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Differential subordination applications to a class of meromorphic multivalent functions associated with Mittag-Leffler function

Abstract

In this paper, using the principal of differential subordination, we obtain some properties of certain class of p-valent meromorphic functions, which are defined by Mittag-Leffler function.

Introduction

Denote by Σp,m the class of analytic meromorphic multivalent functions of the form:

$$ f(z)=\frac{1}{z^{p}}+\sum\limits_{k=m}^{\infty }a_{k}z^{k}\ \left(p\in \mathbb{N}=\{1,2,\ldots\};m>-p\right), $$
(1)

where \(\mathbb {U}^{\ast }=\{z\in \mathbb {C} \) and \(0<|z|<1\}=\mathbb {U}\backslash \{0\}.\ \)We note that Sigmap,1−p=Σp.

For two functions f(z) and g(z), analytic in \(\mathbb {U}, f(z)\) is subordinate to g(z)(f(z)g(z)) in \(\mathbb {U}\), if there exists a function ω(z), analytic in \(\mathbb {U}\) with ω(0)=0 and \(\left \vert \omega (z)\right \vert <1, f(z)=g(\omega (z)) (z\in \mathbb {U})\) and if g(z) is univalent in \(\mathbb {U}\), then (see for details [1] and also [2])

$$f(z)\prec g(z)\Longleftrightarrow f(0)=g(0)\text{ and }f(\mathbb{U})\subset g(\mathbb{U}). $$

The Hadamard product of f(z) and g(z) given by

$$g(z)=\frac{1}{z^{p}}+\sum\limits_{k=m}^{\infty }b_{k}z^{k}\ $$

is defined by

$$ \left(f\ast g\right) \left(z\right) =\frac{1}{z^{p}}+\sum\limits_{k=m}^{ \infty }a_{k}b_{k}z^{k}=\left(g\ast f\right) \left(z\right). $$
(2)

The Mittag-Leffler function Eα(z) (z\(\in \mathbb {U}^{\ast }\)) ([3] and [4]) see also ([5, 6] and [7]) is defined by

$$E_{\alpha }(z)=\sum\limits_{k=0}^{\infty }\frac{1}{\Gamma (k\alpha +1)}z^{k},\alpha \in \mathbb{C},\Re(\alpha)>0. $$

For \(\alpha,\beta,\gamma \in \mathbb {C}\), (α)>0, max { 0, (c)−1} and (c)>0, Srivastava and Tomovski [8] generalized Mittag-Leffler function by the function

$$ E_{\alpha,\beta }^{\gamma,c}(z)=\sum\limits_{k=0}^{\infty }\frac{(\gamma)_{kc}}{ \Gamma (k\alpha +\beta)k!}z^{nk} $$
(3)

and proved that it is an entire function in the complex z-plane, where

$$(\gamma)_{\theta }=\frac{\Gamma (\gamma +\theta)}{\Gamma (\gamma)}\left\{ \begin{array}{c} 1,\ \ \ \ \ \ \ \ \theta =0 \\ \gamma (\gamma +1)\ldots(\gamma +\theta -1),\ \theta \neq 0 \end{array} \right.. $$

Mostafa and Aouf [9] (see also [10]) used the function \(E_{\alpha,\beta }^{\gamma,c}(z)\) and defined the meromorphic function

$$\begin{array}{@{}rcl@{}} \mathcal{M}_{p,\alpha,\beta }^{\gamma,c}(z) &=&z^{-p}\Gamma (\beta)E_{\alpha,\beta }^{\gamma,c}(z) \\ &=&z^{-p}+\sum\limits_{k=m}^{\infty }\frac{\Gamma (\beta)\Gamma \lbrack \gamma +(k+p)c]}{\Gamma (\gamma)\Gamma \lbrack \beta +(k+p)\alpha ](k+p)!} z^{k}, \\ &&\left(\Re \left(\alpha \right) =0\ \text{when\ }\Re \left(c\right) =1\ \text{with}\ \beta \neq 0\right), \end{array} $$
(4)

and for f(z)Σp,m, they defined the operator

$$\begin{array}{@{}rcl@{}} \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z) &=&\mathcal{M}_{p,\alpha,\beta }^{\gamma,c}(z)\ast f(z) \\ &=&z^{-p}+\sum\limits_{k=m}^{\infty }\frac{\Gamma (\beta)\Gamma \lbrack \gamma +(k+p)c]}{\Gamma (\gamma)\Gamma \lbrack \beta +(k+p)\alpha ](k+p)!} a_{k}z^{k}. \end{array} $$
(5)

