Some Reciprocal Classes of Close-to-Convex and Quasi-Convex Analytic Functions

Abstract

The present paper comprises the study of certain functions which are analytic and defined in terms of reciprocal function. The reciprocal classes of close-to-convex functions and quasi-convex functions are defined and studied. Various interesting properties, such as sufficiency criteria, coefficient estimates, distortion results, and a few others, are investigated for these newly defined sub-classes.

1. Introduction

We denote by $\mathcal{A}$ the class of analytic functions on the unit disc $\mathbb{U}=\left\{z\in \mathbb{C}:\left|z\right|<1\right\}$ having the following taylor series representation:

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

(1)

The analytic function f will be subordinate to an analytic function g, if there exists an analytic function $w,$ known as a Schwarz function, with $w\left(0\right)=0$ and $\left|w\left(z\right)\right|<\left|z\right|$, such that $f\left(z\right)=g\left(w\right(z\left)\right)$. Moreover, if the function g is univalent in $\mathbb{U}$, then we have the following (see [1,2]):

$$\begin{array}{cccccc}\hfill f\left(z\right)\prec g\left(z\right),& z\in \mathbb{U}& ⟺\hfill & f\left(0\right)=g\left(0\right)& \mathrm{and}& f\left(\mathbb{U}\right)\subset g\left(\mathbb{U}\right).\end{array}$$

Uralegaddi et al. [3] introduced the reciprocal classes $\mathcal{M}\left(\gamma \right)$ of starlike and $\mathcal{N}\left(\gamma \right)$ of convex functions for $1\le \gamma \le {\displaystyle \frac{4}{3}}$, which were further studied by Owa et al. [4,5,6] for the values $\gamma \ge 1$. The classes $\mathcal{M}\left(\gamma \right)$ of starlike functions and $\mathcal{N}\left(\gamma \right)$ of reciprocal order convex functions $\gamma ,$ $\left(\gamma >1\right)$ are defined as follows:

$$\begin{array}{ccc}\hfill \mathcal{M}\left(\gamma \right)& =& \left\{f\in \mathcal{A}:\mathfrak{Re}{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}<\gamma ,z\in \mathbb{U}\right\},\hfill \\ \hfill \mathcal{N}\left(\gamma \right)& =& \left\{f\in \mathcal{A}:\mathfrak{Re}\left\{1+{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\right\}<\gamma ,z\in \mathbb{U}\right\}.\hfill \end{array}$$

Using the same concept, together with the idea of k-uniformly starlike and $\gamma$ ordered convex functions, Nishiwakiand Owa [7] defined the reciprocal classes of uniformly starlike $\mathcal{MD}\left(k,\gamma \right)$ and convex functions $\mathcal{ND}\left(k,\gamma \right)$. The class $\mathcal{MD}\left(k,\gamma \right)$ denotes the subclass of $\mathcal{A}$ consisting of functions f satisfying the inequality

$$\mathfrak{Re}{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}<k\left|{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}-1\right|+\gamma ,\left(z\in \mathbb{U}\right),$$

for some $\gamma \left(\gamma >1\right)$ and $k\left(k\le 0\right)$ and the class $\mathcal{ND}\left(k,\gamma \right)$ denotes the subclass of $\mathcal{A}$ consisting of functions $f\left(z\right)$ satisfying the inequality

$$\mathfrak{Re}{\displaystyle \frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{f^{\prime }\left(z\right)}}<\gamma +k\left|{\displaystyle \frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{f^{\prime }\left(z\right)}}-1\right|,\left(z\in \mathbb{U}\right),$$

for some $\gamma \left(\gamma >1\right)$ and $k\left(k\le 0\right)$. They also proved that the well-known Alexander relation holds between $\mathcal{MD}\left(k,\gamma \right)$ and $\mathcal{ND}\left(k,\gamma \right).$ This means that

$$\begin{array}{ccc}\hfill f\in \mathcal{ND}\left(k,\gamma \right)& \iff & z{f}^{\prime }\in \mathcal{MD}\left(k,\gamma \right).\hfill \end{array}$$

For a more detailed and recent study on uniformly convex and starlike functions, we refer the reader to [8,9,10,11,12].

Considering the above defined classes, we introduce the following classes.

Definition 1.

Let $f$ belong to $\mathcal{A}$. Then, it will belong to the class $\mathcal{KD}\left(\beta ,\gamma \right)$ if there exists $g\in \mathcal{MD}\left(\gamma \right)$ such that

$$\begin{array}{cc}\hfill \mathfrak{Re}\left\{{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}\right\}<\beta ,& (z\in \mathbb{U}),\end{array}$$

(2)

for some $\beta ,\gamma >1$.

Definition 2.

Let $f$ belong to $\mathcal{A}$. Then, it will belong to the class $\mathcal{QD}\left(\beta ,\gamma \right)$ if there exists $g\in \mathcal{ND}\left(\gamma \right)$ such that

$$\begin{array}{cc}\hfill \mathfrak{Re}\left\{{\displaystyle \frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{g^{\prime }\left(z\right)}}\right\}<\beta ,& (z\in \mathbb{U}),\end{array}$$

(3)

for some $\beta ,\gamma >1$.

It is clear, from (2) and (3), that

$$\begin{array}{ccc}\hfill f\left(z\right)\in \mathcal{QD}\left(\beta ,\gamma \right)& \iff & z{f}^{\prime }\left(z\right)\in \mathcal{KD}\left(\beta ,\gamma \right).\hfill \end{array}$$

Definition 3.

