1 Introduction

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

$$\begin{aligned} f(z)=z^{p}+\sum _{n=1}^{\infty }a_{p+n}z^{p+n},\ \ (p\in {\mathbb {N}} =\{1,2,3,...\},z\ \mathbf {\in }\ {\mathbb {U}}\mathbf {=\{}z\,\mathbf {\in {\mathbb {C}}:}\left| z\right| <1\}). \end{aligned}$$
(1.1)

Aouf [1] defined the class \({\textbf{P}}_{k}(p,\gamma )\) of functions g(z) satisfying \(g(0)=p\) and

$$\begin{aligned} \int \limits _{0}^{2\pi }\left| \frac{\text {Re} \{g(z)\}-\gamma }{p-\gamma }\right| d\theta \le k\pi { \ \ }(z=re^{i\theta };k\ge 2;0\le \gamma <p), \end{aligned}$$
(1.2)

which generalizes the classes:

  1. (i)

    \({\textbf{P}}_{k}(1,\gamma )={\textbf{P}}_{k}(\gamma )\) (see [11, 13] and [7] who proved that it is a convex set);

  2. (ii)

    \({\textbf{P}}_{k}(1,0)={\textbf{P}}_{k}\) (see [15]);

  3. (iii)

    \({\textbf{P}}_{2}(p,\gamma )={\textbf{P}}(p,\gamma )\) is the class in which \(\text {Re} \{g(z)\}>\gamma \) \((0\le \gamma <p);\)

  4. (iv)

    \({\textbf{P}}_{2}(1,\gamma )={\textbf{P}}(\gamma )\) is the class in which \( \text {Re} \{g(z)\}>\gamma \) \((0\le \gamma <1);\)

  5. (v)

    \({\textbf{P}}_{2}(1,0)={\textbf{P}}\) is the class in which \(\text {Re} \{g(z)\}>\gamma \) .

From (1.2), we have \(g\in {\textbf{P}}_{k}(p,\gamma )\) if and only if there exist \(g_{1},g_{2}\in {\textbf{P}}(p,\gamma )\) such that

$$\begin{aligned} g(z)=\left( \frac{k}{4}+\frac{1}{2}\right) g_{1}(z)-\left( \frac{k}{4}-\frac{ 1}{2}\right) g_{2}(z),{ \ \ }(z\in {\mathbb {U}}). \end{aligned}$$
(1.3)

Shams et al. [19], defined the integral operator \(I_{p}^{\alpha }:{\mathbb {A}} (p)\rightarrow {\mathbb {A}}(p)\) by:

$$\begin{aligned} I_{p}^{\alpha }f(z)= & {} \left\{ \begin{array}{cc} \dfrac{\left( p+1\right) ^{\alpha }}{z\Gamma \left( \alpha \right) } \int \nolimits _{0}^{z}\left( \log \dfrac{z}{t}\right) ^{\alpha -1}f(t)dt{ \ } &{} \left( \alpha >0\right) \\ f(z) &{} {\ }(\alpha =0) \end{array} \right. \nonumber \\= & {} z^{p}+\sum _{n=1}^{\infty }\left( \dfrac{p+1}{n+p+1}\right) ^{\alpha }a_{n+p}z^{n+p}{ \ \ \ \ \ }(\alpha \ge 0). \end{aligned}$$
(1.4)

Note that \(I_{1}^{\alpha }=I^{\alpha }\) was defined by Jung et al. [5].

Taking

$$\begin{aligned} G_{p}^{\alpha }(z)=z^{p}+\sum _{n=1}^{\infty }\left( \dfrac{p+1}{n+p+1} \right) ^{\alpha }z^{n+p} \end{aligned}$$
(1.5)

we define the function \(G_{p}^{\alpha *}(z)\) by

$$\begin{aligned} G_{p}^{\alpha }(z)*G_{p}^{\alpha *}(z)=\frac{z^{p}}{(1-z)^{p+\delta } }{ \ }(\delta >-p). \end{aligned}$$
(1.6)

Using the function \(G_{p}^{\alpha *}(z)\), we define the new linear operator \(N_{p}^{\alpha ,\delta }:{\mathbb {A}}(p)\rightarrow {\mathbb {A}}(p)\) by

$$\begin{aligned} N_{p}^{\alpha ,\delta }f(z)= & {} G_{p}^{\alpha *}(z)*f(z) \nonumber \\= & {} z^{p}+\sum _{n=1}^{\infty }\left( \dfrac{n+p+1}{p+1}\right) ^{\alpha } \frac{\left( p+\delta \right) _{n}}{\left( 1\right) _{n}}a_{n+p}z^{n+p}, \end{aligned}$$
(1.7)

where \(\left( v\right) _{n}\) given by

$$\begin{aligned} \left( v\right) _{n}=\frac{\Gamma \left( v+n\right) }{\Gamma \left( v\right) }=\left\{ \begin{array}{cc} 1\text { } &{} (n=0) \\ v\left( v+1\right) ...\left( v+n-1\right) &{} (n\in {\mathbb {N}} ) \end{array} \right. . \end{aligned}$$

