A Criterion for Subfamilies of Multivalent Functions of Reciprocal Order with Respect to Symmetric Points

Abstract

In the present research paper, our aim is to introduce a new subfamily of p-valent (multivalent) functions of reciprocal order. We investigate sufficiency criterion for such defined family.

1. Introduction

Let us suppose that ${\mathcal{A}}_p$ represents the class of p-valent functions $f\left(z\right)$ that are holomorphic (analytic) in the region $\mathbb{E}=\left\{z:\left|z\right|<1\right\}$ and has the following Taylor series representation:

$$f\left(z\right)=z^p+{\displaystyle \sum }_{k=1}^{\infty }{a}_{p+k}{z}^{p+k}.$$

(1)

Two points p and $p^{\prime }$ are said to be symmetrical with respect to o if o is the midpoint of the line segment $p{p}^{\prime }$.

If $f\left(z\right)$ and $g\left(z\right)$ are analytic in $\mathcal{E},$ we say that $f\left(z\right)$ is subordinate to $g\left(z\right),$ written as $f\left(z\right)\prec g\left(z\right),$ if there exists a Schwarz function, $w\left(z\right),$ which is analytic in $\mathcal{E}$ with $w\left(0\right)=0$ and $\left|w\left(z\right)\right|<1$ such that $f\left(z\right)=g\left(w\right(z\left)\right).$ Furthermore, if the function $g\left(z\right)$ is univalent in $\mathcal{E},$ then we have the following equivalence, see [1].

$$\begin{array}{ccc}\hfill f\left(z\right)\prec g\left(z\right)(z\in \mathcal{E})⟺f\left(0\right)=g\left(0\right)& \mathrm{and}& f\left(\mathcal{E}\right)\subset g\left(\mathcal{E}\right).\hfill \end{array}$$

Let ${\mathcal{N}}_{\alpha }$ denotes the class of starlike functions of reciprocal order $\alpha$$\left(\alpha >1\right)$ and is given below

$${\mathcal{N}}_{\alpha }:=\left\{f\left(z\right)\in \mathcal{A}:\mathrm{Re}\left(\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\right)<\alpha ,\left(z\in \mathbb{E}\right)\right\}.$$

(2)

This class was introduced by Uralegaddi et al. [2] amd further studied by the Owa et al. [3]. After that Nunokawa and his coauthors [4] proved that $f\left(z\right)\in {\mathcal{N}}_{\alpha }$, $0<\alpha <\frac{1}{2},$ if and only if the following inequality holds

$$\left|\frac{2\alpha z{f}^{\prime }\left(z\right)}{f\left(z\right)}-1\right|<1,\left(z\in \mathbb{E}\right).$$

Later on, Owa and Srivastava [5] in 2002 generalized this idea for the classes of multivalent convex and starlike functions of reciprocal order $\alpha$$\left(\alpha >p\right)$, and further studied by Polatoglu et al. [6]. For more details of the related concepts, see the article of Dixit et al. [7], Uyanik et al. [8], and Arif et al. [9].

For $-1\le t<s\le 1$ with $s\ne 0\ne t,$$0<\alpha <1,$ and $p\in \mathbb{N},$ we introduce a subclass of ${\mathcal{A}}_p$ consisting of all analytic p-valent functions of reciprocal order $\alpha$, denoted by ${\mathcal{N}}_{\alpha }^p\mathcal{S}\left(s,t\right)$ and is defined as

$${\mathcal{N}}_{\alpha }^p\mathcal{S}\left(s,t\right)=\left\{f\left(z\right)\in {\mathcal{A}}_p:\mathrm{Re}\left(\frac{\left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}{f\left(sz\right)-f\left(tz\right)}\right)<\frac{p}{\alpha },\left(z\in \mathbb{E}\right)\right\},$$

(3)

or equivalently

$$\left|\frac{\left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}{f\left(sz\right)-f\left(tz\right)}-\frac{p}{2\alpha }\right|\le \frac{p}{2\alpha }.$$

(4)

Many authors studied sufficiency conditions for various subclasses of analytic and multivalent functions, for details see [4,10,11,12,13,14,15,16,17].

