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Journal of Function Spaces and Applications
Volume 2012 (2012), Article ID 207410, 8 pages
http://dx.doi.org/10.1155/2012/207410
Research Article

On Univalence Criteria for a General Integral Operator

1Department of Mathematics, University of Piteşti, Târgu din Vale Street, No. 1, 110040 Piteşti, Romania
2Department of Mathematics, “1 Decembrie 1918” University of Alba Iulia, Nicolae Iorga Street, No. 11–13, 510009 Alba Iulia, Romania

Received 5 February 2012; Revised 19 May 2012; Accepted 19 May 2012

Academic Editor: Gestur Ólafsson

Copyright © 2012 Vasile Marius Macarie and Daniel Breaz. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We consider a new general integral operator, and we give sufficient conditions for the univalence of this integral operator in the open unit disk of the complex plane. Several consequences of the main results are also shown.

1. Introduction

Let 𝑈={𝑧|𝑧|<1} be the unit disk of the complex plane and let 𝒜 be the class of functions 𝑓 of the form 𝑓(𝑧)=𝑧+𝑚=2𝑎𝑚𝑧𝑚,(1.1) that are analytic in 𝑈 and satisfy the usual normalization conditions 𝑓(0)=𝑓(0)1=0. We denote by 𝑆 the subclass of 𝒜 consisting of all univalent functions 𝑓 in 𝑈 and consider the class 𝑃 of functions that are analytic in 𝑈 and that satisfy (0)=1 and Re(𝑧)>0 for all 𝑧𝑈.

In the present paper we obtain sufficient conditions for the following general integral operator to be in the class 𝑆 (The univalent functions are of importance in geometric functions theory and may have some applications in fluid mechanics and physics): 𝐻𝛼1,,𝛼𝑛,𝛽,𝑔1,,𝑔𝑛,1,,𝑝𝛽(𝑧)=𝑧0𝑢𝑛𝛽1𝑖=1𝑔𝑖(𝑢)𝛼𝑖𝑝𝑗=1𝑗(𝑢)𝑑𝑢1/𝛽,(1.2) where 𝛼𝑖,𝛽 are complex numbers, 𝛽0, the functions 𝑔𝑖(𝑢)𝒜 for all 𝑖=1,2,,𝑛 and 𝑗(𝑢)𝑃 for all 𝑗=1,2,,𝑝, where 𝑛,𝑝 are positive integers.

For proving our main results we need the following theorems.

Theorem 1.1 (see [1]). Let 𝛼 be a complex number, let Re𝛼>0; and let 𝑓𝒜. If 1|𝑧|2Re𝛼||||Re𝛼𝑧𝑓(𝑧)𝑓||||(𝑧)1,(1.3) for all 𝑧𝑈, then for any complex number 𝛽 with Re𝛽Re𝛼, the function 𝐹𝛽𝛽(𝑧)=𝑧0𝑢𝛽1𝑓(𝑢)𝑑𝑢1/𝛽(1.4) is in the class S.

Theorem 1.2 (see [2]). If the function 𝑔(𝑧) is regular in 𝑈 and |𝑔(𝑧)|<1 in 𝑈, then for all 𝜉𝑈 and 𝑧𝑈, the following inequalities hold: ||||𝑔(𝜉)𝑔(𝑧)1||||||||𝑔(𝑧)𝑔(𝜉)𝜉𝑧1||||,||𝑔𝑧𝜉(1.5)||||||(𝑧)1𝑔(𝑧)21|𝑧|2.(1.6) These are equalities if and only if 𝑔(𝑧)=𝜀(𝑧+𝑢)/(1+̄𝑢𝑧), where |𝜀|=1 and |𝑢|<1.

Remark 1.3 (see [2]). For 𝑧=0, from inequality (1.5) we have ||||𝑔(𝜉)𝑔(0)1||||||𝜉||,𝑔(0)𝑔(𝜉)(1.7) and hence, ||||||𝜉||+||||𝑔(𝜉)𝑔(0)||||||𝜉||1+𝑔(0).(1.8) Writing 𝑔(0)=𝑎 and letting 𝜉=𝑧, we get ||||𝑔(𝑧)|𝑧|+|𝑎|,1+|𝑎||𝑧|(1.9) for all 𝑧𝑈.

2. Main Results

Theorem 2.1. Let 𝛿>0. For 𝑖=1,2,,𝑛, let 𝛼𝑖 be a complex number, let 𝑀𝑖>0, and let 𝑔𝑖𝒜. For 𝑗=1,2,,𝑝, let 𝑁𝑗>0 and 𝑗𝑃. If ||||𝑧𝑔𝑖(𝑧)𝑔𝑖||||(𝑧)𝑀𝑖||||,𝑖=1,2,,𝑛(𝑧𝑈),(2.1)𝑧𝑗(𝑧)𝑗||||(𝑧)𝑁𝑗,𝑗=1,2,,𝑝(𝑧𝑈),(2.2)𝑛𝑖=1||𝛼𝑖||𝑀𝑖+𝑝𝑗=1𝑁𝑗𝛿,(2.3) then for every complex number 𝛽 with Re𝛽𝛿>0, the integral operator 𝐻𝛼1,,𝛼𝑛,𝛽,𝑔1,,𝑔𝑛,1,,𝑝(𝑧) given by (1.2) is in the class 𝑆.

