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# Fekete-Szegö Type Problems and Their Applications for a Subclass of q-Starlike Functions with Respect to Symmetrical Points

by 1,2,3, 4,*,
1
Department of Mathematics and Statistics, University of Victoria, Victoria, BC V8W 3R4, Canada
2
Department of Medical Research, China Medical University Hospital, China Medical University, Taichung 40402, Taiwan
3
Department of Mathematics and Informatics, Azerbaijan University, 71 Jeyhun Hajibeyli Street, AZ1007 Baku, Azerbaijan
4
Department of Mathematics Abbottabad University of Science and Technology, Abbottabad 22010, Pakistan
5
Department of Mathematical Sciences, Faculty of Science and Technology, Universiti Kebangsaan Malaysia, Bangi 43600, Selangor, Malaysia
6
7
*
Author to whom correspondence should be addressed.
Mathematics 2020, 8(5), 842; https://doi.org/10.3390/math8050842
Received: 15 April 2020 / Revised: 26 April 2020 / Accepted: 27 April 2020 / Published: 22 May 2020
(This article belongs to the Special Issue Mathematical Analysis and Analytic Number Theory 2020)

## Abstract

In this article, by using the concept of the quantum (or q-) calculus and a general conic domain $Ω k , q$ , we study a new subclass of normalized analytic functions with respect to symmetrical points in an open unit disk. We solve the Fekete-Szegö type problems for this newly-defined subclass of analytic functions. We also discuss some applications of the main results by using a q-Bernardi integral operator.

