Certain Properties of a Class of Close-to-Convex Functions Related to Conic Domains

Abstract

We aim to de…fine a new class of close-to-convex functions which is related to conic domains. Many interesting properties such as sufficiency criteria, inclusion results, and integral preserving properties are investigated here. Some interesting consequences of our results are also observed.

1. Introduction

Let 𝒜 be the class of functions 𝑓𝑓(𝑧)=𝑧+^∞_𝑛=2𝑎_𝑛𝑧^𝑛, (1) which are analytic in the open unit disc 𝐸={𝑧∈ℂ∶|𝑧|<1}. Let 𝑓 and 𝑔 be analytic in 𝐸, and we say that 𝑓 is subordinate to 𝑔, written as 𝑓(𝑧)≺𝑔(𝑧) if there exists a Schwarz function 𝑤, which 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); see [1].

Kanas and Wisniowska [2, 3] introduced and studied the classes of 𝑘-uniformly convex functions denoted by 𝑘-𝒰𝒞𝒱 and the corresponding class 𝑘-𝒮𝒯 related by the Alexander-type relation. Later, the class 𝑘-uniformly close-to-convex functions denoted by 𝑘-𝒰𝒦 defined as 𝑘-𝒰𝒦=𝑓∈𝒜∶Re𝑧𝑓^′(𝑧)𝑔(𝑧)>𝑘𝑧𝑓^′(𝑧)𝑔(𝑧),𝑔(𝑧)−1∈𝑘-𝒮𝒯,𝑧∈𝐸 (2) was considered by Acu [4]; for study details on these classes, we refer to [5–7]. All these above mentioned classes were generalized to the classes 𝒮𝒟(𝑘,𝛿), 𝒞𝒟(𝑘,𝛿), and 𝒦𝒟(𝑘,𝛽,𝛿) by Shams et al. [8] and Srivastava et al. [9], respectively. The classes 𝒮𝒟(𝑘,𝛿) and 𝒞𝒟(𝑘,𝛿) are defined as 𝒮𝒟(𝑘,𝛿)=𝑓∈𝒜∶Re𝑧𝑓^′(𝑧)𝑓(𝑧)>𝑘𝑧𝑓^′(𝑧)𝑓(𝑧),()=−1+𝛿,𝑧∈𝐸𝒞𝒟𝑘,𝛿𝑓∈𝒜∶Re1+𝑧𝑓^′′(𝑧)𝑓^′(𝑧)>𝑘𝑧𝑓^′′(𝑧)𝑓^′(𝑧)+𝛿,𝑧∈𝐸, (3) where 𝑘≥0, 0≤𝛽, 𝛿<1. The class 𝒦𝒟(𝑘,𝛽,𝛿) known as 𝑘-uniformly close-to-convex functions of order 𝛽 type 𝛿 is the class of all those functions 𝑓∈𝒜 which satisfies the following condition: 𝑓Re^′(𝑧)𝑔^′(𝑧)𝑓≥𝑘^′(𝑧)𝑔^′(𝑧),−1+𝛽,𝑘≥0,0≤𝛽,𝛿<1;𝑧∈𝐸 (4) for some 𝑔∈𝒞𝒟(𝑘,𝛿).

Motivated by the work of Noor et al. [10–13], we define the following.

Definition 1. Let 𝑓∈𝒜. Then, 𝑓 is in the class 𝒯𝒟(𝑘,𝛼,𝛽,𝛾,𝛿) if and only if, for 𝛼≥0, 0≤𝛽,𝛾,𝛿<1, 𝐻𝐻,(),Re𝛼,𝛽,𝛾;𝑓,𝑔≥𝑘𝛼,𝛽,𝛾;𝑓,𝑔−1𝑘≥0;𝑧∈𝐸 (5) for some 𝑔∈𝒞𝒟(𝑘,𝛿), where 𝐻𝛼,𝛽,𝛾;𝑓,𝑔=1−𝛼𝑓1−𝛽^′(𝑧)𝑔^′(𝑧)+𝛼−𝛽1−𝛾𝑧𝑓^′(𝑧)^′𝑔^′(𝑧)−𝛾. (6)

