Notes on a New Class of Univalent Starlike Functions with Respect to a Boundary Point

Abstract

This article presents a newly defined subclass of univalent functions that are starlike with respect to a boundary point, closely related to the Robertson class and specifically associated with a vertical strip domain. Additionally, this study derives generalized coefficient estimates for these classes, as well as for the Robertson class linked to the Nephroid domain and the Lemniscate of Bernoulli.

1. Introduction

Let us define $\mathcal{H}$ as the set of all analytic functions that are defined within the open unit disk $\mathbb{D}=\left\{u\in \mathbb{C}:\left|u\right|<1\right\}$. Let $\mathcal{A}$ represent the collection of all analytic functions $f\in \mathcal{H}$ that are normalized by the conditions $f\left(0\right)=0$ and $f^{\prime }\left(0\right)=1$. Hence, the power series expansion of f is of the form

$$f\left(u\right)=u+\sum _{n=2}^{\infty }{a}_n{u}^n, u\in \mathbb{D}.$$

(1)

Further, let $\mathcal{S}$ be the subclass of $\mathcal{A}$ whose elements are univalent in $\mathbb{D}$. Let the class consisting of all analytic functions that are starlike of the order $\lambda ,0\le \lambda <1$ be analytically defined as

$${\mathcal{S}}^{*}\left(\lambda \right)=\left\{f\in \mathcal{A}:\Re \left({\displaystyle \frac{u{f}^{\prime }\left(u\right)}{f\left(u\right)}}\right)>\lambda , u\in \mathbb{D}\right\}.$$

A function $f\in \mathcal{A}$ is considered subordinate to a function $g\in \mathcal{A}$, denoted as $f\left(u\right)\prec g\left(u\right)$, if there exists a function $h\in \mathcal{H}$, referred to as a Schwarz function, satisfying $h\left(0\right)=0$ and $\left|h\right(u\left)\right|\le \left|u\right|$ such that $f\left(u\right)=g\left(h\right(u\left)\right)$ for all $u\in \mathbb{D}$. Suppose that g is univalent in $\mathbb{D}$. Then f is subordinate to g if and only if $f\left(0\right)=g\left(0\right)$ and $f\left(\mathbb{D}\right)\subseteq g\left(\mathbb{D}\right)$. Let $S^{*}(A,B)$ be the class of Janowski starlike functions, defined as

$$S^{*}(A,B)=\left\{f\in \mathcal{H}:{\displaystyle \frac{u{f}^{\prime }\left(u\right)}{f\left(u\right)}}\prec {\displaystyle \frac{1+Au}{1+Bu}},-1\le B<A\le 1, u\in \mathbb{D}\right\}.$$

Kuroki and Owa [1] introduced the class $S(\eta ,\beta )$, defined by

$$S(\eta ,\beta )=\left\{f\in \mathcal{H}:\eta <\Re \left({\displaystyle \frac{u{f}^{\prime }\left(u\right)}{f\left(u\right)}}\right)<\beta ,u\in \mathbb{D}\right\}$$

for some real number $\eta (\eta <1)$ and some real number $\beta (\beta >1)$. Also, they found the necessary and sufficient condition for $S(\eta ,\beta )$ and examined the coefficient estimates for $S(\eta ,\beta )$. For given $\eta$ and $\beta$ ($0\le \eta <1<\beta$), the function $f\left(u\right)\in S(\eta ,\beta )$ if and only if f satisfies the following two subordination equations:

$${\displaystyle \frac{u{f}^{\prime }\left(u\right)}{f\left(u\right)}}\prec {\displaystyle \frac{1+(1-2\eta )u}{1-u}}\hspace{1em}\hspace{1em}and\hspace{1em}\hspace{1em}{\displaystyle \frac{u{f}^{\prime }\left(u\right)}{f\left(u\right)}}\prec {\displaystyle \frac{1+(1-2\beta )u}{1-u}}.$$

If we let ${\varphi }_{\eta }\left(u\right):={\displaystyle \frac{1+(1-2\eta )u}{1-u}}, 0\le \eta <1$, ${\varphi }_{\eta }\left(\mathbb{D}\right)$ is a domain that lies in the right half-plane with a real part greater than $\eta$, and the function ${\varphi }_{\beta }\left(u\right):={\displaystyle \frac{1+(1-2\beta )u}{1-u}}, \beta >1$, ${\varphi }_{\beta }\left(u\right)$, maps the unit disk onto a domain in the right half-plane with a real part smaller than $\beta$.

Sim and Kwon [2] proved the inclusion relation between Janowski starlike functions $S^{*}(A,B)$ and the class $S(\eta ,\beta )$.

Let $\mathcal{P}$ be a subclass of $\mathcal{H}$ consisting of all functions of the form

$$g\left(u\right)=1+\sum _{n=1}^{\infty }{d}_n{u}^n,g\left(0\right)=1$$

(2)

such that $\Re \left(g\right(u\left)\right)>0$. This class is also known as the Carathéodory class. Let ${\mathcal{P}}^{*}\left(1\right)$ be a subclass of $\mathcal{P}$ of all functions $h\left(u\right)$ such that $h\left(0\right)=1,h^{\prime }\left(0\right)>0,h$ is univalent in $\mathbb{D},$ and $h\left(\mathbb{D}\right)$ is a set consisting of symmeric functions with respect to a real axis and starlike with respect to 1.

While the class ${\mathcal{S}}^{*}\left(\lambda \right)$ had been widely examined by several authors over a long period, not much attention had been given to the class of analytic functions $g\left(u\right),$ $u\in \mathbb{D}$ that map the open unit disk $\mathbb{D}$ onto a starlike domain with respect to a boundary point. Then, Robertson [3] gave the constructive idea for starlike functions with respect to a boundary point. For this, Robertson introduced two classes, $\mathcal{G}$ and ${\mathcal{G}}^{*}$, defined as follows:

The class ${\mathcal{G}}^{*}$ contains all the analytic functions in $\mathbb{D}$, normalized so that $g\left(0\right)=1$ and $g\left(1\right)={lim}_{r\to 1^{-}}g\left(r\right)=0$ and for some real $\lambda$, $\Re \left[e^{i\lambda }g\left(u\right)\right]>0,u\in \mathbb{D}$. Also let $g\left(u\right)$ map $\mathbb{D}$ univalently on a domain starlike with respect to $g\left(1\right)$, and the class $\mathcal{G}$ is defined as the family of functions g of the form (2), which are analytic and non-vanishing in $\mathbb{D}$ such that

$$\Re \left\{2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+{\displaystyle \frac{1+u}{1-u}}\right\}>0, u\in \mathbb{D}.$$

Robertson also proved that the class $\mathcal{G}$ is a subset of the class ${\mathcal{G}}^{*}$ and conjectured that the class ${\mathcal{G}}^{*}$ is a subset of the class $\mathcal{G}$. Later this conjecture was settled by Lyzzaik [4] so that ${\mathcal{G}}^{\ast }=\mathcal{G}.$ Also, it should be noted that the class $\mathcal{G}$ is closely connected with the class ${\mathcal{S}}^{\ast }\left({\displaystyle \frac{1}{2}}\right)$. In the same article, Robertson proved that if $g\in \mathcal{G}$ and $g\ne I$, the identity function, then g is close to convex and univalent in $\mathbb{D}$. In [5], Todorov established a structural formula and coefficient estimates for the class $\mathcal{G}$ with a functional ${\displaystyle \frac{g\left(u\right)}{1-u}}$ for $u\in \mathbb{D}$.