We note that:

$$\begin{aligned} N_{p}^{0,1-p}f(z)= & {} f(z); \\ N_{p}^{\alpha ,1-p}f(z)= & {} N_{p}^{\alpha }f(z)=z^{p}+\sum _{n=1}^{\infty }\left( \dfrac{n+p+1}{p+1}\right) ^{\alpha }a_{n+p}z^{n+p}; \\ N_{1}^{\alpha ,\delta }f(z)= & {} N^{\alpha ,\delta }f(z)=z+\sum _{n=1}^{\infty }\left( \dfrac{n+2}{2}\right) ^{\alpha }\frac{\left( 1+\delta \right) _{n}}{ \left( 1\right) _{n}}a_{n+1}z^{n+1}. \end{aligned}$$

It is readily verified from (1.7) that, this linear operator has the two

recurrence relation:

$$\begin{aligned}{} & {} z(N_{p}^{\alpha ,\delta }f(z))^{^{\prime }}=(p+1)N_{p}^{\alpha +1,\delta }f(z)-N_{p}^{\alpha ,\delta }f(z), \end{aligned}$$
(1.8)
$$\begin{aligned}{} & {} z(N_{p}^{\alpha ,\delta }f(z))^{^{\prime }}=(p+\delta )N_{p}^{\alpha ,\delta +1}f(z)-\delta N_{p}^{\alpha ,\delta }f(z). \end{aligned}$$
(1.9)

By using each of the recurrence relations (1.8) and (1.9), we obtain different results.

For \(0\le \gamma ,\) \(\beta <p,\) \(p\in {\mathbb {N}} \) and \(k\ge 2\). Seoudy [18]. (see also Noor [8 with \(p=1\)]), defined the following classes by:

$$\begin{aligned} {\mathcal {R}}_{k}(p,\gamma )= & {} \left\{ f\in {\mathbb {A}}(p):\frac{zf^{^{\prime }}(z)}{f(z)}\in {\textbf{P}}_{k}(p,\gamma )\right\} , \end{aligned}$$
(1.10)
$$\begin{aligned} V_{k}(p,\gamma )= & {} \left\{ f\in {\mathbb {A}}(p):\frac{\left( zf^{^{\prime }}(z)\right) ^{^{\prime }}}{f^{^{\prime }}(z)}\in {\textbf{P}}_{k}(p,\gamma )\right\} , \end{aligned}$$
(1.11)
$$\begin{aligned} T_{k}\left( p,\gamma ,\beta \right)= & {} \left\{ f\in {\mathbb {A}}(p):g(z)\in {\mathcal {R}}_{2}(p,\gamma ),\frac{zf^{^{\prime }}(z)}{g(z)}\in {\textbf{P}} _{k}(p,\beta )\right\} , \end{aligned}$$
(1.12)
$$\begin{aligned} T_{k}^{*}\left( p,\gamma ,\beta \right)= & {} \left\{ f\in {\mathbb {A}} (p):g(z)\in V_{2}(p,\gamma ),\frac{\left( zf^{^{\prime }}(z)\right) ^{^{\prime }}}{g^{^{\prime }}(z)}\in {\textbf{P}}_{k}(p,\beta )\right\} . \end{aligned}$$
(1.13)

We can easily see that:

$$\begin{aligned} f(z)\in V_{k}(p,\gamma )\Leftrightarrow \frac{zf^{^{\prime }}(z)}{p}\in {\mathcal {R}}_{k}(p,\gamma ) \end{aligned}$$
(1.14)

and

$$\begin{aligned} f(z)\in T_{k}^{*}\left( p,\gamma ,\beta \right) \Leftrightarrow \frac{ zf^{^{\prime }}(z)}{p}\in T_{k}\left( p,\gamma ,\beta \right) . \end{aligned}$$
(1.15)

We note that, for special choices for the parameters k and \(\gamma \) involved in the above classes, we can obtain the well-known subclasses:

  1. (i)

    \({\mathcal {R}}_{2}(p,\gamma )=S_{p}^{*}(\gamma )\) (see [2, 12, 14] and [4]),

  2. (ii)

    \(V_{2}(p,\gamma )=C_{p}(\gamma )\) (see [2, 12] and [4]),

  3. (iii)

    \(T_{2}\left( p,\gamma ,\beta \right) =T_{p}\left( \gamma ,\beta \right) \) (see [3]),

  4. (iv)

    \(T_{2}^{*}\left( p,\gamma ,\beta \right) =T_{p}^{*}\left( \gamma ,\beta \right) \) (see [9, 10] and [20]).

