Properties for certain subclasses of meromorphic functions defined by a multiplier transformation

Abstract

Some inclusion and convolution properties of certain subclasses of meromorphic functions associated with a family of multiplier transformations, which are defined by means of the Hadamard product (or convolution), are investigated. We also obtain closure properties for certain integral operators.

MSC:30C45, 30C80.

1 Introduction

Let $\mathcal{A}$ denote the class of analytic functions f in the open unit disk $\mathbb{U}=\{z\in \mathbb{C}:|z|<1\}$ with the usual normalization $f(0)=f^{\mathrm{\prime }}(0)-1=0$. Let ${\mathcal{S}}^{\ast }(\alpha )$ and $\mathcal{K}(\alpha )$ denote the subclasses of $\mathcal{A}$ consisting of starlike and convex functions of order α ($0\le \alpha <1$) and let ${\mathcal{S}}^{\ast }(0)={\mathcal{S}}^{\ast }$ and $\mathcal{K}(0)=\mathcal{K}$. If f and g are analytic in $\mathbb{U}$, we say that f is subordinate to g in $\mathbb{U}$, written as $f\prec g$ or $f(z)\prec g(z)$, if there exists a Schwarz function w such that $f(z)=g(w(z))$ ($z\in \mathbb{U}$).

A function $f\in \mathcal{A}$ is said to be prestarlike of order α in $\mathbb{U}$ if

$$\frac{z}{{(1-z)}^{2(1-\alpha )}}\ast f(z)\in {\mathcal{S}}^{\ast }(\alpha )(0\le \alpha <1),$$

where $f\ast g$ denotes the familiar Hadamard product (or convolution) of two analytic functions f and g in $\mathbb{U}$. We denote this class by $\mathcal{R}(\alpha )$ (see, for details, [1]). We note that $\mathcal{R}(0)=\mathcal{K}$ and $\mathcal{R}(1/2)={\mathcal{S}}^{\ast }(1/2)$.

Let $\mathcal{N}$ be the class of all functions h which are analytic and univalent in $\mathbb{U}$ and for which $h(\mathbb{U})$ is convex with $h(0)=1$.

Let ℳ denote the class of functions of the form

$$f(z)=\frac{1}{z}+\sum _{k=0}^{\mathrm{\infty }}{a}_k{z}^k$$

which are analytic in the punctured open unit disk $\mathbb{D}=\mathbb{U}\setminus \{0\}$.

For any $n\in {\mathbb{N}}_0=\{0,1,2,\dots \}$, we denote the multiplier transformations $D_{\lambda }^n$ of functions $f\in \mathcal{M}$ by

$$D_{\lambda }^{n}f(z)=\frac{1}{z}+\sum _{k=0}^{\mathrm{\infty }}{\left(\frac{k+1+l}{\lambda }\right)}^n{a}_k{z}^k(\lambda >0;z\in \mathbb{D}).$$

Obviously, we have

$$D_{\lambda }^s(D_{\lambda }^{t}f(z))=D_{\lambda }^{s+t}f(z)$$

for all nonnegative integers s and t. The operators $D_{\lambda }^n$ and $D_1^n$ are the multiplier transformations introduced and studied by Sarangi and Uraligaddi [2] and Uralegaddi and Somanatha [3, 4], respectively. Analogous to $D_{\lambda }^n$, we here define a new multiplier transformation $I_{\lambda ,\mu }^n$ as follows.

Let $f_n(z)=1/z+{\sum }_{k=0}^{\mathrm{\infty }}{((k+1+\lambda )/\lambda )}^n{z}^k$, $n\in {\mathbb{N}}_0$, and let $f_{n,\mu }^{†}$ be such that

$$f_n(z)\ast f_{n,\mu }^{†}(z)=\frac{1}{z}+\sum _{k=0}^{\mathrm{\infty }}\frac{{(\mu )}_{k+1}}{{(1)}_{k+1}}{z}^k(\mu >0;z\in \mathbb{D}),$$

where ${(\nu )}_k$ is the Pochhammer symbol (or the shifted factorial) defined (in terms of the gamma function) by

$${(\nu )}_k:=\frac{\mathrm{\Gamma }(\nu +k)}{\mathrm{\Gamma }(\nu )}=\{\begin{array}{cc}1\hfill & \text{if }k=0\text{ and }\nu \in \mathbb{C}\setminus \{0\},\hfill \\ \nu (\nu +1)\cdots (\nu +k-1)\hfill & \text{if }k\in \mathbb{N}:=\{1,2,\dots \}\text{ and }\nu \in \mathbb{C}.\hfill \end{array}$$

