Abstract

Let 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛾 , 𝛼 ) ( − 1 ≤ 𝐵 < 1 , 𝐵 < 𝐴 , 0 < 𝛾 ≤ 1 and 𝛼 > 0 ) denote the class of functions of the form ∑ 𝑓 ( 𝑧 ) = 𝑧 + ∞ 𝑘 = 𝑛 + 1 𝑎 𝑘 𝑧 𝑘 ( 𝑛 ∈ 𝑁 = { 1 , 2 , 3 , … } ) , which are analytic in the open unit disk 𝑈 and satisfy the following subordination condition 𝑓 ( 𝑧 ) + 𝛼 𝑧 𝑓 ( 𝑧 ) ≺ ( ( 1 + 𝐴 𝑧 ) / ( 1 + 𝐵 𝑧 ) ) 𝛾 , for ( 𝑧 ∈ 𝑈 ; 𝐴 ≤ 1 ; 0 < 𝛾 < 1 ) , ( 1 + 𝐴 𝑧 ) / ( 1 + 𝐵 𝑧 ) , for ( 𝑧 ∈ 𝑈 ; 𝛾 = 1 ) . We obtain sharp bounds on R e 𝑓 ( 𝑧 ) , R e 𝑓 ( 𝑧 ) / 𝑧 , | 𝑓 ( 𝑧 ) | , and coefficient estimates for functions 𝑓 ( 𝑧 ) belonging to the class 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛾 , 𝛼 ) . Conditions for univalency and starlikeness, convolution properties, and the radius of convexity are also considered.

1. Introduction

Let 𝐴 𝑛 denote the class of functions of the form 𝑓 ( 𝑧 ) = 𝑧 + ∞ 𝑘 = 𝑛 + 1 𝑎 𝑘 𝑧 𝑘 ( 𝑛 ∈ 𝑁 = { 1 , 2 , 3 , … } ) , ( 1 . 1 ) which are analytic in the open unit disk 𝑈 = { 𝑧 ∶ 𝑧 ∈ 𝐶 a n d | 𝑧 | < 1 } . Let 𝑆 𝑛 and 𝑆 ∗ 𝑛 denote the subclasses of 𝐴 𝑛 whose members are univalent and starlike, respectively.

For functions 𝑓 ( 𝑧 ) and 𝑔 ( 𝑧 ) analytic in 𝑈 , we say that 𝑓 ( 𝑧 ) is subordinate to 𝑔 ( 𝑧 ) in 𝑈 and we write 𝑓 ( 𝑧 ) ≺ 𝑔 ( 𝑧 ) ( 𝑧 ∈ 𝑈 ) , if there exists an analytic function 𝑤 ( 𝑧 ) in 𝑈 such that | | | | 𝑤 ( 𝑧 ) ≤ | 𝑧 | , 𝑓 ( 𝑧 ) = 𝑔 ( 𝑤 ( 𝑧 ) ) ( 𝑧 ∈ 𝑈 ) . ( 1 . 2 ) Furthermore, if the function 𝑔 ( 𝑧 ) is univalent in 𝑈 , then 𝑓 ( 𝑧 ) ≺ 𝑔 ( 𝑧 ) ( 𝑧 ∈ 𝑈 ) ⟺ 𝑓 ( 0 ) = 𝑔 ( 0 ) , 𝑓 ( 𝑈 ) ⊂ 𝑔 ( 𝑈 ) . ( 1 . 3 ) Throughout our present discussion, we assume that 𝑛 ∈ 𝑁 , − 1 ≤ 𝐵 < 1 , 𝐵 < 𝐴 , 𝛼 > 0 , 𝛽 < 1 , 0 < 𝛾 ≤ 1 . ( 1 . 4 ) We introduce the following subclass of 𝐴 𝑛 .

Definition 1.1. A function 𝑓 ( 𝑧 ) ∈ 𝐴 𝑛 is said to be in the class 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛾 , 𝛼 ) if it satisfies 𝑓 ( 𝑧 ) + 𝛼 𝑧 𝑓 ( 𝑧 ) ≺ ℎ ( 𝑧 ) ( 𝑧 ∈ 𝑈 ) , ( 1 . 5 ) where ⎧ ⎪ ⎨ ⎪ ⎩ ℎ ( 𝑧 ) = 1 + 𝐴 𝑧 1 + 𝐵 𝑧 𝛾 , ( 𝐴 ≤ 1 ; 0 < 𝛾 < 1 ) , 1 + 𝐴 𝑧 1 + 𝐵 𝑧 , ( 𝛾 = 1 ) . ( 1 . 6 ) The classes 𝑇 1 ( 1 − 2 𝛽 , − 1 , 1 , 1 ) = 𝑅 ( 𝛽 ) ( 𝛽 = 0 o r 𝛽 < 1 ) , 𝑇 1 𝑅 ( 𝐴 , 0 , 1 , 𝛼 ) = ( 𝛼 , 𝐴 ) ( 𝐴 > 0 ) ( 1 . 7 ) have been studied by several authors (see [1–5]). Recently, Gao and Zhou [6] showed some mapping properties of the following subclass of 𝐴 1 : 𝑅 𝑓 ( 𝛽 , 𝛼 ) = ( 𝑧 ) ∈ 𝐴 1 𝑓 ∶ R e ( 𝑧 ) + 𝛼 𝑧 𝑓 ( 𝑧 ) > 𝛽 ( 𝑧 ∈ 𝑈 ) . ( 1 . 8 ) Note that 𝑅 ( 𝛽 , 1 ) = 𝑅 ( 𝛽 ) , 𝑇 1 ( 1 − 2 𝛽 , − 1 , 1 , 𝛼 ) = 𝑅 ( 𝛽 , 𝛼 ) . ( 1 . 9 ) For further information of the above classes (with 𝛾 = 1 ) and related analytic function classes, see Srivastava et al. [7], Yang and Liu [8], Kim [9], and Kim and Srivastava [10].

In this paper, we obtain sharp bounds on R e 𝑓 ( 𝑧 ) , R e ( 𝑓 ( 𝑧 ) / 𝑧 ) , | 𝑓 ( 𝑧 ) | , and coefficient estimates for functions 𝑓 ( 𝑧 ) belonging to the class 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛾 , 𝛼 ) . Conditions for univalency and starlikeness, convolution properties, and the radius of convexity are also presented. One can see that the methods used in [6] do not work for the more general class 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛾 , 𝛼 ) than 𝑅 ( 𝛽 , 𝛼 ) .

