Toeplitz Matrices for a Class of Bazilevič Functions and the λ-Pseudo-Starlike Functions

Abstract

In the present paper, we define and study a new family of holomorphic functions which involve the Bazilevič functions and the $\lambda$-pseudo-starlike functions. We establish coefficient estimates for the first four determinants of the symmetric Toeplitz matrices $T_2\left(2\right)$, $T_2\left(3\right)$, $T_3\left(1\right)$ and $T_3\left(2\right)$ for the functions in this family. Further, we investigate several special cases and consequences of our results.

1. Introduction

Let $\mathcal{A}$ stand for the family of functions f of the form

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

(1)

which are analytic in the open unit disk $U=\left\{z\in \mathbb{C}:\left|z\right|<1\right\}$. Let S indicate the class of all functions in $\mathcal{A}$ which are univalent in U.

We can say that $f\in \mathcal{A}$ is called a starlike function in U if

$$\Re \left({\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\right)>0,(z\in U)$$

and a function $f\in \mathcal{A}$ is called a convex function in U if

$$\Re \left({\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}+1\right)>0,(z\in U).$$

A function $f\in \mathcal{A}$ is called a Bazilevič function in U if (see [1])

$$\Re \left({\displaystyle \frac{z^{1-\beta }{f}^{\prime }\left(z\right)}{{\left(f\left(z\right)\right)}^{1-\beta }}}\right)>0(\forall z\in U;\beta \ge 0).$$

Otherwise, a function $f\in \mathcal{A}$ is called a $\lambda$-pseudo-starlike function in U if (see [2])

$$\Re \left({\displaystyle \frac{z{\left(f^{\prime }\left(z\right)\right)}^{\lambda }}{f\left(z\right)}}\right)>0(\forall z\in U;\lambda \ge 1).$$

The estimation of the bounds of Hankel matrices is very much in the focus of the univalent function theory. These matrices and determinants have an important role in many fields of mathematics and have several applications [3]. Toeplitz determinants and Hankel determinants are closely related. Hankel matrices have constant entries along the reverse diagonal, while Toeplitz matrices have constant entries along the diagonal. The symmetric Toeplitz determinant $T_q\left(n\right)$ for $f\in \mathcal{A}$ is defined by

$$T_q\left(n\right)=\left|\begin{array}{cccc}{a}_n& a_{n+1}& \dots & a_{n+q-1}\\ a_{n+1}& a_n& \dots & a_{n+q-2}\\ ⋮& ⋮& ⋮& ⋮\\ a_{n+q-1}& a_{n+q-2}& \dots & a_n\end{array}\right|,$$

where $q\ge 1$, $n\ge 1$ and $a_1=1$. In particular,

$$T_2\left(2\right)=\left|\begin{array}{cc}{a}_2& a_3\\ a_3& a_2\end{array}\right|,T_2\left(3\right)=\left|\begin{array}{cc}{a}_3& a_4\\ a_4& a_3\end{array}\right|,$$

and

$$T_3\left(1\right)=\left|\begin{array}{ccc}1& a_2& a_3\\ a_2& 1& a_2\\ a_3& a_2& 1\end{array}\right|,T_3\left(2\right)=\left|\begin{array}{ccc}{a}_2& a_3& a_4\\ a_3& a_2& a_3\\ a_4& a_3& a_2\end{array}\right|.$$

Toeplitz determinants appear in all branches of pure and applied mathematics, including statistics and probability, image processing, quantum mechanics, queuing networks, signal processing and time series analysis. Also, Toeplitz matrices have an important role in functional analysis, applied mathematics, physics and also technical sciences (see [3]). In recent years, several authors have established estimates of the Toeplitz determinant $\left|T_q\left(n\right)\right|$ for various families of univalent functions (see [4,5,6,7,8,9,10,11,12,13,14,15,16,17]).

