


Proof. Let us define the mapping P (., .) : [a, b] → R given by

P (x, t) =

(b− a)(t− a)2

2− (t− a)3

3, t ∈ [a, x),

(a− b)(t− b)2

2− (t− b)3

3, t ∈ [x, b].

Integrating by parts we have

∫ b

a

P (x, t)f ′(t)dt =

∫ x

a

[(b− a)

(t− a)2

2− (t− a)3

3

]f ′(t)dt

+

∫ b

x

[(a−b)

(t−b)2

2− (t−b)3

3

]f ′(t)dt=−

∫ b

a

w(t)f(t)dt

+

b−a

2

[(x−a)2+(b−x)2

]− 1

2

[(x−a)3+(b−x)3

]f(x)

=−∫ b

a

w(t)f(t)dt +(b− a)3

6f(x).

We also have

∫ b

a

P (x, t)dt =

∫ x

a

[(b−a)

(t−a)2

2− (t−a)3

3

]dt+

∫ b

x

[(a−b)

(t−b)2

2− (t−b)3

3

]dt

=

[(b−a)

6(t−a)3− (t−a)4

12

]∣∣∣∣x

a

−[b−a

6(t−b)3+

(t−b)4

12

]∣∣∣∣b

x

=1

6(b− a)3

(x− a + b

2

).

13


Using the relations

∫ b

a

|P (x, t)|dt =

∫ x

a

[(b−a)

(t−a)2

2− (t−a)3

3

]dt+

∫ b

x

[(b−a)

(t−b)2

2+

(t−b)3

3

]dt

=

[b−a

6(t−a)3− (t−a)4

12

]∣∣∣∣x

a

+

[b−a

6(t−b)3+

(t−b)4

12

]∣∣∣∣b

x

=b−a

6

[(x−a)3+(b−x)3

]− 1

12

[(x−a)4+(b−x)4

]

=(b− a)4

6

[1

4+ 3

(x− a+b

2

)2

(b− a)2

]

− (b− a)4

6

1

16+

3

2

(x− a+b

2

)2

(b− a)2+

(x− a + b

2

)4

(b− a)4

=(b− a)4

6

[3

16+

3

2

(x− a+b

2

)2

(b− a)2−

(x− a+b

2

)4

(b− a)4

],

∣∣∣∣f ′(t)−Γ + γ

2

∣∣∣∣ ≤Γ− γ

2

and

∫ b

a

P (x, t)

(f ′(t)− Γ + γ

2

)dt = −

∫ b

a

w(t)f(t)dt +(b− a)3

6f(x)

− Γ + γ

12(b− a)3

(x− a + b

2

).

we obtain

∣∣∣∣∫ b

a


6f(x)+

Γ+γ

12(b−a)3

(x− a+b

2

)∣∣∣∣

≤∫ b

a

|P (x, t)|∣∣∣∣f ′(t)−

Γ + γ

2

∣∣∣∣ dt ≤ Γ− γ

2

∫ b

a

|P (x, t)| dt

=1

12(Γ− γ)(b− a)4

[3

16+

3

2

(x− a+b

2

)2

(b− a)2−

(x− a+b

2

)4

(b− a)4

].

14


We have

∫ b

a

P (x, t) (f ′(t)−γ) dt=−∫ b

a

w(t)f(t)dt+(b−a)3

6f(x)−1

6(b−a)3

(x− a+b

2

)γ.

Since ∫ b

a

P (x, t) (f ′(t)− γ) dt ≤ max |P (x, t)| (S − γ)(b− a),

max |P (x, t)|= max

b−a

2(x−a)2− (x−a)3

3,b−a

2(b−x)2− (b−x)3

3

,

=1

12(b−a)3

1+

∣∣∣∣∣3x− a+b

2

b−a−4

(x− a+b

2

)3

(b−a)3

∣∣∣∣∣

=

1

12(b−a)3M.

we obtain

∫ b

a


6f(x) ≥ (b−a)4

12

[γ

(M−2

x− a+b2

b−a

)−MS

]

In a similar way we can prove that

∫ b

a


6f(x) ≤ (b− a)4

12

[Γ

(M − 2

x− a+b2

b− a

)−MS

]

Remark 1. Putting x =a + b

2in relations (5) and (6) from Theorem 4,

we have the following quadrature formulae

(7)∫ b

a

w(t)f(t)dt =(b− a)3

6f

(a + b

2

)+R1[f ], |R1[f ]| ≤ 1

64(Γ− γ)(b− a)4

(8)∫ b

a

w(t)f(t)dt=(b−a)3

6f

(a+b

2

)+R2[f ],

(b−a)4

12(γ−S)≤R2[f ]≤ (b−a)4

12(Γ−S).

15


3. THE MEAN VALUE THEOREMS AND INEQUALITY OF

OSTROWSKI TYPE

In [6], Pompeiu derive a variant of Lagrange ′s mean value theorem:

Theorem 5.[6] For every real valued function f differentiable on an interval

[a, b] not containing 0 and for all pairs x1 6= x2 in [a, b], there exist a point

ξ in (x1, x2) such that

x1f(x2)− x2f(x1)

x1 − x2

= f(ξ)− ξf ′(ξ).

Using Pompeiu ′s mean value theorem, S.S. Dragomir obtained in [4] the

following result

Theorem 6.[4] Let f : [a, b] → R be continuous on [a, b] and differentiable

on (a,b) with [a, b] not containing 0. If w : [a, b] → R is nonnegative

integrable on [a, b], then for each x ∈ [a, b], we have the inequality:

∣∣∣∣∫ b

a

f(t)w(t)dt− f(x)

x

∫ b

a

tw(t)dt

∣∣∣∣(9)

≤ ‖f − lf ′‖∞[sgn(x)

(∫ x

a

w(t)dt−∫ b

x

w(t)dt

)

+1

|x|(∫ b

x

tw(t)dt−∫ x

a

tw(t)dt

)],

where l(t) = t, t ∈ [a, b].

Lemma 1. For every real valued function f differentiable on an interval

[a, b] not containing 0 and for all pairs x1 6= x2 in [a, b] there exist a point

16


ξ in (a, b) such that

(x2 − a)(b− x2)x21f(x2)− (x1 − a)(b− x1)x

22f(x1)

x21 − x2

2

(10)

=

(a + b

2ξ − ab

)f(ξ)− ξ

2(ξ − a)(b− ξ)f ′(ξ).

Proof. Let F , G a real functions defined on the interval

[1

b,1

a

]by

F (u) = (1− au)(bu− 1)f

(1

u

),

G(u) = u2.

Since F and G are differentiable on

(1

b,1

a

)and

F ′(u) = 2

(a + b

2− abu

)f

(1

u

)− (1− au)(bu− 1)

u2f ′

(1

u

),

G′(u) = 2u,

then applying Cauchy ′s mean value theorem to F and G on the interval

[x, y] ⊂[1

b,1

a

], it follows that exist η ∈ (x, y) such that

F (x)− F (y)

G(x)−G(y)=

F ′(η)

G′(η),

namely

(1− ax)(bx− 1)f

(1

x

)− (1− ay)(by − 1)f

(1

y

)

x2 − y2

=1

2η

2

(a + b

2− abη

)f

(1

η

)− (1− aη)(bη − 1)

η2f ′

(1

η

).

If we choose x2 =1

x, x1 =

1

yand ξ =

1

η, then we obtain the relation (10).

17



on (a, b) with [a, b] not containing 0. Then for any x ∈ [a, b], we have the

inequality

∣∣∣∣b3 − a3

3x2(x− a)(b− x)f(x)−

∫ b

a

(t− a)(b− t)f(t)dt

∣∣∣∣

≤(

b− a

x

)2

|a + b| · ‖h‖∞

1

4+

(x− a+b

2

b− a

)2

+4

3

∣∣∣∣∣x− a+b

2

b− a

∣∣∣∣∣

3 ,

where h(t) =

(a + b

2t− ab

)f(t)− t(t− a)(b− t)

2f ′(t).

Proof. Applying Lemma 1, for x, t ∈ [a, b], there is a ξ between x and t

such that

(x− a)(b− x)t2f(x)− (t− a)(b− t)x2f(t)

t2 − x2

=

(a + b

2ξ − ab

)f(ξ)− ξ

2(ξ − a)(b− ξ)f ′(ξ),

namely

|t2(x− a)(b− x)f(x)− x2(t− a)(b− t)f(t)|≤ sup

ξ∈[a,b]

∣∣∣∣(

a+b

2ξ−ab

)f(ξ)− ξ

2(ξ−a)(b−ξ)f ′(ξ)

∣∣∣∣·∣∣t2−x2

∣∣=‖h‖∞ ·∣∣t2−x2

∣∣ .

Integrating over t ∈ [a, b], we obtain

∣∣∣∣b3−a3

3(x−a)(b−x)f(x)−x2

∫ b

a

(t−a)(b−t)f(t)dt

∣∣∣∣≤‖h‖∞∫ b

a

∣∣t2−x2∣∣ dt.

18


For a < b and a > 0, we obtain

∫ b

a

∣∣t2−x2∣∣dt =

∫ x

a

(x2 − t2)dt +

∫ b

x

(t2 − x2)dt

=

∫ x

a

[(x− a + b

2

)2

−(

t− a + b

2

)2

− (a + b)(t− x)

]dt

+

∫ b

x

[(t− a + b

2

)2

−(

x− a + b

2

)2

+ (a + b)(t− x)

]dt

=4

3

(x− a + b

2

)3

+ (a + b)

[(x− a + b

2

)2

+(b− a)2

4

]

= |a + b| (b− a)2

1

4+

(x− a+b

2

b− a

)2

+4

3· b− a

a + b

(x− a+b

2

b− a

)3

≤ |a + b| (b− a)2

1

4+

(x− a+b

2

b− a

)2

+4

3·∣∣∣∣∣x− a+b

2

b− a

∣∣∣∣∣

3 .

For a < b and b < 0, we obtain

∫ b

a

∣∣t2−2∣∣dt =

∫ x

a

(t2 − x2)dt +

∫ b

x

(x2 − t2)dt

=

∫ x

a

[(t− a + b

2

)2

−(

x− a + b

2

)2

+ (a + b)(t− x)

]dt

+

∫ b

x

[(x− a + b

2

)2

−(

t− a + b

2

)2

− (a + b)(t− x)

]dt

=−4

3

(x− a + b

2

)3

− (a + b)

[(x− a + b

2

)2

+(b− a)2

4

]

= |a + b| (b− a)2

1

4+

(x− a+b

2

b− a

)2

+4

3· b− a

a + b

(x− a+b

2

b− a

)3

≤ |a + b| (b− a)2

1

4+

(x− a+b

2

b− a

)2

+4

3·∣∣∣∣∣x− a+b

2

b− a

∣∣∣∣∣

3 .

19


This complete the proof.

In a similar way we can obtain the following results.


on (a, b) with [a, b] not containing 0. Then for any x ∈ [a, b], we have the

inequality

∣∣∣∣x−a

x· a+b

2f(x)− 1

b−a

∫ b

a

(t−a)f(t)dt

∣∣∣∣ ≤1

|x|(b−a)·‖h‖∞·1

4+

(x− a+b

2

b−a

)2 ,

where h(t) = af(t) + (t− a)tf ′(t), t ∈ [a, b].

Remark 2. If we choose x =a + b

2in Theorem 8, we obtain

(11)

∣∣∣∣(b− a)2

2f

(a + b

2

)−

∫ b

a

(t− a)f(t)dt

∣∣∣∣ ≤(b− a)2

2 |a + b| · ‖h‖∞ .

Remark 3. If in Theorem 6 we choose x =a + b

2and w(t) = t−a to obtain

the following inequality∣∣∣∣∫ b

a

(t− a)f(t)dt− f

(a + b

2

)(b− a)2

3(a + b)(2b + a)

∣∣∣∣(12)

≤ ‖f − lf ′‖∞[−sgn

(a + b

2

)· (b− a)2

4+

b(b− a)2

2 |a + b|]

.

Now, we show that (11) can be better than (12). For that purpose, we

choose f(t) =1

t, a = 1, b = 3. Thus, the right-hand side of (11) and (12)

become

R.H.S.(11) =1

2, R.H.S.(12) = 1.

If we choose f(t) = e−t, a = 1, b = 3 we obtain

R.H.S.(11) =1

2e, R.H.S.(12) =

1

e

therefore, the right-hand side of (11) is better then (12).

20


References

[1] A. Branga, Spline functions with applications to optimal approximation,

General Mathematics, Vol. 14, no. 4, pp 135-146, 2006.

[2] P. Cerone, S.S. Dragomir, Midpoint type rules from an inequalities point

of view, in Analytic-Compuational Methods in Applied Mathematics,

G.A. Anastassiou (Ed), CRC Press, New York, 2000,135-200.

[3] P. Cerone, S.S. Dragomir, Trapezoidal type rules from an inequalities

point of view, in Analytic-Compuational Methods in Applied Mathe-

matics, G.A. Anastassiou (Ed), CRC Press, New York, 2000, 65-134.

[4] S.S. Dragomir, An inequality of Ostrowski type via Pompeiu ′s mean

value theorem, JIPAM, Volume 6, Issue 3, Article 83, 2005.

[5] A. Ostrowski, Uber die asolutabweichung einer differencienbaren func-

tionen von ihren integral mittelwert, Comment. Math. Hel, 10, 1938,

226-227.

[6] D. Pompeiu, Sur une proposition analogue au theoreme des accroisse-

ments finis, Mathematica, 22 (1946), 143-146.

[7] E.C. Popa, An inequality of Ostrowski type via a mean value theorem,

General Mathematics Vol. 15, No. 1, 2007, 93-100.

[8] F. Sofonea, Analiza numerica si teoria aproximarii, Editura Univer-

sitatii din Bucuresti, 2006.

21


[9] N. Ujevic, Some double integral inequalities and applications, Acta Math.

Univ. Comeniancae, Vol. LXXI, 2(2002), 189-199.

[10] N.Ujevic, A generalization of Ostrovski ′s inequality and applications in

numerical integration, Appl. Math. Lett., 17(2), 2004, 133-137.

Ana Maria Acu, Mugur Acu

University ”Lucian Blaga” of Sibiu


Str. Dr. I. Ratiu, No. 5-7, 550012 - Sibiu, Romania

E-mail: [email protected],


Arif Rafiq

COMSATS Institute of Information Technology


Defense Road, Off Raiwind Road, Lahore - Pakistan

E-mail: [email protected]

22



SOME INTERESTING ELEMENTARY INEQUALITIES

Dumitru Acu

Abstract

In this paper we present the generalizations for some elementary

inequalities.

2000 Mathematics Subject Classification: 26D07

Key words and phrases: elementary inequalities

1. The aim of this paper is to obtain some interesting elementary in-

equalities.

The starting point and initial inspiration was the results obtained in [1],

problem 3.9.31, [2] and [3].

It is proved in [1] that if m and n, with n < m, are natural numbers and

we consider the function

(1) f(x) =1 + x + . . . + xm

1 + x + . . . + xn,

then:

1) f is strictly increasing on (0, +∞),

2) 1 < f(x) <m + 1

n + 1, 0 < x < 1,

23

D. Acu - Some interesting elementary inequalities

(2) f(x) >m + 1

n + 1xm−n, 0 < x < 1,

f(x) >m + 1

n + 1, 1 < x < ∞

1 < f(x) <m + 1

n + 1xn−m, 1 < x < ∞.

In [2], using the functions which generalize the function (1), we obtain

new inequalities by the type (2).

For example, if we consider the function

(3) f : [0, +∞) → [0, +∞), f(x) =P (x) + xn+1Q(x)

xp+1P (x),

where P (x) and Q(x) are two polynomials of the degree n and p respectively

having real positive coefficients such that for all real x 6= 0

(4) P (x) = xnP

(1

x

)and Q(x) = xpQ

(1

x

),

then the following inequalities are valid

(5) 1 < f(x) < 1 +Q(1)

P (1), f(x) > 1 +

Q(1)

P (1)xn+p+2, for x ∈ (0, 1),

f(x) > 1 +Q(1)

P (1), 1 < f(x) < 1 +

Q(1)

P (1)xn+p+2, for x > 1.

In [3], the authors replace the polynomials (4) by the functions

P (x) =

p∑i=0

aixαi , Q(x) =

q∑i=0

bixβi ,

where: 0 = α0 < α1 < . . . < αp, 0 = β0 < β1 < . . . < βq, p, q ∈ N and

ai, i = 0, p and bi, i = 0, q are nonnegative coefficients such that ao > 0, ap >

0, bq > 0.

24


The function (3) is replaced by the functions

fk : [0, +∞) → [0, +∞), fk =P (x) + xαp+kQ(x)

P (x),

where k ≥ 0.

In [4] the following inequality is given: if a, b ∈ R− 0, then we have

(6)1

3≤ a2 − ab + b2

a2 + ab + b2≤ 3. (Problem 1.60, p.6 )

This double inequality is equivalent

(7)1

3≤ t2 − t + 1

t2 + t + 1≤ 3.

with t =a

b. The inequalities (7) are verified for all real numbers t and they

have the form (2).

In this work we generalize the inequalities (6) and (7) and we obtain

another interesting results.

2. In order to generalize the inequalities (6) and (7), we shall make use

of the following elementary result.

Lemma 1. Let the function Tn be defined on R2 − (0, 0) by

Tn(a, b) = an + an−1b + . . . + abn−1 + bn, n ∈ N∗.

i) If n is even, then Tn(a, b) > 0 for any real numbers a, b, with a 6= 0

or b 6= 0.

ii) If n is odd , then

Tn(a, b) > 0, if a + b > 0

and

Tn(a, b) < 0, if a + b < 0.

25


Proof. We have

(8) (a− b)Tn(a, b) = an+1 − bn+1.

If n is even, then n + 1 is odd and, the a− b on an+1− bn+1, a 6= b, have the

same sign and it results Tn(a, b) > 0.

For a = b 6= 0 we have Tn(a, b) = nan > 0.

If n is odd, then n + 1 is even, n + 1 = 2p, p ∈ N∗. Now, we obtain

(9) (a− b)Tn(a, b) = a2p − b2p = (a2 − b2)S2p(a, b)

with S2p(a, b) > 0, for any (a, b) ∈ R2 − (0, 0), and a 6= b.

From (9), we find

(10) Tn(a, b) = (a + b)S2p(a, b),

where Tn(a, b) > 0 if a + b > 0 and Tn(a, b) < 0 if a + b < 0.

For a+b = 0 we have Tn(a, b) = 0, when n is odd. The proof is complete.

Corollary 1. Let the function Un defined on R by

Un(t) = tn + tn−1 + . . . + t + 1, n ∈ N∗

i) If n is even, then Un(t) > 0, for any real number t ∈ R.

ii) If n is odd, then

Un(t) > 0, for any real number t > −1

and

Un(t) < 0, for any real number t < −1.

26


Proof. Putting a = t and b = 1 in Lema 1 we obtain the Corollary 1.

Theorem 1. For any real number a, b ∈ R∗, we have the double inequality

(11)1

2n + 1≤ a2n − a2n−1b + a2n−2b2 − . . .− ab2n−1 + b2n

a2n + a2n−1b + a2n−2b2 + . . . + ab2n+1 + b2n≤ 2n + 1,

n ∈ N∗.

Proof. Setting t =a

b, the double inequality (11) takes the form

(12)1

2n + 1≤ t2n − t2n−1 + t2n−3 − . . . + t2 − t + 1

t2n + t2n−1 + . . . + t2 + t + 1≤ 2n + 1,

with t ∈ R.

We consider the function f : R→ R,

f(t) =U2n(−t)

U2n(t)=

t2n − t2n−1 + . . . + t2 − t + 1

t2n + t2n−1 + . . . + t2 + t + 1.

We obtain f ′(t) =g(t)

U22n(t)

, where

g(t) = [2nt2n−1 − (2n− 1)t2n−2 + . . . + 2t− 1]U2n(t)

− [2nt2n−1 + (2n− 1)t2n−2 + . . . + 2t + 1]U2n(−t)

= 2t4n−2+4t4n−4+. . .+(2n−2)t2n+2+2nt2n−2nt2n−2−(2n−2)t2n−4

− . . .− 8t6 − 6t4 − 4t2 − 2

= 2(t4n−2−1)+4t2(t4n−6−1)+. . .+(2n−2)22n−4(t6−1)+2nt2n−2(t2−1)

= (t2 − 1)V4n−4(t),

with V4n−4(t) > 0 for any t ∈ R.

Thus, the derivative f ′(t) vanishes in t = −1 and t = 1. For t ∈(−∞,−1] and t ∈ [1,∞) the function f increases and for (−1, 1) it de-

creases. This means that at t = −1 the function has a maximum, max f =

27


f(−1) = 2n+1 while at t = 1 the function has a minimum, minf = f(1) =1

2n + 1. Since f(−∞) = f(+∞) = 1, it results

1

2n + 1≤ f(t) ≤ 2n + 1

for any t ∈ R that is (12). The theorem is proved.

For n = 2 from (11) we obtain (6).

Corollary 2. If x1, x2, . . . , xn are real positive numbers, n ∈ N, n ≥ 2 and

k ∈ N∗, then

(13)

1

2k + 1

n∑i=1

xi ≤n∑

i=1

x2k+1i

x2ki + x2n−1

i xi+1 + . . . + xix2k−1i+1 + x2k

i+1

≤ (2k+1)n∑

i=1

xi,

with xn+1 = x1.

Proof. We have

n∑i=1

x2k+1i − x2k+1

i+1

x2ki + x2n−1

i xi+1 + . . . + xix2k−1i+1 + x2k

i+1

=n∑

i=1

(xi − xi+1) = 0,

wheren∑

i=1

x2k+1i

T2k(xi, xi+1)=

1

2

n∑i=1

x2k+1i + x2k+1

i+1

T2k(xi, xi+1).

