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 Σ_p,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 Sigma_p,1−p=Σ_p.

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 E_α(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 $\alpha,\beta,\gamma \in \mathbb {C}$, ℜ(α)>0, max { 0, ℜ(c)−1} and ℜ(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)∈Σ_p,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 f∈Σ_p,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 Σ_p(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 ℜ{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 ([16–18] 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, ℜ(c)−1}, ℜ(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 ℜ{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 f∈Σ_p,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)∈Σ_p,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 Σ_p,m, Kumar and Shukla [21] defined the integral operator г_μ,p:Σ_p,m→Σ_p,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 г_μ,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 г_μ,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)∈Σ_p,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 г_μ,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 Σ_p,m. Also, let g(z)∈Σ_p,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)=z^p+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 $\mathbb {U}$, ℜ{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)∈Σ_p,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)∈Σ_p,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.