Coefficient Bounds for Certain Subclasses of q-Starlike Functions

Abstract

By making use of q-calculus, we define and investigate several new subclasses of bi-univalent mappings related to the q-Noor integral operator. The coefficient bounds $|u_2|$, $|u_3|$ and the Fekete–Szegő problem $\left|u_3-\mu u_2^2\right|$ for mappings belonging to these classes are derived.

1. Introduction and Preliminaries

Let ${\chi }_{\circ }:=\left\{v\in \mathbb{C}:\left|v\right|<1\right\}$, and $\mathcal{A}$ be the class of mappings l which are analytic in ${\chi }_{\circ }$ and normalized by the conditions $l\left(0\right)=l^{{}^{\prime }}\left(0\right)-1=0$. Thus, every l from $\mathcal{A}$ has the following series representation:

$$l\left(v\right)=v+\sum _{n=2}^{\infty }{u}_n{v}^n.$$

(1)

A mapping $l\in \mathcal{A}$ is said to be in the class $\mathcal{S}$ if l is univalent in ${\chi }_{\circ }$. It is well-known that every univalent mapping $l\in \mathcal{A}$ has an inverse $l^{-1}$ which is defined by

$$l^{-1}\left(l\left(v\right)\right)=v\left(v\in {\chi }_{\circ }\right),$$

and

$$l\left(l^{-1}\left(\varpi \right)\right)=\varpi \left(\left|\varpi \right|<r_0\left(l\right);r_0\left(l\right)\ge \frac{1}{4}\right).$$

Indeed, the inverse mapping $l^{-1}$ is specified by

$$m\left(\varpi \right)=l^{-1}\left(\varpi \right)=\varpi -u_2{\varpi }^2+(2u_2^2-u_3){\varpi }^3-(5u_2^3-5u_2{u}_3+u_4){\varpi }^4+\cdots .$$

(2)

A mapping $l\in \mathcal{A}$ is bi-univalent in ${\chi }_{\circ }$ if both l and $l^{-1}$ are univalent in ${\chi }_{\circ }$. Let $\mathsf{\Sigma }$ denote the class of bi-univalent mappings in ${\chi }_{\circ }$ specified by the Taylor–Maclaurin series expansion (1). Many researchers calculated the second coefficient and obtained the values of coefficients for different subclasses of bi-univalent mappings. Levin [1] proved that $\left|u_2\right|<1.51$. Branan et al. [2] showed that $\left|u_2\right|\le \sqrt{2}$, and Netanyahu [3] proved that $max\left|u_2\right|=4/3.$ Also, Srivastava et al. [4], Frasin and Aouf [5] estimated the coefficients $|u_2|$ and $\left|u_3\right|$ for certain subclasses of bi-univalent mapping. For more recent investigations on bi-univalent mappings, one can refer to [1,6,7,8,9,10,11,12,13].

Let l and m be analytic in ${\chi }_{\circ }$. Then l is subordinate to m, if there exists a Schwarz mapping s, which is analytic in ${\chi }_{\circ }$ with $s\left(0\right)=0$ and $\left|s\right(v\left)\right|<1$ for $v\in {\chi }_{\circ }$ such that $l\left(v\right)=m\left(s\right(v\left)\right)$. Furthermore, if the mapping m is univalent in ${\chi }_{\circ }$, then we obtain the following relationship (see [14,15]):

$$l\left(v\right)\prec m\left(v\right)⟺l\left(0\right)-m\left(0\right)=0\mathrm{and}l\left({\chi }_{\circ }\right)\subset m\left({\chi }_{\circ }\right).$$

For two analytic mappings

$$l\left(v\right)=v+\sum _{n=2}^{\infty }{u}_n{v}^n\mathrm{and}m\left(v\right)=v+\sum _{n=2}^{\infty }{b}_n{v}^n,$$

the convolution of l and m is defined by

$$l\left(v\right)\ast m\left(v\right)=v+\sum _{n=2}^{\infty }{u}_n{b}_n{v}^n.$$

We now recall several essential definitions involving $\overline{q}$-calculus which will be utilized in the present paper. Throughout this paper, we will consider $\overline{q}$ to be a fixed number in $(0,1)$.

Definition 1

([16]). For $\varrho \in \mathbb{N}$ and $\overline{q}\in \left(0,1\right)$, the number $\left[\varrho \right]$ is defined by

$$\left[\varrho \right]=\left\{\begin{array}{c}\frac{1-{\overline{q}}^{\varrho }}{1-\overline{q}}\left(\left[0\right]=0\right),\\ {\displaystyle \sum _{k=0}^{n-1}}{\overline{q}}^k=1+\overline{q}+{\overline{q}}^2+\dots +{\overline{q}}^{n-1}.\end{array}\right.$$

Definition 2.

