Sharp Coefficient Estimates for Analytic Functions Associated with Lemniscate of Bernoulli

Abstract

The main purpose of this work is to study the third Hankel determinant for classes of Bernoulli lemniscate-related functions by introducing new subclasses of star-like functions represented by $S_{L\lambda }^{*}$ and $R{L}_{\lambda }$. In many geometric and physical applications of complex analysis, estimating sharp bounds for problems involving the coefficients of univalent functions is very important because these coefficients describe the fundamental properties of conformal maps. In the present study, we defined sharp bounds for function-coefficient problems belonging to the family of $S_{L\lambda }^{*}$ and $R{L}_{\lambda }$. Most of the computed bounds are sharp. This study will encourage further research on the sharp bounds of analytical functions related to new image domains.

1. Introduction and Preliminaries

Analytic classes in geometric function theory are important for understanding and describing the geometric properties of functions, particularly in complex analysis. Geometric function theory is a mathematical field that analyzes functions from a geometric viewpoint. Geometric function theory, a privileged mathematical field, first and foremost focuses on intervening analytical functions from a geometric perspective. The study of analytical functions is a high appeal in the domain of mathematical analysis. In geometric function theory, the notion of analytical classes is directly linked with understanding the numerous functional aspects, particularly the ones of complex domains. Scholars have proposed imperative contributions for the sake of defining, explaining, and exploring several analytical classes such as holomorphic functions, univalent functions (injective), Quasi conformal mappings (which preserve local angles), and many others. Instigative efforts in the initial foundation were rendered by Cauchy, Riemann, and Weierstrass. The concept of conformal mapping, which is today an enlightened concern in the mathematical field, emerged as an imperative study area at the early age of the 20th century. Louis de Branges and Stefan Bergman made contributions to the Bieberbach conjecture, which was eventually proven by Louis de Branges [1]. Due to advancements in complex dynamics and its use in a number of fields, such as mathematical physics, computer science, and engineering, geometric function theory has experienced ongoing development in the 21st century. Univalent functions include subclasses that include star-like and convex functions. Researchers have frequently talked about how to illustrate and focus on innovative geometrical structures as image domains, as well as how to define the holomorphic functions that go along with them. In 1991, Goodman illuminated uniform convexity and established the concept of the parabolic domain. In the middle of the 20th century, the idea of star-like functions developed as a logical extension of convex functions. The investigation of functions that exhibit characteristics similar to stars in the realm of complex analysis has a long and distinguished history, and it has emerged as an important area of study within the domain of univalent functions. Star-like functions make up a subset of univalent functions, and they play a vital role in understanding the geometric attributes exhibited by analytic functions. The exploration and examination of the analytical class of star-like functions have been the subject of numerous studies.

Alexander [2] was the first mathematician to successfully discover an intriguing relationship between star-like and convex functions. Robertson [3] further advanced the theory of convex and star-like functions, also introducing a specific order for this class. Since the early 1900s, coefficient problems have played a major role in the study of geometric theory and analytic functions. The coefficient problem of analytic classes has been addressed in numerous papers. The upper bounds of the third Hankel determinant have been studied in several papers, see [4,5,6,7,8,9,10,11,12].

To understand the fundamental concepts used in our primary results, we need to start with a few basic concepts. An analytic function $f\left(z\right)$ in the open unit disk $U=\left\{z\in \mathbb{C}:\left|z\right|<1\right\}$, which fulfills the conditions $f\left(0\right)=0$ and $f^{\prime }\left(0\right)=1$, where $\mathbb{C}$ indicates the complex plane, belong to class A and have the following Taylor expansion:

$$f\left(z\right)=z+\sum _{n=2}^{\infty }{a}_n{z}^{n}z\in U.$$

(1)

The class S is the set of all such univalent functions satisfying the normalization criteria $f\left(0\right)=0$ and $f^{\prime }\left(0\right)=1$. This class S became the foundation of current studies in this field. Let S be the class of functions that are univalent in A.

For two functions $\varphi$ and $\mathsf{\Psi }\in A,$ the function $\varphi$ is subordinate to the function $\mathsf{\Psi }$, and is indicated by $\varphi \prec \mathsf{\Psi }$. If there is an analytic function ℏ with the properties $\hslash \left(0\right)=0$ and $|\hslash (z\left)\right|<1$ for all $z\in U$, then the following applies:

$$\varphi \left(z\right)=\mathsf{\Psi }(\hslash (z\left)\right),z\in U.$$

Polynomial functions are usually associated with the class of analytic functions P. The class $p\left(z\right)$ of normalized analytic functions contained in U is denoted by class P and has the following expansion:

$$p\left(z\right)=1+\sum _{n=1}^{\infty }{c}_n{z}^n$$

(2)

such that,

$$\mathbf{P}=\left\{p:p\left(0\right)=1\mathrm{and}\Re \left(p\left(z\right)\right)>0z\in U\right\}.$$

The Carathéodory functions, also referred to as functions with positive real parts, hold immense significance in the domain of geometric functions and exhibit connections to nearly all subclasses of univalent functions.

Several subclasses of univalent functions have been developed based on image domains from a geometric perspective. The subfamilies $S^{*}$ and K represent star-like and close-to-convex univalent functions, respectively. The following are the definitions for each family:

$$S^{*}:=\left\{f:f\in S\mathrm{and}\Re \left({\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\right)>0\left(z\in U\right)\right\},$$

$$K:=\left\{f:f\in S\mathrm{and}\Re \left({\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}\right)>0\left(g\in S^{*};z\in U\right)\right\}.$$

A subfamily $S_L^{*}$ of set S was defined by Sokół and Stankiewicz [13]. It is defined as follows:

$$S_L^{*}:=\left\{f:f\in S\mathrm{and}\left|{\left({\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\right)}^2-1\right|<1\left(z\in U\right)\right\}.$$

The geometric meaning of $f\in S_L^{*}$ is that for each $z\in U$, the ratio ${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}$ falls in the area limited by the right half of Bernoulli’s lemniscate $\left|{\omega }^2-1\right|<1.$ Equivalently, by using the familiar concept of subordination, a function $f\in SL$ satisfies the following relationship:

$${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\prec \sqrt{1+z}.$$

This set has been studied further by different researchers, see [13,14,15,16,17]. Inspired by these sub-families, consider the following set of univalent functions f of form (1) as follows:

$$S_{L\lambda }^{*}=\left\{f\in \mathbf{S}:{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\prec {\left(1+z\right)}^{\lambda }z\in U\mathrm{and}\lambda \in \left(0,1\right)\right\}.$$

A subclass $KL$ of univalent functions f of the form (1) is now defined as follows:

$$KL:=\left\{f:f\in S\mathrm{and}\left|{\left({\displaystyle \frac{\left(z{f}^{\prime }\left(z\right)\right)}{g\left(z\right)}}\right)}^2-1\right|<1\left(g\in SL;z\in U\right)\right\}.$$

In particular, consider the subclass of $KL$ denoted by $RL$, where $g\left(z\right)=z$. This subclass can be defined as follows:

$$RL:=\left\{f:f\in S\mathrm{and}{f}^{\prime }\left(z\right)\prec \sqrt{1+z}\left(z\in U\right)\right\}.$$

The subfamilies $S^{*}$ and R were initially considered by Babalola and Zaprawa, see [18,19]. These classes lead us to define the subclass $R{L}_{\lambda }$ of univalent functions f of the form (1) as follows:

$$R{L}_{\lambda }=\left\{f\in \mathbf{S}:f^{\prime }\left(z\right)\prec {\left(1+z\right)}^{\lambda }\left(z\in U\mathrm{and}\lambda \in \left(0,1\right)\right)\right\}.$$

The study of Hankel determinants within these classes offers a unique perspective from which we can understand functions’ fundamental patterns and behaviors that have significant implications in both theoretical and applied mathematics. For the given parameters j, $n\in N$, Pommerenke [20,21] defined the Hankel determinant $H_{j,n}\left(f\right)$ for a function $f\in S$ of form (1) as follows:

$$H_{j,n}\left(f\right)=\left|\begin{array}{cccc}{a}_n& a_{n+1}& \dots & a_{n+j-1}\\ a_{n+1}& a_{n+2}& \dots & a_{n+j}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n+j-1}& a_{n+j-2}& \dots & a_{n+2j-2}\end{array}\right|j,n\in \mathbb{N}.$$

For $j=2$ and $n=1;$

$$H_{2,1}\left(f\right)=\left|\begin{array}{cc}{a}_1& a_2\\ a_2& a_3\end{array}\right|=a_1{a}_3-a_2^2.$$

The determinant $H_{2,1}$ is noted as a Fekete–Szegö functional—see [22].

For $j=2$ and $n=2;$

$$H_{2,2}\left(f\right)=\left|\begin{array}{cc}{a}_2& a_3\\ a_3& a_4\end{array}\right|=a_2{a}_4-a_3^2.$$

To determine the third-order Hankel determinant, use the following mathematical expression:

$$\begin{array}{ccc}\hfill H_{3,1}\left(f\right)& =& \left|\begin{array}{ccc}{a}_1& a_2& a_3\\ a_2& a_3& a_4\\ a_3& a_4& a_5\end{array}\right|,\hfill \\ & =& a_5(a_3-a_2^2)-a_4(a_4-a_2{a}_3)+a_3(a_2{a}_4-a_3^2).\hfill \end{array}$$

This implies that

$$\left|H_{3,1}\left(f\right)\right|\le \left|a_5\right|\left|a_3-a_2^2\right|+\left|a_4\right|\left|a_4-a_2{a}_3\right|+\left|a_3\right|\left|H_{2,2}\left(f\right)\right|.$$

(3)

This study makes a valuable contribution to the field by deriving the third Hankel determinants for the classes $R{L}_{\lambda }$ and ${\mathbf{S}}_{L\lambda }^{*}$ of bounded turning functions associated with the lemniscate Bernoulli.

2. A Set of Lemmas

We require the following results to determine the bounds of Hankel determinants.

Lemma 1.

If $p\in \mathbf{P}$, then there exists x, β, and z with $\left|x\right|\le 1,\left|\beta \right|\le 1$, and $\left|z\right|\le 1$ such that

$$2c_2=c_1^2+xt,$$

(4)

$$4c_3=c_1^3+2c_{1}tx-c_{1}t{x}^2+2t\left(1-{\left|x\right|}^2\right)\beta ,$$

(5)

$$8c_4=c_1^4+tx\left(c_1^2\left(x^2-3x+3\right)+4x\right)-4t\left(1-{\left|x\right|}^2\right)\left(c_1\left(x-1\right)y+\overline{x}{y}^2-\left(1-{\left|x\right|}^2\right)z\right),$$

(6)

where $t:=(4-c_1^2).$

We refer the interested reader to [23,24,25] for the results in (4)–(6), respectively.

Lemma 2

([24]). If $p\in \mathbf{P}$, then the following inequalities hold

$$\left|c_m\right|\le 2,\mathit{for}m\ge 1,$$

(7)

$$\left|c_2-{\displaystyle \frac{\upsilon }{2}}{c}_1^2\right|\le \left\{\begin{array}{cc}2,& 0\le \upsilon \le 2,\\ 2\left|\nu -1\right|,& \mathit{elsewhere}.\end{array}\right.$$

(8)

Lemma 3

([26]). For $p\in \mathbf{P}$, as given by (2), let ${\alpha }_1$, ${\alpha }_2$, ${\alpha }_3$, and ${\alpha }_4$ satisfy the inequalities $0<{\alpha }_2<1$, $0<{\alpha }_1<1$ and

$$\begin{array}{ccc}& & 8{\alpha }_1\left(1-{\alpha }_1\right)\left[{\left({\alpha }_2{\alpha }_3-2{\alpha }_4\right)}^2+{\left({\alpha }_2\left({\alpha }_1+{\alpha }_2\right)-{\alpha }_3\right)}^2\right]\hfill \\ & & +{\alpha }_2\left(1-{\alpha }_2\right){\left({\alpha }_3-2{\alpha }_1{\alpha }_2\right)}^2-4{\alpha }_1{\alpha }_2^2{\left(1-{\alpha }_2\right)}^2\left(1-{\alpha }_1\right)\hfill \\ & \le & 0.\hfill \end{array}$$

Then,

$$\left|{\alpha }_4{c}_1^4+{\alpha }_1{c}_2^2+2{\alpha }_2{c}_1{c}_3-{\displaystyle \frac{3}{2}}{\alpha }_3{c}_2{c}_1^2-c_4\right|\le 2.$$

(9)

Lemma 4

([27]). If $p\left(z\right)=1+{\sum }_{n=1}^{\infty }{c}_n{z}^n\in \mathbf{P}$, then

$$\left|c_2-\mu c_1^2\right|\le \left\{\begin{array}{cc}-4\mu +2,\hfill & \hfill \mu \le 0,\\ 2,\hfill & \hfill 0\le \mu \le 1,\\ 4\mu -2,\hfill & \hfill \mu >1.\end{array}\right.$$

(10)

3. Bound of $\left|H_{3,1}\left(f\right)\right|$ for the Class $S_{L\lambda }^{*}$

In this section, we have determined the upper bound of $\left|H_{3,1}\left(f\right)\right|$ for the Class $S_{L\lambda }^{*}.$

Theorem 1.

