Journal of Inequalities and Applications
Volume 2009 (2009), Article ID 813687, 12 pages
doi:10.1155/2009/813687
Research Article

On Bounded Boundary and Bounded Radius Rotations

Department of Mathematics, COMSATS Institute of Information Technology, 44000 Islamabad, Pakistan

Received 6 January 2009; Revised 6 March 2009; Accepted 19 March 2009

Academic Editor: Narendra Kumar Govil

Copyright © 2009 K. I. Noor et al. 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 establish a relation between the functions of bounded boundary and bounded radius rotations by using three different techniques. A well-known result is observed as a special case from our main result. An interesting application of our work is also being investigated.

1. Introduction

Let 𝐴 be the class of functions 𝑓 of the form

𝑓 ( 𝑧 ) = 𝑧 + 𝑛 = 2 𝑎 𝑛 𝑧 𝑛 , ( 1 . 1 ) which are analytic in the unit disc 𝐸 = { 𝑧 | 𝑧 | < 1 } . We say that 𝑓 𝐴 is subordinate to 𝑔 𝐴 , written as 𝑓 𝑔 , if there exists a Schwarz function 𝑤 ( 𝑧 ) , which (by definition) is analytic in 𝐸 with 𝑤 ( 0 ) = 0 and | 𝑤 ( 𝑧 ) | < 1 ( 𝑧 𝐸 ) , such that 𝑓 ( 𝑧 ) = 𝑔 ( 𝑤 ( 𝑧 ) ) . In particular, when 𝑔 is univalent, then the above subordination is equivalent to 𝑓 ( 0 ) = 𝑔 ( 0 ) and 𝑓 ( 𝐸 ) 𝑔 ( 𝐸 ) .

For any two analytic functions

𝑓 ( 𝑧 ) = 𝑛 = 0 𝑎 𝑛 𝑧 𝑛 , 𝑔 ( 𝑧 ) = 𝑛 = 0 𝑏 𝑛 𝑧 𝑛 ( 𝑧 𝐸 ) , ( 1 . 2 ) the convolution (Hadamard product) of 𝑓 and 𝑔 is defined by

( 𝑓 𝑔 ) ( 𝑧 ) = 𝑛 = 0 𝑎 𝑛 𝑏 𝑛 𝑧 𝑛 ( 𝑧 𝐸 ) . ( 1 . 3 ) We denote by 𝑆 ( 𝛼 ) , 𝐶 ( 𝛼 ) , ( 0 𝛼 < 1 ) , the classes of starlike and convex functions of order 𝛼 , respectively, defined by

𝑆 ( 𝛼 ) = 𝑓 𝐴 : R e 𝑧 𝑓 ( 𝑧 ) , 𝑓 ( 𝑧 ) > 𝛼 , 𝑧 𝐸 𝐶 ( 𝛼 ) = 𝑓 𝐴 : 𝑧 𝑓 ( 𝑧 ) 𝑆 . ( 𝛼 ) , 𝑧 𝐸 ( 1 . 4 ) For 𝛼 = 0 , we have the well-known classes of starlike and convex univalent functions denoted by 𝑆 and 𝐶 , respectively.

Let 𝑃 𝑘 ( 𝛼 ) be the class of functions 𝑝 ( 𝑧 ) analytic in the unit disc 𝐸 satisfying the properties 𝑝 ( 0 ) = 1 and

0 2 𝜋 | | | R e 𝑝 ( 𝑧 ) 𝛼 | | | 1 𝛼 𝑑 𝜃 𝑘 𝜋 , ( 1 . 5 ) where 𝑧 = 𝑟 𝑒 𝑖 𝜃 , 𝑘 2 , and 0 𝛼 < 1 . For 𝛼 = 0 , we obtain the class 𝑃 𝑘 introduced in [1]. Also, for 𝑝 𝑃 𝑘 ( 𝛼 ) , we can write 𝑝 ( 𝑧 ) = ( 1 𝛼 ) 𝑞 1 ( 𝑧 ) + 𝛼 , 𝑞 1 𝑃 𝑘 . We can also write, for 𝑝 𝑃 𝑘 ( 𝛼 ) ,

1 𝑝 ( 𝑧 ) = 2 𝜋 0 2 𝜋 1 + ( 1 2 𝛼 ) 𝑧 𝑒 𝑖 𝑡 1 𝑧 𝑒 𝑖 𝑡 𝑑 𝜇 ( 𝑡 ) , 𝑧 𝐸 , ( 1 . 6 ) where 𝜇 ( 𝑡 ) is a function with bounded variation on [ 0 , 2 𝜋 ] such that