Silverman and Silvia [6] and Obradović and Owa [7] independently defined another class, $\mathcal{G}\left(\beta \right)$, that is contained in ${\mathcal{G}}^{*}$. The class $\mathcal{G}\left(\beta \right)$, $0\le \beta <1$, is defined as follows:

$$\mathcal{G}\left(\beta \right)=\left\{g\in \mathcal{H},g\ne 0:\Re \left(u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+(1-\beta ){\displaystyle \frac{1+u}{1-u}}\right)>0,u\in \mathbb{D}\right\}.$$

Note that $\mathcal{G}\left({\displaystyle \frac{1}{2}}\right)={\mathcal{G}}^{*}$

Recently, Dhurai et al. [8] introduced and investigated the new class ${\mathcal{G}}_m$, which is closely related to the Robertson class $\mathcal{G}$, containing the family of all analytic functions g of the form (2) that satisfy

$$\Re \left\{2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+{\left({\displaystyle \frac{1+u}{1-u}}\right)}^m\right\}>0, u\in \mathbb{D},0<m\le 2$$

and observed that ${\mathcal{G}}_1=\mathcal{G}.$

Recently, Kavitha [9] defined a class which is a Robertson class associated with the Nephroid domain ${\mathcal{Q}}_N\left(u\right)=1+u-{\displaystyle \frac{u^3}{3}}$, denoted by ${\mathcal{G}}_N$, with the family of functions g of the form (2) in ${\mathcal{G}}_N$ satisfying the following inequality:

$$\Re \left\{2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+{\mathcal{Q}}_N\left(u\right)\right\}>0, u\in \mathbb{D}.$$

(3)

The author of [9] explored the representation theorem and growth and distortion theorem and investigated various implications related to differential subordination.

In [10], the class $\mathcal{G}\left(E\right),E>1$, consisting of all analytic and non-vanishing functions g of the form (2), such that

$$\Re \left(2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+u{\displaystyle \frac{P^{\prime }(u;E)}{P(u;E)}}\right)>0, u\in \mathbb{D},$$

which is a function closely related to the class $\mathcal{G}$, was introduced by Jakubowski [10]. Here,

$$P(u;E)={\displaystyle \frac{4u}{{\left(\sqrt{{(1-u)}^2+\frac{4u}{E}}+1-u\right)}^2}}, u\in \mathbb{D}$$

is the Pick function.

For $-1<A\le 1$ and $-A<B\le 1$, Jakubowski and Włodarczyk [11] defined the class $\mathcal{G}(A,B)$ of all g of the form (2), satisfying ${\displaystyle \Re \left(2u\frac{g^{\prime }\left(u\right)}{g\left(u\right)}+\frac{1+Au}{1-Bu}\right)>0, u\in \mathbb{D}}$ (also see the work of Sivasubramanian [12]).

Let us define the functions $p\left(u\right)$ and $p_1\left(u\right)$ as follows:

$$p\left(u\right):=2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+{\displaystyle \frac{1+u}{1-u}}$$

and

$$p_1\left(u\right):=2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+{\left({\displaystyle \frac{1+u}{1-u}}\right)}^m, 0<m\le 2.$$

By using Ma and Minda’s idea, Mohd and Darus, in [13], defined the class ${\mathcal{S}}^{*}\left(\varphi \right)$ containing all $g\in \mathcal{H}$ of the form (2) such that

$${\mathcal{S}}^{*}\left(\varphi \right):=\left\{f\in \mathcal{H}:p\left(u\right)\prec \varphi \left(u\right)\right\}.$$

Note that the function $\sqrt{1+u}$ maps the open unit disk $\mathbb{D}$ onto a Bernoulli Lemniscate, ${(x^2+y^2)}^2-2(x^2-y^2)=0$, denoted by ${\varphi }_L\left(u\right):=\sqrt{1+u}$, and ${\varphi }_L\left(u\right)$ is symmetric with respect to a real axis. Also, the function $u+\sqrt{1+u^2}$ maps $\mathbb{D}$ onto a ’crescent’-shaped domain bounded by two circular arcs:

$${\gamma }_1\subset \mathbb{T}(1,\sqrt{2})=\{z\in \mathbb{C}:|z-1|=\sqrt{2}\}$$

and

$${\gamma }_2\subset \mathbb{T}(1,\sqrt{2})=\{z\in \mathbb{C}:|z+1|=\sqrt{2}\}.$$

Both the circles $\mathbb{T}(1,\sqrt{2})$ and $\mathbb{T}(-1,\sqrt{2})$ pass through the points $z=i$ and $z=-i$ and also the arcs ${\gamma }_1$ and ${\gamma }_2$ lying in the closed right half-plane with common end points at i and $-i.$ We denote this function by ${\varphi }_C\left(u\right):=u+\sqrt{1+u^2}$. The univalence of ${\varphi }_C\left(u\right)$ has been presented in [14]. Gruszecki et al. [15] addressed two classes, which related to the Robertson formula associated with the crescent-shaped domain and Lemniscate of Bernoulli. The classes ${\mathcal{G}}_C$ and ${\mathcal{G}}_L$ contain the family of functions g of the form (2) in ${\mathcal{G}}_C$ and ${\mathcal{G}}_L$ satisfying the subordinations

$${\mathcal{G}}_C=\left\{g\in \mathcal{H}:p\left(u\right)\prec {\varphi }_C\left(u\right), u\in \mathbb{D}\right\}$$

and

$${\mathcal{G}}_L=\left\{g\in \mathcal{H}:p\left(u\right)\prec {\varphi }_L\left(u\right), u\in \mathbb{D}\right\}$$

(4)

respectively. Also, Lecko [16] et al. introduced a class of the Robertson type associated with the exponential fucnction ${\varphi }_e\left(u\right)=e^u$; the class is denoted by ${\mathcal{G}}_e$, consisting of a family of functions g of the form (2) satsifying

$${\mathcal{G}}_e=\left\{g\in \mathcal{H}:p\left(u\right)\prec {\varphi }_e\left(u\right), u\in \mathbb{D}\right\}.$$

Motivated by the above-mentioned works, in the current article we are interested in the function

$${\varphi }_{\eta ,\beta }\left(u\right)=1+{\displaystyle \frac{\beta -\eta }{\pi }}ilog\left({\displaystyle \frac{1-e^{2\pi i\frac{1-\eta }{\beta -\eta }} u}{1-u}}\right), u\in \mathbb{D}$$

for some real number $\eta (\eta <1)$ and for some real $\beta (\beta >1).$ This function, ${\varphi }_{\eta ,\beta }\left(u\right)$, maps $\mathbb{D}$ onto the strip domain between $\eta$ and $\beta$. We define

$${\mathcal{P}}_{\eta ,\beta }:=\left\{p\in \mathcal{H}:p\prec {\varphi }_{\eta ,\beta }\left(u\right)\right\}$$

which implies that

$${\mathcal{P}}_{\eta ,\beta }:=\left\{p\in \mathcal{H}:\eta <\Re \left(p\right)<\beta \right\}.$$

In this article, we are interested in introducing two new subclasses that are closely related to starlike functions with respect to a boundary point in the vertical strip domain. For these new classes, we find general coefficient estimates. The classes are defined as follows.

2. Main Results

Definition 1.

Let $\mathcal{G}(\eta ,\beta )$ consist of the class function $g\in \mathcal{H}$ of the form (2) that satisfies the inequality

$$\eta <\Re \left\{2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+{\displaystyle \frac{1+u}{1-u}}\right\}<\beta , u\in \mathbb{D},$$

(5)

for some real number $\eta (\eta <1)$ and some real number $\beta (\beta >1)$.