From (5) it is easy to have

$$ cz(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z))^{\prime }=\gamma \mathcal{ H}_{p,\alpha,\beta }^{\gamma +1,c}f(z)-(\gamma +pc)\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z) \left(c>0\right) $$
(6)

and

$$ \alpha z\left(\mathcal{H}_{p,\alpha,\beta +1}^{\gamma,c}f\left(z\right) \right)^{^{\prime }}=\beta \ \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f\left(z\right) -\left(p\alpha +\beta \right) \ \mathcal{H}_{p,\alpha,\beta +1}^{\gamma,c}f\left(z\right),\alpha \neq 0. $$
(7)

We note that:

(i) \(\mathcal {H}_{p,0,\beta }^{1,1}f(z)=f(z);\)

(ii) \(\mathcal {H}_{p,0,\beta }^{2,1}f(z)=\left (p+1\right) f(z)+zf^{^{\prime }}(z);\)

(iii) \(\mathcal {H}_{1,0,\beta }^{2,1}f(z)=2f(z)+zf^{^{\prime }}(z).\)

Using the operator \(\mathcal {H}_{p,\alpha,\beta }^{\gamma,c}f(z)\), we have the following definition.

Definition 1.

For fixed A and B (−1≤B<A≤1), we say that a function fΣp,m is in the class \(\Sigma _{p,m}^{\gamma,c}\left (\alpha,\beta ;A,B\right)\) if it satisfies

$$ -\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{p}\prec \frac{1+Az}{1+Bz}. $$
(8)

In view of the definition of differential subordination, (8) is equivalent to

$$ \left\vert \frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}+p}{Bz^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}+pA}\right\vert <1. $$
(9)

We note that:

(i)

$$\begin{array}{@{}rcl@{}} \Sigma_{p,1}^{1,1}\left(0,1;A,B\right) &=&\Sigma_{p}\left(A,B\right) \left(-1\leq B<A\leq 1;\mathbb{U}^{\ast }\right) \\ &=&\left\{ f\in \Sigma_{p}:-\frac{z^{p+1}f^{^{\prime }}(z)}{p}\prec \frac{ 1+Az}{1+Bz}\right\}, \end{array} $$

the class Σp(A,B) was introduced and studied by Mogra [11].

(ii)

$$\begin{array}{@{}rcl@{}} \Sigma_{p,m}^{\gamma,c}\left(\alpha,\beta ;1-\frac{2\eta }{p},-1\right) &=&\Sigma_{p,m}^{\gamma,c}\left(\alpha,\beta,\eta \right) \left(0\leq \eta < p\right) \\ &=&\left\{ f\in \Sigma_{p,m}:\Re\{-z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}\}>\eta \right\}. \\ && \end{array} $$
(10)

Preliminary results

The following lemmas will be required in our investigation.

Lemma 1

[12]. Let h be a convex (univalent) function in \(\mathbb {U}\) with h(0)=1. Also let

$$ \phi (z)=1+d_{p+m}z^{p+m}+d_{p+m+1}z^{p+m+1}+\ldots\text{,} $$
(11)

be analytic in \(\mathbb {U}.\) If

$$ \phi (z)+\frac{z\phi^{^{\prime }}(z)}{\tau }\prec h(z)\ \left(\Re \left(\tau \right) \geq 0;\ \tau \neq 0;\ z\in \mathbb{U}\right), $$
(12)

then

$$ \phi (z)\prec \Psi (z)=\frac{\tau }{p+m}z^{-\frac{\tau }{p+m} }\int\limits_{0}^{z}t^{\frac{\tau }{p+m}-1}h(t)dt. $$
(13)

