Some geometric properties of certain meromorphically multivalent functions associated with the first-order differential subordination

1.   Introduction

Throughout our present discussion, we assume that

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

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

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

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 $\mathcal{G}_n(A, B, \lambda)$ if it satisfies the first-order differential subordination:

Recently, several authors (see, e.g., ^{[1,2,3,4,5,6,7,8,10,11,12,13,14,15,16]} and the references cited therein) introduced and studied various subclasses of meromorphically multivalent functions. Certain properties such as distortion bounds, inclusion relations and coefficient estimates are given. In this note we obtain inclusion relation, coefficient estimate and sharp bounds on $\mathrm{Re}\left(z^pf(z)\right)$ for functions $f(z)$ belonging to the class $\mathcal{G}_n(A, B, \lambda)$. Furthermore, we investigate a new problem, that is, to find

where $f(z)$ varies in the class

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

Lemma ^[9]. 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 $\mathcal{G}_n(A, B, \lambda)$

Theorem 1. Let $0 < \alpha_1 < \alpha_2$. Then $\mathcal{Q}_n(A, B, \alpha_2) \subset \mathcal{Q}_n(A, B, \alpha_1)$.

Proof. Suppose that

for $f(z)\in \mathcal{Q}_n(A, B, \alpha_2)$. Then the function $g(z)$ is analytic in $\mathbb{U}$ with $g(0) = p$. By using (1.3) and (2.1), we have

An application of the above Lemma yields

By noting that $0 < \frac{\alpha_1}{\alpha_2} < 1$ and that the function $\frac{1+Az}{1+Bz}$ is convex univalent in $\mathbb{U}$, it follows from (2.1)–(2.3) that

This shows that $f(z)\in \mathcal{Q}_n(A, B, \alpha_1)$. The proof of Theorem 1 is completed.

Theorem 2. Let $f(z)\in \mathcal{Q}_n(A, B, \alpha)$. Then, for $|z| = r < 1$,

and

All the bounds are sharp for the function $f_n(z)$ given by

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

and

Let $f(z)\in \mathcal{Q}_n(A, B, \alpha)$. Then we can write

where $w(z) = w_nz^n+w_{n+1}z^{n+1}+\cdots$ is analytic and $|w(z)| < 1$ for $z\in \mathbb{U}$. By the Schwarz lemma, we know that $|w(z)|\leq |z|^n$ $(z\in \mathbb{U})$. It follows from (2.11) that

which implies that

After integration we arrive at

Since

we get from (2.12) and left-hand inequality in (2.10) that, for $|z| = r < 1$,

and, for $z\in \mathbb{U}$,

Similarly, by using (2.12) and the right-hand inequality in (2.10), we have (2.6) and (2.7) (with $B\neq -1$).

Furthermore, for the function $f_n(z)$ given by (2.8), we find that $f_n(z)\in \mathcal{A}_n(p)$,

and

Hence $f_n(z)\in \mathcal{Q}_n(A, B, \alpha)$ and, from (2.14), we conclude that the inequalities (2.4) to (2.7) are sharp. The proof of Theorem 2 is completed.

Corollary. Let $f(z)\in \mathcal{Q}_n(A, B, \alpha)$. If

then $f(z)$ is $p$-valent close-to-convex in $\mathbb{U}$.

Proof. Let $f(z)\in \mathcal{Q}_n(A, B, \alpha)$ and (2.15) be satisfied. Then, by using (2.5) in Theorem 2, we see that

This shows that $f(z)$ is $p$-valent close-to-convex in $\mathbb{U}$. The proof of the corollary is completed.

Theorem 3. Let $f(z)\in \mathcal{Q}_n(A, B, \alpha)$. Then, for $|z| = r < 1$,

and

All of the above bounds are sharp.

Proof. It is obvious that

Making use of (2.4) in Theorem 2, it follows from (2.19) that

which gives (2.16).

Similarly, we deduce from (2.6) in Theorem 2 and (2.19) that (2.17) holds true.

Also, with the help of (2.13), we find that

From this and (2.19), we obtain (2.18).

Furthermore, it is easy to see that the inequalities (2.16)–(2.18) are sharp for the function $f_n(z)$ given by (2.8). Now the proof of Theorem 3 is completed.

Theorem 4. Let $f(z)\in \mathcal{Q}_n(A, B, \alpha)$ and $AB\leq 1$. Then, for $|z| = r < 1$,

and

The above bounds are sharp.

Proof. Since $AB\leq 1$, it follows from (2.9) that

By virtue of (2.12) and (2.22), we have, for $|z| = r < 1$,

By noting that

we deduce from (2.23) and (2.24) that the desired inequalities hold true.

