On Classes of Meromorphic Functions Defined by Subordination and Convolution

Abstract

For $p\in {\mathbb{N}}^{*}$, let ${\Sigma }_p$ denote the class of meromorphic p-valent functions. We consider an operator for meromorphic functions denoted by $T_b^n$, which generalizes some previously studied operators. We introduce some new subclasses of the class ${\Sigma }_p$, associated with subordination using the above operator, and we prove that these classes are preserved regarding the operator $J_{p,\gamma }$, so we have symmetry when we look at the form of the class in which we consider the function g and at the form of the class of the image $J_{p,\gamma }\left(g\right)$, where $J_{p,\gamma }\left(g\right)\left(z\right)={\displaystyle \frac{\gamma -p}{z^{\gamma }}{\int }_0^{z}g\left(t\right)t^{\gamma -1}dt}$, $\gamma \in \mathbb{C}$ with $\mathrm{Re}\gamma >p$.

1. Introduction and Preliminaries

Let $U=\{z\in \mathbb{C}:|z|<1\}$ be the unit disc in the complex plane, $\dot{U}=U\setminus \left\{0\right\}$, $\mathcal{H}\left(U\right)=\{f:U\to \mathbb{C},f\mathrm{holomorphic}\mathrm{in}U\}$, $\mathbb{Z}=\{\dots ,-1,0,1,2,\dots \}$ the set of integer numbers, $\mathbb{N}=\{0,1,2,\dots \}$ and ${\mathbb{N}}^{*}=\mathbb{N}\setminus \left\{0\right\}$.

For $p\in {\mathbb{N}}^{*}$, let ${\Sigma }_p$ denote the class of meromorphic functions of the form

$$g\left(z\right)=\frac{a_{-p}}{z^p}+a_0+a_{1}z+\cdots ,z\in \dot{U},a_{-p}\ne 0.$$

To introduce the next meromorphic function subclasses, we need to know what is a subordination.

Definition 1 

([1] (p. 4)). Let f and F be members of $\mathcal{H}\left(U\right)$. The function f is said to be subordinate to F, written as $f\prec F$ or $f\left(z\right)\prec F\left(z\right)$, if there exists a function w analytic in U, with $w\left(0\right)=0$ and $\left|w\right(z\left)\right|<1$, and such that $f\left(z\right)=F\left(w\right(z\left)\right).$

For $p\in {\mathbb{N}}^{*}$ and $h\in \mathcal{H}\left(U\right)$ with $h\left(0\right)=p,$ we use the following notations (defined in previous papers as [2,3,4]):

$${\Sigma }_{p,0}=\{g\in {\Sigma }_p:a_{-p}=1\},$$

$$\Sigma S_p\left(h\right)=\left\{{\displaystyle g\in {\Sigma }_p:-\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\prec h\left(z\right)}\right\},$$

$$\Sigma K_p\left(h\right)=\left\{g\in {\Sigma }_p:-\left[{\displaystyle 1+\frac{z{g}^{″}\left(z\right)}{g^{\prime }\left(z\right)}}\right]\prec h\left(z\right)\right\},$$

$$\Sigma K_{p,0}\left(h\right)=\Sigma K_p\left(h\right)\cap {\Sigma }_{p,0}.$$

For $\phi \in \Sigma K_{p,0}\left(h\right)$, $h_1\in \mathcal{H}\left(U\right)$ with $h_1\left(0\right)=1$, we have

$$\Sigma {\mathcal{C}}_{p,0}(h_1;\phi ,h)=\left\{{\displaystyle g\in {\Sigma }_{p,0}:\frac{g^{\prime }\left(z\right)}{\phi {}^{\prime }\left(z\right)}\prec h_1\left(z\right)}\right\}.$$

$$\Sigma {\mathcal{C}}_{p,0}(h_1;h)=\left\{{\displaystyle g\in {\Sigma }_{p,0}:\mathrm{there}\mathrm{exists}\phi \in \Sigma K_{p,0}\left(h\right)\mathrm{such}\mathrm{that}\frac{g^{\prime }\left(z\right)}{\phi {}^{\prime }\left(z\right)}\prec h_1\left(z\right)}\right\}.$$

Inspired by the convolution of two analytic functions f and g, of the form

$$f\left(z\right)=z+\sum _{k=2}^{\infty }{a}_k{z}^k,f\left(z\right)=z+\sum _{k=2}^{\infty }{b}_k{z}^k,z\in U,$$

that is denoted by $f\ast g$ and is the analytic function ${\displaystyle (f\ast g)\left(z\right)=z+\sum _{k=2}^{\infty }{a}_k{b}_k{z}^k}$, we consider an extension of it on the class of meromorphic functions, thus defining a new operator denoted by $T_b^n$, as it follows:

For $p\in {\mathbb{N}}^{*},n\in \mathbb{N}$ and $b={\left(b_k\right)}_{k\ge 0}$, a sequence of complex numbers, such that ${\displaystyle \sum _{k=0}^{\infty }{b}_k{z}^k}$ is convergent for $z\in U$, we consider the new operator for meromorphic functions, denoted by $T_b^n:{\Sigma }_p\to {\Sigma }_p$, as

$$T_b^n\left(g\right)\left(z\right)=\frac{a_{-p}}{z^p}+\sum _{k=0}^{\infty }{b}_k^n{a}_k{z}^k,z\in \dot{U},$$

where $g\in {\Sigma }_p$ is $g\left(z\right)={\displaystyle \frac{a_{-p}}{z^p}+\sum _{k=0}^{\infty }{a}_k{z}^k,\mathrm{with}z\in \dot{U},a_{-p}\ne 0.}$

We will denote by $SC\left(U\right)$ the set of complex numbers sequences, $b={\left(b_k\right)}_{k\ge 0}$, such that ${\displaystyle \sum _{k=0}^{\infty }{b}_k{z}^k}$ is convergent for $z\in U$. It is obvious that every sequence satisfying the condition ${\displaystyle \underset{k\to \infty }{lim}\frac{|b_{k+1}|}{|b_k|}\le 1}$ belongs to $SC\left(U\right)$.