0 2 𝜋 𝑑 𝜇 ( 𝑡 ) = 2 𝜋 , 0 2 𝜋 | | | | 𝑑 𝜇 ( 𝑡 ) 𝑘 𝜋 . ( 1 . 7 ) For (1.6) together with (1.7), see [2]. Since 𝜇 ( 𝑡 ) has a bounded variation on [ 0 , 2 𝜋 ] , we may write 𝜇 ( 𝑡 ) = 𝐴 ( 𝑡 ) 𝐵 ( 𝑡 ) , where 𝐴 ( 𝑡 ) and 𝐵 ( 𝑡 ) are two non-negative increasing functions on [ 0 , 2 𝜋 ] satisfying (1.7) . Thus, if we set 𝐴 ( 𝑡 ) = ( ( 𝑘 / 4 ) + ( 1 / 2 ) ) 𝜇 1 ( 𝑡 ) and 𝐵 ( 𝑡 ) = ( ( 𝑘 / 4 ) ( 1 / 2 ) ) 𝜇 2 ( 𝑡 ) , then (1.6) becomes

𝑘 𝑝 ( 𝑧 ) = 4 + 1 2 1 2 𝜋 0 2 𝜋 1 + ( 1 2 𝛼 ) 𝑧 𝑒 𝑖 𝑡 1 𝑧 𝑒 𝑖 𝑡 𝑑 𝜇 1 𝑘 ( 𝑡 ) 4 1 2 1 2 𝜋 0 2 𝜋 1 + ( 1 2 𝛼 ) 𝑧 𝑒 𝑖 𝑡 1 𝑧 𝑒 𝑖 𝑡 𝑑 𝜇 2 ( 𝑡 ) . ( 1 . 8 ) Now, using Herglotz-Stieltjes formula for the class 𝑃 ( 𝛼 ) and (1.8), we obtain

𝑘 𝑝 ( 𝑧 ) = 4 + 1 2 𝑝 1 𝑘 ( 𝑧 ) 4 1 2 𝑝 2 ( 𝑧 ) , 𝑧 𝐸 , ( 1 . 9 ) where 𝑃 ( 𝛼 ) is the class of functions with real part greater than 𝛼 and 𝑝 𝑖 𝑃 ( 𝛼 ) , for 𝑖 = 1 , 2 .

We define the following classes:

𝑅 𝑘 ( 𝛼 ) = 𝑓 : 𝑓 𝐴 a n d 𝑧 𝑓 ( 𝑧 ) 𝑓 ( 𝑧 ) 𝑃 𝑘 , 𝑉 ( 𝛼 ) , 0 𝛼 < 1 𝑘 ( 𝛼 ) = 𝑓 : 𝑓 𝐴 a n d 𝑧 𝑓 ( 𝑧 ) 𝑓 ( 𝑧 ) 𝑃 𝑘 . ( 𝛼 ) , 0 𝛼 < 1 ( 1 . 1 0 ) We note that

𝑓 𝑉 𝑘 ( 𝛼 ) 𝑧 𝑓 𝑅 𝑘 ( 𝛼 ) . ( 1 . 1 1 ) For 𝛼 = 0 , we obtain the well-known classes 𝑅 𝑘 and 𝑉 𝑘 of analytic functions with bounded radius and bounded boundary rotations, respectively. These classes are studied by Noor [35] in more details. Also it can easily be seen that 𝑅 2 ( 𝛼 ) = 𝑆 ( 𝛼 ) and 𝑉 2 ( 𝛼 ) = 𝐶 ( 𝛼 ) .

Goel [6] proved that 𝑓 𝐶 ( 𝛼 ) implies that 𝑓 𝑆 ( 𝛽 ) , where

4 𝛽 = 𝛽 ( 𝛼 ) = 𝛼 ( 1 2 𝛼 ) 4 2 2 𝛼 + 1 1 , 𝛼 2 , 1 1 2 l n 2 , 𝛼 = 2 , ( 1 . 1 2 ) and this result is sharp.

In this paper, we prove the result of Goel [6] for the classes 𝑉 𝑘 ( 𝛼 ) and 𝑅 𝑘 ( 𝛼 ) by using three different methods. The first one is the same as done by Goel [6] , while the second and third are the convolution and subordination techniques.