Proof. Let us define the function: 𝑓(𝑧)=𝑧0𝑛𝑖=1𝑔𝑖(𝑢)𝛼𝑖𝑝𝑗=1𝑗(𝑢)𝑑𝑢,(2.4) with 𝑔𝑖𝒜, for all 𝑖=1,2,,𝑛 and 𝑗𝑃, for all 𝑗=1,2,,𝑝, and, thus, we obtain 𝑓(𝑧)=𝑛𝑖=1𝑔𝑖(𝑧)𝛼𝑖𝑝𝑗=1𝑗(𝑧).(2.5) The function 𝑓 is regular in 𝑈 and 𝑓(0)=𝑓(0)1=0. We have 𝑧𝑓(𝑧)𝑓=(𝑧)𝑛𝑖=1𝛼𝑖𝑧𝑔𝑖(𝑧)𝑔𝑖+(𝑧)𝑝𝑗=1𝑧𝑗(𝑧)𝑗(𝑧)(𝑧𝑈).(2.6) From (2.1), (2.2), and (2.6) we obtain 1|𝑧|2𝛿𝛿||||𝑧𝑓(𝑧)𝑓||||(𝑧)1|𝑧|2𝛿𝛿𝑛𝑖=1||𝛼𝑖||𝑀𝑖+𝑝𝑗=1𝑁𝑗(𝑧𝑈),(2.7) and by (2.3), we have 1|𝑧|2𝛿𝛿||||𝑧𝑓(𝑧)𝑓||||(𝑧)1,𝑧𝑈.(2.8) Using (2.8), by Theorem 1.1, it results that the integral operator 𝐻𝛼1,,𝛼𝑛,𝛽,𝑔1,,𝑔𝑛,1,,𝑝(𝑧) given by (1.2) is in the class 𝑆.

Letting 𝑝=1 in Theorem 2.1, we have the following.

Corollary 2.2. Let 𝛿>0,  let 𝑁>0, and let 𝑃. For 𝑖=1,2,,𝑛, let 𝛼𝑖 be a complex number, let 𝑀𝑖>0, and let 𝑔𝑖𝒜. If ||||𝑧𝑔𝑖(𝑧)𝑔𝑖||||(𝑧)𝑀𝑖||||,𝑖=1,2,,𝑛(𝑧𝑈),𝑧(𝑧)||||(𝑧)𝑁(𝑧𝑈)𝑛𝑖=1||𝛼𝑖||𝑀𝑖+𝑁𝛿,(2.9) then for every complex number 𝛽 with Re𝛽𝛿>0, the integral operator 𝐹𝛼1,,𝛼𝑛,𝛽,𝑔1,,𝑔𝑛,𝛽(𝑧)=𝑧0𝑢𝑛𝛽1𝑖=1𝑔𝑖(𝑢)𝛼𝑖(𝑢)𝑑𝑢1/𝛽(2.10) is in the class 𝑆.

Letting 𝑛=1 in Theorem 2.1, we have the following.

Corollary 2.3. Let 𝛿>0, let 𝑀>0, let 𝑔𝒜, and let 𝛼 be a complex number. For 𝑗=1,2,,𝑝, let 𝑁𝑗>0 and let 𝑗𝑃. If ||||𝑧𝑔(𝑧)𝑔||||||||(𝑧)𝑀(𝑧𝑈),𝑧𝑗(𝑧)𝑗||||(𝑧)𝑁𝑗,𝑗=1,2,,𝑝(𝑧𝑈),|𝛼|𝑀+𝑝𝑗=1𝑁𝑗𝛿,(2.11) then for every complex number 𝛽 with Re𝛽𝛿>0, the integral operator 𝐺𝛼,𝛽,𝑔,1,,𝑝𝛽(𝑧)=𝑧0𝑢𝛽1𝑔(𝑢)𝛼𝑝𝑗=1𝑗(𝑢)𝑑𝑢1/𝛽(2.12) is in the class 𝑆.

For 𝑀1=𝑀2==𝑀𝑛=𝑀 and 𝑁1=𝑁2==𝑁𝑝=𝑁 in Theorem 2.1, we have the following.