If $f\in S_{L\lambda }^{*}$, $\lambda \in \left(0,1\right)$ has the series representation of the form as given by (1), then

$$\begin{array}{ccc}\hfill \left|a_2\right|& \le & \lambda ,\hfill \end{array}$$

(11)

$$\begin{array}{ccc}\hfill \left|a_3\right|& \le & {\displaystyle \frac{1}{2}}\lambda ,\hfill \end{array}$$

(12)

$$\begin{array}{ccc}\hfill |a_4|& \le & {\displaystyle \frac{1}{3}}\lambda ,\hfill \end{array}$$

(13)

$$\begin{array}{ccc}\hfill |a_5|& \le & {\displaystyle \frac{1}{4}}\lambda .\hfill \end{array}$$

(14)

These results are sharp.

Proof. 

Let $f\in S_{L\lambda }^{*}$ and $\lambda \in \left(0,1\right)$; according to the definition of subordination, there exists a Schwartz function $\hslash \left(z\right)$ with the following properties:

$$\hslash \left(0\right)=0\mathrm{and}|\hslash (z\left)\right|<1.$$

such that,

$${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}={\left(1+\hslash \left(z\right)\right)}^{\lambda }.$$

(15)

Also, if the function $p\in P$, then

$$p\left(z\right)={\displaystyle \frac{1+\hslash \left(z\right)}{1-\hslash \left(z\right)}}=1+c_{1}z+c_2{z}^2+c_3{z}^3\cdots ,$$

and this further gives

$$\hslash \left(z\right)={\displaystyle \frac{p\left(z\right)-1}{1+p\left(z\right)}},$$

$$\hslash \left(z\right)={\displaystyle \frac{c_{1}z+c_2{z}^2+c_3{z}^3+\cdots }{2+c_{1}z+c_2{z}^2+c_3{z}^3+\cdots }}.$$

On expanding, we have

$${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}=1+a_{2}z+\left(2a_3-a_2^2\right)z^2+\left(3a_4-a_2{a}_3+\left(-2a_2{a}_3+a_2^3\right)z^3\right)\cdots .$$

(16)

Also,

$${\left(1+\hslash \left(z\right)\right)}^{\lambda }=1+{\displaystyle \frac{1}{2}}\lambda c_{1}z+\left(\lambda \left({\displaystyle \frac{1}{2}}{c}_2-{\displaystyle \frac{1}{4}}{c}_1^2\right)+{\displaystyle \frac{1}{8}}\lambda c_1^2\left(\lambda -1\right)\right)z^2\cdots .$$

(17)

The comparison of coefficients $z$, $z^2$, $z^3$, and $z^4,$ along with precise computation and also using (16) and (17), we have

$$\begin{array}{ccc}\hfill a_2& =& {\displaystyle \frac{c_1}{2}}\lambda ,\hfill \end{array}$$

(18)

$$\begin{array}{ccc}\hfill a_3& =& {\displaystyle \frac{c_2}{4}}\lambda -\left({\displaystyle \frac{3}{16}}\lambda -{\displaystyle \frac{3}{16}}{\lambda }^2\right)c_1^2,\hfill \end{array}$$

(19)

$$\begin{array}{ccc}\hfill a_4& =& \left({\displaystyle \frac{5}{24}}{\lambda }^2-{\displaystyle \frac{1}{4}}\lambda \right)c_1{c}_2+\left({\displaystyle \frac{7}{72}}\lambda -{\displaystyle \frac{5}{32}}{\lambda }^2+{\displaystyle \frac{17}{288}}{\lambda }^3\right)c_1^3+{\displaystyle \frac{1}{6}}\lambda c_3,\hfill \end{array}$$

(20)

and

$$\begin{array}{ccc}\hfill a_5& =& \left({\displaystyle \frac{19}{1152}}{\lambda }^4-{\displaystyle \frac{5}{64}}{\lambda }^3+{\displaystyle \frac{277}{2304}}{\lambda }^2-{\displaystyle \frac{15}{256}}\lambda \right)c_1^4+\left(-{\displaystyle \frac{3}{32}}\lambda +{\displaystyle \frac{1}{16}}{\lambda }^2\right)c_2^2\hfill \\ & & +\left({\displaystyle \frac{7}{48}}{\lambda }^2-{\displaystyle \frac{3}{16}}\lambda \right)c_1{c}_3+\left({\displaystyle \frac{5}{48}}{\lambda }^3-{\displaystyle \frac{5}{16}}{\lambda }^2+{\displaystyle \frac{7}{32}}\lambda \right)c_1^2{c}_2+{\displaystyle \frac{1}{8}}\lambda c_4.\hfill \end{array}$$

(21)

Using first inequality of Lemma 2 in (18), we have

$$\left|a_2\right|\le \lambda .$$

Using second inequality of Lemma 2 in (19), we have

$$\begin{array}{ccc}\hfill \left|a_3\right|& =& {\displaystyle \frac{\lambda }{4}}\left|c_2-\left({\displaystyle \frac{3}{4}}-{\displaystyle \frac{3}{4}}\lambda \right)c_1^2\right|,\hfill \\ & \le & {\displaystyle \frac{\lambda }{2}},\hfill \end{array}$$

where $\upsilon ={\displaystyle \frac{3}{4}}-{\displaystyle \frac{3}{4}}\lambda$ for $0\le \upsilon \le 2$ and $\lambda \in \left(0,1\right)$.

Now consider the following:

$$\left|a_4\right|=\left|\left({\displaystyle \frac{5}{24}}{\lambda }^2-{\displaystyle \frac{1}{4}}\lambda \right)c_1{c}_2+\left({\displaystyle \frac{7}{72}}\lambda -{\displaystyle \frac{5}{32}}{\lambda }^2+{\displaystyle \frac{17}{288}}{\lambda }^3\right)c_1^3+{\displaystyle \frac{1}{6}}\lambda c_3\right|.$$

Now, using (4), assuming $c\in \left(0,2\right)$ and $\lambda \in \left(0,1\right),$ and applying triangular inequality, we have

$$|a_4|\le \left({\displaystyle \frac{1}{72}}\lambda -{\displaystyle \frac{5}{96}}{\lambda }^2+{\displaystyle \frac{17}{288}}{\lambda }^3\right)c^3+\left({\displaystyle \frac{5}{48}}{\lambda }^2-{\displaystyle \frac{1}{24}}\lambda \right)c\left|x\right|t+{\displaystyle \frac{1}{24}}\lambda c{\left|x\right|}^{2}t+{\displaystyle \frac{1}{12}}\lambda t\left(1-{\left|x\right|}^2\right)\left|\beta \right|.$$

Using $\left|x\right|=\rho$, $\left|\beta \right|\le 1$, we obtained

$$\begin{array}{ccc}\hfill |a_4|& \le & \left({\displaystyle \frac{1}{72}}\lambda -{\displaystyle \frac{5}{96}}{\lambda }^2+{\displaystyle \frac{17}{288}}{\lambda }^3\right)c^3+\left({\displaystyle \frac{5}{48}}{\lambda }^2-{\displaystyle \frac{1}{24}}\lambda \right)c\rho t+{\displaystyle \frac{1}{24}}\lambda c{\rho }^{2}t+{\displaystyle \frac{1}{12}}\lambda t\hfill \\ & :& =Ƒ\left(c,\rho \right).\hfill \end{array}$$

Suppose,

$$\begin{array}{ccc}\hfill A_1& =& \left({\displaystyle \frac{1}{72}}\lambda +{\displaystyle \frac{5}{96}}{\lambda }^2+{\displaystyle \frac{17}{288}}{\lambda }^3\right)c^3,\hfill \\ \hfill A_2& =& \left({\displaystyle \frac{5}{48}}{\lambda }^2-{\displaystyle \frac{1}{24}}\lambda \right)ct\hfill \\ \hfill A_3& =& {\displaystyle \frac{1}{24}}\lambda ct,\hfill \\ \hfill A_4& =& {\displaystyle \frac{1}{12}}\lambda t,\hfill \end{array}$$

where $t=4-c^2.$ So, we can write

$$\xi \left(c,\rho \right)=A_1+A_2\rho +A_3{\rho }^2+A_4.$$

Differentiating the above expression with respect to $\rho$, we have

$${\displaystyle \frac{\partial \xi }{\partial \rho }}=A_2+{\chi }_1\rho ,$$

with

$${\chi }_1={\displaystyle \frac{1}{12}}\lambda ct.$$

By calculating ${\displaystyle \frac{\partial \xi }{\partial \rho }}>0,$ it is clear that $\xi \left(c,\rho \right)$ is a function of $\rho$ that increases on $\left[0,1\right]$. Consequently, we determined that $Ƒ\left(c,\rho \right)$ reaches its maximum at $\rho =1.$

$$\xi \left(c,1\right)=\zeta \left(c\right)=\left({\displaystyle \frac{1}{72}}\lambda -{\displaystyle \frac{5}{96}}{\lambda }^2+{\displaystyle \frac{17}{288}}{\lambda }^3\right)c^3+{\displaystyle \frac{5}{48}}{\lambda }^{2}ct+{\displaystyle \frac{1}{12}}\lambda t.$$

Using the value of $t=4-c^2,$ after some simplification, we have

$$\zeta \left(c\right)=\left(-{\displaystyle \frac{13}{288}}{\lambda }^3-{\displaystyle \frac{5}{96}}{\lambda }^2+{\displaystyle \frac{1}{72}}\lambda \right)c^3-{\displaystyle \frac{1}{12}}\lambda c^2+{\displaystyle \frac{5}{12}}{\lambda }^{2}c+{\displaystyle \frac{1}{3}}\lambda ,$$

$$\zeta \left(c\right)=B_1{c}^3+B_2{c}^2+B_{3}c+B_4,$$

where

$$\begin{array}{ccc}\hfill B_1& =& \left(-{\displaystyle \frac{13}{288}}{\lambda }^3-{\displaystyle \frac{5}{96}}{\lambda }^2+{\displaystyle \frac{1}{72}}\lambda \right),\hfill \\ \hfill B_2& =& -{\displaystyle \frac{1}{12}}\lambda \hfill \\ \hfill B_3& =& {\displaystyle \frac{5}{12}}{\lambda }^2,\hfill \\ \hfill B_4& =& {\displaystyle \frac{1}{3}}\lambda .\hfill \end{array}$$

Differentiating the above expression with respect to c, we have

$${\displaystyle \frac{\partial \zeta }{\partial c}}=3B_1{c}^2+2B_{2}c+B_3,$$

Differentiating the above expression with respect to c, we have

$${\zeta }^{″}\left(c\right)={\beta }_{1}c-{\beta }_2,$$

with

$$\begin{array}{ccc}\hfill {\beta }_1& =& \left(-{\displaystyle \frac{13}{48}}{\lambda }^3-{\displaystyle \frac{5}{16}}{\lambda }^2+{\displaystyle \frac{1}{12}}\lambda \right),\hfill \\ \hfill {\beta }_2& =& -{\displaystyle \frac{1}{6}}\lambda .\hfill \end{array}$$