Lemma 2

[13]. Let μbe a positive measure on the unit interval [0,1]. Let g(z,t)be a complex valued function defined on \(\mathbb {U}\times \left [ 0,1\right ] \) such that g(.,t) is analytic in \(\mathbb {U}\) for each t[0,1] and such that g(z,.) is μ integrable on [0,1] for all \(z\in \mathbb {U}.\) In addition, suppose that {g(z,t)}>0,g(−r,t)is real and

$$\Re \left\{ \frac{1}{g(z,t)}\right\} \geq \frac{1}{g(-r,t)} \left(\left\vert z\right\vert \leq r<1;t\in \left[ 0,1\right] \right). $$

If the function G is defined by

$$G(z)=\int\limits_{0}^{1}g(z,t)d\mu (t), $$

then

$$\Re \left\{ \frac{1}{G(z)}\right\} \geq \frac{1}{G(-r)} \left(\left\vert z\right\vert \leq r<1\right). $$

Each of the identities (asserted by Lemma 2) is fairly well known (cf., e.g., [[8], ch. 14]).

Lemma 3

[14]. For real or complex numbers a, b, and c (c≠0,−1,−2,…)

$$ \int\limits_{0}^{1}t^{b-1}\left(1-t\right)^{c-b-1}\left(1-tz\right)^{-a}dt=\frac{\Gamma \left(b\right) \Gamma \left(c-b\right) }{\Gamma \left(c\right) }_{2}F_{1}\left(a,b;c;z\right) \ \left(\Re \left(c\right) <\Re \left(b\right) >0\right) ; $$
$$ _{2}F_{1}\left(a,b;c;z\right) =\left(1-z\right)_{\text{ \ \ \ \ } 2}^{-a}F_{1}\left(a,c-b;c;\frac{z}{z-1}\right) \left(z\neq 1\right) $$
(14)

and

$$ _{2}F_{1}\left(a,b;c;z\right) =_{2}F_{1}\left(b,a;c;z\right). $$
(15)

Lemma 4

[15]. Let Φbe analytic in \(\mathbb {U}\) with

$$\Phi (0)=1\text{ and }\Re \{\Phi (z)\}>\frac{1}{2}. $$

Then, for any function F, analytic in \(\mathbb {U}, \left (\Phi \ast F\right) \left (\mathbb {U}\right) \) is contained in the convex hull of \( F\left (\mathbb {U}\right).\)

We used the technique used by ([1618] and [19]).

Main inclusion relationships

Unless otherwise mentioned, we assume throughout this paper that \(-1\leq B<A\leq 1,\alpha,\beta,\gamma \in \mathbb {C}, \Re (\alpha)>0\), max { 0, (c)−1}, (c)>0,δ>0,f(z) given by (1) and \(z\in \mathbb {U}^{\ast }.\)

Theorem 1

Let γ≠0 and f(z) satisfy:

$$ \mathcal{-}\frac{\left(1-\delta \right) z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}+\delta z^{p+1}\left(\mathcal{H }_{p,\alpha,\beta }^{\gamma +1,c}f(z)\right)^{^{\prime }}}{p}\prec \frac{ 1+Az}{1+Bz}, $$
(16)

then

$$ -\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{p}\prec \Psi (z)\prec \frac{1+Az}{1+Bz}, $$
(17)

where

$$ \Psi (z)=\left\{ \begin{array}{cc} \frac{A}{B}+\left(1-\frac{A}{B}\right) \left(1+Bz\right)_{\text{ \ \ \ } 2}^{-1}F_{1}\left(1,1;\frac{\gamma }{\delta c\left(p+m\right) }+1;\frac{Bz }{1+Bz}\right), & B\neq 0 \\ 1+\frac{\gamma }{\gamma +\delta c\left(p+m\right) }Az, & B=0. \end{array} \right. $$
(18)