Corollary 2.4. Let 𝛿>0, let 𝑀>0, and let 𝑁>0. For 𝑖=1,2,,𝑛, let 𝛼𝑖 be a complex number and let 𝑔𝑖𝒜. For 𝑗=1,2,,𝑝, let 𝑗𝑃. If ||||𝑧𝑔𝑖(𝑧)𝑔𝑖||||||||(𝑧)𝑀,𝑖=1,2,,𝑛(𝑧𝑈),𝑧𝑗(𝑧)𝑗||||𝑀(𝑧)𝑁,𝑗=1,2,,𝑝(𝑧𝑈),𝑛𝑖=1||𝛼𝑖||+𝑝𝑁𝛿,(2.13) then for every complex number 𝛽 with Re𝛽𝛿>0, the integral operator 𝐻𝛼1,,𝛼𝑛,𝛽,𝑔1,,𝑔𝑛,1,,𝑝(𝑧) given by (1.2) is in the class 𝑆.

Theorem 2.5. Let 𝛿>0. For 𝑖=1,2,,𝑛, let 𝛼𝑖 be a complex number, let 𝑀𝑖>0, and let 𝑔𝑖𝒜,𝑔𝑖(𝑧)=𝑧+𝑎𝑖2𝑧2+𝑎𝑖3𝑧3+. For 𝑗=1,2,,𝑝, let 𝑁𝑗>0 and 𝑗𝑃 with 𝑗(0)=0, and 𝑐=𝑛𝑖=1𝛼𝑖𝑎𝑖2/|𝑛𝑖=1𝛼𝑖|. If ||||𝑔𝑖(𝑧)𝑔𝑖||||(𝑧)<𝑀𝑖||||,𝑖=1,2,,𝑛(𝑧𝑈),(2.14)𝑗(𝑧)𝑗||||(𝑧)<𝑁𝑗,𝑗=1,2,,𝑝(𝑧𝑈),(2.15)𝑛𝑖=1||𝛼𝑖||𝑀𝑖+𝑝𝑗=1𝑁𝑗||𝑛𝑖=1𝛼𝑖|||||||<1,(2.16)𝑛𝑖=1𝛼𝑖|||||1max|𝑧|<11|𝑧|2𝛿/𝛿|𝑧|((|𝑧|+2|𝑐|)/(1+2|𝑐||𝑧|)),(2.17) then for every complex number 𝛽 with Re𝛽𝛿>0, the integral operator 𝐻𝛼1,,𝛼𝑛,𝛽,𝑔1,,𝑔𝑛,1,,𝑝(𝑧) given by (1.2) is in the class 𝑆.

Proof. We define the function: 𝑓(𝑧)=𝑧0𝑛𝑖=1𝑔𝑖(𝑢)𝛼𝑖𝑝𝑗=1𝑗(𝑢)𝑑𝑢(𝑧𝑈),(2.18) with 𝑔𝑖𝒜, for all 𝑖=1,2,,𝑛 and 𝑗𝑃, for all 𝑗=1,2,,𝑝.
We consider the function 1𝐾(𝑧)=||𝑛𝑖=1𝛼𝑖||𝑓(𝑧)𝑓(𝑧),𝑧𝑈.(2.19) We have 𝑓(𝑧)𝑓=(𝑧)𝑛𝑖=1𝛼𝑖𝑔𝑖(𝑧)𝑔𝑖+(𝑧)𝑝𝑗=1𝑗(𝑧)𝑗(𝑧)(𝑧𝑈).(2.20) From (2.14), (2.15), and (2.20) we obtain 1||𝑛𝑖=1𝛼𝑖||||||𝑓(𝑧)𝑓||||(𝑧)𝑛𝑖=1||𝛼𝑖||𝑀𝑖+𝑝𝑗=1𝑁𝑗||𝑛𝑖=1𝛼𝑖||(𝑧𝑈).(2.21) From (2.16), (2.19), and (2.21) we obtain |𝐾(𝑧)|<1 for all 𝑧𝑈.
We have 𝐾(0)=2𝑛𝑖=1𝛼𝑖𝑎𝑖2/|𝑛𝑖=1𝛼𝑖|=2𝑐, and, using Remark 1.3 we get ||||𝐾(𝑧)|𝑧|+2|𝑐|1+2|𝑐||𝑧|(𝑧𝑈).(2.22) From (2.19) and (2.22), we obtain 1|𝑧|2𝛿𝛿||||𝑧𝑓(𝑧)𝑓|||||||||(𝑧)𝑛𝑖=1𝛼𝑖|||||max|𝑧|<11|𝑧|2𝛿𝛿|𝑧||𝑧|+2|𝑐|1+2|𝑐||𝑧|,(2.23) for all 𝑧𝑈.
From (2.17) and (2.23), we have 1|𝑧|2𝛿𝛿||||𝑧𝑓(𝑧)𝑓||||(𝑧)1(𝑧𝑈).(2.24) So, applying Theorem 1.1, we obtain that the integral operator 𝐻𝛼1,,𝛼𝑛,𝛽,𝑔1,,𝑔𝑛,1,,𝑝(𝑧) given by (1.2) is in the class 𝑆.