Upon computation, it can be observed that ${\zeta }^{″}\left(c\right)<0$, since $c\in \left[0,2\right]$ and $\lambda \in \left(0,1\right)$. This indicates the maximum value of $\zeta \left(c\right)$ at $c=0$. Thus, we obtain

$$|a_4|\le {\displaystyle \frac{1}{3}}\lambda .$$

Now, consider the following:

$$\begin{array}{ccc}\hfill a_5& =& \left({\displaystyle \frac{19}{1152}}{\lambda }^4-{\displaystyle \frac{5}{64}}{\lambda }^3+{\displaystyle \frac{277}{2304}}{\lambda }^2-{\displaystyle \frac{15}{256}}\lambda \right)c_1^4+\left(-{\displaystyle \frac{3}{32}}\lambda +{\displaystyle \frac{1}{16}}{\lambda }^2\right)c_2^2\hfill \\ & & -\left(-{\displaystyle \frac{7}{48}}{\lambda }^2+{\displaystyle \frac{3}{16}}\lambda \right)c_1{c}_3-\left(-{\displaystyle \frac{5}{48}}{\lambda }^3+{\displaystyle \frac{5}{16}}{\lambda }^2-{\displaystyle \frac{7}{32}}\lambda \right)c_1^2{c}_2+{\displaystyle \frac{1}{8}}\lambda c_4.\hfill \end{array}$$

we can write

$$\begin{array}{ccc}\hfill a_5& =& {\displaystyle \frac{-\lambda }{8}}\left[-\left({\displaystyle \frac{38{\lambda }^3-180{\lambda }^2+277\lambda -135}{288}}\right)c_1^4+\left({\displaystyle \frac{3}{4}}-{\displaystyle \frac{1}{2}}\lambda \right)c_2^2+\left(-{\displaystyle \frac{7}{6}}\lambda +{\displaystyle \frac{3}{2}}\right)c_1{c}_3\right.\hfill \\ & & \left.-\left({\displaystyle \frac{21-30\lambda +10{\lambda }^2}{12}}\right)c_1^2{c}_2-c_4\right],\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill {\alpha }_1& =& \left({\displaystyle \frac{3}{4}}-{\displaystyle \frac{1}{2}}\lambda \right),\hfill \\ \hfill {\alpha }_2& =& -{\displaystyle \frac{7}{12}}\lambda +{\displaystyle \frac{3}{4}},\hfill \\ \hfill {\alpha }_3& =& {\displaystyle \frac{7}{6}}-{\displaystyle \frac{5}{3}}\lambda +{\displaystyle \frac{5}{9}}{\lambda }^2,\hfill \\ \hfill {\alpha }_4& =& -\left({\displaystyle \frac{38{\lambda }^3-180{\lambda }^2+277\lambda -135}{288}}\right).\hfill \end{array}$$

We see that $0<{\alpha }_2<1$, $0<{\alpha }_1<1$ for $\lambda \in \left(0,1\right)$ and

$$\begin{array}{ccc}& & 8{\alpha }_1\left(1-{\alpha }_1\right)\left[{\left({\alpha }_2{\alpha }_3-2{\alpha }_4\right)}^2+{\left({\alpha }_2\left({\alpha }_1+{\alpha }_2\right)-{\alpha }_3\right)}^2\right]\hfill \\ & & +{\alpha }_2\left(1-{\alpha }_2\right){\left({\alpha }_3-2{\alpha }_1{\alpha }_2\right)}^2-4{\alpha }_1{\alpha }_2^2{\left(1-{\alpha }_2\right)}^2\left(1-{\alpha }_1\right)\hfill \\ & :& =\phi \left(\lambda \right),\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill \phi \left(\lambda \right)& =& -{\displaystyle \frac{169}{23,328}}{\lambda }^8+{\displaystyle \frac{949}{23,328}}{\lambda }^7+{\displaystyle \frac{1661}{46,656}}{\lambda }^6-{\displaystyle \frac{253}{864}}{\lambda }^5+{\displaystyle \frac{19,955}{124,416}}{\lambda }^4\hfill \\ & & +{\displaystyle \frac{2635}{10,368}}{\lambda }^3-{\displaystyle \frac{881}{10,368}}{\lambda }^2-{\displaystyle \frac{353}{3456}}\lambda -{\displaystyle \frac{9}{512}}.\hfill \end{array}$$

Upon computation, it can be observed that $\phi \left(\lambda \right)<0$ for $\lambda \in \left(0,1\right).$ By using Lemma 3, we have

$$|a_5|\le {\displaystyle \frac{\lambda }{4}}.$$

The sharpness of the results can be obtained by taking the function $f_m:U\to \mathbb{C}$ given by

$$f_m\left(z\right)=zexp\left({\int }_0^z{\displaystyle \frac{{\left(1+t^n\right)}^{\lambda }-1}{t}}\right)dtm=1,2,3,4.$$

Then,

$${\displaystyle \frac{z{f}_m^{\prime }\left(z\right)}{f_m\left(z\right)}}={\left(1+z^n\right)}^{\lambda },m=1,2,3,4.$$

Hence, $f_m\in S_{L\lambda }^{*}$ and

$$f_1\left(z\right)=zexp\left({\int }_0^z{\displaystyle \frac{{\left(1+t\right)}^{\lambda }-1}{t}}\right)dt=z+\lambda z^2+\left({\displaystyle \frac{3{\lambda }^2}{4}}-{\displaystyle \frac{\lambda }{4}}\right)z^3-\cdots ,$$

(22)

$$f_2\left(z\right)=zexp\left({\int }_0^z{\displaystyle \frac{{\left(1+t^2\right)}^{\lambda }-1}{t}}\right)dt=z+{\displaystyle \frac{\lambda }{2}}{z}^3+\cdots ,$$

(23)

$$f_3\left(z\right)=zexp\left({\int }_0^z{\displaystyle \frac{{\left(1+t^3\right)}^{\lambda }-1}{t}}\right)dt=z+{\displaystyle \frac{\lambda }{3}}{z}^4+\cdots ,$$

(24)

and

$$f_4\left(z\right)=zexp\left({\int }_0^z{\displaystyle \frac{{\left(1+t^4\right)}^{\lambda }-1}{t}}\right)dt=z+{\displaystyle \frac{\lambda }{4}}{z}^4+\cdots .$$

(25)

For $\lambda ={\displaystyle \frac{1}{2}}$ in the above result, the following results have been proven in [16]. □

Corollary 1.

If $f\in S_L^{*}$ has the series form as given by (1), then

$$|a_2|\le {\displaystyle \frac{1}{2}},$$

$$|a_3|\le {\displaystyle \frac{1}{4}},$$

$$|a_4|\le {\displaystyle \frac{1}{6}},$$

$$|a_5|\le {\displaystyle \frac{1}{8}}.$$

These estimations are sharp.

Sokół [13] obtained the first three bounds, and a similar method is used to establish the bound for $|a_5|$.

Conjecture 1.

If $f\in S_{L\lambda }^{*}$, $\lambda \in \left(0,1\right)$ has the series representation of the form as given by (1), then

$$\left|a_{n+1}\right|\le {\displaystyle \frac{\lambda }{n}}.\left(n\ge 1\right).$$

Theorem 2.

If $f\in S_{L\lambda }^{*}$, $\lambda \in \left(0,1\right)$ has the series representation of the form as given by (1), then

$$|a_2{a}_3-a_4|\le {\displaystyle \frac{\lambda }{3}}.$$

(26)

Proof. 

From (18) to (20), we have

$$a_2{a}_3-a_4=\left(-{\displaystyle \frac{1}{12}}{\lambda }^2+{\displaystyle \frac{1}{4}}\lambda \right)c_1{c}_2+c_1^3\left({\displaystyle \frac{5}{144}}{\lambda }^3+{\displaystyle \frac{1}{16}}{\lambda }^2-{\displaystyle \frac{7}{72}}\lambda \right)-{\displaystyle \frac{1}{6}}{\lambda }_1{c}_3.$$

Using Lemma 1, applying triangle inequality and $c:=c_1\in [0,2]$ after simplification, the following is obtained:

$$\begin{array}{ccc}\hfill {Ƒ}_1\left(c,x\right)& =& \left(-{\displaystyle \frac{1}{72}}\lambda +{\displaystyle \frac{1}{48}}{\lambda }^2+{\displaystyle \frac{5}{144}}{\lambda }^3\right)c^3+\left({\displaystyle \frac{1}{24}}\lambda -{\displaystyle \frac{1}{24}}{\lambda }^2\right)c\rho t\hfill \\ & & +{\displaystyle \frac{1}{24}}\lambda ct{\rho }^2+{\displaystyle \frac{1}{12}}\lambda t,\hfill \end{array}$$

where $t=4-c^2,$ and we are assuming $\left|x\right|=\rho$ and $\left|\beta \right|\le 1$.

Partially differentiating with respect to $\rho$, we obtain

$${\displaystyle \frac{\partial {Ƒ}_1\left(c,\rho \right)}{\partial \rho }}=\left({\displaystyle \frac{1}{24}}\lambda -{\displaystyle \frac{1}{24}}{\lambda }^2\right)ct+{\displaystyle \frac{1}{12}}\lambda ct\rho .$$

Clearly, ${\displaystyle \frac{\partial {Ƒ}_1\left(c,x\right)}{\partial \rho }}>0$, since $\rho \in \left[0,1\right]$, ${Ƒ}_1\left(c,\rho \right)$ attains maximum value at $\rho =1$, and we have

$$\begin{array}{ccc}\hfill {Ƒ}_1\left(c,1\right)& =& \left(-{\displaystyle \frac{1}{72}}\lambda +{\displaystyle \frac{1}{48}}{\lambda }^2+{\displaystyle \frac{5}{144}}{\lambda }^3\right)c^3+\left({\displaystyle \frac{1}{24}}\lambda -{\displaystyle \frac{1}{24}}{\lambda }^2\right)ct\hfill \\ & & +{\displaystyle \frac{1}{24}}\lambda ct+{\displaystyle \frac{1}{12}}\lambda t.\hfill \end{array}$$

Substituting t with $4-c^2$ in the above equation, we obtain a new expression

$$\begin{array}{ccc}\hfill {Ƒ}_1\left(c,1\right)& =& G_1\left(c\right)=\left(-{\displaystyle \frac{1}{72}}\lambda +{\displaystyle \frac{1}{16}}{\lambda }^2+{\displaystyle \frac{5}{144}}{\lambda }^3\right)c^3-{\displaystyle \frac{1}{12}}\lambda c^2+\left(-{\displaystyle \frac{1}{6}}{\lambda }^2+{\displaystyle \frac{1}{3}}\lambda \right)c\hfill \\ & & +{\displaystyle \frac{1}{3}}\lambda .\hfill \end{array}$$

Suppose,

$$G_1\left(c\right)=K_1{c}^3+K_2{c}^2+K_{3}c+K_4,$$

where

$$\begin{array}{ccc}\hfill K_1& =& {\displaystyle \frac{5}{144}}{\lambda }^3+{\displaystyle \frac{1}{16}}{\lambda }^2-{\displaystyle \frac{7}{72}}\lambda ,\hfill \\ \hfill K_2& =& -{\displaystyle \frac{1}{12}}\lambda ,\hfill \\ \hfill K_3& =& \left(-{\displaystyle \frac{1}{6}}{\lambda }^2+{\displaystyle \frac{1}{3}}\lambda \right),\hfill \\ \hfill K_4& =& {\displaystyle \frac{1}{3}}\lambda .\hfill \end{array}$$

Differentiating $G_1\left(c\right)$ with respect to c, after some simplification, we have

$$G_1^{\prime }\left(c\right)={\tau }_1{c}^2+{\tau }_{2}c+K_3,$$

with

$$\begin{array}{ccc}\hfill {\tau }_1& =& {\displaystyle \frac{5}{48}}{\lambda }^3+{\displaystyle \frac{3}{16}}{\lambda }^2-{\displaystyle \frac{7}{24}}\lambda ,\hfill \\ \hfill {\tau }_2& =& -{\displaystyle \frac{1}{6}}\lambda .\hfill \end{array}$$

Upon computation, it can be observed that $G_1^{\prime }\left(c\right)<0$, so $G_1\left(c\right)$ is a decreasing function and that reaches its maximum value at $c=0$; therefore, we concluded

$$|a_2{a}_3-a_4|\le {\displaystyle \frac{\lambda }{3}}.$$

For $\lambda ={\displaystyle \frac{1}{2}}$ in the above result, the following results have been proven in [16]. □

Corollary 2.

If $f\in S_L^{*}$ has the series form as given by (1), then

$$|a_2{a}_3-a_4|\le {\displaystyle \frac{1}{6}}.$$

Theorem 3.

