A class of analytic functions involving in the Dziok-Srivastava operator

Abstract

Let $\mathcal{A}$ be a class of functions $f(z)$ of the form

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

(0.1)

which are analytic in the open unit disk $\mathbb{U}$. By means of the Dziok-Srivastava operator, we introduce a new subclass

$${\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )(l\le m+1,l,m\in \mathbb{N}\cup \{0\},-\frac{\pi }{2}<\alpha <\frac{\pi }{2},\mu >-cos\alpha )$$

of $\mathcal{A}$. In particular, ${\mathcal{S}}_0^1(2,0,0)$ coincides with the class of uniformly convex functions introduced by Goodman. The order of starlikeness and the radius of α-spirallikeness of order β ($\beta <1$) are computed. Inclusion relations and convolution properties for the class ${\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$ are obtained. A special member of ${\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$ is also given. The results presented here not only generalize the corresponding known results, but also give rise to several other new results.

MSC:30C45.

1 Introduction

Let $\mathcal{A}$ be a class of functions $f(z)$ of the form

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

(1.1)

which are analytic in the open unit disk $\mathbb{U}=\{z:|z|<1\}$. For $\beta <1$, a function $f(z)\in \mathcal{A}$ is said to be starlike of order β in $\mathbb{U}$ if

$$\mathrm{\Re }\frac{z{f}^{\prime }(z)}{f(z)}>\beta (z\in \mathbb{U}).$$

(1.2)

This class is denoted by ${\mathcal{S}}^{\ast }(\beta )$ ($\beta <1$). For $-\frac{\pi }{2}<\alpha <\frac{\pi }{2}$ and $\beta <1$, a function $f(z)\in \mathcal{A}$ is said to be α-spirallike of order β in $\mathbb{U}$ if

$$\mathrm{\Re }\left\{e^{i\alpha }\frac{z{f}^{\prime }(z)}{f(z)}\right\}>\beta cos\alpha (z\in \mathbb{U}).$$

(1.3)

When $0\le \beta <1$, it is well known that all the starlike functions of order β and α-spirallike functions of order β are univalent in $\mathbb{U}$. A function $f(z)\in \mathcal{A}$ is said to be convex univalent in $\mathbb{U}$ if

$$\mathrm{\Re }\{1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\}>0(z\in \mathbb{U}).$$

(1.4)

We denote this class by $\mathcal{K}$. Also, let $\mathcal{UCV}(\subset \mathcal{K})$ be the class of uniformly convex functions in $\mathbb{U}$ introduced by Goodman [1]. It was shown in [2] that $f(z)\in \mathcal{A}$ is in $\mathcal{UCV}$ if and only if

$$\mathrm{\Re }\{1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\}>\left|\frac{z{f}^{″}(z)}{f^{\prime }(z)}\right|(z\in \mathbb{U}).$$

(1.5)

In [2], Rønning investigated the class ${\mathcal{S}}_p$ defined by

$${\mathcal{S}}_p=\{f(z)\in {\mathcal{S}}^{\ast }(0):f(z)=z{g}^{\prime }(z),g(z)\in \mathcal{UCV}\}.$$

(1.6)

The uniformly convex and related functions have been studied by many authors (see, e.g., [1–10] and the references therein).

If

$$f(z)=z+\sum _{n=2}^{\mathrm{\infty }}{a}_n{z}^n\in \mathcal{A}\text{and}g(z)=z+\sum _{n=2}^{\mathrm{\infty }}{b}_n{z}^n\in \mathcal{A},$$

then the Hadamard product (or convolution) of $f(z)$ and $g(z)$ is given by

$$(f\ast g)(z)=z+\sum _{n=2}^{\mathrm{\infty }}{a}_n{b}_n{z}^n.$$

For

$${\alpha }_j\in \mathbb{C}(j=1,2,\dots ,l)\text{and}{\beta }_j\in \mathbb{C}\mathrm{\setminus }\{0,-1,-2,\dots \}(j=1,2,\dots ,m),$$

the generalized hypergeometric function

$${}_{l}F_m({\alpha }_1,\dots ,{\alpha }_l;{\beta }_1,\dots ,{\beta }_m;z)$$

is defined by the following infinite series:

$$\begin{array}{c}{}_{l}F_m({\alpha }_1,\dots ,{\alpha }_l;{\beta }_1,\dots ,{\beta }_m;z)=\sum _{n=0}^{\mathrm{\infty }}\frac{{({\alpha }_1)}_n\cdots {({\alpha }_l)}_n}{{({\beta }_1)}_n\cdots {({\beta }_m)}_n}\frac{z^n}{n!}\hfill \\ (l\le m+1;l,m\in {\mathbb{N}}_0=\mathbb{N}\cup \{0\};z\in \mathbb{U}),\hfill \end{array}$$

where ${(c)}_n$ is the Pochhammer symbol defined by

$${(c)}_n=\{\begin{array}{ll}1& (n=0),\\ c(c+1)\cdots (c+n-1)& (n\in \mathbb{N}).\end{array}$$

