Hankel and Toeplitz Determinants for a Subclass of q-Starlike Functions Associated with a General Conic Domain

Abstract

By using a certain general conic domain as well as the quantum (or q-) calculus, here we define and investigate a new subclass of normalized analytic and starlike functions in the open unit disk $\mathbb{U}$. In particular, we find the Hankel determinant and the Toeplitz matrices for this newly-defined class of analytic q-starlike functions. We also highlight some known consequences of our main results.

1. Introduction and Definitions

Let the class of functions, which are analytic in the open unit disk

$$\mathbb{U}=\left\{z:z\in \mathbb{C}\mathrm{and}\left|z\right|<1\right\},$$

be denoted by $\mathcal{L}\left(\mathbb{U}\right)$. Also let $\mathcal{A}$ denote the class of all functions $f,$ which are analytic in the open unit disk $\mathbb{U}$ and normalized by

$$f\left(0\right)=0\mathrm{and}{f}^{\prime }\left(0\right)=1.$$

Then, clearly, each $f\in \mathcal{A}$ has a Taylor–Maclaurin series representation as follows:

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

(1)

Suppose that $\mathcal{S}$ is the subclass of the analytic function class $\mathcal{A}$, which consists of all functions which are also univalent in $\mathbb{U}$.

A function $f\in \mathcal{A}$ is said to be starlike in $\mathbb{U}$ if it satisfies the following inequality:

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

We denote by ${\mathcal{S}}^{\ast }$ the class of all such starlike functions in $\mathbb{U}$.

For two functions f and g, analytic in $\mathbb{U}$, we say that the function f is subordinate to the function g and write this subordination as follows:

$$f\prec g\mathrm{or}f\left(z\right)\prec g\left(z\right),$$

if there exists a Schwarz function w which is analytic in $\mathbb{U}$, with

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

such that

$$f\left(z\right)=g\left(w\left(z\right)\right).$$

In the case when the function g is univalent in $\mathbb{U}$, then we have the following equivalence (see, for example, [1]; see also [2]):

$$f\left(z\right)\prec g\left(z\right)(z\in \mathbb{U})⟺f\left(0\right)=g\left(0\right)\mathrm{and}f\left(\mathbb{U}\right)\subset g\left(\mathbb{U}\right).$$

Next, for a function $f\in \mathcal{A}$ given by (1) and another function $g\in \mathcal{A}$ given by

$$g\left(z\right)=z+\sum _{n=2}^{\infty }{b}_n{z}^n\left(z\in \mathbb{U}\right),$$

the convolution (or the Hadamard product) of f and g is defined here by

$$\left(f\ast g\right)\left(z\right):=z+\sum _{n=2}^{\infty }{a}_n{b}_n{z}^n=:\left(g\ast f\right)\left(z\right).$$

(2)

Let $\mathcal{P}$ denote the well-known Carathéodory class of functions p, analytic in the open unit disk $\mathbb{U}$, which are normalized by

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

(3)

such that

$$\Re \left(p\left(z\right)\right)>0\left(z\in \mathbb{U}\right).$$

Following the works of Kanas et al. (see [3,4]; see also [5]), we introduce the conic domain ${\mathsf{\Omega }}_k$ $(k\geqq 0)$ as follows:

$${\mathsf{\Omega }}_k=\left\{u+iv:u>k\sqrt{{\left(u-1\right)}^2+v^2}\right\}.$$

(4)

In fact, subjected to the conic domain ${\mathsf{\Omega }}_k$ $(k\geqq 0)$, Kanas and Wiśniowska (see [3,4]; see also [6]) studied the corresponding class k-$\mathcal{ST}$ of k-starlike functions in $\mathbb{U}$ (see Definition 1 below). For fixed $k,$ ${\mathsf{\Omega }}_k$ represents the conic region bounded successively by the imaginary axis $(k=0),$ by a parabola $(k=1),$ by the right branch of a hyperbola $(0<k<1),$ and by an ellipse $(k>1)$.

For these conic regions, the following functions play the role of extremal functions.

$$p_k\left(z\right)=\left\{\begin{array}{cc}{\displaystyle \frac{1+z}{1-z}}=1+2z+2z^2+\cdots \hfill & \left(k=0\right)\hfill \\ 1+{\displaystyle \frac{2}{{\pi }^2}}{\left[log\left({\displaystyle \frac{1+\sqrt{z}}{1-\sqrt{z}}}\right)\right]}^2\hfill & \left(k=1\right)\hfill \\ 1+{\displaystyle \frac{2}{1-k^2}}{sinh}^2\left[\left({\displaystyle \frac{2}{\pi }}arccosk\right)arctan\left(h\sqrt{z}\right)\right]\hfill & \left(0\leqq k<1\right)\hfill \\ 1+{\displaystyle \frac{1}{k^2-1}}\left[1+sin\left({\displaystyle \frac{\pi }{2K\left(\kappa \right)}{\int }_0^{\frac{u\left(z\right)}{\sqrt{\kappa }}}{\displaystyle \frac{dt}{\sqrt{(1-t^2)(1-{\kappa }^2{t}^2)}}}}\right)\right]\hfill & \left(k>1\right),\hfill \end{array}\right.$$

(5)

where

$$u\left(z\right)={\displaystyle \frac{z-\sqrt{\kappa }}{1-\sqrt{\kappa }z}}\left(z\in \mathbb{U}\right),$$

and $\kappa \in (0,1)$ is so chosen that

$$k=cosh\left(\frac{\pi K^{\prime }\left(\kappa \right)}{4K\left(\kappa \right)}\right).$$

Here $K\left(\kappa \right)$ is Legendre’s complete elliptic integral of first kind and

$$K^{\prime }\left(\kappa \right)=K\left(\sqrt{1-{\kappa }^2}\right),$$

that is, $K^{\prime }\left(\kappa \right)$ is the complementary integral of $K\left(\kappa \right)$ (see, for example, ([7], p. 326, Equation 9.4 (209))). Indeed, from (5), we have

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

(6)

The class k-$\mathcal{ST}$ is defined as follows.

