Closure properties of generalized $\lambda$-Hadamard product for a class of meromorphic Janowski functions

1.   Introduction

Let $\Sigma (a, k)$ be the class of meromorphic functions $f$ of the form

which is analytic in the punctured open unit disk $\mathbb{U}^* = \{z\in \mathbb{C}:0 < |z| < 1\} = \mathbb{U}\setminus \{0\}.$

In 2009, El-Ashwah ^[1] introduced the subclass $\Sigma S _{k}^{*}(a, \beta)$ of generalized meromorphically starlike of order $\beta$ as follows,

An analytic function $g: \mathbb{U} = \{z:|z| < 1\}\rightarrow\mathbb{C}$ is subordinate to an analytic function $h: \mathbb{U}\rightarrow\mathbb{C}$, if there is a function $\omega$ satisfying $\omega(0) = 0\; \; \hbox{and}\; \; |\omega(z)| < 1\; \; (z\in\mathbb{U})$, such that $g(z) = h(\omega(z))(z\in\mathbb{U})$. Note that $g(z)\prec h(z)$. Especially, if $h$ is univalent in $\mathbb{U}$, then the following conclusion is true (see ^[2]):

Using the subordinate relationship, we introduce the following subclass of generalized meromorphic Janowski functions (see ^[3,5,6]).

Definition 1. Let $f(z)\in \Sigma (a, k), a > 0, A, B\in \mathbb{R}, \; |A|\leq 1, |B|\leq 1$ and $A\neq B$. The function $f\in\Sigma S_{k}^{*}(a, A, B)$ if and only if

It is obvious that

According to the subordination relationship, the function $f(z)\in \Sigma S_{k}^{*}(a, A, B)$ if and only if there exists an analytic function $\omega(z)$ in the open unit dick $\mathbb{U}$ satisfying $\omega(0) = 0$ and $|\omega(z)| < 1$, such that

which is equivalent to

Let $T(a, k)$ be the subclass of $\Sigma (a, k)$, the function $f$ belonging to $T(a, k)$ of the following form

Let

Next, we introduce the generalized $\lambda$-Hadamard product of the class $T(a, k)$.

Definition 2. Let $a > 0, p > 0, q > 0, \lambda\geq 0, k\in \mathbb{N}, k\geq2, f_{i}(z) = \frac{a}{z}-\sum_{n = k}^{\infty}|a_{n}, _{j}|z^{n}\in {T}(a, k)\; (i = 1, 2)$. The generalized $\lambda$-Hadamard product $(f_{1} \widetilde{\bigtriangleup} f_{2})_{\lambda}(p, q, a; z)$ of the function $f_1$ and $f_2$ is defined by

where

From (1.4) and (1.5), we get

Selecting different parameters $a, \lambda, p$ and $q$, we can obtain the following three special convolutions:

(ⅰ) For $\lambda = 0,$ $({f_{1}\widetilde{\Delta} f_{2}})_{0}(p, q, a; z)$ is the generalized Hadamard product:

In particular, for $a = 1$, the Hadamard product $({f_{1}\Delta f_{2}}) (p, q, 1;z) = \frac{1}{z}-\sum_{n = k}^{\infty}{|a_{n, 1}|}^{p} {|a_{n, 2}|}^{q} z^{n}$ was studied by Janowski ^[3] and Tang et al. ^[6].

(ⅱ) For $\lambda = 0, \; p = q = 1$, $({f_{1}\widetilde{\Delta} f_{2}})_{0}(1, 1, a; z)$ is the general Hadamard product:

In particular, for $a = 1$, $({f_{1} \ast} f_{2})(1;z)$ is the well-known Hadamard product (see ^[1,4,7]):

For $a = 1, k = 1,$ $f_{1}(z) = z^{-1}+ \sum_{n = 1}^{\infty}\frac{(a)_{n}}{(c)_{n}}z^{n}, f_{2}(z) = \frac{a}{z}-\sum_{n = 1}^{\infty}|a_{n, 2}|z^{n},$ $(a)_{n} = a(a+1)\cdots(a+n-1)(n\in \mathbb{N}),$ $({f_{1} \ast} f_{2})(1;z) = \frac{1}{z}-\sum_{n = 1}^{\infty}\frac{(a)_{n}}{(c)_{n}}{|a_{n, 2}|}z^{n}$ was studied by Liu et al. ^[8] (see also ^[9]).

