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Subordination and inclusion theorems for higher order derivatives of a generalized fractional differintegral operator

Abstract

The main object of this paper is to investigate some subordination results of certain subclasses of multivalent analytic functions which are defined by a generalized fractional differintegral operator. Inclusion relations for functions in the class \(\mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta ;A,B\right) \ \) and the images of these functions by the generalized Bernardi-Libera-Livingston integral operator are also considered.

Introduction

Denote the class consisting of analytic and multivalent functions in the open unit disc \(\mathbb {U}=\{z\in \mathbb {C}:\left \vert z\right \vert <1\}\) of the form:

$$ f(z)=z^{p}+\sum_{n=1}^{\infty}a_{p+n}z^{p+n}\;(p\in\mathbb{N}=\{1,2,...\}), $$
(1)

by \(\mathcal {A}(p).\) We note that \(\mathcal {A}(1)=\mathcal {A}.\)

Consider the first-order differential subordination

$$H(\varphi(z),z\varphi^{\prime}(z))\prec h(z), $$

where the symbol stands for subordination of two analytic functions in \(\mathbb {U\ }\)(see [1, 2]). A univalent function q is called dominant, if φ(z)q(z) for all analytic functions φ that satisfy this differential subordination. A dominant \(\widetilde {q}\) is called the best dominant, if \(\widetilde {q}(z)\prec q(z)\) for all dominant q. For \(f\in \mathcal {A}(p)\), the qth order derivative of f(z) could be written as

$$\begin{array}{*{20}l} {}f^{(q)}(z) & =\delta(p,q)z^{p-q}+\sum_{n=1}^{\infty}\delta(p+n,q)a_{p+n} z^{p+n-q},\;z\in\mathbb{U} \left(p>q,\;q\in\mathbb{N}_{0}:=\mathbb{N}\cup\{0\}\right), \end{array} $$
(2)

where

$$\delta(p,q)=\frac{p!}{(p-q)!}:=\left\{ \begin{array} [c]{lll} p(p-1)\dots(p-q+1), & \text{if} & q\neq0,\\ 1, & \text{if} & q=0. \end{array} \right. $$

Let

$$ {\!~\!}_{p}F_{q}\left(a_{1},...,a_{p};b_{1},...,b_{q};z\right) =\sum\limits_{n=0}^{\infty}\frac{(a_{1})_{n}...(a_{p})_{n}}{(b_{1})_{n}...(b_{q})_{n}(1)_{n}}z^{n}, $$
(3)

be the well-known generalized hypergeometric function for complex parameters \(a_{1},...,a_{q}\,,\ b_{1},...,b_{s}\ \ (b_{j}\notin \mathbb {Z}_{0}^{-}=\{0,-1,-2,...\};\ j=1,2,...,s)\ \)and (λ)ν is the Pochhammer symbol defined by

$$(\lambda)_{\upsilon}:=\left\{ \begin{array} [c]{lll} 1 & & \text{if}\ \upsilon=0,\\ \lambda(\lambda+1)(\lambda+2)...(\lambda+\upsilon-1) & & \text{if } \upsilon\in \mathbb{N}. \end{array} \right. $$

In addition, if we put p=2, q=1, a1=a, a2=b, b1=c in (3), we get the (Gaussian) hypergeometric function 2F1(a,b;c;z)(c≠0,−1,−2,…) which satisfies (see [3])

$$ {}\int_{0}^{1}t^{b-1}(1-t)^{c-b-1}(1-tz)^{-a}dt=\tfrac{\Gamma(b)\Gamma (c-b)}{\Gamma(c)}\;_{2}F_{1}(a,b;c;z)\;(\mathfrak{R}(c)>\mathfrak{R}(b)>0); $$
(4)
$$ {\!~\!}_{2}F_{1}(a,b;c;z)={(1-z)^{-a}}_{\,2}F_{1}\left(a,c-b;c;\frac{z}{z-1}\right) ; $$
(5)

and

$$ {\!~\!}_{2}F_{1}(a,b;c;z){=}_{2}F_{1}(b,a;c;z). $$
(6)

We will recall some definitions which will be used in our paper.

Definition 1

[412]. Assume that 0≤λ<1 and \(\mu,\eta \in \mathbb {R}\). Then, in terms of 2F1, the generalized fractional derivative operator for \(f\in \mathcal {A}(p)\) is defined by

$${}J_{0,z}^{\lambda,\mu,\eta,p}f(z):=\frac{d}{dz}\left[ \frac{z^{\lambda-\mu} }{\Gamma(1-\lambda)}\int_{0}^{z}(z-\zeta)^{-\lambda}f(\zeta)_{2}F_{1}\left(\mu-\lambda,1-\eta;1-\lambda;1-\frac{\zeta}{z}\right) d\zeta\right], $$

where f is an analytic function in a simply-connected region of the complex z-plane containing the origin with the order f(z)=O(|z|ε), z→0 when ε> max{0,μη}−1 and the multiplicity of (zζ)λ is removed by requiring log(zζ)to be real when zζ>0.

Remark 1

We note that

(i) \(J_{0,z}^{\lambda,\mu,\eta,p}\left \{ z^{p+n}\right \} =\frac {\Gamma (p+n+1)\Gamma (p+n+1-\mu +\eta)}{\Gamma (p+n+1-\mu)\Gamma (p+n+1-\lambda +\eta)}z^{p+n-\mu }\ \left (n\geq 1\right),\)

(ii) \(J_{0,z}^{\lambda,\lambda,\eta,p}f(z)=D_{z}^{\lambda }f(z)\ \)(see [13]).