Special Cases (i)(𝑘,0,𝛽,𝛾,𝛿)=𝒦𝒟(𝑘,𝛽,𝛿); see [9].(ii)𝒯𝒟(1,0,𝛽,𝛾,0)=𝒰𝒦(𝛽) and 𝒯𝒟(1,1,𝛽,𝛾,0)=𝒰𝒬(𝛾), the classes of uniformly close-to-convex and quasiconvex functions introduced and investigated in [14].(iii)𝒯𝒟(0,𝛼,0,0,0)=𝒬_𝛼, the class of alpha quasiconvex functions, introduced and studied in [11].(iv)𝒯𝒟(0,0𝛽,𝛾,𝛿)=𝒦(𝛽,𝛿), the class of close-to-convex functions of order 𝛽 type 𝛿, [15].(v)𝒯𝒟(0,1,𝛽,𝛾,𝛿)=𝐶^∗(𝛾,𝛿), the class of quasiconvex functions of order 𝛾 type 𝛿, [16].The conditions 𝑘≥0, 𝛼≥0, 0≤𝛽,𝛾,𝛿<1 on the parameters are assumed throughout the entire paper unless otherwise mentioned.

Geometric Interpretation. A function 𝑓∈𝒜 is in the class 𝒯𝒟(𝑘,𝛼,𝛽,𝛾,𝛿) if and only if the functional 𝐻(𝛼,𝛽,𝛾,𝛿;𝑓,𝑔) takes all the values in the conic domain Ω_𝑘 defined as follows: Ω_𝑘=𝑢+𝜄𝑣;𝑢>𝑘(𝑢−1)²+𝑣². (7) Extremal functions for these conic regions are denoted by 𝑝_𝑘(𝑧), which are analytic in 𝐸 and map 𝐸 onto Ω_𝑘 such that 𝑝_𝑘(0)=1 and 𝑝^′_𝑘(0)>1. These functions are given as: 𝑝_𝑘(𝑧)=1+𝑧21−𝑧,𝑘=0,1+𝜋²√log1+𝑧1−√𝑧²2,𝑘=1,1+1−𝑘²sinh²2𝜋arccos𝑘arctanℎ√𝑧,10<𝑘<1,1+𝑘²𝜋−1sin2𝑅(𝑡)^{√𝑢(𝑧)/𝑡}₀1√1−𝑥²1−(𝑡𝑥)²+1𝑑𝑥𝑘²−1,𝑘>1, (8) where 𝑢(𝑧)=(𝑧−√𝑡)/(1−√𝑡𝑧),𝑡∈(0,1),𝑧∈𝐸, and 𝑧 is chosen such that 𝑘=cosh(𝜋𝑅^′(𝑡)/4𝑅(𝑡)), where 𝑅(𝑡) is Legendre's complete elliptic integral of the first kind and 𝑅^′(𝑡) is complementary integral of 𝑅(𝑡); see [2, 3].

2. Preliminaries Result

We require the following results which are essential in our investigations.

Lemma 2 (see [17, page 70]). Let ℎ be convex function in 𝐸 and 𝑞∶𝐸↦𝐶 with Re𝑞(𝑧)>0, 𝑧∈𝐸. If 𝑝 is analytic in 𝐸 with 𝑝(0)=ℎ(0), then 𝑝(𝑧)+𝑞(𝑧)𝑧𝑝^′(𝑧)≺ℎ(𝑧)𝑖𝑚𝑝𝑙𝑖𝑒𝑠𝑝(𝑧)≺ℎ(𝑧). (9)