Definition 1.

A function $f\in \mathcal{A}$ is said to be in the class k-$\mathcal{ST}$ if and only if

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec p_k\left(z\right)\left(\forall z\in \mathbb{U};k\geqq 0\right).$$

We now recall some basic definitions and concept details of the q-calculus which will be used in this paper (see, for example, ([7], p. 346 et seq.)). Throughout the paper, unless otherwise mentioned, we suppose that $0<q<1$ and

$$\mathbb{N}=\left\{1,2,3\cdots \right\}={\mathbb{N}}_0\backslash \left\{0\right\}\left({\mathbb{N}}_0:=\left\{0,1,2,\cdots \right\}\right).$$

Definition 2.

Let $q\in \left(0,1\right)$ and define the q-number ${\left[\lambda \right]}_q$ by

$${\left[\lambda \right]}_q=\left\{\begin{array}{cc}{\displaystyle \frac{1-q^{\lambda }}{1-q}}\hfill & \left(\lambda \in \mathbb{C}\right)\hfill \\ {\displaystyle \sum _{k=0}^{n-1}}{q}^k=1+q+q^2+\cdots +q^{n-1}\hfill & \left(\lambda =n\in \mathbb{N}\right).\hfill \end{array}\right.$$

Definition 3.

Let $q\in \left(0,1\right)$ and define the q-factorial ${\left[n\right]}_q!$ by

$${\left[n\right]}_q!=\left\{\begin{array}{cc}1\hfill & \left(n=0\right)\hfill \\ \prod _{k=1}^n{\left[k\right]}_q\hfill & \left(n\in \mathbb{N}\right).\hfill \end{array}\right.$$

Definition 4

(see [8,9]). The q-derivative (or q-difference) operator $D_q$ of a function f defined, in a given subset of $\mathbb{C}$, by

$$\left(D_{q}f\right)\left(z\right)=\left\{\begin{array}{cc}{\displaystyle \frac{f\left(z\right)-f\left(qz\right)}{\left(1-q\right)z}}\hfill & \left(z\ne 0\right)\hfill \\ f^{\prime }\left(0\right)\hfill & \left(z=0\right),\hfill \end{array}\right.$$

(7)

provided that $f^{\prime }\left(0\right)$ exists.

From Definition 4, we can observe that

$$\underset{q\to 1-}{lim}\left(D_{q}f\right)\left(z\right)=\underset{q\to 1-}{lim}\frac{f\left(z\right)-f\left(qz\right)}{\left(1-q\right)z}=f^{\prime }\left(z\right)$$

for a differentiable function f in a given subset of $\mathbb{C}$. It is also known from (1) and (7) that

$$\left(D_{q}f\right)\left(z\right)=1+\sum _{n=2}^{\infty }{\left[n\right]}_q{a}_n{z}^{n-1}.$$

(8)

Definition 5.

The q-Pochhammer symbol ${\left[\xi \right]}_{n,q}$ $\left(\xi \in \mathbb{C};n\in {\mathbb{N}}_0\right)$ is defined as follows:

$${\left[\xi \right]}_{n,q}=\frac{{\left(q^{\xi };q\right)}_n}{{\left(1-q\right)}^n}=\left\{\begin{array}{cc}1\hfill & \left(n=0\right)\hfill \\ {\left[\xi \right]}_q{\left[\xi +1\right]}_q{\left[\xi +2\right]}_q\cdots {\left[\xi +n-1\right]}_q\hfill & \left(n\in \mathbb{N}\right).\hfill \end{array}\right.$$

Moreover, the q-gamma function is defined by the following recurrence relation:

$${\mathsf{\Gamma }}_q\left(z+1\right)={\left[z\right]}_q{\mathsf{\Gamma }}_q\left(z\right)and{\mathsf{\Gamma }}_q\left(1\right)=1.$$

Definition 6

(see [10]). For $f\in \mathcal{A},$ let the q-Ruscheweyh derivative operator ${\mathcal{R}}_q^{\lambda }$ be defined, in terms of the Hadamard product (or convolution) given by (2), as follows:

$${\mathcal{R}}_q^{\lambda }f\left(z\right)=f\left(z\right)\ast {\mathcal{F}}_{q,\lambda +1}\left(z\right)\left(z\in \mathbb{U};\lambda >-1\right),$$

where

$${\mathcal{F}}_{q,\lambda +1}\left(z\right)=z+\sum _{n=2}^{\infty }\frac{{\mathsf{\Gamma }}_q\left(\lambda +n\right)}{{\left[n-1\right]}_q!{\mathsf{\Gamma }}_q\left(\lambda +1\right)}{z}^n=z+\sum _{n=2}^{\infty }\frac{{\left[\lambda +1\right]}_{q,n-1}}{{\left[n-1\right]}_q!}{z}^n.$$

We next define a certain q-integral operator by using the same technique as that used by Noor [11].

Definition 7.

