paper 01-18-2009. 94KB Jun 04 2011 12:03:06 AM
ACTA UNIVERSITATIS APULENSIS
No 18/2009
THE INTEGRAL OPERATOR ON THE SH (β) CLASS
Daniel Breaz and Nicoleta Breaz1
Abstract. In this paper we present a convexity condition for a integral
operator F defined in formula (2) on the class SH (β).
2000 Mathematics Subject Classification: 30C45.
Keywords and phrases: Univalent functions, integral operator, convex functions.
1. 1.Introduction
Let U = {z ∈ C, |z| < 1} be the unit disc of the complex plane and denote by H (U ), the class of the olomorphic functions in U. Consider A =
{f ∈ H (U ) , f (z) = z + a2 z 2 + a3 z 3 + ..., z ∈ U } be the class of analytic functions in U and S = {f ∈ A : f is univalent in U }.
Denote with K the class of the olomorphic functions in U with f (0) =
f ′ (0) − 1 = 0, where is convex functions in U , defined by
zf ′′ (z)
+ 1 > 0, z ∈ U .
K = f ∈ H (U ) ; f (0) = f ′ (0) − 1 = 0, Re
f ′ (z)
(
(
)
)
In the paper (3) J. Stankiewicz and A. Wisniowska introduced the class of
univalent functions SH (β), define by the next inequality:
)
(
zf ′ (z)
√
√
√ zf ′ (z)
−
2β
2
−
1
2
+
2β
2
−
1
,
< Re
f (z)
f (z)
for some β > 0 and for all z ∈ U .
1
Supported by the GAR 14/2006.
7
(1)
D.Breaz, N. Breaz - The integral operator on the SH (β) class
We consider the integral operator
F (z) =
Z
f1 (t)
t
z
0
!α1
fn (t)
· ... ·
t
!αn
(2)
dt
and we study your properties.
Remark. We observe that for n = 1 and α1 = 1 we obtain the integral
operator of Alexander.
2.Main results
Theorem 1.Let αi , i ∈ {1, ..., n} the real numbers with the properties αi >
0 for i ∈ {1, ..., n} and
√
n
X
2
√
√ .
(3)
αi ≤
2β
2
−
1
+ 2
i=1
We suppose that the functions fi ∈ SH (β) for i = {1, ..., n} and β > 0. In
this conditions the integral operator defined in (2) is convex.
Proof. We calculate for F the derivatives of the first and second order.
From (2) we obtain:
′
F (z) =
f1 (z)
z
!α1
fn (z)
· ... ·
z
!αn
and
′′
F (z) =
n
X
i=1
αi
fi (z)
z
!αi −1
n
fj (z)
zfi′ (z) − fi (z) Y
zfi (z)
z
j=1
!
!αj
j6=i
F ′′ (z)
zf1′ (z) − f1 (z)
zfn′ (z) − fn (z)
=
α
+
...
+
α
.
1
n
F ′ (z)
zf1 (z)
zfn (z)
!
!
fn′ (z) 1
f1′ (z) 1
F ′′ (z)
+
...
+
α
.
=
α
−
−
n
1
F ′ (z)
f1 (z) z
fn (z) z
!
Multiply the relation (4) with z we obtain:
8
!
(4)
D.Breaz, N. Breaz - The integral operator on the SH (β) class
n
n
n
X
zfi′ (z) X
zfi′ (z)
zF ′′ (z) X
αi
αi
αi .
=
−1 =
−
F ′ (z)
fi (z)
fi (z)
i=1
i=1
i=1
!
(5)
The relation (5) is equivalent with
n
n
X
zfi′ (z) X
zF ′′ (z)
α
αi + 1.
+
1
=
−
i
F ′ (z)
fi (z)
i=1
i=1
√
We multiply the relation (6) with 2 and obtain:
√
n
n √
√
X
zF ′′ (z)
zfi′ (z) √ X
+
1
=
−
αi + 2.
2
2α
2
i
′
F (z)
fi (z)
i=1
i=1
!
(6)
(7)
The equality (7) is equivalent with:
√
√ zF ′′ (z)
√ zf ′ (z)
2 F ′ (z) + 1 = α1 2 f11(z) + 2α1 β
2 − 1 + ...
√
√ ′ (z)
n
+αn 2 zf
+
2α
β
2
−
1
−
n (z)
f√
n√ P
√
Pn
n
− i=1 2αi β
2 − 1 − 2 i=1 αi + 2.
We calculate the real part from both terms of the above equality and obtain:
n√
′′
o
√
√
zf1′ (z)
(z)
+
1
=
α
Re
2Re zF
2
+
2β
2
−
1
+ ...
F ′ (z)
n√1
o f1 (z)√
zfn′ (z)
2
2−1 −
+αn Re
+ 2β
√ fn (z) √ P
√
Pn
n
− i=1 2αi β
2 − 1 − 2 i=1 αi + 2.
