New Subclasses of Multivalent Analytic Functions Associated with a Linear Operator

Abstract

Making use of a linear operator, which is defined here by means of the Hadamard product (or convolution), we consider two subclasses 𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵) and 𝐺_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵) of multivalent analytic functions with negative coefficients in the open unit disk. Some modified Hadamard products, integral transforms, and the partial sums of functions belonging to these classes are studied.

1. Introduction

Let 𝐴_𝑝,𝑛 denote the class of functions of the form 𝑓(𝑧)=𝑧^𝑝+^∞_{𝑘=𝑝+𝑛}𝑎_𝑘𝑧^𝑘,𝑝,𝑛∈𝑁={1,2,3,…}, (1) which are analytic in the open unit disk 𝑈={𝑧∶|𝑧|<1}.

For functions 𝑓(𝑧) and 𝑔(𝑧) analytic in 𝑈, we say that 𝑓(𝑧) is subordinate to 𝑔(𝑧) in 𝑈, written 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). (2) Furthermore, if the function 𝑔(𝑧) is univalent in 𝑈, then 𝑓(𝑧)(𝑧)(0)(0),𝑓(𝑈)(𝑈).≺𝑔)(𝑧∈𝑈⟺𝑓=𝑔⊆𝑔 (3)

In terms of the Pochhammer symbol (𝑏)_𝑛 given by (𝑏)_𝑛=𝑏(𝑏+1)⋯(𝑏+𝑛−1)(𝑛∈𝑁), we define the function 𝜑_𝑝,𝑛(𝑎,𝑐;𝑧) by 𝜑_𝑝,𝑛(𝑎,𝑐;𝑧)=𝑧^𝑝+^∞_{𝑘=𝑝+𝑛}(𝑎)_𝑘−𝑝(𝑐)_𝑘−𝑝𝑧^𝑘(𝑧∈𝑈;𝑐∉{0,−1,−2,…}). (4) Corresponding to 𝜑_𝑝,𝑛(𝑎,𝑐;𝑧), we consider here a linear operator 𝐿_𝑝,𝑛(𝑎,𝑐) on 𝐴_𝑝,𝑛 by the following usual Hadamard product (or convolution): 𝐿_𝑝,𝑛(𝑎,𝑐)𝑓(𝑧)=𝜑_𝑝,𝑛(𝑎,𝑐;𝑧)∗𝑓(𝑧)=𝑧^𝑝+^∞_{𝑘=𝑝+𝑛}(𝑎)_𝑘−𝑝(𝑐)_𝑘−𝑝𝑎_𝑘𝑧^𝑘, (5) for 𝑓(𝑧) given by (1). For 𝑝=𝑛=1, 𝐿_1,1(𝑎,𝑐) on 𝐴_1,1 was first defined by Carlson and Shaffer [1]. Its differential-integral representation can be found in [2]. We remark in passing that a much more general convolution operator than the operator 𝐿_𝑝,1(𝑎,𝑐) was introduced by Dziok and Srivastava [3].

Let 𝑇_𝑝,𝑛 denote the subclass of 𝐴_𝑝,𝑛 consisting of functions of the form 𝑓(𝑧)=𝑧^𝑝−^∞_{𝑘=𝑝+𝑛}𝑎_𝑘𝑧^𝑘,𝑎_𝑘≥0;𝑝,𝑛∈𝑁. (6) We now consider the following two subclasses of the class 𝑇_𝑝,𝑛.

Definition 1. A function 𝑓(𝑧)∈𝑇_𝑝,𝑛 is said to be in the class 𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵) if and only if ()𝐿1−𝜆^𝑝,𝑛(𝑎,𝑐)𝑓(𝑧)𝑧^𝑝𝐿+𝜆^𝑝,𝑛(𝑎,𝑐)𝑓(𝑧)^′𝑝𝑧^𝑝−1≺1+𝐴𝑧(),1+𝐵𝑧𝑧∈𝑈 (7) where 𝑎>0,𝑐>0,𝜆≥0,−1≤𝐵≤0,𝐵<𝐴≤1. (8)

Definition 2. A function 𝑓(𝑧)∈𝑇_𝑝,𝑛 is said to be in the class 𝐺_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵) if and only if 𝑧𝑓^′(𝑧)𝑝∈𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵). (9)

For functions 𝑓_𝑗(𝑧)∈𝑇_𝑝,𝑛 given by 𝑓_𝑗(𝑧)=𝑧^𝑝−^∞_{𝑘=𝑝+𝑛}𝑎_𝑘,𝑗𝑧^𝑘,𝑎_𝑘,𝑗≥0;𝑗=1,2, (10) we denote by (𝑓₁⋅𝑓₂)(𝑧) the modified Hadamard (or quasi-Hadamard) product of 𝑓₁(𝑧) and 𝑓₂(𝑧); that is, 𝑓₁⋅𝑓₂(𝑧)=𝑧^𝑝−^∞_{𝑘=𝑝+𝑛}𝑎_𝑘,1𝑎_𝑘,2𝑧^𝑘=𝑓₂⋅𝑓₁(𝑧). (11)

The class 𝐹_𝑝,𝑛1,1,𝜆,1−2𝛼𝑝,−1=𝐹_𝑝(𝑛,𝜆,𝛼),𝜆≥0;0≤𝛼<𝑝 (12) with 𝑛=1 was introduced and studied earlier by Lee et al. [4] (and was further investigated by Aouf and Darwish [5], Aouf et al. [6], and Yaguchi et al. [7]). The class 𝐺_𝑝,𝑛1,1,𝜆,1−2𝛼𝑝,−1=𝐺_𝑝(𝑛,𝜆,𝛼),𝜆≥0;0≤𝛼<𝑝 (13) with 𝑛=1 was studied by Aouf [8] and Aouf et al. [6]. Recently, Aouf [9] investigated the modified Hadamard products of several functions in the classes 𝐹_𝑝(𝑛,𝜆,𝛼) and 𝐺_𝑝(𝑛,𝜆,𝛼) for 𝑛∈𝑁.

In the present paper, we prove a number of theorems involving the modified Hadamard products, integral transforms, and the partial sums of functions in the classes 𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵) and 𝐺_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵). Some of our results are generalizations of the corresponding results in [4–9].