If $f\in S_{L\lambda }^{*}$, $\lambda \in \left(0,1\right)$ has the series representation of the form as given by (1), then

$$\left|a_3-a_2^2\right|\le {\displaystyle \frac{\lambda }{2}}.$$

(27)

Proof. 

From (18) and (19), consider

$$\left|a_3-a_2^2\right|={\displaystyle \frac{\lambda }{4}}\left|c_2-\left({\displaystyle \frac{3}{4}}+{\displaystyle \frac{1}{4}}\lambda \right)c_1^2\right|.$$

Using the second inequality of Lemma 2, the following is obtained:

$$\left|a_3-a_2^2\right|\le {\displaystyle \frac{\lambda }{2}},$$

with $v={\displaystyle \frac{3}{4}}+{\displaystyle \frac{1}{4}}\lambda$ for $0\le \nu \le 2$ and $\lambda \in \left(0,1\right).$

For $\lambda ={\displaystyle \frac{1}{2}}$ in the above result, the following results have been proven in [16]. □

Corollary 3.

If $f\in S_L^{*}$ has the series form as given by (1), then

$$\left|a_3-a_2^2\right|\le {\displaystyle \frac{1}{4}}.$$

Theorem 4.

If $f\in S_{L\lambda }^{*}$, $\lambda \in \left(0,1\right)$ has the series representation of the form as given by (1), then

$$|H_{2,2}\left(f\right)|\le {\displaystyle \frac{1}{4}}{\lambda }^2,$$

(28)

where $H_{2,2}\left(f\right)=a_2{a}_4-a_3^2$.

Proof. 

From (18) to (20), we obtain

$$\begin{array}{ccc}\hfill H_{2,2}\left(f\right)& =& a_2{a}_4-a_3^2={\displaystyle \frac{1}{12}}{\lambda }^2{c}_1{c}_3-{\displaystyle \frac{1}{16}}{\lambda }^2{c}_2^2-\left(-{\displaystyle \frac{1}{96}}{\lambda }^3+{\displaystyle \frac{1}{32}}{\lambda }^2\right)c_1^2{c}_2\hfill \\ & & +\left({\displaystyle \frac{31}{2304}}{\lambda }^2-{\displaystyle \frac{1}{128}}{\lambda }^3-{\displaystyle \frac{13}{2304}}{\lambda }^4\right)c_1^4.\hfill \end{array}$$

From Lemma 1, replacing $\left|x\right|=\sigma$, $\left|\beta \right|\le 1$, $t:=(4-c^2)$ and $c:=c_1\in [0,2]$, after simplification, we obtain

$$\begin{array}{ccc}\hfill {Ƒ}_3\left(c,\rho \right)& =& \left({\displaystyle \frac{7}{2304}}{\lambda }^2-{\displaystyle \frac{1}{384}}{\lambda }^3-{\displaystyle \frac{13}{2304}}{\lambda }^4\right)c^4+\left({\displaystyle \frac{1}{192}}{\lambda }^3-{\displaystyle \frac{1}{192}}{\lambda }^2\right)c^2\sigma t\hfill \\ & & +{\displaystyle \frac{1}{48}}{\lambda }^2{c}^2{\sigma }^{2}t+{\displaystyle \frac{1}{24}}{\lambda }^{2}ct+{\displaystyle \frac{1}{64}}{\lambda }^2{\sigma }^2{t}^2.\hfill \end{array}$$

By differentiating ${Ƒ}_3\left(c,\sigma \right)$ with respect to $\sigma$, we have

$${\displaystyle \frac{\partial {Ƒ}_3\left(c,\rho \right)}{\partial \rho }}={\displaystyle \frac{1}{192}}{\lambda }^3{c}^{2}t-{\displaystyle \frac{1}{192}}{\lambda }^2{c}^{2}t+{\displaystyle \frac{1}{24}}{\lambda }^2{c}^2\sigma t+{\displaystyle \frac{1}{32}}{\lambda }^2\sigma t^2.$$

We can observe that ${\displaystyle \frac{{Ƒ}_3\left(c,\sigma \right)}{\partial \sigma }}\ge 0$, and we can see that ${Ƒ}_3\left(c,\sigma \right)\le {Ƒ}_3\left(c,1\right)$ if we put $\sigma =1$, we obtain

$$\begin{array}{ccc}\hfill {Ƒ}_3\left(c,1\right)& =& G_3\left(c\right)=\left({\displaystyle \frac{7}{2304}}{\lambda }^2-{\displaystyle \frac{1}{384}}{\lambda }^3-{\displaystyle \frac{13}{2304}}{\lambda }^4\right)c^4+\left({\displaystyle \frac{1}{192}}{\lambda }^3-{\displaystyle \frac{1}{192}}{\lambda }^2\right)c^{2}t\hfill \\ & & +{\displaystyle \frac{1}{48}}{\lambda }^2{c}^{2}t+{\displaystyle \frac{1}{24}}{\lambda }^{2}ct-{\displaystyle \frac{1}{64}}{\lambda }^2{t}^2.\hfill \end{array}$$

Putting $t=(4-c^2)$ in the above equation, we have

$$\begin{array}{ccc}\hfill H_3\left(c\right)& =& -\left(-{\displaystyle \frac{7}{2304}}{\lambda }^2+{\displaystyle \frac{1}{128}}{\lambda }^3+{\displaystyle \frac{13}{2304}}{\lambda }^4\right)c^4-{\displaystyle \frac{1}{24}}{\lambda }^2{c}^3-\left({\displaystyle \frac{1}{16}}{\lambda }^2-{\displaystyle \frac{1}{48}}{\lambda }^3\right)c^2\hfill \\ & & +{\displaystyle \frac{1}{6}}{\lambda }^{2}c+{\displaystyle \frac{1}{4}}{\lambda }^2.\hfill \end{array}$$

Suppose

$$H_3\left(c\right)=S_1{c}^4+S_2{c}^3+S_3{c}^2+S_{4}c+S_5,$$

where

$$\begin{array}{ccc}\hfill S_1& =& -\left(-{\displaystyle \frac{7}{2304}}{\lambda }^2+{\displaystyle \frac{1}{128}}{\lambda }^3+{\displaystyle \frac{13}{2304}}{\lambda }^4\right),\hfill \\ \hfill S_2& =& -{\displaystyle \frac{1}{24}}{\lambda }^2,\hfill \\ \hfill S_3& =& -\left({\displaystyle \frac{1}{16}}{\lambda }^2-{\displaystyle \frac{1}{48}}{\lambda }^3\right),\hfill \\ \hfill S_4& =& {\displaystyle \frac{1}{6}}{\lambda }^2,\hfill \\ \hfill S_5& =& {\displaystyle \frac{1}{4}}{\lambda }^2.\hfill \end{array}$$

Differentiating $H_3\left(c\right)$ with respect to c, we have

$$H_3^{\prime }\left(c\right)={\zeta }_1{c}^3+{\zeta }_2{c}^2+{\zeta }_{3}c+S_4.$$

with

$$\begin{array}{ccc}\hfill {\zeta }_1& =& -\left(-{\displaystyle \frac{7}{576}}{\lambda }^2+{\displaystyle \frac{1}{32}}{\lambda }^3+{\displaystyle \frac{13}{576}}{\lambda }^4\right),\hfill \\ \hfill {\zeta }_2& =& S_2^{\prime }=-{\displaystyle \frac{1}{8}}{\lambda }^2,\hfill \\ \hfill {\zeta }_3& =& -\left({\displaystyle \frac{1}{8}}{\lambda }^2-{\displaystyle \frac{1}{24}}{\lambda }^3\right).\hfill \end{array}$$

After computing, we find that $H_3^{\prime }\left(c\right)$$\le 0$, $H_3\left(c\right)$ is a decreasing function and reaches its maximum value at $c=0$. Finally, we have

$$|H_{2,2}\left(f\right)|\le {\displaystyle \frac{1}{4}}{\lambda }^2.$$

For $\lambda ={\displaystyle \frac{1}{2}}$ in the above result, the following results have been proven in [16]. □

Corollary 4.

If $f\in S_L^{*}$ has the series form as given by (1), then

$$|H_{2,2}\left(f\right)|\le {\displaystyle \frac{1}{16}}.$$

Theorem 5.

If $f\in S_{L\lambda }^{*}$, $\lambda \in \left(0,1\right)$ has the series representation of the form as given by (1), then

$$\left|H_{3,1}\left(f\right)\right|\le {\displaystyle \frac{17{\lambda }^2+9{\lambda }^3}{72}}.$$

(29)

Proof. 

Thus, by considering the fact that $a_1=1$, together with (12)–(14), (26), (27), (28), and (3).

For $\lambda ={\displaystyle \frac{1}{2}}$ in the above result, the following result has been proven in [16]. □

Corollary 5.

If $f\in S_L^{*}$ has the series form as given by (1), then

$$\left|H_{3,1}\left(f\right)\right|\le {\displaystyle \frac{43}{576}}.$$

4. Bound of $\left|{\mathit{H}}_{\mathbf{3},\mathbf{1}}\left(\mathit{f}\right)\right|$ for the Class ${\mathit{RL}}_{\mathit{\lambda }}$

Our initial findings are focused on establishing bounds for the functions f that are part of the $R{L}_{\lambda }$ class.

Theorem 6.

If $f\in R{L}_{\lambda }$, $\lambda \in \left(0,1\right)$ has the series form as given by (1), then

$$\begin{array}{ccc}\hfill \left|a_2\right|& \le & {\displaystyle \frac{1}{2}}\lambda ,\hfill \end{array}$$

(30)

$$\begin{array}{ccc}\hfill \left|a_3\right|& \le & {\displaystyle \frac{1}{3}}\lambda ,\hfill \end{array}$$

(31)

$$\begin{array}{ccc}\hfill |a_4|& \le & {\displaystyle \frac{1}{4}}\lambda ,\hfill \end{array}$$

(32)

$$\begin{array}{ccc}\hfill |a_5|& \le & {\displaystyle \frac{1}{5}}\lambda .\hfill \end{array}$$

(33)

These results are sharp.

Proof. 

Assuming $f\in R_{L\lambda }$ and $\lambda \in \left(0,1\right)$, by utilizing the principle of subordination, we have

$$f\left(z\right)={\left(1+\hslash \left(z\right)\right)}^{\lambda }.$$

(34)

$${\left(1+\hslash \left(z\right)\right)}^{\lambda }=1+{\displaystyle \frac{1}{2}}\lambda c_{1}z+\left(\lambda \left({\displaystyle \frac{1}{2}}{c}_2-{\displaystyle \frac{1}{4}}{c}_1^2\right)+{\displaystyle \frac{1}{8}}\lambda c_1^2\left(\lambda -1\right)\right)z^2\cdots .$$

(35)

Define the function

$$p\left(z\right)={\displaystyle \frac{1+\hslash \left(z\right)}{1-\hslash \left(z\right)}}=1+c_{1}z+c_2{z}^2+c_3{z}^3\cdots ,$$

which implies

$$\hslash \left(z\right)={\displaystyle \frac{p\left(z\right)-1}{1+p\left(z\right)}}.$$

$$\hslash \left(z\right)={\displaystyle \frac{c_{1}z+c_2{z}^2+c_3{z}^3+\cdots }{2+c_{1}z+c_2{z}^2+c_3{z}^3+\cdots }}.$$

On expanding, we have

$$f^{\prime }\left(z\right)=1+2a_{2}z+3a_3{z}^2+\cdots .$$

(36)

From (36) and (35), a comparison of the coefficients of $z$, $z^2$, $z^3$, and $z^4,$ we can obtain

$$\begin{array}{ccc}\hfill a_2& =& {\displaystyle \frac{c_1}{4}}\lambda ,\hfill \end{array}$$

(37)

$$\begin{array}{ccc}\hfill a_3& =& {\displaystyle \frac{c_2}{6}}\lambda -\left({\displaystyle \frac{1}{8}}\lambda -{\displaystyle \frac{1}{24}}{\lambda }^2\right)c_1^2,\hfill \end{array}$$