Corresponding to the function

$$z{\cdot }_l{F}_m({\alpha }_1,\dots ,{\alpha }_l;{\beta }_1,\dots ,{\beta }_m;z),$$

the Dziok-Srivastava operator (see [11])

$$H({\alpha }_1,\dots ,{\alpha }_l;{\beta }_1,\dots ,{\beta }_m):\mathcal{A}\to \mathcal{A}$$

is defined by the following Hadamard product:

$$\begin{array}{c}H({\alpha }_1,\dots ,{\alpha }_l;{\beta }_1,\dots ,{\beta }_m)f(z)=(z{\cdot }_l{F}_m({\alpha }_1,\dots ,{\alpha }_l;{\beta }_1,\dots ,{\beta }_m;z))\ast f(z)\hfill \\ (l\le m+1;l,m\in {\mathbb{N}}_0;z\in \mathbb{U}).\hfill \end{array}$$

If $f(z)\in \mathcal{A}$ is given by (1.1), then we have

$$H({\alpha }_1,\dots ,{\alpha }_l;{\beta }_1,\dots ,{\beta }_m)f(z)=z+\sum _{n=1}^{\mathrm{\infty }}\frac{{({\alpha }_1)}_n\cdots {({\alpha }_l)}_n}{{({\beta }_1)}_n\cdots {({\beta }_m)}_n}\frac{a_{n+1}}{n!}{z}^{n+1}(z\in \mathbb{U}).$$

(1.7)

In order to make the notation simple, we write

$$H_m^l({\alpha }_1)=H({\alpha }_1,\dots ,{\alpha }_l;{\beta }_1,\dots ,{\beta }_m)(l\le m+1;l,m\in {\mathbb{N}}_0).$$

(1.8)

It should also be remarked that the Dziok-Srivastava operator $H_m^l({\alpha }_1)$ is a generalization of several linear operators considered in earlier investigations (see [12–19], also see [20]).

In this paper we introduce and investigate the following subclass of $\mathcal{A}$.

Definition A function $f(z)\in \mathcal{A}$ is said to be in ${\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$ if it satisfies the condition

$$\mathrm{\Re }\left\{e^{i\alpha }\frac{z{(H_m^l({\alpha }_1)f(z))}^{\prime }}{H_m^l({\alpha }_1)f(z)}\right\}+\mu >|\frac{z{(H_m^l({\alpha }_1)f(z))}^{\prime }}{H_m^l({\alpha }_1)f(z)}-1|(z\in \mathbb{U}),$$

(1.9)

where

$$l\le m+1,l,m\in {\mathbb{N}}_0,-\frac{\pi }{2}<\alpha <\frac{\pi }{2}\text{and}\mu >-cos\alpha .$$

(1.10)

Note that $f(z)=z\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$ and that

$${\mathcal{S}}_0^1(1,\alpha ,0)=\{f(z)\in \mathcal{A}:\mathrm{\Re }\left\{e^{i\alpha }\frac{z{f}^{\prime }(z)}{f(z)}\right\}>|\frac{z{f}^{\prime }(z)}{f(z)}-1|(z\in \mathbb{U})\}.$$

(1.11)

Also,

$${\mathcal{S}}_0^1(1,0,0)={\mathcal{S}}_p\text{and}{\mathcal{S}}_0^1(2,0,0)=\mathcal{UCV}.$$

(1.12)

Throughout this paper we assume, unless otherwise stated, that l, m, α and μ satisfy (1.10).

2 Subordination theorem

Let $f(z)$ and $g(z)$ be analytic in $\mathbb{U}$. We say that the function $f(z)$ is subordinate to $g(z)$ in $\mathbb{U}$, and we write $f(z)\prec g(z)$, if there exists an analytic function $w(z)$ in $\mathbb{U}$ such that

$$|w(z)|\le |z|\text{and}f(z)=g(w(z))(z\in \mathbb{U}).$$

If $g(z)$ is univalent in $\mathbb{U}$, then

$$f(z)\prec g(z)\iff f(0)=g(0)\text{and}f(\mathbb{U})\subset g(\mathbb{U}).$$

Theorem 1 A function $f(z)\in \mathcal{A}$ is in ${\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$ if and only if

$$e^{i\alpha }\frac{z{(H_m^l({\alpha }_1)f(z))}^{\prime }}{H_m^l({\alpha }_1)f(z)}\prec h(z)cos\alpha +isin\alpha ,$$

(2.1)

where

$$\begin{array}{rcl}h(z)& =& 1+\frac{2}{{\pi }^2}(1+\frac{\mu }{cos\alpha }){(log\frac{1+\sqrt{z}}{1-\sqrt{z}})}^2\\ =& 1+\frac{8}{{\pi }^2}(1+\frac{\mu }{cos\alpha })\{z+\frac{2}{3}{z}^2+\frac{23}{45}{z}^3+\cdots \}(z\in \mathbb{U}).\end{array}$$

(2.2)

Proof Let us define $w(z)=u+iv$ by

$$e^{i\alpha }\frac{z{(H_m^l({\alpha }_1)f(z))}^{\prime }}{H_m^l({\alpha }_1)f(z)}=w(z)cos\alpha +isin\alpha (z\in \mathbb{U}).$$

(2.3)

Then $w(0)=1$ and the inequality (1.9) can be rewritten as

$$u>\frac{cos\alpha }{2(cos\alpha +\mu )}{v}^2+\frac{1}{2}(1-\frac{\mu }{cos\alpha }).$$

(2.4)

Thus

$$w(\mathbb{U})\subset \mathrm{\Omega }=\{w=u+iv:u\text{ and }v\text{ satisfy }(\text{2.4})\}.$$

It follows from (2.2) that $h(0)=1$. In order to prove the theorem, it suffices to show that the function $w=h(z)$ given by (2.2) maps $\mathbb{U}$ conformally onto the parabolic region Ω.