2. Preliminary Results

We need the following results to obtain our results.

Lemma 2.1. Let 𝑓 𝑉 𝑘 ( 𝛼 ) . Then there exist 𝑠 1 , 𝑠 2 𝑆 ( 𝛼 ) such that 𝑓 𝑠 ( 𝑧 ) = 1 ( 𝑧 ) / 𝑧 ( 𝑘 / 4 ) + ( 1 / 2 ) 𝑠 2 ( 𝑧 ) / 𝑧 ( 𝑘 / 4 ) ( 1 / 2 ) , 𝑧 𝐸 . ( 2 . 1 )

Proof. It can easily be shown that 𝑓 𝑉 𝑘 ( 𝛼 ) if and only if there exists 𝑔 𝑉 𝑘 such that 𝑓 𝑔 ( 𝑧 ) = ( 𝑧 ) 1 𝛼 [ 2 ] , 𝑧 𝐸 , s e e . ( 2 . 2 ) From Brannan [7] representation form for functions with bounded boundary rotations, we have 𝑔 𝑔 ( 𝑧 ) = 1 ( 𝑧 ) 𝑧 𝑘 4 + 1 2 𝑔 2 ( 𝑧 ) 𝑧 𝑘 4 1 2 , 𝑔 𝑖 𝑆 , 𝑖 = 1 , 2 . ( 2 . 3 ) Now, it is shown in [8] that for 𝑠 𝑖 𝑆 ( 𝛼 ) , we can write 𝑠 𝑖 𝑔 ( 𝑧 ) = 𝑧 𝑖 ( 𝑧 ) 𝑧 1 𝛼 , 𝑔 𝑖 𝑆 , 𝑖 = 1 , 2 . ( 2 . 4 ) Using (2.3) together with (2.4) in (2.2), we obtain the required result.

Lemma 2.2 (see [9]). Let 𝑢 = 𝑢 1 + 𝑖 𝑢 2 , 𝑣 = 𝑣 1 + 𝑖 𝑣 2 , and Ψ ( 𝑢 , 𝑣 ) be a complex-valued function satisfying the conditions: (i) Ψ ( 𝑢 , 𝑣 ) is continuous in a domain 𝐷 2 , (ii) ( 1 , 0 ) 𝐷 and R e Ψ ( 1 , 0 ) > 0 , (iii) R e Ψ ( 𝑖 𝑢 2 , 𝑣 1 ) 0 , whenever ( 𝑖 𝑢 2 , 𝑣 1 ) 𝐷 and 𝑣 1 ( 1 / 2 ) ( 1 + 𝑢 2 2 ) .
If ( 𝑧 ) = 1 + 𝑐 1 𝑧 + is a function analytic in 𝐸 such that ( ( 𝑧 ) , 𝑧 ( 𝑧 ) ) 𝐷 and R e Ψ ( ( 𝑧 ) , 𝑧 ( 𝑧 ) ) > 0 for 𝑧 𝐸 , then R e ( 𝑧 ) > 0 in 𝐸 .

Lemma 2.3. Let 𝛽 > 0 , 𝛽 + 𝛾 > 0 , and 𝛼 [ 𝛼 0 , 1 ) , with 𝛼 0 = m a x 𝛽 𝛾 1 , 2 𝛽 𝛾 𝛽 . ( 2 . 5 ) If ( 𝑧 ) + 𝑧 ( 𝑧 ) 𝛽 ( 𝑧 ) + 𝛾 1 + ( 1 2 𝛼 ) 𝑧 1 𝑧 , ( 2 . 6 ) then ( 𝑧 ) 𝑄 ( 𝑧 ) 1 + ( 1 2 𝛼 ) 𝑧 1 𝑧 , ( 2 . 7 ) where 1 𝑄 ( 𝑧 ) = 𝛾 𝛽 𝐺 ( 𝑧 ) 𝛽 , 𝐺 ( 𝑧 ) = 1 0 1 𝑧 1 𝑡 𝑧 2 𝛽 ( 1 𝛼 ) 𝑡 𝛽 + 𝛾 1 𝑑 𝑡 = 2 𝐹 1 ( 2 𝛽 ( 1 𝛼 ) , 1 , 𝛽 + 𝛾 + 1 ; 𝑧 / ( 𝑧 1 ) ) , ( 𝛽 + 𝛾 ) ( 2 . 8 ) 2 𝐹 1 denotes Gauss hypergeometric function. From (2.7), one can deduce the sharp result that 𝑃 ( 𝛽 ) , with 𝛽 = 𝛽 ( 𝛼 , 𝛽 , 𝛾 ) = m i n R e 𝑄 ( 𝑧 ) = 𝑄 ( 1 ) . ( 2 . 9 ) This result is a special case of the one given in [10, page 113].