In our study, we use the following lemmas to derive the desired bounds:

Lemma 1 

([18]). If the function $p\in \mathcal{P}$ is given by the series $p\left(z\right)=1+p_{1}z+p_2{z}^2+p_3{z}^3+\cdots$, then the sharp estimate $\left|p_k\right|\le 2$ $\left(k=1,2,3,\cdots \right)$ holds.

Lemma 2 

([19]). If the function $p\in \mathcal{P}$, then

$$2p_2=p_1^2+\left(4-p_1^2\right)x$$

$$4p_3=p_1^3+2p_1\left(4-p_1^2\right)x-p_1\left(4-p_1^2\right)x^2+2\left(4-p_1^2\right)\left(1-{\left|x\right|}^2\right)z,$$

for some $x,z$ with $\left|x\right|\le 1$ and $\left|z\right|\le 1$.

2. Main Results

First of all, we defined the family $\mathcal{R}(\gamma ,\beta ,\lambda )$ in the following way:

Definition 1. 

A function $f\in \mathcal{A}$ is in the family $\mathcal{R}(\gamma ,\beta ,\lambda )$ if it satisfies the condition

$$\Re \left\{\left(1-\gamma \right){\displaystyle \frac{z^{1-\beta }{f}^{\prime }\left(z\right)}{{\left(f\left(z\right)\right)}^{1-\beta }}}+\gamma {\displaystyle \frac{z{\left(f^{\prime }\left(z\right)\right)}^{\lambda }}{f\left(z\right)}}\right\}>0,$$

where $0\le \gamma \le 1$, $\beta \ge 0$, $\lambda \ge 1$ and $z\in U$.

Theorem 1. 

Let $f\in \mathcal{R}(\gamma ,\beta ,\lambda )$ be given by (1). Then,

$$\left|a_2\right|\le {\displaystyle \frac{2}{J}},$$

$$\left|a_3\right|\le {\displaystyle \frac{2}{S}}+{\displaystyle \frac{4B}{S{J}^2}}$$

and

$$\left|a_4\right|\le {\displaystyle \frac{2}{G}}+{\displaystyle \frac{4A}{JSG}}+{\displaystyle \frac{8N}{3J^{3}SG}},$$

where

$$\begin{array}{cc}\hfill G& =\left(1-\gamma \right)(\beta +3)+\gamma (4\lambda -1),\hfill \\ \hfill J& =\left(1-\gamma \right)(\beta +1)+\gamma (2\lambda -1),\hfill \\ \hfill S& =\left(1-\gamma \right)(\beta +2)+\gamma (3\lambda -1),\hfill \\ \hfill A& =\left(1-\gamma \right)\left(\beta -1\right)(\beta +3)-\gamma (6{\lambda }^2-11\lambda +2),\hfill \\ \hfill B& ={\displaystyle \frac{1}{2}}\left[\left(1-\gamma \right)(\beta -1)(\beta +2)+\gamma \left(2{\lambda }^2-4\lambda +1\right)\right],\hfill \\ \hfill N& ={\displaystyle \frac{1}{2}}\left[\left(1-\gamma \right)\left(\beta -1\right)(\beta +3)-\gamma (6{\lambda }^2-11\lambda +2)\right]\times \hfill \\ \hfill & \times \left[\left(1-\gamma \right)(\beta -1)(\beta +2)+\gamma \left(2{\lambda }^2-4\lambda +1\right)\right]-\left(1-\gamma \right)(\beta +2)-\gamma (3\lambda -1).\hfill \end{array}$$

(2)

Proof. 

