1. Introduction

Let $\mathcal{A}_{p}$ denote the class of $p$-valent functions $f$ which are analytic in the regions $\mathbb{U}=\left \{ z\in %TCIMACRO{\U{2102} }% %BeginExpansion \mathbb{C} %EndExpansion :\left \vert z\right \vert < 1\right \}$ and normalized by

$f(z) = {z^p} + _{k = 1}^\infty {a_{p + k}}{z^{p + k}}.$

(1.1)

We note that $\mathcal{A}_{1}=\mathcal{A}.$ Also let $\mathcal{N}(\alpha)$ and $\mathcal{M}(\alpha)$ denote the usual classes of starlike and convex functions of reciprocal order $\alpha$, $\alpha >1$, and are defined by

$\mathcal{N}(\alpha ) = \left \{ f\left( z\right) \in \mathcal{A}:Re \frac{zf^{\prime }(z)}{f(z)}<\alpha, \text{ }\left( z\in \mathbb{U}\right) \right \},$

(1.2)

$\mathcal{M}(\alpha ) =\left \{ f\left( z\right) \in \mathcal{A}:1+Re% \frac{zf^{\prime \prime }(z)}{f^{\prime }(z)}<\alpha, \text{ }\left( z\in \mathbb{U}\right) \right \} .$

(1.3)

These classes were introduced by Uralegaddi et.al ^[21] in 1994 and then studied by the authors in ^[12]. After that Nunokawa and his coauthors ^[11] proved that for $f\in \mathcal{N}(\alpha),$ $0 < \alpha < % \frac{1}{2},$ if and only if the following inequality holds

$\left \vert \frac{zf^{\prime }\left( z\right) }{f\left( z\right) }-\frac{1}{% 2\alpha }\right \vert <\frac{1}{2\alpha }, \left( z\in \mathbb{U}% \right) .$

In 2002, Owa and Srivastava ^[13] generalized this idea for the classes of $p$-valent starlike and $p$-valent convex functions of reciprocal order $\alpha$ with $\alpha >p,$ and further investigated by Polatoglu et.al ^[14]. Recently in 2011, Uyanik et.al ^[22] extended this idea to the classes of $p$-valently spirallike and $p$-valently Robertson functions and discussed coefficient inequalities and sufficient conditions for the functions of these classes.

The convolution (Hadamard product) of functions $f, g\in \mathcal{A} _{p}$ is defined by

$(f\ast g)\left( z\right) =z^{p}+\sum\limits_{n=1}^{\infty }a_{n+p}b_{k+p}z^{n+p}, z\in \mathbb{U},$

where $f$ is given by (1.1) and

$g(z)=z^{p}+\sum\limits_{n=1}^{\infty }b_{n+p}z^{n+p}, z\in \mathbb{U}.$

The incomplete beta function $\phi _{p}$ defined by

$\phi _{p}(a, c;z)=z^{p}+\sum\limits_{n=1}^{\infty }\frac{(a)_{n}}{(c)_{n}}z^{n+p}, (a\in %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion , c\in %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion \diagdown \left( 0, -1, ...\right), z\in \mathbb{U}),$

where $(\alpha)_{n}$ is the pochhamer symbol defined in terms of Gamma function by

$(\alpha )_{n}=\frac{\Gamma (\alpha +n)}{\Gamma (\alpha )}=\left \{ \begin{array}{ll} \alpha (\alpha +1)...(\alpha +n-1), & \text{if }n\in \mathbb{N} \\ 1, & \text{if }n=0.% \end{array}% \right.$

With the help of incomplete beta function $\phi _{p}$ and concepts of convolution, Saitoh in ^[18,19] introduced the operator $\mathcal{L}_{p}: \mathcal{A}_{p}\rightarrow \mathcal{A}_{p}$ and is defined by

$\notag \\ \mathcal{L}_{p}(a, c)f(z) =\phi _{p}(a, c;z)\ast f(z), \notag \\ \notag \\ \;\;\;\;\;\;=z^{p}+\sum\limits_{n=1}^{\infty }\varphi _{n}(a)\text{ }a_{n+p}\text{ }z^{n+p} 1.4$

(1.4)

with $a>-p$ and

$\varphi _{n}(a)=\frac{\Gamma (a+n)\Gamma (c)}{\Gamma (a)\Gamma (c+n)}.$

(1.5)

This operator is an extension of the familiar Carlson-Shaffer operator, which has been used widely on the space of analytic and univalent functions in $\mathbb{U}$, see ^[3,20]. The following identity can be easily derived

$z\left( \mathcal{L}_{p}(a, c)f(z)\right) ^{\prime }=a\mathcal{L}_{p}\left( a+1, c\right) f(z)-(a-p)\mathcal{L}_{p}(a, c)f(z).$

(1.6)

Motivated from the above mentioned work, we now introduce a new subclass of multivalent functions of reciprocal order using the operator defined in (1.4).

An analytic multivalent function $f$ of the form (1.1) belongs to the class $\mathcal{SC}_{p}^{\lambda }(a, b, c, \beta)$, if and only if

$Re\left \{ \frac{2e^{i\lambda }}{b}\left( \frac{\mathcal{L}% _{p}(a+1, c)f(z)}{\mathcal{L}_{p}(a, c)f(z)}-1\right) \right \} <\left( \beta -1\right) \cos \lambda,$

where $b\in %TCIMACRO{\U{2102} }% %BeginExpansion \mathbb{C} %EndExpansion \diagdown \{0\},$ $\lambda$ is real with $\left \vert \lambda \right \vert < % \frac{\pi }{2},$ $\beta >1.$

It is noticed that, by giving specific values to $a, b, c, \beta$ and $\lambda$ in $\mathcal{SC}_{p}^{\lambda }(a, b, c, \beta),$ we obtain many well-known as well as new subclasses of analytic, univalent and multivalent functions, for example;

(ⅰ) For $\lambda =0,$ $a=1=c$ and $b=2$, we obtain $\mathcal{SC} _{p}^{0}(1, 2, 1, \beta)=\mathcal{M}_{p}(\beta)$ studied in ^[13] and further for $p=1,$ we have the class $\mathcal{M}(\beta)$ introduced and studied in ^[7,21].