Also, if $n\in \mathbb{N}$ and $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right),$ then $b^n={\left(b_k^n\right)}_{k\ge 0}\in SC\left(U\right).$

For different choices of the sequence b, we obtain particular forms of the operator $T_b^n$, forms that were studied in many previous articles.

For instance, if we consider $b_k=k$, then the sum

$$\sum _{k=0}^{\infty }{b}_k^n{a}_k{z}^k=\sum _{k=0}^{\infty }{k}^n{a}_k{z}^k,z\in U,$$

is the well-known Sǎlǎgean differential operator of order n [5].

If we take $b_k={\displaystyle \frac{\lambda -p}{k+\lambda }}$ ($p\in {\mathbb{N}}^{*},\lambda \in \mathbb{C}$ with $\mathrm{Re}\lambda >p$), then $T_b^n$ is the operator $J_{p,\lambda }^n$ introduced in [6].

If we consider $b_k=1+q+\dots +q^k,q\in (0,1)$, then the sum

$$\sum _{k=0}^{\infty }{b}_k^n{a}_k{z}^k=\sum _{k=0}^{\infty }{(1+q+\dots +q^k)}^n{a}_k{z}^k,z\in U,$$

is Wanas operator (see [7,8,9]) and $T_b^n$ is the operator $W_q^n$ introduced in [10].

The study of operators on very different classes of functions to remark their properties is always an updating problem, as we can see in the works [11,12,13,14,15,16].

We have the following proprieties for $T_b^n$:

(1)

${\displaystyle T_b^{0}g\left(z\right)=g\left(z\right)}$;

(2)

${\displaystyle T_b^{1}g\left(z\right)=\frac{a_{-p}}{z^p}+\sum _{k=0}^{\infty }{a}_k{b}_k{z}^k}$;

(3)

${\displaystyle T_b^n\left(\alpha g_1+\beta g_2\right)\left(z\right)=\alpha T_b^n\left(g_1\right)\left(z\right)+\beta T_b^n\left(g_2\right)\left(z\right),\alpha ,\beta \in \mathbb{C},g_1,g_2\in {\Sigma }_p}$;

(4)

${\displaystyle T_b^n\left(T_b^m\left(g\right)\left(z\right)\right)=T_b^m\left(T_b^n\left(g\right)\left(z\right)\right)=T_b^{n+m}\left(g\right)\left(z\right)}$;

(5)

${\displaystyle T_b^n\left(z{g}^{\prime }\left(z\right)\right)=z·{\left(T_b^{n}g\left(z\right)\right)}^{\prime }}$;

(6)

${\displaystyle T_b^n\left(T_c^n\left(g\right)\left(z\right)\right)=T_{bc}^n\left(g\right)\left(z\right)}$.

We make the remark that the properties (1)–(5), of the operator $T_b^n$, are also shared by the above-mentioned operators $W_q^n$ (see [10]) and $J_{p,\lambda }^n$ (see [6]). The advantages of studying the operator $T_b^n$ is that it has a simple form and generalizes other operators defined on the class of meromorphic functions.

Next, using the above operator $T_b^n$, we define some new subclasses of the class ${\Sigma }_p$, associated with subordination, such that, in some particular cases, these new subclasses are the well-known classes of meromorphic starlike, convex or close-to-convex functions that were (and still are) studied by many authors working in the field of geometric function theory. We mention here some of the first papers dealing with these special classes of meromorphic functions [17,18,19].

Definition 2. 

Let $p\in {\mathbb{N}}^{*},n\in \mathbb{N}$, $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right),$ and $h\in \mathcal{H}\left(U\right)$ with $h\left(0\right)=p$. We define

$$\Sigma S_{p,b}^n\left(h\right)=\left\{g\in {\Sigma }_p:-\frac{z{\left(T_b^{n}g\left(z\right)\right)}^{\prime }}{T_b^{n}g\left(z\right)}\prec h\left(z\right)\right\}.$$

$$\Sigma K_{p,b}^n\left(h\right)=\left\{g\in {\Sigma }_p:-\left[{\displaystyle 1+\frac{z{\left(T_b^{n}g\left(z\right)\right)}^{″}}{{\left(T_b^{n}g\left(z\right)\right)}^{\prime }}}\right]\prec h\left(z\right)\right\}.$$

$$\Sigma K_{p,0,b}^n\left(h\right)=\Sigma K_{p,b}^n\left(h\right)\cap {\Sigma }_{p,0}.$$

It is obvious that we have

$$g\in \Sigma S_{p,b}^n\left(h\right)\iff T_b^n\left(g\right)\in \Sigma S_p\left(h\right),$$

$$g\in \Sigma K_{p,b}^n\left(h\right)\iff T_b^n\left(g\right)\in \Sigma K_p\left(h\right).$$

If we consider the function ${\displaystyle g\left(z\right)=\frac{1}{z^p}}$, we have ${\displaystyle T_b^n\left(g\left(z\right)\right)=g\left(z\right)=\frac{1}{z^p}}$ and

$$-\frac{z{\left(T_b^{n}g\left(z\right)\right)}^{\prime }}{T_b^{n}g\left(z\right)}=p,-\left[{\displaystyle 1+\frac{z{\left(T_b^{n}g\left(z\right)\right)}^{″}}{{\left(T_b^{n}g\left(z\right)\right)}^{\prime }}}\right]=p,$$

this meaning that ${\displaystyle g\left(z\right)=\frac{1}{z^p}}$ belongs to the new defined classes $\Sigma S_{p,b}^n\left(h\right)$, respectively, $\Sigma K_{p,b}^n\left(h\right).$

Definition 3. 

