Abstract

The authors introduce two new subclasses denoted by (Ω,𝜆,𝑝,𝛼) and 𝜀(Ω,𝜆,𝑝,𝛼) of the class 𝐴(𝑝,𝑛) of 𝑝-valent analytic functions. They obtain coefficient inequality for the class (Ω,𝜆,𝑝,𝛼). They investigate various properties of classes (Ω,𝜆,𝑝,𝛼) and 𝜀(Ω,𝜆,𝑝,𝛼). Furthermore, they derive partial sums associated with the class 𝜀(Ω,𝜆,𝑝,𝛼).

1. Introduction and Definition

Let 𝐴(𝑝,𝑛) denote the class of functions of the form 𝑓(𝑧)=𝑧𝑝+𝑘=𝑛𝑎𝑝+𝑘𝑧𝑝+𝑘(𝑛,𝑝={1,2,}),(1.1) which are analytic and p-valent in the open unit disc 𝑈={𝑧𝑧,|𝑧|<1}. We write 𝐴(1,1)=𝐴.

A function 𝑓𝐴(𝑝,𝑛) is said to be in the class 𝑆(𝑝,𝑛,𝛼) of p-valently star-like functions of order 𝛼 if it satisfies the conditionRe𝑧𝑓(𝑧)𝑓(𝑧)>𝛼(𝑧;0𝛼<𝑝).(1.2) Furthermore, a function 𝑓𝐴(𝑝,𝑛) is said to be in the class 𝐾(𝑝,𝑛,𝛼) of p-valently convex functions of order 𝛼 if it satisfies the conditionRe1+𝑧𝑓(𝑧)𝑓(𝑧)>𝛼(𝑧𝑈;0𝛼<𝑝).(1.3) The classes 𝑆(𝑝,𝑛,𝛼) and 𝐾(𝑝,𝑛,𝛼) were studied by Owa [1]. The class 𝑆(𝑝,𝛼)=𝑆(𝑝,1,𝛼) was considered by Patil and Thakare [2].

We denote by 𝑇(𝑝,𝑛) the subclass of the class 𝐴(𝑝,𝑛) consisting of functions of the form 𝑓(𝑧)=𝑧𝑝𝑘=𝑛𝑎𝑝+𝑘𝑧𝑝+𝑘𝑎𝑝+𝑘0;𝑛,𝑝(1.4) and define two further classes 𝑇(𝑝,𝑛,𝛼) and 𝐶(𝑝,𝑛,𝛼) by𝑇(𝑝,𝑛,𝛼)=𝑆(𝑝,𝑛,𝛼)𝑇(𝑝,𝑛),𝐶(𝑝,𝑛,𝛼)=𝐾(𝑝,𝑛,𝛼)𝑇(𝑝,𝑛).(1.5) For the classes 𝑇(𝑝,𝛼)=𝑆(𝑝,𝛼)𝑇(𝑝),𝐶(𝑝,𝛼)=𝐾(𝑝,𝛼)𝑇(𝑝).

The following lemmas were given by Owa [3].

Lemma 1.1. Let the function 𝑓 be defined by 𝑓(𝑧)=𝑧𝑝𝑘=1𝑎𝑝+𝑘𝑧𝑝+𝑘𝑎𝑝+𝑘0;𝑝.(1.6) Then, 𝑓 is in the class 𝑇(𝑝,𝛼) if and only if 𝑘=1(𝑝+𝑘𝛼)𝑎𝑝+𝑘𝑝𝛼.(1.7) The result is sharp.

Lemma 1.2. Let the function 𝑓 be defined by (1.6). Then, 𝑓 is in the class 𝐶(𝑝,𝛼) if and only if 𝑘=1(𝑝+𝑘)(𝑝+𝑘𝛼)𝑎𝑝+𝑘𝑝(𝑝𝛼).(1.8) The result is sharp.

For a function 𝑓 defined by (1.6) and in the class 𝑇(𝑝,𝛼), Lemma 1.1 yields𝑎𝑝+1𝑝𝛼𝑝+1𝛼.(1.9) On the other hand, for a function 𝑓 defined by (1.6) and in the class 𝐶(𝑝,𝛼), Lemma 1.2 yields𝑎𝑝+1𝑝(𝑝𝛼)(𝑝+1)(𝑝+1𝛼).(1.10) In view of the coefficient inequalities (1.9) and (1.10), it would seem to be natural to introduce and study here two further classes 𝑇𝜀(𝑝,𝛼) and 𝐶𝜀(𝑝,𝛼) of analytic and p-valent functions, where 𝑇𝜀(𝑝,𝛼) denotes the subclass of 𝑇(𝑝,𝛼) consisting of functions of the form𝑓(𝑧)=𝑧𝑝(𝑝𝛼)𝜀𝑧𝑝+1𝛼𝑝+1𝑘=2𝑎𝑝+𝑘𝑧𝑝+𝑘𝑎𝑝+𝑘0,𝑝,𝑘{1};0𝛼<𝑝;0𝜀<1(1.11) and 𝐶𝜀(𝑝,𝛼) denotes the subclass of 𝐶(𝑝,𝛼) consisting of functions of the form𝑓(𝑧)=𝑧𝑝𝑝(𝑝𝛼)𝜀𝑧(𝑝+1)(𝑝+1𝛼)𝑝+1𝑘=2𝑎𝑝+𝑘𝑧𝑝+𝑘𝑎𝑝+𝑘.0,𝑝,𝑘{1};0𝛼<𝑝;0𝜀<1(1.12) The classes 𝑇𝜀(𝑝,𝛼) and 𝐶𝜀(𝑝,𝛼) are studied by Aouf et al. [4].