Let $f$ belong to $\mathcal{A}$. Then, it will belong to the class $\mathcal{KD}\left(k,\beta ,\gamma \right)$ if there exists $g\in \mathcal{MD}\left(k,\gamma \right)$ such that

$$\begin{array}{cc}\hfill ll\mathfrak{Re}\left\{{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}\right\}<k\left|{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}-1\right|+\beta ,& (z\in \mathbb{U}),\end{array}$$

(4)

for some $k\le 0$ and $\beta ,\gamma >1$.

Definition 4.

Let $f$ belong to $\mathcal{A}$. Then, it is said to be in the class $\mathcal{QD}\left(k,\beta ,\gamma \right)$ if there exists $g\in \mathcal{ND}\left(k,\gamma \right)$ such that

$$\begin{array}{cc}\hfill \mathfrak{Re}\left\{{\displaystyle \frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{g^{\prime }\left(z\right)}}\right\}<k\left|{\displaystyle \frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{g^{\prime }\left(z\right)}}-1\right|+\beta ,& (z\in \mathbb{U}),\end{array}$$

(5)

for some $k\le 0$ and $\beta ,\gamma >1$.

We can see, from (4) and (5), that the well-known relation of Alexander type holds between the classes $\mathcal{KD}\left(k,\beta ,\gamma \right)$ and $\mathcal{QD}\left(k,\beta ,\gamma \right)$, which means that

$$\begin{array}{ccc}\hfill f\left(z\right)\in \mathcal{QD}\left(k,\beta ,\gamma \right)& \iff & z{f}^{\prime }\left(z\right)\in \mathcal{KD}\left(k,\beta ,\gamma \right).\hfill \end{array}$$

2. Preliminary Lemmas

Lemma 1.

For positive integers t and σ, we have

$$\sigma \sum _{j=1}^t{\displaystyle \frac{{\left(\sigma \right)}_{j-1}}{(j-1)!}}={\displaystyle \frac{{\left(\sigma \right)}_t}{(t-1)!}},$$

(6)

where ${\left(\sigma \right)}_t$ is the Pochhammer symbol, defined by

$${\left(\sigma \right)}_t={\displaystyle \frac{\Gamma \left(\sigma +t\right)}{\Gamma \left(\sigma \right)}}=\sigma (\sigma +1)(\sigma +2)(\sigma +3)\cdots (\sigma +t-1).$$

Proof. 

Consider

$$\begin{array}{cc}& \sigma {\displaystyle \sum _{j=1}^t}{\displaystyle \frac{{\left(\sigma \right)}_{j-1}}{(j-1)!}}\hfill \\ =\hfill & \sigma \left(1+{\displaystyle \frac{\sigma }{1}}+{\displaystyle \frac{{\left(\sigma \right)}_2}{2!}}+{\displaystyle \frac{{\left(\sigma \right)}_3}{3!}}+{\displaystyle \frac{{\left(\sigma \right)}_4}{4!}}+\cdots +{\displaystyle \frac{{\left(\sigma \right)}_{t-1}}{(t-1)!}}\right)\hfill \\ =\hfill & \sigma (1+\sigma )\left(1+{\displaystyle \frac{\sigma }{2}}+{\displaystyle \frac{\sigma (\sigma +2)}{2\times 3}}+\cdots +{\displaystyle \frac{\sigma (\sigma +2)\cdots (\sigma +t-2)}{2\times \cdots \times (t-1)}}\right)\hfill \\ =\hfill & \sigma (1+\sigma ){\displaystyle \frac{(\sigma +2)}{2}}\left(1+{\displaystyle \frac{\sigma }{3}}+\cdots +{\displaystyle \frac{\sigma (\sigma +3)\cdots (\sigma +t-2)}{3\times 4\times \cdots \times (t-1)}}\right)\hfill \\ =\hfill & \sigma (1+\sigma ){\displaystyle \frac{(\sigma +2)}{2}}{\displaystyle \frac{(\sigma +3)}{3}}\left(1+{\displaystyle \frac{\sigma }{4}}+\cdots +{\displaystyle \frac{\sigma (\sigma +4)\cdots (\sigma +t-2)}{4\times \cdots \times (t-1)}}\right)\hfill \\ =\hfill & \sigma (1+\sigma ){\displaystyle \frac{(\sigma +2)}{2}}{\displaystyle \frac{(\sigma +3)}{3}}{\displaystyle \frac{(\sigma +4)}{4}}\left(1+{\displaystyle \frac{\sigma }{5}}+\cdots +{\displaystyle \frac{\sigma \cdots (\sigma +t-2)}{5\times 6\times \cdots \times (t-1)}}\right)\hfill \\ =\hfill & \sigma (1+\sigma ){\displaystyle \frac{(\sigma +2)}{2}}{\displaystyle \frac{(\sigma +3)}{3}}{\displaystyle \frac{(\sigma +4)}{4}}\cdots \left(1+{\displaystyle \frac{\sigma }{t-1}}\right)\hfill \\ =\hfill & \sigma (1+\sigma ){\displaystyle \frac{(\sigma +2)}{2}}{\displaystyle \frac{(\sigma +3)}{3}}{\displaystyle \frac{(\sigma +4)}{4}}\cdots \left({\displaystyle \frac{\sigma +(t-1)}{t-1}}\right)\hfill \\ =\hfill & {\displaystyle \frac{{\left(\sigma \right)}_t}{(t-1)!}}.\hfill \end{array}$$

 □

Lemma 2.