Let $p\in {\mathbb{N}}^{*},n\in \mathbb{N}$, $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right)$ and $h\in \mathcal{H}\left(U\right)$ with $h\left(0\right)=p$. For $\phi \in \Sigma K_{p,0,b}^n\left(h\right)$, $h_1\in \mathcal{H}\left(U\right)$ with $h_1\left(0\right)=1$, we have

$$\Sigma {\mathcal{C}}_{p,0,b}^n(h_1;\phi ,h)=\left\{{\displaystyle g\in {\Sigma }_{p,0}:\frac{{\left(T_b^{n}g\left(z\right)\right)}^{\prime }}{{\left(T_b^n\phi \left(z\right)\right)}^{\prime }}\prec h_1\left(z\right)}\right\}.$$

$$\Sigma {\mathcal{C}}_{p,0,b}^n(h_1;h)=\left\{{\displaystyle g\in {\Sigma }_{p,0}:\mathit{there}\mathit{exists}\phi \in \Sigma K_{p,0,b}^n\left(h\right)\mathit{such}\mathit{that}\frac{{\left(T_b^{n}g\left(z\right)\right)}^{\prime }}{{\left(T_b^n\phi \left(z\right)\right)}^{\prime }}\prec h_1\left(z\right)}\right\}.$$

It is obvious that for the above classes, we have

$${\displaystyle g\left(z\right)=\frac{1}{z^p}\in \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;g,h)\cap \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;h).}$$

Since in Mathematics symmetry means that one shape is identical to the other shape when it is moved, rotated, or flipped, we will see that the majority of our results have symmetry when we look at the form of the class of function g and at the form of the image class of $J_{p,\gamma }\left(g\right)$, this meaning that for g in $\Sigma S_{p,b}^n\left(h\right)$, satisfying some conditions, we obtain $J_{p,\gamma }\left(g\right)$ in $\Sigma S_{p,b}^n\left(h\right)$. We also have this symmetry property for all previously defined classes.

It is easy to see that, for $n=0$, the class $\Sigma S_{p,b}^0\left(h\right)$ is the class $\Sigma S_p\left(h\right)$, the class $\Sigma K_{p,b}^0\left(h\right)$ is the class $\Sigma K_p\left(h\right)$ and the class $\Sigma {\mathcal{C}}_{p,0,b}^0(h_1;\phi ,h)$ is the class $\Sigma {\mathcal{C}}_{p,0}(h_1;\phi ,h)$, which were studied in [2,3,4].

In the field of geometric theory of analytic functions, since the beginning of the 20th century, many mathematicians studied different operators applied to classes of analytic functions and their properties, but papers on operators applied to classes of meromorphic functions suffer from a gap. Also, there is still more to say about the property of preserving meromorphic classes and the symmetry properties when we apply an operator.

From 2011, the second author of the present paper began to study integral operators on meromorphic multivalent functions and classes obtained by using the analytic solution of a Briot–Bouquet differential equation

$$q\left(z\right)+\frac{nz{q}^{\prime }\left(z\right)}{\beta q\left(z\right)+\gamma }=h\left(z\right),$$

which was the best $(a,n)$-dominant for a Briot–Bouquet subordination. We mention that a Briot–Bouquet subordination has the form

$$p\left(z\right)+\frac{nz{p}^{\prime }\left(z\right)}{\beta p\left(z\right)+\gamma }\prec h\left(z\right),z\in U,p,h\in \mathcal{H}\left(U\right),p\left(0\right)=h\left(0\right).$$

For a better understanding of the expression “the best dominant” we give the next definition:

Definition 4 

([1] (p. 16)). Let $\psi :{\mathbb{C}}^3\times U\to \mathbb{C}$ and h univalent in U. If $p\in \mathcal{H}\left(U\right)$ satisfies the differential subordination

$$\psi (p\left(z\right),z{p}^{\prime }\left(z\right),z^2{p}^{″}\left(z\right);z)\prec h\left(z\right),$$

(1)

then p is called a solution for (1). The univalent function q is said to be a dominant if we have $p\prec q$ for all p verifying (1).

A dominant $\tilde{q}$ with $\tilde{q}\prec q$ for all dominants q of (1) is called the best dominant of (1).

Briot–Bouquet differential subordinations and their solutions began to be studied intensively by P. J. Eenigenburg, S. S. Miller, P. T. Mocanu and M. O. Reade in 1983 (see [20]). Very important and useful results regarding the Briot–Bouquet differential equations and subordinations were published some years later by S. S. Miller, P. T. Mocanu in [21]. To obtain our result, we turned to the solutions of these so-called Briot–Bouquet differential subordinations.

2. Main Results

First of all, for $p,n,m\in \mathbb{N},p\ne 0$, $n>m$, we give a link between the sets $\Sigma S_{p,b}^n\left(h\right)$ and $\Sigma S_{p,b}^m\left(h\right)$, respectively, $\Sigma K_{p,b}^n\left(h\right)$ and $\Sigma K_{p,b}^m\left(h\right)$.

Proposition 1. 

Let $p,n,m\in \mathbb{N},p\ne 0$, $n>m$, $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right)$ and $h\in \mathcal{H}\left(U\right)$ with $h\left(0\right)=p$. We have

$$g\in \Sigma S_{p,b}^n\left(h\right)\iff T_b^{n-m}\left(g\right)\in \Sigma S_{p,b}^m\left(h\right),$$

respectively,

$$g\in \Sigma K_{p,b}^n\left(h\right)\iff T_b^{n-m}\left(g\right)\in \Sigma K_{p,b}^m\left(h\right).$$

Proof. 