is the best dominant of (17). Furthermore,

$$ \Re \left\{ -\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{p}\right\} >\rho \ \left(0\leq \rho <1\right), $$
(19)

where

$$ \rho =\left\{ \begin{array}{cc} \frac{A}{B}+\left(1-\frac{A}{B}\right) \left(1-B\right)_{\text{ \ \ \ } 2}^{-1}F_{1}\left(1,1;\frac{\gamma }{\delta c\left(p+m\right) }+1;\frac{B}{ B-1}\right), & B\neq 0, \\ 1-\frac{\gamma }{\gamma +\delta c\left(p+m\right) }A, & B=0. \end{array} \right. $$
(20)

The result is the best possible.

Proof

Let

$$ \phi (z)=-\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{p}, $$
(21)

where ϕ is given by (11). Differentiating (21) and using (6), we get

$$\mathcal{-}\frac{\left(1-\delta \right) z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}+\delta z^{p+1}\left(\mathcal{H }_{p,\alpha,\beta }^{\gamma +1,c}f(z)\right)^{^{\prime }}}{p}=\phi (z)+ \frac{\delta cz\phi^{^{\prime }}(z)}{\gamma }\prec \frac{1+Az}{1+Bz}. $$

Now, by using Lemma 1 for \(\tau =\frac {\gamma }{\delta c},\) we get

$$\begin{array}{@{}rcl@{}} -\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{p} &\prec &\Psi (z)=\frac{\gamma }{\delta c\left(p+m\right) } z^{-\frac{\gamma }{\delta c\left(p+m\right) }}\int\limits_{0}^{z}t^{\frac{ \gamma }{\delta c\left(p+m\right) }-1}\left(\frac{1+At}{1+Bt}\right) dt \\ &=&\left\{ \begin{array}{cc} \frac{A}{B}+\left(1-\frac{A}{B}\right) \left(1+Bz\right)_{\text{ \ \ \ } 2}^{-1}F_{1}\left(1,1;\tfrac{\gamma }{\delta c\left(p+m\right) }+1;\tfrac{ Bz}{1+Bz}\right), & B\neq 0 \\ 1+\frac{\gamma }{\gamma +\delta c\left(p+m\right) }Az, & B=0. \end{array} \right. \end{array} $$

This proves (17) of Theorem 1. In order to prove (20), we need to show that

$$ \inf_{\left\vert z\right\vert <1}\left\{ \Re(\Psi(z))\right\} =\Psi (-1). $$
(22)

We have

$$\Re\left\{ \frac{1+Az}{1+Bz}\right\} \geq \frac{1-Ar}{1-Br} \left(\left\vert z\right\vert \leq r<1\right). $$

Putting

$$G\left(z,\zeta \right) =\frac{1+A\zeta z}{1+B\zeta z}\text{ and }dv(\zeta)= \frac{\gamma }{\delta c\left(p+m\right) }\zeta^{\frac{\gamma }{\delta c\left(p+m\right) }-1}d\zeta \left(0\leq \zeta \leq 1\right), $$

which is a positive measure on [0,1], we obtain

$$\Psi (z)=\int\limits_{0}^{1}G\left(z,\zeta \right) dv(\zeta). $$

Then

$$\Re(\Psi (z))\geq \int\limits_{0}^{1}\frac{1-A\zeta r}{1-B\zeta r}dv(\zeta)=\Psi (-r) \left(\left\vert z\right\vert \leq r<1\right). $$

Assuming r→1 in the above inequality, we obtain (22). The result in (19) is the best possible and Ψ is the best dominant of (17). This completes the proof of Theorem 1. □

Theorem 2

Let \(f(z) \in \Sigma _{p,m}^{\gamma,c}\left (\alpha,\beta,\eta \right) \left (0\leq \eta < p\right),\) then

$$ \Re \left\{ -z^{p+1}\left[ \left(1-\delta \right) \left(\mathcal{H} _{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}+\delta \left(\mathcal{H}_{p,\alpha,\beta }^{\gamma +1,c}f(z)\right)^{^{\prime }}\right] \right\} >\eta \left(\left\vert z\right\vert <R\right), $$
(23)

where

$$ R=\left\{ \frac{\sqrt{c^{2}\delta^{2}\left(p+m\right)^{2}+\gamma^{2}} -c\delta \left(p+m\right) }{\gamma }\right\}^{\frac{1}{p+m}}. $$
(24)

The result is the best possible.

