On Subclass of Analytic Univalent Functions Associated with Negative Coefficients

Abstract

M. H. Al-Abbadi and M. Darus (2009) recently introduced a new generalized derivative operator 𝜇_𝜆^𝑛,𝑚_1,𝜆2, which generalized many well-known operators studied earlier by many different authors. In this present paper, we shall investigate a new subclass of analytic functions in the open unit disk 𝑈={𝑧∈ℂ∶|𝑧|<1} which is defined by new generalized derivative operator. Some results on coefficient inequalities, growth and distortion theorems, closure theorems, and extreme points of analytic functions belonging to the subclass are obtained.

1. Introduction and Definitions

Let 𝒜(𝑥) denote the class of functions of the form

𝑓(𝑧)=𝑧+^∞_𝑘=𝑥+1𝑎_𝑘𝑧^𝑘,𝑎_𝑘iscomplexnumber, (1.1) and 𝑥∈ℕ={1,2,3,...}, which are analytic in the open unit disc 𝑈={𝑧∈ℂ∶|𝑧|<1} on the complex plane ℂ; note that 𝒜(1)=𝒜 and 𝒜(𝑥)⊆𝒜(1). Suppose that 𝒮(𝑥) denote the subclass of 𝒜(𝑥) consisting of functions that are univalent in 𝑈. Further, let 𝒮^∗_𝛼(𝑥) and 𝒞_𝛼(𝑥) be the classes of 𝒮(𝑥) consisting of functions, respectively, starlike of order 𝛼 and convex of order 𝛼 in 𝑈, for 0≤𝛼<1. Let 𝒯(𝑥) denote the subclass of 𝒮(𝑥) consisting of functions of the form

𝑓(𝑧)=𝑧−^∞_𝑘=𝑥+1|𝑎_𝑘|𝑧^𝑘,𝑥∈ℕ={1,2,3,…}, (1.2) defined on the open unit disk 𝑈={𝑧∈ℂ∶|𝑧|<1}. A function 𝑓∈𝒯(𝑥) is called a function with negative coefficient and the class 𝒯(1) was introduced and studied by Silverman [1]. In [1] Silverman investigated the subclasses of 𝒯(1) denoted by 𝑆^∗_𝒯(𝛼) and 𝐶_𝒯(𝛼) for 0≤𝛼<1. That are, respectively, starlike of order 𝛼 and convex of order 𝛼. Now (𝑥)_𝑘 denotes the Pochhammer symbol (or the shifted factorial) defined by

(𝑥)_𝑘=Γ(𝑥+𝑘)=Γ(𝑥)1for𝑘=0,𝑥∈ℂ⧵{0},𝑥(𝑥+1)(𝑥+2)…(𝑥+𝑘−1)for𝑘∈ℕ={1,2,…},𝑥∈ℂ. (1.3)

The authors in [2] have recently introduced a new generalized derivative operator 𝜇_𝜆^𝑛,𝑚_1,𝜆2 as follows.

