On Uniformly Bazilevic and Related Functions

Abstract

We introduce a new class of functions analytic in the open unit disc, which contains the class of Bazilevic functions and also generalizes the concept of uniform convexity. We establish univalence criterion for the functions in this class and investigate rate of growth of coefficients, arc length problem, inclusion results, and distortion bounds. Some interesting results are derived as special cases.

1. Introduction

Let 𝐴 be the class functions analytic in the open unit disc 𝐸={𝑧∶|𝑧|<1} and satisfying the conditions 𝑓(0)=0,𝑓′(0)=1. Let 𝑆⊂𝐴 be the class of functions which are univalent, and also let 𝑆^∗(𝑟),𝐶(𝑟) be the subclasses of 𝑆 which consists of starlike and convex functions of order 𝛾(0≤𝛾<1), respectively.

Kanas and Wisniowska [1, 2] studied the classes of 𝑘-uniformly convex functions, denoted by 𝑘-𝑈𝐶𝑉 and the corresponding class 𝑈𝑆𝑇 related with the Alexander-type relation. In [3], the domain Ω_𝑘,𝑘∈[0,∞) is defined as follows: Ω_𝑘=𝑢+𝑖𝑣∶𝑢>𝑘(𝑢−1)²+𝑣². (1.1) For fixed 𝑘,Ω_𝑘 represents the conic region bounded, successively, by the imaginary axis (𝑘=0), the right branch of hyperbola (0<𝑘<1), a parabola (𝑘=1), and an ellipse (𝑘>1). Also, we note that, for no choice of 𝑘(𝑘>1),Ω_𝑘 reduces to a disc.

In this paper, we will choose 𝑘∈[0,1]. Related with Ω_𝑘, we define the domain Ω_𝑘,𝛾, see [4], as follows: Ω_𝑘,𝛾=1−𝛾Ω_𝑘+𝛾,0≤𝛾<1. (1.2) For 𝑘∈[0,1], the following functions denoted by 𝑝_𝑘,𝛾(𝑧) are univalent in 𝐸, continuous as regards to 𝑘 and 𝛾, have real coefficients, and map 𝐸 onto Ω_𝑘,𝛾, such that 𝑝_𝑘,𝛾(0)=1,𝑝^′_𝑘,𝛾(0)>0: 𝑝_𝑘,𝛾(𝑧)=1+1−2𝛾𝑧(),(),21−𝑧𝑘=01+1−𝛾1−𝑘²sinh²2𝜋arccos𝑘arctanh√𝑧,(),20<𝑘<11+1−𝛾𝜋²√log1+𝑧1−√𝑧²,(𝑘=1). (1.3) Let 𝑃 denote the class of the Caratheodory functions of positive real part. We define a subclass of 𝑃 as follows.

Definition 1.1. Let 𝑃(𝑝_𝑘,𝛾)⊂𝑃 be the class consisting of functions 𝑝(𝑧) which are analytic in 𝐸 with 𝑝(0)=1 and which are subordinate to 𝑝_𝑘,𝛾(𝑧) in 𝐸. We write 𝑝∈𝑃(𝑝_𝑘,𝛾) implies 𝑝≺𝑝_𝑘,𝛾, where 𝑝_𝑘,𝛾(𝑧) is the function, given by (1.3), and maps 𝐸 onto Ω_𝑘,𝛾,𝑘∈[0,1],𝛾∈[0,1). That is 𝑝(𝐸)⊂𝑝_𝑘,𝛾(𝐸). We note that 𝑃(𝑝_0,0)=𝑃 and 𝑝∈𝑃(𝑝_0,𝛾)=𝑃(𝛾) implies that Re𝑝(𝑧)>𝛾,𝑧∈𝐸. It is easy to verify that 𝑃(𝑝_𝑘,𝛾) is a convex set, and 𝑃(𝑝_𝑘,𝛾)⊂𝑃(𝛾₁),𝛾₁=(𝑘+𝛾)/(1+𝑘).

The class 𝑃(𝑝_𝑘,𝛾) is extended as follows.

Definition 1.2. Let 𝑝(𝑧) be analytic in 𝐸 with 𝑝(0)=1. Then, 𝑝∈𝑃_𝑚(𝑝_𝑘,𝛾) if and only if, for 𝑚≥2,0≤𝛾<1,𝑘∈[0,1],𝑧∈𝐸, we have 𝑝(𝑧)=𝑚4+12𝑝₁(𝑧)−𝑚4−12𝑝₂(𝑧),𝑝₁,𝑝₂∈𝑃𝑝_𝑘,𝛾. (1.4) When 𝑘=0, we obtain the class 𝑃_𝑚(𝛾) which reduces to the class 𝑃_𝑚 with 𝛾=0, introduced and studied in [5]. Also 𝑃₂(𝑝_𝑘,𝛾)=𝑃(𝑝_𝑘,𝛾).

We now define the following.

Definition 1.3. Let 𝑓∈𝐴 with 𝑓(𝑧)𝑓^′(𝑧)/𝑧≠0 in 𝐸. Let, for 𝛼 real and 𝛽∈[−1/2,1), 𝐽𝛼,𝛽,𝑓(𝑧)=(1−𝛼)1−𝛽𝑧𝑓^′(𝑧)𝑓(𝑧)+𝛼1−𝛽+𝑧𝑓^′′(𝑧)𝑓^′(𝑧). (1.5) Then 𝑓∈𝑘−𝑈𝐵_𝑚(𝛼,𝛽,𝛾) if and only if 𝐽𝛼,𝛽,𝑓(𝑧)∈𝑃_𝑚𝑝_𝑘,𝛾,for𝑧∈𝐸. (1.6) For any real number 𝛼 and 𝛽∈[−1/2,1), we note that the identity function belongs to 𝑘−𝑈𝐵_𝑚(𝛼,𝛽,𝛾) so that 𝑘−𝑈𝐵_𝑚(𝛼,𝛽,𝛾) is not empty.