For $\varrho \in \mathbb{N}$, the $\overline{q}$-number shifted factorial is defined by

$$\left[\varrho \right]!=\left[1\right]\left[2\right]\left[3\right]\dots \left[\varrho \right]\left(\left[0\right]!=1\right).$$

We observe that $\underset{\overline{q}\to 1}{lim}\left[\varrho \right]!=\varrho !$.

Definition 3

([17,18]). The $\overline{q}$-derivative operator for $l\in \mathcal{A}$ is described as:

$${\partial }_{\overline{q}}l\left(v\right)=\frac{l\left(\overline{q}v\right)-l\left(v\right)}{(\overline{q}-1)v}(v\in {\chi }_{\circ }).$$

For $n\in \mathbb{N}$ and $v\in {\chi }_{\circ }$, we see that

$${\partial }_{\overline{q}}{v}^n=\left[n\right]v^{n-1},{\partial }_{\overline{q}}\left\{\sum _{n=1}^{\infty }{u}_n{v}^n\right\}=\sum _{n=1}^{\infty }\left[n\right]u_n{v}^{n-1}.$$

Definition 4.

The $\overline{q}$-generalized Pochhammer symbol for $\varrho \in \mathbb{R}$ and $n\in \mathbb{N}$ is defined by

$${\left[\varrho \right]}_n=\left[\varrho \right][\varrho +1][\varrho +2]\dots [\varrho +(n-1)],$$

and for $\varrho >0$, $\overline{q}$-gamma mapping is described as:

$${\mathsf{\Gamma }}_{\overline{q}}(\varrho +1)=\left[\varrho \right]{\mathsf{\Gamma }}_{\overline{q}}\left(\varrho \right)\mathrm{and}{\mathsf{\Gamma }}_{\overline{q}}\left(1\right)=1.$$

Definition 5

([12]). Let $l\in \mathcal{A}$ and $\lambda >-1$. The $\overline{q}$-Noor integral operator is defined by

$${\left[{\mathcal{F}}_{\overline{q},\lambda +1}\left(v\right)\right]}^{-1}\ast {\mathcal{F}}_{\overline{q},\lambda +1}\left(v\right)=v{\partial }_{\overline{q}}l\left(v\right).$$

Now, we define the operator $I_{\overline{q}}^{\lambda }$ as follows:

$$I_{\overline{q}}^{\lambda }l\left(v\right)=l\left(v\right)\ast {\left[{\mathcal{F}}_{\overline{q},\lambda +1}\left(v\right)\right]}^{-1}(v\in {\chi }_{\circ };\lambda >-1),$$

(3)

where

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

Thus, we have

$$I_{\overline{q}}^{\lambda }l\left(v\right)=v+\sum _{n=2}^{\infty }\frac{\left[n\right]!{\mathsf{\Gamma }}_{\overline{q}}(1+\lambda )}{{\mathsf{\Gamma }}_{\overline{q}}(n+\lambda )}{u}_n{v}^n=v+\sum _{n=2}^{\infty }\frac{\left[n\right]!}{{[\lambda +1]}_{n-1}}{u}_n{v}^n.$$

(4)

Clearly,

$$I_{\overline{q}}^{0}l\left(v\right)=v{\partial }_{\overline{q}}l\left(v\right)\mathrm{and}{I}_{\overline{q}}^{1}l\left(v\right)=l\left(v\right).$$

Note that when $\overline{q}\to 1$, $\overline{q}$-Noor integral operator reduces to Noor integral operator, and the following identity holds:

$$v{\partial }_{\overline{q}}\left(I_{\overline{q}}^{\lambda +1}l\left(v\right)\right)=\left(1+\frac{\left[\lambda \right]}{{\overline{q}}^{\lambda }}\right)I_{\overline{q}}^{\lambda }l\left(v\right)-\frac{\left[\lambda \right]}{{\overline{q}}^{\lambda }}{I}_{\overline{q}}^{\lambda +1}l\left(v\right).$$

(5)

If $\overline{q}\to 1$, the equality (5) implies that

$$v{\left(I^{\lambda +1}l\left(v\right)\right)}^{{}^{\prime }}=\left(1+\lambda \right)I^{\lambda }l\left(v\right)-\lambda I^{\lambda +1}l\left(v\right),$$

which is the well-known recurrent formula for the Noor integral operator.

Furthermore, a mapping $l\in \mathcal{A}$ is called starlike of order $\alpha (0\le \alpha <1)$, if

$$\Re \left(\frac{v{l}^{{}^{\prime }}\left(v\right)}{l\left(v\right)}\right)>\alpha .$$

(6)

We use the notation ${\mathcal{S}}^{\ast }\left(\alpha \right)$ for the class of starlike mappings of order $\alpha .$ In particular, when $\alpha =0,$ the class ${\mathcal{S}}^{\ast }:={\mathcal{S}}^{\ast }\left(0\right)$ denotes the familiar class of starlike mappings.

