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. ☐