On a class of $p$-valent functions with negative coefficients defined by opoola differential operator

1. Introduction

Let $A$ denote the class of all functions, $f(z)$ normalized by which are analytic in the unit disc $U=\{z:|z|<1\}$.

Definition 3. [1] For $f(z)\in A$ , Opoola introduced the following operator:

If $f(z)$ is given by ($\text{1}$), then from (2), we see that $(0\le \mu \le \beta$, $t\ge 0$ and $n\in N_0=N\cup 0)$.

Remark 1.

1. When $\beta =\mu$, $t=1$,$D^n(\mu ,\beta ,t)f(z)=D^{n}f(z)$ by Salagean [2],
2. When $\beta =\mu$,$D^n(\mu ,\beta ,t)f(z)=D_{\lambda }^{n}f(z)$ by Al-Oboudi [3].

Definition 1. Let $A_p$ denote the class of functions of the form:

which are analytic and multivalent in the open unit disc $U=(z\in C:\left|z\right|<1).$ We define the following differential operator for the functions $f(z)\in A_p$ If $f(z)$ is given by ($\text{4}$), then from (5), we see that $(0\le \mu \le \beta$, $t\ge 0$ and $n\in N_0=N\cup 0)$. Let $T_p$ denote the subclass of $A_p$ consisting of functions of the form If $f(z)$ is given by Eq. ($\text{7}$), then from Eq. (5), we get $(n\in N_0,a_k\ge 0,p=1,2,..,0\le \mu \le \beta ,t\ge 0,n\in N_0=N\cup 0).$

Remark 2. When $\beta =\mu$ in ($\text{8}$),$D^n(\mu ,\beta ,t,p)f(z)=D_{\delta ,p}^{n}f(z)$ defined by Bulut in [4]. Now, from ($\text{8}$), it follows that $D^n(\mu ,\beta ,t,p)f(z)$ can be written in terms of Convolution as $$D^n(\mu ,\beta ,t,p)f(z)=(f\ast g)(z),$$ where $f(z)$ is as in ($\text{7}$), while $$g(z)=z^p-\sum _{k=p+1}^{\mathrm{\infty }}{[1+(\frac{k}{p}+\beta -\mu -1)t]}^n{z}^k.$$

Definition 2. A function $f(z)\in T_p$ is in the class $S_p^n(\lambda ,\alpha ,\gamma ,\delta )$ if

for some $0\le \lambda <1$, $0\le \gamma <1$,$0<\alpha \le 1$,$0<\delta <1$, $D^n(\mu ,\beta ,t,p)f(z)$ as defined in ($\text{8}$).

Remark 3. When $\mu =\beta$ in ($\text{9}$),the class $S_p^n(\lambda ,\alpha ,\gamma ,\delta )$ reduces to the class $R_p^n(\alpha ,\beta ,\gamma ,\mu )$ studied by Bulut in [4].

Definition 4. [5, 6] The fractional integral of order $l$ is defined, for function $f(z)$ by

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

Definition 5. [5, 6] The fractional derivative of order $l$ is defined, for function $f(z)$ by

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

Definition 6. [5, 6] Under the hypothesis of Definition 4, the fractional derivative of order $p+l$ is defined for functions $f(z)$, by

It readily follows from ($\text{9}$) and ($\text{10}$) that and

Lemma 1.[7] If $f(z)$ and $g(z)$ are analytic in $U$ with $f(z)\prec g(z)$, then for $\sigma >0$ and $z=r{e}^{i\theta }$, $(0<r<1)$, then $${\int }_0^{2\pi }{|f(z)|}^{\sigma }d\theta \le {\int }_0^{2\pi }{|g(z)|}^{\sigma }d\theta .$$

In this work, several properties of the class $S_p^n(\lambda ,\alpha ,\gamma ,\delta )$ are studied, such as coefficient inequalities, hadamard product, radii of close-to-convex, star-likeness, convexity, extreme points, the integral mean inequalities for the fractional derivatives, and further growth and distortion theorem are given using fractional calculus techniques. For more research on classes of multivalent or p-valent functions, see [7, 8, 9, 10, 11, 12, 13,14, 15, 16, 17, 18]