We have $g\in \Sigma S_{p,b}^n\left(h\right)$ equivalent to $T_b^n\left(g\right)\in \Sigma S_p\left(h\right)$.

Since $T_b^n\left(g\right)=T_b^m\left(T_b^{n-m}g\right)$, we obtain $T_b^m\left(T_b^{n-m}g\right)\in \Sigma S_p\left(h\right)$, which is equivalent to

$$T_b^{n-m}\left(g\right)\in \Sigma S_{p,b}^m\left(h\right).$$

The second equivalence can be proved in the same way. □

From Proposition 1, considering $m=n-1$, we have a link between the sets $\Sigma S_{p,b}^n\left(h\right)$ and $\Sigma S_{p,b}^{n-1}\left(h\right)$, respectively, $\Sigma K_{p,b}^n\left(h\right)$ and $\Sigma K_{p,b}^{n-1}\left(h\right)$, as we can see:

Proposition 2. 

Let $p,n\in \mathbb{N},p\ne 0$, $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right)$ and $h\in \mathcal{H}\left(U\right)$ with $h\left(0\right)=p$. We have

$$g\in \Sigma S_{p,b}^n\left(h\right)\iff T_b\left(g\right)\in \Sigma S_{p,b}^{n-1}\left(h\right),$$

respectively,

$$g\in \Sigma K_{p,b}^n\left(h\right)\iff T_b\left(g\right)\in \Sigma K_{p,b}^{n-1}\left(h\right).$$

Proposition 3. 

Let $p,n\in \mathbb{N},p\ne 0$, $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right)$, $h,h_1\in \mathcal{H}\left(U\right)$ with $h\left(0\right)=p$, $h_1\left(0\right)=1$ and $\phi \in \Sigma K_{p,0,b}^n\left(h\right)$. We have

$$g\in \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;\phi ,h)\iff T_b^n\left(g\right)\in \Sigma {\mathcal{C}}_{p,0}(h_1;T_b^n\left(\phi \right),h).$$

Proof. 

From $g\in \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;\phi ,h)$, we have ${\displaystyle \frac{{\left(T_b^{n}g\left(z\right)\right)}^{\prime }}{{\left(T_b^n\phi \left(z\right)\right)}^{\prime }}\prec h_1\left(z\right)}$. Since $\phi \in \Sigma K_{p,0,b}^n\left(h\right)$, we easily obtain from the definition that $T_b^n\left(\phi \right)\in \Sigma K_{p,0}\left(h\right)$.

Hence, $T_b^n\left(g\right)\in \Sigma {\mathcal{C}}_{p,0}(h_1;T_b^n\left(\phi \right),h).$ □

Next, we give a link between the sets $\Sigma {\mathcal{C}}_{p,0,b}^n(h_1;\phi ,h)$ and $\Sigma {\mathcal{C}}_{p,0,b}^m(h_1;T_b^m\left(\phi \right),h)$.

Proposition 4. 

Let $p,n,m\in \mathbb{N},p\ne 0$, $n>m$, $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right)$, $h,h_1\in \mathcal{H}\left(U\right)$ with $h\left(0\right)=p$, $h_1\left(0\right)=1$ and $\phi \in \Sigma K_{p,0,b}^n\left(h\right)$. We have

$$g\in \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;\phi ,h)\iff T_b^{n-m}\left(g\right)\in \Sigma {\mathcal{C}}_{p,0,b}^m(h_1;T_b^{n-m}\left(\phi \right),h).$$

Proof. 

We have $g\in \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;\phi ,h)$ equivalent to $T_b^n\left(g\right)\in \Sigma {\mathcal{C}}_{p,0}(h_1;T_b^n\left(\phi \right),h)$, from Proposition 3.

Since $T_b^n\left(g\right)=T_b^m\left(T_b^{n-m}g\right)$ and $T_b^n\left(\phi \right)=T_b^m\left(T_b^{n-m}\left(\phi \right)\right),$ we obtain

$$T_b^m\left(T_b^{n-m}g\right)\in \Sigma {\mathcal{C}}_{p,0}(h_1;T_b^m\left(T_b^{n-m}\left(\phi \right)\right),h),$$

which is equivalent to

$$T_b^{n-m}\left(g\right)\in \Sigma {\mathcal{C}}_{p,0,b}^m(h_1;T_b^{n-m}\left(\phi \right),h).$$

□

The next result is a lemma that is needed to prove a theorem, which will help us to obtain functions from the class $\Sigma {\mathcal{C}}_{p,0,b}^n(h_1;\phi ,h)$ when a function from the class $\Sigma K_{p,0,b}^n\left(h\right)$ is already given.

Lemma 1. 

Let $p\in {\mathbb{N}}^{*}$, $\lambda \in \mathbb{C}$, $h\in \mathcal{H}\left(U\right)$ with $h\left(0\right)=p$. If $g\in \Sigma K_{p,0}\left(h\right)$, then the function

$$f_{\lambda ,g}=(1+\lambda p)g+\lambda z{g}^{\prime }\in \Sigma {\mathcal{C}}_{p,0}(h_1;g,h),$$

where $h_1=1-\lambda (h-p).$

Proof. 