Now, it resultsn∑

i=1

x2k+1i

T2k(xi, xi+1)=

1

2

n∑i=1

x2k+1i + x2k+1

i+1

T2k(xi, xi+1)=

1

2

n∑i=1

(xi + xi+1)T2k(xi,−xi+1)

T2k(xi, xi+1).

Using (12), we obtain

1

2(2k + 1)

n∑i=1

(xi + xi+1) ≤n∑

i=1

x2k+1i

T2k(xi, xi+1)≤ 2k + 1

2

n∑i=1

(xi + xi+1),

where we have the double inequality (13). The Corollary 2 is proved.

28


3. Let

P (t) = At2n +Bt2n−1 +At2n−2 +Bt2n−3 + . . .+At2 +Bt+A, A > 0, B > 0

be a polynomial of degree 2n, n ∈ N∗ such that P (t) > 0 for any t ∈ R.

We consider the function

f : R→ R, f(t) =P (−t)

P (t).

We have

Theorem 2. The following double inequality

(14)(n + 1)A− nB

(n + 1)A + nB≤ f(t) ≤ (n + 1)A + nB

(n + 1)A− nB

holds for any t ∈ R

Proof. For the derivative of the function f we obtain

f ′(t) =ab(t2 − 1)V4n−4(t)

P 2(t),

with V4n−4(t) > 0 for any real number t.

The derivative f ′(t) vanishes in t = −1 and t = 1. From here, we obtain

that

max f = f(−1) =(n + 1)A + nB

(n + 1)A− nB

and

min f = f(1) =(n + 1)A− nB

(n + 1)A + nB.

Thus, we obtain the double inequality (14).

Remark 1.For A = 1 and B = 1 we obtain (12).

29


4. Another interesting result is the following

Theorem 3.If a, b ∈ R∗ and n ∈ N∗, then

(15)an + bn

an−1 + an−2b + . . . + abn−2 + bn−1≥ a + b

n, for a + b > 0

and

(16)an + bn

an−1 + an−2b + . . . + abn−2 + bn−1≤ a + b

n, for a + b < 0.

Proof. If a + b > 0, then from Lema 1 we have

Tn(a, b) = an−1 + an−2b + . . . + bn−1 > 0

for any a, b,∈ R∗.Using this result, the inequality (15) is equivalent

(n−1)an+(n−1)bn−an−1b−an−2b2−. . .−abn−1−an−1b−an−2b2−. . .−abn−1 ≥ 0

or

an−1(a− b) + an−2(a2 − b2) + . . . + a(an−1 − bn−1)− b(an−1 − bn−1)

−b2(an−2 − bn−2)− . . .− bn−1(a− b) ≥ 0

or

(a− b)(an−1− bn−1)+ (a2− b2)(an−2− bn−2)+ . . .+(an−1− bn−1)(a− b) ≥ 0

where

(17) (a− b)2[Tn−2(a, b) + Tn−3(a, b)T1(a, b) + . . . + Tn−2(a, b)] ≥ 0.

30


Since Tk(a, b) > 0, n = 1, 2, . . . , (n − 2), it results the inequality (17)

holds.

If a + b < 0, then for n odd we have Tn−1(a, b) > 0 and the inequality

(16) is equivalent.

(18) (a− b)2[Tn−2(a, b) + Tn−3(a, b)T1(a, b) + . . . + Tn−2(a, b)] ≤ 0.

Since the numbers n − k and k − 2 have different parities, it results

Tn−k(a, b)Tk−2(a, b) < 0, k = 2, 3, . . . , n and hence (18) holds.

For n even we have Tn(a, b) < 0 and the inequality (16) is equivalent

(17). Since the numbers n − k and k − 2 have the same parity, it results

Tn−k(a, b)Tk−2(a, b) < 0, k = 2, 3, . . . , n and hence (17) holds.

In conclusion, the Theorem 3 is proved.

Remark 2.For n = 4 and n = 5, a + b > 0, from (15) we obtain the

Problem 2.4 from [4](p. 28).

References

[1] D. S. Mitrinovic, in corporation with P. M. Vasic, Analytic Inequalities,

Springer Varleg, Berlin - Heidelberg - New York, 1970.

[2] D. Acu, Generalization of some inequalities, Univ. Beograd, Publ. Elek-

trotehn. Fak., Ser. Math. Fiz., Nr. 678-715 (1980), 58-62.

[3] D. D. Adamovic, I. E. Pecaric, Some inequalities obtained by elementary

methods, Matematiciki Vesnik, No. 35 (1983), 219-230.

31


[4] L. Panaitopol, V. Bandila, M. Lascu, Inequalities, Ed. GIL, 1995 (in

Romanian)

Dumitru Acu




E-mail: acu [email protected]

32



AN INEQUALITY FOR GENERALIZED SPLINE

FUNCTIONS

Adrian Branga

Abstract

The main result of this paper is an inequality for the generalized

spline functions based on the properties of the spaces, operator and

interpolatory set used in the definition.

2000 Mathematics Subject Classification: 41A15, 41A50, 41A52, 65D07

Key words and phrases: spline functions, best approximation, inequalities in

abstract spaces

1. INTRODUCTION

Definition 1. Let E1 be a real linear space, (E2, ‖.‖2) a normed real linear

space, T : E1 → E2 an operator and U ⊆ E1 a non-empty set. The problem

of finding the elements s ∈ U which satisfy

(1) ‖T (s)‖2 = infu∈U

‖T (u)‖2,

is called the general spline interpolation problem, corresponding to the set

U .

33

A. Branga - An inequality for generalized spline functions

A solution of this problem, provided that exists, is named general spline

interpolation element, corresponding to the set U .

The set U is called interpolatory set.

In the sequel we assume that E1 is a real linear space, (E2, (. , .)2, ‖.‖2)

is a real Hilbert space, T : E1 → E2 is a linear operator and U ⊆ E1 is a

non-empty convex set.

Lemma 1. T (U) ⊆ E2 is a non-empty convex set.

The proof follows directly from the linearity of the operator T , taking

into account that U is a non-empty set.

Theorem 1. (Existence Theorem) If T (U) ⊆ E2 is a closed set, then the

general spline interpolation problem (1) (corresponding to U) has at least a

solution.

The proof is shown in the papers [2, 4].

Theorem 2. (Characterization Theorem) An element s ∈ U is solution of

the general spline interpolation problem (1) (corresponding to U) if and only

if T (s) is the unique element in T (U) of the best approximation for 0E2.

For a proof see the paper [2].

For every element s ∈ U we define the set

(2) U(s) := U − s.

Lemma 2. For every element s ∈ U the set U(s) is non-empty (0E1 ∈U(s)).

34


The result follows directly from the relation (2).

Theorem 3. (Uniqueness Theorem) If T (U) ⊆ E2 is closed set and exists

an element s ∈ U solution of the general spline interpolation problem (1)

(corresponding to U), such that U(s) is linear subspace of E1, then the

following statements are true

i) For any elements s1, s2 ∈ U solutions of the general spline interpola-

tion problem (1) (corresponding to U) we have

(3) s1 − s2 ∈ Ker(T ) ∩ U(s);

ii) The element s ∈ U is the unique solution of the general spline inter-

polation problem (1) (corresponding to U) if and only if

(4) Ker(T ) ∩ U(s) = 0E1.

A proof is presented in the papers [2, 3].

2. MAIN RESULT

Lemma 3. An element s ∈ U , such that U(s) is linear subspace of E1, is

solution of the general spline interpolation problem (1) (corresponding to U)

if and only if

(5) (T (s), T (u))2 = 0, (∀) u ∈ U(s).

A proof is shown in the papers [2, 4].

For every element s ∈ U we consider the set

(6) S(s) := v ∈ E1 | (T (v), T (u))2 = 0, (∀) u ∈ U(s).

35


Proposition 1. For every element s ∈ U the set S(s) has the following

properties

i) S(s) is non-empty set (0E1 ∈ S(s));

ii) S(s) is linear subspace of E1;

iii) Ker(T ) ⊆ S(s).

For a proof see the paper [2].

Lemma 4. An element s ∈ U , such that U(s) is linear subspace of E1, is

solution of the general spline interpolation problem (1) (corresponding to U)

if and only if

(7) s ∈ S(s).

The result is a consequence of Lemma 3.

Lemma 5. For every element s ∈ U the set T (S(s)) has the following

properties

i) T (S(s)) is non-empty set (0E2 ∈ T (S(s)));

ii) T (S(s)) is linear subspace of E2;

iii) T (S(s)) ⊆ (T (U(s)))⊥.

A proof is shown in the paper [2].

36


Theorem 4. If an element s ∈ U , such that U(s) is linear subspace of E1,

is solution of the general spline interpolation problem (1) (corresponding to

U), then the following inequality is true

(8) ‖T (u)− T (s)‖2 ≤ ‖T (u)− T (v)‖2, (∀) u ∈ U, (∀) v ∈ S(s),

with equality if and only if T (v) = T (s), i.e. v − s ∈ Ker(T ).

Proof. Let u ∈ U , v ∈ S(s) be arbitrary elements.

Using the properties of the inner product (. , .)2, we deduce

‖T (u)− T (v)‖22 = ‖(T (u)− T (s)) + (T (s)− T (v))‖2

2 =(9)

= ‖T (u)− T (s)‖22 + 2(T (u)− T (s), T (s)− T (v))2 + ‖T (s)− T (v)‖2

2.

As u ∈ U and s ∈ U it obtains u− s ∈ U(s), therefore

(10) T (u− s) ∈ T (U(s)).

Because s ∈ U , from Proposition 1 ii) it follows that S(s) is linear

subspace of E1. On the other hand, as s ∈ U , such that U(s) is linear

subspace of E1, is solution of the general spline interpolation problem (1)

(corresponding to U), using Lemma 4 we deduce s ∈ S(s). Also, we have

v ∈ S(s). Consequently, it follows that s− v ∈ S(s), hence

(11) T (s− v) ∈ T (S(s)).

Taking into account that s ∈ U and using Lemma 5 iii), the formula

(11) implies that

(12) T (s− v) ∈ (T (U(s)))⊥.

37


From relations (10), (12) and using the property of the orthogonality we

deduce

(13) (T (u− s), T (s− v))2 = 0.

As T is a linear operator, the relation (13) can be written as

(14) (T (u)− T (s), T (s)− T (v))2 = 0.

Substituting the formula (14) in the equality (9), it follows that

(15) ‖T (u)− T (v)‖22 = ‖T (u)− T (s)‖2

2 + ‖T (s)− T (v)‖22.

The relation (15) implies

(16) ‖T (u)− T (s)‖2 ≤ ‖T (u)− T (v)‖2,

with equality if and only if ‖T (s)−T (v)‖2 = 0, equivalent T (v) = T (s), i.e.

v − s ∈ Ker(T ).

References

[1] A. M. Acu, Moment preserving spline approximation on finite intervals

and Chakalov-Popoviciu quadratures, Acta Universitatis Apulensis, Nr.

13/2007, pp 37-56

[2] A. Branga, Contributii la Teoria Functiilor spline, Teza de Doctorat,

Universitatea Babes-Bolyai, Cluj-Napoca, 2002.

[3] Gh. Micula, Functii spline si aplicatii, Editura Tehnica, Bucuresti,

1978.

38


[4] Gh. Micula, S. Micula, Handbook of splines, Kluwer Acad. Publ.,

Dordrecht-Boston-London, 1999.

Adrian Branga




E-mail: adrian [email protected]

39



SOME STARLIKENESS CONDITIONS PROVED BY

INEQUALITIES

Daniel Breaz, Nicoleta Breaz

Abstract

Let U = z ∈ C, |z| < 1 be the unit disc of the complex plane.

Let An =f ∈ H (U) , f (z) = z + an+1z

n+1 + an+2zn+2 + ..., z ∈ U

be the class of analytic functions in U and

S∗ (α) =

f ∈ A, Rezf ′ (z)f (z)

> α, z ∈ U

the class of starlike functions of order α. We consider the integral

operator

(1) FΣ (z) =1 +

∑ki=1 βi

zPk

i=1 βi

z∫

0

(k∏

i=1

fi (t)

)tPk

i=1(βi−1)dt

where fi ∈ An and βi ≥ 0 for i = 1, ..., k is the real numbers and

study its starlikeness properties.

2000 Mathematics Subject Classification: 30C45

Key words and phrases: Integral operator, univalent function, starlike function.

40

D. Breaz, N. Breaz - Some starlikeness conditions proved by inequalities

1. INTRODUCTION

In this paper we derive a 1−a1−b

order starlikeness condition for the integral

operator FΣ.This condition is an extension of the results of Gh. Oros [4].

Observe that for k = 1 we obtain the Bernardi integral operator.

Lemma 1 [2]. Let q be a univalent function in U and let θ and φ be analitic

functions in the domain D ⊂ q (U) with φ (w) 6= 0, for w ∈ q (U) . Let

Q (z) = nzq′ (z) φ [q (z)]

h (z) = θ [q (z)] + Q (z)

and suppose that

i) Q is starlike;

ii)Rezh′ (z)

Q (z)= Re

[θ′ [q (z)]

φ [q (z)]+

zQ′ (z)

Q (z)

]> 0.

If p is analytic in U , with

p (0) = q (0) , p′ (0) = ... = p(n−1) (0) = 0, p (U) ⊂ D

and

θ [p (z)] + zp′ (z) φ [p (z)] ≺ θ [q (z)] + zq′ (z) φ [q (z)]

then p ≺ q, and q is the best dominant.

41


2. MAIN RESULTS

Theorem 1. Let a, b, βi, i = 1, ..., k the real number, with the next

property:∑k

i=1 βi ≥ 0, 0 ≤ a ≤ 1,−1 ≤ b ≤ 0.Let the function

h (z) =1 + az

1 + bz+

n (a− b) z

(1 + bz)(1 +

∑ki=1 βi +

(a + b

∑ki=1 βi

)z) .

If fi ∈ An for all i ∈ 1, ..., k and

z(∏k

i=1 fi (z))′

∏ki=1 fi (z)

≺ h (z)

then

RezF ′

Σ (z)

FΣ (z)>

1− a

1− b

or

FΣ ∈ S∗(

1− a

1− b

),

where FΣ is defined in (1).

Proof. From (1) we obtain that

(2)k∑

i=1

βi · FΣ (z) + zF ′Σ (z) =

(k∑

i=1

βi + 1

)(k∏

i=1

fi (z)

).

If we consider

p (z) =zF ′

Σ (z)

FΣ (z),

then (2) becomes

zp′ (z)

p (z) +∑k

i=1 βi

+ p (z) =z

(∏ki=1 fi (z)

)′∏k

i=1 fi (z),

42


But

z(∏k

i=1 fi (z))′

∏ki=1 fi (z)

≺ h (z)

implieszp′ (z)

p (z) +∑k

i=1 βi

+ p (z) ≺ h (z) .

We apply Lemma 1 to prove that

RezF ′

Σ (z)

FΣ (z)>

1− a

1− b.

We take

q (z) =1 + az

1 + bz, θ (w) = w, φ (w) =

1

w +∑k

i=1 βi

, θ [q (z)] =1 + az

1 + bz,

φ [q (z)] =1 + bz

1 +∑k

i=1 βi +(a + b

∑ki=1 βi

)z,

and

Q (z) = nzq′ (z) φ [q (z)] =n (a− b) z

(1 + bz)(1 +

∑ki=1 βi +

(a + b

∑ki=1 βi

)z) .

h (z) = θ [q (z)]+Q (z) =1 + az

1 + bz+

n (a− b) z

(1 + bz)(1 +

∑ki=1 βi +

(a + b

∑ki=1 βi

)z) .

Since Q is starlike and Re φ [q (z)] > 0, from Lemma 1 we deduce that

p ≺ q ⇔ zF ′Σ (z)

FΣ (z)≺ 1 + az

1 + bz⇒ Re

zF ′Σ (z)

FΣ (z)>

1− a

1− b.

This last relation is equivalent to

FΣ ∈ S∗(

1− a

1− b

).

43


Corollary 1. Let∑k

i=1 βi ≥ 0, 0 < α ≤ 1 and

h (z) =1 + αz

1− αz+

2nαz

(1− αz)(1 +

∑ki=1 βi −

(1−∑k

i=1 βi

)αz

) .

If fi ∈ An for all i ∈ 1, ..., kand

z(∏k

i=1 fi (z))′

∏ki=1 fi (z)

≺ h (z)

then

RezF ′

Σ (z)

FΣ (z)>

1− α

1− α,


Proof. In Theorem 1 take a = α, b = −α.

Corollary 2.Let∑k

i=1 βi ≥ 0, 0 < α ≤ 1 and

h (z) =1

1− αz+

nαz

(1− αz)(1 +

∑ki=1 βi − α

∑ki=1 βiz

) .

If fi ∈ An for all i ∈ 1, ..., kand

z(∏k

i=1 fi (z))′

∏ki=1 fi (z)

≺ h (z)

then

RezF ′

Σ (z)

FΣ (z)>

1

1 + α,


Proof. This corollary is obtained if we take a = 0, b = −α in Theorem 1.

44


Remark 1.If we put α = 1 and k = 1 in Corollary 2 and Corollary 1, we

obtain a result of [4].

3. SPECIAL CASES

Corollary 3. Let

h (z) =4 + 4z (1 + n) + z2

4− z2.

If f ∈ An and

zf ′ (z)

f (z)≺ h (z) ,

then

RezF ′ (z)

F (z)>

1

3,

where F (z) =∫ z

0f(t)

tdt is the Alexander operator.

Proof. In Theorem 1, let k = 1,∑k

i=1 βi = 0, a = 12, b = −1

2.

Corollary 4. Let γ ≥ 0, −1 ≤ b ≤ 0 and

h (z) =1 + z

1 + bz+

n (1− b) z

(1 + bz) (1 + γ + (1 + bγ) z).

If f ∈ An and

zf ′ (z)

f (z)≺ h (z) ,

then

RezF ′ (z)

F (z)> 0,

where

F (z) =1 + γ

zγ

z∫

0

f (t) tγ−1dt

45


is the Bernardi operator.

Proof. In Theorem 1 take a = 1, k = 1, β1 = γ and f1 = f .

References

[1] M. Acu, Operatorul integral Libera-Pascu si proprietatile acestuia cu

privire la functiile uniform stelate, convexe, aproape convexe si α-

uniform convexe, Ed. Univ. ”Lucian Blaga” din Sibiu, 2005.

[2] S.S.Miller and P.T.Mocanu,On some classes of order differential sub-

ordination. Michigan Math. J.(1985),185-195.

[3] P.T.Mocanu,On a class of first-order differential subordinations. Babes-

Bolyai Univ., Fac. of Math. Res. Sem.,Seminar on Mathematical Anal-

ysis, Preprint 7(1991), 37-46.

[4] G.Oros, On starlike images by an integral operator. Mathematica,

42(65),(2000),71-74.

Daniel Breaz, Nicoleta Breaz

” 1 Decembrie 1918 ” University


Alba Iulia, Romania


[email protected]

46



NOTE ON SUBCLASS OF β-STARLIKE AND β-CONVEX

FUNCTIONS WITH NEGATIVE COEFFICIENTS

ASSOCIATED WITH SOME HYPERBOLA

Irina Dorca

Abstract

In this paper we define a subclass of β-starlike and β-convex func-

tions with negative coefficients associated with some hyperbola and

we obtain some properties regarding to these classes.


Key words and phrases: β-starlike functions, β-convex functions, hyperbola,

Libera-Pascu integral operator, Briot-Bouquet differential subordination,

generalized Salagean operator

1. INTRODUCTION

Let H(U) be the set of functions which are regular in the unit disc U ,

A = f ∈ H(U) : f(0) = f ′(0)− 1 = 0

and S = f ∈ A : f is univalent in U.

47

I. Dorca - Note on subclass of β-starlike and β-convex functions ...

In [10] the subfamily T of S consisting of functions f of the form

(1) f(z) = z −∞∑

j=2

ajzj, aj ≥ 0, j = 2, 3, ..., z ∈ U

was introduced.

We recall here the definition of the well - known class of starlike functions

S∗ =

f ∈ A : Re

zf ′(z)

f(z)> 0 , z ∈ U

and the definition of the well - known class of convex functions

Scn(α) =

f ∈ A : Re

Dn+2f(z)

Dn+1f(z)> α, z ∈ U

.

Let consider the Libera-Pascu integral operator La : A → A defined as:

(2) f(z) = LaF (z) =1 + a

za

z∫

0

F (t) · ta−1dt , a ∈ C , Re a ≥ 0.

For a = 1 we obtain the Libera integral operator, for a = 0 we obtain

the Alexander integral operator and in the case a = 1, 2, 3, ... we obtain the

Bernardi integral operator.

Generalization of the Libera-Pascu integral operator was studied by

many mathematicians such P. T. Mocanu in [16], E. Draghici in [8] and

D. Breaz in [7].

Let Dn be the Salagean differential operator (see [9]) defined as:

Dn : A → A , n ∈ N and D0f(z) = f(z)

D1f(z) = Df(z) = zf ′(z) , Dnf(z) = D(Dn−1f(z)).

48


Definition 1. [6] Let β, λ ∈ R, β ≥ 0, λ ≥ 0 and f(z) = z +∞∑

j=2

ajzj. We

denote by Dβλ the linear operator defined by

Dβλ : A → A ,

Dβλf(z) = z +

∞∑j=2

[(1 + (j − 1)λ)β]ajzj .

The purpose of this paper is to define a subclass of β-starlike functions

and another one for β-convex functions with negative coefficients associated

with some hyperbola and to obtain some estimations for the coefficients of

the series expansion and some other properties regarding these classes.

2. PRELIMINARY RESULTS

Definition 2.[4] Let f ∈ T , f(z) = z−∞∑

j=2

ajzj, aj ≥ 0, j = 2, 3, ..., z ∈ U .

We say that f is in the class T ?Lβ(α) if:

ReDβ+1

λ f(z)

Dβλf(z)

> α, α ∈ [0, 1), λ ≥ 0, β ≥ 0, z ∈ U.