Goyal and Prajapat [14] (see also [412]) defined the operator \(\mathcal {M}_{0,z}^{\lambda,\mu,\eta,p}:\mathcal {A}(p)\rightarrow \mathcal {A}(p) \left (0\leq \lambda <1,\;\mu < p+1,\;\eta >\max \{\lambda,\mu \}-p-1\right),\ \)by

$$\begin{array}{*{20}l} {}\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z) & :=\frac{\Gamma(p+1-\mu)\Gamma(p+1-\lambda+\eta)}{\Gamma(p+1)\Gamma(p+1-\mu+\eta)}z^{\mu} J_{0,z}^{\lambda,\mu,\eta,p}f(z)\\ & =z^{p}+\sum\limits_{n=1}^{\infty}\frac{(p+1)_{n}(p+1-\mu+\eta)_{n} }{(p+1-\mu)_{n}(p+1-\lambda+\eta)_{n}}a_{p+n}z^{p+n}\\ & ={z^{p}}_{3}F_{2}\left(1,p+1,p+1-\mu+\eta;p+1-\mu,p+1-\lambda +\eta;z\right) \ast f(z),\\ & \end{array} $$
(7)

where the symbol stands for convolution of two power series and \(f\in \mathcal {A}(p)\). It is easy to check that

$$ z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime} =(p-\mu)\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)+\mu\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z). $$
(8)

In this paper, we define the higher order derivative of \(\mathcal {M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\ \)as follows:

$$\begin{array}{*{20}l} {}\left(\!\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\!\right)^{\left(q\right) }\!\,=\,\delta\left(p,q\right) z^{p-q}\,+\,\sum\limits_{n=1}^{\infty}\!\frac{(p+1)_{n}(p+1-\mu+\eta)_{n}}{(p+1-\mu)_{n}(p+1-\lambda+\eta)_{n}} \delta\!\left(p+n,q\right) a_{p+n}z^{p+n-q}\\ \left(p\in \mathbb{N},\ q\in\mathbb{N}_{0},\ p>q,\ 0\leq\lambda<1,\;\mu< p+1,\;\eta>\max\{\lambda,\mu\}-p-1\right). \end{array} $$
(9)

From (9), we have

$$\begin{array}{*{20}l} z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q\right)} & =(p-\mu)\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta +1,p}f(z)\right)^{\left(q-1\right) }+\left(\mu-q+1\right) \\ & \quad\times\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }\ \left(q\in \mathbb{N} \right). \end{array} $$
(10)

We say that \(f\in \mathcal {A}(p)\) is in the class \(\mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta ;A,B\right) \ \)if

$$ \frac{1}{p-q-\zeta}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q\right) }}-\zeta\right) \prec\frac{1+Az}{1+Bz}, $$
(11)

\(0\leq \lambda <1,\;\mu < p+1,\;\eta >\max \{\lambda,\mu \}-p-1,\ 0\leq \zeta < p-q,\ -1\leq B<A\leq 1,\ p\in \mathbb {N},\ q\in \mathbb {N}_{0}\ {and}\ p>q+\zeta.\ \)Denoting by \(\mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta,\xi \right), \) the class of functions \(f\left (z\right) \in \mathcal {A}(p)\ \)which satisfies

$$ \mathfrak{R}\left\{ \frac{1}{p-q-\zeta}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q\right) }}-\zeta\right) \right\} >\xi\;(\xi<1;\ p\in\mathbb{N};\ z\in\mathbb{U}). $$
(12)

Preliminaries

To prove our main results, we shall need the following definition and lemmas.

Definition 2

[2]. Denote the set of all functions f that are analytic and univalent on \(\overline {\mathbb {U}}\setminus E(f)\ \)by \(\mathcal {Q}\), where

$$E(f):=\{\varsigma\in\partial\mathbb{U}:\underset{z\rightarrow\varsigma}{\lim }f(z)=\infty\}, $$

and are such that f(ς)≠0 for \(\varsigma \in \partial \mathbb {U}\setminus E(f)\).

Lemma 1

[15]. Let h(z) be analytic and convex (univalent) function in \(\mathbb {U}\) with h(0)=1. Also let ϕ given by

$$\phi(z)=1+c_{n}z^{n}+c_{n+1}z^{n+1}+... $$

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

$$ \phi(z)+\frac{z\phi^{\prime}(z)}{\alpha}\prec h(z)\;(\mathfrak{R}(\alpha)\geqslant 0;\ \alpha\neq0), $$
(13)

then

$$\phi(z)\prec\psi(z)=\frac{\alpha}{n}z^{-\frac{\alpha}{n}}\int_{0}^{z} t^{\frac{\alpha}{n}-1}h(t)dt\prec h(z), $$

and ψ is the best dominant of (13).