For $f\in \mathcal{A}$, let the q-integral operator ${\mathcal{F}}_{q,\lambda }$ be defined by

$${\mathcal{F}}_{q,\lambda +1}^{-1}\left(z\right)\ast {\mathcal{F}}_{q,\lambda +1}\left(z\right)=z\left(D_{q}f\right)\left(z\right).$$

Then

$$\begin{array}{ll}{\mathcal{I}}_q^{\lambda }f\left(z\right)& =f\left(z\right)\ast {\mathcal{F}}_{q,\lambda +1}^{-1}\left(z\right)\\ & =z+\sum _{n=2}^{\infty }{\psi }_{n-1}{a}_n{z}^n\left(z\in \mathbb{U};\lambda >-1\right),\end{array}$$

(9)

where

$${\mathcal{F}}_{q,\lambda +1}^{-1}\left(z\right)=z+\sum _{n=2}^{\infty }{\psi }_{n-1}{z}^n$$

and

$${\psi }_{n-1}=\frac{{\left[n\right]}_q!{\mathsf{\Gamma }}_q\left(\lambda +1\right)}{{\mathsf{\Gamma }}_q\left(\lambda +n\right)}=\frac{{\left[n\right]}_q!}{{\left[\lambda +1\right]}_{q,n-1}}.$$

Clearly, we have

$${\mathcal{I}}_q^{0}f\left(z\right)=z\left(D_{q}f\right)\left(z\right)\mathrm{and}{\mathcal{I}}_q^{1}f\left(z\right)=f\left(z\right).$$

We note also that, in the limit case when $q\to 1-,$ the q-integral operator ${\mathcal{F}}_{q,\lambda }$ given by Definition 7 would reduce to the integral operator which was studied by Noor [11].

The following identity can be easily verified:

$$z{D}_q\left({\mathcal{I}}_q^{\lambda +1}f\left(z\right)\right)=\left(1+\frac{{\left[\lambda \right]}_q}{q^{\lambda }}\right){\mathcal{I}}_q^{\lambda }f\left(z\right)-\frac{{\left[\lambda \right]}_q}{q^{\lambda }}{\mathcal{I}}_q^{\lambda +1}f\left(z\right).$$

(10)

When $q\to 1-,$ this last identity in (10) implies that

$$z{\left({\mathcal{I}}^{\lambda +1}f\left(z\right)\right)}^{\prime }=\left(1+\lambda \right){\mathcal{I}}^{\lambda }f\left(z\right)-\lambda {\mathcal{I}}^{\lambda +1}f\left(z\right),$$

which is the well-known recurrence relation for the above-mentioned integral operator which was studied by Noor [11].

In geometric function theory, several subclasses belonging to the class of normalized analytic functions class $\mathcal{A}$ have already been investigated in different aspects. The above-defined q-calculus gives valuable tools that have been extensively used in order to investigate several subclasses of $\mathcal{A}$. Ismail et al. [12] were the first who used the q-derivative operator $D_q$ to study the q-calculus analogous of the class ${\mathcal{S}}^{\ast }$ of starlike functions in $\mathbb{U}$ (see Definition 8 below). However, a firm footing of the q-calculus in the context of geometric function theory was presented mainly and basic (or q-) hypergeometric functions were first used in geometric function theory in a book chapter by Srivastava (see, for details, ([13], p. 347 et seq.); see also [14]).

Definition 8

(see [12]). A function $f\in \mathcal{A}$ is said to belong to the class ${\mathcal{S}}_q^{\ast }$ if

$$f\left(0\right)=f^{\prime }\left(0\right)-1=0$$

(11)

and

$$\left|\frac{z}{f\left(z\right)}\left(D_{q}f\right)z-\frac{1}{1-q}\right|\leqq \frac{1}{1-q}.$$

(12)

It is readily observed that, as $q\to 1-$, the closed disk:

$$\left|w-\frac{1}{1-q}\right|\leqq \frac{1}{1-q}$$

becomes the right-half plane and the class ${\mathcal{S}}_q^{\ast }$ of q-starlike functions reduces to the familiar class ${\mathcal{S}}^{\ast }$ of normalized starlike functions in $\mathbb{U}$ with respect to the origin $(z=0)$. Equivalently, by using the principle of subordination between analytic functions, we can rewrite the conditions in (11) and (12) as follows (see [15]):

$$\frac{z}{f\left(z\right)}\left(D_{q}f\right)\left(z\right)\prec \widehat{p}\left(z\right)\left(\widehat{p}\left(z\right)=\frac{1+z}{1-qz}\right).$$

(13)

The notation ${\mathcal{S}}_q^{\ast }$ was used by Sahoo and Sharma [16].

Now, making use of the principle of subordination between analytic functions and the above-mentioned q-calculus, we present the following definition.

Definition 9.

A function p is said to be in the class k-${\mathcal{P}}_q$ if and only if

$$p\left(z\right)\prec \frac{2p_k\left(z\right)}{\left(1+q\right)+\left(1-q\right)p_k\left(z\right)},$$

where $p_k\left(z\right)$ is defined by (5).

Geometrically, the function $p\left(z\right)\in k$-${\mathcal{P}}_q$ takes on all values from the domain ${\mathsf{\Omega }}_{k,q}$ $(k\geqq 0)$ which is defined as follows:

$${\mathsf{\Omega }}_{k,q}=\left\{w:\Re \left(\frac{\left(1+q\right)w}{\left(q-1\right)w+2}\right)>k\left|\frac{\left(1+q\right)w}{\left(q-1\right)w+2}-1\right|\right\}.$$

The domain ${\mathsf{\Omega }}_{k,q}$ represents a generalized conic region.

It can be seen that

$$\underset{q\to 1-}{lim}{\mathsf{\Omega }}_{k,q}={\mathsf{\Omega }}_k,$$

where ${\mathsf{\Omega }}_k$ is the conic domain considered by Kanas and Wiśniowska [3]. Below, we give some basic facts about the class k-${\mathcal{P}}_q$.

Remark 1.

First of all, we see that

$$k-{\mathcal{P}}_q\subseteq \mathcal{P}\left[\frac{2k}{2k+1+q}\right],$$

where $\mathcal{P}\left[\frac{2k}{2k+1+q}\right]$ is the well-known class of functions with real part greater than $\frac{2k}{2k+1+q}.$ Secondly, we have

$$\underset{q\to 1-}{lim}k-{\mathcal{P}}_q=\mathcal{P}\left(p_k\right),$$

where $\mathcal{P}\left(p_k\right)$ is the well-known function class introduced by Kanas and Wiśniowska [3]. Thirdly, we have

$$\underset{q\to 1-}{lim}0-{\mathcal{P}}_q=\mathcal{P},$$

where $\mathcal{P}$ is the well-known class of analytic functions with positive real part.