2. The bounds on R e 𝑓 ( 𝑧 ) , R e ( 𝑓 ( 𝑧 ) / 𝑧 ) , and | 𝑓 ( 𝑧 ) | in 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛾 , 𝛼 )

In this section, we let 𝜆 𝑚 ⎧ ⎪ ⎨ ⎪ ⎩ ( 𝐴 , 𝐵 , 𝛾 ) = 𝑚 𝑗 = 0 𝛾 𝑗 𝐴 − 𝛾 𝑚 − 𝑗 𝑗 𝐵 𝑚 − 𝑗 , ( 𝐴 ≤ 1 ; 0 < 𝛾 < 1 ) , ( 𝐴 − 𝐵 ) ( − 𝐵 ) 𝑚 − 1 , ( 𝛾 = 1 ) , ( 2 . 1 ) where 𝑚 ∈ 𝑁 and 𝛾 𝑗 = ⎧ ⎪ ⎨ ⎪ ⎩ 𝛾 ( 𝛾 − 1 ) ⋯ ( 𝛾 − 𝑗 + 1 ) 𝑗 ! , ( 𝑗 = 1 , 2 , … , 𝑚 ) , 1 , ( 𝑗 = 0 ) . ( 2 . 2 ) With (2.1), it is easily seen that the function ℎ ( 𝑧 ) given by (1.6) can be expressed as ℎ ( 𝑧 ) = 1 + ∞ 𝑚 = 1 𝜆 𝑚 ( 𝐴 , 𝐵 , 𝛾 ) 𝑧 𝑚 ( 𝑧 ∈ 𝑈 ) . ( 2 . 3 )

Theorem 2.1. Let 𝑓 ( 𝑧 ) ∈ 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛾 , 𝛼 ) . Then, for | 𝑧 | = 𝑟 < 1 , R e 𝑓 ( 𝑧 ) ≥ 1 + ∞ 𝑚 = 1 ( − 1 ) 𝑚 𝜆 𝑚 ( 𝐴 , 𝐵 , 𝛾 ) 𝑟 𝛼 𝑛 𝑚 + 1 𝑛 𝑚 , R e 𝑓 ( 𝑧 ) ≤ 1 + ∞ 𝑚 = 1 𝜆 𝑚 ( 𝐴 , 𝐵 , 𝛾 ) 𝑟 𝛼 𝑛 𝑚 + 1 𝑛 𝑚 . ( 2 . 4 ) The bounds in (2.4) are sharp for the function 𝑓 𝑛 ( 𝑧 ) defined by 𝑓 𝑛 ( 𝑧 ) = 𝑧 + ∞ 𝑚 = 1 𝜆 𝑚 ( 𝐴 , 𝐵 , 𝛾 ) 𝑧 ( 𝑛 𝑚 + 1 ) ( 𝛼 𝑛 𝑚 + 1 ) 𝑛 𝑚 + 1 ( 𝑧 ∈ 𝑈 ) . ( 2 . 5 )

Proof. The analytic function ℎ ( 𝑧 ) given by (1.6) is convex (univalent) in 𝑈 (cf. [11]) and satisfies ℎ ( 𝑧 ) = ℎ ( 𝑧 ) ( 𝑧 ∈ 𝑈 ) . Thus, for | 𝜁 | ≤ 𝜎 ( 𝜁 ∈ 𝐶 a n d 𝜎 < 1 ) , ℎ ( − 𝜎 ) ≤ R e ℎ ( 𝜁 ) ≤ ℎ ( 𝜎 ) . ( 2 . 6 )
Let 𝑓 ( 𝑧 ) ∈ 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛾 , 𝛼 ) . Then, we can write 𝑓 ( 𝑧 ) + 𝛼 𝑧 𝑓 ( 𝑧 ) = ℎ ( 𝑤 ( 𝑧 ) ) ( 𝑧 ∈ 𝑈 ) , ( 2 . 7 ) where 𝑤 ( 𝑧 ) = 𝑤 𝑛 𝑧 𝑛 + 𝑤 𝑛 + 1 𝑧 𝑛 + 1 + ⋯ is analytic and | 𝑤 ( 𝑧 ) | < 1 for 𝑧 ∈ 𝑈 . By the Schwarz lemma, we know that | 𝑤 ( 𝑧 ) | ≤ | 𝑧 | 𝑛 ( 𝑧 ∈ 𝑈 ) . It follows from (2.7) that 𝑧 1 / 𝛼 𝑓 ( 𝑧 ) = 1 𝛼 𝑧 ( 1 / 𝛼 ) − 1 ℎ ( 𝑤 ( 𝑧 ) ) , ( 2 . 8 ) which leads to 𝑓 1 ( 𝑧 ) = 𝛼 𝑧 − 1 / 𝛼 𝑧 0 𝜁 ( 1 / 𝛼 ) − 1 ℎ ( 𝑤 ( 𝜁 ) ) 𝑑 𝜁 ( 2 . 9 ) or to 𝑓 1 ( 𝑧 ) = 𝛼 1 0 𝑡 ( 1 / 𝛼 ) − 1 ℎ ( 𝑤 ( 𝑡 𝑧 ) ) 𝑑 𝑡 ( 𝑧 ∈ 𝑈 ) . ( 2 . 1 0 ) Since | | | | 𝑤 ( 𝑡 𝑧 ) ≤ ( 𝑡 𝑟 ) 𝑛 ( | 𝑧 | = 𝑟 < 1 ; 0 ≤ 𝑡 ≤ 1 ) , ( 2 . 1 1 ) we deduce from (2.6) and (2.10) that 1 𝛼 1 0 𝑡 ( 1 / 𝛼 ) − 1 ℎ ( − ( 𝑡 𝑟 ) 𝑛 ) 𝑑 𝑡 ≤ R e 𝑓 1 ( 𝑧 ) ≤ 𝛼 1 0 𝑡 ( 1 / 𝛼 ) − 1 ℎ ( ( 𝑡 𝑟 ) 𝑛 ) 𝑑 𝑡 . ( 2 . 1 2 ) Now, by using (2.3) and (2.12), we can obtain (2.4).
Furthermore, for the function 𝑓 𝑛 ( 𝑧 ) defined by (2.5), we find that 𝑓 𝑛 ( 𝑧 ) = 1 + ∞ 𝑚 = 1 𝜆 𝑚 ( 𝐴 , 𝐵 , 𝛾 ) 𝑧 𝛼 𝑛 𝑚 + 1 𝑛 𝑚 , 𝑓 ( 2 . 1 3 ) 𝑛 ( 𝑧 ) + 𝛼 𝑧 𝑓 𝑛 ( 𝑧 ) = 1 + ∞ 𝑚 = 1 𝜆 𝑚 ( 𝐴 , 𝐵 , 𝛾 ) 𝑧 𝑛 𝑚 = ℎ ( 𝑧 𝑛 ) ≺ ℎ ( 𝑧 ) ( 𝑧 ∈ 𝑈 ) . ( 2 . 1 4 ) Hence, 𝑓 𝑛 ( 𝑧 ) ∈ 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛾 , 𝛼 ) and from (2.13), we see that the bounds in (2.4) are the best possible.
Hereafter, we write 𝑇 𝑛 ( 𝐴 , 𝐵 , 1 , 𝛼 ) = 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛼 ) . ( 2 . 1 5 )