Proof

Since \(f(z)\in \Sigma _{p,m}^{\gamma,c}\left (\alpha,\beta,\eta \right),\) let

$$ -z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}=\eta +\left(p-\eta \right) u(z), $$
(25)

where u(z) in the form (11) and {u(z)}>0. Differentiating (25) and using (6), we get

$$ -\frac{z^{p+1}\left[ \left(1-\delta \right) \left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}+\delta \left(\mathcal{H} _{p,\alpha,\beta }^{\gamma +1,c}f(z)\right)^{^{\prime }}\right] +\eta }{ p-\eta }=u(z)+\frac{c\delta zu^{^{\prime }}(z)}{\gamma }. $$
(26)

Applying the following estimate [20],

$$\frac{\left\vert zu^{^{\prime }}(z)\right\vert }{\Re \left\{ u(z)\right\} } \leq \frac{2(p+m)r^{p+m}}{1-r^{2(p+m)}} \left(\left\vert z\right\vert =r<1\right) ; $$

in (26), we get

$$\begin{array}{@{}rcl@{}} &&\Re \left\{ -\frac{z^{p+1}\left[ \left(1-\delta \right) \left(\mathcal{H} _{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}+\delta \left(\mathcal{H}_{p,\alpha,\beta }^{\gamma +1,c}f(z)\right)^{^{\prime }}\right] +\eta }{p-\eta }\right\} \\ &\geq &\Re \left(u(z)\right) \left(1-\frac{2c\delta (p+m)r^{p+m}}{\gamma \left(1-r^{2(p+m)}\right) }\right). \end{array} $$
(27)

It is easily seen that the right-hand side of (27) is positive, if r<R, where R is given by (24). In order to show that the bound R is the best possible, we consider the function fΣp,m defined by

$$-z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}=\eta +\left(p-\eta \right) \frac{1+z^{p+m}}{1-z^{p+m}}. $$

Noting that

$$\begin{array}{@{}rcl@{}} &&-\frac{z^{p+1}\left[ \left(1-\delta \right) \left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}+\delta \left(\mathcal{H} _{p,\alpha,\beta }^{\gamma +1,c}f(z)\right)^{^{\prime }}\right] +\eta }{ p-\eta} \\ &=&\frac{\gamma \left(1-z^{2\left(p+m\right) }\right) +2c\delta (p+m)z^{p+m}}{\gamma \left(1-z^{p+m}\right)^{2}}=0, \end{array} $$

for

$$z=R\text{ }\Re \left(\frac{i\pi }{p+m}\right). $$

This completes the proof of Theorem 2. □

Putting δ=1 in Theorem 2, we obtain the following result.

Corollary 1

If \(f(z) \in \Sigma _{p,m}^{\gamma,c}\left (\alpha,\beta,\eta \right) \left (0\leq \eta < p\right),\) then \(f(z) \in \Sigma _{p,m}^{\gamma +1,c}\left (\alpha,\beta,\eta \right) \) for |z|<R, where

$$R^{\ast }=\left\{ \frac{\sqrt{c^{2}\left(p+m\right)^{2}+\gamma^{2}} -c\left(p+m\right) }{\gamma }\right\}^{\frac{1}{p+m}}. $$

The result is the best possible.

Theorem 3

If the function f(z)Σp,m satisfies

$$z^{p}\left[ \left(1-\delta \right) \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)+\delta \mathcal{H}_{p,\alpha,\beta }^{\gamma +1,c}f(z)\right] \prec \frac{1+Az}{1+Bz}, $$

then

$$z^{p}\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\prec \Psi_{1}(z)\prec \frac{1+Az}{1+Bz} $$

and

$$\Re \left\{ z^{p}\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right\} >\rho, $$

where Ψ1(z) is in the form (18) and ρ given by (20). The result is the best possible.