(ⅲ) For $p = q = 1,$ $({f_{1}\widetilde{\Delta} f_{2}})_{\lambda}(1, 1, a; z) = ({f_{1}\overline{{\ast}} f_{2})_{\lambda}}(z)$ is $\lambda-$Hadamard product:

In 1996, Choi et al. ^[4] studied the closure properties of the generalized $\lambda$-Hadamard product for univalent functions. In 2001, Liu et al. ^[8] investigated two new classes of meromorphically multivalent functions by using a linear operator. Based on this, in this paper, we mainly consider the closure properties of the generalized $\lambda$-Hadamard product for meromorphic functions. Firstly, we obtain the necessary and sufficient conditions of the class $\overline{\Sigma S}_{k}^{*}(a, A, B)$ by using subordination relationship. Secondly, we discuss the closure properties of the general Hadamard product, the generalized Hadamard product and $\lambda$-Hadamard product for functions of the class $\overline{\Sigma S}_{k}^{*}(a, A, B)$. Finally, we discuss the closure properties of the generalized $\lambda$-Hadamard product and generalize the known conclusions.

2.   Perliminaries

Unless otherwise noted, the parameters $a, k, A$ and $B$ satisfy the condition: $a > 0, k\in\mathbb{N}, k\geq2, A, B \in \mathbb{R}, |A|\leq 1, |B|\leq 1$ and $A\neq B.$

In order to establish our results, we need the following lemmas.

Lemma 1. Let $f$ be of the form (1.1). If

then $f(z)\in \Sigma S_{k}^*(a, A, B).$

Proof. Let $|z| = 1$. By (2.1), we obtain

Using the principle of maximum modulus, we get

Then we conclude that $f(z)\in \Sigma S_{k}^*(a, A, B).$ Therefore, we complete the proof of Lemma 1.

Lemma 2. Let the function $f$ be of the form (1.3).

(ⅰ) If $0\leq B < A\leq 1$. Then the function $f(z)\in \overline{\Sigma S}_{k}^*(a, A, B)$ if and only if

(ⅱ) If $-1\leq A < B\leq 0$. Then the function $f(z)\in \overline{\Sigma S}_{k}^*(a, A, B)$ if and only if

Proof. It appears from (1.3) that $\overline{\Sigma S}_{k}^*(a, A, B)\subset \Sigma S^*_{k}(a, A, B)$. In view of Lemma 1, we just need prove the necessity part of Lemma 2. Let $f(z)\in \overline{\Sigma S}_{k}^*(a, A, B)$. By (1.2), we have

Since $|\hbox{Re}z|\leq |z|$ holds true for all $z\in\mathbb{C}$. If $0\leq B < A\leq 1$, then

Let $z = r\rightarrow 1^{-}$. From (2.4), we can get (2.2). Using the same method, we can obtain the result of (ⅱ). The estimates of (ⅰ) and (ⅱ) are sharp for the function $f$ given by

So, we complete the proof of Lemma 2.

3.   Main results

First of all, we discuss the closure properties of the general Hadamard product of the class $\overline{\Sigma S}_{k}^*(a, A, B)$.