Theorem 1.[4] Let α ∈ [0, 1), λ ≥ 0 and β ≥ 0. The function f ∈ T of the

form (1) is in the class T ?Lβ(α) iff

(3)∞∑

j=2

[(1 + (j − 1)λ)β(1 + (j − 1)λ− α)]aj < 1− α.

Definition 3.[5] Let f ∈ T , f(z) = z−∞∑

j=2

ajzj, aj ≥ 0, j = 2, 3, ..., z ∈ U .

We say that f is in the class T cLβ(α) if:

ReDβ+2

λ f(z)

Dβ+1λ f(z)

> α, α ∈ [0, 1), λ ≥ 0, β ≥ 0, z ∈ U.

49


Theorem 2.[5] Let α ∈ [0, 1), λ ≥ 0 and β ≥ 0. The function f ∈ T of the

form (1) is in the class T cLβ(α) iff

(4)∞∑

j=2

[(1 + (j − 1)λ)β+1(1 + (j − 1)λ− α)]aj < 1− α.

Definition 4. [12] A function f ∈ S is said to be in the class SH(α) if it

satisfies∣∣∣∣zf ′(z)

f(z)− 2α

(√2− 1

)∣∣∣∣ < Re

√2

zf ′(z)

f(z)

+ 2α

(√2− 1

),

for some α (α > 0) and for all z ∈ U .

Remark 1. Geometric interpretation for starlike functions associated with

some hyperbola: Let Ω(α) =

zf ′(z)

f(z): z ∈ U , f ∈ SH(α)

. Then

Ω(α) =w = u + i · v : v2 < 4αu + u2 , u > 0

.

Note that Ω(α) is the interior of a hyperbola in the right half-plane which

is symmetric about the real axis and has vertex at the origin.

Definition 5.[2] Let f ∈ S and α > 0 . We say that the function f is in

the class SHn(α) , n ∈ N , if∣∣∣∣Dn+1f(z)

Dnf(z)− 2α(

√2− 1)

∣∣∣∣ < Re

√2Dn+1f(z)

Dnf(z)

+ 2α(

√2− 1) , z ∈ U .

Remark 2.Geometric Interpretation: If we denote with pα the analytic

and univalent functions with the properties pα(0) = 1 , p′α(0) > 0 and

pα(U) = Ω(α) (see the pervios remark), then f ∈ SHn(α) iff Dn+1f(z)Dnf(z)

≺pα(z) , where the symbol ≺ denotes the subordination in U . We have pα(z) =

(1 + 2α)√

1+bz1−z

− 2α , b = b(α) = 1+4α−4α2

(1+2α)2and the branch of the square root

√w is chosen so that Im

√w ≥ 0 .

50


Theorem 3.[2] If F (z) ∈ SHn(α) , α > 0 , n ∈ N and f(z) = LaF (z) , where La

is the integral operator defined by (2), then f(z) ∈ SHn(α) , α > 0 , n ∈ N .

Theorem 4.[2] Let n ∈ N and α > 0 . If f(z) ∈ SHn+1(α) , then f(z) ∈SHn(α) .

Definition 6. [3] A function f ∈ A is said to be in the class CV H(α) if it

satisfies

∣∣∣∣zf ′′(z)

f ′(z)− 2α

(√2− 1

)+ 1

∣∣∣∣ < Re

√2

zf ′′(z)

f ′(z)

+ 2α

(√2− 1

),

for some α (α > 0) and for all z ∈ U .

Remark 3. Geometric interpretation for convex functions associated with

some hyperbola: Let Ω(α) = w = u + i · v : v2 < 4αu + u2 , u > 0 . Note

that Ω(α) is the interior of a hyperbola in the right half-plane which is sym-

metric about the real axis and has vertex at the origin. With this notations

we have f(z) ∈ CV H(α) if and only ifzf ′′(z)

f ′(z)+ 1 take all values in the

convex domain Ω(α) contained in the right half-plane.

Definition 7.[3] Let f ∈ A and α > 0 . We say that the function f is in

the class CV Hn(α) , n ∈ N , if

∣∣∣∣Dn+2f(z)

Dn+1f(z)− 2α(

√2− 1)

∣∣∣∣ < Re

√2Dn+2f(z)

Dn+1f(z)

+ 2α(

√2− 1) , z ∈ U .

Remark 4.Geometric Interpretation: If we denote with pα the analytic

and univalent functions with the properties pα(0) = 1 , p′α(0) > 0 and

pα(U) = Ω(α) (see the Remark 2.1), then f ∈ CV Hn(α) iff Dn+2f(z)Dn+1f(z)

≺pα(z) , where the symbol ≺ denotes the subordination in U . We have pα(z) =

51


(1 + 2α)√

1+bz1−z

− 2α , b = b(α) = 1+4α−4α2

(1+2α)2and the branch of the square root

√w is chosen so that Im

√w ≥ 0 . If we consider pα(z) = 1 + C1z + · · · we

have C1 = 1+4α1+2α

.

Theorem 5.[3] If F (z) ∈ CV Hn(α) , α > 0 , n ∈ N and f(z) = LaF (z) , where La

is the integral operator defined by (2), then f(z) ∈ CV Hn(α) , α > 0 , n ∈N .

Theorem 6.[3] Let n ∈ N and α > 0 . If f(z) ∈ CV Hn+1(α) , then f(z) ∈CV Hn(α) .

Theorem 7. [12] Let f ∈ SH(α) and f(z) = z + b2z2 + b3z

3 + ... . Then

(5) |b2| ≤ 1 + 4α

1 + 2α, |b3| ≤ (1 + 4α)(3 + 16α + 24α2)

4(1 + 2α)3.

The next theorem is the result of the so called ”admissible functions

method” due to P.T. Mocanu and S.S. Miller (see [13] , [14] , [15]).

Theorem 8. [2] Let h convex in U and Re[βh(z) + γ] > 0, z ∈ U. If

p ∈ H(U) with p(0) = h(0) and p satisfied the Briot-Bouquet differential

subordination

(6) p(z) +zp′(z)

βp(z) + γ≺ h(z), then p(z) ≺ h(z).

3.MAIN RESULTS

Definition 8. Let f ∈ T ?Lβ(α), f(z) = z −∞∑

j=2

aizj, aj ≤ 0, j ≥ 2,

α ∈ [0, 1), λ ≥ 0 and α1 > 0. We say that the function f is in the class

52


T ?HLβ(α; α1),

β ≥ 0, if∣∣∣∣∣Dβ+1

λ f(z)

Dβλf(z)

− 2α1 · (√

2− 1)

∣∣∣∣∣ < Re

√2 · Dβ+1

λ f(z)

Dβλf(z)

+2α1·(

√2−1), z ∈ U.

Remark 5. Geometric interpretation : If we denote with pα1 the analytic

and univalent functions with the properties pα1(0) = 1, p′α1(0) > 0 and

pα1(U) = Ω(α1) (see Remark 1), then f ∈ T ?HLβ(α; α1) if and only if

Dβ+1λ f(z)

Dβλf(z)

≺ pα1(z), where the symbol ” ≺ ” denotes the subordination in

U . We have pα1(z) = (1 + 2α1)

√1 + bz

1− z− 2α1, b = b(α1) =

1 + 4α1 − 4α21

(1 + 2α1)2

and the branch of the square root sqrtw is chosen so that Im√

w ≥ 0. If we

consider pα1(z) = 1− C1z − ..., we have C1 =1 + 4α1

1 + 2α1

.

Theorem 9. Let f ∈ T ?HLβ(α; α1), β ≥ 0, α ∈ [0, 1), α1 > 0 and f(z) =

z −∞∑

j=2

ajzj, aj ≤ 0, j ≥ 2, then

|a2| ≤ 1

(1 + λ)β(1 + λ− α)· 1 + 4α1

1 + 2α1

,

|a3| ≤ 1

(1 + 2λ)β(1 + 2λ− α)· (1 + 4α1)(3 + 16α1 + 24α2

1)

4(1 + 2α1)3.

Proof. If we denote by Dβλf(z) = g(z), g(z) = z −

∞∑j=2

bjzj, we have:

f ∈ T ?HLβ(α; α1) iff g ∈ T ?HLβ(α; α1). From the above series expansions

we obtain:

|aj| ≤ 1

[(1 + (j − 1)λ)β(1 + (j − 1)λ− α)]|bj|, j ≥ 2.

Using the estimations given in (5), from Theorem 7, we obtain the needed

results.

53


Theorem 10. If F (z) ∈ T ?HLβ(α; α1), β ≥ 0, α ∈ [0, 1), α1 > 0 and

f(z) = LaF (z), where La is the integral operator defined by (2), then f(z) ∈T ?HLβ(α; α1).

Proof. By differentiating (2) we obtain (1 + a)F (z) = af(z) + zf ′(z). By

means of the application of the linear operator Dβλ we obtain

(1 + a)Dβ+1λ F (z) = aDβ+1

λ f(z) + Dβ+1λ (zf ′(z))

or

(1 + a)Dβ+1λ F (z) =

[a +

λ− 1

λ

]·Dβ+1

λ f(z) +1

λ·Dβ+2

λ f(z) .

Similarly, by means of the application of the linear operator Dβλ we

obtain

(1 + a)DβλF (z) =

[a +

λ− 1

λ

]·Dβ

λf(z) +1

λ·Dβ+1

λ f(z) .

Thus

Dβ+1λ F (z)

DβλF (z)

=

Dβ+2λ f(z) +

[a +

λ− 1

λ

]·Dβ+1

λ f(z)

Dβ+1λ f(z) +

[a +

λ− 1

λ

]·Dβ

λf(z)

(7) =

Dβ+2λ f(z)

Dβ+1λ f(z)

· Dβ+1λ f(z)

Dβλf(z)

+ [λ(a + 1)− 1] · Dβ+1λ f(z)

Dβλf(z)

Dβ+1λ f(z)

Dβλf(z)

+ [λ(a + 1)− 1]

.

With notationDβ+1

λ f(z)

Dβλf(z)

= p(z), where p(z) = 1− p1(z)− ... , we have

zp′(z) = z ·(

Dβ+1λ f(z)

Dβλf(z)

)′

54


=z(Dβ+1

λ f(z))′ ·Dβλf(z)−Dβ+1

λ f(z) · z(Dβλf(z))′

(Dβλf(z))2

=Dβ+2

λ f(z) ·Dβλf(z)− (Dβ+1

λ f(z))2

(Dβλf(z))2

and

1

p(z)· λzp′(z) =

Dβ+2λ f(z)

Dβ+1λ f(z)

− Dβ+1λ f(z)

Dβλf(z)

=Dβ+2

λ f(z)

Dβ+1λ f(z)

− p(z) .

From the above we have

Dβ+2λ f(z)

Dβ+1λ f(z)

= p(z) +1

p(z)· λzp′(z) .

Thus from (7) we can deduce the following

(8)Dβ+1

λ F (z)

DβλF (z)

=

p(z) ·(

λzp′(z) · 1

z+ p(z)

)+ [λ(a + 1)− 1] · p(z)

p(z) + [λ(a + 1)− 1]

= p(z) +1

p(z) + [λ(a + 1)− 1]· λzp′(z) .

From Remark 1 we haveDβ+1

λ F (z)

DβλF (z)

≺ pα1(z) and thus, using (8), we notice

that

p(z) +1

p(z) + [λ(a + 1)− 1]· λzp′(z) ≺ pα1(z) .

We can deduce from Remark 1 and from the hypothesis Re(pα1(z) + a) >

0, z ∈ U. In this condition, from Theorem 8, we obtain p(z) ≺ pα1(z) or

Dβ+1λ f(z)

Dβλf(z)

≺ pα1(z). This means that f(z) = LaF (z) ∈ T ?HLβ(α; α1).

Theorem 11. Let a ∈ C, Re a ≥ 0, α ∈ [0, 1), α1 > 0 and β ≥ 0. If F (z) ∈T ?HLβ(α; α1), F (z) = z −

∞∑j=2

ajzj, aj ≤ 0, j ≥ 2, f(z) = LaF (z), f(z) =

55


z −∞∑

j=2

bjzj, where La is the integral operator defined by (2), λ ≥ 0, β ≥ 0,

then

|b2| ≤∣∣∣∣a + 1

a + 2

∣∣∣∣ ·1

(1 + λ)β(1 + λ− α)· 1 + 4α1

1 + 2α1

,

|b3| ≤∣∣∣∣a + 1

a + 3

∣∣∣∣ ·1

(1 + 2λ)β(1 + 2λ− α)· (1 + 4α1)(3 + 16α1 + 24α2

1)

4(1 + 2α1)3.

Proof. From f(z) = LaF (z) we have (1 + a)F (z) = af(z) + zf ′(z). Using

the above series expansions we obtain

(1 + a)z −∞∑

j=2

(1 + a)ajzj = az −

∞∑j=2

(a + j)bjzj + z

and thus bj(a + j) = (1 + a)aj, j ≥ 2 . From the above we have |bj| ≤∣∣∣∣a + 1

a + j

∣∣∣∣ · |aj|, j ≥ 2 . Using the estimations from Theorem 9 we obtain the

needed results.

For a = (j − 1)λ − α, for any j ≥ 2, when the integral operator La

become the Libera integral operator, we obtain from the above theorem:

Corollary 1. Let α ∈ [0, 1), α1 > 0, β ≥ 0. If F (z) ∈ T ?HLβ(α; α1), F (z) =

z −∞∑

j=2

ajzj, aj ≤ 0, j ≥ 2, and f(z) = LF (z), f(z) = z −

∞∑j=2

bjzj, where L

is the Libera integral operator defined by

L(F (z)) =1 + (j − 1)λ− α

z(j−1)λ−α

z∫

0

F (t) · t(j−1)λ−α−1dt,

then

|b2| ≤ 1

(1 + λ)β· 1 + 4α1

(1 + 2α1)(2 + λ− α), (j = 2) ,

|b3| ≤ 1

(1 + 2λ)β· (1 + 4α1)(3 + 16α1 + 24α2

1)

4(1 + 2α1)3(3 + 2λ− α), (j = 3) .

56


Theorem 12. Let β ≥ 0, α ∈ [0, 1) and α1 > 0. If f(z) ∈ T ?HLβ+1(α; α1)

then f(z) ∈ T ?HLβ(α; α1).

Proof. With notationDβ+1

λ f(z)

Dβλf(z)

= p(z) we have (see the proof of Theorem

10):

Dβ+2λ f(z)

Dβ+1λ f(z)

= p(z) +1

p(z)· λzp′(z) .

From f(z) ∈ T ?HLβ+1(α; α1) we obtain (see Remark 5) p(z) +1

p(z)·

λzp′(z) ≺ pα1(z). Using the definition of the function pα1(z) we have Repα1(z) >

0 and from Theorem 8 we obtain p(z) ≺ pα1(z) or f(z) ∈ T ?HLβ(α; α1) .

Remark 6.From the above theorem we obtain T ?HLβ(α; α1) ⊂ T ?HL0(α; α1) =

T ?HL(α; α1) ⊂ T ∗L for β ≥ 0, α ∈ [0, 1), α1 > 0.

Remark 7.In a similarly way we can prove the perviously results for β-

convex functions with negative coefficients named T cHLβ(α; α1), β ≥ 0, α ∈[0, 1), α1 > 0 , making use of Theorem 2 instead of Theorem 1 .

Definition 9. Let f ∈ T cLβ(α), f(z) = z −∞∑

j=2

aizj, aj ≤ 0, j ≥ 2,

α ∈ [0, 1), λ ≥ 0 and α1 > 0. We say that the function f is in the class

T cHLβ(α; α1), β ≥ 0, if∣∣∣∣∣Dβ+2

λ f(z)

Dβ+1λ f(z)

− 2α1 · (√

2− 1)

∣∣∣∣∣ < Re

√2 · Dβ+2

λ f(z)

Dβ+1λ f(z)

+2α1·(

√2−1), z ∈ U.

Theorem 13. Let f ∈ T cHLβ(α; α1), β ≥ 0, α ∈ [0, 1), α1 > 0 and f(z) =

z −∞∑

j=2

ajzj, aj ≤ 0, j ≥ 2, then

|a2| ≤ 1

(1 + λ)β+1(1 + λ− α)· 1 + 4α1

1 + 2α1

,

57


|a3| ≤ 1

(1 + 2λ)β+1(1 + 2λ− α)· (1 + 4α1)(3 + 16α1 + 24α2

1)

4(1 + 2α1)3.

Theorem 14. If F (z) ∈ T cHLβ(α; α1), β ≥ 0, α ∈ [0, 1), α1 > 0 and

f(z) = LaF (z), where La is the integral operator defined by (2), then f(z) ∈T cHLβ(α; α1).

Theorem 15. Let a ∈ C, Re a ≥ 0, α ∈ [0, 1), α1 > 0 and β ≥ 0. If F (z) ∈T cHLβ(α; α1), F (z) = z −

∞∑j=2

ajzj, aj ≤ 0, j ≥ 2, f(z) = LaF (z), f(z) =

z −∞∑

j=2

bjzj, where La is the integral operator defined by (2), λ ≥ 0, β ≥ 0,

then

|b2| ≤∣∣∣∣a + 1

a + 2

∣∣∣∣ ·1

(1 + λ)β+1(1 + λ− α)· 1 + 4α1

1 + 2α1

,

|b3| ≤∣∣∣∣a + 1

a + 3

∣∣∣∣ ·1

(1 + 2λ)β+1(1 + 2λ− α)· (1 + 4α1)(3 + 16α1 + 24α2

1)

4(1 + 2α1)3.

Corollary 2. Let α ∈ [0, 1), α1 > 0, β ≥ 0. If F (z) ∈ T cHLβ(α; α1), F (z) =

z −∞∑

j=2

ajzj, aj ≤ 0, j ≥ 2, and f(z) = LF (z), f(z) = z −

∞∑j=2

bjzj, where L

is the Libera integral operator defined by

L(F (z)) =1 + (j − 1)λ− α

z(j−1)λ−α

z∫

0

F (t) · t(j−1)λ−α−1dt,

then

|b2| ≤ 1

(1 + λ)β+1· 1 + 4α1

(1 + 2α1)(2 + λ− α), (j = 2) ,

|b3| ≤ 1

(1 + 2λ)β+1· (1 + 4α1)(3 + 16α1 + 24α2

1)

4(1 + 2α1)3(3 + 2λ− α), (j = 3) .

Theorem 16. Let β ≥ 0, α ∈ [0, 1) and α1 > 0. If f(z) ∈ T cHLβ+1(α; α1)

then f(z) ∈ T cHLβ(α; α1).

58


Remark 8.From the above theorem we obtain T cHLβ(α; α1) ⊂ T cHL0(α; α1) =

T cHL(α; α1) ⊂ T cL for β ≥ 0, α ∈ [0, 1), α1 > 0.

References

[1] M. Acu, On a subclass of functions with negative coefficients, General

Mathematics, Vol. 10, No. 3-4 (2002), 57-66.

[2] M. Acu, On a subclass of n-starlike functions associated with some

hyperbola, General Mathematics Vol. 13, No. 1 (2005), 9198.

[3] M. Acu, On a subclass of n-convex functions associated with some hy-

perbola, General Mathematics, Vol. 13, No. 3(2005), pp 923-30, 2005.

[4] M. Acu, I. Dorca, S. Owa, On some starlike functions with negative

coefficients (to appear).

[5] M. Acu, I. Dorca, D. Breaz, About some convex functions with negative

coefficients, Acta Universitatis Apulensis, No. 14/2007, pag. 97-108,

Alba Iulia.

[6] M. Acu, S. Owa, Note on a class of starlike functions, Proceeding Of

the International Short Joint Work on Study on Calculus Operators in

Univalent Function Theory - Kyoto 2006, 1-10.

[7] D. Breaz, Operatori integrali pe spatii de functii univalente, Editura

Academiei Romane, Bucuresti 2004.

59


[8] E. Draghici, Elemente de teoria functiilor cu aplicatii la operatori in-

tegrali univalenti, Editura Constant, Sibiu 1996.

[9] G. S. Salagean, On some classes of univalent functions, Seminar of

geometric function theory, Cluj - Napoca, 1983.

[10] G. S. Salagean, Geometria Planului Complex, Ed. Promedia Plus, Cluj

- Napoca, 1999.

[11] H. Silverman, Univalent functions with negative coefficients, Proc.

Amer. Math. Soc. 5(1975), 109-116.

[12] J. Stankiewicz and A. Wisniowska, Starlike functions associated with

some hyperbola, Folia Scientiarum Universitatis Tehnicae Resoviensis

147, Matematyka 19(1996), 117-126.

[13] S. S. Miller and P. T. Mocanu, Differential subordonations and univa-

lent functions, Mich. Math. 28 (1981), 157 - 171.

[14] S. S. Miller and P. T. Mocanu, On some classes of first-order differential

subordinations, Mich. Math. 32(1985), 185 - 195.

[15] S. S. Miller and P. T. Mocanu, Univalent solution of Briot-Bouquet

differential equations, J. Differential Equations 56 (1985), 297 - 308.

[16] P. T. Mocanu, Classes of univalent integral operators, J. Math. Anal.

Appl. 157, 1(1991), 147-165.

60


Irina Dorca




E-mail: ira [email protected]

61



ON SOME INEQUALITY FOR CONVEX FUNCTIONS

Bogdan Gavrea

Abstract

In this paper we present a generalization of a result derived in

[3].

2000 Mathematics Subject Classification: 26D15, 26D20

Key words and phrases: Jensen’s inequality, convex functions, linear positive

functionals

1. INTRODUCTION

Let I = [a, b] (a < b) be a fixed interval of the real axis. If f is a convex

function on I, then the well-known Jensen’s inequality holds:

n∑i=1

pif(xi)− f

(n∑

i=1

pixi

)≥ 0

for any pi ∈ [0, 1], i = 1, n satisfying∑n

i=1 pi = 1 and any xi ∈ I, i = 1, n.