Definition 1.1. For 𝑓∈𝒜=𝒜(1), the generalized derivative operator 𝜇_𝜆^𝑛,𝑚_1,𝜆2∶𝒜→𝒜 is defined by 𝜇_𝜆^𝑛,𝑚_1,𝜆2𝑓(𝑧)=𝑧+^∞_𝑘=2(1+𝜆₁(𝑘−1))^𝑚−1(1+𝜆₂(𝑘−1))^𝑚𝑐(𝑛,𝑘)𝑎_𝑘𝑧^𝑘,(𝑧∈𝑈), (1.4) where 𝑛,𝑚∈ℕ₀={0,1,2,…},𝜆₂≥𝜆₁≥0,and𝑐(𝑛,𝑘)=(_𝑛^{𝑛+𝑘−1})=(𝑛+1)_𝑘−1/(1)_𝑘−1.
(1)Special cases of this operator include the Ruscheweyh derivative operator in the cases 𝜇_𝜆^𝑛,1_1,0≡𝜇^𝑛,𝑚_0,0≡𝜇^𝑛,0_0,𝜆2≡𝑅^𝑛 [3], the Salagean derivative operator 𝜇^0,𝑚+1_1,0≡𝑆^𝑛 [4], the generalized Ruscheweyh derivative operator 𝜇_𝜆^𝑛,2_1,0≡𝑅^𝑛_𝜆 [5], the generalized Salagean derivative operator introduced by Al-Oboudi 𝜇_𝜆^0,𝑚+1_1,0≡𝑆^𝑛_𝛽 [6], and the generalized Al-Shaqsi and Darus derivative operator 𝜇^𝜆,𝑛+1_𝛽,0≡𝐷^𝑛_𝜆,𝛽 where (𝑛=𝜆,𝑚=𝑛+1,𝜆₁=𝛽,and𝜆₂=0) can be found in [7]. It is easily seen that 𝜇_𝜆^0,1_1,0𝑓(𝑧)=𝜇^0,𝑚_0,0𝑓(𝑧)=𝜇^0,0_0,𝜆2𝑓(𝑧)=𝜇_𝜆^1,1_1,1𝑓(𝑧)=𝑓(𝑧),𝜇_𝜆^1,1_1,0𝑓(𝑧)=𝜇^1,𝑚_0,0𝑓(𝑧)=𝜇^1,0_0,𝜆2𝑓(𝑧)=𝜇^0,2_1,0𝑓(𝑧)=𝑧𝑓^′(𝑧), and also 𝜇_𝜆^𝑛−1,0_1,0𝑓(𝑧)=𝜇^{𝑛−1,𝑚}_0,0𝑓(𝑧) where 𝑛=1,2,3,….

By making use of the generalized derivative operator 𝜇_𝜆^𝑛,𝑚_1,𝜆2, the authors introduce a new subclass as follows.

Definition 1.2. For 0≤𝛼<1,(𝑛,𝑚∈ℕ₀={0,1,2,…})and𝜆₂≥𝜆₁≥0, let ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼) be the subclass of 𝒮(𝑥) consisting of functions 𝑓 satisfying Re(𝑧(𝜇_𝜆^𝑛,𝑚_1,𝜆2𝑓(𝑧))^′𝜇_𝜆^𝑛,𝑚_1,𝜆2𝑓(𝑧))>𝛼,(𝑧∈𝑈), (1.5) where 𝜇_𝜆^𝑛,𝑚_1,𝜆2𝑓(𝑧)=𝑧+^∞_𝑘=𝑥+1(1+𝜆₁(𝑘−1))^𝑚−1(1+𝜆₂(𝑘−1))^𝑚𝑐(𝑛,𝑘)𝑎_𝑘𝑧^𝑘, (1.6){𝑥=1,2,3,…,(𝑧∈𝑈)}, and 𝜇_𝜆^𝑛,𝑚_1,𝜆2𝑓(𝑧)≠0.
Further, we define the class 𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼) by 𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼)=ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼)∩𝒯(𝑥),{𝑥=1,2,3,…}, (1.7) for 0≤𝛼<1,(𝑛,𝑚∈ℕ₀={0,1,2,…}), and 𝜆₂≥𝜆₁≥0.

Also note that various subclasses of ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼) and 𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼) have been studied by many authors by suitable choices of 𝑛,𝑚,𝜆₁,𝜆₂, and 𝑥. For example,

𝒯ℋ^0,0_0,𝜆2(1,𝛼)≡𝒯ℋ_𝜆^0,1_1,0(1,𝛼)≡𝒯ℋ^0,𝑚_0,0(1,𝛼)≡𝒯ℋ_𝜆^1,1_1,1(1,𝛼)≡𝑆^∗_𝒯(𝛼), (1.8)

starlike of order 𝛼 with negative coefficients. And

𝒯ℋ^1,0_0,𝜆2(1,𝛼)≡𝒯ℋ_𝜆^1,1_1,0(1,𝛼)≡𝒯ℋ^1,𝑚_0,0(1,𝛼)≡𝒯ℋ^0,2_1,0(1,𝛼)≡𝐶_𝒯(𝛼), (1.9)

