Aspects of univalent holomorphic functions involving multiplier transformation and Ruscheweyh derivative

Abstract

Making use multiplier transformation and Ruscheweyh derivative,we introduce a new class of analytic functions $\mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$ defined on the open unit disc, and investigate its various characteristics. Further we obtain distortion bounds, extreme points and radii of close-to-convexity, starlikeness and convexity and neighborhood property for functions belonging to the class $\mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$.

MSC:30C45, 30A20, 34A40.

1 Introduction

Let denote the class of functions of the form $f(z)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_j{z}^j$, which are analytic and univalent in the open unit disc $U=\{z:z\in \mathbb{C}:|z|<1\}$. is a subclass of consisting of the functions of the form $f(z)=z-{\sum }_{j=2}^{\mathrm{\infty }}|a_j|z^j$. For functions $f,g\in \mathcal{A}$ given by $f(z)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_j{z}^j$, $g(z)=z+{\sum }_{j=2}^{\mathrm{\infty }}{b}_j{z}^j$, we define the Hadamard product (or convolution) of f and g by $(f\ast g)(z)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_j{b}_j{z}^j$, $z\in U$.

Definition 1.1 (Ruscheweyh [1])

For $f\in \mathcal{A}$, $n\in \mathbb{N}$, the operator $R^n$ is defined by $R^n:\mathcal{A}\to \mathcal{A}$,

$$\begin{array}{c}{R}^{0}f(z)=f(z),\hfill \\ R^{1}f(z)=z{f}^{\prime }(z),\dots ,\hfill \\ (n+1)R^{n+1}f(z)=z{(R^{n}f(z))}^{\prime }+n{R}^{n}f(z),z\in U.\hfill \end{array}$$

Remark 1.1 If $f\in \mathcal{A}$, $f(z)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_j{z}^j$, then $R^{n}f(z)=z+{\sum }_{j=2}^{\mathrm{\infty }}\frac{(n+j-1)!}{n!(j-1)!}{a}_j{z}^j$, $z\in U$.

If $f\in \mathcal{T}$, $f(z)=z-{\sum }_{j=2}^{\mathrm{\infty }}{a}_j{z}^j$, then $R^{n}f(z)=z-{\sum }_{j=t+1}^{\mathrm{\infty }}\frac{(n+j-1)!}{n!(j-1)!}{a}_j{z}^j$, $z\in U$.

Definition 1.2 [2, 3]

For $f\in \mathcal{A}$, $n\in \mathbb{N}\cup \{0\}$, $\lambda ,l\ge 0$, the operator $I(n,\lambda ,l)f(z)$ is defined by the following infinite series:

$$I(n,\lambda ,l)f(z):=z+\sum _{j=2}^{\mathrm{\infty }}{\left(\frac{\lambda (j-1)+l+1}{l+1}\right)}^n{a}_j{z}^j.$$

Remark 1.2 [4]

It follows from the above definition that

$$\begin{array}{c}I(0,\lambda ,l)f(z)=f(z),\hfill \\ (l+1)I(n+1,\lambda ,l)f(z)=(l+1-\lambda )I(n,\lambda ,l)f(z)+\lambda z{(I(n,\lambda ,l)f(z))}^{\prime },z\in U.\hfill \end{array}$$

Remark 1.3 The operator $I(n,\lambda ,0)=D_{\lambda }^n$ is the generalized Sălăgean operator introduced by Al-Oboudi [5], and $I(n,1,0)=S^n$ is the Sălăgean differential operator [6].

Definition 1.3 [7, 8]

Let $\gamma ,\lambda ,l\ge 0$, $n\in \mathbb{N}$. Denote by $R{I}_{n,\lambda ,l}^{\gamma }$ the operator given by $R{I}_{n,\lambda ,l}^{\gamma }:\mathcal{A}\to \mathcal{A}$,

$$R{I}_{n,\lambda ,l}^{\gamma }f(z)=(1-\gamma )R^{n}f(z)+\gamma I(n,\lambda ,l)f(z),z\in U.$$

Remark 1.4 If $f\in \mathcal{A}$, $f(z)=z+{\sum }_{j=2}^{\mathrm{\infty }}{a}_j{z}^j$, then

$$R{I}_{n,\lambda ,l}^{\gamma }f(z)=z+\sum _{j=2}^{\mathrm{\infty }}\{\gamma {\left(\frac{1+\lambda (j-1)+l}{l+1}\right)}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}{a}_j{z}^j,z\in U.$$