2. Main results

Theorem 1. A function $f(z)\in T_p$ is in the class $S_p^n(\lambda ,\alpha ,\gamma ,\delta )$ if and only if

for some $0\le \lambda <1$, $0\le \gamma <1$,$0<\alpha \le 1$,$0<\delta <1$. The result is sharp for the function $f(z)$ given by $$\begin{array}{rlr}f(z)=z^p-\frac{\delta (\lambda p+\alpha -\gamma )}{k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )}{z}^k,& & (k\ge p+1).\end{array}$$

Proof Suppose that $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$, then we have from ($\text{9}$) that $$\left|\frac{(D^n(\mu ,\beta ,t,p)f(z){)}^{\prime }-p{z}^{p-1}}{\lambda (D^n(\mu ,\beta ,t,p)f(z){)}^{\prime }+(\alpha -\gamma )}\right|<\delta .$$ By substitution, we have $$\left|\frac{p{z}^{p-1}-\sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n{a}_k{z}^{k-1}-p{z}^{p-1}}{\lambda (p{z}^{p-1}-\sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n{a}_k{z}^{k-1})+(\alpha -\gamma )}\right|<\delta ,$$ $$\left|\frac{\sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n{a}_k{z}^{k-1}}{\lambda (p{z}^{p-1}-\sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n{a}_k{z}^{k-1})+(\alpha -\gamma )}\right|<\delta .$$ Since $\mathrm{\Re }z\le \left|z\right|,$ then $$\mathrm{\Re }\left\{\frac{\sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n{a}_k{z}^{k-1}}{\lambda (p{z}^{p-1}-\sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n{a}_k{z}^{k-1})+(\alpha -\gamma )}\right\}<\delta .$$ If we choose $z$ real and let $z\to 1^{-}$, then we get $$\begin{array}{rl}& \sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n{a}_k\le \delta [\lambda (p-\sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n{a}_k)+(\alpha -\gamma )],\\ & \Rightarrow \sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n{a}_k\le \delta \lambda p-\delta \lambda \sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n{a}_k+\delta (\alpha -\gamma ),\\ & \Rightarrow \sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n{a}_k+\delta \lambda \sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n{a}_k\le \delta (\lambda p+\alpha -\gamma ),\\ & \Rightarrow \sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )a_k\le \delta (\lambda p+\alpha -\gamma ).\end{array}$$ Conversely, suppose that the inequality ($\text{15}$) holds true and that $z\in \mathrm{\partial }U:\{z\in C:\left|z\right|=1\}$ and suppose that $$\begin{array}{rl}& |(D^n(\mu ,\beta ,t,p)f(z){)}^{\prime }-p{z}^{p-1}|–\delta |\lambda (D^n(\mu ,\beta ,t,p)f(z){)}^{\prime }+(\alpha -\gamma )|\\ & \le |(D^n(\mu ,\beta ,t,p)f(z){)}^{\prime }-p{z}^{p-1}-\delta (\lambda (D^n(\mu ,\beta ,t,p)f(z){)}^{\prime }+(\alpha -\gamma ))|\\ & =|-\sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n{a}_k{z}^{k-1}-\delta \lambda p{z}^{p-1}\\ & +\delta \lambda \sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n{a}_k{z}^{k-1}-\delta (\alpha -\gamma )|\\ & =|\sum _{k=p+1}^{\mathrm{\infty }}(\delta \lambda -1)k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n{a}_k{z}^{k-1}-\delta (\lambda p{z}^{p-1}+\alpha -\gamma )|\\ & \le \sum _{k=p+1}^{\mathrm{\infty }}(\delta \lambda -1)k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n{a}_k-\delta (\lambda p+\alpha -\gamma )\\ & \le \sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(\delta \lambda +1)a_k-\delta (\lambda p+\alpha -\gamma )\le 0.\end{array}$$ Since by maximum modulus theorem ,that the maximum modulus of an analytic function cannot be attained inside the domain but on the boundary, implies $$|(D^n(\mu ,\beta ,t,p)f(z){)}^{\prime }-p{z}^{p-1}|-\delta |\lambda (D^n(\mu ,\beta ,t,p)f(z){)}^{\prime }+(\alpha -\gamma )|<0,$$ i.e.,$$|(D^n(\mu ,\beta ,t,p)f(z){)}^{\prime }-p{z}^{p-1}|<\delta |\lambda (D^n(\mu ,\beta ,t,p)f(z){)}^{\prime }+(\alpha -\gamma )|.$$ So, $$\frac{|(D^n(\mu ,\beta ,t,p)f(z){)}^{\prime }-p{z}^{p-1}|}{|\lambda (D^n(\mu ,\beta ,t,p)f(z){)}^{\prime }+(\alpha -\gamma )|}<\delta ,$$ implies $$\left|\frac{(D^n(\mu ,\beta ,t,p)f(z){)}^{\prime }-p{z}^{p-1}}{\lambda (D^n(\mu ,\beta ,t,p)f(z){)}^{\prime }+(\alpha -\gamma )}\right|<\delta .$$ Hence, we have that $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta ).$