Corollary 2.2. Let 𝑓 ( 𝑧 ) ∈ 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛼 ) . Then, for 𝑧 ∈ 𝑈 , R e 𝑓 ( 𝑧 ) > 1 − ( 𝐴 − 𝐵 ) ∞ 𝑚 = 1 𝐵 𝑚 − 1 , 𝛼 𝑛 𝑚 + 1 ( 2 . 1 6 ) R e 𝑓 ( 𝑧 ) < 1 + ( 𝐴 − 𝐵 ) ∞ 𝑚 = 1 ( − 𝐵 ) 𝑚 − 1 𝛼 𝑛 𝑚 + 1 ( 𝐵 ≠ − 1 ) . ( 2 . 1 7 ) The results are sharp.

Proof. For 𝛾 = 1 , it follows from (2.12) (used in the proof of Theorem 2.1) that R e 𝑓 1 ( 𝑧 ) > 𝛼 1 0 𝑡 ( 1 / 𝛼 ) − 1 1 − 𝐴 𝑡 𝑛 1 − 𝐵 𝑡 𝑛 𝑑 𝑡 , R e 𝑓 1 ( 𝑧 ) < 𝛼 1 0 𝑡 ( 1 / 𝛼 ) − 1 1 + 𝐴 𝑡 𝑛 1 + 𝐵 𝑡 𝑛 𝑑 𝑡 ( 𝐵 ≠ − 1 ) , ( 2 . 1 8 ) for 𝑧 ∈ 𝑈 . From these, we have the desired results.

The bounds in (2.16) and (2.17) are sharp for the function 𝑓 𝑛 ( 𝑧 ) = 𝑧 + ( 𝐴 − 𝐵 ) ∞ 𝑚 = 1 ( − 𝐵 ) 𝑚 − 1 𝑧 ( 𝑛 𝑚 + 1 ) ( 𝛼 𝑛 𝑚 + 1 ) 𝑛 𝑚 + 1 ∈ 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛼 ) . ( 2 . 1 9 )

Theorem 2.3. Let 𝑓 ( 𝑧 ) ∈ 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛾 , 𝛼 ) . Then, for | 𝑧 | = 𝑟 < 1 , R e 𝑓 ( 𝑧 ) 𝑧 ≥ 1 + ∞ 𝑚 = 1 ( − 1 ) 𝑚 𝜆 𝑚 ( 𝐴 , 𝐵 , 𝛾 ) 𝑟 ( 𝑛 𝑚 + 1 ) ( 𝛼 𝑛 𝑚 + 1 ) 𝑛 𝑚 , R e 𝑓 ( 𝑧 ) 𝑧 ≤ 1 + ∞ 𝑚 = 1 𝜆 𝑚 ( 𝐴 , 𝐵 , 𝛾 ) 𝑟 ( 𝑛 𝑚 + 1 ) ( 𝛼 𝑛 𝑚 + 1 ) 𝑛 𝑚 . ( 2 . 2 0 ) The results are sharp.

Proof. Noting that 𝑓 ( 𝑧 ) = 𝑧 1 0 𝑓 ( 𝑢 𝑧 ) 𝑑 𝑢 , R e 𝑓 ( 𝑧 ) 𝑧 = 1 0 R e 𝑓 ( 𝑢 𝑧 ) 𝑑 𝑢 ( 𝑧 ∈ 𝑈 ) , ( 2 . 2 1 ) an application of Theorem 2.1 yields (2.20). Furthermore, the results are sharp for the function 𝑓 𝑛 ( 𝑧 ) defined by (2.5).

Corollary 2.4. Let 𝑓 ( 𝑧 ) ∈ 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛼 ) . Then, for 𝑧 ∈ 𝑈 , R e 𝑓 ( 𝑧 ) 𝑧 > 1 − ( 𝐴 − 𝐵 ) ∞ 𝑚 = 1 𝐵 𝑚 − 1 , ( 𝑛 𝑚 + 1 ) ( 𝛼 𝑛 𝑚 + 1 ) R e 𝑓 ( 𝑧 ) 𝑧 < 1 + ( 𝐴 − 𝐵 ) ∞ 𝑚 = 1 ( − 𝐵 ) 𝑚 − 1 . ( 𝑛 𝑚 + 1 ) ( 𝛼 𝑛 𝑚 + 1 ) ( 2 . 2 2 ) The results are sharp for the function 𝑓 𝑛 ( 𝑧 ) defined by (2.19).

Proof. For 𝑓 ( 𝑧 ) ∈ 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛼 ) , it follows from (2.6) and (2.10) (with 𝛾 = 1 ) that 1 𝛼 1 0 𝑡 ( 1 / 𝛼 ) − 1 1 − 𝐴 ( 𝑢 𝑡 ) 𝑛 1 − 𝐵 ( 𝑢 𝑡 ) 𝑛 𝑑 𝑡 < R e 𝑓 1 ( 𝑢 𝑧 ) < 𝛼 1 0 𝑡 ( 1 / 𝛼 ) − 1 1 + 𝐴 ( 𝑢 𝑡 ) 𝑛 1 + 𝐵 ( 𝑢 𝑡 ) 𝑛 𝑑 𝑡 , ( 2 . 2 3 ) for 𝑧 ∈ 𝑈 and 0 < 𝑢 ≤ 1 . Making use of (2.21) and (2.23), we can obtain (2.22).

Theorem 2.5. Let 𝑓 ( 𝑧 ) ∈ 𝑇 1 ( 𝐴 , 𝐵 , 𝛼 ) and 𝑔 ( 𝑧 ) ∈ 𝑇 1 ( 𝐴 0 , 𝐵 0 , 𝛼 0 ) ( − 1 ≤ 𝐵 0 < 1 , 𝐵 0 < 𝐴 0 and 𝛼 0 > 0 ). If 𝐴 0 − 𝐵 0 ∞ 𝑚 = 1 𝐵 0 𝑚 − 1 𝛼 ( 𝑚 + 1 ) 0 ≤ 1 𝑚 + 1 2 , ( 2 . 2 4 ) then ( 𝑓 ∗ 𝑔 ) ( 𝑧 ) ∈ 𝑇 1 ( 𝐴 , 𝐵 , 𝛼 ) , where the symbol ∗ stands for the familiar Hadamard product (or convolution) of two analytic functions in 𝑈 .