First of all, we make the remark that, since $g\in {\Sigma }_{p,0}$, we have g of the form

$${\displaystyle g\left(z\right)=\frac{1}{z^p}+\sum _{k=0}^{\infty }{a}_k{z}^k,\mathrm{so}{g}^{\prime }\left(z\right)=-\frac{p}{z^{p+1}}+\sum _{k=1}^{\infty }k{a}_k{z}^{k-1},}$$

and we find that

$$f_{\lambda ,g}\left(z\right)=(1+\lambda p)g\left(z\right)+\lambda z{g}^{\prime }\left(z\right)=$$

$${\displaystyle (1+\lambda p)\left[{\displaystyle \frac{1}{z^p}+\sum _{k=0}^{\infty }{a}_k{z}^k}\right]+\lambda z\left[{\displaystyle -\frac{p}{z^{p+1}}+\sum _{k=1}^{\infty }k{a}_k{z}^{k-1}}\right]=\frac{1}{z^p}+\sum _{k=0}^{\infty }{b}_k{z}^k,}$$

where $b_k=(1+\lambda p+\lambda k)a_k,k\ge 0.$

Hence, we have $f_{\lambda ,g}\in {\Sigma }_{p,0}$.

From the definition of the function $f_{\lambda ,g}$, we have

$$f_{\lambda ,g}^{\prime }=(1+\lambda p)g^{\prime }+\lambda g^{\prime }+\lambda z{g}^{″},$$

hence

$$\frac{f_{\lambda ,g}^{\prime }}{g^{\prime }}=(1+\lambda p)+\lambda \left(1+\frac{z{g}^{″}}{g^{\prime }}\right).$$

Since $g\in \Sigma K_{p,0}\left(h\right)$, we obtain $-\left({\displaystyle 1+\frac{z{g}^{″}\left(z\right)}{g^{\prime }\left(z\right)}}\right)\prec h\left(z\right),z\in U,$ therefore

$$(1+\lambda p)+\lambda \left(1+\frac{z{g}^{″}}{g^{\prime }}\right)\prec (1+\lambda p)-\lambda h\left(z\right)=h_1\left(z\right),z\in U,$$

this meaning that

$$\frac{f_{\lambda ,g}^{\prime }\left(z\right)}{g^{\prime }\left(z\right)}\prec h_1\left(z\right)\iff f_{\lambda ,g}\in \Sigma {\mathcal{C}}_{p,0}(h_1;g,h).$$

□

We notice that for $\lambda =0$ in Lemma 1, we find that

$$f_{0,g}=g\in \Sigma {\mathcal{C}}_{p,0}(1;g,h)\subset \Sigma {\mathcal{C}}_{p,0}(1;h),$$

so we have the next result:

Corollary 1. 

Let $p\in {\mathbb{N}}^{*}$, $h\in \mathcal{H}\left(U\right)$ with $h\left(0\right)=p$. We have

$$\Sigma K_{p,0}\left(h\right)\subset \Sigma {\mathcal{C}}_{p,0}(1;h).$$

Theorem 1. 

Let $p\in {\mathbb{N}}^{*},n\in \mathbb{N}$, $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right)$, $\lambda \in \mathbb{C}$, $h\in \mathcal{H}\left(U\right)$ with $h\left(0\right)=p$. If $g\in \Sigma K_{p,0,b}^n\left(h\right)$, then the function

$$f_{\lambda ,g}=(1+\lambda p)g+\lambda z{g}^{\prime }\in \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;g,h),$$

where $h_1=1-\lambda (h-p).$

Proof. 

Since $g\in \Sigma K_{p,0,b}^n\left(h\right)$, we have from Definition 2 that $T_b^n\left(g\right)\in \Sigma K_{p,0}\left(h\right)$, so, from Lemma 1 we obtain

$$f_{\lambda ,T_b^n\left(g\right)}=(1+\lambda p)T_b^n\left(g\right)+\lambda z{\left(T_b^n\left(g\right)\right)}^{\prime }\in \Sigma {\mathcal{C}}_{p,0}(h_1;T_b^n\left(g\right),h),$$

(2)

where $h_1=1-\lambda (h-p).$

On the other hand, from Proposition 3, we have

$$f_{\lambda ,g}\in \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;g,h)\iff T_b^n\left(f_{\lambda ,g}\right)\in \Sigma {\mathcal{C}}_{p,0}(h_1;T_b^n\left(g\right),h).$$

By using the properties of the operator $T_b^n$, we see that

$$T_b^n\left(f_{\lambda ,g}\right)=T_b^n\left((1+\lambda p)g+\lambda z{g}^{\prime }\right)=$$

$$=(1+\lambda p)T_b^n\left(g\right)+\lambda T_b^n\left(z{g}^{\prime }\right)=(1+\lambda p)T_b^n\left(g\right)+\lambda z{\left(T_b^n\left(g\right)\right)}^{\prime }=f_{\lambda ,T_b^n\left(g\right)}.$$

Using now (2) we obtain that

$$T_b^n\left(f_{\lambda ,g}\right)\in \Sigma {\mathcal{C}}_{p,0}(h_1;T_b^n\left(g\right),h),$$

therefore

$$f_{\lambda ,g}\in \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;g,h).$$

□

We notice that for $\lambda =0$ in Theorem 1 we obtain that

$$f_{0,g}=g\in \Sigma {\mathcal{C}}_{p,0,b}^n(1;g,h)\subset \Sigma {\mathcal{C}}_{p,0}^n(1;h),$$

so we have the next result:

Corollary 2. 

Let $p\in {\mathbb{N}}^{*}$, $n\in \mathbb{N}$, $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right)$ and $h\in \mathcal{H}\left(U\right)$ with $h\left(0\right)=p$. We have

$$\Sigma K_{p,0,b}^n\left(h\right)\subset \Sigma {\mathcal{C}}_{p,0,b}^n(1;h).$$

We consider now the integral operator $J_{p,\gamma }$ (see [4]), defined on the class of meromorphic function ${\Sigma }_p$,

$$J_{p,\gamma }\left(g\right)\left(z\right)={\displaystyle \frac{\gamma -p}{z^{\gamma }}{\int }_0^{z}g\left(t\right)t^{\gamma -1}dt},$$

where $\gamma \in \mathbb{C}$ with $\mathrm{Re}\gamma >p$. We give the conditions such that $J_{p,\gamma }$ preserves the classes defined in Definition 2, respectively, in Definition 3.