(ⅱ) For $p=c=b=1,$ $\lambda =0$ and $a=2,$ we have $\mathcal{SC} _{1}^{0}(1, 2, 1, \beta)=\mathcal{N}(\beta)$ studied in ^[7,21].

(ⅲ) For $a=1=c$ and $b=2,$ we get the class $\mathcal{SC}% _{p}^{\lambda }(1, 2, 1, \beta)=\mathcal{S}_{p}(\lambda, \beta)$ and for $% a=2,$ $b=c=1,$ we obtain $\mathcal{SC}_{p}^{\lambda }(1, 2, 1, \beta)=% \mathcal{C}_{p}(\lambda, \beta)$, introduced and studied in ^[22].

(ⅳ) For $\lambda =0$, $a=2,$ $b=p=1,$ and $c=2-\alpha$, we obtain the class $\mathcal{SC}_{1}^{0}(2, 1, 2-\alpha, \beta)=\mathcal{P}_{\alpha}(\beta)$ ^[4].

The $q^{th}$ Hankel determinant $H_{q}(n),$ $q\geq 1,$ $n\geq 1,$ stated by Pommerenke ^[15] and furter investigated by Noonan and Thomas ^[10] as

$H_{q}(n)=\left \vert \begin{array}{llll} a_{n} & a_{n+1} & \cdots & a_{n+q+1} \\ a_{n+1} & a_{n+2} & \cdots & a_{n+q} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n+q-1} & a_{n+q} & \cdots & a_{n+2q-2} \end{array} \right \vert$

This Hankel determinant is useful and has also been studied by several authors for details see ^[1,2]. The growth rate of Hankel determinant $H_{q}(n)$ as $n\rightarrow \infty$ was investigated, respectively, when $f$ is a member of certain subclass of analytic functions, such as the class of $p$-valent functions ^[15,10], the class of starlike functions ^[15], the class of univalent functions ^[16], the class of close-to-convex functions ^[9], a new class $V_{k}$ ^[9]. It is well known that the Fekete-Szegö functional is $H_{2}(1)=\left \vert a_{3}-a_{2}^{2}\right \vert.$ Fekete and Szegö further generalized the estimate $\left \vert a_{3}-\mu a_{2}^{2}\right \vert$ where $\mu$ is real and $f$ $\in S,$ the class of univalent functions$.$ For our discussion in this paper we investigate the coeficient bound, the upper bounds of the Hankel determinant $H_{3}(1)$ for a subclass of multivalent functions.

We will need the following lemma's for our work.

Lemma 1.2. ^[17]. If $q$ is an analytic function with $Req(z)>0$ and

$q(z)=1 + _{n = 1}^\infty d_{n}z^{n}, \text{ }z\in \mathbb{U},$

(1.7)

then for} $n\geq 1,$

$\left \vert d_{n}\right \vert \leq 2.$

Lemma 1.3. ^[6]. If $q$ is of the form (1.7) with positive real part, then the following sharp estimate holds

$\left \vert d_{2-}\upsilon d_{1}^{2}\right \vert \leq 2\max \left \{ 1, \left \vert 2\upsilon -1\right \vert \right \}, \text{ for all }\upsilon \in \mathbb{C}.$

Lemma 1.4. ^[5]. If $q$ is of the form (1.7) with positive real part, then

$2d_{2} =d_{1}^{2}+x(4-d_{1}^{2}). \\ 4d_{3} =d_{1}^{3}+2(4-d_{1}^{2})d_{1}x-d_{1}(4-d_{1}^{2})x^{2}+2(4-d_{1}^{2})% \left( 1-\left \vert x\right \vert ^{2}\right) z.$

for some $x,$ $z$ with $\left \vert x\right \vert \leq 1$ and $\left \vert z\right \vert \leq 1.$

---

2. Main Results

Theorem 2.1. If $f\in \mathcal{SC}_{p}^{\lambda }(a, b, c, \beta),$ then

$\left \vert a_{p+1}\right \vert \leq \frac{\left \vert b\right \vert \left \vert \eta \right \vert c}{p},$

and

$\left \vert a_{n+p-1}\right \vert \leq \frac{a\left \vert b\right \vert \left \vert \eta \right \vert }{(n+p-2)\varphi _{n-1}(a)}\prod_{j=1}^{n-2}% \left( 1+\frac{a\left \vert b\right \vert \left \vert \eta \right \vert }{% (p+j)}\right), n\geq 3,$

(2.1)

where $\varphi _{n-1}(\delta)$ is given by (1.5) and

$\eta =(1-\beta )\cos \lambda +i\sin \lambda .$

(2.2)

Proof. Let $f\in \mathcal{SC}_{p}^{\lambda }(a, b, c, \beta).$ Then we have

$Re\left \{ \frac{2e^{i\lambda }}{b}\left( \frac{\mathcal{L}% _{p}(a+1, c)f(z)}{\mathcal{L}_{p}(a, c)f(z)}-1\right) \right \} <\left( \beta -1\right) \cos \lambda, z\in \mathbb{U}.$

Now let us define a function $q$ by

$e^{i\lambda }\left( 1-\frac{2}{b}+\frac{2}{b}\frac{\mathcal{L}_{p}(a+1, c)f(z)% }{\mathcal{L}_{p}(a, c)f(z)}\right) =\left( (1-\beta )q(z)+\beta \right) \cos \lambda +i\sin \lambda .$

where $q$ is analytic in $\mathbb{U}$ with $q(0)=1$ and $Req(z)>0$, $% z\in \mathbb{U}$. Then (2.3) can be written as