One way to generalize the class ${\mathcal{S}}^{\ast }\left(\alpha \right)$ is to replace the derivative in (6) by the $\overline{q}$-difference operator ${\partial }_{\overline{q}}$ and replace the right-half plane $\left\{\varpi :\Re \left(\varpi \right)>\alpha \right\}$ with a suitable domain. The appropriate definition turns out to be the following:

Definition 6

([19]). A mapping $l\in \mathcal{A}$ is said to be in the class ${\mathcal{S}}_{\overline{q}}^{\ast }\left(\alpha \right)$ if

$$\left|\frac{\frac{v\left({\partial }_{\overline{q}}l\right)\left(v\right)}{l\left(v\right)}-\alpha }{1-\alpha }-\frac{1}{1-\overline{q}}\right|<\frac{1}{1-\overline{q}}(0\le \alpha <1).$$

(7)

Observe that as $\overline{q}\to 1^{-}$, the disk

$$\left|\varpi -{\left(1-\overline{q}\right)}^{-1}\right|<{\left(1-\overline{q}\right)}^{-1}$$

becomes the right-half plane and the class ${\mathcal{S}}_{\overline{q}}^{\ast }\left(\alpha \right)$ reduces to ${\mathcal{S}}^{\ast }\left(\alpha \right).$ In particular, when $\alpha =0$, the class ${\mathcal{S}}_{\overline{q}}^{\ast }\left(\alpha \right)$ concides with the class ${\mathcal{S}}_{\overline{q}}^{\ast }:={\mathcal{S}}_{\overline{q}}^{\ast }\left(0\right)$, which is proposed by Ismail et al. [19] (see also Srivastava et al. [20]), it was shown that the relationship in (7) is equivalent to

$$\frac{\frac{v\left({\partial }_{\overline{q}}l\right)\left(v\right)}{l\left(v\right)}-\alpha }{1-\alpha }\prec \frac{1+v}{1-\overline{q}v}.$$

Now, by using the $\overline{q}$-Noor integral operator, we define the following function classes.

Definition 7.

For $l\in \mathsf{\Sigma }$ and $0\le \alpha <1,$ l belongs to the bi-univalent class ${\mathcal{N}}_{\mathsf{\Sigma }}^{(\alpha ,\overline{q})}$ if and only if

$$\frac{\frac{v{\partial }_{\overline{q}}l\left(v\right)}{l\left(v\right)}-\alpha }{1-\alpha }\prec \frac{1+v}{1-\overline{q}v},$$

(8)

and

$$\frac{\frac{\varpi {\partial }_{\overline{q}}m\left(\varpi \right)}{m\left(\varpi \right)}-\alpha }{1-\alpha }\prec \frac{1+\varpi }{1-\overline{q}\varpi },$$

(9)

where $m\left(\varpi \right)=l^{-1}\left(\varpi \right)$ is defined by (2).

Definition 8.

For $l\in \mathsf{\Sigma }$, $0\le \alpha <1$ and $\lambda >-1$, l belongs to the bi-univalent class ${\mathcal{N}}_{\mathsf{\Sigma }}^{(\alpha ,\lambda ,\overline{q})}$ if and only if

$$\frac{\frac{v{\partial }_{\overline{q}}{I}_{\overline{q}}^{\lambda }l\left(v\right)}{I_{\overline{q}}^{\lambda }l\left(v\right)}-\alpha }{1-\alpha }\prec \frac{1+v}{1-\overline{q}v},$$

(10)

and

$$\frac{\frac{\varpi {\partial }_{\overline{q}}{I}_{\overline{q}}^{\lambda }m\left(\varpi \right)}{I_{\overline{q}}^{\lambda }m\left(\varpi \right)}-\alpha }{1-\alpha }\prec \frac{1+\varpi }{1-\overline{q}\varpi },$$

(11)

where $m\left(\varpi \right)=l^{-1}\left(\varpi \right)$ is defined by (2).

Example 1.

The mappings

$$\frac{v}{1-v}and-log(1-v)$$

are examples of the above defined subclasses of bi-univalent mappings.

Lemma 1

([21]). If $p\left(v\right)=1+p_{1}v+p_2{v}^2+p_3{v}^3+\cdots$ is an analytic mapping in ${\chi }_{\circ }$ with positive real parts, then

$$\left|p_2-\frac{p_1^2}{2}\right|\le 2-\frac{{\left|p_1\right|}^2}{2}.$$

(12)

2. Main Results

Firstly, we derive the following result.

Theorem 1.