Theorem 2. 

Let $p\in {\mathbb{N}}^{*},n\in \mathbb{N}$, $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right)$, $\gamma \in \mathbb{C}$ with $\mathrm{Re}\left(\gamma \right)>p$, $h\in \mathcal{H}\left(U\right)$ convex in U with $h\left(0\right)=p$ and $g\in \Sigma S_{p,b}^n\left(h\right)$. If $\mathrm{Re}h\left(z\right)<\mathrm{Re}\left(\gamma \right),z\in U,$ then

$$G=J_{p,\gamma }\left(g\right)\in \Sigma S_{p,b}^n\left(q\right),$$

where q is the univalent solution of the Briot–Bouquet differential equation

$$q\left(z\right)+\frac{pz{q}^{\prime }\left(z\right)}{\gamma -q\left(z\right)}=h\left(z\right),z\in U,q\left(0\right)=p.$$

The function q is the best (p,p)-dominant.

Proof. 

Since $g\in \Sigma S_{p,b}^n\left(h\right)$ we have from Definition 2 that

$$-\frac{z{\left(T_b^{n}g\left(z\right)\right)}^{\prime }}{T_b^{n}g\left(z\right)}\prec h\left(z\right),z\in U,$$

which is equivalent to $T_b^n\left(g\right)\in \Sigma S_p\left(h\right).$ From [2], Corolarry 2.13, since the hypotheses are verified, we find that

$$J_{p,\gamma }\left(T_b^n\left(g\right)\right)\in \Sigma S_p\left(q\right),$$

(3)

where q is the univalent solution of the Briot–Bouquet differential equation

$$q\left(z\right)+\frac{pz{q}^{\prime }\left(z\right)}{\gamma -q\left(z\right)}=h\left(z\right),z\in U,q\left(0\right)=p,$$

and the function q is the best (p,p)-dominant.

We know from [6] that

$$J_{p,\gamma }\left(g\right)\left(z\right)=\frac{a_{-p}}{z^p}+{\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{\gamma -p}{k+\gamma }}{a}_k{z}^k,z\in \dot{U},$$

this meaning that

$$J_{p,\gamma }\left(T_b^n\left(g\right)\right)\left(z\right)=\frac{a_{-p}}{z^p}+{\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{\gamma -p}{k+\gamma }}{b}_k^n{a}_k{z}^k,z\in \dot{U},$$

since

$$T_b^n\left(g\right)\left(z\right)=\frac{a_{-p}}{z^p}+\sum _{k=0}^{\infty }{b}_k^n{a}_k{z}^k,z\in \dot{U}.$$

By using the definition of the operator $T_b^n$, we obtain

$$T_b^n\left(J_{p,\gamma }g\left(z\right)\right)=\frac{a_{-p}}{z^p}+{\displaystyle \sum _{k=0}^{\infty }}{b}_k^n{\displaystyle \frac{\gamma -p}{k+\gamma }}{a}_k{z}^k=J_{p,\gamma }\left(T_b^n\left(g\right)\right)\left(z\right),z\in \dot{U}.$$

(4)

From (3) and (4), we obtain

$$T_b^n\left(J_{p,\gamma }\left(g\right)\right)\in \Sigma S_p\left(q\right),$$

which, from Definition 2, is equivalent to

$$J_{p,\gamma }\left(g\right)\in \Sigma S_{p,b}^n\left(q\right),$$

where q is the univalent solution of the Briot–Bouquet differential equation

$$q\left(z\right)+\frac{pz{q}^{\prime }\left(z\right)}{\gamma -q\left(z\right)}=h\left(z\right),z\in U,q\left(0\right)=p,$$

and the function q is the best (p,p)-dominant. □

Since $q\prec h$, we have $\Sigma S_{p,b}^n\left(q\right)\subset \Sigma S_{p,b}^n\left(h\right)$, so we obtain the next corollary:

Corollary 3. 

Let $p\in {\mathbb{N}}^{*},n\in \mathbb{N}$, $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right)$ and $\gamma \in \mathbb{C}$ with $\mathrm{Re}\left(\gamma \right)>p$. Also let $g\in \Sigma S_{p,b}^n\left(h\right)$ with h convex in U and $h\left(0\right)=p$. If

$$\mathrm{Re}h\left(z\right)<\mathrm{Re}\left(\gamma \right),z\in U,$$

then

$$G=J_{p,\gamma }\left(g\right)\in \Sigma S_{p,b}^n\left(h\right).$$

For the next theorem, we will omit the proof because it is similar to the previous one. For the proof of Theorem 3 we use [3], Corollary 2.3.

Theorem 3. 

Let $p\in {\mathbb{N}}^{*},n\in \mathbb{N}$, $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right)$, $\gamma \in \mathbb{C}$ with $\mathrm{Re}\gamma >p$, and $g\in \Sigma K_{p,b}^n\left(h\right)$ with h convex in U. If

$$\mathrm{Re}[\gamma -h(z\left)\right]>0,z\in U,$$

then

$$J_{p,\gamma }\left(g\right)\in \Sigma K_{p,b}^n\left(q\right),$$

where q is the univalent solution of the Briot–Bouquet differential equation

$$q\left(z\right)+\frac{(p+1)z{q}^{\prime }\left(z\right)}{\gamma -q\left(z\right)}=h\left(z\right),z\in U,$$

with $q\left(0\right)=p$.

The function q is the best (p,p+1)-dominant.

Since for Theorem 3 we have $q\prec h$, we obtain the next corollary:

Corollary 4. 