In [1], S.S. Dragomir proves the following result:

62

B. Gavrea - On some inequality for convex functions

Theorem 1.[S.S. Dragomir, [1]] If f is a convex and differentiable function

on I, then

n∑i=1

pif(xi)− f

(n∑

i=1

pixi

)≤ 1

4(b− a) (f ′(b)− f ′(a))

for any xi ∈ [a, b], pi ∈ [0, 1], i = 1, n such that∑n

i=1 pi = 1.

In [2], S. Simic improves the result from Theorem 1 by removing the differ-

entiability restriction on f . More precisely the following result was proved:

Theorem 2.[S. Simic, [2]] Let pi ∈ [0, 1], i = 1, n,∑n

i=1 pi = 1. Then if f

is convex on I we have that:

(1)n∑

i=1

pif(xi)− f

(n∑

i=1

pixi

)≤ f(a) + f(b)− 2f

(a + b

2

):= Sf (a, b).

In [3] the results of [1] and [2] are generalized for the case of normalized linear

positive functionals. Here, we derive a result that generalizes Theorem 2.3

from [3].

2. MAIN RESULTS

Let A be a linear positive functional satisfying A(e0) = 1. The main

result of our paper is given by the following theorem.

Theorem 3. Let f : [a, b] → R be a convex function on the interval [a, b]

and for n ∈ N, n ≥ 2 let ∆n be the partition of [a, b] given by:

∆n : a = x0 < x1 < ... < xn−1 = b.

Then

A(f)− f(a1) ≤n−1∑i=0

f(xi)− nf

(∑n−1i=0 xi

n

).

63


Proof. Let S∆n(f) denote the linear spline that interpolates the function

f in the nodes of partition ∆n. It is well known that we can write S∆n(f)

in terms of fundamental splines as follows:

S∆n(f) =n−1∑i=0

f(xi)li(x),

where

l0(x) =

x1 − x

x1 − x0

x ∈ [x0, x1]

0 x > x1

ln−1(x) =

x− xn−1

xn − xn−1

x ∈ [xn−2, xn−1]

0 x < xn−2

and lk(x) =

x− xk−1

xk − xk−1

x ∈ [xk−1, xk]

xk+1 − x

xk+1 − xk

x ∈ [xk, xk+1]

0 x < xk−1 or x > xk+1

for 0 < k < n− 1. The following identities are well known

n−1∑i=0

li(x) = 1 ∀x ∈ [a, b](2)

n−1∑i=0

xli(x) = x ∀x ∈ [a, b].(3)

Since f is convex on [a, b], we have

A(f) ≤ A(S∆n(f))

or

A(f)− f(a1) ≤n−1∑i=0

A(ei)f(xi)− f(a1).

64


To establish the result of this theorem it is sufficient to prove the inequality

n−1∑i=0

A(li)f(xi)− f(a1) ≤n−1∑i=0

f(xi)− nf

(∑n−1i=0 xi

n

),

which is equivalent to

(4)n−1∑i=0

f(xi)(1− A(li)) + f(a1) ≥ nf

(∑n−1i=0 xi

n

).

Since

1− A(li) ≥ 0, i = 0, n− 1,

from Jensen’s inequality we obtain

n−1∑i=0

f(xi)(1−A(li))+f(a1) ≥[1 +

n−1∑i=0

(1− A(li))

]f

(a1 +

∑n−1i=0 xi − xiA(li)

1 +∑n−1

i=0 (1− A(li))

)

But by using equations (2) and (3) we obtain

1 +n−1∑i=0

(1− A(li)) = n

andn−1∑i=0

xiA(li) = a1

which gives inequality (4), i.e.,

n−1∑i=0

f(xi)(1− A(li)) + f(a1) ≥ nf

(∑n−1i=0 xi

n

).

This concludes our proof.

Remark. For the case n = 2 one obtains Theorem 2.3. from [3].

65


References

[1] S.S. Dragomir, A converse result for Jensen’s discrete inequality via

Gruss inequality and applications in Information Theory, An. Univ.

Oradea Fasc. Mat., 7(4) (1999-2000), pp. 178-189.

[2] S. Simic, On a global upper bound for Jensen’s inequality, J. Math. Anal.

Appl., 343(1), 2008, pp. 414-419.

[3] B. Gavrea, J. Jaksetic, J. Pecaric , On a global upper bound for Jensen’s

inequality, submitted to Nonlinear Analysis Series A: Theory, Methods

and Applications.

Bogdan Gavrea

Technical University of Cluj-Napoca




66



ON SOME INEQUALITIES FOR CONVEX FUNCTIONS OF

HIGHER ORDER

Ioan Gavrea

Abstract

Our goal is to prove some inequalities for convex functions of

higher order. These inequalities generalize some results obtained by

A. Lupas in [2].

2000 Mathematics Subject Classification: 26D15, 26D20

Key words and phrases: convex functions of higher order, positive linear

functionals, quadrature formulas

1. INTRODUCTION

Let s be a natural number and denote by Ks[a, b] the cone of all functions

f : [a, b] → R which are convex (non-concave) of order s on [a, b], i.e., for

any system x0, x1, ..., xs+1 of distinct points from [a, b], we have

[x0, x1, ..., xs+1; f ] ≥ 0

67

I. Gavrea - On some inequalities for convex functions of higher order

where [x0, x1, ..., xs+1; f ] denotes the divided difference of f at the points

x0, x1, ..., xs+1. It is well known the Hermite-Hadamard inequality which

asserts that

(1) f

(a + b

2

)≤ 1

b− a

∫ b

a

f(x)dx ≤ f(a) + f(b)

2, f ∈ K1[a, b].

S. S. Dragomir and C. Pearce, in their book [1], have studied many func-

tional inequalities of the Hermite-Hadamard type.

A. Lupas ([2]) established Hermite-Hadamard type inequalities for con-

vex functions of second order. The following notations are used:

- ej(t) = tj, j ∈ N and Πn denotes the linear spac of polynomials of

degree ≤ n;

- for x0 ∈ [a, b], δx0 is the evaluation functional at x0, i.e., δx0(f) =

f(x0);

- by A we denote the set of all linear positive functionals A : C[a, b] →R normalized by A(e0) = 1 and such that A is different from the

evaluation functional;

- if A ∈ A, then a1 = A(e1) and fj(t) = (t− a1)j.

In [2] the following theorem was proved:

Theorem 1.[A. Lupas] For A ∈ A and f ∈ K2[a, b] the following inequali-

ties are satisfied

(2) δ2(A; f) ≤ A(f) ≤ ∆2(A; f),

68


where

δ2(A; f) =A(f2)

(a1 − a)2 + A(f2)f(a) +

(a1 − a)2

(a1 − a)2 + A(f2)f(a1 +

A(f2)

a1 − a)

∆2(A; f) =(b− a1)

2

(b− a1)2 + A(f2)f(a1 − A(f2)

b1 − a) +

A(f2)

(b− a1)2 + A(f2)f(b).

In (2) the equality cases hold for any polynomial of degree ≤ 2.

The aim of this paper is to generalize Lupas’s result.

2. MAIN RESULTS

Let A ∈ A. We consider the following quadrature formula

(3) A(f) = αf(a1) +n∑

k=1

Akf(xk) + R(f).

Lemma 1. The quadrature formula (3) has degree of exactness 2n if and

only if xk are the roots of the polynomial P ∈ Πn which satisfies the following

conditions

(4) A[ei(e1 − a)P ] = 0, i = 0, 1, ..., n− 1.

The coefficients Ak, k = 1, 2, ..., n and α are given by

Ak =A[(e1 − a)l2k]

xk − a, k = 1, 2, ..., n(5)

α =A(l2)

l2(a),(6)

where

l(x) =n∏

k=1

(x− xk), lk(x) =l(x)

(x− xk)l′(xk).

69


Proof. Let us suppose that the quadrature formula has degree of exactness

2n. This means that

R(P ) = 0, ∀P ∈ Π2n.

We have (e1 − a)l2k, l2 ∈ Π2n and so

(7) R((e1 − a)l2k) = 0 and R(l2) = 0.

From (3) and (7) we get

(8) A[(e1 − a)l2k] = Ak(xk − a)

and

(9) A(l2) = αl2(a)

From (8) and (9) we get (5) and (6). On the other hand we have

R[ei(e1 − a)l] = 0, i = 0, 1, ..., n− 1.

This means that l is an orthogonal polynomial of degree n relative to the

linear positive functional B defined by

B(f) = A[(e1 − a)f ].

It is well known that two orthogonal polynomials of the same degree have

the same roots.

Lemma 2. For every f ∈ Kn[a, b] we have

(10) A[(x− a)lf ] > 0,

where l is the polynomial from Lemma 1.

70


Proof. Let us denote by Ln(f ; a, x1, ..., xn) the Lagrange polynomial of

degree n which interpolates the function f in the points a, x1, ..., xn, we

have

(11) f(x)− Ln(f ; a, x1, ..., xn)(x) = (x− a)l(x)[x, a, x1, ..., xn; f ].

We multiply both sides of (11) by (x−a)l(x) and then apply the functional

A to get

A[(e1 − a)lf ] = A[(e1 − a)2l2[·, a, x1, ..., xn; f ]

].

Using the fact that f ∈ Kn[a, b] implies [x, a, x1, ..., xn; f ] ≥ 0 and that the

functional A is a positive functional we obtain the desired result.

Lemma 3. Let f ∈ K2n[a, b]. Then the remainder R(f) from (3) is positive.

Proof. From (11) we get

R(f) = R [(e1 − a)l[·, a, x1, ..., xn; f ]] .

The function [·, a, x1, ..., xn; f ] ∈ Kn[a, b] if f ∈ K2n[a, b]. By Lemma 2 we

obtain

R(f) ≥ 0.

We obtain the following result

Theorem 2. Let f ∈ K2n[a, b]. Then

(12) A(f) ≥ αf(a) +n∑

k=1

Akf(xk),

where α,Ak, k = 1, 2, ..., n are given by (5) and (6).

71


Let A ∈ A. We consider the following quadrature formula

(13) A(f) = βf(b) +n∑

k=1

Bkf(yk) + R1(f)

Lemma 4. The quadrature formula (13) has the degree of exactness 2n if

and only if yk are the roots of Q ∈ Πn, where Q satisfies the following

conditions:

A[(b− e1)eiQ] = 0, i = 0, 1, ..., n− 1.

The coefficients Bk, k = 1, n and β are given by

Bk =A[(b− e1)l

2k]

b− xk

, k = 1, 2, ..., n(14)

α =A(l2)

l2(b).(15)

Lemma 5. Let f ∈ K2n[a, b]. Then

R1(f) ≤ 0.

The proofs of Lemma 4 and Lemma 5 are omitted.

Theorem 3. Let f ∈ K2n[a, b]. Then

(16) A(f) ≤ βf(b) +n∑

k=1

βkf(yk),

where βk, β, k = 1, n are given by (14) and (15).

Proof. The result of the Theorem follows from Lemma 5.

Corollary 1. If f ∈ K2n[a, b] and A ∈ A then

αf(a) +n∑

k=1

Akf(xk) ≤ A(f) ≤ βf(b) +n∑

k=1

Bkf(yk).

72


Let A be a linear positive functional, A ∈ A and let us consider the following

quadrature formulas:

(17) A(f) =m∑

k=0

αkf(k)(a) + λf(z) + R2(f), f ∈ C(m)[a, b].

Lemma 6. The quadrature formula 17 has degree of exactness maximum

m + 2 if and only if

z = a +A(e1 − a)m+2

A(e1 − a)m+2(18)

λ =A(e1 − a)m+1

(z − a)m+1(19)

αk =A(e1 − a)k − λ(z − a)k

k!, k = 0,m.(20)

Proof. Let us denote by Hm+1(f ; a(m+1), z) Hermite’s polynomial of degree

m + 1 which satisfies the following conditions:

H(k)(m+1)(f ; a(m+1), z)(a) = f (k)(a), k = 0, 1, ..., m

H(m+1)(f ; a(m+1), z)(z) = f(z).

It is well known that

(21) f(x)−H(m+1)(f ; a(m+1), z)(x) = (x− a)m+1(x− z)[x, a(m+1), z; f ],

where [x, a(m+1), z; f ] is the divided difference with the node a taken m + 1

times. By (21), it follows that if the quadrature formula (17) has degree of

exactness m + 2 then we have

A[(e1 − a)m+1(e1 − z)] = 0.

73


The last equality is equivalent with (18). The relation (19) follows from the

equality

R2[(e1 − a)m+1] = 0

and (20) from

R2[(e1 − a)k] = 0.

Remark 1.Using equality (21) we get

R2(f) ≥ 0,

if f ∈ Km+2[a, b].

Theorem 4. Let f ∈ Km+2[a, b]. Then

(22) A(f) ≥ λf(z) +m∑

k=0

αkf(k)(a),

where z, λ, αk, k = 0, 1, ...,m are given by Lemma 6.

Proof. The proof of the theorem follows from the Remark above.

References

[1] S.S. Dragomir, C.E.M. Pearce, Selected Topics on Hermite-Hadamard

Inequalities and Applications, RGMIA Monographs, Victoria University,

2000, http://rgmia.vu.edu/monographs.html.

[2] A. Lupas, Functional inequalities for convex functions of higher order,

Inequality Theory and Applications, Editors Y.J. Cho, J.K. Kim, S. S.

Dragomir, Nova Science Publishers, 2001, pp. 219–229.

74


Ioan Gavrea





75



A VORONOVSKAYA ESTIMATE WITH SECOND ORDER

MODULUS OF SMOOTHNESS

Heiner Gonska, Ioan Rasa

Abstract

We survey some recent quantitative Voronovskaya-type theorems

and prove a new estimate involving the second order modulus of

smoothness. For Bernstein operators and genuine Bernstein-Durrmeyer

operators it is shown that this inequality implies a higher order of ap-

proximation than previous inequalities in terms of the least concave

majorant of the first order modulus of continuity.

2000 Mathematics Subject Classification: 41A10, 41A15, 41A25, 41A36, 26A15.

Key words and phrases: the quantitative Voronovskaya-type theorems,

Bernstein operators, Bernstein-Durrmeyer operators, the second order modulus

of smoothness

1. INTRODUCTION AND PREVIOUS RESULTS

The present note deals with a new quantitative version of Voronovskaya’s

well-known theorem (see [13]) which we recall here as

76

H. Gonska, I. Rasa - A Voronovskaya estimate with second order modulus

Theorem 1 Let the classical Bernstein operators be given by

Bn(f ; x) :=n∑

k=0

f

(k

n

)pn,k(x) :=

n∑

k=0

f

(k

n

)(n

k

)xk(1− x)n−k,

where n ∈ N, f ∈ [0, 1]R , x ∈ [0, 1].

If f is bounded on [0,1], differentiable in some neighborhood of x and has

second derivative f ′′(x) for some x ∈ [0, 1], then

limn→∞

n · [Bn(f ; x)− f(x)] =x(1− x)

2f ′′(x).

If f ∈ C2[0, 1], the convergence is uniform.

Already in 1985 V.S. Videnskij published in [12] (see Theorem 15.2 on

p. 49) the following quantitative version for f ∈ C2[0, 1] where ω1 is the

first order modulus of continuity:

∣∣∣∣n · [Bn(f ; x)− f(x)]− x(1− x)

2f ′′(x)

∣∣∣∣ ≤ x(1− x) · ω1

(f ′′;

√2

n

).

In the first author’s recent note [2] a general quantitative Voronovskaya-

type theorem was given as follows.

Theorem 2 Let q ∈ N0, f ∈ Cq[0, 1] and L : C[0, 1] → C[0, 1] be a positive

linear operator. Then

∣∣∣∣∣L(f ; x)−q∑

r=0

L((e1 − x)r; x) · f (r)(x)

r!

∣∣∣∣∣

≤ L(|e1 − x|q; x)

q!· ω1

(f (q);

L(|e1 − x|q+1; x)

(q + 1)L(|e1 − x|q; x)

).

77


Here ω1 is the least concave majorant of ω1, satisfying

ω1(f ; ε) ≤ ω1(f ; ε) ≤ 2ω1(f ; ε), ε ≥ 0.

For more details concerning ω1 see, e.g., [5].

Corollary 1 (see [1], [2])

If q ∈ N is even, f ∈ Cq[0, 1], then uniformly in x ∈ [0, 1],

nq2 ·

Bn(f ; x)− f(x)−

q∑r=1

Bn((e1 − x)r; x) · f (r)(x)

r!

→ 0, n →∞.

The proof follows from

0 ≤ Bn((e1 − x)q; x) ≤ Aq · n−q2

andBn(|e1 − x|q+1; x)

Bn((e1 − x)q; x)→ 0.

Another consequence of Theorem 2 (see [6]) is

Corollary 2 For q = 2, f ∈ C2[0, 1], and L such that L(e0; x) = 1 and

L(e1 − x; x) = 0 we have

∣∣∣∣L(f ; x)− f(x)− 1

2· L((e1 − x)2; x) · f ′′(x)

∣∣∣∣

≤ 1

2· L((e1 − x)2; x) · ω1

(f ′′;

1

3· L(|e1 − x|3; x)

L((e1 − x)2; x)

).

78


Example 1 For L = Bn we obtain the following improvement of Viden-

skij’s inequality (see [6]):

∣∣∣∣n · [Bn(f ; x)− f(x)]− x(1− x)

2· f ′′(x)

∣∣∣∣ ≤x(1− x)

2· ω1

(f ′′;

1

3√

n

).

In [2] we showed that the upper bound can be replaced by the better

pointwise expression

x(1− x)

2· ω1

(f ′′;

√1

n2+

x(1− x)

n

).

Example 2 For variation-diminishing operators L = S4n giving piecewise

linear interpolators at equidistant knots in [0,1] the quantitative Voronovskaya

theorem reads as follows:

∣∣∣∣n2[S4n(f ; x)− f(x)]− 1

2f ′′(x) · nx(1− nx)

∣∣∣∣

≤ 1

2nx(1− nx) · ω1(f

′′; 13n

) ≤ 1

8· ω1

(f ′′;

1

3n

).

Here nx := nx− [nx] is the fractional part of nx.

This was shown in [7].

Example 3 In his 1972 dissertation A. Lupas (see [10]) investigated a

Beta-type operator Bn defined by

Bn(f ; x) :=

f(x), for x ∈ 0,1,1

B(nx, n− nx)·

1∫

0

tnx−1(1− t)n−1−nxf(t)dt, for x ∈ (0, 1).

79


with B(a, b) =

1∫

0

ta−1(1− t)b−1dt, a, b > 0.

The composition Un := Bn Bn is a ”genuine Bernstein-Durrmeyer

operator”, given explicitly by

Un(f ; x) = f(0)·pn,0(x)+f(1)·pn,n(x)+(n−1)n−1∑

k=1

pn,k(x)·1∫

0

pn−2,k−1(t)f(t)dt.

For recent results concerning Un see [3]. In this case we have (see [2])

|(n + 1)[Un(f ; x)− f(x)]− x(1− x)f ′′(x)|

≤ x(1− x) · ω1

(f ′′; 4 ·

√1

(n + 1)2+

x(1− x)

n + 1

).

Further recent contributions concerning quantitative Voronovskaya-type

theorems were made by Tachev, Rasa and Gonska (see [11], [9], [8]).

A remark on notation: if not otherwise indicated, ||·|| will always denote

the sup norm on C[a, b]. An exceptional case is the following. For

W2,∞[a, b] := f ∈ C[a, b] : f ′ is absolutely continuous with ||f ′′||L∞[a,b] < ∞,

||f ||L∞[a,b] := vrai sup |f(x)| : x ∈ [a, b],

will be a relevant norm in the following section.

2. A VORONOVSKAYA ESTIMATE INVOLVING ω2

Starting from our presentation at the 2006 conference on Numerical

Analysis and Approximation Theory in Cluj-Napoca (see [6]) the question

80


was asked if in the quantitative Voronovskaya theorem ω1(f′′; ...) could be

replaced by the second order modulus of smoothness ω2(f ; δ) given by

sup|f(x− h)− 2f(x) + f(x + h)|, a + h ≤ x ≤ b− h, 0 ≤ h ≤ δ,

for 0 ≤ δ ≤ 1

2.

The answer to this question is ”yes and no” as will be seen below. In

order to come up with the answer, first we make the following

Remark 1 (the basic decomposition) Let q ∈ N0, f ∈ Cq[a, b] be fixed and

g ∈ Cq+2[a, b] be arbitrary. Then for x ∈ [a, b] fixed and t ∈ [a, b], the

remainder in Taylor’s formula is given by

Rq(f ; x, t) = f(t)−q∑

k=0

1k!

f (k)(x) · (t− x)k,

and can be decomposed as follows:

Rq(f ; x, t) = Rq(f − g; x, t) + (Rq −Rq+1)(g; x, t) + Rq+1(g; x, t).

Lemma 1 If f and g are given as in Remark 1, then

|Rq(f ; x, t)| ≤ 2|x− t|q

q!||f (q)||,

(Rq −Rq+1)(g; x, t) =1

(q + 1)!(t− x)q+1 · g(q+1)(x),

|Rq+1(g; x, t)| ≤ |x− t|q+2

(q + 2)!||g(q+2)||.

We will also need the following auxiliary result (to be used below only for

q = 2).

81


Lemma 2 (see [4]) Let q ∈ N0. If f ∈ Cq[0,1] and 0 < h ≤ 12

are fixed,

then for any ε > 0 there are polynomials p = ph such that

||f (q) − p(q)|| ≤ 3

4· ω2(f

(q); h) + ε,

||p(q+1)|| ≤ 5

h· ω1(f

(q); h),

||p(q+2)|| ≤ 3

2h2· ω2(f

(q); h).