## 1. Introduction and Definitions

Let $A$ denote the class of all functions f which are analytic in the open unit disk
$E = { z : z ∈ C and z < 1 }$
and has the normalized Taylor-Maclaurin series expansion of the following form:
$f ( z ) = z + ∑ n = 2 ∞ a n z n .$
Let $S$ be the subclass of all functions in $A$ that are univalent in $E$ (see [1]): If f and $g ∈ A$, the function f is said to be subordinate to the function g, written as $f ≺ g$, if there exists an analytic function w in $E$, with
$w ( 0 ) = 0 and | w ( z ) | < 1 ( z ∈ E ) ,$
such that $f ( z ) = g w ( z )$. Furthermore, the following equivalence will hold true (see [2]), if g is univalent in $E .$
$f ( z ) ≺ g ( z ) ⟺ f ( 0 ) = g ( 0 ) and f ( E ) ⊂ g ( E ) .$
Let $P$ denote the well-known Carathéodory class of functions p, which are analytic in the open unit disk $E$ with
$ℜ p ( z ) > 0 and p ( 0 ) = 1 .$
If $p ∈ P$, then it has the form given by
$p ( z ) = 1 + ∑ n = 1 ∞ c n z n ,$
where $c n ≦ 2 ( n ∈ N )$.
If f is univalent in $E$ and $f ( E )$ is a star-shaped domain with respect to the origin, then f is called starlike in $E$ with respect to the origin. The analytical condition of a starlike function in $E$ is given as follows:
$ℜ z f ′ ( z ) f ( z ) > 0 ( z ∈ E ) .$
The class of all such functions is denoted by $S *$. A function $f ∈ A$ is said to be starlike with respect to symmetrical points (see [3]) if it satisfies the inequality:
$ℜ z f ′ ( z ) f ( z ) − f ( − z ) > 0 ( z ∈ E ) .$
The class of all functions in $S$ which are starlike with respect to symmetrical points is denoted by $S s *$. Furthermore, we denote two interesting subclasses of $S$ by k-$UCV$ and k-$ST$ ($0 ≦ k < ∞$) of functions which are, respectively, k-uniformly convex and k-starlike in $E$ defined for $0 ≦ k < ∞$ by
$k − UCV = f : f ∈ S and ℜ 1 + z f ′ ′ ( z ) f ′ ( z ) > k z f ′ ′ ( z ) f − ( z ) ( z ∈ E )$
and
$k − ST = f : f ∈ S and ℜ z f ′ ( z ) f ( z ) > k z f ′ ( z ) f ( z ) − 1 ( z ∈ E .$
Kanas et al. (see [4,5]; see also [6]) defined and studied classes of k-starlike functions and k-uniformly convex functions subject to the conic domain $Ω k$$( k ≧ 0 )$, where
$Ω k = u + i v : u 2 > k 2 u − 1 2 + v 2 .$
For this conic domain, the following functions play the role of extremal functions:
$p k ( z ) = A 1 ( z ) ( k = 0 ) A 1 ( z ) ( k = 1 ) A 3 ( z ) ( 0 < k < 1 ) A 4 ( z ) ( k > 1 ) ,$
where
$A 1 ( z ) = 1 + z 1 − z ,$
$A 2 ( z ) = 1 + 2 π 2 log 1 + z 1 − z 2 ,$
$A 3 ( z ) = 1 + 2 1 − k 2 sinh 2 2 π arccos k arctan ( h z ) ,$
$A 4 ( z ) = 1 + 1 k 2 − 1 sin π 2 K ( i ) ∫ 0 u ( z ) t 1 1 − x 2 1 − ( i x ) 2 d x + 1 1 − k 2 ,$
$i ∈ ( 0 , 1 )$, and
$k = cosh π K ′ ( i ) K ( i ) ,$
$K ( i )$ is the first kind of Legendre’s complete elliptic integral (see, for details [4,5]). Indeed, from (4), we have
$p k ( z ) = 1 + P 1 z + P 2 z 2 + P 3 z 3 + …$
The quantum (or q-) calculus is an important tool which is used to study various families of analytic functions and has inspired the researchers due to its applications in mathematics and some other related disciplines. Srivastava (see, for details [7]) was the first who used the basic (or q-) hypergeometric functions in Geometric Functions Theory. The extension of the class of starlike functions in the quantum (or q-) calculus was first introduced in [8] by means of the q-difference operator. After that, some remarkable research work was conducted by many mathematicians, which has played an important role in Geometric Function Theory. In particular, Srivastava et al. [9,10] studied the class of q-starlike functions related with the Janowski functions. Mahmood et al. [11] studied the class of q-starlike functions associated with conic regions. The upper bound of the third Hankel determinant for a class of q-starlike functions was investigated in [12] (see also [9]). Kanas and Raducanu [13] introduced the q-analogue of the Ruscheweyh operator by using the concept of convolution and studied some of its properties (see also [11,14,15,16,17,18,19,20]). Many other q-derivative and q-integral operators can be written by using the idea of convolution (we refer, for details, to [21,22,23,24]). For a comprehensive review of the quantum (or q-)-calculus literature, we refer to a recently-published survey-cum-expository review article by Srivastava [25]. In this article, we will use the conic domain $Ω k , q$ and the quantum (or q-) calculus to define and investigate new subclasses of starlike functions with respect to symmetrical points in the open unit disk $E$. We will investigate the Hankel determinant, the Toeplitz matrices and the Fekete-Szegö inequalities, and discuss some applications of the main results by using the q-Bernardi integral operator.
We first give some basic definitions of the quantum (or q-) calculus that will help us in the upcoming sections. We also provide some notations and concepts used in this investigation.
Definition 1.
Let $q ∈ ( 0 , 1 )$ and the q-factorial $[ n ] q !$ be defined as follows:
$[ n ] q ! = 1 ( n = 0 ) ∏ k = 1 n − 1 [ k ] q ( n ∈ N ) .$
Definition 2.
The generalized q-Pochhammer symbol $[ t ] n , q$$( t ∈ C )$ is defined as follows:
$[ t ] n , q = q t , q n ( 1 − q ) n = 1 ( n = 0 ) q [ t + 1 ] q [ t + 2 ] q … [ t + n − 1 ] q ( n ∈ N ) .$
Definition 3.
The q-Gamma function is defined as follows:
$Γ q ( t + 1 ) = [ t ] q Γ q ( t ) and Γ q ( 1 ) = 1 t > 0 .$
Definition 4.
(see [26]) For $f ∈ A ,$ the q-derivative operator or q-difference operator are defined as follows:
$D q f ( z ) = f ( z ) − f ( q z ) ( 1 − q ) z ( z ∈ E ) .$
From (1) and (7), we have
$D q f ( z ) = 1 + ∑ n = 2 ∞ [ n ] q a n z n − 1 .$
Moreover, for $n ∈ N$ and $z ∈ E ,$ we get
$D q z n = [ n ] q z n − 1 , D q ∑ n = 1 ∞ a n z n = ∑ n = 1 ∞ [ n ] q a n z n − 1 .$
When $q → 1 −$, the q-difference operator $D q$ approaches the ordinary differential operator:
$lim q → 1 − D q f ( z ) = f ′ ( z ) .$
Definition 5.
(see [8]) We say that a function $f ∈ A$ belongs to the class $S q *$ if
$f ( 0 ) = 1 = f ′ ( 0 )$
and
$z D q f ( z ) f ( z ) − 1 1 − q ≦ 1 1 − q .$
By applying the principle of subordination, the conditions (9) and (10) can be written as follows (see [27]):
$z D q f ( z ) f ( z ) ≺ 1 + z 1 − q z .$
Now, making use of the quantum (or q-) calculus and the principle of subordination, we define q-starlike and q-convex functions with respect to symmetrical points as follows.
Definition 6.
An analytic function f is said to be in the class $S s * ( q )$ if
$2 z D q f ( z ) f ( z ) − f ( − z ) − 1 1 − q ≦ 1 1 − q .$
By applying the principle of subordination, the condition (11) can be written as follows:
$2 z D q f ( z ) f ( z ) − f ( − z ) ≺ 1 + z 1 − q z .$
Definition 7.
(see [9]) Let $k ∈ [ 0 , ∞ )$ and $q ∈ 0 , 1 .$ A function p is said to be in the class k-$P q$ if and only if
$p ( z ) ≺ p k , q ( z ) ,$
where
$p k , q ( z ) = 2 p k ( z ) 1 + q + 1 − q p k ( z ) − 1$
and $p k ( z )$ is given by (5).
Geometrically, a function $p ∈ k$-$P q$ takes on all values from the domain $Ω k , q$, which is defined as follows:
$Ω k , q = w : ℜ 1 + q w q − 1 w + 2 > k 1 + q w q − 1 w + 2 − 1 .$
Remark 1.
If $q → 1 − ,$ then $Ω k , q = Ω k$ is given by (3).
Remark 2.
For $q → 1 − ,$ then k-$P q = P ( p k ) ,$ where $P ( p k )$ is defined in [4].
In the present investigation, by using the quantum (or q-) calculus and the general conic domain $Ω k , q$, we focus on the Hankel determinant, the Toeplitz matrices and the Fekete-Szegö problems for the function class $S s * ( q ) .$
Definition 8.
An analytic function f is said to be in the class k-$S s * ( q )$ if
$2 z D q f ( z ) f ( z ) − f ( − z ) ∈ k − P q$
or, equivalently,
$ℜ 1 + q 2 z D q f ( z ) f ( z ) − f ( − z ) q − 1 2 z D q f ( z ) f ( z ) − f ( − z ) + 2 > k 1 + q 2 z D q f ( z ) f ( z ) − f ( − z ) q − 1 2 z D q f ( z ) f ( z ) − f ( − z ) + 2 − 1 .$
Special Case:
For $k = 0$ and $q → 1 − ,$ then the class k-$S s * ( q )$ reduces to $S s *$ (see [3]).
Let $n ∈ N 0$ and $j ∈ N$. The jth Hankel determinant was introduced and studied in [28]:
$H j ( n ) = a n a n + 1 … a n + j − 1 a n + 1 a n + 2 … a n + j − 2 … … … … … … a n + j − 1 a n + j − 2 … a n + 2 j − 2$
where $a 1 = 1$. Several authors have studied $H j ( n )$. In particular, sharp upper bounds on $H 2 ( 2 )$ were obtained in [29,30,31] for several classes. The Hankel determinant $H 2 ( 1 )$ represents a Fekete-Szegö functional $a 3 − a 2 2$. This functional has been further generalized as $a 3 − μ a 2 2$ for some real or complex $μ$ and also the functional $a 2 a 4 − a 3 2$ is equivalent to $H 2 ( 2 )$ [30]. Babalola [32] studied the Hankel determinant $H 3 ( 1 ) .$
The symmetric Toeplitz determinant $T j ( n )$ is defined as follows:
$T j ( n ) = a n a n + 1 … a n + j − 1 a n + 1 … ⋮ ⋮ ⋮ a n + j − 1 … a n ,$
so that
$T 2 ( 2 ) = a 2 a 3 a 3 a 2 , T 2 ( 3 ) = a 3 a 4 a 4 a 3 , T 3 ( 2 ) = a 2 a 3 a 4 a 3 a 2 a 3 a 4 a 3 a 2 ,$
and so on. The problem of finding the best possible bounds for $a n + 1 − a n$ has a long history (see [33]). It is known from [33] that
$a n + 1 − a n < c ,$
for a constant c.
Lemma 1.
(see [31]) If p is analytic in $E$ and of the form (2), then
$2 c 2 = c 1 2 + x ( 4 − c 1 2 )$
and
$4 c 3 = c 1 3 + 2 ( 4 − c 1 2 ) c 1 x − ( 4 − c 1 2 ) c 1 x 2 + 2 ( 4 − c 1 2 ) ( 1 − x 2 ) z$
and, for some $x ,$$z ∈ C ,$ with $z ≦ 1 ,$ and $x ≦ 1 .$
Lemma 2.
(see also [34]) If p is analytic in $E$ and of the form (2), and if $μ ∈ C$$( 1 ≦ k ≦ n − 1 ) ,$ then
$c n − μ c k c n − k ≦ 2 max 1 , 2 μ − 1 .$
Lemma 3.
(see [35]; see also [33]) If the function p given by (2) is analytic in $E ,$ then
$c n ≦ 2 ( n ∈ N ) .$
The above inequality is sharp for the function f given by
$f ( z ) = 1 + z 1 − z .$
Lemma 4.
(see [35]) If p is analytic in $E$ and of the form (2), then
$c 2 − v c 1 2 ≦ − 4 v + 2 ( v < 0 ) 2 ( 0 ≦ v ≦ 1 ) 4 v − 2 ( v > 1 )$
The equality holds true for the function p given by
$p ( z ) = 1 + z 1 − z$
or by one of its rotations, when $v < 0$ or $v > 1$. In addition, the equality holds true for the function p given by
$p ( z ) = 1 + z 2 1 − z 2$
or by one of its rotations, when $0 < v < 1 .$ if $v = 0 ,$ the equality holds true if and only if
$p ( z ) = 1 2 + 1 2 λ 1 + z 1 − z + 1 2 − 1 2 λ 1 − z 1 + z , 0 ≦ λ ≦ 1$
or one of its rotations. If $v = 1 ,$ the equality holds true if and only if $p ( z )$ is the reciprocal of one of the functions such that the equality holds true in the case when $v = 0$. In addition, the above upper bound is sharp and it can be improved as follows when:
$c 2 − v c 1 2 + v c 1 2 ≦ 2 ( 0 < v ≦ 1 2 )$
and
$c 2 − v c 1 2 + ( 1 − v ) c 1 2 ≦ 2 ( 1 2 < v ≦ 1 ) .$