If $f\in \mathcal{T}$, $f(z)=z-{\sum }_{j=2}^{\mathrm{\infty }}{a}_j{z}^j$, then

$$R{I}_{n,\lambda ,l}^{\gamma }f(z)=z-\sum _{j=2}^{\mathrm{\infty }}\{\gamma {\left(\frac{1+\lambda (j-1)+l}{l+1}\right)}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}{a}_j{z}^j,z\in U.$$

Remark 1.5 The operator $R{I}_{n,\lambda ,0}^{\gamma }f(z)=R{D}_{\lambda ,\gamma }^{n}f(z)$ which was introduced in [9] and the operator $R{I}_{n,1,0}^{\gamma }f(z)=L_{\gamma }^{n}f(z)$ which was introduced in [10].

Following the work of Najafzadeh and Pezeshki [11] we can define the class $\mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$ as follows.

Definition 1.4 For $\gamma ,\lambda ,l\ge 0$, $0\le \alpha <1$ and $0<\beta \le 1$, let $\mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$ be the subclass of consisting of functions that satisfying the inequality

$$\left|\frac{R{I}_{n,\lambda ,l}^{\mu ,\gamma }f(z)-1}{2\nu (R{I}_{n,\lambda ,l}^{\mu ,\gamma }f(z)-\alpha )-(R{I}_{n,\lambda ,l}^{\mu ,\gamma }f(z)-1)}\right|<\beta ,$$

(1.1)

where

$$R{I}_{n,\lambda ,l}^{\mu ,\gamma }f(z)=(1-\mu )\frac{R{I}_{n,\lambda ,l}^{\gamma }f(z)}{z}+\mu {(R{I}_{n,\lambda ,l}^{\gamma }f(z))}^{\prime },$$

(1.2)

$0<\nu \le 1$.

Remark 1.6 If $f\in \mathcal{T}$, $f(z)=z-{\sum }_{j=2}^{\mathrm{\infty }}{a}_j{z}^j$, then

$$\begin{array}{rcl}R{I}_{n,\lambda ,l}^{\mu ,\gamma }f(z)& =& 1-\sum _{j=t+1}^{\mathrm{\infty }}[1+\mu (j-1)]\\ \times \{\gamma {\left(\frac{1+\lambda (j-1)+l}{l+1}\right)}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}{a}_j{z}^{j-1},z\in U.\end{array}$$

Remark 1.7 The class $\mathcal{RI}(\gamma ,\lambda ,0,\alpha ,\beta )=\mathcal{RD}(\gamma ,\lambda ,\alpha ,\beta )$ defined and studied in [12] and $\mathcal{RI}(\gamma ,1,0,\alpha ,\beta )=\mathcal{L}(\gamma ,\alpha ,\beta )$ defined and studied in [13].

2 Coefficient bounds

In this section we obtain coefficient bounds and extreme points for functions in $\mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$.

Theorem 2.1 Let the function $f\in \mathcal{T}$. Then $f\in \mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$ if and only if

$$\begin{array}{c}\sum _{j=t+1}^{\mathrm{\infty }}(1+\mu (j-1))[1+\beta (2\nu -1)]\{\gamma {\left(\frac{1+\lambda (j-1)+l}{l+1}\right)}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}{a}_j\hfill \\ <2\beta \nu (1-\alpha ).\hfill \end{array}$$

(2.1)

The result is sharp for the function $F(z)$ defined by

$$F(z)=z-\frac{2\beta \nu (1-\alpha )}{(1+\mu (j-1))[1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{z}^j,j\ge t+1.$$