Let $f\in \mathcal{R}(\gamma ,\beta ,\lambda )$. Then, there exists $p\in \mathcal{P}$ such that

$$\left(1-\gamma \right){\displaystyle \frac{z^{1-\beta }{f}^{\prime }\left(z\right)}{{\left(f\left(z\right)\right)}^{1-\beta }}}+\gamma {\displaystyle \frac{z{\left(f^{\prime }\left(z\right)\right)}^{\lambda }}{f\left(z\right)}}=p\left(z\right),$$

(3)

where p has the following series representation:

$$p\left(z\right)=1+p_{1}z+p_2{z}^2+p_3{z}^3+\cdots .$$

If we equate the coefficients in (3), these relations follow:

$$\left[\left(1-\gamma \right)(\beta +1)+\gamma (2\lambda -1)\right]a_2=p_1,$$

(4)

$$\left[\left(1-\gamma \right)(\beta +2)+\gamma (3\lambda -1)\right]a_3+{\displaystyle \frac{1}{2}}\left[\left(1-\gamma \right)(\beta -1)(\beta +2)+\gamma \left(2{\lambda }^2-4\lambda +1\right)\right]a_2^2=p_2$$

(5)

and

$$\begin{array}{c}\left[\left(1-\gamma \right)(\beta +3)+\gamma (4\lambda -1)\right]a_4-\left[\left(1-\gamma \right)\left(\beta -1\right)(\beta +3)-\gamma (6{\lambda }^2-11\lambda +2)\right]a_2{a}_3\hfill \\ \hfill +{\displaystyle \frac{1}{6}}\left[\left(1-\gamma \right)(\beta -1)(\beta -2)(\beta +3)+\gamma \left({\displaystyle \frac{2}{3}}\lambda (\lambda -2)(2\lambda -5)-1\right)\right]a_2^3=p_3.\end{array}$$

(6)

Considering (4), (5) and (6), after simplifying, we find that

$$a_2={\displaystyle \frac{1}{J}}{p}_2,$$

(7)

$$a_3={\displaystyle \frac{1}{S}}{p}_2-{\displaystyle \frac{B}{S{J}^2}}{p}_1^2$$

(8)

and

$$a_4={\displaystyle \frac{1}{G}}{p}_3+{\displaystyle \frac{A}{JSG}}{p}_1{p}_2-{\displaystyle \frac{N}{3J^{3}SG}}{p}_1^3,$$

(9)

where $G,J,S,A,B$ and N are given by (2). By applying Lemma 1, we obtain

$$\left|a_2\right|\le {\displaystyle \frac{2}{J}},$$

$$\left|a_3\right|\le {\displaystyle \frac{2}{S}}+{\displaystyle \frac{4B}{S{J}^2}}$$

and

$$\left|a_4\right|\le {\displaystyle \frac{2}{G}}+{\displaystyle \frac{4A}{JSG}}+{\displaystyle \frac{8N}{3J^{3}SG}}.$$

□

Theorem 2. 

Let $f\in \mathcal{R}(\gamma ,\beta ,\lambda )$ be given by (1). Then,

$$\left|T_2\left(2\right)\right|\le {\displaystyle \frac{4(J^4-4J^{2}B+4B^2)}{J^4{S}^2}}-{\displaystyle \frac{4}{J^2}},$$

(10)

where $J,B$ are given by (2).

Proof. 

Considering (7), (8) and (2), it is easy to see that

$$\begin{array}{cc}\hfill \left|T_2\left(2\right)\right|& =\left|a_3^2-a_2^2\right|\hfill \\ \hfill & =\left|{\displaystyle \frac{p_2^2}{S^2}}-{\displaystyle \frac{2B{p}_1^2{p}_2}{S^2{J}^2}}+{\displaystyle \frac{B^2{p}_1^4}{S^2{J}^4}}-{\displaystyle \frac{p_1^2}{J^2}}\right|.\hfill \end{array}$$

By applying Lemma 2 to express $p_2$ in terms of $p_1$, we obtain

$$\begin{array}{cc}\hfill & \left|a_3^2-a_2^2\right|\hfill \\ \hfill & =\left|{\displaystyle \frac{(J^4-4J^{2}B+4B^2)p_1^4}{4J^4{S}^2}}-{\displaystyle \frac{p_1^2}{J^2}}+{\displaystyle \frac{(J^2-2B)p_1^{2}x(4-p_1^2)}{2J^2{S}^2}}+{\displaystyle \frac{x^2{(4-p_1^2)}^2}{4S^2}}\right|.\hfill \end{array}$$

