• 検索結果がありません。

SOME CRITERIA FOR UNIVALENCE OF CERTAIN INTEGRAL OPERATORS

N/A
N/A
Protected

Academic year: 2022

シェア "SOME CRITERIA FOR UNIVALENCE OF CERTAIN INTEGRAL OPERATORS"

Copied!
6
0
0

読み込み中.... (全文を見る)

全文

(1)

PII. S0161171204401392 http://ijmms.hindawi.com

© Hindawi Publishing Corp.

SOME CRITERIA FOR UNIVALENCE OF CERTAIN INTEGRAL OPERATORS

VIRGIL PESCAR and SHIGEYOSHI OWA Received 21 January 2004 and in revised form 25 March 2004

We derive some criteria for univalence of certain integral operators for analytic functions in the open unit disk.

2000 Mathematics Subject Classification: 30C45.

1. Introduction. LetᏭbe the class of the functionsf (z)which are analytic in the open unit diskU= {z∈C:|z|<1}andf (0)=f(0)−1=0.

We denote by᏿the subclass ofᏭconsisting of functionsf (z)∈Ꮽwhich are univa- lent inU. Miller and Mocanu [1] have considered many integral operators for functions f (z)belonging to the classᏭ. In this paper, we consider the integral operators

Fα(z)= 1

α z

0

f (u)1/α

u1du α

(z∈U) (1.1)

forf (z)∈Ꮽand for someα∈C. It is well known thatFα(z)∈᏿forf (z)∈ and α >0, where᏿denotes the subclass of᏿consisting of all starlike functionsf (z)inU. 2. Preliminary results. To discuss our integral operators, we need the following theorems.

Theorem2.1[3]. Letαbe a complex number withRe(α) >0andf (z)∈. Iff (z) satisfies

1−|z|2 Re(α) Re(α)

zf(z) f(z)

1, (2.1)

for allz∈U, then the integral operator

Gα(z)=

α z

0uα1f(u)du 1/α

(2.2)

is in the class.

(2)

Theorem2.2[4]. Letαbe a complex number withRe(α) >0andf (z)∈. Iff (z) satisfies (2.1) for allz∈U, then, for any complex numberβwithRe(β)Re(α), the integral operator

Gβ(z)= β

z

0uβ1f(u)du 1/β

(2.3) is in the class.

Example2.3. Defining the functionf (z)by f (z)=

z 0

1+uRe(α) 1−uRe(α)

1/2

du (2.4)

with Re(α) >0, we have that

1−z2 Re(α) Re(α)

zf(z) f(z)

=zRe(α). (2.5)

Thus the functionf (z)satisfies the condition ofTheorem 2.2. Therefore, for Re(β) Re(α),

Gβ(z)= β

z 0

uβ−1

1+uRe(α) 1−uRe(α)

1/2

du 1/β

(2.6) is in the class᏿.

Theorem2.4[2]. If the functiong(z)is regular inU, then, for allξ∈Uandz∈U, g(z)satisfies

g(ξ)−g(z) 1−g(z)g(ξ)

ξ−z

1−zξ

, (2.7)

g(z)1−g(z)2

1−|z|2 . (2.8)

The equalities hold only in the caseg(z)=ε((z+u)/(1+uz)), where|ε| =1and|u|<1.

Remark2.5[2]. Forz=0, from inequality (2.7), g(ξ)−g(0)

1−g(0)g(ξ)

|ξ| (2.9)

and, hence

g(ξ) |ξ|+g(0)

1+g(0)|ξ|. (2.10)

Consideringg(0)=aandξ=z, we see that g(z) |z|+|a|

1+|a||z| (2.11)

for allz∈U.

(3)

Schwarz lemma[2]. If the functiong(z)is regular inU,g(0)=0, and|g(z)| 1, for allz∈U, then

g(z) |z| (2.12)

for allz∈U, and|g(0)| 1. The equality in (2.12) forz≠0holds only in the case g(z)=z, where|| =1.

3. Main results

Theorem3.1. Letαbe a complex number withRe(1/α)=a >0and the function g(z)∈satisfying

zg(z) g(z) 1

1 (z∈U). (3.1)

Then, for

|α| 2

(2a+1)(2a+1)/2a, (3.2)

the integral operator

Fα(z)= 1

α z

0

g(u)1/α

u1du α

(3.3) is in the class.

Proof. Let 1/α=β. Then we have F1/β(z)=

β

z 0

uβ−1 g(u)

u β

du 1/β

. (3.4)

We consider the function

f (z)= z

0

g(u) u

β

du. (3.5)

Then the function

h(z)= 1

|β|

zf(z)

f(z) (3.6)

is regular inUand the constant|β|satisfies the inequality

|β| (2a+1) 2

(2a+1)/2a

. (3.7)

From (3.5) and (3.6), we have that h(z)= β

|β|

zg(z) g(z) 1

. (3.8)

(4)

Using (3.8) and (3.1), we obtain

h(z)1 (z∈U). (3.9)

Noting thath(0)=0 and applying the Schwarz lemma forh(z), we get 1

|β|

zf(z) f(z)

|z| (z∈U) (3.10)

and hence we obtain 1−|z|2a

a

zf(z) f(z)

|β|

1−|z|2a a

|z| (z∈U). (3.11) Because

max|z|≤1

1−|z|2a a |z|

= 2

(2a+1)(2a+1)/2a, (3.12)

from (3.11) and (3.7), we have

1−|z|2a a

zf(z) f(z)

1 (3.13)

forz∈U. From (3.13) andTheorem 2.1, it follows that (2.4) belongs to the class᏿. By means of (2.4) and (3.5), we have that the integral operatorF1/β(z)is in the class

