Daniel_breaz. 44KB Jun 04 2011 12:08:35 AM
TWO STARLIKENESS PROPERTIES FOR THE BERNARDI
OPERATORS
by
Daniel Breaz and Nicoleta Breaz
f is starlike, then F is also starlike.In this paper we extend
this result if f satisfies larger conditions and proving that F is starlike of order 2 3 and 1 3 ,
where F is the Bernardi integral operator.
Abstract. It is well known that if
Let U = {z : z < 1} the unit disc in the complex plane, and let H (U ) the space of
holomorphic functions in U.
Let An denote the class of functions f of the form
Introductions.
f ( z ) = z + a n +1 z n +1 + a n + 2 z n + 2 + ..., z ∈ U
wich are analytic in the unit disc U and A1 = A .
Let the class of univalent functions S = { f : f ∈ H u (U ), f (0 ) = f ′(0 ) − 1 = 0} and let
zf ′(z )
S ∗ (α ) = f : Re
> α , z ∈U
f (z )
be the class of the starlike functions of the order α ,0 ≤ α < 1 , in U. For α = 0 ,
denote S ∗ (0 ) = S ∗ the class of the starlike functions in U.
Consider the following integral operator
where f ∈ An .
F (z ) =
γ +1
zγ
∫ f (t )t
z
γ −1
dt , γ ≥ 0 ,
0
Lemma A.[1] Let q univalent in U and let θ and φ analytic functions in the domain
D ⊆ q(U ) , with φ (w) ≠ 0 , when w ∈ q(U ) .
7
Q(z ) = nzq ′(z )φ [q( z )] h(z ) = θ [q(z )] + Q(z )
Let
and suppose that next condition is satisfy:
i)
Q is starlike
ii)
Re
and
θ ′[q( z )] zQ ′(z )
zh ′( z )
= Re
+
>0.
Q( z )
φ [q( z )] Q( z )
If p is analytic in U, with
p(0 ) = q(0), p ′(0 ) = ... = p (n −1) (0 ) = 0,
and
p(U ) ⊂ D
θ [ p(z )] + zp ′( z )φ [ p( z )] p θ [q( z )] + zq ′(z )φ [q( z )] = h( z )
then p p q and q is the best dominant.
Main results.
Theorem 1. Let
h( z ) =
If f ∈ An and
2
2nz
+
2 − z (2 − z )(2 + 2γ − γz )
zf ′( z )
p h( z ) ,
f (z )
then
Re
where
zF ′( z ) 2
> ,
F (z ) 3
1+ γ
F (z ) = γ
z
∫ f (t )t
z
0
8
γ −1
dt (1)
Proof. The relation (1) implies:
γF ( z ) + zF ′( z ) = (γ + 1) f ( z )
Let
The relation (2) implies
But
implies
zF ′( z )
.
F (z )
p(z ) =
zf ′( z )
zp ′(z )
+ p(z ) =
.
f (z )
p(z ) + γ
zf ′( z )
p h( z )
f (z )
zp ′(z )
+ p ( z ) p h( z ) .
p(z ) + γ
We apply the lemma A for proved that
Re
If we let:
zF ′( z ) 2
> .
F (z ) 3
q( z ) =
2
2− z
θ (w) = w
φ (w) =
1
w+γ
θ [q( z )] =
φ [q(z )] =
2
2− z
2− z
2 + 2γ − γz
9
(2)
Q( z ) = nzq ′( z )φ [q( z )] =
h( z ) = θ [q(z )] + Q( z ) =
(2 − z )(2 + 2γ − γz )
2nz
.
2
2nz
+
.
2 − z (2 − z )(2 + 2γ − γz )
Since Q is starlike and Re φ [q(z )] > 0 , by lemma A we deduce
zF ′(z )
zF ′(z ) 2
2
p
⇒ Re
ppq⇔
> .
3
F (z )
F (z )
2− z
Theorem 2. Let
2+ z
4nz
+
h( z ) =
.
2 − z (2 − z )(2 + 2γ + (1 − γ )z )
If f ∈ An and
zf ′( z )
p h( z ) ,
f (z )
then
zF ′(z ) 1
Re
> ,
F (z ) 3
where
F (z ) =
Proof. We are
Let
We deduce
But
1+ γ
zγ
∫ f (t )t
z
γ −1
dt .