3. Main Results

By using the same method as that of Goel [6], we prove the following result. We include all the details for the sake of completeness.

3.1. First Method

Theorem 3.1. Let 𝑓 𝑉 𝑘 ( 𝛼 ) . Then 𝑓 𝑅 𝑘 ( 𝛽 ) , where 𝛽 = 𝛽 ( 𝛼 ) is given by (1.12). This result is sharp.

Proof. Since 𝑓 𝑉 𝑘 ( 𝛼 ) , we use Lemma 2.1, with relation (1.11) to have 1 + 𝑧 𝑓 ' ' ( 𝑧 ) 𝑓 = 𝑘 ( 𝑧 ) 4 + 1 2 𝑧 𝑠 1 ( 𝑧 ) 𝑠 1 𝑘 ( 𝑧 ) 4 1 2 𝑧 𝑠 2 ( 𝑧 ) 𝑠 2 = 𝑘 ( 𝑧 ) 4 + 1 2 𝑧 𝑓 1 ( 𝑧 ) 𝑓 1 ( 𝑘 𝑧 ) 4 1 2 𝑧 𝑓 2 ( 𝑧 ) 𝑓 2 ( , 𝑧 ) ( 3 . 1 ) where 𝑠 𝑖 𝑆 ( 𝛼 ) and 𝑓 𝑖 𝐶 ( 𝛼 ) , 𝑖 = 1 , 2 .
Therefore, from (2.4), we have 𝑧 𝑓 ( 𝑧 ) = 𝑘 𝑓 ( 𝑧 ) 4 + 1 2 𝑧 𝑔 1 ( 𝑧 ) / 𝑧 1 𝛼 𝑧 0 𝑔 1 ( 𝜙 ) / 𝜙 1 𝛼 𝑘 𝑑 𝜙 4 1 2 𝑧 𝑔 2 ( 𝑧 ) / 𝑧 1 𝛼 𝑧 0 𝑔 2 ( 𝜙 ) / 𝜙 1 𝛼 𝑑 𝜙 , ( 3 . 2 ) that is, 𝑧 𝑓 ( 𝑧 ) 𝑓 = 𝑘 ( 𝑧 ) 4 + 1 2 𝑧 0 𝑧 𝜙 1 𝛼 𝑔 1 ( 𝜙 ) 𝑔 1 ( 𝑧 ) 1 𝛼 𝑑 𝜙 𝑧 1 𝑘 4 1 2 𝑧 0 𝑧 𝜙 1 𝛼 𝑔 2 ( 𝜙 ) 𝑔 2 ( 𝑧 ) 1 𝛼 𝑑 𝜙 𝑧 1 , ( 3 . 3 ) where we integrate along the straight line segment [ 0 , 𝑧 ] , 𝑧 𝐸 .
Writing 𝑧 𝑓 ( 𝑧 ) = 𝑘 𝑓 ( 𝑧 ) 4 + 1 2 𝑝 1 𝑘 ( 𝑧 ) 4 + 1 2 𝑝 2 ( 𝑧 ) , ( 3 . 4 ) and using (3.3) , we have 𝑝 𝑖 ( 𝑧 ) = 𝑧 0 𝑧 𝜙 1 𝛼 𝑔 𝑖 ( 𝜙 ) 𝑔 𝑖 ( 𝑧 ) 1 𝛼 𝑑 𝜙 𝑧 1 , ( 3 . 5 ) where 𝑝 𝑖 ( 0 ) = 1 and hence by [11] we have | | | | 𝑝 𝑖 ( 𝑧 ) 1 + 𝑟 2 1 𝑟 2 | | | | 2 𝑟 1 𝑟 2 , | 𝑧 | = 𝑟 , 𝑧 𝐸 . ( 3 . 