## 2. Main Results

Theorem 1.
Let the function f given by $( 1 )$ belong to the class k-$S s * ( q )$. Then
$a 2 ≦ 1 2 P 1 ,$
$a 3 ≦ 1 2 q P 1 + P 2 − P 1 + ( q − 1 ) P 1 2 2$
and
$a 4 ≦ 1 2 ( 1 + q 2 ) P 1 + 2 P 2 − P 1 + 2 q 2 − 2 q + 1 2 q P 1 2 + P 3 + P 1 − 2 P 2 − 2 q 2 − 2 q + 1 2 q P 1 2 + P 1 P 2 + q − 1 q 2 − q + 1 4 q P 1 3 .$
Proof.
For $f ∈ k$-$S s * ( q ) ,$ we have
$2 z D q f ( z ) f ( z ) − f ( − z ) = h ( z ) ≺ H k ( z ) ,$
where
$H k ( z ) = 2 p k ( z ) 1 + q + 1 − q p k ( z ) − 1$
and $p k ( z )$ is given by (5).
The function $p ( z )$ with $p ( 0 ) = 1$ is given as follows:
$p ( z ) = 1 + H k − 1 ( h ( z ) ) 1 − H k − 1 ( h ( z ) ) = 1 + c 1 z + c 2 z 2 + ⋯ .$
After some computation involving (16), we have
$h ( z ) = H k p ( z ) + 1 p ( z ) − 1 .$
Therefore, we find that
$H k p ( z ) + 1 p ( z ) − 1 = 1 + q + 1 2 P 1 c 1 2 z + P 1 c 2 2 + 1 4 P 2 − P 1 + ( q − 1 ) P 1 2 2 c 1 2 z 2 + P 1 c 3 2 + P 2 2 − P 1 2 + ( q − 1 ) P 1 2 4 c 1 c 2 + P 1 8 − P 2 4 − ( q − 1 ) P 1 2 8 + P 3 8 − 1 8 ( q − 1 ) P 1 P 2 − 1 32 ( q − 1 ) 2 P 1 3 c 1 3 z 3 .$
We also have
$2 z D q f ( z ) f ( z ) − f ( − z ) = 1 + 2 q a 2 z + q 2 q a 3 z 2 + 4 q a 4 − 2 q a 2 a 3 z 3 + ⋯ .$
Comparing the corresponding coefficients in (17) and (18) along with Lemma 3, we obtain the required result. □
Theorem 2.
Let the analytic function $f ∈ A$ be in the class k-$S s * ( q )$. Then
$T 3 ( 2 ) ≦ P 1 2 + 1 2 1 + q 2 Ω 1 + Ω 2 P 1 2 4 + 16 Ω 3 + P 1 2 2 q 2 + 2 P 1 2 Ω 5 2 − Ω 4 Ω 5 P 1 2 ,$
where
$Ω 1 = P 1 + 2 P 2 − 2 P 1 + Ω 6 P 1 2 ,$
and
$Ω 2 = P 3 + P 1 − 2 P 2 − Ω 6 P 1 2 + Ω 7 P 1 P 2 + q − 1 Ω 8 P 1 3 .$
Furthermore, we have
$Ω 3 = 2 P 1 Ω 5 P 3 8 + P 1 8 − P 2 4 − 1 8 Ω 6 P 1 2 + 1 8 Ω 7 P 1 P 2 + 1 8 q − 1 Ω 8 P 1 3 − 1 2 q 2 P 2 4 − P 1 4 + q − 1 P 1 2 8 2 ,$
$Ω 4 = P 1 2 q 2 P 2 4 − P 1 4 + q − 1 P 1 2 8 − 2 P 1 Ω 5 P 2 2 − P 1 2 − 1 4 Ω 6 P 1 2 ,$
$Ω 5 = 1 16 1 + q 2 ,$
$Ω 6 = 2 q 2 − 2 q + 1 2 q ,$
$Ω 7 = 1 − 2 q 2 + 2 q 2 q$
and
$Ω 8 = q 2 − q + 1 4 q .$
Here $P 1$ and $P 2$ are given in $( 5 )$.
Proof.
By comparison of coefficients in (17) and (18), we can obtain
$a 2 = 1 4 P 1 c 1$
$a 3 = 1 2 q 1 2 P 1 c 2 + P 2 4 − P 1 4 + ( q − 1 ) P 1 2 8 c 1 2 a 4 = 1 2 ( 1 + q 2 ) 1 2 P 1 c 3 + P 2 − P 1 2 + 2 q 2 − 2 q + 1 8 q P 1 2 c 1 c 2 + 1 8 P 3 + P 1 − 2 P 2 − 1 2 q 2 q 2 − 2 q + 1 P 1 2$
$+ 1 − 2 q 2 + 2 q 16 q P 1 P 2 + q − 1 q 2 − q + 1 32 q P 1 3 c 1 3 .$
A detailed calculation for $T 3 ( 2 )$ yields
$T 3 ( 2 ) = a 2 − a 4 a 2 2 − 2 a 3 2 + a 2 a 4 .$
Now, if $f ∈ k$-$S s * ( q ) ,$ then we have
$a 2 − a 4 ≦ a 2 + a 4 ,$
$a 2 − a 4 ≦ 1 2 P 1 + 1 2 1 + q 2 Ω 1 + Ω 2 .$
We need to maximize $a 2 2 − 2 a 3 2 + a 2 a 4$ for $f ∈ k$-$S s * ( q )$. Thus, by writing $a 2 , a 3 , a 4$ in terms of $c 1 , c 2 , c 3 ,$ with the help of (19) and (21), we get