Also notice that Inequality (5) is well defined, as the function

$$p\left(u\right):=2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+{\displaystyle \frac{1+u}{1-u}}$$

is analytic in $\mathbb{D}$.

Remark 1.

Let $g\in \mathcal{G}(\eta ,\beta )$. If $\eta =0$, then $g\left(u\right)$ is starlike with respect to the boundary point, which implies that $g\left(u\right)$ is univalent in $\mathbb{D}.$

Definition 2.

Let ${\mathcal{G}}_m(\eta ,\beta )$ consist of the functions $g\in \mathcal{H}$ of the form (2) that satisfy the inequality

$$\eta <\Re \left\{2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+{\left({\displaystyle \frac{1+u}{1-u}}\right)}^m\right\}<\beta , 0<m\le 2,u\in \mathbb{D},$$

(6)

for some real number $\eta (\eta <1)$ and some real number $\beta (\beta >1).$ Inequality (6) is well defined, as the function

$$p_1\left(u\right):=2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+{\left({\displaystyle \frac{1+u}{1-u}}\right)}^m, 0<m\le 2$$

is analytic in $\mathbb{D}$.

Lemma 1.

Let $g\in \mathcal{H}$. Then $g\left(u\right)\in \mathcal{G}(\eta ,\beta )$ if and only if

$$p\left(u\right)\prec {\varphi }_{\eta ,\beta }\left(u\right), u\in \mathbb{D}$$

(7)

where $\eta <1$ and $\beta >1.$

Proof. 

Let us consider the function $G\left(u\right)$ as

$$G\left(u\right)=1+{\displaystyle \frac{\beta -\eta }{\pi }}ilog\left({\displaystyle \frac{1-e^{2\pi i\frac{1-\eta }{\beta -\eta }} u}{1-u}}\right), u\in \mathbb{D}$$

(8)

where $\eta <1$ and $\beta >1.$

Let ${\displaystyle a=\frac{1-e^{2\pi i\frac{1-\eta }{\beta -\eta }} u}{1-u}}$, since ${\displaystyle \frac{1-\eta }{\beta -\eta }}\in \mathbb{R}$ and $\left|a\right|=1,$ so $a\in \partial \mathbb{D}.$ Also, ${\displaystyle \frac{1-au}{1-u}}$ is analytic and non-zero in $\mathbb{D}.$ Equation (8) implies that $\mathcal{G}$ rotates the horizontal strip domain $log\left({\displaystyle \frac{1-au}{1-u}}\right)$ by ${90}^{\circ }$ and scales it by ${\displaystyle \frac{\beta -\eta }{\pi }}$ and translates it by 1.

Therefore, the function $G\left(u\right)$ is analytic and univalent in $\mathbb{D}$ with $G\left(0\right)=1$. Also, $G\left(u\right)$ maps the open unit disk $\mathbb{D}$ onto a strip domain w with $\eta <\Re \left(w\right)<\beta .$ With the information and Remark 1, we can conclude that Subordination (7) is equivalent to Inequality (5), which completes the declaration of Lemma 1. □

Lemma 2.

Let $g\in \mathcal{H}$. Then $g\left(u\right)\in {\mathcal{G}}_m(\eta ,\beta )$ if and only if

$$p_1\left(u\right)\prec {\varphi }_{\eta ,\beta }\left(u\right), u\in \mathbb{D}$$

(9)

where $\eta <1$ and $\beta >1.$

Proof. 

The proof of this lemma is similar to the proof of Lemma 1 and hence we omit the details. □

Now we prove that the classes $\mathcal{G}(\eta ,\beta )$ and ${\mathcal{G}}_m(\eta ,\beta )$ are non-empty.

Example 1.

Let us consider the function $g\left(u\right)$ given by

$$g\left(u\right)=(1-u)exp\left\{{\displaystyle \frac{\beta -\eta }{2\pi }}i{\int }_0^u{\displaystyle \frac{1}{t}}log\left({\displaystyle \frac{1-e^{2\pi i\frac{1-\eta }{\beta -\eta }} t}{1-t}}\right)dt\right\}, u\in \mathbb{D}$$

(10)

with $\eta <1$ and $\beta >1$. Then, we have

$$2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+{\displaystyle \frac{1+u}{1-u}}=1+{\displaystyle \frac{\beta -\eta }{\pi }}ilog\left({\displaystyle \frac{1-e^{2\pi i\frac{1-\eta }{\beta -\eta }} u}{1-u}}\right), u\in \mathbb{D}.$$

According to the proof of Lemma 1, it is evident that the function $g\left(u\right)$ given by (10) satisfies the inequality (5), which implies that $g\left(u\right)\in \mathcal{G}(\eta ,\beta )$.

Example 2.

Let us consider the function $g\left(u\right)$ given by

$$g\left(u\right)=exp\left({\displaystyle \frac{\beta -\eta }{2\pi }}i{\int }_0^u{\displaystyle \frac{1}{t}}log\left({\displaystyle \frac{1-e^{2\pi i\frac{1-\eta }{\beta -\eta }} t}{1-t}}\right)dt-m\left(u+{\displaystyle \frac{u^2}{2}}+{\displaystyle \frac{2m^2+1}{3}}{\displaystyle \frac{u^3}{3}}+{\displaystyle \frac{m^3+2m}{4}}{\displaystyle \frac{u^4}{4}}+\dots \right)\right),$$

(11)

$u\in \mathbb{D}$ with $\eta <1$ and $\beta >1$. Then, we have

$$2{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+{\displaystyle \frac{1}{u}}+2m\left(1+u+{\displaystyle \frac{2m^2+1}{3}}{u}^2+{\displaystyle \frac{m^3+2m}{4}}{u}^3+\dots \right)={\displaystyle \frac{1}{u}}+{\displaystyle \frac{\beta -\eta }{u\pi }}ilog\left({\displaystyle \frac{1-e^{2\pi i\frac{1-\eta }{\beta -\eta }} u}{1-u}}\right), u\in \mathbb{D}$$

implies that,

$$2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+{\left({\displaystyle \frac{1+u}{1-u}}\right)}^m=1+{\displaystyle \frac{\beta -\eta }{\pi }}ilog\left({\displaystyle \frac{1-e^{2\pi i\frac{1-\eta }{\beta -\eta }} u}{1-u}}\right), u\in \mathbb{D}.$$

According to the proof of Lemma 2, the function $g\left(u\right)$ given in (11) satisfies the inequality (6), which implies that $g\left(u\right)\in {\mathcal{G}}_m(\eta ,\beta )$.

Lemma 3

(Miller and Mocanu [17]). Let Ξ be a set in the complex plane $\mathbb{C}$ and let b be a complex number such that $\Re \left(b\right)>0$. Suppose that $\psi :{\mathbb{C}}^2\times \mathbb{D}\to \mathbb{C}$ satisfies the condition

$$\psi \left(i\rho ,\sigma ;u\right)\notin \mathsf{\Xi }$$

for all real ${\displaystyle \rho ,\sigma \le -\frac{{|b-i\rho |}^2}{2\Re \left(b\right)}}$ and all $u\in \mathbb{D}.$ If the function $q\left(u\right)$ defined by $q\left(u\right)=b+b_{1}u+b_2{u}^2+\dots$ is analytic in $\mathbb{D}$ and if

$$\psi \left(q\left(u\right),u{q}^{\prime }\left(u\right);u\right)\in \mathsf{\Xi }$$

then $\Re \left(q\right(u\left)\right)>0$ in $\mathbb{D}$.

Theorem 1.