To simplify the notation, we chose $p_1=p$, and since the function p is in the family $\mathcal{P}$ simultaneously, we supposed that $p\in [0,2]$. Applying the triangle inequality with $P=4-p^2$, we can see that

$$\begin{array}{cc}\hfill \left|a_3^2-a_2^2\right|& \le \left|{\displaystyle \frac{(J^4-4J^{2}B+4B^2)p^4}{4J^4{S}^2}}-{\displaystyle \frac{p^2}{J^2}}\right|+{\displaystyle \frac{(J^2-2B)p^2\left|x\right|P}{2J^2{S}^2}}+{\displaystyle \frac{{\left|x\right|}^2{P}^2}{4S^2}}=:F(p,\left|x\right|).\hfill \end{array}$$

It is obvious that $F^{\prime }(p,\left|x\right|)>0$ on $[0,1]$, and thus $F(p,\left|x\right|)\le F(p,\left|1\right|)$.

We can note that $F\left(\left|x\right|\right)$ has a maximum value at $[0,2]$, when $p=2$. It follows that

$$\left|T_2\left(2\right)\right|=\left|a_3^2-a_2^2\right|\le {\displaystyle \frac{4(J^4-4J^{2}B+4B^2)}{J^4{S}^2}}-{\displaystyle \frac{4}{J^2}}.$$

This concludes the proof. □

Remark 1. 

Taking $\gamma =0$ and $\beta =1$ in Theorem 2, we can obtain the result which was proven by Radhika et al. [9] [Theorem 2].

Theorem 3. 

Let $f\in \mathcal{R}(\gamma ,\beta ,\lambda )$ be given by (1). Then,

$$\left|T_2\left(3\right)\right|=|a_4^2-a_3^2|\le {\displaystyle \frac{4{\psi }_1}{9G^2{J}^6{S}^2}}-{\displaystyle \frac{4(J^4-4J^{2}B+4B^2)}{J^4{S}^2}},$$

where

$${\psi }_1=9J^6{S}^2+36J^{5}SA-24J^{3}SN+36J^4{A}^2+16N^2-48J^{2}AN.$$

Proof. 

Applying (8), (9) and (2) and using Lemma 2, we find

$$\begin{array}{c}|a_4^2-a_3^2|=|{\displaystyle \frac{(J^4-4J^{2}B+4B^2)p_1^4}{4J^4{S}^2}}+{\displaystyle \frac{{\psi }_1{p}_1^6}{144G^2{J}^6{S}^2}}-{\displaystyle \frac{(J^2-2B)p_1^{2}x(4-p_1^2)}{2J^2{S}^2}}+{\displaystyle \frac{{\psi }_2{p}_1^4(4-p_1^2)x}{12G^2{J}^4{S}^2}}\hfill \\ -{\displaystyle \frac{{\psi }_3{p}_1^4(4-p_1^2)x^2}{24G^2{J}^{3}S}}-{\displaystyle \frac{x^2{(4-p_1^2)}^2}{4S^2}}+{\displaystyle \frac{{\psi }_4{p}_1^2{(4-p_1^2)}^2{x}^2}{4G^2{J}^2{S}^2}}-{\displaystyle \frac{(JS+A)p_1^2{(4-p_1^2)}^2{x}^3}{4G^{2}JS}}+{\displaystyle \frac{p_1^2{(4-p_1^2)}^2{x}^4}{16G^2}}\\ +{\displaystyle \frac{{\psi }_3{p}_1^3(4-p_1^2){(1-|x|}^2)z}{12G^2{J}^{3}S}}+{\displaystyle \frac{(JS+A)p_1{(4-p_1^2)}^2{(1-|x|}^2)xz}{2G^{2}JS}}-{\displaystyle \frac{p_1{(4-p_1^2)}^2{(1-|x|}^2)x^{2}z}{4G^2}}\\ \hfill +{\displaystyle \frac{{(4-p_1^2)}^2{(1-|x|}^2{)}^2{z}^2}{4G^2}}|,\end{array}$$