$a 2 2 − 2 a 3 2 + a 2 a 4 ≦ P 1 2 c 1 2 4 + Ω 3 c 1 4 − Ω 4 c 1 2 c 2 − P 1 2 c 2 2 8 q 2 + Ω 5 P 1 2 c 1 c 3 .$
Finally, applying the triangle inequality, Lemma 2 and Lemma 3 along with (22) and (23), we obtained the required result. □
Theorem 3.
If an analytic function $f ∈ A$ is in the class k-$S s * ( q ) ,$ then
$a 2 a 4 − a 3 2 ≦ p 1 2 4 q 2 .$
Proof.
Making use of (19), (20) and (21), we have
$a 2 a 4 − a 3 2 = λ 1 c 1 c 3 + λ 2 c 1 2 c 2 − λ 3 c 2 2 + λ 4 c 1 4 ,$
where
$λ 1 = p 1 2 16 1 + q 2 ,$
$λ 2 = 1 16 q 2 1 + q 2 p 1 2 − 1 16 q 2 1 + q 2 p 1 p 2 + 2 − q 64 q 2 1 + q 2 p 1 3 ,$
$λ 3 = p 1 2 16 q 2$
and
$λ 4 = 1 16 1 + q 2 p 1 p 3 − 1 64 q 2 1 + q 2 p 1 2 + 1 32 q 2 1 + q 2 p 1 p 2 + q − 2 128 q 2 1 + q 2 p 1 3 + 4 q 2 − 4 q 3 − q + 2 128 q 2 1 + q 2 p 1 2 p 2 + q − 1 256 q 2 1 + q 2 p 1 4 − 1 64 q 2 p 2 2 .$
By using Lemma 1, we take
$Υ = 4 − c 1 2 and Z = 1 − x 2 z .$
Without loss of generality, we assume that $c = c 1$$0 ≦ c ≦ 2 ,$ so that
$a 2 a 4 − a 3 2 = 1 4 λ 1 + 2 λ 2 − λ 3 + 4 λ 4 c 4 + 1 2 λ 1 + λ 2 − λ 3 Υ c 2 x − λ 1 4 Υ c 2 x 2 − λ 3 4 Υ 2 x 2 + λ 1 2 c Υ Z .$
Taking the moduli on both sides of (25) and using the triangle inequality, we find that
$a 2 a 4 − a 3 2 ≦ 1 4 λ 1 + 2 λ 2 − λ 3 + 4 λ 4 c 4 + 1 2 λ 1 + λ 2 − λ 3 Υ c 2 x + λ 1 4 Υ c 2 x 2 + λ 3 4 Υ 2 x 2 + λ 1 2 1 − x 2 c Υ .$
This can be written as follows:
$a 2 a 4 − a 3 2 ≦ A λ c 4 + B λ x Υ c 2 + λ 1 4 x 2 Υ c 2 + λ 3 4 x 2 Υ 2 + λ 1 2 1 − x 2 Υ c = G ( x ) ,$
where
$A λ = 1 4 λ 1 + 2 λ 2 − λ 3 + 4 λ 4 B λ = 1 2 λ 1 + λ 2 − λ 3 .$
Now, trivially, we have
$G ′ ( c , x ) > 0$
on the closed interval $[ 0 , 1 ]$, which shows that $G ( c , x )$ is an increasing function in the interval $[ 0 , 1 ]$. Therefore, the maximum value occurs at $x = 1$ and we have
$max { G ( c , 1 ) } = G ( c ) ,$
$G ( c , 1 ) = A λ c 4 + B λ Υ c 2 + λ 1 4 Υ c 2 + λ 3 4 Υ 2$
and
$G ( c ) = A λ c 4 + B λ Υ c 2 + λ 1 4 Υ c 2 + λ 3 4 Υ 2 .$
Hence, by putting $Υ = 4 − c 1 2$ and after some simplification, we have
$G ( c ) = A λ − B λ − λ 1 4 + λ 3 4 c 4 + 4 B λ + λ 1 4 − λ 3 2 c 2 + 4 λ 3 .$
We consider $G ′ ( c ) = 0$, for the optimum value of $G ( c )$, which implies that $c = 0$. Thus, $G ( c )$ has a maximum value at $c = 0$. Hence, the maximum value of $G ( c )$ is given by
$max { G ( c ) } = 4 λ 3 ,$
which occurs at $c = 0$ or
$c 2 = 4 B λ + λ 1 4 − λ 3 2 A λ − B λ − λ 1 4 + λ 3 4 .$
Hence, by putting
$λ 3 = p 1 2 16 q 2$
in (27) and after some simplification, we obtain the desired result. □
For $q → 1 − ,$$k = 0 ,$ and $p 1 = 2$ in Theorem 3, we have the following known result for the class $S s * .$
Corollary 1.
(see [36]) If an analytic function $f ∈ A$ that belongs to the class $S * ,$ then
$a 2 a 4 − a 3 2 ≦ 1 .$