Proof Suppose f satisfies (2.1). Then for $|z|<1$, we have

$$\begin{array}{c}|R{I}_{n,\lambda ,l}^{\mu ,\gamma }f(z)-1|-\beta |2\nu (R{I}_{n,\lambda ,l}^{\mu ,\gamma }f(z)-\alpha )-(R{I}_{n,\lambda ,l}^{\mu ,\gamma }f(z)-1)|\hfill \\ =|-\sum _{j=t+1}^{\mathrm{\infty }}(1+\mu (j-1))\{\gamma {\left(\frac{1+\lambda (j-1)+l}{l+1}\right)}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}{a}_j{z}^{j-1}|\hfill \\ -\beta |2\nu (1-\alpha )-(2\nu -1)\sum _{j=t+1}^{\mathrm{\infty }}\{\gamma {\left(\frac{1+\lambda (j-1)+l}{l+1}\right)}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}\hfill \\ \times [1+\mu (j-1)]a_j{z}^{j-1}|\hfill \\ \le \sum _{j=t+1}^{\mathrm{\infty }}[1+\mu (j-1)]\{\gamma {\left(\frac{1+\lambda (j-1)+l}{l+1}\right)}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}{a}_k-2\beta \nu (1-\alpha )\hfill \\ +\sum _{j=t+1}^{\mathrm{\infty }}\beta (2\nu -1)(1+\mu (j-1))\{\gamma {\left(\frac{1+\lambda (j-1)+l}{l+1}\right)}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}{a}_j\hfill \\ =\sum _{j=t+1}^{\mathrm{\infty }}[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {\left(\frac{1+\lambda (j-1)+l}{l+1}\right)}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}{a}_j\hfill \\ -2\beta \nu (1-\alpha )<0.\hfill \end{array}$$

Hence, by using the maximum modulus Theorem and (1.1), $f\in \mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$. Conversely, assume that

$$\begin{array}{c}\left|\frac{R{I}_{n,\lambda ,l}^{\mu ,\gamma }f(z)-1}{2\nu (R{I}_{n,\lambda ,l}^{\mu ,\gamma }f(z)-\alpha )-(R{I}_{n,\lambda ,l}^{\mu ,\gamma }f(z)-1)}\right|\hfill \\ =\left|\frac{-{\sum }_{j=t+1}^{\mathrm{\infty }}[1+\mu (j-1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}{a}_j{z}^{j-1}}{2\nu (1-\alpha )-{\sum }_{j=t+1}^{\mathrm{\infty }}[1+\mu (j-1)](2\nu -1)\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}{a}_j{z}^{j-1}}\right|\hfill \\ <\beta ,z\in U.\hfill \end{array}$$

Since $Re(z)\le |z|$ for all $z\in U$, we have

$$\begin{array}{c}Re\left\{\frac{{\sum }_{j=t+1}^{\mathrm{\infty }}[1+\mu (j-1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}{a}_j{z}^{j-1}}{2\nu (1-\alpha )-{\sum }_{j=t+1}^{\mathrm{\infty }}[1+\mu (j-1)](2\nu -1)\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}{a}_j{z}^{j-1}}\right\}\hfill \\ <\beta .\hfill \end{array}$$

(2.2)

By choosing choose values of z on the real axis so that $R{I}_{n,\lambda ,l}^{\mu ,\gamma }f(z)$ is real and letting $z\to 1$ through real values, we obtain the desired inequality (2.1). □

Corollary 2.2 If $f\in \mathcal{T}$ is in $\mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$, then

$$a_j\le \frac{2\beta \nu (1-\alpha )}{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}},j\ge t+1,$$

(2.3)

with equality only for functions of the form $F(z)$.

Theorem 2.3 Let $f_1(z)=z$ and

$$\begin{array}{c}{f}_j(z)=z-\frac{2\beta \nu (1-\alpha )}{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{z}^j,\hfill \\ j\ge t+1,\hfill \end{array}$$

(2.4)

for $0\le \alpha <1$, $0<\beta \le 1$, $\gamma ,\lambda ,l\ge 0$ and $0<\nu \le 1$. Then $f(z)$ is in the class $\mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$ if and only if it can be expressed in the form

$$f(z)=\sum _{j=t}^{\mathrm{\infty }}{\omega }_j{f}_j(z),$$

(2.5)

where ${\omega }_j\ge 0$ and ${\sum }_{j=1}^{\mathrm{\infty }}{\omega }_j=1$.

Proof Suppose $f(z)$ can be written as in (2.5). Then

$$f(z)=z-\sum _{j=t+1}^{\mathrm{\infty }}{\omega }_j\frac{2\beta \nu (1-\alpha )}{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{z}^j.$$

Now,

$$\begin{array}{c}\sum _{j=t+1}^{\mathrm{\infty }}\frac{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{2\beta \nu (1-\alpha )}{\omega }_j\hfill \\ \times \frac{2\beta \nu (1-\alpha )}{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}=\sum _{j=t+1}^{\mathrm{\infty }}{\omega }_j=1-{\omega }_1\le 1.\hfill \end{array}$$

Thus $f\in \mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$.