Corollary 1. If $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$, then $$a_{p+1}\le \frac{\delta (\lambda p+\alpha -\gamma )p^n}{(p+1)[p+(p(\beta -\mu )+1)t{]}^n(1+\delta \lambda )}.$$

Theorem 2. The class $S_p^n(\lambda ,\alpha ,\gamma ,\delta )$ is a class of convex functions.

Proof Let the functions

be in the class $S_p^n(\lambda ,\alpha ,\gamma ,\delta )$, then for $0\le j\le 1$ $$h(z)=(1-j)f(z)+jg(z)=z^p-\sum _{k=p+1}^{\mathrm{\infty }}{c}_k{z}^k,$$ where $c_k=(1-j)a_k+j{b}_k\ge 0$, then making use of ($\text{15}$), we see that $$\begin{array}{rl}\sum _{k=p+1}^{\mathrm{\infty }}k& [1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )c_k\\ & =\sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )a_k+\sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )b_k\\ & <(1-j)\delta (\lambda p+\alpha -\gamma )+j\delta (\lambda p+\alpha -\gamma )\\ & =\delta (\lambda p+\alpha -\gamma )-j\delta (\lambda p+\alpha -\gamma )+j\delta (\lambda p+\alpha -\gamma )\\ & =\delta (\lambda p+\alpha -\gamma ),\end{array}$$ implies $h(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$, which completes the proof.

Theorem 3. If each of the functions $f(z)$ and $g(z)$ is in the class $S_p^n(\lambda ,\alpha ,\gamma ,\delta )$, then $(f\ast g)(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\mathrm{\Omega })$, where $\mathrm{\Omega }\ge \frac{{\delta }^2(\lambda p+\alpha -\gamma )}{k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda {)}^2-{\delta }^2\lambda (\lambda p+\alpha -\gamma )}$.

Proof From ($\text{15}$), we have

and We need to find the smallest $\mathrm{\Omega }$ such that From ($\text{18}$) and ($\text{19}$), we find by means of Cauchy-Schwarz inequalities that Thus, it is enough to show that $$\frac{k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\mathrm{\Omega }\lambda )}{\mathrm{\Omega }(\lambda p+\alpha -\gamma )}{a}_k{b}_k\le \frac{k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )}{\delta (\lambda p+\alpha -\gamma )}\sqrt{a_k{b}_k}.$$ That is On the other hand, from ($\text{21}$), we have Therefore, in view of ($\text{22}$) and ($\text{23}$), it is enough to show that $$\frac{\delta (\lambda p+\alpha -\gamma )}{k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )}\le \frac{\mathrm{\Omega }(1+\delta \lambda )}{\delta (1+\mathrm{\Omega }\lambda )},$$ i.e., $$\delta (\lambda p+\alpha -\gamma )\delta (1+\mathrm{\Omega }\lambda )\le kM(1+\delta \lambda )\mathrm{\Omega }(1+\delta \lambda ),$$ where $M=[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n$. So, $${\delta }^2[(\lambda p+\alpha -\gamma )+\mathrm{\Omega }\lambda (\lambda p+\alpha -\gamma )]\le k\mathrm{\Omega }M(1+\delta \lambda {)}^2,$$ implies $${\delta }^2(\lambda p+\alpha -\gamma )+{\delta }^2\mathrm{\Omega }\lambda (\lambda p+\alpha -\gamma )\le k\mathrm{\Omega }M(1+\delta \lambda {)}^2,$$ implies $${\delta }^2(\lambda p+\alpha -\gamma )\le k\mathrm{\Omega }M(1+\delta \lambda {)}^2-{\delta }^2\mathrm{\Omega }\lambda (\lambda p+\alpha -\gamma ).$$ Also $$\mathrm{\Omega }[kM(1+\delta \lambda {)}^2-{\delta }^2\lambda (\lambda p+\alpha -\gamma )]\ge {\delta }^2(\lambda p+\alpha -\gamma ),$$ implies $$\mathrm{\Omega }\ge \frac{{\delta }^2(\lambda p+\alpha -\gamma )}{k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda {)}^2-{\delta }^2\lambda (\lambda p+\alpha -\gamma )}.$$