Then

$$I_{\lambda ,\mu }^{n}f(z)=f_{n,\mu }^{†}(z)\ast f(z).$$

(1.1)

We note that $I_{1,2}^{0}f(z)=z{f}^{\mathrm{\prime }}(z)+2f(z)$ and $I_{1,2}^{1}f(z)=f(z)$. It is easily verified from (1.1) that

$$z{(I_{\lambda ,\mu }^{n+1}f(z))}^{\mathrm{\prime }}=\lambda I_{\lambda ,\mu }^{n}f(z)-(\lambda +1)I_{\lambda ,\mu }^{n+1}f(z)$$

(1.2)

and

$$z{(I_{\lambda ,\mu }^{n}f(z))}^{\mathrm{\prime }}=\mu I_{\lambda ,\mu +1}^{n}f(z)-(\mu +1)I_{\lambda ,\mu }^{n}f(z).$$

(1.3)

The definition (1.1) of the multiplier transformation $I_{\lambda ,\mu }^n$ is motivated essentially by the Choi-Saigo-Srivastava operator [5] for analytic functions, which includes the Noor integral operator studied by Liu [6] (also, see [7–9]).

We also define the function $\varphi (a,c;z)$ by

$$\varphi (a,c;z):=\frac{1}{z}+\sum _{k=0}^{\mathrm{\infty }}\frac{{(a)}_{k+1}}{{(c)}_{k+1}}{z}^k(z\in \mathbb{U};a\in \mathbb{R};c\in \mathbb{R}\setminus {\mathbb{Z}}_0^{-};{\mathbb{Z}}_0^{-}:=\{-1,-2,\dots \}).$$

(1.4)

By using the operator $I_{\lambda ,\mu }^n$, we introduce the following class of analytic functions for $\gamma >0$, $\lambda >0$, $s\in \mathbb{R}$, $\mu >0$ and $h\in \mathcal{N}$:

$${\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h):=\{f\in \mathcal{M}:(1-\gamma )z{I}_{\lambda ,\mu }^{n}f(z)+\gamma z^2{(I_{\lambda ,\mu }^{n}f(z))}^{\prime }\prec h(z)\}.$$

In the present paper, we derive some inclusion relations, convolution properties and integral preserving properties for the class ${\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h)$.

The following lemmas will be required in our investigation.

Lemma 1.1 [[10], Lemma 2, p.192]

Let g be analytic in $\mathbb{U}$ and h be analytic and convex univalent in U with $h(0)=g(0)$. If

$$g(z)+\frac{1}{\gamma }z{g}^{\prime }(z)\prec h(z)(Re\{\gamma \}\ge 0;\gamma \ne 0),$$

(1.5)

then

$$g(z)\prec \tilde{h}(z)=\gamma z^{-\gamma }{\int }_0^z{t}^{\gamma -1}h(t)dt\prec h(z)$$

and $\tilde{h}$ is the best dominant of (1.5).

Lemma 1.2 [[1], Theorem 2.4, p.54]

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

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

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

Lemma 1.3 [[11], Lemma 5, p.656]

Let $0<a\le c$. Then

$$Re\{z\varphi (a,c;z)\}>\frac{1}{2}(z\in \mathbb{U}),$$

where ϕ is given by (1.4).

