Closure properties for second-order $\lambda$-Hadamard product of a class of meromorphic Janowski function

1.   Introduction

Let $\Sigma (c, n)(c > 0, n\geq 2, n\in \mathbb{N})$ denote the class of meromorphic functions $f$ which are analytic in the punctured open unit disk $\hat{\mathcal{U}} = \mathcal{U}\backslash\{0\} = \{z\in \mathbb{C}:0 < |z| < 1 \}$ and has the Taylor series expansion of form

We define the following class $\Sigma S _{n}^{*}(c, \beta)$ of the generalized meromorphically starlike of order $\beta$ ^[1,2],

Let $f(z)$ and $g(z)$ be members of analytic function in $\mathcal{U}$, $f(z)$ is said to be subordinate to $g(z)$, if there exists a Schwarz function $\omega(z),$ analytic in $\mathcal{U}$ with

such that

In such a case, we write $f(z)\prec g(z)$. Furthermore if the function $g(z)$ is univalent in $\mathcal{U}$, then we have ^[3,4],

In this paper, we study a class of meromorphic Janowski functions as follows. ^[5,6]

Definition 1.1. A function $f(z)\in \Sigma (c, n)$ belongs to the class $\Sigma S_{n}^{*}(c, a, b),$ if and only if

It is quite clear that

Using the subordination relationship, $f(z)\in \Sigma S_{n}^{*}(c, a, b),$ if satisfies the following condition

where $\omega(z)$ is analytic in $\mathcal{U}$ with $|\omega(z)| < 1$ and $\omega(0) = 0$.

The above equation can also be equivalent to

Let $T(c, n)$ be a subclass of $\Sigma (c, n)$, and $f \in T(c, n)$ is defined as

In particular, taking

Next, we will introduce the second-order generalized $\lambda$-Hadamard product of the class $T(c, n)$.

Definition 1.2. Let $f_{i}(z) = \frac{c}{z}-\sum_{k = n}^{\infty}|a_{k}, _{i}|z^{k}\in {T}(c, n)\; (i = 1, 2)$. The second-order generalized $\lambda$-Hadamard product $(f_{1} \widetilde{\bigtriangleup} f_{2})_{\lambda}(u, v, c; z)$ of the function $f_1$ and $f_2$ is defined by

where

We can obtain, from (1.3) and (1.4), that

In particular, if $u = v = 1$, then $(f_{1} \widetilde{\bigtriangleup} f_{2})_{\lambda}(1, 1, c; z) = ({f_{1}\overline{{\ast}} f_{2})_{\lambda}}(z)$ is the second-order $\lambda$-Hadamard product as follows:

Note. By choosing different parameters $c, \lambda, u$ and $v$, we can get special convolutions as below:

$(1)$ For $\lambda = 0, \; c = 1,$ $({f_{1}\triangle f_{2}})_0(u, v, 1;z) = ({f_{1}\triangle f_{2}}) (u, v; z)$ is the product defined in ^[7,8].

$(2)$ For $\lambda = 0, \; u = v = 1, c = 1,$ $({f_{1}\widetilde{\triangle} f_{2}})_{0}(1, 1, 1;z) = ({f_{1} \ast} f_{2})(z)$ is the famous Hadamard product defined in ^[1].

For the sake of simplicity, the parameters for the rest of the article are specified below

At the same time, let

In 1996, Choi et al. ^[7] studied the generalized $\lambda$-Hadamard product of univalent functions. In 2021, the authors ^[3] discuss the closure properties of the first-order $\lambda$-Hadamard product of the class $\overline{\Sigma S}_{n}^{*}(c, a, b)$.

In this paper, we will continue to discuss the closed problems of second-order generalized $\lambda$-Hadamard product of the class $\Sigma S_{n}^*(c, a, b)$ and obtain some new results.