Note that $\frac{1}{2}(1-\frac{\mu }{cos\alpha })<1$. Consider the transformations

$$w_1=\sqrt{w-1},w_2=exp\left(\pi w_1\sqrt{\frac{2cos\alpha }{cos\alpha +\mu }}\right),t=\frac{1}{2}(w_2+\frac{1}{w_2}).$$

(2.5)

It is easy to verify that the composite function

$$t=ch\left(\pi \sqrt{\frac{2cos\alpha (w-1)}{cos\alpha +\mu }}\right)=g(w)(\text{say})$$

maps ${\mathrm{\Omega }}^{+}=\mathrm{\Omega }\cap \{w=u+iv:v>0\}$ conformally onto the upper half-plane $Im(t)>0$ so that $w=\mathrm{\Re }(w)\in [\frac{1}{2}(1-\frac{\mu }{cos\alpha }),+\mathrm{\infty })$ corresponds to $t=\mathrm{\Re }(t)\in [-1,+\mathrm{\infty })$ and $w=1$ to $t=1$. With the help of the symmetry principle, the function $t=g(w)$ maps Ω conformally onto the region $G=\{t:|arg(t+1)|<\pi \}$. Since

$$t=2{\left(\frac{1+z}{1-z}\right)}^2-1$$

(2.6)

maps $\mathbb{U}$ onto G, we see that

$$\begin{array}{rl}w& =g^{-1}(t)=1+\frac{1}{2{\pi }^2}(1+\frac{\mu }{cos\alpha }){(log(t+\sqrt{t^2-1}))}^2\\ =1+\frac{2}{{\pi }^2}(1+\frac{\mu }{cos\alpha }){(log\frac{1+\sqrt{z}}{1-\sqrt{z}})}^2\\ =h(z)\end{array}$$

maps $\mathbb{U}$ conformally onto Ω. The proof of the theorem is now completed. □

Corollary 1 Let $f(z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$. Then for $z\in \mathbb{U}$,

$$\left|{\left(\frac{H_m^l({\alpha }_1)f(z)}{z}\right)}^{sec\alpha e^{i\alpha }}\right|\le exp\left\{\frac{2}{{\pi }^2}(1+\frac{\mu }{cos\alpha }){\int }_0^1\frac{1}{\rho }{(log\frac{1+\sqrt{\rho |z|}}{1-\sqrt{\rho |z|}})}^{2}d\rho \right\}$$

(2.7)

and

$$\left|{\left(\frac{H_m^l({\alpha }_1)f(z)}{z}\right)}^{sec\alpha e^{i\alpha }}\right|\ge exp\{-\frac{8}{{\pi }^2}(1+\frac{\mu }{cos\alpha }){\int }_0^1\frac{1}{\rho }{(arctan\sqrt{\rho |z|})}^{2}d\rho \}.$$

(2.8)

The results are sharp.

Proof

From Theorem 1 we have

$$\frac{e^{i\alpha }}{cos\alpha }(\frac{z{(H_m^l({\alpha }_1)f(z))}^{\prime }}{H_m^l({\alpha }_1)f(z)}-1)\prec h(z)-1$$

for $f(z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$ and $h(z)$ given by (2.2). Since the function $h(z)-1$ is univalent and starlike (with respect to the origin) in $\mathbb{U}$, using the result of Suffridge [[21], Theorem 3], we get

$$\frac{e^{i\alpha }}{cos\alpha }{\int }_0^z(\frac{{(H_m^l({\alpha }_1)f(t))}^{\prime }}{H_m^l({\alpha }_1)f(t)}-\frac{1}{t})dt\prec {\int }_0^z\frac{h(t)-1}{t}dt.$$

This implies that

$$\frac{e^{i\alpha }}{cos\alpha }log\frac{H_m^l({\alpha }_1)f(z)}{z}={\int }_0^1\frac{h(\rho w(z))-1}{\rho }d\rho (z\in \mathbb{U}),$$

(2.9)

where $w(z)$ is analytic and $|w(z)|\le |z|$ in $\mathbb{U}$.