Lemma 2

[16]. Let h be a convex functions with

$$\mathfrak{R}\left[ \beta h(z)+\gamma\right] >0\ \left(\beta,\gamma\in \mathbb{C},\ z\in\mathbb{U}\right). $$

If p(z) is analytic in \(\mathbb {U\ }{with}\ p(0)=h(0),\ \)then

$$p(z)+\frac{zp^{\prime}\left(z\right) }{\beta p(z)+\gamma}\prec h(z)\Rightarrow p(z)\prec h(z). $$

The class of star-like (and normalized) functions of order α in \(\mathbb {U},\ \)is

$$\mathcal{S}^{\ast}\left(\alpha\right) =\left\{ f\in\mathcal{A}:\mathfrak{R}\left(\frac{zf^{\prime}\left(z\right) }{f\left(z\right) }\right) >\alpha\ \left(\alpha<1;\ z\in\mathbb{U}\right) \right\}. $$

Also in [17], if β>0 and β+γ>0, for a given \(\alpha \in \left [ -\frac {\gamma }{\beta },1\right),\) we define the order of starlikeness of the class \(\mathrm {I}_{\beta,\gamma }\left [ \mathcal {S}^{\ast }\left (\alpha \right) \right ]\ \)by the largest number 𝜗(α;β,γ) such that\(\ \mathrm {I}_{\beta,\gamma }\left [ \mathcal {S}^{\ast }\left (\alpha \right) \right ] \subset \mathcal {S}^{\ast }\left (\vartheta \right),\ \)where Iβ,γ is given by

$$ \mathrm{I}_{\beta,\gamma}(f)(z)=\left[ \frac{\beta+\gamma}{z^{\gamma}} \int\limits_{0}^{z}f^{\beta}(t)t^{\gamma-1}dt\right]^{\frac{1}{\beta}}. $$
(14)

Lemma 3

[17]. Let β>0, β+γ>0 and consider Iβ,γ defined by (14). If \(\alpha \in \left [-\frac {\gamma }{\beta },1\right),\ \)then the order of starlikeness of the class \(\mathrm {I}_{\beta,\gamma }\left [ \mathcal {S}^{\ast }\left (\alpha \right) \right ] \ \)is given by the number \(\vartheta \left (\alpha ;\beta,\gamma \right) =\inf \left \{ \mathfrak {R}\left (q(z)\right):z\in \mathbb {U}\right \},\ \)where

$$q(z)=\frac{1}{\beta Q\left(z\right) }-\frac{\gamma}{\beta}\ \text{and\ } Q(z)=\int\limits_{0}^{1}\left(\frac{1-z}{1-tz}\right)^{2\beta\left(1-\alpha\right) }t^{\beta+\gamma-1}dt. $$

Moreover, if \(\alpha \in \left [ \alpha _{0},1\right),\ {where}\ \alpha _{0}=\max \left \{ \frac {\beta -\gamma -1}{2\beta };-\frac {\gamma }{\beta }\right \}\ {and}\ g=\mathrm {I}_{\beta,\gamma }\left (f\right)\) with \(f\in \mathcal {S}^{\ast }\left (\alpha \right),\ \)then

$$\mathfrak{R}\left(\frac{zg^{\prime}\left(z\right) }{g\left(z\right) }\right) >\vartheta\left(\alpha;\beta,\gamma\right) \ \left(z\in\mathbb{U}\right), $$

where

$$\vartheta\left(\alpha;\beta,\gamma\right) =\frac{1}{\beta}\left[\frac{\beta+\gamma}{{\!~\!}_{2}F_{1}\left(1,2\beta\left(1-\alpha\right),\beta+\gamma+1;\frac{1}{2}\right) }-\gamma\right]. $$

Subordination and Inclusion theorems involving \(\left (\mathcal {M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left (q\right) }\)

We assume throughout this paper unless otherwise mentioned that\(\ p\in \mathbb {N},\ 0\leq \lambda <1,\;\mu < p,\;\eta >\max \{\lambda,\mu \}-p-1,\ -1\leq B<A\leq 1,\ 0\leq \zeta < p-q,\ \xi <1,\ \sigma >0,\ 0< c\leq 1\) and the powers are considered principal ones.

Theorem 1

Assume that \(1\leq q\leq p\ {and}\ f\left (z\right)\in \mathcal {A}\left (p\right) \) satisfy

$$(1-\sigma)\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}+\sigma \frac{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\prec \sqrt{1+cz}, $$

then

$$ \frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\prec Q(z)\prec \sqrt{1+cz}, $$
(15)

where

$$ Q(z)={\left(1+cz\right)^{\frac{1}{2}}}_{\,2}F_{1}\left(-\frac{1} {2},1;\frac{p-\mu}{\sigma}+1;\frac{cz}{1+cz}\right), $$
(16)

is the best dominant of (15). Furthermore,

$$ \mathfrak{R}\left\{ \frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\right\} >M, $$
(17)

where

$$M={\left(1-c\right)^{\frac{1}{2}}}_{\,2}F_{1}\left(-\frac{1}{2},1;\frac{p-\mu}{\sigma}+1;\frac{c}{c-1}\right). $$

The estimate in (17) is the best possible.