For $0\le \alpha <1,$ if $l\in {\mathcal{N}}_{\mathsf{\Sigma }}^{(\alpha ,\overline{q})},$ then

$$\begin{array}{c}\hfill \left|u_2\right|\le (1+\overline{q})(1-\alpha )\sqrt{\frac{2}{2\overline{q}\left[(1+\overline{q})(1-\alpha )\left(\left[2\right]-1\right)-{\overline{q}}^2\right]}},\end{array}$$

$$\begin{array}{c}\hfill \left|u_3\right|\le \frac{(1-\alpha )(1+\overline{q})}{\left[2\right]{\overline{q}}^2}\left\{\overline{q}+\left[2\right](1+\overline{q})(1-\alpha )\right\},\end{array}$$

and

$$\begin{array}{c}\hfill \left|u_3-2u_2^2\right|\le \frac{(1-\alpha ){(1+\overline{q})}^2}{\left[2\right]{\overline{q}}^2}\left(1+\overline{q}-\alpha \right).\end{array}$$

(13)

Proof. 

For the mapping $l\in {\mathcal{N}}_{\mathsf{\Sigma }}^{(\alpha ,\overline{q})}$ of the form (1), we know that

$$\frac{\frac{v{\partial }_{\overline{q}}l\left(v\right)}{l\left(v\right)}-\alpha }{1-\alpha }=1+\frac{\overline{q}}{1-\alpha }{u}_{2}v+\frac{\overline{q}}{1-\alpha }\left(\left[2\right]u_3-u_2^2\right)v^2+\cdots ,$$

(14)

and for its inverse map $m=l^{-1},$ we have

$$\frac{\frac{\varpi {\partial }_{\overline{q}}m\left(\varpi \right)}{m\left(\varpi \right)}-\alpha }{1-\alpha }=1-\frac{\overline{q}}{1-\alpha }{u}_2\varpi +\frac{\overline{q}}{1-\alpha }\left\{\left[2\right]\left(2u_2^2-u_3\right)-u_2^2\right\}{\varpi }^2.$$

(15)

Since both mappings l and its inverse map $m=l^{-1}$ are in ${\mathcal{N}}_{\mathsf{\Sigma }}^{(\alpha ,\overline{q})},$ by the definition of subordination, there exist two Schwarz mappings $p\left(v\right)={\displaystyle \sum _{n=1}^{\infty }}{c}_n{v}^n,$ and $h\left(\varpi \right)={\displaystyle \sum _{n=1}^{\infty }}{d}_n{\varpi }^n,$ where v, $\varpi \in {\chi }_{\circ }.$ It follows from (14) and (15) that

$$\frac{\frac{v{\partial }_{\overline{q}}l\left(v\right)}{l\left(v\right)}-\alpha }{1-\alpha }=\frac{1+\left(p\right(v\left)\right)}{1-\overline{q}\left(p\left(v\right)\right)}=1+(1+\overline{q})c_{1}v+(1+\overline{q})\left(c_2+\overline{q}{c}_1^2\right)v^2+\cdots ,$$

(16)

and

$$\frac{\frac{\varpi {\partial }_{\overline{q}}m\left(\varpi \right)}{m\left(\varpi \right)}-\alpha }{1-\alpha }=\frac{1+\left(h\right(\varpi \left)\right)}{1-\overline{q}\left(h\left(\varpi \right)\right)}=1+(1+\overline{q})d_1\varpi +(1+\overline{q})\left(d_2+\overline{q}{d}_1^2\right){\varpi }^2+\cdots .$$

(17)

For the coefficients of the Schwarz mappings $p\left(v\right)$ and $h\left(\varpi \right)$, we find from [22] that $\left|c_{n-1}\right|\le 1$ and $\left|d_{n-1}\right|\le 1$. By equating the corresponding coefficients of (14) and (16), we get

$$\begin{array}{c}\hfill \frac{\overline{q}}{1-\alpha }{u}_2=(1+\overline{q})c_1,\end{array}$$

(18)

and

$$\begin{array}{c}\hfill \frac{\overline{q}}{1-\alpha }\left\{\left[2\right]u_3-u_2^2\right\}=(1+\overline{q})\left(c_2+\overline{q}{c}_1^2\right).\end{array}$$

(19)

Similarly, by equating the corresponding coefficients of (15) and (17), we have

$$\begin{array}{c}\hfill -\frac{\overline{q}}{1-\alpha }{u}_2=(1+\overline{q})d_1,\end{array}$$

(20)

and

$$\begin{array}{c}\hfill \frac{\overline{q}}{1-\alpha }\left\{\left[2\right]\left(2u_2^2-u_3\right)-u_2^2\right\}=(1+\overline{q})\left(d_2+\overline{q}{d}_1^2\right).\end{array}$$

(21)

From (18) and (20), we obtain

$$\begin{array}{c}\hfill c_1=-d_1,\end{array}$$

(22)

and

$$\begin{array}{c}\hfill \frac{2{\overline{q}}^2}{{(1-\alpha )}^2}{u}_2^2={(1+\overline{q})}^2\left(c_1^2+d_1^2\right).\end{array}$$

(23)

By means of (19) and (21), we know that

$$\left(\frac{2\left[2\right]\overline{q}}{1-\alpha }-\frac{2\overline{q}}{1-\alpha }\right)u_2^2=(1+\overline{q})(c_2+d_2)+\overline{q}(1+\overline{q})\left(c_1^2+d_1^2\right).$$