Proof. It is obtained by combining Lemma 2.1, Lemma 2.4 and Lemma

4.1 in [4]. Lemmas 2.1 and 2.4 show that for f ∈ C[0,1] and 0 < h ≤ 12

there are functions Sh(f, ·) ∈ W2,∞[0,1] such that

||f − Sh(f, ·)|| ≤ 3

4· ω2(f ; h),

||(Shf)′|| ≤ 5

h· ω1(f ; h),

||(Shf)′′||L∞[0,1] ≤ 3

2h2· ω2(f ; h).

The function Sh(f, ·) is not necessarily in C2[0, 1]. Hence it is shown in

Lemma 4.1 that for each g ∈ W2,∞[0, 1] and each ε > 0, there is a polynomial

p = p(g, ε) such that

||g − p|| < ε, ||p|| ≤ ||g||, ||p′|| ≤ ||g′||, and

||p′′|| ≤ ||g′′||L∞[0,1].

Keeping 0 < h ≤ 1

2fixed we choose g = Sh(f, ·) and obtain polynomials

p such that

||f − Shf + Shf − p||∞ ≤ ||f − Shf ||+ ||Shf − p||≤ 3

4· ω2(f ; h) + ε.

82


Moreover,

||p′|| ≤ ||(Shf)′|| ≤ 5

h· ω1(f ; h),

||p′′|| ≤ ||(Shf)′′||L∞[0,1] ≤ 3

2h2· ω2(f ; h).

If f ∈ Cq[0, 1], q ≥ 1, we apply the same argument to f (q) instead of f

and interprete the polynomial p from above as a q − th derivative of some

other polynomial which we also call p.

Theorem 3 Suppose L : C[0,1] → C[0,1] is a positive linear operator such

that Lei = ei, i = 0,1. If f ∈ C2[0, 1], then for any 0 < h ≤ 1

2the following

inequality holds:∣∣∣∣L(f ; x)− f(x)− 1

2L((e1 − x)2; x) · f ′′(x)

∣∣∣∣ ≤ L((e1 − x)2; x)· |L((e1 − x)3; x)|

L((e1 − x)2; x)· 5

6h· ω1(f

′′; h) +

(3

4+

L((e1 − x)4; x)

L((e1 − x)2; x)· 1

16h2

)· ω2(f

′′; h)

.

Proof. We use Lemma 2 in the case q = 2 and let f ∈ C2[0, 1] and

p = p(f, h, ε) be as in the lemma.

Next we use the basic decomposition of Remark 1 and apply L to both sides

to obtain

|L(R2(f ; x, ·); x)| = |L(f ; x)− f(x)− 1

2· L((e1 − x)2; x) · f ′′(x)|

≤ |L(R2(f − p; x, ·); x)|+ |L((R2 −R3)(p; x, ·); x)|+ |L(R3(p; x, ·); x)|≤ L(|R2(f − p : x, ·)|; x) +

1

3!|p(3)(x) · |L((e1 − x)3; x)|+ 1

4!||p(4)|| · L((e1 − x)4; x)

≤ ||(f−p)′′|| · L((e1−x)2; x)+1

3!||p(3)|| · |L((e1−x)3; x)|+ 1

4!||p(4)|| · L((e1−x)4; x)

≤ L((e1 − x)2; x) ·

3

4· ω2(f

′′; h) + ε

+ |L((e1 − x)3; x)| · 5

6h· ω1(f

′′; h)

+L((e1 − x)4; x) · 1

16h2· ω2(f

′′; h).

83


Letting ε tend to zero first and using L((e1−x)2; x) as a common factor

for all remaining terms implies the inequality of the theorem.

The theorem shows that we need the ”central moments” L((e1 − x)i; x)

for i ∈ 2, 3, 4 to proceed further. The inconvenient quantity L(|e1−x|3; x)

is not required (which was the case before).

Corollary 3 Putting h =

√L((e1 − x)4; x)

L((e1 − x)2; x)and assuming that h > 0, the

inequality in the theorem becomes:∣∣∣∣L(f ; x)− f(x)− 1

2L((e1 − x)2; x) · f ′′(x)

∣∣∣∣ ≤ L((e1 − x)2; x)·

|L((e1 − x)3; x)|√L((e1 − x)2 · L((e1 − x)4; x)

· 5

6· ω1

(f ′′;

√L((e1 − x)4; x)

L((e1 − x)2; x)

)

+13

16· ω2

(f ′′;

√L((e1 − x)4; x)

L((e1 − x)2; x)

).

3. APPLICATIONS

We first reconsider S∆n, i.e., piecewise linear interpolation at equidistant

knots. The following lemma describes the moments of S∆n.

Lemma 3 Let x be such thatk − 1

n≤ x ≤ k

n, 1 ≤ k ≤ n. Then, for m ≥ 1,

S∆n((e1−x)m; x) = n

(x− k − 1

n

)(k

n− x

) [(k

n− x

)m−1

−(

k − 1

n− x

)m−1].

Note that for x ∈

k − 1

n,k

n

one has S∆n((e1 − x)m; x) = 0.

84


For all x ∈ [0,1] we have

S∆n((e1−x)m; x) =1

nnx(1−nx)

[(1− nx

n

)m−1

−(−nx

n

)m−1]

.

We are thus lead to the following

Example 4 For the quantities figuring in Corollary 3 we have successively

S∆n((e1 − x)2; x) =1

n2nx(1− nx) ≤ 1

4n2,

S∆n((e1 − x)3; x)2

S∆n((e1 − x)2; x) · S∆n((e1 − x)4; x)︸︷︷︸≤1

=(1− 2nx)2

1− 3nx+ 3nx2=

1 for x =1

2and n even,

0 for x =1

2and n odd.

S∆n((e1 − x)4; x)

S∆n((e1 − x)2; x)=

1

n2[1−3nx+3nx2]

=1

n2for x =

1

2and n even,

<1

n2for nx ∈ (0,1).

So we get

∣∣∣∣n2[S∆n(f ; x)− f(x)]− 1

2f ′′(x) · nx(1− nx)

∣∣∣∣

≤nx(1−nx)︸︷︷︸≤1/4

·

5

6· ω1

(f ′′;

1

n

)+

13

16· ω2

(f ′′;

1

n

)=

O

(1

n

)for f ∈ C3[0,1],

O

(1

n

)for f ∈ C4[0,1].

85


In conclusion: The use of ω2(f′′; ..) does not help in comparison to that of

ω1(f′′; ...), because the assumption f ∈ C4[0,1] does not yield a better order

of approximation.

We will now reconsider Bn and show that for the Bernstein operators the

situation is different. Some important facts on their moments are collected

in

Lemma 4 For the Bernstein operators we have

Bn((e1 − x)2; x) =x(1− x)

n=:

X

n,

|Bn((e1−x)3; x)|2Bn((e1−x)2; x) ·Bn((e1−x)4; x)

=X2(X′)2

n4· n

X· n3

3(n−2)X2+X=

(X′)2

3(n−2)X+1,

Bn((e1 − x)4; x)

Bn((e1 − x)2; x)=

3(n− 2)X2 + X

n3· n

X=

3(n− 2)X + 1

n2.

This leads to

Theorem 4 ∣∣∣∣n[Bn(f ; x)− f(x)]− x(1− x)

2f ′′(x)

∣∣∣∣

≤X·

|X ′|√3(n−2)X+1

· 5

6ω1

(f ′′;

√3(n−2)X+1

n2

)+

13

16ω2

(f ′′;

√3(n−2)X+1

n2

)

≤

|X ′|n

· ||f ′′′||+ 13

8·√

3(n− 2)X + 1

n2· ||f ′′′|| = O

(1√n

)for f ∈ C3[0,1],

|X ′|n

· ||f ′′′||+ 13

16· 3(n− 2)X + 1

n2· ||f (4)|| = O

(1

n

)for f ∈ C4[0,1].

86


Hence the use of ω2(f′′; ...) does help in case of the Bernstein opera-

tors in the sense that a higher degree of approximation is guaranteed for

f ∈ C4[0,1].

Remark 2 For Bernstein operators it is not possible to have an upper

bound in terms of ω2(f′′, ...) only. Consider the function f(x) = e3(x) = x3.

Then

Bn(e3; x)−e3(x)−1

2·e′′3(x)·Bn((e1−x)2; x) =

x(1− x)(1− 2x)

n26= 0 for x /∈ 0, 1

2, 1,

but ω2(e′′3, h) = 0 for all h ≥ 0.

The situation is the same for the genuine Bernstein-Durrmeyer opera-

tors Un = Bn Bn as will be shown next. We start off again with a lemma

dealing with the relevant moments.

Lemma 5 The operators Un satisfy

Un((e1 − x)2; x) =2X

n + 1,

|Un((e1 − x)3; x)|2Un((e1 − x)2; x) · Un((e1 − x)4; x)

=62X2(X′)2

(n+1)2(n+2)2· n+1

2X· (n+1)(n+2)(n+3)

12(n−7)X2+24X

=3

2· (X ′)2 · n + 3

n + 2· 1

(n− 7)X + 2,

Un((e1 − x)4; x)

Un((e1 − x)2; x)=

12(n− 7)X2 + 24X

(n + 1)(n + 2)(n + 3)· n + 1

2X

=6(n− 7)X + 12

(n + 2)(n + 3.

87


We are thus led to the following

Theorem 5 For the genuine Bernstein-Durrmeyer operators we have the

inequality

|(n + 1)[Un(f ; x)− f(x)]− x(1− x) · f ′′(x)|

≤ 2X ·√

3

2· (X′)2 · n + 3

n + 2· 1

(n− 7)X + 2· 5

6ω1

(f ′′;

√6(n− 7)X + 12

(n + 2)(n + 3)

)

+13

16· ω2

(f ′′;

√6(n− 7)X + 12

(n + 2)(n + 3)

)

≤

3|X ′|n + 2

||f ′′′||+ 13

8·√

6(n− 7)X + 12

(n + 2)(n + 3)· ||f ′′′|| = O

(1√n

)for f ∈ C3[0,1],

3|X ′|n + 2

· ||f ′′′||+ 13

16· 6(n− 7)X + 12

(n + 2)(n + 3)· ||f (4)|| = O

(1

n

)for f ∈ C4[0,1].

References

[1] S.N. Bernstein, Complement a l’article de E. Voronovskaya

”Determination de la forme asymptotique de l’approximation des fonc-

tions par les polynomes de M. Bernstein”, C. R. (Dokl.) Acad. Sci.

URSS A (1932), no.4, 86–92.

[2] H. Gonska, On the degree of approximation in Voronovskaja’s theorem,

Studia Univ. Babes-Bolyai, Mathematica 52 (2007), no. 3, 103–116.

[3] H. Gonska, D. Kacso, I. Rasa, On genuine Bernstein-Durrmeyer oper-

ators, Results Math. 50 (2007), no. 3-4, 213–225.

88


[4] H. Gonska and R. Kovacheva, The second order modulus revisited:

remarks, applications, problems, Conf. Semin. Mat. Univ. Bari 257

(1994), 1-32.

[5] H. Gonska and J. Meier, On approximation by Bernstein-type operators:

best constants. Studia Sci. Math. Hungar. 22 (1987), no. 1-4, 287–297.

[6] H. Gonska, P. Pitul and I. Rasa, On Peano’s form of the Taylor re-

mainder, Voronovskaja’s theorem and the commutator of positive linear

operators, In: ”Numerical Analysis and Approximation Theory” (Proc.

Int. Conf. Cluj-Napoca 2006; ed. by O. Agratini & P. Blaga), 55-80.

Cluj-Napoca: Casa Cartii de Stiinta 2006.

[7] H. Gonska, P. Pitul and I. Rasa, On differences of positive linear op-

erators, Carpathian J. Math., 22 (2006), 65–78.

[8] H. Gonska and I. Rasa, Asymptotic behaviour of differentiated Bern-

stein polynomials, to appear in ”Mat. Vesnik”.

[9] H. Gonska and G. Tachev, A quantitative variant of Voronovskaja’s

theorem, to appear in ”Results Math.”, 53 (2008).

[10] A. Lupas, Die Folge der Betaoperatoren, Dissertation, Universitat

Stuttgart 1972.

[11] G. Tachev, Voronovskaja’s theorem revisited, J. Math. Anal. Appl. 343

(2008), no. 1, 399–404.

[12] V.S.Videnskij, Linear Positive Operators of Finite Rank (Russian),

Leningrad: ”A.I. Gerzen” State Pedagogical Institute 1985.

89


[13] E.V. Voronovskaja, Determination de la forme asymptotique de

l’approximation des fonctions par les polynomes de M. Bernstein (Rus-

sian), C. R. Acad. Sc. URSS (1932), 79–85.

Heiner Gonska

University of Duisburg-Essen


D-47048 Duisburg, Germany


Ioan Rasa

Technical University


RO-400020 Cluj-Napoca, Romania


90



POPOVICIU TYPE INEQUALITIES FOR PSEUDO

ARITHMETIC AND GEOMETRIC MEANS

Vasile Mihesan

Abstract

In this paper we prove Popoviciu type inequalitites for the pseudo

arithmetic and geometric means an and gn, defined by

an =Pn

p1x1 − 1

p1

n∑

i=2

pixi and gn = xPn/p1

1 /

n∏

i=2

xpi/p1

i ,

where xi and pi(i = 1, 2, . . . , n) are positive real numbers and Pn =n∑

i=1pi.

2000 Mathematics Subject Classification: 26D15, 26E60

Key words and phrases: pseudo arithmetic and geometric means, inequalities

1. INTRODUCTION

The classical inequality between the weighted arithmetic and geometric

means

(1) Gn = Gn(y;q) =n∏

i=1

yqi/Qn

i ≤ 1

Qn

n∑i=1

qiyi = An(y;q) = An

is valid for all positive real numbers yi and qi(i = 1, 2, . . . , n) with Qn =∑n

i=1 qi. Equality holds in (1) if and only if y1 = y2 = · · · = yn.

91

V. Mihesan - Popoviciu type inequalities ...

For this inequality, which is probably the most important inequality,

many proofs, extensions, refinements and variants are known, see [2], [3],

[4], [7].

In this paper we denote by an and gn the following expressions which

are closely connected to An and Gn. For positive real numbers xi and

pi(i = 1, 2, . . . , n) with Pn =n∑

i=1

pi we define

(2) an = an(x;p) =Pn

p1

x1 − 1

p1

n∑i=2

pixi

and

(3) gn = gn(x;p) = xPn/p1

1 /

n∏i=2

xpi/p1

i

Although there is no general agreement in literature what constitutes a

mean value, most authors consider the intermediate property as the main

feature. Since an and gn do not satisfy this condition, this means the double-

inequalities

min1≤i≤n

xi ≤ an ≤ max1≤i≤n

xi and min1≤i≤n

xi ≤ qn max1≤i≤n

xi

are not true for all positive xi, we call an and gn pseudo arithmetic and

geometric means.

In 1990, H. Alzer [1] published the following comparison of inequality

(1):

(4) an(x;p) ≤ gn(x;p)

which equality holding if and only if x1 = x2 = · · · = xn. For the special

case

92


p1 = p2 = · · · = pn the inequality (4) was proved by S. Iwamoto, R. J.

Tomkins and C. L. Wang [5].

The aim of this paper is to prove Popoviciu type inequalities for pseudo

arithmetic and geometric means.

2. INEQUALITIES INVOLVING THE RATIO an(x;p)/gn(x;p)

Two well-known extensions of the arithmetic mean-geometric mean in-

equality are the following inequalities of Popoviciu and Rado.

(5)(Gn(y;q)/An(y;q)

)Qn ≤(Gn−1(y;q)/An−1(y;q)

)Qn−1

and

(6) Qn(An(y;q)−Gn(y;q)) ≥ Qn−1(An−1(y;q)−Gn−1(y;q))

Equality holds in (5) if and only if yn = An−1 and in (6) if and only if

yn = Gn−1; see [7], [3].

The next proposition provide an analog of Popoviciu inequality (5) for

pseudo arithmetic and geometric means [1].

Proposition 1.Let xi(i = 1, 2, . . . , n; n ≥ 2) be positive real numbers such

that an(x,p) > 0 and an−1(x,p) > 0. Then we have

(7) an(x;p)/gn(x;p) ≤ an−1(x;p)/gn−1(x;p)

with equality holding if and only if x1 = xn.

Using inequality (6) we obtain an extension of the inequality of Popoviciu

type (7).

93


Theorem 1.For all positive real numbers xi(i = 1, 2, . . . , n; n ≥ 2) such

that an(x;p) > 0 and an−1(x,p) > 0 we have

Pn(1− (an(x;p)/gn(x;p))p1/Pn) ≥≥ Pn−1(1− (an−1(x;p)/gn−1(x;p))p1/Pn−1)

(8)

Proof. If we put in (6) y1 = an(x;p), yi = xi(i = 2, 3, . . . , n) and qi = pi

(i = 1, 2, . . . , n) then we obtain

Qn(An(y;q)−Gn(y;q)) = x1Pn(1− (an(x;p)/gn(x;p))p1/Pn)

which leads to inequality (8).

3. INEQUALITIES INVOLVING THE RATIO an(x;q)/gn(x;p)

The most obvious extension is to allow the means in the Rado and

Popoviciu inequalities to have different weights [3].

((Gn(y;p))Pn/pn/(An(y;q))Qn/qn ≤≤ ((Gn−1(y;p))Pn−1/pn/(An−1(y;q))Qn−1/qn

(9)

and

QnAn(y;q)− qn

pn

PnGn(y;p) ≥

≥ Qn−1An−1(y;q)− qn

pn

Pn−1Gn−1(y;p)(10)

Using this inequalities we obtain generalizations of the inequalities of

Popoviciu type (7) and (8) (see also [6]). Let n ≥ 2 be a fixed integer.

Theorem 2.Let xi(i = 1, 2, . . . , n; n ≥ 2) be positive real numbers such that

an(x;q) > 0 and an−1(x;q) > 0. Then we have

(11) an(x;q)/gn(x;p) ≤ an−1(x;q)/gn−1(x;p)

94


Proof. If we set in (9) y1 = an(x;q), yi = xi(i = 2, 3, . . . , n) then we have

(Gn(y;p))Pn/pn/(An(y;q))Qn/qn

= (an(x;q)/gn(x;p))p1/pn · xPn/pn−Qn/qn

1

and

(Gn−1(y;p))Pn−1/pn/(An−1(y;q))Qn−1/qn

= (an−1(x;q)/gn−1(x;p))p1/pn · xPn−1/pn−Qn−1/qn

1

which leads to inequality (11) because

Pn/pn −Qn/qn = Pn−1/pn −Qn−1/qn.

Theorem 3.For all positive real numbers

xi(i = 1, 2, . . . , n; n ≥ 2) such that an(x;q) > 0 and an−1(x;q) > 0 we have

Pn(1− (an(x;q)/gn(x;p))p1/Pn ≥≥ Pn−1(1− (an−1(x;q)/gn−1(x;p))p1/Pn−1

(12)

Proof. If we put in (10) y1 = an(x;q), yi = xi(i = 2, 3, . . . , n) then we have

QnAn(y;q)− qn

pn

PnGn(y;p) = Qnx1 − qn

pn

Pnx1(an(x;q)/gn(x;p))p1/Pn ,

and

Qn−1An−1(y;q)− qn

pnPn−1Gn−1(y;p)

= Qn−1x1 − qn

pnPn−1x1(an−1(x;q)/gn−1(x;p))p1/Pn−1 .

and

Qn − qn

pnPn(an(x;q)/gn(x;p))p1/Pn ≥

≥ Qn−1 − qn

pnPn−1(an−1(x;q)/gn−1(x;p))p1/Pn−1

which leads to inequality (12).

95


References

[1] H. Alzer, Inequalities for pseudo arithmetic and geometric means,

International Series of Numerical Mathematics, vol. 103, Birkhauser-

Verlag Basel, 1992, 5-16.

[2] E.F. Beckenbach and R. Bellman, Inequalities, Springer Verlang,

Berlin, 1983.

[3] P.S. Bullen, D.S. Mitrinovic and P.M. Vasic, Means and Their Inequal-

ities, Reidel Publ. Co., Dordrecht, 1988.

[4] G.H. Hardy, J.E. Littlewood and G. Polya, Inequalities, Cambridge

Univ. Press, Cambridge, 1952.

[5] S. Iwamoto, R.J. Tomkins and C.L. Wang, Some theorems on reverse

inequalities, J. Math. Anal. Appl. 119(1986), 282-299.

[6] V. Mihesan, Rado and Popoviciu type inequalities for pseudo arithmetic

and geometric means (in press)

[7] D.S. Mitrinovic, Analytic Inequalities, Springer Verlag, New York,

1970.

96


Vasile Mihesan



Str. C.Daicoviciu nr.15, 400020 Cluj-Napoca, Romania


97



RADO TYPE INEQUALITIES FOR WEIGHTED POWER

PSEUDO MEANS

Vasile Mihesan

Abstract

We denote by m[r]n (x,p) the following expression, which is closely

connected to the weighted power means of order r,M[r]n .

Let n ≥ 2 be a fixed integer, p = (p1, p2, . . . , pn), p1 > 0, pi ≥0; (i = 2, . . . , n) and Pn =

n∑i=1

pi. We define the weighted power

pseudo means:

m[r]n (x;p) =

(Pnp1

xr1 − 1

pi

n∑i=1

pixri

)1/r, r 6= 0

xPn/p1

1 /n∏

i=2x

pi/p1

i , r = 0. (x ∈ Rr(p)).

where

Rr(p) = x = (x1, x2, . . . , xn)|xi > 0(i = 1, 2, . . . , n), Pnxr1 >

n∑

i=2

pixri .

In this paper we prove Rado type inequalities for the weighted power

pseudo means m[r]n (x;p).