Throughout this paper, we assume that 𝑘∈[0,1],𝛾∈[0,1),𝑚≥2,𝛼,𝛽∈ℝ,𝛽∈[−1/2,1),𝑧∈𝐸, unless otherwise specified.

We note the following special cases.(i)For 𝑚=2,0≤𝛽<1,𝑘=0, we have a subclass of a class introduced by Mocanu [6]. Also see [7, 8].(ii)It is well known that 0−𝑈𝐵₂(𝛼,𝛽,0)=𝐵₂(𝛼,𝛽) contains the Bazilevic functions with 𝛼>0 and 𝐵₂(𝛼,𝛽)⊂𝑆.(iii)𝑘−𝑈𝐵_𝑚(1,0,𝛾)=𝑘−𝑉_𝑚(𝛾)⊂𝑉_𝑚((𝑘+𝛾)/(1+𝑘))⊂𝑉_𝑚, where 𝑉_𝑚 is the well-known class of functions with bounded boundary rotation, see [9].(iv)𝑘−𝑈𝐵_𝑚(0,0,𝛾)=𝑘−𝑈𝑅_𝑚(𝛾)⊂𝑅_𝑚((𝑘+𝛾)/(1+𝑘))⊂𝑅_𝑚, where 𝑅_𝑚 denotes the class of functions with bounded radius rotation, see [9].(v)𝑘−𝑈𝐵₂(1,0,𝛾)=𝑘−𝑈𝐶𝑉(𝛾) is the class of uniformly convex functions of order 𝛾,0−𝑈𝐶𝑉(𝛾)=𝐶(𝛾), and 1−𝑈𝐶𝑉(𝛾)=𝑈𝐶𝑉(𝛾).

Also 𝑘−𝑈𝐵₂(0,0,𝛾)=𝑈𝑆𝑇(𝛾), and 𝑈𝑆𝑇(𝛾)⊂𝑆^∗(𝛾₁),𝛾₁=(𝑘+𝛾)/(1+𝑘).

Remark 1.4. (i)From Definition 1.3, it can easily be seen that 𝑓∈𝑘−𝑈𝐵_𝑚(𝛼,𝛽,𝛾) if and only if, for 𝛼≠0, there exists a function 𝑔∈𝑘−𝑈𝑅_𝑚(𝛾) such that 𝑓(𝑧)=𝑚₁^𝑧₀𝑡^𝑚1−1𝑔(𝑡)𝑡^{(1−𝛽)/𝛼}𝑑𝑡^1/𝑚1=𝑧+⋯, (1.7) where 𝑚₁()=1+1−𝛼1−𝛽𝛼. (1.8) A simple computation shows that (1.7) can be written as (1−𝛼)𝑧𝑓^′(𝑧)𝑓(𝑧)+𝛼1−𝛽1−𝛽+𝑧𝑓^′′(𝑧)𝑓^′(𝑧)=𝑧𝑔^′(𝑧)𝑔(𝑧), (1.9) with 𝑔∈𝑘−𝑈𝑅_𝑚(𝛾).(ii)Also, for 𝑓∈𝑘−𝑈𝐵_𝑚(𝛼,𝛽,𝛾), it can be verified from (1.5) that 𝑧(𝑓(𝑧)/𝑧)^(1−𝛼)(𝑓′(𝑧))^{𝛼/(1−𝛽)} belongs to 𝑘−𝑈𝑅_𝑚(𝛾) for all 𝑧∈𝐸.

2. Preliminary Results

The following lemma is an easy generalization of a result due to Kanas [3].

Lemma 2.1 (see [10]). Let 𝑘∈[0,∞), and let 𝜎,𝛿₁ be any complex numbers with 𝜎≠0 and 0≤𝛾<Re(𝜎𝑘/(𝑘+1)+𝛿₁). If ℎ(𝑧) is analytic in 𝐸,ℎ(0)=1 and satisfies ℎ(𝑧)+𝑧ℎ^′(𝑧)𝜎ℎ(𝑧)+𝛿₁≺𝑝_𝑘,𝛾(𝑧), (2.1) and 𝑞_𝑘,𝛾(𝑧) is an analytic solution of 𝑞_𝑘,𝛾(𝑧)+𝑧𝑞^′_𝑘,𝛾(𝑧)𝜎𝑞_𝑘,𝛾(𝑧)+𝛿₁=𝑝_𝑘,𝛾(𝑧), (2.2) then 𝑞_𝑘,𝛾(𝑧) is univalent, ℎ(𝑧)≺𝑞_𝑘,𝛾(𝑧)≺𝑝_𝑘,𝛾(𝑧), (2.3) and 𝑞_𝑘,𝛾(𝑧) is the best dominant of (2.1), where 𝑞_𝑘,𝛾(𝑧) is given as 𝑞_𝑘,𝛾(𝑧)=¹₀exp_𝑡^𝑡𝑧𝑝_𝑘,𝛾(𝑢)−1𝑢𝑑𝑢𝑑𝑡⁻¹+𝛿₁𝜎. (2.4)

Lemma 2.2 (see [11]). Let 𝑢=𝑢₁+𝑖𝑢₂ and 𝑣=𝑣₁+𝑖𝑣₂, and let 𝜓(𝑢,𝑣) be a complex-valued function satisfying the following conditions:(i)𝜓(𝑢,𝑣) is continuous in 𝐷⊂ℂ²,(ii)(1,0)∈𝐷 and Re𝜓(1,0)>0,(iii)Re𝜓(𝑖𝑢₂,𝑣₁)≤0, whenever (𝑖𝑢₂,𝑣₁)∈𝐷 and 𝑣₁≤−(1+𝑢²₂)/2.If ℎ(𝑧) is a function analytic in 𝐸 such that (ℎ(𝑧),𝑧ℎ′(𝑧))∈𝐷 and Re𝜓{ℎ(𝑧),𝑧ℎ′(𝑧)}>0 for 𝑧∈𝐸, then Reℎ(𝑧)>0 in 𝐸.

