Subordination Properties for Multivalent Functions Associated with a Generalized Fractional Differintegral Operator

Abstract

Using of the principle of subordination, we investigate some subordination and convolution properties for classes of multivalent functions under certain assumptions on the parameters involved, which are defined by a generalized fractional differintegral operator under certain assumptions on the parameters involved.

1. Introduction and Definitions

Denote by $\mathcal{A}(p)$ the class of analytic and p-valent functions of the form:

$$f(z)=z^p+\sum _{n=1}^{\infty }{a}_{p+n}{z}^{p+n}(p\in \mathbb{N}=\{1,2,\dots \};z\in \mathbb{U}=\{z\in \mathbb{C}:\left|z\right|<1\}).$$

(1)

For functions $f,g$ analytic in $\mathbb{U}$, f is subordinate to g, written $f(z)\prec g(z)$ if there exists a function w, analytic in $\mathbb{U}$ with $w(0)=0$ and $\left|w(z)\right|<1,$ such that $f(z)=g(w(z)),z\in \mathbb{U}.$ If g is univalent in $\mathbb{U},$ then (see [1,2]):

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

If $\phi (z)$ is analytic in $\mathbb{U}$ and satisfies:

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

(2)

then $\phi$ is a solution of (2). The univalent function q is called dominant, if $\phi (z)\prec q(z)$ for all $\phi$. A dominant $\tilde{q}$ is called the best dominant, if $\tilde{q}(z)\prec q(z)$ for all dominants q.

Let ${}_2{F}_1(a,b;c;z)$ $\left(c\ne 0,-1,-2,\cdots \right)$ be the well-known (Gaussian) hypergeometric function defined by:

$${}_2{F}_1(a,b;c;z):=\sum _{n=0}^{\infty }\frac{{(a)}_n{(b)}_n}{{(c)}_n{(1)}_n}{z}^n,z\in \mathbb{U},$$

where:

$${(\lambda )}_n:=\frac{\mathsf{\Gamma }\left(\lambda +n\right)}{\mathsf{\Gamma }\left(\lambda \right)}.$$

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

Definition 1.

For $f(z)\in \mathcal{A}(p),$ the fractional integral and fractional derivative operators of order λ are defined by Owa [3] (see also [4]) as:

$$D_z^{-\lambda }f(z):=\frac{1}{\mathsf{\Gamma }(\lambda )}{\int }_0^z\frac{f(\zeta )}{{(z-\zeta )}^{1-\lambda }}d\zeta (\lambda >0),$$

$$D_z^{\lambda }f(z):=\frac{1}{\mathsf{\Gamma }(1-\lambda )}\frac{d}{dz}{\int }_0^z\frac{f(\zeta )}{{(z-\zeta )}^{\lambda }}d\zeta (0\le \lambda <1),$$

where f is an analytic function in a simply-connected region of the complex z-plane containing the origin, and the multiplicity of ${(z-\zeta )}^{\lambda -1}({(z-\zeta )}^{-\lambda })$ is removed by requiring $log(z-\zeta )$ to be real when $z-\zeta >0$.

Definition 2.

For $f(z)\in \mathcal{A}(p)$ and in terms of ${}_2{F}_1,$ the generalized fractional integral and generalized fractional derivative operators defined by Srivastava et al. [5] (see also [6]) as:

$$I_{0,z}^{\lambda ,\mu ,\eta }f(z):=\frac{z^{-\lambda -\mu }}{\mathsf{\Gamma }(\lambda )}{\int }_0^z{(z-\zeta )}^{\lambda -1}f(\zeta ){}_2{F}_1\left(\mu +\lambda ,-\eta ;\lambda ;1-\frac{\zeta }{z}\right)d\zeta \left(\lambda >0,\mu ,\eta \in \mathbb{R}\right),$$

$$J_{0,z}^{\lambda ,\mu ,\eta }f(z):=\left\{\begin{array}{c}\frac{d}{dz}\left\{\frac{z^{\lambda -\mu }{\int }_0^z{(z-\zeta )}^{-\lambda }f{(\zeta )}_2{F}_1\left(\mu -\lambda ,1-\eta ;1-\lambda ;1-\frac{\zeta }{z}\right)d\zeta }{\mathsf{\Gamma }(1-\lambda )}\right\}(0\le \lambda <1),\hfill \\ \frac{d^n}{d{z}^n}{J}_{0,z}^{\lambda -n,\mu ,\eta }f(z)(n\le \lambda <n+1;n\in \mathbb{N}),\hfill \end{array}\right.$$

where $f(z)$ is an analytic function in a simply-connected region of the complex $z-$ plane containing the origin with the order $f(z)=O({\left|z\right|}^{\epsilon }),z\to 0$ when $\epsilon >max\{0,\mu -\eta \}-1$, and the multiplicity of ${(z-\zeta )}^{\lambda -1}$ $({(z-\zeta )}^{-\lambda })$ is removed by requiring $log(z-\zeta )$ to be real when $z-\zeta >0$.

We note that:

$$I_{0,z}^{\lambda ,-\lambda ,\eta }f(z)=D_z^{-\lambda }f(z)(\lambda >0)\mathrm{and}{J}_{0,z}^{\lambda ,\lambda ,\eta }f(z)=D_z^{\lambda }f(z)(0\le \lambda <1),$$

where $D_z^{-\lambda }f(z)$ and $D_z^{\lambda }f(z)$ are the fractional integral and fractional derivative operators studied by Owa [3].