Because fi ∈ SH (β) for i = {1, ..., n} we apply in the above relation the
inequality (1) and obtain:
√
′
√
zf (z)
+ 1 > α1 f11(z) − 2β
2 − 1 + ...
′
√
n (z)
+αn zf
2
−
1
−
−
2β
fn (z)
√
√ Pn
√
Pn
− i=1 2αi β
2 − 1 − 2 i=1 αi + 2.
2Re
′
zf (z)
zF ′′ (z)
F ′ (z)
Because αi fii(z) − 2β
√
√
2 − 1 > 0 for all i ∈ {1, ..., n}, obtain that
n
n
√ X
√
√
X
zF ′′ (z)
αi + 2.
2Re
2
−
1
−
2
2α
β
+
1
>
−
i
′
F (z)
i=1
i=1
!
9
(8)
D.Breaz, N. Breaz - The integral operator on the SH (β) class
Using the hypothesis (3) in (8) we have:
zF ′′ (z)
Re
+1 >0
F ′ (z)
!
(9)
so, F is the convex function.
Corollary 2. Let α the real numbers with the properties 0 < α ≤
√
2
√ .
2β ( 2−1)+ 2
√
We suppose that the functions
f ∈ SH (β) and β > 0. In this conditions the
R z f (z) α
dt is convex.
integral operator F (z) = 0 z
Proof. In the Theorem 1, we consider n = 1, α1 = α and f1 = f .
References
[1] M. Acu, Close to convex functions associated with some hyperbola, Libertas Mathematica, Tomus XXV (2005), 109-114.
[2] D. Breaz, N. Breaz, Two integral operators, Studia Universitatis Babe¸s
-Bolyai, Mathematica, Cluj Napoca, No. 3-2002,pp. 13-21.
[3] J. Stankiewicz, A. Wisniowska, Starlike functions associated with some
hyperbola, Folia Scientiarum Universitatis Tehnicae Resoviensis 147, Matematyka 19(1996), 117-126.
Authors:
Daniel Breaz
Department of Mathematics
” 1 Decembrie 1918 ” University of Alba Iulia
Address: Alba Iulia, Str. N. Iorga, No. 11-13, 510009, Alba
email:[email protected]
Nicoleta Breaz
Department of Mathematics
” 1 Decembrie 1918 ” University of Alba Iulia
Address: Alba Iulia, Str. N. Iorga, No. 11-13, 510009, Alba
email:[email protected]
10
No 18/2009
THE INTEGRAL OPERATOR ON THE SH (β) CLASS
Daniel Breaz and Nicoleta Breaz1
Abstract. In this paper we present a convexity condition for a integral
operator F defined in formula (2) on the class SH (β).
2000 Mathematics Subject Classification: 30C45.
Keywords and phrases: Univalent functions, integral operator, convex functions.
1. 1.Introduction
Let U = {z ∈ C, |z| < 1} be the unit disc of the complex plane and denote by H (U ), the class of the olomorphic functions in U. Consider A =
{f ∈ H (U ) , f (z) = z + a2 z 2 + a3 z 3 + ..., z ∈ U } be the class of analytic functions in U and S = {f ∈ A : f is univalent in U }.
Denote with K the class of the olomorphic functions in U with f (0) =
f ′ (0) − 1 = 0, where is convex functions in U , defined by
zf ′′ (z)
+ 1 > 0, z ∈ U .
K = f ∈ H (U ) ; f (0) = f ′ (0) − 1 = 0, Re
f ′ (z)
(
(
)
)
In the paper (3) J. Stankiewicz and A. Wisniowska introduced the class of
univalent functions SH (β), define by the next inequality:
)
(
zf ′ (z)
√
√
√ zf ′ (z)
−
2β
2
−
1
2
+
2β
2
−
1
,
< Re
f (z)
f (z)
for some β > 0 and for all z ∈ U .
1
Supported by the GAR 14/2006.
7
(1)
D.Breaz, N. Breaz - The integral operator on the SH (β) class
We consider the integral operator
F (z) =
Z
f1 (t)
t
z
0
!α1
fn (t)
· ... ·
t
!αn
(2)
dt
and we study your properties.
Remark. We observe that for n = 1 and α1 = 1 we obtain the integral
operator of Alexander.
2.Main results
Theorem 1.Let αi , i ∈ {1, ..., n} the real numbers with the properties αi >
0 for i ∈ {1, ..., n} and
√
n
X
2
√
√ .
(3)
αi ≤
2β
2
−
1
+ 2
i=1
We suppose that the functions fi ∈ SH (β) for i = {1, ..., n} and β > 0. In
this conditions the integral operator defined in (2) is convex.
Proof. We calculate for F the derivatives of the first and second order.
