paper 24-18-2009. 121KB Jun 04 2011 12:03:08 AM

ON THE UNIVALENCE OF SOME INTEGRAL OPERATORS

Virgil Pescar

Abstract. _{In view of two integral operators Hγ1,γ2,...,γ}

n and Jγ1,γ2,...,γn

for analytic functionsfj, j = 1, nin the open unit diskU, sufficient conditions for univalence of these integral operators are discussed.

2000 Mathematics Subject Classification: 30C45.

Key Words and Phrases: Integral operator, univalence, starlike functions.

1. Introduction

We consider the unit open disk _U =_{z _∈C_{:|z|<1} and A the class of} the functions f which are analytic in _U and f(0) =f′_{(0)−1 = 0.}

We denote by _S the subclass of the functionsf _{∈ A} which are univalent in_U and _S∗ _{denote the subclass ofS consisting of all starlike functions f in}

U.

We consider the integral operators

Jγ(z) =

1

γ

Z z

0

u−1(f(u))γ1 du

γ

(1) for f _{∈ A},γ be a complex number, γ ₆= 0 and

Hγ1,γ2,...,γn(z) =

Z z

0

f1(u)

u

_γ1 1

...

fn(u)

u

_γn1

du (2)

for fj ∈ A and γj complex numbers,γj 6= 0, j = 1, n.

From (2) for n = 1, f₁ = f and 1

γ1 = α we obtain the integral operator Kim-Merkes, Hα, given by

Hα(z) = Z z

0

f(u)

u

α

du (3)

In [3] Kim-Merkes prove that the integral operator Hα is in the class S for _|α_{| ≤} 1₄ and f _{∈ S}.

Pescar in [8], [9], has obtained univalence sufficient conditions for the integral operator Hα.

Pescar, Owa [12] have studied univalence problems for integral operator

Hα.

In [2], D. Breaz and N. Breaz have studied the univalence of integral operator Hγ1,γ2,...,γn.

In this paper we introduce a general integral operator

Jγ1,γ2,...,γn(z) =

( _n X

j=1

1

γj !

Z z

0

u−1f1(u)

1

γ1...fn(u) 1

γndu

)P_n1

j=1 1

γj

(4)

for fj ∈ A, γj complex numbers, γj 6= 0, j = 1, n, which is a generalization of integral operator Jγ, given by (1).

For n = 1, f₁ =f and γ₁ = γ, from (4) we obtain the integral operator

Jγ.

In the present paper, we obtain some sufficient conditions for the integral operators Hγ1,γ2,...,γn and Jγ1,γ2,...,γn to be in the class S.

2. Preliminary results

We need the following lemmas.

Lemma 2.1. [7]Let α be a complex number, Re α >0 and f _{∈ A}. If

1_{− |}z_|2Re α

Re α

zf′′(z)

f′_(z)

≤

1 (5)

for all z _{∈ U}, then the integral operator Fα defined by

Fα(z) =

α

Z z

0

uα−1f′(u)du

_α1

is in the class S.

Lemma 2.2. [1]If f(z) = z+a2z2 +.. is analytic in U and

(1_{− |}z_|2)

zf′′(z)

f′_(z)

≤

1 (7)

for all z _{∈ U}, then the function f(z) is univalent in _U.

Lemma 2.3. (Schwarz [4]). Let f the function regular in the disk _UR=

{z _∈C_{:|z|< R} with |f(z)| < M, M fixed. If f(z) has in z = 0 one zero}

with multiply _≥m, then

|f(z)_{| ≤} M

Rm|z| m_{, z}

∈ UR (8)

the equality (in the inequality (8) for z ₆= 0) can hold only if f(z) = eiθ M

Rmz m_,

where θ is constant.

