Daniel Breaz. 49KB Jun 04 2011 12:08:35 AM

Proceedings of the International Conference on Theory and Applications of
Mathematics and Informatics – ICTAMI 2003, Alba Iulia

INTEGRAL OPERATORS ON THE UCD( α )-CLASS
by
Daniel Breaz
Abstract. We consider the class of functions f = z + a 2 z 2 + a3 z 3 + ... , that are analytically
and univalent in the unit disk.
We consider the class of functions UCD( α ), α ≥ 0 with the properties
Re f ′( z ) ≥ α zf ′′( z ) (∀)z ∈ U .
In this paper we present some results from integrals operators at this class.

1.Introduction
Theorem A. [4] A sufficient condition for a function f of the form

f (z ) = z + a 2 z 2 + a 3 z 3 + ...

(1)

so that it belongs to the UCD(α ), α ≥ 0 class is

∑ k [1 + α (k − 1)]⋅ a
∞

k =2

k

≤ 1.

(2)

2.Main results

Theorem 1. Let be f ∈ S , f (z ) = z + a 2 z 2 + a3 z 3 + ... , z ∈ U . We suppose that the
coefficients of function f verify the condition

∑ k [1 + α (k − 1)]⋅ a
∞

k =2

We consider the integral operator
F (z ) =

∫
z

0

k

f (t )
dt ,
t

≤ 1.

which is the Alexander operator.
In this conditions, we have F ∈ UCD(α ), α ≥ 0, z ∈ U .

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Proceedings of the International Conference on Theory and Applications of
Mathematics and Informatics – ICTAMI 2003, Alba Iulia

Proof
Let be f of the form (1), verifying the condition (2).
We have:

∫(

∫

)

z


t + a 2 t 2 + a3t 3 + K
a t2 a t3

dt = 1 + a 2 t + a 3 t 2 + K dt = t + 2 + 3 + K =
F (z ) =
2
3
t


0
0
z

a
a
= z + 2 z 2 + 3 z 3 ... .
2
3

Denoting

z

0

bk =

ak
,k ≥ 2,
k
we can write, F (z ) = z + b2 z 2 + b3 z 3 + ... .

In that following, we evaluate the relation (2), for the function F , with the
coefficients bk .

∑
∞

k =2
∞

k [1 + α (k − 1)] ⋅ bk =

∑ [1 + α (k − 1)]⋅ a
k =2

k

≤

∑
∞

k =2
∞

k [1 + α (k − 1)] ⋅

ak
=
k

∑ k [1 + α (k − 1)] ⋅ a
k =2

k

∑ k [1 + α (k − 1)]⋅ a
∞

k =2

k

⋅

1
≤
k

< 1.

According to the relation (2), from the Theorem A, we obtain:
F ∈ UCD(α ), α > 0, z ∈ U .

Theorem 2. Let be f ∈ A , f ( z ) = z + a 2 z 2 + a3 z 3 + ... , α ≥ 0 , z ∈ U .We suppose
that the condition (2) is satisfied and we consider the Libera operator,

∫

2
L( f )( z ) =
f (t )dt .
z0
z

Then L ∈ UCD(α ) , α ≥ 0 , z ∈ U .

Proof
Let be f of the form (1) verifying the condition (2).

We have:

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Proceedings of the International Conference on Theory and Applications of
Mathematics and Informatics – ICTAMI 2003, Alba Iulia
2
F (z ) =
z

=

∫(
z

)


2 t2
t3
t4

+ a3
+ K =
t + a 2 t + a 3 t + K dt =  + a 2
3
4
z 2

2

3


a
a
2z

+ 2 z 3 + 3 z 4 + K =

3
4

z 2

0

z

0

2

=z+

∑

∞
2a k k
2a 2 2 2a3 3
z +
z +K = z +
z .
3
4
k =2 k + 1

Denoting

bk =

2a k
,
k +1

we can write L( f )(z ) = z + b2 z 2 + b3 z 3 + ... .
We evaluate the relation (2), for the function obtained after integration with the
coefficients bk..
Since k ≥ 2 , we have
∞
∞
∞
2a k
2
=
≤
k [1 + α (k − 1)] ⋅ a k ⋅
k [1 + α (k − 1)] ⋅ bk =
k [1 + α (k − 1)] ⋅
k + 1 k =2
k +1
k =2
k =2

∑
≤

∑ k [1 + α (k − 1)]⋅ a

∑

∞

k =2

k

⋅

∑

2 2
≤ < 1.
3 3

L( f ) ∈ UCD(α ) , α ≥ 0 ,
so, the Libera operator preserves this class.
that implies

Theorem 3. Let be f ∈ A , f ( z ) = z + a 2 z 2 + a3 z 3 + ... , z ∈ U . We suppose that the
condition (2) is satisfied and we consider the Bernardi operator,

1+ γ
F (z ) = γ
z

∫ f (t ) ⋅ t
z

Then, we have F ∈ UCD(α ) , α ≥ 0 , z ∈ U .

γ −1

dt , γ ≥ −1 .