Goyal and Prajapat [7] (see also [8]) defined the operator:

$$S_{0,z}^{\lambda ,\mu ,\eta ,p}f(z)=\left\{\begin{array}{c}\frac{\mathsf{\Gamma }(p+1-\mu )\mathsf{\Gamma }(p+1-\lambda +\eta )}{\mathsf{\Gamma }(p+1)\mathsf{\Gamma }(p+1-\mu +\eta )}{z}^{\mu }{J}_{0,z}^{\lambda ,\mu ,\eta }f(z)(0\le \lambda <\eta +p+1;z\in \mathbb{U}),\hfill \\ \frac{\mathsf{\Gamma }(p+1-\mu )\mathsf{\Gamma }(p+1-\lambda +\eta )}{\mathsf{\Gamma }(p+1)\mathsf{\Gamma }(p+1-\mu +\eta )}{z}^{\mu }{I}_{0,z}^{-\lambda ,\mu ,\eta }f(z)(-\infty <\lambda <0;z\in \mathbb{U}).\hfill \end{array}\right.$$

For $f(z)\in \mathcal{A}(p)$, we have:

$$\begin{array}{ccc}\hfill S_{0,z}^{\lambda ,\mu ,\eta ,p}f(z)& =& z_3^p{F}_2(1,1+p,1+p+\eta -\mu ;1+p-\mu ,1+p+\eta -\lambda ;z)\ast f(z)\hfill \\ \hfill & =& z^p+\sum _{n=1}^{\infty }\frac{{(p+1)}_n{(p+1-\mu +\eta )}_n}{{(p+1-\mu )}_n{(p+1-\lambda +\eta )}_n}{a}_{p+n}{z}^{p+n}\hfill \\ \hfill (p& \in & \mathbb{N};\mu ,\eta \in \mathbb{R};\mu <p+1;-\infty <\lambda <\eta +p+1),\hfill \end{array}$$

where “∗” stands for convolution of two power series, and ${}_q{F}_s(q\le s+1;q,s\in {\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\})$ is the well-known generalized hypergeometric function.

Let:

$$\begin{array}{ccc}\hfill G_{p,\eta ,\mu }^{\lambda }(z)& =& z^p+\sum _{n=1}^{\infty }{\displaystyle \frac{{(p+1)}_n{(p+1-\mu +\eta )}_n}{{(p+1-\mu )}_n{(p+1-\lambda +\eta )}_n}}{z}^{p+n}\hfill \\ \hfill (p& \in & \mathbb{N};\mu ,\eta \in \mathbb{R};\mu <p+1;-\infty <\lambda <\eta +p+1).\hfill \end{array}$$

and:

$$G_{p,\eta ,\mu }^{\lambda }(z)\ast {\left[G_{p,\eta ,\mu }^{\lambda }(z)\right]}^{-1}=\frac{z^p}{{(1-z)}^{\delta +p}}(\delta >-p;z\in \mathbb{U}).$$

Tang et al. [9] (see also [10,11,12,13,14,15]) defined the operator $H_{p,\eta ,\mu }^{\lambda ,\delta }:\mathcal{A}(p)\to \mathcal{A}(p),$ where:

$$\begin{array}{ccc}\hfill H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)& =& z^p+\sum _{n=1}^{\infty }{\displaystyle \frac{{(\delta +p)}_n{(p+1-\mu )}_n{(p+1-\lambda +\eta )}_n}{{(1)}_n{(p+1)}_n{(p+1-\mu +\eta )}_n}}{a}_{p+n}{z}^{p+n}\hfill \\ \hfill (p& \in & \mathbb{N},\delta >-p,\mu ,\eta \in \mathbb{R},\mu <p+1,-\infty <\lambda <\eta +p+1).\hfill \end{array}$$

It is easy to verify that:

$$z{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }=(\delta +p)H_{p,\eta ,\mu }^{\lambda ,\delta +1}f(z)-\delta H_{p,\eta ,\mu }^{\lambda ,\delta }f(z),$$

(3)

and:

$$z{\left(H_{p,\eta ,\mu }^{\lambda +1,\delta }f(z)\right)}^{\prime }=(p+\eta -\lambda )H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)-(\eta -\lambda )H_{p,\eta ,\mu }^{\lambda +1,\delta }f(z).$$

(4)

By using the operator $H_{p,\eta ,\mu }^{\lambda ,\delta }$, we introduce the following class.

Definition 3.

For $A,B(-1\le B<A\le 1)$, $f\in \mathcal{A}(p)$ is in the class ${\mathcal{T}}_{p,\eta ,\mu }^{\lambda ,\delta }(A,B)$ if

$$\frac{{(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z))}^{\prime }}{p{z}^{p-1}}\prec \frac{1+Az}{1+Bz}(z\in \mathbb{U};p\in \mathbb{N}),$$

which is equivalent to:

$$\left|\frac{{\displaystyle \frac{{(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z))}^{\prime }}{p{z}^{p-1}}}-1}{B{\displaystyle \frac{{(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z))}^{\prime }}{p{z}^{p-1}}}-A}\right|<1(z\in \mathbb{U}).$$

For convenience, we write ${\mathcal{T}}_{p,\eta ,\mu }^{\lambda ,\delta }\left(1-\frac{2\xi }{p},-1\right)={\mathcal{T}}_{p,\eta ,\mu }^{\lambda ,\delta }(\xi )(0\le \xi <p)$, which satisfies the inequality:

$$\Re \left\{\frac{{(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z))}^{\prime }}{z^{p-1}}\right\}>\xi (0\le \xi <p).$$

In this paper, we investigate some subordination and convolution properties for classes of multivalent functions, which are defined by a generalized fractional differintegral operator. The theory of subordination received great attention, particularly in many subclasses of univalent and multivalent functions (see, for example, [13,15,16,17]).