(38)

$$\begin{array}{ccc}\hfill a_4& =& \left({\displaystyle \frac{1}{16}}{\lambda }^2-{\displaystyle \frac{3}{16}}\lambda \right)c_1{c}_2+\left({\displaystyle \frac{7}{96}}\lambda -{\displaystyle \frac{3}{64}}{\lambda }^2+{\displaystyle \frac{1}{192}}{\lambda }^3\right)c_1^3+{\displaystyle \frac{1}{8}}\lambda c_3,\hfill \end{array}$$

(39)

and

$$a_5=\begin{array}{c}\left({\displaystyle \frac{1}{40}}{\lambda }^2-{\displaystyle \frac{3}{40}}\lambda \right)c_2^2+c_1^4\left({\displaystyle \frac{1}{1920}}{\lambda }^4-{\displaystyle \frac{3}{320}}{\lambda }^3+{\displaystyle \frac{83}{1920}}{\lambda }^2-{\displaystyle \frac{3}{64}}\lambda \right)+c_1{c}_3\left({\displaystyle \frac{{\lambda }^2}{20}}-{\displaystyle \frac{3}{20}}\lambda \right)\\ +c_1^2{c}_2\left({\displaystyle \frac{7}{40}}\lambda -{\displaystyle \frac{9}{80}}{\lambda }^2+{\displaystyle \frac{1}{80}}{\lambda }^3\right)+{\displaystyle \frac{\lambda }{10}}{c}_4.\end{array}$$

(40)

Using the first inequality of Lemma 2 in (19), we have

$$\left|a_2\right|\le {\displaystyle \frac{1}{2}}\lambda .$$

Using the second inequality of Lemma 2, we obtained

$$\begin{array}{ccc}\hfill \left|a_3\right|& =& {\displaystyle \frac{\lambda }{6}}\left|c_2-\left({\displaystyle \frac{3}{4}}-{\displaystyle \frac{1}{4}}\lambda \right)c_1^2\right|\hfill \\ & \le & {\displaystyle \frac{\lambda }{3}},\hfill \end{array}$$

with $v={\displaystyle \frac{3}{4}}-{\displaystyle \frac{1}{4}}\lambda$ for $0\le \nu \le 2$ and $\lambda \in \left(0,1\right)$. Now, consider

$$a_4=\left({\displaystyle \frac{1}{16}}{\lambda }^2-{\displaystyle \frac{3}{16}}\lambda \right)c_1{c}_2+\left({\displaystyle \frac{7}{96}}\lambda -{\displaystyle \frac{3}{64}}{\lambda }^2+{\displaystyle \frac{1}{192}}{\lambda }^3\right)c_1^3+{\displaystyle \frac{1}{8}}\lambda c_3.$$

From Lemma 1, then replacing $\left|x\right|=\varsigma$, $\left|\beta \right|\le 1$, $t:=(4-c^2)$ and $c:=c_1\in [0,2]$, after simplification, we have

$$\begin{array}{ccc}\hfill \left|a_4\right|& \le & \left({\displaystyle \frac{1}{96}}\lambda -{\displaystyle \frac{1}{64}}{\lambda }^2+{\displaystyle \frac{1}{192}}{\lambda }^3\right)c^3+\left({\displaystyle \frac{1}{32}}{\lambda }^2-{\displaystyle \frac{\lambda }{32}}\right)c\rho t+{\displaystyle \frac{1}{32}}\lambda c{\rho }^{2}t+{\displaystyle \frac{1}{16}}\lambda t\hfill \\ & :& =k\left(c,\varsigma \right).\hfill \end{array}$$

Partially differentiating with respect to c, we obtain

$${\displaystyle \frac{\partial k}{\partial \varsigma }}=\left({\displaystyle \frac{1}{32}}{\lambda }^2-{\displaystyle \frac{\lambda }{32}}\right)ct+{\displaystyle \frac{1}{16}}\lambda c\varsigma t.$$

The calculation reveals that ${\displaystyle \frac{\partial k}{\partial \varsigma }}>0,$ indicating that $k\left(c,\varsigma \right)$ is an increasing function of $\varsigma$ on the closed interval $\left[0,1\right]$. Consequently, $k\left(c,\varsigma \right)$ reaches its maximum at $\varsigma =1$. Therefore, we have

$$k\left(c,1\right)=H\left(c\right)=\left({\displaystyle \frac{1}{96}}\lambda -{\displaystyle \frac{1}{64}}{\lambda }^2+{\displaystyle \frac{1}{192}}{\lambda }^3\right)c^3+{\displaystyle \frac{1}{32}}{\lambda }^{2}ct+{\displaystyle \frac{1}{16}}\lambda t.$$

Substituting the value of t with $4-c^2,$ we have

$$H\left(c\right)=H_1{c}^3+H_2{c}^2+H_{3}c+H_4,$$

with

$$\begin{array}{ccc}\hfill H_1& =& \left({\displaystyle \frac{1}{96}}\lambda -{\displaystyle \frac{3}{64}}{\lambda }^2+{\displaystyle \frac{1}{192}}{\lambda }^3\right),\hfill \\ \hfill H_2& =& -{\displaystyle \frac{1}{16}}\lambda ,\hfill \\ \hfill H_3& =& {\displaystyle \frac{1}{8}}{\lambda }^2,\hfill \\ \hfill H_4& =& {\displaystyle \frac{1}{4}}\lambda .\hfill \end{array}$$

Partially differentiating $H\left(c\right)$ with respect to c, we obtain

$${\displaystyle \frac{\partial G}{\partial c}}={\gamma }_1{c}^2+{\gamma }_{2}c+H_3,$$

with

$$\begin{array}{ccc}\hfill {\gamma }_1& =& \left({\displaystyle \frac{1}{32}}\lambda -{\displaystyle \frac{9}{64}}{\lambda }^2+{\displaystyle \frac{1}{64}}{\lambda }^3\right),\hfill \\ \hfill {\gamma }_2& =& -{\displaystyle \frac{1}{8}}\lambda .\hfill \end{array}$$

Calculations show that ${\displaystyle \frac{\partial H}{\partial c}}<0,$ for $c\in$$\left[0,2\right]$ and $\lambda \in \left(0,1\right)$, hence $H\left(c\right)$ attains a maximum at $c=0$, and we obtain

$$|a_4|\le {\displaystyle \frac{1}{4}}\lambda .$$

Now consider,

$$a_5=\begin{array}{c}\left({\displaystyle \frac{1}{40}}{\lambda }^2-{\displaystyle \frac{3}{40}}\lambda \right)c_2^2+c_1^4\left({\displaystyle \frac{1}{1920}}{\lambda }^4-{\displaystyle \frac{3}{320}}{\lambda }^3+{\displaystyle \frac{83}{1920}}{\lambda }^2-{\displaystyle \frac{3}{64}}\lambda \right)+c_1{c}_3\left({\displaystyle \frac{{\lambda }^2}{20}}-{\displaystyle \frac{3}{20}}\lambda \right)\\ +c_1^2{c}_2\left({\displaystyle \frac{7}{40}}\lambda -{\displaystyle \frac{9}{80}}{\lambda }^2+{\displaystyle \frac{1}{80}}{\lambda }^3\right)+{\displaystyle \frac{\lambda }{10}}{c}_4.\end{array}$$

After simplification, we have

$$a_5=\begin{array}{c}-{\displaystyle \frac{1}{10}}\lambda \left[\left(-{\displaystyle \frac{1}{4}}\lambda +{\displaystyle \frac{3}{4}}\right)c_2^2+c_1^4\left(-{\displaystyle \frac{83}{192}}\lambda +{\displaystyle \frac{3}{32}}{\lambda }^2-{\displaystyle \frac{1}{192}}{\lambda }^3+{\displaystyle \frac{15}{32}}\right)+c_1{c}_3\left(-{\displaystyle \frac{\lambda }{2}}+{\displaystyle \frac{3}{2}}\right)\right.\\ \left.-c_1^2{c}_2\left({\displaystyle \frac{7}{4}}-{\displaystyle \frac{9}{8}}\lambda +{\displaystyle \frac{1}{8}}{\lambda }^2\right)-c_4\right].\end{array}$$

We can observe that $0<{\alpha }_2<1$, $0<{\alpha }_1<1$ for $\lambda \in \left(0,1\right)$ and

$$\begin{array}{ccc}& & 8{\alpha }_1\left(1-{\alpha }_1\right)\left[{\left({\alpha }_2{\alpha }_3-2{\alpha }_4\right)}^2+{\left({\alpha }_2\left({\alpha }_1+{\alpha }_2\right)-{\alpha }_3\right)}^2\right]\hfill \\ & & +{\alpha }_2\left(1-{\alpha }_2\right){\left({\alpha }_3-2{\alpha }_1{\alpha }_2\right)}^2-4{\alpha }_1{\alpha }_2^2{\left(1-{\alpha }_2\right)}^2\left(1-{\alpha }_1\right)\hfill \\ & :& =\varphi \left(\lambda \right),\hfill \end{array}$$

where

$$\begin{array}{ccc}\hfill {\alpha }_1& =& -{\displaystyle \frac{1}{4}}\lambda +{\displaystyle \frac{3}{4}},\hfill \\ \hfill {\alpha }_2& =& -{\displaystyle \frac{\lambda }{4}}+{\displaystyle \frac{3}{4}},\hfill \\ \hfill {\alpha }_3& =& {\displaystyle \frac{7}{6}}-{\displaystyle \frac{9}{12}}\lambda +{\displaystyle \frac{1}{12}}{\lambda }^2,\hfill \\ \hfill {\alpha }_4& =& -{\displaystyle \frac{83}{192}}\lambda +{\displaystyle \frac{3}{32}}{\lambda }^2-{\displaystyle \frac{1}{192}}{\lambda }^3+{\displaystyle \frac{15}{32}}.\hfill \end{array}$$

with

$$\begin{array}{ccc}\hfill \varphi \left(\lambda \right)& =& -{\displaystyle \frac{1}{\mathrm{18,432}}}{\lambda }^8+{\displaystyle \frac{7}{9216}}{\lambda }^7-{\displaystyle \frac{55}{\mathrm{18,432}}}{\lambda }^6-{\displaystyle \frac{1}{288}}{\lambda }^5+{\displaystyle \frac{365}{\mathrm{18,432}}}{\lambda }^4\hfill \\ & & +{\displaystyle \frac{187}{9216}}{\lambda }^3-{\displaystyle \frac{187}{6144}}{\lambda }^2-{\displaystyle \frac{25}{512}}\lambda -{\displaystyle \frac{9}{512}}.\hfill \end{array}$$

Upon computation, it can be observed that $\varphi \left(\lambda \right)<0$ for $\lambda \in \left(0,1\right).$ By using Lemma 3, we have

$$|a_5|\le {\displaystyle \frac{\lambda }{5}}.$$

The sharpness of the results can be obtained by taking the function $g_n:U\to \mathbb{C}$ given by

$$g_n\left(z\right)={\int }_0^z{\left(1+t^n\right)}^{\lambda }dt,n=1,2,3,4.$$

Then,

$$f_n^{\prime }\left(z\right)={\left(1+z^n\right)}^{\lambda },n=1,2,3,4.$$

Hence, $g_n\in R{L}_{\lambda }$ and

$$g_1\left(z\right)={\int }_0^z{\left(1+t\right)}^{\lambda }dt=z+{\displaystyle \frac{\lambda }{2}}{z}^2+{\displaystyle \frac{\lambda }{6}}\left(\lambda -1\right)z^3+\cdots ,$$

(41)

$$g_2\left(z\right)={\int }_0^z{\left(1+t^2\right)}^{\lambda }dt=z+{\displaystyle \frac{\lambda }{3}}{z}^3+\cdots ,$$

(42)

$$g_3\left(z\right)={\int }_0^z{\left(1+t^3\right)}^{\lambda }dt=z+{\displaystyle \frac{\lambda }{4}}{z}^4+\cdots ,$$

(43)

$$g_4\left(z\right)={\int }_0^z{\left(1+t^4\right)}^{\lambda }dt=z+{\displaystyle \frac{\lambda }{5}}{z}^5+\cdots .$$

(44)

Putting a value of $\lambda ={\displaystyle \frac{1}{2}}$ in the above results, the following results have been proven in [28]. □

Corollary 6.

If $f\in RL$ has the series form as given (1), then

$$|a_2|\le {\displaystyle \frac{1}{4}},$$

$$|a_3|\le {\displaystyle \frac{1}{6}},$$

$$|a_4|\le {\displaystyle \frac{1}{8}},$$

$$|a_5|\le {\displaystyle \frac{1}{10}}.$$

Conjecture 2.