Definition 10.

A function f is said to be in the class $\mathcal{ST}\left(k,\lambda ,q\right)$ if and only if

$$\frac{z\left(D_q{\mathcal{I}}_q^{\lambda }f\right)\left(z\right)}{f\left(z\right)}\in k-{\mathcal{P}}_q\left(k\geqq 0;\lambda \geqq 0\right),$$

or, equivalently,

$$\Re \left(\frac{\left(1+q\right)\frac{z\left(D_q{\mathcal{I}}_q^{\lambda }f\right)\left(z\right)}{f\left(z\right)}}{\left(q-1\right)\frac{z\left(D_q{\mathcal{I}}_q^{\lambda }f\right)\left(z\right)}{f\left(z\right)}+2}\right)>k\left|\frac{\left(1+q\right)\frac{z\left(D_q{\mathcal{I}}_q^{\lambda }f\right)\left(z\right)}{f\left(z\right)}}{\left(q-1\right)\frac{z\left(D_q{\mathcal{I}}_q^{\lambda }f\right)\left(z\right)}{f\left(z\right)}+2}-1\right|.$$

Remark 2.

First of all, it is easily seen that

$$\mathcal{ST}\left(0,1,q\right)={\mathcal{S}}_q^{\ast },$$

where ${\mathcal{S}}_q^{\ast }$ is the function class introduced and studied by Ismail et al. [12]. Secondly, we have

$$\underset{q\to 1-}{lim}\mathcal{ST}\left(k,1,q\right)=k-\mathcal{ST},$$

where k-$\mathcal{ST}$ is a function class introduced and studied by Kanas and Wiśniowska [4]. Finally, we have

$$\underset{q\to 1-}{lim}\mathcal{ST}\left(0,1,q\right)={\mathcal{S}}^{\ast },$$

where ${\mathcal{S}}^{\ast }$ is the well-known class of starlike functions in $\mathbb{U}$ with respect to the origin $(z=0)$.

Remark 3.

Further studies of the new q-starlike function class $\mathcal{ST}\left(k,\lambda ,q\right),$ as well as of its more consequences, can next be determined and investigated in future papers.

Let $n\in {\mathbb{N}}_0$ and $j\in \mathbb{N}$. The following jth Hankel determinant was considered by Noonan and Thomas [17]:

$${\mathcal{H}}_j\left(n\right)=\left|\begin{array}{cccccc}{a}_n\hfill & a_{n+1}\hfill & .& .& .\hfill & a_{n+j-1}\hfill \\ a_{n+1}\hfill & .\hfill & & & & .\hfill \\ .\hfill & .\hfill & & & & .\hfill \\ .\hfill & .\hfill & & & & .\hfill \\ .\hfill & .\hfill & & & & .\hfill \\ a_{n+j-1}\hfill & .\hfill & .& .& .& a_{n+2\left(j-1\right)}\hfill \end{array}\right|,$$

where $a_1=1.$ In fact, this determinant has been studied by several authors, and sharp upper bounds on ${\mathcal{H}}_2\left(2\right)$ were obtained by several authors (see [18,19,20]) for various classes of functions. It is well-known that the Fekete–Szegö functional $\left|a_3-a_2^2\right|$ can be represented in terms of the Hankel determinant as ${\mathcal{H}}_2\left(1\right)$. This functional has been further generalized as $\left|a_3-\mu a_2^2\right|$ for some real or complex $\mu$. Fekete and Szegö gave sharp estimates of $\left|a_3-\mu a_2^2\right|$ for $\mu$ real and $f\in \mathcal{S}$, the class of normalized univalent functions in $\mathbb{U}$. It is also known that the functional $\left|a_2{a}_4-a_3^2\right|$ is equivalent to ${\mathcal{H}}_2\left(2\right)$ (see [18]). Babalola [21] studied the Hankel determinant ${\mathcal{H}}_3\left(1\right)$ for some subclasses of normalized analytic functions in $\mathbb{U}$. The symmetric Toeplitz determinant ${\mathcal{T}}_j\left(n\right)$ is defined by

$${\mathcal{T}}_j\left(n\right)=\left|\begin{array}{cccccc}{a}_n\hfill & a_{n+1}\hfill & .& .& .& a_{n+j-1}\hfill \\ a_{n+1}\hfill & .\hfill & & & & .\hfill \\ .\hfill & .\hfill & & & & .\hfill \\ .\hfill & .\hfill & & & & .\hfill \\ .\hfill & .\hfill & & & & .\hfill \\ a_{n+j-1}\hfill & .\hfill & .& .& .& a_n\hfill \end{array}\right|,$$

so that

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

and so on.

For $f\in \mathcal{S}$, the problem of finding the best possible bounds for $\left|\left|a_{n+1}\right|-\left|a_n\right|\right|$ has a long history (see, for details, [22]). It is a known fact from [22] that

$$|\left|a_{n+1}\right|-\left|a_n\right||<c$$

for a constant c. However, the problem of finding exact values of the constant c for $\mathcal{S}$ and its various subclasses has proved to be difficult. In a very recent investigation, Thomas and Abdul-Halim [23] succeeded in obtaining some sharp estimates for ${\mathcal{T}}_j\left(n\right)$ for the first few values of n and j involving symmetric Toeplitz determinants whose entries are the coefficients $a_n$ of starlike and close-to- convex functions.

In the present investigation, our focus is on the Hankel determinant and the Toeplitz matrices for the function class $\mathcal{ST}\left(k,\lambda ,q\right)$ given by Definition 10.

2. A Set of Lemmas

In order to prove our main results in this paper, we need each of the following lemmas.