In proving our main results, we need the following lemmas.

Lemma 3 (see [10, 11]). A function 𝑓(𝑧)∈𝑇_𝑝,𝑛 defined by (6) is in the class 𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵) if and only if ^∞_{𝑘=𝑝+𝑛}𝜆(𝑎)𝑝+𝑘−𝑝^𝑘−𝑝(𝑐)_𝑘−𝑝𝑎_𝑘≤𝑝𝐴−𝐵.1−𝐵 (14)

Lemma 4 (see [10, 11]). A function 𝑓(𝑧)∈𝑇_𝑝,𝑛 defined by (6) is in the class 𝐺_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵) if and only if ^∞_{𝑘=𝑝+𝑛}𝑘𝜆(𝑎)𝑝+𝑘−𝑝^𝑘−𝑝(𝑐)_𝑘−𝑝𝑎_𝑘≤𝑝²𝐴−𝐵.1−𝐵 (15)

Making use of Lemmas 3 and 4, we can show the following two results.

Corollary 5. Let 0<𝑎₁≤𝑎₀,0<𝑐₀≤𝑐₁,0≤𝜆₁≤𝜆₀,−1≤𝐵₁≤𝐵₀<𝐴₀≤𝐴₁≤1,𝐵₀≤0. (16) Then 𝐹_𝑝,𝑛𝑎₀,𝑐₀,𝜆₀,𝐴₀,𝐵₀⊆𝐹_𝑝,𝑛𝑎₁,𝑐₁,𝜆₁,𝐴₁,𝐵₁,𝐺_𝑝,𝑛𝑎₀,𝑐₀,𝜆₀,𝐴₀,𝐵₀⊆𝐺_𝑝,𝑛𝑎₁,𝑐₁,𝜆₁,𝐴₁,𝐵₁. (17)

Corollary 6. Let 𝑓(𝑧)∈𝑇_𝑝,𝑛. Then 𝑓(𝑧)∈𝐹_𝑝,𝑛(𝑎₀,𝑐₀,𝜆₀,𝐴,𝐵)(𝐺_𝑝,𝑛(𝑎₀,𝑐₀,𝜆₀,𝐴,𝐵)) if and only if (𝑓∗𝑔)(𝑧)∈𝐹_𝑝,𝑛(𝑎₁,𝑐₁,𝜆₁,𝐴,𝐵)(𝐺_𝑝,𝑛(𝑎₁,𝑐₁,𝜆₁,𝐴,𝐵)), where 𝜆₀≥0,  𝜆₁≥0,𝑎_𝑗>0 and 𝑐_𝑗>0(𝑗=0,1), and 𝑔(𝑧)=𝑧^𝑝+^∞_{𝑘=𝑝+𝑛}𝑝+𝑘−𝑝𝜆₀𝑎_0𝑘−𝑝𝑐_1𝑘−𝑝𝑝+𝑘−𝑝𝜆₁𝑎_1𝑘−𝑝𝑐_0𝑘−𝑝𝑧^𝑘. (18)

If we let 𝑎=𝑐=1,𝐴=1−2𝛼𝑝,0≤𝛼<𝑝,𝐵=−1, (19) then Lemmas 3 and 4 reduce to the following result.

Corollary 7. Let 𝑓(𝑧)∈𝑇_𝑝,𝑛 be defined by (6). Then(i) 𝑓(𝑧) is in the class 𝐹_𝑝(𝑛,𝜆,𝛼) if and only if 𝑝+𝑘−𝑝𝜆^∞_{𝑘=𝑝+𝑛}𝑎_𝑘≤𝑝−𝛼. (20)(ii) 𝑓(𝑧) is in the class 𝐺_𝑝(𝑛,𝜆,𝛼) if and only if ^∞_{𝑘=𝑝+𝑛}𝑘𝑝+𝑘−𝑝𝜆𝑎_𝑘≤𝑝𝑝−𝛼. (21)

2. Modified Hadamard Products

Hereafter in this paper we assume that (8) is satisfied: 0≤𝛼<𝑝,0≤𝛼_𝑗<𝑝,−1≤𝐵_𝑗𝐵≤0,_𝑗<𝐴_𝑗≤1,𝑗=1,2,…,𝑚. (22)

Theorem 8. Let 𝑓_𝑗(𝑧)∈𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴_𝑗,𝐵_𝑗)(𝑗=1,2,…,𝑚) and 𝑎≥𝑐>0. Then 𝑓₁⋅𝑓₂⋅…⋅𝑓_𝑚(𝑧)∈𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴(𝐵),𝐵), (23) where 𝑚≥2 and 𝐴(𝐵)()𝑝(𝑐)=𝐵+1−𝐵^𝑛𝑝+𝑛𝜆(𝑎)_𝑛^𝑚𝑚−1_𝑗=1𝐴_𝑗−𝐵_𝑗1−𝐵_𝑗. (24) The result is sharp; that is, 𝐴(𝐵) cannot be decreased for each 𝐵∈[−1,0].