6 ) Therefore, m i n 𝑓 𝑖 𝐶 ( 𝛼 ) m i n | 𝑧 | = 𝑟 𝑝 R e 𝑖 ( 𝑧 ) = m i n 𝑓 𝑖 𝐶 ( 𝛼 ) m i n | 𝑧 | = 𝑟 | | 𝑝 𝑖 | | ( 𝑧 ) . ( 3 . 7 ) Let 𝑧 = 𝑟 𝑒 𝑖 𝜃 and 𝜙 = 𝑅 𝑒 𝑖 𝜃 , 0 < 𝑅 < 𝑟 < 1 . For fixed 𝑧 and 𝜙 , we have from (2.4) | | | | 𝑔 𝑖 ( 𝜙 ) 𝑔 𝑖 | | | | 𝑅 ( 𝑧 ) 𝑟 1 + 𝑟 1 + 𝑅 2 . ( 3 . 8 ) Now, using (3.8), we have, for a fixed 𝑧 𝐸 , | 𝑧 | = 𝑟 , | | | | 𝑧 0 𝑧 𝜙 1 𝛼 𝑔 𝑖 ( 𝜙 ) 𝑔 𝑖 ( 𝑧 ) 1 𝛼 𝑑 𝜙 𝑧 | | | | 𝑟 0 1 + 𝑟 1 + 𝑅 2 ( 1 𝛼 ) 𝑑 𝑅 𝑟 . ( 3 . 9 ) Let 𝑇 ( 𝑟 ) = 𝑟 0 1 + 𝑟 1 + 𝑅 2 ( 1 𝛼 ) 𝑑 𝑅 𝑟 , ( 3 . 1 0 ) with 𝑅 = 𝑟 𝑡 , 0 < 𝑡 < 1 , we have 𝑇 ( 𝑟 ) = 1 0 1 + 𝑟 1 + 𝑟 𝑡 2 ( 1 𝛼 ) 𝑑 𝑡 . ( 3 . 1 1 ) By differentiating we note that 𝑇 ( 𝑟 ) = 2 ( 1 𝛼 ) 1 0 ( 1 𝑡 ) ( 1 + 𝑟 𝑡 ) 2 1 + 𝑟 1 + 𝑟 𝑡 ( 1 2 𝛼 ) 𝑑 𝑡 > 0 , ( 3 . 1 2 ) and therefore 𝑇 ( 𝑟 ) is a monotone increasing function of 𝑟 and hence m a x 0 𝑟 1 𝑇 ( 𝑟 ) = 𝑇 ( 1 ) = 2 2 ( 1 𝛼 ) 1 0 𝑑 𝑡 ( 1 + 𝑡 ) 2 ( 1 𝛼 ) = 2 4 ( 1 𝛼 ) 1 ( 2 𝛼 1 ) , i f 𝛼 2 1 2 l n 2 , i f 𝛼 = 2 . ( 3 . 1 3 ) By letting | | | | 𝛽 ( 𝛼 ) = m i n 𝑧 0 𝑧 𝜙 1 𝛼 𝑔 𝑖 ( 𝜙 ) 𝑔 𝑖 ( 𝑧 ) 1 𝛼 𝑑 𝜙 𝑧 | | | | 1 , 𝑧 𝐸 , ( 3 . 1 4 ) for all 𝑔 𝑖 ( 𝑧 ) 𝑆 , we obtain the required result from (3.7), (3.13), and (3.14).
Sharpness can be shown by the function 𝑓 0 𝑉 𝑘 ( 𝛼 ) given by 𝑧 𝑓 0 ( 𝑧 ) 𝑓 0 = 𝑘 ( 𝑧 ) 4 + 1 2 1 ( 1 2 𝛼 ) 𝑧 𝑘 1 + 𝑧 4 1 2 1 + ( 1 2 𝛼 ) 𝑧 1 𝑧 . ( 3 . 1 5 ) It is easy to check that 𝑓 0 𝑅 𝑘 ( 𝛽 ) , where 𝛽 is the exact value given by (1.12).