We will need the following lemmas for our work.

Lemma 1

(Jack’s lemma [18]). Let Ψ be a non-constant holomorphic function in $\mathbb{E}$ and if the value of $\left|\mathsf{\Psi }\right|$ is maximum on the circle $\left|z\right|=r<1$ at $z_{\circ }$, then $z_{\circ }{\mathsf{\Psi }}^{\prime }\left(z_{\circ }\right)=k$$\mathsf{\Psi }\left(z_{\circ }\right)$, where k$\ge 1$ is a real number.

Lemma 2

(See [1]). Let $\mathfrak{H}\subset \mathbb{C}$ and let $\mathsf{\Phi }:{\mathbb{C}}^2\times {\mathbb{E}}^{*}\to \mathbb{C}$ be a mapping satisfying $\mathsf{\Phi }\left(ia,b,z\right)\notin$$\mathfrak{H}$ for $a,b\in \mathbb{R}$ such that $b\le -\frac{1+a^2}{2}.$ If $p\left(z\right)=1+c_1{z}^1+c_2{z}^2+\cdots$ is regular in ${\mathbb{E}}^{*}$ and $\mathsf{\Phi }\left(p\left(z\right),z{p}^{\prime }\left(z\right),z\right)\in \mathfrak{H}$∀$z\in {\mathbb{E}}^{*}$, then $\mathrm{Re}\left(p\left(z\right)\right)>0.$

Lemma 3

(See [15]). Let $p\left(z\right)=1+c_{1}z+c_2{z}^2+\cdots$ be analytic in $\mathbb{E}$ and η be analytic and starlike (with respect to the origin) univalent in $\mathbb{E}$ with $\eta \left(0\right)=0.$ If $z{p}^{\prime }\left(z\right)\prec \eta \left(z\right),$ then

$$p\left(z\right)\prec 1+\underset{0}{\overset{z}{{\displaystyle \int }}}\frac{\eta \left(t\right)}{t}dt.$$

This result is the best possible.

2. Main Results

Theorem 1.

Let $f\left(z\right)\in {\mathcal{A}}_p$ and satisfies

$${\displaystyle \sum }_{n=1}^{\infty }\left(\alpha \left(p+n\right)+p\frac{\left(s^{p+n}-t^{p+n}\right)}{\left(s^p-t^p\right)}\right)\left|a_{n+p}\right|\le \frac{p}{2}\left(1-\left|2\alpha -1\right|\right).$$

(5)

Then $f\left(z\right)\in {\mathcal{N}}_{\alpha }^p\mathcal{S}\left(s,t\right).$

Proof. 

Let us assume that the inequality (5) holds. It suffices to show that

$$\left|\frac{2\alpha \left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}{f\left(sz\right)-f\left(tz\right)}-p\right|\le p.$$

(6)

Consider

$$\begin{array}{cc}& \left|\frac{2\alpha \left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}{f\left(sz\right)-f\left(tz\right)}-p\right|\hfill \\ =\hfill & \left|\frac{p\left(2\alpha -1\right)\left(s^p-t^p\right)z^p+{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}(2\alpha \left(p+n\right)\left(s^p-t^p\right)-p\left(s^{n+p}-t^{n+p}\right))a_{n+p}{z}^{n+p}}{\left(s^p-t^p\right)z^p+{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\left(s^{n+p}-t^{n+p}\right)a_{n+p}{z}^{n+p}}\right|\hfill \\ \le \hfill & \frac{p\left|2\alpha -1\right|\left(s^p-t^p\right)+{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}(2\alpha \left(p+n\right)\left(s^p-t^p\right)+p\left(s^{n+p}-t^{n+p}\right))\left|a_{n+p}\right|}{\left(s^p-t^p\right)-{\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}\left(s^{n+p}-t^{n+p}\right)\left|a_{n+p}\right|}\hfill \end{array}$$