Next, by using the linear operator \(N_{p}^{\alpha ,\delta }f(z),\) we introduce the following classes of analytic functions for \(0\le \gamma ,\beta <p\) and \(k\ge 2:\)

$$\begin{aligned} {\mathcal {R}}_{p}^{\alpha ,\delta }(k,\gamma )= & {} \left\{ f\in {\mathbb {A}} (p):N_{p}^{\alpha ,\delta }f(z)\in {\mathcal {R}}_{k}(p,\gamma )\right\} , \end{aligned}$$
(1.16)
$$\begin{aligned} V_{p}^{\alpha ,\delta }(k,\gamma )= & {} \left\{ f\in {\mathbb {A}} (p):N_{p}^{\alpha ,\delta }f(z)\in V_{k}(p,\gamma )\right\} , \end{aligned}$$
(1.17)
$$\begin{aligned} T_{p}^{\alpha ,\delta }\left( k,\gamma ,\beta \right)= & {} \left\{ f\in {\mathbb {A}}(p):N_{p}^{\alpha ,\delta }f(z)\in T_{k}\left( p,\gamma ,\beta \right) \right\} , \end{aligned}$$
(1.18)
$$\begin{aligned} T_{p}^{*(\alpha ,\delta )}\left( k,\gamma ,\beta \right)= & {} \left\{ f\in {\mathbb {A}}(p):N_{p}^{\alpha ,\delta }f(z)\in T_{k}^{*}\left( p,\gamma ,\beta \right) \right\} . \end{aligned}$$
(1.19)

Note that

$$\begin{aligned} f(z)\in V_{p}^{\alpha ,\delta }(k,\gamma )\Leftrightarrow \frac{zf^{^{\prime }}(z)}{p}\in {\mathcal {R}}_{p}^{\alpha ,\delta }(k,\gamma ), \end{aligned}$$
(1.20)

and

$$\begin{aligned} f(z)\in T_{p}^{*(\alpha ,\delta )}\left( k,\gamma ,\beta \right) \Leftrightarrow \frac{zf^{^{\prime }}(z)}{p}\in T_{p}^{\alpha ,\delta }\left( k,\gamma ,\beta \right) . \end{aligned}$$
(1.21)

In particular, we set \({\mathcal {R}}_{1}^{\alpha ,\delta }(k,\gamma )=\mathcal { R}^{\alpha ,\delta }(k,\gamma )\), \(V_{1}^{\alpha ,\delta }(k,\gamma )=V^{\alpha ,\delta }(k,\gamma ),T_{1}^{\alpha ,\delta }\left( k,\gamma ,\beta \right) =T^{\alpha ,\delta }\left( k,\gamma ,\beta \right) \) and \( T_{1}^{*(\alpha ,\delta )}\left( k,\gamma ,\beta \right) =T^{*(\alpha ,\delta )}\left( k,\gamma ,\beta \right) .\)

The following lemma will be required in our investigation.

Lemma 1

[6]. Let \(\phi \left( u,v\right) \) be acomplex valued function,

$$\begin{aligned} \phi :D\rightarrow {\mathbb {C}} \,D\subset {\mathbb {C}} \times {\mathbb {C}} { \ \ \ (} {\mathbb {C}} \ \text {is complex plane),} \end{aligned}$$

and let \(u=u_{1}+i\ u_{2}\,v=v_{1}+iv_{2}.\) Suppose that \(\phi \left( u,v\right) \) satisfies the following conditions:

  1. (i)

    \(\phi \left( u,v\right) \) is continuous in D

  2. (ii)

    \(\ \left( 1,0\right) \in D\) and \(\text {Re} \left\{ \phi \left( 1,0\right) \right\} >0;\)

  3. (iii)

    for all \(\left( iu_{2},v_{1}\right) \in \) D such that \(v_{1}\le \frac{-\left( 1+u_{2}^{2}\right) }{2},\) \(\text {Re} \left\{ \phi \left( iu_{2},v_{1}\right) \right\} \le 0.\)

Let \(h\left( z\right) =1+h_{1}z+h_{2}z^{2}+....\)be regular in \({\mathbb {U}}\). Such that \(\left( h\left( z\right) ,zh^{^{\prime }}\left( z\right) \right) \in D\) for all \(z\in {\mathbb {U}}.\) If

$$\begin{aligned} \text {Re} \left\{ \phi \left( h\left( z\right) ,zh^{^{\prime }}\left( z\right) \right) \right\} >0{ \ \ \ }\left( z\in {\mathbb {U}}\right) , \end{aligned}$$

then \(\text {Re} \left\{ h\left( z\right) \right\} >0\), \(z\in {\mathbb {U}}\) .

Lemma 2

[16]. Let \(p\left( z\right) \) be analytic in \({\mathbb {U}}\) with \(p(0)=a\) and \(\Re \{p(z)\}>0,\) \(z\in {\mathbb {U}}.\) Then for \(s>0\) and \( \mu \in {\mathbb {C}} \backslash \{-1\},\)

$$\begin{aligned} \text {Re} \left\{ p(z)+\frac{szp^{^{\prime }}(z)}{p(z)+\mu }\right\} >0\left( \left| z\right| <r_{0}\right) , \end{aligned}$$
(1.22)

where \(r_{0}\) is given by

$$\begin{aligned} r_{0}=\frac{\left| \mu +1\right| }{\sqrt{A+(A^{2}-\left| \mu ^{2}-1\right| )^{\frac{1}{2}}}},{ \ }A=2(s+1)^{2}+\left| \mu \right| ^{2}-1, \end{aligned}$$
(1.23)

and this radius is the best possible.

