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volume 7, issue 4, article 147, 2006.

Received 19 June, 2006;

accepted 04 September, 2006.

Communicated by:H. Silverman

Abstract Contents

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Journal of Inequalities in Pure and Applied Mathematics

UNIVALENCE CONDITIONS FOR CERTAIN INTEGRAL OPERATORS

VIRGIL PESCAR

"Transilvania" University of Bra¸sov

Faculty of Mathematics and Computer Science Department of Mathematics

2200 Bra¸sov ROMANIA

EMail:virgilpescar@unitbv.ro

c

2000Victoria University ISSN (electronic): 1443-5756 171-06

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Univalence Conditions for Certain Integral Operators

Virgil Pescar

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J. Ineq. Pure and Appl. Math. 7(4) Art. 147, 2006

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Abstract

In this paper we consider some integral operators and we determine conditions for the univalence of these integral operators.

2000 Mathematics Subject Classification:Primary 30C45.

Key words: Univalent function, Integral operator.

Contents

1 Introduction. . . 3 2 Preliminary Results. . . 5 3 Main Results . . . 6

References

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Univalence Conditions for Certain Integral Operators

Virgil Pescar

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1. Introduction

Let U = {z ∈C :|z|<1} be the unit disc in the complex plane. The class A and the class S are defined in [2]: let A be the class of functions f(z) = z +a2z2 +· · · , which are analytic in the unit disk normalized with f(0) = f0(0)−1 = 0; letSthe class of the functionsf ∈Awhich are univalent inU.

In [7] is defined the classS(α).For0< α ≤2,letS(α)denote the class of functionsf ∈Awhich satisfy the conditions:

(1.1) f(z)6= 0 for0<|z|<1 and

(1.2)

z f(z)

00

≤α for allz ∈U.

In [7] is proved the next result. For0 < α≤ 2,the functionsf ∈ S(α)are univalent.

In this work, we consider the integral operators

(1.3) Gα(z) =

α

Z z

0

gα−1(u)du α1

and

(1.4) Hα, γ(z) =

α

Z z

0

uα−1

h(u) u

γ

du α1

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Univalence Conditions for Certain Integral Operators

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forg(z)∈S, h(z)∈Sand for someα, γ ∈C.

Kim - Merkes [1] studied the integral operator

(1.5) Fγ(z) =

Z z

0

h(u) u

γ

du

and obtained the following result

Theorem 1.1. If the functionh(z)belongs to the classS, then for any complex numberγ, |γ| ≤ 14,the functionFγ(z)defined by (1.5) is in the classS.

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Univalence Conditions for Certain Integral Operators

Virgil Pescar

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2. Preliminary Results

In order to prove our main results we will use the lemma due to N.N. Pascu [4]

presented in this section.

Lemma 2.1. Let the functionf ∈Aandαa complex number,Reα >0. If

(2.1) 1− |z|2 Reα

Reα

zf00(z) f0(z)

≤1,

for allz ∈U, then for all complex numbersβ, Reβ ≥Reαthe function

(2.2) Fβ(z) =

β

Z z

0

uβ−1f0(u)du β1

is regular and univalent inU.

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Univalence Conditions for Certain Integral Operators

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3. Main Results

Theorem 3.1. Let α be a complex number, Re α ≥ 0 and the function g ∈ S, g(z) =z+a2z2+· · · .If

(j1) |α−1| ≤ Re α

4 for Re α∈(0,1) or

(j2) |α−1| ≤ 1

4 for Re α∈[1,∞), then the function

(3.1) Gα(z) =

α

Z z

0

gα−1(u)du α1

is in the classS.

Proof. From (3.1) we have

(3.2) Gα(z) =

"

α Z z

0

uα−1

g(u) u

α−1

du

#α1 .

The function g(z) is regular and univalent, hence g(z)z 6= 0 for allz ∈ U. We can choose the regular branch of the function h

g(z) z

iα−1

to be equal to 1 at the origin.

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Univalence Conditions for Certain Integral Operators

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Let us consider the regular function inU, given by

(3.3) p(z) =

Z z

0

g(u) u

α−1

du.

Becauseg ∈S, we obtain (3.4)

z g0(z) g(z)

≤ 1 +|z|

1− |z|

for allz ∈U.

We have

1− |z|2 Reα Re α

z p00(z) p0(z)

= 1− |z|2 Reα Re α

z g0(z) g(z) −1

(3.5)

≤ 1− |z|2 Reα

Reα |α−1|

zg0(z) g(z)

+ 1

.

From (3.5) and (3.4) we obtain (3.6) 1− |z|2 Reα

Re α

z p00(z) p0(z)

≤ 1− |z|2 Reα

Reα |α−1| 2 1− |z|. Now, we consider the cases

i1) 0<Reα <1.

The function

s: (0,1)→ <, s(x) = 1−a2x (0 < a < 1)

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is a increasing function and fora =|z|, z ∈U,we obtain

(3.7) 1− |z|2 Reα ≤ 1− |z|2

for allz ∈U.

From (3.6) and (3.7), we have

(3.8) 1− |z|2 Reα

Re α

zp00(z) p0(z)

≤ 4|α−1|

Re α for allz ∈U.