The last expression is bounded above by p if

$$\begin{array}{cc}& p\left|2\alpha -1\right|\left(s^p-t^p\right)+{\displaystyle \sum }_{n=1}^{\infty }(2\alpha \left(p+n\right)\left(s^p-t^p\right)+p\left(s^{n+p}-t^{n+p}\right))\left|a_{n+p}\right|\hfill \\ <& p\left\{\left(s^p-t^p\right)-{\displaystyle \sum }_{n=1}^{\infty }\left(s^{n+p}-t^{n+p}\right)\left|a_{n+p}\right|\right\}.\hfill \end{array}$$

Hence

$${\displaystyle \sum }_{n=1}^{\infty }\left(\alpha \left(p+n\right)+p\frac{\left(s^{p+n}-t^{p+n}\right)}{\left(s^p-t^p\right)}\right)\left|a_{n+p}\right|\le \frac{p}{2}\left(1-\left|2\alpha -1\right|\right).$$

This shows that $f\left(z\right)\in {\mathcal{NS}}_p\left(s,t,\alpha \right).$ This completes the proof. □

Theorem 2.

If $f\left(z\right)\in {\mathcal{A}}_p$ satisfies the condition

$$\left|1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}-\frac{z{\left(f\left(sz\right)-f\left(tz\right)\right)}^{\prime }}{f\left(sz\right)-f\left(tz\right)}\right|<1-\alpha ,\left(\frac{1}{2}\le \alpha <1\right),$$

(7)

then $f\left(z\right)\in {\mathcal{N}}_{\alpha }^p\mathcal{S}\left(s,t\right).$

Proof. 

Let us set

$$q\left(z\right)=\frac{1-\frac{\alpha \left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}{p\left(f\left(sz\right)-f\left(tz\right)\right)}}{1-\alpha }-1.$$

(8)

Then clearly $q\left(z\right)$ is analytic in $\mathbb{E}$ with $q\left(0\right)=0$. Differentiating logarithmically, we have

$$1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}-\frac{z{\left(f\left(sz\right)-f\left(tz\right)\right)}^{\prime }}{f\left(sz\right)-f\left(tz\right)}=-\frac{\left(1-\alpha \right)z{q}^{\prime }\left(z\right)}{\left(\alpha -\left(1-\alpha \right)q\left(z\right)\right)}.$$

So

$$\left|1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}-\frac{z{\left(f\left(sz\right)-f\left(tz\right)\right)}^{\prime }}{f\left(sz\right)-f\left(tz\right)}\right|=\left|-\frac{\left(1-\alpha \right)z{q}^{\prime }\left(z\right)}{\left(\alpha -\left(1-\alpha \right)q\left(z\right)\right)}\right|.$$

From (7), we have

$$\left|\frac{\left(1-\alpha \right)z{q}^{\prime }\left(z\right)}{\left(\alpha -\left(1-\alpha \right)q\left(z\right)\right)}\right|<1-\alpha .$$

Next, we claim that $\left|q\left(z\right)\right|<1$. Indeed, if not, then for some $z_{\circ }\in \mathbb{E},$ we have

$$\underset{\left|z\right|\le \left|z_0\right|}{\max }\left|q\left(z\right)\right|=\left|q\left(z_0\right)\right|=1.$$

Applying Jack’s lemma to $q\left(z\right)$ at the point $z_0$, we have

$$q\left(z_0\right)=e^{i\theta },\frac{z_0{q}^{\prime }\left(z_0\right)}{q\left(z_0\right)}=k,k\ge 1.$$

Then

$$\begin{array}{ccc}\hfill \left|1+\frac{z_0{f}^{″}\left(z_0\right)}{f^{\prime }\left(z_0\right)}-\frac{z{\left(f\left(s{z}_0\right)-f\left(t{z}_0\right)\right)}^{\prime }}{f\left(s{z}_0\right)-f\left(t{z}_0\right)}\right|& =& \left|\frac{\left(1-\alpha \right)z_0{q}^{\prime }\left(z_0\right)}{\left(\alpha -\left(1-\alpha \right)q\left(z_0\right)\right)}\right|\hfill \\ & =& \left|1-\alpha \right|\left|\frac{z_0{q}^{\prime }\left(z_0\right)}{q\left(z_0\right)}\left(\frac{1}{\left(1-\alpha \right)-\alpha e^{-i\theta }}\right)\right|\hfill \\ & =& \left|1-\alpha \right|\left|\frac{k}{\alpha e^{-i\theta }-\left(1-\alpha \right)}\right|\hfill \\ & \ge & \left|1-\alpha \right|\left|\frac{1}{\left(1-\alpha \right)-\alpha e^{-i\theta }}\right|.\hfill \end{array}$$