Lemma 3

[17]. Let \(\phi \) be convex and let g be starlike in \( {\mathbb {U}}.\) Then, for F analytic in \({\mathbb {U}}\) with \(F(0)=1,\left( \left( \phi *Fg\right) /\left( \phi *g\right) \right) \) is contained in the convex hull of \(F(\mathbb {U)}.\)

In this paper, we obtain several inclusion properties of the classes \( {\mathcal {R}}_{p}^{\alpha ,\delta }(k,\gamma ),\) \(V_{p}^{\alpha ,\delta }(k,\gamma ),\) \(T_{p}^{\alpha ,\delta }\left( k,\gamma ,\beta \right) \) and \( T_{p}^{*(\alpha ,\delta )}\left( k,\gamma ,\beta \right) \) associated with the operator \(N_{p}^{\alpha ,\delta }f(z).\)

2 Main results

Unless otherwise mentioned, we assume throughout this paper that \(k\ge 2,\) \( 0\le \gamma <p,\) \(p\in {\mathbb {N}} \) , \(\delta >-p\) and \(\alpha \mathbb {>}0.\)

Theorem 1

Let \(0\le \gamma _{1}\le \gamma <p\), then

$$\begin{aligned} {\mathcal {R}}_{p}^{\alpha +1,\delta }(k,\gamma )\subset {\mathcal {R}} _{p}^{\alpha ,\delta }(k,\gamma _{1}), \end{aligned}$$
(2.1)

where

$$\begin{aligned} \gamma _{1}=\frac{2\gamma -3+\sqrt{\left( 3-2\gamma \right) ^{2}+8\left( p+2\gamma \right) }}{4}. \end{aligned}$$
(2.2)

Proof

Let \(f(z)\in {\mathcal {R}}_{p}^{\alpha +1,\delta }(k,\gamma )\) and

$$\begin{aligned} \frac{z\left( N_{p}^{\alpha ,\delta }f(z)\right) ^{^{\prime }}}{ N_{p}^{\alpha ,\delta }f(z)}= & {} H(z)=\left( p-\gamma _{1}\right) h(z)+\gamma _{1} \nonumber \\= & {} \left( \frac{k}{4}+\frac{1}{2}\right) \left\{ \left( p-\gamma _{1}\right) h_{1}(z)+\gamma _{1}\right\} \nonumber \\{} & {} -\left( \frac{k}{4}-\frac{1}{2}\right) \left\{ \left( p-\gamma _{1}\right) h_{2}(z)+\gamma _{1}\right\} , \end{aligned}$$
(2.3)

where \(h_{i}\) is analytic in \({\mathbb {U}}\) with \(h_{i}(0)=1,i=1,2.\) Using the identity (1.8) in (2.3) and differentiating the resulting equation we obtain

$$\begin{aligned} \frac{z\left( N_{p}^{\alpha +1,\delta }f(z)\right) ^{^{\prime }}}{ N_{p}^{\alpha +1,\delta }f(z)}=\left\{ \gamma _{1}+\left( p-\gamma _{1}\right) h(z)+\tfrac{\left( p-\gamma _{1}\right) zh^{^{\prime }}(z)}{ \left( p-\gamma _{1}\right) h(z)+\gamma _{1}+1}\right\} \in {\textbf{P}} _{k}(p,\gamma ). \end{aligned}$$
(2.4)

Now, we will show that \(H(z)\in {\textbf{P}}_{k}(p,\gamma _{1})\) or \( h_{i}(z)\in P.\) From (2.3) and (2.4) we have

$$\begin{aligned}{} & {} \frac{z\left( N_{p}^{\alpha +1,\delta }f(z)\right) ^{^{\prime }}}{ N_{p}^{\alpha +1,\delta }f(z)} \\{} & {} \quad =\left( \frac{k}{4}+\frac{1}{2}\right) \left\{ \gamma _{1}+\left( p-\gamma _{1}\right) h_{1}(z)+\dfrac{\left( p-\gamma _{1}\right) zh_{1}^{^{\prime }}(z)}{\left( p-\gamma _{1}\right) h_{1}(z)+\gamma _{1}+1}\right\} \\{} & {} \qquad -\left( \frac{k}{4}-\frac{1}{2}\right) \left\{ \gamma _{1}+\left( p-\gamma _{1}\right) h_{2}(z)+\dfrac{\left( p-\gamma _{1}\right) zh_{2}^{^{\prime }}(z)}{\left( p-\gamma _{1}\right) h_{2}(z)+\gamma _{1}+1}\right\} , \end{aligned}$$

this implies that

$$\begin{aligned} \text {Re} \left( \gamma _{1}-\gamma +\left( p-\gamma _{1}\right) h_{i}(z)+ \dfrac{\left( p-\gamma _{1}\right) zh_{i}^{^{\prime }}(z)}{\left( p-\gamma _{1}\right) h_{i}(z)+\gamma _{1}+1}\right) >0{ \ }\left( i=1,2\right) . \end{aligned}$$

We form the functional \(\phi (u,v)\) by taking \(u=h_{i}(z),\) \( v=zh_{i}^{^{\prime }}(z),\)

$$\begin{aligned} \phi (u,v)=\gamma _{1}-\gamma +\left( p-\gamma _{1}\right) u+\dfrac{\left( p-\gamma _{1}\right) v}{\left( p-\gamma _{1}\right) u+\gamma _{1}+1}. \end{aligned}$$
(2.5)