Theorem 4. If $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$, then $f(z)$ is p-valently close-to-convex of order $\rho$ in $\left|z\right|<r_1(\lambda ,\alpha ,\gamma ,\delta ,\rho )$, where $$\begin{array}{rlr}{r}_1(\lambda ,\alpha ,\gamma ,\delta ,\rho )=\underset{k}{inf}{\left\{\frac{[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )(p-\rho )}{\delta (\lambda p+\alpha -\gamma )}\right\}}^{\frac{1}{k-p}},& & (k\ge p+1).\end{array}$$

Proof Let $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$, then $$|\frac{f^{\prime }(z)}{z^{p-1}}-p|<p-\rho ,$$ implies

Since hence, ($\text{24}$) is true if Solving ($\text{26}$) for $\left|z\right|$, we obtain $$\begin{array}{rlr}\left|z\right|<{\left\{\frac{[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )(p-\rho )}{\delta (\lambda p+\alpha -\gamma )}\right\}}^{\frac{1}{k-p}},& & (k\ge p+1).\end{array}$$ Hence, the proof.

Theorem 5. If $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$,then $f(z)$ is p-valently starlike of order $\rho$ in $\left|z\right|<r_2(\lambda ,\alpha ,\gamma ,\delta ,\rho )$, where $$\begin{array}{rlr}{r}_2(\lambda ,\alpha ,\gamma ,\delta ,\rho )=\underset{k}{inf}{\left\{\frac{k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )(p-\rho )}{\delta (\lambda p+\alpha -\gamma )(k-p)}\right\}}^{\frac{1}{k-p}},& & (k\ge p+1).\end{array}$$

Proof Let $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$, then $$|\frac{z{f}^{\prime }(z)}{f(z)}-p|<p-\rho .$$ The inequality

Since i.e., $$\begin{array}{rl}\sum _{k=p+1}^{\mathrm{\infty }}k{a}_k{\left|z\right|}^{k-p}-\sum _{k=p+1}^{\mathrm{\infty }}p{a}_k{\left|z\right|}^{k-p}& <(p-\rho )(1-\sum _{k=p+1}^{\mathrm{\infty }}{a}_k{\left|z\right|}^{k-p})\\ & =\sum _{k=p+1}^{\mathrm{\infty }}k{a}_k{\left|z\right|}^{k-p}-\sum _{k=p+1}^{\mathrm{\infty }}p{a}_k{\left|z\right|}^{k-p}\end{array}$$ $$\begin{array}{rl}& <p-\sum _{k=p+1}^{\mathrm{\infty }}p{a}_k{\left|z\right|}^{k-p}-\rho +\sum _{k=p+1}^{\mathrm{\infty }}\rho a_k{\left|z\right|}^{k-p}\\ & =\sum _{k=p+1}^{\mathrm{\infty }}k{a}_k{\left|z\right|}^{k-p}-\sum _{k=p+1}^{\mathrm{\infty }}\rho a_k{\left|z\right|}^{k-p}\\ & <p-\rho \\ & =\sum _{k=p+1}^{\mathrm{\infty }}\frac{(k-\rho )a_k{\left|z\right|}^{k-p}}{p-\rho }\\ & <1.\end{array}$$ Since $$\sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )a_k\le \delta (\lambda p+\alpha -\gamma ).$$ This holds true if $$\frac{(k-\rho ){\left|z\right|}^{k-p}}{p-\rho }<\frac{k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )}{\delta (\lambda p+\alpha -\gamma )},$$ $$\begin{array}{rlr}\left|z\right|<{\left\{\frac{k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )(p-\rho )}{(k-p)\delta (\lambda p+\alpha -\gamma )}\right\}}^{\frac{1}{k-p}},& & (k\ge p+1),\end{array}$$ hence, the proof.

