Some subclasses of multivalent spirallike meromorphic functions

Abstract

In the present paper, we introduce and investigate two new subclasses ${\mathcal{MS}}_p(\alpha ,\beta )$ and ${\mathcal{MC}}_p(\alpha ,\beta )$ of meromorphic functions. Such results as integral representations and coefficient inequalities are proved. The results presented here would provide extensions of those given in earlier works.

MSC:30C45, 30C80.

Dedication

Dedicated to Professor Hari M Srivastava

1 Introduction

Let ${\mathrm{\Sigma }}_p$ denote the class of functions f of the form

$$f(z)=z^{-p}+\sum _{n=1-p}^{\mathrm{\infty }}{a}_n{z}^n,$$

(1.1)

which are analytic in the punctured open unit disk

$${\mathbb{U}}^{\ast }:=\{z:z\in \mathbb{C}\text{ and }0<|z|<1\}=:\mathbb{U}\setminus \{0\}.$$

Let denote the class of functions p given by

$$p(z)=1+\sum _{n=1}^{\mathrm{\infty }}{p}_n{z}^n(z\in \mathbb{U}),$$

which are analytic in and satisfy the condition

$$\mathrm{\Re }(p(z))>0(z\in \mathbb{U}).$$

A function $f\in {\mathrm{\Sigma }}_p$ is said to be in the class ${\mathcal{MS}}_p(\alpha )$ of meromorphic p-valent starlike functions of order α if it satisfies the inequality

$$\mathrm{\Re }\left(\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}\right)<-\alpha (z\in \mathbb{U};0\leqq \alpha <p).$$

(1.2)

Moreover, a function $f\in {\mathrm{\Sigma }}_p$ is said to be in the class ${\mathcal{MK}}_p(\alpha )$ of meromorphic p-valent convex functions of order α if it satisfies the inequality

$$\mathrm{\Re }(1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)})<-\alpha (z\in \mathbb{U};0\leqq \alpha <p).$$

(1.3)

It is readily verified from (1.2) and (1.3) that

$$f\in {\mathcal{MK}}_p(\alpha )⟺-\frac{z{f}^{\mathrm{\prime }}}{p}\in {\mathcal{MS}}_p^{\ast }(\alpha ).$$

In [1], Wang et al. introduced and investigated two new subclasses of the class ${\mathrm{\Sigma }}_p$. A function $f\in {\mathrm{\Sigma }}_p$ is said to be in the class ${\mathcal{M}}_p(\beta )$ if it is characterized by the condition

$$\mathrm{\Re }\left(\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}\right)>-\beta (z\in \mathbb{U};\beta >p).$$

Also, a function $f\in {\mathrm{\Sigma }}_p$ is said to be in the class ${\mathcal{N}}_p(\beta )$ if and only if

$$\mathrm{\Re }(1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)})>-\beta (z\in \mathbb{U};\beta >p).$$

Let ${\mathcal{A}}_p$ be the class of functions of the form

$$f(z)=z^p+\sum _{n=p+1}^{\mathrm{\infty }}{a}_n{z}^n$$

which are analytic in . If it satisfies the condition

$$\mathrm{\Re }\left(e^{i\alpha }\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}\right)<\beta (z\in \mathbb{U};-\frac{\pi }{2}<\alpha <\frac{\pi }{2};\beta >pcos\alpha ),$$

then we say that $f\in {\mathcal{S}}_p(\alpha ,\beta )$. Furthermore, let ${\mathcal{C}}_p(\alpha ,\beta )$ denote the subclass of ${\mathcal{A}}_p$ consisting of functions which satisfy the inequality

$$\mathrm{\Re }\left(e^{i\alpha }(1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)})\right)<\beta (z\in \mathbb{U};-\frac{\pi }{2}<\alpha <\frac{\pi }{2};\beta >pcos\alpha ).$$

The function classes ${\mathcal{S}}_p(\alpha ,\beta )$ and ${\mathcal{C}}_p(\alpha ,\beta )$ were introduced and studied recently by Uyanik et al. [2].

Motivated essentially by the above mentioned work, we introduce and investigate the following two subclasses of the class ${\mathrm{\Sigma }}_p$ of meromorphic functions.