Lemma 3 (see [17, page 195]). Let ℎ be convex function in 𝐸 with ℎ(0)=0 and 𝐴≥0. Suppose that 𝑗≥4/ℎ^′(0)  and that 𝐵(𝑧),  𝐶(𝑧), and 𝐷(𝑧) are analytic in 𝐸 and satisfy Re𝐵(𝑧)≥𝐴+|𝐶(𝑧)−1|−Re(𝐶(𝑧)−1)+𝑗𝐷(𝑧), (10) for 𝑧∈𝐸. If 𝑝 is analytic in 𝐸 with 𝑝(𝑧)=𝑎₁𝑧+𝑎₂𝑧²+𝑎₃𝑧³+⋯ and the following subordination relation holds: 𝐴𝑧²𝑝^′′(𝑧)+𝐵(𝑧)𝑧𝑝^′(𝑧)+𝐶(𝑧)𝑝(𝑧)+𝐷(𝑧)≺ℎ(𝑧), (11) then 𝑝(𝑧)≺ℎ(𝑧). (12)

Lemma 4 (see [12]). If 𝑓(𝑧)≺ℎ(𝑧) and 𝑔(𝑧)≺ℎ(𝑧), then for 𝛼∈[0,1], (1−𝛼)𝑓(𝑧)+𝛼𝑔(𝑧)≺ℎ(𝑧). (13)

3. Main Results

First, we prove the following sufficiency criteria for the functions in the class 𝒯𝒟(𝑘,𝛼,𝛽,𝛾,𝛿).

Theorem 5. A function 𝑓∈𝒜 is said to be in the class 𝒯𝒟(𝑘,𝛼,𝛽,𝛾,𝛿), if ^∞_𝑛=2Φ_𝑛𝑘;𝛼,𝛽,𝛾,𝛿<1−𝛽1−𝛾, (14) where Φ_𝑛=()𝑛𝑘;𝛼,𝛽,𝛾,𝛿𝑘+1)(1−𝛼1−𝛾𝑛+𝛼1−𝛽²𝑎_𝑛+𝑘+2−𝛽1−𝛾+𝛼(𝑘+1)𝛾−𝛽𝑛𝑏_𝑛. (15)