2. Preliminaries

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

Lemma 1.

[18]. Let h be an analytic and convex (univalent) function in $\mathbb{U}$ with $h(0)=1$. Additionally, let φ given by:

$$\varphi (z)=1+c_n{z}^n+c_{n+1}{z}^{n+1}+\dots$$

(5)

be analytic in $\mathbb{U}$. If:

$$\varphi (z)+\frac{z{\varphi }^{\prime }(z)}{\sigma }\prec h(z)(\Re (\sigma )⩾0;\sigma \ne 0),$$

(6)

then:

$$\varphi (z)\prec \psi (z)=\frac{\sigma }{n}{z}^{-\frac{\sigma }{n}}{\int }_0^z{t}^{\frac{\sigma }{n}-1}h(t)dt\prec h(z),$$

(7)

and ψ is the best dominant of (6).

Denote by $P(\varsigma )$ the class of functions $\mathsf{\Phi }$ given by:

$$\mathsf{\Phi }(z)=1+c_{1}z+c_1{z}^2+\dots ,$$

(8)

which are analytic in $\mathbb{U}$ and satisfy the following inequality:

$$\Re \left\{\mathsf{\Phi }(z)\right\}>\varsigma (0\le \varsigma <1).$$

Using the well-known growth theorem for the Carathéodory functions (cf., e.g., [19]), we may easily deduce the following result:

Lemma 2.

[19]. If $\mathsf{\Phi }\in$ $P(\varsigma )$. Then

$$\Re \left\{\mathsf{\Phi }(\varsigma )\right\}⩾2\varsigma -1+\frac{2(1-\varsigma )}{1+\left|z\right|}(0\le \varsigma <1).$$

Lemma 3.

[20]. For $0\le {\varsigma }_1,{\varsigma }_2<1,$

$$P({\varsigma }_1)\ast P({\varsigma }_2)\subset P({\varsigma }_3)({\varsigma }_3=1-2(1-{\varsigma }_1)(1-{\varsigma }_2)).$$

The result is the best possible.

Lemma 4.

[21]$.$ Let ϕ be such that $\phi (0)=1$ and $\phi (z)\ne 0$ and $A,B\in \mathbb{C},$ with $A\ne B,\left|B\right|\le 1,\nu \in {\mathbb{C}}^{*}.$

$(i)$ If $\left|\frac{\nu (A-B)}{B}-1\right|\le 1$ or $\left|\frac{\nu (A-B)}{B}+1\right|\le 1,B\ne 0$ and $\phi (z)$ satisfies:

$$1+\frac{z{\phi }^{\prime }(z)}{\nu \phi (z)}\prec \frac{1+Az}{1+Bz},$$

then:

$$\phi (z)\prec {(1+Bz)}^{\nu \left(\frac{A-B}{B}\right)}$$

and this is the best dominant.

$(ii)$ If $B=0$ and $\left|\nu A\right|<\pi$ and if $\phi$ satisfies:

$$1+\frac{z{\phi }^{\prime }(z)}{\nu \phi (z)}\prec 1+Az,$$

then:

$$\phi (z)\prec e^{\nu Az},$$

and this is the best dominant.

Lemma 5.

[2]. Let $\mathsf{\Omega }\subset \mathbb{C}$, $b\in \mathbb{C},$ $\Re \left(b\right)>0$ and $\psi :{\mathbb{C}}^2\times \mathbb{U}\to \mathbb{C}$ satisfy $\psi \left(ix,y;z\right)\notin \mathsf{\Omega }$ for all $x,y\le -\frac{{\left|b-ix\right|}^2}{2\Re \left(b\right)}$ and all $z\in \mathbb{U}$. If $p(z)=1+p_{1}z+p_2{z}^2+\dots ,$ is analytic in $\mathbb{U}$ and if:

$$\psi \left(p(z),z{p}^{\prime }\left(z\right);z\right)\in \mathsf{\Omega },$$

then $\Re \left\{p(z)\right\}>0$ in $\mathbb{U}$.

Lemma 6.

[22]. Let $\psi \left(z\right)$ be analytic in $\mathbb{U}$ with $\psi (0)=1$ and $\psi (z)\ne 0$ for all $z.$ If there exist two points $z_1,z_2\in \mathbb{U}$ such that:

$$-\frac{\pi }{2}{\rho }_1=arg\left\{\psi (z_1)\right\}<arg\left\{\psi (z)\right\}<\frac{\pi }{2}{\rho }_2=arg\left\{\psi (z_2)\right\},$$

(9)

for some ${\rho }_1$ and ${\rho }_2\left({\rho }_1,{\rho }_2>0\right)$ and for all $z\left(\left|z\right|<\left|z_1\right|=\left|z_2\right|\right),$ then:

$$\frac{z_1{\psi }^{\prime }(z_1)}{\psi (z_1)}=-i\left(\frac{{\rho }_1+{\rho }_2}{2}\kappa \right)and\frac{z_2{\psi }^{\prime }(z_2)}{\psi (z_2)}=i\left(\frac{{\rho }_1+{\rho }_2}{2}\kappa \right),$$

(10)

where:

$$\kappa \ge \frac{1-\left|a\right|}{1+\left|a\right|}anda=itan\left(\frac{{\rho }_2-{\rho }_1}{{\rho }_2+{\rho }_1}\right).$$

(11)

3. Properties Involving ${\mathit{H}}_{\mathit{p},\mathit{\eta },\mathit{\mu }}^{\mathit{\lambda },\mathit{\delta }}$

Unless otherwise mentioned, we assume throughout this paper that $p\in \mathbb{N},\delta >-p,\mu ,\eta \in \mathbb{R},\mu <p+1,-\infty <\lambda <\eta +p+1,$ $-1\le B<A\le 1,\theta >0$, and the powers are considered principal ones.