class of convex function of order 𝛼 with negative coefficients. Also

𝒯ℋ^0,0_0,𝜆2(𝑥,𝛼)≡𝒯ℋ_𝜆^0,1_1,0(𝑥,𝛼)≡𝒯ℋ^0,𝑚_0,0(𝑥,𝛼)≡𝒯ℋ_𝜆^1,1_1,1(𝑥,𝛼)≡𝑆^∗_𝒯(𝑥,𝛼),𝒯ℋ^1,0_0,𝜆2(𝑥,𝛼)≡𝒯ℋ_𝜆^1,1_1,0(𝑥,𝛼)≡𝒯ℋ^1,𝑚_0,0(𝑥,𝛼)≡𝒯ℋ^0,2_1,0(𝑥,𝛼)≡𝐶_𝒯(𝑥,𝛼),𝒯ℋ_𝜆^0,2_1,0(𝑥,𝛼)≡𝑃(𝑥,𝜆₁,𝛼)(0≤𝜆₁<1),𝒯ℋ_𝜆^1,2_1,0(𝑥,𝛼)≡𝐶(𝑥,𝜆₁,𝛼)(0≤𝜆₁<1). (1.10) The classes 𝑆^∗_𝒯(𝑥,𝛼) and 𝐶_𝒯(𝑥,𝛼) were studied by Chatterjea [8] (see also Srivastava et al. [9]), whereas the classes 𝑃(𝑥,𝜆₁,𝛼) and 𝐶(𝑥,𝜆₁,𝛼) were, respectively, studied by Altintaş [10] and Kamali and Akbulut [11]. When 𝜆₁=𝜆₂=0 or 𝑚=1,𝜆₂=0, or 𝑚=0,𝜆₁=0 in the class ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼), we have the class 𝑅_𝑛(𝛼) introduced and studied by Ahuja [12]. Finally we note that when 𝑚=2,𝜆₂=0 in the class ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼) we have the class 𝒦^𝑛_𝜆(𝑥,𝛼) introduced and studied by Al-Shaqsi and Darus [13].

2. Coefficient Inequalities

In this section, we provide a necessary and sufficient condition for a function 𝑓 analytic in 𝑈 to be in ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼) and in 𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼).

Theorem 2.1. For 0≤𝛼<1 and 𝜆₂≥𝜆₁≥0, let 𝑓∈𝒮(𝑥) be defined by (1.1). If ^∞_𝑘=𝑥+1(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1(1+𝜆₂(𝑘−1))^𝑚𝑐(𝑛,𝑘)|𝑎_𝑘|≤1−𝛼,𝑥=1,2,…, (2.1) then 𝑓∈ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼), where 𝑛∈ℕ={1,2,…} and 𝑚∈ℕ₀={0,1,2,…}.

Proof. Assume that (2.1) holds true. Then we shall prove condition (1.5). It is sufficient to show that |𝑧(𝜇_𝜆^𝑛,𝑚_1,𝜆2𝑓(𝑧))^′𝜇_𝜆^𝑛,𝑚_1,𝜆2𝑓(𝑧)−1|≤1−𝛼,(𝑧∈𝑈). (2.2) So, we have that |𝑧(𝜇_𝜆^𝑛,𝑚_1,𝜆2𝑓(𝑧))^′𝜇_𝜆^𝑛,𝑚_1,𝜆2𝑓(𝑧)−1|=|𝑧(𝜇_𝜆^𝑛,𝑚_1,𝜆2𝑓(𝑧))^′−𝜇_𝜆^𝑛,𝑚_1,𝜆2𝑓(𝑧)𝜇_𝜆^𝑛,𝑚_1,𝜆2|∑𝑓(𝑧)=|^{∞𝑘=𝑥+1}((𝑘−1)(1+𝜆₁(𝑘−1))^𝑚−1/(1+𝜆₂(𝑘−1))^𝑚)𝑐(𝑛,𝑘)𝑎_𝑘𝑧^𝑘𝑧+∑^∞_𝑘=𝑥+1((1+𝜆₁(𝑘−1))^𝑚−1/(1+𝜆₂(𝑘−1))^𝑚)𝑐(𝑛,𝑘)𝑎_𝑘𝑧^𝑘|≤∑|𝑧|<1,_{∞𝑘=𝑥+1}((𝑘−1)(1+𝜆₁(𝑘−1))^𝑚−1/(1+𝜆₂(𝑘−1))^𝑚)𝑐(𝑛,𝑘)|𝑎_𝑘|1−∑^∞_𝑘=𝑥+1((1+𝜆₁(𝑘−1))^𝑚−1/(1+𝜆₂(𝑘−1))^𝑚)𝑐(𝑛,𝑘)|𝑎_𝑘|, (2.3) and expression (2.3) is bounded by (1−𝛼).
Hence (2.2) holds if ^∞_𝑘=𝑥+1(𝑘−1)(1+𝜆₁(𝑘−1))^𝑚−1(1+𝜆₂(𝑘−1))^𝑚𝑐(𝑛,𝑘)|𝑎_𝑘|≤(1−𝛼)[1−^∞_𝑘=𝑥+1(1+𝜆₁(𝑘−1))^𝑚−1(1+𝜆₂(𝑘−1))^𝑚𝑐(𝑛,𝑘)|𝑎_𝑘|], (2.4) which is equivalent to ^∞_𝑘=𝑥+1(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1(1+𝜆₂(𝑘−1))^𝑚𝑐(𝑛,𝑘)|𝑎_𝑘|≤(1−𝛼), (2.5) by (2.1). Thus 𝑓∈ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼). Note that the denominator in (2.3) is positive provided that (2.1) holds.