Proof. Since 𝑔 ( 𝑧 ) ∈ 𝑇 1 ( 𝐴 0 , 𝐵 0 , 𝛼 0 ) ( − 1 ≤ 𝐵 0 < 1 , 𝐵 0 < 𝐴 0 and 𝛼 0 > 0 ), it follows from Corollary 2.4 (with 𝑛 = 1 ) and (2.24) that R e 𝑔 ( 𝑧 ) 𝑧 𝐴 > 1 − 0 − 𝐵 0 ∞ 𝑚 = 1 𝐵 0 𝑚 − 1 𝛼 ( 𝑚 + 1 ) 0 ≥ 1 𝑚 + 1 2 ( 𝑧 ∈ 𝑈 ) . ( 2 . 2 5 ) Thus, 𝑔 ( 𝑧 ) / 𝑧 has the Herglotz representation 𝑔 ( 𝑧 ) 𝑧 = | 𝑥 | = 1 𝑑 𝜇 ( 𝑥 ) ( 1 − 𝑥 𝑧 𝑧 ∈ 𝑈 ) , ( 2 . 2 6 ) where 𝜇 ( 𝑥 ) is a probability measure on the unit circle | 𝑥 | = 1 and ∫ | 𝑥 | = 1 𝑑 𝜇 ( 𝑥 ) = 1 .
For 𝑓 ( 𝑧 ) ∈ 𝑇 1 ( 𝐴 , 𝐵 , 𝛼 ) , we have ( 𝑓 ∗ 𝑔 ) ( 𝑧 ) + 𝛼 𝑧 ( 𝑓 ∗ 𝑔 ) ( 𝑧 ) = 𝐹 ( 𝑧 ) ∗ 𝑔 ( 𝑧 ) 𝑧 ( 𝑧 ∈ 𝑈 ) , ( 2 . 2 7 ) where 𝐹 ( 𝑧 ) = 𝑓 ( 𝑧 ) + 𝛼 𝑧 𝑓 ( 𝑧 ) ≺ 1 + 𝐴 𝑧 1 + 𝐵 𝑧 ( 𝑧 ∈ 𝑈 ) . ( 2 . 2 8 ) In view of the function ( 1 + 𝐴 𝑧 ) / ( 1 + 𝐵 𝑧 ) is convex (univalent) in 𝑈 , we deduce from (2.26) to (2.28) that ( 𝑓 ∗ 𝑔 ) ( 𝑧 ) + 𝛼 𝑧 ( 𝑓 ∗ 𝑔 ) ( 𝑧 ) = | 𝑥 | = 1 𝐹 ( 𝑥 𝑧 ) 𝑑 𝜇 ( 𝑥 ) ≺ 1 + 𝐴 𝑧 ( 1 + 𝐵 𝑧 𝑧 ∈ 𝑈 ) . ( 2 . 2 9 ) This shows that ( 𝑓 ∗ 𝑔 ) ( 𝑧 ) ∈ 𝑇 1 ( 𝐴 , 𝐵 , 𝛼 ) .

Corollary 2.6. Let 𝑓 ( 𝑧 ) ∈ 𝑇 1 ( 𝐴 , 𝐵 , 𝛼 ) , 𝑔 ( 𝑧 ) ∈ 𝑅 ( 𝛽 , 1 ) and 𝜋 𝛽 ≥ − 2 − 9 1 2 − 𝜋 2 . ( 2 . 3 0 ) Then, ( 𝑓 ∗ 𝑔 ) ( 𝑧 ) ∈ 𝑇 1 ( 𝐴 , 𝐵 , 𝛼 ) .

Proof. By taking 𝐴 0 = 1 − 2 𝛽 , 𝐵 0 = − 1 and 𝛼 0 = 1 , (2.24) in Theorem 2.5 becomes 2 ( 1 − 𝛽 ) ∞ 𝑚 = 1 ( − 1 ) 𝑚 − 1 ( 𝑚 + 1 ) 2 𝜋 = 2 ( 1 − 𝛽 ) 1 − 2 ≤ 1 1 2 2 , ( 2 . 3 1 ) that is, 𝜋 𝛽 ≥ − 2 − 9 1 2 − 𝜋 2 . ( 2 . 3 2 ) Hence, the desired result follows as a special case from Theorem 2.5.

Remark 2.7. R. Singh and S. Singh [4, Theorem 3] proved that, if 𝑓 ( 𝑧 ) and 𝑔 ( 𝑧 ) belong to 𝑅 ( 0 , 1 ) , then ( 𝑓 ∗ 𝑔 ) ( 𝑧 ) ∈ 𝑅 ( 0 , 1 ) . Obviously, for − 𝜋 2 − 9 1 2 − 𝜋 2 ≤ 𝛽 < 0 , ( 2 . 3 3 ) Corollary 2.6 generalizes and improves Theorem 3 in [4].

Theorem 2.8. Let 𝑓 ( 𝑧 ) ∈ 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛾 , 𝛼 ) and 𝐴 𝐵 ≤ 1 . Then, for | 𝑧 | = 𝑟 < 1 , | | | | 𝑓 ( 𝑧 ) ≤ 𝑟 + ∞ 𝑚 = 1 𝜆 𝑚 ( 𝐴 , 𝐵 , 𝛾 ) 𝑟 ( 𝑛 𝑚 + 1 ) ( 𝛼 𝑛 𝑚 + 1 ) 𝑛 𝑚 + 1 . ( 2 . 3 4 ) The result is sharp, with the extremal function 𝑓 𝑛 ( 𝑧 ) defined by (2.5).

Proof. It is well known that for 𝜁 ∈ 𝐶 and | 𝜁 | ≤ 𝜎 < 1 , | | | | 1 + 𝐴 𝜁 − 1 + 𝐵 𝜁 1 − 𝐴 𝐵 𝜎 2 1 − 𝐵 2 𝜎 2 | | | | ≤ ( 𝐴 − 𝐵 ) 𝜎 1 − 𝐵 2 𝜎 2 . ( 2 . 3 5 ) Since 𝐴 𝐵 ≤ 1 , we have 1 − 𝐴 𝐵 𝜎 2 > 0 and so (2.35) leads to | | | | 1 + 𝐴 𝜁 | | | | 1 + 𝐵 𝜁 𝛾 ≤ | | | | 1 − 𝐴 𝐵 𝜎 2 1 − 𝐵 2 𝜎 2 | | | | + ( 𝐴 − 𝐵 ) 𝜎 1 − 𝐵 2 𝜎 2 𝛾 = 1 + 𝐴 𝜎 1 + 𝐵 𝜎 𝛾 | | 𝜁 | | . ≤ 𝜎 < 1 ( 2 . 3 6 ) By virtue of (1.6), (2.10), and (2.36), we have | | 𝑓 | | ≤ 1 ( 𝑢 𝑧 ) 𝛼 1 0 𝑡 ( 1 / 𝛼 ) − 1 | | | | 1 ℎ ( 𝑤 ( 𝑢 𝑡 𝑧 ) ) 𝑑 𝑡 ≤ 𝛼 1 0 𝑡 ( 1 / 𝛼 ) − 1 ℎ ( ( 𝑢 𝑡 | 𝑧 | ) 𝑛 ) 𝑑 𝑡 , ( 2 . 3 7 ) for 𝑧 ∈ 𝑈 and 0 ≤ 𝑢 ≤ 1 . Now, by using (2.3), (2.21) and (2.37), we can obtain (2.34).