Let $g\in \mathcal{H},{\displaystyle \frac{1}{2}}\le \eta <1$ and $\Re \left\{p\left(u\right)\right\}>\eta$. Then,

$$\Re \left\{{\displaystyle \frac{g\left(u\right)}{{(1-u)}^2}}\right\}>\delta \left(\eta \right):={\displaystyle \frac{1}{3-2\eta }}, u\in \mathbb{D}.$$

(12)

Proof. 

We denote $\delta \left(\eta \right):=\delta$ and note that ${\displaystyle \frac{1}{2}}\le \delta <1$ for ${\displaystyle \frac{1}{2}}\le \eta <1$. Let us define $q\left(u\right)$ by

$$q\left(u\right)={\displaystyle \frac{1}{1-\delta }}\left({\displaystyle \frac{g\left(u\right)}{{(1-u)}^2}}-\delta \right).$$

Then q is analytic in $\mathbb{D},$$q\left(0\right)=1$ and

$$2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+{\displaystyle \frac{1+u}{1-u}}=1+{\displaystyle \frac{(1-\delta )u{q}^{\prime }\left(u\right)}{\delta +(1-\delta )q\left(u\right)}}=\psi \left(q\left(u\right),u{q}^{\prime }\left(u\right)\right)$$

where

$$\psi (r,s)=1+{\displaystyle \frac{(1-\delta )s}{\delta +(1-\delta )r}}.$$

(13)

Also,

$$\left\{\psi \left(q\left(u\right),u{q}^{\prime }\left(u\right)\right):u\in \mathbb{D}\right\}\subset \left\{w\in \mathbb{C}:\Re \left(w\right)>\eta \right\}:=h_{\eta }.$$

Now, for every real $\rho ,$$\sigma \le -{\displaystyle \frac{1}{2}}\left(1+{\rho }^2\right)$,

$$\Re \left(\psi \left(i\rho ,\sigma \right)\right)=\Re \left(1+{\displaystyle \frac{(1-\delta )\sigma }{\delta +(1-\delta )i\rho }}\right)$$

$$=1+{\displaystyle \frac{\delta (1-\delta )\sigma }{{\delta }^2+{(1-\delta )}^2{\rho }^2}}$$

$$\le 1-{\displaystyle \frac{1}{2}}\delta (1-\delta ){\displaystyle \frac{1+{\rho }^2}{{\delta }^2+{(1-\delta )}^2{\rho }^2}}.$$

Now, let us take

$$h\left(\rho \right)={\displaystyle \frac{1+{\rho }^2}{{\delta }^2+{(1-\delta )}^2{\rho }^2}}.$$

(14)

Then,

$$h^{\prime }\left(\rho \right)={\displaystyle \frac{2(2\delta -1)\rho }{{\left({\delta }^2+{(1-\delta )}^2{\rho }^2\right)}^2}}.$$

Hence, $h^{\prime }\left(0\right)=0$ occurs only at $\rho =0$ and h satisfies

$$h\left(0\right)={\displaystyle \frac{1}{{\delta }^2}}$$

and

$$\underset{\rho \to \infty }{lim}h\left(\rho \right)={\displaystyle \frac{1}{{(1-\delta )}^2}}.$$

Since, ${\displaystyle \frac{1}{2}}\le \delta <1$, we have

$${\displaystyle \frac{1}{{\delta }^2}}\le h\left(\rho \right)<{\displaystyle \frac{1}{{(1-\delta )}^2}}$$

and hence, we get

$$\Re \left(\psi \left(i\rho ,\sigma \right)\right)\le 1-{\displaystyle \frac{1}{2}}\delta (1-\delta ){\displaystyle \frac{1}{{\delta }^2}}={\displaystyle \frac{3\delta -1}{2\delta }}=\eta .$$

This shows that $\Re \left(\psi \left(i\rho ,\sigma \right)\right)\notin h_{\eta }$. By Lemma 3, we obtain $\Re \left(q\right(u\left)\right)>0$ in $\mathbb{D}$, and this implies that Inequality (12) is valid. This completes the proof of Theorem 1. □

Theorem 2.

Let $g\in \mathcal{H},1<\beta <{\displaystyle \frac{3}{2}}$ and $\Re \left\{p\left(u\right)\right\}<\beta$. Then

$$\Re \left\{{\displaystyle \frac{g\left(u\right)}{{(1-u)}^2}}\right\}<\zeta \left(\beta \right):={\displaystyle \frac{1}{3-2\beta }}, u\in \mathbb{D}.$$

(15)

Proof. 

Denote $\zeta \left(\beta \right):=\zeta ={\displaystyle \frac{1}{3-2\beta }}>1$ for $\beta >1$. Let us define $q\left(u\right)$ by

$$q\left(u\right)={\displaystyle \frac{1}{1-\zeta }}\left({\displaystyle \frac{g\left(u\right)}{{(1-u)}^2}}-\zeta \right).$$

Then q is analytic in $\mathbb{D},$$q\left(0\right)=1$, and

$$2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+{\displaystyle \frac{1+u}{1-u}}=1+{\displaystyle \frac{(1-\zeta )u{q}^{\prime }\left(u\right)}{\zeta +(1-\zeta )q\left(u\right)}}=\psi \left(q\left(u\right),u{q}^{\prime }\left(u\right)\right),$$

where

$$\psi (r,s)=1+{\displaystyle \frac{(1-\zeta )s}{\zeta +(1-\zeta )r}}.$$

(16)

Also,

$$\left\{\psi \left(q\left(u\right),u{q}^{\prime }\left(u\right)\right):u\in \mathbb{D}\right\}\subset \left\{w\in \mathbb{C}:\Re \left(w\right)<\beta \right\}:=h_{\beta }.$$

Now, for every real $\rho ,\sigma \le -{\displaystyle \frac{1}{2}}\left(1+{\rho }^2\right),$

$$\Re \left(\psi \left(i\rho ,\sigma \right)\right)=1-{\displaystyle \frac{1}{2}}\zeta (1-\zeta )h\left(\rho \right)$$

where $h\left(\rho \right)$ is given in (14):

$${\displaystyle \frac{1}{{\zeta }^2}}<h\left(\rho \right)\le {\displaystyle \frac{1}{{(1-\zeta )}^2}}$$

For all $\zeta >1$, we have

$$\Re \left(\psi \left(i\rho ,\sigma \right)\right)\ge {\displaystyle \frac{3\zeta -1}{2\zeta }}=\beta .$$

This shows that $\Re \left(\psi \left(i\rho ,\sigma \right)\right)\notin h_{\beta }$. By Lemma 3, we obtain $\Re \left(q\right(u\left)\right)>0$ in $\mathbb{D}$. Hence, Inequality (15) is valid. This completes the proof of Theorem 2. □

Theorem 3.

Let $g\in \mathcal{H},$${\displaystyle \frac{1}{2}}\le \eta <1<\beta <{\displaystyle \frac{3}{2}}$ and $\eta <\Re \left\{p\left(u\right)\right\}<\beta$ in $\mathbb{D}$. Then

$$\delta \left(\eta \right)<\Re \left\{{\displaystyle \frac{g\left(u\right)}{{(1-u)}^2}}\right\}<\zeta \left(\beta \right), u\in \mathbb{D}$$

where $\delta \left(\eta \right)$ and $\zeta \left(\beta \right)$ are given in (12) and (15).

Proof. 