Lemma 2.3 (see [12]). Let 𝑓∈𝐴 with 𝑓(𝑧)𝑓′(𝑧)/𝑧≠0 in 𝐸. Then 𝑓(𝑧) is a Bazilevic function (hence univalent) in 𝐸 if and only if, for 0≤𝜃₁<𝜃₂≤2𝜋,0<𝑟<1,𝑧=𝑟𝑒^𝑖𝜃, one has ^𝜃2_𝜃1Re1+𝑧𝑓^′′(𝑧)𝑓^′(𝑧)+𝛼₁−1𝑧𝑓^′(𝑧)𝑓(𝑧)−𝛽₁Im𝑧𝑓^′(𝑧)𝑓(𝑧)𝑑𝜃>−𝜋, (2.5) where 𝛼₁>0,𝛽₁ real.

3. Main Results

In the following, we establish the criterion of univalence for the class 𝑘−𝑈𝐵_𝑚(𝛼,𝛽,𝛾) with certain restriction on the upper bound of the value of 𝑚.

Theorem 3.1. Let 𝑓∈𝑘−𝑈𝐵_𝑚(𝛼,𝛽,𝛾),𝛼>0,𝛽∈[0,1). Then 𝑓∈𝑆 for 𝑚≤2{𝛼(1+2𝛽)(𝑘+1)/(1−𝛽)(1−𝛾)+1} in 𝐸.

Proof. Since 𝑓∈𝑘−𝑈𝐵_𝑚(𝛼,𝛽,𝛾)⊂𝐵_𝑚(𝛼,𝛽,𝛾₁),𝛾₁=(𝑘+𝛾)/(1+𝑘), we note that, in (1.9), 𝑔∈𝑅_𝑚(𝛾₁). It is known [13] that there exists 𝑔₁∈𝑅_𝑚 such that, for 𝑧∈𝐸, 𝑔(𝑧)𝑔=𝑧¹(𝑧)𝑧^(1−𝛾1). (3.1) Now from (1.9) and (3.1), we have, for 𝑧=𝑟𝑒^𝑖𝜃,0<𝑟<1,0≤𝜃₁<𝜃₂≤2𝜋, ^𝜃2_𝜃1(Re1−𝛼)1−𝛽𝛼𝑧𝑓^′(𝑧)𝑓(𝑧)+1+𝑧𝑓^′′(𝑧)𝑓^′(𝑧)𝑑𝜃>−𝜋1−𝛽𝛼1−𝛾𝑚1+𝑘2−1−2𝛽. (3.2) We use Lemma 2.3 with 𝛼₁=1/𝛼+𝛽(1−1/𝛼),𝛽₁=0 to have the required result.

As special cases, we note that(i)for 𝑘=0,𝛽=0,𝛼=1,𝑓(𝑧) is univalent for 𝑚≤2(1/(1−𝛾)+1). We observe that, when 𝛾=0, we obtain a well-known result that the class 𝑉_𝑚 of functions with bounded boundary rotation contains univalent functions for 2≤𝑚≤4, see [9];(ii)for 𝑘=1,𝛽=0,𝛼=1,𝛾=0,𝑓∈1−𝑈𝑉_𝑚 is univalent for 𝑚≤6.

Let Γ denote the Gamma function, and let 𝐹(𝑎,𝑏,𝑐;𝑧) be the hypergeometric function which is analytic in 𝐸 and is defined by 𝐹()=Γ(𝑐)𝑎,𝑏,𝑐;𝑧Γ(𝑐)Γ(𝑐−𝑏)¹₀𝑢^𝑏−1(1−𝑢)^{𝑐−𝑏−1}(1−𝑧𝑢)^−𝑎𝑑𝑢, (3.3) where Re𝑏>0,Re(𝑐−𝑏)>0.

Define 𝐾𝑚₁,𝑘₁,𝑟=𝑟𝐹𝑘₁𝑚2+1,𝑚₁,𝑘₁𝑚2−1+𝑚₁+1,𝑟^1/𝑚1, (3.4) where 𝑚₁()=1+1−𝛼1−𝛽𝛼,𝑘₁=1−𝛽𝛼1−𝛾1+𝑘,𝛼₁≠0. (3.5) Also, for 0≤𝜃≤2𝜋, let 𝑓_𝜃𝑚₁,𝑘₁,𝑧=𝑚₁^𝑧₀𝜉^𝑚1−11+𝑒^𝑖𝜃𝜉^{−𝑘1((𝑚/2)+1)}1−𝑒^𝑖𝜃𝜉^{𝑘1((𝑚/2)−1)}𝑑𝜉^1/𝑚1. (3.6) We now consider the distortion problem.

Theorem 3.2. Let 𝑓∈𝑘−𝑈𝐵_𝑚(𝛼,𝛽,𝛾),𝛼≠0,𝛽∈[0,1). Then, for |𝑧|=𝑟,(i)−𝐾(𝑚₁,𝑘₁,−𝑟)≤|𝑓(𝑧)|≤𝐾(𝑚₁,𝑘₁,𝑟), for 𝛼>0,(ii)𝐾(𝑚₁,𝑘₁,𝑟)≤|𝑓(𝑧)|≤−𝐾(𝑚₁,𝑘₁,−𝑟), for 𝛼<0.

