On certain subclasses of p-valent functions with negative coefficients defined by a generalized differential operator

1. Introduction

Let \(\mathcal{A}\) denote the class of all functions of the form which are analytic in the open unit disk \(U=\lbrace z:|z|< 1\rbrace\). For a function \(f\in \mathcal{A}\), Raducanu and Orhan [1] introduced the following operator: \[ D_{\alpha\nu}^{0}f(z)=f(z)\] \[ D_{\alpha\nu}^{1}f(z)=\alpha \nu z^{2}f”(z)+(\alpha -\nu)zf'(z)+{(1-\alpha+\nu)}f(z)\] If \(f\) is given by (1), then from the definition of the operator \(D_{\alpha \nu}^{n}f\), the Equation (2) can be rewritten as: where \((n\in N_{0}=N\cup \lbrace 0\rbrace)\).

Remark 1.

1. When \(\alpha =1, \nu=0\), we get the Sălăgean differential operator introduced by Sălăgean in [2].
2. When \(\nu=0\), we obtain differential operator defined by Al-Oboudi in [3].

Let \(\mathcal{A}_{p}\) denote the class of functions of the form which are analytic and p-valent in the open unit disk \(U=\lbrace z:|z|< 1\rbrace\). We can write the following equalities for the functions \(f\in \mathcal{A}_{p}:\) \[ D_{\alpha \nu}^{0,p}f(z)=f(z)\] If \(f\) is given by Equation 4, then from Equation 5 and Equation 6, we see that

Remark 2.

1. If \(\nu=0\), \( D_{\alpha \nu}^{n,p}f=D_{\alpha,p}^{n}f\) defined by Bulut in [4]
2. If \(p=1\), \( D_{\alpha \nu}^{n,p}f=D_{\alpha \nu}^{n}f\) introduced by Raducanu and Orhan in [1]
3. If \(p=1,\alpha=1,\nu=0\), \(D_{\alpha \nu}^{n,p}f=D^{n}f\) defined by Sălăgean in [2]
4. If \(p=1,\nu=0\), \(D_{\alpha \nu}^{n,p}f=D_{\alpha}^{n}f\) defined by Al-Oboudi in [3].

Let \(\mathcal{T}_{p}\) denote the subclass of \(\mathcal{A}_{p}\) consisting of functions of the form If \(f\) is given by Equation 8, then from Equation 5 and Equation 6, we get

Definition 1. A function \(f\in \mathcal{T}_{p}\) is in the class, \(S_{p}^n(\vartheta,\beta,\gamma,\varphi)\) if and only if

for \( 0\le\nu \le\alpha \le 1,0\le \vartheta< 1,\)\(0\le\gamma< 1,\)\(0< \beta\le1,\)\(0< \varphi< 1,p\in N\),\(D_{\alpha \nu}^{n,p}f(z)\) as in 9.

In this paper, basic properties of the class \(S_{p}^n(\vartheta,\beta,\gamma,\varphi)\) are studied such as: coefficient inequalities, growth and distortion theorem, closure property, \(\delta\)-neighborhoods, extreme points, radii of close-to-convexity, starlikeness and convexity for these subclasses.

Remark 3. If \(\nu=0\), \(\vartheta=\alpha\), \(\varphi=\mu\), the class \(S_{p}^n(\vartheta,\beta,\gamma,\varphi)\) reduces to the class \(R_{p}^n(\alpha,\beta,\gamma,\mu)\) investigated by Bulut [4]

Definition 2. A function \(f\in \mathcal{T}_{p}\) is in the class \(S_{p}^{n,(\delta_0)}(\vartheta,\beta,\gamma,\varphi)\), if there exists a function \(g(z)\in S_{p}^n(\vartheta,\beta,\gamma,\varphi)\) such that $$\left|\frac{f(z)}{g(z)}-1\right|< 1-\delta_0 … (z\in U ,0\le \delta_0 < 1)$$ for \(0\le \vartheta< 1,\) \(0\le\gamma< 1,\) \(0< \beta\le1,\) \(0< \varphi< 1.\)

Definition 3. For a function \(f\in \mathcal{T}_{p}\), \(\delta\ge 0\), \(\delta\)-neighborhood of \(f\) is defined as:

in particular, for a function \(h\in \mathcal{T}_{p}\), given by \(h(z)=z^{p}\) \((p\in N)\), we immediately have

The concept of neighborhoods was first introduced by Goodman [5] and generalized by Ruschewey [6] and Altintas [7] (see also [8, 9].