Definition 1 A function $f\in {\mathrm{\Sigma }}_p$ is said to be in the class ${\mathcal{MS}}_p(\alpha ,\beta )$ if it satisfies the condition

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

(1.4)

for some real α and β, where (and throughout this paper unless otherwise mentioned) the parameters α and β are constrained as follows:

$$|\alpha |<\frac{\pi }{2}\text{and}\beta >pcos\alpha .$$

Furthermore, a function $f\in {\mathrm{\Sigma }}_p$ is said to be in the class ${\mathcal{MC}}_p(\alpha ,\beta )$ if it satisfies the inequality

$$\mathrm{\Re }\left(e^{i\alpha }(1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)})\right)>-\beta (z\in \mathbb{U}).$$

(1.5)

Remark 1 Taking $\alpha =0$, we get the function classes introduced by Wang et al. [1].

Remark 2 We note that $f\in {\mathcal{MS}}_p(\alpha ,\beta )$ if and only if

$$-e^{i\alpha }\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}\prec \frac{p{e}^{i\alpha }-(2\beta -p{e}^{-i\alpha })z}{1-z}.$$

(1.6)

Also, $f\in {\mathcal{MC}}_p(\alpha ,\beta )$ if and only if

$$-e^{i\alpha }(1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)})\prec \frac{p{e}^{i\alpha }-2(\beta -p{e}^{-i\alpha })z}{1-z}.$$

(1.7)

For some investigations of meromorphic functions, see (for example) the works [1, 3–10] and the references cited in.

In the present paper, we aim at proving some interesting properties such as integral representations and coefficient inequalities of the function classes ${\mathcal{MS}}_p(\alpha ,\beta )$ and ${\mathcal{MC}}_p(\alpha ,\beta )$.

2 Main results

We begin by presenting an integral representation of functions belonging to the class ${\mathcal{MS}}_p(\alpha ,\beta )$.

Theorem 1 Let $f\in {\mathcal{MS}}_p(\alpha ,\beta )$. Then

$$f(z)=z^{-p}\cdot exp(2(\beta -pcos\alpha )e^{-i\alpha }{\int }_0^z\frac{\omega (t)}{t(1-\omega (t))}dt)(z\in {\mathbb{U}}^{\ast }),$$

(2.1)

where ω is analytic in with $\omega (0)=0$ and $|\omega (z)|<1$.

Proof For $f\in {\mathcal{MS}}_p(\alpha ,\beta )$, we know that (1.6) holds true. It follows that

$$-e^{i\alpha }\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}=p{e}^{i\alpha }-\frac{2(\beta -pcos\alpha )\omega (z)}{1-\omega (z)},$$

(2.2)

where ω is analytic in with $\omega (0)=0$ and $|\omega (z)|<1$. We next find from (2.2) that

$$\frac{f^{\mathrm{\prime }}(z)}{f(z)}+\frac{p}{z}=\frac{2(\beta -pcos\alpha )e^{-i\alpha }\omega (z)}{z(1-\omega (z))}(z\in {\mathbb{U}}^{\ast }),$$

(2.3)

which, upon integration, yields

$$log(z^{p}f(z))=2(\beta -pcos\alpha )e^{-i\alpha }{\int }_0^z\frac{\omega (t)}{t(1-\omega (t))}dt.$$

(2.4)

The assertion (2.1) of Theorem 1 can be easily derived from (2.4). □

Note that $f\in {\mathcal{MS}}_p(\alpha ,\beta )$ if and only if

$$-\frac{z{f}^{\mathrm{\prime }}(z)}{p}\in {\mathcal{MC}}_p(\alpha ,\beta ),$$

we get the following result.

Corollary 1 Let $f\in {\mathcal{MC}}_p(\alpha ,\beta )$. Then

$$f(z)=-p{\int }_{z_0}^z{u}^{-p-1}\cdot exp(2(\beta -pcos\alpha )e^{-i\alpha }{\int }_0^u\frac{\omega (t)}{t(1-\omega (t))}dt)du(z\in {\mathbb{U}}^{\ast }),$$

where ω is analytic in with $\omega (0)=0$ and $|\omega (z)|<1$.

Next, we discuss the coefficient estimates of functions belonging to the classes ${\mathcal{MS}}_p(\alpha ,\beta )$ and ${\mathcal{MC}}_p(\alpha ,\beta )$. The following lemma will be required in the proof of Theorem 2.