Proof. By (24) we have 𝐵<𝐴(𝐵)≤1 for 𝑎≥𝑐>0. Let 𝑓_𝑗(𝑧)=𝑧^𝑝−^∞_{𝑘=𝑝+𝑛}𝑎_𝑘,𝑗𝑧^𝑘∈𝐹_𝑝,𝑛𝑎,𝑐,𝜆,𝐴_𝑗,𝐵_𝑗𝑎_𝑘,𝑗≥0;𝑗=1,2,…,𝑚. (25) Then Lemma 3 gives 1−𝐵_𝑗𝐴_𝑗−𝐵_𝑗^∞_{𝑘=𝑝+𝑛}𝑝+𝑘−𝑝𝜆(𝑎)_𝑘−𝑝𝑝(𝑐)_𝑘−𝑝𝑎_𝑘,𝑗≤1𝑗=1,2,…,𝑚, (26) and hence ^𝑚_𝑗=11−𝐵_𝑗𝐴_𝑗−𝐵_𝑗×^∞_{𝑘=𝑝+𝑛}𝑝+𝑘−𝑝𝜆(𝑎)_𝑘−𝑝𝑝(𝑐)_𝑘−𝑝^𝑚𝑎_𝑘,1𝑎_𝑘,2⋯𝑎_𝑘,𝑚≤^𝑚_𝑗=11−𝐵_𝑗𝐴_𝑗−𝐵_𝑗^∞_{𝑘=𝑝+𝑛}𝑝+𝑘−𝑝𝜆(𝑎)_𝑘−𝑝𝑝(𝑐)_𝑘−𝑝𝑎_𝑘,𝑗≤1. (27) Also, using Lemma 3, ℎ(𝑧)=𝑓₁⋅𝑓₂⋅…⋅𝑓_𝑚(𝑧)=𝑧^𝑝−^∞_{𝑘=𝑝+𝑛}𝑎_𝑘,1𝑎_𝑘,2⋯𝑎_𝑘,𝑚𝑧^𝑘∈𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵) (28) if and only if 1−𝐵𝐴−𝐵^∞_{𝑘=𝑝+𝑛}𝑝+𝑘−𝑝𝜆(𝑎)_𝑘−𝑝𝑝(𝑐)_𝑘−𝑝𝑎_𝑘,1𝑎_𝑘,2⋯𝑎_𝑘,𝑚≤1. (29)
To prove the result of Theorem 8, it follows from (27) and (29) that we need to find the smallest 𝐴 such that 𝑝+𝑘−𝑝𝜆(𝑎)_𝑘−𝑝(1−𝐵)𝑝(𝑐)_𝑘−𝑝()≤𝜆(𝑎)𝐴−𝐵𝑝+𝑘−𝑝_𝑘−𝑝𝑝(𝑐)_𝑘−𝑝^𝑚𝑚_𝑗=11−𝐵_𝑗𝐴_𝑗−𝐵_𝑗𝑘≥𝑝+𝑛;𝑚≥2; (30) that is, that ()𝑝(𝑐)𝐴≥𝐵+1−𝐵^𝑘−𝑝𝑝+𝑘−𝑝𝜆(𝑎)_𝑘−𝑝^𝑚−1×^𝑚_𝑗=1𝐴_𝑗−𝐵_𝑗1−𝐵_𝑗𝑘≥𝑝+𝑛;𝑚≥2. (31) Since (𝑐)_{𝑘+1−𝑝}(𝑎)_{𝑘+1−𝑝}≤(𝑐)_𝑘−𝑝(𝑎)_𝑘−𝑝𝑘≥𝑝+𝑛;𝑎≥𝑐>0, (32) we see that the right-hand side of (31) is a decreasing function of 𝑘. Consequently, taking 𝑘=𝑝+𝑛 in (31), we have ℎ(𝑧)∈𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴(𝐵),𝐵), where 𝐴(𝐵) is given by (24).
Furthermore, by considering the functions 𝑓_𝑗(𝑧) defined by: 𝑓_𝑗(𝑧)=𝑧^𝑝−𝑝(𝑐)_𝑛𝐴_𝑗−𝐵_𝑗𝑝+𝑛𝜆(𝑎)_𝑛1−𝐵_𝑗𝑧^𝑝+𝑛𝑗=1,2,…,𝑚, (33) we have 𝑓_𝑗(𝑧)∈𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴_𝑗,𝐵_𝑗) and 𝑓₁⋅𝑓₂⋅…⋅𝑓_𝑚(𝑧)=𝑧^𝑝−𝑝(𝑐)_𝑛𝑝+𝑛𝜆(𝑎)_𝑛^𝑚𝑚_𝑗=1𝐴_𝑗−𝐵_𝑗1−𝐵_𝑗𝑧^𝑝+𝑛. (34) Noting that (𝑎)𝑝+𝑛𝜆^𝑛(𝑐)_𝑛𝑝(𝑐)_𝑛𝑝+𝑛𝜆(𝑎)_𝑛^𝑚𝑚_𝑗=1𝐴_𝑗−𝐵_𝑗1−𝐵_𝑗𝐴(𝐵)=𝑝−𝐵,1−𝐵 (35) we conclude that 𝐴(𝐵) cannot be decreased for each 𝐵.

By using Lemma 4 instead of Lemma 3, the following theorem can be proved on the lines of the proof of Theorem 8. We omit the details involved.

Theorem 9. Let 𝑓_𝑗(𝑧)∈𝐺_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴_𝑗,𝐵_𝑗)(𝑗=1,2,…,𝑚) and 𝑎≥𝑐>0. Then 𝑓₁⋅𝑓₂⋅…⋅𝑓_𝑚(𝑧)∈𝐺_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴(𝐵),𝐵), (36) where 𝑚≥2 and 𝐴(𝐵)(𝑝=𝐵+1−𝐵)²(𝑐)_𝑛𝑝+𝑛𝑝+𝑛𝜆(𝑎)_𝑛^𝑚−1×^𝑚_𝑗=1𝐴_𝑗−𝐵_𝑗1−𝐵_𝑗∈(𝐵,1). (37) The result is sharp for the functions 𝑓_𝑗(𝑧) defined by 𝑓_𝑗(𝑧)=𝑧^𝑝−𝑝²(𝑐)_𝑛𝐴_𝑗−𝐵_𝑗𝑝+𝑛𝑝+𝑛𝜆(𝑎)_𝑛1−𝐵_𝑗𝑧^𝑝+𝑛∈𝐺_𝑝,𝑛𝑎,𝑐,𝜆,𝐴_𝑗,𝐵_𝑗𝑗=1,2,…,𝑚. (38)

Theorem 10. Let 𝑓_𝑗(𝑧)∈𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴_𝑗,𝐵_𝑗)(𝑗=1,2,…,𝑚) with 𝑎≥𝑐>0 and 𝜆≥1. Then 𝑓₁⋅𝑓₂⋅…⋅𝑓_𝑚(𝑧)∈𝐺_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴(𝐵),𝐵), (39) where 𝑚≥2 and 𝐴(𝐵)=𝐵+(1−𝐵)𝑝+𝑛𝑝𝑝(𝑐)_𝑛𝑝+𝑛𝜆(𝑎)_𝑛^𝑚−1×^𝑚_𝑗=1𝐴_𝑗−𝐵_𝑗1−𝐵_𝑗. (40) The result is sharp for the functions 𝑓_𝑗(𝑧)(𝑗=1,2,…,𝑚) defined by (33).