where

$$\begin{array}{cc}\hfill {\psi }_2& =3J^4{S}^2+9J^{3}SA-4JSN+6J^2{A}^2-4AN,\hfill \\ \hfill {\psi }_3& =3J^{3}S+6J^{2}A-4N,\hfill \\ \hfill {\psi }_4& =J^2{S}^2+2JSA+A^2.\hfill \end{array}$$

To make the notation simple, we select $p_1=p$, and because the function p is simultaneously in the family $\mathcal{P}$, we can assume that $p\in [0,2]$. Using the triangle inequality with $Z=(1-|x{|}^2)$ and $P=4-r^2$, we obtain

$$\begin{array}{c}|a_4^2-a_3^2|=\left|{\displaystyle \frac{{\psi }_1{p}^6}{144G^2{J}^6{S}^2}}-{\displaystyle \frac{(J^4-4J^{2}B+4B^2)p^4}{4J^4{S}^2}}\right|+{\displaystyle \frac{(J^2-2B)p^2\left|x\right|P}{2J^2{S}^2}}+{\displaystyle \frac{{\psi }_2{p}^{4}P\left|x\right|}{12G^2{J}^4{S}^2}}\hfill \\ +{\displaystyle \frac{{\psi }_3{p}^{4}P{\left|x\right|}^2}{24G^2{J}^{3}S}}+{\displaystyle \frac{{\left|x\right|}^2{P}^2}{4S^2}}+{\displaystyle \frac{{\psi }_4{p}^2{P}^2{\left|x\right|}^2}{4G^2{J}^2{S}^2}}+{\displaystyle \frac{(JS+A)p^2{P}^2{\left|x\right|}^3}{4G^{2}JS}}+{\displaystyle \frac{p^2{P}^2{\left|x\right|}^4}{16G^2}}+{\displaystyle \frac{{\psi }_3{p}^{3}PZ}{12G^2{J}^{3}S}}\\ \hfill +{\displaystyle \frac{(JS+A)p\left|x\right|P^{2}Z}{2G^{2}JS}}+{\displaystyle \frac{{p\left|x\right|}^2{P}^{2}Z}{4G^2}}+{\displaystyle \frac{P^2{Z}^2}{4G^2}}=:F_1(p,\left|x\right|).\end{array}$$

We differentiate $F_1(p,\left|x\right|)$ with regard to $\left|x\right|$ with the use of elementary calculus, and obtain

$$\begin{array}{c}{\displaystyle \frac{\partial F_1(p,\left|x\right|)}{\partial \left|x\right|}}={\displaystyle \frac{(J^2-2B)p^2(4-p^2)}{2J^2{S}^2}}+{\displaystyle \frac{{\psi }_2{p}^4(4-p^2)}{12G^2{J}^4{S}^2}}-{\displaystyle \frac{2{\psi }_3{p}^3(4-p^2)\left|x\right|}{12G^2{J}^{3}S}}+{\displaystyle \frac{{\psi }_3{p}^4(4-p^2)\left|x\right|}{12G^2{J}^{3}S}}\hfill \\ +{\displaystyle \frac{{\psi }_4{p}^2{(4-p^2)}^2\left|x\right|}{2G^2{J}^2{S}^2}}-{\displaystyle \frac{p(JS+A){(4-p^2)}^2{\left|x\right|}^2}{G^{2}JS}}+{\displaystyle \frac{3(JS+A)p^2{(4-p^2)}^2{\left|x\right|}^2}{4G^{2}JS}}-{\displaystyle \frac{p{(4-p^2)}^2{\left|x\right|}^3}{2G^2}}\\ \hfill +{\displaystyle \frac{p^2{(4-p^2)}^2{\left|x\right|}^3}{4G^2}}+{\displaystyle \frac{(JS+A)p{(4-p^2)}^2(1-{\left|x\right|}^2)}{2G^{2}JS}}-{\displaystyle \frac{\left|x\right|{(4-p^2)}^2(1-{\left|x\right|}^2)}{G^2}}+{\displaystyle \frac{p\left|x\right|{(4-p^2)}^2(1-{\left|x\right|}^2)}{2G^2}}.\end{array}$$