The classes 𝑇𝜀(𝛼)=𝑇𝜀(1,𝛼),𝐶𝜀(𝛼)=𝐶𝜀(1,𝛼)(1.13)

were considered earlier by Silverman and Silvia [5].

Now, we give the following equalities for the functions 𝑓(𝑧) belonging to the class 𝐴(𝑝,𝑛):𝐷0𝐷𝑓(𝑧)=𝑓(𝑧),1𝐷𝑓(𝑧)=𝐷𝑓(𝑧)=𝑧0𝑓(𝑧)=𝑧𝑝𝑧𝑝1+𝑘=𝑛(𝑝+𝑘)𝑎𝑝+𝑘𝑧𝑝+𝑘1=𝑝𝑧𝑝+𝑘=𝑛(𝑝+𝑘)𝑎𝑝+𝑘𝑧𝑝+𝑘,𝐷2𝑓𝐷(𝑧)=𝐷(𝐷𝑓(𝑧))=𝑧1𝑓(𝑧)=𝑧𝑝𝑧𝑝+𝑘=𝑛(𝑝+𝑘)𝑎𝑝+𝑘𝑧𝑝+𝑘=𝑝2𝑧𝑝+𝑘=𝑛(𝑝+𝑘)2𝑎𝑝+𝑘𝑧𝑝+𝑘,𝐷Ω𝐷𝑓(𝑧)=𝐷Ω1𝑓(𝑧)=𝑝Ω𝑧𝑝+𝑘=𝑛(𝑘+𝑝)Ω𝑎𝑝+𝑘𝑧𝑝+𝑘.(1.14) We define 𝐴(𝑝,𝑛)𝐴(𝑝,𝑛) such that1(Ω,𝜆,𝑝)=𝑝Ω𝐷𝜆Ω𝜆𝑓(𝑧)+𝑝𝑧𝐷Ω𝑓(𝑧)10𝜆𝑝Ω,Ω{0}.(1.15)

A function 𝑓(𝑧)𝐴(𝑝,𝑛) is said to be in the class (Ω,𝜆,𝑝,𝛼) if it satisfies the inequalityRe𝑧((Ω,𝜆,𝑝))𝑧(Ω,𝜆,𝑝)=Re1/𝑝Ω𝐷+(1/𝑝1)𝜆Ω𝑓(𝑧)𝐷+(𝜆/𝑝)𝑧Ω𝑓(𝑧)1/𝑝Ω𝐷𝜆Ω𝐷𝑓(𝑧)+(𝜆/𝑝)𝑧Ω𝑓(𝑧)>𝛼,(1.16) for some 𝛼(0𝛼<𝑝),0𝜆1/𝑝Ω,  Ω{0} and for all 𝑧𝑈.

If Ω=0 and 𝜆=0, we obtain the condition (1.2). Furthermore, we obtain the condition (1.3) for Ω=0 and 𝜆=1.

We denote by 𝑇(𝑝,𝑛) the subclass of the class 𝐴(𝑝,𝑛) consisting of functions of the form 𝑓(𝑧)=𝑧𝑝𝑘=𝑛𝑎𝑘+𝑝𝑧𝑘+𝑝𝑎𝑘+𝑝0;𝑛,𝑝,(1.17) and define the class (Ω,𝜆,𝑝,𝛼) by(Ω,𝜆,𝑝,𝛼)=(Ω,𝜆,𝑝,𝛼)𝑇(𝑝,𝑛).(1.18)

Furthermore, we denote by 𝜀(Ω,𝜆,𝑝,𝛼) the subclass of (Ω,𝜆,𝑝,𝛼) consisting of functions of the form 𝑓(𝑧)=𝑧𝑝(𝑝𝛼)𝜀((𝑝+𝑛)/𝑝)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑝+𝑛𝑘=𝑛+1𝑎𝑝+𝑘𝑧𝑝+𝑘(0𝜀<1).(1.19) The main object of the present paper is to investigate interesting properties and characteristics of the classes (Ω,𝜆,𝑝,𝛼) and 𝜀(Ω,𝜆,𝑝,𝛼). Also, the partial sums is defined for 𝑓 function defined by (1.19).