By Theorem 1 and Theorem 2, we have

$$\Re \left\{{\displaystyle \frac{g\left(u\right)}{{(1-u)}^2}}\right\}>{\displaystyle \frac{1}{3-2\eta }}\hspace{1em}and\hspace{1em}\Re \left\{{\displaystyle \frac{g\left(u\right)}{{(1-u)}^2}}\right\}<{\displaystyle \frac{1}{3-2\beta }}, u\in \mathbb{D}.$$

This essentially completes the proof of Theorem 3. □

We need the following Lemma for finding the generalized coefficients of the above $G(\eta ,\beta ).$

Lemma 4

([18]). Let $\varphi \left(u\right)={\sum }_{n=1}^{\infty }{C}_n u^n$ be analytic and univalent in $\mathbb{D}$ and suppose that $\varphi \left(u\right)$ maps $\mathbb{D}$ onto a convex domain. If $p\left(u\right)={\sum }_{n=1}^{\infty }{B}_n u^n$ is analytic in $\mathbb{D}$ and satisfies the subordination

$$p\left(u\right)\prec \varphi \left(u\right), u\in \mathbb{D}$$

then

$$|B_n|\le |C_1|, n=1,2,3,\cdots .$$

Theorem 4.

If the analytic function $g\in G\left(\eta ,\beta \right)$ is of the form (2), then

$$|d_n|\le \prod _{k=1}^n{\displaystyle \frac{2k+\frac{2\left(\beta -\eta \right)}{\pi }sin\frac{\pi \left(1-\eta \right)}{\beta -\eta }}{2^n.n!}}, n=1,2,3,\cdots .$$

Proof. 

Let us define $p\left(u\right)$ and $\varphi \left(u\right)$ by

$$p\left(u\right)=2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+{\displaystyle \frac{1+u}{1-u}}, u\in \mathbb{D}$$

(17)

and

$$\varphi \left(u\right)=1+{\displaystyle \frac{\beta -\eta }{\pi }}ilog\left({\displaystyle \frac{1-e^{2\pi i\frac{1-\eta }{\beta -\eta }}u}{1-u}}\right), u\in \mathbb{D}.$$

(18)

By the subordination, (7) can be written as

$$p\left(u\right)\prec \varphi \left(u\right), u\in \mathbb{D}.$$

(19)

Note that the function $\varphi \left(u\right)$ defined by (18) is convex in $\mathbb{D}$ and has the form

$$\varphi \left(u\right)=1+\sum _{n=1}^{\infty }{C}_n{u}^n,$$

where

$$C_n={\displaystyle \frac{\beta -\eta }{n\pi }}i\left(1-e^{2n\pi i{\displaystyle \frac{1-\eta }{\beta -\eta }}}\right), n=1,2,3,\cdots .$$

If we let

$$p\left(u\right)=1+\sum _{n=1}^{\infty }{B}_n{u}^n,$$

(20)

then by Lemma 4, we see that Subordination (19) implies that

$$|B_n|\le |C_1|, n=1,2,3,\cdots$$

(21)

where

$$|C_1|={\displaystyle \frac{\beta -\eta }{\pi }}\left|1-e^{2\pi i{\displaystyle \frac{1-\eta }{\beta -\eta }}}\right|={\displaystyle \frac{2\left(\beta -\eta \right)}{\pi }}sin{\displaystyle \frac{\pi \left(1-\eta \right)}{\beta -\eta }}.$$

(22)

Now, Equality (17) implies that

$$p\left(u\right)g\left(u\right)=2u{g}^{\prime }\left(u\right)+(1+u){(1-u)}^{-1}g\left(u\right).$$

By using power series expansion, we get

$$\sum _{n=1}^{\infty }\left(B_n+B_{n-1}{d}_1+B_{n-2}{d}_2+\dots +B_1{d}_{n-1}+d_n\right)u^n=\sum _{n=1}^{\infty }2n{d}_n{u}^n+1$$

$$+\sum _{n=1}^{\infty }\left(2+2d_1+\dots +2d_{n-1}+d_n\right)u^n.$$

Equating the coefficient of $u^n$ on both sides leads to

$$\left|d_n\right|={\displaystyle \frac{1}{2n}}\left|B_1{d}_{n-1}+\dots +B_{n-1}{d}_1+B_n-2\left(1+d_1+d_2+\dots +d_{n-1}\right)\right|$$

$$\le {\displaystyle \frac{1}{2n}}\left(\left|B_1{d}_{n-1}+\dots +B_{n-1}{d}_1+B_n\right|+2\left|1+d_1+d_2+\dots +d_{n-1}\right|\right)$$

$$\le {\displaystyle \frac{1}{2n}}\left(|B_1\left|\right|d_{n-1}|+\dots +|B_{n-1}\left|\right|d_1|+|B_n|+2\left(1+|d_1|+|d_2|+\dots +|d_{n-1}|\right)\right)$$

$$\le {\displaystyle \frac{2+|C_1|}{2n}}\sum _{k=1}^n\left|d_{k-1}\right|,$$

where $d_0=1$ and $C_1$ is given in (22). To prove the assertion of the theorem, we have to show that

$$|d_n|\le {\displaystyle \frac{2+|C_1|}{2n}}\sum _{k=1}^n\left|d_{k-1}\right|\le \prod _{k=1}^n{\displaystyle \frac{2k+|C_1|}{2^n.n!}}.$$

(23)

Now we use the method of mathematical induction for the proof of the theorem.

By substituting $n=1$ and $n=2$, we get

$$|d_1|\le {\displaystyle \frac{2+|C_1|}{2}}\left|d_0\right|={\displaystyle \frac{2+|C_1|}{2}}$$

and

$$|d_2|\le {\displaystyle \frac{\left(2+|C_1|\right)}{2.2}}\left(d_0+d_1\right)$$

$$\le {\displaystyle \frac{\left(2+|C_1|\right)}{2.2}}\left(1+{\displaystyle \frac{2+|C_1|}{2}}\right)$$

$$\le {\displaystyle \frac{(2+C_1)(4+C_1)}{2^2.2!}},$$

It is clear that the assertion is holds true for $n=1$ and $n=2$. We assume that Equation (23) is true for $n=m.$ Then, for $n=m+1$, a simple calculation gives us that

$$|d_{m+1}|\le {\displaystyle \frac{2+|C_1|}{2(m+1)}}\sum _{k=1}^{m+1}\left|d_{k-1}\right|$$

$$\le {\displaystyle \frac{2+|C_1|}{2(m+1)}}\left(\sum _{k=1}^m|d_{k-1}|+|d_m|\right)$$

$$\le {\displaystyle \frac{2+|C_1|}{2(m+1)}}\left(\sum _{k=1}^m|d_{k-1}|+{\displaystyle \frac{2+|C_1|}{2m}}\sum _{k=1}^m\left|d_{k-1}\right|\right)$$

$$\le {\displaystyle \frac{2+|C_1|}{2(m+1)}}\sum _{k=1}^m\left|d_{k-1}\right|\left(1+{\displaystyle \frac{2+|C_1|}{2m}}\right)$$

$$\le {\displaystyle \frac{2+|C_1|}{2m}}\sum _{k=1}^m\left|d_{k-1}\right|\left({\displaystyle \frac{2(m+1)+|C_1|}{2(m+1)}}\right)$$

$$\le \prod _{k=1}^m{\displaystyle \frac{2k+|C_1|}{2^m.m!}}.{\displaystyle \frac{2(m+1)+|C_1|}{2(m+1)}}$$

$$\le \prod _{k=1}^{m+1}{\displaystyle \frac{2k+C_1}{2^{m+1}.(m+1)!}},$$

which implies that Inequality (23) is true for $n=m+1.$

Therefore, by mathematical induction, we get

$$|d_n|\le \prod _{k=1}^n{\displaystyle \frac{2k+|C_1|}{2^n.n!}}, n=1,2,3,\dots$$

where $C_1$ is given in (22). This completes the proof of Theorem 4. □

Theorem 5.