Theorem 6. If $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$, then $f(z)$ is p-valently convex of order $\rho$ in $\left|z\right|<r_3(\lambda ,\alpha ,\gamma ,\delta ,\rho )$, where $$\begin{array}{rlr}{r}_3(\lambda ,\alpha ,\gamma ,\delta ,\rho )=\underset{k}{inf}{\left\{\frac{[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )p(p-\rho )}{\delta (\lambda p+\alpha -\gamma )(k-p)}\right\}}^{\frac{1}{k-p}},& & (k\ge p+1).\end{array}$$

Proof Let $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$, then $$|1+\frac{zf”(z)}{f^{\prime }(z)}-p|<p-\rho .$$ The inequality $$\begin{array}{rl}& |1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}-p|\\ & =\left|\frac{p{z}^{p-1}-\sum _{k=p+1}^{\mathrm{\infty }}k{a}_k{z}^{k-1}+z(p(p-1)z^{p-2}-\sum _{k=p+1}^{\mathrm{\infty }}k(k-1)a_k{z}^{k-2})-p(p{z}^{p-1}-\sum _{k=p+1}^{\mathrm{\infty }}k{a}_k{z}^{k-1})}{p{z}^{p-1}-\sum _{k=p+1}^{\mathrm{\infty }}k{a}_k{z}^{k-1}}\right|\\ & =\left|\frac{p{z}^{p-1}-\sum _{k=p+1}^{\mathrm{\infty }}k{a}_k{z}^{k-1}+p(p-1)z^{p-1}-\sum _{k=p+1}^{\mathrm{\infty }}k(k-1)a_k{z}^{k-1})-p^2{z}^{p-1}+\sum _{k=p+1}^{\mathrm{\infty }}pk{a}_k{z}^{k-1})}{p{z}^{p-1}-\sum _{k=p+1}^{\mathrm{\infty }}k{a}_k{z}^{k-1}}\right|\\ & =\left|\frac{-\sum _{k=p+1}^{\mathrm{\infty }}k(k-p)a_k{z}^{k-p}}{p-\sum _{k=p+1}^{\mathrm{\infty }}k{a}_k{z}^{k-p}}\right|\\ & \le \frac{\sum _{k=p+1}^{\mathrm{\infty }}k(k-p)a_k{\left|z\right|}^{k-p}}{p-\sum _{k=p+1}^{\mathrm{\infty }}k{a}_k{\left|z\right|}^{k-p}}\\ & <p-\rho .\end{array}$$ So, $$\sum _{k=p+1}^{\mathrm{\infty }}k(k-p)a_k{\left|z\right|}^{k-p}<(p-\rho )(p-\sum _{k=p+1}^{\mathrm{\infty }}k{a}_k{\left|z\right|}^{k-p}),$$ implies $$\sum _{k=p+1}^{\mathrm{\infty }}k(k-p)a_k{\left|z\right|}^{k-p}<p(p-\rho )-(p-\rho )\sum _{k=p+1}^{\mathrm{\infty }}k{a}_k{\left|z\right|}^{k-p}),$$ implies $$\sum _{k=p+1}^{\mathrm{\infty }}k(k-\rho )a_k{\left|z\right|}^{k-p}<p(p-\rho ).$$ Since $\sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )a_k\le \delta (\lambda p+\alpha -\gamma )$. This is true if $$\frac{k(k-\rho ){\left|z\right|}^{k-p}}{p(p-\rho )}<\frac{k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )}{\delta (\lambda p+\alpha -\gamma )},$$ $$\begin{array}{rlr}\left|z\right|<{\left\{\frac{[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )p(p-\rho )}{(k-p)\delta (\lambda p+\alpha -\gamma )}\right\}}^{\frac{1}{k-p}},& & (k\ge p+1),\end{array}$$ hence, the proof.

Theorem 7. Let

then, $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$ if and only if it can be expressed in the form $$f(z)={\lambda }_p{f}_p(z)+\sum _{k=p+1}^{\mathrm{\infty }}{\lambda }_k{f}_k(z),$$ where ${\lambda }_p\ge 0$ and ${\lambda }_p$=$1–\sum _{k=p+1}^{\mathrm{\infty }}{\lambda }_k$.