Therefore

$${\left|1+\frac{z_0{f}^{″}\left(z_0\right)}{f^{\prime }\left(z_0\right)}-\frac{z{\left(f\left(s{z}_0\right)-f\left(t{z}_0\right)\right)}^{\prime }}{f\left(s{z}_0\right)-f\left(t{z}_0\right)}\right|}^2\ge \frac{{\left(1-\alpha \right)}^2}{{\left(1-\alpha \right)}^2+{\alpha }^2-2\alpha \left(1-\alpha \right)\cos \theta }.$$

Now the right hand side has minimum value at $\cos \theta =-1,$ therefore we have

$${\left|1+\frac{z_0{f}^{″}\left(z_0\right)}{f^{\prime }\left(z_0\right)}-\frac{z{\left(f\left(s{z}_0\right)-f\left(t{z}_0\right)\right)}^{\prime }}{f\left(s{z}_0\right)-f\left(t{z}_0\right)}\right|}^2\ge {\left(1-\alpha \right)}^2.$$

But this contradicts (7). Hence we conclude that $\left|q\left(z\right)\right|<1$ for all $z\in \mathbb{E}$, which shows that

$$\left|\frac{1-\frac{\alpha \left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}{p\left(f\left(sz\right)-f\left(tz\right)\right)}}{1-\alpha }-1\right|<1.$$

This implies that

$$\left|\frac{\left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}{p\left(f\left(sz\right)-f\left(tz\right)\right)}-1\right|<\frac{1}{\alpha }-1.$$

(9)

Now we have

$$\begin{array}{ccc}\left|\frac{\left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}{p\left(f\left(sz\right)-f\left(tz\right)\right)}-\frac{1}{2\alpha }\right|\hfill & \le & \left|\frac{\left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}{p\left(f\left(sz\right)-f\left(tz\right)\right)}-1\right|+\left|1-\frac{1}{2\alpha }\right|\hfill \\ & <& \frac{1}{\alpha }-1+1-\frac{1}{2\alpha }\hfill \\ & =& \frac{1}{2\alpha }.\hfill \end{array}$$

This implies that $f\left(z\right)\in {\mathcal{N}}_{\alpha }^p\mathcal{S}\left(s,t\right)$.  □

Theorem 3.

If $f\left(z\right)\in {\mathcal{A}}_p$ satisfies the condition

$$\mathrm{Re}\left(-1-\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}+\frac{z{\left(f\left(sz\right)-f\left(tz\right)\right)}^{\prime }}{f\left(sz\right)-f\left(tz\right)}\right)>\left\{\begin{array}{cc}\frac{\alpha }{2\left(\alpha -1\right)},& 0\le \alpha \le \frac{1}{2}\\ \frac{\alpha -1}{2\alpha },& \frac{1}{2}\le \alpha <1,\end{array}\right.$$

(10)

then $f\left(z\right)\in {\mathcal{N}}_{\alpha }^p\mathcal{S}\left(s,t\right)$ for $0\le \alpha <1.$

Proof. 