Conversely, let $f\in \mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$. Then by using (2.3), setting

$${\omega }_j=\frac{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{2\beta \nu (1-\alpha )}{a}_j,j\ge t+1,$$

and ${\omega }_1=1-{\sum }_{j=2}^{\mathrm{\infty }}{\omega }_j$, we have $f(z)={\sum }_{j=t}^{\mathrm{\infty }}{\omega }_j{f}_j(z)$. This completes the proof of Theorem 2.3. □

3 Distortion bounds

In this section we obtain distortion bounds for the class $\mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$.

Theorem 3.1 If $f\in \mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$, then

$$\begin{array}{c}r-\frac{2\beta \nu (1-\alpha )}{(1+\mu t)[1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda t+l}{l+1})}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}}{r}^{t+1}\hfill \\ \le |f(z)|\le r+\frac{2\beta \nu (1-\alpha )}{(1+\mu t)[1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda t+l}{l+1})}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}}{r}^{t+1}\hfill \end{array}$$

(3.1)

holds if the sequence ${\{{\sigma }_j(\gamma ,\lambda ,l,\beta ,\nu )\}}_{j=t+1}^{\mathrm{\infty }}$ is non-decreasing, and

$$\begin{array}{c}1-\frac{2\beta \nu (1-\alpha )(t+1)}{(1+\mu t)[1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda t+l}{l+1})}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}}{r}^t\hfill \\ \le |f^{\prime }(z)|\le 1+\frac{2\beta \nu (1-\alpha )(t+1)}{(1+\mu t)[1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda t+l}{l+1})}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}}{r}^t\hfill \end{array}$$

(3.2)

holds if the sequence ${\{\frac{{\sigma }_j(\gamma ,\lambda ,l,\beta ,\nu )}{j}\}}_{j=t+1}^{\mathrm{\infty }}$ is non-decreasing, where

$${\sigma }_j(\gamma ,\beta ,\nu )=[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {\left(\frac{1+\lambda (j-1)+l}{l+1}\right)}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}.$$

The bounds in (3.1) and (3.2) are sharp, for $f(z)$ given by

$$f(z)=z-\frac{2\beta \nu (1-\alpha )}{(1+\mu t)[1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda t+l}{l+1})}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}}{z}^{t+1},z=\pm r.$$

(3.3)

Proof In view of Theorem 2.1, we have

$$\sum _{j=t+1}^{\mathrm{\infty }}{a}_j\le \frac{2\beta \nu (1-\alpha )}{(1+\mu t)[1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda t+l}{l+1})}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}}.$$

(3.4)

We obtain

$$|z|-|z{|}^{t+1}\sum _{j=t+1}^{\mathrm{\infty }}{a}_j\le |f(z)|\le |z|+|z{|}^{t+1}\sum _{j=t+1}^{\mathrm{\infty }}{a}_j.$$

Thus

$$\begin{array}{c}r-\frac{2\beta \nu (1-\alpha )}{(1+\mu t)[1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda t+l}{l+1})}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}}{r}^{t+1}\hfill \\ \le |f(z)|\le r+\frac{2\beta \nu (1-\alpha )}{(1+\mu t)[1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda t+l}{l+1})}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}}{r}^{t+1}.\hfill \end{array}$$

(3.5)

Hence (3.1) follows from (3.5). Further,

$$\sum _{j=t+1}^{\mathrm{\infty }}j{a}_j\le \frac{2\beta \nu (1-\alpha )}{(1+\mu t)[1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda t+l}{l+1})}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}}.$$

Hence (3.2) follows from

$$1-r^t\sum _{j=t+1}^{\mathrm{\infty }}j{a}_j\le |f^{\prime }(z)|\le 1+r^t\sum _{j=t+1}^{\mathrm{\infty }}j{a}_j.$$

 □

4 Radius of starlikeness and convexity

The radii of close-to-convexity, starlikeness, and convexity for the class $\mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$ are given in this section.