Proof. Let us assume that relation (6) holds. Now, it is sufficient to show that 𝑘𝐻𝛼,𝛽,𝛾,𝛿;𝑓,𝑔−1−Re𝐻𝛼,𝛽,𝛾,𝛿;𝑓,𝑔−1<1. (16)
First, we consider 𝐻=𝛼,𝛽,𝛾,𝛿;𝑓,𝑔−11−𝛼𝑓1−𝛽^′(𝑧)𝑔^′(𝑧)+𝛼−𝛽1−𝛾𝑧𝑓^′(𝑧)^′𝑔^′(𝑧)=−𝛾−11−𝛼𝑓1−𝛽^′(𝑧)𝑔^′(𝑧)+𝛼1−𝛾𝑧𝑓^′(𝑧)^′𝑔^′(𝑧)−(1−𝛼)𝛽−1−𝛽𝛼𝛾=(𝑓1−𝛾−11−𝛼)1−𝛾^′(𝑧)1−𝛽1−𝛾𝑔^′(𝑧)+𝛼1−𝛽𝑧𝑓^′(𝑧)^′1−𝛽1−𝛾𝑔^′(𝑧)−1−𝛾+𝛼𝛾−𝛽𝑔^′(𝑧)1−𝛽1−𝛾𝑔^′(𝑧). (17) Using (1) and the series 𝑔(𝑧)=𝑧+∑^∞_𝑛=2𝑏_𝑛𝑧^𝑛 in (17), we have 𝐻=()𝛼,𝛽,𝛾,𝛿;𝑓,𝑔−11−𝛼1−𝛾1+∑^∞_𝑛=2𝑛𝑎_𝑛𝑧^𝑛−11−𝛽1−𝛾1+∑^∞_𝑛=2𝑛𝑏_𝑛𝑧^𝑛−1+𝛼1−𝛽1+∑^∞_𝑛=2𝑛²𝑎_𝑛𝑧^𝑛−11−𝛽1−𝛾1+∑^∞_𝑛=2𝑛𝑏_𝑛𝑧^𝑛−1−1−𝛾+𝛼𝛾−𝛽1+∑^∞_𝑛=2𝑛𝑏_𝑛𝑧^𝑛−11−𝛽1−𝛾1+∑^∞_𝑛=2𝑛𝑏_𝑛𝑧^𝑛−1=(1−𝛼)1−𝛾∑^∞_𝑛=2𝑛𝑎_𝑛𝑧^𝑛−11−𝛽1−𝛾1+∑^∞_𝑛=2𝑛𝑏_𝑛𝑧^𝑛−1+𝛼1−𝛽∑^∞_𝑛=2𝑛²𝑎_𝑛𝑧^𝑛−11−𝛽1−𝛾1+∑^∞_𝑛=2𝑛𝑏_𝑛𝑧^𝑛−1−1−𝛾+𝛼𝛾−𝛽∑^∞_𝑛=2𝑛𝑏_𝑛𝑧^𝑛−11−𝛽1−𝛾1+∑^∞_𝑛=2𝑛𝑏_𝑛𝑧^𝑛−1≤(1−𝛼)1−𝛾∑^∞_𝑛=2𝑛𝑎_𝑛|𝑧|^𝑛−11−𝛽1−𝛾1−∑^∞_𝑛=2𝑛𝑏_𝑛|𝑧|^𝑛−1+𝛼1−𝛽∑^∞_𝑛=2𝑛²𝑎_𝑛|𝑧|^𝑛−11−𝛽1−𝛾1−∑^∞_𝑛=2𝑛𝑏_𝑛|𝑧|^𝑛−1+1−𝛾+𝛼𝛾−𝛽∑^∞_𝑛=2𝑛𝑏_𝑛|𝑧|^𝑛−11−𝛽1−𝛾1−∑^∞_𝑛=2𝑛𝑏_𝑛|𝑧|^𝑛−1≤(1−𝛼)1−𝛾∑^∞_𝑛=2𝑛𝑎_𝑛1−𝛽1−𝛾1−∑^∞_𝑛=2𝑛𝑏_𝑛+𝛼1−𝛽∑^∞_𝑛=2𝑛²𝑎_𝑛1−𝛽1−𝛾1−∑^∞_𝑛=2𝑛𝑏_𝑛+1−𝛾+𝛼𝛾−𝛽∑^∞_𝑛=2𝑛𝑏_𝑛1−𝛽1−𝛾1−∑^∞_𝑛=2𝑛𝑏_𝑛. (18) Now, 𝑘𝐻𝐻≤(𝐻≤(()∑𝛼,𝛽,𝛾,𝛿;𝑓,𝑔−1−Re𝛼,𝛽,𝛾,𝛿;𝑓,𝑔−1𝑘+1)𝛼,𝛽,𝛾,𝛿;𝑓,𝑔−1𝑘+1)1−𝛼1−𝛾^∞_𝑛=2𝑛𝑎_𝑛1−𝛽1−𝛾1−∑^∞_𝑛=2𝑛𝑏_𝑛+𝛼1−𝛽∑^∞_𝑛=2𝑛²𝑎_𝑛1−𝛽1−𝛾1−∑^∞_𝑛=2𝑛𝑏_𝑛+1−𝛾+𝛼𝛾−𝛽∑^∞_𝑛=2𝑛𝑏_𝑛1−𝛽1−𝛾1−∑^∞_𝑛=2𝑛𝑏_𝑛. (19) The last inequality is bounded above by 1, if (𝑘+1)(1−𝛼)1−𝛾^∞_𝑛=2𝑛𝑎_𝑛+𝛼1−𝛽^∞_𝑛=2𝑛²𝑎_𝑛+1−𝛾+𝛼𝛾−𝛽^∞_𝑛=2𝑛𝑏_𝑛≤1−𝛽1−𝛾1−^∞_𝑛=2𝑛𝑏_𝑛. (20) Hence, ^∞_𝑛=2Φ_𝑛𝑘;𝛼,𝛽,𝛾,𝛿≤1−𝛽1−𝛾, (21) where Φ_𝑛(𝑘;𝛼,𝛽,𝛾,𝛿) is given by (15). This completes the proof.