Let

$$q\left(z\right)=\frac{\frac{p\left(f\left(sz\right)-f\left(tz\right)\right)}{\left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}-\alpha }{1-\alpha }.$$

Then clearly $q\left(z\right)$ is analytic in $\mathbb{E}$. Applying logarithmic differentiation, we have

$$-1-\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}+\frac{z{\left(f\left(sz\right)-f\left(tz\right)\right)}^{\prime }}{f\left(sz\right)-f\left(tz\right)}=\frac{\left(1-\alpha \right)z{q}^{\prime }\left(z\right)}{\alpha +\left(1-\alpha \right)q\left(z\right)}=\mathsf{\Psi }\left(q\left(z\right),z{q}^{\prime }\left(z\right),z\right),$$

where

$$\mathsf{\Psi }\left(u,v;t\right)=\frac{\left(1-\alpha \right)v}{\alpha +\left(1-\alpha \right)u}.$$

Now for all $x,y\in \mathbb{R}$ satisfying the inequality $y\le -\frac{1+x^2}{2}$, we have

$$\mathsf{\Psi }\left(ix,y,z\right)=\frac{\left(1-\alpha \right)y}{\alpha +\left(1-\alpha \right)ix}.$$

Therefore

$$\begin{array}{ccc}\hfill \mathrm{Re}\left(\mathsf{\Psi }\left(ix,y,z\right)\right)& \le & -\frac{\alpha \left(1-\alpha \right)\left(1+x^2\right)}{2\left({\alpha }^2+{\left(1-\alpha \right)}^2{x}^2\right)},\hfill \\ & \le & \left\{\begin{array}{cc}\frac{\alpha }{2\left(\alpha -1\right)},& 0\le \alpha \le \frac{1}{2},\\ \frac{\alpha -1}{2\alpha },& \frac{1}{2}\le \alpha <1.\end{array}\right.\hfill \end{array}$$

We set

$$\Lambda =\left\{\zeta :\mathrm{Re}\left(\zeta \right)>\left\{\begin{array}{cc}\frac{\alpha }{2\left(\alpha -1\right)},& 0\le \alpha \le \frac{1}{2},\\ \frac{\alpha -1}{2\alpha },& \frac{1}{2}\le \alpha <1.\end{array}\right.\right\}$$

Then $\mathsf{\Psi }\left(ix,y;z\right)\notin \Lambda$ for all real x, y such that $y\le -\frac{1+x^2}{2}$. Moreover, in view of (10), we know that $\mathsf{\Psi }\left(q\left(z\right),z{q}^{\prime }\left(z\right),z\right)\in \Lambda$. So applying Lemma 2, we have

$$\mathrm{Re}\left(q\left(z\right)\right)>0,$$

which shows that the desired assertion of Theorem 6 holds.  □

Theorem 4.

If $f\left(z\right)\in {\mathcal{A}}_p$ satisfies

$$\mathrm{Re}\frac{f\left(sz\right)-f\left(tz\right)}{\left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}\left(1-\beta \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}+\beta \frac{z{\left(f\left(sz\right)-f\left(tz\right)\right)}^{\prime }}{f\left(sz\right)-f\left(tz\right)}\right)>\frac{2\alpha +\beta \left(3\alpha -1\right)}{2p},$$

(11)

then $f\left(z\right)\in {\mathcal{N}}_{\alpha }^p\mathcal{S}\left(s,t\right)$ for $0<\alpha <1$ and $\beta \ge 0.$

Proof. 

Let

$$h\left(z\right)=\frac{\frac{p\left(f\left(sz\right)-f\left(tz\right)\right)}{\left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}-\alpha }{1-\alpha }.$$

where $h\left(z\right)$ is clearly analytic in $\mathbb{E}$ such that $h\left(0\right)=1.$ We can write

$$\frac{p\left(f\left(sz\right)-f\left(tz\right)\right)}{\left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}=\alpha +\left(1-\alpha \right)h\left(z\right).$$

(12)

After some simple computation, we have

$$-\beta \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}+\beta \frac{z{\left(f\left(sz\right)-f\left(tz\right)\right)}^{\prime }}{f\left(sz\right)-f\left(tz\right)}=\beta \frac{\alpha +\left(1-\alpha \right)\left(h\left(z\right)+z{h}^{\prime }\left(z\right)\right)}{\alpha +\left(1-\alpha \right)h\left(z\right)}$$