2. Coefficient inequalities

Theorem 4. A function \(f\in \mathcal{T}_{p}\) is in the class \(S_{p}^n(\vartheta,\beta,\gamma,\varphi)\) if and only if

for \( 0\le\nu \le\alpha \le 1, 0\le \vartheta< 1,\) \(0\le\gamma< 1,\) \(0< \beta\le1,\) \(0< \varphi< 1\), \(n\in N_{0},\) \(p\in N\). Furthermore, the result is sharp for the function given as \begin{align*} f(z)=z^p – \frac{\varphi(\vartheta p+\beta-\gamma)}{k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)}a_k, (k\ge p+1). \end{align*}

Proof. Suppose that \(f\in S_{p}^n(\vartheta, \beta, \gamma, \varphi),\) then from inequality 10, we have \begin{eqnarray*} \left| \frac {(D_{\alpha \nu}^{n,p}f(z))’ -pz^{p-1}}{\vartheta(D_{\alpha \nu}^{n,p}f(z))’+(\beta-\gamma)} \right|&=&\left| \frac {pz^{p-1}-\sum_{k=p+1}^{\infty}k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}a_{k}z^{k-1} -pz^{p-1}}{\vartheta(pz^{p-1}-\sum_{k=p+1}^{\infty}k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}a_{k}z^{k-1})+(\beta-\gamma)} \right|\\ &= &\left| \frac {\sum_{k=p+1}^{\infty}k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}a_{k}z^{k-1}}{\vartheta(pz^{p-1}-\sum_{k=p+1}^{\infty}k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}a_{k}z^{k-1})+(\beta-\gamma)} \right|\\&<&\varphi, (z\in U,n\in N_{0}) \end{eqnarray*} it is well known that \(\Re z\le \left|z\right|\), therefore, we obtain \begin{align*} \Re\left\{\frac {\sum_{K=p+1}^{\infty}k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}a_{k}z^{k-1}}{\vartheta(pz^{p-1}-\sum_{k=p+1}^{\infty}k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}a_{k}z^{k-1})+(\beta-\gamma)}\right\} < \varphi. \end{align*} If we choose \(z\) real and let \(z \rightarrow 1^-,\) then we get \begin{eqnarray*} \sum_{K=p+1}^{\infty}k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}a_{k}&\le& \varphi\{\vartheta(p-\sum_{k=p+1}^{\infty}k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}a_{k})+(\beta-\gamma)\} \end{eqnarray*} which is precisely the assertion 13. On contrary, suppose that the inequality 13 hold true and let \(z\in \delta U=\lbrace z\in C:\left|z\right|=1\rbrace.\) Then, from 10, we have \begin{eqnarray*} &&\left|(D_{\alpha \nu}^{n,p}f(z))' -pz^{p-1} \right|- \varphi\left|\vartheta(D_{\alpha \nu}^{n,p}f(z))'+(\beta-\gamma) \right|\le \sum_{k=p+1}^{\infty}k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}a_{k}\left| z\right|^{k-1}\\&&- \varphi(\vartheta p+\beta-\gamma)+ \varphi \vartheta \sum_{k=p+1}^{\infty}k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}a_{k}\left| z\right|^{k-1}\\ &&=\sum_{k=p+1}^{\infty}k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}a_{k}\left| z\right|^{k-1}(1+\varphi\vartheta)a_{k}- \varphi(\vartheta p+\beta-\gamma)\le 0. \end{eqnarray*} By maximum modulus theorem, we have \(f \in S_{p}^n(\vartheta,\beta,\gamma,\varphi).\)