Lemma 1 Let $p\in \mathbb{N}$. Suppose also that the sequence ${\{A_{p+m}\}}_{m=0}^{\mathrm{\infty }}$ is defined by

$$\{\begin{array}{cc}{A}_p=\frac{\beta -pcos\alpha }{p}\hfill & (m=0),\hfill \\ A_{p+m}=\frac{2(\beta -pcos\alpha )}{2p+m}(1+{\sum }_{k=0}^{m-1}{A}_{p+k})\hfill & (m\in \mathbb{N}).\hfill \end{array}$$

(2.5)

Then

$$\begin{array}{rl}{A}_{p+m}=& \frac{2(\beta -pcos\alpha )}{2\beta +m+2p-2pcos\alpha }\prod _{k=0}^m\frac{2\beta +k+2p-2pcos\alpha }{2p+k}\\ (m\in {\mathbb{N}}_0:=\mathbb{N}\cup \{0\}).\end{array}$$

(2.6)

Proof By virtue of (2.5), we get

$$(2p+m+1)A_{p+m+1}=2(\beta -pcos\alpha )(1+\sum _{k=0}^m{A}_{p+k}),$$

(2.7)

and

$$(2p+m)A_{p+m}=2(\beta -pcos\alpha )(1+\sum _{k=0}^{m-1}{A}_{p+k}).$$

(2.8)

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

$$\frac{A_{p+m+1}}{A_{p+m}}=\frac{2\beta +m+2p-2pcos\alpha }{2p+m+1}(m\in {\mathbb{N}}_0).$$

(2.9)

Thus,

$$\begin{array}{rl}{A}_{p+m}& =\frac{A_{p+m}}{A_{p+m-1}}\cdot \frac{A_{p+m-1}}{A_{p+m-2}}\cdot \cdot \cdot \frac{A_{p+1}}{A_p}\cdot A_p\\ =\frac{2\beta +m-1+2p-2pcos\alpha }{2p+m}\cdot \cdot \cdot \frac{2\beta +2p-2pcos\alpha }{2p+1}\cdot \frac{2\beta -2pcos\alpha }{2p}\\ =\frac{2(\beta -pcos\alpha )}{2\beta +m+2p-2pcos\alpha }\prod _{k=0}^m\frac{2\beta +k+2p-2pcos\alpha }{2p+k}(m\in \mathbb{N}).\end{array}$$

(2.10)

The proof of Lemma 1 is thus completed. □

Theorem 2 Let $f(z)=z^{-p}+{\sum }_{m=0}^{\mathrm{\infty }}{a}_{p+m}{z}^{p+m}\in {\mathcal{MS}}_p(\alpha ,\beta )$. Then

$$|a_{p+m}|\leqq \frac{2(\beta -pcos\alpha )}{2\beta +m+2p-2pcos\alpha }\prod _{k=0}^m\frac{2\beta +k+2p-2pcos\alpha }{2p+k}(m\in {\mathbb{N}}_0).$$

(2.11)

Proof Let

$$h(z):=\frac{\beta +e^{i\alpha }\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}+ipsin\alpha }{\beta -pcos\alpha }(z\in \mathbb{U};f\in {\mathcal{MS}}_p(\alpha ,\beta )).$$

(2.12)

We know that $h\in \mathcal{P}$. It follows that

$$e^{i\alpha }z{f}^{\mathrm{\prime }}(z)=(\beta -pcos\alpha )f(z)h(z)-(\beta +ipsin\alpha )f(z).$$

(2.13)

Suppose that

$$h(z)=1+h_{1}z+h_2{z}^2+\cdots .$$

(2.14)

Then

$$\begin{array}{r}{e}^{i\alpha }(-p{z}^{-p}+p{a}_p{z}^p+(p+1)a_{p+1}{z}^{p+1}+\cdots +(p+m)a_{p+m}{z}^{p+m}+\cdots )\\ =(\beta -pcos\alpha )(z^{-p}+a_p{z}^p+a_{p+1}{z}^{p+1}+\cdots )\times (1+h_{1}z+h_2{z}^2+\cdots )\\ -(\beta +ipsin\alpha )(z^{-p}+a_p{z}^p+a_{p+1}{z}^{p+1}+\cdots +a_{p+m}{z}^{p+m}+\cdots ).\end{array}$$