Theorem 2.2. Let 𝑓 be defined by (1.2) and 1−∑^∞_𝑘=𝑥+1(𝑘(1+𝜆₁(𝑘−1))^𝑚−1/(1+𝜆₂(𝑘−1))^𝑚)𝑐(𝑛,𝑘)|𝑎_𝑘|≥0(𝑥=1,2,…). Then 𝑓∈𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼) if and only if (2.1) is satisfied.

Proof. We only prove the right-hand side, since the other side can be justified using similar arguments in proof of Theorem 2.1. Since 𝑓∈𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼) by condition (1.5), we have that Re(𝑧(𝜇_𝜆^𝑛,𝑚_1,𝜆2𝑓(𝑧))^′𝜇_𝜆^𝑛,𝑚_1,𝜆2)∑𝑓(𝑧)=Re{𝑧−^∞_𝑘=𝑥+1(𝑘(1+𝜆₁(𝑘−1))^𝑚−1/(1+𝜆₂(𝑘−1))^𝑚)𝑐(𝑛,𝑘)|𝑎_𝑘|𝑧^𝑘𝑧−∑^∞_𝑘=𝑥+1((1+𝜆₁(𝑘−1))^𝑚−1/(1+𝜆₂(𝑘−1))^𝑚)𝑐(𝑛,𝑘)|𝑎_𝑘|𝑧^𝑘}>𝛼. (2.6) Choose values of 𝑧 on real axis so that 𝑧(𝜇_𝜆^𝑛,𝑚_1,𝜆2𝑓(𝑧))^′/𝜇_𝜆^𝑛,𝑚_1,𝜆2𝑓(𝑧) is real. Letting 𝑧→1⁻ through real values, we have that 1−∑^∞_𝑘=𝑥+1(𝑘(1+𝜆₁(𝑘−1))^𝑚−1/(1+𝜆₂(𝑘−1))^𝑚)𝑐(𝑛,𝑘)|𝑎_𝑘|𝑧^𝑘1−∑^∞_𝑘=𝑥+1((1+𝜆₁(𝑘−1))^𝑚−1/(1+𝜆₂(𝑘−1))^𝑚)𝑐(𝑛,𝑘)|𝑎_𝑘|𝑧^𝑘>𝛼. (2.7) Thus we obtain ^∞_𝑘=𝑥+1(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1(1+𝜆₂(𝑘−1))^𝑚𝑐(𝑛,𝑘)|𝑎_𝑘|≤1−𝛼, (2.8) which is (2.1). Hence the proof is complete.

The result is sharp with the extremal function 𝑓 given by

𝑓(𝑧)=𝑧−(1−𝛼)(1+𝜆₂𝑥)^𝑚(𝑥+1−𝛼)(1+𝜆₁𝑥)^𝑚−1𝑧𝑐(𝑛,𝑥+1)^𝑥+1,(𝑥=1,2,…). (2.9)

Theorem 2.3. Let the function 𝑓 given by (1.2) be in the class 𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼). Then |𝑎_𝑘|≤(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1,𝑐(𝑛,𝑘) (2.10) where 0≤𝛼<1,𝜆₂≥𝜆₁≥0,𝑘≥𝑥+1, and 𝑥=1,2,3,…. Equality holds for the function given by (2.9).