Theorem 1. Let $f_{i}(z) = \frac{a}{z}- \sum_{n = k}^{\infty}|a_{n}|z^{n}\in \overline{\Sigma S}_{k}^*(a, A, B)\; (i = 1, 2)$. If $A$ and $B$ satisfy the condition

then $\frac{1}{a}(f_{1}\ast f_{2})(a; z)\in \overline{\Sigma S}_{k}^*(a, A, B).$

Proof. Let $0\leq B < A\leq 1$. If $f_{i}(z) \in \overline{\Sigma S}_{k}^*(a, A, B)\; (i = 1, 2)$, from (2.2), we obtain

and

To obtain $\frac{1}{a}(f_{1}\ast f_{2})(a; z)\in \overline{\Sigma S}_{k}^*(a, A, B)$, we just prove

By using Cauchy-Schwarz inequality, from (3.1) and (3.2), we have

According to (3.3) and (3.4), we only need to prove

From (3.4) and the condition $0\leq B < A\leq 1$, we have

Therefore, we conclude that $\frac{1}{a}(f_{1}\ast f_{2})(a; z)\in \overline{\Sigma S}_{k}^*(a, A, B).$

For $-1\leq A < B\leq 0$, using the same method above, we get $\frac{1}{a}(f_{1}\ast f_{2})(a; z)\in \overline{\Sigma S}_{k}^*(a, A, B)$. Therefore, we complete the proof of Theorem 1.

Next, we prove the closure properties of the generalized Hadamard product.

Theorem 2. Let $f_{i}(z) = \frac{a}{z}-\sum_{n = k}^{\infty}|a_{n}|z^{n}\in \overline{\Sigma S}_{k}^*(a, A, B)\; (i = 1, 2)$ and $p > 1$. If anyone of the following conditions holds true:

(ⅰ) For $0\leq B < A\leq 1$, $\hat{B}$ and $a$ satisfy

(ⅱ) For $-1\leq A < B\leq 0$, $-1\leq A < \hat{B}\leq 0$ and $a$ satisfies $a > 0$. Then $\frac{1}{a}(f_{1}\Delta f_{2})\left(\frac{1}{p}, \frac{p-1}{p}; a, z\right)\in \overline{\Sigma S}_{k}^*(a, A, \hat{B}).$

Proof. (ⅰ) For $0\leq B < A \leq 1$, by Lemma 2, we can get

We deduce that

and

According to (1.5), we have

To prove $\frac{1}{a}(f_{1}\ast f_{2})(\frac{1}{p}, \frac{p-1}{p}, a; z)\in \overline{\Sigma S}_{k}^*(a, A, \hat{B})$, we only need to show

Applying Hölder inequality, from (3.5) and (3.6), we obtain

Thus, (3.7) holds if

that is,

Let

and

It is easy to see $F_1(n) > 0$ holds true for $n\geq k$. To obtain (3.8), we only need to prove

Because $\frac{G_1(n)}{F_1(n)}$ is an increasing function with respect to $n$ and the condition (i) holds true, so we have

Therefore, we conclude that $\frac{1}{a}(f_{1}\Delta f_{2})\left(\frac{1}{p}, \frac{p-1}{p}; a, z\right)\in \overline{\Sigma S}_{k}^*(a, A, \hat{B})$.

(ⅱ) It is similar to the proof of (ⅰ). For $-1\leq A < B\leq 0$, we only need to show

that is,

Let

and

It is easy to see $F_2(n) > 0$ holds true for $n\geq k$. To make (3.9) establish, we only need to prove

It is clear that $\frac{G_2(n)}{F_2(n)}$ is an decreasing function with respect to $n$ for $0 < a < 1$ and

then we have

It can be concluded that $\frac{1}{a}(f_{1}\Delta f_{2})\left(\frac{1}{p}, \frac{p-1}{p}; a, z\right)\in \overline{\Sigma S}_{k}^*(a, A, \hat{B})$. Clearly, $\frac{G_2(n)}{F_2(n)}$ is an increasing function with respect to $n$ for $a > 1$ and

Since $A > \frac{aA(1-B)+(B-A)}{a(1-B)+(B-A)}$, we have

Thus, we conclude that $\frac{1}{a}(f_{1}\Delta f_{2})\left(\frac{1}{p}, \frac{p-1}{p}; a, z\right)\in \overline{\Sigma S}_{k}^*(a, A, \hat{B})$. The proof of Theorem 2 is completed.