3.2. Second Method

Theorem 3.2. Let 𝑓 𝑉 𝑘 ( 𝛼 ) . Then 𝑓 𝑅 𝑘 ( 𝛽 ) , where 1 𝛽 = 4 ( 2 𝛼 1 ) + 4 𝛼 2 4 𝛼 + 9 . ( 3 . 1 6 )

Proof. Let 𝑧 𝑓 ( 𝑧 ) 𝑘 𝑓 ( 𝑧 ) = ( 1 𝛽 ) 𝑝 ( 𝑧 ) + 𝛽 = ( 1 𝛽 ) 4 + 1 2 𝑝 1 𝑘 ( 𝑧 ) 4 1 2 𝑝 2 ( 𝑧 ) + 𝛽 ( 3 . 1 7 ) 𝑝 ( 𝑧 ) is analytic in 𝐸 with 𝑝 ( 0 ) = 1 . Then 𝑧 𝑓 ( 𝑧 ) 𝑓 = ( 𝑧 ) ( 1 𝛽 ) 𝑝 ( 𝑧 ) + 𝛽 + ( 1 𝛽 ) 𝑧 𝑝 ( 𝑧 ) ( 1 𝛽 ) 𝑝 ( 𝑧 ) + 𝛽 , ( 3 . 1 8 ) that is, 1 1 𝛼 𝑧 𝑓 ( 𝑧 ) 𝑓 = 1 ( 𝑧 ) 𝛼 1 𝛼 ( 1 𝛽 ) 𝑝 ( 𝑧 ) + 𝛽 𝛼 + ( 1 𝛽 ) 𝑧 𝑝 ( 𝑧 ) = ( 1 𝛽 ) 𝑝 ( 𝑧 ) + 𝛽 ( 𝛽 𝛼 ) + 1 𝛼 ( 1 𝛽 ) 1 𝛼 𝑝 ( 𝑧 ) + ( 1 / ( 1 𝛽 ) ) 𝑧 𝑝 ( 𝑧 ) . 𝑝 ( 𝑧 ) + ( 𝛽 / ( 1 𝛽 ) ) ( 3 . 1 9 ) Since 𝑓 𝑉 𝑘 ( 𝛼 ) , it implies that ( 𝛽 𝛼 ) + 1 𝛼 ( 1 𝛽 ) 1 𝛼 𝑝 ( 𝑧 ) + ( 1 / ( 1 𝛽 ) ) 𝑧 𝑝 ( 𝑧 ) 𝑝 ( 𝑧 ) + ( 𝛽 / ( 1 𝛽 ) ) 𝑃 𝑘 , 𝑧 𝐸 . ( 3 . 2 0 ) We define 𝜑 𝑎 , 𝑏 1 ( 𝑧 ) = 𝑧 1 + 𝑏 ( 1 𝑧 ) 𝑎 + 𝑏 𝑧 1 + 𝑏 ( 1 𝑧 ) 1 + 𝑎 , ( 3 . 2 1 ) with 𝑎 = 1 / ( 1 𝛽 ) , 𝑏 = 𝛽 / ( 1 𝛽 ) . By using (3.17) with convolution techniques, see [5], we have that 𝜑 𝑎 , 𝑏 ( 𝑧 ) 𝑧 𝑘 𝑝 ( 𝑧 ) = 4 + 1 2 𝜑 𝑎 , 𝑏 ( 𝑧 ) 𝑧 𝑝 1 𝑘 ( 𝑧 ) 4 1 2 𝜑 𝑎 , 𝑏 ( 𝑧 ) 𝑧 𝑝 2 ( 𝑧 ) ( 3 . 2 2 ) implies 𝑝 ( 𝑧 ) + 𝑎 𝑧 𝑝 ( 𝑧 ) = 𝑘 𝑝 ( 𝑧 ) + 𝑏 4 + 1 2 𝑝 1 ( 𝑧 ) + 𝑎 𝑧 𝑝 1 ( 𝑧 ) 𝑝 1 𝑘 ( 𝑧 ) + 𝑏 4 1 2 𝑝 2 ( 𝑧 ) + 𝑎 𝑧 𝑝 2 ( 𝑧 ) 𝑝 2 ( 𝑧 ) + 𝑏 . ( 3 . 