Theorem 1.

Let $f\in \mathcal{A}(p)$ satisfy:

$$(1-\theta )\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}{p{z}^{p-1}}+\theta \frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta +1}f(z)\right)}^{\prime }}{p{z}^{p-1}}\prec \frac{1+Az}{1+Bz}.$$

(12)

Then:

$$\Re {\left(\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}{p{z}^{p-1}}\right)}^{\frac{1}{\tau }}>{\left(\frac{\delta +p}{\theta }\underset{0}{\overset{1}{\int }}{u}^{\frac{\delta +p}{\theta }-1}\left(\frac{1-Au}{1-Bu}\right)du\right)}^{\frac{1}{\tau }},\tau \ge 1.$$

(13)

The estimate in (13) is sharp.

Proof. 

Let:

$$\varphi (z)=\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}{p{z}^{p-1}}(z\in \mathbb{U}).$$

(14)

Then, φ is analytic in $\mathbb{U}$. After some computations, we get:

$$(1-\theta )\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}{p{z}^{p-1}}+\theta \frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta +1}f(z)\right)}^{\prime }}{p{z}^{p-1}}=\varphi (z)+\frac{\theta z{\varphi }^{\prime }(z)}{\delta +p}\prec \frac{1+Az}{1+Bz}.$$

Now, by using Lemma 1, we deduce that:

$$\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}{p{z}^{p-1}}\prec \frac{\delta +p}{\theta }{z}^{-\frac{\delta +p}{\theta }}\underset{0}{\overset{z}{\int }}{t}^{\frac{\delta +p}{\theta }-1}\left(\frac{1+At}{1+Bt}\right)dt,$$

(15)

or, equivalently,

$$\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}{p{z}^{p-1}}=\frac{\delta +p}{\theta }\underset{0}{\overset{1}{\int }}{u}^{\frac{\delta +p}{\theta }-1}\left(\frac{1+Auw(z)}{1+Buw(z)}\right)du,$$

and so:

$$\Re \left(\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}{p{z}^{p-1}}\right)>\left(\frac{\delta +p}{\theta }\underset{0}{\overset{1}{\int }}{u}^{\frac{\delta +p}{\theta }-1}\left(\frac{1-Au}{1-Bu}\right)du\right).$$

(16)

Since:

$$\Re \left({\chi }^{\frac{1}{\tau }}\right)\ge {\left(\Re \left(\chi \right)\right)}^{\frac{1}{\tau }}\left(\chi \in \mathbb{C},\Re \left\{\chi \right\}\ge 0,\tau \ge 1\right).$$

(17)

The inequality (13) now follows from (16) and (17). To prove that the result is sharp, let:

$$\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}{p{z}^{p-1}}=\frac{\delta +p}{\theta }\underset{0}{\overset{1}{\int }}{u}^{\frac{\delta +p}{\theta }-1}\left(\frac{1+Auz}{1+Buz}\right)du.$$

(18)

Now, for $f(z)$ defined by (18), we have:

$$(1-\theta )\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}{p{z}^{p-1}}+\theta \frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta +1}f(z)\right)}^{\prime }}{p{z}^{p-1}}=\frac{1+Az}{1+Bz}\left(z\in \mathbb{U}\right),$$

Letting $z\to -1,$ we obtain:

$$\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}{p{z}^{p-1}}\to \frac{\delta +p}{\theta }\underset{0}{\overset{1}{\int }}{u}^{\frac{\delta +p}{\theta }-1}\left(\frac{1-Au}{1-Bu}\right)du,$$

which ends our proof. ☐

Putting $\theta =1$ and using Lemma 1 for Equation (15) in Theorem 1, we obtain the following example.

Example 1.

Let the function $f(z)\in \mathcal{A}(p).$ Then, following containment property holds,

$${\mathcal{T}}_{p,\mu ,\eta }^{\lambda ,\delta +1}(A,B)\subset {\mathcal{T}}_{p,\mu ,\eta }^{\lambda ,\delta }(A,B).$$

Using (4) instead of (3) in Theorem 1, one can prove the following theorem.

Theorem 2.

Let $f\in \mathcal{A}(p)$ satisfy

$$(1-\theta )\frac{{\left(H_{p,\eta ,\mu }^{\lambda +1,\delta }f(z)\right)}^{\prime }}{p{z}^{p-1}}+\theta \frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}{p{z}^{p-1}}\prec \frac{1+Az}{1+Bz}.$$

Then:

$$\Re {\left(\frac{{\left(H_{p,\eta ,\mu }^{\lambda +1,\delta }f(z)\right)}^{\prime }}{p{z}^{p-1}}\right)}^{\frac{1}{\tau }}>{\left(\frac{p+\eta -\lambda }{\theta }\underset{0}{\overset{1}{\int }}{u}^{\frac{p+\eta -\lambda }{\theta }-1}\left(\frac{1-Au}{1-Bu}\right)du\right)}^{\frac{1}{\tau }},\tau \ge 1.$$

(19)

The result is sharp.

Putting $\theta =1$ in Theorem 2, we obtain the following example.