Theorem 2.9. Let 𝑓 ( 𝑧 ) = 𝑧 + ∞ 𝑘 = 𝑛 + 1 𝑎 𝑘 𝑧 𝑘 ∈ 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛾 , 𝛼 ) . ( 2 . 3 8 ) Then, | | 𝑎 𝑘 | | ≤ 𝛾 ( 𝐴 − 𝐵 ) ( 𝑘 ( 𝛼 ( 𝑘 − 1 ) + 1 ) 𝑘 ≥ 𝑛 + 1 ) . ( 2 . 3 9 ) The result is sharp for each 𝑘 ≥ 𝑛 + 1 .

Proof. It is known (cf. [12]) that, if 𝜑 ( 𝑧 ) = ∞ 𝑘 = 1 𝑏 𝑘 𝑧 𝑘 ≺ 𝜓 ( 𝑧 ) ( 𝑧 ∈ 𝑈 ) , ( 2 . 4 0 ) where 𝜑 ( 𝑧 ) is analytic in 𝑈 and 𝜓 ( 𝑧 ) = 𝑧 + ⋯ is analytic and convex univalent in 𝑈 , then | 𝑏 𝑘 | ≤ 1 ( 𝑘 ∈ 𝑁 ) .
By (2.38), we have 𝑓 ( 𝑧 ) + 𝛼 𝑧 𝑓 ( 𝑧 ) − 1 = 𝛾 ( 𝐴 − 𝐵 ) ∞ 𝑘 = 𝑛 + 1 𝑘 ( 𝛼 ( 𝑘 − 1 ) + 1 ) 𝑎 𝛾 ( 𝐴 − 𝐵 ) 𝑘 𝑧 𝑘 − 1 ≺ 𝜓 ( 𝑧 ) ( 𝑧 ∈ 𝑈 ) , ( 2 . 4 1 ) where 𝜓 ( 𝑧 ) = ℎ ( 𝑧 ) − 1 𝛾 ( 𝐴 − 𝐵 ) = 𝑧 + ⋯ ( 2 . 4 2 ) and ℎ ( 𝑧 ) is given by (1.6). Since the function 𝜓 ( 𝑧 ) is analytic and convex univalent in 𝑈 , it follows from (2.41) that 𝑘 ( 𝛼 ( 𝑘 − 1 ) + 1 ) | | 𝑎 𝛾 ( 𝐴 − 𝐵 ) 𝑘 | | ≤ 1 ( 𝑘 ≥ 𝑛 + 1 ) , ( 2 . 4 3 ) which gives (2.39).
Next, we consider the function 𝑓 𝑘 − 1 ( 𝑧 ) = 𝑧 + ∞ 𝑚 = 1 𝜆 𝑚 ( 𝐴 , 𝐵 , 𝛾 ) 𝑧 ( 𝑚 ( 𝑘 − 1 ) + 1 ) ( 𝛼 𝑚 ( 𝑘 − 1 ) + 1 ) 𝑚 ( 𝑘 − 1 ) + 1 ( 𝑧 ∈ 𝑈 ; 𝑘 ≥ 𝑛 + 1 ) . ( 2 . 4 4 ) It is easy to verify that 𝑓 𝑘 − 1 ( 𝑧 ) + 𝛼 𝑧 𝑓 𝑘 − 1 𝑧 ( 𝑧 ) = ℎ 𝑘 − 1 𝑓 ≺ ℎ ( 𝑧 ) ( 𝑧 ∈ 𝑈 ) , 𝑘 − 1 𝛾 ( 𝑧 ) = 𝑧 + ( 𝐴 − 𝐵 ) 𝑧 𝑘 ( 𝛼 ( 𝑘 − 1 ) + 1 ) 𝑘 + ⋯ . ( 2 . 4 5 ) The proof of Theorem 2.9 is completed.

3. The Univalency and Starlikeness of 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛼 )

Theorem 3.1. 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛼 ) ⊂ 𝑆 𝑛 if and only if ( 𝐴 − 𝐵 ) ∞ 𝑚 = 1 𝐵 𝑚 − 1 𝛼 𝑛 𝑚 + 1 ≤ 1 . ( 3 . 1 )

Proof. Let 𝑓 ( 𝑧 ) ∈ 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛼 ) and (3.1) be satisfied. Then, by (2.16) in Corollary 2.2, we see that R e 𝑓 ′ ( 𝑧 ) > 0 ( 𝑧 ∈ 𝑈 ) . Thus, 𝑓 ( 𝑧 ) is close-to-convex and univalent in 𝑈 .
On the other hand, if ( 𝐴 − 𝐵 ) ∞ 𝑚 = 1 𝐵 𝑚 − 1 𝛼 𝑛 𝑚 + 1 > 1 , ( 3 . 2 ) then the function 𝑓 𝑛 ( 𝑧 ) defined by (2.19) satisfies 𝑓 𝑛 ( 0 ) = 1 > 0 and 𝑓 𝑛 𝑟 𝑒 𝜋 𝑖 / 𝑛 = 1 − ( 𝐴 − 𝐵 ) ∞ 𝑚 = 1 𝐵 𝑚 − 1 𝑟 𝛼 𝑛 𝑚 + 1 𝑛 𝑚 ⟶ 1 − ( 𝐴 − 𝐵 ) ∞ 𝑚 = 1 𝐵 𝑚 − 1 𝛼 𝑛 𝑚 + 1 < 0 ( 3 . 3 ) as 𝑟 → 1 . Hence, there exists a point 𝑧 𝑛 = 𝑟 𝑛 𝑒 𝜋 𝑖 / 𝑛 ( 0 < 𝑟 𝑛 < 1 ) such that 𝑓 𝑛 ( 𝑧 𝑛 ) = 0 . This implies that 𝑓 𝑛 ( 𝑧 ) is not univalent in 𝑈 and so the theorem is proved.

