Neighborhoods and partial sums of certain subclass of starlike functions

Abstract

The main purpose of the present paper is to derive the neighborhoods and partial sums of a certain subclass of starlike functions.

MSC: 30C45.

1 Introduction

Let ${\mathcal{A}}_m$ denote the class of functions f of the form

$$f(z)=z+\sum _{k=m+1}^{\mathrm{\infty }}{a}_k{z}^k(m\in \mathbb{N}:=\{1,2,3,\dots \}),$$

(1.1)

which are analytic in the open unit disk

$$\mathbb{U}:=\{z:z\in \mathbb{C}\text{ and }|z|<1\}.$$

A function $f\in {\mathcal{A}}_m$ is said to be in the class ${\mathcal{S}}_m^{\ast }(\beta )$ of starlike functions of order β if it satisfies the inequality

$$\mathrm{\Re }\left(\frac{z{f}^{\prime }(z)}{f(z)}\right)>\beta (z\in \mathbb{U};0\leqq \beta <1).$$

(1.2)

Assuming that $\alpha \geqq 0$, $0\leqq \beta <1$ and $f\in {\mathcal{A}}_m$, we say that a function $f\in {\mathcal{H}}_m(\alpha ,\beta )$ if it satisfies the condition

$$\mathrm{\Re }(\frac{z{f}^{\prime }(z)}{f(z)}+\alpha \frac{z^2{f}^{″}(z)}{f(z)})>\alpha \beta (\beta +\frac{m}{2}-1)+\beta -\frac{m\alpha }{2}(z\in \mathbb{U}).$$

(1.3)

For convenience, throughout this paper, we write

$${\gamma }_m:=\alpha \beta (\beta +\frac{m}{2}-1)+\beta -\frac{m\alpha }{2}.$$

(1.4)

Recently, Ravichandran et al. [1] proved that ${\mathcal{H}}_m(\alpha ,\beta )\subset {\mathcal{S}}_m^{\ast }(\beta )$. Subsequently, Liu et al. [2] derived various properties and characteristics such as inclusion relationships, Hadamard products, coefficient estimates, distortion theorems and cover theorems for the class ${\mathcal{H}}_m(\alpha ,\beta )$ and a subclass of ${\mathcal{H}}_m(\alpha ,\beta )$ with negative coefficients. Furthermore, Singh et al. [3] generalized the class ${\mathcal{H}}_m(\alpha ,\beta )$ and found several sufficient conditions for starlikeness. In the present paper, we aim at proving the neighborhoods and partial sums of the class ${\mathcal{H}}_m(\alpha ,\beta )$.

2 Main results

Following the earlier works (based upon the familiar concept of a neighborhood of analytic functions) by Goodman [4] and Ruscheweyh [5], and (more recently) by Altintaş et al. [6–9], Cǎtaş [10], Frasin [11], Keerthi et al. [12] and Srivastava et al. [13], we begin by introducing here the δ-neighborhood of a function $f\in {\mathcal{A}}_m$ of the form (1.1) by means of the definition

$$\begin{array}{rcl}{\mathcal{N}}_{\delta }(f)& :=& \{g\in {\mathcal{A}}_m:g(z)=z+\sum _{k=m+1}^{\mathrm{\infty }}{b}_k{z}^k\text{ and}\\ \sum _{k=m+1}^{\mathrm{\infty }}\frac{k(1+k\alpha -\alpha )-{\gamma }_m}{1-{\gamma }_m}|a_k-b_k|\leqq \delta (\delta ,\alpha \geqq 0;0\leqq \beta <1;{\gamma }_m<1)\}.\end{array}$$

(2.1)

By making use of the definition (2.1), we now derive the following result.