Example 2.

Let the function $f(z)\in \mathcal{A}(p).$ Then, following inclusion property holds

$${\mathcal{T}}_{p,\mu ,\eta }^{\lambda ,\delta }(A,B)\subset {\mathcal{T}}_{p,\mu ,\eta }^{\lambda +1,\delta }(A,B).$$

For a function $f\in \mathcal{A}(p),$ the generalized Bernardi–Libera–Livingston integeral operator $F_{p,\gamma }$ is defined by (see [23]):

$$\begin{array}{ccc}\hfill F_{p,\gamma }f(z)& =& \frac{\gamma +p}{z^p}\underset{0}{\overset{z}{\int }}{t}^{\gamma -1}f(t)dt\hfill \\ & =& \left(z^p+\sum _{k=1}^{\infty }\frac{\gamma +p}{\gamma +p+k}{z}^{p+k}\right)\ast f(z)(\gamma >-p)\hfill \\ & =& z^p{}_3{F}_2\left(1,1,\gamma +p;1,\gamma +p+1;z\right)\ast f(z).\hfill \end{array}$$

(20)

Lemma 7.

If $f\in \mathcal{A}\left(p\right),$ prove that:

(i) 

$H_{p,\eta ,\mu }^{\lambda ,\delta }\left(F_{p,\gamma }f\right)=F_{p,\gamma }\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f\right),$

(ii) 

$$z{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }{F}_{p,\gamma }f(z)\right)}^{\prime }=(p+\gamma )H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)-\gamma H_{p,\eta ,\mu }^{\lambda ,\delta }{F}_{p,\gamma }f(z).$$

(21)

Proof. 

Since

$$\begin{array}{ccc}\hfill H_{p,\eta ,\mu }^{\lambda ,\delta }\left(F_{p,\gamma }f\right)& =& \left[z^p{}_3{F}_2\left(\delta +p,p+1-\mu ,p+1-\lambda +\eta ;p+1,p+1-\mu +\eta ;z\right)\right]\ast \left(F_{p,\gamma }f\right)\hfill \\ & =& \left[z^p{}_3{F}_2\left(\delta +p,p+1-\mu ,p+1-\lambda +\eta ;p+1,p+1-\mu +\eta ;z\right)\right]\ast \hfill \\ & & \left[z^p{}_3{F}_2\left(1,1,\gamma +p;1,\gamma +p+1;z\right)\ast f(z)\right],\hfill \end{array}$$

and:

$$\begin{array}{ccc}\hfill F_{p,\gamma }\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f\right)& =& z^p{}_3{F}_2\left(1,1,\gamma +p;1,\gamma +p+1;z\right)\ast \left(H_{p,\eta ,\mu }^{\lambda ,\delta }f\right)\hfill \\ & =& z^p{}_3{F}_2\left(1,1,\gamma +p;1,\gamma +p+1;z\right)\ast \hfill \\ & & \left[z^p{}_3{F}_2\left(\delta +p,p+1-\mu ,p+1-\lambda +\eta ;p+1,p+1-\mu +\eta ;z\right)\ast f(z)\right].\hfill \end{array}$$

Now, the first part of this lemma follows. Furthermore,

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

(22)

If we replace $f(z)$ by $H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)$ and using the first part of this lemma, we get (21). ☐

Theorem 3.

Suppose that $p+\gamma >0,f\in$ ${\mathcal{T}}_{p,\eta ,\mu }^{\lambda ,\delta }(A,B)$ and $F_{p,\gamma }$ defined by (20). Then:

$$\Re {\left(\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }{F}_{p,\gamma }f(z)\right)}^{\prime }}{p{z}^{p-1}}\right)}^{\frac{1}{\tau }}>{\left(\left(p+\gamma \right)\underset{0}{\overset{1}{\int }}{u}^{p+\gamma -1}\left(\frac{1-Au}{1-Bu}\right)du\right)}^{\frac{1}{\tau }},\tau \ge 1.$$

(23)

The result is sharp.

Proof. 

Let:

$$\varphi (z)=\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }{F}_{p,\gamma }f(z)\right)}^{\prime }}{p{z}^{p-1}}(z\in \mathbb{U}).$$

(24)

Then, φ is analytic in $\mathbb{U}$. After some calculations, we have:

$$\frac{{(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z))}^{\prime }}{p{z}^{p-1}}=\varphi (z)+\frac{z{\varphi }^{\prime }(z)}{p+\gamma }\prec \frac{1+Az}{1+Bz}.$$

Employing the same technique that was used in proving Theorem 1, the remaining part of the theorem can be proven. ☐

Theorem 4.

Let $-1\le B_i<A_i\le 1(i=1,2).$ If each of the functions $f_i\in \mathcal{A}(p)$ satisfies:

$$(1-\theta )\frac{H_{p,\eta ,\mu }^{\lambda ,\delta }{f}_i(z)}{z^p}+\theta \frac{H_{p,\eta ,\mu }^{\lambda ,\delta +1}{f}_i(z)}{z^p}\prec \frac{1+A_{i}z}{1+B_{i}z}(i=1,2),$$

(25)

then:

$$(1-\theta )\frac{H_{p,\eta ,\mu }^{\lambda ,\delta }F(z)}{z^p}+\theta \frac{H_{p,\eta ,\mu }^{\lambda ,\delta +1}F(z)}{z^p}\prec \frac{1+(1-2\varrho )z}{1-z},$$

(26)

where:

$$F(z)=H_{p,\eta ,\mu }^{\lambda ,\delta }(f_1\ast f_2)(z)$$

(27)

and:

$$\varrho =1-\frac{4(A_1-B_1)(A_2-B_2)}{(1-B_1)(1-B_2)}\left[1-\frac{1}{2}{}_2{F}_1\left(1,1;\frac{\delta +p}{\theta }+1;\frac{1}{2}\right)\right].$$

(28)

The result is possible when $B_1=B_2=-1.$

Proof. 