$1-\frac{2}{b}+\frac{2}{b}\frac{\mathcal{L}_{p}(a+1, c)f(z)}{\mathcal{L}% _{p}(a, c)f(z)}=1+\frac{(1-\beta )\cos \lambda +i\sin \lambda }{e^{i\lambda }}% \sum\limits_{n=1}^{\infty }c_{n}z^{n},$

or equivalently

$2e^{i\lambda }\left( \mathcal{L}_{p}(a+1, c)f(z)-\mathcal{L}% _{p}(a, c)f(z)\right) =b\eta \mathcal{L}_{p}(a, c)f(z)\sum\limits_{n=1}^{\infty }c_{n}z^{n},$

(2.4)

where $\eta$ is given by (2.2). From (2.4) and (1.6) we have

$2e^{i\lambda }\left[z\left( \mathcal{L}_{p}(a, c)f(z)\right) ^{^{\prime }}-p% \mathcal{L}_{p}(a, c)f(z)\right] =ab\eta \mathcal{L}_{p}(a, c)f(z)\sum\limits_{n=1}^{% \infty }c_{n}z^{n},$

that is,

$2e^{i\lambda }\left[\sum\limits_{k=1}^{\infty }(k+p-1)\varphi _{k}(a)a_{k+p}z^{k+p}% \right] =ab\eta \left[z^{p}+\sum\limits_{k=1}^{\infty }\varphi _{k}(a)a_{k+p}z^{k+p}\right] \left( \sum\limits_{k=1}^{\infty }d_{n}z^{n}\right) .$

Comparing the coefficients of $z^{n+p-1}$ on both sides, we obtain

$2e^{i\lambda }(n+p-2)\varphi _{n-1}(a)a_{n+p-1}=ab\eta \left \{ d_{1}a_{n-2}\varphi _{n+p-2}(a)+...+d_{n-1}\right \} .$

(2.5)

Taking absolute on both sides and then applying Lemma 1.1, we have

$\left \vert a_{n+p-1}\right \vert \leq \frac{a\left \vert b\right \vert \left \vert \eta \right \vert }{(n+p-2)\varphi _{n-1}(a)}\left \{ 1+\varphi _{1}(a)\left \vert a_{p+1}\right \vert +...+\varphi _{n-2}(a)\left \vert a_{n+p-2}\right \vert \right \}$

(2.6)

We now apply mathematical induction on (2.6). So for $n=2$

$\left \vert a_{p+1}\right \vert \leq \frac{\left \vert b\right \vert \left \vert \eta \right \vert c}{p}.$

which shows that the result is true for $n=2$. For $n=3$

$\left \vert a_{p+2}\right \vert \leq \frac{a\left \vert b\right \vert \left \vert \eta \right \vert }{(p+1)\varphi _{2}(a)}\left \{ 1+\varphi _{1}(a)\left \vert a_{p+1}\right \vert \right \}$

(2.7)

and using the bound of $\left \vert a_{p+1}\right \vert$ in (2.7), we have

$\left \vert a_{p+2}\right \vert \leq \frac{a\left \vert b\right \vert \left \vert \eta \right \vert }{(p+1)\varphi _{2}(a)}\left \{ 1+\frac{a\left \vert b\right \vert \left \vert \eta \right \vert }{p}\right \} .$

Therefore (2.1) holds for $n=3$.

Assume that (2.1) is true for $n=k,$ that is,

$\left \vert a_{k+p-1}\right \vert \leq \frac{a\left \vert b\right \vert \left \vert \eta \right \vert }{(k+p-2)\varphi _{k-1}(a)}\prod_{j=1}^{k-2}% \left( 1+\frac{a\left \vert b\right \vert \left \vert \eta \right \vert }{% (p+j)}\right) .$

Consider

$\left \vert a_{k+p}\right \vert \leq \frac{a\left \vert b\right \vert \left \vert \eta \right \vert }{(k+p-1)\varphi _{k}(a)}\left \{ \left( 1+% \frac{a\left \vert b\right \vert \left \vert \eta \right \vert }{p}\right) +% \frac{a\left \vert b\right \vert \left \vert \eta \right \vert }{(p+1)}% \left( 1+\frac{a\left \vert b\right \vert \left \vert \eta \right \vert }{p}% \right) \right. \\ \;\;\;\;\;\;\;\;\;\;\;\;\left. +\ldots +\frac{a\left \vert b\right \vert \left \vert \eta \right \vert }{% (p+k-1)}\prod_{j=1}^{k-2}\left( 1+\frac{a\left \vert b\right \vert \left \vert \eta \right \vert }{(p+j)}\right) \right \} \\ \;\;\;\;\;\;=\frac{a\left \vert b\right \vert \left \vert \eta \right \vert }{% (k+p-1)\varphi _{k}(a)}\prod_{j=1}^{k-1}\left( 1+\frac{a\left \vert b\right \vert \left \vert \eta \right \vert }{(p+j)}\right) .$

Therefore, the result is true for $n=k+1$ and hence by using mathematical induction, (2.1) holds true for all $n\geq 3.$

The following corollaries which were proved by owa and Nishawski ^[12] comes as a special case from Theorem 2.1 by varying the parameter $a, b, c, p$ and $\lambda.$

Corollary 2.2. If $f\in \mathcal{M}(\beta),$ then

$\left| {{a_n}} \right| \le _{l = 2}^n\frac{{(l + 2\beta - 4)}}{{\left( {n - 1} \right)!}}, {\rm{for}}\;{\rm{all}}\;n \ge 2.$

Corollary 2.3. If $f\in \mathcal{N}(\beta),$ then

$\left| {{a_n}} \right| \le _{l = 2}^n\frac{{(l + 2\beta - 4)}}{{n!}}, \;{\rm{for}}\;{\rm{all}}\;n \ge {\rm{2}}.$

Theorem 2.4. Let $f\in \mathcal{A}_{p}$ and satisfies

$\sum\limits_{n=1}^{\infty }\left \vert \left( 1+\frac{2n}{ab}\right) e^{i\lambda }-\beta \cos \lambda \right \vert \varphi _{n}(a)\left \vert a_{n+p}\right \vert <\left \vert \beta \cos \lambda -e^{i\lambda }\right \vert .$