Lemma 1

(see [20]). If the function $p\left(z\right)$ given by (3) is in the Carathéodory class $\mathcal{P}$ of analytic functions with positive real part in $\mathbb{U},$ then

$$2c_2=c_1^2+x\left(4-c_1^2\right)$$

and

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

for some $x,z\in \mathbb{C}$ with $\left|x\right|\leqq 1$ and $\left|z\right|\leqq 1.$

Lemma 2

(see [24]). Let the function $p\left(z\right)$ given by (3) be in the Carathéodory class $\mathcal{P}$ of analytic functions with positive real part in $\mathbb{U}$. Also let $\mu \in \mathbb{C}$. Then

$$\left|c_n-\mu c_k{c}_{n-k}\right|\leqq 2max\left(1,\left|2\mu -1\right|\right)\left(1\leqq k\leqq n-1\right).$$

Lemma 3

(see [22]). Let the function $p\left(z\right)$ given by (3) be in the Carathéodory class $\mathcal{P}$ of analytic functions with positive real part in $\mathbb{U}$. Then

$$\left|c_n\right|\leqq 2\left(n\in \mathbb{N}\right).$$

This last inequality is sharp.

3. Main Results

Throughout this section, unless otherwise mentioned, we suppose that

$$q\in \left(0,1\right),\lambda >-1\mathrm{and}k\in \left[0,1\right].$$

Theorem 1.

If the function $f\left(z\right)$ given by (1) belongs to the class $\mathcal{ST}\left(k,\lambda ,q\right),$ where $k\in \left[0,1\right],$ then

$$\left|a_2\right|\leqq \frac{\left(1+q\right)p_1}{2q{\psi }_1},$$

$$a_3\leqq \frac{1}{2q{\psi }_2}\left(p_1+\left|p_2-p_1+\frac{\left(q^2+1\right)p_1^2}{2q}\right|\right)$$

and

$$\begin{array}{ll}{a}_4& \leqq \frac{\left(1+q\right)}{4\left(q+q^2+q^3\right){\psi }_3}\left(2p_1+4\left|p_2-p_1+\frac{\left(2+q^2\right)p_1^2}{4q}\right|\right.\\ & +\left|2p_3+2p_1-4p_2-\frac{\left(2\left(1+q^2\right)-q\right)p_1^2}{q}+\frac{\left(4q^2-3q+2\right)}{q}{p}_1{p}_2\right.\\ & \left.+\left.\frac{\left(q^2+2q-1\right)}{2q^2}{p}_1^3\right|\right),\end{array}$$

(14)

where $p_j$ $(j=1,2,3)$ are positive and are the coefficients of the functions $p_k\left(z\right)$ defined by (6). Each of the above results is sharp for the function $g\left(z\right)$ given by

$$g\left(z\right)=\frac{2p_k\left(z\right)}{\left(1+q\right)+\left(1-q\right)p_k\left(z\right)}.$$

Proof. 

Let $f\left(z\right)\in \mathcal{ST}\left(k,\lambda ,q\right)$. Then, we have

$$\frac{z\left(D_{q}f\right)\left(z\right)}{f\left(z\right)}=\mathfrak{q}\left(z\right)\prec S_k\left(z\right),$$

(15)

where

$$S_k\left(z\right)=\frac{2p_k\left(z\right)}{\left(1+q\right)+\left(1-q\right)p_k\left(z\right)},$$

and the functions $p_k\left(z\right)$ are defined by (6).

We now define the function $p\left(z\right)$ with $p\left(0\right)=1$ and with a positive real part in $\mathbb{U}$ as follows:

$$p\left(z\right)=\frac{1+S_k^{-1}\left(\mathfrak{q}\left(z\right)\right)}{1-S_k^{-1}\left(\mathfrak{q}\left(z\right)\right)}=1+c_{1}z+c_2{z}^2+\cdots .$$

(16)

After some simple computation involving (16), we get

$$\mathfrak{q}\left(z\right)=S_k\left(\frac{p\left(z\right)+1}{p\left(z\right)-1}\right).$$

We thus find that

$$\begin{array}{c}{S}_k\left(\frac{p\left(z\right)+1}{p\left(z\right)-1}\right)\hfill \\ =1+\left(\frac{q+1}{2}\right)\left[\frac{p_1{c}_1}{2}z+\left\{\frac{p_1{c}_2}{2}+\left(\frac{p_2}{4}-\frac{p_1}{4}+\left(\frac{\left(q-1\right)p_1^2}{8}\right)\right)c_1^2\right\}{z}^2\right.\hfill \\ +\left\{\frac{p_1{c}_3}{2}+\left(\frac{p_2}{2}-\frac{p_1}{2}+\left(\frac{\left(q-1\right)p_1^2}{4}\right)\right)c_1{c}_2\right.\hfill \\ \left.\left.+\left(\frac{p_1}{8}-\frac{p_2}{4}-\frac{\left(q-1\right)p_1^2}{8}+\frac{p_3}{8}-\frac{\left(q-1\right)p_1{p}_2}{8}+\frac{{\left(q-1\right)}^2{p}_1^3}{32}\right)c_1^3\right\}{z}^3\right]+\cdots .\hfill \end{array}$$

(17)

Now, upon expanding the left-hand side of (15), we have

$$\begin{array}{c}\frac{z\left(D_q{\mathcal{I}}_q^{\lambda }f\right)\left(z\right)}{f\left(z\right)}=1+q{\psi }_1{a}_{2}z+\left\{\left(q+q^2\right){\psi }_2{a}_3-q{\psi }_1^2{a}_2^2\right\}{z}^2\hfill \\ +\left\{\left(q+q^2+q^3\right){\psi }_3{a}_4-\left(2q+q^2\right){\psi }_1{\psi }_2{a}_2{a}_3+q{\psi }_1^3{a}_2^3\right\}{z}^3+\cdots .\hfill \end{array}$$

(18)

Finally, by comparing the corresponding coefficients in (17) and (18) along with Lemma 3, we obtain the result asserted by Theorem 1.  □

Theorem 2.