2. A Coefficient Inequality for the Class (Ω,𝜆,𝑝,𝛼) and Some Theorems for the Class 𝜀(Ω,𝜆,𝑝,𝛼)

First, we give a coefficient inequality for the class (Ω,𝜆,𝑝,𝛼).

Theorem 2.1. Let the function 𝑓 be defined by (1.17). Then, 𝑓 is in the class (Ω,𝜆,𝑝,𝛼) if and only if 𝑘=n𝑘+𝑝𝑝Ω1+𝜆𝑘𝑝Ω1(𝑘+𝑝𝛼)𝑎𝑝+𝑘𝑝𝛼.(2.1)

Proof. Suppose that 𝑓(𝑧)(Ω,𝜆,𝑝,𝛼). Then, we find from (1.16) that Re𝑧((Ω,𝜆,𝑝))𝑧(Ω,𝜆,𝑝)=Re1/𝑝Ω𝐷+(1/𝑝1)𝜆Ω𝑓(𝑧)𝐷+(𝜆/𝑝)𝑧Ω𝑓(𝑧)1/𝑝Ω𝐷𝜆Ω𝐷𝑓(𝑧)+(𝜆/𝑝)𝑧Ω𝑓(𝑧)=Re𝑝𝑧𝑝𝑘=𝑛(𝑘+𝑝)Ω+11/𝑝Ω𝑎+𝜆𝑘/𝑝𝑝+𝑘𝑧𝑝+𝑘𝑧𝑝𝑘=𝑛(𝑘+𝑝)Ω1/𝑝Ω𝑎+𝜆𝑘/𝑝𝑝+𝑘𝑧𝑝+𝑘>𝛼.(2.2) If we choose 𝑧 to be real and let 𝑧1, we get 𝑝𝑘=𝑛(𝑘+𝑝)Ω+11/𝑝Ω𝑎+𝜆𝑘/𝑝𝑝+𝑘1𝑘=𝑛(𝑘+𝑝)Ω1/𝑝Ω𝑎+𝜆𝑘/𝑝𝑝+𝑘𝛼(2.3) or, equivalently, 𝑝𝑘=𝑛(𝑘+𝑝)Ω+11/𝑝Ω𝑎+𝜆𝑘/𝑝𝑝+𝑘𝛼1𝑘=𝑛(𝑘+𝑝)Ω1/𝑝Ω𝑎+𝜆𝑘/𝑝𝑝+𝑘.(2.4) Thus, we have 𝑘=𝑛(𝑘+𝑝)Ω+11/𝑝Ω+𝜆𝑘/𝑝(𝑘+𝑝)Ω1/𝑝Ω𝑎+𝜆𝑘/𝑝𝑝+𝑘𝑝𝛼(2.5) or 𝑘=𝑛𝑘+𝑝𝑝Ω1+𝜆𝑘𝑝Ω1𝑎(𝑘+𝑝𝛼)𝑝+𝑘𝑝𝛼.(2.6) Conversely, suppose that the inequality (2.1) holds true and let 𝑧𝜕𝑈={𝑧𝑧,|𝑧|=1}.(2.7) Then, we find from the definition (1.4) that ||||𝑧((Ω,𝜆,𝑝))||||=|||||𝑧(Ω,𝜆,𝑝)𝑝1/𝑝Ω𝐷+(1/𝑝1)𝜆Ω𝑓(𝑧)𝐷+(𝜆/𝑝)𝑧Ω𝑓(𝑧)1/𝑝Ω𝐷𝜆Ω𝐷𝑓(𝑧)+(𝜆/𝑝)𝑧Ω𝑓(𝑧)|||||=||||𝑝𝑝𝑧𝑝𝑘=𝑛((𝑘+𝑝)/𝑝)Ω(𝑘+𝑝)1+𝜆𝑘𝑝Ω1𝑎𝑝+𝑘𝑧𝑝+𝑘𝑧𝑝𝑘=𝑛((𝑘+𝑝)/𝑝)Ω1+𝜆𝑘𝑝Ω1𝑎𝑝+𝑘𝑧𝑝+𝑘||||𝑝𝑘=𝑛𝑘((𝑘+𝑝)/𝑝)Ω1+𝜆𝑘𝑝Ω1𝑎𝑝+𝑘1𝑘=𝑛((𝑘+𝑝)/𝑝)Ω1+𝜆𝑘𝑝Ω1𝑎𝑝+𝑘.(2.8) By means of inequality (2.1), we can write 𝑘=𝑛𝑘+𝑝𝑝Ω1+𝜆𝑘𝑝Ω1(𝑘+𝑝𝛼)𝑎𝑝+𝑘𝑝𝛼𝑘=𝑛𝑘𝑘+𝑝𝑝Ω1+𝜆𝑘𝑝Ω1𝑎𝑝+𝑘𝑝𝛼𝑘=𝑛𝑘+𝑝𝑝Ω1+𝜆𝑘𝑝Ω1(𝑝𝛼)𝑎𝑝+𝑘(2.9) or 𝑘=𝑛𝑘𝑘+𝑝𝑝Ω1+𝜆𝑘𝑝Ω1𝑎𝑝+𝑘𝑝𝛼(𝑝𝛼)𝑘=𝑛𝑘+𝑝𝑝Ω1+𝜆𝑘𝑝Ω1𝑎𝑝+𝑘.(2.10) Thus, we obtain |||||𝑧1/𝑝Ω𝐷+(1/𝑝1)𝜆Ω𝑓(𝑧)𝐷+(𝜆/𝑝)𝑧Ω𝑓(𝑧)1/𝑝Ω𝐷𝜆Ω𝐷𝑓(𝑧)+(𝜆/𝑝)𝑧Ω𝑓(𝑧)|||||=||||𝑝𝑘=𝑛𝑘((𝑘+𝑝)/𝑝)Ω1+𝜆𝑘𝑝Ω1𝑎𝑝+𝑘𝑧𝑘1𝑘=𝑛((𝑘+𝑝)/𝑝)Ω1+𝜆𝑘𝑝Ω1𝑎𝑝+𝑘𝑧𝑘||||𝑝𝛼(𝑝𝛼)𝑘=𝑛((𝑘+𝑝)/𝑝)Ω1+𝜆𝑘𝑝Ω1𝑎𝑝+𝑘1𝑘=𝑛((𝑘+𝑝)/𝑝)Ω1+𝜆𝑘𝑝Ω1𝑎𝑝+𝑘=𝑝𝛼.(2.11) This evidently completes the proof of Theorem 2.1.