If the function $g\left(u\right)\in {\mathcal{G}}_m(\eta ,\beta )$ is of the form (2), then

$$|d_1|\le {\displaystyle \frac{1}{2}}\left(\left|{\displaystyle \frac{2(\beta -\eta )}{\pi }}sin{\displaystyle \frac{\pi (1-\eta )}{\beta -\eta }}\right|+2m\right),$$

$$|d_2|\le {\displaystyle \frac{1}{8}}\left({\left|{\displaystyle \frac{2(\beta -\eta )}{\pi }}sin{\displaystyle \frac{\pi (1-\eta )}{\beta -\eta }}\right|}^2+(2+4m)\left|{\displaystyle \frac{2(\beta -\eta )}{\pi }}sin{\displaystyle \frac{\pi (1-\eta )}{\beta -\eta }}\right|+8m^2\right),$$

$$\begin{gathered} |d_3|\le {\displaystyle \frac{1}{144}}\left(9{\left|{\displaystyle \frac{2(\beta -\eta )}{\pi }}sin{\displaystyle \frac{\pi (1-\eta )}{\beta -\eta }}\right|}^3+(12+30m){\left|{\displaystyle \frac{2(\beta -\eta )}{\pi }}sin{\displaystyle \frac{\pi (1-\eta )}{\beta -\eta }}\right|}^2\right) \\ +{\displaystyle \frac{(72m^2+24m+24)}{144}}\left|{\displaystyle \frac{2(\beta -\eta )}{\pi }}sin{\displaystyle \frac{\pi (1-\eta )}{\beta -\eta }}\right|+16m+128m^3, \end{gathered}$$

(24)

and

$$|d_n|\le {\displaystyle \frac{1}{2n}}\sum _{k=1}^n\left(\left|{\displaystyle \frac{2(\beta -\eta )}{\pi }}sin{\displaystyle \frac{\pi (1-\eta )}{\beta -\eta }}\right|+Q(n-k+1)\right)d_{k-1}.$$

Also, for $m=1$

$$|d_1|\le 1+{\displaystyle \frac{\left|\frac{2(\beta -\eta )}{\pi }sin\frac{\pi (1-\eta )}{\beta -\eta }\right|}{2}},$$

$$|d_2|\le {\displaystyle \frac{1}{8}}\left(\left|{\displaystyle \frac{2(\beta -\eta )}{\pi }}sin{\displaystyle \frac{\pi (1-\eta )}{\beta -\eta }}\right|+2\right)\left(\left|{\displaystyle \frac{2(\beta -\eta )}{\pi }}sin{\displaystyle \frac{\pi (1-\eta )}{\beta -\eta }}\right|+4\right)$$

and

$$|d_3|\le {\displaystyle \frac{1}{48}}\prod _{k=1}^3\left(\left|{\displaystyle \frac{2(\beta -\eta )}{\pi }}sin{\displaystyle \frac{\pi (1-\eta )}{\beta -\eta }}\right|+2k\right).$$

For $m=2$,

$$|d_1|\le {\displaystyle \frac{\left|\frac{2(\beta -\eta )}{\pi }sin\frac{\pi (1-\eta )}{\beta -\eta }\right|+4}{2}},$$

$$|d_2|\le {\displaystyle \frac{1}{8}}\left({\left|{\displaystyle \frac{2(\beta -\eta )}{\pi }}sin{\displaystyle \frac{\pi (1-\eta )}{\beta -\eta }}\right|}^2+10\left|{\displaystyle \frac{2(\beta -\eta )}{\pi }}sin{\displaystyle \frac{\pi (1-\eta )}{\beta -\eta }}\right|+32\right),$$

$$|d_3|\le {\displaystyle \frac{1}{48}}\left({\left|{\displaystyle \frac{2(\beta -\eta )}{\pi }}sin{\displaystyle \frac{\pi (1-\eta )}{\beta -\eta }}\right|}^3+18{\left|{\displaystyle \frac{2(\beta -\eta )}{\pi }}sin{\displaystyle \frac{\pi (1-\eta )}{\beta -\eta }}\right|}^2\right)$$

$$+{\displaystyle \frac{8}{3}}\left|{\displaystyle \frac{2(\beta -\eta )}{\pi }}sin{\displaystyle \frac{\pi (1-\eta )}{\beta -\eta }}\right|+{\displaystyle \frac{22}{3}},$$

where

$$Q\left(n\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{m}{n}}\right)+\sum _{r=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{m}{n-r}}\right)\prod _{t=1}^r{\displaystyle \frac{(m+t-1)}{t!}}.$$

Proof. 

Let us define $p\left(u\right)$ by

$$P\left(u\right)=2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+{\left({\displaystyle \frac{1+u}{1-u}}\right)}^m,u\in \mathbb{D}, 0<m\le 2,$$

(25)

and $\varphi \left(u\right)$ be as in (18). Then, Subordination (9) can be written as

$$p\left(u\right)\prec \varphi \left(u\right), u\in \mathbb{D},$$

(26)

Now, Equation (25) is the same as

$$2u{g}^{\prime }\left(u\right)+{\left\{{\displaystyle \frac{1+u}{1-u}}\right\}}^{m}g\left(u\right)=p\left(u\right)g\left(u\right).$$

By expanding the above as a power series, we obtain that

$$\sum _{n=1}^{\infty }2n{d}_n{u}^n+1+\sum _{n=1}^{\infty }\left(d_n+Q\left(1\right)d_{n-1}+\dots +Q(n-1)+Q\left(n\right)\right)u^n$$

$$=1+\sum _{n=1}^{\infty }\left(d_n+B_1{d}_{n-1}+\dots +B_{n-1}{d}_1+B_n\right)u^n$$

where

$$Q\left(n\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{m}{n}}\right)+\sum _{r=1}^n\left({\displaystyle \genfrac{}{}{0pt}{}{m}{n-r}}\right)\prod _{t=1}^r{\displaystyle \frac{(m+t-1)}{t!}},$$

(27)

$$Q\left(1\right)=2m,Q\left(2\right)=2m^2,Q\left(3\right)={\displaystyle \frac{4m^3+2m}{3}}.$$

(28)

By comparing the coefficient of $u^n$ and by simple calculation, we find that

$$|d_n|\le$$

$${\displaystyle \frac{1}{2n}}\left(|B_1\left|\right|d_{n-1}|+|B_2\left|\right|d_{n-2}|+\dots +|B_{n-1}\left|\right|d_1|+|B_n|+Q\left(n\right)+Q(n-1)|d_1|+\dots +Q\left(1\right)|d_{n-1}|\right).$$

This implies that

$$\begin{array}{cc}\hfill |d_n|\le & {\displaystyle \frac{1}{2n}}\left(\left(\right|C_1|+Q\left(n\right))+\left(\right|C_1|+Q(n-1))|d_1|+\cdots +(|C_1|+Q\left(1\right))|d_{n-1}|\right)\hfill \\ \hfill =& {\displaystyle \frac{1}{2n}}\sum _{k=1}^n\left(|C_1|+Q(n-k+1)\right)\left|d_{k-1}\right|.\hfill \end{array}$$