2.   Perliminaries

Lemma 2.1. ^[3] If $f \in \Sigma (c, k)$ satisfies

then $f(z)\in \Sigma S_{n}^*(c, a, b).$

Lemma 2.2. ^[3] Let the function $f\in T(c, k)$, then $f\in \overline{\Sigma S}_{n}^*(c, a, b)$ if and only if

3.   Main results

First, we investigate the closure properties of the second-order $\lambda$-Hadamard product $(f_{1}\overline{\ast} f_{2})_{\lambda}(z).$

Theorem 3.1. If $f_i(z) = \frac{c}{z}-\sum_{k = n}^{\infty} |{a_{k, i}}|z^{k}\in \overline{\Sigma S}_{n}^*(c, a, b)\; (i = 1, 2)$ satisfy one of the following conditions:

(1) $0\leq b < a\leq 1, \; 0 < \lambda < \frac{1}{n+1}$,

(2) $-1\leq a < b\leq 0, \; 0 < \lambda < \frac{1}{n+1}$,

then the second-order $\lambda$-Hadamard product $\frac{1}{c}(f_{1}\overline{\ast} f_{2})_{\lambda}(z)\in \overline{\Sigma S}_{n}^*(c, a, b).$

Proof. Let $f_{1}, f_{2}\in \overline{\Sigma S}_{n}^*(c, a, b),$ which are given by (1.2).

(1) For $0\leq b < a\leq 1$, we can get from (2.2),

and

Since

then, to obtain $\frac{1}{c}(f_{1}\overline{\ast} f_{2})\in\overline{ \Sigma S}_{n}^*(c, a, b)$, we only need to verify the following condition is established.

Using Cauchy-Schwarz inequality, it can be obtained from Eqs (3.1) and (3.2) that

From (3.3) and (3.4), we just need to prove

Since

Thus, we just have to show

For $\lambda < \frac{1}{n+1}$, (3.5) can be simplified to $\lambda > \frac{k(a-b)-[k+1+(a+kb)]}{(a-b)k(k+1)}.$

Therefore, we conclude that if

then $\frac{1}{c}(f_{1}\overline{\ast} f_{2})_{\lambda}(z)\in \overline{\Sigma S}_{n}^*(c, a, b).$

(2) For $-1\leq a < b\leq 0$. We can using the same method above, obtain that if

then $\frac{1}{c}(f_{1}\overline{\ast} f_{2})_{\lambda}(z)\in \overline{\Sigma S}_{n}^*(c, a, b).$ This completes the proof of Theorem 3.1.

Corollary 3.1. Let $u > 1, f_{i}(z)\in \overline{\Sigma S}_{n}^*(c, -1, 0)\; (i = 1, 2)$. If $\; 0 < \lambda < \frac{1}{n+1}$, then

Next, we consider the closure properties of the second-order generalized $\lambda$-Hadamard product.

Theorem 3.2. Let $u > 1, f_{i}(z) = \frac{c}{z}-\sum_{k = n}^{\infty}|a_{k}|z^{k}\in \overline{\Sigma S}_{n}^*(c, a, b)\; (i = 1, 2).$ If one of the following conditions,

(i) $0\leq \hat{b} < b < a\leq 1, \; \; \frac{2n+1}{(n+1)(3n+1)} < \lambda < \mathrm{min}(\frac{c(1+b)+2n(a-b)}{(3n^2+2n)(a-b)}, \frac{1}{n+1}),$ and $0\leq\hat{b} < \mathrm{min} (1, a-\frac{(a-b)[n+1+(a+nb)]n[1-(n+1)\lambda]}{c[n+1+(a+nb)]+(a-b)n^2[1-(n+1)\lambda]});$

(ii) $-1\leq a < b < \hat{b}\leq 0, \; {\rm{a}}nd \; \; 0 < \lambda < \frac{2n+1}{(n+1)(3n+1)}$, holds true, then $\frac{1}{c}(f_{1}\widetilde{\bigtriangleup} f_{2})_{\lambda}\bigg(\frac{1}{u}, \frac{u-1}{u}, c, z\bigg)\in \overline{\Sigma S}_{n}^*(c, a, \hat{b}).$

Proof. Let $f_{i}(z)\in \overline{\Sigma S}_{n}^*(c, a, b)(i = 1, 2).$

(1) Let $0\leq b < a \leq 1$, using (2.2), we have

So we get

and

Since

then, to get $\frac{1}{c}({f_{1}\widetilde{\triangle }f_{2}})(\frac{1}{u}, \frac{u-1}{u}, c; z)\in \overline{\Sigma S}_{n}^*(c, a, \widehat{b})$, we just only need to verify that the following condition:

holds true.