#### 2.1. The Fekete-Szegö Problem

Theorem 4.
Let the function $f ∈ A$ given by $( 1 )$ belong to the class k-$S s * ( q )$. Then
$a 3 − μ a 2 2 ≦ 1 4 q 2 P 2 + ( q ( 1 − μ − 1 ) P 1 2 ) ( μ ≦ δ 1 ) P 1 2 q ( δ 1 ≦ μ ≦ δ 2 ) − 1 4 q 2 P 2 + ( q ( 1 − μ − 1 ) P 1 2 ) ( μ ≦ δ 1 ) ,$
where
$δ 1 = 2 P 2 + P 1 q − 1 P 1 − 2 q P 1 2 , δ 2 = 2 P 2 + P 1 q − 1 P 1 + 2 q P 1 2 .$
Proof.
From (19) and (20), we have
$a 3 − μ a 2 2 = P 1 4 q c 2 − v c 1 2 ,$
where
$v = 1 2 1 − P 2 P 1 − q − 1 P 1 2 + μ q P 1 2 .$
By applying the triangle inequality and Lemma 4, we obtain Theorem 4. □
If we set $k = 0$ and $q → 1 −$ in Theorem 4, we thus obtain the following known result.
Corollary 2.
(see [37]) If an analytic function $f ∈$$S s * ( ϕ ) ,$ then
$a 3 − μ a 2 2 ≦ 1 2 ( P 2 − μ 2 P 1 2 ) ( μ ≦ δ 1 ) P 1 2 ( δ 1 ≦ μ ≦ δ 2 ) − 1 2 ( P 2 − μ 2 P 1 2 ) ( μ ≦ δ 1 ) ,$
where
$δ 1 = 2 P 2 − P 1 P 1 2 , δ 2 = 2 P 2 + P 1 P 1 2 .$
Let $δ 1 ≦ μ ≦ δ 2$. Then, in view of Lemma 4, Theorem 4 can be improved as follows.
Theorem 5.
If the function f given by (1) belongs to the class $S s * ( q )$ and if
$δ 1 ≦ μ ≦ δ 3 = 2 P 2 q P 1 ,$
then
$a 3 − μ a 2 2 + 1 q P 1 2 2 P 2 − P 1 − q − 1 P 1 2 + μ q P 1 2 a 2 2 ≦ P 1 2 q .$
Furthermore, if $δ 3 ≦ μ ≦ δ 2 ,$ then
$a 3 − μ a 2 2 + 1 q P 1 2 2 P 2 + P 1 − q − 1 P 1 2 − μ q P 1 2 a 2 2 ≦ P 1 2 q .$
If we set $k = 0$ and $q → 1 −$, we obtain the following known result.
Corollary 3.
(see [37]) If an analytic function $f ∈ S s * ( ϕ )$ and if
$δ 1 ≦ μ ≦ δ 3 = 2 P 2 P 1 2 ,$
then
$a 3 − μ a 2 2 + 1 q P 1 2 2 P 2 − P 1 + μ P 1 2 a 2 2 ≦ P 1 2 .$
Moreover, if
$δ 3 ≦ μ ≦ δ 2 ,$
then
$a 3 − μ a 2 2 + 1 P 1 2 2 P 2 + P 1 − μ P 1 2 a 2 2 ≦ P 1 2 .$