Proof. Suppose 𝑧₀ is a point on the circumference |𝑧|=𝑟 such that |𝑓(𝑧₀)|=min_{0≤𝜃≤2𝜋}|𝑓(𝑟𝑒^𝑖𝜃)|, and let 𝜏 denote the preimage under 𝑓 on the segment [0,𝑓(𝑧₀)].
Consider first case 𝛼>0.
Distortion results for 𝑔₁∈𝑅_𝑚 are given as follows: 1−𝜉^𝑚/2−11+𝜉^𝑚/2+1≤𝑔₁(𝜉)𝜉≤1+𝜉^𝑚/2−11−𝜉^𝑚/2+1, (3.7) see [9]. In view of (1.7), (3.1), and (3.7), we have 𝑓𝑧₀^𝑚1=𝑚_1𝜏𝜉^𝑚1−1𝑔(𝜉)𝜉^{(1−𝛽)/𝛼}𝑑𝜉,≥𝑚₁^𝑟₀𝑡^𝑚1−1(1−𝑡)^{𝑘1(𝑚/2−1)}(1+𝑡)^{−𝑘1(𝑚/2+1)}𝑑𝑡,=𝑚₁𝑟^𝑚11₀𝑢^𝑚1−1(1+𝑟𝑢)^{−𝑘1(𝑚/2+1)}(1−𝑢)^{𝑘1(𝑚/2−1)}𝑑𝑢. (3.8) Hence 𝑓(𝑧)≥𝑓𝑧₀≥−𝐾𝑚₁,𝑘₁,−𝑟, (3.9) where 𝐾(𝑚₁,𝑘₁,𝑟) is defined in (3.4).
The proof of (i) for the upper bound of |𝑓(𝑧)| can be obtained in the similar manner.
The proof of (ii) for the case 𝛼<0 is analogous.

Remark 3.3. The bounds in Theorem 3.2 are sharp for 𝑓∈𝐵_𝑚(𝛼,𝛽,𝛾₁),𝛾₁=(𝑘+𝛾)/(1+𝑘), and the equality occurs for the function 𝑓_𝜃(𝑚₁,𝑘₁,𝑧) given by (3.6) with 𝜃 suitably chosen.
As an application of Theorem 3.2, we derive bound for initial Taylor coefficient of 𝑓(𝑧) as follows.

Corollary 3.4. Let 𝑓(𝑧)=𝑧+∑^∞_𝑛=2𝑎_𝑛𝑧^𝑛 satisfy the conditions of Theorem 3.2. Then 𝑎₂≤𝑘₁(𝑚/2+1)𝑘₁(𝑚/2−1)+𝑚₁+1, (3.10) where 𝑘₁ and 𝑚₁ are given in (3.4).

Proof. First consider the case 𝛼>0 and assume 𝑎₂ to be real.
We find 𝐾𝑚₁,𝑘₁𝑘,𝑟=𝑟+¹(𝑚/2+1)𝑘₁(𝑚/2−1)+𝑚₁𝑟+1²+𝑂𝑟³,𝑓(𝑟)=𝑟+𝑎₂𝑟²+𝑂𝑟³. (3.11) In view of Theorem 3.2(i), we have 𝑎₂≤𝑘₁(𝑚/2+1)𝑘₁(𝑚/2−1)+𝑚₁+1. (3.12) For 𝛼<0, we proceed in a similar manner and use Theorem 3.2(ii) to complete the proof.

We note that −𝐾(𝑚₁,𝑘₁,−𝑟) and 𝐾(𝑚₁,𝑘₁,𝑟) are increasing functions of 𝑟. Thus letting 𝑟→1 in the left hand side of (i) and (ii) in Theorem 3.2, we have the following.

Corollary 3.5. Let 𝑓 satisfy the conditions of Theorem 3.2. Then _{𝑓∈𝑘−∪𝐵𝑚(𝛼,𝛽,𝛾)}𝑓(𝐸)=𝑤∶|𝑤|<𝑟𝑚₁,𝑘₁, (3.13) where 𝑟𝑚₁,𝑘₁=𝐹𝑘₁𝑚2+1,𝑚₁,𝑘₁𝑚2−1+𝑚₁𝐹𝑘+1;1,for𝛼>0,₁𝑚2+1,𝑚₁,𝑘₁𝑚2−1+𝑚₁+1;−1,for𝛼<0. (3.14)

Using Theorem 3.1 and Corollary 3.4, we have the following covering result.

Corollary 3.6. Let 𝑓∈𝑘−𝑈𝐵_𝑚(𝛼,𝛽,𝛾),𝛼>0,𝛽∈[1,0), and 𝑚≤2{𝛼(1+2𝛽)(𝑘+1)/(1−𝛽)(1−𝛾)+1}. If 𝐷 is the boundary of the image of 𝐸 under 𝑓, then every point of 𝐷 is at distance at least 𝑘₁(𝑚/2−1)+𝑚₁+12𝑘₁(𝑚/2−1)+𝑚₁+1+𝑘₁(𝑚/2+1) (3.15) from the origin.