Theorem 3.2. Let (3.1) in Theorem 3.1 be satisfied. If 𝛼 ≥ 1 and ( 𝛼 − 1 ) 1 − ( 𝐴 − 𝐵 ) ∞ 𝑚 = 1 𝐵 𝑚 − 1 + 𝛼 𝑛 𝑚 + 1 𝑛 𝛼 2 1 − ( 𝐴 − 𝐵 ) ∞ 𝑚 = 1 𝐵 𝑚 − 1 ≥ ( 𝑛 𝑚 + 1 ) ( 𝛼 𝑛 𝑚 + 1 ) 𝐴 − 1 , 1 − 𝐵 ( 3 . 4 ) then 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛼 ) ⊂ 𝑆 ∗ 𝑛 .

Proof. We first show that ∞ 𝑚 = 1 𝐵 𝑚 − 1 ≥ 𝛼 𝑛 𝑚 + 1 ∞ 𝑚 = 1 𝐵 𝑚 − 1 ( ( 𝑛 𝑚 + 1 ) ( 𝛼 𝑛 𝑚 + 1 ) 𝛼 ≥ 1 ) . ( 3 . 5 ) Equation (3.5) is obvious when 𝐵 ≥ 0 . For 0 > 𝐵 ≥ − 1 , we have ∞ 𝑚 = 1 𝐵 𝑚 − 1 − 𝛼 𝑛 𝑚 + 1 ∞ 𝑚 = 1 𝐵 𝑚 − 1 ( 𝑛 𝑚 + 1 ) ( 𝛼 𝑛 𝑚 + 1 ) = 𝜇 1 − 𝜇 2 + 𝜇 3 − 𝜇 4 + ⋯ + ( − 1 ) 𝑚 − 1 𝜇 𝑚 + ⋯ , ( 3 . 6 ) where 𝜇 𝑚 = | | 𝐵 | | 𝑛 𝑚 𝑚 − 1 ( 𝑛 𝑚 + 1 ) ( 𝛼 𝑛 𝑚 + 1 ) > 0 . ( 3 . 7 ) Since | 𝐵 | ≤ 1 and 𝑑 𝑥 𝑑 𝑥 = ( 𝑥 + 1 ) ( 𝛼 𝑥 + 1 ) 1 − 𝛼 𝑥 2 ( 𝑥 + 1 ) 2 ( 𝛼 𝑥 + 1 ) 2 ≤ 0 ( 𝑥 ≥ 1 ; 𝛼 ≥ 1 ) , ( 3 . 8 ) { 𝜇 𝑚 } is a monotonically decreasing sequence. Therefore, the inequality (3.5) follows from (3.6).
Let 𝑓 ( 𝑧 ) ∈ 𝑇 𝑛 ( 𝐴 , 𝐵 , 𝛼 ) . Then, 𝑓 R e ( 𝑧 ) + 𝛼 𝑧 𝑓 > ( 𝑧 ) 1 − 𝐴 1 − 𝐵 ( 𝑧 ∈ 𝑈 ) . ( 3 . 9 ) Define 𝑝 ( 𝑧 ) in 𝑈 by 𝑝 ( 𝑧 ) = 𝑧 𝑓 ( 𝑧 ) . 𝑓 ( 𝑧 ) ( 3 . 1 0 ) In view of (3.1) in Theorem 3.1 is satisfied, the function 𝑓 ( 𝑧 ) is univalent in 𝑈 , and so 𝑝 ( 𝑧 ) = 1 + 𝑝 𝑛 𝑧 𝑛 + 𝑝 𝑛 + 1 𝑧 𝑛 + 1 + ⋯ is analytic in 𝑈 . Also it follows from (3.10) that 𝑓 ( 𝑧 ) + 𝛼 𝑧 𝑓 ( 𝑧 ) = ( 1 − 𝛼 ) 𝑓 ( 𝑧 ) + 𝛼 𝑓 ( 𝑧 ) 𝑧 𝑧 𝑝 ( 𝑧 ) + ( 𝑝 ( 𝑧 ) ) 2 . ( 3 . 1 1 )

We want to prove now that R e 𝑝 ( 𝑧 ) > 0 for 𝑧 ∈ 𝑈 . Suppose that there exists a point 𝑧 0 ∈ 𝑈 such that | | 𝑧 R e 𝑝 ( 𝑧 ) > 0 | 𝑧 | < 0 | | 𝑧 , R e 𝑝 0 = 0 . ( 3 . 1 2 ) Then, applying a result of Miller and Mocanu [13, Theorem 4], we have 𝑧 0 𝑝 𝑧 0 + 𝑝 𝑧 0 2 𝑛 ≤ − 2 𝑧 R e 1 − 𝑝 0 − 𝑧 I m 𝑝 0 2 𝑛 ≤ − 2 . ( 3 . 1 3 ) For 𝛼 ≥ 1 , we deduce from Corollaries 2.2 and 2.4, (3.1), (3.5), (3.11), (3.13), and (3.4) that 𝑓 R e 𝑧 0 + 𝛼 𝑧 0 𝑓 𝑧 0 ≤ ( 1 − 𝛼 ) 1 − ( 𝐴 − 𝐵 ) ∞ 𝑚 = 1 𝐵 𝑚 − 1 − 𝛼 𝑛 𝑚 + 1 𝑛 𝛼 2 1 − ( 𝐴 − 𝐵 ) ∞ 𝑚 = 1 𝐵 𝑚 − 1 ≤ ( 𝑛 𝑚 + 1 ) ( 𝛼 𝑛 𝑚 + 1 ) 1 − 𝐴 . 1 − 𝐵 ( 3 . 1 4 ) But this contradicts (3.9) at 𝑧 = 𝑧 0 . Therefore, we must have R e 𝑝 ( 𝑧 ) > 0 ( 𝑧 ∈ 𝑈 ) and the proof of Theorem 3.2 is completed.

Remark 3.3. In [6, Theorem 4(ii)], the authors gave the following: if 0 < 𝛼 < 1 and 𝛽 1 is the solution of the equation 1 − 3 𝛼 2 = 𝛽 + ( 1 − 𝛽 ) ∞ 𝑚 = 2 ( − 1 ) 𝑚 − 1 𝛼 + 2 ( 𝛼 − 1 ) 𝑚 , 𝑚 ( 𝛼 ( 𝑚 − 1 ) + 1 ) ( 3 . 1 5 ) then 𝑅 ( 𝛽 , 𝛼 ) ⊂ 𝑆 ∗ 1 for 𝛽 ≥ 𝛽 1 . However, this result is not true because the series in (3.15) diverges.