If the function $f\left(z\right)$ given by (1) belongs to the class $\mathcal{ST}\left(k,\lambda ,q\right),$ then

$$\begin{array}{ll}{\mathcal{T}}_3\left(2\right)\leqq & \left[\left(\frac{1+q}{2q{\psi }_1}\right)p_1^2+\left(\frac{1+q}{4\left(q+q^2+q^3\right){\psi }_3}\right)\left[{\mathsf{\Omega }}_1+{\mathsf{\Omega }}_2\right]\right]\\ & ·\left[4\left(\frac{{\left(1+q\right)}^2}{16q^2{\psi }_1^2}\right)p_1^2+16\left|{\mathsf{\Omega }}_3\right|+\frac{p_1^2}{4q^2{\psi }_2^2}+2{\mathsf{\Omega }}_5{p}_1^2\left|2-\frac{{\mathsf{\Omega }}_4}{{\mathsf{\Omega }}_5{p}_1^2}\right|\right],\end{array}$$

where

$$\begin{array}{l}{\mathsf{\Omega }}_1=2p_1+4\left|p_2-p_1+\frac{\left(2+q^2\right)}{4q}{p}_1^2\right|,\\ {\mathsf{\Omega }}_2=|2p_3+2p_1-4p_2-\left(2\left(1+q^2\right)-q\right)p_1^2\\ +\left(\frac{4q^2-3q+2}{q}\right)p_1{p}_2+\left(\frac{q^2+q+1}{2q^2}{p}_1^3\right)|,\\ {\mathsf{\Omega }}_3=\frac{1}{2q^2{\psi }_2^2}{\left(\frac{p_2}{4}-\frac{p_1}{4}+\frac{\left(q^2+1\right)p_1^2}{8q}\right)}^2-{\mathsf{\Omega }}_5·[\frac{p_3}{4}+\frac{p_1}{4}-\frac{p_2}{2}\\ -\frac{\left[2\left(1+q^2\right)-q\right]p_1^2}{8q}+\frac{4q^2-3q+2}{8q}{p}_1{p}_2+\left(\frac{q^2+2q-1}{16q^2}\right)p_1^3],\\ {\mathsf{\Omega }}_4=\frac{p_1}{2q^2{\psi }_2^2}\left(\frac{p_2}{4}-\frac{p_1}{4}+\frac{\left(q^2+1\right)p_1^2}{8q}\right)-{\mathsf{\Omega }}_5{p}_1\left(p_2-p_1+\frac{\left(2+q^2\right)p_1^2}{4q}\right),\\ {\mathsf{\Omega }}_5=\frac{{\left(1+q\right)}^2}{16q^2\left(1+q+q^2\right){\psi }_1{\psi }_3}\end{array}$$

and $p_j$ $(j=1,2)$ are positive and are the coefficients of the functions $p_k\left(z\right)$ defined by (6).

Proof. 

Upon comparing the corresponding coefficients in (17) and (18), we find that

$$\begin{array}{ll}{a}_2& =\frac{\left(1+q\right)p_1{c}_1}{4q{\psi }_1},\end{array}$$

(19)

$$\begin{array}{ll}{a}_3& =\frac{1}{2q{\psi }_2}\left[\frac{p_1{c}_2}{2}+\left(\frac{p_2}{4}-\frac{p_1}{4}+\frac{\left(q^2+1\right)p_1^2}{8q}\right)c_1^2\right],\end{array}$$

(20)

$$\begin{array}{ll}{a}_4=& \frac{\left(1+q\right)}{4\left(q+q^2+q^3\right){\psi }_3}\left[p_1{c}_3+\left(p_2-p_1+\frac{\left(2+q^2\right)p_1^2}{4q}\right)c_1{c}_2\right.\\ & +\left(\frac{p_3}{4}+\frac{p_1}{4}-\frac{p_2}{2}-\frac{\left(2\left(1+q^2\right)-q\right)p_1^2}{8q}+\frac{\left(4q^2-3q+2\right)}{8q}{p}_1{p}_2\right.\\ & \left.+\left.\frac{\left(q^2+2q-1\right)}{16q^2}{p}_1^3\right)c_1^3\right].\end{array}$$

(21)

By a simple computation, ${\mathcal{T}}_3\left(2\right)$ can be written as follows:

$${\mathcal{T}}_3\left(2\right)=\left(a_2-a_4\right)\left(a_2^2-2a_3^2+a_2{a}_4\right).$$

Now, if $f\in \mathcal{ST}\left(k,\lambda ,q\right),$ then it is clearly seen that

$$\begin{array}{ll}\left|a_2-a_4\right|& \leqq \left|a_2\right|+\left|a_4\right|\\ & \leqq \left(\frac{1+q}{2q{\psi }_1}\right)p_1^2+\left(\frac{1+q}{4\left(q+q^2+q^3\right){\psi }_3}\right)\left({\mathsf{\Omega }}_1+{\mathsf{\Omega }}_2\right).\end{array}$$

We need to maximize $\left|a_2^2-2a_3^2+a_2{a}_4\right|$ for a function $f\in \mathcal{ST}\left(k,\lambda ,q\right)$. So, by writing $a_2,$ $a_3$, and $a_4$ in terms of $c_1,$ $c_2$, and $c_3,$ with the help of (19)–(21), we get

$$\begin{array}{c}\left|a_2^2-2a_3^2+a_2{a}_4\right|\hfill \\ =\left|\left(\frac{{\left(1+q\right)}^2}{16q^2{\psi }_1^2}\right)p_1^2{c}_1^2-{\mathsf{\Omega }}_3{c}_1^4-{\mathsf{\Omega }}_4{c}_1^2{c}_2-\frac{p_1^2}{8q^2{\psi }_2^2}{c}_2^2+{\mathsf{\Omega }}_5{p}_1^2{c}_1{c}_3\right|.\hfill \end{array}$$

(22)

Finally, by applying the trigonometric inequalities, Lemmas 2 and 3 along with (22), we obtain the result asserted by Theorem 2.  □

As an application of Theorem 2, we first set ${\psi }_{n-1}=1$ and $k=0$ and then let $q\to 1-$. We thus arrive at the following known result.