Noting that $h(z)$ maps the disk $|z|<\rho$ ($0<\rho \le 1$) onto a region which is convex and symmetric with respect to the real axis, we know that

$$h(-\rho |z|)\le \mathrm{\Re }\left\{h(\rho w(z))\right\}\le h(\rho |z|)(z\in \mathbb{U}).$$

(2.10)

Now (2.2), (2.9) and (2.10) lead to

$$log\left|{\left(\frac{H_m^l({\alpha }_1)f(z)}{z}\right)}^{sec\alpha e^{i\alpha }}\right|\le \frac{2}{{\pi }^2}(1+\frac{\mu }{cos\alpha }){\int }_0^1\frac{1}{\rho }{(log\frac{1+\sqrt{\rho |z|}}{1-\sqrt{\rho |z|}})}^{2}d\rho$$

and

$$\begin{array}{rl}log\left|{\left(\frac{H_m^l({\alpha }_1)f(z)}{z}\right)}^{sec\alpha e^{i\alpha }}\right|& \ge \frac{2}{{\pi }^2}(1+\frac{\mu }{cos\alpha }){\int }_0^1\frac{1}{\rho }{(log\frac{1+i\sqrt{\rho |z|}}{1-i\sqrt{\rho |z|}})}^{2}d\rho \\ =-\frac{8}{{\pi }^2}(1+\frac{\mu }{cos\alpha }){\int }_0^1\frac{1}{\rho }{(arctan\sqrt{\rho |z|})}^{2}d\rho \end{array}$$

for $z\in \mathbb{U}$. Hence we have (2.7) and (2.8).

Furthermore, for

$${\alpha }_j\in \mathbb{C}\setminus \{0,-1,-2,\dots \}(j=1,\dots ,l),$$

it is easy to see that the function $f_0(z)$ in ${\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$, defined by

$$\begin{array}{r}{H}_m^l({\alpha }_1)f_0(z)\\ =zexp\{\frac{2}{{\pi }^2}(1+\frac{\mu }{cos\alpha })cos\alpha e^{-i\alpha }{\int }_0^z\frac{1}{t}{(log\frac{1+\sqrt{t}}{1-\sqrt{t}})}^{2}dt\}(z\in \mathbb{U}),\end{array}$$

(2.11)

shows that the estimates (2.7) and (2.8) are sharp. □

Corollary 2 Let $f(z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$, where

$${\alpha }_j\in \mathbb{C}\mathrm{\setminus }\{0,-1,-2,\dots \}(j=1,2,\dots ,l).$$

Then

$$\begin{array}{rl}f(z)=& zexp\left\{\frac{2cos\alpha e^{-i\alpha }}{{\pi }^2}(1+\frac{\mu }{cos\alpha }){\int }_0^1\frac{1}{\rho }{(log\frac{1+\sqrt{\rho w(z)}}{1-\sqrt{\rho w(z)}})}^{2}d\rho \right\}\\ \ast \{z+\sum _{n=1}^{\mathrm{\infty }}\frac{n!{({\beta }_1)}_n\cdots {({\beta }_m)}_n}{{({\alpha }_1)}_n\cdots {({\alpha }_l)}_n}{z}^{n+1}\}(z\in \mathbb{U}),\end{array}$$

(2.12)

where $w(z)$ is analytic in $\mathbb{U}$ with $w(0)=0$ and $|w(z)|<1$ ($z\in \mathbb{U}$).

Proof From (2.9) and (2.2), we have

$$\begin{array}{r}{H}_m^l({\alpha }_1)f(z)\\ =zexp\left\{\frac{2cos\alpha e^{-i\alpha }}{{\pi }^2}(1+\frac{\mu }{cos\alpha }){\int }_0^1\frac{1}{\rho }{(log\frac{1+\sqrt{\rho w(z)}}{1-\sqrt{\rho w(z)}})}^{2}d\rho \right\}(z\in \mathbb{U}).\end{array}$$

(2.13)

For

$${\alpha }_j\in \mathbb{C}\mathrm{\setminus }\{0,-1,-2,\dots \}(j=1,2,\dots ,l),$$

from (2.13) and (1.7), we obtain (2.12). □

3 Properties of the class ${\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$

Theorem 2 Let $f(z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$. Then

$$H_m^l({\alpha }_1)f(z)\in {\mathcal{S}}^{\ast }\left(\frac{1}{2}(1-\frac{\mu }{cos\alpha })\right)$$

(3.1)

and the order $\frac{1}{2}(1-\frac{\mu }{cos\alpha })$ is sharp.

Proof Let $h(z)$ be given by (2.2). It follows from the proof of Theorem 1 that

$$\partial h(\mathbb{U})=\{w=u+iv:u=\frac{cos\alpha }{2(cos\alpha +\mu )}{v}^2+\frac{1}{2}(1-\frac{\mu }{cos\alpha })\}.$$

(3.2)

By using (3.2), we find that

$$\underset{|z|=1(z\ne 1)}{min}\mathrm{\Re }\left\{e^{-i\alpha }(h(z)cos\alpha +isin\alpha )\right\}=\underset{v\in (-\mathrm{\infty },+\mathrm{\infty })}{min}g(v)cos\alpha +{sin}^2\alpha ,$$

where

$$g(v)=\frac{{cos}^2\alpha }{2(cos\alpha +\mu )}{v}^2+\frac{cos\alpha -\mu }{2}+vsin\alpha (-\mathrm{\infty }<v<+\mathrm{\infty }).$$