The bounds in (2.20) and (2.21) are sharp with the extremal function $f_n(z)$ given by (2.8). The proof of Theorem 4 is completed.

Theorem 5. Let $f(z)\in \mathcal{Q}_1(A, B, \alpha)$ and

If

then $(f*g)(z)\in \mathcal{Q}_1(A, B, \alpha),$ where the symbol $*$ denotes the familiar Hadamard product of two analytic functions in $\mathbb{U}$.

Proof. Since $g(z)\in \mathcal{Q}_1(A_0, B_0, \alpha_0)$, we find from the inequality (2.18) in Theorem 3 and (2.25) that

Thus the function $\frac{g(z)}{z^p}$ has the following Herglotz representation:

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

For $f(z)\in \mathcal{Q}_1(A, B, \alpha)$, we have

and

Thus

where

In view of the fact that the function $\frac{1+Az}{1+Bz}$ is convex univalent in $\mathbb{U}$, it follows from (2.26) to (2.28) that

This shows that $(f*g)(z)\in \mathcal{Q}_1(A, B, \alpha)$. The proof of Theorem 5 is completed.

Theorem 6. Let

Then

The result is sharp for each $k\geq n$.

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 $|b_j|\leq 1$ $(j\in \mathbb{N})$.

By using (2.29), we have

In view of the fact that the function $\frac{z}{1+Bz}$ is analytic and convex univalent in $\mathbb{U}$, it follows from (2.31) that

which gives (2.30).

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

Since

and

for each $k\geq n$, the proof of Theorem 6 is completed.

Theorem 7. Let $f(z)\in \mathcal{Q}_n(A, B, 0)$. Then, for $|z| = r < 1$,

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

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

where

The above results are sharp.

Proof. Equality in (2.32) occurs for $z = 0$. Thus we assume that $0 < |z| = r < 1$.

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

where $\varphi(z)$ is analytic and $|\varphi(z)|\leq 1$ in $\mathbb{U}$. It follows from (2.35) that

By using the Carathéodory inequality:

we obtain

Put $\frac{f'(z)}{pz^{p-1}} = u+iv\quad (u, v\in \mathbb{R})$. Then (2.36) and (2.37), together, yield

We note that

Combining (2.38) and (2.39), we have

Also, (2.10) and (2.35) imply that

We now calculate the minimum 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, we get

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

Since

we see that

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

(i) $u_n\leq\frac{1-Ar^n}{1-Br^n}$, that is, $M_n(A, B, \alpha, r)\geq 0$ (see (2.34)). In view of $\psi_n'(u_n) = 0$ and (2.41), the function $\psi_n(u)$ is increasing on the segment $\left[\frac{1-Ar^n}{1-Br^n}, \frac{1+Ar^n}{1+Br^n}\right]$. Therefore, we deduce from (2.40) that, if $M_n(A, B, \alpha, r)\geq 0$, then

This proves (2.32).

Next we consider the function $f(z)$ given by

It is easy to find that

which shows that the inequality (2.32) is sharp.

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

In view of (2.34), $\psi_n(u)$ in (2.40) can be written as follows:

Therefore, if $M_n(A, B, \alpha, r)\leq 0$, then it follows from (2.42), (2.44) and (2.45) that

To show that the inequality (2.33) is sharp, we take

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

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

from the above argument we obtain that

The proof of Theorem 7 is completed.

3.   Conclusions

In this paper, we have introduced and investigated some geometric properties of the class $\mathcal{G}_n(A, B, \lambda)$ which is defined by using the principle of first-order differential subordination. For this function class, we have derived the sharp upper bound on $|z| = r < 1$ for the following functional:

over the class $\mathcal{G}_n(A, B, 0)$. We have also obtained other properties of the function class $\mathcal{G}_n(A, B, \lambda)$.

Motivated by a recently-published survey-cum-expository review article by Srivastava ^[15], the interested reader's attention is drawn toward the possibility of investigating the basic (or $q$-) extensions of the results which are presented in this paper. However, as already pointed out by Srivastava, their further extensions using the so-called ($p$, $q$)-calculus will be rather trivial and inconsequential variations of the suggested extensions which are based upon the classical $q$-calculus, the additional paremeter $p$ being redundant or superfluous (see, for details, [15, p. 340]).

Acknowledgments

The authors would like to express sincere thanks to the referees 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).

Conflicts of interest

The authors agree with the contents of the manuscript, and there are no conflicts of interest among the authors.