Now, we give a characterization theorem for the class 𝜀(Ω,𝜆,𝑝,𝛼).

Theorem 2.2. Let the function 𝑓 be defined by (1.19). Then, 𝑓 is in the class 𝜀(Ω,𝜆,𝑝,𝛼) if and only if 𝑘=𝑛+1𝑘+𝑝𝑝Ω1+𝜆𝑘𝑝Ω1(𝑘+𝑝𝛼)𝑎𝑝+𝑘1(𝑝𝛼)(1𝜀)0𝛼<𝑝,0𝜀<1,0𝜆𝑝Ω.(2.12) The result is sharp for the function 𝑓 given by 𝑓(𝑧)=𝑧𝑝(𝑝𝛼)𝜀((𝑝+𝑛)/𝑝)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑝+𝑛(𝑝𝛼)(1𝜀)((𝑝+𝑘)/𝑝)Ω1+𝜆𝑘𝑝Ω1(𝑧𝑘+𝑝𝛼)𝑝+𝑘(𝑘=𝑛+1,𝑛+2,;𝑛).(2.13)

Proof. Using inequality (2.1), we have 𝑛+𝑝𝑝Ω1+𝜆𝑛𝑝Ω1(𝑛+𝑝𝛼)𝑎𝑝+𝑛+𝑘=𝑛+1𝑘+𝑝𝑝Ω1+𝜆𝑘𝑝Ω1(𝑘+𝑝𝛼)𝑎𝑝+𝑘𝑝𝛼(2.14) or 𝑘=𝑛+1𝑘+𝑝𝑝Ω1+𝜆𝑘𝑝Ω1(𝑘+𝑝𝛼)𝑎𝑝+𝑘𝑝𝛼𝑛+𝑝𝑝Ω1+𝜆𝑛𝑝Ω1(𝑛+𝑝𝛼)𝑎𝑝+𝑛.(2.15) Thus, by setting 𝑎𝑝+𝑛=(𝑝𝛼)𝜀((𝑝+𝑛)/𝑝)Ω1+𝜆𝑛𝑝Ω1(𝑛+𝑝𝛼),(2.16) we obtain 𝑘=𝑛+1𝑘+𝑝𝑝Ω1+𝜆𝑘𝑝Ω1(𝑘+𝑝𝛼)𝑎𝑝+𝑘𝑝𝛼𝑛+𝑝𝑝Ω1+𝜆𝑛𝑝Ω1(𝑛+𝑝𝛼)(𝑝𝛼)𝜀((𝑝+𝑛)/𝑝)Ω1+𝜆𝑛𝑝Ω1(𝑛+𝑝𝛼)(2.17) or 𝑘=𝑛+1𝑘+𝑝𝑝Ω1+𝜆𝑘𝑝Ω1(𝑘+𝑝𝛼)𝑎𝑝+𝑘(p𝛼)(1𝜀).(2.18) A closure theorem for the class 𝜀(Ω,𝜆,𝑝,𝛼) is given by the following.