If $f\left(z\right)\in \mathcal{MD}\left(k,\gamma \right)$, then

$$f\left(z\right)\in \mathcal{MD}\left({\displaystyle \frac{\gamma -k}{1-k}}\right).$$

Proof. 

Using the definition, we write

$$\begin{array}{ccc}\hfill \mathfrak{Re}{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}& <& k\left|{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}-1\right|+\gamma \hfill \\ & \le & k\mathfrak{Re}{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}+\gamma -k,\hfill \end{array}$$

which implies that

$$\left(1-k\right)\mathfrak{Re}{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}<\gamma -k.$$

After simplification, we obtain

$$\mathfrak{Re}{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}<{\displaystyle \frac{\gamma -k}{1-k}},\left(k\le 0,\gamma >1\right).$$

As ${\displaystyle \frac{\gamma -k}{1-k}}>1,$ we have $f\left(z\right)\in \mathcal{MD}\left({\displaystyle \frac{\gamma -k}{1-k}}\right)$. With this, we obtain the required result. □

Lemma 3.

If $f$ belongs to the class $\mathcal{MD}\left(k,\gamma \right)$, then

$$|a_n|\le {\displaystyle \frac{{\left({\delta }_{k,\gamma }\right)}_{n-1}}{\left(n-1\right)!}},$$

(7)

where

$${\delta }_{k,\gamma }={\displaystyle \frac{2(\gamma -1)}{1-k}}.$$

(8)

Proof. 

Let us define a function

$$p\left(z\right)={\displaystyle \frac{\left(\gamma -k\right)-\left(1-k\right)\left({\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\right)}{\gamma -1}},$$

(9)

where $p\in \mathcal{P}$, the class of Caratheodory functions (see [1]). One may write

$${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}={\displaystyle \frac{\left(\gamma -k\right)-\left(\gamma -1\right)p\left(z\right)}{1-k}},$$

(10)

or

$$z{f}^{\prime }\left(z\right)=\left({\displaystyle \frac{\gamma -k}{1-k}}-{\displaystyle \frac{\gamma -1}{1-k}}p\left(z\right)\right)f\left(z\right).$$

(11)

Let us write $p\left(z\right)$ as $p\left(z\right)=1+{\displaystyle \sum _{n=1}^{\infty }}{p}_n{z}^n$ and let f have the series form, as in (1). Then, (11) can be written as

$$\sum _{n=1}^{\infty }n{a}_n{z}^n=\left(\sum _{n=1}^{\infty }{a}_n{z}^n\right)\left({\displaystyle \frac{\gamma -k}{1-k}}-{\displaystyle \frac{\gamma -1}{1-k}}\left(1+\sum _{n=1}^{\infty }{p}_n{z}^n\right)\right),a_1=1$$

which reduces to

$$\begin{array}{ccc}\hfill \sum _{n=1}^{\infty }n{a}_n{z}^n& =& \left(\sum _{n=1}^{\infty }{a}_n{z}^n\right)\left(1-{\displaystyle \frac{\gamma -1}{1-k}}\sum _{n=1}^{\infty }{p}_n{z}^n\right)\hfill \\ & =& \sum _{n=1}^{\infty }{a}_n{z}^n-{\displaystyle \frac{\gamma -1}{1-k}}\left(\sum _{n=1}^{\infty }{a}_n{z}^n\right)\left(\sum _{n=1}^{\infty }{p}_n{z}^n\right).\hfill \end{array}$$

This implies that

$$\sum _{n=1}^{\infty }\left(n-1\right)a_n{z}^n=-{\displaystyle \frac{\gamma -1}{1-k}}\sum _{n=1}^{\infty }\left(\sum _{j=0}^{n-1}{a}_j{p}_{n-j}\right)z^n.$$

After comparing the $n^{th}$ term’s coefficients, appearing on both sides, combined with the fact that $a_0=0$, we obtain

$$a_n={\displaystyle \frac{-\left(\gamma -1\right)}{\left(n-1\right)\left(1-k\right)}}\sum _{j=1}^{n-1}{a}_j{p}_{n-j}.$$

Now, we take the absolute value and then apply the triangle inequality to get

$$|a_n|\le {\displaystyle \frac{\gamma -1}{\left(n-1\right)(1-k)}}\sum _{j=1}^{n-1}\left|a_j\right|\left|p_{n-j}\right|.$$

Applying the coefficient estimates, such that $\left|p_n\right|\le 2$ $(n\ge 1)$ for Caratheodory functions [1], we obtain

$$|a_n|\le {\displaystyle \frac{2(\gamma -1)}{\left(n-1\right)(1-k)}}\sum _{j=1}^{n-1}\left|a_j\right|.$$

$$|a_n|\le {\displaystyle \frac{{\delta }_{k,\gamma }}{n-1}}\sum _{j=1}^{n-1}\left|a_j\right|,$$

(12)

where ${\delta }_{k,\gamma }={\displaystyle \frac{2(\gamma -1)}{1-k}}$. We prove (7) by induction on n. Thus, first for $n=2$, we obtain the following from (12):

$$\left|a_2\right|\le {\displaystyle \frac{{\delta }_{k,\gamma }}{1}}={\displaystyle \frac{{\left({\delta }_{k,\gamma }\right)}_{2-1}}{\left(2-1\right)!}}.$$

(13)