Applying the Hölder inequality, from (3.6) and (3.7), we get

In order to obtain inequality (3.8), we only need to prove

that is,

Let

and

After simplification, we can get

Suppose

and

To prove Theorem 3.2, we divide the procedure into two cases.

(a) Because $P_1(k)$ is increasing with respect to $k$ and $P_1(n) = \frac{c[n+1+(a+nb)]}{a-b}+n^2[1-(n+1)\lambda] > 0,$ $P_1(k) > 0$ is true for $k\geq n$. In order to prove (3.9), we only need to show the following inequality

It is easily to verify (3.10) holds true if $\lambda < \frac{(a-b)n^2+a[n+1+(a+nb)]}{(a-b)(n+1)n^2}$.

It is clear that $\overline{P_1(k)}$ is monotonically increasing, so we get $\lambda < \frac{c(1+b)+2k(a-b)}{(3k^2+2k)(a-b)}.$

Since $\overline{Q_1(k)}$ is monotonically decreasing with respect to $k$ if $\lambda > \frac{2k+1}{(k+1)(3k+1)},$ then

and $\overline{Q_1(k)}$ is increasing with respect to $k$ if $\lambda < \frac{2k+1}{(k+1)(3k+1)}$, then

Because of $\widehat{b}\geq 0$, this case is removed.

(b) Because $P_1(k)$ is decreasing with respect to $k$ and $P_1(n) = \frac{c[n+1+(a+nb)]}{a-b}+n^2[1-(n+1)\lambda] < 0,$ $P_1(k) < 0$ is true for $k\geq n$. (3.9) holds true if

Clearly, (3.11) holds true if $\lambda > \frac{(a-b)n^2+c[k+1+(a+kb)]}{(a-b)(k+1)k^2}$. Since $\lambda < \frac{1}{k+1},$ there is no solution for $\lambda$.

Since $P_1(k)$ and $\overline{P_1(k)}$ are decreasing functions, we have $\lambda > \frac{c(1+b)+2k(a-b)}{(3k^2+2k)(a-b)}.$ $\overline{Q_1(k)}$ is monotonically decreasing with respect to $k$ if $\lambda > \frac{2k+1}{(k+1)(3k+1)},$ we have

And $\overline{Q_1(k)}$ is increasing if $\lambda < \frac{2k+1}{(k+1)(3k+1)}$. On the other hand, because $\lambda > \frac{c(1+b)+2k(a-b)}{(3k^2+2k)(a-b)}$, there is no solution for $\lambda$.

If the condition (ⅰ) is satisfied, then (3.9) is true. Therefore, $\frac{1}{c}(f_{1}\widetilde{\bigtriangleup} f_{2})_{\lambda}\bigg(\frac{1}{u}, \frac{u-1}{u}, c, z\bigg)\in \overline{\Sigma S}_{n}^*(c, a, \widehat{b}).$

(2) For $-1\leq a < b \leq 0$, the proof is similar to (1), so we can obtain

Let

and

After simple calculation, we can obtain

Assume

and

We divide the discussion into two cases as follows:

(c) Because $P_2(k)$ is increasing with respect to $k$ and $P_2(n) = \frac{c[n+1+(a+nb)]}{b-a}+n^2[1-(n+1)\lambda] > 0$, $P_2(k) > 0$ is true for $k\geq n$. In order to prove (3.12), we only need the following inequality to be true

It is not hard to verify (3.13) holds true if $\lambda < \frac{(b-a)n^2+c(n+1-(a+nb))}{(b-a)(n+1)n^2}$.