(24)

By using (22) and (23) in (24), we have

$$\left|u_2\right|\le (1+\overline{q})(1-\alpha )\sqrt{\frac{2}{2\overline{q}\left[(1+\overline{q})(1-\alpha )\left(\left[2\right]-1\right)-{\overline{q}}^2\right]}}.$$

Now, in view of (19), (21) and (22), we see that

$$\frac{2\left[2\right]\overline{q}}{1-\alpha }{u}_3=(1+\overline{q})(d_2-c_2)+\frac{2\left[2\right]\overline{q}}{1-\alpha }{u}_2^2.$$

(25)

From (23) and (25), we know that

$$\left|u_3\right|\le \frac{(1-\alpha )(1+\overline{q})}{\left[2\right]{\overline{q}}^2}\left\{\overline{q}+\left[2\right](1+\overline{q})(1-\alpha )\right\}.$$

By means of (21), we get

$$\frac{\overline{q}}{1-\alpha }\left[2\right]\left(2u_2^2-u_3\right)=(1+\overline{q})\left(d_2+\overline{q}{d}_1^2\right)+\frac{\overline{q}}{1-\alpha }{u}_2^2.$$

(26)

It follows from (20) and (26) that

$$2u_2^2-u_3=\frac{(1-\alpha )(1+\overline{q})}{\left[2\right]{\overline{q}}^2}\left\{\overline{q}{d}_2+\left[{\overline{q}}^2+(1-\alpha )(1+\overline{q})\right]d_1^2\right\}.$$

Thus, by virtue of $\left|c_{n-1}\right|\le 1$ and $\left|d_{n-1}\right|\le 1$, we deduce that

$$\left|u_3-2u_2^2\right|\le \frac{(1-\alpha ){(1+\overline{q})}^2}{\left[2\right]{\overline{q}}^2}\left(1+\overline{q}-\alpha \right).$$

□

Theorem 2.

For $0\le \alpha <1$ and $\lambda >-1$, if $l\in {\mathcal{N}}_{\mathsf{\Sigma }}^{(\alpha ,\lambda ,\overline{q})},$ then

$$\left|u_2\right|\le {[\lambda +1]}_1(1+\overline{q})(1-\alpha )\sqrt{\frac{2{[\lambda +1]}_2}{T}},$$

(27)

where

$$\begin{array}{c}\hfill T=\overline{q}\left\{(1+\overline{q})(1-\alpha )\left\{2\left[2\right]\left[3\right]!{\left({[\lambda +1]}_1\right)}^2-2{\left(\left[2\right]!\right)}^2{[\lambda +1]}_2\right\}-Z\right\}\end{array}$$

with

$$\begin{array}{c}\hfill Z=2{\left(\left[2\right]!\right)}^2{\overline{q}}^2,\end{array}$$

$$\left|u_3\right|\le \frac{(1+\overline{q})(1-\alpha ){[\lambda +1]}_2}{\left[2\right]\left[3\right]!{\overline{q}}^2}\left\{\overline{q}+\frac{\left[2\right]\left[3\right]!{\left({[\lambda +1]}_1\right)}^2(1+\overline{q})(1-\alpha )}{{\left(\left[2\right]!\right)}^2{[\lambda +1]}_2}\right\},$$

(28)

and

$$\left|u_3-2u_2^2\right|\le \frac{{[\lambda +1]}_2(1-\alpha ){(1+\overline{q})}^2}{\left[2\right]\left[3\right]!{\overline{q}}^2}\left(1+\overline{q}-\alpha \right).$$

Proof. 

For the mapping $l\in {\mathcal{N}}_{\mathsf{\Sigma }}^{(\alpha ,\lambda ,\overline{q})}$ of the form (1), we have

$$\begin{array}{cc}\hfill & \frac{\frac{v{\partial }_{\overline{q}}{I}_{\overline{q}}^{\lambda }l\left(v\right)}{I_{\overline{q}}^{\lambda }l\left(v\right)}-\alpha }{1-\alpha }\hfill \\ =\hfill & 1+\frac{\left[2\right]!\overline{q}}{\left(1-\alpha \right){[\lambda +1]}_1}{u}_{2}v+\frac{\overline{q}}{1-\alpha }\left\{\frac{\left[2\right]\left[3\right]!}{{[\lambda +1]}_2}{u}_3-{\left(\frac{\left[2\right]!}{{[\lambda +1]}_1}\right)}^2{u}_2^2\right\}{v}^2+\cdots ,\hfill \end{array}$$