Theorem 1 If $f\in {\mathcal{A}}_m$ satisfies the condition

$$\frac{f(z)+\epsilon z}{1+\epsilon }\in {\mathcal{H}}_m(\alpha ,\beta )(\epsilon \in \mathbb{C};|\epsilon |<\delta ;\delta >0),$$

(2.2)

then

$${\mathcal{N}}_{\delta }(f)\subset {\mathcal{H}}_m(\alpha ,\beta ).$$

(2.3)

Proof By noting that the condition (1.3) can be rewritten as follows:

$$\left|\frac{\frac{z{f}^{\prime }(z)}{f(z)}+\alpha \frac{z^2{f}^{″}(z)}{f(z)}-1}{\frac{z{f}^{\prime }(z)}{f(z)}+\alpha \frac{z^2{f}^{″}(z)}{f(z)}-(2{\gamma }_m-1)}\right|<1(z\in \mathbb{U}),$$

(2.4)

we easily find from (2.4) that a function $g\in {\mathcal{H}}_m(\alpha ,\beta )$ if and only if

$$\frac{z{g}^{\prime }(z)+\alpha z^2{g}^{″}(z)-g(z)}{z{g}^{\prime }(z)+\alpha z^2{g}^{″}(z)-(2{\gamma }_m-1)g(z)}\ne \sigma (z\in \mathbb{U};\sigma \in \mathbb{C};|\sigma |=1),$$

which is equivalent to

$$\frac{(g\ast h)(z)}{z}\ne 0(z\in \mathbb{U}),$$

(2.5)

where

$$h(z)=z+\sum _{k=m+1}^{\mathrm{\infty }}{c}_k{z}^k(c_k:=\frac{k+\alpha k(k-1)-1-[k+\alpha k(k-1)-(2{\gamma }_m-1)]\sigma }{2({\gamma }_m-1)\sigma }).$$

(2.6)

It follows from (2.6) that

$$\begin{array}{rcl}|c_k|& =& \left|\frac{k+\alpha k(k-1)-1-[k+\alpha k(k-1)-(2{\gamma }_m-1)]\sigma }{2({\gamma }_m-1)\sigma }\right|\\ \leqq & \frac{k+\alpha k(k-1)-1+[k+\alpha k(k-1)-(2{\gamma }_m-1)]|\sigma |}{2(1-{\gamma }_m)|\sigma |}\\ =& \frac{k(1+k\alpha -\alpha )-{\gamma }_m}{1-{\gamma }_m}(|\sigma |=1).\end{array}$$

If $f\in {\mathcal{A}}_m$ satisfies the condition (2.2), we deduce from (2.5) that

$$\frac{(f\ast h)(z)}{z}\ne -\epsilon (|\epsilon |<\delta ;\delta >0),$$

or, equivalently,

$$\left|\frac{(f\ast h)(z)}{z}\right|\geqq \delta (z\in \mathbb{U};\delta >0).$$

(2.7)

We now suppose that

$$q(z)=z+\sum _{k=m+1}^{\mathrm{\infty }}{d}_k{z}^k\in {\mathcal{N}}_{\delta }(f).$$

It follows from (2.1) that

$$\begin{array}{rcl}\left|\frac{((q-f)\ast h)(z)}{z}\right|& =& |\sum _{k=m+1}^{\mathrm{\infty }}(d_k-a_k)c_k{z}^{k-1}|\\ \leqq & |z|\sum _{k=m+1}^{\mathrm{\infty }}\frac{k(1+k\alpha -\alpha )-{\gamma }_m}{1-{\gamma }_m}|d_k-a_k|<\delta .\end{array}$$

(2.8)

Combining (2.7) and (2.8), we easily find that

$$\left|\frac{(q\ast h)(z)}{z}\right|=\left|\frac{([f+(q-f)]\ast h)(z)}{z}\right|\geqq \left|\frac{(f\ast h)(z)}{z}\right|-\left|\frac{((q-f)\ast h)(z)}{z}\right|>0,$$

which implies that

$$\frac{(q\ast h)(z)}{z}\ne 0(z\in \mathbb{U}).$$

Therefore, we conclude that

$$q(z)\in {\mathcal{N}}_{\delta }(f)\subset {\mathcal{H}}_m(\alpha ,\beta ).$$

We thus complete the proof of Theorem 1. □

Next, we derive the partial sums of the class ${\mathcal{H}}_m(\alpha ,\beta )$. For some recent investigations involving the partial sums in analytic function theory, one can refer to [14–16] and the references cited therein.