When we take  𝛼=0, 𝑘=1, and 𝑔(𝑧)=𝑧 in the above theorem, we obtain the following sufficient condition for the functions to be in the class 𝒰𝒦(𝛽) which is proved in [14].

Corollary 6 (see [14]). A function  𝑓∈𝐴 is said to be in the class 𝒰𝒦(𝛽) if ^∞_𝑛=2𝑛𝑎_𝑛≤1−𝛽2. (22)

Corollary 7 (see [14]). A function  𝑓∈𝐴 is said to be in the class 𝒰𝒬(𝛾) if ^∞_𝑛=2𝑛²𝑎_𝑛≤1−𝛾2. (23)

The above corollary is obtained when we take 𝛼=1, 𝑘=1,  and   𝑔(𝑧)=𝑧 in Theorem 5.

Theorem 8. Let 𝑓∈𝒯𝒟(𝑘,𝛼,𝛽,𝛾,𝛿) and 𝛼≥4(𝛽/(1+3𝛽)). Then, 𝑓∈𝒦𝒟(𝑘,0,𝛿).

Proof. Let 𝑓^′(𝑧)𝑔^′(𝑧)=𝑝(𝑧), (24) where 𝑝(𝑧) is analytic and 𝑝(0)=1. Now differentiating (24), we have 𝑧𝑓^′(𝑧)^′𝑔^′(𝑧)=𝑧𝑝^′(𝑧)+𝑝(𝑧)𝜓(𝑧), (25) where 𝜓(𝑧)=(𝑧𝑔^′(𝑧))^′/𝑔^′(𝑧). Using (24) and (25) in relation (6), we obtain 𝐻(𝑧)=𝛼,𝛽,𝛾,𝛿;𝑓1−𝛼𝑝(𝑧)+𝛼1−𝛽−𝛽1−𝛾𝑧𝑝^′(𝑧)(𝑧)𝜓(𝑧)=𝛼+𝑝−𝛾1−𝛾𝑧𝑝^′(𝑧)+(1−𝛼)1−𝛾+𝛼1−𝛽𝜓(𝑧)𝑝(𝑧)−𝛽(1−𝛽1−𝛾1−𝛼)1−𝛾+𝛼𝛾1−𝛽(𝑧)1−𝛽1−𝛾=𝐵𝑧𝑝^′(𝑧)+𝐶(𝑧)𝑝(𝑧)+𝐷(𝑧), (26) where 𝐵(𝑧)=𝛼(𝑧)=()𝜓(𝑧)1−𝛾,𝐶1−𝛼1−𝛾+𝛼1−𝛽,𝐷(𝑧)𝛽()1−𝛽1−𝛾=−1−𝛼1−𝛾+𝛼𝛾1−𝛽.1−𝛽1−𝛾 (27)
Now, since 𝑓∈𝒯𝒟(𝑘,𝛼,𝛽,𝛾,𝛿), we have 𝐵(𝑧)𝑧𝑝^′(𝑧)+𝐶(𝑧)𝑝(𝑧)+𝐷(𝑧)≺𝑝_𝑘(𝑧). (28)
Replacing 𝑝(𝑧) by 𝑝_∗(𝑧)=𝑝(𝑧)−1 and 𝑝_𝑘(𝑧) by 𝑝^∗_𝑘(𝑧)=𝑝_𝑘(𝑧)−1, the above subordination is equivalent to 𝐵(𝑧)𝑧𝑝^′_∗(𝑧)+𝐶(𝑧)𝑝_∗(𝑧)+𝐷_∗(𝑧)≺𝑝^∗_𝑘(𝑧), (29) where 𝐷_∗(𝑧)=𝐶(𝑧)+𝐷(𝑧)−1. Using Lemma 3 with 𝐴=0, we obtain 𝑝_∗(𝑧)≺𝑝^∗_𝑘(𝑧). (30) This implies that 𝑓^′(𝑧)𝑔^′(𝑧)=𝑝(𝑧)≺𝑝_𝑘(𝑧). (31) Hence, 𝑓∈𝒦𝒟(𝑘,0,𝛿). This completes the proof.