Corollary 5. If \(f \in S_{p}^n(\vartheta,\beta,\gamma,\varphi)\), then \( a_{p+1}\le \frac{\varphi(\vartheta p+\beta-\gamma)}{(p+1)\left[1+(\alpha \nu (p+1)+\alpha-\nu)(\frac{1}{p})\right]^{n}(1+\varphi\vartheta)}.\)

3. Growth and distortion theorem

Theorem 6. For each \(f(z)\in S_{p}^n(\vartheta,\beta,\gamma,\varphi)\), we have \( \left|z\right|^p-\frac{\varphi(\vartheta p+\beta-\gamma)}{\left[{1+(\alpha \nu (p+1)+\alpha-\nu)(\frac{1}{p})}\right]^{n}(1+\varphi \vartheta)(p+1)}\left|z\right|^{p+1} \le\left|f(z)\right| \le\left|z\right|^p+\frac{\varphi(\vartheta p+\beta-\gamma)}{\left[{1+(\alpha \nu (p+1)+\alpha-\nu)(\frac{1}{p})}\right]^{n}(1+\varphi \vartheta)(p+1)}\left|z\right|^{p+1}.\)

Proof. Let \(f(z)\in S_{p}^n(\vartheta,\beta,\gamma,\varphi),z\in U\), the bound on \(f(z)\) is given by

from Theorem 4, we have by using (15) in (14), we obtain again using (15), we have Consequently, combining (16) and (17) we obtain the desired result.

Theorem 7. For each \(f(z)\in S_{p}^n(\vartheta,\beta,\gamma,\varphi)\), we have \( p\left|z\right|^{p-1}-\frac{\varphi(\vartheta p+\beta-\gamma)}{\left[{1+(\alpha \nu (p+1)+\alpha-\nu)(\frac{1}{p})}\right]^{n}(1+\varphi \vartheta)}\left|z\right|^{p} \le\left|f'(z)\right| \le p\left|z\right|^{p-1}+\frac{\varphi(\vartheta p+\beta-\gamma)}{\left[{1+(\alpha \nu (p+1)+\alpha-\nu)(\frac{1}{p})}\right]^{n}(1+\varphi \vartheta)}\left|z\right|^{p}.\)

Proof. Let \(f(z)\in S_{p}^n(\vartheta,\beta,\gamma,\varphi),z\in U\), the bound on the derivative of \(f(z)\) is given by \begin{align*} \left|f'(z)\right|\le p\left|z\right|^{p-1}+ (p+1)\left|z \right|^p\sum_{k=p+1}^\infty a_k, z\in U, \end{align*} and, in the same way as above, we get our desired result.

4. Closure properties

Theorem 8. Let the functions \begin{align*} f(z)=z^{p}-\sum_{k=p+1}^{\infty}a_{k}z^{k} , && (a_k\ge 0)\\ g(z)=z^{p}-\sum_{k=p+1}^{\infty}b_{k}z^{k} , && (b_k\ge 0), \end{align*} be in the class \( S_{p}^n(\vartheta,\beta,\gamma,\varphi)\). Then for \(0\le \lambda\le1,\) the function \(h\) is defined as $$h(z)=(1-\lambda)f(z)+\lambda g(z)=z^{p}-\sum_{k=p+1}^{\infty}c_{k}z^{k},$$ where \( c_k:=(1-\lambda)a_k+\lambda b_k\ge0,\) is also in \( S_{p}^n(\vartheta,\beta,\gamma,\varphi).\)

Proof. Suppose that each of the functions \(f\) and \(g\) is in the class \(S_{p}^n(\vartheta,\beta,\gamma,\varphi)\). Then making use of inequality (13), we have \begin{eqnarray*} &&\sum_{k=p+1}^{\infty}k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)c_{k}\\&&=(1-\lambda)\sum_{k=p+1}^{\infty}k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)a_{k} \\&&+\lambda\sum_{k=p+1}^{\infty}k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)b_{k} \\&& \le (1-\lambda)\varphi(\vartheta p+\beta-\gamma)+\lambda \varphi(\vartheta p+\beta-\gamma)\\&&= \varphi(\vartheta p+\beta-\gamma),\end{eqnarray*} which completes the proof.