4. The Radius of Convexity

Theorem 4.1. Let 𝑓 ( 𝑧 ) belong to the class 𝑇 𝑛 ( 𝛾 ) defined by 𝑇 𝑛 ( 𝛾 ) = 𝑇 𝑛 ( 1 , − 1 , 𝛾 , 0 ) = 𝑓 ( 𝑧 ) ∈ 𝐴 𝑛 ∶ 𝑓 ( 𝑧 ) ≺ 1 + 𝑧 1 − 𝑧 𝛾 , ( 𝑧 ∈ 𝑈 ) , ( 4 . 1 ) 0 < 𝛿 ≤ 1 and 0 ≤ 𝜌 < 1 . Then, 𝑓 R e ( 1 − 𝛿 ) ( 𝑧 ) 1 / 𝛾 + 𝛿 1 + 𝑧 𝑓 ( 𝑧 ) 𝑓 ( 𝑧 ) > 𝜌 | 𝑧 | < 𝑟 𝑛 ( 𝛾 , 𝛿 , 𝜌 ) , ( 4 . 2 ) where 𝑟 𝑛 ( 𝛾 , 𝛿 , 𝜌 ) is the root in ( 0 , 1 ) of the equation ( 1 − 2 𝛿 + 𝜌 ) 𝑟 2 𝑛 − 2 ( 1 − 𝛿 + 𝑛 𝛿 𝛾 ) 𝑟 𝑛 + 1 − 𝜌 = 0 . ( 4 . 3 ) The result is sharp.

Proof. For 𝑓 ( 𝑧 ) ∈ 𝑇 𝑛 ( 𝛾 ) , we can write 𝑓 ( 𝑧 ) 1 / 𝛾 = 1 + 𝑧 𝑛 𝜑 ( 𝑧 ) 1 − 𝑧 𝑛 𝜑 , ( 𝑧 ) ( 4 . 4 ) where 𝜑 ( 𝑧 ) is analytic and | 𝜑 ( 𝑧 ) | ≤ 1 in 𝑈 . Differentiating both sides of (4.4) logarithmically, we arrive at 1 + 𝑧 𝑓 ( 𝑧 ) 𝑓 ( 𝑧 ) = 1 + 2 𝑛 𝛾 𝑧 𝑛 𝜑 ( 𝑧 ) 1 − ( 𝑧 𝑛 𝜑 ( 𝑧 ) ) 2 + 2 𝛾 𝑧 𝑛 + 1 𝜑 ( 𝑧 ) 1 − ( 𝑧 𝑛 𝜑 ( 𝑧 ) ) 2 ( 𝑧 ∈ 𝑈 ) . ( 4 . 5 ) Put | 𝑧 | = 𝑟 < 1 and ( 𝑓 ′ ( 𝑧 ) ) 1 / 𝛾 = 𝑢 + 𝑖 𝑣 ( 𝑢 , 𝑣 ∈ 𝑅 ) . Then, (4.4) implies that 𝑧 𝑛 𝜑 ( 𝑧 ) = 𝑢 − 1 + 𝑖 𝑣 , 𝑢 + 1 + 𝑖 𝑣 ( 4 . 6 ) 1 − 𝑟 𝑛 1 + 𝑟 𝑛 ≤ 𝑢 ≤ 1 + 𝑟 𝑛 1 − 𝑟 𝑛 . ( 4 . 7 ) With the help of the Carathéodory inequality | | 𝜑 | | ≤ | | | | ( 𝑧 ) 1 − 𝜑 ( 𝑧 ) 2 1 − 𝑟 2 , ( 4 . 8 ) it follows from (4.5) and (4.6) that 𝑓 R e ( 1 − 𝛿 ) ( 𝑧 ) 1 / 𝛾 + 𝛿 1 + 𝑧 𝑓 ( 𝑧 ) 𝑓 𝑧 ( 𝑧 ) ≥ ( 1 − 𝛿 ) 𝑢 + 𝛿 + 2 𝑛 𝛿 𝛾 R e 𝑛 𝜑 ( 𝑧 ) 1 − ( 𝑧 𝑛 𝜑 ( 𝑧 ) ) 2 | | | | 𝑧 − 2 𝛿 𝛾 𝑛 + 1 𝜑 ( 𝑧 ) 1 − ( 𝑧 𝑛 𝜑 ( 𝑧 ) ) 2 | | | | ≥ ( 1 − 𝛿 ) 𝑢 + 𝛿 + 𝑛 𝛿 𝛾 2 𝑢 𝑢 − 𝑢 2 + 𝑣 2 + 𝛿 𝛾 2 ( 𝑢 − 1 ) 2 + 𝑣 2 − 𝑟 2 𝑛 ( 𝑢 + 1 ) 2 + 𝑣 2 𝑟 𝑛 − 1 1 − 𝑟 2 𝑢 2 + 𝑣 2 1 / 2 = 𝐹 𝑛 ( 𝑢 , 𝑣 ) ( s a y 𝜕 ) , ( 4 . 9 ) 𝐹 𝜕 𝑣 𝑛 ( 𝑢 , 𝑣 ) = 𝛿 𝛾 𝑣 𝐺 𝑛 ( 𝑢 , 𝑣 ) , ( 4 . 1 0 ) where 0 < 𝑟 < 1 , 0 < 𝛿 ≤ 1 and 𝐺 𝑛 ( 𝑢 , 𝑣 ) = 𝑛 𝑢 𝑢 2 + 𝑣 2 2 + 1 − 𝑟 2 𝑛 𝑟 𝑛 − 1 1 − 𝑟 2 𝑢 2 + 𝑣 2 1 / 2 + 𝑟 2 𝑛 ( 𝑢 + 1 ) 2 + 𝑣 2 − ( 𝑢 − 1 ) 2 + 𝑣 2 2 𝑟 𝑛 − 1 1 − 𝑟 2 𝑢 2 + 𝑣 2 3 / 2 > 0 ( 4 . 1 1 ) because of (4.6) and (4.7). In view of (4.10) and (4.11), we see that 𝐹 𝑛 ( 𝑢 , 𝑣 ) ≥ 𝐹 𝑛 ( 𝑢 , 0 ) = ( 1 − 𝛿 ) 𝑢 + 𝛿 + 𝑛 𝛿 𝛾 2 1 𝑢 − 𝑢 + 𝛿 𝛾 2 𝑟 𝑛 − 1 1 − 𝑟 2 × 1 − 𝑟 2 𝑛 1 𝑢 + 𝑢 − 2 1 + 𝑟 2 𝑛 . ( 4 . 1 2 )