Proof Assume that $f(z)={\lambda }_p{f}_p(z)+\sum _{k=p+1}^{\mathrm{\infty }}{\lambda }_k{f}_k(z)$, then

implies $$f(z)=z^p-\sum _{k=p+1}^{\mathrm{\infty }}{\lambda }_k\left\{\frac{\delta (\lambda p+\alpha -\gamma )}{k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )}\right\}{z}^k.$$ Thus, $$\begin{array}{rl}& \sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda ){\lambda }_k\frac{\delta (\lambda p+\alpha -\gamma )}{k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )}\\ & =\delta (\lambda p+\alpha -\gamma )\sum _{k=p+1}^{\mathrm{\infty }}{\lambda }_k=\delta (\lambda p+\alpha -\gamma )(1-{\lambda }_p)\le \delta (\lambda p+\alpha -\gamma ),\end{array}$$ which shows that $f(z)$ satisfies condition ($\text{15}$) and therefore, $f\in S_p^n(\lambda ,\alpha ,\gamma ,\delta ).$ Conversely, suppose that $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$, since $$\begin{array}{rlr}{a}_k\le \frac{\delta (\lambda p+\alpha -\gamma )}{k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )},& & (k\ge p+1),\end{array}$$ we may set $$\begin{array}{rl}& {\lambda }_k=\frac{k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )}{\delta (\lambda p+\alpha -\gamma )}{a}_k,\\ & {\lambda }_p=1-\sum _{k=p+1}^{\mathrm{\infty }}{\lambda }_k,\end{array}$$ then we obtain from $f(z)=z^p-\sum _{k=p+1}^{\mathrm{\infty }}{a}_k{z}^k$, $$f(z)=({\lambda }_p+\sum _{k=p+1}^{\mathrm{\infty }}{\lambda }_k)z^p-\sum _{k=p+1}^{\mathrm{\infty }}{\lambda }_k\frac{\delta (\lambda p+\alpha -\gamma )}{k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )}{z}^k,$$ i.e., $$f(z)={\lambda }_p{z}^p+\sum _{k=p+1}^{\mathrm{\infty }}{\lambda }_k(z^p-\frac{\delta (\lambda p+\alpha -\gamma )}{k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )}{z}^k),$$ implies $$f(z)={\lambda }_p{z}^p+\sum _{k=p+1}^{\mathrm{\infty }}{\lambda }_k{f}_k(z).$$ This completes the proof.

Corollary 2. The extreme points of $S_p^n(\lambda ,\alpha ,\gamma ,\delta )$ are given by; $$\begin{array}{rlr}& f_p(z)=z^p,& \\ & f_k(z)=z^p–\frac{\delta (\lambda p+\alpha -\gamma )}{k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )}{z}^k,& (k\ge p+1).\end{array}$$

Theorem 8. Let $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$ and suppose that

for some $0\le q\le j$, $0\le l<1$, $(j-q{)}_{q+1}$ denotes the pochhammer symbol defined by $(j-q{)}_{q+1}=(j-q)(j-q+1)\dots j.$ Also, let the function If there exists an analytic function $w(z)$ defined by with $(k\ge q)$ \begin{align*}\Psi (j) =\frac{\Gamma (j-q)}{\Gamma (j+1-l-q)},& & (0\le l0\) and $z=r{e}^{i\theta }$, $(0<r<1)$,

Proof Let $f(z)=z^p-\sum _{j=p+1}^{\mathrm{\infty }}{a}_j{z}^j.$ By means of ($\text{12}$) and Definition 6, we have

where $\mathrm{\Psi }(j)=\frac{\mathrm{\Gamma }(j-q)}{\mathrm{\Gamma }(j+1-l-q)},(0\le l<1,j\ge p+1).$ Since $\psi$ is a decreasing function of $j$, we get $$0<\mathrm{\Psi }(j)\le \mathrm{\Psi }(p+1)=\frac{\mathrm{\Gamma }(p+1-q)}{\mathrm{\Gamma }(2+p-l-q)}.$$ Similarly, from ($\text{32}$), ($\text{14}$), and Definition 6, we have For some $\sigma >0$ and $z=r{e}^{i\theta }$, $(0<r<1)$, we show that so, by applying Lemma 1, it is enough to show that If the above subordination holds true, then we have an analytic function $w(z)$ with $w(0)=0$, $|w(z)|<1$, such that By the condition of the Theorem, we define the function $w(z)$ by which readily yields $w(0)=0$. For such a function $w(z)$, we have By means of the hypothesis of the theorem, the result is proved.

As a special case $q=0$, we have following results from Theorem 8.

Corollary 3. Let $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$ and suppose that

if there exists an analytic function $w(z)$ defined by with \begin{align*}\psi(j)=\frac{\Gamma(j)}{\Gamma(j+1-l)},& & (0\le l0\) and $z=r{e}^{i\theta }$,\;\;\; $(0<r<1)$

Letting $q=1$, we have the following from Theorem 8.