Proof. Obviously, 𝐵<𝐴(𝐵)≤1 for 𝑚≥2,  𝑎≥𝑐>0, and 𝜆≥1. By applying Lemma 4, we know that 𝑓₁⋅𝑓₂⋅…⋅𝑓_𝑚(𝑧)=𝑧^𝑝−^∞_{𝑘=𝑝+𝑛}𝑎_𝑘,1𝑎_𝑘,2⋯𝑎_𝑘,𝑚𝑧^𝑘∈𝐺_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵), (41) if and only if 1−𝐵𝐴−𝐵^∞_{𝑘=𝑝+𝑛}𝑘𝑝+𝑘−𝑝𝜆(𝑎)_𝑘−𝑝𝑝²(𝑐)_𝑘−𝑝𝑎_𝑘,1𝑎_𝑘,2⋯𝑎_𝑘,𝑚≤1. (42) Proceeding as in the proof of Theorem 8, we need to find the smallest 𝐴 such that ()𝑘𝐴≥𝐵+1−𝐵𝑝𝑝(𝑐)_𝑘−𝑝𝑝+𝑘−𝑝𝜆(𝑎)_𝑘−𝑝^𝑚−1×^𝑚_𝑗=1𝐴_𝑗−𝐵_𝑗1−𝐵_𝑗𝑘≥𝑝+𝑛. (43) Defining the function 𝜑(𝑥) by 𝜑(𝑥)=𝑥𝑝+𝑥−𝑝𝜆^𝑚−1𝑥≥𝑝+𝑛, (44) we see that 𝜑^′(𝑥)=𝑝(1−𝜆)−(𝑚−2)𝜆𝑥𝑝+𝑥−𝑝𝜆^𝑚≤0𝑥≥𝑝+𝑛, (45) for 𝑚≥2 and 𝜆≥1. Hence, the right-hand side of (43) is a decreasing function of 𝑘. Thus, we arrive at (𝑓₁⋅𝑓₂⋅…⋅𝑓_𝑚)(𝑧)∈𝐺_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴(𝐵),𝐵), where 𝐴(𝐵) is given by (40).

Sharpness can be verified easily.

By putting 𝑎=𝑐=1,𝐴(𝐵)=1−2𝛼𝑝𝐴,𝐵=−1,_𝑗=1−2𝛼_𝑗𝑝,𝐵_𝑗=−1𝑗=1,2,…,𝑚. (46) Theorem 10 reduces to the following.

Corollary 11. Let 𝑓_𝑗(𝑧)∈𝐹_𝑝(𝑛,𝜆,𝛼_𝑗)(𝑗=1,2,…,𝑚). Then (𝑓₁⋅𝑓₂⋅…⋅𝑓_𝑚)(𝑧)∈𝐺_𝑝(𝑛,𝜆,𝛼), where 𝑚≥2,  𝜆≥1, and 𝛼=𝑝−𝑝+𝑛𝑝𝑝+𝑛𝜆^𝑚𝑚−1_𝑗=1𝑝−𝛼_𝑗. (47) The result is sharp for the functions 𝑓_𝑗(𝑧)=𝑧^𝑝−𝑝−𝛼_𝑗𝑧𝑝+𝑛𝜆^𝑝+𝑛∈𝐹_𝑝𝑛,𝜆,𝛼_𝑗𝑗=1,2,…,𝑚. (48)

Theorem 12. Let 𝑓_𝑗(𝑧)∈𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴_𝑗,𝐵_𝑗)(𝑗=1,2,…,𝑚), 𝑔_𝑗(𝑧)∈𝐹_𝑝,𝑛𝑎,𝑐,𝜆,𝐶_𝑗,𝐷_𝑗,−1≤𝐷_𝑗≤0;𝐷_𝑗<𝐶_𝑗≤1;𝑗=1,2,…,𝑚 (49) and 𝑎≥𝑐>0. Then one has (i)  ∑^𝑚_𝑗=1(𝑓_𝑗⋅𝑔_𝑗)(𝑧)−(𝑚−1)𝑧^𝑝∈𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴(𝐵),𝐵), where 𝐴(𝐵)()(𝑐)=𝐵+1−𝐵𝑚𝑝_𝑛𝐴₀−𝐵₀𝐶₀−𝐷₀𝑝+𝑛𝜆(𝑎)_𝑛1−𝐵₀1−𝐷₀,𝐴₀=max_{1≤𝑗≤𝑚}𝐴_𝑗,𝐵₀=min_{1≤𝑗≤𝑚}𝐵_𝑗,𝐶₀=max_{1≤𝑗≤𝑚}𝐶_𝑗,𝐷₀=min_{1≤𝑗≤𝑚}𝐷_𝑗, (50) provided that 𝐴(𝐵)≤1. (ii)  (1/𝑚)∑^𝑚_𝑗=1(𝑓_𝑗⋅𝑔_𝑗)(𝑧)∈𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴(𝐵),𝐵), where 𝐴(𝐵)()×𝑝(𝑐)=𝐵+1−𝐵_𝑛𝐴₀−𝐵₀𝐶₀−𝐷₀𝑝+𝑛𝜆(𝑎)_𝑛1−𝐵₀1−𝐷₀∈(𝐵,1]. (51)