If $f\in R{L}_{\lambda }$, $\lambda \in \left(0,1\right)$ has the series form as given by (1), then

$$\left|a_n\right|\le {\displaystyle \frac{\lambda }{n}}\left(n\ge 2\right).$$

Theorem 7.

If $f\in R{L}_{\lambda }$, $\lambda \in \left(0,1\right)$ has the series form as given by (1), then

$$\left|a_2{a}_4-a_3^2\right|\le {\displaystyle \frac{{\lambda }^2}{9}}.$$

(45)

The result is sharp.

Proof. 

From (37) to (39), we have

$$a_2{a}_4-a_3^2=\left({\displaystyle \frac{{\lambda }^3}{576}}-{\displaystyle \frac{{\lambda }^2}{192}}\right)c_1^2{c}_2+\left({\displaystyle \frac{{\lambda }^2}{384}}-{\displaystyle \frac{{\lambda }^3}{768}}-{\displaystyle \frac{{\lambda }^4}{2304}}\right)c_1^4+{\displaystyle \frac{{\lambda }^2}{32}}{c}_1{c}_3-{\displaystyle \frac{{\lambda }^2}{36}}{c}_2^2.$$

From Lemma 1, applying a triangle inequality and replacing $\left|x\right|=\rho$, $\left|\beta \right|\le 1$ and $c:=c_1\in [0,2]$, after simplification, we obtain

$$\begin{array}{ccc}\hfill \left|a_2{a}_4-a_3^2\right|& \le & \left({\displaystyle \frac{{\lambda }^2}{1152}}-{\displaystyle \frac{{\lambda }^3}{2304}}-{\displaystyle \frac{{\lambda }^4}{2304}}\right)c^4+\left({\displaystyle \frac{{\lambda }^3}{1152}}-{\displaystyle \frac{{\lambda }^2}{1152}}\right)t\rho c^2+\hfill \\ & & \left({\displaystyle \frac{{\lambda }^2}{128}}\right)t{\rho }^2{c}^2+\left({\displaystyle \frac{{\lambda }^2}{64}}\right)ct+\left({\displaystyle \frac{{\lambda }^2}{144}}\right){\rho }^2{t}^2,\hfill \\ & :& =\psi \left(c,\rho \right).\hfill \end{array}$$

Taking the partial derivative with respect to $\rho$ we obtain,

$${\displaystyle \frac{\partial \psi }{\partial \rho }}=\left({\displaystyle \frac{{\lambda }^3}{1152}}-{\displaystyle \frac{{\lambda }^2}{1152}}\right)t{c}^2+\left({\displaystyle \frac{{\lambda }^2}{64}}\right)t\rho c^2+\left({\displaystyle \frac{{\lambda }^2}{72}}\right)\rho t^2.$$

Calculations give ${\displaystyle \frac{\partial \psi }{\partial \rho }}>0$ and then $\psi \left(c,\rho \right)$ is an increasing function of $\rho$ on the closed interval $\left[0,1\right]$. It follows that $\psi \left(c,\rho \right)$ attains maximum at $\rho =1$. Hence, we obtain

$$\begin{array}{ccc}\hfill \psi \left(c,1\right)& =& \left({\displaystyle \frac{{\lambda }^2}{1152}}-{\displaystyle \frac{{\lambda }^3}{2304}}-{\displaystyle \frac{{\lambda }^4}{2304}}\right)c^4+\left({\displaystyle \frac{{\lambda }^3}{1152}}-{\displaystyle \frac{{\lambda }^2}{144}}\right)t{c}^2\hfill \\ & & +{\displaystyle \frac{{\lambda }^2}{64}}ct+{\displaystyle \frac{{\lambda }^2}{144}}{t}^2,\hfill \\ & :& =h\left(c\right).\hfill \end{array}$$

Putting the value of $t=4-c^2,$ after simplification, we have

$$h\left(c\right)=-\left(-{\displaystyle \frac{{\lambda }^2}{1152}}+{\displaystyle \frac{{\lambda }^3}{768}}+{\displaystyle \frac{{\lambda }^4}{2304}}\right)c^4-{\displaystyle \frac{1}{64}}{\lambda }^2{c}^3-\left(-{\displaystyle \frac{1}{288}}{\lambda }^3+{\displaystyle \frac{1}{36}}{\lambda }^2\right)c^2+{\displaystyle \frac{1}{16}}{\lambda }^{2}c+{\displaystyle \frac{1}{9}}{\lambda }^2,$$

$$h\left(c\right)=T_1{c}^4+T_2{c}^3+T_3{c}^2+T_{4}c+T_5,$$

where

$$\begin{array}{ccc}\hfill T_1& =& -\left(-{\displaystyle \frac{{\lambda }^2}{1152}}+{\displaystyle \frac{{\lambda }^3}{768}}+{\displaystyle \frac{{\lambda }^4}{2304}}\right),\hfill \\ \hfill T_2& =& -{\displaystyle \frac{1}{64}}{\lambda }^2,\hfill \\ \hfill T_3& =& -\left(-{\displaystyle \frac{1}{288}}{\lambda }^3+{\displaystyle \frac{1}{36}}{\lambda }^2\right),\hfill \\ \hfill T_4& =& {\displaystyle \frac{1}{16}}{\lambda }^2,\hfill \\ \hfill T_5& =& {\displaystyle \frac{1}{9}}{\lambda }^2.\hfill \end{array}$$

Differentiating $h\left(c\right)$ with respect to c, we have

$${\displaystyle \frac{\partial h}{\partial c}}={\kappa }_1{c}^3+{\kappa }_2{c}^2+{\kappa }_{3}c+T_4^{\prime },$$

with

$${\kappa }_1=-\left(-{\displaystyle \frac{{\lambda }^2}{288}}+{\displaystyle \frac{{\lambda }^3}{192}}+{\displaystyle \frac{{\lambda }^4}{576}}\right),$$

$${\kappa }_2=-{\displaystyle \frac{3}{64}}{\lambda }^2,$$

$${\kappa }_3=-\left(-{\displaystyle \frac{1}{144}}{\lambda }^3+{\displaystyle \frac{1}{18}}{\lambda }^2\right).$$

Upon computation, it can be observed that ${\displaystyle \frac{\partial h}{\partial c}}<0.$ Thus, $h\left(c\right)$ is a decreasing function of c on the closed interval $\left[0,2\right]$. It follows that $h\left(c\right)$ attains a maximum at $c=0$. Hence, we obtain

$$\left|a_2{a}_4-a_3^2\right|\le {\displaystyle \frac{{\lambda }^2}{9}}.$$

The result is sharp for the given function $g_2$ in (42).

For $\lambda ={\displaystyle \frac{1}{2}},$ this above result has been shown in [28]. □

Corollary 7.

If $f\in RL$ has the series form as given by (1), then

$$\left|a_2{a}_4-a_3^2\right|\le {\displaystyle \frac{1}{36}}.$$

Theorem 8.

If $f\in R{L}_{\lambda }$, $\lambda \in \left(0,1\right)$ has the series form as given by (1), then

$$\left|a_3-a_2^2\right|\le {\displaystyle \frac{\lambda }{3}}.$$

(46)

The result is sharp.

Proof. 

Using (37) and (38), we have

$$a_3-a_2^2={\displaystyle \frac{1}{6}}\lambda c_2-{\displaystyle \frac{\lambda }{8}}{c}_1^2-{\displaystyle \frac{{\lambda }^2}{48}}{c}_1^2.$$

After simplification, we obtain

$$\left|a_3-a_2^2\right|={\displaystyle \frac{1}{6}}\lambda \left|c_2-c_1^2\left({\displaystyle \frac{3}{4}}+{\displaystyle \frac{\lambda }{8}}\right)\right|.$$

Using Lemma 4, we obtain

$$\left|a_3-a_2^2\right|\le {\displaystyle \frac{1}{3}}\lambda ,$$

with $\mu =\left({\displaystyle \frac{3}{4}}+{\displaystyle \frac{\lambda }{8}}\right)$ for $\lambda \in \left(0,1\right)$ and $0\le \mu \le 1.$ The result is sharp for the given function $g_2$ in (42).

For $\lambda ={\displaystyle \frac{1}{2}},$ the above result reduces to the following, as proven in [28]. □

Corollary 8.

If $f\in RL$ has the series form as given by (1), then

$$\left|a_3-a_2^2\right|\le {\displaystyle \frac{1}{6}}.$$

Theorem 9.

If $f\in R{L}_{\lambda }$, $\lambda \in \left(0,1\right)$ has the series form as given by (1), then

$$\left|H_{3,1}\left(f\right)\right|\le {\displaystyle \frac{1}{16}}{\lambda }^2.$$

Proof. 

${\mathrm{H}}_{3,1}\left(f\right)=2a_2{a}_3{a}_4-a_3^3-a_4^2+a_3{a}_5-a_2^2{a}_5$. Using (37)–(40) in (4), we have

$$\begin{array}{cc}\hfill H_{3,1}\left(f\right)& ={\displaystyle \frac{{\lambda }^2}{\mathrm{552,960}}}\left[\left(1728\lambda +\mathrm{12,096}\right)c_1{c}_2{c}_3+\left(-1296+288{\lambda }^2+504\lambda +24{\lambda }^3\right)c_1^4{c}_2\right.\hfill \\ \hfill & +\left(288+432\lambda +144{\lambda }^2\right)c_1^3{c}_3+\left(1872-1152\lambda -336{\lambda }^2\right)c_1^2{c}_2^2+9216c_2{c}_4\hfill \\ \hfill & +\left(-1152\lambda -6912\right)c_1^2{c}_4-8640c_3^2-c_2^3\left(6912+256\lambda \right)\hfill \\ \hfill & \left.+c_1^6\left(300-108\lambda -105{\lambda }^2-18{\lambda }^3-{\lambda }^4\right)\right].\hfill \end{array}$$

(47)

Now applying Lemma 1 in (47), we obtain

$$\begin{array}{cc}\hfill H_{3,1}\left(f\right)& ={\displaystyle \frac{{\lambda }^2}{\mathrm{552,960}}}\left[72{\lambda }^2{c}^{3}t\left(1-{\left|x\right|}^2\right)y+72\lambda c^{3}t\left(1-{\left|x\right|}^2\right)y\right.\hfill \\ \hfill & +1008cx{t}^2\left(1-{\left|x\right|}^2\right)y-1152c^{3}t\left(1-{\left|x\right|}^2\right)yx-144\lambda c^{4}t{x}^3\hfill \\ \hfill & +1152c^{2}t\left(1-{\left|x\right|}^2\right)y^2\overline{x}-144x^2{t}^{2}c\left(1-{\left|x\right|}^2\right)y-1080c^2{t}^2{x}^3\hfill \\ \hfill & -2304x{t}^2\left(1-{\left|x\right|}^2\right)y^2\overline{x}+432\lambda cx{t}^2\left(1-{\left|x\right|}^2\right)y+2304x^3{t}^2\hfill \\ \hfill & +576\lambda c^{3}t\left(1-{\left|x\right|}^2\right)yx+576\lambda c^{2}t\left(1-{\left|x\right|}^2\right)y^2\overline{x}+468c^2{t}^2{x}^2\hfill \\ \hfill & -2160t^2{\left(1-{\left|x\right|}^2\right)}^2{y}^2-864x^3{t}^3-1152x^{2}t{c}^2+360c^{4}t{x}^2\hfill \\ \hfill & -864x^3{t}^3-1152x^{2}t{c}^2-2160t^2{\left(1-{\left|x\right|}^2\right)}^2{y}^2-84{\lambda }^2{c}^2{t}^2{x}^2\hfill \\ \hfill & +\left(12+4\lambda -9{\lambda }^2-6{\lambda }^3-{\lambda }^4\right)c^6-72c^{4}tx-288c^{4}t{x}^3-32\lambda x^3{t}^3\hfill \\ \hfill & +12{\lambda }^3{c}^{4}tx+48{\lambda }^2{c}^{4}tx-36{\lambda }^2{c}^{4}t{x}^2-576\lambda c^{2}t{x}^2+108\lambda c^{4}t{x}^2\hfill \\ \hfill & -144c^{3}t\left(1-{\left|x\right|}^2\right)y-576\lambda c^{2}t\left(1-{\left|x\right|}^2\right)\left(1-{\left|y\right|}^2\right)z+36x^4{t}^2{c}^2\hfill \\ \hfill & -1152c^{2}t\left(1-{\left|x\right|}^2\right)\left(1-{\left|y\right|}^2\right)z+2304x{t}^{2}t\left(1-{\left|x\right|}^2\right)\left(1-{\left|y\right|}^2\right)z\hfill \\ \hfill & \left.+48\lambda c^2{t}^2{x}^2-216\lambda c^2{t}^2{x}^3.\right]\hfill \end{array}$$