2 Inclusion relations

Theorem 2.1 If $0\le {\gamma }_1<{\gamma }_2$, then

$${\mathcal{M}}_{\lambda ,\mu }^n({\gamma }_2;h)\subset {\mathcal{M}}_{\lambda ,\mu }^n({\gamma }_1;h).$$

Proof

Let

$$g(z)=z{I}_{\lambda ,\mu }^{n}f(z)(f\in {\mathcal{M}}_{\lambda ,\mu }^n({\gamma }_2;h):z\in \mathbb{U}).$$

(2.1)

Then the function g is analytic in $\mathbb{U}$ with $g(0)=1$. Differentiating both sides of (2.1), we have

$$(1+{\gamma }_2)z{I}_{\lambda ,\mu }^{n}f(z)+{\gamma }_2{z}^2{(I_{\lambda ,\mu }^{n}f(z))}^{\prime }=g(z)+{\gamma }_{2}z{g}^{\prime }(z)\prec h(z).$$

(2.2)

Hence an application of Lemma 1.1 with $\mu =1/{\gamma }_2$ yields

$$g(z)\prec h(z).$$

(2.3)

Since $0\le {\gamma }_1/{\gamma }_2<1$ and h is convex univalent in U, it follows from (2.1), (2.2) and (2.3) that

$$\begin{array}{c}(1+{\gamma }_1)z{I}_{\lambda ,\mu }^{n}f(z)+{\gamma }_1{z}^2{(I_{\lambda ,\mu }^{n}f(z))}^{\prime }\hfill \\ =\frac{{\gamma }_1}{{\gamma }_2}[(1-{\gamma }_2)z{I}_{\lambda ,\mu }^{n}f(z)+{\gamma }_2{z}^2{(I_{\lambda ,\mu }^{n}f(z))}^{\prime }]+(1-\frac{{\gamma }_1}{{\gamma }_2})g(z)\hfill \\ \prec h(z).\hfill \end{array}$$

Therefore $f\in {\mathcal{M}}_{\lambda ,\mu }^n({\gamma }_1;h)$, and so we complete the proof of Theorem 2.1. □

Theorem 2.2 If $0<{\mu }_1\le {\mu }_2$, then

$${\mathcal{M}}_{\lambda ,{\mu }_2}^n(\gamma ;h)\subset {\mathcal{M}}_{\lambda ,{\mu }_1}^n(\gamma ;h).$$

Proof Let $f\in {\mathcal{M}}_{\lambda ,{\mu }_2}^n(\gamma ;h)$. Then

(2.4)

In view of Lemma 1.3, we see that the function $z\varphi ({\mu }_1,{\mu }_2;z)$ has the Herglotz representation

$$z\varphi ({\mu }_1,{\mu }_2;z)={\int }_{|x|=1}\frac{d\mu (x)}{1-xz}(z\in \mathbb{U}),$$

(2.5)

where $\mu (x)$ is a probability measure defined on the unit circle $|x|<1$ and

$${\int }_{|x|=1}d\mu (x)=1.$$

Since h is convex univalent in $\mathbb{U}$, it follows from (2.4) and (2.5) that

$$(1+\gamma )z{I}_{\lambda ,{\mu }_1}^{n}f(z)+\gamma z^2{(I_{\lambda ,{\mu }_1}^{n}f(z))}^{\prime }={\int }_{|x|=1}h(xz)d\mu (x)\prec h(z),$$

which completes the proof of Theorem 2.2. □

Theorem 2.3 If $\mu >0$, then

$${\mathcal{M}}_{\lambda ,\mu +1}^n(\gamma ;\tilde{h})\subset {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h),$$

where

$$\tilde{h}(z)=\mu z^{-\mu }{\int }_0^z{t}^{\mu -1}h(t)dt\prec h(z).$$

Proof

Let

$$g(z)=(1+\gamma )z{I}_{\lambda ,\mu }^{n}f(z)+\gamma z^2{(I_{\lambda ,\mu }^{n}f(z))}^{\prime }(f\in \mathcal{M};z\in \mathbb{U}).$$

(2.6)

Then from (1.4) and (2.6), we have

$$z^{-1}g(z)=\gamma \mu I_{\lambda ,\mu +1}^{n}f(z)+(1-\gamma \mu )I_{\lambda ,\mu }^{n}f(z).$$

(2.7)

Differentiating both sides of (2.6) and using (1.4), we obtain

(2.8)

By a simple calculation with (2.7) and (2.8), we get

$$g(z)+\frac{z{g}^{\prime }(z)}{\mu }=(1+\gamma )\frac{I_{\lambda ,\mu +1}^{n}f(z)}{z}+\gamma {(I_{\lambda ,\mu +1}^{n}f(z))}^{\prime }.$$