Proof. It is clear that −1≤𝐵₀≤𝐵_𝑗<𝐴_𝑗≤𝐴₀≤1,  𝐵₀≤0,  −1≤𝐷₀≤𝐷_𝑗<𝐶_𝑗≤𝐶₀≤1,  𝐷₀≤0, 1−𝐵_𝑗𝐴_𝑗−𝐵_𝑗≥1−𝐵₀𝐴₀−𝐵₀>0,1−𝐷_𝑗𝐶_𝑗−𝐷_𝑗≥1−𝐷₀𝐶₀−𝐷₀>0, (52) for 𝑗=1,2,…,𝑚. Let 𝑓_𝑗(𝑧)=𝑧^𝑝−^∞_{𝑘=𝑝+𝑛}𝑎_𝑘,𝑗𝑧^𝑘∈𝐹_𝑝,𝑛𝑎,𝑐,𝜆,𝐴_𝑗,𝐵_𝑗𝑔_𝑗(𝑧)=𝑧^𝑝−^∞_{𝑘=𝑝+𝑛}𝑏_𝑘,𝑗𝑧^𝑘∈𝐹_𝑝,𝑛𝑎,𝑐,𝜆,𝐶_𝑗,𝐷_𝑗,𝑗=1,2,…,𝑚. (53) Then Lemma 3 gives ^𝑚_𝑗=1𝑓_𝑗⋅𝑔_𝑗(𝑧)−(𝑚−1)𝑧^𝑝=𝑧^𝑝−^∞_{𝑘=𝑝+𝑛}^𝑚_𝑗=1𝑎_𝑘,𝑗𝑏_𝑘,𝑗𝑧^𝑘∈𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵), (54) if and only if 1−𝐵𝐴−𝐵^∞_{𝑘=𝑝+𝑛}𝑝+𝑘−𝑝𝜆(𝑎)_𝑘−𝑝𝑝(𝑐)_𝑘−𝑝^𝑚_𝑗=1𝑎_𝑘,𝑗𝑏_𝑘,𝑗≤1. (55) Also, by Lemma 3 and (52), we deduce that 1−𝐵₀𝐴₀−𝐵₀1−𝐷₀𝐶₀−𝐷₀×^∞_{𝑘=𝑝+𝑛}𝑝+𝑘−𝑝𝜆(𝑎)_𝑘−𝑝𝑝(𝑐)_𝑘−𝑝²𝑎_𝑘,𝑗𝑏_𝑘,𝑗≤1−𝐵_𝑗𝐴_𝑗−𝐵_𝑗^∞_{𝑘=𝑝+𝑛}𝑝+𝑘−𝑝𝜆(𝑎)_𝑘−𝑝𝑝(𝑐)_𝑘−𝑝𝑎_𝑘,𝑗×1−𝐷_𝑗𝐶_𝑗−𝐷_𝑗^∞_{𝑘=𝑝+𝑛}𝑝+𝑘−𝑝𝜆(𝑎)_𝑘−𝑝𝑝(𝑐)_𝑘−𝑝𝑏_𝑘,𝑗≤1𝑗=1,2,…,𝑚, (56) and hence 1𝑚1−𝐵₀𝐴₀−𝐵₀1−𝐷₀𝐶₀−𝐷₀×^∞_{𝑘=𝑝+𝑛}𝑝+𝑘−𝑝𝜆(𝑎)_𝑘−𝑝𝑝(𝑐)_𝑘−𝑝^2𝑚_𝑗=1𝑎_𝑘,𝑗𝑏_𝑘,𝑗≤1. (57)
To prove Theorem 12(i), it follows from (55) and (57) that we need to find the smallest 𝐴 such that ()(𝑐)𝐴≥𝐵+1−𝐵𝑚𝑝_𝑘−𝑝𝑝+𝑘−𝑝𝜆(𝑎)_𝑘−𝑝×𝐴₀−𝐵₀1−𝐵₀𝐶₀−𝐷₀1−𝐷₀𝑘≥𝑝+𝑛, (58) for 𝑎≥𝑐>0. This leads to the assertion of Theorem 12(i).
Analogously, we can prove Theorem 12(ii).

In the special case when 𝑎=𝑐=1,𝐴(𝐵)=1−2𝛼𝑝,𝐴(𝐵)2𝛼=1−𝑝,𝐵=−1,𝐴_𝑗=𝐶_𝑗=1−2𝛼_𝑗𝑝,𝐵_𝑗=𝐷_𝑗=−1𝑗=1,2,…,𝑚, (59) Theorem 12 reduces to the following.

Corollary 13. Let 𝑓_𝑗(𝑧)∈𝐹_𝑝(𝑛,𝜆,𝛼_𝑗)(𝑗=1,2,…,𝑚). Then (i)  ∑^𝑚_𝑗=1(𝑓_𝑗⋅𝑓_𝑗)(𝑧)−(𝑚−1)𝑧^𝑝∈𝐹_𝑝(𝑛,𝜆,𝛼), where 𝑚𝛼=𝑝−𝑝+𝑛𝜆𝑝−min_{1≤𝑗≤𝑚}𝛼_𝑗², (60) provided that 𝛼≥0. (ii)  (1/𝑚)∑^𝑚_𝑗=1(𝑓_𝑗⋅𝑓_𝑗)(𝑧)∈𝐹_𝑝(𝑛,𝜆,𝛼), where 1𝛼=𝑝−𝑝+𝑛𝜆𝑝−min_{1≤𝑗≤𝑚}𝛼_𝑗²∈0,𝑝. (61)
Replacing Lemma 3 by Lemma 4 in the proof of Theorem 12, one can prove the following.

Theorem 14. Let 𝑓_𝑗(𝑧)∈𝐺_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴_𝑗,𝐵_𝑗)(𝑗=1,2,…,𝑚), 𝑔_𝑗(𝑧)∈𝐺_𝑝,𝑛𝑎,𝑐,𝜆,𝐶_𝑗,𝐷_𝑗,−1≤𝐷_𝑗≤0;𝐷_𝑗<𝐶_𝑗≤1;𝑗=1,2,…,𝑚 (62) and 𝑎≥𝑐>0. Then (i)  ∑^𝑚_𝑗=1(𝑓_𝑗⋅𝑔_𝑗)(𝑧)−(𝑚−1)𝑧^𝑝∈𝐺_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴(𝐵),𝐵), where 𝐴(𝐵)=𝐵+(1−𝐵)𝑚𝑝²(𝑐)_𝑛𝐴₀−𝐵₀𝐶₀−𝐷₀𝑝+𝑛𝑝+𝑛𝜆(𝑎)_𝑛1−𝐵₀1−𝐷₀, (63) and 𝐴₀,𝐵₀,𝐶₀, and 𝐷₀ are given as in Theorem 12, provided that 𝐴(𝐵)≤1. (ii)  (1/𝑚)∑^𝑚_𝑗=1(𝑓_𝑗⋅𝑔_𝑗)(𝑧)∈𝐺_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴(𝐵),𝐵), where 𝐴(𝐵)()×𝑝=𝐵+1−𝐵₂(𝑐)_𝑛𝐴₀−𝐵₀𝐶₀−𝐷₀𝑛+𝑝𝑝+𝑛𝜆(𝑎)_𝑛1−𝐵₀1−𝐷₀∈(𝐵,1). (64)
As a special case of Theorem 14, one has the following.