It follows from (12) that

$$\begin{array}{cc}& \frac{p\left(f\left(sz\right)-f\left(tz\right)\right)}{\left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}\left(1-\beta \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}+\beta \frac{z{\left(f\left(sz\right)-f\left(tz\right)\right)}^{\prime }}{f\left(sz\right)-f\left(tz\right)}\right)\hfill \\ =& \beta \left(1-\alpha \right)z{h}^{\prime }\left(z\right)+\left(1-\alpha \right)\left(1+\beta \right)h\left(z\right)+\alpha \left(1+\beta \right)\hfill \\ =& \mathsf{\Psi }\left(h\left(z\right),z{h}^{\prime }\left(z\right),z\right)\hfill \end{array}$$

where

$$\mathsf{\Psi }\left(u,v,t\right)=\beta \left(1-\alpha \right)v+\left(1-\alpha \right)\left(1+\beta \right)u+\alpha \left(1+\beta \right).$$

Now for some real numbers x and y satisfying $y\le -\frac{1+x^2}{2},$ we have

$$\begin{array}{ccc}\hfill \mathrm{Re}\left(\mathsf{\Psi }\left(ix,y,z\right)\right)& \le & -\beta \left(1-\alpha \right)\frac{1+x^2}{2}+\alpha \left(1+\beta \right)\hfill \\ & =& \frac{1}{2}\left(2\alpha +\beta \left(3\alpha -1\right)\right).\hfill \end{array}$$

If we set

$$\Lambda =\left\{\zeta :\mathrm{Re}\left(\zeta \right)>\frac{1}{2}\left(2\alpha +\beta \left(3\alpha -1\right)\right)\right\},$$

then $\mathsf{\Psi }\left(ix,y,z\right)\notin \Lambda$ Furthermore, by virtue of (11), we know that $\mathsf{\Psi }\left(h\left(z\right),z{h}^{\prime }\left(z\right),z\right)\in \Lambda$. Thus by using Lemma 2, we have

$$\mathrm{Re}\left(h\left(z\right)\right)>0,$$

which implies that the assertion of Theorem 7 holds true.  □

Theorem 5.

If $f\left(z\right)\in {\mathcal{A}}_p$ satisfies the condition

$$\left|{\left(p-\frac{2\alpha \left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}{\left(f\left(sz\right)-f\left(tz\right)\right)}\right)}^{\prime }\right|\le p\beta {\left|z\right|}^{\gamma },$$

(13)

then $f\left(z\right)\in {\mathcal{N}}_{\alpha }^p\mathcal{S}\left(s,t\right)$ with $0<\alpha <1,$$0<\beta \le \gamma +1$ and $\gamma \ge 0.$

Proof. 

Let we define

$$ϝ\left(z\right)=z\left(p-\frac{2\alpha \left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}{\left(f\left(sz\right)-f\left(tz\right)\right)}\right).$$

(14)

Then $ϝ\left(z\right)$ is regular in $\mathbb{E}$ and $ϝ\left(0\right)=0.$ The condition (14) gives

$$\left|{\left(p-\frac{2\alpha \left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}{\left(f\left(sz\right)-f\left(tz\right)\right)}\right)}^{\prime }\right|=\left|{\left(\frac{ϝ\left(z\right)}{z}\right)}^{\prime }\right|$$

It follows from (13) that

$$\left|{\left(\frac{ϝ\left(z\right)}{z}\right)}^{\prime }\right|\le p\beta {\left|z\right|}^{\gamma }.$$

This implies that

$$\left|\left(\frac{ϝ\left(z\right)}{z}\right)\right|=\left|\underset{0}{\overset{z}{{\displaystyle \int }}}{\left(\frac{ϝ\left(t\right)}{t}\right)}^{\prime }dt\right|\le \underset{0}{\overset{z}{{\displaystyle \int }}}\left|{\left(\frac{ϝ\left(t\right)}{t}\right)}^{\prime }\right|dt\le \frac{p\beta {\left|z\right|}^{\gamma +1}}{\gamma +1},$$

and therefore

$$\left|\left(\frac{ϝ\left(z\right)}{z}\right)\right|<p,$$

which further gives

$$\left|\frac{\left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}{p\left(f\left(sz\right)-f\left(tz\right)\right)}-\frac{1}{2\alpha }\right|<\frac{1}{2\alpha }.$$

Hence $f\left(z\right)\in {\mathcal{N}}_{\alpha }^p\mathcal{S}\left(s,t\right).$ □

Theorem 6.