(2.9)

If $f\in {\mathcal{M}}_{\lambda ,\mu +1}^n(\gamma ;h)$, then it follows from (2.9) that

$$g(z)+\frac{z{g}^{\prime }(z)}{\mu }\prec h(z)(\mu >0).$$

Hence an application of Lemma 1.1 yields

$$g(z)\prec \tilde{h}(z)=\mu z^{-\mu }{\int }_0^z{t}^{\mu -1}h(t)dt\prec h(z),$$

which shows that

$$f\in {\mathcal{M}}_{\lambda ,\mu +1}^n(\gamma ;\tilde{h})\subset {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h).$$

 □

Theorem 2.4 If $s\in \mathbb{R}$ and $\lambda >0$, then

$${\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;\tilde{h})\subset {\mathcal{M}}_{\lambda ,\mu }^{n+1}(\gamma ;h),$$

where

$$\tilde{h}(z)=\lambda z^{-\lambda }{\int }_0^z{t}^{\lambda -1}h(t)dt\prec h(z).$$

Proof By using the same techniques as in the proof of Theorem 2.3 and (1.5), we have Theorem 2.4 and so we omit the detailed proof involved. □

Theorem 2.5 Let $\gamma >0$, $\beta >0$ and $f\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;\beta h+1-\beta )$. If $\beta \le {\beta }_0$, where

$${\beta }_0=\frac{1}{2}{(1-\frac{1}{\gamma }{\int }_0^1\frac{u^{\frac{1}{\gamma }-1}}{1+u}du)}^{-1},$$

(2.10)

then $f\in {\mathcal{M}}_{\lambda ,\mu }^n(0;h)$. The bound ${\beta }_0$ is sharp for the function

$$h(z)=\frac{1}{1-z}(z\in \mathbb{U}).$$

Proof

Let

$$g(z)=z{I}_{\lambda ,\mu }^{n}f(z)(f\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;\beta h+1-\beta );\gamma >0;\beta >0).$$

(2.11)

Then we have

$$\begin{array}{rcl}g(z)+\gamma z{g}^{\prime }(z)& =& (1+\gamma )z{I}_{\lambda ,\mu }^{n}f(z)+\gamma z^2{(I_{\lambda ,\mu }^{n}f(z))}^{\prime }\\ \prec & \beta h(z)+1-\beta .\end{array}$$

Hence an application of Lemma 1.1 yields

$$g(z)\prec \frac{\beta }{\gamma }{z}^{-\frac{1}{\gamma }}{\int }_0^z{t}^{\frac{1}{\gamma }-1}h(t)dt+1-\beta =(h\ast \psi )(z),$$

(2.12)

where

$$\psi (z)=\frac{\beta }{\gamma }{z}^{-\frac{1}{\gamma }}{\int }_0^z\frac{t^{\frac{1}{\gamma }-1}}{1-t}dt+1-\beta .$$

(2.13)

If $0<\beta \le {\beta }_0$, where ${\beta }_0$ is given by (2.10), then from (2.13), we have

$$\begin{array}{rcl}Re\{\psi (z)\}& =& \frac{\beta }{\gamma }{\int }_0^1{u}^{\frac{1}{\gamma }-1}Re\left\{\frac{1}{1-uz}du\right\}+1-\beta \\ >& \frac{\beta }{\gamma }{\int }_0^1\frac{u^{\frac{1}{\gamma }-1}}{1+u}du+1-\beta \\ \ge & \frac{1}{2}.\end{array}$$

By using the Herglotz representation for ψ, it follows from (2.11) and (2.12) that

$$z{I}_{\lambda ,\mu }^{n}f(z)\prec (h\ast \psi )(z)\prec h(z)$$

since h is convex univalent in $\mathbb{U}$. This shows that $f\in {\mathcal{M}}_{\lambda ,\mu }^n(0;h)$.