Clearly, the first two conditions of Lemma 1 are satisfied in the domain \( D\subseteq {\mathbb {C}} \backslash \frac{\left( \gamma _{1}+1\right) }{\gamma _{1}-p}\times {\mathbb {C}} \). Now, we verify condition (iii) as follows:

$$\begin{aligned} \text {Re} \left\{ \phi \left( iu_{2},v_{1}\right) \right\}= & {} \text {Re} \left\{ \left( p-\gamma _{1}\right) iu_{2}+\gamma _{1}-\gamma +\frac{\left( p-\gamma _{1}\right) v_{1}}{\left( p-\gamma _{1}\right) iu_{2}+\gamma _{1}+1} \right\} \nonumber \\\le & {} \gamma _{1}-\gamma -\frac{\left( p-\gamma _{1}\right) \left( \gamma _{1}+1\right) \left( 1+u_{2}^{2}\right) }{2\left[ \left( \gamma _{1}+1\right) ^{2}+\left( p-\gamma _{1}\right) ^{2}u_{2}^{2}\right] }=\frac{ A+Bu_{2}^{2}}{2C}, \end{aligned}$$
(2.6)

where

$$\begin{aligned} A= & {} \left( \gamma _{1}+1\right) \left[ 2\gamma _{1}^{2}+3\gamma _{1}-2\gamma _{1}\gamma -2\gamma -p\right] , \\ B= & {} \left( p-\gamma _{1}\right) \left[ 2\left( p-\gamma _{1}\right) \left( \gamma _{1}-\gamma \right) -\left( \gamma _{1}+1\right) \right] , \\ C= & {} 2\left[ \left( \gamma _{1}+1\right) ^{2}+\left( p-\gamma _{1}\right) ^{2}u_{2}^{2}\right] . \end{aligned}$$

We note that \(\text {Re} \left\{ \phi \left( iu_{2},v_{1}\right) \right\} <0\) if and only if \(A\le 0\) and \(B<0.\) From \(\gamma _{1}\) as given by (2.2), we obtain \(A\le 0\) and from \(0\le \gamma _{1}\le \gamma <p\) we have \(B<0.\) Therefore applying Lemma 1, \(h_{i}\in {\textbf{P}}(i=1,2)\) and consequently \( H(z)\in {\textbf{P}}_{k}(p,\gamma _{1})\) for \(z\in {\mathbb {U}}.\) \(\square \)

Theorem 2

Let \(0\le \gamma _{1}\le \gamma <p\), then

$$\begin{aligned} {\mathcal {R}}_{p}^{\alpha ,\delta +1}(k,\gamma )\subset {\mathcal {R}} _{p}^{\alpha ,\delta }(k,\gamma _{1}), \end{aligned}$$
(2.7)

where

$$\begin{aligned} \gamma _{1}=\frac{(2\gamma -2\delta -1)-\sqrt{\left( 2\delta -2\gamma +1\right) ^{2}+8\left( p+2\gamma \delta \right) }}{4}. \end{aligned}$$
(2.8)

Proof

The proof of Theorem 2 is the same as the proof of Theorem 1 by using (1.9) instead of (1.8). \(\square \)

Theorem 3

Let \(0\le \gamma _{1}\le \gamma <p\) , \( \delta >-p\) and \(k\ge 2\) then

$$\begin{aligned} V_{p}^{\alpha +1,\delta }(k,\gamma )\subset V_{p}^{\alpha ,\delta }(k,\gamma _{1}), \end{aligned}$$
(2.9)

where \(\gamma _{1}\) is given by (2.2).

Proof

Applying (1.20) and Theorem 1, we observe that

$$\begin{aligned} f(z)\in & {} V_{p}^{\alpha +1,\delta }(k,\gamma ) \\\Leftrightarrow & {} \frac{zf^{^{\prime }}(z)}{p}\in {\mathcal {R}}_{p}^{\alpha +1,\delta }(k,\gamma )\Longrightarrow \frac{zf^{^{\prime }}(z)}{p}\in {\mathcal {R}}_{p}^{\alpha ,\delta }(k,\gamma _{1}) \\\Leftrightarrow & {} f(z)\in V_{p}^{\alpha ,\delta }(k,\gamma _{1}), \end{aligned}$$

which evidently prove Theorem 3. \(\square \)

Similarly, we can prove the following result.

Theorem 4

Let \(0\le \gamma _{1}\le \gamma <p\) , \( \delta >-p\) and \(k\ge 2\) then

$$\begin{aligned} V_{p}^{\alpha ,\delta +1}(k,\gamma )\subset V_{p}^{\alpha ,\delta }(k,\gamma _{1}), \end{aligned}$$
(2.10)

where \(\gamma _{1}\) is given by (2.8).