Therefore,

$$H_{3,1}\left(f\right)={\displaystyle \frac{{\lambda }^2}{\mathrm{552,960}}}\left[L_1\left(c,x\right)+L_2\left(c,x\right)y+L_3\left(c,x\right)y^2+\mathsf{\Psi }\left(c,x,y\right)z\right].$$

where $x,y,z\in U$, and

$$\begin{array}{cc}\hfill L_1\left(c,x\right)& =\left[\left(12+4\lambda -9{\lambda }^2-6{\lambda }^3-{\lambda }^4\right)c^6+12{\lambda }^3{c}^{4}tx+48{\lambda }^2{c}^{4}tx\right.\hfill \\ \hfill & -36{\lambda }^2{c}^{4}t{x}^2-84{\lambda }^2{c}^2{t}^2{x}^2+360c^{4}t{x}^2+468c^2{t}^2{x}^2-72c^{4}tx\hfill \\ \hfill & -1080c^2{t}^2{x}^3-144\lambda c^{4}t{x}^3-576\lambda c^{2}t{x}^2+2304x^3{t}^2-864x^3{t}^3\hfill \\ \hfill & -1152x^{2}t{c}^2-288c^{4}t{x}^3+36x^4{t}^2{c}^2-32\lambda x^3{t}^3+12\lambda c^{4}tx\hfill \\ \hfill & +108\lambda c^{4}t{x}^2\left.+48\lambda c^2{t}^2{x}^2-216\lambda c^2{t}^2{x}^3\right],\hfill \end{array}$$

$$\begin{array}{cc}\hfill L_2\left(c,x\right)& =72{\lambda }^2{c}^{3}t\left(1-{\left|x\right|}^2\right)y+72\lambda c^{3}t\left(1-{\left|x\right|}^2\right)\hfill \\ \hfill & +1008cx{t}^2\left(1-{\left|x\right|}^2\right)-1152c^{3}t\left(1-{\left|x\right|}^2\right)x\hfill \\ \hfill & -144x^2{t}^{2}c\left(1-{\left|x\right|}^2\right)+432\lambda cx{t}^2\left(1-{\left|x\right|}^2\right)\hfill \\ \hfill & +576\lambda c^{3}t\left(1-{\left|x\right|}^2\right)x-144c^{3}t\left(1-{\left|x\right|}^2\right),\hfill \end{array}$$

$$\begin{array}{ccc}\hfill L_3\left(c,x\right)& =& 1152c^{2}t\left(1-{\left|x\right|}^2\right)\overline{x}-2304x{t}^2\left(1-{\left|x\right|}^2\right)\overline{x}\hfill \\ & & +576\lambda c^{2}t\left(1-{\left|x\right|}^2\right)\overline{x}-2160t^2{\left(1-{\left|x\right|}^2\right)}^2,\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \mathsf{\Psi }\left(c,x,y\right)& =& -1152c^{2}t\left(1-{\left|x\right|}^2\right)\left(1-{\left|y\right|}^2\right)+2304x{t}^{2}t\left(1-{\left|x\right|}^2\right)\left(1-{\left|y\right|}^2\right)\hfill \\ & & -576\lambda c^{2}t\left(1-{\left|x\right|}^2\right)\left(1-{\left|y\right|}^2\right).\hfill \end{array}$$

We can write

$$\left|H_{3,1}\left(f\right)\right|\le {\displaystyle \frac{{\lambda }^2}{\mathrm{552,960}}}\left[\left|L_1\left(c,x\right)\right|+\left|L_2\left(c,x\right)\right|\beta +\left|L_3\left(c,x\right)\right|{\beta }^2+\left|\mathsf{\Psi }\left(c,x,y\right)\right|\rho \right].$$

By applying the condition $\left|z\right|\le 1$ and using $\left|x\right|=x$, $\left|\beta \right|=y$, we obtain

$$\begin{array}{ccc}\hfill \left|H_{3,1}\left(f\right)\right|& \le & {\displaystyle \frac{{\lambda }^2}{\mathrm{552,960}}}\left[\left|L_1\left(c,x\right)\right|+\left|L_2\left(c,x\right)\right|y+\left|L_3\left(c,x\right)\right|y^2+\left|\mathsf{\Psi }\left(c,x,y\right)\right|\right]\hfill \\ & :& =G\left(c,x,y\right),\hfill \end{array}$$

where

$$G\left(c,x,y\right)={\displaystyle \frac{{\lambda }^2}{\mathrm{552,960}}}\left[g_1\left(c,x\right)+g_2\left(c,x\right)y+g_2\left(c,x\right)y^2+g_4\left(c,x,y\right)\left(1-y^2\right)\right],$$

(48)

with

$$\begin{array}{cc}\hfill g_1\left(c,x\right)& =\left(12+4\lambda +9{\lambda }^2+6{\lambda }^3+{\lambda }^4\right)c^6+12{\lambda }^3{c}^{4}tx+48{\lambda }^2{c}^{4}tx\hfill \\ \hfill & +36{\lambda }^2{c}^{4}t{x}^2+84{\lambda }^2{c}^2{t}^2{x}^2+360c^{4}t{x}^2+468c^2{t}^2{x}^2\hfill \\ \hfill & +72c^{4}tx+1080c^2{t}^2{x}^3+144\lambda c^{4}t{x}^3+576\lambda c^{2}t{x}^2\hfill \\ \hfill & +2304x^3{t}^2+864x^3{t}^3+1152x^{2}t{c}^2+288c^{4}t{x}^3\hfill \\ \hfill & +36x^4{t}^2{c}^2+32\lambda x^3{t}^3+12\lambda c^{4}tx+108\lambda c^{4}t{x}^2\hfill \\ \hfill & +48\lambda c^2{t}^2{x}^2+216\lambda c^2{t}^2{x}^3,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill g_2\left(c,x\right)& =& 72{\lambda }^2{c}^{3}t\left(1-x^2\right)+72\lambda c^{3}t\left(1-x^2\right)+1008cx{t}^2\left(1-x^2\right)\hfill \\ & & +1152c^{3}t\left(1-x^2\right)x+144x^2{t}^{2}c\left(1-x^2\right)+432\lambda cx{t}^2\left(1-x^2\right)\hfill \\ & & +576\lambda c^{3}t\left(1-x^2\right)x+144c^{3}t\left(1-x^2\right),\hfill \end{array}$$

$$\begin{array}{ccc}\hfill g_3\left(c,x\right)& =& 1152c^{2}t\left(1-x^2\right)x-2304x{t}^2\left(1-x^2\right)x\hfill \\ & & +576\lambda c^{2}t\left(1-x^2\right)x-2160t^2{\left(1-x^2\right)}^2,\hfill \end{array}$$

and

$$g_4\left(c,x,y\right)=1152c^{2}t\left(1-x^2\right)+2304x{t}^{2}t\left(1-x^2\right)-576\lambda c^{2}t\left(1-x^2\right).$$

Let us examine the closed cuboid $\left[0,2\right]\times \left[0,1\right]\times \left[0,1\right]$ symbolized by $\mathsf{\Delta }.$ Our goal is to maximize $G\left(c,x,y\right)$ inside, and in the interior of all the 6 faces and the 12 edges of $\mathsf{\Delta }.$

(1)

Assume that $(c,x,y)\in (0,2)\times (0,1)\times (0,1)$. Now, to find points of maxima inside $\mathsf{\Delta }$, we calculate the partial derivative of (48) which is possible if) with respect to y, and we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{\partial G}{\partial y}}& ={\displaystyle \frac{1}{3840}}{\lambda }^2{c}^{3}t\left(1-x^2\right)+{\displaystyle \frac{1}{7680}}{\lambda }^4{c}^{3}t\left(1-x^2\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{7680}}{\lambda }^3{c}^{3}t\left(1-x^2\right)+{\displaystyle \frac{1}{480}}{\lambda }^2{c}^{3}t\left(1-x^2\right)x\hfill \\ \hfill & +{\displaystyle \frac{1}{960}}{\lambda }^3{c}^{3}t\left(1-x^2\right)x+{\displaystyle \frac{1}{1280}}{\lambda }^{3}c{t}^{2}x\left(1-x^2\right)\hfill \\ \hfill & +{\displaystyle \frac{7}{3840}}{\lambda }^{2}c{t}^{2}x\left(1-x^2\right)+{\displaystyle \frac{1}{3840}}{\lambda }^{2}c{t}^2{x}^2\left(1-x^2\right)\hfill \\ \hfill & +2y\left({\displaystyle \frac{1}{480}}{\lambda }^2{c}^{2}t\left(1-x^2\right)x+{\displaystyle \frac{1}{240}}{\lambda }^{2}x{t}^2\left(1-x^2\right)x\right.\hfill \\ \hfill & \left.+{\displaystyle \frac{1}{960}}{\lambda }^3{c}^{2}t\left(1-x^2\right)x+{\displaystyle \frac{1}{256}}{\lambda }^2{t}^2{\left(1-x^2\right)}^2\right)\hfill \\ \hfill & -2y\left({\displaystyle \frac{1}{480}}{\lambda }^2{c}^{2}t\left(1-x^2\right)+{\displaystyle \frac{1}{240}}{\lambda }^2{t}^2\left(1-x^2\right)x\right.\hfill \\ \hfill & \left.+{\displaystyle \frac{1}{960}}{\lambda }^3{c}^{2}t\left(1-x^2\right)\right).\hfill \end{array}$$

For ${\displaystyle \frac{\partial g}{\partial y}}=0$ and $t=4-c^2$, after simplification, we obtain

$$y=-{\displaystyle \frac{1}{4}}\left[{\displaystyle \frac{\left(2c^2+{\lambda }^2{c}^2-\lambda c^2+24x\lambda +2\lambda c^{2}x+56x+2c^{2}x+8x^2-2c^2{x}^2\right)c}{-c^2{x}^2+4x^2+4\lambda c^{2}x-64x+24x{c}^2-23c^2+60-4\lambda c^2}}\right]:=y_o.$$

If $y_o$ is a critical point and $y_o$ lies inside $\mathsf{\Delta }$, it means $y_o\in (0,1)$, which is possible if

$$\begin{array}{c}-\left(2c^2+{\lambda }^2{c}^2-\lambda c^2+24x\lambda +2\lambda c^{2}x\right.\\ \left.+56x+2c^{2}x+8x^2-2c^2{x}^2\right)c\end{array}<\begin{array}{c}4\left(-c^2{x}^2+4x^2+4\lambda c^{2}x\right.\\ \left.-64x+24x{c}^2-23c^2+60-4\lambda c^2\right)\end{array},$$

(49)

and

$${\displaystyle \frac{\left(24x\lambda +56x+8x^2\right)}{\left(2+{\lambda }^2+\lambda +2\lambda x+2x-2x^2\right)}}<c^2.$$

(50)

Suppose,

$$T\left(x\right)={\displaystyle \frac{\left(24x\lambda +56x+8x^2\right)}{\left(2+{\lambda }^2+\lambda +2\lambda x+2x-2x^2\right)}}.$$

Since $T^{\prime }\left(x\right)>0$, for $x\in (0,1)$, $T\left(x\right)$ is increasing in $(0,1).$ Hence, $c^2>{\displaystyle \frac{24\lambda +24}{2+{\lambda }^2+3\lambda }}$ and the computation reveals that there is no critical point of G in $(0,2)\times (0,1)\times (0,1)$. Consequently, (49) does not hold for all $x\in (0,1)$.

(2)

We next consider the case for the interior of the six faces of $\mathsf{\Delta }$.

