proceedings of the fifth international symposium...
TRANSCRIPT
”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
Department of Mathematics
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
”Lucian Blaga” University
from Sibiu, Romania
2Mugur Acu
”Lucian Blaga” University
from Sibiu, Romania
3Dumitru Acu
”Lucian Blaga” University
of Sibiu, Romania
4Adrian Branga
adrian [email protected]
”Lucian Blaga” University
from Sibiu, Romania
5Daniel Breaz
”1 Decembrie 1918” University
of Alba-Iulia, Romania
6Nicoleta Breaz
”1 Decembrie 1918” University
of Alba-Iulia, Romania
7Amelia Bucur
”Lucian Blaga” University
from Sibiu, Romania
8Eugen Constantinescu
”Lucian Blaga” University
from Sibiu, Romania
9Daniela Dicu
Liceul Teoretic ” Gheorghe Lazar”
Avrig, Romania
10Gheorghe Dicu Gr. Sc. Forestier
Curtea de Arges, Romania
11Petrica Dicu
”Lucian Blaga” University
of Sibiu, Romania
12Irina Dorca
”Lucian Blaga” University
of Sibiu, Romania
13Eugen Draghici
”Lucian Blaga” University
from Sibiu, Romania
14Ali Ebadian
Urmia University,
Iran
6
15Bogdan Gavrea
”Tehnical University”
of Cluj-Napoca, Romania
16Ioan Gavrea
”Tehnical University”
of Cluj-Napoca, Romania
17Heiner Gonska
University of Duisburg-Essen
Germania
18Jose Luis Lopez-Bonilla
joseluis.lopezbonilla@gm
Escuela Superior de Ingenieria
Mecanica y Electrica Ins, Mexic
19Vasile Mihesan
”Tehnical University”
of Cluj-Napoca, Romania
20Nicusor Minculete
University ”Dimitrie Cantemir”
of Brasov, Romania
21Shahram Najafzadeh
University of Maragheh,
Iran
22Marian Olaru
”Lucian Blaga” University
from Sibiu, Romania
23Stefan Poka Gheorghe Sincai High Scholl
Cluj-Napoca, Romania
24Emil C. Popa
”Lucian Blaga” University
from Sibiu, Romania
25Ioan Popa
ioanpopa [email protected]
Edmond Nicolau College
Cluj-Napoca, Romania
26Luminita Preoteasa Gr. Sc. Forestier
Curtea de Arges, Romania
27Arif Rafiq
COMSATS Institute of
Information Technology
Lahore, Pakistan
28Sofonea Florin
”Lucian Blaga” University
from Sibiu, Romania
29Gheorghe Sandru Scoala Generala Vistea de Jos
Brasov, Romania
30Doru Stefanescu
University of Bucharest
Romania
31Ioan Tincu
”Lucian Blaga” University
from Sibiu, Romania
32Andrei Vernescu
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
”Mathematical Inequalities” Sibiu, 25 - 27 September 2008
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
A.M. Acu, M. Acu, A. Rafiq - Some inequalities of Ostrowski type...
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
w(t)f(t)dt− (b− a)3
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
A.M. Acu, M. Acu, A. Rafiq - Some inequalities of Ostrowski type...
∫ 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
w(t)f(t)dt− (b− a)3
12(f(a) + f(b)) ≥ 1
12(b− a)4(γ − S).
In a similar way we can prove that
(4)
∫ b
a
w(t)f(t)dt− (b− a)3
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
A.M. Acu, M. Acu, A. Rafiq - Some inequalities of Ostrowski type...
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
A.M. Acu, M. Acu, A. Rafiq - Some inequalities of Ostrowski type...
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
w(t)f(t)dt− (b−a)3
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
A.M. Acu, M. Acu, A. Rafiq - Some inequalities of Ostrowski type...
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
w(t)f(t)dt− (b−a)3
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
w(t)f(t)dt− (b−a)3
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
A.M. Acu, M. Acu, A. Rafiq - Some inequalities of Ostrowski type...
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
A.M. Acu, M. Acu, A. Rafiq - Some inequalities of Ostrowski type...
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
A.M. Acu, M. Acu, A. Rafiq - Some inequalities of Ostrowski type...
Theorem 7. Let f : [a, b] → R be continuous on [a, b] and differentiable
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
A.M. Acu, M. Acu, A. Rafiq - Some inequalities of Ostrowski type...