Theorem 5

Let \(\alpha \ge 0\), \(\delta >-p,\) \(0\le \gamma ,\beta <p\) and \(k\ge 2\) then

$$\begin{aligned} T_{p}^{\alpha +1,\delta }\left( k,\gamma ,\beta \right) \subset T_{p}^{\alpha ,\delta }\left( k,\gamma ,\beta \right) . \end{aligned}$$
(2.11)

Proof

Let \(f(z)\in T_{p}^{\alpha +1,\delta }\left( k,\gamma ,\beta \right) \). Then, in view of the definition of the class \(T_{p}^{\alpha +1,\delta }\left( k,\gamma ,\beta \right) ,\) there exists a function \(g(z)\in \mathcal { R}_{p}^{\alpha +1,\delta }(2,\gamma )\) such that

$$\begin{aligned} \frac{z\left( N_{p}^{\alpha +1,\delta }f(z)\right) ^{^{\prime }}}{ N_{p}^{\alpha +1,\delta }g(z)}\in {\textbf{P}}_{k}(p,\beta ){ \ \ \ \ \ \ }\left( z\in {\mathbb {U}}\right) . \end{aligned}$$

Now let

$$\begin{aligned} \frac{z\left( N_{p}^{\alpha ,\delta }f(z)\right) ^{^{\prime }}}{ N_{p}^{\alpha ,\delta }g(z)}=H(z)=\left( p-\beta \right) h(z)+\beta , \end{aligned}$$
(2.12)

where h(z) is analytic in \({\mathbb {U}}\) with \(h(0)=1.\) Using (1.8) in (2.12), we have

$$\begin{aligned} (p+1)N_{p}^{\alpha +1,\delta }f(z)-N_{p}^{\alpha ,\delta }f(z)=\left[ \left( p-\beta \right) h(z)+\beta \right] N_{p}^{\alpha ,\delta }g(z). \end{aligned}$$
(2.13)

Differentiating (2.13) leads to

$$\begin{aligned}{} & {} (p+1)z\left( N_{p}^{\alpha +1,\delta }f(z)\right) ^{^{\prime }}-z\left( N_{p}^{\alpha ,\delta }f(z)\right) ^{^{\prime }} \nonumber \\{} & {} \quad =\left( p-\beta \right) zh^{^{\prime }}(z)N_{p}^{\alpha ,\delta }g(z)+ \left[ \left( p-\beta \right) h(z)+\beta \right] z\left( N_{p}^{\alpha ,\delta }g(z)\right) ^{^{\prime }}. \end{aligned}$$
(2.14)

Since \(g(z)\in {\mathcal {R}}_{p}^{\alpha +1,\delta }(2,\gamma ),\) by Theorem 1, \(g(z)\in {\mathcal {R}}_{p}^{\alpha ,\delta }(2,\gamma )\), then we have

$$\begin{aligned} \frac{z\left( N_{p}^{\alpha ,\delta }g(z)\right) ^{^{\prime }}}{ N_{p}^{\alpha ,\delta }g(z)}=\left( p-\gamma \right) p(z)+\gamma , \end{aligned}$$

where \(p(z)=1+c_{1}z+c_{2}z^{2}+...\) is analytic in \({\mathbb {U}}\) with \( p(0)=1.\) Then by using (1.8), we have

$$\begin{aligned} (p+1)\frac{N_{p}^{\alpha +1,\delta }g(z)}{N_{p}^{\alpha ,\delta }g(z)} =\left( p-\gamma \right) p(z)+\gamma +1. \end{aligned}$$
(2.15)

From (2.14) and (2.15), we obtain

$$\begin{aligned} \frac{z\left( N_{p}^{\alpha +1,\delta }f(z)\right) ^{^{\prime }}}{ N_{p}^{\alpha +1,\delta }g(z)}=\left( p-\beta \right) h(z)+\beta +\tfrac{ \left( p-\beta \right) zh^{^{\prime }}(z)}{\left( p-\gamma \right) p(z)+\gamma +1}\in {\textbf{P}}_{k}(p,\beta ). \end{aligned}$$
(2.16)

Now, we will show that \(H(z)\in {\textbf{P}}_{k}(p,\beta )\) or \(h_{i}\in {\textbf{P}}(i=1,2).\) From (2.3) and (2.16) we have

$$\begin{aligned} \frac{z\left( N_{p}^{\alpha +1,\delta }f(z)\right) ^{^{\prime }}}{ N_{p}^{\alpha +1,\delta }g(z)}= & {} \left( \frac{k}{4}+\frac{1}{2}\right) \left\{ \left( p-\beta \right) h_{1}(z)+\beta +\frac{\left( p-\beta \right) zh_{1}^{^{\prime }}(z)}{\left( p-\gamma \right) p(z)+\gamma +1}\right\} \\{} & {} -\left( \frac{k}{4}-\frac{1}{2}\right) \left\{ \left( p-\beta \right) h_{2}(z)+\beta +\frac{\left( p-\beta \right) zh_{2}^{^{\prime }}(z)}{\left( p-\gamma \right) p(z)+\gamma +1}\right\} , \end{aligned}$$

this implies that

$$\begin{aligned} \text {Re} \left\{ \left( p-\beta \right) h_{i}(z)+\frac{\left( p-\beta \right) zh_{i}^{^{\prime }}(z)}{\left( p-\gamma \right) p(z)+\gamma +1} \right\} >0{ \ }\left( z\in {\mathbb {U}};i=1,2\right) . \end{aligned}$$