Proof. Since 𝑓∈𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼), then condition (2.1) gives |𝑎_𝑘|≤(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1,𝑐(𝑛,𝑘) (2.11) for each 𝑘=𝑥+1 where 𝑥=1,2,3,….
Clearly the function given by (2.9) satisfies (2.10), and therefore, 𝑓 given by (2.9) is in 𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼) for this function; the result is clearly sharp.

3. Growth and Distortion Theorems

In this section, growth and distortion theorems will be considered and covering property for function in the class will also be given.

Theorem 3.1. Let the function 𝑓 given by (1.2) be in the class 𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼). Then for 0<|𝑧|=𝑟<1, 𝑟−(1−𝛼)(1+𝜆₂𝑥)^𝑚(𝑥+1−𝛼)(1+𝜆₁𝑥)^𝑚−1𝑟𝑐(𝑛,𝑥+1)^𝑥+1≤|𝑓(𝑧)|≤𝑟+(1−𝛼)(1+𝜆₂𝑥)^𝑚(𝑥+1−𝛼)(1+𝜆₁𝑥)^𝑚−1𝑟𝑐(𝑛,𝑥+1)^𝑥+1,𝑥=1,2,…, (3.1) where 0≤𝛼<1,𝑚∈ℕ₀={0,1,2…},and𝑛∈ℕ={1,2,…}.

Proof. We only prove the right-hand side inequality in (3.1), since the other inequality can be justified using similar arguments. Since 𝑓∈𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼) by Theorem 2.2, we have that ^∞_𝑘=𝑥+1(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1(1+𝜆₂(𝑘−1))^𝑚𝑐(𝑛,𝑘)|𝑎_𝑘|≤(1−𝛼). (3.2) Now (𝑥+1−𝛼)(1+𝜆₁𝑥)^𝑚−1(1+𝜆₂𝑥)^𝑚𝑐(𝑛,𝑥+1)(^∞_𝑘=𝑥+1|𝑎_𝑘=|)_{∞𝑘=𝑥+1}(𝑥+1−𝛼)(1+𝜆₁𝑥)^𝑚−1(1+𝜆₂𝑥)^𝑚𝑐(𝑛,𝑥+1)|𝑎_𝑘≤|,_{∞𝑘=𝑥+1}(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1(1+𝜆₂(𝑘−1))^𝑚𝑐(𝑛,𝑘)|𝑎_𝑘|,≤1−𝛼,𝑥=1,2,3,…. (3.3) And therefore, ^∞_𝑘=𝑥+1|𝑎_𝑘|≤(1−𝛼)(1+𝜆₂𝑥)^𝑚(𝑥+1−𝛼)(1+𝜆₁𝑥)^𝑚−1𝑐(𝑛,𝑥+1),𝑥=1,2,…. (3.4) Since 𝑓(𝑧)=𝑧−^∞_𝑘=𝑥+1|𝑎_𝑘|𝑧^𝑘,𝑥=1,2,…, (3.5) then we have that |𝑓(𝑧)|=|𝑧−^∞_𝑘=𝑥+1|𝑎_𝑘|𝑧^𝑘|. (3.6) After that, |𝑓(𝑧)|≤|𝑧|+|𝑧|^∞𝑥+1_𝑘=𝑥+1|𝑎_𝑘||𝑧|^{𝑘−(𝑥+1)}≤𝑟+𝑟^∞𝑥+1_𝑘=𝑥+1|𝑎_𝑘|. (3.7) By aid of inequality (3.4), it yields the right-hand side inequality of (3.1). Thus, this completes the proof.