On the other hand, $\overline{P_2(k)}$ is increasing with respect to $k$, we can get $\lambda < \frac{c(1-b)+2k(b-a)}{(3k^2+2k)(b-a)}$.

It is clear that $\overline{Q_2(k)}$ is decreasing if $\lambda > \frac{2k+1}{(k+1)(3k+1)}$, then

and $\overline{Q_2(k)}$ is increasing with respect to $k$ if $\lambda < \frac{2k+1}{(k+1)(3k+1)}$, then

(d) Because $P_2(k)$ is decreasing with respect to $k$ and$\; P_2(n) = n^2[1-(n+1)\lambda]+\frac{c[n+1-(a+nb)]}{b-a} < 0$, $P_2(k) < 0$ is true for $k\geq n$. In order to prove (3.12), we only need the following inequality to be true

It is not hard to verify (3.14) holds true if $\lambda > \frac{(b-a)n^2+c(n+1-(a+nb))}{(b-a)(n+1)n^2}$.

Also, $\overline{P_2(k)}$ is decreasing with respect to $k$, we can obtain $\lambda > \frac{c(1-b)+2k(b-a)}{(3k^2+2k)(b-a)}$. So, there is no solution for $\lambda$.

If the condition (ⅱ) is satisfied, then (3.12) is true. Therefore, $\frac{1}{c}(f_{1}\widetilde{\bigtriangleup} f_{2})_{\lambda}\bigg(\frac{1}{u}, \frac{u-1}{u}, c, z\bigg)\in \overline{\Sigma S}_{n}^*(c, a, \hat{b}).$

When $b = \frac{1}{2}$ and $a = 1$ in Theorem 3.2, we have the following corollary.

Corollary 3.2. Let $u > 1, f_{i}(z) = \frac{c}{z}-\sum_{k = n}^{\infty} |{a_{k, i}}|z^{k} (i = 1, 2)\in \overline{\Sigma S}_{n}^*(c, 1, \frac{1}{2})$. If

then $\frac{1}{c}(f_{1}\widetilde{\bigtriangleup} f_{2})_{\lambda}\bigg(\frac{1}{u}, \frac{u-1}{u}, c, z\bigg)\in \overline{\Sigma S}_{n}^*(c, 1, \frac{1}{2}).$

By putting $b = -\frac{1}{2}$ and $a = -1$ in Theorem 3.2, we have

Corollary 3.3. Let $u > 1, f_{i}(z) = \frac{c}{z}-\sum_{k = n}^{\infty} |{c_{k, i}}|z^{k} (i = 1, 2)\in \overline{\Sigma S}_{n}^*(c, -1, -\frac{1}{2})\;$.

If $\; -\frac{1}{2} < \hat{b}\leq 0, \; 0 < \lambda < \frac{2n+1}{(n+1)(3n+1)}$, then $\frac{1}{c}(f_{1}\widetilde{\bigtriangleup} f_{2})_{\lambda}\bigg(\frac{1}{u}, \frac{u-1}{u}, c, z\bigg)\in \overline{\Sigma S}_{n}^*(c, -1, -\frac{1}{2}).$

4.   Conclusions

In this paper, we prove the closure properties of the second-order $\lambda$-Hadamard product and the second-order generalized $\lambda$-Hadamard product of the class $\overline{\Sigma S}_{n}^*(c, a, b).$ The results presented in this paper would find further applications for the $\lambda$-Hadamard product of the class of meromorphic Janowski function, which can enrich the research field of Hadamard procduct.

Acknowledgments

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 2022MS01004, 2019MS01023 and 2020MS01011, the Program for Young Talents of Science and Technology in Universities of Inner Mongolia Autonomous Region under Grant NJYT-18-A14, the Higher School Foundation of Inner Mongolia of the People's Republic of China under Grant NJZY22168, the Program for Key Laboratory Construction of Chifeng University (no. CFXYZD202004) and the Research and Innovation Team of Complex Analysis and Nonlinear Dynamic Systems of Chifeng University (no. cfxykycxtd202005).

Conflict of interest

The authors declare no conflict of interest.