We form the functional \(\phi (u,v)\) by choosing \(u=h_{i}(z),\) \( v=zh_{i}^{^{\prime }}(z),\)

$$\begin{aligned} \phi (u,v)=\left( p-\beta \right) u+\frac{\left( p-\beta \right) v}{\left( p-\gamma \right) p(z)+\gamma +1}. \end{aligned}$$

Clearly, conditions (i) and (ii) of Lemma 1 are satisfied in \(D\subseteq {\mathbb {C}} \backslash Q^{*}\times {\mathbb {C}} \), where \(Q^{*}=\left\{ z\in {\mathbb {C}} \text { and }\text {Re} (p(z))=p_{1}>\frac{\gamma +1}{\gamma -p}\right\} \) and \( \ p(z)=p_{1}+ip_{2}.\)

Now, we verify condition (iii) as follows:

$$\begin{aligned} \text {Re} \left\{ \phi (iu_{2},v_{1})\right\}= & {} \text {Re} \left\{ \left( p-\beta \right) iu_{2}+\frac{\left( p-\beta \right) v_{1}}{\left( p-\gamma \right) \left( p_{1}+ip_{2}\right) +\gamma +1}\right\} \\\le & {} -\frac{\left( p-\beta \right) \left[ \left( p-\gamma \right) p_{1}+\gamma +1\right] \left( 1+u_{2}^{2}\right) }{2\left\{ \left[ \left( p-\gamma \right) p_{1}+\gamma +1\right] ^{2}+\left[ p_{2}\left( p-\gamma \right) \right] ^{2}\right\} }<0. \end{aligned}$$

By applying Lemma 1, \(h_{i}(z)\in {\textbf{P}}(i=1,2)\). \(\square \)

Theorem 6

Let \(\alpha \ge 0\) , \(\beta <p,\) \(\delta >-p,0\le \gamma <p\) and \(k\ge 2\) then

$$\begin{aligned} T_{p}^{\alpha ,\delta +1}\left( k,\gamma ,\beta \right) \subset T_{p}^{\alpha ,\delta }\left( k,\gamma ,\beta \right) . \end{aligned}$$
(2.17)

Proof

The proof of Theorem 6 is the same as the proof of Theorem 5 by using (1.9) instead of (1.8). \(\square \)

Theorem 7

Let \(\alpha \ge 0\) , \(\beta <p,\) \(\delta >-p,0\le \gamma <p\) and \(k\ge 2\) then

$$\begin{aligned} T_{p}^{*(\alpha +1,\delta )}\left( k,\gamma ,\beta \right) \subset T_{p}^{*(\alpha ,\delta )}\left( k,\gamma ,\beta \right) . \end{aligned}$$
(2.18)

Proof

By applying (1.21) and Theorem 5, it follows that

$$\begin{aligned} f(z)\in & {} T_{p}^{*(\alpha +1,\delta )}\left( k,\gamma ,\beta \right) \\\Leftrightarrow & {} \frac{zf^{^{\prime }}(z)}{p}\in T_{p}^{\alpha +1,\delta }\left( k,\gamma ,\beta \right) \Longrightarrow \frac{zf^{^{\prime }}(z)}{p} \in T_{p}^{\alpha ,\delta }\left( k,\gamma ,\beta \right) \\\Leftrightarrow & {} f(z)\in T_{p}^{*(\alpha ,\delta )}\left( k,\gamma ,\beta \right) \end{aligned}$$

\(\square \)

Similarly, we can prove the following result.

Theorem 8

Let \(\alpha \ge 0\) , \(\beta <p,\) \(\delta >-p,0\le \gamma <p\) and \(k\ge 2\) then

$$\begin{aligned} T_{p}^{*(\alpha ,\delta +1)}\left( k,\gamma ,\beta \right) \subset T_{p}^{*(\alpha ,\delta )}\left( k,\gamma ,\beta \right) . \end{aligned}$$
(2.19)

Theorem 9

If \(f(z)\in {\mathcal {R}}_{p}^{\alpha ,\delta +1}(k,0),\) for \(z\in {\mathbb {U}},\) then \(f\in {\mathcal {R}}_{p}^{\alpha ,\delta }(k,0)\) for

$$\begin{aligned} \left| z\right| <r_{0}=r_{0}=\frac{\left| \mu +1\right| }{ \sqrt{A+(A^{2}-\left| \mu ^{2}-1\right| )^{\frac{1}{2}}}}, \end{aligned}$$
(2.20)

where \(A=2(s+1)^{2}+\left| \mu \right| ^{2}-1,\) \(\mu \ne -1\) with \( \mu =1\) and \(s=1.\) This radius is the best possible.