2 3 ) Thus, from (3.20) and (3.23), we have ( 𝛽 𝛼 ) + 1 𝛼 ( 1 𝛽 ) 𝑝 1 𝛼 𝑖 ( 𝑧 ) + 𝑎 𝑧 𝑝 𝑖 ( 𝑧 ) 𝑝 𝑖 ( 𝑧 ) + 𝑏 𝑃 , 𝑖 = 1 , 2 . ( 3 . 2 4 ) We now form the functional Ψ ( 𝑢 , 𝑣 ) by choosing 𝑢 = 𝑝 𝑖 ( 𝑧 ) , 𝑣 = 𝑧 𝑝 𝑖 ( 𝑧 ) in (3.24) and note that the first two conditions of Lemma 2.2 are clearly satisfied. We check condition (iii) as follows: 𝜓 R e 𝑖 𝑢 2 , 𝑣 1 = 1 𝑣 1 𝛼 ( 𝛽 𝛼 ) + R e 1 𝑖 𝑢 2 = 1 + ( 𝛽 / ( 1 𝛽 ) ) 𝑣 1 𝛼 ( 𝛽 𝛼 ) + 1 ( 𝛽 / ( 1 𝛽 ) ) 𝑢 2 2 + ( 𝛽 / ( 1 𝛽 ) ) 2 1 1 1 𝛼 ( 𝛽 𝛼 ) 2 1 + 𝑢 2 2 ( 𝛽 / ( 1 𝛽 ) ) 𝑢 2 2 + ( 𝛽 / ( 1 𝛽 ) ) 2 = 𝑢 2 ( 𝛽 𝛼 ) 2 2 + ( 𝛽 / ( 1 𝛽 ) ) 2 1 + 𝑢 2 2 ( 𝛽 / ( 1 𝛽 ) ) 2 𝑢 2 2 + ( 𝛽 / ( 1 𝛽 ) ) 2 = 𝛽 ( 1 𝛼 ) 2 ( 𝛽 𝛼 ) 2 / ( 1 𝛽 ) 2 ( 𝛽 / ( 1 𝛽 ) ) + ( 2 𝛽 2 𝛼 ( 𝛽 / ( 1 𝛽 ) ) ) 𝑢 2 2 2 𝑢 2 2 + ( 𝛽 / ( 1 𝛽 ) ) 2 = ( 1 𝛼 ) 𝐴 + 𝐵 𝑢 2 2 2 𝐶 , 2 𝐶 > 0 , ( 3 . 2 5 ) where 𝛽 𝐴 = ( 1 𝛽 ) 2 [ ] , 1 2 ( 𝛽 𝛼 ) 𝛽 ( 1 𝛽 ) 𝐵 = ( 𝑢 1 𝛽 2 ( 𝛽 𝛼 ) ( 1 𝛽 ) 𝛽 ) , 𝐶 = ( 1 𝛼 ) 2 2 + 𝛽 1 𝛽 2 > 0 . ( 3 . 2 6 ) The right-hand side of (3.25) is negative if 𝐴 0 and 𝐵 0 . From 𝐴 0 , we have 1 𝛽 = 𝛽 ( 𝛼 ) = 4 ( 2 𝛼 1 ) + 4 𝛼 2 4 𝛼 + 9 , ( 3 . 2 7 ) and from 𝐵 0 , it follows that 0 𝛽 < 1 .
Since all the conditions of Lemma 2.2 are satisfied, it follows that 𝑝 𝑖 𝑃 in 𝐸 for 𝑖 = 1 , 2 and consequently 𝑝 𝑃 𝑘 and hence 𝑓 𝑅 𝑘 ( 𝛽 ) , where 𝛽 is given by (3.16). The case 𝑘 = 2 is discussed in [12].