Since

$$g^{\prime }(v)=\frac{{cos}^2\alpha }{cos\alpha +\mu }v+sin\alpha ,g^{″}(v)>0,$$

the function $g(v)$ attains its minimum value at

$$v_0=-\frac{(cos\alpha +\mu )sin\alpha }{{cos}^2\alpha }.$$

Thus

$$\begin{array}{r}\underset{|z|=1(z\ne 1)}{min}\mathrm{\Re }\left\{e^{-i\alpha }(h(z)cos\alpha +isin\alpha )\right\}\\ =g(v_0)cos\alpha +{sin}^2\alpha \\ =-\frac{{sin}^2\alpha (cos\alpha +\mu )}{2cos\alpha }+\frac{cos\alpha (cos\alpha -\mu )}{2}+{sin}^2\alpha \\ =\frac{1}{2}(1-\frac{\mu }{cos\alpha }).\end{array}$$

(3.3)

If $f(z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$, then we deduce from Theorem 1 and (3.3) that

$$\mathrm{\Re }\frac{z{(H_m^l({\alpha }_1)f(z))}^{\prime }}{H_m^l({\alpha }_1)f(z)}>\frac{1}{2}(1-\frac{\mu }{cos\alpha })(z\in \mathbb{U})$$

and the order $\frac{1}{2}(1-\frac{\mu }{cos\alpha })$ in (3.1) is sharp for the function $f_0(z)$ defined by (2.11). □

Theorem 3 Let $f(z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$ and $\frac{1}{2}(1-\frac{\mu }{cos\alpha })\le \beta <1$. Then $H_m^l({\alpha }_1)f(z)$ is α-spirallike of order β in $|z|<\rho$, where

$$\rho =\rho (\beta ,\alpha ,\mu )={(tan\left(\frac{\pi }{4}\sqrt{\frac{2cos\alpha (1-\beta )}{cos\alpha +\mu }}\right))}^2.$$

(3.4)

The result is sharp.

Proof From (3.4) and (2.2) we have

$$0<\rho \le 1(\frac{1}{2}(1-\frac{\mu }{cos\alpha })\le \beta <1)$$

and

$$\begin{array}{rl}h(-\rho )& =1+\frac{2}{{\pi }^2}(1+\frac{\mu }{cos\alpha }){(log\frac{1+i\sqrt{\rho }}{1-i\sqrt{\rho }})}^2\\ =1-\frac{8}{{\pi }^2}(1+\frac{\mu }{cos\alpha }){(arctan\sqrt{\rho })}^2\\ =\beta .\end{array}$$

Hence

$$\underset{|z|<\rho }{inf}\mathrm{\Re }h(z)=h(-\rho )=\beta .$$

(3.5)

Let $f(z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$. Then it follows from Theorem 1 and (3.5) that

$$\mathrm{\Re }\left\{e^{i\alpha }\frac{z{(H_m^l({\alpha }_1)f(z))}^{\prime }}{H_m^l({\alpha }_1)f(z)}\right\}>\beta cos\alpha (|z|<\rho ),$$

that is, $H_m^l({\alpha }_1)f(z)$ is α-spirallike of order β in $|z|<\rho$. Also, the result is sharp for the function $f_0(z)$ defined by (2.11). □

Setting $\beta =\frac{1}{2}(1-\frac{\mu }{cos\alpha })$, Theorem 3 reduces to the following.

Corollary 3 Let $f(z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$. Then $H_m^l({\alpha }_1)f(z)$ is α-spirallike of order $\frac{1}{2}(1-\frac{\mu }{cos\alpha })$ in $\mathbb{U}$. The result is sharp.

For $\beta \le 1$, a function $f(z)\in A$ is said to be prestarlike of order β in $\mathbb{U}$ if

$$\{\begin{array}{ll}\frac{z}{{(1-z)}^{2(1-\beta )}}\ast f(z)\in {\mathcal{S}}^{\ast }(\beta ),& \beta <1,\\ \mathrm{\Re }\frac{f(z)}{z}>\frac{1}{2},& \beta =1,\end{array}$$

(3.6)

(see [20]). We denote this class by $\mathcal{R}(\beta )$ ($\beta \le 1$). The following lemma is due to Ruscheweyh [[20], p.54].

Lemma 1 Let $\beta \le 1$, $f(z)\in \mathcal{R}(\beta )$ and $g(z)\in {\mathcal{S}}^{\ast }(\beta )$. Then, for any analytic function $F(z)$ in $\mathbb{U}$,

$$\frac{f\ast (Fg)}{f\ast g}(\mathbb{U})\subset \overline{co}(F(\mathbb{U})),$$

where $\overline{co}(F(\mathbb{U}))$ denotes the convex hull of $F(\mathbb{U})$.

Applying the lemma, we derive Theorems 4 and 5 below.

Theorem 4 Let

$${\alpha }_1>0\mathit{\text{and}}{\alpha }_1^{\prime }\ge max\{{\alpha }_1,1+\frac{\mu }{cos\alpha }\}.$$

(3.7)

Then

$${\mathcal{S}}_m^l({\alpha }_1^{\prime },\alpha ,\mu )\subset {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu ).$$

(3.8)

Proof

Define

$$\varphi (z)=z+\sum _{n=1}^{\mathrm{\infty }}\frac{{({\alpha }_1)}_n}{{({\alpha }_1^{\prime })}_n}{z}^{n+1}(z\in \mathbb{U})$$

for ${\alpha }_1$ and ${\alpha }_1^{\prime }$ satisfying (3.7). Then $\varphi (z)\in \mathcal{A}$ and

$$\frac{z}{{(1-z)}^{{\alpha }_1^{\prime }}}\ast \varphi (z)=\frac{z}{{(1-z)}^{{\alpha }_1}}(z\in \mathbb{U}).$$