Corollary 15. Let 𝑓_𝑗(𝑧)∈𝐺_𝑝(𝑛,𝜆,𝛼_𝑗)(𝑗=1,2,…,𝑚). Then (i)  ∑^𝑚_𝑗=1(𝑓_𝑗⋅𝑓_𝑗)(𝑧)−(𝑚−1)𝑧^𝑝∈𝐺_𝑝(𝑛,𝜆,𝛼), where 𝛼=𝑝−𝑚𝑝𝑝+𝑛𝑝+𝑛𝜆𝑝−min_{1≤𝑗≤𝑚}𝛼_𝑗², (65) provided that 𝛼≥0. (ii)  (1/𝑚)∑^𝑚_𝑗=1(𝑓_𝑗⋅𝑓_𝑗)(𝑧)∈𝐺_𝑝(𝑛,𝜆,𝛼), where 𝑝𝛼=𝑝−𝑝+𝑛𝑝+𝑛𝜆𝑝−min_{1≤𝑗≤𝑚}𝛼_𝑗²∈0,𝑝. (66)

3. Integral Operator

Theorem 16. Let 0≤𝜆₁≤𝜆₀,0<𝑐₀≤𝑎₀,0<𝑎₁≤𝑐₁𝑜𝑟0<𝑎₁≤𝑎₀𝑎𝑛𝑑0<𝑐₀≤𝑐₁. (67) If 𝑓(𝑧)∈𝐹_𝑝,𝑛(𝑎₀,𝑐₀,𝜆₀,𝐴,𝐵), then the function 𝐼(𝑧) defined by 𝐼(𝑧)=𝜇+𝑝𝑧^𝜇𝑧₀𝑡^𝜇−1𝑓(𝑡)𝑑𝑡𝜇>−𝑝 (68) belongs to 𝐹_𝑝,𝑛(𝑎₁,𝑐₁,𝜆₁,𝐶(𝐷),𝐷), where −1≤𝐷≤0 and 𝐶(𝐷)()×=𝐷+1−𝐷𝑝+𝑛𝜆₁𝑎_1𝑛𝑐_0𝑛𝜇+𝑝(𝐴−𝐵)𝑝+𝑛𝜆₀𝑐_1𝑛𝑎_0𝑛().𝜇+𝑝+𝑛1−𝐵 (69) The result is sharp; that is, the number 𝐶(𝐷) cannot be decreased for each 𝐷.

Proof. Note that 𝐷<𝐶(𝐷)<1. For 𝑓(𝑧)=𝑧^𝑝−^∞_{𝑘=𝑝+𝑛}𝑎_𝑘𝑧^𝑘∈𝐹_𝑝,𝑛𝑎₀,𝑐₀,𝜆₀,𝐴,𝐵, (70) it follows from (68) that 𝐼(𝑧)=𝑧^𝑝−^∞_{𝑘=𝑝+𝑛}𝜇+𝑝𝑎𝜇+𝑘^𝑘𝑧^𝑘𝑧∈𝑈;𝜇>−𝑝. (71)
To prove the result of Theorem 16, we need to find the smallest 𝐶 such that ()×𝜆𝐶≥𝐷+1−𝐷𝑝+𝑘−𝑝₁𝑎_1𝑘−𝑝𝑐_0𝑘−𝑝𝜇+𝑝(𝐴−𝐵)𝑝+𝑘−𝑝𝜆₀𝑐_1𝑘−𝑝𝑎_0𝑘−𝑝(),𝜇+𝑘1−𝐵𝑘≥𝑝+𝑛 (72) where we have used Lemma 3. In view of (67), it is easy to know that the right-hand side of (72) is a decreasing function of 𝑘. Therefore, we conclude that 𝐼(𝑧)∈𝐹_𝑝,𝑛𝑎₁,𝑐₁,𝜆₁,𝐶(𝐷),𝐷, (73) where 𝐶(𝐷) is given by (69).
Furthermore, it can easily be verified that the result is sharp, with the extremal function 𝑓(𝑧)=𝑧^𝑝−𝑝𝑐_0𝑛(𝐴−𝐵)𝑝+𝑛𝜆₀𝑎_0𝑛(1−𝐵)×𝑧^𝑝+𝑛∈𝐹_𝑝,𝑛𝑎₀,𝑐₀,𝜆₀,𝐴,𝐵. (74)

With the aid of Lemma 4 (instead of Lemma 3) and using the same steps as in the proof of Theorem 16, we can prove the following.

Theorem 17. Let (67) in Theorem 16 be satisfied. If 𝑓(𝑧)∈𝐺_𝑝,𝑛(𝑎₀,𝑐₀,𝜆₀,𝐴,𝐵), then the function 𝐼(𝑧) defined by (68) belongs to 𝐺_𝑝,𝑛(𝑎₁,𝑐₁,𝜆₁,𝐶(𝐷),𝐷), where 𝐶(𝐷)(−1≤𝐷≤0) is the same as in Theorem 16. The result is sharp for the function 𝑓(𝑧)=𝑧^𝑝−𝑝²𝑐_0𝑛(𝐴−𝐵)𝑝+𝑛𝑝+𝑛𝜆₀𝑎_0𝑛(1−𝐵)×𝑧^𝑝+𝑛∈𝐺_𝑝,𝑛𝑎₀,𝑐₀,𝜆₀,𝐴,𝐵. (75)

If we let 𝑎₀=𝑐₀=𝑎₁=𝑐₁=1,𝜆₀=𝜆₁=𝜆≥0,𝐴=1−2𝛼𝑝(𝐷)(𝛼),𝐵=𝐷=−1,𝐶=1−2𝛽𝑝, (76) then Theorem 17 yields the following.