This proves that, for $n=2,$ (7) is true. For $n=3$, we obtain

$$\left|a_3\right|\le {\displaystyle \frac{{\delta }_{k,\gamma }}{2}}\left(1+\left|a_2\right|\right)={\displaystyle \frac{{\delta }_{k,\gamma }\left(1+{\delta }_{k,\gamma }\right)}{2}}={\displaystyle \frac{{\left({\delta }_{k,\gamma }\right)}_{3-1}}{\left(3-1\right)!}}.$$

This proves that when $n=3$, (7) holds true. Now, we assume that for $t\le n,$ (7) is true, that means

$$\left|a_t\right|\le {\displaystyle \frac{{\left({\delta }_{k,\gamma }\right)}_{t-1}}{\left(t-1\right)!}}t=1,2,\dots ,n.$$

(14)

Using (12) and (14), we have

$$\left|a_{t+1}\right|\le {\displaystyle \frac{{\delta }_{k,\gamma }}{t}}\sum _{j=1}^t\left|a_j\right|\le {\displaystyle \frac{{\delta }_{k,\gamma }}{t}}\sum _{j=1}^t{\displaystyle \frac{{\left({\delta }_{k,\gamma }\right)}_{j-1}}{\left(j-1\right)!}}.$$

After applying (6), we obtain

$$\left|a_{t+1}\right|\le {\displaystyle \frac{1}{t}}{\displaystyle \frac{{\left({\delta }_{k,\gamma }\right)}_t}{(t-1)!}}={\displaystyle \frac{{\left({\delta }_{k,\gamma }\right)}_t}{t!}}.$$

As a result of mathematical induction, it is shown that (7) is true for all $n\ge 2$. Hence, the required bound is obtained. □

Lemma 4

([13]). Let w be analytic in $\mathbb{U}$ with $w\left(0\right)=0.$ If there exists $z_0\in \mathbb{U}$ such that

$$\underset{\left|z\right|\le \left|z_0\right|}{max}\left|w\left(z\right)\right|=\left|w\left(z_0\right)\right|,$$

then

$$z_0{w}^{\prime }\left(z_0\right)=cw\left(z_0\right),$$

where c is real and $c\ge 1$.

3. Main Results

Theorem 1.

If $f\left(z\right)\in \mathcal{KD}\left(k,\beta ,\gamma \right)$, then

$$f\left(z\right)\in \mathcal{KD}\left({\displaystyle \frac{\beta -k}{1-k}},\gamma \right).$$

Proof. 

If $f\left(z\right)\in \mathcal{KD}\left(k,\beta ,\gamma \right)$, then $k\le 0$, $\beta >1$, and so we obtain

$$\begin{array}{ccc}\hfill \mathfrak{Re}\left\{{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}\right\}& <& k\left|{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}-1\right|+\beta \hfill \\ & \le & \beta +k\mathfrak{Re}\left\{{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}-1\right\},\hfill \end{array}$$

which leads to

$$\mathfrak{Re}\left\{{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}\right\}-k\mathfrak{Re}\left\{{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}\right\}<-k+\beta .$$

After simplification, we obtain

$$\mathfrak{Re}\left\{{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}\right\}<{\displaystyle \frac{\beta -k}{1-k}},\left(k\le 0,\beta >1\right).$$

(15)

This completes the proof. □

In a similar way, one can easily prove the following important result.

Theorem 2.

If $f\in \mathcal{QD}\left(k,\beta ,\gamma \right)$, then

$$f\in \mathcal{QD}\left({\displaystyle \frac{\beta -k}{1-k}},\gamma \right).$$

Theorem 3.

If $f\left(z\right)\in \mathcal{KD}\left(k,\beta ,\gamma \right)$, then

$$|a_n|\le {\displaystyle \frac{{\left({\delta }_{k,\gamma }\right)}_{n-1}}{n!}}+{\displaystyle \frac{\left|{\delta }_{k,\beta }\right|}{n}}\sum _{j=1}^{n-1}{\displaystyle \frac{{\left({\delta }_{k,\gamma }\right)}_{j-1}}{\left(j-1\right)!}},$$

where ${\delta }_{k,\gamma }$ is given by (8) and

$${\delta }_{k,\beta }={\displaystyle \frac{2\left(\beta -1\right)}{1-k}}.$$

(16)

Proof. 

If f is in the class $\mathcal{KD}(k,\beta ,\gamma ),$ then there exists $g\left(z\right)\in \mathcal{MD}\left(k,\gamma \right)$ such that the function

$$p\left(z\right)={\displaystyle \frac{\left(\beta -k\right)-\left(1-k\right)\left({\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}\right)}{\beta -1}}$$

(17)

belongs to $\mathcal{P}.$ Therefore, we write

$$z{f}^{\prime }\left(z\right)={\displaystyle \frac{\beta -k}{1-k}}g\left(z\right)-{\displaystyle \frac{\beta -1}{1-k}}g\left(z\right)p\left(z\right).$$

(18)

Let us write $p\left(z\right)$ as $p\left(z\right)=1+{\sum }_{n=1}^{\infty }{p}_n{z}^n$, $g\left(z\right)$ as $g\left(z\right)=z+{\sum }_{n=2}^{\infty }{b}_n{z}^n$, and let $f\left(z\right)$ have the series form as in (1). Then, (18) can be written as

$$z+\sum _{n=2}^{\infty }n{a}_n{z}^n={\displaystyle \frac{\beta -k}{1-k}}\left(z+\sum _{n=2}^{\infty }{b}_n{z}^n\right)-{\displaystyle \frac{\beta -1}{1-k}}\left(1+\sum _{n=1}^{\infty }{p}_n{z}^n\right)\left(z+\sum _{n=2}^{\infty }{b}_n{z}^n\right).$$