#### 2.2. Applications of the Main Results

In this section, firstly we recall that the Bernardi integral operator $F β$ given in [38] as follows:
$F β ( f ( z ) ) = 1 + β z β ∫ 0 z t β − 1 f ( t ) d t f ∈ A , β > − 1 .$
The q-integral of the function f on $[ 0 , z ]$ is defined as follows (see, for example [39]):
$∫ 0 z f ( t ) d q t = 1 − q z ∑ k = 0 ∞ q k f ( q k z ) ,$
and q-integral of the function $z n$ is given by
$∫ 0 z z n d q t = z n + 1 n + 1 q ,$
where $n ≠ − 1$ and for $q → 1 − ,$ Equation (29) becomes
$∫ 0 z h 1 ( t ) d t = z n + 1 n + 1 .$
Noor [39] introduced the q-Bernardi integral operator $B q ( z )$ as follows:
$B q ( z ) = F β ( f ( z ) ) = 1 + β q z β ∫ 0 z t β − 1 f ( t ) d q t ( β > − 1 ) .$
Let $f ∈ A$. Then, by using Equations (29) and (8), we obtain the following power series for the function $B q ( z )$ in the open unit disk $E$ as follows:
$B q ( z ) = z + ∑ n = 2 ∞ 1 + β q n + β q a n z n .$
Clearly, $B q ( z )$ is analytic in the open unit disk $E$.
Let
$B n = 1 + β q n + β q ( n ≧ 1 ) .$
Applying Theorem 1 on Equation (31), we obtain the following result.
Theorem 6.
If the function $B q ( z )$ given by $( 31 )$ belongs to the class k-$S s * ( q ) ,$ where $k ∈ [ 0 , 1 ] ,$ then
$a 2 ≦ 1 2 B 2 P 1 ,$
$a 3 ≦ 1 + q 2 3 q B 3 − 1 P 1 + P 2 − P 1 + ( q − 1 ) P 1 2 2$
and
$a 4 ≦ 1 + q 2 4 q B 4 P 1 + 2 P 2 − P 1 + 2 3 q B 3 − 1 q − 1 + 1 + q 2 3 q B 3 − 1 P 1 2 + P 3 + P 1 − 2 P 2 − 2 3 q B 3 − 1 q − 1 + 1 + q 2 3 q B 3 − 1 P 1 2 + P 1 P 2 + q − 1 4 q − 1 3 q B 3 − 1 + ( 1 + q ) 3 q B 3 − 1 P 1 3 ,$
where $B 2 ,$$B 3$ and $B 4$ are given in $( 32 )$.
Applying Theorem 2 to Equation (31), we obtain the following result.
Theorem 7.
If the function $B q ( z )$ given by $( 31 )$ belongs to the class k-$S s * ( q ) ,$ then
$T 3 ( 2 ) ≦ P 1 2 B 2 + 1 + q 2 4 q B 4 Ω 10 + Ω 11 P 1 2 4 B 2 2 + 16 Ω 12 + 1 + q 2 P 1 2 2 3 q B 3 − 1 2 + 2 P 1 2 Ω 14 2 − Ω 13 P 1 2 Ω 14 ,$
where
$Ω 10 = P 1 + 2 P 2 − 2 P 1 + Ω 15 P 1 2 , Ω 11 = P 3 + P 1 − 2 P 2 + Ω 15 P 1 2 + Ω 16 P 1 P 2 + q − 1 Ω 17 P 1 3 , Ω 12 = 2 P 1 Ω 14 P 3 8 + P 1 8 − P 2 4 − 1 8 Ω 15 P 1 2 + 1 8 Ω 16 P 1 P 2 + 1 8 q − 1 Ω 17 P 1 3 − 1 + q 2 2 3 q B 3 − 1 2 P 2 4 − P 1 4 + q − 1 P 1 2 8 2 , Ω 13 = 1 + q 2 P 1 2 3 q B 3 − 1 2 P 2 4 − P 1 4 + q − 1 P 1 2 8 + 2 P 1 Ω 14 P 2 2 − P 1 2 + 1 4 Ω 15 P 1 2 , Ω 14 = 1 + q 16 B 2 B 4 4 q , Ω 15 = 2 3 q B 3 − 1 q − 1 + 1 + q 2 3 q B 3 − 1 ,$
$Ω 16 = 1 + q − 2 3 q B 3 − 1 q − 1 2 3 q B 3 − 1 , Ω 17 = 3 q B 3 − 1 q − 1 + 1 + q 4 3 q B 3 − 1 ,$
where $P 1$ and $P 2$ are given in $( 31 )$.
Applying Theorem 3 to Equation (31), we obtain the following result.
Theorem 8.
If the function $B q ( z )$ given by $( 31 )$ belongs to the class k-$S s * ( q ) ,$ then
$a 2 a 4 − a 3 2 ≦ 1 + q 2 p 1 2 4 3 q B 3 − 1 2 .$
For $q → 1 − ,$$k = 0 ,$$β = 0$ and $p 1 = 2$ in Theorem (8), we have the following known result for the class $S s * .$
Corollary 4.
(see [36]) Let $f ∈ S s *$ be of the form $( 1 )$. Then
$a 2 a 4 − a 3 2 ≦ 1 .$
Theorem 9.
If the function $B q ( z )$ given by $( 31 )$ belongs to the class k-$S s * ( q ) ,$ then
$a 3 − μ a 2 2 ≦ 1 + q 4 3 q B 3 − 1 2 P 2 + q − 1 − μ 3 q B 3 − 1 1 + q B 2 2 P 1 2 ( μ ≦ δ 1 ) 1 + q P 1 4 3 q B 3 − 1 , ( δ 1 ≦ μ ≦ δ 2 ) − 1 + q P 1 4 3 q B 3 − 1 2 P 2 + q − 1 − μ 3 q B 3 − 1 1 + q B 2 2 P 1 2 ( μ ≦ δ 1 ) ,$
where
$δ 1 = 1 + q B 2 2 3 q B 3 − 1 2 P 2 + P 1 q − 1 P 1 − 2 P 1 2$
and
$δ 2 = 1 + q B 2 2 3 q B 3 − 1 2 P 2 + P 1 q − 1 P 1 + 2 P 1 2 .$
If we set $k = 0 ,$$β = 0$ and $q → 1 −$ in Theorem 9, we obtain the following known result.
Corollary 5.
(see [37]) If an analytic function $f ∈ A$ belongs to the class $S s * ( ϕ ) ,$ then
$a 3 − μ a 2 2 ≦ 1 2 ( P 2 − μ 2 P 1 2 ) ( μ ≦ δ 1 ) P 1 2 ( δ 1 ≦ μ ≦ δ 2 ) − 1 2 P 2 − μ 2 P 1 2 ( μ ≧ δ 2 ) ,$
where
$δ 1 = 2 P 2 − P 1 P 1 2$
and
$δ 2 = 2 P 2 + P 1 P 1 2 .$