5. \(\delta\)-Neighborhoods

Theorem 9. If

then \( S_{p}^n(\vartheta,\beta,\gamma,\varphi)\subset N_{\delta}^{p}(h,g).\)

Proof. For a function \(f(z)\in S_{p}^n(\vartheta,\beta,\gamma,\varphi)\) of the form (8), Theorem 4 immediately yields $$ (p+1)\left[1+(\alpha \nu(p+1)+\alpha-\nu)(\frac{1}{p})\right]^{n}(1+\varphi \vartheta)\sum_{k=p+1}^{\infty}a_{k}\le \varphi(\vartheta p+\beta-\gamma),$$ therefore,

On the other hand, we also find from (13) that that is which completes the proof.

Theorem 10. If \(g(z)\in S_{p}^n(\vartheta,\beta,\gamma,\varphi)\) and

then \(N_{\delta}^{p}(f,g)\subset S_{p}^{n,(\delta_0)}(\vartheta,\beta,\gamma,\varphi).\)

Proof. Suppose that \(f\in N_{\delta}^p(f,g)\), then by Definition 3, we have $$\sum_{k=p+1}^{\infty}k|a_k-b_k|\le \delta,$$ which readily implies the coefficient inequality given by $$\sum_{k=p+1}^{\infty}|a_k-b_k|\le \frac{\delta}{p+1} (p\in N).$$ Next, since \(g\in S_{p}^n(\vartheta,\beta,\gamma,\varphi)\), we have from inequality (13) that \begin{align*} \sum_{k=p+1}^{\infty}b_{k}\le \frac{\varphi(\vartheta p+\beta-\gamma)}{(p+1)\left[1+(\alpha \nu(p+1)+\alpha-\nu)(\frac{1}{p})\right]^{n}(1+\varphi \vartheta)}, \end{align*} so from the definition of the class, we have \begin{eqnarray*} \left|\frac{f(z)}{g(z)}-1\right|&<&\frac{\sum_{k=p+1}^{\infty}|a_k-b_k|}{1-\sum_{k=p+1}^\infty b_k}\\ &\le& \frac{\delta}{p+1}\frac{(p+1)\left[1+(\alpha \nu(p+1)+\alpha-\nu)(\frac{1}{p})\right]^{n}(1+\varphi \vartheta)}{(p+1)\left[1+(\alpha \nu(p+1)+\alpha-\nu)(\frac{1}{p})\right]^{n}(1+\varphi \vartheta)-\varphi(\vartheta p+\beta-\gamma)}\\ &= &1-\delta_0,\end{eqnarray*} provided that \(\delta_0\) is given precisely by (22). Thus, by the definition, \(f\in S_{p}^{n,\delta_0}(\vartheta,\beta,\gamma,\varphi)\) for \(\delta_0\) given by (22), this completes our proof.

6. Extreme points

Theorem 11. If \( f_{p}(z)=z^{p} , f_{k}(z)=z^{p}-\frac{\varphi(\vartheta p+\beta-\gamma)}{k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)}z^{k} (k\ge p+1) \) then, \(f\in S_{p}^n(\vartheta,\beta,\gamma,\varphi)\) if and only if it can be expressed in the form \(f(z)=\lambda_{p}f_{p}(z)+\sum_{k=p+1}^{\infty}\lambda_{k}f_{k}(z),\) where \(\lambda_k\ge 0\) and \(\lambda_{p}\)=\(1 – \sum_{k=p+1}^{\infty}\lambda_{k}\).