For $h(z)=1/(1-z)$ and $f\in \mathcal{M}$ defined by

$$z{I}_{\lambda ,\mu }^{n}f(z)=\frac{\beta }{\gamma }{z}^{-\frac{1}{\gamma }}{\int }_0^z\frac{t^{\frac{1}{\gamma }-1}}{1-t}dt+1-\beta ,$$

it is easy to verify that

$$(1+\gamma )z{I}_{\lambda ,\mu }^{n}f(z)+\gamma z^2{(I_{\lambda ,\mu }^{n}f(z))}^{\prime }=\beta h(z)+1-\beta .$$

Thus $f\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;\beta h+1-\beta )$. Furthermore, for $\beta >{\beta }_0$, we have

$$Re\{z{I}_{\lambda ,\mu }^{n}f(z)\}\to \frac{\beta }{\gamma }{\int }_0^1\frac{u^{\frac{1}{\gamma }-1}}{1+u}du+1-\beta <\frac{1}{2}(z\to -1),$$

which implies that $f\notin {\mathcal{M}}_{\lambda ,\mu }^n(0;h)$. Hence the bound ${\beta }_0$ cannot be increased when $h(z)=1/(1-z)$ ($z\in \mathbb{U}$). □

3 Convolution properties

Theorem 3.1 If $f\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h)$ and

$$Re\{zg(z)\}>\frac{1}{2}(g\in \mathcal{M};z\in \mathbb{U}),$$

then

$$f\ast g\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h).$$

Proof Let $f\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h)$ and $g\in \mathcal{M}$. Then we have

$$(1+\gamma )z{I}_{\lambda ,\mu }^n(f\ast g)(z)+\gamma z^2{(I_{\lambda ,\mu }^n(f\ast g)(z))}^{\prime }=zg(z)\ast \psi (z),$$

where

$$\psi (z)=(1+\gamma )z\frac{I_{\lambda ,\mu }^{n}f(z)}{+}\gamma z^2{(I_{\lambda ,\mu }^{n}f(z))}^{\prime }\prec h(z).$$

The remaining part of the proof of Theorem 3.1 is similar to that of Theorem 2.2, and so we omit the details involved. □

Corollary 3.1 Let $f\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h)$ be given by (1.1). Then the function

$${\sigma }_m(z)={\int }_0^{1}t{S}_m(tz)dt(z\in \mathbb{U}),$$

where

$$S_m(z)=\frac{1}{z}+\sum _{n=1}^{m-1}{a}_n{z}^{n-1}(m\in \mathbb{N}\setminus \{1\};z\in \mathbb{U}),$$

is also in the class ${\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h)$.

Proof

We have

$${\sigma }_m(z)=\frac{1}{z}+\sum _{n=1}^{m-1}\frac{a_n}{n-1}{z}^{n+1}=(f\ast g_m)(z)(m\in \mathbb{N}\setminus \{1\}),$$

(3.1)

where

$$f(z)=\frac{1}{z}+\sum _{n=1}^{\mathrm{\infty }}{a}_n{z}^{n-1}\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h)$$

and

$$g_m(z)=\frac{1}{z}+\sum _{n=1}^{m-1}\frac{z^n}{n-1}\in \mathcal{M},$$

while, it is known [4] that

$$Re\{z{g}_m(z)\}=Re\{1+\sum _{n=1}^{m-1}\frac{z^n}{n+1}\}>\frac{1}{2}(m\in \mathbb{N}\setminus \{1\};z\in \mathbb{U}).$$

(3.2)

In view of (3.1) and (3.2), an application of Theorem 3.1 leads to ${\sigma }_m\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h)$. □

Theorem 3.2 If $f\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h)$ and

$$z^{2}g(z)\in R(\alpha )(g\in \mathcal{M};z\in \mathbb{U}),$$

then

$$(f\ast g)\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h).$$

Proof

By using a similar method as in the proof of Theorem 3.1, we have

$$(1+\gamma )z{I}_{\lambda ,\mu }^n(f\ast g)(z)+\gamma z^2{(I_{\lambda ,\mu }^n(f\ast g)(z))}^{\prime }=\frac{z^{2}g(z)\ast (z\psi (z))}{z^{2}g(z)\ast z}(z\in \mathbb{U}),$$

(3.3)

where

$$\psi (z)=(1+\gamma )z{I}_{\lambda ,\mu }^{n}f(z)+\gamma z^2{(I_{\lambda ,\mu }^{n}f(z))}^{\prime }\prec h(z).$$

Since h is convex univalent in $\mathbb{U}$, it follows from (3.3) and Lemma 1.2 that Theorem 3.2 holds true. □

If we take $\alpha =0$ and $\alpha =1/2$ in Theorem 3.2, we have the following corollary.