Using the condition (j1) and (3.8) we get

(3.9) 1− |z|2 Reα

Re α

zp00(z) p0(z)

≤1 for allz ∈U.

i2) Re α ≥1.

We observe that the function

q : [1,∞)→ <, q(x) = 1−a2x

x (0< a < 1)

is a decreasing function, and that, if we takea =|z|, z ∈U,then

(3.10) 1− |z|2 Reα

Reα ≤1− |z|2 for allz ∈U.

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Univalence Conditions for Certain Integral Operators

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From (3.6) and (3.10) we obtain

(3.11) 1− |z|2 Reα

Re α

zp00(z) p0(z)

≤4|α−1|.

From (3.11) and (j2), we have

(3.12) 1− |z|2 Reα

Re α

zp00(z) p0(z)

≤1 for allz ∈U.

Using (3.9), (3.12) and because p0(z) =

g(z) z

α−1

, from Lemma 2.1 for α =β it results that the functionGα(z)is in the classS.

Theorem 3.2. If α is a real number, α ∈ 4

5, 54

and the function g ∈ S(α), then the function

(3.13) Gα(z) =

α

Z z

0

gα−1(u)du α1

is in the classS.

Proof. Ifg ∈S(α),theng ∈ Sand by Theorem3.1forα ∈4

5, 54

,we obtain the functionGα(z)in the classS.

Theorem 3.3. Letα, γ be a complex numbers and the functionh∈S, h(z) = z+a2z2+· · · .

If

(p1) |γ| ≤ Reα

4 for Reα∈(0,1)

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Univalence Conditions for Certain Integral Operators

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or

(p2) |γ| ≤ 1

4 for Reα∈[1,∞) then the function

(3.14) Hα, γ(z) =

α Z z

0

uα−1

h(u) u

γ

du α1

is regular and univalent inU.

Proof. Let us consider the regular function inU,defined by

(3.15) f(z) =

Z z

0

h(u) u

γ

du.

For the functionh∈S, we obtain (3.16)

z h0(z) h(z)

≤ 1 +|z|

1− |z|

for allz ∈U.

We obtain

(3.17) 1− |z|2 Reα Re α

z f00(z) f0(z)

≤ 1− |z|2 Reα Reα |γ|

zh0(z) h(z)

+ 1

.

From (3.17) and (3.16), we have (3.18) 1− |z|2 Reα

Reα

z f00(z) f0(z)

≤ 1− |z|2 Reα

Re α |γ| 2 1− |z|

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Univalence Conditions for Certain Integral Operators

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We consider the cases j1) 0<Re α <1.

In this case we obtain

(3.19) 1− |z|2 Reα ≤ 1− |z|2 for allz ∈U.

From (3.18) and (3.19), we get

(3.20) 1− |z|2 Reα

Re α

zf00(z) f0(z)

≤ 4|γ|

Re α for allz ∈U.

By (3.20) and (p1) we have

(3.21) 1− |z|2 Reα

Re α

zf00(z) f0(z)

≤1 for allz ∈U.

j2) Reα ≥1.

For this case we obtain

(3.22) 1− |z|2 Reα

Reα ≤1− |z|2 for allz ∈U.

From (3.18) and (3.22) we have

(3.23) 1− |z|2 Reα

Re α

zf00(z) f0(z)

≤4|γ|.

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Univalence Conditions for Certain Integral Operators

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From (3.23) and (p2), we get

(3.24) 1− |z|2 Reα

Re α

zf00(z) f0(z)

≤1 for allz ∈U.

From (3.21), (3.24) and becausef0(z) =

h(z) z

γ

, from Lemma2.1forα = β it results that the functionHα,γ(z)is in the classS.

Remark 1. Forα = 1,from Theorem3.3we obtain Theorem1.1, the result due to Kim-Merkes.

Theorem 3.4. Letγ be a complex number and the functionh∈S(a).

If

(3.25) |γ| ≤ α

4 for α∈(0,1) or

(3.26) |γ| ≤ 1

4 for α∈[1,2]

then the functionHα, γ(z)defined by (3.14) is in the classS.

Proof. Because h(z) ∈S(α), 0 < α≤2,thenh(z)∈ Sand by Theorem3.3 the functionHα, γ(z)belongs to the classS.

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Univalence Conditions for Certain Integral Operators

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References

[1] Y.J. KIMANDE.P. MERKES, On an integral of powers of a spirallike func- tion, Kyungpook Math. J., 12(2) (1972), 249–253.

[2] P.T. MOCANU, T. BULBOAC ˘A ANDG.St. S ˘AL ˘AGEAN, The Geometric Theory of Univalent Functions, Cluj, 1999.

[3] Z. NEHARI, Conformal Mapping, Mc Graw-Hill Book Comp., New York, 1952 (Dover. Publ. Inc., 1975).

[4] N.N. PASCU, An improvement of Becker’s univalence criterion, Proceed- ings of the Commemorative Session Simion Stoilov, Bra¸sov, (1987), 43–48.

[5] V. PESCAR, New univalence criteria, "Transilvania" University of Bra¸sov, Bra¸sov, 2002.

[6] C. POMMERENKE, Univalent Functions, Vanderhoeck Ruprecht, G˝ottin- gen, 1975.

[7] D. YANG AND J. LIU, On a class of univalent functions, IJMMS, 22(3) (1999), 605–610.

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