3. Main results

Theorem 3.1 Let γj be complex numbers, γj 6= 0, Mj real positive

num-bers, j = 1, n, Pn_j=1Re_γ1

j = 1 and fj ∈ A, fj(z) =z+a2jz

2_+a

3jz3+..., j = 1, n. If

zf_j′(z)

fj(z) − 1

≤

Mj, j = 1, n (9)

and

n X

j=1

Mj

|γj| ≤ 3√3

2 (10)

then the integral operatorsJγ1,γ2,...,γn given by (4) and Hγ1,γ2,...,γn given by (2)

are in the class _S.

Proof. We observe that

=

( _n X j=1 1 γj ! Z z 0 u Pn j=1 1

γj−1

f1(u)

u

_γ1 1

. . .

fn(u)

u

_γn1

du

)P_n1

j=1 1

γj

(11)

Let us consider the function

Hγ1,γ2,...,γn(z) =

Z z

0

f₁(u)

u

_γ1 1

. . .

fn(u)

u

_γn1

du (12)

for fj ∈ A, j = 1, n. The function Hγ1,γ2,...,γn(z) is regular in U.

We define the function pbyp(z) = zHγ′′1,γ2,...,γn(z) H′

γ1,γ2,...,γn(z). The function psatisfies

p(0) = 0 and

|p(z)_{| ≤} n X

j=1

1

|γj|

zf_j′(z)

fj(z) − 1

, z_{∈ U} (13)

From (9) and (13) we have

|p(z)_{| ≤} n X

j=1

Mj

|γj|

(14)

for all z _{∈ U}.

Applying Lemma 2.3 we obtain

|p(z)_{| ≤} n X

j=1

Mj

|γj||

z_|, z _{∈ U} (15)

and hence, we get 1_{− |}z_|2

zH_{γ1,γ2,...,γ}′′ _n(z)

H′

γ1,γ2,...,γn(z)

≤

(1_{− |}z_|2)_|z_|

n X

j=1

Mj

|γj|

(16)

for all z _{∈ U}. We have

max

|z|≤1

(1_{− |}z_|2)_|z_| = 2 3√3 and from (10) and (16) we obtain

1_{− |}z_|2

zH_γ1′′_,γ2,...,γ

n(z)

H′

γ1,γ2,...,γn(z)

≤

for all z _{∈ U}.

From (17) and by Lemma 2.2, we obtain that the integral operator

Hγ1,γ2,...,γn is in the class S.

Because H_{γ1,γ2,...,γ}′ _n(z) = f1_z(z) 1

γ1

. . .fn(_zz)

1

γn

and

Re α=

n X

j=1

Re1 γj

= 1,

from (17) and by Lemma 2.1 it results that the integral operator Jγ1,γ2,...,γn

belongs to class _S.

Corollary 3.2 Let γ be a complex number, Re1_γ = 1 and f _{∈ A}, f(z) = z+a21z2+a31z3 +. . .

If

zf′_(z) f(z) −1

≤

3√3

2 |γ| (18)

for all z _{∈ U}, then the integral operators Jγ and Hα, α= 1_γ, are in the class

S.

Proof. From Theorem 3.1, for n= 1, γ₁ =γ, α = 1

γ, f1 =f it results that

Jγ ∈ S and Hα ∈ S.

Corollary 3.3. If f _{∈ A} and

zf′(z)

f(z) −1 ≤

3√3

2 , z ∈ U (19)

then the integral operator of Alexander H given by

H(z) = Z z

0

f(u)

u du (20)

is in the class _S.

Proof. From Theorem Theorem 3.1, forn = 1, γ1 = 1, f1 =f, H =H1 we

obtain H _{∈ S}.

Theorem 3.4. Let γj be complex numbers and fj ∈ S, fj(z) = z +

If

n X

j=1

1

|γj| ≤ 1

4 (21)

then the integral operator Hγ1,γ2,...,γn given by (2) is in the class S.