Corollary 3.2 If $f\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h)$ and $g\in \mathcal{M}$ satisfies one of the following conditions:

• (i) $z^{2}g(z)$ is convex univalent in $\mathbb{U}$

or

• (ii) $z^{2}g(z)\in S^{\ast }(\frac{1}{2})$,

then $(f\ast g)\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h)$.

4 Integral operators

Theorem 4.1 If $f\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h)$, then the function F defined by

$$F(z)=\frac{c-1}{z^c}{\int }_0^z{t}^{c-1}f(t)dt(Re\{c\}>1)$$

(4.1)

is in the class ${\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;\tilde{h})$, where

$$\tilde{h}(z)=(c-1)z^{-(c-1)}{\int }_0^z{t}^{c}h(t)dt\prec h(z).$$

Proof Let $f\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h)$. Then from (4.1), we obtain

$$(c-1)f(z)=z{F}^{\prime }(z)+cF(z).$$

(4.2)

Define the function G by

$$z^{-1}G(z)=(1+\gamma )I_{\lambda ,\mu }^{n}F(z)+\gamma z{(I_{\lambda ,\mu }^{n}F(z))}^{\prime }(z\in \mathbb{D}).$$

(4.3)

Differentiating both sides of (4.3) with respect to z, we get

$$z{G}^{\prime }(z)-G(z)=(1+\gamma )z{I}_{\lambda ,\mu }^n(z{F}^{\prime }(z))+\gamma z^2{\left(I_{\lambda ,\mu }^n(z{F}^{\prime }(z))\right)}^{\prime }.$$

(4.4)

Furthermore, it follows from (4.2), (4.3) and (4.4) that

(4.5)

Since $f\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h)$, from (4.5), we have

$$G(z)+\frac{1}{c-1}z{G}^{\prime }(z)\prec h(z)(Re\{c\}>1),$$

and so an application of Lemma 1.1 yields

$$G(z)\prec \tilde{h}(z)=\frac{c-1}{z^{c-1}}{\int }_0^z{t}^{c}h(t)dt\prec h(z).$$

Therefore we conclude that

$$F\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;\tilde{h})\subset {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h).$$

 □

Theorem 4.2 If $f\in \mathcal{M}$ and F are defined as in Theorem  4.1, if

$$(1-\alpha )z{I}_{\lambda ,\mu }^{n}F(z)+\alpha z{I}_{\lambda ,\mu }^{n}f(z)\prec h(z)(\alpha >0),$$

(4.6)

then $F\in {\mathcal{M}}_{\lambda ,\mu }^n(0;\tilde{h})$, where

$$\tilde{h}(z)=\frac{c-1}{\alpha }{z}^{-\frac{\alpha }{c-1}}{\int }_0^z{t}^{\frac{c-1}{\alpha }-1}h(t)\prec h(z)(Re\{c\}>1).$$

Proof

Let

$$G(z)=z{I}_{\lambda ,\mu }^{n}F(z)(z\in \mathbb{D}).$$

(4.7)

Then G is analytic in $\mathbb{U}$ with $G(0)=1$ and

$$z{G}^{\prime }(z)=z^2{(I_{\lambda ,\mu }^{n}F(z))}^{\prime }+G(z).$$

(4.8)

It follows from (4.2), (4.6), (4.7) and (4.8) that

$$\begin{array}{c}(1-\alpha )z{I}_{\lambda ,\mu }^{n}F(z)+\alpha z{I}_{\lambda ,\mu }^{n}f(z)\hfill \\ =(1-\alpha )z\frac{I_{\lambda ,\mu }^{n}F(z)}{+}\frac{\alpha }{c-1}[cz{I}_{\lambda ,\mu }^{n}F(z)+z^2{(I_{\lambda ,\mu }^{n}F(z))}^{\prime }]\hfill \\ =G(z)+\frac{\alpha }{c-1}z{G}^{\prime }(z)\prec h(z)(Re\{c\}>1;\alpha >0).\hfill \end{array}$$