(2.8)

Then $f\in \mathcal{SC}_{p}^{\lambda }(a, b, c, \beta).$

Proof. To prove that $f$ belongs to $\mathcal{SC}_{p}^{\lambda }(a, b, c, \beta)$ we need to prove that

$\left \vert \frac{e^{i\lambda }\left( 1-\frac{2}{b}+\frac{2}{b}\frac{% \mathcal{L}_{p}(a+1, c)f(z)}{\mathcal{L}_{p}(a, c)f(z)}\right) -1}{e^{i\lambda }\left( 1-\frac{2}{b}+\frac{2}{b}\frac{\mathcal{L}_{p}(a+1, c)f(z)}{\mathcal{L% }_{p}(a, c)f(z)}\right) -\left( 2\beta \cos \lambda -1\right) }\right \vert <1.$

(2.9)

For this consider the left hand side of (2.9), we have

$LHS =\left \vert \frac{e^{i\lambda }\left( 1-\frac{2}{b}+\frac{2}{b}\frac{% \mathcal{L}_{p}(a+1, c)f(z)}{\mathcal{L}_{p}(a, c)f(z)}\right) -1}{e^{i\lambda }\left( 1-\frac{2}{b}+\frac{2}{b}\frac{\mathcal{L}_{p}(a+1, c)f(z)}{\mathcal{L% }_{p}(a, c)f(z)}\right) -\left( 2\beta \cos \lambda -1\right) }\right \vert \\ \;\;\;\;\;\;=\left \vert \frac{\left( e^{i\lambda }-1\right) bz^{p}+\sum\limits_{n=1}^{\infty }\left( \left( b+\frac{2n}{a}\right) e^{i\lambda }-1\right) \varphi _{n}(a)a_{n+p}z^{n+p}}{\left( be^{i\lambda }-2b\beta \cos \lambda +b\right) +\sum\limits_{n=1}^{\infty }\left( \left( b+\frac{2n}{a}\right) e^{i\lambda }-2b\beta \cos \lambda +b\right) \varphi _{n}(a)a_{n+p}z^{n+p}}\right \vert \\ \;\;\;\;\;\;\leq \frac{\left \vert \left( e^{i\lambda }-1\right) \right \vert +\sum\limits_{n=1}^{\infty }\left \vert \left( 1+\frac{2n}{ab}\right) e^{i\lambda }-1\right \vert \varphi _{n}(a)\left \vert a_{n+p}\right \vert }{\left \vert 2\beta \cos \lambda -e^{i\lambda }-1\right \vert -\sum\limits_{n=1}^{\infty }\left \vert \left( 1+\frac{2n}{ab}\right) e^{i\lambda }-2\beta \cos \lambda +1\right \vert \varphi _{n}(a)\left \vert a_{n+p}\right \vert }.$

The last expression is bounded above by $1$ if

$\left \vert e^{i\lambda }-1\right \vert +\sum\limits_{n=1}^{\infty }\left \vert \left( 1+\frac{2n}{ab}\right) e^{i\lambda }-1\right \vert \varphi _{n}(a)\left \vert a_{n+p}\right \vert \leq \left \vert 2\beta \cos \lambda -e^{i\lambda }-1\right \vert\\ \;\;\;\;\;\;-\sum\limits_{n=1}^{\infty }\left \vert \left( 1+\frac{2n}{ab}\right) e^{i\lambda }-2\beta \cos \lambda +1\right \vert \varphi _{n}(a)\left \vert a_{n+p}\right \vert$

which is equivalent to the condition (2.8) and so $f\in \mathcal{SC}_{p}^{\lambda }(a, b, c, \beta).$

If we set $p=1, \lambda =0$, $a=1,$ $b=2$ and $c=1,$ in above Theorem we have the following result by ^[12].

Corollary 2.5. If $f\in A$ satisfies

$\sum\limits_{n=2}^{\infty }\left \{ \left( n-1\right) +\left \vert n-2\beta +1\right \vert \right \} \left \vert a_{n}\right \vert \leq 2(\beta -1)$

for some $\beta (\beta >1)$, then $f\in \mathcal{M}(\beta)$.

Corollary 2.6. ^[4]. A function $f\in \mathcal{P}_{\alpha }(\beta)$ if and only if

$\sum\limits_{n=2}^{\infty }\frac{\Gamma (n+1)\Gamma (2-\alpha )}{\Gamma (n+1-\alpha )}(n-\beta )\left \vert a_{n}\right \vert \leq (\beta -1).$

The result is sharp.

Theorem 2.7. Let $f\in \mathcal{SC}_{p}^{0}(a, b, c, \beta)$ and of the form (1.1). Then

$\left \vert a_{p+2}-\mu a_{p+1}^{2}\right \vert \leq \frac{\left \vert b\right \vert c(c+1)(1-\beta )}{(a+1)(p+1)}\max \left \{ 1, \left \vert 2\upsilon -1\right \vert \right \},$

(2.10)

where

$\upsilon =\frac{\mu bc(a+1)(p+1)(1-\beta )}{2p^{2}(c+1)}-\frac{ab(1-\beta )}{% 2p}.$

(2.11)

Proof. Let $f\in \mathcal{SC}_{p}^{0}(a, b, c, \beta).$ Then from (2.5) with $\lambda =0$, we have

$a_{p+1} =\frac{bc(1-\beta )}{2p}d_{1} \\ a_{p+2} =\frac{bc(c+1)(1-\beta )}{2(a+1)(p+1)}\left \{ d_{2}+\frac{ ab(1-\beta )}{2p}d_{1}^{2}\right \} .$

For any complex number $\mu,$ we have

$a_{p+2}-\mu a_{p+1}^{2} =\frac{bc(c+1)(1-\beta )}{2(a+1)(p+1)}\left[d_{2}-% \frac{b(1-\beta )}{2p}\left \{ \frac{\mu c(a+1)(p+1)}{p(c+1)}-a\right \} d_{1}^{2}\right] \\ \;\;\;\;\;\;=\frac{bc(c+1)(1-\beta )}{2(a+1)(p+1)}\left[d_{2}-\upsilon d_{1}^{2}% \right],$

where $\upsilon$ is given by (2.11)

Taking modulus on both sides and applying Lemma 1.2, we have

$\left \vert a_{p+2}-\mu a_{p+1}^{2}\right \vert =\left \vert \frac{ bc(c+1)(1-\beta )}{2(a+1)(p+1)}\right \vert \left \vert d_{2}-\upsilon d_{1}^{2}\right \vert \\ \;\;\;\;\;\;\leq \frac{\left \vert b\right \vert c(c+1)(1-\beta )}{(a+1)(p+1)}\max \left \{ 1, \left \vert 2\upsilon -1\right \vert \right \} .$

This proves the required result.