Proof. We consider the function p(z) = zHγ′′1,γ2,...,γn(z) H′

γ1,γ2,...,γn(z), where Hγ1,γ2,...,γn is

defined by (2). We obtain

|p(z)_{| ≤} n X

j=1

1

|γj|

zf′

j(z)

fj(z) − 1

, z _{∈ U} (22)

Because fj ∈ S we have

zf′

j(z)

fj(z)

≤

1+|z|

1−|z|, z ∈ U, j = 1, n and

zf_j′(z)

fj(z) −1 ≤

zf_j′(z)

fj(z)

+ 1_≤ 2

1_{− |}z_|, j = 1, n, z ∈ U (23)

From (21), (23) and (22) we obtain

|p(z)_{| ≤} 1

2(1_{− |}z_|), z∈ U (24)

and hence, we have

(1_{− |}z_|2)

zH_γ1,γ2′′ _,...,γ

n(z)

H′

γ1,γ2,...,γn(z)

≤

1 (25)

for all z _{∈ U}.

By Lemma 2.2 we have Hγ1,γ2,...,γn is in the class S.

Corollary 3.5. Letγ be a complex number andf _{∈ S}, f =z+a21z2+. . .

If

|γ_{| ≤} 1

4 (26)

then the integral operator Hγ ∈ S.

Proof. In Theorem Theorem 3.4 for n = 1,_γ11 =γ, f₁ =f it results that

Theorem 3.6. Let γj be complex numbers, γj = 06 , fj ∈ A, fj(z) =

z+a2jz2+a3jz3+. . . , j = 1, n and a= Pn

j=1Re 1

γj >0. If

zf′

j(z)

fj(z) − 1

≤

(2a+ 1)2a2+1a

2n |γj|, j = 1, n (27) then the integral operator Jγ1,γ2,...,γn is in the class S.

Proof. The integral operator Jγ1,γ2,...,γn has the form (11).

We consider the function

p(z) = zH

′′

γ1,γ2,...,γn(z)

H′

γ1,γ2,...,γn(z)

, z _{∈ U} (28)

where Hγ1,γ2,...,γn(z) is define by (2).

The function p satisfies p(0) = 0 and from (28) we obtain

|p(z)_{| ≤} n X

j=1

1

|γj|

zf_j′(z)

fj(z) −1

, z _∈U (29)

From (27) and (29) we have

|p(z)_{| ≤} (2a+ 1) 2a+1

2a

2 (30)

for all z _{∈ U}.

Applying Lemma 2.3 we obtain

|p(z)_{| ≤} (2a+ 1) 2a+1

2a

2 |z|, z ∈ U (31)

From (28) and (31) we have 1_{− |}z_|2a

a

zH′′

γ1,γ2,...,γn(z)

H′

γ1,γ2,...,γn(z)

≤

(2a+ 1)2a2+1a

2

1_{− |}z_|2a

a |z| (32)

for all z _{∈ U}. Because

max

|z|≤1

1_{− |}z_|2a

a |z|

= 2

(2a+ 1)2a2+1a

from (32) we obtain

1_{− |}z_|2a

a

zH′′

γ1,γ2,...,γn(z)

H′

γ1,γ2,...,γn(z)

≤

1 (33)

for all z _{∈ U}. From (2) we have

H_γ1,γ2′ _,...,γ

n(z) =

f₁(z)

z

_γ1 1

. . .

fn(z)

z

_γn1

, z _{∈ U} (34)

and from (33) by Lemma 2.1 it results that the integral operator Jγ1,γ2,...,γn

belongs to class _S.

Remark 3.7. From Theorem 3.6 for n= 1, γ₁ =γ, f₁ =f, a=Re _γ1 = 1

we obtain the condition

zf′(z)

f(z) −1 ≤

3√3

2 |γ| (35)

and, hence, we obtain Corollary 3.2.

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Author:

Virgil Pescar

Department of Mathematics

”Transilvania” University of Bra¸sov Faculty of Science

Bra¸sov, Romania