(3.9)

In view of ${\alpha }_1^{\prime }\ge {\alpha }_1>0$, it follows from (3.9) that

$$\frac{z}{{(1-z)}^{{\alpha }_1^{\prime }}}\ast \varphi (z)\in {\mathcal{S}}^{\ast }(1-\frac{{\alpha }_1}{2})\subset {\mathcal{S}}^{\ast }(1-\frac{{\alpha }_1^{\prime }}{2}),$$

which implies that

$$\varphi (z)\in \mathcal{R}(1-\frac{{\alpha }_1^{\prime }}{2}).$$

(3.10)

Also, for $f(z)\in \mathcal{A}$, (3.9) leads to

$$\{\begin{array}{l}{H}_m^l({\alpha }_1)f(z)=\varphi (z)\ast H_m^l({\alpha }_1^{\prime })f(z),\\ z{(H_m^l({\alpha }_1)f(z))}^{\prime }=\varphi (z)\ast (z{(H_m^l({\alpha }_1^{\prime })f(z))}^{\prime }).\end{array}$$

(3.11)

Let $f(z)\in {\mathcal{S}}_m^l({\alpha }_1^{\prime },\alpha ,\mu )$. Then, by Theorems 1 and 2, we have

$$\{\begin{array}{l}F(z)=\frac{z{(H_m^l({\alpha }_1^{\prime })f(z))}^{\prime }}{H_m^l({\alpha }_1^{\prime })f(z)}\prec e^{-i\alpha }(h(z)cos\alpha +isin\alpha ),\\ H_m^l({\alpha }_1^{\prime })f(z)\in {\mathcal{S}}^{\ast }(\frac{1}{2}(1-\frac{\mu }{cos\alpha }))\subset {\mathcal{S}}^{\ast }(1-\frac{{\alpha }_1^{\prime }}{2})\end{array}$$

(3.12)

for $h(z)$ given by (2.2) and ${\alpha }_1^{\prime }\ge 1+\frac{\mu }{cos\alpha }$. Since the function $e^{-i\alpha }(h(z)cos\alpha +isin\alpha )$ is convex univalent in $\mathbb{U}$, from (3.10), (3.11), (3.12) and the lemma, we deduce that

$$\begin{array}{rl}\frac{z{(H_m^l({\alpha }_1)f(z))}^{\prime }}{H_m^l({\alpha }_1)f(z)}& =\frac{\varphi (z)\ast (z{(H_m^l({\alpha }_1^{\prime })f(z))}^{\prime })}{\varphi (z)\ast H_m^l({\alpha }_1^{\prime })f(z)}\\ =\frac{\varphi (z)\ast (F(z)H_m^l({\alpha }_1^{\prime })f(z)}{\varphi (z)\ast H_m^l({\alpha }_1^{\prime })f(z)}\\ \prec e^{-i\alpha }(h(z)cos\alpha +isin\alpha ).\end{array}$$

Therefore, by Theorem 1, $f(z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$ and (3.8) is proved. □

Theorem 5 Let $f(z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$ and $g(z)\in \mathcal{R}(\frac{1}{2}(1-\frac{\mu }{cos\alpha }))$. Then

$$(f\ast g)(z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu ).$$

(3.13)

Proof Let $f(z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$. According to Theorems 1 and 2, we have

$$F(z)=\frac{z{(H_m^l({\alpha }_1)f(z))}^{\prime }}{H_m^l({\alpha }_1)f(z)}\prec e^{-i\alpha }(h(z)cos\alpha +isin\alpha )$$

and

$$H_m^l({\alpha }_1)f(z)\in {\mathcal{S}}^{\ast }\left(\frac{1}{2}(1-\frac{\mu }{cos\alpha })\right).$$

(3.14)

If we put $\varphi (z)=(f\ast g)(z)$, then

$$\begin{array}{rl}\frac{z{(H_m^l({\alpha }_1)\varphi (z))}^{\prime }}{H_m^l({\alpha }_1)\varphi (z)}& =\frac{g(z)\ast (z{(H_m^l({\alpha }_1)f(z))}^{\prime })}{g(z)\ast H_m^l({\alpha }_1)f(z)}\\ =\frac{g(z)\ast (F(z)H_m^l({\alpha }_1)f(z))}{g(z)\ast H_m^l({\alpha }_1)f(z)}(z\in \mathbb{U})\end{array}$$

(3.15)

for $g(z)\in \mathcal{R}(\frac{1}{2}(1-\frac{\mu }{cos\alpha }))$.

In view of (3.14) and (3.15), an application of the lemma leads to

$$\frac{z{(H_m^l({\alpha }_1)\varphi (z))}^{\prime }}{H_m^l({\alpha }_1)\varphi (z)}\prec e^{-i\alpha }(h(z)cos\alpha +isin\alpha ).$$

Consequently, by applying Theorem 1, $\varphi (z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$ and the proof of (3.13) is completed. □

Note that $\mathcal{R}(\frac{1}{2})={\mathcal{S}}^{\ast }(\frac{1}{2})$. Since $\mathcal{R}({\beta }_1)\subset \mathcal{R}({\beta }_2)$ for ${\beta }_1\le {\beta }_2\le 1$ (see [[15], p.49], we have

$$\mathcal{K}=\mathcal{R}(0)\subset \mathcal{R}\left(\frac{1}{2}(1-\frac{\mu }{cos\alpha })\right)(-cos\alpha <\mu \le cos\alpha ).$$

Thus Theorem 5 yields the following.