Comparing the $nth$ term’s coefficients on both sides, we obtain

$$n{a}_n=b_n-{\displaystyle \frac{\beta -1}{1-k}}\left[p_{n-1}+p_{n-2}{b}_2+p_{n-3}{b}_3+\dots +p_1{b}_{n-1}\right].$$

By taking the absolute value, we get

$$\begin{array}{ccc}\hfill n|a_n|& =& \left|b_n-{\displaystyle \frac{\beta -1}{1-k}}\left[p_{n-1}+p_{n-2}{b}_2+p_{n-3}{b}_3+\dots +p_1{b}_{n-1}\right]\right|\hfill \\ & \le & \left|b_n\right|+{\displaystyle \frac{\beta -1}{1-k}}\left|p_{n-1}+p_{n-2}{b}_2+p_{n-3}{b}_3+\dots +p_1{b}_{n-1}\right|.\hfill \end{array}$$

Applying the triangle inequality, we obtain

$$n|a_n|\le \left|b_n\right|+{\displaystyle \frac{\beta -1}{1-k}}\left\{\left|p_{n-1}\right|+\left|p_{n-2}{b}_2\right|+\left|p_{n-3}{b}_3\right|+\dots +\left|p_1{b}_{n-1}\right|\right\}.$$

(19)

As $\mathfrak{Re}\left\{p\left(z\right)\right\}>0$ in $\mathbb{U}$, we have $\left|p_n\right|\le 2(n\ge 1)$ (see [1]). Then, from (19), we have

$$n|a_n|\le \left|b_n\right|+{\displaystyle \frac{2\left(\beta -1\right)}{1-k}}\sum _{j=1}^{n-1}\left|b_j\right|,$$

where $b_1=1$. Using Lemma (3), we obtain

$$n|a_n|\le {\displaystyle \frac{{\left({\delta }_{k,\gamma }\right)}_{n-1}}{\left(n-1\right)!}}+{\delta }_{k,\beta }\sum _{j=1}^{n-1}{\displaystyle \frac{{\left({\delta }_{k,\gamma }\right)}_{j-1}}{\left(j-1\right)!}},$$

where ${\delta }_{k,\beta }=$ ${\displaystyle \frac{2\left(\beta -1\right)}{1-k}}$ and ${\delta }_{k,\gamma }$ is defined by (8). This can be written as

$$|a_n|\le {\displaystyle \frac{{\left({\delta }_{k,\gamma }\right)}_{n-1}}{n!}}+{\displaystyle \frac{{\delta }_{k,\beta }}{n}}\sum _{j=1}^{n-1}{\displaystyle \frac{{\left({\delta }_{k,\gamma }\right)}_{j-1}}{\left(j-1\right)!}}.$$

This completes the proof. □

From Definition 4 and Theorem 2, we immediately get the following corollary.

Corollary 1.

If $f\left(z\right)\in \mathcal{QD}(k,\beta ,\gamma )$, then

$$|a_n|\le {\displaystyle \frac{1}{n}}\left[{\displaystyle \frac{{\left({\delta }_{k,\gamma }\right)}_{n-1}}{n!}}+{\displaystyle \frac{{\delta }_{k,\beta }}{n}}\sum _{j=1}^{n-1}{\displaystyle \frac{{\left({\delta }_{k,\gamma }\right)}_{j-1}}{\left(j-1\right)!}}\right],$$

where ${\delta }_{k,\beta }$ and ${\delta }_{k,\gamma }$ are given by (16) and (8), respectively.

By taking $k=0$ in the above results, we obtain the coefficient inequality for the classes $\mathcal{KD}(\beta ,\gamma )$ and $\mathcal{QD}(\beta ,\gamma ).$

Theorem 4.

If a function $f\in \mathcal{KD}(k,\beta ,\gamma )$, then there exists $g\in \mathcal{MD}(k,\gamma )$ such that

$${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}\prec 1+2\left({\beta }_1-1\right)-{\displaystyle \frac{2\left({\beta }_1-1\right)}{1-z}},(z\in \mathbb{U}),$$

(20)

where

$${\beta }_1={\displaystyle \frac{\beta -k}{1-k}}.$$

(21)

Proof. 

Let $f\left(z\right)\in \mathcal{KD}(k,\beta ,\gamma )$. Then, there exists $g\left(z\right)$ in $\mathcal{MD}(k,\gamma )$ and a Schwarz function $w\left(z\right)$ such that

$${\displaystyle \frac{{\beta }_1-\left({\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}\right)}{{\beta }_1-1}}={\displaystyle \frac{1+w\left(z\right)}{1-w\left(z\right)}},$$

(22)

as $w\left(z\right)$ is analytic $\mathbb{U}$ with $w\left(0\right)=0$ and

$$\Re e\left({\displaystyle \frac{1+w\left(z\right)}{1-w\left(z\right)}}\right)>0,(z\in \mathbb{U}).$$