3.3. Third Method

Theorem 3.3. Let 𝑓 𝑉 𝑘 ( 𝛼 ) . Then 𝑓 𝑅 𝑘 ( 𝛽 ) , where 𝛽 = 𝛽 1 ( 𝛼 , 1 , 0 ) = 2 𝛼 1 2 2 2 ( 1 𝛼 ) 1 , i f 𝛼 2 , 1 1 2 l n 2 , i f 𝛼 = 2 . ( 3 . 2 8 )

Proof. Let 𝑧 𝑓 ( 𝑧 ) 𝑘 𝑓 ( 𝑧 ) = 𝑝 ( 𝑧 ) = 4 + 1 2 𝑧 𝑠 1 ( 𝑧 ) 𝑠 1 𝑘 ( 𝑧 ) 4 1 2 𝑧 𝑠 2 ( 𝑧 ) 𝑠 2 ( 𝑧 ) , ( 3 . 2 9 ) and let 𝑧 𝑠 𝑖 ( 𝑧 ) 𝑠 𝑖 ( 𝑧 ) = 𝑝 𝑖 ( 𝑧 ) , 𝑖 = 1 , 2 . ( 3 . 3 0 ) Then 𝑝 , 𝑝 𝑖 are analytic in 𝐸 with 𝑝 ( 0 ) = 1 , 𝑝 𝑖 ( 0 ) = 1 , 𝑖 = 1 , 2 .
Logarithmic differentiation yields 𝑧 𝑓 ( 𝑧 ) 𝑓 ( 𝑧 ) = 𝑝 ( 𝑧 ) + 𝑧 𝑝 ( 𝑧 ) = 𝑘 𝑝 ( 𝑧 ) 4 + 1 2 𝑧 𝑠 1 ( 𝑧 ) 𝑠 1 𝑘 ( 𝑧 ) 4 1 2 𝑧 𝑠 2 ( 𝑧 ) 𝑠 2 = 𝑘 ( 𝑧 ) 4 + 1 2 𝑝 1 ( 𝑧 ) + 𝑧 𝑝 1 ( 𝑧 ) 𝑝 1 𝑘 ( 𝑧 ) 4 1 2 𝑝 2 ( 𝑧 ) + 𝑧 𝑝 2 ( 𝑧 ) 𝑝 2 . ( 𝑧 ) ( 3 . 3 1 ) Since 𝑓 𝑉 𝑘 ( 𝛼 ) , it follows that ( 𝑧 𝑠 𝑖 ) / 𝑠 𝑖 𝑃 ( 𝛼 ) , 𝑧 𝐸 , or 𝑠 𝑖 𝐶 ( 𝛼 ) for 𝑧 𝐸 . Consequently, 𝑝 𝑖 ( 𝑧 ) + 𝑧 𝑝 𝑖 ( 𝑧 ) 𝑝 𝑖 ( 𝑧 ) 𝑃 ( 𝛼 ) , ( 3 . 3 2 ) where 𝑧 𝑠 𝑖 ( 𝑧 ) / 𝑠 𝑖 ( 𝑧 ) = 𝑝 𝑖 ( 𝑧 ) , 𝑖 = 1 , 2 . We use Lemma 2.3 with 𝛾 = 0 , 𝛽 = 1 > 0 , 𝛼 [ 0 , 1 ) , and = 𝑝 𝑖 in (3.32), to have 𝑝 𝑖 𝑃 ( 𝛽 ) , where 𝛽 is given in (3.28) and this estimate is best possible, extremal function 𝑄 is given by 𝑄 ( 𝑧 ) = ( 1 2 𝛼 ) 𝑧 ( 1 𝑧 ) 1 ( 1 𝑧 ) 1 2 𝛼 1 , i f 𝛼 2 , 𝑧 1 ( 𝑧 1 ) l o g ( 1 𝑧 ) , i f 𝛼 = 2 , ( 3 . 3 3 ) see [10]. MacGregor [13] conjectured the exact value given by (3.28). Thus 𝑠 𝑖 𝑆 ( 𝛽 ) and consequently 𝑓 𝑅 𝑘 ( 𝛽 ) , where the exact value of 𝛽 is given by (3.28).

3.4. Application of Theorem 3.3

Theorem 3.4. Let 𝑔 and belong to 𝑉 𝑘 ( 𝛼 ) . Then 𝐹 ( 𝑧 ) , defined by 𝐹 ( 𝑧 ) = 𝑧 0 𝑔 ( 𝑡 ) 𝑡 𝜇 ( 𝑡 ) 𝑡 𝜂 𝑑 𝑡 , ( 3 . 3 4 ) is in the class 𝑉 𝑘 ( 𝛿 ) , where 0 𝜇 < 𝜂 1 , 𝛿 = 𝛿 ( 𝛼 ) = ( 1 ( 𝜇 + 𝜂 ) ( 1 𝛽 ) ) , and 𝛽 ( 𝛼 ) is given by (1.12).

Proof. From (3.34), we can easily write ( 𝑧 𝐹 ( 𝑧 ) ) 𝐹 ( 𝑧 ) = 𝜇 𝑧 𝑔 ( 𝑧 ) 𝑔 ( 𝑧 ) + 𝜂 𝑧 ( 𝑧 ) ( 𝑧 ) + 1 ( 𝜇 + 𝜂 ) . ( 3 . 3 5 ) Since 𝑔 and belong to 𝑉 𝑘 ( 𝛼 ) , then, by Theorem 3.3, 𝑧 𝑔 ( 𝑧 ) / 𝑔 ( 𝑧 ) and 𝑧 ( 𝑧 ) / ( 𝑧<