Proof. Let 𝑓(𝑧)≠𝑤₀,𝑤₀≠0. Then 𝑓₁ given by 𝑓₁(𝑧)=𝑤₀𝑓(𝑧)/(𝑤₀−𝑓(𝑧)) is univalent, since 𝑓 is univalent by Theorem 3.1. Writing 𝑓(𝑧)=𝑧+∑^∞_𝑛=2𝑎_𝑛𝑧^𝑛, we have 𝑓₁(𝑧)=𝑤₀𝑓(𝑧)𝑤₀(𝑧)𝑎−𝑓=𝑧+₂+1𝑤₀𝑧²+⋯, (3.16) and since 𝑓₁∈𝑆, it follows that 𝑎₂+1𝑤₀≤2. (3.17) Now using Corollary 3.4, 1𝑤₀𝑘≤2+¹((𝑚/2)+1)𝑘₁((𝑚/2)−1)+𝑚₁+1, (3.18) and this proves the result.

As a special case, for 𝛼=1,𝛽=0,𝛾=0,𝑘=0, and 𝑚=2,𝑓 is convex, and in this case, |𝑤₀|>1/4.

We now proceed to investigate some inclusion properties.

Theorem 3.7. Let 𝑓∈𝑘−𝑈𝐵_𝑚(𝛼,𝛽,𝛾),𝛼>0,𝛽∈[1,0). Then 𝑓∈𝑘−𝑈𝑅_𝑚(𝛾) for 0≤𝛾<{(1−𝛽+𝛼𝛽)𝑘/𝛼(𝑘+1)},𝑧∈𝐸. In particular, for 𝛼>0,𝛽∈[1,0), one has 𝑘−𝑈𝐵_𝑚𝛼,𝛽⊂𝑘−𝑈𝑅_𝑚. (3.19)

Proof. Since 𝑓∈𝑘−𝑈𝐵_𝑚(𝛼,𝛽,𝛾), we can write from (1.7) (1−𝛼)1−𝛽𝑧𝑓^′(𝑧)𝑓(𝑧)+𝛼1−𝛽+𝑧𝑓^′′(𝑧)𝑓^′(𝑧)=1−𝛽𝑧𝑔^′(𝑧)𝑔(𝑧), (3.20) or (1−𝛼)𝑧𝑓^′(𝑧)𝑓(𝑧)+𝛼1−𝛽1−𝛽+𝑧𝑓^′′(𝑧)𝑓^′(𝑧)=𝑧𝑔^′(𝑧)𝑔(𝑧)∈𝑃_𝑚𝑝_𝑘,𝛾. (3.21) Let 𝑧𝑓′(𝑧)/𝑓(𝑧)=𝑝(𝑧). Then 𝑝(𝑧) is analytic in 𝐸 with 𝑝(0)=1. Then we can write (3.21) as 1−𝛽+𝛼𝛽𝑝(𝑧)+𝛼1−𝛽1−𝛽+𝛼𝛽𝑧𝑝^′(𝑧)𝑝(𝑧)−𝛼𝛽1−𝛽+𝛼𝛽=ℎ₀(𝑧)=𝑧𝑔^′(𝑧)𝑔(𝑧), (3.22) or 𝑝(𝑧)+1/𝑚₁𝑧𝑝^′(𝑧)𝑝(𝑧)1−𝛽𝛼𝑚₁ℎ₀(𝑧)+𝛽𝑚₁, (3.23) where 𝑚₁=1+(1−𝛼)(1−𝛽)/𝛼.
Since ℎ₀∈𝑃_𝑚(𝑝_𝑘,𝛾) and 𝑃_𝑚(𝑝_𝑘,𝛾) is a convex set, see [4], it follows that {((1−𝛽)/𝛼𝑚₁)ℎ₀(𝑧)+(𝛽/𝑚₁)𝑝₀(𝑧)}, with 𝑝₀(𝑧)=1, belong to 𝑃_𝑚(𝑝_𝑘,𝛾) in 𝐸.
Define 𝜙_𝑚1(𝑧)=𝑧(1−𝑧)^1/𝑚1(1−(𝑧/2))(1−𝑧)². (3.24) Writing 𝑝(𝑧)=𝑚4+12𝑝₁(𝑧)−𝑚4−12𝑝₂(𝑧), (3.25) we use convolution technique, see [4], to have 𝑝(𝑧)∗𝜙_𝑚1(𝑧)𝑧=𝑝(𝑧)+1/𝑚₁𝑧𝑝^′(𝑧)𝑝(𝑧), (3.26) where ∗ denotes convolution (Hadamard product).
Therefore, from (3.23), (3.25), and (3.26), we have for 𝑖=1,2𝑝_𝑖(𝑧)+𝑧𝑞^′_𝑖(𝑧)𝑚₁𝑝(𝑧)∈𝑃𝑝_𝑘,𝛾. (3.27) We now use Lemma 2.1 with 𝜎=𝑚₁,𝛿₁=0,Re(𝑚₁𝑘/(𝑘+1))>𝛾 to obtain 𝑝_𝑖(𝑧)≺𝑝_𝑘,𝛾(𝑧),𝑖=1,2, and consequently 𝑝∈𝑃_𝑚(𝑝_𝑘,𝛾) in 𝐸. This proves that 𝑓∈𝑘−𝑈𝑅_𝑚(𝛾) in 𝐸.

For 𝑘=0, we can improve this result by restricting 𝛼,𝛽 suitably as follows.