Proof. Assume that \(f(z)=\lambda_{p}f_{p}(z)+\sum_{k=p+1}^{\infty}\lambda_{k}f_{k}(z)\), then \begin{eqnarray*} f(z)&=&(1 – \sum_{k=p+1}^{\infty}\lambda_{k})z^{p}+\sum_{k=p+1}^{\infty}\lambda_{k}\left\{ z^{p}-\frac{\varphi(\vartheta p+\beta-\gamma)}{k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)}z^{k}\right\}\\&=&z^{p}-\sum_{k=p+1}^{\infty}\lambda_{k}\left\{ \frac{\varphi(\vartheta p+\beta-\gamma)}{k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)}z^{k}\right\}.\end{eqnarray*} Thus, \begin{eqnarray*}&&\sum_{k=p+1}^{\infty}k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta) \lambda_{k} \frac{\varphi(\vartheta p+\beta-\gamma)}{k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)}\\&&= \varphi(\vartheta p+\beta-\gamma)\sum_{k=p+1}^{\infty}\lambda_{k}= \varphi(\vartheta p+\beta-\gamma)(1-\lambda_{p}) \le \varphi(\vartheta p+\beta-\gamma),\end{eqnarray*} which shows that \(f\) satisfies condition (13) and therefore,\(f\in S_{p}^n(\vartheta,\beta,\gamma,\varphi).\) Conversely, suppose that \(f\in S_{p}^n(\vartheta,\beta,\gamma,\varphi)\), since \[ a_{k} \le \frac{\varphi(\vartheta p+\beta-\gamma)}{k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)}, (k\ge p+1),\] we may set \[ \lambda_{k}=\frac{k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)}{\varphi(\vartheta p+\beta-\gamma)}a_{k}, \text{ and } \lambda_{p}=1- \sum_{k=p+1}^{\infty}\lambda_{k},\] then we obtain from \begin{eqnarray*}f(z)&=& z^{p}-\sum_{k=p+1}^{\infty}a_{k}z^{k}\\&=&(\lambda_{p}+\sum_{k=p+1}^{\infty}\lambda_{k})z^{p} – \sum_{k=p+1}^{\infty}\lambda_{k} \frac{\varphi(\vartheta p+\beta-\gamma)}{k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)}z^k\\ &=&\lambda_{p}z^{p}+ \sum_{k=p+1}^{\infty}\lambda_{k}(z^{p}- \frac{\varphi(\vartheta p+\beta-\gamma)}{k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)}z^k)\\&=& \lambda_{p}z^{p} + \sum_{k=p+1}^{\infty}\lambda_{k}f_{k}(z),\end{eqnarray*} which completes the proof.

Corollary 12. The extreme points of \(S_{p}^n(\vartheta,\beta,\gamma,\varphi)\) are given by $$f_{p}(z)=z^{p} , f_{k}(z)=z^{p}-\frac{\varphi(\vartheta p+\beta-\gamma)}{k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)}z^{k} (k\ge p+1)$$

7. Radii of close-to-convexity,starlikeness and convexity

A function \(f\in \mathcal{T}_p\) is said to be \(p\)-valently close-to-convex of order \(\rho\) if it satisfies $$\Re \left\{ f'(z)\right\} >\rho$$ for some \(\rho (0\le\rho

\rho,$$ for some \(\rho (0\le\rho

\rho,$$ for some \(\rho (0\le\rho< p)\) and for all \(z\in U.\)

Theorem 13. If \(f\in S_{p}^n(\vartheta,\beta,\gamma,\varphi)\) then \(f\) is \(p\)-valently close-to-convex of order \(\rho\) in \(\left|z\right|< r_1(\vartheta,\beta,\gamma,\varphi,\rho)\), where \begin{align*} r_1(\vartheta,\beta,\gamma,\varphi,\rho)=\inf_{k}\left\{\frac{\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)a_{k}(p-\rho)}{\varphi(\vartheta p+\beta-\gamma)} \right\}^{\frac{1}{k-p}} k\ge p+1. \end{align*}