Proof

Putting

$$ \phi(z)=\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\ (z\in \mathbb{U}), $$
(18)

then ϕ(z) is analytic in \(\mathbb {U}\). After some computations, we get

$$(1-\sigma)\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}+\sigma \frac{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}} $$
$$=\phi(z)+\left(\frac{\sigma}{p-\mu}\right) z\phi^{\prime}(z)\prec \sqrt{1+cz}. $$

where the influence of \(h(z) = \sqrt {1+cz}\) under certain values of c is illustrated by Fig. 1. To apply Lemma 1, it suffies to show that h(z) is convex, therefore for z=reiθ, r(0,1), θ[−π,π], we have

$$1+\frac{zh^{\prime\prime}}{h^{\prime}}=1-\frac{cz}{2\left(1+cz\right) }=\frac{2+cz}{2\left(1+cz\right) }, $$

and

$$\begin{array}{*{20}l} \mathfrak{R}\left(1+\frac{zh^{\prime\prime}}{h^{\prime}}\right) & =\frac{2+3cr\cos\theta+c^{2}r^{2}}{\left\vert 1+cre^{i\theta}\right\vert^{2}}\geq\frac{2-3cr+c^{2}r^{2}}{\left\vert 1+cre^{i\theta}\right\vert^{2}}\\ & =\frac{\left(2-cr\right) \left(1-cr\right) }{\left\vert 1+cre^{i\theta}\right\vert^{2}}>0. \end{array} $$
Fig. 1
figure1

Influence of \(h(z)=\sqrt {1+cz}\) for different values of c

This implies that h is convex in \(\mathbb {U}\).

Now, by using Lemma 1 (with n=1) and making a change of variables followed by the use of (4) and (5), we deduce that

$$\begin{array}{*{20}l} \frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}} & \prec Q(z)=\frac {p-\mu}{\sigma}z^{-\frac{p-\mu}{\sigma}}\int_{0}^{z}t^{\frac{p-\mu}{\sigma} -1}\left(1+ct\right)^{\frac{1}{2}}dt\\ & ={\left(1+cz\right)^{\frac{1}{2}}}_{\,2}F_{1}\left(-\frac{1}{2},1;\frac{p-\mu}{\sigma}+1;\frac{cz}{1+cz}\right), \end{array} $$

this proves (15). Next, it is enough to show that

$$\inf_{\left\vert z\right\vert <1}\left\{ \mathfrak{R}(Q(z)\right\} =Q(-1). $$

Indeed

$$\mathfrak{R}\left\{ \sqrt{1+cz}\right\} \geqslant\sqrt{1-cr}\ \left(\left\vert z\right\vert \leq r<1\right). $$

Setting

$$G(z,s)=\sqrt{1+czs}\;\text{and\ }d\nu(s)=\frac{p-\mu}{\sigma}s^{\frac{p-\mu }{\sigma}-1}ds\;(0\leq s\leq1), $$

which is a positive measure on the closed interval [0,1], we get

$$Q(z)=\int\limits_{0}^{1}G(z,s)d\nu(s), $$

so that

$$\mathfrak{R}\left\{ Q(z)\right\} \geqslant\int_{0}^{1}\sqrt{1-cr}d\nu(s)=Q(-r)\;(\left\vert z\right\vert \leq r<1). $$

Letting r→1 in the above inequality, we obtain (17). To show that the result in (17) is sharp, let us suppose that

$$\mathfrak{R}\left\{ \frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\right\} >M_{1}, $$

that is

$$\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\prec\frac{1+\left(1-2M_{1}\right) z}{1-z}. $$

From (15), we have

$$\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\prec\frac{1+\left(1-2M\right) z}{1-z}, $$

and so

$$\frac{1+\left(1-2M\right) z}{1-z}\prec\frac{1+\left(1-2M_{1}\right) z}{1-z}, $$

which implies that MM1, that is, M cannot be decreased and the estimate in (17) is the best possible. □

For \(f\in \mathcal {A}(p)\) the generalized Bernardi-Libera-Livingston integeral operator Fp,υ is defined by (see [18]):

$$\begin{array}{*{20}l} F_{p,\upsilon}f(z) & =\frac{\upsilon+p}{z^{p}}\int\limits_{0}^{z} t^{\upsilon-1}f(t)dt\\ & =\left(z^{p}+\sum\limits_{n=1}^{\infty}\frac{\upsilon+p}{\upsilon +p+n}z^{p+n}\right) \ast f(z)\\ & ={z^{p}}_{\,3}F_{2}\left(1,1,\upsilon+p;1,\upsilon+p+1;z\right) \ast f(z)\ (\upsilon>-p). \end{array} $$
(19)

Lemma 4

If \(f\in \mathcal {A}\left (p\right),\ \)then (i) \(\mathcal {M}_{0,z}^{\lambda,\mu,\eta,p}\left (F_{p,\upsilon }f\right) =F_{p,\upsilon }\left (\mathcal {M}_{0,z}^{\lambda,\mu,\eta,p}f\right),\)

(ii)

$$ z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\right)^{\prime}=(p+\upsilon)\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)-\upsilon \mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z), $$
(20)

(iii)

$$ {}z\left(\! \mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\!\right)^{\left(q\right) }\!=(p+\upsilon)\left(\! \mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\!\right)^{\left(q-1\right) }-\left(\upsilon+q-1\right) \left(\! \mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\!\right)^{\left(q-1\right) }. $$
(21)