We find that $(\partial F_1(p,\left|x\right|)/\partial |x\left|\right)\ge 0$ for $\left|x\right|\in [0,1]$ and fixed $p\in [0,2]$. As a result, $F_1(p,\left|x\right|)$ is an increasing function of $\left|x\right|$. Therefore, $F_1(p,\left|x\right|)\le F_1(p,\left|1\right|)$. Therefore,

$$\begin{array}{c}|a_4^2-a_3^2|\le \left|{\displaystyle \frac{{\psi }_1{p}^6}{144G^2{J}^6{S}^2}}-{\displaystyle \frac{(J^4-4J^{2}B+4B^2)p^4}{4J^4{S}^2}}\right|+{\displaystyle \frac{(J^2-2B)p^2(4-p^2)}{2J^2{S}^2}}+{\displaystyle \frac{{(4-p^2)}^2}{4S^2}}\hfill \\ \hfill +{\displaystyle \frac{(2{\psi }_2+A_{3}JS)p^4(4-p^2)}{24G^2{J}^4{S}^2}}+{\displaystyle \frac{(4{\psi }_4+4(JS+A)JS+J^2{S}^2)p^2{(4-p^2)}^2}{16G^2{J}^2{C}^2}}.\end{array}$$

Now, on $[0,2]$ at $P=2$, we find

$$|a_4^2-a_3^2|\le {\displaystyle \frac{4{\psi }_1}{9G^2{J}^6{S}^2}}-{\displaystyle \frac{4(J^4-4J^{2}B+4B^2)}{J^4{S}^2}}.$$

□

Remark 2. 

Taking $\gamma =0$ and $\beta =1$ in Theorem 3, we can obtain the same result as derived by Radhika et al. [9] [Theorem 3].

Theorem 4. 

Let $f\in \mathcal{R}(\gamma ,\beta ,\lambda )$ be given by (1). Then,

$$|T_3\left(1\right)|=|1+2a_2^2(a_3-1)-a_3^2|\le 1+{\displaystyle \frac{4{\psi }_5}{J^4{S}^2}}-{\displaystyle \frac{8}{J^2}},$$

(11)

where

$${\psi }_5=4J^{3}S-8SB-J^4+4J^{2}B-4B^2.$$

(12)

Proof. 

From (7), (8) and (2), by applying Lemma 2 and some calculations, we obtain

$$\begin{array}{c}\hfill |T_3\left(1\right)|=|1+{\displaystyle \frac{p_1^4}{JS}}+{\displaystyle \frac{p_1^{2}x(4-p_1^2)}{JS}}-{\displaystyle \frac{2B{p}_1^4}{J^{4}S}}-{\displaystyle \frac{2p_1^2}{J^2}}-{\displaystyle \frac{(J^4-4J^{2}B+4B^2)p_1^4}{4J^4{S}^2}}\\ \hfill -{\displaystyle \frac{(J^2-2B)p_1^{2}x(4-p_1^2)}{2J^2{S}^2}}-{\displaystyle \frac{x^2{(4-p_1^2)}^2}{4S^2}}|.\end{array}$$