Corollary 9. Let 𝑓∈𝒯𝒟(𝑘,𝛼,0,0,0)=𝒬_𝛼. Then, 𝑓∈𝒦𝒟(0,0,0)=𝒦. That is, 𝒬_𝛼⊂𝒦,  𝛼≥0.

The above result is well-known inclusion proved in [11].

For 𝑓∈𝐴, consider the following integral operator defined by 𝐹(𝑧)=𝐼_𝑚𝑓=𝑚+1𝑧^𝑚𝑧₀𝑡^𝑚−1𝑓(𝑡)𝑑𝑡,𝑚=1,2,3…. (32) This operator was given by Bernardi [18] in 1969. In particular, the operator 𝐼₁ was considered by Libera [19]. Now let us prove the following.

Theorem 10. Let 𝑓∈𝒯𝒟(𝑘,𝛼,𝛽,𝛾,𝛿). Then, 𝐼_𝑚[𝑓]∈𝒦𝒟(𝑘,0,𝛿).

Proof. Let the function 𝑔 be such that (6) is satisfied. It can easily be seen that according to [4], the function 𝐺=𝐼_𝑚[𝑓]∈𝒞𝒟(𝑘,𝛿), and from (32), we deduce (1+𝑚)𝑓^′(𝑧)=(1+𝑚)𝐹^′(𝑧)+𝑧𝐹^′′(𝑧),(1+𝑚)𝑔^′(𝑧)=(1+𝑚)𝐺^′(𝑧)+𝑧𝐺^′′(𝑧). (33)(34) If we let 𝑝(𝑧)=𝐹^′(𝑧)/𝐺^′(𝑧) and 𝑞(𝑧)=1/(𝑚+1+(𝑧𝐺^′′(𝑧)/𝐺^′)), then simple computations yield us 𝑓^′(𝑧)𝑔^′(𝑧)=(1+𝑚)𝐹^′(𝑧)+𝑧𝐹^′′(𝑧)(1+𝑚)𝐺^′(𝑧)+𝑧𝐺^′′(𝑧)=𝑧𝑝(𝑧)𝑔^′(𝑧)+𝑧𝑝(𝑧)𝑔^′′(𝑧)+𝑧𝑔^′(𝑧)𝑝^′(𝑧)(1+𝑚)𝐺^′(𝑧)+𝑧𝐺^′′(𝑧)=𝑝(𝑧)+𝑞(𝑧)𝑧𝑝^′(𝑧). (35) Let 𝑓^′(𝑧)𝑔^′(𝑧)=𝑝(𝑧)+𝑧𝑝^′(𝑧)𝑞(𝑧)=ℎ(𝑧), (36) where ℎ(𝑧) is analytic and 𝑝(0)=1. From (36), we have 𝑧𝑓^′(𝑧)^′𝑔^′(𝑧)=𝜓(𝑧)ℎ(𝑧)+𝑧ℎ^′(𝑧), (37) where 𝜓(𝑧)=(𝑧𝑔^′(𝑧))^′/𝑔^′(𝑧). Using (36) and (37) in (6), we have 𝐻(𝑧)=𝛼,𝛽,𝛾,𝛿;𝑓1−𝛼ℎ(𝑧)+𝛼1−𝛽−𝛽1−𝛾𝑧ℎ^′(𝑧)(𝑧)𝜓(𝑧)=𝛼+ℎ−𝛾1−𝛾𝑧ℎ^′(𝑧)+(1−𝛼)1−𝛾+𝛼1−𝛽𝜓(𝑧)ℎ(𝑧)−𝛽(1−𝛽1−𝛾1−𝛼)1−𝛾+𝛼𝛾1−𝛽(𝑧)1−𝛽1−𝛾=𝐵𝑧ℎ^′(𝑧)+𝐶(𝑧)ℎ(𝑧)+𝐷(𝑧), (38) where 𝐵(𝑧)=𝛼(𝑧)=()𝜓(𝑧)1−𝛾,𝐶1−𝛼1−𝛾+𝛼1−𝛽,𝐷(𝑧)𝛽()1−𝛽1−𝛾=−1−𝛼1−𝛾+𝛼𝛾1−𝛽.1−𝛽1−𝛾 (39) Now proceeding in the similar manner as in the proof of Theorem 5 and using Lemma 3 with 𝐴=0, we obtain 𝑓^′(𝑧)𝑔^′(𝑧)=ℎ(𝑧)≺𝑝_𝑘(𝑧). (40) From (36), it implies that 𝑝(𝑧)+𝑧𝑝^′(𝑧)𝑞(𝑧)≺𝑝_𝑘(𝑧). (41) By employing Lemma 2, we immediately obtain the desired result.