Theorem 4.1 Let the function $f\in \mathcal{T}$ belong to the class $\mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$, Then $f(z)$ is close-to-convex of order δ, $0\le \delta <1$, in the disc $|z|<r$, where

$$r:=\underset{j\ge t+1}{inf}{\left[\frac{(1-\delta )[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{2\beta \nu j(1-\alpha )}\right]}^{\frac{1}{t}}.$$

(4.1)

The result is sharp, with extremal function $f(z)$ given by (2.3).

Proof For given $f\in \mathcal{T}$ we must show that

$$|f^{\prime }(z)-1|<1-\delta .$$

(4.2)

By a simple calculation we have

$$|f^{\prime }(z)-1|\le \sum _{j=t+1}^{\mathrm{\infty }}j{a}_j|z{|}^t.$$

The last expression is less than $1-\delta$ if

$$\sum _{j=t+1}^{\mathrm{\infty }}\frac{j}{1-\delta }{a}_j|z{|}^t<1.$$

We use the fact that $f\in \mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$ if and only if

$$\sum _{j=t+1}^{\mathrm{\infty }}\frac{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{2\beta \nu (1-\alpha )}{a}_j\le 1.$$

Equation (4.2) holds true if

$$\frac{j}{1-\delta }|z{|}^t\le \sum _{j=t+1}^{\mathrm{\infty }}\frac{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{2\beta \nu (1-\alpha )}.$$

Or, equivalently,

$$|z{|}^t\le \sum _{j=t+1}^{\mathrm{\infty }}\frac{(1-\delta )[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{2\beta \nu j(1-\alpha )},$$

which completes the proof. □

Theorem 4.2 Let $f\in \mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$. Then

• 1. f is starlike of order δ, $0\le \delta <1$, in the disc $|z|<r_1$, where

  $$r_1=\underset{j\ge t+1}{inf}{\left\{\frac{(1-\delta )[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{2\beta \nu (1-\alpha )(j-\delta )}\right\}}^{\frac{1}{t}}.$$

• 2. f is convex of order δ, $0\le \delta <1$, in the disc $|z|<r_2$ where,

  $$r_2=\underset{j\ge t+1}{inf}{\left\{\frac{(1-\delta )[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{2\beta \nu j(j-1)(1-\alpha )}\right\}}^{\frac{1}{t}}.$$

Each of these results is sharp for the extremal function $f(z)$ given by (2.5).

Proof 1. For $0\le \delta <1$ we need to show that

$$|\frac{z{f}^{\prime }(z)}{f(z)}-1|<1-\delta .$$

(4.3)

We have

$$|\frac{z{f}^{\prime }(z)}{f(z)}-1|\le \left|\frac{{\sum }_{j=t+1}^{\mathrm{\infty }}(j-1)a_j|z{|}^t}{1-{\sum }_{j=t+1}^{\mathrm{\infty }}{a}_j|z{|}^t}\right|.$$

The last expression is less than $1-\delta$ if

$$\sum _{j=t+1}^{\mathrm{\infty }}\frac{(j-\delta )}{1-\delta }{a}_j|z{|}^t<1.$$

We use the fact that $f\in \mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$ if and only if

$$\sum _{j=t+1}^{\mathrm{\infty }}\frac{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{2\beta \nu (1-\alpha )}{a}_j<1.$$

Equation (4.3) holds true if

$$\frac{j-\delta }{1-\delta }|z{|}^t<\frac{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{2\beta \nu (1-\alpha )}.$$

Or, equivalently,

$$|z{|}^t<\frac{(1-\delta )[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{2\beta \nu (1-\alpha )(j-\delta )},$$

which yields the starlikeness of the family.

• 2. Using the fact that f is convex if and only $z{f}^{\prime }$ is starlike, we can prove (2) with a similar way of the proof of (1). The function f is convex if and only if

  $$|z{f}^{″}(z)|<1-\delta .$$

  (4.4)

We have

$$\begin{array}{c}|z{f}^{″}(z)|\le |\sum _{j=t+1}^{\mathrm{\infty }}j(j-1)a_j|z{|}^{t-1}|<1-\delta ,\hfill \\ \sum _{j=t+1}^{\mathrm{\infty }}\frac{j(j-1)}{1-\delta }{a}_j|z{|}^{t-1}<1.\hfill \end{array}$$

We use the fact that $f\in \mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$ if and only if

$$\sum _{j=t+1}^{\mathrm{\infty }}\frac{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{2\beta \nu (1-\alpha )}{a}_j<1.$$

Equation (4.4) holds true if

$$\frac{j(j-1)}{1-\delta }|z{|}^{t-1}<\frac{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{2\beta \nu (1-\alpha )},$$

or, equivalently,

$$|z{|}^{t-1}<\frac{(1-\delta )[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{2\beta \nu j(j-1)(1-\alpha )},$$

which yields the convexity of the family. □

5 Neighborhood property

In this section we study neighborhood property for functions in the class $\mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$.