(29)

and for its inverse map $m=l^{-1},$ we get

$$\begin{array}{cc}\hfill & \frac{\frac{\varpi {\partial }_{\overline{q}}{I}_{\overline{q}}^{\lambda }m\left(\varpi \right)}{I_{\overline{q}}^{\lambda }m\left(\varpi \right)}-\alpha }{1-\alpha }\hfill \\ =\hfill & 1-\frac{\left[2\right]!\overline{q}}{\left(1-\alpha \right){[\lambda +1]}_1}{u}_2\varpi +\frac{\overline{q}}{1-\alpha }\left\{\frac{\left[2\right]\left[3\right]!}{{[\lambda +1]}_2}\left(2u_2^2-u_3\right)-{\left(\frac{\left[2\right]!}{{[\lambda +1]}_1}\right)}^2{u}_2^2\right\}{\varpi }^2+\cdots .\hfill \end{array}$$

(30)

By noting that both mappings l and its inverse map $m=l^{-1}$ are in the class ${\mathcal{N}}_{\mathsf{\Sigma }}^{(\alpha ,\lambda ,\overline{q})}$, by the definition of subordination, there exist two Schwarz mappings $p\left(v\right)={\displaystyle \sum _{n=1}^{\infty }}{c}_n{v}^n$ and $h\left(\varpi \right)={\displaystyle \sum _{n=1}^{\infty }}{d}_n{\varpi }^n,$ where v, $\varpi \in {\chi }_{\circ }$, it follows that

$$\frac{\frac{v{\partial }_{\overline{q}}{I}_{\overline{q}}^{\lambda }l\left(v\right)}{I_{\overline{q}}^{\lambda }l\left(v\right)}-\alpha }{1-\alpha }=\frac{1+\left(p\right(v\left)\right)}{1-\overline{q}\left(p\left(v\right)\right)}=1+(1+\overline{q})c_{1}v+(1+\overline{q})\left(c_2+\overline{q}{c}_1^2\right)v^2+\cdots ,$$

(31)

and

$$\frac{\frac{\varpi {\partial }_{\overline{q}}{I}_{\overline{q}}^{\lambda }m\left(\varpi \right)}{I_{\overline{q}}^{\lambda }m\left(\varpi \right)}-\alpha }{1-\alpha }=\frac{1+\left(h\right(\varpi \left)\right)}{1-\overline{q}\left(h\left(\varpi \right)\right)}=1+(1+\overline{q})d_1\varpi +(1+\overline{q})\left(d_2+\overline{q}{d}_1^2\right){\varpi }^2+\cdots .$$

(32)

By observing that the coefficients of the Schwarz mappings $p\left(v\right)$ and $h\left(\varpi \right)$ satisfy the conditions $\left|c_{n-1}\right|\le 1$ and $\left|d_{n-1}\right|\le 1,$ and by equating the corresponding coefficients of (29) and (31), we see that

$$\begin{array}{c}\hfill \frac{\left[2\right]!\overline{q}}{\left(1-\alpha \right){[\lambda +1]}_1}{u}_2=(1+\overline{q})c_1,\end{array}$$

(33)

and

$$\begin{array}{c}\hfill \frac{\overline{q}}{1-\alpha }\left\{\frac{\left[2\right]\left[3\right]!}{{[\lambda +1]}_2}{u}_3-{\left(\frac{\left[2\right]!}{{[\lambda +1]}_1}\right)}^2{u}_2^2\right\}=(1+\overline{q})\left(c_2+\overline{q}{c}_1^2\right),\end{array}$$

(34)

Similarly, by comparing the corresponding coefficients of (30) and (32), we get

$$\begin{array}{c}\hfill -\frac{\left[2\right]!\overline{q}}{\left(1-\alpha \right){[\lambda +1]}_1}{u}_2=(1+\overline{q})d_1,\end{array}$$

(35)

and

$$\begin{array}{c}\hfill \frac{\overline{q}}{1-\alpha }\left\{\frac{\left[2\right]\left[3\right]!}{{[\lambda +1]}_2}\left(2u_2^2-u_3\right)-{\left(\frac{\left[2\right]!}{{[\lambda +1]}_1}\right)}^2{u}_2^2\right\}=(1+\overline{q})\left(d_2+\overline{q}{d}_1^2\right).\end{array}$$

(36)

From (33) and (35), we obtain

$$\begin{array}{c}\hfill c_1=-d_1,\end{array}$$

(37)

and

$$\begin{array}{c}\hfill 2{\left(\frac{\left[2\right]!\overline{q}}{(1-\alpha ){[\lambda +1]}_1}\right)}^2{u}_2^2={(1+\overline{q})}^2\left(c_1^2+d_1^2\right).\end{array}$$

(38)

It follows from (34) and (36) that

$$\frac{2\overline{q}}{1-\alpha }\left\{\frac{\left[2\right]\left[3\right]!}{{[\lambda +1]}_2}-{\left(\frac{\left[2\right]!}{{[\lambda +1]}_1}\right)}^2\right\}{u}_2^2=(1+\overline{q})(c_2+d_2)+\overline{q}(1+\overline{q})\left(c_1^2+d_1^2\right).$$

(39)

In view of (37), (38) and (39), we get the required estimate (27).