(2.15)

By evaluating the coefficient of $z^{p+m}$ on both sides of (2.15), we get

$$\begin{array}{rl}{e}^{i\alpha }(p+m)a_{p+m}=& (\beta -pcos\alpha )(h_{2p+m}+a_p{h}_m+a_{p+1}{h}_{m-1}+\cdots +a_{p+m})\\ -(\beta +ipsin\alpha )a_{p+m}.\end{array}$$

(2.16)

On the other hand, it is well known that

$$|h_k|\leqq 2(k\in \mathbb{N}).$$

(2.17)

From (2.16) and (2.17), we easily get

$$|a_p|\leqq \frac{\beta -pcos\alpha }{p}$$

(2.18)

and

$$|a_{p+m}|\leqq \frac{2(\beta -pcos\alpha )}{2p+m}(1+\sum _{k=0}^{m-1}|a_{p+k}|).$$

(2.19)

Suppose that $p\in \mathbb{N}$. We define the sequence ${\{A_{p+m}\}}_{m=0}^{\mathrm{\infty }}$ as follows:

$$\{\begin{array}{cc}{A}_p=\frac{\beta -pcos\alpha }{p}\hfill & (m=0),\hfill \\ A_{p+m}=\frac{2(\beta -pcos\alpha )}{2p+m}(1+{\sum }_{k=0}^{m-1}{A}_{p+k})\hfill & (m\geqq 1).\hfill \end{array}$$

(2.20)

In order to prove that

$$|a_{p+m}|\leqq A_{p+m}(m\in {\mathbb{N}}_0),$$

(2.21)

we use the principle of mathematical induction. It is easy to verify that

$$|a_p|\leqq A_p=\frac{\beta -pcos\alpha }{p}.$$

(2.22)

Thus, assuming that

$$|a_{p+j}|\leqq A_{p+j}(j=0,1,\dots ,m;m\in {\mathbb{N}}_0),$$

(2.23)

we find from (2.19) and (2.23) that

$$\begin{array}{rl}|a_{p+m+1}|& \leqq \frac{2(\beta -pcos\alpha )}{2p+m+1}(1+\sum _{k=0}^m|a_{p+k}|)\\ \leqq \frac{2(\beta -pcos\alpha )}{2p+m+1}(1+\sum _{k=0}^m|A_{p+k}|)\\ =A_{p+m+1}(m\in {\mathbb{N}}_0).\end{array}$$

(2.24)

Therefore, by the principle of mathematical induction, we have

$$|a_{p+m}|\leqq A_{p+m}(m\in {\mathbb{N}}_0).$$

(2.25)

By means of Lemma 1 and (2.20), we know that

$$A_{p+m}=\frac{2(\beta -pcos\alpha )}{2\beta +m+2p-2pcos\alpha }\prod _{k=0}^m\frac{2\beta +k+2p-2pcos\alpha }{2p+k}(m\in {\mathbb{N}}_0).$$

(2.26)

Combining (2.25) and (2.26), we readily get the coefficient estimates (2.11) asserted by Theorem 2. □

From Theorem 2, we easily get the following result.

Corollary 2 Let $f(z)=z^{-p}+{\sum }_{m=0}^{\mathrm{\infty }}{a}_{p+m}{z}^{p+m}\in {\mathcal{MC}}_p(\alpha ,\beta )$. Then

$$|a_{p+m}|\leqq \frac{2p(\beta -pcos\alpha )}{(p+m)(2\beta +m+2p-2pcos\alpha )}\prod _{k=0}^m\frac{2\beta +k+2p-2pcos\alpha }{2p+k}(m\in {\mathbb{N}}_0).$$

Remark 3 By setting $\alpha =0$ in Theorem 2, we get the corresponding result due to Wang et al. [1].

Theorem 3 If $f\in {\mathcal{MS}}_p(\alpha ,\beta )$, then

$$\frac{pcos\alpha -(2\beta -pcos\alpha )r}{1-r}\leqq \mathrm{\Re }(-e^{i\alpha }\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)})\leqq \frac{pcos\alpha +(2\beta -pcos\alpha )r}{1+r}$$

(2.27)

for $|z|=r<1$.