So, from (22), we obtain

$$\begin{array}{ccc}\hfill \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}& =& {\beta }_1-\left({\beta }_1-1\right)\left(\frac{1+w\left(z\right)}{1-w\left(z\right)}\right)\hfill \\ & =& \frac{{\beta }_1\left(1-w\left(z\right)\right)-\left({\beta }_1-1\right)\left(1+w\left(z\right)\right)}{1-w\left(z\right)}\hfill \\ & =& \frac{1+w\left(z\right)-2{\beta }_{1}w\left(z\right)}{1-w\left(z\right)}\hfill \\ & =& \frac{1-w\left(z\right)-2\left({\beta }_1-1\right)w\left(z\right)}{1-w\left(z\right)}\hfill \\ & =& \frac{1-w\left(z\right)+2\left({\beta }_1-1\right)-2\left({\beta }_1-1\right)w\left(z\right)-2\left({\beta }_1-1\right)}{1-w\left(z\right)}\hfill \\ & =& \frac{1-w\left(z\right)+2\left({\beta }_1-1\right)\left(1-w\left(z\right)\right)-2\left({\beta }_1-1\right)}{1-w\left(z\right)}.\hfill \end{array}$$

This implies that

$${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}=1+2\left({\beta }_1-1\right)-{\displaystyle \frac{2\left({\beta }_1-1\right)}{1-w\left(z\right)}},$$

and hence

$${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}\prec 1+2\left({\beta }_1-1\right)-{\displaystyle \frac{2\left({\beta }_1-1\right)}{1-z}},(z\in \mathbb{U}),$$

which is as required in (20). □

Corollary 2.

If $f\in \mathcal{QD}(k,\beta ,\gamma )$, then there exists $g\in \mathcal{ND}\left(k,\gamma \right)$ such that

$${\displaystyle \frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{g^{\prime }\left(z\right)}}\prec 1+2\left({\beta }_1-1\right)-{\displaystyle \frac{2\left({\beta }_1-1\right)}{\left(1-z\right)}},(z\in \mathbb{U}),$$

(23)

where ${\beta }_1$ is given by (21).

Theorem 5.

If $f\in \mathcal{KD}(k,\beta ,\gamma )$, then there exists a function $g\in \mathcal{MD}\left(k,\gamma \right)$ such that

$${\displaystyle \frac{1-(2{\beta }_1-1)r}{1-r}}\le \mathfrak{Re}{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}\le {\displaystyle \frac{1+(2{\beta }_1-1)r}{1+r}},$$

(24)

where $\left|z\right|=r<1$ and ${\beta }_1$ is given by (21).

Proof. 

Using Theorem 4, we define the function $\varphi$ as follows

$$\varphi \left(z\right)=1+2\left({\beta }_1-1\right)+{\displaystyle \frac{2\left(1-{\beta }_1\right)}{1-z}},\left(z\in \mathbb{U}\right).$$

Letting $z=r{e}^{i\theta }(0\le r<1)$, we observe that

$$\mathfrak{Re}\varphi \left(z\right)=1+2\left({\beta }_1-1\right)+{\displaystyle \frac{2\left(1-{\beta }_1\right)\left(1-rcos\theta \right)}{1+r^2-2rcos\theta }}.$$

Let us define

$$\psi \left(t\right)={\displaystyle \frac{1-rt}{1+r^2-2rt}},\left(t=cos\theta \right).$$

As ${\psi }^{\prime }\left(t\right)={\displaystyle \frac{r\left(1-r^2\right)}{{\left(1+r^2-2rt\right)}^2}}\ge 0$ (since $r<1$), we get

$$1+2\left({\beta }_1-1\right)+{\displaystyle \frac{2\left(1-{\beta }_1\right)}{1-r}}\le \mathfrak{Re}\varphi \left(z\right)\le 1+2\left({\beta }_1-1\right)+{\displaystyle \frac{2\left(1-{\beta }_1\right)}{1+r}}.$$

After simplification, we have

$${\displaystyle \frac{1-(2{\beta }_1-1)r}{1-r}}\le \mathfrak{Re}\varphi \left(z\right)\le {\displaystyle \frac{1+(2{\beta }_1-1)r}{1+r}}.$$

With the fact that ${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}\prec \varphi \left(z\right),$ $\left(z\in \mathbb{U}\right)$ and as $\varphi$ is univalent in $\mathbb{U}$, by using (22), we get the required result. □

Corollary 3.

If $f\in \mathcal{QD}(k,\beta ,\gamma )$, then there exists $g\in \mathcal{ND}\left(k,\gamma \right)$ such that

$${\displaystyle \frac{1-(2{\beta }_1-1)r}{1-r}}\le \mathfrak{Re}{\displaystyle \frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{g^{\prime }\left(z\right)}}\le {\displaystyle \frac{1+(2{\beta }_1-1)r}{1+r}},$$

(25)

where $\left|z\right|=r<1$ and ${\beta }_1$ is given by (21).

Theorem 6.

Assume that a function $f\in \mathcal{A}$ satisfies

$$\mathfrak{Re}\left({\displaystyle \frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}}-{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\right)>{\displaystyle \frac{{\beta }_1+1}{2{\beta }_1}},(z\in \mathbb{U}),$$

(26)

for some $g\left(z\right)\in \mathcal{MD}\left(k,\gamma \right)$ and for real ${\beta }_1$ given by (21). If

$$\varphi \left(z\right)={\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}$$

is analytic in $\mathbb{U}$ and $\varphi \left(z\right)\ne 0$ and $\varphi \left(z\right)\ne 2{\beta }_1-1$ in $\mathbb{U}$, then $f\in \mathcal{KD}(k,{\beta }_1)$.