Let $p\in {\mathbb{N}}^{*},n\in \mathbb{N}$, $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right)$, $\gamma \in \mathbb{C}$ with $\mathrm{Re}\gamma >p$, and $g\in \Sigma K_{p,b}^n\left(h\right)$ with h convex in U. If

$$\mathrm{Re}[\gamma -h(z\left)\right]>0,z\in U,$$

then

$$J_{p,\gamma }\left(g\right)\in \Sigma K_{p,b}^n\left(h\right).$$

Theorem 4. 

Let $p\in {\mathbb{N}}^{*},n\in \mathbb{N}$, $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right)$ and $\gamma \in \mathbb{C}$ with $\mathrm{Re}\gamma >p$. Let $h_1$, h be convex functions in U with $h_1\left(0\right)=1,h\left(0\right)=p$ such that we have $\mathrm{Re}h\left(z\right)<\mathrm{Re}\left(\gamma \right),z\in U.$ If $\phi \in \Sigma K_{p,0,b}^n\left(h\right)$ and $g\in \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;\phi ,h)$, then

$$J_{p,\gamma }\left(g\right)\in \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;J_{p,\gamma }\left(\phi \right),q),$$

where q is the univalent solution of the Briot–Bouquet differential equation

$$q\left(z\right)+\frac{(p+1)z{q}^{\prime }\left(z\right)}{\gamma -q\left(z\right)}=h\left(z\right),z\in U,$$

with $q\left(0\right)=p.$

The function q is the best (p,p+1)-dominant.

Proof. 

We have $\phi \in \Sigma K_{p,0,b}^n\left(h\right)\iff T_b^n\left(\phi \right)\in \Sigma K_{p,0}\left(h\right)$ and

$$g\in \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;\phi ,h)\iff T_b^n\left(g\right)\in \Sigma {\mathcal{C}}_{p,0}(h_1;T_b^n\left(\phi \right),h).$$

Using now [4], Theorem 2.2, we obtain that

$$J_{p,\gamma }\left(T_b^n\left(g\right)\right)\in \Sigma {\mathcal{C}}_{p,0}(h_1;J_{p,\gamma }\left(T_b^n\left(\phi \right)\right),q),$$

where q is the univalent solution of the Briot–Bouquet differential equation

$$q\left(z\right)+\frac{(p+1)z{q}^{\prime }\left(z\right)}{\gamma -q\left(z\right)}=h\left(z\right),z\in U,$$

with $q\left(0\right)=p.$

The function q is the best (p,p+1)-dominant.

We use now the fact that $J_{p,\gamma }\left(T_b^n\left(g\right)\right)=T_b^n\left(J_{p,\gamma }\left(g\right)\right)$ and $J_{p,\gamma }\left(T_b^n\left(\phi \right)\right)=T_b^n\left(J_{p,\gamma }\left(\phi \right)\right),$ therefore we obtain

$$T_b^n\left(J_{p,\gamma }\left(g\right)\right)\in \Sigma {\mathcal{C}}_{p,0}(h_1;T_b^n\left(J_{p,\gamma }\left(\phi \right)\right),q)\iff J_{p,\gamma }\left(g\right)\in \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;J_{p,\gamma }\left(\phi \right),q).$$

It is clear from Theorem 3 that we have $J_{p,\gamma }\left(\phi \right)\in \Sigma K_{p,0,b}^n\left(q\right)$, since the function $\phi \in \Sigma K_{p,0,b}^n\left(h\right)$, this meaning that the class $\Sigma {\mathcal{C}}_{p,0,b}^n(h_1;J_{p,\gamma }\left(\phi \right),q)$ is well-defined.

Hence, we find that

$$J_{p,\gamma }\left(g\right)\in \Sigma {\mathcal{C}}_{p,0.b}^n(h_1;J_{p,\gamma }\left(\phi \right),q),$$

where q is the univalent solution of the Briot–Bouquet differential equation

$$q\left(z\right)+\frac{(p+1)z{q}^{\prime }\left(z\right)}{\gamma -q\left(z\right)}=h\left(z\right),z\in U,$$

with $q\left(0\right)=p.$

The function q is the best (p,p+1)-dominant. □

It is easy to see from the proof of Theorem 4 that we also have the next result:

Corollary 5. 

Let $p\in {\mathbb{N}}^{*},n\in \mathbb{N}$, $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right)$, $\gamma \in \mathbb{C}$ with $\mathrm{Re}\gamma >p$. Let $h_1$ and h be convex functions in U with $h_1\left(0\right)=1,h\left(0\right)=p$ and $\mathrm{Re}h\left(z\right)<\mathrm{Re}\left(\gamma \right),z\in U.$

If $g\in \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;h)$, then

$$J_{p,\gamma }\left(g\right)\in \Sigma {\mathcal{C}}_{p,0}^n(h_1;q),$$

where q is the univalent solution of the Briot–Bouquet differential equation

$$q\left(z\right)+\frac{(p+1)z{q}^{\prime }\left(z\right)}{\gamma -q\left(z\right)}=h\left(z\right),z\in U,$$

with $q\left(0\right)=p.$

The function q is the best (p,p+1)-dominant.

Since we know that $q\prec h$, we obtain the next corollaries:

Corollary 6. 

Let $p\in {\mathbb{N}}^{*}n\in \mathbb{N}$, $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right)$, and $\gamma \in \mathbb{C}$ with $\mathrm{Re}\gamma >p$. Let $h_1$ and h be convex functions in U with $h_1\left(0\right)=1,h\left(0\right)=p$ and $\mathrm{Re}h\left(z\right)<\mathrm{Re}\gamma ,z\in U$. If $\phi \in \Sigma K_{p,0,b}^n\left(h\right)$ and $g\in \Sigma {\mathcal{C}}_{p,0}^n(h_1;\phi ,h),$ then

$$J_{p,\gamma }\left(g\right)\in \Sigma {\mathcal{C}}_{p,0}^n(h_1;J_{p,\gamma }\left(\phi \right),h).$$

Corollary 7. 

