articol Acta 4. 47KB Jun 04 2011 12:08:22 AM

THE STARLIKENES PROPERTIES FOR INTEGRAL OPERATORS

by

Daniel Breaz and Nicoleta Breaz

Abstract. In this paper we prove a starlikeness property for the Bernardi operator concerning

( )

α

S -class and two particular properties for Libera and Alexander operators.

INTRODUCTION.

Let U=

{

z:z <1

}

the unit disc in the complex plane and let An the class of

functions f of the form f

( )

z z a z an zn z U

n

n + + ∈

+

= +

+ +

+ 2 2 ...,

1

1 , which are analytic

in the unit disc U and we denote A₁ = A.

Let the class of univalent functions S =

{

f:f∈Hn

( ) ( ) ( )

U , f 0 = f′0 −1=0

}

and let

( )

( )

( )

   



 ′ _{> ∈}

=

∗ _{z U}

z f

z f z f

S α :Re α,

the class of starlike functions of the order α,0≤α<1, in U. For α =0, denote

( )

∗

∗ _=S

S 0 the class of the starlike functions in U. Consider the following integral operator

( )

= +

∫

z

( )

c− ≥ c

c f tt dt c

z c z F

0

1 _{, 0}

1

, where f ∈An, where is the Bernardi operator.

For 0γ = obtained the Alexander operator, and for γ =1 is the operator of Libera.

PRELIMINARY RESULTS.

Definition [1] For 0<α ≤2, we define S

( )

α as the class of functions f ∈A, which satisfy the conditions f

( )

z ≠0, for 0< z <1and

( )

z U

z f

z

∈ ≤ ″   

Theorem A. [1] Let f∈A,f

( )

z ≠0,for 0< z <1. If

( )

z U

z f

z

∈ ≤ ″   

 _{2, , then f is}

univalent in U. Corollary B. [1] Let

( )

A

z b z z

f

n n n

∈ +

=

∑

∞₌₁

1 and

( )

∑

∞₌ − ≤

2

2 1

n

n

b n

n .

Then f is univalent in U.

Remark [1] All the functions from S

( )

α , 0<α ≤2 which satisfy theorem A is univalent in U.

Theorem C. [1] Let f

( )

z =z+a₂z2 +...∈S

( )

α , with 0<α ≤2. Then for any z∈U we have:

( )

z − ≤ za + z

f z

2

1 ₂ α ;

( )

≤ +  +  ≤

   



 ₊

− z a z

z f

z z

a z

2 1

Re 2

1 ₂ α ₂ α ;

( )

2 2

2

1 z a z

z z

f

α

+ +

≥ .

The equalities exist if:

( )

_α

( )

α ,

(

0 2α

)

2

1 2

≤ ≤ ∈

+ ±

= S a

z az

z z

f .

Theorem D. [1] Let f∈S

( )

α with the following form f

( )

z =z+a₃z3+a₄z4 +....

a) If 2

5 2

≤

≤α , then f is starlike for

4 ₅

1 2

⋅ < _α

b) If 3−1≤α ≤2, then Re f′

( )

z >0, for

α 1 3−

<

z .

Corollary E. [1] Let f ∈S

( )

α with the following form f

( )

z =z+a₃z3 +a₄z4 +.... a)If

5 2

0<α ≤ , then f is starlike in U.

b) If 0<α ≤ 3−1, then Ref′

( )

z >0, for z∈U.

Lemma F [2] Let γ >0 and β <1. If f ∈An satisfy inequality

( )

( )

{

f' z +γzf" z

}

>β

Re , z∈U ,

then

( )

{ }

'

(

1

) ( )

{

2 , 1

}

Re f z >β + −β δ n γ − , z∈U , where

( )

∫

+ =1

01

, _γ

ρ ρ γ

δ n

d

n .

MAIN RESULTS.

Theorem 1. Let θ =0,911621907, −1<c≤4,741187 and

( )

{

₂

( )

_{,1 1}

}

1 1 , 2 1

1 1 1 , 2 1

3 1

1 2 2

− 

  



 _

  

 

+ −

−    

 

+ <

+ − +

n c

c n tg

c c c

δ δ

δ

θ .