Definition 5.1 For functions f belong to of the form and $\epsilon \ge 0$, we define $(\eta -\epsilon )$-neighborhood of f by

$$N_{\epsilon }^{\eta }(f)=\{g(z)\in \mathcal{A}:g(z)=z+\sum _{j=2}^{\mathrm{\infty }}{b}_j{z}^j,\sum _{j=2}^{\mathrm{\infty }}{j}^{\eta +1}|a_j-b_j|\le \epsilon \},$$

where η is a fixed positive integer.

By using the following lemmas we will investigate the $(\eta -\epsilon )$-neighborhood of function in $\mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$.

Lemma 5.1 Let $-1\le \beta <1$, if $g(z)=z+{\sum }_{j=2}^{\mathrm{\infty }}{b}_j{z}^j$ satisfies

$$\sum _{j=2}^{\mathrm{\infty }}{j}^{\rho +1}|b_j|\le \frac{2\beta \nu (1-\alpha )}{1+\beta (2\nu -1)}$$

then $g(z)\in \mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$.

Proof By using of Theorem 2.1, it is sufficient to show that

$$\frac{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^{\rho }+(1-\gamma )\frac{(\rho +j-1)!}{\rho !(j-1)!}\}}{2\beta \nu (1-\alpha )}=\frac{j^{\rho +1}}{2\beta \nu (1-\alpha )}[1+\beta (2\nu -1)].$$

But

$$\frac{[1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^{\rho }+(1-\gamma )\frac{(\rho +j-1)!}{\rho !(j-1)!}\}}{2\beta \nu (1-\alpha )}\le \frac{j^{\rho +1}}{2\beta \nu (1-\alpha )}[1+\beta (2\nu -1)].$$

Therefore it is enough to prove that

$$Q(j,\rho )=\frac{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^{\rho }+(1-\gamma )\frac{(\rho +j-1)!}{\rho !(j-1)!}}{j^{\rho +1}}\le 1,$$

the result follows because the last inequality holds for all $j\ge t+1$. □

Lemma 5.2 Let $f(z)=z-{\sum }_{k=2}^{\mathrm{\infty }}{a}_k{z}^k\in \mathcal{T}$, $\gamma ,\lambda ,l\ge 0$, $0\le \alpha <1$, $0<\beta \le 1$ and $\epsilon \ge 0$. If $\frac{f(z)+ϵz}{1+ϵ}\in \mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$, then

$$\sum _{j=t+1}^{\mathrm{\infty }}{j}^{\rho +1}{a}_j\le \frac{2\beta \nu (1-\alpha )(1+ϵ){(t+1)}^{\rho +1}}{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda t+l}{l+1})}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}},$$

where either $\rho =0$ or $\rho =1$. The result is sharp with the extremal function

$$f(z)=z-\frac{2\beta \nu (1-\alpha )(1+ϵ)}{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda t+l}{l+1})}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}}{z}^{t+1},z\in U.$$

Proof Letting $g(z)=\frac{f(z)+ϵz}{1+ϵ}$ we have

$$g(z)=z-\sum _{j=t+1}^{\mathrm{\infty }}\frac{a_j}{1+ϵ}{z}^j,z\in U.$$

In view of Theorem 2.3, $g(z)={\sum }_{j=1}^{\mathrm{\infty }}{\eta }_j{g}_j(z)$ where ${\eta }_j\ge 0$, ${\sum }_{j=1}^{\mathrm{\infty }}{\eta }_j=1$,

$$g_1(z)=z$$

and

$$g_j(z)=z-\frac{2\beta \nu (1-\alpha )(1+ϵ)}{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{z}^j,j\ge t+1.$$