Theorem 3.2. Let the function 𝑓 given by (1.2) be in the class 𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼). Then for 0<|𝑧|=𝑟<1, 1−(𝑥+1)(1−𝛼)(1+𝜆₂𝑥)^𝑚(𝑥+1−𝛼)(1+𝜆₁𝑥)^𝑚−1𝑟𝑐(𝑛,𝑥+1)^𝑥≤|𝑓^′(𝑧)|≤1+(𝑥+1)(1−𝛼)(1+𝜆₂𝑥)^𝑚(𝑥+1−𝛼)(1+𝜆₁𝑥)^𝑚−1𝑟𝑐(𝑛,𝑥+1)^𝑥,𝑥=1,2,…, (3.8) where 0≤𝛼<1,𝜆₂≥𝜆₁≥0,𝑚∈ℕ₀={0,1,2,…},and𝑛∈ℕ={1,2,…}.

Proof. Since 𝑓∈𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼), by Theorem 2.2, we have that ^∞_𝑘=𝑥+1(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1(1+𝜆₂(𝑘−1))^𝑚𝑐(𝑛,𝑘)|𝑎_𝑘|≤1−𝛼,𝑥=1,2,3,…. (3.9) Now (𝑥+1−𝛼)(1+𝜆₁𝑥)^𝑚−1(1+𝜆₂𝑥)^𝑚𝑐(𝑛,𝑥+1)(^∞_𝑘=𝑥+1𝑘|𝑎_𝑘=|)_{∞𝑘=𝑥+1}(𝑥+1−𝛼)(1+𝜆₁𝑥)^𝑚−1(1+𝜆₂𝑥)^𝑚𝑐(𝑛,𝑥+1)𝑘|𝑎_𝑘|≤(𝑥+1)^∞_𝑘=𝑥+1(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1(1+𝜆₂(𝑘−1))^𝑚𝑐(𝑛,𝑘)|𝑎_𝑘|≤(𝑥+1)(1−𝛼),(𝑘≥𝑥+1,𝑥=1,2,3,…). (3.10) Hence ^∞_𝑘=𝑥+1𝑘|𝑎_𝑘|≤(𝑥+1)(1−𝛼)(1+𝜆₂𝑥)^𝑚(𝑥+1−𝛼)(1+𝜆₁𝑥)^𝑚−1𝑐(𝑛,𝑥+1),𝑥=1,2,…. (3.11) Since 𝑓^′(𝑧)=1−^∞_𝑘=𝑥+1𝑘|𝑎_𝑘|𝑧^𝑘−1,𝑥=1,2,…, (3.12) then we have that 1−|𝑧|^𝑥∞_𝑘=𝑥+1𝑘|𝑎_𝑘||𝑧|^{𝑘−(𝑥+1)}≤|𝑓^′(𝑧)|≤1+|𝑧|^𝑥∞_𝑘=𝑥+1𝑘|𝑎_𝑘||𝑧|^{𝑘−(𝑥+1)}, (3.13) and therefore, 1−𝑟^𝑥∞_𝑘=𝑥+1𝑘|𝑎_𝑘|≤|𝑓^′(𝑧)|≤1+𝑟^𝑥∞_𝑘=𝑥+1𝑘|𝑎_𝑘|,𝑥=1,2,…. (3.14) By using the inequality (3.11) in (3.14), we get Theorem 3.2. This completes the proof.

4. Extreme Points

The extreme points of the class 𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼) are given by the following theorem.

Theorem 4.1. Let 𝑓_𝑥(𝑧)=𝑧 and 𝑓_𝑘(𝑧)=𝑧−(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1𝑧𝑐(𝑛,𝑘)^𝑘, (4.1) where 0≤𝛼<1,𝜆₂≥𝜆₁≥0,(𝑛,𝑚∈ℕ₀),𝑘=𝑥+1,𝑥+2,…, and 𝑥=1,2,3,….
Then 𝑓∈𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼) if and only if it can be expressed in the form 𝑓(𝑧)=^∞_𝑘=𝑥𝛿_𝑘𝑓_𝑘(𝑧), (4.2) where 𝛿_𝑘≥0 and ∑^∞_𝑘=𝑥𝛿_𝑘=1.