Suppose that $f_i\in \mathcal{A}(p)(i=1,2)$ satisfy the condition (25). Setting:

$$p_i(z)=(1-\theta )\frac{H_{p,\eta ,\mu }^{\lambda ,\delta }{f}_i(z)}{z^p}+\theta \frac{H_{p,\eta ,\mu }^{\lambda ,\delta +1}{f}_i(z)}{z^p}(i=1,2),$$

(29)

we have:

$$p_i(z)\in P({\varsigma }_i)\left({\varsigma }_i=\frac{1-A_i}{1-B_i},i=1,2\right).$$

Thus, by making use of the identity (3) in (29), we get:

$$H_{p,\eta ,\mu }^{\lambda ,\delta }{f}_i(z)=\frac{\delta +p}{\theta }{z}^{p-\frac{\delta +p}{\theta }}{\int }_0^z{t}^{\frac{\delta +p}{\theta }-1}{p}_i(t)dt(i=1,2),$$

(30)

which, in view of F given by (27) and (30), yields:

$$H_{p,\eta ,\mu }^{\lambda ,\delta }F(z)=\frac{\delta +p}{\theta }{z}^{p-\frac{\delta +p}{\theta }}{\int }_0^z{t}^{\frac{\delta +p}{\theta }-1}F(t)dt,$$

(31)

where:

$$F(z)=(1-\theta )\frac{H_{p,\eta ,\mu }^{\lambda ,\delta }F(z)}{z^p}+\theta \frac{H_{p,\eta ,\mu }^{\lambda ,\delta +1}F(z)}{z^p}=\frac{\delta +p}{\theta }{z}^{-\frac{\delta +p}{\theta }}{\int }_0^z{t}^{\frac{\delta +p}{\theta }-1}(p_1\ast p_2)(t)dt.$$

(32)

Since $p_i(z)\in P({\varsigma }_i)(i=1,2),$ it follows from Lemma 3 that:

$$(p_1\ast p_2)(z)\in P({\varsigma }_3)({\varsigma }_3=1-2(1-{\varsigma }_1)(1-{\varsigma }_2)).$$

(33)

Now, by using (33) in (32) and then appealing to Lemma 2, we have:

$$\begin{array}{ccc}\hfill \Re \left\{F(z)\right\}& =& \frac{\delta +p}{\theta }{\int }_0^1{u}^{\frac{\delta +p}{\theta }-1}\Re \left\{(p_1\ast p_2)(uz)\right\}du\hfill \\ & ⩾& \frac{\delta +p}{\theta }{\int }_0^1{u}^{\frac{\delta +p}{\theta }-1}\left(2{\varsigma }_3-1+\frac{2(1-{\varsigma }_3)}{1+u\left|z\right|}\right)du\hfill \\ & >& \frac{\delta +p}{\theta }{\int }_0^1{u}^{\frac{\delta +p}{\theta }-1}\left(2{\varsigma }_3-1+\frac{2(1-{\varsigma }_3)}{1+u}\right)du\hfill \\ & =& 1-\frac{4(A_1-B_1)(A_2-B_2)}{(1-B_1)(1-B_2)}\left[1-\frac{\delta +p}{\theta }{\int }_0^1{u}^{\frac{\delta +p}{\theta }-1}{(1+u)}^{-1}du\right]\hfill \\ & =& 1-\frac{4(A_1-B_1)(A_2-B_2)}{(1-B_1)(1-B_2)}\left[1-\frac{1}{2}{}_2{F}_1\left(1,1;\frac{\delta +p}{\theta }+1;\frac{1}{2}\right)\right]=\varrho .\hfill \end{array}$$

When $B_1=B_2=-1,$ we consider the functions $f_i(z)\in \mathcal{A}(p)(i=1,2)$, which satisfy (25), are defined by:

$$H_{p,\eta ,\mu }^{\lambda ,\delta }{f}_i(z)=\frac{\delta +p}{\theta }{z}^{p-\frac{\delta +p}{\theta }}{\int }_0^z{t}^{\frac{\delta +p}{\theta }-1}\left(\frac{1+A_{i}t}{1-t}\right)dt(i=1,2).$$

Thus, it follows from (32) that:

$$F(z)=\frac{\delta +p}{\theta }{\int }_0^1{u}^{\frac{\delta +p}{\theta }-1}\left[1-(1+A_1)(1+A_2)+\frac{(1+A_1)(1+A_2)}{(1-uz)}\right]du$$

$$=1-(1+A_1)(1+A_2)+(1+A_1)(1+A_2){(1-z)}^{-1}{}_2{F}_1\left(1,1;\frac{\delta +p}{\theta }+1;\frac{z}{z-1}\right)$$

$$\to 1-(1+A_1)(1+A_2)+\frac{1}{2}(1+A_1)(1+A_2){}_2{F}_1\left(1,1;\frac{\delta +p}{\theta }+1;\frac{1}{2}\right)\mathrm{as}z\to -1,$$

which evidently ends the proof. ☐

Theorem 5.