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
A.M. Acu, M. Acu, A. Rafiq - Some inequalities of Ostrowski type...
This complete the proof.
In a similar way we can obtain the following results.
Theorem 8. Let f : [a, b] → R be continuous on [a, b] and differentiable
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
A.M. Acu, M. Acu, A. Rafiq - Some inequalities of Ostrowski type...
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
A.M. Acu, M. Acu, A. Rafiq - Some inequalities of Ostrowski type...
[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
Department of Mathematics
Str. Dr. I. Ratiu, No. 5-7, 550012 - Sibiu, Romania
E-mail: [email protected],
Arif Rafiq
COMSATS Institute of Information Technology
Department of Mathematics
Defense Road, Off Raiwind Road, Lahore - Pakistan
E-mail: [email protected]
22
Proceedings of the Fifth International Symposium
”Mathematical Inequalities” Sibiu, 25 - 27 September 2008
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
D. Acu - Some interesting elementary inequalities
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
D. Acu - Some interesting elementary inequalities
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
D. Acu - Some interesting elementary inequalities
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
D. Acu - Some interesting elementary inequalities
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
D. Acu - Some interesting elementary inequalities
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
D. Acu - Some interesting elementary inequalities
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
D. Acu - Some interesting elementary inequalities
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
D. Acu - Some interesting elementary inequalities
[4] L. Panaitopol, V. Bandila, M. Lascu, Inequalities, Ed. GIL, 1995 (in
Romanian)
Dumitru Acu
University ”Lucian Blaga” of Sibiu
Department of Mathematics
Str. Dr. I. Ratiu, No. 5-7, 550012 - Sibiu, Romania
E-mail: acu [email protected]
32
Proceedings of the Fifth International Symposium
”Mathematical Inequalities” Sibiu, 25 - 27 September 2008
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
A. Branga - An inequality for generalized spline functions
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
A. Branga - An inequality for generalized spline functions
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
A. Branga - An inequality for generalized spline functions
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
A. Branga - An inequality for generalized spline functions
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
A. Branga - An inequality for generalized spline functions
[4] Gh. Micula, S. Micula, Handbook of splines, Kluwer Acad. Publ.,
Dordrecht-Boston-London, 1999.
Adrian Branga
University ”Lucian Blaga” of Sibiu
Department of Mathematics
Str. Dr. I. Ratiu, No. 5-7, 550012 - Sibiu, Romania
E-mail: adrian [email protected]
39
Proceedings of the Fifth International Symposium
”Mathematical Inequalities” Sibiu, 25 - 27 September 2008
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
D. Breaz, N. Breaz - Some starlikeness conditions proved by inequalities
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
D. Breaz, N. Breaz - Some starlikeness conditions proved by inequalities
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
D. Breaz, N. Breaz - Some starlikeness conditions proved by inequalities
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,
where FΣ is defined in (1).
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,
where FΣ is defined in (1).
Proof. This corollary is obtained if we take a = 0, b = −α in Theorem 1.
44
D. Breaz, N. Breaz - Some starlikeness conditions proved by inequalities
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
D. Breaz, N. Breaz - Some starlikeness conditions proved by inequalities
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
Department of Mathematics
Alba Iulia, Romania
E-mail: [email protected],
46
Proceedings of the Fifth International Symposium
”Mathematical Inequalities” Sibiu, 25 - 27 September 2008
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.