Corollary 18. Let 𝑓(𝑧)∈𝐺_𝑝(𝑛,𝜆,𝛼). Then the function 𝐼(𝑧) defined by (68) belongs to 𝐺_𝑝(𝑛,𝜆,𝛽(𝛼)), where 𝛽(𝛼)=𝛼𝜇+𝑝+𝑛𝑝.𝜇+𝑝+𝑛 (77) The number 𝛽(𝛼) cannot be increased for each 𝛼∈[0,𝑝).

4. Partial Sums

In this section, we let 𝑓(𝑧)∈𝑇_𝑝,𝑛 be given by (6) and define the partial sums 𝑠₁(𝑧) and 𝑠_𝑚(𝑧) by 𝑠₁(𝑧)=𝑧^𝑝,𝑠_𝑚(𝑧)=𝑧^𝑝−^{𝑝+𝑛+𝑚−2}_{𝑘=𝑝+𝑛}𝑎_𝑘𝑧^𝑘,(𝑚∈𝑁⧵{1}). (78) Also we make the notation simple by writing 𝛽_𝑘=𝑝+𝑘−𝑝𝜆(𝑎)_𝑘−𝑝(1−𝐵)𝑝(𝑐)_𝑘−𝑝().𝐴−𝐵𝑘=𝑝+𝑛,𝑝+𝑛+1,… (79)

Theorem 19. Let 𝑓(𝑧)∈𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵) and 𝑎≥𝑐>0. Then for 𝑧∈𝑈, one has the following. 𝑓(𝑧)Re𝑠_𝑚(𝑧)1>1−𝛽_{𝑝+𝑛+𝑚−1}(),𝑠𝑚∈𝑁Re^𝑚(𝑧)𝑓(𝑧)>𝛽_{𝑝+𝑛+𝑚−1}1+𝛽_{𝑝+𝑛+𝑚−1}(𝑚∈𝑁). (80)(81) The results are sharp for each 𝑚∈𝑁.

Proof. Let 𝑎≥𝑐>0 and 𝛽_𝑘 be given by (79). Then 𝛽_𝑘+1≥𝛽_𝑘≥1 for 𝑘≥𝑝+𝑛, and so it follows from Lemma 3 that ^{𝑝+𝑛+𝑚−2}_{𝑘=𝑝+𝑛}𝑎_𝑘+𝛽_{𝑝+𝑛+𝑚−1}⋅^∞_{𝑘=𝑝+𝑛+𝑚−1}𝑎_𝑘≤^∞_{𝑘=𝑝+𝑛}𝛽_𝑘𝑎_𝑘≤1(𝑚≥2), (82) for 𝑓(𝑧)∈𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵).
If we put 𝑝₁(𝑧)=1+𝛽_{𝑝+𝑛+𝑚−1}𝑓(𝑧)𝑠_𝑚(𝑧)𝛽−1=1−^{𝑝+𝑛+𝑚−1}⋅∑^∞_{𝑘=𝑝+𝑛+𝑚−1}𝑎_𝑘𝑧^𝑘−𝑝1−∑^{𝑝+𝑛+𝑚−2}_{𝑘=𝑝+𝑛}𝑎_𝑘𝑧^𝑘−𝑝(𝑧∈𝑈;𝑚≥2), (83) then 𝑝(0)=1 and 𝑝₁(𝑧)−1𝑝₁(𝑧)≤𝛽+1_{𝑝+𝑛+𝑚−1}⋅∑^∞_{𝑘=𝑝+𝑛+𝑚−1}𝑎_𝑘2−2∑^{𝑝+𝑛+𝑚−2}_{𝑘=𝑝+𝑛}𝑎_𝑘−𝛽_{𝑝+𝑛+𝑚−1}⋅∑^∞_{𝑘=𝑝+𝑛+𝑚−1}𝑎_𝑘≤1(𝑧∈𝑈;𝑚≥2) (84) because of (82). Hence, we have Re𝑝₁(𝑧)>0 for 𝑧∈𝑈, which implies that (80) holds true for 𝑚≥2.
Similarly, by setting 𝑝₂(𝑧)=1+𝛽_{𝑝+𝑛+𝑚−1}𝑠_𝑚(𝑧)𝑓(𝑧)−𝛽_{𝑝+𝑛+𝑚−1}(𝑧∈𝑈), (85) it follows from (82) that 𝑝₂(𝑧)−1𝑝₂(𝑧)≤+11+𝛽_{𝑝+𝑛+𝑚−1}⋅∑^∞_{𝑘=𝑝+𝑛+𝑚−1}𝑎_𝑘2−2∑^{𝑝+𝑛+𝑚−2}_{𝑘=𝑝+𝑛}𝑎_𝑘−𝛽_{𝑝+𝑛+𝑚−1}−1⋅∑^∞_{𝑘=𝑝+𝑛+𝑚−1}𝑎_𝑘≤1(𝑧∈𝑈;𝑚≥2). (86) Therefore, we see that Re𝑝₂(𝑧)>0 for 𝑧∈𝑈, that is, that (81) holds for 𝑚≥2.
For 𝑚=1, replacing (82) by 𝛽^∞_{𝑝+𝑛𝑘=𝑝+𝑛}𝑎_𝑘≤^∞_{𝑘=𝑝+𝑛}𝛽_𝑘𝑎_𝑘≤1 (87) and proceeding as the above, we know that (80) and (81) are also true.
Furthermore, by taking the function 𝑓(𝑧)=𝑧^𝑝−𝑧^{𝑝+𝑛+𝑚−1}𝛽_{𝑝+𝑛+𝑚−1}, (88) we find that 𝑠_𝑚(𝑧)=𝑧^𝑝, 𝑓(𝑧)Re𝑠_𝑚(𝑧)1⟶1−𝛽_{𝑝+𝑛+𝑚−1}𝑠as𝑧⟶1,Re^𝑚(𝑧)𝑓(𝑧)⟶𝛽_{𝑝+𝑛+𝑚−1}1+𝛽_{𝑝+𝑛+𝑚−1}as𝑧⟶exp𝜋𝑖.𝑛+𝑚−1 (89) The proof of Theorem 19 is thus completed.

By virtue of Theorem 19 and Definition 2, we easily have the following.