Let $\upsilon \in {\mathbb{C}}^{\ast }$, and let $A,B\in \mathbb{C}$ with $A\ne B$ and $\left|B\right|\le 1.$ Suppose that:

$$\begin{array}{cc}\left|\frac{\upsilon (\delta +p)(A-B)}{B}-1\right|\le 1or\left|\frac{\upsilon (\delta +p)(A-B)}{B}+1\right|\le 1\hfill & ifB\ne 0,\hfill \\ \left|\upsilon (\delta +p)A\right|\le \pi \hfill & ifB=0.\hfill \end{array}$$

If $f\in \mathcal{A}(p)$ with $H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\ne 0$ for all $z\in {\mathbb{U}}^{\ast }=\mathbb{U}\setminus \left\{0\right\},$ then:

$$\frac{H_{p,\eta ,\mu }^{\lambda ,\delta +1}f(z)}{H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)}\prec \frac{1+Az}{1+Bz},$$

implies:

$${\left(\frac{H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)}{z^p}\right)}^{\upsilon }\prec q(z),$$

where:

$$q(z)=\left\{\begin{array}{cc}{(1+Bz)}^{\upsilon (\delta +p)(A-B)/B}\hfill & ifB\ne 0,\hfill \\ e^{\upsilon (\delta +p)Az}\hfill & ifB=0,\hfill \end{array}\right.$$

is the best dominant.

Proof. 

Putting:

$$\Delta (z)={\left(\frac{H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)}{z^p}\right)}^{\upsilon }(z\in \mathbb{U}).$$

(34)

Then, $\Delta$ is analytic in $\mathbb{U}$, $\Delta (0)=1$ and $\Delta (z)\ne 0$ for all $z\in \mathbb{U}.$ Taking the logarithmic derivatives on both sides of (34) and using (3), we have:

$$1+\frac{z{\Delta }^{\prime }(z)}{\upsilon (\delta +p)\Delta (z)}=\frac{H_{p,\eta ,\mu }^{\lambda ,\delta +1}f(z)}{H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)}\prec \frac{1+Az}{1+Bz}.$$

Now, the assertions of Theorem 5 follow by Lemma 4. ☐

Theorem 6.

Let $0\le \alpha \le 1,\zeta >1.$ If $f(z)\in \mathcal{A}\left(p\right)$ satisfies:

$$\Re \left(\left(1-\alpha \right)\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta +2}f(z)\right)}^{\prime }}{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta +1}f(z)\right)}^{\prime }}+\alpha \frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta +1}f(z)\right)}^{\prime }}{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}\right)<\zeta ,$$

(35)

then:

$$\Re \left(\frac{H_{p,\eta ,\mu }^{\lambda ,\delta +1}f(z)}{H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)}\right)<\beta ,$$

where $\beta \in \left(1,\infty \right)$ is the positive root of the equation:

$$2\left(\delta +p+\alpha \right){\beta }^2-\left[2\zeta \left(\delta +p+1\right)-\left(1-\alpha \right)\right]\beta -\left(1-\alpha \right)=0.$$

(36)

Proof. 

Let:

$$\frac{H_{p,\eta ,\mu }^{\lambda ,\delta +1}f(z)}{H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)}=\beta +\left(1-\beta \right)\phi (z).$$

(37)

Then, φ is analytic in $\mathbb{U}$, $\phi (0)=1$ and $\phi (z)\ne 0$ for all $z\in \mathbb{U}.$ Taking the logarithmic derivatives on both sides of (37) and using the identity (3), we have:

$$\left(\delta +p+1\right)\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta +2}f(z)\right)}^{\prime }}{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta +1}f(z)\right)}^{\prime }}-\left(\delta +p\right)\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta +1}f(z)\right)}^{\prime }}{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}=1+\frac{\left(1-\beta \right)z{\phi }^{\prime }(z)}{\beta +\left(1-\beta \right)\phi (z)},$$

and so:

$$\left(1-\alpha \right)\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta +2}f(z)\right)}^{\prime }}{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta +1}f(z)\right)}^{\prime }}+\alpha \frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta +1}f(z)\right)}^{\prime }}{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}=\alpha \beta +\frac{\left(1-\alpha \right)\left(\delta +p\right)\beta }{\delta +p+1}$$

$$+\frac{\left(1-\beta \right)\left[\alpha +\left(1-\alpha \right)\left(\delta +p\right)\right]}{\delta +p+1}\phi (z)+\frac{\left(1-\alpha \right)\left(1-\beta \right)}{\left[\beta +\left(1-\beta \right)\phi (z)\right]\left(\delta +p+1\right)}z{\phi }^{\prime }\left(z\right).$$

Let:

$$\begin{array}{ccc}\hfill \mathsf{\Psi }\left(r,s;z\right)& =& \alpha \beta +\frac{\left(1-\alpha \right)\left(\delta +p\right)\beta }{\delta +p+1}+\frac{\left(1-\beta \right)\left[\alpha +\left(1-\alpha \right)\left(\delta +p\right)\right]}{\delta +p+1}r\hfill \\ & & +\frac{\left(1-\alpha \right)\left(1-\beta \right)}{\left[\beta +\left(1-\beta \right)\phi (z)\right]\left(\delta +p+1\right)}s,\hfill \end{array}$$

and:

$$\mathsf{\Omega }=\left\{w\in \mathbb{C}:\Re (w)<\zeta \right\}.$$

Then, for $x,y\le -\frac{1+x^2}{2},$ we have:

$$\begin{array}{ccc}\hfill \Re \left\{\mathsf{\Psi }\left(ix,y;z\right)\right\}& =& \alpha \beta +\frac{\left(1-\alpha \right)\left(\delta +p\right)\beta }{\delta +p+1}+\frac{\left(1-\alpha \right)\left(1-\beta \right)\beta y}{\left[{\beta }^2+{\left(1-\beta \right)}^2{x}^2\right]\left(\delta +p+1\right)}\hfill \\ & \ge & \alpha \beta +\frac{\left(1-\alpha \right)\left(\delta +p\right)\beta }{\delta +p+1}-\frac{\left(1-\alpha \right)\left(1-\beta \right)}{2\beta \left(\delta +p+1\right)}=\zeta ,\hfill \end{array}$$

where $\beta$ is the positive root of Equation (36). Suppose that:

$$R\left(\beta \right)=2\left(\delta +p+\alpha \right){\beta }^2-\left[2\zeta \left(\delta +p+1\right)-\left(1-\alpha \right)\right]\beta -\left(1-\alpha \right)=0.$$

For $\beta =0,R\left(0\right)=-\left(1-\alpha \right)\le 0$ and for $\beta =1,R\left(1\right)=2\left(\delta +p\right)\left(1-\zeta \right)+2\left(\alpha -\zeta \right)\le 0.$ This proves that $\beta \in \left(1,\infty \right).$ Thus, for $z\in \mathbb{U}$, $\mathsf{\Psi }\left(ix,y;z\right)\notin \mathsf{\Omega }$, and so, we obtain the required result by an application of Lemma 5. ☐

Theorem 7.

Suppose that $0<{\epsilon }_1,{\epsilon }_2\le 1.$ If:

$$-\frac{\pi }{2}{\epsilon }_1<arg\left\{(1-\theta )\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}{p{z}^{p-1}}+\theta \frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta +1}f(z)\right)}^{\prime }}{p{z}^{p-1}}\right\}<\frac{\pi }{2}{\epsilon }_2,$$

(38)

then:

$$-\frac{\pi }{2}{\xi }_1<arg\left(\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}{p{z}^{p-1}}\right)<\frac{\pi }{2}{\xi }_2,$$

(39)

where:

$${\epsilon }_1={\xi }_1+\frac{2}{\pi }arctan\left(\frac{\left({\xi }_1+{\xi }_2\right)\theta }{2\left(\delta +p\right)}\frac{1-\left|a\right|}{1+\left|a\right|}\right),{\epsilon }_2={\xi }_2+\frac{2}{\pi }arctan\left(\frac{\left({\xi }_1+{\xi }_2\right)\theta }{2\left(\delta +p\right)}\frac{1-\left|a\right|}{1+\left|a\right|}\right).$$

(40)

Proof. 

Let:

$$\varphi (z)=\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}{p{z}^{p-1}}(z\in \mathbb{U}).$$

Then, from Theorem 1, we have:

$$(1-\theta )\frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta }f(z)\right)}^{\prime }}{p{z}^{p-1}}+\theta \frac{{\left(H_{p,\eta ,\mu }^{\lambda ,\delta +1}f(z)\right)}^{\prime }}{p{z}^{p-1}}=\varphi (z)+\frac{\theta z{\varphi }^{\prime }(z)}{\delta +p}.$$

Let $U(z)$ be the function that maps $\mathbb{U}$ onto the domain:

$$\left\{w\in \mathbb{C}:-\frac{\pi }{2}{\epsilon }_1<arg(w)<\frac{\pi }{2}{\epsilon }_2\right\},$$

with $U(0)=1,$ then:

$$\varphi (z)+\frac{\theta z{\varphi }^{\prime }(z)}{\delta +p}\prec U(z).$$

Assume that $z_1,z_2$ are two points in $\mathbb{U}$ such that the condition (9) is satisfied, then by Lemma 6, we obtain (10) under the constraint (11). Therefore,

$$\begin{array}{ccc}\hfill arg\left[\left(\delta +p\right)\varphi (z_1)+\theta z_1{\varphi }^{\prime }(z_1)\right]& =& arg\varphi (z_1)\left[\left(\delta +p\right)+\theta \frac{z_1{\varphi }^{\prime }(z_1)}{\varphi (z_1)}\right]\hfill \\ & =& arg\varphi (z_1)+arg\left[\left(\delta +p\right)+\theta \frac{z_1{\varphi }^{\prime }(z_1)}{\varphi (z_1)}\right]\hfill \\ & =& -\frac{\pi }{2}{\xi }_1+arg\left[\left(\delta +p\right)-i\theta \frac{\left({\xi }_1+{\xi }_2\right)\kappa }{2}\right]\hfill \\ & =& -\frac{\pi }{2}{\xi }_1-arctan\left[\frac{\left({\xi }_1+{\xi }_2\right)\theta \kappa }{2\left(\delta +p\right)}\right]\hfill \\ & \le & -\frac{\pi }{2}{\xi }_1-arctan\left[\frac{\left({\xi }_1+{\xi }_2\right)\theta }{2\left(\delta +p\right)}\frac{1-\left|a\right|}{1+\left|a\right|}\right],\hfill \end{array}$$

and:

$$arg\left[\left(\delta +p\right)\varphi (z_2)+\theta z_2{\varphi }^{\prime }(z_2)\right]\ge \frac{\pi }{2}{\xi }_2+arctan\left[\frac{\left({\xi }_1+{\xi }_2\right)\theta }{2\left(\delta +p\right)}\frac{1-\left|a\right|}{1+\left|a\right|}\right].$$

which contradicts the assumption (38). This evidently completes the proof of Theorem 7. ☐