Corollary 4. Let $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$ and suppose that

if there exists an analytic function $w(z)$ define by with \begin{align*} \psi(j)=\frac{\Gamma(j-1)}{\Gamma(j-l)},& & (0\le l0\) and $z=r{e}^{i\theta }$, $(0<r<1)$

Theorem 9. If $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$, then we have $$|D_z^{-l}f(z)|\le \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+l+1)}{\left|z\right|}^{p+l}[1+\frac{\delta (\lambda p+\alpha -\gamma )p^n}{[p+(p(\beta -\mu )+1)t{]}^n(1+\delta \lambda )(p+l+1)}\left|z\right|],$$ and

Proof Suppose that $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$, using Theorem 1, we find that $$\sum _{k=p+1}^{\mathrm{\infty }}k[1+(\frac{k}{p}+\beta -\mu -1)t{]}^n(1+\delta \lambda )a_k\le \delta (\lambda p+\alpha -\gamma ),$$ implies

i.e., From ($\text{7}$), if $f(z)=z^p-\sum _{k=p+1}^{\mathrm{\infty }}{a}_k{z}^k$, $$D_z^{-l}f(z)=\frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+l+1)}{z}^{p+l}-\sum _{k=p+1}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(k+1)}{\mathrm{\Gamma }(k+l+1)}{a}_k{z}^{k+l},$$ implies where Clearly, $\psi$ is a decreasing function of $k$ and we get $$0<\mathrm{\Psi }(k)\le \mathrm{\Psi }(p+1)=\frac{p+1}{p+l+1}.$$ Using ($\text{51}$) and ($\text{53}$), we obtain, $$\begin{array}{rl}|\frac{\mathrm{\Gamma }(p+l+1)}{\mathrm{\Gamma }(p+1)}{z}^{-l}{D}_z^{-l}f(z)|& \le {\left|z\right|}^p+\psi (p+1){\left|z\right|}^{p+1}\sum _{k=p+1}^{\mathrm{\infty }}{a}_k\\ & \le {\left|z\right|}^p+\frac{\delta (\lambda p+\alpha -\gamma )p^n}{[p+(p(\beta -\mu )+1)t{]}^n(1+\delta \lambda )(p+l+1)}{\left|z\right|}^{p+1},\end{array}$$ which is equivalent to assertion ($\text{49}$) and $$\begin{array}{rl}|\frac{\mathrm{\Gamma }(p+l+1)}{\mathrm{\Gamma }(p+1)}{z}^{-l}{D}_z^{-l}f(z)|& \ge {\left|z\right|}^p-\psi (p+1){\left|z\right|}^{p+1}\sum _{k=p+1}^{\mathrm{\infty }}{a}_k\\ & \ge {\left|z\right|}^p-\frac{\delta (\lambda p+\alpha -\gamma )p^n}{[p+(p(\beta -\mu )+1)t{]}^n(1+\delta \lambda )(p+l+1)}{\left|z\right|}^{p+1},\end{array}$$ which completes the proof.

Theorem 10. If $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$, then we have $$|D_z^{l}f(z)|\le \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p-l+1)}{\left|z\right|}^{p-l}[1+\frac{\delta (\lambda p+\alpha -\gamma )p^n}{[p+(p(\beta -\mu )+1)t{]}^n(1+\delta \lambda )(p-l+1)}\left|z\right|],$$ and

Proof If $f(z)=z^p-\sum _{k=p+1}^{\mathrm{\infty }}{a}_k{z}^k$, then $$D_z^{l}f(z)=\frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p-l+1)}{z}^{p-l}-\sum _{k=p+1}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(k+1)}{\mathrm{\Gamma }(k-l+1)}{a}_k{z}^{k-l},$$ implies