## 3. Conclusions

We have made use of the general conic domain $Ω k , q$ and the quantum (or q-) calculus to introduce and investigate several new subclasses of q-starlike functions with respect to symmetrical points in open unit disk $E$. We have studied some interesting results such as the Hankel determinant, the Toeplitz matrices, and the Fekete-Szegö inequalities. We have also discussed some applications of our main results by using a q-Bernardi integral operator.
For further investigation, we can easily follow a known relationship between the q-analysis and $( p , q )$-analysis (see [25] (p. 340, Equations (9.1), (9.2) and (9.3))) and the results for the q-analogues, which we have included in this paper for $0 < q < 1 ,$ can then be easily transformed into the related results for the $( p , q )$-analogues with $0 < q < p ≦ 1$ by adding a rather redundant (or superfluous) parameter p (see, for details [25] (p. 340)).

## Author Contributions

Conceptualization, H.M.S.; Formal analysis, M.D.; Investigation, S.K.; Methodology, N.K.; Validation, Q.Z.A. and H.M.S.; Visualization, S.H.; Writing—Review and Editing, H.M.S. All authors have read and agreed to the published version of the manuscript.

## Funding

This work is supported by Universiti Kebangsaan Malaysia, grant number FRGS/1/2019/STG06/UKM/01/1.

## Conflicts of Interest

The authors declare no conflict of interest.

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