On second-order differential subordination for certain meromorphically multivalent functions

1.   Introduction

Throughout our present investigation, we assume that

Let $\Sigma_n(p)$ denote the class of functions of the form

which are analytic in the punctured open unit disk $\mathbb{U}^* = \{z : 0 < |z| < 1\}$ with a pole at $z = 0$. The class $\Sigma_n(p)$ is closed under the Hadamard product (or convolution)

where

For functions $f(z)$ and $g(z)$ analytic in $\mathbb{U} = \{z:|z| < 1\}$, we say that $f(z)$ is subordinate to $g(z)$ and write $f(z)\prec g(z)$ $(z\in \mathbb{U})$, if there exists an analytic function $w(z)$ in $\mathbb{U}$ such that

Furthermore, if the function $g(z)$ is univalent in $\mathbb{U}$, then

In this paper we introduce and investigate the following subclass of $\Sigma_n(p)$.

Definition. A function $f(z)\in \Sigma_n(p)$ is said to be in the class $\mathcal{R}_n(A, B, \lambda)$ if it satisfies the second-order differential subordination:

Recently, several authors (see, e.g., ^{[1,2,3,4,5,6,7,8,9,11,12,13,14,15]} and the references cited therein) introduced and investigated various subclasses of meromorphically multivalent functions. Some properties such as distortion bounds, inclusion relations and coefficient estimates were given. In this note we obtain inclusion relation and coefficient estimate for functions $f(z)$ in the class $\mathcal{R}_n(A, B, \lambda)$. Furthermore, we investigate a new problem. It is to find

where $f(z)$ varies in the class:

We need the following lemma in order to derive our main results for the class $\mathcal{R}_n(A, B, \lambda)$.

Lemma ^[10]. Let $g(z)$ be analytic in $\mathbb{U}$ and $h(z)$ be analytic and convex univalent in $\mathbb{U}$ with $h(0) = g(0)$. If

where $\mathrm{Re}\mu\geq 0$ and $\mu\neq0$, then $g(z)\prec h(z)$.

2.   Geometric properties of functions in class $\mathcal{R}_n(A, B, \lambda)$

Theorem 1. Let $\lambda_2 < \lambda_1 < 0$. Then

Proof. Suppose that

for $f(z)\in \mathcal{R}_n(A, B, \lambda_2)$. Then $g(z)$ is analytic in $\mathbb{U}$ with $g(0) = p$. By using (1.3) and (2.1), we have

Hence an application of Lemma with $\mu = -\frac{p+1}{\lambda_2} > 0$ yields

Note that $0 < \frac{\lambda_1}{\lambda_2} < 1$ and that the function $\frac{1+Az}{1+Bz}$ is convex univalent in $\mathbb{U}$, then it follows from (2.1), (2.2) and (2.3) that

Thus $f(z)\in \mathcal{R}_n(A, B, \lambda_1)$. The proof of Theorem 1 is completed.

Theorem 2. Let

Then

The result is sharp for each $k\geq n$ $(k\neq p)$.

Proof. It is known that, if

where $\varphi(z)$ is analytic in $\mathbb{U}$ and $\psi(z) = z+\cdots$ is analytic and convex univalent in $\mathbb{U}$, then $|c_j|\leq 1$ $(j\in \mathbb{N})$.

By (2.4) we have

Because the function $\frac{z}{1+Bz}$ is analytic and convex univalent in $\mathbb{U}$, it follows from (2.6) that

which gives (2.5).

Next we consider the function $f_k(z)$ defined by

Since

and

for each $k\geq n\ (k\neq p)$, the proof of Theorem 2 is completed.

Theorem 3. Let $f(z)\in \mathcal{R}_n(A, B, \lambda)$, $g(z)\in \Sigma_n(p)$ and

Then $(f*g)(z)\in \mathcal{R}_n(A, B, \lambda)$.