If $f\left(z\right)\in {\mathcal{A}}_p$ satisfies

$$\left|\frac{\left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}{f\left(sz\right)-f\left(tz\right)}\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}-\frac{z{\left(f\left(sz\right)-f\left(tz\right)\right)}^{\prime }}{f\left(sz\right)-f\left(tz\right)}\right)\right|<p\left(\frac{1-\alpha }{\alpha }\right),$$

(15)

then $f\left(z\right)\in {\mathcal{N}}_{\alpha }^p\mathcal{S}\left(s,t\right),$ where $\frac{p}{p+1}<\alpha <1.$

Proof. 

Let

$$q\left(z\right)=\frac{p\left(f\left(sz\right)-f\left(tz\right)\right)}{\left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}.$$

(16)

Then $q\left(z\right)$ is clearly analytic in $\mathbb{E}$ such that $q\left(0\right)=1$. After logarithmic differentiation and some simple computation, we have

$$z{\left(\frac{1}{q\left(z\right)}\right)}^{\prime }q\left(z\right)=1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}-\frac{z{\left(f\left(sz\right)-f\left(tz\right)\right)}^{\prime }}{f\left(sz\right)-f\left(tz\right)}.$$

(17)

From (16) and (17), we find that

$$z{\left(\frac{1}{q\left(z\right)}\right)}^{\prime }=\frac{\left(s^p-t^p\right)z{f}^{\prime }\left(z\right)}{p\left(f\left(sz\right)-f\left(tz\right)\right)}\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}-\frac{z{\left(f\left(sz\right)-f\left(tz\right)\right)}^{\prime }}{f\left(sz\right)-f\left(tz\right)}\right).$$

Now by condition (15), we have

$$z{\left(\frac{1}{q\left(z\right)}\right)}^{\prime }\prec p\left(\frac{1-\alpha }{\alpha }\right)z=\mathsf{\Theta }\left(z\right),$$

where $\mathsf{\Theta }\left(0\right)=0.$ Applying Lemma 3, we have

$$\frac{1}{q\left(z\right)}\prec 1+\underset{0}{\overset{z}{{\displaystyle \int }}}\frac{\mathsf{\Theta }\left(t\right)}{t}dt=\frac{\alpha +p\left(1-\alpha \right)z}{\alpha },$$

which implies that

$$q\left(z\right)\prec \frac{\alpha }{\alpha +p\left(1-\alpha \right)z}=H\left(z\right).$$

(18)

We can write

$$\begin{array}{ccc}\hfill \mathrm{Re}\left(1+\frac{z{H}^{″}\left(z\right)}{H^{\prime }\left(z\right)}\right)& =& \mathrm{Re}\left(\frac{\alpha -p\left(1-\alpha \right)z}{\alpha +p\left(1-\alpha \right)z}\right)\hfill \\ & \ge & \frac{\alpha -p\left(1-\alpha \right)}{\alpha +p\left(1-\alpha \right)}.\hfill \end{array}$$

Now since $\frac{p}{1+p}<\alpha <1,$ therefore we have

$$\mathrm{Re}\left(1+\frac{z{H}^{″}\left(z\right)}{H^{\prime }\left(z\right)}\right)>0.$$

This shows that H is convex univalent in $\mathbb{E}$ and $H\left(\mathbb{E}\right)$ is symmetric about the real axis, therefore

$$\mathrm{Re}\left(H\left(z\right)\right)\ge H\left(1\right)\ge 0.$$

(19)

Combining (16), (18), and (19), we deduce that

$$\mathrm{Re}\left(q\left(z\right)\right)>\alpha ,$$

which implies that $f\left(z\right)\in {\mathcal{N}}_{\alpha }^p\mathcal{S}\left(s,t\right).$ □