Corollary 3.8. Let 𝑓∈𝐵_𝑚(𝛼,𝛽,𝛾),𝛼∈(0,1],𝛽∈[0,1). Then, 𝑓∈𝑅_𝑚(𝛾₀), where 𝛾₀=22𝑚₁𝛾−1+2𝑚₁𝛾−1²+8𝑚₁. (3.28)

Proof. Writing 𝑝_𝑖(𝑧)=(1−𝛾₀)ℎ_𝑖(𝑧)+𝛾₀ and proceeding as in Theorem 3.7, it follows from (3.25) and (3.27) that for 𝑖=1,21−𝛾₀ℎ_𝑖(𝑧)+1−𝛾₀𝑧ℎ^′_𝑖(𝑧)𝑚₁1−𝛾₀ℎ_𝑖(𝑧)+𝑚₁𝛾₀+𝛾₀−𝛾∈𝑃,𝑧∈𝐸. (3.29) Constructing the functional 𝜓(𝑢,𝑣) with 𝑢=ℎ_𝑖(𝑧),𝑣=𝑧ℎ^′_𝑖(𝑧), we note that the first two conditions of Lemma 2.2 are easily verified. For condition (iii), we proceed as follows: Re𝜓𝑖𝑢₂,𝑣₁=𝛾₀+𝑚−𝛾¹𝛾₀1−𝛾₀𝑣₁𝑚²₁𝛾²₀+𝑚²₁1−𝛾₀²𝑢²₂,𝑣₁≤−1+𝑢²₂2,≤2𝛾₀−𝛾𝑚²₁𝛾²₀+𝑚²₁1−𝛾₀²𝑢²₂−𝑚₁𝛾₀1−𝛾₀1+𝑢²₂2𝑚²₁𝛾²₀+𝑚²₁1−𝛾₀²𝑢²₂,=𝐴+𝐵𝑢²₂,2𝐶 (3.30) where 𝐴=2(𝛾₀−𝛾)(𝑚²₁𝛾²₀)+𝑚₁𝛾₀(1−𝛾₀),  𝐵=2(𝛾₀−𝛾)𝑚²₁(1−𝛾₀)²−𝑚₁𝛾₀(1−𝛾₀),𝐶=[𝑚²₁𝛾²₀+𝑚²₁(1−𝛾₀)²𝑢²₂]>0.
The right-hand side of (3.30) is less than or equal to zero if 𝐴≤0 and 𝐵≤0, and condition (iii) is satisfied. Form 𝐴≤0, we obtain 𝛾₀ as given by (3.28), and B≤0 ensures that 𝛾₀∈[0,1).
We now use Lemma 2.2 to have ℎ_𝑖∈𝑃,𝑖=1,2, and this implies 𝑝_𝑖∈𝑃(𝛾₀),𝑖=1,2.
Consequently 𝑝∈𝑃_𝑚(𝛾₀) in 𝐸, and this completes the proof.

We note that 𝑘−𝑈𝐵_𝑚(𝛼,𝛽,𝛾)⊂𝐵_𝑚(𝛼,𝛽,𝛾₁) with 𝛾₁=(𝑘+𝛾)/(1+𝑘). Thus the above result can be restated in a general form as follows.

Corollary 3.9. Let 𝑓∈𝑘−𝑈𝐵_𝑚(𝛼,𝛽,𝛾),𝛼∈(0,1],𝛽∈[0,1). Then 𝑓∈𝑅_𝑚(𝛾₂) in 𝐸, where 𝛾₂=22𝑚𝛾₁−1+2𝑚𝛾₁−1²+8𝑚₁. (3.31)

By taking 𝛼=1,𝛽=0,𝑘=1, we have the following.

Corollary 3.10. Let 𝑓∈1−𝑈𝐵_𝑚(1,0,𝛾). Then 𝑓∈𝑅_𝑚(𝜎₁), where 0≤𝜎₁<1 with 𝜎₁=¹₀2exp1−𝛾𝜋²¹_𝑡1𝑥√log1+𝑥1−√𝑥²𝑑𝑥𝑑𝑡⁻¹. (3.32) When 𝛾=0, one has 𝜎₁=0.73719⋯, and this value of 𝜎₁ is sharp, see [14].

Proof. In Theorem 3.7, with 𝑘=1,𝛼=1,𝛽=0, we have 𝑚₁=1 and therefore, from (3.27), it follows that 𝑝_𝑖(𝑧)+𝑧𝑝^′_𝑖(𝑧)𝑝_𝑖(𝑧)≺𝑝_1,𝛾(𝑧),𝑖=1,2. (3.33) By taking 𝑝_𝑖(𝑧)=𝑧𝑓^′_𝑖(𝑧)/𝑓_𝑖(𝑧), it implies that Re1+𝑧𝑓_𝑖^′′(𝑧)𝑓^′_𝑖(𝑧)>𝑧𝑓_𝑖^′′(𝑧)𝑓^′_𝑖(𝑧)+𝛾,𝑧∈𝐸,𝑖=1,2. (3.34) That is, 𝑓_𝑖∈𝑈𝐶𝑉(𝛾), and it has been proved in [14] that every function in the class 𝑈𝐶𝑉(𝛾) is starlike of order 𝜎₁ where this order is exact and is given by (3.32).
Now, using the argument given in Theorem 3.7, we have 𝑝(𝑧)=(𝑧𝑓(𝑧)/𝑓(𝑧))∈𝑃_𝑚(𝜎₁),𝑧∈𝐸, and the proof is complete.

We now prove the following.