For ease of notation, we select $p_1=p$, and because the function p is simultaneously in the family $\mathcal{P}$, we can assume that $p\in [0,2]$ without losing generality. The result is that, using the triangle inequality with $P=4-r^2$, we obtain

$$|T_3\left(1\right)|\le \left|1+{\displaystyle \frac{{\psi }_5{p}^4}{4J^4{S}^2}}-{\displaystyle \frac{2p^2}{J^2}}\right|+{\displaystyle \frac{(J^2-2B)p^2(4-p^2)}{2J^2{S}^2}}+{\displaystyle \frac{{(4-p^2)}^2}{4S^2}}.$$

Hence, at $p=2$, we find

$$|T_3\left(1\right)|\le 1+{\displaystyle \frac{4{\psi }_5}{J^4{S}^2}}-{\displaystyle \frac{8}{J^2}}.$$

□

Remark 3. 

Taking $\gamma =0$ and $\beta =1$ in Theorem 4, we can obtain the result which was derived by Radhika et al. [9] [Theorem 4].

Theorem 5. 

Let $f\in \mathcal{R}(\gamma ,\beta ,\lambda )$ be given by (1). Then,

$$|T_3\left(2\right)|=|(a_2-a_4)(a_2^2-2a_3^2+a_2{a}_4)|\le \left[{\displaystyle \frac{2}{J}}-{\displaystyle \frac{8(3J^{2}S+6JA-4N)p^3}{12G{J}^{2}S}}\right]\left[{\displaystyle \frac{4}{J^2}}-{\displaystyle \frac{4{\psi }_6}{3J^4{S}^{2}G}}\right],$$

(13)

where

$${\psi }_6=6G(J^4-4J^{2}B+4B^2)-3J^{2}S(JS+2A)+4SN.$$

Proof. 

From (7), (9) and (2) and by applying Lemma 2, we obtain

$$|a_2-a_4|=\left|{\displaystyle \frac{p_1}{J}}-{\displaystyle \frac{p_1^3}{4G}}-{\displaystyle \frac{p_1(4-p_1^2)x}{2G}}+{\displaystyle \frac{p_1(4-p_1^2)x^2}{4G}}-{\displaystyle \frac{(4-p_1^2){(1-|x|}^2)z}{2G}}-{\displaystyle \frac{A{p}_1^3}{2JSG}}-{\displaystyle \frac{A{p}_1(4-p_1^2)x}{2JSG}}+{\displaystyle \frac{N{p}_1^3}{3J^{2}SG}}\right|.$$

Applying triangle inequality and $p_1=p$, we find

$$|a_2-a_4|\le \left|{\displaystyle \frac{p}{J}}-{\displaystyle \frac{(3J^{2}S+6JA-4N)p^3}{12G{J}^{2}S}}\right|+{\displaystyle \frac{p(JS+A)\left|x\right|P}{JS}}+{\displaystyle \frac{{p\left|x\right|}^{2}P}{4G}}+{\displaystyle \frac{PZ}{2G}}+{\displaystyle \frac{Ap\left|x\right|P}{2JSG}}.$$

Using the same methods as in Theorems 2 and 3, we obtain

$$|a_2-a_4|\le {\displaystyle \frac{2}{J}}-{\displaystyle \frac{8(3J^{2}S+6JA-4N)p^3}{12G{J}^{2}S}}.$$

(14)

Also, by using (7), (8), (9) and (2), applying Lemma 2 and taking $p_1=p\in [0,2]$, we find

$$\begin{array}{c}|a_2^2-2a_3^2+a_2{a}_4|\le \left|{\displaystyle \frac{p^2}{J^2}}-{\displaystyle \frac{{\psi }_6{p}^4}{12J^4{S}^{2}G}}\right|+{\displaystyle \frac{{\psi }_7{p}^2(4-p^2)\left|x\right|}{2J^2{S}^{2}G}}+{\displaystyle \frac{p^2(4-p^2){\left|x\right|}^2}{4JG}}+{\displaystyle \frac{{(4-p^2)}^2{\left|x\right|}^2}{2S^2}}\hfill \\ \hfill +{\displaystyle \frac{p(4-p^2){(1-|x|}^2)}{2JG}}:=F_2(p,\left|x\right|),\end{array}$$