For n = 1,2,3,

$$|d_1|\le \left(|C_1|+Q\left(1\right)\right)\left|d_0\right|.$$

$$|d_2|\le {\displaystyle \frac{1}{4}}\left(\left(|C_1|+Q\left(2\right)\right)|d_0|+\left(|C_1|+Q\left(1\right)\right)\left|d_1\right|\right).$$

$$|d_3|\le {\displaystyle \frac{1}{6}}\left(\left(|C_1|+Q\left(3\right)\right)|d_0|+\left(|C_1|+Q\left(2\right)\right)|d_1|+\left(|C_1|+Q\left(1\right)\right)\left|d_2\right|\right).$$

By using (28), we get

$$|d_1|\le {\displaystyle \frac{1}{2}}\left(|C_1|+2m\right),$$

$$|d_2|\le {\displaystyle \frac{1}{8}}\left(|C_1{|}^2+(2+4m)\left|C_1\right|+8m^2\right),$$

$$|d_3|\le {\displaystyle \frac{1}{144}}\left(9|C_1{|}^3+(12+30m)|C_1{|}^2+(72m^2+24m+24)\left|C_1\right|+16m+128m^3\right).$$

The choice of $m=1$ gives

$$|d_1|\le 1+{\displaystyle \frac{|C_1|}{2}},$$

$$|d_2|\le {\displaystyle \frac{1}{8}}\left(\right|C_1|+2)\left(\right|C_1|+4),$$

$$|d_3|\le {\displaystyle \frac{1}{48}}\prod _{k=1}^3\left(\right|C_1|+2k).$$

Similarly, the choice of $m=2$ gives

$$|d_1|\le {\displaystyle \frac{|C_1|+4}{2}},$$

$$|d_2|\le {\displaystyle \frac{1}{8}}\left(|C_1{|}^2+10\left|C_1\right|+32\right),$$

and

$$|d_3|\le {\displaystyle \frac{1}{48}}\left(|C_1{|}^3+18|C_1{|}^2+128\left|C_1\right|+352\right).$$

Substituting $|C_1|$ as in (22) in the above equations completes the proof of the Theorem 5. □

Theorem 6.

If the analytic function $g\left(u\right)\in {\mathcal{G}}_L$ is of the form (2), then

$$|d_n|\le \prod _{k=1}^n{\displaystyle \frac{1+4k}{4k}}, n=1,2,3,\cdots .$$

Proof. 

Let us define $p\left(u\right)$ and ${\varphi }_L\left(u\right)$ by

$$p\left(u\right)=2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+{\displaystyle \frac{1+u}{1-u}}, u\in \mathbb{D}$$

(29)

and

$${\varphi }_L\left(u\right)=\sqrt{1+u}.$$

(30)

Then, Subordination (4) can be written as

$$p\left(u\right)\prec {\varphi }_L\left(u\right), u\in \mathbb{D}.$$

(31)

Note that the function ${\varphi }_L\left(u\right)$ defined by (30) is convex in $\mathbb{D}$ and has the form

$${\varphi }_L\left(u\right)=1+{\displaystyle \frac{u}{2}}-{\displaystyle \frac{u^2}{8}}+\dots .$$

By (20), and then by Lemma 4, we see that Subordination (31) implies that

$$|B_n|\le {\displaystyle \frac{1}{2}}.$$

(32)

Now, Equality (29) implies that

$$p\left(u\right)g\left(u\right)=2u{g}^{\prime }\left(u\right)+(1+u){(1-u)}^{-1}g\left(u\right).$$

By using power series expansion, we get

$$\sum _{n=1}^{\infty }\left(B_n+B_{n-1}{d}_1+B_{n-2}{d}_2+\dots +B_1{d}_{n-1}+d_n\right)u^n$$

$$=\sum _{n=1}^{\infty }2n{d}_n{u}^n+1+\sum _{n=1}^{\infty }\left(2+2d_1+\dots +2d_{n-1}+d_n\right)u^n.$$

Comparing the coefficients of $u^n$ on both sides leads to

$$\left|d_n\right|={\displaystyle \frac{1}{2n}}\left|B_1{a}_{n-1}+\cdots +B_{n-1}{a}_1+B_n-2\left(1+d_1+d_2+\cdots +d_{n-1}\right)\right|$$

$$\le {\displaystyle \frac{1}{2n}}\left(\left|B_1{a}_{n-1}+\cdots +B_{n-1}{a}_1+B_n\right|+2\left|1+d_1+d_2+\cdots +d_{n-1}\right|\right)$$

$$\le {\displaystyle \frac{1}{2n}}\left(|B_1\left|\right|d_{n-1}|+\cdots +|B_{n-1}\left|\right|d_1|+|B_n|+2\left(1+|d_1|+|d_2|+\cdots +|d_{n-1}|\right)\right)$$

$$\le {\displaystyle \frac{5}{4n}}\sum _{k=1}^n\left|d_{k-1}\right|,$$

where $d_0=1$. To prove the assertion of the theorem, we have to show that

$$|d_n|\le {\displaystyle \frac{5}{4n}}\sum _{k=1}^n\left|d_{k-1}\right|\le \prod _{k=1}^n{\displaystyle \frac{1+4k}{4k}}.$$

(33)

Now we use the method of mathematical induction for the proof of the theorem. By substituting $n=1$ and $n=2$, we have

$$|d_1|\le {\displaystyle \frac{5}{4}}\left|d_0\right|={\displaystyle \frac{5}{4}}.$$

and

$$|d_2|\le {\displaystyle \frac{5}{8}}\left(d_0+d_1\right)$$

$$\le {\displaystyle \frac{45}{32}}={\displaystyle \frac{5}{4}}{\displaystyle \frac{9}{8}}.$$

It is clear that the assertion is valid for $n=1$ and $n=2$. We assume that the proposition is true for $n=m.$ Then, for $n=m+1$, a simple calculation gives us that

$$|d_{m+1}|\le {\displaystyle \frac{5}{4(m+1)}}\sum _{k=1}^{m+1}\left|d_{k-1}\right|$$

$$\le {\displaystyle \frac{5}{4(m+1)}}\left(\sum _{k=1}^m|d_{k-1}|+|d_m|\right)$$

$$\le {\displaystyle \frac{5}{4(m+1)}}\sum _{k=1}^m|d_{k-1}|+{\displaystyle \frac{5}{4(m+1)}}{\displaystyle \frac{5}{4m}}\sum _{k=1}^m\left|d_{k-1}\right|$$

$$\le {\displaystyle \frac{5}{4(m+1)}}\sum _{k=1}^m\left|d_{k-1}\right|\left({\displaystyle \frac{5+4m}{4m}}\right)$$

$$\le \prod _{k=1}^m{\displaystyle \frac{1+4k}{4k}} · {\displaystyle \frac{1+4(m+1)}{4(m+1)}}$$

$$=\prod _{k=1}^{m+1}{\displaystyle \frac{1+4k}{4k}}.$$

This implies that Inequality (33) is true for $n=m+1.$

Hence, by mathematical induction,

$$|d_n|\le \prod _{k=1}^n{\displaystyle \frac{1+4k}{4k}}, n=1,2,3,\cdots ,$$

which completes the proof of Theorem 6. □

Theorem 7.

If the analytic function $g\left(u\right)\in {\mathcal{G}}_N$ is of the form (2), then ${\displaystyle |d_i|\le \frac{49}{36}=1.3611,i=1,2,3,\cdots .}$

Proof. 