Proof. 

Let us define a function $w\left(z\right)$ by

$$w\left(z\right)={\displaystyle \frac{\varphi \left(z\right)-1}{\varphi \left(z\right)+\left(1-2{\beta }_1\right)}},z\in \mathbb{U}.$$

Then, $w\left(z\right)$ is analytic in $\mathbb{U}$ as $\varphi \left(z\right)\ne 2{\beta }_1-1$ and

$$\varphi \left(z\right)={\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}={\displaystyle \frac{1+\left(1-2{\beta }_1\right)w\left(z\right)}{1-w\left(z\right)}}.$$

(27)

Because $\varphi \left(z\right)\ne 0$, we use logarithmic differentiation to get

$${\displaystyle \frac{1}{z}}+{\displaystyle \frac{f^{″}\left(z\right)}{f^{\prime }\left(z\right)}}-{\displaystyle \frac{g^{\prime }\left(z\right)}{g\left(z\right)}}={\displaystyle \frac{\left(1-2{\beta }_1\right)w^{\prime }\left(z\right)}{1+\left(1-2{\beta }_1\right)w\left(z\right)}}+{\displaystyle \frac{w^{\prime }\left(z\right)}{1-w\left(z\right)}},$$

which further yields

$${\displaystyle \frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}}-{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}=1-{\displaystyle \frac{\left(1-2{\beta }_1\right)z{w}^{\prime }\left(z\right)}{1+\left(1-2{\beta }_1\right)w\left(z\right)}}-{\displaystyle \frac{z{w}^{\prime }\left(z\right)}{1-w\left(z\right)}}.$$

(28)

Then, we note that w is analytic in open unit disk and $w\left(0\right)=0$. Therefore, from (28), we obtain

$$\begin{array}{ccc}\hfill \mathfrak{Re}\left({\displaystyle \frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}}-{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\right)& =& \mathfrak{Re}\left(1-{\displaystyle \frac{\left(1-2{\beta }_1\right)z{w}^{\prime }\left(z\right)}{1+\left(1-2{\beta }_1\right)w\left(z\right)}}-{\displaystyle \frac{z{w}^{\prime }\left(z\right)}{1-w\left(z\right)}}\right)\hfill \\ & >& {\displaystyle \frac{{\beta }_1+1}{2{\beta }_1}}.\hfill \end{array}$$

Suppose there exists a point $z_0\in \mathbb{U}$ such that

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

then, by Lemma 4, we can write $w\left(z_0\right)=e^{i\theta }$ and $z_0{w}^{\prime }\left(z_0\right)=c{e}^{i\theta }$ for a point $z_0$, and we have

$$\begin{array}{cc}& \mathfrak{Re}\left({\displaystyle \frac{z_0{g}^{\prime }\left(z_0\right)}{g\left(z_0\right)}}-{\displaystyle \frac{z_0{f}^{″}\left(z_0\right)}{f^{\prime }\left(z_0\right)}}\right)\hfill \\ =& \mathfrak{Re}\left(1-{\displaystyle \frac{\left(1-2{\beta }_1\right)c{e}^{i\theta }}{1+\left(1-2{\beta }_1\right)e^{i\theta }}}-{\displaystyle \frac{c{e}^{i\theta }}{1-e^{i\theta }}}\right)\hfill \\ =& \mathfrak{Re}\left(1-{\displaystyle \frac{c\left(1-2{\beta }_1\right)\left(e^{i\theta }+\left(1-2{\beta }_1\right)\right)}{1+{\left(1-2{\beta }_1\right)}^2+2\left(1-2{\beta }_1\right)cos\theta }}+{\displaystyle \frac{c\left(1-e^{i\theta }\right)}{2\left(1-cos\theta \right)}}\right)\hfill \\ =& 1+{\displaystyle \frac{c\left(2{\beta }_1-1\right)\left[cos\theta +\left(1-2{\beta }_1\right)\right]}{1+{\left(1-2{\beta }_1\right)}^2+2\left(1-2{\beta }_1\right)cos\theta }}+{\displaystyle \frac{c}{2}}\hfill \\ \le & 1-{\displaystyle \frac{c\left(2{\beta }_1-1\right)}{2{\beta }_1}}+{\displaystyle \frac{c}{2}}\hfill \\ =& 1-{\displaystyle \frac{c\left({\beta }_1-1\right)}{2{\beta }_1}}\hfill \\ \le & 1-{\displaystyle \frac{{\beta }_1-1}{2{\beta }_1}},\mathrm{as}c<1\hfill \\ =& {\displaystyle \frac{{\beta }_1+1}{2{\beta }_1}},\hfill \end{array}$$

which gives that

$$\mathfrak{Re}\left\{{\displaystyle \frac{z_0{g}^{\prime }\left(z_0\right)}{g\left(z_0\right)}}-{\displaystyle \frac{z_0{f}^{″}\left(z_0\right)}{f^{\prime }\left(z_0\right)}}\right\}\le {\displaystyle \frac{{\beta }_1+1}{2{\beta }_1}},$$

which is the contradiction to the supposed condition (26). Hence, there is no $z_0\in \mathbb{U}$ such that $\left|w\left(z_0\right)\right|=1$. This implies that $\left|w\left(z\right)\right|<1,(z\in \mathbb{U})$ and, therefore, by (27), we have

$${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}\prec {\displaystyle \frac{1-(2{\beta }_1-1)z}{1-z}}$$

or

$$\mathfrak{Re}\left\{{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}\right\}<{\beta }_1,z\in \mathbb{U}.$$

Hence, we conclude that $f\left(z\right)\in \mathcal{KD}(k,{\beta }_1)$. □

Theorem 7.