From (2) we obtain:
′
F (z) =
f1 (z)
z
!α1
fn (z)
· ... ·
z
!αn
and
′′
F (z) =
n
X
i=1
αi
fi (z)
z
!αi −1
n
fj (z)
zfi′ (z) − fi (z) Y
zfi (z)
z
j=1
!
!αj
j6=i
F ′′ (z)
zf1′ (z) − f1 (z)
zfn′ (z) − fn (z)
=
α
+
...
+
α
.
1
n
F ′ (z)
zf1 (z)
zfn (z)
!
!
fn′ (z) 1
f1′ (z) 1
F ′′ (z)
+
...
+
α
.
=
α
−
−
n
1
F ′ (z)
f1 (z) z
fn (z) z
!
Multiply the relation (4) with z we obtain:
8
!
(4)
D.Breaz, N. Breaz - The integral operator on the SH (β) class
n
n
n
X
zfi′ (z) X
zfi′ (z)
zF ′′ (z) X
αi
αi
αi .
=
−1 =
−
F ′ (z)
fi (z)
fi (z)
i=1
i=1
i=1
!
(5)
The relation (5) is equivalent with
n
n
X
zfi′ (z) X
zF ′′ (z)
α
αi + 1.
+
1
=
−
i
F ′ (z)
fi (z)
i=1
i=1
√
We multiply the relation (6) with 2 and obtain:
√
n
n √
√
X
zF ′′ (z)
zfi′ (z) √ X
+
1
=
−
αi + 2.
2
2α
2
i
′
F (z)
fi (z)
i=1
i=1
!
(6)
(7)
The equality (7) is equivalent with:
√
√ zF ′′ (z)
√ zf ′ (z)
2 F ′ (z) + 1 = α1 2 f11(z) + 2α1 β
2 − 1 + ...
√
√ ′ (z)
n
+αn 2 zf
+
2α
β
2
−
1
−
n (z)
f√
n√ P
√
Pn
n
− i=1 2αi β
2 − 1 − 2 i=1 αi + 2.
We calculate the real part from both terms of the above equality and obtain:
n√
′′
o
√
√
zf1′ (z)
(z)
+
1
=
α
Re
2Re zF
2
+
2β
2
−
1
+ ...
F ′ (z)
n√1
o f1 (z)√
zfn′ (z)
2
2−1 −
+αn Re
+ 2β
√ fn (z) √ P
√
Pn
n
− i=1 2αi β
2 − 1 − 2 i=1 αi + 2.
Because fi ∈ SH (β) for i = {1, ..., n} we apply in the above relation the
inequality (1) and obtain:
√
′
√
zf (z)
+ 1 > α1 f11(z) − 2β
2 − 1 + ...
′
√
n (z)
+αn zf
2
−
1
−
−
2β
fn (z)
√
√ Pn
√
Pn
− i=1 2αi β
2 − 1 − 2 i=1 αi + 2.
2Re
′
zf (z)
zF ′′ (z)
F ′ (z)
Because αi fii(z) − 2β
√
√
2 − 1 > 0 for all i ∈ {1, ..., n}, obtain that
n
n
√ X
√
√
X
zF ′′ (z)
αi + 2.
2Re
2
−
1
−
2
2α
β
+
1
>
−
i
′
F (z)
i=1
i=1
!
9
(8)
D.Breaz, N. Breaz - The integral operator on the SH (β) class
Using the hypothesis (3) in (8) we have:
zF ′′ (z)
Re
+1 >0
F ′ (z)
!
(9)
so, F is the convex function.
Corollary 2. Let α the real numbers with the properties 0 < α ≤
√
2
√ .
2β ( 2−1)+ 2
√
We suppose that the functions
f ∈ SH (β) and β > 0. In this conditions the
R z f (z) α
dt is convex.
integral operator F (z) = 0 z
Proof. In the Theorem 1, we consider n = 1, α1 = α and f1 = f .
References
[1] M. Acu, Close to convex functions associated with some hyperbola, Libertas Mathematica, Tomus XXV (2005), 109-114.
[2] D. Breaz, N. Breaz, Two integral operators, Studia Universitatis Babe¸s
-Bolyai, Mathematica, Cluj Napoca, No. 3-2002,pp. 13-21.
[3] J. Stankiewicz, A. Wisniowska, Starlike functions associated with some
hyperbola, Folia Scientiarum Universitatis Tehnicae Resoviensis 147, Matematyka 19(1996), 117-126.
Authors:
Daniel Breaz
Department of Mathematics
” 1 Decembrie 1918 ” University of Alba Iulia
Address: Alba Iulia, Str. N. Iorga, No. 11-13, 510009, Alba
email:[email protected]
Nicoleta Breaz
Department of Mathematics
” 1 Decembrie 1918 ” University of Alba Iulia
Address: Alba Iulia, Str. N. Iorga, No. 11-13, 510009, Alba
email:[email protected]
10