Taking $\mu =1$, we obtain the following result.

Corollary 2.8. Let $f\in \mathcal{SC}_{p}^{0}(a, b, c, \beta)$ and of the form (1.1). Then

$\left \vert a_{p+2}-a_{p+1}^{2}\right \vert \leq \frac{\left \vert b\right \vert c(c+1)(1-\beta )}{(a+1)(p+1)}\max \left \{ 1, \left \vert 2\upsilon -1\right \vert \right \},$

where

$\upsilon =\frac{bc(a+1)(p+1)(1-\beta )}{2p^{2}(c+1)}-\frac{ab(1-\beta )}{2p}.$

Theorem 2.9. Let $f\in \mathcal{SC}_{p}^{\lambda }(a, b, c, \beta).$ Then

$\left \vert a_{p+1}a_{p+3}-a_{p+2}^{2}\right \vert \leq \left[\frac{4\left \vert b\right \vert c(c+1)(1-\beta )}{(p+1)(a+1)}\right] ^{2}.$

(2.12)

Proof. Let $f\in \mathcal{SC}_{p}^{\lambda }(a, b, c, \beta)$. Then from (2.5) we have

$a_{p+1} =\frac{ab(1-\beta )}{2p\varphi _{1}(a)}d_{1}.$

(2.13)

$a_{p+2} =\frac{ab(1-\beta )}{2(p+1)\varphi _{2}(a)}\left \{ d_{2}+\frac{ ab(1-\beta )}{2p}d_{1}^{2}\right \} .$

(2.14)

$a_{p+3} =\frac{ab(1-\beta )}{2(p+2)\varphi _{3}(a)}\left \{ d_{3}+\frac{ ab(1-\beta )}{2(p+1)}d_{2}^{2}+\frac{\left \{ ab(1-\beta )\right \} ^{2}}{ 4p(p+1)}d_{1}^{2}d_{2}+\frac{ab(1-\beta )}{2p}d_{1}^{2}\right \} .$

(2.15)

From (2.13), (2.14) and (2.15) we obtain

$\left \vert a_{p+1}a_{p+3}-a_{p+2}^{2}\right \vert =\frac{\left \{ ab(1-\beta )\right \} ^{2}}{4p(p+2)\varphi _{1}(a)\varphi _{3}(a)}\times \\ \;\;\;\;\;\;\left \{ d_{3}+\frac{ab(1-\beta )}{2(p+1)}d_{2}^{2}+\frac{\left \{ ab(1-\beta )\right \} ^{2}}{4p(p+1)}d_{1}^{2}d_{2}+\frac{ab(1-\beta )}{2p} d_{1}^{2}\right \} \\ \;\;\;\;\;\;-\frac{\left \{ ab(1-\beta )\right \} ^{2}}{4(p+1)^{2}\varphi _{2}^{2}(a)} \left \{ d_{2}^{2}+\frac{\left \{ ab(1-\beta )\right \} ^{2}}{4p^{2}} d_{1}^{4}+\frac{ab(1-\beta )}{p}d_{1}^{2}d_{2}\right \} .$

After some simplification we have

$\left \vert a_{p+1}a_{p+3}-a_{p+2}^{2}\right \vert =\frac{\left \vert A\right \vert ^{2}}{4}\left[Bd_{1}d_{3}+Cd_{1}d_{2}^{2}+\right. \\ \;\;\;\;\;\;\left. Ed_{1}^{3}d_{2}+Fd_{1}^{3}-Gd_{2}^{2}-Hd_{1}^{4}-Kd_{1}^{2}d_{2}, \right]$

(2.16)

where

$\begin{array}{lll} A=ab(1-\beta ), & B=\frac{1}{p(p+1)\varphi _{1}(a)\varphi _{3}(a)}, & C= \frac{A}{2p(p+1)(p+2)\varphi _{1}(a)\varphi _{3}(a)}, \\ E=\frac{A^{2}}{4p^{2}(p+1)(p+2)\varphi _{1}(a)\varphi _{3}(a)}, & F=\frac{A}{ 2p^{2}(p+1)\varphi _{1}(a)\varphi _{3}(a)}, & G=\frac{1}{(p+1)^{2}\varphi _{2}^{2}(a)}, \\ H=\frac{A^{2}}{4p^{2}(p+1)^{2}\varphi _{2}^{2}(a)}, & K=\frac{A}{ p(p+1)^{2}\varphi _{2}^{2}(a)}. & \end{array}$