Theorem 2.3. Let 𝑓𝑠(𝑧)=𝑧𝑝(𝑝𝛼)𝜀((𝑝+𝑛)/𝑝)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑝+𝑛𝑘=𝑛+1𝑎𝑝+𝑘,𝑠𝑧𝑝+𝑘𝑎𝑝+𝑘,𝑠.0;𝑝,𝑛;0𝜀<1;𝑠=1,2,3,,𝑚(2.19) If 𝑓𝑠𝜀(Ω,𝜆,𝑝,𝛼), then the function 𝑔 given by 𝑔(𝑧)=𝑧𝑝(𝑝𝛼)𝜀((𝑝+𝑛)/𝑝)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑝+𝑛k=𝑛+1𝑏𝑝+𝑘𝑧𝑝+𝑘,(2.20) with 𝑏𝑝+𝑘1=𝑚𝑚𝑠=1𝑎𝑝+𝑘,𝑠0,(2.21) is also in the class 𝜀(Ω,𝜆,𝑝,𝛼).

Proof. Since 𝑓𝑠𝜀(Ω,𝜆,𝑝,𝛼) for 𝑠=1,2,,𝑚, it follows from Theorem 2.2 that 𝑘=𝑛+1𝑘+𝑝𝑝Ω1+𝜆𝑘𝑝Ω1(𝑘+𝑝𝛼)𝑎𝑝+𝑘,𝑠(𝑝𝛼)(1𝜀)(𝑠=1,2,,𝑚).(2.22) By applying (2.22) and the definition (2.21), we write 𝑘=𝑛+1𝑘+𝑝𝑝Ω1+𝜆𝑘𝑝Ω1(𝑘+𝑝𝛼)𝑏𝑝+𝑘=𝑘=𝑛+1𝑘+𝑝𝑝Ω1+𝜆𝑘𝑝Ω1(1𝑘+𝑝𝛼)𝑚𝑚𝑠=1𝑎𝑝+𝑘,𝑠=1𝑚𝑚𝑠=1𝑘=𝑛+1𝑘+𝑝𝑝Ω1+𝜆𝑘𝑝Ω1(𝑘+𝑝𝛼)𝑎𝑝+𝑘,𝑠1𝑚𝑚𝑠=1(𝑝𝛼)(1𝜀)=(𝑝𝛼)(1𝜀).(2.23)

Theorem 2.4. Let 𝑓𝑝+𝑛(𝑧)=𝑧𝑝(𝑝𝛼)𝜀((𝑝+𝑛)/𝑝)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑝+𝑛,𝑓𝑝+𝑘(𝑧)=𝑧𝑝(𝑝𝛼)𝜀((𝑝+𝑛)/𝑝)Ω1+𝜆𝑛𝑝Ω1(𝑧𝑛+𝑝𝛼)𝑝+𝑛(𝑝𝛼)(1𝜀)((𝑝+𝑘)/𝑝)Ω1+𝜆𝑘𝑝Ω1𝑧(𝑘+𝑝𝛼)𝑝+𝑘,(2.24) where 𝑘{𝑛+1,𝑛+2,}.
Then, 𝑓 is in the class 𝜀(Ω,𝜆,𝑝,𝛼) if and only if it can be expressed in the form 𝑓(𝑧)=𝑘=𝑛𝜂𝑝+𝑘𝑓𝑝+𝑘𝜂(𝑧)𝑝+𝑘0;𝑘=𝑛𝜂𝑝+𝑘=1.(2.25)

Proof. Suppose that 𝑓 is given by (2.25), so that we find from (2.24) that 𝑓(𝑧)=𝑧𝑝(𝑝𝛼)𝜀((𝑝+𝑛)/𝑝)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑝+𝑛𝑘=𝑛+1(𝑝𝛼)(1𝜀)((𝑝+𝑘)/𝑝)Ω1+𝜆𝑘𝑝Ω1(𝜂𝑘+𝑝𝛼)𝑝+𝑘𝑧𝑝+𝑘,(2.26) where the coefficients 𝜂𝑝+𝑘 are given with 𝑘=𝑛𝜂𝑝+𝑘=1,  𝜂𝑝+𝑘0. Then, since, 𝑘=𝑛+1𝑘+𝑝𝑝Ω1+𝜆𝑘𝑝Ω1(𝑘+𝑝𝛼)(𝑝𝛼)(1𝜀)((𝑘+𝑝)/𝑝)Ω1+𝜆𝑘𝑝Ω1𝜂(𝑘+𝑝𝛼)𝑝+𝑘=(𝑝𝛼)(1𝜀)𝑘=𝑛+1𝜂𝑝+𝑘=(𝑝𝛼)(1𝜀)1𝜂𝑝+𝑛(𝑝𝛼)(1𝜀),(2.27) we conclude from Theorem 2.2 that 𝑓𝜀(Ω,𝜆,𝑝,𝛼).
Conversely, let us assume that the function 𝑓 defined by (1.19) is in the class 𝜀(Ω,𝜆,𝑝,𝛼). Then, 𝑎𝑝+𝑘(𝑝𝛼)(1𝜀)((𝑘+𝑝)/𝑝)Ω1+𝜆𝑘𝑝Ω1(𝑘+𝑝𝛼),(2.28) which follows readily from (2.12) for 𝑘{𝑛+1,𝑛+2,}.
For 𝑘{𝑛+1,𝑛+2,}, setting 𝜂𝑝+𝑘=((𝑘+𝑝)/𝑝)Ω1+𝜆𝑘𝑝Ω1(𝑘+𝑝𝛼)𝑎(𝑝𝛼)(1𝜀)𝑝+𝑘(2.29) and 𝜂𝑝+𝑛=1𝑘=𝑛+1𝜂𝑝+𝑘, we thus arrive at (2.25). This completes the proof of Theorem 2.4.