Proof Consider the function φ defined by

$$\phi (z):=\frac{p{e}^{i\alpha }-(2\beta -p{e}^{-i\alpha })z}{1-z}(z\in \mathbb{U}).$$

(2.28)

Let $z=r{e}^{i\theta }$ ($0<r<1$), we see that

$$\mathrm{\Re }(\phi (z))=pcos\alpha -\frac{2(\beta -pcos\alpha )r(cos\theta -r)}{1+r^2-2rcos\theta }.$$

(2.29)

Suppose

$$\psi (t):=pcos\alpha -\frac{2(\beta -pcos\alpha )r(t-r)}{1+r^2-2rt}(t:=cos\theta ),$$

(2.30)

we easily find that

$${\psi }^{\mathrm{\prime }}(t)=-2(\beta -pcos\alpha )\cdot \frac{1-r^2}{{(1+r^2-2rt)}^2}>0.$$

(2.31)

This implies

$$pcos\alpha -\frac{2(\beta -pcos\alpha )r}{1-r}\leqq \mathrm{\Re }(\phi (z))\leqq pcos\alpha +\frac{2(\beta -pcos\alpha )r}{1+r},$$

(2.32)

which is equivalent to

$$\frac{pcos\alpha -(2\beta -pcos\alpha )r}{1-r}\leqq \mathrm{\Re }(\phi (z))\leqq \frac{pcos\alpha +(2\beta -pcos\alpha )r}{1+r}.$$

(2.33)

Noting that $-e^{i\alpha }\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}\prec \phi (z)$ and $\phi (z)$ is univalent in , we prove the inequality (2.27). □

Taking $\alpha =0$ in Theorem 3, we have the following corollary.

Corollary 3 If $f\in {\mathcal{MS}}_p(0,\beta )$, then

$$\frac{p-(2\beta -p)r}{1-r}\leqq \mathrm{\Re }\left(\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}\right)\leqq \frac{p+(2\beta -p)r}{1+r}$$

for $|z|=r<1$.

Similar to the proof of Theorem 3, we get the following result.

Corollary 4 If $f\in {\mathcal{MC}}_p(\alpha ,\beta )$, then

$$\frac{pcos\alpha -(2\beta -pcos\alpha )r}{1-r}\leqq \mathrm{\Re }(-e^{i\alpha }(1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)}))\leqq \frac{pcos\alpha +(2\beta -pcos\alpha )r}{1+r}$$

for $|z|=r<1$.

Corollary 5 If $f\in {\mathcal{MC}}_p(0,\beta )$, then

$$\frac{p-(2\beta -p)r}{1-r}\leqq \mathrm{\Re }(1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)})\leqq \frac{p+(2\beta -p)r}{1+r}$$

for $|z|=r<1$.

Now, we present some sufficient conditions for functions belonging to the classes ${\mathcal{MS}}_p(\alpha ,\beta )$ and ${\mathcal{MC}}_p(\alpha ,\beta )$.

Theorem 4 If $f\in {\mathcal{MS}}_p(\alpha ,\beta )$ satisfies the condition

$$\sum _{n=1-p}^{\mathrm{\infty }}\left(\right|n{e}^{i\alpha }+\lambda |+|n{e}^{i\alpha }+2\beta -\lambda \left|\right)|a_n|\leqq |p{e}^{i\alpha }-2\beta +\lambda |-|p{e}^{i\alpha }-\lambda |$$

(2.34)

for some real α, β and λ ($0\leqq \lambda \leqq pcos\alpha$), then $f\in {\mathcal{MS}}_p(\alpha ,\beta )$.

Proof To prove $f\in {\mathcal{MS}}_p(\alpha ,\beta )$, it suffices to show that

$$\left|\frac{e^{i\alpha }\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}+\lambda }{e^{i\alpha }\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}+(2\beta -\lambda )}\right|<1(z\in \mathbb{U};0\leqq \lambda \leqq pcos\alpha ).$$

(2.35)

From (2.34), we know that

$$|p{e}^{i\alpha }-2\beta +\lambda |-\sum _{n=1-p}^{\mathrm{\infty }}|n{e}^{i\alpha }+2\beta -\lambda ||a_n|\geqq |p{e}^{i\alpha }-\lambda |+\sum _{n=1-p}^{\mathrm{\infty }}|n{e}^{i\alpha }+\lambda ||a_n|>0.$$