0
γF ( z ) + zF ′( z ) = (γ + 1) f ( z )
p(z ) =
zF ′( z )
.
F (z )
zf ′( z )
zp ′(z )
+ p(z ) =
.
f (z )
p(z ) + γ
zf ′( z )
p h( z )
f (z )
10
(3)
zp ′(z )
+ p ( z ) p h( z ) .
p(z ) + γ
and obtained
In lemma we consider:
q(z ) =
2+ z
2−z
θ (w ) = w
φ (w) =
1
w+γ
θ [q( z )] =
φ [q(z )] =
2− z
.
2 + 2γ + (1 − γ )z
Q( z ) = nzq ′( z )φ [q( z )] =
h( z ) = θ [q(z )] + Q( z ) =
2+ z
2− z
2nz
(2 − z )(2 + 2γ + (1 − γ )z )
2+ z
2nz
+
.
(
)
(
2 − z 2 − z 2 + 2γ + (1 − γ )z )
Sience Q is starlike and Re φ [q(z )] > 0 , by lemma A obtained
zF ′(z ) 1 + z
zF ′(z ) 1
ppq⇔
p
⇒ Re
>
F (z ) 1 − z
F (z ) 3
REFERENCES
[1].Miller, S.S. and Mocanu P.T., On some classes of first differential subordinations,
Michigan Math. J. 32, pp. 185-195, 1985.
[2].Mocanu, P.T., On a class of first-order differential subordinations, Babes-Bolyai
Univ., Fac. Of Math. Res. Sem., Seminar on Mathematical Analysis, Preprint 7, pp.
37-46, 1991.
11
[3].Oros, Gh. On starlike images by an integral operator, Mathematica, Tome 42(65),
No 1, 2000, pp. 71-74.
[4].Breaz, D., Breaz, N. Starlikeness conditions for the Bernardi operator, to be
appear.
Authors:
Daniel Breaz- „1 Decembrie 1918” University of Alba Iulia, str. N. Iorga, No. 13,
Departament of Mathematics and Computer Science, email: [email protected]
Nicoleta Breaz- „1 Decembrie 1918” University of Alba Iulia, str. N. Iorga, No. 13,
Departament of Mathematics and Computer Science, email: [email protected]
12
OPERATORS
by
Daniel Breaz and Nicoleta Breaz
f is starlike, then F is also starlike.In this paper we extend
this result if f satisfies larger conditions and proving that F is starlike of order 2 3 and 1 3 ,
where F is the Bernardi integral operator.
Abstract. It is well known that if
Let U = {z : z < 1} the unit disc in the complex plane, and let H (U ) the space of
holomorphic functions in U.
Let An denote the class of functions f of the form
Introductions.
f ( z ) = z + a n +1 z n +1 + a n + 2 z n + 2 + ..., z ∈ U
wich are analytic in the unit disc U and A1 = A .
Let the class of univalent functions S = { f : f ∈ H u (U ), f (0 ) = f ′(0 ) − 1 = 0} and let
zf ′(z )
S ∗ (α ) = f : Re
> α , z ∈U
f (z )
be the class of the starlike functions of the order α ,0 ≤ α < 1 , in U. For α = 0 ,
denote S ∗ (0 ) = S ∗ the class of the starlike functions in U.
Consider the following integral operator
where f ∈ An .
F (z ) =
γ +1
zγ
∫ f (t )t
z
γ −1
dt , γ ≥ 0 ,
0
Lemma A.[1] Let q univalent in U and let θ and φ analytic functions in the domain
D ⊆ q(U ) , with φ (w) ≠ 0 , when w ∈ q(U ) .
7
Q(z ) = nzq ′(z )φ [q( z )] h(z ) = θ [q(z )] + Q(z )
Let
and suppose that next condition is satisfy:
i)
Q is starlike
ii)
Re
and
θ ′[q( z )] zQ ′(z )
zh ′( z )
= Re
+
>0.
Q( z )
φ [q( z )] Q( z )
If p is analytic in U, with
p(0 ) = q(0), p ′(0 ) = ... = p (n −1) (0 ) = 0,
and
p(U ) ⊂ D
θ [ p(z )] + zp ′( z )φ [ p( z )] p θ [q( z )] + zq ′(z )φ [q( z )] = h( z )
then p p q and q is the best dominant.