Theorem 2.5. Let 𝑓 be given by (1.19) and define the partial sums 𝑠𝑚(𝑧) by 𝑠𝑚𝑧(𝑧)=𝑝𝑝Ω(𝑝𝛼)𝜀(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑝+𝑛𝑧,𝑚=𝑛,𝑝𝑝Ω(𝑝𝛼)𝜀(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1(𝑧𝑛+𝑝𝛼)𝑝+𝑛𝑚𝑘=𝑛+1𝑎𝑝+𝑘𝑧𝑝+𝑘,𝑚=𝑛+1,𝑛+2,.(2.30) Suppose also that 𝑘=𝑛+1𝑑𝑝+𝑘𝑎𝑝+𝑘𝑝1Ω(𝑝𝛼)𝜀(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1,(𝑛+𝑝𝛼)where𝑑𝑝+𝑘=(𝑝+𝑘)Ω1+𝜆𝑘𝑝Ω1(𝑘+𝑝𝛼)𝑝Ω1(𝑝𝛼)(1𝜀);0𝛼<𝑝,0𝜀<1,0𝜆𝑝Ω.(2.31) Then, for 𝑚𝑛+1, one has Re𝑓(𝑧)𝑠𝑚𝑝(𝑧)>1Ω(𝑝𝛼)(1𝜀)(𝑝+𝑚+1)Ω1+𝜆(𝑚+1)𝑝Ω1,𝑠(𝑝+𝑚+1𝛼)(2.32)Re𝑚(𝑧)>𝑓(𝑧)(𝑝+𝑚+1)Ω1+𝜆(𝑚+1)𝑝Ω1(𝑚+𝑝+1𝛼)𝑝Ω(𝑝𝛼)(1𝜀)+(𝑝+𝑚+1)Ω1+𝜆(𝑚+1)𝑝Ω1(𝑚+𝑝+1𝛼).(2.33) Each of the bounds in (2.32) and (2.33) is the best possible.