Proof

Since \(f(z)\in \mathcal {A}(p)\), then

$$\begin{array}{*{20}l} {}\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}\left(F_{p,\upsilon}f\right) &\,=\,\left[ {z^{p}}_{3}F_{2}\left(1,p+1,p+1-\mu+\eta;p+1-\mu,p+1-\lambda +\eta;z\right) \right] \ast\left(F_{p,\upsilon}f\right) \\ & \,=\,\left[ {z^{p}}_{3}F_{2}\left(1,p+1,p+1-\mu+\eta;p+1-\mu,p+1-\lambda +\eta;z\right) \right] \\ &\quad\ast \left[ {z^{p}}_{3}F_{2}\left(1,1,\upsilon+p;1,\upsilon+p+1;z\right) \ast f(z)\right], \end{array} $$

and

$$\begin{array}{*{20}l} {}F_{p,\upsilon}\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f\right) &={z^{p}}_{3}F_{2}\left(1,1,\upsilon+p;1,\upsilon+p+1;z\right) \ast\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f\right) \\ & ={z^{p}}_{3}F_{2}\left(1,1,\upsilon+p;1,\upsilon+p+1;z\right) \ast\\ & \left[ {z^{p}}_{3}F_{2}\left(1,p+1,p+1-\mu+\eta;p+1-\mu,p+1-\lambda+\eta;z\right) \ast f(z)\right]. \end{array} $$

Now, the first part of this lemma follows. Also, the recurrence relation of Fp,υ is given by

$$ z\left(F_{p,\upsilon}f(z)\right)^{\prime}=(p+\upsilon)f(z)-\upsilon F_{p,\upsilon}f(z). $$
(22)

If we replace f(z) by \(\mathcal {M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\ \)and using the first part of this lemma, we get (20). If we differentiate (20) q-times, we obtain (21). □

Theorem 2

Suppose that \(1\leq q\leq p\ {and}\ f\left (z\right)\in \mathcal {A}(p)\) satisfy

$$(1-\sigma)\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon }f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1} }+\sigma\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\prec \sqrt{1+cz}, $$

where Fp,υ defined by (19), then

$$ \frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\prec \varphi(z)\prec\sqrt{1+cz}, $$
(23)

where φ(z) given by

$$\varphi(z)={\left(1+cz\right)^{\frac{1}{2}}}_{\,2}F_{1}\left(-\frac{1} {2},1;\frac{\upsilon+p}{\sigma}+1;\frac{cz}{1+cz}\right), $$

is the best dominant of (23). Further,

$$ \mathfrak{R}\left\{ \frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon }f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1} }\right\} >L, $$
(24)

where

$$L={\left(1-c\right)^{\frac{1}{2}}}_{\,2}F_{1}\left(-\frac{1}{2},1;\frac{\upsilon+p}{\sigma}+1;\frac{c}{c-1}\right). $$

The result is the best possible.

Proof

Taking

$$ \Theta(z)=\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\ (z\in\mathbb{U)}, $$
(25)

then Θ is analytic in \(\mathbb {U}\). After some calculations, we have

$$\begin{array}{*{20}l} & (1-\sigma)\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}+\sigma\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q-1\right) }}{\delta\left(p,q-1\right) z^{p-q+1}}\\ & =\Theta(z)+\left(\frac{\sigma}{p+\upsilon}\right) z\Theta^{\prime }\left(z\right) \prec\sqrt{1+cz}. \end{array} $$

By employing the same technique that was used in proving Theorem 1, the remaining part of the theorem can be proved. □

Theorem 3

Let\(\ q\in \mathbb {N}_{0}\ {and}\ p>q+\zeta.\ \)If \(f\left (z\right) \in \mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta,\xi \right),\ \)then \(f\left (z\right) \in \mathcal {S}_{p,q}^{\lambda +1,\mu +1,\eta +1}\left (\zeta,\xi \right) \ \)for |z|<R(p,q,μ,ζ,ξ) where

$$ R\left(p,q,\mu,\zeta,\xi\right) =\min\left\{ r>0:t\left(r\right) =0\right\}, $$
(26)

and

$$t\left(r\right) =1-\frac{2r}{\left(p-q-\zeta\right) \left\vert \left(1-\xi\right) \left(1-r\right)^{2}-\left\vert \xi+\frac{q+\zeta-\mu }{p-q-\zeta}\right\vert \left(1-r^{2}\right) \right\vert }. $$

Proof

Assume that \(f\left (z\right) \in \mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta,\xi \right) \) and

$$ u\left(z\right) =\frac{1}{p-q-\zeta}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q\right) }}-\zeta\right), $$
(27)

then, u(z) is analytic in \(\mathbb {U}\) with \(u(0)=1,\ \mathfrak {R}\left \{ u\left (z\right) \right \} >\xi.\) After some computations, we have

$$ {}u\left(z\right) +\frac{zu^{\prime}\left(z\right) }{\left(p-q-\zeta\right) u(z)+\left(q+\zeta-\mu\right) }\,=\,\frac{1}{p-q-\zeta}\!\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q\right) }}-\zeta\right). $$
(28)