Proof. It is sufficient to show that \(\left|\frac{f'(z)}{z^{p-1}}-p\right|< p-\rho.\) Since \(\left|\frac{pz^{p-1}-\sum_{k=p+1}^\infty ka_Kz^{k-1}}{z^{p-1}}-p\right|< p-\rho,\) which implies that $$\left|\frac{f'(z)}{z^{p-1}}-p\right|\le \sum_{k=p+1}^{\infty}ka_k\left|z\right|^{k-p}< p-\rho,$$ implies

and by applying the result of Theorem 4, we get $$ \sum_{k=p+1}^{\infty}a_k \le\frac{\varphi(\vartheta p+\beta-\gamma)}{k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)a_{k}}.$$ Hence,(23) is true if solving (24) for \(z\) we obtain $$\left|z\right|\le \left\{\frac{\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)(p-\rho)}{\varphi(\vartheta p+\beta-\gamma)}\right\}^{\frac{1}{k-p}}$$ which completes the proof.

Theorem 14. If \(f\in S_{p}^n(\vartheta,\beta,\gamma,\varphi)\) then \(f\) is \(p\)-valently starlike of order \(\rho\) in \(\left|z\right|< r_2(\vartheta,\beta,\gamma,\varphi,\rho)\), where \begin{align*} r_2(\vartheta,\beta,\gamma,\varphi,\rho)=\inf_{k}\left\{\frac{k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)(p-\rho)}{\varphi(\vartheta p+\beta-\gamma)(k-\rho)} \right\}^{\frac{1}{k-p}} k\ge p+1. \end{align*}

Proof. In order to prove, it suffices to show that \(\left|\frac{zf'(z)}{f(z)}-p\right| < p-\rho. \)

and by using inequality (13), we get $$ \sum_{k=p+1}^{\infty}a_k \le\frac{\varphi(\vartheta p+\beta-\gamma)}{k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)a_{k}},$$ so, (25) holds true if $$\frac{(k-\rho)\left|z\right|^{k-\rho}}{p-\rho} \le \frac{k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)}{\varphi(\vartheta p+\beta-\gamma)},$$ and then \(f\) is starlike of order \(\rho.\)

Theorem 15. If \(f\in S_{p}^n(\vartheta,\beta,\gamma,\varphi)\), then \(f\) is \(p\)-valently convex of order \(\rho\) in \(\left|z\right|< r_3(\vartheta, \beta, \gamma, \varphi, \rho)\), where \begin{align*} r_3(\vartheta,\beta,\gamma,\varphi,\rho)=\inf_{k}\left\{\frac{\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)p(p-\rho)}{\varphi(\vartheta p+\beta-\gamma)(k-\rho)} \right\}^{\frac{1}{k-p}} k\ge p+1. \end{align*}

Proof. To prove this, it suffices to show that \(\left|1+\frac{zf”(z)}{f'(z)}-p\right| < p-\rho .\) Since

it implies that \begin{align*} \left|1+\frac{zf”(z)}{f'(z)}-p\right|=\left|\frac{-\sum_{k=p+1}^{\infty}k(k-p)a_kz^{k-p}}{p-\sum_{k=p+1}^{\infty}ka_kz^{k-p}}\right|\le\frac{\sum_{k=p+1}^{\infty}k(k-p)a_k\left| z\right| ^{k-p}}{p-\sum_{k=p+1}^{\infty}ka_k\left| z\right| ^{k-p}} < p-\rho \end{align*} and by applying the result in Theorem 4, we get $$ \sum_{k=p+1}^{\infty}a_k \le\frac{\varphi(\vartheta p+\beta-\gamma)}{k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)a_{k}}$$ so, (26) holds true if $$\frac{k(k-\rho)\left|z\right|^{k-p} }{p(p-\rho)} \le \frac{k\left[{1+(\alpha \nu k+\alpha-\nu)\left(\frac{k}{p}-1\right)}\right]^{n}(1+\varphi \vartheta)}{\varphi(\vartheta p+\beta-\gamma)}$$ and then \(f\) is convex of order \(\rho.\)

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Competing Interests

The author(s) do not have any competing interests in the manuscript.