Let

$$p_N\left(u\right)=2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+1+u-{\displaystyle \frac{u^3}{3}},$$

and

$$p_N\left(u\right)=1+p_{1}u+p_2{u}^2+\cdots +p_n{u}^n+\cdots .$$

This implies that

$$2u{g}^{\prime }\left(u\right)+\left(1+u-{\displaystyle \frac{u^3}{3}}\right)g\left(u\right)=p_N\left(u\right)g\left(u\right).$$

By using (2) and $p_N\left(u\right)$, we find that

$$\sum _{n=1}^{\infty }2n{d}_n{u}^n+1+\sum _{n=1}^{\infty }\left(d_n+d_{n-1}-{\displaystyle \frac{d_{n-3}}{3}}\right)u^n=\sum _{n=1}^{\infty }\left(d_n+p_1{d}_{n-1}+\dots +p_{n-1}{d}_1+p_n\right)u^n.$$

By comparing the coefficients of $u^n$, we get

$$|d_n|\le {\displaystyle \frac{1}{2n}}\left|p_n+p_{n-1}{d}_1+\dots +p_1{d}_{n-1}-d_{n-1}+{\displaystyle \frac{d_{n-3}}{3}}\right|$$

$$\le {\displaystyle \frac{1}{2n}}\left(|p_n|+|p_{n-1}\left|\right|d_1|+\dots +|p_1\left|\right|d_{n-1}|+|d_{n-1}|+\left|{\displaystyle \frac{d_{n-3}}{3}}\right|\right).$$

By virtue of the widely known fact that if $p_N\left(u\right)=1+p_{1}u+p_2{u}^2+\cdots \in \mathcal{P}$, then $|p_i|\le 2,i=1,2,3,\cdots ,$ we have

$$|d_n|\le {\displaystyle \frac{1}{n}}\left(1+|d_1|+|d_2|+\cdots +|d_{n-4}|+{\displaystyle \frac{7}{6}}|d_{n-3}|+|d_{n-2}|+{\displaystyle \frac{3}{2}}\left|d_{n-1}\right|\right)$$

$$\le {\displaystyle \frac{1}{n}}\left(\sum _{k=1}^{n-3}|d_{k-1}|+{\displaystyle \frac{7}{6}}|d_{n-3}|+|d_{n-2}|+{\displaystyle \frac{3}{2}}\left|d_{n-1}\right|\right).$$

From the above, by letting $d_0=1,$ we get the following inequalities:

$$|d_1|\le 1,$$

$$|d_2|\le {\displaystyle \frac{1}{2}}\left(|d_0|+|d_1|\right)\le 1,$$

and

$$|d_3|\le {\displaystyle \frac{1}{3}}\left(|d_0|+|d_1|+|d_2|\right)\le 1.$$

Continuing this up to $n-3$ terms, we get

$$|d_{n-3}|\le {\displaystyle \frac{1}{n-3}}\left(|d_0|+|d_1|+\dots +|d_{n-4}|\right)\le 1$$

$$|d_{n-2}|\le {\displaystyle \frac{1}{n-2}}\left(|d_0|+|d_1|+\dots +|d_{n-4}|+{\displaystyle \frac{7}{6}}\left|d_{n-3}\right|\right)$$

$$\le {\displaystyle \frac{1}{n-2}}\left(n-3+{\displaystyle \frac{7}{6}}\right)$$

$$={\displaystyle \frac{6n-11}{6n-12}}.$$

Similarly,

$$|d_{n-1}|\le {\displaystyle \frac{6n-11}{6n-12}}$$

and

$$|d_n|\le {\displaystyle \frac{(2n+1)(6n-11)}{2n(6n-12)}}.$$

From the above $d_i^{\prime }s$, we can conclude that only one ${\displaystyle |d_i|=\frac{49}{36}}$ and the remaining ${\displaystyle |d_i|\le \frac{49}{36}=1.3611,i=1,2,3\dots .}$ □

In the next theorem, we give the general coefficient estimates for the class ${\mathcal{G}}_N$ by a different method.

Theorem 8.

If the function $g\left(u\right)\in {\mathcal{G}}_N$ is of the form (2), the subordination condition is satisfied:

$$p_N\left(u\right)=2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+1+u-{\displaystyle \frac{u^3}{3}}\prec {\displaystyle \frac{1+u}{1-u}},$$

(34)

and then ${\displaystyle |d_i|\le \frac{49}{36}=1.3611,i=1,2,3,\cdots .}$

Proof. 

Let us define $p\left(u\right)$ and $\chi \left(u\right)$ by

$$p_N\left(u\right)=2u{\displaystyle \frac{g^{\prime }\left(u\right)}{g\left(u\right)}}+1+u-{\displaystyle \frac{u^3}{3}}, u\in \mathbb{D}$$

(35)

and

$$\chi \left(u\right)={\displaystyle \frac{1+u}{1-u}}.$$

(36)

Then, Subordination (34) can be written as

$$p_N\left(u\right)\prec \chi \left(u\right), u\in \mathbb{D}.$$

(37)

Note that the function $\chi \left(u\right)$ defined by (36) is convex in $\mathbb{D}$ and has the form

$$\chi \left(u\right)=1+2u+2u^2+\cdots ,$$

and using (20), and then Lemma 4, we see that Subordination (37) implies that

$$|B_n|\le 2.$$

(38)

Now, Equality (35) implies that

$$2u{g}^{\prime }\left(u\right)+\left(1+u-{\displaystyle \frac{u^3}{3}}\right)g\left(u\right)=p_N\left(u\right)g\left(u\right).$$

By using (2) and $p_N\left(u\right)$, we find that

$$\sum _{n=1}^{\infty }2n{d}_n{u}^n+1+\sum _{n=1}^{\infty }\left(d_n+d_{n-1}-{\displaystyle \frac{d_{n-3}}{3}}\right)u^n=\sum _{n=1}^{\infty }\left(d_n+B_1{d}_{n-1}+\cdots +B_{n-1}{d}_1+B_n\right)u^n.$$

By comparing the coefficients of $u^n$, we get

$$|d_n|\le {\displaystyle \frac{1}{2n}}\left|B_n+B_{n-1}{d}_1+\dots +B_1{d}_{n-1}-d_{n-1}+{\displaystyle \frac{d_{n-3}}{3}}\right|$$

$$\le {\displaystyle \frac{1}{2n}}\left(|B_n|+|B_{n-1}\left|\right|d_1|+\dots +|B_1\left|\right|d_{n-1}|+|d_{n-1}|+\left|{\displaystyle \frac{d_{n-3}}{3}}\right|\right).$$

By using (38),

$$|d_n|\le {\displaystyle \frac{1}{n}}\left(1+|d_1|+|d_2|+\dots +|d_{n-4}|+{\displaystyle \frac{7}{6}}|d_{n-3}|+|d_{n-2}|+{\displaystyle \frac{3}{2}}\left|d_{n-1}\right|\right)$$

$$\le {\displaystyle \frac{1}{n}}\left(\sum _{k=1}^{n-3}|d_{k-1}|+{\displaystyle \frac{7}{6}}|d_{n-3}|+|d_{n-2}|+{\displaystyle \frac{3}{2}}\left|d_{n-1}\right|\right).$$

The proof of Theorem 8 is thus completed by using Theorem 26. □

3. Conclusions

In this article, newly defined subclasses of univalent functions are presented, characterized by their starlike nature with respect to a boundary point, along with a class that is intricately connected to the Robertson class, specifically pertaining to a vertical strip domain. Furthermore, this research produces generalized coefficient estimates for these categories, including the Robertson class associated with the Nephroid domain and the Lemniscate of Bernoulli.

This study also indicates that by using q-calculus for values of $0<q<1$, it is possible to establish a functional class. The outcomes derived from this study create new possibilities for subsequent research.

In addition, we can extend these types of studies to incorporate bounded boundary rotation and bounded radius rotation.