Letting \(v(z)=\frac {u(z)-\xi }{1-\xi },\ \)then, v(0)=1 with \(\mathfrak {R}\left \{ v\left (z\right) \right \} >0.\ \)Substituting in (28), we obtain

$$\begin{array}{*{20}l} & \frac{1}{p-q-\zeta}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q\right) }}-\zeta\right) -\xi\\ & =\left(1-\xi\right) \left[ v\left(z\right) +\frac{zv^{\prime}\left(z\right) }{\left(p-q-\zeta\right) \left[ \left(1-\xi\right) v(z)+\xi\right] +\left(q+\zeta-\mu\right) }\right], \end{array} $$

and so

$$\begin{array}{*{20}l} & \mathfrak{R}\left\{ \frac{1}{p-q-\zeta}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q\right)}}-\zeta\right) -\xi\right\} \\ & \geq\left(1-\xi\right) \ \left[ \mathfrak{R}\left\{ v\left(z\right)\right\} -\frac{\left\vert zv^{\prime}\left(z\right) \right\vert }{\left(p-q-\zeta\right) \left\vert \left(1-\xi\right) \left\vert v\left(z\right) \right\vert -\left\vert \xi+\frac{q+\zeta-\mu}{p-q-\zeta}\right\vert\, \,\right\vert }\right] \\ & \geq\left(1-\xi\right) \ \left[ \mathfrak{R}\left\{ v\left(z\right)\right\} -\frac{\left\vert zv^{\prime}\left(z\right) \right\vert }{\left(p-q-\zeta\right) \left\vert \left(1-\xi\right) \ \mathfrak{R}\left\{ v\left(z\right) \right\} -\left\vert \xi+\frac{q+\zeta-\mu}{p-q-\zeta}\right\vert\, \,\right\vert }\right]. \end{array} $$

Applying the following well-known estimate [19]:

$$\mathfrak{R}\left\{ v(z)\right\} \geq\frac{1-r}{1+r}\ \text{and\ }\frac{\left\vert zv^{\prime}(z)\right\vert }{\mathfrak{R}\left\{ v(z)\right\} }\leq\frac{2nr^{n} }{1-r^{2n}}\ (\left\vert z\right\vert =r<1), $$

for n=1, we get

$$\mathfrak{R}\left\{ \frac{1}{p-q-\zeta}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q\right)}}-\zeta\right) -\xi\right\} \geq\left(1-\xi\right) t\left(r\right)\mathfrak{R}\left\{ v(z)\right\}. $$

It is easily seen that t(r) is positive, if |z|<R(p,q,μ,ζ,ξ), where R is given by (26). □

Theorem 4

Let \(f(z)\in \mathcal {A}(p),\ p>\mu,\ \gamma >0\ \)and

$$ \mathfrak{R}\left(\frac{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\prime}}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime}}-\frac{\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta +1,p}f(z)}{\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)}\right) <\frac{\gamma}{p-\mu}, $$
(29)

then

$$\mathfrak{R}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime}}{\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)}\right)^{-\frac{1}{2\gamma}}>\frac{1}{2}. $$

The result is sharp.

Proof

From (8), (29) may be written as

$$\mathfrak{R}\left(1+\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime\prime}}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime}}-\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime}}{\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)}\right) <\gamma, $$

or equivalently,

$$ 1+\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime\prime}}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime}}-\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime}}{\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)}\prec-\frac{2\gamma z}{1-z}. $$
(30)

Letting

$$\digamma(z)=\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime}}{\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)}\right)^{-\frac{1}{2\gamma}}, $$

then, we can express (30) as

$$ z\left(\log\digamma(z)\right)^{\prime}\prec z\left(\log\frac{1} {1-z}\right)^{\prime}. $$
(31)

Fom [20], (31) implies to

$$\digamma(z)\prec\frac{1}{1-z}, $$

or equivalently,

$$\mathfrak{R}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\prime}}{\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)}\right)^{-\frac {1}{2\gamma}}>\frac{1}{2}\ \left(z\in\mathbb{U}\right). $$

To show that the result is sharp, let

$$K(z)=z^{p}+\sum\limits_{n=1}^{\infty}\frac{(p+1)_{n}(p+1-\mu+\eta)_{n} }{(p+1-\mu)_{n}(p+1-\lambda+\eta)_{n}}\frac{2\gamma\left(2\gamma-1\right)...\left(2\gamma-n+1\right) }{n!}z^{p+n}, $$

and so

$$\begin{array}{*{20}l} \mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}K(z) & =z^{p}+\sum\limits_{n=1} ^{\infty}\frac{2\gamma\left(2\gamma-1\right)...\left(2\gamma-n+1\right) }{n!}z^{p+n}\\ & =z^{p}\left(1+z\right)^{2\gamma}. \end{array} $$

It is easy to check that K(z) satisfies (29) and

$$\mathfrak{R}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}K(z)\right)^{\prime}}{\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}K(z)}\right)^{-\frac {1}{2\gamma}}\rightarrow\frac{1}{2} $$

as z→1. This ends our proof. □

Theorem 5

Consider that\(\ q\in \mathbb {N}_{0},\ p>q+\zeta \ {and}\ \)

$$ \left(p-q-\zeta\right) \left(1-A\right) +\left(q+\zeta-\mu\right) \left(1-B\right) \geq0. $$
(32)

(i) Suppose that \(\left (\mathcal {M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left (q\right) }\neq 0\ \)for all \(z\in \mathbb {U}^{\ast }:=\mathbb {U}\backslash \{0\},\ \)then

$$\mathcal{S}_{p,q}^{\lambda+1,\mu+1,\eta+1}\left(\zeta;A,B\right) \subset\mathcal{S}_{p,q}^{\lambda,\mu,\eta}\left(\zeta;A,B\right). $$