2000 Mathematics Subject Classification: 30C45
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
I. Dorca - Note on subclass of β-starlike and β-convex functions ...
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
I. Dorca - Note on subclass of β-starlike and β-convex functions ...
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
I. Dorca - Note on subclass of β-starlike and β-convex functions ...
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
I. Dorca - Note on subclass of β-starlike and β-convex functions ...
(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
I. Dorca - Note on subclass of β-starlike and β-convex functions ...
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
I. Dorca - Note on subclass of β-starlike and β-convex functions ...
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
I. Dorca - Note on subclass of β-starlike and β-convex functions ...
=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
I. Dorca - Note on subclass of β-starlike and β-convex functions ...
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
I. Dorca - Note on subclass of β-starlike and β-convex functions ...
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
I. Dorca - Note on subclass of β-starlike and β-convex functions ...
|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
I. Dorca - Note on subclass of β-starlike and β-convex functions ...
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
I. Dorca - Note on subclass of β-starlike and β-convex functions ...
[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
I. Dorca - Note on subclass of β-starlike and β-convex functions ...
Irina Dorca
University ”Lucian Blaga” of Sibiu
Department of Mathematics
Str. Dr. I. Ratiu, No. 5-7, 550012 - Sibiu, Romania
E-mail: ira [email protected]
61
Proceedings of the Fifth International Symposium
”Mathematical Inequalities” Sibiu, 25 - 27 September 2008
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
B. Gavrea - On some inequality for convex functions
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
B. Gavrea - On some inequality for convex functions
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
B. Gavrea - On some inequality for convex functions
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
Department of Mathematics
Cluj-Napoca, Romania
E-mail: [email protected]
66
Proceedings of the Fifth International Symposium
”Mathematical Inequalities” Sibiu, 25 - 27 September 2008
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
I. Gavrea - On some inequalities for convex functions of higher order
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
I. Gavrea - On some inequalities for convex functions of higher order
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
I. Gavrea - On some inequalities for convex functions of higher order
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
I. Gavrea - On some inequalities for convex functions of higher order
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
I. Gavrea - On some inequalities for convex functions of higher order
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
I. Gavrea - On some inequalities for convex functions of higher order
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
I. Gavrea - On some inequalities for convex functions of higher order
Ioan Gavrea
Technical University of Cluj-Napoca
Department of Mathematics
Cluj-Napoca, Romania
E-mail: [email protected]
75
Proceedings of the Fifth International Symposium
”Mathematical Inequalities” Sibiu, 25 - 27 September 2008
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
H. Gonska, I. Rasa - A Voronovskaya estimate with second order modulus
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
H. Gonska, I. Rasa - A Voronovskaya estimate with second order modulus
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
H. Gonska, I. Rasa - A Voronovskaya estimate with second order modulus
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
H. Gonska, I. Rasa - A Voronovskaya estimate with second order modulus
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
H. Gonska, I. Rasa - A Voronovskaya estimate with second order modulus
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.
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H. Gonska, I. Rasa - A Voronovskaya estimate with second order modulus
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
H. Gonska, I. Rasa - A Voronovskaya estimate with second order modulus
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
H. Gonska, I. Rasa - A Voronovskaya estimate with second order modulus
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
H. Gonska, I. Rasa - A Voronovskaya estimate with second order modulus
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
H. Gonska, I. Rasa - A Voronovskaya estimate with second order modulus
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
H. Gonska, I. Rasa - A Voronovskaya estimate with second order modulus
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
H. Gonska, I. Rasa - A Voronovskaya estimate with second order modulus
[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
H. Gonska, I. Rasa - A Voronovskaya estimate with second order modulus
[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
Department of Mathematics
D-47048 Duisburg, Germany
E-mail: [email protected]
Ioan Rasa
Technical University
Department of Mathematics
RO-400020 Cluj-Napoca, Romania
E-mail: [email protected]
90
Proceedings of the Fifth International Symposium
”Mathematical Inequalities” Sibiu, 25 - 27 September 2008
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
V. Mihesan - Popoviciu type inequalities ...
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
V. Mihesan - Popoviciu type inequalities ...
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
V. Mihesan - Popoviciu type inequalities ...
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
V. Mihesan - Popoviciu type inequalities ...
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
V. Mihesan - Popoviciu type inequalities ...
Vasile Mihesan
Technical University of Cluj-Napoca
Department of Mathematics
Str. C.Daicoviciu nr.15, 400020 Cluj-Napoca, Romania
E-mail: [email protected]
97
Proceedings of the Fifth International Symposium
”Mathematical Inequalities” Sibiu, 25 - 27 September 2008
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
V. Mihesan - Rado type inequalities for weighted power pseudo means
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
V. Mihesan - Rado type inequalities for weighted power pseudo means
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
V. Mihesan - Rado type inequalities for weighted power pseudo means
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
V. Mihesan - Rado type inequalities for weighted power pseudo means
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
V. Mihesan - Rado type inequalities for weighted power pseudo means
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
V. Mihesan - Rado type inequalities for weighted power pseudo means
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
V. Mihesan - Rado type inequalities for weighted power pseudo means
[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
Technical University of Cluj-Napoca
Department of Mathematics
Str. C.Daicoviciu nr.15, 400020 Cluj-Napoca, Romania
E-mail: [email protected]
106
Proceedings of the Fifth International Symposium
”Mathematical Inequalities” Sibiu, 25 - 27 September 2008
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
N. Minculete - Several inequalities about arithmetic functions...
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
N. Minculete - Several inequalities about arithmetic functions...
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
N. Minculete - Several inequalities about arithmetic functions...
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
N. Minculete - Several inequalities about arithmetic functions...
≥ 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).