Corollary 1

(see [25]). If the function $f\left(z\right)$ given by (1) belongs to the class ${\mathcal{S}}^{\ast },$ then

$${\mathcal{T}}_3\left(2\right)\leqq 84.$$

Theorem 3.

If the function $f\left(z\right)$ given by (1) belongs to the class $\mathcal{ST}\left(k,\lambda ,q\right),$ then

$$\left|a_2{a}_4-a_3^2\right|\leqq \frac{1}{4q^2{\psi }_2^2}{p}_1^2,$$

(23)

where $k\in \left[0,1\right]$ and $p_j$ $(j=1,2,3)$ are positive and are the coefficients of the functions $p_k\left(z\right)$ defined by (6).

Proof. 

Making use of (19)–(21), we find that

$$\begin{array}{ll}{a}_2{a}_4-a_3^2& =\frac{A\left(q\right)}{16q^2{\psi }_1{\psi }_3}{p}_1^2{c}_1{c}_3+\left(\frac{A\left(q\right){\psi }_2^2-{\psi }_1{\psi }_3}{16q^2{\psi }_1{\psi }_2^2{\psi }_3}{p}_1{p}_2-\frac{A\left(q\right){\psi }_2^2-{\psi }_1{\psi }_3}{16q^2{\psi }_1{\psi }_2^2{\psi }_3}{p}_1^2\right.\\ & \left.+\frac{A\left(q\right)\left(2+q^2\right){\psi }_2^2-2\left(1+q^2\right){\psi }_1{\psi }_3}{64q^2{\psi }_1{\psi }_3}{p}_1^3\right)c_1^2{c}_2+\frac{1}{16q^2{\psi }_2^2}{p}_1^2{c}_2^2\\ & +\left[\frac{A\left(q\right)}{64q^2{\psi }_1{\psi }_3}{p}_1{p}_3+\left(\frac{A\left(q\right){\psi }_2^2-{\psi }_1{\psi }_3}{64q^2{\psi }_1{\psi }_2^2{\psi }_3}\right)p_1^2+\left(\frac{{\psi }_1{\psi }_3-A\left(q\right){\psi }_2^2}{32q^2{\psi }_1{\psi }_2^2{\psi }_3}\right)p_1{p}_2\right.\\ & +\left(\frac{2\left(1+q^2\right){\psi }_1{\psi }_3-\left(2\left(1+q^2\right)-q\right)A\left(q\right){\psi }_2^2}{128q^3{\psi }_1{\psi }_2^2{\psi }_3}\right)p_1^3\\ & +\left(\frac{A\left(q\right)\left(4q^2-3q+2\right){\psi }_2^2-2\left(1+q^2\right){\psi }_1{\psi }_3}{128q^3{\psi }_1{\psi }_2^2{\psi }_3}\right)p_1^2{p}_2\\ & +\left.\left(\frac{A\left(q\right)\left(q^2+2q-1\right){\psi }_2^2-{\left(1+q^2\right)}^2{\psi }_1{\psi }_3}{256q^4{\psi }_1{\psi }_2^2{\psi }_3}\right)p_1^4-\frac{1}{64q^2{\psi }_2^2}{p}_2^2\right]c_1^4,\end{array}$$

(24)

where

$$A\left(q\right)=\frac{{\left(1+q\right)}^2}{1+q+q^2}.$$

We substitute the values of $c_2$ and $c_3$ from the above Lemma and, for simplicity, take $Y=4-c_1^2$ and $Z=(1-|x{|}^2)z$. Without loss of generality, we assume that $c=c_1$ $(0\leqq c\leqq 2)$, so that

$$\begin{array}{c}{a}_2{a}_4-a_3^2=\left[\frac{q\left(1-q\right)A\left(q\right){\psi }_2^2}{128q^2{\psi }_1{\psi }_3}{p}_1^3+\frac{A\left(q\right)}{64q^2{\psi }_1{\psi }_3}{p}_1{p}_3\right.\hfill \\ +\left(\frac{A\left(q\right)\left(4q^2-3q+2\right){\psi }_2^2-2\left(1+q^2\right){\psi }_1{\psi }_3}{128q^3{\psi }_1{\psi }_2^2{\psi }_3}\right)p_1^2{p}_2\hfill \\ \left.+\left(\frac{A\left(q\right)\left(q^2+2q-1\right){\psi }_2^2-{\left(1+q^2\right)}^2{\psi }_1{\psi }_3}{256q^4{\psi }_1{\psi }_2^2{\psi }_3}\right)p_1^4-\frac{1}{64q^2{\psi }_2^2}{p}_2^2\right]c^4\hfill \\ +\left[\frac{A\left(q\right){\psi }_2^2-{\psi }_1{\psi }_3}{32q^2{\psi }_1{\psi }_2^2{\psi }_3}{p}_1{p}_2+\frac{A\left(q\right)\left(2+q^2\right){\psi }_2^2-2\left(1+q^2\right){\psi }_1{\psi }_3}{128q^2{\psi }_1{\psi }_3}{p}_1^3\right]c^{2}xY\hfill \\ ·\left[-\frac{A\left(q\right)}{64q^2{\psi }_1{\psi }_3}{p}_1^2{c}^{2}Y{x}^2-\frac{1}{64q^2{\psi }_2^2}{p}_1^2{x}^2{Y}^2+\frac{A\left(q\right)}{32q^2{\psi }_1{\psi }_3}{p}_1^{2}cYZ\right].\hfill \end{array}$$

(25)