Therefore, by Lemma 1.1, we conclude that Theorem 4.2 holds true as stated. □

Theorem 4.3 Let $F\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h)$. If the function f is defined by

$$F(z)=\frac{c-1}{z^c}{\int }_0^z{t}^{c-1}f(t)dt(c>1),$$

(4.9)

then

$$\sigma f(\sigma z)\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h),$$

where

$$\sigma =\sigma (c)=\frac{\sqrt{1+{(c-1)}^2}-1}{c-1}.$$

(4.10)

The bound σ is sharp for the function

$$h(z)=\beta +(1-\beta )\frac{1+z}{1-z}(\beta \ne 1;z\in \mathbb{U}).$$

(4.11)

Proof We note that for $F\in \mathcal{M}$,

$$F(z)=F(z)\ast \frac{1}{z(1-z)}\text{and}z{F}^{\prime }(z)=F(z)\ast (\frac{1}{z{(1-z)}^2}-\frac{2}{z^2(1-z)}).$$

Then from (4.9), we have

$$f(z)=\frac{cF(z)+z{F}^{\prime }(z)}{c-1}=(F\ast g)(z)(c>1;z\in \mathbb{D}),$$

(4.12)

where

$$g(z)=\frac{1}{c-1}((c-2)\frac{1}{z(1-z)}+\frac{1}{z{(1-z)}^2})\in \mathcal{M}.$$

(4.13)

Next, we show that

$$Re\{zg(z)\}>\frac{1}{2}(|z|<\sigma ),$$

(4.14)

where $\sigma =\sigma (c)$ is given by (4.10). Letting

$$\frac{1}{1-z}=R{e}^{i\theta }(|z|=r<1;R>0),$$

we see that

$$cos\theta =\frac{1+R^2(1-r^2)}{2R}\text{and}R\ge \frac{1}{1+r}.$$

(4.15)

Then for (4.13) and (4.15), we have

$$\begin{array}{rcl}2Re\{zg(z)\}& =& \frac{2}{c-1}[(c-2)Rcos\theta +R^2(2{cos}^2\theta -1)]\\ =& \frac{R^2}{c-1}[c(1-r^2)+R^2{(1-r^2)}^2-2]+1\\ \ge & \frac{R^2}{c-1}[c-1-2r-(c-1)r^2]+1.\end{array}$$

This evidently gives (4.14), which is equivalent to

$$Re\{\sigma zg(\sigma z)\}>\frac{1}{2}(z\in \mathbb{U}).$$

(4.16)

Let $F\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h)$. Then, by using (4.12) and (4.16), an application of Theorem 3.1 yields

$$\sigma f(\sigma z)=F(z)\ast \sigma g(\sigma z)\in {\mathcal{M}}_{\lambda ,\mu }^n(\gamma ;h).$$

For h given by (4.11), we consider the function $F\in \mathcal{M}$ defined by

$$(1+\gamma )z{I}_{\lambda ,\mu }^{n}F(z)+\gamma z^2{(I_{\lambda ,\mu }^{n}F(z))}^{\prime }=\beta +(1-\beta )\frac{1+z}{1-z}(\beta \ne 1;z\in \mathbb{U}).$$

(4.17)

Then from (4.3), (4.5) and (4.17), we find that

$$\begin{array}{c}(1+\gamma )z{I}_{\lambda ,\mu }^{n}f(z)+\gamma z^2{(I_{\lambda ,\mu }^{n}f(z))}^{\prime }\hfill \\ =\beta +(1-\beta )\frac{1+z}{1-z}+\frac{z}{c-1}{(\beta +(1-\beta )\frac{1+z}{1-z})}^{\prime }\hfill \\ =\beta +\frac{(1-\beta )(c-1+2z-(c-1)z^2)}{(c-1){(1-z)}^2}\hfill \\ =\beta (z=-\sigma ).\hfill \end{array}$$

Therefore we conclude that the bound $\sigma =\sigma (c)$ cannot be increased for each c ($c>1$). □