2000 Mathematics Subject Classification: 26D15, 26E60

Key words and phrases: Rado type inequality, the weighted power pseudo means

98

V. Mihesan - Rado type inequalities for weighted power pseudo means

1. INTRODUCTION

Let y = (y1, y2, . . . , yn) and q = (q1, q2, . . . , qn) be positive n-tuples. If r

is a real number, then the r-th power means of y with weights q,M[r]n (y;q)

is defined by

(1) M [r]n (y; q) =

(1

Qn

n∑i=1

qiyri

)1/r

; r 6= 0

( n∏i=1

yqi

i

)1/Qn

; r = 0

where Qn =n∑

i=1

qi If r, s ∈ R, r ≤ s then [4]

(2) M [r]n (y;q) ≤ M [s]

n (y;q)

is valid for all positive real number yi and qi(i = 1, 2, . . . , n). For r = 0 abd

s = 1 we obtain the classical inequality between the weighted arithmetic

and geometric means.

(3) Gn = Gn(y;q) =( n∏

i=1

yqi

i

)1/Qn ≤ 1

Qn

n∑i=1

qiyi = An(y;q) = An.

In this paper we denote by m[r]n (x;p) the following expression which is

closely connected to M[r]n (x;p).

Let n ≥ 2 be an integer (considered fixed throughout the paper),

p = (p1, p2, . . . , pn), p1 > 0, pi ≥ 0(i = 2, . . . , n) and Pn =∑n

i=1 pi. We

define

(4) m[r]n (x;p) =

(Pn

p1xr

1 − 1p1

n∑i=2

pixri

)1/r

, r 6= 0

xPn/x1

1 /n∏

i=2

xpi/p1

i , r = 0

(x ∈ Rr(p))

99


where Rr(p) = x = (x1, x2, . . . , xn)|xi > 0(i = 1, 2, . . . , n), Pnxr1 >

n∑i=2

pixri.

Although there is no general agreement in literature what constitutes a

mean values most authors consider the intermediate property as the main

feature. Since m[r]n (x;p) do not satisfy this condition, this mean the double

inequalities

min1≤i≤n

xi ≤ m[r]n (x;p) ≤ max

1≤i≤nxi

are not true for all positive xi, we call m[r]n (x;p) weighted power pseudo

means of order r.

For r = 1 we obtain the pseudo arithmetic means an(x; p) and for r = 0

the pseudo geometric means gn(x;p), see [2].

(5) an(x;p) =Pn

p1

x1 − Pn

p1

n∑i=2

pixi, gn(x;p) = xPn/p1

1 /

n∏i=2

xpi/p1

i

H. Alzer [2] published the following companion of inequality (3):

(6) an(x;p) ≤ gn(x;p)

For the special case p1 = p2 = · · · = pn the inequality (6) was prove by

S. Iwamoto, R.J. Tomkins and C.L. Wang [6].

Rado and Popoviciu type inequalities for pseudo arithmetic and geomet-

ric means where given in [2], [9], [10].

We note that inequality (6) is an example of so called reverse inequality.

One of the first reverse inequalities was published by J. Aczel [1] who proved

the following intriguing variant of the Cauchy-Schwarz inequality:

if xi and yi(i = 1, . . . , n) are real number with x21 >

n∑i=2

x2i and y2

1 >n∑

i=2

y2i ,

then

(7) (x1y1 −n∑

i=2

xiyi)2 ≥ (x2

1 −n∑

i=2

x2i )(y

21 −

n∑i=2

y2i )

100


Further interesting reverse inequalities where given in [3], [5], [6], [7], [8],

[12], [13].

The aim of this paper is to prove Rado type inequalities for the weighted

power pseudo means m[r]n (x;p).

2. RADO TYPE INEQUALITIES FOR WEIGHTED POWER

PSEUDO MEANS

The well-known extension of the arithmetic mean-geometric mean in-

equality (3) is the following inequality of Rado [12].

(8) Qn(An(y;q)−Gn(y;q)) ≥ Qn−1(An−1(y;q)−Gn−1(y;q))

The next proposition provide an analog of Rado inequality (8) for pseudo

arithmetic and geometric means [2].

Proposition 1.For all positive real number xi(i = 1, 2, . . . , n; n ≥ 2) we

have

(9) gn(x;p)− an(x;p) ≥ gn−1(x;p)− an−1(x;p)

The most obvious extension is to allow the means in the Rado inequality to

have different weight [4].

(10) QnAn(y;q)− qn

pn

PnGn(y;p) ≥ Qn−1An−1(y;q)− qn

pn

Pn−1Gn−1(y;p)

Using this inequality we obtain the following generalization of the inequality

(9) [10].

101


Proposition 2.For all positive real number xi(i = 1, 2, . . . , n; n ≥ 2) we

have

(11) gn(x;p)− an(x;q) ≥ gn−1(x;p)− an−1(x;q)

An extension of the Rado inequality (10) for weighted power means is the

following inequality [4]. If r, s, t ∈ R such that −∞ < r/t ≤ 1 ≤ s/t < ∞then

(12)

Qn(M [s]n (y;q))t−qn

pn

Pn(M [r]n (y;p))t ≥ Qn−1(M

[s]n−1(y;q))t−qn

pn

Pn(M[r]n−1(y;p))t

Using the inequality (12) we obtain generalization of the inequality of

Rado type (11) for the weighted power pseudo means.

Theorem 1.If r ≤ 1,x ∈ Rr(p) and xr1 ≤ xr

n then

(13) m[r]n (x;p)− an(x;q) ≥ m

[r]n−1(x;p)− an−1(x;q)

If s ≥ 1,x ∈ Rs(q) and x1 ≤ xn then

(14) an(x;p)−m[s]n (x;q) ≥ an−1(x;p)−m

[s]n−1(x;q)

Proof. To prove (13) we put in (12) s = t = 1 and we obtain for r ≤ 1 the

inequality

(15) QnAn(y;q)− qn

pn

PnM [r]n (y;p) ≥ Qn−1An−1(y;q)− qn

pn

Pn−1M[r]n−1(y;p)

If we set in (15) y1 = m[r]n (x;p), yi = xi(i = 2, 3, . . . , n) then we obtain

QnAn(y;q)− qn

pn

PnM[r]n (y;p) = q1(m

[r]n (x;p)− an(x;q)) + (Qn − qn

pn

Pn)x1

102


and

Qn−1An−1(y;q)−qn

pn

Pn−1M[r]n−1(y;p)=q1(m

[r]n−1(x;p)−an−1(x;q))+(Qn−1−qn

pn

Pn−1)x1

which leads to inequality (13), because equality

(16) Qn − qn

pn

Pn = Qn−1 − qn

pn

Pn−1

holds.

We observe that for r ≤ 1,x ∈ Rr(p) and xr1 ≤ xr

n we have

0 < Pnxr1 −

n∑i=2

pixri ≤ Pn−1x

r1 −

n∑i=2

pixri and m

[r]n−1(x;p) exist.

To prove (14) we set in (12) r = t = 1 and we obtain for s ≥ 1 the

inequality

(17) QnM [s]n (y;q)− qn

pn

PnAn(y;p) ≥ Qn−1M[s]n−1(y;q)− qn

pn

Pn−1An−1(y;p)

If we put in (17) y1 = m[s]n (x;q), yi = xi(i = 2, 3, . . . , n) then we have

QnM[s]n (y;q)− qn

pn

PnAn(y;p) = (Qn− qn

pn

Pn)x1 +qn

pn

p1(an(x;p)−m[s]n (x;q))

and

Qn−1M[s]n−1(y;q)−qn

pn

Pn−1An−1(y;p)=(Qn−1−qn

pn

Pn−1)x1+qn

pn

p1(an−1(x;p)−m[s]n−1(x;q))

which leads to inequality (14), because equality (16) holds. For s ≥ 1,x ∈Rs(p) and x1 ≤ xn,m

[s]n−1(x;p) exist.

Theorem 2.If 0 < r ≤ s,x ∈ Rr(p) ∩Rs(q) and x1 ≤ xn then

(18) (m[r]n (x;p))s − (m[s]

n (x;q))s ≥ (m[r]n−1(x;p))s − (m

[s]n−1(x;q))s

and

(19) (m[r]n (x;p))r − (m[s]

n (x;q))r ≥ (m[r]n−1(x;p))r − (m

[s]n−1(x;q))r.

103


Proof. To prove (18) we put in (12) t = s and we obtain for 0 < r ≤ s the

inequality

(20)

Qn(M [s]n (y;q))s−qn

pn

Pn(M [r]n (y;p))s ≥ Qn−1(M

[s]n−1(y;q))s−qn

pn

Pn−1(M[r]n−1(y;p))s

If we set in (20) y1 = m[r]n (x;p), yi = xi(i = 2, 3, . . . , n) then we have

Qn(M [s]n (y;q))s−qn

pn

Pn(M [r]n (y;p))s =q1((m

[r]n (x;p))s−((m[r]

n (x;q))s)+xs1(Qn−qn

pn

Pn)

and

Qn−1(M[s]n−1(y;q))s− qn

pn

Pn−1(M[r]n−1(y;p))s = q1((m

[r]n−1(x;p))s−((m

[r]n−1(x;q))s)

+ xs1(Qn−1− qn

pn

Pn)

which leads to inequality (18), because equality (16) holds. For 0 < r ≤s,x ∈ Rr(p) ∩Rs(q) and x1 ≤ xn,m

[r]n−1(x;p) and m

[s]n−1(x;q) exist.

To prove (19) we set in (12) t = r and we obtain for 0 < r ≤ s the

inequality

(21)

Qn(M [s]n (y;q))r−qn

pn

Pn(M [r]n (y;p))r ≥ Qn−1(M

[s]n−1(y;q))r−qn

pn

Pn−1(M[r]n−1(y;p))r.

If we put in (21) y1 = m[s]n (x; q), yi = xi(i = 2, 3, . . . , n) then we obtain

Qn(M [s]n (y;q))r−qn

pn

Pn(M [r]n (y;p))r =x1(Qn−qn

pn

Pn)+p1qn

pn

((m[r]n (x;p))r−(m[s]

n (x;q))r)

and

Qn−1(M[s]n−1(y;q))r − qn

pn

Pn−1(M[r]n−1(y;p))r = x1(Qn−1 − qn

pn

Pn−1)

+p1qn

pn

((m[r]n−1(x;p))r − (m

[s]n−1(x;q))r).

which leads to inequality (19) because equality (16) holds.

For p = q we obtain the results of [11].

104


References

[1] J. Aczel, Some general methods in the theory of functional equations in

one variable.New applications of functional equations (Russian), Uspehi

Mat. Nauk (N.S.) 11,No.3 (69)(1956), 3-58.

[2] H. Alzer, Inequalities for pseudo arithmetic and geometric means, In-

ternational Series of Numerical Mathematics, Vol. 103, Birkhauser-

Verlag Basel, 1992, 5-16.

[3] R. Bellman, On an inequality concerning an indefinite form, Amer.

Math. Monthly 63 (1956), 108-109.

[4] P.S. Bullen, D.S. Mitrinovic and P.M. Vasic, Means and Their Inequal-

ities, Reidel Publ. Co., Dordrecht, 1988.

[5] Y.J. Cho, M. Matic, J. Pecaric, Improvements of some inequalities of

Aczel’s type, J. Math. Anal. Appl. 256(2001), 226-240.

[6] S. Iwamoto, R.J. Tomkins and C.L. Wang, Some theorems on reverse

inequalities, J. Math. Anal. Appl. 119(1986), 282-299.

[7] L. Losonczi, Inequalities for indefinite forms, J. Math. Anal. Appl.

285(1997),148-156.

[8] V. Mihesan, Applications of continuous dynamic programing to inverse

inequalities, General Mathematics 2(1994), 53-60.

[9] V. Mihesan, Popoviciu type inequalities for pseudo arithmetic and ge-

ometric means (in press)

105


[10] V. Mihesan, Rado and Popoviciu type inequalities for pseudo arithmetic

and geometric means (in press)

[11] V. Mihesan, Inequalities for weighted power pseudo means (in press)

[12] D.S. Mitrinovic, Analytic Inequalities, Springer Verlag, New York,

1970.

[13] X.H. Sun, Aczel-Chebyshev type inequality for positive linear functional,

J. Math.Anal.Appl.245(2000), 393-403.

Vasile Mihesan



Str. C.Daicoviciu nr.15, 400020 Cluj-Napoca, Romania


106



SEVERAL INEQUALITIES ABOUT ARITHMETIC

FUNCTIONS WHICH USE THE E-DIVISORS

Nicusor Minculete

Abstract

The integer d =r∏

i=1

pbii is called an exponential divisor (or e-

divisor) of n =r∏

i=1

paii > 1 if bi|ai for every i = 1, r. Let σ(e)(n)

denote the sum of the exponential divisors of n and τ (e)(n) denote

the number of the exponential divisors of n. The purpose of this pa-

per is to present several inequalities about the arithmetical functions

σ(e) and τ (e). Among these, we have the following:σ(e)(n)τ (e)(n)

≥r∏

i=1

pσ(ai)

τ(ai)

i , (∀)n ∈ N∗, σ(e)(n) >γ(n)ϕ(n)

n

r∏

i=1

σ(ai), (∀)n ∈

N, n ≥ 2 andσ(e)(n)

[τ (e)(n)]2≥ ϕ(n)

n, (∀)n ∈ N∗, where ϕ is Euler’s to-

tient, τ(n) is the number of the natural divisors of n, σ(n) is the sum

of the natural divisors of n and γ is the ”core” of n.

2000 Mathematics Subject Classification: 11A25, 11A99

Key words and phrases: exponential divisors, arithmetical functions, the sum of

the exponential divisors of n, the number of the exponential divisors of n,

Euler’s totient ϕ, the number of the natural divisors of n, the sum of the

natural divisors of n.

107

N. Minculete - Several inequalities about arithmetic functions...

1. INTRODUCTION

The notions of ”exponential divisors” was introduced by M.V. Subbarao

in [10]. Let n > 1 be an integer of canonical form n = pa11 pa2

2 . . . parr . The

integer d =r∏

i=1

pbii is called an exponential divisor (or e-divisor) of n =

r∏i=1

paii > 1, if bi|ai for every i = 1, r. We note d|(e)n. Let σ(e)(n) denote the

sum of the exponential divisors of n. By convention, 1 is an exponential

divisor of itself so that τ (e)(1) = σ(e)(1) = 1. We notice that 1 is not

an exponential divisor of n > 1, the smallest exponential divisor of n =

pa11 pa2

2 . . . parr > 1 is p1p2 . . . pr, where p1p2 . . . pr = γ(n) is called the ”core”

of n (γ(1) = 1).

For example, the exponential divisors of the number p10 are p, p2, p5 and

p10, so τ (e)(p10) = τ(10) = 4 and σ(e)(n) = p + p2 + p5 + p10.

We notice that if n is a squarefree number, then σ(e)(n) = n, and τ (e)(n) = 1.

It is easy to see that τ (e)(n) =∑

d|(e)n1 and σ(e)(n) =

∑

d|(e)nd are multiplicative

functions, and hence

(1) τ (e)(n) = τ(a1)τ(a2) · . . . · τ(ar),

(2) σ(e)(n) =r∏

i=1

σ(e)(paii ) =

r∏i=1

(∑

bi|ai

pbii ).

E.G. Straus and M.V. Subbarao in [9] obtained several results concerning

e-perfect numbers (n is an e-perfect number if σ(e)(n) = 2n), including the

nonexistence of odd e-perfect numbers, and show that the set

σ(e)(n)

n

is

108


dense in [1,∞).

To study the problem of the maximal order, Erdos (see[9]) obtained the

following result:

limn→∞

suplog τ (e)(n) · log log n

log n=

log 2

2

In [1] J. Fabrykovski and M. V. Subbarao showed that

limn→∞

supσ(e)(n)

eγn log log n=

6

π2,

and∑n≤x

σ(e)(n) = Bx2 + O(x1+ε),

where B is an absolute constant ≈ 0.568285.

Let T (n) denote the product of all divisors of n, and let T (e)(n) denote the

product of all exponential divsors of n.

In [6], J. Sandor showed that

(3) T (e)(n) = (t(n))τ(e)(n)

2 ,

where t(1) = 1 and t(n) = pσ(a1)τ(a1)

1 · pσ(a2)τ(a2)

2 · . . . · pσ(ar)τ(ar)r for n = pa1

1 pa22 . . . par

r .

We remarked that

(4) τ (e)(n) ≤ τ(n)

and

(5) t(n) ≤ n.

In [8], J. Sandor showed that, if n is a perfect square, then

(6) 2ω(n) ≤ τ (e)(n) ≤ 2Ω(n),

109


where ω(n) and Ω(n) denote the number of distinct prime factors of n, and

the total number of prime factors of n, respectively. It is easy to see that

for n = pa11 pa2

2 . . . parr > 1, we have ω(n) = r and Ω(n) = a1 + a2 + . . . + ar.

In [7], J. Sandor introduced the arithmetic function t1(n) = p2√

a1

1 · p2√

a2

2 ·. . . · p2

√ar

r with t1(1) = 1, and proved that

(7) t1(n) ≥ t(n) ≥ nγ(n),

for all n ≥ 1.

In [2], N. Minculete showed that

(8) σ(e)(n) ≤ ψ(n) ≤ σ(n),

(9) τ(n) ≤ σ(e)(n)

τ (e)(n),

for all integers n ≥ 1, where ψ is Dedekind’s function.

We note with σ(e)k (n) the sum of kth powers of the exponential divisors of n,

so, σ(e)k (n) =

∑

d|(e)ndk, whence we obtain the following equalities: σ

(e)1 (n) =

σ(e)(n) and σ(e)0 (n) = τ (e)(n)- the number of the exponential divisors of n.

In [3], N. Minculete showed that

(10)σ

(e)k (n)

τ (e)(n)≤

(nk + 1

2

)2

,

for all integers n ≥ 1 and k ≥ 0

2. MAIN RESULTS

We will present three theorems containing some properties of the above

functions.

110


Theorem 1.If σ(e)(n) is the sum of the exponential divisors of n, τ (e)(n)

denote the number of the exponential divisors of n, γ is the ”core” of n, ϕ

is Euler’s function, and σ(n) is the sum of the natural divisors of n, then

(11)σ(e)(n)

τ (e)(n)≥

r∏i=1

pσ(ai)

τ(ai)

i and

(12) σ(e)(n) >γ(n)ϕ(n)

n

r∏i=1

σ(ai),

for all n ≥ 2.

Proof. Let a ∈ N∗ and p be a prime number, and d1, d2, . . . , dr, divisors of

a; it follows that

σ(e)(pa) = pd1 + pd2 + . . . + pdk .

Since the function f : R → (0, +∞), f(x) = px, p ≥ 2, is convex, from

Jensen’s Inequality, we deduce that

pd1 + pd2 + . . . + pdk

k≥ p

d1+d2+...+dkk ,

soσ(e)(pa)

τ(a)≥ p

σ(a)τ(a)

therefore

σ(e)(pa) ≥ τ(a)pσ(a)τ(a) .

Hence, from the fact that the functions σ(e) and τ (e) are multiplicative, we

obtain

σ(e)(n) = σ(e)(pa11 · pa2

2 · . . . · parr ) = σ(e)(pa1

1 )σ(e)(pa22 ) · . . . · σ(e)(par

r ) ≥

111


≥ τ(a1)pσ(a1)τ(a1 · τ(a2)p

σ(a2)τ(a2) · . . . · τ(ar)p

σ(ar)τ(ar) = τ (e)(n)

r∏i=1

pσ(ai)

τ(ai)

i ,

soσ(e)(n)

τ (e)(n)≥

r∏i=1

pσ(ai)

τ(ai)

i .

It is easy to see that

(13) xa > a(x− 1), for all real numbers a ≥ 1 and x > 1,

because, using Bernoulli‘s generalized Inequality, we have

xa = (1 + x− 1)a ≥ 1 + a(x− 1) > a(x− 1).

In the case n > 1, we have n = pa11 pa2

2 . . . parr , and, by applying inequality

(13), for x = pi and a = σ(ai)τ(ai)

, we deduce the inequality

pσ(ai)

τ(ai)

i >σ(ai)

τ(ai)(pi − 1) = pi

σ(ai)

τ(ai)

(1− 1

pi

)

sor∏

i=1

pσ(ai)

τ(ai)

i >

r∏i=1

piσ(ai)

τ(ai)

(1− 1

pi

)=

γ(n)ϕ(n)

nτ (e)(n)

r∏i=1

σ(ai).

From inequality (11), we have

σ(e)(n)

τ (e)(n)≥

r∏i=1

pσ(ai)

τ(ai)

i ,

therefore,σ(e)(n)

τ (e)(n)>

γ(n)ϕ(n)

nτ (e)(n)

r∏i=1

σ(ai).


denote the number of the exponential divisors of n, τ(n) is the number of

the natural divisors of n and ϕ is Euler’s function, then

(14) σ(e)(n) ≥ γ(n)ϕ(n)τ(n)

n,

for all n ≥ 1.

112


Proof. We know that

σ(n) ≥ n + 1, for all n ≥ 1.

Using this inequality and inequality (12), we obtain another inequality,

namely, in the case n > 1, we have n = pa11 pa2

2 . . . parr , so

σ(e)(n) >γ(n)ϕ(n)

n

r∏i=1

σ(ai) ≥ γ(n)ϕ(n)

n

r∏i=1

(ai + 1) =γ(n)ϕ(n)τ(n)

n

In case that n = 1, we deduce σ(e)(1) = γ(1)ϕ(1)τ(1)1

= 1,

therefore σ(e)(n) ≥ γ(n)ϕ(n)τ(n)

nfor all n ≥ 1


denote the number of the exponential divisors of n and ϕ is Euler’s function,

then

(15)σ(e)(n)

[τ (e)(n)]2≥ ϕ(n)

n,

for all n ≥ 1.

Proof. From [4] and [5], we remark the inequalities

τ(n) ≤ 2√

n, for all n ≥ 1, and σ(n) ≥ τ(n)√

n, for all n ≥ 1, and we say

that

τ (e)(n) =r∏

i=1

τ(ai) ≤r∏

i=1

2√

ai = 2r

r∏i=1

√ai ≤ γ(n)

r∏i=1

√ai.