Proof. For $f(z)\in \mathcal{R}_n(A, B, \lambda)$ and $g(z)\in \Sigma_n(p)$, we have

where

From (2.7), we can see that the function $z^pg(z)$ has Herglotz representation:

where $\mu(x)$ is a probability measure on the unit circle $|x| = 1$ and $\int_{|x| = 1}d\mu(x) = 1$.

Because the function $\frac{1+Az}{1+Bz}$ is convex univalent in $\mathbb{U}$, it follows from (2.8)–(2.10) that

This shows that $(f*g)(z)\in \mathcal{R}_n(A, B, \lambda)$. The proof of Theorem 3 is completed.

Theorem 4. Let $f(z)\in \mathcal{R}_n(A, B, 0)$. Then for $|z| = r < 1$,

(ⅰ) if $M_n(A, B, \lambda, r)\geq 0$, we have

(ⅱ) if $M_n(A, B, \lambda, r)\leq 0$, we have

where

The results are sharp.

Proof. Equality in (2.11) occurs for $z = 0$. Thus we assume that $0 < |z| = r < 1$.

For $f(z)\in \mathcal{R}_n(A, B, 0)$, we can write

where $\varphi(z)$ is analytic and $|\varphi(z)|\leq 1$ in $\mathbb{U}$. It follows from (2.14) that

Using the Carathéodory inequality:

we have

Put $-\frac{z^{p+1}f'(z)}{p} = u+iv\quad (u, v\in \mathbb{R})$. Note that $\lambda < 0$, then (2.15) and (2.16) provide

Further, we can see that

Combining (2.17) and (2.18), we have

It is known that for $|\xi|\leq \sigma$ $(\sigma < 1)$,

From (2.20) and(2.14) we have

Now we calculate the maximal value of $\psi_n(u)$ on the segment $\left[\frac{1-Ar^n}{1-Br^n}, \frac{1+Ar^n}{1+Br^n}\right]$. Obviously,

and $\psi_n'(u) = 0$ if and only if

where $L_n$ and $K_B$ are given by (2.13). From (2.13) and (2.18) one can see that $K_B > 0$ and $L_n < 0$.

Since

we see that

But $u_n$ is not always less than $\frac{1+Ar^n}{1+Br^n}$. The following two cases arise.

(ⅰ) $u_n\geq\frac{1+Ar^n}{1+Br^n}$, that is, $M_n(A, B, \lambda, r)\geq 0$ (see (2.13)). In view of $\psi_n'(u_n) = 0$ and (2.21), the function $\psi_n(u)$ is increasing on the segment $\left[\frac{1-Ar^n}{1-Br^n}, \frac{1+Ar^n}{1+Br^n}\right]$. Thus we deduce from (2.19) that, if $M_n(A, B, \lambda, r)\geq 0$, then

This proves (2.11).

Next we consider the function $f(z)\in \mathcal{R}_n(A, B, 0)$ defined by

It is easy to find that

which shows that the inequality (2.11) is sharp.

(ⅱ) $u_n\leq\frac{1+Ar^n}{1+Br^n}$, that is, $M_n(A, B, \lambda, r)\leq 0$. In this case we easily obtain

In view of (2.13), $\psi_n(u)$ in (2.19) can be written as

Therefore, if $M_n(A, B, \lambda, r)\leq 0$, then it follows from (2.22), (2.24) and (2.25) that

To show that the inequality (2.12) is sharp, we consider the function $f(z)$ defined by

where $c_n\in \mathbb{R}$ is determined by

Clearly, $-1 < \varphi(r)\leq 1$, $-1\leq c_n < 1$, $|\varphi(z)|\leq 1$ $(z\in \mathbb{U})$, and so $f(z)\in \mathcal{R}_n(A, B, 0)$. Since

from the above argument we obtain that

Now the proof of Theorem 4 is completed.

Acknowledgments

The authors would like to express sincere thanks to the referee for careful reading and suggestions which helped us to improve the paper. This work was supported by National Natural Science Foundation of China (Grant No.11571299).

Conflict of interest

The authors declare no conflicts of interest.