Upon setting $Z=(1-|x{|}^2)z$ and taking the moduli in (25) and using trigonometric inequality, we find that

$$\begin{array}{c}\left|a_2{a}_4-a_3^2\right|\leqq \left|{\lambda }_1\right|c^4+\left|{\lambda }_2\right|\left|x\right|Y{c}^2+\frac{A\left(q\right)}{64q^2{\psi }_1{\psi }_3}{p}_1^{2}Y{\left|x\right|}^2{c}^2\hfill \\ +\frac{1}{64q^2{\psi }_2^2}{p}_1^2{\left|x\right|}^2{Y}^2+\frac{A\left(q\right)}{32q^2{\psi }_1{\psi }_3}{p}_1^2{c}^{2}Y\left(1-{\left|x\right|}^2\right)\hfill \\ =\mathsf{\Lambda }\left(c,\left|x\right|\right),\hfill \end{array}$$

(26)

where

$$\begin{array}{ll}{\lambda }_1& =\frac{q\left(1-q\right)A\left(q\right){\psi }_2^2}{128q^2{\psi }_1{\psi }_3}{p}_1^3+\frac{A\left(q\right)}{64q^2{\psi }_1{\psi }_3}{p}_1{p}_3\\ & +\left(\frac{A\left(q\right)\left(4q^2-3q+2\right){\psi }_2^2-2\left(1+q^2\right){\psi }_1{\psi }_3}{128q^3{\psi }_1{\psi }_2^2{\psi }_3}\right)p_1^2{p}_2\\ & +\left(\frac{A\left(q\right)\left(q^2+2q-1\right){\psi }_2^2-{\left(1+q^2\right)}^2{\psi }_1{\psi }_3}{256q^4{\psi }_1{\psi }_2^2{\psi }_3}\right)p_1^4-\frac{1}{64q^2{\psi }_2^2}{p}_2^2\\ {\lambda }_2& =\frac{A\left(q\right){\psi }_2^2-{\psi }_1{\psi }_3}{32q^2{\psi }_1{\psi }_2^2{\psi }_3};p_1{p}_2+\frac{A\left(q\right)\left(2+q^2\right){\psi }_2^2-2\left(1+q^2\right){\psi }_1{\psi }_3}{128q^2{\psi }_1{\psi }_3}{p}_1^3.\end{array}$$

Now, trivially, we have

$${\mathsf{\Lambda }}^{\prime }\left(\left|x\right|\right)>0$$

on $\left[0,1\right]$, and so

$$\mathsf{\Lambda }\left(\left|x\right|\right)\leqq \mathsf{\Lambda }\left(1\right).$$

Hence, by puting $Y=4-c_1^2$ and after some simplification, we have

$$\begin{array}{ll}\left|a_2{a}_4-a_3^2\right|& =\left(\left|{\lambda }_1\right|-\left|{\lambda }_2\right|+\frac{{\psi }_1{\psi }_3-A\left(q\right){\psi }_2^2}{64q^2{\psi }_1{\psi }_3}{p}_1^2\right)c^4\\ & +\left(4\left|{\lambda }_2\right|+\left(\frac{A\left(q\right){\psi }_2^2-{\psi }_1{\psi }_3}{16q^2{\psi }_1{\psi }_3}{p}_1^2\right)\right)c^2+\frac{1}{4q^2{\psi }_2^2}{p}_1^2\\ & =G\left(c\right).\end{array}$$

(27)

For optimum value of $G\left(c\right)$, we consider $G^{\prime }\left(c\right)=0,$ which implies that $c=0$. So $G\left(c\right)$ has a maximum value at $c=0$. We therefore conclude that the maximum value of $G\left(c\right)$ is given by

$$\frac{1}{4q^2{\psi }_2^2}{p}_1^2,$$

which occurs at $c=0$ or

$$c^2=-\frac{128\left|{\lambda }_2\right|q^2{\psi }_1{\psi }_3+4A\left(q\right){\psi }_2^2-2{\psi }_1{\psi }_3{p}_1^2}{\left(64q^2\left(\left|{\lambda }_1\right|-\left|{\lambda }_2\right|\right){\psi }_1{\psi }_3+{\psi }_1{\psi }_3-A\left(q\right){\psi }_2^2{p}_1^2\right)}.$$

This completes the proof of Theorem 3.  □

If we put ${\psi }_{n-1}=1$ and let $q\to 1-$ in Theorem 3, we have the following known result.

Corollary 2

(see [26]). If the function $f\left(z\right)$ given by (1) belongs to the class k-$\mathcal{ST},$ where $k\in \left[0,1\right],$ then

$$\left|a_2{a}_4-a_3^2\right|\leqq \frac{p_1^2}{4}.$$

If we put

$$p_1=2\mathrm{and}{\psi }_{n-1}=1,$$

by letting $q\to 1-$ in Theorem 3, we have the following known result.

Corollary 3

(see [18]). If $f\in {\mathcal{S}}^{\ast },$ then

$$\left|a_2{a}_4-a_3^2\right|\leqq 1.$$

By letting $k=1,$ ${\psi }_{n-1}=1,$ $q\to 1-$ and

$$p_1=\frac{8}{{\pi }^2},p_2=\frac{16}{3{\pi }^2}\mathrm{and}{p}_3=\frac{184}{45{\pi }^2}$$

in Theorem 3, we have the following known result.

Corollary 4

(see [27]). If the function $f\left(z\right)$ given by (1) belong to the class $\mathcal{SP},$ then

$$\left|a_2{a}_4-a_3^2\right|\leqq \frac{16}{{\pi }^4}.$$

4. Concluding Remarks and Observations

Motivated significantly by a number of recent works, we have made use of a certain general conic domain and the quantum (or q-) calculus in order to define and investigate a new subclass of normalized analytic functions in the open unit disk $\mathbb{U}$, which we have referred to as q-starlike functions. For this q-starlike function class, we have successfully derived several properties and characteristics. In particular, we have found the Hankel determinant and the Toeplitz matrices for this newly-defined class of q-starlike functions. We also highlight some known consequences of our main results which are stated and proved as theorems and corollaries.