Assume that $k\le 0$ and $\beta >1$. If $f\in \mathcal{A}$ and if there exists $g\in \mathcal{MD}\left(k,\gamma \right)$ such that

$$\left|{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}-1\right|<{\displaystyle \frac{\beta -1}{1-k}}z\in \mathbb{U},$$

(29)

then $f\in \mathcal{KD}(k,\beta ,\gamma )$.

Proof. 

We have

$$\begin{array}{cc}& \left|{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}-1\right|<{\displaystyle \frac{\beta -1}{1-k}}\hfill \\ \Rightarrow & (1-k)\left|{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}-1\right|+1<\beta \hfill \\ \Rightarrow & \left|{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}-1\right|+1<k\left|{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}-1\right|+\beta \hfill \\ \Rightarrow & \mathfrak{Re}{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}<k\left|{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}-1\right|+\beta \hfill \\ \Rightarrow & f\in \mathcal{KD}(k,\beta ,\gamma ).\hfill \end{array}$$

 □

Corollary 4.

Let $f\in \mathcal{A}$ have the form (1). Assume that $g=z+b_2{z}^2+\cdots$ belongs to the class $\mathcal{MD}\left(k,\gamma \right)$ and satisfies

$$\left|{\displaystyle \frac{{\sum }_{n=2}^{\infty }(n{a}_n-b_n)z^{n-1}}{1+{\sum }_{n=2}^{\infty }{b}_n{z}^{n-1}}}\right|<{\displaystyle \frac{\beta -1}{1-k}}z\in \mathbb{U},$$

(30)

for some $k\left(k\le 0\right)$, $\beta \left(\beta >1\right)$.

Then, $f\left(z\right)\in \mathcal{KD}(k,\beta ,\gamma )$.

Proof. 

We have

$$\begin{array}{cc}& \left|{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{g\left(z\right)}}-1\right|\hfill \\ =& \left|{\displaystyle \frac{z+{\displaystyle \sum _{n=2}^{\infty }}n{a}_n{z}^n}{z+{\displaystyle \sum _{n=2}^{\infty }}{b}_n{z}^n}}-1\right|\hfill \\ =& \left|{\displaystyle \frac{{\sum }_{n=2}^{\infty }(n{a}_n-b_n)z^{n-1}}{1+{\sum }_{n=2}^{\infty }{b}_n{z}^{n-1}}}\right|\hfill \\ <& {\displaystyle \frac{\beta -1}{1-k}},\hfill \end{array}$$

and hence (29) follows immediately from (30). □

Theorem 8.

Let $f\in \mathcal{A}$ have the form (1) and let $g=z+{\sum }_{n=2}^{\infty }{b}_n{z}^n$, belonging to the class $\mathcal{MD}\left(k,\gamma \right)$, satisfy

$$1+\sum _{n=2}^{\infty }\left(n\left|a_n\right|+y\left|b_n\right|\right)<yz\in \mathbb{U},$$

(31)

for some $k\left(k\le 0\right)$, $\beta \left(\beta >1\right)$ and where

$$y={\displaystyle \frac{(\beta -1)}{(1-k)}}>0.$$

Then, $f\left(z\right)\in \mathcal{KD}(k,\beta ,\gamma )$.

Proof. 

Consider

$$\begin{array}{cc}& 1+\sum _{n=2}^{\infty }\left(n\left|a_n\right|+y\left|b_n\right|\right)<y\hfill \\ \Rightarrow & 1+\sum _{n=2}^{\infty }n\left|a_n\right|<y-y\sum _{n=2}^{\infty }\left|b_n\right|\hfill \\ \Rightarrow & 0<y-y\sum _{n=2}^{\infty }\left|b_n\right|\hfill \\ \Rightarrow & 0<y-y\sum _{n=2}^{\infty }|b_n\left|\right|z^{n-1}|\hfill \end{array}$$

(32)

$$\begin{array}{cc}\Rightarrow & 0<y\left|1+\sum _{n=2}^{\infty }{b}_n{z}^{n-1}\right|.\hfill \end{array}$$

(33)

We have

$$\begin{array}{cc}& 1+\sum _{n=2}^{\infty }\left(n\left|a_n\right|+y\left|b_n\right|\right)<y\hfill \\ \Rightarrow & 1+\sum _{n=2}^{\infty }n\left|a_n\right|<y-y\sum _{n=2}^{\infty }\left|b_n\right|\hfill \\ \Rightarrow & 1+\sum _{n=2}^{\infty }n\left|a_n\right||z^{n-1}|<y-y\sum _{n=2}^{\infty }|b_n\left|\right|z^{n-1}|\hfill \\ \Rightarrow & \left|1+\sum _{n=2}^{\infty }n{a}_n{z}^{n-1}\right|<y\left|1+\sum _{n=2}^{\infty }{b}_n{z}^{n-1}\right|\hfill \\ \Rightarrow & \left|{\displaystyle \frac{1+{\sum }_{n=2}^{\infty }n{a}_n{z}^{n-1}}{1+{\sum }_{n=2}^{\infty }{b}_n{z}^{n-1}}}\right|<y,\hfill \end{array}$$

from (33). By (30), it follows that $f\in \mathcal{KD}(k,\beta ,\gamma )$. □