Proof

The proof follows by taking the same lines as in the proof of Theorem 1 and taking \(\phi (z)=z^{p}\mathcal {H}_{p,\alpha,\beta }^{\gamma,c}f(z)\) in (21). □

For the function f(z) in the class Σp,m, Kumar and Shukla [21] defined the integral operator гμ,p:Σp,mΣp,m as follows:

$$\begin{array}{@{}rcl@{}} \digamma_{\mu,p}(f)(z) &=&\frac{\mu }{z^{\mu +p}}\int\limits_{0}^{z}t^{\mu +p-1}f(t)dt \\ &=&z^{-p}+\sum\limits_{k=m}^{\infty }\frac{\mu }{k+p+\mu }a_{k}z^{k}\ \left(\mu >0\right). \end{array} $$
(28)

From (28), we get

$$ z\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z)\right)^{^{\prime }}=\mu \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)-\left(\mu +p\right) \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z). $$
(29)

Theorem 4

Let the function f(z)given by (1) be in the class \( \Sigma _{p,m}^{\gamma,c}\left (\alpha,\beta ;A,B\right) \) and гμ,p(f)(z) defined by (28). Then

$$ -\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z)\right)^{^{\prime }}}{p}\prec \Theta (z)\prec \frac{1+Az}{ 1+Bz}, $$
(30)

where

$$ \Theta (z)=\left\{ \begin{array}{cc} \frac{A}{B}+\left(1-\frac{A}{B}\right) \left(1+Bz\right)_{\text{ \ \ \ } 2}^{-1}F_{1}\left(1,1;\frac{\mu }{\left(p+m\right) }+1;\frac{Bz}{1+Bz} \right), & B\neq 0 \\ 1+\frac{\mu }{\mu +p+m}Az, & B=0. \end{array} \right. $$
(31)

is the best dominant of (31). Furthermore,

$$ \Re\left\{ -\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z)\right)^{^{\prime }}}{p}\right\} >\sigma \text{ } \left(0\leq \sigma <1\right), $$
(32)

where

$$ \sigma =\left\{ \begin{array}{cc} \frac{A}{B}+\left(1-\frac{A}{B}\right) \left(1-B\right)_{\text{ \ \ \ } 2}^{-1}F_{1}\left(1,1;\frac{\mu }{\left(p+m\right) }+1;\frac{B}{B-1} \right), & B\neq 0 \\ 1-\frac{\mu }{\mu +p+m}A, & B=0. \end{array} \right. $$
(33)

The result is the best possible.

Proof

Let

$$ L(z)=-\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z)\right)^{^{\prime }}}{p}, $$
(34)

where L in the form (11). Differentiating (34) and using (29), we get

$$-\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{p}=L(z)+\frac{z}{\mu }L^{^{\prime }}(z)\prec \frac{1+Az}{1+Bz}. $$

Now the remaining part of Theorem 4 follows by using the technique used in proving Theorem 1. □

Theorem 5

Let the function гμ,p(f)(z) defined by (28) satisfy:

$$ z^{p}\left[ \left(1-\delta \right) \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z)+\delta \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right] \prec \frac{1+Az}{1+Bz}, $$
(35)

then

$$ \Re\left\{ z^{p}\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z)\right\} >\theta, $$
(36)

where

$$\theta =\left\{ \begin{array}{cc} \frac{A}{B}+\left(1-\frac{A}{B}\right) \left(1-B\right)_{\text{ \ \ \ } 2}^{-1}F_{1}\left(1,1;\frac{\mu }{\delta \left(p+m\right) }+1;\frac{B}{B-1} \right), & B\neq 0 \\ 1-\frac{\mu }{\mu +\delta \left(p+m\right) }A, & B=0. \end{array} \right. $$

The result is the best possible.