In case that n > 1, we have n = pa11 pa2

2 . . . parr ,

hence

σ(e)(n) >γ(n)ϕ(n)

n

r∏i=1

σ(ai) ≥ γ(n)ϕ(n)

n

r∏i=1

τ(ai)√

ai =

113


=ϕ(n)τ (e)(n)

nγ(n)

r∏i=1

√ai ≥ ϕ(n)[τ (e)(n)]2

n,

which means thatσ(e)(n)

[τ (e)(n)]2>

ϕ(n)

n. For n = 1, we obtain

σ(e)(1)

[τ (e)(1)]2=

ϕ(1)

1= 1. Consequently,

σ(e)(n)

[τ (e)(n)]2≥ ϕ(n)

n, for all n ≥ 1.

References

[1] Fabrykowski, J. and Subbarao, M. V., The maximal order and the

average order of multiplicative function ϕ(e)(n), Theorie des Nombres

(Quebec, PQ, 1987), 201-206, de Gruyter, Berlin-New York, 1989.

[2] Minculete, N., Concerning some inequalities about arithmetic functions

which use the exponential divisors.(to appear)

[3] Minculete, N., Considerations concerning some inequalities of the

arithmetic functions ϕ(e)k and τ (e). (to appear)

[4] Panaitopol, L. and Gica, Al. , O introducere ın aritmetica si teoria

numerelor, Editura Universitatii din Bucuresti, 2001.

[5] Panaitopol, L. and Gica, Al. , Probleme de aritmetica si teoria nu-

merelor, Editura GIL, Zalau, 2006.

[6] Sandor, J., On exponentially harmonic numbers, Scientia Magna, Vol.

2 (2006), No. 3, 44-47.

[7] Sandor, J., A Note on Exponential Divisors and Related Arithmetic

Functions, Scientia Magna, Vol.1 (2006), No. 1.

114


[8] Sandor, J., On an exponential totient function, Studia Univ. Babes-

Bolyai, Math., Vol. 41 (1996), No. 3, 91-94.

[9] Straus, E. G. and Subbarao, M. V., On exponential divisors, Duke

Math. J. 41 (1974), 465-471.

[10] Subbarao, M. V., On some arithmetic convolutions in The Theory

of Arithmetic Functions, Lecture Notes in Mathematics, New York,

Springer-Verlag, 1972.

[11] Toth, L., On Certain Arithmetic Functions Involving Exponential Di-

visors, Annales Univ. Sci. Budapest., Sect. Comp. 24 (2004), 285-294.

[12] - http:// www.mathworld.wolfram.com.

Nicusor Minculete

University ”Dimitrie Cantemir” of Brasov

Department of REI

Str. Bisericii Romane, Nr. 107, Brasov, Romania


115



INEQUALITIES BETWEEN SOME ARITHMETIC

FUNCTIONS

Nicusor Minculete, Petrica Dicu

Abstract

The objective of this paper is to present several inequalities be-

tween some arithmetical functions. In these inequalities the follow-

ing arithmetical functions will appear: Euler’s totient ϕ, the func-

tions τ and σ respectively, where τ(n) is the number of natural di-

visors of n, and σ(n) is the sum of natural divisors of n. Among

the main results we have ϕ(mn)σ(mn) ≥ ϕ(m)ϕ(n)σ(m)σ(n) and

nσ(n) ≥ ϕ(n)τ2(n), (∀)m,n ∈ N∗.

2000 Mathematics Subject Classification: 11A25

Key words and phrases: prime number, arithmetical functions, Euler‘s totient ϕ,

the number of the natural divisors of n, the sum of the natural divisors of n.

1. INTRODUCTION

Let n be a positive integer, n ≥ 1. We note with ϕ(n) the number of

positive integers less than or equal to n, that are comprime to n.

Hence

ϕ(n) = cardk|(k, n) = 1, k ≤ n.

116

N. Minculete, P. Dicu - Inequalities between some arithmetic functions

The function ϕ so defined is the totient function. C. Jordan has intro-

duced the function ϕk(n) as the number of k-tuples (a1, a2, ...ak) with all

the components between 1 and n satisfying (a1, a2, ..., ak, n) = 1. This is

a generalization of Euler‘s totient function ϕ, so, ϕ1(n) = ϕ(n). It should

also be recalled from [9] that the function ϕk(n) can be represented in the

following form:

ϕk(n) = nk∏

p|n

(1− 1

pk

).

If n > 1 and n =r∏

i=1

pαii n then we note with ω(n) the number of distinct

prime factors of n, so that ω(n) = r(ω(1) = 0).

Consider σk(n) the sum of kth powers of divisors of n,so,σk(n) =∑

d|n dk,

whence we obtain the following equalities: σ1(n) = σ(n) and σ0(n) = τ(n)-

the number of divisors of n.

Also, ζ(k) =∞∑

n=1

n−k =∏

p prim

1

(1− 1pk )

, where k > 1, denotes the zeta func-

tion of Riemann.

An important point in the study of the arithmetical functions consists in

the establishing of several inequalities between these arithmetical function.

In order to do that, we shall review some properties met in several papers,

Let us remark two properties of the totient function, namely, Gauss‘s iden-

tity

(1)∑

d/n

φ(d) = n;

for n > 1 and n = pα11 pα2

2 ...pαrr we have the relation

(2) ϕ(n) = n

(1− 1

p1

)(1− 1

p2

). . .

(1− 1

pr

).

117


In [2], T.M. Apostol shows that

(3)σ(n)

n<

n

ϕ(n)≤ π2

6· σ(n)

n, (∀)n ∈ N \ 0, 1.

L. Panaitopol and A. Gica [7] show that, for any n > 31, we have

(4) ϕ(n) > τ(n).

In 1972, S. Porubsky proves the inequality

(5) ϕ(n)τ 2(n) ≤ n2, (∀)n ∈ N∗, m 6= 4,

and in 1993, J. Sandor [10] finds the inequality

(6) σk(n)σl(n) ≤ τ(n)σk+1(n), (∀)n ∈ N∗, (∀)k, l ∈ N∗.

J. Sandor, in paper [11], proves that:

(7) ϕ(n)[ω(n) + 1] ≥ n, (∀)n ∈ N,

and

(8) nτ(n) ≥ ϕ(n) + σ(n), (∀)n ∈ N \ 0, 1.

R. Sivaramakrishnan [12] establishes the following inequality:

(9) ϕ(n)τ(n) ≥ n, (∀)n ∈ N∗.

Making an inventory of the arithmetical functions, we can mention some

of them and we can also create the basis for the extension and improvements

of some of them.

118


2. MAIN RESULTS

We first establish several inequalities between the arithmetical functions

ϕ and σ, where tho variable interfere.

Theorem 1.For every m,n ∈ N∗ the following inequality

(10) ϕ(mn)σ(mn) ≥ ϕ(m)ϕ(n)σ(m)σ(n) holds.

Proof. We consider m =r∏

i=1

pαii

∏j

qβj

j , and n =r∏

i=1

pα′ii

∏

k

rγk

k , where pi

are the common prime factors of m and n, it follows that

ϕ(m)σ(m)

m2=

r∏i=1

(1− 1

pαi+1i

) ∏j

(1− 1

qβj+1j

),

ϕ(n)σ(n)

n2=

r∏i=1

(1− 1

pα′i+1i

)∏

k

(1− 1

rγk+1k

),

ϕ(mn)σ(mn)

m2n2=

r∏i=1

(1− 1

pαi+α′i+1i

)∏j

(1− 1

qβj+1j

) ∏

k

(1− 1

rγk+1k

).

Therefore, from

1− 1

pαi+α′i+1i

≥(

1− 1

pαii

) (1− 1

pα′ii

),

taking the product, the following inequalities implies:

ϕ(mn)σ(mn) ≥ ϕ(m)ϕ(n)σ(m)σ(n), (∀)m,n ∈ N∗.

Theorem 2.For every m,n ∈ N∗ the following inequality

(11) ϕ2(mn)σ2(mn) ≤ ϕ2(m)ϕ2(n)σ2(m)σ2(n) holds.

119


Proof. Let us now consider m =r∏

i=1

pαii

∏j

qβj

j and n =r∏

i=1

pα′ii

∏

k

rγk

k ,

where pi are the common prime factors of m and n; it follows that

ϕ(n)σ(n)

n2=

r∏i=1

(1− 1

pα′i+1i

)∏

k

(1− 1

rγk+1k

),

ϕ(m)σ(m)

m2=

r∏i=1

(1− 1

pαi+1i

) ∏j

(1− 1

rβj+1j

).

Hence, we obtain the following relations:

ϕ(m2)σ(m2)

m4=

r∏i=1

(1− 1

p2αi+1i

) ∏j

(1− 1

r2βj+1j

),

ϕ(n2)σ(n2)

n4=

r∏i=1

(1− 1

p2α′i+1i

) ∏

k

(1− 1

r2γk+1j

),

ϕ(mn)σ(mn)

m2n2=

r∏i=1

(1− 1

pαi+α′i+1i

)∏j

(1− 1

qβj+1j

)∏

k

(1− 1

rγk+1k

).

Since the inequalities

(1− 1

pα+1

)2

≤ 1− 1

p2α+1,

(1− 1

pα+α′+1

)2

≤ (1− 1

p2α+1)(1− 1

p2α′+1)

are true, then using the previous relations and taking the product, we deduce

inequality (11).

Theorem 3.For every m,n ∈ N∗ and k ∈ N the following inequalities

(12) ϕk(mn)σk(mn) ≥ ϕk(m)ϕk(n)σk(m)σk(n),

(13) ϕ2k(mn)σ2

k(mn) ≤ ϕ2k(m)ϕ2

k(n)σ2k(m)σ2

k(n),

120


hold.(ϕ0(m) = 1)

Reasoning as in the proofs of Theorems 1 and 2, we can prove that relations

(12) and (13) hold.

Theorem 4.For every n ∈ N∗, the following inequality

(14) nσ(n) ≥ ϕ(n)τ 2(n) holds.

n = pα11 pα2

2 ...pαrr , then

ϕ(n) = n

r∏i=1

(1− 1

pi

).

Hencen

ϕ=

r∏i=1

1

1− 1pi

,

but1

1− 1pi

= 1 +1

pi

+1

p2i

+ . . . ≥ 1 +1

pi

+1

p2i

+ . . . +1

pαii

,

so thatr∏

i=1

1

1− 1pi

≥r∏

i=1

(1 +1

pi

+1

p2i

+ . . . +1

pαii

) =

τ(m)∑i=1

1

di

.

Since σ(n) =

τ(n)∑i=1

di = n

τ(n)∑i=1

1

di

and using the Cauchhy-Buniakowski-Schwarz

Inequality

τ(n)∑i=1

di

τ(n)∑i=1

1

di

≥ τ 2(n), we deduce the following inequality:

nσ(n) ≥ ϕ(n)τ 2(n), (∀)n ∈ N∗.

Corollary 1.For every n, k ∈ N∗, the following inequality

(15) nkσk(n) ≥ ϕk(n)τ 2(n) holds.

121


Reasoning as in proof in Theorem 4, we can prove that relation(15) is true.

Theorem 5.For any two natural numbers n and k, with n ≥ 1 and k ≥ 2,

we have the sequence of inequalities

(16)σk(n

2)

[σk(n)]2≥ nk

σk(n)≥ ϕk(n)

nk≥

∏p prim

(1− 1

pk

)=

1

ζ(k),

Proof. We evaluate the ratio of σk(n2) and [σk(n)]2, so

σk(m2)

[σk(m)]2=

k∏i=1

1− 1

pki

(p

k(αi+1)i −pk

i

pk(αi+1)i −1

)2 ≥

r∏i=1

[1− 1

pki

(pk

i (αi+1)−pki

pk(αi+1)i −1

)]=

r∏i=1

pαiki (pk

i −1)

pk(αi+1)i − 1

≥r∏i

(1− 1

pki

)≥

∏p prim

(1− 1

pk

),

which proves the sequence of inequalities from statement.

Corollary 2 For every n.k ∈ N∗, k 6= 1 we have the following inequalities:

(17)ϕk(n)

σk(n)≥ ζ2(k),

(18)n2k

σ2k(n)

≥ ϕk(n)

nk

1

ζ(k),

(19)nkζ(k)

σk(n)≥ σk(n)ϕk(n)

n2k.

Using inequality (16), these inequalities are verified.

122


Theorem 6.For every n, k ∈ N∗, we have the following inequalities:

(20) ϕ3k(n) ≥ 2ω(n)nkϕk(n),

(21) ϕ2k(n) ≥ 2ω(n)√

nϕk(n),

(22) ϕk(n) ≥ nk−1ϕ(n)

Proof. From the relation

1− 1

p3k=

(1− 1

pk

)(1 +

1

pk+

1

p2k

)≥

(1− 1

pk

)· 3

pk,

taking the product, we infer the inequality

∏

p|n

(1− 1

p3k

)≥

∏

p|n

(1− 1

pk

)· 3

pk≥

∏

p|n

(1− 1

pk

)· 3

pkα, (∀)α ≥ 1.

Therefore

ϕ3k(n) ≥ 3ω(n)nkϕk(n).

To prove the inequality (21) it is sufficient to show that

∏

p|n

(1− 1

p2k

)=

∏

p|n

(1− 1

pk

)(1 +

1

pk

)≥

∏

p|n

(1− 1

pk

)· 2√

pk

≥∏

p|n

(1− 1

pk

)· 2√

pkα, (∀)α ≥ 1.

Hence

ϕ2k(n) ≥ 2ω(n)√

nϕk(n), (∀)n, k ∈ N∗.

Inequality (22) simply results from

ϕk(n) = nk∏

p|n

(1− 1

pk

)≥ nk

∏

p|n

(1− 1

p

)≥ nk−1ϕ(n).

123


References

[1] D. Acu, Aritmetica si teoria numerelor, Editura Universitatii ”Lucian

Blaga” din Sibiu 1999.

[2] Apostol, T-M., Introduction to Analytic Number theory, Springer-

Verlag, New York, 1976.

[3] Creanga, I. and col., Introducere in teoria numerelor, Editura Didactica

si Pedagogica, Bucuresti, 1965.

[4] Gauss, C. Fr., Cercetari aritmetice, Editura Armacord, Timisoara,

1999.

[5] Jordan, C., Traite de substitutions et des equations algebriques, Gau-

thier Villars et Cie Editeurs, Paris, 1957.

[6] Minculete N. si Dicu P.,Concerning the Euler totient, General Mathe-

matics, vol. 16, No. 1 (2008), 85-91.

[7] Panaitopol, L. and Gica, Al., Probleme celebre de teoria numerelor,

Editura Universitatii din Bucuresti, 1998.

[8] Panaitopol, L. and Gica, Al., O introducere in aritmetica si teoria

numerelor, Editura Universitatii din Bucuresti, 2001.

[9] Panaitopol, L. and Gica, Al., Probleme de aritmetica si teoria nu-

merelor, Editura GIL, Zalau, 2006.

[10] Sandor, J.,On Jordan′s arithmetical function, Gazeta Matematica Seria

B, nr. 2-3/1993.

124


[11] Sandor, J., Some Diophantiene Equation for Particular Arithmetic

Functions, Univ. Timisoara, Seminarul de teoria structurilor no

53(1989), p.1-10.

[12] Sivaramakrishnan, R., Classical Theory of Arithmetic Function, Marcel

Dekker, In., New York, 1989.

Nicusor Minculete

University ”Dimitrie Cantemir” of Brasov

Department of REI

Str. Bisericii Romane, Nr. 107, Brasov, Romania


Petrica Dicu



Str. Dr. I. Ratiu, Nr. 5–7, 550012 - Sibiu, Romania


125



AN INTEGRAL INEQUALITY FOR CONVEX FUNCTIONS

OF THREE ORDER

Olaru Ion Marian

Abstract

Using the characterization of convex functions of three order with

the divided difference,we obtain an inequality for the operator

F (f) =1

b− a

b∫

a

f(x)dx.

2000 Mathematics Subject Classification: 34K10, 47H10

Key words and phrases: convex functions, divided difference

1. INTRODUCTION

In this paper we denote by F the following functional

F (f) =1

b− a

b∫

a

f(x)dx.

Clearly, if F is in such a manner defined then F (1) = 1. Sometime instead

of F we write Fx in order to put in evidence the corresponding variable.

For instance

Fx(f) =1

b− a

b∫

a

f(x)dx.

126

I. M. Olaru - An integral inequality for convex functions of three order

Definition 1. We say that a function f : [a, b] → R is convex of order three

if for all distinct points x, y, z, t ∈ [a, b] we have the divided difference

[x, y, z, t; f ] ≥ 0.

2. MAIN RESULTS

The result of this paper is given in

Proposition 1.If f, g : [a, b] → R are convex functions by three order on

the interval [a, b], then:

[F (e2)− F (e)2][8F (fg)− 7F (f)F (g)]− 5F (e2)F (f)F (g) ≥

≥ 5[F (e)(F (f)F (eg) + F (g)F (ef))− F (ef)F (eg)]

where e(x) = x, x ∈ [a, b].

Proof. Under our conditions, for all distinct points x, y, z, t ∈ [a, b] we have

[x, y, z, t; f ][x, y, z, t; g] ≥ 0,

which is equivalence with :

[f(x)(y − z) + f(y)(z − t) + f(z)(t− x) + f(t)(x− y)]

(1) ·[(y − z)g(x) + g(y)(z − t) + g(z)(t− x) + g(t)(x− y)] ≥ 0.

We can no make use of the fact that F is a linear positive functional; by

applying successively on (1) the functionals Fx , Fy , Fz , Ft we obtain the

inequality. We remark that

FtFzFyFx(f(x)g(x)(y − z)2) = 2[F (e2)− F (e)2]F (fg)

127

I. M. Olaru - An integral inequality for convex functions of three order

FtFzFyFx(f(x)g(y)(y − z)(z − t)) = [F (e)2 − F (e2)]F (fg)

FtFzFyFx(f(x)g(z)(y − z)(t− x)) = F (e2)F (f)F (g)−

−F (e)[F (ef)F (g) + F (f)F (eg)] + F (ef)F (eg)

References

[1] A. Lupas, An integral inequalitiy for convex functions,Publications de

la Faculte d’Electrotehnique de L’Universite a Belgrad ,No 384(1972),

pp 17-19.

[2] D. S. Mitrinovic, Analytic Inequalities, Beograd (1970)

Ioan Marian Olaru





128



ON A PROBLEM OF A. SHAFIE

Emil. C. Popa

Abstract

In this paper we establish some properties for the convex functions

and we apply our results to the Andersson’s inequality.

2000 Mathematics Subject Classification: 26D15, 26D10.

Key words and phrases: convex functions, Andersson’s inequality.

1. INTRODUCTION

We consider the following question

PROBLEM. Suppose the function f : R → R be twice differentiable,

and for all x ∈ R we have

f ′′(x) > 0, 0 ≤ f(x) ≤ |x|

Then |f ′(x)| < 1 for all x ∈ R. (RGMIA, 08.2008, A. Shafie)

A. Kechriniotis obtain a proof of this problem using the fact that f ′ is

integrable on all [x0, x] ⊂ R. (RGMIA, 09.2008)

Next we consider a convex function f on R with some conditions and

we proof a general property for f in this case.

129

E. C. Popa - On a problem of A. Shafie

2. THE MAIN RESULTS

Lemma 1.If 0 ≤ k1 < k2 and f : [0,∞) → R a convex function with

k1x ≤ f(x) ≤ k2x for all x ∈ [0,∞). Then there exists

limx→∞

f(x)

x= l ∈ (0,∞).

Proof. We have k1 ≤ f(x)

x≤ k2, x > 0 and f(0) = 0. The function

f(x)

xis increasing, hence exists

limx→∞

f(x)

x= l ∈ (0,∞).

Lemma 2.Suppose 0 ≤ k1 ≤ k2 and f : [0,∞) → R a convex function with

k1x ≤ f(x) ≤ k2x for all x ∈ [0,∞). If c ∈ (0,∞) and f is differentiable in

c with f ′(c) 6= l, then there exists xc ∈ (c∞) such that

f ′(x) =f(xc)

xc

.

Proof. Let

g(x) = f(x)− f(c)− (x− c)f ′(c), x ≥ 0.

If y(x)− f(c) = (x− c)f ′(c) we have evidently f(x) ≥ y(x), x ∈ [0,∞).

Hence g(x) ≥ 0 for all x ≥ 0.

But

g(x) = x

[f(x)

x− f ′(c)

]− f(c) + cf ′(c), x > 0

and hence limx→∞

g(x) = ∞.

Now

g(c) = 0 ≤ g(0) < ∞ = limx→∞

g(x).

130


But f is a convex function hence f is continuous and g is continuous on

(0,∞). Using the Darboux property we have a point xc ∈ (c,∞) such that

g(xc) = g(0). Hence

f(xc)− f(c)− (xc − c)f ′(c) = −f(c) + cf ′(c)

and finally

f(xc) = xcf′(c).

By analogy we have

Lemma 3.If 0 ≤ k1 < k2 and f : (−∞, 0] → R is a convex function with

−k1x ≤ f(x) ≤ −k2x for all x ≤ 0, then there exists

limx→∞

f(x)

x= l′ ∈ (−∞, 0)

Lemma 4.If 0 ≤ k1 < k2 and let f : (∞, 0] → R be a convex function with

−k1x ≤ f(x) ≤ −k2x for all x ≤ 0. If c′ ∈ (−∞, 0) and f is differentiable

in c′ with f ′(c′) 6= l′, then there exists xc′ ∈ (−∞, c′) such that

f ′(c′) =f(xc′)

xc′.

We have now the following result

Theorem 1.If 0 ≤ k1 < k2 and f : R → R is a convex function such that

k1|x| ≤ f(x) ≤ k2|x| for all x ∈ R, then we have

(1) −k2 ≤ f ′s(x) ≤ f ′d(x) ≤ k2

for all x ∈ R

131


Proof. We consider x0 > 0 and c > x0 with the conditions of the Lemma

2. Hencef(xc)

xc

= f ′(c).