Substituting the values of $d_{2}$ and $d_{3}$ frome Lemma 1.3 in (2.16) we have

$\left \vert \;\;\;\;\;\;Bd_{1}d_{3}+Cd_{1}d_{2}^{2}+Ed_{1}^{3}d_{2}+Fd_{1}^{3}-Gd_{2}^{2}-Hd_{1}^{4}-Kd_{1}^{2}d_{2}\right \vert\\ \;\;\;\;\;\;=\left \vert \frac{1}{4}Bd_{1}\left \{ d_{1}^{3}+2d_{1}(4-d_{1}^{2})x-d_{1}(4-d_{1}^{2})x^{2}+2(4-d_{1}^{2})\left( 1-\left \vert x\right \vert ^{2}\right) z\right \} \right. \\ \;\;\;\;\;\;+\frac{1}{4}Cd_{1}\left \{ d_{1}^{4}+2d_{1}^{2}(4-d_{1}^{2})x+(4-d_{1}^{2})^{2}x^{2}\right \} +\frac{1}{% 2}Ed_{1}^{3}\left \{ d_{1}^{2}+(4-d_{1}^{2})x\right \} \\ \;\;\;\;\;\;+Fd_{1}^{3}-\frac{1}{4}G\left \{ d_{1}^{4}+2d_{1}^{2}(4-d_{1}^{2})x+(4-d_{1}^{2})^{2}x^{2}\right \} -Hd_{1}^{4} \\ \;\;\;\;\;\;\left. -\frac{1}{2}Kd_{1}^{2}\left \{ d_{1}^{2}+(4-d_{1}^{2})x\right \} \right \vert .$

Simple computation gives

$4\left \vert Bd_{1}d_{3}+Cd_{1}d_{2}^{2}+Ed_{1}^{3}d_{2}+Fd_{1}^{3}-Gd_{2}^{2}-Hd_{1}^{4}-Kd_{1}^{2}d_{2}\right \vert =\left \vert \left( C+2E\right) d_{1}^{5}+\right.\\ \;\;\;\;\;\;\left( B-G-H-2K\right) d_{1}^{4}+Fd_{1}^{3}+\left( 2B+2Cd_{1}+2Ed_{1}-2G-2K\right) d_{1}^{2}\left( 4-d_{1}^{2}\right) \left \vert x\right \vert\\ \;\;\;\;\;\;\left. +2Bd_{1}\left( 4-d_{1}^{2}\right) \left( 1-\left \vert x\right \vert ^{2}\right) \left \vert z\right \vert +\left \{ Cd_{1}\left( 4-d_{1}^{2}\right) -Bd_{1}^{2}-G\left( 4-d_{1}^{2}\right) \right \} \left( 4-d_{1}^{2}\right) \left \vert x\right \vert ^{2}\right \vert$

(2.17)

Applying triangle inequality and replacing $\left \vert x\right \vert$ by $\rho$ in (2.17) we have

$4\left \vert Bd_{1}d_{3}+Cd_{1}d_{2}^{2}+Ed_{1}^{3}d_{2}+Fd_{1}^{3}-Gd_{2}^{2}-Hd_{1}^{4}-Kd_{1}^{2}d_{2}\right \vert \leq \left( \left \vert C\right \vert +2\left \vert E\right \vert \right) d_{1}^{5}+\left \vert F\right \vert d_{1}^{3}+\\ \;\;\;\;\;\;\left( B-G+\left \vert H\right \vert +2\left \vert K\right \vert \right) d_{1}^{4}+\left \{ 2B+2\left \vert C\right \vert d_{1}+2\left \vert E\right \vert d_{1}-2G+2\left \vert K\right \vert \right \} d_{1}^{2}\left( 4-d_{1}^{2}\right) \rho \notag \\ \;\;\;\;\;\;+2Bd_{1}\left( 4-d_{1}^{2}\right) \left( 1-\rho ^{2}\right) +\left \{ \left \vert C\right \vert d_{1}\left( 4-d_{1}^{2}\right) -Bd_{1}^{2}-G\left( 4-d_{1}^{2}\right) \right \} \left( 4-d_{1}^{2}\right) \rho ^{2} \notag \\ =F(d_{1}, \rho ).$

(2.18)

Taking partial derivative of $F(d_{1}, \rho)$ with respect to $\rho$, we have

$\frac{\partial F(d_{1}, \rho )}{\partial \rho } =\left \{ 2B+2\left \vert C\right \vert d_{1}+2\left \vert E\right \vert d_{1}-2G+2\left \vert F\right \vert \right \} d_{1}^{2}\left( 4-d_{1}^{2}\right) -4Bd_{1}\left( 4-d_{1}^{2}\right) \rho \\ \;\;\;\;\;\;+2\left \{ \left \vert C\right \vert d_{1}\left( 4-d_{1}^{2}\right) -Bd_{1}^{2}-G\left( 4-d_{1}^{2}\right) \right \} \left( 4-d_{1}^{2}\right) \rho$

Clearly $\frac{\partial F(d_{1}, \rho)}{\partial \rho }>0,$ for $0 < \rho < 1$ and $0 < d_{1} < 2.$ Therefore, $F(d_{1}, \rho)$ is an increasing function of $% \rho.$ Also for a fixed $d_{1}\in \lbrack 0, 2],$ we have

$\max_{0\leq \rho \leq 1}F(d_{1}, \rho )=F(d_{1}, \rho )=J(d_{1}).$

Therefore by putting $\rho =1$ in (2.18) we have

$J(d_{1}) =\left \{ \left \vert C\right \vert +2\left \vert E\right \vert \right \} d_{1}^{5}+\left \{ B-G+\left \vert H\right \vert +2\left \vert K\right \vert \right \} d_{1}^{4}+\left \vert F\right \vert d_{1}^{3} \\ \;\;\;\;\;\;+\left \{ 2B+2\left \vert C\right \vert d_{1}+2\left \vert E\right \vert d_{1}-2G+2\left \vert K\right \vert \right \} d_{1}^{2}\left( 4-d_{1}^{2}\right) \\ \;\;\;\;\;\;+\left \{ \left \vert C\right \vert d_{1}\left( 4-d_{1}^{2}\right) -Bd_{1}^{2}-G\left( 4-d_{1}^{2}\right) \right \} \left( 4-d_{1}^{2}\right)$