Let $p\in {\mathbb{N}}^{*},n\in \mathbb{N}$, $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right)$, $\gamma \in \mathbb{C}$ with $\mathrm{Re}\gamma >p$. Let $h_1,h$ be convex functions in U with $h_1\left(0\right)=1,h\left(0\right)=p$ and let $g\in \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;h)$. If $\mathrm{Re}h\left(z\right)<\mathrm{Re}\gamma ,z\in U$, then

$$J_{p,\gamma }\left(g\right)\in \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;h).$$

Using the previous Corollary for m times ($m\in {\mathbb{N}}^{*}$), we find that we have

$$J_{p,\gamma }^m\left(g\right)\in \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;h),$$

when $g\in \Sigma {\mathcal{C}}_{p,0,b}^n(h_1;h)$. The operator $J_{p,\gamma }^m$ was previously used in the paper [6].

The last result will present the class of the image of the function $f_{\lambda ,g},$ when we have $g\in \Sigma K_{p,0,b}^n\left(h\right),$ through the operator $J_{p,\gamma }$.

Proposition 5. 

Let $p\in {\mathbb{N}}^{*},n\in \mathbb{N}$, $b={\left(b_k\right)}_{k\ge 0}\in SC\left(U\right)$, $\gamma \in \mathbb{C}$ with $\mathrm{Re}\gamma >p$ and let $g\in \Sigma K_{p,0,b}^n\left(h\right),$ for $h\in \mathcal{H}\left(U\right)$ convex in U, with $h\left(0\right)=p$ and $\mathrm{Re}h\left(z\right)<\mathrm{Re}\gamma ,z\in U.$ Then,

$$J_{p,\gamma }\left(f_{\lambda ,g}\right)\in \Sigma {\mathcal{C}}_{p,0,b}^n(q_1;J_{p,\gamma }\left(g\right),q),$$

where $q_1=1-\lambda (q-p)$ and q is the univalent solution of the Briot–Bouquet differential equation

$$q\left(z\right)+\frac{(p+1)z{q}^{\prime }\left(z\right)}{\gamma -q\left(z\right)}=h\left(z\right),z\in U,$$

with $q\left(0\right)=p$.

The function q is the best (p,p+1)-dominant.

Proof. 

Since $g\in \Sigma K_{p,0,b}^n\left(h\right),$ we have from Theorem 3 that $J_{p,\gamma }\left(g\right)\in \Sigma K_{p,0,b}^n\left(q\right)$, where q is the univalent solution of the Briot–Bouquet differential equation

$$q\left(z\right)+\frac{(p+1)z{q}^{\prime }\left(z\right)}{\gamma -q\left(z\right)}=h\left(z\right),z\in U,$$

with $q\left(0\right)=p$ and the function q is the best (p,p+1)-dominant.

Using now Theorem 1 for $J_{p,\gamma }\left(g\right)\in \Sigma K_{p,0,b}^n\left(q\right)$, we find that

$$f_{\lambda ,J_{p,\gamma }\left(g\right)}\in \Sigma {\mathcal{C}}_{p,0,b}^n(q_1;g,q),$$

(5)

where $q_1=1-\lambda (q-p)$.

Since we have $f_{\lambda ,g}=(1+\lambda p)g+\lambda z{g}^{\prime }$, it is easy to see that

$$f_{\lambda ,J_{p,\gamma }\left(g\right)}=J_{p,\gamma }\left(f_{\lambda ,g}\right).$$

(6)

From (5) and (6), we obtain

$$J_{p,\gamma }\left(f_{\lambda ,g}\right)\in \Sigma {\mathcal{C}}_{p,0,b}^n(q_1;g,q),$$

where $q_1=1-\lambda (q-p)$. □

3. Conclusions

In this paper, we first introduce a new operator on the classes of meromorphic multivalent functions denoted by $T_b^n$ using the well-known convolution. This operator, for different choices of the sequence b, becomes an operator that was previously studied, so the operator $T_b^n$ is a generalization of some operators, which verify some special properties, studied before. The advantages of studying the operator $T_b^n$ is that it has a simple form and generalizes other operators, sharing also the same basic properties. Then we build new classes of meromorphic functions, using the operator $T_b^n$ and the subordination, denoted by $\Sigma S_{p,b}^n\left(h\right)$, $\Sigma K_{p,b}^n\left(h\right)$, $\Sigma {\mathcal{C}}_{p,0,b}^n(h_1;h)$. It is obvious that classes of starlike meromorphic functions, convex and close-to-convex meromorphic functions are obtained from the above-defined classes when $n=0$ and for a specific function $h.$ Some interesting preserving properties, concerning these classes, are discussed in theorems and corollaries, when we apply the well-known integral operator $J_{p,\gamma }$. To obtain our results, we also turned to the solutions of so-called Briot–Bouquet differential subordinations and to the best dominant for the considered subordination. We mention also here that a Briot–Bouquet differential subordination is a differential subordination that has the form

$$p\left(z\right)+\frac{nz{p}^{\prime }\left(z\right)}{\beta p\left(z\right)+\gamma }\prec h\left(z\right),z\in U,p,h\in \mathcal{H}\left(U\right),\mathrm{with}p\left(0\right)=h\left(0\right).$$

We make the remark that $\Sigma S_{p,b}^n\left(h\right)$-like classes may be defined using also the superordination and the preservation of such a class, following the application of the operator $J_{p,\gamma }$ can be investigated in future works.