Proof

Let

$$ K(z)=z^{p}\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z), $$
(37)

where K in the form (11). Differentiating (37) and using (29) and (35), we get

$$K(z)+\frac{\delta z}{\mu }K^{^{\prime }}(z)\prec \frac{1+Az}{1+Bz}. $$

Now the remaining part of Theorem 5 follows by using the technique used in proving Theorem 1. □

Theorem 6

Let the function f(z)Σp,m satisfy:

$$\mathcal{-}\frac{z^{p+1}\left[ \left(1-\delta \right) \left(\mathcal{H} _{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z)\right)^{^{\prime }}+\delta \left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}\right] }{p}\prec \frac{1+Az}{1+Bz}, $$

then

$$\Re\left\{ -\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\digamma_{\mu,p}(f)(z)\right)^{^{\prime }}}{p}\right\} >\theta, $$

where гμ,p(f)(z) is given by (28) and θ is given as in Theorem 5. The result is the best possible.

Proof

The proof follows by taking the same lines as in Theorem 5. □

Theorem 7

Let f(z)be in the class Σp,m. Also, let g(z)Σp,m satisfy:

$$\Re\left\{ z^{p}\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}g(z)\right\} >0. $$

If

$$\left\vert \frac{\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)}{\mathcal{H} _{p,\alpha,\beta }^{\gamma,c}g(z)}-1\right\vert <1, $$

then

$$ \Re \left\{ -\frac{z\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)} \right\} >0\text{ }\left(\left\vert z\right\vert <R_{0}\right), $$
(38)

where

$$ R_{0}=\frac{\sqrt{9(p+m)^{2}+4p(2p+m)}-3(p+m)}{2(2p+m)}. $$
(39)

Proof

Let

$$ \phi (z)=\frac{\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)}{\mathcal{H} _{p,\alpha,\beta }^{\gamma,c}g(z)}-1=e_{p+m}z^{p+m}+e_{p+m+1}z^{p+m+1}+\ldots \text{,} $$
(40)

we note that ϕ is analytic in \(\mathbb {U}\), with ϕ(0)=0 and |ϕ(z)|≤|z|p+m. Then, by applying the familiar Schwarz Lemma [22], we have ϕ(z)=zp+mΨ(z) is analytic in \(\mathbb {U}\) and \(\left \vert \Psi (z)\right \vert \leq 1 \left (z\in \mathbb {U}\right).\) Therefore, (40) leads to

$$ \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)=\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}g(z)\left(z^{p+m}\Psi (z)+1\right). $$
(41)

Differentiating (41), we obtain

$$ \frac{z\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)}=\frac{z\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}g(z)\right)^{^{\prime }}}{ \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}g(z)}+\frac{z^{p+m}\left[ \left(p+m\right) \Psi (z)+z\Psi^{^{\prime }}(z)\right] }{1+z^{p+m}\Psi (z)}. $$
(42)

Letting \(\chi (z)=z^{p}\mathcal {H}_{p,\alpha,\beta }^{\gamma,c}g(z),\) we see that the function χ is of the form (11) and is analytic in \( \mathbb {U}\), {χ(z)}>0 and

$$\frac{z\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}g(z)\right)^{^{\prime }}}{\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}g(z)}=\frac{z\chi^{^{\prime }}(z)}{\chi (z)}-p, $$

so, we find from (42) that

$$ \Re \left\{ -\tfrac{z\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)} \right\} \geq p-\left\vert \frac{z\chi^{^{\prime }}(z)}{\chi (z)} \right\vert -\left\vert \tfrac{z^{p+m}\left[ \left(p+m\right) \Psi (z)+z\Psi^{^{\prime }}(z)\right] }{1+z^{p+m}\Psi (z)}\right\vert. $$
(43)