Now, by means of (34), (36) and (37), we obtain

$$\frac{2\overline{q}}{1-\alpha }\frac{\left[2\right]\left[3\right]!}{{[\lambda +1]}_2}{u}_3=(1+\overline{q})(d_2-c_2)+\frac{2\left[2\right]\left[3\right]!\overline{q}}{(1-\alpha ){[\lambda +1]}_2}{u}_2^2.$$

(40)

By virtue of (38), (40), $\left|c_{n-1}\right|\le 1$ and $\left|d_{n-1}\right|\le 1,$ we know that

$$\left|u_3\right|\le \frac{(1+\overline{q})(1-\alpha ){[\lambda +1]}_2}{\left[2\right]\left[3\right]!{\overline{q}}^2}\left\{\overline{q}+\frac{\left[2\right]\left[3\right]!{\left({[\lambda +1]}_1\right)}^2(1+\overline{q})(1-\alpha )}{{\left(\left[2\right]!\right)}^2{[\lambda +1]}_2}\right\}.$$

(41)

From (36), we get

$$\frac{\overline{q}}{1-\alpha }\left(\frac{2\left[2\right]\left[3\right]!}{{[\lambda +1]}_2}{u}_2^2-\frac{\left[2\right]\left[3\right]!}{{[\lambda +1]}_2}{u}_3\right)=(1+\overline{q})\left(d_2+\overline{q}{d}_1^2\right)+\frac{\overline{q}}{1-\alpha }{\left(\frac{\left[2\right]!}{{[\lambda +1]}_1}\right)}^2{u}_2^2.$$

(42)

By using (35) on right hand side of Equation (42), we see that

$$2u_2^2-u_3=\frac{(1-\alpha )(1+\overline{q}){[\lambda +1]}_2}{\left[2\right]\left[3\right]!{\overline{q}}^2}\left\{\overline{q}{d}_2+\left[{\overline{q}}^2+(1-\alpha )(1+\overline{q})\right]d_1^2\right\}.$$

(43)

By means of (43), $\left|c_{n-1}\right|\le 1$ and $\left|d_{n-1}\right|\le 1,$ we conclude that

$$\left|u_3-2u_2^2\right|\le \frac{{[\lambda +1]}_2(1-\alpha ){(1+\overline{q})}^2}{\left[2\right]\left[3\right]!{\overline{q}}^2}\left(1+\overline{q}-\alpha \right).$$

□

Theorem 3.

For $0\le \alpha <1,$ if $l\in {\mathcal{N}}_{\mathsf{\Sigma }}^{(\alpha ,\overline{q})}$ and $\mu \in \mathbb{R}$, then

$$\left|u_3-\mu u_2^2\right|\le \left\{\begin{array}{c}\frac{(1+\overline{q})(1-\alpha )}{\left[2\right]\overline{q}}\left(0\le \left|h_1\left(\mu \right)\right|\le \frac{1}{2\left[2\right]\overline{q}}\right),\\ 4(1-\alpha )(1+\overline{q})\left|h_1\left(\mu \right)\right|\left(\left|h_1\left(\mu \right)\right|\ge \frac{1}{2\left[2\right]\overline{q}}\right),\end{array}\right.$$

where

$$h_1\left(\mu \right)=\left(\frac{\left[2\right]}{2}-\mu \right)\frac{(1+\overline{q})(1-\alpha )}{\overline{q}\left\{(1+\overline{q})(1-\alpha )\left(\left[2\right]\left[2\right]-2\left[1\right]\right)-2{\overline{q}}^2\right\}}.$$

Proof. 

Let the mapping l given by (1) be in the class ${\mathcal{N}}_{\mathsf{\Sigma }}^{(\alpha ,\overline{q})}.$ From (23) and (24), we know that

$$u_2^2=\frac{{(1+\overline{q})}^2{(1-\alpha )}^2(c_2+d_2)}{\overline{q}\left\{(1+\overline{q})(1-\alpha )\left(\left[2\right]\left[2\right]-2\left[1\right]\right)-2{\overline{q}}^2\right\}}.$$

(44)

Also, by means of (25), we obtain

$$u_3=\frac{(1+\overline{q})(1-\alpha )(d_2-c_2)}{2\left[2\right]\overline{q}}+\frac{\left[2\right]}{2}{u}_2^2.$$

(45)

It follows from (44) and (45) that

$$u_3-\mu u_2^2=(1+\overline{q})(1-\alpha )\left\{\left(h_1\left(\mu \right)+\frac{1}{2\left[2\right]\overline{q}}\right)d_2+\left(h_1\left(\mu \right)-\frac{1}{2\left[2\right]\overline{q}}\right)c_2\right\},$$

where

$$h_1\left(\mu \right)=\left(\frac{\left[2\right]}{2}-\mu \right)\frac{(1+\overline{q})(1-\alpha )}{\overline{q}\left\{(1+\overline{q})(1-\alpha )\left(\left[2\right]\left[2\right]-2\left[1\right]\right)-2{\overline{q}}^2\right\}}.$$