Corollary 20. Let 𝑓(𝑧)∈𝐺_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵) and 𝑎≥𝑐>0. Then we have 𝑓Re^′(𝑧)𝑠^′_𝑚(𝑧)1>1−𝛽_{𝑝+𝑛+𝑚−1}(),𝑠𝑧∈𝑈;𝑚∈𝑁Re^′𝑚(𝑧)𝑓^′(𝑧)>𝛽_{𝑝+𝑛+𝑚−1}1+𝛽_{𝑝+𝑛+𝑚−1}(𝑧∈𝑈;𝑚∈𝑁). (90) The results are sharp for each 𝑚∈𝑁.

Theorem 21. Let 𝑓(𝑧)∈𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵) with 𝑎≥𝑐>0 and 𝜆≥1. Then one has 𝑓Re^′(𝑧)𝑠^′_𝑚(𝑧)>1−𝑝+𝑛+𝑚−1𝑝𝛽_{𝑝+𝑛+𝑚−1}(),𝑠𝑧∈𝑈;𝑚∈𝑁Re^′𝑚(𝑧)𝑓^′(𝑧)>𝑝𝛽_{𝑝+𝑛+𝑚−1}𝑝+𝑛+𝑚−1+𝑝𝛽_{𝑝+𝑛+𝑚−1}(𝑧∈𝑈;𝑚∈𝑁). (91)(92) The results are sharp for each 𝑚∈𝑁.

Proof. Let 𝑎≥𝑐>0, 𝜆≥1 and 𝛽_𝑘 be given by (79). Then it is easy to verify that 𝛽_𝑘+1≥𝛽𝑘+1𝑘^𝑘≥1𝑝𝑘≥𝑝+𝑛, (93) and hence we deduce from Lemma 3 that ^{𝑝+𝑛+𝑚−2}_{𝑘=𝑝+𝑛}𝑘𝑝𝑎_𝑘+𝛽_{𝑝+𝑛+𝑚−1}⋅𝑝+𝑛+𝑚−1^∞_{𝑘=𝑝+𝑛+𝑚−1}𝑘𝑎_𝑘≤^∞_{𝑘=𝑝+𝑛}𝛽_𝑘𝑎_𝑘≤1(𝑚≥2), (94) for 𝑓(𝑧)∈𝐹_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵).
Defining the function 𝑞₁(𝑧) by 𝑞₁(𝑧)=1+𝑝𝛽_{𝑝+𝑛+𝑚−1}𝑓𝑝+𝑛+𝑚−1^′(𝑧)𝑠^′_𝑚(𝑧)−1(𝑧∈𝑈), (95) it follows from (94) that 𝑞₁(𝑧)−1𝑞₁(𝑧)≤+1𝑝𝛽_{𝑝+𝑛+𝑚−1}⋅𝑝+𝑛+𝑚−1^∞_{𝑘=𝑝+𝑛+𝑚−1}𝑘𝑎_𝑘×2𝑝−2^{𝑝+𝑛+𝑚−2}_{𝑘=𝑝+𝑛}𝑘𝑎_𝑘−𝑝𝛽_{𝑝+𝑛+𝑚−1}⋅𝑝+𝑛+𝑚−1_{∞𝑘=𝑝+𝑛+𝑚−1}𝑘𝑎_𝑘⁻¹≤1(𝑧∈𝑈;𝑚≥2). (96) This leads to the inequality (91) for 𝑚≥2.
Similarly, for the function 𝑞₂(𝑧) defined by 𝑞₂(𝑧)=1+𝑝𝛽_{𝑝+𝑛+𝑚−1}𝑠𝑝+𝑛+𝑚−1^′𝑚(𝑧)𝑓^′(𝑧)−𝑝𝛽_{𝑝+𝑛+𝑚−1},𝑝+𝑛+𝑚−1 (97) we deduce from (94) that 𝑞₂(𝑧)−1𝑞₂(𝑧)≤+11+𝑝𝛽_{𝑝+𝑛+𝑚−1}⋅𝑝+𝑛+𝑚−1^∞_{𝑘=𝑝+𝑛+𝑚−1}𝑘𝑎_𝑘×2𝑝−2^{𝑝+𝑛+𝑚−2}_{𝑘=𝑝+𝑛}𝑘𝑎_𝑘−𝑝𝛽_{𝑝+𝑛+𝑚−1}⋅𝑝+𝑛+𝑚−1−1_{∞𝑘=𝑝+𝑛+𝑚−1}𝑘𝑎_𝑘⁻¹≤1(𝑧∈𝑈;𝑚≥2). (98) This yields the inequality (92) for 𝑚≥2.
For 𝑚=1, replacing (94) by 𝛽_𝑝+𝑛𝑝+𝑛^∞_{𝑘=𝑝+𝑛}𝑘𝑎_𝑘≤^∞_{𝑘=𝑝+𝑛}𝛽_𝑘𝑎_𝑘≤1, (99) we know that (91) and (92) are also true.
Furthermore, the bounds in (91) and (92) are the best possible for the function 𝑓(𝑧) defined by (88).

Finally, Theorem 21 yields the following.

Corollary 22. Let 𝑓(𝑧)∈𝐺_𝑝,𝑛(𝑎,𝑐,𝜆,𝐴,𝐵) with 𝑎≥𝑐>0 and 𝜆≥1. Then 𝑓Re^′(𝑧)+𝑧𝑓^′′(𝑧)𝑠^′_𝑚(𝑧)+𝑧𝑠_𝑚^′′(𝑧)>1−𝑝+𝑛+𝑚−1𝑝𝛽_{𝑝+𝑛+𝑚−1}(),𝑠𝑧∈𝑈;𝑚∈𝑁Re^′𝑚(𝑧)+𝑧𝑠_𝑚^′′(𝑧)𝑓^′(𝑧)+𝑧𝑓^′′(𝑧)>𝑝𝛽_{𝑝+𝑛+𝑚−1}𝑝+𝑛+𝑚−1+𝑝𝛽_{𝑝+𝑛+𝑚−1}(𝑧∈𝑈;𝑚∈𝑁). (100) The results are sharp for each 𝑚∈𝑁.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.