Theorem 2 Let $f\in {\mathcal{A}}_m$ be given by (1.1) and define the partial sums $f_n(z)$ of f by

$$f_n(z)=z+\sum _{k=m+1}^n{a}_k{z}^k(n\in \mathbb{N};n\geqq m+1).$$

(2.9)

If

$$\sum _{k=m+1}^{\mathrm{\infty }}\frac{k(1+k\alpha -\alpha )-{\gamma }_m}{1-{\gamma }_m}|a_k|\leqq 1(\alpha \geqq 0;0\leqq \beta <1;{\gamma }_m<1),$$

(2.10)

then

(1) $f\in {\mathcal{H}}_m(\alpha ,\beta )$;

(2)

$$\mathrm{\Re }\left(\frac{f(z)}{f_n(z)}\right)\geqq \frac{n(1+\alpha +n\alpha )}{(n+1)(1+n\alpha )-{\gamma }_m}(n\in \mathbb{N};n\geqq m+1;z\in \mathbb{U})$$

(2.11)

and

$$\mathrm{\Re }\left(\frac{f_n(z)}{f(z)}\right)\geqq \frac{(n+1)(1+n\alpha )-{\gamma }_m}{(n+1)(1+n\alpha )+1-2{\gamma }_m}(n\in \mathbb{N};n\geqq m+1;z\in \mathbb{U}).$$

(2.12)

The bounds in (2.11) and (2.12) are sharp.

Proof (1) Suppose that $f_1(z)=z$. We know that $z\in {\mathcal{H}}_m(\alpha ,\beta )$, which implies that

$$\frac{f_1(z)+\epsilon z}{1+\epsilon }=z\in {\mathcal{H}}_m(\alpha ,\beta ).$$

From (2.10), we easily find that

$$\sum _{k=m+1}^{\mathrm{\infty }}\frac{k(1+k\alpha -\alpha )-{\gamma }_m}{1-{\gamma }_m}|a_k-0|\leqq 1,$$

which implies that $f\in {\mathcal{N}}_1(z)$. In view of Theorem 1, we deduce that

$$f\in {\mathcal{N}}_1(z)\subset {\mathcal{H}}_m(\alpha ,\beta ).$$

(2) It is easy to verify that

$$\begin{array}{rcl}\frac{(n+1)[1+(n+1)\alpha -\alpha ]-{\gamma }_m}{1-{\gamma }_m}& =& \frac{(n+1)(1+n\alpha )-{\gamma }_m}{1-{\gamma }_m}\\ >& \frac{n(1+n\alpha -\alpha )-{\gamma }_m}{1-{\gamma }_m}>1(n\in \mathbb{N}).\end{array}$$

Therefore, we have

$$\sum _{k=m+1}^n|a_k|+\frac{(n+1)(1+n\alpha )-{\gamma }_m}{1-{\gamma }_m}\sum _{k=n+1}^{\mathrm{\infty }}|a_k|\leqq \sum _{k=m+1}^{\mathrm{\infty }}\frac{k(1+k\alpha -\alpha )-{\gamma }_m}{1-{\gamma }_m}|a_k|\leqq 1.$$

(2.13)

We now suppose that

$$\begin{array}{rcl}\psi (z)& =& \frac{(n+1)(1+n\alpha )-{\gamma }_m}{1-{\gamma }_m}(\frac{f(z)}{f_n(z)}-\frac{n(1+\alpha +n\alpha )}{(n+1)(1+n\alpha )-{\gamma }_m})\\ =& 1+\frac{\frac{(n+1)(1+n\alpha )-{\gamma }_m}{1-{\gamma }_m}{\sum }_{k=n+1}^{\mathrm{\infty }}{a}_k{z}^{k-1}}{1+{\sum }_{k=m+1}^n{a}_k{z}^{k-1}}.\end{array}$$

(2.14)

It follows from (2.13) and (2.14) that

$$\left|\frac{\psi (z)-1}{\psi (z)+1}\right|\leqq \frac{\frac{(n+1)(1+n\alpha )-{\gamma }_m}{1-{\gamma }_m}{\sum }_{k=n+1}^{\mathrm{\infty }}|a_k|}{2-2{\sum }_{k=m+1}^n|a_k|-\frac{(n+1)(1+n\alpha )-{\gamma }_m}{1-{\gamma }_m}{\sum }_{k=n+1}^{\mathrm{\infty }}|a_k|}\leqq 1(z\in \mathbb{U}),$$

which shows that

$$\mathrm{\Re }(\psi (z))\geqq 0(z\in \mathbb{U}).$$

(2.15)

Combining (2.14) and (2.15), we deduce that the assertion (2.11) holds true.