0

Proof
Let be f of the form (1), verifying the condition (2).
We have

1+ γ
F (z ) = γ
z

∫ (t + a t
z

2

0

2

)

+ a3 t + K ⋅ t
3

γ −1

1+ γ
dt = γ
z

63

∫ (t
z

0

γ

)

+ a 2 t γ +1 + a3 t γ + 2 + K dt =

Proceedings of the International Conference on Theory and Applications of
Mathematics and Informatics – ICTAMI 2003, Alba Iulia
1+ γ
= γ
z

 t γ +1

1+ γ
t γ +2

+
+ K = γ
a
2
 γ +1
γ +2
z

0
z

 z γ +1 a 2 z γ + 2 a 3 z γ +3 


 γ + 1 + γ + 2 + γ + 3 K =



∞
γ +1 2
γ +1 3
γ +1 k
⋅z .
z + a3
z + K = z + ∑ ak ⋅
γ +2
γ +3
γ +k
k =2
γ +1
.
Let be bk = a k ⋅
γ +k

= z + a2

We evaluate the relation (2) for the function F (z ) = z + b2 z 2 + b3 z 3 + ...
We have:
∞
∞
∞
γ +1
γ +1
≤
=
k [1 + α (k − 1)] ⋅ bk =
k [1 + α (k − 1)] ⋅ a k
k [1 + α (k − 1)] ⋅ a k ⋅
γ + k k =2
γ +k
k =2
k =2

∑
≤

∑
∞

∑

k [1 + α (k − 1)] ⋅ a k ⋅

F ∈ UCD(α ) .
k =2

∑

γ +1 ∞
≤
k [1 + α (k − 1)] ⋅ a k ≤ 1 , so:
γ +1 ∑
k =2

Theorem 4. Let

1+ γ
F (z ) = γ

∫ f (t ) ⋅ t
z

γ −1

dt .

If f is of the form (1), f ∈ S , α > 0 and Re f ′( z ) ≥ α zf ′′( z ) (∀)z ∈ U , then
z

0

(1 + γ )F ′(z ) + zF ′′(z ) ≥ α z[(2 + γ )F ′′ + zF ′′′].

Proof

F (z ) =

1+ γ
zγ

∫
z

f (t ) ⋅ t γ −1 dt ⇔

0

zγ
F (z ) =
1+ γ

∫ f (t ) ⋅ t
z

0

γ −1

dt .

After successive derivations, we obtain:
zγ
zF ' ( z )
γ
γ
⋅ z γ −1 ⋅ F ( z ) +
⋅ F ' ( z ) = f ( z ) ⋅ z γ −1 ⇔
⋅ F (z ) +
= f (z ) ⇒
1+ γ
1+ γ
1+ γ
1+ γ
F ′( z ) zF ′′( z )
γ
1
= f ' ( z ) ⇔ F ′(z ) +
⋅ z ⋅ F ′′( z ) = f ' ( z ) ⇔
+
⋅ F ' (z ) +
1+ γ
1+ γ
1+ γ
1+ γ
zF ′′′( z )
1
⇔ F ′′(z ) +
F ′′( z ) +
= f ′′(z ) .
1+ γ
1+ γ
2+γ
1
So, we have f ′′(z ) =
F ′′(z ) +
zF ′′′(z ) .
1+ γ
1+ γ

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Proceedings of the International Conference on Theory and Applications of
Mathematics and Informatics – ICTAMI 2003, Alba Iulia
If Re f ′(z ) ≥ α zf ′′( z ) (∀)z ∈ U , α > 0 , then
f ∈ UCD(α ) ⇒ f ′( z ) ≥ α zf ′′(z ) .

In this inequality, we put the expressions of
F ′( z ) +

⇔

1
zF ′′( z ) ≥ α
1+ γ

2+γ
z 
 1+ γ

f ′ and


1
 F ′′(z ) +
zF ′′′(z ) ⇔
1+ γ


f ′′ , and we obtain:

1
(1 + γ )F ′(z ) + zF ′′(z ) ≥ α z (2 + γ )F ′′(z ) + z 2 F ′′′(z ) ⇔
1+ γ
1+ γ

[

]

⇔ (1 + γ )F ' (z ) + zF " ( z ) ≥ α z (2 + γ )F " (z ) + z 2 F ′′′(z ) .

Remark 5. If f is of the form (1), f ∈ S , α > 0 and Re f ′( z ) ≥ α zf ′′( z ) (∀)z ∈ U ,
then
1
log f ' (z ) ≤
.
αz
Proof

According to the Theorem 4 we have:
2+γ 
1
1 2
 F ′′(z ) +
F ′( z ) +
zF ′′( z ) ≥ α z 
z F ′′′( z ) ⇔
γ
γ
+
1+ γ
1
1
+



2+γ
1
′
F ′′( z ) +
zF ′′′(z )
 

1
1+ γ
1+ γ
′
′
′
= α ⋅ z ⋅ log F (z ) +
⇔1≥α ⋅ z ⋅
zF ( z ) ⇔
1
1+ γ

 
F ′(z ) +
zF ′′( z )
1+ γ

′
′
⇔ 1 ≥ α ⋅ z ⋅ [log f ′(z )] ⇔ [log f ′(z )] ≤

α⋅z
1

.

References

1.Miller S.S., Mocanu P.T.- Integral operators on certain classes of analytic functions,
Univalent Functions, Fractional Calculus, and their Applications, Halstead Pres, John
Wiley (1989), 153-166.
2. Miller S.S., Mocanu P.T., Reade M.O.- Starlike integral operators, Pacific Journal
of Mathematics, vol. 79, No. 1/1978, 157-168.

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Proceedings of the International Conference on Theory and Applications of
Mathematics and Informatics – ICTAMI 2003, Alba Iulia
3. Miller S.S., Mocanu P.T.- Differential subordinations and univalent functions,
Michigan, Math. J. 28(1981), 157-171.
4. Rosy T., Stephen A.B., Subramanian K.G., Silverman H.- Classes of convex
functions, Internat. J. Math.& Math. Sci., vol 23, No. 12(2000), 819-825.
5. Silverman H.- Univalent functions with negative coefficients, Proc. Amer. Math.
Soc. 51(1975), 109-116.
Author:
Daniel Breaz, “1 Decembrie 1918” University of Alba Iulia, Romania, dbreaz@uab.ro

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