where Clearly, $\mathrm{\Psi }$ is a decreasing function of $k$ and we get $$0<\mathrm{\Psi }(k)\le \mathrm{\Psi }(p+1)=\frac{p+1}{p-l+1}.$$ Using ($\text{51}$) and ($\text{56}$), we obtain, $$\begin{array}{rl}|\frac{\mathrm{\Gamma }(p-l+1)}{\mathrm{\Gamma }(p+1)}{z}^l{D}_z^{l}f(z)|& \le {\left|z\right|}^p+\psi (p+1){\left|z\right|}^{p+1}\sum _{k=p+1}^{\mathrm{\infty }}{a}_k\\ & \le {\left|z\right|}^p+\frac{\delta (\lambda p+\alpha -\gamma )p^n}{[p+(p(\beta -\mu )+1)t{]}^n(1+\delta \lambda )(p-l+1)}{\left|z\right|}^{p+1},\end{array}$$ which is equivalent to assertion ($\text{54}$). Similarly, $$\begin{array}{rl}|\frac{\mathrm{\Gamma }(p-l+1)}{\mathrm{\Gamma }(p+1)}{z}^l{D}_z^{l}f(z)|& \ge {\left|z\right|}^p-\mathrm{\Psi }(p+1){\left|z\right|}^{p+1}\sum _{k=p+1}^{\mathrm{\infty }}{a}_k\\ & \ge {\left|z\right|}^p-\frac{\delta (\lambda p+\alpha -\gamma )p^n}{[p+(p(\beta -\mu )+1)t{]}^n(1+\delta \lambda )(p-l+1)}{\left|z\right|}^{p+1},\end{array}$$ which completes the proof.

Corollary 5. If $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$, then we have

Proof From Definition 4, we have $$\underset{l\to 0}{lim}{D}_z^{-l}f(z)=f(z).$$ Therefore, setting $l=0$ in ($\text{49}$), we obtain $$|D_z^{0}f(z)|\le \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+0+1)}{\left|z\right|}^{p+0}[1+\frac{\delta (\lambda p+\alpha -\gamma )p^n}{[p+(p(\beta -\mu )+1)t{]}^n(1+\delta \lambda )(p+0+1)}\left|z\right|],$$ and $$\begin{array}{r}|D_z^{0}f(z)|\ge \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+0+1)}{\left|z\right|}^{p+0}[1-\frac{\delta (\lambda p+\alpha -\gamma )p^n}{[p+(p(\beta -\mu )+1)t{]}^n(1+\delta \lambda )(p+0+1)}\left|z\right|],\end{array}$$ i.e., $$\begin{array}{rl}{\left|z\right|}^p-\frac{\delta (\lambda p+\alpha -\gamma )p^n}{[p+(p(\beta -\mu )+1)t{]}^n(1+\delta \lambda )(p+1)}{\left|z\right|}^{p+1}& \le |f(z)|\\ & \le {\left|z\right|}^p+\frac{\delta (\lambda p+\alpha -\gamma )p^n}{[p+(p(\beta -\mu )+1)t{]}^n(1+\delta \lambda )(p+1)}{\left|z\right|}^{p+1},\end{array}$$ which is (57).

Corollary 6. If $f(z)\in S_p^n(\lambda ,\alpha ,\gamma ,\delta )$, then we have

Proof From Definition 4, we have $$\underset{l\to 1}{lim}{D}_z^{l}f(z)=f^{\prime }(z).$$ Therefore, setting $l=1$ in ($\text{54}$), we obtain $$|D_z^{1}f(z)|\le \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p)}{\left|z\right|}^{p-1}[1+\frac{\delta (\lambda p+\alpha -\gamma )p^n}{[p+(p(\beta -\mu )+1)t{]}^n(1+\delta \lambda )(p)}\left|z\right|],$$ and $$\begin{array}{r}|D_z^{1}f(z)|\ge \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p)}{\left|z\right|}^{p-1}[1-\frac{\delta (\lambda p+\alpha -\gamma )p^n}{[p+(p(\beta -\mu )+1)t{]}^n(1+\delta \lambda )(p)}\left|z\right|],\end{array}$$ i.e., $$p{\left|z\right|}^{p-1}-\frac{\delta (\lambda p+\alpha -\gamma )p^n}{[p+(p(\beta -\mu )+1)t{]}^n(1+\delta \lambda )}{\left|z\right|}^p\le |f^{\prime }(z)|\le p{\left|z\right|}^{p-1}+\frac{\delta (\lambda p+\alpha -\gamma )p^n}{[p+(p(\beta -\mu )+1)t{]}^n(1+\delta \lambda )}{\left|z\right|}^p,$$ which is ($\text{58}$).

Acknowledgments :

The authors acknowledge the management of the University of Ilorin for providing us with a suitable research laboratory and library to enable us carried out this research.

Conflicts of Interest:

”The author declares no conflict of interest.”

Data Availability:

All data required for this research is included within this paper.

Funding Information:

No funding is available for this research.