Corollary 4

• (i) If $f(z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,0)$ and $g(z)\in {\mathcal{S}}^{\ast }(\frac{1}{2})$, then

  $$(f\ast g)(z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,0).$$

• (ii) If $f(z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$ with $-cos\alpha <\mu \le cos\alpha$ and $g(z)\in \mathcal{K}$, then

  $$(f\ast g)(z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu ).$$

Theorem 6 The function $f(z)\in \mathcal{A}$ defined by

$$H_m^l({\alpha }_1)f(z)=\frac{z}{{(1-bz)}^{2cos\alpha e^{-i\alpha }}}(z\in \mathbb{U})$$

(3.16)

belongs to the class ${\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$, where

$${\alpha }_j\in \mathbb{C}\mathrm{\setminus }\{0,-1,-2,\dots \}(j=1,2,\dots ,l),$$

b is complex and

$$|b|\le \{\begin{array}{ll}\frac{cos\alpha +\mu }{3cos\alpha -\mu }& (-cos\alpha <\mu <\frac{cos\alpha }{3}),\\ \sqrt{\frac{\mu }{cos\alpha +\mu }}& (\mu \ge \frac{cos\alpha }{3}).\end{array}$$

(3.17)

The result is sharp, that is, $|b|$ cannot be increased.

Proof For $f(z)\in \mathcal{A}$ defined by (3.16) and

$${\alpha }_j\in \mathbb{C}\mathrm{\setminus }\{0,-1,-2,\dots \}(j=1,2,\dots ,l),$$

we easily have

$$e^{i\alpha }\frac{z{(H_m^l({\alpha }_1)f(z))}^{\prime }}{H_m^l({\alpha }_1)f(z)}=\frac{1+bz}{1-bz}cos\alpha +isin\alpha (z\in \mathbb{U}).$$

(3.18)

Hence, by Theorem 1, $f(z)\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$ if and only if

$$\frac{1+bz}{1-bz}\prec h(z),$$

(3.19)

where $h(z)$ is given by (2.2). Clearly, (3.19) is equivalent to

$$\{w:|w-\frac{1+{|b|}^2}{1-{|b|}^2}|<\frac{2|b|}{1-{|b|}^2}\}\subset h(\mathbb{U})$$

(3.20)

for $0<|b|<1$. Let

$$\delta =min\left\{\right|w-\frac{1+{|b|}^2}{1-{|b|}^2}|:w\in \partial h(\mathbb{U})\},$$

(3.21)

where $\partial h(\mathbb{U})$ is given by (3.2). Then we have

$$\{\begin{array}{l}\delta =min\{\sqrt{g(u)}:u\ge \frac{1}{2}(1-\frac{\mu }{cos\alpha })\},\\ g(u)={(u-\frac{1+{|b|}^2}{1-{|b|}^2})}^2+2(1+\frac{\mu }{cos\alpha })(u-\frac{cos\alpha -\mu }{2cos\alpha })(u\ge \frac{cos\alpha -\mu }{2cos\alpha }).\end{array}$$

(3.22)

Note that

$$\frac{1}{2}(1-\frac{\mu }{cos\alpha })<\frac{1+{|b|}^2}{1-{|b|}^2},g^{\prime }(u)=2(u-(\frac{2{|b|}^2}{1-{|b|}^2}-\frac{\mu }{cos\alpha })).$$

(3.23)

(i)If

$$-cos\alpha <\mu <\frac{cos\alpha }{3}\text{and}|b|=\frac{cos\alpha +\mu }{3cos\alpha -\mu },$$

(3.24)

then

$$\frac{1-|b|}{1+|b|}=\frac{1}{2}(1-\frac{\mu }{cos\alpha }),{|b|}^2={\left(\frac{cos\alpha +\mu }{3cos\alpha -\mu }\right)}^2<\frac{cos\alpha +\mu }{5cos\alpha +\mu },$$

and so

$$\frac{2{|b|}^2}{1-{|b|}^2}-\frac{\mu }{cos\alpha }<\frac{1}{2}(1-\frac{\mu }{cos\alpha }).$$

(3.25)

From (3.22), (3.23) and (3.25), we have $g^{\prime }(u)>0(u\ge \frac{1}{2}(1-\frac{\mu }{cos\alpha }))$, and hence

$$\delta =\sqrt{g\left(\frac{1}{2}(1-\frac{\mu }{cos\alpha })\right)}=\frac{1+{|b|}^2}{1-{|b|}^2}-\frac{1}{2}(1-\frac{\mu }{cos\alpha })=\frac{2|b|}{1-{|b|}^2}.$$

(3.26)