Putting $A = 1, B = 0$ in Theorem 2, we obtain the following corollary.

Corollary 1. Let $f_{i}(z) = \frac{a}{z}-\sum_{n = k}^{\infty} |{a_{n, i}}|z^{n}\in \overline{\Sigma S}_{k}^*(a, 1, 0)\; (i = 1, 2)$ and $p > 1$. If $\hat{B}$ and $a$ satisfy

Then $\frac{1}{a}(f_{1}\bigtriangleup f_{2})\left(\frac{1}{p}, \frac{p-1}{p}; a, z\right)\in \overline{\Sigma S}_{k}^*(a, A, \hat{B})$. Finally, we consider the closure properties of $\lambda$-Hadamard product $(f_{1}\overline{\ast} f_{2})_{\lambda}(z)$ and the generalized $\lambda$-Hadamard product $(f_{1}\widetilde{\Delta} f_{2})_{\lambda}(p.q, a; z)$ as follows.

Theorem 3. Let $f_i(z) = az-\sum_{n = k}^{\infty} |{a_{n, i}}|z^{n}\in \overline{\Sigma S}_{k}^*(a, A, B)\; (i = 1, 2)$. If $A, B$ and $\lambda$ satisfy

Then $\frac{1}{a}(f_{1}\overline{\ast} f_{2})_{\lambda}(z)\in \overline{\Sigma S}_{k}^*(a, A, B).$

Proof. Let $f_{i}(z)\in \overline{\Sigma S}_{k}^*(a, A, B)\; (i = 1, 2)$. If $0\leq B < A\leq 1$, from Lemma 2, we obtain

and

By Definition 2, we have

To show $\frac{1}{a}(f_{1}\bar{\ast} f_{2})_\lambda(z)\in\overline{ \Sigma S}_{k}^*(a, A, B),$ we just prove

Applying Cauchy-Schwarz inequality to (3.11) and (3.12), we get

According to (3.13) and (3.14), we only need to prove

Since $\sqrt{|a_{n, 1}||a_{n, 2}|}\leq \frac{a(A-B)}{(n+1)+(A+nB)},$ we only need to prove

Let

Clearly, $M(n)$ is an increasing function for $n\geq k$. This implies that (3.15) holds true if

For $\lambda < \frac{1}{k+1}$, the above formula can be expressed as $\lambda \geq \frac{(A-B)-(k+1-A-nB)}{(A-B)(k+1)}.$

To summarize, we conclude that $\frac{1}{a}(f_{1}\overline{\ast} f_{2})_{\lambda}(z)\in \overline{\Sigma S}_{k}^*(a, A, B)$ for $\frac{(A-B)-(k+1-A-nB)}{(A-B)(k+1)} < \lambda < \frac{1}{k+1}.$ Using the same method, we can get $\frac{1}{a}(f_{1}\overline{\ast} f_{2})_{\lambda}(z)\in \overline{\Sigma S}_{k}^*(a, A, B)$ for

Thus, we complete the proof of Theorem 3.

Theorem 4. Let $f_{i}(z) = \frac{a}{z}-\sum_{n = k}^{\infty}|a_{n}|z^{n}\in \overline{\Sigma S}_{k}^*(a, A, B)\; (i = 1, 2)$ and $p > 1$.

(1) For $0\leq B < A\leq 1, 0\leq \hat{B} < A$. If anyone of the following conditions holds true.

