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

[4–12]. 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,μ−η}−1 and the multiplicity of (z−ζ)^−λ is removed by requiring log(z−ζ)to be real when z−ζ>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 [4–12]) 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 I_β,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 I_β,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=re^iθ, 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

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 M≤M₁, 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 F_p,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 F_p,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 F_p,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.