Theorem 3.11. Let 0<𝛼₂<𝛼₁≤1,𝛽∈[0,1). Then 𝑘−𝑈𝐵_𝑚𝛼₁,𝛽,𝛾⊂𝑘−𝑈𝐵_𝑚𝛼₂,𝛽,𝛾,𝑧∈𝐸. (3.35)

Proof. Let 𝑓∈𝑘−𝑈𝐵_𝑚(𝛼₁,𝛽,𝛾). Then, for 𝑧∈𝐸1−𝛼₁𝑧𝑓^′(𝑧)𝑓(𝑧)+𝛼₁1−𝛽1−𝛽+𝑧𝑓^′′(𝑧)𝑓^′(𝑧)=𝐻₁(𝑧),𝐻₁∈𝑃_𝑚𝑝_𝑘,𝛾. (3.36) Also, by Theorem 3.7, we have 𝑧𝑓^′(𝑧)𝑓(𝑧)=𝐻₂(𝑧),𝐻₂∈𝑃_𝑚𝑝_𝑘,𝛾. (3.37) Now 1−𝛼₂𝑧𝑓^′(𝑧)𝑓(𝑧)+𝛼₂1−𝛽1−𝛽+𝑧𝑓^′′(𝑧)𝑓^′(𝑧)=𝛼₂𝛼₁𝐻₁(𝑧)+𝛼1−²𝛼₁𝐻₂(𝑧)=𝐻(𝑧), (3.38) and since 𝑃_𝑚(𝑝_𝑘,𝛾) is convex set, 𝐻∈𝑃_𝑚(𝑝_𝑘,𝛾) in 𝐸. This proves the result.

Let 𝑀(𝑟)=max_{0≤𝜃≤2𝜋}|𝑓(𝑟𝑒^𝑖𝜃)|,𝐿(𝑟) the length of the curve 𝐶,𝐶=𝑓(𝑟𝑒^𝑖𝜃),0≤𝜃≤2𝜋, and 𝐴(𝑟) the area of the region bounded by 𝐶.

We will now study the arc length problem for the class 𝐵_𝑚(𝛼,𝛽,𝛾) as follows.

Theorem 3.12. For 𝑘=0, let 𝑓∈B_𝑚(𝛼,𝛽,𝛾). Then, for 0<𝑟<1,(i)𝐿(𝑟)≤(𝜋𝑀(𝑟)/𝛼(1−𝛽))[𝑚{(1−𝛾)+(𝛼−1)(1−𝛾₀)}+2𝛼𝛽],𝛼≥2, where 𝛾₀ is given by (3.28).(ii)𝐿(𝑟)≤(𝜋𝑀(𝑟)/𝛼)[𝑚(2−𝛼)+2𝛼𝛽/(1−𝛽)],0<𝛼<2.

Proof. We prove (i), and proof of (ii) follows on similar lines.
Solving (1.7) for 𝑓′, we obtain a formal representation as 𝑧𝑓^′(𝑧)=𝑔(𝑧)^{(1−𝛽)/𝛼}𝑧^𝛽𝑓(𝑧)^{−(1−𝛼)(1−𝛽)/𝛼},(𝛼≠0), (3.39) with 𝑧=𝑟𝑒^𝑖𝜃. 𝐿(𝑟)=₀^2𝜋𝑧𝑓^′(𝑧)𝑑𝜃=₀^2𝜋𝑧𝑓^′(𝑧)𝑒^{−𝑖arg(𝑧𝑓′(𝑧))}𝑑𝜃. (3.40) Integration by parts gives us 𝐿(𝑟)=₀^2𝜋𝑓(𝑧)𝑒^{−𝑖arg(𝑧𝑓′(𝑧))}Re𝑧𝑓^′(𝑧)^′𝑓^′(𝑧)≤𝑀(𝑟)𝑑𝜃𝛼₀^2𝜋Re𝐽𝛼,𝛽,𝑓(𝑧)+(𝛼−1)𝑧𝑓^′(𝑧)𝑓(𝑧)+𝛼𝛽1−𝛽𝑑𝜃, (3.41) where we have used Logarithmic differentiation of (3.39).
Now, from Corollary 3.8, it follows that 𝑧𝑓′/𝑓∈𝑃_𝑚(𝛾₀), where 𝛾₀ is given by (3.28). Also, since 𝑓∈𝐵_𝑚(𝛼,𝛽,𝛾), we have Re𝐽𝛼,𝛽,𝑓(₀^2𝜋𝑧)𝑑𝜃≤𝑚1−𝛾𝜋. (3.42) Using these observations in (3.41), we prove part (i).

Remark 3.13. In Theorem 3.12, we have, for 𝛼>0, 𝐿(𝑟)=𝑂(1)𝑀(𝑟), (3.43) where 𝑂(1) is a constant depending only on 𝛼,𝛽,𝑚, and 𝛾. This can be improved by using the univalence criterion of 𝑓∈𝐵_𝑚(𝛼,𝛽,𝛾) as follows.

Corollary 3.14. Let 𝑓∈𝐵_𝑚(𝛼,𝛽,𝛾),𝛼>0,𝛽∈[0,1), and 𝑚≤2{𝛼(1+2𝛽)/(1−𝛽)(1−𝛾)+1}. Then 𝐿(𝑟)=𝑂(1)𝐴(𝑟)1log1−𝑟². (3.44)

Proof. In view of the univalence of 𝑓 by Theorem 3.1, we have 𝑀(𝑟)≤(4/𝑟)𝑀(𝑟²), see [9], and 𝑀𝑟²=^∞_𝑛=1𝑎_𝑛𝑟^2𝑛=^∞_𝑛=1𝑛^1/2𝑎_𝑛𝑟^𝑛𝑛^−1/2𝑟^𝑛. (3.45) Making now use of the area theorem and the Schwarz inequality, we obtain the required result.

Corollary 3.15. Let 𝑓∶𝑓(𝑧)=𝑧+∑^∞_𝑛=2𝑎_𝑛𝑧^𝑛 satisfy the condition of Theorem 3.12. Then 𝑎_𝑛(1)𝑀))=𝑂((𝑛−1/𝑛𝑛,(𝑛⟶∞). (3.46)

Proof is immediately since by Cauchy’s Theorem.