Theorem 2.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, 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
N. Minculete - Several inequalities about arithmetic functions...
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
Theorem 3.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 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
N. Minculete - Several inequalities about arithmetic functions...
=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
N. Minculete - Several inequalities about arithmetic functions...
[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
E-mail: [email protected]
115
Proceedings of the Fifth International Symposium
”Mathematical Inequalities” Sibiu, 25 - 27 September 2008
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
N. Minculete, P. Dicu - Inequalities between some arithmetic functions
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
N. Minculete, P. Dicu - Inequalities between some arithmetic functions
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
N. Minculete, P. Dicu - Inequalities between some arithmetic functions
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
N. Minculete, P. Dicu - Inequalities between some arithmetic functions
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
N. Minculete, P. Dicu - Inequalities between some arithmetic functions
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
N. Minculete, P. Dicu - Inequalities between some arithmetic functions
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
N. Minculete, P. Dicu - Inequalities between some arithmetic functions
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
N. Minculete, P. Dicu - Inequalities between some arithmetic functions
[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
E-mail: [email protected]
Petrica Dicu
University ”Lucian Blaga” of Sibiu
Department of Mathematics
Str. Dr. I. Ratiu, Nr. 5–7, 550012 - Sibiu, Romania
E-mail: [email protected]
125
Proceedings of the Fifth International Symposium
”Mathematical Inequalities” Sibiu, 25 - 27 September 2008
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
University ”Lucian Blaga” of Sibiu
Department of Mathematics
Str. Dr. I. Ratiu, Nr. 5–7, 550012 - Sibiu, Romania
E-mail: [email protected]
128
Proceedings of the Fifth International Symposium
”Mathematical Inequalities” Sibiu, 25 - 27 September 2008
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
E. C. Popa - On a problem of A. Shafie
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
E. C. Popa - On a problem of A. Shafie
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
E. C. Popa - On a problem of A. Shafie
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
E. C. Popa - On a problem of A. Shafie
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
E. C. Popa - On a problem of A. Shafie
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
University ”Lucian Blaga” of Sibiu
Department of Mathematics
Str. Dr. I. Ratiu, Nr. 5–7, 550012 - Sibiu, Romania
E-mail: [email protected]
135
Proceedings of the Fifth International Symposium
”Mathematical Inequalities” Sibiu, 25 - 27 September 2008
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
F. Sofonea, A.M. Acu, A. Rafiq - An error analysis for a family ...
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
F. Sofonea, A.M. Acu, A. Rafiq - An error analysis for a family ...
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
F. Sofonea, A.M. Acu, A. Rafiq - An error analysis for a family ...
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
F. Sofonea, A.M. Acu, A. Rafiq - An error analysis for a family ...
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).
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F. Sofonea, A.M. Acu, A. Rafiq - An error analysis for a family ...
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
F. Sofonea, A.M. Acu, A. Rafiq - An error analysis for a family ...
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).
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F. Sofonea, A.M. Acu, A. Rafiq - An error analysis for a family ...
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
F. Sofonea, A.M. Acu, A. Rafiq - An error analysis for a family ...
Florin Sofonea, Ana Maria Acu
University ”Lucian Blaga” of Sibiu
Department of Mathematics
Str. Dr. I. Ratiu, No. 5-7, 550012 - Sibiu, Romania
E-mail: [email protected],
Arif Rafiq
COMSATS Institute of Information Technology
Department of Mathematics
Defense Road, Off Raiwind Road, Lahore - Pakistan
E-mail: [email protected]
145
Proceedings of the Fifth International Symposium
”Mathematical Inequalities” Sibiu, 25 - 27 September 2008
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.
2000 Mathematics Subject Classification: 33C05
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
I. Tincu, G. Sandru - A proof of an inequality
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.
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I. Tincu, G. Sandru - A proof of an inequality
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
University ”Lucian Blaga” of Sibiu
Department of Mathematics
Str. Dr. I. Ratiu, Nr. 5–7, 550012 - Sibiu, Romania
E-mail: [email protected]
Gheorghe Sandru
Scoala Generala Vistea de Jos
Brasov, Romania
149