Differentiating with respect to $d_{1}$, we have

$J^{\prime }(d_{1}) =5\left \{ \left \vert C\right \vert +2\left \vert E\right \vert \right \} d_{1}^{4}+4\left \{ B-G+\left \vert H\right \vert +2\left \vert K\right \vert \right \} d_{1}^{3}+3\left \vert F\right \vert d_{1}^{2} \\ \;\;\;\;\;\;+4\left \{ B-G+2\left \vert K\right \vert \right \} d_{1}\left( 4-d_{1}^{2}\right) -4\left \{ B-G+2\left \vert K\right \vert \right \} d_{1}^{3} \\ \;\;\;\;\;\;+6\left \{ \left \vert C\right \vert +\left \vert E\right \vert \right \} d_{1}^{2}\left( 4-d_{1}^{2}\right) -4\left \{ \left \vert C\right \vert +\left \vert E\right \vert \right \} d_{1}^{4}+\left \vert C\right \vert d_{1}^{2}\left( 4-d_{1}^{2}\right) ^{2} \\ \;\;\;\;\;\;-4\left \vert C\right \vert d_{1}^{2}\left( 4-d_{1}^{2}\right) -2Bd_{1}\left( 4-d_{1}^{2}\right) +2Bd_{1}^{3}-4Gd_{1}\left( 4-d_{1}^{2}\right)$

Again differentiating with respect to $d_{1}$ we have

$J^{\prime \prime }(d_{1}) =20\left \{ \left \vert C\right \vert +2\left \vert E\right \vert \right \} d_{1}^{3}+12\left \{ B-G+\left \vert H\right \vert +2\left \vert K\right \vert \right \} d_{1}^{2} \\ \;\;\;\;\;\;+6\left \vert F\right \vert d_{1}+4\left \{ B-G+2\left \vert K\right \vert \right \} \left( 4-d_{1}^{2}\right) -8\left \{ B-G+2\left \vert K\right \vert \right \} d_{1}^{2} \\ \;\;\;\;\;\;-12\left \{ B-G+2\left \vert K\right \vert \right \} d_{1}^{2}+126\left \{ \left \vert C\right \vert +\left \vert E\right \vert \right \} \left( 4-d_{1}^{2}\right) -126\left \{ \left \vert C\right \vert +\left \vert E\right \vert \right \} d_{1}^{3} \\ \;\;\;\;\;\;-16\left \{ \left \vert C\right \vert +\left \vert E\right \vert \right \} d_{1}^{3}+2\left \vert C\right \vert d_{1}\left( 4-d_{1}^{2}\right) ^{2}-4\left \vert C\right \vert d_{1}^{3}\left( 4-d_{1}^{2}\right) ^{2}-8\left \vert C\right \vert d_{1}\left( 4-d_{1}^{2}\right) \\ \;\;\;\;\;\;+8\left \vert C\right \vert d_{1}^{3}-2B\left( 4-d_{1}^{2}\right) +4Bd_{1}^{2}+6Bd_{1}^{2}-4G\left( 4-d_{1}^{2}\right) +8Gd_{1}^{2}.$

For maximum value of $J(d_{1}),$ clearly $J^{\prime }(d_{1})=0$ for $d_{1}=0$ and $J^{\prime \prime }(0) < 0,$ so $J(d_{1})$ has maximum value at $d_{1}=0$ hence

$\left \vert a_{p+1}a_{p+3}-a_{p+2}^{2}\right \vert \leq \left[\frac{4\left \vert b\right \vert c(c+1)(1-\beta )}{(p+1)(a+1)}\right] ^{2}.$

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3. Subordination Results for the Function Class $\mathcal{SC}_{p}^{\lambda }(a, b, c, \beta)$

Given functions $f, g\in A,$ $f$ is said to subordinate to $g$ denoted by $f\prec g,$ $z\in \mathbb{U},$ if there exist a function $w\in V,$ where

$V=\left \{ w\in A:w(0)=0, \text{ }\left \vert w(z)\right \vert <1, \text{ } z\in \mathbb{U}\right \}$

such that $f(z)=g(w(z)).$

Lemma 3.1. ^[18]. Let $q(z)$ be convex in $\mathbb{U}$ and $Re\left(\mu _{1}q(z)+\mu _{2}\right) >0,$ where $\mu _{1}, \mu _{2}\in %TCIMACRO{\U{2102} }% %BeginExpansion \mathbb{C} %EndExpansion \diagdown \left \{ 0\right \}$, $z\in \mathbb{U}.$ If $h(z)$ is analytic in $\mathbb{U}$ with $q(0)=h(0)$ and

$h(z)+\frac{zh^{^{\prime }}(z)}{\mu _{1}h(z)+\mu _{2}}\prec q(z), z\in \mathbb{U},$

then $h(z)\prec q(z).$

Lemma 3.2. A function $f\in \mathcal{SC}_{p}^{\lambda }(a, b, c, \beta)$, if and only if

$e^{i\lambda }\left( 1-\frac{2}{b}+\frac{2}{b}\frac{\mathcal{L}_{p}(a+1, c)f(z) }{\mathcal{L}_{p}(a, c)f(z)}\right) \prec q(z), z\in \mathbb{U},$

where

$q(z)=\frac{\cos \lambda -\left \{ 2\beta \cos \lambda +i\sin \lambda -\cos \lambda \right \} z}{1-z}.$

(3.1)

for some real $\lambda (\left \vert \lambda \right \vert < \frac{\pi }{2})$ and $\beta >p.$

The proof of above lemma is similier to that of Theorem 1 in (22) so we omit the proof.