(ii) Also, assuming that

$$ \frac{1-A}{1-B}\geq\frac{1}{p-q-\zeta}\max\left\{ \frac{p-2q-2\zeta+\mu-1} {2},-\left(q+\zeta-\mu\right) \right\}, $$
(33)

then

$$\mathcal{S}_{p,q}^{\lambda+1,\mu+1,\eta+1}\left(\zeta;A,B\right) \subset\mathcal{S}_{p,q}^{\lambda,\mu,\eta}\left(\zeta,\xi\right). $$

where the bound

$$ \xi\left(A,B\right) =\frac{1}{p-q-\zeta}\left[ \frac{p-\mu}{{\!~\!}_{2} F_{1}\left(1,\frac{2\left(p-q-\zeta\right) \left(A-B\right) } {1-B},p-\mu+1;\frac{1}{2}\right) }-\left(q+\zeta-\mu\right) \right], $$
(34)

is the best possible

Proof

Let \(f\left (z\right) \in \mathcal {S}_{p,q}^{\lambda +1,\mu +1,\eta +1}\left (\zeta ;A,B\right) \ \)and

$$ G\left(z\right) =z\left(\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q\right) }}{\delta\left(p,q\right) z^{p-q} }\right)^{\frac{1}{p-q-\zeta}}. $$
(35)

Since \(\left (\mathcal {M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left (q\right) }\neq 0\ \)for all \(z\in \mathbb {U}^{\ast },\ \) then G(z) is analytic in \(\mathbb {U}\) with G(0)=0 and G(0)=1. Differentiating both sides of (35) logarithmically, we get

$$ \Psi\left(z\right) =\frac{zG^{\prime}\left(z\right) }{G\left(z\right)}=\frac{1}{p-q-\zeta}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q\right) }}-\zeta\right). $$
(36)

Using (10) in (36), we have

$$ \left(p-\mu\right) \frac{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q\right) }}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q\right) }}=\left(p-q-\zeta\right) \Psi\left(z\right) +\left(q+\zeta-\mu\right). $$
(37)

Differentiating both sides of (37) logarithmically, we get

$${}\frac{1}{p-q-\zeta}\!\left(\!\frac{z\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda+1,\mu+1,\eta+1,p}f(z)\right)^{\left(q\right) }}\,-\,\zeta\!\right)\! \!=\Psi\left(z\right) +\frac{z\Psi^{\prime}\left(z\right)}{\left(p-q-\zeta\right) \Psi\left(z\right) +\left(q+\zeta-\mu\right)}. $$

Combining this identity together with \(f\left (z\right)\in \mathcal {S}_{p,q}^{\lambda +1,\mu +1,\eta +1}\left (\zeta ;A,B\right),\ \)we obtain

$$\Psi\left(z\right) +\frac{z\Psi^{\prime}\left(z\right) }{\left(p-q-\zeta\right) \Psi\left(z\right) +\left(q+\zeta-\mu\right) } \prec\frac{1+Az}{1+Bz}\equiv h(z). $$

We will use Lemma 2 for \(\widetilde {\beta }=\left (p-q-\zeta \right),\ \widetilde {\gamma }=\left (q+\zeta -\mu \right).\ \)Since h(z) is a convex function in \(\mathbb {U\ }\)and

$$\mathfrak{R}\left[ \left(p-q-\zeta\right) \frac{1+Az}{1+Bz}+\left(q+\zeta -\mu\right) \right] >0, $$

whenever (32) holds. Then \(f(z)\in \mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta ;A,B\right) \ \)from Lemma 2. This completes the proof of (i). To prove (ii), we assume that (33) holds, then all the assumptions of Lemma 3 are satisfied for the above values of \(\widetilde {\beta },\ \widetilde {\gamma }\ \)and \(\widetilde {\alpha }=\frac {1-A}{1-B}.\ \)It follows that \(\mathcal {S}_{p,q}^{\lambda +1,\mu +1,\eta +1}\left (\zeta ;A,B\right) \subset \mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta,\xi \right) \) where ξ(A,B) given by (34) is the best possible. □

Theorem 6

Assume that\(\ q\in \mathbb {N}_{0},\ p>q+\zeta \ {and}\ \)

$$ \left(p-q-\zeta\right) \left(1-A\right) +\left(q+\zeta+\upsilon\right) \left(1-B\right) \geq0. $$
(38)

(i) Suppose that \(\left (\mathcal {M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon }f(z)\right)^{\left (q\right) }\neq 0\ \)for all \(z\in \mathbb {U}^{\ast },\ \)then

$$\mathcal{S}_{p,q}^{\lambda,\mu,\eta}\left(\zeta;A,B\right) \subset F_{p,\upsilon}\left(\mathcal{S}_{p,q}^{\lambda,\mu,\eta}\left(\zeta;A,B\right) \right). $$

(ii) Also, assuming that

$$ \frac{1-A}{1-B}\geq\frac{1}{p-q-\zeta}\max\left\{ -\frac{q+2\zeta+\upsilon +1}{2},-\left(p+\zeta+\upsilon\right) \right\}, $$
(39)

then

$$\mathcal{S}_{p,q}^{\lambda,\mu,\eta}\left(\zeta;A,B\right) \subset \mathcal{S}_{p,q}^{\lambda,\mu,\eta}\left(\zeta;\tau\left(A,B\right) \right). $$

where the bound

$$ {}\tau\left(A,B\right) =\frac{1}{p-q-\zeta}\left[ \frac{2p-q+\upsilon}{{\!~\!}_{2}F_{1}\left(1,\frac{2\left(p-q-\zeta\right) \left(A-B\right)}{1-B},2p-q+\upsilon;\frac{1}{2}\right) }-\left(p+\zeta+\upsilon\right) \right], $$
(40)

is the best possible.