(ⅰ) $a < 1, \lambda < \mathrm{min}\bigg\{\frac{1}{2(n+1)}, \frac{a(1+B)+(A-B)}{(2n+1)(A-B)}, \frac{k(A-B)+a(k+1+A+kB)}{k(k+1)(A-B)}\bigg\},$

(ⅱ) $a > 1, \frac{1}{2(k+1)} < \lambda < \mathrm{min}\bigg\{\frac{a(1+B)+(A-B)}{(2n+1)(A-B)}, \frac{k(A-B)+a(k+1+A+kB)}{k(k+1)(A-B)}\bigg\},$

(ⅲ) $a < 1, \mathrm{max}\bigg\{\frac{a(1+B)+(A-B)}{(2k+1)(A-B)}, \frac{k(A-B)+a(k+1+A+kB)}{k(k+1)(A-B)}\bigg\} < \lambda < \frac{1}{2(n+1)},$

(ⅳ) $a > 1, \mathrm{max}\bigg\{\frac{1}{2(k+1)}, \frac{a(1+B)+(A-B)}{(2k+1)(A-B)}, \frac{k(A-B)+a(k+1+A+kB)}{k(k+1)(A-B)}\bigg\} < \lambda.$

Then the generalized $\lambda$-Hadamard product: $\frac{1}{a}(f_{1}\widetilde{\Delta} f_{2})_{\lambda}\bigg(\frac{1}{p}, \frac{p-1}{p}, a; z\bigg)\in \overline{\Sigma S}_{k}^*(a, A, B).$

(2) For $-1\leq A < B\leq 0.$ If anyone of the following conditions holds true.

(ⅴ) $a < 1, \lambda < \mathrm{min}\bigg\{\frac{1}{2(n+1)}, \frac{(B-A)+a(1-B)}{(2k-1)(B-A)}, \frac{k(B-A)+a(k+1-A-kB)}{k(k+1)(B-A)}\bigg\}$,

$\mathrm{max}(A, \frac{aA[(k+1)-(A+kB)]+[1-(k+1)\lambda](k+1+A)(B-A)}{a[(k+1)-(A+kB)]-k[1-(k+1)\lambda](B-A)})\leq \hat{B} \leq 0,$

(ⅵ) $a > 1, \frac{1}{2(k+1)} < \lambda < \mathrm{min}\bigg\{\frac{(B-A)+a(1-B)}{(2k-1)(B-A)}, \frac{k(B-A)+a(k+1-A-kB)}{k(k+1)(B-A)}\bigg\},$

$\mathrm{max} (A, \frac{aA[(k+1)-(A+kB)]+[1-(k+1)\lambda](k+1+A)(B-A)}{a[(k+1)-(A+kB)]-k[1-(k+1)\lambda](B-A)})\leq \hat{B} \leq 0,$

(ⅶ) $a < 1, \mathrm{max}\bigg\{\frac{(B-A)+a(1-B)}{(2k-1)(B-A)}, \frac{k(B-A)+a(k+1-A-kB)}{k(k+1)(B-A)}\bigg\} < \lambda < \frac{1}{2(n+1)}, A\leq \hat{B} < 0,$

(ⅷ) $a > 1, \mathrm{max}\bigg\{\frac{1}{2(k+1)}, \frac{(B-A)+a(1-B)}{(2k-1)(B-A)}, \frac{k(B-A)+a(k+1-A-kB)}{k(k+1)(B-A)}\bigg\} < \lambda, A\leq \hat{B} < 0.$

Then the generalized $\lambda$-Hadamard product: $\frac{1}{a}(f_{1}\widetilde{\Delta} f_{2})_{\lambda}\bigg(\frac{1}{p}, \frac{p-1}{p}, a; z\bigg)\in \overline{\Sigma S}_{k}^*(a, A, B).$

Proof. Let $f_{i}(z)\in \overline{\Sigma S}_{k}^*(a, A, B)(i = 1, 2)$. The method for the proof of Theorem 4 is similar to that of Theorem 2. If $-1\leq A < B \leq 0,$ we just need to prove

that is

Let

and

The following discussion can be divided into two cases:

(a) It is easy to see that $P(n) > 0$ if $\lambda < \frac{(B-A)+a(1-B)}{(B-A)(2n-1)}$. To prove that (3.16) holds true, we need only to prove the following inequality