Theorem 11. For 𝛼>𝛼₁≥0, 𝒯𝒟𝑘,𝛼,𝛽,𝛾,𝛿⊆𝒯𝒟𝑘,𝛼₁,𝛽,𝛾,𝛿. (42)

Proof. Let 𝑓∈𝒯𝒟(𝑘,𝛼,𝛽,𝛾,𝛿). Then, consider 𝐻𝛼₁,𝛽,𝛾,𝛿;𝑓,𝑔=1−𝛼₁𝑓1−𝛽^′(𝑧)𝑔^′(𝑧)+𝛼−𝛽₁1−𝛾𝑧𝑓^′(𝑧)^′𝑔^′(𝑧)−𝛾. (43) After some simple computations, we have 𝐻𝛼₁=1,𝛽,𝛾,𝛿;𝑓,𝑔𝛼1−𝛽1−¹𝛼𝑓^′(𝑧)𝑔^′(𝑧)+𝛼−𝛽₁𝛼1−𝛼𝑓1−𝛽^′(𝑧)𝑔^′(𝑧)+𝛼−𝛽1−𝛾𝑧𝑓^′(𝑧)^′𝑔^′(𝑧)=𝛼−𝛾1−¹𝛼𝐻+𝛼0,𝛽,𝛾,𝛿;𝑓,𝑔¹𝛼𝐻𝛼,𝛽,𝛾,𝛿;𝑓,𝑔. (44) Now, since 𝑓∈𝒯𝒟(𝑘,𝛼,𝛽,𝛾,𝛿), we have 𝐻(𝛼,𝛽,𝛾,𝛿;𝑓,𝑔)≺𝑝_𝑘(𝑧). Also, Theorem 8 gives us that 𝐻(0,𝛽,𝛾,𝛿;𝑓,𝑔)≺𝑝_𝑘(𝑧). The use of Lemma 4 leads us to the required relation; that is, 𝐻(𝛼₁,𝛽,𝛾,𝛿;𝑓,𝑔)≺𝑝_𝑘(𝑧). This completes the proof.

Acknowledgment

The authors would like to thank the reviewer of this paper for his/her valuable comments on the earlier version of this paper. They would also like to acknowledge Prof. Dr. Ehsan Ali, VC AWKUM, for the financial support on the publication of this article.