Proof

Let \(f\left (z\right) \in \mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta ;A,B\right) \ \)and

$$ H\left(z\right) =z\left(\frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\right)^{\left(q\right) }}{\delta\left(p,q\right) z^{p-q}}\right)^{\frac{1}{p-q-\zeta}}. $$
(41)

Since \(\left (\mathcal {M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon }f(z)\right)^{\left (q\right) }\neq 0\ \)for all \(z\in \mathbb {U}^{\ast },\ \)then H(z) is analytic in \(\mathbb {U}\) with H(0)=0 and H(0)=1. Differentiating both sides of (41) logarithmically, we get

$$ \Phi\left(z\right) =\frac{zH^{\prime}\left(z\right) }{H\left(z\right)}=\frac{1}{p-q-\zeta}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\right)^{\left(q\right) }}-\zeta\right). $$
(42)

Using (21) in (42), we have

$$ \left(p+\upsilon\right) \frac{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q\right) }}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}F_{p,\upsilon}f(z)\right)^{\left(q\right) }}=\left(p-q-\zeta\right) \Phi\left(z\right) +\left(q+\zeta+\upsilon\right). $$
(43)

Differentiating both sides of (43) logarithmically, we get

$${}\frac{1}{p-q-\zeta}\left(\frac{z\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q+1\right) }}{\left(\mathcal{M}_{0,z}^{\lambda,\mu,\eta,p}f(z)\right)^{\left(q\right) }}-\zeta\right) =\Phi\left(z\right) +\frac{z\Phi^{\prime}\left(z\right) }{\left(p-q-\zeta\right) \Phi\left(z\right) +\left(q+\zeta+\upsilon\right) }. $$

Combining this identity together with \(f\left (z\right) \in \mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta ;A,B\right),\ \)we obtain

$$\Phi\left(z\right) +\frac{z\Phi^{\prime}\left(z\right) }{\left(p-q-\zeta\right) \Phi\left(z\right) +\left(q+\zeta+\upsilon\right) }\prec\frac{1+Az}{1+Bz}\equiv h(z). $$

We will use Lemma 2 for \(\widetilde {\beta }=\left (p-q-\zeta \right),\ \overline {\gamma }=\left (q+\zeta +\upsilon \right).\ \)Since h(z) is a convex function in \(\mathbb {U\ }\)and

$$\mathfrak{R}\left[ \left(p-q-\zeta\right) \frac{1+Az}{1+Bz}+\left(q+\zeta+\upsilon\right) \right] >0, $$

whenever (38) holds. Then \(f(z)\in F_{p,\upsilon }\left (\mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta ;A,B\right) \right) \ \)from Lemma 2. This proves (i). To prove (ii), we assume that (39) holds, then all the assumptions of Lemma 3 are satisfied for \(\widetilde {\beta },\ \overline {\gamma }\ \) which stated above and \(\widetilde {\alpha }=\frac {1-A}{1-B}.\ \)It follows that

$$\mathcal{S}_{p,q}^{\lambda,\mu,\eta}\left(\zeta;A,B\right) \subset\mathcal{S}_{p,q}^{\lambda,\mu,\eta}\left(\zeta;\tau\left(A,B\right)\right), $$

where τ(A,B) given by (40) is the best possible □

Conclusion

In our present investigation, we have derived some subordination results of certain subclasses of multivalent analytic functions which are defined by a generalized fractional differintegral operator. We have also successfully considered inclusion relations for functions in the class \(\mathcal {S}_{p,q}^{\lambda,\mu,\eta }\left (\zeta ;A,B\right)\) and the images of these functions by the generalized Bernardi-Libera-Livingston integral operator.

Abbreviations

\(\mathcal {A}(p)\) :

The class of analytic and multivalent functions in the open unit disc \(\mathbb{U}=\{z\in \mathbb{C}|z| <1\}\)

:

Convolution of two power series

:

Subordination of two analytic functions in \(\mathbb {U\ }\)

2 F 1(a,b;c;z)(c≠0,−1,−2,…):

The well-known (Gaussian) hypergeometric function

\(J_{0,z}^{\lambda,\mu,\eta,p}f(z)\) :

The generalized fractional derivative operator for \(f\in \mathcal {A}(p)\)

F p, υ :

The generalized Bernardi-Libera-Livingston integeral operator

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Acknowledgements

The author would like to express her sincere thanks to Hillal Elshehabey from Mathematics department, Faculty of Science, South Valley university, for his help in plotting the function h(z), as well as the referees for their valuable suggestions and comments, which enhanced the paper.

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Zayed, H.M. Subordination and inclusion theorems for higher order derivatives of a generalized fractional differintegral operator. J Egypt Math Soc 27, 17 (2019). https://doi.org/10.1186/s42787-019-0020-2

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Keywords

  • Differential subordination
  • Multivalent functions
  • Higher order derivatives
  • Generalized fractional differintegral operator

Mathematics Subject Classification

  • 30C45
  • 30C50.