᏿, and hence we conclude that the integral operatorFα(z)is in the class᏿. Example3.2. If we take the functiong(z)=zezandα=1/a >0, then

g(z)=z+a2z2+a3z3+··· (3.14) is analytic inUand

zg(z) g(z) 1

= |z|<1 (z∈U). (3.15) Since the functiong(z)satisfies the condition ofTheorem 3.1, we have

Tα(z)= 1

α z

0

eu/αu1/α−1du α

. (3.16)

Theorem3.3. Letα,βbe complex numbers withRe(β)Re(α) >0and the function g(z)∈satisfying

zg(z)−g(z) zg(z)

1 (z∈U). (3.17)

Then, for

|α| max

|z|1

1−|z|2 Re(α) Re(α)

|z|

|z|+a2 1+a2|z|

, (3.18)

(5)

the integral operator

Fα,β(z)=

β z

0

g(u)1/α

uβ1/α1du 1/β

(3.19) is in the class.

Proof. We have

Fα,β(z)= β

z 0uβ1

g(u) u

1/α

du 1/β

. (3.20)

We consider the function

f (z)= z

0

g(u) u

1/α

du, (3.21)

which is regular inU. The function

p(z)= |α|f(z)

f(z), (3.22)

where the constant|α|satisfies inequality (3.18), is regular inU. From (3.22) and (3.21), we obtain

p(z)=|α| α

zg(z)−g(z) zg(z)

(3.23) and using (3.17), we have

p(z)<1 (z∈U) (3.24)

and|p(0)| = |a2|. ApplyingRemark 2.5, we obtain αf(z)

f(z)

|z|+a2

1+a2|z| (z∈U). (3.25) It follows that

1−|z|2 Re(α) Re(α)

zf(z) f(z)

1

|α|

1−|z|2 Re(α) Re(α)

|z|

|z|+a2 1+a2|z|

(3.26)

for allz∈U. We consider the function Q(x)=1−x2 Re(α)

Re(α)

x

x+a2 1+a2x

x= |z|;x∈[0,1]

. (3.27)

BecauseQ(1/2) >0,Q(x)satisfies

xmax[0,1]Q(x) >0. (3.28)

(6)

Using this fact, (3.26) gives us that 1−|z|2 Re(α)

Re(α)

zf(z) f(z)

1

|α|max

|z|1

1−|z|2 Re(α) Re(α)

|z|

|z|+a2 1+a2|z|

. (3.29) From (3.29) and (3.18), we obtain (2.1). Using (2.1) andTheorem 2.2, we obtain that the integral operator (2.4) belongs to the class᏿. Therefore, it follows from (2.4) and (3.21) thatFα,β(z)is in the class᏿.

Corollary3.4. Letαbe a complex number withRe(α) >0and the functiong(z)∈satisfying (3.18). Then, for

max|z|1

1−|z|2 Re(α) Re(α)

|z|

|z|+a2 1+a2|z|

|α| 1, (3.30)

the integral operator (3.3) is in the class.

Proof. FromTheorem 3.3forβ=1/α, the condition Re(β)Re(α) >0 is identical with|α|<1 and we haveFα,β(z)=Fα(z).

Acknowledgment. The authors thankfully acknowledge the kind and helpful ad- vice of Professor H. M. Srivastava on the paper.

References

[1] S. S. Miller and P. T. Mocanu,Differential Subordinations. Theory and applications, Mono- graphs and Textbooks in Pure and Applied Mathematics, vol. 225, Marcel Dekker, New York, 2000.

[2] Z. Nehari,Conformal Mapping, McGraw-Hill, New York, 1952.

[3] N. N. Pascu,On a univalence criterion. II, Itinerant Seminar on Functional Equations, Approx- imation and Convexity (Cluj-Napoca, 1985), Preprint, vol. 85, Universitatea “Babe¸s- Bolyai”, Cluj-Napoca, 1985, pp. 153–154.

[4] ,An improvement of Becker’s univalence criterion, Proceedings of the Commemora- tive Session: Simion Stoïlow (Bra¸sov, 1987), University of Bra¸sov, Bra¸sov, 1987, pp. 43–

48.

Virgil Pescar: Department of Mathematics, Faculty of Mathematics and Computer Science, Transilvania University of Bra¸sov, 2200 Bra¸sov, Romania

E-mail address:[email protected]

Shigeyoshi Owa: Department of Mathematics, Kinki University, Higashi-Osaka, Osaka 577- 8502, Japan

E-mail address:[email protected]

参照

関連したドキュメント

Mitsuo Izuki, Department of Mathematics, Faculty of Science, Hokkaido University, Kita 10 Nishi 8, Kita-ku, Sapporo, Hokkaido 060-0810, Japan. E-mail

Masayuki Horiguchi Department of Mathematics, Faculty of Science, Kanagawa University Address: Tsuchiya 2946, Hiratsuka City, Kanagawa Prefecture, 259‐1293, Japan

Nalinee Wechwiriyakul: Department of Mathematics, Faculty of Science, The University of the Thai Chamber of Commerce, Bangkok 10400 , Thailand. E-mail

Lenbury: Department of Mathematics, Faculty of Science, Mahidol University, 272 Rama 6 Road, Bangkok 10400, Thailand. E-mail

Amir Akbary: Department of Mathematics and Computer Science, University of Lethbridge, 4401 University Drive West, Lethbridge, AB, Canada T1K 3M4. E-mail

Kim: Institute of Science Education, Kongju National University, Kongju 314-701, Korea E-mail address: [email protected]. Jang: Department of Mathematics and Computer Science,