Theorem 3.3. Let $\beta >p,$ $b\in %TCIMACRO{\U{2102} }% %BeginExpansion \mathbb{C} %EndExpansion \diagdown \{0, -1\}.$ Then

$\mathcal{SC}_{p}^{0}(a+1, b, c, \beta )\subset \mathcal{SC}_{p}^{0}(a, b+1, c, \beta _{1}),$

where

$\beta _{1}=\frac{b(a+1)}{a(b+1)}\beta -\frac{b-a}{a(b+1)}.$

Proof. Suppose $f\in \mathcal{SC}_{p}^{0}(a+1, b, c, \beta)$ and set

$1-\frac{2}{b+1}+\frac{2}{b+1}\frac{\mathcal{L}_{p}(a+1, c)f(z)}{\mathcal{L}% _{p}(a, c)f(z)}=h(z),$

(3.2)

where $h$ is analytic in $\mathbb{U}$ and $h(0)=1.$

Logarithmic differentiation of (3.2), gives

$\frac{z\left( \mathcal{L}_{p}(a+1, c)f(z)\right) ^{^{\prime }}}{\mathcal{L}% _{p}(a+1, c)f(z)}-\frac{z\left( \mathcal{L}_{p}(a, c)f(z)\right) ^{^{\prime }}% }{\mathcal{L}_{p}(a, c)f(z)}=\frac{\left( b+1\right) zh^{^{\prime }}(z)}{% (b+1)\left \{ h(z)-1\right \} +2}.$

Using the identity (1.6) we have

$1-\frac{2}{b}+\frac{2}{b}\frac{\mathcal{L}_{p}(a+2, c)f(z)}{\mathcal{L} _{p}(a+1, c)f(z)}=1-\frac{a}{a+1}\frac{b+1}{b}+\frac{a}{a+1}\frac{b+1}{b}h(z)+ \frac{2}{b(a+1)}\frac{zh^{^{\prime }}(z)}{h(z)-1+\frac{2}{b+1}}.$

(3.3)

Let

$1-\frac{a}{a+1}\frac{b+1}{b}+\frac{a}{a+1}\frac{b+1}{b}h(z)=H(z),$

where $H$ is analytic in $\mathbb{U}$ and $H(0)=1$. From (3.3) we have

$1-\frac{2}{b}+\frac{2}{b}\frac{\mathcal{L}_{p}(a+2, c)f(z)}{\mathcal{L} _{p}(a+1, c)f(z)}=H(z)+\frac{zH^{^{\prime }}(z)}{\mu _{1}H(z)+\mu _{2}},$

where $\mu _{1}=\frac{b\left(a+1\right) }{2}$ and $\mu _{2}=\frac{2a-ab-b}{2 }.$ Since $f(z)\in VD_{p}^{0}(a+1, b, c, \beta),$ so from Lemma 3.2 we have

$H(z)+\frac{zH^{^{\prime }}(z)}{\mu _{1}H(z)+\mu _{2}}=1-\frac{2}{b}+\frac{2}{ b}\frac{\mathcal{L}_{p}(a+2, c)f(z)}{\mathcal{L}_{p}(a+1, c)f(z)}\prec q(z),$

where $q$ is given by

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

Applying Lemma 3.1 we have

$H(z)\prec q(z)$

or equivalently

$h(z)\prec \frac{1-\left( 2\beta _{1}-1\right) z}{1-z},$

where

$\beta _{1}=\frac{b(a+1)}{a(b+1)}\beta -\frac{b-a}{a(b+1)}.$

This complete the proof.

Theorem 3.4. Let $\ f\in \mathcal{SC}_{p}^{0}(a, b, c, \beta).$ Then $F\in \mathcal{SC} _{p}^{0}(a, b, c, \beta),$ where $F$ is Bernardi integral operator defined by

$F(z)=\frac{p+1}{z^{c}}\int_{0}^{z}t^{c-1}f(t)dt, c>-1.$

(3.5)

Proof. Suppose

$1-\frac{2}{b}+\frac{2}{b}\frac{\mathcal{L}_{p}(a+2, c)F(z)}{\mathcal{L}% _{p}(a+1, c)F(z)}=h(z),$

(3.6)

where $h$ is analytic in $\mathbb{U}$ and $h(0)=1$.

Now differentiating (3.5) we have

$(c+p)f(z)=cF(z)+zF^{^{\prime }}(z)$

Applying the operator $\mathcal{L}_{p}(a, c)$ we have

$(c+p)\mathcal{L}_{p}(a, c)f(z)=c\mathcal{L}_{p}(a, c)F(z)+\mathcal{L} _{p}(a+1, c)F(z)$

(3.7)

and

$(c+p)\mathcal{L}_{p}(a+1, c)f(z)=c\mathcal{L}_{p}(a+1, c)F(z)+\mathcal{L}% _{p}(a+2, c)F(z)$

(3.8)

From (3.7) and (3.8) we have

$\frac{\mathcal{L}_{p}(a+1, c)f(z)}{\mathcal{L}_{p}(a, c)f(z)}=\frac{c\frac{% \mathcal{L}_{p}(a+1, c)f(z)}{\mathcal{L}_{p}(a, c)f(z)}+\frac{\mathcal{L}% _{p}(a+2, c)f(z)}{\mathcal{L}_{p}(a+1, c)f(z)}\frac{\mathcal{L}_{p}(a+1, c)f(z)% }{\mathcal{L}_{p}(a, c)f(z)}}{c+\frac{\mathcal{L}_{p}(a+1, c)f(z)}{\mathcal{L}% _{p}(a, c)f(z)}}.$

(3.9)

Logarithmic differentiation of (3.6), totgether with (1.6) and (3.9) we have

$1-\frac{2}{b}+\frac{2}{b}\frac{\mathcal{L}_{p}(a+1, c)f(z)}{\mathcal{L}% _{p}(a, c)f(z)}=h(z)+\frac{zh^{^{\prime }}(z)}{\mu _{3}h(z)+\mu _{4}},$

where $\mu _{3}=\frac{ab}{2}$ and $\mu _{4}=c+p-\frac{ab}{2}$

Since $f\in \mathcal{SC}_{p}^{0}(a, b, c, \beta),$ so from Lemma 3.2 we have

$h(z)+\frac{zh^{^{\prime }}(z)}{\mu _{3}h(z)+\mu _{4}}=1-\frac{2}{b}+\frac{2}{ b}\frac{\mathcal{L}_{p}(a+1, c)f(z)}{\mathcal{L}_{p}(a, c)f(z)}\prec q(z),$

where $q$ is given by (3.4)

Applying Lemma 3.1 we have

$h(z)\prec q(z),$

which implies that $F\in \mathcal{SC}_{p}^{0}(a, b, c, \beta).$

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Conflict of Interest

All authors declare no conflicts of interest in this paper.

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