(2.36)

Now, by the maximum modulus principle, we deduce from (1.1) and (2.36) that

$$\begin{array}{rl}\left|\frac{e^{i\alpha }\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}+\lambda }{e^{i\alpha }\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}+(2\beta -\lambda )}\right|& =\left|\frac{(-p{e}^{i\alpha }+\lambda )+{\sum }_{n=1-p}^{\mathrm{\infty }}(n{e}^{i\alpha }+\lambda )a_n{z}^{n+p}}{(-p{e}^{i\alpha }+2\beta -\lambda )+{\sum }_{n=1-p}^{\mathrm{\infty }}(n{e}^{i\alpha }+2\beta -\lambda )a_n{z}^{n+p}}\right|\\ <\frac{|p{e}^{i\alpha }-\lambda |+{\sum }_{n=1-p}^{\mathrm{\infty }}|n{e}^{i\alpha }+\lambda ||a_n|}{|p{e}^{i\alpha }-2\beta +\lambda |-{\sum }_{n=1-p}^{\mathrm{\infty }}|n{e}^{i\alpha }+2\beta -\lambda ||a_n|}\\ \leqq 1.\end{array}$$

(2.37)

Therefore, if f satisfies the coefficient estimate (2.34), then we know that f satisfies the inequality (2.35). This completes the proof of Theorem 4. □

Corollary 6 If $f\in {\mathcal{MC}}_p(\alpha ,\beta )$ satisfies the inequality

$$\sum _{n=1-p}^{\mathrm{\infty }}|n|\left(\right|n{e}^{i\alpha }+\lambda |+|n{e}^{i\alpha }+2\beta -\lambda \left|\right)|a_n|\leqq p\left(\right|p{e}^{i\alpha }-2\beta +\lambda |-|p{e}^{i\alpha }-\lambda \left|\right)$$

for some real α, β and λ ($0\leqq \lambda \leqq pcos\alpha$), then $f\in {\mathcal{MC}}_p(\alpha ,\beta )$.

We need the following lemma to prove our next theorem.

Lemma 2 (See [11])

Let φ be a nonconstant regular function in . If $|\phi |$ attains its maximum value on the circle $|z|=r<1$ at $z_0$, then

$$z_0{\phi }^{\mathrm{\prime }}(z_0)=k\phi (z_0),$$

where $k\geqq 1$ is a real number.

Theorem 5 If $f\in {\mathcal{MS}}_p(0,\beta )$ satisfies

$$|1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)}-\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}|<\frac{\beta -p}{2\beta }(z\in \mathbb{U})$$

(2.38)

for some real $\beta >p$, then $f\in {\mathcal{MS}}_p(0,\beta )$.

Proof Let us define the function ϕ by

$$\varphi (z):=\frac{\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}+p}{\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}+2\beta -p}(z\in \mathbb{U}),$$

(2.39)

then we see that ϕ is analytic in and $\varphi (0)=0$. It follows from (2.39) that

$$\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}=\frac{-p+(2\beta -p)\varphi (z)}{1-\varphi (z)}.$$

(2.40)

Differentiating both sides of (2.40) logarithmically, we obtain

$$1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)}-\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}=\frac{(2\beta -p)z{\varphi }^{\mathrm{\prime }}(z)}{-p+(2\beta -p)\varphi (z)}+\frac{z{\varphi }^{\mathrm{\prime }}(z)}{1-\varphi (z)}.$$

(2.41)

By virtue of (2.38) and (2.41), we find that

$$|1+\frac{z{f}^{\mathrm{\prime }\mathrm{\prime }}(z)}{f^{\mathrm{\prime }}(z)}-\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}|=\left|\frac{2(\beta -p)z{\varphi }^{\mathrm{\prime }}(z)}{[-p+(2\beta -p)\varphi (z)][1-\varphi (z)]}\right|<\frac{\beta -p}{2\beta }.$$

(2.42)