Thus, in view of (12), we deduce that

$$\left|u_3-\mu u_2^2\right|\le \left\{\begin{array}{c}\frac{(1+\overline{q})(1-\alpha )}{\left[2\right]\overline{q}}\left(0\le \left|h_1\left(\mu \right)\right|\le \frac{1}{2\left[2\right]\overline{q}}\right),\\ 4(1-\alpha )(1+\overline{q})\left|h_1\left(\mu \right)\right|\left(\left|h_1\left(\mu \right)\right|\ge \frac{1}{2\left[2\right]\overline{q}}\right).\end{array}\right.$$

(46)

□

Theorem 4.

For $0\le \alpha <1$ and $\lambda >-1$, if $l\in {\mathcal{N}}_{\mathsf{\Sigma }}^{(\alpha ,\lambda ,\overline{q})}$ and $\mu \in \mathbb{R}$, then

$$\left|u_3-\mu u_2^2\right|\le \left\{\begin{array}{c}\frac{(1+\overline{q})(1-\alpha ){[\lambda +2]}_2}{\left[2\right]\left[3\right]!\overline{q}}\left(0\le \left|h_2\left(\mu \right)\right|\le \frac{{[\lambda +2]}_2}{2\left[2\right]\left[3\right]!\overline{q}}\right),\\ 4(1-\alpha )(1+\overline{q})\left|h_2\left(\mu \right)\right|\left(\left|h_2\left(\mu \right)\right|\ge \frac{{[\lambda +2]}_2}{2\left[2\right]\left[3\right]!\overline{q}}\right),\end{array}\right.$$

where

$$h_2\left(\mu \right)=\left(\frac{\left[2\right]}{2\left[2\right]\left[3\right]!\overline{q}}-\mu \right)T_1$$

with

$$T_1=\frac{{[\lambda +2]}_2{\left({[\lambda +1]}_1\right)}^2(1+\overline{q})(1-\alpha )}{\overline{q}\left\{(1+\overline{q})(1-\alpha )\left({\left({[\lambda +1]}_1\right)}^2\left[2\right]\left[2\right]\left[3\right]!-2\left[1\right]{\left(\left[2\right]!\right)}^2{[\lambda +2]}_2\right)-2{\overline{q}}^2\right\}}.$$

(47)

Proof. 

Let the mapping l given by (1) be in the class ${\mathcal{N}}_{\mathsf{\Sigma }}^{(\alpha ,\lambda ,\overline{q})}.$ In view of (38) and (39), we know that

$$u_2^2=\frac{{[\lambda +2]}_2{\left({[\lambda +2]}_1\right)}^2{(1+\overline{q})}^2{(1-\alpha )}^2(c_2+d_2)}{\overline{q}\left\{(1+\overline{q})(1-\alpha )\left({\left({[\lambda +2]}_1\right)}^2\left[2\right]\left[2\right]\left[3\right]!-2\left[1\right]{\left(\left[2\right]!\right)}^2{[\lambda +2]}_2\right)-2{\overline{q}}^2{\left(\left[2\right]!\right)}^2\right\}},$$

(48)

and from (40), we have

$$u_3=\frac{(1+\overline{q})(1-\alpha ){[\lambda +2]}_2(d_2-c_2)}{2\left[2\right]\left[3\right]!\overline{q}}+\frac{\left[2\right]}{2}{u}_2^2.$$

(49)

It follows from (48) and (49) that

$$u_3-\mu u_2^2=(1+\overline{q})(1-\alpha )\left\{\left(h_2\left(\mu \right)+\frac{{[\lambda +2]}_2}{2\left[2\right]\left[3\right]!\overline{q}}\right)d_2+\left(h_2\left(\mu \right)-\frac{{[\lambda +2]}_2}{2\left[2\right]\left[3\right]!\overline{q}}\right)c_2\right\},$$

(50)

where

$$h_2\left(\mu \right)=\left(\frac{\left[2\right]}{2}-\mu \right)T_1$$

with $T_1$ given by (47). Thus, by virtue of (12) and (50), we conclude that the desired inequality holds. □

3. Conclusions

The main purpose of this article is to achieve several attentiveness and constructive manipulations of $\overline{q}$-calculus in Geometric Function Theory. By utilizing the $\overline{q}$-calculus theory, we initiate and examine a new subclass of bi-univalent functions in open unit disk ${\chi }_{\circ }$, and we find the second and the third Taylor–Maclaurin coefficients of mappings as well as find the Fekete–Szegő problem $\left|u_3-\mu u_2^2\right|$ in this function class. These results are improvements on the estimates obtained in the recent studies. In particular, we discuss the applications of $\overline{q}$-calculus by using the $\overline{q}$-Noor integral operator.