Proof. From (2.31) and Theorem 2.2, we have that 𝑓𝜀(Ω,𝜆,𝑝,𝛼). By definition 𝑑𝑝+𝑘, we can write 𝑑𝑝+𝑘=((𝑝+𝑘)/𝑝)Ω1+𝜆𝑘𝑝Ω1(𝑘+𝑝𝛼)=𝑘(𝑝𝛼)(1𝜀)1+𝑝Ω1+𝜆𝑘𝑝Ω1𝑝𝛼+𝑘=𝑘(𝑝𝛼)(1𝜀)1+𝑝Ω1+𝜆𝑘𝑝Ω11+𝑘1𝜀(𝑝𝛼)(1𝜀)>1.(2.34) Under the hypothesis of this theorem, we can see from (2.31) that 𝑑𝑝+𝑘+1>𝑑𝑝+𝑘>1, for 𝑘=𝑛+1,𝑛+2. Therefore, we have 𝑚𝑘=𝑛+1𝑎𝑝+𝑘+𝑑𝑝+𝑚+1𝑘=𝑚+1𝑎𝑝+𝑘𝑘=𝑛+1𝑑𝑝+𝑘𝑎𝑝+𝑘𝑝1Ω(𝑝𝛼)𝜀(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1(𝑝+𝑛𝛼).(2.35) Using (1.19) and (2.30), we can write 𝑓(𝑧)𝑠𝑚=𝑧(𝑧)𝑝𝑝Ω(𝑝𝛼)𝜀/(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑝+𝑛𝑘=𝑛+1𝑎𝑝+𝑘𝑧𝑝+𝑘𝑧𝑝𝑝Ω(𝑝𝛼)𝜀/(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑝+𝑛𝑚𝑘=𝑛+1𝑎𝑝+𝑘𝑧𝑝+𝑘=1𝑘=𝑚+1𝑎𝑝+𝑘𝑧𝑝+𝑘𝑧𝑝𝑝Ω(𝑝𝛼)𝜀/(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑝+𝑛𝑚𝑘=𝑛+1𝑎𝑝+𝑘𝑧𝑝+𝑘.(2.36) Set Ψ1(𝑧)=𝑑𝑝+𝑚+1𝑓(𝑧)𝑠𝑚1(𝑧)1𝑑𝑝+𝑚+1𝑑=1𝑝+𝑚+1𝑘=𝑚+1𝑎𝑝+𝑘𝑧𝑘𝑝1Ω(𝑝𝛼)𝜀/(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑛𝑚𝑘=𝑛+1𝑎𝑝+𝑘𝑧𝑘.(2.37) By applying (2.35) and (2.37), we find that ||||𝜓1(𝑧)1𝜓1||||=||||||||||(𝑧)+1𝑑𝑝+𝑚+1𝑘=𝑚+1𝑎𝑝+𝑘𝑧𝑘22𝑝Ω(𝑝𝛼)𝜀(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑛2𝑚𝑘=𝑛+1𝑎𝑝+𝑘𝑧𝑘𝑑𝑝+𝑚+1𝑘=𝑚+1𝑎𝑝+𝑘𝑧𝑘||||||||||𝑑𝑝+𝑚+1𝑘=𝑚+1𝑎𝑝+𝑘𝑧𝑝+𝑘2𝑝1Ω(𝑝𝛼)𝜀(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1(𝑛+𝑝𝛼)𝑚𝑘=𝑛+1𝑎𝑝+𝑘𝑑𝑝+𝑚+1𝑘=𝑚+1𝑎𝑝+𝑘1(𝑧𝑈;𝑚𝑛+1),(2.38) which shows that ReΨ1(𝑧)>0. Thus, we obtain ReΨ1𝑑(𝑧)=Re𝑝+𝑚+1𝑓(𝑧)𝑠𝑚1(𝑧)1𝑑𝑝+𝑚+1>0Re𝑓(𝑧)𝑠𝑚1(𝑧)>1𝑑𝑝+𝑚+1(2.39) or Re𝑓(𝑧)𝑠𝑚𝑝(𝑧)>1Ω(𝑝𝛼)(1𝜀)(𝑝+𝑚+1)Ω1+𝜆(𝑚+1)𝑝Ω1(𝑝+𝑚+1𝛼).(2.40) Let 𝑓(𝑧)=𝑧𝑝𝑝Ω(𝑝𝛼)𝜀(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑝+𝑛1𝑝Ω(𝑝𝛼)𝜀/(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1(𝑛+𝑝𝛼)𝑑𝑝+𝑚+1𝑧𝑝+𝑚+1.(2.41) Then, 𝑓(𝑧) satisfies the condition (2.31) and 𝑓𝜀(Ω,𝜆,𝑝,𝛼). Thus, we can write 𝑓(𝑧)𝑠𝑚=𝑝(𝑧)1Ω(𝑝𝛼)𝜀/(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑛𝑝1Ω(𝑝𝛼)𝜀/(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑛𝑝1Ω(𝑝𝛼)𝜀/(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1(𝑛+𝑝𝛼)/𝑑𝑝+𝑚+1𝑧𝑚+1𝑝1Ω(𝑝𝛼)𝜀/(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑛𝑝=11Ω(𝑝𝛼)𝜀/(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1(𝑛+𝑝𝛼)/𝑑𝑝+𝑚+1𝑧𝑚+1𝑝1Ω(𝑝𝛼)𝜀/(𝑝+𝑛)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑛(2.42) and taking as 𝑧111𝑑𝑝+𝑚+1𝑝=1Ω(𝑝𝛼)(1𝜀)(𝑝+𝑚+1)Ω1+𝜆(𝑚+1)𝑝Ω1(𝑝+𝑚+1𝛼),(2.43) which shows that the bound in (2.32) is the best possible.
By using definitions of 𝑓(𝑧) and 𝑠𝑚(𝑧), we can write 𝑠𝑚(𝑧)=𝑧𝑓(𝑧)𝑝(𝑝𝛼)𝜀/((𝑝+𝑛)/𝑝)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑝+𝑛𝑚𝑘=𝑛+1𝑎𝑝+𝑘𝑧𝑝+𝑘𝑧𝑝(𝑝𝛼)𝜀/((𝑝+𝑛)/𝑝)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑝+𝑛𝑘=𝑛+1𝑎𝑝+𝑘𝑧𝑝+𝑘=𝑧𝑝(𝑝𝛼)𝜀/((𝑝+𝑛)/𝑝)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑝+𝑛𝑧𝑝(𝑝𝛼)𝜀/((𝑝+𝑛)/𝑝)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑝+𝑛𝑘=𝑛+1𝑎𝑝+𝑘𝑧𝑝+𝑘𝑘=𝑛+1𝑎𝑝+𝑘𝑧𝑝+𝑘+𝑘=𝑚+1𝑎𝑝+𝑘𝑧𝑝+𝑘𝑧𝑝(𝑝𝛼)𝜀/((𝑝+𝑛)/𝑝)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑝+𝑛𝑘=𝑛+1𝑎𝑝+𝑘𝑧𝑝+𝑘=1+𝑘=𝑚+1𝑎𝑝+𝑘𝑧𝑘1(𝑝𝛼)𝜀/((𝑝+𝑛)/𝑝)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑛𝑘=𝑛+1𝑎𝑝+𝑘𝑧𝑘.(2.44) If we put 𝜓2(𝑧)=1+𝑑𝑝+𝑚+1𝑠𝑚(𝑧)𝑓𝑑(𝑧)𝑝+𝑚+11+𝑑𝑝+𝑚+1=1+1+𝑑𝑝+𝑚+1𝑘=𝑚+1𝑎𝑝+𝑘𝑧𝑘1(𝑝𝛼)𝜀/((𝑝+𝑛)/𝑝)Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑛𝑘=𝑛+1𝑎𝑝+𝑘𝑧𝑘(2.45) and make use of (2.35), we can deduce that ||||𝜓2(𝑧)1𝜓2||||=|||||||||||||||||(𝑧)+11+𝑑𝑝+𝑚+1𝑘=𝑚+1𝑎𝑝+𝑘𝑧𝑘22(𝑝𝛼)𝜀𝑝+𝑛𝑝Ω1+𝜆𝑛𝑝Ω1𝑧(𝑛+𝑝𝛼)𝑛2𝑚𝑘=𝑛+1𝑎𝑝+𝑘𝑧𝑘+𝑑𝑝+𝑚+11𝑘=𝑚+1𝑎𝑝+𝑘𝑧𝑘|||||||||||||||||1+𝑑𝑝+𝑚+1𝑘=𝑚+1𝑎𝑝+𝑘22(𝑝𝛼)𝜀𝑝+𝑛𝑝Ω1+𝜆𝑛𝑝Ω1(𝑛+𝑝𝛼)2𝑚𝑘=𝑛+1𝑎𝑝+𝑘𝑑𝑝+𝑚+11𝑘=𝑚+1𝑎𝑝+𝑘=1+𝑑𝑝+𝑚+1𝑘=𝑚+1𝑎𝑝+𝑘21(𝑝𝛼)𝜀𝑝+𝑛𝑝Ω1+𝜆𝑛𝑝Ω1(𝑛+𝑝𝛼)𝑚𝑘=𝑛+1𝑎𝑝+𝑘𝑑𝑝+𝑚+11𝑘=𝑚+1𝑎𝑝+𝑘1+𝑑𝑝+𝑚+1𝑘=𝑚+1𝑎𝑝+𝑘2𝑑𝑝+𝑚+1𝑘=𝑚+1𝑎𝑝+𝑘𝑑𝑝+𝑚+11k=𝑚+1𝑎𝑝+𝑘=1,(2.46)|(𝜓2(𝑧)1)/(𝜓2(𝑧)+1)|1 requires that Re{𝜓2(𝑧)}>0. Thus, we obtain ΨRe2=(𝑧)1+𝑑𝑝+𝑚+1𝑠Re𝑚(𝑧)𝑓𝑑(𝑧)𝑝+𝑚+11+𝑑𝑝+𝑚+1>0(2.47) or 𝑠Re𝑚(𝑧)𝑓>𝑑(𝑧)𝑝+𝑚+11+𝑑𝑝+𝑚+1.(2.48) It follows from the last inequality that assertion (2.33) of the Theorem 2.5 holds.
The bound in (2.33) is sharp with the extremal function given by (2.41). The proof of the theorem is thus completed.