Letting 𝑝=1 in Theorem 2.5, we have the following.

Corollary 2.6. Let 𝛿>0, let 𝑁>0, and let 𝑃 with (0)=0. For 𝑖=1,2,,𝑛, let 𝛼𝑖 be a complex number, 𝑀𝑖>0 and 𝑔𝑖𝒜,𝑔𝑖(𝑧)=𝑧+𝑎𝑖2𝑧2+𝑎𝑖3𝑧3+, and 𝑐=𝑛𝑖=1𝛼𝑖𝑎𝑖2/|𝑛𝑖=1𝛼𝑖|. If ||||𝑔𝑖(𝑧)𝑔𝑖||||(𝑧)<𝑀𝑖||||,𝑖=1,2,,𝑛(𝑧𝑈),(𝑧)||||(𝑧)<𝑁(𝑧𝑈),𝑛𝑖=1||𝛼𝑖||𝑀𝑖+𝑁||𝑛𝑖=1𝛼𝑖|||||||<1,𝑛𝑖=1𝛼𝑖|||||1max|𝑧|<11|𝑧|2𝛿,/𝛿|𝑧|((|𝑧|+2|𝑐|)/(1+2|𝑐||𝑧|))(2.25) then for every complex number 𝛽 with Re𝛽𝛿>0, the integral operator 𝐹𝛼1,,𝛼𝑛,𝛽,𝑔1,,𝑔𝑛,𝛽(𝑧)=𝑧0𝑢𝑛𝛽1𝑖=1𝑔𝑖(𝑢)𝛼𝑖(𝑢)𝑑𝑢1/𝛽(2.26) is in the class 𝑆.

Letting 𝑛=1 in Theorem 2.5, we have the following.

Corollary 2.7. Let 𝛿>0, 𝑀>0,  𝑔𝒜,𝑔(𝑧)=𝑧+𝑎2𝑧2+𝑎3𝑧3+ and let 𝛼 be a complex number. For 𝑗=1,2,,𝑝, let 𝑁𝑗>0 and 𝑗𝑃 with 𝑗(0)=0. If ||||𝑔(𝑧)𝑔||||||||(𝑧)<𝑀(𝑧𝑈),𝑗(𝑧)𝑗||||(𝑧)<𝑁𝑗,𝑗=1,2,,𝑝(𝑧𝑈),|𝛼|M+𝑝𝑗=1𝑁𝑗1<|𝛼|max|𝑧|<11|𝑧|2𝛿||𝑎/𝛿|𝑧||𝑧|+22||/||𝑎1+22||,|𝑧|(2.27) then for every complex number 𝛽 with Re𝛽𝛿>0, the integral operator 𝐺𝛼,𝛽,𝑔,1,,𝑝𝛽(𝑧)=𝑧0𝑢𝛽1𝑔(𝑢)𝛼𝑝𝑗=1𝑗(𝑢)𝑑𝑢1/𝛽(2.28) is in the class 𝑆.

For 𝑀1=𝑀2==𝑀𝑛=𝑀 and 𝑁1=𝑁2==𝑁𝑝=𝑁 in Theorem 2.5, we have the following.

Corollary 2.8. Let 𝛿>0, 𝑀>0, 𝑁>0. For 𝑖=1,2,,𝑛, let 𝛼𝑖 be a complex number and 𝑔𝑖𝒜,𝑔𝑖(𝑧)=𝑧+𝑎𝑖2𝑧2+𝑎𝑖3𝑧3+. For 𝑗=1,2,,𝑝, let 𝑗𝑃 with 𝑗(0)=0, and 𝑐=𝑛𝑖=1𝛼𝑖𝑎𝑖2/|𝑛𝑖=1𝛼𝑖|.
If ||||𝑔𝑖(𝑧)𝑔𝑖||||||||(𝑧)<𝑀𝑖=1,2,,𝑛(𝑧𝑈),𝑗(𝑧)𝑗||||𝑀(𝑧)<𝑁𝑗=1,2,,𝑝(𝑧𝑈),𝑛𝑖=1||𝛼𝑖||+𝑝𝑁||𝑛𝑖=1𝛼𝑖|||||||<1,𝑛𝑖=1𝛼𝑖|||||1max|𝑧|<11|𝑧|2𝛿,/𝛿|𝑧|((|𝑧|+2|𝑐|)/(1+2|𝑐||𝑧|))(2.29) then for every complex number 𝛽 with Re𝛽𝛿>0, the integral operator 𝐻𝛼1,,𝛼𝑛,𝛽,𝑔1,,𝑔𝑛,1,,𝑝(𝑧) given by (1.2) is in the class 𝑆.

References

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