If f ∈A_n, satisfies inequality

( )

( )

( )

{

( )

}



  

  −    

 

   

 

+ −

− + +

   

  − + + −

> ′

θ δ

δ

θ

2 2 2

2

3 1 1 1 1 , 1 1 , 1 4 1 2

3 1 1 1 2 1

Re

tg c c

n n

c

tg c c

z

f (1)

then

*

S Fc∈ ,

Proof. For this proof we need the following result proved by Yong Chan Kim and H.M. Srivastava in [4] and which we are presented below.

We consider δ

( )

n,γ defined as Lemma F and we have the following conditions satisfied :

911621907 ,

0

=

θ , 17418γ ≥0,

( )

( )

( )

{

1 ,

} ( )

{

2 ,1 1

}

1 , 2 3 1 2 2 − − − < − − n n n tg δ γ δ δ γ θ γγ γ .

If f ∈An satisfies inequality:

( )

( )

{

}

( )

( )

{

−

}

{

−

( )

} ( )

_ − − _ +     ₋ − + − > + θ γγ γ δ δ γ θ γγ γ γ 2 2 2 2 2 2 3 1 1 , 1 1 , 1 4 2 3 1 1 2 1 " ' Re tg n n tg z f z

f , z∈U, (2)

then f ∈S*.

We go back to our proof. One easily observe that

( )

( )

z F

( )

z c

z F z

f c ⋅ c″

+ + ′ = 1 1 ' .

It is known the fact that f ∈An then Fc∈An.

Concerning the relation (1) and (2) above and consider

1 1

+ =

c

γ we obtain:

( )

( )

( )

>

      _⋅ ″ + + ′

= z F z

c z F z

f _{c c}

1 1 Re ' Re

( )

( )

{

( )

}

₃1 0

1 1 1 , 1 1 , 1 4 1 2 3 1 1 1 2 1 2 2 2 2 >       −       _      + − − + +       − + + − > θ δ δ θ tg c c n n c tg c c .

Concerning the result proved by Yong Chan Kim and H.M. Srivastava, this last inequality implies the fact that Fc∈S*.

( )

(

)

₃1 0,02531...,

1 2 ln 1 8 4

3 1 5 1

Re

2 2

2

− ≈    

  − −

+

− −

> ′

θ θ

tg tg z

f z∈U ,

then

( )

( )

*

0 1

2

S dt t f z z F

z

∈

=

∫

,

where F₁ is the operator of Libera.

Proof. This corollary is obtained from theorem 1, for c=n=1. So we obtain

( )

( )

( )

(

)

₃1 0,02531...,

1 2 ln 1 8 4

3 1 5 1

2 1 Re

Re

2 2

2

1

1 ≈−

   

  − −

+

− −

>    



 ′ _{+ ⋅} ″

= ′

θ θ

tg tg z

F z z F z

f

condition which implies the starlikenes of Libera’s operator.

Corollary 3. Let θ =0,911621907. If f ∈A₁ satisfy the inequality

( )

0,417352...

2 ln 4 2 ln 4 6

2 ln 4 2 ln 4 3 Re

2 2

≈ +

− +

− > ′ z f

then

( )

( )

*

0

0 dt S

t t f f F

z

∈

=

∫

,

where F₀ is Alexander’s operator.

Proof. We considerc=0,n=1 in theorem 1. So we obtain δ

( )

1,1 =ln2and

( )

{

( )

( )

}

0,417352...

2 ln 4 2 ln 4 6

2 ln 4 2 ln 4 3 Re

Re ₂

2 0

0 ≈

+ −

+ − > ″ ⋅ + ′ =

′ z F z z F z

f ,

condition which implies the starlikenes of Alexander’s operator.

REFERENCES

[2] Nunokava M., Obradovic M., Owa S.- One criterion for univalency, Proc. Amer. Math. Soc. 106(1989), no.4. 1035-1037.

[3] Nunokava M. - On the theory of multivalent functions, Tsukuba J. Math. 11(1987), no.2, 273-286.

[4] Kim Y.C., Srivastava H.M.-Some applications of a differential subordination, Internat. J. Math.&Math. Sci. Vol. 22. No. 3.(1999), 649-654.

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