Clearly, $\frac{Q(n)}{P(n)}$ is an increasing function with respect to $n$ for $a < 1, \lambda < \frac{1}{2(n+1)}$ or $a > 1, \lambda > \frac{1}{2(k+1)}.$ If $\lambda < \frac{k(B-A)+a(k+1-A-kB)}{k(k+1)(B-A)},$ then we have

Therefore, $\frac{1}{a}(f_{1}\widetilde{\bigtriangleup} f_{2})_{\lambda}\bigg(\frac{1}{p}, \frac{p-1}{p}, a, z\bigg)\in \overline{\Sigma S}_{k}^*(a, A, B)$ if the conditions (v) or (vi) are satisfied.

(b) It is easy to see that $P(n) < 0$ if $\lambda > \frac{(B-A)+a(1-B)}{(B-A)(2n-1)}$ and $P(k) = \frac{a[(k+1)-(A+kB)]}{B-A}+k[1-(k+1)\lambda] > 0$. To prove that (3.16) holds true, we need only to prove the following inequality

Clearly, $\frac{Q(n)}{P(n)}$ is a decreasing function with respect to $n$ for $a < 1, \lambda < \frac{1}{2(n+1)}$ or $a > 1, \lambda > \frac{1}{2(k+1)}.$

Therefore, $\frac{1}{a}(f_{1}\widetilde{\bigtriangleup} f_{2})_{\lambda}\bigg(\frac{1}{p}, \frac{p-1}{p}, a, z\bigg)\in \overline{\Sigma S}_{k}^*(a, A, B)$ if the conditions (vii) or (viii) are satisfied.

To summarize, we conclude that the conclusion (2) is established. Using the same method to that in the proof of (2), the conclusion (1) holds true. Thus, we complete the proof of Theorem 4.

Let $A = 1$ and $B = 0$ in Theorem 4, we have the following corollary.

Corollary 2. Let $f_{i}(z)\in \overline{\Sigma S}_{k}^*(a, 1, 0)\; (i = 1, 2)$ and $p > 1$. If anyone of the following conditions is satisfied.

(ⅰ) $a < 1, \lambda < \mathrm{min}\bigg\{\frac{1}{2(n+1)}, \frac{a+1}{(2n+1)}, \frac{k+a(k+2)}{k(k+1)}\bigg\},$

(ⅱ) $a > 1, \frac{1}{2(k+1)} < \lambda < \mathrm{min}\bigg\{\frac{a+1}{(2n+1)}, \frac{k+a(k+2)}{k(k+1)}\bigg\},$

(ⅲ) $a < 1, \mathrm{max}\bigg\{\frac{a+1}{(2k+1)}, \frac{k+a(k+2)}{k(k+1)}\bigg\} < \lambda < \frac{1}{2(n+1)},$

(ⅳ) $a > 1, \mathrm{max}\bigg\{\frac{1}{2(k+1)}, \frac{a+1}{(2k+1)}, \frac{k+a(k+2)}{k(k+1)}\bigg\} < \lambda.$

Then the generalized $\lambda$-Hadamard product $(f_{1}\widetilde{\Delta} f_{2})_{\lambda}\left(\frac{1}{p}, \frac{p-1}{p}, a; z\right)\in \ \overline{\Sigma S}_{k}^*(a, 1-2\beta, \hat{B}).$

Acknowledgments

This work is supported by the Natural Science Foundation of the People's Republic of China under Grant 11561001, the Natural Science Foundation of Inner Mongolia of the People's Republic of China under Grants 2018MS01026, 2019MS01023 and 2020MS01011, the Program for Young Talents of Science and Technology in Universities of Inner Mongolia Autonomous Region under Grant NJYT-18-A14 and the Higher-School Science Foundation of Inner Mongolia of the People's Republic of China under Grants NJZY20198 and NJZY20200.

Conflict of interest

All authors declare no conflict of interest in this paper.