Butf(xc)

xc

≤ k2 and we have

f ′s(t) ≤ f ′d(t) ≤ f ′(c) ≤ k2

for all t < c.

Next we consider c′ < 0, f is differentiable in c′. Using Lemma 3 and

Lemma 4 then there exists xc′ < c′ < 0 such that

f(xc′)

xc′= f ′(c′).

Butf(xc′)

xc′≥ −k2 and hence

f ′d(t) ≥ f ′s(t) ≥ f ′(c′) ≥ −k2

for all t > c′.

In conclusion

−k2 ≤ f ′s(t) ≤ f ′d(t) ≤ k2

for all t ∈ (c′, c)

and in particular

−k2 ≤ f ′s(x0) ≤ f ′d(x0) ≤ k2.

Finally

−k2 ≤ f ′s(x) ≤ f ′d(x) ≤ k2

for all x > 0.

By analogy

−k2 ≤ f ′s(x) ≤ f ′d(x) ≤ k2

132


for all x < 0.

But for x′ < 0 < x we have

f ′s(x′) ≤ f ′d(x

′) ≤ f ′s(0) ≤ f ′d(0) ≤ f ′s(x) ≤ f ′d(x)

and hence

−k2 ≤ f ′s(x) ≤ f ′d(x) ≤ k2

for all x ∈ R.

3. An Application

B.X. Andersson [1] showed that if the functions fk are convex and in-

creasing in [0, 1] with fk(0) = 0 then

(2)

∫ 1

0

n∏

k=1

fk(x)dx ≥ 2n

n + 1

n∏

k=1

∫ 1

0

fk(x)dx.

An interesting special case of Andersson’s inequality is obtained by

taking all the fk to the same function f , when we get

(3)

∫ 1

0

fn(x)dx ≥ 2n

n + 1

(∫ 1

0

f(x)dx

)n

.

We observe that in the conditions of the PROBLEM of A. Shafie, the

Andersson’s inequality (3) is true.

We consider now 0 ≤ k1 < k2 and f ∈ C1(−∞,∞) is a convex function

with k1|x| ≤ f(x) ≤ k2|x| for all x ∈ (−∞,∞). Using the Theorem 1 we

have |f ′(x)| ≤ k2 for all x ∈ [0, 1].

We observe that 0 ≤ f ′(x) ≤ k2, x ∈ [0, 1] and

∫ 1

0

fn(x)f ′(x)dx ≤ k2

∫ 1

0

fn(x)dx.

133


But ∫ 1

0

fn(x)f ′(x)dx =1

n + 1f (n+1)(1),

and hence

(4)

∫ 1

0

fn(x)dx ≥ 1

k2(n + 1)fn+1(1),

(5)(n + 1)k2

f (n+1)(1)

∫ 1

0

fn(x)dx ≥ 1.

Now ∫ 1

0

f(x)dx ≤ k2

2

and hence

(6)2m

km2

(∫ 1

0

f(x)dx

)m

≤ 1.

Of (5) and (6) we obtain:

(n + 1)k2

fn+1(1)

∫ 1

0

fn(x)dx ≥ 2m

km2

(∫ 1

0

f(x)dx

)m

,

∫ 1

0

fn(x)dx ≥ 2m

n + 1· fn+1(1)

km+12

(∫ 1

0

f(x)dx

)m

.

We have hence the following property

Theorem 2.Let f : [0, 1] → R+ be a convex function with f ∈ C1[0, 1] and

0 ≤ f(x) ≤ kx with k > 0. If m,n ∈ N, m 6= n, then

(7)

∫ 1

0

fn(x)dx ≥ 2m

n + 1· fn+1(1)

km+1

(∫ 1

0

f(x)dx

)m

We observe that (7) is an inequality related with Andersson’s inequality (3).

134


References

[1] Andersson B.X., An inequality for convex functions, Nordisk Mat. Tidsk

6(1958), 25-26.

[2] Fink A.M., Andersson’s inequality and best possible inequalities, JIPAM,

4(3), Art 54, 9p. (2003).

[3] Fink A.M. Andersson’s inequality, Math. Ineq.& Appl., Vol. 6, No.2

(2003) 241-245.

[4] Siretki Gh., Calculus, Ed. II-a, Bucuresti(1982), (in romanian).

Emil C. Popa





135



AN ERROR ANALYSIS FOR A FAMILY OF FOUR-POINT

QUADRATURE FORMULAS

Florin Sofonea, Ana Maria Acu, Arif Rafiq

Abstract

Various error inequalities for a family of four point quadrature

rules are established.

2000 Mathematics Subject Classification: 65D30 , 65D32

Key words and phrases: Quadrature rule, Error inequalities, Numerical

integration.

1. INTRODUCTION

Definition 1.It is called a quadrature formula or formula of numerical in-

tegration, the following formula

I[f ] =

∫ b

a

f(x)dx =m∑

i=0

Aif(xi) +R[f ],

where xi ∈ [a, b], respectively Ai, i = 0,m are called the nodes, respectively

the coefficients of the quadrature formula, and R[f ] is the remainder term.

136

F. Sofonea, A.M. Acu, A. Rafiq - An error analysis for a family ...

We consider the space L2[a, b] with the inert product 〈·, ·〉 defined by

〈f, g〉 =1

b− a

∫ b

a

f(t)g(t)dt.

Denote by ‖f‖ =√〈f, f〉 the norm in this space. The Chebyshev functional

is defined by

T (f, g) = 〈f, g〉 − 〈f, e〉〈g, e〉,

where f, g ∈ L2[a, b] and e = 1. This functional satisfies the pre-Gruss

inequality

(1) T (f, g)2 ≤ T (f, f)T (g, g).

Denote by σ(f ; a, b) =√

(b− a)T (f, f).

In [4], N. Ujevic obtain an optimal 2-point quadrature formula of open

type ∫ 1

−1

f(t)dt = f(√

6− 3)

+ f(3−

√6)

+R[f ],

and establish some error inequalities for this formula.

Theorem 1.[4] Let f : [−1, 1] → R be a function such that f ′ ∈ L1[−1, 1].

If there exists a real number γ1 such that γ1 ≤ f ′(t), t ∈ [−1, 1], then

|R[f ]| ≤ 2(3−√

6)(S − γ1),

and if there exist a real number Γ1 such that f ′(t) ≤ Γ1, t ∈ [−1, 1], then

|R[f ]| ≤ 2(3−√

6)(Γ1 − S),

where S =f(1)− f(−1)

2. If there exist real numbers γ1, Γ1 such that γ1 ≤

f ′(t) ≤ Γ1, t ∈ [−1, 1], then

|R[f ]| ≤(

25

2− 5

√6

)(Γ1 − γ1) .

137


Theorem 2.[4] Let f : [−1, 1] → R be an absolutely continuous function

such that f ′ ∈ L2[−1, 1]. Then

(2) |R[f ]| ≤√

74

3− 10

√6 σ(f ′;−1, 1).

The inequality (2) is sharp in the sense that the constant

√74

3− 10

√6

cannot be replaced by a smaller one.

In [5], F. Zafar and N.A. Mir obtain some similar results for a family

four-point quadrature rules.

∫ 1

−1

f(t)dt =[hf(−1) + (1− h)f(−4 + 4h + 2

√3− 6h + 4h2)(3)

+ (1− h)f(4− 4h− 2√

3− 6h + 4h2) + hf(1)]

+R[f ],

In proof our results we will need the following lemma.

Lemma 1. Let

f(t) =

f1(t), t ∈ [a, x1],

f2(t), t ∈ (x1, x2],

f(3)(t), t ∈ (x2, b],

where x1, x2 ∈ [a, b], x1 < x2, f1 ∈ C1[a, x1], f2 ∈ C1[x1, x2], f3 ∈ C1[x2, b].

If f1(x1) = f2(x2) and f2(x2) = f3(x3) then f is an absolutely continuous

function.

2. MAIN RESULTS

Our goal is to obtain similar error inequalities for a new family of four-

point quadrature rules. The set of values of nodes from quadrature formula

138


obtained in this paper include the set of values of nodes from quadrature

formula (3) obtained by F. Zafar and N.A. Mir.

Theorem 3.If f ∈ [−1, 1] → R is a function such that f ′ ∈ L1[−1, 1], and

h ∈(−∞,

1

3

], then

∫ 1

−1

f(t)dt = hf(−1)+(1−h)f

(−

√3h−1

3(h−1)

)+(1−h)f

(√3h−1

3(h−1)

)(4)

+ hf(1) +R[f ],

R[f ] = −∫ 1

−1

p1(t)f′(t)dt,

where

(5) p1(t) =

t− h + 1, t ∈[−1,−

√3h− 1

3(h− 1)

],

t, t ∈(−

√3h− 1

3(h− 1),

√3h− 1

3(h− 1)

),

t− 1 + h, t ∈[√

3h− 1

3(h− 1), 1

].

If there exist real numbers γ, Γ such that γ ≤ f ′(t) ≤ Γ, t ∈ [−1, 1], then

(6) |R[f ]| ≤ K1(h)Γ− γ

2,

where

K1(h)=

1

3(h−1)

[−6h2+15h−5+2

√3(h−1)2 ·

√3h−1

3(h−1)

], h ∈ (−∞, 0]

1

3(h−1)

[6h3−12h2+15h−5+2

√3(h−1)2

√3h−1

3(h−1)

], h ∈

(0,

1

3

]

139


If there exist a real numbers γ such that γ ≤ f ′(t), t ∈ [−1, 1], then

(7) |R[f ]| ≤ 2K2(h)(S − γ),

and if there exist a real number Γ, such that f ′(t) ≤ Γ, t ∈ [−1, 1], then

(8) |R[f ]| ≤ 2K2(h)(Γ− S),

where S =f(1)− f(−1)

2and

K2(h) =

−√

3h− 1

3(h− 1)− h + 1, h ∈ (−∞, c1] ∪ [c2, 1/3],

√3h− 1

3(h− 1), h ∈ (c1, c2),

c1 ∈ (−1/2,−1/3), c2 ∈ (1/5, 1/4).

Proof. The relation (4) can be obtained integrating by parts

R[f ] = −∫ 1

−1

f ′(t)p1(t)dt. Since

(9)

∫ 1

−1

p1(t)dt = 0,

we obtain

(10)

∫ 1

−1

(f ′(t)− Γ + γ

2

)p1(t)dt =

∫ 1

−1

f ′(t)p1(t)dt = −R[f ].

From relation (10) we have

|R[f ]| =

∣∣∣∣∫ 1

−1

(f ′(t)− Γ + γ

2

)p1(t)dt

∣∣∣∣

≤∥∥∥∥f ′ − Γ + γ

2

∥∥∥∥∞·∫ 1

−1

|p1(t)| dt ≤ K1(h)Γ− γ

2,

140


since ∥∥∥∥f ′ − Γ + γ

2

∥∥∥∥∞≤ Γ− γ

2

and

∫ 1

−1

|p1(t)| dt =

∫ h−1

−1

(−t+h−1)dt+

∫ −q

3h−13(h−1)

h−1

(t−h+1)dt+2

∫ q3h−1

3(h−1)

0

tdt

+

∫ 1−h

q3h−1

3(h−1)

(−t + 1− h)dt +

∫ 1

1−h

(t− 1 + h)dt

=1

3(h−1)

[6h3−12h2+15h−5+2

√3(h−1)2

√3h−1

h−1

], for h∈

(0,

1

3

]

∫ 1

−1

|p1(t)| dt =

∫ −q

3h−13(h−1)

−1

(t−h+1)dt+2

∫ q3h−1

3(h−1)

0

tdt+

∫ 1

q3h−1

3(h−1)

(−t+1−h)dt

=1

3(h−1)

[−6h2+15h−5+2

√3(h−1)2

√3h−1

h−1

], for h ∈ (−∞, 0] .

From relation (9) we have

|R[f ]|=∣∣∣∣∫ 1

−1

f ′(t)p1(t)dt

∣∣∣∣=∣∣∣∣∫ 1

−1

(f ′(t)−γ)p1(t)dt

∣∣∣∣≤∫ 1

−1

|f ′(t)−γ| |p1(t)| dt

≤ supt∈[−1,1]

|p1(t)| ·∫ 1

−1

(f ′(t)− γ) dt = 2(S − γ) · supt∈[−1,1]

|p1(t)| .

Since

supt∈[−1,1]

|p1(t)| =

−√

3h− 1

3(h− 1)− h + 1, h ∈ (−∞, c1] ∪ [c2, 1/3],

√3h− 1

3(h− 1), h ∈ (c1, c2),

where c1 ∈ (−1/2,−1/3), c2 ∈ (1/5, 1/4), we obtain the relation (7).

In a similar way we can obtain the relation (8).

141


Theorem 4.Let f : [−1, 1] → R be an absolutely continuous function such

that f ′ ∈ L2[−1, 1]. Then

(11) |R[f ]| ≤√

∆(h)σ(f ′;−1, 1),

where ∆(h) = −2

3

√3

√3h− 1

h− 1(h−1)2 +2h2−4h+

4

3. The inequality (11) is

sharp in the sense that the constant√

∆(h) cannot be replaced by a smaller

one.

Proof. Let p1be defined by (5). We have

〈p1, f′〉 =

1

2

∫ 1

−1

p1(t)f′(t)dt = −1

2R[f ].

On the other hand, we have 〈p1, f′〉 = T (f ′, p1), since 〈p1, e〉 = 0.

From relation (1) it follows

|T (f ′, p1)|≤√

T (f ′, f ′)√

T (p1, p1)=1

2‖p1‖2 σ(f ′;−1, 1)=

1

2

√∆(h)σ(f ′;−1, 1),

namely |R[f ]| =√

∆(h)σ(f ′;−1, 1).

We have to prove that this inequality is sharp. For that purpose, we

define the function

f(t)=

1

2t2 − th + t +

1

2− h, t ∈

[−1,−

√3h− 1

3(h− 1)

]

1

6

(−6h+2

√3(h−1)

√3h−1

h−1+3t2+3

), t∈

(−√

3h−1

3(h−1),

√3h−1

3(h−1)

]

1

2t2 + th− t +

1

2− h, t ∈

(√3h− 1

3(h− 1), 1

]

We remark that the function f ′(t) = p1(t). From Lemma 1 we see that

the function f is an absolutely continuous function. For this function the

142


left-hand side of (11) become

L.H.S.(11) = ∆(h).

The right-hand side of (11) becomes

R.H.S(11) = ∆(h).

We see that L.H.S.(11) = R.H.S(11). Thus, (11) is sharp.

Remark 1.For h = 0 we obtain the Gauss two point quadrature formula

∫ 1

−1

f(t)dt = f

(−√

3

3

)+ f

(√3

3

)+R[f ],

where

(12) |R[f ]| ≤(−2

3

√3 +

4

3

)σ(f ′;−1, 1) ' 0.1786 σ(f ′;−1, 1).

Remark 2.For h =1

6we get Lobatto four-point quadrature rule as follows

∫ 1

−1

f(t)dt =1

6

[f(−1) + 5f

(−√

5

5

)+ 5f

(√5

5

)+ f(1)

]+R[f ],

where

(13) |R[f ]| ≤(− 5

18

√5 +

13

18

)σ(f ′;−1, 1) ' 0.1011 σ(f ′;−1, 1).

Remark 3.For h =1

4we get 3/8 Simpson ′ s rule as follows:

∫ 1

−1

f(t)dt =1

4

[f(−1) + 3f

(−1

3

)+ 3f

(1

3

)+ f(1)

]+R[f ],

where

(14) |R[f ]| ≤ 1

12σ(f ′;−1, 1) ' 0.0833 σ(f ′;−1, 1).

143


Remark 4.The estimates (12), (13), respectively (14) may be considered

estimates of Gauss two-point, Simpson ′s 3/8, respectively Lobatto four-point

quadrature rule for the class of functions

H = f : [−1, 1] → R, fabsolutely continuous function, f ′ ∈ L2[−1, 1].

References

[1] A. M. Acu, A. Babos, An error analysis for a quadrature formula,

The 14-the International Conference The Knowledge-based organization

Technical Sciences Computer Science, Modelling & Simulation and E-

learning Technologies. Physics, Mathematics and Chemistry. Conference

Proceedings 8, ISSN: 1843-6722, 290-298

[2] Adrian Branga, Development in series of orthogonal polynomials with

applications in optimization, General Mathematics, vol. 15, no. 1, pp

101-110, 2007.

[3] A. Lupas , C. Manole , Capitole de Analiza Numerica , Universitatea din

Sibiu , Colectia Facultatii de Stiinte-Seria Matematica 3 , Sibiu 1994.

[4] N. Ujevic, Error inequalities for a quadrature formula of open type, Re-

vista Colombiana de Matematicas, Volumen 37 (2003), 93-105.

[5] F. Zafar, N.A.Mir, Some generalized error inequalities and applica-

tions, Journal of Inequalities and Applications, Volume 2008, Article

ID 845934, 15 pages.

144


Florin Sofonea, Ana Maria Acu





[email protected]

Arif Rafiq

COMSATS Institute of Information Technology


Defense Road, Off Raiwind Road, Lahore - Pakistan


145



A PROOF OF AN INEQUALITY

Ioan Tincu, Gheorghe Sandru

Abstract

In this paper we show that if f ∈ Πn and |f(x)| ≤ A for all

x ∈ [−1, 1], then

|f(x)| ≤ A|Tn(x)| for all x ∈ R \ [−1, 1],

where Tn is Cebısev polynomial of first kind.


Key words and phrases: Jacobi, Chebychev, interpolation.

1. INTRODUCTION

Let us use the following notation:

- R(α,β)n (x) =

n∑

k=0

(−1)k

n

k

(n + α + β + 1)k

(α + 1)k

(1− x

2)k is Jacobi poly-

nomial where α, β > −1, and (x)k = x(x + 1) . . . (x + k − 1) is Pochmaier

symbol.

- Tn(x) = R(− 1

2,− 1

2)

n (x) is Chebychev polynomial of first kind.

- Un(x) = R( 12, 12)

n (x) denotes Chebychev polynomial of second kind.

146

I. Tincu, G. Sandru - A proof of an inequality

- The points xk = cos kπn

, k = 0, 1, ..., n are the roots of polynomial

ω(x) = (x2 − 1)Un−1(x)

- Ln(x0, x1, . . . , xn; f |x) =n∑

k=0

ω(x)

(x− xk)ω′(xk)f(xk) denotes Lagrange

polynomial which interpolates f : [−1, 1] → R at points xk ∈ [−1, 1], k =

0, 1, . . . , n.

2. PRINCIPAL RESULTS

Proposition 1.If f is a polynomial of degree ≤ n such that

|f(x)| ≤ A,A > 0, (∀)x ∈ [−1, 1]

then

|f(x)| ≤ A|Tn(x)|, (∀)x ∈ R\[−1, 1]

Proof. If f is an arbitrary polynomial of degree ≤ n according to Lagrange

interpolation formula we have

f(x) = Ln(x0, x1, ..., xn; f |x) =n∑

k=0

ω(x)

(x− x0)ω′(xk)f(xk)

We have:

ω′(xk) = (x2k − 1)U ′

n−1(xk), k = 1, 2, ..., n− 1,

U ′n−1(xk) = (−1)k

x2k−1

,

ω′(1) = 2, ω′(−1) = 2 · (−1)n,

(1)

f(x) = ω(x)

[n−1∑

k=1

f(xk)

(x− xk)(x2k − 1)U ′

n−1(xk)+

f(1)

2(x− 1)+

(−1)nf(−1)

2(x + 1)

]

147


The polynomials Tn, Un−1 verify:

(2)

T ′n(x) = n2Un−1(x),

(1− x2)T ′′n (x)− xT ′

n(x) + n2Tn(x) = 0

Tn(x) = xUn−1(x) + (x2 − 1)U ′n−1(x)

From (1) obtain:

(3) |f(x)| ≤ A|(x2 − 1)Un−1(x)|[

n−1∑

k=1

1

|x− xk| +1

2|1− x| +1

2|1 + x|

]

Let x > 1. Then, from (3) results

|f(x)| ≤ A(x2 − 1)Un−1(x)

[n−1∑

k=1

1

x− xk

+x

x2 − 1

]

= A(x2 − 1)Un−1(x) ·[U ′

n−1(x)

Un−1(x)+

x

x2 − 1

],

|f(x)| ≤ A[(x2 − 1)U ′n−1(x) + xUn−1(x)].

From (2) results

|f(x)| ≤ ATn(x), (∀)x > 1.

Let x < −1. In (3), for x = −t, t > 1 we obtain

|f(−t)| ≤ A(t2 − 1)|(−1)nUn−1(t)

[n−1∑

k=1

1

t + xk

+t

t2 − 1

]

= A(t2 − 1)Um−1(t)

[−U ′

n−1(−t)

Un−1(−t)+

t

t2 − 1

]

= A[(t2 − 1)U ′n−1(t) + tUn−1(t)] = ATn(t) = A(−1)nTn(−t),

|f(x)| ≤ A(−1)nTn(x) = ATn(|x|)An = A|Tn(x)| for x < −1,

which completes the proof.

By means of a linear transformation we find as a corollary following.

148


Proposition 2.If f is a polynomial of degree ≤ n such that

|f(x)| ≤ A, A > 0, (∀)x ∈ [0, 1]

then |f(x)| ≤ A|Tn(2x− 1)|, (∀)x ∈ R\[0, 1].

References

[1] A. Lupas, C. Manole, Capitole de Analiza Numerica, Ed. Univ.”Lucian

Blaga”, Sibiu, 1994

[2] G. Szego, Orthogonal Polynomials, American Mathematical Colloquium

Publications, New York, 1939

Ioan Tincu





Gheorghe Sandru

Scoala Generala Vistea de Jos

Brasov, Romania

149

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