We can write 𝑛𝑎_𝑛=12𝜋𝑟^𝑛₀^2𝜋𝑧𝑓^′(𝑧)𝑒^{−𝑖𝑛𝜃}=1𝑑𝜃2𝜋𝑟^𝑛𝐿(𝑟). (3.47) Now taking 𝑟=(1−1/𝑛) in Theorem 3.12, we prove the result.

We study the arc length problem and corresponding rate of growth of coefficient for the class 𝑘−𝑈𝐵_𝑚(𝛼,𝛽,𝛾).

Theorem 3.16. Let, for 𝛼≥1,−1/2≤𝛽<1,𝑓∈𝑘−𝑈𝐵_𝑚(𝛼,𝛽,𝛾). Then for 𝑚>2(1−𝑘₁)/𝑘₁, one has 𝐿(𝑟)=𝑂(1)𝑀(𝑟)^1−𝑚111−𝑟^{{𝑘1(𝑚/2+1)−1}},(𝑟⟶1), (3.48) where 𝑂(1) depends only on 𝑚₁,𝑘₁, and 𝑚 and 𝑚₁,𝑘₁ are as given by (3.4).

Proof. From (1.7), we can write 𝑧𝑓^′(𝑧)𝑓(𝑧)^{(1−𝛼)(1−𝛽)/𝛼}=𝑔(𝑧)^{(1−𝛽)/𝛼},𝑔∈𝑘−𝑅_𝑚𝛾. (3.49) Now, for 𝑧=𝑟𝑒^𝑖𝜃,𝛼≥1, 𝐿(𝑟)=_|𝑧|=𝑟𝑓^′(𝑧)|𝑑𝑧|=₀^2𝜋𝑧𝑓^′(𝑧)𝑒^{−𝑖arg𝑧𝑓′(𝑧)}≤𝑑𝜃_02𝜋𝑔(𝑧)^{(1−𝛽)/𝛼}𝑧^𝛽𝑓(𝑧)^{−(1−𝛼)(1−𝛽)/𝛼}𝑒^{−𝑖arg𝑧𝑓′}𝑑𝜃. (3.50) Since 𝑔∈𝑘−𝑈𝑅_𝑚(𝛾), we use a result proved in [4] and write 𝑔(𝑧)=𝑔₁(𝑧)^𝑚/4+1/2𝑔₂(𝑧)^𝑚/4−1/2,𝑔₁,𝑔₂∈𝑘−𝑈𝑅₂𝛾. (3.51) Since 𝑘−𝑈𝑅₂(𝛾)⊂𝑆^∗((𝑘+𝛾)/(1+𝑘)), see [4].
We can write 𝑔_𝑖(𝑧)𝑠=𝑧^𝑖(𝑧)𝑧^{1−((𝑘+𝛾)/(1+𝑘))},𝑠_𝑖∈𝑆^∗,𝑠=𝑧^𝑖(𝑧)𝑧^{(1−𝛾)/(1+𝑘)},𝑖=1,2. (3.52) Using (3.51), (3.52), and distortion results for the class 𝑆^∗ of starlike functions, we obtain from (3.50) 𝐿(𝑟)≤𝐶₁(𝑀(𝑟))^1−𝑚1₀^2𝜋𝑟1−𝑟𝑒^{𝑖𝜃2𝑘1(𝑚/4+1/2)}(1𝑀(𝑟𝑑𝜃,=𝑂)())^1−𝑚111−𝑟^{𝑘1(𝑚/4+1/2)−1},(𝑟⟶1), (3.53) where 𝑘₁(𝑚/2+1)>1.

Remark 3.17. (i)For the case 0<𝛼<1, we can solve the arc length problem in a similar manner. We define 𝑚(𝑟)=min_{0≤𝜃≤2𝜋}|𝑓(𝑧)| and proceed as before to obtain 𝐿(𝑟)=𝑂(1)(𝑚(𝑟))^1−𝑚111−𝑟^{𝑘1(𝑚/2+1)−1},(𝑟⟶1). (3.54)(ii)For 𝛼<0, we have 𝐿(𝑟)=𝑂(1)(𝑀(𝑟))^1−𝑚111−𝑟^{𝑘1(𝑚/2−1)−1}1,for𝑚>𝑘₁+1. (3.55)

We can derive the following result of order of growth of coefficients from Theorem 3.16 with the same method used before.

Corollary 3.18. Let 𝑓∶𝑓(𝑧)=𝑧+∑^∞_𝑛=2𝑎_𝑛𝑧^𝑛∈𝑘−𝑈𝐵_𝑚(𝛼,𝛽,𝛾),𝛽∈[0,1),𝛾∈[0,1),𝛼≥1. Then 𝑎_𝑛=𝑂(1)𝑀^(1−𝑚1)𝑛^{{𝑘1(𝑚/2+1)−2}},(𝑛⟶∞), (3.56) where 𝑂(1) depends only on 𝑚,𝑚₁,𝑘₁ and 𝛾.
The exponent {𝑘₁(𝑚/2+1)−2} is best possible for 𝑘=0,𝛾=0. The extremal function satisfies the equation 𝑓(𝑧)𝑧^1−𝛼𝑓^′(𝑧)^{𝛼/(1−𝛽)}=𝑔₀(𝑧)𝑧, (3.57) with 𝑔₀(𝑧)=1(𝑚+2)1+𝑧1−𝑧^𝑚/2+1−1∈𝑅_𝑚. (3.58)

Acknowledgment

The authors are grateful to Dr. S. M. Junaid Zaidi, Rector, COMSATS Institute of Information Technology, Islamabad, Pakistan, for providing excellent research facilities and environment.