Main results.
Theorem 1. Let
h( z ) =
If f ∈ An and
2
2nz
+
2 − z (2 − z )(2 + 2γ − γz )
zf ′( z )
p h( z ) ,
f (z )
then
Re
where
zF ′( z ) 2
> ,
F (z ) 3
1+ γ
F (z ) = γ
z
∫ f (t )t
z
0
8
γ −1
dt (1)
Proof. The relation (1) implies:
γF ( z ) + zF ′( z ) = (γ + 1) f ( z )
Let
The relation (2) implies
But
implies
zF ′( z )
.
F (z )
p(z ) =
zf ′( z )
zp ′(z )
+ p(z ) =
.
f (z )
p(z ) + γ
zf ′( z )
p h( z )
f (z )
zp ′(z )
+ p ( z ) p h( z ) .
p(z ) + γ
We apply the lemma A for proved that
Re
If we let:
zF ′( z ) 2
> .
F (z ) 3
q( z ) =
2
2− z
θ (w) = w
φ (w) =
1
w+γ
θ [q( z )] =
φ [q(z )] =
2
2− z
2− z
2 + 2γ − γz
9
(2)
Q( z ) = nzq ′( z )φ [q( z )] =
h( z ) = θ [q(z )] + Q( z ) =
(2 − z )(2 + 2γ − γz )
2nz
.
2
2nz
+
.
2 − z (2 − z )(2 + 2γ − γz )
Since Q is starlike and Re φ [q(z )] > 0 , by lemma A we deduce
zF ′(z )
zF ′(z ) 2
2
p
⇒ Re
ppq⇔
> .
3
F (z )
F (z )
2− z
Theorem 2. Let
2+ z
4nz
+
h( z ) =
.
2 − z (2 − z )(2 + 2γ + (1 − γ )z )
If f ∈ An and
zf ′( z )
p h( z ) ,
f (z )
then
zF ′(z ) 1
Re
> ,
F (z ) 3
where
F (z ) =
Proof. We are
Let
We deduce
But
1+ γ
zγ
∫ f (t )t
z
γ −1
dt .
0
γF ( z ) + zF ′( z ) = (γ + 1) f ( z )
p(z ) =
zF ′( z )
.
F (z )
zf ′( z )
zp ′(z )
+ p(z ) =
.
f (z )
p(z ) + γ
zf ′( z )
p h( z )
f (z )
10
(3)
zp ′(z )
+ p ( z ) p h( z ) .
p(z ) + γ
and obtained
In lemma we consider:
q(z ) =
2+ z
2−z
θ (w ) = w
φ (w) =
1
w+γ
θ [q( z )] =
φ [q(z )] =
2− z
.
2 + 2γ + (1 − γ )z
Q( z ) = nzq ′( z )φ [q( z )] =
h( z ) = θ [q(z )] + Q( z ) =
2+ z
2− z
2nz
(2 − z )(2 + 2γ + (1 − γ )z )
2+ z
2nz
+
.
(
)
(
2 − z 2 − z 2 + 2γ + (1 − γ )z )
Sience Q is starlike and Re φ [q(z )] > 0 , by lemma A obtained
zF ′(z ) 1 + z
zF ′(z ) 1
ppq⇔
p
⇒ Re
>
F (z ) 1 − z
F (z ) 3
REFERENCES
[1].Miller, S.S. and Mocanu P.T., On some classes of first differential subordinations,
Michigan Math. J. 32, pp. 185-195, 1985.
[2].Mocanu, P.T., On a class of first-order differential subordinations, Babes-Bolyai
Univ., Fac. Of Math. Res. Sem., Seminar on Mathematical Analysis, Preprint 7, pp.
37-46, 1991.
11
[3].Oros, Gh. On starlike images by an integral operator, Mathematica, Tome 42(65),
No 1, 2000, pp. 71-74.
[4].Breaz, D., Breaz, N. Starlikeness conditions for the Bernardi operator, to be
appear.
Authors:
Daniel Breaz- „1 Decembrie 1918” University of Alba Iulia, str. N. Iorga, No. 13,
Departament of Mathematics and Computer Science, email: [email protected]
Nicoleta Breaz- „1 Decembrie 1918” University of Alba Iulia, str. N. Iorga, No. 13,
Departament of Mathematics and Computer Science, email: [email protected]
12