When $c=0$ (48) has the form

$$\begin{array}{cc}\hfill h_1\left(x,y\right)& =g\left(0,x,y\right)=-{\displaystyle \frac{7}{120}}{\lambda }^2{x}^2{y}^2-{\displaystyle \frac{1}{240}}{\lambda }^2{x}^4{y}^2-{\displaystyle \frac{1}{15}}{\lambda }^{2}x{y}^2\hfill \\ \hfill & +{\displaystyle \frac{1}{10}}{\lambda }^2{x}^3+{\displaystyle \frac{1}{15}}{\lambda }^2{x}^3{y}^2+{\displaystyle \frac{1}{15}}{\lambda }^{2}x+{\displaystyle \frac{1}{270}}{\lambda }^3{x}^3+{\displaystyle \frac{1}{16}}{\lambda }^2{y}^2.\hfill \end{array}$$

(51)

Since $h_1\left(x,y\right)$ has no critical point in $(0,1)\times (0,1)$. Since

$${\displaystyle \frac{\partial h_1}{\partial y}}=-{\displaystyle \frac{7}{60}}{\lambda }^2{x}^{2}y-{\displaystyle \frac{1}{120}}{\lambda }^2{x}^{4}y+{\displaystyle \frac{2}{15}}{\lambda }^{2}xy+{\displaystyle \frac{2}{15}}{\lambda }^2{x}^{3}y+{\displaystyle \frac{1}{8}}{\lambda }^{2}y\ne 0x,y\in \left(0,1\right).$$

When $c=2$, (48) reduces to

$$h_2=g\left(2,x,y\right)={\displaystyle \frac{1}{720}}{\lambda }^2+{\displaystyle \frac{1}{2160}}{\lambda }^3+{\displaystyle \frac{1}{960}}{\lambda }^4+{\displaystyle \frac{1}{1440}}{\lambda }^5+{\displaystyle \frac{1}{8640}}{\lambda }^6.$$

(52)

When $x=1$, (48) reduces to

$$\begin{array}{ccc}\hfill h_3\left(c\right)& =& g\left(c,1,y\right)={\displaystyle \frac{{\lambda }^2}{\mathrm{552,960}}}\left[\left(12-28\lambda +9{\lambda }^2-6{\lambda }^3+{\lambda }^4\right)c^6\right.\hfill \\ & & +\left(-336{\lambda }^2+48{\lambda }^3-1248\lambda +1728\right)c^4+92,160\hfill \\ & & +\left(1344{\lambda }^2+4992\lambda -29952\right)c^2\left.+2048\lambda \right].\hfill \end{array}$$

(53)

Solving ${\displaystyle \frac{\partial h_3}{\partial c}}=0,$ we obtain the optimal point at $c=0$. Thus, $h_3$ achieves its maximum at $c=0$, which is as follows:

$$h_3\left(0\right)\le \left({\displaystyle \frac{{\lambda }^3}{270}}+{\displaystyle \frac{{\lambda }^2}{6}}\right).$$

When $x=0$, (48) reduces to

$$\begin{array}{cc}\hfill h_4\left(c,y\right)& =g\left(c,0,y\right)={\displaystyle \frac{{\lambda }^2}{\mathrm{552,960}}}\left[\left(\lambda +3\right)\left({\lambda }^3+3{\lambda }^2+4\right)c^6-72y\left(2+\lambda +{\lambda }^2\right)c^5\right.\hfill \\ \hfill & +\left(-1152+3312y^2-576\lambda y^2+576\lambda \right)c^4+288y\left(2+\lambda +{\lambda }^2\right)c^3\hfill \\ \hfill & +\mathrm{34,560}{y}^2\left.+\left(-\mathrm{21,888}{y}^2+4608-2304\lambda +2304\lambda y^2\right)c^2\right].\hfill \end{array}$$

(54)

On $y=1$, a numerical method shows that there is no solution in $(0,2)\times (0,1)$ for the system of equations ${\displaystyle \frac{\partial g\left(c,x\right)}{\partial c}}=0$ and ${\displaystyle \frac{\partial g\left(c,x\right)}{\partial c}}=0$.

When $y=0,$ there does not exist any solution for the system of equations ${\displaystyle \frac{\partial g\left(c,x\right)}{\partial c}}=0$ and ${\displaystyle \frac{\partial g\left(c,x\right)}{\partial c}}=0$ in $(0,2)\times (0,1)$.

(3)

Now, we are going to find the maxima of $G(c,x,y)$ on the edges of $\mathsf{\Delta }$.

By taking $y=1$ in (51), we have

$$h_5\left(x\right)=g\left(0,x,1\right)={\displaystyle \frac{1}{16}}{\lambda }^2{\left(1-x^2\right)}^2+{\displaystyle \frac{1}{15}}{\lambda }^2{x}^2\left(1-x^2\right)+{\displaystyle \frac{1}{6}}{\lambda }^2{x}^3+{\displaystyle \frac{1}{270}}{\lambda }^3{x}^3.$$

Since $h_5^{\prime }\left(x\right)>0$ for $x\in \left(0,1\right)$, we can see that $h_5\left(x\right)$ is increasing in $[0,1]$ and hence attains its maximum value at $x=1$. Thus, we have

$$h_5\left(1\right)\le {\displaystyle \frac{1}{6}}{\lambda }^2+{\displaystyle \frac{1}{270}}{\lambda }^3.$$

By putting $y=0$ in (54), we have

$$\begin{array}{ccc}\hfill h_6\left(c\right)& =& g_1\left(c,0,0\right)=c^6\left({\displaystyle \frac{1}{\mathrm{46,080}}}{\lambda }^2+{\displaystyle \frac{1}{\mathrm{138,240}}}{\lambda }^3+{\displaystyle \frac{1}{\mathrm{61,440}}}{\lambda }^4+{\displaystyle \frac{1}{\mathrm{92,160}}}{\lambda }^5+{\displaystyle \frac{1}{\mathrm{552,960}}}{\lambda }^6\right)\hfill \\ & & +\left({\displaystyle \frac{1}{120}}{\lambda }^2+{\displaystyle \frac{1}{240}}{\lambda }^3\right)c^2-\left({\displaystyle \frac{1}{480}}{\lambda }^2+{\displaystyle \frac{1}{960}}{\lambda }^3\right)c^4-{\displaystyle \frac{1}{480}}{\lambda }^2{c}^4.\hfill \end{array}$$

We can observe that $h_6^{\prime }\left(c\right)>0$ for $c\in [0,2]$; we see that $h_6\left(c\right)$ is increasing in $[0,2]$ and hence attains its maximum value at $c=2$. Thus, we have

$$h_6\left(2\right)\le {\displaystyle \frac{1}{720}}{\lambda }^2+{\displaystyle \frac{1}{2160}}{\lambda }^3+{\displaystyle \frac{1}{960}}{\lambda }^4+{\displaystyle \frac{1}{1440}}{\lambda }^5+{\displaystyle \frac{1}{8640}}{\lambda }^6.$$

By putting $y=1$ in (54), we have

$$\begin{array}{cc}\hfill h_7\left(c\right)& =g\left(c,0,1\right)={\displaystyle \frac{{\lambda }^2}{\mathrm{552,960}}}\left[\left(\lambda +3\right)\left({\lambda }^3+3{\lambda }^2+4\right)c^6-72y\left(2+\lambda +{\lambda }^2\right)c^5\right.\hfill \\ \hfill & +\left(-1152+3312-576\lambda +576\lambda \right)c^4+288\left(2+\lambda +{\lambda }^2\right)c^3\hfill \\ \hfill & \left.+\left(-\mathrm{21,888}+4608-2304\lambda +2304\lambda \right)c^2+\mathrm{34,560}\right].\hfill \end{array}$$

Clearly, $h_7^{\prime }\left(c\right)<0,$ thus $h_7\left(c\right)$ is decreasing in $[0,2]$ and hence attains its maximum value at $c=0$. Thus, we have

$$h_7\left(0\right)\le {\displaystyle \frac{1}{16}}{\lambda }^2.$$

As we can see that (53) is independent of y, we have

$$\begin{array}{cc}\hfill h_8\left(c\right)& =g\left(c,1,1\right)=g\left(c,1,0\right)={\displaystyle \frac{{\lambda }^2}{\mathrm{552,960}}}\left[\left(12-28\lambda +9{\lambda }^2-6{\lambda }^3+{\lambda }^4\right)c^6\right.\hfill \\ \hfill & +\left(-336{\lambda }^2+48{\lambda }^3-1248\lambda +1728\right)c^4+\left(1344{\lambda }^2+4992\lambda -\mathrm{29,952}\right)c^2\hfill \\ \hfill & \left.+2048\lambda +92160\right].\hfill \end{array}$$

We can observe that $h_8^{\prime }\left(c\right)<0$ for $c\in [0,2]$, and we can see that $h_8\left(c\right)$ is decreasing in $[0,2].$ Hence, it attains its maximum value at $c=0$. Thus, we have

$$h_8\left(0\right)\le {\displaystyle \frac{{\lambda }^3}{270}}+{\displaystyle \frac{{\lambda }^2}{6}}.$$

By putting $c=0$ in (53), we have

$$g\left(0,1,y\right)={\displaystyle \frac{1}{6}}{\lambda }^2+{\displaystyle \frac{1}{270}}{\lambda }^3$$

$$g\left(2,0,y\right)=g\left(2,1,y\right)={\displaystyle \frac{1}{720}}{\lambda }^2+{\displaystyle \frac{1}{2160}}{\lambda }^3+{\displaystyle \frac{1}{960}}{\lambda }^4+{\displaystyle \frac{1}{1440}}{\lambda }^5+{\displaystyle \frac{1}{8640}}{\lambda }^6.$$

Assuming that $y=0$ in (51), we obtain

$$h_9\left(x\right)=g\left(0,x,0\right)={\displaystyle \frac{1}{10}}{\lambda }^2{x}^3+{\displaystyle \frac{1}{270}}{\lambda }^3{x}^3+{\displaystyle \frac{1}{15}}{\lambda }^{2}x.$$

We can observe that $h_9^{\prime }\left(x\right)>0$ for $c\in [0,2]$; we can observe that $h_9\left(x\right)$ is increasing in $[0,1].$ Thus, it reaches its maximum value at $x=1$. Consequently, we have

$$h_9\left(1\right)\le {\displaystyle \frac{1}{10}}{\lambda }^2+{\displaystyle \frac{1}{270}}{\lambda }^3+{\displaystyle \frac{1}{15}}{\lambda }^2.$$

When we put $y=1$ and $y=0$ in (52), we obtain

$$g\left(2,x,1\right)=g\left(2,x,0\right)={\displaystyle \frac{1}{720}}{\lambda }^2+{\displaystyle \frac{1}{2160}}{\lambda }^3+{\displaystyle \frac{1}{960}}{\lambda }^4+{\displaystyle \frac{1}{1440}}{\lambda }^5+{\displaystyle \frac{1}{8640}}{\lambda }^6.$$

By putting $c=0$ and $x=0$ in (48), we have

$$g\left(0,0,y\right)={\displaystyle \frac{1}{16}}{\lambda }^2{y}^2\le {\displaystyle \frac{1}{16}}{\lambda }^2.$$

A simple calculation gives

$$g\left(0,0,y\right)\le {\displaystyle \frac{1}{16}}{\lambda }^2.$$

From the above cases, we concluded the following:

$$\left|H_{3,1}\left(f\right)\right|\le {\displaystyle \frac{1}{16}}{\lambda }^2.$$

The result is sharp for the $g_4$ given in (44). □

For $\lambda ={\displaystyle \frac{1}{2}},$ the above result reduces to the following, proven in [28].

Corollary 9.

If $f\in RL$ has the series form as given by (1), then

$$\left|H_{3,1}\left(f\right)\right|\le {\displaystyle \frac{1}{64}}.$$

5. Conclusions

Many researchers have dedicated their efforts to defining the subclasses of analytic functions. In this context, we have specifically focused on the subclasses of star-like functions, which are denoted by $S_{L\lambda }^{*}$ and $R{L}_{\lambda }$, associated with Bernoulli’s lemniscate. We derived an initial coefficient estimate and then established the bounds of the third-order Hankel determinant for these classes. All the estimations are proven to be sharp. The Fekete–Szegö functional and the second and third-order Hankel determinants represent noteworthy findings in our research. The results we obtained from our analysis showed a substantial improvement when compared to the results for simpler versions of these already well-known functions. This suggests a notable enhancement in the effectiveness of the studied functions. Many other fascinating dimensions remain unexplored, and we firmly believe this work can serve as a reference point for many more to come.