If Ω=0, 𝜆=0, and 𝑛=1 are taken in Theorem 2.5, the following result is obtained given by Liu [6].

Corollary 2.6. Let 𝑓 be given by (1.19) and define the partial sums 𝑠𝑚(𝑧) by 𝑠𝑚𝑧(𝑧)=𝑝(𝑝𝛼)𝜀𝑧𝑝+1𝛼𝑝+1𝑧,𝑚=1,𝑝(𝑝𝛼)𝜀𝑧(𝑛+𝑝𝛼)𝑝+1𝑚𝑘=2𝑎𝑝+𝑘𝑧𝑝+𝑘,𝑚=2,3,.(2.49) Suppose also that 𝑘=2𝑑𝑝+𝑘𝑎𝑝+𝑘1(𝑝𝛼)𝜀𝑝+1𝛼where𝑑𝑝+𝑘=𝑘+𝑝𝛼(𝑝𝛼)(1𝜀);0𝛼<𝑝,0𝜀<1.(2.50) Then, for 𝑚2, one has Re𝑓(𝑧)𝑠𝑚>(𝑧)(𝑝𝛼)𝜀+𝑚+1𝑠𝑝+𝑚+1𝛼Re𝑚(𝑧)>𝑓(𝑧)𝑝+𝑚+1𝛼(.𝑝𝛼)(2𝜀)+𝑚+1(2.51) Each of the bounds in (2.51) is the best possible.