where

$${\psi }_7=2G(J^2-2B)-J{S}^2-SA.$$

Next, we may find the maximum value of $F_2(p,\left|x\right|)$ on the closed area $[0,2]\times [0,1]$. We can note that a maximum of $[0,2]\times [0,1]$ exists at an interior point $(p_0,|x\left|\right)$. Using differentiation $F_2(p,\left|x\right|)$ with regard to $\left|x\right|$, we obtain

$${\displaystyle \frac{\partial F_2(p,\left|x\right|)}{\partial \left|x\right|}}={\displaystyle \frac{{\psi }_7{p}^2(4-p^2)}{2J^2{S}^{2}G}}+{\displaystyle \frac{p^2(4-p^2)\left|x\right|}{2JG}}+{\displaystyle \frac{{(4-p^2)}^2\left|x\right|}{S^2}}-{\displaystyle \frac{p(4-p^2)\left|x\right|)}{JG}}.$$

If $p=0$,

$$F_2(0,\left|x\right|)={\displaystyle \frac{8}{S^2}}{\left|x\right|}^2\le {\displaystyle \frac{8}{S^2}}$$

If $p=2$,

$$F_2(2,\left|x\right|)={\displaystyle \frac{4}{J^2}}-{\displaystyle \frac{4{\psi }_6}{3J^4{S}^{2}G}}$$

If $\left|x\right|=0$,

$$F_2(p,0)=\left|{\displaystyle \frac{p^2}{J^2}}-{\displaystyle \frac{{\psi }_6{p}^4}{12J^4{S}^{2}G}}\right|+{\displaystyle \frac{p(4-p^2)}{2JG}}$$

which has the highest possible value

$${\displaystyle \frac{4}{J^2}}-{\displaystyle \frac{4{\psi }_6}{3J^4{S}^{2}G}}$$

on $[0,2]$. Also, if $\left|x\right|=1$, we obtain

$$F_2(p,1)=\left|{\displaystyle \frac{p^2}{J^2}}-{\displaystyle \frac{{\psi }_6{p}^4}{12J^4{S}^{2}G}}\right|+{\displaystyle \frac{(2{\psi }_7+J{S}^2)p^2(4-p^2)}{4J^2{S}^{2}G}}+{\displaystyle \frac{{(4-p^2)}^2}{2S^2}}$$

which has the highest possible value

$${\displaystyle \frac{4}{J^2}}-{\displaystyle \frac{4{\psi }_6}{3J^4{S}^{2}G}}$$

on $[0,2]$. Therefore,

$$|T_3\left(2\right)|=|(a_2-a_4)(a_2^2-2a_3^2+a_2{a}_4)|\le \left[{\displaystyle \frac{2}{J}}-{\displaystyle \frac{8(3J^{2}S+6JA-4N)p^3}{12G{J}^{2}S}}\right]\left[{\displaystyle \frac{4}{J^2}}-{\displaystyle \frac{4{\psi }_6}{3J^4{S}^{2}G}}\right].$$

□

Remark 4. 

Taking $\gamma =0$ and $\beta =1$ in Theorem 5, we can obtain the result which was obtained by Radhika et al. [9] [Theorem 5].

3. Conclusions

The primary objective was to define a new family $\mathcal{R}(\gamma ,\beta ,\lambda )$ of holormorphic functions, associating the Bazilevič functions and the $\lambda$-pseudo-starlike functions. We generated Taylor–Maclaurin coefficient estimates for the first four determinants of the Toeplitz matrices $T_2\left(2\right)$, $T_2\left(3\right)$, $T_3\left(2\right)$ and $T_3\left(1\right)$ for the functions belonging to this newly introduced family. Moreover, the results obtained may provide an opportunity for researchers to find the determinants of the Toeplitz matrices for functions of other families.