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

$$\underset{|z|\leqq |z_0|}{max}|\varphi (z)|=|\varphi (z_0)|=1.$$

Then, Lemma 2 gives us that $\varphi (z_0)=e^{i\theta }$ and $z_0{\varphi }^{\mathrm{\prime }}(z_0)=k{e}^{i\theta }$ ($k\geqq 1$). For such a point $z_0$, we have that

$$\begin{array}{rl}|1+\frac{z_0{f}^{\mathrm{\prime }\mathrm{\prime }}(z_0)}{f^{\mathrm{\prime }}(z_0)}-\frac{z_0{f}^{\mathrm{\prime }}(z_0)}{f(z_0)}|& =\left|\frac{2(\beta -p)k{e}^{i\theta }}{[-p+(2\beta -p)e^{i\theta }][1-e^{i\theta }]}\right|\\ =\frac{2(\beta -p)k}{\sqrt{p^2+{(2\beta -p)}^2-2p(2\beta -p)cos\theta }\sqrt{2-2cos\theta }}\\ \geqq \frac{\beta -p}{2\beta }.\end{array}$$

(2.43)

This contradicts our condition (2.38). Therefore, there is no $z_0\in \mathbb{U}$ such that $|\varphi (z_0)|=1$. This implies that $|\varphi (z)|<1$ ($z\in {\mathbb{U}}^{\ast }$), that is,

$$\left|\frac{\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}+p}{\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)}+(2\beta -p)}\right|<1(z\in \mathbb{U}).$$

Thus, we conclude that $f\in {\mathcal{MS}}_p(0,\beta )$. □

Theorem 6 If $f\in {\mathcal{MS}}_p(0,\beta )$ for some real $p<\beta \leqq p+\frac{1}{2}$, then

$$\mathrm{\Re }\left(\frac{1}{z^{p}f(z)}\right)>\frac{1}{1-2\beta +2p}(z\in \mathbb{U}).$$

(2.44)

Proof Consider the function η such that

$$\frac{1}{z^{p}f(z)}=\frac{1+(1-2\gamma )\eta (z)}{1-\eta (z)}$$

(2.45)

for $\gamma =\frac{1}{1-2\beta +2p}$ and $f(z)\in {\mathcal{MS}}_p(0,\beta )$. Then we know that

$$\mathrm{\Re }(-\frac{z{f}^{\mathrm{\prime }}(z)}{f(z)})=\mathrm{\Re }(p+\frac{(1-2\gamma )z{\eta }^{\mathrm{\prime }}(z)}{1+(1-2\gamma )\eta (z)}+\frac{z{\eta }^{\mathrm{\prime }}(z)}{1-\eta (z)})<\beta .$$

(2.46)

Since $\eta (z)$ is analytic in and $\eta (0)=0$, we suppose that there exists a point $z_0\in \mathbb{U}$ such that

$$\underset{|z|\leqq |z_0|}{max}|\eta (z)|=|\eta (z_0)|=1.$$

Then, applying Lemma 2, we can write that $\eta (z_0)=e^{i\theta }$ and $z_0{\eta }^{\mathrm{\prime }}(z_0)=k{e}^{i\theta }$ ($k\geqq 1$). This gives us that

$$\begin{array}{rl}\mathrm{\Re }(-\frac{z_0{f}^{\mathrm{\prime }}(z_0)}{f(z_0)})& =\mathrm{\Re }(p+\frac{(1-2\gamma )k{e}^{i\theta }}{1+(1-2\gamma )e^{i\theta }}+\frac{k{e}^{i\theta }}{1-e^{i\theta }})\\ \geqq p-\frac{(1-2\gamma )k}{2\gamma }-\frac{k}{2}\\ \geqq p+\frac{\gamma -1}{2\gamma }=\beta ,\end{array}$$

(2.47)

which contradicts the inequality (2.46). Therefore, there is no $z_0\in \mathbb{U}$ such that $|\eta (z_0)|=1$. This means that $|\eta (z)|<1$, and that

$$\mathrm{\Re }\left(\frac{1}{z^{p}f(z)}\right)>\frac{1}{1-2\beta +2p}(z\in \mathbb{U}).$$

(2.48)

The proof of Theorem 6 is thus completed. □

In view of Theorem 6, we get the following result.

Corollary 7 If $f\in {\mathcal{MC}}_p(0,\beta )$ for some real $p<\beta \leqq p+\frac{1}{2}$, then

$$\mathrm{\Re }\left(\frac{p}{z^{p+1}{f}^{\mathrm{\prime }}(z)}\right)>\frac{1}{1-2\beta +2p}(z\in \mathbb{U}).$$