Using the following known estimates [23] (see also [20]),

$$\left\vert \frac{\chi^{^{\prime }}(z)}{\chi (z)}\right\vert \leq \frac{ 2(p+m)r^{p+m-1}}{1-r^{2(p+m)}}\text{ and }\left\vert \frac{\left(p+m\right) \Psi (z)+z\Psi^{^{\prime }}(z)}{1+z^{p+m}\Psi (z)}\right\vert \leq \frac{p+m }{1-r^{p+m}}\left(\left\vert z\right\vert =r<1\right), $$

in (43), we have

$$\Re \left\{ -\frac{z\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)} \right\} \geq \frac{p-3(p+m)r^{p+m}-(2p+m)r^{2(p+m)}}{1-r^{2(p+m)}}, $$

which is certainly positive, provided that r<R0,R0 given by (39). □

Theorem 8

Let the function f(z)Σp,m satisfy:

$$z^{p}\left[ \left(1-\delta \right) \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)+\delta \mathcal{H}_{p,\alpha,\beta }^{\gamma +1,c}f(z)\right] \prec \frac{1+Az}{1+Bz}, $$

then

$$\Re\left\{ \left(z^{p}\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{\frac{1}{q}}\right\} >\epsilon^{\frac{1}{q}}\text{ }\left(q\in \mathbb{N} \right), $$

where ε in the form (20). The result is the best possible.

Proof

Let

$$ \phi (z)=z^{p}\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z), $$
(44)

where ϕ in the form (11). Differentiating (44) and using (6), we have

$$z^{p}\left[ \left(1-\delta \right) \mathcal{H}_{p,\alpha,\beta }^{\gamma,c}f(z)+\delta \mathcal{H}_{p,\alpha,\beta }^{\gamma +1,c}f(z)\right] =\phi (z)+\frac{\delta cz\phi^{^{\prime }}(z)}{\gamma }\prec \frac{1+Az}{1+Bz}. $$

Now the remaining part of Theorem 8 follows by using the technique used in proving Theorem 1, and using the inequality:

$$\Re (w^{\frac{1}{q}})\geq \left(\Re (w)\right)^{\frac{1}{q}}\text{ }\left(\Re (w)>0;q\in \mathbb{N}\right), $$

we have the result asserted by Theorem 8. □

Theorem 9

Let the function \(f(z)\in \Sigma _{p,m}^{\gamma,c}\left (\alpha, \beta ;A,B\right)\) and let g(z)Σp,m satisfy:

$$\Re\left(z^{p}g(z)\right)>\frac{1}{2}. $$

Then,

$$\left(f\ast g\right) (z)\in \Sigma_{p,m}^{\gamma,c}\left(\alpha,\beta ;A,B\right). $$

Proof

We have

$$-\frac{z^{p+1}\left(\mathcal{H}_{p,\alpha,\beta }^{\gamma,c}\left(f\ast g\right) (z)\right)^{^{\prime }}}{p}=-\frac{z^{p+1}\left(\mathcal{H} _{p,\alpha,\beta }^{\gamma,c}f(z)\right)^{^{\prime }}}{p}\ast z^{p}g(z). $$

Since

$$\Re \left(z^{p}g(z)\right) >\frac{1}{2} $$

and \(\frac {1+Az}{1+Bz}\) is convex in \(\mathbb {U}\), it follows from (8) and Lemma 4 that \(\left (f\ast g\right) (z)\in \Sigma _{p,m}^{\gamma,c}\left (\alpha,\beta ;A,B\right),\) which completes the proof of Theorem 9. □

Remark 1

For different value of γ, c, α, β, and p in the above results, we obtain results corresponding to the functions given in the introduction.

Availability of data and materials

All data generated or analyzed during this study are included in this published article.

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Mostafa, A.O., El-hawsh, G.M. Differential subordination applications to a class of meromorphic multivalent functions associated with Mittag-Leffler function. J Egypt Math Soc 27, 31 (2019). https://doi.org/10.1186/s42787-019-0028-7

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Keywords

  • Analytic function
  • Convex function
  • Meromorphic multivalent functions
  • Subordination and Mittag-Leffler function

mathematics subject classification

  • 30C45
  • 30C50
  • 30C55