Moreover, if we put

$$f(z)=z+\frac{1-{\gamma }_m}{(n+1)(1+n\alpha )-{\gamma }_m}{z}^{n+1}(n\in \mathbb{N}\setminus \{1,2,\dots ,m-1\};m\in \mathbb{N}),$$

(2.16)

then for $z=r{e}^{i\pi /n}$, we have

$$\frac{f(z)}{f_n(z)}=1+\frac{1-{\gamma }_m}{(n+1)(1+n\alpha )-{\gamma }_m}{z}^n\to \frac{n(1+\alpha +n\alpha )}{(n+1)(1+n\alpha )-{\gamma }_m}(r\to 1^{-}),$$

which implies that the bound in (2.11) is the best possible for each $n\in \mathbb{N}\setminus \{1,2,\dots ,m-1\}$.

Similarly, we suppose that

$$\begin{array}{rcl}\phi (z)& =& \frac{(n+1)(1+n\alpha )+1-2{\gamma }_m}{1-{\gamma }_m}(\frac{f_n(z)}{f(z)}-\frac{(n+1)(1+n\alpha )-{\gamma }_m}{(n+1)(1+n\alpha )+1-2{\gamma }_m})\\ =& 1-\frac{\frac{(n+1)(1+n\alpha )+1-2{\gamma }_m}{1-{\gamma }_m}{\sum }_{k=n+1}^{\mathrm{\infty }}{a}_k{z}^{k-1}}{1+{\sum }_{k=m+1}^{\mathrm{\infty }}{a}_k{z}^{k-1}}.\end{array}$$

(2.17)

In view of (2.13) and (2.17), we conclude that

$$\left|\frac{\phi (z)-1}{\phi (z)+1}\right|\leqq \frac{\frac{(n+1)(1+n\alpha )+1-2{\gamma }_m}{1-{\gamma }_m}{\sum }_{k=n+1}^{\mathrm{\infty }}|a_k|}{2-2{\sum }_{k=m+1}^n|a_k|-\frac{n(1+\alpha +n\alpha )}{1-{\gamma }_m}{\sum }_{k=n+1}^{\mathrm{\infty }}|a_k|}\leqq 1(z\in \mathbb{U}),$$

which implies that

$$\mathrm{\Re }(\phi (z))\geqq 0(z\in \mathbb{U}).$$

(2.18)

Combining (2.17) and (2.18), we readily get the assertion (2.12) of Theorem 2. The bound in (2.12) is sharp with the extremal function f given by (2.16).

The proof of Theorem 2 is thus completed. □

Finally, we turn to ratios involving derivatives. The proof of Theorem 3 below is much akin to that of Theorem 2, we here choose to omit the analogous details.

Theorem 3 Let $f\in {\mathcal{A}}_m$ be given by (1.1) and define the partial sums $f_n(z)$ of f by (2.9). If the condition (2.10) holds, then

$$\mathrm{\Re }\left(\frac{f^{\prime }(z)}{f_n^{\mathrm{\prime }}(z)}\right)\geqq \frac{(n+1)(n\alpha +{\gamma }_m)-{\gamma }_m}{(n+1)(1+n\alpha )-{\gamma }_m}(n\in \mathbb{N};n\geqq m+1;z\in \mathbb{U})$$

(2.19)

and

$$\mathrm{\Re }\left(\frac{f_n^{\mathrm{\prime }}(z)}{f^{\prime }(z)}\right)\geqq \frac{(n+1)(1+n\alpha )-{\gamma }_m}{(n+1)(2+n\alpha -{\gamma }_m)-{\gamma }_m}(n\in \mathbb{N};n\geqq m+1;z\in \mathbb{U}).$$

(2.20)

The bounds in (2.19) and (2.20) are sharp with the extremal function given by (2.16).

Remark By setting $\alpha =0$ and $m=1$ in Theorems 2 and 3, we get the corresponding results obtained by Silverman [16].