Let us now calculate the minimum value of 𝐹 𝑛 ( 𝑢 , 0 ) on the closed interval [ ( 1 − 𝑟 𝑛 ) / ( 1 + 𝑟 𝑛 ) , ( 1 + 𝑟 𝑛 ) / ( 1 − 𝑟 𝑛 ) ] . Noting that 1 − 𝑟 2 𝑛 𝑟 𝑛 − 1 1 − 𝑟 2 [ 8 ] ) ≥ 𝑛 ( s e e ( 4 . 1 3 ) and (4.7), we deduce from (4.12) that 𝑑 𝐹 𝑑 𝑢 𝑛 ( 𝑢 , 0 ) = 1 − 𝛿 + 𝛿 𝛾 2 1 − 𝑟 2 𝑛 𝑟 𝑛 − 1 1 − 𝑟 2 − 1 + 𝑛 𝑢 2 1 − 𝑟 2 𝑛 𝑟 𝑛 − 1 1 − 𝑟 2 − 𝑛 ≥ 1 − 𝛿 + 𝛿 𝛾 2 1 − 𝑟 2 𝑛 𝑟 𝑛 − 1 1 − 𝑟 2 − + 𝑛 1 + 𝑟 𝑛 1 − 𝑟 𝑛 2 1 − 𝑟 2 𝑛 𝑟 𝑛 − 1 1 − 𝑟 2 − 𝑛 = 1 − 𝛿 + 2 𝛿 𝛾 𝐼 𝑛 ( 𝑟 ) ( 1 − 𝑟 𝑛 ) 2 , ( 4 . 1 4 ) where 𝐼 𝑛 𝑛 ( 𝑟 ) = 2 1 + 𝑟 2 𝑛 − 𝑟 1 + 𝑟 2 + ⋯ + 𝑟 2 𝑛 − 2 . ( 4 . 1 5 ) Also 𝐼 𝑛 ( 𝑟 ) = 𝑛 2 𝑟 2 𝑛 − 1 − 1 + 3 𝑟 2 + ⋯ + ( 2 𝑛 − 1 ) 𝑟 2 𝑛 − 2 ( 4 . 1 6 ) and 𝐼 1 ( 𝑟 ) = 𝑟 − 1 < 0 . Suppose that 𝐼 𝑛 ( 𝑟 ) < 0 . Then, 𝐼 𝑛 + 1 ( 𝑟 ) = ( 𝑛 + 1 ) 2 𝑟 2 𝑛 + 1 − ( 2 𝑛 + 1 ) 𝑟 2 𝑛 − 1 + 3 𝑟 2 + ⋯ + ( 2 𝑛 − 1 ) 𝑟 2 𝑛 − 2 < 𝑛 2 𝑟 2 𝑛 − 1 + 3 𝑟 2 + ⋯ + ( 2 𝑛 − 1 ) 𝑟 2 𝑛 − 2 < 𝐼 𝑛 ( 𝑟 ) < 0 . ( 4 . 1 7 ) Hence, by virtue of the mathematical induction, we have 𝐼 𝑛 ( 𝑟 ) < 0 for all 𝑛 ∈ 𝑁 and 0 ≤ 𝑟 < 1 . This implies that 𝐼 𝑛 ( 𝑟 ) > 𝐼 𝑛 ( 1 ) = 0 ( 𝑛 ∈ 𝑁 ; 0 ≤ 𝑟 < 1 ) . ( 4 . 1 8 ) In view of (4.14) and (4.18), we see that 𝑑 𝐹 𝑑 𝑢 𝑛 ( 𝑢 , 0 ) > 0 1 − 𝑟 𝑛 1 + 𝑟 𝑛 ≤ 𝑢 ≤ 1 + 𝑟 𝑛 1 − 𝑟 𝑛 . ( 4 . 1 9 ) Further it follows from (4.9), (4.12), and (4.19) that 𝑓 R e ( 1 − 𝛿 ) ( 𝑧 ) 1 / 𝛾 + 𝛿 1 + 𝑧 𝑓 ( 𝑧 ) 𝑓 ( 𝑧 ) − 𝜌 ≥ 𝐹 𝑛 1 − 𝑟 𝑛 1 + 𝑟 𝑛 , 0 − 𝜌 = ( 1 − 𝛿 ) 1 − 𝑟 𝑛 1 + 𝑟 𝑛 + 𝛿 1 − 2 𝑛 𝛾 𝑟 𝑛 − 𝑟 2 𝑛 1 − 𝑟 2 𝑛 = 𝐽 − 𝜌 𝑛 ( 𝑟 ) 1 − 𝑟 2 𝑛 , ( 4 . 2 0 ) where 0 ≤ 𝜌 < 1 and 𝐽 𝑛 ( 𝑟 ) = ( 1 − 2 𝛿 + 𝜌 ) 𝑟 2 𝑛 − 2 ( 1 − 𝛿 + 𝑛 𝛿 𝛾 ) 𝑟 𝑛 + 1 − 𝜌 . ( 4 . 2 1 ) Note that 𝐽 𝑛 ( 0 ) = 1 − 𝜌 > 0 and 𝐽 𝑛 ( 1 ) = − 2 𝑛 𝛿 𝛾 < 0 . If we let 𝑟 𝑛 ( 𝛾 , 𝛿 , 𝜌 ) denote the root in ( 0 , 1 ) of the equation 𝐽 𝑛 ( 𝑟 ) = 0 , then (4.20) yields the desired result (4.2).

To see that the bound 𝑟 𝑛 ( 𝛾 , 𝛿 , 𝜌 ) is the best possible, we consider the function 𝑓 ( 𝑧 ) = 𝑧 0 1 − 𝑡 𝑛 1 + 𝑡 𝑛 𝛾 𝑑 𝑡 ∈ 𝑇 𝑛 ( 𝛾 ) . ( 4 . 2 2 ) It is clear that for 𝑧 = 𝑟 ∈ ( 𝑟 𝑛 ( 𝛾 , 𝛿 , 𝜌 ) , 1 ) , 𝑓 ( 1 − 𝛿 ) ( 𝑟 ) 1 / 𝛾 + 𝛿 1 + 𝑟 𝑓 ( 𝑟 ) 𝑓 𝐽 ( 𝑟 ) − 𝜌 = 𝑛 ( 𝑟 ) 1 − 𝑟 2 𝑛 < 0 , ( 4 . 2 3 ) which shows that the bound 𝑟 𝑛 ( 𝛾 , 𝛿 , 𝜌 ) cannot be increased.

Setting 𝛿 = 1 , Theorem 4.1 reduces to the following result.

Corollary 4.2. Let 𝑓 ( 𝑧 ) ∈ 𝑇 𝑛 ( 𝛾 ) and 0 ≤ 𝜌 < 1 . Then, 𝑓 ( 𝑧 ) is convex of order 𝜌 in ( | 𝑧 | < 𝑛 𝛾 ) 2 + ( 1 − 𝜌 ) 2 1 / 2 − 𝑛 𝛾 1 − 𝜌 1 / 𝑛 . ( 4 . 2 4 ) The result is sharp.