Proof. Suppose that 𝑓 can be expressed as in (4.2). Our goal is to show that 𝑓∈𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼).
By (4.2), we have that 𝑓(𝑧)=^∞_𝑘=𝑥𝛿_𝑘𝑓_𝑘(𝑧)=𝛿_𝑥𝑓_𝑥(𝑧)+^∞_𝑘=𝑥+1𝛿_𝑘𝑓_𝑘(𝑧)=𝛿_𝑥𝑓_𝑥(𝑧)+^∞_𝑘=𝑥+1𝛿_𝑘[𝑧−(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1𝑧𝑐(𝑛,𝑘)^𝑘]=^∞_𝑘=𝑥𝛿_𝑘𝑧−^∞_𝑘=𝑥+1𝛿_𝑘(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1𝑧𝑐(𝑛,𝑘)^𝑘=𝑧−^∞_𝑘=𝑥+1𝛿_𝑘(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1𝑧𝑐(𝑛,𝑘)^𝑘. (4.3) Now 𝑓(𝑧)=𝑧−^∞_𝑘=𝑥+1|𝑎_𝑘|𝑧^𝑘=𝑧−^∞_𝑘=𝑥+1𝛿_𝑘(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1𝑧𝑐(𝑛,𝑘)^𝑘, (4.4) so that |𝑎_𝑘𝛿|=^𝑘(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1.𝑐(𝑛,𝑘) (4.5) Now, we have that ^∞_𝑘=𝑥+1𝛿_𝑘=1−𝛿_𝑥≤1,𝑥=1,2,3,…. (4.6) Setting ^∞_𝑘=𝑥+1𝛿_𝑘=^∞_𝑘=𝑥+1𝛿_𝑘(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1𝑐(𝑛,𝑘)(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1𝑐(𝑛,𝑘)(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚≤1, (4.7) we arrive to ^∞_𝑘=𝑥+1(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚𝑐(𝑛,𝑘)|𝑎_𝑘|≤1. (4.8) And therefore, ^∞_𝑘=𝑥+1(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1(1+𝜆₂(𝑘−1))^𝑚𝑐(𝑛,𝑘)|𝑎_𝑘|≤1−𝛼,𝑥=1,2,3,…. (4.9) It follows from Theorem 2.2 that 𝑓∈𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼).
Conversely, let us suppose that 𝑓∈𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼); our goal is, to get (4.2). From (4.2) and using similar last arguments, it is easily seen that 𝑓(𝑧)=𝑧−^∞_𝑘=𝑥+1|𝑎_𝑘|𝑧^𝑘=𝑧−^∞_𝑘=𝑥+1𝛿_𝑘(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1𝑧𝑐(𝑛,𝑘)^𝑘, (4.10) which suffices to show that |𝑎_𝑘𝛿|=^𝑘(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1.𝑐(𝑛,𝑘) (4.11) Now, we have that 𝑓∈𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼), then by previous Theorem 2.3, |𝑎_𝑘|≤(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1.𝑐(𝑛,𝑘) (4.12) That is (𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1𝑐(𝑛,𝑘)|𝑎_𝑘|(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚≤1. (4.13) Since ∑^∞_𝑘=𝑥𝛿_𝑘=1, we see 𝛿_𝑘≤1, for each 𝑘=𝑥,𝑥+1,𝑥+2,…, and 𝑥=1,2,3,….
We can set that 𝛿_𝑘=(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1𝑐(𝑛,𝑘)|𝑎_𝑘|(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚. (4.14) Thus, the desired result is that |𝑎_𝑘𝛿|=^𝑘(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1.𝑐(𝑛,𝑘) (4.15) This completes the proof of the theorem.

Corollary 4.2. The extreme points of 𝒯ℋ_𝜆^𝑛,𝑚_1,𝜆2(𝑥,𝛼) are the functions 𝑓_𝑥𝑓(𝑧)=𝑧,_𝑘(𝑧)=𝑧−(1−𝛼)(1+𝜆₂(𝑘−1))^𝑚(𝑘−𝛼)(1+𝜆₁(𝑘−1))^𝑚−1𝑧𝑐(𝑛,𝑘)^𝑘, (4.16) where 0≤𝛼<1,(𝑛,𝑚∈ℕ₀={0,1,2,…}),𝜆₂≥𝜆₁≥0, and 𝑘=𝑥+1,𝑥+2,…,(𝑥=1,2,3,…).

Acknowledgments

This work is fully supported by UKM-GUP-TMK-07-02-107, Malaysia. The authors are also grateful to the referee for his/her suggestions which helped us to improve the contents of this article.