So we obtain

$$\begin{array}{rcl}g(z)& =& {\eta }_{1}z+\sum _{j=t+1}^{\mathrm{\infty }}{\eta }_j[z-\frac{2\beta \nu (1-\alpha )(1+ϵ)}{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{z}^j]\\ =& z-\sum _{j=t+1}^{\mathrm{\infty }}{\eta }_k\frac{2\beta \nu (1-\alpha )(1+ϵ)}{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}{z}^j.\end{array}$$

Since ${\eta }_j\ge 0$ and ${\sum }_{j=2}^{\mathrm{\infty }}{\eta }_j\le 1$, it follows that

$$\sum _{j=t+1}^{\mathrm{\infty }}{a}_k\le \underset{j\ge t+1}{sup}{j}^{\rho +1}\frac{2\beta \nu (1-\alpha )(1+ϵ)}{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}.$$

Since whenever $\rho =0$ or $\rho =1$ we conclude that

$$W(j,\rho ,\gamma ,\alpha ,\beta ,ϵ)=j^{\rho +1}\frac{2\beta \nu (1-\alpha )(1+ϵ)}{[1+\mu (j-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda (j-1)+l}{l+1})}^n+(1-\gamma )\frac{(n+j-1)!}{n!(j-1)!}\}}$$

is a decreasing function of j, the result will follow. The proof is complete. □

Theorem 5.1 Let $\rho =0$ or $\rho =1$ and suppose $0\le \beta <1$ and

$$\begin{array}{rcl}-1& \le & \theta \\ <& \frac{[1+\mu (t-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda t+l}{l+1})}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}-2\beta \nu (1-\alpha )(1+ϵ){(t+1)}^{\eta +1}}{[1+\mu (t-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda t+l}{l+1})}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}},\end{array}$$

$f(z)\in \mathcal{T}$ and $\frac{f(z)+ϵz}{1+ϵ}\in \mathcal{RI}(\gamma ,\lambda ,l,\alpha ,\beta )$, then the $(\eta -\epsilon )$-neighborhood of f is the subset of $\mathcal{RI}(\lambda ,\lambda ,l,\alpha ,\beta )$, where

$$\begin{array}{rcl}\epsilon & \le & 2(1-\alpha )\{\theta \gamma [1+\mu (t-1)][1+\beta (2\nu -1)]\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}\\ -\beta \gamma [1+\theta (2\nu -1)](1+ϵ){(t+1)}^{\eta +1}\}/([1+\theta (2\nu -1)][1+\mu (t-1)]\\ \times [1+\beta (2\nu -1)]\{\gamma {\left(\frac{1+\lambda t+l}{l+1}\right)}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}).\end{array}$$

The result is sharp.

Proof For $f(z)=z-{\sum }_{j=2}^{\mathrm{\infty }}|a_j|z^j$, let $g(z)=z+{\sum }_{j=2}^{\mathrm{\infty }}{b}_j{z}^j$ be in $N_{\epsilon }^{\eta }(f)$. So by Lemma 5.2, we have

$$\begin{array}{rcl}\sum _{j=2}^{\mathrm{\infty }}{j}^{\eta +1}|b_j|& =& \sum _{j=2}^{\mathrm{\infty }}{j}^{\eta +1}|a_j-b_j-a_j|\\ \le & \epsilon +\frac{2\beta \nu (1-\alpha )(1+ϵ){(t+1)}^{\eta +1}}{[1+\mu (t-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda t+l}{l+1})}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}}.\end{array}$$

By using Lemma 5.1, $g(z)\in \mathcal{L}(\gamma ,\alpha ,\beta )$ if

$$\epsilon +\frac{2\beta \nu (1-\alpha )(1+ϵ){(t+1)}^{\eta +1}}{[1+\mu (t-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda t+l}{l+1})}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}}\le \frac{2\theta \nu (1-\alpha )}{1+\theta (2\nu -1)},$$

that is, $\epsilon \le \frac{2(1-\alpha )\{\theta \gamma [1+\mu (t-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda t+l}{l+1})}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}-\beta \gamma [1+\theta (2\nu -1)](1+ϵ){(t+1)}^{\eta +1}\}}{[1+\theta (2\nu -1)][1+\mu (t-1)][1+\beta (2\nu -1)]\{\gamma {(\frac{1+\lambda t+l}{l+1})}^n+(1-\gamma )\frac{(n+t)!}{n!t!}\}}$, and the proof is complete. □

Author’s contributions

The author drafted the manuscript, read and approved the final manuscript.