Proof

We begin by setting

$$\begin{aligned} \frac{z\left( N_{p}^{\alpha ,\delta }f(z)\right) ^{^{\prime }}}{ N_{p}^{\alpha ,\delta }f(z)}= & {} p(z) \nonumber \\= & {} \left( \frac{k}{4}+\frac{1}{2}\right) p_{1}(z)-\left( \frac{k}{4}-\frac{1 }{2}\right) p_{2}(z), \end{aligned}$$
(2.21)

where \(h_{i}\) is analytic in \({\mathbb {U}}\) with \(h_{i}(0)=1,i=1,2.\) Using a similar argument as in Theorem 1, we obtain

$$\begin{aligned} \frac{z\left( N_{p}^{\alpha +1,\delta }f(z)\right) ^{^{\prime }}}{ N_{p}^{\alpha +1,\delta }f(z)}= & {} p(z)+\frac{zp^{^{\prime }}(z)}{p(z)+1} \nonumber \\= & {} \left( \frac{k}{4}+\frac{1}{2}\right) \left( p_{1}(z)+\frac{ zp_{1}^{^{\prime }}(z)}{p_{1}(z)+1}\right) \nonumber \\{} & {} -\left( \frac{k}{4}-\frac{1}{2}\right) \left( p_{2}(z)+\frac{ zp_{2}^{^{\prime }}(z)}{p_{2}(z)+1}\right) . \end{aligned}$$
(2.22)

Applying Lemma 2, we get

$$\begin{aligned} \text {Re} \left\{ p_{i}(z)+\frac{zp_{i}^{^{\prime }}(z)}{p_{i}(z)+1}\right\} >0\text { for }\left| z\right| <r_{0}, \end{aligned}$$
(2.23)

where \(r_{0}\) is given by (2.20). \(\square \)

Theorem 10

Let \(\Psi \) be a convex function and \(f(z)\in \mathcal {R }_{p}^{\alpha ,\delta }(2,\gamma ).\) Then \(G\in {\mathcal {R}}_{p}^{\alpha ,\delta }(2,\gamma ),\) where \(G=\Psi *f.\)

Proof

Let \(G=\Psi *f\) , where f(z) is given by (1.1) and

$$\begin{aligned} \Psi (z)=z^{p}+\sum _{n=1}^{\infty }b_{n+p}z^{n+p}. \end{aligned}$$
(2.24)

Then

$$\begin{aligned} N_{p}^{\alpha ,\delta }G(z)= & {} N_{p}^{\alpha ,\delta }\left[ z^{p}+\sum _{n=1}^{\infty }a_{n+p}b_{n+p}z^{n+p}\right] \nonumber \\= & {} z^{p}+\sum _{n=1}^{\infty }\left( \dfrac{n+p+1}{p+1}\right) ^{\alpha } \frac{\left( p+\delta \right) _{n}}{\left( 1\right) _{n}} a_{n+p}b_{n+p}z^{n+p} \nonumber \\= & {} \left( \Psi *N_{p}^{\alpha ,\delta }f\right) \left( z\right) . \end{aligned}$$
(2.25)

Also, \(f(z)\in {\mathcal {R}}_{p}^{\alpha ,\delta }(2,\gamma ).\) Therefore \( N_{p}^{\alpha ,\delta }f(z)\in {\mathcal {R}}_{2}(p,\gamma ).\) By logarithmic differentiation of (2, 25), we obtain

$$\begin{aligned} \frac{z\left( N_{p}^{\alpha ,\delta }G(z)\right) ^{^{\prime }}}{ pN_{p}^{\alpha ,\delta }G(z)}=\frac{\Psi (z)*F(z)N_{p}^{\alpha ,\delta }f(z)}{\Psi (z)*N_{p}^{\alpha ,\delta }f(z)}, \end{aligned}$$
(2.26)

where \(F=z\left( N_{p}^{\alpha ,\delta }G(z)\right) ^{^{\prime }}/p\left( N_{p}^{\alpha ,\delta }G(z)\right) \) is analytic in \({\mathbb {U}}\) and \( F(0)=1. \) From Lemma 3, we can see that \(z\left( N_{p}^{\alpha ,\delta }G(z)\right) ^{^{\prime }}/p\left( N_{p}^{\alpha ,\delta }G(z)\right) \) is contained in the convex hull of \(F({\mathbb {U}}).\) Since \(z\left( N_{p}^{\alpha ,\delta }G(z)\right) ^{^{\prime }}/p\left( N_{p}^{\alpha ,\delta }G(z)\right) \) is analytic in \({\mathbb {U}}\) and

$$\begin{aligned} F(\mathbb {U)=}\ \Omega \ \mathbb {=}\ \left\{ \omega :\frac{z\left( N_{p}^{\alpha ,\delta }\omega (z)\right) ^{^{\prime }}}{p\left( N_{p}^{\alpha ,\delta }\omega (z)\right) }\in {\textbf{P}}_{2}(\gamma )\right\} , \end{aligned}$$
(2.27)

then \(z\left( N_{p}^{\alpha ,\delta }G(z)\right) ^{^{\prime }}/p\left( N_{p}^{\alpha ,\delta }G(z)\right) \) lies in \(\Omega ,\) this implies that \( G=\Psi *f\in {\mathcal {R}}_{p}^{\alpha ,\delta }(2,\gamma ).\)

\(\square \)

Remark

Specealizing the parameters \(\delta ,\alpha \) and p in the above results, we obtain results concerning the operators \(N_{p}^{\alpha }f(z)\) and \(N^{\alpha ,\delta }f(z)\) given in the introduction.