(ii) If

$$-cos\alpha <\mu <\frac{cos\alpha }{3}\text{and}\frac{cos\alpha +\mu }{3cos\alpha -\mu }<|b|<\sqrt{\frac{cos\alpha +\mu }{5cos\alpha +\mu }},$$

(3.27)

then

$$\frac{1-|b|}{1+|b|}<\frac{1}{2}(1-\frac{\mu }{cos\alpha })\text{and}{g}^{\prime }(u)>0(u\ge \frac{1}{2}(1-\frac{\mu }{cos\alpha })).$$

Hence

$$\delta =\sqrt{g\left(\frac{1}{2}(1-\frac{\mu }{cos\alpha })\right)}<\frac{2|b|}{1-{|b|}^2}.$$

(3.28)

(iii) If

$$\mu \ge \frac{cos\alpha }{3}\text{and}|b|=\sqrt{\frac{\mu }{cos\alpha +\mu }},$$

(3.29)

then

$${|b|}^2=\frac{\mu }{cos\alpha +\mu }\ge \frac{cos\alpha +\mu }{5cos\alpha +\mu },$$

and so

$$\frac{2{|b|}^2}{1-{|b|}^2}-\frac{\mu }{cos\alpha }\ge \frac{1}{2}(1-\frac{\mu }{cos\alpha }).$$

Thus $g(u)$ attains its minimum value at

$$u_0=\frac{2{|b|}^2}{1-{|b|}^2}-\frac{\mu }{cos\alpha }$$

and

$$\delta =\sqrt{g(u_0)}=2|b|\sqrt{\frac{cos\alpha +\mu }{cos\alpha (1-{|b|}^2)}}=\frac{2|b|}{1-{|b|}^2}.$$

(3.30)

(iv) If

$$\mu \ge \frac{cos\alpha }{3}\text{and}\sqrt{\frac{\mu }{cos\alpha +\mu }}<|b|<1,$$

(3.31)

then from (iii) we easily have

$$\delta =\sqrt{g(u_0)}<\frac{2|b|}{1-{|b|}^2}.$$

(3.32)

Now, by virtue of (3.19), (3.20), (3.21), and (i)-(iv), we have proved the theorem. □

Theorem 7 Let

$$f(z)=z+\sum _{n=2}^{\mathrm{\infty }}{a}_n{z}^n\in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu ),$$

where

$${\alpha }_j\in \mathbb{C}\setminus \{0,-1,-2,-3,\dots \}(j=1,2,\dots ,l).$$

Then

$$|a_2|\le \frac{8(cos\alpha +\mu )}{{\pi }^2}\left|\frac{{\beta }_1\cdots {\beta }_m}{{\alpha }_1\cdots {\alpha }_l}\right|.$$

(3.33)

The result is sharp.

Proof It can be easily verified that, for $z\in \mathbb{U}$,

$$\frac{z{(H_m^l({\alpha }_1)f(z))}^{\prime }}{H_m^l({\alpha }_1)f(z)}=1+\frac{{\alpha }_1\cdots {\alpha }_l}{{\beta }_1\cdots {\beta }_m}{a}_{2}z+\cdots$$

(3.34)

and

$$\begin{array}{rl}h(z)& =1+\frac{8z}{{\pi }^2}(1+\frac{\mu }{cos\alpha }){\left(\sum _{n=1}^{\mathrm{\infty }}\frac{z^{n-1}}{2n-1}\right)}^2\\ =1+\frac{8}{{\pi }^2}(1+\frac{\mu }{cos\alpha })\sum _{n=1}^{\mathrm{\infty }}\left(\frac{1}{n}\sum _{\nu =0}^{n-1}\frac{1}{2\nu +1}\right)z^n\\ =1+\frac{8}{{\pi }^2}(1+\frac{\mu }{cos\alpha })z+\cdots ,\end{array}$$

(3.35)

where

$$f(z)=z+a_2{z}^2+\cdots \in {\mathcal{S}}_m^l({\alpha }_1,\alpha ,\mu )$$

and $h(z)$ is given by (2.2). From (3.34), (3.35) and Theorem 1, we obtain

$$\begin{array}{rl}\frac{{\pi }^2{e}^{i\alpha }}{8(cos\alpha +\mu )}(\frac{z{(H_m^l({\alpha }_1)f(z))}^{\prime }}{H_m^l({\alpha }_1)f(z)}-1)& =\frac{{\pi }^2{e}^{i\alpha }{\alpha }_1\cdots {\alpha }_l}{8(cos\alpha +\mu ){\beta }_1\cdots {\beta }_m}{a}_{2}z+\cdots \\ \prec \frac{{\pi }^{2}cos\alpha }{8(cos\alpha +\mu )}(h(z)-1)\in \mathcal{K}.\end{array}$$

(3.36)

It is the well-known Rogosinski result (cf. [[22], p.195]) that if

$$g(z)=\sum _{n=1}^{\mathrm{\infty }}{b}_n{z}^n$$

is analytic in $\mathbb{U}$, $g(z)\prec \varphi (z)$ and $\varphi (z)\in \mathcal{K}$, then $|b_n|\le 1$ ($n\in \mathbb{N}$). Hence (3.33) follows from (3.36) at once. □

The estimate (3.33) is sharp since equality is attained for the function $f_0(z)$ defined by (2.11).